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DIFFERENTIATION DIFFERENTIATION.............................................................. ............................................................................................ ............................................................. .............................................................. ...................................... ....... 4 IMPLICIT DIFFERENTIATION DIFFERENTIATION ............................................................. ............................................................................................ ............................................................. ........................................... ............. 5 PARAMETRIC PARAMETRIC DIFFERENTIATION DIFFERENTIATION ......................................................... ........................................................................................ .............................................................. ...................................... ....... 7 TRIGONOMETRIC TRIGONOMETRIC DIFFERENTIATION DIFFERENTIATION ............................................................ .......................................................................................... .......................................................... ............................ 9 DIFFERENTIATION DIFFERENTIATION OF INVERSE TRIGONOMETRIC TRIGONOMETRIC FUNCTIONS FUNCTIONS ....................................... ................................................................ ......................... 10 DIFFERENTIATION DIFFERENTIATION OF EXPONENTIAL EXPONENTIAL FUNCTIONS FUNCTIONS .......................................................... ......................................................................................... ...............................11 DIFFERENTIATION DIFFERENTIATION OF NATURAL LOGARITHMS LOGARITHMS ........................................................... .......................................................................................... .................................... ..... 13 PARTIAL DERIVATIVES DERIVATIVES .......................................................... ......................................................................................... ............................................................. ........................................................ ..........................15 First Partial Derivative ............................................................ ........................................................................................... ............................................................. ................................................... ..................... 15 Second PARTIAL DERIVATIVE DERIVATIVE ............................................................ ........................................................................................... .............................................................. .................................... .....16 INTEGRATION INTEGRATION RESULTS.................................................................... .................................................................................................. ............................................................. .............................................. ............... 17 Even powers of Odd powers on Even powers of
and
............................................................ ........................................................................................... .............................................................. .................................... ..... 19
and
............................................................ ........................................................................................... .............................................................. .................................... ..... 20
........................................................... .......................................................................................... ............................................................. ........................................................ ..........................21
INTEGRATION INTEGRATION BY PARTS....................................................................... ..................................................................................................... ............................................................. ......................................... ..........22 Reduction Formulae ............................................................ ........................................................................................... ............................................................. ........................................................ ..........................25 PARTIAL FRACTIONS FRACTIONS.......................................................... ......................................................................................... ............................................................. ............................................................. ...............................28 Denominator with Linear Factors ....................................... ...................................................................... .............................................................. ................................................... .................... 28 Denominator with unfactorizable quadratic factor............................................................. ...................................................................................... ..........................29 Denominator with a repeated factor ........................................................... ......................................................................................... ........................................................ ..........................30 Improper Fractions Fractions (degree of numerator
≥
degree of denominator)................................. denominator)................................................ ............... 31
TRAPEZIUM RULE (NUMERICAL INTEGRATION)................. INTEGRATION)................................................ .............................................................. .............................................. ............... 33 COMPLEX NUMBERS........................................................... .......................................................................................... ............................................................. ............................................................. ...............................34 SQUARE ROOT OF NEGATIVE NUMBERS ................................................................................ .......................................................................................................... ..........................34 Operations on Complex Numbers ...................................................... ..................................................................................... .............................................................. .................................... ..... 34 Adding and Subtracting Complex Numbers.............................. .......................................................................................... 34 Multiplying Complex Numbers ................................................................................................................................................. 34 34 Dividing C omplex Numb ers ....................................................................................................................................................... 34
Square Roots of Complex Numbers ..................................... ................................................................... ............................................................. ................................................... .................... 35 Quadratic Equations ............................................................ ........................................................................................... ............................................................. ........................................................ ..........................35 Equations with Real Coefficients .................................................................................................................................. ...........35 Equations with Complex Coeffic ients .................................................................................................................................... 35
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Roots of Equations...................................... ...................................................................... .............................................................. ............................................................. ................................................... ....................36 Argand Diagram...................................... Diagram..................................................................... .............................................................. .............................................................. ........................................................ .........................37
– – De Moivre’s Theorem
Representing Sums and Differences on Argand Diagrams .......................... ................................................................ 37
Modulus Argument Form .................................................................... .................................................................................................. ............................................................. .................................... ..... 37 The Modulus of a Complex Number N umber ....................................................................................................................................... 37 The Argument Ar gument of Complex Number N umber ........................................................................................................................................ 38
Modulus Argument Form ...................................................................................................................... .................................. 39
.......................................................... ......................................................................................... ............................................................. ........................................................ ..........................41
Multiples of Sine and a nd Cosine ...................................................................................................................................................... 43
The Exponential Exponential Form of a Complex Number ............................................... ............................................................................. ................................................... ..................... 43 Locus on the Argand diagram ................................ .............................................................. ............................................................. ............................................................. .................................... ...... 44 SEQUENCES............................................................. ............................................................................................ .............................................................. ............................................................. ................................................... ..................... 49 SEQUENCES.............................................................. ............................................................................................ ............................................................. ............................................................. .............................................. ................ 50 Types of Sequences ......................................................................... ....................................................................................................... ............................................................. .............................................. ............... 51 Convergent Sequences S equences ...................................................................................................................................................... ...........51 Divergent Sequences ..................................................................................................................................................................... 51
Convergence of a Sequence ....................................................................................... ..................................................................................................................... .............................................. ................ 52 Recurrence Relations .......................................................... ......................................................................................... ............................................................. ........................................................ ..........................52 SERIES ............................................................ ............................................................................................ ............................................................... ............................................................. ............................................................. ...............................54 SERIES ............................................................ ............................................................................................ ............................................................... ............................................................. ........................................................ ..........................55 Using Sigma Notation.......................................................................... ........................................................................................................ ............................................................. ......................................... ..........55 Sum of a Series .................................................................. ................................................................................................. ............................................................. ............................................................. ...............................55 Mathematical Induction .......................................................... ......................................................................................... ............................................................. ................................................... .....................56 Method of Differences .................................................... ................................................................................... ............................................................. ............................................................. ............................... 57 ARITHMETIC ARITHMETIC PROGRESSIONS PROGRESSIONS............................................................. ............................................................................................ .............................................................. .................................... ..... 58 GEOMETRIC PROGRESSIONS PROGRESSIONS.................................................................................... .................................................................................................................. .............................................. ................ 60
MACLAURIN’S SERIES PASCAL’S TRIANGLE
............................................................. ............................................................................................ ............................................................. ........................................................ ..........................63
TAYLOR SERIES ........................................................... ......................................................................................... ............................................................. .............................................................. .................................... ..... 66
BINOMIAL THEOREM.............................................................. ............................................................................................ ............................................................. .............................................................. ...............................68 ........................................................... .......................................................................................... ............................................................. ............................................................. ...............................69
FACTORIALS ............................................................ .......................................................................................... ............................................................. ............................................................. .............................................. ................ 69 THE BINOMIAL THEOREM ......................................................................... ....................................................................................................... ............................................................. .................................... ..... 71 Extension of the Binomial Expansion ................................. ............................................................... ............................................................ ................................................... ..................... 72
ROOTS OF EQUATIONS EQUATIONS ........................................................... .......................................................................................... ............................................................. ............................................................. ...............................76
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THE INTERMEDIATE INTERMEDIATE VALUE THEOREM ............................................................. ........................................................................................... ................................................... ..................... 77 DETERMINING DETERMINING THE ROOTS OF AN EQUATION EQUATION .................................................................... ................................................................................................... ...............................78 BISECTION METHOD................................................................................ .............................................................................................................. ............................................................. .................................... ..... 78 LINEAR INTERPOLATION INTERPOLATION ........................................................... .......................................................................................... ............................................................. .............................................. ................78 NEWTON RAPHSON................................................................... ................................................................................................. ............................................................ ................................................... ..................... 79 DERIVING AN ITERATIVE ITERATIVE FORMULA ................................................ .............................................................................. ............................................................. .................................... ..... 80 MATRICES........................................................... .......................................................................................... .............................................................. ............................................................. ........................................................ ..........................81 MATRICES........................................................... .......................................................................................... .............................................................. ............................................................. ................................................... ..................... 82 Matrix Multiplication................................................................. ............................................................................................... ............................................................ ................................................... .....................82 THE DETERMINANT OF A
×
MATRIX ........................................................... ......................................................................................... .............................................. ................ 83
The Transpose of a Matrix........................................................... .......................................................................................... ............................................................. .............................................. ................ 84 Finding the inverse of A Matrix (Cofactor Method) ............................................................................ ...................................................................................... .......... 85 SYSTEMS OF EQUATIONS EQUATIONS............................................................ ........................................................................................... ............................................................. .............................................. ................86 ROW REDUCTION REDUCTION ............................................................ ........................................................................................... ............................................................. ............................................................. ...............................87 Row Reduction and Systems of Equations .......................................................................................................................... 87
DIFFERENTIAL EQUATIONS EQUATIONS........................................................... .......................................................................................... ............................................................. ................................................... ..................... 90 DIFFERENTIAL DIFFERENTIAL EQUATIONS EQUATIONS ........................................................... .......................................................................................... ............................................................. .............................................. ................ 91 Separable Differential Equations.................................... Equations................................................................... .............................................................. ........................................................ .........................91 The Integrating Integrating Factor.................................................................................. ................................................................................................................ ............................................................. ...............................92 Linear Differential Equations with Constant Coefficients........................................................... ........................................................................... ................ 93
–
Homogenous Differential Equations ...................................................................................................................................... 93 Non Homogeneous Differential Equations ...................................................................................................................... 94
Differential Equations Requiring a Substitution......................................................... ........................................................................................ .................................... ..... 98 Mathematical Modelling.......................................................... ......................................................................................... ............................................................. ................................................ .................. 101
DIFFERENTIATION
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At the end of this section students should be able to:
ln
–
,,
1. find the derivative of , where is is a differentiable function of ; 2. find the derivative of (to include functions of (to polynomials or trigonometric); trigonometric); 3. apply the chain rule to obtain gradients and equations of tangents and normals to curves given by their parametric equations; 4. use the concept of implicit differentiation, with the assumption that one of the variables is a function of the other; 5. differentiate differentiate any combinations of polynomials, trigonometric, trigonometric, exponential and logarithmic functions; 6. differentiate differentiate inverse trigonometric functions 7. obtain second derivatives, , of the functions in 3, 4, 5 above; . 8. find the first and second partial derivatives of
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IMPLICIT DIFFERENTIATION
0 5
A function which is written in the form is is called an explicit function: is stated explicitly in terms of . However, functions such as or
are implicit
LESSON 4 4
functions.
LESSON 1 1
Differentiate
2 SOLUTION
LESSON 2 determine SOLUTION
SOLUTION
Determine
3 3 5 22 3 0 33322 SOLUTION
3 3 5
6 6 2 3 6 2 3 2 2 62 0 2 2 6 2 2 2 622 2 6 2 2 62 6 2 636 0 2 3 31,13, 3 1 1 0 1 2611262 32 1 LESSON 5
Find the equations of the
tangents at the points where
Use implicit differentiation to
1 for
Determine
for
with respect to .
.
SOLUTION
.
1 1 2 2 0 2 22 2 5 0 5 0 4 1110 10 0 1 4 LESSON 3 3
410 1
When
for
.
Gradient at (6,
Equation of line: Using (6,
)
)
on the curve
.
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1 32 6 10 3 2 10 6232 62 32 3 32 6 12 3 2 12 22 6 22 6 2 2 2 0 2 1 212 2 2 2 1 1 0 2221 0 2 2 0 6 2 22 60 6 0 33,33, 2 2 2 0 3,3,3,333,2 2,2, 2 2 2 2221 11 22 222 Gradient at (6, 3)
Equation of line Using (6, 3)
LESSON 6
Find and classify the stationary
points on the curve
.
SOLUTION
Stationary points occur when
Sub.
into
and and
.
We have already stated that
2 2 2220 1 2 3,3,2331 1 2 1 1 2332 1 25 → Maximum 2,2, 2 2 222 1 5 → Minimum For
For
0
and
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× 5 3 2 5 1 2 2 3 × 2 3 31 22 3 3 2 22 3 1 44 1 1 √ 2 7 7 1 4 11 4 122 7 7 2 √ 21 7 2 2 7 7−−2 2 7 7
PARAMETRIC DIFFERENTIATION
Given that and and a parameter, then
where where is is called
LESSON 1 Find the gradient of the stated curve at the point defined. ; where SOLUTION
when
LESSON 2 Find the gradient of the stated curve at the point defined. ;
SOLUTION
when
÷ 122 72− 266 1√ √ 22 7 1 26261 112 1 7 2 301 , 22 11 8 8−− 24 24 24 1 4× × 24 6 1 88 1 8 2 222 2 1 1 7 when
LESSON 3 to the curve
Find the equation of the normal at the point where
the curve crosses the line SOLUTION
when
when
(1, 7)
.
