Bird - Higher Engineering Mathematics - 5e - Solutions Manual.pdf

© 2006 John Bird. All rights reserved. Published by Elsevier.

HIGHER

ENGINEERING

MATHEMATICS

5TH EDITION

JOHN BIRD

SAMPLE OF WORKED SOLUTIONS

TO EXERCISES

© 2006 John Bird. All rights reserved. Published by Elsevier. ii

INTRODUCTION

In ‘Higher Engineering Mathematics 5th Edition’ are some 1750 further problems arranged

regularly throughout the text within 250 Exercises. A sample of solutions for over 1000 of these

further problems has been prepared in this document. The reader should be able to cope with

the remainder by referring to similar worked problems contained in the text.

CONTENTS

Page Chapter 1 Algebra 1

Chapter 2 Inequalities 13

Chapter 3 Partial fractions 19

Chapter 4 Logarithms and exponential functions 25

Chapter 5 Hyperbolic functions 41

Chapter 6 Arithmetic and geometric progressions 48

Chapter 7 The binomial series 55

Chapter 8 Maclaurin’s series 65

Chapter 9 Solving equations by iterative methods 71

Chapter 10 Computer numbering systems 85

Chapter 11 Boolean algebra and logic circuits 94

Chapter 12 Introduction to trigonometry 110

Chapter 13 Cartesian and polar co-ordinates 131

Chapter 14 The circle and its properties 135

Chapter 15 Trigonometric waveforms 144

Chapter 16 Trigonometric identities and equations 155

Chapter 17 The relationship between trigonometric and hyperbolic functions 163

Chapter 18 Compound angles 168

Chapter 19 Functions and their curves 181

Chapter 20 Irregular areas, volumes and mean values of waveforms 197

© 2006 John Bird. All rights reserved. Published by Elsevier. iii

Chapter 21 Vectors, phasors and the combination of waveforms 202

Chapter 22 Scalar and vector products 212

Chapter 23 Complex numbers 219

Chapter 24 De Moivre’s theorem 232

Chapter 25 The theory of matrices and determinants 238

Chapter 26 The solution of simultaneous equations by matrices and determinants 246

Chapter 27 Methods of differentiation 257

Chapter 28 Some applications of differentiation 266

Chapter 29 Differentiation of parametric equations 281

Chapter 30 Differentiation of implicit functions 287

Chapter 31 Logarithmic differentiation 291

Chapter 32 Differentiation of hyperbolic functions 295

Chapter 33 Differentiation of inverse trigonometric and hyperbolic functions 297

Chapter 34 Partial differentiation 306

Chapter 35 Total differential, rates of change and small changes 312

Chapter 36 Maxima, minima and saddle points for functions of two variables 319

Chapter 37 Standard integration 327

Chapter 38 Some applications of integration 332

Chapter 39 Integration using algebraic substitutions 350

Chapter 40 Integration using trigonometric and hyperbolic substitutions 356

Chapter 41 Integration using partial fractions 365

Chapter 42 The t = tan θ/2 substitution 372

Chapter 43 Integration by parts 376

Chapter 44 Reduction formulae 384

Chapter 45 Numerical integration 390

Chapter 46 Solution of first order differential equations by separation of variables 398

Chapter 47 Homogeneous first order differential equations 410

Chapter 48 Linear first order differential equations 417

Chapter 49 Numerical methods for first order differential equations 424

Chapter 50 Second order differential equations of the form 2

2

d y dya b cy 0dxdx

+ + = 435

Chapter 51 Second order differential equations of the form 2

2

d y dya b cy f (x)dxdx

+ + = 441

Chapter 52 Power series methods of solving ordinary differential equations 458

© 2006 John Bird. All rights reserved. Published by Elsevier. iv

Chapter 53 An introduction to partial differential equations 474

Chapter 54 Presentation of statistical data 489

Chapter 55 Measures of central tendency and dispersion 497

Chapter 56 Probability 504

Chapter 57 The binomial and Poisson distributions 508

Chapter 58 The normal distribution 513

Chapter 59 Linear correlation 523

Chapter 60 Linear regression 527

Chapter 61 Sampling and estimation theories 533

Chapter 62 Significance testing 543

Chapter 63 Chi-square and distribution-free tests 553

Chapter 64 Introduction to Laplace transforms 566

Chapter 65 Properties of Laplace transforms 569

Chapter 66 Inverse Laplace transforms 575

Chapter 67 The solution of differential equations using Laplace transforms 582

Chapter 68 The solution of simultaneous differential equations using Laplace transforms

590

Chapter 69 Fourier series for periodic functions of period 2π 595

Chapter 70 Fourier series for a non-periodic functions over period 2π 601

Chapter 71 Even and odd functions and half-range Fourier series 608

Chapter 72 Fourier series over any range 616

Chapter 73 A numerical method of harmonic analysis 623

Chapter 74 The complex or exponential form of a Fourier series 627


1

CHAPTER 1 ALGEBRA

EXERCISE 1 Page 2

2. Find the value of 2 35pq r when p = 25

, q = -2 and r = -1

( ) ( )2 32 3 2 25 5 2 1 5 4 15 5

pq r ⎛ ⎞= − − = × × ×− =⎜ ⎟⎝ ⎠

-8

5. Simplify ( )( )2 3 3 2x y z x yz and evaluate when x = 12

, y = 2 and z = 3

( )( )2 3 3 2x y z x yz = 2 3 3 1 1 2 5 4 3x y z x y z+ + + =

When x = 12

, y = 2 and z = 3, 5 4 3x y z = ( ) ( )5 4 3 3 3

4 35 5 4

1 2 3 3 3 272 32 2 2 2 2−

×⎛ ⎞ = = = = =⎜ ⎟⎝ ⎠

13 12

6. Evaluate 3 1 1 12 2 2 2a bc a b c

− −⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

when a = 3, b = 4 and c = 2

1 1 13 1 1 1 21

3 3 1 2 22 3 22 2 2 22

a ba bc a b c a b c a b cc

⎛ ⎞ ⎛ ⎞+ −− ⎜ ⎟ ⎜ ⎟− − + −⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞

= = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

When a = 3, b = 4 and c = 2, ( )2 2

2 2

9 23 42 4

a bc

±= = = ± 4 1

2

8. Simplify ( )

( )

1 1 13 2 2 3

3

a b c ab

a b c

−⎛ ⎞⎜ ⎟⎝ ⎠

( )

( )

1 1 13 2 2 1 11 13

1 3 1 1 1 1 18 2 9 1 33 3 32 2 3 13 2 2 3 2 2 6 3 2

3 132 2

a b c aba b c a b a b c a b c

a b c a b c

−− + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + − − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠ = = =

= 11 1 36 3 2a b c

− or

6 11 3

3

a bc


2

EXERCISE 2 Page 3

3. Remove the brackets and simplify: ( ) ( ) 24 2 3 5 2 2 3p p q p q q⎡ ⎤− − − + +⎣ ⎦

( ) ( ) 24 2 3 5 2 2 3p p q p q q⎡ ⎤− − − + +⎣ ⎦ = 24 2 15 3 2 4 3p p q p q q− − − − +⎡ ⎤⎣ ⎦

= [ ]24 30 6 4 8 3p p q p q q− − − − +

= [ ]24 26 11p p q− − = 24p – 26p + 11q = 11q – 2p

6. Simplify 2 4 6 3 4 5y y y+ ÷ + × −

2 4 6 3 4 5y y y+ ÷ + × − = 4 22 3 4 5 2 12 56 3

y y y yy y

+ + × − = + + − = 2 3 123

yy− +

8. Simplify 2 3 2 6a ab a b ab− × ÷ +

2 3 2 6a ab a b ab− × ÷ + =

22 2 2 22 63

6 6a a ba ab ab a ab a a abb b

− × + = − + = − + = ab


3

EXERCISE 3 Page 4

3. Solve the equation: 1 1 03 2 5 3a a

+ =− +

1 1

3 2 5 3a a= −

− + from which, (5a + 3) = -(3a – 2)

i.e. 5a + 3 = -3a + 2

and 5a + 3a = 2 – 3

Thus, 8a = -1 and a = - 18

4. Solve the equation: 3 61

tt= −

−

If 3 61

tt= −

− then ( )3 6 1t t= − −

i.e. 3 6 6t t= − + from which, 6 6 3 3t t t= − =

Hence, if 6 = 3 t then 6 23

t = = and t = 22 = 4

6. Make l the subject of 2 ltg

π=

If 2 ltg

π= then 2t l

gπ= and

2

2t l

gπ⎛ ⎞ =⎜ ⎟⎝ ⎠

from which, 2

2tl gπ

⎛ ⎞= ⎜ ⎟⎝ ⎠

or 2

24gtlπ

=

7. Transpose LmL rCRµ

=+

for L

If LmL rCRµ

=+

then ( )m L rCR Lµ+ = i.e. mL mrCR Lµ+ =


4

from which, ( )mrCR L mL L mµ µ= − = − and mrCRLmµ

=−

8. Make r the subject of the formula 2

2

11

x ry r

+=

−

If 2

2

11

x ry r

+=

− then ( ) ( )2 21 1x r y r− = +

from which, 2 2x xr y yr− = +

and ( )2 2 2x y yr xr r y x− = + = +

Thus, 2 x yrx y−

=+

and x yrx y

⎛ ⎞−= ⎜ ⎟+⎝ ⎠


5

EXERCISE 4 Page 5

2. Solve the simultaneous equations 5 1 32 4 0a bb a= −+ + =

5a + 3b = 1 (1)

a + 2b = -4 (2)

5 × (2) gives: 5a + 10b = -20 (3)

(1) – (3) gives: - 7b = 21 from which, b = 217−

= -3

Substituting in (1) gives: 5a + 3(-3) = 1

from which, 5a = 1 + 9 = 10 and a = 102

= 2

3. Solve the simultaneous equations

2 49 (1)5 3 153 5 0 (2)7 2 7

x y

x y

+ =

− + =

15 × (1) gives: 3x + 10y = 49 (3)

14 × (2) gives: 6x – 7y = -10 (4)

2 × (3) gives: 6x + 20y = 98 (5)

(5) – (4) gives: 27y = 108 from which, y = 10827

= 4

Substituting in (3) gives: 3x + 40 = 49 and 3x = 49 – 40 = 9

from which, x = 3

4.(b) Solve the quadratic equation by factorisation: 28 2 15 0x x+ − =

If 28 2 15 0x x+ − = then (4x – 5)(2x + 3) = 0

hence, 4x – 5 = 0 i.e. 4x = 5 i.e. x = 54

and 2x + 3 = 0 i.e. 2x = -3 i.e. x = 32

−


6

5. Determine the quadratic equation in x whose roots are 2 and -5

If roots are x = 2 and x = -5 then (x – 2)(x + 5) = 0 i.e. 2 2 5 10 0x x x− + − =

i.e. 2 3 10 0x x+ − =

6.(a) Solve the quadratic equation, correct to 3 decimal places: 22 5 4 0x x+ − =

If 22 5 4 0x x+ − = then 25 5 4(2)( 4) 5 (25 32) 5 572(2) 4 4

x⎡ ⎤− ± − − − ± + − ±⎣ ⎦= = =

Hence, x = 5 574

− + = 0.637 or x = 5 574

− − = -3.137


7

EXERCISE 5 Page 8

3. Determine ( ) ( )210 11 6 2 3x x x+ − ÷ +

5x - 2 22 3 10 11 6x x x+ + −

210 15x x+ - 4x - 6 - 4x - 6

Hence, 210 11 62 3

x xx+ −+

= 5x - 2

5. Divide ( )3 2 2 33 3x x y xy y+ + + by (x + y)

2 22x xy y+ +

3 2 2 33 3x y x x y xy y+ + + +

3 2x x y+

2 22 3x y xy+ 2 22 2x y xy+

2 3xy y+ 2 3xy y+

Hence, 3 2 2 33 3x x y xy y

x y+ + +

+ = 2 22x xy y+ +

6. Find ( )25 4 ( 1)x x x− + ÷ −

5x + 4 21 5 4x x x− − +

25 5x x− 4x + 4 4x - 4 8

Hence, 25 4

1x x

x− +−

= 5x + 4 + 81x −


8

8. Determine ( )4 35 3 2 1 ( 3)x x x x+ − + ÷ −

3 25 18 54 160x x x+ + + 4 33 5 3 2 1x x x x− + − +

4 35 15x x−

318x 3 218 54x x− 254 2x x− 254 162x x− 160x + 1 160x - 480 481

Hence, 4 35 3 2 1

3x x x

x+ − +

− = 3 2 4815 18 54 160

3x x x

x+ + + +

−


9

EXERCISE 6 Page 9

2. Use the factor theorem to factorise 3 2 4 4x x x+ − −

Let f(x) = 3 2 4 4x x x+ − −

If x = 1, f(x) = 1 + 1 – 4 – 4 = -6

x = 2, f(x) = 8 + 4 – 8 – 4 = 0 hence, (x – 2) is a factor

x = 3, f(x) = 27 + 9 – 12 – 4 = 20

x = -1, f(x) = -1 + 1 + 4 – 4 = 0 hence, (x + 1) is a factor

x = -2, f(x) = -8 + 4 + 8 – 4 = 0 hence, (x + 2) is a factor

Thus, 3 2 4 4x x x+ − − = (x + 1)(x + 2)(x – 2)

4. Use the factor theorem to factorise 3 22 16 15x x x− − +

Let f(x) = 3 22 16 15x x x− − +

If x = 1, f(x) = 2 – 1 – 16 + 15 = 0 hence, (x – 1) is a factor

x = 2, f(x) = 16 – 4 – 32 +15 = -5

x = 3, f(x) = 54 – 9 – 48 + 15 = 12

x = -1, f(x) = – 1 – 1 + 16 + 15 = 29

x = -2, f(x) = -16 – 4 + 32 + 15 = 27

x = -3, f(x) = -54 – 9 + 48 + 15 = 0 hence, (x + 3) is a factor

3 2 3 2

2

2 16 15 2 16 15( 1)( 3) 2 3

x x x x x xx x x x− − + − − +

=− + + −

2x - 5 2 3 22 3 2 16 15x x x x x+ − − − +

3 22 4 6x x x+ − 25 10 15x x− − + 25 10 15x x− − +

Hence, 3 22 16 15x x x− − + = (x – 1)(x + 3)(2x – 5)


10

6. Solve the equation 3 22 2 0x x x− − + =

Let f(x) = 3 22 2x x x− − +

If x = 1, f(x) = 1 – 2 – 1 + 2 = 0 hence, (x – 1) is a factor

x = 2, f(x) = 8 – 8 – 2 + 2 = 0 hence, (x – 2) is a factor

x = 3, f(x) = 27 – 18 – 3 + 2 = 8

x = -1, f(x) = -1 – 2 + 1 + 2 = 0 hence, (x + 1) is a factor

Hence, 3 22 2x x x− − + = (x – 1)(x – 2)(x + 1)

If 3 22 2 0x x x− − + = then (x – 1)(x – 2)(x + 1) = 0

from which, x = 1, x = 2, or x = -1


11

EXERCISE 7 Page 11

2. Determine the remainder when 3 26 5x x x− + − is divided by (a) (x + 2) (b) (x – 3)

(a) Remainder is 3 2ap bp cp d+ + + where a = 1, b = -6, c = 1, d = -5 and p = -2

Hence, remainder = 3 21( 2) 6( 2) 1( 2) 5− − − + − − = -8 – 24 – 2 – 5 = -39

(b) When p = 3, remainder = 3 21(3) 6(3) 1(3) 5− + − = 27 – 54 + 3 – 5 = -29

4. Determine the factors of 3 27 14 8x x x+ + + and hence solve the cubic equation

3 27 14 8 0x x x+ + + =

Remainder is 3 2ap bp cp d+ + + where a = 1, b = 7, c = 14, d = 8

Let p = 1, then remainder = 3 21(1) 7(1) 14(1) 8+ + + = 30

Let p = -1, then remainder = 3 21( 1) 7( 1) 14( 1) 8− + − + − + = -1 + 7 – 14 + 8 = 0, hence (x + 1) is a

factor

Let p = -2, then remainder = 3 21( 2) 7( 2) 14( 2) 8− + − + − + = -8 + 28 – 28 + 8 = 0, hence (x + 2) is a

factor

Let p = -3, then remainder = 3 21( 3) 7( 3) 14( 3) 8− + − + − + = -27 + 63 – 42 + 8 = 2

Let p = -4, then remainder = 3 21( 4) 7( 4) 14( 4) 8− + − + − + = -64 + 112 – 56 + 8 = 0, hence (x + 4) is

a factor

Hence, 3 27 14 8x x x+ + + = (x + 1)(x + 2)(x + 4)

If 3 27 14 8 0x x x+ + + = then (x + 1)(x + 2)(x + 4) = 0

from which, x = -1, x = -2 or x = -4


12

6. Using the remainder theorem, solve the equation 3 22 7 6 0x x x− − + =

Remainder is 3 2ap bp cp d+ + + where a = 2, b = -1, c = -7, d = 6

Let p = 1, then remainder = 3 22(1) ( 1)(1) ( 7)(1) 6+ − + − + = 2 - 1 – 7 + 6 = 0, hence (x - 1) is a

factor

Let p = 2, then remainder = 3 22(2) ( 1)(2) ( 7)(2) 6+ − + − + = 16 - 4 – 14 + 6 = 4

Let p = -1, then remainder = 3 22( 1) ( 1)( 1) ( 7)( 1) 6− + − − + − − + = -2 - 1 + 7 + 6 = 10

Let p = -2, then remainder = 3 22( 2) ( 1)( 2) ( 7)( 2) 6− + − − + − − + = -16 - 4 + 14 + 6 = 0, hence (x + 2)

is a factor

Let p = -3, then remainder = 3 22( 3) ( 1)( 3) ( 7)( 3) 6− + − − + − − + = -54 - 9 + 21 + 6 = -36

The third root can be found by division, i.e. ( )

3 2 3 2

2

2 7 6 2 7 61)( 2 2

x x x x x xx x x x− − + − − +

=− + + −

2x - 3 2 3 22 2 7 6x x x x x+ − − − +

3 22 2 4x x x+ − 23 3 6x x− − + 23 3 6x x− − +

Hence, 3 22 7 6x x x− − + = (x – 1)(x + 2)(2x – 3)

If 3 22 7 6 0x x x− − + = then (x – 1)(x + 2)(2x – 3) = 0

from which, x = 1, x = -2 or x = 1.5


13

CHAPTER 2 INEQUALITIES EXERCISE 8 Page 13

2. Solve the following inequalities: (a) 2x > 1.5 (b) x + 2 ≥ 5

(a) 2x > 1.5 i.e. x > 2(1.5) i.e. x > 3

(b) x + 2 ≥ 5 i.e. x ≥ 5 – 2 i.e. x ≥ 3

4. Solve the following inequalities: (a) 7 24

k−≤ 1 (b) 3z + 2 > z + 3

(a) 7 24

k−≤ 1 i.e. 7 – 2k ≤ 4 i.e. 7 – 4 ≤ 2k i.e. 3 ≤ 2k and k ≥ 3

2

(b) 3z + 2 > z + 3 i.e. 3z – z > 3 – 2 i.e. 2z >1 and z > 12

5. Solve the following inequalities: (a) 5 – 2y ≤ 9 + y (b) 1 - 6x ≤ 5 + 2x

(a) 5 – 2y ≤ 9 + y i.e. 5 – 9 ≤ y + 2y i.e. -4 ≤ 3y i.e. 43

− ≤ y or y ≥ 43

−

(b) 1 - 6x ≤ 5 + 2x i.e. 1 – 5 ≤ 2x + 6x i.e. -4 ≤ 8x i.e. 48

− ≤ x or x ≥ - 12


14

EXERCISE 9 Page 14 1. Solve the inequality: 1t + < 4 If 1t + < 4 then -4 < t + 1 < 4 -4 < t + 1 becomes -5 < t i.e. t > -5 t + 1 < 4 becomes t < 3 Hence, -5 < t < 3 3. Solve the inequality: 2 1x − < 4 If 2 1x − < 4 then -4 < 2x – 1 < 4

-4 < 2x – 1 becomes -3 < 2x and 32

− < x

2x – 1 < 4 becomes 2x < 5 and x < 52

Hence, 32

− < x < 52

5. Solve the inequality: 1 k− ≥ 3 1 k− ≥ 3 means 1 – k ≥ 3 and 1 – k ≤ -3 i.e. 1 – 3 ≥ k and 1 + 3 ≤ k i.e. k ≤ -2 and k ≥ 4


15

EXERCISE 10 Page 15

2. Solve the inequality: 2 45

tt+−

> 1

If 2 45

tt+−

> 1 then 2 45

tt+−

- 1 > 0 i.e. 2 4 55 5

t tt t+ −

−− −

> 0 and ( ) ( )2 4 55

t tt

+ − −−

> 0

i.e. 95

tt+−

> 0 Hence, either (i) t + 9 > 0 and t – 5 > 0

or (ii) t + 9 < 0 and t – 5 < 0 (i) t > -9 and t > 5 and both inequalities are true when t > 5 (ii) t < -9 and t < 5 and both inequalities are true when t < -9

Hence, 2 45

tt+−

> 1 is true when t > 5 or t < -9

3. Solve the inequality: 3 45

zz−+

≤ 2

If 3 45

zz−+

≤ 2 then 3 45

zz−+

- 2 ≤ 0 i.e. 3 4 2( 5)5 ( 5)

z zz z− +

−+ +

≤ 0 and (3 4) 2( 5)5

z zz

− − ++

≤ 0

i.e. 3 4 2 105

z zz

− − −+

≤ 0 i.e. 145

zz−+

≤ 0

Hence, either (i) z - 14 ≤ 0 and z + 5 > 0 or (ii) z – 14 ≥ 0 and z + 5 < 0 (i) z ≤ 14 and z > -5 i.e. -5 < z ≤ 14 (ii) z ≥ 14 and z ≥ -5 Both of these inequalities are not possible to satisfy.

Hence, 3 45

zz−+

≤ 2 is true when -5 < z ≤ 14


16

EXERCISE 11 Page 16 3. Solve the inequality: 22x ≥ 6

22x ≥ 6 i.e. 2x ≥ 3 hence, x ≥ 3 or x ≤ - 3 4. Solve the inequality: 23 2k − ≤ 10

23 2k − ≤ 10 i.e. 23k ≤ 12 and 2k ≤ 4 Hence, - 4 ≤ k ≤ 4 i.e. -2 ≤ k ≤ 2 6. Solve the inequality: ( )21t − ≥ 36 ( )21t − ≥ 36 then (t – 1) ≥ 36 or (t – 1) ≤ - 36 i.e. (t – 1) ≥ 6 or (t – 1) ≤ -6 i.e. t ≥ 7 or t ≤ -5 8. Solve the inequality: ( )24 5k + > 9 ( )24 5k + > 9 then 4k + 5 > 9 or 4k + 5 < - 9 i.e. 4k + 5 > 3 or 4k + 5 < -3 i.e. 4k > -2 or 4k < -8

and k > 12

− or k < -2


17

EXERCISE 12 Page 17 1. Solve the inequality: 2 6x x− − > 0

2 6x x− − > 0 thus (x – 3)(x + 2) > 0 Either (i) x – 3 > 0 and x + 2 > 0 or (ii) x – 3 < 0 and x + 2 < 0 (i) x > 3 and x > -2 i.e. x > 3 (ii) x < 3 and x < -2 i.e. x < -2 3. Solve the inequality: 22 3 2x x+ − < 0

22 3 2x x+ − < 0 thus (2x – 1)(x + 2) < 0 Either (i) 2x – 1 > 0 and x + 2 < 0 or (ii) 2x – 1 < 0 and x + 2 > 0

(i) 2x > 1 i.e. x > 12

and x < -2 both of which are not possible

(ii) 2x < 1 i.e. x < 12

and x > -2 thus -2 < x < 12

5. Solve the inequality: 2 4 4z z+ + ≤ 4

2 4 4z z+ + ≤ 4 i.e. 2 4z z+ ≤ 0 i.e. z(z + 4) ≤ 0 Either (i) z ≤ 0 and z ≥ -4 i.e. -4 ≤ z ≤ 0 or (ii) z ≥ 0 and z ≤ -4 both of which are not possible 7. Solve the inequality: 2 4 7t t− − ≥ 0 Since 2 4 7t t− − ≥ 0 then ( )22t − - 7 – 4 ≥ 0


18

i.e. ( )22t − ≥ 11 and t – 2 ≥ 11 or t – 2 ≤ - 11 thus, t ≥ ( )2 11+ or t ≤ ( )2 11−


19

CHAPTER 3 PARTIAL FRACTIONS

EXERCISE 13 Page 20

2. Resolve 2

4( 4)2 3

xx x

−− −

into partial fractions.

Let 2

4( 4) 4 16 ( 3) ( 1)2 3 ( 1)( 3) ( 1) ( 3) ( 1)( 3)

x x A B A x B xx x x x x x x x

− − − + +≡ = + =

− − + − + − + −

Hence, 4x – 16 = A(x – 3) + B(x + 1)

If x = -1, -20 = -4A from which, A = 5

If x = 3, 12 – 16 = 4B from which, B = -1

Hence, 2

4( 4) 5 12 3 ( 1) ( 3)

xx x x x

−= −

− − + −

4. Resolve 23(2 8 1)

( 4)( 1)(2 1)x x

x x x− −

+ + − into partial fractions.

Let

23(2 8 1) ( 1)(2 1) ( 4)(2 1) ( 4)( 1)( 4)( 1)(2 1) ( 4) ( 1) (2 1) ( 4)( 1)(2 1)

x x A B C A x x B x x C x xx x x x x x x x x

− − + − + + − + + +≡ + + =

+ + − + + − + + − Hence, 26 24 3x x− − = A(x + 1)(2x – 1) + B(x + 4)(2x – 1) + C(x + 4)(x + 1)

If x = -4, 96 + 96 -3 = A(-3)(-9) from which, 189 = 27A and A = 7

If x = -1, 6 + 24 -3 = B(3)(-3) from which, 27 = -9B and B = -3

If x = 0.5, 1.5 - 12 -3 = C(4.5)(1.5) from which, -13.5 = 6.75C and C = -2

Hence, 23(2 8 1) 7 3 2

( 4)( 1)(2 1) ( 4) ( 1) (2 1)x x

x x x x x x− −

= − −+ + − + + −

5. Resolve 2

2

9 86

x xx x+ ++ −


Since the numerator is of the same degree as the denominator, division is firstly required.


20

1 2 26 9 8x x x x+ − + +

2 6x x+ − 8x + 14

Hence, 2

2 2

9 8 8 1416 6

x x xx x x x+ + +

= ++ − + −

Let 2

8 14 8 14 ( 2) ( 3)6 ( 3)( 2) ( 3) ( 2) ( 3)( 2)

x x A B A x B xx x x x x x x x

+ + − + += ≡ + =

+ − + − + − + −

Hence, 8x + 14 = A(x – 2) + B(x + 3)

If x = -3, -24 + 14 = -5A form which, -10 = -5A and A = 2

If x = 2, 16 + 14 = 5B from which, 30 = 5B and B = 6

Hence, 2

2

9 8 2 616 ( 3) ( 2)

x xx x x x+ +

= + ++ − + −

7. Resolve 3 23 2 16 20( 2)( 2)

x x xx x− − +− +


3x - 2 2 3 24 3 2 16 20x x x x− − − +

33 12x x−

22 4 20x x− − + 22 8x− +

- 4x + 12

Hence, 3 2

2

3 2 16 20 12 43 2( 2)( 2) 4

x x x xxx x x− − + −

≡ − +− + −

Let 2

12 4 12 4 ( 2) ( 2)4 ( 2)( 2) ( 2) ( 2) ( 2)( 2)x x A B A x B x

x x x x x x x− − + + −

= ≡ + =− − + − + − +

Hence, 12 – 4x = A(x + 2) + B(x - 2)

If x = 2, 4 = 4A from which, A = 1

If x = -2 20 = -4B from which, B = -5

Hence, 3 23 2 16 20 1 53 2( 2)( 2) ( 2) ( 2)

x x x xx x x x− − +

≡ − + −− + − +


21

EXERCISE 14 PAGE 22

2. Resolve 2

2

7 3( 3)

x xx x+ +

+ into partial fractions.

Let 2 2

2 2 2

7 3 ( )( 3) ( 3)( 3) ( 3) ( 3)

x x A B C A x x B x Cxx x x x x x x+ + + + + +

≡ + + =+ + +

Hence, 2x + 7x + 3 = A(x)(x + 3) + B (x + 3) + C 2x

If x = 0 3 = 3B from which, B = 1

If x = -3 9 – 21 + 3 = 9C i.e. -9 = 9C f rom which, C = -1

Equating 2x coefficients: 1 = A + C from which, A = 2

Hence, 2

2 2

7 3 2 1 1( 3) ( 3)

x xx x x x x+ +

= + −+ +

4. Resolve 2

2

18 21( 5)( 2)

x xx x+ −− +


Let 2 2

2 2 2

18 21 ( 2) ( 5)( 2) ( 5)( 5)( 2) ( 5) ( 2) ( 2) ( 5)( 2)

x x A B C A x B x x C xx x x x x x x+ − + + − + + −

≡ + + =− + − + + − +

Hence, 2 218 21 ( 2) ( 5)( 2) ( 5)x x A x B x x C x+ − = + + − + + −

If x = 5 18 + 105 – 25 = 49A i.e. 98 = 49A from which, A = 2

If x = -2 18 – 42 – 4 = -7C i.e. -28 = -7C from which, C = 4

Equating 2x coefficients: -1 = A + B from which, B = -3

Hence, 2

2 2

18 21 2 3 4( 5)( 2) ( 5) ( 2) ( 2)

x xx x x x x+ −

= − +− + − + +


22

EXERCISE 15 PAGE 23

1. Resolve ( )( )

2

2

137 2

x xx x

− −+ −


Let ( )( ) ( )

( )( )

22

2 2 2

( )( 2) 713( 2)7 2 7 7 ( 2)

Ax B x C xx x Ax B Cxx x x x x

+ − + +− − +≡ + =

−+ − + + −

Hence, ( )2 213 ( )( 2) 7x x Ax B x C x− − = + − + +

If x = 2, 4 – 2 –13 = 11C i.e. -11 = 11C from which, C = -1

Equating 2x coefficients: 1 = A + C from which, A = 2

Equating constant terms: -13 = -2B + 7C = -2B – 7 i.e. 2B = 13 – 7 = 6 from which, B = 3

Hence, ( ) ( ) ( )2

2 2

13 2 3 1( 2)7 2 7

x x xxx x x

− − += −

−+ − +

4. Resolve ( )

3 2

2 2

4 20 7( 1) 8

x x xx x+ + −− +


Let ( ) ( )

( ) ( )( )

2 2 23 2

22 2 2 2 2

( 1) 8 8 ( )( 1)4 20 7( 1) ( 1)( 1) 8 8 ( 1) 8

A x x B x Cx D xx x x A B Cx Dx xx x x x x

− + + + + + −+ + − +≡ + + =

− −− + + − +

Hence, ( ) ( )3 2 2 2 24 20 7 ( 1) 8 8 ( )( 1)x x x A x x B x Cx D x+ + − = − + + + + + −

( ) ( )2 2 2( 1) 8 8 ( )( 2 1)A x x B x Cx D x x= − + + + + + − +

If x = 1, 1 + 4 + 20 – 7 = 9B i.e. 18 = 9B from which, B = 2

Equating 3x coefficients: 1 = A + C (1’)

Equating 2x coefficients: 4 = -A + B – 2C + D (2’)

Equating x coefficients: 20 = 8A + C – 2D (3’)

Since B = 2, A + C = 1 (1)

-A – 2C + D = 2 (2)


23

8A + C – 2D = 20 (3)

2 × (2) gives: -2A – 4C + 2D = 4 (4)

(3) + (4) gives: 6A – 3C = 24 (5)

3 × (1) gives: 3A + 3C = 3 (6)

(5) + (6) gives: 9A = 27 from which, A = 3

From (1): 3 + C = 1 from which, C = -2

From (2): -3 + 4 + D = 2 from which, D = 1

Hence, ( ) ( )3 2

22 2 2

4 20 7 3 2 1 2( 1) ( 1)( 1) 8 8

x x x xx xx x x

+ + − −= + +

− −− + +

5. When solving the differential equation 2

22 6 10 20 td d e

dt dtθ θ θ− − = − by Laplace transforms, for

given boundary conditions, the following expression for Λ θ results:

Λ θ = ( )( )

3 2

2

394 42 4022 6 10

s s s

s s s s

− + −

− − +

Show that the expression can be resolved into partial fractions to give:

Λ ( ) ( )2

2 1 5 32 2 2 6 10

ss s s s

θ −= − +

− − +

Let ( )( ) ( )( ) ( )

( )

3 2

2 2

2 2

2

394 42 402

( 2)2 6 10 6 10

( 2) 6 10 ( ) 6 10 ( )( )( 2)

( 2) 6 10

s s s A B Cs Ds ss s s s s s

A s s s B s s s Cs D s s

s s s s

− + − +≡ + +

−− − + − +

− − + + − + + + −=

− − +

Hence, ( ) ( )3 2 2 2394 42 40 ( 2) 6 10 ( ) 6 10 ( )( )( 2)2

s s s A s s s B s s s Cs D s s− + − = − − + + − + + + −

( ) ( )3 2 3 2 28 2 20 6 10 ( )( 2 )A s s s B s s s Cs D s s= − − − + − + + + −

If s = 0, -40 = A(-20) from which, A = 2

If s = 2, 32 – 78 + 84 – 40 = B (8 – 24 + 20) i.e. -2 = 4B from which, B = 12

−


24

Equating 3s coefficients: 4 = A + B + C i.e. 4 = 2 - 12

+ C from which, C = 52

Equating 2s coefficients: 392

− = -8A – 6B – 2C + D i.e. 392

− = -16 + 3 – 5 + D

from which, D = 32

−

Hence, ( )( ) ( )

3 2

2 2

39 1 5 34 42 40 22 2 2 2( 2)2 6 10 6 10

s s s s

s ss s s s s s

− + − − −≡ + +

−− − + − +

i.e. Λ ( ) ( )2

2 1 5 32 2 2 6 10

ss s s s

θ −= − +

− − +


25

CHAPTER 4 LOGARITHMS AND EXPONENTIAL FUNCTIONS

EXERCISE 16 Page 26 2. Evaluate: 2log 16 Let x = 2log 16 then 42 16 2x = = from which, x = 4 Hence, 2log 16 = 4

4. Evaluate: 21log8

Let x = 21log8

then 33

1 12 28 2

x −= = = from which, x = -3

Hence, 21log8

= -3

7. Evaluate: 4log 8 Let x = 4log 8 then 4 8x = i.e. ( )2 32 2

x= i.e. 2 32 2x =

from which, 2x = 3 and x = 32

Hence, 4log 8 = 1 12

11. Solve the equation: 41log 22

x = −

If 41log 22

x = − then x = 52

5 552

1 1 14244

−= = = ± =

132

±


26

12. Solve the equation: lg 2x = −

If lg x = -2 then 10log 2x = − and x = 22

11010

− = = 1100

or 0.01

16. Write in terms of log 2, log 3 and log 5 to any base: 416 5log

27⎛ ⎞×⎜ ⎟⎜ ⎟⎝ ⎠

1

1 144 44 3 4 34 4

3

16 5 2 5log log log 2 5 3 log 2 log5 log327 3

− −

⎛ ⎞⎛ ⎞ ⎛ ⎞× ×⎜ ⎟= = × × = + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎜ ⎟⎝ ⎠ ⎝ ⎠

= 4 log 2 + 14

log 5 – 3 log 3

17. Write in terms of log 2, log 3 and log 5 to any base: 4

34

125 16log81

⎛ ⎞×⎜ ⎟⎜ ⎟⎝ ⎠

4

34

125 16log81

⎛ ⎞×⎜ ⎟⎜ ⎟⎝ ⎠

= ( )3

3 3 3 33

5 2log log 5 2 3 log 5 log 2 log 33

− −⎛ ⎞×= × × = + +⎜ ⎟

⎝ ⎠

= 3 log 5 + log 2 – 3 log 3 19. Simplify: log 64 + log 32 – log 128

log 64 + log 32 – log 128 = ( )6 5

6 5 7 6 5 7 47

2 2log 2 log 2 log 2 log log 2 log 22

+ −⎛ ⎞×+ − = = =⎜ ⎟

⎝ ⎠ = 4 log 2

20. Evaluate:

1 1log16 log82 3

log 4

−

1 1log16 log82 3

log 4

− =

( ) ( )2

1 1114 32 3 232

2 2 2 2

2loglog 2 log 2 2log16 log8 log 2 log 2 log 2log 2 log 2 log 2 log 2 2log 2

⎛ ⎞⎜ ⎟−− − ⎝ ⎠= = = = = 1

2


27

22. Solve the equation: 3log 2 log log16 logt t t− = +

3log 2 log log16 logt t t− = +

i.e. ( )32log log 16t t

t⎛ ⎞

=⎜ ⎟⎝ ⎠

i.e. ( ) ( )2log 2 log 16t t=

from which, 22 16t t= i.e. 22 16 0t t− = i.e. 2t(t – 8) = 0

Hence, t = 8 (note that t = 0 is not a valid solution to the equation)


28

EXERCISE 17 Page 27 1. Solve the equation 3 6.4x = correct to 4 significant figures If 3 6.4x = then 10 10log 3 log 6.4x = and

x = 10

10

log 6.4 0.80617997... 1.689675...log 3 0.47712125...

= = = 1.690, correct to 4 significant figures.

3. Solve the equation 1 2 12 3x x− −= correct to 4 significant figures If 1 2 12 3x x− −= then 10 10( 1) log 2 (2 1) log 3x x− = − i.e. 10 10 10 10log 2 log 2 2 log 3 log 3x x− = − i.e. ( )10 10 10 10 10 10log 3 log 2 2 log 3 log 2 2log 3 log 2x x x− = − = −

Hence, x = 10 10

10 10

log 3 log 22log 3 log 2

−−

= 0.2696, correct to 4 significant figures.

5. Solve the equation 25.28 4.2x= correct to 4 significant figures If 25.28 4.2x= then 10 10log 25.28 log 4.2x=

from which, x = 10

10

log 25.28log 4.2


6. Solve the equation 2 1 24 5x x− += correct to 4 significant figures If 2 1 24 5x x− += then 10 10(2 1) log 4 ( 2) log 5− = +x x i.e. 10 10 10 102 log 4 log 4 log 5 2log 5x x− = + i.e. 10 10 10 102 log 4 log 5 2log 5 log 4x x− = + i.e. 10 10 10 10(2 log 4 log 5) 2log 5 log 4x − = +

from which, x = 10 10

10 10

2 log 5 log 42log 4 log 5

+−



29

8. Solve the equation 0.027 3.26x = correct to 4 significant figures If 0.027 3.26x = then 10 10log 0.027 log 3.26x =

from which, x = 10

10

log 3.26log 0.027

= -0.3272, correct to 4 significant figures.

9. The decibel gain n of an amplifier is given by: n = 210

1

P10logP

⎛ ⎞⎜ ⎟⎝ ⎠

where 1P is the power input

and 2P is the power output. Find the power gain 2

1

PP

when n = 25 decibels.

When n = 25 then: 25 = 210

1

P10logP

⎛ ⎞⎜ ⎟⎝ ⎠

from which, 210

1

P25 log10 P

⎛ ⎞= ⎜ ⎟

⎝ ⎠ i.e. 2.5 = 2

101

PlogP

⎛ ⎞⎜ ⎟⎝ ⎠

Thus, 2.52

1

P 10P

= i.e. power gain, 2

1

PP

= 316.2


30

EXERCISE 18 Page 29

3. Evaluate, correct to 5 significant figures: ( )1.72952.1127 2.1127

2.1347 3.6817

4 15.6823( ) ( ) ( )2

ee ea b ce e

−−

−

−−

Using a calculator:

2.1347

5.6823( )ae− = 48.04106, correct to 5 decimal places.

2.1127 2.1127

( )2

e eb−− = 4.07482, correct to 5 decimal places.

( )1.7295

3.6817

4 1( )

ec

e

− − = -0.08286, correct to 5 decimal places.

4. The length of a bar, l , at a temperature θ is given by 0l l eαθ= , where l and α are constants.

Evaluate l , correct to 4 significant figures, when 0l = 2.587, θ = 321.7 and α = 41.771 10−×

Using a calculator, ( )4321.7 1.771 100 (2.587)l l e eαθ

−× ×= = = 2.739, correct to 4 significant figures.


31

EXERCISE 19 Page 31 2. Use the power series for xe to determine, correct to 4 significant figures, (a) 2e (b) 0.3e− and

check your result by using a calculator.

(a) 2 3 4

1 ....2! 3! 4!

x x x xe x= + + + + +

When x = 2, 2e = 2 3 4 5 62 2 2 2 21 2 ...

2! 3! 4! 5! 6!+ + + + + + +

= 1 + 2 + 2 + 1.33333 + 0.66666 + 0.26666 + 0.08888 + 0.02540 + 0.00635

+ 0.00141 + 0.00028 + 0.00005 + …

= 7.389, correct to 4 significant figures, which may be checked with a calculator.

(b) When x = -0.3, 0.3e− = 2 3 4 5( 0.3) ( 0.3) ( 0.3) ( 0.3)1 0.3 ...

2! 3! 4! 5!− − − −

− + + + + +

= 1 – 0.3 + 0.04500 – 0.00450 + 0.00034 – 0.00002 + …


3. Expand (1 – 2x) 2xe to six terms.

(1 – 2x) 2xe = (1 – 2x)2 3 4(2 ) (2 ) (2 )1 2 ...

2! 3! 4!x x xx

⎛ ⎞+ + + + +⎜ ⎟

⎝ ⎠

= (1 – 2x) 2 3 44 21 2 23 3

x x x x⎛ ⎞+ + + +⎜ ⎟⎝ ⎠

= 2 3 4 2 3 44 2 81 2 2 2 4 4 ...3 3 3

x x x x x x x x+ + + + − − − − −

= 2 3 481 2 23

x x x− − −


32

EXERCISE 20 Page 32 1. Plot a graph of y = 3 0.2xe over the range x = -3 to x = 3. Hence determine the value of y when

x = 1.4 and the value of x when y = 4.5

Figure 1

From Figure 1, when x = 1.4, y = 3.95 and when y = 4.5, x = 2.05 4. The rate at which a body cools is given by 0.05250 teθ −= where the excess temperature of a body

above its surroundings at time t minutes is Cθ° . Plot a graph showing the natural decay curve

for the first hour of cooling. Hence determine (a) the temperature after 25 minutes, and (b) the

time when the temperature is 195 C°

From Figure 2 on page 33,

(a) after t = 25 minutes, temperature θ = 70 C°

(b) when the temperature is 195 C° , time t = 5 minutes


33

Figure 2


34

EXERCISE 21 Page 34 2. Evaluate, correct to 5 significant figures:

1.76 0.1629

1.41

2.946ln 5 ln 4.8629 ln 2.4711( ) ( ) ( )lg10 2ln 0.00165 5.173

e ea b c− −

Using a calculator,

1.76

1.41

2.946ln( )lg10

ea = (2.946)(1.76)(1.41)


0.16295( )2 ln 0.00165

eb−


ln 4.8629 ln 2.4711( )5.173

c − = 0.13087, correct to 5 significant figures.

4. Solve, correct to 4 significant figures: 1.77.83 2.91 −= xe

If 1.77.83 2.91 −= xe then 1.7 7.832.91

e− = and 1.7 7.83ln ln2.91

xe− ⎛ ⎞= ⎜ ⎟⎝ ⎠

i.e. -1.7x = 7.83ln2.91

⎛ ⎞⎜ ⎟⎝ ⎠

and x = 1 7.83ln1.7 2.91

⎛ ⎞− ⎜ ⎟⎝ ⎠


5. Solve, correct to 4 significant figures: 216 24 1t

e−⎛ ⎞

= −⎜ ⎟⎝ ⎠

If 216 24 1t

e−⎛ ⎞

= −⎜ ⎟⎝ ⎠

then 216 124

t

e−

= −

from which, 2 16124

t

e−= −

and 16ln 12 24t ⎛ ⎞− = −⎜ ⎟

⎝ ⎠

and t = 162ln 124

⎛ ⎞− −⎜ ⎟⎝ ⎠



35

7. Solve, correct to 4 significant figures: 1.593.72ln 2.43x

⎛ ⎞ =⎜ ⎟⎝ ⎠

If 1.593.72ln 2.43x

⎛ ⎞ =⎜ ⎟⎝ ⎠

then 1.59 2.43ln3.72

⎛ ⎞ =⎜ ⎟⎝ ⎠x

from which, 2.433.721.59 e

x

⎛ ⎞⎜ ⎟⎝ ⎠=

and x = 2.433.72

2.433.72

1.59 1.59ee

⎛ ⎞−⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

= = 0.8274, correct to 4 significant figures.

8. The work done in an isothermal expansion of a gas from pressure 1p to 2p is given by:

10

2

pw w lnp

⎛ ⎞= ⎜ ⎟

⎝ ⎠

If the initial pressure 1p = 7.0 kPa, calculate the final pressure 2p if 0w 3w=

If 0w 3w= then 03w = 10

2

pw lnp

⎛ ⎞⎜ ⎟⎝ ⎠

i.e. 3 = 1

2

plnp

⎛ ⎞⎜ ⎟⎝ ⎠

and 3 1

2 2

p 7000ep p

= =

from which, final pressure, 32 3

7000p 7000ee

−= = = 348.5 Pa


36

EXERCISE 22 Page 37

2. The voltage drop, v volts, across an inductor L henrys at time t seconds is given by 200RtLv e

−

= ,

where R = 150 Ω and L = 12.5 310−× H. Determine (a) the voltage when t = 6160 10−× s, and

(b) the time for the voltage to reach 85 V.

(a) Voltage 200RtLv e

−

= = ( )( )6

3

150 160 101.9212.5 10200 200

−

−

×−

−× =e e = 29.32 volts

(b) When v = 85 V, 3150

12.5 1085 200 −−×=

t

e from which, 3150

12.5 1085200

−−×=

t

e

and 3

150 85ln12.5 10 200

t−

⎛ ⎞− = ⎜ ⎟× ⎝ ⎠

Thus, time t = 312.5 10 85ln

150 200

−× ⎛ ⎞− ⎜ ⎟⎝ ⎠

= 671.31 10 s−×

4. A belt is in contact with a pulley for a sector θ = 1.12 radians and the coefficient of friction

between the two surfaces is µ = 0.26. Determine the tension on the taut side of the belt, T

newtons, when tension on the slack side 0T = 22.7 newtons, given that these quantities are

related by the law 0T T eµθ= . Determine also the value of θ when T = 28.0 newtons.

Tension 0T T eµθ= = ( )( )0.26 1.12 0.291222.7 22.7e e= = 30.4 N

When T = 28.0 N, 28.0 = 22.7 0.26e θ from which, 0.2628.022.7

e θ=

Thus, 28.00.26 ln22.7

θ ⎛ ⎞= ⎜ ⎟⎝ ⎠

and θ 1 28.0ln0.26 22.7

⎛ ⎞= ⎜ ⎟⎝ ⎠

= 0.807 rad

5. The instantaneous current i at time t is given by: 10t

CRi e−

= when a capacitor is being charged.

The capacitance C is 67 10−× F and the resistance R is 60.3 10× ohms. Determine:

(a) the instantaneous current when t is 2.5 seconds, and


37

(b) the time for the instantaneous current to fall to 5 amperes.

Sketch a curve of current against time from t = 0 to t = 6 seconds.

(a) Current, i = 6 62.5

7 10 0.3 1010−−

× × ×e = 3.04 A

(b) When i = 5, 2.15 10−

=t

e from which, 2.1510

−=

t

e

Thus, 5ln10 2.1

t⎛ ⎞ = −⎜ ⎟⎝ ⎠

and time, t = (2.1) ln 0.5− = 1.46 s

Time t 0 1 2 3 4 5 6

Current i 10 6.21 3.86 2.40 1.49 0.92 0.57 A graph of current against time is shown in Figure 3.

Figure 3

7. The current i flowing in a capacitor at time t is given by: 12.5 1t

CRi e−⎛ ⎞

= −⎜ ⎟⎝ ⎠

where resistance R is 30 kilohms and the capacitance C is 20 microfarads. Determine: (a) the current flowing after 0.5 seconds, and (b) the time for the current to reach 10 amperes.


38

(a) Current, 12.5 1−⎛ ⎞

= −⎜ ⎟⎝ ⎠

tCRi e = 6 3

0.520 10 30 1012.5 1 −−× × ×

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠e = 7.07 A

(b) When i = 10 A, 0.610 12.5 1−⎛ ⎞

= −⎜ ⎟⎝ ⎠

t

e from which, 0.610 112.5

−= −

t

e

Thus, 0.6 10112.5

−= −

t

e and 10ln 10.6 12.5t ⎛ ⎞− = −⎜ ⎟

⎝ ⎠

i.e. time, t = 100.6 ln 112.5

⎛ ⎞− −⎜ ⎟⎝ ⎠

= 0.966 s


39

EXERCISE 23 Page 40 2. At particular times, t minutes, measurements are made of the temperature, Cθ° , of a cooling

liquid and the following results are obtained:

Temperature Cθ° 92.2 55.9 33.9 20.6 12.5

Time t minutes 10 20 30 40 50

Prove that the quantities follow a law of the form 0kteθ θ= , where 0θ and k are constants, and

determine the approximate value of 0θ and k.

Since 0

kteθ θ= then ( )0 0ln ln ln lnk t k te eθ θ θ= = +

i.e. 0ln lnktθ θ= +

where gradient = k and intercept on vertical axis = 0lnθ

Using log-linear graph paper, a graph of lnθ against t is shown in Figure 4

Figure 4 Since the graph is a straight line the quantities do follow a law of the form 0

kteθ θ=


40

Gradient, k = ln 92.2 ln12.510 50

ABBC

−=

− = -0.05

At point A in Figure 4, 92.2 Cθ = ° , t = 10 min Substituting in 0

kteθ θ= gives: ( )( )0.05 10092.2 eθ −=

from which, 0θ0.5

0.5

92.2 92.2ee−= = = 152


41

CHAPTER 5 HYPERBOLIC FUNCTIONS

EXERCISE 24 Page 42 1.(a) Evaluate sh 0.64 correct to 4 significant figures.

(a) sh 0.64 = ( )0.64 0.6412

e e−− = 0.6846, correct to 4 significant figures.

Alternatively, using a scientific calculator, using hyp, sin 0.64 = 0.6846 2.(b) Evaluate ch 2.4625 correct to 4 significant figures.

(b) ch 2.4625 = ( )2.4625 2.462512

e e−+ = 5.910, correct to 4 significant figures.

Alternatively, using a scientific calculator, using hyp, cos 0.72 = 5.910 3.(a) Evaluate th 0.65 correct to 4 significant figures.

(a) th 0.65 = 0.65 0.65

0.65 0.65

1.39349505...2.4375866...

e ee e

−

−

−=

+ = 0.5717, correct to 4 significant figures.

Alternatively, using a scientific calculator, using hyp, tan 0.65 = 0.5717 4.(b) Evaluate cosech 3.12 correct to 4 significant figures.

(b) cosech 3.12 = 13.12sh

= 0.08849, correct to 4 significant figures, using a calculator.

5.(a) Evaluate sech 0.39 correct to 4 significant figures.

(a) sech 0.39 = 10.39ch



42

6.(b) Evaluate coth 1.843 correct to 4 significant figures.

(b) coth 1.843 = 11.843th


7. A telegraph wire hangs so that its shape is described by 5050yy ch= . Evaluate, correct to 4

significant figures, the value of y when x = 25.

When x = 0.25, 5050yy ch= = 2550

50ch = 50 ch 0.50 = 56.38, correct to 4 significant figures.

8. The length l of a heavy cable hanging under gravity is given by l = 2 ( / 2 )c sh L c . Find the value

of l when c = 40 and L = 30.

l = 2 ( / 2 )c sh L c = 2(40) sh 30 3802(40) 8

sh⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

= 30.71

9. 2 0.55 tanh(6.3 / )V L d L= is a formula for velocity V of waves over the bottom of shallow water,

where d is the depth, and L is the wavelength. If d = 8.0 and L = 96, calculate the value of V.

2 0.55 tanh(6.3 / )V L d L= = (6.3)(8.0)0.55(96) tanh

96⎡ ⎤⎢ ⎥⎣ ⎦

= 52.8 tanh 0.525 = 25.425829…

Hence, V = 25.425829... = 5.042


43

EXERCISE 25 Page 46 2. Prove the following identities: (a) coth x ≡ 2 cosech 2x + th x (b) ch 2θ - 1 ≡ 2 sh 2 θ

(a) R.H.S. = 2 cosech 2x + th x = 22 2 1 1

2 2sh x sh x sh x sh x

sh x ch x sh xch x ch x sh x ch x ch x sh xch x+

+ = + = + =

= 2

cothch x ch x xsh xch x sh x

= = = L.H.S

(b) R.H.S. = 2 sh 2 θ = ( )( )2

2 22 122 4 2

e e e e e e e e e e e eθ θ

θ θ θ θ θ θ θ θ θ θ−

− − − − −⎛ ⎞− ⎡ ⎤= − − = − − +⎜ ⎟ ⎣ ⎦⎝ ⎠

= 2 2

2 0 0 2 2 21 1 222 2 2 2 2

e ee e e e e eθ θ

θ θ θ θ−

− −⎡ ⎤ ⎡ ⎤− − + = − + = + −⎣ ⎦ ⎣ ⎦

= 2 2

1 2 12

e e chθ θ

θ−+

− = − = L.H.S.

4. Prove the following identities: (a) sh(A + B) ≡ sh A ch B + ch A sh B

(b) 2 2

42 2

1 tanh2 cothsh x ch x x

ch x x+ −

≡

(a) R.H.S. = sh A ch B + ch A sh B = 2 2 2 2

A A B B A A B Be e e e e e e e− − − −⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞− + + −+⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

= 14

A B A B A B A B A B A B A B A Be e e e e e e e+ − − + − − + − − + − −⎡ ⎤+ − − + − + −⎣ ⎦

= ( )( ) ( )1 2 2 ( )

4 2

+ − +− ++ −⎡ ⎤− = = +⎣ ⎦

A B A BA BA B e ee e sh A B = L.H.S

(b) L.H.S. = 2 2 2 2

2 2 22

2

12 coth

2

sh x ch x sh x sh xch x x ch xch x

sh x

+ − +=

⎛ ⎞⎜ ⎟⎝ ⎠

since 2 21ch x sh x− =

= 2 2 4

42 2 4

2 tanh2

sh x sh x sh x xch x ch x ch x

⎛ ⎞= =⎜ ⎟

⎝ ⎠ = R.H.S.


44

5. Given 6 2x xP e Qe ch x sh x−− ≡ − , find P and Q

6 2x xP e Qe ch x sh x−− ≡ − = ( ) ( )6 2 32 2

x x x xx x x xe e e e e e e e

− −− −⎛ ⎞ ⎛ ⎞+ −

− = + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 3 3x x x xe e e e− −+ − + i.e. 2 4x x x xP e Qe e e− −− = + from which, P = 2 and Q = -4 6. If 5 4x xe e Ash x Bch x−− ≡ + , find A and B

5 42 2 2 2 2 2

x x x xx x x x x xe e e e A A B Be e Ash x B ch x A B e e e e

− −− − −⎛ ⎞ ⎛ ⎞− +

− ≡ + = + = − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 2 2 2 2

x xA B A Be e−⎛ ⎞ ⎛ ⎞+ − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

i.e. 5 42 2

x x x xA B A Be e e e− −+ −⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Hence, 52

A B+= i.e. A + B = 10 (1)

and 42

A B−= i.e. A – B = 8 (2)

(1) + (2) gives: 2A = 18 from which, A = 9 From (1) B = 1


45

EXERCISE 26 Page 48 2. Solve 2 ch x = 3, correct to 4 decimal places.

