Maths Quest Specialist 12 Textbook 4E TI-Nspire CAS Companion

MATHS QUEST 12

Specialist Mathematics

4TH EDITION

TI-NSPIRE C AS C ALCULATOR COMPANION

VCE M AT H EM AT I CS U N I T S 3 & 4

MATHS QUEST 12

Specialist Mathematics RAYMOND ROZEN | PAULINE HOLLAND | BRIAN HODGSON HOWARD LISTON | JENNIFER NOLAN | GEOFF PHILLIPS

4TH EDITION

TI-NSPIRE C AS C ALCULATOR COMPANION

First published 2013 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10/12 pt Times LT Std © John Wiley & Sons Australia, Ltd 2013 The moral rights of the authors have been asserted. ISBN: 978 1 118 31811 9 978 1 118 31809 6 (flexisaver) 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. Cover and internal design images: © vic&dd/Shutterstock.com Typeset in India by Aptara Illustrated by Aptara and Wiley Composition Services Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1 Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their permission to reproduce copyright material in this book. Images Texas Instruments: Screenshots from TI-Nspire reproduced with permission of Texas Instruments Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to rectify any error or omission in subsequent editions will be welcome. In such cases, please contact the Permissions Section of John Wiley & Sons Australia, Ltd.

Contents Introduction

vi

CHAPTER 6

Integral calculus

CHAPTER 1

Coordinate geometry

1

15 21

Relations and regions of the complex plane 27 CHAPTER 5

29

53

CHAPTER 9

Vectors

CHAPTER 4

Differential calculus

CHAPTER 8

Kinematics

CHAPTER 3

Complex numbers

CHAPTER 7

Differential equations

CHAPTER 2

Circular functions

37

63

CHAPTER 10

Vector calculus CHAPTER 11

Mechanics

79

71

47

Introduction This booklet is designed as a companion to Maths Quest 12 Specialist Mathematics Fourth Edition. It contains worked examples from the student text that have been re-worked using the TI-Nspire CX CAS calculator with Operating System v3. The content of this booklet will be updated online as new operating systems are released by Texas Instruments. The companion is designed to assist students and teachers in making decisions about the judicious use of CAS technology in answering mathematical questions. The calculator companion booklet is also available as a PDF file on the eBookPLUS under the preliminary section of Maths Quest 12 Specialist Mathematics Fourth Edition.

vi

Introduction

Chapter 1

Coordinate geometry Worked example 4

Sketch the graph of y =

x2 + 2 including all asymptotes and intercepts. 3x

think 1

On a Graphs page, complete the entry line as: x2 + 2 f 1( x ) = 3x then press ENTER ·.

2

To determine the equation of the oblique asymptote,

Write/display

x2 + 2 into two functions. 3x To do this on a Calculator screen press: • MENU b • 2: Number 2 • 7: Fraction Tools 7 • 1: Proper Fraction 1 Complete the entry line as: propfrac(f1(x)) then press ENTER ·.

divide y =

3

Write the equations of the asymptotes.

The equations of the asymptotes are: 1 x = 0 and y = x . 3

Chapter 1 • Coordinate geometry

1

4

To determine the turning points, press: • MENU b • 3: Algebra 3 • 1: Solve 1 • MENU b • 4: Calculus 4 • 1: Derivative 1 Complete the entry line as:  d  solve  ( f 1( x )) = 0, x   dx  then press ENTER ·.

5

Solving

To find the y-coordinates of the stationary points by substitution, complete the entry lines as:

x=

f 1( 2 )

f 1( 2 ) Press ENTER · after each entry. Describe the nature and coordinates of the stationary points, as deduced from the graph. Sketch the graph of y =

−

2 or x = 2.

The coordinates of the stationary points are:  2 2 Local minimum  2,   3 

−

6

d  x2 + 2   = 0 for x gives dx  3 x 

− −2 2 Local maximum  2, 3  

x2 + 2. 3x

y 10

x = 0 (Asymptote)

5

−10

−5

0 −5 −10

2

Maths Quest 12 Specialist Mathematics

5

1 y = – x (Asymptote) 3 x 10

Worked example 12

Sketch the graph of

( x − 1) 2 ( y − 2) 2 + = 1. 25 9

think 1

Write/display

Compare

( x − 1)2 ( y − 2)2 + = 1 with 25 9

( x − h) 2 ( y − k ) 2 + = 1. a2 b2

h = 1, k = 2 and so the centre is (1, 2). a2 = 25 b2 = 9 a=5 b=3

2

The major axis is parallel to the x-axis as a > b.

3

The extreme points (vertices) parallel to the x-axis for the ellipse are: (−a + h, k) (a + h, k)

Vertices are: (−5 + 1, 2) = (−4, 2)

(5 + 1, 2) = (6, 2)

The extreme points (vertices) parallel to the y-axis for the ellipse are: (h, −b + k) (h, b + k)

and (1, −3 + 2) = (1, −1)

(1, 3 + 2) = (1, 5)

4

5

Find the x- and y-intercepts. On a Calculator page, complete the entry lines as:  ( x − 1)2 ( y − 2)2  + = 1, x  y = 0 solve   25  9  ( x − 1)2 ( y − 2)2  + = 1, y  x = 0 solve   25  9 Press ENTER · after each entry.

The x-intercepts are: 3−5 5 3+5 5 , x = 3 3 The y-intercepts are: x=

y= 6

10 − 6 6 10 + 6 6 , y = 5 5

To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 4: Ellipse 4

( x − h )2 ( y − k )2

+ =11 a2 b2 Complete as shown: ( x − 1)2 ( y − 2 )2 + =1 52 32 Press ENTER · after the entry, the graph is shown.

• 1:


3

7

Sketch the graph of the ellipse.

+ 6√6 0, 10 ––––––– 5

y 6 4

3 − 5√5 , –––––– 3

−6

0

(−4, 2)

−4 5

4


(1, 2)

0 −2

(y − 2)

2

+ –––––– =1 9

(1, 5)

2

−2

− 6√6 0, 10 –––––––

(x − 1)2 –––––– 25

2

(6, 2)

4

6 x

(1, −1) 3 + 5√5 , –––––– 3

0

Worked example 13

Sketch the graph of

( x − 2) 2 ( y + 4) 2 + = 1. 9 16

think

Write/display

( x − 2)2 ( y + 4)2 + = 1 with 9 16 ( x − h) 2 ( y − k ) 2 + = 1. a2 b2

h = 2, k = −4 So the centre is (2, −4). a2 = 9 b2 = 16 a=3 b=4

1

Compare

2

The major axis is parallel to the y-axis as b > a.

3


Vertices are: (−3 + 2, −4) = (−1, −4)

4


and (2, −4 − 4) = (2, −8)

5

(3 + 2, −4) = (5, −4) (2, 4 − 4) = (2, 0)

Find the x- and y-intercepts. On a Calculator page, complete the entry lines as:  ( x − 2)2 ( y + 4)2  solve  + = 1, x  y = 0  9  16  ( x − 2)2 ( y + 4)2  + = 1, y  x = 0 solve   9  16 Press ENTER · after each entry.

The x-intercept is x = 2. The y-intercepts are: y= 6

− 12 − 4

3

5

,

y=

−

12 + 5 . 3

To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 4: Ellipse 4 • 1: Complete the entry line as:

( x − 2 )2

2 y − − 4) ( + =1

32 42 then press ENTER ·. Note that the viewing window has been changed.


5

7

y


2 −12 + 4√5

0, –––––––– 3

−1 −2 −4

−12 − 4√5 0, –––––––– 3

6


−6

0 (2, 0) 123456 x (−1, −4) (2, −4)

−8 (2, −8) −10

(5, −4) (x − 2)2 –––––– 9

2

(y + 4)

+ –––––– =1 16

Worked example 14

Sketch the graph of 5 x2 + 9(y − 2)2 = 45. think

Write/display

5x2 + 9(y − 2)2 = 45

1

Rearrange and simplify by dividing both sides by 45 to make the RHS = 1.

2

Simplify by cancelling.

3

Compare

4

Major axis is parallel to the x-axis as a > b.

5


Vertices are: (−3 + 0, 2) = (−3, 2)


and (0, 5 + 2) or (0, 2 − 5 )

6

7

x 2 ( y − 2)2 + = 1 with 9 5 ( x − h) 2 ( y − k ) 2 + = 1. a2 b2

5 x 2 9( y − 2)2 45 + = 45 45 45 2 2 x ( y − 2) + =1 9 5 h = 0, k = 2 and so the centre is (0, 2). b2 = 5 as a, b > 0 a2 = 9 a=3 b= 5

(3 + 0, 2) = (3, 2)

−

(0, 5 + 2) (0, 2 + 5 )

≈ (0, −0.24)

≈ (0, 4.24)

Find the x-intercepts. On a Calculator page, complete the entry line as: solve(5 x 2 + 9( y − 2)2 = 45, x ) | y = 0 then press ENTER ·.

x= 8

−3

5

5

,

x=

3 5 5

To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 6: Conic 6 • 1: Complete the entry line as: 5 x 2 + 0 xy + 9 y 2 + 0 x + − 36 y + − 9 = 0 then press ENTER ·.


7

9


y 6 (0, 2 + 5 ) 5x2 + 9(y − 2)2 = 45 4 2 (0, 2)

(−3, 2) −4 −3 −2 −3√5 –––– , 5

8


−1 0

(3, 2)

0 1

2

−2 (0, 2 − 5 )

3

4x

3√5 –––– , 5

0

Worked Example 15

Sketch the graph of the relation described by the rule: 25x2 + 150x + 4y2 − 8y + 129 = 0. Think 1

Write/Display

To locate the intercepts, on a Calculator page, complete the entry lines as: solve ( 25 x 2 + 150 x + 4 y 2 − 8 y + 129 = 0, x ) |y=0 solve ( 25 x 2 + 150 x + 4 y 2 − 8 y + 129 = 0, y ) |x=0 Make a record of the intercepts.