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6 2 At (1, 7):
6 16 3 7 31 10916 16 163 10916 44 1 1 5 444 1 1 3 5 3 × 4 32 332 × 14 8 3 1 6 3 36 1 6 3 Gradient of normal is
LESSON 4 4
Find
and
for the
parametric equations
and
SOLUTION
LESSON 5 5
Find
and
for the
parametric equations and . Hence find and classify the stationary point(s). SOLUTION
2 6 × 23636 1333 3 33 3 33399× 1 1 6 9 6 0 33 3 0 3 3 0 3333 6 31 32626 1212 26,26,1212 613 1621
Stationary points occur when
Minimum point
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TRIGONOMETRIC DIFFERENTIATION
sec 22 33 sec tan sec s ec 22 3 tan s e c ′cos cos sec s ec os 2 3 t a n sec si n sseinc t a n ≡ 2 1 s e c csccot cscsect cotacotn 1 tan sec secccsc tann sec 1 sec 2sec 2secsect sectann 2se 2 sec c 2sec 2 sec tan tan csc 1tan 2se 2 sec c 1t s e c 1 t a n cotcosec4t 5 21tan 1tan 1 tan 1tan 3cs+ 3+csc−−12 12 2 tan 1 tan n 22 tan tan1 tann 22 1 sec4 4sec4t 4sec4 tan4 an4 cot5csc5 5 3cs33csc6612 12 csc12 cot12 1 2 1 2 18 csc12 1 2 cot12 1 2 −− +112csc 2csc4 2csc4 −2 csc 4 − − 4 4csc 4 csc 4 4 cot 4 4 2csc4 4 4 cot44 4 4csc2csc (e)
LESSON 2
SOLUTION
LESSON 1
(a) (b) (c) (d) (e)
SOLUTION (a)
(b)
(c)
(d)
Differentiate the following w.r.t
Recall:
Given that
, show that
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DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS LESSON 1 1 PROOF Let then
Differentiate
sin−
.
sisinsin−, – < < sisin 1 cos cos1 cos sisi n 1 coscos 1 1si 1sisinn 1 √ 1 1 sin− √ 1 1 − cos coscoscos,− , 0 ≤ ≤ cos cos si n 1 1 cossinsisi n 1 ssiinn 1 1cos 1coscos 1 coscos− √ 1 1 − t a n tantan−, < < tann sec 1 1 sescec 1 tan 11 tan 1 1 But
LESSON 2 2 PROOF Let then
Differentiate
But
LESSON 3 3 PROOF Let then
But
Differentiate
.
−− cos− 3 3 cos LESSON 4
Show that if
cos− 3
, then
SOLUTION
iff
cos 33 sincos 3 si3n sin 11 1cos 1cos 3 3 coscos− 33 1 3 1 3 3 31 9 √ 19 sin− √ 1 1 . ′′ cos− √ 1 1 . ′′ tan− 1 1 . ′′
In general, given that is a function of , we have
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DIFFERENTIATION OF EXPONENTIAL FUNCTIONS FUNCTION
DERIVATIVE
LESSON 1 (a) (b) (c) (d) (e)
Differentiate the following
2− + 1 2−− 223cos3 663cos3 cos3 1 [2]
SOLUTION (a)
(b)
(c)
(d)
2 12 1 2 + 2− 1 1 −1 2− 121− − 1 1 2 1 −−2221121 1 13
131 55 15 151 5 5 1 3 2 15 3 2 15 33 22 15 1355 2 sin 2 2 0 sisin cosos sisin c cos sisin co coss coscos s sin 2 cos (e)
LESSON 2 2
Determine
for
SOLUTION
LESSON 3 that
Given that
, prove
SOLUTION
2 2 20 coscos 2 2coscos sisi n 2 sin
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LESSON 4 4
Determine
defined parametrically by
for the equation
1 sin− 2 1 2 sin− 2 1 2 1 22 × 21 4 × 21 √ 14 11 4 √ 14 and
SOLUTION
.
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DIFFERENTIATION OF NATURAL LOGARITHMS
2×2 ×+1 1 2l n 1 1 2212122llnn n 211lnn 1 l n 2 1 1 3 3 2 llnn22 11 11 2l lnn33332222 221 1 2 332 2 22321 1223662222 1 6 24 1 11233 6 2 221911331 2 22 13 2 2 +− l n 1 ≡ 2 ln11 1ln1 11 1 1 1 1 2 1 1 2 . 1 2 (vii)
LESSON 1 following. (ii) (iii) (iv) (v) (vi)
1 Differentiate each of the
l3lnn377112 llnnsisin 4411 ln + ln22 133 2 ln33 1 1 3 1 3l33ln7772 2 7217 2 2
(vii)
(viii)
SOLUTION (ii)
(iii)
(iv)
(v)
(vi)
ln 2 1 11 1 ln 4siscoscoins444 4co4sicotnt44
22lnlnn 1 22 lnn 2l2ln 1
(viii)
LESSON 2
that
SOLUTION
Given that
, show
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LESSON 3 3
Differentiate
SOLUTION
3 l1n ln 3 ln3 3l nln33
3 w.r.t
ln − ln ln √ 3 l1n l n1 l1ln2 3 1 2 3 3 3 3 3 3 3 3 Sub 3 33 3 3 3 In general, if LESSON 4
SOLUTION
then
If
, find
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PARTIAL DERIVATIVES For partial derivatives we differentiate with respect to one variable and treat the other variable(s) as constants.
,, 4ln ,, 44 ln 2 4ln 4 4ln 8 ,, 442ln 48 LESSON 3
Given that
, determine
(i)
(ii)
SOLUTION
LESSON 1
Given that
,, 22 ,, ,, 2 52 3 2 52 ,, 2 2 − 4 25 − 4 5 3 2 3 2 8843333 233 266 22 3 2 88433 23 222 22 , evaluate
(i)
(ii)
SOLUTION
(i) For
we differentiate
with with respect to
, treating as a constant.
(ii)
LESSON 2
(i)
(ii)
LESSON 4
determine (i) (ii) (iii)
Given that
, determine
SOLUTION (i)
(i)
(ii)
(ii)
SOLUTION
(i)
(ii)
Given that
(iii)
2 4 3
,
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2 ,, 2,2, 1 ,, 3 2 2 2,2, 1 321 12 LESSON 5 5
For
determine
.
,
SOLUTION
LESSON 1
Given that
,, 4 4 ,, 3 444 22 −− 6 6 ,, 4 42 3 4 2 3 8 ,, 844− 3 8 2
, determine
(i)
(ii)
(iii)
SOLUTION
(i)
(ii)
(iii)
NB:
LESSON 2
Given that
ln sinn (i)
, determine
(ii) (iii) (iv) (v) (vi)
SOLUTION
ln1 sin 48 ln 12 sin 2 222 3−coscos3− 44 322 cos 23 32 322 sin 23 4 32 cos23 94 sin23 ln2 sin 2− 2 2 3cos2 3 2 cos3 3 22 − cos − 2 34 cos23 32 32sin23 22 34 cos23 49 sin 23 (i)
(ii)
(iii)
(iv)
(v)
(vi)
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INTEGRATION RESULTS SOLUTION
+ ′ √′
FUNCTION
LESSON 1
(i) (ii)
(v) (vi)
ln|| ln| | 1 1 ++ 1 + llnn||cos| csec| ossec|| 1 lnsecsec ln|sesectanttan| lntan2 ln|−sisin|| sin 1 tan−
(vii)
(i)
(ii)
(iii)
or
Determine
∫ ∫ − ∫ −+ ∫∫ tanan+− ∫ 12∫ 332l2l n5
(iii) (iv)
INTEGRAL
(iv)
(v)
(vi)
∫ 1 22 l2ln n| | ∫−ln|44 1| ∫ −5 2 2 12 +52 ln|12| 1 2| ∫ 1+− 4 6 2 2 6 9 12 ln|22 6 9| ∫ tan sin cos sicosn ∫ 12ln|cos| c33os | 5 3 3 5 2 ln ∫ 12ln 33 2 ln 2 ln4 32ln
(vii)
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LESSON 2 (i) (ii) (iii) (iv) (v) (vi)
Evaluate each of the following.
∫∫ 2− ∫ 2 ∫ ++ ∫∫ 3 √ 6 + ∫∫ csc
(v)
(vii)
(viii)
(vi)
SOLUTION (i)
(ii)
(iii)
(iv)
∫ 1 2 − ∫ 2 − 2 2 22 11−− 2 ∫ 22 4 ∫ ++ 5−1 −−− 25 2−1 −−−−− 52 1 1 52 1 5−−− 21 −
∫ √ 6 6 14 4 6 14 36+ 2 16 6 ∫3 ∫ + ∫csc ∫ −− ∫∫ −− + −− ∫ −− 1 √ 2 2 sin− 2 ∫ −− 1 + 3 2 22 1 23 3 222 2 1 1−− 2 3 22
(vii)
(viii)
LESSON 3 3 (a) (b) (c)
SOLUTION (a)
(b)
Determine
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(c)
12 1222 1−− 32113 tan−223 4 22 1−− 2 tan− 3 ∫ −−
Even powers of
and
sin sin csions2 11cos2 2 sisin 2 2 sin 1 cos2 2sin22 2 4 cos cos cos co s 2 2co 2 cos s 1 1 1 cos2 cos 2 2 cos 1 cos2 2sin22 2 4 LESSON 1 1
Determine
SOLUTION
Using the identity
LESSON 2
Determine
SOLUTION
Using the identity
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sin sinsin 1si ncossisi n sinsin sin 1cos 1cos s sinn sincn cos sinn cos sisin cos co coss 3 cos cos co coss cos cos cos 1si 1sin c cos sin cos sin sin 3 sin 2 sin 2 sin2sisi2 n 2 sisi n 22 1cos 1cos 22 1sisi n 2 cos sin22cos 2cos 2 2 cos2 6 LESSON 1
Determine
SOLUTION
Since
LESSON 2 2
Determine
SOLUTION
LESSON 3 3
SOLUTION
Determine
cos sin cos sin co coss coscos sin cos cos 1si 1si n sin scoincos s sisisnin cos sin 5 7 LESSON 4 4
SOLUTION
Determine
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tan tansec 1sec se tcan 1 1 t a n 1si n cos1 sin cos cos lsen|seseseccttttaann| ln|sec| | s ec | | ln sec sec sec t tan tan 2 tan 2 1se secc 2 1 2 tan 2 LESSON 1
Determine
SOLUTION
Since
LESSON 2 2
SOLUTION
Determine
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–– –– –
INTEGRATION BY PARTS
When choosing we use the following acronym. Logarithms
Inverses Algebra
Trigonometric Ratios
Exponentials
LESSON 1 1
SOLUTION
Determine
1 LESSON 2 2 SOLUTION
Determine
121 2 3 3 1681 2 3 3 1681 2 3 31414 2 2 3 1681 2 3 31212 3 3 56 2 3 34 1 1 1 2 1 √ 312 1 2 1 33 11 22− √ 312 3 1 − 3 3 12 121 1 22 √ 12 1 2 3 1 11 2 2 3 312 1 2 3 1 11 2 2 1 2 2 1 2 23 1 1 2 12 1 2 2 2 c cos cos 1 coscos sisin c cos sin sin s sin cos cos LESSON 3 3
SOLUTION
LESSON 4 4
∫ 22 3
2 3 3 1 1 2 3 3 12 2 3 3 2 3 3 12 1 2 3 3 121 22 3
Evaluate
SOLUTION
Evaluate
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LESSON 5 5
Determine
ln
SOLUTION
lnn 1.ln ln 1 ln1 lnn 1 1 lnn LESSON 6 6
SOLUTION
SOLUTION
Evaluate
sisin
Determine
lnn ln 1 3 lnn 3 ln 3 13 ln 19 9 3l3ln 1
LESSON 7 7
lnn
sin
2 sisin coscos sin coscos 2 cos 22 2 22cosco s cos sisin 2222 ssisiin— nn — 22sicosconns sisin 2cos 2cos— — cocoss 022 2cos 2co s0sisin 2 coscos 2 4 2 sisin sisisisinn coscos 1 12 1 coscosisisn 2 sisin 2 cos sisin 12 LESSON 8 8
SOLUTION
Evaluate
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1sin 1 1 1 2 sisin 2 12 coscos 12 sisin 1 5 sisin 2 1 sisin 4 1 coscos 4 sisin 4 sisinsisin 2 2 sisinsisin 1 4 cocoss 5 5
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LESSON 9
Establish a reduction formula
that could be used to find
∫
and use it
4 − − − 4 −− 44 3 44 12121212 2 44 12121212 24242424 4 1212 24 24 24 4 1212 24 24 44 12121212 24 24 24 24 24 24 ≡ cos ∫ sincos∫ cos− − −− when
.
SOLUTION
LESSON 10
If
show that
.
Hence, find
.
SOLUTION
STEP 1: Write the integral in the product form
co coss cos− cos− 1 1 cos− sisin STEP 2: Integrate by parts or an appropriate
method
coscos sisin sin coscos−− 1 1 cos− sin
STEP 3: Simplify
sincos− 1 1 cos− 1 co coss sin coscos−− 1 1 cos− 1cos 1cos
sincos−− 1 1 cos− 1 1 cos
sin cos sincos 1−1− − 111−1− 1sin coscos−− 1111−− sincos −−
STEP 4: Apply the derived formula to the rest of the question
51 4 5 sincos 5 31 2 3 sincos 3
STEP 5: When you have reduced your integral to its lowest form, go back to the original integral &
1 c cos sisin 15 sincos 45 13 sincos 23 sin ∫ 1 1 2 1 1 −− 1 1 plug in the final value of .