2 ch x = 3 from which, 32

ch x =

i.e. 32 2

x xe e−+= or 3 0x xe e−+ − =

i.e. ( )2

3 0x x x xe e e e−+ − = or ( )23 1 0x xe e− + =

Hence, ( ) ( ) ( )( )

( )

23 3 4 1 1 3 52 1 2

xe⎡ ⎤− − ± − − ±⎣ ⎦= = = 2.61803 or 0.381966

Thus, x = ln 2.61803 or x = ln 0.381966 i.e. x = 0.9624 or x = -0.9624 i.e. x = ± 0.9624 3. Solve 3.5 sh x + 2.5 ch x = 0, correct to 4 decimal places. 3.5 sh x + 2.5 ch x = 0

i.e. 3.5 2.5 02 2

x x x xe e e e− −⎛ ⎞ ⎛ ⎞− ++ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

i.e. 1.75 1.75 1.25 1.25 0− −− + + =x x x xe e e e and 3 0.5 0x xe e−− = or 3 0.5x xe e−=

i.e. 0.53

x

x

ee− = i.e. 2 0.5

3xe =

Hence, 2x = ln 0.53

from which, x = 1 0.5ln2 3

= -0.8959

5. Solve 4 th x - 1 = 0, correct to 4 decimal places.

4 th x - 1 = 0 i.e. 4 1x x

x x

e ee e

−

−

⎛ ⎞−=⎜ ⎟+⎝ ⎠


46

i.e. ( )4 x x x xe e e e− −− = + and 4 4 0x x x xe e e e− −− − − = Hence, 3 5 0x xe e−− = and 3 5x xe e−=

Thus, 53

x

x

ee− = from which, 2 5

3xe =

i.e. 2x = ln 53

and x = 1 5ln2 3

= 0.2554

6. A chain hangs so that its shape is of the form 56 ( / 56)y ch x= . Determine, correct to 4

significant figures, (a) the value of y when x is 35, and (b) the value of x when y is 62.35

(a) When x = 35, 3556 ( / 56) 5656

y ch x ch⎛ ⎞= = ⎜ ⎟⎝ ⎠

= 67.30, using a calculator.

(b) When, y = 62.35, then 62.35 = 56 ( / 56)ch x

Thus, 62.3556 56

xch= or 56 56 62.35

2 56

x x

e e−

+=

i.e. 56 56 62.35256

x x

e e− ⎛ ⎞+ = ⎜ ⎟

⎝ ⎠ = 2.22679

Thus, 2

56 56 56 562.22679 0x x x x

e e e e−⎛ ⎞ ⎛ ⎞⎛ ⎞

+ − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

i.e. 2

56 562.22679 1 0x x

e e⎛ ⎞

− + =⎜ ⎟⎝ ⎠

from which, ( ) ( ) ( )( )

( )

2

562.22679 2.22679 4 1 1 2.22679 0.95859...

2 1 2

⎡ ⎤− − ± − − ±⎣ ⎦= =x

e

= 1.60293 or 0.623857

Hence, ln1.6029356x= or ln 0.623857

56x=

i.e. x = 56 ln 1.60293 = 26.42 or x = 56 ln 0.623857 = -26.42, which is not possible.


47

EXERCISE 27 Page 49 3. Expand the following as a power series as far as the term in 5x : (a) sh 3x (b) ch 2x

(a) sh 3x = ( ) ( ) ( )3 53 33 ...

3! 5!x x

x + + + = 3 527 3 3 3 3 336 5 4 3 2 1

x x x× × × ×+ +

× × × ×

= 3 59 8132 40

x x x+ + as far as the term in 5x

(b) ch 2x = ( ) ( )2 42 21 ...

2! 4!x x

+ + + = 2 4161 2 ...24

x x+ + +

= 2 421 23

x x+ + as far as the term in 4x

4. Prove the identity: 3 57 3126 120

sh shθ θ θ θ θ− ≡ + + as far as the term in 5θ only.

L.H.S. = ( ) ( )3 5 3 52 22 2 ... ...

3! 5! 3! 5!sh sh

θ θ θ θθ θ θ θ⎛ ⎞ ⎛ ⎞

− ≡ + + + − + + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= 3 5 3 58 32 1 12 ... ...6 120 6 120

θ θ θ θ θ θ⎛ ⎞ ⎛ ⎞+ + + − + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ( ) 3 58 1 32 126 6 120 120

θ θ θ θ⎛ ⎞ ⎛ ⎞− + − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

as far as the term in 5θ only

= 3 57 316 120

θ θ θ+ + = R.H.S.

5. Prove the identity: 2 3 4 5

2 12 2 8 24 384 1920− ≡ − + − + − +sh chθ θ θ θ θ θθ as far as the term in 5θ only.

L.H.S. = ( ) ( ) ( ) ( )3 5 2 4/ 2 / 2 / 2 / 22 2 ... 1 ...

2 2 2 3! 5! 2! 4!sh ch

θ θ θ θθ θ θ⎛ ⎞ ⎛ ⎞− ≡ + + + − + + +⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 3 5 2 45

2 2 1 1148 2 (120) 8 (16)(24)

⎛ ⎞ ⎛ ⎞+ + − + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠θ θ θ θ θ

= 2 3 4 5

18 24 384 1920θ θ θ θθ− + − + + + as far as the term in 5θ only


48

CHAPTER 6 ARITHMETIC AND GEOMETRIC PROGRESSIONS

EXERCISE 28 Page 52 1. Find the 11th term of the series 8, 14, 20, 26, … The 11th term of the series 8, 14, 20, 26,… is given by: a + (n – 1)d where a = 8, n = 11 and d = 6 Hence, the 11th term is: 8 + (11 – 1)(6) = 8 + 60 = 68 3. The seventh term of a series is 29 and the eleventh term is 54. Determine the sixteenth term. The n’th term of an arithmetic progression is: a + (n – 1)d The 7th term is: a + 6d = 29 (1) The 11th term is: a + 10d = 54 (2)

(2) – (1) gives: 4d = 25 from which, d = 254

Substituting in (1) gives: a + 2564

⎛ ⎞⎜ ⎟⎝ ⎠

= 29

i.e. a + 37.5 = 29 from which, a = 29 – 37.5 = -8.5

Hence, the 16th term is: -8.5 + (16 – 1) 254

⎛ ⎞⎜ ⎟⎝ ⎠

= -8.5 + 93.75 = 85.25

5. Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13.6, … 29 = 7 + (n – 1)(2.2) from which, 29 – 7 = 2.2(n – 1)

i.e. 222.2

= n – 1 i.e. 10 = n – 1 and n = 11 i.e. 29 is the 11th term of the series.

7. Determine the sum of the series 6.5, 8.0, 9.5, 11.0, …, 32


49

In the series: 6.5, 8.0, 9.5, 11.0, …, 32, a = 6.5 and d = 1.5 The n’th term is 32, hence, a + (n – 1)d = 32 i.e. 6.5 + (n - 1)(1.5) = 32

from which, 32 – 6.5 = (n – 1)(1.5) and 25.51.5

= n – 1 i.e. 17 = n - 1 and n = 18

Sum of series, [ ] [ ] [ ]nn 18S 2a (n 1)d 2(6.5) (18 1)(1.5) 9 13 25.52 2

= + − = + − = + = 346.5


50

EXERCISE 29 Page 53

2. Three numbers are in arithmetic progression. Their sum is 9 and their product is 20 14

.

Determine the three numbers. Let the three numbers be (a – d), a and (a + d) Thus, (a – d) + a + ( a + d) = 9 i.e. 3a = 9 and a = 3 Also, a(a – d)(a + d) = 20.25 Since, a = 3, then 3 ( )29 d− = 20.25

i.e. 2 20.259 d 6.753

− = =

and 9 – 6.75 = 2d from which, 2d = 2.25 and d = 2.25 = 1.5 Hence, the three numbers are: (a – d) = 3 – 1.5 = 1.5, a = 3 and (a + d) = 3 + 1.5 = 4.5 4. Find the number of terms of the series 5, 8, 11,… of which the sum is 1025

Sum of n terms is given by: [ ]nnS 2a (n 1)d2

= + −

i.e. 1025 = [ ]n 2(5) (n 1)(3)2

+ −

i.e. [ ]2 1025 n 10 3(n 1)× = + − Hence, 2050 = [ ] [ ] 2n 10 3n 3 n 7 3n 7n 3n+ − = + = + i.e. 23n 7n 2050 0+ − =

This is a quadratic equation, hence n = 27 7 4(3)( 2050) 7 24649 7 157

2(3) 6 6− ± − − − ± − ±

= =

i.e. number of terms, n = 25 (the negative answer having no meaning)


51

6. The first, tenth and last terms of an arithmetic progression are 9, 40.5 and 425.5 respectively. Find (a) the number of terms, (b) the sum of all the terms, and (c) the 70th term. (a) a = 9 and the 10th term is: a + (10 – 1)d = 40.5 i.e. 9 + 9d = 40.5 and 9d = 40.5 – 9 = 31.5

hence 31.5d 3.59

= =

Last term is given by: a + (n – 1)d i.e. 9 + (n – 1)(3.5) = 425.5 i.e. (n – 1)(3.5) = 425.5 – 9 = 416.5

and n – 1 = 416.5 1193.5

=

Hence, the number of terms, n = 120

(b) Sum of all the terms, [ ] [ ] [ ]nn 120S 2a (n 1)d 2(9) (120 1)(3.5) 60 18 416.52 2

= + − = + − = +

= 26070 (c) The 70th term is: a + (n – 1)d = 9 + (70 – 1)(3.5) = 9 + 69(3.5) = 250.5 8. An oil company bores a hole 80 m deep. Estimate the cost of boring if the cost is £30 for drilling the first metre with an increase in cost of £2 per metre for each succeeding metre. The series is: 30, 32, 34, … to 80 terms, i.e. a = 30, d = 2 and n = 80

Thus, total cost, [ ] [ ] [ ]nn 80S 2a (n 1)d 2(30) (80 1)(2) 40 60 158 40(218)2 2

= + − = + − = + = = £8720


52

EXERCISE 30 Page 55 1. Find the 10th term of the series 5, 10, 20, 40, … The 10th term of the series 5, 10, 20, 40, … is given by: n 1a r − where a = 5, r = 2 and n = 10 i.e. the 10th term = n 1a r − = ( )( )10 1 95 2 (5)(2) 5(512)− = = = 2560 3. The first term of a geometric progression is 4 and the 6th term is 128. Determine the 8th and 11th terms.

The 6th term is given by: 5a r 128= i.e. 54r 128= and 5 128r 324

= =

Thus, r = 5 32 = 2 Hence, the 8th term is: n 1a r − = 8 1 74(2) 4(2) 4(128)− = = = 512 and the 11th term is: 11 1 104(2) 4(2) 4(1024)− = = = 4096

4. Find the sum of the first 7 terms of the series 2, 5, 12 12

,… (correct to 4 significant figures).

Common ratio, r = ar 5a 2= = 2.5 (Also,

2ar 12.5ar 5

= = 2.5)

Sum of 7 terms, ( )( )

( )( )

( )n 7

n

a r 1 2 2.5 1 2 610.35 1S

r 1 2.5 1 1.5− − −

= = =− −

= 812.5, correct to 4 significant

figures.

6. Find the sum to infinity of the series 1 1 52 , 1 , ,.....2 4 8

−

The series is a G.P. where r = 1.252.5

− = -0.5 and a = 2.5

Hence, sum to infinity, a 2.5 2.5 5S1 r 1 ( 0.5) 1.5 3∞ = = = =− − −

= 1 23


53

EXERCISE 31 Page 57 1. In a geometric progression the 5th term is 9 times the 3rd term, and the sum of the 6th and 7th terms is 1944. Determine (a) the common ratio, (b) the first term, and (c) the sum of the 4th to 10th terms inclusive. (a) The 5th term of a geometric progression is: 4ar and the 3rd term is: 2ar

Hence, 4ar = 9 2ar from which, 4

2 9rr

= i.e. 2 9r =

from which, the common ratio, r = 3 (b) The 6th term is 5ar and the 7th term is 6ar Hence, 5ar + 6ar = 1944 Since r = 3, 243a + 729a = 1944

i.e. 972a = 1944 and first term, a = 1944972

= 2

(c) Sum of the 4th to 10th terms inclusive is given by:

( )( )

( )( )

( ) ( )( )

10 3 10 3

10 3

1 1 2 3 1 2 3 11 1 (3 1) 3 1

a r a rS S

r r− − − −

− = − = −− − − −

= ( ) ( )10 3 10 33 1 3 1 3 3 59049 27− − − = − = − = 59022 4. If the population of Great Britain is 55 million and is decreasing at 2.4% per annum, what will be the population in 5 years time? G.B. population now = 55 million, population after 1 year = 0.976 × 55 million, population after 2 years = ( )20.976 55× million Hence, population after 5 years = ( )50.976 55× = 48.71 million 6. If £250 is invested at compound interest of 6% per annum, determine (a) the value after 15 years, (b) the time, correct to the nearest year, it takes to reach £750.


54

(a) First term, a = £250, common ratio, r = 1.06

Hence, value after 15 years = ( )1515 (250) 1.06ar = = £599.14

(b) When £750 is reached, 750 = nar

i.e. 750 = 250 ( )1.06 n

and 750250

= 1.06n i.e. 3 = 1.06n

Taking logarithms gives: lg 3 = ( )lg 1.06 lg1.06n n=

from which, n = lg3lg1.06

= 18.85

Hence, it will take 19 years to reach more than £750 7. A drilling machine is to have 8 speeds ranging from 100 rev/min to 1000 rev/min. If the speeds form a geometric progression determine their values, each correct to the nearest whole number. First term, a = 100 rev/min

The 8th term is given by: 8 1 1000ar − = from which, 7 1000 10100

r = = and 7 10r = = 1.3895

Hence, 1st term is 100 rev/min 2nd term is ar = (100)(1.3895) = 138.95 3rd term is 2 2(100)(1.3895) 193.07ar = = 4th term is 3 3(100)(1.3895) 268.27ar = = 5th term is 4 4(100)(1.3895) 372.76ar = = 6th term is 5 5(100)(1.3895) 517.96ar = = 7th term is 6 6(100)(1.3895) 719.70ar = = 8th term is 7 7(100)(1.3895) 1000ar = = Hence, correct to the nearest whole numbers, the eight speeds are: 100, 139, 193, 268, 373, 518, 720 and 1000 rev/min


55

CHAPTER 7 THE BINOMIAL SERIES

EXERCISE 32 Page 59 1. Expand ( )52 3+a b using Pascal’s triangle From page 58 of textbook, ( )5 5 4 3 2 2 3 4 5a x a 5a x 10a x 10a x 5a x x+ = + + + + + Thus, replacing a with 2a and x with 3b gives: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )5 5 4 3 2 2 3 4 52 3 2 5 2 3 10 2 3 10 2 3 5 2 3 3+ = + + + + +a b a a b a b a b a b b = + + + + +5 4 3 2 2 3 4 532a 240a b 720a b 1080a b 810ab 243b


56

EXERCISE 33 Page 61 1. Use the binomial theorem to expand ( )42+a x

( )42+a x = ( ) ( )( ) ( ) ( )( )( ) ( ) ( )2 3 44 3 24 3 4 3 2a 4a 2x a 2x a 2x 2x

2! 3!+ + + +

= 4 3 2 2 3 4a 8a x 24a x 32a x 16x+ + + + 3. Expand ( )42 3−x y

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( ) ( )4 4 3 2 2 3 44 3 4 3 22 3 2 4 2 3 2 3 2 3 3

2! 3!− = + − + − + − + −x y x x y x y x y y

= 4 3 2 2 3 416 96 216 216 81− + − +x x y x y xy y

4. Determine the expansion of 522⎛ ⎞+⎜ ⎟

⎝ ⎠x

x

( ) ( ) ( )( ) ( ) ( )( )( ) ( )

( )( )( )( ) ( )

5 2 35 4 3 2

4 5

5 4 5 4 32 2 2 22 2 5 2 2 22! 3!

5 4 3 2 2 224!

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x x x x xx x x x

xx x

= 5 33 5

320 160 3232 160 320+ + + + +x x xx x x

6. Determine the sixth term of 13

33

⎛ ⎞+⎜ ⎟⎝ ⎠

qp

The 6th term of ( )+ na x is ( )( )( )

( 1) 11 2 ..... ( 1)1 !

− − −− − −−

n r rn n n to r termsa x

r

Hence, the 6th term of 13

33

⎛ ⎞+⎜ ⎟⎝ ⎠

qp is: ( )( )( )( )( ) ( )5

13 513 12 11 10 93

5! 3− ⎛ ⎞⎜ ⎟⎝ ⎠

qp = ( )5

81287 33

⎛ ⎞⎜ ⎟⎝ ⎠

qp

= 8 534749 p q


57

9. Use the binomial theorem to determine, correct to 5 significant figures: (a) ( )70.98 (b) ( )92.01 (a)

( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( ) ( )( )( )( ) ( )7 7 2 3 47 6 7 6 5 7 6 5 40.98 1 0.02 1 7 0.02 0.02 0.02 0.02 ..

2! 3! 4!= − = + − + − + − + − +

= 1 – 0.14 + 0.0084 – 0.00028 + 0.0000056 - … = 0.86813, correct to 5 significant figures.

(b) ( ) ( ) ( )9

9 9 99 90.012.01 2 0.01 2 1 2 1 0.0052

⎛ ⎞= + = + = +⎜ ⎟⎝ ⎠

= ( )( ) ( )( )( )9 2 39 8 9 8 72 1 9(0.005) (0.005) (0.005) ...

2! 3!⎡ ⎤+ + + +⎢ ⎥

⎣ ⎦

= ( )9 92 1 0.045 0.0009 0.0000105 ... 2 (1.0459105)+ + + + = = 535.51, correct to 5 significant figures.


58

EXERCISE 34 Page 63

2. Expand ( )2

11+ x

in ascending powers of x as far as the term in 3x , using the binomial theorem.

Sate the limits of x for which the series is valid.

( )( ) ( )( ) ( ) ( )( )( ) ( )2 2 3

2

2 3 2 3 41 1 1 2 ...2! 3!1

− − − − − −= + = − + + +

+x x x x

x

= 2 21 2 3 4 ...− + − +x x x and x < 1

3. Expand ( )3

12+ x



( )3

12+ x

= ( ) ( )( ) ( )( )( )3 2 33 3 3 3 4 3 4 5

2 2 1 2 1 3 ...2 2 2! 2 3! 2

−− − −

⎡ ⎤− − − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + = − + + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

x x x xx

= 2 33

1 3 3 51 ...2 2 2 4

⎡ ⎤− + − +⎢ ⎥⎣ ⎦x x x

= 2 31 3 3 51 ...8 2 2 4⎡ ⎤− + − +⎢ ⎥⎣ ⎦

x x x

The series is true provided 2x < 1 i.e. x < 2

5. Expand 11 3+ x



11 3+ x

= ( ) ( ) ( )1 2 32

1 3 1 3 51 2 2 2 2 21 3 1 (3 ) 3 3 ...2 2! 3!

−

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞− − − − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠+ = + − + +⎜ ⎟⎝ ⎠

x x x x

= 2 33 27 13512 8 16

− + −x x x as far as the term in 3x

The series is true provided 3x < 1 i.e. x < 13


59

7. When x is very small show that : (a) ( ) ( )2

1 5121 1

≈ +− −

xx x

(b) ( )( )4

1 21 10

1 3−

≈ +−

xx

x (c)

( )3

1 5 19161 2

+≈ +

−

x xx

(a) ( ) ( )

( ) ( ) ( )122

2

1 1 1 1 2 121 1

− − ⎛ ⎞= − − ≈ + +⎜ ⎟⎝ ⎠− −

xx x xx x

≈ 1 22

+ +x x ignoring the 2x term and above

≈ 512

+ x

(b) ( )( )

( )( ) ( )( )44

1 21 2 1 3 1 2 1 12

1 3−−

= − − ≈ − +−

xx x x x

x

≈ 1 + 12x – 2x ignoring the 2x term and above ≈ 1 + 10x

(c) ( )

( ) ( )1 12 3

3

1 5 5 21 5 1 2 1 12 31 2

−+ ⎛ ⎞⎛ ⎞= + − ≈ + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠−

x x x x xx

≈ 2 513 2

+ +x x ignoring the 2x term and above

≈ 1916

+ x 5 2 15 4 192 3 6 6

+⎛ ⎞+ = =⎜ ⎟⎝ ⎠

9. Express the following as power series in ascending powers of x as far as the term in 2x . State in each case the range of x for which the series is valid.

(a) 11−⎛ ⎞

⎜ ⎟+⎝ ⎠xx

(b) ( ) ( )

( )

23

2

1 1 3

1

− −

+

x x

x

(a) ( ) ( ) ( )1 1 22 2

1 1 1 31 2 2 2 21 1 1 11 2 2! 2 2!

−

⎡ ⎤ ⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞− − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎢ ⎥−⎛ ⎞ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥ ⎢ ⎥= − + ≈ − + − − +⎜ ⎟+⎝ ⎠ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x x xx x xx

as far as the term

in 2x


60

= 2 2 2 2 23 31 1 1

2 8 2 8 2 8 2 4 8⎛ ⎞⎛ ⎞− − − + ≈ − + − + −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

x x x x x x x x x

≈ 2

12

− +xx as far as the term in 2x

The series is valid if x < 1

(b) ( ) ( )

( )( )( ) ( )

23 122 23

2

1 1 31 1 3 1

1

−− −= + − +

+

x xx x x

x

≈ ( ) ( )2

2

2 13 31 1 2 3 .. 1 ..

2! 2

⎡ ⎤⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎢ ⎥ ⎛ ⎞⎝ ⎠⎝ ⎠⎢ ⎥+ − + − + − +⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦

xx x x as far as the term in 2x

≈ ( )( )2

21 1 2 12

⎛ ⎞+ − − −⎜ ⎟

⎝ ⎠

xx x x

≈ ( )2

2 21 2 2 12

⎛ ⎞− − + − −⎜ ⎟

⎝ ⎠

xx x x x as far as the term in 2x

≈ ( )2 2

2 21 3 1 1 32 2

⎛ ⎞− − − ≈ − − −⎜ ⎟

⎝ ⎠

x xx x x x neglecting 3x terms and above

≈ 2712

− −x x

The series is valid provided 3x < 1 i.e. x < 13


61

EXERCISE 35 Page 65

2. Kinetic energy is given by 212

mv . Determine the approximate change in the kinetic energy

when mass m is increased by 2.5% and the velocity v is reduced by 3% New mass = 1.025 m = (1 + 0.025)m and new velocity = 0.97 v = (1 – 0.03)v

New kinetic energy = 2 2 21 1(1 0.025) (1 0.03) (1 0.025)(1 0.06)2 2

+ − ≈ + −m v mv

≈ 2 21 1(1 0.025 0.06) (0.965)2 2

+ − =mv mv

i.e. 96.5% of the original kinetic energy. Thus, the approximate change in kinetic energy is a reduction of 3.5% 3. An error of +1.5% was made when measuring the radius of a sphere. Ignoring the products of small quantities determine the approximate error in calculating (a) the volume, and (b) the surface area.

(a) Volume of sphere, V = 343

rπ .

New volume = 3 3 3 3 34 4 4 4(1.015 ) (1 0.015) [1 3(0.015)] (1 0.045)3 3 3 3

= + ≈ + = +r r r rπ π π π

= 1.045 V i.e. the volume has increased by 4.5% (b) Surface area of sphere, A = 24 rπ New surface area = ( )2 2 2 24 1 0.015 4 [1 2(0.015)] 4 (1 0.03)+ ≈ + = +r r rπ π π = 1.03 A i.e. the surface area has increased by 3%


62

6. The electric field strength H due to a magnet of length 2 l and moment M at a point on its axis

distance x from the centre is given by: ( ) ( )2 2

1 12

⎧ ⎫⎪ ⎪= −⎨ ⎬− +⎪ ⎪⎩ ⎭

MHl x l x l

Show that if l is very small compared with x, then H ≈ 3

2Mx

( ) ( )

2 2

2 2 2 2 22 2

1 1 1 1 1 12 2 2

1 1

M M M l lHl l x l x xx l x l l lx x

x x

− −

⎧ ⎫⎪ ⎪⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎛ ⎞ ⎛ ⎞= − = − = − − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠− + ⎛ ⎞ ⎛ ⎞ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎩ ⎭⎩ ⎭ − +⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

2 2

2 2 41 12 2M l l M lx l x x x l x

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎧ ⎫≈ + − − ≈⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎩ ⎭⎩ ⎭

≈ 3

2Mx

7. The shear stress τ in a shaft of diameter D under a torque T is given by: τ = 3

k TDπ

. Determine

the approximate percentage error in calculating τ if T is measured 3% too small and D is 1.5% too large. New value of T = (1 – 0.03)T and new value of D = (1 + 0.015)D

Hence, new shear stress = ( )( ) 33 3 3

(1 0.03) 1 0.03 1 0.015(1 0.015)k T kT

D Dπ π−− ⎡ ⎤= − +⎣ ⎦+

( )( ) [ ]3 31 0.03 1 0.045 1 0.03 0.045kT kTD Dπ π

≈ − − ≈ − −⎡ ⎤⎣ ⎦

( )3 1 0.075 (1 0.075)kTD

τπ

≈ − = −

i.e. the new torque has decreased by 7.5% 9. In a series electrical circuit containing inductance L and capacitance C the resonant frequency is

given by: 12

=rf LCπ. If the values of L and C used in the calculation are 2.6% too large and

0.8% too small respectively, determine the approximate percentage error in the frequency.


63

New value of inductance = (1 + 0.026)L and new value of capacitance = (1 – 0.008)C

Hence, new resonant frequency = 1 12 (1 0.026) (1 0.008) 2 (1 0.026) (1 0.008)L C L Cπ π

=+ − + −

= ( )1 1 112 2 22

1 1 0.026 (1 0.008)2

L Cπ

− − −−+ −

( )( ) ( )1 12 2

1 11 0.013 1 0.004 1 0.013 0.00422 LCL C ππ

≈ − + ≈ − +⎡ ⎤⎣ ⎦

( )1 0.009rf≈ − i.e. the new resonant frequency is 0.9% smaller

10. The viscosity η of a liquid is given by: η =4k rlν

, where k is a constant. If there is an error in r

of +2%, in ν of +4% and l of -3%, what is the resultant error in η? New value of r = (1 + 0.02)r, new value of ν = (1 + 0.04) ν and new value of l = (1 – 0.03) l

Hence, new value of viscosity = ( ) ( ) ( )4 4 4

4 1 1(1 0.02) 1 0.02 1 0.04 1 0.03(1 0.04) (1 0.03)

k r k rl lν ν

− −+ ⎡ ⎤= + + −⎣ ⎦+ −

( )( )( )4

1 0.08 1 0.04 1 0.03k rlν

≈ + − +⎡ ⎤⎣ ⎦

( ) ( )4

1 0.08 0.04 0.03 1 0.07≈ + − + ≈ +k r

lη

ν

i.e. the viscosity increases by 7%

12. The flow of water through a pipe is given by: G = ( )53d HL

. If d decreases by 2% and H by

1%, use the binomial theorem to estimate the decrease in G. New value of d = (1 - 0.02)d and new value of H = (1 – 0.01)H


64

Hence, new value of G = ( )5 1 5 5 1 1

5 5 52 2 2 2 2 2

1 12 2

3 3 3 (1 0.02) (1 0.01)− −= =

d H d H d HL L L

5 15 2 2

12

3 5 11 (0.02) 1 (0.01)2 2

⎡ ⎤⎛ ⎞⎛ ⎞≈ − −⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

d H

L

( )( ) ( )5 5(3 ) (3 )1 0.05 1 0.005 1 0.05 0.005≈ − − ≈ − −⎡ ⎤⎣ ⎦

d H d HL L

(1 0.055)G≈ − i.e. the flow G has decreased by 5.5%


65

CHAPTER 8 MACLAURIN’S SERIES

EXERCISE 36 Page 70 1. Determine the first four terms of the power series for sin 2x using Maclaurin’s series. Let f(x) = sin 2x f(0) = sin 0 = 0 f ′(x) = 2 cos 2x f ′(0) = 2 cos 0 = 2 f ′′(x) = -4 sin 2x f ′′(0) = -4 sin 0 = 0 f ′′′(x) = -8 cos 2x f ′′′(0) = -8 cos 0 = - 8 ( )ivf x = 16 sin 2x (0)ivf = 16 sin 0 = 0 ( )vf x = 32 cos 2x (0)vf = 32 cos 0 = 32 ( )vif x = -64 sin 2x (0)vif = -64 sin 0 = 0 ( )viif x = -128 cos 2x (0)viif = -128 cos 0 = - 128

Maclaurin’s series states: f(x) = f(0) + x f ′(0) + 2

2!x f ′′(0) +

3

3!x f ′′′(0) + ….

=2 3 4 5 6 7

0 (2) (0) ( 8) (0) (32) (0) ( 128)2! 3! 4! 5! 6! 7!x x x x x xx+ + + − + + + + −

i.e. f(x) = 3 5 78 32 1282

6 120 5040x x xx − + −

i.e. sin 2x = 3 5 74 4 823 15 315

x x x x− + −

3. Use Maclaurin’s series to determine the first three terms of the power series for ( )ln 1 xe+ Let f(x) = ( )ln 1 xe+ f(0) = ( )0ln 1 e+ = ln 2

f ′(x) = 1

x

x

ee+

f ′(0) = 0

0

11 2

ee

=+


66

f ′′(x) = ( ) ( )

( )2

1

1

x x x x

x

e e e e

e

+ −

+ f ′′(0) =

( ) ( )( )

0 0 0 0

2 20

1 2 1 12 41

e e e e

e

+ − −= =

+


2!x f ′′(0) +

3

3!x f ′′′(0) + ….

=21 1ln 2 ...

2 2! 4xx ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

i.e. ( )ln 1 xe+ = 2

ln 22 8x x

+ +

6. Develop, as far as the term in 4x , the power series for sec 2x. Let f(x) = sec 2x f(0) = sec 0 = 1 f ′(x) = 2 sec 2x tan 2x f ′(0) = 0 f ′′(x) = ( )( ) ( )( )22sec 2 2sec 2 tan 2 4sec 2 tan 2x x x x x+ = 2 2 2 24sec 2 sec 2 tan 2 4sec 2 sec 2 sec 2 1⎡ ⎤ ⎡ ⎤+ = + −⎣ ⎦ ⎣ ⎦x x x x x x = 2 34sec 2 2sec 2 1 8sec 2 4sec 2x x x x⎡ ⎤− = −⎣ ⎦ f ′′(0) = 8 – 4 = 4 f ′′′(x) = ( )224sec 2 2sec 2 tan 2 8sec 2 tan 2x x x x x− = 348sec 2 tan 2 8sec 2 tan 2x x x x− f ′′′(0) = 0 ( )ivf x = ( )( ) ( )( )3 2 248sec 2 2sec 2 tan 2 144sec 2 (2sec 2 tan 2 )x x x x x x⎡ ⎤+⎣ ⎦

( )( ) ( )( )28sec 2 2sec 2 tan 2 16sec 2 tan 2x x x x x⎡ ⎤− +⎣ ⎦

(0)ivf = 96 + 0 – 16 – 0 = 80


2!x f ′′(0) +

3

3!x f ′′′(0) + ….

= 2 3 4

1 (0) (4) (0) (80)2! 3! 4!x x xx+ + + +

= 2 4801 224

x x+ +

i.e. sec 2x = 2 4101 23

x x+ + as far as the term in 4x


67

7. Expand 2 cos3e θ θ as far as the term in 2θ using Maclaurin’s series. Let f(θ) = 2 cos3e θ θ f(0) = 0 cos 0e = 1 f ′(θ) = ( )( ) ( )( )2 23sin 3 cos3 2e eθ θθ θ− +

= ( )2 2cos3 3sin 3e θ θ θ− f ′(0) = ( )0 2cos 0 3sin 0e − = 2 f ′′(θ) = ( )( ) ( )( )2 26sin 3 9cos3 2cos3 3sin 3 2e eθ θθ θ θ θ− − + − f ′′(0) = - 9 + 4 = -5

Maclaurin’s series states: f(θ) = f(0) + θ f ′(0) + 2

2!θ f ′′(0) +

3

3!θ f ′′′(0) + ….

= 2

1 (2) ( 5)2!θθ+ + −

i.e. 2 25cos 3 1 22

e θ θ θ θ= + − as far as the term in 2θ

8. Determine the first three terms of the series for 2sin x by applying Maclaurin’s theorem. Let f(x) = 2sin x f(0) = sin 0 = 0 f ′(x) = 2 sin x cos x f ′(0) = 2 sin 0 cos 0 = 0 f ′′(x) = (2 sin x)(-sin x) + (cos x)(2 cos x) = ( )2 2 2 22sin 2cos 2 cos sin 2cos 2x x x x x− + = − = f ′′(0) = 2 cos 0 = 2 f ′′′(x) = -4 sin 2x f ′′′(0) = -4 sin 0 = 0 ( )ivf x = -8 cos 2x (0)ivf = -8 cos 0 = -8 ( )vf x = 16 sin 2x (0)vf = 16 sin 0 = 0 ( )vif x = 32 cos 2x (0)vif = 32 cos 0 = 32


2!x f ′′(0) +

3

3!x f ′′′(0) + ….

= 2 3 4 5 6

0 (0) (2) (0) ( 8) (0) (32) ...2! 3! 4! 5! 6!x x x x xx+ + + + − + + +

i.e. 2 2 4 61 2sin3 45

x x x x= − + to three terms


68

EXERCISE 37 Page 72

1. Evaluate 0.6 sin

0.23e dθ θ∫ , correct to 3 decimal places, using Maclaurin’s series.

Let f(θ) = sin3e θ f(0) = sin 03e = 3 f ′(θ) = ( )( )sin3 cose θ θ f ′(0) = ( )( )sin 03 cos 0e = 3 f ′′(θ) = ( )( ) ( )( )sin sin3 sin cos 3 cose eθ θθ θ θ− + = ( )sin 23 cos sine θ θ θ− f ′′(0) = ( )sin 0 23 cos 0 sin 0e − = 3 f ′′′(θ) = ( )( ) ( )( )sin 2 sin3 2cos sin cos cos sin 3 cose eθ θθ θ θ θ θ θ− − + − f ′′′(0) = ( )( ) ( )( )sin 0 2 sin 03 2cos 0sin 0 cos 0 cos 0 sin 0 3 cos 0e e− − + − = (3)(-1) + (1)(3) = 0

Maclaurin’s series states: f(θ) = f(0) + θ f ′(0) + 2

2!θ f ′′(0) +

3

3!θ f ′′′(0) + ….

= 233 3 02

θ θ+ + +

Hence, 0.6 sin

0.23e dθ θ∫ =

0.630.6 2 2

0.20.2

3 33 3 32 2 2

d θθ θ θ θ θ⎡ ⎤⎛ ⎞+ + = + +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

∫

= ( ) ( ) ( ) ( )3 32 20.6 0.23 33(0.6) 0.6 3(0.2) 0.2

2 2 2 2

⎡ ⎤ ⎡ ⎤+ + − + +⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= (1.8 + 0.54 + 0.108) – (0.6 + 0.06 + 0.004) = 2.448 – 0.664 = 1.784, correct to 3 decimal places 4. Use Maclaurin’s theorem to expand ln( 1)x x + as a power series. Hence evaluate, correct to 3

decimal places, 0.5

0ln(1 )x x dx+∫

From page XX, 2 3 4

ln( 1) ....2 3 4x x xx x+ = − + − +

Hence, 1 2 3 4 50.5 0.52

0 0ln(1 ) ...

2 3 4 5x x x xx x dx x x dx

⎛ ⎞+ = − + − + +⎜ ⎟

⎝ ⎠∫ ∫


69

5 7 9 11 133 2 2 2 2 20.52

0...

2 3 4 5 6x x x x xx

⎛ ⎞⎜ ⎟= − + − + −⎜ ⎟⎜ ⎟⎝ ⎠

∫

=

0.55 7 9 11 13 152 2 2 2 2 2

0

...5 7 9 11 13 152 3 4 5 62 2 2 2 2 2

x x x x x x⎡ ⎤⎢ ⎥⎢ ⎥− + − + − +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= ( ) ( ) ( ) [ ]11 135 7 92 22 2 2

2 1 2 1 20.5 0.5 0.5 (0.5) (0.5) .. 05 7 27 22 65⎡ ⎤

− + − + + −⎢ ⎥⎣ ⎦

= 0.07071 – 0.0126 + 0.00327 – 0.00100 + 0.00034 - … = 0.06072… = 0.061, correct to 3 decimal places.


70

EXERCISE 38 Page 74

1. Determine 3

31

2 1lim2 3 5x

x xx x→

⎧ ⎫− +⎨ ⎬+ −⎩ ⎭

3

31

2 1lim2 3 5x

x xx x→

⎧ ⎫− +⎨ ⎬+ −⎩ ⎭

= 2

21

3 2 3 2lim6 3 6 3x

xx→

⎧ ⎫− −=⎨ ⎬+ +⎩ ⎭

= 19

4. Determine 2

20

sin 3lim3x

x xx x→

⎧ ⎫−⎨ ⎬+⎩ ⎭

2

20

sin 3lim3x

x xx x→

⎧ ⎫−⎨ ⎬+⎩ ⎭

= 0

2 3cos3 3lim3 2 3x

x xx→

− −⎧ ⎫ =⎨ ⎬+⎩ ⎭ = -1

5. Determine 30

sin coslimθ

θ θ θθ→

−⎧ ⎫⎨ ⎬⎩ ⎭

30

sin coslimθ

θ θ θθ→

−⎧ ⎫⎨ ⎬⎩ ⎭

= ( )( )

2 20 0

cos sin cos sinlim lim3 3θ θ

θ θ θ θ θ θθ θ→ →

⎧ ⎫− − +⎡ ⎤⎪ ⎪ ⎧ ⎫⎣ ⎦ =⎨ ⎬ ⎨ ⎬⎩ ⎭⎪ ⎪⎩ ⎭

= 0 0

cos sin ( sin ) cos (1) cos 1 1 2lim lim6 6 6 6θ θ

θ θ θ θ θ θ θθ→ →

+ − + + +⎧ ⎫ ⎧ ⎫= = =⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭

= 13

7. Determine 30

sinh sinlimx

x xx→

−⎧ ⎫⎨ ⎬⎩ ⎭

30

sinh sinlimx

x xx→

−⎧ ⎫⎨ ⎬⎩ ⎭

= 20 0 0

cosh cos sinh sin cosh coslim lim lim3 6 6x x x

x x x x x xx x→ → →

− + +⎧ ⎫ ⎧ ⎫ ⎧ ⎫= =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭ ⎩ ⎭

= 1 1 26 6+

= = 13

8. Determine 2

sin 1limln sinπθ

θθ→

−⎧ ⎫⎨ ⎬⎩ ⎭

2

sin 1limln sinπθ

θθ→

−⎧ ⎫⎨ ⎬⎩ ⎭

= 2 2

coslim lim sin sin1 2cos

sinπ πθ θ

θ πθθ

θ→ →

⎧ ⎫⎪ ⎪⎪ ⎪ = =⎨ ⎬⎛ ⎞⎪ ⎪⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

= 1


71

CHAPTER 9 SOLVING EQUATIONS BY ITERATIVE METHODS

EXERCISE 39 Page 80 1. Find the positive root of the equation 2 3 5 0x x+ − = , correct to 3 significant figures, using the

method of bisection, Let f(x) = 2 3 5x x+ − then, using functional notation:

f(0) = - 5

f(1) = 1 + 3 - 5 = - 1

f(2) = 4 + 6 - 5 = + 5

Since there is a change of sign from negative to positive there must be a root of the equation between

x = 1 and x = 2.

The method of bisection suggests that the root is at 1 22+ = 1.5, i.e. the interval between 1 and 2 has been

bisected.

Hence f(1.5) = (1.5) 2 + 3(1.5) - 5 = 1.75

Since f(1) is negative, f(1.5) is positive, and f(2) is also positive, a root of the equation must lie between

x = 1 and x = 1.5, since a sign change has occurred between f(1) and f(1.5)

Bisecting this interval gives 1 1.52+ i.e. 1.25 as the next root.

Hence f(1.25) = (1.25) 2 + 3(1.25) - 5 = 0.3125

Since f(1) is negative and f(1.25) is positive, a root lies between x = 1 and x = 1.25

Bisecting this interval gives 1 1.252

+ i.e. 1.125

Hence f(1.125) = (1.125) 2 + 3(1.125) - 5 = - 0.359375

Since f(1.125) is negative and f(1.25) is positive, a root lies between x = 1.125 and x = 1.25

Bisecting this interval gives 1.125 1.252+ i.e. 1.1875


72

Hence f(1.1875) = (1.1875) 2 + 3(1.1875) - 5 = - 0.02734375

Since f(1.1875) is negative and f(1.25) is positive, a root lies between x = 1.1875 and x = 1.25

Bisecting this interval gives 1.1875 1.252+ i.e. 1.21875

Hence f(1.21875) = (1.21875) 2 + 3(1.21875) - 5 = 0.1416016

Since f(1.21875) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.21875

Bisecting this interval gives 1.1875 1.218752+ = 1.203125

Hence f(1.203125) = 0.056885

Since f(1.203125) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.203125


Hence f(1.1953125) = 0.0147095

Since f(1.1953125) is positive and f(1.1875) is negative, a root lies between x = 1.1953125 and

x = 1.1875


Hence, f(1.191406) = - 0.006334

Since f(1.191406) is negative and f(1.1953125) is positive, a root lies between x = 1.191406 and

x = 1.1953125


The last two values obtained for the root are 1.1914… and 1.1934….

The last two values are both 1.19, correct to 3 significant figures. We therefore stop the iterations here.

Thus, correct to 3 significant figures, the positive root of x 2 + 3x - 5 = 0 is 1.19

2. Using the bisection method solve 2xe x− = , correct to 4 significant figures


73

Let f(x) = 2xe x− − then f(0) = 1 – 0 – 2 = -1 f(1) = 1 1 2 0.28e − − = − f(2) = 2 2 2 3.38e − − = Hence, a root lies between x = 1 and x = 2. Let the root be x = 1.5 f(1.5) = 1.5 1.5 2 0.98169e − − = Hence, a root lies between x = 1 and x = 1.5 due to a sign change.

Bisecting this interval gives 1 1.52+ = 1.25

f(1.25) = 1.25 1.25 2 0.240343e − − = Hence, a root lies between x = 1 and x = 1.25 due to a sign change.

Bisecting this interval gives 1 1.252

+ = 1.125

f(1.125) = 1.125 1.125 2 0.04478e − − = − Hence, a root lies between x = 1.125 and x = 1.25 due to a sign change.


f(1.1875) = 0.091374 Hence, a root lies between x = 1.1875 and x = 1.125 due to a sign change.




f(1.140625) = -0.011902 Hence, a root lies between x = 1.140625 and x = 1.15625 due to a sign change.


74




f(1.14453125) = -0.0035626 Hence, a root lies between x = 1.14453125 and x = 1.1484375 due to a sign change.



Bisecting this interval gives 1.14648 1.144531252

+ = 1.14551

The last two values are both 1.146, correct to 4 significant figures. We therefore stop the iterations here.

Thus, correct to 4 significant figures, the positive root of 2xe x− = is 1.146

4. Solve x – 2 – ln x = 0 for the root nearest to 3, correct to 3 decimal places using the bisection

method Let f(x) = x – 2 – ln x f(2) = 2 – 2 – ln 2 = - 0.693 f(3) = 3 – 2 – ln 3 = - 0.0986 f(4) = 4 – 2 – ln 4 = 0.61371 Hence, the root lies between x = 3 and x = 4 because of the sign change. Table 1 shows the results in tabular form.


75

Table 1 1x 2x 1 2

3 2x xx +⎛ ⎞= ⎜ ⎟

⎝ ⎠ ( )3f x

3 3 3 3.125 3.125 3.15625 3.140625 3.14844 3.1445325 3.146486

4 3.5 3.25 3.25 3.1875 3.125 3.15625 3.140625 3.14844 3.1445325

3.5 3.25 3.125 3.1875 3.15625 3.140625 3.14844 3.145325 3.146486 3.14550925

0.24724 0.071345 -0.014434 0.028263 0.0068654 -0.003797 0.0015312 -0.0011327 0.0001999

Correct to 3 decimal places, the solution of x – 2 – ln x = 0 is 3.146


76

EXERCISE 40 Page 83 2. Use an algebraic method of successive approximation to solve 3 2 14 0x x− + = , correct to 3

decimal places. Let f(x) = 3 2 14x x− +

f(0) = 14

f(1) = 1 – 2 + 14 = 13

f(2) = 8 – 4 + 14 = 18 (There are no positive values of x)

f(-1) = -1 + 2 + 14 = 15

f(-2) = -8 + 4 + 14 = 10

f(-3) = -27 + 6 + 14 = -7 Hence a root lies between x = -2 and x = -3 First approximation Let the first approximation be -2.6 Second approximation Let the true value of the root, x2 , be (x1 + δ1) Let f(x1 + δ1) = 0, then since x1 = -2.6, (-2.6 + δ1)3 - 2(-2.6 + δ1) + 14 = 0 Neglecting terms containing products of δ1 and using the binomial series gives: [(-2.6)3 + 3(-2.6)2 δ1] + 5.2 - 2δ1 + 14 ≈ 0 -17.576 + 20.28δ1 + 5.2 - 2 δ1 + 14 ≈ 0 18.28δ1 ≈ 17.576 – 5.2 - 14

δ1 ≈ 17.576 5.2 14 0.0888418.28− −

≈ −

Thus x2 ≈ -2.6 - 0.08884 = -2.6888 Third approximation Let the true value of the root, x3 , be (x2 + δ2) Let f(x2 + δ2) = 0, then since x2 = -2.6888,


77

(-2.6888 + δ2)3 - 2(-2.6888 + δ2) + 14 = 0 Neglecting terms containing products of δ2 gives: ( ) ( )3 2

2 22.6888 3 2.6888 5.3776 2 14 0− + − δ + − δ + ≈ -19.439 + 21.6888δ2 + 5.3776 - 2δ2 + 14 ≈ 0 19.6888δ2 ≈ 19.439 – 5.3776 - 14

δ2 ≈ 19.439 5.3776 1419.6888− − ≈ 0.003119

Thus x3 ≈ (x2 + δ2) = -2.6888 + 0.003119 ≈ -2.6857 Fourth approximation Let the true value of the root, x4 , be (x3 + δ3) Let f(x3 + δ3) = 0, then since x3 = -2.6857, (-2.6857 + δ3)3 - 2(-2.6857 + δ3) + 14 = 0 Neglecting terms containing products of δ3 gives: ( ) ( )3 2

3 32.6857 3 2.6857 5.3714 2 14 0− + − δ + − δ + ≈ -19.3719 + 21.63895δ3 + 5.3714 - 2δ3 + 14 ≈ 0 19.63895δ2 ≈ 19.3719 – 5.3714 - 14

δ2 ≈ 19.3719 5.3714 1419.63895− − ≈ 0.00002546

Thus x4 ≈ (x3 + δ3) = -2.6857 + 0.000025 ≈ -2.6857 Since x3 and x4 are the same when expressed to the required degree of accuracy, then the required root is -2.686, correct to 3 decimal places. 3. Use an algebraic method of successive approximation to solve 4 33 7 5.5 0x x x− + − = , correct to

3 significant figures. Let f(x) = 4 33 7 5.5x x x− + −

f(0) = -5.5


78

f(1) = 1 – 3 + 7 – 5.5 = -0.5

f(2) = 16 – 24 + 14 – 5.5 = 0.5

Hence a root lies between x = 1 and x = 2 First approximation Let the first approximation be 1.5 Second approximation Let the true value of the root, x2 , be (x1 + δ1)

Let f(x1 + δ1) = 0, then since x1 = 1.5,

(1.5 + δ1) 4 - 3(1.5 + δ1) 3 + 7(1.5 + δ1) - 5.5 = 0

Neglecting terms containing products of δ1 and using the binomial series gives:

[(1.5) 4 + 4(1.5) 3 δ1] – 3[(1.5) 3 + 3(1.5) 2 δ1] + 10.5 + 7δ1 - 5.5 ≈ 0

5.0625 + 13.5δ1 – 10.125 – 20.25δ1 + 10.5 + 7δ1 – 5.5 ≈ 0

0.25δ1 ≈ 0.0625

δ1 ≈ 0.0625 0.250.25

≈

Thus x2 ≈ 1.5 + 0.25 = 1.75 Third approximation Let the true value of the root, x3 , be (x2 + δ2) Let f(x2 + δ2) = 0, then since x2 = 1.75,

(1.75 + δ2) 4 - 3(1.75 + δ2)3 + 7(1.75 + δ2) – 5.5 = 0

Neglecting terms containing products of δ2 gives:

( ) ( ) ( ) ( )4 3 3 22 2 21.75 4 1.75 3[ 1.75 3 1.75 ] 7(1.75 ) 5.5 0+ δ − + δ + + δ − ≈

9.3789 + 21.4375δ2 – 16.078125 – 27.5625δ2 + 12.25 + 7δ2 – 5.5 ≈ 0

0.875δ2 ≈ -0.050775

δ2 ≈ 0.050775 0.058030.875

−≈ −

Thus x3 ≈ (x2 + δ2) = 1.75 – 0.05803 ≈ 1.692 Fourth approximation Let the true value of the root, x4 , be (x3 + δ3) Let f(x3 + δ3) = 0, then since x3 = 1.692,


79

(1.692 + δ3) 4 - 3(1.692 + δ3)3 + 7(1.692 + δ3) – 5.5 = 0 Neglecting terms containing products of δ3 gives: ( ) ( ) ( ) ( )4 3 3 2

3 3 31.692 4 1.692 3[ 1.692 3 1.692 ] 7(1.692 ) 5.5 0+ δ − + δ + + δ − ≈ 8.19599 + 19.37586δ3 – 14.5318977 – 25.76578δ3 + 11.844 + 7δ3 – 5.5 ≈ 0 0.61008δ3 ≈ -0.0080923

δ3 ≈ 0.0080923 0.01326430.61008

−≈ −

Thus x4 ≈ (x3 + δ3) = 1.692 - 0.0132643 ≈ 1.6787 Fifth approximation Let the true value of the root, x5 , be (x4 + δ4) Let f(x4 + δ4) = 0, then since x4 = 1.6787, (1.6787 + δ4) 4 - 3(1.6787 + δ4)3 + 7(1.6787 + δ4) – 5.5 = 0 Neglecting terms containing products of δ4 gives: ( ) ( ) ( ) ( )4 3 3 2

4 4 41.6787 4 1.6787 3[ 1.6787 3 1.6787 ] 7(1.6787 ) 5.5 0+ δ − + δ + + δ − ≈ 7.941314 + 18.9225δ4 – 14.1919 – 25.3623δ4 + 11.7509 + 7δ4 – 5.5 ≈ 0 0.5602δ4 ≈ -0.000314

δ4 ≈ 0.000314 0.000560.5602

−≈ −

Thus x5 ≈ (x4 + δ4) = 1.6787 - 0.00056 ≈ 1.6781 Since x4 and x5 are the same when expressed to the required degree of accuracy, then the required root is 1.68, correct to 3 decimal places. From earlier, f(x) = 4 33 7 5.5x x x− + − f(0) = -5.5 f(-1) = 1 + 3 – 7 – 5.5 = -8.5

f(-2) = 16 + 24 – 14 – 5.5 = 20.5 Hence, a root lies between x = -1 and x = -2.

This root may be found in exactly the same way as for the positive root above.


80

EXERCISE 41 Page 85 2. Use Newton’s method to solve 33 10 14x x− = , correct to 4 significant figures. Let f(x) = 33 10 14x x− − f(0) = -14 f(1) = 3 – 10 – 14 = -21 f(2) = 24 – 20 – 14 = -10 f(3) = 81 – 30 – 14 = 36 Hence, a root lies between x = 2 and x = 3. Let 1r 2.2=

A better approximation is given by: ( )12 1

1

f rr r

f '(r )= − 2f '(x) 9x 10x= −

Hence, 3

2 2

3(2.2) 10(2.2) 14 4.056r 2.2 2.2 2.38819(2.2) 10(2.2) 21.56

− − −= − = − =

−

3f (2.3881) 2.9771577r 2.3881 2.3881 2.2796f '(2.3881) 27.44619

= − = − =

4f (2.2796) 1.257655r 2.2796 2.2796 2.3321f '(2.2796) 23.973184

−= − = − =

5f (2.3321) 0.7297097r 2.3321 2.3321 2.3036f '(2.3321) 25.627214

= − = − =

60.3633356r 2.3036 2.3183

24.7231566−

= − =

70.1962136r 2.3183 2.310525.187634

= − =

80.10180935r 2.3105 2.314624.94069

−= − =

90.05452775r 2.3146 2.312425.070358

= − =

100.02944745r 2.3124 2.3136

25.0007438−

= − =

110.0163322r 2.3136 2.312925.0387046

= − =

120.01037987r 2.3129 2.313

25.0165577−

= − =

Hence, correct to 4 significant figures, x = 2.313


81

4. Use Newton’s method to solve 4 33 4 7 12 0x x x− + − = , correct to 3 decimal places. Let f(x) = 4 33 4 7 12x x x− + − f(0) = -12 f(1) = 3 – 4 + 7 – 12 = -6 f(2) = 48 – 32 + 14 – 12 = 18 Hence, a root lies between x = 1 and x = 2 There are no further positive roots since the 4x term predominates. f(-1) = 3 + 4 – 7 – 12 = -12 f(-2) = 48 + 32 – 14 – 12 = 54 Hence, a root lies between x = -1 and x = -2 There are no further negative roots since, once again, the 4x term predominates. For the root between x = 1 and x = 2: Let 1r 1.25= 3 2f '(x) 12x 12x 7= − +

( ) ( )( ) ( )

4 31

2 1 3 21

3 1.25 4 1.25 7(1.25) 12f (r ) 3.73828r r 1.25 1.25 1.56985f '(r ) 11.687512 1.25 12 1.25 7

− + − −= − = − = − =

− +

31.734046r 1.56985 1.49715

23.8522585= − =

40.12925743r 1.49715 1.4908120.3720909

= − =

50.000994417r 1.49081 1.49076

20.089988= − =

Hence, correct to 3 decimal places, the positive root is: x = 1.491 For the root between x = -1 and x = -2: Let 1r 1.2= −

( ) ( )( ) ( )

4 31

2 1 3 21

3 1.2 4 1.2 7( 1.2) 12f (r ) 7.2672r r 1.2 1.2 1.4343f '(r ) 31.01612 1.2 12 1.2 7

− − − + − − −= − = − − = − − = −

−− − − +

32.4589815r 1.4343 1.388053.094585

= − − = −−

40.11488764r 1.3880 1.3856

48.207045= − − = −

−


82

50.00051387r 1.3856 1.3856

47.960999= − − = −

−

Hence, correct to 3 decimal places, the negative root is: x = -1.386

7. Use Newton’s method to solve 2300 62

e θ θ− + = , correct to 3 significant figures.

Let f(θ) = 2300 62

e θ θ− + − f(0) = 300 – 6 = 294 f(1) = 35.1 f(2) = 0.495

f(3) = -3.756 Hence, a root lies between x = 2 and x = 3, very close to x = 2. There are no further positive roots since the 2300e θ− term predominates. There are no negative roots since f(x) will always be positive. Let 1r 2= 2f '(x) 600e 0.5− θ= − +

4

12 1 4

1

f (r ) 300e 1 6 0.494692r r 2 2 2.0492f '(r ) 600e 0.5 10.489383

−

−

+ −= − = − = − =

− + −

30.00436387r 2.0492 2.04979.45952774

−= − =

−

Hence, correct to 3 significant figures, x = 2.05 9. A Fourier analysis of the instantaneous value of a waveform can be represented by:

1sin sin 34 8

y t t tπ⎛ ⎞= + + +⎜ ⎟⎝ ⎠

Use Newton’s method to determine the value of t near to 0.04, correct to 4 decimal places, when the amplitude, y, is 0.880.

When y = 0.88, then 10.88 sin sin 34 8

t t tπ⎛ ⎞= + + +⎜ ⎟⎝ ⎠

or 1sin sin 3 0.88 04 8

t t tπ⎛ ⎞+ + + − =⎜ ⎟⎝ ⎠

Let f(t) = 1sin sin 3 0.884 8

t t tπ⎛ ⎞+ + + −⎜ ⎟⎝ ⎠

Let 1r = 0.04 3f '(t) 1 cos t cos3t8

= + +


83

12 1

1

10.04 sin 0.04 sin 3(0.04) 0.88f (r ) 0.00035154 8r r 0.04 0.043f '(r ) 2.3715033451 cos0.04 cos3(0.04)8

π+ + + −

= − = − = −+ +

= 0.03985

30.000004203r 0.03985 0.03985

2.371529496−

= − =

Hence, correct to 4 decimal places, t = 0.0399 10. A damped oscillation of a system is given by the equation: 0.57.4 sin 3ty e t= − Determine the value of t near to 4.2, correct to 3 significant figures, when the magnitude y of the oscillation is zero. Let f(t) = 0.57.4 sin 3− te t ( )( ) ( )( )0.5t 0.5t 0.5t 0.5tf ' (t) 7.4e 3cos3t sin 3t 3.7e 22.2e cos3t 3.7e sin 3t= − + − = − − = ( )0.5te 22.2cos3t 3.7sin 3t− + Let 1r 4.2=

( )

2.11

2 1 2.11

f (r ) 7.4e sin12.6 2.03182922r r 4.2 4.2 4.189f '(r ) e 22.2cos12.6 3.7sin12.6 182.2023833

− −= − = − = − =

− + −

(Note, sin 12.6 is sin 12.6 rad)

3f (4.189) 0.037825r 4.189 4.189 4.189f '(4.189) 180.3134965

−= − = − =

−

Hence, correct to 3 significant figures, t = 4.19 11. The critical speeds of oscillation, λ, of a loaded beam are given by the equation: 3 23.250 0.063 0λ λ λ− + − = Determine the value of λ which is approximately equal to 3.0 by Newton’s method, correct to 4 decimal places.


