Maths Quest 11 Advanced General Mathematics (Spec) Classpad Edition

VCE mATHEmATiCS UNiTS 1 & 2

mATHS QUEST 11 Advanced General mathematics C A S i O C L A S S PA D E D i T i O N

mARK BARNES RUTH BAKOGiANiS KYLiE BOUCHER TRACY HERFT ROBYN WiLLiAmS JENNiFER NOLAN

mARK DUNCAN GEOFF PHiLLiPS

CONTRiBUTiNG AUTHORS iAN WEBSTER | DOUGLAS SCOTT | ELENA iAmPOLSKY | SONJA STAmBULiC ROSS ALLEN | DAViD PHiLLiPS | mURRAY ANDERSON | ROBERT CAHN | KATHY CHAmBERS miCHAEL HALFPENNY | CAROLYN mEWS | JOHN SHORT | NiCK SimPSON BORiS SmOLYAR | JENNY WATSON | BARBARA WOODS SUPPORT mATERiAL JOHN DOWSEY | DENNiS FiTZGERALD | EmiLY HUi | ViNOD NARAYAN SimONE RiCHARDSON | PETER SWAiN | DAViD TYNAN

First published 2010 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10/12pt Times LT © John Wiley & Sons Australia, Ltd 2010 The moral rights of the authors have been asserted. National Library of Australia Cataloguing-in-Publication entry

Title: Edition: ISBN:

Notes: Target Audience: Subjects: Other Authors/ Contributors: Dewey Number:

Maths Quest 11 Advanced General Mathematics / Mark Barnes . . . [et al.]. Casio ClassPad edition. 978 1 7421 6222 5 (student book.) 978 1 7421 6290 4 (teacher book.) 978 1 7421 6232 4 (student ebook.) 978 1 7421 6297 3 (teacher ebook.) Includes index. For secondary school age. Mathematics — Textbooks. Mathematics — Problems, exercises, etc. Barnes, Mark. 510.712

Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Screenshots © Casio. ClassPad is a registered trademark of Casio Computer Co., Ltd Illustrated by Paul Lennon, Aptara and the Wiley Art Studio Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1

Contents CHAPTER 3

Introduction vii About eBookPLUS ix Acknowledgements x

Relations and functions 71 3a Relations 71

Exercise 3A 73

CHAPTER 1

3b Functions 74

Number systems: real and complex 1 1a Review of set notation 1

Exercise 1A 9 1B Subsets of the set of real numbers 1C 1d 1e

1f

1G

10

Exercise 1B 13 Properties of surds 14 Exercise 1C 20 The set of complex numbers 22 Exercise 1D 24 Multiplication and division of complex numbers 24 Exercise 1E 28 Representing complex numbers on an Argand diagram 29 Exercise 1F 31 Factorising quadratic expressions and solving quadratic equations over the complex number field 31 Exercise 1G 36

Summary 38

Exercise 3B 77 3c Inverse functions 77 Exercise 3C 80 Summary 81 Chapter review 82 eBookPLUS activities 85

Exam Practice 1 Based on Chapters 1–3 86

CHAPTER 4

Algebra 88 4a Review of index laws 88

Exercise 4A 91 4B Standard form and significant figures 92 4c 4d

Chapter review 40 eBookPLUS activities 43

CHAPTER 2

Transformations 44 2a Translations of points and graphs 44

Exercise 2A 49 2B Reflections of points and graphs 50

Exercise 2B 52 2C Dilations from axes 53 Exercise 2C 55 2D The ellipse and the hyperbola 56 Exercise 2D 61 2E Successive transformations 62 Exercise 2E 63 Summary 65 Chapter review 67 eBookPLUS activities 70

4e 4f 4g

Exercise 4B 94 Transposition 96 Exercise 4C 99 Solving linear equations and simultaneous linear equations 101 Exercise 4D 111 Applications 114 Exercise 4E 119 Algebraic fractions 122 Exercise 4F 125 Linear literal equations 126 Exercise 4G 128

Summary 129 Chapter review 131 eBookPLUS activities 135

CHAPTER 5

Trigonometric ratios and their applications 136 5a Trigonometry of right-angled triangles 136

Exercise 5A 141 5B Elevation, depression and bearings 143 Exercise 5B 146

5C The sine rule 148 5D 5E 5F 5G 5H

Exercise 5C 152 The cosine rule 154 Exercise 5D 158 Area of triangles 159 Exercise 5E 162 Trigonometric identities 163 Exercise 5F 165 Radian measurement 165 Exercise 5G 166 Arcs, sectors and segments 167 Exercise 5H 170


7G Transformation of data 259

Exercise 7G 261 Summary 263 Chapter review 264 eBookPLUS activities 269


CHAPTER 8

Further algebra 272 8a Polynomial identities 272

Exercise 8A 275 8B Partial fractions 276

Exercise 8B 282 CHAPTER 6

Sequences and series 178 6a Describing sequences 178

Exercise 6A 184 6B Arithmetic sequences 186

Exercise 6B 191 6c Arithmetic series 192

Exercise 6C 194 6d Geometric sequences 195 Exercise 6D 199 6e Geometric series 200 Exercise 6E 205 6F Applications of sequences and series 206 Exercise 6F 208 Summary 211 Chapter review 213 eBookPLUS activities 217

CHAPTER 7

Variation 218 7a Direct variation 218

Exercise 7A 224 7B Further direct variation 227 7C 7D 7E 7F

iv

Exercise 7B 231 Inverse variation 235 Exercise 7C 241 Further inverse variation 244 Exercise 7D 246 Joint variation 249 Exercise 7E 252 Part variation 255 Exercise 7F 257

Contents

8c Simultaneous equations 283

Exercise 8C 287 Summary 289 Chapter review 290 eBookPLUS activities 292

CHAPTER 9

Algebra and logic 293 9a Statements (propositions), connectives and

9B 9c 9d 9E

truth tables 293 Exercise 9A 298 Valid and invalid arguments 300 Exercise 9B 307 Techniques of proof 310 Exercise 9C 314 Sets and Boolean algebra 317 Exercise 9D 324 Digital logic 325 Exercise 9E 331


CHAPTER 10

Linear and non-linear graphs 340 10a The circle 340

Exercise 10A 342 10b The ellipse 343

Exercise 10B 346 10C The parabola 347

Exercise 10C 350 10d The hyperbola 351 Exercise 10D 356

10E Polar coordinates 357 10F 10G 10h

10i

Exercise 10E 359 Polar equations 360 Exercise 10F 362 Polar graphs 363 Exercise 10G 365 Review of complex numbers and polar form of complex numbers 367 Exercise 10H 372 Addition of ordinates, reciprocals and squares of simple graphs 372 Exercise 10I 378


CHAPTER 11

Linear programming 389 11a Graphs of linear inequations 389 11b

11c 11d

11e 11f

Exercise 11A 393 Graphs of simultaneous linear inequations 394 Exercise 11B 398 Graphs of systems of linear inequations 399 Exercise 11C 401 Maximising and minimising linear functions 402 Exercise 11D 408 Solving linear programming problems 409 Exercise 11E 411 Further applications of linear programming 415 Exercise 11F 417


CHAPTER 12

Coordinate geometry 429 12a Distance between two points 429

Exercise 12A 431 12B Midpoint of a line segment 432 Exercise 12B 434 12C Dividing a line segment internally in the ratio a : b 435 Exercise 12C 438 12D Dividing a line segment externally in the ratio a : b 439 Exercise 12D 442

12e Parallel lines 442

Exercise 12E 444 12f Perpendicular lines 445 Exercise 12F 446 12G Applications 447 Exercise 12G 451 Summary 454 Chapter review 456 eBookPLUS activities 459


CHAPTER 13

Vectors 462 13a Introduction to vectors 462

Exercise 13A 465 13B Operations on vectors 466

Exercise 13B 470 13C Magnitude, direction and components

of vectors 471 Exercise 13C 474 13D i , j notation 475   Exercise 13D 478 13E Applications of vectors 480 Exercise 13E 483 Summary 487 Chapter review 488 eBookPLUS activities 491

CHAPTER 14

Statics of a particle 492 14a Force and tension 492

Exercise 14A 494 14b Newton’s first law of motion 495

Exercise 14B 497 14C Equilibrium — forces at an angle 498

Exercise 14C 501 14d Connected bodies in equilibrium 502

Exercise 14D 505 Summary 507 Chapter review 508 eBookPLUS activities 510

CHAPTER 15

Kinematics 511 15A Introduction to kinematics 511

Exercise 15A 518

Contents

v

15B Velocity–time graphs and acceleration–time

graphs 521 Exercise 15B 529 15C Constant acceleration formulas 532 Exercise 15C 535 15D Instantaneous rates of change 538 Exercise 15D 546 Summary 549 Chapter review 551 eBookPLUS activities 555

CHAPTER 16

Geometry in two and three dimensions 556 16a Review of basic geometry 556

Exercise 16A 560 16B Geometric constructions 562 Exercise 16B 567 16c Polygons 567 Exercise 16C 575 16D Three-dimensional geometry 576 Exercise 16D 578

vi

Contents

16e Circle geometry 579

Exercise 16E 582 16F Tangents, chords and circles 584 Exercise 16F 589 16G Geometry in architecture, design and art 592 Exercise 16G 597 Summary 599 Chapter review 601 eBookPLUS activities 605


CHAPTER 17

Univariate data — Available on eBookPlus CHAPTER 18

Bivariate data — Available on eBookPlus Answers 608 Index 673

Introduction Maths Quest 11 Advanced General Mathematics is specifically designed for the VCE General Mathematics course, for students on the Specialist Mathematics pathway, and based on the award-winning Maths Quest series. This resource contains: • a student textbook with accompanying eBookPLUS • a teacher edition with accompanying eGuidePLUS • a solutions manual.

Student textbook Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos and to make navigation through the text easier. Clear, concise theory sections contain worked examples and highlighted important text and remember boxes. Icons appear for the eBookPLUS to indicate that interactivities, eLessons and digital documents are available online to help with the teaching and learning of particular concepts. Worked examples in a Think/Write format provide clear explanation of key steps and suggest presentation of solutions. Many worked examples have eBookPLUS icons to indicate that a Tutorial is available to elucidate the concepts being explained. Worked examples also contain CAS calculator instructions and screens to exemplify judicious use of the calculator. Exercises contain many carefully graded skills and application problems, including multiple choice questions. Cross-references to relevant worked examples appear with the first ‘matching’ question throughout the exercises. Exercises also contain questions from past VCE examination papers as well as Exam tips. A selection of questions are tagged as technology-free to indicate to students that they should avoid using their calculators or other technologies to assist them in finding a solution. Exam practice sections contain exam style questions, including time and mark allocations for each question. Fully worked solutions are available on the eBookPLUS for students. Each chapter concludes with a summary and chapter review exercise containing examinationstyle questions (multiple choice, short answer and extended response), which help consolidate students’ learning of new concepts. Also included are questions from past VCE exams along with relevant Exam tips. Technology is fully integrated (in line with VCE recommendations).

Student website — eBookPLUS The accompanying eBookPLUS contains the entire student textbook in HTML plus additional exercises. Students may use the eBookPLUS on laptops, school or home computers, and cut and paste material for revision, assignments or the creation of notes for exams. Career profiles and History of mathematics place mathematical concepts in context. Investigations, often suggesting the use of technology, provide further discovery learning opportunities.

Introduction

vii

WorkSHEET icons link to editable Word documents, and may be completed on screen or printed and completed by hand. SkillSHEET icons link to printable pages designed to help students revise required concepts, and contain additional examples and problems. Interactivity icons link to dynamic animations which help students to understand difficult concepts. eLesson icons link to videos or animations designed to elucidate concepts in ways other than what the teacher can achieve in the classroom. Tutorial icons link to one-way engagement activities which explain the worked examples in detail for students to view at home or in the classroom. Test yourself tests are also available and answers are provided for students to receive instant feedback.

Teacher website — eGuidePLUS The accompanying eGuidePLUS contains everything in the eBookPLUS and more. Two tests per chapter, fully worked solutions to WorkSHEETs, the work program and other curriculum advice in editable Word format are provided. Maths Quest is a rich collection of teaching and learning resources within one package. Maths Quest 11 Advanced General Mathematics provides ample material, such as exercises, analysis questions, investigations, worksheets and technology files, from which teachers may set school assessed coursework (SAC).

viii

Introduction

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About eBookPLUS

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Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their assistance and for permission to reproduce copyright material in this book.

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x

Acknowledgements

1

1a 1b 1c 1D 1E

number systems: real and complex areaS oF STudy

• Review of properties and computation with natural numbers, integers, and rational numbers • Forms of representation including, for example, equivalence of decimal and fractional forms (terminating and infinitely recurring decimals for rational numbers) • Geometric representation of natural numbers, integers and rational numbers on a number line • Application of rational number arithmetic to computation in practical situations involving small and large numbers using scientific notation • Definition and properties of real and complex number systems

Review of set notation Subsets of the set of real numbers Properties of surds The set of complex numbers Multiplication and division of complex numbers 1F Representing complex numbers on an Argand diagram 1G Factorising quadratic expressions and solving quadratic equations over the complex number field

• Irrational numbers and complex numbers as roots of quadratic equations • Geometric representation of irrational numbers that are roots of quadratic equations on the real number line and representation of complex numbers on an Argand diagram • Operations with irrational numbers of the form a + b n , where a and b are rational numbers and n is a positive integer that is not a perfect square • Operations with complex numbers of the form a + bi, where a and b are rational numbers and i2 = -1

eBook plus Digital doc

10 Quick Questions

1a

review of set notation Sets contain elements. In this chapter the elements are numbers. For example, the following are six elements: 1, 2, 3, 4, 5, 6. ξ is the universal set — the set of all elements under consideration. So, in this example, ξ = {1, 2, 3, 4, 5, 6}. ∅ is the empty or null set. This set contains no elements. ∅ = {}. An upper case letter, such as A, represents a subset of ξ. In our example, A = {1, 3, 5} and B = {1, 2, 3, 4}. ∈ is read as ‘is an element of’. For example, 3 ∈ A. ∉ is read as ‘is not an element of’. For example, 2 ∉ A. ⊂ is read as ‘is a subset of’. For example, {1, 3} ⊂ A. ⊃ is read as ‘is a superset of’. For example, A ⊃ {1, 3}. Related symbols, such as ⊇, ⊆ and ⊄, are also used.

Chapter 1

number systems: real and complex

1

A′ is the complement of A. This set contains all the elements not in A that are in ξ. For example, given ξ = {1, 2, 3, 4, 5, 6}, if A = {1, 3, 5}, then A′= {2, 4, 6}. A ∪ B is the union of A and B. This set contains all the elements in sets A and B. For the example above, A ∪ B = {1, 2, 3, 4, 5}. A ∩ B is the intersection of A and B. This set contains all the elements in both A and B. For the example above, A ∩ B = {1, 3}. C \ D is read as ‘C slash D’. This set contains all the elements in C that are not in D. If C = {1, 2, 5, 6} and D = {2, 5}, then C \ D = {1, 6}. This notation is particularly useful in modifying a given set to exclude a small number of elements. A Venn diagram may be used to illustrate set notation. ξ

A

B 1

2

3

4

5 6

Worked Example 1

ξ = {2, 4, 6, 8, 10, 12}. C = {4, 8, 12}. D = {2, 6, 10, 12}. a Illustrate these sets on a Venn diagram. Then state: b C′ c C ∪ D d C ∩ D e (C ∪ D)′ f C′ ∩ D′ Think a Draw a Venn diagram, and enter the elements in the

appropriate region.

g C′\{2}.

Write a

C

ξ

D

4

2 12

8

b The set C′ is the complement of C and contains all

elements not in the set C.

10

b C′ = {2, 6, 10}

c The set C ∪ D is the union of C and D and contains

c C ∪ D = {2, 4, 6, 8, 10, 12}

d The set C ∩ D is the intersection of C and D and

d C ∩ D = {12}

e The set (C ∪ D)′ is the complement of the union of

e (C ∪ D)′= ∅

all elements in sets C and D.

contains elements common to sets C and D.

sets C and D. It contains elements not in the union of sets C and D. In this case, there are no elements not in the union of sets C and D.

f The set C′∩ D′ is the intersection of C′ and D′. It

contains elements common to the sets C′ and D′. There are no common elements to C′ and D′. g The set C′\{2} is the set of C′ without the element 2.

It contains all the elements of the set C′ but not 2.

2

6

f C′ ∩ D′ = ∅

g C′\{2} = {6, 10}

Maths Quest 11 Advanced General Mathematics for the Casio ClassPad

The set of real numbers Non-signed numbers — natural numbers and fractions Numbers were invented to quantify objects in the environment. Hunter–gatherers used counting or natural numbers to communicate how many of a particular animal were seen on a hunting trip. The set of natural numbers is given as N = {1, 2, 3, . . .}.

The set of fractions is given as D. D includes 1 , 1 , 3 , 7 , 121 , . . . , where fractions greater than 1 2 3 4 3

16

may be expressed as an improper fraction or as a mixed number. As 3 or 9 are also fractions, 3 3 N ⊂ D. Fractions were invented to divide an object into equal parts or a number of objects into equal parts. For practical applications in the ancient world, the set of fractions (D) was all that was needed. Mathematical oddities occurred, such as a proof that 2 could not be expressed as a fraction, but these were considered more of a philosophical or mystical argument than a practical use of numbers.

Signed numbers — integers and rational numbers The systematic consideration of the concept of number in algebra, and the numbers required to solve equations of the form x + 2 = 0 and 3x + 1 = 0, resulted in the invention of integers and rational numbers. The set of integers is given by Z = {. . . , -3, -2, -1, 0, +1, +2, +3, . . .}. Note: When we are dealing with signed numbers the positive signs are usually omitted, that is, Z = {. . . , -3, -2, -1, 0, 1, 2, 3, . . .}. Z - is the set of negative integers: Z - = {. . . , -3, -2, -1}. Z + is the set of positive integers: Z + = {1, 2, 3, . . .}. Therefore, Z = Z - ∪ {0} ∪ Z +. p The set of rational numbers is given by Q. These are numbers of the form , where p ∈ Z and q q ∈ Z \{0}. Consistent with the definition of Q, Z ⊂ Q. Q- is the set of negative rational numbers. Q+ is the set of positive rational numbers. Therefore, Q = Q- ∪ {0} ∪ Q+.

Chapter 1 Number systems: real and complex

3

Was Q the largest set of numbers that could be mapped on a number line? The Greeks had shown that 2, one of the solutions of x2 = 2, was not rational, and further evidence that Q was not the largest set of numbers that could be mapped onto a number line came from writing fractions in decimal form. Rational numbers in their simplest form with denominators such as 2, 4, 5, 8, 10 produce terminating decimals. Some examples include: 3 7 89 123 = 0.375, = 0.4375, = 0.712, = 1.9921875 8 16 125 64 Rational numbers in their simplest form with denominators such as 3, 6, 7, 9, 11, 13, 14, 15, 17 produce non-terminating recurring or repeating decimals. Some examples include: 1 3 17 99

= 0.333… = 0.3 , 1 = 0.1666… = 0.16 , 6

5 12

= 0.42666… = 0.426

  , 3 = 0.428571428571… = 0.428571   , 17 = 1.307692   = 0.171717… = 0.17 13 7

Division by zero Are 2 and 0 rational numbers? 2 is the solution of 0x = 2, for which there are no solutions. 0 is 0

0

0

0

the solution of 0x = 0, for which there are an infinite number of solutions. As neither case gives a unique rational number solution, then division by zero is not defined. The 00 case is sometimes referred to as the indeterminate case. Worked Example 2

Using a calculator, express the following rational numbers in decimal form. a

5 16

4 7

b c

17 12

Think

Write

a Since the denominator is 16, expect a terminating

a

5 16

1

Since the denominator is 7, expect a non-terminating, repeating decimal.

b

4 7

= 0.5714285714…

2

Indicate the repeating sequence using dot notation.

4 7

  = 0.571428

1

Since the denominator is 12, expect a nonterminating, repeating decimal.

2

Indicate the repeating sequence using dot notation.

decimal.

b

c

c

= 0.3125

17 12

= 0.41666…

17 12

= 0.416

Irrational numbers Irrational numbers are given by I. They are numbers that can be placed on a number line and may be expressed as non-terminating, non-recurring decimals. For example: -

1 3 3 , 5 + 1, 4 3, 5 4, π . 2 p Irrational numbers cannot be written in the form , where p ∈ Z and q ∈ Z \{0}. q -

4

2,


Many irrational numbers in decimal form, such as 2 and π, have digits that have no pattern. For these numbers it is impossible to predict the next digit from the preceding digits. However, other irrational numbers can be constructed with a pattern; for example: 0.10110011100011110000… and 0.01011011101111… There are two important subsets of the set of irrational numbers: the set of algebraic numbers and the set of transcendental numbers. Algebraic numbers are those that are the solution of an algebraic polynomial equation of the form: an x n + an - 1 x n - 1 + … + a2 x 2 + a1 x + a0, 1

where a0, a1, a2, …, an - 1, an ∈ Z. For example, algebraic numbers include 3 3 from one of the 3

solutions of x3 - 3 = 0 and 2 4 from x4 - 8 = 0. Transcendental numbers occur in the evaluation of some functions, such as trigonometric functions. For example, sin (32°4′) and p are transcendental numbers. The functions that produce these numbers are often called transcendental functions.

Real numbers Finally, the set of real numbers is given as R. R includes all numbers that can be put on a number line, where R = Q ∪ I. The Venn diagram shows the relationships between R, Q, I, Z and N. ξ = R (Real numbers)

Q (Rational numbers) Z (Integers) I (Irrational numbers)

N (Natural numbers)

Worked Example 3

For each of the numbers below, using R, Q, I, Z and N, state all the sets for which they are a member. - 17   a -5 b c 3 2 d 27.179 e 4.153 3

f 17.1354… Think a b

-5 - 17 3

is an integer. is a rational number, as it can be written as a

fraction. 3 2 is an irrational number. c   is a rational number, as it is a recurring d 27.179 decimal. e 4.153 is a rational number, as it is a terminating

decimal.

1

g 1.011011101111…

h 32 5

1

i 17 4

Write a b

-5

is a negative integer (Z -). It is also a rational number (Q) and a real number (R). - 17 3

is a rational number (Q) and a real

number (R). c 3 2 is an irrational number (I) and a real number (R).   is a rational number (Q) and a real d 27.179 number (R). e 4.153 is a rational number (Q) and a real

number (R).


5

f 17.1354… is an irrational number as there is no

f 17.1354… is an irrational number (I ) and a

indication that there is a recurring pattern.

real number (R).

g 1.011011101111… is an irrational number.

g 1.011011101111… is an irrational number (I)

and a real number (R).

1

1

h 32 5 can be simplified to 2 and is therefore a

h 32 5 is a natural number (N). It is also an

natural number.

integer (Z ), a rational number (Q) and a real number (R).

1

1

i 17 4 is an irrational number.

i 17 4 is an irrational number (I ) and a real

number (R).

Worked example 4

eBook plus

a Express each of the following in the form , where a ∈ Z and b ∈ Z \{0}. b    a 0.6 b 0.23 c 0.41 d 2.1234 Think a

b

c

d

6

Tutorial

int-1211 Worked example 4

WriTe

1

Write 0.6 in expanded form.

2

Multiply [1] by 10.

3

Subtract [1] from [2].

4

State the simplest answer.

1

  in the expanded form. Write 0.23

2

Multiply [1] by 100.

3


4


1

Write 0.41 in the expanded form.

2

Multiply [1] by 10.

3


4


1 2

  in the expanded form. Write 2.1234 Multiply [1] by 1000.

3


4


a

0.6 = 0.666666… 10 × 0.6 = 6.66666…

[1] [2]

9 × 0.6 = 6 6 0.6 = 9 2 = 3 b

  = 0.232323… 0.23   = 23.232323… 100 × 0.23

[1] [2]

  = 23 99 × 0.23

c

  = 23 0.23 99 0.41 = 0.41111… 10 × 0.41 = 4.11111… 9 × 0.41 = 3.7

3.7 37 = 9 90   2.1234 = 2.1234234… d   = 2123.423423… 1000 × 2.1234

[1] [2]

0.41 =

  = 2121.3 999 × 2.1234   = 2121.3 2.1234 999 21213 = 9999 2357 = 1111

maths Quest 11 advanced General mathematics for the Casio Classpad

[1] [2]

Note: The CAS calculator can perform some of these calculations for you. On the Main screen, tap: • Action • Transformation • toFrac Complete the entry line as: 0.6666666666667 Then press E.

The basic properties of number are assumed to be true if a counterexample cannot be found. For example, the statement ‘the product of two integers is an integer’ is accepted as true because a counterexample has not been found, but the statement ‘the quotient, when defined, of two integers is an integer’ is false because a counterexample  2  is not an integer. 3

Worked Example 5

Determine counterexamples for the following. a The product of two irrational numbers is irrational. b The sum of two irrational numbers is irrational. Think a

b

Write

Take a simple irrational number such as 2. Multiply by another irrational number, say 2. State your answer. 1

2

Take two irrational numbers such as 0.101100111000… and 0.010011000111…. Add these numbers. State your answer.

a Because

2 × 2 = 2, which is a rational number, the statement ‘the product of two irrational numbers is irrational’ is shown to be false.

b 0.101100111000… + 0.010011000111… =

0.111111111111…

Because 0.111111111111… is a rational number, the statement ‘the sum of two irrational numbers is irrational’ has been shown to be false.

Standard form or scientific notation Very large or very small numbers are conveniently expressed in standard form, a × 10b, where a ∈ R and 1 ≤ a < 10 and b ∈ Z. For example, 1234111 = 1.234111 × 106 and 0.000000000045 = 4.5 × 10-11.

Precision of answers — significant figures and decimal places In everyday life, the exact answer to a problem is not always required. For instance, if you were asked how long it took you to get to school that morning, you would not be expected to give the answer accurate to hundredths of a second. A reasonable answer to such a question would be in minutes.


7

The numerical answer to a calculation may be required to be given correct to a set number of decimal places, and this is done through a process of rounding. To determine the number of decimal places contained in a number, count the number of digits after the decimal point. For example, 0.35 has 2 decimal places. For numbers expressed to a given number of decimal places, remember to round up if the next digit is 5 or more. For example, rounded to 2 decimal places, 2.234 becomes 2.23 and 2.236 becomes 2.24. To determine the number of significant figures contained in a number, count the number of digits from the first non-zero digit. For example, 0.035 contains 2 significant figures. Any zeros at the end of a number after a decimal point are considered to be significant. For example, 1.40 has 3 significant figures. The trailing zeros at the end of a number are not considered to be significant. For example, 24 000 has 2 significant figures. For numbers expressed to a given number of significant figures, remember to round. For example, rounded to 2 significant figures, 2.234 becomes 2.2 and 2.236 also becomes 2.2. Some examples are shown in the following table. 2 significant figures

3 significant figures

2 decimal places

3 decimal places

471 860.237 8

470 000

472 000

471 860.24

471 860.238

1.238 9

1.2

1.24

1.24

1.239

1.006 8

1.0

1.01

1.01

1.007

0.016 78

0.017

0.016 8

0.02

0.017

0.00

0.002

0.20

0.199

Number

0.001 556 0.199 1

0.0016 0.001 56 0.20

0.199

Worked Example 6

Calculate the following products and quotients without using a calculator, expressing your answer in scientific notation correct to 1 significant figure. -10

a 8 × 1024 × 3 × 10

b

7 × 1017 8 × 10 10

Think a

b

8

Write

1

Multiply the terms by using the properties of indices: an × am = an + m.

a 8 × 1024 × 3 × 10-10 = 24 × 1014

2

Write the answer in standard form, correct to 1 significant figure.

1

Multiply the terms by using the properties of indices: an ÷ am = an - m.

24 × 1014 = 2.4 × 10 × 1014 = 2.4 × 1015 = 2 × 1015 17 b 7 × 10- = 0.875 × 10 27 8 × 10 10

2

Write the answer in standard form, correct to 1 significant figure.

0.875 × 1027 = 0.9 × 1027 or = 9 × 1026


REMEMBER

1. Recall and apply the definitions of ξ, ∅, ∈, ⊂, A′, A ∪ B, A ∩ B and C \D, and how to illustrate these on a Venn diagram. 2. Recall and apply the definitions of the following sets of numbers: N, D, Z, Q, I and R and their relationships. 3. Rational numbers, expressed as decimals, are either terminating or non-terminating recurring. Irrational numbers, expressed as decimals, are non-terminating and non-recurring. 4. Division by zero is not defined. 5. Given a non-terminating, non-repeating decimal, use the process to express the number p in the form of a rational number. q 6. Do simple calculations in scientific notation by hand and more complex calculations using a calculator. 7. Express answers to the stated number of significant figures or decimal places. Exercise

1A

Review of set notation 1 WE 1 If ξ = {1, 2, 3, 4, 5, 6}, A = {1, 2} and B = {2, 3}, show these on a Venn diagram, and then state the following sets. a A′ b A ∪ B c A ∩ B d A\{2} e A ∪ B′ f A′ ∪ B′ g (A ∩ B)′ 2 WE2 Use a calculator to express the following rational numbers in decimal form. 213 15 16 a b c 64 44 13 3 WE3 For each number below, using J, Q, I and R state all the sets in which they are a member. - 2,

1 1 16 21 - 2   , 4.135218976…, 4.232332333… , , 3 , 6 3 , 16 4 , 5 5 , π , 21.72, 2.567 8 16 7

4 WE4  a 0.24

Express the following in their simplest rational form.    b 1.123 c 0.123

  d 2.1123

5 WE5 Find a counter-example, if possible, for the following statements. If a counter-example is found, the statement is false (F). If a counter-example is not found, accept the statement to be true (T). a The product of two integers is an integer. b The division of an integer by an integer is a rational number. c The difference of two irrational numbers is irrational. d The sum of an irrational number and a rational number is irrational. 6 Copy the Venn diagram at right and then shade the region represented by each of the following sets. A B ξ a A′ b A ∪ B c A ∩ B d (A ∪ B) \ (A ∩ B) e A′ ∩ B f A′ ∩ B′ g (A ∪ B)′


9

7 Complete the following table. Number



2 decimal places

3 decimal places

1267.1066 7.6699 8.000 56 0.999 87 0.076 768 0.000 174 95 Calculate the following products and quotients without using a calculator, 8 We6 expressing your answer in scientific notation to 1 significant figure. a 1.5 × 1016 × 4 × 1012

b 1.2 × 1024 × 3 × 10-10

d 7 × 1014 × 9 × 10-8

e

8 × 1017 4 × 10 10

c 3.2 × 1025 × 2 × 1015 f

2.5 × 1012 × 8 × 10 5 × 108

-7

9 Calculate the following products and quotients using a calculator, expressing your answer in scientific notation to 3 significant figures. b 8.2583 × 1025 × 9.2527 × 10-7

a 1.4574 × 1021 × 3.6677 × 109 c

-

5.7789 × 1017 4.6999 × 1010

d

-

2.578 × 1012 (8.775 × 10 7 + 7.342 × 10 6 ) 5.878 × 1013

10 mC The smallest subset of R in which 4 - 2 3 27 belongs is: a Z+ b Zc Q+ D Q-

E I

9 4

  belongs is: 11 mC The smallest subset of R in which 4.4567 a Z+ b Zc Q+ D Q-

E I

12 mC If ξ = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 4} and B = {5}, then A′\ B is: c ∅ a {1, 2, 3, 4, 5} b {5, 6, 7, 8} D {6. 7. 8} E {1, 2, 3, 4, 5, 6, 7, 8} 13 mC 3.0102 and 92457 to 4 significant figures are: a 3.01 and 92450 b 3.010 and 92450 c 3.01 and 92460 D 3.010 and 92460 E 3.0102 and 92457   and 0.232 233 222 333… respectively belong to the following sets. 14 mC 0.23, 0.23 a Z, Z, I b Q, Q, I c Q, I, I D Z, Q, I E Q, Q, Q

1B

Subsets of the set of real numbers


WorkSHEET 1.1

notation Mathematics is a form of communication. While the mathematical concepts can be unambiguously defined, there are often difficulties in communicating these concepts because of differences in notation. There is no international body that defines notation that is to be used worldwide. There can be significant differences between nations and regions. The development of notations in mathematics may be likened to the chaotic international development of the

10


English language. Just as with the English language, mathematical notations change with time and geography. At the secondary-school level, there is some attempt to standardise the notations used. As a consequence, students are exposed only to some commonly used notations. At universities, however, there is a proliferation of notations because international references are used. How big is the problem? It is more an annoyance than a problem. Consider 2.4, 2⋅4 and 2,4. All represent the same decimal number but in some countries 2⋅4 = 2 × 4. Then again, others use 2.4 = 2 × 4.

Some notations for subsets of the set of real numbers There are different forms of notation for representing subsets. 1. Describing, for example, {integers between and including 1 and 5} 2. Listing, for example, {1, 2, 3, 4, 5} 3. Set builder notation, for example, {x: x ∈ Z, 1 ≤ x ≤ 5}, which is read as ‘the set of numbers x such that x is an element of the set of integers and x is greater than or equal to 1 and less than or equal to 5’. If x ∈ R, it is not necessary to include the nature of x. For example, {x: x ≥ 2} represents the set of real numbers greater than or equal to 2. The two sets above may be represented on a number line as follows. −2 −1 0 1 2 3 4 5 6 7 x ∈ z −2 −1 0 1 2 3 4 5 6 x

If x ∈ Q, the graph on the number line appears to look like the corresponding graph for x ∈ R because the number line appears to be continuous (although all irrational numbers are missing). For example, {x: x ∈ Q, x ≥ 2} would appear to be identical to the graph of {x: x ≥ 2} shown above. If individual numbers are excluded from a given set, indicate this on a number line by an open point. For example, {x: x ≥ 2}\{3} is represented on a number line below. −3 −2 −1 0 1 2 3 4 x

A given set can be stated in more than one way using set builder notation. For example, {1, 2, 3, 4, 5} can be written as {x: x ∈ Z, 0 < x < 6}, {x: 1 ≤ x ≤ 5} or {x: x ∈ Z +, x ≤ 5}. 4. Interval notation is used for subsets of the set of real numbers. The definition of interval notation is shown below. These sets are also illustrated on a number line. a

b

x

[a, b] = {x : a ≤ x ≤ b}

a

b

x

[a, b) = {x : a ≤ x < b}

a

b

x

(a, b) = {x : a < x < b}

a

x

(−∞, a] = {x : x ≤ a}

a

x

(a, ∞) = {x : x > a}


11

Notes 1. If a terminal point is included, a closed dot is shown on the number line. 2. If the terminal point is not included, an open dot is shown. 3. If the terminal point is ±∞, an arrowhead is shown and the notation uses a round bracket, for example, [a, ∞). Worked Example 7

List the following sets and then express each set using set builder notation. Illustrate each set on a number line. a {Integers between -3 and 4} b {Integers less than 2} Think a

b

Write

1

This set involves integers. List the set of integers. Express the set using set builder notation.

2

Draw a number line showing arrowheads on each end. Ensure that the numbers from -3 to 4 are shown, using an appropriate scale. Since, the set of integers is to be represented, do not join the dots.

1

This set involves integers. List the set of integers. Express the set using set builder notation.

2

Draw a number line showing arrowheads on each end. Ensure that the numbers from -3 to 4 are shown, using an appropriate scale. Since, the set of integers is to be represented, do not join the dots, but show an arrow on the left side of -3.

a {-2, -1, 0, 1, 2, 3} = {x: x ∈ Z, -3 < x < 4}

−4 −3 −2 −1 0 1 2 3 4 5 6 x

b {…, -2, -1, 0, 1} = {x: x ∈ Z, x < 2}

−3 −2 −1 0 1 2 3 4 x ∈ z

Worked Example 8

Use set builder notation to represent the following sets. a {Rational numbers greater than 27} b {Integers between and including both 100 and 300, except for 200} c {Positive integers less 9 and greater than 50} d {Real numbers from 1 to 3, including 3} e {Real numbers that are less than 7 and greater than 2} f {Positive real numbers that are less than 2 or greater than 7} Think

Write

a The numbers in this set belong to the set of rational

a {x: x ∈ Q, x > 27}

b The numbers in this set belong to the set of integers,

b {x: x ∈ Z, 100 ≤ x ≤ 300}\{200}

c The numbers in this set belong to the set of positive

c {x: x ∈ Z +, x < 9} ∪ {x: x ∈ Z +, x > 50}

numbers, Q.

Z. Exclude 200. Z +.

integers, Express the set of positive integers less 9 and greater than 50 as the union of two sets.

12


d The numbers in this set belong to the set of real

d {x: 1 < x ≤ 3}

e The numbers in this set belong to the set of real

e {x: x < 7} ∩ {x: x > 2} or, more simply,

f The numbers in this set belong to the set of positive

f {x: x < 2} ∪ {x: x > 7} or

numbers, R, from 1 to 3, including 3.

numbers, R, that are less than 7 and greater than 2. real numbers, R+, that are less than 2 or greater than 7.

{x: 2 < x < 7}

R\{x: 2 ≤ x ≤ 7}

Worked Example 9

Use interval notation to represent the following sets. a {x: -2 < x ≤ 3} b {x: x ≤ 4} c {x: 3 < x ≤ 5} ∪ {x: 4 ≤ x < 7} d {x: 3 < x ≤ 5} ∩ {x: 4 ≤ x < 7} Think

Write

a x ∈ R. Only the end point 3 is included; therefore,

a (-2, 3]

b x ∈ R. Negative infinity is always preceded by a

b (-∞, 4]

c x ∈ R. Only the inner end points are included.

c (3, 5] ∪ [4, 7)

d x ∈ R. Only inner end points are included.

d (3, 5] ∩ [4, 7) = [4, 5]

use a square bracket.

round bracket when using interval notation.

REMEMBER

1. Set builder notation. For example, {x: x ∈ Q, -3 < x ≤ 17}. If x ∈ R, the set does not have to be stated. For example, {x: -3 < x ≤ 17}. 2. Interval notation. For example, (-3, 17] = {x: -3 < x ≤ 17}, [-3, ∞) = {x: x ≥ -3}. This notation can only be used if x ∈ R. 3. All subsets of R can be illustrated on a number line. Exercise

1B

Subsets of the set of real numbers 1 WE7 List the following sets and then express each set using set builder notation. Then illustrate each set on a number line. a {Integers between -6 and 1} b {Integers from -3 to 4} c {Integers greater than -6 and less than or equal to 4} d {Positive integers less than 5} e {Integers less than 5} f {Integers greater than 2} g {Negative integers greater than -5} 2 WE8 a b c d

Use set builder notation to represent the following sets.

{Rational numbers greater than 5} {Rational numbers greater than 5 and less than or equal to 20} {Positive rational numbers less than 20} {Integers between 5 and 20, except for 8 and 9}


13

e f g h i

{Positive integers less than 100, except for integers between 40 and 50} {Real numbers from 2 to 5, including 2} {Real numbers which are less than 5 and greater than 3} {Real numbers which are less than 3 and greater than 7} {Positive real numbers which are less than 3 and greater than 7}

3 WE9 Use interval notation to represent the following sets; then illustrate the sets on a number line. b {x: x < 2} a {x: -3 ≤ x ≤ 1} c {x: -2 < x < 1} d {x: x ≥ 2} e {x: 2 ≤ x < 5} ∪ {x: 4 ≤ x < 6} f {x: x < 5} ∪ {x: 4 ≤ x < 6} g {x: 2 ≤ x < 5} ∩ {x: 4 < x ≤ 6} h {x: x > 5} ∩ {x: 4 < x ≤ 6} 4 MC Which of the following sets is an incorrect representation of the set {all integers from 1 to 5}? a {1, 2, 3, 4, 5} b {x: x ∈ Z, 1 ≤ x ≤ 5} c {x: x ∈ Z, 1 ≤ x < 6} E [1, 5] D Z +\{x: x ∈ Z, x ≥ 6} 5 MC For the set illustrated on the given number line, which of the following cannot be true? −5 −4 −3 −2 −1 0 1 2 3 4 5 x

a (-5, 5] b {x: -5 < x ≤ 5} D {Real numbers from 5 to 5, not including -5}

1C

c {x: x ∈ Q, -5 < x ≤ 5} e [-5, 5]

Properties of surds A surd is any irrational number of the form n a , where a > 0 and n ∈ Z +. In this section we will focus on the surds of the form a , where a ∈ Q. For example, 21 is a surd, but 36 = 6 is a rational number and not a surd.

Simplifying surds 2 cannot be simplified because it does not have a perfect square factor, but

8 can be

simplified since 8 = 4 × 2 = 4 × 2 = 2 × 2 = 2 2 . A surd is not simplified until all perfect square factors are removed, so the simplified version of 32 is not 2 8 but 4 2. Worked Example 10

Simplify the following surds. (Assume that x and y are positive real numbers.) a

384 b 3 405 c

-1 8

175

Think a

Write

384 = 64 × 6

1

Express 384 as a product of two factors where one factor is the largest possible perfect square.

2

Express 64 × 6 as the product of two surds.

= 64 × 6

3

Simplify the square root from the perfect

=8 6

a

square (that is, 64 = 8). 14


b

c

b 3 405 = 3 81 × 5

1

Express 405 as a product of two factors, one of which is the largest possible perfect square.

2

Express 81 × 5 as a product of two surds.

= 3 81 × 5

3

Simplify 81.

= 3×9 5

4

Multiply together the whole numbers outside the root (3 and 9).

= 27 5

1

Express 175 as a product of two factors where one factor is the largest possible perfect square.

2

Express 25 × 7 as a product of 2 surds.

=

3

Simplify 25.

=

4

Multiply together the numbers outside the square root sign.

=

c

-1

175 =

8

-1 8 -1 8 -1 8 -5 8

25 × 7 × 25 × 7 ×5 7 7

Addition and subtraction of surds Only like surds may be added or subtracted. Like surds, in their simplest form, have the same number under the square root sign. For example, 5 3 + 7 3 = (5 + 7) 3 = 12 3 and 5 3 - 7 3 = (5 - 7) 3 = - 2 3 .

Worked Example 11

Simplify each of the following expressions involving surds. Assume that a and b are positive real numbers. a 3 6 + 17 6 - 2 6 b 5 3 + 2 12 - 5 2 + 3 8 c Think a

b

100 a 3 b2 + ab 36 a - 5 4 a2 b

Write

All three terms are like, since they contain the same surd ( 6 ), so group the like terms together and simplify. 1

1 2

Simplify the surds where possible.

a 3 6 + 17 6 - 2 6 = (3 + 17 - 2) 6

= 18 6 b 5 3 + 2 12 - 5 2 + 3 8

= 5 3+2 4×3-5 2 +3 4×2 = 5 3 +2×2 3 -5 2 +3×2 2 = 5 3+4 3-5 2 +6 2 2

c

1

Collect the like terms. Simplify the surds where possible.

=9 3+ 2 c

1 2

100 a3 b 2 + ab 36a - 5 4 a 2 b = 1 × 10 a 2 × a × b 2 + ab × 6 a - 5 × 2 × a b 2

1 2

= × 10 × a × b a + ab × 6 a - 5 × 2 × a b = 5ab a + 6ab a - 10 a b 2

Add the like terms.

= 11ab a - 10 a b


15

multiplication of surds Using the property a × b = ab , where a, b ∈ R+, 2 × 6 = 12 = 4 × 3 = 2 3 . Using the distributive property a(b + c) = ab + ac, 2 ( 3 + 6 ) = 2 × 3 + 2 × 6 = 6 + 12 = 6 + 2 3 . Using an extension of the distributive property, ( 3 + 1) ( 3 - 2) = 3 × 3 - 2 3 + 3 - 2 = 3 - 3 - 2 = 1 - 3 . When appropriate, the expansion of a perfect square may be used; that is, (a + b)2 = a2 + 2ab + b2 and (a - b)2 = a2 - 2ab + b2. For example, ( 3 - 2 )2 = 3 - 2 3 × 2 + 2 = 5 - 2 6

definition of the conjugate The conjugate of a + b is a - b The conjugate of 3 - 2 5 is 3 + 2 5 . The product of a conjugate pair is rational if the numbers under the square root are rational. For example, ( 3 + 2 )( 3 - 2 ) = 3 × 3 - 3 × 2 + 2 × 3 - 2 × 2 = 3 - 6 + 6 - 2 = 1. This is a special case of the difference of perfect squares expansion, (a + b)(a - b) = a2 - b2.

Worked example 12

eBook plus

Multiply the following surds, expressing answers in simplest form.

Tutorial

a 6 12 × 2 6 3 5

b

70 × 14 10

Think a


1

WriTe

Simplify 12 .

a 6 12 × 2 6 = 6 4 × 3 × 2 6

=6×2 3 ×2 6 = 12 3 × 2 6 2

Multiply the coefficients and multiply the surds.

= 24 18

3

Simplify the product surd.

= 24 9 × 2 = 24 × 3 2 = 72 2

b

1

2

Multiply the coefficients and multiply the surds.

b

3 5

1

3

1

70 × 4 10 = 5 × 4 × 70 × 10

Simplify the product surd.

3

= 20 700 3

= 20 100 × 7 3

= 20 × 10 7 3

16

Simplify by dividing both 10 and 20 by 10 (cross-cancel).


3 = 2 7 or 3 7 2

Worked Example 13

Expand and simplify the following where possible. a

7 ( 18 - 3) b - 2 3 ( 10 - 5 3 ) c ( 5 + 3 6 )( 2 3 - 2 )

Think

Write

Method 1: Using the rule a

b

7 ( 18 - 3)

1

Write the expression.

2

Simplify 18.

= 7 (3 2 - 3)

3

Expand the bracket.

= 7 × 3 2 + 7 × -3

4

Simplify.

a

= 3 14 - 3 7 -2

3 ( 10 - 5 3 )

1


2

(a) Expand the brackets.

= - 2 3 × 10 - 2 3 × -5 3

(b) Be sure to multiply through with the negative.

= -2 30 × 10 9

Simplify.

= -2 30 + 10 × 3

3

b

= -2 30 + 30 c

1


2

Expand the brackets.

c ( 5 + 3 6 ) (2 3 -

= 5 ×2 3 + 5 × - 2 +3 6 ×2 3 + 3 6×

3

Simplify.

2)

-

2

= 2 15 - 10 + 6 18 - 3 12 = 2 15 - 10 + 6 × 3 2 - 3 × 2 3 = 2 15 - 10 + 18 2 - 6 3

Method 2: Using a CAS calculator a,b 1 & c

On the Main screen complete the entry lines as: 7 ( 18 - 3) 3 ( 10 - 5 3 )

-2

( 5 + 3 6 )(2 3 - 2 ) Press E after each entry.

2

Write the answers.

a 3( 2 - 1) 7 b 30 - 2 30 c (2 3 -

2 ) 5 - 6 3 + 18 2


17

Division of surds a b

=

a 6 6 , where a, b ∈ R+. For example, = = 3 b 2 2

a b ab × = , where a and b are rational, we can express b b b b answers with rational denominators. For example, Using the property

a

=

2

=

2

6

×

=

12 2 3 3 = = 6 6 3

6 6 6 Using the property of conjugates, binomial surds in the denominator may be rationalised. For example, 7-2 2 7+ 2

7-2 2

=

7+ 2

×

By multiplying the original surd by

7- 2 7- 2

=

7 - 14 - 2 14 + 2 × 2 11 - 3 14 = 7-2 5

7- 2

, we are multiplying by 1, so the number is 7- 2 unchanged but is finally expressed in its rational denominator form. Worked Example 14

Express the following in their simplest form with a rational denominator. Assume that x and y are positive real numbers. a

9 88 6 99

b

6 13

c

1 2 6- 3

+

1 3 6 +2 3

Think a

b

Write

=

9 88 6 99

Simplify the fraction under the root.

=

9 8 6 9

3

Simplify the surds.

=

9×2 2 6×3

4

Multiply the whole numbers in the numerator and those in the denominator.

= 2

1

Write the fraction.

2

Multiply both the numerator and the denominator by the surd 13 or 1.

1

Rewrite the surds, using

2

a b

=

a . b

a

b

9 88 6 99

6 13 = =

c

18

1

Write the first fraction.

2

Multiply the numerator and the denominator by the conjugate of the denominator.

c

= =


6 13

×

13 13

78 13 1 2 6- 3 1 2 6- 3

×

2 6+ 3 2 6+ 3

2 6+ 3 (2)2 × 6 - 3

3

Expand the denominator.

=

4

Simplify the denominator.

=

5

Write the second fraction.

6

Multiply the numerator and the denominator by the conjugate of the denominator.

=

7

Expand the denominator.

=

8

Simplify the denominator.

=

9

Add the two fractions together. Find the lowest common denominator first.

2 6+ 3 3 6-2 3 + 21 42

2 6+ 3 21 1

3 6+2 3 1 3 6+2 3

×

3 6-2 3 3 6-2 3

3 6-2 3 × 6 - 22 × 3

32

3 6-2 3 42

=

2 6 + 3  2 3 6 - 2 3 ×  +  2 21 42

=

4 6 +2 3 3 6 -2 3 + 42 42

10

Add the numerators.

=

7 6 42

11

Simplify where appropriate.

=

6 6

REMEMBER

1. To simplify a surd, take out all perfect square factors from the number under the root sign. 2. Like surds have the same number under the root sign when expressed in its simplest form. 3. Only like surds can be added and subtracted. 4. Multiplication properties: a × b = ab , a ( b + c ) = ab + ac , and ( a + b )( c + d ) = ac + ad + bc + bd 5. The conjugate of a + b is a - b . 6. Special expansions: (a) Expansion of perfect squares: (a + b)2 = a2 + 2ab + b2 and (a - b)2 = a2 - 2ab + b2 (b) Difference of perfect squares: (a + b)(a - b) = a2 - b2 7. Division properties: ( a + b) ( c + d)

=

a b

( a + b) ( c + d)

= ×

a a , = b b ( c - d) ( c - d)

a b ab × = b b b =

ac - ad + bc - bd , using the conjugate c-d

of c + d .


19

Exercise

1c

Properties of surds 1 WE10 a f

Simplify the following surds. b

24

g 7 80

300

2 WE11

c

56

h

d

125 128 4

i

e

98 2 18 5

48 -

j

3 50 10

Simplify the following expressions.

a 7 2 + 4 3 - 5 2 - 6 3

b 2 + 5 7 - 6 - 4 7

c 3 5 - 6 3 + 5 5 - 4 2 - 8 5

d 18 - 12 + 75 + 27

e

50 - 72 + 80 + 45

g

2 3 3 2 5 3 5 2 + 4 8 8 4

3 WE12

f 3 12 - 5 18 + 4 27 + 5 98 h

2 27 3 32 5 48 5 2 + 5 5 3 2

Express the following surds in their simplest form.

a 6 × 15

b 2 3 × 5 7

c 4 7 × 3 14

d

20 15 × 3 4

4 WE13 Expand, giving your answers in their simplest form. b

-

c 2 3 (3 3 + 2 )

d

-4

e ( 5 - 3 )( 5 - 2 )

f (3 3 - 2)(2 5 + 3)

g ( 18 - 12 )( 3 - 2 2 )

h ( 5 - 3 )2

i ( 5 + 7 )2

j (2 3 - 3 2 )2

k (2 12 + 3 18 )2

l ( 5 - 3 )( 5 + 3 )

m (2 5 - 3 )(2 5 + 3 )

n (2 3 + 3 2 )(2 3 - 3 2 )

a 3 ( 5 - 2 )

2( 7 - 6) 3 ( 32 - 3 12 )

o (5 5 - 10)(5 5 + 10) 5 WE14 Express the following surds in their simplest form with a rational denominator. a e i m

18

b

3 4 3

f

7 5 3

j

3- 2 12 - 8

n

12 + 8

2 24

c

3 3 2 8

g

3 12 2 2

k

2 5+3 2 2 5- 3

o

5-2 3

3 88

d

22 5

h

4 2 5- 3

l

5+ 3 3 5-4

p

5+4

5 3 1 5- 3 5+ 3 5- 3 2 18 - 24 3 8 - 54

6 Express the following surds in their simplest form with a rational denominator. 1 1 1 1 a b + 2 2 -3 2 2 +3 3 2 -2 3 2 2 +3 3 c

20

3 5 3 2-2 3

-

2 5 -1 2 2 +3 3

d

2 5+3 3 3 5 -2 3


-

3 5 -2 3 2 5+3 3

e f

4 2 +3 3 4 2-2 3 2 5+3 3 3 5-4 3

× ÷

5 2-2 3 Exam tip When subtracting fractions, be careful of the signs. Make sure you subtract the whole second fraction. Use brackets to help.

6 2 +3 3 3 5+4 3

[Authors’ advice]

2 5+3 3

7 Given that x = 2 - 3 2, find each of the following, giving the answer in surd form with a rational denominator. 1 1 x2 - 2x a x + b x c x x x+2 d

x2 + 2x x+3

e x2 - 4x - 14

f 2x2 - 2x - 9

g Using your answers to e and f, state if 2 - 3 2 is a solution of x2 - 4x - 14 = 0 and 2x2 - 2x - 9 = 0. 8 Show that 5 - 2 3 is a solution of one of the following equations: x2 - 13x + 10 = 0 or x2 - 10x + 13 = 0. 9 Show that 2 + 1 is a solution of both of the following equations: x 2 - 2 2 x + 1 = 0 and x 2 - (2 2 + 3) x + 4 + 3 2 = 0. 10 MC Expressed in its simplest form, a 3

b

-

3

11 MC Expressed in its simplest form, a 2a

3 2 1 75 27 48 = 5 3 2 D 0 c 7 3 14 a3 b 2 7ab 2

=

c 2ab

b 2a

E - 3 3

E

D 2a 2 b 0

2a 2

12 MC Expressed in its simplest form, (3 3 - 4 8 )(2 3 - 3 8 ) = a 114 - 34 6

b 120 - 34 6

D 18 - 24 2

E - 18 - 34 6

13 MC Expressed in its simplest form, a

5 3 2 2

b

15 21 6 14

=

c 5 3 2

5 6 2 2

14 MC Expressed in its simplest form,

2 5+ 3 5- 3

a 13 + 3 15 2

b 12 - 15 2

D 12 + 3 15 2

E 13 + 15 2

15 MC Expressed in its simplest form, a 3 5 D 52 - 30 5

c - 78 - 17 24

E 5 6 2

= c 18 - 3 15 2

3 5-5 3 5+5

b - 30 5

D 5 6 4

-

3 5+5 3 5-5

= c 30 5

-

E 3 5


21

1d

The set of complex numbers

eBook plus

The need to invent further numbers became clear when equations such as x2 = -1 and x2 = -9 were considered. Clearly there are no real solutions, so imaginary numbers were invented, using the symbol i, where i2 = -1. The equation x2 = -1 has two solutions, x = -i and x = i. As

Digital doc

WorkSHEET 1.2

-9

= 9 × - 1 = 9 × - 1 = 3 × i 2 = 3i, x 2 = - 9 has the solutions x = ±3i. Quadratic equations such as x2 - 4x + 5 = 0 were investigated further. Using the general formula for the solution of a quadratic equation, that is, if ax2 + bx + c = 0, then b 2 - 4 ac 4 ± 16 - 20 4 ± - 4 4 ± 2i ,x= = = = 2 ± i. If the discriminant, b2 - 4ac is 2a 2 2 2 negative, the equation has no real solutions, but it does have two complex solutions. A complex number is any number of the form x + yi, where x, y ∈ R. C is the set of complex numbers where C = {x + yi : x, y ∈ R} Just as x is commonly used in algebra to represent a real number, z is commonly used to represent a complex number, where z = x + yi. ξ C If x = 0, z = yi is a pure imaginary number. If y = 0, z = x R is a real number, so that. R ⊂ C. This is represented on the Q Venn diagram at right. x=

-b ±

Z N

I

notation If z = a + bi, the real component of z, Re(z) = a, and the imaginary component of z, Im(z) = b. For example, if z = - 2 - 2 3i, Re(z) = -2 and Im(z) = - 2 3 (not - 2 3i). Similarly, Re(- 2 - 2 3i) = - 2 and Im(- 2 - 2 3i) = - 2 3 )

equality of complex numbers If a + bi = c + di, then a = c and b = d. For two complex numbers z1 and z2 to be equal, their real and imaginary components must be equal. Worked example 15

If (x + 2) + (y - 4)i = (2x + y) + xi, find x and y. Think

22

1

Let the real parts be equal and the imaginary parts be equal to form two equations.

2

Rearrange the linear simultaneous equations.

3

Add equations [3] and [4] to solve for x.

4

Substitute x = -1 into equation [2] to find the value of y.

5

State the answer.

WriTe

Re: x + 2 = 2x + y Im: y - 4 = x -x -x

- y = -2 +y=4 -2x

=2 x = -1

y - 4 = -1 y=3


x = -1, y = 3

[1] [2] [3] [4]

multiplication of a complex number by a real constant If z = a + bi, then kz = k(a + bi) = ka + kbi. For example, if z = -2 + 3i, then -3z = -3(-2 + 3i) = 6 - 9i.

adding and subtracting complex numbers If z1 = a + bi and z2 = c + di, then z1 + z2 = (a + c) + (b + d )i and z1 - z2 = (a - c) + (b - d)i. Worked example 16

eBook plus

-3

If z1 = 2 − 3i and z2 = + 4i, find a z1 + z2 b z1 - z2

Tutorial

c 3z1 − 4z2.

Think


WriTe

Method 1: Using the rule a Use the definition for addition of complex numbers: z1 + z2 = (a + c) + (b + d)i b Use the definition for subtraction of complex numbers: z1 − z2 = (a − c) + (b − d)i c First multiply each complex number by the constant and then use the definition for subtraction of complex numbers to answer the question.

a z1 + z2 = (2 - 3) + (-3 + 4)i

= -1 + i

b z1 - z2 = (2 + 3) + (-3 - 4)i

= 5 - 7i

c 3z1 - 4z2 = 3(2 - 3i) - 4(-3 + 4i)

= 6 - 9i - (-12 + 16i) = 18 - 25i

Method 2: Using a CAS calculator After setting the calculator to Cplx mode, on the Main screen complete the entry lines as: 2 - 3i W x -3 + 4i W y x+y x-y 3x - 4y Press E after each entry. Note: x and y are substituted for z1 and z2.

a,b 1 & c

2

Write the answers.

a z1 + z2 = -1 + i b z1 - z2 = 5 - 7i c 3z1 - 4z2 = 18 - 25i

rememBer

1. Complex numbers are of the form a + bi where i2 = -1. 2. If z = a + bi, then Re(z) = a and Im(z) = b. 3. If z1 = a + bi and z2 = c + di then z1 + z2 = (a + c) + (b + d)i and z1 - z2 = (a - c) + (b - d)i. 4. If a + bi = c + di, then a = c and b = d.

Chapter 1


23

Exercise

1d

The set of complex numbers 1

Express the following in terms of i. a

-

b

d

-

e

2

16 10 + 10

-

c 2 +

7

-

20

- 28

12

State the values of Re(z) and Im(z) for the following. a 3 + 4i d 8 -

3 a c e g

-

40

WE15

b - 2 + 2i

c ( 2 - 1) + ( 2 + 1)i

e -6

f 13i

Solve to find x and y in the following.

(x + 1) + ( y - 1)i = 2 + 3i (2x + i) + (3 − 2yi) = x + 3i (2x + 3yi) + 2(x + 2yi) = 3 + 2i (2x − 3i) + (−3 + 2y)i = y − xi

4 WE16

b (x + 4) − (3 + yi) = 2 + 5i d (x + 2i) + 2(y + xi) = 7 − 4i f (x + i) + (2 + yi) = 2x + 3yi

If z1 = 3 − 4i and z2 = 2 − 3i, evaluate the following.

a z1 + z2

b z1 - z2

d 2z1 - 3z2

e

2 z1 + 2 2 z2

c 2z1 + 3z2 f

2 z1 + 3z2

5 Find the following components. a Re(2 + 3i + 3(4 - 2i)) b Re( 3 + 2 2i + 2 (- 3 - 3i )) c Im(2(2 - 3i) - 3(4 - 2i)) d Im(2 3 - 2 2i + 2 (- 3 - 6i )) 6 MC If z1 = 2 - i and z2 = 3 - 2i, then Re(2z1 - 3z2) = a 13 b -13 d -5 e 4

c 5

7 MC If z1 = 2 - i and z2 = 3 - 2i, then Im(2z1 - 3z2) = b 4 a 4i d -8 e -8i

c -4

8 MC If (2 + xi) + (4 - 3i) = x + 3yi, then the respective values of x and y are: a 6, 1 b 3, 6 c 6, -3 d 6, 3 e 1, 6

1e

Multiplication and division of complex numbers Multiplication of complex numbers If z1 = a + bi and z2 = c + di where a, b, c, d ∈ R, then z1z2 = (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac – bd) + (ad + bc)i Note that this is an application of the distributive property.

24


Worked Example 17

Simplify a 2i(2 - 3i) b (2 - 3i)(-3 + 4i). Think

Write a 2i(2 - 3i) = 4i - 6i2

a Expand the brackets.

= 6 + 4i

b Expand the brackets as for binomial

expansion and simplify.

b (2 -

3i)(-3

+ 4i) = -6 + 8i + 9i - 12i2 = 6 + 17i

Alternatively, on the Main screen, complete the entry lines as: 2i(2 - 3i) (2 - 3i) (−3 + 4i) Press E after each entry.

The conjugate of a complex number If z = a + bi, then its conjugate, z , is z = a - bi . The sum of a complex number and its conjugate z + z = a + bi + a - bi = 2a, which is a real number. The product of a complex number and its conjugate zz = (a + bi)(a - bi) = a2 - (bi)2 = a2 + b2, which is a real number. \

Worked Example 18

If z1 = 2 + 3i and z2 = -4 - 5i, find a z1 + z2

b z1 + z2

c z1 z2

d z1 z2 .

Think

Write


1

Determine the conjugate of each complex numbers using the definition: if z = a + bi, then its conjugate, z , is z = a - bi .

2

Evaluate z1 + z2 .

a

z1 = 2 + 3i ⇒ z1 = 2 - 3i z2 = -4 - 5i ⇒ z2 = - 4 + 5i z1 + z2 = (2 - 3i) + ( - 4 + 5i) = - 2 + 2i


25

b

1

To evaluate z1 + z2 , first evaluate z1 + z2.

2

Evaluate z1 + z2 .

b z1 + z2 = (2 + 3i) + (-4 - 5i)

= -2 - 2i

z1 + z2 = - 2 + 2i

c Evaluate z1 z2 using binomial expansion.

c

z1 z2 = (2 - 3i)(-4 + 5i) = -8 + 10i + 12i - 15i2 = 7 + 22i

d Evaluate z1z2 first and then evaluate z1z2 .

d

z1z2 = (2 + 3i)(-4 - 5i) = -8 - 10i - 12i - 15i2 = 7 - 22i z1z2 = 7 + 22i

Method 2: Using a CAS calculator On the Main screen, complete the entry lines as: 2 + 3i W x -4 − 5i W y conjg(x) + conjg(y) conjg(x + y) conjg(x) × conjg(y) conjg(x × y) Press E after each entry. Note: ‘conjg’ can be typed directly onto the screen or can be found by tapping: • Action • Complex • conjg

division of complex numbers If z1 = a + bi and z2 = c + di where a, b, c, d ∈ R, then using the conjugate z1 a + bi = z2 c + di a + bi c - di × c + di c - di ac - adi + bci - bdi 2 = c 2 - (di)2 (ac + bd ) - (ad - bc)i = c2 + d 2 ac + bd ad - bc = 2 i c + d 2 c2 + d 2 =

Worked example 19

eBook plus

Express each of the following in a + bi form. 4-i a 2

26

3 - 4i b 3i

c (3 - 2i)-1

2 - 3i d 2+i


Tutorial


Think

Write

Method 1: Using the rule a Divide each term of the numerator by 2.

a

b Multiply the numerator and denominator

b

by i and then divide each term of the numerator by 3. Write the answer in the required form a + bi.

c

1

Express (3 - 2i)-1 as a reciprocal.

2

Multiply the numerator and the denominator by the complex conjugate of the denominator.

4-i 1 = 2- i 2 2 3 - 4i i 3i - 4 × = 3i i 3 4 = -i 3

c (3 - 2i )-1 =

1 3 + 2i × 3 - 2i 3 + 2i 3 + 2i = 9 - (2i)2 3 + 2i = 9 - 4i 2 3 + 2i = 13

=

d

3

Write the answer in the required form a + bi.

1

Multiply the numerator and the denominator by the complex conjugate of the denominator.

d

Write the answer in the required form a + bi.

=

3 2 + i 13 13

2 - 3i 2 - 3i 2 - i = × 2+i 2+i 2-i

2

1 3 - 2i

4 - 2i - 6i + 3i 2 4 - i2 1 - 8i = 5 1 8 = - i 5 5 =

Method 2: Using a CAS calculator On the Main screen, complete the entry lines as: 4 -i 2 3 - 4i 3i (3 - 2i)−1 2 - 3i 2+i Press E after each entry.


27

REMEMBER

1. Multiplication of complex numbers: If z1 = a + bi and z2 = c + di where a, b, c, d ∈ R, then z1z2 = (a + bi)(c + di)

= ac + adi + bci + bdi2

= (ac - bd) + (ad + bc)i

2. If z = a + bi, then its conjugate, z , is z = a - bi , where the sum and product are both real. z + z = 2a and z z = a 2 - (bi)2 = a2 + b2 3. Division of complex numbers: If z1 = a + bi and z2 = c + di where a, b, c, d ∈ R, then using the conjugate z1 a + bi = z2 c + di

= a + bi × c - di c + di c - di ac - adi + bci - bdi 2 = c 2 - (di)2 (ac + bd ) - (ad - bc)i c2 + d 2 ac + bd ad - bc = i c2 + d 2 c2 + d 2 =

Exercise

1e

Multiplication and division of complex numbers 1 WE17 & 19

Evaluate the following, giving the answer in its simplest a + bi form.

a 2i(2 + 3i)

b (2 - 3i)(1 + i)

d (2 - 3i)2

e (6 + 7i)(6 - 7i)

g

3 + 4i 3 - 4i

h -1

j (3 + 2i) m

(3 - 2i)2 (2 - i ) 2

p

2-i 3+i 3 + 2i 2 - i

1 + 2i 2+i

c (-2 - i)(1 - 3i) 3 - 4i f 5i (2 + i ) 2 1 + 2i 1 1 l + 2 - 3i 2 + 3i i

-2

k (3 + 2i)

n 2 + 3i +

1 2 + 3i

o

3-i 3+i

+

3+i 3-i

2 WE18 If z1 = 4 - 3i and z2 = 3 - 4i, evaluate the following, giving the answer in its simplest a + bi form. a z1 d

28

z12

b z1z2

c z1 z2

e 2iz2

f ( z1 + z2 )2


-

g z1 1 j

 z1   z 

2

z  i  1  z2 

z1 z2

h

k z1 +

2

1 z1

-1

l

z1 z2 z1 z2

c

z1  z1  = z2  z2 

If z1 = a + bi and z2 = c + di, prove that:

3

b z1 + z2 = z1 + z2

a z1 z2 = z1z2

4 Find x and y in each of the following. a (x + yi)(2 + i) = 3 + 6i

x + yi = 1+ i 1 + 2i

b

5 Solve for z.

b (2 - 3i)z = -3 - 2i

a (4 + 3i)z = 2 - i

-

-

6 For each of the following, state z and find z 1, then state z 1 in terms of z. a z = 4 + 5i

b z = a + bi

7 mC Expressed in a + bi form, (2 3 - 3i)(3 3 - 2i) = a 24 - 13 3i

b 12 - 13 3i

D (6 3 - 6) - 13 3i

E 12 - 5 3i (2 3 - 3i)

8 mC Expressed in a + bi form,

(3 3 - 2i)

a

24 31

-

5 3 31

i

b

12 5 3 - 31 31

D

24 23

-

5 3 i 31

E

212 23

-

c (6 3 + 6) - 13 3i

= c

i

5 3 23

24 31

-

13 3 31

i

i -

9 mC Expressed in a + bi form, (1 + i)2 + (1 + i) 2 = a

5 i 2

D

9 4

9

-4i

b

9 4

7

E

3 i 2

-4i

c 0

10 mC If (2x + yi)(3 - 2i) = 4 + 5i, then the respective values of x and y are:

1F

a

2 13

D

1 5

and and

23 13 23 5

b

1 13

E

2 5

and and

23 13

c

23 5

7 13

and

23 13


WorkSHEET 1.3

representing complex numbers on an argand diagram

eBook plus Interactivity

int-0968 Complex sums and differences

Whereas real numbers can be represented on a number line, complex numbers with their real and imaginary components require a plane.

Chapter 1


29

The Argand diagram or Argand plane has a horizontal axis Re(z) and a vertical axis Im(z). The complex number z = a + bi is represented by the point (a, b). Because of the similarities with the Cartesian plane a + bi is referred to as the Cartesian or rectangular form. The complex numbers 2 + 3i, 4, –3i and –2 – 4i are shown on the Argand plane below. Im(z) 4

2 + 3i

2 −4 −2

2

−2

4 4 Re(z)

−3i

−2 − 4i −4

Worked Example 20 a Express the following in their simplest form: i0, i, i2, i3, i4, i5. b Use the pattern in these results to find the simplest form for: i8, i21 and i 63. c Illustrate the points from part a on an Argand plane, and state their distance from the origin and

the angle of rotation about the origin to rotate from one power of i to the next.

Think

Write/Draw

a Use the fact that i2 = −1 and your knowledge of

a i0 = 1

b The pattern repeats as shown in part a .

i1 = i i2 = -1 i3 = i2 × i = -i i4 = i2 × i2 = 1 i5 = i4 × i = i b i8 = (i4)2 =1 i21 = (i5)4 × i =i i 63 = (i 15)4 × i3 3 =i = -i

c

c

index laws to simplify each term.

1

Rule up a pair of labelled, scaled axes for the Argand plane. Place each of the points from part a onto the plane and label them.

Im(z) 2 i2 −2 −1

5 1 i, i 0 i , i4

−1 i3

1

2 Re(z)

−2

30

2

Determine the distance of each point from the origin.

All points are 1 unit from the origin.

3

State the angle of rotation about the origin to rotate from one power of i to the next.

The angle of rotation about the origin to rotate from one power of i to the next is 90° in an anticlockwise direction.


REMEMBER

1. Complex numbers are represented as an ordered pair on a complex plane or Argand diagram. 2. For an Argand plane, the horizontal axis is Re(z), and the vertical axis is Im(z). Exercise

1f

Representing complex numbers on an Argand diagram 1 Plot the following points on an Argand plane. a 2 + 3i d -2 - 3i

c -2 + 3i f 2

b 2 - 3i e -3i

2 WE20a Give the following in their simplest form. a i7 d i 15 g (2i)9 3 a b c d

b i37 e (2i)6 h -(-2i) 9

-

c i 4 f (-2i)8

-

If z = 3 + 2i, state z and calculate z 1. Plot z, z , and z 1 on an Argand plane. What transformation plots z onto z 1? What is the relationship between the origin and the points representing z and z 1?

4 a If z = 2 + 3i, calculate iz, i2z, i3z and i4z. b Plot z, iz, i2z, i3z and i4z on an Argand plane. c State a transformation that will plot the point inz onto in + 1z for n ∈ Z +. -

-

5 a If z = 1 + i, calculate z 2, z 1, z0, z2, z3 and z4. b Plot z 2, z 1, z0, z, z2, z3 and z4 on an Argand plane. c State the rotation required and the change in distance from the origin required to plot the point zn onto zn + 1 for n ∈ Z. d State the rotation required and the distance from the origin required to plot the point z0 onto zn for n ∈ Z +. e State, using the results above, the following powers of z in their simplest Cartesian form. i z5 ii z 3 iii z10 iv z17 v z 13.

1G

Factorising quadratic expressions and solving quadratic equations over the complex number field In the introduction to complex numbers, a quadratic equation of the form ax2 + bx + c = 0, -

b ± b 2 - 4 ac . The 2a expression under the square root sign is called the discriminant, and is represented by ∆, where ∆ = b2 - 4ac. The discriminant can be used to determine the nature of the solutions. It can also be used to determine possible methods for factorising a quadratic expression. The table overleaf gives the method for factorising a quadratic expression and the nature of the solutions of a quadratic equation for given values of ∆, where a, b, c ∈Q\{0}. where a ∈ R\{0}, b, c ∈R, was solved using the quadratic formula x =


31

Values of the discriminant

Factorising methods for an expression

Nature of solution(s) of an equation

∆=0

A perfect square. State the answer.

One rational solution

∆ is a perfect square

Factorise over Q or complete the square.

Two rational solutions

∆ > 0 and is not a perfect square

Complete the square.

Two irrational solutions

∆<0

Complete the square.

Two complex solutions

Factorising quadratic expressions over the complex number field Just as in previous years, where factorising over R implies that all the coefficients must be real numbers, factorising over C implies that all the coefficients must be complex numbers. As factors over C are required in this section, the variable label will be z. In worked example 21 below, the factors for the expressions in parts a and b are factors over both R and C, but the factors for the expression in part c are factors over C only. It is still correct to say that 2z2 + 3 does not factorise over R. If c = 0 and a, b ∈R\{0}, then factorise az2 + bz by taking out the common factor z(az + b). If b = 0 and a, c ∈R\{0}, then factorise az2 + c using the difference of squares to factorise.

Worked Example 21

Factorise each of the following quadratic expressions over C. a 2z2 + 6z b 2z2 - 6 c 2z2 + 3 Think

Write

Method 1: Using the rule a Factorise 2z2 + 6z by taking out the

highest common factor.

b

Factorise 2z2 - 6 by taking out the highest common factor. Factorise further using the difference of two squares.

b

1

Factorise 2z2 + 3 by taking out the common factor of 2.

c

2

Factorise further using the difference of two squares. Let 3 = 3 i 2. 2 2

1 2

c

32

a 2z2 + 6z = 2z(z + 3)

2z2 - 6 = 2(z3 - 3) = 2( z - 3 )( z + 3 ) 3  2z 2 + 3 = 2  z 2 +   2 3   = 2  z2 - i2   2   3  3  = 2 z i  z + i  2  2 


3

 6  6  = 2 z i  z + i    2 2 

Rationalise the denominators by multiplying the relevant terms by

2 2

.

Method 2: Using a CAS calculator a , b On the Main screen, tap: & c • Action

• Transformation • rFactor Complete the entry lines as: rFactor(2z2 + 6z) rFactor(2z2 – 6) rFactor(2z2 + 3) Press E after each entry.

Worked Example 22

Factorise each of the following quadratic expressions over C. a z2 - 6z + 9 b z2 - 4z - 60 c 2z2 - 6z - 6 d -2z2 - 3z - 5 Think a

b

c

1

Write

Calculate the value of the discriminant to determine the nature of the factors.

2

Since ∆ = 0, the expression is a perfect square.

1


2

Since ∆ = 256, which is a perfect square, then the factors will be rational.

1


2

Since ∆ = 84, which is not a perfect square but is positive, use the difference of two squares to find two factors over R.

a ∆ = b2 - 4ac

= (-6)2 - 4(1)(9) =0 2 z - 6z + 9 = (z - 3)2

b ∆ = b2 - 4ac

= (-4)2 - 4(1)(-60) = 256

z2 - 4z - 60 = (z - 10)(z + 6) c ∆ = b2 - 4ac

= (-6)2 - 4(2)(-6) = 84 2z2 - 6z - 6 = 2(z2 - 3z - 3) 2  3  = 2   z 2 - 3z + 2  - 3  

()

( )-

 3 = 2 z - 2 

(

3

=2 z- 2d

1


2

21 2

( ) 

21  4 

)( z -

3 2

+

21 2

3 2

2

)

d ∆ = b2 - 4ac

= (-3)2 - 4(-2)(-5) = -31


33

2

Since ∆ = -31 which is not a perfect square but is negative, use the difference of two squares to find two factors over C.

- 2z 2

- 3z - 5 = - 2 z 2 + 2 z + 2

5

)

( )  +

5 2

9 - 16  

+

31 i 4

(

3

 3 = - 2  z2 + 4 z + 

( )+  2(z + ) 

 3 = - 2 z + 4  =-

3 4

3 4

2

2

31  16  

2

31 2  16 i 

(

= - 2 z + 43 -

31 i 4



)( z +

3 4

)

Solving quadratic equations over the complex number field Two methods can be used to solve quadratic equations over the field complex number: 1. Factor first and use the null factor property to state solutions 2. Use the formula for the solution of a quadratic equation. The null factor property states that if ab = 0, then a = 0 or b = 0 or a = b = 0. From worked example 22 d,   3 31  - 2 z 2 - 3z - 5 = - 2 z + 3 - 31 i - 3 z+ + i ,    4 4  4 4  so the solutions of 2z - 3z - 5 = 0 are from  z+

3 31 3 31 3 31 i = 0 and z + + i = 0. The solutions are z = ± i. 4 4 4 4 4 4

If az2 + bz + c = 0, where a ∈C \{0}, b, c ∈ C, the formula for the solution of the quadratic equation is z =

-

b ± b 2 - 4 ac . 2a

Worked example 23

eBook plus

Solve the following using the formula for the solution of a quadratic equation. a 2z2 + 4z + 5 = 0 b 2iz2 + 4z - 5i = 0 Think

a

Use the quadratic formula -

b ± b 2 - 4 ac to solve over C, 2a where a = 2, b = 4, c = 5.

z=

a z=

=

2

34

Express the answer in the form a + bi.

-

4 ± 16 - 40 4

-

4±

-

24

4 -

4± 4×-6 = 4 =

-

4 ± 2 6i 2 4

= -1 ±


WriTe

Method 1: Using the rule 1

Tutorial

6 i 2


b

1

Use the quadratic formula -

b ± b 2 - 4 ac to solve over C, 2a where a = 2i, b = 4, c = -5i. z=

b z=

= =

-

4 ± 16 - 4 × -10i 2 4i

-

4 ± 16 - 40 4i

-

4 ± - 24 4i

-

4 ± 4 × -6 i × i 4i ( - 4 ± 2 6i)i = 4 =

2

Express the answer in the form a + bi.

=i± =

Method 2: Using a CAS calculator a On the Main screen, tap: & • Action b • Advanced • solve Complete the entry lines as: solve(2z2 + 4z + 5 = 0, z) solve(2iz2 + 4z - 5i = 0, z) Press E after each entry.

6 2

- 6 6 + i or +i 2 2

REMEMBER

1. Quadratic expressions can be factorised. Quadratic equations can be solved. 2. Quadratic expressions of the form az2 + c can be factorised using the difference of squares method. Quadratic expressions of the form az2 + bz can be factorised by taking out z as a common factor. Quadratic expressions of the form az2 + bz + c can be factorised by completing the squares method. 3. Quadratic equations of the form az2 + bz + c = 0 can be solved by either factorising and using the null factor property or by using the quadratic formula. 4. The formula for the solution of az2 + bz + c = 0, where a ∈ C\{0}, b, c ∈C, is -b ±

b 2 - 4 ac . 2a 5. The discriminant, ∆ = b2 - 4ac, can be used to determine the number (one or two) and the nature of the solutions, particularly if they are rational, irrational or complex only. 6. Real solutions of a quadratic equations can be represented on a number line, whereas complex solutions can be represented on an Argand diagram. z=


35

Exercise

1g

Factorising quadratic expressions and solving quadratic equations over the complex number field 1 WE 21

2

Factorise the following quadratic expressions over C.

a 2z2 - 6 1 d 2z2 + 2

b 2z2 - 3

c 3z2 + 6

e z2 - 4z

f 6z2 - 2z

g 2 2 z 2 - 2 z

h -4z2 - 3z

Factorise the following quadratic expressions over C without using the completion of the square method. a z2 + 8z + 16 b 2z2 - 8z + 8 c 2z2 + 3z - 2 2 2 d z + 2z - 3 e 2z - 2z - 24 f -12z2 + 10z + 12

3 WE 22 Factorise the following quadratic expressions over C using the completion of the square method. a z2 + 4z + 14 b z2 + 10z + 16 c 2z2 + 5z - 3 d z2 + z - 3 e z2 + 8z + 16 f z2 + 2z + 3 2 2 g 2z - 5z + 2 h 2z + 8z + 16 i -2z2 + 5z + 4 2 j 4z + 4z - 1 4 WE 22 Factorise the following quadratic expression, and then solve the given quadratic equation. a 3z2 - 2 = 0 b 2z2 + 5 = 0 c 2z2 - 7z = 0 d z2 - 6z + 5 = 0 e z2 - 5z + 6 = 0 f 2z2 - 5z + 3 = 0 g z2 - 4z + 2 = 0 h 2z2 + 5z + 4 = 0 2 i z - 6z + 5 = 0 j -3z2 - 2z - 1 = 0 5 WE 23 Solve the following quadratic equations over C using the formula for the solution of a quadratic equation. a z2 - 10z + 25 = 0 b z2 - 10z + 5 = 0 2 + 4z + 7 = 0 d 2z2 - 7z + 6 = 0 c z 2 e 3z - 7z + 7 = 0 f -2z2 + 4z - 6 = 0 6 Expand the following. a (z - (2 + 3i)) (z - (2 - 3i)) b (z - (2 + 3i))2 c (z - 2 + 3i) (z - 3 - 2i) 7 Solve the following quadratic equations over C, using the formula for the solution of a quadratic equation. a iz2 - 6z + 5i = 0 b (2 + i)z2 - iz - (2 - i) = 0 c -3iz2 - (1 + i)z + 5i = 0 8 MC Using the smallest set from Q, I and C, the solutions of 2z2 - 5z + 6 = 0 and 5z2 - 11z + 5 = 0, respectively, belong to the sets: a C, C b C, Q c C, I D I, I E I, Q

36


9 MC The factors of z2 + 6z + 11 and 2z2 - 4z + 3, respectively, are:  2  2  a ( z + 3 - 2i)( z + 3 + 2i), 2  z - 1 i  z - 1 + i 2  2    2  2  B ( z + 3 - 2i)( z + 3 + 2i),  z - 1 i  z - 1 + i 2  2    2  2  C ( z - 3 - 2i)( z - 3 + 2i), 2  z - 1 i  z - 1 + i 2  2    2  2  D ( z + 3 - 2i)( z + 3 + 2i),  z + 1 i  z + 1 + i 2  2    2  2  E ( z - 3 - 2i)( z - 3 + 2i), 2  z + 1 i  z + 1 + i 2  2  


37

Summary Review of set notation

• Recall and apply the definitions of ξ, ∅, ∈, ⊂, A′, A ∪ B, A ∩ B, and C \ D, and how to illustrate these on a Venn diagram. • Recall and apply the definitions of the following sets of numbers: N, D, Z, Q, I, and R and their relationships. • Rational numbers, expressed as a decimal, are either terminating or non-terminating, recurring. • Irrational numbers, expressed as a decimal, are non-terminating and non-recurring. • Division by zero is not defined. p • Given a non-terminating, non-repeating decimal, use the process to express the number in the form of a q rational number. • Do simple calculations in scientific notation by hand and more complex calculations using a calculator. • Express answers to the stated number of significant figures or decimal places. Subsets of the set of real numbers

• Set builder notation. For example, {x: x ∈ Q, -3 < x ≤ 17}. If x ∈ R, the set does not have to be stated. For example, {x: -3 < x ≤ 17}. • Interval notation. For example, (-3, 17] = {x: -3 < x ≤ 17}, [-3, ∞) ={x: x ≥ -3}. This notation can only be used if x ∈ R. • All subsets of R can be illustrated on a number line. Properties of surds

• To simplify a surd, take out all perfect square factors from the number under the root sign. • Like surds have the same number under the root sign when expressed in its simplest form. • Only like surds can be added and subtracted. • Multiplication properties:

a × b = ab , a ( b + c ) = ab + ac and

( a + b )( c + d ) = ac + ad + bc + bd • The conjugate of a + b is a - b • Special expansions: 1. Expansion of perfect squares: (a + b)2 = a2 + 2ab + b2 and (a - b)2 = a2 - 2ab + b2 2. Difference of perfect squares: (a + b)(a - b) = a2 - b2 • Division properties: ( a + b)

•

( c + d)

=

a b

( a + b) ( c + d)

a , b

= ×

a b

a b ab × = b b b

=

( c - d) ( c - d)

=

ac - ad + bc - bd , using the conjugate of c-d


• • • •

Complex numbers are of the form a + bi where i2 = –1. If z = a + bi, then Re(z) = a and Im(z) = b. If z1 = a + bi and z2 = c + di then z1 + z2 = (a + c) + (b + d )i and z1 − z2 = (a − c) + (b − d )i. If a + bi = c + di, then a = c and b = d. Multiplication and division of complex numbers

• Multiplication of complex numbers: 1. If z1 = a + bi and z2 = c + di where a, b, c, d ∈ R, then z1z2 = (a + bi)( c + di) = ac + adi + bci + bdi2 = (ac – bd) + (ad + bc)i 38


c + d.

2. If z = a + bi, then its conjugate, z , is z = a - bi where the sum and product are both real. z + z = 2a and zz = a2 - (bi)2 = a2 + b2 • Division of complex numbers: If z1 = a + bi and z2 = c + di where a, b, c, d ∈ R, then using the conjugate z1 a + bi = z2 c + di

= a + bi × c - di c + di c - di ac - adi + bci - bdi 2 = c 2 - (di)2 (ac + bd ) - (ad - bc)i = c2 + d 2 ac + bd ad - bc = i c2 + d 2 c2 + d 2 Representing complex numbers on an Argand diagram

• Complex numbers are represented as an ordered pair on a complex plane or Argand diagram. • For an Argand plane, the horizontal axis is Re(z), and the vertical axis is Im(z). Factorising quadratic expressions and solving quadratic equations over the complex number field

• • • • •

Quadratic expressions can be factorised. Quadratic equations can be solved. Quadratic expressions of the form az2 + c can be factorised using the difference of squares method. Quadratic expressions of the form az2 + bz can be factorised by taking out z as a common factor. Quadratic expressions of the form az2 + bz + c can be factorised by completing the square method. Quadratic equations of the form az2 + bz + c = 0 can be solved by either factorising and using the null factor property or by using the quadratic formula. • The formula for the solution of az2 + bz + c = 0, where a ∈ C \{0}, b, c ∈ C, is -b ±

b 2 - 4 ac . 2a • The discriminant, ∆ = b2 - 4ac, can be used to determine the number (one or two) and the nature of the solutions, particularly if they are rational, irrational or complex only. • Real solutions of a quadratic equation can be represented on a number line, whereas complex solutions can be represented on an Argand diagram. z=


39

chapter review Short answer

1 For each of the following, state all the sets (from Z, Q, I, R and C) in which they are a member. a

-4

b 1

e 27 3

f

-

4

-2 7

16 3

c

5

g 2p

8

d 3 3 h -3.221

i 3.21683947. . . (no pattern) j

-3 + 2

5

  k 1.1234

l i2

2 Express each of the following repeating decimals in its simplest rational number form.     a 0.24 b 1.123 c 1.123 8 × 10 4 × 6 × 10 5 , giving your answer in 3 × 10 8 scientific notation, correct to 1 significant figure.

3 Calculate

4 Use interval notation to represent each of the following sets, then plot each on a separate number line. a {x: x > 2} b {x: 3 < x ≤ 4} c {x: 4 ≤ x ≤ 10} ∪ {x: 8 < x < 12} d {x: 4 ≤ x ≤ 10} ∩ {x: 8 < x < 12} 5 Use set builder notation to represent each of the following sets, then plot each on a separate number line. a {7, 8, 9, 10, 11, 12} b {integers between 3 and 8} c {rational numbers between 1 and 2} d {rational numbers greater than 5, except for 10} e {real numbers greater than 4 and less than 5} f {real numbers less than 4 or greater than 6} a

80 + 72 - 50 - 45

b

5 (2 5 + 10 )

c (3 3 - 2 2 )(3 3 + 2 2 )

4 3- 2 2 3+ 2

f

2 3+ 2 3- 2

-

3+2 2 3+ 2

Exam tip Be careful with the signs. Make sure you subtract the whole second fraction. [Authors’ advice]

7 By substituting for x, show that 3 - 1 is a solution z 2 - 4 3z + 6 = 0 . 8 Find each of the following components of a complex number. a Im(2 +

-8)

b Re((2 - i) + ( 3 + 2i)) c Im( - 2 3 (2 - 6i )) 9 Find x and y. a (3 + 2 yi) + (1 - i ) = x + yi b ( x + 2 y ) + (2 x + y )i = 9 + 6i 10 If z1 = 2 - 4i and z2 = 2 + 3i, express each of the following complex numbers in a + bi form. a 2 z1 + 3z2 b z1 c z1z2 -1 z z z d 1 e z1 + z1 1 f 1 - 2 z1 z2 z1 11 Find z in its simplest form. z 2-i b a = 3 - 3i = 3 - 3i 2-i z 12 Plot the following complex numbers on an Argand diagram. a 2 − 3i b i 29 c i 29 2 + 3i d i(2 + 3i) e i 13 Factorise the following quadratic expressions over C. a z2 + 7 b 2z2 + 3 2 c 4z + 5z d 16z2 + 4z - 6 2 e z + 4z - 2 f z2 + 3z + 3 2 g 2z + 3z + 4 Exam tip Make sure you take the coefficient of the squared term first.

d (3 3 - 2 2 )2 40

of z 2 - 2 3z + 2 = 0 , but is not a solution of

m 2 + - 5 n 0.172117722111777222. . . (pattern continues) o 3 - 3 p 4i q 3 + 7i r 4 + 0i

6 Simplify the following surds.

e


[Authors’ advice]

14 For each of the following quadratic equations, factorise the corresponding quadratic expression, then solve the equation. a 2z2 + 5 = 0 b z2 + 4z + 5 = 0 c 3z2 + 5z + 2 = 0 15 Solve each of the following quadratic equations using the formula for the solution of a quadratic equation. a z2 + 6z + 10 = 0 b 2z2 + 4z + 5 = 0 c 3z2 - 5z + 4 = 0

2 Which of the following is a rational number?  4 0.14 a b   0 1.167 c

0 0

d -

e

5 {x: 4 ≤ x ≤ 6} ∪ {x: 5 < x < 7} and {x: 4 ≤ x ≤ 6} ∩ {x: 5 < x < 7} can also be written as: a (5, 6], [4, 7) b [4, 7], (5, 6) c [4, 7), (5, 6] d [5, 6), (4, 7] e R, ∅

15 15 c + 3 5

d

5 3 2

3

+

5

15

2 15 15

a 3 7 - 12

b 7 3 + 4 6

c 3 3 - 12 6

d 4 3 - 7 6

e 7 3 - 4 6 8 In its simplest form, 3 3 + 3 (1 + 3 ) +

3 3

+

4

is:

3 -1

a 4 + 5 3

b 2 + 3 3

c 3 + 5 3

d 5 + 7 3

e 4 + 3 9 In its simplest form, ( 3 - 2 )( 3 + 2 ) + ( 3 + 2 )2 +

5( 3 + 2 ) 3- 2

is:

4 9

4 R\[3, 6) can also be written as: a (-∞, 3] ∩ (6, ∞) b (-∞, 3) ∩ [6, ∞) c (-∞, 3) ∪ [6, ∞) d R\{x: 3 < x ≤ 6} e (-∞, 3] ∪ (6, ∞)

b

7 In its simplest form, 2 54 - 3 27 + 4 48 - 5 24 is:

7 11

3 The numbers 999, 0.000105 and 0.496 expressed to 2 significant figures are: A 1000, 0.00011 and 0.5 B 1000, 0.00010 and 0.5 C 99, 0.00011 and 0.49 D 99, 0.00010 and 0.49 E 1000, 0.00011 and 0.50

a 5.3 + 0.6

e

Multiple choice

1 The numbers 27.366, 0.027356 and 273460.123 expressed to 3 significant figures are: A 27.4, 0.0274 and 273000 B 27.3, 0.0273 and 273000 C 27.4, 0.027 and 273000 D 27.37, 0.027 and 274000 E 27.37, 0.027 and 273460.12

5 3 + = 3 5

6 In its simplest form,

a 31 + 4 6

b 31

c 11 + 2 6

d 9 + 4 6

e 31 + 12 6 10 If z1 = 2 - i and z2 = 1 - 2i, then Re (z1z2) + Im (z1z2) = e 5 a 4 b -5 d -4 c -5i z  z  11 If z1 = 2 + i and z2 = -i, then Re  1  + Im  1  =  z2   z2  b -1

a 0 12

in + 1

+ is: a 1 in

+

in + 2

+

b -1

c 1 in + 3,

d -3

e 3

n ∈ Z, in its simplest form

c 0

d i

13 When the complex numbers 1 + 2i,

e -i

5 , 1 - 2i

1 + 7i 1 and (2 + i)(4 + 3i) are plotted 3+i 5 on an Argand diagram, the number of distinct points shown would be: a 1 b 2 c 3 d 4 e 5

i 3 ( - 2 + i),


41

14 The equation z2 - 2z - 3 + 4i = 0 is best described as having: a two rational solutions b two real solutions c two complex solutions D one complex solution E one real solution 15 The equation 3z2 + 2z + a = 0, where a ∈ R, has two distinct real solutions if: a a<3 b a>3 c a≤3 1 1 E a> D a< 3 3

16 The equation 3z2 + az + 2 = 0, where a ∈ R, has two distinct real solutions if: a a>2 6 b 0 2 6 E

-2

6
exTended reSponSe

1 a b c d e f g

Plot z = 1 + 3i on an Argand plane. If O is the origin, show using trigonometry, that the angle made by Oz with the positive real axis is 60°. Calculate the distance Oz. Calculate z 1, z0, z, z2 and z3, then plot them on the same Argand diagram used in a. State the rotation about the origin and the change in distance from the origin to plot zn + 1 given zn. State the rotation about the origin and the distance from the origin, to plot zn given z0 = 1. State, using the results above, the following powers of z in their simplest form. i z4 ii z6 iii z9 iv z 2 v z9

2 If the solutions of a quadratic equation are -1 and 2, then a quadratic equation giving these solutions is (z + 1)(z - 2) = 0. A quadratic expression that produces these factors is (z - 2)(z + 1) = z2 - z - 2, but this is not the only expression whose corresponding quadratic equation has the solutions -1 and 2. a If the coefficients of a quadratic equation with solutions -1 and 2 are integers, give an example of another quadratic expression whose corresponding quadratic equation would have the solutions -1 and 2. State all quadratic expressions that satisfy the conditions above. b In each case, find a quadratic expression with the coefficient of z2 = 1 that, when expressed as a quadratic equation, gives the solutions: i z = -2, z = 3 ii z = 4 + 3 , z = 4 - 3 iii z = 4 + 2i, z = 4 - 2i iv z = 3 + 2 3 , z = 2 - 3 v z = 3 + 2i, z = 2 - i c Note that the quadratic expressions in b ii and b iii have integer coefficients. If the coefficients of a quadratic equation are integers and a and b below are integers, then state the other solution. i z = a+b 3 ii z = a + bi d z = 2 + 3i is a solution of z2 + bz + c = 0 where b, c ∈ Z. i State the other solution, then find the values of b and c. ii Substitute z = 2 + 3i into z2 + bz + c = 0, then find the values of b and c. e If z2 - 2z + 3 = 0, find the solutions using the formula for the solution of a quadratic equation, then factorise z2 - 2z + 3. f If 2z2 + 4z + 3 = 0, find the solutions using the formula for the solution of a quadratic equation, then factorise 2z2 + 4z + 3. eBook plus

Digital doc

Test Yourself Chapter 1

42


eBook plus

aCTiviTieS

chapter opener Digital doc

• 10 Quick Questions: Warm up with ten quick questions on the set of complex numbers. (page 1) 1a

Review of set notation

Tutorial

• We4 int-1211: Watch how to express a recurring decimal fraction. (page 6) 1b

Subsets of the set of real numbers

Digital doc

• WorkSHEET 1.1: Identify irrational numbers and simplify surds. (page 10) 1c

Properties of surds

Tutorial

• We12 int-1027: Watch how to multiply surds and express answers in their simplest form. (page 16) 1D


Digital doc

• WorkSHEET 1.2: Multiply and divide surds and represent complex numbers on an Argand diagram. (page 22) Tutorial

• We16 int-1028: Watch how to add and subtract complex numbers. (page 23) 1E

Multiplication and division of complex numbers

Tutorial

• We19 int-1029: Watch how to express the quotient of complex numbers in the form a + bi. (page 26) Digital doc

• WorkSHEET 1.3: Identify subsets of complex numbers using mathematical notation and add and simplify surds. (page 29)

1F

Representing complex numbers on an argand diagram

Interactivity

• Complex sums and differencs int-0968: Consolidate your understanding of how to represent complex numbers on an Argand diagram. (page 29) 1G

Factorising quadratic expressions and solving quadratic equations over the complex number field

Tutorial

• We23 int-1030: Watch how to solve quadratic equations over the complex number field. (page 34) chapter review Digital doc

• Test Yourself: Take the end-of-chapter test to test your progress. (page 42) To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 1


43

2

2a 2b 2c 2d 2E

Translations of points and graphs Reflections of points and graphs Dilations from axes The ellipse and the hyperbola Successive transformations

Transformations areas oF sTudy

• Sketching relations in the Cartesian plane from descriptions, equations or formulas and identifying their key features

• Graphical representation of circles, ellipses, parabolas and hyperbolas, and sketching their graphs • Identifying asymptotes eBook plus Digital doc

10 Quick Questions

2a

Translations of points and graphs introduction to transformations Under a transformation of the Cartesian plane, each point (x, y) maps onto its image point (x ′, y ′). In this chapter, the transformation Tr will be defined by the rule (x, y) → (x ′, y ′) and the following transformations will be considered: translations, reflections and dilations. For each transformation and combination of transformations, points and their images as well as rules and their image rules will be considered.

some simple relations The following relations with the given rules and properties will be dealt with in this and subsequent chapters. 1. Linear (straight lines) • y = mx + c, where m is the gradient and (0, c) is the y-intercept • x = a, a vertical line with the gradient undefined

y y = mx + c

2. Quadratic (parabolas) • y = ax2 + bx + c • x-intercepts (if they exist) are found by solving ax2 + bx + c = 0. • (0, c) is the y-intercept. − b • The equation of the axis of symmetry is: x = . 2a − b • The turning point has an x-coordinate of x = and 2a the y-coordinate is found by substitution of the x-coordinate into the rule. • If a > 0 the shape is ∪ (upright) and if a < 0 the shape is ∩ (inverted). 44

maths Quest 11 advanced General mathematics for the Casio ClassPad

x=a (0, c) a

x

y y = ax2 + bx + c

x y = −ax2 + bx + c

y

3. Inverse (hyperbolas) 1 • y = x • Horizontal asymptote y = 0, the x-axis • Vertical asymptote x = 0, the y-axis

asymptote x=0

y=

1 x

x y=0 asymptote

y

4. Circle • x2 + y2 = a2 • Centre (0, 0) and radius a

a x2 + y2 = a2 −a

a

0

x

−a y

5. Exponential • y = ax, a > 0 • Horizontal asymptote y = 0 • Common point of (0, 1) for any a value

y = ax (0, 1) y=0 x asymptote

Translations Under a translation given by Th, k, (x, y) → (x + h, y + k), x ′ = x + h, y ′ = y + k: h is a translation in the x direction (horizontally) k is a translation in the y direction (vertically). T1, 2 represents a translation of 1 unit to the right and 2 units up for all points in a plane. T−1, −2 represents a translation of 1 unit to the left and 2 units down for all points in a plane.

Points under translations Worked Example 1

The point A (3, 1) maps onto A′ under the translation T1, 2 . Find the coordinates of A′. Think

The translation T1, 2 indicates that A is moved 1 unit to the right and 2 units up. Write this using mapping notation to find the coordinates of A′.

Write

(x, y) → (x + 1, y + 2) (3, 1) → (4, 3) A′ (4, 3)

Worked Example 2

Find a translation that maps A(3, -1) onto A′(2, 3). Think

Write

1

Use mapping notation to determine the translations.

(3, -1) → (2, 3) ⇒ (3, -1) → (3 - 1, -1 + 4)

2

State the answer.

A is translated 1 unit to the left and 4 units up, that is, T−1, 4.

Chapter 2 Transformations

45

Worked examPle 3

A translation is defined by the rule (x, y) → (x − 2, y + 3). If the image point is A′(5, 6), find the coordinates of the original point A. Think

WriTe

1

State the image equations. 5 is the image of x under translation of −2 units. 6 is the image of y under translation of 3 units.

x′ = x − 2 =5 y′ = y + 3 =6

2

Solve for x and y.

x = 7, y = 3

3

State the answer.

If the image point is A′(5, 6), and the translation is defined by the rule (x, y) → (x − 2, y + 3), the original point is A(7, 3).

rules under translations Worked examPle 4

eBook plus

Find the image rule for each of the following, given the original rule and translation. Tutorial a y = x, T−2, −3 int-1032 b y = 2x2, T− 4, 5 Worked example 4 c y = f (x), T h, k Think a

b

46

WriTe a x′ = x − 2

1

State the image equations.

2

Find x and y in terms of x′ and y′.

3

Substitute into y = x.

y=x ⇒ y′ + 3 = x′ + 2 ⇒ y′ = x′ − 1

4

Express the answer without using the primes.

Given y = x under translation T−2, −3, the equation of the image (or image rule) is y = x − 1.

5

Illustrate these transformations on a graph.

1


y′ = y − 3 x = x′ + 2 y = y′ + 3

b x′ = x − 4

y′ = y + 5


c

x = x′ + 4 y = y′ - 5

2


3

Substitute into y = 2x2.

y = 2x2 ⇒ y′ - 5 = 2(x′ + 4)2 ⇒ y′ = 2(x′ + 4)2 + 5

4

Express the answer without using the primes. Note: In first form of the answer, the turning point is (-4, 5), which was the answer expected as (0, 0) → (-4, 5).

Given y = 2x2 under translation T-4, 5, the equation of the image (or image rule) is y = 2(x + 4)2 + 5 or y = 2x2 + 16x + 37

5

Illustrate these transformations on a graph.

1


2


3

Substitute into y = f (x).

y′ - k = f (x′ - h) ⇒ y′ = f (x′ - h) + k

4


Given y = f (x) under translation Th, k, the equation of the image (or image rule) is y = f (x - h) + k.

c x′ = x + h

y′ = y + k x = x′ - h y = y′ - k

Worked Example 5

Find the image of x2 + y2 = 1 under Th, k. Think

Write

1

x2 + y2 = 1 is a circle with centre (0, 0) and a radius of 1. State the image equations.

x′ = x + h y′ = y + k

2


x = x′ - h y = y′ - k

3

Substitute into x2 + y2 = 1.

(x′ - h)2 + (y′ - k)2 = 1

4


Given x2 + y2 = 1 under translation Th, k, the equation of the image (or image rule) is (x - h)2 + (y - k)2 = 1.


47

Worked Example 6

Given the rule and its image rule under a translation, state a possible translation and its abbreviated version in the form Ta, b. a y = x, y = x + 1 b y = x2, y = (x - 2)2 + 1 c y = x2 + 1, y = x2 + 2x - 4 Think a

b

1

The original rule is y = x and its image is y = x + 1. We need a point on each graph for comparison. Substitute a value for x, say x = 0, into both rules.

a y=x

When x = 0, y = 0 ⇒ (0, 0) y=x+1 When x = 0, y = 0 + 1 ⇒ (0, 1) ⇒ (0, 0) → (0, 1)

2

State the required translations in the form Ta, b. That is, no translation on the x-axis and translation of 1 unit up on the y-axis.

1

The original rule is y = x2 and its image is y = (x - 2)2 + 1. We need a point on each graph for comparison. Determine the turning points of each equation. State the required translations in the form Ta, b. That is, translation of 2 units to the right on the x-axis and translation of 1 unit up on the y-axis.

b y = x2 has a turning point at (0, 0).

The original rule is y = x2 + 1 and its image is y = x2 + 2x - 4. We need a point on each graph for comparison. Determine the turning points for each equation.

c y = x2 + 1 has a turning point at (0, 1).

2

c

Write

1

2

State the required translations in the form Ta, b. That is, translation of 1 unit left on the x-axis and translation of 6 units down on the y-axis.

T0, 1: no translation on the x-axis and translation of 1 unit up on the y-axis.

y = (x - 2)2 + 1 has a turning point of (2, 1). ⇒ (0, 0) → (2, 1) T2, 1: translation of 2 units to the right on the x-axis and translation of 1 unit up on the y-axis.

y = x2 + 2x - 4 = (x + 1)2 - 5 The turning point is at (-1, -5). ⇒ (0, 1) → (-1, -5)

T-1, -6: translation of 1 unit left on the x-axis and translation of 6 units down on the y-axis.

REMEMBER

1. The rules, graphs and basic properties of lines, parabolas, hyperbolas, circles and exponentials as listed under the heading ‘Some simple relations’ are starting points for the rules in this chapter. 2. Translations: Th, k, where x ′ = x + h, y ′ = y + k • Th, k is a translation h in the x direction (horizontally) and a translation k in the y direction (vertically). Note: Th, k is an abbreviated version of a translation; however, your description of the translation must be explained in words. 3. Use x ′ = x + h, y ′ = y + k to find: (i) image points and image rules after a translation, given a point or a rule and the translation (ii) original points and original rules before a translation, given an image point or an image rule and the translation (iii) the translation, given a point and its image or a rule and its image.

48


Exercise

2A

Translations of points and graphs 1 WE 1

Given the coordinates of A and the translation, find the coordinates of the image, A′.

a A(1, 2), T3, 4 c A(-1, 2), T3, -4 2 WE 2

Given a point and its image, state the possible translation in words as well as using

Ta, b notation. a A(2, 3), A′(4, 5) c A(-1, -4), A ′(-3, 1) 3 WE 3

b A(1, -2), T-3, 4 d A(-1, -2), T-3, -4

b A(2, -3), A ′(4, -2) d A(-2, -4), A ′(-1, -7)

Given the translation and the coordinates of the image, A′, find the coordinates of A.

a T2, 3, A ′(4, 1) c T3, -2, A ′(-2, -3)

b T-1, 2, A ′(-2, 1) d T-2, -3, A ′(-5, -2)

Find the image rule for each of the following, given the original rule and translation. 4 WE 4 For parts a to e, sketch the graphs on the same axes using the original rule and its image. a y = x, T1, -3 b y = 2x, T1, -2 c y = -x, T2, -2 d y = 2x + 1, T-1, -2 2 2 e y = x , T-1, 2 f y = 2x , T2, -1 g y = x2 + 2, T-3, 4 h y = -2x2, T-2, 3 2 i y = f (x), T -3, 2 j y = f (x), T -1, -2 k y = 2x , Th, k l y = -3x, Th, k 5 WE 5 a Sketch on the same set of axes, x2 + y2 = 4 and its image under the translation T1, -1, translated 1 unit in the positive x direction and 1 unit in the negative y direction. 1 b Sketch on the same set of axes, y = and its image under the translation T2, -1 (2 units x right and 1 unit down). State the equations of the asymptotes. 6 WE 6

Given a rule and its image rule under a translation, state a possible translation.

a y = 2x, y = 2x + 3

b y = -x2, y = -(x + 4)2 - 3

c y = x2 + 2x + 1, y = x2 + 4x - 1

7 Given a rule and its image rule under a translation and a point to consider, state a possible translation. a x2 + y2 = 9, (x - 1)2 + (y + 3)2 = 9. Consider the centre (h, k). b (x + 1)2 + (y - 2)2 = 4, x2 + y2 - 2x + 2y - 2 = 0. Consider the centre (h, k). Complete the square on both x and y for the image circle. 1 1 c y = , y = + 2. Consider the point of intersection of the asymptotes. x x −1 8 Using the property that under a translation Th, k, y = f (x) → y - k = f (x - h), state a possible translation in words and in the form Ta, b, given a rule and its image rule. a y = 2x, y = 2x + 3 b y = 2x, y = 2x + 1 + 7 x x + 3 c y = 2 + 1, y = 2 -5 d y = 2x + 3 - 4, y = 2x - 2 + 1 9 MC Under T2, -1, the image of (2, 3) and the point whose image is (2, 3) are: a (4, 2), (0, 4) b (0, 4), (4, 2) c (4, 2), (4, 2) d (4, 2), (6, 1) E (0, 4), (0, 4) 10 MC Under T-2, 1, the image equation of y = x2 and the equation whose image equation is y = x2 are: A y = (x + 2)2 + 1, y = (x - 2)2 - 1 B y = (x - 2)2 - 1, y = (x + 2)2 + 1 C y = (x - 2)2 + 1, y = (x + 2)2 - 1 D y = (x + 2)2 - 1, y = (x - 2)2 + 1 E y = (x + 2)2 + 1, y = (x + 2)2 - 1


49

11 MC The translations which map (3, 4) onto (2, -4) and y = x2 + 1 onto y = (x - 2)2 + 2 are, respectively: a T-1, -8 and T2, 1 b T1, 8 and T2, 1 c T-1, -8 and T2, 2 d T1, 8 and T2, 2 E T-1, -8 and T-2, 1 1 1 12 MC The translations which map y = 2x onto y = 2x - 1 + 1 and y = onto y = − 1 are, x x +1 respectively: a T-1, -1 and T1, 1 d T1, -1 and T-1, 1

2B

b T1, 1 and T-1, -1 E T-1, -1 and T-1, 1

c T-1, 1 and T1, -1

Reflections of points and graphs On the diagram, A′ is the image of the point A under reflection in the line L. L is called the mediator. My = 0 represents reflection in the line y = 0, the x-axis. Mx = 0 represents reflection in the line x = 0, the y-axis.

y x

L

A y C(−x, y)

To determine the rule for each reflection, consider the diagram at right. My = 0 : (x, y) → (x, -y) gives point B. Mx = 0 : (x, y) → (-x, y) gives point C. Similarly, for reflection in the lines y = x and y = -x, consider the diagram at right. My = x : (x, y) → (y, x) gives point E. My = -x : (x, y) → (-y, -x) gives point F.

A′

−x

E(y, x)

y

−y

A(x, y) x x B(x, −y)

y y=x y

D(x, y) y

y

x x

−y −x F(−y, −x)

D(x, y) x x y = −x

Note: M is used to represent reflections since a reflection is a mirror image about its specified axis or line. Worked Example 7

Find the coordinates of the image of (2, -3) under the following reflections. a My = 0 (reflection in the x-axis) b Mx = 0 (reflection in the y-axis) c My = x (reflection in the line y = x) d My = -x (reflection in the line y = −x) Think

Write

a A reflection in the x-axis, My = 0, means (x, y) → (x, -y).

a (x, y) → (x, -y)

b A reflection in the y-axis, Mx = 0, means (x, y) → (-x, y).

b (x, y) → (-x, y)

c A reflection in the line y = x, My = x, means

c (x, y) → (y, x)

d A reflection in the line y = -x, My = -x, means

d (x, y) → (-y, -x)

(x, y) → (y, x). (x, y) →

50

(-y, -x).

(2, -3) → (2, 3)

(2, -3) → (-2, -3)

(2, -3) → (-3, 2)


(2, -3) → (3, -2)

Worked examPle 8

eBook plus

a Find the image equation of y = 2x + 1 under the following reflections.

i My = 0 (reflection in the x-axis)

ii My = −x (reflection in the line y = −x)

b Find the image equation of y = 4(x − 1)2 under the following reflections.

i Mx = 0 (reflection in the y-axis)

ii My = x (reflection in the line y = x)

Think a

i

ii

Tutorial

int-1212

Worked example 8

WriTe

1

A reflection in the x-axis, My = 0, means (x, y) → (x, −y).

2

Transpose to make x and y the subjects.

3

Substitute to find the image equations.

4

State the image equation without the primes.

1

A reflection in the line y = −x, My = −x, means (x, y) → (−y, −x).

2


3

Substitute to find the image equations. Make y′ the subject.

a

i x′ = x,

y′ = −y

x = x′ y = −y′ y = 2x + 1 ⇒ −y′ = 2x′ + 1 y = −2x − 1 ii x′ = −y,

y′ = −x

x = −y′ y = −x′ y = 2x + 1 ⇒ −x′ = −2y′ + 1 1 1 x′ + 2 2 1 1 y= x + 2 2

⇒ y′ =

b

i

ii

4


1

A reflection in the y-axis, Mx = 0, means (x, y) → (−x, y).

2


3


4


1

A reflection in the line y = x, My = x, means (x, y) → (y, x).

b

i x′ = −x,

y′ = y

x = −x′ y = y′ y = 4(x − 1)2 ⇒ y′ = 4(−x′ − 1)2 y = 4(−x − 1)2 ii x′ = y

y′ = x x = y′ y = x′

2


3


y = 4(x − 1)2 ⇒ x′ = 4(y′ − 1)2

4


The image equation is x = 4(y − 1)2. If the equation is required with y as the subject, x ( y − 1)2 = , 4 y −1= ±

x 1 or y = ± x +1 4 2

Chapter 2

Transformations

51

REMEMBER

1. Reflections: My = f (x), where y = f (x) is the equation of the mediator. 2. Use simple diagrams to find the rules of Mx = 0, My = 0, My = x, and My = -x, where Mx = 0 is a reflection in the y-axis, My = 0 is a reflection in the x-axis, My = x is a reflection in the line y = x and My = -x is a reflection in the line y = −x. Note: The above notation is an abbreviated version of a reflection; however, your description of reflections must be explained in words. 3. Use the image rules to find: (i) image points and image rules after a reflection, given a point or a rule and the reflection (ii) original points and original rules before a reflection, given an image point or an image rule and the reflection (iii) the reflection, given a point and its image or a rule and its image.

Exercise

2B

Reflections of points and graphs Find the coordinates of the image of (-2, 4) under the following reflections.

1 WE 7 a b c d

My = 0 (reflection in the x-axis) Mx = 0 (reflection in the y-axis) My = x (reflection in the line y = x) My = -x (reflection in the line y = −x)

2 WE 8

Find the image rules of the given rules under the following reflections. In each

case, sketch, on the same set of axes, the graphs of the relations before and after the reflection. a y = 3x, My = x b y = -2x, My = -x c y = 2x + 3, My = 0 d y = 2x2, My = 0 g y = x2 + 1, My = -x j y =

1 + 1, M x = 0 x

m y = 2x, My = 0 p y =

−

−

1 h y = , Mx = 0 x 1 k y = + 1, M y = x x −1

1 2 x , My = x 4 1 i y = , My= x x +1 2 l y = + 3, M y = − x x −1

n y = 2x + 3, Mx = 0

o y = x , M y = x

e y = -3x2, Mx = 0

f y =

x + 2, M y = − x

3 MC Under My = 0, the image of (2, 3) and the point whose image is (2, 3) are, respectively: a (2, -3) and (2, -3) b (2, -3) and (-2, 3) c (-2, 3) and (-2, 3) d ( 2, 3) and (2, 3) E ( 2, 3) and ( 2, 3) − 1 4 MC Under My = x, the image equations of y = x and y = are, respectively: x −1 1 1 a y = x 2 ( x ≥ 0) and y = + 1 b y = x 2 ( x ≤ 0) and y = + 1 x x 1 1 c y = x 2 ( x ≥ 0) and y = − 1 d y = x 2 ( x ≤ 0) and y = x x +1 1 E y = x 2 ( x ≤ 0) and y = − 1 x

52


dilations from axes

2C

A dilation of factor k from the y-axis (or parallel to the x-axis) is y represented by Dk, 1, with a rule (x, y) → (kx, y). As shown on the diagram, y the x-coordinate has been transformed by a dilation factor k, k ∈ R+, while the y-coordinate is unchanged. x kx x Similarly, D1, k , represents a dilation of factor k from the x-axis (or parallel to the y-axis), with a rule (x, y) → (x, ky). D2, 3, represents a dilation of factor 2 from the y-axis and a dilation of factor 3 from the x-axis, with a rule (x, y) → (2x, 3y). Worked examPle 9

Find the image of (4, −2) under the following dilations. a D2, 1 (dilation factor of 2 from the y-axis) 1 b D 1 (dilation factor of 2 from the x-axis) 1,

c D1 2

2

,3

(dilation factor of

1 2

from the y-axis and 3 from the x-axis)

Think

WriTe a x′ = 2x

a A dilation factor of 2 from the y-axis, D2, 1, means

(x, y) → (2x, y).

b A dilation factor of

(x, y) → (x, 1 y).

1 2

y′ = y (x, y) → (2x, y) (4, −2) → (8, −2)

b x′ = x

from the x-axis, D 1 , means 1,

2

y′ = 1 y

2

2

(x, y) → (x, 1 y) 2

(4, c A dilation factor of

1 2

→ (4, −1)

1 c x′ = 2 x,

from the y-axis and 3 from the

x-axis, D1 , means (x, y) → ( 1 x, 3y). 2 2

−2)

y′ = 3y (x, y) → ( 1x, 3y)

,3

2

(4,

−2)

→ (2, −6)

Worked examPle 10

Find the image equation of y = x2 under the following dilations. a D2, 1 (dilation factor of 2 parallel to the x-axis) 1 b D1 (dilation factor of 2 parallel to the x-axis and 3 parallel to the y-axis) 2

,3

Think a

1

eBook plus Tutorial


WriTe

A dilation factor of 2 parallel to the x-axis (i.e. from the y-axis), D2, 1, means (x, y) → (2x, y).

a x′ = 2x

y′ = y

Chapter 2

Transformations

53

2


3


1 x′ 2 y = y′ x=

y = x2  x′  ⇒ y′ =    2

4

b

1

State the image equation without the primes. A dilation factor of 12 parallel to the x-axis and 3 parallel to the y-axis (i.e. from the x-axis), D1 , means (x, y) → ( 1 x, 3y). 2

,3

2

The image equation is y =

1 2 x . 4

1 x 2 y′ = 3y

b x′ =

2

2


x = 2x′ 1 y = y′ 3

3


y = x2 1 ⇒ y ′ = (2 x ′) 2 3

4


The image equation is y = 12x2.

Worked Example 11

Find the original point if the image point under D2, 3 (dilation factor of 2 from the y-axis and 3 from the x-axis) is (6, -9). Think 1

A dilation factor of 2 from the y-axis and 3 from the x-axis, D2, 3, means (x, y) → (2x, 3y).

2

Transpose to make x and y the subjects. This means that the original point will have an 1 1 x-value 2 that of the image point and a y-value 3 that of the image point.

3

Multiply the x-value of the image point by 2 and the y-value of the image point by 3 to find the original point.

Write

x′ = 2x y′ = 3y 1 x = x′ 2 1 y = y′ 3 1 1  (x, y) →  x, y  2 3  (6, 9) → (3, -3) The original point is (3, -3).

Worked Example 12

Find the original equation, if the image equation under the dilation D3, 1 (dilation factor of 3 parallel to the x-axis) is y = 3x2 + 1. Think 1

54

A dilation factor of 3 parallel to the x-axis (i.e. from the y-axis), D3, 1, means (x, y) → (3x, y).

Write

x′ = 3x y′ = y


2

The image equation is y = 3x2 + 1. Reintroduce the primes and substitute the values for x′ and y′ to find the original equation.

y′ = 3x′2 + 1 ⇒ y = 3(3x)2 + 1

3

Simplify and state the original equation.

The original equation is y = 27x2 + 1.

REMEMBER

1. Dilations: x′ = mx, y′ = ny under Dm, n, with a dilation factor of m from the y-axis (or parallel to the x-axis) and a dilation factor of n from the x-axis (or parallel to the y-axis). Note: Dm, n is an abbreviation for the dilations; however, you must express the dilations in words. 2. Use the image rules to find: (i) image points and image rules after a dilation, given a point or a rule and the dilation (ii) original points and original rules before a dilation, given an image point or an image rule and the dilation (iii) the dilation, given a point and its image or a rule and its image. Exercise

2C

Dilations from axes 1 WE 9

Find the coordinates of the image point for the given point under the given dilation.

a (2, 3), D2, 1 d (-2, -6), D1 2

2 WE 10 a d g j

,2

c (4, -3), D2, 3

, 3 4

For the given equation under the given dilation, find the image equation.

y = x, D2, 1 y = 0, D1, 2 x = 2, D1, 5 y = x2, D1, 2

m y = 2x, D2, 3 p y = (x - 1)2, D2, 3 s y = 2-x, D2, 3 3 WE 11

b (-2, 5), D1, 3 e (-6, 4), D 2 3

y = x + 1, D1, 2 y = 0, D1, 4 x = 0, D2, 4 y = x2 - 1, D2, 3 1 n y = , D1, 2 x q y = (x + 2)2 + 1, D2, 2 t y = f (x), Da, b b e h k

y = x + 1, D2, 1 x = 2, D2, 1 y = x2, D2, 1 y = 2x, D1, 2 1 o y = , D2, 2 x r y = 3(x - 1)2 + 2, D3, 2 c f i l

Find the coordinates of the original point, given the image point under the given

dilation. a (3, 5), D1, 2 (dilation factor of 2 from the x-axis) b (3, 5), D2, 1 (dilation factor of 2 parallel to the x-axis) c (-2, 3), D2, 2 (dilation factor of 2 from the x- and y-axes) d (-2, -3), D2, 3 (dilation factor of 2 parallel to the x-axis and 3 parallel to the y-axis) 4 WE 12 Find the original equation, given the image equation under the given dilation. a y = x, D2, 1 (dilation factor of 2 parallel to the x-axis) b y = x2, D1, 2 (dilation factor of 2 from the x-axis) c y = 2x, D2, 2 (dilation factor of 2 from the x- and y-axes) 1 d y = , D2, 3 (dilation factor of 2 parallel to the x-axis and 3 parallel to the y-axis) x


55

5 mC Under D3, 2 (dilation factor of 3 from the y-axis and 2 from the x-axis), the image of (3, −2) and the point whose image is (3, −2) are, respectively: b (9, −4) and (1, −1) c (−1, 1) and (9, −4) a (9, −4) and (−1, 1) −1) and (9, −4) E (9, 4) and (1, 1) d (1, 6 mC Under D2, 3 (dilation factor of 2 parallel to the x-axis and 3 parallel to the y-axis), the image rule of y = f (x) and the rule whose image rule is y = f (x) are, respectively: 1  x a y = 3 f   and y = f (2 x) 2 3 1  x c y = f   and y = 3 f (2 x) 3 2

1  x f (2 x) and y = 3 f    2 3 1  x d y = 3 f (2 x) and y = f   3 2 b y=

eBook plus

 x E y = 3 f   and y = 3 f (2 x ) 2

2d

Digital doc

WorkSHEET 2.1

The ellipse and the hyperbola Note: Further work on the ellipse and the hyperbola is presented in chapter 10.

The ellipse y An ellipse is a conic section that is closely related to the circle. b Consider the image of the circle x2 + y2 = 1 under the x′ y′ 1 dilation Da, b , x′ = ax, y′ = by, so x = , y = a b −1 −a a x −1 1 2 2 x y  ′  ′   x 2 + y 2 = 1 →   +   = 1, therefore the image −b  a  b x2 y2 equation is 2 + 2 = 1. a b This is the standard form or general equation for an ellipse, with centre at (0, 0), x-intercepts (±a, 0) and y intercepts (0, ±b). Pronumeral a is called the semi-major axis and b the semi-minor axis. The ellipse has two axes of symmetry. The longer axis is called the major axis and its length is equal to 2a. Hence, a is the length of the semi-major axis. The shorter axis of symmetry is called the minor axis and its length is equal to 2b. Hence, b is the length of the semi-minor axis.

The hyperbola Consider the image of the hyperbola y = the dilation D 1 2

,

1 2

. x′ =

1 2

x, y ′ =

1 2

1 under x

y, so

x = 2 x ′, y = 2 y ′. 1 1 y = → 2y ′ = , therefore the image rule is y = 1 . x 2x ′ 2x  1 1  1 , The point  and 1 unit from the origin.  is on the line y = 2 2 2x 1   1 1 , is rotated clockwise through 45°,   maps 2 2 2x onto (1, 0) and the asymptotes x = 0 and y = 0 become y = x and y = −x. If y =

56

y


−1

1

x

y y = −ba x y = ab x −a

a

x

This rectangular hyperbola (rectangular because the asymptotes are at right angles) has the equation x2 - y2 = 1. x2 y2 This equation under the dilation Da, b becomes 2 − 2 = 1. This is the standard form a b or general equation for the hyperbola with centre (0, 0), x-intercepts or vertices (±a, 0) ±b and asymptotes y = x. a ( x − h) 2 ( y − k ) 2 − = 1, If the centre of the hyperbola is at (h, k) then the equation becomes a2 b2 ± b which has vertices (h ± a, k) and asymptotes ( y − k ) = ( x − h). a

Worked Example 13

x 2 y2 + = 1 under the following transformations. In each case, 9 4 sketch the graph of the original ellipse and its image. a T2, -3 (translation 2 units right and 3 units down) b My = x (reflection in the line y = x) Find the image equation of

Think a

1

2

3

4

5

b

1 2

Write/draw

Translation of 2 units right and 3 units down, T2, -3, means (x, y) → (x + 2, y - 3). Transpose to make x and y the subjects.

a x′ = x + 2

y′ = y - 3 x = x′ - 2 y = y′ + 3

x2 y2 + = 1 is 9 4 an ellipse with centre (0, 0). Substitute to find the image equations.

x2 y2 + =1 9 4 ( x ′ − 2)2 ( y ′ + 3)2 ⇒ + =1 9 4

State the image equation without the primes. The image is an ellipse with centre (2, −3). Sketch the original ellipse by plotting key points. Sketch the image graph by plotting key points. Some key points are: (0, 0) → (2, -3) (-3, 0) → (-1, -3) (3, 0) → (5, -3) (0, -2) → (2, -5) (0, 2) → (2, -1)

The image equation is

The original equation

Reflection in the line y = x, My = x, means (x, y) → (y, x). Transpose to make x and y the subjects.

( x − 2)2 ( y + 3)2 + = 1. 9 4

Centre: (2, -3). y 6 4 2 −6 −4 −2 −2 −4 −6

x2 9

y2 4

+

2

4

=1 6 x

(2, −3) (x − 2)2 9

+

(y + 3)2 4

=1

b x′ = y

y′ = x y = x′ x = y′


57

3


x2 y2 + = 1 is an 9 4 ellipse with centre (0, 0). Substitute to find the image equations.

x2 y2 + =1 9 4 ( y ′) 2 ( x ′) 2 ⇒ + =1 9 4

4

State the image equation without the primes. The image is an ellipse with centre (0, 0).

The image equation is

5

Sketch the original ellipse by plotting key points as for part a , then sketch the image ellipse using key points. Key points are: (0, 0) → (0, 0) (−3, 0) → (0, −3) (3, 0) → (0, 3) (0, −2) → (−2, 0) (0, 2) → (2, 0)

x2 y2 + = 1. 4 9

Centre: (0, 0). y y=x 3 x2 9

2 −3

+

−2 −3

x2 4

+

y2 9

=1

eBook plus

x2

y2

− = 1 under the following transformations. 4 9 In each case, find the equation of the asymptotes and sketch on separate axes the graph of the original hyperbola and its image showing the asymptotes. a T−2, 4 (translation left 2 units and 4 units up) b My = x (reflection about the line y = x) Think a

58

1

=1 x

3

Worked examPle 14

Find the image equation of

y2 4

Tutorial


WriTe

Translation of 2 units left and 4 units up, T−2, 4, means (x, y) → (x − 2, y + 4).

2


3


x2 y2 − = 1 is 4 9 an hyperbola with asymptotes ±3 y= x and vertices at (±2, 0). 2 Substitute to find the image equations and the image asymptotes and image vertices (h ± a, k).

a x′ = x − 2

y′ = y + 4 x = x′ + 2 y = y′ − 4

The image equation: x2 y2 − =1 4 9 ( x ′ + 2)2 ( y ′ − 4)2 ⇒ − =1 4 9 The equations of the image asymptotes: ±3 y= x 2 ±3 y′ − 4 = ( x ′ + 2) 2 The image vertices: (h ± a, k) = (−2 ± 2, 4) = (−4, 4) and (0, 4)


4

State the image equation and its asymptotes without the primes.

The image equation: ( x + 2)2 ( y − 4)2 − =1 4 9 Asymptotes: ±3 y−4= ( x + 2) 2 ⇒y= y=

5

Sketch the original graph by plotting ±3 x2 y2 x − = 1 with asymptotes y = 4 9 2 and vertices (±2, 0).

Sketch the image graph by plotting ( x + 2)2 ( y − 4)2 − = 1, with asymptotes 4 9 ±3 y= ( x + 2) + 4 and vertices 2 (-4, 4) and (0, 4).

2

( x + 2) + 4

−3 3 x + 7 or y = x + 1 2 2

y = −32 x

−2

6

±3

y = 32 x

−1

1

y = −32 x + 1

y

x

2

y = 32 x + 7

8 6

(−4, 4)

4 (0, 4) 2

−6 −4 −2

2

−2

4

6

x

−4 −6 −8 b

1

2

3

Reflection in the line y = x, My = x, means (x, y) → (y, x). Transpose to make x and y the subjects. x2 y2 − = 1 is an 4 9 ±3 hyperbola with asymptotes y = x and 2 vertices at (±2, 0). Substitute to find the image equations and the image asymptotes and image vertices (h ± a, k). The original equation

b x′ = y

y′ = x x = y′ y = x′

The image equation: x2 y2 − =1 4 9 ( y ′) 2 ( x ′) 2 =1 − 4 9 The equation of the image asymptotes: ±3 y= x 2 ±3 x′ = y′ 2 The image vertices: (0, ±2) ⇒


59

4

State the image equation and its asymptotes without the primes.

The image equation: y2 x2 − =1 4 9 −

x2 y2 =1 + 9 4

Asymptotes: ±3 x= y 2 ⇒y= 5

Sketch the original graph by plotting ±3 x2 y2 x and − = 1 with asymptotes y = 4 9 2 vertices (±2, 0).

±2

3

y = −32 x

x y 4

y = 32 x

2 −2

−2

x

2

−4

6

Sketch the image graph by plotting − x2 ±2 y2 = 1 with asymptotes y = x + 9 4 3 and vertices (0, ±2).

y y = −23 x

y = 32 x

4 2

−4 −2

2

4

x

−2 −4

REMEMBER

x2 y2 + = 1, centre (0, 0), x-intercepts (±a, 0), y-intercepts (0, ±b). a > b where a2 b2 a is the semi-major axis and b is the semi-minor axis.

1. Ellipse:

( x − h) 2 ( y − k ) 2 + = 1, centre (h, k). a2 b2 x2 y2 3. Hyperbola: 2 − 2 = 1, centre (0, 0), vertices (±a, 0), a b 2. Ellipse:

±

b x. a ( x − h) 2 ( y − k ) 2 − = 1, centre (h, k), vertices (h ± a, k), asymptotes 4. Hyperbola: a2 b2 ± b (y − k) = ( x − h). a asymptotes y =

60


Exercise

2D

The ellipse and the hyperbola 1

For each of the following ellipses, state the coordinates of the centre and the values of the semi-major and semi-minor axes. Sketch the graph of the ellipse. ( x − 5)2 ( y + 3)2 + =1 25 9 ( x + 2)2 ( y − 2)2 d + =1 4 16

x2 y2 + =1 25 9 y2 c x 2 + =1 4 a

b

2 For each of the following hyperbolas, state the coordinates of the centre and the vertices. Find the equations of the asymptotes; then sketch the graph. x2 y2 − =1 9 16 ( x + 2)2 ( y − 4)2 e − =1 4 16 − 2 x y2 h + =1 9 16 − ( x + 2)2 ( y − 4)2 k + =1 4 16

a x2 - y2 = 1

b

d (x - 3)2 - (y + 2)2 = 1 g -x2 + y2 = 1 j

-(x

+ 4)2 + (y - 2)2 = 1

x2 − y2 = 1 4 ( x + 2)2 ( y − 4)2 f − =1 16 9 y2 x2 i − =1 9 16 − ( x + 2)2 ( y − 4)2 l + =1 16 4 c

x2 y2 + = 1 under the following transformations. In each 16 9 case, sketch the graph of the image ellipse. a T-2, -4 (translation 2 units left and 4 units down) b My = x (reflection about the line y = x) c D 1 (dilation factor of 2 from y-axis and 1 from the x-axis)

3 WE 13 Find the image equation of

2,

3

3

x2

y2

− = 1 under the following transformations. In each 9 36 case, find the equation of the asymptotes and sketch on separate axes the graph of the image showing the asymptotes. a T-1, 2 b My = x c D1 1

4 WE 14 Find the image equation of

, 3 2

−

x2 y2 5 Find two pairs of values of a and b for which 2 + 2 = 1 has the same asymptotes as a b x2 y2 − = 1. 9 4 ( x − 1)2 ( y + 2)2 6 MC + = 1 is defined for: 9 4 a -2 ≤ x ≤ 4, -4 ≤ y ≤ 0 b -4 ≤ x ≤ 2, 0 ≤ y ≤ 4 c -1 ≤ x ≤ 3, -5 ≤ y ≤ 1 d 3 ≤ x ≤ 1, 1 ≤ y ≤ 5 E 3 ≤ x ≤ 3, 2 ≤ y ≤ 2 7 MC The equations of the asymptotes of a y = 2 x − 4 − 2 2 , y = c y = 2 x + 4 + 2 2 , y =

− −

( x − 2)2 ( y + 4)2 − = 1 are: 4 8

2x − 4 + 2 2

2 b y = 2 x − 4 − 2 , y =

2x + 4 − 2 2

d y = 2x - 8, y = -2x

−

2 x−4+ 2 2

e y = 2x + 8, y = -2x


61

successive transformations

2e

eBook plus

Finding the final image rule under successive transformations Worked examPle 15

a

b

int-0969 Successive transformations

eBook plus

Under the translation T−2, 3 followed by the reflection My = −x: a find the coordinates of the final image of the point (2, −3) b find the final image equation of y = x2. Think

Interactivity

Tutorial


WriTe

T−2, 3 M y = −x ( x, y ) →( x − 2, y + 3)  →( − y − 3, − x + 2)

1

Determine a single rule that replaces the successive transformations.

a

2

Use this rule to state the image point.

1

Set up the image equations.

2

Transpose to make x and y the subjects. Substitute the values of x and y to find the image equation. Transpose to make y the subject of the image equation.

(x, y) → (−y − 3, −x + 2) (2, −3) → (−(−3) − 3, −2 + 2) (2, −3) → (0, 0) b x′ = −y − 3 y′ = −x + 2 y = −x′ − 3 x = −y′ + 2 y = x2 ⇒ −x − 3 = (−y + 2)2 As (−y + 2)2 = (y − 2)2 therefore the image equation is (y − 2)2 = −x − 3 or y = 2 ± − x − 3

3 4

Worked examPle 16

Find a transformation for each of the following. a y = f (x) → y = 2 f (x) b y = f (x) → y = f (2x) c y = f (2x) → y = f (2x − 2) d y = f (2x) → y = − f (2x) e y = f (−x) → y = f (−x + 2) − 2 Think

WriTe

a x′ = x, y′ = 2y b 2x′ = x or x′ =

a D1, 2 (dilation factor of 2 from the x-axis)

1 x, y′ = y 2

b D1 2

,1

(dilation factor of 1 from the y-axis) 2

c 2x = 2x′ − 2 or x′ = x + 2, y′ = y

c T2, 0 (translation 2 units right)

d x′ = x, y′ = −y

d My = 0 (reflection about the x-axis)

e

−x′

+ 2 = −x or x′ = x + 2, y′ = y − 2

e T2, −2 (translation 2 units right and 2 units down)

rememBer

Find a single rule to replace successive transformations and use it to find final image points or equations.

62


Exercise

2E

Successive transformations 1

Find the rule of a single transformation which replaces the following successive abbreviated transformations, given the original is (x, y). a D2, 1, then T3, 4 b T3, 4, then D2, 1 c My = x, then T3, 4 d T3, 4, then My = x e D2, 3, then T3, 4 f T3, 4, then D2, 3 g T3, 4, then T-4, 3 h D2, 1, then D3, 2 i My = x, then Mx = 0 j My = x, then My = -x, then My = 0 k T2, -3, then D3, 2, then My = 0 l My = 0, then T2, -3, then D3, 2 m D3, 2, then My = 0, then T2, -3

2

Under a transformation, Tr1, (x, y) → (2x + 1, -3y + 2). Under a transformation, Tr 2, (x, y) → (-2y + 2, -3x + 2). Find the rule of a single transformation that replaces the following successive transformations. a Tr1, then Tr2 b Tr2, then Tr1 c Tr1, then Tr1 d Tr 2, then Tr 2

3 WE 15

Find the final image of (-2, 3) under the following successive transformations.

a D2, 3, then T2, 4 c T-2, -3, then D2, 3 4 WE 15

b Mx = 0, then T1, -2 d T1, -2, then D2, 2, then My = -x

Find the final image equation for y =

transformations. a T2, 4, then D2, 3 c D2, 3, then T-2, -3 5

1 under the following successive x

b T1, -2, then Mx = 0 d D2, 2, then T1, -2, then My = -x

Find the transformation for each of the following. a y = x2 → y = (x - 2)2 + 3 d y = 3x → y = 3-x g y =

− 1 1 →y= x x

j x 2 + y 2 = 1 →

b y = (x - 1)2 → y = 4(x - 1)2 e y = 3-x → y = 3-x + 4 h y =

1 1 →y= 2x 2x − 5

x2 y2 + =1 4 25

l x2 - y2 = 1 → (x + 4)2 - (y - 3)2 = 1 n y = 4x2 → y = (2x - 1)2 6 WE 16

c y = 2x2 → y = -2x2 f y = 32x → y = 32x + 1 - 2 i y =

5 5 →y= x +1 x+2

x2 y2 + = 1 → x2 + y2 = 1 4 25 x2 y2 m − = 1 → x2 − y2 = 1 9 16 − 2 x y2 o − x 2 + y 2 = 4 → + =1 9 16 k

Find a transformation for each of the following.

a y = f (x) → y = -f (x) c y = -f (-x) → y = -f (-x + 1) e y = -2 f (-2x + 1) → y = -2 f (2(-x + 1))

b y = -f (x) → y = -f (-x) d y = -f (-x + 1) → y = -2 f (-x + 1) f y = -2 f (2(-x + 1)) → y = -2 f (-2x + 5) + 4

7 State the successive transformations starting with the first equation and finishing with the second. There is more than one correct order for the transformations. a y = x → y = -2x + 4 b y = x2 → y = -2(x + 1)2 - 3 1 2 +4 d y = x2 → y = (2x + 3)2 c y = → y = x x−3 e y = 2x → y = -2 4x + 3 f y = 2x → y = 3(23x - 5)


63

h y = f (x) → y = 2f (−2x − 1) − 4

g y = f (x) → y = − 4 f (2x − 1) + 3 i y = x6 → y = − (−2x + 4)6 + 7

8 Sketch the graphs of the following using successive transformations, starting with the first equation. a y = x2, y = −2(x − 2)2 − 1 b y = x2, y = (2x − 1)2 − 2 c y = 1, y = +4 x 2x − 3 e y = 2x → y = 3(22x + 3)

d y = 2x → y = −2x + 3 + 4 f y = 2x → y = −3(22x + 3) + 1

9 mC Under successive transformations, y = f (x) → y = f (2x) → y = f (−2x) → y = f (−2x − 2). The transformations, in order, are: a D2, 1, My = 0, T−2, 0 b D1 , Mx = 0, T−2, 0 c D1 , Mx = 0, T−1, 0 2

d D1 , My = 0, T1, 0 2

,1

,1

2

,1

E D1 , Mx = 0, T−1, 0 2

,1

10 mC Under the sequence of transformations, T1, 2, My = x, D2, 3 the final image rule 1 for y = is: x 6 −6 x−4 1 d y = 3( x + 4) − 1 a y=

6 −3 x+4 1 E y = 3( x + 4) + 1 b y=

c y=

6 +3 x−4 eBook plus Digital doc

WorkSHEET 2.2

64


Summary Translations of points and graphs

• The rules, graphs and basic properties of lines, parabolas, hyperbolas, circles and exponentials as listed under ‘Some simple relations’ on page 44 are starting points for the rules in this chapter. • Translations: Th, k, where x ′ = x + h, y ′ = y + k – Th, k is a translation h in the x direction (horizontally) and a translation k in the y direction (vertically). Note: Th, k is an abbreviated version of a translation; however, your description of the translation must be explained in words. • Use x ′ = x + h, y ′ = y + k to find: 1. image points and image rules after a translation, given a point or a rule and the translation 2. original points and original rules before a translation, given an image point or an image rule and the translation 3. the translation, given a point and its image or a rule and its image. Reflections of points and graphs

• Reflections: My = f (x), where y = f (x) is the equation of the mediator. • Use simple diagrams to find the rules of Mx = 0, My = 0, My = x, My = -x, where Mx = 0 is a reflection in the y-axis, My = 0 is a reflection in the x-axis, My = x is a reflection in the line y = x and My = -x is a reflection in the line y = −x. Note: The above notation is an abbreviated version of a reflection; however, your description of reflections must be explained in words. • Use the image rules to find: 1. image points and image rules after a reflection, given a point or a rule and the reflection 2. original points and original rules before a reflection, given an image point or an image rule and the reflection 3. the reflection, given a point and its image or a rule and its image. Dilations from axes

• Dilations: x ′ = mx, y ′ = ny under Dm, n, with a dilation factor of m from the y-axis (or parallel to the x-axis) and a dilation factor of n from the x-axis (or parallel to the y-axis). Note: Dm, n is an abbreviation for the dilations; however, you must express the dilations in words. • Use the image rules to find: 1. image points and image rules after a dilation, given a point or a rule and the dilation 2. original points and original rules before a dilation, given an image point or an image rule and the dilation 3. the dilation, given a point and its image or a rule and its image. The ellipse and the hyperbola

x2 y2 + = 1, centre (0, 0), x-intercepts (±a, 0), y-intercepts (0, ±b). a > b where a is the semi-major a2 b2 axis and b is the semi-minor axis.

• Ellipse:

• Ellipse:

( x − h) 2 ( y − k ) 2 + = 1, centre (h, k). a2 b2


65

±b x2 y2 − 2 = 1, centre (0, 0), vertices (±a, 0), asymptotes y = x. 2 a b a ( x − h) 2 ( y − k ) 2 • Hyperbola: − = 1, centre (h, k), vertices (h ± a, k), asymptotes a2 b2

• Hyperbola:

(y − k) =

±b

a

( x − h)

Successive transformations

• Find a single rule to replace successive transformations and use it to find final image points or equations.

66



1 A translation that maps (-2, 3) onto (4, 1) also maps y = x2 onto its image. Find the equation of the image. 2 Given a rule and its image rule, state the translation in the form Ta, b and express it in words. a y = x2, y = (x + 2)2 - 4 1 1 b y = , y = +1 x x +1 −

c

( x − 4)2 ( y + 2)2 + =1 4 16

d

( x − 1)2 ( y + 2)2 + =1 2 3

8 For each of the following hyperbolas, state the coordinates of the centre and the vertices. Find the equations of the asymptotes; then sketch the graph. Exam tip Sketch the centre, asymptotes and vertices; then make sure you sketch the graph in the [Author’s advice] correct quadrants.

−

c y = x, y = x+2 d x2 + y2 = 4, (x – 2)2 + (y + 5)2 = 4 e y = 5x, y = 5x + 3 - 4 f y = x2, y = x2 + 4x - 2 3 Find the image of the given point under the given reflection. a (-2, 0), Mx = 0 b (0, -3), My = 0 c (2, 3), My = x d ( -3, 2), My = -x 4 Find the image rule for the given rule under the given reflection. Sketch the graphs of the given rule and the image rule on the same set of axes. a y = 2x, Mx = 0 b y = 2x2, My = 0 1 c y = 4x2, My = x d y = , My = −x x +1 e y = 2x + 1, My = 0 f x2 + (y + 1)2 = 9, My = 0 g y = x , M y = x

h y =

−

x, Mx =0

5 Find the coordinates of the original point, given the coordinates of its image and the dilation. a (2, 3), D2, 1 b (-6, 3), D1, 3 c ( − 4, 4), D1

1 , 2 4

6 For the given equation under the given dilation, find the image equation. 1 a y = x2, D2, 2 b y = , D2, 3 x d y = (x + 1)2 + 2, D3, 2 c y = x , D4,1 7 For each of the following ellipses, state the coordinates of the centre, the semi-major and semiminor axes; then sketch the graph. x2 a + y2 = 1 4 ( x − 2)2 ( y + 4)2 b + =1 9 4

a c

x2 y2 − =1 4 16 − 2 x

4

+

b ( x − 2)2 −

y2 =1 16

d

−

( y − 1)2 =1 4

( x + 3)2 y 2 + =1 4 4

9 Given a point and an equation, find the coordinates of the image point and the image equation under the given successive transformations. a (x, y) → (2x - 2, 3y - 3), (2, -3), y = x2 b (x, y) → (2x - 2, 3y + 3), (-3, 2), y = x2 1 c (x, y) → (-2x + 2, -3y - 3), (-2, -1), y = x 1 d (x, y) → (-y - 2, -2x + 4), (-2, -3), y = x 10 Find the rule of a transformation that replaces the successive transformations in the given order. a T-2, 3, D2, 1 b D2, 1, T-2, 3 c D2, 2, T2, -3, My = x d My = x, D2, 2, T2, -3 e T2, -3, My = x, D2, 2 11 Find the transformation for each of the successive transformations. 2 3 a y = 3x 1 → y = 3 − x  → y = − 3 − x  → y = − 7(3− x )  → y = − 7(3−2 x ) 4

 → y = − 7(3−2 x + 2 ) 5

− − 1 1 1 2 1 3  →y =  →y =  → x x 2x − 1 y= −1 2x + 1

b y =

1

2

→ y = 2 f ( x )  → y = − 2 f ( x) c y = f ( x )  3

4

 → y = − 2 f (3 x )  → y = − 2 f ( − 3x) 5

 → y = − 2 f ( − 3 x + 2) + 2


67

12 Starting with the first equation, graph the second equation using successive transformations. a y = x → y = -2x + 3 b y = x2 → y = -2(x - 1)2 + 6 c y = x → y = − 2 2 x − 1 + 2 d y = 3x → y = -2(3 2x + 2) e x 2 + y 2 = 1,

( x + 1)2 ( y − 2)2 + =1 4 9

f x 2 − y 2 = 1,

x 2 ( y − 2)2 − =1 4 9

Multiple choice

1 The translation that maps (-2, 4) onto (4, -1) maps (3, 5) onto: A (1, 1) B (9, 0) C (5, 2) D (9, 2) E (5, 0) 2 Which of the following correspond to an equation and its image under the translation T2, -1? A y = x, y = x - 3 B y = 2x, y = 2x - 5 1 1 C y = x2, y = x2 - 5 D y = , y = −1 x x−2 E y = 2x, y = 2x - 1 - 2 1 1 3 Under a translation, y = maps onto y = + 3. x x+2 The equations of the asymptotes for the image are: A x = 0, y = 0 B x = 2, y = -3 C x = 2, y = 0 D x = -2, y = 3 E x = 2, y = 3 4 Under a reflection, y = x2 maps onto y = -x2. The image of (-2, 3) under this reflection is: A (2, 3) B (-3, 2) C (-3, -2) D (2, 3) E ( 2, 3) 5 The image equation of y = x + 1 − 1 under My = x is: A y = x2 + 2x B y = x2 + 2x + 2 2 C y = x D y = x2 + 2x - 2 2 E y = x - 2 6 Under a dilation (x, y) → (ax, by), the image equation of y = f (x) is: A y =

1 b

 x f   a

C y = bf (ax) 1 E y = f (ax ) b

68

 x B y = bf    a  x D y = af    b

7 (6, -4) is the image point under D 1 . The original 3, point is: 2 -2) B (18, 8) C (2, -8) A (2, -2) E ( 8, 2) D (18, ( x + 2)2 ( y + 3)2 8 The translation that maps + =1 9 16 x 2 ( y + 7)2 onto = 1 is: + 9 16 A T0, -7 B T-2, 4 C T2, 10 D T-2, -10 E T2, -4 9 The equations of the asymptotes of ( x − 2)2 − ( y − 1)2 = 1 are: 4 A y = 2x - 3, y = -2x + 5 − 1 1 B y = x − 1, y = x+3 2 2 − 1 1 C y = x + 2, y = x 2 2 − 1 1 D y = x, y = x+2 3 2 E y = 2x + 3, y = -2x - 5 10 The coordinates of the vertices of ( x − 1)2 ( y + 3)2 − = 1 are: 4 9 B (-1, 3), (3, 3) A (-1, -3), (3, -3) C ( 3, 3), (1, 3) D (-3, -3), (-1, -3) E ( 3, 3), ( 1, 3) 11 The rule of a single transformation that replaces T2, 3, then D2, 1, then My = x is (x, y) → A (y - 3, 2x - 4) B (y + 3, 2x + 4) 1   C  y − 3, x − 1   2

D (-y + 3, -2x + 4)

E (y - 3, 2x - 2) 12 Under the transformations Tr1: (x, y) → (2x + 1, -2y + 2) and Tr 2: (x, y) → (-y + 2, 2x), the rule for Tr1 followed by Tr 2 is (x, y) → A (2y + 4, 4x + 2) B (-2y + 5, -4x + 2) C ( 2x + 1, 4y + 4) D (-y + 2, 2x) E (2y, 4x + 2) 13 Under successive transformations, 1

2

y = x  → y = 2 x  → y = 2 x + 1, the transformations 1 and 2 are: A D2, 1 and T-1, 0 B D1 and T− 1


2

,1

2

,0

c D1 and T1 2

,1

2

,0

d D2,1 and T− 1 2

,0

E D1, 2 and T−1, 0 14 Under successive transformations

15 Which of the following sequences of successive transformations does not produce the rule y = 2x → y = −3(22x − 3) + 1? a M y = 0 , D1, 3 , D1 , T3 2

1

y = − f ( − x )  → y = − 2 f (− 2 x )

b M y = 0 , D1 , T3

2

 → y = − 2 f (− 2 x + 2) + 3, the transformations 1 and 2 are: b D1 a D2, 2 and T1, 3 2

c D1 2

E D

,2

2,

1 2

and T1, 3 and T−1, 3

,1

2

,2

and T−1, 3

d D2, 1 and T−1, 0

,3

2

c D1 , M y = 0 , T3 2

,3

2

2

,1

,1

,1

d D1, 3 , M y = 0 , D1 , T3 2

,1

2

,1

E D1 , T3 , M y = 0 2

,3

2

,1

exTended resPonse

1 a State the translation that maps x2 + y2 = 1 onto (x − 1)2 + (y + 2)2 = 1. b By completing the square on both x and y, express x2 + y2 + 2x + 4y − 13 = 0 in the form (x − h)2 + (y − k)2 = r2. State the coordinates of the centre and the radius. c Find the translation that maps x2 + y2 = 1 onto x2 + y2 + 2x − 2y − 7 = 0. x2 ( x + 1)2 2 a State the translation that maps − y 2 = 1 onto − ( y − 3)2 = 1. 4 4 b By completing the square on both x and y, express 9x2 − 4y2 + 18x − 8y − 31 = 0 in the form ( x − h) 2 ( y − k ) 2 − = 1. State the coordinates of the centre and the semi-major and semi-minor axes. a2 b2 c Find the translation that maps 92 − y2 + 2x − 3 = 0 onto x2 − y2 + 4x + 2y − 1 = 0.



Chapter 2

Transformations

69

eBook plus

aCTiviTies


• 10 Quick Questions: Warm up with ten quick questions on transformations. (page 44) 2a

Translations of points and graphs

Tutorial

• We4 int-1032: Watch how to determine the rule for the image achieved after translations using a CAS calculator. (page 46) 2b

Reflections of points and graphs

Digital doc

• We8 int-1212: Watch how to find the image equations following transformations. (page 51) 2c

dilations from axes

Tutorial

• We10 int-1033: Watch how to find the image equation after dilations. (page 53) Digital doc

• WorkSHEET 2.1: Sketch and interpret nonlinear graphs. (page 56) 2d

The ellipse and the hyperbola

Tutorial

• We14 int-1034: Watch how to find the image equation of an ellipse after transformations. (page 58) 2E

Successive transformations

Tutorial

• We15 int-1035: Watch how to determine the coordinates and image equations after various transformations. (page 62) Digital doc

• WorkSHEET 2.2: Sketch graphs and identify transformations. (page 64) chapter review Digital doc


Interactivity

• Successive transformations int-0969: Consolidate your understanding of the effects of successive transformations. (page 62)

70


3

3a Relations 3b Functions 3c Inverse functions

relations and functions AreAs oF sTudy

• Sketching relations in the Cartesian plane from descriptions, equations or formulas and identifying their key features

• Sketching the graph of an inverse function from the graph of a simple function

eBook plus

3A

Digital doc

relations

10 Quick Questions

A relation is a set of ordered pairs. The following are examples of relations in listed notation. A = {(1, 2), (2, 3), (3, 4), (4, 5)} and B = {(1, 1), (1, 2), (2, 1), (3, 2)} The set of first elements from each ordered pair is the domain. A typical domain element is often represented by x. The set of second elements from each ordered pair is the range. A typical range element is often represented by y. For relation A, the domain = {1, 2, 3, 4} and the range = {2, 3, 4, 5}. Alternatively, using set builder notation, the domain = {x: x ∈ Z, 1 ≤ x ≤ 4}, which reads as ‘the set of x-values, such that x is an integer between 1 and 4 inclusive’, and the range = {y: y ∈ Z, 2 ≤ y ≤ 5}, which reads ‘the set of y-values, such that y is an integer between 2 and 5 inclusive’. For relation B, the domain = {1, 2, 3} and the range = {1, 2}. Some relations have a rule that relates the domain elements with the range elements. For relation A, the rule is y = x + 1 and using set builder notation A = {(x, y): x ∈ Z, y = x + 1}, which reads as ‘the set of coordinate pairs, (x, y), such that the domain is an element of integer numbers and the rule for y is y = x + 1’. Note that a relation may be uniquely defined using the domain and rule. For relation B, there is no obvious rule, so listing is the only method for representing this relation.

Graphs of relations Worked exAMPle 1

eBook plus

Graph the following relations and state the range. a {(x, y): x ∈ Z +, y = x + 1} b {(x, y): x ∈ [−1, 2), y = x2} 2 2 c {(x, y): x + y = 4}

Tutorial


Chapter 3

relations and functions

71

Think a

1

Write

The graph of y = x + 1 is linear. The domain of the relation is x ∈ Z +, which is all positive integers. Create a table of values, starting with the smallest value of x. A small number of points will suffice.

2

Determine the range.

3

Draw a pair of axes, showing an appropriate scale and plot the points. Do not join the dots as x ∈ Z +. Place an arrow on the last point to indicate that the pattern continues in that direction.

a The smallest positive integer is 1.

x+1 1+1 2+1 3+1 4+1

x 1 2 3 4

The range is y ≥ 2, where y ∈ Z +. y 6 4

y=x+1

2 0

b

1

2

3

c

1

The graph of y = x2 is parabolic. The domain of the relation is x ∈ [−1, 2), so the graph is continuous over the interval. The square bracket indicates that there should be a closed circle at x = −1 and the round bracket indicates that there should be an open circle at x = 2. Create a table of values to determine the points on the graph. Draw a pair of axes, showing an appropriate scale and plot the points. Join the dots.

Determine the range. The smallest y-value is 0 and the highest is 4, which has an open circle and is therefore not included. x2 + y2 = 4 is a circle with centre (0, 0) and a radius of 2. The domain is not explicitly stated and therefore x ∈ [−2, 2] is implied. Draw a pair of axes, showing an appropriate scale and sketch the circle as described.

y 2 3 4 5

b

2

4

6

x

x

x2

y

-1

(-1)2

1

0

(0)2

0

1

(1)2

1

2

2

4

(2)

y 5 4 y = x2 3 2 1 −1 0 1 2 3 4 5 x

The range is: y ∈ [0, 4). y

c

−2

2

x2 + y2 = 4

0

2

x

−2 2

Determine the range.

The range is: y ∈ [−2, 2].

REMEMBER

1. A relation is a set of ordered pairs. 2. The domain of a relation is the set of first elements of a set of ordered pairs. 3. The range of a relation is the set of second elements of a set of ordered pairs.

72


Exercise

3a

Relations 1 For each of the following relations, state the domain and range. a {(2, 3), (4, 7), (6, 8)} b {(2, 1), (3, 1), (4, 1)} c {(2, 2), (2, 3), (2, 5)} d {(1, 2), (1, 3), (2, 2), (3, 2)} f {(x, y): x ∈ Z, x ≥ 5, y = -x} e {(x, y): x ∈ Z +, y = 2x - 1} g {(x, y): x ∈ Z, 4 ≤ x < 21, y = x - 4} h {(x, y): x ∈[2, 7), y = x + 1} i {(x, y): x ∈[2, 4), y = x2} j {(x, y): x ∈(-1, 1), y = x2} k {(x, y): x ∈(-2, -1), y = x2} l {(x, y): y = x2 + 3} -x2 + 3} n {(x, y): x2 + y2 = 9} m {(x, y): y = 2 2 o {(x, y): (x - 2) + y = 4} p {(x, y): x ∈ [0, 3), x2 + y2 = 9} 2 WE1 Sketch the graph of each of the following relations. State the domain and range. a {(1, 2), (1, 3), (2, 2), (3, 2)} b {(2, 2), (2, 3), (2, 5)} c {(x, y): x ∈ Z, x ≤ 5, y = x + 2} d {(x, y): x ∈[-2, 3], y = -x2} 2 e {(x, y): y = (x - 1) + 3} f {(x, y): y = -2(x - 1)2 - 2} 2 g {(x, y): x ∈ ( 2, 1], y = 2(x + 1) + 3} h {(x, y): (x - 2)2 + (y + 1)2 = 9} 2 2 i {(x, y): (x + 2) + (y - 2) = 5} j {(x, y): x ∈[0, 2), x2 + y2 = 9} k {(x, y): x ∈[0, 1), (x - 2)2 + (y + 1)2 = 9} l {(x, y): y ∈[0, 3), x2 + y2 = 9} 2 2 m {(x, y): x ∈[0, 2), y ≥ 0, x + y = 9} n {(x, y): y ≥ 3, (x + 2)2 + (y - 2)2 = 9} 3

For each of the following graphs, state the domain and range. a

b

−10 −5

c

y

y

y

10

10

10

5

5

5

5

−5

10

x

−10 −5

−10

−5

5

10

x

−10

d

y

y

10

10

10

5

5

5

−5

5

10

x

−10 −5

−5

5

10

x

−10 −5

−5

h y

y

10

10

10

5

5

5

−10

5

10

x

5

10

5

10

x

x

i

y

−5

10

−10

−10

g

5

f

y

−10

−10 −5

−5

−10

e

−10 −5

−10 −5

−10 −5

−5

−10

5

10

x

−10 −5

−5

x

−10

Chapter 3 Relations and functions

73

j

k

−10 −5

l

y

y

y

10

10

10

5

5

5

−5

5

10

x

−10 −5

−10

4

x

−10 −5

−5

5

10

x

−10

( y − 3)2 = 1 are, respectively: 4 b [−2, 0] and [1, 5] c [−2, 0] and [−5, −1] − e [0, 2] and [ 5, 1]

MC The domain and range of ( x + 1)2 +

MC The domain of

a R\(−1, 3) c R\[−1, 3) e R\{−1, 3}

3B

10

−10

a [−1, 1] and [−2, 2] d [0, 2] and [1, 5] 5

−5

5

( x − 1)2 ( y + 2)2 + = 1 is: 4 16 b R\(−1, 3] d R\[−1, 3]


WorkSHEET 3.1

Functions A function is a relation that does not repeat the first element in any of its ordered pairs. That is, for any x-value, there is only one y-value. So, a function is a relation that is either one-to-one (1–1) or many-to-one (many–1). A one-to-one function has one x-value for a given y-value, whereas a many-to-one function has more than one x-value for a given y-value. In general terms, a relation can be A–B. To test for A, you can use a horizontal-line test. If the line crosses the graph once, A = 1; if it crosses more than once, A = many. To test for B, you can use a vertical-line test; again, if the line crosses once, B = 1, and if it crosses more than once, B = many. For a relation to be a function, B must always equal 1. Graphs of one-to-one and many-to-one functions are shown below. y

y Horizontal line test (A = 2) 1

1 1 Vertical line test (B = 1)

x Horizontal line test (A = 1)

1−1 function

2 x 1 Vertical line test (B = 1)

Many−1 function

A graph of a relation also represents a function if a vertical line does not cut the graph more than once. We have seen that listing or set builder notation can be used to state a function, but there is an additional mapping notation for functions only. f: X → Y, f (x) = .............. For this mapping notation, f is the label for the mapping or function. X is always clearly expressed as the domain. Y, the co-domain, is a set that is large enough to contain the range, so the range ⊆ co-domain. f (x) or y is the image of x under the mapping. f (x) = ............... or y = f (x) is the rule for f. 74


Worked exAMPle 2

Explain why each of the following is an invalid use of mapping notation. 1 a f: R → R, f ( x ) = x b f: R+ → R+, f (x) = x − 1 c f: R → R, f (x) = ± x Think

WriTe a The function not defined when x = 0

a The domain is not specified.

and this needs to be stated clearly in the mapping notation.  1

−1

b The range is not a subset of the co-domain.

b f = . The range ⊄ co-domain.  2 2

c f (x) is not a function.

c The graph is not a function since it

is one-to-many. Mapping notation cannot be used if the relation is not a function.

Worked exAMPle 3

eBook plus

Find the range for the following functions. a f: R+ → R, f (x) = 4x − 1 b f: R → R, f (x) = −x2 − 4x + 5 c f: R → R, f (x) = 2x − 1 Think a

b

1

Tutorial


WriTe

f (x) = 4x − 1 is linear. The domain is x ∈ R+ or x ∈ (0, ∞).

2

f (0) = −1, but (0, −1) is not included and therefore this lower end of the range must be represented using a round bracket. State the range.

1

f (x) = −x2 − 4x + 5 is an inverted parabola over the set of real − b numbers. Use x = to 2a determine the x-value of the turning point, as this can be used to indicate the maximum y-value of the graph.

a When x = 0, f (0) = 4(0) − 1

= −1

The range: y ∈ (−1, ∞).

b x=

− − ( 4) −2

= −2

2

Substitute this x-value into f (x) to determine the maximum y-value.

f (−2) = −4 + 8 + 5 = 9

3

State the range.