2

To sketch the graph of the ellipse, on a Graphs page press: • MENU b • 3: Graph Entry/Edit 3 • 2: Equation 2 • 6: Conic 6 • 1: Complete the entry line as: 25 x 2 + 0 xy + 4 y 2 + 0 x + − 8 y + 129 = 0 then press ENTER ·.

3

Write the x-intercepts.

x= x=

4

− 15 − 4

6

5 − 15

+4 6 5


y (−3, 6)

(−5, 1) −15 − 4√6 , –––––––– 5

0

−5

(−3, 1)

6 (−1, 1)

−3 −1 0 −4 (−3, −4)

−15 + 4√6 , –––––––– 5

x 0

Chapter 1 • Coordinate geometry 9

Worked Example 16

Determine the Cartesian equation of the curve with parametric equations x = 2 + 3 sin (t) and y = 1 − 2 cos (t ) where t ∈ R. Describe the graph and state its domain and range. Think

Write/Display

1

Use a CAS calculator to sketch the graph in a Graphs page, in parametric mode, by completing the entry line as:  x1(t ) = 2 + 3 sin (t )  y1(t ) = 1 − 2 cos (t )  Then press ENTER ·.

2

Rewrite the parameters by isolating cos (t) and sin (t).

y −1 x−2 = sin (t) and − = cos (t) 2 3

3

Square both sides of each equation then add.

( x − 2)2 ( y − 1)2 = sin2 (t) + cos2 (t) + 9 4 =1

4

Describe the relation.

This represents an ellipse with centre (2, 1).

5

The domain is the range of the parametric equation x = 2 + 3 sin (t).

Domain is [2 − 3, 2 + 3] = [−1, 5]

6

The range is the range of the parametric equation y = 1 – 2 cos (t).

Range is [1 − 2, 1 + 2] = [−1, 3]

10 Maths Quest 12 Specialist Mathematics

Worked Example 25

Express each of the following as partial fractions. a

5 x 2 + 10 x − 52 2 x3 − 5 x2 + 3 x + 7 b ( x − 2)( x + 4) x2 − x − 2

Think

a & b

1

Write/Display

On a Calculator page, press: • MENU b • 3: Algebra 3 • 3: Expand 3 Complete the entry lines as:  5 x 2 + 10 x − 52  expand   ( x − 2) × ( x + 4)   2 x 3 − 5x 2 + 3x + 7  expand    x2 − x − 2 Press ENTER · after each entry.

2

Write the answers.

a

5 x 2 + 10 x − 52 2 2 = − +5 ( x − 2) × ( x + 4) x + 4 x − 2

b

2 x 3 − 5x 2 + 3x + 7 1 3 + 2x − 3 = + x2 − x − 2 x +1 x − 2


Worked Example 27

Sketch the graph of the function y =

x 2 − 5x + 6 . x−4

Think 1

Use a CAS calculator to express the rational function as partial fractions by completing the following steps. Press: • MENU b • 2: Number 2 • 7: Fraction Tools 7 • 1: Proper Fraction 1 Complete the entry line as:  x 2 − 5x + 6  propFrac   x − 4  then press ENTER ·.

2

Express the function as partial fractions.

3

Sketch the graphs of y1 = x − 1 2 (asymptote) and y2 = on the same x − 4 axes.

Write/Display

y=

2 + x −1 x−4 y y1 = x − 1 2 y2 = x—— −4

0

x

1 2 3 4

−1

x=4 4

Determine any x-intercepts.

y = 0, x2 − 5x + 6 = 0 (x − 2)(x − 3) = 0 ⇒ x = 2 and x = 3

5

Determine the y-intercept.

x = 0, y =

6

Add the two graphs by addition of ordinates x 2 − 5x + 6 . to obtain the graph of y = x−4

6 4 −3 y= 2 −

y

(3, 0)

(2, 0)

y1 = x − 1 2 y2 = x—— −4

0 (0,

−1

− 3–2 )

1 2 3 4

2 y = x − 1 + x—— −4

x=4


x

7

Open a Graphs page, and complete the entry lines as; f 1( x ) = x − 1 2 x−4 f 3 ( x ) = f 1( x ) + f 2 ( x ) f 2( x ) =

then press ENTER ·.


ChapTer 2

Circular functions Worked example 2

If cosec (x) = 43 and, 0 ≤ x ≤ 90°, find x (to the nearest tenth of a degree). Think 1

express the equation cosec (x) = in terms of sin (x).

2

On a Calculator page, press: • Menu b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as

WriTe 4 3

cosec ( x ) =

1 4 = sin ( x ) 3

4   1 solve  = , x | 0 ≤ x ≤ 90  sin( x ) 3  Then press enTeR ·. Alternatively, the three reciprocal functions are built into the TI-nspire. They can be accessed by the µ key, or through the catalogue, or you can simply use the letter keys and enter csc, sec or cot as needed. 3

Write the solution.

Solving cosec (x) = 43 for x ∈[0,90°], x = 48.5904°

4

Round off the answer to 1 decimal place.

x = 48.6°

ChapTer 2 • Circular functions

15

Worked Example 12

Solve cosec (x) = 1.8 over the interval 0 ≤ x ≤ 4π. Give your answer(s) correct to 2 decimal places. Think

Write

1

On a Calculator page, press: • Menu b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as solve (csc(x) = 1.8, x) | 0 ≤ x ≤ 4π Then press ENTER ·.

2

Write the solution.

Solving cosec (x) = 1.8 over the interval 0 ≤ x ≤ 4π gives

3

Round the answers to 2 decimal places.

x = 0.59, 2.55, 6.87, 8.84


Worked Example 13

a Expand, and simplify where possible, each of the following. i sin (x − 2y) ii cos (x + 30°) b Simplify the expression sin (2x) cos (y) + cos (2x) sin (y). Think

a

i On a Calculator page, press: & • MENU b ii • 3: Algebra 3

Write

a

i & ii

• B: Trigonometry B • 1: Expand 1 Complete then entry lines as shown, then press ENTER ·.

b

1

Write the appropriate compoundangle formula.

2

Substitute A = 2x and B = y to reveal the answer.

3

Alternatively, on a Calculator page, press: • Menu b • 3: Algebra 3 • B: Trigonometry B • 2: Collect 2 Complete the entry line as tCollect(sin(2x)cos(y) + cos(2x)sin(y)) Then press ENTER ·.

Write the solution.

b

sin (A) cos (B) + cos (A) sin (B) = sin (A + B) sin (2x) cos (y) + cos (2x) sin (y) = sin (2x + y)

sin (2x) cos (y) + cos (2x) sin (y) = sin (2x + y)

Chapter 2 • Circular functions 17

Worked Example 15

Simplify: a sin (270 − C °) b sec

 π −θ . 2 

Think

a & b

1

On a Calculator page, press: • Menu b • 3: Algebra 3 • 3: Expand 3 Complete the entry lines as: expand(sin (270 − c), c)   π − θ  , θ expand  sec 2   

Write

a & b

Press ENTER · after each entry. Note: The calculator should be in degree mode for the first expansion above, and radian mode for the second. 2

Express the answer as a reciprocal function.


π sec  − θ  = cosec(θ ) 2 

Worked example 16

5π  Find the exact value of cot  .  12  Think

WriTe

express

2

express cot in terms of its reciprocal,

3

use the appropriate compound-angle formula to expand the denominator.

4

5

5π  π π cot   = cot  +   4 6  12 

π π 5π as the sum of and . 6 4 12

1

express in simplest fraction form.

Simplify.

1 . tan

=

=

=

=

=

=

=

1 π π  tan +  4 6 1 π  tan   + tan  π    4  6    π π   1 − tan   tan      4  6 

π π 1 − tan   tan    4  6 π π tan   + tan    4  6  1  3  1

1 − (1)  1+ 1− 1+

3

1 3 1 3

 3 −1  3   3 +1  3 

3 −1 3 +1

6

Rationalise the denominator.

=

( 3 − 1) ( 3 − 1) ( 3 + 1) ( 3 − 1)

7

Simplify.

=

3− 2 3 +1 3−1

4−2 3 2 = 2− 3 =

ChapTer 2 • Circular functions

19

Note: It is possible to check the answer using a calculator. On a Calculator page, complete the entry line as: 5π cot    12  then press enTeR ·.

20


ChapTer 3

Complex numbers Worked example 1

Using the imaginary number i, write down an expression for: −16

a

b

−5

.

Think

a & b

WriTe

1

Change the document settings to Rectangular mode. To do this, press: • HOME c • 5: Settings 5 • 2: Settings 2 • 2: Document Settings 2 Tab down to Real or Complex and select Rectangular.

2

On a Calculator page, complete the entry lines as: −

a & b

16

−

5 Press ENTER · after each entry.

ChapTer 3 • Complex numbers

21

Worked Example 4

Simplify z = i4 − 2i2 + 1 and w = i6 − 3i4 + 3i2 − 1. Think

Write

1

On a Calculator page, complete the entry lines as: i4 − 2i2 + 1 i6 − 3i4 + 3i2 − 1 Press ENTER · after each entry.