84

Let f(λ) = 3 23.250 0.063− + −λ λ λ 2f '( ) 3 6.5 1λ = λ − λ + Let 1r 3.0=

( ) ( )( ) ( )

3 2

2 2

3.0 3.250 3.0 3.0 0.063f (3.0) 0.687r 3.0 3.0 3.0 2.91918f '(3.0) 8.53 3.0 6.5 3.0 1

− + −= − = − = − =

− +

30.0370604r 2.91918 2.914307.5901656

= − =

40.00015139r 2.91430 2.914287.5364835

= − =

Hence, correct to 4 decimal places, λ = 2.914


85

CHAPTER 10 COMPUTER NUMBERING SYSTEMS

EXERCISE 42 Page 86 2. Convert the following binary numbers to denary numbers:

(a) 10101 (b) 11001 (c) 101101 (d) 110011 (a) 4 3 2 1 0

210101 1 2 0 2 1 2 0 2 1 2= × + × + × + × + ×

= 16 + 0 + 4 + 0 + 1 = 1021 (b) 4 3 2 1 0

211001 1 2 1 2 0 2 0 2 1 2= × + × + × + × + ×

= 16 + 8 + 0 + 0 + 1 = 1025 (c) 5 4 3 2 1 0

2101101 1 2 0 2 1 2 1 2 0 2 1 2= × + × + × + × + × + ×

= 32 + 0 + 8 + 4 + 0 + 1 = 1045 (d) 5 4 3 2 1 0

2110011 1 2 1 2 0 2 0 2 1 2 1 2= × + × + × + × + × + ×

= 32 + 16 + 0 + 0 + 2 + 1 = 1051 4. Convert the following binary numbers to denary numbers:

(a) 11010.11 (b) 10111.011 (c) 110101.0111 (d) 11010101.10111 (a) 4 3 2 1 0 1 2

211010.11 1 2 1 2 0 2 1 2 0 2 1 2 1 2− −= × + × + × + × + × + × + ×

= 16 + 8 + 0 + 2 + 0 + 1 12 4+ = 1026.75

(b) 4 3 2 1 0 1 2 3

210111.011 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2− − −= × + × + × + × + × + × + × + ×

= 16 + 0 + 4 + 2 + 1 + 1 14 8+ = 1023.375

(c) 5 4 3 2 1 0 1 2 3 4

2110101.0111 1 2 1 2 0 2 1 2 0 2 1 2 0 2 1 2 1 2 1 2− − − −= × + × + × + × + × + × + × + × + × + ×

= 32 + 16 + 0 + 4 + 0 + 1 + 1 1 14 8 16+ + = 3


86

(d) 7 6 5 4 3 2 1 0 1 2211010101.10111 1 2 1 2 0 2 1 2 0 2 1 2 0 2 1 2 1 2 0 2− −= × + × + × + × + × + × + × + × + × + ×

+ 3 4 51 2 1 2 1 2− − −× + × + ×

= 128 + 64 + 16 + 4 + 1 + 1 1 1 12 8 16 32+ + + = 10213.71875


87

EXERCISE 43 Page 88 2. Convert the following denary numbers into binary numbers: (a) 31 (b) 42 (c) 57 (d) 63 (a) 2 31 Remainder 2 15 1 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 231 11111= (b) 2 42 Remainder 2 21 0 2 10 1 2 5 0 2 2 1 2 1 0 0 1 Hence, 10 242 101 010= (c) 2 57 Remainder 2 28 1 2 14 0 2 7 0 2 3 1 2 1 1 0 1 Hence, 10 257 111 001= (d) 2 63 Remainder 2 31 1 2 15 1 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 263 111 111=


88

4. Convert the following denary numbers into binary numbers:

(a) 47.40625 (b) 30.8125 (c) 53.90625 (d) 661.65625

(a) 2 47 Remainder 2 23 1 2 11 1 2 5 1 2 2 1 2 1 0 0 1 Hence, 10 247 101111= 0.40625 × 2 = 0.8125 0.8125 × 2 = 1.625 0.625 × 2 = 1.25 0.25 × 2 = 0.50 0.50 × 2 = 1.00 Hence, 10 20.40625 0.01101= Thus, 10 247.40625 101 111.011 01= (b) 2 30 Remainder 2 15 0 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 230 11110= 0.8125 × 2 = 1.625 0.625 × 2 = 1.25 0.25 × 2 = 0.50 0.50 × 2 = 1.00 Hence, 10 20.8125 0.1101= Thus, 10 230.8125 11 110.110 1= (c) 2 53 Remainder 2 26 1 2 13 0 2 6 1 2 3 0 2 1 1 0 1 Hence, 10 253 110 101= 0.90625 × 2 = 1.8125 0.8125 × 2 = 1.625 0.625 × 2 = 1.25 0.25 × 2 = 0.50 0.50 × 2 = 1.00 Hence, 10 20.90625 0.111 01= Thus, 10 253.90625 110 101.111 01=


89

(d) 2 61 Remainder 2 30 1 2 15 0 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 261 111101= 0.65625 × 2 = 1.3125 0.3125 × 2 = 0.625 0.625 × 2 = 1.25 0.25 × 2 = 0.50 0.50 × 2 = 1.00 Hence, 10 20.65625 0.101 01= Thus, 10 261.65625 111 101.101 01=


90

EXERCISE 44 Page 89 3. Convert the following denary numbers to binary numbers, via octal: (a) 247.09375 (b) 514.4375 (c) 1716.78125 (a) 8 247 Remainder 8 30 7 8 3 6 0 3 Hence, 10 8 2247 367 011110 111= = from Table 10.1, page 88 0.09375 × 8 = 0.75 0.75 × 8 = 6.00 Hence, 10 8 20.09375 0.06 0.000110= = from Table 10.1, page 88 Hence, 10 8 2247.09375 763.06 11 110 111.000 11= = (b) 8 514 Remainder 8 64 2 8 8 0 8 1 0 Hence, 10 8 2514 1002 001 000 000 010= = from Table 10.1, page 88 0 1 0.4375 × 8 = 3.50 0.50 × 8 = 4.00 Hence, 10 8 20.4375 0.34 0.011100= = from Table 10.1, page 88 Hence, = =10 8 2514.4375 1002.34 1000000 010.0111 (c) 8 1716 Remainder 8 214 4 8 26 6 8 3 2 Hence, 10 8 21716 3264 011 010 110 100= = from Table 10.1, page 88 0 3 0.78175 × 8 = 6.25 0.25 × 8 = 2.00 Hence, 10 8 20.78125 0.62 0.110 010= = from Table 10.1, page 88 Hence, 10 8 21716.79125 3264.62 11010110 100.11001= = 4. Convert the following binary numbers to denary numbers, via octal: (a) 111.0111 (b) 101001.01 (c) 1110011011010.0011


91

(a) 111.0111 = 111.011 100 2

= 7 . 3 84 = 0 1 22

3 47 8 3 8 4 8 78 8

− −× + × + × = + + = 107.4375

(b) 101001.01 = 101 001.010 2

= 5 1 . 82 = 1 0 1 25 8 1 8 2 8 40 18

−× + × + × = + + = 1041.25

(c) 1110011011010.0011 = 001 110 011 011 010.001 100 2 = 1 6 3 3 2 . 1 84 = 4 3 2 1 0 1 21 8 6 8 3 8 3 8 2 8 1 8 4 8− −× + × + × + × + × + × + ×

= 4096 + 3072 + 192 + 24 + 2 + 1 48 64+ = 107386.1875


92

EXERCISE 45 Page 92 2. Convert the hexadecimal number 162C into decimal.

1 0162C 2 16 C 16 32 12= × + × = + = 1044

4. Convert the hexadecimal number 162F1 into decimal

2 1 0 2 1 0162F1 2 16 F 16 1 16 2 16 15 16 1 16= × + × + × = × + × + ×

= 512 + 240 + 1 = 10753 6. Convert the decimal number 10200 into hexadecimal 16 200 Remainder 16 12 8 0 12 (i.e. C) Hence, 10 16200 C8= 8. Convert the decimal number 10238 into hexadecimal 16 238 Remainder 16 14 14 (i.e. E) 0 14 (i.e. E) Hence, 10 16238 EE= 10. Convert the binary number 211101010 into hexadecimal

211101010 = 1110 1010 grouping in 4’s = E A16 from Table 10.2, page 90.

i.e. 2 1611101010 EA=


93

12. Convert the binary number 210100101 into hexadecimal

210100101 = 1010 0101 grouping in 4’s = A 516 from Table 10.2, page 90.

i.e. 2 1610100101 A5= 14. Convert the hexadecimal number 16ED into binary

16ED = 1110 1101 2 from Table 10.2, page 90. 16. Convert the hexadecimal number 16A21 into binary

16A21 = 1010 0010 0001 2 from Table 10.2, page 90.


94

CHAPTER 11 BOOLEAN ALGEBRA AND LOGIC CIRCUITS

EXERCISE 46 Page 97 1. Determine the Boolean expression and construct a truth table for the following switching circuit:

For the circuit to function, Z = C AND [(A AND B) OR ( A AND B)]

i.e. Z = ( )C. A.B A.B+

The truth table is shown below:

A B C A.B A A .B A.B + A .B Z = C.( A.B + A .B) 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 0 0 0 0 0 1 1

1 1 1 1 0 0 0 0

0 0 1 1 0 0 0 0

0 0 1 1 0 0 1 1

0 0 0 1 0 0 0 1

2. Determine the Boolean expression and construct a truth table for the following switching circuit:

For the circuit to function, Z = C AND [(A AND B ) OR ( A )]

i.e. Z = ( )C. A.B A+



95

A B C A B A. B A. B + A Z = C.( A. B + A ) 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

1 1 1 1 0 0 0 0

1 1 0 0 1 1 0 0

0 0 0 0 1 1 0 0

1 1 1 1 1 1 0 0

0 1 0 1 0 1 0 0

4. Determine the Boolean expression and construct a truth table for the following switching circuit:

For the circuit to function, Z = C AND [(B AND C AND A ) OR (A AND (B OR C )]

i.e. Z = ( )C. B.C.A A.(B C)+ +


A B C A

C B.C. A B + C A.( B + C ) B.C. A + A.( B + C ) Z

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

1 1 1 1 0 0 0 0

1 0 1 0 1 0 1 0

0 0 0 1 0 0 0 0

1 0 1 1 1 0 1 1

0 0 0 0 1 0 1 1

0 0 0 1 1 0 1 1

0 0 0 1 0 0 0 1

The remaining problems in this exercise have solutions given.


96

EXERCISE 47 Page 100 3. Simplify: ( )F.G F.G G. F F+ + +

( )F.G F.G G. F F+ + + = F.G F.G G(1) F.G F.G G+ + = + + from 10, Table 11.8,

= ( )G. F F G+ + = G.1 G G G+ = + from 10, Table 11.8,

= G from 9, Table 11.8

4. Simplify: ( )F.G F. G G F.G+ + +

( )F.G F. G G F.G+ + + = F.G F.1 F.G F.G F F.G+ + = + + from 10, Table 11.8,

= ( )F. G G F F.1 F+ + = +

= F + F = F from 9 and 10, Table 11.8 6. Simplify: F.G.H F.G.H F.G.H+ +

F.G.H F.G.H F.G.H+ + = ( )F.H. G G F.G.H+ +

= F.H F.G.H+ from 10, Table 11.8,

= ( )H. F F.G+

8. Simplify: P.Q.R P.Q.R P.Q.R+ +

P.Q.R P.Q.R P.Q.R+ + = ( )Q.R. P P P.Q.R+ +

= Q.R P.Q.R+ from 10, Table 11.8 9. Simplify: F.G.H F.G.H F.G.H F.G.H+ + +

F.G.H F.G.H F.G.H F.G.H+ + + = ( ) ( )F.G. H H F.G H H+ + +

= F.G F.G+ from 10, Table 11.8,

= ( )G. F F+

= G from 10, Table 11.8


97

12. Simplify: ( ) ( )R. P.Q P.Q P.Q P. Q.R Q.R+ + + +

( ) ( )R. P.Q P.Q P.Q P. Q.R Q.R+ + + + = ( ) ( )R.P.Q R.P. Q Q P.R. Q Q+ + + +

= R.P.Q R.P P.R+ + from 10, Table 11.8,

= ( )R.P.Q P. R R+ +

= R.P.Q P+ from 10, Table 11.8,

= P R.Q+ from 17, Table 11.8


98

EXERCISE 48 Page 101 2. Simplify: (A B.C ) (A.B C)+ + +

(A B.C ) (A.B C)+ + + = ( ) ( )A .B.C A B C+ + + by de Morgan’s law,

= A.B.C A B C+ + +

= C.(A.B 1) A B+ + +

= C A B+ + or A B C+ + from 8, Table 11.8

3. Simplify: (A.B B.C). A.B+

(A.B B.C). A.B+ = ( ) ( ) ( )A.B . B.C . A B+ by de Morgan’s law,

= ( ) ( ) ( )A B . B C . A B+ + +

= ( ) ( ) ( )A B . B C . A B+ + +

= ( ) ( )A.B A.C B.B B.C . A B+ + + +

= ( ) ( )A.B A.C B B.C . A B+ + + + from 13, Table 11.8,

= ( ) ( )A.B A.C B . A B+ + + from 15, Table 11.8,

= ( ) ( )B. A 1 A.C . A B⎡ ⎤+ + +⎣ ⎦

= ( ) ( )B A.C . A B+ + from 8, Table 11.8,

= A.B B.B A.C.A A.B.C+ + +

= A.B 0 0 A.B.C+ + + from 14, Table 11.8,

= A.B A.B.C+ 5. Simplify: (P.Q P.R).(P.Q.R)+

(P.Q P.R).(P.Q.R)+ = ( ) ( ) ( )P Q P R .P. Q R⎡ ⎤+ + + +⎢ ⎥⎣ ⎦ by de Morgan’s law,

= ( )( )P Q P R P.Q P.R+ + + +


99

= ( ) ( )1 Q R . P.Q P.R+ + +

= P.Q P.R Q.P.Q Q.P.R R .P.Q R .P.R+ + + + +

= P.Q P.R 0 Q.P.R R .P.Q 0+ + + + + from 14 and 9, Table 11.8,

= ( )P. Q R Q.R R .Q+ + +

= ( )( )P. Q R R Q Q+ + +

= ( )P. Q R R+ + from 10, Table 11.8,

= ( )P . Q R+ from 9, Table 11.8


100

EXERCISE 49 Page 105 3. Use Karnaugh map techniques to simplify: (P.Q).(P.Q)

P.Q corresponds to P = 0, Q = 0, i.e. the top left hand cell of the Karnaugh map, shown as a 1.

P.Q corresponds to P = 0, Q = 1, i.e. the bottom left hand cell, hence P.Q corresponds to each of

the other three cells, shown as 2’s.

Only one cell has both a 1 and a 2 in it, i.e. P = 0, Q = 0

Hence, (P.Q).(P.Q) = P .Q 4. Use Karnaugh map techniques to simplify: A.C A.(B C) A.B.(C B)+ + + +

If a Boolean expression contains brackets it is often easier to remove them, using the laws and rules

of Boolean algebra, before plotting the function on a Karnaugh map.

Thus, A.C A.(B C) A.B.(C B)+ + + + = A.C A.B A.C A.B.C A.B.B+ + + +

= A.C A.B A.C A.B.C A.0+ + + +

= A.C A.B A.C A.B.C+ + +

A.C corresponds to A = 1, C = 0, shown as 1’s in the two right hand cells in the top row

A.B corresponds to A = 0, B = 1, shown as 1’s in the two cells in the second column

A.C corresponds to A = 0, C = 1, shown as 1’s in the two left hand cells in the bottom row

A.B.C corresponds to A = 1, B = 1, C = 1, shown as a 1 in the cell in the third column, bottom row

A 4-cell couple and two 2-cell couples are formed as shown by the broken lines.

The only variable common to the 4-cell couple is B = 1, i.e. B.

The variable common to the 2-cell couple on the top right of the map is A = 1 and C = 0, i.e. A.C


101

The variables common to the 2-cell couple on the bottom left of the map is A = 0 and C = 1, i.e.

A.C

Thus, A.C A.(B C) A.B.(C B)+ + + + simplifies to B A.C A.C+ +

6. Use Karnaugh map techniques to simplify: P.Q.R P.Q.R P.Q.R P.Q.R+ + +

P.Q.R , i.e. P = 0, Q = 0 and R = 0, is shown on the Karnaugh map as a 1




Two 2-cell couples are formed as shown.

For the cell containing the 2 and the 3, the variables common are P = 1, Q = 1, i.e. P.Q

For the cell containing the 3 and the 4, the variables common are P = 1, R = 1, i.e. P.R

Hence, P.Q.R P.Q.R P.Q.R P.Q.R+ + + may be simplified to: P.Q + P.R + P.Q.R

i.e. P.Q.R P.Q.R P.Q.R P.Q.R+ + + simplifies to: P.(Q + R) + P.Q .R 8. Use Karnaugh map techniques to simplify: A.B.C.D A.B.C.D A.B.C.D+ +

A.B.C.D , i.e. A = 0, B = 0, C = 1 and D = 1, is shown as a 1 on the four variable matrix

A.B.C.D , i.e. A = 0, B = 0, C = 1 and D = 0, is shown as a 2


Two 2-cell couples are formed as shown.


102

The variables common to the vertical couple is A = 0, B = 0, C = 1, i.e. A.B.C

The variables common to the horizontal couple is B = 0, C = 1, D = 0, i.e. B.C.D

Hence, A.B.C.D A.B.C.D A.B.C.D+ + simplifies to: A.B.C + B.C.D or ( )B .C. A D+

10. Use Karnaugh map techniques to simplify: A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D+ + + +

A.B.C.D , i.e. A = 0, B = 0, C = 0 and D = 1, is shown as a 1 on the four variable matrix below.





A 4-cell couple is formed as shown and the variables common to it are A = 1, D = 0, i.e. A.D

Hence, A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D+ + + + simplifies to:

A.D A .B .C.D+

11. Use Karnaugh map techniques to simplify: A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D+ + + + + +

The Karnaugh map for the given expression is shown below. A 4-cell couple and three 2-cell couples are formed as shown.

The variables common to the 4-cell couple are A = 0 and C = 1, i.e. A.C


103

The variables common to the 2-cell couple on the far right of the top row are A = 1, C = 0 and

D = 0, i.e. A.C.D

The variables common to the 2-cell couple on the far left and far right of the top row are B = 0,

C = 0 and D = 0, i.e. B.C.D

The variables common to the 2-cell couple at the top and bottom of the first column are A = 0,

B = 0 and D = 0, i.e. A.B.D

Hence, A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D+ + + + + +

simplifies to: A.C + A.C.D + B.C.D + A.B.D

i.e. ( )A .C A .C.D B .D. A C+ + +


104

EXERCISE 50 Page 109 5. Simplify the expression given in column 4 of the truth table below and devise a logic circuit to meet the requirements of the simplified expression.

From column 4, 1Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +

The Karnaugh map is shown below.

The vertical 2-cell couple corresponds to: A.B

The horizontal 4-cell couple corresponds to: C

Hence, 1Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +

simplifies to: 1Z A.B C= +

A logic circuit to meet these requirements is shown below.

7. Simplify the expression given in column 6 of the truth table of question 5 above and devise a logic circuit to meet the requirements of the simplified expression. From column 6, 3Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +

The Karnaugh map is shown below.


105

The horizontal 2-cell couple corresponds to: A.C

The 4-cell couple corresponds to: B

Hence, 3Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +

simplifies to: 3Z B A.C= +


9. Simplify the Boolean expression: P.Q.R P.Q.R P.Q.R+ + and devise a logic circuit to meet the

requirements of the simplified expression. The Karnaugh map for the Boolean expression: P.Q.R P.Q.R P.Q.R+ + is shown below.

The 2-cell couple on the far right of the map corresponds to: P.R

The other 2-cell couple corresponds to: Q.R

Hence, P.Q.R P.Q.R P.Q.R+ +

simplifies to: ( )P .R Q.R or R. P Q+ +



106

11. Simplify the Boolean expression:

A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D+ + + +

and devise a logic circuit to meet the requirements of the simplified expression.

The Karnaugh map for the Boolean expression:

A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D+ + + +

is shown below.

The 2-cell couple on the bottom row of the map corresponds to: A.C.D

The 4-cell couple corresponds to: B.D

Hence, A.B.C.D A.B.C.D A.B.C.D A.B.C.D A.B.C.D+ + + +

simplifies to: ( )A.C.D B.D or D. A .C B+ +


12. Simplify the Boolean expression: ( ) ( )P.Q.R . P Q.R+ and devise a logic circuit to meet the

requirements of the simplified expression.

The Karnaugh map for the Boolean expression: ( ) ( )P.Q.R . P Q.R+ is shown below.

P.Q.R is shown with a 1

P.Q.R is shown with 2’s


107

P + Q.R is shown with 3’s

P Q.R+ is shown with 4’s

Hence, ( ) ( )P.Q.R . P Q.R+ is represented by the cells containing both 2’s and 4’s

The 2-cell vertical couple of the map corresponds to: P.Q

The 2-cell horizontal couple corresponds to: P.R

Hence, ( ) ( )P.Q.R . P Q R+ +

simplifies to: P .Q P.R+ or ( )P. Q R+



108

EXERCISE 51 Page 112 The solutions to questions 1 to 6 are shown on page 112 and 113. 7. In a chemical process, three of the transducers used are P, Q and R, giving output signals of

either 0 or 1. Devise a logic system to give a 1 output when:

(a) P and Q and R all have 0 outputs, or when

(b) P is 0 and (Q is 1 or R is 0).

The Boolean expression to meet the requirements is:

( )P.Q.R P. Q R+ + = P.Q.R P.Q P.R+ +

= ( )P.R . Q 1 P.Q+ +

= P.R P.Q+

= ( )P . Q R+

A logic circuit to satisfy this Boolean expression is shown below:

8. Lift doors should close, (Z), if:

(a) the master switch, (A), is on and either

(b) a call, (B), is received from any other floor, or

(c) the doors, (C), have been open for more than 10 seconds, or

(d) the selector push within the lift, (D), is pressed for another floor.

Devise a logic circuit to meet these requirements.


Z = A.(B + C + D)



109

9. A water tank feeds three separate processes. When any two of the processes are in operation at

the same time, a signal is required to start a pump to maintain the head of water in the tank.

Devise a logic circuit using nor-gates only to give the required signal.


Z = A.(B + C) + B.(A + C) + C.(A + B)

= A.B + A.C + A.B + B.C + A.C + B.C

= A.B + A.C + B.C

i.e. Z = A.(B + C) + B.C


10. A logic signal is required to give an indication when:

(a) the supply to an oven is on, and

(b) the temperature of the oven exceeds 210°C, or

(c) the temperature of the oven is less than 190°C.

Devise a logic circuit using nand-gates only to meet these requirements.


Z = A.B +A.C

i.e. Z = A.(B + C)



110

CHAPTER 12 INTRODUCTION TO TRIGONOMETRY

EXERCISE 52 Page 115 2. Triangle PQR is isosceles, Q bring a right angle. If the hypotenuse is 38.47 cm find (a) the

lengths of sides PQ and QR, and (b) the value of ∠QPR

(a) Since triangle PQR in the diagram below is isosceles, PQ = QR From Pythagoras, 2 2 2 2(38.47) (PQ) (QR) 2(PQ)= + =

from which, ( )2

2 38.47PQ2

= and PQ = 238.47 38.47

2 2= = 27.20 cm

Hence, PQ = QR = 27.20 cm

(b) Since triangle PQR is isosceles, ∠P = ∠R and since ∠Q = 90°, then ∠P + ∠R = 90° Hence, ∠QPR = 45° (=∠QRP) 3. A man cycles 24 km south and then 20 km due east. Another man, starting at the same time as

the first man, cycles 32 km due east and then 7 km due south. Find the distance between the two

men.

With reference to the diagram below, AB = 32 – 20 = 12 km and BC = 24 – 7 = 17 km

Hence, distance between the two men, AC = ( )2 212 17+ = 20.81 km by Pythagoras.


111

4. A ladder 3.5 m long is placed against a perpendicular wall with its foot 1.0 m from the wall.

How far up the wall (to the nearest centimetre) does the ladder reach? If the foot of the ladder is

now moved 30 cm further away from the wall, how far does the top of the ladder fall?

Distance up the wall, AB = ( )2 23.5 1.0− = 3.35 m by Pythagoras.

( ) ( ) ( )2 2 2 2A 'B A 'C ' BC' 3.5 1.30 3.25m⎡ ⎤= − = − =⎣ ⎦

Hence, the amount the top of the ladder has moved down the wall, given by AA′ = 3.35 – 3.25 = 0.10 m or 10 cm


112

EXERCISE 53 Page 117

2. If cos A = 1517

find sin A and tan A, in fraction form.

Since adjacentcos inehypotenuse

= then the sides 15 and 17 are as shown in the diagram.

By Pythagoras, BC = ( )2 217 15− = 8

Hence, sin A = opposite BChypotenuse AC

= = 817

and tan A = opposite BCadjacent AB

= = 815

4. Point P lies at co-ordinates (-3, 1) and point Q at (5, -4). Determine (a) the distance PQ, (b) the

gradient of the straight line PQ, and (c) the angle PQ makes with the horizontal.

(a) From the diagram below, PQ = ( )2 25 8+ = 9.434 by Pythagoras

(b) Gradient of PQ = 1 4 5

3 5 8− −

=− − −

= - 0.625

(c) 5Tan8

θ = from which, the angle PQ makes with the horizontal, θ = 1 5tan8

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 32°


113

EXERCISE 54 Page 119 2. Solve triangle DEF

By Pythagoras, FE = ( )2 24 3+ = 5 cm

Tan E = 43

from which, ∠E = 1 4tan3

− = 53.13° or 53°8′

Hence, ∠F = 180° - 90° - 53.13° = 36.87° or 36°52′ 4. Solve triangle JKL and find its area

Sin 51° = 6.7JL

from which, JL = 6.7sin 51°

= 8.62 cm

Tan 51° = 6.7KL

from which, KL = 6.7tan 51°

= 5.43 cm

∠J = 180° - 90° - 51° = 39° (Checking: JL = ( )2 26.7 5.43 8.62cm+ = )

Area of triangle JKL = 12

(KL)(JK) = 12

(5.43)(6.7) = 18.19 2cm

6. Solve triangle PQR and find its area


114

By Pythagoras, 2 2 2PR 3.69 8.75+ = from which, PR = ( )2 28.75 3.69− = 7.934 m

Sin R = 3.698.75

from which, ∠R = 1 3.69sin8.75

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 24.94° or 24°57′

∠Q = 180° - 90° - 24.94° = 65.06° or 65°3′

Area of triangle PQR = 12

(PQ)(PR) = 12

(3.69)(7.934) = 14.64 2m

7. A ladder rest against the top of the perpendicular wall of a building and makes an angle of 73°

with the ground. If the foot of the ladder is 2 m from the wall, calculate the height of the building.

The ladder is shown in the diagram below, where BC is the height of the building.

Tan 73° = BC2

from which, height of building, BC = 2 tan 73° = 6.54 m


115

EXERCISE 55 Page 121 2. From the top of a vertical cliff 80.0 m high the angle of depression of two buoys lying due west

of the cliff are 23° and 15°, respectively. How far are the buoys apart?

In the diagram below, the two buoys are shown as A and B.

Tan 15° = 80

AC from which, AC = 80

tan15° = 298.56 m

Tan 23° = 80BC

from which, BC = 80tan 23°

= 188.47 m

Hence, distance apart, AB = AC – BC = 298.56 – 188.47 = 110.1 m 4. A flagpole stands on the edge of the top of a building. At a point 200 m from the building the

angles of elevation of the top and bottom of the pole are 32° and 30° respectively. Determine the

height of the flagpole.

In the diagram below, the flagpole is shown as AB.

Tan 32° = AC

200 from which, AC = 200 tan 32° = 124.97 m

Tan 30° = BC200

from which, BC = 200 tan 30° = 115.47 m

Hence, height of flagpole, AB = AC – BC = 124.97 – 115.47 = 9.50 m 6. From a window 4.2 m above horizontal ground the angle of depression of the foot of a building

across the road is 24° and the angle of elevation of the top of the building is 34°. Determine,

correct to the nearest centimetre, the width of the road and the height of the building.


116

In the diagram below, D is the window, the width of the road is AB and the height of the building across the road is BC.

In the triangle ABD, ∠D = 90° - 24° = 66°

Tan 66° = AB4.2

hence, width of road, AB = 4.2 tan 66° = 9.43 m

From triangle DEC, tan 34° = CE CE CEDE AB 9.43

= = from which, CE = 9.43 tan 34° = 6.36 m

Hence, height of building, BC = CE + EB = CE + AD = 6.36 + 4.2 = 10.56 m 7. The elevation of a tower from two points, one due east of the tower and the other due west of it

are 20° and 24°, respectively, and the two points of observation are 300 m apart. Find the height

of the tower to the nearest metre.

In the diagram below, the height of the tower is AB and the two observation points are at C and D.

Tan 20° = AB

BC from which, AB = BC tan 20°

Tan 24° = AB300 BC−

from which, AB = (300 - BC) tan 24°

i.e. BC tan 20° = (300 - BC) tan 24° = 300 tan 24° - BC tan 24° i.e. 0.36397 BC = 133.57 – 0.44523 BC i.e. 0.8092 BC = 133.57

and BC = 133.570.8092

= 165.06 m

Tan 20° = AB165.06

from which, height of tower, AB = 165.06 tan 20° = 60 m, to the nearest metre


117

EXERCISE 56 Page 123 4.(a) Evaluate correct to 4 decimal places: secant 73°

secant 73° = 1cos 73°

= 3.4203

5.(b) Evaluate correct to 4 decimal places: cosecant 15.62°

cosecant 15.62° = 1sin15.62°

= 3.7139

6.(c) Evaluate correct to 4 decimal places: cotangent 321°23′

cotangent 321°23′ = 1 1 123tan 321 23' tan 321.38333tan 32160

= =° °°

= -1.2519

7.(a) Evaluate correct to 4 decimal places: sine 23π

Note that sine 23π means sine 2

3π radians = 0.8660

2 2 180 2rad 120 hence, sin sin1203 3 3π π ° π⎡ ⎤= × = ° ≡ °⎢ ⎥π⎣ ⎦

8.(c) Evaluate correct to 4 decimal places: cot 2.612

cot 2.612 = 1tan 2.612 rad

= -1.7083

12. Determine the acute angle 1sec 1.6214− in degrees (correct to 2 decimal places), degrees and

minutes, and in radians (correct to 3 decimal places).

Using a calculator, 1sec 1.6214− = 1 1cos1.6214

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 51.92° or 55°55′ or 51.92 rad180π

× = 0.906 rad


118

13. Determine the acute angle 1cosec 2.4891− in degrees (correct to 2 decimal places), degrees and

minutes, and in radians (correct to 3 decimal places).

Using a calculator, 1cosec 2.4891− = 1 1sin2.4891

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 23.69° or 23°41′ or 23.69 rad180π

× = 0.413 rad

16. Evaluate, correct to 4 significant figures: 11.5 tan 49 11' sin 903cos 45

° − °°

11.5 tan 49 11' sin 90 11.5 tan 49.18333 sin 90 13.315 1

3cos 45 3cos 45 2.2132° − ° °− ° −

= =° °

= 5.805

18. Evaluate, correct to 4 significant figures: 6.4cosec 29 5' sec812cot12

° − °°

1 16.4

6.4cosec 29 5' sec81 13.1665 6.39245sin 29.08333 cos8112cot12 9.409262

tan12

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟° − ° −° °⎝ ⎠ ⎝ ⎠= =° ⎛ ⎞

⎜ ⎟°⎝ ⎠

= 0.7199

20. If tan x = 1.5276, determine sec x, cosec x and cot x. (Assume x is an acute angle) If tan x = 1.5276, then x = 1tan 1.5276− = 56.79°

sec x = sec 56.79° = 1cos56.79°

= 1.8258

cosec x = cosec 56.79° = 1sin 56.79°

= 1.1952

cot x = cot 56.79° = 1tan 56.79°

= 0.6546

21. Evaluate, correct to 4 significant figures: ( )( )( )

sin 34 27 ' cos69 2 '2 tan 53 39 '° °

°

( )( )

( )( )( )sin 34 27 ' cos69 2 ' sin 34.45 cos69.03333

2 tan 53 39 ' 2 tan 53.65° ° ° °

=° °

= 0.07448


119

23. Evaluate, correct to 4 significant figures: cos ec 27 19 ' sec 45 29 '1 cosec 27 19 'sec 45 29 '

° + °− ° °

1 1

cos ec 27 19 ' sec 45 29 ' 2.179086 1.426296sin 27.31666 cos 45.483331 11 cosec 27 19 'sec 45 29 ' 1 (2.179086)(1.426296)1

sin 27.31666 cos 45.48333

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟° + ° +° °⎝ ⎠ ⎝ ⎠= =− ° ° −⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟° °⎝ ⎠⎝ ⎠

= 3.605382.10802−

= -1.710

25. Evaluate, correct to 5 significant figures: (a) cosec(-143°) (b) cot(-252°) (c) sec(-67°22′)

Using a calculator: (a) cosec(-143°) = 1sin( 143 )− °

= -1.6616

(b) cot(-252°) = 1tan( 252 )− °

= -0.32492

(c) sec(-67°22′) = ( )

1cos 67.36666− °

= 2.5985


120

EXERCISE 57 Page 126 2. Use the sine rule to solve triangle ABC, given B = 71°26′, C = 56°32′ and b = 8.60 cm, and find

its area.

Triangle ABC is shown below.

∠A = 180° - 71°26′ - 56°32′ = 52°2′

From the sine rule, 8.60 csin 71 26 ' sin 56 32 '

=° °

from which, c = 8.60sin 56 32 'sin 71 26 '

°°

= 7.568 cm

Also from the sine rule, a 8.60sin 52 2 ' sin 71 26 '

=° °

from which, a = 8.60sin 52 2 'sin 71 26 '

°°

= 7.152 cm

Area = 1 1a csin B (7.152)(7.568)sin 71 26 '2 2

= ° = 25.65 2cm

4. Use the sine rule to solve triangle DEF, given d = 32.6 mm, e = 25.4 mm and D = 104°22′, and

find its area.

Triangle DEF is shown below.

From the sine rule, 32.6 25.4

sin104 22 ' sin E=

° from which, sin E = 25.4sin104 22 '

32.6° = 0.75477555

and E = 1sin 0.75477555− = 49.0° or 49°0′ Hence, ∠F = 180° - 104°22′ - 49°0′ = 26°38′

From the sine rule, 32.6 fsin104 22 ' sin 26 38'

=° °

from which, f = 32.6sin 26 38'sin104 22 '

°°

= 15.09 mm


121

Area = 1 1d esin F (32.6)(25.4)sin 26 38'2 2

= ° = 185.6 2mm

5. Use the sine rule to solve triangle JKL, given j = 3.85 cm, k = 3.23 cm and K = 36° and find its

area.

Triangle JKL is shown below.

From the sine rule, 3.23 3.85

sin 36 sin J=

° from which, sin J = 3.85sin 36

3.23° = 0.7006109

and J = 1sin 0.7006109− = 44.476° = 44°29′ or J = 180° - 44°29′ = 135°31′ Case 1: When J = 44°29′, ∠L = 180° - 36° - 44°29′ = 99°31′

From the sine rule, l 3.23sin 99 31' sin 36

=° °

from which, l = 3.23sin 99 31'sin 36

°°

= 5.420 cm

Area = 1 1l jsin K (5.420)(3.85)sin 362 2

= ° = 6.132 2cm

Case 2: When J = 135°31′, ∠L = 180° - 135°31′ - 36° = 8°29′

From the sine rule, l 3.23sin8 29 ' sin 36

=° °

from which, l = 3.23sin8 29 'sin 36

°°

= 0.811 cm

Area = 1 1j k sin L (3.85)(3.23)sin8 29 '2 2

= ° = 0.917 2cm


122

EXERCISE 58 Page 127 2. Use the cosine and sine rules to solve triangle PQR, given q = 3.25 m, r = 4.42 m and P = 105°,

and find its area

Triangle PQR is shown below.

By the cosine rule, 2 2 2p 4.42 3.25 2(4.42)(3.25)cos105= + − ° = 19.5364 + 10.5625 – (-7.4359) = 37.5348 and p = 37.5348 = 6.127 m

From the sine rule, 6.127 4.42sin105 sin R

=°

from which, sin R = 4.42sin1056.127

° = 0.696816

and R = 1sin 0.696816− = 44.172° or 44°10′ ∠Q = 180° - 105° - 44°10′ = 30°50′

Area = 1 (4.42)(3.25)sin1052

° = 6.938 2m

4. Use the cosine and sine rules to solve triangle XYZ, given x = 21 mm, y = 34 mm and

z = 42 mm, and find its area

Triangle XYZ is shown below.

By the cosine rule, 2 2 221 42 34 2(42)(34)cos X= + −

from which, cos X = 2 2 242 34 21 0.8679972(42)(34)+ −

=


123

and ∠X = 1cos (0.867997)− = 29.77° or 29°46′

From the sine rule, 21 34sin 29 46 ' sin Y

=°

from which, sin Y = 34sin 29 46 ' 0.803807021

°=

and ∠Y = 1sin 0.8038070− = 53.495° or 53°30′ Hence, ∠Z = 180° - 29°46′ - 53°30′ = 96°44′

Area = 1 (21)(34)sin 96 44 '2

° = 355 2mm


124

EXERCISE 59 Page 129 2. Two sides of a triangular plot of land are 52.0 m and 34.0 m, respectively. If the area of the plot

is 620 2m find (a) the length of fencing required to enclose the plot, and (b) the angles of the

triangular plot.

The triangular plot of land ABC is shown below.

(a) Area = 620 = 1

2(52.0)(34.0)sin A from which, sin A = 620 0.7013571 (52.0)(34.0)

2

=

and ∠A = 1sin 0.701357− = 44.54° or 44°32′ By the cosine rule, 2 2 2BC (52.0) (34.0) 2(52.0)(34.0)cos 44.54= + − ° = 1339.677 and BC = 1339.677 = 36.6 m Hence, length of fencing required = AB + BC + CA = 52.0 + 36.6 + 34.0 = 122.6 m

(b) Area = 620 = 12

(52.0)(36.6)sin B from which, sin B = 2(620)(52.0)(36.6)

and ∠B = 1 2(620)sin(52.0)(36.6)

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 40°39′

∠A = 44°32′ hence, ∠C = 180° - 44°32′ - 40°39′ = 94°49′ 3. A jib crane is shown below. If the tie rod PR is 8.0 m long and PQ is 4.5 m long determine (a)

the length of jib RQ, and (b) the angle between the jib and the tie rod.


125

(a) Using the cosine rule on triangle PQR shown below gives: 2 2 2RQ 8.0 4.5 2(8.0)(4.5)cos130= + − ° = 130.53 and jib, RQ = 130.53 = 11.43 m = 11.4 m, correct to 3 significant figures

(b) From the sine rule, 4.5 11.43

sin R sin130=

° from which, sin R = 4.5sin130 0.3015923

11.43°=

and the angle between the jib and the tie rod, ∠R = 1sin 0.3015923− = 17.553° or 17°33′ 4. A building site is in the form of a quadrilateral as shown below, and its area is 1510 2m .

Determine the length of the perimeter of the site.

The quadrilateral is split into two triangles as shown in the diagram below.

Area = 1510 = 12

(52.4)(28.5)sin 72° + 12

(34.6)(x)sin 75°

i.e. 1510 = 710.15 + 16.71 x


126

from which, x = 1510 710.1516.71− = 47.87 m

Hence, perimeter of quadrilateral = 52.4 + 28.5 + 34.6 + 47.9 = 163.4 m 5. Determine the length of members BF and EB in the roof truss shown below.

Using the cosine rule on triangle ABF gives: 2 2 2BF 2.5 5 2(2.5)(5)cos50= + − °= 15.18 from which, BF = 15.18 = 3.9 m Using the sine rule on triangle ABF gives:

3.9 2.5sin 50 sin B

=°

from which, sin B = 2.5sin 50 0.4910543.9

°=

and ∠ABF = 1sin 0.491054− = 29.41° Assuming ∠ABE = 90°, then ∠FBE = 90° - 29.41° = 60.59° Using the sine rule on triangle BEF gives:

4 3.9sin 60.59 sin E

=°

from which, sin E = 3.9sin 60.59 0.84934994

°=

and ∠E = 1sin 0.8493499− = 58.14° Thus, ∠EFB =180° - 58.14° - 60.59° = 61.27° Using the sine rule on triangle BEF again gives:

BE 4sin 61.27 sin 60.59

=°

from which, BE = 4sin 61.27sin 60.59

°°

= 4.0 m


127

EXERCISE 60 Page 131 1. PQ and QR are the phasors representing the alternating currents in two branches of a circuit.

Phasor PQ is 20.0 A, and is horizontal. Phasor QR (which is joined to the end of PQ to form

triangle PQR) is 14.0 A and is at an angle of 35° to the horizontal. Determine the resultant

phasor PR and the angle it makes with phasor PQ.

Phasors PQ and QR are shown in the phasor diagram below.

Using the cosine rule, 2 2 2PR (20.0) (14.0) 2(20.0)(14.0)cos145= + − ° = 1054.725 from which, resultant phasor, PR = 1054.725 = 32.48 A

Using the sine rule, 14.0 32.48sin P sin145

=°

from which, sin P = 14.0sin145 0.24723132.48

°=

and ∠P = 1sin 0.247231− = 14.31° or 14°19′ 4. An idler gear, 30 mm in diameter, has to be fitted between a 70 mm diameter driving gear and a

90 mm diameter driven gear, as shown below. Determine the value of angle θ between the centre

lines.

The triangle involving angle θ is shown below, where AB = 45 mm radius + 15 mm radius = 60 mm and BC = 35 mm radius + 15 mm radius = 50 mm


128

Using the cosine rule gives: 2 2 299.78 60 50 2(60)(50)cos= + − θ

from which, 2 2 260 50 99.78cos 0.642675

2(60)(50)+ −

θ = = −

and angle, θ = 1cos ( 0.642675)− − = 130° 5. A reciprocating engine mechanism is shown below. The crank AB is 12.0 cm long and the

connecting rod BC is 32.0 cm long. For the position shown determine the length of AC and the

angle between the crank and the connecting rod.

The mechanism is shown below with the measurements marked.

Using the sine rule, 32.0 12.0

sin 40 sin C=

° from which, sin C = 12.0sin 40 0.241045

32.0°=

and ∠C = 1sin 0.241045− = 13.95° The angle between the crank and the connecting rod, ∠ABC =180° - 40° - 13.95° = 126.05° or 126°3′

Using the sine rule gives:

AC 32.0sin126.05 sin 40

=° °

from which, AC = 32.0sin126.05sin 40

°°

= 40.25 cm

Alternatively, using the cosine rule, 2 2 2AC 12.0 32.0 2(12.0)(32.0)cos126.05 1619.9611= + − ° = from which, AC = 1619.9611 = 40.25 cm


129

6. In the diagram in question 5 above, determine how far C moves, correct to the nearest millimetre

when angle CAB changes from 40° to 160°, B moving in an anticlockwise direction.

A diagram showing the position of the crank and connecting rod when angle CAB is 160° is shown below.

Using the sine rule, 32.0 12.0

sin160 sin C=

° from which, sin C = 12.0sin160 0.1282576

32.0°=

and ∠C = 1sin 0.1282576− = 7.37° Hence, ∠AB′C′ =180° - 7.37° - 160° = 12.63° Using the sine rule again gives:

AC' 32.0sin12.63 sin160

=° °

from which, AC′ = 32.0sin12.63sin160

°°

= 20.46 cm

Hence, the distance that C moves, i.e. CC′ = AC - AC′ = 40.25 – 20.46 = 19.8 cm

8. An aeroplane is sighted due east from a radar station at an elevation of 40° and a height of

8000 m, and later at an elevation of 35° and height 5500 m in a direction E 70° S. If it is

descending uniformly, find the angle of descent. Determine also the speed of the aeroplane in

km/h if the time between the two observations is 45 s.

From the sketch of the aeroplanes flight shown below.


130

Tan 40° = 8000OA

from which, OA = 8000tan 40°

= 9534.03 m

Tan 35° = 5500OB

from which, OB = 5500tan 35°

= 7854.81 m

From the cosine rule, 2 2 2AB (9534.03) (7854.81) 2(9534.03)(7854.81)cos 70= + − ° from which, AB = 10068.24 m From the view shown below, XY = 8000 – 5500 = 2500 m

Hence, angle of descent, θ = 1 2500tan10068.24

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 13.95° or 13°57′

Flight distance of aeroplane between observations, XZ = ( )2 210068.24 (2500)⎡ ⎤+⎣ ⎦

= 10373.98 m or 10.374 km

Hence, speed of descent = dis tan ce 10.374 km 10.374 60 6045time 45h

60 60

× ×= =

×

km/h = 829.9 km/h


131

CHAPTER 13 CARTESIAN AND POLAR CO-ORDINATES

EXERCISE 61 Page 134 2. Express (6.18, 2.35) as polar co-ordinates, correct to 2 decimal places, in both degrees and

radians. From the diagram, 2 2r 6.18 2.35= + = 6.61

and 1 1y 2.35tan tanx 6.18

− −θ = = = 20.82° or 20.82180π

× rad = 0.36 rad

Hence, (6.18, 2.35) in Cartesian co-ordinates corresponds to (6.61, 20.82°) or (6.61, 0.36 rad) in polar co-ordinates. 4. Express (-5.4, 3.7) as polar co-ordinates, correct to 2 decimal places, in both degrees and


and 1 1y 3.7tan tanx 5.4

− −α = = = 34.42°

Thus, θ = 180° - α = 145.58° or 145.58180π

× rad = 2.54 rad

Hence, (-5.4, 3.7) in Cartesian co-ordinates corresponds to (6.55, 145.58°) or (6.55, 2.54 rad) in polar co-ordinates.


132

5. Express (-7, -3) as polar co-ordinates, correct to 2 decimal places, in both degrees and

radians. From the diagram, 2 2r 7 3= + = 7.62

and 1 3tan7

−α = = 23.20°

Thus, θ = 180° + 23.20° = 203.20° or 203.20180π

× rad = 3.55 rad

Hence, (-7, -3) in Cartesian co-ordinates corresponds to (7.62, 203.20°) or (7.62, 3.55 rad) in polar co-ordinates. 8. Express (9.6, -12.4) as polar co-ordinates, correct to 2 decimal places, in both degrees and


and 1 12.4tan9.6

−α = = 52.25°

Thus, θ = 360° - 52.25° = 307.75° or 307.75180π

× rad = 5.37 rad

Hence, (9.6, -12.4) in Cartesian co-ordinates corresponds to (15.68, 307.75°) or (15.68, 5.37 rad) in polar co-ordinates.


133

EXERCISE 62 Page 136 1. Express (5, 75°) as Cartesian co-ordinates, correct to 3 decimal places.

In the diagram, x = 5 cos 75° = 1.294 and y = 5 sin 75° = 4.830 Hence, (5, 75°) in polar form corresponds to (1.294, 4.830) in Cartesian form. 4. Express (3.6, 2.5 rad) as Cartesian co-ordinates, correct to 3 decimal places. x = 3.6 cos 2.5 rad = -2.884 y = 3.6 sin 2.5 rad = 2.154 Hence, (3.6, 2.5 rad) in polar form corresponds to (-2.884, 2.154) in Cartesian form. 5. Express (10.8, 210°) as Cartesian co-ordinates, correct to 3 decimal places. x = 10.8 cos 210° = -9.353 y = 10.8 sin 210° = -5.400 Hence, (10.8, 210°) in polar form corresponds to (-9.353, -5.400) in Cartesian form. 8. Express (6, 5.5 rad) as Cartesian co-ordinates, correct to 3 decimal places. x = 6 cos 5.5 rad = 4.252 y = 6 sin 5.5 rad = -4.233 Hence, (6, 5.5 rad) in polar form corresponds to (4.252, -4.233) in Cartesian form. 9. The diagram below shows 5 equally spaced holes on an 80 mm pitch circle diameter. Calculate

their co-ordinates relative to axes 0x and 0y in (a) polar form, (b) Cartesian form.

Calculate also the shortest distance between the centres of two adjacent holes.


134

(a) In the diagram below, hole A is at an angle of 90° Hence, in polar form, hole A is 40∠90°.

The holes will be equally displaced, 3605° i.e. 72° apart.

Thus, in polar form the holes are at 40∠90°, 40∠(90° + 72°), i.e. 40∠162°, 40∠(162° + 72°), i.e. 40∠234°, 40∠(234° + 72°), i.e. 40∠306°, and 40∠(306° + 72°), i.e. 40∠378° or 40∠18°. Summarising, the holes are at 40∠18°, 40∠90°, 40∠162°, 40∠234°, 40∠306°

(b) 40∠18° = (40 cos 18°, 40 sin 18°) = (38.04 + j12.36) in Cartesian form 40∠90° = (40 cos 90°, 40 sin 90°) = (0 + j40) in Cartesian form 40∠162° = (40 cos 162°, 40 sin 162°) = (-38.04 + j12.36) in Cartesian form 40∠234° = (40 cos 234°, 40 sin 234°) = (-23.51 – j32.36) in Cartesian form 40∠306° = (40 cos 306°, 40 sin 306°) = (23.51 - j32.36) in Cartesian form In triangle ABC in the above diagram, AC = 40 – 12.36 = 27.64, and BC = 38.04

Thus, by Pythagoras’ theorem, AB = ( )2 227.64 38.04+ = 47.02 mm

i.e. the shortest distance between the centres of two adjacent holes is 47.02 mm.


135

CHAPTER 14 THE CIRCLE AND ITS PROPERTIES

EXERCISE 63 Page 138 1. If the radius of a circle is 41.3 mm, calculate the circumference of the circle. Circumference, c = 2πr = 2π(41.3) = 259.5 mm 2. Find the diameter of a circle whose perimeter is 149.8 cm. If perimeter, or circumference, c = πd, then 149.8 = πd

and diameter, d = 149.8π

= 47.68 cm

3. A crank mechanism is shown below, where XY is a tangent to the circle at point X. If the circle

radius 0X is 10 cm and length 0Y is 40 cm, determine the length of the connecting rod XY.

If XY is a tangent to the circle, then ∠0XY = 90° Thus, by Pythagoras, 2 2 20Y 0X XY= +

from which, 0Y = ( )2 2 2 20Y 0X 40 10 1500− = − = = 38.73 cm


136

EXERCISE 64 Page 139 1. Convert to radians in terms of π: (a) 30° (b) 75° (c) 225°

(a) 30° = 30180π

× rad = 6π rad

(b) 75° = 75180π

× rad = 512π rad

(c) 225° = 225180π

× rad = 45 15rad rad36 12π π

= = 54π rad

2. Convert to radians: (a) 48° (b) 84°51′ (c) 232°15′

(a) 48° = 48180π

× rad = 0.838 rad

(b) 84°51′ = 5184 84.9560 180 180

π π⎛ ⎞× = ×⎜ ⎟⎝ ⎠

rad = 1.481 rad

(c) 232°15′ = 232.25180π

× rad = 4.054 rad

3. Convert to degrees: (a) 5 rad6π (b) 4 rad

9π (c) 7 rad

12π

(a) 5 rad6π = 5 180

6π °×

π = 5 × 30 = 150°

(b) 4 rad9π = 4 180

9π °×

π = 4 × 20 = 80°

(c) 7 rad12π = 7 180

12π °×

π = 7 × 15 = 105°

4. Convert to degrees and minutes: (a) 0.0125 rad (b) 2.69 rad (c) 7.241 rad

(a) 0.0125 rad = 1800.0125 °×

π = 0.716° or 0°43′


137

(b) 2.69 rad = 1802.69 °×

π = 154.126° or 154°8′

(c) 7.241 rad = 1807.241 °×

π = 414.879° or 414°53′


138

EXERCISE 65 Page 140 2. If the angle subtended at the centre of a circle of diameter 82 mm is 1.46 rad, find the lengths of

the (a) minor arc, (b) major arc

If diameter d = 82 mm, radius r = 822

= 41 mm

(a) Minor arc length, s = rθ = (41)(1.46) = 59.86 mm (b) Major arc length = circumference – minor arc = 2π(41) – 59.86 = 257.61 – 59.86 = 197.8 mm 3. A pendulum of length 1.5 m swings through an angle of 10° in a single swing. Find, in

centimetres, the length of the arc traced by the pendulum bob.

Arc length of pendulum bob, s = rθ = (1.5) 10180π⎛ ⎞×⎜ ⎟

⎝ ⎠ = 0.262 m or 26.2 cm

5. Determine the angle of lap, in degrees and minutes, if 180 mm of a belt drive are in contact with

a pulley of diameter 250 mm.

Arc length, s = 180 mm, radius, r = 2502

= 125 mm

Since s = rθ, the angle of lap, θ = s 180r 125= = 1.44 rad = 1801.44×

π = 82.506° or 82°30′

6. Determine the number of complete revolutions a motorcycle wheel will make in travelling 2 km,

if the wheel’s diameter is 85.1 cm If wheel diameter = 85.1 cm, then circumference, c = πd = π(85.1) cm = 267.35 cm = 2.6735 m

Hence, number of revolutions of wheel in travelling 2000 m = 20002.6735

= 748.08

Thus, number of complete revolutions = 748


139

8. Determine (a) the shaded area in the diagram below, (b) the percentage of the whole sector that

the area of the shaded portion represents.

(a) Shaded area = 2 21 1(50) (0.75) (38) (0.75)2 2

− = 2 21 (0.75) 50 382

⎡ ⎤−⎣ ⎦ = 396 2mm

(b) Percentage of whole sector = 2

396 100%1 (50) (0.75)2

× = 42.24%


140

EXERCISE 66 Page 142 1. Determine the radius, and the co-ordinates of the centre of the circle given by the equation

2 2x y 6x 2y 26 0+ + − − =

Method 1: The general equation of a circle 2 2 2(x a) (y b) r− + − = is 2 2x y 2ex 2fy c 0+ + + + =

where co-ordinate, a = - 2e2

, co-ordinate, b = - 2f2

and radius, r = 2 2a b c+ −

Hence, if 2 2x y 6x 2y 26 0+ + − − = then a = - 2e 62 2= − = -3, b = - 2f 2

2 2−

= − = 1

and radius, r = 2 2( 3) (1) ( 26) (9 1 26) 36⎡ ⎤− + − − = + + =⎣ ⎦ = 6

i.e. the circle 2 2x y 6x 2y 26 0+ + − − = has centre at (-3, 1) and radius 6, as shown below.

Method 2: 2 2x y 6x 2y 26 0+ + − − = may be rearranged as: 2 2(x 3) (y 1) 36 0+ + − − = i.e. 2 2 2(x 3) (y 1) 6+ + − = which has a radius of 6 and centre at (-3, 1)

2. Sketch the circle given by the equation 2 2x y 6x 4y 3 0+ − + − = Method 1: The general equation of a circle 2 2 2(x a) (y b) r− + − = is 2 2x y 2ex 2fy c 0+ + + + =

where co-ordinate, a = - 2e2

, co-ordinate, b = - 2f2

and radius, r = 2 2a b c+ −

Hence, if 2 2x y 6x 4y 3 0+ − + − = then a = - 2e 62 2

−= − = 3, b = - 2f 4

2 2= − = -2


141

and radius, r = 2 2(3) ( 2) ( 3) (9 4 3) 16⎡ ⎤+ − − − = + + =⎣ ⎦ = 4

i.e. the circle 2 2x y 6x 4y 3 0+ − + − = has centre at (3, -2) and radius 4, as shown below.

Method 2: 2 2x y 6x 4y 3 0+ − + − = may be rearranged as: 2 2(x 3) (y 2) 16 0− + + − = i.e. 2 2 2(x 3) (y 2) 4− + + = which has a radius of 4 and centre at (3, -2)

4. Sketch the curve 2yx 6 1

6⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

If 2yx 6 1

6⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦ then

2x y16 6

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

and 2 2x y1

6 6⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

i.e. 2 2

2 2

x y 16 6

+ = and 2 2 2x y 6+ =

which is a circle of radius 6 and co-ordinates of centre at (0, 0), as shown below.