LESSON 11
that
SOLUTION
If
, show
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1 1 1 11−− 1 2 −− 1 2 1 1 11 . 2 12 1 1 2 11−− 12 1 1 2 −− 2 1 1 −− ∫lnln , ≥ 0.0. ≥ 1,1, 12 12 −− ln 22ln ln −−. 1 ln 22ln ln −− 22ln ln ln −− . . 2 ln ln −− ln 1 1 ln 2 ln 1 2 −− 112 32 −− 12 32 1 1 3 3 2 1 2 32 1 1 2 4 2 4 2 2 2 LESSON 12
It is given that By By considering
or otherwise, show that for or
Hence, find SOLUTION
in terms of
METHOD 1
Rearrange the equation
Take integrals of both sides
14 34 34 12 34 ln 1 21 34312 3 12 12 83 8 8 8 METHOD 2
ln ln 1ln −−. 1 1 2 2 ln 1 2 ln −− 112 ln 1 2 −− 2 2 −− is found the same way
4 sisin
LESSON 13 By using the substitution , find the exact value of
1 √ 16 1 6
− 16 1− √16 1 6 16√16
Show that
∫1616 −− 11 −,− 2≥√ 23 44cos4cosisin s 2;2;4cos 4cos 6
Deduce, or prove otherwise, that if for for
SOLUTION
, then
. Hence find
.
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0; 0 11 6 √ 16 1 16 4 cos 16 4si 4 si n 4cos √ 1616si 1 616sin 164cos 16 1 sisi n cos √ 1si 1 sin cos cos 1 1 0 6 − 16 11 6 −. 2 −16 1 6. 22 16 1 6 1 1− − 2√ 216 1 1 1 6 16 1 11616 − √ 16 1 6 1 1√ 16 116166− √ 16 1 6 1616 1√ 11616− 1 1 √ 16 √ 16 √ 16 1 6 1 6 1 6 166√ 16 1161− √ 16 1 6 − 16 1− √16 1 6 166√16
√ 16 1 6
− 1616 1 √16 − 16 1 6 2
1616 1 1−− −− 16 1 6 0 1616 1 1−− 2 1√ 1212 1616 1 1−− 2. 2 . √ √ 4.4. √ √3 2 161616 121√ −3− 2 √ 2 2 16 4√ 3 11 6 sin− 4 20 6 √ 16 2 4 83 4√ 3 3 2√ 3
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PARTIAL FRACTIONS LESSON 2
Denominator with Linear Factors LESSON 1 1 fractions.
Express
−−−
in partial
2 27 2227 1 1 ≡ 2 2 1 1 2 7 1 1 1 2 2 2 3 3 9 3 13 1 2 13 1 2 27 2 2 72 21 2 13 1 ln|112|2|3l n| 1 1| ln 2 2 233 111333 3 1 3 3 2 11113 33 33 1 3 3 3 33 1 24 112 2 0 3 13 33 33 3 2 2 12 2 233 111333 1 2311 1 213 3 SOLUTION
When
(a) Express
(b) Hence determine SOLUTION (a)
Comparing coefficients of
in partial fractions
∫ +−−
++ −−
132 19195 6 1113132219 3 3 13 After factorizing the denominator we get
When
13 19 1313 1 19 119 223232 1 331 22 1133
3 121 2 2 453 15 3 202 10 1 616 132 19195 6 11 1 23 2 32 3 13 +−− ∫ ++ ln|11111 |3l n|223 22| 2l n|33233| When
When
When
When
When
++ −− +
(b)
P a g e | 29
LESSON 3 3
++ +++++ +
Express
++ + +++ ++
in partial
5 2 2 6 22 5 6 2 5 4 5 5341 3 3 3331 333 1 5 2 26 2 25 2225 2 5 2 2 2 2 5 2 3 3 3 3 3 2 5 6 3 3 1 5 254 6 3 321 5 5 2 2 5 8 10 36 4 4 2 5 2 3 3 1 5 4 3 2 2 5 6 2 3 1 3 3 3 1 3 3 222 32 4 5 2 2 2 5 −+ + 3 1 1 7 l n 2 2 −+ 2 5 ∫ + 1 7 3 21 l n 2 2 5 2 5 2 5 2 2 51 2 1 5 27 5 32 2 21 l n 2 2 2 51 2 1 ln 2 2 32 ln| 2 5| 7 1 11 4 2 5 2 1 ln 2 2 32 ln| 2 5| 7 1 11 2 3 7 − | | 1 1 l n 2 2 l n 2 5| 5 t a n 25 2 2 2 22 0 2 5 2 2 5 1− ln 5t5 tan LESSON 1 1
Express
fractions.
fractions and hence determine
in partial
SOLUTION
SOLUTION
LESSON 2 2
Express
and hence determine SOLUTION
in partial fractions .
P a g e | 30
LESSON 1 1
Express
fractions
−−−+
in partial
63 99 36 999 3339 3339 3 3 33 393 3 3 330 93 3 91 9 11 1 2 63 99 1 32 3 333 SOLUTION
When
When
Equating coefficients of
LESSON 2 2
Express
+++ −−
in partial
fractions and hence determine
5 2 2 6 22 5 9 24 2 99 24 2 24 2 49 3218 2 2 3 3 2 2 293243 2323 3322333 3 2 2 3 2 1535 502 25 22 2 02 9 24 2 3 4 3 18 2 2 3 3 2 4 932184 SOLUTION
When
Equating coefficients of
When
22 2 3 3 3 3−− 2ln| 2 2| 33 3
P a g e | 31
≥
+−− + + 2 31 1 4 2 31 1 4 1 1 2 3 4 1 1 1 1 1 24 4 4 1111 3 3 23 3 2 3 2 31 1 4 22 3 4 12 1 2 31 1 4 2 2 3 4− 12 1 22 l ln 4 2ln| 1 1| + + +− −+ ∫ ++−− If for
,
has has degree and
has has degree
, then quotient has degree
LESSON 1
Express
in partial
fractions and hence determine
SOLUTION
LESSON 2 2
2 3 32 1 122222 12221 1 2222 2 1 1 2 1414 5 11 5 4 2 5 31110 2 → 2 3 11 14 3 2 1 31 222 2 101022 1 5 2 2 2 1 1 2 2 32 1120 221 1 145 21 2 32 1120 ln|22 1| 145 ln| 2 2| SOLUTION
Express
fractions and hence determine
in in partial
When
When
Equating coefficients of
−−− ++ ++ ++ + 1919 44 21
LESSON 3 3
Express
in
partial fractions and hence determine
3 3 2 2 11 33 1922 441 121 3 3 2 2 1 1 19 44 21 3 3 2 2 1 1 193 3 4242211 1 2 2 1 1 3 33 1 1 3 3222 63 2 11 1 42 2 SOLUTION
When
When
When
P a g e | 32
1 11 666 5 3 31919 22 4444 1 1221 5 33 3 21 2 12 1 3 31919 22 1441 121 1 1 5 5 3 3 2 2 Equating coefficients of Equating coefficients of
3 3 2 2
1 1
2 5 3ln| 3 3| ln| 2 2| 2l n| 1 1|
P a g e | 33
TRAPEZIUM RULE (NUMERICAL INTEGRATION) Introduction
1
The area under the curve equal width (as shown above)
can be estimated by finding the sum of the areas of trapeziums of
− ℎ2 ℎ2 ⋯ ℎ2 − ℎℎ2 ⋯− 2 2 ⋯ ⋯ − The area of a trapezium with parallel sides The width of each trapezium is
and
where is the number of trapeziums.
Thus we have area under curve is
≈ 2 2 ⋯− LESSON 1
1 − 0.4
Using 5 trapeziums, estimate
0.00.4 00. 41 11 1.1616 0.1.01..82 0.1.82 11 1.2.64446444 1.12.6 21. 6 1 15 3.5656 2 0 1 ≈ 25 1 5 5 21.1.161.642.443.566 ≈ 4.72 SOLUTION
ℎ
and width is given by the formula
Width of each trapezium is
P a g e | 34
COMPLEX NUMBERS INTRODUCTION
̅
If then then is its conjugate and vice versa. It is also important to note that the product of a complex number and its conjugate is which is always a Real number. The conjugate is also denoted .
∗ √ 11 1 + : 2 → ̅ 22 ,, 3 2 2 6 32 62 31 3 5 6 5 5 √ 1616 111616 √ 1√ 1√ 1616 4 + − √ 1818 9211 3√ 22 37 52 552 2 1355 7272414152 5 2 1 451 214 ± ±± ±± 1 2929 4129 62i 6 2i 54i 5 4i 116 5 25 2 4 1 1 82 8 2 18 18 12 12 9 22334 36 8861 616 6 6 8 32 3 254 5 4 15151515 81212 12121010 10 810 8 23 23 2 2 Complex numbers are written in the form where and are real numbers and is is the imaginary unit such that or
Sometimes the letter is used to denote a complex number, . A complex number can also be written as an ordered pair of its real numbers, . is also known as the real part i.e.
(i)
First we find the conjugate of the denominator
Multiply the numerator and the denominator by the conjugate
is also known as the imaginary part i.e.
SQUARE ROOT OF NEGATIVE NUMBERS
Thus complex numbers can be used to find the square roots of negative numbers. Examples
(ii) Express
With this extension of the number system we can now solve equations which we once unsolvable.
For example, (i)
(ii)
(i) (ii) (iii)
(iv)
in in the form
P a g e | 35
SOLUTION
11 0 ±√ ±±√ 11 33±33 0 413 3±3 ± 3231 3±3 ±2√ √ 33 4 22 11 4 2221 ± 022 441 2 ± 1212 24 √ 8 2 ±4 2 ±±2√ 43383 11 1 8√ 3 4± 4 22 2 2 1515 10 10 0 22 222 2 1510 1 510 0 2 22 2 1 5 1 0 0 1510 2 2 ± 2221 44111510 2 2 ± 22 2 222 2 2 60 40 22 2 2 ± √ 28 6 60 40 22 2 2 ±2 √ 6032 6032 22 √ 606060323232 (a)
LESSON 1 1
Find
√ 1515 88
SOLUTION We assume that the square root of a complex number is a complex number
√ 1515 8 15 8 15 8 22 15 2 4 8 4 15 16 15 1515 16 0 1616 1 0 16 1 ±1 ±4 √ 15 8 4 or 4 ±±4
Invalid since is real
Thus we see that a complex number has 2 square roots, which are complex numbers.
LESSON 1 (a) (b) (c)
Solve the following equations
1330 0 4 2 1
(b)
(c)
LESSON 2 2
SOLUTION
Determine such that
P a g e | 36
6060 3232 60 22 16 → 16 25616 60 6060 60256256 0 46464 4 0 ±2 2 82 162 8 2 82 216 8 2 2 ±2 82 82 22 2 2 2 82 82 5 22 2 2 2 82 82 3 2 22 0 ±√ ± √ 2 44 4,4, > 0 −±√ −− 2 ± 2 When
When
For the equation
Letting
, we have
As a result we can conclude that if a quadratic equation has complex roots they occur in conjugates.
2 2
In general, if a polynomial has complex roots they occur in conjugate pairs. For example, if is is the root of a polynomial equation then is is also a root of the same equation. Recall:
0 sum of ro ot s prproduct of roots 0 0 ,, andand 543 5 5 5 555 5525 0 25 0 4 3 4 3 3 43 4 34 334 3 25 8 8 25 25 0 15 12 0 If
, then
and
where and are the roots of the equation i.e.
Also, if then
,
where
and
are the roots of the
equation.
LESSON 1
Given one root find the equation
(i) (ii)
SOLUTION
(i) Let
then then
Equation is
(ii) Let
then then
Equation is
LESSON 2 equation roots.
Given that
is is a root of the , find the other 2
SOLUTION Since complex roots occur in conjugate pairs and a cubic polynomial has 3 roots one root must be real.
21, 1 2, ∈ ℝ 12 1 2 112 2 3 1 1 2 1 Let
P a g e | 37
INTRODUCTION
,,
A complex number can can be represented on a diagram called an Argand diagram as (i) a point with coordinates
(ii) a vector
5 8 2 58 2 7 9 58 2 3 7 LESSON 1 1
Find
and and
and
and for . Hence, represent on Argand diagrams.
SOLUTION
(i) (ii)
| | || √ 13 4 511 √ 33 | | 1 1 1 √ 2
The modulus of a complex number, , is a measure of the magnitude of , and is written as .
Thus modulus LESSON 1
(a) (b)
(c) (d)
SOLUTION
(a)
.
Determine the modulus of
P a g e | 38
| | |334 344 | 33 4 5 11 √ 33 | | 51 1 √ 33 11 √ √ 33 2 || 5 | | 3 4 22 3,3,4 2,2,11 (b)
SOLUTION (a)
(c)
(d)
LESSON 2 what is
If ?
and and
,
(b)
SOLUTION
We are trying to find the distance between and . In other words, what is the distance between the points and and on on the Argand Diagram?