2

Write the answer.

z=4 w = −8

Worked Example 5

Evaluate each of the following.  1 − 3 i − i 2 − i3  a Re(7 + 6i) b Im(10) c Re(2 + i − 3i3) d Im    2 Think

Write

a , b , 1 On a Calculator page, press: • Menu b c & • 2: Number 2 d

• 9: Complex Number Tools 9 • 2: Real Part 2 or • 3: Imaginary Part 3 Complete the entry lines as: real(7 + 6i) imag(10) real(2 + i − 3i2)  1 − 3i − i 2 − i 3  imag    2 Press ENTER · after each entry. 2

Write the answers.

a Re(7 + 6i) = 7 b Im(10) = 0 c Re(2 + i − 3i3) = 2

 1 − 3i − i 2 − i3  −  = 1  2

d Im 


Worked Example 10

Determine Re(z2w) + Im(zw2) for z = 4 + i and w = 3 − i. Think 1

On a Calculator page, complete the entry lines as: Define z = 4 + i Define w = 3 – i Press ENTER · after each entry. Then press: • Menu b • 2: Number 2 • 9: Complex Number Tools 9 • 2: Real Part 2 Complete the entry line as: real(z2 × w) + imag(z × w2) Then press ENTER ·. Note: The imaginary part can be found in the same menu as the real part.

2

Write the answer.

Write

Re(z2 w) + Im(zw2) = 37

Worked Example 12

Write down the conjugate of each of the following complex numbers. a 8 + 5i b −2 − 3i Think

a & b

1

On a Calculator page, press: • Menu b • 2: Number 2 • 9: Complex Number Tools 9 • 1: Complex Conjugate 1 Complete the entry lines as: conj (8 + 5i) conj (−2 − 3i) Press ENTER · after each entry.

2

Write the answers.

Write

a 8 − 5i b −2 + 3i

Chapter 3 • Complex numbers 23

Worked Example 16

If z = a + bi, find a and b such that Think 1

5 z − 15 = 4 − 3 i. z−1 Write

On a Calculator page, press: • Menu b • 3: Algebra 3 • C: Complex C • 1: Solve 1 Complete the entry line as:  5z − 15  = 4 − 3i, z  cSolve    z −1 Then press ENTER ·.

2

a is the real part of z, b is the imaginary part.

a = 2, b = −3

Worked Example 17

Find the modulus of the complex number z = 8 − 6i. Think 1

On a Calculator page, press: • Menu b • 2: Number 2 • 9: Complex Number Tools 9 • 5: Magnitude 5 Complete the entry line as: |8 − 6i| Then press ENTER ·.

2

Write the answer.


Write

| z | = |8 − 6i| = 10

Worked Example 23

Express each of the following in polar form, r cis (θ ), where θ = arg(z), −π < θ ≤ π. a z = 1 + i b z = 1 − 3 i Think

a & b

1

On a Calculator page, complete the entry line as: 1+i Then press: • Menu b • 2: Number 2 • 9: Complex Number Tools 9 • 6: Convert to Polar 6 Then press ENTER ·.

2

Write the answer.

3

Use the relationship reiθ = r cos (θ ) + ir sin (θ ) to express the answer in the required form. The calculator always gives θ in principle valued form.

Write

a & b

π For a , 1 + i = 2 cis    4

Key in 1 − 3i and repeat the above procedure. 4

Write the answer.

For b , 1 − 3i = 2e

−i

π 3

−π  = 2 cis   3

Chapter 3 • Complex numbers 25

Worked Example 36

a If f (z) = z3 + 7z2 + 16z + 10, find all factors of f (z) over C. b Factorise P(z) = z3 − (3 −i)z2 + 2z − 6 + 2i. Think

a & b

1

On a Calculator page, press: • Menu b • 3: Algebra 3 • C: Complex C • 2: Factor 2 Complete the entry lines as cFactor (z3 + 7z2 + 16z + 10, z) cFactor (z3 − (3 − i)z2 + 2z − 6 + 2i, z) Press ENTER · after each entry.

2

Write the answers in the required form.


Write

a & b

For a, the three factors of P(z) are (z + 1), (z + 3 − i) and (z + 3 + i) For b, P( z ) = ( z − 3 + i)( z + 2i)( z − 2i)

CHAPTER 4

Relations and regions of the complex plane WORKED EXAMPLE 16

Express each of the following expressions in Cartesian form. a Re(z + 5) b Im(z − 2 − 3i) c | z − 4 + 2i | THINK

a , 1 On a Calculator page, complete the entry lines as: b Define z = x + yi & Then press ENTER ·. c To answer part a press:

2

• MENU b • 2: Number 2 • 9: Complex Number Tools 9 • 2: Real Part 2 or • 3: Imaginary Part 3 or • 5: Magnitude 5 Complete the entry line as: real(z + 5) imag(z − 2 − 3i) | z − 4 + 2i | Press ENTER · after each entry. Write the answers.

WRITE

a, b & c

For a , Re(z + 5) = x + 5. For b , Im(z − 2 − 3i) = y − 3. For c , | z − 4 + 2i | = x 2 − 8 x + y 2 + 4 y + 20 .

CHAPTER 4 • Relations and regions of the complex plane

27

ChapTer 5

Differential calculus Worked example 1

Differentiate the following expressions with respect to x.  4x a y = tan (6 x) b y = 2 tan   3  Think

a & b

1

On a Calculator page, press: • Menu b • 4: Calculus 4 • 1: Derivative 1 Complete the entry lines as: d (tan (6 x )) dx

WriTe

a & b

d  4x   2 tan    dx 3 Press enTeR · after each entry.

2

Write the solutions.

For a,

d 6 [tan (6 x )] = dx [cos (6 x )]2

For b,

d dx

  2 tan

 4x  =    3  3  cos 

8  4x    3  

2

ChapTer 5 • Differential calculus

29

Worked Example 4

Find the equation of the tangent to the curve y = 3x + cos (2x) + tan (x) where x = Think 1

π . 4

Write

On a Calculator page, press: • Menu b • 4: Calculus 4 • 9: Tangent Line 9 Complete the entry lines as:

π tangentLine  3 x + cos (2 x ) + tan ( x ), x ,   4 Then press ENTER ·.

2

Write the solution.


Equation of the tangent is y = 3x + 1

Worked example 6

Find, using calculus, f ″( x) if f ( x) is equal to: a ecos (2x) + loge (x)

b

sin ( x ) . x

Think

a & b

1

On a Calculator page, complete the entry line as: f (x) : = ecos (2x) + ln(x) Then press enTeR ·. Note: The syntax used here is another way of defining a function or variable. You can use the Define or Store methods if you prefer.

2

Complete the entry line as:

WriTe

a & b

d2 ( f ( x )) dx 2 Then press enTeR ·.

3

Write the solution.

The second derivative, f '' ( x ) = [4 sin 2 (2 x ) − 4 cos (2 x )]e cos (2 x ) −

4

1 x2

On a Calculator page, complete the entry line as: sin ( x ) f (x) : = x Complete the entry line as: d2 ( f ( x )) dx 2 Press enTeR · after each entry.

ChapTer 5 • Differential calculus

31

5

Write the solution.

6

You may rearrange the answer to a form similar to that given in the solution obtained manually as follows. Press: • Menu b • 3: Algebra 3 • 2: Factor 2 Then select and paste the previous answer to obtain the entry line:

The second derivative,   cos ( x ) 3 1  sin ( x ) − f ′′( x ) =  5 − 3 x  2  x2 4x

   3 1 cos ( x )   sin ( x ) − factor   5 − 3 x      4x 2 x2  Then press enTeR ·. 7

Write the solution.

The second derivative: (3 − 4 x 2 ) sin ( x ) − 4 x cos ( x ) f ′′( x ) = 5 4x 2

Worked example 16

Find the equation of the normal to the curve with equation: x y = 2 cos −1   at the point where x = 3 .  2 Think

32

1

On a calculator page, press: • Menu b • 4: Calculus 4 • A: normalLine A Complete the entry line as: x   normalLine  2 cos−1   , x , 3    2 Then press enTeR ·.

2

Write your solution in an appropriate form.


WriTe

The equation of the normal is y =

x 3 π − + . 2 2 3

Worked Example 18

Find the antiderivative for each of the following expressions: −3 20 1 b a c . 2 2 16 + x 2 25 − x 49 − x Think

Write

a , 1 On a Calculator page, press: • Menu b b • 4: Calculus 4 & • 3: Integral 3 c

Complete the entry lines as:

∫ ∫ ∫

  1  dx  2 25 − x  −3    dx  2 49 − x 

 20   dx  16 + x 2 

Press ENTER · after each entry. Note: The calculator finds the second form of the antiderivative in part b . Also, it does not include the constant. You will have to do that yourself. 2

Write your solutions, remembering to include the constant of integration.

a b c

∫ ∫ ∫

1 25 −

x2

x dx = sin −1   + c  5

x2

x dx = − 3 sin −1   + c  7

−3

49 −

20 x dx = 5 tan −1   + c  4 16 + x 2

Chapter 5 • Differential calculus 33

Worked Example 21

Differentiate the equation y2 + 3x2 = 4 to find Think 1

On a Calculator page, press: • Menu b • 4: Calculus 4 • E: Implicit Differentiation E Complete the entry line as: impDif (y2 + 3x2 = 4, x, y) Then press ENTER ·.

2

Substitute for y as in part a (which is preferable in this straightforward equation) or continue to use the calculator to make y the subject in the equation. Press: • Menu b • 3: Algebra 3 • 1: Solve 1 Complete the entry line as: Solve (y2 + 3x2 = 4, y) Then press ENTER ·.

3

Express the domain, 3x2 − 4 ≤ 0 shown in the screen in a more appropriate form. Take care to change ≤ to < as y is in the denominator in the solution.

4

Write your solution, remembering to include the domain.


dy in terms of x. dx Write

−2

3 3

2 3 3

−2 3 dy 3x 2 3 =± ;
Worked Example 23

dy For x2y2 + y = 2, find the gradient of the tangent, , at the point (1, −2). Hence, determine the dx equation of the tangent at this point. Think 1

2

Write

The gradient of the implicit function can be found using a CAS calculator. To do this, on a Calculator page, press: • Menu b • 4: Calculus 4 • E: Implicit Differentiation E Complete the entry line as: impDif (x2 × y2 + y = 2, x,y) | x = 1 and y = −2 Then press ENTER ·.