142

EXERCISE 67 Page 143 2. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm.

Determine the linear velocity of a point on the rim of one of the wheels of the bicycle, and the

angular velocity of the wheels.

Linear velocity, v = 36 km/h = 36 10003600× m/s = 10 m/s

(Note that changing from km/h to m/s involves dividing by 3.6)

Radius of wheel, r = 5002

= 250 mm = 0.25 m

Since, v = ωr, then angular velocity, ω = v 10r 0.25= = 40 rad/s

3. A train is travelling at 108 km/h and has wheels of diameter 800mm.

(a) Determine the angular velocity of the wheels in both rad/s and rev/min.

(b) If the speed remains constant for 2.70 km, determine the number of revolutions made by a

wheel, assuming no slipping occurs.

(a) Linear velocity, v = 108 km/h = 1083.6

m/s = 30 m/s

Radius of wheel = 8002

= 400 mm = 0.4 m

Since, v = ωr, then angular velocity, ω = v 30r 0.4= = 75 rad/s

rev 60s75rad / s 75rad / s2 rad min

= × ×π

= 716.2 rev/min

(b) Linear velocity, v = st

hence, time, t = s 2700mv 30m / s= = 90 s = 90

60 = 1.5 minutes

Since a wheel is rotating at 716.2 rev/min, then in 1.5 minutes it makes 716.2 rev/min × 1.5 min = 1074 rev/min


143

EXERCISE 68 Page 145 2. Calculate the centripetal force acting on a vehicle of mass 1 tonne when travelling around a bend

of radius 125 m at 40 km/h. If this force should not exceed 750 N, determine the reduction in

speed of the vehicle to meet this requirement.

Centripetal force = 2mv

r where mass, m = 1 tonne = 1000 kg, radius, r = 125 m and

velocity, v = 40 km/h = 403.6

m/s

Hence, centrifugal force =

240(1000)3.6

125

⎛ ⎞⎜ ⎟⎝ ⎠ = 988 N

If centrifugal force is limited to 750 N, then 2(1000)v750

125=

from which, velocity, v = (750)(125)1000

⎛ ⎞⎜ ⎟⎝ ⎠

= 9.6825 m/s

= 9.6825 × 3.6 = 34.89 km/h

Hence, reduction in speed is 40 – 34.89 = 5.14 km/h

3. A speed-boat negotiates an S-bend consisting of two circular arcs of radii 100 m and 150 m. If

the speed of the boat is constant at 34 km/h, determine the change in acceleration when leaving

one arc and entering the other.

Speed of the boat, v = 34 km/h = 343.6

m/s

For the first bend of radius 100 m, acceleration =

2

22

1

34v 3.6 0.892 m / sr 100

⎛ ⎞⎜ ⎟⎝ ⎠= =

For the second bend of radius 150 m, acceleration =

2

2

343.6 0.595m / s150

⎛ ⎞⎜ ⎟⎝ ⎠= − = − , the negative sign

indicating a change in direction Hence, change in acceleration is 0.892 – (-0.595) = 1.49 2m / s


144

CHAPTER 15 TRIGONOMETRIC WAVEFORMS

EXERCISE 69 Page 151 2. Determine the angles between 0° and 360° whose cosecant is 2.5317 Cosecant, and thus sine, is positive in the 1st and 2nd quadrants.

If cosec θ = 2.5317, then 1 1 1cos ec (2.5317) sin 23.265 or 23 16 '2.5317

− − ⎛ ⎞θ = = = ° °⎜ ⎟⎝ ⎠

.

From the diagram, the two values of θ between 0° and 360° are: 23°16′ and 180° - 23°16′ = 156°44′ 4. Solve the equation: 1cos ( 0.5316) t− − = Cosine is negative in the 2nd and 3rd quadrants.

1cos (0.5316)− = 57.886° or 57°53′ as shown in the diagram below.

From the diagram, t = 180° - 57°53′ = 122°7′ and t = 180° + 57°53′ = 237°53′ 6. Solve the equation: 1tan 0.8314− = θ


145

Tangent is positive in the 1st and 3rd quadrants.

1tan 0.8314−θ = = 39.74° or 39°44′

From the diagram, the two values of θ between 0° and 360° are: 39°44′ and 180° + 39°44′ = 219°44′


146


2. State the amplitude and period of the waveform y = 2 5xsin2

and sketch the curve between 0°

and 360°.

If y = 2 5xsin2

, amplitude = 2 and period = 36052

° = 144°

A sketch y = 2 5xsin2

is shown below.

4. State the amplitude and period of the waveform y = 3 cos2θ and sketch the curve between 0°

and 360°.

If y = 3 cos2θ , amplitude = 3 and period = 360

12

° = 720°

A sketch y = 3 cos2θ is shown below.

6. State the amplitude and period of the waveform y = 6 sin(t - 45°) and sketch the curve between 0° and 360°.


147

If y = 6 sin(t - 45°), amplitude = 6 and period = 3601° = 360°

A sketch y = 6 sin(t - 45°) is shown below.

7. State the amplitude and period of the waveform y = 4 cos(2θ + 30°) and sketch the curve between 0° and 360°.

If y = 4 cos(2θ + 30°), amplitude = 4 and period = 3602° = 180°

A sketch y = 4 cos(2θ + 30°) is shown below.

(Note that y = 4 cos(2θ + 30°) leads y = 4 cos 2θ by 302° = 15°)

9. State the amplitude and period of the waveform y = 5 2 3cos2θ and sketch the curve between 0°

and 360°.

If y = 5 2 3cos2θ , amplitude = 5 and period = 180

32

° = 120°


148

A sketch y = 5 2 3cos2θ is shown below.


149

EXERCISE 71 Page 159 1. Find the amplitude, periodic time, frequency and phase angle (stating whether it is leading or

lagging A sin ωt) of the alternating quantity: i = 40 sin(50πt + 0.29) mA If i = 40 sin(50πt + 0.29) mA, then amplitude = 40 mA,

ω = 50π rad/s = 2πf from which, frequency, f = 502ππ

= 25 Hz,

periodic time, T = 1 1f 25= = 0.040 s or 40 ms,

and phase angle = 0.29 rad leading or 1800.29 °×

π = 16.62° leading or 16°37′ leading.

4. A sinusoidal voltage has a maximum value of 120 V and a frequency of 50 Hz. At time t = 0, the

voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form

v = A sin(ωt ± α). Let v = A sin(ωt ± α) = 120 sin(2πft + φ) = 120 sin(100πt + φ) volts, since f = 50 Hz. (a) When t = 0, v = 0 hence, 0 = 120 sin(0 + φ), i.e. 0 = 120 sin φ from which, sin φ = 0 and φ = 0 Hence, if v = 0 when t = 0, then v = 120 sin 100πt volts (b) When t = 0, v = 50 V hence, 50 = 120 sin(0 + φ)

from which, 50 sin120

= φ and 1 50sin 24.624 24.624 0.43rad120 180

− π⎛ ⎞φ = = °= × =⎜ ⎟⎝ ⎠

Hence, if v = 50 when t = 0, then v = 120 sin(100πt + 0.43)volts 5. An alternating current has a periodic time of 25 ms and a maximum value of 20 A. When time

t = 0, current i = -10 amperes. Express the current i in the form i = A sin(ωt ± α).

If periodic time T = 25 ms, then frequency, 3

1 1fT 25 10−= =

× = 40 Hz

Angular velocity, ω = 2πt =2π(40) = 80π rad/s


150

Hence, current i = 20 sin(80πt + φ) When t = 0, i = -10, hence -10 = 20 sin φ

from which, sin φ = 10 0.520

− = − and 1sin ( 0.5) 30 or rad6

− πφ = − = − ° −

Thus, i 20sin 80 t A6π⎛ ⎞= π −⎜ ⎟

⎝ ⎠

7. The current in a.c. circuit at any time t seconds is given by: i = 5 sin(100πt – 0.432) amperes.

Determine (a) the amplitude, periodic time, frequency and phase angle (in degrees), (b) the value

of current at t = 0, (c) the value of current at t = 8 ms, (d) the time when the current is first a

maximum, (e) the time when the current first reaches 3 A. Sketch one cycle of the waveform

showing relevant points. (a) If i = 5 sin(100πt - 0.432) mA, then amplitude = 5 A,

ω = 100π rad/s = 2πf from which, frequency, f = 1002ππ

= 50 Hz,

periodic time, T = 1 1f 50= = 0.020 s or 20 ms,

and phase angle = 0.432 rad lagging or 1800.432 °×

π = 24.75° lagging or 24°45′ lagging.

(b) When t = 0, i = 5 sin(- 0.432) = -2.093 A (note that -0.432 is radians) (c) When t = 8 ms, i = 5 ( )3sin 100 8 10 0.432−⎡ ⎤π × −⎣ ⎦ = 5 sin (2.081274) = 4.363 A

(d) When the current is first a maximum, 5 = 5 sin(100πt – 0.432) i.e. 1 = sin(100πt – 0.432) and 100πt – 0.432 = 1sin 1 1.5708− = (again, be sure your calculator is on radians)

from which, time t = 1.5708 0.432100+π

= 0.006375 s or 6.375 ms

(e) When i = 3 A, 3 = 5 sin(100πt – 0.432)

i.e. 35

= sin(100πt – 0.432)


151

and 100πt – 0.432 = 1 3sin 0.64355

− =

from which, time t = 0.6435 0.432100+π

= 0.003423 s or 3.423 ms

A sketch of one cycle of the waveform is shown below.

Note that since phase angle φ = 24.75°, in terms of time tφ then

t24.75360 20

φ≡ from which, tφ = 1.375 ms

Alternatively, tφ = 0.432100

φ=

ω π = 1.375 ms, as shown in the sketch.


152

EXERCISE 72 Page 165 1. A complex current waveform i comprises a fundamental current of 50 A r.m.s. and frequency

100 Hz, together with a 24% third harmonic, both being in phase with each other at zero time.

(a) Write down an expression to represent current i. (b) Sketch the complex waveform of current

using harmonic synthesis over one cycle of the fundamental.

(a) Fundamental current: r.m.s. = 12

× maximum value

from which, maximum value = 2 r.m.s. 2 50× = × = 70.71 A Hence, fundamental current is: 1i = 70.71 sin 2π(100)t = 70.71 sin 628.3t A Third harmonic: amplitude = 24% of 70.71 = 16.97 A Hence, third harmonic current is: 3i = 16.97 sin 3(628.3)t = 16.97 sin 1885t A Thus, current i = 1i + 3i = 70.71 sin 628.3t + 16.97 sin 1885t amperes. (b) The complex waveform for current i is shown sketched below:

2. A complex voltage waveform v is comprised of a 212.1 V r.m.s. fundamental voltage at a

frequency of 50 Hz, a 30% second harmonic component lagging by π/2 rad, and a 10% fourth

harmonic leading by π/3 rad. (a) Write down an expression to represent voltage v. (b) Sketch the

complex voltage waveform using harmonic synthesis over one cycle of the fundamental

waveform.


153

(a) Voltage,

v = 212.1 212.1 212.1sin 2 (50)t (0.30) sin 2(2 )(50)t (0.1) sin 4(2 )(50)t0.707 0.707 2 0.707 3

π π⎡ ⎤ ⎡ ⎤π + π − + π +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ volts

i.e. v = 300 sin 314.2 t + 90sin 628.3t 30sin 1256.6t2 3π π⎛ ⎞ ⎛ ⎞− + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ volts

(b) The complex waveform representing v is shown sketched below:

5. A voltage waveform is described by:

v = 200 sin 377t + 80sin 1131t 20sin 1885t4 3π π⎛ ⎞ ⎛ ⎞+ + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠volts

Determine (a) the fundamental and harmonic frequencies of the waveform, (b) the percentage

third harmonic, and (c) the percentage fifth harmonic. Sketch the voltage waveform using

harmonic synthesis over one cycle of the fundamental.


154

(a) From the fundamental voltage, 377 = 1 12 fω = π i.e. fundamental frequency, 1377f2

=π

= 60 Hz

From the 3rd harmonic voltage, 1131 = 3 32 fω = π

i.e. 3rd harmonic frequency, 31131f2

=π

= 180Hz

From the 5th harmonic voltage, 1885 = 5 52 fω = π

i.e. 5th harmonic frequency, 51885f

2=

π = 300Hz

(b) Percentage 3rd harmonic = 80 100%200

× = 40%

(c) Percentage 5th harmonic = 20 100%200

× = 10%

The complex waveform representing v is shown sketched below:

© 2006 John Bird. All rights reserved. Published by Elsevier. 155

CHAPTER 16 TRIGONOMETRIC IDENTITIES AND EQUATIONS


2. Prove the trigonometric identity: ( )2

1 cos ec1 cos

= θ− θ

L.H.S. = ( ) 22

1 1sin1 cos

=θ− θ

(since 2 2sin cos 1θ+ θ = )

= 1sinθ

= cosec θ = R.H.S.

4. Prove the trigonometric identity: 3cos x cos x sin x cos x

sin x−

=

L.H.S. = 3 2 2cos x cos x cos x(1 cos x) cos x sin x

sin x sin x sin x− −

= = = cos x sin x = sin x cos x = R.H.S.

5. Prove the trigonometric identity: ( ) ( )2 2 21 cot 1 cot 2cos ec+ θ + − θ = θ L.H.S. = ( ) ( )2 2 2 21 cot 1 cot 1 2cot cot 1 2cot cot+ θ + − θ = + θ+ θ+ − θ+ θ

= ( )2 22 2cot 2 2 cos ec 1+ θ = + θ−

= 2 22 2cos ec 2 2cos ec+ θ− = θ = R.H.S.

6. Prove the trigonometric identity: ( )2sin x sec x cos ec x1 tan x

cos x tan x+

= +

L.H.S. = ( )2 2

21 1 sin x cos xsin x sin xsin x sec x cos ec x cos x sin x cos x sin x

sin xcos x tan x sin xcos xcos x

+⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟+ ⎝ ⎠ ⎝ ⎠= =⎛ ⎞⎜ ⎟⎝ ⎠

= sin x cos x sin x cos xsin xcos x sin x cos x

+ +⎛ ⎞ =⎜ ⎟⎝ ⎠

= sin x cos x tan x 1 1 tan xcos x cos x

+ = + = + = R.H.S.


EXERCISE 74 Page 169 2. Solve: 3 cosec A + 5.5 = 0 for angles between 0° and 360°

Since 3 cosec A + 5.5 = 0 then 3 cosec A = -5.5 and cosec A = 5.53

−

i.e. 1 5.5sin A 3

= − or sin A = 35.5

−

from which, A = 1 3sin5.5

− ⎛ ⎞−⎜ ⎟⎝ ⎠

= -33.056° or -33°3′

Since sine is negative, the angle 33°3′ occurs in the 3rd and 4th quadrants as shown in the diagram

below.

Hence, the two angles for A between 0° and 360° whose sine is 3

5.5− are:

180° + 33°3′ = 213°3′ and 360° - 33°3′ = 326°57′ 3. Solve: 4(2.32 – 5.4 cot t) = 0 for angles between 0° and 360° Since 4(2.32 – 5.4 cot t) = 0 then 2.32 – 5.4 cot t = 0 and 2.32 = 5.4 cot t

i.e. cot t = 2.325.4

from which, tan t = 5.42.32

Hence, t = 1 5.4tan 66.75 or 66 45'2.32

− ⎛ ⎞ = ° °⎜ ⎟⎝ ⎠

Since tan is positive, the angle 66°45′ occurs in the 1st and 3rd quadrants as shown in the diagram

below.


Hence, the two angles for t between 0° and 360° whose tan is 5.4

2.32 are:

66°45′ and 180° + 66°45′ = 246°45′


EXERCISE 75 Page 170 1. Solve: 25sin y 3= for angles between 0° and 360°

Since 25sin y 3= then 2 3sin y 0.605

= = and sin y 0.60 0.774596...= = ±

and y = 1sin (0.774596...) 50 46 '− = ° Since sine y is both positive and negative, a value for y occurs in each of the four quadrants as shown in the diagram below.

Hence the values of y between 0° and 360° are: 50°46′, 180° - 50°46′ = 129°14′, 180° + 50°46′ = 230°46′ and 360° - 50°46′ = 309°14′ 2. Solve: 25 3cos ec D 8+ = for angles between 0° and 360° Since 25 3cosec D 8+ = then 23cosec D 8 5 3= − = i.e. 2cos ec D 1=

Hence, 2

1 1sin D

= and 2sin D 1= from which, sin D = 1 1= ±

and D = ( )1sin 1− ±

There are two values of D between 0° and 360° which satisfy this equation, as shown in the

sinusoidal waveform below

Hence, D = 90° and 270°


EXERCISE 76 Page 170 2. Solve: 28 tan 2 tan 15θ+ θ = for angles between 0° and 360° Since 28 tan 2 tan 15θ+ θ = then 28 tan 2 tan 15 0θ+ θ− = i.e. ( )( )4 tan 5 2 tan 3 0θ− θ+ =

i.e. 4 tan θ - 5 = 0 from which, tan θ = 5 1.254= and 1tan 1.25−θ = = 51°20′

and 2 tan θ + 3 = 0 from which, tan θ = 3 1.52

− = − and 1tan 1.5−θ = − = -56°19′

From the diagram below, the four values of θ between 0° and 360° are: 51°20′, 180° - 56°19′ = 123°41′, 180° + 51°20′ = 231°20′ and 360° - 56°19′ = 303°41′

3. Solve: 22cosec t 5cosec t 12− = for angles between 0° and 360° Since 22cosec t 5cosec t 12− = then 22cosec t 5cosec t 12 0− − = and (2 cosec t + 3)(cosec t – 4) = 0

i.e. 2 cosec t + 3 = 0 from which, cosec t = 32

− and sin t = 23

− from which, t = -41°49′

and cosec t - 4 = 0 from which, cosec t = 4 and sin t = 14

from which, t = 14°29′

From the diagram below, the four values of θ between 0° and 360° are:


14°29′, 180° - 14°29′ = 165°31′, 180° + 41°49′ = 221°49′ and 360° - 41°49′ = 318°11′


EXERCISE 77 Page 171 1. Solve: 212sin 6 cosθ− = θ for angles between 0° and 360° Since 212sin 6 cosθ− = θ then ( )212 1 cos 6 cos− θ − = θ

i.e. 212 12cos 6 cos− θ− = θ

i.e. 212cos cos 6 0θ+ θ− =

Factorising gives: (4 cos θ + 3)(3 cos θ - 2) = 0

i.e. 4 cos θ + 3 = 0 from which, cos θ = 3 0.754

− = − and θ = 1cos ( 0.75)− − = -41°25′

and 3 cos θ - 2 = 0 from which, cos θ = 23

and θ = 1 2cos3

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 48°11′

From the diagram below, the four values of θ between 0° and 360° are: 48°11′, 180° - 41°25′ = 138°35′, 180° + 41°25′ = 221°25′ and 360° - 48°11′ = 311°49′

3. Solve: 24cot A 6cosecA 6 0− + = for angles between 0° and 360° Since 24cot A 6cosecA 6 0− + = then ( )24 cosec A 1 6cosecA 6 0− − + =

and 24cosec A 6cosecA 2 0− + =

Factorising gives: (2 cosec A – 1) (2 cosec A – 2) = 0

i.e. 2 cosec A - 1 = 0 from which, cosec A = 12= and sin A = 2 which has no solutions

and 2 cosec A - 2 = 0 from which, cosec A = 1 and sin A = 1, which has only one solution

between 0° and 360°, i.e. A = 90°


5. Solve: 22.9cos a 7sin a 1 0− + = for angles between 0° and 360° Since 22.9cos a 7sin a 1 0− + = then 22.9(1 sin a) 7sin a 1 0− − + =

i.e. 22.9 2.9sin a 7sin A 1 0− − + =

and 22.9sin a 7sin a 3.9 0+ − =

Hence, sin a = 27 7 4(2.9)( 3.9) 7 94.242(2.9) 5.8

⎡ ⎤− ± − − − ±⎣ ⎦ = = 0.46685 or -2.88064, which has

no solution

Thus, a = 1sin (0.46685)− = 27°50′

and, from the diagram below, 180° - 27°50′ = 152°10′


CHAPTER 17 THE RELATIONSHIP BETWEEN

TRIGONOMETRIC AND HYPERBOLIC FINCTIONS EXERCISE 78 Page 174 2. Verify the following identity by expressing in exponential form:

cos j(A – B) = cos jA cos jB + sin jA sin jB

L.H.S. = cos j(A – B) = ch(A – B) from (5), page 173

R.H.S. = cos jA cos jB + sin jA sin jB

= ch A ch B + j sh A j sh B from (5) and (6), page 173

= ch A ch B + 2j sh A sh B

= ch A ch B – sh A ah B

= ch(A – B)

Hence, L.H.S. = R.H.S. i.e. cos j(A – B) = cos jA cos jB + sin jA sin jB 3. Verify the following identity by expressing in exponential form:

cos j2A = 1 – 2 2sin jA

L.H.S. = cos j2A = ch 2A from (5), page 173

R.H.S. = 1 – 2 2sin jA = 1 – 2(sin jA)(sin jA) = 1 – 2(j sh A)(j sh A)

= 1 - 2 2 2 2j sh A 1 2sh A= +

= ch 2A from Table 5.1, page 45

Hence, L.H.S. = R.H.S. i.e. cos j2A = 1 – 2 2sin jA 5. Verify the following identity by expressing in exponential form:

sin jA – sin jB = A B A B2cos j sin j2 2+ −⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

L.H.S. = sin jA – sin jB = j sh A – j sh B = j(sh A – sh B)

R.H.S. = A B A B2cos j sin j2 2+ −⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= A B A B2c h j s h2 2+ −⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= A B A B2 j c h s h2 2+ −⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


= A B A B A B A B2j ch ch sh sh sh ch ch sh2 2 2 2 2 2 2 2

⎡ ⎤ ⎡ ⎤+ −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

=

A B A B A B A B A B A Bch ch sh ch ch ch ch sh sh sh sh ch2 2 2 2 2 2 2 2 2 2 2 22j

A B A Bsh sh ch sh2 2 2 2

⎡ ⎤− +⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

= 2 2 2 2A A B A B B A B B B A A2j ch sh ch ch ch sh sh sh ch sh sh ch2 2 2 2 2 2 2 2 2 2 2 2

⎡ ⎤− + −⎢ ⎥⎣ ⎦

= 2 2 2 2A A B B B B A A2j sh ch ch sh sh ch sh ch2 2 2 2 2 2 2 2

⎡ ⎤⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= 2 2 2 2A A B B B B A A2j sh ch ch sh sh ch ch sh2 2 2 2 2 2 2 2

⎡ ⎤⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= ( ) ( )A A B B2j sh ch 1 sh ch 12 2 2 2

⎡ ⎤−⎢ ⎥⎣ ⎦ from Table 5.1, page 45

= A A B Bj 2sh ch 2sh ch2 2 2 2

⎡ ⎤−⎢ ⎥⎣ ⎦

= A Bj sh 2 sh 22 2

⎡ ⎤⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ from Table 5.1, page 45

= j [ sh A – sh B ] = L.H.S


EXERCISE 79 Page 175 1. Use the substitution A = jθ to obtain the hyperbolic identity corresponding to the trigonometric identity: 2 21 tan A sec A+ = Let A = jθ then 2 21 tan A sec A+ =

becomes: 1 + ( ) ( )2 2tan j sec jθ = θ

i.e. ( ) ( )2 21 tan j sec j+ θ = θ

i.e. ( )2

2 11 j tanhcos j

⎛ ⎞+ θ = ⎜ ⎟θ⎝ ⎠

i.e. 2

2 2 11 j tanhch

⎛ ⎞+ θ = ⎜ ⎟θ⎝ ⎠

i.e. 2 21 tanh sec h− θ = θ 3. Use the substitution A = jθ and B = jφ to obtain the hyperbolic identity corresponding to the trigonometric identity: sin(A – B) = sin A cos B – cos A sin B Substituting A = jθ and B = jφ in sin(A – B) = sin A cos B – cos A sin B gives: sin(jθ – jφ) = sin jθ cos jφ – cos jθ sin jφ i.e. j sinh (θ – φ) = j sinh θ cosh φ – j cosh θ sinh φ i.e. j sinh (θ – φ) = j (sinh θ cosh φ – cosh θ sinh φ) i.e. sinh (θ – φ) = sinh θ cosh φ – cosh θ sinh φ 4. Use the substitution A = jθ to obtain the hyperbolic identity corresponding to the

trigonometric identity: 2

2 tan Atan 2A1 tan A

=−

Substituting A = jθ in 2

2 tan Atan 2A1 tan A

=−

gives: 2

2 tan jtan 2j1 tan j

θθ =

− θ


i.e. ( )2 2 2

2 j tanh 2jtanhj tanh 21 j tanh1 jtanh

θ θθ = =

− θ− θ

i.e. 2

2 j tanhj tanh 21 tanh

θθ =

+ θ

i.e. 2

2 tanhtanh 21 tanh

θθ =

+ θ

6. Use the substitution A = jθ to obtain the hyperbolic identity corresponding to the

trigonometric identity: 3 3 1sin A sin A sin 3A4 4

= −

Substituting A = jθ in 3 3 1sin A sin A sin 3A4 4

= −

gives: 3 3 1sin j sin j sin 3j4 4

θ = θ− θ

i.e. ( )3 3 3 1j sin h j sin h j sin h 34 4

θ = θ− θ

Dividing by j gives: ( )2 3 3 1j sin h sin h sin h 34 4

θ = θ− θ

i.e. ( )3 3 1sin h sin h sin h 34 4

− θ = θ− θ

i.e. 3 1 3sinh sinh 3 sinh4 4

θ = θ − θ

7. Use the substitution A = jθ to obtain the hyperbolic identity corresponding to the trigonometric identity: ( )2 2cot A sec A 1 1− = Substituting A = jθ in ( )2 2cot A sec A 1 1− =

gives: ( )2 2cot j sec j 1 1θ θ− =

i.e. 2 2

1 1 1 1tan j cos j

⎛ ⎞− =⎜ ⎟θ θ⎝ ⎠

i.e. ( ) ( )2 2

1 1 1 1tan j cos j

⎛ ⎞− =⎜ ⎟

⎜ ⎟θ θ⎝ ⎠


i.e. ( )2 2

1 1 1 1chjth

⎛ ⎞− =⎜ ⎟θ⎝ ⎠θ

and ( )22 2

1 sec h 1 1j th

θ− =θ

i.e. ( )2 2coth sec h 1 1− θ θ− =

or ( )2 2coth 1 sech 1θ − θ =


CHAPTER 18 COMPOUND ANGLES


3. Show that (a) 2sin x sin x 3 cos x3 3π π⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(b) 3sin cos2π⎛ ⎞− −φ = φ⎜ ⎟

⎝ ⎠

(a) L.H.S. = 2 2 2sin x sin x sin x cos cos x sin sin x cos cos x sin3 3 3 3 3 3π π π π π π⎛ ⎞ ⎛ ⎞+ + + = + + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= 1 3 1 3sin x cos x sin x cos x2 2 2 2

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞+ + − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= 32 cos x2

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 3 cos x = R.H.S.

The diagram below shows an equilateral triangle ABC of side 2 with each angle 60°. Angle A is

bisected. By Pythagoras, AD = 2 22 1 3− = . Hence, sin3π = sin 60° = 3

2 and cos 60° = 1

2

(b) L.H.S. = 3 3 3sin sin cos cos sin2 2 2π π π⎛ ⎞ ⎡ ⎤− − φ = − φ− φ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

= [ ]( 1)cos (0)sin− − φ− φ = cos φ = R.H.S.

4. Prove that: (a) 3sin sin 2 (sin cos )4 4π π⎛ ⎞ ⎛ ⎞θ + − θ− = θ+ θ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(b) ( )( )

cos 270tan

cos 360° + θ

= θ°− θ


(a) L.H.S. = 3 3 3sin sin sin cos cos sin sin cos cos sin4 4 4 4 4 4π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞θ+ − θ− = θ + θ − θ − θ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 1 1 1 1sin cos sin cos2 2 2 2

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞θ + θ − θ − − θ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

= [ ] ( )1 2sin cos sin cos sin cos2 2

θ+ θ+ θ+ θ = θ+ θ

= 2 (sin cos )θ+ θ = R.H.S. The diagram below shows an isosceles triangle where AB = BC = 1 and angles A and C are both

45°. By Pythagoras, AC = 2 21 1 2+ = . Hence, sin 4π = sin 45° = cos 45°= 1

2

(b) L.H.S. = ( )( )

cos 270 cos 270 cos sin 270 sin 0 ( 1)sincos 360 cos360 cos sin 360 sin (1)cos 0

° + θ ° θ− ° θ − − θ= =

°−θ ° θ+ ° θ θ+

= sin tancos

θ= θ

θ = R.H.S.

7. Solve the equation: 4 sin(θ - 40°) = 2 sin θ for values of θ between 0° and 360° 4 sin(θ - 40°) = 2 sin θ

i.e. 4[sin θ cos 40° - cos θ sin 40°] = 2 sin θ

i.e. 3.064178 sin θ - 2.57115 cos θ = 2 sin θ

Hence, 1.064178 sin θ = 2.57115 cos θ

sin 2.57115 2.4160901cos 1.064178

θ= =

θ

i.e. tan θ = 2.4160901 and θ = 1tan (2.4160901)− = 67°31′ and 247°31′ (see diagram below)


EXERCISE 81 Page 181 2. Change the function: 4 sin ωt – 3 cos ωt into the form R sin(ωt ± α) Let 4 sin ωt – 3 cos ωt = R sin(ωt + α)

= R[sin ωt cos α + cos ωt sin α]

= (R cos α) sin ωt + (R sin α) cos ωt

Hence, 4 = R cos α from which, cos α = 4R

and -3 = R sin α from which, sin α = 3R

−

There is only one quadrant where both cosine is positive and sine is negative, i.e. the 4th quadrant,

as shown in the diagram below.

R = ( )2 24 3+ = 5 and α = 1 3tan4

− = 0.644 rad (make sure your calculator is on radians)

Hence, 4 sin ωt – 3 cos ωt = 5 sin(ωt – 0.644)

3. Change the function: -7 sin ωt + 4 cos ωt into the form R sin(ωt ± α) Let -7 sin ωt + 4 cos ωt = R sin(ωt + α)



Hence, -7 = R cos α from which, cos α = 7R

−

and 4 = R sin α from which, sin α = 4R

There is only one quadrant where both cosine is negative and sine is positive, i.e. the 2nd quadrant,



R = ( )2 27 4+ = 8.062 and φ = 1 4tan7

− = 0.519 rad

Thus, in the diagram, α = π - 0.519 = 2.622

Hence, -7 sin ωt + 4 cos ωt = 8.062 sin(ωt + 2.622)

5. Solve the following equations for values of θ between 0° and 360°:

(a) 2 sin θ + 4 cos θ = 3 (b) 12 sin θ - 9 cos θ = 7 (a) Let 2 sin θ + 4 cos θ = R sin(θ + α)

= R[sin θ cos α + cos θ sin α]

= (R cos α) sin θ + (R sin α) cos θ



There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st

quadrant, as shown in the diagram below.

R = ( )2 24 2+ = 4.472 and α = 1 4tan

2− = 63.43° or 63°26′

Hence, 2 sin θ + 4 cos θ = 4.472 sin(θ + 63°26′)

Thus, since 2 sin θ + 4 cos θ = 3 then 4.472 sin(θ + 63°26′) = 3

i.e. sin(θ + 63°26′) = 3 0.670844.472

=


and θ + 63°26′ = 1sin 0.67084− = 42°8′ or 180° - 42°8′ = 137°52′

Thus, θ = 42°8′ - 63°26′ = -21°18′ ≡ 360° -21°18′ = 338°42′

or θ = 137°52′ - 63°26′ = 74°26′

i.e. θ = 74°26′ and 338°42′ satisfies the equation 2 sin θ + 4 cos θ = 3

(b) Let 12 sin θ - 9 cos θ = R sin(θ + α)

= R[sin θ cos α + cos θ sin α]

= (R cos α) sin θ + (R sin α) cos θ


and -9 = R sin α from which, sin α = 9R

−

There is only one quadrant where both cosine is positive and sine is negative, i.e. the 4th

quadrant, as shown in the diagram below.

R = ( )2 212 9+ = 15 and α = 1 9tan

12− = 36°52′

Hence, 12 sin θ - 9 cos θ = 15 sin(θ - 36°52′)

Thus, since 12 sin θ - 9 cos θ = 7 then 15 sin(θ - 36°52′) = 7

i.e. sin(θ - 36°52′) = 715

and θ - 36°52′ = 1 7sin15

− = 27°49′ or 180° - 27°49′ = 152°11′

Thus, θ = 27°49′ + 36°52′ = 64°41′ or θ = 152°11′ + 36°52′ = 189°3′

i.e. θ = 64°41′ and 189°3′ satisfies the equation 12 sin θ - 9 cos θ = 7 7. The third harmonic of a wave motion is given by: 4.3 cos 3θ - 6.9 sin 3θ.

Express this in the form R sin(3θ ± α)


Let 4.3 cos 3θ - 6.9 sin 3θ = R sin(3θ + α)

= R[sin 3θ cos α + cos 3θ sin α]

= (R cos α) sin 3θ + (R sin α) cos 3θ

Hence, -6.9 = R cos α from which, cos α = 6.9R

−

and 4.3 = R sin α from which, sin α = 4.3R

There is only one quadrant where both cosine is negative and sine is positive, i.e. the 2nd quadrant,


R = ( )2 26.9 4.3+ = 8.13 and φ = 1 4.3tan6.9

− = 31°56′

and α = 180° - 31°56′ = 148°4′ = 2.584 rad

Hence, 4.3 cos 3θ - 6.9 sin 3θ = 8.13 sin(3θ + 2.584)

8. The displacement x metres of a mass from a fixed point about which it is oscillating is given by:

x = 2.4 sin ωt + 3.2 cos ωt, where t is the time in seconds. Express x in the form R sin(ωt + α). Let x = 2.4 sin ωt + 3.2 cos ωt = R sin(ωt + α)



Hence, 2.4 = R cos α from which, cos α = 2.4R

and 3.2 = R sin α from which, sin α = 3.2R

There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st quadrant, as

shown in the diagram below.


R = ( )2 22.4 3.2+ = 4 and α = 1 3.2tan2.4

− = 53.13° or 0.927 rad

Hence, x = 2.4 sin ωt + 3.2 cos ωt = 4 sin(ωt + 0.927) m



2. Prove the following identities: (a) 22

cos 21 tancos

φ− = φ

φ (b) 2

2

1 cos 2t 2cot tsin t+

=

(c) ( )( )tan 2x 1 tan x 2tan x 1 tan x

+=

−

(d) 2 cosec 2θ cos 2θ = cot θ - tan θ

(a) L.H.S. = 2 2 2 2

2 2 2 2

cos 2 cos sin cos sin1 1 1cos cos cos cos

⎛ ⎞ ⎛ ⎞φ φ− φ φ φ− = − = − −⎜ ⎟ ⎜ ⎟φ φ φ φ⎝ ⎠ ⎝ ⎠

= 1- ( )21 tan− φ = 2tan φ = R.H.S.

(b) L.H.S. = ( )2 2

22 2 2

1 2cos t 11 cos 2t 2cos t 2cot tsin t sin t sin t

+ −+= = = = R.H.S.

(c) L.H.S. = ( )( ) ( )( )( )( )( )2

2 tan x 1 tan x2 tan x 1 tan xtan 2x 1 tan x 1 tan x 1 tan x1 tan xtan x tan x tan x

+⎛ ⎞ +⎜ ⎟+ − +−⎝ ⎠= =

= ( )( )

2 tan x1 tan x 2 tan x 2

tan x tan x 1 tan x 1 tan x−

= =− −

= R.H.S.

(d) L.H.S. = 2 cosec 2θ cos 2θ = 2

2 2 2(cos 2 ) 2cot 2 2 tansin 2 tan 21 tan

⎛ ⎞ θ = θ = =⎜ ⎟ θθ θ⎝ ⎠− θ

= ( )2 22 1 tan 1 tan 1 tan cot tan

2 tan tan tan− θ − θ

= = − θ = θ− θθ θ θ

= R.H.S.

3. If the third harmonic of a waveform is given by 3V cos3θ , express the third harmonic in terms of

the first harmonic cos θ, when 3V = 1. When 3V = 1, 3V cos3θ = cos 3θ = cos(2θ + θ) = cos 2θ cos θ - sin 2θ sin θ = ( ) ( )22cos 1 (cos ) 2sin cos (sin )θ− θ − θ θ θ = 3 22cos cos 2cos sinθ− θ− θ θ


= ( )3 22cos cos 2cos 1 cosθ− θ− θ − θ = 3 32cos cos 2cos 2cosθ− θ− θ+ θ = 34cos 3cosθ− θ = R.H.S.


EXERCISE 83 Page 184 2. Express as a sum or difference: cos 8x sin 2x

cos 8x sin 2x = [ ]1 sin(8x 2x) sin(8x 2x)2

+ − − from (2), page 183

= [ ]1 sin10x sin 6x2

−

4. Express as a sum or difference: 4 cos 3θ cos θ

4 cos 3θ cos θ = [ ]14 cos(3 ) cos(3 )2

⎧ ⎫θ+ θ + θ−θ⎨ ⎬⎩ ⎭

from (3), page 183

= 2[cos 4θ + cos 2θ]

6. Determine: 2sin 3t cos t dt∫

2 sin 3t cost = [ ]12 sin(3t t) sin(3t t)2

⎧ ⎫+ + −⎨ ⎬⎩ ⎭

from (1), page 183

= sin 4t + sin 2t

Hence, ( )2sin 3t cos t dt sin 4t sin 2t dt= +∫ ∫ = 1 1cos 4t cos 2t c4 2

− − +

8. Solve the equation: 2 sin 2φ sin φ = cos φ in the range φ = 0° to φ = 180°. 2 sin 2φ sin φ = cos φ i.e. 2(2 sin φ cos φ) sin φ = cos φ i.e. 24sin cos cosφ φ = φ i.e. 24sin cos cos 0φ φ− φ = and ( )2cos 4sin 1 0φ φ− = Hence, cos φ = 0 from which, φ = 1cos 0− = 90°


and 24sin 1φ = from which, 2 1sin4

φ = and sin φ = 1 0.54= ±

Hence, φ = 1sin 0.5− = 30° and 150° (see diagram below) and φ = 1sin ( 0.5)− − = 210° and 330°

Since the range is from φ = 0° to φ = 180°, then the only values of φ to satisfy: 2 sin 2φ sin φ = cos φ are: φ = 30°, 90° and 150°


EXERCISE 84 Page 185 1. Express as a product: sin 3x + sin x

sin 3x + sin x = 3x x 3x x2sin cos2 2+ −⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

from (5), page 184

= 2 sin 2x cos x

3. Express as a product: cos 5t + cos 3t

cos 5t + cos 3t = 5t 3t 5t 3t2cos cos2 2+ −⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

from (7), page 184

= 2 cos 4t cos t

5. Express as a product: 1 cos cos2 3 4

π π⎛ ⎞+⎜ ⎟⎝ ⎠

1 1 3 4 3 4cos cos 2cos cos2 3 4 2 2 2

⎧ π π π π ⎫⎛ ⎞ ⎛ ⎞+ −⎪ ⎪⎜ ⎟ ⎜ ⎟π π ⎪ ⎪⎛ ⎞+ = ⎨ ⎬⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎪ ⎪⎜ ⎟ ⎜ ⎟

⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

from (7), page 184

=

712 12cos cos2 2

π π⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 7cos cos24 24π π

6. Show that: (a) sin 4x sin 2x tan xcos 4x cos 2x

−=

+ (b) 1 sin(5x ) sin(x ) cos3x sin(2x )

2−α − +α = −α

(a) L.H.S. =

4x 2x 4x 2x2cos sinsin 4x sin 2x 2 2

4x 2x 4x 2xcos 4x cos 2x 2cos cos2 2

+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠=

+ −+ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 2cos3x sin x sin x tan x2cos3x cos x cos x

= = = R.H.S.

(b) L.H.S. = 1 sin(5x ) sin(x )2

−α − +α


= [ ]1 (sin 5x cos cos5x sin ) (sin x cos cos x sin )2

α − α − α + α

= [ ]1 cos (sin 5x sin x) sin (cos5x cos x)2

α − − α +

= 1 5x x 5x x 5x x 5x xcos 2cos sin sin 2cos cos2 2 2 2 2⎡ ⎤⎧ + − ⎫ ⎧ + − ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞α − α⎨ ⎬ ⎨ ⎬⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭ ⎩ ⎭⎣ ⎦

= [ ]1 2cos (cos3x sin 2x) 2sin (cos3x cos 2x)2

α − α

= cos 3x (cos α sin 2x – sin α cos 2x)

= cos 3x (sin 2x cos α – cos 2x sin α)

= cos 3x sin(2x - α) = R.H.S.


CHAPTER 19 FUNCTIONS AND THEIR CURVES

EXERCISE 85 Page 199 1. Sketch y = 3x - 5

2. Sketch y = -3x + 4

3. Sketch y = 2x 3+

4. Sketch y = ( )2x 3−


5. Sketch y = ( )2x 4 2− +

6. Sketch y = x - 2x

7. Sketch y = 3x 2+


8. Sketch y = 1 + cos 3x

9. Sketch y = 3 - 2sin x4π⎛ ⎞+⎜ ⎟

⎝ ⎠

10. Sketch y = 2 ln x



EXERCISE 86 Page 201 1. Determine whether the following functions are even, odd or neither even nor odd: (a) 4x (b) tan 3x (c) 3t2e (d) 2sin x (a) Let f(x) = 4x . Since f(-x) = f(x) then 4x is an even function and is symmetrical about the f(x)

axis as shown below:

(b) Let f(x) = tan 3x. Since f(-x) = - f(x) then tan 3x is an odd function and is symmetrical about

the origin as shown below:

(c) Let f(t) = 3t2e . The function is neither even not odd, and is as shown below:

(d) Let f(x) = 2sin x . Since f(-x) = f(x) then 2sin x is an even function and is symmetrical about

the f(x) axis as shown below:


3. State whether the following functions, which are periodic of period 2π, are even or odd:

(a) , when 0

f ( ), when 0θ − π ≤ θ ≤⎧

θ = ⎨−θ ≤ θ ≤ π⎩ (b)

x, when x2 2f (x)

30, when x2 2

π π⎧ − ≤ ≤⎪⎪= ⎨ π π⎪ ≤ ≤⎪⎩

(a) A sketch of f(θ) against θ is shown below. Since the function is symmetrical about the f(θ) axis,

it is an even function.

(b) A sketch of f(x) against x is shown below. Since the function is symmetrical about origin, it is

an odd function.


EXERCISE 87 Page 203 2. Determine the inverse function of f(x) = 5x - 1 If y = f(x), then y = 5x - 1

Transposing for x gives: x = y 15+

Interchanging x and y gives: y = x 15+

Hence, if f(x) = 5x – 1, then 1 1f (x) (x 1)5

− = +

4. Determine the inverse function of 1f (x) 2x

= +

If y = f(x), then y = 1 2x+

Transposing for x gives: x = 1y 2−

Interchanging x and y gives: y = 1x 2−

Hence, if 1f (x) 2x

= + , then 1 1f (x)x 2

− =−

6. Determine the principal value of the inverse function 1cos 0.5−

Using a calculator (set on radians), 1cos 0.5− = 1.0472 rad or 3π rad

8. Determine the principal value of the inverse function 1cot 2−

Using a calculator (set on radians), 1cot 2− = 1 1tan2

− = 0.4636 rad

10. Determine the principal value of the inverse function 1sec 1.5−

Using a calculator (set on radians), 1sec 1.5− = 1 1cos1.5

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 0.8411 rad


12. Evaluate x, correct to 3 decimal places: x = 1 1 11 4 8sin cos tan3 5 9

− − −+ −

Using a calculator (set on radians), x = 1 1 11 4 8sin cos tan3 5 9

− − −+ −

= 0.3398 + 0.6435 – 0.7266 = 0.257 13. Evaluate y, correct to 4 significant figures: y = 1 1 13sec 2 4cos ec 2 5cot 2− − −− + Using a calculator (set on radians), y = 1 1 13sec 2 4cos ec 2 5cot 2− − −− +

= 1 1 11 1 13cos 4sin 5 tan22 2

− − −⎛ ⎞ ⎛ ⎞− + ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= 2.3562 – 3.1416 + 2.3182 = 1.533



2. Determine the asymptotes parallel to the x- and y-axes for 2 xyx 3

=−

2 xy

x 3=

− hence, 2y (x 3) x− =

and 2y (x 3) x 0− − = (1) i.e. 2 2y x 3y x 0− − = i.e. ( )2 2x y 1 3y 0− − = (2) From equation (1), equating highest power of y to zero gives: x – 3 = 0, i.e. x = 3 From equation (2), equating highest power of x to zero gives: 2y – 1 = 0, i.e. y = ± 1 Hence, asymptotes parallel to the x- and y-axes occur at x = 3, y = 1 and y = -1

3. Determine the asymptotes parallel to the x- and y-axes for x(x 3)y(x 2)(x 1)

+=

+ +

x(x 3)y

(x 2)(x 1)+

=+ +

hence, y(x + 2)(x + 1) = x(x + 3) (1)

i.e. ( )2 2y x 3x 2 x 3x 0+ + − − = and 2 2yx 3yx 2y x 3x 0+ + − − = i.e. 2x (y 1) 3xy 2y 3x 0− + − − = (2) From equation (1), equating highest power of y to zero gives: (x + 2)(x + 1) = 0, i.e. x = -2 and x = -1 From equation (2), equating highest power of x to zero gives: y – 1 = 0, i.e. y = 1 Hence, asymptotes parallel to the x- and y-axes occur at x = -2, x = -1 and at y = 1 5. Determine all the asymptotes for ( )2 2x y 16 y− = Equating highest power of x to zero gives: 2y 16− = 0, i.e. y = ± 4


Since ( )2 2x y 16 y− = then 2 2 2x y 16x y 0− − = Equating highest power of y to zero gives: 2x = 0, i.e. x = 0 Let y = mx + c, then ( )22x mx c 16 mx c⎡ ⎤+ − = +⎣ ⎦

i.e. 2 2 2 2x m x 2mxc c 16 mx c⎡ ⎤+ + − = +⎣ ⎦ i.e. 2 4 3 2 2 2m x 2mcx c x 16x mx 1 0+ + − − − = Equating coefficient of highest power of x to zero gives: 2m 0= , i.e. m = 0 Equating next coefficient of highest power of x to zero gives: 2mc = 0, i.e. c = 0 Hence, the only asymptotes occur at y = 4, y = -4 and at x = 0 7. Determine the asymptotes and sketch the curve for 2 2xy x y 2x y 5− + − = 2 2xy x y 2x y 5− + − = (1) Equating the highest power of y to zero gives: x = 0, which is an asymptote. Equating the highest power of x to zero gives: -y = 0, i.e. y = 0, which is an asymptote. Letting y = mx + c in equation (1) gives: ( ) ( )2 2x mx c x mx c 2x (mx c) 5+ − + + − + = i.e. ( )2 2 2 3 2x m x 2mcx c mx cx 2x mx c 5 0+ + − − + − − − = and 2 3 2 2 3 2m x 2mcx c x mx cx 2x mx c 5 0+ + − − + − − − = i.e. ( ) ( ) ( )2 3 2 2m m x 2mc c x x c 2 m c 5 0− + − + + − − − = Equating the coefficient of the highest power of x to zero gives: 2m m 0− = , i.e. m(m – 1) = 0 i.e. m = 0 or m = 1 Equating the coefficient of the next highest power of x to zero gives: 2mc – c = 0 When m = 0, c = 0 and when m = 1, 2c – c = 0, i.e. c = 0 Hence, y = mx + c becomes y = x, which is an asymptote. Thus, asymptotes occur at x = 0, y = 0 and at y = x


A sketch of the curve 2 2xy x y 2x y 5− + − = , together with its asymptotes is shown below:



1. Sketch the graphs of (a) 2 7y 3x 9x4

= + + (b) 2y 5x 20x 50= − + +

(a) 2 7y 3x 9x4

= + + dy 6x 9 0dx

= + = for a turning point

from which, x = 96

− = -1.5

When x = -1.5, y = 23( 1.5) 9( 1.5) 1.75− + − + = -5 Hence, a turning point occurs at (-1.5, -5)

2

2

d y 6dx

= , which is positive, hence, (-1.5, -5) is a minimum point.

A sketch of the graph 2 7y 3x 9x4

= + + is shown below.

(b) 2y 5x 20x 50= − + + dy 10x 20 0dx

= − + = for a turning point

from which, 20 = 10x and x = 2 When x = 2, 2y 5(2) 20(2) 50= − + + = 70 Hence, a turning point occurs at (2, 70)

2

2

d y 10dx

= − , which is negative, hence, (2, 70) is a maximum point.

A sketch of the graph 2y 5x 20x 50= − + + is shown below.


4. Sketch the curve depicting: 2

2 x 16y4−

=

Since 2

2 x 16y4−

= then 2 24y x 16= −

i.e. 2 216 x 4y= −

i.e. 2 2x 4y 1

16 16− =

and 2 2

2 2

x y 14 2

− = which is a hyperbola, symmetrical about the x- and y-

axes, distance between vertices being 2(4), i.e. 8 units along the x-axis.

A sketch of 2

2 x 16y4−

= , i.e. 2 2

2 2

x y 14 2

− = is shown below.

5. Sketch the curve depicting: 2 2y x5

5 2= −

Since 2 2y x5

5 2= − then

2 2x y 52 5+ =


and 2 2x y 1

10 25+ =

i.e. ( )

2 2

2 2

x y 1510

+ =

which is an ellipse, centre (0, 0), major axis, AB = 2(5) = 10 units along y-axis, and minor axis,

CD = 2 10 along the x-axis.

A sketch of the curve 2 2y x5

5 2= − , i.e.

( )2 2

2 2

x y 1510

+ = is shown below.

7. Sketch the curve depicting: 2 2x y 9=

Since 2 2x y 9= then 22

9yx

= and 3yx

=

which is a rectangular hyperbola, lying in the 1st and 3rd quadrants only, as shown in the sketch below.

9. Sketch the circle given by the equation 2 2x y 4x 10y 25 0+ − + + =


Since 2 2x y 4x 10y 25 0+ − + + = then ( ) ( )2 2x 2 y 5 4 0− + + − = i.e. ( ) ( )2 2 2x 2 y 5 2− + + = which is a circle of centre (2, -5) and radius 2, as shown in the sketch below.

11. Describe the shape of the curve represented by the equation ( )2y 3 x 1⎡ ⎤= −⎣ ⎦

Since ( )2y 3 x 1⎡ ⎤= −⎣ ⎦ then ( )2 2y 3 x 1= −

and 2 2y 3x 3= −

i.e. 2 23 3x y= −

i.e. 2

2 y1 x3

= −

i.e. ( ) ( )

2 2

2 2x y 11 3

− =

which is a hyperbola, symmetrical about the x- and y-axes, with vertices 2(1) = 2 units apart

along the x-axis.

14. Describe the shape of the curve represented by the equation ( )12y 3x=

Since ( )12y 3x= then ( )y 3x= or 2y 3x=

which is a parabola, vertex at (0, 0) and symmetrical about the x-axis.


15. Describe the shape of the curve represented by the equation 2 2y 8 2x− = − Since 2 2y 8 2x− = − then 2 2y 2x 8+ =

and 2 2y 2x 1

8 8+ =

i.e. ( )

2 2

2 2

y x 1(2)8

+ =

which is an ellipse, centre (0, 0), with major axis 2 8 units along the y-axis, and minor axis

2(2) = 4 units along the x-axis.


CHAPTER 20 IRREGULAR AREAS, VOLUMES AND MEAN

VALUES OF WAVEFORMS EXERCISE 90 Page 218 1. Plot a graph of 2y 3x x= − by completing a table of values of y from x = 0 to x = 3. Determine

the area enclosed by the curve, the x-axis and ordinates x = 0 and x = 3 by (a) the trapezoidal

rule, (b) the mid-ordinate rule and (c) by Simpson’s rule. A table of values is shown below.

x 0 0.5 1.0 1.5 2.0 2.5 3.0 2y 3x x= − 0 1.25 2.0 2.25 2.0 1.25 0

A graph of 2y 3x x= − is shown below.

(a) Using the trapezoidal rule, with 6 intervals each of width 0.5 gives:

area ( ) 0 00.5 1.25 2.0 2.25 2.0 1.252

⎡ + ⎤⎛ ⎞≈ + + + + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ = (0.5)(8.75) = 4.375 square units

(b) Using the mid-ordinate rule, with 6 intervals, with mid-ordinates occurring at 0.25 0.75 1.25 1.75 2.25 2.75

where the y-values are: 0.6875 1.6875 2.1875 2.1875 1.6875 0.6875 area ≈ (0.5)[0.6875 + 1.6875 + 2.1875 + 1.6875 + 0.6875] = (0.5)(9.125) = 4.563 square units (c) Using Simpson’s rule, with 6 intervals each of width 0.5 gives:


area ( ) ( ) ( ) [ ]1 1(0.5) 0 0 4 1.25 2.25 1.25 2 2.0 2.0 (0.5) 0 19 83 3

≈ + + + + + + = + +⎡ ⎤⎣ ⎦

= 1 (0.5)(27)3

= 4.5 square units

Simpson’s rule is considered the most accurate of the approximate methods. An answer of 4.5 square units can be achieved with the other two methods if more intervals are taken. 3. The velocity of a car at one second intervals is given in the following table:

time t(s) 0 1 2 3 4 5 6

velocity v(m/s) 0 2.0 4.5 8.0 14.0 21.0 29.0

Determine the distance travelled in 6 seconds (i.e. the area under the v/t graph) using Simpson’s

rule. Using Simpson’s rule with 6 intervals each of width 1 s gives:

area ( ) ( ) ( ) [ ]1 1(1) 0 29.0 4 2.0 8.0 21.0 2 4.5 14.0 29.0 124 373 3

≈ + + + + + + = + +⎡ ⎤⎣ ⎦

= 1 (190)3

= 63.33 square units

5. The deck of a ship is 35 m long. At equal intervals of 5 m the width is given by the following

table: Width (m) 0 2.8 5.2 6.5 5.8 4.1 3.0 2.3

Estimate the area of the deck. Using the trapezoidal rule with 7 intervals each of width 5 m gives:

area ( ) [ ]0 2.35 2.8 5.2 6.5 5.8 4.1 3.0 (5) 1.15 27.42

⎡ + ⎤⎛ ⎞≈ + + + + + + = +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= (5)(28.55) = 143 2m (To use Simpson’s rule needs an even number of intervals, so could not be used in this question).


EXERCISE 91 Page 219 1. The areas of equidistantly spaced sections of the underwater form of a small boat are as follows:

1.76 2.78 3.10 3.12 2.61 1.24 0.85 2m

Determine the underwater volume if the sections are 3 m apart.

Underwater volume = [ ]3 (1.76 0.85) 4(2.78 3.12 1.24) 2(3.10 2.61)3

+ + + + + +

= 2.61 + 28.56 + 11.42 = 42.59 3m 3. The circumference of a 12 m long log of timber of varying circular cross-section is measured at

intervals of 2 m along its length and the results are:

Distance from one end (m) 0 2 4 6 8 10 12

Circumference (m) 2.80 3.25 3.94 4.32 5.16 5.82 6.36

Estimate the volume of the timber in cubic metres.

If circumference c = 2πr then radius, r = c2π

Cross-sectional area = 2 2

2 c cr2 4

⎛ ⎞π = π =⎜ ⎟π π⎝ ⎠

Hence, the cross-sectional areas are: 2

22.80 0.6239m4

=π

, 2

23.25 0.8405m4

=π

, 1.2353 2m ,

1.4851 2m , 2.1188 2m , 2.6955 2m , 3.2189 2m Hence, volume of timber

( ) ( ) ( )2 0.6239 3.2189 4 0.8405 1.4851 2.6955 2 1.2353 2.11883

≈ + + + + + +⎡ ⎤⎣ ⎦

= ( ) ( )2 23.8428 20.0844 6.7082 30.63543 3

+ + =

= 20.42 3m


EXERCISE 92 Page 222 1. Determine the mean value of the periodic waveforms shown over half a cycle.