The range: y ∈ (−∞, 9].

Chapter 3


75

c

1

Use a CAS calculator to draw this graph. f (x) = 2x - 1 is an exponential graph over the set of real numbers. As x → -∞, 2x → -1, y = -1 is an asymptote.

2

Use the graph and the information described above to state the range.

c

The range is: y ∈ (-1, ∞).

Worked Example 4

If f: R → R, f (x) = 2x2 - 4x + 1, find a f (x2) b f (2x + 1) Think

Write

f (x2),

x2

a

To find substitute and simplify.

b

To find f (2x + 1), substitute 2x + 1 for x and simplify.

for x

a

f (x) = 2x2 - 4x + 1 f (x2) = 2(x2)2 - 4(x2) + 1 = 2x4 - 4x2 + 1

b

f (x) = 2x2 - 4x + 1 f (2x + 1) = 2(2x + 1)2 - 4(2x + 1) + 1 = 2[4x2 + 4x + 1] - 8x - 4 + 1 = 8x2 + 8x + 2 - 8x - 4 + 1 = 8x2 - 1

Alternatively, on the Main screen, complete the entry line as: Define f (x) = 2x2 - 4x + 1 f (x2) f (2x + 1) Press E after each entry.

REMEMBER

1. A function is a relation that does not repeat the first element in any of its ordered pairs. That is, for any x-value, there is one y-value. 2. A function can be one-to-one or many-to-one. 3. The graph of a function cannot be crossed more than once by any vertical line.

76


4. Mapping notation for functions: f: X → Y, f (x) = ..............., where X is the domain, Y is the co-domain (range ⊆ Y) and f (x) is the image of x. 5. When required, the range can be determined from the domain and the rule. exerCise

3B

Functions 1

We2

Explain why the following are an invalid use of mapping notation.

a f : R → R, f ( x ) = x + 5 d f : R + → R, f ( x ) = 2

±

b f : R → R, f ( x ) =

x2 + x x

c f: [1, 5] → R+, f (x) = 4 − x

e f: R+ → R+, f (x) = x3 − 3x

x

f f: R → R, x = 2

We3 Find the range for the following functions.

a c e g i

f: [2, 5) → R, f (x) = 3x − 2 f: R → R, f (x) = 3x + 2 f: R → R, f (x) = −2x2 + 8x + 1 f: (1, 3) → R, f (x) = x2 + 4x + 2 f: [1, 3) → R, f (x) = −2x + 2

k f : [4, ∞ ) → R, f ( x ) = x + 5 m f : ( − ∞ , 2) → R, f ( x ) = o f : R\{0} → R, f ( x ) =

x2

1 2− x

f: R → R, f (x) = x2 + 2x + 3 f: (2, 4] → R, f (x) = 3 − 2x f: R → R, f (x) = 4 − 2x f: (1, 3) → R, f (x) = x2 − 4x + 2 f: R → R, f (x) = 3−x + 2 2 l f : R\{1} → R, f ( x ) = x −1 −1 n f : R → R, f ( x ) = x 2 +1

b d f h j

+x x

3 For f (x) = x2 − 5x + 6, find: a the factors of f (x) and the solutions of f (x) = 0

b We4 f (x2 − 1) in its simplest form.

4 The solutions of f (x) = 0 are 2 and 6. Find the solutions of: a f (x2 − x) = 0 b f (2x − 2) = 0. 5

MC Which of the following is a valid use of mapping notation?

a f : R → R, f ( x ) =

1 x −1

b f : R+ → R+, f ( x) =

±

c f : [4, ∞ ) → R, f ( x ) = x − 4 1 e f : R → R+, f ( x) = 2 x +1 6

3C

1 −2 x +1

d f: R → R, x = 2

MC The ranges for f : [4, ∞ ) → R, f ( x ) =

a [3, ∞), (−∞, 0)

b [ 5 , ∞), ( − ∞, 0)

d [ 5 , ∞ ), (0, ∞ )

e [ 5 , ∞ ), (2, ∞ )

inverse functions

1 are: x−2 c [3, ∞), (2, ∞)

x + 5 and f : ( − ∞ , 2) → R, f ( x ) =

eBook plus −1

Under a mapping f, X maps onto Y. Under an inverse mapping f , Interactivity int-0970 Y maps onto X. The following statements result: −1 Inverse functions 1. f can only exist if f is a one-to-one function. − 2. Domain f 1 = range f. −1 3. Range f = domain f. − 4. If x and y are interchanged, the rule for f 1 is obtained from the rule for f.

Chapter 3


77

-

5. If x and y are interchanged, the graph of f 1 can be obtained from the graph of f by reflecting in the line y = x. y 3 (−1, 2)

y 3

y=x

2

2

1

1

−3 −2 −1 −1

1

2

3x

−3 −2 −1 −1

(2, −1)

−2

y=x

−

f 1(x)

1

2

f (x)

3x

−2

−3

−3

It is possible to consider inverse relations. For example, the rule for the inverse of y = x2, a many-to-one function is x = y2, a one-to-many relation. However, if we are to start with a function f and finish with a function f 1, then f has to be a one-to-one function. Worked Example 5 -

If f: (-1, 2] → R, f (x) = 2x + 4, find the domain, range and rule of f 1(x), and sketch the graphs of f and f 1 on the same set of axes. Think

Write -

-

1

First, determine if the inverse function, f 1(x), exists.

f (x) is linear and a 1-1 function, so f 1(x) exists.

2

Determine the range of f (x). The domain is x ∈ (-1, 2], so substitute the end values of x to determine the range.

f (-1) = -2 + 4 = 2 f (2) = 4 + 4 = 8 Therefore, the range of f (x) is y ∈ (2, 8].

3

The domain of f (x) = the range of f 1(x). The range of f (x) = the domain of f 1(x). State the domain and range of f 1(x).

4

To determine the rule of f 1(x), let f (x) = y and interchange x and y. Then make y the subject.

5

Fully define the rule for the inverse function.

f 1: (2, 8] → R, f 1 ( x ) =

6

Sketch the graphs over the required domains, showing the line y = x.

y f(x) = 2x + 4 8 y=x 6 − 4 f 1(x) = 12 x − 2 2 2 0 −8 −6 −4 −2 4 6 8 x −2

-

-

Domain f (x): x ∈ (-1, 2] ⇒ range of f 1(x) is y ∈ (-1, 2] Range of f (x): y ∈ (2, 8] ⇒ Domain of f 1(x) is (2, 8] Let y = 2x + 4 ⇒ x = 2y + 4 ⇒ -2y = 4 - x 1 ⇒ y= x−2 2 −

−

−4 −6 −8

78


1 x−2 2

Worked exAMPle 6

f: (−∞, −1] −1

eBook plus

If → R, f (x) = + 2x + 2, find the domain, range and rule − of f (x), and sketch the graphs of f and f 1 on the same set of axes. x2

Think 1


WriTe

First, determine if the inverse − function, f 1(x), exists. Since 2 f (x) = x + 2x + 2 is an upright parabola, it is necessary to locate the turning point using x =

Tutorial

−b

2a

−

2 − = 1 2 Since the turning point occurs at x = −1, and the domain is − x ∈ (−∞, −1], f (x) is a 1–1 function and f 1(x) exists. x=

.

Determine the range of f (x). The domain is x ∈ (−∞, −1], so substitute the end value of x to determine the range (this value is at the turning point).

x = −1 ⇒ f( = 1 − 2 + 2 =1 The point (−1, 1) is the minimum point on the graph.

3

The domain of f (x) = the range − of f 1(x). The range of f (x) = the − domain of f 1(x). State the domain and range − of f 1(x).

Domain f (x): x ∈ (−∞, −1] − ⇒ range of f 1(x) is y ∈ (−∞, −1] Range of f (x): y ∈ [1, ∞) − ⇒ Domain of f 1(x) is [1, ∞)

4

To determine the rule of f 1(x), let f (x) = y and interchange x and y. Then make y the subject.

2

−

−1)

Let y = x2 + 2x + 2 ⇒ y = (x + 1)2 + 1 Interchange x and y ⇒ x = (y + 1)2 + 1 ⇒ x − 1 = (y + 1)2 ⇒ y = ± x −1 −1

5

Since f (x) = x2 + 2x + 2, and x ∈ (−∞, −1] (left side of the − parabola), then f 1 should be

−

−

f 1: [1, ∞ ) → R, f 1 ( x ) =

−

x −1 −1

−

y = x − 1 − 1. Fully define the rule for the inverse function. 6

Sketch the graphs over the required domains, showing the line y = x.

f(x) = x2 + 2x + 2

y 5 4 3 2 1

0 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

y=x

1 2 3 4 5 x − f 1(x) = −1− √x − 1

Chapter 3


79

7

To view the graph and its inverse on a CAS calculator, open the Graph & Tab screen. Complete the entry lines as: y1 = x2 + 2x + 2 | x ≤ −1 y2 = −1 − x −1 y3 = x Tick each of the equations and tap $.

reMeMBer −

1. The inverse function of f, f 1, exists if f is a one-to-one function. − 2. For f 1, domain = range of f, range = domain of f and the rule is found by interchanging x and y in the rule for f. − − 3. To graph f 1 from the graph of f, or f from the graph of f 1, reflect in the line y = x. exerCise

3C

inverse functions 1

2

− We5 For each of the following functions f, determine the domain, range and rule of f 1. −

Sketch the graphs of f and f 1 on the same set of axes. a f: [0, ∞) → R, f (x) = 3x − 2 b f: [1, 3) → R, f (x) = 2x − 4 c f: (−1, 2] → R, f (x) = −2x + 4 d f: (−∞, 2] → R, f (x) = 2x + 1 e f: (−2, 2] → R, f (x) = −2x − 2 f f: [−4, 2) → R, f (x) = −x − 3

− We6 For each of the following functions f, determine the domain, range and rule of f 1. −

Sketch the graphs of f and f 1 on the same set of axes. a f: (1, ∞) → R, f (x) = x2 − 2x + 2 b f: (−∞, −2]→ R, f (x) = x2 + 4x + 5 c f: (1, ∞) → R, f (x) = −x2 + 2x + 3

d f : [ 12 , ∞ ) → R, f ( x ) = − 2 x 2 + 2 x + 3

e f : ( − 2, ∞) → R, f ( x ) = x + 2

f

g f : ( − 2, ∞) → R, f ( x) =

−

x + 2 +1

f : ( 12 , ∞ ) → R, f ( x ) = 2 x − 1 + 1

h f : ( − ∞ , 3 ) → R, f ( x ) =

−

3− x + 2

− f 1

3 For each of the following functions f, determine if exists. For those that exist, find the − domain, range and rule of f 1. 1 1 a f : R\{0} → R, f ( x ) = b f : R\{0} → R, f ( x ) = 2 x x 1 1 c f : R\ 12 → R, f ( x ) = d f : (2, ∞) → R, f ( x) = ( x − 2)2 2x − 1 − 4 MC For f: [1, ∞) → R, f (x) = 2x2 − 4x + 4, the domain and rule of f 1 are:

{}


WorkSHEET 3.2

80

− − a [2, ∞ ), f 1 ( x ) = 2 − x − 1 b [3, ∞ ), f 1 ( x ) = 1 + x − 2 2 2 − d [3, ∞ ), f 1 ( x ) = 1 − x − 1 2

− e [2, ∞ ), f 1 ( x ) = 1 + x − 2 2


−

c [2, ∞ ), f 1 ( x ) = 2 − x − 1

Summary Relations

• A relation is a set of ordered pairs. • The domain of a relation is the set of first elements of a set of ordered pairs. • The range of a relation is the set of second elements of a set of ordered pairs. Functions

• A function is a relation that does not repeat the first element in any of its ordered pairs. That is, for any x-value, there is one y-value. • A function can be one-to-one or many-to-one. • The graph of a function cannot be crossed more than once by any vertical line. • Mapping notation for functions: f : X → Y, f (x) = ............... where X is the domain, Y is the co-domain (range ⊆ Y) and f(x) is the image of x. • When required, the range can be determined from the domain and the rule. Inverse functions

-

• The inverse function of f, f 1, exists if f is a one-to-one function. • For f 1; domain = range of f, range = domain of f and the rule is found by interchanging x and y in the rule for f. • To graph f 1 from the graph of f, or f from the graph of f 1, reflect in the line y = x.


81

chapter review 10

1 For each of the following relations, state the domain and range. Sketch the graph. a {(2, 3), (2, 4), (3, 3)} b {(x, y): x ∈ Z, y = -x} c {(x, y): x ∈ Z +, y = 2x} d {(x, y): (x - 1)2 + (y - 2)2 = 4} e {( x, y ): f {( x, y ):

y

d

Short answer

( x − 1)2 ( y − 2)2 + = 1} 4 16

5 −10 −5

e

−10 −5

−5

10

−10 −5

−5

5

10

x

−10

y 10

x

−10

10

5

f 5

5

x

y

y 5

10

10

2 For the graphs of the relations below, state the domain and range. 10

5

−10

( x − 1)2 ( y − 2)2 − = 1} 4 16

a

−5

5 −10 −5

−5

x

−10

b

y

3 Explain why each of the following are an invalid use of the mapping notation.

10 5 −10 −5

−5

5

10

x

c f : R → R, f ( x ) = 2 ± 3 4 Find the range for each of the following functions. a f : [3, 4) → R, f (x) = x - 4 b f : R → R, f (x) = (x - 4)(x - 2) c f : R → R, f (x) = e x - 4

y 10

d f : [4, ∞) → R, f ( x ) = x + 5

5 −10 −5

−5

−10

82

x2 + 1 x2

b f : R+ → R+, f (x) = x2 - 1

−10

c

a f : R → R, f ( x ) =

5

10

x

e f : [ − 2, 2] → R, f ( x ) = 4 − x 2 f f : R → R, f (x) = -2x2 + 4x - 1 3 g f : R\{2} → R, f ( x ) = +1 x−2


−

h f : (2, ∞) → R, f (x) = log2(x) 1 i f : R → R, f ( x ) = 2 x +1 Exam tip Make sure you state the range taking the domain restriction into account.

5 For f (x) = x2 - 3x + 2: a find the factors of f (x) and the solutions of f (x) = 0 b find f (x2 + 1), the factors of f (x2 + 1) and the solutions of f (x2 + 1) = 0. Multiple choice

  x2 y2 + = 1 and 1 The ranges of ( x, y ): x ∈[ − 2, 0], 4 9   2 2   x y − = 1 are: ( x, y ): 4 9   a [-3, 3], [-3, 3] b [-3, 3], R c [-3, 0], R e [ 3, 0], [ 3, 3] d [0, 3], R 2 The implied domains for {(x, y): x = 2y} and {( x, y ): y = 25 − x 2 } are: a (0, ∞), [-5, 5] b [0, ∞), (-∞, -5] ∪ [5, ∞) c R, [-5, 5] d [0, ∞), [-5, 5] e (0, ∞), (-∞, -5] ∪ [5, ∞) 3 Which of the following is not a proper use of mapping notation? a f : [2, 3] → [3, 4], f (x) = x + 1 b f : (2, 3) → [3, 4], f (x) = x + 1 1 c f : (2, 3] → (7, 8], f ( x ) = x + 6 2 d f : [2, 3] → R-, f (x) = -x + 1 e f : [2, 3] → R+, f (x) = -x + 3

4 The ranges of f : R + → R, f ( x ) = x + 4 and f: R → R, f (x) = x2 + 2x + 1 are: a (-∞, 2), [0, ∞) b (-∞, -2), [0, ∞) c ( ∞, 0], [0, ∞) d (-∞, 2), R+ + e ( ∞, 2), R 5 If under the mapping f : R- → R, 1 1 f ( x ) = 2 , and f ( x ) = , then x is: 3 x −1 −9 1 b a c -2 24 8 d 2 e ±2 6 Which of the following equations represents a function for which an inverse function exists?   x2 y2 + = 1 a ( x, y ): x ∈[0, 2], 4 9     x2 y2 + = 1 b ( x, y ): y ∈[0, 3], 4 9     x2 y2 − = 1 c ( x, y ): x ∈[2, ∞ ), 4 9   2   x y2 − = 1 d ( x, y ): y ∈[0, ∞ ), 4 9     x2 y2 − = 1 e ( x, y ): x ∈[2, ∞ ), y ∈( − ∞ , 0], 4 9   2 7 For f : ( ∞, 2] → R, f (x) = x - 4x + 1, the domain and rule of f 1 are: −

a ( − ∞ , 2], f 1 ( x ) = 2 − x + 3 −

b [ − 3, ∞ ], f 1 ( x ) = 2 − x + 3 −

c [ − 3, ∞ ], f 1 ( x ) = 2 + x + 3 −

d (3, ∞ ), f 1 ( x ) = 2 − x − 3 −

e (3, ∞ ), f 1 ( x ) = 2 + x − 3

Extended response -

-

1 For each of the following functions f, find the domain, range and rule of f 1. Sketch the graphs of f and f 1 on the same set of axes. a f : [0, ∞) → R, f (x) = 2x + 1 b f : (-2, 4] → R, f (x) = -2x + 1 Exam tip Make sure you write the equation of c f : [1, ∞) → R, f (x) = x2 - 2x + 3 the asymptotes. d f : (-∞, 0) → R, f (x) = x2 - 2x - 1 e f : [2, ∞) → R, f ( x ) = x − 2 − f f : ( − ∞, 1] → R, f ( x ) = 1 − x g f : [ − 3, 0] → R, f ( x ) = 2 1 − h f : [ + 3, ∞) → R, f ( x ) = 2

x2 9

x2 −1 9


83

2 The perimeter for a new seal enclosure is to have a maximum side length of 8 m. The width is to be twice the length (x). a Draw a diagram of the enclosure and label the sides. b Define a rule that gives the perimeter, P, of the new enclosure. c What is the largest value that x can be? d State the domain and range. e Write in function notation the rule for the perimeter. f Define a function for the area of the enclosure, A(x). g If the maximum area allowed is 18 m2, find the dimension of the enclosure. eBook plus Digital doc


84


eBook plus

ACTiviTies


• 10 Quick Questions: Warm up with ten quick questions on relations and functions. (page 71) 3a

Relations

Tutorial

• We1 int-1036: Watch how to graph relations and state their range. (page 71) Digital doc

• WorkSHEET 3.1: Use set notation to identify the domain and range of relations and functions. (page 74) 3b

Functions

Tutorial

• We3 int-1037: Watch how to determine the determine the range for functions given in function notation. (page 75) 3c

Inverse functions

chapter review Digital doc


Interactivity

• Inverse functions int-0970: Consolidate your understanding of graphs of inverse functions. (page 77) Tutorial

• We6 int-1038: Watch how to find the domain and range of an inverse function. (page 79) Digital doc

• WorkSHEET 3.2: Identify types of relations and their domain and range. (page 80)

Chapter 3


85

EXAM PRACTICE 1 SHORT ANSWER

20 minutes

1 Simplify 3 72 + 4 12 − 300 . 2 Sketch the graph of f : f (x) = [2x + 1].

[−3,

2 marks

3] → R, where 3 marks

3 Let f : R → R, where f (x) = x – 2 and let 1 g: R\{2} → R, where g( x ) = . f ( x) Sketch the graphs of f (x) and g(x) on the same set 4 marks of axes. Label all key features. 4 Show that if z1 = a + bi and z2 = c - di, then equals

ac − bd + (bc + ad )i . c2 + d 2

z1 z2 3 marks

MULTIPLE CHOICE

Chapters 1 TO 3

Which point represents z1 - z2? A H B J D L E M

C K

4 The graph of {(x, y): y = | x - 2 |} is transformed by a dilation of factor 2 from the x-axis. The image of this transformation can be represented by the rule: A {(x, y): y = | 2x - 2 |} 1   B ( x, y ): y = | x − 2 |  2     1 C ( x, y ): y = x − 2  2   D {(x, y): y = | x - 4 |} E {(x, y): y = 2 | x - 2 |} 5 The graph of f is shown below. y

10 minutes

Each question is worth 1 mark. 1 If z1 = 4 - 3i and z2 = 1 + 2i, then z1z2 equals: A 5 - i B 4 - 6i C 2 - 5i D 10 + 5i E −2 + 5i a  2 Point A (x, y) is translated by   and then reflected b  in the y-axis to the image B. The coordinates of point B are: A (−x + a, y + b) B (−x − a, y + b) C (x + a, −y + b) D (x + a, −y - b) E (−x - a, −y − b)

x

−

Which one of the following represents f 1? A

b

y

x

c

x

Im (z)

e

y

K z2 J

86

M

z1 H

Re (z)


x

d

y

3 The points z1 and z2 are shown on the Argand diagram below. L

y

x

y

x

exTended resPonse

30 minutes

1 Graphic designer Rhonda has been contracted by Wacky World to design a new logo for their T-shirts. Her first design is shown below.

To allow Rhonda to have the flexibility to change her designs, she determines a function that models her design. For her design, 1 cm represents 1 unit on the Cartesian plane. y y2

C 1 cm

B y1 O

A

x

Graph 1

She finds that part of her design follows the rule y = | 2x − 3 |. a For the rule y1 = | 2x − 3 |, determine the coordinates of: i A ii B iii C 1 + 1 + 2 = 4 marks The second part of the design is defined by y2, which is determined by transforming y1 onto the image of y2 as shown in Graph 1 above. b Describe a transformation that maps the graph of y1 to y2. 1 mark c Use set notation to determine the rule for y2. 3 marks d In the context of this problem, determine the feasible domain and range. Write your answer using function notation. 1 + 1 = 2 marks 2 Rhonda inserts another ‘W’ in the design. The second ‘W’ is smaller than the first. Rhonda uses the method of finding a rule to describe part of the ‘W’ and then applies transformations to complete the design. Using Graph 1, her new design for ‘W’ will include points AB′C′. a The coordinates of B′ are (0, 1). Describe a transformation that maps B to the image B′. 2 marks The second ‘W’ includes the original point A. b Determine a rule that describes the line that passes through points A and B′. 3 marks  4 c The point C′ lies on the line found in part c. If C′ has coordinates  x,  , find the exact value of x.  3 2 marks

d Rhonda is going to place the second smaller ‘W’ above the first ‘W’. The position of B′ will become (−1, 3). Write down in matrix form, the combination of transformations that map B′ to its new position.

2 marks


Solutions Exam practice 1

exam practice 1

87

4

4a 4b 4c 4d

Review of index laws Standard form and significant figures Transposition Solving linear equations and simultaneous linear equations 4e Applications 4F Algebraic fractions 4G Linear literal equations

Algebra AreAs of sTudy

• Applications of arithmetic involving natural numbers, integers, rational numbers and real numbers • Substitution and transposition in linear relations, such as temperature conversion • The solution of linear equations, including literal linear equations • Developing formulas from word descriptions, and substitution of values into formulas

• The construction of tables of values from a given formula using technology • Linear relations defined recursively and simple applications • The algebraic and graphical solution of simultaneous linear equations in two variables • Solution of worded problems involving a linear equation or simultaneous linear equations in two variables • The construction of a table of values from a given formula using technology eBook plus Digital doc

10 Quick Questions

4A

review of index laws Index, power or exponent 23 Base number

You learned in earlier years that 23 means ‘multiply two by itself three times’, that is, 2 × 2 × 2 = 8. The index indicates the number of times the base is multiplied by itself. In general,

34 3 81 = 3× × 3×3 =  Index form Expanded form Basic numeral a m = a× a× ×a  …  m times

In summary, the index laws are: Multiplication: Division: Raising to a power: Raising to the power of zero: Raising to the power of one:

88

am × an = am + n am am ÷ an = n = am - n a m n (a ) = am × n a0 = 1 a1 = a

maths Quest 11 Advanced General mathematics for the Casio Classpad

n

an  a (a × b)m = am × bm;   = n  b b 1 a m= m a

Products and quotients: Negative powers:

m

Fractional powers: a n = n am = ( n a ) When simplifying expressions and equations with indices, we generally use a combination of these laws. m

Worked Example 1 4 3 2 2 Simplify 3 x y × 2 x y z . 5 x 8 yz

Think

write

1

Add the indices of x and the indices of y and multiply the constants together in the numerator.

2

Subtract the indices of x, y and z in the denominator from those in the numerator.

3

Simplify to write answer with positive indices and use the rule z0 = 1.

3x 4 y3 × 2 x 2 y 2 z 6 x6 y 5z = 5 x8 yz 5 x8 yz

= 65 x - 2 y 4 z 0

=

6y4 5x 2

When numbers with an index are then raised to another index, the indices are multiplied.

Worked Example 2

Simplify -

-

a x 3y2 × (2x2y 1)3 b ( a4 b3 )

-2

÷

a 7 b2 - . a3 b 4

Think a

b

1

Write

Remove the brackets by multiplying the indices.

-

-

a x 3y2 × (2x2y 1)3

-

-3

= x 3y2 × 23x6y -1

2

Add the indices of x and y.

= 8x3y

3

Simplify to write answer with a positive index.

=

1


b (a 4 b 3 )

-2

÷

8x3 y

a7b2 a3b

-4 -

= a 8b

2

Change the division sign to a multiplication sign and write the reciprocal of the second term.

=

-

a 8b 1

-6

-6

÷ ×

a7b2 a3 b

-4

-

a3b 4 a7b2

Chapter 4 Algebra

89

-

-

a 5b 10 a7b2

3

Add the indices of a and b in the numerators.

=

4

Subtract the indices of a and b in the denominator from those in the numerator and simplify to positive indices.

=a

-12 -8 b

=

1 a12 b8

Worked exAmple 3

Simplify a 2n - 1 × 62n × 3n + 1

eBook plus

Think a

1

b

Tutorial

b 2n × 41 - n × 162n - 1.


WriTe

Change the 6 into 2 × 3.

a 2n - 1 × 62n × 3n + 1

= 2n - 1 × (2 × 3)2n × 3n + 1

2


= 2n - 1 × 22n × 32n × 3n + 1

3

Add indices of numbers with base 2 and add indices of numbers with base 3.

= 23n - 1 33n + 1

1

Change all numbers to a base of 2.

2


= 2n × 22 - 2n × 28n - 4

3

Simplify by adding indices of numbers with base 2.

= 27n - 2

b 2n × (22)1 - n × (24)2n - 1

Worked exAmple 4

Simplify each of the following, expressing the answer with a positive index. 7

a

128 × 4 64

b

3

x 2 y6 ÷ x 3 y5

Think a

WriTe a

7

128 × 4 64

1


2

Write using fractional indices.

= 128 7 × 64 4

3

Write 128 and 64 in index form with a base of 2.

= ( 2 7 ) 7 × ( 26 ) 4

4

Multiply the powers.

= 21 × 2 4

5

Simplify and write the answer.

= 21 × 2 2

1

1

1

1

6

3

5

= 22 b

1


2

Write the expression using index notation.

b

3

x 2 y6 ÷ x3 y 5 1

90

1

= (x 2 y6 ) 3 ÷ (x3 y 5 ) 2


2

Remove the brackets by multiplying the powers.

4

Collect terms with the same base by subtracting the powers.

= x3

=x

5

Simplify the powers.

6

Rewrite with positive powers.

3 5

= x 3 y2 ÷ x 2 y 2

3

2 3 5 22y 2

=

- 5 -1 6y 2

1 5 1 x6y2

REMEMBER

A combination of the laws below can be used to simplify indicial equations. 1 a m= m a0 = 1 a1 = a a am × an = am + n m an

Exercise

4A

am ÷ a n =

= n am = ( n a ) m

am = a m - n an

(am)n = am × n

(a × b)m = am × bm

Review of index laws 1 WE 1 Simplify the following expressions. a d

12a 2 b 3a

b

3 1

- 3a 3 b 2

e 2a 2 b 4 × 3a 2 b

-2

6a b

2 WE 2

8a3b 4 2ab -1

- a2b- 4

2ab3 1

-1

-

5

f 8 3 a 2 b3 ÷ 4 a 2 b 2

Simplify the following expressions.

a

( x 3 y 2 )2 2x4 y2

b

d

2( x 2 y 2 )3 ( 2 x 3 y )3

3 -2 2 e (m n ) 2mn

g

(5m 2 n)2 3m 1 -2 2 3 × (3m n ) n2

j

9w2 ( - 3v1w 2 )- 3 -1 2 ÷ 2 (2v w ) ( 2v 3 ) 2

3 ab 2

(2 xy 2 )3 4 x 2 y3 -

c

(3 x 3 y 2 )2 4( x 2 y 2 ) 4

f

( 2m 1n - 2 ) 3 2(m 2 n 2 )4

i

(4 v 1w 2 ) 2 2w2 -2 2 3 ÷ 2 -2 2 ( 3m n ) (v w )

-

-

3 MC Expressing a

c

h

( - 2 v 3 w )3 2 w 2 × (5v 2 w 1 )3 (vw)2

-

-

-

ab in index form gives: b ab

-3

-3

c (ab)

1

1

d ab 8

e (ab) 8

2

Chapter 4 Algebra

91

4

Simplify.

We3

a 2n - 1 × 4n + 1 × 16n d 5

27 4 n + 2 ÷ 16 3n -1

e

6n 2 n + 23 n

c

5n - 3 × 3n + 1 75n

f 72m - 2 × 4 × 32m

We4 Simplify each of the following, expressing your answer with positive indices. 3

9 × 81

a d

3

( xy 3 ) ÷ ( x 2 y ) 2

g j

4B

b 3n + 2 × 9n - 1 × 27

×

e

5 24

×4

3

( 4 x 3 ) 3 × ( 3 x 4 )8 ( x + 1)

b

2 x3

2

x +1

h k

1 x6 -1 2

×8

1

-1

-2

f 27

3

4 (64 m6 ) 3

4m

( )

5

4

c x2 ÷ x3

i

-2

x+2+

x x+2

4

2

× 93 × 3

-5 4

1 x

-4

l ( y - 4) y - 4

standard form and significant figures standard form Standard form (or scientific notation) involves a practical use of indices. A very large or very small number can be expressed in standard form as a more convenient way of writing it. This notation involves expressing the number as a number between one and ten multiplied by a power of 10. a × 10n, where 1 ≤ a < 10 A computer may complete a basic operation in approximately 0.000 000 000 8 seconds. It is easier to write a number such as this in standard form as 8 × 10 10 seconds. Likewise, a light year is a measure of distance equal to 9 460 528 400 000 000 kilometres. In standard form, this number can be expressed as 9.460 528 4 × 1015 Standard form is not only a more economical means of expressing these numbers, it also makes calculations involving these numbers easier through the use of index laws. To write a number in standard form : • Move the decimal point so that the number appears to be between 1 and 10 • Count the number of decimal places the decimal point has been moved (positive if moving left or negative if moving right) • Multiply by the power of 10 equal to this number. • The sign of the power will be positive if the magnitude of the original number is greater than 10 • The sign of the power will be negative if the magnitude of the original number is between 0 and 1.

Worked exAmple 5

eBook plus

350 000 × 0.04 Solve and express as a basic numeral. 70 Think 1

92

Express the problem in standard form.

Tutorial

int-1040

WriTe

Worked example 5

350 000 × 0.04 70 3.5 × 10 5 × 4 × 10 2 = 7 × 10


14 × 103 7 × 10

2

Simplify the numerator using index laws where possible.

=

3

Divide using index laws where possible.

= 2 × 102

4

Express as a basic numeral.

= 200

Significant figures Often we will be interested in all the figures in a particular number. • Significant figures are counted from the first non-zero digit (1–9). For example, 0.0092 has two significant figures (9 and 2). • Any zeros at the end of the number after the decimal point are considered to be significant. For example, 0.250 has three significant figures (2, 5 and 0), whereas 0.025 has two significant figures (2 and 5). • The trailing zeros at the end of a number are not considered significant. For example, 1200 has two significant figures (1 and 2). • All zeros between two non-zero digits are always significant. For example, 102.587 has 6 significant figures (1, 0, 2, 5, 6 and 7).

Worked Example 6

State the number of significant figures in the following numbers. a 3.205 60 b 20.01 c 0.0034 d 35 000 Think

Write

a Significant figures are counted from the first non-

a 3.205 60 has 6 significant figures.

zero digit (1–9). There are two zeros after the decimal point, which are to the right of a non-zero digit and so all digits are significant. b Significant figures are counted from the first non-

b 20.01 has 4 significant figures.

zero digit (1–9). All zeros between two non-zero digits are always significant. All digits in this case are significant. c Significant figures are counted from the first non-

c 0.0034 has 2 significant figures.

zero digit (1–9). The first non-zero digit in this case is 3. Only 3 and 4 are significant. d The trailing zeros at the end of a number are

d 35000 has 2 significant figures.

not considered significant. Only 3 and 5 are significant.

Calculations involving significant figures When performing calculations associated with significant figures, the following rules apply. • When adding or subtracting numbers, count the number of decimal places to determine the number of significant figures. The answer cannot contain more places after the decimal point than the least number of decimal places in the numbers being added or subtracted. • When multiplying or dividing numbers, count the number of significant figures. The answer cannot contain more significant figures than the number being multiplied or divided with the least number of significant figures.

Chapter 4 Algebra

93

Worked Example 7

Evaluate, expressing your answer to the appropriate number of significant figures: a 345.87 + 20.1 b 23.020 × 0.023. Think a

1

2 3 4

b

1

2 3

Write

This question involves addition; therefore, find the least number of decimal places of the numbers in the question.

a 345.87 has 2 decimal places.

Add the numbers. Round the answer to 1 decimal place. Interpret this answer.

347.87 + 20.1 = 367.97 368.0 The answer has 1 decimal place and 4 significant figures. b 23.020 has 5 significant figures. 0.023 has 2 significant figures. The answer will have 2 significant figures. 23.020 × 0.023 = 0.529 46 0.53

The question involves multiplication; therefore, find the least number of significant figures of the numbers in the question. Multiply the numbers. Express the answer to 2 significant figures.

20.1 has 1 decimal place. The answer will have 1 decimal place.

REMEMBER

1. To write a number in standard form: • Move the decimal point so that the number appears to be between 1 and 10. • Count the number of decimal places the decimal point has been moved (left: +; right: –). • Multiply by the power of 10 equal to this number. 2. The number of significant figures in a number can be determined by considering each of the following rules: • Significant figures are counted from the first non-zero digit (1–9). • Any zeros at the end of the number after the decimal point are considered to be significant. • The trailing zeros at the end of a number are not considered significant. • All zeros between two non-zero digits are always significant. 3. When performing calculations associated with significant figures, the following rules apply: • When adding or subtracting numbers, count the number of decimal places to determine the number of significant figures. The answer cannot contain more places after the decimal point than the least number of decimal places in the numbers being added or subtracted. • When multiplying or dividing numbers, count the number of significant figures. The answer cannot contain more significant figures than the number being multiplied or divided with the least number of significant figures. Exercise

4b

Standard form and significant figures 1 Express the following in standard form. a 360 400 b 213.457 d 0.0324 e 0.000 100 31

94


c 1 023.98 f 570 201 009

2 WE5 Solve by expressing the numbers in standard form and simplifying using index laws. Express your answer as a basic numeral. 28 000 420 000 11 200 000 a b c 350 1400 2800 d

80 000 000 16 000

e

3 100 000 1550

f

7 500 000 1500

g

0.000 24 0.3

h

0.000 018 0.06

i

0.000 056 0.0350

j

0.000 84 0.0021

k

5 800 000 0.02

l

130 000 0.0026

m

0.0066 11 000

n

0.000 095 190 000

o

18 000 × 0.0045 900

p

4900 × 0.001 75 35

q

750 000 00 × 0.000 025 1250

r

25 600 000 × 0.000 000 004 0.0064

3 MC 10.0673 expressed in standard form is: a 1.006 73 × 10 b 10.0673 × 10 4 2 d 0.100 673 × 10 e 1.006 73 × 10 1 4 WE6 a d g j

Specify the number of significant figures in the following:

0.023 210.50 0.120 10 0.000 002

5 WE 7

-2

c 0.100 673 × 10

b e h k

10.21 10.10 7620 4730.90

c f i l

3045 34 700.002 190.00 2 800 000

Calculate the following to the correct number of significant figures.

a 2.456 + 0.9 d 0.2507 - 0.120 g 403.5 ÷ 5.1

b 12.340 + 1.02 e 1.903 × 230.576 h 2.01 ÷ 0.05080

c 120.350 - 2.04 f 28.1 × 2.1020

6 MC The solution to 130.70 − 28.9913 with the correct number of significant figures is: a 101.71 b 101.7090 c 101.7 d 101 e 101.709 7 MC The solution to 32.3695 ÷ 1.870 with the correct number of significant figures is: a 17.3 b 17 c 17.309 d 17.31 e 17.3100 8 Complete the following calculations, expressing your answer to the appropriate number of significant figures. a It is 1.35 kilometres from Jane’s house to school. Her average step length is 0.7 metres. How many steps does it take for Jane to walk to school. b If a container of sugar cubes has a mass of 250 g when full (excluding the mass of the container), how many sugar cubes would be required to fill the container if they each have a mass of 3.24 g? 9 The outer ‘skin’ of a human cell, the cell membrane, is approximately 0.000 000 008 4 metres thick. If the radius of the cell (including the cell membrane) is 0.000 004 2 metres, what fraction of the radius does the cell membrane constitute? 10 An Olympic size swimming pool contains 2 500 000 litres of water. The average daily water usage for a family of four is 625 litres. How long would it take for a family to use the volume of water equivalent to an Olympic size pool.

Chapter 4 Algebra

95

11 The Earth has a mass of approximately 5970 yottagrams (where a yottagram, Yg, is 1021 kg). The mass of the Moon is 73500 zettagrams (where a zettagram, Zg, is 1018 kg). What percentage of the mass of the Earth is the mass of the Moon?

4c

Transposition A formula is an equation or a rule that defines the relationship between two or more variables. If a formula describes a relationship between two variables, both of which are to the power of 1, and does not contain terms that include a product or quotient of those variables, then such a relation is said to be linear. The graph that represents a linear relation is a straight line, which is where the term linear is derived from. For example: x − 4y − 7 = 0 and y = −3x + 6 are linear relations, whereas x x + y − xy = 3 or x2 + y = 29 or =7 y are not (as explained previously). Linear relations are often found in practical situations. For example, the formula for the circumference of a circle, C = πD, and the formula for the conversion of temperature from 9 degrees Celsius to degrees Fahrenheit, F = 5 C + 32, both describe linear relations. If we wanted to find many values of C given various values of F, it would be more convenient to have the corresponding formula — the formula that would have C on one side and everything else on the other side of the equals sign. The variable that is by itself is called the subject of the formula (that is, a formula describes its subject in terms of all other variables). In the formula I = 20R, I is the subject. To make R the subject, we need to rearrange the formula. Such a rearrangement is called transposition.

96


I = 20R I 20 R divide both sides of the equation by 20. = 20 20 I Simply. =R 20 I Write the subject on the left-hand side. R= 20 To rearrange or transpose a formula, we need to perform the same inverse operations to both sides of the equation until the desired result is achieved. To transpose the equation

Worked exAmple 8

Transpose the formula 4x = 2y - 3 to make y the subject. Think

WriTe

4x = 2y - 3

1

Write the given formula.

2

Add 3 to both sides of the equation.

3

Divide each term on both sides of the equation by 2.

4 x 3 2y + = 2 2 2

4

Simplify both sides of the equation.

2x +

4x + 3 = 2y - 3 + 3 4x + 3 = 2y

3 =y 2 y = 2x +

3 2

To transpose the above formula, we use the same methods as those employed for solving linear equations. The only difference is that in the end we do not obtain a unique (or specific) numerical value for the required variable, but rather an expression in terms of other variables. Most of the relations that describe real-life situations are non-linear. Consider, for example, any formula for area or volume. A few examples are the area of a circle A = π r2 (non-linear, 1 since it contains r to the power of 2); the area of a triangle A = 2 bh (non-linear, since it contains the product of two variables); and the volume of a cube V = s3 (non-linear, since s is cubed). Non-linear formulas can be transposed by performing identical inverse operations to both sides of the equations. The inverse of x2 is x, the inverse of x is x2 and so on. Worked exAmple 9

eBook plus

Transpose each of the following formulas to make the pronumerals indicated in brackets the subject. ab - ac 4 a A = 3 π r2 (r) b P= ( a) c m = pq - rs ( s) d Think a

Tutorial


WriTe

1

Write the equation.

2

Multiply both sides of the equation by 3.

a

4

A = 3 πr2 4

3 × A = 3 πr2 × 3 3A = 4πr2

Chapter 4

Algebra

97

3

4

b

3 A 4π r 2 = 4π 4π 3A = r2 4π

Divide both sides by 4π.

3A = r2 4π

Take the square root of both sides. Note: From an algebraic point of view we should write ± in front of the root. However, since r represents a physical quantity (radius of a sphere in this case), it can take only positive values.

1

On the Main screen, tap: • Action • Advanced • solve Complete the entry line as: a×b-a×c   , a solve  p =   d Then press E.

2

Write the answer.

r=

3A 4π

b

a=

dP b-c

Note: Capital P should be used in the answer. c

1

Write the equation.

2

The inverse of x is x2 so square both sides.

3

Subtract pq from both sides.

4

Divide both sides by −r.

c

m=

pq - rs

m2 = pq − rs m2 − pq = pq − rs − pq m2 − pq = −rs m 2 - pq - rs = -r r m 2 - pq -r pq - m 2 s= r s=

5

Multiply the numerator and denominator by −1 (optional).

REMEMBER

1. Transposition is the rearrangement of terms within a formula. 2. Relations can be transposed by performing identical inverse operations to both sides of the equation that describes the relation. 3. The subject of the formula is the variable that is by itself on one side of the equals sign while all other variables are on the other side.

98


Exercise

4c

Transposition 1 WE8 Transpose each of the following formulas to make the pronumeral indicated in brackets the subject. (Where two pronumerals are indicated, perform a separate transposition for each.) a 5y + 4x = 20 (x, y) b 3x − 4y + 12 = 0 (y) c m = 3a − 14 (a) d 5p = 2 − 3k (k) 1

3

e 2 a = 4 b

(a, b)

f 10 − 3a = 2a − b

g a = 3b − 0.5c

(c)

h

i 5(3 − 2d) = 6(f + 4)

(d, f )

3a 2(b + 3a) + = 1 2 3

(a, b)

k

(a, b)

2(a - 3) = b 5 7(a - 4 b) 5(b - 2a) j = 3 4 2x 3x - 6 y l -6= 5 10

(a) (a, b) (x, y)

2 WE9 Transpose each of the following formulas to make the pronumerals indicated in brackets the subject. (Where two pronumerals are indicated, perform a separate transposition for each.) a v2 = u2 + as (a, u) b S = 4πr2 (r) 1 1 1 r   + ( R, R1 ) c = (r ) d A = A0  1 + R R1 R2  100  t e s = (u + v ) (t, u) 2

f T = 2π

L g

g C = a 2 + b 2

h s = ut + 2 at2 (a)

( L , g)

1

(b)

j R = mv2 - mv1 t

i P = I2R (I, R)

(m, v1 )

Questions 3 to 6 refer to the following information. A gardener charges a $40 fixed fee for each visit plus $12 per hour of work. 3 MC Which of the following graphs represents the above information, where C represents the total cost of a visit and t the time the gardener worked (in hours)? b

a C

40

12 0

d C

40 0

40 0

t

(5, 100)

t

t

e

(4, 76)

c C

40 C

0

t

t 12

0

40 C

4 MC Which of the following represents the relationship between t and C? a C + 40 − 12t = 0 b 12t + C = 40 c 12t + 40 − C = 0 d t = 12C + 40 e 40 + 12t + C = 0

Chapter 4 Algebra

99

5 MC When the relationship between t and C is transposed to make t the subject, it is then written as: C + 40 C - 12 a t = b 12t + 40 = C c t = 12 40 C 3 C d t = + e - 3 13 = t 40 10 12 6 MC If the total bill came to $79, for how long did the gardener work? a 3 h b 3 h 15 min c 3 h 30 min d 3 h 45 min e 4 h Questions 7–10 refer to the following information. The volume of a square-based pyramid 1

with the side of the base s and the height h is given by the formula V = 3s2h. 7 MC The side length of the base of a square-based pyramid with the height h and volume V is given by: a s = 3

v h

b s =

3h V

c s =

h 3V

d s =

V 3h

e s =

3V h

8 MC The height of a square-based pyramid with the side of the base 5 cm and volume 75 cm3 is: a 8 cm b 9 cm c 10 cm D 11 cm E 12 cm 9 MC If both the side of the base and the height are doubled, the volume is: A doubled b tripled C increased by 4 times D increased by 6 times E increased by 8 times 10 MC If the side of the base of a pyramid is doubled, but its volume remains unchanged, the height: 1

A becomes twice as large

B becomes 2 of the original size

1

C becomes 4 of the original size

D becomes 4 times as large

3

E becomes 4 of the original size 11 The sum of the interior angles of a regular polygon is given by S = (n − 2) × 180°, where n is the number of sides. a Transpose the formula to make n the subject. b Use the appropriate formulas to complete the following table: Polygon

Number of sides (n)

Triangle

3

Hexagon

6

Dodecagon

12

Nonagon

9

Heptagon

7

Sum of interior angles (S)

1080° 540° 360° 1440°

100



WorkSHEET 4.1

4d

n 12 The sum of n terms of an arithmetic sequence is given by the formula S = [2a + ( n - 1)d ], 2 where a is the first number of the sequence and d is the common difference. a Transpose the formula to make a the subject and hence find the first term in a sequence that has n = 26, d = 3 and S = 1079. b Transpose the formula to make d the subject and hence find the common difference of an arithmetic sequence with 20 terms, a = 18 and S = -20.

solving linear equations and simultaneous linear equations A linear equation is an equation that contains a pronumeral (unknown value) raised to the power of 1. Such an equation may also be called an equation of the first degree. Examples of linear or first degree equations include: x+5 2x - 4 = 8, y = 7x - 12 and y = . 3 Equations of the type: 1 y = , y = x , 2x2 - 4 = 8, x2 + y2 = 4 and y = x3 - 8 x are not linear since they contain pronumerals that are raised to powers other than 1; in these 1

cases, -1, 2 , 2, 2 and 3, respectively. A linear equation is an equation that contains a pronumeral raised to the power of 1. It may also be called an equation of the first degree.

solving linear equations When we are asked to solve an equation, we are to find the value of the pronumeral so that when it is substituted into the original equation, it will make the equation a true statement. Equations are solved by performing a number of inverse operations to both sides of the equation until the value of the unknown is found. When solving equations, the order of operations process, BODMAS (i.e. Brackets Of Division, Multiplication, Addition, Subtraction) is reversed. We may therefore apply the SAMDOB process (BODMAS in reverse). This means that the operations of subtraction and addition are taken care of first, followed by multiplication and division. Brackets are dealt with last. Worked exAmple 10

Solve the following equations. 3x a 2x - 3 = 4 b 10 =5 2 Think a

WriTe

1

Write the given equation.

2

(Optional step.) Rule up a table with two columns to the side of the equation. In the first column, note each of the operations performed on x in the correct order. In the second column, write the corresponding inverse operation. The arrows indicate which operation to begin with.

a 2x - 3 = 4

Operation ↓

Inverse

×2

÷2

-3

+3 ↑

Chapter 4

Algebra

101

b

2x − 3 = 4 2x − 3 + 3 = 4 + 3 2x = 7

3

Solve the equation by making x the subject. Add 3 to both sides of the equation.

4

Divide both sides of the equation by 2.

5

Simplify.

1


2

(Optional step.) As in part a above.

2x 7 = 2 2 1

x = 3 2 (or 3.5)

b

10 -

Operation

Inverse

×-3

÷-3

÷2

×2

+ 10

- 10

↓ 3

Solve the equation by making x the subject. Subtract 10 from both sides of the equation.

3x =5 2

3x =5 2 3x 10 - 10 = 5 - 10 2 10 -

- 3x

2 4

- 3x

Multiply both sides of the equation by 2.

2

5

-3x

Divide both sides of the equation by −3.

Simplify.

= -5

× 2 = -5× 2 -3x

-3

6

↑

= - 10 =

x=

-10 -3

10 3

= 3 13

Step 2 in worked example 10 is an optional step that may be used initially to help you become familiar with the process of solving equations. The answers may be checked by substituting the values obtained back into the original equation or using a CAS calculator. If the pronumeral appears in the equation more than once, we must collect terms containing the unknown on one side of the equation and all other terms on the other side. Worked Example 11

Solve for x in the equation: 2x − 4 = 4x + 6. Think

102

1


2

Transpose 4x to the LHS of the equation by subtracting it from both sides of the equation.

Write

2x − 4 = 4x + 6 2x − 4x − 4 = 4x − 4x + 6 −2x − 4 = 6


3


4


5

Simplify.

−2x

−4+4=6+4 −2x = 10 -2x -2

=

10 -2

x = −5

If the equation contains brackets, they should be expanded first. In some cases, however, both sides of the equation can be divided by the coefficient in front of the brackets instead of expanding. Worked Example 12

Solve for x in 2(x + 5) = 3(2x - 6). Think

Write



2(x + 5) = 3(2x − 6)

2

Expand each of the brackets on both sides of the equation.

2x + 10 = 6x − 18

3

Transpose 6x to the LHS of the equation by subtracting it from both sides of the equation.

4

Subtract 10 from both sides of the equation.

5


2x − 6x + 10 = 6x − 6x − 18 −41x + 10 = −18 −4x

+ 10 − 10 = −18 − 10 −4x = −28 -4 x -4

6

Simplify.

=

- 28 -4

x=7

Method 2: Using a CAS calculator 1

On the Main screen, tap: • Action • Advanced • solve Complete the entry line as: solve (p = 2(x + 5) = 3(2x - 6), x) Then press E.

2

Write the answer.

Solving 2(x + 5) = 3(2 x − 6) for x, gives x = 7.

Chapter 4 Algebra

103

If an equation contains a fraction, we should first remove the denominators by multiplying each term of the equation by the lowest common denominator (LCD).

Worked Example 13

Find the value of x that will make the following a true statement: Think

x+2 x = 5- . 3 2

Write

Method 1: Using the rule x+2 x = 53 2

1


2

Determine the LCD of 2 and 3.

3

Multiply each term of the equation by the LCD.

4


5

Expand the bracket on the LHS of the equation.

2x + 4 = 30 − 3x

6

Add 3x to both sides of the equation.

2x + 3x + 4 = 30 − 3x + 3x 5x + 4 = 30

7

Subtract 4 from both sides of the equation.

8


9

Simplify.

LCD of 2 and 3 is 6. x+2 x ×6 = 5×6- ×6 3 2 6( x + 2) 6x = 30 3 2 2(x + 2) = 30 − 3x

5x + 4 − 4 = 30 − 4 5x = 26 5 x 26 = 5 5 1

x = 5 5 (or 5.2)

Method 2: Using a CAS calculator

104

1

On the Main screen, tap: • Action • Advanced • solve Complete the entry line as: x  x+2 solve  = 5 - , x  3 2  Then press E.

2

Write the answer.

Solving

26 x+2 x 1 or 5 5. = 5 - for x, gives x = 5 3 2


Sometimes in equations containing fractions, a pronumeral appears in the denominator. Such equations are solved in the same manner as those in the previous examples. However, care must be taken to identify the value (or values) for which the pronumeral will cause the denominator to be zero. If in the process of obtaining the solution the pronumeral is found to take such a value, it should be discarded. Even though the process of identifying the value of the pronumeral that causes the denominator to be zero is at this stage merely a precaution, this process should be practised as it will prove useful in future chapters.

Worked exAmple 14

Solve the following equation for x:

eBook plus

2 3 1 . + = x 2x x - 1

Tutorial


Think

WriTe


Identify the values of x that will cause the denominator to be zero. Note: Once the equation has been solved, values that cause the denominator to be zero will be discarded.

First fraction: x=0 Second fraction: 2x = 0 x=0 Third fraction: x - 1 = 0 x=1 x cannot assume the values of 0 and 1, since this will cause the fraction to be undefined.

2


2 3 1 + = x 2x x - 1

3

Determine the LCD of x, 2x and x - 1.

LCD of x, 2x and x - 1 is 2x(x - 1).

4

Multiply each term of the equation by the LCD.

2 3 1 × 2 x ( x - 1) + × 2 x ( x - 1) = × 2 x ( x - 1) x 2x x -1

5


6

Expand the brackets on the LHS of the equation.

7

Collect like terms onto the LHS by subtracting 2x from both sides of the equation.

7x - 2x - 7 = 2x - 2x 5x - 7 = 0

8


5x - 7 + 7 = 0 + 7 5x = 7

9


10

Simplify. Note: The value of 1.4 is a valid solution.

4 x ( x - 1) 6 x ( x - 1) 2 x ( x - 1) + = x 2x x -1 4(x – 1) + 3(x - 1) = 2x 4x - 4 + 3x - 3 = 2x 7x - 7 = 2x

5x 7 = 5 5 2

x = 1 5 (or 1.4)

Chapter 4

Algebra

105


On the Main screen, tap: • Action • Advanced • solve Complete the entry line as: 1 2 3  solve  + ,x =  x 2 x x - 1  Then press E.

2

Write the answer.

Solving

7 2 3 1 2 for x, gives x = or 1 5 . + = 5 x 2x x - 1

Simultaneous equations It is impossible to solve one linear equation with two unknowns. There must be two equations with the same two unknowns for a solution to be found. Such equations are called simultaneous equations.

Graphical solution of simultaneous equations If two straight lines intersect, the point of their intersection belongs to both lines and hence the coordinates of that point (x, y) will represent the solution of two simultaneous equations that define the lines. When we are solving simultaneous equations graphically, the accuracy of the solution is highly dependent on the quality of the graph. Therefore, all graphs must be drawn on graph paper as accurately as possible. It is good practice to verify any answer obtained from a graph by substituting it into the original equations, or to check using a CAS calculator. Worked Example 15

Solve the following pair of simultaneous equations graphically: a x + 2y = 4 b y + 3x = 17 x - y = 1 2x - 3y = 4 Think a

1 2

3

106

Write

Rule up a set of axes. Label the origin and the x and y axes. Find the x-intercept for the equation x + 2y = 4, by making y = 0.

Find the y-intercept for the equation x + 2y = 4, by making x = 0. Divide both sides of the equation by 2.

a (See graph at step 7 on page 107.)

x-intercept: when y = 0, x + 2y = 4 x+2×0=4 x=4 The x-intercept is at (4, 0). y-intercept: when x = 0, x + 2y = 4 0 + 2y = 4 2y = 4 y=2 The y-intercept is at (0, 2).


4

Plot the points on graph paper and join them with the straight line. Label the graph.

(Refer to the graph at step 7.)

5

Find the x-intercept for the equation x – y = 1, by making y = 0.

x-intercept: when y = 0, x−y=1 x−0=1 x=1 The x-intercept is at (1, 0).

6

Find the y-intercept for the equation x − y = 1, by making x = 0. Multiply both sides of the equation by −1.

y-intercept: when x = 0, x−y=1 0−y=1 −y = 1 − − y × 1 = 1 × −1 y = −1 The y-intercept is at (0, −1).

7

Plot the points on graph paper and join them with the straight line. Label the graph.

y x–y=1 2 1 0 −1

b

x + 2y = 4 (2, 1) 1

2

4

x

8

From the graph, read the coordinates of the point of intersection.

The point of intersection between the two graphs is (2, 1).

9

Verify the answer by substituting the point of intersection into the original equations.

Substitute x = 2 and y = 1 into x + 2y = 4. LHS = 2 + 2 × 1 RHS = 4 =2+2 =4 LHS = RHS Substitute x = 2 and y = 1 into x − y = 1 LHS = 2 − 1 RHS = 1 =1 LHS = RHS In both cases LHS = RHS; therefore, the solution set (2, 1) is correct.

1

Rearrange both equations to make y the subject. To do this, on the Main screen, complete the entry lines as: solve(y + 3x = 17, y) solve(2x - 3y = 4, y) Press E after each entry.

b

Chapter 4 Algebra

107

2

On the Graph & Tab screen, complete the function entry lines as: y1 = 2( x - 2) 3 y2 = 17 - 3x Tick the equation boxes and tap $.

3

To find the point of intersection, tap: • Analysis • G-Solve • Intersect

4

Write the answer.

The point of intersection between the two graphs is (5, 2).

Parallel lines If two equations have the same gradient, they represent parallel lines. Such lines will never meet and so never have a point of intersection (that is, there is no solution). y The following pair of equations, y = 2x + 3 and y = 2x + 5 define two 5 parallel lines; hence, there is no solution. The graph at right demonstrates 5 that the straight lines never intersect. + y=

2x

y=

Coincidental lines If two lines coincide, then there are an infinite number of solutions. For example, consider the two straight lines given by the equations y = 2x + 1 and 4x − 2y = −2. Rearranging the second equation we obtain the same line.

108

4x − 2y = −2 4x − 4x − 2y = −4x − 2 −2y = −4x − 2 -2y -4 x 2 = --2 -2 2 y = 2x + 1


+ 2x

3 3

x

0 y

1 + 2x −2 y= y= 2 − 4x

0

x

The two equations when graphed represent the same line — they coincide. Therefore, every point on the line will represent the solution as there is not one unique point that satisfies both equations.

Algebraic solution of simultaneous equations When using algebra to solve simultaneous equations, the aim is to obtain one equation with one unknown from two equations with two unknowns by various algebraic manipulations. This can be done in two ways — substitution and elimination — as outlined below. Substitution method The method of substitution is easy to use when at least one of the equations represents one unknown in terms of the other. To solve simultaneous equations using the method of substitution: 1. Check that one of the equations is transposed so that one of the unknowns is expressed in terms of the other. 2. Substitute the transposed equation into the second equation. 3. Solve for the unknown variable. Worked Example 16

Use the method of substitution to solve the simultaneous equations y = 2x + 3 and 4x - y = 5. Think

Write

y = 2x + 3 4x − y = 5

1

Write the equations one under the other, and number them.

2

Substituting (2x + 3) into [2]: 4x − (2x + 3) = 5

3

Substitute the expression for y (2x + 3) from equation [1] into equation [2]. Note: By substituting one equation into the other, we are left with one equation and one unknown. Solve for x. (a) Expand the brackets on the LHS of the equation. (b) Simplify the LHS of the equation by collecting like terms. (c) Add 3 to both sides of the equation. (d) Divide both sides of the equation by 2.

4

Substitute 4 in place of x into [1] to find the value of y.

Substituting x = 4 into [1]: y=2×4+3

5

Evaluate.

6

Answer the question.

Solution: x = 4, y = 11 or solution set (4, 11).

7

Verify the answer by substituting the point of intersection into the original equations or use a CAS calculator.

The answer was checked using a CAS calculator and found to be correct.

[1] [2]

4x − 2x − 3 = 5 2x − 3 = 5 2x − 3 + 3 = 5 + 3 2x = 8 2x 8 = 2 2 x=4

=8+3 = 11

If neither of the equations give one unknown in terms of the other, we can still use a method of substitution by first transposing one of the equations.

Chapter 4 Algebra

109

Elimination method As the name suggests, the idea of the elimination method is to eliminate one of the variables. This is done in the following way. 1. Choose the variable you want to eliminate. 2. Make the coefficients of that variable equal in both equations. 3. Eliminate the variable by either adding or subtracting the two equations. Once this is done, the resulting equation will contain only one unknown which then can be easily found. Worked Example 17

Use the elimination method to solve the following simultaneous equations. 2x + 3y = 4 x - 3y = 2 Think

Write

2x + 3y = 4 x − 3y = 2

1

Write the equations one under the other, and number them.

2

Add equations [1] and [2] to eliminate y. Note: y is eliminated since the coefficients of y in both equations are equal in magnitude and opposite in sign.

[1] + [2]: 2x + 3y = 4 + (x − 3y = 2) 3x = 6

3


4

Substitute the value of x into equation [2]. Note: x = 2 may be substituted in either equation. Solve for y. (a) Subtract 2 from both sides of the equation. (b) Divide both sides of the equation by −3.

5

[1] [2]

3x 6 = 3 3 x=2 Substituting x = 2 into [2]: x − 3y = 2 2 − 3y = 2 2 − 2 − 3y = 2 − 2 −3y = 0 -3y 0 = -3 3 y=0

6


Solution: x = 2, y = 0 or solution set (2, 0).

7

Verify the answer by substituting the point of intersection into the original equations or using a CAS calculator.

The answer was checked using a CAS calculator and found to be correct.

If there is no pair of equal coefficients, we can make them the same by multiplying or dividing one or both equations by an appropriate number. Worked Example 18

Solve the following simultaneous equations using a CAS calculator. 2x + 3y = 4 3x + 2y = 10

110


Think

Write

1

On the Main screen, tap: • ) • {N Complete the entry line as shown. Then press E.

2


Solution: x =

22 , 5

y=

-8 5

 22 - 8  , . 5 5

or 

REMEMBER

1. A linear equation is an equation that contains a pronumeral raised to the power of 1. It may also be called an equation of the first degree. 2. Linear equations are solved by using inverse operations. When solving linear equations the order of operations process, BODMAS, is reversed. 3. If the pronumeral appears more than once, the terms containing the unknown are collected onto one side of the equation and the numbers onto the other. 4. If the equation contains brackets, either expand, or divide both sides by the coefficient in front of the bracket. 5. If an equation contains fractions, multiply each term of the equation by the LCD. Simultaneous linear equations can be solved either graphically or algebraically. 6. Graphical method: – Draw the straight lines representing the equations and find the coordinates of the point of intersection. 7. Algebraic methods: (a) Substitution: Transpose one of the equations so that one of the unknowns is expressed in terms of the other and substitute into the second equation. (b) Elimination: Equate the coefficients of one unknown and eliminate it by either adding or subtracting the two equations.

Exercise

4d

Solving linear equations and simultaneous linear equations 1 WE 10 Solve the following equations. a x + 3 = −16 b 3 − 2x = 10 - 6x x 1 d = -5 e = 4 2 7

c −0.2x = 10 3x + 11 = 20 f 4

Chapter 4 Algebra

111

g

- 2x

+3=-7

4 11x + 2 j =7 5

2 WE 11 Solve for x. a 2x = 7 + 9x d 3x + 4 = x − 6 g 2x − 7 = −2x + 1 j 13 − 3x = 4 − 6x 3 WE 12 Solve for x. a 4(x − 20) = 16 c −2(2x − 7) = 3x e 4(x + 3) = 2(7 − 4x) + 5

h

x-9 =5 7

k 6 b e h k

6- x =5 13 2x l - 17 = 20 3 i

x = 15 2

15x + 22 = −7x + 2 = 3x + 18 15x + 2 = 13x − 10 −9 + 11x = −3 − x −5x

c 12 − 17x = −5x f 5x − 3 = −3 − 5x i 8 − 3x = 4 − x

b −(5 + 6x) = 13 d 8(2x + 1) = −2(7 − 3x) f 5(2x − 4) − 3 + 7(2 − x) = 0

4 WE 13 For each of the following, find the value of x that will make the statement true. x + 4 2x + 1 x a b - x = 12 = 3 2 7 2x 7 x - 1 2 - 3x c x + -3= 0 d = 6 4 5 7 x 3(2 - 3 x ) 5 x 3x - 2 7 - x 2 x + 1 1 e f = + + = 3 2 8 4 3 2 6 7- x 2(3 x - 6) 4(6 - 2 x ) 3( x + 7) g h -4= = 55 3 9 6 5 WE 14 For each of the following: a state which value (or values) of x will cause the equation to be undefined b solve for x. 2 1 4 3 4 2 i ii - = + = x +1 x x x - 1 x 3x 2(3 x - 6) 1 4 5 7 iv =1 iii ( x - 1)( x + 1) + x + 1 = x - 1 2x 4 x

v

3(7 x - 4) = 4 x-2

vi

-3 15 11 - = 4 x 8x x - 3

6 WE 15 Solve the following pairs of simultaneous equations graphically. Verify your answer with a CAS calculator. a 3x + y = 6, x − 2y = 2 b −x + 3y = 3, 2x + 3y = 12 c x = y + 2, 2y − x = 0 d 3x + 2y = −6, y = 1 − x 7 MC The pair of simultaneous equations y = 5 − x and y = −x − 1 has: A 1 solution B 2 solutions C no solutions D an infinite number of solutions E none of these 8 MC The pair of simultaneous equations 2x − 2y = 4 and y − x + 2 = 0 has: A 1 solution B 2 solutions C no solutions D an infinite number of solutions E none of these 9 Complete the following statements. a If two lines with different gradients go through the origin, then the solution to the pair of simultaneous equations defining those lines is _______________ (give coordinates). b If two lines have the same gradients but different y-intercepts, then the pair of simultaneous equations defining such lines will have _______________ solution(s).

112


c If two lines are defined by the equation y = mx + c and have the same value of c but different values of m, then the solution to the pair of simultaneous equations will be _______________ (give coordinates). 10 WE 16 Solve the following pairs of simultaneous equations by the method of substitution. a y = 2x + 3 b x = y c y = 3x − 6 6x + y = 11 6x − 2y = 10 y = 16 + 5x d x = −y e 2y − 6 = x f x = 5 − 4y 3x − 6y = 36 7x + 3y = −25 2y − 3x = 13 Use the method of elimination to solve each of the following: 11 WE 17 a 2x + y = 3 b x − y = 7 c 3x − 2y = −1 − − 4x − y = 9 y+x= 1 3x − 6y = −9 e 2x + 4y = 24 f 2x − y = 0 d x + 3y = 7 −6x + 4y = 8 2x − 4y = −9 5x − 3y = −19 12 MC Nathan is solving a pair of simultaneous equations 2x − 3y = 5 [1] and 3x + 4y = 10 [2] using the elimination method. To eliminate one of the variables, he could multiply equation [1] and equation [2] by: A 2 and 3 respectively B 3 and 4 respectively C 3 and 2 respectively D 5 and 10 respectively E 4 and 2 respectively 13 WE 18 Solve each of the following pairs of equations using the elimination method. b x − 2y = −4

a 2x − 3y = 8 3x + 4y = −5

e 5y − 2x = 4

x + 3y = −5

6x + y = 4

1

2x − y = -3

3x + y = 9

d 2y − x = −10

1 x 5

f

3

+ 5y = 5

2y + x = 20

4 2 1 + = we know that x will not be equal to: x - 3 x +1 x

14 MC Without solving the equation B −1

A 3

2

c 3 x + 3y = 5

C 0

D 0 or −1

E −1 or 3 or 0

3( x - 1) 4 - 2x , each term of the equation could be multiplied by: -5= 2 3 A 2 B 3 C 4 D 5 E 6 2x 16 MC To solve the equation = 4, the operations that must be performed are: 3 2 A × both sides by 2, then ÷ by 3 B × both sides by 3

15 MC To solve

2

C ÷ both sides by 3

D × both sides by 3, then ÷ by 4

E × both sides by 4, then ÷ by 2 Question 17 refers to the diagram at right.

y

[1]

[2] 2 1 0 −1 −2

1

3

x

17 MC a The equation of line [1] is: 2

A y = 3 x − 2

B y = 2x − 2

D y = 2x + 1

E y = 2 + 2x

1

C y = 2 x − 2

Chapter 4 Algebra

113

b The equation of line [2] is: 2

2

a y = 3x + 2

b y = 3x + 3

c y=

-2 3

x-2

d 3y = 2x + 6 e 3y + 2x = 6 c The point of intersection of the two lines has the coordinates: a

(, ) 3 2

1 2

1  b  2, 2 

c (2, 1)

d

3   , 1 2

e

3   2,  2

18 Find the value of z, such that the solution to the following equation is x = 1. 3 z 8x = x - 2 x + 1 ( x - 2)( x + 1) 19 Solve the following equation. 5 4 6 = + 2x - 2 x - 1 x - 2 20 Find the values of x, y and z in the following three simultaneous equations with three unknowns. 2x + 3y - z = -7 3x + 2y + z = 2 x - 4y + 2z = 15 eBook plus Digital doc

Investigation Comparing production costs

4e

Applications Linear equations can often be used to help us in problem solving. This is usually done in the following way. 1. Identify the unknown and choose any convenient pronumeral (usually x) to represent it. 2. Use the information given in the problem to compose an equation in terms of the pronumeral. 3. Solve the equation to find the value of the pronumeral. 4. Interpret your result by relating the answer back to the problem.

Worked exAmple 19

If the sum of twice a certain number and 5 is multiplied by 3 and then divided by 7, the result is 9. Find the number. Think

WriTe

1

Assign the pronumeral x to the unknown value.

2

Build the equation according to the information given. (a) Twice the number — this means × 2, so write this.

2x

(b) The sum of twice the number and 5 — this means 2x + 5, so add this on.

2x + 5

(c) The sum is multiplied by 3 — this means 3(2x + 5). Add this on. Note: We include brackets to indicate the order of operations.

3(2x + 5)

(d) The result is divided by 7 — this means Add this on.

3(2 x + 5) . 7

(e) The result is 9 — which means that all of the previous computations will equal 9. Write this.

114

Let x = the unknown number.

3(2 x + 5) 7 3(2 x + 5) =9 7


3

Solve for x. (a) Multiply both sides of the equation by 7.

(b) Divide both sides of the equation by 3 since they are both divisible by 3. (c) Subtract 5 from both sides of the equation.

3(2 x + 5) ×7= 9×7 7 3(2x + 5) = 63 3(2 x + 5) 63 = 3 3 2x + 5 = 21 2x + 5 − 5 = 21 − 5 2x = 16 2 x 16 = 2 2

(d) Divide both sides of the equation by 2.

x=8

(e) Simplify. 4


The unknown number is 8.

Sometimes the problem contains more than one unknown. In such cases one of the unknowns is called x and the other unknowns are then expressed in terms of x. Worked Example 20

Sarah is buying tulip bulbs. Red tulip bulbs cost $5.20 each, while yellow tulip bulbs cost $4.70 each. If 22 bulbs cost Sarah $107.40, how many of each type did she buy? Think

Write

1

Define the variables. Note: Since there are 22 bulbs altogether, the number of yellow tulip bulbs is 22 − the number of red tulip bulbs; that is, 22 − x.

Let x = the number of red tulip bulbs. Let 22 − x = the number of yellow tulip bulbs.

2

Write an expression for the cost of the red tulips. Note: One red tulip costs $5.20; therefore x red tulips cost 5.20 × x.

Total cost of red tulip bulbs = 5.20 × x = 5.2x

Chapter 4 Algebra

115

3

Write an expression for the cost of the yellow tulips. Note: One yellow tulip costs $4.70; therefore 22 − x tulips cost 4.70 × (22 − x).

Total cost of yellow tulip bulbs = 4.70 × (22 − x) = 4.7(22 − x)

4

Formulate an equation relating the total cost of the red and yellow tulips and the expressions obtained in steps 2 and 3.

The total cost of the red and yellow tulip bulbs is $107.40. Also the total cost of red and yellow tulip bulbs is 5.2x + 4.7(22 − x). Therefore, 5.2x + 4.7(22 − x) = 107.4

5

Solve the equation. (a) Expand the brackets on the LHS of the equation.

5.2x + 103.4 − 4.7x = 107.4 0.5x + 103.4 = 107.4

(b) Collect the like terms on the LHS of the equation. (c) Subtract 103.4 from both sides of the equation.

0.5x + 103.4 − 103.4 = 107.4 − 103.4 0.5x = 4

(d) Divide both sides of the equation by 0.5.

0.5 x 4 = 0.5 0.5 x=8 There are 8 red and 14 (that is, 22 − 8) yellow tulip bulbs.

6

(e) Simplify. Interpret the answer obtained.

7


Sarah purchased 8 red and 14 yellow tulip bulbs.

Worked Example 21

A train (denoted as train 1) leaves station A and moves in the direction of station B with an average speed of 60 km/h. Half an hour later another train (denoted as train 2) leaves station A and moves in the direction of the first train with an average speed of 70 km/h. Find: a the time needed for the second train to catch up with the first train b the distance of both trains from station A at that time. Think

116

Write

1

Define the variables. Note: Since the first train left half an hour earlier, the time taken for it to reach the meeting point will be x + 0.5.

Let x = the time taken for train 2 to reach train 1. Therefore, the travelling time, t, for each train is: Train 1: t1 = x + 0.5 Train 2: t2 = x

2

Write the speed of each train.

Train 1: Train 2:

v1 = 60 v2 = 70

3

Write the distance travelled by each of the trains from station A to the point of the meeting. (Distance = speed × time.)

Train 1: Train 2:

d1 = 60(x + 0.5) d2 = 70x

4

Equate the two expressions for distance. Note: When the second train catches up with the first train, they are the same distance from station A — that is, d1 = d2.

When the second train catches up with the first train, d1 = d2.


5

Solve the equation. On the Main screen, tap: • Action • Advanced • solve Then complete the entry line as: solve(60(x + 0.5) = 70x, x) Then press E.

6

Substitute 3 in place of x into either of the two expressions for distance, say into d2.

Substitute x = 3 into d2 = 70x d2 = 70 × 3

7

Evaluate.

8

Answer the questions.

a The second train will catch up with the first train

= 210 3 hours after leaving station A.

b Both trains will be 210 km from station A.

Simultaneous equations are used to solve a variety of problems containing more than one unknown. Here is a simple algorithm that can be applied to any of them: 1. Identify the variables. 2. Set up simultaneous equations by transforming written information into algebraic sentences. 3. Solve the equations by using the substitution, elimination or graphical methods. 4. Interpret your answer by referring back to the original problem. Worked Example 22

Find two consecutive numbers that add up to 99. Think 1

Define the two variables.

2

Formulate two equations from the information given and number them. Note: Consecutive numbers follow one another and differ by 1. Hence, if x is the first number, the next number will be x + 1 — that is, y = x + 1. Substitute the expression (x + 1) from equation [2] for y into equation [1].

3 4

Solve for x. (a) Simplify the LHS of the equation by collecting like terms. (b) Subtract 1 from both sides of the equation. (c) Divide both sides of the equation by 2.

Write

Let x = the first number. Let y = the second number. x + y = 99 y = x + 1

[1] [2]

Substituting (x + 1) into [1]: x + x + 1 = 99 2x + 1 = 99 2x + 1 − 1 = 99 − 1 2x = 98 2 x 98 = 2 2 x = 49

Chapter 4 Algebra

117

5

Substitute 49 in place of x into equation [1] to find the value of y.

Substituting x = 49 into equation [2]: y=x+1

6

Evaluate.

7

Verify the answer by checking that the two values are consecutive and that they sum 99.

8


y = 49 + 1 = 50 49 and 50 are consecutive numbers. 49 + 50 = 99 The obtained values satisfy the problem. The two consecutive numbers that add up to 99 are 49 and 50.

Worked Example 23

Two hamburgers and a packet of chips cost $8.20, while one hamburger and two packets of chips cost $5.90. Find the cost of a packet of chips and a hamburger. Think

Write

1

Define the two variables.

2

Formulate an equation from the first sentence and call it [1]. Note: One hamburger costs $x, two hamburgers cost $2x. Thus, the total cost of cost of two hamburgers and one packet of chips is 2x + y and it is equal to $8.20. Formulate an equation from the second sentence and call it [2]. Note: One packet of chips costs $y, two packets cost $2y. Thus, the total cost of two packets of chips and one hamburger is x + 2y and it is equal to $5.90.

3

4

Solve for the simultaneous equations. On the Main screen, tap: • ) • {N Complete the entry line as shown. Then press E.

5

Answer the question and include appropriate units.

Let x = the cost of one hamburger. Let y = the cost of a packet of chips. 2x + y = 8.20

[1]

x + 2y = 5.90

[2]

A hamburger costs $3.50 and a packet of chips costs $1.20.

It is extremely important to be consistent with the use of units while setting up equations. For example, if the cost of each item is expressed in cents, then the total cost must also be expressed in cents.

118


REMEMBER

To solve worded problems using linear equations, follow these steps: 1. Identify the variables. 2. Set up an equation by transforming the written information into an algebraic statement or statements. 3. Solve the equation. 4. Interpret the result by relating the answer back to the original problem. To solve problems involving simultaneous equations, follow these steps: 5. Identify and define the variables. 6. Transform written information into algebraic statements. 7. Solve the pair of equations graphically or algebraically using the methods of substitution or elimination. 8. Interpret the result by relating the answer back to the problem. 9. Always make sure the numbers in the equations are in the same units. Exercise

4e

Applications 1 WE 19 The average of three consecutive odd numbers is −3. Find the largest number. 2 Half of a certain number is subtracted from 26 and the result is then tripled, and the answer is 18. Find the number. 3 The sum of one-third of a number and 5 is 27. Find the number. 4 WE20 Fiona is buying tulip bulbs. Red tulip bulbs cost $6.40 each, while yellow tulip bulbs cost $5.20 each. If 28 bulbs cost Fiona $167.20, how many of each type did she buy? 5 A rectangle is 2.5 times as long as it is wide. Find the dimensions of the rectangle if its perimeter is 56 cm. 6 In an isosceles triangle, two sides of equal length are together 8 cm longer than the third side. If the perimeter of the triangle is 32 cm, what is the length of each side? 7 In a scalene triangle the first angle is 3 times as large as the second, while the third angle is 20° smaller than the second. Find the size of each angle; hence, name the triangle according to its angles’ sizes. 8 All items at a clothing store have been reduced by 15%. If Stephanie purchased a shirt at the reduced price of $84.15, what was its original price?

Chapter 4 Algebra

119

9 MC a If 7 times a number subtracted from 52 gives 3, then the number is: 6

A −7 B 7 C 8 D 6 E 7 7 b The sum of one-quarter of a number and 10 is 15. The value of the number is: A 100 B 50 C 40 D 20 E 10 1

10 a I am 3 times as old as my cousin Carla, who is 33 times as old as my daughter Nina. If our total ages are 43 years, how old is my cousin? b Another cousin, Zara, is Carla’s older sister. Zara is as many times as old as my daughter Nina as the number of years that she is older than Carla. How old is my other cousin? 11 Simon is only 16 years old, but he has already lived in four different countries because of his father’s job. He was born and spent a few years of his early childhood in the USA, then the family moved to Germany, where he stayed one year longer than he had in the USA. After that, he lived in London for twice as long as he had in Germany. Finally, they came to live in Melbourne. So far, he has been in Australia for 2 years less than he lived in America. a At what age did Simon leave his country of birth? b For how long did Simon live in each country? 12 In the storeroom of a fruit shop there were two boxes of apples, one of Golden Delicious and the other of Jonathans, which were to be sold at $2.80 and $3.50/kg, respectively. The apples, however, were accidentally mixed together and, instead of sorting them out, the owner decided to sell them as they were. So as not to make a loss, he sold the mixed apples at $3.10/kg. How many kilograms of each type of apple were there if together they weighed 35 kg? 13 WE21 Alex and Nat are going for a bike ride. Nat can ride at 10 km/h, while Alex can develop a maximum speed of 12 km/h if he needs to. Nat leaves home at 10 am, while Alex stays behind for 15 minutes and then sets out to catch up with Nat. When will Alex reach Nat, assuming that both of them are riding at their maximum speed?

14 Samuel is paddling with a constant speed towards a certain place he has marked on his map. With the aid of a current (which has a speed of 2 km/h) it takes him only 1 h 20 min to reach his destination. However, on the way back he has to paddle against the current and it then takes him 4 h to reach his starting point. Find Samuel’s speed on the still water.

15 One administrative assistant can type 1.5 times as fast as another. If they both work together, they can finish a certain job in 6 hours. However, the slower one working alone will need 15 hours to finish the same job. How many hours will the quicker assistant alone need to complete the job?

120


16 Maya needs to renovate her house. She has enough money to pay a plumber for 28 days or a carpenter for 21 days. For how many days can she pay the renovators if they both work at the same time? If Maya’s next pay cheque will come in 2 weeks, can she afford to hire both specialists until then? 17 In a particular school a number of VCE students obtained a tertiary entrance score higher than 99.4 and 15% more students obtained a score higher than 99.0, but lower than 99.4. If there were 43 students whose tertiary entrance scores were above 99.0, how many of those obtained a score above 99.4? 18 WE22 Find two consecutive numbers that add up to 89. 19 When three times the first number is added to twice the second number, the result is 13. Four times the difference of those numbers is 44. Find the numbers. 20 Half of the sum of two numbers is 6 less than the first number. One-third of their difference is one less than the second number. Find the numbers. 21 Five times the first number is twice as large as four times the second number. When the difference of the two numbers is multiplied by 20, the result is 3. Find the numbers. 22 A rectangle’s length is 2 cm more than its width. If the perimeter of a rectangle is 24 cm, find its dimensions and, hence, its area. (2x − 1) cm 23 In the rectangle at right, find the values of x and y. Hence, determine the perimeter.

(x + 2) cm

10 cm (5 + y) cm

24 The sides of an equilateral triangle have the following lengths: (x + y) cm, (2x − 3) cm and (3y − 1) cm. Find the perimeter of the triangle. 25 The perimeter of a rhombus ABCD is 10 cm longer than the perimeter of an isosceles triangle ABC. Find the length of AC, the diagonal of a rhombus, if it is 2 cm smaller than its side. 26 A table consists of 2 columns and 5 rows. Each of its cells is a rectangle with length x cm and width y cm. The perimeter of the table is 70 cm and the total length of interior lines is 65 cm. a Draw a diagram to represent the above information. b Find the dimensions of each cell and comment on its shape. 27 Phuong conducts a survey. She asks 72 people whether or not they use the internet at home. There were three times as many people who answered ‘Yes’ as those who answered ‘No’. Find the number of people in each category and hence help Phuong to complete the following statement: ‘According to the survey _______________ (insert fraction) of the population uses the internet at home.’ 28 WE23 At the end of the day, two shop assistants compare their sales. One sold 5 toasters and 2 sandwich-makers for a total of $149.65, while the other sold 3 of each for a total value of $134.70. Find the price of each item. 29 In an aquatic centre, ‘pool and spa’ entry is $3.50, while ‘pool, spa, sauna and steam room’ entry is $5.20. At the end of the day, a cashier finds that she sold 193 tickets altogether and her takings are 40c short of $800. How many of each type of ticket were sold? 30 Spiro empties his piggy bank. He has 42 coins, some of which are 5c coins and some of which are 10c coins, to the total value of $2.50. How many 5c coins and how many 10c coins does he have? 31 Maya and Rose are buying meat for a picnic. Maya’s family likes lamb more than pork, so she buys 3 kg of lamb and only half as much pork. Rose’s family have different tastes, so she buys 4.5 kg of pork and one-third as much lamb. If Maya spends $13.50, which is $8.25 less than Rose spends, what is the cost of 1 kg of each type of meat?

Chapter 4 Algebra

121

32 Bella and Boris are celebrating their 25th wedding anniversary. Today, their combined age is exactly 100. If Boris is 4 years older than Bella, how old was his bride on the day of their wedding? 33 Interpreting Pty Ltd translates each English text into both French and Japanese. It takes a French interpreter 0.6 hours to translate a page of any scientific text and 1 hour to translate a page of fiction. A Japanese interpreter needs 0.95 hours to translate scientific text and 1.5 hours for fiction. If the French interpreter works 8 hours a day, while the Japanese interpreter is prepared to take some of her work home and spend up to 12.5 hours per day altogether, what is the maximum number of pages of each type of text that can be translated each day by Interpreting Pty Ltd? eBook plus Digital doc

WorkSHEET 4.2

34 Sasha is making dim sims and spring rolls for his guests. He is going to prepare everything first and then cook. On average it takes 0.2 hours to prepare one portion of dim sims and 0.25 hours to prepare one portion of spring rolls. He needs 0.05 hours and 0.15 hours to cook each portion of dim sims and spring rolls respectively. If he spends 2 hours on preparation and 51 minutes on cooking, how many portions of dim sims and spring rolls does Sasha make?

Algebraic fractions

4f

Algebraic fractions are fractions that contain a pronumeral. Performing operations involving these fractions involves the same rules as those for numerical fractions.

Addition and subtraction Fractions can only be added or subtracted when they have a common denominator.

Worked exAmple 24

Simplify x x a + 2 5

b

x 3 + . 2 x

Think a

1

2

WriTe

Find the lowest common denominator. 5

Multiply the first term by 5 and the second 2

term by 2 .

b

Add the numerators.

1

Find the lowest common denominator.

2

Multiply the first term by 2

3

122

Add the numerators.

Lowest common denominator = 10 5 x 2 x × + × 5 2 2 5 5x 2 x = + 10 10 7x 10

3

term by 2 .

a 2 × 5 = 10

x and the second x

b Lowest common denominator = 2x

x2 6 + 2x 2x x2 + 6 2x


Worked exAmple 25

Simplify 3 2 a x x-1

eBook plus

b

Tutorial

2a 2 . x+3 x-3

Think


WriTe


1


2

Multiply the first term by the second term by x . x

b

a Lowest common denominator = x(x - 1)

x -1 and x -1

3( x - 1) 2x x ( x - 1) x ( x - 1)

3

Add the numerators.

3( x - 1) - 2 x x ( x - 1)

4

Expand and simplify the numerator.

x-3 x ( x - 1)

1


2

Multiply the first term by the second term by x + 3 . x+3

b Lowest common denominator = (x + 3)(x - 3), or

x2 - 9 2a( x - 3) 2( x + 3) + 2 x2 - 9 x -9

x-3 and x-3

3

Add the numerators.

2a( x - 3) + 2( x + 3) x2 - 9

4

Expand the numerator.

2ax - 6a + 2 x + 6 x2 - 9

Method 2: Using a CAS calculator a & b

1

On the Main screen, tap: • Action • Transformation • combine Complete the entry lines as: combine  3 - 2   x x - 1

a & b

combine  2a - 2   x + 3 x - 3 Press E after each entry.

2

Write the answer.

3 2 x-3 = 2 x x -1 x - x 2a 2 2ax - 2 x - 6a - 6 b = x+3 x-3 x2 - 9 a

Chapter 4

Algebra

123

Multiplication and division When multiplying, cancelling vertically or diagonally helps to simplify expressions before multiplying top and bottom. This may involve factorising expressions to identify common factors. Worked Example 26

Simplify a

3 x 2 20 × 4 9x

b

x2 + 4 2y . × 2 6 y2 5 x + 20

Think

Write


b

1

Cancel common factors between numerators and denominators then simplify.

a Common factors: 3, x, 4

3 x 2 20 x 5 × = × 4 9x 1 3 5x = 3

2

Multiply numerators together and denominators together.

1

Factorise the denominator.

2

Cancel common factors and simplify.

1 1 × 3y 5

3

Multiply numerators together and denominators together.

1 15y

b

x2 + 4 2y × 2 6y 5( x 2 + 4)

Method 2: Using a CAS calculator a & b

124

1

On the Main screen, complete the entry lines as: 3 x 2 20 × 4 9x 2 x +4 2y × 2 6y2 5 x + 20 Press E after each entry.

2

Write the answers.

a & b

3 x 2 20 5 x × = 4 9x 3 x2 + 4 2y 1 b × 2 = 2 6y 5 x + 20 15 y a


REMEMBER

1. To add or subtract algebraic fractions, first find a common denominator and solve by adding or subtracting the numerator. 2. If the pronumeral is in the denominator, you can generally find the common denominator by multiplying the denominators together. 3. When multiplying or dividing, cancel down before multiplying through numerators and denominators. Exercise

4f

Algebraic fractions 1 WE24 Simplify the following expressions. a 3a 2b 5b a + b + 7 4 3 4 d

3g 4 g 2 3

e

h 4 + 5 h

g

m 2 7 m

h

1 2n 3n 3

2

c

d 4d 5 3

f

2k 5 + 3 2

Simplify the following expressions. a

2 3 + p ( p + 2)

b

3 5 + 2q (q + 5)

c

4 3 + (r + 1) (r - 2)

d

1 7 (s - 3) (s + 4)

e

3 2 (2t + 3) (t - 1)

f

3 5v + (2v - 3) 8

g

3w 5 2 ( w - 2)

h

5( x - 1) ( x + 3) + 3 2

3 WE25 Simplify the following expressions. a

3 7 ( y + 3) ( y - 3)

b

1 5 + ( z + 2) ( z - 2)

c

1 4 + (3 - 2 x ) ( x - 2)

d

3 2 + (1 - y ) ( y + 3)

e

2 5 + (a + 3)2 (a + 3)

f

1 7 (3b - 2) (3b - 2)2

2 2 4 MC When simplifying the expression , the lowest common denominator 2 3( x - 3) ( x - 3) is: a x - 3 b 3(x - 3) c (x - 3)2 2 3 d 3(x - 3) e 3(x - 3) 5 MC Simplifying

2a 2 - a 4 a 2 - 4 a + 1 gives: ÷ 2a 3 b 3 (2a 2 b) 2

2a 2 b(2a - 1)

b

a d

- 2a 2

b(2a - 1)

e

4a2 b(2a - 1)

c

2a 2 b (2a - 1)

- 2a 2

b(2a + 1)

Chapter 4 Algebra

125

9 - e 2 2(e + 3) gives: ÷ e2 e3 3- e 3- e e(3 - e) 3+ e a b c d 2 2e 2 2 7 We26 Simplify the following expressions. 3b 2(b + 5) 2 x 3 2( x + 2) a b × × 2 3(b + 5) 8b 2 x+2 10 x 6

mC Simplifying

d 8

9 - e2 2e × (3 + e) e2

3+ e 2e

d2 + 5 12d × 2 3 3d 2d + 10 7h(h + 2) 12h 2 f × 3 2 7h + 14 h 2 c

g 2 (2 - g) 3g × 6 4 - g2

Simplify the following expressions. a

( j - 3)( j + 2) 12( j - 3)( j + 7) × 3( j + 7)2 2( j + 2)2

2m 2 - m - 3 3m × 6(m - 1) 2m(2m - 3)2 p + 3 3( p + 3) e ÷ 7 21 p c

g

4G

e

e

2(3r 2

3r 2 15r 3 ÷+ 5r + 2) (3r + 2)

b

2(k + 1)2 (k - 2) 15(k + 5)(k - 2) × 5(k + 5)3 3(k + 1)3 - ( n + 1) 2

9n -1 (q - 2)(q + 3) (q - 2)2 f ÷ q+5 3q + 15 d

h

6 n2

×

n2

3(s 2 - 4) 2s - 4 ÷ 18s - 27 - (4 s 2 - 9)

linear literal equations


Literal equations are those that are written in terms of pronumerals int-0971 such that no unique numerical solution will be possible, but rather Linear literal equations an expression containing these pronumerals. An equation such as mx - n = p could be described as a linear literal equation in x, as it is linear and contains pronumerals rather than numbers. (Note, that in this case x is defined as the variable.) A solution to a literal equation can be determined algebraically by the use of inverse operations just as a numerical equation can. The difference is that the solution will be a general one — that is, in terms of the pronumerals. In the example above, the solution to this equation will be obtained by isolating x as the subject as follows: mx - n = p (add n to both sides of the equation) mx = p + n (divide by m on both sides of the equation) p+ n x= m Note that literal equations always contain at least one pronumeral (apart from the variable), but they may also contain numerals. Worked exAmple 27

Solve for x. ax a -c= d b

b

m 3n = ( x - a) x

Think a

126

1

WriTe

Add c to both sides.

a

ax -c = d b ax =d+c b


b

ax = b(d + c)

2

Multiply both sides by b.

3

Divide both sides by a.

1

Multiply both sides by x - a.

2

Multiply both sides by x.

3 5

Expand. Collect x terms. Factorise.

mx = 3nx - 3na 3nx - mx = 3na x(3n - m) = 3na

6

Divide both sides by 3n - m.

x=

4

b ( d + c) a m 3n b ( x - a) = x 3n( x - a) m= x mx = 3n(x - a) x=

3na 3n - m

Solving simultaneous literal equations requires the same method as numerical linear equations, namely, substitution or elimination methods. The solutions will be in terms of the pronumerals. Worked exAmple 28

Solve for x and y.

eBook plus Tutorial

ax - by = -4 2ax - 3by = 6

Think


WriTe

1

Assign a number to each equation.

ax - by = -4 2ax - 3by = 6

[1] [2]

2

Multiply equation [1] by 2. Assign a number to the new equation. Subtract equation [2] from equation [3] to eliminate the x term.

2ax - 2by = -8

[3]

3 4

Solve for y.

5

Substitute this value of y into equation [1] and solve for x.

6

State the solution.

by = -14 y=

-

14 b

ax + 14 = -4 ax = -18 18 x= a x=

-

18 14 , y= a b

rememBer

1. Solve linear literal equations by using inverse operations to obtain an equation with the variable as the subject. 2. Solve simultaneous literal equations using elimination or substitution methods. 3. The solutions for literal equations will be in terms of the pronumerals.

Chapter 4

Algebra

127

Exercise

4g

Linear literal equations 1 WE27

Solve for x.

x =c b x+r d = 3t s 1 g ax + b = c 1 m - =m x n 1 1 +e= m x+d f (x + d ) j

p x -

2( x - m) =p n 3k f +l = 4 x+l

2x =y w d e - f = g x

a

b

c

h 2bx - c = 4a

i a(b - x) = b - a

k r ( x - s) =

1 b

l nx - p(x - q) = n(x + p)

bx cx + =n n m

o

c+ x d e + = x 2 x 3x

x y - = c in terms of x is: a b a(c - a) b b ac(c + ay ) e b

c

ac(c - ay ) b

n

dx = x- f e

2 MC The solution to abc - ya b a(cb + y ) d b a

3 WE28 Solve the following simultaneous equations. a ax + by = a2 + b2 b ax + by = a2 - ab + 2b2 2 2 ax - by = a - b ax - by = a2 - ab - 2b2 c

e

x y + =1 a b x y - =3 a b x a + b2 + by = a a x a2 + b2 +y= b ab

g a 2 x + by = ab - 2b 2 y b 2 - 2b bx + = a a

d

x y a+b + = +1 a b a x y a+b + = 2a b 2a y =4 b y (b - a) x + =0 3b

f (a - b) x +

h

x b - by = a a x a-b + ay = b b

n 4 The sum of n terms of an arithmetic sequence is given by the formula S = [2a + ( n - 1)d ], 2 where a is the first number of the sequence and d is the common difference. a Transpose the formula to make a the subject and hence find the first term in a sequence which has n = 26, d = 3 and S = 1079. b Transpose the formula to make d the subject and hence find the common difference of an arithmetic sequence with 20 terms, a = 18 and S = −20.

128


Summary Review of index laws

a0 = 1 a1 = a a- m =

1 am

am × an = am + n am ÷ an =

am = am - n an

(am)n = am × n m

a n = n am = ( n a )

m

(a × b)m = am × bm Standard form and significant figures

• The number of significant figures in a number can be determined by considering each of the following rules: 1. Significant figures are counted from the first non-zero digit (1–9). 2. Any zeros at the end of the number after the decimal point are considered to be significant. 3. The trailing zeros at the end of a number are not considered significant. 4. All zeros between two non-zero digits are always significant. • When performing calculations associated with significant figures, the following rules apply: 1. When adding or subtracting numbers, count the number of decimal places to determine the number of significant figures. The answer cannot contain more places after the decimal point than the least number of decimal places in the numbers being added or subtracted. 2. When multiplying or dividing numbers, count the number of significant figures. The answer cannot contain more significant figures than the number being multiplied or divided with the least number of significant figures. Transposition

• Transposition is the rearrangement of terms within a formula. • The subject of the formula is the variable that is by itself on one side of the equation, while all other variables are on the other side. Solving linear equations and simultaneous linear equations

• Linear equations can be solved by using inverse operations. When solving linear equations, the order of operations process, BODMAS, is reversed. • Simultaneous linear equations can be solved either graphically or algebraically. 1. Graphical method: Draw the straight lines representing the equations and find the coordinates of the point of intersection. 2. Algebraic methods: (a) Substitution: Transpose one of the equations so that one of the unknowns is expressed in terms of the other and substitute into the second equation. (b) Elimination: Equate the coefficients of one unknown and eliminate it by either adding or subtracting the two equations.

Chapter 4 Algebra

129

Applications using linear equations

• To solve worded problems using linear equations, follow these steps: 1. Identify the variables. 2. Set up an equation by transforming the written information into an algebraic statement or statements. 3. Solve the equation. 4. Interpret the result by relating the answer back to the original problem. Applications using simultaneous equations

1. Identify and define the variables. 2. Transform written information into algebraic statements. 3. Solve the pair of equations graphically or algebraically using the methods of substitution or elimination. 4. Interpret the result by relating the answer back to the problem. 5. Always make sure the numbers in the equations are in the same units. Algebraic fractions

• To add or subtract algebraic fractions, first find a common denominator and solve by adding or subtracting the numerator. • If the pronumeral is in the denominator, you can generally find the common denominator by multiplying the denominators together. • When multiplying or dividing, cancel down before multiplying through numerators and denominators. Linear literal equations

• Linear literal equations are equations that have a variable, such as x, with constants and coefficients that are either numbers or pronumerals. • To solve linear literal equations, use the inverse operations to obtain an equation with the variable as the subject in terms of the other pronumerals. • The solution will be in terms of the pronumerals. • Simultaneous linear literal equations can be solved using elimination or substitution methods. The solutions will be in terms of the pronumerals.

130


chapter review Short answer 2 -4 3

1 Simplify 5 x 5 y

× (3 x

-1

3 y 2 )2 .

2 Write the following in standard form and simplify: a 450 000 ÷ 120 000 000 b 0.000012 × 0.34 c 245 × 17 000 3 Calculate the following to the correct number of significant figures. a 3.2418 + 103.27 b 1.0065 × 1 200 4 Transpose each of the following formulas to make the pronumeral, indicated in brackets, the subject. (If two pronumerals are indicated, make a separate transposition for each.) a 6x − 12y + 15 = 0 (x) 7(3 - 4 d ) 8(e + 7) b (d, e) = - 3 5 3 5 Solve for x. 2 x - 3 6 - 3x 7x = 2+ 5 4 3 6 a Find the equations of the two lines shown on the diagram. y 4 2 −1 0

8

x

b Find the coordinates of the point of intersection (the diagram is not drawn to scale). 7 Solve the following simultaneous equations. a 6x + 2y = 12 x − 2y = 2 b 8y − 24 = 4x 7x + 3y = −25 c 15 − 3x − 3y = 30 2x + y = −4 8 Before opening the store, a cashier makes sure that his register contains at least $5 in change. He counts a number of 10c coins, twice as many 5c coins and 4 times as many 20c coins to the total value of exactly $5. How many coins of each type does he count?

9 A building company charges a $2300 set fee plus $500 a day while it is working on a project within the time limits that are specified by a contract. If the project is completed earlier than the set time, the company will still charge $500 for each of the remaining days. However, if the project is not completed by the due date, the company will pay a $135 penalty for each extra day until the work is done. From the given information, construct a set of formulas for the total cost of work, T, the number of days it takes to complete the job according to the contract, n, and the number of extra days, e. 10 Jessica is 3 years older than Rebecca. In 5 years she will be 3 times as old as Rebecca was 2 years ago. Find the girls’ present ages. 11 Simplify the following: 5m - 4 m + 6 1 4 a b 2 3 3x - 4 x 3 2 + c ( x + 1)( x - 2) ( x - 3)( x + 1) 12 Simplify

3( g - 2)( g + 3)2 12( g - 1)2 ( g + 3)5 × . ( g - 2) 4( g - 1)4

25w - 15 w2 - 9 ÷ . 2w - 8 4( w 2 - 16) 14 Solve for x. b - ax a b px - w + x = pk = mn g 3 k 13 Simplify

15 Solve the following simultaneous equations for x and y: y 5b a 2ax - by = 1 + 2b2 b bx - = a 2a Multiple choice 3

 3m 2  1  is equal to:  n4  a

9m 5 n7

b

27m6 n12

D

6m 5 n7

E

9m6 n12

c

3m3 n3

2 The number of significant figures in 20.034 is: a 2 b 3 c 4 D 5 E 6

Chapter 4 Algebra

131

3 The solution to 1303.45 + 23000 with the correct number of significant figures is: a 24303.5 b 24303.50 c 24300 D 24303 E 24000 4 The solution to 25.69 × 2.5040 with the correct number of significant figures is: a 64.3 b 64.33 c 64 D 64.373 E 64.3730 2B + 3 5 If A = is transposed to make B the subject, 4 then: 4A + 3 3 a B = 2A − 2 b B = 2 c B = 2A − 3 E B = 4A +

D B = 4A −

3 2

3 2

Questions 6 and 7 refer to the shape at right.

22 , the perimeter of a certain shape is 7 11x given by P = x + x + x + . When transposed 7 to make x the subject, x is: 7P 32 a b 32 7P 7P c 7P − 14 D 14 7( P - 3) E 11

6 Using π =

7 If the perimeter of the above shape is 8 cm, then x is equal to: a

4 7

D 3

cm

b 42 cm

2 11

E 1.75 cm

cm

c 4 cm

2x 8 The solution to the equation - 5 = - 1 is: 3 a 1 b 2 c 3 D 5 E 6 3x = 6, the following operations 4 could be performed to both sides of the equation: A Add 12; multiply by 4; divide by −3 B Multiply by 4; divide by −3; subtract 12 C Multiply by −4; divide by 3; subtract 12

9 To solve 12 -

132

D Subtract 12; multiply by 4; divide by −3 E Multiply by 4; subtract 12; divide by −3 10 An equation that is the same as 2(3x − 1) = 5x + 3 is: a 6x = 5x + 1 b 11x = 5 c x − 2 = 3 D −2 = 11x + 3 E 11x − 2 = 3 11 The value for x that satisfies the equation 1 - 2x x + = 2 is: 3 4 a 4

b 3

D −4

E −3

c

3 4

12 The value for x that satisfies the equation 6 8 4 - = is: x +1 x x a 12 b 6 c 2 D −2 − E 6 13 The sum of solutions of the pair of simultaneous equations y + x = 12 and 2y − x = 6 is: a 36 b 12 c 20 D 24 E 18 14 If y = 3x − 4 and y = 5 + 4x, then the values of x and y, respectively, are: b 9 and 31 c −31 and −9 a −9 and −31 E 9 and -31 D −9 and 31 15 The point of intersection of the two lines in the graph at right is:

y 4 2 −2 −10

1

a (1, 3) 1

1

b (1 2 , 32) 1

D (1 9 , 3 9 )

5

1

x

1

c (1 5 , 3 5 )

E (2, 3)

16 Which statement below is not true for the pair of simultaneous equations y + x = 22 and 3x − y = 26? A The sum of the numbers is 22. B Three times the first number is 26 larger than the second number. C Three times one number is 26 smaller than the other number. D The difference between 3 times one number and the other is 26. E When one number is subtracted from 22, the other number is obtained.


17 If 3 times a number subtracted from 6 gives 9, then the number is: c 1 a 5 b −1 D

1 3

E 3

18 The perimeter of a regular hexagon is 12.6 cm more than the perimeter of a square with the same side length. The length of the side of a hexagon is: a 2.1 cm b 3.15 cm c 1.26 cm D 12.6 cm E 6.3 cm 19 When half a number is subtracted from 8, the result is the same as adding double that number to 2. The equation that matches this information is: x x a 2 x - 8 = + 2 b 8 - 2 x = 2 + 2 2 x x c 8 - = 2 + 2 x D + 8 = 2 x + 2 2 2 x E - 8 = 2 x + 2 2 20 The sum of two numbers is 42 and their difference is 4. The smaller of the numbers is: a 23 b 17 c 18 D 19 E 24

21 Ben is 1 year short of being twice as old as Esther. If their ages total 20 years, Ben is: a 11 b 12 c 13 D 14 E 15 5( x - 1) ( x + 3) 22 When is expressed as 3 2x ax 2 + bx + c , then a, b and c, respectively, are: 6x b 10, −13 and 9 D 10, −13 and −9

a 10, 13 and 9 c −10, 13 and −9 E 10, 13 and −9 23 Simplifying a

3- e 2

9 - e 2 2(e + 3) gives: ÷ e2 e3 3- e e(3 - e) b c 2e 2

D

3+ e 2

E

3+ e 2e mx 24 The solution to x = x - 3 p in terms of x is: 2 a

6 m

b

D

-6p

m

6p m 6 E p

c

-6

m

Extended response

1 Adrian has begun a new job as a car salesperson. His fortnightly wage is calculated in two parts: a set amount of $600 plus 2% of sales made each fortnight. a Write the rule describing Adrian’s fortnightly wage. b How much can Adrian expect to earn if his sales in a particular fortnight are: i $20 000 ii $65 000 iii $100 000? c How much must Adrian make in sales to obtain a fortnightly wage of: i $1300 ii $1800 iii $2400? Brett, also a salesperson in the motor vehicle industry, is paid a fortnightly salary of $860 regardless of sales made. d Compare Adrian’s fortnightly wage to Brett’s fortnightly salary. e Write the rule describing Brett’s fortnightly salary. f How much would Adrian have to make in sales in one fortnight to obtain the same amount as Brett earns? 2 Joseph has $15 000 to invest. He does not want to ‘keep all of his eggs in the one basket’, so he decides to split the money in the following ways. He puts some of his money in the bank, which offers an interest rate of 6% p.a., and the remainder into a building society, which offers an interest rate of 11% p.a. If Joseph plans to take a trip to Queensland, costing $1200, and he wants to pay for the trip using only the interest earned from his investments after 1 year, how must he split his $15 000?

Chapter 4 Algebra

133

3 Michael wishes to rent a car for a long weekend. The cost, C, of renting a Toyota Corolla from company A is given by C = 25 + 0.08n, and the cost of renting from company B C = 40 + 0.05n, where n is the number of kilometres travelled. Cost ($)

C

[1] 40 25 0

a b c d e f g h i j

[2]

Number of km

n

Which company, A or B, does line [1] represent? What could the numbers 25 and 40 represent? What does the point of intersection of lines [1] and [2] represent? Find the coordinates of the point of intersection. If Michael decides to travel along the Great Ocean Road, which is about 350 km each way, from which company, A or B, should he rent so that he pays less? Next long weekend, Michael is planning to go to Phillip Island, which is about 150 km each way. From which company should he rent this time? Explain to Michael how he can decide from which company to rent, if he knows the approximate distance he intends to travel, without doing any calculations. Write the formula for d, the difference between the cost of renting the car from the two companies (A or B). Write the difference equation which corresponds to the equation in part h. Use the difference equation to generate a table of values for distances from 0 to 1000 km inclusive, with increments of 100 km. Hence, find the distance for which the cost of renting from company A will exceed the cost of renting from company B by more than $10.

4 Novak Novelties manufactures a variety of children’s 3-D puzzles. The director of the company has asked his assistants Caitlin, Bridget and Emese to prepare a report on production costs, expenses and returns on the puzzles. Each puzzle costs the company $15 to produce. In addition, the company has monthly overheads of $21 000. The selling price of each puzzle is $45. a Write an equation describing the expenses; that is, the total cost, C, of producing n puzzles each month. b Write an equation describing the selling price of n puzzles. c Plot and label the graph of the equation obtained in part a. Does it commence at the origin? Explain. d Plot and label the graph of the equation obtained in part b on the same axis. Does it commence at the origin? Explain. e The point of intersection of the two lines on your graph is called the break-even point. Explain what this means in terms of the given problem. f Find the coordinates of the break-even point (point of intersection). g Shade the portion between the two lines to the left of the break-even point. Explain what this portion represents. h Shade the portion between the two lines to the right of the break-even point. Explain what this portion represents. Profit may be defined as the selling price minus the total cost. i Write an equation describing the profit obtained, P, after selling n puzzles. j Determine whether a profit or loss is made when: i 400 ii 600 iii 800 iv 1000 puzzles are sold in a particular month. eBook plus Digital doc


134


eBook plus

ACTiviTies


• 10 Quick Questions: Warm up with ten quick questions on algebra. (page 88) 4a

Review of index laws

Tutorial

• We3 int-1039: Watch how to simplify expressions involving indices. (page 90) 4b

Standard form and significant figures

Tutorial

• We5 int-1040: Watch how to simplify an expression involving the product and quotient of numbers in standard form. (page 92) 4c

Transposition

Tutorial

• We9 int-1041: Watch how to solve equations for specific pronumerals. (page 97) Digital doc

• WorkSHEET 4.1: Transpose simple equations, and use transposition to solve worded problems. (page 101) 4d

Solving linear equations and simultaneous linear equations

Tutorial

• We14 int-1042: Watch how to solve equations involving fractions. (page 105) Digital doc

• Investigation: Comparing production costs. (page 114) 4e

applications

4F

algebraic fractions

Tutorial

• We25 int-1043: Watch how to simplify algebraic expressions involving fractions. (page 123) 4G

Linear literal equations

Interactivity

• Linear literal equations int-0971: Consolidate your understanding of how to solve linear literal equations. (page 126) Tutorial

• We28 int-1044: Watch how to solve literal simultaneous equations. (page 127) chapter review Digital doc


Digital doc

• WorkSHEET 4.2: Transpose equations with algebraic fractions and apply this method to more complex worded problems. (page 122)

Chapter 4

Algebra

135

5 Trigonometric ratios and their applications areas oF sTudy

• Right-angled triangles and solutions to problems involving right-angled triangles using sine, cosine and tangent • The relationships sin2 (θ ) + cos2 (θ ) = 1, cos (θ ) = sin (90° − θ ) and sin (θ ) = cos (90° − θ) • Two-dimensional applications including angles of depression and elevation • Exact values of sine, cosine and tangent for 30°, 45° and 60°

5A 5B 5C 5D 5E 5F 5G 5H

Trigonometry of right-angled triangles Elevation, depression and bearings The sine rule The cosine rule Area of triangles Trigonometric identities Radian measurement Arcs, sectors and segments

• Solution of triangles by the sine and cosine rules • Areas of triangles, including the formula A = s ( s − a ) ( s − b ) ( s − c) • Circle mensuration: radian measure, arc length, areas of sectors and segments • Applications, for example, navigation and surveying in simple contexts eBook plus Digital doc

5a

10 Quick Questions

Trigonometry of right-angled triangles

Trigonometry, derived from the Greek words trigon (triangle) and metron (measurement), is the branch of mathematics that deals with the relationship between the sides and angles of a triangle. It involves finding unknown angles, side lengths and areas of triangles. The principles of trigonometry are used in many practical situations such as building, surveying, navigation and engineering. In previous years you will have studied the trigonometry of right-angled triangles. We will review this material before considering non–right-angled triangles. sin (θ ) = cos (θ ) =

opposite side O hypotenuse which is abbreviated to sin (θ ) = H adjacent side A hypotenuse which is abbreviated to cos (θ ) = H

opposite side O tan (θ ) = adjacent side which is abbreviated to tan (θ ) = A

B Opposite (O) C

Hypotenuse (H )

θ (A) Adjacent

A

The symbol θ (theta) is one of the many letters of the Greek alphabet used to represent the angle. Other symbols include α (alpha), β (beta) and γ (gamma). Non-Greek letters may also be used.

136


Writing the mnemonic SOH–CAH–TOA each time we perform trigonometric calculations will help us to remember the ratios and solve the problem.

Pythagoras’ theorem For specific problems it may be necessary to determine the side lengths of a right-angled triangle before calculating the trigonometric ratios. In this situation, Pythagoras’ theorem is used. Pythagoras’ theorem states: In any right-angled triangle, c2 = a2 + b2.

c

a b

Worked Example 1

Determine the value of the pronumerals, correct to 2 decimal places.

a

b 4

x

7

50° Think a

1

Write

Label the sides, relative to the marked angles.

a

4

x O

b

24° 25′ h

H

50°

2

Write what is given.

Have: angle and hypotenuse (H)

3

Write what is needed.

Need: opposite (O) side

4

Determine which of the trigonometric ratios is required, using SOH–CAH–TOA.

5

Substitute the given values into the appropriate ratio.

6

Transpose the equation and solve for x.

7

Round the answer to 2 decimal places.

1

Label the sides, relative to the marked angle.

sin (θ ) =

O H

sin (50°) =

x 4

4 × sin (50°) = x x = 4 × sin (50°) = 3.06 b

A 7 24° 25′ h H

2


Have: angle and adjacent (A) side

3


Need: hypotenuse (H)

4


cos (θ ) =

A H

5


cos (24°25′ ) =

7 h

Chapter 5 Trigonometric ratios and their applications

137

6

Solve for h. On the Main screen, complete the entry line as: 7   solve  cos(dms(24, 25) = , h  h  Then press E.

7


= 7.69

Worked Example 2

Find the angle θ, giving the answer in degrees and minutes. θ 12 18 Think 1

Write

Label the sides, relative to the marked angles. θ

A 12

O 18 2


3


4


tan (θ ) = O

5


tan (θ ° ) =

6

To calculate tan−1, on the Main screen, complete the entry lines as:

Have: opposite (O) and adjacent (A) sides Need: angle

A 18 12

 −1  18    tan  12   toDMS( Press E after each entry line. Note: toDMS can be located by tapping: • Action • Transformation • toDMS

138


7

q ° = tan

Write the answer to the nearest minute.

−1

= 56° 19′

 18   12 

Exact values Most of the trigonometric values that we will deal with in this chapter are only approximations. However, angles of 30°, 45° and 60° have exact values of sine, cosine and tangent. Consider an equilateral triangle, ABC, of side length 2 units. If the triangle is perpendicularly bisected, then two congruent triangles, ABD and CBD, are obtained. From triangle ABD it can be seen that BD creates a right-angled triangle with angles of 60° and 30° and base length (AD) of 1 unit. The length of BD is obtained using Pythagoras’ theorem. Using triangle ABD and the three trigonometric ratios the following exact values are obtained:

1 sin (30°) = 2 3 cos (30°) = 2 tan (30°) =

1 3

B

2

cos (45°) =

1

3

2

60° A

C

D 2

3 2 1 cos (60°) = 2 sin (60°) =

or

3 3

tan (60°) =

3 or 3 1

Consider a right-angled isosceles triangle EFG whose equal sides are of 1 unit. The hypotenuse EG is obtained by using Pythagoras’ theorem. (EG)2 = (EF)2 + (FG)2 = 12 + 12 =2 EG = 2 Using triangle EFG and the three trigonometric ratios, the following exact values are obtained: 1 2 or sin (45°) = 2 2

30°

or

G

E

2

1

45° 1

F

2

2 2 1 tan (45°) = or 1 1

Worked Example 3

Determine the height of the triangle shown in surd form. h 60° 8 cm


139

Think 1

Write

Label the sides relative to the marked angle. h O 60° A 8 cm

2


Have: angle and adjacent (A) side

3


4


Need: opposite (O) side O tan (θ ) = A

5


6

tan (60°) =

h 8

Substitute exact values where appropriate.

3=

h 8

7

Transpose the equation to find the required value.

h=8 3

8

State the answer.

The triangle’s height is 8 3 cm.

REMEMBER B

1. For any right-angled triangle:

O A O Opposite cos (θ ) = tan (θ ) = (O) H H H 2. To determine which trigonometric ratio to use when C solving a right-angled triangle, follow these steps: (a) Label the diagram using the symbols θ, O, A, H. (b) Write what is given. (c) Write what is needed. (d) Determine which of the trigonometric ratios is required, using SOH–CAH–TOA. a (e) Substitute the given values into the rule and solve. 3. Pythagoras’ theorem, c2 = a2 + b2, may also be used to solve right-angled triangles. 4. Angles of 30°, 45° and 60° have exact values for sine, cosine and tangent. sin (θ ) =

θ

45°

sin (θ )

1 2

1

cos (θ )

3 2

1

tan (θ )

140

30°

1 3

=

2 2 3 3

60°

=

2 2

3 2

=

2 2

1 2

1

3


Hypotenuse (H )

θ (A) Adjacent

c

b

A

exerCise

5a

Trigonometry of right-angled triangles 1

Copy and label the sides of the following right-angled triangles using the words

eBook plus

hypotenuse, adjacent, opposite and the symbol θ.

Digital doc

a

b

θ

SkillSHEET 5.1

c

Adjacent

d

Opposite

θ

Labelling rightangled triangles

eBook plus

2 We 1 Find the value of the pronumerals, correct to 2 decimal places. a

b

10

Digital doc

SkillSHEET 5.2

x

c

7.5

x

d

47°8' 17

40°

Using trigonometric ratios

684

x

32°14'

62°38' x

eBook plus

e

f

1.03

g 78°

Digital doc

x

SkillSHEET 5.3

504

3.85

14°25'

Degrees and minutes

x

h

27°47'

x

17 x

y 38°48'

3 We 2 Find the angle θ, giving the answer in degrees and minutes. a b c 5 θ

10

θ

28

7

20

12

θ

d

e θ

2.1

f

4.2

30

11.7

48 θ

θ

6.8

g

h 53.2

θ

θ

1.74

3.26

78.1

Chapter 5

Trigonometric ratios and their applications

141

4

Digital doc Composite shapes 1

An isosceles triangle has a base of 12 cm and equal angles

of 30°. Find, in surd form: 30° 30° a the height of the triangle 12 cm b the area of the triangle c the perimeter of the triangle, giving your answers in simplest surd form.

eBook plus SkillSHEET 5.4

We 3

5

14

Find the perimeter of the composite shape at right, in surd form. The length measurements are in metres. 26

60°

6 A ladder 6.5 m long rests against a vertical wall and makes an angle of 50° to the horizontal ground. a How high up the wall does the ladder reach? b If the ladder needs to reach 1 m higher, to the nearest minute, what angle should it make to the ground? 7 A 400-m-long road goes straight up a slope. If the road rises 50 m vertically, what is the angle that the road makes with the horizontal? 8 An ice-cream cone has a diameter of 6 cm and a sloping edge of 15 cm. Find the angle at the bottom of the cone. 9 A vertical flagpole is supported by a wire attached from the top of the pole to the horizontal ground, 4 m from the base of the pole. Joanne measures the angle the wire makes with the ground and finds this is 65°. How tall is the flagpole? 10 A stepladder stands on a floor, with its feet 1.5 m apart. If the angle formed by the legs is 55°, how high above the floor is the top of the ladder? eBook plus Digital doc

SkillSHEET 5.5 Composite shapes 2

11

The angle formed by the diagonal of a rectangle and one of its shorter sides is 60°. If the diagonal is 8 cm long, find the dimensions of the rectangle, in surd form.

12 In the figure at right, find the value of the pronumerals, correct to 2 decimal places.

d a

7 b

50°

13 In the figure at right, find the value of the pronumerals, correct to 2 decimal places.

30° c

48° b a 14

14 In the figure at right, find the value of the pronumeral x, correct to 2 decimal places.

33°

15 An advertising balloon is attached to a rope 120 m long. The rope makes an angle of 75° to level ground. How high above the ground is the balloon?

58° 6 x

16 An isosceles triangle has sides of 17 cm, 20 cm and 20 cm. Find the magnitude of the angles.

142


70°

17 A garden bed at right is in the shape of a trapezium. What volume of garden mulch is needed to cover it to a depth of 15 cm?

12 m 120° 4m

18 A gable roof has sloping sides of 8.3 m. It rises to a height of 2.7 m at the centre. a What is the angle of slope of the two sides? b How wide is the roof at its base?

8.3 m

8.3 m 2.7 m

19 A ladder 10 m long rests against a vertical wall at an angle of 55° to the horizontal. It slides down the wall, so that it now makes an angle of 48° with the horizontal. a Through what vertical distance did the top of the ladder slide? b Does the foot of the ladder move through the same distance? Justify your answer.

Elevation, depression and bearings Trigonometry is especially useful for measuring distances and heights which are difficult or impractical to access. For example, two important applications of right-angled triangles involve: 1. angles of elevation and depression, and 2. bearings.

Angles of elevation and depression

ne

of

sig

ht

Angles of elevation and depression are employed when dealing with directions which require us to look up and down respectively. An angle of elevation is the angle between the horizontal and an object which is higher than the observer (for example, the top of a mountain or flagpole). Li

5B

θ Angle of elevation

An angle of depression is the angle between the horizontal and an object which is lower than the observer (for example, a boat at sea when the observer is on a cliff).

Angle of depression

t

ne

Li

of

h sig

θ

Unless otherwise stated, the angle of elevation or depression is measured and drawn from the horizontal. Angles of elevation and depression are each measured from the horizontal. When solving problems involving angles of elevation and depression, it is best always to draw a diagram. The angle of elevation is equal to the angle of depression since they are alternate ‘Z’ angles.

D

E D and E are alternate angles ∴∠D=∠E


143

Worked Example 4

From a cliff 50 metres high, the angle of depression of a boat at sea is 12°. How far is the boat from the base of the cliff? Think 1

Write

Draw a diagram and label all the given information. Include the unknown length, x, and the angle of elevation, 12°.

12° 50 m 12°

x

2


Have: angle and opposite side

3


4

Determine which of the trigonometric ratios is required (SOH–CAH–TOA).

Need: adjacent side O tan (θ ) = A

5


6


tan (12° ) =

x × tan (12° ) = 50 x=

7


8


50 x 50 tan (12)

= 235.23 The boat is 235.23 m away from the base of the cliff. N

Bearings Bearings measure the direction of one object from another. There are two systems used for describing bearings. True bearings are measured in a clockwise direction, starting from north (0° T).

150° T

Compass bearing equivalent is S30°E N

Conventional or compass bearings are measured: first, relative to north or south, and second, relative to east or west.

20° W

E

20°

W

E

S S N20°W S70°E True bearing equivalent True bearing equivalent is 340° T is 110° T

The two systems are interchangeable. For example, a bearing of 240° T is the same as S60°W. When solving questions involving direction, always start with a diagram showing the basic compass points: north, south, east and west.

144

N


N

N W

240° T S

E

W

E 60° S S60°W

Worked examPle 5

A ship sails 40 km in a direction of N52°W. How far west of the starting point is it? Think 1

WriTe/draW

Draw a diagram of the situation, labelling each of the compass points and the given information.

N x

40

km 52°

W

E S

2

Write what is given for the triangle.

Have: angle and hypotenuse

3

Write what is needed for the triangle.

4


5


Need: opposite side O sin (θ ) = H x sin (52° ) = 40

6


7


8


40 × sin (52° ) = x x = 40 × sin (52° ) = 31.52 The ship is 31.52 km west of the starting point.

Worked examPle 6

eBook plus

A ship sails 10 km east, then 4 km south. What is its bearing from its starting point? Think 1

Tutorial


WriTe

Draw a diagram of the situation, labelling each of the compass points and the given information.

N 10 km

θ

4 km

S 2

Write what is given for the triangle.

Have: adjacent and opposite sides

3

Write what is needed for the triangle.

4


Need: angle O tan (θ ) = A

5


6

Transpose the equation and solve for θ, using the inverse tan function.

7

Convert the angle to degrees and minutes.

tan (θ ) =

4 10

−  4  θ = tan 1    10 

= 21.801 409 49° = 21°48′

Chapter 5


145

8

9

Express the angle in bearings form. The bearing of the ship was initially 0° T; it has since rotated through an angle of 90° and an additional angle of 21°48′. To obtain the final bearing these values are added. Answer the question.

earing = 90° + 21°48′ B = 111°48′ T

The bearing of the ship from its starting point is 111°48′ T.

REMEMBER

1. Angles of elevation and depression are each measured from the horizontal. 2. The angle of elevation is equal to the angle of depression since they are alternate ‘Z’ angles. 3. True bearings are measured in a clockwise direction, starting from north (0° T). 4. Conventional or compass bearings are measured first, relative to north or south, and second, relative to east or west. 5. Whenever solving problems involving angles of elevation and depression or bearings, you should always draw a diagram and label all the given information. 6. Set up a compass as the basis of your diagram for bearings questions. Exercise

5B

Elevation, depression and bearings 1 WE 4 From a vertical fire tower 60 m high, the angle of depression to a fire is 6°. How far away, to the nearest metre, is the fire? 2 A person stands 20 m from the base of a building, and measures the angle of elevation to the top of the building as 55°. If the person is 1.7 m tall, how high, to the nearest metre, is the building? 3 An observer on a cliff top 57 m high observes a ship at sea. The angle of depression to the ship is 15°. The ship sails towards the cliff, and the angle of depression is then 25°. How far, to the nearest metre, did the ship sail between sightings? 4 Two vertical buildings, 40 m and 62 m high, are directly opposite each other across a river. The angle of elevation of the top of the taller building from the top of the smaller building is 27°. How wide is the river? (Give the answer to 2 decimal places.) 5 To calculate the height of a crane which is on top of a building, Denis measures the angle of elevation to the bottom and top of the crane. These were 62° and 68° respectively. If the building is 42 m high find, to 2 decimal places: a how far Denis is from the building b the height of the crane. 6 A new skyscraper is proposed for the Melbourne Docklands region. It is to be 500 m tall. What would be the angle of depression, in degrees and minutes, from the top of the building to the island on Albert Park Lake, which is 4.2 km away? 7 From a rescue helicopter 2500 m above the ocean, the angles of depression of two shipwreck survivors are 48° (survivor 1) and 35° (survivor 2). a Draw a labelled diagram which represents the situation. b Calculate how far apart the two survivors are. 8 A lookout tower has been erected on top of a mountain. At a distance of 5.8 km, the angle of elevation from the ground to the base of the tower is 15.7° and the angle of elevation to the observation deck (on the top of the tower) is 15.9°. How high, to the nearest metre, is the observation deck above the top of the mountain?

146


9

From a point A on level ground, the angle of elevation of the top of a building 50 m high is 45°. From a point B on the ground and in line with A and the foot of the building, the angle of elevation of the top of the building is 60°. Find, in simplest surd form, the distance from A to B.

10 Express the following conventional bearings as true bearings. a N35°W b S47°W c N58°E

d S17°E

11 Express the following true bearings in conventional form. a 246° T b 107° T c 321° T

d 074° T

12 mC a A bearing of S30°E is the same as: A 030° T B 120° T C 150° T b A bearing of 280° T is the same as: A N10°W B S10°W C S80°W

D 210° T

E 240° T

D N80°W

E N10°E

13 We 5 A pair of canoeists paddle 1800 m on a bearing of N20°E. How far north of their starting point are they, to the nearest metre? 14 A yacht race consists of four legs. The first three legs are 4 km due east, then 5 km south, followed by 2 km due west. a How long is the final leg, if the race finishes at the starting point? b On what bearing must the final leg be sailed? 15 We 6 A ship sails 20 km south, then 8 km west. What is its bearing from the starting point? 16 A cross-country competitor runs on a bearing of N60°W for 2 km, then due north for 3 km. a How far is he from the starting point? b What is the true bearing of the starting point from the runner? 17 Two hikers set out from the same campsite. One walks 7 km in the direction 043° T and the other walks 10 km in the direction 133° T. a What is the distance between the two hikers? b What is the bearing of the first hiker from the second? 18 A ship sails 30 km on a bearing of 220°, then 20 km on a bearing of 250°. Find: a how far south of the original position it is b how far west of the original position it is c the true bearing of the ship from its original position, to the nearest degree. 19 The town of Bracknaw is due west of Arley. Chris, in an ultralight plane, starts at a third town, Champton, which is due north of Bracknaw, and flies directly towards Arley at a speed of 40 km/h in a direction of 110° T. She reaches Arley in 3 hours. Find: a the distance between Arley and Bracknaw b the time to complete the journey from Champton to Bracknaw, via Arley, if she increases her speed to 45 km/h between Arley and Bracknaw. 20 From a point, A, on the ground, the angle of elevation of the top of a vertical tower due north of A is 46°. From a point B, due east of A, the angle of elevation of the top of the tower is 32°. If the tower is 85 m high, find: a the distance from A to the foot of the tower b the distance from B to the foot of the tower c the true bearing of the tower from B.

Chapter 5


147

21 A bird flying at 50 m above the ground was observed at noon from my front door at an angle of elevation of 5°. Two minutes later its angle of elevation was 4°. a If the bird was flying straight and level, find the horizontal distance of the bird: i from my doorway at noon ii from my doorway at 12.02 pm. b Hence, find: i the distance travelled by the bird in the two minutes ii its speed of flight in km/h.

5C

The sine rule When working with non–right-angled triangles, it is usual B to label the angles A, B and C, and the sides a, b and c, so that side a c a is the side opposite angle A, side b is the side opposite angle B and side c is the side opposite angle C. A C b In a non–right-angled triangle, a perpendicular line, h, can be drawn from the angle B to side b. B h Using triangle ABD we obtain sin (A) = . Using triangle CBD c h we obtain sin (C) = . c a h a Transposing each equation to make h the subject, we A C obtain: h = c × sin (A) and h = a × sin (C). Equate to get D b c × sin (A) = a × sin (C). h– h– c = sin (A) and a = sin (C) Transpose to get c a = sin (C ) sin ( A) In a similar way, if a perpendicular line is drawn from angle A to side a, we get b c = sin (B ) sin (C ) From this, the sine rule can be stated. In any triangle ABC:

B c

a b c = = sin ( A) sin ( B) sin (C )

A

a b

Notes 1. When using this rule, depending on the values given, any combination of the two equalities may be used to solve a particular triangle. 2. To solve a triangle means to find all unknown side lengths and angles. The sine rule can be used to solve non–right-angled triangles if we are given: 1. two angles and one side length 2. two side lengths and an angle opposite one of these side lengths. Worked Example 7

In the triangle ABC, a = 4 m, b = 7 m and B = 80°. Find A, C and c. Think 1

Write B

Draw a labelled diagram of the triangle ABC and fill in the given information.

c A

148

b=7


80° a = 4 C

C

2

Check that one of the criteria for the sine rule has been satisfied.

The sine rule can be used since two side lengths and an angle opposite one of these side lengths have been given.

3

Write the sine rule to find A.

To find angle A: a b = sin ( A) sin (B ) 4 7 = sin ( A) sin (80)

4

Substitute the known values into the rule.

5

Transpose the equation to make sin (A) the subject.

4 × sin (80°) = 7 × sin (A) 4 × sin (80) = sin ( A) 7 4 × sin (80) sin ( A) = 7

6

Evaluate.

 −  4 × sin (80 )  A = sin 1   7 −1 = sin (0.562 747 287) = 34.246 004 71°

8

Round the answer to degrees and minutes. Determine the value of angle C using the fact that the angle sum of any triangle is 180°.

= 34°15′ C = 180° − (80° + 34°15′) = 65°45′

9

Write the sine rule to find c.

To find side length c: c b = sin (C ) sin (B )

10


c 7 = sin (65  45′ ) sin (80 )

11

Transpose the equation to make c the subject.

12

Evaluate. Round the answer to 2 decimal places and include the appropriate unit.

7

c=

7 × sin (65  45′ ) sin (80 )

7 × 0.911762 043 0.984 807 753 6.382 334 305 = 0.984 807 753 = 6.480 792 099 = 6.48 m =

The ambiguous case When using the sine rule there is one important issue to consider. If we are given two side lengths and an angle opposite one of these side lengths, then two different triangles may be drawn. For example, if a = 10, c = 6 and C = 30°, two possible triangles could be created. B

B c=6 A

a = 10 30°

a = 10

c=6 C

A

30°

C


149

In the first case, angle A is an acute angle, while in the second case, angle A is an obtuse angle. The two values for A will add to 180°. The ambiguous case does not work for each example. It would be useful to know, before commencing a question, whether or not the ambiguous case exists and, if so, to then find both sets of solutions. The ambiguous case exists if C is an acute angle and a > c > a × sin (C), or any equivalent statement; for example, if B is an acute angle and a > b > a × sin (B), and so on.

Worked examPle 8

eBook plus

In the triangle ABC, a = 10 m, c = 6 m and C = 30°. a Show that the ambiguous case exists. b Find two possible values of A, and hence two possible values of B and b. Think

Tutorial


WriTe

Method 1: Using the rules a

1

2

Check that the conditions for an ambiguous case exist, i.e. that C is an acute angle and that a > c > a × sin (C).

a C = 30° so C is an acute angle.

sin (C) = sin (30°) = 0.5 a > c > a × sin (C) 10 > 6 > 10 sin (30°) 10 > 6 > 5 This is correct. This is an ambiguous case of the sine rule.

State the answer.

Case 1 b

1


b

B a = 10

c=6

30°

A

150

2


3


4


5

Evaluate angle A, in degrees and minutes.

6

Determine the value of angle B, using the fact that the angle sum of any triangle is 180°.

7

Write the sine rule to find b.

C

To find angle A: a c = sin ( A) sin (C ) 10 6 = sin ( A) sin (30) 10 × sin (30°) = 6 × sin (A) 10 × sin (30) = sin ( A) 6 10 × sin (30) sin ( A) = 6  −  10 × sin (30 )  A = sin 1    6 A = 56°27′ B = 180° − (30° + 56°27′) = 93°33′ To find side length b: b c = sin (B) sin (C )


8


9

Transpose the equation to make b the subject and evaluate.

b 6 =  sin (93 33′) sin (30 ) 6 × sin (93 33′) sin (30 ) = 11.98 m

b=

Case 2 b

1


b

B a = 10

c=6 A

30°

C

2

Write the alternative value for angle A. Subtract the value obtained for A in Case 1 from 180°.

To find the alternative angle A: If sin (A) = 0.8333, then A could also be: A = 180° − 56°27′ = 123°33′

3

Determine the alternative value of angle B, using the fact that the angle sum of any triangle is 180°.

B = 180° − (30° + 123°33′) = 26°27′

4

Write the sine rule to find the alternative b.

To find side length b: b c = sin (B) sin (C )

5


b 6 =  sin (26 27 ′) sin (30 )

6

Transpose the equation to make b the subject and evaluate.

b=

6 × sin (26  27 ′ ) sin (30 ) = 5.35 m



B c=6 A

2

In part a it was shown that the ambiguous case of the sine rule exists. Therefore, on the Main screen, complete the entry line as: 6  10  = , a | 0 ≤ a ≤ 180 solve   sin(a) sin(30)  Then press E.

3

Convert the angles to degrees and minutes.

a = 10 30°

C

A = 56°27′ or A = 123°33′


151

If A = 56°27′, B = 180 - (30 + 56°27′) = 93°33′ If A = 123°33′, B = 180 - (30 + 123°33′) = 26°27′

4

Calculate the size of the angle B given each angle A.

5

To find the side length b, on the Main screen, complete the entry line as: b 6  solve   sin(dms(93, 33)) = sin(30) , b b 6  solve   sin(dms(26, 27)) = sin(30) , b Press E after each entry.

6

If B = 93°33′, b = 11.98 m If B = 26°27′, b = 5.35 m

Write the answers.

Hence, for this example there were two possible solutions as shown by the diagram below. B

B a = 10

c=6 A

30°

c=6 C

A

a = 10 30°

C

REMEMBER

1. The sine rule states that for any triangle ABC: a b c = = sin ( A) sin ( B) sin (C ) 2. When using this rule it is important to note that, depending on the values given, any combination of the two equalities may be used to solve a particular triangle. 3. The sine rule can be used to solve non–right-angled triangles if we are given: (a) two angles and one side length (b) two side lengths and an angle opposite one of these side lengths. 4. The ambiguous case exists if C is an acute angle and a > c > a × sin (C). Exercise

5C

The sine rule 1 WE 7 In the triangle ABC, a = 10, b = 12 and B = 58°. Find A, C and c. 2 In the triangle ABC, c = 17.35, a = 26.82 and A = 101°47′. Find C, B and b. 3 In the triangle ABC, a = 5, A = 30° and B = 80°. Find C, b and c. 4 In the triangle ABC, c = 27, C = 42° and A = 105°. Find B, a and b.

152


5 In the triangle ABC, a = 7, c = 5 and A = 68°. Find the perimeter of the triangle. 6 Find all unknown sides and angles for the triangle ABC, given A = 57°, B = 72° and a = 48.2. 7 Find all unknown sides and angles for the triangle ABC, given a = 105, B = 105° and C = 15°. 8 Find all unknown sides and angles for the triangle ABC, given a = 32, b = 51 and A = 28°. 9 Find the perimeter of the triangle ABC if a = 7.8, b = 6.2 and A = 50°. 10 MC In a triangle ABC, A = 40°, C = 80° and c = 3. The value of b is: A 2.64 B 2.86 C 14 D 4.38 E 4.60 11 WE 8 In the triangle ABC, a = 10, c = 8 and C = 50°. Find two possible values of A, and hence two possible values of b. 12 In the triangle ABC, a = 20, b = 12 and B = 35°. Find two possible values for the perimeter of the triangle. 13 Find all unknown sides and angles for the triangle ABC, given A = 27°, B = 43° and c = 6.4. 14 Find all unknown sides and angles for the triangle ABC, given A = 100°, b = 2.1 and C = 42°. 15 Find all unknown sides and angles for the triangle ABC, given A = 25°, b = 17 and a = 13. 16 To calculate the height of a building, Kevin measures the angle of elevation to the top as 48°. He then walks 18 m closer to the building and measures the angle of elevation as 64°. How high is the building? 17 A river has parallel banks which run directly east–west. Kylie takes a bearing to a tree on the opposite side. The bearing is 047° T. She then walks 10 m due east, and takes a second bearing to the tree. This is 305° T. Find: a her distance from the second measuring point to the tree b the width of the river, to the nearest metre. 18 A ship sails on a bearing of S20°W for 14 km, then changes direction and sails for 20 km and drops anchor. Its bearing from the starting point is now N65°W. a How far is it from the starting point? b On what bearing did it sail the 20 km leg? 19 A cross-country runner runs at 8 km/h on a bearing of 150° T for 45 mins, then changes direction to a bearing of 053° T and runs for 80 mins until he is due east of the starting point. a How far was the second part of the run? b What was his speed for this section? c How far does he need to run to get back to the starting point? 20 From a fire tower, A, a fire is spotted on a bearing of N42°E. From a second tower, B, the fire is on a bearing of N12°W. The two fire towers are 23 km apart, and A is N63°W of B. How far is the fire from each tower? 21 MC A boat sails on a bearing of N15°E for 10 km, then on a bearing of S85°E until it is due east of the starting point. The distance from the starting point to the nearest kilometre is, then: A 10 km B 38 km C 110 km D 113 km E 114 km 22 MC A hill slopes at an angle of 30° to the horizontal. A tree which is 8 m tall is growing at an angle of 10° to the vertical and is part-way up the slope. The vertical height of the top of the tree above the slope is: A 7.37 m B 8.68 m C 10.84 m D 15.04 m E 39.89 m


153

23 A cliff is 37 m high. The rock slopes outward at an angle of 50° to the horizontal, then cuts back at an angle of 25° to the vertical, meeting the ground directly below the top of the cliff. Carol wishes to abseil from the top of the cliff to the ground as shown in the diagram. Her climbing rope is 45 m long, and she needs 2 m to secure it to a tree at the top of the cliff. Will the rope be long enough to allow her to reach the ground?

50° 25° rope

rock 37 m


WorkSHEET 5.1

5d

The cosine rule In any non–right-angled triangle ABC, a perpendicular line can be drawn from angle B to side b. Let D be the point where the perpendicular line meets side b, and the length of the perpendicular line be h. Let the length AD = x units. The perpendicular line creates two right-angled triangles, ADB and CDB. B Using triangle ADB and Pythagoras’ theorem, we obtain: c2 = h2 + x2 [1] a c h Using triangle CDB and Pythagoras’ theorem, we obtain: a2 = h2 + (b – x)2 [2] A C D Expanding the brackets in equation [2]: b−x x b a2 = h2 + b2 – 2bx + x2 Rearranging equation [2] and using c2 = h2 + x2 from equation [1]: a2 = h2 + x2 + b2 – 2bx = c2 + b2 – 2bx = b2 + c2 – 2bx From triangle ABD, x = c × cos (A), therefore a2 = b2 + c2 − 2bx becomes a2 = b2 + c2 – 2bc × cos (A) This is called the cosine rule and is a generalisation of Pythagoras’ theorem. In a similar way, if the perpendicular line was drawn from angle A to side a or from angle C to side c, the two right-angled triangles would give c2 = a2 + b2 – 2ab × cos (C) and b2 = a2 + c2 – 2ac × cos (B) respectively. From this, the cosine rule can be stated: In any triangle ABC a2 = b2 + c2 − 2bc cos (A) b2 = a2 + c2 − 2ac cos (B) c2 = a2 + b2 − 2ab cos (C)

154


B c A

a b

C

The cosine rule can be used to solve non–right-angled triangles if we are given: 1. three sides of the triangle 2. two sides of the triangle and the included angle (the angle between the given sides). Worked Example 9

Find the third side of triangle ABC given a = 6, c = 10 and B = 76°, correct to 2 decimal places. Think

Write



B c = 10 A

76°

a=6 C

b

2

Check that one of the criteria for the cosine rule has been satisfied.

Yes, the cosine rule can be used since two side lengths and the included angle have been given.

3

Write the appropriate cosine rule to find side b.

To find side b: b2 = a2 + c2 − 2ac cos (B)

4

Substitute the given values into the rule.

5

Evaluate.

6


= 62 + 102 − 2 × 6 × 10 × cos (76°) = 36 + 100 − 120 × 0.241 921 895 = 106.969 372 5 b = 106.969 372 5 = 10.34 correct to 2 decimal places



B c = 10 A

2

Write the appropriate cosine rule to find side b.

3

On the Main screen, complete the entry line as: solve(b2 = 62 + 102 - 2×6×10×cos(76), b) Then press E.

4

Since b represents the side length of a triangle, then b > 0.

76° b

a=6 C

b2 = a2 + c2 - 2ac cos (B)

b = 10.34, correct to 2 decimal places.


155

Note: Once the third side has been found, the sine rule could be used to find other angles if necessary. If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject. b2 + c 2 − a 2 cos (A) = 2 bc a 2 + c 2 − b2 cos (B) = 2 ac a 2 + b2 − c 2 cos (C) = 2 ab

Worked examPle 10

eBook plus

Find the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm. Think

Tutorial

int-1213

WriTe

Worked example 10


Draw a labelled diagram of the triangle, call it ABC and fill in the given information. Note: The smallest angle will be opposite the smallest side.

B c=7 A

a=4 C

b=9

Let a = 4 b=7 c=9 2

Check that one of the criteria for the cosine rule has been satisfied.

3

Write the appropriate cosine rule to find angle A.

4

Substitute the given values into the rearranged rule.

5

Evaluate.

6

Transpose the equation to make A the subject by taking the inverse cos of both sides.

7

Round the answer to degrees and minutes.

The cosine rule can be used since three side lengths have been given. b2 + c2 − a2 cos (A) = 2bc 72 + 92 − 4 2 2×7×9 49 + 81 − 16 = 126 114 = 126  114  A = cos−1   126  = 25.208 765 3° =

= 25°13′



B c=7 A

2

156

Write the appropriate cosine rule to find the angle A.

a=4 b=9

C

a2 = b2 + c2 − 2bc cos (A)


3

On the Main screen, complete the entry line as: solve(42 = 92 + 72 − 4×9×7×cos(a), a) | 0 ≤ a ≤ 180 Then press E.

4


A = 25.2088° = 25°13′

Worked examPle 11

eBook plus

Two rowers set out from the same point. One rows N70°E for 2000 m and the other rows S15°W for 1800 m. How far apart are the two rowers? Think 1

Tutorial

int-1047

WriTe

Worked example 11

N

Draw a labelled diagram of the triangle, call it ABC and fill in the given information.

2000 m 70° C

A

15° 1800 m B 2 3

Check that one of the criteria for the cosine rule has been satisfied. Write the appropriate cosine rule to find side c.

The cosine rule can be used since two side lengths and the included angle have been given. To find side c: c2 = a2 + b2 − 2ab cos (C) = 20002 + 18002 − 2 × 2000 × 1800 × cos (125°)

4

Substitute the given values into the rule.

5

Evaluate.

6


= 40 000 000 + 3 240 000 − 7 200 000 × −0.573 576 436 = 11 369 750.342 c = 11 369 750.342 = 3371.906 04 = 3371.91

7


The rowers are 3371.91 m apart.

rememBer

1. In any triangle ABC:

a2 = b2 + c2 − 2bc cos (A) b2 = a2 + c2 − 2ac cos (B) c2 = a2 + b2 − 2ab cos (C)

Chapter 5


157

2. The cosine rule can be used to solve non–right-angled triangles if we are given: (a) three sides of the triangle (b) two sides of the triangle and the included angle (that is, the angle between the two given sides). 3. If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject. b2 + c2 − a2 cos (A) = 2bc a2 + c2 − b2 cos (B) = 2ac 2 a + b2 − c2 cos (C) = 2ab

Exercise

5D

The cosine rule 1 WE 9 Find the third side of triangle ABC given a = 3.4, b = 7.8 and C = 80°. 2 In triangle ABC, b = 64.5 cm, c = 38.1 cm and A = 58°34′. Find a. 3 In triangle ABC, a = 17, c = 10 and B = 115°. Find b, and hence find A and C. 4 WE 10 Find the smallest angle in the triangle with sides 6 cm, 4 cm and 8 cm. 5 In triangle ABC, a = 356, b = 207 and c = 296. Find the largest angle. 6 In triangle ABC, a = 23.6, b = 17.3 and c = 26.4. Find the size of all the angles. 7

In triangle DEF, d = 3 cm, e = 7 cm and F = 60°. Find f in exact form.

8 WE 11 Two rowers set out from the same point. One rows N30°E for 1500 m and the other rows S40°E for 1200 m. How far apart are the two rowers? 9 Maria cycles 12 km in a direction N68°W, then 7 km in a direction of N34°E. a How far is she from her starting point? b What is the bearing of the starting point from her finishing point? 10 A garden bed is in the shape of a triangle, with sides of length 3 m, 4.5 m and 5.2 m. a Calculate the smallest angle. b Hence, find the area of the garden. (Hint: Draw a diagram, with the longest length as the base of the triangle.) 11 A hockey goal is 3 m wide. When Sophie is 7 m from one post and 5.2 m from the other, she shoots for goal. Within what angle, to the nearest degree, must the shot be made if it is to score a goal? 12 An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly? 13 A plane flies in a direction of N70°E for 80 km, then on a bearing of S10°W for 150 km. a How far is the plane from its starting point? b What direction is the plane from its starting point? 14 Ship A is 16.2 km from port on a bearing of 053° T and ship B is 31.6 km from the same port on a bearing of 117° T. Calculate the distance between the two ships. 15 A plane takes off at 10.00 am from an airfield, and flies at 120 km/h on a bearing of N35°W. A second plane takes off at 10.05 am from the same airfield, and flies on a bearing of S80°E at a speed of 90 km/h. How far apart are the planes at 10.25 am?

158


16 Three circles of radii 5 cm, 6 cm and 8 cm are positioned so that they just touch one another. Their centres form the vertices of a triangle. Find the largest angle in the triangle. 8

17 For the given shape at near right, determine: a the length of the diagonal b the magnitude (size) of angle B c the length of x.

5 cm

6 cm

x

150° B 7 60° 10

8 cm

18 From the top of a vertical cliff 68 m high, an observer notices a yacht at sea. The angle of depression to the yacht is 47°. The yacht sails directly away from the cliff, and after 10 minutes the angle of depression is 15°. How fast does the yacht sail?

5E

Area of triangles

1

The area of any triangle is given by the rule Area = 2 bh where b is the base length and h is the perpendicular height of the triangle.

h b

However, often the perpendicular height is not given directly and needs to be calculated first. In the triangle ABC, b is the base length and h is the perpendicular height of the triangle.

B c

Using the trigonometric ratio for sine: sin (A) =

h c

A

h

a

b

C

Transposing the equation to make h the subject, we obtain: h = c × sin (A) Therefore, the area of triangle ABC becomes: 1

Area = 2 bc sin (A) Depending on how the triangle is labelled, the formula could read: 1

1

1

Area = 2 ab sin (C) Area = 2 ac sin (B) Area = 2 bc sin (A) The area formula may be used on any triangle provided that two sides of the triangle and the included angle (that is, the angle between the two given sides) are known. Worked Example 12

Find the area of the triangle shown.

Think 1


7 cm

120°

9 cm

Write/draw B c = 7 cm 120° A

a = 9 cm C

Let a = 9 cm, c = 7 cm, B = 120° 2

Check that the criterion for the area rule has been satisfied.

The area rule can be used since two side lengths and the included angle are known.


159

3 4 5

Write the appropriate rule for the area. Substitute the known values into the rule.

1

Area = 2 ac sin (B) 1

= 2 × 9 × 7 × sin (120°) = 27.28 cm2


Note: If you are not given the included angle, you will need to find it in order to calculate the area. This may involve using either the sine or cosine rule.

Worked examPle 13

eBook plus

A triangle has known dimensions of a = 5 cm, b = 7 cm and B = 52°. Find A and C and hence the area. Think 1

Tutorial


WriTe


B a=5

52° A

C

b=7

Let a = 5, b = 7, B = 52° 2

Check whether the criterion for the area rule has been satisfied.

The area rule cannot be used since the included angle has not been given.

3


To find angle A: a b = sin ( A) sin ( B)

4


5


6

Evaluate.

7


8

Determine the value of the included angle, C, using the fact that the angle sum of any triangle is 180°.

9

Write the appropriate rule for the area.

10 11

5 7 = sin ( A) sin (52) 5 × sin (52°) = 7 × sin (A) 5 × sin (52) = sin ( A) 7 5 × sin (52) sin ( A) = 7 −1

A = sin

 5 × sin (52)    7

= 34.254 151 87°

160

= 34°15′ C = 180° − (52° + 34°15′) = 93°45′ Area =

1 2

ab sin (C)


=

1 2

× 5 × 7 × sin (93°45′)


= 17.46 cm2.


Heron’s formula If we know the lengths of all the sides of the triangle but none of the angles, we could use the 1 cosine rule to find an angle, then use 2 bc sin (A) to find the area. Alternatively, we could use Heron’s formula to find the area. Heron’s formula states that the area of a triangle is: Area = s( s − a)( s − b)( s − c) where s is the semi-perimeter of the triangle; that is, 1

s = 2 (a + b + c) The proof of this formula is beyond the scope of this course. Worked Example 14

Find the area of the triangle with sides of 4 cm, 6 cm and 8 cm. Think 1

Write


C 4 cm

6 cm

B

8 cm

A

Let a = 4, b = 6, c = 8 2

Determine which area rule will be used.

Since three side lengths have been given, use Heron’s formula.

3

Write the rule for Heron’s formula.

Area = s(s − a)(s − b)(s − c)

4

Write the rule for s, the semi-perimeter of the triangle. Substitute the given values into the rule for the semi-perimeter.

5

1

s = 2 (a + b + c) =

1 2

(4 + 6 + 8)

1

= 2 (18) =9

6

Substitute all of the known values into Heron’s formula.

7

Evaluate.

Area = 9(9 − 4)(9 − 6)(9 − 8) = 9 × 5 × 3 ×1 = 135 = 11.618 950 04

8

Round the answer to 2 decimal places and include the appropriate unit.

= 11.62 cm2

REMEMBER

1. If two sides of any triangle and the included angle (that is, the angle between the two given sides) are known, the following rules may be used to determine the area of that triangle. 1 1 1 Area = 2 ab sin (C) Area = 2 ac sin (B) Area = 2 bc sin (A)


161

2. Alternatively, if the lengths of three sides of a triangle are known, Heron’s formula may be used to find the area of the triangle: Area = s(s − a) (s − b) (s − c) where s is the semi-perimeter of the triangle; that is, 1

s = 2 (a + b + c) Exercise

5E

Area of triangles 1 WE 12 Find the area of the triangle ABC with a = 7 cm, b = 4 cm and C = 68°. 2 Find the area of the triangle ABC with a = 7.3 cm, c = 10.8 cm and B = 104°40′. 3 Find the area of the triangle ABC with b = 23.1 m, c = 18.6 m and A = 82°17′. 4 Find the exact area of the triangle DEF with d = 6, e = 9 and F = 60°. 5 Find the exact area of the triangle QPR with p = 12, r = 10 and Q = 45°. 6 WE 13 MC In a triangle, a = 15 m, b = 20 m and B = 50°. The area of the triangle is: A 86.2 m2 B 114.9 m2 C 149.4 m2 2 2 D 172.4 m E 181.7 m 7 WE 14 Find the area of the triangle with sides of 5 cm, 6 cm and 8 cm. 8 Find the area of the triangle with sides of 40 mm, 30 mm and 5.7 cm. 9 Find the area of the triangle with sides of 16 mm, 3 cm and 2.7 cm. 10 Find the area of the equilateral triangle with sides 4 cm. Leave your answer in simplified surd form. 11 MC A triangle has sides of length 10 cm, 14 cm and 20 cm. The area of the triangle is: B 65 cm2 C 106 cm2 A 41 cm2 2 2 D 137 cm E 1038 cm 12 A triangle has a = 10 cm, c = 14 cm and C = 48°. Find A and B and hence the area. 13 A triangle has a = 17 m, c = 22 m and C = 56°. Find A and B and hence the area. 14 A triangle has b = 32 mm, c = 15 mm and B = 38°. Find A and C and hence the area. 15 A piece of metal is in the shape of a triangle with sides of length 114 mm, 72 mm and 87 mm. Find its area using Heron’s formula. 16 A triangle has the largest angle of 115°. The longest side is 62 cm and another side is 35 cm. Find the area of the triangle. 17 A triangle has two sides of 25 cm and 30 cm. The angle between the two sides is 30°. Find: a its area b the length of its third side c its area using Heron’s formula. 18 The surface of a fish pond has the shape shown in the diagram at right. How many goldfish can the pond support if each fish requires 0.3 m2 surface area of water? 19 Find the area of this quadrilateral.

3.5 m

2m

5m 4m

8m

4m

60° 5m

162

1m


20 A parallelogram has diagonals of length 10 cm and 17 cm. An angle between them is 125°. Find: a the area of the parallelogram b the dimensions of the parallelogram. 21 A lawn is to be made in the shape of a triangle, with sides of length 11 m, 15 m and 17.2 m. How much grass seed, to the nearest kilogram, is needed if it is sown at the rate of 1 kg per 5 m2? 22 A bushfire burns out an area of level grassland shown in the diagram. What is the area, in hectares, of the land that is burnt?

400 m

23 An earth embankment is 27 m long, and has a cross-section shown in the diagram. Find the volume of earth needed to build the embankment. 24 mC A parallelogram has sides of 14 cm and 18 cm, and an angle between them of 72°. The area of the parallelogram is: A 86.2 cm2 B 118.4 cm2 C 172.4 cm2 D 239.7 cm2 eBook plus Digital doc

WorkSHEET 5.2

5F

km

2 km River

1.8

200 m Road

100° 2m

130° 50° 5m

80°

E 252 cm2

25 mC An advertising hoarding is in the shape of an isosceles triangle, with sides of length 15 m, 15 m and 18 m. It is to be painted with two coats of purple paint. If the paint covers 12 m2 per litre, the amount of paint needed, to the nearest litre, would be: A 9L B 18 L C 24 L D 36 L E 42 L

Trigonometric identities An identity is a relationship that holds true for all values of a pronumeral or pronumerals. The sine and cosine functions are related functions and the following identities exist between them.

The Pythagorean identity Think of a triangle within the unit circle. We know that the hypotenuse is 1 unit.

y 1

a 1 a = sin (θ )

sin (θ ) =

θ a

0 b

b 1 b = cos (θ )

cos (θ ) =

Chapter 5

x


163

So the triangle formed has a height of sin (θ ) and a base length of cos (θ ). Pythagoras’ theorem then tells us that a2 + b2 = 12 ∴ sin2 (θ ) + cos2 (θ) = 1 Note: sin2 (θ ) = (sin (θ ))2 and cos2 (θ ) = (cos (θ ))2 The Pythagorean identity is sin2 (θ ) + cos2 (θ ) = 1.

Worked Example 15

Find the value of sin (θ ) given cos (θ ) =

5 and 0° < θ < 90°. 13

Think

Write

1

Write the Pythagorean identity.

2

Substitute the known value cos (θ ) =

3

Solve to find the required value.

4

Write the final answer.

sin2 (θ ) + cos2 (θ ) = 1 5 . 13

2

 5 sin2 (θ ) +   = 1 13 25 =1 169 144 sin2 (θ ) = 169 12 sin (θ ) = ± 13

sin2 (θ ) +

As θ is in the first quadrant sin (θ ) =

Complementary angles

12 . 13

α In the diagram at right we can see that: b a a c cos (θ ) = sin (θ ) = c c a b cos (α ) = sin (α ) = θ c c b So for our diagram cos (θ ) = sin (α ) and sin (θ ) = cos (α ). We also know that θ + α = 90°, so α = 90° - θ. By substituting this into cos (θ ) = sin (α ) and cos (α ) = sin (θ ) we get cos (θ ) = sin (90° - θ ) and sin (θ ) = cos (90° - θ ). Worked Example 16

Find the value of cos (70°) given sin (20°) = 0.342. Think 1 2 3

164

Write the equation with the required complementary angle formula. Identify the value of θ. Substitute the angle into the equation and simplify.

Write

cos (θ ) = sin (90° - θ )

θ = 70° cos (70°) = sin (90° - 70°) = sin (20°) = 0.342


REMEMBER

1. The Pythagorean identity is sin2 (θ ) + cos2 (θ ) = 1. 2. cos (θ ) = sin (90° - θ ) and sin (θ ) = cos (90° - θ ). Exercise

5F

Trigonometric identities 1

4 and 0° < θ < 90°. 5

5

12 and 0° < θ < 90°. 13 6 Find the value of cos (θ ) given sin (θ ) = and 0° < θ < 90°. 10 2 Find the value of sin (θ ) given cos (θ ) = and 0° < θ < 90°. 7 Use your knowledge of exact values to show that the Pythagorean identity is true for θ = 30°.

6

WE 16 Find the value of sin (12°) given cos (78°) = 0.208.

7

Find the value of cos (42°) given sin (48°) = 0.743.

2 3 4

5G

WE 15 Find the value of sin (θ ) given cos (θ ) =

Find the value of cos (θ ) given sin (θ ) =

Radian measurement In all of the trigonometry tasks covered so far, the unit for measuring angles has been the degree. There is another commonly used measurement for angles, the radian. This is used in situations involving length and areas associated with circles. Consider the unit circle, a circle with a radius of 1 unit. OP is the radius. If OP is rotated θ ° anticlockwise, the point P traces a path along the circumference of the circle to a new point, P1. The arc length PP1 is a radian measurement, symbolised by θ c. Note: 1c is equivalent to the angle in degrees formed when the length of PP1 is 1 unit; in other words, when the arc is the same length as the radius. If the length OP is rotated 180°, the point P traces out half the circumference. Since the circle has a radius of 1 unit, and C = 2πr, the arc PP1 has a length of π. The relationship between degrees and radians is thus established. 180° = π c This relationship will be used to convert from one system to another. Rearranging the basic conversion factor gives: 180° = π π 1° = 180 

O

P OP = 1 unit P1

O

θc

θ°

P

OP = 1 unit 1– 2

circumference 180°

P1

O

P

π To convert an angle in degrees to radian measure, multiply by . 180   180 Also, since π = 180°, it follows that 1c = . π 180  To convert an angle in radian measure to degrees, multiply by . π Chapter 5 Trigonometric ratios and their applications

165

Where possible, it is common to have radian values with π in them. It is usual to write radians without any symbol, but degrees must always have a symbol. For example, an angle of 25° must have the degree symbol written, but an angle of 1.5 is understood to be 1.5 radians. Worked examPle 17

Convert 135° to radian measure, expressing the answer in terms of π. Think 1

2

WriTe

To convert an angle in degrees to radian measure, π multiply the angle by . 180

135° = 135° ×

π 180 

135π 180 3π = 4

=

Simplify, leaving the answer in terms of π.

Worked examPle 18

Convert the radian measurement

4π to degrees. 5

Think

WriTe

1

To convert radian measure to an angle in degrees, 180 multiply the angle by . π

2

Simplify. Note: The π cancels out.

4π 4π 180  = × 5 5 π 720  = 5 = 144°

If the calculation does not simplify easily, write the answers in degrees and minutes, or radians to 4 decimal places. If angles are given in degrees and minutes, convert to degrees only before converting to radians. rememBer

1. 180° = πc π 2. To convert an angle in degrees to radian measure, multiply by . 180  180  3. To convert an angle in radian measure to degrees, multiply by . π exerCise

5G

radian measurement 1

answers in terms of π. a 30° b 60° e 225° f 270° i 72° j 200°

166

eBook plus

We 17 Convert the following angles to radian measure, expressing

Digital doc

SkillSHEET 5.6

c 120° g 315°


d 150° h 480°

Changing degrees to radians

2

We 18 Convert the following radian measurements into degrees.

π 4 17π f 6 a

3π 2 π g 12 b

7π 6 13π h 10

5π 3 11π i 8

c

d

e

7π 12

j 8π

3 Convert the following angles in degrees to radians, giving answers to 4 decimal places. a 27° b 109° c 243° d 351° e 7° f 63°42′ g 138°21′ h 274°8′ i 326°53′ j 47°2′ 4 Convert the following radian measurements into degrees and minutes. a 2.345 b 0.6103 c 1 d 1.61 e 3.592 f 7.25 g 0.182 h 5.8402 i 4.073 j 6.167

5h

arcs, sectors and segments

eBook plus

arc length

Interactivity

An arc is a section of the circumference of a circle. The length of the arc is proportional to the angle subtended at the centre. For example, an 1 angle of 90° will create an arc which is 4 the circumference. We have already defined an arc length as equivalent to θ radians if the circle has a radius of 1 unit.

Therefore, a simple dilation of the unit circle will enable us to calculate the arc length for any sized circle, as long as the angle is expressed in radians. If the radius is dilated by a factor of r, the arc length is also dilated by a factor of r.

int-0972 Sectors

θ° r=1

θ°

θc

rθ

c

r

Dilation by factor of r

Therefore, l = rθ, where l represents the arc length, r represents the radius and θ represents an angle measured in radians. Worked examPle 19

Find the length of the arc which subtends an angle of 75° at the centre of a circle with radius 8 cm. Think 1

2

WriTe/draW

Draw a diagram representing the situation and label with the given values.

75° r=8

Convert the angle from 75° to radian measure by π multiplying the angle by . 180 

75° = 75° ×

Chapter 5

=

l = rθ

π 180 

75π 180


167

= 1.3090

3

Evaluate to 4 decimal places.

4

Write the rule for the length of the arc.

5

Substitute the values into the formula.

= 8 × 1.3090

6

Evaluate to 2 decimal places and include the appropriate unit.

= 10.4720 = 10.47 cm

l = rθ

Note: In order to use the formula for the length of the arc, the angle must be in radian measure.