(c)
(d)
| | |34 2 | |32 32 41 41| 32 32 41 41 √ 5050 arg t–
The angle is called the argument of ( ) where is the angle the vector representing the complex number on the Argand diagram makes with the positive real axis. Thus
. To
avoid complications we use and this is known as the principal argument of . LESSON 1
(a) (b)
(c) (d)
Determine the argument of
arg 1 tan− 11 4 arg 34 2. 2t1 an− 43 arg 1 1√ 33 2 tan− √ 13 3 arg 5
P a g e | 39
5c 5 c os o s s s i n n 2 3 2 3 cosco s sisin 6 5c5 cosos6si n ∴ cosco s sisin || 5c55cos os 6 sisin 6 cos sinn ar g 6 1 – co c o s s i n √ 2cos 2cos4 sinn4 c o s s i n – coscos coscos s in sin 113 4 511 √ 33 coscoscossinncosscions sinn sisinncos si n coscocscossisincoscos sisisisin ncossisin √11 2cos 2cos sin 4 4 coscos sin 53coscos42.2.211 sin2.2.211 + + 2cos 11 √ 33 2 sinn 2 c+ o+ s s s i n c o s s i n c o s s i n n c o s s i n n 55cos s3inn 3 scocoss co cosssisicoscosnnsisi sisi nnco ncos s sinsin cosccoosco os coss sisi n sisinn sisincos cossin coscos sin c c os o s sisi n cos5cosisisn 5cos si sinn ̅ c o s s i n n cos c os s i n || cosco1s si sinn | | | | | 3 | | ar g 5sisin55cos arg || argarg argg arg arg arg 5sin 3cos3 If
Therefore, is
has has modulus and argument then and
in in modulus argument form
LESSON 3 and
LESSON 1 Write the following in modulus argument form
(a)
(a) (b)
SOLUTION
(c) (d)
(a)
Prove that for
(b)
SOLUTION
(i)
(ii)
(iii)
SOLUTION
(b)
(iv)
LESSON 2 Find the modulus and argument of the following
NB:
If
(i)
(ii)
SOLUTION
(i)
Furthermore, we can conclude
(a)
(b)
(ii)
(c)
(d)
then then
P a g e | 40
LESSON 4
11
1 √ 33
Given that , determine (i)
argument form.
and (ii)
and and
in modulus
SOLUTION
|| 1 √ 33 2 arg tan− √ 13 3 || 1 1 1 √ 2 arg t tan− 1 4 2√ 2cos 2cos7 si7n 2√ 2 2 2 c os s i n √ 12 12 √ cos si n √ 22cos 12 sin 12 (i)
(ii)
–
P a g e | 41
INTRODUCTION
ccosos sisin coscos sisi n 2 2 sincn coss coscos 2 s sin22 then then
This can be extended to give
LESSON 1
SOLUTION
cos cos sin Use De Moivre’s Theorem to prove the fol owingidentities coscos 4 ≡ 8cos 8 coscos 1 1
4 cos4 sin4 cos sinn cos 4 coscos sinn 6co 6coss sin 4c4 coss sin sin cos 4 4cosos sin 6 6 cos sin 4cosos sisin sisi n cos4 cos 6 cossin sisi n cos 6 6 cos 1 co coss 1cos 1cos 1 cosos cos 6 6 cos 6co6 coss co coss 2 cos 1 8co8 coss 8 coscos 1 1 Use de Moivre’s theorem to show that sisin 5 co coss sinn cos sin s sin ,, 5 cos5 s i n 5 c o s s i n n 1co1 coss 5co 5 coss sin 10cos sin 10cos sin 5cos os sin 1sin cos 5cos sin 10cos sin 10cos sin 5cos sin sin sisin 5 5co5 coss sisin 10co 10coss sin sisi n 5, 10, 10, 1 When
Equating real parts
LESSON 2
where
and are integers determined.
SOLUTION
When
Equating Imaginary Parts
P a g e | 42
Use de Moivre’s theorem to show 3 t a n t a n tan3 13tan cos3 s i n 3 c o s s i n n cos 3co 3 coss sin 3cos 3cos sin sin LESSON 3 that
(ii)
−√ √
1 √ √ 3 – 1t√ 33tan − √ 133 √ √ 33 2 By DeMoi3 vre’s Theorem 3− cos 3 3cosos sisinn 3co 3cosssisin sin 11 √ 33− −2 cos os 3si n 3 sin3 3 2 cos cos 3 cos3 cos 3 3 cos sin 118 cos sinn 1 1 8 sisin 3 3co3 coss sisin sisi n 18 sin3 3 – √ 3√ 3 tan3 cos3 cos3 coscos 3s3incossss ssiinn 3 coscocoss si n cossin 1 2 3 3 3 √ √ √ coscos 3cossi n cos 1 − t a n an t tan √ 3 6 3 t13tan √ 3 2 cos6 s sinn6 By De Moivre’s Theorem cos si n √ 3 3 2cos6si n 6 −√ √ 6 cos106 sisinn10 2 1 0 si n 5 5 cos 1024cos s i n n 3 3 BycosDeMoisivren’sTheor e m 12 √ 23 1024 4 4 cos12× sin12× 1coscos 34sin3 4 51512 512√ 12√ 33 Rewriting form
in Modulus Argument
SOLUTION
Equating Real Parts:
Equating Imaginary Parts
LESSON 5 5
Express
argument form. Hence, find . SOLUTION
LESSON 4
(i)
(ii)
SOLUTION
(i)
Find the value of
in in the modulus
in the form
P a g e | 43
INTRODUCTION
sin cos 1 cos 1 sisin coscos sisin cos sinnco. coscos s sinsisin ∴ 1 cos sin cos sinn 2c 2 c o s 1 cos sinn cos sinn 22 sisin By De Moivre’s Theorem 1 coscos sisin ,so , so that c1os os sisin 1 2co2 coss and 2sin cos 1 4. 1 6. 1 4..1 1 4 6 4 1 14 1 6 cos sin,n , 2co2 coss ∴ 2cos 2cos 2cos2cos 4 42cos2 2cos2 6 16cos 1 2cos2cos 4 8 coscos 2 6 cos 8 coscos 4 4 cos 2 3 3
Expressions for powers of and in terms of sines and cosines of multiples of can be derived using the following results
If
then
LESSON 1 1 Express cosines of multiples of . SOLUTION
If
in terms of
s i n 1 3. 1 3..1 1 3 3 1 1 3 1 cos sin,n , 2sin ∴ 2sin 2sin 3 3 322sin 8sin 1 2sin336sin sin 4 sin3 4 sin LESSON 2 2 multiples of .
Express
in terms of sines of
SOLUTION
If
From Maclaurin’s Theor em cos 1 2!2! 4!4! 6!6! ⋯ sin 3!3! 5!5! ⋯ ! ! ! ! c o s si n 1 1 ⋯ coscos sisi n 1 1 ! ! ! ! ⋯ 1 11 1 Then
This series the expansion of
appears to be similar to
i.e.
Looking at the powers of
P a g e | 44
111 1 Now let’s try the expans ion 11 2!2! 3!3! 4!4! 5!5! ⋯ 11 2!2! 3!3! 4!4! 5!5! ⋯ 11 2 4 ⋯ 3!3! 5!5! ⋯ cos sisin cos sinn 11+ √ −− |11| 1 1 √ 2 arg tan−1 4 √ +22−− √ 1 √ 3 arg | | 1 1 √ 2 4 || −√ 33 1 11 2 arg t tan √ 3 6 || √2 2 5 arg √ 2 arg arg 4 6 12 2 Grouping Imaginary and Real terms
NB: If
∗ − then
We will be using the notation
LESSON 1 If the point in the complex plane corresponds to the complex number , find the locus of in each of the following situations.
|||2 23| 4 | 3 3 | 2 || 3 0 ,0, 0 ,, ∴ | 0 0 0| 3 | | 9 3
(a) (b) (c) SOLUTION (a) The distance between the point and and the point representing the complex number is is 3 CARTESIAN FORM
LESSON 1 Express the following complex numbers in the form . (a)
i.e. a circle with centre at (0, 0) and radius 3
(b)
SOLUTION
(a)
(b)
Let
and
(b)
| 2 2| 4 2,0 2,2, 0 ,, || 2| 2 4 0| 4 | 2 22 | 4 4 2 2 16
Circle with centre and and radius 4 The distance between the point and and the point representing the complex number is is 4 CARTESIAN FORM
P a g e | 45
77 11 227 0 2 4 , 00 Circle with centre
(c)
| 3 3 | 2 3,3,1 3, 3, 1 ,, || 3 33 1 1| | 2 2 3 3 1 1 2
Circle with centre and radius 2 The distance between the point and and the point representing the complex number is is 2. CARTESIAN FORM
LESSON 2 Determine the Cartesian equation of the locus of points satisfying the following conditions. (a) (b)
2−|− 3 √| 3 || + 2| 3| || 22|| 33 33| | | ||| 44 43 32424 3636 3 3 8242412 36360 0 4 4 40,0, 4 −+− √ 3 || 1 1| √1| 3 | √2 32|| 2| | 1 11 | √ 33|222 | 2 2221 1 1414 11 113 012 123 12 3
SOLUTION (a)
CARTESIAN FORM
Circle with centre (b)
and and radius 2
and radius
√
.
,,
LESSON 3 Sketch the locus of the point representing the complex number , given that . Write down the Cartesian equation of the locus.
| 3| | 2 5| | 3| | 2 5| | 3 3| | 25 | 25 ,, 0, 0 , 3 ,, ⊥ 2, 2,55 | 0 3| | 25 25| | 3 3| | 2 2 55| 3 3 2 2 5 5 6 9 4 4 1010 2 25 1616 4 4 20 0 4 5 5 0 arargg 3 33 2 SOLUTION
Rewriting
The distance between the point , representing the complex number , and the point is equal to the distance between and the point . Therefore, we are finding the bisector of .
LESSON 4 Describe and sketch the locus of the points satisfying the following conditions.
(a)
(b)
CARTESIAN FORM
SOLUTION (a)
arg 3 3
P a g e | 46
arg 30 3 0 4 arg− 3 3 4 tan 3 3 4 3 3 t3an>43 1 3,3, 0 ; This is the half line starting at
SOLUTION
| 2| 1 | 0 2 2| 1
Circle with centre
, not
including (3, 0), making an angle of with the
positive real axis. (b)
arargg32 32 32 3 ar 2g2 3 3 2 2 3 3 23 t√a 3n333 √ 3 √ 3 √ 3– 2 3√ 3 > 33,3,2 ; The half line starting at
| 2| ≤ 1
LESSON 6 Shade on an Argand diagram the region in which .
0,0, 2
and and radius 1.
LESSON 7 (a) Sketch on one Argand diagram: (i) the locus of points satisfying
, exclusive,
which makes an angle of with the positive
| | | 2 2| arg 4 | | ≤ | 2 2| – ≤ arargg ≤ ||| |022|| | 20 2 0| 0 ,0, 1 ar2g,2, 0 arg– 0 0 ,0, 1 4
(ii) the locus of points satisfying
real axis.
(b) Shade on your diagram the region in which and and
SOLUTION (a) (i)
2 1 21 1 1 11 1 2,2, 1 LESSON 5 where
Describe and sketch the locus of
SOLUTION
Using vectors
This is the line passing through the point and parallel to the vector
, i.e
This is the perpendicular bisector of the line segment joining the points and and
(ii)
Half line starting at
, excluding,
making an angle of with the positive axis.
P a g e | 47
(b)
(ii) LESSON 8
arg sin−
LESSON 9 (a) Sketch on an Argand diagram the locus of points satisfying the equation
| 6 6| 3
| 6 6| 3 | | 1 < ≤ 1
(b) It is given that satisfies the equation . (i) Write down the greatest possible value of . (ii) Find the greatest possible value of , giving your answer in the form , where . SOLUTION
(a) Circle with centre
arg
0,0, 6
(a) On the same Argand diagram, sketch the loci of points satisfying: (i) (ii)
|arg3 3 33| 5
(b) (i) From your sketch, explain why there is only one complex number satisfying both equations. (ii) Verify that this complex number is
74 –
SOLUTION and and radius 3
3,3,3,3,110
(a) (i) Circle with centre (ii) Half line, starting at making an angle of real axis.
(b) (i) 9 is the largest possible value of
| |
and and radius 5 , exclusive,
with the positive
.
(b) (i) There is only one complex number satisfying both equations since there is only one point of intersection due to the
P a g e | 48
73 | |743 | 4 3 3 | 4434 | 33 5 ar garg744 444 3 tan− 44 34
half-line which starts within the circle. (ii) If is the point of intersection it is must satisfy both conditions.
SEQUENCES
P a g e | 49
At the end of this section, students should be able to:
{}
1. define the concept of a sequence of of terms as a function from the positive integers to the real numbers; 2. write a specific term from the formula for the th term, or from a recurrence relation; 3. describe the behaviour of convergent and divergent sequences, through simple examples; 4. apply mathematical induction to establish properties properties of sequences.
P a g e | 50
SEQUENCES INTRODUCTION A sequence is a list of numbers which obey a particular pattern. Each number in the sequence is called a term of the sequence. These are usually denoted where is the first term, is the second term and is the term. In some cases the sequence can be defined by a formula an expression for the term.
, , , … , −, – 4+ 1 11++ + 441 1 1 3 4423 11 71111 4445 11 15191159 3, 7, +11, 15, 19, …. 12 1 11 23 3 2 1 42 4 3 1 53 5 4 1 64 3 54 55 6 2,2, 2, 3, 4, 5, … 1 1 21 12 21 14 21 81 2 16
(d)
LESSON 1 Write down the first 5 terms of the following sequences: (a) (b) (c) (d)
SOLUTION (a)
(b)
(c)
1 121 1 3211 1 2, 4 , 118,++16, +32, … 11 1 12 1 12 2 11 2 3 1 3 3 11 3 4 1 4 4 11 4 5 1 5 5 1 112 35 14 56 2 , 3, 4 , 5, 6, … 5, 8, 11, 14, …. 2, … , , , , , … 1,1,1, , ,, , , , , …, … × , ×, × , × , × , …. 2,2, , , , …
LESSON 2 For each of the following sequences determine an expression for the term, . (a) (b) 8, (c) (d) (e) (f) (g)
6,
4,
2,
0,
SOLUTION
3 3 2 2 2 10 2 +
(a) Consecutive terms differ by 3 therefore we try . To create the right formula we add 2 i.e. (b) Consecutive terms differ by therefore we try . To create the correct expression we need to add 10 i.e. (c) The numerators are the natural numbers and the denominators are two more than the numerator i.e.