(a) Over half a cycle, mean value = 3

3

area under curve (2 10 10 )Aslength of base 10 10 s

−

−

× ×=

× = 2 A

(b) Over half a cycle, mean value = ( )3

3

1 5 10 (100)Vs2

5 10 s

−

−

×

× = 50 V

(c) Over half a cycle, mean value = ( )( )3

3

1 15 10 5 As2

15 10 s

−

−

×

× = 2.5 A

3. An alternating current has the following values at equal intervals of 5 ms.

Time (ms) 0 5 10 15 20 25 30

Current (A) 0 0.9 2.6 4.9 5.8 3.5 0

Plot a graph of current against time and estimate the area under the curve over the 30 ms period

using the mid-ordinate rule and determine its mean value. A graph of current against time is shown plotted below


Mid-ordinates are shown by the broken lines in the above diagram. The mid-ordinate values are: 0.4, 1.6, 3.8, 5.7, 4.9 and 2.2 area [ ]3(5 10 ) 0.4 1.6 3.8 5.7 4.9 2.2−≈ × + + + + + = [ ]3(5 10 ) 18.6−× = 393 10−× As = 0.093 As

Mean value = 3

3

area 93 10 Aslength of base 30 10 s

−

−

×=

× = 3.1 A

5. An indicator diagram of a steam engine is 12 cm long. Seven evenly spaced ordinates, including

the end ordinates, are measured as follows:

5.90 5.52 4.22 3.63 3.32 3.24 3.16 cm

Determine the area of the diagram and the mean pressure in the cylinder if 1 cm represents 90 kPa

Area ( ) ( ) ( ) ( )1 2 5.90 3.16 4 5.52 3.63 3.24 2 4.22 3.323

≈ + + + + + +⎡ ⎤⎣ ⎦

= [ ]1 1(2) 9.06 49.56 15.08 (2)(73.7)3 3

+ + = = 49.13 2cm

Mean value = 2

349.13cm Pa90 1012cm cm

× × = 368.5 kPa


CHAPTER 21 VECTORS, PHASORS AND THE COMBINATION

OF WAVEFORMS EXERCISE 93 Page 228 2. Forces A, B and C are coplanar and act at a point. Force A is 12 kN at 90°, B is 5 kN at 180° and

C is 13 kN at 293°. Determine graphically the resultant force. The forces are shown in diagram (a) below. Using the ‘nose-to-tail’ method, the vector diagram is

shown in Figure (b). The 12 kN force is drawn first, then the 5 kN force is ‘added’ to the end of the

12 kN force. Finally, the 13 kN force is ‘added’ to the end of the 5 kN force. Since the nose of the

13 kN force actually touches the tail of the 12 kN force then the resultant of the three forces is

zero.

(a) (b) 4. Three forces of 2 N, 3 N and 4 N act as shown below. Calculate the magnitude of the resultant

force and its direction relative to the 2 N force.


Total horizontal component = 3 cos 0° + 4 cos 60° + 2 cos 300° = 6

Total vertical component = 3 sin 0° + 4 sin 60° + 2 sin 300° = 1.732

From the diagram below, R = ( )2 26 1.732+ = 6.24 N and 1 1.732tan6

− ⎛ ⎞θ = ⎜ ⎟⎝ ⎠

= 16.10°

Hence the direction of the resultant relative to the 2 N force is 16.10° + 60° = 76.10°. Thus, the resultant is 6.24 N at an angle of 76.10° to the 2 N force. 6. The acceleration of a body is due to four component, coplanar accelerations. These are 2 2m / s

due north, 3 2m / s due east, 4 2m / s to the south-west and 5 2m / s to the south-east. Calculate the

resultant acceleration and its direction. The space diagram is shown below.

Total horizontal component = 3 cos 0° + 2 cos 90° + 4 cos 225° + 5 cos 315° = 3.707

Total vertical component = 3 sin 0° + 2 sin 90° + 4 sin 225° + 5 sin 315° = -4.364

From the diagram below, R = ( )2 23.707 4.364+ = 5.7 2m / s

and 1 4.364tan 503.707

− ⎛ ⎞θ = = °⎜ ⎟⎝ ⎠

correct to 2 significant figures.

Hence, the resultant acceleration is 5.7 2m / s at 310° (i.e. at E 50° S)


8. A ship heads in a direction of E 20° S at a speed of 20 knots while the current is 4 knots in a

direction of N 30° E. Determine the speed and actual direction of the ship. The vector diagram is shown below.

Total horizontal component = 4 cos 60° + 20 cos 340° = 20.794 knots

Total vertical component = 4 sin 60° + 20 sin 340° = -3.376 knots

From the diagram below, R = ( )2 220.794 3.376+ = 21.07 knots

and 1 3.376tan 9.2220.794

− ⎛ ⎞θ = = °⎜ ⎟⎝ ⎠

Hence, the speed of the ship is 21.07 knots and its actual direction is E 9.22° S


EXERCISE 94 Page 231 2. Calculate the resultant of (a) 1v + 2v - 3v (b) 3v - 2v + 1v when 1v = 15 m/s at 85°,

2v = 25 m/s at 175° and 3v = 12 m/s at 235° (a) Total horizontal component of 1v + 2v - 3v = 15 cos 85° + 25 cos 175° - 12 cos 235° = -16.715

Total vertical component of 1v + 2v - 3v = 15 sin 85° + 25 sin 175° - 12 sin 235° = 26.952

From the diagram below, R = ( )2 216.715 26.952+ = 31.71 m/s,

1 26.952tan 58.1916.715

− ⎛ ⎞α = = °⎜ ⎟⎝ ⎠

and thus θ = 180° - 58.19° = 121.81°

i.e. the resultant of 1v + 2v - 3v is 31.71 m/s at angle of 121.81° (b) Total horizontal component of 3v - 2v + 1v = 12 cos 235° - 25 cos 175° + 15 cos 85° = 19.329

Total vertical component of 3v - 2v + 1v = 12 sin 235° - 25 sin 175° + 15 sin 85° = 2.934

From the diagram below, R = ( )2 219.329 2.934+ = 19.55 m/s,

and 1 2.934tan 8.6319.329

− ⎛ ⎞θ = = °⎜ ⎟⎝ ⎠

i.e. the resultant of 3v - 2v + 1v is 19.55 m/s at angle of 8.63°


EXERCISE 95 Page 232 1. A car is moving along a straight horizontal road at 79.2 km/h and rain is falling vertically

downwards at 26.4 km/h. Find the velocity of the rain relative to the driver of the car. The space diagram is shown in diagram (a). The velocity diagram is shown in diagram (b) and the velocity of the rain relative to the driver is given by vector rc where rc = re + ec

rc = ( )2 279.2 26.4+ = 83.5 km/h and 1 79.2tan 71.626.4

− ⎛ ⎞θ = = °⎜ ⎟⎝ ⎠

(a) (b)

i.e. the velocity of the rain relative to the driver is 83.5 km/h at 71.6° to the vertical. 2. Calculate the time needed to swim across a river 142 m wide when the swimmer can swim at

2 km/h in still water and the river is flowing at 1 km/h. At what angle to the bank should the

swimmer swim? The swimmer swims at 2 km/h relative to the water, and as he swims the movement of the water

carries him downstream. He must therefore aim against the flow of the water – at an angle θ shown

in the triangle of velocities shown below where v is the swimmers true speed.

v = 2 22 1 3− = km/h = 1000360

⎛ ⎞⎜ ⎟⎝ ⎠

m/min = 28.87 m/min

Hence, if the width of the river is 142 m, the swimmer will take 14228.87

= 4.919 minutes

= 4 min 55 s


In the above diagram, sin θ = 12

from which, θ = 30°

Hence, the swimmer needs to swim at an angle of 60° to the bank (shown as angle α in the

diagram.

3. A ship is heading in a direction N 60° E at a speed which in still water would be 20 km/h. It is

carried off course by a current of 8 km/h in a direction of E 50° S. Calculate the ship’s actual

speed and direction. In the triangle of velocities shown below (triangle 0AB), 0A represents the velocity of the ship in

still water, AB represents the velocity of the water relative to the earth, and 0B is the velocity of the

ship relative to the earth.

Total horizontal component of v = 20 cos 30° + 8 cos 310° = 22.46

Total vertical component of v = 20 sin 30° + 8 sin 310° = 3.87

Hence, v = ( )2 222.46 3.87+ = 22.79 km/h,

and 1 3.87tan 9.7822.46

− ⎛ ⎞θ = = °⎜ ⎟⎝ ⎠

Hence, the ships actual speed is 22.79 km/h in a direction E 9.78° N



2. Two alternating voltages are given by 1v 10sin t= ω volts and 2v 14sin t volts3π⎛ ⎞= ω +⎜ ⎟

⎝ ⎠. By

plotting 1v and 2v on the same axes over one cycle obtain a sinusoidal expression for

(a) 1v + 2v (b) 1v - 2v

(a) 1v 10sin t= ω , 2v 14sin t volts3π⎛ ⎞= ω +⎜ ⎟

⎝ ⎠ and 1v + 2v are shown sketched below:

1v + 2v leads 1v by 36° = 36180π

× = 0.63 rad

Hence, by measurement, 1v + 2v = 20.9 sin(ωt + 0.63) volts


(b) 1v 10sin t= ω , 2v 14sin t volts3π⎛ ⎞= ω +⎜ ⎟

⎝ ⎠ and 1v - 2v are shown sketched below:

1v - 2v lags 1v by 78° = 78180π

× = 1.36 rad

Hence, by measurement, 1v - 2v = 12.5 sin(ωt – 1.36) volts

4. Express 7sin t 5sin t4π⎛ ⎞ω + ω +⎜ ⎟

⎝ ⎠ in the form ( )Asin tω ±α using phasors.

The space diagram is shown in (a) below and the phasor diagram is shown in (b).

(a) (b)

Using the cosine rule: 2 2 2R 7 5 2(7)(5)cos135 123.497= + − ° = from which, R 123.497= = 11.11


Using the sine rule: 5 11.11sin sin135

=θ °

from which, 5sin135sin 0.3182311.11

°θ = =

and 1sin 0.31823−θ = = 18.56° or 0.324 rad

Hence, in sinusoidal form, 7sin t 5sin t4π⎛ ⎞ω + ω +⎜ ⎟

⎝ ⎠ = 11.11sin( t 0.324)ω +

6. Express i 25sin t 15sin t3π⎛ ⎞= ω − ω +⎜ ⎟

⎝ ⎠ in the form ( )A sin tω ±α using phasors.

The relative positions of currents 1i and 2i are shown in diagram (a) below. Phasor 2i is shown

reversed in diagram (b) to give - 2i . The phasor diagram for i = 1i - 2i is shown in diagram (c)

(a) (b) (c)

Using the cosine rule: 2 2 2i 25 15 2(25)(15)cos60 475= + − ° = from which, i 475= = 21.79

Using the sine rule: 15 21.79sin sin 60

=θ °

from which, 15sin 60sin 0.596221.79

°θ = =

and 1sin 0.5962−θ = = 36.60° or 0.639 rad

Hence, in sinusoidal form, i 25sin t 15sin t3π⎛ ⎞= ω − ω +⎜ ⎟

⎝ ⎠ = 21.79sin( t 0.639)ω −

8. Express 3x 9sin t 7sin t3 8π π⎛ ⎞ ⎛ ⎞= ω + − ω −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ in the form ( )A sin tω ±α using phasors.

180rad 60

3 3π π °

= × = °π

and 3 3 180rad 67.58 8π π °

= × = °π

The relative positions of currents 1x and 2x are shown in diagram (a) below. Phasor 2x is shown


reversed in diagram (b) to give - 2x . The phasor diagram for x = 1x - 2x is shown in diagram (c)

(a) (b) (c) Using the cosine rule: 2 2 2x 9 7 2(9)(7)cos(60 67.5 ) 206.704= + − °+ ° = from which, x 206.704 14.377= =

Using the sine rule: 7 14.377sin sin127.5

=α °

from which, 7sin127.5sin 0.386314.377

°α = =

and 1sin 0.3863−α = = 22.72° Measured to the horizontal, θ = 60° + 22.72° = 82.72° or 1.444 rad

Hence, in sinusoidal form, 3x 9sin t 7sin t3 8π π⎛ ⎞ ⎛ ⎞= ω + − ω −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ = 14.38sin( t 1.444)ω +


CHAPTER 22 SCALAR AND VECTOR PRODUCTS EXERCISE 97 Page 241 2. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine the quantities (a) p q⋅ (b) p r⋅ (a) p q⋅ = (2)(0) + (-3)(4) + (0)(-1) = -12 (b) p r⋅ = (2)(1) + (-3)(2) + (0)(-3) = 2 – 6 = -4 4. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine the quantities (a) p (b) r (a) p = ( )2 2 2(2) ( 3) (0)+ − + = 13

(b) r = ( )2 2 2(1) (2) ( 3)+ + − = 14

5. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine the quantities (a) ( )p q r⋅ +

(b) 2r ⋅ (q – 2p) (a) ( )p q r⋅ + = (2i – 3j)( 4j – k + i + 2j – 3k) = (2i – 3j)( i + 6j – 4k)

= (2)(1) + (-3)(6) + (0)(-4) = 2 – 18 + 0 = -16 (b) 2r ⋅ (q – 2p) = (2i + 4j – 6k) ⋅ (4j – k – 2(2i – 3j)) = (2i + 4j – 6k) ⋅ (– 4i + 10j – k)

= (2)(-4) + (4)(10) + (-6)(-1)

= -8 + 40 + 6 = 38 7. If p = 2i – 3j, q = 4j – k and r = i + 2j – 3k find the angle between (a) p and q (b) q and r

(a) From equation (4), page 239, 1 1 2 2 3 32 2 2 2 2 2

1 2 3 1 2 3

a b a b a bcosa a a b b b

+ +θ =

+ + + +

= 2 2 2 2 2 2

(2)(0) ( 3)(4) (0)( 1) 1213 172 ( 3) 0 0 4 ( 1)

+ − + − −=

+ − + + + − = -0.8072

from which, θ = 1cos ( 0.8072)− − = 143.82°


(b) cos θ = 2 2 2 2 2 2

(0)(1) (4)(2) ( 1)( 3) 11 0.71302417 140 4 ( 1) 1 2 ( 3)

+ + − −= =

+ + − + + −

from which, θ = 1cos (0.713024)− = 44.52° 8. If p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine the direction cosines of (a) p (b) q

(c) r

(a) For p, 2 2

2 2cos132 ( 3)

α = =+ −

= 0.555, 3cos13−

β = = -0.832 and 0cos13

γ = = 0

(b) For q, 2 2

0 0cos174 ( 1)

α = =+ −

= 0, 4cos17

β = = 0.970 and 1cos17−

γ = = -0.243

(c) For r, 2 2 2

1 1cos141 2 ( 3)

α = =+ + −

= 0.267, 2cos14

β = = 0.535 and 3cos14−

γ = = -0.802

10. Find the angle between the velocity vector 1v = 5i + 2j + 7k and 2v = 4i + j - k

cos θ = 2 2 2 2 2 2

(5)(4) (2)(1) (7)( 1) 15 0.4003278 185 2 7 4 1 ( 1)

+ + −= =

+ + + + −

from which, θ = 1cos (0.40032)− = 66.40° 11. Calculate the work done by a force F = (-5i + j + 7k) when its point of application moves from

point (-2i – 6j + k) m to the point (i – j + 10k) m. Work done = F ⋅ d where d = (i – j + 10k) - (-2i – 6j + k) = 3i + 5j + 9k Hence, work done = (-5i + j + 7k) ⋅ (3i + 5j + 9k) = -15 + 5 + 63 = 53 Nm


EXERCISE 98 Page 244 1. When p = 3i + 2k, q = i – 2j + 3k and r = -4i + 3j – k, determine (a) p q× (b) q p×

(a) p q× = i j k3 0 21 2 3−

= i 0 22 3−

- j 3 21 3

+ k 3 01 2−

= 4i – 7j – 6k

(b) q p× = i j k1 2 33 0 2

− = i 2 3

0 2−

- j 1 33 2

+ k 1 23 0

− = -4i + 7j + 6k

2. When p = 3i + 2k, q = i – 2j + 3k and r = -4i + 3j – k, determine (a) p r× (b) r q×

(a) p r× = ( )( ) ( )2p p r r p r⋅ ⋅ − ⋅ where p ⋅ p = (3)(3) + (2)(2) = 13,

r ⋅ r = (-4)(-4) +(3)(3) + (-1)(-1) = 26

and p ⋅ r = (3)(-4) + (0)(3) + (2)(-1) = -14

Hence, p r× = 2(13)(26) ( 14) (338 196) 142⎡ ⎤− − = − =⎣ ⎦ = 11.92

(b) r q× = = ( )( ) ( )2r r q q r q⋅ ⋅ − ⋅ where r ⋅ r = (-4)(-4) + (3)(3) + (-1)(-1) = 26, q ⋅ q = (1)(1) +(-2)(-2) + (3)(3) = 14 and r ⋅ q = (-4)(1) + (3)(-2) + (-1)(3) = -13

Hence, r q× = 2(26)(14) ( 13) (364 169) 195⎡ ⎤− − = − =⎣ ⎦ = 13.96

4. When p = 3i + 2k, q = i – 2j + 3k and r = -4i + 3j – k, determine (a) p × (r × q) (b) (3p × 2r) × q

(a) r × q = i j k4 3 1

1 2 3− −

− = 7i + 11j + 5k


Hence, p × (r × q) = i j k3 0 27 11 5

= (-22)i – (1)j + (33)k = -22i - j + 33k

(b) 3p × 2r = i j k9 0 68 6 2− −

= (-36)i – (-18 + 48)j + (54)k = -36i - 30j + 54k

(3p × 2r) × q = i j k36 30 541 2 3

− −−

= (-90 + 108)i – (-108 - 54)j + (72 + 30)k

= 18i + 162j + 102k

6. For vectors a = -7i + 4j + 12

k and b = 6i – 5j – k find (i) a b⋅ (ii) a b× (iii) a b×

(iv) b a× and (v) the angle between the vectors.

(i) a b⋅ = (-7)(6) + (4)(-5) +( 12

)(-1) = - 62 12

(ii) a b× =

i j k17 42

6 5 1

−

− −

= i 542

⎛ ⎞− +⎜ ⎟⎝ ⎠

- j(7 – 3) + k(35 – 24) = 11 i 4j 11k2

− − +

(iii) a ⋅ a = (-7)(-7) + (4)(4) + 1 12 2

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 65.25

b ⋅ b = (6)(6) + (-5)(-5) + (-1)(-1) = 62

a ⋅ b = (-7)(6) + (4)(-5) + ( )1 12

⎛ ⎞ −⎜ ⎟⎝ ⎠

= - 62.5

a b× = ( )( ) ( ) ( )265.25 62 62.5 4045.5 3906.25 139.25⎡ ⎤− − = − =⎣ ⎦ = 11.80

(iv) b a× = i j k6 5 1

17 42

− −

−

= i 5 42

⎛ ⎞− +⎜ ⎟⎝ ⎠

- j(3 – 7) + k(24 – 35) = 11 i 4j 11k2

+ −

(v) cos θ = 2

2 2 2 2 2

1( 7)(6) (4)( 5) ( 1)62.52 0.9826388

65.25 621( 7) (4) (6) ( 5) ( 1)2

⎛ ⎞− + − + −⎜ ⎟ −⎝ ⎠ = = −⎛ ⎞− + + + − + −⎜ ⎟⎝ ⎠

from which, θ = 1cos ( 0.9826388)− − = 169.31°


8. A force of (2i – j + k) newton’s acts on a line through point P having co-ordinates (0, 3, 1)

metres. Determine the moment vector and its magnitude about point Q having co-ordinates

(4, 0, -1) metres. Position vector, r = (0i + 3j +k) – (4i + 0j – k) = -4i + 3j + 2k

Moment, M = r × F where M = i j k4 3 2

2 1 1−

− = (3 + 2)i – (-4 - 4)j + (4 - 6)k

= (5i + 8j - 2k) Nm

Magnitude of M, M = ( ) ( ) ( )2r F r r F F r F× = ⋅ ⋅ − ⋅ where r ⋅ r = (-4)(-4) + (3)(3) +(2)(2) = 29

F ⋅ F = (2)(2) + (-1)(-1) +(1)(1) = 6

r ⋅ F = (-4)(2) + (3)(-1) +(2)(1) = -9

i.e. M = ( )( ) ( )229 6 9− − = 9.64 Nm 10. Calculate the velocity vector and its magnitude for a particle rotating about the z-axis at an

angular velocity of (3i – j + 2k) rad/s when the position vector of the particle is at

(i – 5j + 4k) m.

Velocity vector, v = ω × r = (3i – j + 2k) × (i – 5j + 4k) = i j k3 1 21 5 4−

= (-4 + 10)i – (12 - 2)j + (-15 + 1)k

= 6i - 10j - 14k

Magnitude of v, v = ( ) ( ) ( )2r r rω⋅ω ⋅ − ⋅ω

where ω⋅ω = (3)(3) + (-1)(-1) + (2)(2) = 14

r⋅r = (1)(1) + (-5)(-5) + (4)(4) = 42

r⋅ω = (1)(3) + (-5)(-1) + (4)(2) = 16

Hence, v = ( )( ) ( )214 42 16− = 18.22 m/s


EXERCISE 99 Page 246 1. Find the vector equation of the line through the point with position vector 5i – 2j + 3k which is

parallel to the vector 2i + 7j – 4k. Determine the point on the line corresponding to λ = 2 in the

resulting equation. Vector equation of the line, r = a + λb = (5i – 2j + 3k) + λ(2i + 7j – 4k) i.e. r = (5 + 2λ)i + (7λ - 2)j + (3 - 4λ)k When λ = 2, r = 9i + 12j – 5k 2. Express the vector equation of the line in problem 1 in standard Cartesian form. The vector equation of a straight line in standard Cartesian form is:

31 2

1 2 3

z ax a y ab b b

−− −= = = λ

Since a = 5i – 2j + 3k, 1 2 3a 5, a 2 and a 3= = − = and b = 2i + 7j – 4k, 1 2 3b 2, b 7 and b 4= = = −

then the Cartesian equations are: x 5 y 2 z 32 7 4− + −

= = = λ−

or x 5 y 2 3 z2 7 4− + −

= = = λ

4. Express the straight line equation 1 4y 3z 12x 15 4− −

+ = = in vector form.

2x 1 1 4y 3z 11 5 4+ − −

= = i.e.

11 1 zx y32 4

1 5 42 4 3

−+ −= =

−

i.e. 1 2 31 1 1a , a and a2 4 3

= − = =

and 1 2 31 5 4b , b and b2 4 3

= = − =

Hence, in vector form the equation is:

r = ( )1 1a b+ λ i + ( )2 2a b+ λ j + ( )3 3a b+ λ k


= 1 12 2

⎛ ⎞− + λ⎜ ⎟⎝ ⎠

i + 1 54 4

⎛ ⎞− λ⎜ ⎟⎝ ⎠

j + 1 43 3

⎛ ⎞+ λ⎜ ⎟⎝ ⎠

k

or r = ( )1 12

λ − i + ( )1 1 54

− λ j + ( )1 1 43

+ λ k


CHAPTER 23 COMPLEX NUMBERS EXERCISE 100 Page 250 1. Solve the quadratic equation 2x 25 0+ = Since 2x 25 0+ = then 2x 25= − i.e. x 25 ( 1)(25) 1 25 j 25= − = − = − = from which, x = ± j 5 2. Solve the quadratic equation 22x 3x 4 0+ + = Since 22x 3x 4 0+ + = then

23 3 4(2)(4) 3 ( 1)(23) 3 ( 1) (23) 3 j (23)3 23x2(2) 4 4 4 4

⎡ ⎤− ± − − ± − − ± − − ±− ± −⎣ ⎦= = = = =

= - 3 23j4 4± or (-0.750 ± j1.199)

4. Evaluate (a) 8j (b) 7

1j

− (c) 13

42j

(a) 8j = ( )42j = ( )41− = 1

(b) ( ) ( )3 37 6 2j j j j j j 1 j= × = × = × − = −

Hence, 7 2

1 1 1 j j j jj j j j( j) j ( 1) 1

− − − −− = − = = = = =

− − − − − = -j

(c) ( )613 12 2 6j j j j j j ( 1) j= × = × = × − =

Hence, 13

42 j

= 2

2 2( j) j2 j2j j( j) j 1

− − −= = =

− − = -j2


EXERCISE 101 Page 253 1. Evaluate (a) (3 + j2) + (5 – j) and (b) (-2 + j6) – (3 – j2) and show the results on an Argand

diagram. (a) (3 + j2) + (5 – j) = (3 + 5) + j(2 – 1) = 8 + j (b) (-2 + j6) – (3 – j2) = -2 + j6 – 3 + j2 = (-2 – 3) + j(6 + 2) = -5 + j8 (8 + j) and (-5 + j8) are shown on the Argand diagram below.

3. Given 1Z = 1 + j2, 2Z = 4 – j3, 3Z = -2 + j3 and 4Z = -5 – j, evaluate in a + jb form:

(a) 1 2 3Z Z Z+ − (b) 2 1 4Z Z Z− + (a) 1 2 3Z Z Z+ − = 1 + j2 + 4 – j3 – (-2 + j3) = 1 + j2 + 4 – j3 + 2 – j3 = (1 + 4 + 2) + j(2 – 3 – 3) = 7 – j4 (b) 2 1 4Z Z Z− + = (4 – j3) – (1 + j2) + (-5 – j) = 4 – j3 – 1 – j2 – 5 – j = (4 – 1 – 5) + j(-3 – 2 – 1) = -2 - j6 5. Given 1Z = 1 + j2, 2Z = 4 – j3, 3Z = -2 + j3 and 4Z = -5 – j, evaluate in a + jb form:

(a) 1 3 4Z Z Z+ (b) 1 2 3Z Z Z (a) 1 3 4Z Z Z+ = (1 + j2)(-2 + j3) + (-5 – j) = -2 + j3 – j4 + 2j 6 - 5 – j

= -2 + j3 – j4 - 6 – 5 - j = -13 – j2 (b) 1 2 3Z Z Z = (1 + j2)(4 – j3)(-2 + j3) = (4 – j3 + j8 - 2j 6)(-2 + j3) = ( 10 + j5)(-2 + j3)

= -20 + j30 – j10 + 2j 15 = -20 + j30 –j10 – 15 = -35 + j20


7. Given 1Z = 1 + j2, 2Z = 4 – j3, 3Z = -2 + j3 and 4Z = -5 – j, evaluate in a + jb form:

(a) 1 3

1 3

Z ZZ Z+

(b) 12 3

4

ZZ ZZ

+ +

(a) 1 3

1 3

Z ZZ Z+

= 2(1 j2)( 2 j3) 2 j3 j4 j 6 8 j

(1 j2) ( 2 j3) 1 j5 1 j5+ − + − + − + − −

= =+ + − + − + − +

= 2

2 2

( 8 j)( 1 j5) 8 j40 j j 5 3 j41( 1 j5)( 1 j5) 1 5 26− − − − + + + +

= =− + − − +

= 3 41j26 26

+

(b) 12 3

4

ZZ ZZ

+ + = (4 – j3) + 1 j25 j+

− − + (-2 + j3) = 4 – j3 + 2 2

(1 j2)( 5 j)5 1

+ − ++

- 2 + j3

= 4 – j3 + 25 j j10 j 2

26− + − + - 2 + j3

= 4 – j3 + 7 j926

− − - 2 + j3 = 2 - 7 9j26 26

−

= 52 7 9j26 26 26

− − = 45 9j26 26

−

9. Show that 25 1 j2 2 j5 57 j242 3 j4 j

⎛ ⎞− + −− = +⎜ ⎟+ −⎝ ⎠

2

2 2

1 j2 (1 j2)(3 j4) 3 j4 j6 j 8 11 j2 11 2j3 j4 3 4 25 25 25 25+ + − − + − +

= = = = ++ +

2

2

2 j5 (2 j5)( j) j2 j 5 5 j2 5 j2j j( j) j 1

− − − += = = = +

− − −

L.H.S. = 25 1 j2 2 j5 25 11 2 25 11 2j (5 j2) 5 j 22 3 j4 j 2 25 25 2 25 25

⎛ ⎞− + − ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − + − + = − − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥+ − ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎝ ⎠

= 25 11 125 2 50 25 114 48j j2 25 25 2 25 25⎡ − − ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤− + = − − −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎣ ⎦

= 25 114 25 48j2 25 2 25⎛ ⎞ ⎛ ⎞− − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 57 + j24 = R.H.S.



2. Solve the complex equation: 2 j j(x jy)1 j+

= +−

2 j j(x jy)1 j+

= +−

hence, (2 j)(1 j) j(x jy)(1 j)(1 j)+ +

= +− +

i.e. 2

22 2

2 j2 j j jx j y1 1

+ + += +

+

i.e. 1 j3 jx y2+

= −

i.e. 1 3j2 2+ = -y + jx

Hence, x = 32

and y = 12

−

3. Solve the complex equation: (2 j3) (a jb)− = + (2 j3) (a jb)− = +

Squaring both sides gives: ( )22 j3 a jb− = +

(2 – j3)(2 – j3) = a + jb

i.e. 4 – j6 – j6 + 2j 9 = a + jb

i.e. -5 – j12 = a + jb

Hence, a = -5 and b = -12

5. If Z = R + jωL + 1j Cω

, express Z in (a + jb) form when R = 10, L = 5, C = 0.04 and ω = 4

Z = R + jωL + 1j Cω

= 10 + j(4)(5) + 1j(4)(0.04)

= 10 + j20 + 6.25j

= 10 + j20 + 6.25( j)j( j)

−−

= 10 + j20 - 2

6.25j−

= 10 + j20 – j6.25

= 10 + j13.75


EXERCISE 103 Page 256 2. Express the following Cartesian complex numbers in polar form, leaving answers in surd form:

(a) 2 + j3 (b) -4 (c) -6 + j (a) 2 + j3 From the diagram below, r = 2 22 3 13+ =

and 1 3tan 56.31 or 56 19 '2

− ⎛ ⎞θ = = ° °⎜ ⎟⎝ ⎠

Hence, 2 + j3 = 13 56 19'∠ ° in polar form (b) -4 = -4 + j0 and is shown in the diagram below, where r = 4 and θ = 180°

Hence, -4 = 4 180∠ ° in polar form (c) -6 + j From the diagram below, r = 2 26 1 37+ =

and 1 1tan 9.46 or 9 28'6

− ⎛ ⎞α = = ° °⎜ ⎟⎝ ⎠

thus θ = 180° - 9°28′ = 170°32′

Thus, -6 + j = 37 170 32'∠ °


3. Express the following Cartesian complex numbers in polar form, leaving answers in surd form:

(a) - j3 (b) ( )32 j− + (c) 3j (1 j)−

(a) - j3 From the diagram below, r = 3 and θ = -90°

Hence, - j3 = 3∠-90° in polar form

(b) ( )32 j− + = (-2 + j)(-2 + j)(-2 + j) = (4 – j2 – j2 + 2j )(-2 + j)

= (3 – j4)(-2 + j) = -6 + j3 + j8 - 2j 4 = -2 + j11

From the diagram below, r = 2 22 11 125+ = and 1 11tan 79.70 or 79 42 '2

− ⎛ ⎞α = = ° °⎜ ⎟⎝ ⎠

and θ = 180° - 79°42′ = 100°18′

Hence, ( )32 j− + = -2 + j11 = 125 100 18'∠ ° in polar form

(c) 3j (1 j)− = (j)( 2j )(1 – j) = -j(1 – j) = -j + 2j = -1 – j

From the diagram below, r = 2 21 1 2+ = and 1 1tan 451

− ⎛ ⎞α = = °⎜ ⎟⎝ ⎠

and θ = 180° - 45° = 135°

Hence, 3j (1 j)− = -1 – j = 2 135∠− °


5. Convert the following polar complex numbers into (a + jb) form giving answers correct to 4

significant figures: (a) 6∠125° (b) 4∠π (c) 3.5∠-120° (a) 6∠125° = 6 cos 125° + j 6 sin 125° = -3.441 + j4.915 (b) 4∠π = 4 cos π + j sin π = - 4.000 + j0 (Note that π is radians) (c) 3.5∠-120° = 3.5 cos(-120°) + j 3.5 sin(-120°) = -1.750 – j3.031 6. Evaluate in polar form: (a) 3 20 15 45∠ °× ∠ ° (b) 2.4 65 4.4 21∠ °× ∠− ° (a) 3 20 15 45∠ °× ∠ ° = 3 15 (20 45 )× ∠ °+ ° = 45∠65° (b) 2.4 65 4.4 21∠ °× ∠− ° = 2.4 4.4 (65 21 )× ∠ °+ − ° = 10.56∠44° 7. Evaluate in polar form: (a) 6.4 27 2 15∠ °÷ ∠− ° (b) 5 30 4 80 10 40∠ °× ∠ °÷ ∠− °

(a) 6.4 27 2 15∠ °÷ ∠− ° = 6.4 27 6.4 27 152 15 2

∠ °= ∠ °− − °

∠− ° = 3.2∠42°

(b) 5 30 4 80 10 40∠ °× ∠ °÷ ∠− ° = 5 30 4 80 5 4 (30 80 40 )10 40 10∠ °× ∠ ° ×

= ∠ °+ °− − °∠− °

= 2∠150°

8. Evaluate in polar form: (a) 4 36 8π π

∠ + ∠ (b) 2 120 5.2 58 1.6 40∠ °+ ∠ °− ∠− °

(a) 4 36 8π π

∠ + ∠ = 4cos j4sin 3cos j3sin6 6 8 8π π π π⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ = (3.464 + j2) + (2.772 + j1.148)

= 6.236 + j3.148

From the diagram below, r = 2 26.236 3.148 6.986+ =

and 1 3.148tan 26.79 or 26 47 ' or 0.467rad6.236

− ⎛ ⎞θ = = ° °⎜ ⎟⎝ ⎠


Hence, 4 36 8π π

∠ + ∠ = 6.986∠26°47′ or 6.986∠0.467 rad

(b) 2 120 5.2 58 1.6 40∠ °+ ∠ °− ∠− °

= (2 cos 120° + j 2 sin 120°) + (5.2 cos 58° + j 5.2 sin 58°) – (1.6 cos(-40°) + j 1.6 sin(-40°))

= (-1 + j 1.732) + (2.756 + j4.410) – (1.226 – j1.028)

= -1 + j1.732 + 2.756 + j4.410 – 1.226 + j1.028

= 0.530 + j7.170

From the diagram below, r = 2 20.530 7.170 7.190+ =

and 1 7.170tan 85.77 or 85 46 '0.530

− ⎛ ⎞θ = = ° °⎜ ⎟⎝ ⎠

Hence, 2 120 5.2 58 1.6 40∠ ° + ∠ ° − ∠ − ° = 7.190∠85°46′


EXERCISE 104 Page 259 1. Determine the resistance R and series inductance L (or capacitance C) for each of the following

impedances assuming the frequency to be 50 Hz:

(a) (3 + j8) Ω (b) (2 – j3) Ω (c) j14 Ω (d) 8 60∠− °Ω (a) If Z = (3 + j8) Ω then resistance, R = 3 Ω and inductive reactance, LX = 8 Ω (since the j

term is positive)

LX = 2πfL = 8 hence, inductance, L = 8 82 f 2 (50)

=π π

= 0.0255 H or 25.5 mH

(b) If Z = (2 – j3) Ω then resistance, R = 2 Ω and capacitive reactance, CX = 3 Ω (since the j

term is negative)

C1X

2 fC=

π = 3 hence, capacitance, C = 3 61 1 1.061 10 or 1061 10

2 f (3) 2 (50)(3)− −= = × ×

π π

= 1061 µF

(c) If Z = j14 Ω i.e. Z = (0 + j14) Ω then resistance, R = 0 Ω and LX = 14 Ω

i.e. 2πfL = 14 hence, inductance, L = 142 (50)π

= 0.04456 H or 44.56 mH

(d) If Z = 8 60∠− °Ω = 8 cos(-60°) + j 8 sin(-60°) = (4 – j6.928) Ω

Hence, resistance, R = 4 Ω and CX = 6.928 Ω

i.e. 1 6.9282 fC

=π

and capacitance, C = 61 459.4 102 (50)(6.928)

−= ×π

= 459.4 µF

2. Two impedances, 1Z (3 j6)= + Ω and 2Z (4 j3)= − Ω are connected in series to a supply

voltage of 120 V. Determine the magnitude of the current and its phase angle relative to the

voltage. In a series circuit, total impedance, TOTAL 1 2Z Z Z= + = (3 + j6) + (4 – j3) = (7 + j3) Ω

= 2 2 1 37 3 tan7

− ⎛ ⎞+ ∠ ⎜ ⎟⎝ ⎠

= 7.616∠23.20° Ω or 7.616∠23°12′ Ω

Since voltage V = 120∠0° V, then current, I = V 120 0Z 7.616 23 12 '

∠ °=

∠ ° = 15.76∠-23°12′ A

i.e. the current is 15.76 A and is lagging the voltage by 23°12′


3. If the two impedances in Problem 2 are connected in parallel determine the current flowing and

its phase relative to the 120 V supply voltage. In a parallel circuit shown below, the total impedance, TZ is given by:

2 2 2 2T 1 2

1 1 1 1 1 3 j6 4 j3 3 6 4 3j jZ Z Z 3 j6 4 j3 3 6 4 3 45 45 25 25

− += + = + = + = − + +

+ − + +

i.e. TT

1 admit tan ce,YZ

= = 0.22667 – j0.01333 = 0.2271∠3°22′ siemen

Current, I = TT

V VY (120 0 )(0.2271 3 22 ') 27.25 3 22 'Z

= = ∠ ° ∠ ° = ∠ ° A

i.e. the current is 27.25 A and is leading the voltage by 3°22′

5. For the circuit shown, determine the current I flowing and its phase relative to the applied

voltage.

2 2 2 2T 1 2 3

1 1 1 1 1 1 1 30 j20 40 j50 1Z Z Z Z 30 j20 40 j50 25 30 20 40 50 25

+ −= + + = + + = + +

− + + +

= 30 20 40 50 1j j1300 1300 4100 4100 25

+ + − +

i.e. TT

1 admit tan ce,YZ

= = 0.07283 + j0.00319 = 0.0729∠2°30′ S


Current, I = TT

V VY (200 0 )(0.0729 2 30 ') 14.6 2 30 'Z

= = ∠ ° ∠ ° = ∠ ° A

i.e. the current is 14.6 A and is leading the voltage by 2°30′

7. A delta-connected impedance AZ is given by: 1 2 2 3 3 1A

2

Z Z Z Z Z ZZZ

+ +=

Determine AZ in both Cartesian and polar form given 1Z (10 j0)= + Ω , 2Z (0 j10)= − Ω and

3Z (10 j10)= + Ω .

1 2 2 3 3 1A

2

Z Z Z Z Z Z (10 j0)(0 j10) (0 j10)(10 j10) (10 j10)(10 j0)ZZ (0 j10)

+ + + − + − + + + += =

−

= 2j100 j100 j 100 100 j100 200 j100 200 j100j10 j10 j10 j10

− − − + + −= = −

− − − −

= j200 10 (10 j20)10

+ = + Ω

From the diagram below, r = 2 210 20+ = 22.36 and 1 20tan 63.4310

− ⎛ ⎞θ = = °⎜ ⎟⎝ ⎠

Hence, AZ (10 j20) 22.36 63.43= + Ω = ∠ °Ω 8. In the hydrogen atom, the angular momentum, p, of the de Broglie wave is given by:

( )jhp jm2

⎛ ⎞Ψ = − ± Ψ⎜ ⎟π⎝ ⎠

Determine an expression for p.

If ( )jhp jm2

⎛ ⎞Ψ = − ± Ψ⎜ ⎟π⎝ ⎠ then p = ( ) ( )( )2jh jm jh hjm j m

2 2 2± Ψ⎛ ⎞⎛ ⎞ ⎛ ⎞− = − ± = − ±⎜ ⎟⎜ ⎟ ⎜ ⎟π Ψ π π⎝ ⎠⎝ ⎠ ⎝ ⎠

= ( )h m2

±π

= mh2

±π


10. Three vectors are represented by P, 2∠30°, Q, 3∠90° and R, 4∠-60°. Determine in polar form

the vectors represented by (a) P + Q + R (b) P – Q – R (a) P + Q + R = 2∠30° + 3∠90° + 4∠-60° = (1.732 + j1) + (0 + j3) + (2 – j3.464)

= (3.732 + j0.536) = 3.770∠8.17° (b) P – Q – R = 2∠30° - 3∠90° - 4∠-60° = (1.732 + j1) - (0 + j3) - (2 – j3.464)

= (-0.268 + j1.464))

From the diagram below, r = 1.488 and 1 1.464tan 79.630.268

− ⎛ ⎞α = = °⎜ ⎟⎝ ⎠

and 180 79.63 100.37θ = ° − ° = °

Hence, P – Q – R = 1.488∠100.37°

11. In a Schering bridge circuit, XX X CZ (R jX )= − ,

22 CZ jX= − , ( )( )( )

3

3

3 C3

3 C

R jXZ

R jX

−=

− and 4 4Z R= ,

where C1X

2 fC=

π. At balance: ( )( ) ( )( )X 3 2 4Z Z Z Z= .

Show that at balance 3 4X

2

C RRC

= and 2 3X

4

C RCR

=

Since ( )( ) ( )( )X 3 2 4Z Z Z Z=

then ( )( ) ( )( )3

X 2

3

3 CX C C 4

3 C

R jX(R jX ) jX R

R jX

⎧ ⎫−⎪ ⎪− = −⎨ ⎬−⎪ ⎪⎩ ⎭

Thus, ( )( )( )

( )( )3 2

X

3

3 C C 4X C

3 C

R jX jX R(R jX )

R jX

− −− =

−

i.e. ( )( ) ( )( )

3 22

X

3 3

2C C 43 C 4

X C3 C 3 C

j X X RjR X R(R jX )

R jX R jX

−− = +

− −


i.e. ( )

2 2

X

3

C 4 C 4X C

C 3

X R X R(R jX )

X R ( j)− = −

−

= 2 2

3

C 4 C 4

C 3

X R X RX jR

+

i.e. XX C(R jX )− = 2 2

3

C 4 C 4

C 3

X R X Rj

X R−

Equating the real parts gives: 2

3

4C 4 32

X 4C 2

3

1 RX R 2 fC2 fCR R1X 2 fC2 fC

ππ= = =

ππ

i.e. 3 4X

2

C RR

C=

Equating the imaginary parts gives: 2

X

C 4C

3

X RX

R− = −

i.e. 4

2 4

X 3 2 3

1 R2 fC R1

2 fC R 2 fC Rπ

= =π π

from which, 2 3X

4

C RC

R=


CHAPTER 24 DE MOIVRE’S THEOREM EXERCISE 105 Page 261 2. Determine in polar and Cartesian forms (a) [ ] 43 41∠ ° (b) ( )52 j− − (a) [ ] 4 43 41 3 4 41∠ ° = ∠ × ° = 81∠164° = 8 cos 164° + j 8 sin 164° = -77.86 + j22.33

(b) ( ) ( )5552 j 5 153.435 5 5 153.435⎡ ⎤− − = ∠− ° = ∠ ×− °⎣ ⎦

= 55.90∠-767.175° = 55.90∠-47°10′

= 55.90 cos -47°10′ + j 55.90 sin -47°10′

= 38 – j41 3. Convert (3 – j) into polar form and hence evaluate ( )73 j− , giving the answer in polar form.

(3 – j) = 2 2 1 13 1 tan3

− ⎛ ⎞+ ∠ −⎜ ⎟⎝ ⎠

= 10 18 26'∠− °

Hence, ( ) ( )7773 j 10 18 26 ' 10 7 18 26 '⎡ ⎤− = ∠− ° = ∠ ×− °⎣ ⎦ = 3162∠-129°2′

5. Express in both polar and rectangular forms: ( )53 j8−

( ) ( )5553 j8 73 69.444 73 5 69.444⎡ ⎤− = ∠− ° = ∠ ×− °⎣ ⎦ = 45530∠-347.22° = 3162∠12°47′

= 45530 cos 12°47′ + j45530 sin 12°47′

= 44400 + j10070

7. Express in both polar and rectangular forms: ( )616 j9− −

From the diagram below, r = 2 216 9 337+ = and 1 9tan 29.35816

− ⎛ ⎞α = = °⎜ ⎟⎝ ⎠

and 180 29.358 209.358θ = °+ ° = °

Hence, ( ) ( )66616 j9 337 209.358 337 6 209.358⎡ ⎤− − = ∠ ° = ∠ × °⎣ ⎦


= 638.27 10 1256.148× ∠ ° = 6(38.27 10 ) 176 9'× ∠ °

( )6 6(38.27 10 ) 176 9 ' 10 38.27 cos176 9 ' j38.27sin176 9 '× ∠ ° = ° + °

= 610 ( 38.18 j2.570)− +


EXERCISE 106 Page 263 2. Determine the two square roots of the following complex numbers in Cartesian form and show

the results on an Argand diagram: (a) 3 – j4 (b) -1 – j2

(a) [ ] [ ]123 j4 5 53.13 5 53.13− = ∠− ° = ∠− °

The first root is: 12 15 53.13 2.236 26.57 (2 j1)

2∠ ×− ° = ∠− ° = −

and the second root is: 52.236 ( 26.57 180 ) ( 2 j1)∠ − °+ ° = − +

Hence, (3 j4) (2 j)− = ± − as shown in the Argand diagram of Figure (a) below.

(a) (b)

(b) 121 j2 5 243.435 5 243.435⎡ ⎤ ⎡ ⎤− − = ∠ ° = ∠ °⎣ ⎦ ⎣ ⎦

The first root is: ( )12 15 243.435 1.495 121.72 ( 0.786 j1.272)

2∠ × ° = ∠ ° = − +

and the second root is: 1.495 (121.72 180 ) 1.495 58.28 (0.786 j1.272)∠ °− ° = ∠− = −

Hence, ( 1 j2) (0.786 j1.272)− − = ± − as shown in the Argand diagram if Figure (b) above.

4. Determine the modulus and argument of the complex number: ( )133 j4+

( ) ( )11 1

333 313 j4 5 53.13 5 53.13 5 17.71 1.710 17 43'3

+ = ∠ ° = ∠ × ° = ∠ ° = ∠ °

Hence, the modulus is 1.710, and the arguments are 17°43′, 17°43′ + 3603° = 137°43′,

and 137°43′ + 3603° = 257°43′ since the three roots are equally displaced by 120°.

5. Determine the modulus and argument of the complex number: ( )142 j− +


( ) ( )111444

12 j 5 153.435 5 153.435 1.223 38 22 '4

⎡ ⎤− + = ∠ ° = ∠ × ° = ∠ °⎣ ⎦

There are 4 roots equally displaced 3604° , i.e. 90° apart.

Hence, the modulus is 1.223, and the arguments are: 38°22′, 38°22′ + 90° = 128°22′,

128°22′ + 90° = 218°22′ and 218°22′ + 90° = 308°22′

7. Determine the modulus and argument of the complex number: ( )234 j3 −−

( ) [ ] ( )22 233 3

24 j3 5 36.87 5 36.87 0.3420 24 35'3

−− −− = ∠− ° = ∠− ×− ° = ∠ °

There are 3 roots equally displaced 3603° , i.e. 120° apart.

Hence, the modulus is 0.3420, and the arguments are: 24°35′, 24°35′ + 120° = 144°35′,

and 144°35′ + 120° = 264°35′ 8. For a transmission line, the characteristic impedance 0Z and the propagation coefficient γ are

given by:

0R j LZG j C

⎛ ⎞+ ω= ⎜ ⎟+ ω⎝ ⎠

and ( )( )R j L G j Cγ = + ω + ω⎡ ⎤⎣ ⎦

Given R = 25 Ω, L = 35 10−× H, G = 680 10−× siemens, C = 60.04 10−× F and ω = 2000π rad/s,

determine, in polar form, 0Z and γ.

( )3R j L 25 j(2000 ) 5 10 25 j31.416 40.15 51.49−+ ω = + π × = + = ∠ °

( )6 6 6 6G j C 80 10 j(2000 ) 0.04 10 10 (80 j251.33) 263.755 10 72.34− − − −+ ω = × + π × = + = × ∠ °

Hence, ( )0 6

R j L 40.15 51.49Z 152224.6 20.85G j C 263.755 10 72.34−

⎛ ⎞+ ω ∠ °⎛ ⎞= = = ∠− °⎜ ⎟ ⎜ ⎟+ ω × ∠ °⎝ ⎠⎝ ⎠

= ( )1152224.6 20.852

∠ − ° = 390.2∠-10.43° Ω

( )( ) ( )( )6R j L G j C 40.15 51.49 263.755 10 72.34−⎡ ⎤γ = + ω + ω = ∠ ° × ∠ °⎡ ⎤⎣ ⎦ ⎣ ⎦

= ( ) 10.01058976 123.83 0.01058976 123.832

∠ ° = ∠ × °

= 0.1029∠61.92°


EXERCISE 107 Page 265 2. Convert (-2.5 + j4.2) into exponential form. (-2.5 + j4.2) = 4.89 120.76 or 4.89 2.11rad∠ ° ∠

and 4.89∠2.11 ≡ 4.89 j 2.11e

4. Convert 1.2 j 2.51.7e − into rectangular form.

( )( )1.2 j 2.5 1.2 j 2.5 1.21.7e 1.7e e 1.7e 2.5rad− −= = ∠− = 1.2 1.21.7e cos( 2.5) j1.7e sin( 2.5)− + −

= -4.52 – j3.38

6. If j 2.1z 7e= , determine ln z (a) in Cartesian form, and (b) in polar form. (a) If j 2.1z 7e= then ln z = ( )j 2.1 j 2.1ln 7e ln 7 ln e= + = ln 7 + j2.1 in Cartesian form (b) ln 7 + j2.1 = 2.86∠47.18° or 2.86∠0.82 rad 8. Determine in polar form (a) ln(2 + j5) (b) ln(-4 – j3) (a) ln(2 + j5) = ( ) ( )j1.19 j1.19ln 29 1.19 ln 29 e ln 29 ln e∠ = = +

= ln 29 j1.19 1.6836 j1.19+ = +

= 2.06∠35.25° or 2.06∠0.615 rad (b) ln(-4 – j3) = ( ) ( ) ( )j3.785ln 5 216.87 ln 5 3.785 ln 5e∠ ° = ∠ =

= ln 5 j3.785 1.6094 j3.785+ = +

= 4.11∠66.96° or 4.11∠1.17 rad 9. When displaced electrons oscillate about an equilibrium position the displacement x is given by

the equation:

( )24mf hht j t2m 2m a

x A e

⎧ ⎫−⎪ ⎪− +⎨ ⎬−⎪ ⎪⎩ ⎭=

Determine the real part of x in terms of t, assuming ( )24mf h− is positive.