Worked Example 20

Find the angle subtended by a 17 cm arc in a circle of radius 14 cm: a in radians b in degrees. Think a

b

Write

1

Write the rule for the length of the arc.

2


3

Transpose the equation to make θ the subject.

4

Evaluate to 4 decimal places and include the appropriate unit.

1

To convert radian measure to an angle in 180  degrees, multiply the angle by . π Evaluate.

2 3

a

l = rθ 17 = 14θ

θ=

17 14

= 1.214 285 714 = 1.2143c 180  b 1.2143c = 1.2143 × π

Convert the angle to degrees and minutes.

= 69.573 446 55°

= 69°34′

Area of a sector In the diagram at right, the shaded area is the minor sector AOB, and the unshaded area is the major sector AOB. The area of the sector is proportional to the arc length. For 1 1 example, an area of 4 of the circle contains an arc which is 4 of the circumference. area of sector arc length Thus, in any circle: = area of circle circumference of circle A rθ where θ is measured in radians. = 2 πr 2 πr rθ × π r 2 A= 2π r 1

= 2 r 2 θ The area of a sector is:

168

1

A = 2 r 2 θ


Major sector O

A

B Minor sector

Worked examPle 21

A sector has an area of 157 cm2, and subtends an angle of 107°. What is the radius of the circle? Think 1

WriTe

107° = 107° ×

Convert the angle from 107° to radian measure by π multiplying the angle by 180  .

107π 180 = 1.8675

π 180 

=

2


3

Write the rule for the area of a sector.

A=

1 2

r2θ

4


157 =

1 2

× r2 × 1.8675

5

Transpose the equation to make r 2 the subject.

6

Take the square root of both sides of the equation. Evaluate to 2 decimal places and include the appropriate unit.

7

2 × 157 = r2 1.8675 r2 = 168.139 016 5 r = 12.966 842 97 = 12.97 cm

area of a segment A segment is that part of a sector bounded by the arc and the chord. As can be seen from the diagram at right: Area of segment = area of sector − area of triangle 1

1

A = 2 r2θ − 2 r 2 sin (θ ° )

1

= 2 r2 (θ − sin (θ °)) Note: θ is in radians and θ ° is in degrees.

The area of a segment:

r

θ

Segment

1

A = 2 r2 (θ − sin (θ°))

Worked examPle 22

eBook plus

Find the area of the segment in a circle of radius 5 cm, subtended by an angle of 40°. Think

WriTe

π 40° = 40° × 180 40π = 180 = 0.6981

Tutorial


1

Convert the angle from 40° to radian measure π by multiplying the angle by . 180

2


3

Write the rule for the area of a segment.

A = 2 r2 (θ − sin (θ °))

4

Identify each of the variables.

r = 5, θ = 0.6981, θ ° = 40°

5


A = 2 × 52 (0.6981 − sin (40°))

6

Evaluate.

1

1 1

= 2 × 25 × 0.0553 = 0.691 25

7

Round to 2 decimal places and include the appropriate unit.

= 0.69 cm2

Chapter 5


169

rememBer

1. Arc length: 2. Area of a sector:

l = rθ 1 A = 2 r2θ 1

3. Area of a segment: A = 2 r2 (θ − sin ((θ ° )) where r = radius, θ = angle (measured in radians) and θ ° = angle (measured in degrees).

exerCise

5h

arcs, sectors and segments 1 We 19 Find the length of the arc which subtends an angle of 65° at the centre of a circle of radius 14 cm. 2 Find the length of the arc which subtends an angle of 153° at the centre of a circle of radius 75 mm. 3

Find the length of the arc which subtends an angle of 135° at the centre of a circle of

radius 10 cm. Leave answer in terms of π. 4 An arc of a circle is 3.5 cm long, and subtends an angle of 41° at the centre of the circle. What is the radius of the circle? 5 An arc of a circle is 27.8 cm long, and subtends an angle of 205° at the centre of the circle. What is the radius of the circle? 6

An arc of a circle is 4 cm long and subtends an angle of 60° at the centre of the circle. What is the radius of the circle? Write your answer in terms of π.

7 We 20 Find the angle subtended by a 20 cm arc in a circle of radius 75 cm: a in radians b in degrees. 8 Find the angle subtended by an 8 cm arc in a circle of radius 5 cm: a in radians b in degrees. 9 An arc of length 8 cm is marked out on the circumference of a circle of radius 13 cm. What angle does the arc subtend at the centre of the circle? 10 An arc of length 245 mm is marked out on the circumference of a circle of radius 18 cm. Find the angle that the arc subtends at the centre of the circle. 11 The minute hand of a clock is 35 cm long. How far does the tip of the hand travel in 20 minutes? 12 A child’s swing is suspended by a rope 3 m long. What is the length of the arc it travels if it swings through an angle of 42°? 13 Find the area of the sector of a circle of radius 17 cm with an angle of 56°. 14 Find the area of the sector of a circle of radius 6.2 cm with an angle of 256°. 15

Find the area of a sector of a circle of radius 6 cm with an angle of 100°. Write your answer in terms of π.

16 We21 A sector has an area of 825 cm2, and subtends an angle of 70°. What is the radius of the circle? 17 A sector with an area of 309 cm2 is part of a circle of radius 18.2 cm. Find the angle in the sector. 170


18 Find the area of a sector of a circle of radius 30 cm if the sector has an arc length of 18 cm. 19 A garden bed is in the form of a sector of a circle of radius 4 m. The arc of the sector is 5 m long. Find: a the area of the garden bed b the volume of mulch needed to cover the bed to a depth of 10 cm. 20 The minute hand on a clock is 62 cm long. What area does the hand sweep through in 40 minutes? 21 A sector whose angle is 150° is cut from a circular piece of cardboard whose radius is 12 cm. The two straight edges of the sector are joined so as to form a cone. a What is the surface area of the cone? b What is the radius of the cone? 22 WE 22 Find the area of the segment in a circle of radius 25 cm subtended by an angle of 100°. 23 Find the area of the segment of a circle of radius 4.7 m that subtends an angle of 85°20′ at the centre. 24 A segment of a circle subtends an angle of 75° at the centre. The area of the segment is 100 cm2. Find the radius of the circle. 25 In a circle of radius 15 cm, a sector has an area of 100 cm2. Find the angle subtended by the sector. 26 Two circles of radii 3 cm and 4 cm have their centres 5 cm apart. Find the area of the intersection of the two circles. 27 MC The angle subtended by a 28 cm arc in a circle of radius 20 cm in radians is: A 0.71 B 40.93 C 80.21 D 1.4 E 0.4 28 MC The area of the segment in a circle of radius 12 cm, subtended by an angle of 60° is: A 6.52 cm2 B 30.31 cm2 C 26.08 cm2 D 15.24 cm2 E 13.04 cm2 29 Two irrigation sprinklers spread water in circular paths with radii of 7 m and 4 m. If the sprinklers are 10 m apart, find the area of crop that receives water from both sprinklers. 30 MC The length of the arc which subtends an angle of 50° at the centre of a circle with radius 10 cm is: A 8.73 cm B 0.87 cm C 10.43 cm D 6.25 cm E 0.63 cm


171

Summary Trigonometry of right-angled triangles

• For any right-angled triangle: O A O sin (θ ) = cos (θ ) = tan (θ ) = H H A • Pythagoras’ theorem, c2 = a2 + b2 may also be used to solve right-angled triangles. c

a

B Hypotenuse (H )

Opposite (O) C

θ (A) Adjacent

A

b

• Angles of 30°, 45° and 60° have exact values of sine, cosine and tangent.

θ

30°

45°

sin (θ )

1 2

1

cos (θ )

3 2

1

tan (θ )

1 3

=

2 2 3 3

60°

=

2 2

3 2

=

2 2

1 2

1

3

Elevation, depression and bearings

• Angles of elevation and depression are each measured from the horizontal. • The angle of elevation is equal to the angle of depression since they are alternate ‘Z’ angles. • True bearings are measured in a clockwise direction, starting from north (0° T). The sine rule

• The sine rule states that for any triangle ABC: a b c = = sin ( A) sin ( B) sin (C ) When using this rule, it is important to note that, depending on the values given, any combination of the two equalities may be used to solve a particular triangle. • The sine rule may be used to solve non–right-angled triangles if we are given: (a) two angles and one side length (b) two side lengths and an angle opposite one of these side lengths. • The ambiguous case exists if C is an acute angle and a > c > a × sin (C). The cosine rule

• In any triangle ABC: a2 = b2 + c2 − 2bc cos (A) b2 = a2 + c2 − 2ac cos (B) c2 = a2 + b2 − 2ab cos (C) • The cosine rule can be used to solve non–right-angled triangles if we are given: (a) three sides of the triangle (b) two sides of the triangle and the included angle (that is, the angle between the two given sides).

172


• If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject. b2 + c2 − a2 cos (A) = 2bc a2 + c2 − b2 cos (B) = 2ac a2 + b2 − c2 cos (C) = 2ab Area of triangles

• If two sides of any triangle and the included angle (that is, the angle between the two given sides) are known, the following rules may be used to determine the area of that triangle. Area =

1 2

ab sin (C)

1

Area = 2 ac sin (B) 1

Area = 2 bc sin (A) • Alternatively, if three side lengths of a triangle are known, Heron’s formula may be used to find the area of a triangle: Area = s(s − a)(s − b)(s − c) where s is the semi-perimeter of the triangle; that is, 1

s = 2 (a + b + c) Trigonometric identities

• An identity is a relationship that holds true for all values of a pronumeral or pronumerals. • The Pythagorean Identity states that sin2 (θ ) + cos2 (θ ) = 1. • Sine and cosine are called complementary functions since: cos (θ ) = sin (90° − θ ) and sin (θ ) = cos (90° − θ ) Radian measurement

• 180° = πc π • To convert an angle in degrees to radian measure, multiply by . 180  180  • To convert an angle in radian measure to degrees, multiply by . π Arcs, sectors and segments

• Arc length:

l = rθ

• Area of a sector:

A = 2 r2θ

• Area of a segment:

A = 2 r2 (θ − sin (θ°))

1

1

Where r = radius, θ = angle (measured in radians) and θ ° = angle (measured in degrees).


173


1 A stepladder stands on a floor with its feet 2 m apart. If the angle formed by the legs with the floor is 60°, how high above the floor is the top of the ladder? 2 Two buildings, 15 m and 27 m high, are directly opposite each other across a river. The angle of depression of the top of the smaller building from the top of the taller one is 30°. How wide is the river? 3 In the triangle shown at right find the exact length of side m.

x

12 cm

30°

45°

4 A triangle has sides of length 12 m, 15 m and 20 m. If Q is the largest angle find cos (q ). 5 A triangle has two sides of 20 cm and 25 cm. The angle between the two sides is 45°. Find its area. 6 A triangular garden area is bound by three straight edges of lengths 4 m, 5 m and 7 m. Find the exact area of the garden. 3 7 Find the value of cos (θ ) given sin (θ ) = and 8 0° < θ < 90°. 8 a Convert the following angles to radian measure, expressing answers in terms of π . i 80° ii 125° iii 640° b Convert the following radian measurement into degrees.

i

π 20

ii

15π 8

iii 7π

9 A paddock is in the shape of a sector with radius of 75 m and an angle of 60°. Find: a the amount of fencing needed to enclose the paddock b the area enclosed by the paddock. Multiple choice

1 In the triangle, the value of θ, to the nearest degree, is: A 37° B 39° C 51° D 52° E 53°

174

4

θ 5

2 A ladder 4.5 m long rests against a vertical wall, with the foot of the ladder 2 m from the base of the wall. The angle the ladder makes with the wall, to the nearest degree, is: A 24° B 26° C 35° D 64° E 66° 3 A person stands 18 m from the base of a building, and measures the angle of elevation to the top of the building as 62°. If the person is 1.8 m tall, how high is the building, to the nearest metre? A 11 m B 18 m C 36 m D 22 m E 34 m 4 A bearing of 310° T is the same as: A N40°W B N50°W C S50°W D S50°E E N50°E 5 In triangle ABC, a = 10, b = 7 and B = 40°. A possible value for C, to the nearest degree, is: A 37° B 52° C 68° D 73° E 113° 6 Two boats start from the same point. One sails due north for 10 km and the other sails south east for 15 km. Their distance apart is: A 10.62 km B 14.83 km C 17.35 km D 21.38 km E 23.18 km 7 A triangle has sides measuring 5 cm, 8 cm and 10 cm. The largest angle in the triangle, to the nearest degree, is: A 52° B 82° C 98° D 128° E 140° 8 The area of the triangle with a = 10 m, b = 8 m and C = 72° is: B 76.08 m2 C 10.15 m2 A 12.36 m2 2 2 D 38.04 m E 123.10 m 9 A garden bed is in the shape of a triangle, with sides of length 4 m, 5.2 m and 7 m. The volume of topsoil needed to cover the garden to a depth of 250 mm is: A 2.32 m3 B 2.57 m3 C 2.81 m3 3 3 D 3.17 m E 3.76 m 10 When 75° is converted to radian measure, the value of the angle, expressed in terms of π, is: 12π π 5π A B C 5 24 12 5π 7π D E 12 12


11 When 5.321 is converted to degrees and minutes, the value of the angle is: A 305°27′ B 304°52′ C 5°19′ D 1°42′ E 152°26′ 12 An arc in a circle of radius 5 cm is 3.5 cm long. The angle, to the nearest degree, subtended at the centre by the arc is: A 35° B 40° C 50° D 68° E 82° 13 A sector has an area of 40 cm2, and an angle of 30°. The arc length of the sector, to 2 decimal places, is: A 1.64 cm B 2.66 cm C 4.83 cm D 6.47 cm E 12.36 cm

14 The area of the shaded region in the figure at right to the nearest cm2 is: B 846 cm2 A 800 cm2 2 C 898 cm D 952 cm2 2 E 983 cm

40 cm 120°

15 A clock has a minute hand 75 cm long. The area that it sweeps when passing through 48 minutes, to 2 decimal places, is: B 1.35 m2 A 0.90 m2 2 C 1.41 m D 1.88 m2 E 2.01 m2

exTended resPonse

1 Three circles of radii 2 cm, 3 cm and 4 cm are placed so that they just touch each other. A triangle is formed by joining their three centres. Find: a the three angles of the triangle b the area of the triangle, correct to 3 decimal places c the shaded area correct to 3 decimal places.

2 A farmer owns a large triangular area of flat land, bounded on one side by an embankment to a river flowing NE, on a second side by a road which meets the river at a bridge where the angle between river and road is 105°, and on the third side by a long fence. Find: a the length of the river frontage, correct to 3 decimal places b the area of the land correct to 3 decimal places. The farmer decides to divide the land into two sections of equal area, by running a fence from the bridge to a point on the opposite side. c On what bearing must the fence be built? d What is the length of the fence, correct to 3 decimal places?

Chapter 5

N River 45°

33° Fence

105° Road

42°

3.2 km


175

3 a A four-wheel-drive vehicle leaves a camp site and travels across a flat sandy plain in a direction of S65°E, for a distance of 8.2 km. It then heads due south for 6.7 km to reach a waterhole. i How far is the waterhole from the camp site? ii What is the bearing of the waterhole from the camp site? b A search plane sets off to find the vehicle. It is on a course that takes it over points A and B, two locations on level ground. At a certain time, from point A, the angle of elevation to the plane is 72°. From point B, the angle of elevation is 47°. If A and B are 3500 m apart, find the height of the plane off the ground. 4 Christopher lives on a farm. He has decided that this year B he will plant a variety of crops in his large but unusually 124 m shaped vegetable garden. He has divided the vegetable garden 56 m into six triangular regions, which he will fence off as shown in 2 1 the diagram at right. Christopher needs to calculate the C A 52 m perimeter and area of each region so he can purchase the 95º 64º 38º correct amount of fencing material and seedlings. 80º 6 58º a Separate each of the regions into single triangles and label 3 85 m each with the information provided. 68 m b Use the appropriate rules to determine all unknown 5 43 m lengths and relevant angles. F c How much fencing material is required to section off the 4 six regions? d If fencing material is $4.50 per metre (and only sold by D the metre) what will the cost be? E e Calculate the area of each region and hence determine the total area available for planting.



176


eBook plus

aCTiviTies

Chapter opener Digital doc

• 10 Quick Questions: Warm up with ten quick questions on trigonometric ratios and their applications. (page 136) 5A

Trigonometry of right-angled triangles

Digital docs

• SkillSHEET 5.1: Practise labelling right-angled triangles. (page 141) • SkillSHEET 5.2: Practise using trigonometric ratios. (page 141) • SkillSHEET 5.3: Practise degrees and minutes. (page 141) • SkillSHEET 5.4: Practise composite shapes 1. (page 142) • SkillSHEET 5.5: Practise composite shapes 2. (page 142) 5B

Elevation, depression and bearings

Tutorial

• We6 int-1045: Watch how to determine the bearing of a ship from its starting point. (page 145) 5C

The sine rule

Tutorial

• We8 int-1046: Watch how to show the ambiguous case of the sine rule exists and apply it. (page 150) Digital doc

• WorkSHEET 5.1: Use trigonometry to find two unknowns in right-angled triangles; solve worded problems of elevation, depression and bearings. (page 154) 5D

The cosine rule

Tutorials

• We 10 int-1213: Watch how to find the smallest angle in a triangle. (page 156) • We 11 int-1047: Watch how to calculate the distance between two rowers. (page 157) 5E

Digital doc

• WorkSHEET 5.2: Solve more complex right-angled triangle problems with two unknowns, worded problems of elevation and depression and apply the sine and cosine rules to non-right angled triangles. (page 163) 5G

Radian measurement

Digital doc

• SkillSHEET 5.6: Practise changing degrees to radians. (page 166) 5H

Arcs, sectors and segments

Interactivity

• Sectors int-0972: Consolidate your understanding of how to calculate the area of a sector. (page 167) Tutorial

• We 22 int-1049: Watch how to find the area of a segment. (page 169) Chapter review Digital doc


Area of triangles

Tutorial

• We 13 int-1048: Watch how to find the area of a triangle given two side lengths and an angle. (page 160)

Chapter 5


177

6

6A 6b 6c 6D 6E 6F

Describing sequences Arithmetic sequences Arithmetic series Geometric sequences Geometric series Applications of sequences and series

Sequences and series areaS oF STudy

• Sequences and series as maps between the natural numbers and the real numbers, and the use of technology to generate sequences and series and their graphs • Sequences generated by recursion: arithmetic (tn + 1 = tn + d ), geometric (tn + 1 = rtn) and fixed

point iteration (for example, t1 = 2, tn + 1 = tn 2, t1 = 0.5, tn + 1 = 0.8tn(1 − tn)) • Practical applications of sequences and series, such as financial arithmetic, population modelling and musical scales

eBook plus

6a

describing sequences

Digital doc

10 Quick Questions

Sequences of numbers play an important part in our everyday life. For example, the following sequence: 2.25, 2.37, 2.58, 2.57, 2.63, . . . gives the end-of-day trading price (for 5 consecutive days) of a share in an electronics company. It looks like the price is on the rise, but is it possible to accurately predict the future price per share of the company? The following sequence is more predictable: 10 000, 9000, 8100, . . . This is the estimated number of radioactive decays of a medical compound each minute after administration to a patient. The compound is used to diagnose tumours. In the first minute, 10 000 radioactive decays are predicted; during the second minute, 9000, and so on. Can you predict the next number in the sequence? You’re correct if you said 7290. Each successive term here is 90% of, or 0.90 times, the previous term. Sequences are strings of numbers. They may be finite in number or infinite. Number sequences may follow an easily recognisable pattern or they may not. A great deal of recent mathematical work has gone into deciding whether certain strings follow a pattern (in which case subsequent terms could be predicted) or whether they are random (in which case subsequent terms cannot be predicted). This work forms the basis of chaos theory, speech recognition software for computers, weather prediction and stock market forecasting, to name but a few uses. The list is almost endless.

178


Sequences which follow a pattern can be described in a number of different ways. They may be listed in sequential order, they may be described as a functional definition, or they may be described in an iterative definition.

1 Listing in sequential order Consider the sequence of numbers t: {5, 7, 9, . . .}. The numbers in sequential order are firstly 5 then 7 and 9 with the indication that there are more numbers to follow. The symbol t is the name of the sequence and the first three terms in the sequence shown are t1 = 5, t2 = 7 and t3 = 9. The fourth term, t4 if the pattern were to continue, would be the number 11. In general, tn is the nth term in the sequence. In this example, the next term is simply the previous term with the number 2 added to it, with the first term being the number 5. Another possible sequence is t: {5, 10, 20, 40, . . .}. In this case it appears that the next term is twice the previous term. The fifth term here, if the pattern continued, would be t5 = 80. It can be difficult to determine whether or not a pattern exists in some sequences. Can you find the next term in the following sequence? t: {1, 1, 2, 3, 5, 8, . . .} Here the next term is the sum of the previous two terms, hence the next term would be 5 + 8 which is equal to 13, and so on. This sequence is called the Fibonacci sequence and is named after its discoverer, Leonardo Fibonacci, a thirteenth century mathematician. Here is another sequence; can you find the next term here? t: {7, 11, 16, 22, 29, . . .} In this sequence the difference between successive terms increases by 1 for each pair. The first difference is 4, the next difference is 5 and so on. The sixth term is thus 37 which is 8 more than 29.

2 Functional definition A functional definition is expressed in the form: tn = 2n − 7, n ∈ {1, 2, 3, 4, . . .} Using this definition the nth term can be readily calculated. For this example t1 = 2 × 1 − 7 = −5, t2 = 2 × 2 − 7 = −3, t3 = 2 × 3 − 7 = −1 and so on. We can readily calculate the 100th term, t100 = 2 × 100 − 7 = 193, simply by substituting the value n = 100 into the expression for tn. Look at the following example: dn = 4.9n2, n ∈ {1, 2, 3, . . .} For this example, in which the sequence is given the name d, d1 = 4.9 × 12 = 4.9, d2 = 4.9 × 22 = 19.6. Listing the sequence would yield d: {4.9, 19.6, 44.1, 78.4, . . .}. The 10th term would be 4.9 × 102 = 490. Here is another example: cn = cos (nπ) + 1, n ∈ {1, 2, 3, . . .} Here the sequence would be c: {0, 2, 0, 2, . . .}.

3 Iterative definition An iterative definition is expressed in the form: tn + 1 = 3tn − 2; t1 = 6 This definition looks complicated, but is actually straightforward. You may have already come across this idea on a spreadsheet. The word iteration means the calculation of the next term from the previous term using the same procedure. The symbol tn + 1 simply means the next term after the term tn. In the above example the first term, t1, is 6 (this is given in the definition) and so the next term, t2, is 3 × 6 − 2 = 16, and the following term is 3 × 16 − 2 = 46. In each and all cases the next term is found by multiplying the previous term by 3 and then subtracting 2. We could write the sequence out as a table:

Chapter 6 Sequences and series

179

n 1

Comment

tn t1 = 6

Given in the definition

t2 = 3t1 − 2

Using t1 to find the next term, t2

= 3 × 6 − 2 = 16 2

t3 = 3t2 − 2


= 3 × 16 − 2 = 46 3

t4 = 3t3 − 2


= 3 × 46 − 2 = 136 An example of this sequence using notation found in a spreadsheet would be: A1 = 6 (the first term is equal to 6) A2 = 3 × A1 − 2 (the next term is 3 times the previous term minus 2). You could then apply the Fill Down option in the Edit menu of the spreadsheet from cell A2 downwards to generate as many terms in the sequence as required. This would result in the next cell down being three times the previous cell, less 2. The iterative definition finds a natural use in a spreadsheet environment and consequently much use is made of it. A drawback is that you cannot find the nth term directly as in the functional definition, but the advantage is that more complicated systems can be successfully modelled using iterative descriptions and hence are more interesting and relevant. Worked Example 1 7

a Find the next three terms in the sequence, b: {14, 7, 2 , . . .}. b Find the 4th, 8th and 12th terms in the following sequence: en = n2 − 3n, n ∈ {1, 2, 3, . . .}. c Find the 2nd, 3rd and 5th terms for the following sequence: kn + 1 = 2kn + 1, k1 = −0.50. Think a

180

Write

1

In this example the sequence is listed and a simple pattern is evident. From inspection, the next term is half the previous term and so the sequence would be 7 7 7 7 14, 7, 2 , 4 , 8 , 16 .

2

On the Main screen, complete the entry lines as: 14 ans × 0.5 Press E repeatedly to generate the sequence.

a The next three terms are 7 , 7 , 7 . 4 8 16


b

b en = n2 − 3n

1

This is an example of a functional definition. The nth term of the sequence is found simply by substitution into the expression en = n2 − 3n.

2

Find the 4th term by substituting n = 4.

e4 = 42 − 3 × 4 =4

3


e8 = 82 − 3 × 8 = 40

4


e12 = 122 − 3 × 12 = 108

5

On the Main screen, tap: • Action • List – Create • seq Complete the entry line as: seq(n2 - 3n, n, 1, 12, 1) Then press E. Note: Scroll through the numbers in the sequence to find the 4th, 8th and 12th terms.

An alternative method to the one above for generating a sequence is shown. On the Sequence screen, complete the sequence as shown. To create the table, tap #.

6

To find the required terms, tap r.


181

c

c kn + 1 = 2kn + 1,

1

This is an example of an iterative definition. We can find the 2nd, 3rd and 5th terms for the sequence kn + 1 = 2kn +1, k1 = −0.50 by iteration.

2

Substitute k1 = −0.50 into the formula to find k2.

k2 = 2 × −0.50 + 1 =0

3

Continue the process until the value of k5 is found.

k3 = 2 × 0 + 1 =1 k4 = 2 × 1 + 1 =3 k5 = 2 × 3 + 1 =7

4

Write the answer.

Thus k2 = 0, k3 = 1 and k5 = 7.

5

Again, on the Sequence screen, complete the sequence as shown, tapping V to type the sequence. To create the table, tap #.

k1 = −0.50

To enlarge the table, tap r.

Logistic equation The logistic equation is a model of population growth. It gives the rule for determining the population in any year, based on the population in the previous year. Since we need the previous term in order to be able to generate the next term of the sequence, then the logistic equation is an example of an iterative definition. It is of the general form: tn + 1 = atn(1 - tn), where 0 < t0 < 1 and a is a constant. 182


Depending on the value of a, sequences generated by use of the logistic equation could be convergent, divergent, or oscillating. A string of numbers that converges to (settles at) a certain fixed value is called a convergent sequence. Sequence tn can converge to only one possible number, x, called the limit of the sequence. This can be written as tn → x. (The symbol → is read as ‘tends to’, or ‘approaches’.) A sequence whose terms grow further and further apart is called divergent. That is, a sequence is divergent if tn → ∞, or tn → −∞ as n → ∞. Finally, a sequence whose terms tend to fluctuate between two (or more) values is called oscillating. An oscillating sequence is neither convergent nor divergent. Worked exampLe 2

eBook plus

Given that a = 2 and t0 = 0.7, use the logistic equation to generate a sequence Tutorial of 6 terms, and state whether the sequence is convergent, divergent, or oscillating. int-1051 If the sequence is convergent, state its limit. Worked example 2 ThInk

WrITe

Method 1: Using the rule tn + 1 = atn(1 – tn) = 2tn(1 – tn)

1

Write the logistic equation, replacing a with its given value (that is, 2).

2

To find t1, substitute the value of t0 (that is, 0.7) in place of tn and evaluate.

t1 = 2t0(1 – t0) = 2 × 0.7 × (1 – 0.7) = 0.42

3

To find the next term, t2, substitute the value of t1 (that is, 0.42) in place of tn and evaluate.

t2 = 2t1(1 – t1) = 2 × 0.42 × (1 – 0.42) = 0.4872

4

Continue the iterative process four more times, each time substituting the value of the previous term into the logistic equation to find the next term.

t3 = 2t2(1 – t2) = 2 × 0.4872 × (1 – 0.4872) = 0.499 672 3 t4 = 2t3(1 – t3) = 2 × 0.499 672 3 × (1 – 0.499 672 3) = 0.499 999 8 t5 = 2t4(1 – t4) = 2 × 0.499 999 8 × (1 – 0.499 999 8) = 0.5 t6 = 2t5(1 – t5) = 2 × 0.5 × (1 – 0.5) = 0.5

5

The terms of the sequence are growing closer and closer to each other, finally settling at 0.5.

The sequence is convergent; the limit of the sequence is 0.5.

Method 2: Using technology 1

On the Sequence screen, complete the sequence as shown, tapping V to type the sequence.

Chapter 6

Sequences and series

183

2

To create the table, tap #. To find the required terms, tap r. Note: The first term is when n = 2, thus the terms are n - 1.

Note that instead of saying ‘the limit of the sequence is 0.5’ in the previous example, we could simply write tn → 0.5. REMEMBER

1. A sequence is a string of numbers or expressions which may follow a recognisable pattern. 2. A sequence can be described in a number of ways. (a) As a list — for example: tn: {1, 7, 2, 6, . . .} (Note: t3 = 2) (b) As a function — for example: tn = 2n − n2, n ∈ {1, 2, 3, . . .} (Note: t5 = 2 × 5 − 52 = −15) (c) As a recursive or iterative formula — for example: tn + 1 = 2tn − 3, t1 = 6 (Note: t2 = 2 × 6 − 3 = 9) 3. The logistic equation is a model of population growth and is an example of an iterative definition. It is of the general form: tn + 1 = atn(1 – tn), where 0 < t0 < 1 and a is a constant. 4. A sequence that converges to (settles at) a certain fixed number, x (the limit of the sequence) is called convergent. This can be written as tn → x. A sequence whose terms grow further and further apart is called divergent. That is, a sequence is divergent if tn → ∞, or tn → −∞ as n → ∞. A sequence whose terms fluctuate between two (or more) values is called oscillating. An oscillating sequence is neither convergent nor divergent. Exercise

6A

Describing sequences 1 WE 1a For each of the following sequences, write a rule for obtaining the next term in the sequence and hence evaluate the next three terms. a {1, 4, 7, . . .} b {1, 0, −1, −2, . . .} c {1, 4, 16, 64, . . .} e g i k

184

{2, −5, 8, −11, 14, . . .} {3, 4, 7, 11, 18, . . .} {1, 0, −1, 0, 1, . . .} {1024, −512, 256, −128, . . .}

3 3

d {3, 2 , 4 , . . .} f {2, 5, 9, 14, 20, . . .} h {2a − 5b, a − 2b, b, −a + 4b, . . .} j {1.0, 1.1, 1.11, . . .}


2 WE 1b Find the first, fifth and tenth terms in the following sequences. a tn = 2n − 5, n ∈ {1, 2, 3, . . .} b tn = 4 × 3n − 2, n ∈ {1, 2, 3, . . .} n c tn = , n ∈ {1, 2, 3, . . .} d tn = 17 − 3.7n, n ∈ {1, 2, 3, . . .} n +1 e tn = 5 ×

 1   2

n

, n ∈ {1, 2, 3, . . .}

f tn = 5 ×  1  2 − 3n2 n,

(−1)n

(3 − n)

, n ∈ {1, 2, 3, . . .}

g tn = + n, n ∈ {1, 2, 3, . . .} h tn = n ∈ {1, 2, 3, . . .} i tn = n2 − n + 41, n ∈ {1, 2, 3, . . .} j tn = a + (n − 1)d, n ∈ {1, 2, 3, . . .} k tn = arn − 1, n ∈ {1, 2, 3, . . .} 3 WE 1c Using a CAS calculator, or other method, find the third, eighth and tenth terms in the following sequences. a un + 1 = un + 2, u1 = 3

b un + 1 = un − 2, u1 =

c un + 1 = 3un, u1 = 0.85

d un + 1 = −2un, u1 = −3

3

e un + 1 = 4 un, u1 =

4 3

1 2

f un + 1 = un − 7, u1 = 14

g un + 1 = −un + 2, u1 = 3

h un + 1 = un + (−1)nun, u1 = 3

i un = 2un − 1, u1 =

j un + 1 = aun + a, u1 = a

−1 4

k un + 2 = un + 1 + un, u1 = 1, u2 = 1

l un + 1 = −un2 − 2, u1 = 3

4 WE2 Given the following values of a and t0, use the logistic equation to generate a sequence of six terms. State whether the sequence is convergent, divergent, or oscillating. If the sequence is convergent, state its limit. a a = 0.8, t0 = 0.5 b a = 0.4, t0 = 0.6 c a = 1.1, t0 = 0.9 d a = 1.9, t0 = 0.4 e a = 2.1, t0 = 0.5 f a = 2.5, t0 = 0.3 g a = 3, t0 = 0.2 h a = 3.4, t0 = 0.7 i a = 4.2, t0 = 0.1 j a = 4.5, t0 = 0.8 5 Study the pattern in each of the following sequences and where possible write the next two terms in the sequence, describing the pattern that you use. a 5, 6, 8, 11, . . . b 4, 9, 12, 13, 12, 9, . . . c 9, 8, 9, 0, . . . 1

d 6, 12, 12, 6, 1 2 , . . .

e 5, 8, 13, 21, . . .

f 1, 3, 7, 15, . . .

g 1, 3, 2, 4, 3, . . . 6

MC a Which of the following functional definitions could be used to describe a

sequence {3, 1, −1, . . .}? A tn = n − 2, n ∈ {1, 2, 3, . . .} b tn = 2n − 5, n ∈ {1, 2, 3, . . .} c tn = 5n − 2, n ∈ {1, 2, 3, . . .} d tn = 5 − 2n, n ∈ {1, 2, 3, . . .} e tn = 2(5 − n), n ∈ {1, 2, 3, . . .} b Which of the following recursive definitions could be used to describe a sequence {20, −10, 5, . . .}? t t A tn + 1 = tn − 30, t1 = 20 B tn + 1 = n , t1 = −20 C tn + 1 = tn − n , t1 = 20 2 2 − tn D tn + 1 = tn − 10, t1 = 20 E tn + 1 = , t1 = 20 2 6 n2 − 12 c Which of the following sequences is generated by the definition tn = , 2 n ∈ {1, 2, 3, . . .}? A {−3, 6, 15, . . .} D {−3, 6, 12, . . .}

B {−3, 6, −12, . . .} E {−3, 6, 18, . . .}

C {−3, 6, 21, . . .}

7 Write the iterative definition for each of the following sequences. a {7, 5, 3, 1, −1, . . .} b {12, 6, 3, 1.5, . . .} c {12, 12.6, 13.2, . . .} f {2, 4, 16, 256, . . .} d {2, 11, 56, 281, . . .} e {4, −12, 36, . . .}


185

8 In the township of Grizabella, the population of stray cats in any given year is given as pn + 1. This can be calculated using the formula pn + 1 = 1.3pn(1 – pn), where pn is the number of cats (in hundreds) in the preceding year. If in 2005 there were 28 stray cats in Grizabella township, calculate: a the expected number of stray cats for 2006 and 2007 b the limiting number of stray cats that Grizabella township can sustain.


History of mathematics Leonardo Fibonacci

6B

9 In the neighbouring township of Macavity, the size of the population of stray cats follows the logistic equation pn + 1 = 0.3pn(1 − pn), where pn + 1 and pn refer to the population size (in hundreds) in any given year and in the preceding year respectively. It is known that in 2005, there were 62 stray cats in the township. By generating and examining the sequence of numbers using the above equation, decide what will happen in the long run to the population size of stray cats in Macavity township. (That is, will the population of cats keep increasing, decreasing, or settle at a particular value?)

arithmetic sequences At a racetrack a new prototype racing car unfortunately develops an oil leak. Each second, a drop of oil hits the road. The driver of the car puts her foot on the accelerator and the car increases speed at a steady rate as it hurtles down the straight. The diagram below shows the pattern of oil drops on the road with the distances between the drops labelled.

10 metres

18 metres

186

26 metres

42 metres

34 metres


The sequence of distances travelled in metres each second is {10, 18, 26, 34, 42, . . .}. The first term in the sequence, t1, is 10 and as you can see, each subsequent term is 8 more than the previous term. This type of sequence is given a special name — an arithmetic sequence. An arithmetic sequence is a sequence where there is a common difference between any two successive terms. We can list the sequence in a table as in table A. From this table we can see that it is possible to write a functional definition for the sequence in terms of the first term, 10, and the common difference, 8, and thus:

tn = 10 + (n − 1) × 8 = 2 + 8n, n ∈ {1, 2, 3, . . .}

We can readily get a general formula for the nth term of an arithmetic sequence whose first term is a and whose common difference is d (see table B). Table A n 1 2 3 4 n

tn 10 + 0 × 8 10 + 1 × 8 10 + 2 × 8 10 + 3 × 8 10 + (n − 1) × 8 = 10 + 8n − 8 = 2 + 8n

Table B n 1 2 3 4 n

tn 10 18 26 34 2 + 8n

tn a+0×d a+1×d a+2×d a+3×d a + (n − 1) × d = (a − d) + dn

In general then: The nth term of an arithmetic sequence is given by tn = a + (n − 1) × d = (a − d ) + nd, n ∈ {1, 2, 3, . . .} where a is the first term and d is the common difference. If we consider three successive terms in an arithmetic sequence, namely x, y and z, then since y − x = the common difference, d, and z − y = d, it follows that: z+x y−x=z−y⇒y= 2 The middle term of any three consecutive terms in an arithmetic sequence is called an arithmetic mean and is the average of the outer two. z+ x That is, y = for any 3 consecutive terms, x, y, z of the arithmetic sequence. 2

Worked Example 3

Show that the following sequences are arithmetic. − 7 − 11 a 4 , 8 , −1, . . . b x2 − 4x, 3x2 − 7x, 5x2 − 10x, . . . Think a

1

Write

To show that a sequence is arithmetic you need to show that the difference between any two successive terms is a constant. Find the difference between the first and the second terms.

a t2 − t1

= = =

− 11 8

- 

− 7  4

− 11 + 14

8 3 8


187

2

3

b

1

2

3

Find the difference between the second and the third terms.

Compare the differences and draw your conclusion.

t3 − t2

= -1 - 

=3

− 11  8

8

t2 − t1 = t3 − t2 =

3 8

∴ The sequence is arithmetic.

Find the difference between the first two terms.

b t2 − t1

= 3x2 − 7x − (x2 − 4x) = 2x2 − 3x

Find the difference between the second and the third terms.

t3 − t2

= 5x2 − 10x − (3x2 − 7x) = 2x2 − 3x

Compare the differences and draw your conclusion.

t2 − t1 = t3 − t2 = 2x2 − 3x ∴ The sequence is arithmetic.

Worked Example 4

State which of the following are arithmetic sequences by finding the difference between successive terms. For those which are arithmetic, find the next term in the sequence, t4, and consequently find the functional definition for the nth term for the sequence, tn. a t: {4, 9, 15, . . .} b t: {−2, 1, 4, . . .} Think a

b

1

To check that a sequence is arithmetic, see if a common difference exists.

2

There is no common difference, as 5 ≠ 6.

1

To check that a sequence is arithmetic, see if a common difference exists.

a 9−4=5

15 − 9 = 6 Since there is no common difference the sequence is not arithmetic.

b 1 − −2 = 3

4 − 1 = 3

2

The common difference is 3.

The sequence is arithmetic with the common difference d = 3.

3

The next term in the sequence, t4, can be found by adding 3 to the previous term, t3.

t4 = t3 + 3 =4+3 =7

4

To find the functional definition, write the formula for the nth term of the arithmetic sequence.

tn = a + (n − 1) × d = (a − d) + nd

5

Identify the values of a and d.

a = −2 and d = 3

tn = (−2 − 3) + n × 3 tn = 3n − 5

6

188

Write

Substitute a = and simplify.

−2

and d = 3 into the formula


Worked exampLe 5

eBook plus

Find the missing terms in this arithmetic sequence: {41, a, 55, b, . . .}. ThInk

Tutorial

int-1052

WrITe

Worked example 5

x+z 2

1

The first three successive terms are 41, a, 55. Write the rule for the middle term of the three successive terms of an arithmetic sequence.

For x, y, z: y =

2

Identify the variables.

x = 41; y = a; z = 55

3

Substitute the values of x, y and z into the formula in step 1 and evaluate.

a=

4

Find the common difference. (The second term is now known.)

d = t2 − t1 = 48 − 41 =7

5

Find the value of b by adding the common difference to the preceding term.

b = 55 + 7 = 62

6

State your answer.

So a = 48, b = 62

41 + 55 2 = 48

Worked exampLe 6

Find the 16th and nth terms in an arithmetic sequence with the 4th term 15 and 8th term 37. ThInk

WrITe


Write the formula for the nth term of the arithmetic sequence.

tn = a + (n − 1) × d

2

Substitute n = 4 and t4 = 15 into the formula and label it equation [1].

t4: a + 3d = 15

[1]

3

Substitute n = 8 and t8 = 37 into the formula and label it equation [2].

t8: a + 7d = 37

[2]

4

Solve the simultaneous equations: subtract equation [1] from equation [2] to eliminate a.

5

Divide both sides by 4.

[2] − [1]: a + 7d − a − 3d = 37 − 15 4d = 22 22 d= 4 1

= 52 6

1

Substitute d = 5 2 into equation [1], and solve for a.

1

Substituting d = 5 2 into [1]: 1

a + 3 × 5 2 = 15 1

a = −1 2

Chapter 6


189

7

To find the nth term of the arithmetic sequence, substitute the values of a and d into the general formula and simplify.

1

= =

8

To find the 16th term, substitute n = 16 into the formula, established in the previous step and evaluate.

1

tn = −1 2 + (n − 1) × 5 2 −3 2

+ (n - 1)

11 2

− 3 + 11n − 11

2 11n − 14 tn = , n ∈ {1, 2, 3, . . .} 2 11 × 16 − 14 If n = 16, t16 = 2 = 81


Write the two equations that represent t4 and t8.

2

To solve equations [1] and [2] simultaneously, on the Main screen complete the entry line as shown. Then press E.

3

Write the answer.

t4: a + 3d = 15 t8: a + 7d = 37

[1] [2]

−3

11 and d = , 2 2 11n − 14 tn = 2

If a =

t16 = 81

REMEMBER

1. An arithmetic sequence is one where successive terms have a common difference. This common difference is given the symbol d. Thus tn + 1 − tn = d for all values of n. The first term in the sequence is given the symbol a. 2. If x, y, z are successive terms in an arithmetic sequence then y is called an arithmetic x+z . mean and is given by y = 2 That is, the middle term is the average of the outer two terms. 3. An arithmetic sequence can be written as a, a + d, a + 2d, . . . and so the nth term tn is: tn = a + (n − 1)d using the function notation, or tn + 1 = tn + d, t1 = a using the iterative notation.

190


exerCISe

6B

arithmetic sequences 1 We3 Show that the following sequences are arithmetic. a {−12, −7, −2, . . .} b {−0.12, 3.48, 7.08, . . .} −3 3 9

d {2.3, −1.7, −5.7, . . .}

c { 8 , 8 , 8 , . . .} 5 −1 −7 , 9, 9

. . .}

f {18, −18, −54, . . .}

13

h {x + 9, 2x + 7, 3x + 5, . . .}

e {9,

2

4

g {5 3 , 715, 815, . . .}

i {3x2 − 4x, 5x2 − 2x, 7x2, . . .} j {3(2 − x), 2(2 − x), 2 − x, . . .} 2 We4 State which of the following are arithmetic sequences by finding the difference between successive terms. For those which are arithmetic, find the next term in the sequence, t4, and consequently find the functional definition for the nth term for the sequence, tn. a tn: {3, 5, 7, . . .} b tn: {4, 7, 11, . . .} d tn: {−3, 0, 3, . . .} c tn: {3, 6, 12, . . .} e tn: {−2, −6, −10, . . .} 3 3 3

3 3 9

g tn: { 4 , 2 , 1 , . . .} 1 − 3 13 , −4, 2

i tn: { 4 ,

2 11 9

f tn: { 7 , 14 , 7 , . . .}

. . .}

h tn: { 4 , 2 , 4 , . . .} j tn: {2π + 3, 4π + 1, 6π − 1, . . .}

3 Find the term given in brackets for each of the following arithmetic sequences. a {4, 9, 14, . . .}, (t21) b {−2, 10, 22, . . .}, (t58) − − c { 27, 12, 3, . . .} (t100) d {2, −11, −24, . . .} (t2025) 4

Find the functional definition for the nth term of the following arithmetic sequences: a where the first term is 5 and the common difference is −3 1 b where the first term is 2.5 and the common difference is 2 − c where the first term is 3 and the common difference is 3 d where the first term is 2x and the common difference is 5x. 5 Find the nth term in the arithmetic sequence where the first term is 6 and the third term is 10. 6 Find the nth term in the arithmetic sequence where the first term is 3 and the third term is 13. 7 We5 Find the missing terms in this arithmetic sequence: {16, m, 27, n} 8 Find the missing terms in the arithmetic sequence below. x − 3y, _______________, −3x + 5y, _______________, . . . 9 We6 Find the 4th term and nth term in the arithmetic sequence whose first term is 6 and whose 7th term is −10.

10 If t10 = 100 and t15 = 175, find the first term, the common difference and hence the nth term for the arithmetic sequence. −1

3

11 If t10 = 2 and t13 = 4, find the first term, the common difference and hence the nth term for the arithmetic sequence.


SkillSHEET 6.1 Using elimination to solve simultaneous equations

12 Insert four evenly spaced numbers between 8 and 36.

13 For the arithmetic sequence {22, m, n, 37, . . .}, find the values for m and n. 14 For the following arithmetic sequences, find the iterative definition and use it in a CAS calculator to generate the first 50 numbers in the sequence. a tn: {3, 7, 11, . . .} b tn: {−3, 0, 3, . . .} c tn: {−2, −6, −10 . . .} 3 3 9

e tn: { 4 , 2 , 4 , . . .}

2 11 9

d tn: { 7 , 14 , 7 , . . .} 1 − 3 − 13 , 4, 2

f tn: { 4 ,

. . .}

g tn: {2π + 3, 4π + 1, 6π − 1, . . .}

Chapter 6


191

15 The first three terms in an arithmetic sequence are 37, 32, 27 and the kth term is −3. Find the value for k. 16 Find the value of x such that the following forms an arithmetic progression: . . . x, 3x + 4, 10x − 7 . . . 17 MC For the following sequence t: {4, 11, 18, . . .}, the difference between the 4th and the 10th term is: A 35 B 42 C 49 D 56 E 63 18 MC The tenth term in an arithmetic sequence is 12 and the third term is −2. The first term in the sequence is: A −7 B −3 C −5 D −8 E −6 3

19 The ratio between the first term and the second term in an arithmetic sequence is 4 . The ratio between the second term and the third term is 4 . 5 a Calculate the ratio of the third term to the fourth term. b Find the ratio of the nth and the nth + 1 term in the sequence.

6c

Arithmetic series Often we have a sequence of numbers and we wish to know their sum. For an example, we return to the oil drops on the racetrack from the start of the previous section on arithmetic sequences. The distance covered by the car each second illustrated the concept of an arithmetic sequence. The total distance covered by the car is the sum of the individual distances covered each second. So after one second the car has travelled 10 m, after 2 seconds the car has travelled 10 + 18 m = 28 m, after three seconds the car has travelled a total distance of 10 + 18 + 26 m = 54 m, and so on. 10 metres

18 metres

28 m

26 metres

54 m

34 metres

42 metres

88 m

130 m

A series, Sn, is the sum of a sequence of n terms t1 + t2 + t3 + . . . + tn. Thus:

S1 = t1 S2 = t1 + t2 S3 = t1 + t2 + t3 Sn = t1 + t2 + t3 + . . . + tn − 2 + tn − 1 + tn.

For an arithmetic sequence, the sum of the first n terms, Sn, can be written in two ways: 1. The first term in the arithmetic sequence is a, the common difference is d, and the last term — that is, the nth term — in the sequence is l.

Sn = a + (a + d ) + (a + 2d ) + . . . (a + (n − 3)d ) + (a + (n − 2)d + a + (n − 1)d = a + (a + d ) + (a + 2d ) + . . . (l − 3d ) + (l − 2d ) + (l − d ) + l

[1]

2. We can write the sum Sn in reverse order starting with the nth term and summing back to the first term a: Sn = l + (l − d ) + (l − 2d ) + . . . + (a + 2d ) + (a + d ) + a. [2]

192


If we add equation [1] and equation [2] together and recognise that there are n terms each of which equal (a + l ) we get: 2Sn = (a + l ) + (a + l ) + . . . n times = n(a + l ) n and so: S n = (a + l ) 2 n or since l is the nth term, l = a + (n − 1)d, so Sn = [a + a + ( n − 1)d ] 2 n Sn = [2a + ( n − 1)d ] 2 The sum of the first n terms in the arithmetic sequence is given by n Sn = ( a + l ) 2 where a is the first term and l is the last term; or alternatively, since l = a + (n − 1)d, by n Sn = ( 2 a + ( n − 1) d ) 2 where a is the first term and d is the common difference. If we know the first term, a, the common difference, d, and the number of terms, n, that we wish to add together we can calculate the sum directly without having to add up all the individual terms. It is worthwhile also to note that Sn + 1 = Sn + tn + 1. This tells us that the next term in the series Sn + 1 is the present sum Sn plus the next term in the sequence tn + 1. This result is useful in spreadsheets where one column gives the sequence and an adjacent column is used to give the series. Worked exampLe 7

eBook plus

Find the sum of the first 20 terms in the sequence tn: {12, 25, 38, . . .}. ThInk

Write the formula for the sum of the first n terms in the arithmetic sequence.

2

Identify the variables.

3

Substitute values of a, d and n into the formula and evaluate.

int-1053

WrITe


Tutorial

Sn =

Worked example 7

n (2a + ( n − 1)d ) 2

a = 12, d = 25 − 12 = 13, n = 20 20 (2 × 12 + 19 × 13) 2 S20 = 2710

S20 =


On the Sequence screen, complete the sequence as shown, tapping V to type the sequence. To create the table, tap #.

Chapter 6


193

2

To find the required terms, tap r.

3

Write the answer.

S20 = 2710

REMEMBER

n (2a + ( n − 1)d ) and 2 so to find Sn the values for a and d need to be found for the sequence whose series is required. 2. In general, tn + 1 = Sn + 1 − Sn. 1. The sum of the first n terms of an arithmetic sequence is Sn =

Exercise

6C

Arithmetic series 1 WE 7 Consider the following sequences and find the sums of the terms as indicated. a tn: {1, 2, 3, . . .}. Find S10, S50, S100. b tn: {1, 3, 5, . . .}. Find S5, S10, S20. c tn = 3n + 7, n ∈ {1, 2, 3, . . .}. Find S5, S10, Sn. d tn = −4n + 15, n ∈ {1, 2, 3, . . .}. Find S5, S10, Sn. e tn + 1 = tn + 5.5, t1 = 2.5. Find S5, S10, S20. f tn + 1 = tn + π, t1 = 2π. Find S5, S10, S20. g The first term is 4 and the common difference is 3. Find S4, S16, S64. 1 h The first term is 14 and the common difference is −3 2 . Find S4, S9, S14. i The first term is 50 and the 10th term is −40. Find S10. j The 5th term is 10 and the 8th term is 16. Find S5, S50, S500. 2 a Find the sum of the first 50 positive integers. b Find the sum of the first 100 positive integers. 3 a Find the sum of all the half-integers between 0 and 100. 1 1 1 1 Note: The sequence of half-integers is { 2 , 12 , 2 2 , 3 2 , . . .} b Compare your answer with that for question 2b. 4 Find the sum of the first 12 terms of an arithmetic sequence in which the second term is 8 and thirteenth term is 41. 5 A sequence of numbers is defined by tn: {15, 9, 3, −3, . . .}. a Find the sum of the first 13, 16 and 19 terms in the sequence. b Find the sum of all the terms between and including t10 and t15.

194


6 A sequence of numbers is defined by tn = 2n − 7, n ∈ {1, 2, 3, . . .}. Find a the sum of the first 20 terms b the sum of all the terms between and including t21 and t40 c the average of the first 40 terms. Hint: You need to find the sum first. 7 Find the equation that gives the sum of the first n positive integers. 8 a Show that the sum of the first n odd integers is equal to the perfect square n2. b Show that the sum of the first n even integers is equal to n2 + n. 9 A sequence is 5, 7, 9, 11, . . . How many consecutive terms need to be added to obtain 357? 10 Consider the sum of the first n integers. For what value of n will the sum first exceed 1000? 11 a Find the sum of all integers divisible by 3 which lie between 200 and 400. b Find the sum of all integers divisible by 6 which lie between 200 and 400.


WorkSHEET 6.1

6d

12 The first term in an arithmetic sequence is 5 and the sum of the first 20 terms is 1240. Find the common difference, d. 13 The sum of the first four terms of an arithmetic sequence is 58, and the sum of the next four terms is twice that number. Find the sum of the following four terms. 14 The sum of a series is given by Sn = 4n2 + 3n. Use the result that tn + 1 = Sn + 1 − Sn to prove that the sequence of numbers, tn, whose series is Sn = 4n2 + 3n is arithmetic. Find both the functional and iterative equations for the sequence, tn.

Geometric sequences A farmer is breeding worms which he hopes to sell to local shire councils for use in the decomposition of waste at rubbish dumps. Worms reproduce readily and the farmer expects a 10% increase per week in the mass of worms that he is farming. A 10% increase per week would mean that the mass of worms would increase by a constant factor of 10 (1 + 100 ) or 1.1. He starts off with 10 kg of worms. By the beginning of the second week he will expect 10 × 1.1 = 11 kg of worms, by the start of the third week he would expect 11 × 1.1 = 10 × (1.1)2 = 12.1 kg of worms, and so on. This is an example of a geometric sequence. A geometric sequence is the sequence where each term is obtained by multiplying the preceding term by a certain constant factor. The first term is 10 and the common factor here is 1.10 which represents a 10% increase on the previous term. We can put the results of the above example into a table.

Chapter 6


195

From this table we can see that t2 = 1.1 × t1, t3 = 1.1 × t2 1 10 × (1.1)0 10 and so on. In general: 2 10 × (1.1)1 11 tn + 1 = 1.1 × tn The common factor or common ratio 12.1 3 10 × (1.1)2 whose value is 1.1 for this example can be 4 10 × (1.1)3 13.31 found by dividing any two successive tn +1 10 × (1.1)n − 1 n 10 × (1.1)n − 1 terms: . tn A geometric sequence, t, can be written in terms of the first term, a, and the common ratio, r. Thus: t: {a, ar, ar2, ar3, . . . , arn − 1, . . .} The first term t1 = a, the second term t2 = ar, the third term t3 = ar2 and consequently the nth term, tn is arn − 1. n

tn

tn

For a geometric sequence: tn = arn − 1 where a is the first term and r the common ratio, given by tn + 1 r= tn If we consider three consecutive terms in a geometric sequence, x, y and z, then y z =r= x y where r is the common factor. Thus the middle term, y, called the geometric mean, can be calculated in terms of the outer two terms, x and z. For a geometric sequence . . . , x, y, z, . . . :

y2 = xz

Worked Example 8

State whether the sequence is geometric by finding the ratio of successive terms: tn: {2, 6, 18, . . .}. If it is geometric, find the next term in the sequence, t4, and the nth term for the sequence, tn. Think

196

Write

1

Find the ratio

t2 . t1

t2 6 = t1 2 =3

2

Find the ratio

t3 . t2

t3 18 = t2 6 =3

3

Compare the ratios and make your conclusion.

4

Since the sequence is geometric, to find the fourth term, multiply the preceding (third) term by the common ratio.

t4 = t3 × r = 18 × 3 = 54

5

Write the general formula for the nth term of the geometric sequence.

tn = arn − 1

t 2 t3 = = 3, the sequence is t1 t2 geometric with the common ratio r = 3. Since


6

Identify the values of a and r.

a = 2; r = 3

7

Substitute the values of a and r into the general formula.

tn = 2 × 3n − 1

Worked exampLe 9

Find the nth term and the 10th term in the geometric sequence, where the first term is 3 and the third term is 12. ThInk

WrITe

1

Write the general formula for the nth term in the geometric sequence.

tn = arn − 1

2

State the value of a (the first term in the sequence) and the value of the third term.

a = 3; t3 = 12

3

Substitute all known values into the general formula.

12 = 3 × r3 − 1 = 3 × r2

4

Solve for r (note that there are two possible solutions).

r2 = 3 =4 r= ± 4 = ±2

5

Substitute the values of a and r into the general equation. Since there are two possible values for r, you must show both expressions for the nth term of the sequence.

So tn = 3 × 2n − 1, or tn = 3 × (−2)n − 1

6

Find the 10th term by substituting n = 10 into each of the two expressions for the nth term.

When n = 10, t10 = 3 × 210 − 1 (using r = 2) = 3 × 29 = 1536 or t10 = 3 × (−2)10 − 1 (using r = −2) = 3 × (−2)9 = −1536

12

Worked exampLe 10

eBook plus

The fifth term in a geometric sequence is 14 and the seventh term is 0.56. Find the common ratio, r, the first term, a, and the nth term for the sequence. ThInk

Tutorial


WrITe

1

Write the general rule for the nth term of the geometric sequence.

tn = arn − 1

2

Use the information about the 5th term to form an equation. Label it [1].

When n = 5, tn = 14 14 = a × r5 − 1 14 = a × r4

[1]

When n = 7, tn = 0.56 0.56 = a × r 7 − 1 0.56 = a × r6

[2]

3

Similarly, use information about the 7th term to form an equation. Label it [2].

Chapter 6


197

4

Solve equations simultaneously: Divide equation [2] by equation [1] to eliminate a.

[2] ar 6 0.56 gives = [1] 14 ar 4

5

Solve for r.

r2 = 0.04 r = ± 0.04 = ± 0.2

6

Since there are two solutions, we have to perform two sets of computations. Consider the positive value of r first. Substitute the value of r into either of the two equations, say equation [1], and solve for a.

If r = 0.2 Substitute r into [1]: a × (0.2)4 = 14 0.0016a = 14 a = 14 ÷ 0.0016 = 8750

7

Substitute the values of r and a into the general equation to find the expression for the nth term.

The nth term is: tn = 8750 × (0.2)n − 1

8

Now consider the negative value of r.

If r = −0.2

9

Substitute the value of r into either of the two equations, say equation [1], and solve for a. (Note that the value of a is the same for both values of r.)

Substitute r into [1] a = (-0.2)4 = 14 0.0016a = 14 a = 14 ÷ 0.0016 = 8750

10

Substitute the values of r and a into the general formula to find the second expression for the nth term of the sequence.

The nth term is: tn = 8750 × (−0.2)n − 1

11

Write the two equations that represent t5 and t7.

t5: 14 = a × r4 t7: 0.56 = a × r6

12

To solve equations [1] and [2] simultaneously, on the Main screen complete the entry line as shown. Then press E.

13

Write the answer.

When r = -0.2 and a = 8750, tn = 8750 × (-0.2)n − 1 When r = 0.2 and a = 8750, tn = 8750 × (0.2)n − 1

198


[1] [2]

rememBer

1. A geometric sequence is one where each successive term is obtained by multiplying the preceding term by the constant number. This number is called the common ratio t and is given the symbol r. Thus n + 1 = r for all values of n. tn The first term in the sequence is given the symbol a. 2. If x, y, z are successive terms in the geometric sequence then y is called a geometric mean and is given by y2 = xz. 3. A geometric sequence can be written as a, ar, ar2, . . . and so the nth term tn is: tn = ar n − 1 using the function notation, or tn + 1 = rtn, t1 = a using the iterative notation. exerCISe

6d

Geometric sequences

eBook plus

1 We8 State which of the following are geometric sequences by finding Digital doc the ratio of successive terms. For those which are geometric, find the Spreadsheet 036 next term in the sequence, t4 and the nth term for the sequence, tn. Fibonacci sequences a tn: {3, 6, 9, . . .} b tn: {4, 12, 36, . . .} c tn: {3, 6, 12, . . .}

d tn: {4, 6, 9, . . .}

f tn: {2, −6, 18, . . .}

g tn: { 7 , 14 , 14 , . . .}

3 3 9

i tn: { 4 , 2 , 4 , . . .}

2

6

1 −3 , 2

j tn: { 4 ,

e tn: {−3, 1,

−1

3

, . . .}

9

h tn: { 4 , 2 , 1, . . .}

3 3 3

9, . . .}

k tn: {2π, 4π 2, 8π 3, . . .}

2 For each of the following: i show that the sequence is geometric ii find the nth term and consequently the 6th and the 10th terms. a t: {5, 10, 20, . . .} b t: {2, 5, 12.5, . . .} c t: {1, −3, 9, . . .} d t: {2, −4, 8, . . .} e t: {2.3, 3.45, 5.175, . . .} 1

1

g t: {3 , 12,

1 , 48

. . .}

1

f t: { 2 , 1, 2, . . .} 3 −1 1 , , 5 15

h t: { 5,

. . .}

1 2 4 j t : { , 2 , 3 , . . .} x x x 3 We9 Find the nth term and the 10th term in the geometric sequence where: a the first term is 2 and the third term is 18 (Why are there two possible answers?) b the first term is 1 and the third term is 4 (Why are there two possible answers?) c the first term is 5 and the fourth term is 40 d the first term is −1 and the second term is 2 1 e the first term is 9 and the third term is 81. (Why are there two possible answers?) i t: {x, 3x4, 9x7, . . .}

4 Find the 4th term in the geometric sequence where the first term is 6 and the 7th term is 3 . 32 5 Find the nth term in the geometric sequence where the first term is 3 and the fourth term is 6 2. 6 For the geometric sequence 3, m, n, 192, . . . , find the values for m and n. 7 Consider the geometric sequence t: {16, m, 81, n, . . .}. Find the values of m and n, if it is known that both are positive numbers.

Chapter 6


199

8 For the geometric sequence a, 15, b, 0.0375, . . . , find the values of a and b, given that they are positive numbers. 9 WE 10 The third term in a geometric sequence is 100 and the fifth term is 400. Find the common ratio, r, the first term, a, and the nth term for the sequence. 1

27

10 If t2 = 2 and t5 = 16 , find the first term, a, the common factor, r, and hence the nth term for the geometric sequence. 11 Find the value of x such that the following sequence forms a geometric progression: x − 1, 3x + 4, 6x + 8. 12 Insert three terms in between 8, _, _, _,

1 32

such that the sequence of numbers is geometric.

13 The difference between the first term and the second term in a geometric sequence is 6. The difference between the second term and the third term is 3. a Calculate the difference between the third term and the fourth term. b Find the nth term in the sequence. 14 The first two terms in a geometric sequence are 120, 24, and the kth term is 0.0384. Find the value for k.

6e

Geometric series When we add up or sum the terms in a sequence we get the series for that sequence. If we look at the geometric sequence {2, 6, 18, 54, . . .} where the first term t1 = a = 2 and the common ratio is 3 we can quickly calculate the first few terms in the series of this sequence.

S1 = t1 = 2 S2 = t1 + t2 = 2 + 6 = 8 S3 = t1 + t2 + t3 = 2 + 6 + 18 = 26 S4 = t1 + t2 + t3 + t4 = 2 + 6 + 18 + 54 = 80

In general the sum of the first n terms is:

Sn = t1 + t2 + t3 + . . . + tn − 2 + tn − 1 + tn.

For a geometric sequence the first term is a, the second term is ar, the third term is ar 2 and so on up to the nth term which is arn − 1. Thus:

Sn = a + ar + ar 2 + . . . + ar n − 3 + ar n − 2 + ar n − 1

[1]

If we multiply equation [1] by r we get:

rSn = ar + ar 2 + ar 3 + . . . ar n − 2 + ar n − 1 + ar n

Note that on the right-hand side of equations [1] and [2] all but two terms are common, namely the first term in equation [1], a, and the last term in equation [2], arn. If we take the difference between equation [2] and equation [1] we get: rSn − Sn = ar n − a [2] − [1] ∴ (r − 1)Sn = a(r n − 1) a(r n − 1) ∴ Sn = ; r ≠1 (r cannot equal 1) r −1 We now have an equation which allows us to calculate the sum of the first n terms of a geometric sequence. The sum of the first n terms of a geometric sequence is given by: a( r n − 1) Sn = ;r ≠1 r −1 where a is the first term of the sequence and r is the common ratio.

200


[2]

Worked Example 11

Find the sum of the first 5 terms (S5) of these geometric sequences. −1 1 a tn: {1, 4, 16, . . .} b tn = 2 (2)n − 1, n ∈ {1, 2, 3, . . .} c tn + 1 = 4 tn, t1 = 2 Write

Think


a Sn =

a(r n − 1) r −1

1

Write the general formula for the sum of the first n terms of the geometric sequence.

2

Write the sequence.

tn: {1, 4, 16, . . .}

3

Identify the variables: a is the first term; r can be established by finding the ratio; n is known from the question.

a = 1; r = 1 = 4; n = 5

4

Substitute the values of a, r and n into the formula and evaluate.

S5 =

4

1(4 5 − 1) 4 −1 1024 − 1 = 3 = 341

Method 2: Using technology a

1

On the Spreadsheet screen, type the initial value of 1 in cell A1. Complete the entry line in cell A2 as: = 4 × A1 Then press E.

2

To find the terms in the sequence highlight A2 to A5 and tap: • Edit • Fill Range


201

b

3

Tap OK.

4

To sum the sequence, in cell B1, tap: • Action • List-Calculation • sum

5

Highlight cells A1 to A5 and then press E. The answer will appear in cell B1.

6

Write the answer. Write the sequence. Compare the given rule with the general formula for the nth term of the geometric sequence tn = ar n − 1 and identify values of a and r; the value of n is known from the question.

1 2

3

202

Substitute values of a, r and n into the general formula for the sum and evaluate.

If tn: {1, 4, 16, . . .} then, S5 = 341.

b tn = 2(2)n − 1, n ∈ {1, 2, 3, . . .}

a = 2; r = 2; n = 5

2(25 − 1) 2 −1 2(32 − 1) = 1 = 62

S5 =


c

1

c tn + 1 = 4 tn, t1 =

1

Write the sequence.

2

This is an iterative formula, so the coefficient of tn is our r; a = t1; n is known from the question.

3

1

r = 4; a =

S5 =

Substitute values of a, r and n into the general formula for the sum and evaluate.

= =

−1 2

−1 2

;n=5

 − 1  1  5 − 1  2  4   1 −1 4 −1  1 × − 1 2  1024 −3 4 − 341 512

The infinite sum of a geometric sequence where r < 1 When the constant ratio, r, is less than 1 or greater than −1, that is, {r: −1 < r < 1}, each successive term in the sequence gets closer to zero. This can readily be shown with the following two examples. 1 −1 , 4

g: {2, −1, 2 , 1

. . .} where a = 2 and r =

1

−1 2 1

h: {40, 2 , 160 , . . .} where a = 40 and r = 80 In both the examples, successive terms approach zero as n increases. In the second case the approach is more rapid than in the first and the first sequence alternates positive and negative. A simple investigation with a spreadsheet will quickly reveal that for geometric sequences with the size or magnitude of r < 1 the series eventually settles down to a near constant value. We say that the series converges to a value S∞ which is the sum to infinity of all terms in the geometric sequence. We can find the value S∞ by recognising that as n → ∞ the term r n → 0, provided r is between −1 and 1. We write this technically as −1 < r < 1 or | r | < 1. The symbol | r | means the magnitude or size of r. Using our equation for the sum of the first n terms: a(r n − 1) Sn = ;r ≠1 r −1 Taking −1 as a common factor from the numerator and denominator: a(1 − r n ) Sn = 1− r n n As n → ∞, r → 0 and hence 1 − r → 1. Thus the top line or numerator will equal a when n → ∞: a S∞ = ; |r | < 1 1− r We now have an equation which allows us to calculate the sum to infinity, S∞ of a geometric sequence. The sum to infinity S∞ of the geometric sequence is given by: a ; | r| < 1 1− r where a is the first term of the sequence and r is the common ratio whose magnitude is less than one. S∞ =


203

Worked exampLe 12

eBook plus

a Find the sum to infinity for the sequence tn: {10, 1, 0.1, . . .}. b Find the fourth term in the geometric sequence whose first term is 6

Tutorial

int-1055

and whose sum to infinity is 10.

Worked example 12

ThInk a

b

WrITe a tn = ar n − 1

1

Write the formula for the nth term of the geometric sequence.

2

From the question we know that the first term, a, is 10 and r = 0.1.

a = 10,

3

Write the formula for the sum to infinity.

S∞ =

a ; |r | < 1 1− r

4

Substitute a = 10 and r = 0.10 into the formula and evaluate.

S∞ =

10 1− 0.1

S∞ =

10 0.9

b S∞ =

r = 0.1

=

100 9

1

= 11 9

a ; |r| < 1 1− r

1

Write the formula for the sum to infinity.

2

From the question it is known that the infinite sum is equal to 10 and that the first term a is 6. Write down this information.

a = 6; S∞ = 10

3

Substitute known values into the formula.

10 =

4

Solve for r.

10(1 − r) = 6 10 − 10r = 6 10r = 4 r = 0.4

5

Write the general formula for the nth term of the geometric sequence.

tn = ar n − 1

6

To find the 4th term substitute a = 6, n = 4 and r = 0.4 into the formula and evaluate.

t4 = 6 × (0.4)3 = 0.384

6 1− r

rememBer

1. The sum of the first n terms in a geometric sequence is:

or

Sn =

a(1 − r n ) with r ≠ 1 1− r

Sn =

a(r n − 1) with r ≠ 1 r −1

2. When the magnitude of r is less than one, that is, −1 < r < 1, the sum of a geometric sequence to infinity, S∞ is given by: a S∞ = 1− r

204


exerCISe

6e

Geometric series

eBook plus

1 We 11 Consider the following sequences and find the terms indicated. a tn: {1, 2, 4, . . .}. Find S5, S10, S20. b tn: {1, 3, 9, . . .}. Find S5, S10, S20. c tn = 3(−2)n − 1, n ∈ {1, 2, 3, . . .}. Find S5, S10, S20. d tn = −4(1.2)n − 1, n ∈ {1, 2, 3, . . .}. Find S1, S10, S20.

Digital doc

Spreadsheet 036 Fibonacci series

3

e tn + 1 = 2tn, t1 = 2 . Find S1, S5, S10. 1

f tn + 1 = 2 tn, t1 =

−2 3

. Find S1, S5, S10.

g The first term is 3000 and the common ratio is 1.05. Find S4, S16, S64. h The first term is 1400 and the common ratio is −1.1. Find S4, S9, S14. i The first term is 20; every other term is obtained by multiplying the preceding term by 5. Find S5, S10. −1 j The first term is −2; every other term is obtained by multiplying the preceding term by 2 . Find S5, S10. 2 Consider the following geometric sequences and find the terms indicated. a The first term is 440 and the 12th term is 880. Find S6. b The 5th term is 1 and the 8th term is 8. Find S1, S10, S20. 3 Find the sum of the first 12 terms of a geometric sequence in which the 8 second term is 3 and the fifth term is 9. 1 4 What minimum number of terms of the series 2 + 3 + 4 2 + . . . must be taken to give a sum in excess of 100? 5 The sum of the first four terms of a geometric sequence is 312, and the sum of the next four terms is 625 times that number. Find the sum of the following four terms. 6 Find the sum of all powers of 2 between 500 and 50 000. 7 Find the sum of all powers of 4 between 500 and 50 000. 8 We 12a

Find the sum to infinity for the following geometric sequences. b tn: {1,

1 1

d tn: {1, 3, 9 , . . .}

−2 4 −8 , , 9, 3 9

. . .}

2 4

c tn: {1, 3, 9 , . . .} e tn: {1,

−1 1 −1 , , , 2 4 8

1 1

a tn: {1, 2 , 4 , . . .}

. . .} 1 1 1

9 For the infinite geometric sequence { 2 , 4 , 8 , . . .}, find the sum to infinity. Consequently, find what proportion each of the first three terms contributes to this sum as a percentage. 1

1

3

9

10 For the infinite geometric sequence {1, 4 , 16 , . . .}, find the sum to infinity. Consequently, find what proportion each of the first three terms contributes to this sum as a percentage. 11 For the infinite geometric sequence {1, 4 , 16 , . . .}, find the sum to infinity. Consequently, find what proportion each of the first three terms contributes to this sum as a percentage.  1

n −1

12 A sequence of numbers is defined by tn = 3  2  , n ∈ {1, 2, 3, . . .}. a Find the sum of the first 20 terms. b Find the sum of all the terms between and including t21 and t40. c Find the sum to infinity, S∞. 13 A sequence of numbers is defined by tn: {9, −3, 1, . . .}. a Find the sum of the first 9 terms. b Find the sum of all the terms between and including t10 and t15. c Find the sum to infinity, S∞. Chapter 6


205

14 The first term of the geometric sequence is 5 and the fourth term is 0.078 125. Find the sum to infinity. 15 The sum of the first four terms of a geometric sequence is 30 and the sum to infinity is 32. Find the first three terms of the sequence. 16 For the geometric sequence 5 + 3 , 5 − 3, . . ., find the common factor, r, and the sum of the infinite series, S∞. 2

eBook plus

17 If 1 + 3x + 9x2 + . . . = 3 , find the value of x.

Digital doc

18 We 12b The first term in a geometric sequence is 4 and S∞ = 6. Find the common factor, r.

WorkSHEET 6.2

19 If the common ratio for a geometric sequence is 0.99 and the sum to infinity is 100, what is the value of the first and second terms in the sequence? 20 Show that xn − 1 always has a factor (x − 1) for n ∈ {1, 2, 3, . . .}. 21 A student stands at one side of a road 10 metres wide, and walks half-way across. The student then walks half of the remaining distance across the road, then half the remaining distance again and so on. a Will the student ever make it past the other side of the road? b Does the width of the road affect your answer?

6F

applications of sequences and series


int-0973 Applications of sequences and series eLesson

This section consists of a mixture of problems where the work covered in the first five exercises is applied to a variety of situations. eles-0080 The following general guidelines can assist you in solving the The Fibonacci sequence problems. 1. Read the question carefully. 2. Decide whether the information suggests an arithmetic or geometric sequence. Check to see if there is a constant difference between successive terms or a constant ratio. If there is neither, look for a simple number pattern such as the difference between successive terms changing in a regular way. 3. Write the information from the problem using appropriate notation. For example, if you are told that the 5th term is 12, write t5 = 12. If the sequence is arithmetic, you then have an equation to work with, namely: a + 4d = 12. If you know the sequence is geometric, then ar4 = 12. 4. Define what you have to calculate and write an appropriate formula or formulas. For example, if you have to find the 10th number in a sequence which you know is geometric, you have an equation: t10 = ar9. This can be calculated if a and r are known or can be established. 5. Use algebra to find what is required in the problem. Worked exampLe 13

In 1970 the cost of 1 megabyte of computer memory was $2025. In 1980 the cost for the same amount of memory had reduced to $45 and by 1990 the cost had dropped to $1. a What was the cost of 1 megabyte of memory in the year 2000? b How much memory, in megabytes, could you buy for $10 in the year 2010 based on the current trend?

206


ThInk a

1

2

3

WrITe

Present the given information in a table.

a

Year

1970

1980

1990

2000

2010

Cost ($)

2025

45

1

?

?

1

1

Study the table. The information suggests a geometric sequence for the cost at each ten-year interval. Verify this by checking for a constant ratio between successive terms.

45 ÷ 2025 = 45 and 1 ÷ 45 = 45 so the three terms form a geometric sequence with

To find the cost in the year 2000, find the fourth term in the sequence by multiplying the preceding (third) term by the common ratio.

t4 = t3 × r

common ratio r =

t4 = 1 × =

1 . 45

1 45

1 45

= 0.022 . . .

b

4

Interpret the result and clearly answer the question.

1

If the cost of 1 megabyte can be found in the year 2010 then the amount of memory purchased for $10 can be determined. To find the predicted cost in the year 2010 the fifth term in the sequence needs to be determined.

In the year 2000 one would have paid about 2 cents for a megabyte of memory. b t5 = t4 × r

=

1 45

=

1 2025

×

1 45

of a dollar per megabyte

2

Take the reciprocal of t5 to get the amount of memory per dollar.

The amount of memory per dollar is 2025 megabytes.

3

Find the amount of memory that can be purchased for $10.

So $10 would buy 10 × 2025 = 20 250 megabytes.

Worked exampLe 14

eBook plus

Express the recurring decimal 0.131 313 13 . . . as a proper fraction.

Tutorial

int-1056

ThInk 1

Express the given number as a geometric series.

2

State the values of a and r.

3

Find the sum to infinity, S∞. Write the formula for the sum to infinity. Substitute values of a and r into the formula and simplify.

4

5

Multiply both numerator and denominator by 100 to get rid of the decimal point.

Worked example 14

WrITe

0.131 313 . . . = 0.13 + 0.001 3 + 0.000 013 . . . a = 0.13 and r =

0.0013 0.13

= 0.01

a 1− r 0.13 S∞ = 1 − 0.01 0.13 S∞ = 0.99 13 S∞ = 99 S∞ =

Chapter 6


207

rememBer

To solve problems, use the following guidelines. 1. Identify the type of the sequence by checking whether there is a common difference, or a common ratio. 2. Translate given information into mathematical statements, using appropriate notation. 3. Define what you have to find and write appropriate formula(s). 4. Use algebra to find what is required. exerCISe

6F

applications of sequences and series 1 We 13 In 1970 the Smith family purchased a small house for $60 000. Over the following years, the value of their property rose steadily. In 1975 the value of the house was $69 000 and in 1980 it reached $79 350. a Assuming that the pattern continues through the years, find (to the nearest dollar) the value of the Smiths’ house in i 1985, ii 1995. b By what factor will the value of the house have increased by the year 2010, compared to the original value? 2 An accountant has been working with the same company for 15 years. She commenced on a salary of $28 000 dollars and has received a $2500 increase each year. a What type of sequence of numbers does her annual income follow? b How much did she earn in her 15th year of employment? c How much has she earned from the company altogether? d What was her percentage increase at the end of i her first and ii her fourteenth year of employment? 3 A chemist has been working with the same company for 15 years. He commenced on a salary of $28 000 dollars and has received a 4% increase each year. a What type of sequence of numbers does his annual income follow? b How much did he earn in his 15th year of employment? c How much has he earned from the company altogether? d What was his increase in salary at the end of i his first and ii his fourteenth year of employment? 4 A biologist is growing a tissue culture in a Petri dish. The initial mass of the culture was 20 milligrams. By the end of the first day the culture was a mass of 28 milligrams. a Assuming that the daily growth is arithmetic, find the mass of the culture after the second, third, tenth and nth day. b On what day will the culture mass first exceed 200 milligrams? c Assuming that the daily growth is geometric, find the mass of the culture after the second, third, tenth and nth day. d On what day will the culture mass first exceed 200 milligrams? 5 Logs of wood can be stacked so that there is one more log on each descending layer than on the previous layer. The top row has 6 logs and there are 20 rows. a How many logs are in the stack altogether?

208


b The logs are to be separated into two equal piles. They are separated by removing logs from the top of the pile. How many rows down will workers take away before they remove half the stack? 6 As I was going to St Ives I met a man with seven wives. Every wife had seven sacks, Every sack had seven cats, Note: This is a variation on the original Every cat had seven kits. riddle, which asks ‘How many were Kits, cats, sacks and wives, going to St Ives’. How many were coming from St Ives? 7 Thoughtful Frank has 100 movie tickets to give away to people at a local shopping centre. He gives the first person one ticket, the next person two tickets, the third person three tickets and so on until he can no longer give the nth person n tickets. How many tickets did the last lucky person receive? How many tickets did Frank have left? 8 Kind-hearted Kate has 200 movie tickets to give away to people at the shopping centre. She gives the first person one ticket, the next person two tickets, the third person four tickets and so on following a geometric progression until she can no longer give the nth person 2(n − 1) tickets. How many tickets did the last lucky person receive? How many tickets did Kate have left? 9 The King of Persia, so the story goes, offered Xanadu any reward to secure the safety of his kingdom. As his reward, Xanadu requested a chessboard with one grain of rice on the first square, two grains on the second, four on the third and so on until the 64th square had its share of rice deposited. a Find the total number of grains of rice that the king needed to supply. b If each grain of rice weighs 0.10 grams, how many kilograms of rice does this represent? (Note: There are 103 grams in 1 kilogram.) 10 As legend has it, the King of Constantinople offered Xanadu’s cousin Yittrius any reward to secure the safety of his city. This Yittrius accepted: she requested a chessboard with one grain of rice on the first square, three grains of rice on the second square, five grains of rice on the third square and so on until the 64th square had its share of rice deposited. a Find the total number of grains of rice that the king needed to supply. b If each grain of rice weighs 0.10 gram, how many kilograms of rice does this represent? (Note: There are 103 grams in 1 kilogram.) 11 A student is 3.0 m from the door to a classroom and decides that he will take a 1.0 m step followed by a step of half that distance, and half again and so on until he gets to the classroom door. Show that he will never get any closer than one metre from the door. 2

12 A hiker walks 36 km on the first day and 3 that distance on the second. Every day thereafter 2 she walks 3 of the distance she walked on the day before. Will the hiker cover the distance of 100 km to complete the walk and on what day will she complete the task? 13 WE 14 Recurring decimals can be expressed as rational numbers. Find the fraction equivalent of the following recurring decimal numbers by writing the decimal number as a sum of infinite terms. a 0.1111 . . . = 0.1 + 0.01 + 0.001 + . . . b 0.333 333 333 . . . c 0.5757 . . . d 2.343 434 . . . e 3.142 142 142 . . . f 21.2121 . . . g 16.666 . . . 14 In 1990, 100 students enrolled for a hypocorisma subject at a local university. Each subsequent year for the next decade the enrolment increased by 20%. a Find the number of students enrolled in hypocorisma in 1995. b Over the course of the decade find the total number of students who had enrolled in hypocorisma.


209

15 For tax purposes, the value of a computer used for a business depreciates by 8.5% of the initial cost each year. For economic reasons the business sells its computers when they first depreciate to less than half their initial value. After how many years will a computer used by this business be sold? 16 The side lengths of a right-angled triangle form the successive terms of an arithmetic sequence. The perimeter of the triangle is 72 m. What are the side lengths of the triangle? 17 A circular board is divided into a series of concentric circles of radius 1 cm, 2 cm, 3 cm and 4 cm as shown at right. a Find the areas of each of the successive shaded regions and show that they form an arithmetic progression. b A dart is fired at the board at random and hits the board. What is the probability of striking each of the four regions of the board? (Note: The probability of striking a region = area of region ÷ total area.)

4 cm 3 cm 1 cm

2 cm

18 A bullet is fired vertically up into the air. In the first second it has an average speed of 180 m/s; that is, it travels 180 m up into the air during the first second. Each second its speed diminishes by 12 m/s. Thus during the 2nd second the bullet has an average speed only 168 m/s and accordingly travels 168 m further up into the air. a Find an equation for the average speed of the bullet for the nth second that it is in the air. b Find the time when the average speed of the bullet is equal to zero. c Find the maximum height of the bullet above where it was fired. 19 Coffee cools according to Newton’s Law of Cooling in which the temperature of the coffee above room temperature drops by a constant fraction each unit of time. The table below shows the temperature of a cup of coffee in a room at 20 °C each minute after it was made. Remember to subtract the room temperature from the temperature of the coffee before you do your calculations. Time (min)

Temperature (°C)

1

80.0

2

74.0

3

68.6

The person who made the coffee will drink it only if it has a temperature in excess of 50 °C. What is the minimum time after the cup of coffee has been made before it becomes undrinkable?

20 Two arithmetic sequences, tn and un, are multiplied together. That is, each term is multiplied by the other to form a new term. tn = 2n − 3, n ∈ {1, 2, 3, . . .} and un = 3n, n ∈ {1, 2, 3, . . .} Show that the new sequence of numbers t1 × u1, t2 × u2, t3 × u3, . . . is an arithmetic series and hence find the arithmetic sequence for that new series. (Hint: For a sequence, an, with a series An, an = An − An − 1) 210


Summary Describing sequences

• A sequence is a string of numbers or expressions. It may contain a finite or infinite number of terms and may or may not follow a recognisable pattern. • A sequence can be described in a number of ways. 1. As a list tn: {1, 7, 2, 6, . . .} (note that t3 = 2) 2. As a function: tn = 2n − n2, n ∈ {1, 2, 3, . . .} (note that t5 = 2 × 5 − 52 = −15) 3. As a recursive or iterative formula: tn + 1 = 2tn − 3, t1 = 6 (note that t2 = 2 × 6 − 3 = 9) • The logistic equation is a model of population growth of the general form: tn + 1 = atn(1 - tn), where 0 < t0 < 1 and a is a constant. • A convergent sequence is a sequence whose terms settle at a certain fixed number, x, called the limit of the sequence. This can be written as tn → x. A sequence whose terms grow further and further apart is called divergent. That is, a sequence is divergent if tn → ∞, or tn → −∞ as n → ∞. A sequence whose terms fluctuate between two (or more) values is called oscillating. Arithmetic sequences

• An arithmetic sequence is one whose successive terms have a common difference. This common difference is given the symbol d. Thus tn + 1 − tn = d for all values of n. The first term in the sequence is given the symbol a. • If x, y, z are successive terms in an arithmetic sequence then the middle term ( y) is called an arithmetic mean and is equal to the average of the two outer terms (x and z): x+z y= 2 • An arithmetic sequence can be written as a, a + d, a + 2d, . . . and so the nth term, tn, is: tn = a + (n − 1)d using the function notation, or tn + 1 = tn + d, t1 = a using the iterative notation. Arithmetic series

• The sum of the first n terms of the arithmetic sequence is given by n Sn = (2a + ( n − 1)d ) 2 • In general, tn + 1 = Sn + 1 − Sn. Geometric sequences

• A geometric sequence is one in which each successive term is obtained by multiplying the preceding term by t a constant number. This number is called the common ratio and is given the symbol r. Thus n + 1 = r for all tn values of n. The first term in the sequence is given the symbol a. • If x, y, z are successive terms in an arithmetic sequence then y is called a geometric mean and is given by y2 = xz. • A geometric sequence can be written as a, ar, ar2, . . . and so the nth term, tn, is tn = ar n − 1 using the function notation, or tn + 1 = rtn, t1 = a using the iterative notation.


211

Geometric series

• The sum of the first n terms in a geometric sequence is given by a(1 − r n ) with r ≠ 1 1− r a(r n − 1) or Sn = with r ≠ 1 r −1 • When the magnitude of r is less than one, that is, −1 < r < 1, the sum of a geometric sequence to infinity S∞ is a given by S∞ = . 1− r

Sn =

Applications of sequences and series

• To solve problems, use the following guidelines. 1. Identify the type of sequence by checking whether there is a common difference, or a common ratio. 2. Translate given information into mathematical statements, using appropriate notation. 3. Define what you have to find and write appropriate formula(s). 4. Use algebra to find what is required.

212



1 Write the iterative definition for each of the following sequences: a {7, 11, 19, 35, 67, . . .} b {−2, 5, 26, 677, . . .} 2 For the arithmetic sequence where t3 = 10 and t6 = 478, find a the functional rule for the nth term in the sequence b the iterative rule for the sequence. 3 A car at a racetrack starts from rest and travels 0.5 m in the 1st second and 1.0 m in the 2nd second following an arithmetic progression in the distances covered each subsequent second. a How far will it travel during the 10th second? b After 10 seconds of motion, how far will it have travelled in total? c To the nearest whole second, how long will it take to travel 1000 m (1 km)? 4 At Bugas Heights a radiation leak in a waste disposal tank potentially exposes staff to a 1000 milli-rem h dose on the first day of the accident, a 800 milli-rem h dose on the second day after the accident and a 640 milli-rem h dose on the third day following the accident. a Assuming a geometric sequence, find the amount of potential exposure dose by the 10th day. b Find the total potential exposure dose in the first 5 days. 5 The infinite sum of a geometric sequence is 99 and the first term is 10. Find the common ratio for the sequence. 6 Find the sum of the following expressions: 1 4

a 1 + +

1 16

+

1 64

. . .

2 3

4 9

b 1 − + -

8 27

. . .

7 Find the fraction equivalent of the following recurring decimals: a 0.222 222 b 2.454 545 454 Multiple choice

1 Consider the sequence tn + 1 = 2tn + 4; t3 = 12. The second term in the sequence is: A 10 B 6 C 28 D 4 E 8

2 A series is listed as 3, 10, 21, 36, . . . The next term in the series is: A 51 B 52 C 53 D 54 E 55 3 The 23rd term in the sequence of numbers {7, 3, −1, . . .} is: A −88 B −81 C −74 −83 E 90 D 4 Consider the arithmetic sequence 52, a, 41, b. The numerical value of the expression a − 3b is: A −60 D

−71

B −64 12 E

C −67 1

2

72 12

5 Consider the arithmetic sequence x − 2y, 3x − 4y, 4x − 7y, . . . An expression for y in terms of x is: A y = x B y = −x − C y = 2x D y = 2x E y = −3x 6 A car is accelerating such that in the 1st second it travels 2.0 metres, in the 2nd second it travels 3.5 metres, in the 3rd second it travels 5.0 metres, and so on for a total of 15 seconds. The total distance travelled by the car is: A 630 m B 93.75 m C 187.5 m D 375 m E 315 m 7 The sum of the first four terms in an arithmetic sequence is 70. The sum of the first six terms is 63. The sixth term of the sequence is equal to: B −7 A −14 C 0 D 7 E 14 8 For a geometric sequence, the 4th term is 5 and the 7th term is −625. The second term in the sequence is: A −2.5 B −1.25 C 0.25 D −0.25 E 0.20 9 The sum of an infinite geometric sequence is 5.6 with the common ratio equal to 0.20. The sum of the first four terms of the geometric sequence is closest to: A 5.0 B 5.2 C 5.4 D 5.6 E 5.8


213

10 The sum of the first 10 terms of a geometric sequence is 400. The next term in the sequence is 3 times the previous term. The first term in the sequence is:

11 The sequence tn = 81 ×  1  3

n−1

and the series

t Sn = 1 − 0.1n are combined to form the ratio n . sn When n = 9 the value of the ratio is:

A

17 731

A

1 27

B

400 1473

B

1 270

C

100 7381

C

1 243

D

200 781

1 D 81

E

10 387

E

10 81

EXtended response

1 Consider a square of side length 2 units. a What is the perimeter of the square? b Each of the four midpoints form the vertices of a new square inscribed within the original square. Find the perimeter of this new square. c Repeat the process to find the perimeter of a third square inscribed within the second. d Give an expression for the perimeter of the nth square. 2 Consider the following iterative definitions: a tn + 1 = tn - 3 , t1 = 1 4 8 b tn + 1 = atn, t1 = b2 c tn + 1 = 3tn2 - 1.5, t1 = 0.5 If each of these definitions is used to generate a sequence of numbers: i decide whether the sequence is arithmetic, geometric or neither, and ii find its fourth term. 3 In January 2004, Rachel and Nathan inherited a small trout farm from their Uncle Michael. They were told that in any given year the trout population, pn + 1, could be easily calculated using the formula pn + 1 = 0.5pn(1 – pn), where pn is the number of trout (in thousands) in the preceding year. They were also told that on the day of their inheritance the farm housed 800 fish. a Use the above formula to predict (to the nearest whole number) the size of the fish population on Rachel and Nathan’s farm for the next three years; that is, for i January 2005 ii January 2006 iii January 2007. b What will happen to the size of the fish population if it continues to change according to the above formula? How long will it take? After extensive research, Rachel and Nathan decided to modernise their newly acquired farm. A new feeding system and other improvements were installed, and were completed within the first two years (that is, by January 2006). As a result, in any given year the trout population, pn + 1, could now be calculated using the formula pn + 1 = 1.6pn(1 – pn), where pn is the number of trout (in thousands) in the preceding year. c Use this new formula, and the figure obtained in part a ii (that is, the trout population in January 2006) as your starting point to predict the size of the trout population for January 2007 and January 2008. d Do Rachel and Nathan still run the risk of losing all of their trout stock? Explain your answer. e Will the size of the trout population ever reach and increase beyond the initial number of 800 fish? Give reasons for your answer.

214


4 On an island in the Pacific Ocean the population of a species of insect (species A) is increasing geometrically with a population of 10 000 in 1990 and an annual growth rate of 12.0%. Another species of insect (species B) is also increasing its population, but arithmetically with numbers 15 000 in 1990 and an annual increment of 1000 per annum. a Using a spreadsheet or other method, determine the difference in the numbers of the two species during the last decade of the twentieth century (that is, up to 1999). b In what year will the first species be greater in number than the second species, assuming that growth rates remain fixed? A scientist has a mathematical model where the species can cohabit provided that they have equal numbers in the year 2000. c If the growth rate in species A is to remain unchanged, what would the annual increment in species B need to be to achieve this? d If the annual increment in species B is to remain unchanged, what would the growth rate in species A need to be to achieve this? 5 a A series is given by the equation Sn = 2n2 + 3n. Show that the sequence is arithmetic and give the expression for the nth term in the sequence, tn. b A series is given by the equation Sn = an2 + bn. Show that the sequence is arithmetic and give the expression for the nth term in the sequence, tn, in terms of a and b. 6 a Australian Heating is a company that produces heating systems. The number of heating systems produced annually is modelled by an increasing geometric sequence. The number of heating systems produced in each of the first 3 years is shown in Table 1. Table 1: Annual production of heating systems. Year Number of heating systems produced

1

2

3

2000

2200

2420

i ii iii iv

Show that the common ratio r of this sequence is 1.1. What is the annual percentage increase in the number of heating systems produced each year? How many heating systems will be produced in year 5? The number of heating systems produced annually continues to follow this pattern. In total, how many heating systems will they produce in the first 10 years of operation? v The geometric sequence in Table 1 can also be produced using an iterative definition of the form Pn + 1 = bPn + c, where P1 = 2000 and Pn is the number of heating systems produced in the nth year. Determine the values of b and c. b The purchase and installation of a basic heating system with five outlets costs $3500. Each additional outlet costs an extra $80. i Determine the cost of installing a heating system with eight outlets. ii A customer has $4400 to spend on a heating system and outlets. Determine the greatest number of outlets that can be bought with this heating system.

Chapter 6


215

iii Australian Heating recommends that a house with 20 squares of living area should have 12 heating outlets. Using this recommended ratio, determine the cost of installing a heating system for a house having 35 squares of living area. c The number, Sn, of heating systems sold in the nth year is defined by the iterative definition: Sn = 1.2Sn − 1 − 200, where n ≤ 5 and S3 = 2224 i Use this definition to determine how many heating systems were sold in the first year. ii What percentage (correct to 1 decimal place) of heating systems produced during the first three years was sold within the three years? eBook plus Digital doc

Test Yourself

exam TIp Remember an iterative formula of the form: Tn = aTn − 1 + b will have a value of b = 0 if the sequence is geometric and a = 1 if the sequence is arithmetic.

Chapter 6

[Assessment report 2004]

[©VcAA 2004]

216


eBook plus

aCTIvITIeS


• 10 Quick Questions: Warm up with ten quick questions on sequences and series. (page 178) 6A

Describing sequences

Tutorial

• We 2 int-1051: Watch how to determine the behaviour of the sequence using a CAS calculator. (page 183) Digital doc

• History of mathematics: Learn about the mathematician Leonardo Fibonacci. (page 186) 6b

Arithmetic sequences

Tutorial

• We 5 int-1052: Watch how to find the missing terms in an arithmetic sequence. (page 189) Digital doc

• SkillSHEET 6.1: Practise using elimination to solve simultaneous equations. (page 191) 6c

Arithmetic series

Tutorial

• We 7 int-1053: Watch how to find the sum of the first 20 terms in an arithmetic sequence. (page 193) Digital doc

• WorkSHEET 6.1: Complete and determine arithmetic series. (page 195) 6D

Geometric sequences

• Spreadsheet 036: Investigate Fibonacci series. (page 205) • WorkSHEET 6.2: Solve more complex problems with arithmetic series, complete and determine geometric series and apply geometric series theory to a worded problem. (page 206) 6F

Applications of sequences and series

Interactivity

• Applications of sequences and series int-0973: Consolidate your understanding of sequences and series. (page 206) eLesson

•

eles-0080: The

Fibonacci sequence (page 206)

Tutorial

• We 14 int-1056: Watch how to express a recurring decimal as a proper fraction. (page 207) chapter review

Tutorial

• We 10 int-1054: Watch how to find a specific term in a geometric sequence. (page 197) Digital doc

• Spreadsheet 036: Investigate Fibonacci sequences. (page 199) 6E

Digital docs

Geometric series

Digital doc


Tutorial

• We 12 int-1055: Watch how to find the sum to infinity of a geometric sequence and a term given the sum to infinity and the first term. (page 204)

Chapter 6


217

7

7A 7B 7C 7D 7E 7F 7G

Direct variation Further direct variation Inverse variation Further inverse variation Joint variation Part variation Transformation of data

Variation areas oF sTudY

• Numerical, graphical and algebraic approaches to direct, inverse and joint variation • Transformation of data to establish relationships 1 between variables, for example, x2, to linear x

• Modelling of given nonlinear data using the relationships y = kkxx 2 + c, c, y =

k + c, where k is a x

positive real number eBook plus Digital doc

7a

10 Quick Questions

direct variation

We often need to look at variables related to an everyday situation, and then work out the relationship between them. This is called mathematical modelling. The variables may be recorded in a table or illustrated graphically. Once a relationship has been formulated, it is interesting to see how one variable changes with respect to changes in the other. This change is called a variation and depends on the relationship that exists between the variables. In this chapter we shall consider four types of variation: direct, inverse, joint and part variation. We shall also look at the transformation of data in order to establish relationships between variables. Consider the following example. Leon has agreed to sponsor his friend who is participating in the ‘40-hour famine’. He will sponsor his friend $1.20 per hour. The amount Leon has to pay depends on the number of hours his friend goes without food. This information may be represented graphically, or placed in a table, as shown. n, number of hours without food C, total cost of sponsorship ($)

1

2

3

4

5

1.20

2.40

3.60

4.80

6.00

The table shows the amount to be paid for up to 5 hours. (If necessary, it could be extended to 40 hours, assuming that Leon can pay for each hour his friend goes without food and his friend can go without food for the specified time.) From the table, we can see that as the number of hours increases, the total cost of sponsorship also increases (as expected). The ratio between the cost, C, and the number of hours, n, is the

218


same for each pair of values, that is: 1.20 2.40 3.60 4.80 6.00 = = = = = 1.20. 1 2 3 4 5 Hence, the relationship between the two variables can be written as: C = 1.20 or C = 1.20 n n C = 1.20n is an equation of a straight line with gradient 1.20. The line passes through the origin. Graphically, the relationship can be represented as shown at right. Summarising our observations, we can say that the following is true for the given information.

C($) 6.00 4.80 3.60 2.40 1.20

1. The ratio between any pair of corresponding values is 0 1 2 3 4 5 n (hours) constant. 2. The graph which represents the data is a straight line passing through the origin. 3. The gradient of the line is equal to the ratio between the variables. 4. As one variable increases, the other variable also increases. In cases like this, we can say that one of the variables is directly proportional to (or varies directly as) the other. The ratio between any two corresponding variables is constant and is called the constant of proportionality (or constant of variation) and is denoted by the symbol k. Hence, for the example on the previous page: C is directly proportional to n or C varies directly as n Using mathematical notation this is written as: C∝n where the symbol ∝ means is directly proportional to or varies directly as. The ratio between corresponding values of C and n is constant and in our example is equal C  to 1.20  = 1.20 . So the constant of proportionality (or constant of variation), k, is 1.20. The n  relationship between variables can be expressed as C = 1.20n. Generally, for any two variables x and y, where y varies directly as x, (that is, y ∝ x), there y exists a relationship between them such that = k or y = kx, where k is a constant. The graph x of the relationship is a straight line passing through the origin with the gradient k — the constant of variation. If y∝x then y = kx where k is the constant of variation and k ∈ R. From the definition, it follows that the existence of the direct variation between two variables can be established either numerically or graphically, as shown in the following example.

Worked Example 1

For the given data, establish whether direct variation exists between x and y using: a a numerical approach (clearly specify k, the constant of variation, if applicable) and b a graphical approach.

Chapter 7 Variation

219

c Confirm your result using a CAS calculator.

x

4

7

8

10

y y x

5.2

9.1

10.4

13

THINK a

1

WRITE/Draw

y for each of the x four pairs of values. One variable varies directly as the other if the ratio between any two corresponding values is constant. Find the ratio

a Ratio =

y x

5.2 = 1.3 4 9.1 = 1.3 7 10.4 = 1.3 8 13 = 1.3 10

First pair: Second pair: Third pair: Fourth pair:

b

2

Compare each of the four ratios and answer the question.

3

Copy and complete the table.

1 2

3

220

Plot the information from the table onto a set of axes. Join the given points and see if a straight line is obtained. Note: For a direct variation to exist between two variables x and y, a straight line passing through the origin (0, 0) must be obtained.

Calculate the gradient using any two points on the straight line. Answer the question. Note: The gradient of the straight line will equal k, the constant of variation, if direct variation exists between the variables x and y.

Since all four ratios are the same (that is, 1.3), y varies directly as x.

b

x

4

7

8

10

y y x

5.2

9.1

10.4

13

1.3

1.3

1.3

1.3

y 13 10.4 9.1 5.2 0

4

7 8 10

x

The given points fit perfectly on a straight line. If the straight line is extended it will pass through the origin. Therefore y varies directly as x. Let (x1, y1) = (4, 5.2) and let (x2, y2) = (10, 13) y -y m= 2 1 x2 - x1 13 - 5.2 10 - 4 7.8 = 6 = 1.3 The gradient of the straight line is equal to k, the constant of variation. =


c

1

On the Statistics screen, enter the data into the table. Label the columns x and y.

2

To draw the scatterplot of the data, tap: • SetGraph • Settings Set: Type: Scatter XList: main\x YList: main\y Freq: 1 Mark: square • Set • y

3

To check the linear relationship, tap: • Calc • Linear Reg Set: XList: main\x YList: main\y Freq: 1 • OK

c

The relationship is one of direct variation.

If we know that one variable varies directly as the other, it is possible to establish the value of k, the constant of variation, and hence determine the value of any variable given its corresponding value.

Chapter 7 Variation

221

Worked Example 2

If it is known that m ∝ n, find: a the constant of variation b the missing values. n

3

m

6 10

THINK a

b

222

9 18

20

WRITE

m n

1

Write the rule for k. Since m ∝ n, then m m = kn and hence k = . n

2

Obtain a pair of values where both m and n are known.

Select n = 9, m = 18

3

Substitute these values into the given equation and obtain a value for k.

k=

4

Answer the question. Note: To obtain k, we can use only corresponding values; that is, those in which one value is underneath the other in the table.

The constant of variation is 2. Therefore m = 2 or m = 2n. n

1

Find the unknown value by substituting its given corresponding pair into the m equation = 2. n

2

First pair: substitute n = 3. Transpose the equation to make m the subject.

3

Second pair: substitute m = 10. Transpose the equation to make n the subject.

4

Third pair: substitute n = 6. Transpose the equation to make m the subject.

5

Final pair: substitute m = 10. Transpose the equation to make n the subject.

a k=

18 9 =2

b From part a

m =2 n

m First pair = 2 3 m=2×3 =6 10 Second pair = 2 n 10 = 2 × n 2n = 10 10 n= 2 =5 m Third pair = 2 6 m=2×6 = 12 20 Final pair = 2 n 20 = 2 × n 2n = 20 20 n= 2 = 10


6


n

3

5

6

9

10

m

6

10

12

18

20

A similar approach may be applied to solving worded problems which involve direct variation. Worked example 3

eBook plus

If the distance, d km, travelled by a person varies directly as the time, t hours, and Tutorial it is known that the person travelled 12 km while walking for 2.5 hours, find: int-1057 Worked example 3 a how far he will travel in 3 hours b how long he must walk in order to travel 6.72 km. THINk a

b

WrITe

1

Write the rule for k. Since d ∝ t, then d d = kt and hence k = . t

2

Substitute the given values for d and t into the equation and solve for k.

3

Substitute t = 3 into the equation d = 4.8. t

4

Transpose the equation to make d the subject.

5

Answer the question and include the appropriate unit.

1

Substitute d = 6.72 into the equation d obtained in part a ; that is, = 4.8. t

k=

a

d t

12 2.5 = 4.8 The constant of variation is 4.8. Therefore, d = 4.8 t d = 4.8 When t = 3, 3 k=

d = 4.8 × 3 = 14.4 He will travel 14.4 km in three hours. b From part a :

When d = 6.72,

d = 4.8 t 6.72 = 4.8 t 6.72 = 4.8 × t 4.8t = 6.72 6.72 t= 4.8 = 1.4

2

Transpose the equation to make t the subject.

3

Convert the answer to hours and minutes by multiplying the decimal part of the answer by 60.

1.4 hours = 1 h + (0.4 × 60) min = 1 h 24 min

4


In order to travel 6.72 km, he must walk for 1 h 24 min.

Although we were not actually asked to find the constant of variation, k, it was a necessary step in order to solve the problem.

Chapter 7

Variation

223

REMEMBER

1. For any two variables x and y, where y varies directly as x (or y is directly proportional to x), the following properties exist: (a) the ratio between any pair of corresponding values is constant and equal to k, the constant of proportionality (or constant of variation) (b) the graph, which represents the variables x and y, is a straight line passing through the origin with the gradient equal to k (c) as one variable increases, the other variable also increases. 2. The notation used to express y varies directly as x is given by: y∝x or y = kx where k ∈ R 3. When solving problems involving direct variation follow these steps: (a) establish a ratio relating the given variables and determine the value of k, the constant of proportionality (b) use the ratio and k to determine the value of any variable given its corresponding value.

Exercise

7A

Direct variation 1 WE 1 For the given data in each of the following tables, establish whether direct variation exists using: i a numerical approach (clearly specify k, the constant of variation, if applicable) and ii a graphical approach. a

x

c

e

2

3

4

b

5

y y x

3

4.5

6

7.5

p

1

4

6

8

r r p

2

8

14

16

s

1

2

5

9

t t s

-3

-6

-15

-18

d

f

a

3

4

5

6

b b a

1

4 3

2 13

2

m

2

3

8

10

n n m

1

1.5

4

5

u

-5

-4

7

8

v v u

-10

-8

14

16

2 MC Which of the following represents direct variation between x and y? i y

0

224

ii y

x

0

iii y

x


0

x

iv y

v y

x

0

A ii only 3

vi y

x

0

B ii and iv

x

0

C iii only

D i, v and vi

E None of the above

Sam and Nicholas collected some data for their statistics project and arranged them in a table as follows. n

0

1

2

7

10

18

19

20

C

20

35

50

125

170

290

305

320

Nicholas wants to see if direct variation exists between the two variables n and C and suggests using either a numerical or graphical method. However, Sam argues that by simple inspection of the table, he knows that C is not directly proportional to n. a Is Sam right about his conclusion? b Explain how he could arrive at his conclusion without plotting the data, or without finding ratios of corresponding values of C and n. 4 WE2 For each of the following, it is known that y ∝ x. Find: a the constant of variation b the missing values. i x ii x 4 5 10 4 y iii

v

x

−2

y

−2.2

x y

vii

23

3

5

iv 9.9 vi

6

10.4 20.8

41.6 72.8

2 −1

5

viii

9

2

9

y

27.6

−1.1

4

x y

18.4

7

18

x

−3

y

−7.5

x

36

−2

y x

2

2.5 12.5 30 5

6

10 3

3

8

4

y

4.5

45

1 −2.5

3 2 3

40

1

1.5

7 4

Questions 5 to 7 refer to the following information. Given that n ∝ m: m

4

9 1 23

n

12

3

5 MC The gradient of the line which represents the relationship between m and n is: A 2 1 3

B 3

C

1 3

D 4

6 MC The relationship between m and n cannot be written as: 1 m n 1 A = 3 B = C 3n = m D m = n 3 n m 3 7 MC The missing values of n when m = 4 and m = 12 are: A 12, 36 respectively

B 4 , 4 respectively

D 16, 48 respectively

E

3

16 , 3

E

4 3

1 E n = m 3

C 1, 3 respectively

16 respectively

Chapter 7 Variation

225

8 mC a ∝ b and a = 21 when b = 15. When a = 49, b is equal to: A 68.6 B 70 C 34.3 D 35

E 17.5

9 We3 A long-distance truck driver finds that when he is driving from Melbourne to Sydney, the distance he travels is directly proportional to time. It takes him 2 hours and 36 minutes to travel 234 km. Find: a how far he will travel in 3 hours b the time taken to travel 117 km c the speed of the truck.

10 The directions on the ‘Anti-Bacterial Clean & More’ bottle recommend that you dilute half a cup of the cleaner in 5 litres of warm water. a Complete the following table. Volume of water (L)

1

2

3

4

5

10

15

20

30

1 2

Amount of cleaner (cups)

b The cup on the bottle has only one graduation (at 1 cup). Explain which amounts of water 2 will be convenient to use with this detergent. 11 The perimeter, P, of a certain shape is directly proportional to the side length, s, and P = 12 cm when s = 4 cm. a Find the perimeter of the shape when its side length is 12.3 cm. b Find the length of the side when the perimeter of the shape is 88.2 cm. c Name the shape.

226


rri

do r

Bathroom Bedroom 2

Bedroom 1

2m

Laundry

Backyard 2m

3m

Co

12 Mila is going to polish all the floors in her unit, except for the kitchen, laundry and bathroom, where she has tiles. The plan of Mila’s unit is shown. a Find the area of the floor that is to be polished. b A particular type of varnish is sold in 3 L cans. If one can covers 34.5 square metres of floor, how many litres are necessary to polish the floors twice? How many 3 L cans should Mila purchase? c If each can of varnish costs $12.60, how much money will be wasted? (That is, find the cost of the left-over varnish.)

Living Entrance 10 m

3m

Kitchen

Dining

4m

3m

13 The graph at right shows the relationship between the acceleration, F (N) a (in m/s2), of a certain body and the force, F (in newtons), acting on 1600 that body. a Show that force varies directly as acceleration. b Find the constant of proportionality of the two variables. c Write the equation of the relationship. d Find the force necessary to produce an acceleration of 4 m/s2. e Find the acceleration which is produced when a force of 1000 newtons is acting on the body.

(2,1600)

2

a (m/s2)

14 Mark is reading a book about world famous explorers and travellers. In one of the chapters, he comes across a distance shown as 26 miles and wishes to convert this distance to kilometres. Mark remembers that 5 miles is approximately equivalent to 8 km. a Find the appropriate equivalent of 26 miles in kilometres. In reality, 5 miles is equivalent to 8.045 km. b Now, find the new equivalent of 26 miles in kilometres. c Compare the difference between the two values. d Express the difference as a percentage of the real distance. 15 If (2, a) and (a, 12.5) belong to a direct proportional relationship, what are the values of a and hence what is the constant of proportionality.

7B

Further direct variation In the previous section, we considered cases of direct variation between two variables raised to the power of 1. Graphically, those relationships were represented by straight lines. In other words, we dealt only with examples of direct linear variation. However, variation is not necessarily linear. There are many cases of direct variation where variables are raised to powers other than 1. In such cases, instead of writing y ∝ x we write the variation statement according to the information provided in the given problem. For example, if it is known that y varies directly as the square of x, we write y ∝ x2. If y varies directly as the square root of x, we write y ∝ x . After the variation statement is written, we proceed in the same manner as with cases of direct linear variation. 1. Write the equation of variation. 2. Substitute known values to find the constant of proportionality, k. 3. Find all unknown values as required in the given problem.

Worked Example 4

For the following: a fill in the x values of the table b establish whether y varies directly as the square root of x by: y i calculating the ratio x ii plotting the graph of y versus x.

x

1

4

9

16

2.5

5

7.5

10

x y y x

THINK a

1

WRITE

Evaluate each of the x values.

a

1 =1

4=2

9=3

16 = 4

Chapter 7 Variation

227

2

Substitute the x values into the table.

x x y

1

4

9

16

1

2

3

4

2.5

5

7.5

10

y x b

i

1

Find the ratio

y

x 4 pairs of values.

b

for each of the

i Ratio =

y

x First pair:

2.5 1

= 2.5

5 2

= 2.5

Third pair:

7.5 3

= 2.5

Fourth pair:

10 4

= 2.5

Second pair:

2

Compare each of the 4 ratios and answer the question.

3


Since all 4 ratios are the same (that is, 2.5) y varies directly as x. x

1

4

9

16

1

2

3

4

2.5

5

7.5

10

2.5

2.5

2.5

2.5

x y y x ii

1

ii

Plot the information from the table onto a set of axes. Join the given points and see if a straight line is obtained. Answer the question. Note: For a direct variation to exist between two variables x and y, a straight line passing through the origin (0, 0) must be obtained.

2

y 10 7.5 5 2.5 0 1 2 3 4

x

The given points fit perfectly on a straight line. If the straight line is extended it will pass through the origin. Therefore y varies directly as x.

Worked Example 5

For the given data, establish the rule relating the variables x and y then graph the relationship using a CAS calculator.

228

x

3

6

7

10

y

28.8

115.2

156.8

320


THINK 1

On the Statistics screen, enter the data into the table. Label the columns x and y.

2

Find the rule that fits the data. To do this, tap: • Calc • Power Reg Set: XList: main\x YList: main\y Freq: 1 Copy Formula: y1 • OK

3

Write down the rule for x and y.

4

Graph the rule. On the Graph & Tab screen, equate the rule to y1, and tap $. Adjust the window accordingly.

WRITE/DISPLAY

y = 3.2x2

Worked Example 6

The volume of a sphere, V, is directly proportional to the cube of its radius, r. If a sphere with radius 3 cm has volume 113.04 cm3, find: a the constant of proportionality, to 3 decimal places b the volume of a sphere with radius 10 cm c the radius of a sphere which has a volume of 33.493 cm3, to 3 decimal places.

Chapter 7 Variation

229

THINk a

b

c

WrITe

V ∝ r3

a

1

Write the variation statement.

2

Write the equation of variation.

3

Substitute V = 113.04 and r = 3 into the equation to find the constant of proportionality, k.

4

Transpose the equation to make k the subject.

5

Evaluate. Round the answer to 3 decimal places.

6

Rewrite the equation of variation by substituting the value for k.

1

Write the equation obtained in part a .

2

Substitute r = 10 into the equation.

V = 4.187 × 103

3

Evaluate and include the appropriate unit.

V = 4.187 × 1000 = 4187 cm3

1


2

Substitute V = 33.493 into the equation.

3

Transpose the equation to make r3 the subject.

4

Solve for r.

r=

5

Evaluate. Round your answer to 3 decimal places and include the appropriate unit.

r = 2.000 cm

V = kr3 113.04 = k × (3)3 113.04 = 27k k=

113.04 27

= 4.187 V = 4.187r3 b

V = 4.187r3 When r = 10,

V = 4.187r3

c

When V = 33.493, 33.493 = 4.187r3 33.493 r3 = 4.187 3

33.493 4.187

Sometimes we are interested in knowing how the change in one variable (for example doubling or tripling) will affect the other. Worked example 7

eBook plus

The area, A, of a square varies directly as the square of its side, s. Find the effect on the area when the side of the square is: a doubled b halved. THINk a

230

WrITe

1


2


3

If we double the side of the square, it will be twice as large as the original size (that is 2s). Therefore, substitute 2s into the equation in place of s and simplify. Note: Anew is used to make it distinct from the original A.

a

A ∝ s2 A = ks2 Anew = k(2s)2 = k × 4s2 = 4ks2


Tutorial


b

Anew = 4A

4

Rewrite the equation in terms of the original area A.

5

Interpret the result.

1

Repeat steps 1 and 2 from part a .

2

If we halve the side of the square, the

When the side of a square is doubled, the area is quadrupled.

S

A ∝ s2 A = ks2

b

Anew = k    2 S

S

new side will be . Substitute into the 2 2 equation of variation in place of s. Note: Anew is used to make it distinct from the original A. 3

Simplify.

=k×

2

S2

4

1 4

= ks2 4 5

Rewrite the equation in terms of the original area A.

Anew = 1 A 4


When the side of a square is halved, the area is divided by 4 (that is, one quarter of the original area).

rememBer

There are many cases of direct variation, where variables are raised to powers other than 1. To solve these problems, follow these steps: 1. Write the statement of variation. 2. Write the equation of variation. 3. Substitute known values to find the constant of proportionality, k. 4. Find all unknown values as required in the given problem. exerCIse

7B

Further direct variation

eBook plus

1 We4 For each of the following:

Digital doc

Spreadsheet 133

a Fill in the x values of the table.

Transforming data

b Establish whether y varies directly as the square root of x by y calculating the ratio and plotting the graph of y versus x. x i x 1 4 9

16

x y y x

3

6

9

12

Chapter 7

Variation

231

ii

x

1

9

25

49

5

15

25

35

4

16

36

64

1

2

3

7

x y y x iii

x x y y x

2

For each of the following: a Fill in the x2 values of the table. y b Establish whether y varies directly as the square of x by calculating the ratio 2 and x plotting the graph of y versus x2. i

x

1

2

3

4

y y x2

2

8

18

32

x

1

3

5

6

y y x2

0.5

4.5

12.5

18

x

2

4

6

8

y y x2

1

4

9

12

x

1

2

4

6

1 6

2 3

8 3

6

2

x

ii

x2

iii

x2

iv

x2 y y x2

3 For each of the following: a Fill in the x3 values of the table. b Find the ratio of y and hence deduce whether or not y ∝ x3. x3

232


i

x

1

2

3

4

5

y y x3

1 4

2

7

16

32

x

2

3

5

y y x3

24

81

375

x

2

4

6

2 3

16 3

18

1

2

4

6

5

40

300

1080

x3

ii

x3

iii

x3 y y x3 iv

x x3 y y x3

4 WE5 Determine whether y varies directly as x, square of x or square root of x using a CAS calculator: x

1

4

16

36

y

0.3

0.6

1.2

1.8

1

4

5 If n ∝ m : m

1 2

n

25 3 4

1

a find the constant of variation b write the relationship as an equation c find the missing values. 6 MC a ∝ b2 and a = 16 when b = 8. When a = 1, b = A 1

B 2

C 4

1

D 4

E 16

7 MC m ∝ n and m = 3 when n = 4. When n = 64, m = A 32 B 24 C 16 D 12

E 8

8 MC p ∝ s3 and s = 6 when p = 72. When p = 9, s = A 1 B 2 C 3 D 4

E 5

Chapter 7 Variation

233

9 For each of the following write: a the variation statement b the equation of variation (use k as the constant of proportionality). i The total surface area of a sphere, A, varies directly as the square of its radius, r. ii The volume of a cube, V, is directly proportional to the cube of the length of its side, s. iii The kinetic energy, E, of a moving body is directly proportional to the square of its velocity, v. iv The radius of a circle, r, varies directly as the square root of its area, A. v The distance to the horizon, d, varies directly as the square root of the height, h, above the sea level. 10 We 6 The total surface area of a spherical balloon is directly proportional to its radius squared. If a balloon with a radius of 10 cm has a total surface area of 1256 cm2, what is: a the constant of proportionality, to 2 decimal places? b the total surface area of the balloon when it has a radius of 12.5 cm? c the radius of the balloon when it has a total surface area of 2826 cm2? 5 mm 5 mm 11 A set of measuring cups consists of 4 cylindrical cups that nestle inside each other. The height of all 4 cups is the same and their radii 6 cm differ by 5 mm. The capacity of each cup is directly proportional to the square of its radius. The smallest cup has a radius of 3 cm and holds 113.04 mL of water. Find: a the constant of proportionality 5 mm b the capacities of the other 3 cups. (Note: 1 mL is equivalent to 1 cm3.) 12 In a clock, the period, T, of the motion of the pendulum — that is, the time the pendulum takes to swing back and forth — is directly proportional to the square root of its length, l. If a 0.9 m long pendulum completes one full cycle of motion in 1.9 seconds, what is the length of the pendulum with a period of 2.5 seconds? What is the period of the motion of a pendulum which is exactly 2 m long? (Express the answers correct to 2 decimal places.) 13 For a particular type of square-based pyramid, its volume, V, varies directly as the cube of the side of its base, s. A square-based pyramid has a volume of 9 cm3 and the side of its base is 3 cm. Find: a the formula connecting the two variables b the volume of a pyramid with a base area of 25 cm2, correct to 2 decimal places c the length of the side of its base, if the volume is 576 cm3. d From studying measurement we know that the volume of any square-based pyramid with a base of size s and a height h is given by the formula 1 V = 3 s2h. Compare this formula with the one obtained in part a and comment on a particular feature of the pyramids in question.

234


14 WE7 For a constant resistance, R, the voltage of an electric circuit, V, is directly proportional to current, I. Find the effect on the voltage when the current is: a doubled b tripled c halved. 15 For each of the following, find the effect on m, when n is: i doubled ii tripled iii quadrupled iv halved. a m ∝ n2 b m ∝ n3 c m ∝ n 16 For each of the following, determine the effect on x when y is: i quadrupled ii divided by 4. a y ∝ x2 b y ∝ x (Hint: In each case write the equation of variation and transpose it to make x the subject first.) 17 The area of a circle varies directly as the square of its radius. a Find the effect on the area if the radius is: i doubled ii tripled iii multiplied by 8 iv multiplied by 10 v halved vi divided by 3. b If a circle with radius 7 cm has an area of 154 cm2, use your answers to part a to find the area of the circles with the following radii: 56 cm, 70 cm, 2 1 cm, 14 cm, 3.5 cm, 21 cm. 3 18 Copy and complete the following statements: a If y ∝ x, when x is multiplied by any number n, y is … b If y ∝ x2, when x is multiplied by any number n, y is … c If y ∝ x , when x is multiplied by any number n, y is … d If y ∝ x3, when x is multiplied by any number n, y is … b  19 (2, b) and  , 4 b belong to the same relation in which the second element varies directly as 3  the square of the first. a Find the value of b, a positive real number. b Find the constant of variation. c Find the value of a in a pair (3, a) which belongs to the same relation.

7C

Inverse variation Consider the following example. Stan used to collect basketball cards. Eventually he became bored with this hobby and decided to give all of his 120 cards to his classmates. If Stan distributed the whole collection between his 2 best friends, Mark and Eugene, they would each receive 60 cards. If he included another friend, Ashley, they would each receive 40 cards and so on. The more people who shared Stan’s collection, the fewer cards each person received. There are 25 people in Stan’s class, including himself. If he were to distribute 120 cards between all of his classmates, each student would receive 5 cards. This information can be represented graphically or as shown in the table. n C

1 120

2 60

3 40

4 30

5 24

6 20

8 15

10 12

12 10

15 8

20 6

24 5

C

(Note that only factors of 120 are included in order to avoid fractional answers). It is obvious that as the number of students, n, who are to share the collection increases, the number of cards, C, that each student receives, decreases. The product of the two variables is constant for each pair and equal to 120 — the size of the collection. That is: 1 × 120 = 2 × 60 = 3 × 40 = 4 × 30 = 5 × 24 = 6 × 20 = 8 × 15 = 10 × 12 = 120 and so on.

120

60 40 30 20 10 123456 8 10 12 15

20 24 n

Chapter 7 Variation

235

Hence, the relationship between two variables can be written as: C × n = 120, or 120 . C= n The graph of the relation is a hyperbola which has the C and n axes as its asymptotes. Summarising our observations, we can say that the following is true for the given information: 1. An increase in one variable causes a decrease in the other. 2. The product of the two corresponding variables is a constant. 3. Neither variable is equal to 0. 4. The graph which represents the data is a hyperbola. 1 If we calculate the values of for each of the values in our table, we will then be able to draw n 1 a graph of C against . n n 1 n C

1

2

3

4

5

6

8

10

12

15

20

24

1

1 2

1 3

1 4

1 5

1 6

1 8

1 10

1 12

1 15

1 20

1 24

120

60

40

30

24

20

15

12

10

8

6

5

1 is a straight line n directed from, but not passing through, the origin. (Note that we exclude the origin itself, hence the open circle at (0, 0), since the number of cards per person when shared between 0 students is undefined.) 1 Hence, we can deduce that C varies directly as , that is, as n the reciprocal of n. As you can see, the graph of C versus

C 120

60 30

In cases like this, we say that one variable varies inversely as 5 (or is inversely proportional to) the other. The product of any 1 1– 1– 1 0— 1 n– 24 4 2 two corresponding variables is constant and is called a constant of proportionality, k. Hence, C is inversely proportional to n (or C varies inversely as n, or directly as the reciprocal 1 of n). It is written as C ∝ . n The product of any two corresponding values of C and n is constant and equal to 120, that is Cn = 120. Therefore the constant of variation k = 120. 120 Therefore, the relationship between the two variables can be written as C = . n 1 Generally, for any two variables x and y, where y varies inversely as x, that is, y ∝ , there x k exists a relationship between them such that y = or yx = k, where k is a constant, called x the constant of proportionality (or the constant of variation). The graph of the relationship is a 1 hyperbola whereas the graph of y against is a straight line directed from, but not passing x through, the origin, and having the gradient k (where x ≠ 0). As with direct variation (Section 7A), the existence of inverse variation can be established either numerically, or graphically. Summarising this: 1 If y∝ x

236


k x where k is the constant of variation and k ∈ R\{0}, x ∈ R\{0}. y=

then

Worked Example 8

For the data represented in the table below, establish whether an inverse variation exists between x and y using: a a numerical approach (clearly specify k, the constant of variation, if applicable) b a graphical approach. x y

1 20

2

3

4

5

10

2 63

5

4

xy THINK a

b

1

WRITE/display

Find the product of xy for each of the 5 pairs of values. Note: One variable varies inversely as the other if the product between any 2 corresponding values is constant.

2

Compare each of the five products and answer the question.

3


1

2

1 Calculate the values of . x Place these values into a table.

a Product = xy

First pair: Second pair: Third pair: Fourth pair: Fifth pair:

1 × 20 = 20 2 × 10 = 20 3 × 6 23 = 20 4 × 5 = 20 5 × 4 = 20

Since the product of the corresponding values is the same in each case (that is 20), y varies inversely as x.

b

x

1

2

3

4

5

y

20

10

6 23

5

4

xy

20

20

20

20

20

x

1

2

3

4

5

1 x

1

1 2

1 3

1 4

1 5

y

20

10

63

2

5

4

1 x and y values into the table. Label the columns x and y. On the Statistics screen, enter the

Chapter 7 Variation

237

3

To find the rule that fits the data, tap: • Calc • Linear Reg Set: XList: main\x YList: main\y Freq: 1 Copy Formula: y1 • OK

4

To graph the rule, open the Graph & Tab screen. Enter the rule as y1. Tap $. Adjust the window accordingly.

5


1 is a straight line x directed from, but not passing through, the origin, hence an open circle at the point (0, 0). 1 Therefore y ∝ . x The graph of y against

As for cases involving direct variation, if we know that one variable varies inversely (or indirectly) as the other, it is possible to establish the value of k, the constant of variation, and hence determine the value of any variable given its corresponding value. Worked Example 9

1 For the data in the table below given that y ∝ , find: x a the constant of proportionality b the missing values. x y THINK a

238

1

2 24

4 8

6

WRITE


a y∝

1 x


b

k x

2


y=

3

Hence, write down the rule for k. Obtain a pair of values where both x and y are known. Substitute these values into the given equation and obtain a value for k. Note: To obtain k, we can use only corresponding values; that is, those in which one value is underneath the other in the table.

k = xy When x = 4, y = 6. k=4×6 = 24

Find the unknown value by substituting its given corresponding pair into the equation xy = 24. First pair: substitute y = 24. Transpose the equation to make x the subject.

b From part a xy = 24

1

2

The constant of variation is 24. 24 Therefore, xy = 24 or y = . x x × 24 = 24 24x = 24 24 x= 24 x=1

First pair:

Second pair: substitute x = 2. Transpose the equation to make y the subject.

Second pair:

Third pair: substitute y = 8. Transpose the equation to make x the subject.

Third pair:


2 × y = 24 2y = 24 24 2 y = 12 y=

x × 8 = 24 8x = 24 24 8 x=3 x=

x

1

2

3

4

y

24

12

8

6

Inverse variation can be applied in many practical situations. The approach to solving word problems involving inverse variation is similar to that for direct variation. Steps to solve word problems involving inverse variation. 1. Write the statement of variation. 2. Write the equation of variation. 3. Use any pair of variables whose values are known to establish the constant of variation. 4. Find all other required values. Worked example 10

Marina is buying prizes for the competitions which are to be held during Maths Week. She has enough money to buy 24 pens valued at $2.50 each. a Pens of better quality are sold at $4 each. How many of these can she buy? b If she wishes to buy enough identical prizes for 100 students, what would be the maximum cost of each prize?

eBook plus Tutorial


Chapter 7

Variation

239

THINK a

b

1

WRITE

Define the variables. Write the statement of variation. This is an example of inverse variation since the amount of money is fixed and as the price increases, the number of prizes which can be bought decreases.

a Let p = the price of each prize.

Let n = the number of prizes. 1 n∝ p k p

2


n=

3

Obtain a pair of values where both p and n are known. Substitute these values into the given equation. Transpose the equation and solve for k.

4

Rewrite the equation of variation by substituting the value of k.

When p = 2.5, n = 24 k 24 = 2.5 24 × 2.5 = k k = 60 60 n= p

5

Substitute p = 4 to find the corresponding value of n.

When p = 4, n =

6


Marina may purchase 15 pens at $4 each.

1

Write down the equation obtained in part a .

2

Substitute n = 100 to find the corresponding value of p.

When n = 100, 100 =

3

Transpose the equation to make p the subject.

4

Evaluate.

100 × p = 60 100p = 60 60 p= 100 = 0.60

5


60 4 = 15

b n = 60

p

60 p

The maximum cost of each item would have to be $0.60 or 60c.

REMEMBER

1. For any variables x and y, where y varies inversely (or indirectly) as x, the following properties exist: (a) one variable increases as the other decreases (b) neither variable is equal to 0 (c) the product of any pair of corresponding values is constant and equal to k (d) the graph which represents the relationship is a hyperbola. 2. The notation used to express y varies inversely as x is given by: 1 y∝ x k or y = where k ∈ R\{0} and where x ∈ R\{0} x

240


Exercise

7c

Inverse variation 1 WE8a

For the data represented in the tables below, establish whether an inverse variation

exists between x and y using a numerical method and give the constant of proportionality. a

x

1

2

3

4

y

12

6

4

3

b

xy c

1 28

2

3

4

14

1 93

7

d

xy e

1

3

6

9

y

18

6

3

2

x

2

4

5

6

y

5

2.5

2

13

xy

x y

x

2

xy

x

4

5

6

10

12

y

15

12

10

8

5

f

xy

x

4 5

2 3

1 2

1 4

2 10

y

5

6

8

16

25

xy

2 WE8b For the data represented in the tables below, establish whether an inverse variation exists between x and y using a graphical method. a

c

e

b

x

2

4

8

10

y

60

30

15

12

x

1 2

1 3

1 4

1 5

y

5

7.5

10

12.5

x

3

5

8

10

y

13

8

5

4

d

f

x

1

4

5

8

y

40

10

8

6

x

1 4

1 2

2

4

y

20

10

5

2.5

x

2 3

3 5

1 4

1

y

18

20

48

12

1 1 3 MC If the relationship between m and n is such that m ∝ and m = 6 when n = , then n 2 which of the following belongs to the same relationship? 1

A (3, 4 )

B (12, 1) C (3, 1) 1 4 MC The expression y ∝ could be represented by: x i y

ii y

0

x iv y

0

A i and iii

iii y

x

0

1

B iii only

0

x

vi y

v y

x–

1 E (12 , 1)

D (3, 6)

1

x–

C ii and vi

0

D iii and v

1

x–

E iii and iv

Chapter 7 Variation

241

1 5 WE9 Given that y ∝ , for each of the following find: x a the constant of proportionality b the missing values. i

iii

x

2

y

6

x

1

2.5

2.4 4

y v

x

1 5

1 4

y

1 53 1 2

20

x

vi

2

20

viii

30 1 60

1 10

x

4

y

30

x

1

10

15

12 3

y

5 8

1 4

iv

6

16

y vii

ii

3

2

13

x

4

y

16

8

x

3

4

y

62 3

1

83

62

32

48

4 30 4 3

2

1 Questions 6 to 8 refer to the following table, representing the relationship u ∝ . v v

4

u

15

5

8 10

7.5

3

6 MC To obtain a straight line graph directed from, but not passing through, the origin, we should plot the values from the table with: A v on the horizontal axis and u on the vertical axis B u on the horizontal axis and v on the vertical axis 1 C v on the horizontal axis and on the vertical axis u 1 D on the horizontal axis and v on the vertical axis u 1 E on the horizontal axis and u on the vertical axis. v 7 MC The gradient of the straight line is: A 4 4 D 15

B 15 E 4 × 15

C

15 4

8 MC Consider the second and third pairs of the relationship and complete the following statement: Compared to the value of v in the third pair, the value of u in the second pair is: A twice as large B twice as small C 4 times as large D 15 times as small E 60 times as large 9 WE 10 Annette is buying prizes for the competitions which are to be held during Science Week. She has enough money to buy 40 items valued at $3.20 each. a Items of better quality are sold at $4.00 each. How many of these can she buy? b If Annette wishes to buy sufficient identical prizes for 80 students, what would be the maximum cost of each prize?

242


10 A team of 5 bricklayers is about to start on a new project. According to the schedule, they have 12 days to complete their job. a If one of the bricklayers gets sick just before they start, how many days would the remaining 4 bricklayers need to complete the job? b It takes the sick bricklayer 5 days to get better. He then goes back to work. In how many days will the team complete their job under these circumstances? c Now imagine that when the bricklayer got sick, his mates invited another workman to help. When the sick bricklayer returned to work 5 days later, they decided to keep working as a team of 6. How long, to the nearest day, will it take to complete the job in this case? 11 A gardener has enough fertiliser to use on a rectangular flower bed measuring 4 m by 9 m. a Would there be enough fertiliser to use on another bed if it is 3 m longer, and 3 m narrower? b Would there be enough fertiliser to use on a bed which is twice as short and twice as wide as the first one? c What is the side length of the square bed which requires exactly the same amount of fertiliser as the first one? 12 The frequency of sound, f, is inversely proportional to the wavelength, l. a A signal generator emits waves of frequency 1000 Hz and wavelength 0.34 m. Find the frequency of waves whose wavelength is: i 200 m ii 800 m. b If the constant of proportionality is the speed of sound, v, state the value of the speed of sound and write the equation which relates the frequency, speed and wavelength of sound. 13 When a force is applied to an object, its acceleration, a, varies inversely with its mass, m. When the mass of the object is 20 kg, the corresponding acceleration is 5 m/s2. a Find the effect on the acceleration when the mass of the object is: i doubled ii tripled iii halved. b The constant of variation is the force, F newtons. Find the magnitude of the force applied to the object in question.

Chapter 7 Variation

243

14 From Physics, we know that the work done, Wd, when a certain object is pushed can be calculated by multiplying the force, F, applied to it by the distance, d, it was pushed; that is, Wd = Fd. Which of the three variables should be fixed as a constant so that the other two would vary inversely? 15 Before use, 1 mL of Betadine sore throat mouthwash must be diluted with water to produce 20 mL of ready-to-use solution. a How many millilitres of water are required? Instead of diluting the mouthwash each time it is needed, Irene decided to prepare a large jar of the solution using the whole bottle. b If a bottle contains 15 mL of concentrated Betadine, how much water is required? c What is the total volume of the solution? At a certain stage the solution contains 190 mL of water. d How many millilitres of Betadine are there in the solution at that stage? e What percentage of the solution is used? 16 In a tape recorder, the speed of rotation of a reel varies inversely as its diameter; that is, the reel with a lot of tape on it rotates slower than the one with the small amount of tape. If the reel with the large amount of tape has a speed of 70 revolutions per minute, what is the speed of the other reel at that moment? 17 The speed of a gear wheel, s, varies directly as the reciprocal of its number of teeth, t. a In the gear train shown, the first gear drives all the others. It has 20 teeth and makes 300 revolutions per minute. Gears 2 and 4 are the same and have 8 teeth each while gear 3 has 40 teeth. Find the speed with which gears 2, 3 and 4 rotate. b If the process were reversed and the fourth gear became the driving one, what would be the speeds of gears 1, 2 and 3, given that the new driving gear rotated with the same speed as the old one; that is, 300 revolutions per minute?


WorkSHEET 7.1

7d

Further inverse variation In the section on further direct variation, we saw that direct variation exists between variables raised to powers other than 1. The same is true for inverse variation. We will concern ourselves mainly with powers of 2, 3 and 1 (square roots). It should be understood that there could be 2 inverse variation relationships with variables raised to any other power.

Worked example 11

The electrical resistance, R, of a wire varies inversely as the square of its diameter, d. If a wire with diameter of 4 mm has the resistance of 4 ohms, find: a the resistance of a wire with diameter 1.2 cm in exact form b the diameter of the wire, correct to 2 decimal places, when the resistance is 8 ohms. THINk a

244

1

WrITe


a

R∝


1 d2

b

2


3

Substitute R = 4 and d = 4 into the equation to find constant of proportionality, k.

4

Transpose the equation to make k the subject and evaluate.

5

Rewrite the equation of variation by substituting the value for k.

6

Convert d = 1.2 cm into mm.

7

Substitute 12 mm into the equation in place of d.

8

Simplify and include the appropriate unit.

1

Rewrite the equation obtained in part a .

2

Substitute R = 8 into the equation.

3

Transpose the equation to make d2 the subject.

4

Solve for d by taking the positive square root of both sides.

5

Simplify and include the appropriate unit.

R=

k d2

When d = 4 and R = 4, k 4= 2 4 k 4= 16 4 × 16 = k k = 64 R=

64 d2

1.2 cm = 1.2 × 10 = 12 mm When d = 12, R =

64 122

4 ohms 9 64 b From part a , R = 2 d =

8=

64 d2

8d2 = 64 64 d2 = 8 =8 d= 8 =2 2 = 2.83 mm

When two variables are inversely proportional, an increase in one variable causes a decrease in the other. The size of the increase/decrease depends on the type of relationship.

Worked example 12

If y ∝

1 x

, m∝

eBook plus

1 1 , p∝ , find the effect on y, m and p, when x is doubled. x2 x

Tutorial


THINk 1

Write the given variation statement for each.

WrITe

1 1 1 y ∝ , m ∝ 2, p ∝ x x x

Chapter 7

Variation

245

2

Write the equation of variation for each. Note: k does not necessarily represent the same numerical value. It is just conventional to denote the constant of variation with k.

3

Substitute 2x into the equations in place of x to find the new values of y, m and p.

4

Rewrite the equations in terms of the original y, m and p.

5


k x k m= 2 x k p= x y=

k 2x 1 k = × 2 x k mnew = (2 x ) 2 k = 2 4x 1 k = × 2 4 x k pnew = 2x k 1 = × x 2 ynew =

1 y 2 1 mnew = m 4 1 pnew = p 2 ynew =

When x is doubled, y is halved, m is divided by 4 (a quarter of the original or is decreased by a factor of 4) and p is divided by 2 (or is decreased by a factor of 2 ).

REMEMBER

1. Inverse variation exists between variables, raised to powers other than 1. 2. When two variables are inversely proportional, an increase in one variable causes a decrease in the other. The size of the increase/decrease depends on the type of relationship.

Exercise

7D

Further inverse variation 1

246

Copy and complete the given tables and hence deduce whether inverse variation exists for each of the following.


a

x

1

2

3

b

4

x3 y

60

20 9

7.5

15 16

y

x3y c

x

2

3

4

5

15

4.4

1.875

0.96

1

3

5

7

343

12 19

2 93

1

x3 x3y

x

2

4

5

d

6

x3

x x3

y

130

41

16

22

25

111

y

64

x3y

27

125

x3y

2 For each of the following, plot y against x and decide whether inverse variation exists between y and the square root of x. a

c

x

1

4

9

25

y

60

30

20

12

x

4

9

16

36

y

45

30

20

15

b

d

x

1

4

9

16

y

30

15

10

7.5

x

4

16

25

49

y

10

5

4

27

6

3 Determine whether m varies inversely as n , n, n2 or n3 by plotting corresponding graphs. A CAS calculator may be used to create appropriate tables of values and then to plot the data. n

16

36

64

m

1296

256

81

4 If p is inversely proportional to the cube of q, find: a the constant of proportionality b the missing values. q p

2

3

6 5 15 8

8

17

4 27

5 Write each of the following as: a a variation statement b an equation with the constant of variation p. i The current, I, is inversely proportional to the resistance, R. ii The force, F, varies inversely as the square of the distance, d, between 2 magnetic poles. iii The resistance, R, varies inversely as the square of the diameter of the wire, d. iv The number, n, of oscillations of a pendulum is inversely proportional to the square root of its length, l. v The time of a journey, t, varies directly as the reciprocal of an average speed, v. vi The light intensity, I, is directly proportional to the reciprocal of the square of the distance from the source of light, d. 6 WE 11 The intensity of the light, I, varies inversely as the square of the distance between the observer and the source of light, d. If I = 5.5 units when d = 7 m, find: a the distance of the observer from the source of light, correct to 2 decimal places, when its intensity is 12 units

Chapter 7 Variation

247

b the intensity of light observed from a distance of 1.75 m.

7 The intensity of sound, I, is inversely proportional to the area of the source of sound. If I = 0.5 W/m2 when the sound passes through the rectangular opening measuring 2 m by 5 m, what is: a the intensity of the sound for i a square opening with a side length of 70 cm? ii a rectangular opening with a length of 3 m and a width that is half of its length? b the length of the side of a square opening if the sound passing through it has an intensity 5 of 9 W/m2?

IB I (or B ), where IA, IB, IC is the intensity of sound IA IC

perceived by A, B and C respectively.

2m

b the ratio

2m

8 The period, T, of the motion of the particle oscillating on the end of an elastic string is inversely proportional to the square root of the stiffness of the string, k. When an object of mass 2 kg is suspended from a string with the stiffness of 10 N/m, it oscillates with the period 2π of seconds. Find: 7 a the period when the same object oscillates on the end of the elastic string with the stiffness of 25 N/m 32π b the stiffness of the string, if the time needed for one oscillation is . 7 9 Three friends A, B and C stand in a straight line 1 m apart from each other. A and C are both 2 m from the speaker. If the intensity of sound varies inversely as the square of the distance of the person from the source of sound, find: a the distance of person B from the sound

1m A

1m B

10 We12 For a certain type of prism with a fixed volume, the height, H, varies inversely as the square of the side of the base, s. Find the effect on the height of the prism if the side of its base is: a doubled b halved c quadrupled d divided by 4. 248


C

1 and y = 12 when x = 4, what is the effect on y when x is tripled? x A y is tripled. B y is divided by 3. C y is quadrupled. D y is divided by 4. E y is unchanged. b The value m varies inversely as the square root of n, and m = 10 when n = 4. When n is halved, m will be: A halved B doubled C multiplied by 2 D divided by 2 E none of these answers. c If p is inversely proportional to the square of s, then for the value of p to be divided by 4, the corresponding value of s should be: A quadrupled B divided by 4 C doubled D halved E squared

11 mC a If y ∝


Investigation The effect of changing one variable on another

7e

Joint variation Up to now, we have considered relationships between two variable quantities. However, in real life there are many situations that involve more than two variables. In such situations, where one variable is directly proportional to the product or quotient of other variables, we say that joint variation takes place. The variation statement is written as before, except that there will be more than one variable on the right-hand side of the statement. For example, if one quantity, x, varies directly as the product of two other quantities, y and z, it is said that x varies jointly as y and z and is written as x ∝ yz, or x = kyz, where k is a constant.

Worked example 13 a From the table below, state whether x varies

jointly as y and z.

b Establish whether x ∝

below.

y

1

2

3

4

y

1

1

2

3

z

2

3

4

5

z

1

2

3

4

x

10

30

60

100

x

5

6

8

9

THINk a

y from the table z

WrITe

1

Write the statement of variation.

2


3


4

Substitute the values x, y and z into the equation to find the constant of proportionality, k.

5

Evaluate k for each corresponding value of x, y and z.

6

Compare each k value.

x ∝ yz

a

x = kyz x kyz = yz yz x k= yz k=

10 30 k= 2 ×1 3×2

10 2 =5

=

30 6 =5

=

k=

60 3×4

60 12 =5

=

k=

100 4×5

100 20 =5

=

k = 5 for each corresponding value of x, y and z.

Chapter 7

Variation

249

b

7


1


2


3


4

Substitute the values of x, y and z into the equation to find the constant of proportionality, k.

5

Evaluate k for each corresponding value of x, y and z.

6

Compare each k value.

7


x varies jointly as y and z, that is x = 5yz. y b x∝ x ky x= z z ky z ×x= × y z y xz =k y xz k= y k=

5 ×1 6×2 k= 1 1

k=

8×3 9×4 k= 2 3

5 12 24 36 = = = = 1 1 2 3 = 5 = 12 = 12 = 12 k is not consistent for each corresponding value of x, y and z. y x is not ∝ . z

Worked Example 14

The volume of a cone, V cm3, varies jointly as the square of its radius, r cm, and the height, h cm. When r = 5 cm and h = 10 cm, V = 261.8 cm3, find: a the volume of a cone, when the radius is doubled and the height remains unchanged, to 2 decimal places b the height of the cone, when r = 12 cm and V = 2714.3 cm3, to 2 decimal places. THINK a

250

WRITE a

V ∝ r2h

1


2 3

Write the equation of variation. Substitute the known values of V, r and h into the equation.

V = kr2h 261.8 = k × 52 × 10 = k × 25 × 10

4

Simplify the RHS of the equation.

5


261.8 = 250k 261.8 250 k = 250 250

6

Evaluate.

7

Rewrite the equation of variation with 1.0472 in place of k.

V = 1.0472r2h

8

Write the values of r and h to be used and substitute them into the given equation.

Substitute r = 10, h = 10 into V. V = 1.0472 × 102 × 10

261.8 250 = 1.0472

k=


b

= 1.0472 × 100 × 10 = 1047.2

9

Evaluate.

10

Give the answer to 2 decimal places and include the appropriate unit.

1

Write the equation found in part a .

2

Write the values of r and V to be used and substitute them into the given equation.

When r = 12 and V = 2714.3, h = ? 2714.3 = 1.0472 × 122 × h

3


4

Transpose the equation to make h the subject. Evaluate.

2714.3 = 1.0472 × 144h = 150.797h 2714.3 150.797h = 150.79 150.797 2714.3 h= 150.797 = 17.999 7

5

Round the answer to 2 decimal places and include the appropriate unit.

V = 1047.20 cm3 b V = 1.0472r2h

h = 18.00 cm

Worked example 15

eBook plus

The electrical resistance, R, of a wire varies jointly as its length, l, and the reciprocal of the square of its diameter, d. Find the percentage change in the resistance if the length of the wire is increased by 25% and its diameter is decreased by 20%. THINk

Tutorial


WrITe

1


R∝

1 d2

2


R=

kl d2

3

Write the values of the new length and diameter. Note: If we treat the original l and d as one (1), then an increase of 25% (that is, 0.25) would make the new value 1.25, whereas a decrease of 20% (that is, 0.20) would make the new value 0.80.

4

Substitute the new values of l and d into the given equation and call the new resistance Rnew.

5


6

Replace

kl with R. d2

lnew = l + 25%l = l + 0.25l = 1.25l

Rnew =

dnew = d - 20%d = d - 0.20d = 0.80d

k × 1.25l (0.80 d )2

k × 1.25l = 0.64 d 2 1.953 125kl = d2 Rnew = 1.953 125R

Chapter 7

Variation

251

7

Convert the decimal to a percentage by multiplying it by 100%.

Rnew = 195.3125% of R

8


The resistance has increased by 95.3125%, which is almost double the original R.

rememBer

1. Joint variation occurs when one variable is directly proportional to the product, or quotient, of other variables. 2. Regardless of the number of variables in the joint variation relationship, there is only one constant of variation, k. exerCIse

7e

eBook plus

Joint variation

Digital doc

Spreadsheet 060

1 We 13a For each of the following state whether x varies jointly as y and z. a

c

y

1

2

3

4

z

2

3

4

x

4

12

y

2

z x

y

2

3

4

5

5

z

1

4

4

5

24

40

x

6

36

48

75

3

5

6

y

4

6

8

10

1

2

3

4

z

2

4

5

6

1

3

8

12

x

2

6

10

15

c

d

y

1

2

3

4

z

2

4

6

x

2

2

y

2

z x

b

y

1

2

3

4

8

z

5

4

3

2

2

2

x

4

10

20

40

4

6

8

y

2

3

5

6

1

2

3

5

z

2

3

4

5

6

6

6

6

x

3

4

5

6

d

3 If it is known that m ∝ n2p: a find the constant of variation b fill in the table at right. 4 If v ∝

u

Joint variation

y For each of the following establish whether x ∝ . z

2 We 13b a

b

and v = 2, when u = 4 and w w = 36, what is k, the constant of variation? Hence, fill in the table.

n

3

p

3

m

6

18

u

3

4

w v

5 6

10 9

126 5

4 9

6

450 10

16 5

6

6

5 Write the equation of variation for each of the following. a Power, P, varies directly as the square of the voltage, V, and inversely as the resistance, R. b Power, P, varies jointly as the resistance, R, and the square of the current, I. 252


c Kinetic energy, E, is directly proportional to the square of the velocity, v, and the mass, m. d Force of a circular motion, F, varies directly as the mass, m, and inversely as the radius, R. e Frequency of a sound, f, varies directly as the square root of the tension, t, and inversely as the square root of the mass per unit length, m. 6 We14 The volume of a cylinder, V cm3, varies jointly as the square of its radius, r cm, and the height, h cm. When r = 12 cm and h = 15 cm, V = 6785.84 cm3, find: a the volume of the cylinder, when the radius is tripled and the height remains unchanged, to 2 decimal places b the height of the cylinder, when r = 7 cm and V = 1328.7 cm3, to 2 decimal places. 7 The change in potential energy, ∆U, varies jointly as the mass of the object, m, and the change in the height, ∆h. When an object which weighs 10 kg is lifted from the floor to a vertical distance of 1.4 m, the object gains 140 J of potential energy. a Find the gain in the potential energy when an object twice as heavy is lifted to the height of 1 m. b Find the mass of the object which gains 220.5 J of potential energy when lifted to the height of 3.5 m. c To what height should an object of mass 250 g be lifted, so that the change in its potential energy reaches 75 J? 8 The energy expenditure, in joules, for various activities varies jointly as the weight, in kilograms, of the person partaking in the activity and the time, in minutes, spent on the activity. For instance, a girl weighing 52 kg will expend 615 J of energy when playing badminton for half an hour. a Alex weighs 102 kg. If he plays badminton for 2 hours, how much energy will he use? b If Alex wants to expend 2091 J of energy, what length of time must he play for? c Alex diets for a few months and as a result weighs considerably less. If he still wishes to expend the same amount of energy as in part b, should his play time be longer or shorter than before? d Select the correct alternative in the following sentences. i If two people engage in the same activity for the same period of time, the heavier person will expend (more/less) energy. ii If two people of different weight expend the same amount of energy while playing badminton, the heavier person must play for a (longer/shorter) time. 9 We15 The variable m varies directly as the square root of n and inversely as the square of p. When n = 16 and p = 3, then m = 2.4. a Write the equation which describes the relationship between m, n and p. b Find the value of m when n = 9 and p = 6. c Find the value of p, which is a positive number, when n = 441 and m = 4.536. 10 The resistance of any wire, R ohms, is directly proportional to its length, l m, and inversely proportional to its cross-section area, A m2. A copper cable which is 800 m long and has a

Chapter 7

Variation

253

-

cross-section area of 5 mm2 (5 × 10 6 m2), has a resistance of 2.848 ohms. a The constant of proportionality is called the resistivity of the material. Find the resistivity of the copper cable. b Find the resistance of a 4 km long copper cable which has a cross-section area of 16 mm2. Questions 11 to 13 refer to the following information. The volume of a right cone, V, varies jointly as the square of its radius, r, and its height, h. The original cone has r = 10 cm and h = 20 cm. 11 mC A second cone has a radius which is double that of the first cone and a height which is half that of the first cone. The volume of the second cone compared to the volume of the first cone will be: A the same B halved C doubled D divided by 4 E quadrupled 12 mC If the volume of the original cone is approximately 2094 cubic units, then the constant of variation is close to: A but less than 1 B but greater than 1 C 10 D but less than 10 E but greater than 100 13 mC If the radius of the original cone is halved, what should the height be, so that the volume remains unchanged? A 5 cm B 10 cm C 20 cm D 40 cm E 80 cm 14 mC The frequency of a string of a musical instrument, F, varies jointly as the square root of the tension, t, in the string, the reciprocal of the length of the string, l, and the reciprocal of the square root of the mass per unit length, m. The equation that describes the relationship between the 4 variables could not be written as: A F=

k l

F = k l Fl E = k C


SkillSHEET 7.1 Percentage increase and decrease

254

t m t

B Fl = k D

Fk = l

t m t m

m t m 15 We 15 The electrical resistance, R, of a wire varies jointly as its length, l, and the reciprocal of the square of its diameter, d. Find the percentage change in the resistance if the length of the wire is increased by 30% and its diameter is decreased by 15%. 16 When Casey rides his motorbike in circles, the force of circular motion, F, varies directly as the square of its velocity, v, and inversely as R, the radius of the circle. Find: a the percentage change in force when, while riding along the same track, Casey: i increases his velocity by 10% ii decreases his velocity by 10% b the percentage change in force when Casey is riding around a circle whose radius is 25% larger than that of the original c the percentage increase in the radius, required to maintain the size of the original force, when the velocity is increased by 15%.



WorkSHEET 7.2

7F

17 Coulomb’s Law states that the force between two charges at rest, F, is directly proportional to the product of the charges, q1 and q2, and inversely proportional to the square of the distance between the charges, r. a Write the equation which represents this relationship. b What effect will the following changes have on the size of the force, F? i The distance between the charges is doubled. ii The distance between the charges is halved. iii One of the charges is doubled. iv Both charges are doubled.

part variation So far in this chapter, we have considered relationships in which one quantity varied as a product or quotient of other quantities. However, relationships may also consist of two or more parts added together. In this situation, we say that part variation takes place. When part variation occurs, each of the parts will have its own constant of variation. If the relationship between two variables x and y is such that y varies partly as x and is partly constant, it is written as y = ax + b and is called part linear variation.

Worked example 16

My telephone bill consists of 2 parts: a fixed charge of $32 (paid whether any calls are made or not) and a charge proportional to the number of calls made. Last quarter I made 296 calls and my bill was $106. a Find the equation of variation. b Find the amount to be paid when 300 calls are made. THINk a

1

WrITe/dIsplaY

Define each variable to be used.

2


3

Substitute the values for A and n into the equation.

4

To solve the equation for k, on the Main screen, tap: • Action • Advanced • solve Complete the entry line as: solve(106 = 296k + 32, k) Then press E.

5

Rewrite the equation substituting 1 in 4 place of k.

a Let A = the total amount to be paid, in dollars

Let n = the number of calls A = k n + 32

When n = 296 and A = 106, 106 = 296k + 32

So

A = 1 n + 32 4

Chapter 7

Variation

255

b

1

2

Substitute n = 300 into the given equation.

b

When n = 300, A = 1 (300) + 32 4

= 1 × 300 + 32

Evaluate.

4

= 75 + 32 = 107 3

Answer the question and include the appropriate unit.

The amount to be paid when 300 calls are made is $107.

Worked example 17

eBook plus

The variable y varies as the sum of two quantities, one of which varies directly as x and the other inversely as x2. When x = 1, y = -17 and when x = 2, y = 1. a Find the equation for y in terms of x. b Find y when x = 5. THINk a

y∝x+

a

2

Write the equation of variation. Let a represent the constant of variation of x and let b represent the constant of 1 variation of 2 . x

3

Substitute the first set of values for x and y into the equation and label it [1].

When x = 1, y = -17. -17 = a + b

Substitute the second set of values for x and y into the equation and label it [2].

When x = 2, y = 1. b 1 = 2a + 4

To obtain the values of a and b, solve the equations simultaneously.

y = ax +

b x2

(a) Multiply equation [2] by 4 and label it equation [3].

[2] × 4: 4 = 8a + b

(b) Subtract equation [1] from equation [3].

[3] - [1]: 4 = 8a + b -(-17 = a + b) 21 = 7a 21 7a = 7 7 3=a a=3

(c) Divide both sides of the equation by 7.

256

1 x2


5


WrITe

1

4

Tutorial

6

Substitute a = 3 into equation [1].

Substituting a = 3 into [1]: -17 = 3 + b

7

Solve for b. Subtract 3 from both sides of the equation.

-17

−3=3−3+b =b b = -20

-20


[1]

[2]

[3]

b

8

Substitute the values of a and b into the equation of variation.

9

Verify the answer obtained using a CAS calculator.

1


2

Substitute x = 5 into the given equation.

3

Evaluate.

4


b

y = 3x −

20 x2

y = 3x −

20 x2

When x = 5, y = 3(5) −

20 52

20 25 = 15 − 0.8 = 14.2 = 15 −

When x = 5, y = 14.2.

REMEMBER

1. Part variation occurs if the relationship consists of two or more parts, added together. When part variation occurs, each of the parts will have its own constant of variation. 2. If the relationship between two variables x and y is such that y varies partly as x and is partly constant, it is written as y = ax + b and is called linear variation. Exercise

7F

Part variation 1 Write an equation defining each of the following relationships. a y varies as the sum of two quantities, the square of x and a constant. b l varies partly as the square root of m and partly as the cube root of n. c y varies partly as the reciprocal of x and partly as a constant. d y varies as the sum of two quantities, the cube of x and the square of z. 2 WE 16 My telephone bill consists of two parts: a fixed charge of $48 (paid whether any calls are made or not) and a charge proportional to the number of calls made. Last quarter I made 400 calls and my bill was $232. a Find the equation of variation b Find the amount to be paid when 440 calls are made. 3 Lana and Michael are planning their engagement party. They found that the cost of the party consists of two parts: a fixed charge for renting the reception hall and hiring the band, and a charge for food proportional to the number of people who are invited. They were told that if 100 people were invited, the cost would be $4500, while if 150 people were invited, the cost would be $5750. a Determine the cost of the fixed charge. b Find the equation which relates the total cost of the party, C, and the number of people, n. c Find the cost if 120 people are invited. d Represent the situation graphically. 4 Janus knows that the cost of producing French bread at home (using a breadmaker) consists of two parts: a fixed part, which represents the cost of the breadmaker, and the cost of the ingredients, which is proportional to the number of loaves made. It costs him $243 to make 20 loaves and $299 to make 100 loaves.

Chapter 7 Variation

257

a b c d

Find the cost of the breadmaker. Find the cost of flour and other ingredients necessary to produce each loaf. Find the cost of making 500 loaves. If Janus sells the French bread to the local cafe at $2.50 per loaf, find the number of loaves that he must produce in order to make a profit.

5 The cost of a health program, not including the cost of vitamins and other supplements, is partly constant and partly varies with the number of weeks a customer wishes to stay on the program. It costs $501 for 12 weeks and $633 for 18 weeks. a Find the cost of being on the program for 16 weeks. b Find the number of weeks a customer was on the program, if she paid a total of $721. c If every new client must pay an initial joining fee and then pay a health consultant for each weekly visit, state the amount of the joining fee and the amount charged by the consultant for each visit. 6 mC The relationship between two variables, m and n, is described by a linear variation. If m = 10.5 when n = 3, such a relationship could be represented by: 9 1 1 1 A m = n2 + 1 2 B m= +72 C m = n2 + 2 n n 2

6

D n= 5m+ 5

E m - 3 = 2.5n

7 The relationship between the velocity of the body, v, and the time, t, is described by part linear variation. The velocity of the body moving in a straight line with uniform acceleration is 20 m/s after 5 seconds and is 26 m/s 3 seconds later. Find: a the equation of linear variation b the initial velocity c the velocity when t = 28 d the time when the velocity is 34 m/s. 8 We17 The variable y varies as the sum of two quantities, one of which varies directly as the square of x and the other inversely as x. When x = 2, y = 11, and when x = 4, y = 47.5. a Find the equation for y in terms of x. b Find y when i x = 5 ii x = 1 . eBook plus Digital doc

SkillSHEET 7.2 Removing a fraction from a linear equation

3

9 The variable y varies as the sum of two quantities, x2 and a constant. When x = 2, y = 5 and when x = 1, y = 2. a Find the equation for y in terms of x by first finding the values of the two constants. b Find y when x = 1 . 3

10 The variable m varies partly as n and partly as the cube of n. When n = 2, m = 14 and when n = -1, m = -5.5. Find the equation defining this relation and hence find the value of m when n = 4. 11 The variable y varies partly as the reciprocal of x and partly as a constant. When x = 5, y = 6 and when x = 10, y = 4.5. a Find the equation for y in terms of x. b Find y when x = 12. c Find x when y = 30. 12 mC The relationship between two variables p and r is given by the following formula: b . The variable r = 6.25 when p = 4 and r = -17 when p = 1. p a The value of a is: C 20 A 6 B -20 3 D 23 E 3 b The value of b is: A 6 B -20 C 20 3 D 23 E 3 r=a p+

258


c When p = 25, r is equal to: A 30.92 B 32 C 29.08 D 34 E 30 d If r is equal to 59.77, p is equal to: A 36 B 49 C 64 D 81 E 100 13 The owner of a fancy dress store is making Halloween hats of a conical shape. The area of the material needed varies partly as radius of the hat and partly as the square of the radius. To make a hat with a radius of 8 cm, 703.36 cm2 of cardboard is needed, while for a hat of radius 10 cm, 942 cm2 of cardboard is needed. a Find the equation which connects the two variables. b Find the area of the cardboard required to make a Halloween hat of radius: i 9 cm ii 11 cm.

7G

Transformation of data

eBook plus

As shown in the first section of this chapter, when one quantity varies Interactivity directly as the other, the graph representing the relationship is a straight int-0974 Transformation line passing through the origin. We also observed that when graphed, of data the data may sometimes give the curves of a parabola, hyperbola, cubic and so on. However, when data are transformed appropriately, they will produce the graph of a straight line. By analysing the transformation which the values undergo, in order to produce a straight line graph, it is possible to establish relationships between the variables. Worked example 18

The following table shows the values of the total surface area, TSA, of spheres and their corresponding radii, r. Radius (r) (cm)

1

2

3

4

5

(cm2)

12.57

50.27

113.1

201.06

314.16

TSA

Graph the values given in the table and comment on the shape of the graph. Using the graph, or otherwise, find the equation which relates total surface area of the sphere, TSA, and its radius r. THINk 1

WrITe/dIsplaY

On the Statistics screen, enter values and label list1: radius and list2: tsa.