8 The gradient of the tangent at (1, −2) is . Use this information and the general 3 equation of a straight line to determine the equation of the tangent.

y − y1 = m( x − x1 ) 8 y − − 2 = ( x − 1) 3 8x 8 y+2= − 3 3 8x 8 y= − −2 3 3 8 x 14 y= − 3 3

Chapter 5 • Differential calculus 35

ChapTer 6

Integral calculus Worked example 1

Find the antiderivative of the following expressions. 3 x2 + 1 a (x + 3)7 b 4x(2 x2 + 1)4 c x3 + x Think

a & b

1

WriTe

On a Calculator page, press: • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry lines as:

a & b

∫ ( x + 3)7 dx ∫ (4 x × (2 x 2 + 1)4 ) dx

∫

 3x 2 + 1   dx  3 x +x

Press enTeR · after each entry. Note: The calculator cannot determine the solution for part c . It simply returns the input you entered. You will have to do this problem manually. Also, it does not include the constant for any of the antiderivatives. ensure you include the constant with your answers.

c

2

Write your solutions, remembering to include the constant of integration.

1

Recognise that 3x2 + 1 is the derivative of x3 + x. Let u = x3 + x.

2

Find

du . dx

( x + 3)8 +c 8 (2 x 2 + 1)5 ∫ 4 x (2 x 2 + 1)4 dx = 5 + c

∫ ( x + 3)7 dx =

c

Let u = x3 + x. du = 3x 2 + 1 dx

ChapTer 6 • Integral calculus

37

Worked Example 2

Antidifferentiate the following functions with respect to x. x+3 2 3 a f ( x) = 2 b f ( x ) = ( x − 1) cos ( 3 x − x ) ( x + 6 x )3 Think

a

1

Write

On a Calculator page, press: • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry line as: x+3 dx 2 ( x + 6 x )3 Then press ENTER ·. The answer is in an equivalent but more complex form than the solution found manually.

a

∫

2

Collect the terms in factorised form over a common denominator as follows: Press: • Menu b • 3: Algebra 3 • 2: Factor 2 Complete the entry line as  − ( x + 9) 1 1  + − factor  2  432( x + 6) 432 x 144 x 2  Then press ENTER ·.

b

∫ (x

x+3 1 +c dx = − 2 3 4 x ( x + 6)2 + 6x)

3

Write your solution, remembering to include the constant of integration.

1

Express in integral notation.

2

Recognise that x2 − 1 is a multiple of the derivative of 3x − x3.

3

Let u = 3x − x3.

Let u = 3x − x3.

4

Find

du . dx

du = 3 − 3x 2 dx = 3(1 − x2) = −3(x2 − 1)


b

2

∫ ( x 2 − 1) cos (3x − x 3 ) dx

5

Substitute u for 3x − x3 and ( x 2 − 1) =

−1

du . 3 dx

So ∫ ( x 2 − 1) cos (3 x − x 2 ) dx −1

du dx 3 dx − cos u du dx =∫ 3 dx − cos u du = ∫ 3 −1 cos u du = 3∫ = ∫ cos u ×

6

Antidifferentiate with respect to u.

=

7

Replace u with 3x − x3.

=

− sin

3

u

+c

− sin (3 x

3

− x3)

+c

Chapter 6 • Integral calculus 39

Worked Example 7

Find the antiderivative of the following expressions. x  2

a sin2 

x  4

b 2 cos2  Think

a

Write

1


2

x Use identity 1 to change sin2   .  2

a

 x

∫ sin2  2  dx = ∫ 12 [1 − cos ( x )] dx

1

1 2

∫ [1 − cos ( x )] dx

3

Take the factor of 2 to the front of the integral.

=

4

Antidifferentiate by rule.

= 12 [ x − sin ( x )] + c

5

Simplify the answer.

=

6

If you use a calculator for problems such as these, you may find the answer expressed in a form that is different from the ones above. The screen dump shows the result, in the first line, of part a done using a CAS calculator. The compact form shown in the second line can be obtained as follows. Press: • Menu b • 3: Algebra 3 • B: Trigonometry B • 2: Collect 2 Complete the entry line as:

x 1 − sin ( x ) + c 2 2

x x x tCollect  − sin   cos     2  2 2 Then press ENTER ·.

b

 x

∫ sin2  2  dx =

x − sin ( x ) +c 2

7

Write your solution, remembering to include the constant of integration.

1


2

x Use identity 2 to change cos2   .  4

=

∫ 2(

3

Simplify the integral.

=

∫  1 + cos  2   dx

4


x = x + 2 sin   + c  2


b

2

x

∫ 2 cos  4  dx 

1 2

)  1 + cos  2x   dx  x

Worked example 12

For each of the following rational expressions: i express as partial fractions ii antidifferentiate the result. x+7 2x − 3 a b 2 ( x + 2 )( x − 3) x − 3x − 4 Think

WriTe

a i 1 express the rational expression

a

i

as two separate fractions with denominators (x + 2) and (x − 3) respectively.

x+7 a b = + ( x + 2)( x − 3) ( x + 2) ( x − 3)

=

a( x − 3) + b( x + 2) ( x + 2)( x − 3)

2

express the partial fractions with the original common denominator.

3

equate the numerator on the left-hand side with the righthand side.

so

4

Let x = −2 so that a can be evaluated.

Let x = −2, and thus 5 = −5a

5

Solve for a.

6

Let x = 3 so that b can be evaluated.

7

Solve for b.

8

Rewrite the rational expression as partial fractions.

9

A CAS calculator can convert expressions in partial fraction form as follows. Press: • Menu b • 3: Algebra 3 • 3: expand 3 Complete the entry line as:

x + 7 = a(x − 3) + b(x + 2)

a = −1 Let x = 3, and thus 10 = 5b b=2 Therefore

−1 x+7 2 = + ( x + 2)( x − 3) x + 2 x − 3

  x+7 expand   ( x + 2)( x − 3)  Then press enTeR ·.

10

ii 1 express the integral in partial

fraction form.

2

x+7 2 1 = − ( x + 2)( x − 3) x − 3 x + 2

Write the answer.


ii

x+7

∫ (x + 2)(x − 3) dx  1 2  =  ∫  x + 2 + x − 3  dx −

= −loge (| x + 2 |) + 2 loge (| x − 3 |) + c; x ∈R\{−2, 3}


41

3

 ( x − 3)2  − = loge   + c; x ∈R\{ 2, 3} | + | x 2  

Simplify using log laws.

b i 1 Factorise the denominator.

b

i

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

2

express the partial fractions with denominators (x − 4) and (x + 1) respectively.

=

a b + x − 4 x +1

3

express the right-hand side with the original common denominator.

=

a( x + 1) + b( x − 4) ( x − 4)( x + 1)

4

equate the numerators.

So 2x − 3 = a(x + 1) + b(x − 4)

5

Let x = 4 to evaluate a.

Let x = 4,

6

Solve for a.

7

Let x = −1 to evaluate b.

8

Solve for b.

9

Rewrite the rational expression as partial fractions.

ii 1 express the integral in its

partial fraction form.

a=1 Let x = −1, −5 = −5b b=1 Therefore ii

∫

2x − 3 1 1 = + x − 3x − 4 x − 4 x +1 2

2x − 3 dx x − 3x − 4 2

=

42

5 = 5a

∫

 1 + 1  dx   x − 4 x + 1

2


= loge (| x − 4 |) + loge (| x + 1|) + c

3


= loge (| x − 4 |)(| x + 1|) + c; x ∈R\{−1, 4} or loge (| x2 − 3x − 4 |) + c; x ∈R\{−1, 4}


Worked example 16

Evaluate the following definite integrals. a

∫

2

0

x−2 dx x2 + 5 x + 4

b

∫

π 2

cos ( x ) 1 + sin ( x ) dx

0

Think

a

WriTe

x−2

2

∫0 x 2 + 5x + 4 dx

1

Write the integral.

2

Factorise the denominator of the integrand.

3

express in partial fraction form with denominators x + 1 and x + 4.

=

a b + x +1 x + 4

4

express the partial fractions with the original common denominator.

=

a( x + 4) + b( x + 1) x 2 + 5x + 4

5

equate the numerators.

6

Let x = −1 to find a.

Let x = −1, −3 = 3a a = −1

7

Let x = −4 to find b.

Let x = −4, −6 = −3b b=2

8

Rewrite the integral in partial fraction form.

So

a

Consider:

x−2 x−2 = x + 5 x + 4 ( x + 1)( x + 4) 2

x − 2 = a(x + 4) + b(x + 1)

2

∫ ∫ 0

=

x−2 dx x 2 + 5x + 4 2

0

−

1 2 + dx x +1 x + 4

−

Antidifferentiate the integrand.

= [ loge (| x + 1 |) + 2 loge (| x + 4 |)]20

10

evaluate the integral.

= [−loge (3) + 2 loge (6)] − [−loge (1) + 2 loge (4)] = −loge (3) + 2 loge (6) − 2 loge (4)

11


= 2 loge (1.5) − loge (3) = loge (2.25) − loge (3) = loge (0.75) (or approx. −2.88)

12

You may attempt this problem using a CAS calculator. On a Calculator page, press • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry line as:

9

∫

2

0

x−2    ( x 2 + 5 x + 4)  dx

Then press enTeR ·. Pressing Ctrl / enTeR · will give an approximate answer. 13

Write the solution.