( )2

24mf hht j t 4mf hh t h t 22m 2m a j t

2m a2m 2m 4mf hx A e Ae e Ae t2m a

⎧ ⎫−⎪ ⎪− +⎨ ⎬ −−⎪ ⎪ − −⎩ ⎭ −

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ −⎜ ⎟= = = ∠⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

= ht h t2 2

2m 2m4mf h 4mf hAe cos t jAe sin t2m a 2m a

− −⎛ ⎞ ⎛ ⎞− −+⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

Hence, the real part is: ht 22m 4mf hAe cos t

2m a− ⎛ ⎞−

⎜ ⎟⎜ ⎟−⎝ ⎠


CHAPTER 25 THE THEORY OF MATRICES AND

DETERMINANTS EXERCISE 108 Page 270

2. Determine

13 624 7 6

22 4 0 5 73

5 7 4 31 05

⎛ ⎞⎜ ⎟

−⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟− + −⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎜ ⎟⎝ ⎠

⎜ ⎟−⎜ ⎟⎝ ⎠

1 13 6 (4 3) ( 7 6) (6 )2 24 7 6

2 22 4 0 5 7 ( 2 5) (4 ) (0 7)3 3

5 7 4 3 31 0 (5 1) (7 0) ( 4 )5 5

⎛ ⎞ ⎛ ⎞+ − + +⎜ ⎟ ⎜ ⎟−⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + − = − + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎜ ⎟ ⎜ ⎟⎝ ⎠⎜ ⎟ ⎜ ⎟− + − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

17 1 62

13 3 73

24 7 35

⎛ ⎞−⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

4. Determine

1 23 1 1.3 7.42 34 7 1 3 2.5 3.9

3 5

⎛ ⎞⎜ ⎟− −⎛ ⎞ ⎛ ⎞

+ −⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎜ ⎟⎝ ⎠ ⎝ ⎠− −⎜ ⎟⎝ ⎠

1 2 1 2(3 1.3) ( 1 7.43 1 1.3 7.42 3 2 3

4 7 1 3 2.5 3.9 1 3( 4 2.5) (7 3.9)3 5 3 5

⎛ ⎞ ⎛ ⎞+ − − − + −⎜ ⎟ ⎜ ⎟− −⎛ ⎞ ⎛ ⎞+ − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− − − + − − + − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= (3 0.5 1.3) ( 1 0.666 7.4

( 4 0.333 2.5) (7 0.6 3.9)+ + − + −⎛ ⎞

⎜ ⎟− − − − +⎝ ⎠

= 4.8 7.7 3

6.83 10.3

⎛ ⎞−⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

6. Determine

13 624 7 6 3.1 2.4 6.4

22 2 4 0 3 5 7 4 1.6 3.8 1.93

5 7 4 5.3 3.4 4.831 05

⎛ ⎞⎜ ⎟

−⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− + − − − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎜ ⎟⎝ ⎠ ⎝ ⎠

⎜ ⎟−⎜ ⎟⎝ ⎠


13 624 7 6 3.1 2.4 6.4

22 2 4 0 3 5 7 4 1.6 3.8 1.93

5 7 4 5.3 3.4 4.831 05

⎛ ⎞⎜ ⎟

−⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− + − − − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠ ⎝ ⎠

⎜ ⎟−⎜ ⎟⎝ ⎠

= (8 9 12.4) ( 14 18 9.6) (12 1.5 25.6)

( 4 15 6.4) (8 2 15.2) (0 21 7.6)(10 3 21.2) (14 0 13.6) ( 8 1.8 19.2

+ − − + − + −⎛ ⎞⎜ ⎟− + + − − + +⎜ ⎟⎜ ⎟− − + − − + +⎝ ⎠

= 4.6 5.6 12.1

17.4 9.2 28.614.2 0.4 13.0

− −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

8. Determine

1 23 1 2 34 7 1 3

3 5

⎛ ⎞⎜ ⎟−⎛ ⎞

× ⎜ ⎟⎜ ⎟− ⎜ ⎟⎝ ⎠ − −⎜ ⎟⎝ ⎠

1 2 3 1 3( ) (2 )3 1 2 3 2 3 5

4 7 1 3 7 8 21( 2 ) ( )3 5 3 3 5

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟−⎛ ⎞× =⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎜ ⎟ ⎜ ⎟⎝ ⎠ − − − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

5 31 26 51 134 63 15

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠

10. Determine 4 7 6 42 4 0 11

5 7 4 7

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− × −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

4 7 6 4 (16 77 42)2 4 0 11 ( 8 44 0)

5 7 4 7 (20 77 28)

− + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− × − = − − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 135

5285

⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

12. Determine 4 7 6 3.1 2.4 6.42 4 0 1.6 3.8 1.9

5 7 4 5.3 3.4 4.8

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− × − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

4 7 6 3.1 2.4 6.42 4 0 1.6 3.8 1.9

5 7 4 5.3 3.4 4.8

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− × − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

= (12.4 11.2 31.8) (9.6 26.6 20.4) (25.6 13.3 28.8)

( 6.2 6.4 0) ( 4.8 15.2 0) ( 12.8 7.6 0)(15.5 11.2 21.2) (12 26.6 13.6) (32 13.3 19.2)

+ + − + + −⎛ ⎞⎜ ⎟− − + − + + − − +⎜ ⎟⎜ ⎟− − + − − +⎝ ⎠

= 55.4 3.4 10.112.6 10.4 20.416.9 25.0 37.9

⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟−⎝ ⎠



2. Calculate the determinant of

1 22 31 33 5

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠

1 2

1 3 2 1 3 2 27 202 31 3 2 5 3 3 10 9 903 5

− +⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − − − = − + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠− −

= 790

−

4. Evaluate j2 j3

(1 j) j−

+

j2 j3

(1 j) j−

+ = (j2)(j) – (-j3)(1 + j) = 2j 2 + j3(1 + j) = -2 + j3 + 2j 3 = -2 + j3 – 3 = -5 + j3

5. Evaluate 2 40 5 207 32 4 117∠ ° ∠− °∠− ° ∠− °

( )( ) ( )( )2 40 5 20

2 40 4 117 5 20 7 327 32 4 117∠ ° ∠− °

= ∠ ° ∠− ° − ∠− ° ∠− °∠− ° ∠− °

= 8∠-77° - 35∠-52° = (1.800 – j7.795) – (21.548 – j27.580)

= (-19.75 + j19.79)

From the diagram below, r = 2 2(19.75 19.79 )+ = 27.96 and 1 19.79tan 45.0619.75

− ⎛ ⎞α = = °⎜ ⎟⎝ ⎠

and 180 45.06 134.94θ = °− ° = °

Hence, 2 40 5 207 32 4 117∠ ° ∠− °∠− ° ∠− °

= (-19.75 + j19.79) or 27.96∠134.94°



2. Determine the inverse of

1 22 31 33 5

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠

1 2

3 2 2 3 20 27 72 31 3 10 9 9 10 90 903 5

−= − − − = − = = −

− −

Hence, the inverse of

1 22 31 33 5

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠

is:

90 3 90 23 27 5 7 31 5 3

7 1 1 90 1 90 190 3 2 7 3 7 2

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ − − − −− − ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎜ ⎟=⎜ ⎟⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟− − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

=

54 607 730 457 7

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠

=

5 47 87 72 34 67 7

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠

3. Determine the inverse of 1.3 7.4

2.5 3.9−⎛ ⎞⎜ ⎟−⎝ ⎠

The inverse of 1.3 7.4

2.5 3.9−⎛ ⎞⎜ ⎟−⎝ ⎠

is: 3.9 7.412.5 1.3( 1.3)( 3.9) (7.4)(2.5)

− −⎛ ⎞⎜ ⎟− −− − − ⎝ ⎠

=

3.9 7.43.9 7.41 13.43 13.432.5 1.3 2.5 1.313.43

13.43 13.43

⎛ ⎞⎜ ⎟− −⎛ ⎞

= ⎜ ⎟⎜ ⎟− −− ⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠

= 0.290 0.5510.186 0.097⎛ ⎞⎜ ⎟⎝ ⎠



3. Calculate the determinant of 4 7 62 4 0

5 7 4

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

4 7 62 4 0 4( 16) ( 7)(8) 6( 34)5 7 4

−− = − − − + −

− using the top row

= -64 + 56 – 204 = -212

5. Calculate the determinant of 3.1 2.4 6.41.6 3.8 1.9

5.3 3.4 4.8

⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟−⎝ ⎠

3.1 2.4 6.41.6 3.8 1.9 3.1( 18.24 6.46) 2.4(7.68 10.07) 6.4( 5.44 20.14)

5.3 3.4 4.8− − = − + − + + − −

− using the first row

= 3.1(-11.78) – 2.4(17.75) + 6.4(-25.58) = -36.518 – 42.6 – 163.712 = -242.83

7. Evaluate 3 60 j2 1

0 (1 j) 2 300 2 j5

∠ °+ ∠ °

[ ]3 60 j2 1

0 (1 j) 2 30 3 60 j5(1 j) 4 30 j2(0 0) 1(0 0)0 2 j5

∠ °+ ∠ ° = ∠ ° + − ∠ ° − − + + using the top row

= ( ) ( )23 60 j5 j 5 4 30 3 60 j5 5 (3.464 j2)∠ ° + − ∠ ° = ∠ ° − − +

= ( ) ( )3 60 j5 5 3.464 j2 3 60 8.464 j3∠ ° − − − = ∠ ° − +

= ( )3 60 8.98 160.48 26.94 220.48∠ ° ∠ ° = ∠

= 26.94∠-139.52° or (-20.49 – j17.49) 8. Find the eigenvalues λ that satisfy the following equations:

(a) (2 ) 2

01 (5 )−λ

=− −λ

(b) ( )

(5 ) 7 50 (4 ) 1 02 8 3

−λ −−λ − =

− −λ


(a) (2 ) 2

01 (5 )−λ

=− −λ

hence, (2 - λ)(5 - λ) – (-2) = 0

i.e. 10 - 7λ + 2λ + 2 = 0

i.e. 2λ - 7λ + 12 = 0

and (λ - 4)( λ - 3) = 0

from which, λ - 4 = 0 or λ - 3 = 0

Thus, eignevalues, λ = 3 or 4

(b) ( )

(5 ) 7 50 (4 ) 1 02 8 3

−λ −−λ − =

− −λ

hence, ( )( ) [ ](5 ) 4 3 8 7(0 2) 5 0 2(4 ) 0−λ −λ − −λ + − + − − −λ =⎡ ⎤⎣ ⎦

i.e. 2(5 ) 12 8 14 40 10 0⎡ ⎤− λ − −λ + λ + − + − λ =⎣ ⎦

and ( )( )25 4 26 10 0−λ λ −λ − + − λ =

i.e. 2 3 25 5 20 4 26 10 0λ − λ − −λ + λ + λ + − λ =

and 3 26 11 6 0−λ + λ − λ + =

or 3 26 11 6 0λ − λ + λ − =

Let f(λ) = 3 26 11 6λ − λ + λ −

f(0) = -6

f(1) = 1 – 6 + 11 – 6 = 0 hence, (λ - 1) is a factor



Thus, (λ - 1)(λ - 2)(λ - 3) = 0

from which, eigenvalues, λ = 1 or 2 or 3



3. Determine the adjoint of 4 7 62 4 0

5 7 4

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

Matrix of cofactors of 4 7 62 4 0

5 7 4

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

is: 16 8 34

14 46 6324 12 2

− − −⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟− −⎝ ⎠

Adjoint = transpose of cofactors = 16 14 248 46 1234 63 2

− −⎛ ⎞⎜ ⎟− − −⎜ ⎟⎜ ⎟− −⎝ ⎠

5. Find the inverse of 4 7 62 4 0

5 7 4

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

From question 3 above, adjoint = 16 14 248 46 63

34 63 2

− −⎛ ⎞⎜ ⎟− − −⎜ ⎟⎜ ⎟− −⎝ ⎠

4 7 62 4 0 4( 16) 7(8) 6( 34) 2125 7 4

−− = − + + − = −

−

Hence, the inverse of 4 7 62 4 0

5 7 4

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

is: 16 14 24

1 8 46 12212

34 63 2

− −⎛ ⎞⎜ ⎟− − − −⎜ ⎟⎜ ⎟− −⎝ ⎠

6. Find the inverse of

13 62

25 73

31 05

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

Matrix of cofactors of

13 62

25 73

31 05

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

is:

2 2105 33 33 2 65 101 142 18 323 2

⎛ ⎞− − −⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠


Transpose of cofactors = adjoint =

2 3 13 425 5 3

3 110 2 1810 2

2 6 323

⎛ ⎞− −⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟⎜ ⎟⎜ ⎟− − −⎜ ⎟⎝ ⎠

13 62

25 73

31 05

−

−

= ( )2 1 2 6 60 1 18 900 5 9233 6 3 75 2 3 5 1 3 15 15

− − −⎛ ⎞ ⎛ ⎞− − + + − = − − − = = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Hence, the inverse of

13 62

25 73

31 05

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

is:

2 3 13 425 5 3

1 3 110 2 18923 10 215 2 6 32

3

⎛ ⎞− −⎜ ⎟⎜ ⎟⎜ ⎟− −⎜ ⎟

− ⎜ ⎟⎜ ⎟− − −⎜ ⎟⎝ ⎠

=

2 3 13 425 5 3

15 3 110 2 18923 10 2

2 6 323

⎛ ⎞− −⎜ ⎟⎜ ⎟⎜ ⎟− − −⎜ ⎟⎜ ⎟⎜ ⎟− − −⎜ ⎟⎝ ⎠


CHAPTER 26 THE SOLUTION OF SIMULTANEOUS

EQUATIONS BY MATRICES AND DETERMINANTS EXERCISE 113 Page 279 2. Use matrices to solve: 2p + 5q + 14.6 = 0

3.1p + 1.7 q + 2.06 = 0 2p + 5q = -14.6

3.1p + 1.7 q = -2.06

Hence, 2 5 p 14.6

3.1 1.7 q 2.06−⎛ ⎞⎛ ⎞ ⎛ ⎞

=⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

The inverse of 2 5

3.1 1.7⎛ ⎞⎜ ⎟⎝ ⎠

is: 1.7 5 1.7 51 13.1 2 3.1 23.4 15.5 12.1

− −⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟− −− −⎝ ⎠ ⎝ ⎠

Thus, p 1.7 5 14.6 14.521 1q 3.1 2 2.06 41.1412.1 12.1

− − −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −− −⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= 1.23.4

⎛ ⎞⎜ ⎟−⎝ ⎠

i.e. p = 1.2 and q = -3.4

3. Use matrices to solve: x + 2y + 3z = 5

2x – 3y – z = 3

-3x + 4y + 5z = 3 Since x + 2y + 3z = 5

2x – 3y – z = 3

-3x + 4y + 5z = 3

then, 1 2 3 x 52 3 1 y 33 4 5 z 3

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟− − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

Matrix of cofactors is: 11 7 12 14 107 7 7

− − −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

and the transpose of cofactors is: 11 2 77 14 71 10 7

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟− − −⎝ ⎠

1 2 32 3 13 4 5

− −−

= 1(-11) – 2 (7) + 3(-1) = -28


The inverse of 1 2 32 3 13 4 5

⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟−⎝ ⎠

is: 11 2 7

1 7 14 728

1 10 7

−⎛ ⎞⎜ ⎟−⎜ ⎟− ⎜ ⎟− − −⎝ ⎠

Thus, x 11 2 7 5 28 1

1 1y 7 14 7 3 28 128 28

z 1 10 7 3 56 2

− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − −⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i.e. x = 1, y = -1 and z = 2

5. Use matrices to solve: p + 2q + 3r + 7.8 = 0

2p + 5q – r – 1.4 = 0

5p – q + 7r – 3.5 = 0 Since p + 2q + 3r = - 7.8

2p + 5q – r = 1.4

5p – q + 7r = 3.5

from which, 1 2 3 p 7.82 5 1 q 1.45 1 7 r 3.5

−⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟− =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

Matrix of cofactors is: 34 19 2717 8 1117 7 1

− −⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟−⎝ ⎠

and the transpose of cofactors is: 34 17 1719 8 727 11 1

− −⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟−⎝ ⎠

1 2 32 5 15 1 7

−−

= 1(34) – 2 (19) + 3(-27) = -85

The inverse of 1 2 32 5 15 1 7

⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

is: 34 17 17

1 19 8 785

27 11 1

− −⎛ ⎞⎜ ⎟− −⎜ ⎟− ⎜ ⎟−⎝ ⎠

Thus, p 34 17 17 7.8 348.5 4.1

1 1q 19 8 7 1.4 161.5 1.985 85

r 27 11 1 3.5 229.5 2.7

− − − −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i.e. p = 4.1, q = -1.9 and r = -2.7


8. In a mechanical system, acceleration x , velocity x and distance x are related by the

simultaneous equations: 3.4 x 7.0 x 13.2x 11.39+ − = −

6.0 x 4.0 x 3.5x 4.98− + + =

2.7 x 6.0 x 7.1x 15.91+ + =

Use matrices to find the values of x , x and x.

Since 3.4 x 7.0 x 13.2x 11.39+ − = −

6.0 x 4.0 x 3.5x 4.98− + + =

2.7 x 6.0 x 7.1x 15.91+ + =

then,

x3.4 7.0 13.2 11.396.0 4.0 3.5 x 4.98

2.7 6.0 7.1 15.91x

⎛ ⎞− −⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Matrix of cofactors is: 7.4 52.05 46.8128.9 59.78 1.577.3 67.3 55.6

−⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠

and the transpose of cofactors is: 7.4 128.9 77.3

52.05 59.78 67.346.8 1.5 55.6

−⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎝ ⎠

3.4 7.0 13.26.0 4.0 3.5

2.7 6.0 7.1

−− = 3.4(7.4) – 7.0 (-52.05) + (-13.2)(-46.8) = 1007.27

The inverse of 3.4 7.0 13.26.0 4.0 3.5

2.7 6.0 7.1

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠

is: 7.4 128.9 77.2

1 52.05 59.78 67.31007.27

46.8 1.5 55.6

−⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟− −⎝ ⎠

Thus,

x 7.4 128.9 77.3 11.39 503.635 0.51 1x 52.05 59.78 67.3 4.98 775.5979 0.77

1007.27 1007.2746.8 1.5 55.6 15.91 1410.178 1.4x

⎛ ⎞ − −⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

i.e. x = 0.5, x = 0.77 and x = 1.4


EXERCISE 114 Page 282 1. Use determinants to solve the simultaneous equations: 3x – 5y = -17.6

7y – 2x – 22 = 0 Since 3x – 5y + 17.6 = 0

– 2x + 7y – 22 = 0

then x y 15 17.6 3 17.6 3 5

7 22 2 22 2 7

−= =

− −− − − −

i.e. x y 113.2 30.8 11

−= =

− −

from which, x = 13.211

− = -1.2 and y = 30.811

= 2.8

3. Use determinants to solve the simultaneous equations: 3x + 4y + z = 10

2x – 3y + 5z + 9 = 0

x + 2y – z = 6 Since 3x + 4y + z – 10 = 0

2x – 3y + 5z + 9 = 0

x + 2y – z – 6 = 0

then x y z 14 1 10 3 1 10 3 4 10 3 4 13 5 9 2 5 9 2 3 9 2 3 5

2 1 6 1 1 6 1 2 6 1 2 1

− −= = =

− − −− − −

− − − − − −

i.e. x y z 14( 21) 1(0) 10( 7) 3( 21) 1( 21) 10( 7) 3(0) 4( 21) 10(7) 3( 7) 4( 7) 1(7)

− −= = =

− − − − − − − − − − − − − − − +

i.e. x y z 114 28 14 14

− −= = =

−

Hence, x = 1414

= 1, y = 2814

= 2 and z = 1414− = -1


7. Applying mesh-current analysis to an a.c. circuit results in the following equations:

(5 – j4) 1I - (-j4) 2I = 100∠0°

(4 + j3 – j4) 2I - (-j4) 1I = 0

Solve the equations for 1I and 2I .

(5 – j4) 1I - (-j4) 2I - 100∠0° = 0

- (-j4) 1I + (4 + j3 – j4) 2I + 0 = 0

i.e. (5 – j4) 1I + j4 2I - 100 = 0

j4 1I + (4 – j) 2I + 0 = 0

Hence, 1 2I I 1j4 100 (5 j4) 100 (5 j4) j4

(4 j) 0 j4 0 j4 (4 j)

−= =

− − − −− −

i.e. ( )

1 22

I I 1100(4 j) j400 (5 j4)(4 j) j4

−= =

− − − −

i.e. 1 2I I 1400 j100 j400 32 j21

= =− − −

Thus, 1I = 400 j100 412.31 14.0432 j21 38.275 33.27

− ∠− °=

− ∠− ° = 10.77 19.23 A∠ °

and 2I = j400 400 9032 j21 38.275 33.27− ∠− °

=− ∠− °

= 10.45 56.73 A∠− °

9. The forces in three members of a framework are 1F , 2F and 3F . They are related by the

simultaneous equations shown below.

1.4 1F + 2.8 2F + 2.8 3F = 5.6

4.2 1F – 1.4 2F + 5.6 3F = 35.0

4.2 1F + 2.8 2F – 1.4 3F = -5.6

Find the values of 1F , 2F and 3F using determinants.

1.4 1F + 2.8 2F + 2.8 3F - 5.6 = 0

4.2 1F – 1.4 2F + 5.6 3F - 35.0 = 0

4.2 1F + 2.8 2F – 1.4 3F + 5.6 = 0


Hence, 31 2 FF F 12.8 2.8 5.6 1.4 2.8 5.6 1.4 2.8 5.6 1.4 2.8 2.81.4 5.6 35.0 4.2 5.6 35.0 4.2 1.4 35.0 4.2 1.4 5.62.8 1.4 5.6 4.2 1.4 5.6 4.2 2.8 5.6 4.2 2.8 1.4

− −= = =

− − −− − − − − −

− − −

i.e. 1 2F F2.8( 17.64) 2.8(90.16) 5.6( 13.72) 1.4( 17.64) 2.8(170.52) 5.6( 29.4)

−=

− − − − − − − −

= 3F 11.4(90.16) 2.8(170.52) 5.6(17.64) 1.4( 13.72) 2.8( 29.4) 2.8(17.64)

−=

− − − − − +

i.e. 31 2 FF F 1225.008 337.512 450.016 112.504

− −= = =

− − −

Thus, 1F = 225.008112.504

= 2 2F = 337.512112.504− = -3 and 3F = 450.016

112.504 = 4

10. Mesh-current analysis produces the following three equations: 1 220 0 (5 3 j4)I (3 j4)I∠ ° = + − − − 2 1 310 90 (3 j4 2)I (3 j4)I 2I∠ ° = − + − − − 3 215 0 10 90 (12 2)I 2I− ∠ °− ∠ ° = + − Solve the equations for the loop currents 1 2 3I , I and I Rearranging gives: (8 – j4) 1I - (3 – j4) 2I + 0 3I - 20 = 0

-(3 – j4) 1I + (5 – j4) 2I - 2 3I - j10 = 0

0 1I - 2 2I + 14 3I + (15 + j10) = 0 Hence,

31 2 II I(3 j4) 0 20 (8 j4) 0 20 (8 j4) (3 j4) 20

(5 j4) 2 j10 (3 j4) 2 j10 (3 j4) (5 j4) j102 14 (15 j10) 0 14 (15 j10) 0 2 (15 j10)

−= =

− − − − − − − − −− − − − − − − − − − −− + + − +

= 1(8 j4) (3 j4) 0(3 j4) (5 j4) 2

0 2 14

−− − −

− − − −−

i.e.

[ ] [ ] [ ] [ ]

1 2I I(3 j4) 2(15 j10) j140 20 14(5 j4) 4 (8 j4) 2(15 j10) j140 20 14(3 j4)

−=

− − − + + − − − − − + + − − −

= [ ] [ ] [ ]

3I(8 j4) (5 j4)(15 j10) j20) (3 j4) (3 j4)(15 j10) 20 2(3 j4)− − + − + − − − + − −

= [ ] [ ]

1(8 j4) 14(5 j4) 4 (3 j4) 14(3 j4)

−− − − + − − −


i.e. [ ] [ ] [ ] [ ]

1 2I I(3 j4) 30 j120 20 66 j56 (8 j4) 30 j120 20 42 j56)

−=

− − − + − − − − + − − +

= [ ] [ ]

3I(8 j4) 115 j30 (3 j4) 85 j30 40(3 j4)− − + − − + − −

= [ ] [ ]

1(8 j4) 66 j56 (3 j4) 42 j56

−− − + − − +

i.e. 1 2I I( 90 j360 j120 480) 1320 j1120 240 j960 j120 480 840 j1120

−=

− − + + + − + − + + + + −

= 3I920 j240 j460 120 255 j430 120 120 j160− − − − + + − +

= 1528 j448 j264 224 126 j168 j168 224

−− − − − + + +

i.e. 31 2 II I 1( 1710 j640) (1080 j40) (545 j110) (402 j376)

− −= = =

− + − − −

Hence, 1I = ( 1710 j640) 1825.84 20.52(402 j376) 550.44 43.09

− − + ∠− °=

− ∠− ° = 3.317 22.57 A∠ °

2I = (1080 j40) 1080.74 2.12(402 j376) 550.44 43.09

− ∠− °=

− ∠− ° = 1.963 40.97 A∠ °

3I = (545 j110) 555.99 11.41 555.99 191.41(402 j376) 550.44 43.09 550.44 43.09− − − ∠− ° ∠− °

= =− ∠− ° ∠− °

= 1.010 148.32 A∠− °


EXERCISE 115 Page 283 1. Q(3), Exercise 113

Use Cramers rule to solve: x + 2y + 3z = 5

2x – 3y – z = 3

-3x + 4y + 5z = 3

x =

5 2 33 3 13 4 5 5( 11) 2(18) 3(21) 281 2 3 1( 11) 2(7) 3( 1) 282 3 13 4 5

− −

− − + −= =

− − + − −− −

−

= 1

y =

1 5 32 3 13 3 5 1(18) 5(7) 3(15) 28

28 28 28

−− − +

= =− − −

= -1

z =

1 2 52 3 33 4 3 1( 21) 2(15) 5( 1) 56

28 28 28

−− − − + − −

= =− − −

= 2

1. Q(7), Exercise 113

Use Cramers rule to solve: s + 2 v + 2a = 4

3s – v + 4a = 25

3s + 2v – a = -4

s =

4 2 225 1 4

4 2 1 4( 7) 2( 9) 2(46) 821 2 2 1( 7) 2( 15) 2(9) 413 1 43 2 1

−− − − − − +

= =− − − +

−−

= 2

v =

1 4 23 25 43 4 1 1( 9) 4( 15) 2( 87) 123

41 41 41− − − − − + − −

= = = -3


a =

1 2 43 1 253 2 4 1( 46) 2( 87) 4(9) 164

41 41 41

−− − − − +

= = = 4

2. Q(8), Exercise 114 Use Cramers rule to solve: 1 2 3i 8i 3i 31+ + = − 1 2 33i 2i i 5− + = − 1 2 32i 3i 2i 6− + =

1i =

31 8 35 2 1

6 3 2 31( 1) 8( 16) 3(27) 2401 8 3 1( 1) 8(4) 3( 5) 483 2 12 3 2

−− −

− − − − − += =

− − + − −−−

= -5

2i =

1 31 33 5 12 6 2 1( 16) 31(4) 3(28) 192

48 48 48

−−

− + += =

− − − = -4

3i =

1 8 313 2 52 3 6 1( 27) 8(28) 31( 5) 96

48 48 48

−− −− − − − − −

= =− − −

= 2


EXERCISE 116 Page 285 1. In a mass-spring-damper system, the acceleration , velocity and displacement x m are related by

the following simultaneous equations:

6.2 x 7.9 x 12.6x 18.0+ + =

7.5x 4.8x 4.8x 6.39+ + =

13.0 x 3.5x 13.0x 17.4+ − = −

By using Gaussian elimination, determine the acceleration, velocity and displacement for the

system, correct to 2 decimal places.

6.2 x 7.9 x 12.6x 18.0+ + = (1)

7.5x 4.8x 4.8x 6.39+ + = (2)

13.0 x 3.5x 13.0x 17.4+ − = − (3)

(2) - 7.56.2

× (1) gives: 0 – 4.7565 x - 10.442 x = -15.384 (2′)

(3) - 13.06.2

× (1) gives: 0 – 13.065 x - 39.419 x = -55.142 (3′)

(3′) - 13.0654.7565−−

× (2′) gives: 0 + 0 – 10.737 x = -12.886

from which, x = 12.88610.737−−

= 1.2

From (3′), -13.065 x - 39.419(1.2) = -55.142

i.e. -13.065 x = -55.142 + 39.419(1.2)

and x = 55.142 39.419(1.2)13.065

− +−

= 0.60

From (1), 6.2 x 7.9(0.60) 12.6(1.2) 18.0+ + =

6.2 x 18.0 4.74 15.2 1.86= − − = −

and x = 1.866.2− = -0.30


2. The tensions, 1 2 3T , T and T in a simple framework are given by the equations:

1 2 35T 5T 5T 7.0+ + =

1 2 3T 2T 4T 2.4+ + =

1 24T 2T 4.0+ =

Determine 1 2 3T , T and T using Gaussian elimination.

1 2 35T 5T 5T 7.0+ + = (1)

1 2 3T 2T 4T 2.4+ + = (2)

1 2 34T 2T 0T 4.0+ + = (3)

(2) - 15

× (1) gives: 2 30 T 3T 1.0+ + = (2′)

(3) - 45

× (1) gives: 2 30 2T 4T 1.6− − = − (3′)

(3′) - 21−⎛ ⎞

⎜ ⎟⎝ ⎠

× (2′) gives: 32T 0.4=

from which, 3T 0.2=

In (3′) 22T 4(0.2) 1.6− − = −

22T 1.6 0.8 0.8− = − + = −

and 2T 0.4=

In (1) 15T 5(0.4) 5(0.2) 7.0+ + =

i.e. 15T 7.0 2.0 1.0 4.0= − − =

and 1T 0.8=


CHAPTER 27 METHODS OF DIFFERENTIATION EXERCISE 117 Page 291

1.(c) Find the differential coefficient with respect to x of 1x

If y = 11 xx

−= then 2dy 1xdx

−= − = 2

1x

−

2.(a) Find the differential coefficient with respect to x of 2

4x−

If y = 22

4 4xx

−−= − then ( )3 3dy ( 4) 2x 8x

dx− −= − − = or 3

8x

3. Find the differential coefficient with respect to x of (a) 2 x (b) 3 53 x (c) 4x

(a) If y = 122 x 2x= then

1 12 2

12

dy 1 1(2) x xdx 2 x

− −⎛ ⎞= = =⎜ ⎟

⎝ ⎠ = 1

x

(b) If y = 5

3 5 33 x 3x= then 2 23 3dy 5(3) x 5x

dx 3⎛ ⎞

= =⎜ ⎟⎝ ⎠

= 3 25 x

(c) If y = 12

12

4 4 4xx x

−= = then

3 32 2

32

dy 1 2(4) x 2xdx 2 x

− −⎛ ⎞= − = − = −⎜ ⎟

⎝ ⎠ =

3

2x

−

4.(a) Find the differential coefficient with respect to x of 3

3x−

If y = 13

133

3 3 3xx x

−− = − = − then

4 43 3

43

dy 1 1( 3) x xdx 3 x

− −⎛ ⎞= − − = =⎜ ⎟

⎝ ⎠ =

3 4

1x

5.(c) Find the differential coefficient with respect to x of 5x

3e

If y = 5x5x

3 3ee

−= then ( )5x 5xdy (3) 5e 15edx

− −= − = − = 5x

15e

−

7. Find the gradient of the curve 4 3y 2t 3t t 4= + − + at the points (0, 4) and (1, 8).

If 4 3y 2t 3t t 4= + − + , then gradient, 3 2dy 8t 9t 1dt

= + −


At (0, 4), t = 0, hence gradient = 3 28(0) 9(0) 1+ − = -1

At (1, 8), t = 1, hence gradient = 3 28(1) 9(1) 1+ − = 16

8. Find the co-ordinates of the point on the graph 2y 5x 3x 1= − + where the gradient is 2.

If 2y 5x 3x 1= − + , then gradient = dy 10x 3dx

= −

When the gradient is 2, 10x – 3 = 2 i.e. 10x = 5 and 1x2

=

When 1x2

= , 21 1 5 3 3y 5 3 1 1

2 2 4 2 4⎛ ⎞ ⎛ ⎞= − + = − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Hence, the co-ordinates of the point where the gradient is 2 is 1 3,2 4

⎛ ⎞⎜ ⎟⎝ ⎠

9. (a) Differentiate 2 3

2 2y 2ln 2 2(cos5 3sin 2 )e θ= + θ− θ+ θ −

θ

(b) Evaluate dydθ

in part (a) when 2π

θ = , correct to 4 significant figures.

(a) 2 3

2 2y 2ln 2 2(cos5 3sin 2 )e θ= + θ− θ+ θ −

θ

= 2 32 2ln 2 2cos5 6sin 2 2e− − θθ + θ− θ− θ−

Hence, ( )3 3dy 24 2( 5sin 5 ) 6(2cos 2 ) 2 3ed

− − θ= − θ + − − θ − θ − −θ θ

= 3 3

4 2 610sin 5 12cos 2e θ− + + θ − θ +

θ θ

(b) When 2π

θ = , dydθ

= 33

2

4 2 5 2 610sin 12cos2 2

e22

π⎛ ⎞⎜ ⎟⎝ ⎠

π π− + + − +

π⎛ ⎞π⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= -1.032049 + 1.2732395 + 10 + 12 + 0.0538997



EXERCISE 118 Page 293 1. Differentiate 32x cos3x with respect to x.

If y = 32x cos3x , then ( )( ) ( )( )3 2dy 2x 3sin 3x cos3x 6xdx

= − +

= 3 26x sin 3x 6x cos3x− + = ( )26x cos 3x xsin 3x−

3. Differentiate 3te sin 4t with respect to x.

If y = 3te sin 4t , then ( )( ) ( )( )3t 3tdy e 4cos 4t sin 4t 3edt

= +

= ( )3te 4cos 4t 3sin 4t+

5. Differentiate te ln t cos t with respect to x.

If y = te ln t cos t , then ( )( ) ( ) ( ) ( )( )t t tdy 1e ln t sin t cos t e ln t edt t

⎡ ⎤⎛ ⎞= − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= t cos te ln t sin t cos t ln tt

⎧ ⎫− + +⎨ ⎬⎩ ⎭

= t 1e ln t cos t ln t sin tt

⎧ ⎫⎛ ⎞+ −⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

6. Evaluate didt

, correct to 4 significant figures, when t = 0.1, and i = 15t sin3t.

Since i = 15t sin3t, then di (15t)(3cos3t) (sin 3t)(15)dt

= +

= 45t cos 3t + 15 sin 3t

When t = 0.1, didt

= 45(0.1) cos 0.3 + 15 sin 0.3 (note 0.3 is radians)

= 4.2990 + 4.4328




3. Differentiate the quotient 33

2sin 2θθ

with respect to x.

If y =

33 23 3

2sin 2 2sin 2θ θ

=θ θ

, then ( ) ( )

( )

1 32 2

2

92sin 2 3 4cos 22dy

d 2sin 2

⎛ ⎞ ⎛ ⎞θ θ − θ θ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=

θ θ

=

1 32 2

2

9 sin 2 12 cos 24sin 2

θ θ− θ θθ

=

2

3 3sin 2 4 cos 24sin 2

θ θ − θ θθ

4. Differentiate the quotient ln 2tt

with respect to x.

If y = ln 2tt

, ( ) ( )

( )

1 1 12 12 2

2

2

1 1 1t ln 2t t t t ln 2tt 2dy t 12 1 ln 2tdt t t 2t

−− −

−⎛ ⎞⎛ ⎞ − ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎛ ⎞⎝ ⎠= = = −⎜ ⎟

⎝ ⎠

= 32

1 11 ln 2t2t

⎛ ⎞−⎜ ⎟⎝ ⎠

= 3

1 11 ln 2t2t

⎛ ⎞−⎜ ⎟⎝ ⎠

6. Find the gradient of the curve 2

2xyx 5

=−

at the point (2, - 4)

If 2

2xyx 5

=−

then gradient,

( )( ) ( ) ( )

2 2 2 2

2 2 22 2 2

x 5 (2) (2x)(2x)dy 2x 10 4x 10 2xdx x 5 x 5 x 5

− − − − − −= = =

− − −

At the point (2, - 4), x = 2, hence gradient = ( )

2

2 22

10 2(2) 10 8 18(4 5) 12 5

− − − − −= =

−− = -18

7. Evaluate dydx

at x = 2.5, correct to 3 significant figures, given 22x 3y

ln 2x+

=


If 22x 3y

ln 2x+

= then ( )

( )

2

2

1ln 2x (4x) (2x 3)dy xdx ln 2x

⎛ ⎞− + ⎜ ⎟⎝ ⎠=

When x = 2.5, ( )

( )

2

2

1ln 5 (10) [2(2.5) 3]dy 16.09438 6.22.5dx 2.59029ln 5

⎛ ⎞− + ⎜ ⎟ −⎝ ⎠= = = 3.82, correct to 3

significant figures.


EXERCISE 120 Page 296 1. Find the differential coefficient of ( )532x 5x− with respect to x.

If y = ( )532x 5x− then ( ) ( )43 2dy 5 2x 5x 6x 5dx

= − −

4. Find the differential coefficient of ( )53

1

x 2x 1− + with respect to x.

If y = ( )

( ) 5353

1 x 2x 1x 2x 1

−= − +

− + then ( ) ( ) ( )

( )

263 2

63

5 3x 2dy 5 x 2x 1 3x 2dx x 2x 1

− − −= − − + − =

− +

= ( )

( )

2

63

5 2 3x

x 2x 1

−

− +

6. Find the differential coefficient of ( )22cot 5t 3+ with respect to x.

If y = cot x = cos xsin x

then

( )2 2

22 2 2

sin x cos xdy (sin x)( sin x) (cos x)(cos x) 1 cosec xdx sin x sin x sin x

− +− − −= = = = −

Thus, if y = ( )22cot 5t 3+ , then ( ) ( ) ( )2 2dy 2 cosec 5t 3 10tdt

⎡ ⎤= − +⎣ ⎦ = ( )2 220t cosec 5t 3− +

8. Find the differential coefficient of tan2e θ with respect to θ.

If y = tan2e θ , then ( )tan 2dy 2e secd

θ= θθ

= 2 tan2sec e θθ

9. Differentiate sin3π⎛ ⎞θ θ−⎜ ⎟

⎝ ⎠ with respect to θ, and evaluate, correct to 3 significant figures, when

2π

θ =


If y = sin3π⎛ ⎞θ θ−⎜ ⎟

⎝ ⎠ then ( ) ( )dy cos sin 1

d 3 3π π⎛ ⎞ ⎛ ⎞= θ θ− + θ−⎜ ⎟ ⎜ ⎟θ ⎝ ⎠ ⎝ ⎠

When 2π

θ = , dy cos sin cos sind 2 2 3 2 3 2 6 6

π π π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟θ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 1.360 + 0.5 = 1.86, correct to 3




2. (a) Given f(t) = 23

2 1 3t t 15 t t

− + − + determine f ′′(t).

(b) Evaluate f ′′(t) when t = 1.

(a) f(t) = 1

2 2 3 1 23

2 1 3 2t t 1 t t 3t t 15 t t 5

− −− + − + = − + − +

f ′(t) = 1

4 2 24 1t 3t 3t t5 2

−− −+ − −

f ′′(t) = 3

5 3 24 112t 6t t5 4

−− −− + + = 5 3 3

4 12 6 15 t t 4 t− + +

(b) When t = 1, f ′′(t) = 5 3 3

4 12 6 1 4 112 65 (1) (1) 5 44 1− + + = − + + = -4.95

4. Find the second differential coefficient with respect to x of (a) 22cos x (b) ( )42x 3−

(a) If y = 22cos x , dy 4cos x( sin x) 4sin x cos xdx

= − = −

and 2

2 22

d y ( 4sin x)( sin x) (cos x)( 4cos x) 4sin x 4cos xdx

= − − + − = −

= ( )2 24 sin x cos x−

(b) If y = ( )42x 3− , 3 3dy 4(2x 3) (2) 8(2x 3)dx

= − = −

and 2

22

d y 24(2x 3) (2)dx

= − = ( )248 2x 3−

5. Evaluate f ′′(θ) when θ = 0 given f(θ) = 2 sec 3θ If f(θ) = 2 sec 3θ, then f ′(θ) = 6 sec 3θ tan 3θ

and f ′′(θ) = ( )( ) ( )( )26sec3 3sec 3 tan 3 18sec3 tan 3θ θ + θ θ θ

= 3 218sec 3 18sec3 tan 3θ+ θ θ

When θ = 0, f ′′(0) = 2

3

18 18 tan 0 18 18(0)cos 0 cos0 1 1

+ = + = 18


7. Show that, if P and Q are constants and y = P cos(ln t) + Q sin(ln t), then

2

22

d y dyt t y 0dt dt

+ + =

y = P cos(ln t) + Q sin(ln t)

dy P Q Psin(ln t) Qcos(ln t)sin(ln t) cos(ln t)dt t t t

− += − + =

[ ]2

2 2

P cos(ln t) Qsin(ln t)(t) Psin(ln t) Qcos(ln t) (1)d y t tdt t

−⎡ ⎤− − − +⎢ ⎥⎣ ⎦=

= 2

P cos(ln t) Qsin(ln t) Psin(ln t) Qcos(ln t)t

− − + −

Hence, 2

22

d y dyt t ydt dt

+ + = ( )22

P cos(ln t) Qsin(ln t) Psin(ln t) Qcos(ln t)tt

− − + −

+ (t) Psin(ln t) Qcos(ln t)t

− + + P cos(ln t) + Q sin(ln t)

= P cos(ln t) Qsin(ln t) Psin(ln t) Qcos(ln t) Psin(ln t) Qcos(ln t)− − + − − +

+ P cos(ln t) + Q sin(ln t)

= 0


CHAPTER 28 SOME APPLICATIONS OF DIFFERENTIATION EXERCISE 122 Page 299 1. An alternating current, i amperes, is given by i = 10 sin 2πft, where f is the frequency in hertz

and t the time in seconds. Determine the rate of change of current when t = 20 ms, given that

f = 150 Hz.

Current, i = 10 sin 2πft

Rate of change of current, 3di (10)(2 f )cos 2 ft (10)(2 150)cos 2 150 20 10dt

−⎡ ⎤= π π = π× π× × ×⎣ ⎦

= 3000π cos 6π = 3000π A/s

3. The voltage across the plates of a capacitor at any time t seconds is given by v = Vt

CRe−

, where

V, C and R are constants. Given V = 300 volts, C = 60.12 10−× F and 6R 4 10= × Ω find (a) the

initial rate of change of voltage and (b) the rate of change of voltage after 0.5 s.

(a) If v = Vt

CRe−

, then t

CRdv 1V edt CR

−⎛ ⎞= −⎜ ⎟⎝ ⎠

Initial rate of change of voltage, (i.e. when t = 0), 06 6

dv 1(300) edt 0.12 10 4 10−

⎛ ⎞= −⎜ ⎟× × ×⎝ ⎠

= - 3000.48

= -625 V/s

(b) When t = 0.5 s, t 0.5

CR 0.48dv V 300e edt CR 0.48

− −⎛ ⎞= − = −⎜ ⎟⎝ ⎠

= -220.5 V/s

4. The pressure p of the atmosphere at height h above ground level is given by p =hc

0p e−

, where 0p

is the pressure at ground level and c is a constant. Determine the rate of change of pressure with

height when 0p = 51.013 10× pascals and c = 46.05 10× at 1450 metres.

Pressure, p =hc

0p e−

Rate of change of pressure with height, ( ) 41450h

5c 6.05 100 4

dp 1 1(p ) e 1.013 10 edh c 6.05 10

−−×

⎛ ⎞⎛ ⎞= − = × −⎜ ⎟⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

= -1.635 Pa/m


EXERCISE 123 Page 301 1. A missile fired from ground level rises x metres vertically upwards in t seconds and

225x 100t t2

= − . Find (a) the initial velocity of the missile, (b) the time when the height of the

missile is a maximum, (c) the maximum height reached, (d) the velocity with which the missile

strikes the ground.

(a) Distance, 225x 100t t2

= −

Initial velocity, (i.e. when t = 0), dx 100 25t 100 25(0)dt

= − = − = 100 m/s

(b) When height is a maximum, velocity = 0, i.e. dx 100 25t 0dt

= − =

from which, 100 = 25t and time t = 4 s

(c) When t = 4 s, maximum height, x = 225100(4) (4) 400 2002

− = − = 200 m

(d) When x = 0 (i.e. on the ground), 2250 100t t2

= −

i.e. 25t 100 t 02

⎛ ⎞− =⎜ ⎟⎝ ⎠

Hence, either t = 0 (at the start) or 25 25100 t 0 i.e. 100 t2 2

− = =

and 200t25

= = 8 s

Velocity, i.e. dxdt

, when t = 8 s is given by dxdt

= 100 – 25t = 100 – 25(8)

= 100 – 200 = -100 m/s (negative indicating

reverse direction to the starting velocity)

3. The equation 210 24t 3tθ = π+ − gives the angle θ, in radians, through which a wheel turns in t

seconds. Determine (a) the time the wheel takes to come to rest, (b) the angle turned through in

the last second of movement.

Angle, 210 24t 3tθ = π+ −


(a) When the wheel comes to rest, angular velocity = 0, i.e. d 0dtθ=

Hence, d 24 6t 0dtθ= − = from which, time, t = 4 s

(b) Distance moved in the last second of movement = (distance after 4 s) – (distance after 3 s)

= 2[10 24(4) 3(4) ]π+ − - 2[10 24(3) 3(3) ]π+ −

= [10 96 48]π+ − - [10 72 27]π+ −

= 96 – 48 -72 + 27 = 3 rad

4. At any time t seconds the distance x metres of a particle moving in a straight line from a fixed

point is given by x = 4t + ln(1 – t). Determine (a) the initial velocity and acceleration (b) the

velocity and acceleration after 1.5 s, (c) the time when the velocity is zero.

(a) Distance, x = 4t + ln(1 – t)

Velocity, v = ( ) 1dx 14 ( 1) 4 1 tdt 1 t

−= + − = − −−

Initial velocity, i.e. when t = 0, v = 141

− = 3 m/s

Acceleration, a = 2

22 2

d x 1(1 t) ( 1)dt (1 t)

−= − − = −−

Initial acceleration, i.e. when t = 0, a = 11

− = -1 2m / s

(b) After 1.5 s, velocity, v = 1 1 14 4 4 4 2(1 t) (1 1.5) ( 0.5)

− = − = − = +− − −

= 6 m/s

and acceleration, a = 2 2 2

1 1 1 1(1 t) (1 1.5) ( 0.5) 0.25

− = − = − = −− − −

= -4 2m / s

(c) When the velocity is zero, 14 0(1 t)

− =−

i.e. 14(1 t)

=−

i.e. 4 – 4t = 1 and 4 – 1 = 4t

from which, time, t = 3 s4


EXERCISE 124 Page 305 2. Find the turning point on x (6 )= θ −θ and determine its nature.

2x (6 ) 6= θ −θ = θ−θ

dx 6 2 0d

= − θ =θ

for a turning point, from which, θ = 3

When θ = 3, 2 2x 6 6(3) 3 18 9 9= θ−θ = − = − =

Hence, (3, 9) are the co-ordinates of the turning point 2

2

d x 2d

= −θ

, which is negative, hence a maximum occurs at (3, 9)

3. Find the turning point on 3 2y 4x 3x 60x 12= + − − and distinguish between them.

3 2y 4x 3x 60x 12= + − −

2dy 12x 6x 60 0dx

= + − = for a turning point

i.e. 22x x 10 0+ − = i.e. (2x + 5)(x – 2) = 0

from which, 2x + 5 = 0 i.e. x = -2.5

and x – 2 = 0 i.e. x = 2

When x = -2.5, 3 2y 4( 2.5) 3( 2.5) 60( 2.5) 12 94.25= − + − − − − =

When x = 2, 3 2y 4(2) 3(2) 60(2) 12 88= + − − = −

Hence, (-2.5, 94.25) and (2, -88) are the co-ordinates of the turning points

2

2

d x 24x 6d

= +θ

When x = -2.5, 2

2

d xdθ

is negative, hence (-2.5, 94.25) is a maximum point.

When x = 2, 2

2

d xdθ

is positive, hence (2, -88) is a minimum point.

5. Find the turning point on xy 2x e= − and determine its nature.


xy 2x e= − hence, xdy 2 e 0dx

= − = for a turning point.

i.e. x2 e= and x = ln 2 = 0.6931

When x = 0.6931, 0.6931y 2(0.6931) e 0.6136= − = −

Hence, (0.6931, -0.6136) are the co-ordinates of the turning point.

2x

2

d y edx

= − When x = 0.6931, 2

2

d ydx

is negative, hence (0.6931, -0.6136) is a maximum point.

8. Determine the maximum and minimum values on the graph y = 12 cos θ - 5 sin θ in the range

θ = 0 to θ = 360°. Sketch the graph over one cycle showing relevant points.

y = 12 cos θ - 5 sin θ

dy 12sin 5cos 0d

= − θ− θ =θ

for a maximum or minimum value

i.e. -12 sin θ = 5 cos θ from which, sin 5cos 12

θ= −

θ i.e. tan θ = 5

12−

Hence, 1 5tan12

− ⎛ ⎞θ = −⎜ ⎟⎝ ⎠

= - 22°37′

Tangent is negative in the 2nd and 4th quadrants as shown in the diagram below.

Hence, θ = 180° - 22°37′ = 157°23′ and 360° - 22°37′ = 337°23′

When θ = 157°23′, y = 12 cos 157°23′ - 5 sin 157°23′ = -13

When θ = 337°23′, y = 12 cos 337°23′ - 5 sin 337°23′ = 13

Hence, (157°23′, -13) and (337°23′, 13) are the co-ordinates of the turning points.

2

2

d y 12cos 5sindx

= − θ+ θ


When θ = 157°23′, 2

2

d ydx

is positive, hence (157°23′, -13) is a minimum point

When θ = 337°23′, 2

2

d ydx

is negative, hence (337°23′, 13) is a maximum point

A sketch of y = 12 cos θ - 5 sin θ is shown below. (When y = 0, 12 cos θ - 5 sin θ = 0 and

12 cos θ = 5 sin θ; hence, sin 12cos 5

θ=

θ from which, tan θ = 2.4 and θ = 1tan 2.4 67.4− = ° and

247.4°; also, at θ = 0, y = 12 cos 0 – 5 sin 0 = 12).

9. Show that the curve 32y (t 1) 2t(t 2)3

= − + − has a maximum value of 23

and a minimum value

of -2.

3 3 22 2y (t 1) 2t(t 2) (t 1) 2t 4t3 3

= − + − = − + −

2dy 2(t 1) 4t 4 0dx

= − + − = for a turning point.

i.e. ( )22 t 2t 1 4t 4 0− + + − =

i.e. 22t 4t 2 4t 4 0− + + − =

i.e. 22t 2 0− = from which, 2t 1= and t = ±1

When t = 1, 32y (1 1) 2(1 2) 23

= − + − = −


When t = -1, 32 16 2y ( 1 1) 2( 1)( 1 2) 63 3 3

= − − + − − − = − + =

2

2

d y 4(t 1) 4dt

= − + When t = 1, 2

2

d ydt

is positive, hence (1, -2) is a minimum point.

When t = -1, 2

2

d ydt

is negative, hence 21,3

⎛ ⎞−⎜ ⎟⎝ ⎠

is a maximum point.

Hence, the maximum value of 32y (t 1) 2t(t 2)3

= − + − is 23

and the minimum value is -2


EXERCISE 125 Page 309 1. The speed, v, of a car (in m/s) is related to time t s by the equation: 2v 3 12t 3t= + − . Determine

the maximum speed of the car in km/h.

Speed, 2v 3 12t 3t= + −

dv 12 6t 0dt

= − = for a maximum value, from which, 12 = 6t and t = 2 s

When t = 2, 2v 3 12(2) 3(2) 3 24 12= + − = + − = 15 m/s

2

2

d v 6dt

= − , which is negative, hence indicating that v = 15 m/s is the maximum speed.

15 m/s = 60 60 s / h15m / s 15 3.61000 m / km

×× = × = 54 km/h = maximum speed.

3. A shell is fired vertically upwards and its vertical height, x metres, is given by x = 24t - 23t ,

where t is the time in seconds. Determine the maximum height reached.

Height, x = 24t - 23t

dx 24 6t 0dt

= − = for a maximum value, from which, 24 = 6t and t = 4 s

2

2

d x 6dt

= − , which is negative – hence a maximum value.

Maximum height = 24(4) - 23(4) = 96 – 48 = 48 m

4. A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area

of the metal for which the volume of the box is 3.5 3m

A lidless box with square ends is shown in the diagram below, having dimensions x by x by y.

Volume of box, V = 2 3x y 3.5m= (1)

Area of metal, A = 2 22

3.52x 3xy 2x 3xx

⎛ ⎞+ = + ⎜ ⎟⎝ ⎠

from equation (1)


i.e. A = 2 12x 10.5x−+

2dA 4x 10.5x 0

dx−= − = for a maximum or minimum value.

i.e. 32

10.5 10.54x i.e. x 2.625x 4

= = = from which, 3x 2.625= = 1.3795

23

2

d A 4 21xdx

−= + When x = 1.3795, 2

2

d Adx

is positive – hence a minimum value.

Minimum or least area of metal = 2 210.5 10.52x 2(1.3795)x 1.3795

+ = + = 11.42 2m

6. Calculate the height of a cylinder of maximum volume which can be cut from a cone of height

20 cm and base radius 80 cm.

A sketch of a cylinder within a cone is shown below. Cylinder volume, V = 2r hπ

A section is shown below.

By similar triangles, 20 h

80 80 r=

− from which, h = 20(80 r) 80 r r20

80 4 4− −

= = −


Hence, cylinder volume, V = 3

2 2r rr 20 20 r4 4

π⎛ ⎞π − = π −⎜ ⎟⎝ ⎠

2dV 3 r40 r 0dr 4

π= π − = for a maximum or minimum value.

i.e. 23 r 3r 16040 r i.e. 40 and r

4 4 3π

π = = =

2

2

d V 6 r40dr 4

π= π− When r = 160

3,

2

2

d Vdr

is negative, hence a maximum value.

Hence, height of cylinder, h = 20 -

160r 40320 204 4 3= − = − = 6.67 cm

7. The power developed in a resistor R by a battery of emf E and internal resistance r is given by:

( )

2

2E RP

R r=

+. Differentiate P with respect to R and show that the power is a maximum when

R = r.

( )

2

2E RP

R r=

+ hence

2 2 2

4

dP (R r) (E) E R(2)(R r) 0dR (R r)

+ − += =

+for a maximum value

Thus, 2 2E (R r) 2R(R r) 0⎡ ⎤+ − + =⎣ ⎦

i.e. 2 2 2R 2Rr r 2R 2Rr 0+ + − − =

i.e. 2 2r R 0− =

and R = r

2 2 2

4

E r RdPdR (R r)

⎡ ⎤−⎣ ⎦=+

and [ ]4 2 2 2 2 32

2 8

(R r) E 2R E r R 4(R r)d PdR (R r)

⎡ ⎤+ − − − +⎣ ⎦=+

When R = r, 2

2

d PdR

is negative, hence power is a maximum when R = r.

9. Resistance to motion, F, of a moving vehicle, is given by F = 5 100xx+ . Determine the

minimum value of resistance.


F = 5 100xx+ = 15x 100x− +

2

2

dF 55x 100 100 0dx x

−= − + = − + = for a maximum or minimum value.

i.e. 100 = 2

5x

and 2 5x 0.05 and x 0.05 0.2236100

= = = =

2

32 3

d F 1010xdx x

−= = which is positive when x = 0.2236, hence x = 0.2236 gives a minimum value of

resistance.

Maximum resistance to motion, F = 5 100xx+ = 5 100(0.2236)

0.2236+ = 44.72

11. The fuel economy E of a car, in miles per gallon, is given by:

E = 21 + 2 2 6 42.10 10 v 3.80 10 v− −× − ×

where v is the speed of the car in miles per hour. Determine, correct to 3 significant figures,

the most economical fuel consumption, and the speed at which it is achieved.

E = 21 + 2 2 6 42.10 10 v 3.80 10 v− −× − ×

2 6 3dE 4.20 10 v 4(3.80) 10 v 0dv

− −= × − × = for a maximum (most economical fuel consumption).

i.e. 2 6 34.20 10 v 4(3.80) 10 v− −× = ×

i.e. 2

26

4.20 10v 2763.1584(3.80) 10

−

−

×= =

×

from which, speed, v = 2763.158 = 52.6 m.p.h

22 6 2

2

d E 4.20 10 12(3.80) 10 vdv

− −= × − × and when v = 52.6, 2

2

d Edv

is negative, hence v is the

maximum speed. Maximum fuel economy, E = 21 + 2 2 6 42.10 10 v 3.80 10 v− −× − × = 21 + 2 2 6 42.10 10 (52.6) 3.80 10 (52.6)− −× − × = 50.0 miles/gallon


EXERCISE 126 Page 311 2. For the curve y = 23x 2x− find (a) the equation of the tangent, and (b) the equation of the

normal at the point (2, 8)

(a) y = 23x 2x− Gradient, m = dy 6x 2dx

= −

At the point (2, 8), x = 2, hence, m = 6(2) – 2 = 10

Equation of tangent is: 1 1y y m(x x )− = −

i.e. y – 8 = 10(x – 2)

i.e. y – 8 = 10x – 20

and y = 10x - 12

(b) Equation of normal is: 1 11y y (x x )m

− = − −

i.e. 1y 8 (x 2)10

− = − −

i.e. 10(y – 8) = -x + 2

i.e. 10y – 80 = -x + 2

and 10y + x = 82

4. For the curve y = 21 x x+ − find (a) the equation of the tangent, and (b) the equation of the

normal at the point (-2, -5)

(a) y = 21 x x+ − Gradient, m = dy 1 2xdx

= −

At the point (-2, -5), x = -2, hence, m = 1 – 2(-2) = 5

Equation of tangent is: 1 1y y m(x x )− = −

i.e. y – (-5) = 5(x – -2)

i.e. y + 5 = 5x + 10

and y = 5x + 5

(b) Equation of normal is: 1 11y y (x x )m

− = − −


i.e. 1y 5 (x 2)5

− − = − − −

i.e. 5(y + 5) = -x - 2

i.e. 5y + 25 = -x - 2

and 5y + x + 27 = 0

5. For the curve 1t

θ = find (a) the equation of the tangent, and (b) the equation of the normal at

the point 13,3

⎛ ⎞⎜ ⎟⎝ ⎠

(a) 11 tt

−θ = = Gradient, m = 22

d 1tdt t

−θ= − = −

At the point 13,3

⎛ ⎞⎜ ⎟⎝ ⎠

, t = 3, hence, m = 2

1 13 9

− = −

Equation of tangent is: 1 1m(t t )θ−θ = −

i.e. 1 1 (t 3)3 9

θ− = − −

i.e. 9θ - 3 = -t + 3

and 9θ + t = 6

(b) Equation of normal is: 1 11 (t t )m

θ−θ = − −

i.e. 1 1 (t 3)139

θ− = − −⎛ ⎞−⎜ ⎟⎝ ⎠

i.e. 1 9(t 3)3

θ− = −

i.e. 1 9t 273

θ− = −

and 29t 263

θ = − or 3θ = 27t - 80


EXERCISE 127 Page 312 2. The pressure p and volume v of a mass of gas are related by the equation pv = 50. If the pressure

increases from 25.0 to 25.4, determine the approximate change in the volume of the gas. Find

also the percentage change in the volume of the gas.

pv = 50 i.e. v = 150 50pp

−= and 22

dv 5050pdp p

−= − = −

Approximate change in volume, 2 2

dv 50 50v p (25.4 25.0) (0.4)dp p 25.0

⎛ ⎞ ⎛ ⎞δ ≈ ⋅δ = − − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

-0.032

Percentage change in volume = 0.032 3.2 3.2100%50 50 2p 25.0

− − −× = = = -1.6%

4. The radius of a sphere decreases from 6.0 cm to 5.96 cm. Determine the approximate change in

(a) the surface area, and (b) the volume.

(a) Surface area of sphere, A = 24 rπ and dA 8 rdr

= π

Approximate change in surface area, ( )dAA r 8 r (6.0 5.96) 8 (6.0)( 0.04)dr

δ ≈ ⋅δ = π − = π −

= -6.03 2cm

(b) Volume of sphere, V = 34 r3π and 2dV 4 r

dr= π

Approximate change in volume, ( )2 2dVV r 4 r ( 0.04) 4 (6.0) ( 0.04)dr

δ ≈ ⋅δ = π − = π − = -18.10 3cm

5. The rate of flow of a liquid through a tube is given by Poiseuilles’s equation as: 4p rQ

8 Lπ

=η

where Q is the rate of flow, p is the pressure difference between the ends of the tube, r is the

radius of the tube, L is the length of the tube and η is the coefficient of viscosity of the liquid. η

is obtained by measuring Q, p, r and L. If Q can be measured accurate to ±0.5%, p accurate to

±3%, r accurate to ±2% and L accurate to ±1%, calculate the maximum possible percentage error

in the value of η.

4p rQ

8 Lπ

=η

from which, 4p r

8 LQπ

η =


4 4

2

d p r p rQ ( 0.005Q) ( 0.005)dQ 8LQ 8LQ

⎛ ⎞ ⎛ ⎞η π πδη ≈ ⋅δ = − ± = ±⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

4 4d r p rp ( 0.03p) ( 0.03)dp 8LQ 8LQ

⎛ ⎞ ⎛ ⎞η π πδη ≈ ⋅δ = ± = ±⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3 4d 4p r p rr ( 0.02r) ( 0.08)dr 8LQ 8LQ

⎛ ⎞ ⎛ ⎞η π πδη ≈ ⋅δ = ± = ±⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

4 4

2

d p r p rL ( 0.01L) ( 0.01)dL 8L Q 8LQ

⎛ ⎞ ⎛ ⎞η − π πδη ≈ ⋅δ = ± = ±⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Maximum possible percentage error ≈ 0.5% + 3% + 8% + 1% = 12.5%


CHAPTER 29 DIFFERENTIATION OF PARAMETRIC EQUATIONS


2. A parabola has parametric equations: x = 2t , y = 2t. Evaluate dydx

when t = 0.5

If x = 2t , then dx 2tdt

=

If y = 2t, then dy 2dt

=

Hence,

dydy 2 1dt

dxdx 2t tdt

= = =

When t = 0.5, dy 1dx 0.5

= = 2

3. The parametric equations for an ellipse are x = 4 cos θ, y = sin θ. Determine (a) dydx

(b) 2

2

d ydx

(a) If x = 4 cos θ, then dx 4sind

= − θθ

If y = sin θ, then dy cosd

= θθ

Hence,

dydy cosd

dxdx 4sind

θθ= =− θ

θ

= 1 cot4

− θ

(b) ( )2

2

2 2 3

d dy d 1 1cot cosecd y 1 1 1 1d dx d 4 4dxdx 4sin 4sin 16 sin sin 16 sind

⎛ ⎞ ⎛ ⎞− θ − − θ⎜ ⎟ ⎜ ⎟θ θ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠= = = = − = −⎜ ⎟ ⎜ ⎟− θ − θ θ θ θ⎝ ⎠ ⎝ ⎠θ

= 31 cosec16

− θ

4. Evaluate dydx

at 6π

θ = radians for the hyperbola whose parametric equations are x = 3 sec θ,

y = 6 tan θ


If x = 3 sec θ, then dx 3sec tand

= θ θθ

If y = 6 tan θ, then 2dy 6secd

= θθ

Hence, 2

dy 1dy 6sec 2sec 2d cos2dx sindx 3sec tan tan sin

d cos

⎛ ⎞⎜ ⎟θ θθ θ= = = = =⎜ ⎟θθ θ θ θ⎜ ⎟

θ θ⎝ ⎠

When 6π

θ = , dy 2 2dx 0.5sin

6

= =π

= 4

6. The equation of a tangent drawn to a curve at point ( )1 1x , y is given by: ( )11 1

1

dyy y x xdx

− = − .