(d) Ignoring the signs, each numerator is 1 and the denominators are the natural numbers . Since the signs alternate between positive and
P a g e | 51
11++
11++ + 1 1 ++ 1 1 +
negative, starting with positive, we use . Therefore
(e) Each numerator is 1 and the denominators are powers of 2 i.e.
(f) Each numerator is 1 and the first number of the denominator is and the second is . Therefore
(g) The numerators are the natural numbers but they begin with 2, i.e. and the denominators are the square numbers. Therefore
The sequence above diverges since it does not converge to any specific value.
A sequence can be classified as being convergent, divergent, oscillating or periodic.
Convergent sequences as the name suggests converge to a definite limit.
→lim
This oscillating sequence above is divergent.
The sequence above is convergent because it is tending to a value. This divergent sequence is PERIODIC as it consists of a set of values which are constantly repeated. The repeating pattern of the sequence consists of three values therefore the sequence is said to have a period of 3.
The sequence above is OSCILLATING and converges.
Divergent sequences are sequences which are not convergent.
P a g e | 52
LESSON 1 Determine which of the following functions is convergent or divergent. If the sequence is convergent, determine the limit of the sequence. (a) (b) (c) (d)
+ − − √ ++ lim 3 → →lim 31 1 →lim 1 →lim 1 3 1 3 lim → →lim 7 →lim 7 1 →lim 1 7 0 lim √ +−+ 12 → 1 2 →lim √ 33 4 →lim 443 44 1 4 4 1 →lim 3 4
SOLUTION (a)
4 1 4 →lim 3 4 4 1 4 →lim 3 4 l→im →lim 1 By L’Hopital 0→lim 3 DOES NOT EXIST Not convergent
(d)
Convergent and converges to 0.
LESSON 1 following
A sequence is given by the
4 + 3
converges and it converges to 3
(b)
is convergent and it converges to 0.
(c)
Write down the first four terms of the sequence. SOLUTION
4 + 3 7 + 3 1010 + 3 1313 1,1, + 33 1, 1 ∈1ℤ+ 1,1, ∈ ℤ+ 1 1 1 1 1 1 LESSON 2 defined by
A sequence of positive integers is
Prove by induction that
SOLUTION
When
,
P a g e | 53
1 1 1 + 1 1 1 + 33 1 11 1 133 1 13 1 3 2 1 2 1 1 1 1 1 + 1 1 1, ∈ ℤ+ 5 + 2++ 1 2 2+ 1 2 1 5 + 2 + 1 + 2 1 ++ 2++ 2 + 12 22+2 1 2 1 + Therefore
is true
Assume true for
Now,
Therefore, is true when Mathematical Induction
LESSON 3 and
is true. Hence, by
A sequence is defined by . Prove by induction that
.
SOLUTION
Therefore
Assume
is true
is true for
Now,
Therefore
is ture whenever
is true.
Hence by mathematical oinduction
2+ 1
.
P a g e | 54
SERIES At the end of this section, students should be able to:
1. 2. 3. 4.
5. 6. 7. 8.
Ʃ
use the summation notation; notation; define a series, as the sum of the terms of a sequence; identify the th term of a series, in the summation notation; define the th partial sum as the sum of the first terms of the sequence, that is,
= ;
apply mathematical induction to establish properties of series; find the sum to infinity of a convergent series; apply the method of differences to appropriate series, series, and find their sums; use the Maclaurin theorem for the expansion of series; 9. use the Taylor theorem for the expansion of series .
P a g e | 55
SERIES
2 = 2
INTRODUCTION
, , , , … , ⋯
Given the sequence corresponding series is
, the
⋯ is the
11++
partial sum where:
the first partial sum the second partial sum
the
are powers of 2 i.e.
partial
sum
LESSON 1 Write each of the following series using sigma notation. (a) (b) (c) (d) (e) (f) (g)
581114⋯ 86420 86420 2 2 ⋯ ⋯ ⋯ 1 1 ⋯ ⋯ ×2 × × ⋯× × 33 2 3 2 = 2 2 102 102 =
1 ++ 1 1 1 = 1 = 2−
(v) Each numerator is 1 and the denominators
the third partial sum
(iv) Ignoring the signs, each numerator is 1 and the denominators are the natural numbers . Since the signs alternate between positive and negative, starting with positive, we use . Therefore
1 1 1 = 11 1 1 1 1 =
(vi) Each numerator is 1 and the first number of the denominator is and the second is . Therefore
(vii) The numerators are the natural numbers but they begin with 2, i.e. and the denominators are the square numbers. Therefore
SOLUTION
(i) Consecutive terms differ by 3 therefore we try . To create the right formula we add 2 i.e.
(ii) Consecutive terms differ by therefore we try . To create the correct expression we need to add 10 i.e.
(iii) The numerators are the natural numbers and the denominators are two more than the numerator i.e.
+
The following standard results can be used to find the sum of various series.
1 1 , 1 1 2 2 1 , 2 6 = = = 4 1 ∑∑== 1 1 LESSON 1
Find each of the following sums
(a) (b)
SOLUTION
(a)
∑== 1 1
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= = = 4
(b)
406 4 1 124 1 42 4 1 ∑= = = 16146 471161 1 61 94 9 1 1 ∑∑====11221 1 ∑== 1 1 1 1 = 1 = = 1 611112222116 1 1226 1 6 22 3616 22 36 156 22 65 1 1 ∑= 26 2 = 2 = 2= 4 1 1 2 6 1122 1
LESSON 2 Express each of the following in a factorized form. (a) (b)
SOLUTION (a)
(b)
31 11133 441121 141222111 1 133 123 8 4 12 LESSON 1 that
Prove by mathematical induction
1 = 1 1 12 1 33 2
1 : = 1 1 12 1 33 2 : 11 1 1 121 11 131 2 0 0 1 : = 1 1 12 1 3 2 2 + 1 +: = 1 12 1 1 1 1 3 1 2 121 1 1 2233 5 121 1 1 2 23 5 5 1 1 for all positive integers . SOLUTION
Therefore, Assume
Now,
is true.
is true for
+
term term
+ 121 13 2 2 1 1 11 1 1
121 1 1 1 133 2 1212121 1 121 1 1 1 13 2 2 12 1
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121 1 133 2 12 12 121 1 133 1111 10 10 121 1 1 2 233 5 + 1 1 133 2 1 12 = 1 1 = {={ 1 }
Therefore
is true whenever
is true.
Hence by mathematical induction
:1: 1: 111112 2111 21211 1 = 1 1 3 1 1 22 ∑= = 1 1 ∑== ∑== (iii)
for all positive integers .
If
, then
LESSON 1
LESSON 2
(i) Express
in partial fractions. (ii) Use the method of differences to show that
(i) Show that
1 1 2 2 1 1 1 ≡ 33 1 1
(iii)Write (iii) Write down the limit to which
(ii) Hence use the method of differences to find an expression for
1 1 =
∑∑=== 1 ∑1= . 1111222 2 11111 1 33 1 111 2 ∑= 1: 1 1 12131 120211 1 2:3: 233445 231234 4: 456 435
converges as tends to infinity. (iv) Find
(iii) Show that you can obtain the same expression for using the standard results for and
giving your answer to 3 significant figures.
SOLUTION
SOLUTION (i)
(ii)
16 1 12 1 1 2 3 1 1 16 1 122 1 6 1 1 16 1 12 1 3 16 1 122 4 3 1 1 2 2 4 1 1 2 2 4 3 2 1 = 1 1 2 2 2 1 1 2 2 4 = 1 1 2 2 4 = 1 1 2 2
(i)
+ + 4+ +++ + 1 1 2 2 2 2
1 1 0 44 02 12 2 1 4 11 1112 12 When
When
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3 3 2 42 222 222 1 1 4 2 3 1 ++1+1+2 2 1 1 2 2 ∑= + 4 2 3 1 11:1 2 22 3 111 2 2 1 2 3 2 3 2: 22 33 141 3: 32 43 51 4: 4 5 6 1: 2 12 1 3 3 111 1 : 4 1 1 2 2 = 1 1 2 2 21 32 22 11 1 13 1 21 2 When
(ii)
32 12 1 21 2
→ ∞4 3 2 1 = 1 1 2 2 2 1 2 32 1 2 1 1 1 2 32 ∑= +++++ + + = 141 2 2 = 141 2 2 0.3210410012 10021 32 4912 4921 (iii) As
(iv)
ARITHMETIC ARITHME TIC PROGRESSIONS PROGRESSIONS
, , , … −, , … − 2 1 1 for every > 1
INTRODUCTION A sequence
or there exists a constant , called the such that
is called an if
That is
Therefore,
LESSON 1 Find the common difference for each of the following arithmetic progressions.
3, 5, 7,2 9,7,11, … … 2, 5, 8, 11, … 3,3, 5,5, 53 5 37,7, …2 8,8, 3,3,22 2,23 , 5 5 5 2,2 , 5,511, 88,8, 3 11,… 11, 15, … is an arithmetic − 4 1 − 4 1 1 1 44 5 5 − 4 1 44 55 4 4 (a) (b) 8, (c)
3,
,
SOLUTION
(a)
(b)
(c)
NB: Any pair of consecutive terms can be used. LESSON 2
Prove that the sequence 3, 7, progression.
SOLUTION We need to show that
Therefore,
is a constant.
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5 2 2
LESSON 3 The sum, , of the first terms of a sequence is given by . Show that the sequence is an arithmetic progression with common difference 10.
− 5 2 2 1 15 1 1 2 5 2 1 15 5 2 5 2 1 155 7 5 2 55 7 55 7 7 5 2 55 12 7 5 2 5 1212 7 1010 7 7 − 1010 1 1 7 1010 10 7 10 17 − 1010 7 7 1010 17 17 10 SOLUTION
502 1111 120120 502 21 11 1353 5 70 36 36 25050 3636 12 30602
LESSON 2 The last term of an arithmetic progression of 20 terms is 295 and the common difference is 4. Calculate the sum of the progression. SOLUTION
20,20, 295, 4 201 201 141 295 219 220 2219219 2020 14 5140 We need to determine
LESSON 3 The sum of the first 6 terms of an arithmetic progression is 54.75 and the sum of the next 6 terms is 63.75. Find the common difference and the first term. SOLUTION
62 2254. 756 1 1 54.75 6 15 54. 54 . 7 5 5, 9, 13 54. 75 7 5 63. 75 7 5 118. 11 8. 5 12 55 101 1 011 14 1222 66 121 121118.118.5 118.5 2 41 6 15 54. 54 . 7 5 12 66 118. 11 8. , , , … , ⋯ ×2: 5 109.118.10119.8.55 30 12 12 66 36 2 22 1 1 14 9 8.5 250 120
LESSON 4 If the first three terms of an arithmetic progression are 5, 9, and 13, what is the value of the 10 th term? SOLUTION
(1)
The common difference, , is 4
(2)
Solving (1) and (2) simultaneously
Sum Formulae for Finite Arithmetic Sequence If is a finite arithmetic sequence, then the corresponding series is called a finite arithmetic series. The sum of the first terms of the series, which we denote , would be stated as LESSON 1 Find the sum of the even numbers from 50 to 120 inclusive. SOLUTION
(1)
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GEOMETRIC PROGRESSIONS
, , , … , , …
INTRODUCTION A sequence
is called a
or there exists a nonzero constant , called the such that
if
− , > 1 27, … is a Geometric Progression. 33−, −− 3−, 3, 3 − 33− 3 − 33−−− −−− − 3 Therefore,
or
LESSON 1
Prove that the sequence 1, 3, 9,
SOLUTION
We need to show that
is a
constant.
LESSON 2 The first and fourth terms of a geometric progression are 6 and 20.25 respectively. Determine the 8 th term of the progression. SOLUTION
6 20.25 √ 23.30..632755 3.1.3575 63 6561 2 64
Since length cannot be negative
Sum of a Geometric Progression The sum of the first terms of a G.P is given by
LESSON 1 The fourth term of a geometric progression is 6 and the seventh term is . Calculate (i) the common ratio, (ii) the first term, (iii) the sum of the first eleven terms. SOLUTION (i)
486 2 486 8 8 8 6 6 34−− −3 4 1 2222 1 512. 25 1 51 −1 5 11− 4 5 5 13 5 45 1−3 4 5 413 5 41−4 3 5 44 313− 114 3 13 (ii)
(iii)
LESSON 2
LESSON 3 The lengths of the sides of a triangle are in geometric progression and the longest side has a length of 36 cm. Given that the perimeter is 76 cm, find the length of the shortest side. SOLUTION Let longest side be and shortest side be .
36 76 3636 36363636 7676 369 9363610 40 0
3 52 53 2 2 0 3 3636 23 16 1 , 1 ><1 1 or 1 1 1 , 48
Given that
, find
and prove that this sequence is a Geometric Progression. SOLUTION
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56 13−− − 556 131− − 65 31− 136 3
(ii)
1 12 14 18 161 ⋯
If we think about it we should realise that the sum appears to be 2. Since the sum appears to tend towards a specific number as it goes on indefinitely we refer to this series as a CONVERGENT series. The sum of this series can be given using the formula
1 , 1< 1< 1 1 1 12 2
therefore
and
,
Thus we see that our intuitive answer is indeed correct. LESSON 1 The first and fourth terms of a geometric progression are 500 and 32 respectively. Find (i) the values of second and third terms (ii) the sum to infinity of the progression
500 3232 500 32 250032 1258 5 500500 2 200 5
SOLUTION (i)
LESSON 2 The first term of a geometric progression is and the common ratio is . Given that and that the sum to infinity is 4, calculate the third term. SOLUTION
Sum to Infinity What would be the sum of the infinite series
For our series above we have
2 8085005000 5 −500 1 2500 25 3 12 121 44 41 1212 4 116 4 121212 1 3 4 31 34 16
LESSON 3 The first term of a geometric series is 120. The sum to infinity of the series is 480. Given that the sum of the first terms is greater than 300, determine the smallest possible value of . SOLUTION
1 120 4804804801 1 120 1 1 3 14 4 1 > 3001
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1201334 > 300 143 12011 4 > 300 4 3 4801 > 300 4 3 134 3 > 58 43 < 8 3 ln 43 < ln 83 ln43< ln 8 > llnn 834 ln 34 is negative > 43.4 1 = 2 1 < < 1 1 1 1 1 1 ⋯ = 12 2 4 8 16 412 12 11< < 1 1 2 12 1
LESSON 4 Determine whether the geometric series
is convergent. If it converges, determine its sum. SOLUTION We need to show that .