∫

2

0

x−2  3 dx = loge    4 ( x 2 + 5 x + 4) ≈ − 2.88


43

b

1

Write the integral.

b

∫

π 2

0

cos ( x ) 1 + sin ( x ) dx

2

Let u = 1 + sin (x) to antidifferentiate.

3

Find

4

cos ( x ) dx =

du dx dx

cos ( x ) dx =

5

Change terminals by finding u π when x = 0 and x = . 2

When x = 0,

Simplify the integrand.

So

6

Let u = 1 + sin (x) du = cos ( x ) dx

du . dx

When x =

∫

π 2

0

π π , u = 1 + sin    2 2 =1+1 =2

2 1 u2 1 2 1 u2 1

du dx dx du

3 2

7


=  23 u 2  1 

8

evaluate the integral.

= 23 × 2 2 − 23 × 12

3

= or 9

using CAS, press • Menu b • 4: Calculus 4 • 3: Integral 3 Complete the entry line as:

∫

π 2

0

cos ( x ) 1 + sin ( x ) dx

Then press enTeR ·.

44

u = 1 + sin (0) =1

cos ( x ) 1 + sin ( x ) dx

∫ = ∫ =

du dx dx


4 2 2 − 3 3 4 2−2 3

3

Worked Example 19

Find the area bounded by the curves y = x2 − 2 and y = 2x + 1. Think 1

Write

On a Calculator page, complete the entry line as: 3

∫ 1 ( x 2 − 2 − (2 x + 1)) dx −

Then press ENTER ·.

2

State the solution.

2

The area bounded by the two curves is 10 3 square units.

Chapter 6 • Integral calculus 45

ChapTer 7

Differential equations Worked example 7

Find the general solution to

dy 1 = 2x + . dx 1 − x2

Think 1

WriTe

On a Calculator page, press: • Menu b • 4: Calculus 4 • D: Differential equation Solver D Complete the entry line as:   1 deSolve  y′ = 2 x + , x , y , 2   1− x then press enTeR ·.

2

− The general solution is y = sin 1 (x) + x2 + c.

Write the solution.

Worked example 8

Find the particular solution to h′(t) =

t t2

+9

Think 1

where h(4) = 1. WriTe

On a Calculator page, press: • Menu b • 4: Calculus 4 • D: Differential equation Solver D. Complete the entry line as:   t deSolve  h' = and h(4) = 1, t , h  , 2   t +9 then press enTeR ·.

2

Write the solution.

The particular solution is h(t ) = t 2 + 9 − 4.

ChapTer 7 • Differential equations

47

Worked Example 10 x

Find the general solution to f ′′( x ) = e 2 − 2. Think 1

Write

On a Calculator page, press: • Menu b • 4: Calculus 4 • D: Differential Equation Solver D Complete the entry line as: x   deSolve  y′′ = e 2 − 2, x , y  ,   then press ENTER ·.

2

Write the solution.

The general solution is x

y = 4e 2 − x 2 + cx + d .

Worked Example 11

Find the particular solution to y″(x) = 4 sin (2x + π) given that y′(0) = 1 and y(0) = 4. Think 1

On a Calculator page, press: • Menu b • 4: Calculus 4 • D: Differential Equation Solver D Complete the entry line as: deSolve(y″ = 4 sin (2x + π) and y′(0) = 1 and y(0) = 4, x, y), then press ENTER ·.

2

Write the solution. Note that it is different from but equivalent to the solution found manually.


Write

The particular solution is y = sin (2x) − x + 4.

Worked example 13

Find general solutions for each of the following differential equations. dy =y a dx b y′(x) = tan (y) Think

a

b

WriTe

1

On a Calculator page, press: • Menu b • 4: Calculus 4 • D: Differential equation Solver D Complete the entry line as: deSolve(y′ = y, x, y), then press enTeR ·.

2

Write the solution, making sure to write the integration constant in a conventional form.

1

express the derivative as

2

Invert both sides of the differential equation.

3

Simplify

4

Let u = sin (y) to antidifferentiate using the ‘derivative present’ method.

5

Find

6

du into Substitute u and cos ( y) = dy the integral.

7

dy . dx

1 . tan ( y)

du . dy

a

The general solution is y = Aex.

b

y′(x) = tan (y) dx = tan ( y) dy dx 1 = ,y≠0 dy tan ( y) =

cos ( y) sin ( y)

∫

cos ( y) dy sin ( y) Let u = sin (y) x=

du = cos (y) dy x=

∫

1 du × dy u dy

Simplify the integrand.

=

∫

du u

8


= loge (| u |) + c

9

Replace u by sin ( y).

x = loge (| sin (y) |) + c

10

Subtract c from both sides.

x − c = loge (| sin (y) |)

11

express the equation in exponential form.

e x − c = | sin (y) |

ChapTer 7 • Differential equations

49

50

12

Write the constant in a more suitable form.

| sin (y) | = e−cex

13

Remove the modulus sign and take sin−1 of both sides to make y the subject.

sin (y) = ±e−cex y = sin−1 (Ae x), where A = ±e−c

14

State the solution.

Therefore, the general solution is y = sin−1 (Ae x).


Worked Example 14 1

a Find the particular solution of y′ = 1 − y 2 if y(0) = 2 .

b State the largest domain for which the solution applies. Think

a & b

1

On a Calculator page, complete the entry line as: 1   deSolve y′ = 1 − y 2 and y(0) = , x , y  ,   2 then press ENTER ·.

2

The solution does not have y as the subject. You can do this yourself manually to obtain the solution.

3

Alternatively, you can use the solve feature of the calculator as follows. Press: • Menu b • 3: Algebra 3 • 1: Solve 1. Complete the entry line as: π − solve  sin 1 ( y) − = x , y  ,   6 then press ENTER ·.

4

The domain, as given by the calculator, has to be re-worked into − 2π π the form: ≤x≤ . 3 3 However, although this is a valid domain for the solution of the trigonometric equation, the end points of this domain are not valid for the differential equation due to the 1 − y 2 term in the denominator. Putting all of this together, write down the correct domain.

Write

a & b

π  The particular solution is y = sin  x +  . 6

Domain is

−

2π π
Chapter 7 • Differential equations 51

Worked Example 24

Use the numerical integration method to find the value of y when x = 15 if at x = 0. Think

dy = 2 x , given that y = 1 dx

Write 15

1

Substitute appropriate values and expressions, a = 0, b = 1 and f (x) = 2x, in the formula. On a Calculator page, press: • Menu b • 4: Calculus 4

y(15) =

∫ 2 x dx + 1 0

• F: Numerical Calculations F • 3: Numerical Integral 3 Complete the entry line as: nInt(2x,x,0,15) + 1, then press ENTER ·.

2

Write the answer.

The value of y when x = 15 is 226.

3

Find the antiderivative of 2x.

y = x2 + c

4

Evaluate the constant by applying the initial condition y(0) = 1.

y = x2 + 1

5

Substitute x = 15 and compare the new answer to the previous one.

y = 225 + 1 = 226 The answers are the same.


Chapter 8

Kinematics Worked example 3

The position of a particle moving in a straight line is given by: x(t) = 2t3 + t loge (t) − 4, t > 0. Find: a the velocity at any time t b the acceleration at any time t. think

a

1

To determine the velocity at any time, differentiate x with respect to t, using a CAS calculator by completing the following steps. Define the function x(t). To do this, on a Calculator page, press: • MENU b • 1: Actions 1 • 1: Define 1 Complete the entry line as: Define x(t) = 2t3 + t ln(t) − 4 then press ENTER ·.

2

Differentiate x with respect to t. To do this press: • MENU b • 4: Calculus 4 • 1: Derivative 1 Complete the entry line as: d ( x (t )) dt then press ENTER ·.

3

Write the solution using the correct notation and variables.

Write

a

The velocity at any time is given as: v(t) = loge (t) + 6t2 + 1.

Chapter 8 • Kinematics

53

b

1

To determine the acceleration at any time, differentiate the velocity with respect to t. To do this, complete the entry line as: d (ln(t ) + 6 × t 2 + 1) dt then press ENTER ·.

2

Write the solution using the correct notation and variables.

b

See the screen above.

The acceleration at any time is given as: 1 a(t ) = 12t + . t

Worked example 4

Find the acceleration in terms of x if x(t) = sin (2t) − cos (2t). think

54

1

To find the acceleration given the position, on a Calculator page, complete the entry line as: d2 (sin (2t ) − cos (2t )) dt 2 then press ENTER ·. Note: The second derivative template is located in the Maths expression template.

2

Write the solution in terms of x.


Write

a(t) = 4 cos (2t) − 4 sin (2t) a(t) = −4 (sin (2t) − cos (2t)) But x = sin (2t) − cos (2t) a(x) = −4x

Worked Example 6

The acceleration of a particle moving in straight line is given by: dv = 5 e t − 6 t + 4 cm/s2, where v is the velocity at any time. dt If the particle starts at the origin with a velocity of −1 cm/s, find: a the velocity at any time t b the displacement x (t) from the origin at any time t c the displacement from the origin 1 second. Think

a Solving the differential equation, to find

Write

a

the velocity press: • MENU b • 4: Calculus 4 • D: Differential Equation Solver D Complete then entry line as: deSolve ( v ' = 5et − 6t + 4 and v ( 0 ) = − 1, v , t ) then press ENTER ·.

b Solving the differential equation, to find & the displacement press: c • MENU b

b & c

• 4: Calculus 4 • D: Differential Equation Solver D Complete then entry line as: deSolve (x ' = 5et − 3t 2 + 4t − 5 and x ( 0 ) = 0, x , t ) then press ENTER ·.

To find the displacement after 1 second, complete as shown.