Determine the equation of the tangent drawn to the ellipse x = 3 cos θ, y = 2 sin θ at 6π

θ =

( )11 1

1

dyy y x xdx

− = −

At point θ, 1x 3cos= θ and 1dx 3sind

= − θθ

1y 2sin= θ and 1dy 2cosd

= θθ

Hence, 1

1

11

dydy 2cos 2d cotdxdx 3sin 3

d

θθ= = = − θ− θ

θ

The equation of a tangent is: y – 2 sin θ = ( )2 cot x 3cos3

− θ − θ

At 6π

θ = , y – 2 sin 6π = 2 cot x 3cos

3 6 6π π⎛ ⎞− −⎜ ⎟⎝ ⎠

i.e. y – 1 = ( )( )2 1.732 x 2.5983

− −

i.e. y – 1 = -1.155(x – 2.598)

i.e. y – 1 = -1.155x + 3

and y = -1.155x + 4


EXERCISE 129 Page 318 1. A cycloid has parametric equations x = 2(θ - sin θ), y – 2(1 – cos θ). Evaluate, at θ = 0.62 rad,

correct to 4 significant figures, (a) dydx

(b) 2

2

d ydx

(a) If x = 2(θ - sin θ), then dx 2 2cosd

= − θθ

If y = 2(1 – cos θ), then dy 2sind

= θθ

Hence,

dydy 2sin sind

dxdx 2(1 cos ) 1 cosd

θ θθ= = =− θ − θ

θ

When θ = 0.62 rad, dy sin 0.62dx 1 cos0.62

=−


(b)

2 2

2 2 2

2

(1 cos )(cos ) (sin )(sin ) cos cos sind dy d sind y (1 cos ) (1 cos )d dx d 1 cos

dxdx 2(1 cos ) 2(1 cos ) 2(1 cos )d

− θ θ − θ θ θ− θ− θθ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ − θ − θθ θ − θ⎝ ⎠ ⎝ ⎠= = = =

− θ − θ − θθ

= 2 2

3 3

cos (cos sin ) cos 12(1 cos ) 2(1 cos )

θ− θ+ θ θ−=

− θ − θ

When θ = 0.62 rad, 2

2 3

d y (cos 0.62) 1 0.1861215dx 2(1 cos 0.62) 2(0.00644748)

− −= =

− = -14.43, correct to 4 significant

figures.

2. The equation of a normal drawn to a curve at point ( )1 1x , y is given by: ( )1 11

1

1y y x xdydx

− = − − .

Determine the equation of the normal drawn to the parabola 21 1x t , y t4 2

= = at t = 2.

If 21

1x t ,4

= dx 1 tdt 2

=

If 11y t ,2

= dy 1dt 2

=

Hence, 1

1

11

dy 1dy 1dt 2

dx 1dx tt2dt

= = =


Equation of a normal is: ( )1 11

1

1y y x xdydx

− = − −

i.e. 21 1 1y t x t12 4t

⎛ ⎞− = − −⎜ ⎟⎝ ⎠

At t = 2, equation of normal is: y – 1 = -2(x – 1)

i.e. y – 1 = -2x + 2 or y = -2x + 3

4. Determine the value of 2

2

d ydx

, correct to 4 significant figures, at 6π

θ = rad for the cardioid

x = 5(2θ - cos 2θ), y = 5(2 sin θ – sin 2θ).

If x = 5(2θ - cos 2θ), then ( )dx 10 10sin 2 10 1 sin 2d

= + θ = + θθ

If y = 5(2 sin θ – sin 2θ), hence dy 10cos 10cos 2 10(cos cos 2 )d

= θ− θ = θ− θθ

dydy 10(cos cos 2 ) cos cos 2d

dxdx 10(1 sin 2 ) 1 sin 2d

θ− θ θ− θθ= = =+ θ + θ

θ

( )22

2

(1 sin 2 )( sin 2sin 2 ) (cos cos 2 )(2cos 2 )d dy d cos cos 21 sin 2d y d dx d 1 sin 2

dxdx 10(1 sin 2 ) 10(1 sin 2 )d

+ θ − θ+ θ − θ− θ θθ− θ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ + θθ θ + θ⎝ ⎠ ⎝ ⎠= = =

+ θ + θθ

When 6π

θ = rad,

2

2

2

2 2 2 21 sin sin 2sin cos cos 2cos6 6 6 6 6 6

21 sind y 6

2dx 10 1 sin6

π π π π π π⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ − + − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

π⎛ ⎞+⎜ ⎟⎝ ⎠=

π⎛ ⎞+⎜ ⎟⎝ ⎠

=

(1.86603)(1.23205) (0.366025)(1)(3.48205)18.660254

−

= 0.02975, correct to 4 significant

figures.


5. The radius of curvature, ρ, of part of a surface when determining the surface tension of a liquid

is given by:

3/ 22

2

2

dy1dxd ydx

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦ρ =

Find the radius of curvature (correct to 4 significant figures) of the part of the surface having

parametric equations (a) x = 3t, 3yt

= at the point 1t2

=

(b) 3 3x 4cos t, y 4sin t= = at t = rad6π

(a) x 3t= , hence dx 3dt

=

13y 3tt

−= = , hence 22

dy 33tdt t

−= − = −

2

2

dy 3dy 1dt t

dxdx 3 tdt

−= = = − and at 1t

2= , 2

dy 1 4dx 1

2

= − = −⎛ ⎞⎜ ⎟⎝ ⎠

( )2

2 32

2 3

d dy d 1 d td y 2t 2dt dx dt t dtdxdx 3 3 3 3tdt

−−

⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= = = = = and at 1t

2= ,

2

32 3

d y 2 2 16dx 3t 313

2

= = =⎛ ⎞⎜ ⎟⎝ ⎠

Hence, radius of curvature, ( )

3/ 2232

3

2

2

dy1 1 4dx 1716 16d y3 3dx

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟ ⎡ ⎤+ −⎝ ⎠⎢ ⎥ ⎣ ⎦⎣ ⎦ρ = = = = 13.14

(b) 3x 4cos t= , hence 2 2dx 12cos t( sin t) 12cos t sin tdt

= − = −

3y 4sin t= , hence 2dy 12sin t cos tdt

=

2

2

dydy 12sin t cos t sin tdt tan tdxdx 12cos t sin t cos t

dt

= = = − = −−

and at t = rad6π , dy tan 0.57735

dx 6π

= − = −


( )2 2

2 2 2 4

d dy d tan td y sec t 1dt dx dtdxdx 12cos t sin t 12cos t sin t 12cos t sin tdt

⎛ ⎞−⎜ ⎟ −⎝ ⎠= = = =

− −

and at t rad6π ,

2

42

d y 1 0.29630dx

12 cos sin6 6

= =π π⎛ ⎞

⎜ ⎟⎝ ⎠

Hence, radius of curvature, ( )

3/ 2232

3

2

2

dy1 1 0.57735dx 1.333333d y 0.29630 0.29630dx

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟ ⎡ ⎤+ −⎝ ⎠⎢ ⎥ ⎣ ⎦⎣ ⎦ρ = = = = 5.196


CHAPTER 30 DIFFERENTIATION OF IMPLICIT FUNCTIONS EXERCISE 130 Page 320 2. Differentiate the given functions with respect to x:

(a) 5 ln 3t2

(b) 2y 13 e4

+ (c) 2 tan 3y

(a) d 5 5 1 dtln 3tdx 2 2 t dx

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 5 dt2t dx

(b) ( )2y 1 2y 1d 3 3 dye 2edx 4 4 dx

+ +⎛ ⎞ =⎜ ⎟⎝ ⎠

= 2y 13 dye2 dx

+

(c) ( ) ( )2d dy2 tan 3y (2) 3sec 3ydx dx

= = 2 dy6sec 3ydx

4. Differentiate the following with respect to u:

(a) 2(3x 1)+

(b) 3 sec 2θ (c) 2y

(a) 1 2d 2 d dx2(3x 1) 2(3x 1) (3)du 3x 1 du du

− −⎛ ⎞ ⎡ ⎤ ⎡ ⎤= + = − +⎜ ⎟ ⎣ ⎦ ⎣ ⎦+⎝ ⎠ = 2

6 dx(3x 1) du

−+

(b) ( ) ( )d d3sec 2 3 2sec 2 tan 2du du

θθ = θ θ = d6sec 2 tan 2

duθ

θ θ

(c) 1 32 2

32

d 2 d 1 dy 1 dy2y (2) ydu du 2 du duy y

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

= 3

1 dyduy

−



1. Determine ( )2 3d 3x ydx

( )2 3d 3x ydx

= ( ) ( )2 3 3 2d d3x y y 3xdx dx

+ using the product rule

= ( ) ( )( )2 2 3dy3x 3y y 6xdx

⎛ ⎞ +⎜ ⎟⎝ ⎠

= 2 2 3dy9x y 6xydx

+

= 2 dy3xy 3x 2ydx

⎛ ⎞+⎜ ⎟⎝ ⎠

3. Determine d 3udu 4v

⎛ ⎞⎜ ⎟⎝ ⎠

d 3udu 4v

⎛ ⎞⎜ ⎟⎝ ⎠

= ( )( ) ( )

( )2 2 2

dv dv4v 3 3u 4 12v 12u 12 dvdu du v u16v 16v du4v

⎛ ⎞− −⎜ ⎟ ⎛ ⎞⎝ ⎠ = = −⎜ ⎟⎝ ⎠

= 2

3 dvv u4v du

⎛ ⎞−⎜ ⎟⎝ ⎠

5. Determine dzdy

given z = 32x ln y

z = 32x ln y hence, ( ) ( )3 2dz 1 dx2x ln y 6xdy y dy

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= 3

22x dx6x ln yy dy

+ or 2 x dx2x 3ln yy dy

⎛ ⎞+⎜ ⎟

⎝ ⎠



1. Determine dydx

given 2 2x y 4x 3y 1 0+ + − + =

( ) ( ) ( ) ( ) ( )2 2d d d d dx y 4x 3y 1 0dx dx dx dx dx

+ + − + =

dy dy2x 2y 4 3 0 0dx dx

+ + − + =

2x + 4 = ( )dy dy dy3 2y 3 2ydx dx dx

− = −

Hence, dy 2x 4dx 3 2y

+=

−

3. Given 2 2x y 9+ = evaluate dydx

when x = 5 and y = 2

2 2d d d(x ) (y ) (9)dx dx dx

+ =

i.e. 2x + 2y dydx

= 0

i.e. 2y dydx

= -2x and dy xdx y

= −

When x = 5 and y = 2, dy 5dx 2

= −

5. Determine dydx

given 2 23y 2xy 4x 0+ − =

Given 2 23y 2xy 4x 0+ − = then ( ) ( )( )dy dy6y 2x 1 y 2 8x 0dx dx

⎡ ⎤⎛ ⎞+ + − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

( )dy 6y 2x 8x 2ydx

+ = −

and dy 8x 2ydx 6y 2x

−=

+ = 4x y

3y x−+

7. Determine dydx

given 43y 2x ln y y x+ = +

Given 43y 2x ln y y x+ = + then ( ) ( )( ) 3dy 1 dy dy3 2x ln y 2 4y 1dx y dx dx

⎡ ⎤⎛ ⎞+ + = +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦


i.e. 3dy 2x3 4y 1 2ln ydx y

⎛ ⎞+ − = −⎜ ⎟

⎝ ⎠

and 3

dy 1 2ln y2xdx 3 4yy

−=

+ −

8. If 2 2 3 253x 2x y y 04

+ − = evaluate dydx

when x = 12

and y = 1.

Given 2 2 3 253x 2x y y 04

+ − = then ( ) ( )( )2 2 3dy 5 dy6x 2x 3y y 4x y 0dx 2 dx

⎡ ⎤⎛ ⎞+ + − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

i.e. 3 2 25 dy6x 4xy y 6x y2 dx

⎛ ⎞+ = −⎜ ⎟⎝ ⎠

and 3

2 2

dy 6x 4xy5dx y 6x y2

+=

−

When x = 12

and y = 1, ( )

( ) ( )

3

22

1 16 4 1dy 3 2 52 2dx 2.5 1.5 15 11 6 1

2 2

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟ +⎝ ⎠ ⎝ ⎠= = =−⎛ ⎞− ⎜ ⎟

⎝ ⎠

= 5

10. Find the gradients of the tangents drawn to the ellipse 2 2x y 2

4 9+ = at the point where x = 2.

Given 2 2x y 2

4 9+ = then 2x 2y dy 0

4 9 dx+ = and 2y dy 2x

9 dx 4= −

from which,

2xdy 2x 9 9x4

2ydx 4 2y 4y9

= − = − ⋅ = −

If 2 2x y 2

4 9+ = and x = 2, then

24 y 24 9+ = from which,

2y 19=

Hence, 2y 9= and y = ± 3

Thus, dy 9x (9)(2) 3dx 4y 4( 3) 2

= − = − = ±±

or ±1.5


CHAPTER 31 LOGARITHMIC DIFFERENTIATION EXERCISE 133 Page 326

2. Use logarithmic differentiation to differentiate y = 3

2 4

(x 1)(2x 1)(x 3) (x 2)

+ +− +

with respect to x.

If y = 3

2 4

(x 1)(2x 1)(x 3) (x 2)

+ +− +

then ln y = 3

2 4

(x 1)(2x 1)ln(x 3) (x 2)

⎧ ⎫+ +⎨ ⎬− +⎩ ⎭

i.e. = 3 2 4ln(x 1) ln(2x 1) ln(x 3) ln(x 2)+ + + − − − +

i.e. = ln(x 1) 3ln(2x 1) 2ln(x 3) 4ln(x 2)+ + + − − − +

Differentiating w.r.t. x gives: 1 dy 1 3 2 4(2)y dx (x 1) (2x 1) (x 3) (x 2)

= + − −+ + − +

i.e. dy 1 6 2 4ydx (x 1) (2x 1) (x 3) (x 2)

⎧ ⎫= + − −⎨ ⎬+ + − +⎩ ⎭

i.e. 3

2 4

dy (x 1)(2x 1) 1 6 2 4dx (x 3) (x 2) (x 1) (2x 1) (x 3) (x 2)

⎧ ⎫+ += + − −⎨ ⎬− + + + − +⎩ ⎭

4. Use logarithmic differentiation to differentiate y = 2xe cos3x(x 4)−

with respect to x.

If y = 2xe cos3x(x 4)−

then ln y = ( )

2xe cos3xlnx 4

⎧ ⎫⎪ ⎪⎨ ⎬

−⎪ ⎪⎩ ⎭ =

12x 2ln e ln(cos3x) ln(x 4)+ − −

= 2x + ln(cos 3x) - 12

ln(x - 4)

Differentiating w.r.t. x gives:

11 dy ( 3sin 3x) 22y dx cos3x (x 4)

−= + −

−

i.e. dy 1y 2 3tan 3xdx 2(x 4)

⎧ ⎫= − −⎨ ⎬−⎩ ⎭

i.e. 2xdy e cos 3x 12 3tan 3x

dx 2(x 4)(x 4)⎧ ⎫

= − −⎨ ⎬−− ⎩ ⎭

6. Use logarithmic differentiation to differentiate y = 4

2x

2x tan xe ln 2x

with respect to x.

If y = 4

2x

2x tan xe ln 2x

then ln y = 4

4 2x2x

2x tan xln ln 2x ln(tan x) ln e ln(ln 2x)e ln 2x

⎧ ⎫= + − −⎨ ⎬

⎩ ⎭


= ln 2 + ln 4x + ln(tan x) – 2x ln e – ln(ln 2x)

= ln 2 + 4 ln x + ln(tan x) – 2x(1) – ln(ln 2x)

Differentiating w.r.t. x gives: 21 dy 4 1 1 10 (sec x) 2y dx x tan x ln 2x x

⎛ ⎞= + + − − ⎜ ⎟⎝ ⎠

i.e. 2

dy 4 cos x 1 1y 2dx x sin x cos x x ln 2x

⎧ ⎫⎛ ⎞= + − −⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

i.e. 4

2x

dy 2x tan x 4 1 12dx e ln 2x x sin xcos x x ln 2x

⎧ ⎫= + − −⎨ ⎬⎩ ⎭

8. Evaluate dydθ

, correct to 3 significant figures, when 4π

θ = given y = 5

2e sinθ θ

θ

If y = 5

2e sinθ θ

θ then ln y =

52

5

2e sin ln 2 ln e ln(sin ) lnθ

θ⎧ ⎫θ= + + θ − θ⎨ ⎬

θ⎩ ⎭

= 5ln 2 ln(sin ) ln2

+ θ+ θ − θ

Differentiating w.r.t. θ gives: 1 dy 1 5 10 1 (cos )y d sin 2

⎛ ⎞= + + θ − ⎜ ⎟θ θ θ⎝ ⎠

i.e. 5

dy 5 2e sin 5y 1 cot 1 cotd 2 2

θ θ⎧ ⎫ ⎧ ⎫= + θ− = + θ−⎨ ⎬ ⎨ ⎬θ θ θ⎩ ⎭ ⎩ ⎭θ

When 4π

θ = , 4

5

2e sindy 54 1 cotd 4 2

44

π ⎧ ⎫π⎪ ⎪π⎪ ⎪= + −⎨ ⎬πθ ⎛ ⎞⎪ ⎪π⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎪ ⎪⎝ ⎠⎩ ⎭⎝ ⎠

= 5.673935[1 + 1 – 3.18309886]

= - 6.71, correct to 3 significant figures.


EXERCISE 134 Page 328 2. Differentiate y = ( )x2x 1− with respect to x. Since y = ( )x2x 1− then ln y = ln ( )x2x 1− = x ln(2x – 1)

Differentiating w.r.t. x gives: 1 dy 2(x) [ln(2x 1)](1)y dx 2x 1

⎛ ⎞= + −⎜ ⎟+⎝ ⎠ by the product rule

i.e. dy 2xy ln(2x 1)dx 2x 1

⎧ ⎫= + −⎨ ⎬−⎩ ⎭

i.e. ( )xdy 2x2x 1 ln(2x 1)dx 2x 1

⎧ ⎫= − + −⎨ ⎬−⎩ ⎭

3. Differentiate y = x (x 3)+ with respect to x.

y = x (x 3)+ = 1x(x 3)+ and ln y =

1x 1ln(x 3) ln(x 3)

x+ = +

Differentiating w.r.t. x gives: ( )21 dy 1 1 [ln(x 3)] xy dx x x 3

−⎛ ⎞⎛ ⎞= + + −⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠ by the product rule

and 2

dy 1 ln(x 3)ydx x(x 3) x

⎧ ⎫+= −⎨ ⎬+⎩ ⎭

i.e. x2

dy 1 ln(x 3)(x 3)dx x(x 3) x

⎧ ⎫+= + −⎨ ⎬+⎩ ⎭

5. Show that when xy 2x= and x = 1, dy 2dx

=

xy 2x= and ln y = x xln 2x ln 2 ln x ln 2 x ln x= + = +

Differentiating w.r.t. x gives: 1 dy 10 (x) (ln x)(1) 1 ln xy dx x

⎛ ⎞= + + = +⎜ ⎟⎝ ⎠

by the product rule

i.e. xdy y(1 ln x) 2x (1 ln x)dx

= + = +

When x = 1, 1dy 2(1) (1 ln1)dx

= + = 2

6. Evaluate xd (x 2)dx

− when x = 3


Let y = 1xx (x 2) (x 2)− = − then ln y =

1x 1 ln(x 2)ln(x 2) ln(x 2)

x x−

− = − =

Differentiating w.r.t. x gives: 2

1(x) [ln(x 2)](1)1 dy x 2y dx x

⎛ ⎞ − −⎜ ⎟−⎝ ⎠= by the quotient rule

and 2

x ln(x 2)dy (x 2)ydx x

⎧ ⎫− −⎪ ⎪−⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

i.e. x

2

(x 2)dy x ln(x 2)dx x x 2

− ⎧ ⎫= − −⎨ ⎬−⎩ ⎭

When x = 3, 3

2

(3 2)dy 3 1 3ln(3 2) 0dx 3 3 2 9 1

− ⎧ ⎫ ⎛ ⎞⎧ ⎫= − − = −⎨ ⎬ ⎨ ⎬⎜ ⎟−⎩ ⎭ ⎝ ⎠⎩ ⎭ = 1

3

7. Show that if y θ= θ and 2θ = , dydθ


If y θ= θ then ln y = ln lnθθ = θ θ

Differentiating w.r.t. θ gives: ( )1 dy 1 (ln )(1)y d

⎛ ⎞= θ + θ⎜ ⎟θ θ⎝ ⎠ = 1 + ln θ

and dy y(1 ln ) (1 ln )d

θ= + θ = θ + θθ

When 2θ = , 2dy 2 (1 ln 2)d

= +θ



CHAPTER 32 DIFFERENTIATION OF HYPERBOLIC

FUNCTIONS

EXERCISE 135 Page 331 2. Differentiate the following with respect to the variable:

(a) 2 sec h 5x3

(b) 5 tcosech8 2

(c) 2 coth 7θ

(a) ( )d 2 2sec h 5x 5sec h 5x tanh 5xdx 3 3

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 10 sech 5x tanh 5x3

−

(b) d 5 t 5 1 t tcosec h cosec h cothdt 8 2 8 2 2 2⎛ ⎞ ⎛ ⎞⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= 5 t tcosech coth16 2 2

−

(c) ( ) ( )( )2d 2cot h 7 2 7cosec h 7d

θ = − θθ

= 214cosech 7− θ

3. Differentiate the following with respect to the variable:

(a) 2 ln (sh x) (b) 3 ln th4 2

⎛ θ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(a) ( ) ( ) ( )d 12ln(shx) 2 chxdx shx

⎛ ⎞= ⎜ ⎟⎝ ⎠

= 2cot h x

(b) 2

2

2

1

ch chd 3 3 1 1 3 3 12 2ln th sec hdx 4 2 4 2 2 8 8th sh ch sh

2 2 2 2ch

2

θ θ⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎡ ⎤⎛ θ ⎞ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ θ θ θ θ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

θ

= 3 18 ch sh

2 2

⎛ ⎞⎜ ⎟⎜ ⎟θ θ⎜ ⎟⎝ ⎠

= 3 sech cosech8 2 2

θ θ

5. Differentiate the following with respect to the variable:

(a) 3

3sh 4x2x

(b) ch 2tcos 2t


(a) ( )( ) ( )( )

( )

3 2

23 3

2x 12ch 4x 3sh 4x 6xd 3sh 4xdx 2x 2x

−⎛ ⎞ =⎜ ⎟⎝ ⎠

by the quotient rule

= 3 2

6

24x ch 4x 18x sh 4x4x− = 4

12xch 4x 9sh 4x2x−

(b) ( )( ) ( )( )2

cos 2t 2sh 2t ch 2t 2sin 2td ch 2tdx cos 2t cos 2t

− −⎛ ⎞ =⎜ ⎟⎝ ⎠

= ( )

2

2 cos 2t sh 2t ch 2t sin 2tcos 2t

+


CHAPTER 33 DIFFERENTIATION OF INVERSE

TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS


1.(a) Differentiate 1sin 4x− with respect to x.

If y = 1sin 4x− , then 2

dy 4dx 1 (4x)

=⎡ ⎤−⎣ ⎦

= ( )2

4

1 16x−

2.(b) Differentiate 12 xcos3 3

− with respect to x.

If y = 12 xcos3 3

− , then ( )2 2

dy 2 1dx 3 3 x

⎡ ⎤−⎢ ⎥=

⎢ ⎥−⎣ ⎦

= ( )2

2

3 9 x

−

−

3.(a) Differentiate 13 tan 2x− with respect to x.

If y = 13 tan 2x− , then 2

dy 23dx 1 (2x)

⎡ ⎤= ⎢ ⎥+⎣ ⎦

= 2

61 4x+

4.(b) Differentiate 1 3sec x4


If y = 1 3sec x4

− , then ( )2 2 2 2

3dy 1 1 14dx 9x 9x 16 x 9x 163 3x x 1 xx 1 16 164 4 4

= = = =⎡ ⎤ ⎛ ⎞ ⎛ ⎞− −⎛ ⎞ −− ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

= ( )2

4

x 9x 16−

5.(a) Differentiate 15 cos ec2 2

− θ with respect to θ.

If y = 15 cos ec2 2

− θ , then ( )2 2

dy 5 2d 2 2

⎡ ⎤−⎢ ⎥=

⎢ ⎥θ θ θ −⎣ ⎦

= ( )2

5

4

−

θ θ −


6.(b) Differentiate 1 2cot 1− θ − with respect to θ.

If y = 1 2cot 1− θ − , then ( )

( ) ( )

12 2

2 2 22

1 1 2dy 2d 1 1 11 1

−− θ − θ − θ

= =θ ⎡ ⎤ ⎡ ⎤θ − + θ −+ θ − ⎣ ⎦⎢ ⎥⎣ ⎦

= ( )2

1

1

−

θ θ −

7. Show that the differential coefficient of 12

xtan1 x

− ⎛ ⎞⎜ ⎟−⎝ ⎠

is 2

2 4

1 x1 x x

+− +

If y = 12

xtan1 x

− ⎛ ⎞⎜ ⎟−⎝ ⎠

then

( )( ) ( )( )( ) ( )

( )( )

2 2 2

2 22 2 2

2 2 2 4 22 2

2 22

1 x 1 x 2x 1 x 2x1 x 1 xdy 1 x

dx 1 2x x xx 1 x x11 x 1 x

− − − − +

− − += = =

− + +⎛ ⎞ − ++ ⎜ ⎟−⎝ ⎠ −

= ( )2

2 4

1 x1 x x

+− +

8.(b) Differentiate 2 1t sec 2t− with respect to t.

If y = 2 1t sec 2t− then ( )( )

( )( )2 1

2

dy 2t sec 2t 2tdt 2t 2t 1

−

⎛ ⎞⎜ ⎟

= +⎜ ⎟⎡ ⎤−⎜ ⎟⎣ ⎦⎝ ⎠

= ( )

1

2

t 2t sec 2t4t 1

−+−

9.(a) Differentiate ( )2 1 2cos 1−θ θ − with respect to θ.

If y = ( )2 1 2cos 1−θ θ − then ( )( )

( ) ( )2 1 2

22

dy 2 cos 1 2d 1 1

−

⎛ ⎞⎜ ⎟− θ ⎡ ⎤= θ + θ − θ⎜ ⎟ ⎣ ⎦θ ⎜ ⎡ ⎤ ⎟− θ −⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠

= ( )( )

31 2

4 2

22 cos 11 2 1

− θθ θ − −

⎡ ⎤− θ − θ +⎣ ⎦

= ( )( )

31 2

2 4

22 cos 12

− θθ θ − −

θ −θ = ( )

( )3

1 2

2 2

22 cos 12

− θθ θ − −

θ −θ

= ( )( )

31 2

2

22 cos 12

− θθ θ − −

θ −θ = ( )

( )2

1 2

2

22 cos 12

− θθ θ − −

− θ


10.(b) Differentiate 1x cos ec x− with respect to x.

If y = 1x cos ec x− then ( )( )

( )

12

1

2

1 xdy 2x cos ec x 1dx

x x 1

−

−

⎛ ⎞−⎜ ⎟

⎜ ⎟ ⎡ ⎤= + ⎣ ⎦⎜ ⎟⎡ ⎤−⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠

= ( )

1x cos ec x2 x x x 1

−⎛ ⎞− ⎡ ⎤⎜ ⎟ + ⎣ ⎦⎜ ⎟−⎝ ⎠

= 1 1cosec x2 (x 1)

− −−

11.(a) Differentiate 1

2

sin 3xx

−

with respect to x.

If y = 1

2

sin 3xx

−

then

( )( )

( )( )

( )

2 1

2

22

3x sin 3x 2x1 3xdy

dx x

−

⎛ ⎞⎜ ⎟

−⎜ ⎟⎡ ⎤−⎜ ⎟⎣ ⎦⎝ ⎠=

= ( )

( )

( )

21

2

14 4 2

3x 2x sin 3x1 3x x 3x 2sin 3x

x x 1 9x

−

−

⎛ ⎞⎜ ⎟

−⎜ ⎟⎡ ⎤ ⎡ ⎤−⎜ ⎟⎣ ⎦⎝ ⎠ ⎢ ⎥= −

⎢ ⎥−⎣ ⎦

= ( )

13 2

1 3x 2sin 3xx 1 9x

−⎧ ⎫⎪ ⎪−⎨ ⎬

−⎪ ⎪⎩ ⎭


EXERCISE 137 Page 338 1. Use logarithmic equivalents of inverse hyperbolic functions to evaluate the following correct to

4 decimal places: (a) 1 1sinh2

− (b) 1sinh 4− (c) 1sinh 0.9−

2 2

1 x x a xsinh lna a

−⎧ ⎫+ +⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

(a) 1 1sinh2

− = 2 21 2 1 1 5ln ln ln1.618034

2 2

⎧ ⎫ ⎛ ⎞+ + +⎪ ⎪ = =⎜ ⎟⎨ ⎬ ⎜ ⎟⎪ ⎪ ⎝ ⎠⎩ ⎭ = 0.4812

(b) 1 1 4sinh 4 sinh1

− −= = ( )2 24 1 4ln ln 4 17 ln8.123106

1

⎧ ⎫+ +⎪ ⎪ = + =⎨ ⎬⎪ ⎪⎩ ⎭

= 2.0947

(c) 1sinh 0.9− = ( )2 20.9 1 0.9ln ln 0.9 1.81 ln 2.245362

1

⎧ ⎫+ +⎪ ⎪ = + =⎨ ⎬⎪ ⎪⎩ ⎭

= 0.8089

2. Use logarithmic equivalents of inverse hyperbolic functions to evaluate the following correct to

4 decimal places: (a) 1 5cosh4

− (b) 1cosh 3− (c) 1cosh 4.3−

2 2

1 x x x acosh lna a

−⎧ ⎫+ −⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

(a) 1 5cosh4

− = 2 25 5 4 5 3ln ln ln 2

4 4

⎧ ⎫+ − +⎪ ⎪ ⎛ ⎞= =⎨ ⎬ ⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

= 0.6931

(b) 1cosh 3− = ( )2 23 3 1ln ln 3 8 ln 5.828427

1

⎧ ⎫+ −⎪ ⎪ = + =⎨ ⎬⎪ ⎪⎩ ⎭

= 1.7627

(c) 1cosh 4.3− = ( )2 24.3 4.3 1ln ln 4.3 17.49 ln8.482105

1

⎧ ⎫+ −⎪ ⎪ = + =⎨ ⎬⎪ ⎪⎩ ⎭

= 2.1380

3. Use logarithmic equivalents of inverse hyperbolic functions to evaluate the following correct to

4 decimal places: (a) 1 1tanh4

− (b) 1 5tanh8

− (c) 1tanh 0.7−

1 x 1 a xtanh ln

a 2 a x− +⎛ ⎞= ⎜ ⎟−⎝ ⎠


(a) 1 1tanh4

− = 1 4 1 1 5ln ln2 4 1 2 3

+⎛ ⎞ =⎜ ⎟−⎝ ⎠ = 0.2554

(b) 1 5tanh8

− = 1 8 5 1 13ln ln2 8 5 2 3

+⎛ ⎞ =⎜ ⎟−⎝ ⎠ = 0.7332

(c) 1tanh 0.7− = 1 1 0.7 1 1.7ln ln2 1 0.7 2 0.3

+⎛ ⎞ =⎜ ⎟−⎝ ⎠ = 0.8673


EXERCISE 138 Page 341 1.(b) Differentiate 1sinh 4x− with respect to x.

If y = 1sinh 4x− then ( )2

dy 4dx 4x 1

=⎡ ⎤+⎣ ⎦

= ( )2

4

16x 1+

2.(a) Differentiate 1 t2cosh3

− with respect to t.

If y = 1 t2cosh3

− then 2 2

dy 12dt t 3

⎡ ⎤= ⎢ ⎥

−⎣ ⎦ =

( )2

2

t 9−

3.(b) Differentiate 13 tanh 3x− with respect to x.

If y = 13 tanh 3x− then ( )2

dy 33dx 1 3x

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦ = 2

91 9x−

4.(a) Differentiate 1 3xsech4


If y = 1 3xsech4

− then ( )2 2 2 2

3dy 1 1 14

xdx 9x 16 9x3x 3x 16 9xx 1 x1 416 164 4

− − − −= = = =

⎡ ⎤ ⎛ ⎞ ⎛ ⎞− −⎛ ⎞ −− ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

= ( )2

4

x 16 9x

−

−

5.(b) Differentiate 11 cos ec h 4x2


If y = 11 cos ec h 4x2

− then ( )2

dy 1 4dx 2 4x 4x 1

⎡ ⎤−⎢ ⎥= ⎢ ⎥

⎡ ⎤+⎢ ⎥⎣ ⎦⎣ ⎦

= ( )2

1

2x 16x 1

−

+

6.(a) Differentiate 1 2xcoth7



If y = 1 2xcoth7

− then 2 2 2 2

2 2 2 2 (49)dy 7 7 7 74x 49 4xdx 49 4x2x 11 49 497

= = = =− −⎛ ⎞ −− ⎜ ⎟

⎝ ⎠

= 2

1449 4x−

7.(b) Differentiate ( )1 21 cosh x 12

− + with respect to x.

If y = ( )1 21 cosh x 12

− + then ( )

( ) ( ) ( )

12 2

2 2 22

1 x 1 (2x)dy 1 x2dx 2 2 x 1 x 1 1x 1 1

−⎡ ⎤⎢ ⎥+⎢ ⎥= =⎢ ⎥⎡ ⎤ + + −+ −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

= ( ) ( ) ( )22 2

x x

2x x 12 x 1 x=

++ =

( )2

1

2 x 1+

8.(a) Differentiate 1sech (x 1)− − with respect to x.

If y = 1sech (x 1)− − then 22

dy 1 1dx (x 1) 1 (x 2x 1)(x 1) 1 (x 1)

− −= =

− − − +⎡ ⎤− − −⎣ ⎦

2

1(x 1) (2x x )

−=

− − =

[ ]1

(x 1) x(2 x)−

− −

9.(b) Differentiate 1coth (cos x)− with respect to x.

If y = 1coth (cos x)− then ( )2 2 2

dy sin x sin x sin x 1dx 1 cos x sin x sin x1 cos x

− − − −= = = = =

−− = cosec x−

10.(a) Differentiate 1sinh−θ θ with respect to θ.

If y = 1sinh−θ θ then ( )

( )( )1

2

dy 1 sinh 1d 1

−⎛ ⎞⎜ ⎟= θ + θ⎜ ⎟θ ⎜ ⎟θ +⎝ ⎠

= ( )

1

2sinh

1−θ

+ θθ +

11.(b) Differentiate ( )

1

2

tanh x1 x

−



If y = ( )

1

2

tanh x1 x

−

− then

( ) ( )( )

( )

22

22

11 x tanh x 2xdy 1 xd 1 x

⎛ ⎞− − −⎜ ⎟−⎝ ⎠=θ −

= ( )

1

22

1 2x tanh x

1 x

−+

−

12. Show that 1d x cosh (cosh x) 2xdx

−⎡ ⎤ =⎣ ⎦

( )( )

( )1 1

2

d sinh xx cosh (cosh x) x cosh (cosh x) (1)dx cosh x 1

− −⎡ ⎤⎢ ⎥⎡ ⎤ = +⎣ ⎦ ⎢ ⎥−⎣ ⎦

= ( )2

x sinh x xsinh x

+ since 1cosh (cosh x) x− =

= x sinh x xsinh x

+

= x + x = 2x

13.(b) Determine ( )2

3 dx4x 25+

∫

( ) ( )2 22 222

3 1 1 3 1dx 3 dx 3 dx54x 25 2x 5 2x2x 15 1

55

= = =⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ ⎛ ⎞+ ⎛ ⎞⎛ ⎞⎣ ⎦ ++⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎝ ⎠⎣ ⎦

∫ ∫ ∫ ∫

= 2 2

2 23 5 35 5dx dx5 2 22x 2x1 1

5 5

⎛ ⎞ =⎜ ⎟⎝ ⎠ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

∫ ∫

= 13 2xsinh c2 5

− +


1 dtt 5−

∫

( ) 2 22 2

11 1 1 1 1 5 5dt dt dt dt

15 5t 5 t t t5 1 1 15 5 5

⎛ ⎞= = = ⎜ ⎟⎜ ⎟⎡ ⎤ ⎡ ⎤ ⎡ ⎤− ⎛ ⎞ ⎝ ⎠⎛ ⎞ ⎛ ⎞− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

∫ ∫ ∫ ∫

= 2

15 dt

t 15

⎡ ⎤⎛ ⎞ −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

∫ = 1 tcosh c5

− +



3 dx16 2x−∫

( ) ( ) ( ) ( )2 22 2 2 2

3 3 3 1 3 1 8dx dx dx dx2 216 2x 2 8 x 88 x 8 x

⎛ ⎞= = = ⎜ ⎟− − ⎝ ⎠− −∫ ∫ ∫ ∫ = 13 xtanh c

2 8 8− +


CHAPTER 34 PARTIAL DIFFERENTIATION EXERCISE 139 Page 345

2. Find zx∂∂

and xy∂∂

given z = x3 – 2xy + y2

If z = 3 2x 2xy y− + then ( ) ( ) ( )3 2 2 2z d d dx (2y) (x) y (1) 3x 2y(1) y (0)x dx dx dx∂

= − + = − +∂

23x 2y= −

and ( )3 2 3x d d d(x ) (1) (2x) (y) y (x )(0) 2x(1) 2yy dy dy dy∂

= − + = − +∂

2x 2y= − +

5. Find zx∂∂

and xy∂∂

given z = 3 22

y 1x yx y

− +

If z = 3 2 3 2 2 12

y 1x y x y yx yx y

− −− + = − + then ( )( ) ( )( )2 2 3z y 3x y 2x 0x

−∂= − − +

∂2 2

3

2y3x yx

= +

and ( )( ) ( )3 2 2x x 2y 1 x yy

− −∂= − −

∂ 3

2 2

1 12x yx y

= − −

6. Find zx∂∂

and xy∂∂

given z = cos 3x sin 4y

If z = cos 3x sin 4y then ( ) ( ) ( )( )z dsin 4y cos3x sin 4y 3sin 3xx dx∂

= = −∂

= 3sin 3xsin 4y−

and ( ) ( ) ( )( )x dcos3x sin 4y cos3x 4cos 4yy dy∂

= =∂

= 4cos 3xcos4y

8. The resonant frequency fr in a series electrical circuit is given by fr =1

2 LCπ. Show that

r3

f 1L 4 CL∂ −

=∂ π

fr =1 12 21 1 L C

22 LC− −

=ππ

and 1 3

r 2 21 3 32 2

f 1 1 1 1C LL 2 2 4 C L4 C L

− −⎛ ⎞⎛ ⎞∂= − = − = −⎜ ⎟⎜ ⎟∂ π π⎝ ⎠⎝ ⎠ π

= 3

14 CL

−

π


9. An equation resulting from plucking a string is: n n b n by sin x k cos t csin tL L Lπ ⎧ π π ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

Determine yt

∂∂

and yx∂∂

n n b n b n n b n n by sin x k cos t csin t k sin x cos t csin x sin tL L L L L L Lπ ⎧ π π ⎫ π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

Hence, y n n b n b n n b n bk sin x sin t csin x cos tt L L L L L L

∂ ⎡ π ⎤ ⎡ π ⎤ π ⎡ π ⎤ ⎡ π ⎤ π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= n b n n b n bsin x ccos t k sin tL L L L

⎧ ⎫π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞−⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

and y n n n b n n n bk cos x cos t c cos x sin tx L L L L L L∂ π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= n n n b n bcos x k cos t csin tL L L L

⎧ ⎫π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

10. In a thermodynamic system, k =T S H

R TAe∆ −∆

where R, k and A are constants. Find

(a) kT∂∂

(b) AT∂∂

(c) ( S)T

∂ ∆∂

(d) ( H)T

∂ ∆∂

(a) ( )( ) ( )( )T S HR T

2 2

RT S T S H Rk AeT R T

∆ −∆⎡ ⎤ ∆ − ∆ −∆⎡ ⎤∂= ⎢ ⎥ ⎢ ⎥∂ ⎢ ⎥ ⎣ ⎦⎣ ⎦

= T S H

R T2 2

RT S RT S R HAeR T

∆ −∆⎡ ⎤ ⎡ ⎤∆ − ∆ + ∆⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥⎣ ⎦

= T S H

R T2 2

R HAeR T

∆ −∆⎡ ⎤ ⎡ ⎤∆⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥⎣ ⎦ =

T S HR T

2

HAeR T

∆ −∆⎡ ⎤ ⎡ ⎤∆⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥⎣ ⎦

= T S H

RT2

A H eR T

∆ −∆∆

(b) Since k = T S H

R TAe∆ −∆

then T S H

R TH T S

RTA k e k e∆ −∆

− ∆ − ∆

= =

Thus, ( )( ) ( )( )S T SR T

2 2

RT S H T S RA k eT R T

∆ − ∆⎛ ⎞ −∆ − ∆ − ∆⎛ ⎞∂= ⎜ ⎟⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

= S T SR T

2 2

R T S R H RT Sk eR T

∆ − ∆⎛ ⎞⎛ ⎞− ∆ − ∆ + ∆⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= S T SR T

2 2

R Hk eR T

∆ − ∆⎛ ⎞⎛ ⎞− ∆⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= H T S

RT2

k H eR T

∆ − ∆∆−


(c) If k = T S H

R TAe∆ −∆

then T S H

R Tk eA

∆ −∆

= and k T S HlnA R T

∆ −∆⎛ ⎞ =⎜ ⎟⎝ ⎠

i.e. kR T ln T S HA

⎛ ⎞ = ∆ −∆⎜ ⎟⎝ ⎠

(1)

i.e. kT S H R T lnA

⎛ ⎞∆ = ∆ + ⎜ ⎟⎝ ⎠

and 1H k kS R ln H T R lnT A A

−∆ ⎛ ⎞ ⎛ ⎞∆ = + = ∆ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Hence, ( )2( S) H T 0T

−∂ ∆= ∆ − +

∂ = 2

HT∆

−

(d) From equation (1), k kH T S R T ln T S R lnA A

⎧ ⎫⎛ ⎞ ⎛ ⎞∆ = ∆ − = ∆ −⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎩ ⎭

Hence, ( H)T

∂ ∆∂

= kS R lnA

⎛ ⎞∆ − ⎜ ⎟⎝ ⎠



2. If z 2 ln xy= find (a) 2

2

zx∂∂

(b) 2

2

zy∂∂

(c) 2z

x y∂∂ ∂

(d) 2z

y x∂∂ ∂

(a) 1z 2 2(y) 2xx xy x

−⎛ ⎞∂= = =⎜ ⎟∂ ⎝ ⎠

and 2

2

zx∂∂

= ( )1 22x 2xx

− −∂= −

∂= 2

2x

−

(b) 1z 2 2(x) 2yy xy y

−⎛ ⎞∂= = =⎜ ⎟∂ ⎝ ⎠

and 2

2

zy∂∂

= ( )1 22y 2yy

− −∂= −

∂= 2

2y

−

(c) ( )2

1z 2yx y x

−∂ ∂=

∂ ∂ ∂ = 0

(d) ( )2

1z 2xy x y

−∂ ∂=

∂ ∂ ∂ = 0

3. If ( )( )x y

zx y−

=+

find (a) 2

2

zx∂∂

(b) 2

2

zy∂∂

(c) 2z

x y∂∂ ∂

(d) 2z

y x∂∂ ∂

(a) 22 2 2

z (x y)(1) (x y)(1) x y x y 2y 2y(x y)x (x y) (x y) (x y)

−∂ + − − + − += = = = +

∂ + + +

2

32

z 4y(x y) (1)x

−∂= − +

∂ = 3

4y(x 3)

−+

(b) 22 2 2

z (x y)( 1) (x y)(1) x y x y 2x 2x(x y)y (x y) (x y) (x y)

−∂ + − − − − − − + −= = = = − +

∂ + + +

2

32

z ( 2x) 2(x y) (1)y

−∂ ⎡ ⎤= − − +⎣ ⎦∂ = 3

4x(x 3)+

(c) 2

2 3 23 2

z 4x 22x(x y) ( 2x) 2(x y) (x y) ( 2)x y x (x y) (x y)

− − −∂ ∂ ⎡ ⎤ ⎡ ⎤= − + = − − + + + − = −⎣ ⎦ ⎣ ⎦∂ ∂ ∂ + +

= 3 3

4x 2(x y) 2x 2y(x y) (x y)− + −

=+ +

= 3

2(x y)(x y)

−+

(d) 2

2 3 23 2

z 4y 22y(x y) (2y) 2(x y) (x y) (2)y x y (x y) (x y)

− − −∂ ∂ −⎡ ⎤ ⎡ ⎤= + = − + + + = +⎣ ⎦ ⎣ ⎦∂ ∂ ∂ + +

= 3 3

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

− + + − + +=

+ +

= 3

2x 2y(x y)

−+

= 3

2(x y)(x y)

−+


5. If 2z x sin(x 2y)= − find (a) 2

2

zx∂∂

(b) 2

2

zy∂∂

Show also that 2z

x y∂∂ ∂

= 2z

y x∂∂ ∂

= 22x sin(x 2y) 4x cos(x 2y)− − −

(a) ( ) [ ]2 2z x cos(x 2y) sin(x 2y) (2x) x cos(x 2y) 2x sin(x 2y)x∂

= − + − = − + −∂

( )[ ] [ ] [ ]2

22

z x sin(x 2y) cos(x 2y) (2x) (2x)cos(x 2y) sin(x 2y) (2)x∂

= − − + − + − + −∂

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

= ( )22 x sin(x 2y) 4xcos(x 2y)− − + −

(b) [ ]2 2z x cos(x 2y) ( 2) 2x cos(x 2y)y∂

= − − = − −∂

2

2

zy∂∂

= ( )[ ]22x sin(x 2y) ( 2)− − − −

= 24x sin(x 2y)− −

( )[ ] [ ]2

2 2z 2x cos(x 2y) 2x sin(x 2y) cos(x 2y) ( 4x)x y x∂ ∂ ⎡ ⎤= − − = − − − + − −⎣ ⎦∂ ∂ ∂

= 22x sin(x 2y) 4xcos(x 2y)− − −

( )[ ] [ ]2

2 2z x cos(x 2y) 2x sin(x 2y) x sin(x 2y) ( 2) (2x) cos(x 2y) ( 2)y x y∂ ∂ ⎡ ⎤= − + − = − − − + − −⎣ ⎦∂ ∂ ∂

= 22x sin(x 2y) 4xcos(x 2y)− − −

7. Given 3xzy

⎛ ⎞= ⎜ ⎟

⎝ ⎠ show that

2zx y∂∂ ∂

= 2z

y x∂∂ ∂

and evaluate 2

2

zx∂∂

when 1x2

= and y = 3

1 12 23xz 3x y

y−⎛ ⎞

= =⎜ ⎟⎝ ⎠

1 1 1 12 2 2 2z 1 33 y x y x

x 2 2− − − −⎛ ⎞∂

= =⎜ ⎟∂ ⎝ ⎠

1 3 1 32 2 2 2z 1 33 x y x y

y 2 2− −⎛ ⎞⎛ ⎞∂

= − = −⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

1 3 3 122 2 2 2

3 12 2

z 3 3 1 3x y y xx y x 2 2 2 4y x

− − −⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ −= − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= ( )3

3

4 xy

−


1 1 1 322 2 2 2

1 32 2

z 3 3 1 3y x x yy x y 2 2 2 4x y

− − − −⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ −= = − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= ( )3

3

4 xy

−

Thus, 2z

x y∂∂ ∂

= 2z

y x∂∂ ∂

( )1 1 1 322 2 2 2

2 3

z 3 3 1 3y x y xx x 2 2 2 4 yx

− − − −⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ −= = − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ⎝ ⎠⎝ ⎠ ⎝ ⎠

When 1x2

= and y = 3,

( )( )2

2 3

3

2 4z 3 3 1 8 2 4 2(2) 2x 4 4 4 4 21 11 4 3 44 (3) 2 82

−∂ − − − − − −= = = = = = = = −

∂ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

= 1 or 0.70712

− −

8. An equation used in thermodynamics is the Benedict-Webb-Rubine equation of state for the

expansion of a gas. The equation is:

22

0 V0 0 2 2 3 6 2 3

C 1CRT 1 1 A 1Vp B RT A (bRT a) e

V T V V V T V

γ−

γ⎛ ⎞+⎜ ⎟α⎛ ⎞ ⎛ ⎞⎝ ⎠= + − − + − + + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Show that 22

V02 2 4 2

p 6 C 1 e CT V T V V

γ−⎧ ⎫∂ γ⎪ ⎪⎛ ⎞= + −⎨ ⎬⎜ ⎟∂ ⎝ ⎠⎪ ⎪⎩ ⎭

( ) 22 2 V0 0 0 2 3 6 2 3

RT 1 1 A 1p B RT A C T (bRT a) T C 1 eV V V V V V

γ−

− −α γ⎛ ⎞⎛ ⎞= + − − + − + + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

23

30 0 V2 2 3 2 3

B R 2C Tp R bR 12T C 1 eT V V V V V V

γ− −−∂ γ⎛ ⎞⎛ ⎞= + + − − +⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

242

40 V2 2 2 3

6C Tp 16T C 1 eT V V V

γ− −−−∂ γ⎛ ⎞⎛ ⎞= + +⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

= 2V02 4 2

6 C 1 e CV T V V

γ−⎧ ⎫γ⎪ ⎪⎛ ⎞+ −⎨ ⎬⎜ ⎟

⎝ ⎠⎪ ⎪⎩ ⎭


CHAPTER 35 TOTAL DIFFERENTIAL, RATES OF CHANGE

AND SMALL CHANGES EXERCISE 141 Page 350 2. Find the total differential, dz, given z = 2xy – cos x

Since z = 2xy – cos x then dz = z zdx dyx y∂ ∂

+∂ ∂

= (2y + sin x)dx + 2x dy

3. Find the total differential, dz given z = x yx y−+

Since z = x yx y−+

then 2 2

z (x y)(1) (x y)(1) 2yx (x y) (x y)∂ + − −

= =∂ + +

and 2 2

z (x y)( 1) (x y)(1) 2xy (x y) (x y)∂ + − − − −

= =∂ + +

Thus, dz = z zdx dyx y∂ ∂

+∂ ∂

= 2 2

2y 2xdx dy(x y) (x y)

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

5. Find the total differential, dz, given z = xy + xy

- 4

Since z = xy + xy

- 4 then

121 xz 12y y

x y 2y x

−

∂= + = +

∂

and ( )22

z xx x y xy y

−∂= − = −

∂

Thus, dz = z zdx dyx y∂ ∂

+∂ ∂

= 2

1 xy dx x dyy2y x

⎛ ⎞⎛ ⎞+ + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

7. Given u = ln sin(xy) show that du = cot(xy)(y dx + x dy) Since u = ln sin(xy) then


du = u udx dyx y∂ ∂

+∂ ∂

= 1 1y cos xy dx x cos xy dysin(xy) sin(xy)

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞+⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

= y cot(xy) dx + x cot(xy) dy

i.e. du = cot(xy)[ y dx + x dy ]


EXERCISE 142 Page 352 1. The radius of a right cylinder is increasing at a rate of 8mm/s and the height is decreasing at a

rate of 15 mm/s. Find the rate at which the volume is changing in cm3/s when the radius is 40 mm

and the height is 150 mm.

Volume of cylinder, V = 2r hπ

Rate at which volume is changing, dV V dr V dhdt r dt h dt

∂ ∂= +∂ ∂

= ( ) ( )2dr dh2 rh rdt dt

π + π

= [ ] 22 (40)(150) (8) (40) ( 15)⎡ ⎤π + π −⎣ ⎦

= 96 000π - 24 000π

= 72 000π 3mm / s

= 72π 3cm / s = 226.2 3cm / s

3. Find the rate of change of k, correct to 4 significant figures, given the following data:

k = f(a, b, c); k = 2b ln a + c2 ea; a is increasing at 2 cm/s; b is decreasing at 3 cm/s; c is

decreasing at 1 cm/s; a = 1.5 cm, b = 6 cm and c = 8 cm.

Since k = 2b ln a + c2 ea

then rate of change of k, dk k da k db k dcdt a dt b dt c dt

∂ ∂ ∂= + +∂ ∂ ∂

= ( ) ( )2 a a2b da db dcc e 2ln a 2cea dt dt dt

⎛ ⎞+ + +⎜ ⎟⎝ ⎠

= ( ) ( )2 1.5 1.5(2)(6) 8 e (2) 2 ln1.5 ( 3) 2(8)e ( 1)1.5

⎛ ⎞+ + − + −⎜ ⎟⎝ ⎠

= 589.656 – 2.433 – 71.707

= 515.5 cm/s

5. Find the rate of change of the total surface area of a right circular cone at the instant when the

base radius is 5 cm and the height is 12 cm if the radius is increasing at 5 mm/s and the height is

decreasing at 15 mm/s.


Total surface area of a cone, A = πrl + 2rπ = ( )2 2 2r r h rπ + +π (see diagram of cone below)

Rate of change of surface area,

dA A dr A dhdt r dt h dt

∂ ∂= +∂ ∂

= ( ) ( ) ( ) ( ) ( )1 1

2 2 2 2 2 22 21 dr 1 dhr r h (2r) r h ( ) 2 r r r h (2h)2 dt 2 dt

− −⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎛ ⎞⎪ ⎪ ⎪ ⎪π + + + π + π + π +⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎝ ⎠⎩ ⎭ ⎩ ⎭

= ( )

( )( )

22 2

2 2 2 2

r dr rh dhr h 2 rdt dtr h r h

⎧ ⎫ ⎧ ⎫π π⎪ ⎪ ⎪ ⎪+ π + + π +⎨ ⎬ ⎨ ⎬+ +⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

= ( )

( )( )

22 2

2 2 2 2

(5) (5)(12)5 12 2 (5) (0.5) ( 1.5)5 12 5 12

⎧ ⎫ ⎧ ⎫π π⎪ ⎪ ⎪ ⎪+ π + + π + −⎨ ⎬ ⎨ ⎬+ +⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

in centimetre units

= 25 6013 10 (0.5) ( 1.5)13 13π π⎛ ⎞ ⎛ ⎞+ π+ π + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= (78.298)(0.5) + (14.50)(-1.5)

= 39.149 – 21.75

= 17.4 2cm / s


EXERCISE 143 Page 354 2. An equation for heat generated H is H = i2Rt. Determine the error in the calculated value of H if

the error in measuring current i is +2%, the error in measuring resistance R is –3% and the error

in measuring time t is +1%

Since H = i2Rt then H H HH i R ti R t

∂ ∂ ∂δ ≈ δ + δ + δ

∂ ∂ ∂

( ) ( ) ( )2 22iRt (0.02i) i t ( 0.03R) i R (0.01t)≈ + − +

2 2 20.04i Rt 0.03i Rt 0.01i Rt≈ − +

≈ (0.04 – 0.03 + 0.01) H

≈ 0.02 H

i.e. the error in H is + 2%

3. fr = 12 LCπ

represents the resonant frequency of a series connected circuit containing

inductance L and capacitance C. Determine the approximate percentage change in fr when L is

decreased by 3% and C is increased by 5%

Since fr = 1 12 21 1 L C

22 LC− −

=ππ

then r rr

f ff L CL C∂ ∂

δ ≈ δ + δ∂ ∂

3 1 3 12 2 2 21 1L C C L

2 2L C2 2

− − − −⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟

≈ δ + δ⎜ ⎟ ⎜ ⎟π π⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3 1 3 12 2 2 2L C C L( 0.03L) (0.05C)

4 4

− − − −⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟≈ − −⎜ ⎟ ⎜ ⎟π π⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1 1 1 12 2 2 20.03L C 0.05C L

4 4

− − − −

≈ −π π

1 10.015 0.0252 LC 2 LC

⎛ ⎞ ⎛ ⎞≈ −⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

i.e. r r rf (0.015 0.025)f 0.01fδ ≈ − ≈ −

i.e. the approximate percentage change in fr is – 1%


4. The second moment of area of a rectangle about its centroid parallel to side b is given by

I =3bd

12 . If b and d are measured as 15 cm and 6 cm respectively, and the measurement errors

are +12 mm in b and –1.5 mm in d, find the error in the calculated value of I.