Since
, the series converges.
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MACLAURIN’S MACLAURIN’S SERIES INTRODUCTION
0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯
must be differentiable must
0 0 Us– e Maclaurin’s Theorcosem to find coscos0.0.2 coscos 0 coscos0 1 s sin 0 sisi n0 0 cos cos 0 co coss 0 1 sin 0 sin0 0 coscos 0 coscos 0 1 sin 0 sisi n0 0 c cos 0 co coss0 1 0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯ 1 0 112!2! 0 3!3! 1 4!4! 0 5!5! 1 1 6!6! 11 2!2! 4!4! 6!6! ⋯ 0. 0 . 2 0. 0 . 2 0. 0 . 2 0.0.2 1 2 24 720 0.98 must must exist at
The derivatives of
must must exist at
Only within specific values of is the series valid.
LESSON 1 the first four non zero terms for determine an approximation for SOLUTION
, hence .
1 cos LESSON 2
Find the Maclaurin expansion for up to and including the term in .
SOLUTION
1 cos 1 2 cos 1 2 11 2!2! . using result Ques tion 1 11 2!2! 2 ⋯ 11 2 12 Fi n d t h e Macl a ur i n ’ s s e r i e s f o r tan LESSON 3 up to
.
SOLUTIONhttp://sirhunte.teachable.com/courses /93027/lectures/2211764
0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯
tan sec 2tanse n sec 2se2secc 4sec tan
0 tan0 0 0 sec 0 1 0 2tann0 sec 0 0 2se2sec0c0 4sec 0tan 0 2
0 0 2!2! 2 3!3! ⋯
P a g e | 64
23!3! + Obtain the Maclaurin’s series expansion for 0.0.1 LESSON 4
A function is defined as
.
(a)
up up to and including the term in . (ii) Hence, estimate to four decimal places. to SOLUTION
++ 0 2 + 0 2 4 + 0 4 8 + 0 8 16 0 16 0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯ 2 4 4 2!2! 84 3!3! 161624!4! 0.0.1 2 2 22 20.02.1 22 0.03.1 30.1 0.1 3.3201 −− ,′ 13 13−− 0 1 −− −− 22 13 1 3 33 6 13 1 3 (i)
Valid for
LESSON 6
l n 1 ln1 – ln +− ln 3+− 3
(i) Use the Maclaurin series for and and to obtain the first three non zero to terms in the Maclaurin series for
You are given that
Find and and Maclaurin series for
(ii) Find the value of for which
. Hence
find an approximation to , giving your answer to three decimal places.
SOLUTION (i) .
. Hence obtain the as as far as the term in .
By considering the equivalent binomial expansion, give the set of values of for which the Maclaurin series is valid. SOLUTION
.
State the range of validity for this series.
(ii)
LESSON 5
0 6 −− 18 1813 1 3 33 545413 1 3−− 0 54 −− 96 9612 1 2 33 28828813 1 3−− 0 288 0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯ 1 6 36 4 288 6 16 1 6 9 4848 1 < 3 < 1 → < <
ln11 1 1 1
0 ln11 0 0 1 0 1 0 1 1 10
1 2 20 1 020 1 6 0 661 60
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1 24 0 241 240
0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯ 1 0 2!2! 2 3!3! 6 4!4! 2424 5!5! ⋯ 2 3 4 5 ⋯ ln1 0 ln1 0 1 1 10 1 1 0 1 1 10 1 10 1 2 20 1 20 6 0 6 6 1 1 0 24 024 24 1 1 0 0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯ 0 2!2! ⋯2 3!3! 6 4!4! 2424 5!5! 2 3 4 5
ln 11 ln1 ln1 2 3 4 5 ⋯ 2 3 4 5 2 23 25 1 < < 1 +− 3 14 2 3 33 12 1 ln3 ln 11 212 1 2 1 2 1 221.0296 3 2 5 2
(ii)
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INTRODUCTION
Maclaurin’s Series is: 0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯ For Taylor’s Series, we let 2!2! 3!3! ⋯ ! ⋯ 1. 2.
The function The function
has has to be infinitely differentiable has to be defined in a region near the value has
Furthermore, replacing with
.
we get
2!2! 3!3! ⋯ ! ⋯
= !!
– ln 3 3
LESSON 1 Find the first three non zero terms of the Taylor expansion of . SOLUTION
LESSON 2
+
–
Find the first four non zero
1 + 3 3−− 31 3 1 14 313 1 161 323 1 321 363 1 1283 terms for the Taylor expansion of .
with centre
ln ln 3 3 → 3 l n1 3 l n1 3 3 3 1 3 19 2!2! 3!3! ⋯ ! ⋯ ln 3 3 ln3 n3 13 19 2!2!1 2!2! ln3 n3 13 181 3!3! ⋯ ! ⋯ SOLUTION
Let
1 3 3 14 161 1 1 321 2!2!1 1 1 1283 3!3!1 1 1
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2 2 2 3 2 3 √ √ √ sin16 2 2√ 2 163 3 4 16 12 16 0.1920
14 161 1 1 641 1 1 2563 1 1
– sin
LESSON 3 (i) Obtain the first four non zero terms of the Taylor Series expansion of in ascending powers of
.
(ii) Hence, calculate an approximation to SOLUTION
sin 3!3! ⋯ 2!2!! 4 sin √ 2 4 sin4 2 coscos 4 cos4 √2 2 s sin 4 sisin 4 √ 22 cos cos co s cos √ 2 4 2 sin √2 2 √ 22 √ 241 √ 22 2!2!1 4 2 3!3! 4 √2 2 √ 22 4√ 2√ 42 4 12 4
(i) let
(ii)
sin
sin → 3 4 16 4 16
.
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BINOMIAL THEOREM At the end of this section, students should be able to:
∈ ℚ
1. explain the meaning and use simple properties of 2. recognise that
that is ,
! , ∈ ℤ and and
, that is,
where
, is the number of ways in which objects may be chosen
from distinct objects; 3. expand for ; 4. apply the Binomial Theorem to real-world problems, for example, in mathematics of finance, science.
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PASCAL’S TRIANGLE INTRODUCTION
For any positive integer :
1 11 1 1 2 1 1 1 3 3 3 1 1 4 6 6 3 1 ! 1 1 2 2 3 3 … 321 8! 8×7× 8× 7×6×6×5×5×4×4×3×3× 2×1 2× 1 0!0! !! FACTORIALS For example,
Is Is defined as 1.
LESSON 1
Simplify
SOLUTION
!
9!9!6!6! 9 ×8×7×6×5×4×3×2×1 6×5×4×3×2 6×5×4 9 ×3×2 × 8 × 7 ×1 504 !! 2 2! ! 2 2! 1 1 2 2 3 3 … 321 2 2! 11 2 2! 2 2! 2 2! 1 1 1 2 2! 11 −−! − −! 72 13 13 24 24 35 35 …… 332211 72 1 1 2 2 72 LESSON 2
Simplify
SOLUTION
LESSON 3
SOLUTION
Solve the equation
.
PASCAL’S TRIANGLE 1
1
1
1 1 1
2
1
3 4
3 6
1 4
1
3 2 7272 3 70 0 10 10 7 7 0 10 7 since
LESSON 4
in invalid
Show that
1 2 2! 31 31! 32 3! 1 2 2! 3131! 3 3332 2! 31 31! 33 33! 3131! 3331! 1 3 32 3! SOLUTION
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LESSON 5
+ +!!
(a) Express
and
in in the form
are integers.
+ +! + +!
, where
(b) Hence find
1 = 2 2!! + + !! + + ! + + !
SOLUTION (a)
11 11 212!! 2 2 2 2 1 1! 2 2! 1 1 2 2 1 2 1 2 1 1 1 1 1 2 2!! 1 1! 2 2! 1 1 1 = 2 2!! = 11! = 2 2! 1: 2!2!1 3!3!1 2: 3!3!1 4!4!1 3: 4!4!1 5!5!1 Equating coefficients of :
Equating constants:
(b)
1: !1 11 1! : 11 1! 21 2! 1 1 1 = 2 2!! 2 2 2!
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THE BINOMIAL THEOREM INTRODUCTION
We now look at an alternative to Pascal’s Triangle using!Factorials. ! !! For any positive integer :
= − 0 1 − 2 − 3 − ⋯ 1 1 − 32 3 2 LESSON 1
Determine the expansion
.
32 3 25 5 5 5 5 5 0 3 1 3 22 2 322 3 3 22 43 22 5 22 72440 240 243810 24243 581811080 22 1010 2727720 10241009 883232 531616 13232 2 3 3 23 2 3 5 0 105 23 3 1959552 2522523232243 13 1 32 13 1 3 16 1 6 9 2 64 2 65 2 66 6015154 1 26 2 1 1 6 9…60 …60 12 1 727266 12 540 12540 96060 46969 SOLUTION
LESSON 2
Find the 6th term of the expansion
SOLUTION
The 6th term of the expansion
begins with
LESSON 3
.
occurs when
since the summation index
.
Find the coefficient of
in the expansion of
SOLUTION
We only need the terms which will result in a
term after multiplication.
We isolate the multiplications which would create an
term.
.
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LESSON 4
Find the term independent of in the expansion
3
.
SOLUTION
86 33 12 1 28286399 64 16
The term independent of :
1 11 2!2! 1 1 13!3! 2 2 ⋯ 1 < < 1 1 1 1 1 2!2! 1 1 13!3! 2 2 ⋯ 1 < < 1 → < < 12 1 2 12 1 23 1 2 23 3 1 3 3 3 1 1 2 22 2 2 121 22 2 2 12 2321 22 1 3 32 12 12 4 32 12 12 312 8 13 1 3 32 12 1 < 2 < 1 12 < < 12 For any real number
provided that
LESSON 1
SOLUTION
Find the binomial expansion of
up to and including the term in
.
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LESSON 2
Find the first three terms of the expansion of
SOLUTION
−
.
423 423 2 3−− 421 421 32 − 424 2− 11 32 − − 3 21 21 2 3 1 3 221 11 2 1111 11 21 2 212 1 32 94 ⋯ 23 2 3 92 + ++ + + + + ++ + 9 1 25 2 5 + ++ + + + 99 2225 25 5 1 5 2 0 5 3 9 69 3 6 1 2 5 5 1 25 +331 −− 33331 11 11 1111 11 211 1111 1112 12 3121 33 3 3 3 3 LESSON 3
(i) Express
in in the form
where
(ii) Hence, or otherwise, find the expansion of
and
are integers.
as a power series in ascending order up to and as
including the term in . (iii)Find (iii) Find the range of values of for which the series expansion of
SOLUTION
(i)
Equating constants:
Equationg coefficients of :
(ii)
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625 6625 2 55−− − 621 21 2 − 62− 11 −52 31 31 52 5 1 5 1 5 31 31 511 252 1251111 11 21 2 1111 1112 12 321 2 313 115 2 754 3758 33 2 4 8 31 256 933 3 363 3 3513 3 152 754 3758 2 4 8 1 < < 1 1 < < 1 → < 2< 2 5 < < 5 √ 1 √ 8282 √ 1 1 11 14 14 14 1211 14 14 114 23211 11 14 323 1287 82 8 1 1 181 1 81 811 811 811 1 81 11 81 (iii) Valid for
LESSON 4
and
Use the expansion of
SOLUTION
This works for small .
to setimate
to four decimal places.
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1 31 31 81 1 1 3 1 7 1 331 4 81 32 81 128 81 31.1.003 3.009 16 186 6 8 6 LESSON 5
(a) Find the binomial expansion of
√ 144144 1 6 6 14 11 4664 1 66 86 8 6 8 1 8 11 68 4141 618 4141 4118 418 42 4 2 4 81441446 1122 12 6 24 3 2 1 2 √ 144 144 ≅ 423 4 3 ≅ 497
(b) Find the binomial expansion of
(a)
(b)
(c)
up to and including the term in
(c) Hence, find an estimate for the the value of SOLUTION
up to and inclusing the term in
.
.
in the form where and are integers.
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ROOTS OF EQUATIONS At the end of this section, students should be able to:
0
1. test for the existence of a root of where f is continuous using the Intermediate Value Theorem; 2. use interval bisection to find an approximation for a root in a given interval; 3. use linear interpolation to find an approximation for a root in a given interval; 4. explain, in geometrical terms, the working of the Newton-Raphson method; 5. use the Newton-Raphson method to find successive approximations to the roots of , where is differentiable; differentiable; 6. use a given iteration to determine a root of an equation to a specified degree of accuracy.
0
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,, 0 2 1 1 1 21 1 1 11 2 2 22 2 1 1 1 × 2 1
,
THE INTERMEDIATE VALUE THEOREM
If in
is a continuous function on the closed interval such such that .