Chapter 8 • Kinematics 55

Worked Example 7

The acceleration of a body travelling in a straight line is given by: a(t) = 6t − 2 m/s2; when t = 0, x = 0 and v = −1. a Find the displacement at any time t. b Find the distance travelled in the first 3 seconds. Think

a

1

Write

To find the displacement at any time t, first find the velocity by antidifferentiating a with respect to t, using a CAS calculator by completing the following steps. To do this, on a Calculator page, complete the entry lines as: Define a(t) = 6t − 2

a

∫ (a(t)) dt

Press ENTER · after each entry. To find the constant of integration complete the entry lines as: Define v(t) = 3t2 − 2t + c solve(v(0) = −1,c) Press ENTER · after each entry. 2

State the velocity at any time.

3

Antidifferentiate v to find the displacement x by completing the entry line as:

∫ (3t

2

v(t) = 3t2 − 2t − 1

− 2t − 1) dt

then pressing ENTER ·. The constant of integration can be found by completing the entry lines as: Define x(t) = t3 − t2 − t + d solve(x(0) = 0,d) Press ENTER · after each entry.

b

4

State the displacement at any time t.

1

To find when the object might have changed direction, find t when v = 0. Complete the entry lines as: Define x(t) = t3 − t2 − t Define v(t) = 3t2 − 2t − 1 solve(v(t) = 0,t) Press ENTER · after each entry. −1 Since t ≠ , t = 1 3 Complete the entry lines as: x(0) x(1) x(3) Press ENTER · after each entry.


x(t) = t3 − t2 − t b

Therefore the body travels 16 m in the first three seconds of motion.

2

Write the solution given that Distance = | x(3) − x(0) | + | x(1) − x(0) |.

Solving v(t) = 0, for t gives − 1 t = 0 as t ≠ 3 x(0) = 0 x(1) = −1 x(3) = 15 Distance = | x(3) − x(0) | + | x(1) − x(0) | = 16 Therefore the body travels 16 m in the first 3 seconds of motion.

Chapter 8 • Kinematics 57

Worked Example 9

A ball is thrown vertically upward from a platform 16 metres above the ground and has an initial velocity of 24.5 m/s. Find: a the maximum height above the platform that the ball reaches b the time taken to reach the maximum height c the time taken for the ball to fall to the ground from the maximum height d the total time the ball is in the air e the speed of the ball when it returns to the level of the platform f the speed of the ball when it hits the ground. Think

a

1

Sketch a diagram of the motion of the ball. Consider up to be positive and down to be negative.

Write

a

v=0 a = −9.8 m/s2

Platform t=0 16 m u = 24.5 Ground

b

2

At its maximum height the velocity of the ball, v, is 0.

v = 0 at the ball’s maximum height.

3

List what is given and what has to be found.

Given: u = 24.5 v = 0 a = −9.8 Require: s = ?

4

Select an appropriate formula.

v2 = u2 + 2as

5

Using the CAS calculator to solve the equation for s, the displacement, complete the entry line as: solve(02 = 24.52 + 2 × −9.8 × s,s) then press ENTER ·.

6

Write the solution.

1

List the given information and what has to found.

2



Solving 02 = 24.52 + 2 × (−9.8) × s for s gives s = 30.625 m. Therefore, the maximum height reached above the platform is 30.625 metres. b

u = 24.5 a = −9.8 v=0 t=? v = u + at

c

3

Using the CAS calculator solve the equation for t. Complete the entry line as: solve(0 = 24.5 − 9.8 × t,t) then press ENTER ·.

4

State the solution.

1

Sketch another diagram for the ball falling from its maximum height.

Solving 0 = 24.5 − 9.8 × t for t gives t = 2.5. Therefore, the ball takes 2.5 seconds to reach its maximum height. c

Max. height u=0 30.625 m Platform a = −9.8 m/s2 16 m Ground

2

List the given information and what has to be found. Both a and s are negative as their direction is downwards.

3


4

Using a CAS calculator to solve the equation for t, complete the entry line as: 1 solve(−46.625 = 0 × t + 2 × −9.8 × t2,t)

Given: u = 0 a = −9.8 s = −(30.625 + 16) = −46.625 Require: t = ? s = ut + 12 at 2


5

Write the solution.

1

Solving −46.625 = 0 × (t) + 2 × (−9.8) × t2 for t gives t = 3.085. Therefore, it takes approximately 3.085 seconds for the ball to fall to the ground from its maximum height.


59

d

e

f

60

1

Add the time travelling up to the time travelling down.

2

State the answer.

1

List the given information and what has to be found.

2


3

Using the CAS calculator to solve the equation for v, complete the entry line as: solve(v2 = 02 + 2 × −9.8 × −30.625, v) Then press ENTER ·.

4

Write the solution.

1

List the information and what has to be found.

2


3

Use the CAS calculator to solve for v. Complete the entry line as: solve(v2 = 02 + 2 × −9.8 × −46.625, v) then press ENTER ·.

4

Write the solution.


d

Total time = 2.5 + 3.085 = 5.585 Therefore, the ball is in the air for approximately 5.585 seconds.

e

u=0 a = −9.8 s = −30.625 v=? v2 = u2 + 2as

Solving v2 = 02 + 2 × −9.8 × −30.625 for v gives v = −24.5 or v = 24.5. Since the object is travelling downward, v = −24.5. Therefore, the speed of the ball when it returns to the level of the platform is 24.5 m/s. f

u=0 a = −9.8 s = −46.625 v=? v2 = u2 + 2as

Solving v2 = 02 + 2 × −9.8 × −46.625 for v gives v = −30.23 or v = 30.23. Therefore, the ball hits the ground with an approximate speed of 30.23 m/s.

Worked example 16

The acceleration of a particle travelling in a straight line is given by velocity at any time t. At time t = 0, the body is at rest at the origin. Find, as a function of time: a the velocity b the acceleration c the position. think

a

1

Write the acceleration as given.

2

Invert both sides of the differential equation.

3

Express t in integral form.

4

Antidifferentiate the integrand, using the CAS calculator. Complete the entry line as: 1 dv v +1 then press ENTER ·. Given that t = 0 when v = 0, to find the constant of integration, complete the entry line as: solve(0 = ln(| 0 + 1 |) + c,c) then press ENTER ·. Rearrange the equation to make v the subject by completing the entry line as: solve(t = ln(| v + 1 |),v) then press ENTER ·.

dv = v + 1, where v is the dt

Write

a

dv = v +1 dt dt 1 or = dv v + 1 t=

∫

1 dv v +1

∫

5

Write the solution.

t=

∫

1 dv = loge (| v + 1 |) + c v +1

Solving 0 = loge (| 0 + 1 |) + c for c gives c = 0. So, t = loge (| v + 1 |) Then, solving t = loge (| v + 1 |) for v gives v = −et − 1 or v = et − 1. Since v = 0 when t = 0 Then v = et − 1 b

c

1

Differentiate v with respect to t to find the acceleration (or we could substitute v into the original equation for acceleration).

2

State the answer.

1

Express x as the antiderivative of v.

2


b

dv dt d (et − 1) a= dt a=

a = et c

x=

∫

(et − 1) dt

x = et − t + c


61

62

3

Substitute t = 0 and x = 0, the initial condition given.

When t = 0, x = 0 0 = e0 − 0 + c

4

Solve for c.

c = −1

5

State the answer.

x = et − t − 1


Chapter 9

Vectors Worked example 9

Find the unit vector in the direction of u ~:

a

a in the figure at right

~u

b in u = 2 i − 3 j + 2 3 k .

think

a

1

Express the vector in component form.

2

Compute the magnitude of the vector, u.

3

Divide each component of the original vector by the magnitude to get uˆ.

4

Comfirm that uˆ has a magnitude of 1.

3j ~

6i~

Write

a

C (6, 3)

u = 6i + 3 j u = 6 2 + 32 = 45 =3 5 6 3 uˆ = i+ j 3 5 3 5 =

2 5 5 i+ j 5 5

u = x 2 + y2 20 5 = + 25 25 =

25 25

=1

Chapter 9 • Vectors

63

b

1

To find the unit vector in the direction of u by using a CAS calculator, complete the following steps. Define the vector u. To do this press: • MENU b • 1: Actions 1 • 1: Define 1 Complete the entry line as:

b

Define u =  2, − 3,2 3  Then press ENTER ·. Find the unit vector in the direction of u . To do this press: • MENU b • 7: Matrix & Vector 7 • C: Vectors C • 1: Unit Vector 1 Complete the entry line as: unitV(u) Then press ENTER ·. 2

64

Write the unit vector in the direction of u.


2 3 2 3 uˆ = i − j + k 5 5 5

Worked Example 15

Let u = 5 i + 2 j . Find a vector parallel to u such that the dot product is 87. Think

1

Let the required vector v = ku.

3

Find the dot product of u . v. Simplify.

4

Equate the result to the given dot product 87.

5

Solve for k.

6

Substitute k = 3 into vector v. An alternative method is to use a CAS calculator. Let the required vector v = ku. Using a CAS calculator, define u and v by completing the following steps. On a Calculator page, press: • MENU b • 1: Actions 1 • 1: Define 1 Complete the entry line as: Define u = [5,2] ENTER ·. then press Repeat for v = k [5,2]

2

7

8

9

10

Write

Let v = k (5i + 2 j) = 5ki + 2 kj u . v = (5i + 2 j) . (5ki + 2 kj) = 25k + 4k = 29k 29k = 87 k=3 v = 15i + 6 j u = 5i + 2 j So v = k (5i + 2 j)

Using a CAS calculator solve u . v = 87 for Press: k, by completing the followingsteps. • MENU b • 3: Algebra 3 • 1: Solve 1 After ‘solve’, then press: • MENU b • 7: Matrix and Vector 7 • C: Vector C • 3: Dot Product 3 Complete the entry line as: solve(dotP(u, v ) = 87, k ) then press ENTER ·. Write the solution.