Chapter 7

Variation

259

260

2

To find the rule of the data and to graph it, tap: • Calc • Power Reg Set: XList: main\radius YList: main\tsa Freq: 1 Copy Formula: y1 • OK Open the Graph & Tab screen and tap $.

3

Comment on the graph obtained.

The graph is not a straight line, passing through the origin, so direct variation does not exist between the two variables. Hence, there is no direct variation between the radius and the total surface area of the sphere.

4

Make assumptions about the graph obtained.

The graph resembles a parabola, so it is reasonable to assume that area is directly proportional to the square of the radius.

5

Write the variation statement for the assumption made.

TSA ∝ r2

6

Write the variation equation.

7


TSA = kr2 TSA k= 2 r

8

Test the assumption by finding the values of r2 TSA and check that the ratio 2 is constant. r On the Spreadsheet screen, label column A: r, column B: tsa and column C: tsa / r2. Enter the data in columns A and B. In cell C2, complete the entry line: = B2/(A2)^2 Then press E. To fill down, highlight C2 to C6 and tap: • Edit • Fill Range • OK

9

Comment on the result obtained.

10

Alternatively, once the values of r2 have been calculated, rule up a table of values and plot TSA versus r2.

The ratio is constant for each corresponding pair (when rounded to 2 decimal places). Hence, TSA ∝ r2 TSA = kr2 TSA = 12.57r2 r2 TSA

1

4

9

12.57

50.27

113.1


16

25

201.06 314.16

TSA 300 200 100 0 14 9 16 25 r2 11

Comment on the graph obtained.

The graph is a straight line, passing through the origin.

12

Establish the value of k by substituting any pair of values from the table into the equation of variation and write the equation relating the two variables.

TSA = kr2 When r = 1, TSA = 12.57, k = ? 12.57 = k × 1 12.57 = k k = 12.57 TSA = 12.57r2

rememBer

1. If the data when plotted give a curve, aim to transform them in such a way that a straight line is produced. 2. Once this has been achieved, establish the relationship between the variables by analysing the transformation. exerCIse

7G

Transformation of data 1

We 18 The data in the table at right follow a x 1 2 3 4 5 particular variation relationship. y 1.76 7.04 15.84 28.16 44 a Graph the values given in the table and comment on the shape of the graph. b Using the graph, or otherwise, find the equation which relates the two variables, x and y.

2

The data in the table below follow a particular variation relationship. x

1

2

4

5

10

12.5

y

100

50

25

20

10

8

Digital doc

Spreadsheet 133

a Plot the graph of y versus x and comment on its shape. 1 1 b Plot the graph of y versus and 2 and hence establish x x the relationship between the two variables. c Find the equation which relates the two variables, x and y. 3

eBook plus


The following table shows the volumes, V, of 6 spheres, to the nearest cm3, with various diameters, D. D (cm) V

(cm3)

22

26

30

34

38

42

5575

9203

14 137

20 580

28 731

38 792

a Graph the values given in the table and comment on the shape of the graph.

Chapter 7

Variation

261

b Using the graph, or otherwise, find the equation which relates the volume of a sphere, V, to its diameter, D. (Give the value of the constant of variation correct to 4 decimal places.) 4 The data in the table below follows a particular variation relationship. x

1

2

3

4

5

6

7

8

9

10

y

5

5.60

5.95

6.20

6.40

6.56

6.69

6.81

6.91

7

a Graph the values given in the table and comment on its shape. b If the relationship between the two variables x and y is given by the formula y = a log10 (x) + b, determine the values of the constants a and b. 5 MC The graph represents the relationship between m and n. It is likely that: A n ∝ m B n ∝ m2 C n ∝ m3 1 D n ∝ E n ∝ m m 6 MC If s ∝ t2, the relationship between s and t could be shown by: A s B s C t

t

0

D s

t

0

0

m

s

0

t

0

n

E s

0

t

7 As part of her Science project, Julia tested the reaction times of 10 people (the ‘subjects’), using the experiment outlined below. Julia holds a 30-cm ruler from the end labelled 30 cm. The subject, whose reaction time is being recorded, keeps his or her fingers open at the other end of the ruler, level with the 0-cm mark of the scale. Subjects close their fingers on the ruler as soon as they see that it has been released. The distance, d cm, that the ruler falls is noted and the reaction time, t, is calculated according to a certain formula. The table shows the reaction times calculated for the 10 subjects, using the distances the ruler fell before each had closed their fingers on it. Person

1

2

3

4

5

6

7

8

9

10

d (cm)

3

7

15

5.5

12

21

4.5

6

11

17

t (s)

0.0779 0.1191 0.1743 0.1055 0.1559 0.2062 0.0955 0.1102 0.1492 0.1855

a Use a CAS calculator to plot values of t against d and comment on the result. b Use a CAS calculator to plot the graphs of t versus d2 and t versus d and comment on the results. c Analyse the graphs obtained in part b and hence find the rule which relates reaction time, t, with the distance, d. d Use the equation from part c to calculate the reaction time if the distance the ruler falls is 20 cm.

262


Summary Direct variation

• For any two variables x and y, where y varies directly as x: y 1. The ratio between any pair of corresponding values, , is constant and equal to k, the constant of k proportionality (or constant of variation). 2. The graph, which represents the variables x and y, is a straight line passing through the origin with the gradient equal to k. 3. As one variable increases, the other variable also increases. • The notation used to express y varies directly as x is given by: y ∝ x. • There are many cases of direct variation, where variables are raised to powers other than one, that is, y ∝ x n. Inverse variation

• For any two variables x and y, where y varies inversely as x: 1. One variable increases as the other decreases. 2. Neither variable is equal to 0. 3. The product of any pair of corresponding values, xy, is constant and equal to k. 4. The graph which represents the relationship is a hyperbola. k • The notation used to express y varies inversely as x is given by y = , where k ∈ R\{0} and x ∈ R\{0}. x 1 • Inverse variation exists between variables raised to powers other than one, y ∝ . xn Joint variation

• Joint variation occurs when one variable is directly proportional to the product, or quotient, of other variables. • Regardless of the number of variables in the joint variation relationship, there is only one constant of variation, k. Part variation

• Part variation occurs if the relationship consists of two or more parts, added together. When part variation b occurs, each of the parts will have its own constant of variation. For example, y = ax 2 + , where a and b x are constants. • If the relationship between two variables x and y is such that y varies partly as x and is partly constant, it is written as y = ax + b, where a and b are constants and is called a part linear variation. Solving variation problems

• To solve any type of variation problem, follow these steps: 1. Write the statement of variation. 2. Write the equation of variation. 3. Substitute known values to find the constant of proportionality, k. 4. Find all unknown values as required in the given problem. Transformation of data

• If data when plotted give a curve, aim to transform them in such a way that a straight line is produced. Once this is achieved, establish the relationship between the variables by analysing the transformation.

Chapter 7 Variation

263


1 Work done, W joules, on a mass by a certain force is directly proportional to the distance, d m, the mass is moved. When a 3 kg mass is moved a distance of 4 m, the work done is equal to 84 joules. Find: a the constant of proportionality b the work done, when the same mass is moved a distance of 12 m c the distance the mass was moved, if the work done on the 3 kg mass is 136.5 joules. 2 The total surface area, TSA cm2, of a cube is directly proportional to the square of its length, l m. If a cube of length 4 cm has a total surface area of 96 cm2, find: a the constant of proportionality b the total surface area of a cube of length 12 cm c the length of a cube which has a total surface area of 384 cm2. 1 3 The table below represents the relationship y ∞ . x Find: x y

5 19.2

6

12

8

a the constant of proportionality b the missing values. 4 The intensity of sound, I, is inversely proportional to the area of the source of sound. If I = 0.6 W/m2 when the sound passes through the rectangular opening measuring 4 m by 8 m, find: a the intensity of the sound for: i a square opening with a side length of 2 m ii a rectangular opening with a length of 7 m by 3 m b the length of the side of a square opening, if the sound passing through it has an intensity of 1.2 W/m2. 5 Power, P, is the rate of doing work. It varies directly as the work done, W, and inversely as time, t. a What type of variation is involved? b Write the statement of variation. c What will be the effect on P, when time during which the work is done, is: i doubled? ii tripled?

264

d From question 1 we know that W varies directly as the distance, d. What would be the effect on P if the mass is moved over twice the distance in: i twice the time? ii half the time? Multiple choice

1 If m ∝ n and (n1, m1) and (n2, m2) represent corresponding values, which of the following equalities is not true? m1 m2 = A n1m1 = n2m2 B n1m2 = n2m1 C n1 n2 n1 m1 m2 n2 = = E D n2 m2 m1 n1 2 Variable a is directly proportional to b, and a = 12 when b = 48. When b = 10, a will equal: A 40 B 5 C 2.5 D 2 E 1 3 If y is directly proportional to x and x is quadrupled, then the value of y is: A doubled B tripled C quadrupled D halved E divided by 4 4 Which of the following tables represent the relationship between x and y, such that y ∝ x ? A

C

E

x

4

9

y

6

12

x

9

16

y

12

16

x

9

25

y

9

20

B

D

x

4

16

y

6

16

x

4

25

y

8

10

5 Which of the following relationships could be represented by this graph? m

0

n

A m ∝ n B m ∝ n C m ∝ n2 D m ∝ n and m ∝ n2 E none of these


6 The total surface area, TSA, of a particular 3-dimensional shape is directly proportional to the square of the side, s, and TSA = 24 when s = 2. When s = 11, the total surface area is: A 66 B 726 C 132 D 121 E 847 7 If a ∝ b3 and b is doubled, then the value of a will be: 1 A multiplied by 2 B unchanged C multiplied by 2 D multiplied by 4 E multiplied by 8 8 The graphs which show an inverse variation between the variables a and b are: ii b i b

B s2

A s

C

1 s

1 1 E s2 s Questions 12 and 13 refer to the table below. D

m

4

9

16

p

6

4

3

12 The table of values represents the variation relationship: 1 A p ∝ m B p ∝ C p ∝ m2 m 1 1 D p ∝ 2 E p ∝ m m 13 The gradient of the straight line shown in the diagram is equal to: p

0

a

1–

a

iii b

1

m

0

a

A i only B ii only C i and ii D ii and iii E iii only 9 If m varies inversely as n, which of the following statements is false? A The graph of m versus n is a hyperbola. B The product of any 2 corresponding values of m and n is constant. m C The quotient of any two corresponding n values is constant. 1 D The graph of m against is a straight line. n n m E For any corresponding values 1 = 2 . n2 m1 10 If y varies inversely as x and x is tripled, then the value of y is: A multiplied by 3 B divided by 3 C divided by 1 D multiplied by 9 3 E divided by 9 11 The variable r varies inversely as the square of s and is represented by the vertical axis. For the resulting graph to be a straight line, directed from the origin, the horizontal axis must be represented by:

3 A 3 × 16 B 3 × 4 C 16 3 D E none of these 4 14 The variable y varies jointly as x, the square root of z and as the reciprocal of the square of w. The equation which does not describe the above variation relationship is: kx z B kx z = w 2 y w2 yw 2 yw 2 C kx = D z = kx z A y =

E w 2 = kxy z 15 The variable x varies jointly as the square of b and the cube of c. When b is doubled and c is halved, x is: A halved B doubled C divided by 8 D quadrupled E multiplied by 8 16 The height of a cylinder, h, is directly proportional to its volume, V, and inversely proportional to the square of its radius, r. When V = 15.7 and r = 1, the height is equal to 5. The constant of proportionality k is closest to: A 3 D

1 9

B

1 3

C 9

E 5

Chapter 7 Variation

265

17 Which of the following graphs represents a part linear variation between x and y? i

should be applied to the data, in order to produce a straight line passing through the origin is:

ii y

y

t

x

x

iii y

0

0

x

A i only B ii only C iii only D i and iii E i, ii and iii 18 m and n are related by a part nonlinear variation. The equation which could represent such a relationship is: A m = 2 - 3n B m = 6 + n + 2n C m + n = 12.5 D m = n - 0.5n2 3n 2 +n E m= n 19 The table below shows the corresponding values b of x and y, such that y = ax + 2 . x x

1

2

y

-11

6

The respective values of a and b are: A 3, -20 B 20, -3 C 5, -16 D 5, 16 E 16, 5 20

A B C D E

squaring each value of p squaring each value of t taking the square root of each value of p taking the reciprocal of each value of p none of the above transformations

21 For a particular set of data, plotting y against x produced a straight line passing through the origin. Before the transformation, the graph representing the data would have been: A

0

C

B

y

D

x

E

p

2

7

10

5

t

12

147

300

75

y

exTeNded respoNse

1 A compound microscope contains 2 lenses: an objective lens and an eyepiece lens. The angular magnification, M, of such a microscope varies directly as the distance between the two lenses, L, and inversely as the product of the focal length of the objective lens, fo and focal length of the eyepiece lens, fe. M = 2.508 when L = 12 cm, fo = 52 cm and fe = 2.3 cm. a Find, correct to the nearest whole number, the constant of variation, k, and hence deduce the formula which relates the 4 variables. b Use the formula to find the angular magnification of a microscope with L = 10 cm, fo = 48 cm and fe = 3.5 cm, correct to 3 decimal places.


x

y

0

x

y

When plotted, the data from this table produce the graph shown above right. The transformation which

266

p

x

y

x

2 Solids expand when they are heated. The length of a steel rod, l, after it has been heated is partly constant and partly varies directly as the change in temperature, ∆T , that took place during the heating process. A 20 m long rod expands by 4.8 mm after it has been exposed to a 20 degrees change in temperature. a State the length of the rod, in metres, after it has been heated. b Find the equation which relates the length of the expanded steel rod, l, to the change in temperature, ∆T. c If the constant of variation represents the product of the length of the rod before it has been heated and the coefficient of thermal expansion, λ, find the value of λ for the steel. Write the value obtained in scientific notation. 3

x

1

4

y

3 20

25

9

2

3 12 20

16

25

38 25

93 43

For the data in the above table, plot: i y versus x ii y versus x2 iii y versus x . a From the plotted data determine which variation is being represented. b Use the table of values or the appropriate graph from part a to find the relationship between x and y. 4 The data in the table below follow a particular variation relationship. p

1

2

4

6

8

10

V

114

57

28.5

19

14.25

11.4

a Plot the graph of V versus p and comment on its shape. b Using the graph or otherwise, find the equation which relates the two variables p and V. c Use the result obtained in part b to complete the following sentence which describes this particular variation relationship, also known as Boyle’s Law. For a given mass of gas at constant temperature, the volume, V, varies _____________ with the pressure, p. 5 The magnification, M, produced by a lens is related jointly to the distance of the object from the lens, u cm, and the distance of the image from the lens, v cm. Table A shows the values of magnification of the image of the object, when u varies and v is fixed. Table B shows the values of magnification of the image of the object, when v varies and u is fixed. Table A u

1

2

4

5

M

−40

−20

−10

−8

Table B v

6

10

12

M

-1 2

-5 6

−1

20 -

2

13

24 −2

a Plot the values from table A and comment on the shape of the graph. b Use the graph to select the most appropriate relationship between the variables M and u, from the 1 following: M ∝ u2, M ∝ u or M ∝ . u c Test your assumption either numerically or graphically. If your assumption proved to be incorrect, repeat b and c; otherwise proceed to d. d Plot the values from table B and comment on the shape of the graph.

Chapter 7 Variation

267

e Complete the following statement: M varies ______ as v. f If it is known that M, u and v are related by joint variation and the constant of variation is -1, deduce the formula which connects the 3 variables. (Hint: Make magnification, M, the subject.) 6 The kinetic energy, E, of a car is related to the speed, v, the car travels and the mass, m. Table A shows values of kinetic energy, E, for several cars of different mass for a certain fixed value of speed. Table A E (joules) m (kg)

695 556

772 840

850 124

927 408

1 000 692

1800

2000

2200

2400

2600

1 . E From your graphs decide on the type of variation that exists between E and m. b Prove your answer to part a using a numerical method. Table B shows values of speed, v, for the cars with different mass, m, for a certain fixed value of kinetic energy. a Plot the graphs of m versus E, m versus E2 and m versus

Table B v (m/s)

29.3

27.8

26.5

25.4

24.4

m (kg)

1800

2000

2200

2400

2600

c Plot the graph of m versus v. What type of variation does the shape of the graph suggest? 1 1 1 d Plot the graphs of m versus , m versus 2 , m versus . v v v From your graphs, write the variation statement for the relationship between m and v. e Complete the following statement: E varies _____ as m and _____ as _____ of v. f If m = 2000 kg when v = 30.2 m/s and E = 912 040 J, find the constant of variation, k, and hence write the formula relating m, v and E. g Use your formula to find the value of kinetic energy when the car weighing 2100 kg travels with the speed of 135 km/h. (Hint: Convert the speed into metres per second first.) 7 The electrical resistance, R, of a wire varies jointly as its length, l, and the reciprocal of the square of its diameter, d. a Write the statement of variation. b Find the constant of proportionality and complete the table of values below. R

12

l

0.8

d

4

3

15

20

0.5

3.75

2

5

1

k c Use the results obtained in parts a and b to find the equation which relates the electrical resistance of the wire to its length and diameter. d i If l is doubled and d remains the same, what happens to the value of R? ii If l remains the same and d is doubled, what happens to the value of R? iii If both l and d are doubled, what happens to the value of R? e i If l is decreased by 20% and d is decreased by 60%, determine whether the value of R increases or decreases and by what amount. eBook plus ii If l is decreased by 10% and d is increased by 20%, determine whether the value Digital doc of R increases or decreases and by what amount. Test Yourself f If l is decreased by 75%, what change must be made to d for R to remain unchanged? Chapter 7

268


eBook plus

aCTIVITIes


• 10 Quick Questions: Warm up with ten quick questions on variation. (page 218) 7A

Direct variation

Tutorial

• We3 int-1057: Watch how to calculate distance travelled and the travel time of a person whose distance travelled varies directly with time. (page 223) 7B

Further direct variation

Tutorial

• We7 int-1058: Watch how to calculate the effect on the area of a square when the sides are doubled and halved. (page 230) Digital doc

• Spreadsheet 133: Investigate transforming data. (page 231) 7C

Inverse variation

Tutorial

• We10 int-1059: Watch how to use variation to determine costs of prizes. (page 239) Digital doc

• WorkSHEET 7.1: Use provided data to establish direct variation, provide data from a proposed direct variation, solve worded problems, establish direct variation graphically, use provided data to establish inverse variation and provide data from a proposed inverse variation. (page 244) 7D

Further inverse variation

Digital docs

• Spreadsheet 060: Investigate joint variation. (page 252) • SkillSHEET 7.1: Practise percentage increase and decrease. (page 254) • WorkSHEET 7.2: Determine both direct and inverse variation numerically and graphically and solve more complex worded problems. (page 255) 7F

Part variation

Tutorial

• We17 int-1062: Watch how to find the equation given how one variable varies with another. (page 256) Digital doc

• SkillSHEET 7.2: Practise removing a fraction from a linear equation. (page 258) 7G


Interactivity

• Transformation of data int-0974: Consolidate your understanding of transformation of data. (page 259) Digital doc

Tutorial

• Spreadsheet 133: Investigate transforming data. (page 261)

Digital doc

Digital doc

• We12 int-1060: Watch how to find the effect on dependent variables when independent variables are doubled. (page 245) • Investigation: The effect of changing one variable on another. (page 249) 7E Tutorial

Joint variation

• We15 int-1061: Watch how to find the percentage change in resistance if wire length is varied. (page 251)

Chapter review


Chapter 7

Variation

269


1 Simplify

20 minutes

2 5 . (3 x - 1) (2 x + 1)

3 marks

2 Solve the following pair of simultaneous equations for x and y. ax + by = c x – y = d 3 marks 1 1 3 and are the first two terms in an ( 2) - 1 ( 2) + 2 arithmetic sequence. Determine the value of the common difference. 3 marks 4 A circular pendant is being made for a necklace. It is to be made using two different materials.

Copper plate

θ Gold metal

The diagram above shows the two different sections for the pendant. If the area of gold metal π is A = 2π - 3 3 cm2 and the angle θ = and 3 3 sin (θ ) = , show that the radius is 2 3 . 2 4 marks MULTIPLE CHOICE

10 minutes

Each question is worth 1 mark. 1 2x° 3.4

5.2

The value of x° in the triangle above is closest to: A 20.42 B 24.58 C 40.83 D 49.17 E 98.34 270

Chapters 4 TO 7

2 The relationship between m and n is such that 1 1 m ∝ , when n = the value of m = 7.2. If n 3 m = 9.8, the value of n will be closest to: A 0.245 B 0.454 C 2.204 D 3.267 E 4.083 3

25 cm 7π 6

The diagram above shows the dimensions of a decorative glass insert for a door. The glass panel is divided into four equal sections. Each section of glass is stained a different colour. The area, in cm2, of each section is closest to: A 204.53 B 286.34 C 490.87 D 818.12 E 1145.37 4 Tina starts her new fitness schedule. On the first day she walks 2 km. The next day she walks 3 km. On day 3, she walks 4.5 km and on day 4 she walks 6 km. If Tina continues to increase the distance she walks in this pattern for 7 more days, the total distance, in km, she would have walked would be: A 54.5 B 87.5 C 97.5 D 99.5 E 104.5 5 During her weekly grocery shopping, Sian buys 2 loaves of bread and 3 litres of milk for $10.50. The next week she buys 3 loaves of bread and 4 litres of milk for $14.95. The next week, Sian is having a barbecue for her friends. The amount of money it will cost Sian to buy 5 loaves of bread will be: A $8.00 B $10.50 C $10.60 D $14.25 E $22.25 6 The volume of a cylinder varies jointly as the square of the radius and the height. The radius is decreased by 20% and the height is increased by 10%. The percentage change in the volume is a: A 4.4% increase B 12.0% decrease C 29.6% decrease D 70.4% increase E 88.0% increase


exTeNded respoNse

30 minutes

1 Theo is raising money for a children’s charity. He plans to cycle from Melbourne to Sydney along the coastal road, a total distance of 1110 km. Before he begins, Theo plans his rest stops each day. On the first day he plans to cycle for 50 km before stopping for a rest. He then plans to stop for a rest after every 25 km he cycles. a If he continues to follow this plan for the total distance between Melbourne and Sydney, determine the number of rest stops he will need. 2 marks Theo plans to cycle 100 km each day. After 500 km, he suffers an injury to his foot and can only manage to cycle 85% of the previous day’s total distance. b If the daily distance cycled by Theo continues to follow this pattern, will he reach Sydney? Justify your answer with appropriate calculations. 3 marks 2 After his first 300 km, Theo has a rest day. He decides to walk to the lookout on top of the hill behind the hotel he is staying in. He leaves his hotel and sets out on a bearing of 030°T and walks the 3.5 km to the lookout. a Determine how far north Theo is from the hotel. Write your answer in exact form. 2 marks After walking 3.5 km to the lookout, Theo decides to follow another path down the hill. He sets out on a bearing of 135°T and walks 4.5 km to the road.

Lookout 135°

θ°

3.5 km

4.5 km

30° Hotel

b Show that θ ° equals 75°. 1 mark c Determine the minimum distance, in km, Theo has to walk to return to the hotel. Write your answer correct to 2 decimal places. 2 marks Theo would like to determine the vertical distance the lookout is from the hotel. He takes some measurements during his hike. He determines the angle of elevation from his point A to the top of the lookout is 15.20°. He walks 2 km towards the hill to point B and determines the angle of elevation to the top of the lookout to be 32.35°. The diagram below represents Theo’s measurements. Lookout y km A

32.35°

15.20° 2 km

B

x km

d If the distance from point B to the vertical base of the lookout is defined as x and the vertical height of the lookout is defined as y, show that y = 2 tan (15.2°) + x tan (15.2°). 2 marks e Determine the height, in m, of the lookout. Write your answer correct to the nearest metre. 3 marks eBook plus Digital doc


exam practice 2

271

8

8A Polynomial identities 8B Partial fractions 8C Simultaneous equations

Further algebra areas oF sTudy

• The solution of simultaneous equations arising from the intersection of a line with a parabola, circle or rectangular hyperbola using algebra eBook plus

8a

polynomial identities

Digital doc

10 Quick Questions

Before discussing the definition of a polynomial identity, it is important to remember some basic definitions. • An algebraic expression is made up of terms. • In the term axn, a is referred to as the coefficient of xn. • A constant is a term with no variable beside it. For example 2x3 + 3 is an algebraic expression made up of two terms. The coefficient of 3 x is 2. The constant is 3. A polynomial identity is an identity of the form: kn xn + kn - 1xn - 1 + kn - 2x n - 2 + . . . . . + k1x + k0, n ∈ N where kn , kn - 1. . . are constants and n is an element of the set of natural numbers N. The degree of a polynomial is given by the highest value of n. Hence a polynomial of degree 1 is linear, of degree 2 is a quadratic, of degree 3 is a cubic, of degree 4 is a quartic and so on. Worked example 1

Which of the following are polynomials? Give reasons for your answers. 1 a x3 + 2x2 + 1 b x+ c (2x + 6)5 x Think a In order for x3 + 2x2 + 1 to be a polynomial,

the powers must all be greater than or equal to 0, which they are. The highest power of x is 3. 1 b In order for x + to be a polynomial, the x powers must all be greater than or equal to 0, which they are not.

272

WriTe a x3 + 2x2 + 1 is a polynomial of degree 3 since it has

descending powers of x and these powers are all greater than or equal to zero, i.e. n ∈ N.

b This is not a polynomial since the second term has

a power of -1.


c In order for (2x + 6)5 to be a polynomial, the

powers must all be greater than or equal to 0, which they are. The highest power of x is 5.

c This is a polynomial of degree 5, since when

expanded, it has n ∈ N.

Two polynomials are said to be equal if each x-value generates the same y-value. Polynomials are identical if they are of the same degree and corresponding coefficients are equal. Therefore, if: ax3 + bx2 + cx + d = 2x3 - 4x + 8 then a = 2, b = 0, c = -4 and d = 8. If two polynomials are known to be equal, then the process of equating coefficients can be used to solve problems. Worked Example 2

If 5x3 + 2x2- 7x + 1 = (2 a + b)x3 - ax2 -(b - c) x + 1, then find the values of a, b and c. Think

Write

Method 1: Technology-free 1

If 5x3 + 2x2- 7x + 1 = (2a + b)x3 - ax2 - (b - c)x + 1, then the each corresponding term must be equal. Equate the terms.

5x3 = (2a + b)x3 ⇒ 5 = 2a + b

[1]

2x2 = -ax2 ⇒ 2 = -a ⇒ -2 = a

[2]

-7x

= -(b - c)x = -(b - c) ⇒7 = b - c

⇒-7

[3] -2

2

Solve these equations using substitution.

Substituting a = into equation [1] gives 2(-2) + b = 5 b=9 Substituting b = 9 into equation [3] gives 9-c=7 c=2

3

Write the answer.

a = -2, b = 9 and c = 2

Method 2: Technology-enabled 1

On the Main screen, using the soft keyboard, tap: • ) • {N Enter the equations as shown. Then press E.

2

Write the answer.

a = -2, b = 9 and c = 2

Chapter 8 Further algebra

273

Worked example 3

Determine values of a and b if m4 + 4 = (m2 + am + 2)(m2 + bm + 2). Think 1

The right-hand side must first be expanded.

2

Equate the coefficients. The coefficients of m3, m2, and m are zero.

3

Solve for a and b.

4

Write the answer.

WriTe

m4 + 4 = m4 + bm3 + 2m2 + am2 + abm2 + 2am + 2m2 + 2bm + 4 = m4 + (b + a)m3 + (4 + ab)m2 + (2a + 2b)m + 4 3 0m = (b + a)m3 ⇒0=b+a [1] 0m2 = (4 + ab)m2 ⇒ 0 = 4 + ab

[2]

0m = (2a +2b)m ⇒ 0 = 2a + 2b From equation [1], b = -a Substitute b = -a into equation [2] 0 = 4 - a2 a2 = 4 a = ±2 a=2 or a = -2 and b = -2 b=2 When a = 2, b = -2 and when a = -2, b = 2.

Worked example 4

eBook plus

If x - 4 is a factor of x3 - 6x2 + 2x + 24, find the other factor. Think

[3]

Tutorial

WriTe



Since the expression x3 - 6x2 + 2x + 24 is cubic, the other factor must be a quadratic, hence it is of the form ax2 + bx + c.

x3 - 6x2 + 2x + 24 = (x - 4)(ax2 + bx + c) RHS = ax3 + bx2 + cx - 4ax2 - 4bx - 4c = ax3 + (b - 4a)x2 + (c - 4b)x - 4c

2

Equate the coefficients.

x3 = ax3 ⇒1=a

[1]

-6x2

= (b - 4a)x2 ⇒ = b - 4a

[2]

2x = (c - 4b)x ⇒ 2 = c - 4b

[3]

-6

3

274

Solve for a, b and c.

Substitute a = 1 into equation [2] = b - 4a b = -2 Substitute b = -2 into equation [3] 2 = c - 4b c = -6

-6


4

Substitute the values for a, b and c into ax2 + bx + c and write the answer.

When a = 1, b = -2 and c = -6 then the quadratic factor of x3 - 6x2 + 2x + 24 is x2 - 2x - 6.


On the Main screen, tap: • Action • Transformation • factor Complete the entry line as: factor(x3 - 6x2 + 2x + 24) Then press E.

2

Write the answer.

The quadratic factor of x3 - 6x2 + 2x + 24 is x2 - 2x - 6.

REMEMBER

1. A polynomial identity is an identity of the form knxn + kn - 1 xn - 1 + kn - 2 xn - 2 + . . . . . + k1x + k0, n ∈ N where kn, kn - 1 . . . are constants and n is an element of the set of natural numbers N. 2. The degree of a polynomial is given by the highest value of n. 3. Polynomials are identical if they are of the same degree and corresponding coefficients are equal. 4. If two polynomials are known to be equal, then the process of equating coefficients can be used to solve problems. Exercise

8A

Polynomial identities 1 WE 1 For each of the following expressions: i state whether or not it is a polynomial ii if yes to i then give its degree. 2 x3 + 2x a 2x4 + 1 b + x2 + 3 c (3x2 + 2)3 d 3 x x 2 WE 2 Find the values of a, b and c if (2a + b)x3 + (b - c)x2 + (a + 2c)x + 5 = 3x3 - 5x2 + 10x + 5. 3

Find the values of a, b and c if x3 + 2x2 - 13x + 16 = (x - 2)(ax2 + bx + c) + 6.

4

Find constants a, b given that 2x3- 5x2 + 10 = (x - 2)2 (2x + a) + bx + c.

5 WE 3 Determine the values of a and b if m4 + 25 = (m2 + am + 5)(m2 + bm + 5). 6

If x2 = a(x + 1)2 + b(x + 1) + c, find the values of a, b and c.

7

If ax3 + bx2 + cx + d = (2x - 1)2 (mx + n), express b in terms of c and d.


275

8 We 4 If x - 2 is a factor of x3 + 3x2 - 16x + 12, find the other factor. 9 10

8B

If x + 1 is a factor of x3 - x2 + x + 3, find the other factor. If 2x + 1 is a factor of 2x3 + 7x2 - 7x - 5, find the other factor.

partial fractions


int-0975 When a function is expressed as one polynomial divided by another, Partial fractions g ( x ) f(x) = , it is often desirable to express this using partial fractions. h( x ) This enables the function to be graphed more easily and also helps with the process of integration (which you will learn about in Mathematical Methods CAS).

proper fractions If g(x)and h(x) are both linear functions, then the function can be expressed as a proper fraction in the form: b f(x) = A + . h( x ) Worked example 5

Express

b 4x + 5 in the form A + . x-3 x-3

Think

WriTe


Express the numerator as 4(x - 3) + b; the value of b must be 17.

2

Write the answer in the form A+

b . x-3

4 x + 5 4( x - 3) + 17 = x-3 x-3 17 = 4+ x-3

Method 2: Technology-enabled

276

1

On the Main screen, tap: • Action • Transformation • propFrac Complete the entry line as:  4 x + 5 propFrac   x - 3  Then press E.

2

Write the answer in the form b . A+ x-3

4x + 5 17 = 4+ x-3 x-3


Consider the case where g(x) is a polynomial of degree 1 and h(x) is a polynomial of degree 2. g( x ) In this case the function, f ( x ) = , is a proper fraction, since the numerator has a smaller h( x ) power than the denominator. For every linear factor (ax + b) in the denominator, there will be a partial fraction of the form A . f ( x) = ax + b For every repeated linear factor of the form (ax + b)2 in the denominator, then the partial A B fractions will be of the form f ( x ) = . On occasions when it is impossible + (ax + b) (ax + b)2 A B to express the partial fractions in the form f ( x ) = , they can be written as + (ax + b) (ax + b)2 A B C . f ( x) = + + (ax + b) (ax + b) (ax + b)2 Worked Example 6

Express

x+3 in partial fraction form. x 2 - 3 x - 40

Think

Method 1: Technology-free

Write

x+3 x+3 , x ∈ R\{-5, 8} = - 3 x - 40 ( x - 8)( x + 5) x+3 A B = + ( x - 8)( x + 5) x - 8 x + 5

1

Factorise the denominator x2 - 3x - 40.

2

The denominator has two linear factors so there will be two partial fractions of the A B form + . ( x - 8) ( x + 5)

3

Express the sum of the two fractions on the right as a single fraction.

x+3 A( x + 5) + B( x - 8) = ( x - 8)( x + 5) ( x - 8)( x + 5)

4

Equate the numerators and simplify.

x + 3 = A(x + 5) + B(x - 8) x + 3 = Ax + 5A + Bx - 8B x + 3 = (A + B)x + 5A - 8B

5

Equate the coefficients to solve for A and B.

x = (A + B)x ⇒1=A+B ⇒ 1 - B = A

[1]

3 = 5A - 8B

[2]

x2

Substitute equation [1] into equation [2]. ⇒ 3 = 5(1 - B) - 8B ⇒ 13B = 2 2 ⇒B= 13 11 ⇒ A= 13 6

Substitute the values for A and B and write the answer in the form A B . + ( x - 8) ( x + 5)

x+3 11 2 , = + - 3 x - 40 13( x - 8) 13( x + 5) x ∈R\{-5, 8} x2


277


On the Main screen, complete the entry line as: x+3 2 x - 3 x - 40 Highlight the equation and tap: • Interactive • Transformation • expand • Partial Fraction • OK

2

Write the answer in the form A B + . ( x - 8) ( x + 5)

x2

x+3 11 2 = + , - 3 x - 40 13( x - 8) 13( x + 5)

x ∈R \ {-5, 8}

Worked example 7

eBook plus

2x - 1 Express in partial fractions. ( x - 2 )( x + 1)2 Think

Tutorial

int-1064

WriTe

Worked example 7


2

2x - 1 A B C , = + + 2 ( x 2 ) ( x + 1 ) ( x - 2)( x + 1) ( x + 1)2

The denominator has one linear factor and one repeated linear factor so there will be three partial fractions of the form A B C + + . ( x - 2) ( x + 1) ( x + 1)2

x ∈R\{-1, 2}.

Express the sum of the three fractions on the right as a single fraction.

2x - 1 = ( x - 2)( x + 1)2 A( x + 1)2 + B( x - 2)( x + 1) + C ( x - 2) ( x - 2)( x + 1)2

3


2x -1 = A(x2 + 2x + 1) + B (x2 - x - 2) + C (x - 2) 2x -1 = Ax2 + 2Ax + A + Bx2 - Bx - 2B + Cx - 2C 2x -1 = (A + B)x2 + (2A - B + C)x + A - 2B - 2C

4

Equate the coefficients to solve for A, B and C.

0x2 = (A + B)x2 ⇒ 0=A+B ⇒ A = -B

[1]

2x = (2A - B + C)x ⇒ 2 = 2A - B + C

[2]

-1

278

= A - 2B - 2C


[3]

Substitute equation [1] into equations [2] and [3].

-3B

6

Solve these equations simultaneously.

7

Substitute the values for A, B and C and write the answer in the form A B C + + . ( x - 2) ( x + 1) ( x + 1)2

3C = 3 ⇒C=1 Hence, -3B + 1 = 2 ⇒ -3B = 1 ⇒B= 1 3 1 ⇒ A= 3 2x - 1 1 1 1 = + , 2 3 ( x 2 ) 3 ( x + 1 ) ( x - 2)( x + 1) ( x + 1)2

5

-3B

+C=2 - 2C = -1

x ∈R\{-1, 2}


On the Main screen, complete the entry line as: 2x - 1 ( x - 2)( x + 1)2 Highlight the equation and tap: • Interactive • Transformation • expand • Partial Fraction • OK

2

Write the answer in the form

2x - 1 1 1 1 = + , 2 3 ( x 2 ) 3 ( x + 1 ) ( x - 2)( x + 1) ( x + 1)2

A B C . + + ( x - 2) ( x + 1) ( x + 1)2

x ∈R\{-1, 2}

Sometimes the denominator may consist of an irreducible quadratic (a quadratic which cannot be factorised using real numbers). These types of functions need to be expressed in partial fractions of the form: f ( x) =

A Bx + C . + 2 ax + b cx + dx + e

Worked Example 8 2 Express 5 x + 9 x + 10 in partial fractions. x3 - 8

Think


Factorise the denominator.

Write

5 x 2 + 9 x + 10 5 x 2 + 9 x + 10 = x3 - 8 ( x - 2)( x 2 + 2 x + 4)


279

2

3

5 x 2 + 9 x + 10 A Bx + C = + x - 2 x2 + 2x + 4 x3 - 8

The denominator has a linear factor and an irreducible quadratic factor so the partial fractions will be of the form A Bx + C . + 2 x - 2 x + 2x + 4

x ∈R\{2}.

Express the sum of the two fractions on the right as a single fraction.

5 x 2 + 9 x + 10 = x3 - 8 A( x 2 + 2 x + 4) + ( Bx + C )( x - 2) ( x - 2)( x 2 + 2 x + 4)

4


5x2 + 9x + 10 = A(x2 + 2x + 4) + (Bx + C) (x - 2) = Ax2 + 2Ax + 4A + Bx2 - 2Bx + Cx - 2C) = (A + B)x2 + (2A - 2B + C)x + 4A -2C

5

Equate the coefficients to solve for A, B and C.

5x2 = (A + B)x ⇒A+B=5 ⇒ B = 5 - A

[1]

9x = (2A - 2B + C)x ⇒ 9 = 2A - 2B + C

[2]

10 = 4A - 2C

[3]

6

Substitute equation [1] into equation [2] and then subtract equation [3] to solve for A, B and C.

7

Substitute the values for A, B and C and write the answer in the form A Bx + C . + x - 2 x2 + 2x + 4

Substituting [1] into [2] 2A - 2(5 - A) + C = 9 4A - 10 + C = 9 4A + C = 9 Subtracting equation [3] 3C = 9 C=3 A=4 B=1 5 x 2 + 9 x + 10 4 x+3 = + , x ∈ R \{2} x - 2 x2 + 2x + 4 x3 - 8


280

On the Main screen, complete the entry line as: 5 x 2 + 9 x + 10 x3 - 8 Highlight the equation and tap: • Interactive • Transformation • expand • Partial Fraction • OK


Write the answer in the form A Bx + C . + x - 2 x2 + 2x + 4

2

5 x 2 + 9 x + 10 4 x+3 = + , x ∈ R \{2} x - 2 x2 + 2x + 4 x3 - 8

improper fractions

g( x ) is an improper h( x ) fraction. In this case, division of polynomials needs to be performed first either by long division or synthetic division.

In the case where g(x) has a higher power than h(x) the function f ( x ) =

Worked example 9

eBook plus

2 Express x + 5 x - 2 as a partial fraction. x-1

Think

Tutorial

int-1065

WriTe

Worked example 9


The degree of the denominator is less than the degree of the numerator, so division must be performed first.

2

Divide the numerator by the denominator using long division.

3

Express the answer as partial fractions.

x - 1 is the divisor. x+6 x - 1 x 2 + 5x - 2

)

x2 - x 6x - 2 6x - 6 4 2 x + 5x - 2 4 = x+6+ , x ∈ R \{1} x -1 x -1


On the Main screen, complete the entry line as: x 2 + 5x - 2 x -1 Highlight the equation and tap: • Interactive • Transformation • expand • Partial Fraction • OK

2

Write the answer.

x 2 + 5x - 2 4 = + x + 6, x ∈R\{1} x -1 x -1

Chapter 8

Further algebra

281

rememBer

g( x ) : h( x ) • If g (x) and h (x) are both linear functions, then the function can be expressed in the For rational functions of the form f ( x ) =

b . h( x ) • Where the numerator is a linear function and the denominator is a quadratic which can A B be factorised, then the partial fraction will be of the form f ( x) = . + ax + b cx + d • When the denominator has repeated linear factors of the form (ax + b)2 then the partial A B fractions will be of the form f ( x ) = + . (ax + b) (ax + b)2 On occasions when it is impossible to express the partial fractions in the form A B f ( x) = + , they can be written as (ax + b) (ax + b)2 A B C f ( x) = + + . (ax + b) (ax + b) (ax + b)2 • When the denominator contains an irreducible quadratic then the partial fractions will form f ( x ) = A +

A Bx + C . + 2 ax + b cx + dx + e • In the case where g (x) has a higher power than h (x) the function is an improper fraction so division of polynomials needs to be performed either by long division or synthetic division. be of the form f ( x ) =

exerCise

8B

partial fractions 1

a 2

Digital doc

4

5

We8

4x + 7 x-2

b

x-5 x 2 - 5x + 6

x2

c

x+7 2x - 1

c

2x - 1 x 2 + 8x - 9

Express each of the following as partial fractions. 2 x + 14 x-4 b c 2 ( x - 1)( x + 3)2 x - 6x + 9

d

3x - 4 2x + 2

d

3x + 2 2x2 - 9x + 7

d

3x - 5 ( x - 2)( x + 1)2

Express each of the following as partial fractions.

x 2 + 3 x + 18 x2 + 5 b 2 2 ( x + 1)( x - 2 x + 5) x ( x + 3 x + 1)

We9

a

282

x+6 ( x + 1)( x - 4)

We7

a

b

Express each of the following as partial fractions.

x -1 a ( x + 2)2

eBook plus WorkSHEET 8.1

2x - 3 x +1

We6

a 3

Express each of the following as the sum of two terms.

We5

c

(x2

2 2 d x + 5x - 1 - 5 x + 5)( x - 2) x 3 - 27

Express each of the following as partial fractions. + 3x + 1 x+2

3 b x + 2x - 3 x-4

3 2 c 3x + 2 x - 4 x + 5 x2 + x + 6


3 d x +3 2x - 1

8c

Simultaneous equations It is impossible to solve one linear equation with two unknown variables. There must be two equations with the same two unknowns for a solution to be found. Such equations are called simultaneous equations. There are several different ways to solve simultaneous equations. In this section we consider algebraic solutions of simultaneous equations arising from the intersection of a line with a parabola, circle or rectangular hyperbola.

Worked Example 10

Solve simultaneously: y = x and y = x2 + 3x + 1. Think

Write


Write the equations and label them [1] and [2].

y = x y = x2 + 3x + 1

2

Substitute equation [1] into equation [2].

Substituting [1] into [2]: x2 + 3x + 1 = x

3

Transpose to make the RHS equal 0 and simplify.

x2 + 3x + 1 − x = 0 x2 + 2x + 1 = 0

4

Factorise.

5

Solve for x.

6

Substitute −1 instead of x into equation [1].

Substituting −1 into [1]: y = −1

7

Write the answer.

Solution set: (−1, −1)

[1] [2]

(x + 1)2 = 0 x+1=0 x = −1



2

Write the answer.

Solving y = x and y = x2 + 3x +1 for x and y gives x = -1 and y = -1. That is, (-1, -1).


283

Worked example 11

eBook plus

Solve simultaneously: y = x + 1 and x2 + y2 = 4. Think

Tutorial

WriTe




y=x+1 x2 + y2 = 4

2


Substituting [1] into [2]: x2 + (x + 1)2 = 4

3

Expand (x + 1)2, using the perfect square identity and transpose to make the RHS = 0.

x2 + x2 + 2x + 1 - 4 = 0 2x2 + 2x - 3 = 0

4

Solve for x, using the quadratic formula.

a = 2, b = 2, c = -3 x= = = = =

5

Write the two values of x separately.

6

Substitute

7

Substitute

-1 +

7

-1 -

7

x1 = y1 =

instead of x into

y2 =

2 equation [1] and simplify.

± 22 - 4 × 2 × (-3) 2×2 - ± 2 4 + 24 4 2 ± 28 4 2±2 7 4 -1 ± 7 2 -1 +

7

2

, x2 =

-1 +

-1 -

7

+1 2 1- 7 2 = + 2 2 1- 7 + 2 = 2 =

284

-2

7 +1 2 1+ 7 2 = + 2 2 1+ 7 + 2 = 2 1+ 7 = 2

instead of x into

2 equation [1] and simplify.

[1] [2]

1- 7 2


-1 -

2

7

8

Write the answer. (Make sure the values of x and y are matched properly; that is, x1 is placed with y1 and x2 with y2.)

 -1 + 7 1 + 7  Solution set:  2 , 2  ,    -1 - 7 1 - 7   2 , 2   



2

Write the answer.

Solving y = x + 1 and x2 + y2 = 4 for x and y gives x= x=

-(

7 + 1) and y = ( 7 - 1) or 2 2

7 -1 and y = 2

7 +1 2

That is,  - ( 7 + 1) - ( 7 - 1)  7 - 1 7 + 1 ,   or  2 , 2  . 2 2    

Worked Example 12

Solve simultaneously: y = 2x − 1 and y =

2 . x-3

Think

Write


2


y = 2x - 1 2 y= x-3


Substituting [1] into [2]: 2 2x - 1 = x-3

[1] [2]


285

3

Solve for x: (a) Multiply both sides of the equation by (x − 3). (b) Expand and make the RHS = 0. (c) Identify the values of a, b and c. (d) Substitute the values of a, b and c into the quadratic formula and simplify.

(2x − 1)(x − 3) = 2 2x2 − 7x + 1 = 0 a = 2, b = − 7, c = 1 7 ± (-7)2 - 4 × 2 × 1 2×2 7 ± 49 - 8 = 4 7 ± 41 = 4

x=

7 + 41 7 - 41 , x2 = 4 4

4

Write the two values of x separately.

x1 =

5

Substitute 7 + 41 into [1] and simplify. 4

 7 + 41  y1 = 2 ×  -1 4   7 + 41 2 2 2 5 + 41 = 2 =

6

Substitute

7 - 41 into [1] and simplify. 4

 7 - 41  y2 = 2 ×  -1 4   7 - 41 2 2 2 5 - 41 = 2 =

7

Write the answer (leave it in surd form).

 7 + 41 5 + 41  , Solution set:  , 4 2    7 - 41 5 - 41  ,   4 2  


286



2

Solving y = 2 x - 1 and y =

Write the answer.

x=

2 for x and y gives x-3

-(

or x =

-( 41 - 5) 41 - 7) and y = 4 4 41 + 5 41 + 7 and y = 4 4

That is,  - ( 41 - 7) - ( 41 - 5)   41 + 7 41 + 5  . or  , ,   4 4 4 4    

rememBer

Simultaneous equations, arising from the intersection of a line with a parabola, circle or a rectangular hyperbola, can be solved using algebra as follows: 1. Transpose one of the equations (it is better to choose a linear equation) to make either x or y the subject and substitute into the other equation. 2. Simplify the resulting equation (if properly simplified, it will result in a quadratic equation). 3. Solve the quadratic equation to find the value(s) of one variable. 4. Substitute the value(s) of the first variable into either of the two equations (preferably into the transposed one) and solve for the second variable. 5. Write the solution set.

exerCise

8C

eBook plus

simultaneous equations 1

a y = x, y = x2 + 5x + 4 c y = 2x, y = x2 + 4x + 1 e y = -2x, y = x2 - 2x - 1

g y + x = 1, y = 3x2 + 2x - 1 i 2x - 4y = 12, y = -2x2 + x + 6 2

3

Digital doc

We 10 Solve each of the following simultaneously.

SkillSHEET 8.1

Using substitution b y = -x, y = x2 + 3x + 4 to solve simultaneous 2 d y = 3x, y = x + 8x + 6 equations f y = x + 5, y = x2 − 4x + 11 x2 -4 h 2x + 3y = 6, y = 2 j 2y + 6x = 4, y = 4 - 3x2

We 11 Solve each of the following simultaneously. a y = x, y2 + x2 = 1 b y = -x, y2 + x2 = 1 2 + y2 = 4 d y = 1 - x, 4 = y2 + x2 c y = 2x, x 2 2 e x + y = 2, x + y - 9 = 0 f y - x - 3 = 0, (x + 3)2 + y2 = 16 x h y + 2 = 3x, (x - 1)2 + (y + 3)2 = 4 g y = 1 - , (x - 2)2 + y2 - 1 = 0 2 i 2 x + 4y = 4, x2 + (y + 1)2 = 25 j 6x - 3y = 12, (x - 2)2 + (y - 1)2 - 36 = 0 We 12 Solve each of the following simultaneously.

a y = 2x, y =

2 x -1

c y = 3x - 4, y =

4 1 - 2x

4 2- x 1 d y = 1 - 2x, y = -1 x+2 b y = x + 1, y =

Chapter 8

Further algebra

287

e 2y - x = 6, y - 3 =

2 x -1

f 2x + 4y - 8 = 0, y + 1 =

3 2- x

2 -1 - y = 0 3x 1 1 =y+2 j x + y = 5, =y i 4x - 3y = 12, 4 4 - 3x 3x - 2 4 mC Which of the following represent the solution to the pair of simultaneous equations x + y = 6 and y = 3x2 + 12x + 10? g x - 3 = 2y, y =

i

(

-1 3

1

,63

)

A i only

2 x -1

ii (-1, 7) B i and ii

h x = 2y - 4,

iii (-1, 1)

iv (-4, 10)

C ii and iv

D ii and iii

5 Buttons are to be attached to a shirt as shown on the diagram at right. If we draw a set of axes through the centre of the button, the position of the two holes can be described as the points of intersection of the line y = x with the circle x2 + y2 = 1. The other two holes are positioned at the points of intersection of the line y = -x with the same circle. Find the coordinates of the four holes. Give the answer correct to 2 decimal places. eBook plus Digital doc

WorkSHEET 8.2

288


E i and iv

y = −x

y2 + x2 = 1

y=x

Summary Polynomials

• A polynomial identity is an identity of the form: kn xn + kn - 1xn - 1 + kn - 2 x n - 2 + . . . . . + k1x + k0, n ∈ N where kn , kn - 1. . . are constants and n is contained within the set of natural numbers N. • The degree of a polynomial is given by the highest value of n. • Polynomials are identical if they are of the same degree and corresponding coefficients are equal. • If two polynomials are known to be equal, then the process of equating coefficients can be used. Partial fractions

g( x ) : h( x ) b • If g(x) and h(x) are both linear functions, then the function can be expressed in the form f ( x ) = A + . h( x )

For rational functions of the form f ( x ) =

• Where the numerator is a linear function and the denominator is a quadratic which can be factorised, then the A B partial fraction will be of the form f ( x ) = + . ax + b cx + d • When the denominator has repeated linear factors of the form (ax + b)2 then the partial fractions will be of A B the form f ( x ) = + . On occasions when it is impossible to express the partial fractions in (ax + b) (ax + b)2 the form f ( x ) =

A B A B C + + + , they can be written as f ( x ) = . (ax + b) (ax + b)2 (ax + b) (ax + b) (ax + b)2

• When the denominator contains an irreducible quadratic then the partial fractions will be of the form A Bx + C f ( x) = + . (ax + b) cx 2 + dx + e • In the case where g(x) has a higher power than h(x), the function is then an improper fraction so division of polynomials needs to be performed either by long division or synthetic division. Simultaneous equations

Simultaneous equations, arising from the intersection of a line with a parabola, circle or a rectangular hyperbola, can be solved using algebra as follows: • Transpose one of the equations (it is better to choose a linear equation) to make either x or y the subject and substitute into the other equation. • Simplify the resulting equation (if properly simplified, it will result in a quadratic equation). • Solve the quadratic equation to find the value(s) of one variable. • Substitute the value(s) of the first variable into either of the two equations (preferably into the transposed one) and solve for the second variable. • Write the solution set.


289


1 Determine the values of a and b where x4 + 7x3 +12x2 + x - 1 = (x2 + ax + 1)(x2 + bx - 1). 2 x - 2 is a factor of x3 + x2 - 11x + 10; find the other factor. x - 22 3 Express as partial fractions. x 2 - 8 x - 20 x3 - 2x 2 + x - 1 4 Express as partial fractions. x2 - 2 5 Find the coordinates of the points of intersection of 15 . the line y = 5x with the hyperbola y = x-2 6 Find the coordinates of the points of intersection of the line y = x with the parabola y = 4 - x2. Multiple choice

1 If 2x3 + 7x2 - 16x + 6 = (2x - 1)(ax2 + bx + c), then the values of a, b and c are: A a = 1, b = 4, c = 6 B a = -1, b = -4, c = 6 C a = 1, b = 4, c = -6 D a = 1, b = -4, c = 6 E a = 1, b = 4, c = -5 2 If 2x2 - 5x - 3 = a(x - b)(x - c) then the values of a, b and c are A a = 2, b = 3, c = 1 B a = 1, b = 3, c = 1 1 D a = 2, b = 3, c = 1 C a = 2, b = 3, c = 2 2 1 E a = 1, b = 3, c = 2

3 If

4x + 2 A B , then: = + ( x + 2)( x - 4) x + 2 x - 4

A A = 1, B = 3 C A = 1, B = 4 E A = 2, B = 3

B A = 3, B = 1 D A = 4, B = 1

5 x 2 + x + 12 A Bx + C , then the = + x3 + 2x 2 + 4 x x x 2 + 2x + 4 values of A, B and C would be: A A = 1, B = 2, C = 3 B A = 1, B = 3, C = 2 C A = 1, B = 2, C = 5 D A = -3, B = 2, C = 5 E A = 3, B = 2, C = 5

4 If

5 A solution to the pair of simultaneous equations 23 y = 1.5x and y = + 2.71 is: 20( x - 4) A  5 , 3 

B

2 2

D

 3 9  ,  2 4

C  3 , 15 

 5 9  ,  2 4

2 4

E none of these

x and (y − 1)2 + (x − 3)2 = 9 4 are solved simultaneously. When one of the equations is substituted into the other and the resultant equation is transposed to the form ax2 + bx + c = 0, the values of a, b and c are:

6 The equation y = 1 -

A

17 , 6, 6

18

1

D 16, -6, 9

B

17 , 6, 6

0

E

17 , 6, 16

9

C

1 , 6, 16

0

extended response

1 Find the coordinates of the points of intersection of the line y = x with: 2 a the hyperbola y = +6 3x - 1 b the circle (y + 1)2 + x2 − 4 = 0. In each case give the answer correct to 2 decimal places. 2 Consider the design, shown on the diagram at right: If we take the point of intersection of the straight lines to be an origin, the design can be described by the following system of equations: y=x y = −x 1 y = x -1 y = x y=0 x=0 x2 + y2 = 4 290


C B A T

E

F

G

I

D

H

S

M

R

QP O

N

J K L

As can be seen from the diagram, there are 20 points of intersection (not counting the centre point). a What is the radius of a circle described by the equation x2 + y2 = 4? b Using the answer to a, state the coordinates of points A, F, K and P. c Find the coordinates of point I by solving an appropriate pair of simultaneous equations algebraically. Leave the answer in a surd form. d State the coordinates of points C, R and N using symmetry. e Find the coordinates of point H by solving algebraically an appropriate pair of simultaneous equations. f Using the symmetry of the design and your answer to part e, write the coordinates of points D, S and M. 1 g State the points of intersection of the hyperbola y = and the circle. x h Find the coordinates of the points in question g by solving an appropriate pair of equations graphically, using a table of values or one of the iteration methods. Give the answer correct to 2 decimal places. -1 i State the points of intersection of the hyperbola y = and the circle. x j Choose a method and use it to find the coordinates of the points in question i. Give the answer correct to 2 decimal places. 3 A section of a roller coaster track is shown at right. It consists of three parts with the following equations: 3 2 d 16

AB:

h=-

+ 10

BC:

h = 163 d 2 - 3d + 16

Height (m)

CDE: h = 0.02d 3 - 1.25d 2 + 25d - 147.56 where h is the height of the track above the ground level and d is the horizontal distance from A. a Find the coordinates of point B, by solving a pair of simultaneous equations algebraically. b The track is closest to the ground when it is 8 metres horizontally from A. What is its height at that point? c Find the horizontal distance(s) from A, when the car is 6 metres above ground level. d Use a CAS calculator to find the coordinates of point C. h e By using a table of values or otherwise, find the E coordinates of point D. D f Point E is 30 metres horizontally from A and is the highest point of this section of the track. Find the A maximum height of the track. B g The track runs alongside the amusements pavilion. The C roof of the pavilion follows the rule h = 0.4d + 4. As seen from the diagram, the car, while on this section of the track, will be level with the roof four times. Find the d Distance height of the car above the ground at each of these four points. eBook plus Digital doc


Chapter 8

Further algebra

291

eBook plus

aCTiviTies


• 10 Quick Questions: Warm up with ten quick questions on further algebra. (page 272) 8A

Polynomial identities

Tutorial

• We4 int-1063: Watch how to find the quadratic factor of a cubic given the linear factor. (page 274) 8B

Partial fractions

Tutorials

• We7 int-1064: Watch how to express a linear function divided by a cubic as a partial fraction. (page 278) • We9 int-1065: Watch how to express a cubic divided by a linear function as a partial fraction. (page 281) Digital doc

• WorkSHEET 8.1: Use the bisection and secant methods as well as the null factor law to solve simple and complex simultaneous equations, and apply learning to worded problems. (page 282)

Chapter review Digital doc


Interactivity

• Partial fractions int-0975: Consolidate your understanding of how to determine partial fractions. (page 276) 8C

Simultaneous equations

Tutorial

• We11 int-1066: Watch how to solve simultaneously a linear and an elliptical equation. (page 284) Digital docs

• SkillSHEET 8.1: Practise using substitution to solve simultaneous equations. (page 287) • WorkSHEET 8.2: Practise finding solutions to linear and non-linear simultaneous equations. (page 288)

292


9

9A Statements (propositions), connectives and truth tables 9B Valid and invalid arguments 9C Techniques of proof 9D Sets and Boolean algebra 9E Digital logic

algebra and logic areaS oF STudy

• Propositions, connectives and truth tables • Tautologies, validity and proof patterns • The application of proof to number patterns and algebra

• Electronic gates and circuits • Laws and properties of Boolean algebra • Boolean algebra and its application to circuit simplification eBook plus

9a

Statements (propositions), connectives and truth tables

Digital doc

10 Quick Questions

Older than calculus (17th century), algebra (9th century) and even geometry (300 years before Christ) is the study of logic. Some of the material described in this section was developed by Aristotle, one of the most famous of the ancient Greek philosophers, yet it is still used today by people as diverse as mathematicians, lawyers, engineers and computer scientists. All of our modern digital technology owes its birth to the application of the principles of logic; every meaningful computer program ever written has relied on the principles you will learn in this chapter. Furthermore, logic can be seen as the study of argument. You will be able to analyse logically the arguments of teachers, politicians and advertisers to determine if they should convince you of their ideas, programs and products.

Statements A statement is a sentence which is either true or false. For example, ‘This book is about Mathematics’ is a true (T) statement, while ‘The capital of Australia is Perth’ is a false (F) statement. Some sentences are not statements at all. ‘Go to the store’ is an instruction, ‘How old are you?’ is a question, ‘See you later!’ is an exclamation, ‘You should see the latest Spielberg movie’ is a suggestion. To determine whether a sentence is a statement, put the expression ‘It is true that . . .’ (or ‘It is false that . . .’) at the front of the sentence. If it still makes sense, then it is a statement.

Chapter 9

algebra and logic

293

Beware of some ‘near-statements’ such as ‘I am tall’, or ‘She is rich’, because these are relative sentences; they require more information to be complete. They can be turned into statements by saying ‘I am tall compared to Mary’ or ‘She is rich compared to Peter’. In some books, statements are called propositions. Worked Example 1

Classify the following sentences as either statements, instructions, suggestions, questions, exclamations or ‘near-statements’. a Germany won World War II. b Would you like to read my new book? c The most money that Mary can earn in one day is $400. d When it rains, I wear rubber boots. e Hello! f You will need to purchase a calculator in order to survive Year 11 Mathematics. g Do not run in the hallways. h You should read this book. i I am short. Think a Put the phrase ‘It is true that . . .’ in front of the

Write a This is a (false) statement.

sentence. If the new sentence makes sense, it is classed as a statement. b Put the phrase ‘It is true that . . .’ in front of the

b This is a question.

sentence. If the new sentence makes sense, it is classed as a statement. c Put the phrase ‘It is true that . . .’ in front of the

c This is a statement. We cannot at this time

sentence. If the new sentence makes sense, it is classed as a statement.

determine if it is true or false without further information.

d Put the phrase ‘It is true that . . .’ in front of the

d This is a (presumably true) statement.

sentence. If the new sentence makes sense, it is classed as a statement. e Put the phrase ‘It is true that . . .’ in front of the

e This is an exclamation.

sentence. If the new sentence makes sense, it is classed as a statement. f Put the phrase ‘It is true that . . .’ in front of the

f This is a (true) statement.

sentence. If the new sentence makes sense, it is classed as a statement. g Put the phrase ‘It is true that . . .’ in front of the

g This is an instruction.

sentence. If the new sentence makes sense, it is classed as a statement. h Put the phrase ‘It is true that . . .’ in front of the

h This is a suggestion.

sentence. If the new sentence makes sense, it is classed as a statement. i Put the phrase ‘It is true that . . .’ in front of the

sentence. If the new sentence makes sense, it is classed as a statement.

294

i This is a near-statement because it

requires additional information to be complete. It can be turned into a statement by saying ‘I am shorter than Karen’.


Connectives and truth tables Two (or more) statements can be combined into compound statements using a connective. For example, the statement ‘The book is new and about Mathematics’ is a compound of the single statements ‘The book is new’, ‘The book is about Mathematics’. Notice the connective ‘and’ which is used to join them. Two main connectives — ‘and’ and ‘or’ — are used in compound sentences. Other connectives are ‘not’, ‘if . . . then . . .’, ‘if . . . and only if . . .’. The truth of a compound statement is determined by the truth of the separate single statements. Considering this example, there are 4 cases; Case 1: ‘The book is new’ is true. ‘The book is about Mathematics’ is true. Case 2: ‘The book is new’ is true. ‘The book is about Mathematics’ is false. Case 3: ‘The book is new’ is false. ‘The book is about Mathematics’ is true. Case 4: ‘The book is new’ is false. ‘The book is about Mathematics’ is false. This list can be summarised using a truth table. Let p = ‘The book is new’ and q = ‘The book is about Mathematics’. What about the third column? This represents the truth value of the compound statement ‘p and q’. To determine this truth value we need to examine the logical definition of the connective ‘and’. For the compound statement to be true, both single statements must be true. If either is false then the whole statement is false. Therefore, we can complete the truth table for ‘and’ (using the common symbol ∧ to represent ‘and’). Similarly, the truth table for ‘or’, using the symbol ∨, is shown. The implication here is that it takes only one (or both) of the statements to be true for a statement such as ‘Mary went to the store or the library’ to be true. If she went to the store then certainly she went to the store or the library. Similarly, if she went to the library in this example she could have gone to both.

p and q

p

q

T

T

T

F

F

T

F

F

p

q

p∧q

T

T

T

T

F

F

F

T

F

F

F

F

p

q

p∨q

T

T

T

T

F

T

F

T

T

F

F

F

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There are some compound statements where it is not possible for both statements to be true at the same time. For example: ‘John is fifteen or sixteen years old.’ Clearly, in this case John cannot be both 15 and 16. This is an example of ‘exclusive-or’. Also be careful not to confuse the logical use of ‘and’ with the English common usage. For example, the sentence ‘Boys and girls are allowed in the swimming pool after 6.00 pm’ is made up of the compound sentences ‘Boys are allowed . . .’ and ‘Girls are allowed . . .’. In reality, what is being said is that either boys or girls or both are allowed, so logically, the sentence should be ‘Both boys and girls are allowed in the swimming pool after 6.00 pm’. In some textbooks ‘and’ is called the conjunction and ‘or’ is called the disjunction.

Worked Example 2

Determine the truth table for the compound statement: ‘The suspect wore black shoes or was a female wearing a skirt’. Think

Write

1

Identify and label the individual statements.

p = ‘The suspect wore black shoes.’ q = ‘The suspect was female.’ r = ‘The suspect wore a skirt.’

2

Form a compound statement. Clearly p is separate from q and r.

p or (q and r) p ∨ (q ∧ r)

3

Create a truth table. Since there are three statements and each can have two values (T or F), there are 2 × 2 × 2 = 8 rows in the table. The (q ∧ r) column is completed by looking at the q and r columns.

p

q

r

(q ∧ r)

p ∨ (q ∧ r)

T

T

T

T

T

T

T

F

F

T

T

F

T

F

T

The last column is completed by just looking at the p column and at the (q ∧ r) column.

T

F

F

F

T

F

T

T

T

T

F

T

F

F

F

F

F

T

F

F

F

F

F

F

F

4

Note: Use brackets to indicate the separation.

As can be observed in the last column, p ‘dominates’ the table. Regardless of the truth of q and r, the entire statement is true if p is true (rows 1–4). Otherwise, if p is false then both q and r must be true (row 5). Negation Another connective is the negation, or ‘not’, and is denoted by the symbol ~. This is merely the opposite of the original statement. If p = ‘It is raining’, then ~p = ‘It is not raining’. Be careful when negating English sentences. For example, the negation of ‘I am over 21’ isn’t ‘I am under 21’, but ‘I am not over 21’. Can you see the difference?

Worked Example 3

Complete the truth table for the compound statement p ∨ ~p. Think

296

Write


Set up a truth table. Since there is only one statement here (p), we need only two rows, either p or not p.

p

~p

p ∨ ~p

T

F

T

F

T

T

Note: The compound statement in worked example 3 is always true! An English sentence equivalent to this statement could be ‘I will be there on Monday or I will not be there on Monday’. Equivalent statements Two statements are equivalent if their truth tables are identical. Each row of the truth tables must match. If there is even one difference then the statements are not equivalent. The symbol ⇔ is used to indicate equivalence, as in p ⇔ q. This is read as ‘p is true if and only if q is true’. Worked example 4

eBook plus

By completing truth tables, show that ~( p ∧ q) ⇔ (~p ∨ ~q).

Tutorial

int-1067

Think 1

Worked example 4

WriTe

Set up a truth table. Since there are two statements, we need 2 × 2 = 4 rows.

p

q

~p

~q

( p ∧ q)

T

T

F

F

T

F

F

Complete the ~p, ~q columns by negating p and q separately.

T

F

F

T

F

T

T

F

T

T

F

F

T

T

3

Complete the (p ∧ q) column.

F

F

T

T

F

T

T

4

Negate the (p ∧ q) column.

5

Find (~p ∨ ~q) using columns 3 and 4.

6

Observe that the final two columns are equal in every row.

2

Note that the equivalence operators, p & q, have a truth table of their own, as shown at right. This clearly demonstrates that p ⇔ q is true when the truth value of p equals the truth value of q. i.e. if both p & q are true or if both p & q are false

~( p ∧ q) (~p ∨ ~q)

p

q

p⇔q

T

T

T

T

F

F

F

T

F

F

F

T

rememBer

1. A statement (sometimes called a proposition) is a sentence which is either true or false. 2. Logical connectives are ‘and’, ‘or’, ‘not’, ‘if . . . then’, and ‘equals’ (see the table at the end of the chapter). 3. Each logical connective has a truth table (see the table at the end of the chapter). 4. A compound statement is made up of two or more statements connected with logical connectives. 5. Two (compound) statements are equivalent if they have identical truth tables.

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Exercise

9a

Statements (propositions), connectives and truth tables 1 WE 1 Classify the following sentences as statements (propositions), instructions, suggestions, exclamations, opinions or ‘near-statements’. If they are statements, then indicate whether they are true (T), false (F) or indeterminate without further information (T/F). a That was the best Hollywood movie of 2009. b That movie won the most Oscar nominations in 2009.

c d e f g h i j 2

When the power fails, candles are a good source of light and heat. Why did you use that candle? Collingwood hasn’t won a Grand Final for almost 15 years. Collingwood hasn’t won a Grand Final because they have too few good players. Please go to the store before it closes. The store closes at 6.00 pm. A dingo is considered to be a native Australian mammal. Mary is tall for her age. Break up the following compound statements into individual single statements.

a b c d e f

The car has 4 seats and airconditioning. The Departments of Finance and Defence were both over budget in 2006. Bob, Carol, Ted and Alice went to the hotel. To be a best-seller a novel must be interesting and relevant to the reader. Either Sam or Nancy will win the trophy. You can choose from ice-cream or fruit for dessert. We have vanilla or strawberry ice-cream. g There are some statements which cannot be proved to be true or false. h Most of my friends studied Mathematics, Physics, Engineering or Law and Arts. 298


3

Convert the following pairs of simple sentences into a compound sentence. Be sure to use ‘and’ and ‘or’ carefully. a John rode his bicycle to school. Mary rode her bicycle to school. b The book you want is in row 3. The book you want is in row 4. c The weather is cold. The weather is cloudy. d Many people read novels. Many people read history. e In a recent poll, 45% preferred jazz. In a recent poll, 35% preferred classical music.

f Two is an even number. Two is a prime number. 4 mC For the compound statement p ∧ q (p and q), the number of different ways for this statement to be true is: A 0 B 1 C 2 D 3 E 4 5 mC For the compound statement p ∨ q (p or q), the number of different ways for this statement to be true is: A 0 B 1 C 2 D 3 E 4 6

As you saw in worked example 4, if there is a compound statement with two single statements, p and q, then there are 2 × 2 = 4 rows in the truth table. List all the different rows for compound statements made up of: a 3 single statements b 4 single statements c 5 single statements. You should be able to develop a pattern of completing the Ts and Fs in a logical sequence.

7 We2 Write the following compound sentences in symbolic form (p, q, r), and determine the truth table. a The Sydney flight was on time; the Perth flight was fully booked. b John, Zia and David passed General Mathematics.

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c Either Alice and Renzo, or Carla, will have to do the dishes. (Note use of commas.) d The committee requires two new members. One must be a female, the other must be either a student or a professor. 8

Find the truth table for the compound statement p ∧ ~p.

9 We3 Complete the truth tables for the following compound statements. a p ∧ ~q b ~p ∧ ~q c ( p ∧ q) ∧ r d p ∨ ~q e ~p ∨ ~q f ( p ∨ q) ∨ r 10

Let p = ‘It is raining’, q = ‘I bring my umbrella’. Write a sentence for the following compound statements. a p∧q b p∨q c ~p ∧ q

11

Let p = ‘Peter likes football’, and q = ‘Quentin likes football’. Write a sentence for the following compound statements. a p∧q b p∨q c p ∨ ~q

12 We4 By completing truth tables show that ~( p ∨ q) ⇔ (~p ∧ ~q). 13

9B

Determine if the compound statement ~( p ∨ q) is equivalent to ~p ∨ ~q.

14

Determine if the following compound statement pairs are equivalent. a ( p ∧ q) ∨ ~p b ( p ∨ q) ∨ ~p ( p ∨ q) ∧ ~p p ∨ ~p

15

Determine if the brackets in an expression alter the truth table by comparing: ( p ∧ q) ∨ r with p ∧ (q ∨ r ).

16

Repeat question 15 with the following statement pairs. a Compare ( p ∧ q) ∧ r with p ∧ (q ∧ r). b Compare ( p ∨ q) ∨ r with p ∨ (q ∨ r). c Based upon the results of questions 15 and 16, what might you conclude about the effect of brackets on a compound expression?

Valid and invalid arguments The purpose of the logical connectives ‘and’, ‘or’, and ‘not’ is to form statements, true or false, in order to evaluate the truth, or otherwise, of something called an argument. An argument is a set of one or more propositions (statements). Before we can evaluate arguments we need one more connective: the implication (or conditional) statement.

implication Consider the following ‘classical’ statement: ‘If it is raining then I bring my umbrella’. This is the combination of the two statements ‘It is raining’ and ‘I bring my umbrella’, connected by two words: ‘If’ and ‘then’. Each of the two statements has individual truth values; either could be true or false. The first statement is called the antecedent, the second is called the consequent, and in symbolic form this is written as p ⇒ q. This is called implication because the first statement implies the second; it is also called conditional, because the outcome of the second statement is conditional on the first. How can we determine the truth table of p ⇒ q? This is not as simple as employing a mere definition.

300


Referring to our example, consider the question ‘Under what conditions would p ⇒ q be a lie?’. 1. If it is indeed raining and I bring my umbrella then, clearly p ⇒ q is true. 2. If it is raining and I don’t bring my umbrella, then I lied to you! Thus, p ⇒ q is false. 3. What if it is not raining? I have told you nothing about what p q p⇒q I would do in that case. I might either bring my umbrella, T T T or I might not. In either case you cannot say I lied to you, so p ⇒ q is true. T F F To summarise, then, the truth table at right may be constructed: F T T This leads us immediately to ask the question: Is p ⇒ q the same F F F as q ⇒ p? Worked Example 5

Determine by using truth tables if ( p ⇒ q) ⇔ (q ⇒ p). Think

Write

1

Set up a truth table for p, q and p ⇒ q. This is shown in the 3rd column.

2

Exchange the roles of p and q to determine the truth table for q ⇒ p. This is shown in the last column.

3

Clearly, they are not equivalent.

p

q

p⇒q

q⇒p

T

T

T

T

T

F

F

T

F

T

T

F

F

F

T

T

This is a most important result; it is a result that people (who think they are arguing logically) often mistake for a valid statement. Again, referring to the original implication; ‘If I bring my umbrella, then it is raining’ says a much different thing from the original statement and is called its converse. In this example, it seems to be making the argument that my bringing the umbrella can control the weather!

Converse, contrapositive and inverse As we have just seen, there are alternative forms of p ⇒ q, such as the converse. These, and their relationship to p ⇒ q are shown in the table below. Symbol

Relationship to p ⇒ q

Implication

p⇒q

(assumed) True

Converse

q⇒p

False

Contrapositive

~q ⇒ ~p

True

Inverse

~p ⇒ ~q

False

Name

Often the contrapositive is a more realistic way of stating an implication than the original statement is. Be careful, however, not to use the converse or inverse as they are (generally) false when p ⇒ q is true.

Arguments An argument is a series of statements divided into two parts — the premises and the conclusion. The premises are a series of statements intended to justify the conclusion. For example, consider the following argument: A terrier is a breed of dog. Premise Rover is a terrier. Premise Therefore, Rover is a dog. Conclusion Generally, an argument will have only one conclusion and (usually) two premises.


301

Conclusion and premise indicators To help identify the conclusion, look for words or phrases like: therefore, accordingly, hence, thus, consequently, it must be so that, so, it follows that, implies that What follows one of these conclusion indicators is the conclusion; by default everything else is a premise. There are also premise indicators: because, given that, since, seeing that, may be inferred from, owing to, for, in that Worked Example 6

Identify the premises and conclusions for each of these arguments. a A Commodore is a model of a Holden car. My car is a white Commodore. Therefore, my car is a Holden. b Military defence depends upon adequate government funding. Adequate government funding depends on a healthy economy. A healthy economy depends upon an intelligent fiscal policy. Military defence depends upon an intelligent fiscal policy. c Pregnant mothers should not smoke. Cigarettes can harm the foetus. d The weather report on Channel 9 is accurate. I will bring an umbrella tomorrow, because the weather report on Channel 9 predicts rain for tomorrow. Think

Write

a Examine each sentence looking for

the conclusion indicators, or examine the sequence of the sentences. b Note how the sequence of statements connects one with the next. The last is therefore the conclusion.

c In this case the sentences have been

reversed. This is a common mistake. d Again, the order of statements

is somewhat confused. Can you re-write this argument in a more logical fashion?

a A Commodore is a model of a Holden car.

My car is a white Commodore. Therefore, my car is a Holden. b Military defence depends upon adequate government funding. Adequate government funding depends on a healthy economy. A healthy economy depends upon an intelligent fiscal policy. Military defence depends upon an intelligent fiscal policy. c Pregnant mothers should not smoke. Cigarettes can harm the foetus. d The weather report on Channel 9 is accurate. I will bring an umbrella tomorrow. The weather report on Channel 9 predicts rain for tomorrow.

Premise Premise Conclusion Premise Premise Premise Conclusion Conclusion Premise Premise Conclusion Premise

In some books, statements are called propositions and arguments are called inferences.

Categorical propositions and the deductive argument The standard argument consists of two premises and a conclusion: All dogs are mammals. Premise Rover is a dog. Premise Therefore, Rover is a mammal. Conclusion

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Note: Observe the use of the key word ‘All’. (Beware of arguments that use the key word ‘some’, as in ‘Some journalists are hard-working’. This is a weaker form of argument, the study of which is beyond the scope of this course.) The first premise is called a categorical statement or proposition, and this form of argument can be called the classical deductive argument. However, as we shall see, there are many cases where we will not have a valid deductive argument, even if everything looks correct: these situations are called fallacies. As an example, consider the following argument: All dogs are mammals. Premise Rover is a mammal. Premise Therefore, Rover is a dog. Conclusion Clearly, no one should be convinced by this argument. Both premises might be true, but the conclusion does not follow logically from them and we would say that this is an invalid argument. This is an example of a formal, or structural fallacy. Some categorical propositions can be turned into implications. For instance, the statement ‘All dogs are mammals’ can be written as ‘If it is a dog, then it is a mammal’. This says exactly the same thing. Beware of certain statements such as ‘If it is sunny tomorrow, I will go to the beach’. This is not the same as saying ‘On all sunny days I will go to the beach’. The key word here is ‘tomorrow’ — this restricts the statement so that the key word ‘all’ cannot be used. However, the implication can still be used in a valid argument: If it is sunny tomorrow, I will go to the beach (after checking the weather tomorrow). It is sunny. I will go to the beach. This is certainly a valid argument. At this point, we can define a symbolic form for this kind of deductive argument: p⇒q p q In other words, we start with an implication, which we assert to be true, then follow by stating that the antecedent is true, and argue that the conclusion is true. Can you see how the ‘Rover’ argument fits into this pattern? Note that this is only one form of (potentially) valid argument. Worked Example 7

Determine if the following arguments are valid. a All mathematics books are interesting. b This is a book about mathematics. Therefore, this book is interesting. c Some history books are boring. d This book is about history. Therefore, this book is boring. Think a

1

If I study hard, I will pass Physics. I passed Physics. I must have studied hard. If I don’t study, I will fail Physics. I didn’t study. I will fail Physics. Write

Change the first statement to: ‘If . . . then . . .’.

a If it is a mathematics book, then it is

interesting. 2

3

(b) Put the argument into symbolic form.

p = It is a mathematics book. q = It is interesting. p⇒q p q

Determine if it is a valid form.

Yes, this is a valid form for an argument.

(a) Assign each statement a symbol.


303

b

1

(a) Assign each statement a symbol. (b) Put the argument into symbolic form.

2


‘some’.

1

c The use of the word ‘some’ means that

the statement cannot be put into this form. Thus, the entire argument is not valid.

(a) Assign each statement a symbol. (b) Put the argument into symbolic form.

2

q = I will pass Physics. p⇒q q p No, this is not a valid form for an argument.

c Consider the first statement. Note the use of the word

d

b p = I study hard.


d p = I don’t study.

q = I will fail Physics. p⇒q p q Yes, this is a valid form for an argument.

Note: Even if the statements are expressed in negative form; ‘I don’t study’ … ‘I will fail Physics’, it is still possible to have a valid argument. Can you devise a ‘positive’ argument which is the equivalent to the one in part d? It is important to note that an argument may be valid even if the truth of the component statements cannot be established. Consider the following (nonsense) argument: All fribbles are granches. An hommie is a fribble. Therefore, an hommie is a granch. We certainly cannot establish the truth of the two premises (let alone know what fribbles, granches or hommies are), but presuming they are true, the argument is valid. Furthermore, consider the argument: If it is a dog then it can do algebra. Rover is a dog. Therefore Rover can do algebra. This is a valid form of argument, but one (or more) of the premises is (are) false. In this case we do not have a sound argument and would certainly not use it to convince anyone of the mathematical ability of dogs. In other words, ‘All sound arguments are valid, but all valid arguments are not necessarily sound’.

Valid forms of argument There are many valid forms of argument. We shall limit our discussion to the most important ones, tabulating five of them below. Argument form and name

304

Example

p⇒q p q

If Mary is elected, then she must be honest. Mary was elected. Mary must be honest.

Modus ponens

This is our standard form.


Argument form and name

Example

p∨ q ∼p q

Either John or Jemma was born in Canada. John was not born in Canada. Jemma was born in Canada.

Disjunctive syllogism

Note that the roles of p and q can be interchanged here.

p⇒q q⇒r p⇒r

If it is raining I will bring my umbrella. If I bring my umbrella then I will not get wet. If it is raining I will not get wet.

Hypothetical syllogism

Many statements (p, q, r, …) can be linked together this way to form a valid argument.

p⇒q ∼q ∼p

If I study hard I will pass Physics. I did not pass Physics. I did not study hard.

Modus tollens

This is a valid form of a negative argument.

p⇒q∧r ⇒s p∧r q∧s

If we holiday in France we will have to practise speaking French, and if we holiday in Germany, we will have to practise German. We will holiday in France and Germany. We will have to practise speaking French and German.

Constructive dilemma

There are several other forms more complex than these which are beyond the scope of this course.

Proving the validity of an argument form It may not be satisfactory to merely declare that the five arguments in the previous table are automatically valid. There is a way to mathematically establish their validity using a truth table. The procedure is as follows. Step 1. Set up a single truth table for all the premises and for the conclusion. Step 2. Examine the row (or rows) in the table where all the premises are true. Step 3. If the conclusion is true in each of the cases of step 2, then the argument is valid. Otherwise it is invalid. Worked Example 8

Establish the validity of the modus ponens argument, namely: p ⇒ q p q Think 1

Set up a truth table for each of the premises, namely p and p ⇒ q and the conclusion q. Note that p and q are set up first, in the usual way, and that p ⇒ q is completed from them in the usual way.

Write

p

q

p⇒q

T

T

T

T

F

F

F

T

T

F

F

T


305

2

Find the rows where all the premises are true.

The premises are all true in the 1st row only.

3

Compare with the conclusion column (q).

The conclusion is also true, so the argument is valid.

Worked example 9

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Show that the following argument is invalid. p⇒q q p Think 1

Tutorial


WriTe

Set up a truth table, for each of the premises, namely q and p ⇒ q and the conclusion p. Note that p and q are set up first, in the usual way, and that p ⇒ q is completed from them in the usual way.

p

q

p⇒q

T

T

T

T

F

F

F

T

T

F

F

T

2

Find the rows where all the premises are true.

The premises are all true in the 1st row and 3rd row.

3

Compare with the conclusion column ( p).

The conclusion is true in the 1st row, but false in the 3rd, so the argument is invalid.

This is a common error in logical argument, and is called affirming the consequent. In conclusion, if an argument fits exactly one of the five given forms, then it is immediately assumed to be valid; otherwise it must be established to be valid using truth tables. rememBer

1. The implication p ⇒ q has the following alternative forms: (a) implication (assumed true) (b) converse (False) (c) contrapositive (True) (d) inverse (False). 2. An argument consists of one or more statements called premises and a statement called a conclusion. 3. An argument is valid if the conclusion is true when all the premises are true. 4. A valid argument is sound if all the premises are true. 5. An invalid argument (sometimes referred to as a fallacy) occurs when the premises might be true but the conclusion does not follow logically from them. 6. Among others, there are five valid deductive argument forms: (a) modus ponens (b) disjunctive syllogism (c) hypothetical syllogism (d) modus tollens (e) constructive dilemma.

306


exerCiSe

9B

Valid and invalid arguments 1

We5 Establish the validity of the contrapositive; namely, determine using truth tables if ( p ⇒ q) ⇔ (~q ⇒ ~p).

2

Establish the truth table for the inverse; namely, show that (p ⇒ q) is not equivalent to (~p ⇒ ~q).

Let p = ‘It is bread’ and q = ‘It is made with flour’. Write out the implication, converse, contrapositive and inverse in sentences. 4 mC The contrapositive of the statement ‘If a child is playing quietly, then it is doing something bad’ is: A If a child is playing quietly then it is not doing something bad. B If a child is not playing quietly then it is not doing something bad. C If a child is not doing something bad then it is not playing quietly. D If a child is doing something bad then it is not playing quietly. E If a child is doing something bad then it is playing quietly. 5 mC The inverse of the statement ‘If you are not careful, then you will get hurt’ is: A If you are careful, then you will not get hurt. B If you do not get hurt, then you are careful. C If you get hurt then you are not careful. D If you do not get hurt then you are not careful. E If you are not careful, then you will not get hurt. 3

6

7

We6 Identify the premises and conclusion in the following arguments. a All cats are fluffy. My pet is a cat. My pet is fluffy. b Two is the only even prime number. Prime numbers are divisible by themselves and 1. All even numbers are divisible by themselves and by 2. c Growing apples depends on good water. Growing apples depends on good irrigation. Good water depends on good irrigation. We7 Determine which of the following are valid arguments.

a If you are a mathematician, you can do algebra. b You are a mathematician. You can do algebra. c If it is a native Australian mammal, then it is a marsupial. A wombat is a native Australian mammal. A wombat is a marsupial. e d Some TV shows are boring. Neighbours is a TV show. Neighbours is boring. 8

All footballers are fit. David is not a footballer. David is not fit.

All musicians can read music. Louise can read music. Louise is a musician.

Look again at the arguments in question 7 which were not valid. If possible, turn them into valid arguments. Assume that the first statement in each argument is always correct.

9 mC Which of the following statements about logic is false? A An argument must have a conclusion. B An argument consists of premises and a conclusion. C An ‘If . . . then . . .’ compound statement is called a conditional statement. D A, B and C are all false. E A, B and C are all true.

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10 WE8 Establish the validity of the disjunctive syllogism argument, namely p ∨ q ∼p q 11

Establish the validity of the three remaining valid forms of argument, namely: a hypothetical syllogism

b modus tollens

c constructive dilemma

12

p⇒q q⇒r p⇒r p⇒q ∼q ∼p p⇒q∧r ⇒s p∧r q∧s

The following are valid arguments. Determine which of the five forms of argument were used. a Either you clean up your room or you will not watch any television tonight. You did not clean up your room. Therefore you will not watch any television tonight. b If you help your mother with the dishes, I will take you to the football game tomorrow. I didn’t take you to the football game. Therefore you didn’t help your mother with the dishes. c If you study statistics, then you will understand what standard deviation means. You studied statistics. Therefore you will understand what standard deviation means.

13 MC Consider the following valid argument. If John plays for us on Saturday, then we will win. If we win on Saturday, then we will come in first place on the ladder. If we come in first place on the ladder, then we play our first final at home. Therefore, if John plays for us on Saturday, then we play our first final at home.

308


This is an example of: A modus ponens C hypothetical syllogism E constructive dilemma

B disjunctive syllogism D modus tollens

14 We9 a Determine the validity of the following argument, using truth tables. p⇒q ∼p ∼q b Show that the following is an example of this argument. If elected with a majority, my government will introduce new tax laws. My government was not elected with a majority. Therefore, my government will not introduce new tax laws. 15

A common argument is of the form: If you work hard, then you will become rich. You don’t work hard. Therefore, you will not become rich. a Put this argument in symbolic form. b Show that it is an invalid form of argument. (This is called ‘denying the antecedent’.)

16

Determine the validity of the following arguments. b ∼p ∧ ∼q a p⇒q r⇒p r ⇒ ∼q r p ⇒ ∼r

17

c ∼p ⇒ ∼q p q

Determine the validity of the following arguments. a All dogs have five legs. All five-legged creatures are called chickens Therefore, all dogs are chickens. b All dogs have five legs. All chickens have five legs. Therefore, all dogs are chickens. c If you deposit money in the bank, then you will earn interest. You didn’t earn any interest. Therefore, you didn’t deposit any money in the bank. d If I wanted an easy course to study, I would choose Human Development and if I wanted an interesting course to study, I would choose General Mathematics. I can choose an easy course, and an interesting one. Therefore, I will study Human Development and General Mathematics. e If the team plays well, either the offence was good or the defence was good. The defence wasn’t good and the team did not play well. eBook plus Therefore, the offence wasn’t good. Digital doc

WorkSHEET 9.1

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9c

Techniques of proof As mentioned at the beginning of this chapter, logic is not only used to establish the validity of arguments, but its techniques are used to establish the truth (or otherwise) of mathematical statements. For example, it is not satisfactory to say that Pythagoras’ theorem is true; it must be proved to be true as well. The tools of logic are also the tools of proof.

Tautologies A tautology is a compound statement which is always true. While this might seem like a useful thing, in arguments such statements are to be avoided, as they contribute nothing towards an argument. An example of a tautology is: ‘The game will be won or lost in the last 30 minutes’. A statement like this, while perhaps favoured by sports announcers, is of no value in establishing the truth of whether a particular team will win. A tautology is more useful as a technique of proof. To establish if a compound statement is a tautology, construct a truth table from its component parts. If the compound statement is always true, then it is a tautology. Worked Example 10

Consider the statement: ‘If John and Jim are qualified lawyers, then John is a qualified lawyer’. Establish whether or not this statement is a tautology. Think

Write

1

Define symbols for each part of the statement.

Let p = John is a qualified lawyer. Let q = Jim is a qualified lawyer.

2

Convert the statement to its symbolic form.

( p ∧ q) ⇒ p

3

Set up a truth table.

4

Establish whether or not the statement is a tautology.

p

q

p∧q

(p ∧ q) ⇒ p

T

T

T

T

T

F

F

T

F

T

F

T

F

F

F

T

The last column shows all ‘T’, therefore the statement is a tautology.

As can be seen, the original statement would contribute very little to any argument as to the qualifications of either John or Jim as lawyers. The ‘opposite’ of a tautology is a selfcontradictory statement; one which is always false. It too has little use in arguments. However, the concept of a tautology can be used in establishing validity or, in mathematical language, proving arguments.

Proof using tautologies An argument is valid under the following condition: ‘If all the premises are true then the conclusion is true’. Let p and q be the premises of an argument and r be the conclusion. If ( p ∧ q) ⇒ r is a tautology, then the argument is valid, or proved. Note that p, q and r can be, themselves, compound statements. In fact, this method is exactly the same as that presented in the previous section, but is a more mechanical technique.

310


Worked Example 11

Establish the validity of the modus tollens argument, namely p ⇒ q ~q ~p Think

Write

1

Set up a standard truth table for p and q.

2

Determine the truth table for p ⇒ q.

3

Form the truth table of the ‘and’ of both statements, namely; p ⇒ q ∧ ~q (column 4).

4

5

Form the truth table of the implication using the 1st two statements along with the conclusion, namely: (p ⇒ q ∧ ~q) ⇒ ~p (column 5). Determine the validity of the argument.

p

q

p⇒q

p ⇒ q ∧ ~q

(p ⇒ q ∧ ~q) ⇒ ~p

T

T

T

F

T

T

F

F

F

T

F

T

T

F

T

F

F

T

T

T

Since the last column is always true, this is a tautology and the original argument is valid.

The other valid forms of argument can also be established using the same technique. This is left as an exercise.

Mathematical proofs using valid argument forms Any of the valid forms of argument can be used to prove theorems in mathematics. While this can be a tedious way of proving things, it certainly will establish them beyond doubt. Worked Example 12

Prove the following: ‘If two straight lines have equal gradients, then they do not intersect’. Think

Write

1

Find a property of lines with equal gradients.

If lines have equal gradient then they are parallel. This is a well-known result from linear graphs.

2

Find a property of parallel lines.

If lines are parallel then they do not intersect. This is a well-known result from geometry.

3

Define p, q and r from compound statements.

Let p = Two lines have equal gradients. Let q = Two lines are parallel. Let r = Two lines do not intersect. Step 1: If p then q, or p ⇒ q. Step 2: If q then r, or q ⇒ r.

Create a symbolic form of steps 1 and 2. 4

Determine a conclusion from the valid argument form (hypothetical syllogism).

p⇒q q⇒r p⇒r


311

5

Write the conclusion.

If two straight lines have equal gradients, then they do not intersect. QED

Whenever a theorem is proved, state quod erat demonstrandum (QED). This means ‘It is demonstrated’, but could also mean ‘quite easily done’!

proof by contrapositive As mentioned in the section on logical connectives, an alternative to p ⇒ q is its contrapositive ~q ⇒ ~p. If we can establish that the contrapositive statement is true, then the original implication is true. Worked example 13

Prove, using the contrapositive, that: ‘If n3 is odd, then n is odd (where n is any integer)’. Think

WriTe

1

Write the statement in symbolic form.

Let p = ‘n3 is odd’; q = ‘n is odd’.

2

Write the contrapositve as a statement.

~q ⇒ ~p; ‘If n is not odd then n3 is not odd’.

3

Prove the truth of the contrapositive.

If n is not odd, it is even (or 0) and can be represented by 2x, where x is any integer. Let n = 2x then n3 = (2x)3 = 8x3.

This relies on the fact that multiplying anything by 2 results in an even number.

But 8x3 = 2 × 2 × 2 × x3 and, hence, must be even. Since 8x3 = n3 is even it is not odd. Since the contrapositive statement is shown to be true, the original argument, namely that if n3 is odd, then n is odd, is also proved. QED

proof by contradiction Another method of proof involves assuming the statement that we are trying to prove is false. Then this leads to an apparent contradiction, so we assume that the statement is true. Worked example 14

eBook plus

Prove, by contradiction, that the product of two negative numbers (non-zero) is positive. Think

312

WriTe

Tutorial


1

Assume that the statement is false.

Let a and b be two negative numbers. Assume that a × b is negative.

2

Determine the magnitude of the product. Invoke the assumption that the product is negative.

The magnitude of a × b = | ab | = | a | × | b | ab = −| a | × | b |


3

Consider the case c = −a, so that c is positive, and a positive number is being multiplied by a negative number.

cb = −| c | × | b | | c | = | a | cb = −| a | × | b |

4

Now from steps 2 and 3, the two expressions are equal.

−| a |

5

This is a contradiction, since c = −a.

Contradiction implies that the original statement is true. QED

since c is < 0 and b > 0 since c = −a substitution

× | b | = ab = cb from steps 2 and 3 a = c divide both sides by b

The contradiction must be as a direct result of the assumption of the original statement being false. In the above case, this occurred in steps 2 and 3, leading to two different expressions for the same thing (−|a| × |b|). This proof technique is based upon the logical argument form: (p ∧ ~q) ⇒ (r ∧ ~r) ⇔ (p ⇒ q) where p ⇒ q (or more specifically q) is the statement you are trying to prove, and (r ∧ ~r) is the contradiction that arises by assuming q was false (~q). This method of proof is also called indirect proof, or reductio ad absurdum (reducing to an absurdity).

Proof by counter-example Perhaps the simplest method of proof is that by assuming the statement to be true, an example arises which shows that the statement is false. Therefore the original statement cannot be true. Worked Example 15

Prove, by counter-example, that the statement ‘the square root of x2 is x’ is false. Think

Write

1

Consider a single case; let x = 8.

x2 = 64

2

Invoke the rule for square roots.

The square root of 64 is ±8.

3

Substitute back.

The square root of x2 is ± x, thus the original statement is false.

Remember, you need only a single example where the statement is false and hence, by extension, the entire statement is false. In the above example, the proof relied on the fact that there were two answers, not one as implied in the statement. There are many other methods of proof, but the ones you have seen in this chapter will provide you with a toolbox of techniques for proving a large number of mathematical statements.

Proof by mathematical induction Mathematical induction is a method of proof, although it is considered less effective than the ones you have already learned. It is used to prove formulas, results and similar things where there is a sequence of results for different values. ( n)( n + 1) For example, the sum of the series 1 + 2 + 3 + . . . + n = has different results depending 2 on the value of n. We will use the method of induction on the above proof. Step 1 Show that the result is true for n = 1 (or in some proofs n = 0).

If n = 1, then the sum = 1 and ( n)( n + 1) 1 × 2 = =1 2 2


313

Step 2 Assume it is true for any value n. Step 3 Prove it true for n + 1 (add one more term to the series). This step is as a result of adding with common denominators. This step is as a result of the common factor of (n + 1) in the numerator. Let m = n + 1, so m + 1 = n + 2 Step 4 Clearly, this is the same formula as the one we assumed true.

Assume: 1 + 2 + 3 + … + n =

( n)( n + 1) 2

(1 + 2 + 3 + . . . + n) + (n + 1) = ( n)( n + 1) + ( n + 1) 2 ( n )( n + 1) + 2( n + 1) = 2 ( n + 1 )( n + 2) = 2 = (m)(m + 1) 2 (m)(m + 1) 2 = 1 + 2 + 3 + . . . m = 1 + 2 + 3 + . . . (n + 1)

The logic behind induction is that we can keep on increasing the value of n by one at a time, until all (possible) values have been ‘proved’. Thus the statement is proved for all values! REMEMBER

1. A tautology is a compound statement which is always true. 2. An argument with premises p, q, r… and conclusion c is valid if the compound statement p ∧ q ∧ r . . . ⇒ c is a tautology (that is, is always true). 3. A mathematical argument can be proved using (among others): (a) a valid deductive argument form (b) contrapositive (prove ~q ⇒ ~p, which proves p ⇒ q) (c) contradiction (prove that the opposite statement is false) (d) counter-example (prove false by finding an example) (e) mathematical induction. Exercise

9c

Techniques of proof 1 WE 10 Determine if the following statement is a tautology. ‘If she plays well she will win, or she will lose.’ 2

Modify the sentence in question 1 to: ‘If she plays well she will win, or if she plays poorly she will lose’. Determine if this sentence is a tautology.

3

Modify the sentence in question 1 to: ‘Either she plays well and wins or she will lose’. Determine if this sentence is a tautology.

4 WE 11 Using tautology, establish the validity of the hypothetical syllogism argument, namely: p⇒q q⇒r p⇒r

314


5

Using tautology, establish the validity or otherwise of the following arguments. b p ⇒ ∼q c ∼p ⇒ ∼q d ∼p ⇒ q a ∼p ⇒ ∼q p q ⇒ ∼r ∼q ⇒ r q ∼q p ⇒ ∼r ∼p ⇒ r p

6 a

b c

d

Express the following arguments in symbolic form. If you are a loser, then you didn’t train hard enough. If you didn’t train hard enough, you were distracted. Therefore, if you are a loser, then you were distracted. If it is not raining, then I will wash my car. It is raining. Therefore, I will not wash my car. If it is cloudy, then I do not bring my umbrella. I did bring my umbrella. Therefore, it is not cloudy. If the hard drive isn’t working, then the program will not work. If the program is not working, then the printer will not work. Therefore, if the hard drive isn’t working the printer will not work.

7 mC The missing truth values in each column in the following table are (from left to right): p

q

p⇒q

q⇒p

~p ⇒ ~q

~q ⇒ ~p

T

T

T

T

T

T

T

F

F

F

T

F

F

A F, T, T, F

F

T B F, F, T, T

F

F

T

T

C T, F, T, F

D T, T, T, T

T E T, T, F, T

8 We12 Using a valid argument, prove the following mathematical result. If a number, x, is even, then x2 is even. 9

Using a valid argument, prove that 242 is even.

10 We13 Prove, using the contrapositive, that: If n2 is odd then n is odd. 11

Prove the following results using the contrapositive. a If ax = bx and x ≠ 0, then a = b. b If n2 > 4, then n > 2. c If n2 is divisible by 2, then n is divisible by 2.

12 We14 Prove, by contradiction, that the product of a negative number and a positive number is negative. 13

Prove, by contradiction, that there is no smallest positive real number (a, such that a > 0).

14

Prove, by contradiction, that 2 is irrational.

15

Prove, by contradiction, that there is no largest positive integer (n, such that n > 0).

16 We15

Prove, by counter-example, the statement that ‘x2 = 4 has only one solution’ is false.

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315

17 Prove, by counter-example, the fact that all prime numbers are odd. 18

19

Consider the formula p = n2 + n + 11. Let n = 1, 2. . . n = 1, p = 13 n = 2, p = 17 n = 3, p = 23 n = 4, p = 31 It seems that: If n is a positive integer, then p is a prime number. Prove, or disprove, this statement.

Consider the two lines PQ, and RS shown below. A third line crossing both of them, called a transversal, forms two angles a and b. Investigate a proof for the statement that: If a + b = 180°, then PQ is parallel to RS. P

Q

a b

R

20

n = 5, p = 41

S

Consider any triangle ABC as shown in the figure below. Let the longest side of the triangle be labelled x and the other two sides be labelled y and z. Investigate a proof for the statement that: x ≤ y + z. B y A

z x

C

21 Examine some well-known results in mathematics. Can you prove them by induction? Here are some ideas to get you started. Some will work, some won’t. a The odd number series 1 + 3 + 5 + . . . + (2n – 1) = n2. Hint: How many odd numbers are there in this series? b (1 + x)n ≥ (1 + xn) for n ≥ 1 and x > 0 c Pythagoras’ theorem d Prove (n)(n + 1) is an even number for any integer value of n. e 2n ≥ n2 for all integers ≥ 4 f Let a, b and c be three consecutive integers. Prove (a + b + c)3 is always divisible by 3. g Prove the quadratic formula. Given ax2 + bx + c = 0, prove that the roots are −b ±

b 2 − 4 ac . 2a h Prove that a polygon of n sides can be composed of n – 2 triangles. Hint: Start with n = 3. i 1 + 2 + 4 + 8 + . . . + 2n = 2n + 1 – 1 j 4n ≥ n4 for all integers ≥ 2 x=

22 a Can you (reasonably) predict what type of proof is likely to be solvable by induction? b What are the limitations of the proofs, in terms of values of n which can be used? c In some scientific research, particularly medicine, the concept of induction is used as follows: If I treat Patient 1 with Drug X, she is cured. If I treat Patient 2 with Drug X, she is cured. If I treat Patient n with Drug X, she is cured. d What conclusion(s) is the researcher trying to draw? Comment on its (their) validity. 316


23 Express your thoughts on the usefulness or otherwise of proof by induction, paying attention not only to mathematical proofs but those used in areas such as science, commerce and politics. eBook plus

9d

Digital doc

Sets and Boolean algebra

WorkSHEET 9.2

Many of the rules of logic that we have seen thus far can be collected into a single set of rules and procedures called Boolean algebra, named after the 19th century English mathematician, George Boole. Boole is also responsible for the introduction of sets into mathematics.

Sets and their properties A set is a collection of objects (or members) that have something in common. Sets can be numbers, such as the set of integers from 1 to 10, A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} or people, such as the set of Australian Prime Ministers, B = {Barton, Deakin, . . ., Keating, Howard, Rudd} or things, such as the set of suits in a deck of cards, C = {Diamonds, Hearts, Clubs, Spades} Sets can be finite, containing a fixed number of members, such as set A above, with 10 members, or infinite, such as the set of positive integers, N = {1, 2, 3, . . .}. Finally, a set with no members is called a null set. An example is the set of female Australian Prime Ministers, F = { } or φ. Implicit in sets is the concept that there are objects in the set and objects not in the set. If an object x is in set A, we could write x ∈ A, while if object y is not in set A we would write y ∉ A. ε Sets can be displayed visually using a Venn diagram as shown at right. The area inside the circle represents the set A with its members A = {2, 4, 6, 8}. The white area outside the 2 6 circle represents all objects not in the set. In future we will not 8 4 generally show the members in the set, but state its ‘rule’. What could be the rule for the set in this figure? The rectangle itself represents the universal set, the set of all possible members (some are in A, some are not) and is denoted by the symbol ε. In this example the universal set could be all the integers. As in arithmetic, there are a series of operations and properties which enable us to manipulate sets. Consider two sets, A and B, and the possible operations on them. Intersection: The area in common between two sets is known as the intersection and is shown here in grey. Symbol:

A∩B ‘A intersection B’ or ‘in both A and B’

Union:

The area in either A or B is the union, and is shown here in red.

Symbol:

ε A

B

A

B

ε

A∪B ‘A union B’ or ‘in either A or B or both’ (Continued )

Chapter 9

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317

Negation:

The area not in A is the negation or complement, and is shown here in blue.

Symbol:

ε A

A′ ‘Complement of A’ or ‘A-prime’ or ‘not in A’

Given these operations, we can now look at the rules of sets, comparing them to the rules of arithmetic. For some laws we will need three sets. Name

Symbolic form

Commutative Law

1. A ∪ B = B ∪ A

Identity sets

1. A ∪ ∅ = A

a+b=b+a

a+0=a

2. A ∩ ε = A

The null set has no effect on ‘union’; the universal set has no effect on ‘intersection’.

1. A ∪ A′ = ε

Inverse

a + (−a) = 0 1 a× =1 a

The placement of brackets has no effect on the final result when the operations are the same.

a + (b + c) = (a + b) + c

Bracketed expressions can be expanded when different operations are involved.

a × (b + c) = a × b + a × c (Note that there is only one of these laws in arithmetic.)

Performing operations on a set will create a result which still belongs to the same class of sets (S).

If a and b are real numbers, then: a + b is a real number a × b is a real number.

2. A ∩ A′ = ∅ Associative Law

1. A ∪ (B ∪ C) = (A ∪ B) ∪ C 2. A ∩ (B ∩ C) = (A ∩ B) ∩ C

Distributive Law

1. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) 2. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Closure

Corresponding arithmetic

Order of a single operation is not important.

2. A ∩ B = B ∩ A

Complements

Description

Consider sets A, B and S. If A, B ⊂ S, then A∪B⊂S A∩B⊂S

a×b=b×a

a×1=a

a × (b × c) = (a × b) × c

It is important to note that union (∪) acts similarly to addition, while intersection (∩) is similar to multiplication, except in the complements, where their roles are reversed. While the commutative laws are self-evident, the remaining laws can be demonstrated using Venn diagrams. Closure is a concept which, for now, will have to be taken for granted. For example, closure applies for integers with the operations of addition and multiplication. It 1 does not apply for division (for example 2 ), as the result (0.5) is not an integer even though 1 and 2 are. Worked Example 16

Using Venn diagrams, establish the complements rules, namely: a A ∪ A′ = ε b A ∩ A′ = ∅ Think

318

Write


a

1

Draw a Venn diagram, indicating A and A′.

ε

a

A'

b

2

Demonstrate A ∪ A′ = ε.

1

Refer to the Venn diagram in part a

2

Demonstrate A ∩ A′ = ∅.

A

Since ∪ means either, it is clear that a member is either in A or in A′ which is by definition equal to ε. 1

b Since ∩ means both, it is clear that there

.

are no members in both A and A′, so the set is null or ∅.


History of mathematics George Boole

Boolean algebra

By replacing the set symbols with Boolean ones, we get the laws of Boolean algebra, which are exactly the same as those for sets. Set name

Set symbol

Boolean name

Boolean symbol

Intersection

∩

and

•

Union

∪

or

+

Complement

′

not

′

Universal set

ε

‘everything’

I

Null set

∅

‘nothing’

O

Thus the set laws become: Name

Set law

Boolean law

Commutative Laws

1. A ∪ B = B ∪ A 2. A ∩ B = B ∩ A

A+B=B+A A•B=B•A

Identity

1. A ∪ ∅ = A 2. A ∩ ε = A

A+O=A A•I=A

Complements

1. A ∪ A′ = ε 2. A ∩ A′ = ∅

A + A′ = I A • A′ = O

Associative

1. A ∪ (B ∪ C) = (A ∪ B) ∪ C 2. A ∩ (B ∩ C) = (A ∩ B) ∩ C

A + (B + C) = (A + B) + C A • (B • C) = (A • B) • C

Distributive

1. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) 2. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

A + (B • C) = (A + B) • (A + C)

Closure

A • (B + C) = A • B + A • C

Whatever applies to sets also applies to Boolean algebra.

Only the first distributive law may require some explanation. Do not confuse the Boolean ‘+’ sign with addition!

Chapter 9

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319

Worked Example 17

Establish the distributive law, namely, A + (B • C) = (A + B) • (A + C), using a Venn diagram. Think 1

Write

Consider the left-hand side term (B • C ), which is the intersection of B and C.

A

B

C 2

Now, create the union with A, namely A + (B • C). In this figure, the red shading shows the ‘new’ area added. The final result is the region which has either colour.

A

B

C 3

Now, consider the 1st ‘term’ of the right-hand side, namely (A + B).

A

B

C 4

Now, consider the 2nd term of the right-hand side, namely (A + C ).

A

B

C 5

Now, consider the intersection of the two regions in steps 3 and 4, which produces the region (A + B) • (A + C). The purple area is the resultant region.

A

B

C 6

Compare the two results.

Clearly the area in step 2 equals the area in step 5, thus A + (B • C) = (A + B) • (A + C ).

deMorgan’s Laws and additional results There are two further important results in Boolean algebra involving the negation of the union and intersection operations. These rules, called deMorgan’s Laws, can be proved using the results from Boolean algebra, or can be demonstrated using Venn diagrams.

320


deMorgan’s First Law states: (A + B)′ = A′ • B′ deMorgan’s Second Law states: (A • B)′ = A′ + B′ These laws can be interpreted as saying that ‘the complement of union is intersection’ and ‘the complement of intersection is union’.

Worked example 18

eBook plus

Prove the first of deMorgan’s Laws; namely, that the complement of the union of 2 sets is the intersection of their complements using: a the rules of Boolean algebra and b Venn diagrams. Think a

1

2

Tutorial


WriTe

State the requirements of proof in Boolean algebra terms. Since (A + B)′ is the complement of (A + B), then A′ • B′ must satisfy both the Complement Laws.

Simplify the left side of equation [1]. (a) This is as a result of the 1st Distributive Law. (b) This is as a result of the 1st Commutative Law. (c) This is as a result of the 1st Complement Law. Note: The term (I + B) represents the union of B with I, which is ‘everything’. Similarly, the term (A + I ) represents the union of A with I, which is ‘everything’.

a If (A + B)′ = A′ • B′ then the two

complement laws must be satisfied. Therefore, we must show that: (A + B) + (A′ • B′) = I 1st Complement Law [1] (A + B) . (A′ • B′) = O 2nd Complement Law [2] 1st Complement Law LHS = (A + B) + (A′ • B′) = (A + B + A′) • (A + B + B′) = (A + A′ + B) • (A + B + B′) = (I + B) • (A + I ) =I•I =I = RHS

QED

3

Complete the simplification.

4

Simplify the left side of equation [2]. (a) This is as a result of the 2nd Commutative Law. (b) This is as a result of the 2nd Distributive Law. (c) This is as a result of the 2nd Commutative Law. Note: The term A′ • A is the intersection of A and its complement, which is ‘nothing’ or O. Similarly, B′ • B = O.

2nd Complement Law LHS = (A + B) • (A′ • B′) = (A′ • B′) • (A + B) = A′ • B′ • A + A′ • B′ • B

Complete the simplification. Note that the intersection and the union of O with any set must be O, since there is nothing in O.

5

= A′ • A • B′ + A′ • B′ • B

= O • B′ + A′ • O =O+O =O = RHS

Chapter 9

QED

algebra and logic

321

b

1

2

Draw a Venn diagram representing the lefthand side of the equation; that is (A + B)′. (a) Draw a rectangle with two large, partly intersecting circles. Label one of the circles as A the other as B. (b) Identify the portion required. Note: A + B, that is, A ∪ B, represents the portion inside the two circles. Therefore, its complement (A + B)′ is represented by the portion outside the two circles. (c) Shade the required portion.

b (A+B)' A

A

B

ε A

A′

ε B′

B

ε A

Comment on the Venn diagrams obtained.

B

ε

Draw a Venn diagram representing the righthand side of the equation; that is, A′ • B′. (a) Draw a rectangle with two large partly intersecting circles. Label one of the circles as A the other as B. (b) Identify the portion required. Note: A′, the complement of A, represents the portion outside the two circles and the non-intersecting part of circle B • B′, the complement of B, represents the portion outside the two circles and the nonintersecting part of circle A. A′ • B′ (that is A′ ∩ B′), is represented by the common shaded portion. A′ • B′ is represented by the portion outside the two circles. (c) Shade the required portion.

3

ε

B

A′ B′

The Venn diagrams obtained are identical, therefore, deMorgan’s first Law, (A + B)′ = A′ • B′, holds true.

The results above establish the first of deMorgan’s Laws. The 2nd Law can be proved in a similar fashion. Based upon the rules for Boolean algebra, some important additional results can be tabulated. Rule

322

Explanation

A+A=A

The union of any set with itself must still be itself.

A•A=A

The intersection of any set with itself must still be itself.

A+I=I

The union of any set with ‘everything’ must be ‘everything’ — I.

A•O=O

The intersection of any set with ‘nothing’ must be ‘nothing’ — O.


Rule

Explanation

A • (A + B) = A

Consider that only the part of (A + B) which intersects with A must be just A itself.

A + (A • B) = A

Consider the fact that A • B is within A, if B ⊂ A, or is A if A′ ⊂ B so that its union with A must be just A itself.

These results are easily established with Venn diagrams, and are left as an exercise. At this point it is worth noting that the key operations of sets and Boolean algebra are intimately related to those of deductive logic. These can be summarised by adding columns to an earlier table.

Set symbol

Logic name

Logic symbol

Boolean name

Boolean symbol

Intersection

∩

and

∧

and

•

Union

∪

or

∨

or

+

Complement

′

not

~

not

′

Universal set

ε

‘everything’

I

Null set

∅

‘nothing’

O

Set name

There are no logical equivalents to ‘everything’ or ‘nothing’. Let us use the rules of Boolean algebra to prove an earlier result.

Worked Example 19

The following pair of logical statements were established to be equivalent: ( p ∨ q) ∨ ~p p ∨ ~p Establish this fact using Boolean algebra. Think

Write

1

Write the first logic statement and equate it with its corresponding statement using Boolean algebra.

( p ∨ q) ∨ ~p = (P + Q) + P′

2

Simplify the right-hand side of the equation.

3

Write the second logic statement and equate it with its corresponding statement using Boolean algebra.

p ∨ ~p = P + P′

4

Simplify the right-hand side of the equation.

5

Comment on the results obtained.

The two statements are both equal to I and therefore equivalent to each other. QED

= (P + P′) + Q 1st Commutative Law = (I) + Q = I + Q Identity Law =I Complements

=I


323

REMEMBER

1. A set is a collection of objects with some attribute in common. 2. Sets can be related using ‘intersection’, ‘union’, and ‘complement’ (see the table at the end of the chapter). 3. Sets follow the following mathematical laws (see the table on page 319): (a) Commutative (b) Identity and complements (c) Associative (d) Distributive (e) Closure. 4. Boolean algebra uses all the rules and properties of sets, but replaces the symbols (see the table at the end of the chapter). 5. deMorgan’s Laws relate various operations as follows: (A + B)′ = A′ • B′ (A • B)′ = A′ + B′ Exercise

9D

Sets and Boolean algebra 1 WE 16 Demonstrate the 2nd Associative Law, namely: A ∩ (B ∩ C) = (A ∩ B) ∩ C using Venn diagrams. 2 WE 17 Demonstrate the 2nd Distributive Law, namely: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) or A • (B + C ) = A • B + A • C using Venn diagrams. 3 MC Which of the following statements about sets is false? A The symbol ∪ represents the union of two or more sets. B All sets must have at least one member. C Some sets can have an infinite number of members. D A member cannot be in both set A and set A′ at the same time. E The set represented by the symbol ∅ has no members. 4 a b c d e f 5

324

Write the following sets using the notation A = {. . .}. A = the set of all even positive integers less than 20 B = the set of all positive integers divisible by 4 C = the set of all even prime numbers D = the set of court cards in a deck of playing cards E = the set of integers, less than 0, which are square numbers F = the set of integers less than 10 Which of the sets in question 4 are finite?

6

Demonstrate, using a Venn diagram, the intersection of the following two sets: A = the set of two-digit positive odd numbers B = the set of two-digit square numbers. List the members of the intersection on the diagram.

7

Demonstrate, using a Venn diagram, the intersection of the following two sets: A = the set of two-digit positive even numbers B = the set of two-digit palindromes (numbers which are the same backwards and forwards). List the members of the intersection on the diagram.


8

Demonstrate on a Venn diagram the regions defined by: a A ∩ B′ b A′ ∩ B′

c A′ ∩ (B ∩ C ).

9

The laws of sets can be demonstrated with specific sets. Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, B = {2, 4, 6, 8, 10}, C = {1, 4, 9} Consider the 1st Distributive Law: A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C ). a Find the set represented by the expression (B ∩ C). b Find the set represented by A ∪ (B ∩ C ). c Find the set represented by (A ∪ B). d Find the set represented by (A ∪ C ). e Find the set represented by (A ∪ B) ∩ (A ∪ C ) and show that this is the same set as that in the answer to part b. 10 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, B = {2, 4, 6, 8, 10}, C = {1, 4, 9} a Find the set represented by the expression (B ∪ C ). b Find the set represented by A ∩ (B ∪ C ). c Find the set represented by (A ∩ C ). d Find the set represented by (A ∩ B). e Find the set represented by (A ∩ B) ∪ (A ∩ C ) and show that this is the same set as that in the answer to part b. 11 We18a Using the rules for Boolean algebra, prove the 2nd of deMorgan’s Laws: (A • B)′ = A′ + B′ 12

Simplify the following logical expressions, using the rules of Boolean algebra. a A + A′ • B + A • B b (A + B + A′) + B′ c A + A′ • B d A • B • (A + C)

13 We18b Show, using Venn diagrams, that: a (A + B) • A = A b (A + B) • B′ = A c A + B • A′ = A + B. 14 We19 Determine, using Boolean algebra, if the following two statements are equivalent. ( p ∧ q) ∧ ~p p ∧ ~p 15 Prove the following using Boolean algebra or Venn diagrams. a A + B + A′ + B′ = I b (A + B) • A′ • B′ = O c (A + B) • (A + B′) = A d A • B + C • (A′ + B′) = A • B + C Hint: Use the results from question 13 to shorten your proofs.

9e eBook plus Interactivity

int-0976 Truth tables

digital logic The contribution of logic and Boolean algebra to the design of digital computers is immense. All digital circuits rely on the application of the basic principles we have learned in this chapter. Computer software is constructed using logic gates based on some of the rules of logic laid down by Aristotle.

digital truth values Digital circuits consist of electrical current flowing through wires which connect the various components. The computer recognises the presence of electricity as ‘True’ and the absence of electricity as ‘False’. Furthermore, it is the accepted convention that we denote the presence of electricity by 1 and the absence by 0. (In some systems the value of

Chapter 9

algebra and logic

325

1 is given to positive electricity and 0 to negative electricity.) Thus we have the basic conversion rule which we will apply here as shown in the table. The so-called ‘on–off’ values come from the notion of a switch: if the current flows through, the switch is on; otherwise it is off — as with a light switch.

Logical value

Digital value

Spoken value

False

0

‘Off’

True

1

‘On’

Gates A gate is an electrical component that controls the flow of On electricity in some way. It is similar to a gate on a farm, which sometimes lets the sheep through and sometimes doesn’t. The simplest possible gate is the switch itself. It has two states, on Off and off, as shown in the figures at right. When drawing a switch on a diagram it is conventional to show the ‘off’ position. By combining switches in certain configurations, we can create simple logic circuits. Worked Example 20

Consider the pair of switches arranged (in parallel) as shown in the figure at right. Assume there is electricity at P. What positions of the two switches, x and y, will allow a current to flow through?

x P y

Think 1

2

Q

Write

List the possible positions for each switch. Switch x can be either off or on (0 or 1) independently of y, so there are 2 × 2 = 4 possible positions. Consider x = 0, y = 0. There will be no current at Q. Otherwise, if x = 1 there will be a current at Q. Similarly, if y = 1 there will be a current at Q. If both x = 1 and y = 1 there will be a current at Q. Note: We can consider this as the ‘truth table’ for this circuit.

x

y

Q

0

0

0

0

1

1

1

0

1

1

1

1

Because of the similarity of this truth table to the Boolean operator ‘+’ (‘or’), we can symbolise this circuit as Q = x + y. In theory, a computer could be constructed from nothing more than thousands (millions, billions . . .) of switches. However, the design of a logic circuit would be a long, time-consuming process. Furthermore, it is not clear ‘who’ turns the switches on or off. Hence, logic gates were constructed as ‘black box’ components which could be combined, quickly, to perform relatively complex operations. A gate consists of one or two inputs and one output. The inputs are ‘wires’ which are either off (0) or on (1). Similarly, the output is either 0 or 1. Inputs require a continuous source of electricity in order to remain at either 0 or 1. The following table shows the gates we will use. Note that inputs are always on the left, output on the right. Name NOT

326

Symbol

Truth table Input

Output

0

1

1

0


Comments Equivalent to Boolean ‘not’.

Name OR

NOR

AND

NAND

Symbol

Truth table

Comments

A

Input A

Input B

Output

B

0

0

0

0

1

1

1

0

1

1

1

1

A

Input A

Input B

Output

B

0

0

1

0

1

0

1

0

0

1

1

0

A

Input A

Input B

Output

B

0

0

0

0

1

0

1

0

0

1

1

1

A

Input A

Input B

Output

B

0

0

1

0

1

1

1

0

1

1

1

0

Equivalent to Boolean ‘or’.

Equivalent to Boolean ‘or’ followed by a ‘not’.

Equivalent to Boolean ‘and’.

Equivalent to Boolean ‘and’ followed by a ‘not’.

NAND and NOR gates, although without equivalent Boolean expressions, are convenient ways of combining AND (or OR) with NOT. For example, a NAND gate is equivalent to the combination shown at right. Very sophisticated circuits can be constructed from combinations of these 5 gates, and the truth table of the output for all possible inputs can be determined.

Worked Example 21

Determine the truth table for the output Q, in terms of the inputs a, b and c.

Think 1

Working from left to right, determine the truth table for the output d in terms of inputs a and b.

a b c

Q

Write a b c

d Q


327

2

3

Use the truth table for an AND gate.

Now consider the output d to be the input to the OR gate, combined with c to determine the truth table at Q. Note that the 1st four rows correspond to step 2 for the case of c = 0, while the 2nd four rows correspond to step 2 for the case of c = 1.

a

b

d

0

0

0

0

1

0

1

0

0

1

1

1

a

b

d

c

Q

0

0

0

0

0

0

1

0

0

0

1

0

0

0

0

1

1

1

0

1

0

0

0

1

1

0

1

0

1

1

1

0

0

1

1

1

1

1

1

1

An alternative approach is to start with all inputs (a, b and c) and lay out a ‘blank’ truth table for these three inputs. Add columns for each gate as required.

A ‘blank’ truth table for three inputs.

a

b

c

a

b

c

d

Q

0

0

0

0

0

0

0

0

0

0

1

0

0

1

0

1

0

1

0

0

1

0

0

0

0

1

1

0

1

1

0

1

1

0

0

1

0

0

0

0

1

0

1

1

0

1

0

1

1

1

0

1

1

0

1

1

1

1

1

1

1

1

1

1

The completed truth table.

It should be clear that this truth table is equivalent to the one in step 3 of worked example 21, with the rows in different order. Furthermore, this circuit of an AND and an OR gate is logically equivalent to the statement (a ∧ b) ∨ c, or in Boolean algebra terms (a • b) + c.

An application — Burglar alarms Logic circuits can be used to design a burglar alarm. Typically, the owner turns a switch on to arm the alarm and a burglar entering the building effectively turns another switch on or off setting the alarm ringing. Consider the alarm circuit in the figure at right. The first s1 a switch (S1) is set by the owner, while the second (S2) is Q accidentally set by the burglar. s2 b

328


Worked example 22

eBook plus

Determine the truth table for the burglar alarm above, and describe the alarm’s mechanism. Think

Define the inputs, S1 and S2. Both switches are shown in the off position in the figure.

2

Set up the truth table for S1, S2, a, b and Q. Note that a = S1 and b = ~S2.

3

Describe the mechanism by examining each row of the truth table. Note that the logical equivalent of this circuit is S1 ∧ ~S2 or, in Boolean terms, S1 • S2′.


WriTe

1

Tutorial

S1: off = 0, on = 1 S2: off = 0, on = 1 S1 S2

a

b

Q

0

0

0

1

0

0

1

0

0

0

1

0

1

1

1

1

1

1

0

0

1st row: S1 = 0 implies that the alarm is ‘off’ or disabled. 2nd row: S1 = 0, the alarm is still disabled. 3rd row: S1 = 1 and S2 = 0 sets off the alarm (Q = 1). 4th row: S1 = 1 and S2 = 1, the alarm is ‘on’ or enabled. The owner sets the alarm mechanism by closing S1; the burglar sets off the alarm by opening S2.

This is the most common form of burglar alarm, with a wire running through, say, a window. When the burglar opens or breaks the window, current stops flowing through the wire effectively turning off S2. Several windows can be protected with a single wire going from window to window, or with a separate mechanism for each window.

Simplifying logic circuits In some cases an apparently complex circuit can be reduced to a simpler one. Worked example 23

Find a circuit equivalent to the one shown at right.

a

c

Q

b

Think 1

Determine the truth table of the circuit. Start by determining the output at c. Note that the inputs to the AND gate are ‘inverted’ by the two NOT gates.

WriTe

a

b

c

0

0

1

0

1

0

1

0

0

1

1

0

Boolean expression = (a′ • b′)

Chapter 9

algebra and logic

329

2

Complete the truth table by determining the output at Q. This is just the negation of c.

a

b

c

Q

0

0

1

0

0

1

0

1

1

0

0

1

1

1

0

1

3

Write out the Boolean expression for Q, by working backwards from Q.

Q = c′ (but c = a′ • b′) = (a′ • b′)′

4

Simplify, using the rules for Boolean algebra.

Q = a″ + b″ = a + b

5

Create the equivalent circuit. In this case it is a single OR gate.

2nd deMorgan’s Law ‘double’ negative

a

Q

b

The original, more complicated circuit might have been used because of availability or cost of components. Otherwise, it would be advantageous to use the circuit in step 5. Often, one has to design the logic circuit, given a Boolean expression. Worked Example 24

Determine the logic circuit for the Boolean expression Q = (a + b′) • (a + c′). Think

Write

1

Determine the number of ‘independent’ inputs.

There are three inputs: a, b and c.

2

Reduce the original Boolean expression to simpler, component parts.

Let u = b′ and v = c′ Q = (a + u) • (a + v)

This last expression is as simple as possible.

Let w = a + u Let x = a + v Q=w•x

3

4

5

w

Begin with the last, ‘simplest’ expression. This is an AND gate with w and x as inputs, Q as output. (a) Using w = a + u, add an OR gate with a and u as inputs, w as output.

u

(b) Using x = a + v, add an OR gate with a and v as inputs, x as output.

v

Note that input a has been ‘duplicated’ for each OR gate. (a) Using u = b′, and v = c′, add two NOT gates to complete the circuit. (b) The two a inputs must be connected.

w x

a a

b

c


Q

u w x

a

What would be the truth table for Q?

330

Q

x

v

Q

rememBer

1. In digital logic, the truth values are replaced with the presence or absence of electricity. 2. The presence of electricity = 1 = True; the absence of electricity = 0 = False. 3. Gates implement the logical connectives AND, OR, NOT as well as NOR, NAND (see the table in the summary at the end of the chapter). 4. All the rules of logic, sets and Boolean algebra can be used to design and simplify digital circuits. exerCiSe

9e

digital logic 1

2

We20 Consider the pair of switches arranged (in series) as shown at right. Assuming that there is electricity at P, when is there current at Q for various positions of the switches x and y?

4

x

y

Q

y

Q

mC The Boolean equivalent of the circuit in question 1 is:

A Q=x+y D Q = x′ + y 3

P

B Q=x–y E Q = (x • y)′

C Q=x•y

Consider the three switches arranged as shown at right. a Assuming that there is electricity at P, when is there current at Q, for various positions of the switches x, y and z? b Write a Boolean expression equivalent to this circuit.

P

x z

Consider the circuit depicted at right which represents a light fixture in a hallway connected to two switches, x and y. The light is ‘on’ whenever there is a direct connection between P and Q. a Determine the truth table for this circuit. b What would be an application for this?

P

x

y

Q

5

mC A Boolean expression for Q in question 4 in terms of x and y is: A x+y B x•y C (x + y) • (x′ + y′) D x • y + x′ • y′ E (x + y) • (x + y)′

6

Modify the circuit in question 4 so that the light comes on only when either (or both) of the two switches is in the ‘on’ position. a

eBook plus

7

Truth tables

Determine the truth table for the output Q, in

terms of the inputs a, b and c for thecircuit at right.

Digital doc

Spreadsheet 137

We21

8

Determine the truth table for the output Q, in terms of the inputs a, b and c for the circuit at right.

9 mC The Boolean equivalent to the circuit in question 8 is: A [a • ( b + c)]′ B a′ + (b′ • c′) D Both A and B are connect E A, B and C are all connect 10 We22 a Determine the truth table for the output Q, in terms of the inputs a, b and c. b Hence, show that this circuit is equivalent to the one in question 8.

Q

b c a b c

Q

C (a′ + b′) • (a′ + c′) b a

Q

c

Chapter 9

algebra and logic

331

11

12

Determine the truth table for the output Q, in terms of the inputs a, b, c and d.

Burglar alarms revisited: Consider the modification to the alarm system in worked example 22, used to protect a safe. If there is a ‘1’ at R, the alarm rings. If there is a ‘1’ at Q, the safe can be unlocked. a Determine the truth table for this circuit. b Hence, describe the operation of this circuit.

a b c d

Q

s1

a b a

s2

c

13 We23 a Use deMorgan’s laws to show that a • b = (a′ + b′)′. b Hence, construct a logic circuit equivalent to an AND gate. 14

Show how a single NAND gate can be the equivalent of a NOT gate.

15 We24 Determine the logic circuit for the Boolean expression Q = a • (b + c)′.

332

16

The designer of the circuit in question 15 does not have any NOT gates available. Re-design the circuit using NOR and/or NAND gates to replace the NOT gate.

17

Design a logic circuit for the Boolean expression Q = a • b + a′ • c, without using any NOT gates.

18

A conditional circuit: Up until now, we have not seen a b a⇒b a digital equivalent, or even a Boolean equivalent, of the 0 0 1 important logical expression a ⇒ b. The truth table for the conditional statement is shown at right. 0 1 1 a From the following list of statements, find the one which 1 0 0 has the same truth table as a ⇒ b. 1 1 1 i a • b′ ii a′ • b iii a′ + b iv a + b′ • a • b′ • b b Design a logic circuit equivalent to a ⇒ b. c Design a logic circuit equivalent to b ⇒ a. d Find a Boolean statement equivalent to (a ⇒ b) • a. e Find a Boolean statement equivalent to the modus ponens argument, namely: a⇒b eBook plus a Digital docs b Investigation and simplify, as much as possible, using the result A computer storage device from part d. Investigation Binary addition f Design a circuit equivalent to the Boolean statement from part e, and show that the output is always 1. Thus you have established the validity of the modus ponens argument.


Q R

Summary Statements (propositions), connectives and truth tables

• • • • •

A statement (sometimes called a proposition) is a sentence which is either true or false. Logical connectives are ‘and’, ‘or’, ‘not’, ‘If . . . then’, and ‘equals’ (see the table at the end of this summary). Each logical connective has a truth table (see the table at the end of the chapter). A compound statement is made up of two or more statements connected with logical connectives. Two (compound) statements are equivalent if they have identical truth tables. Valid and invalid arguments

• The implication p ⇒ q has the following alternative forms: 1. Implication (assumed true) 2. Converse (False) 3. Contrapositive (True) 4. Inverse (False) • An argument consists of one or more statements called premises and a statement called a conclusion. • An argument is valid if the conclusion is true when all the premises are true. • A valid argument is sound if all the premises are true. • An invalid argument (sometimes referred to as a fallacy) occurs when the premises might be true but the conclusion does not follow logically from them. • Among others, there are 5 valid deductive argument forms: 1. Modus ponens 2. Disjunctive syllogism 3. Hypothetical syllogism 4. Modus tollens 5. Constructive dilemma. Techniques of proof

• A tautology is a compound statement which is always true. • An argument with premises p, q, r. . . and conclusion c is valid if the compound statement p ∧ q ∧ r . . . ⇒ c is a tautology (that is, is always true). • A mathematical argument can be proved using (among others): 1. A valid deductive argument form 2. Contrapositive (prove ~q ⇒ ~p, which proves p ⇒ q) 3. Contradiction (prove that the opposite statement is false) 4. Counter-example (prove false by finding an example) 5. Mathematical induction. Sets and Boolean algebra

• A set is a collection of objects with some attribute in common. • Sets can be related using intersection, union, complement (see the table on the next page). • Sets follow the following mathematical laws (see also, the table on page 319): 1. Commutative 2. Identity and Complements 3. Associative 4. Distributive 5. Closure. • Boolean algebra uses all the rules and properties of sets, but replaces the symbols (see the table on page 334). • deMorgan’s Laws relate various operations as follows: (A + B)′ = A′ • B′ (A • B)′ = A′ + B′


333

Digital logic

• In digital logic, the truth values are replaced with the presence or absence of electricity. • The presence of electricity = 1 = True; the absence of electricity = 0 = False. • Gates implement the logical connectives AND, OR, NOT as well as NOR, NAND, XOR (see the table below). • All the rules of logic, sets and Boolean algebra can be used to design and simplify digital circuits. Logic name ‘And’

‘Or’

‘Not’

‘If . . . then . . .’

‘Equal’

NAND

334

Logic Set symbol symbol ∧

∨

~

⇒

⇔

n/a

∩

∪

′

n/a

=

n/a

Set name

Boolean symbol

Intersection

•

Union

Complement

n/a

Equivalent

n/a

+

′

n/a

=

n/a

Truth table p

q

p∧q

T

T

T

T

F

F

F

T

F

F

F

F

p

q

p∨q

T

T

T

T

F

T

F

T

T

F

F

F

p

~p

T

F

F

T

p

q

p⇒q

T

T

T

T

F

F

F

T

T

F

F

F

p

q

p⇔q

T

T

T

T

F

F

F

T

F

F

F

T

p

q

p NAND q

T

T

F

T

F

T

F

T

T

F

F

T


Digital gate

n/a

n/a

Logic name NOR

XOR

Logic Set symbol symbol n/a

n/a

n/a

n/a

Set name

Boolean symbol

n/a

n/a

n/a

n/a

Truth table

Digital gate

p

q

p NOR q

T

T

F

T

F

F

F

T

F

F

F

T

p

q

p XOR q

T

T

F

T

F

T

F

T

T

F

F

F


335


1 Write the following compound statements in symbolic form. a Melpomeni and Jacques purchased new bicycles. b Either it is cold or it is warm and sunny. c The dinner was late, expensive and poorly cooked.

Multiple choice

1 The sentences ‘The capital of Australia is Canberra’, ‘Australia is part of the Southern Hemisphere’ and ‘Australia’s population is over 20 million’ are examples of: A statements B instructions C suggestions D exclamations E near-statements

2 Determine the truth table for (p ∧ q) ∨ (~p ∧ ~q). 3 Write the converse, contrapositive and inverse of the following statement: If a politician is intelligent, she sends her children to good schools. 4 Establish the validity of the following argument: If the bicycle is not red then it is an Italian bicycle. If a bicycle is not an Italian bicycle then it is green. My bicycle is red. Therefore the bicycle is not Italian. −

5 Prove that x 1 =

1 , provided x ≠ 0. x

6 Prove, by contradiction, that x > x, when 0 < x < 1. 7 Let A = the set of all positive prime numbers less than 100. Let B = the set of all positive two-digit numbers with the digit 1 in them. Let C = the set of all positive two-digit numbers whose sum of digits = 7. List the following sets: a A ∩ B b A ∪ (B ∩ C) c A ∩ B ∩ C. 8 Prove, using the rules of Boolean algebra, that (A + A′ • B) • (B + B • C) = B. 9 Design a logic circuit equivalent to the Boolean expression Q = [A • (B′ • C′)] + [A • (B • C)]. 10 By simplifying the expression for Q in question 9, design a circuit with fewer components. (Hint: Consider using NAND or NOR gates.)

336

2 If there is a compound statement with 6 single statements; p, q, r, s, t and v, then how many rows will there be in the truth table? A 6 B 8 C 12 D 36 E 64 3 The truth table at right represents A p ∧ q B p ∨ q C p ∨ ~q D ~p ∨ q E p ∧ ~q

p

q

?

T

T

T

T

F

T

F

T

F

F

F

T

4 The sentence ‘I like either ham or steak with eggs for breakfast’ can be symbolised as: A h ∨ (s ∨ e) B h ∧ (s ∧ e) C h ∧ (s ∨ e) D h ∨ (s ∧ e) E h ⇔ (s ∧ e) 5 The inverse statement to: ‘If I buy a new coat then I am happy’ is: A If I don’t buy a new coat then I am not happy. B If I am not happy then I won’t buy a new coat. C If I am happy then I will buy a new coat. D If I don’t buy a new coat then I am happy. E If I am happy then I won’t buy a new coat.


6 The following argument: If a lawyer is honest then you should hire him. You shouldn’t hire John Smith. John Smith is not honest. is an example of which valid form? A Modus ponens B Disjunctive syllogism C Hypothetical syllogism D Modus tollens E Constructive dilemma

10 The shaded area in the figure at right represents: A A∪Β B A′ ∪ Β C A ∪ Β′ D A ∩ Β′ E (A ∪ Β)′

ε A

B

11 If I = the ‘universal set’, O = the ‘empty set’ and A = any set, then A • I is: A A B I C O E none of these D A′

7 The following argument is an example of which valid form? If I study hard I will pass my exams. a 12 The Boolean I did not pass my exams. expression equivalent bc Q I did not study hard. to the circuit at right is: d B Disjunctive syllogism A Modus ponens A Q = (a • b ) + (c • d) B Q = (a • b) • (c + d) C Hypothetical syllogism D Modus tollens C Q = [(a + b) • c + d] D Q = [(a • b) + c] + d E Constructive dilemma E Q = a • [b + (c + d)] 8 The statement ‘If she sinks that putt she will either 13 The Boolean equivalent to the circuit shown below win or lose the tournament’ is a tautology because: is: A it makes no sense B it is always true A (x + y) + z x C it is always false D it is no use in an argument B (x + y) • z z E none of these C x • (y + z) y

9 By proving ~q ⇒ ~p, you have proved p ⇒ q is an example of proof by: A counter-example B contradiction C contrapositive D deduction E induction

D x • (y • z) E x • (y + z)

exTended reSponSe

1 Other valid argument forms: In exercise 9B you learned five valid forms of argument. There are several others, including the destructive dilemma. Consider the following argument. If we want to reduce greenhouse gases, we should use more nuclear power, and if we wish to reduce nuclear accidents we should use conventional power. We will either not use nuclear power or not use conventional power. Therefore, we will either not reduce greenhouse gases or we will not reduce the risk of nuclear accidents. a Put each statement into symbolic form. b Set up the truth table for the three statements. c Determine if the argument is valid by finding the rows in the truth table where all premises are true and comparing them with the conclusion. d Use this technique to determine the validity of the following argument. a⇒b b⇒c ∼d ⇒ a ∼c d e Can you devise an example of this argument form?

Chapter 9

algebra and logic

337

2 Implication versus equivalence: The statements a ⇒ b (implication) and a ⇔ b (equivalence) are quite different logically. Implication is ‘If . . . then . . .’, while equivalence is ‘If and only if . . . then . . .’. a Write the truth tables for each of the statements. b Determine the two implications that must be true for the equivalence to be true. c Consider the following implication: If a positive integer is even, then it is divisible by 2. Determine if this is also an equivalence. d Determine if all equivalences are also implications. e Consider the following implications. Determine if they are also equivalences. i If a positive integer is odd, then it is not divisible by 2. ii If a number is positive then its square root is also positive. iii If a number is positive then its square is also positive. iv If a triangle has three equal sides, then it has three equal angles of 60° each. v If a polygon is a square then it is a rectangle. vi If x is odd, then x3 is also odd. vii If two statements p and q are true, then p ∨ q is true. viii If two statements p and q are true, then p ∧ q is true. ix If two statements p and q are true, then p ⇒ q is true. x If an argument is sound then it is also valid. 3 Consider the logic circuit at right. w a How many inputs are there? Name them. x b How many outputs are there? Name them. c Write a truth table for the given logic circuit. d i What occurs when w = 1, x = 0 and y = 1? y ii What occurs when w = 0, x = 0, y = 1 and z = 0? e Does input z have any effect on the other inputs? z f i Write the Boolean expression for Q. ii Simplify the expression obtained in part i using the rules for Boolean algebra. g i Write the Boolean expression for R. ii Simplify the expression obtained in part i using the rules for Boolean algebra. h Use parts f and g to find a simplified circuit to the original one given.

Q

R

4 Consider the logic circuit at right. It consists of two inputs, a Q a and b, and four outputs, Q, R, S, and T. a Determine the outputs when a = b = 0. R b b Determine the outputs when a = 0 and b = 1. S c Repeat for the remaining possible values of a and b. d Show that Q = 1 only when a = b = 0. T e Can you describe the pattern for this circuit, which is called a 2-bit decoder? f The circuit has two NOT gates and four AND gates for the 2-bit decoding of a and b. How many gates would be required for a 3-bit, 4-bit and n-bit decoder? eBook plus Digital doc


338


eBook plus

aCTiViTieS


• 10 Quick Questions: Warm up with ten quick questions on algebra and logic. (page 293) 9A

Statements (propositions), connectives and truth tables

Tutorial

• We4 int-1067: Watch how to complete the truth table. (page 297) 9B

Valid and invalid arguments

Tutorial

• We9 int-1068: Watch how to use a truth table to show an argument is valid. (page 306) Digital doc

• WorkSHEET 9.1: Draw truth tables, explain and establish valid and invalid arguments and solve a worded problem. (page 309) 9C

Techniques of proof

Tutorial

• We14 int-1069: Watch how to prove by contradiction that the product of two negative numbers is positive. (page 312) Digital doc

• WorkSHEET 9.2: Complete more complex truth tables, solve worded problems and apply understanding of tautologies, valid and invalid arguments. (page 317) 9D

Sets and Boolean algebra

Digital doc

• History of mathematics: Learn about mathematician George Boole. (page 319)

9E

Digital logic

Interactivity

• Truth tables int-0976: Consolidate your understanding of truth tables using logic. (page 325) Tutorial

• We22 int-1071: Watch how to determine a truth table for a burglar alarm. (page 329) Digital docs

• Spreadsheet 137: Investigate truth tables. (page 331) • Investigation: A computer storage device. (page 332) • Investigation: Binary addition. (page 332) Chapter review Digital doc


Tutorial

• We18 int-1070: Watch how to prove the first of deMorgan’s laws. (page 321)

Chapter 9

algebra and logic

339

10 Linear and non-linear graphs areas oF sTudy

• Sketching relations in the Cartesian plane from descriptions, equations or formulas and identifying their key features • Sketching relations in the Cartesian plane from rules and tables of values • Polar coordinates and polar graphs

10a 10b 10C 10D 10E 10F 10G 10H

The circle The ellipse The parabola The hyperbola Polar coordinates Polar equations Polar graphs Review of complex numbers and polar form of complex numbers 10I Addition of ordinates, reciprocals and squares of simple graphs

• Graphical representation of circles, ellipses, parabolas and hyperbolas; sketching graphs, including focus-directrix properties • Sketching graphs by addition of ordinates; identifying asymptotes • Sketching the graph of reciprocal and square relations from the graph of a simple relation eBook plus

10a

Digital doc

The circle

10 Quick Questions

The circle belongs to the family of conics. That is, a circle is a curve produced by the intersection of a plane with a cone. A circle is the path traced out by a point at a constant distance (the radius) from a fixed point (the centre). Consider the circles shown below right. The first circle has its centre at the origin and radius r. Let P (x, y) be a point on the circle. By Pythagoras: x2 + y2 = r2. y

The equation of a circle with centre (0, 0) and radius r is: x2 + y2 = r 2

If the circle is translated h units to the right, parallel to the x-axis and k units upwards, parallel to the y-axis, then: The equation of a circle with centre (h, k) and radius r is: (x − h)2 + (y − k)2 = r 2


x

x

y y k

P(x, y) (y − k) (x − h) h

340

P(x, y) y

r

x x

Worked Example 1

Sketch the graph of 4x2 + 4y2 = 25, stating the centre and radius. Think 1

Express the equation in standard form by dividing both sides by 4.

Write

x2 + y 2 = r2 4x2 + 4y2 = 25 x2 + y2 =

25 4  5

x2 + y2 =  2  2

State the coordinates of the centre.

3

Find the length of the radius by taking the square root of both sides.

2

Centre (0, 0) r2 =  5 

2

2

r=

5 2

Radius = 2.5 units 4

Sketch the graph.

y 2.5 −2.5

2.5

x

−2.5

Worked Example 2

Sketch the graph of (x − 2)2 + (y + 3)2 = 16, clearly showing the centre and radius. Think

Write

1

Express the equation in standard form by expressing 16 as 42.

(x − h)2 + (y − k)2 = r2 (x − 2)2 + (y + 3)2 = 42

2


Centre (2, −3)

3

State the length of the radius.

r2 = 42 r=4 Radius = 4 units

4

Sketch the graph.

y 1 −2 −3

2

4

6 x

−7

Chapter 10 Linear and non-linear graphs

341

Worked exampLe 3

Sketch the graph of the circle x2 + 2x + y2 − 6y + 6 = 0. Think

WriTe

Express the equation in standard form using the ‘completing the square’ method twice.

(x − h)2 + (y − k)2 = r2 + 2x + y2 − 6y + 6 = 0 2 (x + 2x + 1) − 1 + (y2 − 6y + 9) − 9 + 6 = 0 (x + 1)2 + (y − 3)2 − 4 = 0 (x + 1)2 + (y − 3)2 = 4 (x + 1)2 + (y − 3)2 = 22

2


Centre (−1, 3)

3

State the length of the radius.

r2 = 22 r=2 Radius = 2 units

4

Sketch the graph.

1

x2

y 5 3 1 −3 −1 1

x

rememBer

Circle graphs: 1. x2 + y2 = r2 2. (x − h)2 + (y − k)2 = r2 exerCise

10a

a + = 49 c x2 + y2 = 36 e 2x2 + 2y2 = 50

Digital doc

2

y2

b x2 + y2 = 42 d x2 + y2 = 81 f 9x2 + 9y2 = 100

Sketch the graphs of the following, clearly showing the centre and the radius.

We2

a (x − + (y − 2)2 = 52 2 c (x + 3) + (y − 1)2 = 49 e x2 + (y + 3)2 = 4 1)2

3

b (x + 2)2 + (y + 3)2 = 62 d (x − 4)2 + (y + 5)2 = 64 f (x − 5)2 + y2 = 100

We3 Sketch the graphs of the following circles.

a b c d e f 342

Sketch the graphs of the following, stating the centre and radius of each.

We1

x2

eBook plus

Completing the square

radius r radius r

The circle 1

SkillSHEET 10.1

centre (0, 0) centre (h, k)

x2 + 4x + y2 + 8y + 16 = 0 x2 − 10x + y2 − 2y + 10 = 0 x2 − 14x + y2 + 6y + 9 = 0 x2 + 8x + y2 − 12y − 12 = 0 x2 + y2 − 18y − 19 = 0 2x2 − 4x + 2y2 + 8y − 8 = 0


4

The graph of (x − 2)2 + (y + 5)2 = 4 is: b y y

mC

a

C

y

5

2

x

5 −5

x

−2

x

−2

D

E

y 2

y 2

x −5

x

−5

5

mC The centre and radius of the circle (x + 1)2 + (y − 3)2 = 4 is: a (1, −3), 4 b (−1, 3), 2 C (3, −1), 4 D (1, −3), 2 E (−1, 3), 16

6 Find the equation representing the outer edge of the galaxy as shown in the photo at right, using the astronomical units provided.

10B eBook plus eLesson

eles-0079 Elliptical paths

The ellipse Another member of the family of conics is the ellipse. A conic may be generally defined as a curve, where the ratio of the distance from any point, P (x, y) on the curve to a fixed point (the focus, F), to its distance from a fixed straight line (the directrix) is a constant (the eccentricity, e). The ellipse opposite has: 1. a centre (0, 0) 2. eccentricity: 0 < e < 1 y 3. two foci at F (ae, 0) and F´(−ae, 0) b a a 4. two directrices with equations x = and x = − e F' e a −a −ae 5. a semi-major axis, a – −e 6. a semi-minor axis, b (when a > b). −b From the definition: FP = e PD FP = ePD (FP)2 = e2 (PD)2 (square both sides)

Chapter 10

P(x, y) F ae

a

a– e

D x

Linear and non-linear graphs

343

a  (ae − x)2 + (0 − y)2 = e 2  x −  e 



2

2 a  a  − 2 x +  e   − 2aex + + = e   a2e2 − 2aex + x2 + y2 = e2x2 − 2aex + a2 x2 − e2x2 + y2 = a2 − a2e2 x2(1 − e2) + y2 = a2(1 − e2)

a2e2

x2

x2 +

y2

e2  x

2

y2 = a2 1 − e2

x2 y2 =1 = a 2 a 2 (1 − e 2 ) x2 y2 + = 1 where b2 = a2(1 − e2) a2 b2

The equation of an ellipse with centre (0, 0), foci (±ae, 0), directrices, x = ± a and 0 < e < 1 is: e x 2 y2 where b2 = a2 (1 − e2) + = 1 a 2 b2 If the ellipse is translated h units to the right, parallel to the x-axis, and k units upwards, parallel to the y-axis, then: The equation of an ellipse with centre (h, k), foci (±ae + h, k), directrices ±a x= + h and 0 < e < 1 is: e

y k+b k

F'

F

k−b h–a h h+a x (−ae+h, k) (ae+h, k) x = − a–e + h x = a–e + h

( x − h) 2 ( y − k ) 2 = = 1 where b2 = a2 (1 − e2) a2 b2

Worked Example 4

Sketch the ellipse directrices.

( x + 1)2 ( y − 2 )2 + = 1, giving the coordinates of the foci and the equations of the 25 16

Think 1

Express the equation in standard form.

Write

( x − h) 2 ( y − k ) 2 + =1 a2 b2 ( x + 1)2 ( y − 2)2 + =1 25 16 ( x + 1)2 ( y − 2)2 + =1 52 42

2

344


Centre (h, k) h = −1 and k = 2 Hence, centre is (−1, 2).


3

State the values of a and b.

a = 5 and b = 4

4

Calculate the eccentricity.

Since b2 = a2(1 − e2) then 42 = 52(1 − e2) 16 = 25(1 − e2) 16 = 25 − 25e2 25e2 = 9 9 e2 = 25 3 5

e= 5

Find the coordinates of the foci.

Foci (±ae + h, k) ae = 5 × 3 5 ae = 3 ae + h = 3 + (−1) =2 −ae + h = −3 + (−1) = −4 Hence, foci are (2, 2) and (−4, 2). ±a +h Directrices are x = e

6

Find the equations of the directrices.

a  3 = 5÷  e  5 a 25 = e 3 25 − a +h= + ( 1) e 3 22 = 3

− and a + h = e

=

− 25

3

+ (−1)

− 28

3 Hence, equations of directrices are x= and x = 7

22 1 or 7 3 3 − 28

3

1

or − 9 3 . y 6

Sketch the graph. F' −9 1–3

2

−6 −4 −1 −2

F 2 4

7 1–3 x


345

REMEMBER

Equations Ellipse (0 < e < 1) x2 y2 + =1 a2 b2 where b2 = a2(1 − e2)

( x − h) 2 ( y − k ) 2 + =1 a2 b2 where b2 = a2(1 − e2)

Exercise

10b

Features Major axis length 2a Minor axis length 2b Centre (0, 0) Foci (±ae, 0) Directrices ±a x= e Centre (h, k) Foci (±ae + h, k) Directrices ±a x= +h e

Graph y b − a–e

F' −ae

−a

P(x, y) F ae

a

a– e

D x

−b y k+b F'

k

F

k−b h−a h h+a x (−ae+h, k) (ae+h, k) x = − a–e + h x = a–e + h

The ellipse 1 Sketch the following ellipses, showing the coordinates of the foci and the equations of the directrices. 2 2 a x + y = 1 100 49

2 2 b x + y = 1 25 4

2 2 c x + y = 1 64 36

2 2 d x + y = 1 121 81

2 e x + y 2 = 1 16

f x2 + 4y2 = 4

2 WE4 Sketch the following ellipses, showing the coordinates of the foci and the equations of the directrices. 2 2 a ( x − 1) + ( y + 2) = 1 9 4

2 2 b ( x + 5) + ( y − 2) = 1 25 16

2 2 c ( x + 5) + ( y + 1) = 1 49 25

d

e

( x − 5)2 + y2 = 1 36

( x − 2)2 ( y − 3)2 + =1 169 25

f x2 + 9(y + 2)2 = 9

3 MC Which of the following statements is not true of the graph of a The centre is at (0, 0). b a = 2, b = 1 346


x2 + y 2 = 1? 4

C e=

3 2 −

D The foci are ( 3 , 0) and (

3, 0).

−4 3 4 3 and y = 3 3 4 mC The equation of the ellipse at right is:

E The directrices are y =


WorkSHEET 10.1

10C

y 2

a

( x + 3)2 + ( y − 2)2 = 1 9

b

( x + 3)2 y 2 + =1 9 2

C

( x + 3)2 y 2 + =1 9 4

D

( x − 3)2 + y2 = 1 9

E

( x + 3)2 y 2 − =1 9 4

0

−6

x

−3 y

5 In order to program a gemstone cutting machine, a jeweller requires an equation for the edge of the stone based on the coordinate system shown at right (centred at the stone’s centre). What is the equation required?

1.4 x 1.0

The parabola The parabola is also a conic, as shown at right. The eccentricity, e, for a parabola is equal to one. Hence, the distance from any point P (x, y) on the curve to the focus is equal to the distance of that point from the directrix. The parabola opposite has: 1. a vertex (0, 0) y 2. eccentricity: e = 1 D 3. focus at F (a, 0) P(x, y) 4. directrix with equation x = −a F From the definition:

−a

a

FP = e PD FP = 1 ∴ PD FP = PD (FP)2 = (PD)2 (Square both sides.) (a − x)2 + (0 − y)2 = (x − −a)2 (a − x)2 + (0 − y)2 = (x + a)2 a2 − 2ax + x2 + y2 = x2 + 2ax + a2 −2ax2 + y2 = 2ax y2 = 4ax The equation of a parabola with vertex (0, 0), focus (a, 0), directrix, x = −a and e = 1 is: y2 = 4ax

x

y

k

F(h + a, k)

h−ah h+a

Chapter 10

x


347

If the parabola is translated h units to the right (parallel to the x-axis) and k units upwards (parallel to the y-axis), then: The equation of a parabola with vertex (h, k), foci (a + h, k), directrix, x = −a + h and e = 1 is: (y − k)2 = 4a(x − h)

Worked exampLe 5

Sketch the parabola y2 = 8x showing the vertex, the coordinates of the focus and the equation of the directrix. Think

WriTe

1


y2 = 4ax y2 = 8x

2

Give the coordinates of the vertex.

Vertex (0, 0)

3

Find the value of a.

4a = 8 a=2

4

Find the coordinates of the focus.

Focus (a, 0) Focus at (2, 0)

5

Find the equation of the directrix.

Directrix at x = −a x = −2

6

Sketch the graph.

y

−2 x = −2

F 2

(2, 0) x

Worked exampLe 6

Sketch the parabola (y − 1)2 = 12 (x + 2) showing the vertex, the coordinates of the focus and the equation of the directrix. Think 1

348


WriTe

(y − k)2 = 4a(x − h) (y − 1)2 = 12 (x + 2) Hence h = −2 and k = 1.


2


3


4


Focus (a + h, k) a + h = 3 + (−2) =1 Focus at (1, 1)

5


Directrix at x = −a + h −a + h = −3 + (−2) = −5 x = −5

6

Sketch the graph.

Vertex (h, k) Hence, vertex is (−2, 1) 4a = 12 a=3

y F(1, 1)

(−2, 1) −5

−2 −1

1

x

x = −5

Worked exampLe 7

eBook plus −12

Sketch the parabola = (x − 5) showing the vertex, the coordinates of the focus and the equation of the directrix. y2

Think

Tutorial


WriTe

1

Write the equation.

y2 = −12(x − 5)

2


(y − k)2 = 4a(x − h) Hence h = 5 and k = 0.

3


Vertex (h, k) Hence, the vertex is (5, 0).

4


4a = −12 a = −3

5


Focus (a + h, k) a + h = −3 + 5 =2 Focus at (2, 0).

6

7


y

Directrix at x = −a + h −a + h = 3 + 5 =8 x=8

F (2, 0)

5 8x x=8

Sketch the graph.

Note: The parabola is facing in the opposite direction to that in the previous worked example. Why?

Chapter 10


349

REMEMBER

Equations Parabola (e = 1) y2 = 4ax

(y − k)2 = 4a(x − h)

Features

Graph y

Vertex (0, 0) Focus (a, 0) Directrix x = −a

D −a

P(x, y) F a

x

y

Vertex (h, k) Focus (a + h, k) Directrix x = −a + h

F(h + a, k)

k

h−ah h+a

x

The distance between the vertex and the focus = a. The distance between the focus and the directrix = 2a. Exercise

10C

The parabola 1 WE 5 Sketch the following parabolas, showing the vertex, the coordinates of the focus and the equation of the directrix. a y2 = 4x d y2 = 24x g y2 = 9x

b y2 = 16x e y2 = 6x h y2 = 7x

c y2 = 20x f y2 = 10x

2 WE 6 Sketch the following parabolas, showing the vertex, the coordinates of the focus, and the equation of the directrix. a d g j

y2 = 8(x − 3) y2 = 16(x + 4) (y + 3)2 = 20(x + 1) (y − 7)2 = 12x

b e h k

y2 = 4(x − 1) (y − 1)2 = 4(x − 2) (y + 6)2 = 8(x + 3) (y + 5)2 = 6x

c f i l

y2 = 12 (x + 1) (y − 3)2 = 12 (x − 5) (y − 2)2 = 4x (y + 4)2 = 10x

3 WE 7 Sketch the following parabolas, showing the vertex, the coordinates of the focus, and the equation of the directrix. a c e g i k

y2 = −4x y2 = −10x y2 = −12 (x + 1) y2 = −14(x − 2) (y − 1)2 = −8(x + 3) (y + 2)2 = −18(x − 1)

b d f h j l

y2 = −8x y2 = −6x y2 = −20(x + 3) y2 = −2 (x − 4) (y − 2)2 = −16(x + 5) (y + 5)2 = −10(x + 2)

4 MC The equation of the parabola at right is: A y2 = 4x B y2 = 4(x − 1) 2 C y = 4(x + 1) D y2 = 8(x + 1) 2 E y = 12 (x − 1)

350


y

−2

F 46

x

5 MC Which one of the following statements is not true of the graph of the parabola (y + 1)2 = −4x? B The vertex is at (0, −1). A The focus is to the left of the y-axis. C The focus has coordinates (−1, 0). D The directrix is at x = 1. E It has the same directrix as y2 = −4x. 6 Match each of the following graphs with its equation. a y y c b

y F

2 −3

x

−6 −3

i y2 = −12(x − 2)

ii y2 = 12x

x

2

−3

y

d

F −1

x

5

x

iii (y − 2)2 = −12x

iv (y − 2)2 = 12(x + 3)

7 Find the Cartesian equation of each of the following parabolas and sketch each graph. a Focus at (2, 0) Directrix x = −2 b Focus at (2, 0) Directrix x = −3 c Focus at (4, 0) Directrix x = 2 d Focus at (6, 0) Directrix x = −2 − − e Focus at ( 2, 0) Directrix x = 4 f Focus at ( 4, 0) Directrix x = −1 − − g Focus at (1, 2) Directrix x = 1 h Focus at ( 2, 3) Directrix x = 2

10d

The hyperbola Another member of the family of conics is the hyperbola. However, its eccentricity is greater than 1 (see the diagram below). y b F' −a −ae − a–e

y = b–a x

D

a– e

a

F ae

P(x, y)

x

−b x=

− a–e

x = a–e

y = − b–a x

The hyperbola above has: 1. a centre (0, 0) and vertices at (±a, 0) 2. eccentricity: e > 1 3. two foci at F (ae, 0) and F´ (−ae, 0) −a a 4. two directrices with equations x = and x = e e − bx bx 5. two asymptotes with equations y = and y = where b2 = a2(e2 − 1). a a From the definition:

FP =e PD FP = ePD (FP)2 = e2(PD)2 [Square both sides] a  (ae − x)2 + (0 − y)2 = e 2  x −   e

2


351

− 2aex +

x2

+

y2

2  2  x2 − 2 a x +  a   e =  e  e  

a2e2

a2e2 − 2aex + x2 + y2 = e2x2 − 2aex + a2 a2e2 − a2 = e2x2 − x2 − y2 a2(e2 − 1) = x2(e2 − 1) − y2 a2 = x 2 −

y2 [Divide both sides by (e2 − 1)] −1

e2

x2 y2 = 1 [Divide both sides by a2] − a 2 a 2 (e 2 − 1)

x2 y2 = 1 where b2 = a2(e2 − 1) − a2 b2 The equation of a hyperbola with centre (0, 0), vertices (± a, 0), foci (± ae, 0), vertices (± a, 0), ± b ± directrices, x = a , asymptotes, y =   x and e > 1 is:  a e

x 2 y2 − 2 = 1 where b2 = a2(e2 − 1) 2 a b If the hyperbola is translated h units to the right, parallel to the x-axis, and k units upwards, parallel to the y-axis, then: y

The equation of a hyperbola with centre (h, k), vertices at (± a + h, k), foci (± ae + h, k), directrices, x=

±a

e

+ h, asymptotes, y − k =

±b

a

x = − a–e + h y = b–a(x − h) + k a x = –e + h

k+b

( x − h) and

k k−b

e > 1 is:

F'

F h −a

h − ae

( x − h) 2 ( y − k ) 2 − = 1 where b2 = a2 (e2 − 1) a2 b2

h

h+a

h + ae y = − b–a(x − h) + k

Worked Example 8

x 2 y2 − = 1 indicating the coordinates of the centre, vertices and foci, 9 4 the equations of the directrices, and the equations of the asymptotes.