LESSON 1
and and the product
< 0
then there exists
Use the Intermediate Value Theorem to show that has a root between 1 and 2.
SOLUTION
is a polynomial and therefore continuous on the is
By the Intermediate Value Theorem there must be some root between 1 and 2.
1 3 4 1 0 3 4 1 0 1 1 31 41 1 6
interval
∈ 1,1, 2 0 such such that
1,1, 2
. Therefore there is a
LESSON 2 Use the Intermediate Value Theorem to verify that there is a root of the equation between 0 and 1.
3 4
SOLUTION
Let
0,0, 1 0 × 1 6 ∈ 0,0, 1 0
is a polynomial and therefore continuous on the interval is
By the Intermediate Value Theorem there must be some root between 0 and 1.
.
such such that
.
. Therefore there is a
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DETERMINING THE ROOTS OF AN EQUATION
3 1
LESSON 1 Given that has a root between 2 and 3, find this root to 1 decimal place using the bisection method.
LESSON 1 Use Linear Interpolation, twice over, to determine the root of the equation in the interval .
3 1 3 1 3 1 0 SOLUTION
SOLUTION
3 1 1 3 1 0 3 1 2 2 32 2 1 33 3 3 33 3 1 2 – 2.2.5 2.5 3 2.2.5 2.2.5 1 1.1. 625625 – 2.2.755 2.75 32.2.755 2.2.7575 1 0.140625 – 2.2.87575 2.875 3 2.2.87575 2.875 1 0.8418 2.75 – 2.2.8125125 2.8125 3 2.2.8125125 2.8125 1 0.33
2,2, 3
Let
Mid point of interval is 2.5
Since there is a sign change between 2.5 and 3 the root occurs within this interval. We therefore now bisect this interval. Mid point is 2.75
Due to sign changes the root must be in the interval 2.75 and 3. Mid point is 2.875
Due to sign changes the root is between 2.875
and
Mid point is 2.8125
3 1 2 2 32 2 1 3 3 3 33 31 3 1 2 23 3 2 2 2 2 2 33 2 4 9 3 5 1133 5 2.6 Let
Using similar triangles
Therefore root lies between 2.75 and 2.8125 and since to 1 decimal place both limits are the same the root is approximately 2.8.
13 13 13 5 5 3 5 135 1 1.1. 1
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21.1 32.2. 6 2 2.2. 6 1.13 2 5.5.2 3.3.3 1.1.11 3.1 8.5 2.74
INTRODUCTION
4 10900900 0 – LESSON 1
The equation has exactly one real root, . Taking as the first approximation to , use the Newton Raphson method to find a second approximation, , to . Give your answer to four significant figures.
4 900900 3 2 4 10 1010 103 101100 2410110100 4900900 1010 28440 9.859 2 5 3 13 0 – 1<<2 2 1.55 3 13 0 1,1, 2 1 2 12 551 331 1313 33 2 21 ×522 < 032 13 2929 1,1,2 2 5 3 13 1. 56 10 3 1.1.5 1921.1.56 1.1.55 1.1.51010 1.1.351.1.53 13 1.1.5 63 SOLUTION
Let
LESSON 2
The equation has exactly one real root
.
(i) Show that lies in the interval . (ii) Using the Newton Raphson method with initial estimate to estimate the root of the equation in the interval , correct to 2 decimal places. SOLUTION
(i) Let
Since , by the Intermediate Value Theorem there is a root in the interval
(ii)
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1.1984 1.1984984 21.1.198498461.1.1 98498451.1.1 98498410101.1 .198498431.1.19849843 13 1.1.114679840.0516
1.1467 21.1.1467467 51.1.1467467 31.1.1467467 13 10101.1.1467 6 1. 1 . 1 467 467 467 3 1.1984984 0.001357 01357
1.1453 1.15 + + 2 4 1 0 1.2
Since and correct to 2 decimal places are both equal to 1.15, .
DERIVING AN ITERATIVE FORMULA EXAMPLE 1
Show that
is an
approximate iterative formula for finding the root of . Apply the iterative formula with initial approximation , to obtain an approximation of the root to 2 decimal places. SOLUTION
2 4 1 0 2 4 1 42 1 442 1 + 442 1 1 41.1.22 1 1.2386 41.1.23863862 1 1.2551
41.1.25515512 1 1.2621 1.26 Therefore, the approximation is
decimal places.
correct to 2
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MATRICES At the end of this section, students should be able to: 1. 2. 3. 4. 5. 6.
××
2,2, 3
reduce a system of linear equations to echelon form; ; row-reduce the augmented matrix of an system of linear equations, determine whether the system is consistent, and if so, how many solutions it has; find all solutions of a consistent system; invert a non-singular matrix; solve a system of linear equations, having a non-singular coefficient matrix, matrix, by using its inverse.
3×3 3×3
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MATRICES
A matrix is a rectangular array of elements enclosed in brackets. A matrix is defined: number of rows
×
Two matrices are equal if they contain the same corresponding elements.
The addition and subtraction of matrices is only possible if the matrices are of the same size.
number of columns (this is the size/order of a matrix).
A square matrix contains the same number of rows and columns.
o
Matrix addition is commutative and associative.
Two matrices can be multiplied if the number of columns in the first matrix is equal to the number of rows in the second. o
Matrix multiplication is not commutative but it is associative.
2 × 2 10 01 3 × 3 11 1 222 311 210 111 101 − The identity matrix for for
LESSON 1
If
matrices is
and for
matrices it is
100 010 001
and
(i) find
(ii) deduce SOLUTION (i)
(ii)
11 1 222 131 210 111 101
1 2 2 1 1 1 0 1 1 1 2 1 1 1 1 1 1 2 0 1 1 1 1112 2 22211 13300 111111122211 1311 1111 1 22200 1311 4 0 0 00 40 04 − 4 14 2 1 1 − 10 11 01 Therefore
.
P a g e | 83
×
(i)
ℎ ℎ ℎ | | 0 | || | | | 0 | | × | | ∈ ℝ | | || || | | | 0 – det− 311 121 213 – INTRODUCTION
The determinant of
is
.
o
If the matrix contains a row with ALL zeros then
.
o
For square matrices
o
If contains two identical rows or
columns then
o
.
.
Interchanging two rows or columns of a
matrix changes the sign of the
determinant.
o
For
,
where where is a
matrix.
o
If a row or a column of a matrix is
multiplied by a constant, , then the
determinant of the matrix is multiplied by .
o
Adding a multiple of one row to another does not affect the determinant.
o
o
11 21 | 323| 13121 13111311 11312 2 1 1 1 2 11335 12 211 || ≠ 0 – 2 31 14 20 3 22 310 143 0 2 10 43 32 4312 10 0 2639 32334 28 0122 0 9 3322 9 36 61 105 925 SOLUTION
(ii) Since
LESSON 3
,
is non singular.
Find the value of for which the
matrix
is singular. SOLUTION
LESSON 4
Determine
by by
factoring.
A square matrix is singular if
SOLUTION
otherwise it is non singular.
be obtained by factorizing the matrix. This is done
o
LESSON 2
The matrix
is given by
.
(i) Find the value of the determinant of
(ii) State, giving a reason, whether or non singular.
.
is singular
The determinant of a matrix can
by factoring out the common factor from each row or column.
3 213 621 1055 1 6 3 35523 21 01
factoring factoring out 3 from
factoring factoring out 5 from
To find the determinant it is easier to use the
elements of
because it contains a zero.
1515 3 232 1212 1 16 1515377 1313 120 31 3 112 221 | | 5 , – 6 4 4 1 9 3 1 6 3 5 5 9 4 26 81 5 7 9 3 1 6 54 26 2 2 3 3 2 3 3 2 | | 1 1 213 213 1 51122 11212 19 | | 4 69 38 2575 38 1575 69 4436590 90 241 41 187 87 55 16 52 46 4 2 1 16 52 46 57 9 6 38 567 6788 249 5×4 320 The transpose of a matrix is created by
interchanging the rows and columns. The
LESSON 5
The matrix
.
transpose of a matrix is denoted
.
o
(i) Show that
.
(ii) Matrix is changed to form new matrices and
. Write down the determinant of EACH
o
A square matrix is symmetric if
o
A square matrix is called skew
symmetric is
.
.
of the new matrices, giving a reason for your
For example, if
answers in EACH case.
(a) Matrix is formed by interchanging column 1 and column 2 of matrix
then
and
then interchanging row 1 and row 2 of the
resulting matrix.
LESSON 6
The matrices and are given
by
and
(b) Column 1 of matrix is formed by adding column 2 to column 1 of matrix
. The
.
other columns remain unchanged.
(c) Matrix
is formed by multiplying each
Calculate
element of matrix by 4.
SOLUTION
(a) the determinant of
(i)
(b)
SOLUTION
(ii) (a) interchanging columns will change the
(a)
sign of the determinant to 5 and then
interchanging the rows will change the sign of the determinant to
(b) The determinant remains as
.
(b)
since
adding a multiple of a row does not affect
the determinant.
(c) Since each element is multiplied by 4 the value of the determinant is
− || adj
Step 4: Use the relation
to to find
− 10 11 13 4 12 2 × 2 10 11 13 4 1 2 1 1 3 1 1 1 | | 1 1 4 1 2 1 3 ±1 1 1 5 5 4 2 2 3 11 3 0 3 0 1 111 12 141 21 141 11 1 2 4 2 4 1 1 1 1 1 1 1 1 1 1 ( ) 1 3 0 3 0 1 5 12 4 3 6 3 | | , ∆ 2 3 1 | | 512 63 23 41 53 31 2 − 3 124 36 31 | | ( ) .
If
then
,
and and
LESSON 7
are called the minors of are
,
and
respectively. A minor of an element is obtained by deleting the row and column containing that
SOLUTION
element and finding the determinant of the
Let
matrix which remains.
The cofactors of a matrix are determined by multiplying the minor by
in the following
order
Step 1: Find the determinant of the matrix
Determinant of , det ,
The elements of any row or column can their
corresponding cofactors can be used to determine
the determinant. i.e.
using using
the first row and its corresponding cofactors. Step 2: Write the matrix, say
Step 3: Write the matrix,
the adjoint of ( adj)
, of cofactors.
. This matrix is called
Find the inverse of
3×1 1 0 3 73 52 01 – − 10 128 | | 1 52 01 073 01373 52 1215 311 520 013 173 301 731 520 02 31 31 13 13 02 ( 55 0 7 117 0 7 5 ) 156 821 25 75 86 2115 1 1 5 26 515 − 2 71 82 215 − − 1−5 6 15 10 2 71 82 215 128 15 212188 12 17517101010110100 612128228 1211212215 5 8 12 226 113 LESSON 8
and are
are related by the equation
is non singular.
Find (a)
(b)
, when
SOLUTION
(a)
(b)
matrices and , where
LESSON 9
Solve the equations
4 554519129 5 5 2020 , 1 5 5 2 41 55 41 2019 1 −15 30 45 − 150 28 5 06 2156 − 1501 158 306 4516 192 1 23005 0 2 5 20 150 450450 23 3 SOLUTION
Step 1: Write the system in the form
where
and are matrices
Step 2: Find
Step 3: Multiply both sides of equation by
This equation is said to be consistent since it has a solution.
→ 104 111 132 100 100 001 1 1 1 1 0 0 4 → 00 13 36 04 10 01 – → 10 11 1310 10 00 0 13 0 6 24 10 11 0 → 00 31 36 04 10 01 3 × 3 1 3 → 01 10 32 10 11 00 00 10 1 1 1 00 0 23 14 31 01 10 1 3 → 00 10 31 043 11 013 000 1 1 0 1 0 2 4 2 1 3 → 0 1 0 4 1 – 0 0 1 53 1 32 1 1 0 0 3 2 → (00 10 0144 21 131) 3 3 → 10 5 3 2 1 − ℎ 0 ℎ 3 124 36 33
The following elementary row operations can be performed on a matrix. -
The interchanging of rows
-
The multiplication of a row by a non
zero scalar.
-
The adding of the multiple of a row to
another row.
These operations can convert a
matrix (or
any other matrix) to row echelon form
or
This can be done by 1.
Beginning with the left most column and use row operations to make the first element in this column a 1 and the elements below it zeros.
2.
Ignore the row and the column with the 1 created from step 1 and repeat step 1 on
100 ℎ → 100 10
the remaining matrix.
3.
Repeat this process until the desired matrix is obtained.
1 0 11 13 4 12 LESSON 10
SOLUTION
Find the inverse of
A system of equations can either be consistent or inconsistent. A consistent system can have
a unique solution
infinitely many solutions.
An inconsistent system has no solution. .
The use of row reduction greatly assists in To obtain the inverse of a matrix
−| | 104 111 132100 010 001
we can use row operations to convert the
augmented matrix
to to
.
determining the consistency of a system of equations. An upper triangle matrix unique solution.
10 1 0 0 1
indicates a
100 10 00 100 10 0 3 333 4 2 2 6
Matrices of the form
Since the system has a unique solution it is said to
infinitely many solutions. Matrices of the form solution. LESSON 11
be consistent. indicate no
A system of three equations is
(i) Write the augmented matrix for the system. (ii) Use row reduction to solve the system of equations.
1 0 11 1333 41 11 12 36 04 11 3236 3 1 1 1 → 04 11 32 36 4 → 100 131 136 363 1 1 1 3 → 00 31 36 63 → 001 310 362 603 1 0 2 0 3 → 00 10 3333 13 → 010 100 321 103 : 3 13 3 3 3011 3 2 2 3
SOLUTION (i)
(ii)
From From
:
: :
From
:
2 211 20
indicate
LESSON 12
Determine the general solutions
of the system of equations.