Solving u . v = 87 for k gives k = 3 Therefore, v = 15i + 6 j

Chapter 9 • Vectors 65

Worked Example 19

Determine the value of p if the following vectors are linearly dependent. a = i + pj − k, b = i + 3 j + 8 k and c = i − j + 2 k Think

1

Let a = mb + nc.

2

To solve this equation for m, n and p define vectors a, b and c as shown previously. Then press: • MENU b • 3: Algebra 3 • 1: Solve 1

Write

Let a = mb + nc Solve a = mb + nc for m, n and p.

Complete the entry line as: solve(a = m × b + n × c,{m, n, p}) Then press ENTER ·.

3

Write the solution.

Solving a = m × b + n × c for m, n and p gives: m = − 0.5 n = 1.5 p = −3


Worked example 21

Let u = − 2 i + 3 j + k and v = 3 i + 2 j − k. Find: a the scalar resolute of v on u b the vector resolute of v parallel to u, namely v to u, namely c the vector resolute of v perpendicular v⊥ . think

a

1

To find the scalar resolute of v on u, find uˆ. Using a CAS calculator, define v and u as shown previously and find uˆ by completing the entry line as: unitV(u) Then press ENTER ·.

2

Write uˆ .

3

4

To find uˆ . v (the scalar resolute of v use CAS to on u), first define w = uˆ . the dot product uˆ . v, Then, to find complete the entry line as: dotP(w,v) Then press ENTER ·.

Write the scalar resolute of v on u.

Write

a

uˆ =

−

uˆ . v =

14 3 14 14 i+ j+ k 7 14 14

−

14 14

Chapter 9 • Vectors

67

b

1

To find the vector resolute of v parallel to u (that is, (uˆ . v )uˆ) using CAS, complete the entry line as follows: dotP(w,v) × w Then press ENTER ·.

2

Write the vector resolute of v parallel to u.

c To find the vector resolute of v

perpendicular to u (that is, v − (uˆ . v )uˆ) the entry line as: using CAS, complete  1 v−  7

−

3 14

−

1   14 

then press ENTER ·. Write the solution in an appropriate form.

68


b

1 1 3 (uˆ . v )uˆ = i − j− k 7 14 14 c

20 31 13 v − (uˆ . v )uˆ = i+ j− k 7 14 14

Worked Example 25

Let a particle’s position as a function of time be given by u = 2 cos ( ti ) + 3 sin ( t j ). Find the equation of the path and sketch its graph. Think

Write

1

Express the i and j components of u in terms of their functions.

2

In this case, first eliminate the constants in front of the trigonometric functions.

x = 2 cos (t) y = 3 sin (t) x = cos (t) 2 y = sin (t) 3

3

Square both sides of the equation.

x2 = cos2 (t) 4 y2 = sin2 (t) 9

4

Add the 2 equations.

5

Use the trigonometric identity cos2 (θ ) + sin2 (θ ) = 1.

6

Use CAS to sketch the graph using the parametric equations. On a Graphs page, press: • MENU b • 3: Graph Entry/Edit 3 • 3: Parametric 3 Complete the entry line as shown: x1(t ) = 2 cos (t )

x 2 y2 + = cos2 (t) + sin2 (t) 4 9 x 2 y2 + =1 4 9

y1(t ) = 3sin (t ) 0 ≤ t ≤ 6.28 tstep = 0.13 then press ENTER ·.

7

y

This is the equation of an ellipse of centre (0, 0) where a = 2 and b = 3.

3

−2

0

2

x

−3

Chapter 9 • Vectors 69

Chapter 10

Vector calculus Worked example 3

An object has a position vector, in metres, r ( t ) = (3 t 2 − sin (π t )) i + ( e− t + 2 t ) j ; t ≥ 0 seconds. a Find the velocity vector v( t ). b Find the velocity of the object at t = 2 s. c Find the speed of the object at t = 2 s. d Find the average velocity in the first 2 seconds of the body’s motion. think

a

b

1

Define the position vector, r ( t ). To do this press: • MENU b • 3: Actions 3 • 1: Define 1 Then complete as shown, then press ENTER ·. To find v ( t ). To do this press: b • MENU • 4: Calculus 4 • 1: Derivative 1 Complete the entry line as: d ( r ( t )) dt

2

Write the solution using correct vector notation.

1

Substitute t = 2

2


Write

a

v ( t ) = ( 6t − π cos (π t )) i + ( −e − t + 2 ) j b

1 v ( 2 ) = (12 − π ) i +  2 − 2  j e  

Chapter 10 • Vector calculus

71

c

d

72

1

To find the speed, find the norm of the vector.

2

State the speed correct to 2 decimal places.

1

The average velocity in the first r ( 2) − r ( 0 ) 2 seconds, is 2−0

2

State the average velocity


c

The speed of the object is 9.05 m/s. d

vav = 6 i + 1.567 j

Worked example 5

A body moves in such a way that its position vector in metres at an instant t seconds is given by: 1 1     r ( t ) =  15 t − t 2 + 1 i +  12 t − t 3  j, t ≥ 0.    2 3 

Find: a the velocity vector v( t ) b the acceleration vector a( t ) c the angle between the velocity vector and the acceleration vector of the body at a time t = 1 s, to the nearest degree d the time when the body has an acceleration of magnitude 9.8 m/s2. think

a

b

Write

1

Differentiate r, the position vector, with a respect to t, using a CAS calculator by completing the following steps. Define the position function r (t ). To do this press: • MENU b • 1: Actions 1 • 1: Define 1 Complete the entry line as: 1 1 Define r (t ) = 15t − t 2 + 1,12t − t 3  2 3   then press ENTER ·. To find v (t ) differentiate r with respect to t, press: • MENU b • 4: Calculus 4 • 1: Derivative 1 Complete the entry line as: d (r (t )) dt then press ENTER ·.

2


1

To find the acceleration vector a(t ) complete the following steps. Define the velocity function v (t ), by completing the entry line as:

dr v (t ) = dt v (t ) = (15 − t ) i + (12 − t 2 ) j b

Define v (t ) = [15 − t , 12 − t 2 ] Then press ENTER ·. Differentiate v (t ) with respect to t, by entry line as: completing the d ( v (t )) dt Then press ENTER ·.

2


a(t ) = a(t ) =

dv dt − i − 2tj


73

c

1

To find the angle between the velocity and acceleration vectors, complete the following steps.

c

Define the acceleration function a(t ) = [ − 1, − 2t ] Calculate the velocity vector when t = 1, by finding v(1) and the acceleration vector when t = 1, by finding a(1). Calculate the magnitude of vectors v(1) and a(1) by completing the entry lines as: norm([14, 11]) norm([–1, –2]). Press ENTER · after each entry. Calculate the dot product a(1) . v (1) (Refer to the previous chapter, Vectors, to review the dot product.) 2

The angle, θ, between the two vectors v(1) and a(1) is found using the equation a(1) . v (1) = a(1) × v (1) × cos (θ ) This can be done by hand or using the calculator. Write the solution using correct vector notation.

v (1) = 14 i + 11 j a(1) = − i − 2 j a(1) . v (1) cos (θ ) = a(1) × v (1) v(1) = 317 a(1) = 5 a(1) . v (1) = − 36 − 36 cos (θ ) = 317 5

θ = 155° to the nearest degree. The angle between the two vectors is approximately 155°.

74


d State the acceleration a(t ) at any time, t.

Find the magnitude ofthe acceleration at any time, t. Use the CAS calculator to solve for t when the acceleration is equal to 9.8 m/s2. Complete the entry line as:

solve

(

d

a(t ) = − i − 2tj a(t ) = 1 + 4t 2

)

1 + 4t 2 = 9.8, t Then press ENTER ·. Note: Only positive solutions for t are allowed.

Solving 1 + 4t 2 = 9.8 for t gives t ≈ 4.87, t ≥ 0. When t = 4.87 s the magnitude of the acceleration is 9.8 m/s2.

Chapter 10 • Vector calculus 75

Worked Example 10

An object is thrown off a building (t = 0 s) on a windy day. The acceleration of the object in m/s2 − t

1 i − 9.8 e 10 j . At a time, t = 1, the object has a velocity in m/s of v(1) = 2 i − 3 j . is given by a( t ) = 12 The building is 50 m above the ground and hence the initial position of the object is taken to be r = 0 i + 50 j . What is the initial acceleration of the object? a b Determine the velocity vector v( t ) for all times t ≥ 0. c Determine the position vector r ( t ) for all times t ≥ 0.

Think

a

b

1

Define the acceleration function a(t ), by completing the entry line as: −t 1  Define a(t ) =  , − 9.8e 10   12  Then press ENTER ·. Substitute t = 0 into a(t ) to find the initial acceleration.

2

Write the solution

1

The velocity vector is found by integrating the acceleration a(t ) with respect to time.

2

The vector constant of integration must be found. To do this using the CAS calculator, define the velocity vector v (t ) by completing the entry line as:

Write

a

1 a(0) = i − 9.8 j 2 b

−t  t  Define v (t ) =  + m, 98e 10 + n  12   Then press ENTER ·.

3

State the integrand and the velocity vector v (t ) with the vector constant of integration expressed as mi + nj


v (t ) = ∫ a(t ) dt + c −t t 10 v (t ) = i + 98e j + mi + nj 12

4

Evaluate the vector constant of integration using the fact that v (1) = 2 i − 3 j . Complete the entry line as: solve( v (1) = [2, − 3],{m, n}) Then press ENTER ·.

c

5

State the values of m and n and the velocity vector.

1

Redefine the velocity vector v (t ) with the correct values of m and n.

2

The position vector r (t ) is found by integrating the vector v (t ) with respect to time.