Sketch the hyperbola

Think

352

1

Express the equation in standard form by expressing 9 and 4 as perfect squares.

2


Write

x2 y2 − =1 a2 b2 x2 y2 − =1 9 4 x2 y2 − =1 32 2 2 Centre (0, 0)


x

3 4

State the values of a and b and give the coordinates of the vertices. Calculate the eccentricity by using b2 = a2(e2 − 1).

a = 3 and b = 2, vertices (±3, 0) Since b2 = a2(e2 − 1) then 22 = 32(e2 − 1) 4 = 9(e2 − 1) 4 = 9e2 − 9 9e2 = 13 e2 =

13 9 13 3

e= 5

Find the coordinates of the foci.

Foci are (± ae, 0) 13 3

ae = 3 ×

= 13 Hence, foci are (± 13, 0). 6

Find the equations of the directrices.

Directrices are at x =

±a

e

.

 13  a = 3÷  e  3  =

9 13

9 13 13 Hence, the equations of the directrices are =

x= 7

Find the equations of the asymptotes.

±9

13 . 13

Asymptotes y=

±b

x a ±2 = x 3 Hence, the equations of the asymptotes are y= 8

−2 2 and x y= x. 3 3

y

Sketch the graph.

y = 2–3 x

2 F' − 13 −3

3

F 13

x

−2 x=

9 13 − —— 13

9 13 x = —— 13

y = − 2–3 x


353

eBook plus

Worked exampLe 9

Sketch the hyperbola

( x + 1)2 ( y − 2 )2 − = 1. 16 9

Think 1


2


3

State the values of a and b and give the coordinates of the vertices.

4

Calculate the eccentricity by using b2 = a2(e2 − 1).

5

Find the coordinates of the foci by substituting a, e, h and k into (± ae + h, k).

6

Find the equations of the directrices by substituting ±a a, e and h into x = = h. e

Tutorial


WriTe

( x − h) 2 ( y − k ) 2 − =1 a2 b2 ( x + 1)2 ( y − 2)2 − =1 16 9 ( x + 1)2 ( y − 2)2 − =1 42 32 Centre (h, k) h = −1 and k = 2 Hence, the centre is (−1, 2). a = 4 and b = 3, vertices (−5, 2) and (3, 2) Since b2 = a2(e2 − 1) then 32 = 42(e2 − 1) 9 = 16(e2 − 1) 9 = 16e2 − 16 16e2 = 25 25 e2 = 16 5 e= 4 Foci (± ae + h, k) 5 ae = 4 × 4 =5 ae + h = 5 + (−1) and −ae + h = −5 + (−1) =4 = −6 − Hence, the foci are (4, 2) ( 6, 2). Directrices, x = ± a + h e a  5  16 = 4÷  =  4 5 e 11 16 a + h = + (− 1) = 5 5 e − 21 16 − + ( 1) = 5 e 5 Hence, the equations of the directrices are − 21 11 x= and x = . 5 5 Equations of asymptotes: ±b y−k = ( x − h) a

and

7

354

Find the equations of the asymptotes by substituting ±b a, e, h and k into y − k = ( x − h) . a

−a


+h=

−

±3

[ x − (− 1)] 4 ±3 = ( x + 1) 4 3 y − 2 = ( x + 1) and 4 − 3 y−2= ( x + 1) 4 4(y − 2) = 3(x + 1) and 4(y − 2) = −3(x + 1) 4y − 8 = 3x + 3 and 4y − 8 = −3x − 3 3x − 4y + 11 = 0 and 3x + 4y − 5 = 0 Hence, the equations of the asymptotes are 3x − 4y + 11 = 0 and 3x + 4y − 5 = 0. y−2=

8

Sketch the graph.

3x − 4y + 11 = 0

y 5 2

F'

F

x −1 1 3 4 −1 x = − 21–5 x = 21–5 3x + 4y − 5 = 0

−6 −5

rememBer

Equations Hyperbola (e > 1) x2 y2 − =1 a2 b2 where b2 = a2(e2 − 1)

Features Vertices (± a, 0) Foci (± ae, 0) Directrices ±a x= e Asymptotes ±b y= x a

Graph y b F' −a −ae − a–e

Vertices (± a + h, k) Foci (± ae + h, k) Directrices ±a x= +h e Asymptotes ±b y−k = ( x − h) a

a– e

a

F ae

P(x, y)

x

−b x=

( x − h) 2 ( y − k ) 2 − =1 a2 b2 where b2 = a2(e2 − 1)

y = b–a x

D

y

− a–e

y = − b–a x

x = a–e

x = − a–e + h y = b–a (x − h) + k a x = –e + h

k+b k

F'

F h−a

k−b

h − ae

h

h+a

x

h + ae y = − b–a (x − h) + k

a = distance between vertices ÷ 2

Chapter 10


355

Exercise

10D

The hyperbola 1 WE 8 Sketch the following hyperbolas, showing the coordinates of the centre, vertices and foci, the equations of the directrices and asymptotes. 2 2 a x − y = 1 16 9

d

2 2 b x − y = 1 144 25

x2 y2 − =1 9 9

e

2 2 c x − y = 1 4 4

x2 y2 − =1 64 36

f 4x2 − 9y2 = 36

2 WE 9 Sketch the following hyperbolas. 2 2 a ( x − 1) − y = 1 16 9

2 2 b ( x + 3) − y = 1 144 25

2 2 c x − ( y + 2) = 1 9 9

d x2 − (y − 3)2 = 4

2 2 e ( x + 1) − ( y − 2) = 1 64 36

f 4(x − 5)2 − 9(y + 3)2 = 36

3 For the following hyperbolas, find: i the eccentricity ii the foci iii the equations of the directrices iv the equations of the asymptotes. a

y

b

y

1

2 −3

−2

3

x

y

c

2

−5

5 −1

x

−5

x

5 −2

1

4 MC Given that a hyperbola has vertex points of (4, 0) and (−4, 0), with asymptotes of y = 2 x − and y = 1 x, the equation of the hyperbola is: 2

2 2 A x − y = 1 4 2

2 2 b x + y = 1 4 2

2 2 D x − y = 1 16 4

2 2 e y − x = 1 2 4

2 2 c x − y = 1 64 16

5 MC The equations of the asymptotes of the hyperbola with foci (±2.5, 0) and a distance of 4 units between the two vertices are: A y =

±1 4

x

B y =

±3 4

D y =

±5 4

x

E y = ± x

C y =

x

6 MC The equation of the graph at right is: 2 2 A x − y = 1 64 9

2 2 B x − y = 1 16 12

2 2 C x − y = 1 16 9

2 2 D x + y = 1 16 9

e 3x2 − 4y2 = 12 356


±8 5x

y

F' −5 −4

F 4 5

x

2 and foci ( ± 3 2 , 0).

7 Find the equation of the hyperbola with eccentricity

8 Find the equation of the hyperbola with vertices (±5, 0) and directrices at x = ±4. 9 A boomerang manufacturer’s specifications for a particular model of boomerang appear at right. Find an equation for the dashed curve drawn through the boomerangs.

y

F'

F –20

–30

10E

20

Polar coordinates

y

In the Cartesian coordinate system, a point, P, is located by using (x, y) coordinates. The same point can be located by stating the distance of the point from the origin, the radius, r, and the angle, θ, it makes with the positive x direction. These are known as polar coordinates. We write the polar coordinates of point P as [r, θ]. Note: θ may be given in degrees or radians.

x

30

P(x, y) r

θ x

y x

Worked Example 10

Plot the following polar coordinates. a [2, 60°] b [ − 3,

2π ] 3

Think a

Write

1

Draw the positive x direction.

2

Rotate 60° anticlockwise.

3

Extend the line 2 units.

a

[2, 60°]

y 2

60° x

b

1

Draw the positive x direction.

2

2π Rotate anticlockwise. 3

3

Extend the line 3 units in the opposite direction.

b

y π 2— 3

x 2π [−3, —3 ]

Note: [−3, 2π ] is the same as [3, 5π ]. Why? 3 3 Can you find another set of coordinates for the same point?


357

From the initial polar coordinates diagram, by trigonometry: x = r cos (θ ) y = r sin (θ ) Hence, we can convert polar coordinates to Cartesian coordinates.

Worked Example 11

Convert [2, 2π ] to Cartesian coordinates. 3 Think 1

Write

Find the x-coordinate.

x = r cos (θ )  2π  = 2 cos    3 =2×− x = −1

2

Find the y-coordinate.

1 2

y = r sin (θ ) 2π = 2 sin    3

[2, 2—π ]

=2× 3 2 y = 3 3

State the Cartesian coordinates.

Hence, the Cartesian coordinates are (−1, 3).

4

Alternatively, on the Main screen, using the soft keyboard, tap: • ) • Complete the entry line as shown. Then press E. Note: The angle symbol can be found in 9 K.

From the initial polar coordinates diagram, by trigonometry: y tan (θ ) = x By applying Pythagoras’ theorem: r = x 2 + y 2 Hence, we can convert Cartesian coordinates to polar coordinates.

358


3

y 2

3 2π — 3

x

Worked exampLe 12

eBook plus

Convert (3, −4) to polar coordinates.

Tutorial

Think 1

Find θ.

=

32 + ( − 4 ) 2

=

9 + 16

=

25 y x −4 = (θ in 4th quadrant) 3

tan (θ ) =

Note that as tan (θ ) is negative, θ is in the fourth quadrant.

3

Worked example 12

x2 + y2

r =

Find r.

r =5 2

int-1074

WriTe

State the polar coordinates (choose θ = 306°52′ in this case).

θ = tan

−1  − 4    3

so θ = −53°8′ or (360 − 53°8′) θ = −53°8′ or 306°52′ Hence, the polar coordinates are [5, 306°52′]. y

−4

θ r

3

x

rememBer

y

Cartesian coordinates (x, y) Polar coordinates [r, θ] where x = r cos (θ ) and y = r sin (θ ) y and r = x 2 + y 2 and tan (θ ) = . x

P(x, y) r

θ x

exerCise

10e eBook plus Digital doc

SkillSHEET 10.2 Converting radians to degrees

y x

polar coordinates 1

We 10 Plot the following polar coordinates on the same graph. a [2, 0°] b [5, 180°] c [0.5, 270°]

d [3, 90°]

2 Plot the following polar coordinates on the same graph.  π a 1,   3

 −π  b 1,   3

 2π  c 1,   3 

3 Locate each of the following points on the same graph. a [3, 45°] b [4, 100°] c [1, 300°]

Chapter 10

 5π  d 1,   3  d [2.5, −30°]


359


WorkSHEET 10.2

10F

4

We11 Convert the following to Cartesian coordinates. (See below for a graphics calculator approach.) a [2, 45°] b [5, 30°] c [3, 60°] d [2.7, 90°] π  h [7.8, π] e [1.5, 120°] f [12, 210°] g  2.6,  2  5π   π  k [25, 5.8c] l [1.6, 4c] i 10,  j  9.1, 3 3    

5

We12 Convert the following Cartesian coordinates into polar coordinates. (See below for a CAS calculator approach.) Express θ in parts a−e in degrees and the rest in radians. a (5, 0) b (0, 4.3) c (−30, 0) d (0, −9) e (6, 6) f (−4, 3) g (−5, −12) h (6, −8)

i (−1,

3)

j (−2, −2)

k (2, 3, −2)

l (5, 6)

polar equations A polar equation is an equation written in terms of r and/or θ. Using the conversions for x and y into polar coordinates, x = r cos (θ ) y = r sin (θ ) we can change Cartesian equations into polar equations.

Worked exampLe 13

Convert the following Cartesian equations into polar equations. a x2 + y2 = 25 b y + 2x c 2x − 3y = 5 d x2 + y2 + 6x − 8y = 0 Think a

2 2 e x + y =1 16 9

WriTe

1

Substitute the polar expressions for x and y.

2

Expand and simplify. (Use the identity cos2 (θ ) + sin2 (θ ) = 1.)

3

Alternatively, since x2 + y2 = 25 represents a circle of radius 5 units then the polar equation must be r = 5.

a x2 + y2 = 25

Since x = r cos (θ ) and y = r sin (θ ) (r cos (θ ))2 + (r sin (θ ))2 = 25 r2 cos2 (θ ) + r2 sin2 (θ ) = 25 r2 (cos2 (θ ) + sin2 (θ )) = 25 r2 = 25 r=5 y 5

5 x

−5 −5 b

1

2

360


Divide both sides by r cos (θ ) and sin (θ ) recall the identity = tan (θ ) . cos (θ )

b y = 2x

Since x = r cos (θ ) and y = r sin (θ ) then r sin (θ ) = 2r cos (θ ) sin (θ ) =2 cos (θ ) tan (θ ) = 2 since sin (θ ) = tan (θ ) cos (θ )


c

d

e

3

Isolate θ.

1


2

Simplify.

1


−

θ = tan 1(2) θ = 63°26′ or 1.107° c 2x − 3y = 5 Since x = r cos (θ ) And y = r sin (θ ) then 2r cos (θ ) − 3r sin (θ ) = 5 r (2 cos (θ ) − 3 sin (θ )) = 5 5 r= 2 cos (θ ) − 3 sin (θ ) d x2 + y2 + 6x − 8y = 0

Since x = r cos (θ ) and y = r sin (θ ) r2 cos2 (θ ) + r2 sin2 (θ ) + 6r cos (θ ) − 8r sin (θ ) = 0 r2 (cos2 (θ ) + sin2 (θ )) + r(6 cos (θ ) − 8 sin (θ )) = 0

2

Note that cos2 (θ ) + sin2 (θ ) = 1.

r2 + r(6 cos (θ ) − 8 sin (θ )) = 0

3

Divide both sides by r.

r + 6 cos (θ ) − 8 sin (θ ) = 0 Hence, r = 8 sin (θ ) − 6 cos (θ ).

1


2

Simplify.

e

x2 y2 + =1 16 9 Since x = r cos (θ ) and y = r sin (θ ) (r cos (θ ))2 (r sin (θ ))2 + =1 16 9 r 2 cos2 (θ ) r 2 sin 2 (θ ) + =1 16 9 9r 2 cos2 (θ ) + 16r 2 sin 2 (θ ) =1 144 r2 (9 cos2 (θ ) + 16 sin2 (θ )) = 144 144 r2 = 2 9 cos (θ ) + 16 sin 2 (θ )

Similarly, polar equations can be changed to Cartesian form.

Worked Example 14

Convert the following polar equations into Cartesian equations. 2 a r = 4 cos (θ ) b tan (θ ) = 2 c r = 1 + sin (θ ) Think a

Write

1

Find r2 by multiplying both sides of the equation by r.

2

Substitute the Cartesian expressions for r and θ.

a r = 4 cos (θ )

r2 = 4r cos (θ ) Since r2 = x2 + y2 and x = r cos (θ ) then x2 + y2 = 4x


361

3

Simplify by ‘completing the square’.

Note: This is the equation of a circle of radius 2 units and centre (2, 0). b 1 Substitute tan (θ ) = y . x

c

x2 − 4x + y2 = 0 2 x − 4x + 4 − 4 + y2 = 0 (x − 2)2 + y2 = 4 b tan (θ ) = 2

As tan (θ ) = y , y = 2 x x y = 2x

2

Simplify by making y the subject.

1

Simplify the equation by multiplying both sides of the equation by (1 + sin (θ )).

2

Substitute the Cartesian expressions for r and θ.

Since y = r sin (θ ) r+y=2

3

Make r the subject.

4

Find r2 by squaring both sides.

5

Substitute for r2.

6

Expand and simplify.

Since r2 = x2 + y2 2 x + y2 = (2 − y)2 = 4 − 4y + y2 2 x = 4 − 4y 4y = 4 − x2 x2 y = 1− 4

Note: This is the equation of a parabola.

c

2 1 + sin (θ ) r(1 + sin (θ )) = 2 r + r sin (θ ) = 2 r=

r=2−y r2 = (2 − y)2

REMEMBER

Cartesian coordinates (x, y) Polar coordinates [r, θ] where x = r cos (θ ) and y = r sin (θ ) y also r = x 2 + y 2 and tan (θ ) = . x

y

P(x, y) r

θ x

Exercise

10F

Polar equations 1 WE 13 Convert the following Cartesian equations into polar equations. a x = 3 b y = 2 c x2 + y2 = 9 d x2 + y2 = 36 e y = 5x f y = x g 3x − 4y = 1 h 5x + y = 7 i x2 + y2 − 10x + 6y = 0 j x2 + y2 + 6x + 8y = 0 2 2 k x + y − 12y = 0 l x2 + y2 − 2x = 0 2 2 2 2 m x + y = 1 n x + y = 1 9 4 4 25

362


y x

2 WE 14 Convert the following polar equations into Cartesian equations. a r = 2 b r = 5 c r = 6 sin (θ ) d r = 2 cos (θ ) f tan (θ ) = -4 e tan (θ ) = 3 3π g θ = π h θ = 4 4 i r cos (θ ) = 4

j r sin (θ ) = -1

k r = 4 sin (θ ) − 2 cos (θ )

l r = 6 sin (θ ) + 8 cos (θ )

m r =

10G

3 1 + sin (θ )

n r =

4 1 − cos (θ )

Polar graphs Polar equations can be graphed using polar coordinates. This is often a better alternative than converting polar equations to the sometimes more complicated Cartesian equation form. When using polar equations, θ is assumed to be measured in radians.

Worked Example 15

Sketch the graph of the ‘Spiral of Archimedes’, r = θ for 0 ≤ θ ≤ 2π. The ‘Spiral of Archimedes’ curve is named after the Greek mathematician who first investigated its properties. Think 1

Write

Construct a table of values for 0 ≤ θ ≤ 2π π using increments of and find the 6 corresponding r values to 2 decimal places.

Sketch the graph using a protractor and ruler to plot each of the points from the table. Remember r is the distance from the centre (the origin). The circular protractor on a Mathomat is ideal for this.

π 5— 6

π 3

2π 3

5π 6

r

0

0.52 1.05 10.57 2.09 2.62

θ

π

7π 6

π 2— π 3 3— 4

4π 3

3π 2

5π 3

11π 6

2π

3.14 3.67 4.19 4.71 5.24 5.76 6.28 π — 2

π — 3

π — 4

π — 6

π 7π — 6

π 2

0

r 2

π 6

θ

O π 5— π 4 4— 3

π 3— 2

π 7— π 4 5— 3

π 11 — 6


363

Worked exampLe 16

Sketch the graph of r = θ for 0 < θ < 4π using a CAS calculator. Think

WriTe

On the Geometry screen, tap: • Function • Polar Complete the entry line as: r: θ

θ min = 0 θ max = 4π • OK Adjust the window size if required. Compare the graphs of worked examples 15 and 16.

Worked exampLe 17

eBook plus

Sketch the graph of r = 8 for 0 ≤ θ ≤ 2π.

Tutorial


Think 1

2

Construct a table of values for 0 ≤ θ ≤ 2π and find the corresponding r values to 2 decimal places.

WriTe/dispLay

π 2 8

π

7π 6

4π 3

3π 2

5π 3

11π 6

2π

8

8

8

8

8

8

8

r

8

θ r

π — 2

8

π −8

8

−8 π 3— 2

364

5π 6 8

π 3 8

0

Sketch the graph using a protractor and ruler to plot each of the points from the table. Remember r is the distance from the centre (the origin).

2π 3

π 6 8

θ


8

3

On the Graph & Tab screen, tap: • Type • r = Type Complete the entry line as: r1 = 8 | 0 ≤ θ ≤ 2π Tick the r1 box and then tap $.

Worked Example 18

Sketch the graph of r = 2 sin (θ ) for 0 ≤ θ ≤ 2π using a CAS calculator. Think

Write

On the Graph & Tab screen, tap: • Type • r = Type Complete the entry line as: r1 = 2sin(θ ) | 0 ≤ θ ≤ 2π Tick the r1 box and then tap $.

REMEMBER

Polar graphs may be plotted by using a table of values and sketching by hand, or by using a CAS calculator. Exercise

10G

Polar graphs 1 WE 15, 16 Using a graphics calculator or other method, sketch the graph of each of the following polar equations for 0 ≤ θ ≤ 4π. − a r = θ b r = −θ c r = 2θ d r = 12 θ 2 WE 17 Using a graphics calculator or other method, sketch the graph of each of the following polar equations for 0 ≤ θ ≤ 2π. d r = -2 a r = 2 b r = 4 c r = 1.5


365

3 WE 18 Using a CAS calculator or other method, sketch the graph of each of the following polar equations for 0 ≤ θ ≤ 2π. a r = sin (θ ) b r = 1.5 sin (θ ) c r = -3 sin (θ ) d r = 4 sin (θ ) e Hence, predict the graph of r = −5 sin (θ ). Check your prediction on the CAS calculator. 4 Using a CAS calculator or other method, sketch the following polar equations for 0 ≤ θ ≤ 2π. d r = -4 cos (θ ) a r = cos (θ ) b r = 2 cos (θ ) c r = 3 cos (θ ) e Hence, predict the graph of r = −0.5 cos (θ ). Check your prediction on the CAS calculator. 5 Using a CAS calculator or other method, sketch r = 3 sin (θ ) + 4 cos (θ ) for 0 ≤ θ ≤ 2π. a Comment on the shape of the curve. b State: i the y-intercepts ii the x-intercepts iii the length of the diameter iv the length of the radius v the coordinates of the centre vi the Cartesian equation of the curve. 6 Using a CAS calculator or other method, sketch r = 5 sin (θ ) + 12 cos (θ ) for 0 ≤ θ ≤ 2π. a Comment on the shape of the curve. b State: i the y-intercept(s) ii the x-intercept(s) iii the length of the diameter iv the length of the radius v the coordinates of the centre vi the Cartesian equation of the curve. 7 Using a CAS calculator, plot the graph of r = 1 − cos (θ ) for 0 ≤ (θ ) ≤ 2π. Hence, investigate the graphs: a r = cos (θ ) − 1 b r = 1 − sin (θ ) c r = 4(sin (θ ) − 1). 8 Using a CAS calculator, plot the graph of r = 1 + 2 sin (θ ) for 0 ≤ θ ≤ 2π. Hence, investigate the graphs of the type r = b + a sin (θ ) where a > b. a r = 1 + 3 sin (θ )

b r = 1 + 6 sin (θ )

c r = 2 + 6 sin (θ )

d r = 1 + 2 cos (θ )

e r = 1 − 2 sin (θ )

f Hence, predict the graph of r = 1 − 3 cos (θ )

9 Using a CAS calculator, plot the graph r = sin (2θ ) for 0 ≤ θ ≤ 2π. Hence, investigate the graphs: a r = sin (3θ ) b r = sin (4θ ) c r = cos (2θ ) e r =

−2

sin (3θ )

d r = 2 sin (3θ ) f Hence, predict the graph of r = 4 cos (3θ )

10 Compare and comment on the graphs of the polar equations r = 2 − 4 sin (θ ) and r = −2 − 4 sin (θ ) 366


11 Use a graphing program (or CAS calculator) to investigate graphs of the following polar equations. Replace a and b with actual values (such as 1, 2, 3, 4 and so on) and hence comment on the effect of a and b on each graph. Equations

Name of graph

a r = a + b sin (θ )

Limacon

b r = a sin (bθ )

Four leaf rose

c r = 2a(1 + cos (θ )) a d r = θ a cos (2θ ) e r = cos (θ )

Cardioid

f r2 = a2 cos (2θ ) You may need to plot two graphs here, namely

Hyperbolic spiral Right strophoid Lemniscate of Bernoulli

r = a cos (2θ ) and r = −a cos (2θ ) g r =

10h

2a θ π sin (θ )

Quadratix of Hippias

Review of complex numbers and polar form of complex numbers Review of complex numbers Points to remember • A complex number is made up of real and imaginary parts. • Addition and subtraction is performed by adding or subtracting the real and imaginary parts separately. • Multiplication and division by a constant is performed by multiplying or dividing the real and imaginary parts by that constant. • When multiplying two complex numbers, proceed as if you were multiplying two binomial brackets; to square a complex number, use the perfect square identity. • The conjugate of a complex number is obtained by changing the sign of the imaginary component; the product of a conjugate pair is a real number.

Worked Example 19

If w = 1 - 3i and z = 2 + 4i write down 1 a w + z b z - w c 2w d z e wz f z2 g z h w • w. 2 Think

Write

a Add the real and imaginary parts separately.

a w + z = 1 - 3i + 2 + 4i

b Simplify and add the real and imaginary parts

b z - w = 2 + 4i - (1 - 3i)

separately.

=3+i

= 1 + 7i


367

c When multiplying complex numbers by a constant,

multiply the real and imaginary parts by that constant. d Simplify by multiplying the real and imaginary parts

c 2w = 2(1 - 3i)

= 2 - 6i

d

by the constant. e When multiplying two complex numbers, proceed as

if you were multiplying two binomial brackets.

e wz = (1 - 3i)(2 + 4i)

= 2 + 4i - 6i + 12 = 14 - 2i

f z2 = (2 + 4i)2

f To square a complex number, use the perfect square

= 4 + 16i - 16 = -12 + 16i

identity. g The conjugate of a complex number is obtained by

1 1 z = (2 + 4i ) 2 2 = 1 + 2i

g z = 2 − 4i

changing the sign of the imaginary component. h The product of a conjugate pair is a real number.

h w • w = (1 − 3i )(1 + 3i )

= 1+ 9 = 10

Polar form of complex numbers As seen previously, for the Cartesian coordinates (x, y) the polar coordinates are [r, θ] where x = r cos (θ ) and y = r sin (θ ) Hence, for the complex number in Cartesian form z = x + yi, then z = r cos (θ ) + r sin (θ )i or z = r(cos (θ ) + i sin (θ )) This is abbreviated to z = rcis (θ ). I P(x + iy) This can be represented on the complex number plane as shown at right. By Pythagoras’ theorem, r = x 2 + y 2 ; this is represented by z which is the absolute value or the modulus of z.

r

θ

Worked Example 20

Find z for each of the following. a z = 3 + i b z = 2 - 3i c z = -1 - i Think

368

Write

a

z = x2 + y2

a

z = 32 + 12 = 10

b

z = x2 + y2

b

z = 22 + (− 3)2 = 13

c

z = x2 + y2

c

z = (− 1)2 + (− 1)2 = 2


y x

R

Worked Example 21

Express the following complex numbers in polar form. a z = 1 + i b z =

−

3+i

Think

Write

Method 1: Technology-free a

z is in the first quadrant

I

a θ=

(1, 1)

π So, z = 2 cis    4

r

θ b

π and z = 2 4

R

z is in the second quadrant I i

(− 3 , i) θ − 3

b z =

( 3) +1 −

2

= 3+1 =2

π π 5π θ= + = 2 3 6

R

 5π  z = 2 cis    6

Method 2: Technology-enabled a & b

1

On the Main screen, tap: • Action • Vector • toPol Complete the entry lines as: toPol([1 1]) − toPol( [ 3 1] ) Press E after each entry.

2

Write the answers.

a & b

π a z = 1 + i in polar form is z = 2 cis   .  4 bz=

−

3 + i in polar form is

 5π  z = 2 cis   .  6


369

Worked Example 22

 5π  Express 3 cis   in Cartesian form.  6  Think

Write

Method 1: Technology-free Use the rule r cis (θ ) = r(cos (θ ) + i sin (θ )).

 5π   5π   5π  3 cis   = 3 (cos   + i sin   )  6  6  6 −

 3 1  = 3 + i 2   2 =

−

3 3 + i 2 2


On the Main screen, using the soft keyboard, tap: • ) • Complete the entry line as shown. Then press E. Note: The angle symbol can be found in 9 K.

2

Write the answer.

3  5π  − 3 + 3 cis   = i  6 2 2

Multiplying and dividing in polar form It can be easily proved by expansion that cis (θ ) × cis (α) = cis (θ + α). That is, (cos (θ ) + i sin (θ ))(cos (α ) + i sin (α )) = cos (θ ) cos (α ) − sin (θ )sin (α ) + i sin (θ ) cos (α ) + i cos (θ )sin (α ) = cos (θ ) cos (α ) − sin (θ )sin (α ) + i(sin (θ ) cos (α ) + cos (θ )sin (α )) = cos (θ + α ) + i sin (θ + α ) = cis (θ + α ) Hence, r1cis (θ ) × r2cis (α) = r1r2cis (θ + α) It can similarly be proved that

370

r1cis (θ ) r1 = cis (θ − α ) r2 cis (α ) r2


Worked Example 23

Simplify π  2π  × 6 cis   b a 3 cis    3  3

 4π  8 cis    3 .  2π  2 cis    3

Think

Write

Method 1: Technology-free a Use the rule

r1cis (θ ) × r2cis (α) = r1r2cis (θ + α).

 2π  π  2π π  + 3 cis   × 6 cis   = 18 cis   3  3  3 3  = 18 cis (π ) = 18(cos (π ) + i sin (π )) = 18(−1 + 0) = −18

b

Use the rule r1cis (θ ) r1 = cis (θ − α ). r2 cis (α ) r2

 4π  8 cis    3   4π 2π  = 4 cis  −  3 3   2π  2 cis    3  2π  = 4 cis    3   2π   2π   = 4  cos   + i sin     3  3    −1 3 = 4 + i 2   2 = − 2 + 2 3i

REMEMBER

1. For the complex number in Cartesian form, z = x + yi, then z = r cos (θ ) + r sin (θ )i or z = r (cos (θ ) + i sin (θ )) This is abbreviated to z = r cis (θ ). 2. r = x 2 + y 2 and is represented by | z | which is the absolute value or modulus of z. 3. r1cis (θ ) × r2cis (α) = r1r2cis (θ + a) 4. r1cis (θ ) = r1 cis (θ − α ) r2 cis (α ) r2


371

exerCise

10h

review of complex numbers and polar form of complex numbers 1

2

We 19 If w = 3 + i and z = 2 − 3i find:

a 3w

b 2z

e w − 4z i ww

f wz j zz

d 2w + z

c w = 3(−2 − i)

d w=

h w

We 20 Find | w | for each of the following:

a w = 2 + 2i 3

c 1w 3 g w2

b w=3−i

We 21 Express each of the following complex numbers in polar form.

a 1−i b 2i 1 1 e 2 3 + 2i d + i 2 2 4 We 22 Express each of the following in Cartesian form. π b −3 cis (π) a 2 cis    4 d 5

π 3 cis    6

c

−3

f 1 − 3i  2π  c 3 cis    3

π e cis    2

f

5π   6 

− 2 cis 

We 23 Express each of the following in Cartesian form.

 2π  π a 2 cis   × 3 cis    5  10    π  d  cis      6 

g

10i

1 (1 + i) 3

2

 5π  12 cis    3 π 20 cis    6

π π b 2 cis   × 4 cis    6  12   2π  3 cis    5 e  9π  2 cis    10  h

c

f

− 4 cis  3π 

 7π   5  × 3 cis  5 

π 8 cis    6  − 5π  32 cis   6 

 −π  6 cis    4  −π  18 cis   2 

addition of ordinates, reciprocals and squares of simple graphs


int-0977 Addition of ordinates

addition of ordinates Addition of ordinates is a method that can be used to sketch the graph of a function whose rule can be thought of as the sum of two functions. For example, by selecting suitable x-values, the 1 graph of y = x + x can be sketched by adding the ordinates (y-values) of the graph y = x with 1 those of the graph y = . Suitable x-values would include: x

372


1. the endpoints of the two graphs 2. the points of intersection of the two graphs 3. the x-intercepts of either of the two graphs. Of course, if the rules of the two functions are known, additional points can be obtained by direct substitution.

Worked Example 24

Sketch the graph that results from the addition 1 of ordinates of the functions y = x and y = , x whose graphs are shown at right.

y 3

y = 1–x

y=x

2 1 −3 −2 −1 0 −1

1

2

3

x

−2 −3 Think 1

Add the ordinates at the LHS end points of the graph: A small negative value and a large negative value should give a slightly larger negative value. Mark this point on the set of axes.

Write y 3

y = 1–x

y=x

2 1 −3 −2 −1 0 −1

1

2

3

x

−2 −3

2

Add the ordinates at the first point of intersection. Because they are the same we simply double the y-value of this point. Mark the point on the set of axes.

1 y y = –x 3

y=x

2 1 −3 −2 −1 0 −1

1

2

3

x

−2 −3

3

At its x-intercept, the straight line graph y = x has an ordinate of zero while the non-linear graph has an undefined ordinate. This indicates that our graph will also be undefined at the point where x = 0, giving a vertical asymptote.

Vertical asymptote at x = 0


373

4

Add the ordinates at the second point of intersection. Mark the point on the set of axes.

1 y y = –x 3

y=x

2 1 −3 −2 −1 0 −1

1

2

3

x

−2 −3 5

Add the ordinates at the RHS end points of the graph: A small positive value and a large positive value should give a slightly larger positive value. Mark this point on the set of axes.

1 y y = –x y = x 3

2 1 −3 −2 −1 0 −1

1

2

3

x

−2 −3 6

Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero, and toward the line y = x as x approaches positive or negative infinity.

y 3

y = x + 1–x

2 1 −3 −2 −1 0 −1

1

2

3

x

−2 −3 7

To perform the addition of ordinates on a CAS calculator, open the Graph & Tab screen. Complete the entry lines as: y1 = x 1 y2 = x 1 y3 = x + x Tick the y1, y2 and y3 boxes and then tap $.

Note: When we are merely sketching, we judge the distances by eye, so the values of y obtained by adding ordinates are approximate rather than exact. In addition we recognise that as the ordinates of one graph approach zero, the y-values obtained approach those of the second graph

374


and an asymptote results. Likewise, as the ordinates of one graph approach infinity, the y-values obtained through addition of ordinates also approach infinity resulting in a vertical asymptote.

Reciprocals and squares of simple graphs

1 When finding the square or reciprocal of simple graphs such as y = x2, y = x and y = , we obtain x a new graph that is distorted or transformed in a predictable way.

Sketching the reciprocal of a graph A graph of the reciprocal of a function can be drawn by sketching the original function and marking the reciprocal of the ordinates for each x-value. For reciprocal graphs, points with an ordinate of 1 will be unchanged as the reciprocal of 1 is 11 , which of course equals 1. Points with an ordinate of −1 will be unchanged for the same reason. The reciprocal of each ordinate remains on the same side of the x-axis. For ordinates greater than 1, the reciprocal will be smaller than 1. The larger the ordinate, the closer its reciprocal will be to the x-axis (approaches 0). For ordinates between 0 and 1, the reciprocals will be greater than 1. The same also applies for ordinates that are negative. In general, the further an ordinate is from the x-axis, the closer its reciprocal will be to the x-axis; the closer the ordinate is to the x-axis, the further its reciprocal will be from the x-axis.

Worked Example 25 y y = x2 3

Sketch the reciprocal of the graph of the function y = x2 shown at right.

2 1 −2 −1 0 −1 Think 1

Write y

Sketch the given function and draw the horizontal line y = 1 on the same axes.

2

Mark any points where the ordinates of the graph are equal to 1.

3

Select several points on the original graph with ordinates greater than 1 (above the line). Estimate their value and mark their reciprocals below the line.

2 x

1

3

2

2

2 1

1– –1 2 3

−2

3

3

−1

1–

2 –1 3

0 −1

1

2

x

−2 4

Select several points on the original graph with ordinates between 0 and 1 (between the line and the x-axis). Estimate their value and mark their reciprocals above the line.

y 3

3 3

2 2 2 1– –1 2 3

−2

1

1– 1 2 – 3

−1

0 −1

3

3

22 1 1– –2 3

1– 2 1– 3

1

2

x


375

5

Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero, and toward the x-axis as x approaches positive or negative infinity.

y 3

3 3

2 2 2 1

1– 1 2 – 3

1– –1 2 3

−2

6

−1

0 −1

y = x1–2 3

3

22 1 1– –2 3

1– 2 1– 3

1

2

x

Alternatively, on the Graph & Tab screen, complete the entry lines as: y1 = x2 1 y2 = 2 x Tick the y1 and y2 boxes and then tap $.

Sketching the square of a graph The graph of the square of a function can be drawn by sketching the original function, and marking the square of the ordinates for each x-value. Squaring the ordinates of a graph will have two effects. Firstly, the resulting values are always positive so we need to consider marking only the new points above the x-axis. Secondly, the new points become further away from the x-axis in the case of ordinates greater than 1 or less than −1, and closer to the x-axis in the case of ordinates between −1 and 1. Of course, ordinates with a value of 1 or −1 will become equal to 1 when they are squared.

Worked Example 26

Sketch the square of the graph of the function y = 2 shown below. x y 3 2 1 −3 −2 −1 0 −1 −2 −3

376

y = 2–x

1

2

3 x


Think 1

2

Sketch the given function and draw the horizontal lines y = 1 and y = −1 on the same axes Mark any points where the ordinates of the graph are equal to 1 and change any ordinates of −1 to 1. (Squaring 1 results in 1.)

Write y 3 2 1 1

1

−3 −2 −1 −1

1

3

x

−2 −3

3

2

Select, estimate, and square several ordinates between the horizontal lines y = −1 and y = 1 marking these positive values closer to the x-axis.

y 4 3 2 1 1

–1 4

1

−4 −3 −2 −1 −1 1

1

–1 1 2 – 4

2

3

4

x

–—

2

−2 −3

4

−4

Select, estimate, and square several ordinates outside the horizontal lines y = −1 and y = 1 marking these positive values further away from the x-axis.

1 –1 4

y

9 9 9 8 7 6 5 4 4 4 3 3 2 2 1 1

−4 −3 −2 –1 0 −1 1 –— 2 −2 −2 −3 −3 −4 5

1

2

–1 1 2 – 4

3

4

x

y

Join the points with a smooth curve noting that the graph will asymptote toward the y-axis as x approaches zero.

–1 4

9 9 9 4 y =— 8 x2 7 6 5 4 4 4 3 3 2 2 1 1 1

–4 –3 –2 –1 0 –1 1 –— 2 –2 –2 –3 –3 –4

1

2

–1 –1 2 4

3

4

x


377

6

Alternatively, on the Graph & Tab screen, complete the entry lines as: 2 y1 = x 2  2 y2 =    x Tick the y1 and y2 boxes and then tap $.

REMEMBER

1. Addition of ordinates is a method that can be used to sketch the graph of a function whose rule can be thought of as the sum of two functions. It involves adding the ordinates (y-values) of the two functions for each value of x. 2. Suitable points at which to add ordinates are: (a) the LHS and RHS end points (b) the points of intersection of the two graphs (c) the x-intercepts of either of the two functions. 3. A graph of the reciprocal of a function can be drawn by sketching the original function and marking the reciprocal of the ordinates for each x-value. 4. The graph of the square of a function can be drawn by sketching the original function and marking the square of the ordinates for each x-value. Exercise

10i

Addition of ordinates, reciprocals and squares of simple graphs y 1 WE 24 Sketch the graph that results from the addition of ordinates of the functions y = 2x 1 and y = , whose graphs are shown at right. x 2 Sketch the graph that results from the addition of ordinates of the functions y = x2 and y = 3x. 3 Sketch the graph that results from the addition of ordinates of 1 the functions y = 2x − 4 and y = . x 4 WE 25 Sketch the reciprocal of the graph of the function y = 3x2 shown at right.

3

y = 2x

2 y = 1x–

1 −3 −2 −1 0 −1

1


3 x

−2 −3 y

y = 3x 2

3

−1 0

378

2

1

x

5 Sketch the reciprocal of the graph of the function y = 2x − 3. 6 Sketch the reciprocal of the graph of the function −1 y= . 2x 7 WE 26 Sketch the square of the graph of the 1 function y = shown at right. x 8 Sketch the square of the graph of the function y = x2. 9 Sketch the square of the graph of the function y = −x.

y

y = 1–x

1 −1 0 −1

1


x

379

Summary y

• Cartesian coordinates (x, y) • Polar coordinates [r, (θ )] where x = r cos (θ ) and y = r sin (θ ) y and r = x 2 + y 2 and tan (θ ) = . x

P(x, y) r

θ x

Equations

Features

Circle x2 + y2 = r2

Graph y

Centre (0, 0), radius r

r

P(x, y) y x

x

(x − h)2 + (y − k)2 = r2

y y

Centre (h, k), radius r

P(x, y) (y − k)

k

(x − h) h

Ellipse (0 < e < 1) x2 y2 + =1 a2 b2 where

b2

=

a2(1

−

e2)

Major axis length 2a Minor axis length 2b Centre (0, 0) Foci (± ae, 0) ±a Directrices x = e

x x

y b − a–e

F' −ae

−a

P(x, y) F ae

a

a– e

D x

−b

( x − h) 2 ( y − k ) 2 + =1 a2 b2

Centre (h, k) Foci (± ae + h, k)

where b2 = a2(1 − e2)

Directrices x =

Parabola (e = 1) y2 = 4ax

380

y k+b

±a

e

F'

k

+h

F

k−b h−a h h+a x (−ae+h, k) (ae+h, k) x = − a–e + h x = a–e + h

Vertex (0, 0) Focus (a, 0) Directrix x = −a


y D −a

P(x, y) F a

x

y x

Equations (y − k)2 = 4a(x − h)

Features

Graph y

Vertex (h, k) Focus (a + h, k) Directrix x = −a + h

F(h + a, k)

k

x

h−ah h+a

Hyperbola (e > 1)

b2 = a2(e2 − 1)

x2 y2 − =1 a2 b2

Vertices (± a, 0) Foci (± ae, 0)

where b2 = a2(e2 − 1)

Directrices x =

y b

±a

Asymptotes y =

e ±b

a

F' −a −ae − a–e

x

y = b–a x

D

a– e

a

F ae

P(x, y)

x

−b x = − a–e

( x − h) 2 ( y − k ) 2 − =1 a2 b2 where b2 = a2(e2 − 1)

Vertices (± a + h, k) Foci (± ae + h, k) ±a Directrices x = +h e Asymptotes ±b y−k = ( x − h) a

y

y = − b–a x

x = a–e

x = − a–e + h y = b–a (x − h) + k x = a–e + h

k+b k k−b

F'

F h−a

h − ae

h

h+a

x

h + ae y = − b–a (x − h) + k

Polar graphs

• Polar graphs may be plotted by using a table of values and sketching by hand, or by using a CAS calculator. Polar form of complex numbers

• For the complex number in Cartesian form z = x + yi: z = (r cos (q ) + i sin (q )) = r cis (q ) • r = x 2 + y 2 = |z| which is the absolute value or modulus of z. • r1 cis (q) × r2 cis (a) = r1r2 cis (q + a) •

r1cis (θ ) r1 = cis (θ − α ) r2 cis (α ) r2 Addition of ordinates, reciprocals and squares of simple graphs

• Addition of ordinates is a method that can be used to sketch the graph of a function whose rule can be thought of as the sum of two functions. It involves adding the ordinates (y-values) of the two functions for each value of x.


381

• Suitable points at which to add ordinates are: 1. the LHS and RHS end points 2. the points of intersection of the two graphs 3. the x-intercepts of either of the two functions. • A graph of the reciprocal of a function can be drawn by sketching the original function and marking the reciprocal of the ordinates for each x-value. • The graph of the square of a function can be drawn by sketching the original function and marking the square of the ordinates for each x-value.

382



1 Sketch graphs of the following, clearly showing the key features. a x2 + y2 = 9 b (x − 2)2 + (y + 1)2 = 16 c

x2 + y2 = 1 4

e y2 = 8x g

x2 y2 − =1 64 36

( x + 3)2 ( y + 1)2 + =1 9 4 f (y + 2)2 = 16(x − 5) d

h

( x − 4)2 ( y − 1)2 − =1 32 18

x2 y2 + = 1, find: 25 16 i the length of the major axis ii the length of the minor axis iii the coordinates of the x- and y-intercepts iv the eccentricity v the coordinates of the foci vi the equation of the directrices. Hence, sketch the graph.

2 a For the equation b

3 a For the equation y2 = 4(x + 1), find: i the coordinates of the vertex ii the coordinates of the focus iii the equation of the directrix. b Sketch the graph. x2 y2 4 a For the equation − = 1, find: 144 25 i the eccentricity ii the coordinates of the vertices iii the coordinates of the foci iv the equations of the directrices v the equations of the asymptotes b Hence, sketch the graph. 5 Plot the following polar coordinates. π a [2, ] b [3, 120°] c [−1, π] 6 5π d [5, −50°] e [1, ] 4 6 Convert the following Cartesian coordinates to polar coordinates. a (1, 1) b (0, 3) c ( 3 , −1) d (−2, −1) e (−5, 12)

7 Transform the following polar equations into Cartesian equations. π a r = 2 b θ = − 4 3 c r cos (θ ) = 6 d r = 1 + sin (θ ) 8 Transform the following Cartesian equations into polar form. a x2 + y2 = 25 b x = 2 c y = 2x d 2x − 3y = 5 9 Sketch the graphs of the following polar equations. π a r = 3 b θ = 6 1 2

c r = 3 − 3 cos (θ )

d r =

sin (3θ )

e r = 3 cos (2θ ) θ g r = 2

f r = 1 + 4 sin (θ ) h r = 5 cos (θ )

 −π  10 Convert to 2 cis  to Cartesian form.  4  11 Express z1 =

−

3 + i in polar form

Exam tip Note that this is a second quadrant

angle not a fourth quadrant angle. incorrect answer here for q.

−π

6

is a common

[Assessment report 2007]

[© VCAA 2007]

12 For the equation y = 4 − x, sketch the graph on the same set of axes as: a its reciprocal b its square. Multiple choice

1 The centre of the circle with equation (x − 1)2 + (y + 3)2 = 25 is: A (−1, 3) B (3, −1) C (1, −3) − D (1, 3) E ( 3, 1) 2 The equation of the y circle at right is: A x2 + y2 = 16 B (x − 1)2 + y2 = 9 3 x −1 C x2 + (y − 1)2 = 9 2 2 D (x + 1) + y = 16 E (x + 1)2 + y2 = 9


383

3 The eccentricity of the ellipse with equation x2 y2 + = 1 is equal to: 16 4 A

3 4

B

3 2

C

1 4

5 2

D

A y =

±25 x 9

B y =

±9 x 25

D y =

±3 x 5

E y =

±5 x 3

8 The graph of

C y =

x2 y2 − = 1 is: 16 9 y 9

A −16

16 x

−9 y

B

9 −16 −9

16

y 3

C −4

4 x −3 y

D

4 −3

384

3 −4

x

x

3 −4

E 4

x2 y2 4 The foci of the ellipse with equation + =1 25 9 are located at: B (±5, 0) C (±4, 3) A (±4, 0) D (±5, 3) E (0, ±3) 5 The vertex of the parabola with equation (y − 1)2 = 12x is at: A (1, 3) B (0, −1) C (0, 1) D (1, 0) E (3, 1) 6 The equation of the y parabola shown at right is: A (y + 2)2 = 4(x + 2) 3 B (y − 3)2 = 8(x − 2) C (y − 3)2 = −4(x − 2) 2 4 x 2 − D (y − 3) = 8(x − 2) E (y − 3)2 = 4(x − 2)2 x2 y2 7 The hyperbola with equation − = 1 has asymptotes with equations: 25 9 ±4 5

y

E

−3

4

x

x2 y2 9 The hyperbola with equation − = 1 has directrices with equations: 16 9 A x = D x =

±16 5 ±4

B x = E x =

±9 5 ±5

10 The Cartesian coordinates of [6, A (3 3 , 3) D

 1 3   , 2 2 

B (3, 3 3 )

C x =

±16 9

π ] are: 3 C (1, 3 )

E (3 2 , 3 2 )

11 The polar coordinates of (−3, 0) are: 3π A [3, ] B [−3, π] C [3, π] 2 − D [ 3, 2π] E [3, 0] 12 The polar equation r = 3 sin (θ ) can be expressed in Cartesian form as: A x2 + y2 = 9 B x2 + y2 − 3x = 0 2 2 C x + y + 3y = 0 D x2 + y2 − 3y = 0 E y = 3 13 The Cartesian equation x + y = 0 can be expressed in polar form as: −π π A r = 1 B θ = C θ = 4 4 D r = 45° E r = θ 14 The graph of the polar equation r = 3 cos (4θ ) is a ‘rose’. The number of leaves is: A 4 B 3 C 2 D 12 E 8 15 The maximum value of r when r = −2 + 3 sin (θ ) is: A 1 B 5 C 2 D 3 E 6 16 If z = 2 - 3i then |z| = C 13 A 2 B -3 E 1 D 13 π 17 Let u = 7 cis   and v = acis (b) where a and b are  4 real constants. π If uv = 42 cis   , then  20  −π −π B a = 35 and b = A a = 6 and b = 5 5 π 1 D a = 35 and b = C a = 6 and b = 5 5 1 E a = 6 and b = 5 [© VCAA 2005]


18 Which of the following graphs results from the addition of ordinates of the functions y = −x and 1 y = whose graphs are shown below? x y

y

C

y = 1–x

y = −x

x x

A

y

D

y

x

x

B

y

y

E

x x

Extended response

1 Find the exact coordinates of the points of intersection of the parabolas shown. y A

−6

0

_ _9 2

1– 2

3

x

B


385

2 For the hyperbola shown in the figure at right: a state the gradient of each asymptote b find the equations of each asymptote x2 y2 c given that the equation of the hyperbola is − = 1, a2 b2 state the relationship between a and b d find the eccentricity e find the equation of the hyperbola.

y (1, 1)

3 A rectangle is bounded by the straight lines with equations x = 2, x = 2a, y = b, and y = −b, where a > b + 1. a Sketch the rectangle on the Cartesian plane. b Find the equation of the largest ellipse that can be enclosed within the rectangle. c Find the value of the eccentricity of the enclosed ellipse.

0

x

4 A football field designed in the shape of an ellipse (e = 0.8 ) has a semi-major axis of 100 metres. Player P is located on the boundary line level with the goalposts and kicks to his teammate F, a distance of 30 metres away who happens to mark the ball directly in front of goal at the focal point. y

P B

x

F Goalposts

a Find the distance of player P from the closest directrix D. b Find the distance of player F from the boundary line B, directly in front of him. c Hence, find the minimum distance player F must kick the ball in order to score. 5 A racetrack is to be designed so that its inner and outer boundaries are elliptical. The directrices of the inner ellipse (e = 0.6 ) touch the ellipse that forms the outer edge of the track, while its foci are 300 metres apart. y Q 300 metres D′�

F′

O

F

P

D

x

Find the width of the track PD given that FP = 30 metres. Find the length of the semi-major axis OP. Find the length of the semi-minor axis OQ. Find the equation of the inner ellipse (assuming it is centred at the origin). Given that the track has a constant width of 50 metres, will the eccentricity of the outer ellipse be greater than, less than, or equal to 0.6? f Find the eccentricity of the outer ellipse. g Find the equation of the outer ellipse. a b c d e

386


6 Use a CAS calculator to plot the graph of the polar equation r = 3 sin (θ) where 0 ≤ (θ) ≤ 2π. a Explain why the graph is the same when 0 ≤ (θ ) ≤ π. b By experimenting with different values for θ step, develop and justify an argument that polar graphs with the general equation r = a sin (θ ) are not circular. c Determine the relationship between θ step and the shape of the graph obtained. d Given that polar graphs with the general equation r = a sin (θ ) are in fact circular, what assumptions are made when an equation is determined only by inspection of its graph? 7 Find the Cartesian equation, and sketch the graph, of the smallest ellipse that passes through the point (6, 0) and encloses the circle with the polar equation r = 4 cos (θ). Indicate the coordinates of the foci and the equations of the directrices. It may be helpful to use a CAS calculator to first plot the graph of the circle. eBook plus Digital doc


Chapter 10


387

eBook plus

aCTiviTies


• 10 Quick Questions: Warm up with ten quick questions on linear and nonlinear graphs. (page 340) 10a

The circle

Digital doc

• SkillSHEET 10.1: Practise completing the square. (page 342) 10b

The ellipse

eLesson

• Elliptical paths eles-0079: Watch an eLesson on objects that travel in elliptical paths. (page 343) Digital doc

• WorkSHEET 10.1: Sketch graphs of circles and ellipses, noting important features. (page 347) 10C

The parabola

Tutorial

• We 7 int-1072: Watch how to sketch a conical parabola. (page 349) 10D

The hyperbola

Tutorial

• We 9 int-1073: Watch how to sketch a conical hyperbola. (page 354) 10E

Polar coordinates

Tutorial

• We 12 int-1074: Watch how to convert Cartesian coordinates to polar coordinates. (page 359) Digital docs

• SkillSHEET 10.2: Practise converting radians to degrees. (page 359) • WorkSHEET 10.2: Sketch graphs of circles, ellipses, parabolas and hyperbolas; noting important features. Convert between polar and Cartesian coordinates. (page 360)

388

10G

Polar graphs

Tutorial

• We 17 int-1076: Watch how to sketch a polar graph on the calculator. (page 364) 10I

addition of ordinates, reciprocals and squares of simple graphs

Interactivity

• Addition of ordinates int-0977: Consolidate your understanding of the technique of addition of ordinates. (page 372) Chapter review Digital doc



11

11A Graphs of linear inequations 11B Graphs of simultaneous linear inequations 11C Graphs of systems of linear inequations 11D Maximising and minimising linear functions 11E Solving linear programming problems 11F Further applications of linear programming

Linear programming AreAs oF sTudy

• Graphs of linear equations and linear inequalities • The solution of simultaneous equations by algebraic, numerical and graphical methods

• Graphical approach to solving simple optimisation problems using linear programming

eBook plus

11A

Graphs of linear inequations

Digital doc

10 Quick Questions

What is linear programming? Linear programming was developed during the late 1940s to assist in the organisation of military supplies. Today it is extensively used in fields such as business, engineering, industry and social sciences. Linear programming is a mathematical technique used to solve real-life situations in which a particular quantity is to be maximised or minimised — for example time, money, profit — subject to given constraints or restrictions. A specific example would be maximising the profit a company makes, subject to: 1. the number of employees 2. the number of hours they can work 3. the cost of producing goods and so on. Linear implies that the restrictions and the quantity to be maximised or minimised follow linear patterns. Programming means that it follows a systematic plan. The problem below is an example of a linear programming problem.

Problem A local manufacturer makes two types of deck chair: the deluxe and the standard. Each chair goes through two processes: assembling and finishing. It takes 4 hours to assemble each deluxe chair and 3 hours for each standard chair. The finishing process takes 2 hours for each chair. The profit on each deluxe chair is $15 and on each standard chair, $12. If employees spend at most 48 hours on assembling and 28 hours on finishing each week, how many chairs of each type should be made weekly to maximise the manufacturer’s profit?

Chapter 11

Linear programming

389

Mathematical formulation of the problem 1. Define the variables: (a) Let d represent the number of deluxe chairs. (b) Let s represent the number of standard chairs. (c) Let P represent the profit. 2. Write the constraints as inequations in terms of the variables: 4d + 3s ≤ 48 time available for assembling chairs 2d + 2s ≤ 28 time available for finishing chairs d ≥ 0  Since the number of chairs can not be negative  s ≥ 0  the variables d and s must be positive or zero. 3. Determine what must be maximised: the maximum profit needs to be obtained. 4. Write the function that needs to be maximised in terms of the variables: P = 15d + 12s From this point the problem can be solved by graphical methods, which we use later in the chapter. To solve linear programming problems, it is important to be able to sketch graphs of inequalities and determine points of intersection between graphs. Once we understand this, we can then begin to maximise and minimise linear functions and start to solve linear programming problems like the one above.


y=

x

When a straight line is drawn on a plane, it divides the plane into three sets of points: the points above the line, on the line and below the line. The diagram on the right shows the graph of y = x. The points on y y>x the line itself satisfy the equation y = x. The region above the line represents the points that satisfy the inequation y > x, while the region below the line represents the points that satisfy the x inequation y < x. y< x To indicate which part of the plane we want, shading is used. Note: Throughout this chapter the following format will be used: 1. The area that is not required will be shaded. 2. The area that is required (region required) will remain unshaded. To sketch the graph of an inequation, the following algorithm can be used: 1. For the boundary, replace the inequality sign with an equals sign and sketch the graph of the equation thus formed. (a) If the inequality sign was < or >, use a dotted line (to indicate that the points on the line itself are not included). (b) If the inequality sign was ≤ or ≥, use a solid line (to indicate that the points on the line are included in the region required). 2. To determine the side of the line where the required region lies: (a) Choose any point on one side of the line (a test point). (b) Substitute the coordinates of the test point into the inequation. (c) If the result is a true statement, the point that was chosen belongs to the required region. (d) If the result is not a true statement, the selected point belongs to the region that is not required. 3. Shade the region that is not required. 4. Add the legend to your graph: Region required The following worked examples illustrate this concept.

390


Worked Example 1

Sketch the graph of a y ≥ 0 b x < 6 and leave the region required unshaded. Think a

Write/DRAW a

1

Rule a labelled set of axes including the origin.

2

For the boundary, replace ≥ with = and sketch the equation y = 0 (this is a horizontal line which coincides with the x-axis). Since the inequation contains ≥, a solid line must be drawn.

Boundary equation is y = 0.

3

Choose a test point on the y-axis, say y = 6.

Test point: (0, 6)

4

Substitute 6 into y ≥ 0 to see if it satisfies the inequation; that is, is 6 ≥ 0? (Correct.)

Is 6 ≥ 0? Yes

5

Since the inequation is correct, shade the side of the line that does not contain the point.

6

Add the legend to the graph, indicating that the region required is not shaded.

y

0

x

Region required

b

b

1

Rule a labelled set of axes, including the origin.

2

Sketch the equation x = 6 (a vertical line that passes through 6 on the x-axis). Since the inequation contains <, a broken line must be drawn.

Boundary equation is x = 6.

3

Choose a test point on the x-axis, say x = 0.

Test point: (0, 0)

4

Substitute 0 into x < 6 to see if it satisfies the inequation; that is, is 0 < 6? (Correct.)

5


6

y

0

6 x

Region required

Add the legend. The graph in part a above is a closed half-plane defined by y ≥ 0. It is called ‘closed’ since it includes the equation y = 0. It includes the points on or above the line y = 0. The graph in part b is an open half-plane defined by x < 6 (‘open’ since it excludes the equation x = 6). It is the set of points to the left of the line x = 6. Sometimes the inequation has to be transposed before sketching to make x (or y) the subject. While transposing, keep in mind that multiplying or dividing both sides of an inequation by a negative number changes the direction of the sign of inequality to its opposite.

Chapter 11 Linear programming

391

Worked Example 2

Sketch the graph of −y + 3 < 5 and leave the required region unshaded. Think 1

Write/DRAW

To make y the subject, subtract 3 from both sides of the inequation.

−y

−y

+3<5 +3−3<5−3 −y < 2

2

Multiply both sides of the inequation by −1.

3

Change the direction of the inequality sign.

4


5

For the boundary, sketch the equation y = −2 (a horizontal line, passing through −2 on the y-axis). Since the inequation contains >, a broken line must be drawn.

Boundary equation is y = −2.

6

Choose a point on the y-axis, say y = 0.

Test point: (0, 0)

7

−1

× −y < −1 × 2 y > −2

Is 0 > −2? Yes

−y

Substitute 0 into + 3 < 5 to see if it satisfies the inequation; that is, is 0 > −2? (Correct.)

y

8


9

Add the legend.

x

0 −2 Region required

In the following worked example, we consider the graphing of linear inequations with two variables, x and y. Worked Example 3

Sketch the graph of the inequation y − 4x ≤ 8 and indicate the required region. Think

Write/Draw


Replace ≤ with = to find the x- and y-intercepts of the boundary.

Boundary equation is y − 4x = 8 x-intercept: y = 0

2

To determine the x-intercept, let y = 0.

3

4


5

To sketch the graph of the equation y − 4x = 8, mark the x- and y-intercepts on the set of axes and join them with the straight line. Since the inequation contains ≤, a solid line must be drawn.

6

392

To determine the y-intercept, let x = 0.

To determine the region required, choose a test point on one side of the line, say (0, 0).

0 − 4x = 8 −4x = 8 x = −2

(−2, 0)

y-intercept: x = 0 y−4×0=8 y = 8 Test point: (0, 0) Is 0 ≤ 8? Yes

y 8

−2 0


(0, 8)

x

7

8 9

Substitute the coordinates of the test point into the inequation to see if (0, 0) satisfies it; that is, is 0 ≤ 8? Since the inequation is correct, shade the side of the line that does not contain the point.

y Region required

8

−2 0

x

Add the legend.


Make y the subject of the inequation by adding 4x to both sides.

2

On the Graph & Tab screen, tap: • d • X (to select the inequality symbol) Complete the entry line as: y1 ≤ 8 + 4x Tick the box and tap $.

3

Indicate the required region. Note: The calculator shades the required region.

y − 4x ≤ 8 y ≤ 8 + 4x

The required region.

reMeMber

1. The graph of an inequation containing ≤ or ≥ is a closed half-plane; ≤ or ≥ indicates that a solid line is drawn and it is included in the required region. 2. The graph of an inequation containing < or > is the open half-plane; < or > indicates that a dotted line is drawn and it is excluded from the region required. 3. The required region is normally left unshaded; this is stated by adding the legend to the graph: Region required. Calculators tend to shade the required region. 4. Before sketching the inequation, it must be transposed to make the pronumeral the subject. 5. Multiplying or dividing both sides of an inequation by a negative number changes the direction of the sign of inequality to its opposite. 6. The origin (0, 0) is the most convenient test point to use when determining the region required, unless the straight line passes through it. exerCise

11A eBook plus Digital doc

SkillSHEET 11.1 Solving linear inequations

Graphs of linear inequations 1

We 1

unshaded. a y≥2 e y>0 i x > −2

Sketch graphs to represent the following inequations. Leave the required region b y≤0 f y ≤ −6 j x<1

c y > −1 g x≥4 k x>0

d y<6 h x≤0 l x ≤ −1 Chapter 11

Linear programming

393

2 WE2 Sketch graphs to represent the following inequations. Leave the required region unshaded. b −y ≥ 2 c −x > 3 d −x ≥ −5 a −x < 7 − − − e y > 4 f y ≤ 3 g x + 2 > 0 h y − 3 ≤ 0 i x − 8 < 0 j −x + 4 ≥ 0 k y − 2 < 3 l −y − 6 ≤ 2 3 MC The expression y > 8 is best represented by the following graph: y y y y y B C D E A

0

8

0

x

Region required

Region required

8

8

8

8 x

x

0

0 x

x

0

Region required

Region required

Region required

4 WE3 Sketch graphs to represent these inequations. Leave the required region unshaded. a y − 2x ≤ 4 b y + x ≥ −1 c 4x + 4y ≤ 16 d y − x ≤ 0 e y ≥ −x + 2 f y > 2x − 14 g y < 6x − 24 h y ≥ −7x + 21 i x + y > 0 j y ≥ x + 7 k x > y − 2 l y > 12x − 24 m 3y ≤ x + 12 n −2y ≥ 4x + 6 o x − y < 10 p y < x − 4 q 2y > 4x − 8 r 4x − 2y ≤ 8 s 2x − y > −1 t y − x − 4 < 0 u y + 2x − 6 ≥ 0 v 2y + 8x + 4 ≤ 0 w 9x + 9y + 9 ≥ 0 x 5x + 2y − 10 > 0 The expression y ≥ −2x is best represented by the following graph:

5 MC A

B

y

0

x

0

6

5

x

Region required

11b

−12x

y 0

0

x

Region required

Region required

6 MC The expression B A y

C

y

D

y x

0

Region required

y

E

y

x

0

x

0

Region required

Region required

+ 10y − 60 < 0 is best represented by the following graph: y y C D E y 6

6 x

−5

Region required

−5

0

6 x

Region required

−5

0

0 x

Region required

6 x

−5 Region required

Graphs of simultaneous linear inequations The graph of a linear equation is a straight line and the solution to a pair of simultaneous linear equations is a point of intersection of the two lines. The graph of a linear inequation is a half plane and the solution to a pair of simultaneous linear inequations is the area that is common to both half planes; that is, the area of their intersection. To find the graphical solution to a pair of simultaneous linear inequations, the following algorithm can be used: 1. On the same set of axes sketch each of the inequations, shading the regions that are not required. 2. The solution is represented by the area that remains unshaded, so specify this by adding a legend to the graph (that is, Region required).

394


Worked Example 4

Sketch the following pair of simultaneous linear inequations, determine the point of intersection and leave the required region (that is, the solution) unshaded. x ≥ 2, x + 2y ≤ 0 Think 1

2

Sketch the graph of x ≥ 2. (a) The graph of the equation x = 2 is a vertical line which intersects the x-axis at 2 (there is no y-intercept). Since the inequation contains ≥, draw a solid line. (b) Substitute x = 3 into x ≥ 2 to see if it satisfies the inequation; that is, is 3 ≥ 2? (c) Since the inequation is correct, shade the side of the line that does not contain the test point.

Write/DRAW

Boundary equation for x ≥ 2 is x=2 Test point: (3, 0) Is 3 ≥ 2? Yes y 0

2

x

Sketch the graph of x + 2y ≤ 0.

Boundary equation for x + 2y ≤ 0 is

(a) To sketch the graph of x + 2y = 0, first determine the x-intercept by letting y = 0.

x + 2y = 0 x-intercept: y = 0 x+0=0 x = 0 y-intercept: x = 0 0 + 2y = 0 y = 0 If x = 4, 4 + 2y = 0 2y = −4 y = −2

(b) Determine the y-intercept by letting x = 0. (c) Since x- and y-intercepts coincide, an alternative point must be chosen. To obtain the alternative point, let x or y equal any number other than 0. Say, let x = 4. (d) To sketch the graph of x + 2y = 0, mark the points (0, 0) and (4, −2) on the set of axes and join them with a straight line. (Use a solid line, since the inequation contains the ≤ sign.)

(0, 0)

(0, 0)

(4, −2)

y 0

2

4x (4, −2)

(e) Select any point on one side of the line, say (1, 1) and substitute its coordinates into x + 2y ≤ 0 to see if it satisfies the inequation; that is, is 3 ≤ 0? (f) Since the inequation is false, shade the side of the line that contains the test point. 3

Find the coordinates of the point of intersection of the two lines. (a) Write the given inequations as equations and label them [1] and [2]. (b) Substitute x = 2 into equation [2]. (c) Solve for y.

Test point: (1, 1) Is 3 ≤ 0? No

y 0

2

4x (4, −2)

x = 2 x + 2y = 0 Substituting [1] into [2]: 2 + 2y = 0 2y = −2 y = −1

[1] [2]


395

The solution set is (2, −1).

(d) State the coordinates of the point of intersection. 4

Write the coordinates of the point of intersection on the graph and add the legend.

y Region required

0 −1

(2, −1)

2

4x (4, −2)

Worked exAMPLe 5

eBook plus

Sketch the following pair of simultaneous linear inequations, determine the point of intersection and indicate the required region. 2x + 3y ≤ 6 x−y≥3 Think

Tutorial


WriTe/drAW


Sketch the graph of 2x + 3y ≤ 6.

Boundary of 2x + 3y ≤ 6 is

(a) To sketch the graph of 2x + 3y = 6, first determine the x-intercept by letting y = 0.

2x + 3y = 6 x-intercept: y = 0 2x + 0 = 6 x=3 y-intercept: x = 0 0 + 3y = 6 y=2

(b) Determine the y-intercept by letting x = 0.

(c) On the set of axes mark x- and y-intercepts and join them with a solid straight line (since the inequation contains the ≤ sign). (d) Substitute the coordinates of the point (0, 0) into 2x + 3y ≤ 6 to see if it satisfies the inequation; that is, is 0 ≤ 6? (e) Since the inequation is correct, shade the side of the line that does not contain the test point.

2

(0, 2)

Test point: (0, 0) Is 0 ≤ 6? Yes y 2 0

3

x

Sketch the graph of x – y ≥ 3.

Boundary of x − y ≥ 3 is

(a) To sketch x − y = 3, first determine the x-intercept by letting y = 0.

x−y=3 x-intercept: y = 0 x−0=3 x=3

(3, 0)

y-intercept: x = 0 0−y=3 y = −3

(0, −3)


(c) Mark the x- and y-intercepts and join them with a solid straight line.

396

(3, 0)

Test point: (0, 0) Is 0 ≥ 3? No


y

(d) Substitute the coordinates of the point (0, 0) into x − y ≥ 3 to see if it satisfies the inequation; that is, is 0 ≥ 3?

2

(e) Since the inequation is false, shade the side of the line that contains the test point. 3

4

0

3

x

−3

To find the point of intersection of the two lines. (a) Write the given inequations as equations and label them [1] and [2]. (b) To equate the coefficients of y, multiply equation [2] by 3 and label the resulting equation [3].

2x + 3y = 6 x − y = 3 [2] × 3: 3x − 3y = 9

(c) Eliminate y by adding equations [1] and [3].

[1] + [3]: 2x + 3y + 3x − 3y = 6 + 9

(d) Solve for x.

(e) Substitute x = 3 into either equation, say [2], and solve for y.

Substituting x = 3 into [2]: 3−y=3 y=0

(f) State the coordinates of the point of intersection.

Solution set is (3, 0).

[1] [2] [3]

5x = 15 x=3

Add to the graph the coordinates of the point of intersection and the legend.

y Region required

2

(3, 0)

0

3

x

−3


Make y the subject of the inequations.

2

On the Graph & Tab screen, tap: • d • X (to select the inequality symbol) Complete the entry lines as: 6 − 2x y1 ≤ 3 y2 ≤ x − 3 Tick the boxes and tap $.

3

Indicate the required region.

2x + 3y ≤ 6 3y ≤ 6 - 2x 6 − 2x y≤ 3 x-y≥3 -y ≥ 3 - x y ≤ x - 3

[1]

[2]

The required region.


397

4

To determine the point of intersection between the lines, tap: • Analysis • G-Solve • Intersect

State the point of intersection.

The point of intersection between y ≤

6 − 2x and 3

y ≤ x − 3 is (3, 0).

reMeMber

1. The graphical solution to a pair of simultaneous linear inequations is the region common to both inequations. 2. To find the graphical solution of simultaneous linear inequations: (a) Sketch each of the inequations on the same set of axes. (b) Find the coordinates of the point of intersection of the two lines that form the boundaries and add it to the graph. (c) Add the legend to the graph to indicate that the solution (region required) is the area that remains either unshaded or is shaded. exerCise

11b eBook plus

Graphs of simultaneous linear inequations 1

Sketch the graphs of the following pairs of simultaneous inequations, determine the point of intersection and leave the required region unshaded. a x≥0 b y≥2 c x≤1 x + 4y ≤ 0 x+y≤1 x−y≤2 d y ≤ 2x e y≤6 f y ≥ −3 − y ≤ 3x 5x + 10y ≤ 20 3x − 4y ≤ −24

2

We5 Sketch the graphs of the following pairs of simultaneous inequations, determine the point of intersection and leave the required region unshaded. a 2x − 3y ≤ 0 b 2x + 4y ≥ 8 c 4x + 3y ≤ 12 x + 2y ≤ 0 3x + y ≤ 3 x + 4y ≥ 4 d x + y ≥ 10 e 5x + 4y ≥ 20 f 3x + 2y ≥ 6 x − y ≤ 10 x−y≤5 3x − 2y ≥ 6 g 5x + 2y ≥ 15 h 4x − 6y ≤ 12 i 7x − y ≤ 14 3x + 6y ≤ 18 2x + 2y ≤ 10 3x + 4y ≥ 9 j 4x − y ≥ 8 k 2x + 2y ≥ 6 l −6x + y ≥ 12 14x + 2y ≥ 14 x − y ≤ −4 6x − 3y ≥ 6 m 4x − y ≤ 2 n 3x + 3y ≤ 3 o x − 5y ≥ 10 4x + y ≤ 2 2x + y ≥ −1 4x + 2y ≤ 12

Digital doc

SkillSHEET 11.2 Simultaneous equations

398

We4


x y − − ≥ 2 2 3 x y + ≤2 3 2 3 MC a For the simultaneous inequations y ≥ 3 and x + 2y ≤ 6, the figure showing the correct graphs and required region (unshaded) is: y y B C y D y E A y p 6x − 3y ≤ −3 3x + 4y ≥ 4

q 8x + 4y ≤ −8 y x − ≤1 2

3

3

0 eBook plus Digital doc

WorkSHEET 11.1

6

x

Region required

6

0

Region required

3

3

3 6

x

Region required

0

6

x

Region required

x

0

6

Region required

b For the simultaneous inequations 24x − 12y ≤ −12 and 12x + 16y ≥ 10, the figure showing the correct graphs and required region (unshaded) is: y y y y y B C D E A 5– 8 − 1– 2

1

5– 8

0

5– 6

x

Region required

11C

x

0

r

− 1– 2

1

5– 8

0

5 – 6

x

− 1– 2

Region required

1

5– 8

0

5– 6

x

Region required

1

− 1– 2

5– 8

0

5– 6

x

Region required

− 1– 2

1 0

5– 6

x

Region required

Graphs of systems of linear inequations In the previous section we discussed how to sketch the graphs of pairs of simultaneous linear inequations. The number of inequations to be graphed simultaneously can be extended. We refer to the groups that contain more than two inequations as systems. The solution to a system of linear inequations is the area, common to all half-planes, representing those inequations. Graphically, the solution is given by the region that remains unshaded.

Worked exAMPLe 6

Sketch the following system of linear inequations and leave the required region unshaded. (Do not calculate the coordinates of the points of intersection of the straight lines.) x + 2y ≥ 4 [1] 2x − y ≥ 3 [2] x≤5 [3] y≥1 [4] Think 1

Sketch the graph of x + 2y ≥ 4. (a) To sketch x + 2y = 4, first determine the x-intercept by letting y = 0. (b) Determine the y-intercept by letting x = 0.

WriTe/drAW

Boundary equation for x + 2y ≥ 4 is x + 2y = 4 x-intercept: y = 0 x+2×0=4 x=4 (4, 0) y-intercept: x = 0 1 × 0 + 2y = 4 2y = 4 y=2 (0, 2)

Chapter 11

Linear programming

399

(c) On the set of axes mark the x- and y-intercepts and join them with a solid straight line (since the inequation contains a ≥ sign). (d) Substitute the coordinates of the point (0, 0) into x + 2y ≥ 4 to see if it satisfies the inequation; that is, is 0 ≥ 4?

y

Test point: (0, 0) Is 0 ≥ 4? No 2 1 x

4

(e) Since the inequation is false, shade the side of the line that contains the test point. 2

Sketch the graph of 2x − y ≥ 3. (a) To sketch 2x − y = 3, first determine the x-intercept by letting y = 0.


3

4

5

(c) Mark the x- and y-intercepts on the set of axes and join them with a solid straight line (since the inequation contains a ≥ sign). (d) Substitute the coordinates of the point (0, 0) into 2x − y ≥ 3 to see if it satisfies the inequation; that is, is 0 ≥ 3? (e) Since the inequation is false, shade the side of the line that contains the test point. Sketch the graph of x ≤ 5. (a) Sketch the graph of x = 5 (a vertical straight line passing through 5 on the x-axis). Use a solid line, as the inequation contains a ≤ sign. (b) Substitute x = 0 into x ≤ 5 to see if it satisfies the inequation; that is, is 0 ≤ 5? (c) Since the inequation is correct, shade the side of the line that does not contain the test point. Sketch the graph of y ≥ 1. (a) Sketch the graph of y = 1 (a horizontal line, passing through 1 on the y-axis). Use a solid line, as the inequation contains the ≥ sign. (b) Substitute y = 0 into y ≥ 1 to see if it satisfies the inequation; that is, is 0 ≥ 1? (c) Since the inequation is false, shade the side of the line that contains the test point. The unshaded polygon is the region required. Label the vertices of the polygon A, B and C and add the legend to the graph.

Boundary equation for 2x − y ≥ 3 is 2x − y = 3 x-intercept: y = 0 2x − 0 = 3 2x = 3 x = 1.5

(1.5, 0)

y-intercept: x = 0 0−y=3 y = −3

(0, −3)

y

Test point: (0, 0) Is 0 ≥ 3? Νο

2 1 −3

11–2

Boundary equation for x ≤ 5 is x = 5 y

Test point: (0, 0) Is 0 ≤ 5? Yes

2 1 −3

11–2

4

5 x

Boundary equation for y ≥ 1 is y = 1 y


2 1 −3

11–2

4

y

A

1 −3


11–2

5 x

C

Region required

2

400

x

4

4

B 5 x

In worked example 6 a boundary polygon, ABC, has been formed. The terminology below will be used in the sections that follow. 1. The unshaded region and the polygon are together called the feasible region. They represent all the points that satisfy the system of linear inequalities. 2. Points A, B and C are vertices of the feasible region and can be determined by finding the points of intersection of the relevant lines (that is, solving simultaneous equations where necessary). The vertices are also referred to as feasible points or corner points.

REMEMBER

1. A group of simultaneous linear inequations is referred to as a system. 2. The graphical solution to the system of linear inequations is the area, common to all half-planes, representing those inequalities. It is the region that remains unshaded. 3. The unshaded area together with the boundary polygon, formed as a result of sketching the system of simultaneous linear inequations, is called a feasible region. 4. The vertices of the polygon are called feasible (or corner) points.

Exercise

11c

Graphs of systems of linear inequations 1 WE6 Sketch the graphs of the following systems of inequations and leave the required region unshaded. (Do not calculate the coordinates of the points of intersection of the straight lines.) a x ≥ 0 b x ≥ 0 c x ≤ 0 d 2x + 3y ≤ 3 y≥0 y≥0 y≤0 x≥2 x≥3 x≥2 x≤5 y≤2 y≤4 y≥6 y ≥ −7 e 6x + 5y ≤ 30 f −2x − 4y ≥ 8 g −4x + y ≤ 4 h 4x + 6y ≥ 12 − x≥ 1 x≤4 2x + y ≤ 4 3x + y ≤ 9 y ≥ −3 y ≤ −2 y≥0 x≥0 i −8x − 4y ≤ 4 j x + y ≥ 1 k 2x − y ≤ 2 l x − 3y ≤ 1 y≥x x−y≥1 x + 2y ≥ 1 2x ≥ y x≥0 x≥2 x≤3 x ≥ 12 y≥6 m 4x + 3y ≥ 12 2x + 5y ≤ 10 x≥1 y≤1 q 3x + 3y ≤ 3 2x + y ≥ −1 x≤1 y≥1

y≥2 n 2x − 3y ≤ 0 x + 2y ≤ 0 x≤5 y≤0 r 6x + 3y ≥ 12 −4x + 2y ≤ 16 x≥0 y≥0

y≤1 o 3x + 2y ≥ 6 3x − 2y ≤ 6 x≥0 y≥0 s 5x + 10y ≤ 15 6x − 2y ≤ 9 x≥0 y≥0

u 3x + 5y ≥ 15 x+y≤8

v 6x + 3y ≥ 18 3x − y ≥ −6

w x + y ≤ 9 8x − 3y ≤ 24

x≥5

x≤4

x≥0 y≥0

x≥0 y≥0

y ≤ 12 x

x≥0 y≥0

y≥0 p 4x − 2y ≤ 2 4x + y ≥ 2 x≤2 y≥2 t x + 2y ≤ 16 2x + 5y ≥ 15 x≤5 x≥0 y≥0 − x 7x + 3y ≥ 21 y ≥ −3x y≤7 x≤0 y≤0


401

2

MC The required region for the system of inequations x ≥ 0, y ≥ 0, 7x + 5y ≤ 35 and 2x + y ≥ 8 is:

A

8 7

y

0

B

x

45

y

0

Region required

3

8 7

C

45

8 7

x

y

0

Region required

D

45

8 7

x

y

0

Region required

E

y

8 7

x

45

0

Region required

45

x

Region required

MC The required region for the system of inequations x ≥ 0, y ≥ 0, 3x − 8y ≥ 35 and

x + 12 y ≥ 4 is: A

B

y

8

−43–8

0 4

11–3 x 2

Region required

4

5

11d

C

y

8

3

−4–8

0 4

2x

11–3

3

−4–8

Region required

D

y

8

0 4

2x

11–3

Region required

E

y

8

3

−4–8

0 4

2x

11–3

y

8

3

−4–8

Region required

Region required

MC The system of inequations which best describes the graph at right is:

A B C D E

x ≥ 0, y ≥ 0, x + y ≥ 9, x + 2y ≤ 12 x ≤ 0, y ≥ 0, x + y ≥ 9, x + 2y ≥ 12 x ≥ 0, y ≤ 0, x + y ≥ 12, 2x + y ≥ 9 x ≥ 0, y ≥ 0, x + y ≤ 9, x + 2y ≥ 12 x ≥ 0, y ≥ 0, x + y ≤ 9, 2x + y ≥ 12

MC The system of inequations which best describes the graph at right is:

A B C D E

x ≥ 0, y ≥ 0, 2x + 3y ≥ 24, 2x − 2y ≥ −12 x ≥ 0, y ≥ 0, 2x + 3y ≤ 24, 2x − 2y ≥ −12 x ≥ 0, y ≥ 0, 2x + 3y ≤ 24, 2x − 2y ≤ −12 x ≤ 0, y ≤ 0, 2x + 3y ≤ 24, 2x − 2y ≥ −12 x ≥ 0, y ≥ 0, 3x + 2y ≤ 24, 2x − 2y ≥ −12

Maximising and minimising linear functions

x

0 4 112– 3

y 9 6 x

0

9 12

Region required

y 8 6 −6

0

x 12

Region required


int-0978

Maximising and minimising Linear programming is employed to maximise or minimise linear functions linear functions subject to the constraints given by a system of linear inequations. In this section we learn to maximise and minimise linear functions using two methods: the sliding-line method and the corner-point method.

sliding-line method To maximise/minimise linear functions using the sliding-line method: 1. Sketch the feasible region. 2. Determine the coordinates of all corner points. 3. Graph the linear function to be maximised or minimised. 4. (a) To maximise the linear function, slide the line up and find the last point the line touches in the feasible region. (b) To minimise the linear function, slide the line down and find the last point the line touches in the feasible region.

402


While maximising/minimising a linear function, you will at one stage need to graph it. The function is usually expressed in terms of x and y and needs to be transposed first to make y the subject. Consider the function c = 4x + 2y. c When transposed, it will give: y = − 2 x + . 2 c The linear function has a gradient of −2 and a y-intercept of . 2 y c 3– If values of −1, 0, 1, 2, 3, . . . are assigned to c, a series of parallel y = −2x + 2– 2 lines will be formed (that is, lines with the same gradient, but 1 different y-intercepts), as shown in the diagram at right. 1– 2 For maximising/minimising linear functions using the sliding-line 0 method, any one of the lines shown in the diagram at right can be c=3 x 1– c=2 −2 selected to be ‘the sliding line’. The following example illustrates c=1 c=0 the concept. c = −1 Worked Example 7 a Sketch the system of linear inequations given by:

6x + 8y ≤ 24, x ≤ 2, x ≥ 0, y ≥ 0 and leave the required region unshaded. b Determine the coordinates of the vertices of the feasible region. c Determine the maximum and minimum values of R = 2x + 2y subject to the constraints above, using the sliding-line method. Think a

1

Write/DRAW

Sketch the graph of 6x + 8y ≤ 24. (a) To sketch the graph of 6x + 8y = 24, first determine the x-intercept by letting y = 0.


(c) Mark the x- and y-intercepts on the set of axes and join them with a straight line. Since the inequation contains the ≤ sign, a solid line must be drawn. (d) Substitute the coordinates of the point (0, 0) into 6x + 8y ≤ 24 to see if it satisfies the inequation; that is, is 0 ≤ 24? (e) Since the inequation is correct, shade the side of the line that does not contain the test point. 2

a Boundary equation of 6x + 8y ≤ 24

is 6x + 8y = 24 x-intercept: y = 0 6x + 8 × 0 = 24 6x = 24 x = 4 y-intercept: x = 0 6 × 0 + 8y = 24 8y = 24 y = 3 Test point: (0, 0) Is 0 ≤ 24? Yes y

(4, 0)

(0, 3)

Region required

3

0

4

x

Sketch the graph of x ≤ 2. (a) Sketch the graph of x = 2 (a vertical straight line, passing through the 2 on the x-axis). Use a solid line, since the inequation contains a ≤ sign.

Boundary equation of x ≤ 2 is x = 2


403

(b) Substitute x = 0 into x ≤ 2 to see if it satisfies the inequation; that is, is 0 ≤ 2? (c) Since the inequation is correct, shade the side of the line that does not contain the test point. 3

4

b

c

Sketch the graph of y ≥ 0. (a) Sketch y = 0 (which is actually the x-axis), using a solid line, as the inequation contains a ≥ sign. (b) Choose a point on the y-axis, say y = 1, and check that it satisfies y ≥ 0; that is, is 1 ≥ 0? (c) Since the inequation is correct, shade the side of the line that does not contain the test point.

Boundary equation for x ≥ 0 is x = 0 Test point: (1, 0) Is 1 ≥ 0? Yes

Boundary equation of y ≥ 0 is y = 0 y Region required Test point: (0, 1) Is 1 ≥ 0? Yes 3

0

C O

B (2, 1.5) A 2

4

x

5

Transfer all of the above information onto a graph. Label the vertices of the feasible region A, B, C and O.

1

Determine the coordinates of the corner points: read the coordinates of points O, A and C from the graph.

2

The coordinates of point B can be determined by finding the point of intersection of the lines 6x + 8y = 24 and x = 2. Substitute x = 2 into 6x + 8y = 24 and solve for y.

6x + 8y = 24 x = 2

3

State the coordinates of point B.

The solution set is B (2, 1.5).

1

Transpose the equation that needs to be maximised or minimised, to make y the subject.

2

404

Sketch the graph of x ≥ 0. (a) Sketch x = 0 (which is actually the y-axis), using a solid line, since the inequation contains a ≥ sign. (b) Choose a point on the x-axis, say x = 1, and check that it satisfies the inequation x ≥ 0. That is, is 1 ≥ 0? (c) Since the inequation is correct, shade the side of the line that does not contain the test point.

Test point: (0, 0) Is 0 ≤ 2? Yes

Select any value of R, say, 0.

b O (0, 0), A (2, 0) and C (0, 3)

[1] [2]

Substituting [2] into [1]: 6 × 2 + 8y = 24 12 + 8y = 24 8y = 12 y = 1.5

c

R = 2x + 2y 2y = −2x + R y = −x + R 2

Let R = 0.

y = −x +

y = −x


0 2

3

Sketch the linear function y = −x.

y 3

Region required

C

B (2, 1.5)

O A 0 2 y = −x

4

Treat the linear function as a sliding line. (a) To maximise the linear function, slide the line up and find the last corner point the line touches in the feasible region. (This can be done easily by placing a ruler along the line and sliding it up, parallel to the line, until it touches the last corner point.) (b) To minimise the linear function, slide the line down and find the last corner point the line touches in the feasible corner.

y 3

0

4

x

Region required

Maximum C O

B (2, 1.5) A 2

4

x

Minimum

5

Observe from the graph at which point the maximum and minimum intercepts occur within the feasible region.

Maximum occurs at B (2, 1.5). Minimum occurs at O (0, 0).

6

Substitute the coordinates of point B into equation R = 2x + 2y, to determine the maximum value of R.

Rmax = 2 × 2 + 2 × 1.5 =4+3 =7

7

Substitute the coordinates of point O into R = 2x + 2y to find the minimum value of R.

Rmin = 2 × 0 + 2 × 0 =0+0 =0

Corner-point method To maximise/minimise linear functions using the corner-point method: 1. Sketch the feasible region. 2. Determine the coordinates of all corner points. 3. Apply the corner point method by substituting coordinates of each corner point into the linear function which is to be maximised or minimised. 4. Select maximum and minimum values. Using worked example 9, substitute the values of each corner point into the equation R = 2x + 2y and then select the maximum and minimum values. Corner point

R = 2x + 2y

R

O (0, 0)

2×0+2×0

0

A (2, 0)

2×2+2×0

4

2 × 2 + 2 × 1.5

7

2×0+2×3

6

B (2, 1.5) C (0, 3)

(Minimum)

(Maximum)


405

Worked exAMPLe 8

eBook plus

a Sketch the following system of linear inequations and indicate the required region.

x + y ≤ 10, y ≥ x − 4, y ≤ 2x + 1, x ≥ 0, y ≥ 0 b Determine the coordinates of the vertices of the feasible region. c Determine the maximum and minimum values of z = 3x − y subject to the above constraints, using the corner-point method. Think

Tutorial


WriTe


1

2

Determine the intercepts for [1], [2] and [3].

3

Sketch all graphs.

4

Find the required region for each inequation by using a test point. Transfer all of the information onto the graph. Label the vertices of the feasible region A, B, C, D and O.

5

b

For the boundary equations, replace the inequality signs with an = sign in each inequation and label the resulting equations [1], [2], [3], [4] and [5].

1

Read the coordinates of points O, A and D from the graph.

2

Find point B. (a) The coordinates of point B can be determined by solving equations [1] and [3] simultaneously: substitute [3] into [1]. (b) Solve for x. (c) Substitute x = 3 into equation [3] to find the value of y. (d) State the coordinates of the point B.

3

406

Find point C. (a) To obtain the coordinates of the point C, solve equations [1] and [2] simultaneously. (b) Substitute [2] into [1].

a x + y = 10

y=x−4 y = 2x + 1 x=0 y=0 x + y = 10 x-intercept: y = 0, x = 10 y-intercept: x = 0, y = 10 y=x−4 x-intercept: y = 0, x = 4 y-intercept: x = 0, y = −4 y = 2x + 1 x-intercept: y = 0, x = −0.5 y-intercept: x = 0, y = 1 y 10

[1] [2] [3] [4] [5] (10, 0) (0, 10) (4, 0) (0, −4) (−0.5, 0) (0, 1)

Region B required

1 A D − 1–2 O 4 −4

C 10 x

b O (0, 0), A (0, 1) and D (4, 0)

x + y = 10 y = 2x + 1 Substituting [3] into [1]: x + 2x + 1 = 10 3x + 1 = 10 3x = 9 x=3 Substituting x = 3 into [3]: y=2×3+1 y=7 The solution set is B (3, 7). x + y = 10 y=x−4 Substituting [2] into [1]: x + x − 4 = 10


[1] [3]

[1] [2]

(d) Substitute x = 7 into equation [2] to find the value of y.

Substituting x = 7 into [2]: y=7−4 y=3 The solution set is C (7, 3).

(e) State the coordinates of the point C. c

2x − 4 = 10 2x = 14 x=7

(c) Solve for x.

c

1

Substitute the coordinates of each corner point into the linear function, z, to determine its maximum and minimum values.

2

Select maximum and minimum values of z.

zmin = −1 zmax = 18

Method 2: Using a CAS calculator a 1 Make y the subject of the first inequation and write the others.

a x + y ≤ 10

b

2

On the Graph & Tab screen, tap: • d (to change to the appropriate inequality) Complete the entry lines as: y1 ≤ 10 − x y2 ≥ x - 4 y3 ≤ 2x + 1 x4 ≥ 0 y5 ≥ 0 Tick the boxes and tap $.

3

Indicate the required region.

1

To determine the points of intersection between the lines, tap: • Analysis • G-Solve • Intersect Select the required equations by using the arrow up/down and then E. Repeat for each pair of lines and state the points of intersection.

Corner point O (0, 0) A (0, 1) B (3, 7) C (7, 3) D (4, 0)

z = 3x − y 3×0−0 3×0−1 3×3−7 3×7−3 3×4−0

z 0 −1 2 18 12

(vertex A) (vertex C)

y ≤ 10 - x [1] x ≥ 0 [4] y ≥ x - 4 [2] y ≥ 0 [5] y ≤ 2x + 1 [3]

The required region. b

The points of intersection are: (3, 7), (7, 3), (4, 0).


407

c

2

To determine the points of intersection between the line x = 0 and y = 2x + 1, substitute the value x = 0 into the equation y = 2x + 1.

1

To determine the maximum and minimum values, on the Main screen, complete the entry lines as: 3x − y W z z|x=0|y=0 z|x=0|y=1 z|x=3|y=7 z|x=7|y=3 z|x=4|y=0 Press E after each entry.

2

Write the maximum and minimum values of z.

y = 2x + 1 x = 0 ⇒ y = 2(0) + 1 ⇒y=1 That is the point (0, 1). From the graph, (0, 0) is also a corner point. c

zmin = −1 zmax = 18

at (0, 1) at (7, 3)

The linear function to be maximised or minimised is often referred to as the objective function. The maximum or minimum values of the objective function always occur at a corner point (vertex) of the feasible region. reMeMber

1. To maximise/minimise a linear (objective) function: (a) Sketch the feasible region. (b) Determine the coordinates of all corner points. (c) Apply the sliding-rule method by drawing a graph of the objective function and moving a ruler up or down along it. The last corner point of the feasible region to be passed gives the maximum or minimum value of the objective function. 2. Alternatively, apply the corner-point method by substituting the coordinates of each corner point into the objective function and selecting minimum and/or maximum value(s). exerCise

11d eBook plus Digital docs

SkillSHEET 11.3 Vertices of feasible regions

SkillSHEET 11.4 Sliding-line method

408

Maximising and minimising linear functions 1

We7 For each of the following systems of inequations:

i Sketch the system of inequations and indicate the required region. ii Determine the coordinates of the vertices (corner points) of the feasible region. iii Determine the maximum or minimum value (as specified) of the objective function for the given constraints, using the sliding-line method. a Maximise z = x − y subject to x ≥ 0, y ≥ 0, x ≤ 4, y ≤ 6. b Minimise z = x + 3y subject to x ≥ 0, y ≥ 0, y ≥ x, y ≤ 7. c Maximise z = x + 2y subject to x ≥ 0, y ≥ 0, y ≤ x, x ≤ 5. d Maximise z = 4x + 6y subject to x ≥ 0, y ≥ 0, x + y ≤ 4, 3x + 8y ≤ 24. e Minimise z = 3x − 6y subject to x ≥ 0, y ≥ 0, 2x + 2y ≤ 8, 6x + 8y ≤ 30.


f g h i j eBook plus

2

Digital docs

Spreadsheet 068 Linear programming

SkillSHEET 11.5 Corner-point method

Maximise z = 0.8x + 1.2y subject to x ≥ 0, y ≥ 0, x − 4y ≤ 10, 2x + 7y ≤ 28. Minimise z = 9x + 3y subject to x ≥ 0, y ≥ 0, 2x + 3y ≥ 18, 3x + 4y ≤ 30. Maximise z = −3x + y subject to x ≥ 0, y ≥ 0, 5x + 2y ≥ 20, x ≤ 3, y ≤ 9. Minimise z = x + y subject to x ≥ 0, y ≥ 0, −2x + y ≤ 3, 6x − 3y ≤ 12, 3x + 3y ≤ 15. Maximise z = 3x + 4y subject to x ≥ 0, y ≥ 0, −2x + y ≤ 9, 3x − 5y ≤ 12, x ≤ 6, y ≤ 10.

We8 For each of the following systems of inequations:

i Sketch the system of inequations and indicate the required region. ii Determine the coordinates of the vertices (corner points) of the feasible region. iii Determine the maximum or minimum value (as specified) of the objective function for the given constraints, using the corner-point method. a Minimise z = 2x + y subject to x ≥ 0, y ≥ 0, −x + y ≤ 3, 4x + 7y ≤ 28, 2x + 8y ≥ 0, x ≤ 6. b Maximise z = 2x + y subject to x ≥ 0, y ≥ 0, x + y ≤ 8, y ≤ 2. c Minimise z = 5x − y subject to x ≥ 0, y ≥ 0, x ≤ 6, x − y ≥ −8. d Minimise z = 3x + 4y subject to x ≥ 0, y ≥ 0, x + y ≥ 4, x − y ≥ −4, x ≤ 8. e Maximise z = 1.8x + 2.2y subject to x ≥ 0, y ≥ 0, x ≤ 10, y ≤ 7, 1 x + y ≥ 8. 2

f g h i j 3


WorkSHEET 11.2

11e

Minimise z = 0.7x − 0.3y subject to x ≥ 0, y ≥ 0, 2x − y ≥ −4, 3x + 4y ≤ 36, y ≥ 5. Maximise z = 1.5x + 2.7y subject to x ≥ 0, y ≥ 0, 5x − 6y ≥ −30, x + y ≤ 10, y ≥ 6. Minimise z = 3.2x − 1.4y subject to x ≥ 0, y ≥ 0, −7x + 4y ≤ 28, 4x + 2y ≥ 16, x + y ≤ 14. Maximise z = 7x − 3y subject to x ≥ 0, y ≥ 0, 2x + y ≤ 9, −2x + 6y ≤ 18, x − y ≤ 3. Minimise z = 9.2x − 5.1y subject to x ≥ 0, y ≥ 0, 15x + 6y ≥ 30, −6x − 4y ≥ −36, y ≤ 7, x ≤ 4.

a The minimum value of the equation z = x − y subject to x ≥ 0, y ≥ 0, x ≤ 7, y ≤ 8 is: D −1 E 1 A 0 B 7 C −8 b The maximum value of the equation z = 2x + 5y subject to x ≥ 0, y ≥ 0, y ≤ 4, 2x + 4y ≤ 24 is: A 40 B 16 C 28 D 20 E 0 c The maximum value of the equation z = 5x − 7y subject to x ≥ 0, y ≥ 0, x + y ≤ 4, 3x + 9y ≤ 24 is: A −18.9 B −28 C −4 D 20 E 32 − d The minimum value of the equation z = 10x − 4y subject to x ≥ 0, y ≥ 0, 7x + 4y ≤ 14, −8x − 8y ≤ −16, x ≤ 3 is: A −5 B −14 C −28 D 20 E −20 MC

solving linear programming problems The techniques developed earlier in this chapter can now be employed to solve linear programming problems. For the purpose of this course, all problems will involve exactly two variables. It is important to note that linear programming problems involve only ≤ and ≥ signs; therefore, when graphing inequalities, only solid lines will be used. To solve linear programming problems, use the following algorithm: 1. Define the variables. 2. Write the inequations (constraints) in terms of the variables. 3. Determine what must be maximised or minimised. This is called the objective function. 4. Write the objective function in terms of the variables. 5. Draw the graph of each constraint and obtain the feasible region (unshaded region). 6. Obtain the coordinates of the corner points. 7. Employ the sliding-line method or the corner point method to obtain the maximum or minimum value of the objective function.

Chapter 11

Linear programming

409

WORKED EXAMPLE 9

eBook plus

Bright Spark Enterprises Pty Ltd produces two types of computer games, A and B. The company is contracted to produce at least 20 type-A games and at least 60 type-B games each week. The factory can produce a maximum of 120 games per week. The profit on type-A games is $10 and the profit on type-B games is $15. How many of each game should be produced each week so as to make the greatest weekly profit, assuming all games produced can be sold? THINK 1

Define the variables.

2

Write the inequations (constraints) in terms of the variables: (a) The number of type-A games produced should be 20 or more. (b) The number of type-B games produced should be 60 or more. (c) The total production can not exceed 120 games.

Tutorial


WRITE

Let x = number of type-A computer games produced. Let y = number of type-B computer games produced.

x ≥ 20 y ≥ 60 x + y ≤ 120

3

Determine what must be maximised or minimised.

4

Express the objective function in terms of the variables x and y.

5

Find the required region for x + y ≤ 120. (a) To sketch the graph of x + y = 120, determine the x-intercept and the y-intercept. (b) Substitute the coordinates of the point (0, 0) into x + y ≤ 120 to see if it satisfies the inequation; that is, is 0 ≤ 120?

Boundary for x + y ≤ 120 is x + y = 120 x-intercept: y = 0, x = 120 y-intercept: x = 0, y = 120 Test point: (0, 0) Is 0 ≤ 120? Yes

6

Find the required region for x ≥ 20.

Test point for x ≥ 20: (0, 0) Is 0 ≥ 20? No

7

Find the required region for y ≥ 60.

Test point for y ≥ 60: (0, 0) Is 0 ≥ 60? No

8

Transfer all of the above information onto the graph. Label the corner points.

Maximum profit (P) required. P = 10x + 15y

y 120

C

60

A

Region required B

0 20

410


120 x

(120, 0) (0, 120)

9

Determine the points of the feasible region. (a) Point A: Read the coordinates of the point of intersection of the horizontal and vertical lines from the graph.

A (20, 60)

(b) Point B: Solve x + y = 120 and y = 60 simultaneously; substitute y = 60 into x + y = 120 and solve for x.

x + y = 120 y = 60 Substituting [2] into [1]: x + 60 = 120 x = 60 B (60, 60) x + y = 120 x = 20 Substituting [4] into [3]: 20 + y = 120 y = 100

(c) State the coordinates of point B. (d) Point C: solve x + y = 120 and x = 20 simultaneously; substitute x = 20 into x + y = 120 and solve for y.

[1] [2]

[3] [4]

(e) State the coordinates of point C.

C (20, 100)

10

Employing the corner-point method, substitute the coordinates of the corner points into the objective function P.

P = 10x + 15y At A (20, 60) At B (60, 60) At C (20, 100)

11

Select the maximum value of P.

Pmax = 1700 at C (20, 100)

12

Relate the answer to the original question.

A maximum profit of $1700 will be obtained when 20 type-A and 100 type-B games are produced.

P = 10 × 20 + 15 × 60 = 1100 P = 10 × 60 + 15 × 60 = 1500 P = 10 × 20 + 15 × 100 = 1700

reMeMber

To solve linear programming problems: 1. define the variables 2. write the constraints in terms of the variables 3. define the objective function 4. draw the graphs of the constraints to obtain the feasible region (use solid lines only) 5. find the coordinates of the corner points 6. use either the sliding-line or the corner-point method to find the maximum or minimum value of the objective function.

exerCise

11e eBook plus Digital doc


solving linear programming problems 1

We9 For a semester project, Cathy’s Business Management Team produces two styles of sundial clock. Each clock is made from an old vinyl record. The team is able to produce up to 24 clocks weekly. A minimum of 5 style-A and 3 style-B clocks are ordered each week. The profit on style-A clocks is $2 and the profit on style-B clocks is $3. a To obtain a maximum weekly profit, how many of each style of clock should be produced each week? b Assume all clocks produced can be sold. What is the maximum weekly profit?

Chapter 11

Linear programming

411

2 Katrina and Erin design two types of tracksuit for Right-on-Track sportswear. Design A requires 2 m of material while design B requires 3 m of material. The total amount of material available each day is 60 m. At least 3 of design-A and at least 4 of design-B tracksuits must be produced each day to satisfy orders. Design-A tracksuits are sold at a profit of $5 while design-B tracksuits are sold at a profit of $6.50. a To obtain a maximum daily profit, how many tracksuits of each design should be produced each day? b Assume all tracksuits produced can be sold. What is the maximum daily profit? 3 Active-8 Enterprises hires out rollerblades and bicycles along the bay. Each day the company supplies at least 15 pairs of rollerblades, and a minimum of 5 but no more than 25 bicycles. No more than 40 pairs of rollerblades and bicycles are hired on any particular day. The profit on hiring out a pair of roller-blades is $4 and the profit on hiring out a bicycle is $3. a To obtain a maximum daily profit, how many pairs of rollerblades and how many bicycles should be hired out each day? b What is the maximum daily profit? 4 Squeaky Clean soap manufacturers produce two brands of liquid soap. To meet demand, at least 20 litres of brand A and at least 14 litres of brand B must be produced. Due to other factors the manufacturers are able to produce a maximum of 80 litres. The profit on brand A is $20 and the profit on brand B is $24. a To obtain a maximum profit, how much of brand A and B should be produced? b What is the maximum profit? 5 The Sweat It Out Gymnasium offers its participants aerobic classes and circuit classes. At least 25 aerobic classes and at least 9 circuit classes must be held each week. The gym is able to offer a maximum of 45 classes per week. Aerobic classes produce a profit of $6 while circuit classes produce a profit of $4. a In order to obtain a maximum profit, how many aerobic and circuit classes should be held each week? b What is the maximum profit? 6 Jillaroo’s Adventures factory manufactures two types of tent: a 2-person tent and a 3-person tent. To meet demand, the factory manufactures at least eighteen 2-person and at least eighteen 1 3-person tents each week. The 2-person tent takes 1 2 hours and the 3-person tent takes 1

2 2 hours to make. The equipment needed to produce the tents can be used for a maximum of 75 hours per week. Two-person tents return a profit of $24 while 3-person tents return a profit of $28. a How many of each type of tent should be produced weekly to obtain the maximum profit? b What is the maximum profit? 7 It-Will-Print manufactures bubble jet and laser printers. To meet demand the company must produce a minimum of 5 laser printers, and the total of printers must be, at most, 25 each week. A bubble jet printer takes 2 hours to make and a laser printer takes 3 hours. Due to power restrictions the manufacturing plant can operate for only 60 hours per week. Bubble jet printers return a profit of $12 while laser printers return a profit of $15. a How many of each type of printer should be produced weekly to obtain the maximum profit? b What is the maximum profit? 412


8 A farmer decides to divide his land into two sections and plant corn and peas. He has 80 hectares of land available and must devote at least 10 hectares to peas and at least 10 hectares to corn. At harvest time it takes 1 hour per hectare to collect the corn and 3 hours per hectare to collect the peas. The maximum time available for collecting the crops is 120 hours. He can make a profit of $180 per hectare of corn and $160 per hectare of peas. a How much of each crop should be sown to obtain a maximum profit? b What is the maximum profit? 9 Peter’s task as the new assistant manager at Sureway supermarket is to decide how to achieve maximum profit from sales on two brands of fruit juice. He notes that, on any given day, fruit juice sales are greater than 70 litres but less than 90 litres. Peter also notices that Nature’s Own fruit juice sales are equal to or greater than generic brand sales, and that at least 10 litres of generic brand juice is sold daily. a How many litres of each type of juice should be sold for maximum profit if the profit on Nature’s Own fruit juice is $2.50/L while the profit on the generic brand is $1.20/L? b What is the maximum profit? 10 Sandra and Loreta’s Shantai resort has been redesigned specifically to cater for tourists and people attending business conventions. It is able to cater for a minimum of 360 guests and a maximum of 510 on a monthly basis. The number of tourists is always greater than the number of people at the conventions but never double the people attending the conventions. The profit made per tourist per month is $15 while the profit made per businessperson per month is $18. a How many tourists and businesspeople should the hotel cater for each month to make a maximum profit? b What is the maximum profit? 11 Mathematically Minded Limited produces 3-D puzzles and logic games. It takes 3 hours to produce the parts of a batch of twenty 3-D puzzles and 2 hours to produce the parts of a batch of twenty logic games. The minimum time available for production of these items is 12 hours. A batch of 3-D puzzles takes 1 hour to assemble while a batch of logic puzzles takes 2 hours to assemble. A maximum of 10 hours is allocated to assembling the games. The packaging of a batch of 3-D puzzles and a batch of logic games requires 1 and 2 hours respectively. The minimum time allocated for the packing of these items is 8 hours. a How many of each item should be made to minimise costs if the overhead cost of each 3-D puzzle is $1.80 while the overhead cost of each logic game is $1.25? b What is the minimum cost? 12 A refrigerator manufacturer makes two models of refrigerator: Arctic Snow and Cool Breeze. The manufacturers are able to produce up to 40 Arctic Snow and 50 Cool Breeze models per fortnight, and production must not exceed 80 models (per fortnight). Each Arctic Snow model requires 12 hours to make and each Cool Breeze model takes 10 hours. The factory is able to operate for a maximum of 840 production hours per fortnight. Arctic Snow is able to generate a profit of $200 and Cool Breeze is able to generate a profit of $110. a How many of each model should be manufactured to obtain a maximum profit? b What is the maximum profit? c If the profit generated by Arctic Snow was $140 (and the profit on Cool Breeze remained the same), would the number of each model made still produce the maximum profit?


413

13 Zorko Industries has produced two new cement products: CP1 and CP2. Each 50-kg bag of the cement products consists of specific amounts of substances a, b and c (in units per bag) according to the table shown below: Product

a

b

c

CP1

20

20

15

CP2

25

15

10

The amounts of substances a, b and c available are 400, 300 and 210 units, respectively. Each 50-kg bag of CP1 yields a profit of $45 and each 50-kg bag of CP2 yields a profit of $50. a Let x represent the number of bags of CP1 and let y represent the number of bags of CP2. Explain why. b Given that 20x + 25y ≤ 400, write two similar constraints on x and y. c There are two other constraints. What are they? d What is the objective function? e What is the maximum profit (to the nearest dollar)? 14 MC Elio is commissioned to paint still-lifes and abstracts for the local gallery. He must produce a minimum of 20 pieces for an upcoming exhibition. It takes him on average 10 hours to paint a still-life and 6 hours to paint an abstract, and the maximum time he has to spend on his paintings is 180 hours. If s represents still-lifes and a represents abstracts, the inequations for this information are: A a ≥ 0, s ≥ 0, s + a ≥ 20, 10s + 6a ≤ 180 B a ≥ 0, s ≥ 0, s + a ≥ 20, s ≥ 10, a ≥ 6, s + a ≤ 180 C a ≥ 0, s ≥ 0, s + a ≤ 180, 10s + 6a ≥ 20 D s ≥ 10, a ≥ 6, s + a ≤ 180, 10s + 6a ≥ 20 E s ≥ 10, a ≥ 6, s + a ≥ 20, 10s + 6a ≤ 180 15 MC The feasible region (unshaded region) for the previous problem can best be defined by the graph: A

B a

a 30 20 20 0 18

C

30 20

s

Region required

a

D

a

30

Region required

a

30

20

6 0 10 18 20 s

E

30

20 20

0 18

s

Region required

20 0 18

20 s

Region required

20 0 18

Region required

16 Sonic Boom Sound Systems has developed a new product and needs to organise a research team to run a series of tests. The team is to comprise experienced engineers (e) and training technicians (t). The team is to consist of no more than 8 people, and at least 2 engineers but no more than 6 engineers, and at least 1 but no more than 5 technicians. The number of engineers must be greater than the number of technicians. A minimum of 18 tests need to be conducted on the product in a week. Engineers are able to conduct 5 tests per week and technicians 3. Engineers are paid $450 per week while technicians are paid $250. a How many engineers and technicians should be chosen to keep wages to a minimum? b What is the minimum weekly wage bill?

414


s

11f

Further applications of linear programming In this section we consider more-complex linear programming problems. Although you might find it harder to write the constraints and to define the objective function, the technique of solving these problems is exactly the same as the one discussed in the previous section.

Worked Example 10

The dietitian of the local football club purchases two types of powdered food products for her team. The nutritional contents of the two products per 250 g are listed in the table below: Component

Product A

Product B

Carbohydrates

25 g

30 g

Fat

2g

4g

Protein

15 g

10 g

The team’s minimum daily requirements of carbohydrates, fat and protein are 30 g, 4 g and 15 g respectively. If product A costs 50 cents per 250 g and product B costs 60 cents per 250 g, how much of each type should be used to supply the team’s daily nutritional requirements at the least cost? Think

Write

1

Define the variables.

Let x = the amount of product A (in units of 250 g) y = the amount of product B (in units of 250 g)

2

Write the inequations (constraints) in terms of the variables.

25x + 30y ≥ 30 2x + 4y ≥ 4 15x + 10y ≥ 15 x≥0 y≥0

3

Determine what must be maximised or minimised. Express the objective function in terms of the variables.

Minimum cost required. C = 0.50x + 0.60y

For boundary equations, replace the ≥ sign in the constraints with the = sign and label the resulting equations [1] to [5].

Boundary equations: 25x + 30y = 30 2x + 4y = 4 15x + 10y = 15 x = 0 y = 0

Determine the intercepts for [1], [2] and [3].

25x + 30y = 30 x-intercept: y = 0, x = 1.2 y-intercept: x = 0, y = 1 2x + 4y = 4 x-intercept: y = 0, x = 2 y-intercept: x = 0, y = 1 15x + 10y = 15 x-intercept: y = 0, x = 1 y-intercept: x = 0, y = 1.5

4 5

6

[1] [2] [3] [4] [5] (1.2, 0) (0, 1) (2, 0) (0, 1) (1, 0) (0, 1.5)


415

7

8

Sketch all graphs. (a) Sketch the graphs of the equations [1], [2] and [3] by marking their respective x- and y-intercepts on the set of axes and joining them with a straight line. (b) Sketch the graphs of the equations [4] and [5] — these are the x- and y-axes. Find the required region for 25x + 30y ≥ 30. (a) Substitute the coordinates of the point (0, 0) into 25x + 30y ≥ 30 to see if it satisfies the inequation; that is, is 0 ≥ 30?

For 25x + 30y ≥ 30, Test point: (0, 0) Is 0 ≥ 30? No

(b) Since the inequation is incorrect, shade the side of the line that contains the test point. 9

Find the required region for 2x + 4y ≥ 4. (a) Substitute the coordinates of the point (0, 0) into 2x + 4y ≥ 4 to see if it satisfies the inequation; that is, is 0 ≥ 4?

For 2x + 4y ≥ 4, Test point: (0, 0) Is 0 ≥ 4? Νο


Find the required region for 15x + 10y ≥ 15.

For 15x + 10y ≥ 15,

(a) Substitute the coordinates of the point (0, 0) into 15x + 10y ≥ 15 to see if it satisfies the inequation; that is, is 0 ≥ 15?



12

Find the required region for x ≥ 0.

For x ≥ 0,

(a) Choose a point on the x-axis, say x = 1, and check that it satisfies the inequation x ≥ 0. That is, is 1 ≥ 0?

Test point: (1, 0) Is 1 ≥ 0? Yes

(b) Since the inequation is correct, shade the side of the line that does not contain the test point. Find the required region for y ≥ 0. (a) Choose a point on the y-axis, say y = 1, and check that it satisfies the inequation y ≥ 0. That is, is 1 ≥ 0? (b) Since the inequation is correct, shade the side of the line that does not contain the test point.

13

Transfer all of the above information onto a graph. Label the corner points.

14

Determine the coordinates of the corner points of the feasible region. (a) Read the coordinates of the points A and C from the graph. (b) The coordinates of point B can be determined by solving equations [2] and [3] simultaneously. Write equations [2] and [3].

416

For y ≥ 0,

y

Region required Test point: (0, 1)

1.5 A B 1

Is 1 ≥ 0? Yes

C 0

1 1.2 2

x

A (0, 1.5) and C (2, 0) 2x + 4y = 4 15x + 10y = 15


[2] [3]

Multiply equation [2] by 2.5 and call the resultant equation [6]. Subtract equation [6] from [3] and solve for x.

Substitute x = 0.5 into equation [2] and solve for y.

[2] × 2.5: (2x + 4y = 4) × 2.5 5x + 10y = 10

[6]

[3] − [6]: 15x + 10y − (5x + 10y) = 15 − 10 10x = 5 x = 0.5 Substituting x = 0.5 into [2]: 2 × 0.5 + 4y = 4 1 + 4y = 4 4y = 3 y = 0.75

Write the coordinates of point B.

Β (0.5, 0.75)

15

Employing the corner point method, substitute the coordinates of the corner points into the objective function C.

C = 0.50x + 0.60y At A (0, 1.5) C = 0.50 × 0 + 0.60 × 1.5 = 0.90 At B (0.5, 0.75) C = 0.50 × 0.5 + 0.60 × 0.75 = 0.70 At C (2, 0) C = 0.50 × 2 + 0.60 × 0 = 1.00

16

Select the minimum value of C.

Cmin. = 0.70 at B (0.5, 0.75)

17

Relate the answer to the original question: (a) We used x to denote the amount of product A in 250 g units. Find the amount of product A in grams. (b) We used y to denote the amount of product B in 250 g units. Find the amount of product B in grams. (c) Write the answer to the problem in words.

Amount of product A = 0.5 × 250 = 125 g Amount of product B = 0.75 × 250 = 187.5 g A minimum cost of $0.70 will be spent by using 125 grams of product A and 187.5 grams of product B.

reMeMber

In practical problems involving time, number of items and so on, variables cannot take negative values. Therefore, in linear programming, variables such as x and y are always positive or 0; that is, x ≥ 0 and y ≥ 0.

exerCise

11F eBook plus Digital doc


Further applications of linear programming Solve the following linear programming problems, applying skills learned in the previous exercises. 1

We10 A rug cleaning service has designed a revolutionary treatment which involves two chemicals, A and B. Each of these chemicals contains (among other components) different amounts (in units per kg) of substances a, b and c as shown in the table on the next page.

Chapter 11

Linear programming

417

Number of units (per kg) Substance

Chemical A

Chemical B

a

10

5

b

1

2

c

2

12

The minimum amounts of substances a, b and c required are 20, 6 and 18 units, respectively. One kilogram of chemical A costs the company $16, while 1 kg of chemical B costs the company $22. What is the minimum possible cost (to the nearest dollar) of the treatment if both chemicals must be used? 2 Smelter’s Steel Works manufactures two types of steel rod: type A and type B. Steel rod A takes 2 hours to make while steel rod B takes 4 hours to make. For optimal plant utilisation, the machine press used to make the rods must operate for a minimum of 56 hours over a 1-week period. At least 6 of each type of rod must be made weekly but no more than 16 of steel rod A and no more than 10 of steel rod B can be made per week. a If the profit on steel rod A is $300 and the profit on steel rod B is $900, how many rods of each type must be manufactured to obtain a maximum profit? b What will be the maximum profit? 3 A clothing manufacturer makes two styles of uniform: style A and style B. Each uniform needs to be sewn, pressed and packaged. Each style-A uniform requires 5 minutes for sewing, 6 minutes for pressing and 3 minutes for packaging. Each style-B uniform requires 8 minutes for sewing, 12 minutes for pressing and 3 minutes for packaging. The profit on each style-A uniform is $7 and $12 on each style-B uniform. The times required for the sewing, pressing and packaging, at most, are 480, 600 and 450 minutes respectively. a Specify the variables. b Write the 5 constraints. c Specify the objective function. d Determine how many uniforms of each style should be made each day to maximise the manufacturer’s daily profit. e What is the maximum daily profit? 4 The members of a local football team require a diet which provides them with the daily minimum requirements of essential vitamins A, B and C. The daily minimum requirements are 36 units of A, 12 units of B and 8 units of C (per kilogram). These requirements could be met if two products, Zest and Boom, were combined. The amounts of vitamins (in units per kilogram) are shown in the table below: Product

A

B

C

Zest

12

3

1

Boom

6

4

8

a How many kilograms of products Zest and Boom should be used to keep costs to a minimum, given that product Zest costs $4.80 per kilogram and product Boom costs $3.50 per kilogram? b What is the minimum cost? 418


5 Luxurious Limousine Services offer two types of chauffeured limousine package: the Gold Pass and the Classic. The number of Gold Pass packages ranges from 140 to 200 while the number of Classic packages ranges from 80 to 120. Gold Pass packages are at least twice as popular as Classic packages. A profit of $20 is made on each Gold Pass ride while a profit of $10 is made on each Classic ride. a Which combination of chauffeur-driven rides will yield a maximum profit? b What is the maximum profit? 6 Let It Grow industries have been developing a new type of fertiliser in their two production plants. The fertiliser requires 3 ingredients: I1, I2 and I3. The amounts of these ingredients (in units per tonne) available at each plant are provided in the table below: Production plant

I1

I2

I3

Plant A

5

9

5

Plant B

3

3

10

A minimum amount of 15 units of I1 is available while a maximum of 27 and 50 units of I2 and I3, respectively, are available. Plant A yields a profit of $270 per day and plant B yields a profit of $500. a How many tonnes of fertiliser should be produced daily at each plant to yield a maximum profit? b What is the maximum profit?


419

Summary Graphs of linear inequations

• The graph of an inequation containing a ≤ or ≥ sign is a closed half-plane; a solid line indicates that the points on the line are included in the region required. • The graph of an inequation containing a < or > sign is an open half-plane. The points on the line are not included in the region required. This is indicated by using a dotted line. • To find which side of the line contains points that make the inequation a true statement, a test point is used. • The required region may be shaded or unshaded and each graph must include a legend. Graphs of simultaneous linear inequations

• The graphical solution to a pair of simultaneous linear inequations is given by the intersection of the two half-planes which represent those inequations. Graphs of systems of linear inequations

• A group of simultaneous linear inequations is called a system. • The graphical solution to the systems of linear inequations is given by the area common to all half-planes representing those inequations. • The region required together with the boundary polygon is called the feasible region. • The vertices of the polygon (feasible region) are referred to as feasible points or corner points. Solving linear programming problems

• To solve linear programming problems: 1. Define the variables. 2. Write the constraints (inequations) in terms of the variables. 3. Define the objective function (the linear function to be minimised or maximised). 4. Obtain the feasible region by sketching graphs of the constraints. 5. Find the coordinates of the corner points. 6. Use either the sliding-line or corner-point method to find the maximum/minimum value of the objective function. • Only solid lines are used when sketching the constraints. • In linear programming, variables always take positive values or 0.

420



1 Sketch the graphs to represent the following inequations. Leave the required region unshaded. a 3 − x ≥ 0 b y + 1 < −2 c 3x ≥ 2y − 6 Exam tip Use a point either side of the line to

find the required region.

2 Sketch the graphs of the following pair of simultaneous inequations and leave the required region unshaded. 2y + 5x ≥ 6 1 x − 2y ≤ 4 3 The unshaded region ABC on the graph below is the solution set for a system of three simultaneous linear inequations. Find the three inequations. y

A 400

7 a Maximise the objective function D = 4x + 5y, subject to the system of inequations shown in the graph below. y

A (0, 3) 0

0

400 −200

x

Region required

4 The unshaded region OABCD on the graph below is the solution set for a system of five simultaneous linear inequations. Find the five inequations. y 600 B 400 A C O D −400 0 300 600x −300 Region required

5 Graph the region required for the following systems of inequations and specify all corner points. a x ≥ 0, y ≥ 0, 3y − 2x ≤ 7 and −2y + 5x ≤ 10 b x ≥ 0, y ≥ 0, 6x + 8y ≥ 24 and 8x + 12y ≤ 48 6 Graph the region required for the following system of inequations: x ≥ 0, y ≥ 0, 2y − 6x ≤ 12, 2y + 4x ≤ 14 and 2y − 2x ≥ −2 a Specify each of the corner points. b If the profit is given by P = 5x + 4y, determine the maximum profit subject to the above constraints.

(8, 2) C x

Region required Exam tip Use the outer points to substitute into the equation D = 4x + 5y to find the maximum value.

b Minimise the objective function C = 6x − 2y, subject to the system of inequations shown in the graph below. y

B C

B (6, 6)

(2, 6) B A (1, 5)

0

C

(6, 2)

D (4, 0)

x

Region required

8 For which values of x and y will the objective function P = 10x + 12y be a minimum, subject to: 2x + 6y ≤ 24 2x + 4y ≤ 18 x≥0 y ≥ 0? 9 For which values of x and y will the objective function C = 100x + 1200y be a maximum, subject to: 7x + 5y ≤ 35 5x + 10y ≤ 34 x≤2 x≥0 y ≥ 0? 10 A local factory produces runners and walking shoes. It is able to produce a minimum of 400 pairs of runners and 350 pairs of walking shoes and must meet the weekly demand of up to 900 pairs of shoes altogether. The profit on a pair of runners is $12.50 and on a pair of walking shoes, $10. a Specify the variables. b Write the three constraints. c If we need to maximise profit, what is the objective function?


421

c

Multiple choice

1 The region required (unshaded region) for the inequation x > −6 is represented by: y y B A

0

−6

−6

x

c

Region required

Region required

y

e

0

3

x

Region required

2 The region required (unshaded region) for the inequation y < −x is represented by: y y A B x

0

Region required

y 0

Region required

6 Region required

4 The region required (unshaded region) for the graph at right can be defined by the inequation: y A y < 2x + 4 B y ≤ −2x + 4 4 C y > 2x + 4 D y > −2x + 4 E y ≥ −2x + 4 0 2

9

9

y

0

c

6

Region required

x

Region required

d

y

6

1

6

x

0

5 9 x

Region required

y 1

7–2

6 0

7 –2

7 –2

0

e

x

y

Region required Region required

6

1

9

0

1

7–2

0

1 7–2

y 5

x

5

x

x

Region required

d

Region required

e

x

0

5 9 x

Region required Region required

3 The region required (unshaded region) for the inequation 2x + y ≤ 6 is represented by: y y B A 6

0

6

3

x

Region required

422

x

5 The graph which best displays the required (unshaded) region of the simultaneous inequations 3x + 2y ≥ 18, 4x + 6y ≥ 30 is shown in: y A B y 5

c

x

3

Region required

−6

0

0

x

0

0 x

−6

0 x

−6

x

3

y 6

Region required

e y

y

d

d

0

Region required

Region required

y

x

0

y

6

0

3

x

Region required

6 The unshaded region which best represents the system of inequations x ≥ 0, y ≥ 0, x + y ≤ 8 and y − 2x ≤ 4 is shown in: A y B y 8

8

4

4

−2 0

8 x

Region required


−2 0

8 x

Region required

c

y 8

d

4 8 x

−2 0

Region required

e

y 8

D x ≥ 300, y ≥ 500, x + y ≤ 900, x + y ≤ 1400 E x ≥ 300, y ≥ 500, x + y ≥ 900, x + y ≥ 1400

4

11 The unshaded region which best represents the system of inequations is shown in: y B 1400 y A

8 x

−2 0

1400

Region required

500

500

2 −4

900

900

y 8

0

x 0

c

Region required

7 The region required for a system of inequations is y given by the graph at right. 8 If the revenue (in dollars) is (6, 4) given by R = 4x + 3y, the maximum revenue will be: 2 x A $8 B $32 0 2 8 C $34 D $36 E $38 Region required

9 The minimum value of the objective function C, where C = 3x − 2y, for the feasible region is: A −6 B 2 C 10 D 4.5 E −5

y

(0, 1)

(3, 3 —12 ) (4, 1) x

0 Region required

y (2, 6) (8, 7) (9, 4) (1, 2) (8, 2) x

0 Region required

Questions 10–12 refer to the following information. A drink manufacturer produces two types of sports drink. Each month, at least 300 litres of type A sports drink and 500 litres of type B sports drink must be produced to meet demand. The factory must produce at least 900 litres of sports drink but no more than 1400 litres. 10 If x represents the amount in litres of sports drink A, and y represents the amount in litres of sports drink B, then the system of inequations for the above information is: A x ≤ 300, y ≤ 500, x + y ≥ 900, x + y ≤ 1400 B x ≥ 300, y ≥ 500, x + y ≥ 900, x + y ≤ 1400 C x ≥ 300, y ≥ 500, x + y ≤ 900, x + y ≥ 1400

1400

Region required

d 1400 y

y

900

900

500

x 300 900 1400

0

Region required

e

1400

x 300 900 1400

0

Region required

8

8 The maximum value of the objective function S, where S = 5y − x, for the feasible region is: A 9 B 14.5 C 27 D 11 E 2

x 300 900 1400

500 0

x 300 900 1400 Region required

y

900 500 0

x 300 900 1400 Region required

12 Given that the profit on a litre of sports drink A is $1.20 and the profit on a litre of sports drink B is $1.00, the maximum profit possible is: A $1580 B $1500 C $1460 D $1620 E $960 13 In a linear programming problem involving animal management on a farm: • x represents the number of cows on the farm • y represents the number of sheep on the farm. The feasible region (with boundaries included) for the problem is indicated by the shaded region on the diagram below. y 100 80 60 40 20 0

20 40 60 80 100

x

One of the constraints defining the feasible region indicates that: A there must be 20 cows and 60 sheep. B there must be 40 cows and 40 sheep. C the number of sheep cannot exceed 40.


423

D the number of cows must be at least 60. E the total number of cows and sheep cannot exceed 80. [© VCAA 2006] 14 The cost of manufacturing a number of frying pans consists of a fixed cost of $400 plus a cost of $50 per frying pan. The manufacturer could break even by selling: A 10 frying pans at $90 each. B 10 frying pans at $45 each. C 15 frying pans at $60 each. D 15 frying pans at $30 each. E 20 frying pans at $50 each. [© VCAA 2006] 15 The four inequalities below were used to construct the feasible region for a linear programming problem. x≥0 y≥0 x+y≤9 1 y≤ x 2 A point that lies within this feasible region is: A (4, 4) B (5, 3) C (6, 2) D (6, 4) E (7, 3) [© VCAA 2006] 16 Russell is a wine producer. He makes both red and white wine. Let x represent the number of bottles of red wine he makes and y represent the number of bottles of white wine he makes.

This year he plans to make at least twice as many bottles of red wine as white wine. An inequality representing this situation is: A y ≤ x + 2 B y ≤ 2x C y ≥ 2x D x ≤ 2y E x ≥ 2y [© VCAA 2007] 17 The following five constraints apply to a linear programming problem. x ≥ 0, y ≥ 0, x + y ≥ 50, x + y ≤ 100, y ≤ x In the diagram below, the shaded region (with boundaries included) represents the feasible region for this linear programming problem. y

(50, 50) (25, 25) 0

(50, 0)

(100, 0)

x

The aim is to maximise the objective function Z = 2x + ky. If the maximum value of Z occurs only at the point (100, 0), then a possible value for k is: A 1 B 2 C 3 D 4 E 5 [© VCAA 2007]

Extended response

1 Nick intends to sow o hectares of oats and w hectares of wheat on his farm. He has 35 hectares of land available on which to sow crops. Oats requires 3 hours of labour and wheat requires 4 hours of labour per hectare, and a total of 120 hours of labour is available. a If the profit on oats per hectare is $200 and on wheat $240, how much of each crop must be sown to obtain the maximum profit? b What is the maximum profit Nick can make? c Would this change if the profit on oats per hectare became $250 and the profit on wheat remained unchanged? 2 A company manufactures two products, A and B. Each product must undergo three chemical processes for the number of hours specified in the table at right: Process A B The minimum amount of time required for processes 1, 2 and 3 is P1 10 5 70, 28 and 36 hours respectively. a Using this information, construct the three constraints. P2 2 4 Let x represent the number of items of product A produced. P3 3 3 Let y represent the number of items of product B produced.

424


b Are any other constraints assumed in this situation? c Sketch the 5 constraints, leaving the region required (feasible region) unshaded. The costs associated with the chemical processes on product A and product B are $300 and $200 per item respectively. d If we want to minimise cost, what is the objective function? e Find the minimum cost of the chemical processes associated with manufacturing the products in question. 3 Vicki’s Vitamin Company sells two Vitamin C products, C1 Macro-nutrients and C2, each boosted with a certain amount of macronutrients a α, β and γ (units per kilogram) as shown in the table at right. Scott Scurvy has been advised by his dietitian to combine the b 2 products so that the resulting mixture provides at least 50 units γ of α, 28 units of β and 60 units of γ. If C1 costs $3.50 per kilogram and C2 costs $5.00 per kilogram, a find how Scott should combine the products to achieve minimum cost, and b state this minimum.

C1

C2

10

5

4

4

6

15

4 Sklo, a glass and crystalware company, is about to launch a new line of products called Spring Blooms. All products in this line (cake platters, cheese platters and fruit bowls) feature the same flower design. The company plans to produce a trial batch of 2100 items. Based on previous experience, they know that the number of cake platters should be at least double the number of cheese platters and that the demand for fruit bowls will not exceed 600. They also need to produce at least 200 cheese platters and no more than 1600 cake platters. a Let x be the number of cake platters and let y be the number of cheese platters produced. Write the constraints to represent the information. b Sketch the constraints on a set of axes. Leave the feasible region unshaded. c Find the coordinates of the corner points of the feasible region. Assuming that every item produced will be sold, the company can make a profit of $24 on every cake platter, $18 on every cheese platter and $21.50 on every fruit bowl sold. d If the Sklo company wishes to maximise their profit, write the objective function. e Find the quantities of each type of merchandise that need to be produced and sold in order to maximise the profit. f State the maximum profit. 5 Chantelle, a discerning cat, likes two types of dry food produced by the Superior Cat Food company: Chicken Bites and Fish Bites. Each of these two products contains three main nutrients (A, B and C), essential for healthy teeth and shiny fur. Chicken Bites contain 3% of nutrient A, 5% of nutrient B and 5% of nutrient C; Fish Bites contain 5% of nutrient A, 8% of nutrient B and 2.5% of nutrient C. Chantelle needs at least 80 g of dry food every day, and her minimum daily requirements of the A, B and C nutrients are 2.5 g, 4.5 g and 3 g respectively. a Let x represent the quantity (in grams) of Chicken Bites and let y represent the quantity (in grams) of Fish Bites fed to Chantelle on a daily basis. Write 6 constraints to represent the information. b Sketch the constraints, leaving the feasible region unshaded. c Find the coordinates of the corner points of the feasible region. Alan, the local vet, sells both of Chantelle’s favourite Superior Cat Food products at the following prices: $12 for a 1-kg packet of Chicken Bites and $16 for a 1-kg packet of Fish Bites.


425

d Elena, Chantelle’s owner, wants to minimise the cost of her cat’s food. Write the objective function that can be used to help Elena achieve her goal. e Use the objective function from part d to find the daily quantities of each type of food that Chantelle needs to be fed in order to minimise the cost. f What is the minimum cost? During Chantelle’s annual health-check, Alan told Elena that Fish Bites are now available in 1.5-kg packets and can be purchased from him at $21 per packet. Elena quickly calculated that it would be cheaper to buy these new, larger packets of Fish Bites. g Construct a new function for the cost of pet food. h Calculate the new amounts of each type of food needed to minimise the cost. i Find the new minimum daily cost of Chantelle’s food.

6 Harry offers dog washing and dog clipping services. Let x be the number of dogs washed in one day and y be the number of dogs clipped in one day. It takes 20 minutes to wash a dog and 25 minutes to clip a dog. There are 200 minutes available each day to wash and clip dogs. This information can be written as Inequalities 1 to 3. Inequality 1: x ≥ 0 Inequality 2: y ≥ 0 Inequality 3: 20x + 25y ≤ 200 a Draw the line that represents 20x + 25y = 200. In any one day the number of dogs clipped is at least twice the number of dogs washed. b Write an inequality to describe this information in terms of x and y. Inequality 4: c i On the previous graph, draw and clearly indicate the boundaries of the region represented by Inequalities 1 to 4. ii On a day when exactly five dogs are clipped, what is the maximum number of dogs that could be washed? The profit from washing one dog is $40 and the profit from clipping one dog is $30. Let P be the total profit obtained in one day from washing and clipping dogs. d Write an equation for the total profit, P, in terms of x and y. e i Determine the number of dogs that should be washed and the number of dogs that should be clipped in one day in order to maximise the total profit.

426


ii What is the maximum total profit that can be obtained from washing and clipping dogs in one day? [© VCAA 2006]

7 Gas is generally cheaper than petrol. The car must run on petrol for some of the driving time. Let x be the number of hours driving using gas and y be the number of hours driving using petrol. Inequalities 1 to 5 below represent the constraints on driving a car over a 24-hour period. Explanations are given for Inequalities 3 and 4. Inequality 1: x ≥ 0 Inequality 2: y ≥ 0 1 Inequality 3: y ≤ x The number of hours driving using petrol must not 2 exceed half the number of hours driving using gas. 1 Inequality 4: y ≥ x The number of hours driving using petrol must be at 3 least one third the number of hours driving using gas. Inequality 5: x + y ≤ 24 a Explain the meaning of Inequality 5 in terms of the context of this problem. The lines x + y = 24 and y =

1 x are drawn on the graph below. 2 y

25

24

20 15 10

8

5 0

5

10 1516 20

24

25

x

b On the graph above: 1 x 3 ii clearly shade the feasible region represented by Inequalities 1 to 5. On a particular day, the Goldsmiths plan to drive for 15 hours. They will use gas for 10 of these hours. c Will the Goldsmiths comply with all constraints? Justify your answer. On another day, the Goldsmiths plan to drive for 24 hours. Their car carries enough fuel to drive for 20 hours using gas and 7 hours using petrol. d Determine the maximum and minimum number of hours they can drive using gas while satisfying all constraints. Maximum = __________ hours Minimum = __________ hours i draw the line y =

[© VCAA 2007]



Chapter 11

Linear programming

427

eBook plus

ACTiviTies


• 10 Quick Questions: Warm up with ten quick questions on linear programming. (page 389) 11A


Digital doc

• SkillSHEET 11.1: Practise solving linear inequations. (page 393) 11B

Graphs of simultaneous linear inequations

Tutorial

• We5 int-1159: Watch how to sketch a pair of inequations and indicate the region required. (page 396) Digital doc

• SkillSHEET 11.2: Practise simultaneous equations. (page 398) • WorkSHEET 11.1: Sketch linear inequations and simultaneous linear inequations, showing the required regions. (page 399) 11D

Maximising and minimising linear functions

Interactivity

• Maximising and minimising linear functions int-0978: Consolidate your understanding of how to maximise and minimise linear functions. (page 402) Tutorial

• We8 int-1160: Watch how to sketch a system of linear equations and determine the feasible region and maximise the objective function. (page 406) Digital doc

• SkillSHEET 11.3: Practise vertices of feasible regions. (page 408) • SkillSHEET 11.4: Practise the sliding-line method. (page 408) • Spreadsheet 068: Investigate linear programming. (page 409) • SkillSHEET 11.5: Practise the corner-point method. (page 409)

428

• WorkSHEET 11.2: Sketch linear inequations, solve linear equations graphically and algebraically, sketch systems of linear inequations and determine maximum and minimum values of the objective function. (page 409) 11E

Solving linear programming problems

Tutorial

• We9 int-1161: Watch how to determine the number of two types of games to produced to maximise profit. (page 410) Digital doc

• Spreadsheet 068: Investigate linear programming. (page 411) 11F

Further applications of linear programming

Digital doc

• Spreadsheet 068: Investigate linear programming. (page 417) Chapter review Digital doc



12

12a Distance between two points 12b Midpoint of a line segment 12c Dividing a line segment internally in the ratio a :b 12d Dividing a line segment externally in the ratio a :b 12e Parallel lines 12F Perpendicular lines 12G Applications

Coordinate geometry areaS oF STudy

• Finding equations of straight lines (including vertical lines) from given information • Cutting a line segment internally and externally in a given ratio • Application of coordinate geometry: for example, design, orienteering, navigation and geometrical proofs

• Pythagoras’ theorem and its application to finding the distance between two points • Calculation of coordinates of the midpoint of a line segment • Gradients of parallel and perpendicular lines

eBook plus

12a

Digital doc

distance between two points A B

AC = x2 − x1 BC = y2 − y1 By Pythagoras’ theorem: AB2 = AC2 + BC2 = (x2 − x1)2 + (y2 − y1)2 Hence

10 Quick Questions

Coordinate geometry is a branch of Mathematics with many practical applications. The distance between two points can be calculated easily using Pythagoras’ theorem. It is particularly useful when trying to find a distance that is difficult to measure directly. For example, finding the distance from a point on one side of a lake to a point on the other side. Let A(x1, y1) and B(x2, y2) be two points on the Cartesian plane as shown below. Triangle ABC is a right-angled triangle. y y2 y1

B(x2, y2) A

C

(x1, y1)

AB = ( x2 − x1 )2 + ( y 2 − y1 )2

x1

x2

x

The distance between two points A(x1, y1) and B(x2, y2) is: AB = ( x2 − x1 )2 + ( y2 − y1 )2

Chapter 12

Coordinate geometry

429

Worked exaMple 1

Find the distance between the points A and B in the figure at right.

y 4 A

Think

WriTe

1

−3

1

From the graph find points A and B.

A(−3, 1) and B(3, 4)

2

Let A have coordinates (x1, y1).

Let (x1, y1) = (−3, 1)

3

Let B have coordinates (x2, y2).

Let (x2, y2) = (3, 4)

4

Find the length AB by applying the formula for the distance between two points.

AB = ( x2 − x1 )2 + ( y2 − y1 )2

B

3

x

= [3 − ( − 3)]2 + (4 − 1)2 = (6)2 + (3)2 = 36 + 9 = 45 =3 5 = 6.71 (correct to 2 decimal places)

Worked exaMple 2

Find the distance between the points P(−1, 5) and Q(3, −2). Think

WriTe

1

Let P have coordinates (x1, y1).

Let (x1, y1) = (−1, 5)

2

Let Q have coordinates (x2, y2).

Let (x2, y2) = (3, −2)

3

Find the length PQ by applying the formula for the distance between two points.

PQ = ( x2 − x1 )2 + ( y2 − y1 )2 = [3 − (− 1)]2 + (− 2 − 5)2 = (4)2 + (− 7)2 = 16 + 49 = 65 = 8.06 (correct to 2 decimal places)

Worked exaMple 3

Prove that the points A(1, 1), B(3, isosceles triangle.

eBook plus −1)

and

C(−1, −3)

Think 1

Plot the points. Note: For triangle ABC to be isosceles, two sides must have the same magnitude.

are the vertices of an


WriTe y

A

1 −1 C

430

Tutorial

1

−3

Maths Quest 11 advanced General Mathematics for the Casio Classpad

3 B

From the diagram, AC appears to have the same x length as BC.

2

Find the length AC.

AC = [1 − (− 1)]2 + [1 − (− 3)]2 = (2)2 + (4)2 = 20

3

Find the length BC.

BC = [3 − (− 1)]2 + [− 1 − (− 3)]2 = (4)2 + (2)2 = 20

4

Find the length AB.

AB = [3 − (1)]2 + [− 1 − (1)]2 = (2)2 + (− 2)2 = 4+4 = 8 =2 2

5

State your proof.

Since AC = ΒC, triangle ABC is an isosceles triangle.

reMeMBer

The distance between two points A(x1, y1) and B(x2, y2) is: AB = ( x2 − x1 )2 + ( y2 − y1 )2 exerCiSe

12a

distance between two points 1 2


Spreadsheet 021 Distance between two points

We 1

Find the distance between each pair of points shown

at right. We2 Find the distance between the following pairs of points. a (2, 5), (6, 8) b (−1, 2), (4, 14) − − − c ( 1, 3), ( 7, 5) d (5, −1), (10, 4) e (4, −5), (1, 1) f (−3, 1), (5, 13) − g (5, 0), ( 8, 0) h (1, 7), (1, −6) − j (−a, 2b), (2a, −b) i (a, b), (2a, b)

y

O 6 B 5 4 P 3 C 2A E H N L 1 −6 −5 −4−3−2−1 0 1 2 3 4 5 6 x −1 F −2 M −3 I J −4 D −5 −6 G

K

3 We3 Prove that the points A(0, −3), B(−2, −1) and C(4, 3) are the vertices of an isosceles triangle. 4

The points P(2, −1), Q(−4, −1) and R(−1, 3 3 − 1) are joined to form a triangle. Prove that triangle PQR is equilateral.

5 Prove that the quadrilateral with vertices A(−1, 3), B(5, 3), C(1, 0) and D(−5, 0) is a parallelogram. 6 Prove that the triangle with vertices D(5, 6), E(9, 3) and F(5, 3) is a right-angled triangle. 7

The vertices of a quadrilateral are A(1, 4), B(−1, 8), C(1, 9) and D(3, 5). a Find the lengths of the sides. b Find the lengths of the diagonals. c What type of quadrilateral is it?

Chapter 12

Coordinate geometry

431

8 MC If the distance between the points (3, b) and (−5, 2) is 10 units, then the value of b is: b −4 c 4 a −8 d 0 e 2 9 MC A rhombus has vertices A(1, 6), B(6, 6), C(−2, 2) and D(x, y). The coordinates of D are: b (2, 3) c (−2, 3) a (2, −3) − d (3, 2) e (3, 2) 10 A rectangle has vertices A(1, 5), B(10.6, z), C(7.6, −6.2) and D(−2, 1). Find: a the length of CD b the length of AD c the length of the diagonal AC d the value of z. 11 Show that the triangle ABC with coordinates A(a, a), B(m, −a) and C(−a, m) is isosceles.

12B

Midpoint of a line segment We can determine the coordinates of the midpoint of a line segment by applying the midpoint formula shown below.

Midpoint formula y Consider the line segment connecting the points B(x2, y2) A(x1, y1) and B(x2, y2). (y2 − y) Let P(x, y) be the midpoint of AB. AC is parallel to PD. P(x, y) (x2 − x) D PC is parallel to BD. (y − y1) AP is parallel to PB (collinear). A Hence triangle APC is similar to triangle PBD. (x1, y1) (x − x1) C But AP = PB (since P is the midpoint of AB). x Hence, triangle APC is congruent to triangle PBD. Therefore x − x1 = x2 − x 2x = x1 + x2 x + x2 x= 1 2 y + y2 Similarly it can be shown that y = 1 . 2 In general, the coordinates of the midpoint of a line segment joining the points (x1, y1) and (x2, y2) can be found by averaging the x- and y y-coordinates of the end points, respectively. (x2, y2) M

The coordinates of the midpoint of the line segment joining  x + x2 y1 + y2  (x1, y1) and (x2, y2) are:  1 ,  2 2 

1

(x1, y1)

Worked Example 4

Find the coordinates of the midpoint of the line segment joining (−2, 5) and (7, 1). Think 1

432

Label the given points (x1, y1) and (x2, y2).

Write

Let (x1, y1) = (−2, 5) and (x2, y2) = (7, 1)


+ x _____ y +y (x_____, 2 2 ) 2

1

2

x

2

Find the x-coordinate of the midpoint.

x= = =

x1 + x2 2 −2+7

2 5 2 1

=22 3

Find the y-coordinate of the midpoint.

y=

y1 + y2 2

=

5+1 2

=

6 2

=3 4

Give the coordinates of the midpoint.

1

Hence, the coordinates of the midpoint are (2 2 , 3).

Worked Example 5

The coordinates of the midpoint, M, of the line segment AB are (7, 2). If the coordinates of A are (1, −4), find the coordinates of B. Think 1

Label the start of the line segment (x1, y1) and the midpoint (x, y).

2

Find the x-coordinate of the end point.

3

Find the y-coordinate of the end point.

4

Give the coordinates of the end point.

5

Check that the coordinates are feasible.

Write

Let (x1, y1) = (1, −4) and (x, y) = (7, 2) x1 + x2 2 1 + x2 7= 2 14 = 1 + x2 x2 = 13 y +y y= 1 2 2 −4+ y 2 2= 2 4 = -4 + y2 y2 = 8 x=

Hence, the coordinates of the point B are (13, 8). y 8 2 −4

B (13, 8) M (7, 2) 1 7 A (1, −4)

13

x

Chapter 12 Coordinate geometry

433

reMeMBer

The coordinates of the midpoint of the line segment joining (x1, y1) and (x2, y2) are:

y (x2, y2)

 x1 + x2 y1 + y2   2 , 2 

M

1

(x1, y1)

exerCiSe

12B eBook plus Digital doc

Spreadsheet 075 Midpoint of a segment

+ x _____ y +y (x_____, 2 2 ) 2

1

2

Midpoint of a line segment 1 We4 Use the formula method to find the coordinates of the midpoint of the line segment joining the following pairs of points. a (−5, 1), (−1, −8) b (4, 2), (11, −2) c (0, 4), (−2, −2) − − − d (3, 4), ( 3, 1) e (a, 2b), (3a, b) f (a + 3b, b), (a − b, a − b) 2 We5 The coordinates of the midpoint, M, of the line segment AB are (2, −3). If the coordinates of A are (7, 4), find the coordinates of B. 3

Find: a the coordinates of the centre of a square with vertices A(0, 0), B(2, 4), C(6, 2) and D(4, −2) b the side length c the length of the diagonals.

4 MC The midpoint of the line segment joining the points (−2, 1) and (8, −3) is: a (6, −2) b (5, 2) c (6, 2) d (3, −1) e (5, −2) 5 MC If the midpoint of AB is (−1, 5) and the coordinates of B are (3, 8), then A has coordinates: a (1, 6.5) b (2, 13) c (−5, 2) d (4, 3) e (7, 11) 6 a The vertices of a triangle are A(2, 5), B(1, −3) and C(−4, 3). Find: i the coordinates of P, the midpoint of AC ii the coordinates of Q, the midpoint of AB iii the length of PQ iv the length of BC. b Hence show that BC = 2PQ. 7

a A quadrilateral has vertices A(6, 2), B(4, −3), C(−4, −3) and D(−2, 2). Find: i the midpoint of the diagonal AC ii the midpoint of the diagonal BD. b Comment on your finding.

8 a The points A(−5, 3.5), B(1, 0.5) and C(−6, −6) are the vertices of a triangle. Find: i the midpoint, P, of AB ii the length of PC iii the length of AC iv the length of BC. b Describe the triangle. What could PC represent? 9 Find the equation of the straight line that passes through the midpoint of A(−2, 5) and B(−2, 3), and has a gradient of −3. 10 Find the equation of the straight line that passes through the midpoint of A(−1, −3) and 2 B(3, −5), and has a gradient of 3.

434

x


12C

Dividing a line segment internally in the ratio a :b We can also determine the coordinates of a point dividing a line segment internally in a given ratio either by plotting the given coordinates and using a first-principles approach or by applying a given formula.

Worked Example 6

First-principles method Find the coordinates of the point, P, that divides the line segment joining the points A(2, 3) and B(6, 11) internally in the ratio 3:1. Think 1

Write

Show the end points (x1, y1) and (x2, y2) on a sketch graph and show an estimated position of the internal point, P.

y 11 3

1 B(6, 11) P

A(2, 3)

3 2 2

3

6

x 3

Find the x-coordinate of P. Since P divides AB in the ratio 3:1 then 3 P is located 4 of the length of the line segment AB from A, (P is 3 parts from A and B is 4 parts from A).

The x-coordinate of P is 4 of the way between where x = 2 and x = 6. 3 x = 2 + 4 (6 − 2)

Find the y-coordinate of P.

Similarly, the y-coordinate of P is 4 of the way between y = 3 and y = 11. 3 y = 3 + 4(11 − 3)

3

=2+4×4 =2+3 =5 3

3

=3+4×8 =3+6 =9 4

Give the coordinates of the point.

Hence, the coordinates of the point P are (5, 9).

General formula Consider the line segment connecting the points A(x1, y1) and B(x2, y2). Let P(x, y) be the point on AB that divides it in the ratio a:b as shown at right. AC is parallel to PD. PC is parallel to BD. AP is parallel to PB (collinear). Hence, triangle APC is similar to triangle PBD.

y

B(x2, y2) b P(x, y) a

(y2 − y)

(x2 − x) D (y − y1)

A (x1, y1) (x − x1) C

x


435

AB = PB AC = PD AC = PD x − x1 = x2 − x

Given that then but so

a a a  Note: The ratio a:b may be written in fractional form as . b b b PC AP a = = BD PB b x − x1 x2 − x a b

b(x − x1) = a(x2 − x) bx − bx1 = ax2 − ax ax + bx = ax2 + bx1 x(a + b) = ax2 + bx1

ax2 + bx1 a+b Similarly it can be shown that

x=

ay2 + by1 a+b The coordinates of the point that divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio a : b are:

y=

 ax2 + bx1 ay2 + by1   a + b , a + b  Note: When a = b the formula simplifies to that for the midpoint of a line segment as described earlier.

Worked Example 7

Formula method Find the coordinates of the point, P, that divides the line segment joining A(2, 3) and B(6, 11) internally in the ratio 3 : 1. Think

436

Write

1

Label the end points (x1, y1) and (x2, y2).

Let (x1, y1) = (2, 3) and (x2, y2) = (6, 11)

2

Find a and b.

3

Use the formula to find the x-coordinate and the y-coordinate of P.

a:b = 3:1 Hence a = 3, b = 1 ax + bx1 x= 2 a+b 3(6) + 1(2) = 3+1 18 + 2 = 4 20 = 4 = 5

4


ay2 + by1 a+b 3(11) + 1(3) = 3+1 33 + 3 = 4 36 = 4 =9

y=

Hence, the coordinates of the point dividing the line segment in the ratio 3:1 are (5, 9).


Worked exaMple 8

eBook plus

−4)

If P(3, is the point that divides the line segment AB internally in the ratio 1 : 2, find the coordinates of point A if the coordinates of point B are (5, 8). Think

Tutorial


WriTe

1

Label the end point (x2, y2) and the point P(x, y).

Let (x2, y2) = (5, 8) and (x, y) = (3, −4)

2

Find a and b.

a:b = 1:2 Hence a = 1, b = 2

3

Find the x1-coordinate.

4

Find the y1-coordinate.

Let A be (x1, y1). Hence, if: ax + bx1 x= 2 a+b then 1(5) + 2 x1 5 + 2 x1 3= = 1+ 2 3 9 = 5 + 2x1 2x1 = 4 x1 = 2 ay + by1 y= 2 a+b −

4=

1(8) + 2 y1 1+ 2

8 + 2 y1 3 −12 = 8 + 2y 1 2y1 = −20 y1 = −10 =

Hence, the coordinates of the point A are (2, −10).

5

Give the coordinates of A.

6


y 8

−4 −10

B 2 2 3 5 x P 1 A

reMeMBer

The coordinates of the point P that divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio a:b are:

y

 ax2 + bx1 ay2 + by1   a + b , a + b 

b a

B (x2, y2)

P (x, y)

A (x1, y1) x

Chapter 12

Coordinate geometry

437

exerCiSe

12C eBook plus

dividing a line segment internally in the ratio a : b 1

line segment joining the following pairs of points internally in the given ratio. a (1, 7), (4, 1) 1:2 b (1, 7), (4, 1) 2:1 c (5, −1), (3, 3) 1:1 d (2, 13), (6, 9) 1:1

Digital doc

Spreadsheet 022 Dividing a segment internally

2

We 7 Use the formula method to find the coordinates of the point that divides the line segment joining the following pairs of points internally in the given ratio. a (−3, 5), (1, −5) 3:1 b (−3, 5), (1, −5) 1:3 c (2, 8), (7, 3) 2:3 d (2, 8), (7, 3) 3:2 e (2, −9), (−2, −5) 5:3 f (2, −9), (−2, −5) 3:5

3

If P(6, 1) is the point that divides the line segment AB internally in the ratio 2:5, find the coordinates of A if the coordinates of B are (1, 7).


SkillSHEET 12.1 Dividing a line in a given ratio

Digital doc

WorkSHEET 12.1

We 8

4 The point P(5, 4) divides the line segment joining A(2, −5) and B(c, d) in the ratio 3:2. Find c and d. 5

eBook plus

Use the first-principles method to find the coordinates of the point that divides the

We 6

MC a The point, P, divides the line segment AB internally in the ratio 1:4. If A is (−3, 2) and

B is (7, −3), then the coordinates of P are: b (−1, 1) c (−4, −1) a (5, −2)

d (10, 5)

e (2, 0.5)

b Point Q(3, 10) divides the line segment joining the points A(7, 4) and B(1, 13) internally in the ratio: a 1:2 b 1:1 c 2:1 d 3:1 e 1:3 y c Points P and Q are the points of trisection of AB in the diagram 5 at right. The coordinates P and Q respectively are: B Q

a (0, −1), (2, 3) 1

1

b (1, 0), (3, 4) 2

c (0, 1 3), (6 3, 3 3)

d(0, 1), (4, 3)

1

−4 A

P −1

8 x

e (0, 2), (5, 3 2 ) 6 a Triangle ABC has vertices A(3, 15), B(6, 9) and C(−3, −6). Find: i the coordinates of L, the midpoint of BC ii the coordinates of M, the midpoint of AC iii the coordinates of N, the midpoint of AB. b AL, BM and CN are the medians of triangle ABC. A median is a line drawn from the vertex of a triangle to the midpoint of the opposite side. Find: i the coordinates of the point on AL that divides it in the ratio 2:1 ii the coordinates of the point on BM that divides it in the ratio 2:1 iii the coordinates of the point on CN that divides it in the ratio 2:1. c Comment on your findings from b i, ii and iii. d The three medians are concurrent. Their common point, usually labelled G, is called the centroid of the triangle. Graph the triangle ABC and draw the medians AL, BM and CN. Mark the centroid. 7 Triangle PQR has vertices P(−6, 3), Q(−3, 9) and R(3, 12). Find: a the midpoints U, V and W of QR, PR and PQ respectively b the coordinates of the centroid, G c the ratio PG:GU and PG:PU.

438


12D

Dividing a line segment externally in the ratio a : b We can also determine the coordinates of a point dividing a line segment externally in a given ratio either by plotting the given coordinates and using a first principles approach, or by applying a given formula.

Worked Example 9

First-principles method Find the coordinates of the point, P, that divides the line segment joining A(2, 3) and B(6, 11) externally in the ratio 3:1. Think 1

Write

Show the end points A(x1, y1) and B(x2, y2) on a sketch graph and an estimated position of the external point P(x, y).

y 1

3

11

B(6, 11)

A(2, 3)

3 2 2

3

P(x, y)

x

6

3 2

Find the x-coordinate of P. Since P divides AB externally in the ratio 3 3:1 then P is located 2 of the length of the line segment AB from A. (P is 3 parts from A and B is 2 parts from A.)

The x-coordinate of P is x = 2 and where x = 6. 3 x = 2 + 2 (6 − 2)

Find the y-coordinate of P.

Similarly, the y-coordinate of P is y = 3 and y = 11.

of the way between where

3

=2+ 2×4 =2+6 =8 3 2

of the way between

3

y = 3 + 2 (11 − 3) =3+

3 2

×8

= 3 + 12 = 15 4


Hence, the coordinates of the point P are (8, 15).

5


y 15 3

11

1

P(8, 15)

B(6, 11)

A(2, 3)

3 2

6 8

x

General formula Consider the line segment connecting the points A(x1, y1) and B(x2, y2).


439

Let P(x, y) be an external point on the extension of AB that divides it in the ratio a:b as shown at right. AD is parallel to BC. PD is parallel to PC (collinear). AP is parallel to BP (collinear). Hence, triangle APD is similar to triangle BPC. Given that AP = a BP b AD PD AP a then = = = BC PC BP b AD x − x1 but = BC x − x2 a x − x1 so = b x − x2 a(x − x2) = b(x − x1) ax − ax2 = bx − bx1 ax − bx = ax2 − bx1 x(a − b) = ax2 − bx1 ax − bx1 x= 2 a−b Similarly it can be shown that: ay − by1 y= 2 a−b

y

The coordinates of the point P that divides the line segment joining the points (x1, y1) and (x2, y2) externally in the ratio a:b are:  ax2 − bx1 ay2 − by1   a − b , a − b 

P(x, y)

b

a

(x2, y2)B x − x 2 A (x1, y1)

x − x1

y − y2 C y − y1 D x

y

P(x, y) a

b B(x2, y2)

A(x1, y1)

Worked Example 10

Formula method Find the coordinates of the point that divides the line segment joining the points (2, 3) and (6, 11) externally in the ratio 3:1. Think 1 2

440

Label the end points A(x1, y1) and B(x2, y2). Find a and b.

3

Use the formula to find the x-coordinate and the y-coordinate of the required point.

4


Write

Let (x1, y1) = (2, 3) and (x2, y2) = (6, 11) a:b = 3:1 Hence a = 3, b = 1 ax − bx1 ay − by1 x= 2 y= 2 a−b a−b 3(6) − 1(2) 3(11) − 1(3) = = 3−1 3−1 18 − 2 33 − 3 = = 2 2 16 30 = = 2 2 = 8 = 15 Hence, the coordinates of the point dividing the line externally in the ratio 3:1 are (8, 15).


x

Worked exaMple 11

eBook plus

If P(3, −4) is the point that divides the line segment AB externally in the ratio 1:2, find the coordinates of A if the coordinates of B are (5, 8). Think

Tutorial


WriTe

1

Label the end point (x2, y2) and the point P(x, y).

Let (x2, y2) = (5, 8) and (x, y) = (3, −4)

2

Find a and b.

a:b = 1:2 Hence a = 1, b = 2

3

Find the x1-coordinate.

Let A be (x1, y1). Hence, if: ax − bx1 x= 2 a−b 1(5) − 2 x1 then 3 = 1− 2 5 − 2x 3= − 1 1 −3

= 5 − 2x1 2x1 = 5 + 3 =8 x1 = 4 4

Find the y1-coordinate.

5

Give the coordinates of A.

ay2 − by1 a−b − 4 = 1(8) − 2 y1 1− 2 8 − 2 y1 = −1 4 = 8 − 2y1 2y1 = 8 − 4 =4 y1 = 2 y=

Hence, the coordinates of the point A are (4, 2).

reMeMBer

The coordinates of the point, P, that divides the line segment joining y the points (x1, y1) and (x2, y2) externally in the ratio a:b are:  ax2 − bx1 ay2 − by1   a − b , a − b 

P(x, y) a

b B(x2, y2)

A(x1, y1) x

Chapter 12

Coordinate geometry

441

exerCiSe

12d eBook plus

dividing a line segment externally in the ratio a :b 1

We 9 Use the first-principles method to find the coordinates of the point that divides the line segment joining the following pairs of points externally in the given ratio. a (1, 7), (4, 1) 1:2 b (1, 7), (4, 1) 2:1 c (−3, 5), (1, −5) 3:1 d (−3, 5), (1, −5) 1:3

2

We 10 Use the formula method to find the coordinates of the point that divides the line segment joining the following pairs of points externally in the given ratio. a (2, 8), (7, 3) 2:3 b (2, 8), (7, 3) 3:2 c (2, −9), (−2, −5) 5:3 d (2, −9), (−2, −5) 3:5 − e (5, 1), (3, 3) 4:1 f (5, −1), (3, 3) 1:4

Digital doc

Spreadsheet 023 Dividing a segment externally

3

We 11 The point P(6, 1) is the point that divides the line segment AB externally in the ratio 2:5. Find the coordinates of A if the coordinates of B are (1, 7). 4 The point P(5, 4) divides the line segment joining A(2, −5) and B(c, d) externally in the ratio 3:2. Find c and d. 5 MC P is the point that divides the line segment AB externally in the ratio 4:1. If A is (−2, 3) and B is (7, −3) then the coordinates of P are: 2 a (10, −5) b (−5, 5) c (8 3 , −3) −

d ( 1 , 3) 3

6

e (6, 3)

MC Point Q(−5, 22) divides the line segment joining the points A(7, 4) and B(1, 13)

externally in the ratio: a 1:2 b 2:1 d 1:1 e 1:3 7 A give way sign has the shape of an equilateral triangle with a side length of 87 cm. The sign is attached in two places to a metal pole. a How far from the top of the sign should the holes be drilled if the top hole divides the vertical height of the sign in the ratio 1:9 and the bottom hole in the ratio 8:9? b How high is the top of the sign from the ground if the distance to the base of the pole from the top and bottom of the sign is in the ratio 7:6?