22222 343 2 4 2 8 11 22 12 34 2 4 21 82 2 3 → 02 44 32 78 1 2 2 3 2 → 00 48 36147 14 → 10 212 234374 0 81 62 142 3 8 → 00 10 034 740 0 0 0 0 SOLUTION
The row of zeros indicates that
,
therefore we can use a parameter for one of the
variables and express the other variables in terms of this parameter. Furthermore, the row of zeros indicates that the system has infinitely many solutions.
343 774 43 47 4 4 Let
From row 2:
From row 1:
2 23 2 7 3 23 4 7 4 2 3 2 1 2 1 2 3 2 2 1 1 1 0 1 0 Therefore, if we say that
then
and
So
.
would be one solution for the
system.
LESSON 13
Discuss the solutions of the given
4 8 3 33 410 2 2 2 13 2 132 121 410 4 1 1 1 Interchange and → 3 2 32 21 10 32 → 100 160 140 422 8 04 0 0 8 44 4 08 0 0 8 8 0 equations when (i)
(ii)
SOLUTION
(i) When
, we have
This result indicates that when
, the
system of equations is inconsistent and has no
solution.
(ii) When
, we have that
This result indicates that when
8
, the
system of equations has infinitely many solutions.
66 44 2222 131 23 11 2 44 31 13 4 3 3 Let
From
:
From
:
DIFFERENTIAL DIFFERENT IAL EQUA EQUATIONS TIONS At the end of this section, students should be able to:
0 a, ,, ∈ ℝ
1. solve first order linear differential equations using an integrating factor, using given that is a real constant or a function of , and is a function; 2. solve first order linear differential equations given boundary conditions; 3. solve second order ordinary differential equations with constant coefficients of the form , , where and is: is: (a) a polynomial, (b) an exponential function, (c) a trigonometric function; and the complementary function may consist of (a) real and distinct root, (b) 2 equal roots, (c) 2 complex roots; 4. solve second order ordinary differential equation given boundary conditions; 5. use substitution to reduce a second order ordinary differential equation to a suitable form.
DIFFERENTIAL EQUATIONS A differential equation is an equation which contains derivatives of a function or functions. For a first order differential equation the highest derivative is the first derivative. For a second order differential equation the highest derivative is the second derivative.
sin where
The solution of this type of equation can be achieved by separating the variables and integrating both sides of the equation with respect
These solutions are called general solutions of the differential equation because the value of the
constant is unknown.
to the relative variables. LESSON 3 LESSON 1
SOLUTION
5 3
Solve the differential equations
5 3 5 3 55 3 2 53 3 103 6 coscos tan cost o s t a n 1tan coscos ln cossin coscos lnsisin sisin+ sisin LESSON 2
SOLUTION
Solve the differential equation
Find the particular solution of the
csc 4,4, 1 cscsc;c1 ; 3 when 4 csc1 − csc sisin − coscos 1 1 cos1 31 4 1 2 14 1 → 4 coscos 4 differential equation
when
SOLUTION
.
INTRODUCTION
∫
Linear differential equations of the form
can be solved by multiplying throughout by the Integrating Factor,
LESSON 4
.
Solve the following differential
3 7 3 7 . ∫3 . ∫− − − − 3− 7− − 3− 7− 27 equations
SOLUTION
Step 1: Write the DE in the form
Step 2: Calculate the I.F using
Step 3: Multiply each term in the equation by I.F
L.H.S resembles the product rule where I.F is
actually
Step 4: Integrate both sides of the equation w.r.t.
72 cotcot 2 coscos 2 cotcot 2 coscos cotcot cos cotcot sin lnsin . sicoscosn sin sin.n . sin . 2sincn cos sin cos cos 2c2 cosss sin si nsinsin csc + + 1 1 2 2 sin2 sin2 2 1 1 s1in cs csc LESSON 5
given that
Solve the differential equation
when
.
SOLUTION
when
LESSON 6
Determine the particular solution
of the differential equation
− 4 4 3 0 4 4 − 4 − given that
SOLUTION
when
.
I.F ∫ 4 − 4 − 3, 0 32 − 2 When
0 andand d 0 ,,
are first and second order homogenous equations where
and are constants. The solution of
these equations is called the complementary function (C.F).
LESSON 7
5 2 2 1 5 2 1 5 2 5 ln 25 + SOLUTION
5 2 0
Solve the differential equation
In general, the solution (complementary function)
0 −
of a first order differential equation is of the form
is
Auxiliary Quadratic Equation Given the equation
0
If the auxiliary quadratic equation has a repeated root , ,
The complementary function is determined by the roots of the nature of the roots of the quadratic
is the general solution of the differential equation.
auxiliary equation which is
0 5 4 0 5 4 0 1,15,4 4 0 LESSON 8
Solve the equation
SOLUTION
Auxiliary equation
If the auxiliary quadratic equation has real and distinct roots, and then
is the general solution of the differential equation.
4 4 0 4 4 0 4 1 4 1 0 2twicee LESSON 9
Solve the equation
SOLUTION
Auxiliary equation
2 5 0 2 5 0 12± 25 0 cos2 cos 2 sin 22 ± cos cos s sin LESSON 10
Solve the equation
SOLUTION
Auxiliary equation
If the auxiliary quadratic equation has complex roots of the form ,
is the general solution of the differential equation.
–
– – , ≠ 0
Non homogeneous first order and second order differential equation are of the form
The particular integral is any solution of these types of differential equations.
is a Polynomial is
5 6 → → 0 0665 5 5 66 6 11 6 55 6 5 6 0 36 LESSON 11
SOLUTION
Solve the equation
It seems sensible to think that the
solution is of the form
Substituting into original equation
So
is a solution of the given equation,
but it cannot be the complete solution since it does not contain any arbitrary constant. However, it must be part of the complete solution, and is called a particular integral (P.I). The remainder of the solution can be found by considering the
5 6 0 16 365
simpler differential equation
4 3 2 1 ±24 0 − SOLUTION
Auxiliary equation
C.F:
Particular Integral:
→ 22 → 2
24 4 4 44 22 4 3 3 2221 1 1 4 3 → 341 4 2 → 2 5 2 4 1 → 8 − 34 12 58 Substituting into original equation
P.I is
General solution:
is a Trigonometric Function is
whose solution is
LESSON 13
Thus the complete solution is
differential equation
which is the combination of the complementary
SOLUTION
function and the particular integral. LESSON 12
4 3 2 1
Solve the differential equation
Find the complete solution of the
4 5 cos sisi n 4 5 cos s sin 44 1 1511 1 0 0 sisi n cos
Auxiliary equation
Trial Solution
14 , 1 cos cos s sinn sin coscos coscos sisin cos csos sisinsinncoscos5sisisnin c 4co cos 44 5 5 coscoscosos sisi44n5 5 sin 333 5 55 11 1741 17 174 coscos 171 sin 174 coscos 171 sisin C.F is
Substituting into original equation
Particular Integral is
The complete solution is
LESSON 14 (i) Solve the D.E
4 3 65sin2
3, 7 when 0
(ii) Hence, find the particular solution for which
SOLUTION
Auxiliary equation
1,4 33 0
− − sisi n cos sisi n 2 coscos 2 22 coscos 2 2sin2 44sisin 2 4cos2 334sisisinn22 4co cos2 4cos s 2 4 2 c o s 2 2si 2si n 2 cos2 6565 sisin 2 C.F is
Trial Solution
Substituting into original equation
465si4 8 n82 33 sin2 44 8 8 33 cos2
88 8 si65n2 88 coscos 2 65sisi65 n 2 8 0 → 1si& n 2 8 8 coscos 2 − − sisin 2 8 cos 2 7 0, 3, − −− −sisin 2 8 cos 2 33 2 coscos 2 16sin 2 33 8 sisin0 8cos8cos0 11 3 2cos 77 3 3 2co s 0 16si 16 si n 0 3 2 3 3 9 Particular Integral is General solution is
When
3 3 119 → 10 & 21 21− 1010− sin2 8 cos 2 2 1010 26 5 11 0 22±10102 0411010 21 1 ± 3 − cos3 cos 3 sin 33 Particular Solution is
is an Exponential Function is
LESSON 15
given that
Solve the differential equation
and
when
your answer in the form
.
SOLUTION
Auxiliary equation:
Complementary function:
Particular Integral:
. Give
53 cos0cos 0 sin 0 2 11;11; 0 − cos3 cos 3− sin 33 2 33 sisin 3 3cos3 3cos3 cos0 3si 11111 cos 0 s i n 0 3si n 03cos0 03cos0 2 1 3 1 3 3 2 12 43 −3 cos 3 4 sisin 3 2 2 3 22− 3 → 0 → ∞ 0 3 3231 1 0 0 1,1, 3− − − −− − LESSON 16
Solve the differential equation
given that
as
and that
when
.
SOLUTION
Auxiliary equation:
Trial Solution
Both the trial and and the original have have the same form. Coefficients differ.
2 10 26 10 26 13 22 − cos3 cos 3 sin 33 2 5; 0 General solution:
Particular solution:
C.F:
In this case, the particular integral would be of the form
but since this is already included in
the complementary function we have to use .
Trial Solution
Both the trial and and the original have have the same form. Coefficients differ.
− − − − − 2 4 2 2−− 2−− 2− − 3− 2− 4 12 2 12 − − 12 − →→ ∞0 − → →∞0 3 0 → 0:0: 0 3,3, 0 0 − 12 − 12 − 3 5 12 12 0 2 52 − 12 −
LESSON 17 (i) Show that by using the substitution
may be written in the form
Particular Integral:
(ii) Find the general solution of
General Solution:
as
and that
Therefore we have
From general solution (
Particular solution:
and hence find the general solution of
when
:
):
2 2 2 2
differential equation
SOLUTION (i)
× 1 × 2 1 1 1 2 2 2 2 0 2
(ii) Auxiliary equation:
Complementary function:
Particular Integral Let
, the
∴ 2 2 2 22 2 1 1 2 12 2 0 14 1 → 1 1 12 14 1 4 4 2 1 4 4 2 1 ≠0 233 1 333 2 2 18 6 9 18 233 1 333 2 2 18 Particular Integral: Complete solution:
LESSON 18 It is given that the differential equation
(a) Show that the substitution this differential equation into
, satisfies
transforms
(b) Hence find the general solution of the differential equation
giving your answer in the form
SOLUTION (a)
.
2 23 1 1 3333 2 18 6 2 99 6 6 1818 2 6 99 1818 6 9 1818 6 9 0 3,3 − 0 0 66 9 18 6 9 9 9 9 18 9 18 2 6 9 0 9 124 3 2 43
(b) Auxiliary equation
Complementary function:
Particular Integral Let
− 2 − 2 43 − 2 34 General solution:
(b)
LESSON 19
, > 0,0, > 0 2 4 2 88 1 1212 12 4 3 3 88 1 1212 12 × 12 − → 2 2 2
(a) Given that function of , show that:
and is a
(i)
(ii)
(b) Hence show that the substitution transforms the differential equation
into
(c) Hence find the general solution of the differential equation
giving your answer in the form
SOLUTION (a) (i)
Now,
(ii)
− 2 − 2 2 2 4 88 1 1212 12 88 1 12 12 4 2 88 −2 12 12 4 2 16 2 12 1212 4 1616 12 1212 4 3 3 4 3 0 1,1, 3 0 0 44 3 3 44 3 3 3 3 33 3 3 1 44 3 3 0 4 3 0 3 4
.
(c) Auxiliary equation
Complementary function:
43 43 , 4 3 General Solution: Since
LESSON 20 (a) A pond is initially empty and is then filled
5√ 41 5 5
gradually with water. After minutes, minutes, the depth of the water, metres, satisfies the differential equation
Solve this differential equation to find in
terms of .
(b) Another pond is gradually filling with water,
after minutes, minutes, the surface of the water forms
a circle of radius metres. The rate of change of the radius is inversely proportional to the area of the surface of the water. (i)
Write down a differential equation, in the variables and and and a constant of proportionality, which represents how the radius of the surface of the water is changing with time. (You are not required to solve your differential equation.)
(ii)
When the radius of the pond is 1 metre, the radius is increasing at a rate of 4.5 metres per second. Find the radius of the pond when the
radius is increasing at a rate of 0.5 metres per second. SOLUTION (a)
√ + + ++
1√ 45 1 4 5 − 51 1 −− 45 4 5 5 1 45 4551 15 1 −− 2 25 √ 4 5 15 1 1 0, 0 25 4 50 15 1 1 0 12 1 1 5 √ 4 5 5 1 1 5 11 √ 4 5 2 2 1 4 5 52 211 5 52 211 4 15 52 211 45 9 1, 4.5 29 1 2 9 2 29 When
(b) (i)
(ii)
When
12 29 39 LESSON 2 The number of bacteria in a liquid culture is observed to grow at a rate proportional to the number of cells present. At the beginning of the experiment there are 10,000 cells and after three hours there are 500,000. How many will there be after one day of growth if this unlimited growth continues? What is the of the bacteria? SOLUTION
1 ln1 + 0, 50050 000 10 000 ln50 1ln350 ≅ 1.304 103 000. 22 .. ln2 ln1.2304 ≅ 0.532 hourhourss 1.304
Let represent the number of bacteria present at time . Then the rate of change is
where is the constant of proportionality
where When Also, Therefore,
(initial population)
Using the information from the problem we now determine
Doubling time refers to the amount of time for the bacteria to double in number from its original number.