3

State the integrand and the position vector r (t ), with the vector constant of integration expressed as pi + qj

4

The vector constant of integration must be found. To do this using the CAS calculator, define the position vector r (t ) by completing the entry line as: Define

23 and n = −91.674 12 −t   t 23 v (t ) =  +  i +  98e 10 − 91.674  j   12 12  

m=

c

r (t ) = ∫ v (t ) dt + d −t 2  23 t   r (t ) =  + t + p i +  − 980e 10 − 91.674t + q  j  24 12   

−t  t 2 23  r (t ) =  + t + p, − 980e 10 − 91.674t + q  24 12  



77

78

5

Use r (0) = 0 i + 50 j to solve the vector do this, complete constants p and q. To the entry line as: solve(r (0) = [0, 50],{p, q}) then press ENTER ·.

6

State the values of p and q and the position vector.


p = 0 and q = 1030 −t   t 2 23   r (t ) =  + t  i +  − 980e 10 − 91.674t + 1030  j  24 12   

CHAPTER 11

Mechanics WORKED EXAMPLE 5

In a science laboratory a 1.0-kg mass is suspended by two taut strings as shown at right. The tension forces in string 1 and string 2 are A and B respectively. vector diagram showing all three forces which a Draw a force act on the 1.0-kg mass. b By resolving vectors into i and j components find the magnitudes of A and B respectively. THINK

a

b

1

The mass has three forces acting on it: the weight force vertically downwards and the two tension forces. One tension force acts horizontally the second acts at an angle of 42° to the horizontal. Draw the force vector diagram.

1

The weight vector can be written as −g j. The first tensile force can be written as Ai .

3

4 5

String 1 42°

j ~ ~i

1.0-kg mass

WRITE

2

2

String 2

a

~B

~i

b

A ~

42°

j ~

W ~ = g~

W = −g j A = Ai

The second tensile force can be resolved as B = − B cos (42 ) i + B sin (42 ) j Express the resultant force, R, as the sum of the three forces.

B = − B cos (42 ) i + B sin (42) j

Set the sum of the three vectors to zero in accordance with Newton’s third law.

But R = 0 i + 0 j

R = ( A − B cos (42 )) i + ( B sin (42 ) − g) j

CHAPTER 11 • Mechanics

79

80

6

Use CAS to solve the vector equation. First define the vector R, by completing the entry line as: Define r = [a − b × cos (42), b × sin (42) − 9.8] Then press ENTER ·.

7

To solve the equation complete the entry line as: solve(r = [0, 0], {a, b}) Then press ENTER ·.

8

Write the solution correct to 1 decimal place, showing the correct units.


A = 10.9 N B = 14.6 N

WORKED EXAMPLE 6

© John Wiley & Sons Australia/Jennifer Wright

j A car of mass 800 kg is parked in a street which has an angle of elevation ~ of 15°. The i direction is parallel down the street and the j direction is to the street. perpendicular ~i 15° The car is subject to three forces, namely its weight, W , the normal applied force of the reaction force, N , of the road acting on the car and the brake (this is actually a static friction force) F. forces, W , N and F, acting on the car, taking the car a Draw a vector diagram indicating the three as a particle. b What is the magnitude of the resultant force R? express it as a vector using i − j notation. c Resolve the weight, W , into its components and d Calculate the magnitude of N, the normal contact force, and the magnitude of the applied force of the brake, F.

j ~

15°

~i

15° THINK

a

b

1

2

Draw the force vector diagram.

1

The car is in equilibrium since it is stationary.

2

Apply Newton’s First Law of Motion: the resultant force, R, must be zero. Therefore, the magnitude of the resultant force, R, is zero.

3

c

A stationary car parked on a street will have a vertical weight force, a normal reaction force and a static frictional force resisting its sliding or rolling down the street.

1

Draw a diagram showing the resolution of the weight, W , into j. components parallel to i and

WRITE

a

N ~

j ~ ~i

~F 15° W ~

b

R=0 R=0N c

j ~ ~i

15° 15° Wy W ~

2

The component of W parallel to i , Wx , is W sin (15°).

Wx

F = 207g


81

d

82

3

Substitute W = 800g and evaluate.

4

The component of W parallel to j, Wy is W cos (15°).

5

Substitute W = 800g and evaluate.

6

Express W in vector notation.

1

The components of the net forces parallel to the i and j vectors are both zero.

2

Use CAS to solve for F and N by completing the entry line: solve([207 × g − f, n − 773 × g] = [0, 0], { f, n}) Then press ENTER ·.

3

Write the solutions for F and N.


= 800g sin (15°)i ≈ 207gi Wy = W cos (15°) = −800g cos (15°) j ≈ −773g j W = Wx i + Wy j W = 207gi − 773g j d (207 g − F ) i + ( N − 773g) j = 0 i + 0 j

F = 207g N = 773g

WORKED EXAMPLE 11

A skier of mass 60 kg begins a ski run from the top of a mountain down an 18° slope. A constant friction force of 37 N acts on the skier as he descends. a Draw a vector force diagram depicting the three forces which act on the skier. b Write the equation of motion using Newton’s Second Law of Motion. c Determine the magnitude of the normal contact force and the acceleration of the skier. d If there were no frictional force present, find the acceleration of the skier. THINK

a Three significant forces act on the skier: the

WRITE

a

weight force W , the normal reaction force N friction F . and the force of

j ~

~N ~F

Mass = 60 kg ~i 18° W ~

b

c

1

Use Newton’s second law.

2

The vector W is to be resolved into componentsparallel to i and j.

3

We know that m = 60 and ay = 0.

1

But: W = 60 × 9.8 = 584N F = 37N θ = 18°

2

Use CAS to solve the vector equation, R = 60 ai + 0 j , by completing the entry as: line solve([60 × 9.8 × sin (18) − 37, n − 60 × 9.8 × cos (18)] = [60 × a, 0], {n, a}) Then press ENTER ·.

3

Write the solutions correct to 2 decimal places, with the correct units.

b

c

R = ma R = (W sin (θ ) − F ) i + ( N − W cos (θ )) j R = max i + ma y j R = 60 ax i + 0 j W = 584N F = 37N θ = 18°

N = 559.22 N a = 2.41 m/s2


83

d

84

1

To find the acceleration when no d frictional force is present, edit the resultant force equation by completing the entry line: solve([60 × 9.8 × sin (18), n − 60 × 9.8 × cos (18)] = [60 × a, 0], {n, a}) Then press ENTER ·.

2

Write the solution, with the correct units.


a = 3.03 m/s2

WORKED EXAMPLE 15

35kg 25kg Two carts connected by a light, inextensible rope are being accelerated Tractor across a tarmac by a tractor. The leading card (CartL) has a mass of 35 kg; the trailing cart (CartT) has a mass of 25 kg. The coefficient CartT CartL of friction on the tarmac is μ = 0.20. The tractor exerts a pulling force of 190 N on the cart of mass 35 kg, as shown at right. The breaking tension in the rope is 150 N. a Draw a force vector diagram for each of the two masses. b Calculate the magnitude of the acceleration of the two carts. c Calculate the tension in the rope which connects the two masses. d If the two masses are to be accelerated at a higher rate, what is the greatest acceleration of the carts before the rope breaks? THINK

a

1

2

3

b

Each cart can be represented as a particle with its own set of forces acting.

WRITE

a

N ~1 (CartT) T ~

F1 ~

The trailing cart (CartT) will have four forces acting on it. These forces are the normal contact force, N1; the weight force, W1; the force of friction, F1; and the tensile force, T , acting to right. the

W ~1 Mass 1 = 25 kg

The leading cart (CartL) will have five forces, namely the normal contact force, N 2; the weight force, W2; the frictionand tensile forces, F2 and T , acting to the left; and the force of the tractor, A, acting towards the right.

N2 ~ ~F2

A ~ (Cart ) L

~T W ~2 Mass 2 = 35 kg

R = (T − F1 ) i + ( N1 − 25g) j = 25ai + 0 j N1 − 25g = 0

1

Use Newton’s second law to write the equation of motion for CartT.

2

Equate the j components of the equation.

3

Solve for N1.

N1 = 25g

4

Find the friction force acting on CartT using F1 = μ N1.

F1 = µ N1 = 0.2(25g) = 5g

5

Equate the i components of the equation ofmotion.

6

Substitute F1 = 5g into the equation and call it equation 1.

T − 5g = 25a

7

Use Newton’s second law to write the equation of motion for CartL.

CartL:

8

Equate the j component of the equation. Solve for N2.

9 10

Find the friction force F2 acting on CartL.

b

CartT:

T − F1 = 25a [1]

R = (190 − F2 − T ) i + ( N 2 − 35g) j = 35ai + 0 j N2 − 35g = 0 N2 = 35g F2 = 0.2(35g) = 7g


85

11

Equate the i components of the equation ofmotion for CartL.

190 − F2 − T = 35a

12

Substitute F2 = 7g into the equations and call it equation 2.

190 − 7g − T = 35a

13

Use CAS to solve equations 1 and 2, by completing the entry line as: solve(t − 5 × 9.8 = 25 × a and 190 − 7 × 9.8 − t = 35 × a, {t, a}) then pressing ENTER ·.

14

Write the solution.

c Write the solution. d

86

1

To find the maximum acceleration substitute T = 150 into equation 1.

2

Solve for a.

3

State the solution.


[2]

a = 1.2 m/s2, correct to 1 decimal place. c

T = 79 N correct to the nearest whole number. Therefore the tension in the rope is 79 N. Substitute T = 150 into equation 1: 150 − 5g = 25a a = 4.04 The maximum acceleration of the carts before the rope breaks is 4.04 m/s2.

Maths Quest Specialist 12 Textbook 4E TI-Nspire CAS Companion

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