12e

c 3:1 87 cm

parallel lines In a previous chapter, ‘Linear and non-linear graphs’ (Chapter 10), we investigated linear graphs and equations. We are now going to investigate further properties of straight lines. The equation of a straight line may be expressed in the form: y = mx + c where m is the gradient of the line, and c is the y-intercept. The gradient can be calculated if two points, (x1, y1) and (x2, y2) are given. y −y m= 2 1 x2 − x1 An alternative form for the equation of a straight line is: ax + by + c = 0 where a, b and c are constants. Another alternative form is: y − y1 = m(x − x1) where m is the gradient and (x1, y1) is a point on the line.

442


Worked exaMple 12

Show that AB is parallel to CD given that A has coordinates (−1, −5), B has coordinates (5, 7), C has coordinates (−3, 1), and D has coordinates (4, 15). Think

WriTe

1

Find the gradient of AB.

2

Find the gradient of CD.

Let A(−1, −5) = (x1, y1) and B(5, 7) = (x2, y2) y −y m= 2 1 Since x2 − x1 7 − (− 5) mAB = 5 − (− 1) 12 = 6 =2 Let C(−3, 1) = (x1, y1) and D(4, 15) = (x2, y2) 15 − 1 mCD = 4 − (− 3) =

14 7

=2 3

Compare the gradients to determine if they are parallel. (Note: || means ‘is parallel to’.)

Since parallel lines have the same gradient and mAB = mCD = 2, then AB||CD.

Collinear points lie on the same straight line. Worked exaMple 13

eBook plus

Show that the points A(2, 0), B(4, 1) and C(10, 4) are collinear. Think 1


Tutorial

int-1165

WriTe

Let A(2, 0) = (x1, y1) and B(4, 1) = (x2, y2) y −y Since m= 2 1 x2 − x1 1− 0 mAB = 4−2 =

2

Find the gradient of BC.

Worked example 13

1 2

Let B(4, 1) = (x1, y1) and C(10, 4) = (x2, y2) mBC = 4 − 1 10 − 4

3

Show that A, B and C are collinear.

=

3 6

=

1 2

Since mAB = mBC =

1 2

then AB||BC Since B is common to both line segments, A, B and C must lie on the same straight line. That is, A, B and C are collinear.

Chapter 12

Coordinate geometry

443

Worked exaMple 14

Find the equation of the straight line that passes through the point (2, 6) and is parallel to the line y = 3x + 1. Think 1

2

WriTe

In order to find the equation of a straight line, we need to know the gradient and a point on the line. One point is given and since the line is parallel to y = 3x + 1, the gradients will be the same. Use the formula y − y1 = m(x − x1) and substitute the coordinates of the point and the gradient to find the equation of the line.

Point on the line: (2, 5) Gradient: m = 3.

y − y1 = m(x − x1) y − 5 = 3(x − 2) y − 5 = 3x − 6 y = 3x − 1

reMeMBer

1. The equation of a straight line may be expressed in the form: y = mx + c where m is the gradient of the line and c is the y-intercept, or y − y1 = m(x − x1) where m is the gradient and (x1, y1) is a point on the line. 2. The gradient can be calculated if two points, (x1, y1) and (x2, y2) are given by using y −y m= 2 1 x2 − x1 3. Parallel lines have the same gradient. 4. Collinear points lie on the same straight line.

exerCiSe

12e eBook plus

parallel lines 1

a b c d e f

Digital doc

Spreadsheet 046 Gradient


2

SkillSHEET 12.2

Find if AB is parallel to CD given the following coordinates.

A(4, 13), B(2, 9), C(0, −10), D(15, 0). A(2, 4), B(8, 1), C(−6, −2), D(2, −6). A(−3, −10), B(1, 2), C(1, 10), D(8, 16). A(1, −1), B(4, 11), C(2, 10), D(−1, −5). A(1, 0), B(2, 5), C(3, 15), D(7, 35). A(1, −6), B(−5, 0), C(0, 0), D(5, −4). Which pairs of the following straight lines are parallel?

a 2x + y + 1 = 0 c 2y − x = 3 x e y = −1 2 g 3y = x + 4

Expressing the equation of a straight line in the form y = mx + c

3

444

We 12

We 13

b y = 3x − 1 d y = 4x + 3 f 6x − 2y = 0 h 2y = 5 − x

Show that the points A(0, −2), B(5, 1) and C(−5, −5) are collinear.


4 Show that the line that passes through the points (−4, 9) and (0, 3) also passes through the point (6, −6). 5 In each of the following, show that ABCD is a parallelogram. a A(2, 0), B(4, −3), C(2, −4), D(0, −1) b A(2, 2), B(0, −2), C(−2, −3), D(0, 1) c A(2.5, 3.5), B(10, −4), C(2.5, −2.5), D(−5, 5)


Spreadsheet 085

6 In each of the following, show that ABCD is a trapezium. a A(0, 6), B(2, 2), C(0, −4), D(−5, −9) b A(26, 32), B(18, 16), C(1, −1), D(−3, 3) c A(2, 7), B(1, −1), C(−0.6, −2.6), D(−2, 3) 7

a (4, 3)

Parallel checker

8 eBook plus Digital doc

WorkSHEET 12.2

12F

MC The line that passes through the points (0, −6) and (7, 8) also passes through:

9

b (5, 4)

MC The point as (−1, 5) is: a (2, 9)

(−1,

c (−2, 10)

d (1, −8)

e (1, 4)

5) lies on a line parallel to 4x + y + 5 = 0. Another point on the same line b (4, 2)

d (−2, 3)

c (4, 0)

e (3, −11)

We 14 Find the equation of the straight line given the following conditions: a passes through the point (−1, 3) and parallel to y = −2x + 5. b passes through the point (4, −3) and parallel to 3y + 2x = −3.

perpendicular lines In this section, we shall examine some of the properties of perpendicular lines. Observing the graphs can be very useful in investigating these properties. Consider the diagram below, where the line segment AB is perpendicular to the line segment BC. Line AC is parallel to the x-axis. Line BD is the height of the resulting triangle ABC. Let mAB = m1

Let

=a b = tan(θ ) mBC = m2 − = a c = −tan(α) − = b a =

Hence

y

B

α θ a

A

θ b

D

c

α C

x

−

1 m1

m2 =

−1

m1 −1

or m1m2 = Hence, if two lines are perpendicular to each other, then the product of their gradients is −1. Two lines are perpendicular if and only if: m1m2 = −1 − 1 m2 = or m1

Chapter 12

Coordinate geometry

445

Worked exaMple 15

Show that the lines y = to one another.

eBook plus −5x

+ 2 and 5y − x + 15 = 0 are perpendicular

Tutorial


Think 1

WriTe

Find the gradient of equation 1. Hence

2

Find the gradient of equation 2.

5y − x + 15 = 0 Rewrite in the form y = mx + c 5y = x − 15 y= x −3 5 Hence

3

Test for perpendicularity. (The two lines are perpendicular if the product of their gradients is −1.)

y = −5x + 2 m1 = −5

m2 =

1 5 1

m1m2 = −5 × 5 = −1 Hence, the two lines are perpendicular to each other.

reMeMBer

Two lines are perpendicular if and only if: m1m2 = −1 − 1 . or m2 = m1 exerCiSe

12F eBook plus

perpendicular lines 1 We 15 Show that the lines y = 6x − 3 and x + 6y − 6 = 0 are perpendicular to one another. 2

Digital doc

Determine if AB is perpendicular to CD, given the following coordinates. a b c d e f

Spreadsheet 085 Perpendicular checker

3

A(1, 6), B(3, 8), C(4, −6), D(−3, 1) A(2, 12), B(−1, −9), C(0, 2), D(7, 1) A(1, 3), B(4, 18), C(−5, 4), D(5, 0) A(1, −5), B(0, 0), C(5, 11), D(−10, 8) A(−4, 9), B(2, −6), C(−5, 8), D(10, 14) A(4, 4), B(−8, 5), C(−6, 2), D(3, 11) Determine which pairs of the following straight lines are perpendicular.

a c e g

x + 3y − 5 = 0 y=x y = 3x + 2 2x + y = 6

b d f h

y = 4x − 7 2y = x + 1 x + 4y − 9 = 0 x+y=0

4 Show that the following sets of points form the vertices of a right-angled triangle. a A(1, −4), B(2, −3), C(4, −7) b A(3, 13), B(1, 3), C(−4, 4) c A(0, 5), B(9, 12), C(3, 14) 5 Prove that the quadrilateral ABCD is a rectangle when A is (2, 5), B(6, 1), C(3, −2) and D(−1, 2).

446


6 Find the equation of the straight line that cuts the x-axis at 3 and is perpendicular to the equation 3y − 6x = 12. 7 Calculate the value of m for which the following pairs of equations are perpendicular to each other. a 2y − 5x = 7 and 4y + 12 = mx b 5x − 6y = −27 and 15 + mx = −3y 8 Prove that the quadrilateral ABCD is a rhombus, given A(2, 3), B(3, 5), C(5, 6) and D(4, 4). Hint: The diagonals of a rhombus intersect at right angles. 9 MC The gradient of the line perpendicular to the line with equation 3x − 6y = 2 is: a 3

b

−6

c 2

10 MC Triangle ABC has a right angle at B. The vertices value of z is: 1

a 84

12G

b 4

1 2 are A(−2,

d

e

9), B(2, 8) and C(1, z). The

3

c 12

−2

d 74

e

applications

−4


In this section we look at two important applications: the equation of a straight line, and equations of horizontal and vertical lines.

int-0979 Applications of coordinate geometry

The equation of a straight line

The equation of a straight line can be determined by two methods. The y = mx + c method requires the gradient, m, and a given point to be known, in order to establish the value of c. Note: Since the value of c represents the y-intercept, it can be substituted directly if known. Worked exaMple 16

Find the equation of the straight line that passes through the point (3, −1) and is parallel to the straight line with equation y = 2x + 1. Think

WriTe

1

Write the general equation.

2

Find the gradient of the given line.

3

Substitute for m in the general equation.

4

Substitute the given point to find c.

5

Substitute for c in the general equation.

y = mx + c y = 2x + 1 has a gradient of 2 Hence m = 2 so y = 2x + c (x, y) = (3, −1) ∴ −1 = 2(3) + c =6+c c = −7 y = 2x − 7 or 2x − y − 7 = 0

The alternative method comes from the gradient definition. y − y1 m= 2 x2 − x1 Hence m(x2 − x1) = y2 − y1 Chapter 12

Coordinate geometry

447

Using the general point (x, y) instead of the specific point (x2, y2) gives the general equation: y − y1 = m(x − x1) This requires the gradient, m, and a given point (x1, y1) to be known. Worked Example 17

Find the equation of the line that passes through the point (0, 3) and is perpendicular to a straight line with a gradient of 5. Think

Write

m=5 −1 m1 = 5

1

Find the gradient of the perpendicular line.

Given

2

Substitute for m and (x1, y1) in the general equation.

Since y − y1 = m(x − x1) and (x1, y1) = (0, 3) then y − 3 =

−1 5

(x − 0)

− = x 5 5(y − 3) = −x 5y − 15 = −x x + 5y − 15 = 0

Horizontal and vertical lines For horizontal lines the gradient is equal to zero, and so the equation y = mx + c becomes y = c. Notice that x does not appear in the equation because there is no x-intercept. Horizontal lines are parallel with the x-axis. In the case of vertical lines, the gradient is infinite or undefined. The general equation for a vertical line is given by x = a. In this case, just as the equation suggests, a represents the x-intercept. Notice that y does not appear in the equation because there is no y-intercept. Vertical lines are parallel with the y-axis. The graphs of y = 4 and x = −3 are shown below to highlight this information. y

y

−3

0

y=4

4

0

x = −3

x

Worked Example 18

Find the equation of: a the vertical line that passes through the point (2, −3) b the horizontal line that passes through the point (−2, 6). Think a For a vertical line, there is no y-intercept so y does not

appear in the equation. The x-coordinate of the point is 2. b For a horizontal line, there is no x-intercept so x does not appear in the equation. The y-coordinate of the point is 6.

448

Write a x=2 b y=6


x

Worked Example 19

Find the equation of the perpendicular bisector of the line joining the points (0, −4) and (6, 5). Think 1

Find the gradient of the line joining the given points using the general equation.

Write

Let (0, −4) = (x1, y1) Let (6, 5) = (x2, y2) y − y1 m= 2 x2 − x1 m=

2

Find the gradient of the perpendicular line.

Find the midpoint of the line joining the given points.

=

9 6

=

3 2

For lines to be perpendicular, m2 = −2

m1 = 3

5 − ( − 4) 6−0

−

1 m1

3

x1 + x2 2 0+6 = 2

x=

=3 y + y2 y= 1 2 = =

−4

+5 2

1 2 1

Hence (3, 2 ) are the coordinates of the midpoint. 4

Substitute for m and (x1, y1) in the general equation.

Since y − y1 = m(x − x1) −2 1 and (x1, y1) = (3, 2 ) and m1 = 3

5

Simplify by removing the fractions.

then y −

(a) Multiply both sides by 3.

3(y − 3 2

(b) Multiply both sides by 2.

1 2

=

−2

3 1 − ) = 2(x 2 −2x + 6

(x − 3) − 3)

3y − = 6y − 3 = −4x + 12 4x + 6y − 15 = 0 y 5

Note: The diagram at right shows the geometric situation.

(6, 5)

1

2 –2 1– 2

−4

3

6 x

−4


449

Worked exaMple 20

eBook plus

ABCD is a parallelogram. The coordinates of A, B and C are (1, 5), (4, 2) and (2, −2) respectively. Find: a the equation of AD b the equation of DC c the coordinates of D. Think a

b

1


WriTe

Draw the parallelogram ABCD. Note: The order of the lettering of the geometric shape determines the links in the diagram. For example: ABCD means join A to B to C to D to A. This avoids any ambiguity.

2

Find the gradient of BC.

3

State the gradient of AD.

4

Using the given coordinates of A and the gradient of AD find the equation of AD.

1


2

State the gradient of DC.

3

Using the given coordinates of C and the gradient of DC find the equation of DC.

c Solve simultaneously to find D, the point of intersection

of the equations AD and DC.

Note: Alternatively, a CAS calculator could be used to determine the point of intersection of AD.

450

Tutorial

a

y 5

A

2 B D x −1 1 2 4 −2 C −

2−2 2−4 −4 = − 2 =2 Since mBC = 2 and AD||BC then mAD = 2 y = 2x + c Let (x, y) = (1, 5) 5 = 2(1) + c c=3 Hence, the equation of AD is y = 2x + 3. 2−5 b mAB = 4 −1 −3 = 3 = −1 Since mAB = −1 and DC||AB then mDC = −1 mBC =

y = −x + c Let (x, y) = (2, −2) −2 = −(2) + c c=0 Hence, the equation of DC is y = −x. y = 2x + 3 [1] Equation of DC: y = −x [2] [1] – [2]: 0 = 3x + 3 3x = −3 x = −1 Substituting x = −1 in [2]: y = −(−1) =1 Hence, the coordinates of D are (−1, 1).

c Equation of AD:


reMeMBer

The equation of a straight line can be determined by two methods: 1. The y = mx + c method. This requires the gradient, m, and a given point to be known, in order to establish the value of c. If the y-intercept is known, then this can be directly substituted for c. 2. Alternative method: y − y1 = m(x − x1) This requires the gradient, m, and a given point (x1, y1) to be known. 3. The general equation for a vertical line is given by x = a and a horizontal line is given by y = c. exerCiSe

12G eBook plus Digital doc

Spreadsheet 029 Equation of a straight line

applications 1 2

We 16 Find the equation of the straight line that passes through the point (4, −1) and is parallel

to the straight line with equation y = 2x − 5.

We 17 Find the equation of the line that passes through the point (−2, 7) and is perpendicular 2

to a line with a gradient of 3. 3 Find the equations of the following straight lines. a Gradient 3 and passing through the point (1, 5). b Gradient −4 and passing through the point (2, 1). c Passing through the points (2, −1) and (4, 2). d Passing through the points (1, −3) and (6, −5). e Passing through the point (5, −2) and parallel to x + 5y + 5 = 0. f Passing through the point (1, 6) and parallel to x − 3y − 2 = 0. g Passing through the point (−1, −5) and perpendicular to 3x + y + 2 = 0. 4 Find the equation of the line which passes through the point (−2, 1) and is: a parallel to the straight line with equation 2x − y − 3 = 0 b perpendicular to the straight line with equation 2x − y − 3 = 0. 5 Find the equation of the line that contains the point (1, 1) and is: a parallel to the straight line with equation 3x − 5y = 0 b perpendicular to the straight line with equation 3x − 5y = 0. 6

7

We 18 Find the equation of: a the vertical line that passes through the point (1, −8) b the horizontal line that passes through the point (−5, −7). MC a The vertical line passing through the point (3, −4) is given by:

a y = −4 b x=3 c y = 3x − 4 d y = −4x + 3 e x = −4 b Which of the following points does the horizontal line given by the equation y = −5 pass through? a (−5, 4) b (4, 5) c (3, −5) − d (5, 4) e (5, 5) c Which of the following statements is true? a Vertical lines have a gradient of zero. b The y-coordinates of all points on a vertical line are the same. c Horizontal lines have an undefined gradient. d The x-coordinates of all points on a vertical line are the same. e A horizontal line has the general equation x = a.

Chapter 12

Coordinate geometry

451

d Which of the following statements is false? A Horizontal lines have a gradient of zero. B The straight line joining the points (1, −1) and (−7, −1) is vertical. C Vertical lines have an undefined gradient. D The straight line joining the points (1, 1) and (−7, 1) is horizontal. E A horizontal line has the general equation y = c. 8 The triangle ABC has vertices A(9, −2), B(3, 6) and C(1, 4). a Find the midpoint, M, of BC. b Find the gradient of BC. c Show that AM is the perpendicular bisector of BC. d Describe triangle ABC. 9 WE 19 Find the equation of the perpendicular bisector of the line joining the points (1, 2) and (−5, −4). 10 Find the equation of the perpendicular bisector of the line joining the points (−2, 9) and (4, 0). 11 WE 20 ABCD is a parallelogram. The coordinates of A, B and C are (4, 1), (1, −2) and (−2, 1) respectively. Find: a the equation of AD b the equation of DC c the coordinates of D. 12 The map shows the proposed course for a yacht race. Buoys y Scale: 1 unit ⇔ 1 km N have been positioned at A(1, 5), B(8, 8) and C(12, 6), but 11 the last buoy’s placement, D(10, w), is yet to be finalised. 10 9 a How far is the first stage of the race, that is, from the Buoy B 8 start, O, to buoy A? 7 Buoy b The race marshall boat, M, is situated halfway between 6 A Buoy C M 5 buoys A and C. What are the coordinates of the boat? 4 c Stage 4 of the race (from C to D) is perpendicular to E 3 stage 3 (from B to C). What is the gradient of CD? Buoy D 2 H 1 d Find the linear equation that describes stage 4. O (Start) 1 2 3 4 5 6 7 8 9 10 11 12 x e Hence determine the exact position of buoy D. 2 f An emergency boat is to be placed at point E, 3 of the way from buoy A to buoy D. Into what internal ratio does point E divide the distance from A to D? g Determine the coordinates of the emergency boat. h How far is the emergency boat from the hospital, located at H, 2 km North of the start? 13 MC a The equation of the line passing through the point (4, 3) and parallel to the line 2x − 4y + 1 = 0 is: A x − 2y + 2 = 0 B 2x − y − 5 = 0 C 2x − y − 10 = 0 D 2x − y − 11 = 0 E 2y + x + 2 = 0 b The equation of the perpendicular bisector of the line segment AB where A is (−3, 5) and B is (1, 7) is: A 2y = x + 13 B y = 2x − 8 C 2y = x + 11 D y = −2x + 4 E y = 2x − 4 c The coordinates of the centroid of triangle ABC with vertices A(1, 8), B(9, 6) and C(−1, 4) are: A (4, 5) B (0, 6) C (3, 6) D (5, 7) E (2, 7) 14 To supply cities with water when the source is a long distance away, artificial channels, called aqueducts, may be built. More than 2000 years after it was built, a Roman aqueduct still stands in southern France. It brought water from a source in Uzès to the city of Nîmes. The aqueduct does not follow a direct route between these two locations as there is a mountain range

452


between them. The table shows the approximate distance from Uzès along the aqueduct to each town (or in the case of Pont du Gard, a bridge) and the aqueduct’s height above sea level at each location. Distance from Uzès (km)

Height of aqueduct above sea level (m)

Uzès

0

76

Pont du Gard (bridge)

16

65

St. Bonnet

25

64

St. Gervasy

40

61.5

Nîmes

50

59

Location

a Show the information in the table as a graph with the distance from Uzès along the horizontal axis. Join the plotted points with straight lines. b Calculate the gradient of the steepest part of the aqueduct (in m/km). c Suppose the aqueduct started at Uzès and ended at Nîmes, but had a constant gradient. Write a linear equation to describe its course. d Using the equation found in part 3, calculate the height of the aqueduct at the Pont du Gard. This calculated height is higher than the actual height. How much higher? e Why do you think the Romans made the first part of the aqueduct steeper than the rest?


453

Summary Distance between two points

• The distance between two points A(x1, y1) and B(x2, y2) is: AB = ( x2 − x1 )2 + ( y2 − y1 )2 Midpoint of a line segment

• The coordinates of the midpoint of the line segment joining (x1, y1) and (x2, y2) are:

y (x2, y2)

 x1 + x2 y1 + y2   2 , 2 

M

+ x _____ y +y (x_____, 2 2 ) 1

(x1, y1)

2

1

2

x

Dividing a line segment internally in the ratio a:b

• The coordinates of the point that divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio a:b are:

y

B(x2, y2)

b

P(x, y)

a

 ax2 + bx1 ay2 + by1   a + b , a + b 

A(x1, y1) x

Dividing a line segment externally in the ratio a:b

• The coordinates of the point that divides the line segment joining the points (x1, y1) and (x2, y2) externally in the ratio a:b are:

 ax2 − bx1 ay2 − by1   a − b , a − b 

y

P(x, y) a

b B(x2, y2)

A(x1, y1) x

Parallel lines

• Parallel lines have the same gradient. • Collinear points lie on the same straight line. Perpendicular lines

• Two lines are perpendicular if and only if: m1m2 = −1 or m2 =

−

1 m1

Applications

• Equations of a straight line: 1. Gradient and y-intercept form: y = mx + c y2 − y1 where m (the gradient) = and c is the y-intercept x2 − x1 2. General form: ax + by + c = 0 454


• To find the equation of a straight line: 1. Given gradient and y-intercept y = mx + c 2. Given gradient and a point y − y1 = m(x − x1) or y = mx + c method 3. Given two points Find m, then use: y − y1 = m(x − x1) or y = mx + c method • The general equation for a vertical line is given by x = a and a horizontal line is given by y = c.


455


1 Find the distance between the points (1, 3) and (7, −2). 2 Prove that triangle ABC is isosceles given A(3, 1), B(−3, 7) and C(−1, 3). 3 Show that the points A(1, 1), B(2, 3) and C(8, 0) are the vertices of a right-angled triangle. 4 The midpoint of the line segment AB is (6, −4). If B has coordinates (12, 10), find the coordinates of A. 5 Find the coordinates of the point which divides the line joining the point A(−2, 6) and the point B(2, −4) internally in the ratio 3:1. 6 Find the coordinates of the point which divides the line joining the points (2, 8) and (5, 2) externally in the ratio 1:2. 7 Show that the points A(3, 1), B(5, 2) and C(11, 5) are collinear. 8 Show that the lines y = 2x − 4 and x + 2y − 10 = 0 are perpendicular to one another. 9 Find the equation of the straight line passing through the point (6, −2) and parallel to the line x + 2y − 1 = 0. Exam tip To find the straight line you need a point on the line and its gradient.

10 Find the equation of the line perpendicular to 3x − 2y + 6 = 0 and having the same y-intercept. 11 Find the equation of the perpendicular bisector of the line joining the points (−2, 7) and (4, 11). 12 Find the equation of the straight line joining the point (−2, 5) and the point of intersection of the straight lines with equations y = 3x − 1 and y = 2x + 5. 13 Using the information given in the diagram. a Find: y i the gradient of AD B(4, 9) ii the gradient of AB 9 iii the equation of BC C 4 A iv the equation of DC O D v the coordinates of C. x b Describe quadrilateral 45 9 ABCD.

456

14 In triangle ABC, A is (1, 5), B is (−2, −3) and C is (8, −2). a Find: i the gradient of BC ii the midpoint, P, of AB iii the midpoint, Q, of AC. b Hence show that: i PQ is parallel to BC ii PQ is half the length of BC. 15 Triangle ABC has vertices A(a, b), B(−3, 6) and C(5, −2). The centroid, G, of the triangle has coordinates (−2, −1). a Find: i the midpoint, M, of BC ii the coordinates of A iii the gradient of BC iv the gradient of AM v the length of AB vi the length of AC. b Describe triangle ABC. Multiple choice

1 The distance between the points (1, 5) and (6, − 7) is: A 53

B

29

C 13

E 12 D 193 2 The midpoint of the line segment joining the points (−4, 3) and (2, 7) is: B (−2, 10) C (−6, 4) A (−1, 5) − − D ( 2, 4) E ( 1, 2) 3 If the midpoint of the line segment joining the points A(3, 7) and B(x, y) has coordinates (6, 2), then the coordinates of B are: A (15, 3) B (0, −6) C (9, −3) − D (4.5, 4.5) E ( 9, 3) 4 C is a point that divides the line segment AB internally in the ratio 1:2. If A is the point (−4, 1) and B is the point (2, −5), then the coordinates of C are: 1 1 B (−2, −1) A (0, −3) C (−33, 23) 2 2 E (−1, −2) D (2 3 , −3 3 ) The following information refers to questions 5, 6, 7 and 8. Triangle ABC has vertices A(1, 5), B(4, −1) and C(−6, −3).


5 The median from A meets the line segment, BC, at M. The coordinates of M are: 1

A (2 2 , 2) −4)

1

B (−3 2 , 1)

C (−1, −2)

E (10, 2) D (2, 6 The centroid divides a median in the ratio 2 :1. The coordinates of the centroid are: 1

1

2

A (0, 1 2 )

B (3, 2 3 )

D

E ( 2 , 1)

−1 1 ( 3 , 3)

C (0, 3)

1

7 The gradient of the median, AM, is: 1

A 3 2

2

B 7 C undefined −2 D 2 E 8 The equation of the median, AM, is: A 2x − 7y − 12 = 0 B 7x − 2y + 3 = 0 C 7x − 2y − 3 = 0 D x + 1 = 0 E 2x − 7y + 12 = 0 9 D is a point that divides the line segment AB externally in the ratio 2 : 3. If A is the point (2, 3) and B is the point (7, 6), then the coordinates of D are: 1 A (−8, −3) B (8, −3) C (4, 4 5 ) D (17, 12) E (8, 3) 10 The gradient of the line joining the points (2, 7) and (5, −8) is: A −5 B −1 C 5 −3 E 1 D 7 11 If the points (−6, −11), (2, 1) and (x, 4) are collinear, then the value of x is: 1 A 4 B 3.2 C 4 5

D 16

E 3

12 The gradient of the line perpendicular to 3x − 4y + 7 = 0 is: −4 3 4 C 3 A 4 B 3 D 3 E −4

13 The equation of the line perpendicular to 2x + y − 1 = 0 and passing through the point (1, 4) is: A 2x + y − 6 = 0 B 2x + y − 2 = 0 C x − 2y + 7 = 0 D x + 2y + 9 = 0 E x − 2y = 0 The following information refers to questions 14, 15, 16 and 17. The diagram at y right shows a square inscribed in a circle. (a, b) A (1, 4) D The square has coordinates A(1, 4), x B(2, −3), C(−5, -4) and D(a, b). B C (2, −3)

(−5, −4)

14 The circle has a radius of: A 10 units B 7.07 units C 6 units D 5 units E 12 units 15 The coordinates of the centre are: A (−4, 0) B (−2, 0) C (0, −2) D (−1, 1) −4) E (0, 16 The gradient of the diagonal, BD, is: −3 A −1 C 1 B 4 D

−3

5

E

4 3

17 The coordinates of the point D are: A (−3, 6) B (3, −6) − C ( 6, 2) D (−2, 4) − E ( 6, 3)

Extended response

1 ABCD is a quadrilateral with vertices A(4, 9), B(7, 4), C(1, 2) and D(a, 10). Given that the diagonals are perpendicular to each other, find: a the equation of the diagonal AC b the equation of the diagonal BD B c the value of a. 2 The centroid, G, of a triangle ABC divides the medians internally in the ratio 2:1. For example: AG:GD = 2:1 where D is the midpoint of BC. A′B′C′ is a triangle with coordinates A′(5, 4), B′(−2, 5) and C′(6, 9). Find the coordinates of the centroid, G′.

F

G

D 1

2 A

E

C


457

3 The course for a car rally is planned so that each participating team must y pass four checkpoints in order to complete the course at the point where it North first begins. The first checkpoint, A, is located 8 km south and 5 km east of the start, while the third checkpoint, C, is 15 km north of A and 8 km east of the start. Start a Find the distance to checkpoint A. b Find the coordinates of checkpoint C.

C B 5

D m

8

x

−8

A c Checkpoint B is located 3 of the way from A to C. Find the coordinates of B. d Find the distance from A to C. e Checkpoint D is located m km directly east of the start. One of the teams realises that their car is very low on fuel and decides to drive directly from B to a service station at D. They know that this will save them 5.10 km of travel. Write an expression for the distance from B to D. f Write an expression for the distance from B to D travelling via checkpoint C. y g Find the coordinates of checkpoint D. h Find the total distance of the course (without taking any shortcut). 8m 2

C T

4 An architect decides to design a building with a 14-metre-square base such that the external walls are initially vertical to a height of 50 metres, but taper so that their separation is 8 metres at its peak height of 90 metres. A profile of the building is shown with the point (0, 0) marked as a reference at the centre of the base. a Write the equation of the vertical line connecting A and B. b Write the coordinates of B and C. c Find the length of the tapered section of wall from B to C. d The top floor of the building is on a level with point T that divides BC internally in the ratio 9:1. Find the height of the top floor of the building. 5 In a game of lawn bowls, the object is to bowl a biased ball so that it gets as close as possible to a smaller white ball called a ‘jack’. During a game, a player will sometimes bowl a ball quite quickly so that it travels in a straight line in order to displace an opponents ‘guard balls’. In a particular game, player X has 2 guard balls close to the jack. The coordinates of the jack are (0, 0) and the coordinates of the guard balls 4

B 90 m 50 m

A

0 14 m

x y

S(−30, 24)

24

—) B(−1–2 , 57 40

57 – 40

− 1 57

are A(−1, 5 ) and B( 2, 40 ). Player Y bowls a ball so that it travels in a straight line toward the jack. The ball is bowled from the position S, with the coordinates (−30, 24). −30 a Will player Y displace one of the guard balls? If so, which one? b Due to bias, the displaced guard ball is knocked so that it begins to travel in a straight line (at right angles to the path found in part a). Find the equation of the line of the guard ball. c Show that guard ball A is initially heading directly toward guard ball B. d Given its initial velocity, guard ball A can travel in a straight line for 1 metre before its bias affects it path. Calculate and explain whether guard ball A will collide with guard ball B. eBook plus Digital doc


458


A(−1, 4–5 )

4– 5

−1 −1– 2

(Not to scale)

x

eBook plus

aCTiviTieS


• 10 Quick Questions: Warm up with ten quick questions on coordinate geometry. (page 429) 12a

distance between two points

Tutorial

• We3 int-1162: Watch how to prove that three points are vertices of an isosceles triangle. (page 430) Digital doc

• Spreadsheet 021: Investigate the distance between two points. (page 431) 12b

Midpoint of a line segment

Digital doc

• Spreadsheet 075: Investigate the midpoint of a segment. (page 434) 12c

dividing a line segment internally in the ratio a:b

Tutorial

• We8 int-1163: Watch how to find the coordinates of a point of a line segment given it is divided internally in a ratio 1:2. (page 437) Digital docs

• Spreadsheet 022: Investigate dividing a segment internally. (page 438) • SkillSHEET 12.1: Practise dividing a line in a given ratio. (page 438) • WorkSHEET 12.1: Solve problems of distance between two points, locate the coordinates of the midpoint of a segment and determine coordinates of the segment internally by dividing a line using first principles and ratios. (page 438) 12d

dividing a line segment externally in the ratio a:b

Tutorial

• We11 int-1164: Watch how to find the coordinates of a point of a line segment given it is divided externally in a ratio 1:2. (page 441) Digital doc

• Spreadsheet 023: Investigate dividing a segment externally. (page 442) 12e

Parallel lines

Tutorial

• We13 int-1165: Watch how to show that three points are collinear. (page 443) Digital docs

• Spreadsheet 046: Investigate gradients. (page 444)

• SkillSHEET 12.2: Practise expressing the equation of a straight line in the form y = mx + c. (page 444) • Spreadsheet 085: Investigate parallel lines. (page 445) • WorkSHEET 12.2: Solve more complex problems of distance between two points, locate the coordinates of the midpoint of a segment, determine coordinates of the segment by internally and externally dividing a line using ratios and demonstrate understanding of parallel lines and collinear points. (page 445) 12F

Perpendicular lines

Tutorial

• We15 int-1166: Watch how to show that two straight lines are perpendicular. (page 446) Digital doc

• Spreadsheet 085: Investigate perpendicular lines. (page 446) 12G

applications

Interactivity

• Applications of coordinate geometry int-0979: Apply your knowledge of coordinate geometry by using the interactivity. (page 447) Tutorial

• We20 int-1167: Watch how to determine equations of two sides of a parallelogram and find the coordinates of a vertex. (page 450) Digital doc

• Spreadsheet 029: Investigate the equation of a straight line. (page 451) chapter review Digital doc


Chapter 12

Coordinate geometry

459

EXAM PRACTICE 3

Chapters 8 TO 12

4 p

q

r

Truth value

F

T

T

T

F

T

F

F

3 marks

T

F

F

F

3 The partial fraction can be resolved as follows: 3x + 4 A B C ≡ + + ( x − 1)( x 2 − 2 x + 1) x − 2 ax + b (ax + b)2 a Determine the values of a and b. 2 marks b Show that C = -7. 2 marks c Hence, determine the values of A and B. 3 marks

T

F

T

F

SHORT ANSWER

20 minutes

 4π  1 Convert 2 3 cis   to exact Cartesian  3  coordinates. 2 Prove by induction that ‘n (n + 1)’ is always divisible by 2 for all integer values of n.

MULTIPLE CHOICE

3 marks

10 minutes

Each question is worth 1 mark. 1 When converted to polar form the circle with equation (x - 3)2 + (y + 1)2 = 10 would become which one of the following: A r = 18 B r = 2(3 cos (θ ) - sin (θ )) C r = 2(sin (θ ) - 3 cos (θ )) 10 sin (θ ) + cos (θ ) 2 + 10 E r = cos (θ ) + sin (θ )

D r =

2 A perpendicular line PQ is drawn so that is cuts the line segment AB in the ratio of 1:3. The coordinates of A and B are respectively (1, 3) and (5, 11). The equation PQ would be: A y - 2x = 1 B y + 2x = 17 C 2y + x = 12 D 2y + x = 21 E 2y + x = 22 3 Expressed in the form x + yi the polar equation  − 5π  z = 4 cis  would be which one of the  6 

Which one of the following compound statements represents the truth value for the truth table above? A (p ∨ q) ∨ r B p ∧ (r ∧ q) C (p ∨ q) ∧ r D (p ∨ r) ∧ q E ∼ p ∨ (r ∧ q) 5 y 6 4 2 O

C 2 + 2 3i D 2 3 − 2i E

460

−2

3 − 2i

4

6

x

The graph above shows the feasible region of a set of linear inequalities. If the objective function z = 2y - ax is maximised at the point (3, 4) then the value of a could be: A -2 B -1 c 0 D 1 E 2 6 The line y = -3x + 4 intersects the parabola (y + 2)2 - 3(x + 1) = 0 at points P and Q. If the coordinates of P are (2a, 4b) then the values of a and b would be respectively: A

13 + 37 ( 37 ) − 5 and 12 8

B

13 + 37 ( − 37 ) − 5 and 12 8

C

13 + 37 ( − 37 ) − 5 and 6 2

D

13 + 37 ( 37 ) − 5 and 6 2

E

13 + 37 and 2 3

following? A 2 − 2 3i B − 2 − 2 3i

2


(

−(

37 ) − 5

)

exTended reSponSe

35 minutes

1 The shape of a shade sail can be modelled using part of a rectangular hyperbola and two linear equations. a If the equation of the rectangular hyperbola is given by 8x2 − y2 − 16 = 0, y show that e = ±3. 2 marks To model the shape of the shade sail, the domain is restricted to R+ ∪ {0}. The rectangular hyperbola that follows this constraint is shown on the set of axes at right. 0

x

√2

y

b On the graph above right, clearly label the foci point and the asymptotes. 3 marks

y1

To complete the shape of the shade sail, two line segments (y1 and y2) are drawn as shown on the graph at right. The line segments begin at (0, 0) and end at the point of intersections with the hyperbola. Line y2 is found by reflecting y1 in the x-axis. 4 c Explain why the gradient of y1 must lie between 0 and . 2 marks 2 d If the length of y1 is 2 5 , show that the point of intersection of y1 and the rectangular hyperbola is (2, 4). e If the length of y2 is also 2 5 , determine the equation of y2.

x

0 y2

3 marks 2 marks

2 The shade sails are made by Cover All Shades. There are two types of material that shade sails can be made from. Each type of material is described by the amount of ultraviolet (UV) rays that can penetrate through the fibres. The two materials are defined as light protection (l) and maximum protection (m). The cost per metre of the light protection fabric is $12.75 and the cost per metre of the maximum protection fabric is $21.50. Cover All Shades keeps 100 metres of light protection material and 100 metres of maximum protection material in stock. Each sail requires 15 metres of material to construct. A maximum of 14 sails are made each week. At least 4 light protection sails are made each week. The information above can be represented by the following inequalities: Inequality 1: l ≥ 0 Inequality 2: m ≥ 0 Inequality 3: l + m ≤ 210 Inequality 4: l ≥ 60 a An additional Inequality 5 is described as m ≥ 2l. In the context of this problem, explain Inequality 5. b Lines 1 to 5 are shown on the graph at right. Clearly show the 1 mark feasible region. c Write down the coordinates of all boundary points for the feasible region found in part b. 2 marks d Determine the objective function, S, for this problem. Write your answer as an equation in terms of S, l and m. 2 marks e Determine the maximum weekly sales, in dollars, that Cover All Shades can expect. Write your answer to the nearest cent. 2 marks f Outdoors Living places an order for 6 light protection and 9 maximum protection sails. Explain, in the context of this problem, if Cover All Shades is able to deliver this order within a week.

2 marks

m 210

120

60

210

l 2 marks



exam practice 3

461

13

13A Introduction to vectors 13B Operations on vectors 13C Magnitude, direction and components of vectors 13D i , j notation  13E Applications of vectors

Vectors aReaS oF STudy

• Concept of the position vector of a point in the Cartesian plane • The representation of plane vectors as ordered pairs (a, b) • Plane vectors as directed line segments • The representation of a vector (a, b) in the form ai + bj where i and j are the standard  unit vectors   orthogonal

• The magnitude of a plane vector (a, b) and its calculation • Addition of plane vectors, using components or the parallelogram rule • Simple vector algebra (addition, subtraction, multiplication by a scalar) • Applications of vectors: for example, geometric proofs, orienteering, navigation and statics eBook plus

13a

introduction to vectors

Digital doc

10 Quick Questions

A scalar quantity is one which is specified by size, or magnitude, only. Distance is an example of a scalar quantity; it needs only a number to specify its size or magnitude. Time, length, volume, temperature and mass are scalars. A vector quantity is specified by both magnitude and direction. Displacement measures the final position compared to the starting position and requires both a magnitude (eg. distance 800 m) and a direction (eg. 230°T). Displacement is an example of a vector quantity. Force, velocity and acceleration are also vectors. They all require a size and a direction to be specified completely.

Representation of vectors Vectors can be represented by directed line segments. For example, if north is straight up the page and a scale of 1 cm = 20 m is used, then a displacement of 100 m south is represented by a 5 cm line straight down the page. We place an arrow on the line to indicate the direction of the vector, as shown at right. The start and end points of a vector can be labelled with capital letters. For example, the vector shown at right can have the starting point, or tail, labelled S and the end point, or head, labelled F: → This vector can then be referred to as SF. The vector can also be represented by a lower-case letter over a tilde, for example, s~ . 462


S

N W

E S

100 m

F

1 cm = 20 m

Representing vectors as ordered pairs (a, b) A vector in the x–y plane can be described by an ordered pair (a, b). The values a and b are called components where a gives the change of position relative to the positive x-axis and b gives the change of position relative to the positive y-axis, of the end of the vector compared to the start. For example (2, 4) represents a change of position of 2 units in the positive x direction and 4 units in the positive y direction. Note, the vector represented by (2, 4) doesn’t necessarily start at the origin. It can be in any position on the Cartesian plane.

y 4 3 2 1 −2 −1 0 −1

1

2

3 x

−2 −3

Position vectors A position vector describes a point in the Cartesian plane. Position vectors start at the origin O(0, 0). For → example, for A(3, 1) the position vector OA is shown. Note we can also use (3, 1) to describe any vector that travels three units across and one up, but it is only a position vector if it starts at (0, 0).

y 2 A

1 0O 1

2

3 x

Worked Example 1 y

Write the following vectors in the form (a, b). → → a OC b DA

B

A

C x

O D Think

Write

→

a From O to C, we travel +4 units in the positive x

a OC = (4, 1)

b From D to A, we travel −5 units in the positive x

b DA = ( − 5, 3)

direction and +1 unit in the positive y direction.

direction and +3 units in the positive y direction.

→

Worked Example 2

If we started at (5, −2) where would we end up after a displacement of (3, 2)? Think

Write

1

Write (5, −2) + (3, 2).

(5, −2) + (3, 2)

2

We start at (5, −2) and move +3 units in the positive x direction and +2 units in the positive y direction.

= (8, 0)

3

Write the answer.

We would end up at the point (8, 0).

Chapter 13 Vectors

463

WoRked examPle 3

Draw d, the position vector of (−2, 3), on a set of axes.  Think

1

A position vector must start at (0, 0) and end at the point specified. Make sure the arrow is pointing away from the origin.

2

Label the vector.

dRaW y 3

(−2, 3)

~d x

0

−2

equality of vectors Two vectors are equal if they are: 1. equal in magnitude 2. parallel, and 3. point in the same direction. WoRked examPle 4 y

Which of the following vectors are equal? Think 1

Vectors a and e are of equal length, parallel  same direction. and point in the

a=e  

2

Vectors b and g are of equal length, parallel and point in the same direction.

b=g  

~b

~a

WRiTe

~c f ~ g ~

WoRked examPle 5

~e

x

eBook plus

An aircraft flies 200 km north, then 400 km east. Tutorial Draw a vector diagram to represent the path taken by the aircraft and also the int-1168 displacement of the aircraft from its starting point to its finishing point. Worked example 5 Think 1

Take north as vertically up the page and east to the right.

2

Draw a short, vertical directed line segment to represent a displacement of 200 km north.

3

Draw a horizontal directed line segment with its tail joined to the head of the first. This represents a displacement of 400 km east.

dRaW N E

W

200 km

S 400 km

200 km

464

~d


4

Draw a directed line segment from the tail of the ‘north’ vector (point S) to the head of the ‘east’ vector (point F). This represents the displacement of the aircraft from its starting point to its finishing point.

400 km

F

200 km S

RememBeR

1. A scalar quantity is specified by magnitude or size only. 2. A vector quantity is specified by both magnitude and direction. 3. Vectors can be represented by directed line segments, as in this diagram. → A vector can also be denoted by AB or a.  4. A vector can be represented by an ordered pair (a, b). 5. Position vectors start at the origin. 6. Two vectors are equal if they are: (a) equal in magnitude (b) parallel and (c) point in the same direction. exeRCiSe

13a eBook plus Digital doc

Spreadsheet 144 Vectors introduction

B

~a A

introduction to vectors 1 We 1 Represent the following vectors as a displacement in the form (a, b). → → → → → a AB b AC c AF d BC e BD → → → → → f CD g CA h ED i EF j FE

y

D

−3 −2 −1 O −1 E −2 F

2 We2 If we started at the point (2, −5) where would we end up after each of these displacements? b (−3, 5) c (0, 4) d (2, −5) a (3, −2) 3

Examine the diagram at right. Represent the change of position of each of the vectors shown in the form (a, b).

4 We3 Draw the position vector for each of the following points on the same set of axes. (4, 1) (−3, 2) (0, −3) (−2, −2) 5 We4 Which of the vectors shown in question 3 are equal? 6

3 2 1

e (−2, 5)

y

B A 1 2 3 4 x C

f (6, 3)

g ~ ~c ~b

~a

f ~ ~h

~d

~e x

Represent each of the following vectors on separate diagrams. a the position vector of (2, 3) b the position vector of (0, 5) c the position vector of (−3, 2)

Chapter 13

Vectors

465

d e f g

a displacement of (2, −8) starting from the point (4, 4) a displacement of (−2, 5) starting from the point (3, −6) a displacement of (0, 3) starting from the point (2, 5) the position vector of (4, −2) followed by (3, 5)

7 mC A vector which starts at the point (−2, 1) and finishes at the point (3, −3) is represented by a displacement of: A (4, −5) B (5, −4) C (1, −2) D (−5, 4) E (3, 2) In questions 8 to 11, draw vector diagrams to represent the paths described and the displacement of the finishing point from the starting point. 8 We5 An aeroplane flies 1000 km north from airport A to airport B. It then travels to airport C, which is 1200 km north-east of B. 9

Marcus cycles 20 km in an easterly direction and then travels 30 km due south.


SkillSHEET 13.1 Bearings

10

Bianca rows straight across a river in which a current is flowing at 3.5 km/h. Bianca can row at 11.5 km/h.

11

An aeroplane takes off and flies at an angle of elevation of 25° for 25 km. It then flies horizontally for 300 km.


SkillSHEET 13.2 Angles of elevation and depression

13B

operations on vectors addition of vectors If we travel from A to B and then from B to C, the combined B effect is to start from A and finish at C. We write → → → AB + BC = AC A → Notice that the tail of the second vector BC is joined to the → head of the first vector AB. If the addition is reversed, so that the tail of the first vector is joined to the head of the second vector, the combined effect is → → → → → also a vector equal to AC . So AB + BC = BC + AB This shows that changing the order in which vectors are added does not alter the combined effect of the vectors. This method for adding two vectors is called the triangle rule for vectors. The addition of vectors a and b can be shown by forming a vector from the tail of a to the head of b.  

466


C

~b

~a ~a + ~b

Negative vectors Just as moving −2 units on the x-axis is opposite in direction to moving 2 units along the x-axis, the negative of a given vector is opposite in direction to the original vector.

~b −b ~

The vector −b has the same magnitude as b but is in the opposite  direction. 

Subtraction of vectors Subtraction of vectors can be performed by combining vector addition and negative vectors. a − b = a + ( − b)     For example, if a and b are vectors as shown at right, then we can  find a − b by:    1. expressing it as an addition: a − b = a + ( − b)    

~b ~a

2. reversing the arrow on vector b so that it becomes −b  

−b ~ ~a

3. adding −b to a as shown to form a − b.    

−b ~ ~a

~a − ~b

Worked Example 6

Using d , e and f as shown in the diagram, draw vector diagrams to show:    a d + e b d + e + f c e − f .        Think a

1 2

b

1 2

c

1

f ~ ~d

Draw

Draw the vector d and join the tail of e to  the head of d .   d + e is shown by the vector drawn from  tail  of d to the head of e. the  

a

d + e + f is obtained by joining the head  d + e (from part a ) with the tail of f . of    d + e + f is shown by the vector drawn from the tail of d (or d + e) to the head    of f . 

b

-

Reverse the arrow on f to obtain f and  tail of - f .  join the head of e to the  

~e

~e

~d + ~e

~d

~d

~e

~e

~e

f ~

~d

~d

f ~

~d + ~e +~f

c

−f ~ ~e

Chapter 13 Vectors

467

2

−f ~

e − f is shown by the vector drawn from  tail  of e to the head of − f . the  

~e

~e −~f

WoRked examPle 7

(−5,

eBook plus

(−2,

If a = (1, 4), b = 2) and c = 3) , find each of the following:  b a− c  c a + b + c. a a + b        Think

Tutorial


WRiTe

a Add the corresponding components of each

vector to give the answer for a + b.   b Subtract the corresponding components of each vector to give the answer for a − c .   c a + b + c may be calculated by adding the    corresponding components of a and b and c.   

a a + b = (1, 4) + (−5, 2)



 = (−4, 6)

b a − c = (1, 4) − (−2, 3)



 = (3, 1)

c a + b + c = (1, 4) + (−5, 2) + (−2, 3)





 = (−6, 9)

Scalar multiplication A displacement of (2, 3) followed by another displacement of (2, 3) equals a displacement of (4, 6). We could write this as 2(2, 3) = (4, 6). The vector represented by (2, 3) has been multiplied by the number 2 to give the vector represented by (4, 6). This process is called multiplication by a scalar or scalar multiplication. Scalar multiplication means that the vector is made larger or smaller by a scale factor. In the case above, the scalar is 2. In general, we can say that if k ∈ R: 1. ka is a vector k times as big as a and in the same direction as  a for k > 0.  2. ka is in the opposite direction to a for k < 0.  

~a

WoRked examPle 8

If a = (5, −4) and b = (−3, 2) calculate:   a 2a+ b b 3(b − a ).     Think

a

1 2

b

1

2

468

Multiply each component of a by 2 to  obtain 2a.  Add the components of 2a and b to obtain   2a + b.   Subtract the components of a from b to   obtain b − a.   Multiply the components of a − b by 3 to   obtain 3(b − a).  

WRiTe a

2a = 2(5, −4)  = (10, −8) 2a + b = (10, −8) + (−3, 2)   = (7, −6)

b

b − a = (−3, 2) – (5, −4)   = (−8, 6) 3(b − a) = 3(−8, 6)   = (−24, 18)


2a ~

3a ~

−a ~

−2.5a ~

Worked Example 9

→ ABEF and BCDE are parallelograms with AB represented by a and →  AF represented by b. The length of BC is twice the length of AB. Express  in terms of a and b. the following vectors  → → →  a BC b AC c BD Think a

1

2

b

1 2 3

c

→ → BC and AB are in the same direction and → → BC is twice as big as AB. → Replace AB by a → → → AC = AB + BC using vector addition. → → Replace AB and BC by a and 2a   respectively.

1

Simplify. → → CD = AF since opposite sides of a parallelogram are parallel and the same size.

2

→ → → BD = BC + CD using the triangle rule to add vectors.

3

→ → Replace BC and CD by 2a and b   respectively.

Write

→

E

F

D

~b A

~a B

C

→

a BC = 2AB

= 2a  → → → b AC = AB + BC = a + 2a   = 3a  → c CD = b  → → → BD = BC + CD

= 2a + b  

Worked Example 10

→ → → Simplify the expression AB + BC − EC . Think 1

2

3

→ → AB + BC represents a vector from A to B with the vector from B to C added on. This is the same as the vector from A to C. → → − EC is the same as vector CE since the negative of a vector reverses the direction. → → AC + CE represents a vector starting at A going to C and then from C to E. This is the → same as AE.

Write

→ → → AB + BC − EC → → = AC − EC → → = AC + CE

→ = AE

Chapter 13 Vectors

469

RememBeR

→ → → 1. Vectors are added using the triangle rule, AB + BC = AC. 2. Subtraction of vectors is performed by using a − b = a + ( − b) (−b is a vector which has   to b.)  the same magnitude as b but is in the opposite direction   3. ‘Multiplication of a vector by a scalar’ means that the vector is made larger or smaller by a scale factor. 4. ka is a vector k times as big as a and in the same direction as a, if k > 0; if k < 0,  ka is in the opposite direction  to a, where k ∈ R.  then   exeRCiSe

13B


Spreadsheet 143 Vectors

operations on vectors 1 We6 Using vectors a, b and c as shown, sketch:    c −c b 2b a 3a    d a+b e a+c f b+c ~a       g a + 2b h 2a + 3c i a−c       j b−c k a+b+c l a−b−c ~b         2 Draw two vectors u and v such that u + v = (0, 0).     ~c 3 a Draw two possible representations of u + v = (3, 5).   b Draw two possible representations of u + v = (−3, 2).   4 We 7 If m = (−2, 3), n = (4, 0) and p = (−1, 5), find each of the following.    c n− p d m− n− p a m+n b m+ n+ p           5 We8 Using m, n and p from question 4, calculate the following.    c 2(m + n) a 3n − p b 2m + n − p d 4 p − 3n          F G 6 The figure shows a cube. Write all the vectors that are equal to the following vectors. C → → → a OA b OC c OD E → → → D d GF e OB f AD

B

A O Refer to the cube shown in question 6. → → → Let a = OA, c = OC and d = OD. Write, in terms of a, c and d , the vectors representing:    →  → →  → → a DE b OB c AC d AE e EA → → → → → f EG g DF h OF i AG j DB → → C B 8 ABCDEF is a regular hexagon with vectors OA and OB represented by a and b respectively. Write, in terms of a and b, the     ~b vectors. D A → → → → a a DO b DA c AD d AB O ~ → → → → e BC f AC g CD h ED E F → → i EA j DF → → → → 9 We10 Show that OA + AB + BC = OC .

7 We9

470


→ → → → Express in simplest form AB + BC + DE − DC . → → → → 11 Show that EF + GH − GF − EH = 0. → → → → 12 mC In simplest form, MN − QP + NP + QR equals: → → → A 0 B MR C MQ D QN 10


→ E NR

WorkSHEET 13.1

13C

magnitude, direction and components of vectors magnitude


The magnitude of a vector can be calculated from the length of the line segment representing the vector. The magnitude of a vector a is denoted by a or a.  

direction

The direction of a vector can be found by applying appropriate trigonometric ratios to find a relevant angle. This angle is usually the angle that the vector makes with a given direction such as north, the positive x-axis, the horizontal or vertical and so on.

SkillSHEET 13.3 Using trigonometric ratios

WoRked examPle 11

Find the magnitude and direction, relative to the positive x-axis, of the vector (3, −2). Think 1

dRaW/WRiTe

Draw a diagram of the vector and denote it as a with the angle between a and the positive   x-axis as q.

2

The magnitude of a is the length of the line  the vector. segment representing

3

Use Pythagoras’ theorem to calculate this length.

4

Calculate the angle q using trigonometry.

5

State the solution with the angle down from the positive x-axis given as a negative.

y

θ

3

~a

x 2

a = 32 + 2 2  = 13 2 tan (q ) = 3 q = 33.7° The vector (3, −2) has a magnitude of 13 units and makes an angle of −33.7° with the positive x-axis.

The angle that a vector makes with the positive x-axis can be found using trigonometry. If the vector points in the negative x direction then you will need to add your found angle q to 90° or subtract it from 180° to find the required angle. See the diagram.

y 3

(−3, 3)

180° − θ

θ −3

0

Chapter 13

x

Vectors

471

Upward vectors are expressed as positive angles anticlockwise from the positive x-axis. Downward vectors are expressed as negative angles clockwise from the positive x-axis. In general, if r = (a, b) then the direction of r compared to the positive x-axis is found by   b appropriately adjusting q where tan (q) = . a

Vector components We have seen that two vectors may be added to give one resultant vector. The reverse process may be used to express one vector as the sum of two other vectors. This process is called ‘breaking the vector into two components.’ A vector can be broken into two perpendicular components such as x and y or north and east. It may be convenient to find the effect of a vector in a particular direction. We do this by breaking the vector into two components. ~F V A force F acting as shown will move an object to the right and upwards. ~  The force F can be separated into two component parts; one in the  H, and the other in the vertical direction, V . horizontal direction, H ~   F = H +V    The effect of the force in the horizontal direction is given entirely by H and the effect in the  vertical direction is given by V .  By breaking F into component parts in two perpendicular directions we can analyse the effect  or both of these directions. of the vector in one Worked Example 12

Write the horizontal and vertical components of a vector of magnitude 5 and angle of 120° with the positive x-axis. Think 1

Represent the vector on the Cartesian plane.

Draw/Write y

5

120° x

0 2

Construct a right-angled triangle with the vector as the hypotenuse and the other sides H for horizontal and V for vertical.

y

V

5 120° x

H 0 3

Calculate the angle between the vector and the x-axis and indicate it on the graph.

y 5 V

60° H 0

472


x

4

Calculate V using the sine ratio.

V 5 V = 5 sin (60°)

sin (60°) =

5 3 (or 4.33) 2 H cos (60°) = 5 H = 5 cos (60°) 5 = (or 2.5) 2 =

5

Calculate H using the cosine ratio.

6

State the solution, adding negative signs where necessary.

The vector has a horizontal component of and a vertical component of

5 3 . 2

WoRked examPle 13

a

1

5 2

eBook plus

A car travels 12 km in a direction N30°E. How far: a north b east of its starting point has it travelled?

Think

−

Tutorial


dRaW/WRiTe

Draw a vector diagram representing the motion of the car. Call the vector a and its eastern and northern components e and  n, respectively. 

a

N ~e ~n 30° ~a E

b

2

Calculate n (the magnitude of n) using the  cosine ratio.

3

State the distance travelled by the car to the north.

1

Calculate e (the magnitude of e) using the  sine ratio.

2

State the distance travelled by the car to the east.

n 12 n = 12 cos (30°) = 10.4

cos (30°) =

The car has travelled to approximately 10.4 km north of its starting point. e b sin (30°) = 12 e = 12 sin (30°) =6 The car has travelled to 6 km east of its starting point.

Chapter 13

Vectors

473

RememBeR

1. The magnitude of a vector r is denoted by r or r.   2. A vector represented by (a, b) has a magnitude equal to a 2 + b 2 and a direction with b the positive x-axis given by the appropriate adjustment to q, where tan (q ) = . a 3. A vector may be broken into two component parts, usually in perpendicular directions. exeRCiSe

13C eBook plus

magnitude, direction and components of vectors 1

Digital doc

SkillSHEET 13.3 Using trigonometric ratios

474

We11 Calculate the magnitude and direction, relative to the positive x-axis, of the following displacements. c (2, 4) d (1, 1) a (6, 2) b (4, −1) − − e ( 2, 1) f ( 1, 4) g (1, 0) h (−2, −2)

2

Refer to the diagram of the cube shown. If the sides of the F G cube are 1 unit in length, write the magnitudes of these vectors in B C exact form: → → → → E a OA b AB c OB d OD D → → → → e AD f DF g OE h EF A O → → i OF j AG 3 We12 Write the horizontal and vertical components of these vectors. Write your answers in exact form where possible. a Magnitude 2, angle of 60° with the x-axis b Magnitude 3, angle of 150° with the x-axis c Magnitude 10, angle of −60° with the x-axis d Magnitude 2, angle of −120° with the x-axis e Magnitude 20, angle of 45° with the x-axis f Magnitude 4, parallel to the y-axis g Magnitude 12, parallel to the x-axis h A speed of 30 m/s vertically downwards i A move of 10 m to the left at an angle of 30° downwards from the x-axis j A move of 20 m to the right at an angle of 30° upwards from the x-axis k A speed of 50 m/s horizontally to the right l A force of 40 N at an angle of 20° to the horizontal m A force of 98 N vertically downwards n A force of 1250 N at an angle of 15° to the horizontal. 4

A vector has a horizontal component of x (< 0) and a vertical component of y (> 0). Write the magnitude and direction from the positive x-axis of the vector.

5

We13 A yacht sails 32 km in a direction S25°E. How far a south b east of its starting point has it travelled?


6 Justine cycles 8 km in a northerly direction. She then travels 6 km in an easterly direction. Calculate the magnitude and direction of her displacement. eBook plus Digital doc

WorkSHEET 13.2

13d

7 For the following pairs of vectors, calculate the magnitude and direction of a + b and a − b.     a a = 10 km north and b = 6 km north-east   b a = 25 units east and b = 20 units S30°W   c a = 10 units and b = 8 units in the opposite direction   d a = 12 km west and b = 12 km south   e a = 20 km and b = 15 km in the same direction   f a = 50 units in a direction 300 °T and b = 40 units in a direction 30 °T  

i , j notation  unitvectors 1. A unit vector is any vector with a magnitude or length of 1 unit. y 2. The vector i is defined as the unit vector in the positive x direction.  3. The vector j is defined as the unit vector in the positive y direction.  a displacement of d = (2, 5) represents a move of 2 units in the For example, j ~  positive y direction. positive x direction and 5 units in the x O ~i An alternative way of representing this is d = 2i + 5 j    Any vector in two dimensions can be represented as a combination of i and j vectors, the  the coefficient of j coefficient of i representing the magnitude of the horizontal component and   representing the magnitude of the vertical component. y In general we may represent any two-dimensional vector r as:  y r = xi + yj where x, y ∈ R    ~r O

x x

WoRked examPle 14 a Draw a vector to represent a = 3 i − j .







b Find the magnitude and direction of the vector a. Think a

1



Draw axes with i and j as unit vectors in the  x and y directions respectively.

dRaW/WRiTe a

y 2

j ~

1 −2 −1 O −1 2

Represent 3i − j as a vector from 0 that is  3 units in the positive x direction and 1 unit in the negative y direction and mark the angle between a and the x-axis as q. 

1

2

3

~i

x

y 2 1 −2 −1 O −1

j ~

θ ~a

3

~i

x

Chapter 13

Vectors

475

b

a = 32 + (− 1)2  = 10

1

The magnitude of a (that is, a ) may be found   using Pythagoras’ theorem.

2

Find the value of angle q using the tangent ratio.

tan (q) =

Give the direction of vector a relative to the  positive x-axis.

Vector a makes an angle of −18.4° from  the positive x-axis.

3

b

1 3

q = 18.4°

As we have seen, angles are usually given with respect to the positive x direction. We may generalise this procedure: y For any vector, r: 1. r = xi + yj  ~r    yj 2. Magnitude r , r = x 2 + y2 ~ θ   3. The direction from the positive x-axis is given by appropriately x xi~ O y adjusting q where tan (q) = . x Addition, subtraction and multiplication by a scalar for a vector in i , j form follows the rules   of normal arithmetic with each component treated separately. If a = x1 i + y1 j and b = x2 i + y2 j      a + b = ( x1 + x2 ) i + ( y1 + y2 ) j     a − b = ( x1 − x2 ) i + ( y1 − y2 ) j     ka = k x1 i + k y1 j    Worked Example 15

If a = 3 i + j and b = − 2 i + 5 j, express in i , j form         a a + b b 2 a − b.     Think

a Add the i components and j components separately.



b



1

2 a is calculated by multiplying the i and j    components of a by 2. 

2

2a − b is calculated by subtracting the i and j    components of b respectively from 2a.  

Write a a + b = (3 i + j ) + (− 2 i + 5 j )





    = 3i − 2 i + j + 5 j     = i +6j   2a = 2(3i + j) b    = 6 i + 2 j.   2a − b = 6 i + 2 j − ( − 2 i + 5 j )       = 6i + 2 j + 2i − 5 j     = 8i − 3 j  

Worked Example 16

→

→

OA = 3 i + j and OB = − i + 4 j.   →   → a Represent OA and OB on a diagram.

476


→

b Find, in terms of i and j , the vector AB .





→

c If M is the midpoint of AB, find the vector OM in terms of i and j .



Think a



draw/Write

1

Draw axes with i and j as unit vectors in the  x and y directions respectively.

2

Represent OA as 3i + j and OB as − i + 4 j on     the axes.

→

a

→

1

A

−1 O −1 b

c

→ → → may be expressed as AO + OB using the

1

AB triangle rule for adding vectors.

2

Change AO to negative OA .

3

Express this in i , j form.  

4

Simplify.

1

Mark the point M in the middle of AB.

→

→ → →

→ →

= − OA + OB

→

AB = − (3i + j) + ( − i + 4 j)     c

= − 4i + 3 j  

B

y 4 3 2 1

−1 O −1

→

→

2

Express OM as the sum of OA + 12 AB .

3

Express this in i , j form.   Simplify.

4

x

1 2 3

b AB = AO + OB

→

→

→ OA = 3i + j   → − OB = i + 4 j  

y B 4 3 2

M A x

1 2 3

→ → 1→ OM = OA + AB 2 1

= (3i + j) + 2 (− 4 i + 3 j)     = i + 2.5 j  

REMEMBER

1. Any two-dimensional vector may be written in the form r = xi + y j, where i and j are      unit vectors in the x and y directions, respectively. 2. r = x 2 + y 2  3. The angle made by r with the positive x-axis is given by appropriately adjusting θ,  y where tan (θ) = . x 4. Vectors may be added, subtracted or multiplied by a scalar in i , j form by adding, subtracting or multiplying the i and j components separately.   

Chapter 13 Vectors

477

exeRCiSe

13d eBook plus Digital doc

Spreadsheet 143 Vectors

i , j notation 1 We  14a Draw a vector to represent each of the following. a 4i + 3 j b 4i − 3 j c 2i + 2 j       d i− j e 4i + j f 5i      g −6j h − 2i i − 8i − 6 j     − j 5i + 12 j   2 We 14b Calculate the magnitude and direction of each of the vectors in question 1. 3 a d g j

If a = 3i + 2 j, b = i − j and c = − 2 j, find the following in i , j form.           3a b a+b c a−c      2b e a+b+c f 2b − c       3a + 2b + c h 4c i 4c − a       3c − a − b   

4 We 15 a u  d u−v   5

If u = 2 i − 3 j and v = 3i + j, find the following in exact form.       b u+v c 3v   

Represent the following position vectors in the form xi + y j.   y 4 (1, 3) 3 (2, 3) 2 ~b ~a ~h 1

(−4, 3) ~c

−6 −5 −4 −3 −2 −1 0 −1 d e ~ ~ −2 (−6, −2) −3 (−2, −3) −4

6 We 16

→

(5, 1)

1 2 3 4 5 6 x f ~ (1, −3)

g ~ (6, −4)

→

OA = 2 i − j and OB = 4 i + 3 j    

→

→

a Represent OA and OB on a diagram.

→

b Find, in terms of i and j, the vector AB .   → c If M is the midpoint of AB, find the vector OM in terms of i and j .   7

→

→

→

→

c AC

d ON where N is the midpoint of OB

e AB

f MN

→

478

→

OACB is a rectangle in which the vector OA = 4 i and OB = 6 j. Express the following in   terms of i and j .   → → → a OC b OM where M is the midpoint of OA

→


8

The position of the points A, B and C is defined by:

→

→

→

OA = 4 i , OB = 10 i + 2 j and OC = 4 i + 4 j      a Find the vectors representing the three sides of the triangle ABC (that is, find in i , j   → → → form the vectors AB, AC and BC ). b Calculate the magnitude of these three sides. Leave answers in exact form. c What type of triangle is ABC? 9

→

→

→

M, N and P are three points defined by: OM = − i + j, ON = i + 4 j and OP = 5i + 10 j       → → a Find MN and NP .

→

→

b Show that MN and NP are parallel vectors. a = 4 i − 2 j and b = − 3i + j       a Find 3a − 2b and 3a + 4 b.     b Explain why 3a + 4 b is parallel to the y-axis.   11 mC The magnitude of the vector 2 i + 2 j is:   A 2+2 B 2 2

10

D 2

E

C

6

2+2

12 mC If a = 3i − 5 j and b = − 3i − 2 j, then a − 2b equals:         A 9i − j B 9i + j C − 3i − j       D − 3i + j E − 4i − 9 j     13 mC The vectors u = 2 i and v = 6 i − 2 j. The magnitude of u + v is:        A 68 B 60 C 40 + 4 D

32 + 2

E 6

14 mC The angle the vector 3i − 4 j makes with the positive x-axis is nearest to:   53° A 37° B C −53° D −37° E −127° 15

Find the vector a + b, which represents the planned shot of a pool player.   ~b = 1.

6i ~ − 0.4 ~j

8

~a =

2.3 ~i +

3.1 ~j

y


x

Investigation Angle between two vectors in i, j notation  

16

Vector m = 12 i + x j . The magnitude of m is 13. Find the value of x.     Chapter 13

Vectors

479

13e eBook plus Interactivity

int-0980 Applications of vectors

applications of vectors Vectors have a wide range of applications such as in orienteering, navigation, mechanics and engineering. Vectors are applied whenever quantities specified by both magnitude and direction are involved. When solving problems involving vectors: 1. Draw a vector diagram depicting the situation described. 2. Use the appropriate skills to answer the question being asked.

WoRked examPle 17

A boat is being rowed straight across a river at a speed of 6 km/h. The river is flowing at 2 km/h. If i is the unit vector in the direction that the river is flowing and j is the unit vector in the direction   of i and j. Hence, find the straight across the river, represent the velocity of the boat in terms  place magnitude and direction of the velocity of the boat correct to 1 decimal Think 1

dRaW/WRiTe y

Draw a set of axes with i in the direction of the positive x-axis and j in the direction of the  positive y-axis.

j ~

x

O 2 3

Indicate the velocity vector of the boat, v,  6). starting at O and finishing at the point (2,

~i

v = 2i + 6 j   

y

j ~

6

~i

Represent the velocity of the boat in terms of i and j.  

~v

O 4 5 6

The magnitude of v is 22 + 62 .  Evaluate the magnitude correct to 1 decimal place. Draw a right-angled triangle with v as the  hypotenuse and q as the angle between v and  the i direction. 

Express q using the tangent ratio.

x

v = 22 + 62  = 40 ≈ 6.3 y 6 ~v

O 7

2

θ

6

2

tan (q ) =

2

x

6 2

=3 8

Evaluate q correct to 1 decimal place.

9

State the magnitude and direction of the velocity of the boat.

q = 71.6° The velocity of the boat has a magnitude of approximately 6.3 km/h and is directed at approximately 71.6° from the riverbank.

Note: The magnitude of velocity is referred to as speed.

480


WoRked examPle 18

eBook plus

An aircraft is heading north with an airspeed of 500 km/h. Tutorial A wind of 80 km/h is blowing from the south-west. Using i and int-1171  Worked example 18 j as unit vectors in the directions east and north respectively:  Represent the aircraft’s air velocity in terms of i and j. a b Represent the aircraft’s exact ground velocity in terms of i and j.  its ground speed. c Hence, find the direction in which the aircraft is heading and Think

dRaW/WRiTe

a Express a in terms of i and j.

a a = 500 j

b

b

1

2

Indicate the vector representing the wind speed, w, by placing its tail at the head of the firstvector, directed in a direction 45° from the north with a magnitude of 80, since the wind speed is 80 km/h from the south-west.

4

Represent the combined effect of the two speeds with a vector v using the triangle  rule.

5

Express w, exactly, in terms of i and j  trigonometry.   using basic

6

Express the aircraft’s ground velocity, v,  as the sum of a and w.   Express in terms of i and j.   Indicate the angle between v and the  y-axis as q.

1 2

Use the tangent ratio to evaluate q to 1 decimal place. The length of the horizontal component of v is 40 2 . The  length of the vertical component of v is  500 + 40 2 .



N



y

y

80 45° w ~ 500

E

W S

Indicate the vector representing the aircraft’s airspeed, a, starting at O and  (0, 500). finishing at the point

3

7

c

   Draw a set of axes with i in the direction of the positive x-axis and j in the  direction of the positive y-axis.

~a

j ~

~i

O

x

~v

O

x

w = 80 sin (45°) i + 80 cos (45°) j    = 40 2 i + 40 2 j   v = a+w   

c

= 40 2 i + (500 + 40 2 ) j   y

tan (q ) =

500 + 40 2 ≈ 0.1016 q = 5.8°

w ~ ~a O

40 2

~v

θ x

3

Calculate the magnitude of v correct to  1 decimal place.

v = (40 2 )2 + (500 + 40 2 )2  ≈ 559.4

4

State the direction and magnitude of the ground speed of the aircraft.

The aircraft is flying with a ground speed of approximately 559.4 km/h in a N5.8°E direction.

Chapter 13

Vectors

481

Statics When the vector sum of the forces acting on a stationary particle is zero, then the situation is said to be static and the particle will remain stationary. The particle is also said to be in equilibrium. In the case of two forces, we have the situation shown at right. In the case of three forces, we have the situation shown in the diagram P F ~ ~ below, left. Where the three forces are acting so that the particle is in equilibrium, the lines representing the forces can be rearranged into a triangle of forces (diagram below, right) since their vector sum is zero. Hence, problems can be solved using trigonometry (including the sine rule and cosine rule) and sometimes Pythagoras’ theorem. ~F R ~ R ~

~P

~F

~P

+ + =O ~F ~P R ~ ~

Note: The three forces are still acting in the same direction and have the same magnitudes (or lengths) as they did in the ‘real’ situation. Worked Example 19

Three forces are acting on the particle P as shown in the diagram. Force A is vertically up and has a magnitude  of 20 N (20 newtons) while Force B is horizontally to the right  particle is in equilibrium, and has a magnitude of 40 N. If the find the magnitude of the Force C to the nearest tenth of a  nearest tenth of a degree. newton and give its direction to the Think 1

Draw the three forces as a triangle of forces.

2

Label the angle between the forces A and C   as θ.

3

4 5

Calculate C using Pythagoras’ theorem. 

Evaluate C correct to 1 decimal place.  Evaluate θ using the tangent ratio.

A ~ 20 N

P 40 N

~B

C ~

Write A ~ 20

C ~ ~B

40 2

2

2

C = A +B    = 20 2 + 40 2 = 400 + 1600 = 2000 C = 2000  C = 44.7 newtons  40 tan (q ) = 20 −

q = tan 1 (2) q = 63.4° 6

482

State the answer to the question.

The force has a magnitude of 44.7 N and it is acting downwards at an angle of 63.4° from the vertical.


Geometric proofs Vectors can also be used to prove a range of geometric theorems. From earlier in the chapter, you will remember that two vectors are equal if they are equal in magnitude, are parallel and point in the same direction. One important vector property that is useful in geometric proofs is that if a = kb , where k ∈ R (k ≠ 0), then the two vectors, a and b are parallel.   WoRked examPle 20

eBook plus

Show that the line joining the midpoint of two sides of a triangle is parallel to the third side and equal to half of its length.

B

Tutorial

int-1172

M A

Think 1

2

3

C

WRiTe

→ Let side AB represent vector AB and side BC → represent vector BC . Use the symbol a for  → → vector AB, and b for vector BC.  → Let side AC represent vector AC. Express AC in terms of a and b. → Express MN in terms of a and b.  

→ → AB = a and BC = b  

→ AC = a + b   → → → MN = MB + BN 1 → 1 → = 2 AB + 2 BC 1

4

5 6

Worked example 20

N

1

Simplify the expression by taking out 2 as a common factor.

= 2a+ 2b   1 = 2 (a + b)  

→ → Express MN in terms of AC.

1 → = 2 AC

1

→ → → MN is parallel to AC since AC is a → multiple of MN.

→ → Therefore, MN is parallel to AC and its length → is half the length of AC.

RememBeR

When solving problems involving vectors: 1. Draw a vector diagram depicting the situation described. 2. Use the appropriate skills to answer the question being asked. exeRCiSe

13e

applications of vectors 1 We17 A boat is being rowed straight across a river at a speed of 7 km/h. The river is flowing at 2.5 km/h. If i is the unit vector in the direction that the river is flowing and j is the unit   in terms of i vector in the direction straight across the river, represent the velocity of the boat  and j. Hence, find the magnitude and direction of the velocity of the boat.  Chapter 13

Vectors

483

2 A boat is being rowed straight across a river at a speed of 10 km/h. The river is flowing at 3.4 km/h. Find the magnitude and direction of the velocity of the boat. 3 WE18 An aircraft is heading north with an airspeed of 650 km/h. A wind of 60 km/h is blowing from the south-west. Using i and j as unit vectors in the directions east and north   respectively: a represent the aircraft’s airspeed. b represent the aircraft’s ground speed in terms of i and j.   and its ground speed. c hence, find the direction in which the aircraft is heading 4 An aircraft is heading south with an airspeed of 600 km/h. A wind of 50 km/h is blowing in a S30°W direction. Find the direction in which the aircraft is heading and the ground speed. 5 Forces of 3i + 4 j and 2 i + 2 j act simultaneously on an object.   and direction  Find the magnitude of the resultant of the two forces. 6 Forces of 5i − 4 j, 3i − j and −2 i + 3 j act simultaneously on an object.   of the resultant of the three forces.  and direction Find the magnitude 7 A hiker is located at a position given by (8, 6) where the coordinates represent the distances in kilometres east and north of O respectively. If a campsite is at a position given by (3, 2), find the distance and direction of the hiker from the campsite. 8 A hiker is located at a position given by (−5, 3) where the coordinates represent the distances in kilometres east and north of O respectively. If a campsite is at a position given by (3, −2), find the distance and direction of the hiker from the campsite. 9 A bushwalker starts walking at 8.00 am from a campsite at (−4, 8), where the coordinates represent the distances in kilometres east and north of O respectively. After 1 hour she is at (−2, 6.5). → → Take i and j as unit vectors along OX and OY .  a Write, in terms of i and j, her position at the   start and after 1 hour. b Calculate the distance travelled in 1 hour. c She then continues at the same rate and in the same direction. What is her position vector after: i 2 hours? ii 3 hours? d Show that her position t hours after 8.00 am is given by: r 1 = ( − 4 + 2t ) i + (8 − 1.5t ) j    Another bushwalker commences walking from his campsite also at 8.00 am. His position is given by: r 2 = (7.4 − 1.8t ) i + (2 + 0.5t ) j    e What are the coordinates of this bushwalkers campsite?

484


f What is his position after 2 hours of walking? g By equating i and j components, show that the two bushwalkers meet.   each campsite that each bushwalker has travelled when they meet. h Find the distance from 10 The i , j system may be extended to three dimensions  vector k in the z direction. witha unit Take i , j and k as unit vectors in the directions east,  vertically  up respectively.  north and z

x

~i

~k O

y

j ~

Frank travels 2 km in a direction N30°E from O to a point A. He then climbs a 100 m high cliff. → a Write the vector OA in i , j form.   b Calculate how far Frank has travelled to the north of his starting point.

→ → c If T represents the top of the cliff, write down the vectors AT and OT using i , j, k    components. → d Calculate the magnitude of OT.

(1

2t

−

10

t 2) ~j

y

t ~i +

The position vectors for an arrow and a moving target are shown at right, where t is the time in seconds since the target began to move, and h is the height of the target. If the arrow is to hit the target, when must this happen and what must the value of h be for this to occur?

40

11

hj 5)i~+ ~ 3 + (5t

h x

12

Forces of − 2 i + 3 j, 4 i − 5 j, xi + j and 3i − yj act on a particle which is in equilibrium.         Find the values of x and y.

13 WE19 Three forces are acting on the particle P shown. Force A is vertically up and has magnitude of 16 N while force B is horizontally to the right and has a magnitude of 28 N. If  the magnitude of the force C to the nearest tenth of a newton the particle is in equilibrium, find  and give its direction to the nearest tenth of a degree. A ~ 16 N

P 28 N

~B

C ~

Chapter 13 Vectors

485

14 Three forces are acting on the particle P shown. Force A has a magnitude of 35 N while force  B has a magnitude of 40 N. If the particle is in equilibrium, find the magnitude of the force C to the nearest tenth of a newton and give its direction to the nearest tenth of a degree.  C ~ A ~ 35 N

P 120° 40 N ~B

15

WE20 PQR is a triangle in which M is the midpoint of QR. Prove that → → 1 → PM = 2 (PR − QP).

P

Q

16

486

M

R

Prove that if the midpoints E, F, G and H of a rhombus ABCD are joined, then a parallelogram EFGH is formed. (Extension: Show that the parallelogram is, in fact, a rectangle.)


Summary Introduction to vectors

• A scalar quantity is specified by magnitude or size only. • A vector quantity is specified by both magnitude and direction. • Vectors can be represented by directed line segments, as in this diagram. → • A vector can also be denoted by AB or a.  pair (a, b). • A vector can be represented by an ordered • Position vectors start at the origin. • Two vectors are equal if they are: (a) equal in magnitude (b) parallel (c) point in the same direction.

B

~a A

Operations on vectors

• Vectors are added using the triangle rule. • Subtraction of vectors is performed by using a − b = a + (− b).  isin the  opposite  The vector − b has the same magnitude as b but   direction to b.  of a vector by a scalar’ means that the vector is made larger or • ‘Multiplication smaller by a scale factor. • The vector ka is k times as big as a and in the same direction as a   if k > 0; if k < 0, then ka is in the opposite direction to a.  

~b

~a ~a + ~b −b ~

~a

~a − ~b

Magnitude, direction and components of vectors

• The magnitude of a vector r is denoted by r or r.   • A vector represented by (a, b) has a magnitude equal to a 2 + b 2 and a direction with the positive x-axis b given by appropriately adjusting θ where tan (q ) = . a • A vector may be broken into two component parts, usually in perpendicular directions. i , j notation  

• Any two-dimensional vector may be written in the form r = xi + yj , where i and j     are unit vectors in the x and y directions respectively.  • r = x 2 + y 2 

y ~r 0

θ

xi~

yj ~ x

• The angle made by r with the positive x-axis is given by appropriately adjusting θ,  y where tan (q ) = . x • Vectors may be added, subtracted or multiplied by a scalar in i , j form by adding, subtracting or multiplying   the i and j components separately.   Applications of vectors

• When solving problems involving vectors: 1. Draw a vector diagram depicting the situation described. 2. Use the appropriate skills to answer the question being asked.

Chapter 13 Vectors

487


1 On the same set of axes draw the following vectors: a = (3, −2), b = (0, 4) and c = (−2, 5). Calculate:  a+b   a   b 3c − 2b  c − c  → → → → → → 2 Show that CA + EB − ED + DC − DA = DB. 3 Write the horizontal and vertical components of a vector of magnitude 4 which makes an angle of 120° with the positive x-axis. 4 Let d = i + 4 j, e = −2i − 3 j and f = 4 i .        a Calculate the following. ii 3e iii 2e + f i d − e      b Write the magnitude and direction of d , e   and f .  → → 5 OG = 8i + 2 j and OH = − 4 i − 6 j    →  → a Represent the vectors OG and OH on a diagram. → b Find, in terms of i and j, the vector GH.  →  c Calculate the magnitude of GH. → 6 a If M is the midpoint of OG = 8i + 2 j and N is  →  the midpoint of OH = − 4 i − 6 j, what are the   → → vectors OM and ON in terms of i and j?  →  b Show that MN = − 6 i − 4 j.  → → c How are MN and GH related? 7 A distressed yacht is located at a position given by (43, 36) where the coordinates represent the distances in kilometres east and north of a port respectively. If a ship is at a position given by (50, 32), find the distance and direction of the yacht from the ship. 8 A boat is being rowed straight across a river at a speed of 9 km/h. The river is flowing at 3.2 km/h. Find the magnitude and direction of the velocity of the boat.

488

Multiple choice

Questions 1 to 4 refer to the figure below. 1 Start at the point (−4, −2). The coordinates of the point at which we finish after a ~b displacement equal to d is: A (−1, 2) ~a ~c B (−1, −2) − C (1, 2) ~d D (−1, 1) E (1, 2) 2 a − b is equal to:   A (3, 1) B (5, 1) C (5, 5) D (1, 5) E (1, 3) 3 A vector parallel to c :  A (2, 0) B (2, 2) C (10, 5) D (0, 8) E (1, 1) 4 Compared to c the vector -2 c would be represented  segment of: by a directed line A equal length with the arrow pointing up B equal length with the arrow pointing down C equal length with the arrow pointing to the right D double the length with the arrow pointing up E double the length with the arrow pointing down 5 Using the figure at right, y ~ choose the correct statement. B x − y = − z x A x + y = z    ~    ~z C − x + y = z D x + y = − z       E − x − y = z    → → → → 6 Expressed in simplest form, DE + FH − FG + EG equals: → → A 0 B DF C FE → → D DH E DG For questions 7 and 8 consider the position vector of (4, −8). 7 The magnitude of this vector is: A −4 B 4 D 4 3 E 4 5


C − 4 3

8 The angle between this vector and the positive x-axis is nearest to: A 63.4° B 26.6° C −63.4° − D 116.6° E 116.6° Questions 9 to 12 refer to: a = 2 i − j, b = − i − j and c = 4 i + 2 j          9 The magnitude of a is:  A 1 B 3 C 5 D 3 E 2

13 The magnitude of the resultant force is: A 101 B 7 C 109 D 17 E 5

10 The angle vector c makes with the positive x-axis,  θ, is found by solving: 1

A tan (θ) = 2

B tan (θ) = 2

D cos (θ) = 2

E tan (θ) =

1

C sin (θ) = 2

−1 2

11 The vector 2b + c is:   A 0 B parallel to the x-axis C parallel to the y-axis D equal in magnitude to a E a unit vector 12 a − b + c equals:    A 7 i + 2 j B 5i C 5i + 2 j      D 7 i E 7 i + 4 j    The following information applies to questions 13 and 14. Two forces, i − 3 j and 2 i + 7 j , act simultaneously on   an object.  

14 The direction of the resultant force is: A 53.1° clockwise from j  B 73.3° anticlockwise from i  C 53.1° anticlockwise from i  D 73.3° clockwise from j  E 53.1° clockwise from i .  15 A force of magnitude 18 newtons acts on a body at an angle of 150° in the anticlockwise direction to the vector i.  A vector representation of this force could be: − A 9 3i + 9 j B 9 i + 9 3 j C − 9 3i + 9 j       D − 9 i − 9 3 j E 9 3i − 9 j    

Extended response

1 A triangular course has been planned for a yacht race. Point O is the start and y A finish of the race. The race goes from O to A to B to O with the coordinates of A and B being (24, 16) and (36, 10) respectively. The coordinates represent B distances in kilometres east and north of O. Take i and j as unit vectors along the x- and y-axes. → →   a Write (in terms of i and j) the vectors OA and OB. O Land   → b Hence, show that AB = 12 i − 6 j.  → → → c Calculate the magnitudes of OA, AB and OB. d State the distance of the race. → e Write the angle that OA makes with the x-axis. → f Calculate the angle that AB makes with the x-axis and hence show that the bearing of B from A is 116.6°T.

x

Chapter 13 Vectors

489

g While travelling along the third leg of the race (from B to O), the yacht is subjected to a sudden gust of wind of 20 km/h from the north. If the yacht was travelling at 25 km/h towards O, draw a vector diagram to show the velocity, v, of the yacht. 

2 Use a vector method to show that the diagonals of a rectangle bisect each other. 3 A mass of 9.8 kg exerts a force of 98 N vertically down. It is suspended in equilibrium by a 50-cm piece of inextensible string with the ends fixed on the same horizontal level 40 cm apart. Determine the magnitude of the tension force, T, in the string and the  angle the string makes with the vertical. eBook plus Digital doc


490


40 cm T ~

T ~

98 N

eBook plus

aCTiViTieS


• 10 Quick Questions: Warm up with ten quick questions on vectors. (page 462) 13A

Introduction to vectors

Tutorial

• We5 int-1168: Watch how to draw a vector diagram to represent the path take by an aircraft. (page 464) Digital docs

• Spreadsheet 144: Introduction to vectors. (page 465) • SkillSHEET 13.1: Practise bearings. (page 466) • SkillSHEET 13.2: Practise angles of elevation and depression. (page 466) 13B

Operations on vectors

Tutorial

• We7 int-1169: Watch how to perform vector addition and subtraction. (page 468) Digital docs

• Spreadsheet 143: Investigate vectors. (page 470) • WorkSHEET 13.1: Use graphs to find vectors, represent vectors diagrammatically, solve worded problems, and use provided diagrams to create diagrams of vectors. (page 471) 13C

Magnitude, direction and components of vectors

Tutorial

• We13 int-1170: Watch how to calculate the distance north and east from a car’s starting point. (page 473) Digital docs

• SkillSHEET 13.3: Practise using trigonometric ratios. (pages 471 and 474) • WorkSHEET 13.2: Revision of solving and representing vectors, solve problems of magnitude and direction of vectors; apply knowledge of vectors to worded problem. (page 475)

13D

i , j notation  

Digital docs

• Spreadsheet 143: Investigate vectors. (page 478) • Investigation: Angle between two vectors in i, j notation. (page 479) 13E

Applications of vectors

Interactivity

• Applications of vectors int-0980: Apply your knowledge of vectors by using the interactivity. (page 480) Tutorials

• We18 int-1171: Watch how to determine the air and ground velocity, direction and speed of an aircraft. (page 481) • We20 int-1172: Watch how to show properties of a line joining the two midpoints of sides of a triangle. (page 483) Chapter review Digital doc


Chapter 13

Vectors

491

14

14A 14B 14C 14D

Force and tension Newton’s first law of motion Equilibrium — forces at an angle Connected bodies in equilibrium

statics of a particle AREAs oF sTudy

• Inertial mass, momentum, force and weight • Equations of motion using absolute units

• Description of equations of motion from a diagram showing all the forces acting on a body eBook plus

Note: Students should have completed the unit on vectors (Chapter 13) before attempting this chapter.

14A

Digital doc

10 Quick Questions

Force and tension

A force is the measure of the strength of a push or pull exerted on, or by, a body. Most forces are ‘contact forces’; that is, to push we must touch; to pull we usually touch or have a connecting string. However, gravitational and magnetic forces etc. act through space and thus are not contact forces. Mass is the amount of inertia a body possesses. Inertia is the tendency of an object in motion to remain in motion, or an object at rest to remain at rest, unless acted upon by a force. A sack of potatoes has greater inertia than a tomato, hence it has a greater mass, and, by definition, a greater weight. Weight is the gravitational force exerted on a body downwards by the Earth at the Earth’s surface. Weight W (measured in newtons) is calculated as the product of mass m (in kilograms) and gravitational acceleration g (in m/s2), where the value g = 9.8 m/s2 is the acceleration due to gravity. Be aware of the difference between weight and mass. Weight is the force acting on a mass at or near the Earth’s surface. As a body moves away from the Earth, mass remains constant but weight decreases. Thus, a tomato will weigh less at the summit of Mount Everest than at Bondi Beach. Mass is measured in kilograms; weight is measured in kilograms-weight (kg-wt) or newtons (N). (1 newton is the force that accelerates 1 kg at 1 metre per second per second). TN

m kg-wt

mg newtons

It follows that a body of mass m kg has a weight of mg newtons (on or near the Earth’s surface). mg N

492


Tension is the force exerted by a connecting string or cord when taut. For a vertical string suspending a mass, the tension in the string is equal to the weight (downward force) created by the mass. WoRkEd ExAMPlE 1

What is the weight of a mass of 3 kg? ThiNk

WRiTE

1

Mass is inertia.

Mass = 3 kg

2

Weight is mass multiplied by g.

Weight = 3g N

WoRkEd ExAMPlE 2

A mass of 0.5 kg is added to a mass of 200 mg. What is the total weight? ThiNk

WRiTE

1

Mass is inertia.

Mass = 0.5 kg + 0.2 kg = 0.7 kg

2

Weight is mass multiplied by g.

Weight = 0.7g N

Resolving a force Consider a force F making an angle of θ with the positive x-axis. y

j F

F F sin (θ )

θ

θ x

F cos (θ )

i

Just as with a vector, a force can be resolved into components. Resolving a force means splitting the force up into horizontal and vertical components. F = F cos (θ )i + F sin (θ ) j or x = F cos (θ), y = F sin (θ) WoRkEd ExAMPlE 3

eBook plus

Resolve a force of 5 N into two perpendicular components at angles of 30° and 60°. ThiNk 1

Draw a diagram to represent the situation. The two perpendicular components are the horizontal and vertical components.

Tutorial


WRiTE/dRAW j Force = 5 N

60° 30° 5 cos(30°)

5 sin(30°) i

Chapter 14

statics of a particle

493

2

The horizontal component can be calculated.

5 cos (30°)i = 4.330 N

3

The vertical component can be calculated.

5 sin (30°)j = 2.5 N

4

State the answer.

A force of 5 N of two perpendicular components, a horizontal component of 4.330 N and a vertical component of 2.5 N.

REMEMBER

1. A force is the measure of the strength of a push or pull exerted on, or by, a body. 2. Mass is the amount of inertia a body possesses. 3. Weight is the gravitational force exerted on a body downwards by the Earth at the Earth’s surface. Weight W (in newtons) is calculated as the product of mass m (in kilograms) and gravitational acceleration g (in m/s2). 4. Tension is the force exerted by a connecting string or cord when taut. For a vertical string suspending a mass, the tension in the string is equal to the weight (downward force) created by the mass.

Exercise

14A

Force and tension 1

Define mass? In which units is it measured?

2

Define weight? In which units is it measured?

3

Define tension? In which units is it measured?

4 WE 1

What is the weight of a mass of 10 kg?

5

A mass of 0.7 kg is suspended from a string. What is the tension in the string?

6

What is the mass of a stone which weighs 25 N?

7 WE3 A force of 3 N acts at 20° to the horizontal. Resolve the force into horizontal and vertical components. 8 A force of 10 N acts at an angle to the horizontal. Its vertical component is 8 N. Find the angle. 9 What is the horizontal component in question 8? 10 If v = 3i - 4 j represents a force, find its magnitude and the acute angle that the force makes    with the vertical. 11 Use trigonometry in question 10 to prove that the horizontal component of v is 3 and the  vertical component is -4. 12 If p = 2 i + 5 j represents one force and q = - 4 i - j represents a second force, use a vector to find     p - q. Find the angle that this combined method force makes with the horizontal.   13 MC A stone is suspended from a string. The tension in the string is 5 N. The mass of the stone is: 5 5 C N E kg A 5 N D 5 kg B 5g N g g

494


14 MC A force of 20 N acts at an angle to the horizontal. Its horizontal component is 3 N. The angle is: A sin 1 (0.3) B cos 1 (0.3) C tan 1 (0.3) D sin 1 (0.15) E cos 1 (0.15) 15 MC A eBook plus Digital doc

WorkSHEET 14.1

14B

13

If v = 2 i - 3 j represents a force, the magnitude of the force is:    C -1 D 25 B 5

E 5

16 MC If p = - i + 3 j represents a force and q = - 6 i - 4 j represents a second force,       -5p - q equals:   A -11i - 11 j B 11i - 11 j C 11i + 11 j D -11i + 5 j E - 7i - j          

Newton’s first law of motion Newton’s first law of motion applies to both stationary bodies and to bodies moving with constant speed. In this chapter, which deals with statics only, it is appropriate only to consider the former, that is, stationary bodies. ‘A stationary body will remain at rest unless it is acted on by an external force.’ Thus, if the vector total of all forces acting on a body is zero, the body will remain static (stationary). It follows then that: If a body is stationary, the vector total of all forces acting on the body is zero. Or expressed another way, for a body to be stationary: ‘To every action there is an equal but opposite reaction’. This statement implies that the equal but opposite reaction acts along exactly the same line but in the opposite direction. This means that for every ‘shove’ there is a ‘push back’ in return. For a particle to be in equilibrium, all forces acting on a body or on a system of bodies are in balance. The body or system of bodies is static (or moving with constant speed). For the studies covered in this chapter the body or system of bodies is stationary. What are the weaknesses in the system at right? Beam Hook

Consider what might happen: (a) Nothing. (a) In balance, forces are equal. Tension equals weight. Horizontal forces (not shown) are equal.

Light string Weight

(b) The weight falls (b) because: (i) the string breaks (i) If the string breaks, the weight must be too great for the strength of the string. Just as the Earth pulls down on the body, the string balances this by pulling up on the body. This creates a tension in the string. If the weight is greater than the physical ability of the string to create sufficient tension bear this weight, the string must break. A thick string is presumably stronger, certainly visually, than a thin string. A rope is stronger than a thick string, etc. The maximum tension a string can carry before breaking is known as its tensile strength.

Chapter 14


495

(ii) the hook comes out of the beam

(ii) This will happen if the upward force holding the hook into the beam, created by a combination of screw length and thread radius, is less than the downward tension in the string. Note that there must be a downward tension to balance the upward tension to hold the weight of the mass. Since these tensions are equal but opposite forces, they cancel each other out in the ‘big picture’. Consider the ‘big picture’ force diagram below: If the force, F, exerted by the hook is F = mg N, the system will hold. Be aware that the hook may be capable of holding a larger weight but only requires to apply enough force to balance the weight. For the hook itself in isolation F = T2 For the mass itself in isolation T1 = mg N C Beam

F = mg N T2 = mg N T1 = mg N Mass of m kg

W = mg N

(iii) the beam falls.

(iii) The beam will fall if the string doesn’t break, the hook doesn’t fall out and the combined forces, C, supporting the beam are less than the weight. i.e. if C ≤ mg

WoRkEd ExAMPlE 4

eBook plus

a Find a force P that is equal but opposite to the force F = 2 i - 5 j .





b Find the magnitude of P.

 c Find a force M such that | M| = 2| P|.    ThiNk

a P is in the same line but acting in

 opposite direction to F. This the  implies that there is a rotation of 180° anticlockwise. Use a graph or another method to find P. 



Tutorial




WRiTE a P=



- 2i



+ 5j 

P ~ −2i + 5j

j

180° i

2i − 5j F ~

496


b c

The magnitude of P is the absolute value of P, |P|.    We know that |P| = 29. Use this  information to determine a force M . 

b P =



c

((- 22 ) + 52 = 29

P = 29  ⇒ 2 P = 2 29  ⇒ 2 P = 116  ⇒ M = 116 = 16 + 100  ⇒ M = - 4 i + 10 j   

REMEMBER

1. A stationary body will remain at rest unless it is acted on by an external force. 2. If a body is stationary the vector total of all forces acting on a body is zero. 3. To every action there is an equal but opposite reaction. This means that for every ‘shove’ there is a ‘push back’ in return. Exercise

14B

Newton’s first law of motion 1

What is a force?

2

In what units is force measured?

3

a What is an equal force? b What happens if an equal force acts in the opposite direction: i in the same line of action ii in a different line of action?

4

A woman is standing on a surface. Explain the three scenarios that may occur. Namely: a she rises from the surface b she falls through the surface c she remains standing on the surface.

5

Mary-Alice is trying to prevent the evil burglar ‘black-hearted Ned’ from entering her house. Ned is pushing the front door to open it, Mary-Alice is pushing against the door to keep it shut. Describe what might happen to the door using such words as force, equal, opposite, greater than, less than, etc. The following information relates to questions 6–9. A stone hangs from an light inextensible string (it cannot stretch) suspended from a beam by a hook. The stone has a mass of 2 kg.

6

What is the tension in the string in Newtons?

7

If the string in has a tensile strength (breaking tension) of 35 N, how many more kilograms of stone can be added to the existing stone before the string breaks?

8

If the hook can exert a maximum force of 30 N, what is the maximum mass of stone that it can hold? Will this weight break the string? Explain.

9

The beam in question is not well supported and it will fall from its position if a force greater than 27 N is applied to it. A total weight of 37 N is suspended from the string. Describe what happens.

Chapter 14 Statics of a particle

497

10

Draw a force diagram for the following situation: a force of 10 N from the left comes up against a force of 9 N from the right. What is the resultant force and in what direction?

11

Mary-Alice is pushing against a door with a force of 160 N. Ned, the burglar, is pushing in the opposite direction with a force of 280 N. Mary-Alice’s niece Christine is able to assist with a force of up to 130 N. What will happen? Write down a suitable inequation showing this.

12

MC A stone is suspended from a string. The tensile strength of the string is g N. The

maximum mass of the stone is: B g2 N

A 1 N 13 14

C

1 N g

D g kg

E 1 kg

A mass of 12 kg is suspended from a string hanging from a hook. What is the tension in the string? MC A mass of g kg is suspended from a string hanging from a hook. The tension in the

string is: B g2 N

A 1 N 15

C

1 N g

D g kg

E 1 kg

What is wrong with the diagram below showing a mass resting on a surface?

10 N 25 N

16

WE 4 Draw the vector u = 4 i - 3 j . If this vector represents a force, state the equal but  opposite force in terms of u. Find the magnitude of u in newtons and state the magnitude of   the equal but opposite force.

17 MC In question 16, the acute angle that u makes with the i axis is:   A 143.13° B 53.13° C 126.87° D 36.87° E -53.13° 18 MC A body hangs from a light inextensible cord. If its mass is m then the tension (T) in the cord is: A T = m B T ≥ m C T ≥ mg D T = mg E T + mg = 0 The following information relates to questions 19 and 20. A force is represented by the vector u = 3i - 2 j N.    19 MC The acute angle that this force makes with the i axis is:  -1 2 -1 2 -1 - 2 - -2 - -3 A tan 3 B sin 3 C cos D sin 1 3 E tan 1 2 3

()

()

( )

( )

( )

20 MC The magnitude of the force is: A 13 kg

14C

B

-

13 kg

C 13 N

D - 13 N

E 1 N

Equilibrium — forces at an angle A body is in equilibrium when the forces acting on it are in balance. If the forces act along different lines of action it is usual to resolve the forces along two suitable perpendicular axes. This is done by using simple trigonometric ratios and the sine or cosine rules.

498


WoRkEd ExAMPlE 5

A mass of 10 kg is suspended from a string. The string is pulled horizontally by a force of 150 N. Find the angle this force makes with the angled part of the string. ThiNk 1

WRiTE/dRAW

Draw a diagram to represent the situation described.

T

j

150 N

α° θ°

i

10g N 2

Resolve the tension horizontally and vertically (i - j) using the acute angle for convenience.

T sin(α)

T

j α ° θ° T cos(α)

150 N i

10g N 3

Equate forces acting opposite to each other in the same line of action. Solve for a °. The angle required, θ, is the supplement of a °.

T cos (a °)i = 150i T sin (a °)j = 10g j T sin (a ) 10 g ⇒ = T cos (a ) 150 10 g ⇒ tan (a ) = 150 -  10 g  a = tan 1   150  ⇒ a = 33.16° Required angle, θ = 180 – 33.16 = 166.84°

⇒

WoRkEd ExAMPlE 6

eBook plus

A mass of 5 kg is suspended from a beam by two strings as shown in the diagram. Find the tension in each string.

10 cm

Tutorial


5 cm T2

T1

Mass of 5 kg

Chapter 14


499

Think 1

Re-draw the right angled triangle and use trigonometric ratios to determine the angles a and b.

Write/draw 10

α

β

5

cos (a ) =

5 10

-  5 ⇒ a = cos 1    10 

cos (α °) = 0.5, thus α = 60° ⇒ b = 90 - 60 = 30° 2

Resolve the tension horizontally and vertically using the acute angles found in the previous step.

T2 sin(60°)

T1 sin(30°)

T2 cos(60°)

T1 cos(30°)

5g 3

Equate forces acting opposite to each other in the same line of action.

T1 cos (30°) = T2 cos (60°) 3 1 T1 = T2 2 2 ⇒ 3T1 = T2 ⇒

[1]

T1 sin (30°) + T2 sin (60°) = 5g ⇒

1 3 T1 + T = 5g 2 2 2

⇒ T1 + 3T2 = 10 g 4

Solve the simultaneous equations [1] and [2] to determine the tension in each string.

Substitute equation [1] into equation [2]. T2 = 3T1 ⇒ T1 + 3 ( 3T1 ) = 10 g ⇒ 4T1 = 10 g 5 g 2 5 ⇒ T1 = × 9.8 N 2 T1 = 24.50 N ⇒ T1 =

T2 = 3T1 ⇒ T2 = 3 × 24.50 ⇒ T2 = 42.44 N

500


[2]

REMEMBER

1. A body is in equilibrium when the forces acting on it are in balance. 2. If the forces act along different lines of action it is usual to resolve the forces along two suitable perpendicular axes. This can be done using simple trigonometry ratios or the sine and cosine rules. Exercise

14C

Equilibrium — forces at an angle 1

What is equilibrium?

2

A bullet is fired from a gun. Is it in equilibrium? Explain.

3

A ball falls off the roof of a tall building. Is it in equilibrium? Explain.

4

A door is pushed with a force of 150 N from the left. What additional force is required to establish equilibrium?

5 WE5 A ball of mass 10 kg suspended from a beam experiences a horizontal pulling force of 2 N to the right. Find the new tension in the string and the angle at which the string makes with the beam. 6 WE6 A mass of 3 kg is suspended from a beam by two strings, 5 cm and 12 cm long. If the strings are perpendicular to each other, find the tension in each string. 7 Two strings 3 metres and 6 metres long are attached to two points on a beam 7 metres apart. The two strings are knotted together and a third string carrying a bob is attached to the knot. If the tension in the 3-metre string is 5 newtons, find the mass of the bob. 8 Two strings 5 metres and 9 metres long are attached to two points. The two strings are knotted together making an angle of 120°. A third string carrying a ball is attached to the knot. If the mass of the ball if 2 kg, find the tensions in the strings. 9 A stone of weight M newtons suspended by a string is pulled sideways by a force of 3 newtons acting horizontally. If the tension in the string is 4 newtons, find M. 10 In the diagram below, the three forces are in equilibrium. Find P. 5N PN

30° 38.68° 4N

11

A stone of weight M newtons, suspended by a string, is pulled sideways by a force, v = 3i newtons. If the tension in the string is t = - ai + 3 j newtons, find a and M.     12 A stone of weight M newtons, suspended by a string, is pulled sideways by a force, v = 2 i - 5 j newtons. If the tension in the string is t = -bi + 9j newtons, find M and b.     13 Two perpendicular strings are attached to two points on a beam. The two strings are knotted together and a third string carrying a bob is attached to the knot. If the tension in the string is t = - 5i + 6 j newtons, find the mass of the bob.    14 A mass of 3 kg is suspended from a beam by two strings, 5 cm and 10 cm long. If the strings are perpendicular to each other find the tension in each string.


501

Find the single force that must be added to balance the forces t = 5i + 3 j newtons and    d = - 4 i - 7 j newtons, which act at the same point.    16 In question 15, what is the magnitude of the single force and in which direction does it act?

15

17 Find P and Q in the diagram below, which represents forces acting on a body in equilibrium. 7N

30°

QN

60°

PN

The diagram below relates to questions 18–20. The figure shows a body in equilibrium under the actions of three forces. RN PN

α°

θ°

MN


WorkSHEET 14.2

14d

18 MC Which one of the following is true? A P + M sin (θ °) = R cos (a °) B P + M cos (θ ° ) = R sin (a °) C R = M cos (θ °) D P+R+M=0 E P + M cos (θ °) = R cos (a °) 19 MC Which one of the following is true? A P+M +R=0 B P-M +R=0 C P+M -R=0          D P-M -R=0 E P+M = R       20 MC Which one of the following is true? A a ° = θ ° B a ° + θ ° = 90° C a ° - θ ° = 90° D a ° + θ ° =180° E none of these

Connected bodies in equilibrium

eBook plus Interactivity int-0981

A complete system in equilibrium may contain more than one body. In these cases, it is often easier to deal with each body separately before looking at the ‘total picture’.


WoRkEd ExAMPlE 7

The forces acting on two bobs suspended from light inextensible strings are shown in the diagram at right. Find T and m.

30° TN mg N

502


T1 N

T2 N 60° TN

5g N

ThiNk 1

WRiTE/dRAW

Separate the two bobs. Draw a diagram representing the forces applied to the right bob.

T1 sin (30°) N

TN

T1 cos (30°) N

5g N 2

T1 sin (30°) = 5 g ( j direction)

Calculate the force T.

⇒ T1 = 10 g T = T1 cos (30°) (i direction) = 5 3g 3

Draw a diagram representing the forces applied to the left bob.

T2 sin (60°) N

T2 cos (60°) N

TN

mg N 4

T2 cos (60°) = T (i-direction)

Calculate the value of m given T = 5 3g.

Since, T = 5 3g ⇒ T2 = 10 3g But, T2 sin (60°) = mg (j direction) ⇒ 10 3g ×

3 = mg 2

⇒ m = 15 5

T = 5 3g and m = 15

State the answers.

WoRkEd ExAMPlE 8

The forces acting on two bobs suspended from light inextensible strings are shown in the diagram at right. If the connecting string is at an angle of 10° to the horizontal find T and m.

eBook plus

T2 N 60° 10°

Tutorial

T1 N 30°

TN


TN mg N

5g N

Chapter 14


503

Think 1

Write/draw

Separate the two bobs. Draw a diagram representing the forces applied to the right bob.

T1 sin (30°)

T cos (10°)

T1 cos (30°)

T sin (10°) 5g 2

Calculate the force, T. Set up two equations. To solve these equations simultaneously, using a CAS calculator, let T1 = x and T = y. On the Main screen, using the soft keyboard, tap: • ) • {N Enter the equations as shown. Then press E. Note: Remember to multiply the answer 12.660 by g = 9.8 to get the answer shown at right.

T1 sin (30°) = 5g + T sin (10°) T cos (10°) = T1 cos (30°)

T = 124.1 N 3

Draw a diagram representing the forces applied to the left bob.

T2 sin (60°) T sin (10°)

T cos (10°)

T2 cos (60°)

mg

504

4

Calculate the value of m given T = 124.1 N.

T cos (10°) = T2 cos (60°) Given T = 124.1 N 124.1 cos (10) ⇒ T2 = cos (60) ⇒ T2 = 244.4 N T2 sin (60°) + T sin (10°) = mg 244.4 sin (60) + 124.1sin (10) ⇒m= 9.8 ⇒ m = 23.80 kg

5

State the answers.

T = 124.1 N m = 23.80 kg


[1] [2]

REMEMBER

A complete system in equilibrium may contain more than one body. In these cases it is often easier to deal with each body separately before looking at the ‘total picture’.

Exercise

14d

Connected bodies in equilibrium 1

WE7 Two masses of 3 kg and 7 kg are suspended from two points on a beam by light

inextensible cords, which are themselves connected by a third inextensible horizontal cord. Write the weights of these masses as forces in vector form. What is the total upward vertical force required to maintain the system in equilibrium? Write your answer in vector form. 2

In question 1, the cord from the beam to the 3 kg makes an angle of 30° with the horizontal. What is the tension in each of the strings?

3

For the situation described in question 1, write the tension in the cord holding the 3 kg mass in vector form.

4 WE8 Two masses of 4 kg and 10 kg are suspended (10 kg lower) from two points on a beam by light inextensible cords which are themselves connected by a third inextensible cord, which makes an angle of 60° to the vertical. Write the weights of these masses as forces in vector form. What is the total upward vertical force required to maintain the system in equilibrium? Write this in vector form. 5 For the situation in question 4, if the cord from the beam to the smaller mass makes an angle of 45° to the vertical, find the tension in each string and the angle that the other cord makes with the vertical. 6 Write the tension in the cord suspending the smaller mass in question 5 in vector form. For questions 7–12, use the diagram at right showing two connected masses of 2 kg and p kg: 7

60°

8 MC

2 kg

B T1 = pg E T1 sin (60°) = pg

C T1 cos (30°) = p

Which one of the following statements is true? B T1 cos (30°) = T2 cos (60°) E T1 + T2 = T

C T1 = T

Which one of the following statements is true?

4 3 3 D T2 = 4g

A T2 =

p kg


A T1 = T2 D T1 sin (30°) = T2 cos (60°) 10 MC

T

B T2 = 2g D T2 cos (60°) = 2g

A T1 = p D T1 cos (60°) = pg 9 MC

30° T

MC Which one of the following statements is true?

A T2 = 2 C T2 cos (60°) = 2 E T2 sin (60°) = 2g

T1

T2

4 3 g 3 E 2g + T2 = T B T2 =

C T2 = 4


505

11 MC


2 3 4g B T = 3 3 2 3 D T = g E pg + T1 = T 3 12 MC Which one of the following statements is true? A T =

2 3 2 3 D p = g 3

A p =

13

C T1 =

2 B p = g 3

C p =

2 3g

E pg − T1 = T

The diagram at right shows two connected weights of 6 N and rg N. Find the value of r.

T2 N

T1 N 20°

TN

50° 15°

TN rg N

6N

14

4 3

Use the diagram at right, in which the connecting string is horizontal, to find M in terms of a and b and m.

T1

α

β

T2

M kg

m kg

15 In the diagram in question 14, if T1 = 3i + 4 j and T2 = - 3i + 5 j find the values of M, m,       a and b. 16 Two masses are suspended between two walls by light inextensible strings as shown at right. Express both weights and all tensions in vector form. Use vectors to establish a relationship between M and m.

Wall T1

α T M kg

506


β T 2 T m kg

Wall

Summary Force and tension

• A force is the measure of the strength of a push or pull exerted on, or by, a body. • Mass is the amount of inertia a body possesses. • Inertia is the tendency of an object in motion to remain in motion, or an object at rest to remain at rest, unless acted upon by a force. • Weight is the gravitational force exerted on a body downwards by the Earth at the Earth’s surface. Weight W (in newtons) is calculated as the product of mass m (in kilograms) and gravitational acceleration g, (in m/s2). • Tension is the force exerted by a connecting string or cord when taut. For a vertical string suspending a mass, the tension in the string is equal to the weight (downward force) created by the mass. Newton’s first law of motion

• A stationary body will remain at rest unless it is acted on by an external force. • If a body is stationary the vector total of all forces acting on a body is zero. • To every action there is an equal but opposite reaction. This means that for every ‘shove’ there is a ‘push back’ in return. Equilibrium — forces at an angle

• A body is in equilibrium when the forces acting on it are in balance. • If the forces act along different lines of action it is usual to resolve the forces along two suitable perpendicular axes. This is done using simple trigonometry ratios or the sine and cosine rules. Connected bodies in equilibrium

• A complete system in equilibrium may contain more than one body. In these cases it is often easier to deal with each body separately before looking at the ‘total picture’.


507

chapter review Short Answer

1 If v = 6 i - 2 j represents a force, find its magnitude and theacute angle that the force makes with the vertical. 2 A mass of 15 kg is suspended from a string hanging from a hook. What is the tension in the string? 3 A mass of 7 kg is suspended from a beam by two strings, 7 cm and 24 cm long. If the strings are perpendicular to each other, find the tension in each string. 4 Two masses of 2 kg and 5 kg are suspended from two points on a beam by light inextensible cords, which are themselves connected by a third inextensible horizontal cord. Write the weights of these masses as forces in vector form. What is the total upward vertical force required to maintain the system in equilibrium? Write this in vector form. Multiple Choice

1 A stone is suspended from a string. The tension in the string is 3 N. The mass of the stone is: 3 A 3 N B 3g N C g N 3 D 3 kg E g kg 2 A force of 200 N acts at an angle to the horizontal. Its horizontal component is 60 N. The angle is: A sin 1 (0.6) B cos 1 (0.6) C tan 1 (0.6) -1 -1 D sin (0.3) E cos (0.3) 3 If v = 5i - 2 j represents a force, the magnitude of the force is: A 29 B 29 C 21 D 21 E -7 4 A stone is suspended from a string. The tensile strength of the string is 3g N. The maximum mass of the stone is: 3 A 3 N B 3g2 N C g N D 3g kg E 3 kg

5 A body hangs from a light inextensible cord. If its mass is 2m then the tension (T ) in the cord is: A T = 2m B T ≥ 2m C T ≥ 2mg D T = 2 mg E T + 2 mg = 0

508

6 A mass of

g

kg is suspended from a string hanging 4 from a hook. The tension in the string is: 1 1 1 A N B g2 N C N 4g 4 4 g 1 D kg E kg 4 4 7 Two strings are attached to two points on a beam. The two strings are knotted together and a third string carrying a bob is attached to the knot. If the tension in the shorter string is t = - 5i + 6 j newtons  and the tension in the longer string isT = 5i + aj    newtons, the mass of the bob is: (6 + a) A kg B 10g N C 10 kg g 10 D kg E (6 + a) kg g 8 A mass of 1 kg is suspended from a beam by two strings, 5 cm and 12 cm long, knotted together. If the strings are perpendicular to each other, then the total upward force acting to balance the weight of the mass is: A 17 N B 17 kg C 1 kg D g N E g kg 9 Find the single force that must be added to balance the forces t = 2 i - 3 j newtons and d = - 5i - 2 j       newtons, which act at the same point. A F = 3i + 5 j B F = - 3i - 5 j       c F = 3i - 5 j D F = - 3i - 5 j       E F = 7 i - j    10 A particle is held in equilibrium by three concurrent coplanar forces P, Q and R. P has magnitude 5 newtons and acts in the east direction. Q has magnitude 5 newtons and acts in the north direction. The magnitude and direction of R are, respectively: A 5 newtons, south-east B 5 2 newtons, north-east C 5 2 newtons, south-west D 10 newtons, south-west E 10 newtons, north-east


[© VCAA 2005]

12 Which of the following statements is true? A T1 = m B T1 = mg C T1 = m cos (a °) D T1 cos (90 − a)° = mg E T1 sin (90 − a)° = mg

Questions 11–13 relate to the diagram below. T2

α

β

M kg

T1

m kg

13 Which of the following statements is true? A T1 = T2 B T1 cos (a °) = T2 cos (b °) C T1 − mg = 0 D T1 sin (a °) = T2 sin (b °) E T1 + T2 = Mg + mg

11 Which of the following statements is true? A T2 = M B T2 = Mg C T2 cos (b °) = M D T2 cos (b °) = Mg E T2 sin (b °) = Mg ExTENdEd REsPoNsE

1 Two strings 2 metres and 5 metres long are attached to two points. The two strings are knotted together making an angle of 100°. A third string carrying a ball is attached to the knot. If the mass of the ball if 20 kg, find the tensions in the strings. 2 A stone of weight P newtons suspended by a string is pulled sideways by a force of 10 newtons acting horizontally. If the tension in the string is 18 newtons, find P. 3 The three forces in the diagram below are in equilibrium. Find P. 7N 20°

3N

52.94°

PN

4 A stone of weight M newtons, suspended by a string, is pulled sideways by a force, v = ai - bj newtons. If the  the string  makes with tension in the string is t = - bi + aj newtons, find M in terms of a. Find the angle that    the vertical. 5 Two perpendicular strings are attached to two points on a beam. The two strings are knotted together and a third string carrying a bob is attached to the knot. If the tension in one string is t = 7 i + j newtons, find the    mass of the bob. 6 Two masses of 2 kg and 5 kg are suspended (5 kg lower) from two points on a beam by light inextensible cords, which are themselves connected by a third inextensible cord that makes an angle of 30° to the horizontal. If the cord from the beam to the smaller mass makes an angle of 15° to the vertical, find the tension in each string and the angle that the other cord makes with the vertical. eBook plus Digital doc


Chapter 14


509

eBook plus

ACTiviTiEs


• 10 Quick Questions: Warm up with ten quick questions on statics of a particle. (page 492) 14A

Force and tension

Tutorial

• WE 3 int-1173: Watch how to resolve a force into two perpendicular components. (page 493) Digital doc

• WorkSHEET 14.1: Calculate mass and force of static objects and resolve forces into perpendicular components. (page 495) 14B

Newton’s first law of motion

Tutorial

• WE 4 int-1174: Watch how to determine an opposing force and its magnitude and multiply it by a scalar quantity. (page 496) 14C

Equilibrium — forces at an angle

Tutorial

• WE 6 int-1175: Watch how to find the tension in two strings suspending a mass. (page 499) Digital doc

• WorkSHEET 14.2: Calculate mass and force of static objects, tension in a string suspending them and the angles made with the horizontal. (page 502)

510

14D


Interactivity

• Connected bodies in equilibrium int-0981: Consolidate your understanding of the forces in connected bodies in equilibrium. (page 502) Tutorial

• WE 8 int-1177: Watch how to find the tension of a string and the mass of a suspended body. (page 503) Chapter review Digital doc



15

15A Introduction to kinematics 15B Velocity–time graphs and acceleration–time graphs 15C Constant acceleration formulas 15D Instantaneous rates of change

Kinematics AreAS oF STuDy

• Diagrammatic and graphical representation of empirical position–time data for a single particle in rectilinear motion, including examples with variable velocity (data may be obtained by a student moving along a 100 metre tape according to a given set of instructions, data logging or previous experimental data) • Graphical modelling and numerical analysis of position–time and velocity–time relationships based on continuous hybrid functions formed by straight–line segments, including consideration of average velocity and distance travelled over an interval • Modelling and analysis of rectilinear motion under constant acceleration, including use of constant acceleration formulas: v = u + at, 1 1 v2 = u2 + 2as, s = (u + v)t and s = ut + at2 2 2 • Qualitative graphical analysis of the relationship between position–time, velocity–time and

acceleration–time graphs for simple cases of rectilinear motion involving variable acceleration • Numerical approximation to instantaneous rate of change of a function f at time t = a by evaluation of the central difference f (a + h) - f (a - h) for small values of h using 2h technology and its application to approximate evaluation of instantaneous velocity and instantaneous acceleration in simple cases of rectilinear motion involving variable velocity and variable acceleration • Approximation of velocity–time relationships by step functions and its application to approximate evaluation of distance travelled in simple cases of rectilinear motion involving variable velocity and variable acceleration, as a sum of areas of rectangles, using technology eBook plus Digital doc

15A

introduction to kinematics

10 Quick Questions

Our lives are perpetually involved in movement. Walking around the house, being transported to school, throwing a ball, riding a bicycle, picking up a pen, climbing stairs, going on a holiday are just a few examples. Most of our movements are routine, and we don’t give them a second thought. However, sometimes we do need to think about what we are doing; for example, understanding motion can be a matter of life and death in situations such as crossing a road safely, deciding when it is prudent to overtake when driving, or calculating where a cyclone is heading. Even in less-dramatic situations like keeping an appointment on time, or judging how and when to throw a ball while playing sport, we give more thought to motion. Then we start to employ questions of judgement: How far is it? How long will it take? How will I get there?

Chapter 15

Kinematics

511

Our interest in analysing motion extends far beyond these examples taken from our daily lives. People have long been fascinated by movement in the world about them: by the motion of the planets and stars, by the flight of birds, by the oscillations of pendulums and by the growth of plants, to name a few. The study of motion is fundamental in all branches of science. The name kinematics is given to the study of the motion of bodies, objects or particles. In this chapter, we consider motion that is only one-dimensional; that is, straight-line motion. This is called rectilinear motion (to distinguish it from curvilinear motion, which deals with curves). Examples of rectilinear motion include a ball travelling along a pool table in a single direction, or an ice-hockey puck that has been hit along the ice. For mathematical convenience, all moving objects that we consider in this chapter will be treated as points; that is, the objects do not rotate or change shape. To look at how we might analyse motion, let’s consider the latest jump by Bill the Bungy jumper. Bill jumps from a bridge that is 120 metres above the ground and is attached to an 80-metre elastic rubber rope. He falls vertically towards the ground. In the first 2 seconds he falls 20 metres and in the next 2 seconds he falls a further 60 metres. After 80 metres the bungy rope starts to stretch, and therefore slows the fall so that Bill travels a further 20 metres in 2 seconds. The stretched bungy rope then pulls him up a distance of 15 metres in 2 seconds, passing what is called the ‘equilibrium position’. (This is the position that Bill would eventually remain in, once he stopped bouncing on the rope.) He continues travelling up a further 10 metres in 2 seconds. Bill continues bouncing until he is lowered safely to ground level.

100 metres

25 metres

512


If we take the starting point, S, to be 0 metres, then the first 10 seconds of Bill’s jump can be displayed as follows. Stage 1 S

0 metres at t = 0 s

A

20 metres at t = 2 s

B


Stage 2

C

Stage 3

E


D



Position The position of a particle moving in a straight line Q P is established by its distance from a fixed reference −6 −5−4 −3 −2 −1 O 1 2 3 4 5 x point on the line. This is usually the origin, O, with positions to the right of O normally being taken as positive. Consider the particles, P and Q which both start from the origin, O. The position of particle P is 4 units to the right, therefore x = 4. Particle Q is 3 units to the left of the origin and therefore has a position of x = -3. We could describe Bill’s motion by noting his S A B position at various times. We show this on a straight line (vertical or O 20 80 x (m) horizontal) by indicating his location relative to a Positive reference point, usually the origin, O. Positions to direction the right of O are normally taken as positive. Point S, at the origin, (actually 120 metres above the ground), shows Bill’s starting position. Taking downwards as positive, point A is at 20 and point B is at 80.

Displacement The displacement of a moving particle is its change in position relative to a fixed point. Displacement gives both the distance and direction that a particle is from E D t=8 a point. t = 10 S A C This can be represented on a t=6 t = 0 t = 2 t = 4 B position–time line (or displacement– time line), as shown at right, for the O 20 75 80 85 100 x (m) first 10 seconds of Bill the bungy jumper’s path. Note: The direction of the motion is indicated by the arrows.

Chapter 15 Kinematics

513

Bill travels from C (100 metres) to E (75 metres). The displacement from C to E is the change in position from C to E. Displacement = final position − initial position = 75 − 100 = −25 metres The distance from C to E is 25 metres but the displacement is −25 metres. Displacement is a vector quantity and has both magnitude and direction. (In this case the magnitude is 25 metres and the direction is negative.) Distance is a scalar quantity and has magnitude only. For the first 10 seconds of Bill’s jump, his displacement is 75 metres (75 − 0). However, the distance Bill has moved is 125 metres. Note: At point C, Bill is momentarily at a stop (his velocity is 0) and his motion changes direction from down to up.

Velocity

Average velocity = change in position change in time final position - initial position = change in time x2 - x1 = t2 - t1

x Position

Velocity is also a vector quantity. The average velocity of a particle is the rate of change of its position with respect to time. This can be shown on a position– time graph. The red line shows the position of the particle, x, at time, t.

x2 x1

Change in position δ x Change in time δ t t1

Time

t2

t

δx δt Bill’s average velocity over the first 10 seconds of his jump can be calculated as follows: x -x Average velocity = 2 1 t2 - t1 =

75 - 0 10 - 0 75 = 10 = 7.5 m/s The commonly used units of velocity are cm/s, m/s or km/h. Note: 1 m/s = 3.6 km/h. The instantaneous velocity is the velocity at a given point of time. That is, it is the gradient of the displacement–time graph at a given point. =

Speed Speed is the magnitude of velocity and so it is a scalar quantity. Average speed = distance travelled time taken Instantaneous speed is the magnitude of instantaneous velocity and is always positive. Bill’s average speed over the first 10 seconds of his jump can be calculated as follows: Average speed =

125 10

= 12.5 m/s (compared to the average velocity of 7.5 m/s). 514


WorKeD exAMPle 1

The following position–time line shows a particle that moves from S to A in 2 seconds then from A to F in 3 seconds. Find: a the starting position, S F b the final position, F S A c the displacement of F from S d the distance travelled from S to F 12 x 4 6 8 10 −2 0 2 e the average velocity from S to F f the average speed from S to F. ThinK

WriTe

a Read the position of point S.

a The position of point S is -2.

b Read the position of point F.

b The position of point F is 2.

c Displacement = final position - initial position

c Displacement = 2 - -2

d Add the distance from S to A to the distance from A to F.

d Distance = 12 + 8 = 20 units

e Average velocity = change in position change in time

= 4 units to the right of S

e Average velocity =

x2 - x1 t2 - t1

=

2 - -2 5- 0 4

f Average speed =

distance travelled time taken

=5 = 0.8 units/second in the positive direction f Average speed =

20 5

= 4 units/second

Constant velocity

The velocity is constant from t = 0 to t = 4.

x

Velocity can be determined by the gradient of a position– time graph. If the position–time graph is a series of connected straight-line sections, then the velocity is constant over the duration of each straight-line section.

0

WorKeD exAMPle 2

The velocity is constant from t = 4 to t = 10.

4

10

t

eBook plus Tutorial


Position (cm)

At Luna Park there is a new game called ‘Hit the duck’. To win, you must knock down a mobile duck that moves back and forth in a straight line on a 5-metre track. You have three shots with small sandbags. The position–time graph shows the position of the duck, x centimetres to the right of its starting point, along the track at x various times, t seconds. 500 a What is the initial position of the duck? 400 b How long did the game last? 300 c What is the final displacement of the duck from its starting position? 200 d Write the times for which the velocity is: 100 i positive ii negative iii zero. 0 e Hence, find the velocity for each of the three time intervals in part d. f What was the average speed of the duck during this game?

5 6 10 Time (s)

Chapter 15

t

Kinematics

515

Think

Write

a The initial position of the duck is when t = 0.

a When t = 0, the initial position of the duck

b The graph finishes when t = 10.

b The game lasted for 10 seconds.

c Displacement = final position − initial position.

c

d

d

i Look for where the gradient slopes upwards

is 200 cm to the right of its starting point.

to the right.

t = 5.

t = 10.

downwards to the right.

iii The gradient is zero from t = 5 to t = 6.

iii Look for where the gradient is horizontal.

change in time

i The gradient is positive from t = 0 to ii The gradient is negative from t = 6 to

ii Look for where the gradient slopes

e Velocity = change in position

Displacement = 100 − 200 = −100 cm

e

i Velocity =

x2 - x1 t2 - t1

400 - 200 5- 0 200 = 5 = 40 cm/s x2 - x1 ii Velocity = t2 - t1 100 - 400 = 10 - 6 - 300 = 4 = −75 cm/s x - x1 iii Velocity = 2 t2 - t1 400 - 400 = 6-5 0 = 1 = 0 cm/s =

f Average speed = distance travelled

time taken

f Average speed =

500 10

= 50 cm/s

Position expressed as a function of time When the position is expressed as a function of time, the position–time graph can be sketched and the motion then analysed. If the position–time graph is curved, then the velocity (or gradient) is always changing and never constant. Worked Example 3

A particle moves in a straight line so that its position, x cm, from a fixed point, O, on the line, at time, t, seconds, is given by the rule: 1 x = 2 (t − 1)2, t ∈ [0, 5]

516


The position–time graph is shown: a Copy and complete the table below. 1

2

3

4

Position (cm)

0

t

x

5

x b What is the initial position of the particle? c What is the significance of the position at t = 1? d Show the movement of the particle on a position–time line. e i What is the displacement of the particle?

8 7 6 5 4 3 2 1 0

1

2 3 4 Time (s)

5 t

ii Hence, determine the average velocity of the particle. f i What is the distance travelled by the particle? ii Hence, determine the particle’s average speed. Think a

1

Write

Substitute each value of t into the rule 1 x = 2 (t − 1)2 and evaluate for x.

1

a When t = 0, x = 2 (0 − 1)2 1

= 2 (−1)2 = 0.5 1 When t = 1, x = 2 (1 − 1)2 1

= 2 (0)2 =0 1 When t = 2, x = 2 (2 − 1)2 1

= 2 (1)2 = 0.5 1 When t = 3, x = 2 (3 − 1)2 1

= 2 (2)2 =2 1 When t = 4, x = 2(4 − 1)2 1

= 2 (3)2 = 4.5 1 When t = 5, x = 2 (5 − 1)2 1

= 2 (4)2 =8 2

Complete the table.

t

0

1

2

3

4

5

x

0.5

0

0.5

2

4.5

8

b State the position of the particle when t = 0.

b The initial position is 0.5 cm from O.

c At t = 1 the particle is at the position x = 0 and

c At t = 1 the particle is changing direction.

d The particle starts at x = 0.5, moves to x = 0

d

the position–time graph shows that the particle is changing direction. then turns and finishes at x = 8.

t=5

t=2 t=1

t=0 0

1

2

3

4 5 cm

6

7

8

x


517

e

i Displacement =

e

final position − initial position

ii Average velocity =

change in position change in time

Displacement = 8 − 0.5 = 7.5 cm x2 - x1 ii Average velocity = t2 - t1 = 8 - 0.5 5- 0 i

=

7.5 5

= 1.5 cm/s f

i Add the distance travelled from t = 0

to t = 1, to the distance travelled from t = 1 to t = 5. ii Average speed = distance travelled time taken

f

i

The distance from t = 0 to t = 1 is 0.5 cm and the distance from t = 1 to t = 5 is 8 cm. The total distance is 8.5 cm.

ii

Average speed =

8.5 5

= 1.7 cm/s

REMEMBER

1. A particle’s position gives its location relative to a reference point, usually the origin, O. 2. A particle’s displacement is the change in its position relative to a fixed point. Displacement gives both the distance and direction that the particle is from a point. Displacement = final position − initial position 3. The average velocity of a particle is the rate of change of its position with respect to time. change in position Average velocity = change in time final position - initial position = change in time distance travelled 4. Average speed = time taken Exercise

15A

Introduction to kinematics 1 WE 1 Each of the following position–time lines shows a particle which moves from S to A in 2 seconds, then from A to F in 3 seconds. In each case, find: i the starting position, S ii the final position, F iii the displacement of F from S iv the distance travelled from S to F v the average velocity from S to F vi the average speed from S to F. F F b a A A S

S

−2−1 0 1 2 3 4 5 6 7 8 x

c

F

S

d A

A

−8−4 0 4 8 12 16 20 24 28 32 x

e

A

S

−4 −2 0 2 4 6 8 10 12 14 16 x S

−4 −2 0 2 4 6 8 10 12 14 16 x

F

−4 −2 0 2 4 6 8 10 12 14 16 x

518

F


2

Represent each of the following situations on a position–time line. a A particle starts at S, 2 units to the left of the origin. It is then displaced 10 units to A and undergoes a final displacement of −5 units to F. b A particle starts at S, 3 units to the left of the origin. It is then displaced −10 units to A and undergoes a final displacement of 8 units to F. c A particle starts at S, 6 units to the right of the origin. It is then displaced −8 units to A and undergoes a final displacement of 7 units to F. d A particle starts at S, 4 units to the left of the origin. It is then displaced −11 units to A and undergoes a final displacement of 6 units to F. e A particle starts at S, 3 units to the left of the origin. It is then displaced 8 units to A, followed by a displacement of −7 units to B and undergoes a final displacement of −5 units to F. f A particle starts at S, 8 units to the right of the origin. It is then displaced 3 units to A, followed by a displacement of −4 units to B and undergoes a final displacement of 2 units to F. Each movement from S to F described in question 2 takes 6 seconds and the measurements are in centimetres. In each case determine: i the displacement of F from S ii the total distance travelled by the particle iii the average velocity iv the average speed. C t= 5 Use the position–time line at right to answer questions 4 to 7. S t= 0 B t= 4

4 MC The displacement of F from S, in cm, is: A -24 5 MC A 24 6 MC A 4.25 7 MC A 2

B 24

F t= 8

A t= 3 −10−8−6−4−2 0 2 4 6 8 10 12 14 16 18 20 22 x

C 32

D 14

E 56

The distance travelled in moving from S to F, in cm is: B 34

C 44

D -34

E 56

The average speed in moving from S to F, in cm/s is: B 7

C 5.5

D -6.8

E 3

The average velocity in moving from A to C, in cm/s is: B 10

C -10

8 WE 2 The position–time graph shows the position of a moving particle, x centimetres to the right of the origin, O, at various times, t seconds. a What is the initial position of the particle? b What is the final displacement of the particle from its starting position? c Write the times for which the velocity is: i positive ii negative iii zero. d Hence, find the velocity for each of the three time intervals in part c. e What was the average speed of the particle?

D -2

E -0.5 x

500 Position (cm)

3

400 300 200 100 0

4

6 Time (s)

12

t


519

x 600 500 Position (cm)

9 The position–time graph shows the position of a moving particle, x centimetres to the right of the origin, O, at various times, t seconds. a What is the initial position of the particle? b What is the final displacement of the particle from its starting position? c Write the times for which the velocity is: i positive ii negative iii zero. d Hence, find the velocity for each of the three time intervals in part c. e What was the average speed of the particle?

400 300 200 100 0

45

8 Time (s)

13

t

10 WE 3 A particle moves in a straight line so that its position, x cm, from a fixed point, O, on the line at time, t seconds, is given by the rule: 1

x = 2 (t − 2)2, t ∈ [0, 8]

Position (cm)

The position–time graph is shown below: x 20 18 16 14 12 10 8 6 4 2 0 12345 678 t Time (s)

a Copy and complete the table below. t x

0

2

4

6

8

b What is the significance of the position at t = 2? c Show the movement of the particle on a position–time line. d Determine the average velocity of the particle. e What is the particle’s average speed? 11 A particle moves in a straight line so that its position, x cm, from a fixed point, O, on the line at time t seconds is given by the rule: x = t2 − 8t + 12, t ∈ [0, 8] a Copy and complete the table below. t x

0

2

4

6

8

b Sketch the position–time graph for the particle. Check your answer using a CAS calculator. c What is the significance of the position at t = 4? d Show the movement of the particle on a position–time line. e Determine the average velocity of the particle. f What is the particle’s average speed? 12 A particle moves in a straight line so that its position, x cm, from a fixed point, O, on the line at time t seconds is given by the rule: x = t2 − 4t − 5, t ∈ [0, 6] a Sketch the position–time graph for the particle. Check your answer using a CAS calculator. b Show the movement of the particle on a position–time line.

520


c Determine the average velocity of the particle. d What is the particle’s average speed? 13 A particle moves in a straight line so that its position, x cm, from a fixed point, O, on the line at time, t seconds, is given by the rule: x = -t2 + 2t + 8, t ∈ [0, 6] a b c d

Velocity–time graphs and acceleration–time graphs


int-0267 Motion graphs (kinematics)

Velocity–time graphs Let us take another look at the position–time line for the bungy jump performed by Bill that was described at the start of the chapter. S

A

t=0

t=2

0

t = 10

E

D t=8 C t=6

B t=4

20

75 80 85

100

x

Metres

This situation can be represented on a position–time graph shown at right. The curve reflects the fact that the change of position over time (velocity) is not constant.

We can calculate the average velocity in each of the stages as follows:

Position (m)

15B

Sketch the position–time graph for the particle. Check your answer using a CAS calculator. Show the movement of the particle on a position–time line. Determine the average velocity of the particle. What is the particle’s average speed?

x 100 90 80 70 60 50 40 30 20 10 0

(6, 100) B

A

(8, 85) D (10, 75) (4, 80) E C

(2, 20)

S 1 2 3 4 5 6 7 8 9 10 t Time (s)

From S to A: Average velocity =

x2 - x1 t2 - t1

From A to B: Average velocity =

x2 - x1 t2 - t1

=

20 - 0 2-0

=

80 - 20 4-2

=

20 2

=

60 2

= 10 m/s From B to C: Average velocity =

x2 - x1 t2 - t1

= 30 m/s From C to D: Average velocity =

=

100 - 80 6-4

=

=

20 2

=

= 10 m/s

x2 - x1 t2 - t1 85 - 100 8-6 - 15 2

= -7.5 m/s

Chapter 15

Kinematics

521

From D to E: Average velocity =

x2 - x1 t2 - t1

= 75 - 85 10 - 8 =

- 10 2 −5

Average velocity (m/s)



= m/s Note: The negative velocities occur when the motion is upwards, since we decided to define downwards as positive. vav We can now represent the motion of Bill’s bungy jump 30 during each stage on a velocity–time graph (or more 20 particularly, an average velocity–time graph). Notice that the graph shows that the velocity is constant 10 during each of the stages (shown as the ‘step formation’ of 0 the graph). This is because we have calculated the average 2 4 6 8 10 t −10 velocity of each stage. If we were to analyse the average velocity over smaller time intervals, we would get more steps Time (s) with smaller widths, as is displayed in the second graph. vav If we allowed these time intervals (step widths) to get 40 closer and closer to zero, then the associated average 30 velocities would effectively become a series of connected points that would collectively produce a velocity–time graph 20 something like the one displayed at right. 10 This is a velocity–time graph as it shows Bill’s velocity at 0 every instance of the first 10 seconds of motion during his 1 2 3 4 5 6 7 8 9 10 t bungy jump. There are no horizontal lines (steps) because −10 the velocity is changing every instant over the course of the −20 motion. This change in velocity over time is called Time (s) acceleration. Acceleration is also a vector quantity. v For the first 4 seconds of motion, the graph is a straight 40 line because Bill is subjected only to acceleration due to gravity, which is constant at 9.8 m/s2. This means that every 30 second, Bill’s velocity increases by 9.8 m/s while he is 20 moving downwards. 10 For the period of time where the bungy rope is stretched, 0 (greater than 80 m) from t = 4 seconds to about t = 9 seconds, 1 2 3 4 5 6 7 8 9 10 t the elasticity of the rope causes the acceleration to continually −10 change according to the tension in the bungy rope. That is −20 why the velocity–time graph is curved during this time. Time (s) From about t = 9 seconds to t = 10 seconds, (where the bungy rope is less than 80 m) the rope is again slack and Bill is subject to acceleration due only to gravity again. At this stage the motion is upwards, but since acceleration due to gravity acts downwards, Bill is slowing down or decelerating. Average acceleration = change in velocity change in time =

v2 - v1 t2 - t1

= δv δt The most common units of acceleration are cm/s2 or m/s2. For the moment we will consider only examples that involve constant acceleration.

522


Worked Example 4

Draw a velocity–time graph to match the following description. An object that is moving in a straight line has an initial velocity of 5 m/s. It accelerates at a constant rate until it reaches a velocity of 10 m/s after 6 seconds. It maintains this velocity for 8 seconds and then decelerates at a constant rate for a further 4 seconds when it comes to rest. Write/Draw

1

The velocity ranges from 0 m/s to 10 m/s.

2

The total time is 6 + 8 + 4 = 18 seconds.

3

Draw a set of axes with velocity on the vertical axis and time on the horizontal axis. Label each axis appropriately. Sketch a straight line from (0, 5) to (6, 10) to show the acceleration in the first stage. Draw a horizontal line from (6, 10) to (14, 10) to show the constant velocity during the second stage. Draw a straight line from (14, 10) to (18, 0) to show the final stage of deceleration.

5 6

5

0

2 4 6 8 1012141618 t Time (s)

Notice that the gradient of each straight-line section of the velocity–time graph gives the acceleration of the object.

Analysing the velocity–time graph The gradient of a velocity–time graph allows us to calculate the acceleration of an object moving in a straight line. In addition to this, the area between the velocity–time graph and the time axis also provides useful information relating to displacement and distance. Earlier, it was shown that: Average velocity = change in position change in time δx or vav = δt where v represents average velocity. av

Rearranging this results in: δx = vav × δt In other words, the signed area between a velocity–time graph and the time axis is equal to the change in position or displacement. When we calculate the signed area, we take the area above the time axis as positive displacement and the area below the time axis as negative displacement. If the distance (rather than the displacement that the particle has travelled) is required, then there is no need to sign the areas. That is, the distance travelled is the total area between the velocity–time graph and the time axis. vav Using the average velocity–time graph describing Bill’s 40 bungy jump from earlier, the information described above 30 can be highlighted as follows. 20 The displacement is equal to the sum of the signed areas of the rectangles. 10 Displacement = 10 × 2 + 30 × 2 + 10 × 2 − 7.5 × 2 − 5 × 2 0 = 20 + 60 + 20 − 15 − 10 2 4 6 8 10 t −10 = 75 metres Average velocity (m/s)

4

v 10 Velocity (m/s)

Think

Time (s)


523

Velocity (m/s)

The distance is equal to the sum of all the unsigned area v of the rectangles. 5 Distance = 10 × 2 + 30 × 2 + 10 × 2 + 7.5 × 2 + 5 × 2 = 20 + 60 + 20 + 15 + 10 Area 1 = 125 metres 9 10 The following can be obtained from the figure shown at Area 2 t 5 7 right. 1. The object is travelling at a constant velocity of 5 m/s until t = 5 s. It slows down until it stops at t = 7 s, before it −5 changes direction and increases its speed to 5 m/s at t = 9 s. Time (s) The object then slows down and stops when t = 10 s. 2. The gradient of the line between t = 0 s and t = 5 s is zero, so the acceleration is 0 m/s2. 10 10 Between t = 5 s and t = 9 s, the gradient is 4 , so the acceleration is 4 m/s2. Between t = 9 s and t = 10 s the gradient is 5, so the acceleration is 5 m/s2. 3. Total displacement = Area 1 - Area 2. 4. Total distance = Area 1 + Area 2. Note: When appropriate, break the area between the velocity–time graph and the time axis into simple shapes; for example rectangles, triangles or trapeziums. Area of a rectangle = L × W -

-

1

Area of a triangle = 2 bh 1

Area of a trapezium = 2 (a + b)h WorKeD exAMPle 5

eBook plus

v 10 Velocity (m/s)

Consider the velocity–time graph obtained in worked example 4 to find: a the acceleration in the first 6 seconds b the acceleration in the last 4 seconds c the total displacement d the total distance travelled.

a Average acceleration =


5

0

ThinK

Tutorial

2 4 6 8 1012141618 t Time (s)

WriTe/DrAW

change in velocity change in time

a Average acceleration =

v2 - v1 t2 - t1

= 10 - 5 6-0 = b Average acceleration =


m/s2

v2 - v1 t2 - t1 = 0 - 10 18 - 14

b Average acceleration =

= = 524

5 6


- 10 4 -2.5

m/s2

1

The displacement is equal to the total signed area under the velocity–time graph.

2

Divide the given graph into two trapeziums, one from t = 0 to t = 6 and the other from t = 6 to t = 18.

c v 10 Velocity (m/s)

c

8

5 5

10 Area 1 6 2

3

10 Area 2

12 6

4

8

10 12 14 16 18 t Time (s)

1

Area 1 = 2 (5 + 10) × 6

Calculate the area of each trapezium.

1

= 2 × 15 × 6 = 45 units2 1

Area 2 = 2 (8 + 12) × 10 1

= 2 × 20 × 10 = 100 units2 Displacement = Area 1 + Area 2 = 45 + 100 = 145 m

Find the displacement.

d The distance is equal to the total unsigned area

under the velocity–time graph.

d The distance is equal to 145 m.

Note: Since the velocity is always positive in this example, the distance is equal to the displacement.

Acceleration–time graphs Just as the gradient of a position–time graph gives the rate of change of position or velocity, the gradient of a velocity–time graph gives the rate of change of velocity or acceleration. Where the velocity is increasing the acceleration is positive. Where the velocity is decreasing the acceleration is negative. Where the velocity is not changing the acceleration is zero. Consider a modified velocity–time graph of the first 10 seconds of motion of Bill’s bungy jump. We will assume the acceleration is constant, but v different through each of the stages of the jump. B 40 change in velocity Since average acceleration = , the 30 change in time acceleration for each stage is: 20 v -v From S to B: Average acceleration = 2 1 t2 - t1 10

= 40 - 0 4-0 40 4

=

= 10 m/s2

Velocity (m/s)

4

0 −10 −20

S 2

4

C 6 8

E 10 t

D Time (s)


525

From B to C: Average acceleration =

v2 - v1 t2 - t1

= 0 - 40 6-4 = =

- 40 2

-20

m/s2 v2 - v1 From C to D: Average acceleration = t2 - t1 = 15 - 0 8-6

=

- 15 2

Acceleration (m/s2)

= -7.5 m/s2 v -v From D to E: Average acceleration = 2 1 t2 - t1 = 0 - 15 10 - 8

=

15 2

a 10 0 −10

2

4

6

8

10 t

−20

Time (s) = 7.5 m/s2 Therefore, the acceleration–time graph would look like the graph above. Note: The signed area under the acceleration–time graph gives the change in velocity. In the graph on the previous page, the area between the graph and the time axis from t = 0 s to t = 4 s is 40, which is verified on the previous velocity–time graph. WorKeD exAMPle 6

eBook plus

Velocity (m/s)

Consider the motion of an elevator, which has its velocity–time graph as shown. Tutorial Take positive values to represent upward motion. int-1180 a In what sections — OA, AB, BC, etc. — is the lift: Worked example 6 i accelerating positively? ii accelerating negatively? v iii travelling at a constant velocity? A B 8 b Determine the acceleration for each section of 4 the lift’s journey. 35 40 C 25 27 c Sketch the acceleration–time graph. O G t D 5 18 20 d If the lift started at ground level, 0 metres, −4 determine its position at: −8 i C ii G. e Determine the average velocity of the lift. −12 E F f How far did the lift travel? Time (s) g What was the lift’s average speed? ThinK a

i Acceleration is positive where the

velocity is increasing.

a

i The acceleration is positive from O to A and

from F to G.

ii Acceleration is negative where the

ii The acceleration is negative from B to C

iii Acceleration is zero where the velocity

iii The acceleration is zero from A to B, from

velocity is decreasing. is not changing.

526

WriTe

and from D to E.

C to D and from E to F.


change in velocity change in time v -v = 2 1 t2 - t1

b Average acceleration =

b From O to A, average acceleration =

=

v2 - v1 t2 - t1 8- 0 5- 0 8

=5 = 1.6 m/s2 From A to B, average acceleration =

8-8 18 - 5 0

= 13 = 0 m/s2 From B to C, average acceleration =

0-8 20 - 18

=

-8

=

−4

From C to D, average acceleration = =

2

m/s2

0-0 25 - 20 0 5

= 0 m/s2 From D to E, average acceleration = 12 - 0 27 - 25

=

- 12 2

= −6 m/s2 - 12 - -12 From E to F, average acceleration = 35 - 27 =

0 8

= 0 m/s2 From F to G, average acceleration = 0 - 12 40 - 35

=

12 5

= 2.4 m/s2 so the acceleration–time graph is a series of horizontal lines (steps).

a

c Acceleration (m/s2)

c The acceleration is constant in each section,

4 2 0 −2

5

18 20 25 27

35 40

t

−4 −6 Time (s)


527

d

i

Since the lift started at position 0 metres, the position at point C is the signed area under the trapezium OABC.

d

i

The position at C is the area of trapezium OABC 1

= 2 (13 + 20) × 8 1

= 2 × 33 × 8 = 132 metres ii

The position at point G is the signed area under the trapezium DEFG plus position at point C.

ii

The position at G is the signed area under the trapezium DEFG plus position at point C = =

−1 2 −1 2

(8 + 15) × 12 + 132 × 23 × 12 + 132

= −138 + 132 = − 6 metres (that is, the lift ends up 6 metres below ground level). e Average velocity =


x2 - x1 t2 - t1

e Average velocity =

=

-6 - 0

=

40 - 0 -6 40

= −0.15 m/s f The total distance travelled by the lift is the

total area between the velocity–time graph and the time axis. g Average speed = distance travelled time taken

f The total distance travelled by the lift is

132 + 138 = 270 metres.

g Average speed =

270 40

= 6.75 m/s

REMEMBER

1. Average velocity = 2. Average speed =



3. Average acceleration =


4. The signed area between a velocity–time graph and the time axis is equal to the change in position or displacement. The area above the time axis is positive displacement and the area below the time axis is negative displacement. 5. The unsigned area between a velocity–time graph and the time axis is equal to the distance travelled. 6. Final position = displacement + initial position

528


Velocity–time graphs and acceleration–time graphs 1 WE4 Draw a velocity–time graph to match each of the following descriptions. a An object, which is moving in a straight line, has an initial velocity of 4 m/s. It accelerates at a constant rate until, after 5 seconds, it reaches a velocity of 9 m/s. It maintains this velocity for 10 seconds and then decelerates at a constant rate for a further 5 seconds, when it comes to rest. b An object, which is moving in a straight line, has an initial velocity of 6 m/s. It accelerates at a constant rate until, after 8 seconds, it reaches a velocity of 12 m/s. It maintains this velocity for 15 seconds and then decelerates at a constant rate for a further 5 seconds until it reaches a velocity of 8 m/s. c An object, which is moving in a straight line, has an initial velocity of −5 m/s. It accelerates at a constant rate until, after 10 seconds, it reaches a velocity of 4 m/s. It maintains this velocity for 12 seconds and then decelerates at a constant rate for a further 9 seconds, when it comes to rest. d An object, which is moving in a straight line, has an initial velocity of 5 m/s. It decelerates at a constant rate until, after 6 seconds, it reaches a velocity of −5 m/s. It maintains this velocity for 4 seconds and then accelerates at a constant rate for a further 6 seconds, when it comes to rest. e An object, which is moving in a straight line, has an initial velocity of −8 m/s. It maintains this velocity for 10 seconds and then accelerates at a constant rate until, after 8 seconds, it reaches a velocity of 4 m/s. It maintains this velocity for 12 seconds and then decelerates at a constant rate for a further 4 seconds, when it reaches a velocity of 2 m/s, which it maintains. 2 WE5 Consider the velocity–time graph shown to find: v a the acceleration in the first 5 seconds 12 b the acceleration in the last 5 seconds c the total displacement d the total distance travelled. Velocity (m/s)

15B

8 4

0

5

10

15

20 t

Time (s)

3 a b c d

Consider the velocity–time graph shown to find: the acceleration in the first 6 seconds the acceleration in the last 6 seconds the total displacement the total distance travelled.

v 16 12 Velocity (m/s)

Exercise

8 4 0

16

6

22 t

Time (s)


529

a b c d

v 9

Consider the velocity–time graph shown to find: the acceleration in the first 6 seconds the acceleration in the last 12 seconds the total displacement the total distance travelled.

Velocity (m/s)

4

6

3

0

6

Time (s)

Use the velocity–time graph at right to answer questions 5 to 7.

6

MC The magnitude of the acceleration is greatest between the points: A A and B B B and C C A and B and D and E D D and E E E and F MC The average velocity from A to

F is equal to: A 3.3 m/s C 4 m/s E −4 m/s 7

v 8

B 2.3 m/s D 2.8 m/s

B D

Velocity (m/s)

5

4

C

A

0

2

4

6

E 8 10 12 14 16 18 20 t

−4

F Time (s)

MC The average speed from A to F is equal to:

A 3.3 m/s

B 2.3 m/s

C 4 m/s

E −4 m/s

D 2.8 m/s

9 Consider the motion of a lift in a high-rise building. The lift’s velocity–time graph is as shown. The lift starts from the twenty-fifth floor, which is 100 metres above ground level. Take positive values to represent upward motion.


Velocity (m/s)

Velocity (m/s)

8 WE6 Consider the motion of an elevator, whose v velocity–time graph is as shown. Take positive A 6 values to represent upward motion. a In what sections, OA, AB, BC, etc. is the lift: i accelerating positively? O 4 ii accelerating negatively? iii travelling at a constant velocity? b Determine the acceleration for each section of −8 the lift’s journey. c Sketch the acceleration–time graph. d If the lift started at ground level, 0 metres, determine its position at: i C ii G. v e Determine the average velocity of the lift. 6 f How far did the lift travel? g What was the lift’s average speed?

530

25 t

13

O

6

−9

A

B C 16 19

27 31 38 41 D G t

Time (s)

E

E

14

B

F

F

C D G 24 30 36 40 45 t

Time (s)

a In what sections, OA, AB, BC, etc. is the lift: i accelerating positively? ii accelerating negatively? iii travelling at a constant velocity? b Determine the acceleration for each section of the lift’s journey. c Sketch the acceleration–time graph. d Determine the lift’s position at: i C ii G. e Determine the average velocity of the lift. f How far did the lift travel? g What was the lift’s average speed?

10 A car is travelling at a constant speed of 108 km/h when it passes a stationary police motorcycle. Four seconds later the motorcycle sets off in pursuit with a constant acceleration of 5 m/s2 until it reaches a speed of 126 km/h, which it then maintains. (1 m/s = 3.6 km/h) a For how long does the motorcycle accelerate? b Sketch a velocity–time graph which represents the motion of both the car and the motorcycle. c How long after the car first passes the motorcycle does it take for the motorcycle to catch up to the car? d How far have they travelled? 11 Polly is leading a 50-kilometre bicycle race when her bicycle gets a puncture 360 metres from the finish line. She changes her tyre, and the instant she takes off again, Molly passes her, travelling at a constant speed of 14 m/s. Polly accelerates at a constant rate for 5 seconds, when she reaches a speed of 16 m/s, which she maintains until the finish. a Sketch a velocity–time graph that represents the motion of both Polly and Molly. b Verify that Polly still wins the race. c How far from the finish line are they when Polly catches up to Molly? d If Molly started to accelerate at a constant rate from the moment that Polly caught up to her, what would her acceleration be if they were to dead heat?


531


WorkSHEET 15.1

15C

12 Max the monkey is climbing a coconut tree in a straight line, to find a coconut for lunch. His motion is described as follows. Max starts from rest at ground level with constant acceleration until he reaches a speed of 1.5 m/s after 4 seconds. He maintains this speed for 8 seconds, when he decelerates to a stop after another 2 seconds. After a further 9 seconds, Max heads back down the tree with constant acceleration, reaching a speed of 2.5 m/s in 2 seconds. He maintains this speed for 5 seconds, when he jumps from the tree. (Take positive as up.) a Draw a velocity–time graph representing the motion of the monkey until he leaves the tree. b At what height did Max leap off the tree? c What was the total distance travelled by Max on the tree? d What was Max the monkey’s average speed: i while on the tree? ii while in motion on the tree? Challenge: When Max begins his descent, a palm leaf falls from the tree at a height of 25 metres. It falls with a constant acceleration of 2 m/s2. e Verify that Max the monkey is still on the tree when the palm leaf hits the ground and determine where Max is at this time.

Constant acceleration formulas

Velocity (m/s)

Acceleration due to gravity is usually 9.8 m/s2. It can vary slightly depending on the distance from the centre of the Earth. This means that a falling object or an object thrown into the air is subject to a constant (or uniform) downward acceleration of 9.8 m/s2. Since acceleration is a vector quantity, when the object is moving upwards, it is subject to an acceleration of -9.8 m/s2; that is, a deceleration or retardation. Consider an object moving in a straight line, which has an initial velocity of u. It accelerates constantly until it reaches a velocity of v after t seconds. v Its velocity–time graph is shown at right. We can use this graph to derive various formulas, which can v be applied to problems involving constant acceleration. Since acceleration, a, is the change in velocity over time, u δv a= δt t 0 t v-u = Time (s) t Multiply both sides by t: at = v − u Make v the subject, so: v = u + at [1] Furthermore, since average velocity is the change in position, s, over time, δs u+v average velocity = or δt 2 s u+v = So, t 2 1 Therefore, s = 2 (u + v)t [2] Substituting v = u + at (equation [1]) into equation [2] 1

s = 2 (u + u + at)t 1

= 2 (2u + at)t 1

= 2 (2ut + at2) 532


1

Therefore, s = ut + 2 at2 From [1], t = Substituting t =

[3]

v-u a

v-u into equation [2]: a 1  v - u s = 2 (u + v)   a  2 2 1v -u  = 2  a 

2as = v2 − u2 Therefore, v2 = u2 + 2as [4] In summary, if u is the initial velocity, v is the final velocity, s is the displacement, a is the constant acceleration and t is the time interval, then the following formulas apply for straight line motion: v = u + at s=

1 (u 2

+ v)t

[2]

1 2 at 2

[3]

+ 2as

[4]

s = ut + v2

=

u2

[1]

Notes 1. ‘At rest’ means the velocity is zero. 2. 1 m/s = 3.6 km/h. (Verify this.) 3. When an object is travelling in one direction, u can be treated as the initial speed, v as the final speed and s as the distance travelled. Worked Example 7

A stone is dropped from a bridge that is 150 metres above a river. Find: a the time taken for the stone to reach the river b the stone’s speed on impact. Give answers to the nearest tenth. Think a

1

2

b

Write

List the given information and what has to be found. 1

Find t using s = ut + 2 at2 by substituting in s = 150, a = 9.8 and u = 0.

a Given: s = 150, a = 9.8 and u = 0

Require: t = ? 1

s = ut + 2 at2 1

150 = 0 × t + 2 × 9.8 × t2

3

Solve the equation for t.

150 = 4.9 t2 30.6122 = t2 t = 30.6122 = 5.533

4

State the solution.

The stone reaches the river after approximately 5.5 seconds.

1

List the given information and what has to be found.

2

Find v using = + 2as by substituting u = 0, a = 9.8 and s = 150. v2

u2

b Given: s = 150, a = 9.8 and u = 0

Require: v = ?

v2 = u2 + 2as = 02 + 2 × 9.8 × 150 Chapter 15 Kinematics

533

3

Solve the equation for v.

v2 = 2940 v = 2940 = 54.22

4

State the solution.

The stone reaches the river at a speed of 54.2 m/s.

Worked Example 8

A driver is forced to suddenly apply the brakes of his car when a dog appears in front of it. The car skids in a straight line, stopping 2 centimetres short of the startled dog. The car skidded a distance of 12 metres for 2 seconds. a At what speed was the car travelling as it began to skid? b What was the acceleration of the car during the skid? Think a

1

2

b

Write


a Given: s = 12, t = 2 and v = 0

Require: u = ?

1 (u 2

1

Find u using s = + v)t by substituting s = 12, t = 2 and v = 0.

s = 2 (u + v)t 1

12 = 2 (u + 0) × 2 1

3

Solve the equation for u.

12 = 2 u × 2 u = 12

4

State the solution.

The initial speed of the car was 12 m/s.

1


b Given: v = 0, u = 12 and t = 2

Require: a = ?

2

Find a using v = u + at by substituting v = 0, u = 12 and t = 2.

v = u + at 0 = 12 + a × 2

3

Solve the equation for a.

-12

4

State the solution.

The acceleration of the car was -6 m/s2.

= 2a a = –6

Worked Example 9

A ball is thrown upwards at 14.7 m/s from a tower that is 50 metres above the ground. a Determine the total time that the ball is in the air before it reaches the ground. b Find the ball’s speed when it first strikes the ground. (Give answers to the nearest tenth.) Think a

1

Let u be up as the positive direction.

2

List the given information and what has to be found. 1 Find t using s = ut + at2 by substituting 2 u = 14.7, a = -9.8, s = -50.

3

534

Write a

Given: u = 14.7, a = -9.8, s = -50 Require: t = ? 1 s = ut + at2 2 -50 = 14.7t + 1 (-9.8)t2 2 50 = 14.7t - 4.9t2 4.9t2 - 14.7t - 50 = 0


4

Solve the quadratic equation by using the quadratic formula.

t= t=

-b±

b 2 - 4 ac 2a

14.7 ± (14.7)2 - 4(4.9)(- 50) 2(4.99)

t = -2.0 and 5.0 t = 5.0 seconds, since time can not be negative b

1


2

Find v using v2 = u2 + 2as by substituting in u = 0, a = 9.8 and s = 61.025.

3

Solve the equation for v.

4

State the solution.

b Given: u = 0, a = 9.8 and s = 61.025

Require: v = ?

v2 = u2 + 2as = 02 + 2 × 9.8 × 61.025 = 1196.09 v = 1196.09 = 34.5845 The ball first strikes the ground at a speed of 34.6 m/s.

REMEMBER

1. If u is the initial velocity, v is the final velocity, s is the displacement, a is the constant acceleration and t is the time interval, then the following formulas apply for straight line motion: (a) v = u + at 1 (b) s = 2 (u + v)t 1

(c) s = ut + 2at2 (d) v2 = u2 + 2as 2. When an object is travelling in one direction, u can be treated as the initial speed, v as the final speed and s as the distance travelled. 3. ‘At rest’ means that the velocity is zero. 4. 1 m/s = 3.6 km/h 5. Acceleration due to gravity is 9.8 m/s2 for falling objects and -9.8 m/s2 for objects travelling upwards.

Exercise

15c

Constant acceleration formulas 1 WE 7 A stone is dropped from a bridge that is 98 metres above a river. Giving answers to the nearest tenth, find: a the time taken for the stone to reach the river b the stone’s speed on impact. 2

A particle moving from rest with constant acceleration reaches a speed of 16 m/s in 4 seconds. Find: a the acceleration b the distance travelled.


535

3

An object travelling at 8 m/s accelerates uniformly over a distance of 20 metres until it reaches a speed of 18 m/s. Find: a the acceleration b the time taken.

4 A parachutist free-falls from an aircraft for 6 seconds. Find: a the speed of the parachutist after 6 seconds b the distance travelled after 6 seconds. 5 A ball is dropped from a tower and reaches the ground in 4 seconds. Find: a the height of the tower b the speed of the ball when it hits the ground. 6 We 8 A driver is forced to suddenly apply the brakes of his car when a cat appears in front of it. The car skids in a straight line stopping 8 cm short of the startled cat. The car skidded a distance of 15 metres for 3 seconds. a At what speed was the car travelling as it began to skid? b What was the acceleration of the car during the skid? 7 How long does it take for: a a car to accelerate on a straight road at a constant 6 m/s2 from an initial speed of 17 m/s to a final speed of 28 m/s? b a downhill skier to accelerate from rest at a constant 2 m/s2 to a speed of 10 m/s? 8 A skateboarder is travelling down a gently sloping path at a speed of 10 m/s when he stops skating. He rolls a further 60 metres before coming to a stop. Assuming the acceleration is uniform, find: a the acceleration b the time it takes to come to a stop. 9 A falcon is hovering in the air when it suddenly dives vertically down to swoop on its prey, which is 150 metres directly below it. If the acceleration is uniform and it takes the falcon 5 seconds to reach its prey, find: a the final speed of the falcon in m/s and km/h b the acceleration of the falcon. 10 A tram is travelling at 16 m/s when the brakes are applied, reducing the speed to 6 m/s in 2 seconds. Assuming the retardation is constant, find: a the acceleration b the distance travelled 2 seconds after the brakes are applied c the braking distance of the tram. 11 MC A train travels a distance of 1800 metres in 90 seconds while accelerating uniformly from rest.

536


a The speed of the train after 90 seconds can be determined using the formula: A v = u + at D s = ut +

1 2

B s= at2

1 2

(u + v)t

C C = 2πr

E v2 = u2 + 2as

b The speed in km/h after 90 seconds is: A 2 B 36 C 144 c The speed in km/h after 45 seconds is: A 72 B 36 C 144 d The distance travelled after 45 seconds is: A 225 m B 900 m C 675 m

D 216

E 40

D 216

E 20

D 1350 m

E 450 m

12 We 9 A ball is thrown upwards at 9.8 m/s from a tower that is 30 metres above the ground. a Determine the total time that the ball is in the air before it reaches the ground. b Find the ball’s speed when it first strikes the ground. (Give answers to the nearest tenth.) 13 A ball is thrown upwards at 20 m/s from a tower that is 80 metres above the ground. a Determine the total time that the ball is in the air before it reaches the ground. b Find the ball’s speed when it first strikes the ground. (Give answers to the nearest tenth.) 14 An object is projected vertically upwards from the top of a building that is 50 metres above the ground. Its initial speed is 28 m/s. If the object then falls to the ground, find: a its maximum height above the ground b the total time taken to reach the ground c the speed of the object when it reaches the ground. 15 A car moving from rest with uniform acceleration takes 12 seconds to travel 144 metres. What is its speed after 6 seconds? 16 A bird’s egg falls from a nest in a tree. If it is initially 39.2 metres above the ground, calculate: a its speed when it is halfway to the ground b its speed on striking the ground c the time taken to reach the ground.

Chapter 15

Kinematics

537

17 A cage is descending into a well at a constant speed of 2 m/s when a stone falls through the wire in the cage. If the stone reaches the water at the bottom of the well 10 seconds before the cage, find the height above the water at which the stone fell out of the cage. 18 A balloon is rising with a speed of 19.6 m/s when a gas cylinder falls off the balloon. If the balloon is 80 metres above the ground when the cylinder falls off, how long will it take the cylinder to reach the ground and what will its speed be then?

15D

Instantaneous rates of change Instantaneous velocity

t

0

1

2

3

x

0

1

8

27

The position–time graph is shown at right.

Position (cm)

As we have discussed previously, the instantaneous velocity at a given time is in fact the gradient of the position–time graph at that time. We have also seen that when the velocity is variable the position–time graph will be curved. Consider a particle moving in a straight line such that its position, x cm, at any time, t seconds, is described by the rule: x(t) = t3, t ∈ [0, 3] x Completing a table of values will give: 30 27 24 21 18 15 12 9 6 3

x(t) = t3

0

538


1

2 3 Time (s)

t

Instead, we can apply the rule: δx Average velocity = δt to estimate the gradient (velocity). This involves taking two points on the curve on either side of t = 2. To ensure that the point at t = 2 is in the middle of the two points chosen, each point must be the same distance, h, either side of t = 2. The gradient of the line that joins the two points on the curve at t = 2 - h and t = 2 + h estimates the gradient at t = 2. Finding the rise and the run between the two points allows us to calculate the gradient as v (2) =

( 2 + h )3 - ( 2 - h )3 . 2h

x

Position (cm)

30 27 24 21 18 15 12 9 6 3

Tangent at t=2 0

1

t

2 3 Time (s)

x

Position (cm)

The velocity at any given time, say at t = 2 seconds, is equal to the gradient of the curve at t = 2. The gradient of a curve at any given point is the gradient of the tangent to the curve at that point. So, the velocity at t = 2 is equal to the gradient of the tangent to the curve at t = 2. To physically determine the gradient of the tangent often leads to inaccurate results. Care needs to be taken, firstly to draw an accurate and smooth curve, then to place the tangent at exactly the right position. There is too much room for error with this process.

30 27 24 21 18 15 12 9 6 3

[(2 + h), 3 (2 + h) ] (2, 8) h h 0

1

[(2 − h), 3 (2 − h) ]

2 3 Time (s)

t

The smaller the value of h, the closer this gradient will be to the true gradient of the tangent. 1 0.1 0.01 h For example, using a calculator to find v(2) when h = 1, 0.1 and 0.01 produces the results 13 12.01 12.0001 v(2) shown in the table. It is quite clear from this table that as h gets smaller and smaller the value of v(2) is approaching 12. If it is not already obvious it becomes even more so if h = 0.001 or 0.0001 and so on. In summary, the instantaneous velocity at t = t0, v(t0), (of a particle moving in a straight line) with its position described as x(t) is found by evaluating: v (t0 ) =

x (t0 + h) - x (t0 - h) 2h

for very small values of h (h > 0). This technique uses the same process to that of differentiating from first principles which was covered in Mathematical Methods (CAS) Units 1 & 2, and thus we can say: dx the derivative of x with respect to t. dt v(t) = x′(t) v (t ) =

or

Worked Example 10

A particle is travelling in a straight line with its position, x cm, at any time, t seconds, given as x(t) = t3 - t, t ∈ [0, 3]. Find the velocity of the particle after 1.5 seconds.


539

Think

Write

1

Given the expression x(t) = t3 - t, we want v(1.5).

x(t) = t3 - t

2

Find the velocity equation by differentiating position, x, with respect to time, t (v(t) = x′(t)).

v(t) = x′(t) v(t) = 3t2 - 1

3

Substitute t = 1.5 seconds.

v(1.5) = 3(1.5)2 - 1 v(1.5) = 5.75

4

State the solution.

The velocity of the particle at t = 1.5 seconds is 5.75 cm/s.

Instantaneous acceleration When the acceleration is variable, then the velocity–time graph is curved. The instantaneous acceleration at a given time is the gradient of the velocity–time graph at that time. So, like the instantaneous velocity: The instantaneous acceleration at t = t0, a(t0), (of a particle moving in a straight line) with its velocity described as v(t) is found by evaluating: a(t0 ) =

v (t0 + h) - v (t0 - h) 2h

for very small values of h (h > 0). Again the technique uses the same process to that of differentiating from first principles, and we can say: dv a(t ) = the derivative of v with respect to t. dt or a(t) = v′(t) Worked Example 11

A particle is travelling in a straight line with its velocity, v cm/s, at any time, t seconds, given as 8 v( t ) = , t ≥ 0. t+1 Find the acceleration of the particle after 1 second. Think 1

2

540

8 Given the expression, v (t ) = we want t +1 a(1).

Write/display

v (t ) =

8 t +1

Find the acceleration equation by differentiating velocity with respect to time (a(t) = v′(t)). To do this, on the Main screen, complete the entry lines as: 8 Define v(t) = t +1 d (v (t )) dt 8 Define a(t) = (t + 1)2 Press E after each entry.


3

Substitute t = 1 second into the formula for a(t).

4

State the solution.

The acceleration of the particle at t = 1 seconds is -2 cm/s2.

Approximating velocity–time graphs We have already seen that the distance travelled by a particle travelling in a straight line is the unsigned area between the velocity–time graph and the time axis. When the acceleration is constant, we calculate the areas of rectangles, triangles or trapeziums.

v The area of the trapezium gives the distance travelled. t

If the acceleration is variable, the velocity–time graph is curved v and so it needs to be approximated by straight-line functions. This will result in the area under the graph comprising either rectangles, triangles or trapeziums. Then the distance travelled can be estimated. One way to approximate the velocity–time curve is to use a series of ‘horizontal steps’ over the required domain or time values. t This can be achieved by first dividing the domain interval into n equally sized time intervals, each h units long. Next, evaluate the velocity at the midpoint of each of these intervals. Each of these velocities can be treated as the average velocity over its corresponding interval. The result will be a ‘step function’ graph something like the figure following. v v4

Note: t4 − t3 = t3 − t2 = t2 − t1 = t1 − 0 = h units

v3 v2 v1 0

t1

t2

t3

t4

t

The unsigned area under this velocity–time graph can be found by determining the sum of each rectangular area (h × vn). This gives an estimate for the distance travelled over a given period of time. As the rectangle width (or interval width), h, gets smaller and smaller, the number of rectangles, n, increases and therefore the estimate gets closer and closer to the exact distance.


541

The following worked example will outline the steps involved, with the aid of graphs.

WorKeD exAMPle 12

eBook plus

A particle is travelling in a straight line with its velocity, v (in m/s), at any time t seconds, given as: v(t) = t2 + t, t ≥ 0 Estimate the distance travelled during the first 4 seconds of its motion by approximating the velocity with step functions each 1 unit wide.

1

Sketch the graph of v(t) = t2 + t over the domain [0, 4].

2

Since h is 1 and the domain is [0, 4], then the intervals are from 0 to 1, 1 to 2, 2 to 3 and 3 to 4.

3

The midpoints of each interval are 0.5, 1.5, 2.5 and 3.5.


WriTe/DrAW v Velocity (m/s)

ThinK

Tutorial

20 16 12 v(t) = t2 + t

8 4 0

1

2

3

4

t

4

t

Time (s)

Evaluate v(0.5), v(1.5), v(2.5) and v(3.5). These represent the height of each rectangle.

5

Sketch the step function graph over the domain [0, 4] as an approximation of the velocity–time relationship.

v(0.5) = 0.52 + 0.5 = 0.75 v(1.5) = 1.52 + 1.5 = 3.75 v(2.5) = 2.52 + 2.5 = 8.75 v(3.5) = 3.52 + 3.5 = 15.75 v Velocity (m/s)

4

20 16 12 8 4 0

1

2

3

Time (s)

542

6

Determine the sum of each rectangular area under the step function.

Area of each rectangle = length × width Total area = (0.75 × 1) + (3.75 × 1) + (8.75 × 1) + (15.75 × 1) = 1(0.75 + 3.75 + 8.75 + 15.75) = 29

7

State the unsigned area as the approximate distance travelled.

The particle travels approximately 29 metres during the first 1 4 seconds. (Compared to the exact distance of 29 3 metres.)


In summary, if the acceleration is variable, then the distance, d, travelled by a particle can be estimated from a velocity–time function by evaluating: d = h[v(t1) + v(t2) + v(t3) + … + v(tn)] where h = step function width (time interval width) n = the number of intervals tn = midpoint of time interval, n v(tn) = velocity at time, tn. The method shown above is an approximation of the displacement (area under the curve), that can be improved by reducing the step function width. However, to calculate the exact displacement (area under the curve), calculus is used. Using your knowledge from Mathematical Methods (CAS) Units 1 & 2, this can be achieved as shown in worked example 13. That is, a formula for the distance travelled by an object, d(t), can be found by finding the antiderivative of the formula for its velocity, v(t), with respect to time. A formula for its velocity, v(t) can be found by finding the antiderivative of its acceleration, a(t), with respect to time. d (t ) = ∫ v (t )dt v (t ) = ∫ a(t )dt Displacement x(t) Differentiate

Antidifferentiate Velocity v(t)

Differentiate

Antidifferentiate Acceleration a(t)

Notes 1. The signed area between a velocity–time curve and the t-axis gives the displacement. 2. If the velocity is positive over the given time interval then the displacement is equal to the distance. WorKeD exAMPle 13

eBook plus

A particle is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as: v(t) = t2 + t, t ≥ 0 Calculate the exact distance travelled during the first 4 seconds of its motion. ThinK

Tutorial


WriTe/DiSPlAy


In order to calculate the distance, d(t), travelled during the first 4 seconds of motion, the antiderivative of the expression, v(t) = t2 + t, from t = 0 to t = 4 needs to be found.

d (t ) = ∫ v (t )dt 4

d (t ) = ∫ (t 2 + t )dt 0

Chapter 15

Kinematics

543

4

2

Antidifferentiate the expression.

3

Substitute the limits and solve.

 t3 t2  d (t ) =  +   3 2 0  43 4 2   03 0 2   d (t ) =  +  - +  2  3 2   3  64 16   d (t ) =  +  - 0    3 2   d (t ) = 29 13

4

State the exact distance travelled.

The exact distance travelled during the first 4 seconds of its 1 motion, is 29 3 metres.


On the Main screen, using the soft keyboard, tap: • ) • • P Complete the entry line as: 4

∫ (t 2 + t )dt 0

Then press E.

2

State the exact distance travelled.

The exact distance travelled during the first 4 seconds of its 1

motion, is 29 3 metres. WorKeD exAMPle 14

eBook plus

m/s2

A car accelerates from rest at 2 for 5 seconds. a Write an equation for the acceleration. b Write the equation for the velocity. c Calculate the distance covered in the first 5 seconds. ThinK

WriTe


1

The acceleration is 2 m/s2.

a a(t) = 2

b

1

The velocity equation is produced by antidifferentiating the formula for acceleration.

b a(t) = 2

2

544

It is given that v = 0 when t = 0. Calculate the constant, c, using this information.

v(t) = ∫ 2 dt v(t) = 2t + c 0 = 2(0) + c c=0 ∴ v(t) = 2t, 0 ≤ t ≤ 5


Tutorial


c

1

2

To calculate the distance, d(t), covered in the first 5 seconds, antidifferentiate the velocity equation with limits 0 and 5.

c d(t) =

5

∫4 (2t)dt 5

d(t) = [t2]0 d(t) = [52 - 02] d(t) = 25

State the distance covered in the first 5 seconds.

The distance travelled in the first 5 seconds is 25 metres.

Method 2: Using a CAS calculator a

1

The acceleration is 2 m/s2.

a a(t) = 2

b

1

To determine the equation for velocity, given a(t) = 2, on the Main screen, using the soft keyboard, tap: • ) • • P Complete the entry line as:

b

∫ (2)dt + c

Then press E.

c

2

Write the equation for velocity.

v(t) = 2t + c

3

It is given that v = 0 when t = 0. Calculate the constant, c, using this information.

0 = 2(0) + c c=0 ∴ v(t) = 2t, 0 ≤ t ≤ 5

1

To calculate the distance, d(t), covered in the first 5 seconds, on the Main screen, complete the entry line as:

c

5

∫ (2t )dt 0

Then press E.

2

State the distance covered in the first 5 seconds.

The distance travelled in the first 5 seconds is 25 metres.


545

REMEMBER

1. The instantaneous velocity at t = t0, v(t0), (of a particle moving in a straight line) with its position described as x(t) is found by evaluating: x (t + h) - x (t0 - h) v (t0 ) = 0 , for very small values of h (h > 0). 2h 2. The instantaneous acceleration at t = t0, a(t0), (of a particle moving in a straight line) with its velocity described as v(t) is found by evaluating: v (t + h) - v (t0 - h) a(t0 ) = 0 , for very small values of h (h > 0). 2h 3. If the acceleration is variable, then the distance, d, travelled by a particle can be estimated from a velocity–time function by evaluating: d = h[v(t1) + v(t2) + v(t3) + . . . + v(tn)] where h = step function width (time interval width) n = the number of intervals tn = midpoint of time interval n v(tn) = velocity at time tn. 4. Displacement x(t) Differentiate


Differentiate


Exercise

15D

Instantaneous rates of change 1 WE 10

A particle is travelling in a straight line with its position, x cm, at any time,

t seconds, given as x(t) = t3 + t, t ∈ [0, 5]. Find the velocity of the particle after 2 seconds. 2

A particle is accelerating in a straight line with its position, x cm, at any time, t seconds, given as x(t) = t4, t ∈ [0, 4]. Find the velocity of the particle after 3.5 seconds.

3 A missile travelling in a straight line has its position, x m, at any time, t seconds, given by x(t) = 2t3 − 4t, t ∈ [0, 6]. Find the velocity of the missile after 4 seconds. 4 A particle is travelling in a straight line with its position, x cm, at any time, t seconds, given as x (t ) =

8 , t ≥ 0. Find the velocity of the particle after 3 seconds. t +1

1 5 The position of a lift, x m, at any time, t seconds, is given as x (t ) = - t 2, t ∈ [0, 8]. t + 2 Find the velocity of the lift after 1.5 seconds.

546


6 WE 11 A particle is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as v (t ) =

16 , t ≥ 0. Find the acceleration of the particle after 2 seconds. t+2

7 An ant is travelling in a straight line with its velocity, v (in cm/s), at any time, t seconds, given as v (t ) =

8 , t ≥ 0. Find the acceleration of the ant after 3.5 seconds. (t + 1)2

8 A lift moves with its velocity, v (in m/s), at any time t seconds, given as v (t ) = t 2 -

4 t ∈ [0, 4]. t +1

Find the acceleration of the lift after 1 second. 9 An object is accelerating in a straight line such that its velocity, v (in cm/s), at any time, t seconds, is given as v(t) = t3 + 2t2 - 3t, t ∈ [0, 6]. Find the acceleration of the object after 2 seconds. Questions 10 and 11 refer to the following information: The position of an object travelling in a straight line is given by x m. At any time, t seconds, its position is x(t) = 2loge (t + 1), t ≥ 0. 10 MC The velocity at t = 3 is nearest to: A 0.92 m/s B 2.77 m/s C 0.51 m/s

D 1.37 m/s

E 0.50 m/s

11 MC The velocity at t = 6 is nearest to: A 1 m/s B 0.29 m/s C 3.74 m/s

D 1.84 m/s

E 5.0 m/s

Questions 12 and 13 refer to the following information: An object is travelling in a straight line such that its velocity, v (in m/s), at any time, t seconds, is given as: v(t) = 3e2 − t, t ≥ 0. 12 MC The acceleration, in m/s2, after 2 seconds is equal to: B −3 C 1.5 D 3 A −1.5

E −6

13 MC Using a step function 0.2 seconds wide to approximate the velocity, the distance travelled after 2 seconds is nearest to: A 10.57 m B 9 m C 18 m D 19.14 m E 16.8 m 14 WE12 A particle is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as: v(t) = t2 + 3t, t ≥ 0. Estimate the distance travelled during the first 6 seconds of its motion by approximating the velocity with step functions each 1 unit wide. 15 An object is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as v(t) = t3 + t, t ≥ 0. Calculate the exact distance travelled during the first 3 seconds of its motion. 16 WE13 A particle is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as v(t) = t2 + 3t, t ≥ 0. Calculate the exact distance travelled during the first 6 seconds of its motion. 17 A particle starts at rest and travels in a straight line with its velocity, v (in m/s), at any time t seconds, given as v(t) = t3 + t, t ≥ 0. a Find the equation for the position of the particle with respect to time. b Calculate the distance covered in the first 3 seconds.


547

18 We14 An object initially starts from rest and accelerates in a straight line, a (in m/s2), at any time, t seconds, given as a(t) = 2t + 1, t ≥ 0. a Find the equation for the velocity of the object with respect to time. b Find the equation for the position of the object with respect to time. c Calculate the distance travelled in the first 4 seconds. eBook plus Digital doc

WorkSHEET 15.2

548

19 An object initially travelling at 15 m/s accelerates in a straight line, a (in m/s2), at any time, t seconds, given as a(t) = 12t2 - 4t + 4, t ≥ 0. a Find the equation for the velocity of the object with respect to time. b Find the equation for the position of the object with respect to time. c Calculate the distance travelled in the first 2 seconds.


Summary Position and velocity

• A particle’s position gives its location relative to a reference point, usually the origin, O. • A particle’s displacement is the change in its position relative to a fixed point. Displacement gives both the distance and direction that a particle is from a point. Displacement = final position − initial position • The average velocity of a particle is the rate of change of its position with respect to time. Average velocity = = • Average speed =

change in position change in time final position - initial position change in time


Velocity–time graphs and acceleration–time graphs

change in velocity change in time • The signed area between a velocity–time graph and the time axis is equal to the change in position or displacement. The area above the time axis is positive displacement and the area below the time axis is negative displacement. • The unsigned area between a velocity–time graph and the time axis is equal to the distance travelled. • Final position = displacement + initial position • Average acceleration =

Constant acceleration formulas

• If u is the initial velocity, v is the final velocity, s is the displacement, a is the constant acceleration and t is the time interval, then the following formulas apply for straight line motion: 1

v = u + at s = 2 (u + v)t 1

s = ut + 2 at2 v2 = u2 + 2as • When an object is travelling in one direction, u can be treated as the initial speed, v as the final speed and s as the distance travelled. • ‘At rest’ means that the velocity is zero • 1 m/s = 3.6 km/h. • Acceleration due to gravity is 9.8 m/s2 for falling objects and -9.8 m/s2 for objects travelling up. Instantaneous rates of change

• The instantaneous velocity at t = t0, v(t0), (of a particle moving in a straight line) with its position described as x(t) is found by evaluating: v (t0 ) =

x (t0 + h) - x (t0 - h) 2h

for very small values of h (h > 0). • The instantaneous acceleration at t = t0, a(t0), (of a particle moving in a straight line) with its velocity described as v(t) is found by evaluating: a(t0 ) = for very small values of h (h > 0).

v (t0 + h) - v (t0 - h) 2h


549

• If the acceleration is variable, then the distance, d, travelled by a particle can be estimated from a velocity–time function by evaluating: d = h[v(t1) + v(t2) + v(t3) + … + v(tn)] where h = step function width (time interval width) n = the number of intervals tn = midpoint of time interval, n v(tn) = velocity at time, tn. Displacement x(t) Differentiate


Differentiate


550


chapter review 4 Consider the velocity–time graph shown to find:

Short answer

v

1 a Represent the following situation on a position– time line. A particle starts at S, 5 units to the left of the origin. It is then displaced 6 units to A, followed by a displacement of -4 units to B. It undergoes a final displacement of 10 units to F. b If the movement described above takes 4 seconds and the measurements are in centimetres, determine: i the displacement of F from S ii the total distance travelled by the particle iii the average velocity iv the average speed.

Velocity (m/s)

15

a b c d

3 5

15 t

10

the acceleration in the first 4 seconds the acceleration in the last 3 seconds the total displacement the total distance travelled.

5 Consider the motion of an elevator, which has its velocity–time graph as shown. Take positive values to represent upward motion.

50 Position (cm)

6

Time (s)

x

40 30

v

20

12

10

A

B

8 2

4

6

8

10 12 14 16 18

t

Time (s)

a What is the initial position of the particle? b What is the final displacement of the particle from its starting position? c Write the times for which the velocity is: i positive ii negative iii zero. d Hence, find the velocity for each of the three time intervals in part c. 3 Draw a velocity–time graph to match the following description. An object, which is moving in a straight line, has an initial velocity of -6 m/s. It accelerates at a constant rate until it reaches a velocity of 6 m/s after 12 seconds. It maintains this velocity for 6 seconds and then decelerates at a constant rate for a further 3 seconds, when it comes to rest.

Velocity (m/s)

0

9

0

2 The position–time graph shows the position of a moving particle, x centimetres to the right of the origin, O, at various times, t seconds. 60

12

4 20 22 24 26 28 30 D Gt 2 4 6 8 10 12 14 16 18 C

O −4 −8 −12

E

F

Time (s)

a Determine the acceleration for each section of the lift’s journey. b Sketch the acceleration–time graph. c If the lift started at ground level, 0 metres, determine its position at: i C ii G. d Determine the average velocity of the lift. e How far did the lift travel?


551

6 An object travelling at 5 m/s accelerates uniformly over a distance of 60 metres until it reaches a speed of 20 m/s. Find: a the acceleration b the time taken.

Multiple choice

Use the position–time line below to answer questions 1 to 4. Ct=6

Ft=8 Bt=4

St=0

7 A ball is dropped from a tower and reaches the At=2 ground in 3 seconds, using a = 10 m/s2. Find: a the height of the tower −10 −8 −6 −4 �2 0 2 4 6 8 10 12 14 16 18 20 22 b the speed of the ball when it hits the ground. 1 The displacement of F from S, in cm, is: 8 A tram is travelling at B 24 C 60 A −20 12 m/s when the D 20 E 12 brakes are applied, 2 The distance travelled in moving from S to F, in reducing the speed to cm, is: 4 m/s in 3 seconds. A −20 B 60 C 48 Assuming the −20 E 28 D retardation is constant, 3 The average speed in moving from S to F, in cm/s, find: is: a the acceleration A 2.5 B 20 c -2.5 b the distance D 4 E 7.5 travelled 4 The average velocity in moving from B to C, in 2 seconds after the cm/s, is: brakes are applied A 10 B 5 C -10 c how long it takes the tram to stop after the -5 E 0.1 D brakes are applied Questions 5 to 7 refer to the following velocity–time d the braking distance of the tram. graph. v 9 The position of a lift, x m, at any time, t seconds, is

Find the velocity of the lift after 5 seconds. 10 A particle is travelling in a straight line with its velocity, v (in m/s), at any time, t seconds, given as v(t) = (2t + 3)2, t ≥ 0. a Find the acceleration of the particle after 2 seconds. b Find the distance travelled during the first 4 seconds of its motion. 11 A passenger jet of mass 48 000 kg moves from rest with constant acceleration along a runway due to a total thrust of 105 600 newtons supplied by its engines. Assume that air resistance and other frictional forces are negligible. The magnitude of the acceleration of the jet is 2.2 m/s2. a How many seconds, correct to one decimal place, does it take the jet to reach its lift-off speed of 70 m/s? b What distance is needed, correct to the nearest metre, for the jet to take off?

552

10

x(t) = (t + 3)2 + 5, t ∈ [0, 8]

[©VCAA 2006]

B

8 Velocity (m/s)

given as:

C

6 4 A 2 0 −2 −4

2

4

6

D

E 8 10 12 t

Time (s)

F

5 The magnitude of the acceleration is greatest between the points: A A and B B B and C C C and D D D and E E E and F 6 The average velocity from A to F is equal to: A 14 m/s B -4 m/s C 4.25 m/s -4.25 m/s E 6 m/s D 7 The average speed is equal to: A 5.25 m/s B -5.25 m/s C 4.25 m/s -4.25 m/s E 6 m/s D


x

Questions 8 to 11 refer to the following information: A drag car travels a distance of 600 metres in 12 seconds while accelerating uniformly from rest.

8 The acceleration of the drag car can be determined using the formula: A v = u + at 1 B s = 2 (u + v)t C C = 2πr 1 D s = ut + 2 at2 E v2 = u2 + 2as 9 The speed, in km/h, after 12 seconds is: A 72 B 360 C 180 D 480 E 320 10 The speed, in km/h, after 6 seconds is: A 36 B 90 C 240 D 180 E 160 11 The distance travelled after 6 seconds is: A 300 m B 200 m C 150 m D 500 m E 450 m

Questions 12 and 13 refer to the following: A particle is travelling in a straight line with its position, x m, at any time, t seconds, given as x(t) = t3 – 4t2, t ≥ 0. 12 The velocity at t = 3 is: B 4.5 m/s C 3 m/s A –3 m/s D 1.5 m/s E 5 m/s 13 The acceleration at t = 3 is: A –9 m/s2 B 10 m/s C 3 m/s2 2 2 D 9 m/s E 10 m/s Questions 14 and 15 relate to the following information: An object is travelling in a straight line such that its velocity, v (in m/s), at any time, t seconds, is given as v(t) = (t + 1)2, t ≥ 0. 14 The acceleration, in m/s2, after 2 seconds is equal to: A 9 B -3 C 6 D 3 E -6 15 The distance travelled from t = 2 to t = 4: A 8 2 m

B 32 2 m

D 16 1 m

E

3

3

3 17 1 3

C 411 m 3

m

Extended response

1 An object moves in a straight line so that its position, x cm, from a fixed point, O, on the line at time, t seconds, is given by the rule: x(t) = –t2 + 4t + 12, t ∈ [0, 6] a Sketch the position–time graph for the particle. b Determine the average velocity of the particle. c Show the movement of the particle on a position–time line. d What is the particle’s average speed? e Find the velocity of the particle at: i t = 1 and ii t = 3 seconds. f During what times is the particle travelling faster than it is at t = 1 and t = 3 seconds? Stuntman Bus

Velocity (m/s)

2 During the filming of a movie, a stuntman has to chase a moving bus v and jump into it. The stuntman is required to stand still until the bus 10 passes him. He must then start chasing it. The velocity–time graph at 8 right describes the motion of the stuntman and the bus from the instant 6 the bus door passes the stuntman. a At what instant did the stuntman reach the same speed as the bus? 4 b What is the acceleration of the stuntman during the first 4 seconds? 2 c At what instant did the stuntman catch up to the bus? d How far did the stuntman run to reach the door of the bus? 0 2 Suppose the bus accelerates after 8 seconds at 1 m/s2 until it reaches 11 m/s and the stuntman maintains his speed of 10 m/s. e How far behind the bus is the stuntman after 8 seconds? f Verify that the stuntman is still behind the bus when the bus stops accelerating. g Explain why the stuntman will never catch the bus.

4

6

8

10 12

t

Time (s)


553

3 A girl at the bottom of a 100-m high cliff throws a tennis ball vertically upwards. At the same instant, a boy at the very top of the cliff drops a golf ball so that it hits the tennis ball while both balls are still in the air. The acceleration of both balls can be taken as 10.0 m/s2 downwards. a If the balls collide when the tennis ball is at the top of its path, what is the position of the tennis ball when it strikes the golf ball? b With what speed is the tennis ball thrown for this to occur? c What is the speed of the golf ball when it strikes the tennis ball? d How long has each ball been in motion when they collide? 4 Alana can accelerate to her maximum speed of 8 m/s in 1.6 seconds. Her sister Lily can accelerate to her maximum speed of 8.2 m/s in 2 seconds. Assume that they both accelerate uniformly and they can maintain their maximum speed once they reach it. Alana challenges Lily to a 100-metre race. a Who will win the race? b What is the winning margin? The girls’ brother Blake has a strong interest in handicap racing. He works out two variations of a handicap that will result in a dead heat. c If both girls run the full distance, how much earlier should the loser have started for a dead heat to result? d If they start at the same time, how much less distance should the loser have to cover for a dead heat to result? Blake trains the loser to accelerate fast enough for a dead heat to result. e Find this acceleration. eBook plus Digital doc


554


eBook plus

ACTiViTieS


• 10 Quick Questions: Warm up with ten quick questions on kinematics. (page 511) 15A

Introduction to kinematics

Tutorial

• We 2 int-1178: Watch how to use a position–time graph to determine values for time, displacement, velocity and average speed. (page 515) 15B

Velocity–time graphs and acceleration– time graphs

Interactivity

• Motion graphs (kinematics) int-0267: Consolidate your understanding of motion graphs using the interactivity. (page 521) Tutorials

• We 5 int-1179: Watch how to use a velocity–time graph to determine values for acceleration, displacement and the distance travelled of an object. (page 524) • We 6 int-1180: Watch how to use a velocity–time graph to determine values for acceleration, displacement and the distance travelled of an object. (page 526) Digital doc

• WorkSHEET 15.1: Interpret and create position– time lines, interpret and create position-time and velocity-time graphs to aid in solving worded problems. (page 532)

15D

Instantaneous rates of change

Tutorials

• We 12 int-1181: Watch how to estimate the distance travelled of a particle moving in a straight line using a velocity–time graph. (page 542) • We 13 int-1182: Watch how to calculate the exact distance travelled of a particle given its velocity as a function of time. (page 543) • We 14 int-1183: Watch how to write the equations for acceleration and velocity of a car. (page 544) Digital doc

• WorkSHEET 15.2: Calculate duration, speed, velocity and acceleration of bodies in motion. (page 548) Chapter review Digital doc


Chapter 15

Kinematics

555

i6 Geometry in two and three dimensions areas oF sTudy

• Angle sum of a triangle and of polygons • Straight edge and compass constructions such as: – construction of a line parallel to a given line – bisecting a given angle – perpendicularly bisecting a line segment – construction of a perpendicular to a given line from a point not on the line – construction of exact angles of 60°, 30° and 45° • Construction and investigation of various regular and star polygons • Construction and investigation of polyhedra, Platonic solids

16a 16b 16c 16D 16e 16F 16G

Review of basic geometry Geometric constructions Polygons Three-dimensional geometry Circle geometry Tangents, chords and circles Geometry in architecture, design and art

• Geometry in art and design, tessellations, patterns, perspective • Theorems relating to angles in a circle, such as: – the angle subtended at the circumference is half the angle subtended at the centre by the same arc – angles in the same segment of a circle are equal (and converse) – opposite angles in a cyclic quadrilateral are supplementary – the alternate segment theorem • Theorems on intersecting chords where the chords intersect inside or outside the circle (as secants), as well as limiting case, where one of the lines is a tangent eBoo k plus eBook

16a

review of basic geometry

Digital doc

10 Quick Questions

Although many of the facts, definitions and theorems that we will use in this chapter may be familiar to you, it is worthwhile to collect them here in one place. We can start with the understanding that a point in space is a location with no dimensions; that is, it is a concept that helps us determine the location of real objects. A point is often labelled with a capital letter.

lines and angles A line segment joins two points. It is a basic postulate (or principal) that only one straight line segment can join two given points. In the figure at right, there is exactly one straight line segment that can A B join points A and B. If we extend the line segment to ‘infinity’ in both directions, B A we have what is properly called a line. If we extend the line segment to ‘infinity’ in one direction only, we have a ray.

556


In the figure at right we have a ray extending from point A. Often A B the terms line segment, line and ray are used interchangeably. The context should help you determine which of the three kinds of ‘line’ is really being used. Three points can determine an angle. In Figure 1, the angle, as indicated A B by the arc of a circle, is named ∠BAC or ∠CAB. The vertex of the angle is at point A and is placed in the middle of the name, so we do not write ∠ABC, or ∠BCA in this example. C A straight angle is formed when the three points of the Figure 1 previous example are all in a D line. In Figure 2, the straight angle is ACB, with the vertex at C. Note that another line C A B segment can be drawn to point D. By convention, a Figure 2 straight angle equals 180 degrees and is written as ∠ACB = 180°. Therefore, it can be seen that ∠ACD + ∠DCB = ∠ACB = 180°. These two angles are called supplementary angles. D If the point D is moved so that ∠ACD = ∠DCB, then we have created a right C A B angle. In Figure 3, both Figure 3 ∠ACD and ∠DCB are right angles. From the previous equation, since ∠ACD = ∠DCB, we can rewrite ∠ACD + ∠DCB = ∠ACB = 180° to ∠ACD + ∠ACD = 180° 2(∠ACD) = 180° ∠ACD = 90° So a right angle = 90°. A small box at the vertex is used to indicate a right angle. See the figure in worked example 1. Worked Example 1

Find the value of x in the given diagram.

D

E 32°

C

A Think 1

Write an equation involving the required angle. ∠ACB is a straight angle.

2

Replace angles with known values. ∠BCD = 90° (right angle).

3

Solve for the missing angle.

x B

Write

∠BCD + x + ∠ECA = 180° 90° + x + 32° = 180° 122° + x = 180° 122° – 122° + x = 180° – 122° x = 58°

Chapter 16 Geometry in two and three dimensions

557

Parallel lines and angles It is a basic postulate of geometry that parallel lines never meet. When parallel lines are intersected (cut) by a transversal (straight line), special angles are formed. We shall now look at the properties of several of these angles. Consider the pair of parallel lines in the figure at right. Note Transversal the use of arrows to indicate that they are parallel. Now consider D E the two angles ∠ABC and ∠BDE, marked with arcs. Since the parallel lines (E and C) never meet, we could easily move one B C A on top of the other without affecting the angles. Thus it stands to reason that the angles are equal. These equal angles are called D E corresponding angles. B Now, consider the two crossing lines shown in the figure at C A right. Since ∠EBD and ∠ABC are both straight angles (180°) we can write: C ∠EBC + ∠CBD = ∠CBD + ∠ABD ∠EBC = ∠ABD

E

D

B A

(the two angles indicated with arcs). These angles are called vertically opposite angles. Next, consider the parallel lines at right, and the two angles D ∠CBD and ∠EDB (indicated with arcs). They are positioned B A ‘inside’ the parallel lines, on the same side (to the right) of the transversal. These angles are called co-interior angles and sum to 180°. D Finally, consider the pair of parallel lines on the right and the B two angles ∠EDB and ∠FBD. F A From the earlier result about parallel lines we know that ∠EDB = ∠CBA, and from the result about crossing lines, that ∠CBA = ∠FBD. Therefore, ∠EDB = ∠FBD (the two angles indicated with arcs). These are called alternate angles. From the above results, we are in a position to perform our first ‘proof’. Worked Example 2

Prove that the sum of the (interior) angles of any triangle is 180 degrees. In other words, prove that: a + b + c = 180°. Think 1

b a

Draw/write

Construct a line parallel to one of the sides.

b a

2

558

Label some additional angles (d and e) which will be needed.

3

The upper parallel line is also a straight angle.

4

But we have some alternate angles.

5

Substitute into the equation in step 3.

c d

b

a

e c

d + b + e = 180° d=a e=c a + b + c = 180° . . . QED


c

E C

E C

QED is Latin for quod erat demonstrandum (‘as has been demonstrated’), but could just as well be thought of as ‘Quite Easily Done’! Remember that, when proving something, you are allowed to use other facts which have already been established, without having to prove them each time. For worked example 2, step 4 we relied on the previous result about alternate angles.

angles in polygons A polygon is a closed figure made up of three or more straight line segments.

This figure is not a polygon because it is not closed.

This figure is not a polygon because it has a curved line.

This figure is a polygon because it is closed and has four (more than two) line segments, all straight.

Now that we know that the sum of angles in a triangle is 180°, we can find the sum of angles in any polygon by dividing it into triangles as shown. A F q

m n o r

s

E

p

x

t

u D

B w v

C

The sum of the angles in a hexagon = 4 × 180° (since it contains 4 triangles) = 720°

m + q + r = 180° n + s + t = 180° o + u + v = 180° p + w + x = 180° Note that there is a general formula used to determine the sum of the angles in any polygon. The development of this formula is discussed in the exercises that follow.

loci The locus (plural loci) of a point is the path traced out by a point when it moves according to a given rule. For example, a circle is formed by the locus of points in a plane that are equidistant from a fixed point. Worked example 3

eBook plus

A sprinkler is placed 3 m east and 1 m north of a tap. If the sprinkler can reach a maximum distance of 5 m, write an equation for the outmost edge of the watering distance if the tap is considered to be the origin. Think 1

The sprinkler will water a circular area. Hence, the equation will be of the form (x - h)2 + (y - k)2 = r2.

Tutorial


WriTe

The centre of the circle is at (3, 1) and the radius is 5. Hence, the equation must be (x - 3)2 + (y - 1)2 = 25.

Chapter 16

Geometry in two and three dimensions

559

rememBer

1. 2. 3. 4. 5. 6. 7.

8. 9. 10. 11. 12. 13. exerCise

16a

A point is a marker for a location in space. Two points connected make a line segment. Extend the line segment to infinity in one direction to construct a ray. Extend the line segment to infinity in both directions to construct a line. Where two lines meet, an angle is formed. Two lines which never meet are parallel. A line intersecting a pair of parallel lines is called a transversal and creates a number of special angles. (i) Corresponding angles are equal in magnitude (size). (ii) Co-interior angles add up to 180°. (iii) Alternate angles are equal in magnitude (size). Vertically opposite angles are equal in magnitude (size). Supplementary angles add up to 180°. Complementary angles add up to 90°. The sum of interior angles in a triangle is 180°. A polygon is a closed figure made up of three or more straight line segments. The locus of a point is the path traced out by a point when it moves according to a given rule.

review of basic geometry 1 We 1

Find the values of the pronumerals in the following figures.

a

eBook plus

b

Digital doc

34°

SkillSHEET 16.1

2 mC

int-0997

122°

128°

b isosceles e right-angled

c complementary

3 mC Which of the following statements is false? a An angle is always formed when two lines meet. b The sum of angles in a right-angled triangle is 180 degrees. c A line segment extended to infinity in both directions is called a ray. D Parallel lines never meet. e Reflex angles are bigger than acute angles. 4

5

560

45° x x

Angles which add up to 180 degrees are called:

a parallel D supplementary

Interactivity Bisecting a line segment

xy

x

Geometry review

eBook plus

c

We 2 Prove that the co-interior (or allied) angles marked with arcs in the figure at right are supplementary. In other words, prove that ∠EDB + ∠DBC = 180°.

From the result in worked example 2 and question 4, show that there are only two different angle values: all angles in that figure must take on one or the other of these values. Furthermore, show that if you know only one of these angles, then all other angles can be determined. Demonstrate this using the value in the figure at right.


E

D F

F

B A 101° D B

C

E A

C

6

Prove the following theorem: ‘The value of the exterior angle of a triangle equals the sum of the other two interior angles’. In other words, in referring to the figure at right: d = a + b.

7

b a

c d

Find the values of the unknown angles in the following figures. a x

b

y

2x

c

y

47°

66°

55° z

y z x

31°

x

8 Find the values of the unknown angles in the following figures. a

75° y x

z

b

c

30°

x

3x

20° x 40°

x + 3°

y

9 Show that the sum of the interior angles of the pentagon at right equals 540 degrees. In other words, show that: ∠AED + ∠EDC + ∠DCB + ∠CBA + ∠BAE = 540°

A

B C

E D

10 Find the value of x in the figure at right.

x − 22°

x x + 15°

x − 20°

x + 10° x + 5° A

11 Prove that the diagonals of a square form a right angle; that is, that angle a in the figure at right equals 90 degrees.

B a

D 12 a Copy and complete the following table relating the sum of the interior angles to the number of sides of the polygon. b Can you establish a general formula for the sum in terms of n, the number of sides?

Number of sides (n) Sum of interior angles

3

4

180°

5

6

7

8

9

10

C

20

540°

13 What shape is the locus of points 3 cm from Point A? 14 Write the equation for the locus of points that are 1 unit from the origin. 15 WE 3 A television broadcasting centre has its location at (-1, 3) on a grid. If the broadcasting range is 50 km, write an equation to represent the outer edge of the broadcast region. 16 What path is formed by the locus of points equidistant from two points P and Q?


561

16B

Geometric constructions A traditional part of any study of geometry includes the skill of producing constructions. In this section, you are to use only a straight edge (you can use a ruler, but no measuring!) and a pair of compasses. Note: The correct term for this instrument is a pair of compasses (not to be confused with the compass used in navigation). For convenience, the pair of compasses will sometimes be referred to only as compasses, and the point as a compass point. Armed with our straight edge and compasses, we are able to construct a variety of geometric figures. We shall now look at constructions that show us how to bisect lines, bisect angles, and draw special angles (for example, 60°). The only new definition required here is bisection. You can bisect a line by dividing it in half, or bisect an angle so that its measure is halved.

Bisecting lines In the following examples, constructions drawn with a ruler and a pair of compasses are shown in grey.

Worked Example 4 a Use a ruler and a pair of compasses to bisect a line, AB. b Use a CAS calculator to bisect a line segment. Think a

b

562

1

Draw/display

(a) Draw a line AB. (b) Place the compass point at A, with any radius (more than half the length of the line). (c) Draw a circle.

2

With the same radius as in step 1, repeat for point B.

3

The two circles will intersect at two points. Join these points with a straight line.

1

On the Geometry screen, select the line segment by tapping: • Draw • Line Segment (or from the drop down menu as shown)

a

A

b


B

2

To draw a line segment, tap the screen to set your first point and then tap the screen again for your second point. A line segment will be drawn between the two points.

3

To bisect the line, tap G. Then tap on the line AB, to select it and tap: • Draw • Construct • Perp. Bisector

With this construction, not only have you divided the original line (AB) in half, but the line that you drew in step 3 is perpendicular (at a 90° angle) to AB. Therefore, this line is called the perpendicular bisector of AB. Why does it work? It is useful to examine the geometry of the construction to help understand why it provides the correct result. The figure at right shows the essential part of the construction. C Points C and D are the intersections of the two circles. Join points A to C and B to C to create two triangles, as shown A E B in the next figure. D Because the two circles had the same radius, AC = BC. The two triangles also share the common side CE. By symmetry, ∠ACE = ∠BCE and ∠CAE = ∠CBE. Therefore, since two of the angles are equal; the third angles are also equal; that ∠AEC = ∠BEC. C From all this it is clear that the two triangles are identical. Therefore, AE = EB and we have effectively bisected AB A E B (divided it into two equal parts).

Worked Example 5 a Use a ruler and compasses to construct a line parallel to a given line. b Use a CAS calculator to construct a line parallel to a given line.


563

Think a

b

Draw/display

1

Pick any two points, A and B, on the given line.

2

From point A, draw a circle of any radius (more than half the distance from A to B).

a

3

With the same radius repeat step 1 at point B.

4

Join the highest points (or lowest points) of the circles with a straight line. This will be parallel to AB.

1

b Construct a straight line segment in the same manner as for worked example 4. To construct a parallel line, tap [ and touch the screen to place point C. Tap G. Then tap the line AB and point C to select them.

2

Then tap: • Draw • Construct • Parallel

A

B

Note: The radius is the distance between the compass points, and this determines the distance between the parallel lines.

Bisecting angles Another important construction is that employed to bisect angles. Worked Example 6 a Use a ruler and compasses to bisect any angle. b Use a CAS calculator to bisect any angle.

564


Think a

1

2

draw/display

With any radius, and the point of the compass at the vertex V, draw an arc of a circle which crosses both arms of the angle. The crossings are labelled A and B.

a

V

With any radius and the point of the compass at A, draw an arc inside the angle. This arc should be long enough so that the line representing half the angle would cross it.

3

With the same radius, repeat step 2, putting the point of the compass at B. The two arcs will cross at point C.

4

Join the vertex V to C. This line bisects the angle, namely: ∠AVB ∠AVC = 2

V

V b

1

On the Geometry screen, construct a straight line segment in the same manner as for worked example 4. Then make another line segment from B to C. To construct an angle bisector, tap G and select lines AB and BC.

2

Then tap: • Draw • Construct • Angle Bisector

B A

B A

C

B A

b


565

When drawn carefully, this construction is an accurate way of halving an angle; it is even more accurate than with a protractor. For example, if the original angle was 68.3°, it would be difficult with a protractor to obtain an angle of 34.15°.

Constructing angles The last group of constructions involves the ‘special angles’ of 30°, 45°, 60° and 90°. You have already seen how to construct a 90° angle (see page 567), so bisecting this will produce a 45° angle. Worked Example 7

Use a ruler and compasses to construct a 60° angle. Think 1

Draw

Draw a line as the base of the angle. Select a point for the vertex, A; put the compass point there and draw a circle of any radius, crossing the line at B. A

2

With the same radius as in step 1, put the compass point at B and draw an arc, crossing the other circle at two places, C and D.

B

D A

B C

3

Join A to C. Angle CAB = 60°.

D

A

B C

There are several other 60° angles in the construction above. Can you find them? (Hint: Use the symmetry of the construction.) A 30° angle can be constructed by bisecting a 60° angle. REMEMBER

1. Geometric constructions are made with a pair of compasses and straight edge only. 2. The perpendicular bisector of a line divides the line in half and generates a single angle. 3. For a line parallel to any given line, the radius (the distance between the compass points) determines the distance between them. 4. The angle bisector divides any angle in half. 5. From a 60° angle and a perpendicular bisector, 30° and 45° angles can be made.

566


exerCise

16B eBook plus Interactivities

int-0998 Bisecting an angle

int-0999 Circumcentre

Geometric constructions C

1 We 4 Using a ruler and compasses, construct perpendiculars AC and BD to the line AB. 2 a Consider the isosceles triangle in the figure at right. Construct perpendicular bisectors of each of the three lines AB, BC, and CA. b What do you notice about these bisectors? 3 a Repeat question 2 for any scalene triangle. b What do you notice?

B

A A

B

D

C

4 We 5 A pair of circular pulleys of the same radius is connected by a closed band of rubber. The distance between the pulleys is equal to their diameter. Make a straight-edge-and-compass construction of this system. 5 Three circular pulleys of the same radius are arranged as follows: Pulley A is directly above Pulley B, at a distance equal to twice the diameter. Pulley C is to the right of Pulley B at a distance equal to three times the diameter. The rubber band connecting them runs on the ‘outside’ of the system. Make a straight-edge-and-compass construction. 6 Sometimes it may be necessary to use a ruler to help with the construction. Make two parallel lines exactly 20 cm apart. (Hint: How can you do this if the compasses won’t open wide enough?) 7 We 4, 6 Using a ruler and compasses only, construct a 45° angle. 8 Using a ruler and compasses only, construct a 30° and then a 15° angle. 9 We 6, 7 Use the following set of instructions to duplicate the angle ABC. Step 1. With your compass point at B, trace an arc cutting AB and BC at D and E respectively. Step 2. Draw a line similar to line BC on another sheet of paper. A Step 3. With the same radius as in step 1, draw an arc on the new line. Step 4. Use your compass to ‘measure’ the distance from D to E. Step 5. Use this radius to draw an arc on the new line, putting the B C compass point where the first arc cut the line. Step 6. Join the vertex of the new line with the point where the two arcs cross. eBook plus Digital doc

WorkSHEET 16.1

16C

10 In worked example 3, a perpendicular line was constructed from a point not on the line. Devise a method of constructing a perpendicular from a point on the line.

A

polygons As we have seen, a polygon is a closed figure with straight sides. In this section, we look at the following aspects of polygons: 1. triangle constructions 2. quadrilaterals 3. regular polygons 4. star polygons.

Triangle constructions Let us investigate the properties of the perpendicular bisectors of each side of any triangle. Worked example 8

eBook plus

Construct the perpendicular bisectors and median bisectors of each side of any triangle and investigate their properties.

Tutorial


Chapter 16


567

Think

568

1

Draw any triangle and add circles centred at each vertex. The radius should be large enough so that perpendicular bisectors can be drawn.

2

Use the construction circles to draw perpendicular bisectors. The black lines join pairs of intersecting arcs. It should be clear that the bisectors all meet at a point. This point is called the circumcentre.

3

Use this point as a centre and draw a circle which just touches each vertex. To do this, put the compass point at the point where the bisectors met and the pencil point at any vertex. You should observe that the resultant circumcircle (or outcircle) just touches each vertex.

4

Alternatively, on the Geometry screen construct a triangle by tapping P. Then tap the screen to draw the triangle. You can adjust the size and position of the triangle by tapping and dragging.

5

Then tap G. Select the line you want the perpendicular bisector to pass through and tap: • Draw • Construct • Perp. Bisector Repeat these steps for the other two sides of the triangle.

DRAW/display


6

To draw the circumcircle, tap U. Then tap the intersection point (circumcentre) and one of the vertices of the triangle.

7

Furthermore, from step 2, we can determine the midpoint of each side from the perpendicular bisectors. Note the use of short bars to indicate the bisection of the sides, at points P, Q and R.

8

Q

R P

Join each midpoint to its opposite vertex. These lines also meet at a single point, called the centroid, which has applications in physics and engineering, as it is effectively the point of symmetry of the triangle.

Q

R P

Imagine that the triangle in the previous example is made out of a thick piece of cardboard. The centroid is the point where you could place a finger and ‘balance’ the triangle. The reason for this is that the median bisectors in step 5 are the three axes of symmetry of the triangle, each axis dividing the triangle into two equal areas. Since they all meet, the centroid is the point of symmetry. The incentre is the intersection of the angle bisectors of a triangle. Worked Example 9 B

Construct the incentre of the triangle shown at right: a by hand b using a CAS calculator. C Think a

1

A

Draw/display

(a) Construct the angle bisectors by drawing arcs centred at each vertex (A, B and C). (b) From the intersection of these arcs and the sides of the triangles, draw intersecting arcs between pairs of sides. In the figure at right this has been done to vertex A only, to keep the drawing uncluttered.

a

B

C

A


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2

Complete the construction of the angle bisectors and observe that they, too, meet at a point — the incentre.

3

By placing the compass point at the incentre and carefully drawing a circle, it is possible to construct the incircle, which just touches each side of the triangle.

B

A

C B

C b

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1

Construct a triangle using the method set out in worked example 8. Tap G and select the two lines you want the angle bisector to go through. Then tap: • Draw • Construct • Angle Bisector

2

Repeat this process for each of the other angles of the triangle.

3

To draw the incircle, tap U. Then tap the intersection point (incentre) and a point to allow the circle to fit inside the triangle.

A

b


Quadrilaterals There are many names for the various kinds of quadrilaterals and it is useful to look at their definitions in terms of their features.

Square The square has: 1. four equal sides 2. four right angles. A square can be constructed easily from a circle by finding perpendicular bisector of the diameter.

Rectangle The rectangle has: 1. opposite sides equal 2. four right angles. A rectangle can be constructed from a circle by using the construction shown to draw two lines parallel to a diameter. The rectangle is formed from the intersection of the parallel lines with the circle.

Parallelogram A parallelogram has: 1. opposite sides equal in length 2. opposite sides parallel 3. opposite angles equal.

Rhombus A rhombus has: 1. all four sides equal in length 2. opposite sides parallel 3. opposite angles equal.

Trapezium A trapezium (or trapezoid) has one pair of opposite sides parallel.

Other All other four-sided figures are generally just called quadrilaterals, even though the above figures are also quadrilaterals. Obviously there is some overlap between these definitions; for example, a square is a kind of rectangle, just as a rhombus is a kind of parallelogram. Worked Example 10

Construct a rhombus using compasses and a straight edge. Think 1

(a) Construct a pair of parallel lines. (b) From any point on one line (A) draw an arc of radius equal to the length of the side of the rhombus. This arc cuts the first line at B and the second line at C.

Draw

A

B C


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2

Using the same radius as in step 1, put the compass point at C and draw an arc cutting the same line at D.

A C

3

Join points A to C and B to D to form the rhombus ABDC.

A

D B

C

D

It is easy to extend this kind of construction for other types of quadrilaterals.

Regular polygons A regular polygon is one with each side the same length and with each interior angle the same size. For triangles, the regular polygon is the equilateral triangle; for quadrilaterals, it is the square. Worked example 7 should give you an idea of how to construct an equilateral triangle; worked example 10 should help you construct a square. The construction of a regular hexagon (6 sides) is particularly easy.

Worked Example 11

Construct a regular hexagon using a straight edge and compasses. Think 1

Draw

Draw a circle whose radius is equal to the length of one side of the hexagon. A

2

Using the same radius as in step 1, and with the compass point anywhere on the circle (A), draw an arc which cuts the circle at point B.

A A B

3

4

572

Repeat step 2 by: (a) putting the compass at B and cutting the circle at C (b) putting the compass at C and cutting the circle at D (c) putting the compass at D and cutting the circle at E (d) putting the compass at E and cutting the circle at F

E

F

D

A C

Join points A–B–C–D–E–F to form a hexagon.


B

Several regular polygons, such as those with 7, 9, 11 and 13 sides, cannot be drawn using just a straight edge and ruler. By dividing up the circle’s 360 degrees, it is possible to work out the 360 angle ‘between’ each side. For example, the hexagon’s 6 sides must be = 60° apart. Since 6 construction of a 60° angle is easy, the hexagon can be constructed. This does not occur with 360 regular polygons such as the heptagon, where the angle would be = 51.428 . . .°. In cases 7 like this, a protractor may be used.

Star polygons The last step of worked example 10 was to join the points, in order, to form the regular polygon. What would happen if you joined points by ‘skipping’ others? Consider the figure at right. Point A has been joined to E and then to C E F and back to A again. A similar pattern has been used starting at point B. In other words one point was ‘skipped’ for each line. In this case, D the polygon formed (after removing the central lines) is the only one A possible with 6 points: the ‘Star of David’. These types of polygon are called star polygons because of the obvious resemblance. C Star polygons can be constructed with the aid of a protractor, B as polygons which are 7-sided, 9-sided and so on have no exact construction method. Worked Example 12

Construct all the star polygons from a regular nonagon (9 sides), using a straight edge, compasses and protractor. Think 1

2

Draw

(a) Draw a circle of any radius, and draw a line from the centre to the circumference (point A). Since there are to be 9 sides, the interior angle 360 will be = 40°. 9 (b) Using a protractor, measure a 40° angle from A to cross the circle at B. Measure a 40° angle from B to cross the circle at C. Repeat for the remaining 7 angles.

B 40°

C

B 40° 40° E 40° 40° A 40° 40° 40° 40° J 40° F D

G 3

Instead of joining A–B–C–D–E–F–G–H–J–A, try skipping one point, joining: A–C–E–G–J–B–D–F–H–A.

A

H C

D

B

E

A

F

J G

H


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4

Remove the construction lines to form the star polygon.

C

D

B A

E J

F G 5

Repeat steps 3 and 4, but skip two points, joining: A–D–G–A and B–E–H–B and C–F–J–C.

H C

D

B

E

A J

F G 6

Repeat steps 3 and 4, but skip three points, joining: A–E–J–D–H–C–G–B–F–A.

H C

D

B

E

A J

F G 7

H

Repeat steps 3 and 4, but skip four points, joining: A–F–B–G–C–H–D–J–E–A. Since this is the same figure as in step 6 with the order of the points reversed, there are no more different star polygons to be drawn. Note: The greater the number of points that are skipped, the sharper the points of the star polygon become. Furthermore, if we skip too many points (as in step 7) the same star polygon is produced; thus, there is a limited number of star polygons for a given regular polygon. REMEMBER

1. The perpendicular bisectors of any triangle meet at a point called the circumcentre. From this centre a circle (the circumcircle) that just touches each vertex of the triangle can be drawn. 2. The centroid is the point where the lines connecting each vertex with the midpoint of the opposite side meet. 3. The incentre is the point where the angle bisectors of each vertex meet. From this centre a circle (the incircle) that just touches each side can be drawn. 4. A quadrilateral can be a square, rectangle, rhombus, parallelogram or trapezium. All these can be constructed using a straight edge and compasses only. 5. A regular polygon has equal sides and equal interior angles. Regular polygons of 3, 4, 5, 6, 8 and 12 sides can be constructed easily. Other regular polygons may require the assistance of a protractor. 6. A star polygon is formed by joining alternate vertices of a regular polygon in a regular pattern. Each regular polygon may have 0 (triangle, quadrilateral), 1 (pentagon, hexagon) or more different star polygons.

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Exercise

16C

Polygons 1 WE 8 Construct a scalene triangle with one angle greater than 90 degrees. Investigate the properties of perpendicular bisectors and side length bisectors. 2 Construct a right-angled triangle (one with an angle of 90 degrees). Investigate the properties of the circumcentre and centroid. 3 Which of the basic constructions of the previous sections might be used to construct an isosceles triangle? 4 An ancient method of getting a right angle is to use the Pythagorean triple: a triangle with sides of 3, 4 and 5 units. Devise a method of constructing such a triangle with compasses and a straight edge only. 5 WE 9 Construct the following triangle. Start with a base side of 6 cm. From the left-hand endpoint of the line draw a circle of radius 7 cm and from the right-hand endpoint draw a circle of 5 cm. Join the two endpoints to the place where the two arcs meet, above the line, to form the triangle. Construct the incentre of this triangle. 6 Construct a scalene triangle and determine the incentre. What did you observe about the properties of the incentre? 7 WE 10 Devise a method of constructing a parallelogram (with unequal sides) similar to that of worked example 9. 8 WE 11 Begin the construction of a regular dodecagon (12-sided polygon) using the following steps: a Construct a regular hexagon. b Join the vertices of the hexagon to the centre of the circle first used in the construction. What should the final three steps be? 9 The construction of a regular pentagon is quite difficult. Use the D following instructions to help in the compass-and-straight-edge construction. The various points defined in the steps are shown below. B Step 1. Draw a circle of any radius, and mark the centre C and the A F C M diameter AB. Step 2. Find the perpendicular bisector of AB meeting the circle at D and E. E Step 3. Find the midpoint (M) of CB and with radius MD draw an arc cutting AB at F. The side length of the pentagon is equal to the distance DF. Using this information, complete the construction of the pentagon. (Hint: This step is similar to the construction of a hexagon.) 10 WE 12 The stars on the Australian flag are 7-pointed (the points represent the 6 states and the Northern Territory). There is no compass- and straight-edge construction, so use a protractor to help you construct the 7-pointed star polygon. 11 Construct all the different star polygons possible from an octagon. What is an easy compass-and-straight-edge construction of the octagon?


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16D

Three-dimensional geometry In this section we shall investigate some facets of solids, as opposed to two-dimensional or plane objects that we have studied thus far. However, we shall limit ourselves to polyhedrons.

Nets of polyhedrons A polyhedron is a solid object made up of plane (flat) polygons. So a cube and a pyramid are examples of polyhedrons, but a cylinder is not. A net of a polyhedron is a ‘flattened’ representation of a solid. Since each face of the polyhedron is itself a polygon, the net (or collection of all faces) must be a collection of polygons, such that, when folded correctly, the solid can be constructed. The place where two faces meet is called an edge, while the place where three or more faces meet is called a vertex. Worked Example 13

Construct a net for the triangular pyramid depicted at right (base is shaded). Note: The diagram is in three dimensions, so is in perspective. Consequently, your two-dimensional net will have sides and shapes which look slightly different. Think

Count the number of faces.

2

Count the number of edges. Edges occur where two faces meet.

3

Count the number of vertices. Vertices occur where three or more faces meet. Use the information from steps 1 to 3 to help construct the net. Since there are four triangular faces, the net must have four triangles. If you folded the net along the lines BC, BD and CD so that the three A points met, you would be able to construct the pyramid.

4

B C

write/Draw

1

A

D

The faces are: • base, or polygon BCD (shaded) • front, or polygon ACD • left, or polygon ABC • right, or polygon ABD. There are four faces. The edges are represented by the lines: BC, BD, CD, AC, AD, AB. There are six edges. The vertices (or corners) are A, B, C, D. There are four vertices. A

A B

C

D

A

Note: There are several possible nets for the same object, but all of them require that one or more vertices are shown more than once. In the above example, vertex A was triplicated. Can you construct another net for the above figure where a different vertex is triplicated? There is a mathematical relationship between the numbers of vertices, edges and faces. This is known as Euler’s formula. Let V be the number of vertices, E be the number of edges and F the number of faces or regions. Then: V = E − F + 2. In worked example 13, we can confirm this: as V = 4, E = 6, F = 4, so substituting in Euler’s formula gives 4 = 6 − 4 + 2. This formula can be used to determine, say, the number of edges, if you know the number of vertices and faces.

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The Platonic solids A Platonic solid (named after the Greek philosopher Plato) is one where all the faces are regular polygons. It turns out there are only five Platonic solids in all. 2. The cube is made up of 6 faces of squares.

1. The tetrahedron is made up of 4 faces of equilateral triangles. This is the same basic object as in worked example 12. 3. The octahedron is made up of 8 faces of equilateral triangles.

4. The dodecahedron is made up of 12 faces of pentagons.

108°

5. The icosahedron is made up of 20 faces of equilateral triangles.

Worked Example 14

Confirm Euler’s formula for the octahedron.

B

C A

Think

Write

E

F

D

1

Count the number of faces.

There are 8 faces, determined by the triangles: ABE, ABC, ACD, ADE, FED, FDC, FCB, FBE.

2

Count the number of edges.

The edges are represented by the lines: AB, AC, AD, AE, BC, CD, DE, EB, FB, FC, FD, FE. There are 12 edges.

3

Count the number of vertices.

There are 6 vertices: A, B, C, D, E, F.

4

Write Euler’s formula.

V=E−F+2

5

Confirm Euler’s formula.

V = 6, E = 12, F = 8 6 = 12 − 8 + 2 ✓

REMEMBER

1. A polyhedron is a solid made up of flat polygonal faces. 2. The net of a polyhedron is a two-dimensional representation of the solid, showing connected edges.


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3. The place where two faces meet is called an edge; the place where three or more faces meet is called a vertex. 4. For any solid, Euler’s formula applies: Vertices = Edges − Faces + 2. 5. A Platonic solid is a polyhedron made from a single regular polygon. There are only five different Platonic solids. Exercise

16d

Three-dimensional geometry 1 WE 13 Construct a net for a rectangular box of dimensions 40 cm by 30 cm by 20 cm. 2 Confirm Euler’s formula for the solid in question 1.

G

3 WE 14 List the faces, edges and vertices for the solid drawn at right. Confirm Euler’s formula for this solid.

A

H F

J

E M

4 MC A cylinder is not a polyhedron because: a it has only 3 faces. b it is not Platonic. D it has faces that are not plane polygons.

K D

B

L

C

c it has no edges. e none of the above.

5 MC A polyhedron has 6 faces and 5 vertices. The number of edges of this solid is: a 3 b 5 c 7 D 9 e 11 6 Draw a pyramid with a square base (similar to those in Egypt). Find the numbers of vertices, edges and faces. Confirm Euler’s formula. 7 MC Which of the following is not a net for a cube? a

b

D

E

c

8 Draw a net for the Platonic solid, the tetrahedron. 9 Draw a net for the Platonic solid, the octahedron. 10 Draw the net for the object in question 3. 11 Consider the Platonic solid, the octahedron, shown at right. Imagine that the bottom gets sliced off horizontally along AB, which is parallel to the plane of PQRS. What would be the shape of the new exposed surface? 12 Construct, with compasses and a straight edge (and scissors and glue), the Platonic solid, the octahedron. 13 Given a cube whose side length is a, as shown at right, determine the length of the diagonal of the base (the line AB), using Pythagoras’ theorem. Similarly, show that the so-called major diagonal (AC) has a length of 3a.

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Q

P

R

S

A

B C

a

a

B a

A

14 Consider the cube shown at near right. Use the midpoints of the lines AB, BC and BD, to ‘cut’ an imaginary subcube out of the original. How many of these little cubes fit inside the big one? 15 Now consider the cube being cut in the way shown far right, again by finding the midpoints of AB, BC and BD. What fraction of the original cube is this object?

16e

D A

A

B

D

B

C

C

Circle geometry Until now, we have avoided the most ‘perfect’ form in geometry, the circle. There are several important properties and theorems of circles which form part of the classical study of geometry.

Review of circle definitions A circle is formed by drawing the curved line which is at a given D E distance from a single point (O). This distance is called the radius. The line itself is called the circumference. A line from a point on the circumference (A), through the centre (O), to another point (B) on the A B O circumference is called a diameter. A similar line (DE) which does not pass through the centre is called a chord. OC is a radius. C Mathematically, it is easy to see that the length of AB is twice that of OC. D E In other words, diameter = 2 × radius. The length of any chord is less than that of the diameter (DE < AB). A chord divides the circle into two regions. In the figure at right, the A B O shaded area is called the minor segment. The rest of the circle’s area is called the major segment. A minor segment’s area is always less than half the area of the circle. The part of the circumference joining points D and E is called an arc. The shorter arc (in the shaded region) is the minor arc; the longer one is the major arc. D E A chord can define (or subtend) certain angles. The angle formed by joining the chord to the centre (∠DOE) is called the angle subtended by OO B the chord at the centre. The angle formed by joining the chord to a third point on the circumference (∠DFE) is called the angle subtended by the C chord at the circumference. F Obviously for a given chord (DE) there is only one angle subtended at the centre, but there are different angles subtended at the circumference as the point F is moved around the circle. A tangent line is one that just touches the circumference at a single point (C) and is always perpendicular to the radius (OC). There is a close connection between this tangent and the one defined in trigonometry.

Circle theorems Circle theorem 1 Consider the figure at right. A chord AB is drawn and a third point on the circumference (C) defines the angle ACB (marked angle y in the figure). From the same chord, lines are drawn to the centre (O), defining the angle x. It can be proven that x = 2y. Circle theorem 1: The angle subtended by the chord at the centre is twice the value of the angle subtended by the same chord at the circumference.

A

B x O y C


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Circle theorem 2 A

Now, imagine moving the point C in the figure anywhere along the circumference to, say, points C1 or C2. In each case ‘new’ angles (y1, y2) are subtended by the same chord. But from Circle theorem 1: x = 2y2 and x = 2y1, therefore y2 = y1 This is true as long as C1 and C2 stay on the same side of AB as C (that is, in the same segment as C).

B x O y2 y y1 C2 C

Circle theorem 2: Angles subtended at the circumference, in the same segment, by the same chord, are equal.

Worked example 15

eBook plus

Find the values of the angles x, y and z. Think 1

2

3

Tutorial

int-1186

WriTe

The angles x and 62° are subtended by the same chord at the circumference. The angle at the centre (62°) is twice the angle at the circumference (x).

∠AOB = 2∠ADB 62° = 2x x = 31°

The angles x and y are subtended by the same chord at the circumference, so they are equal.

∠ADB = ∠ACB x=y 31° = y y = 31°

Triangle ADO is isosceles, and angles on the base of an isosceles triangle are equal, so x = z.

Worked example 15

C y D

x

62° O

A

x=z 31° = z z = 31°

eBook plus

Find the values of the angles x, y and z.

1

580

Use Circle theorem 1 to find x.

B

z

Worked example 16

Think

C1

Tutorial

WriTe


x = ∠AOB = 2∠ACB (Circle theorem 1) x = 2(46°) = 92°

2

Use Circle theorem 2 to find y.

y = ∠ADB = ∠ACB (Circle theorem 2) y = 46°

3

Use properties of isosceles triangle AOB to find z.

x + z + ∠OAB = 180° x + z + z = 180° x + 2z = 180° 2z = 180° - 92° = 88° z = 44°


A

z x O 46° C

B y D

Circle theorem 3

A

B

A

B

B1 A1 Now, imagine moving the chord AB to x x the point A1B1 as shown in Figure 1. B1 A1 Notice that the values of both angles x O O and y are increasing. y y What would happen if A1 and B1 C were moved even further, so that the C chord became a diameter? Figure 2 Figure 1 This is shown in Figure 2. Now, x is a straight angle (180°), and from Circle theorem 1, x = 2y. Therefore, y = 90°. This special case of the theorem can be stated as circle theorem 3:

Circle theorem 3: The angle subtended by a diameter is a right angle (90°).

Circle theorem 4 The figure at right is called a cyclic quadrilateral, because all four vertices touch the circumference, and it encloses the centre. It can be proven that the sum of opposite angles is always 180°; that is, a + c = 180° and b + d = 180°. Circle theorem 4: Opposite angles in a cyclic quadrilateral add up to 180 degrees.

A

d

a O

D

c C

b B

Worked Example 17

Find the values of x and y. Think 1

Write D

ABCD is a cyclic quadrilateral and opposite angles in such a shape add up to 180°.

89° A x

O

103° C y B

2

Angles x and 103° are opposite angles.

x + 103° = 180° x = 77°

3

Angles y and 89° are opposite angles.

y + 89° = 180° y = 91°

REMEMBER

1. A chord is a line drawn connecting any two points on the circumference of a circle. 2. Joining the end points of a chord to a third point creates an angle, subtended by that chord (or arc). 3. A line just touching the circumference and perpendicular to the radius is a tangent line. 4. Joining four points on the circumference with straight lines between neighbouring points forms a cyclic quadrilateral.


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5. Circle theorem 1: The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord (or arc) at the circumference. 6. Circle theorem 2: Angles subtended by the same chord (or arc) are equal. 7. Circle theorem 3: The angle subtended by the diameter is a right angle (90°). 8. Circle theorem 4: The sum of opposite angles in a cyclic quadrilateral equals 180°.

exerCise

16e eBook plus Interactivities

int-1002 Circle theorem 1




int-1006 Centrecircumference angles

int-1007 Angles on the same arc

int-1008

Circle geometry

z

1 We15 Find the values of the angles x, y and z in the figure at right. 2 mC The angle subtended by a chord at the centre is _____________ the angle subtended by the same chord at the _____________. The above sentence can be completed using the words: a half, circumference D equal, circumference

b twice, diameter e twice, edge

88° O

x

c twice, circumference

3 Find the value of the angle x in Figure 1, below. 4 mC In question 3, if angle ∠OAC = 20°, then ∠OBC equals: a 37° b 57° c 59° D 94°

e 131°

5 We16 Find the values of the angles x, y and z in Figure 2, below. 6 Using the result from Circle theorem 2, prove the additional result that x = z in Figure 3, below. A B

A

Cyclic quadrilaterals

x O

x 80° z

37° C

B

x y

34° O

y

z E O

y C

D Figure 2

Figure 1

Figure 3

7 mC In Figure 4 (below), the angle (or angles) which is (are) half of ∠AOC is (are): a ∠ABC only b ∠ADC only c ∠ABC and ∠ADC D ∠ADC and ∠ACD e ∠ACD only 8 The angle subtended at the circumference by a diameter is ________ degrees. 9 Find the values of x and y in Figure 5, below. 10 We17 Find the values of x and y in Figure 6, below. B A

C 50° O

x

O D

Figure 4

582

y

54° O

30° y

Figure 5


y

84° Figure 6

x

11 In the following 5 figures, identify the circle theorem that helps you find the value of the angle labelled x. Then find the actual value of x. a

b

111°

c

102°

x 51°

O

81°

x

O

120°

81°

O x

d

e

41°

22° O

x

x O 110°

51°

12 In Figure 7 (below) find the value of angle y in terms of angle x. (Hint: Use Circle theorems 2 and 4 combined.) 13 Find the values of the angles x, y and z in Figure 8 (below). 14 Prove the general case of the result of question 13, namely that: x=y=z 15 In Figure 9 (below), chords AB and CD are parallel. Find the values of the angles x, y and z. y

A y

x O

40°

z

A

O

z

42°

B

O

x D

Figure 7

B

C

D

Figure 8

y

x

C

Figure 9

16 Prove the general case of the result of question 15, namely that: x=y=z 17 The proof of Circle theorem 1 relied on a result about the external angle of any triangle, as shown in the figure at right. The result is that d = a + b. Prove this result. eBook plus Digital doc

WorkSHEET 16.2

18 The proof of Circle theorem 1 also relied on a particular construction whereby a line joining the vertices of the two angles in question were joined and extended towards the circumference. This is the line COD in worked example 15 and in the figure at right. Go back and refer to this now. Imagine the point C being moved so that it was much closer to B as in the figure. Clearly, it is not possible to form the line COD, so how can we prove the theorem in this case? Step 1. Draw a line from C to O (see the figure at right). Step 2. Using the result about equal angles in isosceles triangles, find some equal angles. There are three isosceles triangles in the diagram. To get you started, ∠EBC = y + ∠ECO. (Why?) Complete the proof, namely that x = 2y.

b c d

a D

A

x O

B

y E

C

y C A

z x O

E

B y C

C

Chapter 16


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16F

Tangents, chords and circles There are many geometric problems which use the four circle theorems to help in their solution. To begin, we need to recall the definition of a tangent to a circle as a line which just touches the circumference and forms a right angle to the radius at the point of contact. The last figure in the section on reviewing circle definitions shows an example of a tangent line.

Worked Example 18

Construct, with a straight edge and compasses, a tangent to a circle at any point a by hand b using a CAS calculator. This task is relatively easy, relying on the earlier construction of a perpendicular bisector. Think a

1

2

3

b

584

1

Draw/Display

(a) Given the circle centred at O, and any point A on the circumference, extend the radius outward. (b) Using the same radius as the circle (OA), draw an arc crossing the extended line at B. The result is that A is the midpoint of OB.

a

Construct the perpendicular bisector of the line OB. You will need a larger radius for the compass than in step 1. Draw arcs above and below the line OB by first placing the compass point at B, then at O. The arcs should cross at C and D. Join the points C and D. Since the line CD is perpendicular to OB, it is also perpendicular to OA. Since it just touches the circle at A, it must also be the tangent.

B A O

C

B A

O

D

C

B A

O

b To construct a circle, on the Geometry screen, tap U. Then tap the screen to select the centre of the circle and again to select a point on the circumference.


D

2

To draw a tangent to the circle, tap: • Draw • Construct • Tangent to Curve Tap the point on the screen where you want the tangent to the curve to be drawn.

3

To add the line AB, tap: • Draw • Infinite Line Tap the centre of the circle and the point of intersection of the tangent and the circle.

The alternate segment theorem Consider the figure shown. Line BC is a tangent to the circle at the point A. A line is drawn from A to anywhere on the circle, point D. The angle ∠BAD defines a segment (the shaded area). The unshaded part of the circle is called the alternate segment to ∠BAD. Now consider angles subtended by the chord AD in the alternate segment such as the angles marked in red and blue. The alternate segment theorem states that these are equal to the angle which made the segment, namely: ∠BAD = ∠AED and ∠BAD = ∠AFD

D

O

B

A

C

E

D

O F

Proof of the alternate segment theorem We are required to prove that ∠BAD = ∠AFD. 1. Construct the diameter from A through O, meeting the circle at G. This can be done simply with a straight edge, as both points A and O are known. Join G to the points D and F. 2. Use Circle theorem 3 to find some right angles. This refers to the property that angles subtended at the circumference by a diameter are right angles. ∠BAG = ∠CAG = 90° (property of tangents) ∠GFA = 90° (Circle theorem 3) ∠GDA = 90° (Circle theorem 3)

B

A

C

G

D

O F

B

A

C


585

3. Consider triangle GDA. We know that ∠GDA = 90°. Solve. ∠GDA + ∠DAG + ∠AGD = 180° 90° + ∠DAG + ∠AGD = 180° ∠DAG + ∠AGD = 90° 4. (a) ∠BAG is also a right angle. ∠BAG = ∠BAD + ∠DAG = 90° (b) Equate the two results. ∠DAG + ∠AGD = ∠BAD + ∠DAG (c) Cancel the equal angles on both sides. ∠AGD = ∠BAD 5. Now consider the fact that both triangles DAG and DAF are subtended from the same chord (DA). ∠AGD = ∠AFD (Circle theorem 2) Equate the two equations. ∠AFD = ∠BAD QED

Chords and tangents in circles We shall now consider three situations: where chords intersect inside a circle, where they meet outside a circle and where one of the chords is a tangent. First, let us consider the case of two chords which meet inside a circle.

Chords meeting inside a circle Worked Example 19

Two chords, AB and CD, meet at a 50° angle at point E. Find the values of the angles x, y and z. Think 1

2

Find the value of x using the sum of angles in a triangle.

Write/draw

x + 50° + 44° = 180° x = 180° − 50° − 44° = 86°

(a) Construct a line joining B to D. (b) Now the angle y and the 44° angle (∠DCB) are subtended at the circumference by the same chord (BD).

C 44° x B 50° E A y z D

C 44° x B 50° E A y z D

(c) Apply Circle theorem 2. 3

y = 44°

(a) Now, construct a line joining C to A.

(Circle theorem 2)

C 44° x B 50° E

(b) This time the angles x and z are subtended by the same chord (AC). A y

z D

(c) Apply Circle theorem 2.

586

z = x = 86°


(Circle theorem 2)

As a consequence of this example, the two triangles AED and CEB have all three angles equal and are therefore similar triangles. This can be stated as a theorem: Two chords which meet inside a circle form two similar triangles.

Chords meeting outside a circle Let us now consider the case where the two chords meet outside the circle. Worked Example 20

Two chords, PA and PB, meeting at P, intersect the circumference of a circle at C and D. Find the values of the angles x, y and z.

A

y C

Think

B

Write

z

88° x 28° D

1

Use the sum of angles in a triangle to find x.

x + 88° + 28° = 180° (sum of angles in a triangle) x = 180° − 88° − 28° = 64°

2

(a) Find z. Consider the straight angle ∠ACP.

∠DCA + 88° = 180° (straight angle = 180°) ∠DCA = 92° ∠DCA + z = 180° (Circle theorem 4) 92° + z = 180° z = 88°

(b) Consider the cyclic quadrilateral ABDC. Use Circle theorem 4 to find z. (Note: z will always equal ∠DCP.) 3

(a) Consider the straight angle ∠BDP. (b) Use Circle theorem 4 to find y. (Note: y will always equal x (∠CDP).)

P

∠BDC + x = 180° (straight angle = 180°) ∠BDC + 64° = 180° (from step 1) ∠BDC = 116° y + ∠BDC = 180° (Circle theorem 4) y + 116° = 180° y = 64°

A chord meeting a tangent Finally, consider what happens when one of the chords becomes a tangent. Worked Example 21

A chord (BD) meets a tangent (AD) at point D. The chord crosses the circumference at C. Construct a pair of similar triangles based on the diagram shown. Think 1

Join points A to B and A to C and label angles a, b, c, d, e, x, y. To find similar triangles, attempt to find equal angles.

B

C

Write/DRAW B

A

D

e x C c y ab d A D


587

2

Since AD is a tangent, and AB and AC are chords, the alternate segment theorem can apply.

y=x b=e

3

There are two straight angles along lines BD and AD.

x + c = 180° (straight angle along AD) y + a + b = 180° (straight angle along BD) x+c=y+a+b (equate two straight angles) c = a + b (result from step 2, x = y)

4

Consider the angles in small triangle ACD and the largest triangle BAD.

ACD angles: b, c, d BAD angles: e, (a + b), d

5

Use the results from steps 2 and 3 (b = e and c = a + b).

6

Since three angles are equal, the triangles are similar.

(alternate segment theorem on AB) (alternate segment theorem on AC)

= b, c, d Hence, triangle ACD and triangle BAD are similar.

So, no matter what the geometry of the two chords (even where one is a tangent), a pair of similar triangles is always formed.

Worked example 22

eBook plus

A plank rests on two cylindrical rollers, as shown in the diagram. The radii of the rollers are 40 cm and 50 cm and their distance apart on the ground is 120 cm. Find the value of the angle x.

Tutorial


50 cm x

40 cm 120 cm

Think 1

WriTe/draW

Redraw the diagram as a similar triangle problem. Label the triangle and transfer all the relevant information onto the diagram.

A C 50 cm

E

40 cm

x 2

D

y

120 cm

2

Write a similarity statement. Note: With any similar figures, the corresponding angles are equal and their corresponding sides are in equal ratio.

∆CED is similar to ∆AEB ∆CED ∼ ∆AEB

3

To find missing side lengths, establish the scale factor by calculating the ratio of two corresponding sides whose lengths are given.

Scale factor =

AB CD

=

50 40

=

5 4

= 1.25 588

B


4

Use the scale factor to find the length of side ED.

EB = 1.25ED y + 120 = 1.25y y − y + 120 = 1.25y − y 120 = 0.25y 0.25y = 120 0.25y 120 = 0.25 0.25

(a) Write the relationship between EB and ED. (b) Substitute the known values into the equation. (c) Transpose the equation to make the letter y the subject. (d) Divide both sides by 0.25. 5

Use triangle CED to find the magnitude of the angle required. (a) Identify the appropriate ratio to use. Note: The opposite and adjacent sides are known; therefore, choose the tangent ratio. (b) Substitute the known values into the ratio and evaluate. (c) Transpose the equation to make the angle x the subject, using the inverse tangent function. 2 (d) Evaluate for x and round the answer correct to 2 decimal places.

6


y = 480 C

E

x 2 480 cm

40 cm D

O A  x 40 tan   =  2  480 tan (θ) =

 x 1 tan   =  2  12 x −  1 = tan 1   2  12  x = 4.7636° 2 x = 2 × 4.7636° = 9.5272° = 9.53° The required angle is 9.53°.

REMEMBER

1. The tangent line to a circle from either a point on the circumference or a point outside the circle can be constructed with a compass and straight edge only. 2. The angle formed by a tangent line and a chord divides the circle into two segments. Angles subtended by the chord in the alternate segment equal the angle formed by the tangent line and chord. This is called the alternate segment theorem. 3. Similar triangles have all three angles equal. Their corresponding sides are in equal ratio. 4. Two chords can meet inside the circle or be extended to meet outside the circle. In both cases similar triangles are formed. 5. An extended chord meeting a tangent line outside the circle also forms similar triangles. Exercise

16F

Tangents, chords and circles 1 WE 18 Draw a circle of any radius. Construct a radius which points to the right (that is, due east). Construct a tangent to the circle at the point where this radius touches the circle.


589

2 Line AB is a tangent to the circle, as shown in Figure 1, on the right. Find the values of the angles labelled x and y.

B y

x

A

O 21°

Figure 1

Questions 3, 4 and 5 refer to Figure 2 on the right. The line MN is a tangent to the circle and EA is a straight line. The circles have the same radius. 3 Find 6 different right angles.

D

M

F O

E

G C

A

B eBook plus

4

mC If ∠DAC = 20°, then ∠CFD and ∠FDG are respectively: a 70° and 50° b 70° and 40° D 70° and 70° e 50° and 50°

c 40° and 70°

5

mC A triangle similar to FDA is: a FDG b FGB D GDE e none of the above

c EDA

Digital doc

SkillSHEET 16.2 Similar triangles

N

Figure 2

A y

6 Find the values of the angles x and y in Figure 3. (Hint: Use the alternate segment theorem.) 7 Show that if the sum of the two given angles in question 6 is 90°, then the line AB must be a diameter.

O B

8 Find the value of x in Figure 4, given that the line underneath the circle is a tangent.

42° 62°

x

Figure 3 x

9 In Figure 5, express x in terms of a and b.

100°

x a

O

20°

O

b Figure 4

Figure 5

10 Two tangent lines to a circle meet at an angle y, as shown in Figure 6 below. Find the values of the angles x, y and z. 11 Solve question 10 in the general case (see Figure 7) and show that y = 2a. This result is important for space navigation (imagine the circle to be the earth) in that an object at y can be seen by people at x and z at the same time. 12 We19 In Figure 8, find the values of the angles x, y and z. a

10° z

Figure 6

590

z 75°

z O

O x

y

y

x Figure 7


O y

20° Figure 8

x

13 a Find two pairs of similar triangles in Figure 9. b Since similar triangles are formed, find the lengths of the sides labelled x and y, as shown in Figure 10 at right.

B

B E

A

3

A

C

O

2.5

7

Ey x 5 O

D

D

Figure 9 A

14 WE 20 a Identify a pair of similar triangles in Figure 11. b Find the values of the angles x and y. c If AC = 10 cm, BC = 4 cm and EC = 11 cm, find ED.

C

Figure 10

x

B 70°

y

C

25°

D

E

Figure 11

15 MC Examine Figure 12 at right. The angles x and y (in degrees) are respectively: a 51 and 99 b 51 and 129 c 39 and 122 D 51 and 122 e 51 and 110

19 x

51 y

O

Figure 12 S

16 WE 21 In Figure 13, chord SQ meets a tangent PQ at point Q. Construct a pair of similar triangles based on the diagram shown.

R P

Q

Figure 13

Questions 17, 18 and 19 refer to Figure 14, at right. The line BA is a tangent to the circle at point B. Chord CD is extended to meet the tangent at A.

C x

y D

O

17 Find the values of the angles x and y.

z 50° 45° A B Figure 14

18 MC The triangle which is similar to triangle BAD is: A COD B CAB C BCD D BDC

E AOB

19 MC The value of the angle z is: A 50° B 85°

E 130°

C 95°

D 100°

20 Find the values of the angles x, y and z in Figure 15 (at right). The line AB is tangent to the circle at B.

O

C

D

A

z

33° y 92° x B Figure 15


591

21 Find the values of the angles x, y and z in Figure 16 (at right). The line AB is tangent to the circle at B. The line CD is a diameter.

C

x

O

y D 25°

z

A

z

A

B Figure 16

22 Solve question 21 in the general case; that is, express angles x, y and z in terms of a (see Figure 17).

C x

O

y D a

B Figure 17

23 We22 A plank rests on two cylindrical rollers, as shown. The radii of the rollers are 10 cm and 25 cm and their distance apart on the ground is 90 cm. Find the value of the angle x.

25 cm x

10 cm 90 cm

eBook plus

24 A plank rests on two cylindrical rollers, as shown. The radii of the rollers are 18 cm and 24 cm. The plank makes an angle of 30° with the horizontal ground. Find the distance, x cm, between the rollers.

Digital doc

Investigation

24 cm

Circle constructions

16G

14º

18 cm x

Geometry in architecture, design and art The most powerful tool in the architect’s arsenal and a crucial one for artists is geometry. Both groups rely on rendering ‘real’ three-dimensional objects onto (usually) a two-dimensional sheet (or canvas). Furthermore, the designs of buildings, sculpture, packaging and even automobiles rely on geometrical constructions. This topic could fill an entire chapter (if not a course) on its own, so we will restrict ourselves to only a small part of it.

perspective Architects often render (construct) a view of a three-dimensional space on a sheet of paper. This gives the viewer some idea of how the object being designed fits into that space. Perspective drawing, although a relatively new technique (about 400 years old), is crucial to forming an accurate understanding of the object.

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To construct a simple perspective drawing we need to understand a few basic concepts: centre of vision, horizon and vanishing point. The horizon is a horizontal line which represents the boundary between ‘earth’ and ‘sky’, as far as the ‘observer’ (you) can see. The centre of vision is the distant horizontal line level where your eyes are looking. The line joining your eyeball to your centre of vision meets it at the vanishing point.


Career profile Rachel Nolan

Worked example 23

eBook plus

Construct a perspective drawing of a road, telegraph poles and some posts. Think 1

Tutorial


WriTe/draW

On a sheet of paper, lightly draw in a centre of vision line. Using the tip of your pencil, mark a selected vanishing point somewhere near the middle of this line.

VP

Chapter 16

CV


593

2

Use your ruler to draw a converging line from each bottom corner of your page to the vanishing point to create a road.

3

(a) Using free hand, draw in a mountain range along the centre of vision line. (b) Using a ruler draw a large vertical line on the left-hand side of the road, near the bottom corner of the page. Use this line to draw the closest telegraph pole.

4

(a) Draw faint lines from the top and bottom of the telegraph pole to the vanishing point. (b) Using these lines as a guide, draw in another five vertical lines (these will become other telegraph poles), becoming closer together as they move towards the vanishing point. Add detail to the telegraph poles. (c) Using the same process, draw in a number of white road side posts on the other side of the road.

H VP

CV

H VP

CV

There are computer programs (for example, AutoCad) which will, when given a mathematical model of a three-dimensional object, produce a perspective drawing, allowing you to move both the horizon and vanishing points and to observe the resulting perspective.

The Golden Ratio The ancient Greeks designed many beautiful structures. One reason for this beauty was the proportions used in such structures in the columns, arches and windows. It was discovered that rectangles in a particular ratio were particularly pleasing to the eye. These rectangles are called Golden Rectangles and the ratio is known as the Golden Ratio. Even today, this ratio can be found in nature, in certain windows, in the framing of photographs (only approximately Golden) and in many other architectural elements. Consider the rectangle ABCD at right. Draw a line EF, which A E B creates a square AEFD. If the remaining rectangle (shaded) EBCF has side lengths in the same ratio as the original ABCD, then both are Golden Rectangles. In other words, AB:BC = BC:CF, or expressing it as a fraction, AB BC . = BC CF D F C Furthermore, since EBCF is now itself a Golden Rectangle, A E B it too can be subdivided into a square and another Golden Rectangle (EBGH). This process can be repeated infinitely, creating smaller and smaller Golden Rectangles. How many are H G there in the figure at right? D

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F

C

Worked example 24

Using a straight edge and compasses, construct a Golden Rectangle. Think 1

WriTe/draW

Construct a square, starting from two points, A and B. (a) From points A and B, construct perpendicular lines (AM and BN). (b) Draw a circle with radius AB and centre A, crossing line AM at D. (c) Draw another circle, with radius AB and centre B, crossing BN at C. Join CD. The line AD will be the short side of the golden rectangle. (d) Find the midpoint, E of AB, by finding the perpendicular bisector of line AB.

N

M

D

A

3

Put your compass point at E and pencil at C. Draw the circle passing through points C and D. (a) Extend lines DC and AB horizontally until they are both outside the circle. (b) Construct the perpendicular to AF at point F and extend this line until it meets DC at G. The rectangle AFGD is a Golden Rectangle. Can you identify the other Golden Rectangle in this drawing?

A

B N

M

B C

D

C

D

A A

2

N

M

E

B

B

D

A

E

Tessellations

C

D

B

A

E

C

G

B

F

eBook plus

Another important element in design and architecture is a tessellation: a repeated series of a polygon or polygons which exactly cover the Digital doc Investigation plane, without leaving gaps. A simple tessellation is the chequerboard, The Golden Ratio comprising 64 squares. The board can be extended infinitely in both horizontal and vertical directions, so it is a tessellation. There are three types of tessellation: regular tessellations, semiregular tessellations and irregular tessellations.

regular tessellations A regular tessellation is built up from copies of a single regular polygon. The tessellation at right is made up of regular hexagons. Note that although the plane is filled, the edges are ragged (look down the left side). The point where the three polygons meet (black dot) is called a vertex. There are only three regular tessellations: equilateral triangles, squares and hexagons.

Chapter 16


595

semiregular tessellations The figure at right is called a semiregular tessellation because: 1. it is made up of two (or more) regular polygons 2. the arrangement of polygons at each vertex is the same. At each vertex there are three triangles and two squares. This vertex can be written as {3, 3, 3, 4, 4} and is called the order of the vertex. The order of a vertex can be determined by closely examining the polygons which surround the vertex. Consider the enlarged version of one vertex of the above tessellation. Determine the number of sides of each polygon and list this number of sides in a clockwise fashion, starting with a polygon of the smallest number. So the order {3, 3, 3, 4, 4} can just as easily be {3, 4, 4, 3, 3} or {3, 3, 4, 4, 3}.

33 3 4 4

irregular tessellations The figure at left may appear to be a semiregular tessellation, made up of triangles and squares, but examine carefully the vertex marked in blue. It has 3 triangles and 2 squares {3, 3, 3, 4, 4} while the vertex marked in red would be {3, 3, 3, 3, 3, 3}. Therefore this is an irregular tessellation. In fact, there are only 8 different semiregular tessellations, so that almost all tessellations are irregular. Irregular tessellations can be as simple as the figure at left or as complex as the one below. A famous artist, M. C. Escher, specialised in extremely complex tessellations. Worked example 25

Consider the tessellation shown. a Name the polygons making up the tessellation. b Determine the order of all the different vertices and, hence, the type of tessellation.

Think

WriTe

a Carefully examine the tessellation.

a The polygon types are regular triangle, quadrilateral

and hexagon. This could possibly be a semiregular tessellation. b

1

2

596

Examine the tessellation, looking for different vertex arrangements. It appears that there are three types, marked red, green and purple in the figure. Evaluate the vertices.

b Red vertex: triangle, square, hexagon, square or

3, 4, 6, 4 Green vertex: triangle, triangle, square, triangle, square or 3, 3, 4, 3, 4 Purple vertex: triangle, square, triangle, triangle, square or 3, 4, 3, 3, 4 Although the green and purple vertices appear different, their order is the same. Thus, there are only two vertex types, making this an irregular tessellation.


rules for creating tessellations A multitude of techniques can be used to generate tessellations; here are a few important ones to get you started: 1. All triangles tessellate. 2. All quadrilaterals tessellate. 3. Pentagons will tessellate if the sum of two nonadjacent angles equals 180° or if two sides are parallel. 4. To create the tessellation from a single polygon you can perform the following transformations: (a) translation — moving the polygon left, right, up, down (b) rotation — rotating the polygon through any angle (usually a multiple of 60°) (c) glide reflection — a mirror image followed by a translation. 5. The sum of the angles of each polygon at any vertex must equal 360°.

rememBer

1. Perspective involves rendering three-dimensional scenes on a two-dimensional surface (paper). (a) The horizon is the farthest visible horizontal line. (b) The vanishing point is the farthest visible point on the centre of vision (usually the horizon) in the direction in which the viewer is looking. (c) All horizontal lines in the direction of the viewer’s eyes meet at the vanishing point. 2. (a) A Golden Rectangle has dimensions such that, when the largest possible square is removed, the remaining rectangle has sides in the same ratio as the original rectangle. (b) The ratio of the longest to shortest side of a Golden Rectangle is called the Golden Ratio. 3. Tessellations are patterns constructed from polygons which completely cover a flat surface without gaps. (a) There are regular (only 3), semiregular (only 8) and irregular tessellations (all others). (b) The order of a vertex in a tessellation indicates the type of polygon used and the number of each type. (c) All triangles and quadrilaterals will tessellate; many other polygons will also tessellate.

exerCise

16G

Geometry in architecture, design and art 1

We23 Construct a perspective drawing that shows a railway line running from the observer to the horizon. There is a series of telegraph poles next to the track.

Chapter 16


597

2 MC The vanishing point is: A the place where all straight lines meet B the place where the horizon meets your position C the place where the horizon meets straight lines extending away from your position D the place where all horizontal lines meet E the place where all parallel lines meet Questions 3 and 4 relate to the two figures at right. They represent perspective drawings of a cube of 1 m on each side. The top of the box is shaded blue, the front is shaded red.

a

b

3 MC Compare the two boxes. Which of the following statements is true? A The viewer is below the cube in figure a. B The viewer is above the cube in figure b. C The viewer is taller in figure b. D The viewer is taller in figure a. E The viewers are of the same height. 4 From the following list, suggest the nature of the viewers: short child, tall adult, baby, ant, eagle. 5 WE 24 Construct a Golden Rectangle whose short side is 5 cm. 6 Estimate the ratio of the longest side to the shortest side for your figure in question 5. 7 Calculating the Golden Ratio C G D The figure AFGD is a golden rectangle. Its longest side is b and its shortest side is a. a a Since ABCD is a square, find the length of CG. b Express the ratio, as a fraction, of the longest to shortest side of B F A AFGD in terms of a and b. c Since BCGF is also a Golden Rectangle, express the ratio, as a b fraction, of its longest to shortest sides. d Equate these two fractions (they are both Golden Ratios), and find an equation involving b and a on the left-hand side and 0 on the right-hand side. e Solve this equation for b in terms of a. (Hint: Use the quadratic formula.) f Now, since a could be anything, set a = 1 to find the Golden Ratio. 8 Giovanni has constructed four Golden Rectangles a, b, c and d. He measures one side of each of them, and finds that they have the following dimensions. a Short side = 4 b Long side = 4 c Short side = 11.23 d Long side = 500.5 Find the other dimensions to 3 decimal places. 9 WE 25 Consider the tessellation shown at right. a Name the polygons making up the tessellation. b Determine the order of all the different vertices and hence state the type of tessellation. 10 Make a copy of each shape below on a piece of stiff paper. Use each as a template to produce a tessellation.

598


Summary Review of basic geometry

• • • • • • •

• • • • • •

A point is a marker for a location in space. Two points connected make a line segment. Extend the line segment to infinity in one direction to construct a ray. Extend the line segment to infinity in both directions to construct a line. Where two lines meet, an angle is formed. Two lines which never meet are parallel. A line intersecting a pair of parallel lines is called a transversal and creates a number of special angles. 1. Corresponding angles are equal in magnitude (size). 2. Co-interior angles add up to 180°. 3. Alternate angles are equal in magnitude (size). Vertically opposite angles are equal in magnitude (size). Supplementary angles add up to 180°. Complementary angles add up to 90°. The sum of interior angles in a triangle is 180°. A polygon is a closed figure made up of three or more straight line segments. The locus of a point is the path traced out by a point when it moves according to a given rule. Geometric constructions

• • • • •

Geometric constructions are made with a pair of compasses and a straight edge only. The perpendicular bisector of a line divides the line in half and generates a right angle. For a line parallel to any given line, the radius of the compasses determines the distance between them. The angle bisector divides any angle in half. From the 60° angle and the perpendicular bisector, the 30° and 45° angles can be made. Polygons

• The perpendicular bisectors of any triangle meet at a point called the circumcentre. From this centre a circle (the circumcircle) which just touches each vertex of the triangle can be drawn. • The centroid is the point where the lines connecting each vertex with the midpoint of the opposite side meet. • The incentre is the point where the angle bisectors of each vertex meet. From this centre a circle (the incircle) which just touches each side can be drawn. • A quadrilateral can be a square, rectangle, rhombus, parallelogram or trapezium. All these can be constructed using a straight edge and compasses only. • A regular polygon has equal sides and equal interior angles. Regular polygons of 3, 4, 5, 6, 8 and 12 sides can be constructed easily. To construct other regular polygons, you may also need to use a protractor. • A star polygon is formed by joining alternate vertices of a regular polygon in a regular pattern. Each regular polygon may have 0 (triangle, quadrilateral), 1 (pentagon, hexagon) or more different star polygons. Three-dimensional geometry

• • • • •

A polyhedron is a solid made up of flat polygon faces. The net of a polyhedron is a two-dimensional representation of the solid, showing connected edges. The line where two faces meet is called an edge; the point where three or more faces meet is called a vertex. For any solid, Euler’s formula applies: Vertices = Edges − Faces + 2. A Platonic solid is a polyhedron made from a single regular polygon. There are only five different Platonic solids.


599

Circle geometry

• • • • •

A chord is a line that connects any two points on the circumference of a circle. Joining the end points of a chord to a third point creates an angle, subtended by that chord (or arc). A line just touching the circumference and perpendicular to the radius is the tangent line. Joining four points on the circumference with straight lines between neighbouring points forms a cyclic quadrilateral. Circle theorem 1: The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord (or arc) at the circumference. Circle theorem 2: Angles subtended by the same chord (or arc) are equal. Circle theorem 3: The angle subtended by the diameter is a right angle (90°). Circle theorem 4: The sum of opposite angles in a cyclic quadrilateral equals 180°.

• • •

Tangents, chords and circles

• The tangent line to a circle from either a point on the circumference or a point outside the circle can be constructed with compasses and a straight edge only. • The angle formed by a tangent line and a chord divides the circle into two segments. Angles subtended by the chord in the alternate segment equal the angle formed by the tangent line and chord. This is called the alternate segment theorem. • Similar triangles have all three angles equal. Their corresponding sides are in equal ratio. • Two chords can meet inside the circle or outside the circle. In both cases similar triangles are formed. • A chord meeting a tangent line outside the circle also forms similar triangles. Geometry in architecture, design and art

• Perspective involves rendering three-dimensional scenes on a two-dimensional surface (paper). 1. The horizon is the farthest visible horizontal line. 2. The vanishing point is the farthest visible point on the horizon in the direction in which the viewer is looking. 3. All horizontal lines (in the direction that the viewer is looking) meet at the vanishing point. • A Golden Rectangle has dimensions such that, when the largest possible square is removed, the remaining rectangle has sides in the same ratio as the original rectangle. • The ratio of the longest to the shortest side of a Golden Rectangle is called the Golden Ratio. • Tessellations are patterns constructed from polygons which completely cover a flat surface without gaps. 1. There are regular (only 3), semiregular (only 8) and irregular tessellations (all others). 2. The order of a vertex in a tessellation indicates the types of polygon used and the number of each type. 3. All triangles and quadrilaterals will tessellate; many other polygons will also tessellate.

600


chapter review 9 Find the values of the lengths marked x and y in Figure 1, below.

Short answer

1 a If x = 40°, find y. b Express the angle y in terms of x.

10 Find the value of x in Figure 2, below.

x y x + 3°

2 Using compasses and a straight edge only, construct a 15° angle. List the steps involved. 3 Using compasses and a straight edge only, construct a 150° angle. List the steps involved.

Tetrahedron

4

Cube

6

Octahedron

8

Dodecahedron

12

Isosahedron

20

8 Find the values of the pronumerals in each diagram below. a b x x

2x + 6° Figure 2

Verification of Euler’s formula

Multiple choice

1 The sum of angles in an 11-sided polygon is: A 360° B 1620° C 1980° D 3240° E 3960° 2 The value of x in the figure below is: x + 4°

z

15° O

10

12 Write down the equation of the locus of points that are 3 units from the point (2, -3).

Vertices

7 Three-dimensional objects which are not polyhedrons also have nets. Determine the net for the: a cylinder b cone.

y

6

36°

11 Which of the following vertex orders can form part of a tessellation? All polygons are regular. a 3, 3, 3, 4, 4 b 4, 4, 5, 5 c 3, 3, 6 d 3, 4, 4, 4 e 3, 3, 3, 3, 3, 3 f 3, 4, 3, 4

6 Copy and complete the following table of vertices, edges and faces for the five Platonic solids. Edges

3

O

Figure 1

5 Consider the hexagon at right, where all the vertices are joined to all other vertices. How many triangles of different shapes can you find? One is shown shaded.

Faces

x O

4

4 a Using compasses, a protractor and a straight edge, construct all the different star polygons from a regular decagon. b What angle is subtended by each side at the centre?

Name

y

62°

88° y

O

x

A 42° B 43° C 45° D 47° E unable to be determined from the given information


601

3 Which of the following angles cannot be obtained from compass-and-straight-edge constructions of 60° and 90° angles? A 15° B 22.5° C 75° D 100° E 120° 4 Which of the following regular polygons cannot be constructed with just compasses and a straight edge? A Equilateral triangle B Square C Pentagon D Hexagon E Heptagon 5 A regular heptagon has how many different star polygons? A None B 1 C 2 D 3 E 7 6 A polyhedron has 7 vertices and 9 edges. How many faces does it have? A 2 B 4 C 5 D 14 E 16 7 In Figure 1, the values of x and y respectively are: A 30° and 50° B 50° and 30° C 50° and 50° D 70° and 30° E 70° and 70°

x

x O

30°

100° y

Figure 1

A

30° z y B O 110°

6

4 x

8 O

y

60° 40°

Figure 2

Figure 3

8 In Figure 2, line AB is a diameter. The values of x, y and z respectively are: A 60°, 60° and 60° B 60°, 60° and 70° C 60°, 50° and 70° D 90°, 20° and 70° E 50°, 60° and 70° 9 In Figure 3, the values of angle x and side y respectively are: A 60° and 3 B 60° and 5.33 C 60° and 12 D 40° and 5.33 E 40° and 12 10 A Golden Rectangle has a short side of 56.44 cm. The longest side is: A 34.9 cm B 90.8 cm C 91.2 cm D 91.3 cm E 91.4 cm

extended response

1 The curved figure at right represents part of an athletics track. It shows an arc derived from two concentric circles (circles having the same centre). It is used to mark a curved line for an athletics event. The sports master wishes to find the radius of the inner arc. (You should approach this problem by first tracing the arc onto a piece of paper. a Join the ends of the arc with the line AB and find its perpendicular bisector, DE. b What kind of line is DE? c Let the distance AC = x and DC = y. Express CE in terms of x and y and the unknown radius r. d Since AB and DE are intersecting chords, write an expression showing the appropriate sides of similar triangles having equal ratios. e Modify the expression in part d to obtain an expression for r in terms of y. 2 Use the diagrams in question 1 to calculate the inner radius of a wooden arch over a doorway. The arch has measurements of x = 30 cm and y = 12 cm. Find the radius.

D A

C

B

E

3 A tunnel 500 m long has a semicircular cross-section. It requires temporary supports, AB and BC, as shown. Point B is vertically above D. B a What is the relationship between ∆ABC and ∆DBC? If AB = 24 m and BC = 7 m, find: b the diameter of the tunnel A C D c the height of B above the floor of the tunnel d i the distance of D from the wall at C ii the distance of D from the wall at A. e While the temporary supports are in place, what should drivers of large trucks entering the tunnel take into account?

602


f What is the maximum height of a 3 m wide truck, so that it can pass under the supports? g Determine whether a truck that is 4 m wide and 5.2 m high will pass through the tunnel safely. 4 Construct a dodecahedron. a Construct a regular pentagon (about 10 cm on each side). b Construct the net of the solid. The figure on the right is a model of a net. Use your pentagon from a as your model. c By folding and taping, you can construct the solid. d A construction model is easier to work with if you add tabs to certain edges so that, when folded, the tabs can be connected more easily than the actual edges. Determine where the tabs need to go in your net. 5 Determine the size of z in the figure below right.

Tab

57° z

6 Find the lengths of the pronumerals in the diagram below.

12.5

8.5

6.5 7.5

y x

7 Determine the values of the unknown angles shown below.

86°

w

34°

y

x

z

8 Find the value of the pronumerals in the diagram below.

a

c

58° b

d e


603

9 Calculate the magnitude of x. D

C 43°

3x − 5

B

E A

10 A plank rests on two cylindrical rollers, as shown. The radii of the rollers are 30 cm and 40 cm. The plank makes an angle of 10° with the horizontal ground. Find the distance, x cm, between the rollers. 30 cm

10°

40 cm

x

11 A tunnel 300 m long has a semicircular cross-section. It requires temporary supports, AB and BC, as shown. B

A

D

C

Point B is vertically above D. If AB = 12 m and BC = 5 m, find: a the diameter of the tunnel b the height of B above the floor of the tunnel c the distance D is from the wall at C d the maximum height a 3-m-wide truck could be, so that it can pass under the supports. 12 Research investigations. Geometry is used in a wide variety of professions and activities. Here are a few research topics where the kind of geometry in this chapter has been useful. You may wish to investigate one or more of these, using your library as a resource. a Navigation at sea, over land or in space b Drafting and industrial design c Perspective in painting and drawing d Architecture and building construction e Aeroplane wing design f Tiling floors, walls and ceilings g Computer graphics Which particular aspects of geometry are used in each of these activities? eBook plus Digital doc


604


eBook plus

aCTiviTies


• 10 Quick Questions: Warm up with ten quick questions on geometry in two and three dimensions. (page 556) 16a

Review of basic geometry

Tutorial

• We 3 int-1184: Watch how to write an equation modelling the outer most edge of water from a sprinkler. (page 559) Digital doc

• SkillSHEET 16.1: Practise basic geometry. (page 560) Interactivity

• Bisecting a line segment int-0997: Consolidate your understanding of bisecting a line segment. (page 560) 16b

Geometric constructions

Digital doc

• WorkSHEET 16.1: Determine the value of angles on parallel lines cut by a transversal, complete and table of sum of interior values, and construct bisectors and parallel lines, labelling important features. (page 567) Interactivities

• Bisecting an angle int-0998: Consolidate your understanding of bisecting an angle. (page 567) • Circumcentre int-0999: Consolidate your understanding of circumcentre. (page 567) 16c

Polygons

Tutorial

• We 8 int-1185: Watch how to construct the perpendicular bisectors and median bisectors of each triangle. (page 567) 16e

circle geometry

Tutorials

• We 15 int-1186: Watch how to use circle theorem 2; determine the magnitude of angles. (page 580) • We 16 int-1187: Watch how to use circle theorems 1 and 2; determine the magnitude of angles. (page 580) Interactivities

• Circle theorem 1 int-1002: Consolidate your understanding of circle theorem 1. (page 582) • Circle theorem 2 int-1003: Consolidate your understanding of circle theorem 2. (page 582) • Circle theorem 3 int-1004: Consolidate your understanding of circle theorem 3. (page 582) • Circle theorem 4 int-1005: Consolidate your understanding of circle theorem 4. (page 582)

• Centre-circumference angles int-1006: Consolidate your understanding of centre-circumference angles. (page 582) • Angles on the same arc int-1007: Consolidate your understanding of angles on the same arc. (page 582) • Cyclic quadrilateral int1-1008: Consolidate your understanding of cyclic quadrilaterals. (page 582) Digital doc

• WorkSHEET 16.2: Review basic geometry, construct more complex shapes such as the hexagon, draw a possible net for a cube and use a provided 3D diagram to prove Euler’s rule. (page 583) 16F

Targets, chords and circles

Tutorial

• We 22 int-1188: Watch how to use similar figures to determine the value of an unknown length. (page 588) Digital docs

• SkillSHEET 16.2: Practise similar triangles. (page 590) • Investigation: Circle constructions. (page 592) 16G

Geometry in architechure, design and art

Digital docs

• Career profile: Learn how Rachel Nolan, an architect, uses mathematics. (page 593) • Investigation: The Golden Ratio. (page 595) Tutorial

• We 23 int-1189: Watch how to construct a perspective drawing. (page 593) chapter review Digital doc


Chapter 16


605


b

10 5

10

a

35 t (s)

A car’s motion is represented by the velocity–time graph shown above. a Determine the average acceleration, in m/s2, between 20 and 35 seconds. 2 marks b Describe the car’s motion during the time interval 10 to a seconds. 1 mark c If the total area travelled by the car is 9545 m: i write an expression for the area in terms of a and b. 3 marks 9095 ii Hence, show that b = . 2 marks (595 - 29a) 3 Tracy is going to swim from Point A to Point B across a narrow river. She can swim at a speed of 3 km/h. The river is flowing at a rate of 1 km/h. Point B

2 + 2 = 4 marks

MULTIPLE CHOICE

10 minutes

Each question is worth 1 mark. 1 If u = 4i - 3 j then the unit vector in the direction of   be which  u would one of the following: A 5(4 i - 3 j) B 1(4 i - 3 j) 5     1 D 4 i - 3 j C 7 ( i - j)     4i - 3 j E   7 → 2 PQR is a triangle. M is the midpoint of PR. If → → → PQ = PR = p and RQ = q, then in terms of p and q,     → RM equals: 1 1 A 2 p + q B 2 p - q     1 D q - p C q - 2 p     E p + q   3 A particle moves in a straight line with acceleration 20t - 3t2 at time t seconds. The particle has initial velocity 1 m/s from a fixed point O. The velocity of the particle, in m/s, when t = 3 seconds is: A 2 B 3 C 34 D 63 E 64

4 r° y° x°

1 km/h (river flow)

Point A

a If i is the unit vector in the direction that the  is flowing and j is the unit vector in the river  the river, represent direction straight across velocity of Tracy’s swimming in terms of 2 marks i and j.   606

b On the diagram below left, draw the vectors that represent this situation. 2 marks c Determine the: i exact direction ii exact magnitude of her velocity, in km/h.

30 minutes

1 PRST is a parallelogram. M is the point on PR such that 2PM = MR, and N is the midpoint of ST. → → PR = p and RS = r.   → a Find MR in terms of p. 1 mark 1 → b Show that MN = r + 6 p. 3 marks   v (m/s) 2

0

Chapters 13 TO 16

p°

Using the circle above, establish which one of the following values of the pronumerals is false: x° and y = 2p° A r ° = 2 r+p B r + p = 180° and y + x = 2 C y + 2r = 360° and p = (180 - r)° D y + p = 180° and r + x = 180° E 2p = (360 - x)° and y = (360 - 2r)°


5

45° 4 kg

-1  4  c cos   2 

°

T2 kg

-  5( 2 ) - 4  e sin 1   4  

Two strings attached to two points in the same horizontal plane suspend a ball of mass 5 kg. The two strings make angles of 45° and θ ° to the horizontal respectively. If the strings have respective tensions of 4 kg wt and T2 kg wt, then the value of θ ° is: -  5( 2) - 4  a tan 1   4  

-  5 2 D sin 1    4 

-  5( 2)  b tan 1    4 

6 A hot-air balloon is travelling upward at a constant speed of 21.2 m/s. When the balloon is 70 metres from the ground, a camera lens that was placed on the edge accidentally falls off. Ignoring air resistance and assuming acceleration due to gravity is -9.8 m/s2, then the time taken, in seconds, for the camera lens to hit the ground would be closest to: a 2.19 b 3.78 c 4.35 D 6.52 e 6.60

exTended response

30 minutes

1 Rhonda and Yorak are travelling along a highway in their car. The car they are in is travelling at a constant speed of 102 km/h. a Convert 102 km/h to km/min. 1 mark They pass a sign informing them that there are road works ahead. They need to slow down to 40 km/h. b If Rhonda and Yorak are travelling exactly at 102km/h before uniformly decelerating to 40 km/h and it takes 50 metres for their car to reach 40 km/h: i determine the time, in seconds, it takes their car to slow down to 40km/h. Write your answer to the nearest second. ii Hence, determine the rate of deceleration, in m/s2. Write your answer correct to 2 decimal places. 3 + 2 = 5 marks

Rhonda and Yorak stop at a roadside store to buy drinks. At the store they meet friends, Beth and Derek, who are travelling on a motorbike. They all decide to meet for lunch at the next town, Remark, which is 204 km down the road. Rhonda and Yorak leave 5 minutes before their friends and accelerate uniformly for 30 seconds until they reach a speed of 95 km/h. Beth and Derek leave and accelerate uniformly until they reach a speed of 105 km/h. v (km/min)

19 12

Rhonda and Yorak

1 2

t (min)

The graph above represents the motion of the car Rhonda and Yorak are travelling in. c On the graph above, represent the motion of the motorbike Beth and Derek are travelling on. 3 marks d If it takes Beth and Derek 1 hour and 57 minutes to reach Remark, determine the time, in seconds, for their motorbike to accelerate from rest to 105 km/h. Write your answer to the nearest second. 3 marks e Determine the amount of time, in minutes, it takes Beth and Derek eBook plus to pass Rhonda and Yorak? Write your answer correct to 2 decimal places. 4 marks Digital doc f Determine the time, in minutes, Beth and Derek have to wait until Solutions Exam practice 4 Rhonda and Yorak arrive at Remark for their planned lunch stop. Write your answer correct to 2 decimal places. 3 marks

exam practice 4

607

Answers d (A ∪ B) / (A ∩ B)

CHAPTER 1

Number systems: real and complex Exercise 1A — Review of set notation 1 ξ a {3, 4, 5, 6} A B b {1, 2, 3} c {2} 4 1 2 3 d {1} e {1, 2, 4, 5, 6} 5 6 f {1, 3, 4, 5, 6} g {1, 3, 4, 5, 6}    769 c 1.230 2 a 3.328 125 b 0.3409

A

B

A

B

A

B

A

B

e A′ ∩ B

3 -2 ⇒ Integer: Z, Q, R 16 8 ⇒ Natural number: N, Z, Q, R

21 ⇒ Rational: Q, R 16 - 3 2 ⇒ Rational: Q, R 7

f A′ ∩ B′

6 3 ⇒ Irrational: I, R 1

16 4 ⇒ Natural number: N, Z, Q, R 1

5 5 ⇒ Irrational: I, R π ⇒ Irrational: I, R -21.72 ⇒ Rational: Q, R   ⇒ Rational: Q, R 2.567 4.135 218 976 ⇒ Irrational: I, R 4.232 332 333 ⇒ Irrational: I, R

4 a

8 33

b

5 6

True True False True A′

e.g.

a b c d a

374 333

c

2 - 2 = 0.

A

b A ∪ B A

B

A

B

c A ∩ B

61 495

g (A ∪ B)′

d

3517 1665

7

3 Sig. 4 Sig. Number fig. fig. 2 D.P. 3 D.P. 1267.1066 1270 1267 1267.11 1267.107 7.6699 7.67 7.670 7.67 7.670 8.000 56 8.00 8.001 8.00 8.001 0.999 87 1.00 1.000 1.00 1.000 0.076 768 0.0768 0.076 77 0.08 0.077 0.000 174 95 0.000 175 0.000 175 0 0.00 0.000

8 a 6 × 1028 b 4 × 1014 c 6 × 1040 d 6 × 107 e 2 × 1027 f 4 × 10 3 30 9 a 5.35 × 10 b 7.64 × 1018 c 1.23 × 107 d 3.60 × 10 7 10 B 11 E 12 D 13 D 14 B Exercise 1B — Subsets of the set of real numbers 1 a {x : x ∈ Z, -6 < x < 1} −6 −5 −4 −3 −2 −1 0 1 x b {x : x ∈ Z, -3 ≤ x ≤ 4} c {x : x ∈ Z, -6 < x ≤ 4} d {x : x ∈ Z, 0 < x < 5}

608

Answers

−3

0

−6

0 0

4 x x

4 5

x

e {x : x ∈ Z, x < 5}

0

f {x : x ∈ Z, x > 2}

0

g {x : x ∈ Z, -5 < x < 0} 2 3

a b c d e f g h i a

1

2

x

5 3

4

−5 −4 −3 −2 −1

0

c {x : -2 < x < 1}

x

e 5 - 10 - 15 + 6

f 6 15 + 9 3 - 4 5 - 6

g 7 6 - 18

h 2(4 - 15 )

i 2(6 + 35)

j 4(5 - 3 6 )

k 2(105 + 36 6 ) m 17 o 25

l 2 n -6

Interval notation (-2, 1)

Interval notation [2, 5) ∪ [4, 6) = [2, 6) −2 0 2 4 6

f {x : x < 5} ∪ {x : 4 ≤ x < 6} g {x : 2 ≤ x < 5} ∩ {x : 4 < x ≤ 6}

x

Interval notation (-∞, 5) ∪ [4, 6) = (-∞, 6) −4 −2 0 2 4 6

b

4 2 3

c 6

d

15 3

e

4 15 35

f

g

5 2 8

h

x

Interval notation [2, 5) ∩ (4, 6] = [2, 6] \ {4, 5}

k 4 - 15

m 5 - 2 6

n

p

Interval notation (5, ∞) ∩ (4, 6] = (5, 6] 0 1 2 3 4 5 6 x

4 E 5 C Exercise 1C — Properties of surds

-

4 - 3 15 7

2 6 9

i 3 + 6 l

14 + 5 3 11

o

16 5 - 31 11

c

- 3(6 + 5

6+2 3 3

6 a 4 2

b

5(9 2 + 4 3 ) 114

c

195 10 + 78 15 - 12 2 + 18 3 114

d

1794 - 464 15 - 42 3 231

e

35 - 7 6 15

−2 0 2 4 5 6 8 x

h {x : x > 5} ∩ {x : 4 < x ≤ 6}

5+ 3 2

j 2 10 - 6

x

−2 −1 0 1 2

e {x : 2 ≤ x < 5} ∪ {x : 4 ≤ x < 6}

6

x

Interval notation [2, ∞)

d {x : x ≥ 2}

6 - 9)

5 a

x

−2 −1 0 1 2

14 + 2 3

-8(2

d

x

−2 −1 0 1 2

-

c 18 + 2 6

Interval notation (-∞, 2)

b {x : x < 2}

b

x

{x : x ∈ Q, x > 5} {x : x ∈ Q, 5 < x ≤ 20} {x : x ∈ Q+, x < 20} {x : x ∈ Z, 5 < x < 20}\{8, 9} {x : x ∈ Z+, x < 100}\{40, 50} {x : 2 ≤ x < 5} {x : 3 < x < 5} {x : x < 3} ∪ {x : x > 7} {x : x ∈ R+, x < 3} ∪ {x : x ∈ R, x > 7} Interval notation [ -3, 1] {x : -3 ≤ x ≤ 1} −3 −2 −1 0 1 2

15 - 6

4 a

f

-

47 - 12 15 3

7 a

26 - 45 2 14

b

30 - 39 2 14

2 - 24 2 7

e 0

2)

f 31 - 18 2

b 2 14

c 5 5

d 7 2

d

e 4 3

f 10 3

g 28 5

h 2 2

g x = 2 - 3 2 is a solution to x2 - 4x - 14. However, since there was no remainder, it is not a solution to 2x2 - 2x - 9. 8 Answers will vary. 9 Answers will vary. 10 B 11 A 12 A 13 D 14 A 15 E

i

6 2 5

-

j

3 2 2

2 a 2 2 - 2 3 - 2(3

c 3 + 2 2) e 7 5 - 2 g

9 3 13 2 8 8

3 a 3 10

b 10 21

b

7-4

d 3( 2 + 2 3 ) f 2(9 3 + 10 2 ) h

118 3 49 2 15 10

c 84 2

d

5 3 6

Exercise 1D — The set of complex numbers c b ± 1 a ±4i 7i 2 ± 2 5i d

10 ± 10i

e

1± 2 7 2

Answers

Answers 1A ➜ 1D

1 a 2 6

609

2 a Re (3 + 4i) = 3

b Re (- 2 + 2i ) = - 2

Im (3 + 4i) = 4

Im (- 2 + 2i) = 2

c Re (z) = 2 - 1

Exercise 1F — Representing complex numbers on an Argand diagram 1 a b Im (z) Im (z)

d Re (z) = 2 2

Im (z) = 2 + 1 e Re (z) = -6 Im (z) = 0 3 a x = 1 b y=4

3

d x = 1 y=3

e

f x = 2 1 y=2

g

2

y=7 x= y=

c

c 12 - 17i

e

i

-1

l

4 13

2

- 5i

f

4 5

j

3 13

c

-5

+ 5i

g

-7 25

2

k

5 + 12 i 119 119

16

n

16 8 3i 7 7

- 13 i

63

m 25 - 25 i -9

24 i 25

+

2 a 4 + 3i

b 24 - 7i

c 24 + 7i

e -8 + 6i

f 28i

g

4 a x =

9

2

5 a z = 5 + 5 i z 6 a z 1 = 41

7 B 8 A 9 E 10 B

Answers

h

0

−2

Re (z)

Re (z)

2

−2

f

Im (z)

4 5

2

0 −2

2

b i

e -64

f

3

+ 5i

3 2i 13 13

-256

c 1

d i

g -512i

h

104 25

d 7 + 24i h -i

78 i 25

l

+

b z = -i z a2 + b2 z or z 1 = zz

b z

-1

=

-i 512

Im (z)

b

z Re (z) z−

1

3

+ 25 i

Re (z)

0 z−1

20

4 25

2

−3i

2 a -i

3 a

0

Re (z)

c • Dilated by a factor of 13 from the Real and Imaginary axis. • Reflected about the Real axis

y - 2x =1 b 5 y=3 x = -1

y=5 1

d -5 - 12i

- 13 i

12 5

2

-

p

527 336 i 24 + 7i j 625 - 625 i k 25 25 3 a Answers will vary. b Answers will vary. c Answers will vary.

610

+ 5i

o 1

13

0

−2

d - 2( 2 + 3 )

c 0

3

Im (z)

2

6 D 7 B 8 A Exercise 1E — Multiplication and division of complex numbers e 85

2 − 3i

2

Im (z)

f (3 2 + 2 3 ) - (4 2 + 3 3 )i

b 5 - i

d

Im (z)

Re (z)

2

−2

−2 − 3i

e 7 2 - 10 2i

1 a -6 + 4i

0

−2

−2

d i

3-3 2

Re (z)

−2 + 3i 2 −2

b 1 - i

b

2

2

−2

6 5 12 5

4 a 5 - 7i

5 a 14

0

−2

f x = 2 1 y=2

x=4

2 + 3i

2

Im (z) = ±2 10 f Re (z) = 0 Im (z) = 13 c x = -3 x=1 y = -5 y = -1

14 25

-

1

d z 1 is 13 of the distance of z from the origin. 4 a

b z = 2 + 3i iz = i(2 + 3i) = -3 + 2i i2z = i(-3 + 2i) = -2 - 3i i3z = i(-2 - 3i) = 3 - 2i i4z = i(3 - 2i) = 2 + 3i

Im (z) iz 0

i 2z

z, i 4z

Re (z) i 3z

c The distance from the origin remains the same with a 90° anticlockwise rotation. 1 1 1 2i - 1 = = =- = i 5 a z 2 = 2 1 + 2i - 1 2i 4 2 (1 + i ) 1 1- i 1- i 1- i 1 i × = = = 1+ i 1- i 1+1 2 2 2 z0 = (1 + i)0 = 1 z1 = 1 + i z2 = (1 + i)2 = 1 + 2i - 1 = 2i z3 = (1 + i)3 = 2i(1 + i) = 2i - 2 = -2 + 2i z4 = (1 + i)4 = (-2 + 2i)(1 + i) = -2 - 2i + 2i - 2 = -4 z

-1

=

b

e (z - 3)(z - 2) = 0 z = 3, z = 2

Im (z) z3

z2

z4 z−2

g ( z - 2 - 2 )( z - 2 + 2 ) = 0

z1 z0

z−1

z = 2± 2

Re (z)

 5 7  5 7 h 2  z +  z + +  =0 4 4  4 4  

c Plotting zn compared to zn + 1 where n ∈ Z, undergoes a rotation 45° anticlockwise and its distance from the origin is increased by a factor of 2. nπ d • Rotated n × 45° or anticlockwise 4 • Distance from the origin increased by n 2 e i -4 - 4i

ii

iv 256 + 256i

v

-1 4 -1 128

1

- 4 i +

1 i 128

1 a 2( z - 3 )( z + 3 )

b ( 2 z - 3 )( 2 z + 3 )

c 3( z - 2i)( z + 2i)

d

e z(z - 4)

f 2z(3z - 1)

2 z (2 z - 1)

2 a (z + c (2z - 1)(z + 2) e (2z + 6)(z - 4) 4)2

1 2

(2z - i)(2z + i)

h -z(4z + 3) b (2z - 4)(z - 2) d (z + 3)(z - 1) f -(4z - 6)(3z + 2)

3 a ( z + 2 - 10i)( z + 2 + 10i) b (z + 2)(z + 8)

( )

c 2 z - 12 ( z + 3)    d  z + 1 - 13   z + 1 + 13   2 2  2 2  e (z + 4)2 f ( z + 1 - 2i)( z + 1 + 2i)

( )

h 2(z + 2)2  5 57   5 57  i - 2  z -  z - +  4 4  4 4   -

( )

4 z - 12

2

4 a ( 3z - 2 )( 3z + 2 ) = 0 ±

6 2 b ( 2 z - 5i)( 2 z + 5i) = 0 z=

z=

±

c z(2z - 7) = 0 7 z = 0, z = 2

d (z - 5)(z - 1) = 0 z = 5, z = 1

 1 2  1 2 z +  z + +  =0 3 3 3 3    -1

z=

3

±

2 3

5 a z = 5

b z = 5 ± 2 5

c z = - 2 ± i 3

d z = 2, z = 2

7 ± i 35 6

f z = 1 ± i 2

e z =

3

6 a z2 - 2(2 - 3i)z + 13 b z2 + 6i - 9 c z2 - 5z + iz - 5i + 12 7 a z = - 3i ± i 14 b z =

1 ± 2 19 + i(2 + 19 ) 10

c z =

(i - 1) ± i 2(i - 30) 6

8 A 9 A Chapter review Short answer 1 a -4 → Z, Q, R, C -2 7 8

b

-16 3

→ Q, R, C

d 3 3 → I, R, C

→ Q, R, C

1

e 27 3 → N, Z, Q, R, C g 2π → I, R, C

4 f 5 → I, R, C h 3.221 → Q, R, C

i 3.21 683 947. . . → I, R, C   → Q, R, C k 1.1234

j

2

-3 + 2

5 → I, R, C l i2 → Z, Q, R, C

m 2 + - 5 → C n 0.172 117 722 111 777 222 → I, R, C o 3 - 3 → I, R, C p 4i → C q 3 + 7i → C r 4 + 0i → Z, Q, R, C b 374 c 556 a 8

3 2 × 4 a

10i 2

-3

c

33

333

495

1018 0 1 2 3 x 4 6 8 10 12 14 16 x

b d

0 1 2 3 4 5 6 x

Answers 1E ➜ 1G

j

j

c

g 2( z - 2) z - 12

-5

7 ± 4 4 i (z - 5)(z - 1) = 0 z = 5, z = 1 z=

iii 32i

Exercise 1G — Factorising quadratic expressions and solving quadratic equations over the complex number field

g

f (2z - 3)(z - 1) = 0 3 z = 2, z = 1

6 8 10 12 14 x

Answers

611

5 a

6 7 8 9 10 11 12 13 x

c

3 4 5 6 7 8 9 x

10i 10i ,z= 2 2 b (z + 2 - i)(z + 2 + i) = 0 z = -2 + i, z = -2 - i

1 2 3 4 5 6 7 8 9 10 11 12 x ∈ Q

e

f

3 4 5 6 7 x

5+ 2

d 35 - 12 6

2 3 4 5 6 7 x

z=

-2 3,

z = -1

c 19

15 a z = -3 ± i

e 13 - 3 6 5

f 9 + 2 6

b z = c z =

8 a 2 2 9 a y = 1, x = 4

b 2 + 3 c 6 2 b x = 1, y = 4

10 a 10 + i

b 2 + 4i

1 20

( )

2 c 3 z + 3 ( z + 1) = 0

b 5(2 + 2 )

7 Answers will vary.

d

14 a ( 2 z - 5i)( 2 z + 5i) = 0 z=

0 1 2 3 x∈Q

d

6 a

b

e

21 - 38i 10

11 a 3(1 - 3i)

b

12 a

b

Im (z)

-8

c

-

f

+ 14i

28 - 231i 130

3+i 9

-

2±i 6 2

5 ± i 23 6

Multiple choice 1 A 2 A 6 C 7 E 11 D 12 C 16 D Extended response 1 a Im (z) 2 1

Im (z)

3 E 8 D 13 A

4 C 9 E 14 C

5 C 10 B 15 D

1 + √3i

0 −2 −1 −1

Re (z)

1 2

−2

0

0

Re (z)

−i

Re (z)

b Answers will vary. c OZ = 2 d z 0 = 1, z = 1 + 3 i, z r = - 2(1 - 3 i), z 3 = -8 Im (z)

2 − 3i

c

d

Im (z)

Im (z) −3

+ 2i

i

z2

0

e

Re (z)

Im (z)

Re (z) 3 − 2i

13 a ( z - 7i)( z + 7i) b ( 2 z - 3i)( 2 z + 3i) c z(4z + 5) d 2(4z + 3)(2z - 1) e ( z + 2 - 6 )( z + 2 + 6 )  3 3i   3 3i  f  z +  z + +  2 2  2 2  

( )( )

g 2 z + 14 z + 45

Answers

Re (z)

z1 z0 z−1

Re (z)

π anticlockwise 3 2. Distance from origin doubles. f zn would nπ 1. Be rotated n × 60° or anticlockwise 3 2. Distance from the origin 1 × (2)n. g i - 8(1 + 3i) ii 64 e 1. Rotate 60° or

0

612

0

z3

1 - 3i iii -512 iv 8 1 v 512 2 a 4z2 - 4z - 8, n2z2 - n2z - 2n2, n ∈ Z b i z2 - z - 6 ii z2 - 8z + 13

iii z2 - 8z + 20 v z2 - (5 + i)z + 8 + i

iv z 2 - (5 + 3 ) z + 3

ii z = a - bi c i z = a - b 3 d i z = 2 - 3i ii b = 4 b = 4 and c = 13 c = 13

5 a

e z = 1 ± i 2

( z - 1 - i 2 )( z - 1 + i 2 ) = 0

f z = - 1 ±

y

i 2 2

−2

 2i   2i   z +1  z +1+  2 2   

(x − 1)2 + (y + 1)2 = 4

b

CHAPTER 2

Transformations Exercise 2A — Translations of points and graphs 1 a A′(4, 6) b A′(-2, 2) d A′(-4, -6) c A′(2, -2) b T2, 1 2 a T2, 2 Translated 2 right and Translated 2 right and 2 upward. 1 upward. d T1, -3 c T-2, 5 Translated 2 left and Translated 1 right and 5 upward. 3 downward. 3 a A(2, -2) b A(-1, -1) d A(-3, 1) c A(-5, -1) b y = 2x - 4 4 a y = x - 4

6 a T0, 3 b T-4, -3 7 a T-1, -3 b T2, -3 8 a Translated: Up 3 T0, 3 b Translated: Up 7 and left 1 T-1, 7 c Translated: Down 6 and left 3 T-3, 6 d Translated: Up 5 and right 5 T-5, 5

y = 2x − 4 −4

−4

c y = -x

9 10 11 12

d y = 2x + 1

y = −x

y

y y = 2x + 1 y = 2x + 1

y = −x 1 x

0

x

0

y =0

Original asymptotes: x = 0, y = 0 Image asymptotes: x = 2, y = -1

x

0

x y = −1

x =0 x =2

y y = 2x

x

y = x 1−2 −1

0 −1.5

y=x−4 0

y y = 1x

y=x

x

0 (1, −1) 2

−2

y

x2 + y2 = 4

2

c T-1, -5 c T1, 2

A A C B

Exercise 2B — Reflections of points and graphs b (2, 4) 1 a (-2, -4) c (4, -2) d (-4, 2) 2 a y y = 3x (1, 3) y=x

e y = (x +

1)2

f y = 2(x -

+2

y = (x + 1)2 + 2

-1

0

3

y = x2

Answers 2A ➜ 2B

y

2)2

(3, 1) x

3

0

b

x

y = −2x

y

(−1, 2)

g y = (x + 3)2 + 6 i y = f (x + 3) + 2 k y = -2(x - h)2 + k

h y = -2(x + 2)2 + 3 j y = f (x + 1) - 2 l y = -3(x - h) + k

0

(2, −1)x − y= x

2

Answers

613

c

y

i

y = 2x + 3

y

3 y = x +1 1

1 x

0

x

0

y = −1

y = 1x − 1

−3

y = −2x − 3 x = −1

d

y

j

y = 2x2

y −1

y = x1 + 1

y= x + 1

y = −1

x

0

x

0

y = −2x2

e

k

y

f

y = x 1− 1 + 1 y= 1

0

y = −3x2

y

x

x

0

y = −3x2

x= 1

l

+

− y = √−4x y

y= 2 + 3

y

x−1

y= 3

x

0

x

0 y= 2 − 1 x+3

y= y

g

−1 2 x

x = −3

4

m

y

x= 1 y = 2x

y = x2 + 1 1

1

x

0 −1

x

−1 0

y = −2x

+ −

y = √−x − 1

n h

− y = x1

y y = 1x

0

614

Answers

y = 2−x + 3 y = 2x + 3 y y=3

x

0

x

y = −1

o

y

Exercise 2D — The ellipse and the hyperbola 1 a b y y

y = x2 y = √x

3 (0, 0)

(1, 1) 0

p

0

−5

x

0

5 x

c

y

(5, −3)

−3 −6 −9

−3

d

y

y 6

2

y = −√ x + 2

0

−2

2

2 a

y = −x

x +1 2 e y = 0 g x = 2 x i y = 4

f x = 4 h x = 0

3x 2 k y = -3 4

l y = 2 × 2x

x

d y1 = 0

2 x

n y =

4 o y = x

3 p y = ( x - 2)2 4

1 q y = ( x + 4)2 + 2 2

2 r y = ( x - 3)2 + 4 3

s y =

c

(3, ) ( 1, ) 5 2

-

3 2

4 a y = 2x c y = 5 B 6 A

1 × 22 x 2

y=x

c

x

1

y = 12 x

−2

y = −2x

y

f

y = 2x + 8

x

3

y = −x + 1 y

y= x − 5

x

3 (4, −2)

− y = 43 x + 52

(−2, 4) (−4, 4)

y = 43 x

y

(2, −2) −2

x

2

− y = 43 x

−3

d

y − y = 12 x

e

b

x

0 −2

y = 34 x + 92

y (−2, 4)

(0, 4)

(2, 4)

(−6, 4)

x

x

 x t y = bf    a b

( , 5) 3 2

d (-1, -1)

g

y = −x

1 6x

h

− y = 34 x

y = 43 x

y

(0, 4)

(0, 1)

x2 b y = 2 d y =

y=x

y

x

0 (0, −1)

x (0, −4)

Answers

Answers 2c ➜ 2D

3 a

−1 0

j y = 2x2

m y = 3 × 2 2

x 3 × 22

−4

y

Exercise 2C — Dilations from axes 1 a (4, 3) b (-2, 15) d (-1, -12) c (8, -9) e ( 4, 3) x 2 a y = b y = 2x + 2 2 c y =

x

1

−2

3 A 4 B

(−2, 2)

(0, 0)

−1 0

x

10 x

5

615

y

i −

4 a y = 34 x

(0, 3)

y = 43 x

(−4, 2) (−1, 2)

3

0 x

0

−3 y = 2x + 4

y

c

y=x+6

y = −2x

y

y = −3x

y = 3x

(−4, 2) x

−1

1

x

y = −x − 2

k

y

5 a = 3, b = 2 6 A 7 A Exercise 2E — Successive transformations 1 a (2x + 3, y + 4) b (2x + 6, y + 4) c (y + 4, x + 3) d (y + 4, x + 3) e (2x + 3, 3y + 4) f (2x + 6, 3y + 12) g (x - 1, y + 7) h (6x, 2y) j (-x, -y) i (y, -x) k (2x + 4, -3y + 9) l (3x + 6, -2y - 6) m (3x + 2, -2y - 3)

y = 2x + 8

y = −2x

(−2, 8) (−2, 4)

(−2, 0) x

l

y = x2 + 5

y (−2, 6) (−2, 4)

− y = 2x + 3 x

(−2, 2)

2 a (9y - 4, -4x - 1) c (4x + 3, 9y - 4)

b (6y + 4, -6x + 5) d (9x - 4, 4y - 2)

3 a (-2, 13) c (-8, 0) 6 + 12 4 a y = x-4

b (3, 1) d (-2, 2) 1 b y = -2 x +1

c y

3 a

5 a

y

b

4

b c

(0, 0) 0

−8

2

x

−3

3 −4

(−2, −4)

x

d e f

c

y

6 4 d y = -3 -1 x+2 x-2 T2, 3 Translated 2 right and 3 upward. D1, 4 Dilation factor 4 from x-axis. My = 0 Reflected about x-axis. Mx = 0 Reflected about y-axis. T4, 0 Translated 4 right. T- 1 2

, -2

1

Translated 2 left and 2 downward. 1 −8

0 −1

8x

g My = 0 Reflected about x-axis. h T5 2

,0 5

Translated 2 right.

616

Answers

y = −x 2

2

(2, 2)

(0, −3)

(−4, 3) (−4, 1)

y y = −x−

x

j

b

y

x

i T-1, 0 Translated 1 left. j D2, 5 Dilation factor 2 from y-axis and 5 from x-axis. k D1 1

h Mx = 0, D1 , T- 1 2

1

i My = 0, x = 0, D1 , T2, 7 2

,1

1

Reflected about x and y-axis. Dilation factor of 2 from y-axis and translated 2 right and 7 upward.

, 3 4

8 a

y = x2

y

1

Dilation factor of 3 from y-axis and 4 from x-axis. 2

1

y-axis and 2 from x-axis. Translated left and 4 downward.

l T-4, 3 Translated 4 left and 3 upward. m D1 1

n T1

, -4

1 2

Dilation factor 2 from y-axis and 5 from x-axis.

1

2

Reflected about y-axis, dilation factor of 2 from

, 2 5

1

,2

(2, −1)

,0

x y = −2(x − 2)2 − 1

1 2

Translated right. o D 3 2

,2

−9

3

Dilation factor of 2 from y-axis and 2 from x-axis. 6 a My = 0 Reflected about x-axis. b Mx = 0 Reflected about y-axis. c T1, 0 Translated 1 right. d D1, 2 Dilation factor of 2 from x-axis. e T1 2

b

y = x2

1

x

−1 2

,0

c

1

Translated 2 right. f T3 2

y = (2x − 1)2 y

y

y = 2x 2− 3 + 4 y= 4

,4

3

Translated 2 right and 4 upward.

x

y = −1x

7 a My = 0, D1 , T0, 4 2

,1

2

,1

2

x = −3

1

Reflected about x-axis, dilation factor of 2 from y-axis and translated 4 upward. b My = 0, D1, 2, T-1, -3 Reflected about x-axis, dilation factor of 2 from x-axis and translated 1 left and 3 downward. c T3, 4, D1, 2 Translated 3 right and 4 upward, dilation factor of 2 from x-axis. d D1 , T- 3

2

d

y = 2x

y

y= 4

1 x

−1

,0

1

e Mx = 0, y = 0, D 1 , T3 ,1

4

e

,0

y

1

Reflected about the x and y-axis, dilation factor of 4

24 y = 2x

3

from y-axis and translated 4 right. f D1 , T5 3

,3

3

Answers 2E ➜ 2E

4

y = −2x+3 + 4

−4

3

Dilation factor of 2 from y-axis and translated 2 left.

,0

1

Dilation factor of 3 from y-axis and 3 from x-axis.

y = 3(22x+3)

5 3

Translated right. g My = 0, D1 , T1 2

,4

2

1

,3 1

Reflected about x-axis and dilation 2 from y-axis and

x

1 2

4 from x-axis. Translated by right and 3 upward.

Answers

617

y

f y= 1

d

y = 2x

y y = x 1+ 1

y = 1x + 1 y=1

1 x

0

x

x = −1

e

y

−23 y = −3(22x+3) + 1 2 y=1

9 C 10 C Chapter review Short answer 1 y = (x - 6)2 - 2 2 a T-2, -4 Translated 2 left and 4 downward. b T-1, 1 Translated 1 left and 1 upward. c T-2, 0 Translated 2 left. d T2, -5 Translated 2 right and 5 downward. e T-3, -4 Translated 3 left and 4 downward. f T-2, -6 Translated 2 left and 6 downward. 3 a (2, 0) b (0, 3) d (-2, 3) c (-3, 2) 4 a y y = −2x

x

0

y

f

x2 + (−y + 1)2 = 9

(0, 1) 0 (0, −1)

x

x2 + (y + 1)2 = 9

g

y

y = x2, x ≥ 0 y = √x (1, 1) x

0

y = 2x y

h x

0

0 y = −√−x

b

y

x y = −√x

y = 2x2

b (-6, 1)

5 a (1, 3) c (-8, 16) x

0

y = −2x2

c

y

0

x2 2

c y =

x 2

b y =

6 x 2

x  d y = 2  + 1 + 4 3  b

y

y

1 x

y = +− √x 2

Answers

6 a y =

7 a

y = 4x2 (1, 1)

618

y = −1

−2

−2 −10

2

x

0 −1 −2 −6

5x

1 (2, −4)

c

y

y

d

2 0 2 −6

8 a

x

6 (4, −2)

(1, −2)

y

y = −2x

y =2x

−2

b

x

0

2

y = −2x + 5 y

x

2

, -1

c 1. D1, 2 Dilation factor of 2 from x-axis (parallel to y-axis). 2. My = 0 Reflection in the x-axis. 1 3. D1 Dilation factor of 3 from y-axis ,1 3 (parallel to x-axis). 4. Mx = 0 Reflection in the y-axis. 2 5. T2 Translation 3 of a unit right and 2 units ,2 up. 3 12 a y = −2x + 3 y

y = 2x − 3 (2, 1)

(1, 1)

11 a 1. Mx = 0 Reflection in the y-axis. 2. My = 0 Reflection in the x-axis. 3. D1, 7 Dilation factor of 7 from x-axis (parallel to y-axis). 1 4. D1 Dilation factor of 2 from y-axis ,1 2 (parallel to x-axis). 5. T1, 0 Translation 1 right (positive x direction). b Mx = 0 or 1. My = 0 Reflection in the y-axis or (x-axis). 1 D 2. 1 Dilation factor of 2 from y-axis ,1 2 (parallel to x-axis). Translation 12 unit left and 1 unit down. 3. T- 1

(3, 1) x

y=x

3

c

y = −2x y

0

y = 2x

x

(4, 0)

b

y (1, 6) y = x2

x (−4, 0) x

0

d

y

y = −x −3

y=x+3

(−3, 2)

y = −2(x − 1)2 + 6

c

y

(−3, 0)

( 12 , 2) 0

x

x

0

(−3, −2)

x  0 a y = f  + 2 + 3 1 2 

 x + 2 +3 b y = f   2 

 y - 2 c x = 2 f  -3  2 

 y + 3 d x = 2 f  +2  2 

d

y

1 0

y = 3x

Answers 2E ➜ 2E

3( x + 3)2 3( x - 3)2 - 2 b (-8, 9) y = -2 2 2 6 2 c (6, 0) y = d (-1, 8) y = -3 +4 x-2 x+2 9 a (2, -12) y =

y  e x = 2 f  - 2 - 3 2 

y = √x

x y = −2(3

−2x

+ 2)

−18

Answers

619

e

y

j Domain: (-1, 1) Range: (0, 1) k Domain: (-2, -1) Range: (1, 4) l Domain: R Range: [3, ∞) m Domain: R Range: (-∞, 3] n Domain: [-3, 3] Range: [-3, 3] o Domain: [0, 4] Range: [-2, 2] p Domain: [0, 3) Range: [-3, 0) ∪ (0, 3] y 2 a

(x +1)2 (y −2)2 + =1 9 4

x

0

x2 + y2 = 1

f

y = −x

− y = 32 x + 2 y = 32 x + 2

(−2, 2)

(2, 2) −1 0

1

x

1 2 3x

0 −3 −2 −1 −1

Multiple choice 1 B 2 D 3 D 6 B 7 C 8 E 11 B 12 E 13 B Extended response 1 a T1, -2 Translated 1 right and 2 down. b Centre (-1, -2)

−2 −3

4 E 9 D 14 C

5 A 10 A 15 E

Radius = 18 = 3 2 c T-1, 1 and D3, 3 Translated 1 unit left and 1 unit up. Dilation factor of 3 from the x and y axis. 2 a T-1, 3 Translated 1 unit left and 3 units up. b Centre: (-1, -1) Semi-major = 3 Semi-minor = 2 c T1, 1 Translated 1 right and 1 up.

Answers

y

d

x

0 1 2 3 4 5 x∈Z

0 −2 −1 −1

y = −x2

−2 −3 −4

(−2, −4)

(3, −9)

e

y

y = (x − 1)2 + 3

f

y y = −2(x − 1)2 − 2 x

0 (1, −2) 4

−4

(1, 3)

Relations and functions

620

1 2 3x

−2 −3 −4

y 4 3 2 1

c

y 5 4 3 2 1 0 −3 −2 −1 −1

CHAPTER 3 Exercise 3A — Relations 1 a Domain: {2, 4, 6} Range: {3, 7, 8} b Domain: {2, 3, 4} Range: {1} c Domain: {2} Range: {2, 3, 5} d Domain: {1, 2, 3} Range: {2, 3} e Domain: {x : x ∈ Z+} Range: {y : y ∈ Z, y ≥ 1} f Domain: {x : x ∈ Z, x ≥ 5} Range: {y : y ∈ Z, y ≤ -5} g Domain: {x : x ∈ Z, 4 ≤ x ≤ 21} Range: {y : y ∈ Z, 0 ≤ y ≤ 17} h Domain: [2, 7) Range: [3, 8) i Domain: [2, 4) Range: [4, 16)

b

3 2 1

x

0

(−1, 3) y

g

(−2, 1)

1

y = −2(x + 1)2 + 3 x

0

(1, −5)

h

y 2 −1

−4

(x − 2)2 + (y + 1)2 = 9

(2, −1)

5 x

i

(−2, 2 + √5) y (x + 2)2 + (y − 2)2 = 5

(−2, 2)

(−2 − √5, 2)

(−2 + √5, 2 )

y

j

x

0

(−2, 2 − √5)

x2 + y2 = 9

3 √5 0

x

2

−√5 −3

k

4 B

(x − 2)2 + (y + 1)2 = 9

y √5 −1

x

−√5 −1

(1, −2√2 − 1) y

l

3

0

−3

m

3

x

3 a Factors: 0 = (x - 3)(x - 2) Solutions: x = 3, x = 2 b Factors: 0 = (x2 - 4)(x2 - 3)

y 3

5 A

Exercise 3B — Functions 1 a Domain: [-5, ∞) not R. b Function not defined when x = 0. c The range [-1, 3] ⊆ Co-domain R +. d Not a function since 1-many. e The range R- ∪ {0} ⊆ Co-domain R +. f Not a function since 1-many. 2 a Range: [4, 13) b Range: [2, ∞) c Range: (2, ∞) d Range: [-5, -1) e Range: (-∞, 9] f Range: (-∞, 4) g Range: (7, 23) h Range: [-2, -1) i Range: (-6, 0] j Range: (2, ∞) k Range: [3, ∞) l Range: R\{0} m Range: R + n Range: (-1, 0) o Range: R\{1}

(1, 2√2 − 1) 0

d Domain: R\{-5} Range: R \{5} e Domain: R\{-5, -1} Range: R \{0} f Domain: R Range: [-2.5, 0) g Domain: (-∞, -5) ∪ (0, ∞) Range: R h Domain: [-8, 8] Range: [-5, 5] i Domain: (-∞, -5) ∪ (5, ∞) Range: R + ∪ {0} j Domain: R\{-5} Range: R\{0} k Domain: R Range: (-∞, -2) ∪ (2, ∞) l Domain: R\{-4, 4} Range: (4, ∞)

±

Solutions: x = ±2, x = 3 4 a x = 2, x = -1, x = 3, x = -2 b x = 2, x = 3

(2, √5)

5 E

x

0

6 A

Exercise 3C — Inverse functions y 1 a b f(x) (−2, 5) y 5

n

(2√2 − 2, 3)

f (x)

−2

f(x)

x

0 2 3

x

−4 y 6 f(x)

c 3 a Domain: [-2, ∞) Range: R b Domain: R Range: R c Domain: R\{0} Range: R \{0}

4x

0

−4

−6

0 −6

y 6

d

−

f 1(x)

6x

−6

0

6x

Answers 3A ➜ 3c

(−2√2 − 2, 3)

−

f 1(x)

−1

2 3

−2

y 4

−6

Answers

621

y 4

e

y

e

2

2 4x f (x)

0

−6

−2

−1

−6

x

2

y f(x) 2

−1

f (x)

−2

g

f(x)

−5

y

0

x

2

y

−

f 1(x)

3

0

−3

f(x)

x

x

3 f(x)

−3

−

f 1(x) 0

−

f 1(x)

−2

4x

0

2 a

0

f(x)

y 5

−4

f(x)

−2

f f

−

f 1(x)

y

h

3

y

b

0

f(x) −3

f(x)

3

x

−3 −

f 1(x)

2 −2

0

x

2

−2

−

f 1(x)

y 4

c −

f 1(x)

4x

0

−4

−4

f(x)

f (x)

−1

f (x)

3

-

0 −3

3

x

f (x) f 1 (x)

f(x)

4 E

622

Answers

1 2

R\{0} f (x) d Inverse function f 1 ( x) =

−3

{}

R\

-1

y

d

3 a Inverse function Domain Range f (x) R\{0} R\{0} R\{0} R\{0} f 1 (x) 1 f 1 ( x ) = , x ∈ R \{0} x b The inverse function does not exist. c Inverse function 1 1 f 1 ( x) = + , x ∈ R \{0} 2x 2 Domain Range

x + 2, x ∈ R + x Domain (2, ∞) R +

R\{0}

{}

R\

1 2

Range R + (2, ∞)

Chapter review Short answer y 1 a 4

b y

b

(−2, 5) (−1, 4) −

y 4

f 1(x)

3 2 1

1 0

1 2 3 4x

0

x∈Z

−7

2

4x

0

−2

-x

1 + , -7 ≤ x ≤ 2 2 Domain Range f (x) (-2, 4] [-7, 5) f 1 (x) [-7, 5) (-2, 4] -

f 1 (x) =

(5, −2) −4

c

y 8 6 4 2

f (x)

y 4

d

−7

(4, −7)

(1, 2)

c + 1 2 3 x∈Z

0

-

−

1

e

y 6

f (−1, 2)

(1, 2)

−1 0

y

3

x

0

−4

x

d

f(x)

−3

x

y

-

f 1 (x) = x + 2 + 1, x > -1 Domain Range f (x) (-∞, 0) (-1, ∞) -1 f (x) ( 1, ∞) (-∞, 0)

e

3x − f 1(x)

0 −1 −1

1

y

f 1(x)

2

f(x)

0

−2

-

f 1 (x) = x2 + 2, x ≥ 0 Domain Range f (x) [2, ∞) [0, ∞) f 1 (x) [0, ∞) [2, ∞)

−

x

2

−2 -

y

f

f 1 (x) = 1 - x2, x ≤ 0 Domain Range f (x) (-∞, 1] (-∞, 0] -1 f (x) ( ∞, 0] (-∞, 1]

1 0 −1 −1

f(x)

x

1

−

f 1(x) y

g

-

f 1 (x) =

2 f(x) 0

−3

1

2

x

−

f 1(x)

−3

y

h

−

f 1(x)

3

−2

0

f (x) f 1 (x)

f(x)

3

x

-

f 1 (x) = f (x) f 1 (x)

x2 ,0≤x≤2 4 Domain Range [-3, 0] [0, 2] [0, 2] [-3, 0]

3 1-

x2 + 1, x ≥ 0 4 Domain Range [3, ∞) [0, ∞) [0, ∞) [3, ∞)

3

Answers 3C ➜ 3C

c 2 ± 3 is not a function of x, and not a function. b Range: [-1, ∞) 4 a Range: [-1, 0) d Range: [3, ∞) c Range: (-4, ∞) e Range: [0, 2] f Range: (-∞, 1] g Range: R\{1} h Range: (1, ∞) i Range: (0, 1] 5 a (x - 2) (x - 1) Factors b x4 - x2 x = 2, x = 1 Solutions 0 = (x2 - 1) (x2) Factors x = ±1, x = 0 Solutions Multiple choice 1 B 2 A 3 E 4 E 5 C 6 E 7 C Extended response y f (x) 1 a x 1 f 1 (x) = - , x ≥ 1 2 2 −1 f (x) Domain Range 1 f (x) [0, ∞) [1, ∞) x 0 1 f 1 (x) [1, ∞) [0, ∞)

3

2 1

−2

2 a Domain: x ∈ R Range: y ∈ R b Domain: x ∈ R\{-4, 2} Range: y ∈ R \{0} c Domain: x ∈ R Range: (-∞, -6] ∪ [2, ∞) d Domain: x ∈ R Range: y ∈ R e Domain: x ∈ R Range: y ∈ R f Domain: (-∞, -3] ∪ (0, ∞) Range: (-∞, -3] ∪ (0, ∞) 3 a Incorrect domain b [-1, ∞) ⊆ R + (Co-domain)

1

−4

(3, 2) 0

f 1 (x) = x + 4 + 1, x ≥ -4 Domain Range f (x) [1, ∞) [-4, ∞) f 1 (x) [-4, ∞) [1, ∞)

f 1(x)

3

1 2 3 4x

−2 −1 0

f(x)

y

−3

Answers

623

2 a

b c d e f g

b Reflection in the y-axis c {(x, y) : y = |-2x - 3|

x

d Domain =

2x P(x) = 6x The largest value of x is 8. 4 Domain: [0, 3 ]; range: [0, 8] 4 {p: [0, 3 ] → R, p(x) = 6x} A(x) = 2x2 Length is 3 m, width is 6 m

Range = [0, 4] 2 a Dilation by a factor of 2

1 3

from the x-axis

b y = 3 x + 1 c x = d

Exam practice 1 Short answer

1 2

 - 1   2 

CHAPTER 4

Algebra

1 = 18 2 − 2 3 2

Exercise 4A — Review of index laws

y 6 5 4 3 2 1

b 4a2b3

1 a 4ab

d

x 0 −3−2.5−2−1.5−1−0.5 −1 0.5 1 1.5 2 2.5 3 −2 −3 −4 −5 −6

- 5 a b

e

z

2 a

1 2

d

y3 4 x3

e

g

25m9 9 n5

h

j

y

3

- 7 7  ,  2 2 

5 ba 2 1 b4

b 2xy3

x2y2

1 2m 7 n5 - 16 w 6

125v17

c

-a

2b 7

f 2 a3b

c

9 4 x2y4

f

n14 16m5

i

-

8v 2 m6 27 w 2 n6

-1

3888v13 w6

3 E 4 a 27n + 1 0 −2

2

x

d

b 33n + 3

3n + 2

e

22 n + 8 7

a + bi c + di 4 LHS = × c - di c + di =

(a + bi )(c + di ) (c - di )(c + di )

j

= RHS

1 2 3

3

1 a i ( 2 ,0) ii (3, 0)

624

7

iii ( 2 , 4)

Answers

y2

e

2

x3 g x

ac + bdi 2 + adi + bci = c 2 - di 2 ac - bd + adi + bci = c2 + d 2 Multiple choice 1 D 2 B 3 A 4 E 5 D Extended response

b x 6

1

d

4 5

5 a 3 3 x=2

32n

3

( x + 1) 2

1 7 24

h 64m10 2( x + 1) 1 k ( x + 2) 2

c 3 × 5 -n - 3 f 34m - 4 × 23m - 4 c f

7

x6 1 2

33

i x2 l

3

( y - 4) 2

Exercise 4B — Standard form and significant figures a 3.604 × 105 - d 3.24 × 10 2 a 80 d 5000 g 0.0008 j 0.4 m 6 × 10 7 p 0.245 A

b e b e h k n q

2.134 57 × 102 1.0031 × 10 4 300 2000 0.0003 2.9 × 108 5 × 10 10 1.5

c 1.029 398 × 103 f 5.702 010 09 × 108 c 4000 f 5000 i 0.0016 l 5 × 107 o 0.09 r 16

4 5

a e i a e

b f j b f

2sf 4sf 4sf 3 438.8

c g k c g

4sf 8sf 1sf 13.4 59.1

d h l d h

4sf 5sf 6sf 118 79

5sf 3sf 2sf 0.131 39.6

1

c a = 3 m +

3

x + 4 b y = 4 x + 3 5

14 3

2

5

d k = 3 - 3 p 1

e a = 2 b or b = 3 a

f a = 2 + 5 b or b = 5a - 10

g c = -2a + 6b

h a = 2 b + 3

3

2

k a =

-4 b 21

or b = 2

- 7 or b =

l x = 60 - 6y or y =

6

3

-2

x + 10 S 4π

v -u , u = v 2 - as b r = s

c R =

 A  RR2 R1R2 - 1 d 100  , R1 = A  0  R2 + R1 R2 - R

e t =

2

2s 2s ,u= -v u+v t

g b = C 2 - a 2 i I =

P P , R= 2 R I

k -18

l 4 1 2

2 a -1 e 2

b -1 f 12

c 1 g 2

d -5 h -6

i 2

j

3 a 24

b -3

7 12

-3

2

 T   2π  f L = g   , g = L    2π   T  h a =

2(s - ut ) t2

j m =

tR tR , v1 = v2 v2 - v1 m

3 D 4 C 5 E 6 B 7 E 8 B 9 E 10 C S 11 a n = +2 180 b Number of Sum of interior Polygon sides (n) angles (S) 3

180°

Hexagon

6

720°

Dodecagon

12

1800°

Nonagon

9

1260°

Heptagon

7

900°

Octagon

8

1080°

Pentagon

5

540°

Quadrilateral

4

360°

Decagon

10

1440°

1 2

d -2.2

b -14

c 2 14

d

f 2

g

f 3

72 149

7 11

h 3 i ii 1, -1 vi 0, 3

i i 0, 1 v 2

b i

-1 2 3

ii

10 19

iii 5 23

iv

3 4

v

4 5

vi

6 a 7 C 8 D 9 a 10 a e 11 a e 12 C 13 a e 14 E 15 E 16 C 17 a 18 2 19

6 5

13 47

(2, 0)

b (3, 2)

c (4, 2)

(0, 0) (1, 5) (-4, 1) (-1, 5) (2, 5)

b b f b f

(1, -2) (0.5, 1)

b (2, 3) f (10, 5)

c (3, 9)

B

b E

c D

57 43

d (-8, 9)

c (0, c) No (2.5, 2.5) c (-11, -39) d (4, -4) (-3, 2) c (1, 2) d (-2, 3) (3, -4) (1.5, 3) d (4, -3)

or 1 15

Answers 4A ➜ 4E

Triangle

2

k

c 2

5 a i 0, 1 iv 0

2 a a =

2

j 3

-59

-

- 21 a 4

-1

d 5 6 h 44

e

58 a 127

5

c -50 g 25

4 a 1 14

6f -9 10 d - 9 i d = or f = 10 6

j a =

1

b - 3 2 f 12

e

5

-

127 b 58

2s - 2 na , d = -2 n( n - 1)

i

Exercise 4C — Transposition -4

b d =

1 a -19 e 2

9 0.002 10 4.0 × 103 days 11 1.23% 5

2s - n( n - 1)d ,a=4 2n

Exercise 4D — Solving linear equations and simultaneous linear equations

6 A 7 D 8 a 2000 steps b 77 sugar cubes

1 a x = 5 - 4 y y =

12 a a =

0 x = 1, y = -2, z = 3 2 Exercise 4E — Applications 1 -1 2 40 3 66 4 18 red, 10 yellow tulips 5 Width = 8 cm Length = 20 cm 6 10 cm, 10 cm, 12 cm 7 40°, 120°, 20°, obtuse angled 8 $99

Answers

625

9 a B b D 10 a 10 b 15 11 a 3 b 3 years in America, 4 years in Germany, 8 years in London, 1 year in Australia. 12 15 Jonathan 20 Golden Delicious 13 Alex will reach Nat in 1.25 h; that is, at 11.30 am. 14 4 km/h 15 10 h 16 12 days, No 17 20 18 44, 45 19 7, -4 20 2 2 23 24 26

17, 5 l = 7 cm w = 5 cm A = 35 cm2 x = 8 cm y = 10 cm P = 50 cm 33 cm a

21

2 1 , 5 4

25

6 cm

x y

c

-9 (2t + 3)(t - 1)

g

3w 2 - 6 w - 20 2( w - 2)

( y + 3)( y - 3) - 7 x + 10 c (3 - 2 x )( x - 2) 5a + 17 e (a + 3)2 4 D 5 A 6 C 2x 7 a 5 2 c 2 d e

- 2( j - 3)2

( j + 7)( j + 2) m +1 4(m - 1)(2m - 3)

g

23b 12 g d 6 4 k + 15 f 6 b

-1 10r (r + 1)

1 b ac a a-b i x = b + a 1 k x = +s br g x =

g

m2 - 14 7m

h

1 - 2 n2 3n

2 a

5p + 4 p( p + 2)

b

13q + 15 2q(q + 5)

c x = 2a y = -b

-17d

15 h 2 + 20 e 5h

Answers

d

- 6s + 25

(s - 3)(s + 4)

9 + 10 v 2 8(2v - 3) 13 x - 1 h 6 f

6z + 8 ( z + 2)( z - 2) y + 11 d (1 - y )( y + 3) b

f

(3b - 9) (3b - 2)2

1 4b 2(3 - e) d e b

f 6h b d

2(k - 2)2 (k + 5)2 (k + 1) - 3( n + 1)

2 n( n - 1)

3(q + 3) f q-2 - (s + 2)(2s + 3) h 6

Exercise 4G — Linear literal equations wy 1 a x = 6c b x = z np c x = d x = 3st - r +m 2 d 3k f x = e x = -l f +g 4-l

1- f -d ef e d o x = - - c 3 2 2 D 3 a x = a, y = b

c

626

3g 3 6(2 + g)

e p

8 Toaster: $19.95 2 Sandwich maker: $24.95 29 ‘Pool & Spa’: 120 ‘Pool, spa, sauna & steam room’: 73 30 5c coins: 34 10c coins: 8 31 Lamb: $2.50 Pork: $4.00 32 23 33 10 science text pages and 2 fiction text pages 34 Dim sims: 5 Spring rolls: 4 Exercise 4F — Algebraic fractions 25a 28

- 4 y - 30

3 a

c

1 a

-t

e

8 a

b 5 cm × 5 cm; square 7 54; Yes 2 3 18; No; 4

7r - 5 (r + 1)(r - 2)

4a + c 2b n j x = m( n + 1) h x =

l x = q - n n2 m mb + cn ef p x = d

m x =

n x =

b x = a - b y = 2b d x = 3a + b y = -b

1 2(b - a) 3b y= 2 b a - 2b g x = h x = z a a - 2b y = 2a - 2b y= 2ab 4 a 4 b -2

Extended response 1 a W = 600 + 0.02s b i $1000 ii $1900 iii $2600 c i $35 000 ii $60 000 iii $90 000 d Brett is paid $860 each fortnight regardless of whether or not he makes any sales. However, the amount Adrain takes home depends on how many sales he can make. He will take home a minimum of $600 each fortnight but must continue to make sales if he wishes to increase this amount. e S = 860 f $13 000 2 9000 for 6%, 6000 for 11% 3 a B b Fixed fee c Same cost d (500, 65) e B f A g Under 500 km — choose A, over 500 km — choose B; = 500 km — either h d = 0.03n - 15 i dn - dn - 1 = 0.03, d0 = -15 j Between 800 and 900 4 a C = 15n + 21 000 b SP = 45n c ($)

Chapter review Short answer -4

-16

45 x 15 y 3 a 0.003 75 b 0.00 000 408 c 4 165 000 a 106.51 b 1200 5 a x = 2y - 2

b d = 5 x =

e=

- 43 21 - 43 10

10

- 21 e 21

- 10 d

-6

Expenses/selling price ($000)

1 2 3 4

71

6 a y = 2x + 2 [1] y = -0.5x + 4 [2] b (0.8, 3.6) 7 a (2, 0) b (-4, 1) c (1, -6) 8 20 of 20c 5 of 10c 10 of 5c 9 T = 2300 + 500n - 135e T - 2300 + 135e n = 500 2300 + 500 n - T e = 135 10 Rebecca: 7 years old Jessica: 10 years old - 11x + 16 13m - 24 b 11 a x (3 x - 4) 6 5x - 5 c ( x + 1)( x + 2)( x - 3) 12

0

5

tC

s Co

=1

(700, 31 500)

P = 45n

n

A C D D

No, the graph does not commence at the origin, but at 21 000. This occurs because of initial costs relating to electricity, materials, rent and so on, prior to any products being manufactured. d Refer to graph in part c. Yes, the graph does commence at the origin. This occurs because if there are no sales, no money is received. e The break-even point occurs when expenses (total cost of manufacturing the puzzles) equal the selling price (money received from sale of puzzles). Therefore, the company is neither making a profit nor running at a loss. f (700, 31 500) g Refer to the graph in part c. This portion (the blue portion) of the graph represents expenses (total cost of manufacturing the puzzles) being greater than the selling price (money received from sale of puzzles). Therefore, the company is making a loss. h Refer to the graph in part c. This portion (the yellow portion) of the graph represents the selling price (money received from sale of puzzles) being greater than the expenses (total cost of manufacturing the puzzles). Therefore, the company is making a profit. i P = 30n - 21 000 j i -9000 (a loss of $9000) ii -3000 (a loss of $3000) iii 3000 (a profit of $3000) iv 9000 (a profit of $9000)

Answers

Answers 4F ➜ 4G

13

5 10 15 20

00

10

2 n+

No. of puzzles

9( g + 3)5 ( g - 1)2

10(5w - 3)( w + 4) ( w + 3)( w - 3) 3 pk 2 + kw b - gmn 14 a x = b x = kp+3 a 1 15 x = 2a y = -2b Multiple choice 1 B 2 D 3 E 4 B 6 A 7 E 8 E 9 D 11 D 12 D 13 B 14 A 16 C 17 B 18 E 19 C 21 C 22 D 23 C 24 B

35 33 31 29 27 25 23 21 1

10 0 20 0 30 0 40 0 50 0 60 0 70 0 80 0 90 0

f x =

e x = a b y= a

627

CHAPTER 5

Trigonometric ratios and their applications Exercise 5A — Trigonometry of right-angled triangles 1 a b Adjacent

Opposite

Adjacent θ

θ

Hypotenuse

Hypotenuse

Opposite

c

d

Opposite

Hypotenuse θ

Adjacent Hypotenuse

Adjacent

Opposite

θ

2 a 6.43 b 11.89 c 24.99 d 354.05 e 4.14 f 18.11 g 445.90 h x = 21.14, y = 27.13 3 a 44°26′ b 67°23′ c 44°25′ d 17°10′ e 68°58′ f 38°41′ g 47°4′ h 61°55′ 4 a 2 3 cm b 12 3 cm 2 c 12 + 8 3 cm 5 26 3 + 54 m 6 a 4.98 m b 66°56′ 7 7°11′ 8 23°4′ 9 8.58 m 10 1.44 m 11 4 and 4 3 12 a = 5.36, b = 4.50, c = 4.78, d = 10.72 13 a = 14.90, b = 20.05 14 x = 13.39 15 115.91 m 16 64°51′, 64°51′, 50°18′ 17 10.91 m3 18 a 18°59′ b 15.7 m 19 a 0.76 m b No, the foot of the ladder moves through a distance of 0.96 m. Exercise 5B — Elevation, depression and bearings 1 571 m 2 30 m 3 91 m 4 43.18 m 5 a 22.33 m b 13.27 m 6 6°47′ 7 a b 1319.36 m Helicopter 35°

S2

48°

2500 m

S1

8 22 m 50 3 9 50 m 3 10 a 325° T b 227° T

628

Answers

c 058° T

d 163° T

11 a S66°W b S73°E c N39°W d N74°E 2 a C 1 b D 13 1691 m 14 a 5.39 km b N21°48′W 15 201°48′ T 16 a 4.36 km b 156°35′ T 17 a 12.2 km b 348 T or N12°W 18 a 29.82 km b 38.08 km c 232° T 19 a 112.76 km b 5 hours 30 minutes 20 a 82.08 m b 136.03 m c 301°7′ T 21 a i 571.5 m ii 715 m b i 143.5 m ii 4.31 km/h Exercise 5C — The sine rule 1 44°58′, 77°2′, 13.79 2 39°18′, 38°55′, 17.21 3 70°, 9.85, 9.4 4 33°, 38.98, 21.98 5 19.12 6 C = 51°, b = 54.66, c = 44.66 7 A = 60°, b = 117.11, c = 31.38 8 B = 48°26′, C = 103°34′, c = 66.26; or B = 131°34′, C = 20°26′, c = 23.8 9 24.17 10 A 11 A = 73°15′, b = 8.73; or A = 106°45′, b = 4.12 12 51.9 or 44.86 13 C = 110°, a = 3.09, b = 4.64 14 B = 38°, a = 3.36, c = 2.28 15 B = 33°33′, C = 121°27′, c = 26.24; or B = 146°27′, C = 8°33′, c = 4.57 16 43.62 m 17 a 6.97 m b 4 m 18 a 13.11 km b N20°47′W 19 a 8.63 km b 6.48 km/h c 9.90 km 20 22.09 km from A and 27.46 km from B. 21 D 22 B 23 Yes, she needs 43 m altogether. Exercise 5D — The cosine rule 1 7.95 2 55.22 cm 3 23.08, 41°53′, 23°7′ 4 28°57′ 5 88°15′ 6 A = 61°15′, B = 40°, C = 78°45′ 7 37 cm 8 2218 m 9 a 12.57 km b S35°1′E 10 a 35°6′ b 6.73 m2 11 23° 12 89.12 m 13 a 130 km b S22°12′E 14 28.5 km 15 74.3 km 16 70°49′ 17 a 8.89 m b 76°59′ c x = 10.07 m 18 1.14 km/h

Exercise 5E — Area of triangles 1 12.98 cm2 2 38.14 cm2 27 3 2 6 C 8 570.03 mm2 10 4 3

3 212.88 m2

4

5 30 2 7 14.98 cm2 9 2.15 cm2 11 B 12 A = 32°4′, B = 99°56′, area = 68.95 cm2 13 A = 39°50′, B = 84°10′, area = 186.03 m2 14 A = 125°14′, C = 16°46′, area = 196.03 mm2 15 3131.41 mm2 16 610.38 cm2 17 a 187.5 cm2 b 15.03 cm c 187.47 cm2

18 17 goldfish 19 22.02 m2 2 20 a Area = 69.63 cm b Dimensions are 12.08 cm and 6.96 cm. 21 17 kg 22 52.2 hectares 24 D 23 174.5 m3 25 B Exercise 5F — Trigonometric identities 3 2 5 1 5 13

3

4 5

3 5 7 6 0.208

4

5 Answers will vary. 7 0.743 Exercise 5G — Radian measurement

π 6 5π e 4 2 i π 5 1 a

2 3 4

a e i a e i a e i

45° 105° 247.5° 0.4712 0.1222 5.7052 134°22′ 205°48′ 233°22′

π 3 3π f 2 10 π j 9

b

b f j b f j b f j

270° 510° 1440° 1.9024 1.1118 0.8209 34°58′ 415°24′ 353°21′

2π 3 7π g 4

5π 6 8π h 3

c

d

c 210° g 15°

d 300° h 234°

c 4.2412 g 2.4147

d 6.1261 h 4.7845

c 57°18′ g 10°26′

d 92°15′ h 334°37′

57.33 m

124 m

56 m

1

A

105.86 m

2 C

95º

64º

52 m

O

O

O

C

52 m O

56 m

80º 3

25º

78.10 m

68 m

68 m 43 m

D 34.25 m

E

D

A

O

105.86 m

58º

85 m

6

65.19 m 5

F

4

38º 85 m

43 m F

72.11 m

O

Answers 5A ➜ 5H

Exercise 5H — Arcs, sectors and segments 1 15.88 cm 2 200.28 mm 15π 4 4.89 cm 3 2 12 cm 5 7.77 cm 6 r = π 7 a 0.2667c b 15°17′ c b 91°40′ 8 a 1.6 9 35°16′ 10 77°59′ 11 73.3 cm 12 2.20 m 14 85.88 cm2 13 141.23 cm2 15 A = 10π cm2 16 36.75 cm 17 106°54′ 18 270 cm2

9 a 10 m2 1 b 1 m3 2 20 8050.85 cm 21 a 188.5 cm2 b 5 cm 2 22 237.66 cm 23 5.44 m2 24 24.14 cm 25 50°56′ 27 D 26 6.64 cm2 28 E 29 2.95 m2 30 A Chapter review Short answer 1 1.73 m 2 20.8 m 3 m =12 2 cm 4 -0.08611 5 125 2 m 2 6 4 3 m 2 55 7 8 4π 25π 32π 8 a i 9 ii 36 iii 9 b i 9° ii 337.5° iii 1260° 9 a 228.54 m b 2945.25 m2 Multiple choice 1 C 2 B 3 C 4 B 5 D 6 E 7 C 8 D 9 B 10 D 11 B 12 B 13 D 14 E 15 C Extended response 1 a 44°25′, 57°7′, 78°28′ b 14.697 cm2 2 c 1.270 cm 2 a 3.931 km b 6.075 km2 c N89°53′E d 2.190 km 3 a i 12.59 km ii S36°10′ E b 2783 m 4 a and b B B

E

Answers

629

c 840.84 m d $3784.50 e Area 1 = 2952.80 m2 (Note: Due to rounding, this answer may vary slightly depending on which side lengths and angles were used.) Area 2 = 1308.64 m2 Area 3 = 1741.14 m2 Area 4 = 617.87 m2 Area 5 = 1549.81 m2 Area 6 = 2769.89 m2 Total area = 10 940.15 m2

CHAPTER 6

Sequences and series Exercise 6A — Describing sequences 1 a Add 3 (to the previous term); 10, 13, 16 b Subtract 1 (from the previous term); −3, −4, −5 c Multiply by 4; 256, 1024, 4096 d Divide by 2; 3, 3 , 3 8 16 32 e Add three to the magnitude then change the sign; −17, 20, −23 f The difference between the terms increases by 1 for each pair; 27, 35, 44 g Add the preceding two terms; 29, 47, 76 h Add 3b − a; −2a + 7b, −3a + 10b, −4a + 13b i Many possible answers — assume the sequence repeats; 0, −1, 0 j Append 1 to the decimal expansion of the preceding term; 1.111, 1.1111, 1.111 11 k Divide by −2; 64, −32, 16 4 2 a -3, 5, 15 b , 108, 26 244 3

c

1 5 10 , , 2 6 11

d 13.3, -1.5, -20

e

5 5 , , 5 2 32 1024

f

5 , 4

g 0, 4, 11

h

3 243 59049 , , 2 32 1024

i 41, 61, 131 k a, ar4, ar9

j a, a + 4d, a + 9d

20, 640

3 a 7, 17, 21 b -3 12 , -13 12 , -17 12 c 7.65, 1858.95, 16 730.55 d −12, 384, 1536 e

11

d 0.456, 0.471 321, 0.473 437, 0.473 659, 0.473 682, 0.473 684; sequence converges to 9 19

630

Answers

g 0.48, 0.749, 0.564, 0.738, 0.581, 0.73; oscillates h 0.714, 0.694, 0.722, 0.683, 0.736, 0.66; oscillates i 0.378, 0.987, 0.052, 0.207, 0.689, 0.901; diverging j 0.72, 0.907, 0.379, 1.059, -0.281, −1.619; diverging a 15, 20; difference between subsequent terms increases by 1. b There are many possible answers. A possible pattern is the addition of 5 then 3 then 1 then −1. The next two terms are 4, −3. Here the difference between successive terms follows an arithmetic sequence. c Many possible answers since there is no obvious pattern. It could be the start of a telephone number. d Each successive term is multiplied by an increasing 1 -1 10 1 1 factor of 2 starting 2 = 2 then 2 = 1 and then 2

5

1

6 7 8 9

3

3

followed by 4 ; 16 , 256 . e 34, 55; Each subsequent term is the sum of the preceding two terms. f 31, 63; Terms are 1 less than powers of 2. g 5, 4; Add 2 to find the next term and then subtract 1 to find the subsequent term and repeat. a D b E c C a tn + 1 = tn − 2, t1 = 7 b tn + 1 = tn ÷ 2, t1 = 12 c tn + 1 = tn + 0.6, t1 = 12 d tn + 1 = tn × 5 + 1, t1 = 2 e tn + 1 = −3tn, t1 = 4 f tn + 1 = (tn)2, t1 = 2 a 26 and 25 b 23 cats The population size will rapidly decrease and by 2009, the stray cat population will be gone. (Happily, they were all taken in by good and loving households.)

Exercise 6B — Arithmetic sequences 2 a Arithmetic, difference = 2; t4 = 9; tn = 1 + 2n b Not arithmetic c Not arithmetic d Arithmetic, difference = 3; t4 = 6; tn = −6 + 3n e Arithmetic, difference = −4; t4 = −14; tn = 2 − 4n 1

3 729 6561 , , 4 4096 65536 0, −35, −49

f g 3, −1, −1 h 0, 0, 0 i −1, −32, −128 j a3 + a2 + a, a8 + a7 + . . . + a, a10 + a9 + . . . + a k 2, 21, 55 l −123, −7.55 × 1066, −3.2 × 10267 a 0.2, 0.128, 0.089, 0.065, 0.049, 0.037; sequence converges to 0 b 0.096, 0.0347, 0.0134, 0.0053, 0.0021, 0.0008; sequence converges to 0 c 0.099, 0.0981, 0.0973, 0.0967, 0.096, 0.0955; sequence converges to 1

4

e 0.525, 0.523 687 5, 0.523 821 7, 0.523 808 3, 0.523 809 6, 0.523 809 5; sequence converges to 11 21 f 0.525, 0.623, 0.587, 0.606, 0.597, 0.602; sequence 3 converges to 5

25

-3

1

f Arithmetic, difference = 2 ; t4 = 14 ; tn = 14 + 2 n g Not arithmetic 3 3 h Arithmetic, difference = 4 ; t4 = 3; tn = 4 n -7

7

i Arithmetic, difference = 4 ; t4 = −5; tn = 2 − 4 n j Arithmetic, difference = 2(π − 1); t4 = 8π − 3; tn = 5 + 2(π − 1)n 3 a 104 b 682 c 1458 d −26 310 4 a tn = 8 − 3n, n = 1, 2, 3, . . . n b tn = 2 + , n = 1, 2, 3, . . . 2 c tn = −6 + 3n, n = 1, 2, 3, . . . d tn = −3x + 5nx, n = 1, 2, 3, . . . 5 tn = 4 + 2n, n = 1, 2, 3, . . . 6 5n − 2 7 m = 21.5, n = 32.5 9 −2 8 −x + y, −5x + 9y

5

2 1 14

d tn + 1 = tn +

5

13 m = 27, n = 32

13.6, 19.2, 24.8, 30.4 a tn + 1 = tn + 4; t1 = 3 b tn + 1 = tn + 3; t1 = −3 c tn + 1 = tn − 4; t1 = −2 1 ;t 2 1

2

11 - 4 14 ; 12 ; - 4 3 + 12 n

10 −35; 15; 15n − 50

=

2 7

e tn + 1 = tn +

3 ;t 4 1

=

3 4

f tn + 1 = tn −

7 ;t 4 1

=4

3

 - 1   3

1

1

, t6 = 3072 , t10 = 7 86 432

n -1

, t6 =

- 1 ,t 405 10

-

=

1 32 805

i tn = x × (3x3)n − 1, t6 = 243x16, t10 = 19 683x28 n -1

i 50 j 30, 2550, 250 500 2 a 1275 b 5050 3 a 5000 1 b Each of the 100 terms is 2 less than its corresponding term in question 3. There are 100 terms and so the answer to this question is 50 less than in question 2. 4 258 b −324 5 a −273, −480, −741 b 1080 c 34 6 a 280 n( n + 1) 2 8 a Various answers b Various answers 9 17 10 45 11 a 20 100 b 9900 12 6 13 174 14 The iterative equation is tn + 1 = tn + 8, t1 = 7. The functional equation is t(n) = 8n − 1, n = 1, 2, 3, . . . Exercise 6D — Geometric sequences 1 a Not geometric b Geometric, ratio = 3; t4 = 108; tn = 4 × 3n − 1 c Geometric, ratio = 2; t4 = 24; tn = 3 × 2n − 1 7

3

1

d Geometric, ratio = 2 ; t4 = 13 2 ; tn = 3n − 123 − n 3

1

; t4 = 9 ; tn = (−3)2 − n

f Geometric, ratio = −3; t4 = −54; tn = 2 × (−3)n − 1 3 ;t 2 4

=

27 ;t 28 n

=

2 7

n -1  3 ×  2  × 2n − 3

h Geometric, ratio = 2; t4 = 6; tn = 3 i Not geometric 1 j Geometric, ratio = −6; t4 = −54; tn = 4 × (−6)n − 1 Geometric, ratio = 2π; t4 = 16π 4; tn = (2π)n tn = 5 × 2n − 1, t6 = 160, t10 = 2560 tn = 2 × 2.5n − 1, t6 = 195.31, t10 = 7629.39 tn = 1 × (−3)n − 1, t6 = −243, t10 = −19683

3

-1 1 or 27 . The nth 27 − (−3)5 − 3n, t10 = ±3 25.

could be tn = 4

±3 4

term is tn = 35 − 3n or

5 3 × 2

( n - 1) 2 729

6 m = 12, n = 48 7 m = 36, n = 4 8 a = 300, b = 0.75 9 t1 = 25, r = ±2, tn = 25 × 2n − 1 or tn = 25 × (−2)n − 1 1

3

10 t1 = 3 , r = 2 , tn = 3n − 221 − n

13 a

1 -1 8

1 1

11 −6

12 2, 2 , 8 , or −2, 2 , 3 2

24 2n

b

4 k = 6 1 Exercise 6E — Geometric series b 121, 29 524, 1.74 × 109 1 a 31, 1023, 1 048 575 c 33, −1023, −1 048 575 d −4, −103.8, −746.8 e

3 , 2

12 930, 70 972, 1 302 280 −309, 2239, −1865 15620, 48 828 120 −1.375, −1.332

g h i j

f

46.5, 1534.5

b

2 a 3108

1 , 16

- 2 - 31 - 341

15

3

,

24

,

256

15

6316 , 66 53516

3 458 6 65 024

4 9 7 21 504

8 a 2

b

2 3

c

5 121 875 000 3 2

d 3

e

3 5

9 1.0; 50%, 25%, 12.5% 4

10 3 ; 75%, 18.75%, 4.6875% 11 4; 25%, 18.75%, 14.06% 20   12 a 6 1 - 1  = 5.999 994 278 2   − b 5.722 × 10 6 c 6

()

Answers

Answers 6A ➜ 6E

-1

1  2 32 512 × , t6 = 6 , t10 = 10 x  x  x x a There are two possible answers because the ratio could be −3 or 3. The nth term is tn = 2 × 3n − 1 or tn = 2 × (−3)n − 1, t10 = ±39 366. b There are two possible answers because the ratio could be −2 or 2. The nth term is tn = 2n − 1 or tn = (−2)n − 1, t10 = ±512. c The nth term is tn = 5 × 2n − 1, t10 = 2560. d The nth term is tn = −1 × (−2)n − 1, t10 = 512. e There are two possible answers because the ratio

j tn =

18 E n+2 b 19 a n+3 Exercise 6C — Arithmetic series 1 a 55, 1275, 5050 b 25, 100, 400 17 n + 3n2 c 80, 235, d 15, −70, n(13 − 2n) 2 e 67.5, 272.5, 1095 f 20π, 65π, 230π g 34, 424, 6304 h 35, 0, −122 12

k a b c

 1

h tn = 5 ×

5 6

2

n -1

1

g tn = 3 ×  4 

g tn + 1 = tn + 2π − 2; t1 = 2π + 3 15 9 16 3 17 B

g Geometric, ratio =

1

f tn = 2 × 2n − 1, t6 = 16, t10 = 256

1

e Geometric, ratio =

d tn = 2 × (−2)n − 1, t6 = −64, t10 = −1024 e tn = 2.3 × (1.5)n − 1, t6 = 17.47, t10 = 88.42

631

3 13 a 6 1 4

( )  = 6.750 343 3

b −3.425 × 10−4 3

c 6 4 2

14 6 3

15 16, 8, 4

16 4 - 15 , 17

( 3 + 5) ( 15 - 3)

=

-1

(4 3 + 3 5 ) 3 18

6

19 1, 0.99

1 3

20 Various answers

21 a Mathematically, the student will never make it past the other side of the road. After each attempt, the distance remaining is halved and this result is the extra distance walked at the next attempt. Thus the distance travelled across the road approaches, but never reaches 10 metres. b As shown in part a, the extra distance travelled at each attempt is equal to half the remaining distance from the previous attempt. Given that there will always be an amount remaining to travel, only half this amount can be achieved on the next attempt, regardless of the width of the road. Exercise 6F — Applications of sequences and series ii $120 681 1 a i $91 253 b 3.059 times b $63 000 c $682 500 2 a Arithmetic d i 8.9% ii 4.1% b $48 487 c $560 660 3 a Geometric d i $1120 ii $1865 4 a 36 mg, 44 mg, 100 mg, 20 + 8n mg b 23rd day c 39 mg, 55 mg, 579 mg, 28 × (1.4)n − 1 or 20 × 1.4n d 7th day 5 a 310 b The workers must remove 12 full rows and 17 logs from the 13th row. 6 2801 7 The last person received 13 tickets and Frank had 9 left. 8 The last person received 64 tickets and Kate had 73 left. 9 a 1.8 × 1019 grains of rice b 1.8 × 1015 kg 10 a 4096 grains of rice b 0.41 kg 11 The student will cover a distance of 2 m and thus will never get closer than 1 m from door. 12 Yes, 7th day 13 a

1 9

b

1 3

c

57 99

e

3139 999

f

700 33

g

50 3

d

232 99

14 a 249 b 2596 5 After 8 years, that is, at the start of the ninth year. 1 16 18 m, 24 m, 30 m 17 a π, 3π, 5π, 7π — arithmetic progression with a = π and d = 2π b

632

1 3 5 7 , , , 16 16 16 16

Answers

a 192 − 12n m/s b During the 16th second c 1440 m After 8 minutes, the coffee has cooled to below 50 °C. The sequence for the arithmetic series tnun is 12n − 15, n ∈ {1, 2, 3, . . .} Chapter review Short answer 1 a tn + 1 = 2tn − 3; t1 = 7 b tn + 1 = (tn)2 + 1; t1 = −2 2 a Functional rule: t(n) = 156n − 458 b Iterative rule: tn + 1 = tn + 156; t1 = −302 3 a 5.0 m b 27.5 m c 63 s 4 a 134.2 milli-rem b 3361.6 milli-rem

18 19 20

-1 9

5

89 99

6 a

4 3

b

3 5

7 a

2 9

b 211

5

Multiple choice 1 D 2 E 6 C 7 B 11 E Extended response

3 B 8 E

4 A 9 D

1 a 8

b 4 2

c 4

d Pn = 8

2 a i Arithmetic 3

1

ii t4 = −28

( ) 2 2

5 B 10 C

n-1

b i Geometric ii t4 = a3b2 c i Neither ii t4 = −1.394 531 25 a i 80 ii 37 iii 18 b The trout population will reach 0 in 9 years. c 57 and 86 d No, as the terms in the sequence will keep increasing e No, the sequence will converge to 0.375, that is, the limiting number is 375 fish.

4 a

Pop B Pop A Annual Growth increment rate = 1000 Difference = 1.12

Year

n

1990

1

10 000

15 000

5000

1991

2

11 200

16 000

4800

1992

3

12 544

17 000

4456

1993

4

14 049

18 000

3951

1994

5

15 735

19 000

3265

1995

6

17 623

20 000

2377

1996

7

19 738

21 000

1262

1997

8

22 107

22 000

−107

1998

9

24 760

23 000

−1760

1999

10

27 731

24 000

−3731

2000

11

31 058

25 000

−6058

2001

12

34 785

26 000

−8785

b During 1997 c Annual increment of 1606 insects d Annual growth rate of 1.095 96 a The common difference is 4, which is a constant; therefore, the sequence is arithmetic with definition tn = 4n + 1. b The common difference is 2a, which is a constant; therefore, the sequence is arithmetic with definition tn = 2an + b − a. 6 a i Answers will vary. ii 10% iii 2928 iv 31 784 v b = 1.1, c = 0 b i $3740 ii 16 iii $4780 c i 1850 ii 92.1%

5

Variation Exercise 7A — Direct variation 3

1

d i Yes, k = 2 e i No f i Yes, k = 2 a–f ii Check graphs with your teacher. 2 C 3 a Yes b When n = 0, C = 20. Since the graph does not pass through the origin, a direct variation relationship does not exist. ii k = 2 4 a i k = 2.3 iii k = 1.1 iv k = 2.5 v k = 5.2

vi k =

1 2

viii k =

2 3 1 4

b i x: 8, 12; y: 9.2, 11.5 ii x: 18, 20, 22.5; y: 8, 14 iii x: −1, 9; y: 3.3, 5.5 iv x: −1, 5, 12; y: −5 v x: 2, 8, 14; y: 31.2 vi x: 1, 12; y: 2, 4 vii x: −2, 4; y: 1, 2.5 1 3

viii x: 6, 7; y: 2 , 4

1, 1.5, 2, 3

iii No

ii x2 = 1, 9, 25, 36 2 a i x2 = 1, 4, 9, 16 2 iii x = 4, 16, 36, 64 iv x2 = 1, 4, 16, 36 b i Yes, k = 2 ii Yes, k = 12 iii No

iv Yes, k = 16

x3 = 1, 8, 27, 64, 125 x3 = 8, 64, 216 No 1 Yes, k = 12

ii iv ii iv

x3 = 8, 27, 125 x3 = 1, 8, 64, 216 Yes, k = 3 No

1 4

9 a i iv b i iv

b n =

1 4

m

c m: 9, 16; n: 14 , 1 14

7 D A ∝ r2 r∝ A A = kr2 r=k A

ii v ii v

8 C V ∝ s3 d∝ h V = ks3 d=k h

iii E ∝ v2 iii E = kv2

10 a 12.56

b 1962.5 cm2

11 a 12.56

b 153.86 mL, 200.96 mL, 254.34 mL

12 1.56 m, 2 2 (≈ 2.83) s 13 a V = 1 s 3 b 41.67 cm3 3

b Multiples of 5

b 29.4 cm 36.9 cm Equilateral triangle b 8 L, 3 cans 46 sq. m $4.20 The graph is a straight line passing through the origin. b 800 c F = 800a d 3200 newtons e 1.25 m/s2

d h = s

c 12 cm

14 a Doubled 5 a i × 4 1

b Tripled c Halved ii × 9 iii × 16 iv × 1

b i × 8

ii × 27

iii × 64

c i × 2

ii 3

iii × 2

17 a i × 4 iv × 100 iv 616 cm2

ii × 9 v × 1

iv ×

1 2

iii × 64 vi × 19

4

v 38.5 cm2

8 a Multiplied by n 1 c Multiplied by n 19 a b = 12

4 1

iv × 8

ii Ηalved 1 ii × 16

16 a i Doubled b i × 16

b i 9856 cm2 ii 15 400 cm2 7 B 8 D b 1.3 h or 1 h 18 min

c 15 cm

1

iii 17 9 cm2 vi 1386 cm2

b Multiplied by n2 d Multiplied by n3

b k = 3

c a = 27

Exercise 7C — Inverse variation 1 a Yes, k = 12 c Yes, k = 28 e No 2 a Yes d No 3 C

b Yes, k = 18 d Yes, k = 10 f No b No e No

c Yes f Yes

Answers 6F ➜ 7C

1 a 1 c 12 a c 13 a

1 1 3 2 , , , , 10 5 10 5

iii x = 2, 4, 6, 8 b i Yes, k = 3 ii Yes, k = 5

6 B

c i No

5 C 6 D 9 a 270 km c 90 km/h

x = 1, 3, 5, 7

x = 1, 2, 3, 4 ii

1 a i

5 a k =

1

b i Yes, k = 3

10 a

Exercise 7B — Further direct variation

4 y varies as the square root of x.

1 a i Yes, k = 2 or 1.5

vii k =

b 41.834 km d 0.56%

15 a = ±5, k = ±2.5

3 a i iii b i iii

CHAPTER 7

14 a 41.6 km c 0.234 km

4 D

Answers

633

5 a i k = 12 iii k = 32 v k = 5

ii k = 60 iv k = 26 vi k = 16

vii k = 1

viii k = 20

2

b i x: 5; y: 4.8, 4 ii x: 2, 5; y: 6 iii x: 2; y: 32, 8 iv x: 2, 4; y: 26 v x: 1; y: 25, 10, 2.5 vi x: 1, 2; y: 1 , 1 2

vii x: 2, 5; y:

1 1 , 16 40

2 3

6 E 9 a 10 a 11 a 12 a

7 E 8 A b $1.60 32 b 13 c 11 15 b Yes c 6 m No i 1.7 ii 0.425 v 340 b k = v = 340, f = or f = λ λ 13 a i Halved ii Divided by 3 iii Multiplied by 2 b 100 N 14 Wd 15 a 19 mL b 285 mL c 300 mL d 10 mL e 1 or 33.33% 3

6 280 revolutions per minute 1 17 a 2 and 4: 750 rev per min; 3: 150 rev per min b s1 = 120, s2 = 300, s3 = 60

4 a 1000

b q: 4, 5; p: 125, 37 27

1 R 1 iii R ∝ 2 d

1 v p b i I = R p iii R = 2 d

v t ∝

v t =

6 7 8

4.74 m i 10.20 W/m2 3m 0.57

a a b a

p v

iv n ∝

1

b 0.0390 625 4 3 1

10 a × 4

b × 4

c × 16

11 a B

b C

c C

1

2

3

4

1

8

27

64

3 a

15 16

4 k = 3, u = 8, w = 1, 9, 25, v = 6

60

5 a P =

60

7.5

x3y

60

60

60

b No

1 2

2

3

4

5

x3

8

27

64

125

y

15

4.4

1.875

0.96

x3y

120

118.8

120

120

kv 2 R

x

2

4

5

6

x3

8

64

125

216

130

1 416

2 2 25

11 1 64

c No

y x3y

1040

65

260

2531 8

d Yes x

1

3

5

7

x3

1

27

125

343

y

343

12 19

2 93

1

x3y

343

343

343

343

27

125

e F = k 6 7 8

a a c a c d

d × 16

d Yes d No

b n = 2, p = 4, 7, m = 75

c E = kmv2 x

l

1 d2 p ii F = 2 d p iv n = l p vi I = 2 d b 88 units ii 1.11 W/m2

b

9 a 1.732

1 d2 1

vi I ∝

x y

Answers

ii F ∝

x3

20 9

d Yes 1

Exercise 7E — Joint variation b Yes c No 1 a Yes b Yes c No 2 a Yes

Exercise 7D — Further inverse variation 1 a Yes

634

c No

5 a i I ∝

3

viii x: 10, 15; y: 5,

2 a Yes b Yes 3 m varies inversely as n2.

b P = kRI2 d F =

km R

t m

610 72.70 cm3 200 J 30 m 4825.4 J Longer i More

5.4 n 9 a m = p2 c 5 − 10 a 1.78 × 10 8 11 C 12 B 13 E 14 D 15 R increased by 79.93% 16 a i Increase by 21% b Decrease by 20%

b 8.63 cm b 6.3 kg b 52 min ii Shorter b 0.45 b 4.45 ohm

ii Decrease by 19% c 32.25%

17 a F =

Chapter review Short answer

kq1q2 r2

b l = a m + b3 n

1 2 3 4

d y = ax3 + bz2

5 a Joint

b P ∝

23 2 a A = 50 n + 48 or A = 0.46n + 48

c i Halved

ii × 3

b $250.40 3 a 2000 c $5000

d i No effect

ii × 4

1

ii × 4

iii × 2

iv × 4

b i × 4

Exercise 7F — Part variation 1 a y = ax2 + b a c y = + b x

b C = 2000 + 25n d C 5000 2000 120

4 5 6 7

a c a c E a c

$229 $579 $589 $237, $22

V = 10 + 2t 66 m/s 2 8 a y = 3 x 2 x

b 10 m/s d 12 s

b

1 19

3

n , 52 2 15 5 b 4.25 c 9 y = +3 x b D c C d E A 2 + 62.8r A = 3.14r ii 1070.74 cm2 i 819.54 cm2

10 m = 5n +

Exercise 7G — Transformation of data

t varies directly as d

c t = 0.045 d d t = 0.2012 seconds

Multiple choice 1 A 2 6 B 7 11 D 12 16 B 17 21 B

W t

1

3 8 13 18

C E E C

4 9 14 19

C C B D

5 10 15 20

C C E C

C B A A

Extended response 1 a k = 25, M =

25 L fo f c

b 1.488

2 a 20.0048 m b l = 20 + 0.000 24 ∆T − c 1.2 × 10 5 3 a y varies as the square of x; that is, y ∝ x2. 3

b y = 20 x2 b V =

114 p

c Inversely 5 a The graph has a hyperbolic shape; there is no direct variation present. 1 b Select M ∝ . u c Mu = −40, −40, −40, −40 d The graph is a straight line through the origin. e Directly -v f M = u 6 a E varies directly with m. 1800 2000 2200 2400 = = = b 695 556 772 840 850 124 927 408 2600 = 2.59 × 10 -3 = 1 004 692 1 c Inverse d m ∝ 2 v e Directly, inversely, the square 2E f k = 2, m = 2 g 1 476 562.5 J v l 7 a R ∝ 2 d b

R

6

12

15

18

20

l

0.8

0.9

0.5

3.75

1

d

4

3

2

5

6

k

120

120

120

120

120

Answers

Answers 7D ➜ 7G

1 a Parabola b y = 1.76x2 2 a Hyperbola b y varies inversely as x. 100 c y = x 3 a There is no direct variation present. b V = 0.5236D3 4 a Logarithmic graph b a = 2, b = 5, y = 2 log10 (x) + 5 5 D 6 B 7 a The calculator shows that t does not vary directly as d. b The calculator shows that: t does not vary directly as d 2

b 252 J c 6.5 m 2 c 8 cm b 864 cm b x = 2.5, 5; y = 9.6, 4 ii 0.91 W/m2

21 6 48 i 4.8 W/m2 4m

4 a Hyperbola 2

9 a y = x2 + 1

2 a 1 13 a b

b 70 cents d 128 loaves b 22 weeks

ii -5 3

b i 74.6

11 a

n

a a a a b

635

c R =

120l d2

Exercise 8B — Partial fractions 5 15 1 a 2 b 4 + x +1 x-2

d i R is doubled.

c

1 15 + 2 2(2 x - 1)

d

3 7 2 2x + 2

2 a

2 1 x - 4 x +1

b

3 2 x-2 x-3

f d must be halved.

c

19 1 + 10( x + 9) 10( x - 1)

d

5 1 2x - 7 x - 1

Exam practice 2

3 a

1 3 x + 2 ( x + 2)2

b

1 1 x - 3 ( x - 3)2

c

1 1 2 x - 1 x + 3 ( x + 3)2

1

ii R is 4 of its original value. iii R is halved. e i Increase, R is 5 times as large. 5

ii Decrease, R is 8 of its original value.

Short answer 7 - 11x 1 (2 x + 1)(3 x - 1) 2 x =

bd + c b+a

d

1 1 8 + 9( x - 2) 9( x + 1) 3( x + 1)2

y =

c - da b+a

4 a

2 8- x + x + 1 x2 - 2x + 5

3

3 c

2

4 r = 2 3 Multiple choice 1 B 2 A 3 A 4 D 5 D 6 C Extended response 1 a n = 43 b S∞ = 566.67 After the injury + 500 before the injury. 1066.7 km; Theo will not reach Sydney. 7 3 4 q ° = 30 + 45 Distance is 4.93 km. y = 2 tan (15.2°) + x tan (15.2°) The vertical height of lookout from the hotel is 952 m.

2 a x =

b c d e

CHAPTER 8 Exercise 8A — Polynomial identities i i i i

Yes No Yes Yes

ii 4 ii 6 ii 2

2 a = 2, b = -1, c = 4 3 a = 1, b = 4, c = -5 4 a = 3, b = 4, c = -2 5 a = 10 , b = - 10 or a = - 10 , b = 10 6 7 8 9 10

636

a = 1, b = -2, c = 1 b = -12d - 4c x2 + 5x - 6 x2 - 2x + 3 x2 + 3x - 5

Answers

5  4 x + 15  x  x 2 + 3 x + 1

-

2 2x - 6 + x - 2 x 2 - 5x + 5

23 4x 26 + + d 27( x - 3) 27( x 2 + 3 x + 9) 9( x 2 + 3 x + 9) 5 a x + 1 -

1 x+2

c 3 x - 1 +

b x 2 + 4 x + 18 +

11 - 21x x2 + x + 6

d

69 x-4

x2 x 1 25 + + + 2 4 8 8(2 x - 1)

Exercise 8C — Simultaneous equations 1 a (−2, −2) c (−1, −2) e (1, −2), (−1, 2)

b (−2, 2) d (−2, −6), (−3, −9) f (2, 7), (3, 8)

 - 3 + 33 9 - 33   - 3 - 33 9 + 33  , , g  ,   6 6   6 6    - 2 + 4 7 22 - 8 7   - 2 - 4 7 22 + 8 7  , , h  ,   3 9 3 9    

Further algebra 1 a b c d

b

i (-2, -4),

(

9 - 15 , 4 8

)

 3 + 33 1 - 33   3 - 33 1 + 33  , , j  ,   2   6 2   6   2 2 2 a  , ,   2    2  2 b  ,  2

-

-

 2 ,   2  

2 , 2 -

-

2  2 

2 2  , 2 2 

 2 5 4 5   -2 5 -4 5  c  , , ,   5   5 5   5

1+ 7 1- 7  1- 7 1+ 7  d  , , ,   2   2 2   2  2 + 14 2 - 14   2 - 14 2 + 14  e  , , ,   2 2   2 2   f (- 3 + 2 2 , 2 2 ), (- 3 - 2 2 , - 2 2 )  10 + 2 5 , g  5 

-

5   10 - 2 5 5 , ,   5   5 5 

 -1 + 6 -13 + 3 6   -1 - 6 -13 - 3 6  h  , , ,   5 5  5   5   4 + 2 109 3 - 109   4 - 2 109 3 + 109  i  , , ,   5 5 5 5      12 + 179 4 + 2 179  , j  , 5 5    12 - 179 4 - 2 179  ,   5 5   1+ 5  1- 5  3 a  , 1 + 5 ,  , 1 - 5 2 2     b No real solutions c No real solutions  -1 + 7   -1 - 7  , 2 - 7 ,  , 2 + 7 d  2 2      1 + 17 13 + 17   1 - 17 13 - 17  , , e  ,   4 4  2   2   f  10 + 4, 

-

10  10    ,  10 + 4,  2   2 

 1+ 5  1- 5 g  2 + 5 ,  ,  2 - 5,  2 2    

 - 9 + 93 3 + 93   - 9 - 93 3 - 93  , , h  ,   3 6   3 6  

 20 - 247 4 - 2 247  ,   6 9    25 + 301 5 - 301   25 - 301 5 + 301  , , j  ,   6 6 6 6     4 E 5 (0.71, 0.71), (−0.71, −0.71), (0.71, −0.71), (−0.71, 0.71)

 -1 + 17 -1 + 17   -1 - 17 -1 - 17  , , 6   or   2 2 2 2     Multiple choice 1 C 2 D 3 A 4 E 5 D 6 B Extended response 1 a (0.22, 0.22), (6.12, 6.12) b (0.82, 0.82), (−1.82, −1.82) 2 a 2 b A(−2, 0), F(0, 2), K(2, 0), P(0, −2) c I( 2 , 2 ) d C(- 2 , 2 ), R(- 2 , - 2 ), N( 2 , - 2 ) e H(1, 1) f D(−1, 1), S(−1, −1), M(1, −1) g G, J, T, Q h G(0.52, 1.93), J(1.93, 0.52), T(−1.93, −0.52), Q(−0.52, −1.93) i B, E, O, L j B(−1.93, 0.52), E(−0.52, 1.93), L(1.93, −0.52), O(0.52, −1.93)

3

a c e g

b 4 m B = (4, 7) d C = (12, 7) 4.73 m and 11.27 m f 17.44 m D = (16.67, 14.48) 5.92 m, 9.05 m, 12.12 m and 15.83 m

CHAPTER 9

Algebra and logic Exercise 9A — Statements (propositions), connectives and truth tables 1 a Opinion b T/F c T d Question e T f Opinion g Instruction h T/F i F j Near-statement 2 a The car has 4 seats. The car has airconditioning. b The Department of Finance was over budget in 2006. The Department of Defence was over budget in 2006. c Bob went to the hotel. Carol went to the hotel. Ted went to the hotel. Alice went to the hotel. d To be a best-seller a novel must be interesting to the reader. To be a best-seller a novel must be relevant to the reader. e Sam will win the trophy. Nancy will win the trophy f You can choose vanilla ice-cream for dessert. You can choose strawberry ice-cream for dessert. You can choose fruit for dessert.

Answers

Answers 8A ➜ 9A

 20 + 247 4 + 2 247  i  , , 6 9  

Chapter review Short answer 1 a = 3, b = 4 2 x2 + 3x - 5 2 1 3 x + 2 x - 10 3x - 5 4 x - 2 + 2 x -2 5 (3, 15), (-1, -5)

637

g There are some statements which cannot be proved to be true. There are some statements which cannot be proved to be false. h Most of my friends studied Mathematics. Most of my friends studied Physics. Most of my friends studied Engineering. Most of my friends studied Law. Most of my friends studied Arts. 3 a John and Mary rode their bicycles to school. b The book you want is in row 3 or 4. c The weather is cold and cloudy. d Many people read novels or history. e In a recent poll 80% preferred jazz or classical music. f Two is an even prime number. Two is the only even prime number (alternative answer). 4 B 5 D 6 a

b

p

q

r

T

T

T

T

T

F

T

F

T

T

F

p

q

p∧q

p = Sydney on time

T

T

T

q = Perth fully booked

T

F

F

p∧q

F

T

F

2 = 4 rows

F

F

F

7 a

2

p

q

r

p∧q∧r

p = John passed

T

T

T

T

q = Zia passed

T

T

F

F

r = David passed

T

F

T

F

p∧q∧r

T

F

F

F

23 = 8 rows

F

T

T

F

F

T

F

F

F

F

T

F

F

F

F

F

b

8 ways

p

q

r

(p ∧ q) ∨ r

p = Alice does dishes

T

T

T

T

F

q = Renzo does dishes

T

T

F

T

T

F

T

T

T

F

F

F

F

T

T

T

c

F

T

T

r = Carla does dishes

F

T

F

(p ∧ q) ∨ r

F

F

T

F

F

F

2 = 8 rows 3

p

q

r

s

p

q

r

s

T

T

T

T

F

T

T

T

T

T

T

F

F

T

T

F

T

T

F

T

F

T

F

T

T

T

F

F

F

T

F

F

T

F

T

T

F

F

T

T

T

F

T

F

F

F

T

F

T

F

F

T

F

F

F

T

T

F

F

F

F

F

F

F

d

p

Answers

F

F

F

T

T

F

F

F

F

q

r

p ∧ (q ∨ r)

p = female member

T

T

T

T

T

T

F

T

r = professor

T

F

T

T

p ∧ (q ∨ r)

T

F

F

F

F

T

T

F

F

T

F

F

F

F

T

F

F

F

F

F

23

= 8 rows

p

q

r

s

t

p

q

r

s

t

p

q

r

s

t

p

q

r

s

t

T

T

T

T

T

T

F

T

T

T

F

T

T

T

T

F

F

T

T

T

T

T

T

T

F

T

F

T

T

F

F

T

T

T

F

F

F

T

T

F

T

T

T

F

T

T

F

T

F

T

F

T

T

F

T

F

F

T

F

T

T

T

T

F

F

T

F

T

F

F

F

T

T

F

F

F

F

T

F

F

T

T

F

T

T

T

F

F

T

T

F

T

F

T

T

F

F

F

T

T

T

T

F

T

F

T

F

F

T

F

F

T

F

T

F

F

F

F

T

F

T

T

F

F

T

T

F

F

F

T

F

T

F

F

T

F

F

F

F

T

T

T

F

F

F

T

F

F

F

F

F

T

F

F

F

F

F

F

F

F

32 ways

638

T

q = student

16 ways 6 c

F F

8

p

~p

p ∧ ~p

T

F

F

F

T

F

9 a

c

d

f

14 a

p

q

p ∧ ~q

p

q

~p ∧ ~q

T

T

F

T

T

F

T

F

T

T

F

F

F

T

F

F

T

F

F

F

F

F

F

T

b

p

q

r (p ∧ q) ∧ r

p

q

r

(p ∧ q) ∧ r

T

T

T

T

F

T

T

F

T

T

F

F

F

T

F

F

T

F

T

F

F

F

T

F

T

F

F

F

F

F

F

F

e

T T

T

F

T

T

F

T

F

F

F

F

T

F

F

T

F

T

T

T

T

F

F

F

T

T

F

F

Not equivalent b p q (p ∨ q) T

~p

(p ∨ q) ∨ ~p

(p ∨ ~p)

F

T

T

T

T

T

F

T

F

T

T

F

T

T

T

T

T

F

F

F

T

T

T

Equivalent 15

p

q

r

(p ∧ q) ∨ r

p ∧ (q ∨ r)

T

T

T

T

T

T

F

T

T

F

T

T

T

p

q

~p ∨ ~q

T

T

T

T

F

T

T

T

F

T

T

F

F

F

F

F

T

T

T

F

F

T

F

F

F

F

F

T

T

F

F

F

F

F

F

p

q

p ∨ ~q

T

T

T

F

F

T

F

F

T

T

F

F

T

F

F

T

p

q

r (p ∨ q) ∨ r

p

q

r (p ∨ q) ∨ r

T

T

T

T

F

T

T

T

T

T

F

T

F

T

F

T

p

q

r

(p ∧ q) ∧ r

p ∧ (q ∧ r)

T

F

T

T

F

F

T

T

T

T

T

T

T

T

F

F

T

F

F

F

F

T

T

F

F

F

T

F

T

F

F

It is raining and I bring my umbrella. It is raining or I bring my umbrella. It is not raining and I bring my umbrella. Peter and Quentin like football. Peter or Quentin like football. Peter likes football or Quentin does not like football.

12

p

q

~(p ∨ q)

(~p ∧ ~q)

T

T

F

F

T

F

F

F

F

T

F

F

F

F

T

T

Not equivalent 16 a

T

F

F

F

F

F

T

T

F

F

F

T

F

F

F

F

F

T

F

F

F

F

F

F

F

Equivalent b

p

q

r

(p ∨ q) ∨ r

p ∨ (q ∨ r)

T

T

T

T

T

T

T

F

T

T

F

T

T

T

p

q

~(p ∨ q)

~p ∨ ~q

T

T

T

F

F

T

F

F

T

T

F

T

T

T

T

T

F

F

T

F

T

F

T

F

F

T

T

Not equivalent

F

T

F

T

T

F

F

T

T

T

F

F

F

F

F

Answers 9A ➜ 9A

10 a b c 11 a b c

13

q (p ∧ q) ~p (p ∧ q) ∨ ~p (p ∨ q) (p ∨ q) ∧ ~p

p

Equivalent

Answers

639

c Brackets have no effect on expressions with a single ∨ or ∧ operator, but have an effect if they are mixed up together. Exercise 9B — Valid and invalid arguments 1 p q ~p ~q (p ⇒ q) (~q ⇒ ~p)

2

T

T

F

F

T

T

T

F

F

T

F

F

F

T

T

F

T

T

F

F

T

T

T

T

q

p∨q

~p

T

T

T

F

T

F

T

F

F

T

T

T

F

F

F

T

Conclusion is true whenever all premises are true (3rd row), thus a valid argument. 11 a p q r p⇒q q⇒r p⇒r

p

q

~p

~q

(p ⇒ q)

(~p ⇒ ~q)

T

T

T

T

F

F

T

T

T

T

F

F

T

F

T

T

F

T

T

F

T

F

T

F

T

F

T

F

T

F F

F

F

T

T

T

3 If it is bread then it is made with flour; If it is made with flour then it is bread; If it is not made with flour then it is not bread; If it is not bread then it is not made with flour. 4 C 5 A 6 a Conclusion: My pet is fluffy. b Conclusion: Two is the only even prime number. c Conclusion: Growing apples depends on good irrigation. 7 a and c are valid. 8 b All footballers are fit. David is not fit. David is not a footballer. d Cannot be made into a valid argument. e All musicians can read music. Louise is a musician. Louise can read music. *11 c

9 D 10 p

T

T

F

T

F

F

F

T

F

T

T

F

F

T

F

T

T

T

T

F

T

F

T

F

T

T

T

T

F

F

T

T

T

T

T

T

T

F

F

T

F

F

T

F

F

T

T

F

T

F

F

T

T

T

Conclusion is true whenever all premises are true (4th row), thus a valid argument. c See table at foot of page*.

p

q

r

s

p⇒q

r⇒s

p⇒q∧r⇒s

p∧r

q∧s

T

T

T

T

T

T

T

T

T

T

T

T

F

T

F

F

T

F

T

T

F

T

T

T

T

F

T

T

T

F

F

T

T

T

F

F

T

F

T

T

F

T

F

T

F

T

F

T

F

F

F

F

T

F

T

F

F

T

F

T

F

F

F

T

F

F

F

F

T

F

F

F

F

T

T

T

T

T

T

F

T

F

T

T

F

T

F

F

F

F

F

T

F

T

T

T

T

F

T

F

T

F

F

T

T

T

F

F

F

F

T

T

T

T

T

F

F

F

F

T

F

T

F

F

F

F

F

F

F

T

T

T

T

F

F

F

F

F

F

T

T

T

F

F

Answers

T

Conclusion is true whenever all premises are true (1st, 5th, 7th and 8th rows). b p q p⇒q ~q ~p

Conclusion is true whenever all premises are true (1st row).

640

T

17 a b c d e

12 a Disjunctive syllogism b Modus tollens c Modus ponens 13 C 14 a b p⇒q p q p ⇒ q ~p ~q ∼p T T T F F ∼q T F F F T F

T

T

T

F

F

F

T

T

T

Valid — hypothetical syllogism p ⇒ q; r ⇒ q; p ⇒ r, invalid Valid – modus tollens Valid – constructive dilemma p = The team plays well q = The offence was good r = The defence was good p ⇒ (q ∨ r) ∼p ∧ ∼p ∼r

p = If elected with a majority q = My government will introduce new tax laws Conclusion is false whenever all premises are true (3rd and 4th rows), thus an invalid argument. 15 a p ⇒ q b p q ~p ~q p⇒q ∼p T T F F T ∼q T F F T F F

T

T

F

T

F

F

T

T

T

onclusion false when premises true C (3rd row), therefore invalid. 16 a

p

q

r

p⇒q

r ⇒ ~q

p ⇒ ~r

T

T

T

T

F

F

T

T

F

T

T

T

T

F

T

F

T

F

T

F

F

F

T

T

F

T

T

T

F

T

F

T

F

T

T

T

F

F

T

T

T

F

F

F

T

T

Valid argument b p q r

c

F

T

T

T

F

F

T

T

F

T

F

T

T

F

F

F

T

F

T

T

F

F

F

T

F

F

F

F

T

F

F

F

Invalid argument

T

T

T

T

F

F

T

T

F

T

T

F

T

T

F

T

T

T

F

F

T

F

F

F

F

F

T

F

T

T

T

T

F

F

F

T

F

T

T

F

T

F

F

T

T

T

T

F

F

F

F

F

T

T

T

Invalid argument

Exercise 9C — Techniques of proof 1 p = She plays well q = She wins p⇒q

(p ⇒ q) ∨ ~q

T

T

T

T

T

T

F

F

T

T

F

T

T

T

F

F

T

T

Therefore, this is a tautology. 2

p

q

p⇒q

T

T

T

T

T

T

F

F

T

T

T

F

T

T

F

T

T

F

F

F

T

T

T

T

T

~p ⇒ ~q (p ⇒ q) ∨ ~p ⇒ ~q

Therefore, this is a tautology. 3

p

q

p∧q

(p ∧ q) ∨ ~q

T

T

T

T

T

F

F

T

F

F

T

F

F

T

F

F

F

T

p

q

~p ⇒ ~q

T

T

T

T

F

T

F

T

F

F

Invalid argument

T

~r

Answers 9b ➜ 9c

T

r

q

r⇒p

T

q

p

~p ∧ ~q

T

(q ∨ r) p ⇒ (q ∨ r) ~q ∧ ~p

p

Therefore, this is not a tautology.

Answers

641

4

p

q

r

p⇒q

q⇒r

(p ⇒ q) ∧ (q ⇒ r) p ⇒ r

(p ⇒ q) ∧ (q ⇒ r) → (p ⇒ r)

T

T

T

T

T

T

T

T

T

T

F

T

F

F

F

T

T

F

T

F

T

F

T

T

T

F

F

F

T

F

F

T

F

T

T

T

T

T

T

T

F

T

F

T

F

F

T

T

F

F

T

T

T

T

T

T

F

F

F

T

T

T

T

T

The last column is always true. Thus, this is a tautology and the argument is valid. 5 a

b

c

d

642

p

q

r

~p ⇒ ~q

~q ⇒ r

(~p ⇒ ~q) ∧ (~q ⇒ r)

~p ⇒ r

(~p ⇒ ~q) ∧ (~q ⇒ r) → (~p ⇒ r)

T

T

T

T

T

T

T

T

T

T

F

T

T

T

T

T

T

F

T

T

T

T

T

T

T

F

F

T

F

F

T

T

F

T

T

F

T

F

T

T

F

T

F

F

T

F

F

T

F

F

T

T

T

T

T

T

F

F

F

T

F

F

F

T

Hence, this is a valid argument. p

q

r

p ⇒ ~q

q ⇒ ~r

(~p ⇒ ~q) ∧ (q ⇒ ~r)

p ⇒ ~r

(~p ⇒ ~q) ∧ (q ⇒ ~r) → (p ⇒ ~r)

T

T

T

F

F

F

F

T

T

T

F

F

T

F

T

T

T

F

T

T

T

T

F

F

T

F

F

T

T

T

T

T

F

T

T

T

F

F

T

T

F

T

F

T

T

T

T

T

F

F

T

T

T

T

T

T

F

F

F

T

T

T

T

T

Hence, this is an invalid argument (row 3). q T

~p ⇒ ~q T

(~p ⇒ ~q) ∧ q T

(~p ⇒ ~q) ∧ q ⇒ p T

T

F

T

F

T

F

T

F

F

T

F

F

T

F

T

p T

Hence, this is a valid argument. q T

~p ⇒ q T

(~p ⇒ q) ∧ p T

(~p ⇒ q) ∧ p ⇒ ~q F

T

F

T

T

T

F

T

T

F

T

F

F

F

F

T

p T

Hence, this is an invalid argument (row 1).

Answers

A

B

C

A

B

C

2

A

A

B

B

C

C

3 B 4 a A = {2, 4, 6, 8, 10, 12, 14, 16, 18} b B = {4, 8, 12, 16, . . .} c C = {2} d D = {Jack, Queen, King} e E = ∅ f F = {9, 8, 7, 6, …} 5 A, C, D, E 6 ε A

25 49 81

B

7 ε A

22 44 66 88

B

8 a ε

b ε A

c

B

A

A

B

B

C

9 a {4} b {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} c {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} d {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} e {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 10 a (B ∪ C) = {1, 2, 4, 6, 8, 9, 10} b A ∩ (B ∪ C) = {1, 2, 4, 6, 8, 9, 10} c (A ∩ C) = {1, 4, 9} d (A ∩ B) = {2, 4, 6, 8, 10} e (A ∩ B) ∪ (A ∩ C) = {1, 2, 4, 6, 8, 9, 10} 11 Part 1: Show that (A • B) + (A′ + B′) = I (A • B) + (A′ + B′) = (A + A′ + B′) • (B + A′ + B′) = (I + B′) • (A′ + I ) = (I ) • (I ) =I

QED art 2: Show that (A • B) • (A′ + B′) = O P (A • B) • (A′ + B′) = A • B • A′ + A • B • B′ =O•B+A•O =O+O = O QED 12 a A + B b I c A + B d A • B 13 Answers will vary. 14 (p ∧ q) ∧ ~p = (p • q) • p′ = (p • p′) • q = O • q = O QED p ∧ ~p = p • p′ = O

Answers

Answers 9D ➜ 9D

6 a p ⇒ ~q b ~p ⇒ q ~q ⇒ r p p ⇒ r ~q c p ⇒ ~q d ~p ⇒ ~q q ~q ⇒ ~r ~p ~p ⇒ ~r 7 D 8 If x2 is even, then write it as 2n. (2n)2 = 4n2 If a number is multiplied by 4 then it is even. Therefore 4n2 is even. 9 If a number, x, is even, then x2 is even. 24 is even, therefore 242 is even. 10 Assume n is not odd, show that n2 is even. 11 a Assume a ≠ b, multiply both sides by x. b Assume n < 2, write it as 2 - x and square it. c Assume n is not divisible by 2, therefore it is odd, therefore write it as (2x + 1), then square it. 12 Assume it is positive and compare it with the product of 2 positive numbers of the same magnitude. 13 Assume that a is the smallest positive real number. a Let x = 2 Since a > 0, then x > 0 and x < a (property of division). This contradicts the assumption that a is the smallest positive number. a 14 Assume 2 is rational, so that 2 = , where a and b b are integers that have no common factors. a2 Therefore 2 = 2 , or a2 = 2b2. b Therefore a2 is a multiple of 2 and therefore a is a multiple of 2 (from a2 = a × a) Since it is a multiple of 2, write a = 2x. Therefore a2 = 4x2 = 2b2. Therefore b2 = 2x2 and is thus a multiple of 2. Therefore both a and b are multiples of 2 and have a common factor of 2. This contradicts our initial statement, so it must be false. 15 Assume n is the largest possible integer. Let x = n + 1. Therefore x > n, which contradicts our initial statement. 16 Let x = 2, then x2 = 4, let x = −2 , then x2 = 4. 17 Demonstrate that 2 is both a prime number and is even. 18 Prove by counter-example. 19 Consider the contrapositive statement. 20 Consider what would happen if x > y + z. This would imply that the shortest distance from A to C is not a straight line! 21 Answers will vary. 22 a Mathematical induction is used particularly for sums of series. b The values of n are restricted to integer values. c Answers will vary. 23 Answers will vary. Exercise 9D — Sets and Boolean algebra 1

643

15 a A + B + A′ + B′ = A + A′ + B + B′ = I + I = I b (A + B) • A′ • B′ = A • A′ • B′ + B • A′ • B′ = OB′ + OA′ = O + O = O c (A + B) • (A + B′) = (A + B) • A + (A + B) • B′ =A+A=A d A • B + C • (A′ + B′) = A • B + C • (A • B)′ =A•B+C

8

Exercise 9E — Digital logic 1 x y Q

c

Output

0

0

0

1

0

0

1

1

0

1

0

1

0

1

1

1

1

0

0

1

1

0

1

0

1

0

0

1

1

0

0

0

0

0

1

0

1

1

0

0

1

1

1

x

4 a

y

9 E 10 a

z

Q

0

0

0

0

0

0

1

1

0

1

0

0

0

1

1

1

1

0

0

0

1

0

1

1

1

1

0

1

1

1

1

1

x

y

Q

0

0

1

0

1

0

1

0

0

1

1

1

b Q = (x y) + z •

x

Q

a

b

c

Output

0

0

0

0

0

0

1

1

0

1

0

1

0

1

1

1

1

0

0

1

0

1

1

1

1

1

Answers

b

c

Output

0

0

0

1

0

0

1

1

0

1

0

1

0

1

1

1

1

0

0

1

1

0

1

0

1

1

0

0

1

1

1

0

0

y

P

a

b Same truth table as question 8. 11 a b c d Output

b Used where there are 2 people who can activate the light separately. 5 D 6

644

b

1

2 C 3 a

7

a

12 a

0

0

0

0

0

0

0

1

0

0

0

1

0

1

0

0

1

1

0

0

1

0

0

0

0

1

0

1

0

0

1

1

0

1

0

1

1

1

0

1

0

0

0

1

1

0

0

1

1

1

0

1

0

1

1

0

1

1

1

1

1

0

0

0

1

1

0

1

0

1

1

1

0

1

1

1

1

1

0

S1

S2

a

b

c

Q

R

1

0

0

0

0

1

0

0

1

0

1

0

1

0

0

0

0

1

1

0

1

0

1

0

1

1

1

1

1

1

1

0

1

0

b When S1 = 0, the system is disabled; the safe can’t be opened. When S1 = 1 and S2 = 0, the alarm rings. When S1 = 1 and S2 = 1, the safe can be opened. 13 a a • b = [(a • b)′]′ a • b = (a′ + b′)′ b

14

a

15

a b c

16

Q

c

17

b a

Q

a' c

18 a iii a′ + b b a

Q

b

c

b

Q

a

d a • b e (a • b)′ + b f a Q

b

a

b

(a • b)′

(a • b)′ + b

0

0

1

1

0

1

1

1

1

0

1

1

1

1

0

1

I

g

~r ⇒ I

~I ⇒ g

r

~I

T

T

T

T

T

T

F

* *

T

T

F

T

T

T

F

T

F

T

T

T

T

T

T

F

F

T

F

T

T

F

T

T

T

T

F

F

F

T

F

T

T

F

F

F

F

T

F

T

F

T

F

F

F

F

F

F

T

b a

Q

10 Q = A • [(B + C)′ + B • C]


b c

b c ∨ (w ∧ s)

c l ∧ e ∧ p

(p ∧ q) (~p ∧ ~q) (p ∧ q) ∨ (~p ∧ ~q)

T

T

T

F

T

T

F

F

F

F

F

T

F

F

F

F

F

F

T

T

3 Converse = If she sends her children to good schools, the politician is intelligent. Contrapositive = If she doesn’t send her children to good schools, the politician is not intelligent. Inverse = If a politician isn’t intelligent she doesn’t send her children to good schools.

a

Q

Multiple choice 1 A 2 E 3 C 4 D 5 A 6 D 7 D 8 B 9 C 10 C 11 A 12 D 13 B Extended response 1 a S1 = (g ⇒ n) ∧ (a ⇒ c) S2 = ~n ∨ ~c S3 = ~g ∨ ~a Premises (S1, S2) are all true in rows 4, 12, 13, 15, 16 Conclusion (S3) is true in rows 3, 4, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 Conclusion is true when premises are true, so this is a valid argument.

Answers

Answers 9E ➜ 9E

1 a m ∧ j 2 p q

r

The conclusion (column 7) is false when the premises (columns 4, 5, 6) are all true (*). Invalid argument 1 5 Let: a = ; multiply both sides by x. x ax = 1 a(x1) = 1 = x0 (Definition of x0) x0 a = 1 = x -1 (Definition of division of indices) x 6 Assume x < x x < x2 (Square both sides) 1 < x (Divide both sides by x) This contradicts the initial condition, so the assumption is wrong. 7 a {11, 13, 17, 19, 31, 41, 61, 71, 91} b {2, 3, 5, 7, 11, 13, 16, 17, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97} c {61} 8 (A + A′ • B) • (B + B • C) = (A + A′ • B) • (B • I + B • C) {B • I = B} = (A + A′ • B) • B(I + C) {Distributive Law} {I + C = I} = (A + A′ • B) • B • I = A • B • I + A′ • B • B • I {Distributive Law} {B • I = B, B • B = B} = A • B + A′ • B {Distributive Law} = (A + A′) • B {A + A′ = I} = I • B = B QED 9 c

Q

a b

4

645

b, c

g

n

a

c

g⇒n

a⇒c

S1

S2 = ~n ∨ ~c

S3 = ~g ∨ ~a

T

T

T

T

T

T

T

F

F

T

T

T

F

T

F

F

T

F

T

T

F

T

T

T

T

F

T

T

T

F

F

T

T

T

T

T

T

F

T

T

F

T

F

T

F

T

F

T

F

F

F

F

T

F

T

F

F

T

F

T

F

T

T

T

F

F

F

F

T

F

T

T

F

T

T

T

T

T

T

F

T

F

T

T

F

T

F

F

T

T

F

T

F

T

T

T

T

F

T

F

T

F

F

T

T

T

T

T

F

F

T

T

T

T

T

T

T

F

F

T

F

T

F

F

T

T

F

F

F

T

T

T

T

T

T

F

F

F

F

T

T

T

T

T

S3 = ~g ∨ ~a Premises S1 and S2 are all true in rows 4, 12, 13, 15 and 16. Conclusion is true when premises are true. Therefore a valid argument. d

a

b

c

d

a⇒b

b⇒c

~d ⇒ a

~c

T

T

T

T

T

T

T

F

T

T

T

F

T

T

T

F

T

T

F

T

T

F

T

T

T

T

F

F

T

F

T

T

T

F

T

T

F

T

T

F

T

F

T

F

F

T

T

F

T

F

F

T

F

T

T

T

T

F

F

F

F

T

T

T

F

T

T

T

T

T

T

F

F

T

T

F

T

T

F

F

F

T

F

T

T

F

T

T

F

T

F

F

T

F

F

T

F

F

T

T

T

T

T

F

F

F

T

F

T

T

F

F

F

F

F

T

T

T

T

T

F

F

F

F

T

T

F

T

Argument is valid because conclusion is true when all premises are true (row 15). e Answers will vary.

646

Answers

2 a

b c d e 3 a b c

CHAPTER 10

a

b

a⇒b

a⇔b

T

T

T

T


T

F

F

F

F

T

T

F

Exercise 10A — The circle b 1 a y

F

F

T

T

a ⇒ b and b ⇒ a Yes Yes, because a ⇔ b implies both a ⇒ b and b ⇒ a. Equivalences: i, iv, vi, viii 4 inputs; w, x, y, z 2 outputs; Q, R w

x

y

Z

Q

R

0

0

0

0

0

0

0

0

0

1

0

1

0

0

1

0

0

0

0

0

1

1

0

1

0

1

0

0

0

0

0

1

0

1

0

1

0

1

1

0

1

1

0

1

1

1

1

1

1

0

0

0

0

0

1

0

0

1

0

1

1

0

1

0

0

1

1

0

1

1

0

1

1

1

0

0

0

0

1

1

0

1

0

1

1

1

1

0

1

1

1

1

1

1

1

1

d i Q = 0 and R = 1 ii Q = 0 and R = 0 e No, input z is independent of w, x, y. f i Q = (w • x + x • y) • (w • y + x • y) ii Q = x • y g i R = w • y + x • y + z ii R = w • x • y + z h x

–7

c

–4

Centre (0, 0); radius 7 y 6

d

6 x

–6

Centre (0, 0); radius 4 y 9 9 x

–9

–6

e

–9

Centre (0, 0); radius 6 y 5

f

y 1 3

5 x

–5

Centre (0, 0); radius 5 2 a

Centre (0, 0); radius

5 (1, 2)

d

y 8 1 –3

4

x

12 x

10

–5

15 x

5

–10

–5

b

a

Q

R

S

T

3 a (x + 2)2 + (y + 4)2 = 22

0

0

1

0

0

0

y

0

1

0

1

0

0

–4 –2 –2

1

0

0

0

1

0

–4

1

1

0

0

0

1

b (x − 5)2 + (y − 1)2 = 42 y 5 1

x

–3 1

–6

c (x − 7)2 + (y + 3)2 = 72

–10

8

y 10

2 x

2

4

–13

f

–1 –3

y –4 –5

y

–3

4 x

3

7

y 4

–2 6 –3

–8

–9

c

R

10 3

y 3

6 x

–3

–2

3

1 –3 – 3

b

y 7

e

1 3– x

1 3

–3 –

–5

–4

Centre (0, 0); radius 9 3–

–6

e This circuit ‘decodes’ or distinguishes the inputs, which are 00, 01, 10 and 11. Only one of the outputs = 1 for each of the 4 possible inputs. f 3-bit: 3 NOT, 8 AND 4-bit: 4 NOT and 16 AND n-bit: n NOT and 2n AND

4 x

–4

5

9

x

d (x + 4)2 + (y − 6)2 = 82 y 14

7 14

x

6 12

–4 –2 4

Answers 10A ➜ 10A

4 a –d

7 x

–7

–10

Q

y w z

y 4

7

x

Answers

647

f (x − 1)2 + (y + 2)2 = 32

e x2 + (y − 9)2 = 102 y 19

1 –2

9 –10 –1

b

y 1

c

4 D 5 B 6 (x – 5)2 + (y – 3)2 = 16

25 21 – — 21

y 6

36— 35 x = 175 – 35

f –3

–6 – 32 7 — 7

x=

x=

d

y 9

–121 — 10 20

F F' –11 – 2 10 2 10 11

x=

x =1212010

e

x=

y 1

– 4—3 3

x= 2 a

F 3 2

x

3

4—3 3

(2 2, –2)

x = 9—42 4 C

b

y F (1, 0) x

–1

c

e

F' –4

1

F

x

4

(1 + 5, –2)

x=

x = –4

F (5, 0) x

5+9 5 — 5

F (6, 0) x

–6

f

y

1

F 1 (12– , 0) x

1 x = –2 2–

h

y 1 –24– 1 x = –2 –4

y 1 –22–

x = –1 2–

g

y

x = –6

1 –12–

5+9 5 — 5

F (4, 0) x

–4

x = –5

x

y

d

y

–5

= 4—33

y

(1 – 5, –2)

Answers

x

–1

–2

648

x

x = –1

F' –2 – 3

5–9 5 — 5

x

1 a

15 x = 16— 15

y 1

= 5 – 59—5

–3

3 E y2 5 x 2 + 2 = 1 1.4

–1

–16 15 — 15

–4 3 —

– = 9—42

3 F

Exercise 10C — The parabola

16—15 15

F 15 4

F' –4 – 15

f

x

–2

–9

–121 10 20

– 16—15 15

x

121 — 10 20

x = 175 +3536— 35

–1

F'

(–2 2, –2)

32— 7 7

x

y

x

8

11

F' 5 – 35 F –1 5 –1 5 + 35

32— 7 7

F 2 7

F' –8 – 2 7

y

x

5

x

–1 x = 1612

1

21 x = 25— 21

c – 32— 7 7

–1 x = –1212

25—21 21

F 21

– 25 21 — 21

x=

F (14, 3) –11–10 14 15 –2 2

e

y 2 F' –5 – 21 –2

1

– 1612

F' –1 (–10, 3) 3 –1212

x

51 x = 100— 51

b

y 8

F 51 10

–100 51 — 51

5

4

d

7

x=

y

–15

–5 – 2 6 –5 + 2 6 x F' –5 F 2 (–5 – 2 6, –1) (–5 + 2 6, –1) –6 x = –15 x=5

y F' –10 – 51 –7

x — x = 10 3

— 3

5

Exercise 10B — The ellipse 1 a

– 40

x=

–5

x

1

3 –3

F' F 2 (–8, 2) (–2, 2) –10 –5 –2

4 x

–2 4 10

y 6

1

–13 –3

F 1 (24– , 0) x

F 1 (22– , 0) x

y 3 –14– 3 x = –1 –4

F 3 (14– , 0) x

2 a

b

y

F 1 (2, 0) x

F 1 3 (5, 0) x x=1

c

d

y F (2, 0) x

F (3, 1)

h

y –1 –6

4

x = –6

i

j

y 2 –1

–11– 2

–5

3 a

b

F (–1, 0) y F 1 (–22–, 0)

e

1 x = 2 –2

F (–4, 0) –1

g

y

h

1

k

–1 1 F (–3–2, –2)

1

e x

1 –2 –2

x

F

1 (–4 –2, –5) –5

13

–

12

5

5

–x y =12

F 13

–5

x

–5 –x y = 12

2 y x= 2

y=x

d

y

2

x = –6 –5

– x = 3—2 y x = 3—2 y = x 2 2

3 2F 2 2 2

3F 3 2

F'–3

x

–3 2

y = –x

x = 6 2– y = 3– x 5 4

1

x = –2

F'–8 2 –10 –6 –5

x

–3

f

– x = 9—13

y

13

6

y

1

— x = 144

y = –x 2

5 –2

x –2 1 x = 5 –2

x

–5 –1 x = –1

l

y

x=

y F (–9, 2)

x

– y — x = 144

2 F'–2 – 2 –2 2 –2

x = 4 –2

1

–3 x = –1

1

4 –2

j

y

x

x = 9—13 y = 2– x 13

3

2 8F 6 2– 10 5

x

–6

F

F' – 13 –3

3 –2

– y = 3– x 4

13 x

Answers 10B ➜ 10D

–1

c

y

x = 5 –2 F (–5, 1)

2 x

–3

4 1

F 4 5

–12 F' –13

y

1 F (3–2 , 0)

2 5 –2 x

3

y = –4 x

3.2

13

1 x = 1 –2

x=2

F 1 (–12– , 0)

i

b

x=2

y

2 x

x = –3.2 x = 3.2 1

F (–8, 0)

x

2

y

–

3 y = –4 x

1 –2 x

f

y

F (–2, 3) 3

x

3 F' –5–4 –3

2 x

F 1 (–12–, 0)

1

2 –2 x

h (y − 3)2 = −8x

Exercise 10D — The hyperbola

x=2

d

x

y

–3.2

F (–2, 0)

x

1

1y –22– F –4 –1

1 4 x

2 –1 1

1 a

y

x=1

c

x F 1– (2 2, –4)

x = –2 –2

y

1

F (1, 2)

1

x

f y2 = −6(x + 2 2 )

y

–4

F 6

–2 2

g (y − 2)2 = 4x

y

–21–

x

(3, 7) x

2

x F 1– (12 , –5)

1 x = –1–2

F

x

d y2 = 16(x − 2)

e y2 = −12(x − 1)

F

x = –3

l

y

F 3 4

–2

7

2

y

–3

x

1

x = –1

k

x F –6 (–1, –6)

F 2

–3 – 1–

y 2

y

(1, 2)

x

y

x

–3

x = –5

F

(8, 3)

y –5

F x (4, –3)

–3

F 2

c y2 = 4(x − 3)

2 58 x=2

x=1

g

F

y

–2

y 3

x

1

b y2 = 10(x + 2 )

y F (0, 0) x

–8 –4

f

1

y

x = –8

y

5 C b iv d i

7 a y2 = 8x

x=0

–4 –1 x = –4

e

4 E 6 a ii c iii

y

– y = 2– x 3

Answers

649

2 a

y

Exercise 10E — Polar coordinates 1 2 y

3x – 4y – 3 = 0

3 –4

6

1

3

x

–11 — x = 21 x= — 5 5

b

3x + 4y – 3 = 0

3

y 5x – 12y + 15 = 0

4

300°

y

x–y–2=0

–5

2

x+y+2=0

2

d

y

x–y+3=0

5

(– 2 2, 3)

( 2 2, 3)

3 1 –2

x

2

x+y–3=0 x= 2

x=– 2

e

y

3x – 4y + 11 = 0

8 (–11, 2)

– 37 x= — 5

f

–1 –4

x

7

3x + 4y – 5 = 0

— x = 27 5

y

– 9 13 + 5 x = —––

13

2 –1

5

2x – 3y – 19 = 0 8 x

–3 (– 13 + 5, –3) –5

( 13 + 5, –3)

9 13 + 5 x = —–– 13

2x + 3y – 1 = 0

± 9 13 ± 13 ii ( 13 , 0) iii x = 3 13

iv y =

±2

3

x

b i

± 25 26 26 ± ii ( 26 , 0) iii x = iv y = 5 26

±

1 x 5

c i

± 25 29 29 ± ii ( 29 , 0) iii x = iv y = 5 29

±

2 x 5

4 D x2 y2 7 = 1 9 9

Answers

5 B

(b), (d)

 5 3 1 , 2 2 b   2 d (0, 2.7)

3 3 ) 4

f ( −6 3 , -6)

g (0, 2.6)

h (−7.8, 0)

i ( 5, 5 3 )

j (4.55,

k (22.14, −11.62)

l (−1.05, −1.21)

-

91 3 20 )

5 a [5, 0°] d [9, 270°]

b [4.3, 90°] e [6, 2 , 45°]

c f

g [13, 4.32c] 5π ] j [2 2 , 4

h [10, 5.36c] 11π k [4, ] 6

i l

[30, 180°] [5, 2.5c] 2π [2, ] 3 [7.81, 0.88c]

Exercise 10F — Polar equations 3 2 1 a r = b r = cos (q ) sin (q ) c r = 3 d r = 6 π f q = e θ = 78°41′ 4 1 7 g r = h r = 3 cos (q ) - 4 sin (q ) 5cos (q ) + sin (q ) i r = 10 cos (θ) − 6 sin (θ) j r = −6 cos (θ) − 8 sin (θ) l r = 2 cos (θ) k r = 12 sin (θ)

(9, 2)

2 –9

x

(a)

3 3 ) 2

e (−0.75,

– x = 3—–2 x = 3—–2

1

π — 3 π — 3

–30° x 1 2.5 (d) (c)

c (1.5,

x (3 2, –2)

–2

(– 3 2, –2)

650

3

4 a ( 2 , 2 )

1

3 a i

5π — 3

–— x = 105 13

c

(a) 1

45°

x

10

5x + 12y + 15 = 0

– –— x = 183 13

(a) 2 x

100°

–3 –5

1

y π 2— 3

(b) y

5 –16

(d)

(b) –5 –0.5 (c)

–3

(c)

6 C x2 y2 8 9x2 − 16y2 = 225 9 =1 400 500

m r 2 =

36 4 cos 2 (q ) + 9 sin2 (q )

n r 2 =

100 25 cos2 (q ) + 4 sin2 (q )

2

a c e g i k

x2 + y2 = 4 x2 + (y − 3)2 = 9 y = 3x y = x x = 4 x2 + y2 + 2x − 4y = 0

m y =

9 - x2 6

x2 + y2 = 25 (x − 1)2 + y2 = 1 y = −4x y = −x y = −1 x2 + y2 − 8x − 6y = 0 y 2 - 16 n x = 8

b d f h j l

π 5— π 2 — 2

π 0

3π

c

2π

π 7— 2

4π

y 20 10 –10 0 –20 –10 5 20 –20

2 a

d

0 –5

–5

2 x

d

2 x

b i 0, 5 ii 0, 12 iv 6.5 v (6, 2.5) vi (x − 6)2 + (y − 2.5)2 = 42.25 7

y

–1

1 x

y

c

1 x

b

–1 y

d

0

–2

–2

0

–1

–1

0 –1

–1

iii 13

y 1 0

x

c 1

0

2 x

1

–1

y –1

y

x

5

–1

5

2 x

a

y 1

1 0

x

10

A circle

–2

b

y 1

y 10

0

0

–1.5

iii 5

4 x

y 2

–2

1.5 x

4 x

2

b i 0, 3 ii 0, 4 iv 2.5 v (2, 1.5) vi (x − 2)2 + (y − 1.5)2 = 6.25 6 a

0

–4

1.5

3 a

0

–4

y

0

5x

y

–2

–1.5

x

A circle

4

0

c

5 10

5 x

y 2

y

b

y 2

–2

x

5 a

y 10 5 0 –10 –5 –5

0

Exercise 10G — Polar graphs 1 a b y

5

–5

–2

x

–2 0

–4

e

8

x

1

x

0

0

–2

b 4 a

b

y

y 1

1 1 x

d

y 1 –1 0 –1

3x

1 2

–4

–2

1

2

0 –1

0

y 0.5 –0.5

0

y 8 4 –2 0

x

e

y

y

–2

e

c

–1 1

d

x

–1 0 1

0 2 3 x

1

–1

c

x

3

x

x

x

y –1

1

2

0 1

x

–1 –3

f

y 2

x

–2 0 x –2

Answers 10E ➜ 10G

0

–1

–1 0 –1

2

y 7 5

–5

y 4 2

3

y

a

y

–0.5

Answers

651

y

9

y 1

a

1 0

–1

x

1

0

–1

1

x

1x

0

–1

2 x

y 2

0

b=2

c Increasing the a value increases the size of the resulting cardioid in all directions.

π

0

2 x

0

–2

–2

d The graph centres a spiral about the centre from the chosen a value. Horizontal asymptote at a value.

y 4

f

π

y 1

e

0

–2

x

–1

–1 y

d

1

–1

c

y 1

b

0

–1

–1

a

4 x

0

–4

π

0

–4

10

r = 2 − 4 sin (θ) y

r = −2 − 4 sin (θ) y

0

0 x

x

–2

–2

–6

–6

he graphs are identical. T 11 a If a = b cardioid results. If a < b, the limacon contains an inner loop. If a > b, there is no inner loop (a dimple results).

π

π

π

f The value of a is the length of one petal (lie on the horizontal axis). a

π

0 a

0

a

0 a=b

e The loop passes through the horizontal axis at the chosen a value.

0

g The a value helps to determine the position of the horizontal asymptote. 4a

π

0 2a

a>b

b a is the radius of each petal b is the number of petals if b is odd. (If b is even, then there will be 2b petals) b=3

π

0

π

0

Exercise 10H — Review of complex numbers and polar form of complex numbers 1 a 9 + 3i b 4 - 6i 1

c 1 + 3 i -5

e + 13i g 8 + 6i i 10

652

Answers

d 8 - i f 9 - 7i h 3 - i j 13

2 a 2 2

c

d

 -π  2cis   4 

3 a

-3cis

c

-

(p)

d

3 3 3 + i 2 2

d f

e

12 3 i 2

-

g

Short answer 1 a

15 i 5

3 3 + i 2 2

_3 _2 _1 _1 _2 _3

c

3

y 3 2 1 _3 _2 __1 0 1 _2 _3

5

_3 _2 __1 0 1 _2 _3

7 y = –1x 2

y –3 –6

2 x

–2

f

y F 2

–2

y

x

–2 1 5

x = –2

g

x

–1

–1

e

1

–3

F (9, –2)x

x=1

2 x = –6 –5

y = 3x 2 y=x

y

x = 6 2– 5

y = 3– x 4

6 8F 10

F'–8 –10

1 2 3 x

x

–6 – y = –3 x 4

h y = 2x _ 4 + _x1

4

y 3 2 1

y = 2x _ 4 y = _x1 x 1 2 3

y = 2x _ 3

_3 _2 __1 0 1 _2 _3

6

y = 2x

8

y 16 2 x = 20 –—

x = 16 2—+ 20 5 y = 3– x – 2

5

y = 3x2

4

1 y=— 2

F'

3x

1 2 3 x

F

1

4–5 2

5 2+4x

4

–

3 y = –4 x + 4

2 a i 10

y 3 2 1

— y = 2x _3 _2 __1 0 1 _2 _3

1 2 3 x

1 2 3 x

_

_1

y = 2x 1_ 3

y 3 2 1

_3 _2 __1 0 1 y = –x1 _2 _3

d

y

1 2 3 x

y = (x2 )2 y 3 y = x2 2 1 _3 _2 __1 0 1 2 3 x _12

iv

3 5

b

ii 8

iii (±5, 0), (0, ±4)

v (±3, 0)

vi x =

± 25 3

y 4 F' — –5 –3 – 25 3 –4

F 3 5

3 a i (−1, 0) b y –1 F –2

x 25 — 3

ii (0, 0)

iii x = −2

x

Answers

Answers 10h ➜ 10I

y 3 2 1

6 x

2

1

_3 _2 __1 0 1 _2 _3

2 3

x

–3

3 2 1

y = –x1

–2 –1

–5

y = 2x

1

3

3-i

Exercise 10I — Addition of ordinates, reciprocals and squares of simple graphs 2 1 y = 2x + –x1 y y = x2+3x

1

y 3

–3

h 1 + i

3 2

b

y 3

1 3 d + i 2 2 1 f 2

-

1 2 3 x

Chapter review

b 4 2 (1 + i)

5 a 6i

-

π cis    4 2

b 3

y = x2

_3 _2 __1 0 1 _2

1

2 (1 + i)

y 3 2 1

y = _x

 -π  f 2cis   3 

e i

c

2 3

π b 2cis    2

π e 4 cis    6 4 a

9

b 10

c 3 5

653

13 12

4 a i

ii (±12, 0)

iv x =

b

± 144

v y =

13

x = 144 — 13

y

−144

x= —

13

–12 F' –13

12

5

±5 12

11

x

-

12 a

−5

b

y

(b)

y 4 3 2 1

y= 4_x

_3 _2 __1 0 1 _2 _3

x

–x y = 12

5

3+i

–5 x y = 12

F 13

–5

10 1 - i

iii (±13, 0)

y 6 5 4 3 2 1

y =4 _ x 3

2

1 (e)

(c)

(a) x

1 5

__1 0 1 (d)

b y = −x

c x = 6

d y =

1 C 6 D 11 C 16 D

 π b  3,   2 d [ 5, 3.61c]

8 a r = 5

9 - x2 6 b r cos (θ) = 2

c θ = 1.11°

d r =

9 a

b

y

x

d

2 x

0.5 x

y 5

–5

h

y 5

654

Answers

5 x

–5

x

–2

–6 –4 –2 –5

2 4 6

4 A 9 A 14 E

c a = b

d e =

5 C 10 B 15 B

2

x

y 2 2 –2

2a x

b Equation is

( x - a - 1)2 y 2 + 2 =1 (a - 1)2 b

b2 (a - 1)2 4 a PD = 37.5 metres b 20 metres c 7.5 metres 5 a 50 metres b 180 metres c 144 metres x2 y2 + =1 d 32 400 20 736 e Less than 0.6 since the outer ellipse becomes more circular (less eccentric) x2 y2 f e = 0.537 g + =1 52 900 37 636 c e = 1 -

–0.5

2

g

B E C C

_b

–0.5

f

2

3 8 13 18

2 a Gradients are 1 and −1 b y = x and y = −x

2

y 0.5

y

–2

D D D A

b

–4

e

2 7 12 17

x2 y2 e Equation: =1 2 2 3 a y

y

x

–2

1 2 3 4 5 x

 - 11 - 5 3  B ,  2   8

π — 6

y 4 2 –6 –4 –2

2

y = (4 _ x)

Extended response  - 11 5 3  1 Points of intersection are: A  ,  2   8

5 2 cos (q ) - 3 sin (q )

3

c

1 2 3 4 5 x

Multiple choice

π  6 a  2 ,  4  π   c  2, 6   e [13, 1.97c] 7 a x2 + y2 = 4

3

1 y = —–– 4_x

4

6 x

6 a The graph is confined to the first 2 quadrants. b Various answers.

c The graph consists of an n-sided figure where π θ step = and n > 2. n d θ step is infinitely small (θ step → 0). 7

( x - 2)2 16

y2

+

k

Region required

2 a

8 3+6 6-8 3 x= 3 3

4

x

6

–2

c

x = 8 3—+ 6

8 3 x = 6 –—

3

3

b

y

–x < 7

x

0

Region required y

0

x

0

–3

1 a

b

y

e

f

y

y –y ≤ 3

4

y

x

0

x

0

x

0

Region required

y≤0

y≥2

g

Region required

h

y

y

Region required

Region required

d

y

3

y

y > –1 x

x

0

–2

6

0 –1

x

0 –3

–y > –4

2

c

Region required

Region required

Exercise 11A — Graphs of linear inequations

x

5

–x ≥ –5

CHAPTER 11

Linear programming

–y ≥ 2

Region required y

d

–x > 3

x

0 –2

F

2

Region required

y

–7

2 F'

x

0

–1

x>0

y

–2

x

0

Foci (2 3 + 2, 0); (2 − 2 3, 0) Directrices x =

y x ≤ –1

=1

4

l

y

x

0 y–3≤0

x+2>0 x

0

Region required

Region required

y<6

i

j

y

y

Region required

e

f

y y>0

y y≤

x

0

0

–6

x

0 –6

Region required

h

y

x –x + 4 ≥ 0

Region required

k

l

–y – 6 ≤ 2

5

y

x

0

0

4

Region required

i

Region required

x≤0

j

Region required

3 E

Region required

y

4 a

y

b

y

y y + x ≥ –1

4 –2

0

x x > –2

Region required

0 x<1

1

Region required

x

–8

0

x

0

x

x

–2

x

0 y – 2x ≤ 4

Region required

–1

x

0 –1

Answers 11A ➜ 11A

x≥4

x

4

y

y–2<3 0

0

Region required

y

Region required

g

8

x–8<0

Region required

Answers

655

c

d

y

y

s

x

4

2 x

2

y≥–x+2

h

y

x

0

Region required

2

n x

0

x – y < 10 10

4

r

2

–4 Region required

x

–2

–3

f

y

x

4

Region required

Region required

b

x

1 0

y 3

3

9– (2–, 3 5) 2 0 1 4

x

d 4 —, — (36 13 13)

3 4

y 10

x

0

(10, 0) 10


x

Region required

y 4

–4

x

0 –3

–8

(0, 0)

0

c

2

y 6

Region required

y

0

x

Region required

y 2 1

x

4x – 2y ≤ 8

2y > 4x – 8

1

2

2 a

Region required

y

0

(0, 0) 0

(−12, −3)

–4

Region required

Answers

(−8, 6)

6

0

0

–10

656

e

x

y

x

y

x

Region required

y
0

q

–2

2 (1, −1)

Region required

p

x

Region required

d

Region required

y

1

2

–3

Region required

o

0

y

0 1

0

–1.5

3y ≤ x + 12

1

x

x

–2y ≥ 4x +6

y

(−1, 2)

Region required

y

4

2

4

Region required

y

b

(0, 0) – 1– 0 1

–24

Region required

Region required

y

c

0 2

x > y –2

–12

1 a

y

x

x

2

6 D

y ≥ x +7

y > 12x –24 0

m

x

0

l

0

–1

Region required

y

Region required y 5 5x + 2y – 10 > 0

Exercise 11B — Graphs of simultaneous linear inequations

y

–7

x

x

5 B

7

x+y> 0

2y + 8x + 4 ≤ 0

–2

Region required

Region required

j

1

–2

x

y ≥ –7x +21

Region required

9x + 9y + 9 ≥ 0 0

–1

0

x

3

Region required y

w

y 3

–24

k

2

0

x

0

– 1–

21

x

0 4

y

x

7

Region required

y

v y + 2x – 6 ≥ 0

–14

y < 6x – 2 4

y–x– 4<0 Region required

y 6

0

Region required

–1

u

y

x

0

–4

2x – y > –1

Region required

y > 2x – 14 0

x

0

2

Region required

f

4

– 1–

y–x≤0

y

i

x

0

Region required

g

y

1

0 4x + 4y ≤ 16

e

t

y

4

Region required

x

e

f

y

y

5 45

0

x

h

1

15 — ( 9–, 4 8)

3 0

0

2

5

x

1

5

y

e

3

y

f

6

x

0 –2

x

5

–1 0

x

5

0 –2

–4

–3

i

Region required y

j

Region required y 7

—, 21 — ( 65 31 31) x 0 23

g

0 1 2


k

1–) ( −1–, 2 32

–4

l

3

x

0

–2

3

(–3 1–, 2 –9)

n

0 1 –

(1–, 2 0)

i

2

p 3 10

x

—, −14 — ( 40 11 11)

–2

Region required

y 1

(0, 1)

0 – 1– 2

4– 3

m

6

x 6

Region required

n

x

y 2 1

p

y

y

x 0 –1

q

3

Region required

x 0

2

Region required

r

y

y 8

6 x

0

Region required

1– 2

2

Region required

1 – 1– 0 1 2 –1 Region required

x

4 –4

0

x 2

Answers 11B ➜ 11C

x 2

–3

4

5

2

0

Exercise 11C — Graphs of systems of linear inequations b 1 a y y

3

Region required

3

b A

x 0

– 1– 2

Region required

Region required

3 a B

1– 1 2

3

y

o

–2 (0, −2) Region required

x 0

– 1–

3

2 1 0 1 3 5

4

–4 0

1

1

4

y —, 60 —) ( −12 13 13

y

x 0

Region required

x

2

2

–2

Region required

x

l

y 1– 2

1

Region required

1

r

y

0

k

x

0 –1

0–1

Region required x

y

2 1

1– 2

Region required

y

j

y

x

0 1 –1

– 1–

2

0

3

Region required

6

1

6

–1

x

1 –2

y

x

Region required

q

0

(–2, 3)

–2

x

0

Region required

Region required

y 2

o

y

2

2

12

Region required

m

x –1 0

Region required

y 4

y 9

4

—,−28 — ( 15 11 11)

–8

h

y

x

4

Region required

Region required

1

2–4

3– 2

Region required

Region required

—, 4–) (21 5 5

x

2

0

–7

y

3 6

y 2

x

x

Region required

y

d

y

–3

Region required 7–2

(2, 0)

0

—, −5 – ( 40 9 9)

–5

g

c

3

Region required

Answers

657

s

t

y 1.5 0

v

8

16

0

x

10

1 –2 –2

iii Maximum value 10.35 x

14

Region required

Region required y

g i, ii

6

y

iii Minimum value 18

7 1– 2

3

8

0

x

5

–2

9

0

3 4

x

y

9

–3 0 1 –3

h i, ii

y 2 10 (–5 , 9) 9

x

2

0

2 B 3 E 4 D 5 B Exercise 11D — Maximising and minimising linear functions (4, 6)

3 4

x

Region required

i i, ii

iii Minimum value 0

y (2– ,13 –)

5 3

iii Maximum value 4

y

iii Maximum 4 value 7 5

(3, 9) (3, 5–)

Region required

Region required

6

9

Region required

y

x

–8

1 a i, ii

10 x 0

7

1 023

6

x

Region required

Region required

w

5 71– 2

Region required y

– 8– (182 15 , 15)

4

0

–4.5

u

y

f i, ii

8 3

x

1.5 3

y

3 3

(3, 2) x

0 2 5 – –3 2

0


x

4

j i, ii

Region required

b i, ii

iii Minimum value 0

y 7

(7, 7)

iii Maximum value 15

(5, 5)

(6, 6–5 ) 12 0 –– 5

x

4 6

Region required

Region required y


2

1

–4 –2

x

0

c i, ii

y (6, 10) 10 9 (1– , 10)

2 a i, ii

y 4 3

iii Minimum value 0

( 7–, 40 –) 11 11

(6, 4– ) 7

0

5

Region required

Region required

d i, ii

iii Maximum value 20 4

y 4 3

( 8–, 12 –)

0

4

5

b i, ii

8

5

(6, 2)

5

8

2 0

x

3 3–

x

8

Region required

iii Minimum 1 value -22 2

y 4


y

Region required

e i, ii

6 7 x

0

–3

x

c i, ii

y

(6, 14)

8

(1, 3)

4

0

4

5

x –8

Region required

658

Answers

0

6

Region required

x

iii Minimum value -8

y

d i, ii

(8, 12)

iii Minimum value 12

4 –4 0

4

x

8

Region required

e i, ii

iii Maximum value 33.4

y 8 7

(2, 7)

(10, 7) (10, 3)

0

x

10 16

Region required

f i, ii

y 9 5 4

iii Minimum value -1.15

20 84

–,–) ( 11 11

( 1– , 5)

–2 0

2

1

(5 –3, 5) 12 x

Region required

g i, ii

y 10 6 5

iii Maximum value 23.72

30 80

–,–) ( 11 11

( 6– , 6)

–6 0

(4, 6)

5

10 x

Region required

h i, ii

14


y 6

5

– , 11 – ) (211 11

8 7

4

7

– , 7 –) ( 15 15

–4 0 4 14 Region required

i i, ii

x


y 9 3 0 –3

–9

18 27

( –7 , –7 ) (4, 1) 3 4 1– 2

x

a Brand A = 20, Brand B = 260 b $1840 a 36 aerobic classes, 9 circuit classes b $252 a 20 2-person tents, 18 3-person tents b $984 a 15 bubble jet printers, 10 laser printers b $330 a 70 hectares of corn, 10 hectares of peas b $14 200 a Nature’s Own = 80 litres, generic brand = 10 litres b $212 a 255 tourists, 255 businesspeople b $8415 a 2 batches of 3-D puzzles, 3 batches of logic games b $147 a 40 Arctic and 40 Cool Breeze b $11 960 c Yes ($9560) a Variables need to be defined in order to determine how many bags of CP1 and CP2 will produce maximum profit. b 20x + 15y ≤ 300, 15x + 10y ≤ 210 c x ≥ 0, y ≥ 0 d P = 45x + 50y e $836 14 A 15 A 16 a 3 engineers and 1 technician b $1600 Exercise 11F — Further applications of linear programming 1 $69 2 a Rod A = 16, rod B = 10 b $13 800 3 a x = the number of type A uniforms produced; y = the number of type B uniforms produced. b 5x + 8y ≤ 480, 6x + 12y ≤ 600, 3x + 3y ≤ 450, x ≥ 0, y ≥0 c P = 7x + 12y d 80 type A uniforms and 10 type B uniforms e $680 2 1 4 a 2 5 kg of Zest, 1 5 kg of Boom b $15.72 (rounded down $15.70) 5 a 200 Gold Pass packages, 100 Classic packages b $5000 6 a 1.6 tonnes at plant A, 4.2 tonnes at plant B b $2532 Chapter review Short answer 1 a b y y

4 5 6 7 8 9 10 11 12 13

Region required

j i, ii

9 7 5

3 a C

b C

x

c

Region required

y

x

c D

x

0 –3

3

d B

Exercise 11E — Solving linear programming problems 1 a Style A = 5, Style B = 19 b $67 2 a Design A = 24, Design B = 4 b $146 3 a 35 rollerblades, 5 bicycles b $155

0

–2

Answers 11D ➜ 11F

2 4 6 Region required

3

Region required

(4, 3)

0

0


y ( 4–3 , 7)

x

Region required

2

y 3 0 6– 5

4

x


Answers

659

3 x + y ≥ 400, y ≤ 400, x − 2y ≤ 400 4 x ≥ 0, y ≥ 0, −x + y ≥ −300, x + y ≤ 600, x − y ≥ −400 5 a b y y 1 (0, 2–3)

2 1–3

B

(4, 5)

3 A (0, 3)

(0, 0)O C (2, 0) 0 2

1 –3 –2

4 B (0, 4)

A

x

0


6 a

y 7 B 6 A

(6, 0) C 4 6 x D (4, 0)

e 1600 cake platters, 200 cheese platters and 300 fruit bowls f $48 450 5 a x ≥ 0; y ≥ 0; x + y ≥ 80; 0.03x + 0.05y ≥ 2.5; 0.05x + 0.08y ≥ 4.5; 0.05x + 0.025y ≥ 3 b y A (0, 120)

Region required

b $27.40

O (0, 0) A (0, 6) B ( 1–5 , 6 3–)

B (40, 40) 2

5

1 C (63 –3, 16 –3)

2– 2– C –2 O D C (2 3, 13 ) 0 1 31– D(1, 0) x –1 2

(831–3 , 0)

6

a 54 b −4 x = 0, y = 0, minimum = 0 x = 0, y = 3.4, maximum = 4080 a x = the number of runners produced, y = the number of walking shoes produced. b x ≥ 400, y ≥ 350, x + y ≤ 900 c P = 12.5x + 10y

7 8 9 10

4 E 11 B

5 B 12 A

6 C 13 E

7 D 14 A

Extended response 1 a 20 hectares of oats, 15 hectares of wheat b $7600 c Yes 2 a 10x + 5y ≥ 70, 2x + 4y ≥ 28, 3x + 3y ≥ 36 b Yes, x ≥ 0 and y ≥ 0 We are dealing with items produced; therefore x ≥ 0 and y ≥ 0. y c 14 A (0, 14) 12 B (2, 10) C (10, 2) 7 D (14, 0) 0 7 12 14 x Region required

d C = 300x + 200y e $2600 3 a 5 kg C1, 2 kg C2 b $27.50 4 a x ≥ 0; y ≥ 0; x ≤ 1600; y ≥ 200; x ≥ 2y; x + y ≥ 1500; x + y ≤ 2100 b y

2200 2000 1800 1600 1400 1200 1000 800 600 400 200

2200 (0, 2100) 2000 1800 1600 (0, 1500) 1400 1200 (1400, 700) 1000 B 800 (1000, 500) C (1600, 500) 600 A 400 D (1600, 200) 200 (1300, 200) E x Region required

c A(1000, 500) B(1400, 700) C(1600, 500) D(1600, 200) E(1300, 200) d P = 2.5x – 3.5y + 45 150

660

Answers

x

Region required

Region required

Multiple choice 1 D 2 D 3 D 8 B 9 A 10 B 15 C 16 E 17 A

D (90, 0)

c d e f h i a

1

2

A(0, 120) B(40, 40) C(633 , 16 3 ) D(90, 0) C = 0.012x + 0.016y 1 2 633 g of Chicken Bites and 16 3 g of Fish Bites per day $1.03 per day g C = 0.012x + 0.014y 1 2 633 g of Chicken Bites and 16 3 g of Fish Bites per day $0.99 per day y 8

(0, 8)

6 4 2 (0, 0)

2

4

6

8

10

x

b y ≥ 2x c i Refer to diagram. ii 2 d P = 40x + 30y e i 2 dogs washed and 6 dogs clipped ii $260 a Total number of driving hours with either petrol or gas can not exceed 24 hours. b i & y ii 25 7

20 15 10 1x y=− 3

5 0

5

10

15

20

25

x

c Yes, they do comply with all constraints since (10, 5) fits within the feasible region. d The maximum number of hours using gas is 18 and the minimum is 17.

CHAPTER 12

Coordinate geometry Exercise 12A — Distance between two points 1 AB = 5, CD = 2 10 or 6.32, EF = 3 2 or 4.24, GH = 2 5 or 4.47, IJ = 5, KL = 26 or 5.10, MN = 4 2 or 5.66, OP = 10 or 3.16

2 a 5 d 7.07

b 13 e 6.71

c 10 f 14.42

a2 + 4b2 g 13 h 13 i 2 2 j 3 a + b 3, 4, 5 and 6 Answers will vary. 7 a AB = 4.47, BC = 2.24, CD = 4.47, DA = 2.24 b AC = 5, BD = 5 c Rectangle 8 B 9 D 10 a 12 b 5 c 13 d −2.2 11 Answers will vary. Exercise 12B — Midpoint of a line segment 1

1

1 a (−3, −3 2 ) 1

1

d (0, 1 2 )

2 (−3, −10) 3 a (3, 1) 4 D

1

e (2a, 2 b)

f (a + b, 2 a)

b 4.47

c 6.32 5 C

1

6 a i (−1, 4) 7 8 9 10

c (−1, 1)

b (7 2 , 0)

ii (1 2 , 1)

iii 3.9

iv 7.8

b Answers will vary. a i (1, −0.5) ii (1, −0.5) b Answers will vary. iii 9.55 a i (−2, 2) ii 8.94 b Isosceles triangle, height. y = -3x - 2 3y - 2x + 14 = 0

iv 9.55

Exercise 12C — Dividing a line segment internally in the ratio a : b 1 a (2, 5) b (3, 3) c (4, 1) d (4, 11) 1 1 d (5, 5) 2 a (0, −2 2 ) b (−2, 2 2 ) c (4, 6) −1

1

e ( 2 , −6 2 ) f ( 1 , −7 1 ) 2 2 2

3 (8, −1 5 )

4 c = 7, d = 10

5 a B 1 1 6 a i (1 2 , 1 2 )

b C 1 ii (0, 4 2 )

c D 1 iii (4 2 , 12)

b i (2, 6) ii (2, 6) iii (2, 6) c The medians of the triangle are concurrent and divide themselves in the ratio 2:1. d

y 18

A (3, 15)

15

1

70

G (2, 6)

3

1

0

60

1

L (1–2 , 1 –2) 1 2 3 4 5 6

x

50 0

−6

10

20

30

40

50

Distance from Uzes (km)

−9 1

1

1

7 a U: (0, 10 2 ), V: (-1 2 , 7 2 ), W: (-4 2 , 6) b (-2, 8) c 2 : 3

b -0.69 m/km c y = -0.34x + 76, where 76 is the height in metres above sea level and x is the distance in km from Uzès d 5.56 m e Check with teacher.

Answers

Answers 12A ➜ 12G

C (–3, –6)

- 18

80

6

−5 −4 −3 −2 −1 −3

-8

7 a m = 5 b m = 5 9 E 10 B Exercise 12G — Applications 1 y = 2x − 9 2 3x + 2y − 8 = 0 3 a y = 3x + 2 b y = −4x + 9 c 3x − 2y − 8 = 0 d 2x + 5y + 13 = 0 e x + 5y + 5 = 0 f x − 3y + 17 = 0 g x − 3y − 14 = 0 4 a 2x − y + 5 = 0 b x + 2y = 0 5 a 3x − 5y + 2 = 0 b 5x + 3y − 8 = 0 6 a x = 1 b y = -7 7 a B b C c D d B 8 a (2, 5) b 1 c Answers will vary. d Isosceles triangle 9 y = −x − 3 10 4x − 6y + 23 = 0 b y = x + 3 c (1, 4) 11 a y = −x + 5 b (6.5, 5.5) c 2 12 a 5.10 km e (10, 2) f 2:1 d y = 2x − 18 g (7, 3) h 7.071 km b D c C 13 a A Height of aqueduct above sea level (m) 14 a

B (6, 9)

9 1

2

3 (4, 3 5 ) 4 c = 3, d = −2 5 A 6 B 7 a 7.53 cm; 35.46 cm b 139.925 m Exercise 12E — Parallel lines 1 a No b Yes c No d No e Yes f No 2 b, f; c, e 3, 4, 5 and 6 Answers will vary. 7 B 8 E 9 a y = -2x + 1 b 3y + 2x +5 = 0 Exercise 12F — Perpendicular lines 1 Answers will vary. 2 a Yes b Yes c No d Yes e Yes f No 3 a, e;- b, f; c, h; d, g 1 3 6 y = 2 x + 2

N (4 1–2, 12)

12

M (0, 4 –2 )

Exercise 12D — Dividing a line segment externally in the ratio a : b 1 a (−2, 13) b (7, −5) c (3, −10) d (−5, 10) b (17, −7) c (−8, 1) 2 a (−8, 18) 1 1 2 1 − e (23 , 43 ) f (5 3 , −23 ) d (8, 15)

661

Chapter review Short answer 1 61 2 and 3 Answers will vary. 4 (0, −18) 6 (−1, 14) 9 x + 2y − 2 = 0 11 3x + 2y − 21 = 0 13 a i 4 5

-3

5 (1, 2 ) 7 and 8 Answers will vary. 10 2x + 3y − 9 = 0 12 3x − 2y + 16 = 0 ii 45

iii 4x + 5y − 61 = 0 v (9, 5) b Square 1 14 a i 10

1

iv 5x − 4y − 25 = 0

x

-

ii ( 1 , 1) 2

1

iii (4 2 , 1 2 )

c The asymptote line’s gradient is

b Answers will vary. 15 a i (1, 2) iii −1 v 194 b Isosceles triangle

−

−

ii ( 8, 7) iv 1 vi 194

2 7 12 17

C D A B

A A C E

Extended response 1 a 7x − 3y − 1 = 0 c −7 2 (3, 6) 3 a 9.43 km c B(7, 2) d 15.30 km 4 5

3 C 8 B 13 C

4 B 9 A 14 D

5 C 10 A 15 B

b C(8, 7)

120

e (m - 7)2 + 4

41

Exam Practice 3 Short answer 3 , -3)

1

(-

2 3

n(n + 1) is divisible by 2 by proof a a = 1 b = -1 b C = -7 -3 c A = 7 B = 2

Answers

m

b 3x + 7y − 49 = 0

b y = 4 x + 20 or 25x − 20y + 41 = 0 5 c Since the gradient of the path AB is 4 , which is the same as the gradient of the known path of travel from the common point A, the direction of travel is toward B. d dAB = 0.8 metres. Yes, guard ball A will collide with guard ball B as it will not be deviated from its linear path under 1 metre of travel.

662

b

210

f 5.10 + (m - 8)2 + 49 g m = 30 km. h Total distance = 77.82 km. a x = −7 b B(−7, 50) C(−4, 90) c 40.1 metres d 86 metres a Since the gradient of SA = the gradient of SO = -0.8, the points S, A and O are collinear. Player Y will displace guard ball A. 5

b 4 = a 2 It needs to have a gradient of greater than 0 because it has a positive slope. d (2,4) must be a point on the line if the distance is the same from (0,0). e y2 = -2x 2 a There are at least twice as many maximum protection sails made than light protection.

Multiple choice 1 6 11 16

Multiple choice 1 B 2 C 3 E 4 D 5 A 6 B Extended response 1 a ±3 b y

60

210

l

c (70, 140) (60, 120) (60, 190) d S = 12.75 l + 21.50 m e $ 4850.00 f No they could not deliver this order because the point (90, 135) is outside the feasible region.

CHAPTER 13

Vectors Exercise 13A — Introduction to vectors 1 a (2, 2) b (1, −2) c (−4, −3) d (−1, −4) e (−6, −2) f (−5, 2) g (−1, 2) h (1, 2) i (1, −1) j (−1, 1) 2 a (5, −7) b (−1, 0) − c (2, 1) d (4, −10) e (0, 0) f (8, −2) 3 a a = (1, 1) b b = (0, 3)   c c = ( 1, 2) d d = (- 1, 2)   e e = (- 1, -1) f f = (2, 4)   g g = (1, 0) h h = (1, 1)  

4

10

y 2 1

B

A

11.5 km/h

−4 −3 −2 −1 0 1 2 3 4 x −1 D −2 −3 C

5 a = h, c = d     6 a y

11

b

y

Exercise 13B — Operations on vectors 1 a b

5 (0, 5) 4 3 2 1

x

1 2

3a ~

2b ~

1 2

c

d

y

x

–3 –2 –1

e

y 5 4 3 2 1

(7, 3)

W

E S

A N

20 km W 30 km F

f

~c a~

~c

~b + ~c ~b

g

h a~ + 2b~ a~

i

a~ a~ – ~c

x

k

2

~v

3 a

2a ~

j

–c ~

–c ~

~b ~b – ~c

~c

a~ + ~b + ~c

~u

3c ~

2a ~ + 3c~

2b ~

l

~b

~a –c ~

–b ~ ~a – ~b – ~c

or any two vectors equal in length but opposite in direction

y (3, 5) 5 v 4 ~ 3 2 u 1 ~

b

y ~v 2 (–3, 2) u ~ –3 –2 –1 O x

E S

or any similar result or any similar result 4 a (2, 3) b (1, 8) c (5, −5) d (−5, −2) c (4, 6) d (−16, 20) 5 a (13, -5) b (1, 1) → → → → → → 6 a CB, DE, GF b AB, DG, EF → → → → → → c AE, CG, BF d DE, CB, OA → → e DF f BG 7 a a b a + c   - a + c c d d   -d - a + c e f    g a + c h a + c + d    - a + c + d i j a + c - d      

Answers

Answers 13A ➜ 13B

N

1000 km

S

a~ + ~c

O 12 3 x

C 1200 km

9

e

(0, 3) (2, 5)

~b

~a

(2, 8)

3 4 5

~a + ~b

(4, 4)

(3, 5)

7 B

B 45°

y 8 7 6 5 4 3 2 1 12

d

–c ~

~a

–3–2–1 –1 1 2 3 4 5 6 7 x –2 –3 (4, –2) –4

8

c

–6–5–4–3–2–1 –1 1 2 3 4 5 6 x (2, –8) –2 –3 –4 –5 (6, –4) –6

f y 6 5 4 3 2 1 12 3 4 5 6 x –6–5–4–3–2–1 –1 (1, –1) –2 –3 (–2, 5) –4 –5 (3, –6) –6

g

x y 6 5 4 3 2 1

3 2 1

(–3, 2)

300 km

25 km 25°

(2, 3)

3 2 1

3.5 km/h

663

a a b 2a - 2a - c d a + b   - e a f 2a + b  - -  g b h a + b    i a + b j 2a - b   →  10 AE 12 B Exercise 13C — Magnitude, direction and components of vectors

c

8

1

2

Magnitude

Direction

a

2 10

18.4°

b

17

−14.0°

c

2 5

63.4°

d

2

45°

e

5

153.4°

f

17

104.0°

g

1

Parallel to x-axis

h

2 2

-135°

a 1 c 2 e 2 g 2 i 3

3 a 1,

b

c 5, - 5 3

O

e

-3

2

3

f h j l n

g

i

O

664

Answers

O 4 x

–3

5 x

O

h

y x

–2 O

j

y O –8

y 12

x

–6

O –5

, 1.5

y

y

x

x

x

–6

4 x 2 + y 2 , angle θ from the positive direction of the y x-axis where tan (180° - q ) = x 5 a 29 b 13.5 km 6 10 km, N36.9°E 7 a 14.86 km, N16.6°E and 7.15 km, N36.4°W b 22.91 units, S40.9°E and 39.05 units, N63.7°E c 2 units in direction of a and 18 units in direction  of a  d 16.97 km, SW and 16.97 km, NW e 35 km in direction of a and 5 km in direction of a   f 64.03 units, 338.7°T and 64.03 units, 261.3°T

3

4

y O

0, 4 0, −30 10 3, 10 37.6, 13.7 1207.4, 323.5

Exercise 13D — i , j notation   b 1 a y

f

y

2

1

O –1

1

d −1, - 3

e 10 2 , 10 2 g 12, 0 i - 5 3, −5 k 50, 0 m 0, −98

y

x

2

O

b 1 d 1 f 2 h 1 j 3

3

d

y 2

4 x

x

2 a 5, 36.9° c 2 2, 45°

b 5, −36.9° d 2 , -45°

e 17, 14° g 6, −90° i 10, −143.1°

f 5, 0° h 2, 180° j 13, 112.6°

3 a 9 i + 6 j   c 3i + 4 j   e 4 i - j   g 11i + 2 j   i - 3i - 10 j   4 a 13 c 3 10

b 4 i + j   d 2 i - 2 j   f 2 i  h - 8 j  j - 4 i - 7 j   b 29 d 17

5 a 2 i + 3 j   c - 4 i + 3 j   e - 2 i - 3 j   g 6 i - 4 j   y 6 a 4 3

b i + 3 j   d - 6 i - 2 j   f i - 3 j   h 5i + j   b 2 i + 4 j   c 3i + j  

2 1 O –1 –2

B

A

3 4 5 x ~j ~i

7 a 4 i + 6 j b   c 6 j d  e - 4 i + 6 j f   8 a 6 i + 2 j, 4 j, 6 i + 2 j b      c Isosceles → → 9 a MN, 2 i + 3 j, NP, 4 i + 6 j     10 a 18i - 8 j, - 2 j b    11 C 12 A 13 A

2i  3j - 2 i + 3 j   2 10 , 4, 2 10

i component is zero  14 C

15 3.9 i + 2.7 j 16 x = 5   Exercise 13E — Applications of vectors 1 2.5i + 7 j; 7.43 km/h; 70.3° from the river bank   71.2° from the river bank 2 10.6 km/h, 3 a 650 j b 30 2 i + (650 + 30 2 ) j   693.7 km/h  c N3.5°E, 4 S2.2°W, 643.8 km/h 5 7.8 units, 50.2° from positive x-axis 6 6.3 units, −18.4° from positive x-axis 7 6.4 km, N51.3°E or 51.3°T 8 9.4 km, N58°W or 302°T 9 a - 4 i + 8 j, - 2 i + 6.5 j b 2.5 km     c i 5 j ii 2 i + 3.5 j e (7.4, 2)   3)  h First 7.5 km, second 5.6 km f (3.8, 10 a i + 3 j b 3 km   c 0.1k , i + 3 j + 0.1k d 2.002     11 1 s, 2 m 12 x = −5, y = −1 13 32.3 N, 60.3° from the vertical 14 37.8 N, 53.4° from the vertical 15 and 16 Teacher to check student proofs Chapter review Short answer 1 (–2, 5) y 5 4 (0, 4) 3 2

O –3 –2 –1 –1 –2

2 3 x (–3, –2)

3 −2, 2 3 4 a i 3i + 7 j  

ii

b

5 a

6i - 9 j  

iii - 6 j 

Magnitude

Direction

d 

17

76.0°

e  f 

13

−123.7°

4

0°

G 2 4 6 8 x j ~ ~i

-

12 i - 8 j c 14.4   → 1→ 6 a 4 i + j, - 2 i - 3 j c MN = 2 GH     7 8.1 km, N60.3°W or 299.7°T 8 9.6 km/h, 70.4° from the river bank b

Multiple choice 1 A 2 B 6 D 7 E 11 B 12 A

3 D 8 C 13 E

E S

20 km/h

~v 10 O

25 km/h 36 x (East)

15.5°

2 Teacher to check 3 81.7 N, 53.1° with the vertical

CHAPTER 14

Statics of a particle Exercise 14A — Force and tension 1 Mass is the amount of inertia a body possesses. kg 2 Weight is the force which attracts a mass to the Earth. Newtons 3 The force in a connector, often a string. Newtons 4 10g N 5 0.7g N 25 6 kg = 2.55 / kg 7 2.819 N; 1.026 N g

4 E 9 C 14 C

5 D 10 A 15 C

Exercise 14B — Newton’s first law of motion 1 Force is push or pull. 2 Newtons 3 a A force with exactly the same number of newtons. b i Nothing. The forces are balanced. ii Rotation. 4 a The force pushing up from the surface is greater than her weight. b The force pushing up from the surface is less than her weight. c The force pushing up from the surface equals her weight. 5 If the force Ned is using is greater than the force MaryAlice is using the door will open and vice-versa. If the forces are equal the door will not open. 6 2g N 7 1.571 kg 8 3.061 kg. No. The string can hold 3.571 kg. 9 Any weight over 35 N will break the string. Thus the hook and beam stay where they are. 9N 10 N 10 Resultant 1 N to the right 1 1 12 13 14 15 16 17 19

The door will remain closed if 160 + C ≥ 280. Christine’s force is 120 ≤ C ≤ 130. E 12g N B The body will fall through the surface. -U = - 4 i + 3 j; 5 N; 5 N    E 18 D A 20 C

Answers

Answers 13C ➜ 14B

y 2 O –4 –2 –2 –4 –6 H

-

W

8 53.13° 9 6 N 10 5 N; 36.87° 11 5 cos(-53.13°) = 3; 5 sin(-53.13°) = -4 12 6 i + 6 j; 45°   13 E 14 E 15 A 16 B

a (3, 2) b (−6, 7) c (2, −5)

Extended response 1 a 24 i + 16 j, 36 i + 10 j c 28.84, 13.42, 37.36   d Approx. 80 km  e 33.7° f −26.6º g N y

665

Exercise 14C — Equilibrium — forces at an angle 1 All forces are balanced. Vector sum = 0. 2 No. The forces acting on it cause it to slow down. 3 No. Gravitational forces cause it to accelerate. 4 150 N from the right 5 88.83°; 98.02 N 6 11.31 N; 27.14 N 7 0.5604 kg 8 21.18 N; 17.50 N; 2g N 9 2.646 N 10 7.453 N 11 3; 3 N 12 4 N, 2 13 1.037 kg 14 26.30 N; 13.15 N 15 - i + 4 j   16 17 N; 104.04° from the positive i axis 17 12.12 N; 7 N 18 E 19 A 20 E Exercise 14D — Connected bodies in equilibrium 1 w1 = - 3g j; w2 = - 7 g j; 10 g N; f = 10 g j    N; 50.92 N   2 58.8 N; 85.43 3 t = - 3 3gi + 3g j    4 w1 = - 4 gj; w2 = - 10 g j; 14 g N; f = 14 g j     N; 64.39°  5 131.2 N; 102.9 N; 107.1 6 t = 92.75 i + 92.75 j    7 E 8 D 9 B 10 B 11 D 12 A 13 0.0403 kg 14 M = m tan (a) cot (b) 5 4 15 ; ; 36.87°; 30.96° g g

Exercise 15A — Introduction to kinematics ii 1 iii 2 units 1 a i -1 2 iv 10 units v units/second 5 vi 2 units/second b i 4 ii 0 iii -4 units -4 iv 24 units v units/second 5 4 vi 4 5 units/second ii 12 iii 20 units c i -8 iv 52 units v 4 units/second 2 vi 10 5 units/second d i 12 ii 6 iii -6 units 1 iv 26 units v 1 5 units/second 1 vi 5 5 units/second e i 10 ii 16 iii 6 units 1 iv 30 units v 1 5 units/second vi 6 units/second 2 a F

b

A

x F

A

S

–14–13–12–11–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 x

c

F S

A

–2 0 2 4 6 8 10

d

e

x F

A –16 –14

S –12

–10

–8

–6

–4

A

S

–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6

f

0 x

–2

B

F

x

F

B S

1 2 10 ; 71.57° 2 15g N 3 19.21 N; 65.86 N

A

0 1 2 3 4 5 6 7 8 9 10 11 12 x

- 5g

j; w2 = j; 7 g N; f = 7 gj 4 w1 =        Multiple choice 1 E 2 E 3 B 4 E 6 B 7 A 8 D 9 A 11 D 12 E 13 D Extended response 1 100.2 N; 186.8 N 2 14.97 3 8.386 4 2a; 45°

Answers

Kinematics

–2 0 2 4 6 8 10


666

CHAPTER 15

S

16 w1 = - mg i ; w2 = - mg i ; T1 = - T1 sin (a ) i + T1 cos (a ) j;       T2 = T2 sin (b ) i + T2 cos (b ) j;   T cos (a ) + T2 cos (b ) M = -m+ 1 g

- 2g

5 5.102 kg 6 24.01 N; 7.174 N; 45.84 N; 7.79°

5 D 10 C

3 a i 5 cm 1 iv 2 2 cm/s

ii 15 cm

iii

b i -2 cm iv 4 cm/s

ii 24 cm

iii

c i -1 cm 1 iv 2 2 cm/s

ii 15 cm

iii

d i -5 cm 5 iv 2 6 cm/s

ii 17 cm

iii

5 6

cm/s

-1 3 -1 6 -5 6

cm/s cm/s cm/s

e i -4 cm 1 iv 3 3 cm/s

ii 20 cm

iii

-2 3

b

cm/s

–9 –7 –5 –3 –1 1 3 5 7 x (cm)

1

-50

d i 66 cm/s ii cm/s e 50 cm/s a 600 cm b -400 cm c i t = 5 to t = 8 ii t = 0 to t = 4 and t = 8 to t = 13 iii t = 4 to t = 5 d i 100 cm/s ii -125 cm/s and -40 cm/s iii 0 cm/s e 76.9 cm/s a t 0 2 4 x

2

0

iii t = 4 to t = 6 iii 0 cm/s

2

b 6

8

8

18

x b

c

e 2.5 cm/s 12

t= 2 t= 1 t= 0 –16–12 –8 –4 0 4 8 12 x -4

2

4

6

8

0

-4

0

12

v 12 10 8 6 4 2 0

Velocity (m/s) 1 2 3 4 5678

t

t

12 6

0

c

23 28 t

8

v 4 Velocity (m/s)

–4 –2 0 2 4 6 8 10 12 x (cm)

f 4 cm/s x = t2 – 4t – 5

1 2 3 4 56

t

2 0 –2

10

31

t

–4 –6

d

22

Answers 14C ➜ 15B

t= 0

t= 2

Position (cm)

20

Time (s)

t= 4

x

10 15

v

Time (s)

v 5 Velocity (m/s)

Position (cm)

x = t – 8t + 12

c The particle changes direction or turns. d t= 6 t= 8

8 6 4 2 0 –2 –4 –6 –8 –10

5

Time (s)

b

2

Time (s)

e 0 cm/s 12 a

1

d 4 3 cm/s

cm/s

x

12 11 10 9 8 7 6 5 4 3 2 1 0 –1 –2 –3 –4

t

t= 6

1 a

0 2 4 6 8 10 12 14 16 18 20 x (cm)

0

1 2 3 4 56

Exercise 15B — Velocity–time graphs and acceleration–time graphs

t=0s

d 2 cm/s 11 a t

x = – t2+ 2t + 8

Time (s)

b The particle changes direction or turns. c t=8s t=2s

d 3 3 cm/s

x

10 8 6 4 2 0 –2 –4 –6 –8 –10 –12 –14 –16 –18

Position (cm)

2 3

1

c 2 cm/s 13 a

Velocity (m/s)

c i t = 6 to t = 12 ii t = 0 to t = 4

t= 6

t= 0

f i 1 cm ii 9 cm iii 6 cm/s 1 iv 1 2 cm/s 4 C 5 E 6 B 7 D 8 a 300 cm b 200 cm 9 10

t= 4 t= 2

0

6

10

16 t

–5 Time (s)

Time (s)

Answers

667

e

d i 44 metres below ground level

Velocity (m/s)

v 8

ii 13 metres above ground level e -1.93 m/s f 201 metres 10 a 7 seconds b v

4 0

10

18

30 34

t

–4

Velocity (m/s)

–8 Time (s)

2 a 1.6 m/s2

b -2.4 m/s2

c 190 m

d 190 m

3 a -2 m/s2

b 1 3 m/s2

c 148 m

d 148 m

1

4 a 0.5

b 0.25

c 159 m

d 159 m

5 C 6 D 7 C

Acceleration (m/s2)

a 2 O

1.5

A 0 4

B C 16 19

D E 27 31

B C

D E

4

12 14 23 25

30

t (s)

−2.5

b 1.5 m c 31.5 m d i 1.05 m/s

ii 9 metres above ground level f 177 metres

m/s

0

F G 38 41 t

d i 93 metres above ground level 9 41

Time (s)

b Polly takes 25 seconds and Molly takes 25.7 seconds. c 80 m d 0.8 m/s2 12 a v (m/s)

A

Time (s)

e

t

5

F G

–2

Polly Molly

0

b OA: 1.5 m/s2 AB: 0 m/s2 BC: -2 m/s2 CD: 0 m/s2 DE: -2 m/s2 EF: 0 m/s2 2 FG: 2 3 m/s2

d 1575 m

16 14

iii From A to B, C to D and E to F

c

t

11

c 52.5 seconds 11 a v

m/s2

8 a i From O to A and F to G ii From B to C and D to E

4

Time (s)

Velocity (m/s)

m/s2

Motorcycle Car

35 30

0

g 4.47 m/s

13

g 4 41 m/s

ii 1.5 m/s e Yes, the Monkey is 6.5 m above the ground.

9 a i From B to C and D to E ii From O to A and F to G iii From A to B, C to D and E to F b OA: -1.5 m/s2 AB: 0 m/s2 BC: 0.9 m/s2 CD: 0 m/s2 DE: 1 m/s2 EF: 0 m/s2 FG: -1.2 m/s2 c Acceleration (m/s2)

a 1.2 1.0 0.8 0.6 0.4 0.2 0 –0.2 –0.4 –0.6 –0.8 –1.0 –1.2 –1.4 –1.6 –1.8

Answers

1 2 3 4 5

a a a a a

b b b b b

4.5 seconds 4 m/s2 6.5 m/s2 58.8 m/s 78.4 m

1

5

6

14

24

30

36

40 45 t

7 a 1 6 m/s

b 5 seconds

-5

8 a 6 m/s2 9 a 60 m/s and 216 km/h 10 a -5 m/s2 11 a B

43.8 m/s 32 m 1.54 seconds 176.4 m 39.2 m/s

b -3 3 m/s

6 a 10 m/s

Time (s)

668

Exercise 15C — Constant acceleration formulas

b 12 seconds b 12 m/s2

b 22 m b C

12 a 3.7 seconds

c 25.6 m c A b 26.2 m/s

d E

13 a 6.6 seconds

b 44.4 m/s

4 a 90 m 1 b 7.1 seconds 15 12 m/s 16 a 19.6 m/s b 27.7 m/s 17 24.04 m 18 6.5 seconds; 44.2 m/s

c i 95 metres above ground level

c 70.0 m/s

c 2.8 seconds

d

t4 t2 + 4 2

3

b d = 24 4 m

t3 t2 b x(t) = + 18 a v(t) = t2 + t 3 2 19 a v(t) = 6t3 - 2t2 + 4t + 15

1

c d = 29 3 m

3t 4 2t 3 + 2t2 + 15t 2 3

b x(t) =

11 30

2

0 a 28 m/s2 1 11 a 31.8 seconds

b 217 3 m b 1114 metres

b i 12 cm ii 20 cm iii 3 cm/s iv 5 cm/s b -10 cm

Multiple choice 1 D 2 B 6 C 7 A 11 C 12 C Extended response x 1 a

iii t = 8 to t = 12 iii 0 cm/s

Velocity (m/s)

6

3 6

–6

9 12 15 18 21 t

Time (s)

b -5 m/s2

4 a 1.5 m/s2

c 190.5 m

d 190.5 m

Acceleration (m/s2)

5 a OA: AB: 0 BC: CD: 0 m/s2 DE: -4 m/s2 EF: 0 m/s2 FG: 4 m/s2 b a 5 4 3 2 1 0 –1 –2 –3 –4 –5

m/s2

-5

m/s2

4 C 9 B 14 C

5 D 10 D 15 B

x(t) = –t2 + 4t + 12

1 2 3 4 56

t

b -2 cm/s c

t= 4

t= 6

m/s2

t= 2

–2 0 2 4 6 8 10 12 14 16 18 x 1

d 3 3 m/s i 2 cm/s e ii -2 cm/s f 0 ≤ t < 1 and 3 < t ≤ 6 2 a 3.2 seconds b 2.5 m/s2 c 10 seconds d 80 m e 4 m f Stuntman has travelled 90 m and bus has travelled 92.5 m (stuntman is still 2.5 m behind the bus) g After t = 11, the bus is travelling faster and since it is already in front of the stuntman, the stuntman cannot catch the bus. 3 a 50 metres above the ground b 10 10 m/s c 10 10 m/s d 10 seconds 4 a Lily b 0.84 m c 0.105 seconds d 0.84 m e 5.76 m/s2

CHAPTER 16 3

10 12

20 23

27 30 t

Geometry in two and three dimensions Exercise 16A — Review of basic geometry

Time (s)

1 a 56° 2 D

b 67.5° 3 C

c x = 38°, y = 32°

Answers

Answers 15C ➜ 16A

1 33

18 16 14 12 10 8 6 4 2 0

3 E 8 D 13 E

t= 0

c i t = 12 to t = 18 ii t = 0 to t = 8 1 d i 3 3 cm/s ii -3.75 cm/s 3 v

0

1

Time (s)

–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 x

2 a 40 cm

2

b 18 3 m d 27 m

A

S

b 4 5 seconds b 30 m/s

8 a -2 3 m/s2 c 4.5 seconds 9 16 m/s

Position (cm)

F

B

4

6 a 3 8 m/s2 7 a 45 m

2

c d = 56 3 m Chapter review Short answer 1 a

e 179 metres

m/s

1

Exercise 15D — Instantaneous rates of change 1 13 cm/s 2 171.5 cm/s 3 92 m/s 5 -3.082 m/s 6 -1 m/s2 4 -0.5 cm/s 2 2 7 0.176 cm/s 8 3 m/s 9 17 cm/s2 10 E 11 B 12 B 3 13 D 14 125.5 m 15 24 4 m 16 126 m 17 a x(t) =

ii 11 metres above ground level

669

4, 5 and 6 Answers will vary. 7 a x = 60°, y = 114° b x = 47°, y = 55°, z = 78° c x = 59°, y = 121°, z = 59° 8 a x = 75°, y = 30°, z = 150° b x = 45°, y = 42° c 60° 9 Answers will vary. 10 122° 11 AB = BC, therefore ABC is isosceles. ∠BAC = ∠BCA = 45° (property of isosceles triangles). In triangle OAB, ∠OAB + ∠ABO + ∠BOA = 180° (sum of angles in a triangle). ∠OAB = ∠ABO = 45°, so ∠BOA = 90°. 12 a Number of sides Sum of interior angles 3

180°

4

360°

5

540°

6

720°

7

900°

8

1080°

9

1260°

10

1440°

20

3240°

b General formula: 180°(n - 2 13 Circle with radius 3 cm 14 x2 + y2 = 1 15 (x + 1)2 + (y - 3)2 = 2500 16 The perpendicular bisector of the line joining the two points P and Q. Exercise 16B — Geometric constructions 1, 2a and 3a Constructions. Check with your teacher. 2 b The bisectors meet at the same point. 3 b The bisectors still meet at a point. 4 r 2r r 5

A r 4r B r

6r

r C

6–9 Constructions. Check with your teacher. 10 Draw an arc with A as the centre, crossing the line at B and C. With B as the centre, draw an arc above the line. Repeat with the same radius at C. Join the points where the arcs meet to point A. Exercise 16C — Polygons 1 Perpendicular bisectors meet outside the triangle. 2 Perpendicular bisectors meet on the hypotenuse. 3 Construct the perpendicular bisector of the base; vary the radius to get different triangles. 4 Answers will vary.

670

Answers

5

6 and 7 Answers will vary. 8 Find the angle bisectors of the angles formed at the centre of the circle. Join them to the circumference to create 12 vertices. Join the 12 vertices to create the dodecagon. 9 and 10 Constructions 11 Create perpendicular bisectors, then angle bisectors to create 45° angles from the centre of circle, and a diameter.

Exercise 16D — Three-dimensional geometry 1 and 2 Answers will vary. 3 Faces = ABCDEF, GHJKLM, ABGM, EFHJ, CDKL, AFHG, DEJK, BCLM. Edges = AB, BC, CD, DE, EF, FA, GM, ML, LK, KJ, JH, HG, AG, FH, EJ, DK, CL, BM. Vertices = A, B, C, D, E, F, G, H, J, K, L, M. V = 12, E = 18, F = 8; 12 = 18 - 8 + 2 4 D 5 D 6 V = 5, E = 8, F = 5 7 C 8

9

10

G H J K H G M LK J H

F A B

CD E F ED A F

11 Square 12 Construction 13 2a 14 8 1 15 32 Exercise 16E — Circle geometry 1 x = 44°, y = 44°, z = 44° 2 C 3 74° 4 B 5 x = 46°, y = 34°, z = 46°

6 AED = BEC (opposite angles) ADE = BCE (Circle theorem 2) Thus, similar triangles, so x = z. 7 B 8 90 9 x = 30°, y = 60° 10 x = 126°, y = 96° 11 a Theorem 3, 90º b Theorem 1, 25.5° c Theorem 4, 99º d Theorem 2, 51° e Theorem 1, 58° x 13 x = y = z = 20° 12 y = 180° - 2 14 Answers will vary. 15 x = y = z = 42° 16 Answers will vary. 17 a + b + c = 180° — sum of angles in a triangle c + d = 180° — straight angle a + b + c = c + d — equating the two equations a + b = d QED 18 y + ECO = EBC — isosceles triangle BEC + EBC + y = 180° — sum of angles in triangle EBC BEC + y + ECO + y = 180° — substitution 2y + ECO = 180° - BEC — rearranging x + OAE + AEO = 180° — sum of angles in triangle OAE AEO = BEC — opposite angles x + OAE = 180° - BEC — rearranging 2y + ECO = x + OAE — equating two equations involving 180° - BEC ECO = OAE — isosceles triangle 2y = x QED Exercise 16F — Tangents, chords and circles 1 Construction 2 x = 42°, y = 132° 3 MAC, NAC, FDA, FBA, EDG, EBG 4 B 5 D 6 x = 42°, y = 62° 7 Answers will vary. 8 60 9 x = 180° - a - b 10 x = 80°, y = 20°, z = 80° 11 Answers will vary. 12 x = 85°, y = 20°, z = 85° 13 a ABE, DCE and ADE, BCE b 4.17, 5.83 14 a CBD, CEA b 85°, 70° c 7.36 cm 15 D 16 S

P

R Q

a b-a

c

e b = 8 9 10

b

b a

d b2 - ab - a2 = 0

a ± 5a 2

f

1+ 5 , approximately 1.618 2

Long side = 6.472 b Short side = 2.472 Long side = 18.170 d Short side = 309.333 Equilateral triangle, square 3, 3, 3, 4, 4 and 3, 3, 3, 3, 3, 3; irregular

a c a b

Chapter review Short answer 1 a 97° b y = 177° - 2x 2 Make a 60° angle, bisect it, then bisect one of the smaller (30°) angles. 3 Make a straight angle, then a 60° angle, then bisect the 60° angle. Its supplement is 150°. 4 a

b Angle subtended by each side at centre = 36° 5 There are 10 differently shaped triangles.

Five different triangles

Three different triangles

6 Name

Verification of Faces Edges Vertices Euler’s formula

Tetrahedron

4

6

4

4=6-4+2

Cube

6

12

8

8 = 12 - 6 + 2

Octahedron

8

12

6

6 = 12 - 8 + 2

Dodecahedron

12

30

20

20 = 30 - 12 + 2

Isosahedron

20

30

12

12 = 30 - 20 + 2

7 a

b

Answers 16B ➜ 16G

7 x = 50°, y = 95 18 B 19 C 1 20 x = 33°, y = 55°, z = 22° 21 x = 25°, y = 65°, z = 40° 22 x = a, y = 90° - a, z = 90° - 2a 23 18.92° 24 48.87 cm Exercise 16G — Geometry in architecture, design and art 1 2 C 3 D

7 a b - a

4 Figure a, tall adult; figure b, short child 5 Construction 6 Approximately 1.6 to 1

Answers

671

8 a x = 15°, y = 75°

b x = 30°, y = 92°, z = 90°

9 x = 28, y = 15.7 10 24° 11 a, e 12 (x - 2)2 + (y + 3)2 = 9 Multiple choice 1 B 2 B 3 D 6 B 7 B 8 E Extended response 1 b Diameter 2r - y x d = or an equivalent ratio x y

Exam practice 4 Short answer

→

1 a MR = 23 p 4 E 9 B

5 C 10 D

c 2r - y e

x2 + y2 2y

2 43.5 cm 3 a ∆ABC and ∆DBC are similar triangles as they both contain a right angle and share ∠C. Therefore the third angle is the same. b 25 m c 6.72 m d i 1.96 m ii 23.04 m e Drivers must take into account that the height and width of the tunnel space will have decreased and therefore larger trucks may not be able to pass through safely. f 5.91 m g Yes, the maximum height is 5.64 m. 4 Constructions. Check with your teacher. 5 z = 16.5° 6 x = 10.4 y = 5.1 7 x = 86° (alternate segment) y = 34° (alternate segment) w = 60° z = 52° 8 a = 90° (tangent to circle) b = 32° (complement) c = 32° (isosceles triangle) d = 60° (equilateral triangle) e = 184° (angle at centre) 9 x = 33° 10 The distance between the rollers is x = 114.3 cm. 11 a Diameter of tunnel is 13 m. b B is 4.62 m above the floor of the tunnel.

672

Answers

c D is 1.92 m from C. d Maximum height of truck is 3.55 m. 12 Answers will vary.

→

b MN = r + 16 p 2 a a = -4m/s2 b Constant speed/ velocity c i A = 5b + (a - 10) b + 30b(20 - a) + 450 9095 ii b = (595 - 29a) 3 a v = i + 3 j    b i √3 j

c i N 30° W ii 2 km/h Multiple choice 1 B 2 C 3 E 4 D 5 A 6 D Extended response 17 1 a = 10 b i t = 3 seconds ii a = -6.79 m/s2 c ν km/min Rhonda and Yorak

19 12

1 2

d e f

5

t min

51 seconds Time to pass 54.62 minutes Area under graph = 204. Let time taken be T. 129.092 min 12.09 min

Index acceleration instantaneous 540–41 variable 538–546 acceleration–time graphs 525–28 change in velocity 526 gradient 525 signed area calculation 526 acceleration, constant average velocity 532 formulas 532–35 straight-line motion 533 algebra algebraic fractions 122–25 applications 114–19 Boolean 317, 319–20 degree of polynomial 272 deMorgan’s Laws 320–23 improper fractions 281 index laws 88–91 linear literal equations 126–27 partial fractions 276–82 polynomial identities 272–75 proper fractions 276–81 significant figures 93–4 simultaneous equations 109–11, 283–87 standard form 92–3 transposition 96–8 algebraic fractions addition 122–23 division 124 multiplication 124 subtraction 122–23 alternate segment theorem 585–86 angles alternate 558 bisecting 564–66 co-interior 558 complementary 164 constructing 566 corresponding 558 depression 143–44 elevation 143 geometry 556–57 loci/locus 559 parallel lines and 558–59 polygons, in 559 subtended by chords in circles 579 vertically opposite 558 arc length 167–68 major and minor 579 area sector 168–69 segment 169 triangle 159–62 Argand diagram representation complex numbers 29–31

arguments 301–02 affirming the consequent 306 antecedent statement 300 categorical propositions 302–04 conclusion 301 conclusion indicators 302 constructive dilemma 305 contradiction 312–13 contrapositive 301, 312 converse 301 deductive 302–04 implication 300–01 indirect proof 313 inverse 301 mathematical induction 313–14 modus tollens 305 premise 301 syllogism 305 techniques of proof 310–14 valid 304–06, 310 arithmetic sequences 186–90 series 192–94 axes dilations from 53–5 bearings 144–46 Boolean algebra 317, 319–20 deMorgan’s Laws 320–23 table of laws 319 centre of vision 593 centroid 569 chords 579 circles, in 584–89 major and minor segment 579 circle alternate segment theorem 585–86 angles subtended by chords 579 arc 579 chords 579, 586–89 circumference 579 conic form 340 diameter 579 geometry 579–82 graphs 340–42 relations 45 tangents 579, 586–89 circle theorems 579–81 circumcentre 568 closed half-plane 391 collinear points 443 complementary angles 164 complex numbers 367–68 addition 23 Argand diagram representation 29–31 conjugate 25–6

Index

673

complex numbers (contd.) division 26–7 equality 22 multiplication 24–5 multiplication by a real constant 23 notation 22 polar form 367–71 quadratic equations, solving 31–2, 34–5 quadratic expressions, factorising 31–34 set 22–3 subtraction 23 components of vectors 472–73 conics circle 340–42 ellipse 343–46 hyperbola 351–55 parabola 347–50 conjugate surds 16 connected bodies in equilibrium 502–05 connectives 295–97 negation 296–97 constant acceleration formulas 532–35 average velocity 532 straight-line motion 533 constant of proportionality 219, 236 constant velocity 515–16 constructive dilemma 305 convergent sequence 183 coordinate geometry collinear points 443 distance between two points 429–31 division of line segment in ratio a:b 435–41 equation of straight line 447–48 horizontal line 448 line segment midpoint 432–34 midpoint formula 432–33 parallel line 442–44 perpendicular line 445–46 vertical line 448 x = a 448 y = c 448 y = mx + c 447 y - y1 = m(x - x1) 448 corner-point method 405–08 cosine rule 154–58 cube 577 data transformation 259–61 deductive arguments 302–04 deMorgan’s Laws 320–23 depression, angle of 143–44 digital logic digital truth values 325–26 gates 326–28 digital truth values 325–26 dilations from axes 53–5 direct variation 218–24 variables raised to powers other than 1 227–31 direction of vectors 471–72 displacement moving particle, of 513–14 divergent sequence 183 dodecahedron 577

674

Index

elevation, angle of 143 ellipse 56 conic form 343 graphs 343–46 equations linear 101–06 linear literal 126–27 logistic equation 182–84 polar 360–62 quadratic and complex numbers 34–5 simultaneous 106–11, 283–87 simultaneous linear 394–98 straight-line 442, 447–48 equilibrium 498–501 connected bodies 502–05 equivalent statements 297 Euler’s formula 576 exact values 139–40 final image rule successive transformations 62 force equilibrium 498–501 inertia 492 mass 492 resolving into components 493–94 tension 493 weight 492 fractions algebraic 122–25 improper 281 non-signed numbers, as 3 partial 276–82 proper 276–81 functional definition 179 functions 74–7 inverse 77–80 linear 402–08 position as a function of time 516–18 gates logic 326 table of NAND and NOR gates 326–27 geometric constructions 562–66 mean 196 proofs 483 sequences 195–99 series 200–04 sum to infinity S ∞ of geometric sequence 203 geometric constructions bisecting angles 564–66 bisecting lines 562–64 centroid 569 circumcentre 569 constructing angles 566 incentre 569 geometric mean 196 geometric proofs 483 geometric series infinite sum of sequence where r < 1 203–04 geometry alternate segment theorem 585–86 angles 556–57, 558–59

architecture, design and art, in 592–97 chords 584–89 circle 579–82, 584–89 Golden Ratio 594–95 lines 556–558 parallel lines 558–59 perspective 592–94 tangents 584–89 tessellations 595–97 three-dimensional 576–78 two-dimensional 556–74 Golden Ratio 594–95 Golden Rectangles 594 graphs acceleration–time 525–28 circle 340–42 ellipse 343–46 hyperbola 351–55 linear 340–82 linear inequations 390–93 non-linear 340–82 parabola 347–50 polar 363–65 position–time graph 538 reflections 50–2 relations 44–5, 71–2, 74 simultaneous equations 106–09 simultaneous linear equations 106–09 simultaneous linear inequations 394–98 systems of linear inequations 399–401 translations 44–8 velocity–time 521–25, 541–45 Heron’s formula 161 horizon 593 hyperbola conic form 351 graph 351–55 transformations 56–7, 58–60 icosahedron 577 i, j notation 475–77 angles 476–77 unit vectors 475 improper fractions 281 incentre 569 index laws fractional powers 89 negative power 89 products 89 quotient 89 raising to a power 88 infinite sum of geometric sequence where r < 1 203–04 instantaneous acceleration 540–41 instantaneous rates of change 538–46 acceleration 540–41 approximating velocity–time graphs 541–45 position–time graph 538 unsigned area calculation 523–24 velocity–time 521–25, 541–45 instantaneous velocity 514, 538–40 integers signed numbers, as 3 inverse functions 77–80

inverse relations 78 inverse variation 235–40 variables raised to powers other than 1 244–46 irrational numbers 4–5 irreducible quadratic 279 iterative definition 179–80 joint variation 249–52 kinematics 511–18 acceleration–time graphs 525–28 constant acceleration formulas 532–35 constant velocity 515–16 curvilinear motion 512 displacement 513–14 instantaneous acceleration 540–41 instantaneous rates of change 538–46 instantaneous velocity 514, 538–40 position 513 position as a function of time 516–18 position–time graph 538 rectilinear motion 512 speed 514–15 straight-line motion 512, 533 velocity 514 velocity–time graphs 521–25, 541–45 line segment 556 external division in ratio a:b 439–41 internal division in ratio a:b 435–37 midpoint 432–34 linear equations 101–06 literal 126–27 simultaneous 106–11 linear functions corner-point method 405–08 maximising and minimising 402–08 sliding-line method 402–05 linear graphs 340–82 linear inequations graphs 390–93 simultaneous, graphs of 394–98 systems of, graphs of 399–401 linear programming 389–90 applications 415–17 closed half-plane 391 corner-point method 405–08 graphs of linear inequations 390–93 maximising and minimising linear functions 402–08 open half-plane 391 problem solving methods 409–11 sliding-line method 402–05 lines bisecting 562–64 coincidental 108 horizontal 448 parallel 108–09, 442–44, 558–59 perpendicular 445–46 vertical 448 x = a 448 y = c 448 y = mx + c 447 y - y1 = m(x - x1) 448 loci/locus 559

Index

675

logic arguments 300–06 circuits 328–30 digital 325–31 gates 326–28 proof 310–14 logic circuits burglar alarms 328–29 simplification 329–30 logistic equation 182–84 magnitude of vectors 471 mathematical induction 313–14 mathematical proof methods contradiction 312–13 contrapositive 312 counter-example 313–14 mathematical induction 313–14 valid argument forms 311–12 midpoint formula 432–33 modus tollens 305 motion kinematics 512 negation 296–97 nets of polyhedrons 576 Newton’s first law of motion 495–97 tensile strength 495 non-linear graphs 340–82 notation interval 11 set builder notation 11 standard/scientific form 7–9 number systems complex numbers 24–35 division by zero 4 fractions 3 integers 3 irrational numbers 4–5 natural numbers 3 non-signed numbers 3 notations for subsets of set of real numbers 11–3 rational numbers 3–4 real numbers 5–7 set of complex numbers 22–3 set of real numbers 3–7 signed numbers 3–4 subsets of set of real numbers 10–14 surds 14–21 octahedron 577 open half-plane 391 ordinates addition 372–75 oscillating sequence 183 parabola conic form 347 graph 347–50 relations 44 parallel lines 442–44 angles and 558–59 collinear points 443

676

Index

parallelogram quadrilateral form 571 part linear variation 255 part variation 255–57 partial fractions improper fractions 281 proper fractions 276–81 particle statics connected bodies in equilibrium 502–05 equilibrium 498–501 force 492–94 Newton’s first law of motion 495–97 tensile strength 495 tension 493 perpendicular line bisector 563 line 445–46 perspective 592–94 centre of vision 593 horizon 593 vanishing point 593 Platonic solids 577 points reflections 50–2 translations 44–8 polar coordinates 357–59 division in polar form 370–71 equations 360–63 graphs 363–65 polar form complex numbers 368–69 multiplication in polar form 370–71 polygons angles 559 parallelogram 571 quadrilaterals 571–72 rectangle 571 regular 572–73 rhombus 571 square 571 star 573–74 trapezium 571 triangle construction 567–70 polyhedrons Euler’s formula 576 nets 576 polynomial identities 272–75 degree of polynomial 272 identical 273 position function of time, as 516–18 moving particle 513 position–time graph 538 premise of argument 301 indicators 302 proof techniques contradiction 312–13 contrapositive 312 counter-example 313 tautology 310–11 valid argument forms 311–12

proper fractions 276–81 irreducible quadratic 279 Pythagoras’ theorem 137–38 Pythagorean identity 163–64 quadrilaterals 571–72 radian measurement 165–66 rational numbers signed numbers, as 3–4 real number systems notation forms 11–3 subsets of the set of real numbers 10–3 real numbers 5–7 reciprocals sketching in simple graphs 375–76 rectangle quadrilateral form 571 relations 71–2 circle 45 domain 71 exponential 45 inverse/hyperbolas 45, 78 linear 44 quadratic/parabolas 44 range 71 rhombus quadrilateral form 571 right-angled triangles exact values 139–40 Pythagoras’ theorem 137–39 trigonometry 136–40 scalar multiplication vectors 468–69 sectors area 168–69 segments alternate theorem 585–86 area 169 major and minor 579 sequences 178–84 applications 206–08 arithmetic 186–90 convergent 183 divergent 183 functional definition 179 geometric 195–99 infinite sum of geometric sequence where r < 1 203–04 iterative definition 179–80 limit 183 logistic equation 182–84 oscillating 183 sequential order 179 sum to infinity S∞ of geometric sequence 203 sequential order 179 series applications 206–08 arithmetic 192–94 common difference 193 geometric 200–04 set notation 1–9 non-signed numbers 3 signed numbers 3–4 standard/scientific form notation 7–9

sets 317 intersection 317 negation 318 real numbers 3–7 table of laws 319 table of rules 318 union 317 significant figures 7–8, 93–4 calculations 93–4 simultaneous equations 106–11, 283–87 algebraic solution 109–11 elimination 110–11 graphs 106–09 substitution 109 simultaneous linear equations coincidental lines 108 graphical solution 106–09 parallel lines 108–09 simultaneous linear inequations graphs 394–98 sine rule 148–52 ambiguous case 149–52 sliding-line method 402–05 speed 514–15 square quadrilateral form 571 sketching in simple graphs 376–78 standard form 92–3 decimal places 7–8 set notation 7–9 significant figures 7–8 statements/propositions 293–94 antecedent 300 conditional 300 consequent 300 equivalent 297 implication 300–01 statics 482 surds addition 15 definition of conjugate 16 division 18–9 multiplication 16–7 simplifying 14–5 subtraction 15 syllogistic argument disjunctive 305 hypothetical 305 systems of linear inequations graphs 399–401 tangents 579 circles, in 584–89 tautologies 310 proof of arguments 310–11 techniques of proof tautology 310–11 tensile strength 495 tension 493 tessellations 595–97 irregular 596 regular 595 rules for creation 597 semi regular 596

Index

677

tetrahedron 577 three-dimensional geometry nets of polyhedrons 576 platonic solids 577 transformation of data 259–61 transformations dilations from axes 53–5 ellipse 56, 57 final image rule 62 hyperbola 56–7, 58–60 reflections of graphs and points 50–2 relations 44–5 successive 62 translations of graphs 44–8 translations of points 45–6 translations graphs 44–5 points 45–6 rules 46–8 transposition of formulas 96–8 trapezium quadrilateral form 571 triangles area 159–62 geometric construction 567–70 Heron’s formula 161 right-angled 137–40 trigonometric identities 163–65 complementary angles 164 Pythagorean identity 163–64 trigonometric ratios ambiguous case 149–52 angle of depression 143–44 angle of elevation 143 arc length 167 area of sector 168–69 area of segment 169 area of triangle 159–62 bearings 144–46 cosine rule 154–58 Heron’s formula 161 identities 163–65 radian measurement 165–66 sine rule 148–52 trigonometry of right-angled triangles 136–40

678

Index

truth tables 295–97 truth value 295 unit vectors 475 valid arguments 310 proving 305–06 vanishing point 593 variable acceleration 538–546 variation constant of proportionality 219, 236 direct 218–24, 227–31 inverse 235–40, 244–46 joint 249–52 part 255–57 part linear variation 255 transformation of data 259–61 variables raised to powers other than 1 227–31, 244–46 vectors addition 466 applications 480–83 components 472–73 directed line segments 462 direction 471–72 equality 464 geometric proofs 483 i, j notation 475–77 magnitude 471 negative 467 operations 466–70 ordered pairs, as 463 position 463 scalar multiplication 468–69 statics 482 subtraction 467 unit 475 velocity 514 velocity constant 515–16 instantaneous 514, 538–40 moving particle 514 position–time graph 538 velocity–time graphs 521–25 approximating 541–45 gradient 523 signed area calculation 523–24 unsigned area calculation 541

Maths Quest 11 Advanced General Mathematics (Spec) Classpad Edition

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