Since I =3bd

12 then I II b d

b d∂ ∂

δ ≈ δ + δ∂ ∂

3 2d 3bdb d

12 12⎛ ⎞ ⎛ ⎞

≈ ∂ + ∂⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3 26 3(15)(6)(1.2) ( 0.15)12 12⎛ ⎞ ⎛ ⎞

≈ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

≈ 21.6 – 20.25 = 1.35

i.e. error in I = + 1.35 4cm

5. The side b of a triangle is calculated using b2 = a2 + c2 – 2ac cos B. If a, c and B are measured as

3 cm, 4 cm and π/4 radians respectively and the measurement errors which occur are +0.8 cm,

- 0.5 cm and +π/9 radians respectively, determine the error in the calculated value of b.

b2 = a2 + c2 – 2ac cos B from which, ( )1

2 2 2b a c 2accos B= + −

Approximate error in b, b b bb a c Ba c B∂ ∂ ∂

δ ≈ δ + δ + δ∂ ∂ ∂

( )1

2 2 21 a c 2accos B (2a 2ccos B a2

−⎡ ⎤≈ + − − ∂⎢ ⎥⎣ ⎦

+ ( )1

2 2 21 a c 2accos B (2c 2a cos B c2

−⎡ ⎤+ − − ∂⎢ ⎥

⎣ ⎦

+ ( )1

2 2 21 a c 2accos B (2acsin B B2

−⎡ ⎤+ − ∂⎢ ⎥

⎣ ⎦

122 21 3 4 2(3)(4)cos (6 8cos (0.8)

2 4 4

−⎡ ⎤π π⎛ ⎞⎢ ⎥≈ + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

+ 122 21 3 4 2(3)(4)cos (8 6cos ( 0.5)

2 4 4

−⎡ ⎤π π⎛ ⎞⎢ ⎥+ − − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

+ 122 21 3 4 2(3)(4)cos (24sin

2 4 4 90

−⎡ ⎤π π π⎛ ⎞ ⎛ ⎞⎢ ⎥+ −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦


≈ (0.17645)(0.343146)(0.8) + (0.17645)(3.75736)(-0.5)

+ (0.17645(16.97056)90π⎛ ⎞

⎜ ⎟⎝ ⎠

i.e. approximate error in b ≈ 0.04844 – 0.3315 + 0.10453 = - 0.179 cm

7. The rate of flow of gas in a pipe is given by: v =6 5

C dT

, where C is a constant, d is the diameter

of the pipe and T is the thermodynamic temperature of the gas. When determining the rate of

flow experimentally, d is measured and subsequently found to be in error by +1.4%, and T has

an error of –1.8%. Determine the percentage error in the rate of flow based on the measured

values of d and T.

Flow rate, v =5162

6 5

C d Cd TT

−=

Approximate error in flow rate, v vv d Td T∂ ∂

δ ≈ δ + δ∂ ∂

12 11

66 5

1(C) d 52 (0.014d) C d T ( 0.018T)6T

−

−

⎛ ⎞⎜ ⎟ ⎛ ⎞⎛ ⎞

≈ + − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎜ ⎟⎝ ⎠

6 65 5

C d C d 5(0.007) (0.018)6T T

⎛ ⎞ ⎛ ⎞⎛ ⎞≈ +⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

≈ (0.007 + 0.015) v

≈ 0.022 v

i.e. the percentage error in the rate of flow is + 2.2%


CHAPTER 36 MAXIMA, MINIMA AND SADDLE POINTS FOR

FUNCTIONS OF TWO VARIABLES EXERCISE 144 Page 359 2. Find the maxima, minima and saddle points for the following functions:

(a) f(x, y) = 2 2x y 2x 4y 8+ − + +

(b) f(x, y) = 2 2x y 2x 4y 8− − + +

(c) f(x, y) = 2 22x 2y 2xy 2x y 4+ − − − +

(a) Let f(x, y) = z = 2 2x y 2x 4y 8+ − + +

(i) z 2x 2x∂

= −∂

and z 2y 4y∂

= +∂

(ii) For stationary points, 2x - 2 = 0 from which, x = 1

and 2y + 4 = 0 from which, y = -2

(iii) The co-ordinates of the stationary point is (1, -2)

(iv) 2

2

z 2x∂

=∂

2

2

z 2y∂

=∂

( )2z 2y 4 0

x y x∂ ∂

= + =∂ ∂ ∂

(v) When x = 1, y = -2, 2

2

z 2x∂

=∂

2

2

z 2y∂

=∂

and 2z 0

x y∂

=∂ ∂

(vi) 22z 0

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

(vii) 22 2 2

(1, 2) 2 2

z z zx y x y−

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) ( )( )20 2 2 4= − = − which is negative

(viii) Since ∆ < 0 and 2

2

zx∂∂

> 0 then (1, -2) is a minimum point.

(b) Let f(x, y) = z = 2 2x y 2x 4y 8− − + +

(i) z 2x 2x∂

= −∂

and z 2y 4y∂

= − +∂

(ii) For stationary points, 2x - 2 = 0 from which, x = 1

and - 2y + 4 = 0 from which, y = 2

(iii) The co-ordinates of the stationary point is (1, 2)


(iv) 2

2

z 2x∂

=∂

2

2

z 2y∂

= −∂

( )2z 2y 4 0

x y x∂ ∂

= − + =∂ ∂ ∂

(v) When x = 1, y = 2, 2

2

z 2x∂

=∂

2

2

z 2y∂

= −∂

and 2z 0

x y∂

=∂ ∂

(vi) 22z 0

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

(vii) 22 2 2

(1,2) 2 2

z z zx y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) ( )( )20 2 2 4= − − = which is positive

(viii) Since ∆ > 0 then (1, 2) is a saddle point.

(c) Let f(x, y) = z = 2 22x 2y 2xy 2x y 4+ − − − +

(i) z 2 2y 4xx∂

= − −∂

and z 2 2x 2yy∂

= − −∂

(ii) For stationary points, 2 – 2y – 4x = 0

i.e. 1 – y – 2x = 0 (1)

and 2 – 2x – 2y = 0

i.e. 1 – x – y = 0 (2)

From (1), y = 1 – 2x

Substituting in (2) gives: 1 – x –(1 – 2x) = 0

i.e. 1 – x – 1 + 2x = 0 from which, x = 0

When x = 0 in equations (1) and (2), y = 1

(iii) The co-ordinates of the stationary point is (0, 1)

(iv) 2

2

z 4x∂

= −∂

2

2

z 2y∂

= −∂

( )2z 2 2x 2y 2

x y x∂ ∂

= − − = −∂ ∂ ∂

(v) When x = 0, y = 1, 2

2

z 4x∂

= −∂

2

2

z 2y∂

= −∂

and 2z 2

x y∂

= −∂ ∂

(vi) 22

2z ( 2) 4x y

⎛ ⎞∂= − =⎜ ⎟∂ ∂⎝ ⎠

(vii) 22 2 2

(0,1) 2 2

z z zx y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

( )( )4 4 2 4= − − − = − which is negative


2

zx∂∂

< 0 then (0, 1) is a maximum point.

4. Locate the stationary points of the function z = 2 212x 6xy 15y+ +


(i) z 24x 6yx∂

= +∂

and z 6x 30yy∂

= +∂

(ii) For stationary points, 24x + 6y = 0 (1)

and 6x + 30y = 0 (2)

(iii) From (1), 6y = -24x i.e. y = -4x

Substituting in (2) gives: 6x + 30(-4x) = 0

i.e. 6x = 120x i.e. x = 0

When x = 0, y = 0, hence the co-ordinates of the stationary point is (0, 0)

(iv) 2

2

z 24x∂

=∂

2

2

z 30y∂

=∂

( )2z 6x 30y 6

x y x∂ ∂

= + =∂ ∂ ∂

(v) When x = 0, y = 0, 2

2

z 24x∂

=∂

2

2

z 30y∂

=∂

and 2z 6

x y∂

=∂ ∂

(vi) 22z 36

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

(vii) 22 2 2

(0,0) 2 2

z z zx y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) ( )( )26 24 30= − which is negative


2

zx∂∂

> 0 then (0, 0) is a minimum point

5. Find the stationary points of the surface z = 3 3x xy y− + and distinguish between them.

(i) 2z 3x yx∂

= −∂

and 2z x 3yy∂

= − +∂

(ii) For stationary points, 23x - y = 0 (1)

and 2x 3y− + = 0 (2)

(iii) From (1), y = 23x

Substituting in (2) gives: - x + ( )223 3x = 0

- x + 427x = 0

and ( )3x 27x 1 0− =

i.e. x = 0 or 327x 1 0− = i.e. 3 1x27

= and 31 1x27 3

⎛ ⎞= =⎜ ⎟⎝ ⎠

Hence, x = 0 or 1x3

=


From (1), when x = 0, y = 0

and when 1x3

= , y = 23x = 21 13

3 3⎛ ⎞ =⎜ ⎟⎝ ⎠

Hence, the stationary points occur at (0, 0) and 1 1,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

(iv) 2

2

z 6xx∂

=∂

2

2

z 6yy∂

=∂

( )2

2z x 3y 1x y x∂ ∂

= − + = −∂ ∂ ∂

(v) For (0, 0), 2

2

z 0x∂

=∂

2

2

z 0y∂

=∂

and 2z 1

x y∂

= −∂ ∂

For 1 1,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

, 2

2

z 2x∂

=∂

2

2

z 2y∂

=∂

and 2z 1

x y∂

= −∂ ∂

(vi) For (0, 0), 22z 1

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

For 1 1,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

, 22z 1

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

(vii) ( )( )22 2 2

(0,0) 2 2

z z z 1 0 0 1x y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

which is positive

( )( )1 1,3 3

1 2 2 3⎛ ⎞⎜ ⎟⎝ ⎠

∆ = − = − which is negative

(viii) Since ( )0,0∆ > 0 then (0, 0) is a saddle point

1 1,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

∆ < 0 and 2

2

zx∂∂

> 0 then 1 1,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

is a minimum point


EXERCISE 145 Page 363 1. The function z = 2 2x y xy 4x 4y 3+ + + − + has one stationary value. Determine its co-ordinates and its nature.

(i) z 2x y 4x∂

= + +∂

and z 2y x 4y∂

= + −∂

(ii) For stationary points, 2x + y + 4 = 0 (1)

and 2y + x – 4 = 0 (2)

(iii) (1) + (2) gives: 3x + 3y = 0 from which, y = -x

Substituting in (1), 2x – x + 4 = 0 i.e. x = -4, thus y = +4

Hence, the stationary point occurs at (-4, 4)

(iv) 2

2

z 2x∂

=∂

2

2

z 2y∂

=∂

( )2z 2y x 4 1

x y x∂ ∂

= + − =∂ ∂ ∂

(v) When x = -4, y = 4, 2

2

z 2x∂

=∂

2

2

z 2y∂

=∂

and 2z 1

x y∂

=∂ ∂

(vi) 22z 1

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

(vii) 22 2 2

( 4,4) 2 2

z z zx y x y−

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠



2

zx∂∂

> 0 then (-4, 4) is a minimum point

3. Determine the stationary values of the function f(x, y) = 4 2 2 2 2x 4x y 2x 2y 1+ − + − and

distinguish between them.

Let f(x, y) = z = 4 2 2 2 2x 4x y 2x 2y 1+ − + −

(i) 3 2z 4x 8xy 4xx∂

= + −∂

and 2z 8x y 4yy∂

= +∂

(ii) For stationary points, 3 24x 8xy 4x 0+ − = (1)

and 28x y 4y 0+ = (2)

(iii) From (2), ( )24y 2x 1 0− = from which, y = 0

From (1), if y = 0, 34x 4x 0− = i.e. ( )24x x 1 0− =

from which, x = 0 or x = ± 1


Hence, the stationary points occur at (0, 0) and (1, 0) and (-1, 0)

(iv) 2

2 22

z 12x 8y 4x∂

= + −∂

2

22

z 8x 4y∂

= +∂

( )2

2z 8x y 4y 16xyx y x∂ ∂

= + =∂ ∂ ∂

(v) For (0, 0), 2

2

z 4x∂

= −∂

2

2

z 4y∂

=∂

and 2z 0

x y∂

=∂ ∂

For (1, 0), 2

2

z 8x∂

=∂

2

2

z 12y∂

=∂

and 2z 0

x y∂

=∂ ∂

For (-1, 0), 2

2

z 8x∂

=∂

2

2

z 12y∂

=∂

and 2z 0

x y∂

=∂ ∂

(vi) For all three stationary points, 22z 0

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

(vii) 22 2 2

(0,0) 2 2

z z zx y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) ( )( )20 4 4 32= − − = which is positive

22 2 2

(1,0) 2 2

z z zx y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠


22 2 2

( 1,0) 2 2

z z zx y x y−

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠


(viii) Since (0,0)∆ > 0, then (0, 0) is a saddle point.

Since (1,0)∆ < 0 and 2

2

zx∂∂

> 0 then (1, 0) is a minimum point.

Since ( 1,0)−∆ < 0 and 2

2

zx∂∂

> 0 then (-1, 0) is a minimum point.

4. Determine the stationary points of the surface f(x, y) = 3 2 2x 6x y− − Let f(x, y) = z = 3 2 2x 6x y− −

(i) 2z 3x 12xx∂

= −∂

and z 2yy∂

= −∂

(ii) For stationary points, 23x 12x− = 0 (1)

and -2y = 0 (2)

(iii) From (1) 3x(x 4)− = 0 from which, x = 0 or x = 4

From (2), y = 0

Hence, the stationary points occurs at (0, 0) and (4, 0)


(iv) 2

2

z 6x 12x∂

= −∂

2

2

z 2y∂

= −∂

( )2z 2y 0

x y x∂ ∂

= − =∂ ∂ ∂

(v) At (0, 0), 2

2

z 12x∂

= −∂

2

2

z 2y∂

= −∂

and 2z 0

x y∂

=∂ ∂

At (4, 0), 2

2

z 12x∂

=∂

2

2

z 2y∂

= −∂

and 2z 0

x y∂

=∂ ∂

(vi) For both points, 22z 0

x y⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

(vii) 22 2 2

(0,0) 2 2

z z zx y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) ( )( )20 12 2 24= − − − = − which is negative

(4,0)∆ ( ) ( )( )20 12 2 24= − − = which is positive

(viii) Since (0,0)∆ < 0 and 2

2

zx∂∂

< 0 then (0, 0) is a maximum point.

Since (4,0)∆ >0 then (4, 0) is a saddle point. 6. A large marquee is to be made in the form of a rectangular box-like shape with canvas covering

on the top, back and sides. Determine the minimum surface area of canvas necessary if the

volume of the marquee is to be 250 m3.

A sketch of the marquee is shown above.

Volume of marquee, V = xyz = 250 (1)

Surface area, S = xy + yz + 2xz (2)

From (1), z = 250xy

Substituting in (2) gives: S = 1 1250 250 250 500xy y 2x xy xy 250x 500yxy xy x y

− −⎛ ⎞ ⎛ ⎞+ + = + + = + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠


2

S 250yx x∂

= −∂

and 2

S 500xy y∂

= −∂

For a stationary point, 2

S 250y 0x x∂

= − =∂

from which, 2

250yx

= or 2yx 250= (3)

and 2

S 500x 0y y∂

= − =∂

from which, 2

500xy

= or 2xy 500= (4)

Dividing equation (3) by equation (40 gives:

2

2

yx 250xy 500

= i.e. x 1y 2= and y = 2x

Substituting y = 2x in equation (3) gives: 32x 250= and x = 3 125 = 5 m

and y = 2x = 10 m

From equation (1), xyz = 250 i.e. (5)(10)z = 250 from which, z = 5 m

2

2 3

S 750x x∂

=∂

2

2 3

S 1000y y∂

=∂

and 2S 1

x y∂

=∂ ∂

When x = 5 and y = 10, 2

2

S 6x∂

=∂

2

2

S 1y∂

=∂

and 2S 1

x y∂

=∂ ∂

22 2 2

2 2

S S Sx y x y

⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂∆ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠


Since ∆ < 0 and 2

2

zx∂∂

> 0 then the surface area is a minimum.

Minimum surface area, S = xy + yz + 2xz

= (5)(10) + (10)(5) + (2)(5)(5)

= 50 + 50 + 50 = 150 2m


CHAPTER 37 STANDARD INTEGRATION


3. Determine (a) 23x 5x dxx

⎛ ⎞−⎜ ⎟⎝ ⎠∫ (b) 2(2 ) d+ θ θ∫

(a) ( )2 23x 5x 3x 5xdx dx 3x 5 dxx x x

⎛ ⎞ ⎛ ⎞−= − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ ∫ ∫ =

23x 5x c2

− +

(b) ( )2 2(2 ) d 4 4 d+ θ θ = + θ+ θ θ∫ ∫ = 3

24 2 c3θ

θ + θ + +

4.(b) Determine 4

3 dx4x∫

4 1 3

4 34

3 3 3 x 3 x 1dx x dx c c x c4x 4 4 4 1 4 3 4

− + −− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = + = + = − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− + −⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

∫ ∫ = 3

1 c4x

− +

5.(b) Determine 4 51 x dx4∫

5 915 94 4

4 5 4 41 1 1 x 1 x 1 4x dx x dx c c x c5 94 4 4 4 4 914 4

+⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = + = + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

∫ ∫ = 4 91 x c9

+

6.(b) Determine 5 4

3 dx7 x∫

4 114 15 55 5

45 45

3 3 1 3 3 x 3 x 3 5dx dx x dx c c x c4 17 7 7 7 7 17 x 1x5 5

− +− ⎛ ⎞⎛ ⎞= = = + = + = +⎜ ⎟⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎝ ⎠⎝ ⎠− +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∫ ∫ ∫ = 515 x c7

+

8. Determine (a) 23 sec 3x dx4∫ (b) 22cosec 4 dθ θ∫

(a) 23 3 1sec 3x dx tan 3x c4 4 3

⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫ = 1 tan 3x c

4+

(b) ( )2 12cosec 4 d 2 cot 4 c4

⎛ ⎞θ θ = − θ +⎜ ⎟⎝ ⎠∫ = 1 cot 4 c

2− θ +

9. Determine (a) 5 cot 2t cosec2t dt∫ (b) 4 sec 4t tan 4t dt3∫


(a) ( ) 15 cot 2t cosec2t dt 5 cosec2t c2

⎛ ⎞= − +⎜ ⎟⎝ ⎠∫ = 5 cosec2t c

2− +

(b) 4 4 1sec 4t tan 4t dt sec 4t c3 3 4

⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫ = 1 sec4t c

3+

10.(b) Determine 5x

2 dx3 e∫

5x 5x 5x

5x

2 dx 2 2 1 2e dx e c e c3 e 3 3 5 15

− − −⎛ ⎞⎛ ⎞= = + = − +⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠∫ ∫ = 5x

2 c15e−

+

11.(b) Determine 2u 1 duu

⎛ ⎞−⎜ ⎟⎝ ⎠∫

2 2u 1 u 1 1du du u duu u u u

⎛ ⎞ ⎛ ⎞− ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

∫ ∫ ∫ = 2u lnu c

2− +

12. Determine (a) 2(2 3x) dx

x+

∫ (b) 21 2t dt

t⎛ ⎞+⎜ ⎟⎝ ⎠∫

(a) 1 1 32 2 22 2 2

1 1 12 2 2

(2 3x) 4 12x 9x 4 12x 9xdx dx dx 4x 12x 9x dxx x x x x

−⎛ ⎞ ⎛ ⎞+ + + ⎜ ⎟= = + + = + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠∫ ∫ ∫ ∫

=

1 3 52 2 24x 12x 9x c1 3 5

2 2 2

+ + + = 3 5188 x 8 x x c5

+ + +

(b) ( )2

2 2 22

1 1 1 12t dt 2t 2t dt 4 4t dt t 4 4t dtt t t t

−⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞+ = + + = + + = + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠∫ ∫ ∫ ∫

= 1 3t 4t4t c1 3

−

+ + +−

= 31 4t4t c

t 3− + + +



2. Evaluate (a) ( )2 2

13 x dx

−−∫ (b) ( )3 2

1x 4x 3 dx− +∫

(a) ( ) ( )2 33 32 2

11

1x 2 8 13 x dx 3x 3(2) 3( 1) 6 33 3 3 3 3−

−

⎡ ⎤−⎡ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞− = − = − − − − = − − − − −⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦∫

= 1 23 23 3

⎛ ⎞− −⎜ ⎟⎝ ⎠

= 6

(b) ( ) ( )33 23 2

11

x 4x 1x 4x 3 dx 3x 9 18 9 2 33 2 3

⎡ ⎤ ⎛ ⎞− + = − + = − + − − +⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦∫ = ( ) 10 1

3⎛ ⎞− ⎜ ⎟⎝ ⎠

= 113

−

4. Evaluate (a) /3

/ 62sin 2 d

π

πθ θ∫ (b)

2

03sin t dt∫

(a) [ ]/3 /3

/ 6/ 6

2 2 22sin 2 d cos 2 cos cos2 3 6

π π

ππ

π π⎡ ⎤θ θ = − θ = − −⎢ ⎥⎣ ⎦∫ (note that 2 2and3 6π π are in radians)

= -[-0.5 – 0.5] = -[-1] = 1

(b) [ ]2 2

003sin t dt 3 cos t 3[cos 2 cos0]= − = − −∫ (note that 2 is 2 radians)

= -3[-0.41615 – 1] = 4.248

6. Evaluate, correct to 4 significant figures, (a) 2 2

1cosec 4tdt∫ (b) ( )

/ 2

/ 43sin 2x 2cos3x dx

π

π−∫

(a) [ ] [ ]2 22

11

1 1 1 1 1cosec 4tdt cot 4t cot 8 cot 44 4 4 tan8 tan 4

⎡ ⎤= − = − − = − −⎢ ⎥⎣ ⎦∫

= ( )1 0.147065 0.8636914

− − − = 0.2527

(b) ( )/ 2

/ 2

/ 4/ 4

3 23sin 2x 2cos3x dx cos 2x sin 3x2 3

ππ

ππ

⎡ ⎤− = − −⎢ ⎥⎣ ⎦∫

= 3 2 2 3 3 2 2 3cos sin cos sin2 2 3 2 2 4 3 4

π π π π⎛ ⎞ ⎛ ⎞− − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 3 2 20 (0.707107)2 3 3

⎛ ⎞ ⎛ ⎞+ − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 2.638

8. Evaluate, correct to 4 significant figures, (a) 3

2

2 dx3x∫ (b)

23

1

2x 1dxx+

∫

(a) [ ] ( )3 3 3

22 2

2 2 1 2 2dx dx ln x ln 3 ln 23x 3 x 3 3

= = = −∫ ∫ = 0.2703


(b) 2 23 3 3 32

11 1 1

2x 1 2x 1 1dx dx 2x dx x ln xx x x x

⎛ ⎞+ ⎛ ⎞ ⎡ ⎤= + = + = +⎜ ⎟ ⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠∫ ∫ ∫ = (9 + ln 3) – (1 + ln 1)

= 9 + ln 3 – 1 = 8 + ln 3 = 9.099

9. The entropy change ∆S, for an ideal gas is given by: ∆S = 2 2

1 1

T V

vT V

dT dVC RT V

−∫ ∫

where T is the thermodynamic temperature, V is the volume and R = 8.314. Determine the

entropy change when a gas expands from 1 litre to 3 litres for a temperature rise from 100 K to

400 K given that: Cv = 45 + 6 × 10-3 T + 8 × 10-6 T2

∆S = ( )400 33 6 2

100 1

dT dV45 6 10 T 8 10 T 8.314T V

− −+ × + × −∫ ∫

= 400 33 6

100 1

45 dV6 10 8 10 T dT 8.314T V

− −⎛ ⎞+ × + × −⎜ ⎟⎝ ⎠∫ ∫

= [ ]4006 2

331

100

8 10 T45ln T 6 10 T 8.314 ln V2

−−⎡ ⎤×

+ × + −⎢ ⎥⎣ ⎦

= 6 2 6 2

3 38 10 (400) 8 10 (100)45ln 400 6 10 (400) 45ln100 6 10 (100)2 2

− −− −⎡ ⎤⎛ ⎞ ⎛ ⎞× ×

+ × + − + × +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

- 8.314[ln3 – ln 1]

= (269.616 + 2.4 + 0.64) – (207.233 + 0.6 + 0.04) – 9.134

= 272.656 – 207.873 – 9.134

= 55.65

10. The p.d. between boundaries a and b of an electric field is given by: V = b

a0 r

Q dr2 rπ ε ε∫

If a = 10, b = 20, Q = 2 × 10-6 coulombs, ε0 = 8.85 × 10-12 and εr = 2.77, show that V = 9 kV.

V = b

a0 r

Q dr2 rπ ε ε∫ = [ ]

b b

aa0 r 0 r 0 r

Q 1 Q Qdr ln r (ln b ln a)2 r 2 2

= = −πε ε πε ε πε ε∫

= ( )( )

6

12

2 10 (ln 20 ln10)2 8.85 10 2.77

−

−

×−

π × = 9000 V or 9 kV


11. The average value of a complex voltage waveform is given by:

( )AV 0

1V 10sin t 3sin 3 t 2sin 5 t d( t)π

= ω + ω + ω ωπ ∫

Evaluate VAV correct to 2 decimal places.

( )AV 0

1V 10sin t 3sin 3 t 2sin 5 t d( t)π

= ω + ω + ω ωπ ∫

= 0

1 3 210cos t cos3 t cos5 t3 5

π⎡ ⎤− ω − ω − ω⎢ ⎥π ⎣ ⎦

= ( ) ( )1 10cos cos3 0.4cos5 10cos 0 cos0 0.4cos0− π− π− π − − − −⎡ ⎤⎣ ⎦π

= ( ) ( )1 10 1 0.4 10 1 0.4+ + − − − −⎡ ⎤⎣ ⎦π

= ( ) ( )1 22.811.4 11.14− − =⎡ ⎤⎣ ⎦π π = 7.26


CHAPTER 38 SOME APPLICATIONS OF INTEGRATION

EXERCISE 148 Page 375 2. Sketch the curves y = x2 + 3 and y = 7 – 3x and determine the area enclosed by them.

The two curves intersect when x2 + 3 = 7 – 3x

i.e. x2 + 3x – 4 = 0

i.e. (x + 4)(x – 1) = 0

i.e. when x = -4 and x = 1

The two curves are shown below.

Area enclosed by curves =

( ) ( ) ( )1 1 1 12 2 2

4 4 4 4(7 3x)dx x 3 dx (7 3x) x 3 dx 4 3x x dx

− − − −− − + = − − + = − −∫ ∫ ∫ ∫

= 12 3

4

3x x4x2 3

−

⎡ ⎤− −⎢ ⎥

⎣ ⎦

= 3 1 644 16 242 3 3

⎛ ⎞ ⎛ ⎞− − − − − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 1 2 1 22 18 2 186 3 6 3

⎛ ⎞ ⎛ ⎞− − = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 5206

square units

3. Determine the area enclosed by the three straight lines y = 3x, 2y = x and y + 2x = 5.


y = -2x + 5 and y = 3x intersect when -2x + 5 = 3x i.e. when 5 = 5x i.e. when x = 1

y = -2x + 5 and y = x2

intersect when -2x + 5 = x2

i.e. when 5 = 2.5x i.e. when x = 2

The three straight lines are shown below.

Shaded area = 1 2

0 1

x x3x dx ( 2x 5) dx2 2

⎛ ⎞− + − + −⎜ ⎟⎝ ⎠∫ ∫

= ( )1 22 2 2

2

0 1

3x x x 3 1 1x 5x (0) 4 10 1 1 52 4 4 2 4 4

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− + − + − = − − + − + − − − + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

= ( )1 3 1 11 5 3 1 14 4 4 4

⎡ ⎤⎛ ⎞ ⎛ ⎞+ − = +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= 122

square units


EXERCISE 149 Page 377 2. The distance of points y from the mean value of a frequency distribution are related to the variate

x by the equation y = x + 1x

. Determine the standard deviation (i.e. the r.m.s. value), correct to

4 significant figures for values of x from 1 to 2.

Standard deviation = the r.m.s. value = 2

2 2

1 1

1 1 1 1x dx x x dx2 1 x x x

⎛ ⎞ ⎛ ⎞⎛ ⎞+ = + +⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎝ ⎠ ⎝ ⎠⎝ ⎠∫ ∫

= ( )2 22 2 221 1

1x 2 dx x 2 x dxx

−⎛ ⎞+ + = + +⎜ ⎟⎝ ⎠∫ ∫

= 2 23 1 3

1 1

x x x 12x 2x3 1 3 x

−⎡ ⎤ ⎡ ⎤+ + = + −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

= 8 1 14 2 1 4.83333 2 3

⎡ ⎤⎛ ⎞ ⎛ ⎞+ − − + − =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ = 2.198

3. The current i = 25 sin 100πt mA flows in an electrical circuit. Determine, using integral calculus,

its mean and r.m.s. value each correct to 2 decimal places over the range t = 0 to t = 10 ms.

Mean value = ( )3

3 10 1010 10

3 00

1 2525sin100 t dt 100 cos100 t10 10 0 100

−− ×

×

−

⎡ ⎤π = − π⎢ ⎥× − π⎣ ⎦∫

= 3100(25) cos(100 10 10 ) cos 0100

−⎡ ⎤− π× × −⎣ ⎦π

= [ ] [ ]25 25 50cos cos 0 1 1− π− = − − − =π π π

= 15.92 mA

r.m.s. value = 3 310 10 10 102 2 2

3 0 0

1 125 sin 100 t dt (100)(25) (1 cos 200 t)dt10 10 0 2

− −× ×

− π = − π× − ∫ ∫

since cos 2A = 1 – 22sin A from which, 2 1sin A (1 cos 2A)2

= − ,

= 310 102 2

3

0

(100)(25) sin 200 t (100)(25) sin 2t 10 10 (0 sin 0)2 200 2 200

−×−π ⎡ π ⎤⎡ ⎤ ⎛ ⎞− = × − − −⎜ ⎟⎢ ⎥⎢ ⎥π π⎣ ⎦ ⎝ ⎠⎣ ⎦

= 2 2

3(100)(25) 25 2510 102 2 2

−⎡ ⎤× = =⎣ ⎦ = 17.68 mA


4. A wave is defined by the equation: v = 1 3E sin t E sin 3 tω + ω

where 1E , 3E and ω are constants. Determine the r.m.s. value of v over the interval 0 t π≤ ≤

ω.

r.m.s. value = ( )21 30

1 E sin t E sin 3 t d( t)0

πω ω + ω ω

π−

ω

∫

= ( )2 2 2 21 1 3 30

E sin t 2E E sin t sin 3 t E sin 3 t d( t)πωω

ω + ω ω + ω ωπ ∫

2

0 00

1 cos 2 t t sin 2 tsin t dt dt 0 (0 0)2 2 4 2 2

ππ π ωω ω − ω ω ⎡ π ⎤ π⎡ ⎤ ⎛ ⎞ω = = − = − − − =⎜ ⎟⎢ ⎥⎢ ⎥ω ω ω⎣ ⎦ ⎝ ⎠⎣ ⎦∫ ∫

2

0 00

1 cos 6 t t sin 6 tsin 3 t dt dt 0 (0 0)2 2 12 2 2

ππ π ωω ω − ω ω ⎡ π ⎤ π⎡ ⎤ ⎛ ⎞ω = = − = − − − =⎜ ⎟⎢ ⎥⎢ ⎥ω ω ω⎣ ⎦ ⎝ ⎠⎣ ⎦∫ ∫

( )

[ ]

0 0 0

0

1sin t sin 3 t dt sin 3 t sin t dt cos 4 t cos 2 t dt (see page183of textbook)2

1 sin 4 t sin 2 t 1 (0 0) (0 0) 02 4 2 2

π π πω ω ω

πω

ω ω = ω ω = − ω − ω

ω ω⎡ ⎤= − − = − − − − =⎢ ⎥ω ω⎣ ⎦

∫ ∫ ∫

Hence, r.m.s. value = 22

2 2 311 3

EEE E2 2 2 2

⎛ ⎞ω π π⎡ ⎤+ = +⎜ ⎟⎢ ⎥π ω ω⎣ ⎦ ⎝ ⎠

= 2 2

1 3E E2

⎛ ⎞+⎜ ⎟⎝ ⎠



2. The area between 2

yx

= 1 and y + x2 = 8 is rotated 360° about the x-axis. Find the volume

produced.

2y x= and y = 8 - 2x intersect when 2x = 8 - 2x i.e. 2 2x = 8 and 2x = 4

i.e. when x = 4 2= ± . A sketch of the two curves is shown below.

Volume of solid of revolution = ( ) ( )2 22 22 2

2 28 x dx x dx

− −π − − π∫ ∫

= ( ) ( )2 2 22 4 4 2

2 2 264 16x x dx x dx 64 16x dx

− − −π − + − π = π −∫ ∫ ∫

= 23

2

16x 128 12864x 128 1283 3 3

−

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞π − = π − − − +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎣ ⎦

= 256 768 256 5122563 3 3

−⎡ ⎤ ⎡ ⎤ ⎛ ⎞π − = π = π⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎝ ⎠ = 2170

3π cubic units

3. The curve y = 2x2 + 3 is rotated about (a) the x-axis between the limits x = 0 and x = 3, and

(b) the y-axis, between the same limits. Determine the volume generated in each case.

(a) The curve is shown below.

( ) ( )353 322 4 2 3

x axis 0 00

4xVolume 2x 3 dx 4x 12x 9 dx 4x 9x5−

⎡ ⎤= π + = π + + = π + +⎢ ⎥

⎣ ⎦∫ ∫

= π[(329.4) – (0)] = 329.4π cubic units


(b)

21 2y axis 3

Volume x dy− = π∫ Since y = 2x2 + 3, then 2x2 = y – 3 and 2 y 3x2−

=

Hence, 212 221

y axis 33

y 3 y 21 9volume dy 3y 63 92 2 2 2 2 2−

⎡ ⎤⎡ ⎤ ⎛ ⎞− π π⎛ ⎞ ⎛ ⎞= π = − = − − −⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠⎣ ⎦∫

= ( ) ( ) 162157.5 4.52 2π π

− − =⎡ ⎤⎣ ⎦

= 81π cubic units

4. The profile of a rotor blade is bounded by the lines x = 0.2, y = 2x, xy e−= , x = 1 and the x-axis.

The blade thickness t varies linearly with x and is given by: t = (1.1 – x)K, where K is a

constant.

(a) Sketch the rotor blade, labelling the limits.

(b) Determine, using an iterative method, the value of x, correct to 3 decimal places, where

2x = xe−

(c) Calculate the cross-sectional area of the blade, correct to 3 decimal places.

(d) Calculate the volume of the blade in terms of K, correct to 3 decimal places.

(a) A sketch of the rotor blade is shown below.


(b) Using the Newton-Raphson method (from chapter 9),

let f(x) = 2x - xe−

f(0) = 0 - 0e = -1

f(1) = 2 - 1e− = 1.632 hence a root lies between x = 0 and x = 1

Let 1r = 0.4

xf '(x) 2 e−= +

12 1

1

f (r ) 0.12967995r r 0.4 0.35144f '(r ) 2.670320046

= − = − =

23 2

2

f (r ) ( 0.00079407)r r 0.35144 0.35173f '(r ) 2.70367407

−= − = − =

34 3

3

f (r ) ( 0.000010033)r r 0.35173 0.35173f '(r ) 2.703470033

−= − = − =

Hence, correct to 3 decimal places, x = 0.352

(c) Cross-sectional area of blade = 0.352 1 x

0.2 0.3522x dx e dx−+∫ ∫

= 0.352 12 x

0.2 0.352x e (0.0839) ( 0.3354)−⎡ ⎤ ⎡ ⎤− = − −⎣ ⎦ ⎣ ⎦

= 0.419 square units

(d) Volume of the blade = 0.352 1 x

0.2 0.3522x(1.1 x)K dx e (1.1 x)K dx−− + −∫ ∫

= ( ) ( )0.352 12 x x

0.2 0.352K 2.2x 2x dx K 1.1e xe dx− −− + −∫ ∫

= 0.3522 3 1x x x

0.3520.2

2.2x 2xK K 1.1e xe e2 3

− − −⎡ ⎤ ⎡ ⎤− + − − − −⎢ ⎥ ⎣ ⎦⎣ ⎦ (the latter

integral using integration by parts – see page 418 of textbook)

= 0.3522 3 1x x x

0.3520.2

2.2x 2xK K 1.1e xe e2 3

− − −⎡ ⎤⎡ ⎤− + − + +⎢ ⎥ ⎣ ⎦

⎣ ⎦

= [ ] [ ]K 0.107218 0.038667 K 0.331091 0.177227− + −

= 0.068551K + 0.153864K

= 0.222K, correct to 3 decimal places.


EXERCISE 151 Page 380 2. Find the position of the centroid of the area bounded by y = 5x2, the x-axis and ordinates x = 1

and x = 4. A sketch of the area is shown below.

( )44

4 4 4 4 42 3

1 1 1 14 4 4 432 2 3 3

1 1 1

1

5x 5 54 1 (255)xy dx x 5x dx 5x dx 4 318.754 4x 5 5 1055xy dx 5x dx 5x dx 4 1 (63)3 33

⎡ ⎤⎢ ⎥ ⎡ ⎤−⎣ ⎦⎣ ⎦

= = = = = = =⎡ ⎤ ⎡ ⎤−⎣ ⎦⎢ ⎥⎣ ⎦

∫ ∫ ∫∫ ∫ ∫

= 3.036

( )4 4 22 2 4541 1 4 5 54 1

11

1 1y dx 5x dx 1 1 25x 5 52 2y 25x dx 4 1 (1023)105 210 210 5 210 210ydx

⎡ ⎤⎡ ⎤= = = = = − =⎢ ⎥ ⎣ ⎦

⎣ ⎦

∫ ∫∫

∫

= 24.36

Hence, the centroid lies at (3.036, 24.36)

3. Determine the position of the centroid of a sheet of metal formed by the curve y = 4x – x2 which

lies above the x-axis. y = 4x – x2 = x(4 – x) i.e. when y = 0, x = 0 and x = 4. The area of the sheet of metal is shown sketched below.

By symmetry, x 2=


( )( )

( )( )

43 4 54 4 422 2 2 3 4

0 0 0 04 4 4 432 2

20 0 0

0

1 16x 8x x1 1 1y dx 4x x dx 16x 8x x dx 2 3 4 52 2 2yxy dx 4x x dx 4x x dx 2x3

⎡ ⎤− +⎢ ⎥− − + ⎣ ⎦

= = = =⎡ ⎤− − −⎢ ⎥⎣ ⎦

∫ ∫ ∫∫ ∫ ∫

= ( )1 341.333 512 204.8 (0) 17.06652

64 10.66632 (0)3

− + −⎡ ⎤⎣ ⎦=

⎛ ⎞− −⎜ ⎟⎝ ⎠

= 1.6

Hence, the co-ordinates of the centroid are (2, 1.6)

5. Sketch the curve y2 = 9x between the limits x = 0 and x = 4. Determine the position of the

centroid of this area. The curve y2 = 9x is shown sketched below.

By symmetry, y 0=

( )

452

34 4 4 5 520 0 0 0

4 4 1 4 3 34 320 0 20

0

3x5 6 6 6x 4 0 (32)xydx x 3 x dx 3x dx 2 5 5 5x

2(8)2 x 2 4 0ydx 3 x dx 3x dx 3x32

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ ⎡ ⎤ ⎡ ⎤−⎣ ⎦ ⎣ ⎦ ⎣ ⎦= = = = = = =

⎡ ⎤ ⎡ ⎤⎡ ⎤ −⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∫ ∫ ∫∫ ∫ ∫

= 38.416

= 2.4

Hence, the centroid is at (2.4, 0)


EXERCISE 152 Page 382 2. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid of a

metal template in the form of a quadrant of a circle of radius 4 cm. (The equation of a circle,

centre 0, radius r is x2 + y2 = r2).

(a) A sketch of the template is shown below.

Using Pappus, volume, V = ( )( )area 2 yπ

i.e. ( )( )3 21 4 1r r 2 y2 3 4⎛ ⎞π = π π⎜ ⎟⎝ ⎠

from which, ( )

3

2

2 r 4r 4(4)3y3 3r 2

4

π= = =

π π⎛ ⎞ππ⎜ ⎟

⎝ ⎠

= 1.70 cm

By symmetry, x 1.79cm=

Hence, the centroid of the template is at (1.70, 1.70)

(b) ( )( )( )

( )( )

( )

( )

432 2

144 42 2 200 0 0

4 4 4 422 2 2 21 2 2

0 0 0

0

1 16 x2

3x 16 x dxxydx x 16 x dx 2

x4 x xydx 4 x dx 4 x dx sin 4 x2 4 2

−

⎡ ⎤− −⎢ ⎥⎢ ⎥⎢ ⎥− − ⎢ ⎥⎣ ⎦

= = = =⎡ ⎤− − + −⎢ ⎥⎣ ⎦

∫∫ ∫∫ ∫ ∫

(the numerator being an algebraic substitution - see chapter 39), and the denominator

being a sin θ substitution - see chapter 40)

i.e. ( )

( ) ( )

32

1 1

1 640 163 3x

12.5668sin 1 2(0) 8sin 0 0(4)− −

⎡ ⎤⎡ ⎤⎛ ⎞− −⎢ ⎥⎢ ⎥⎜ ⎟

⎢ ⎥⎢ ⎥⎝ ⎠⎣ ⎦= =⎢ ⎥⎡ ⎤+ − +⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

= 1.70


By symmetry, y 1.70cm=

Hence, the centroid lies on the centre line OC (see diagram), at co-ordinates (1.70, 1.70).

The distance from 0 is given by ( )2 21.70 1.70+ = 2.40 cm

3. (a) Determine the area bounded by the curve y = 5x2, the x-axis and the ordinates x = 0 and

x = 3.

(b) If this area is revolved 360° about (i) the x-axis, and (ii) the y-axis, find the volume of the

solids of revolution produced in each case.

(c) Determine the co-ordinates of the centroid of the area using (i) integral calculus, and (ii) the

theorem of Pappus. (a) The area is shown in the sketch below.

Shaded area = ( )333 2 3

00

5x 55x dx 3 03 3

⎡ ⎤= = −⎢ ⎥⎣ ⎦

∫ = 45 square units

(b)(i) ( )353 3 322 2 4 5

x axis 0 0 00

xVolume y dx 5x dx 25 x dx 25 5 3 05−

⎡ ⎤⎡ ⎤= π = π = π = π = π −⎢ ⎥ ⎣ ⎦

⎣ ⎦∫ ∫ ∫

= 1215π cubic units

(ii) y axisVolume − = (volume generated by x = 3) – (volume generated by y = 5x2)

= 45245 45 452

0 0 00

y y y(3) dy dy 9 dy 9y5 5 10

⎡ ⎤⎛ ⎞ ⎛ ⎞π − π = π − = π −⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦∫ ∫ ∫

= 2459(45) (0) [405 202.5]

10⎡ ⎤⎛ ⎞

π − − = π −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

= 202.5π cubic units

(c)(i) ( ) ( )

343 3 3 42 3

0 0 0 03

0

5x 5 3 0xydx x 5x dx 5x dx 4 4x45 45 45 45ydx

⎡ ⎤⎢ ⎥ −⎣ ⎦

= = = = =∫ ∫ ∫∫

= 2.25


353 32 4 5

0 0 03

0

1 25x1 1 5y dx 25x dx 3 02 52 2 2y45 45 45ydx

⎡ ⎤⎢ ⎥ ⎡ ⎤−⎣ ⎦⎣ ⎦

= = = =∫ ∫∫

= 13.5

Hence, (2.25, 13.5) are the co-ordinates of the centroid.

(ii) Using Pappus, volume generated when the shaded area is revolved about 0y = (area) ( )2 xπ

i.e. 202.5π = (45) ( )2 xπ

from which, 202.5x(45)(2 )

π=

π = 2.25

Similarly, volume generated when the shaded area is revolved about 0x = (area) ( )2 yπ

i.e. 1215π = (45) ( )2 yπ

from which, 1215y(40)(2 )

π=

π = 13.5

Hence, (2.25, 13.5) are the co-ordinates of the centroid.

4. A metal disc has a radius of 7.0 cm and is of thickness 2.5 cm. A semicircular groove of

diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine the volume of

metal removed using Pappus’ theorem and express this as a percentage of the original volume of

the disc. Find also the mass of metal removed if the density of the metal is 7800 kg m-3. A side view of the rim of the disc is shown below.

When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid

of the semicicrcular area removed is at a distance of 4r3π

from its diameter, from problem 2 above,


i.e. 4(1.0)3π

= 0.424 cm from PQ.

Distance of centroid from XX = 7.0 – 0.424 = 6.576 cm.

Distance moved in 1 revolution by the centroid = 2π(6.576) cm

Area of semicircle = 2 2

2r (1.0) cm2 2 2π π π

= =

By Pappus, volume generated = area × distance moved by the centroid

i.e. volume of metal removed = ( )2 (6.576)2π⎛ ⎞ π⎜ ⎟

⎝ ⎠ = 364.90cm

Volume of disc = ( ) ( )22 3r h 7.0 2.5 384.845cmπ = π =

Thus, percentage of metal removed = 64.90 100%384.845

× = 16.86%

Mass of metal removed = density × volume

= 7800 3

kgm

6 364.90 10 m−× × = 0.5062 kg or 506.2 g


EXERCISE 153 Page 389 2. Determine the second moment of area and radius of gyration for the triangle shown below about

(a) axis DD (b) axis EE and (c) an axis through the centroid of the triangle parallel to axis DD.

From Table 38.1, page 385:

(a) Second moment of area about DD, ( )( )33

DD

12.0 9.0b hI12 12

= = = 729 4cm

Radius of gyration about DD, DDh 9.0k6 6

= = = 3.67 cm

(b) Second moment of area about EE, ( )( )33

EE

12.0 9.0b hI4 4

= = = 2187 4cm

Radius of gyration about EE, EEh 9.0k2 2

= = = 6.36 cm

(c) Second moment of area about axis through centoid, ( )( )33 12.0 9.0b h36 36

= = = 243 4cm

Radius of gyration about axis through centoid, h 9.018 18

= = = 2.12 cm

5. For each of the areas shown below determine the second moment of area and radius of gyration

about axis LL, by using the parallel axis theorem.


(a) Second moment of area,3 3

2 2 2LL GG

bl (3.0)(5.0)I I Ad Ad (3.0)(5.0)(2.5 2.0)12 12

= + = + = + +

= 31.25 + 303.75 = 335 4cm

2LL LLI Ak= from which, radius of gyration, LL

LLI 335k

area 15.0⎛ ⎞= = ⎜ ⎟⎝ ⎠

= 4.73 cm

(b) Second moment of area, 2LL GGI I Ad= +

3 3

4GG

bh (18)(12)I 864cm36 36

= = = where h = 2 215 9− = 12 cm, as shown in the diagram below.

Hence, area of triangle, A = 1 (18)(12)

2 = 108 2cm

Thus, 2

2 2LL GG

12I I Ad 864 108 10 864 108(14)3

⎛ ⎞= + = + + = +⎜ ⎟⎝ ⎠

= 22032 4cm = 22032 4cm



LLI 22032k

area 108⎛ ⎞= = ⎜ ⎟⎝ ⎠

= 14.3 cm

(c) Second moment of area, ( )24 4

2 2 2 2LL GG

r 4 (2.0)I I Ad r 5 (2.0) (7)4 2 4π π⎛ ⎞= + = + π + = + π⎜ ⎟

⎝ ⎠

= 12.57 + 615.75 = 628 4cm


LL 2

I 628karea (2.0)

⎛ ⎞= = ⎜ ⎟π⎝ ⎠

= 7.07 cm

8. A circular cover, centre 0, has a radius of 12.0 cm. A hole of radius 4.0 cm and centre X, where

0X = 6.0 cm, is cut in the cover. Determine the second moment of area and the radius of gyration

of the remainder about a diameter through 0 perpendicular to 0x. Second moment of area about diameter, i.e. axis CC in the diagram below

4 4

2CC DD GG

r rI I I Ad4 4π π ⎡ ⎤= − = − +⎣ ⎦


= 4 4

2 2(12) (4.0) (4.0) (6.0)4 4

⎡ ⎤π π− + π⎢ ⎥⎣ ⎦

= 16286 – [201 + 1810] = 14275 = 14280 4cm , correct

to 4 significant figures.

2

CC CCI Ak= from which, radius of gyration,

CCCC 2 2

I 14275 14275karea (12.0) (4.0) 128

⎛ ⎞= = =⎜ ⎟π − π π⎝ ⎠

= 5.96 cm

9. For the sections shown below, find the second moment of area and the radius of gyration about

axis XX.

(a) (b) (a) For rectangle A in the diagram below,

second moment of area about 3 3

4A

bl (18.0)(3.0)C 40.5m12 12

= = =


Hence, A

2 2XXI 40.5 Ad 40.5 (3.0)(18.0)(12.0 1.5) 40.5 9841.5= + = + + = + = 9882 4mm

For rectangle B, second moment of area about 3 3

4B

bl (4.0)(12.0)C 576m12 12

= = =

Hence, B

2XXI 576 (4.0)(12.0)(6.0) 576 1728= + = + = 2304 4mm

Thus, total second moment of area about XX, TXXI 9882 2304 12186= + = = 12190 4mm ,


Radius of gyration about XX, TXXXX

I 12186karea 54 48

= =+

= 10.9 mm

(b) Rectangle A (see diagram below):

Second moment of area about 3 3

4A

bl (6.0)(2.0)C 4cm12 12

= = =

A

2XXI 4 (2.0)(6.0)(6.0) 4 432= + = + = 436 4cm

Rectangle B:


4B

bl (2.5)(3.0)C 5.625cm12 12

= = =

B

2XXI 5.625 (2.5)(3.0)(3.5)= + = 97.5 4cm

Rectangle C:


4C

bl (6.0)(2.0)C 4cm12 12

= = =

C

2XXI 4 (6.0)(2.0)(1.0)= + = 16 4cm

Hence, total second moment of area about axis XX, IXXI = 436 + 97.5 + 16 = 549.5 4cm

Radius of gyration about XX, IXXXX

I 549.5karea 12 7.5 12

= =+ +

= 4.18 cm


11. Find the second moment of area and radius of gyration about the axis XX for the beam section

shown below.

Second moment of area about axis XX,

A B CXX XX XX XXI I I I= + +

= ( )( ) ( )( ) ( )( )3 3 32 2 26.0 1.0 2.0 8.0 10.0 2.0

(6.0)(10.5) (16)(6.0) (20)(1.0)12 12 12

⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ + + + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= 662 + 661.33 + 26.66 = 1350 4cm

Radius of gyration about XX, XXXX

I 1350karea 6 16 20

= =+ +

= 5.67 cm


CHAPTER 39 INTEGRATION USING ALGEBRAIC SUBSTITUTIONS


4. Integrate ( )61 5x 32


Let u = 5x – 3 then du 5dx

= and dx = du5

Hence, 7

6 61 du 1 1 uu u du c2 5 10 10 7

⎛ ⎞= = +⎜ ⎟

⎝ ⎠∫ ∫ = 71 (5x 3) c

70− +

5. Integrate 3(2x 1)−−

with respect to x.

Let u = 2x – 1 then du 2dx

= and dx = du2

Hence, 3 3 du 3 1 3dx du ln u c(2x 1) u 2 2 u 2− −

= = − = − +−∫ ∫ ∫ = 3 ln(2x 1) c

2− − +

7. Evaluate ( )1 5

03x 1 dx+∫ correct to 4 significant figures.

Let u = 3x + 1 then du 3dx

= and dx = du3

Hence, 6

5 5 6du 1 1 u 1u u du c (3x 1) c3 3 3 6 18

⎛ ⎞= = + = + +⎜ ⎟

⎝ ⎠∫ ∫

Thus, ( ) ( )11 5 6 6 6

0 0

1 13x 1 dx 3x 1 4 118 18

⎡ ⎤ ⎡ ⎤+ = + = −⎣ ⎦⎣ ⎦∫ = 227.5

8. Evaluate ( )2 2

0x 2x 1 dx+∫ correct to 4 significant figures.

Let u = 22x 1+ then du 4xdx

= i.e. dudx4x

=

Hence, ( ) ( )3

1 32 32 2 22 2du 1 1 u 1 1x (2x 1) dx x u u du c 2x 1 c 2x 1 c34x 4 4 6 62

⎛ ⎞⎜ ⎟

+ = = = + = + + = + +⎜ ⎟⎜ ⎟⎝ ⎠

∫ ∫ ∫

Thus, ( ) ( ) [ ]2

2 32 2

00

1 1x 2x 1 dx 2x 1 27 16 6⎡ ⎤+ = + = −⎢ ⎥⎣ ⎦∫ = 4.333


9. Evaluate /3

02sin 3t dt

4π π⎛ ⎞+⎜ ⎟

⎝ ⎠∫ correct to 4 significant figures.

3/3

00

2 2 32sin 3t dt cos 3t cos cos4 3 4 3 3 4 4

π

π π ⎡ π ⎤ ⎡ π π π⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = − + = − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦∫ (note angles are in radians)

= [ ]2 0.70711 0.707113

− − −

= 0.9428


EXERCISE 155 Page 394 2. Integrate 55cos t sin t with respect to t.

Let u = cos t then du sin tdt

= − and dudtsin t

=−

65 5 5du u5cos t sin t dt 5u sin t 5 u du 5 c

sin t 6⎛ ⎞

= = − = − +⎜ ⎟− ⎝ ⎠∫ ∫ ∫ = - 65 cos t c

6+

3. Integrate 23sec 3x tan 3x with respect to x.

Let u = tan 3x then 2du 3sec 3xdx

= i.e. 2

dudx3sec 3x

=

Hence, 2

2 22

du u3sec 3x tan 3x dx 3sec 3x (u) u du c3sec 3x 2

= = = +∫ ∫ ∫ = 21 tan 3x c2

+

Alternatively, let u = sec 3x then du 3sec3x tan 3xdx

= i.e. dudx3sec3x tan 3x

=

Hence, 2 2

2 2 du u u3sec 3x tan 3x dx 3u tan 3x du du u du3sec3x tan 3x sec3x u

= = = =∫ ∫ ∫ ∫ ∫

= 2u c

2+ = 21 sec 3x c

2+

5. Integrate lnθθ

with respect to θ.

Let u = ln θ then du 1d

=θ θ

and dθ = θdu

Hence, ( )2ln u ud du u du c

2θ

θ = θ = = +θ θ∫ ∫ ∫ = ( )21 ln c

2θ +

6. Integrate 3 tan t with respect to t.

sin 2t3 tan 2t dt 3 dtcos 2t

=∫ ∫ Let u = cos 2t then du 2sin 2tdt

= − i.e. dudt2sin 2t−

=

Hence, sin 2t sin 2t du 3 1 33 dt 3 du ln u ccos 2t u 2sin 2t 2 u 2

−⎛ ⎞⎛ ⎞= = − = − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫ ∫ ∫


= ( ) 13 3ln cos 2t c ln cos 2t c2 2

−− + = +

= 3 lnsec 2t c2

+

8. Evaluate ( )22x 11

03x e dx

−

∫ correct to 4 significant figures.

Let u = 22x 1− then du 4xdx

= i.e. dudx4x

=

Hence, ( ) ( )2 22x 1 2x 1u u udu 3 3 33xe dx 3xe e du e c e c4x 4 4 4

− −= = = + = +∫ ∫ ∫

( ) ( )2

2 12x 11 2x 1 1 1

0 0

3 33x e dx e e e4 4

− − −⎡ ⎤ ⎡ ⎤= = −⎣ ⎦⎢ ⎥⎣ ⎦∫ = 1.763

10. Evaluate ( )

1

50 2

3x dx4x 1−


Let u = 24x 1− then du 8xdx

= i.e. dudx8x

=

Hence, ( ) ( )

45

5 45 42 2

3x 3x du 3 3 u 3 3dx u du c c cu 8x 8 8 4 32u4x 1 32 4x 1

−− ⎛ ⎞⎛ ⎞= = = + = − + = − +⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠− −

∫ ∫ ∫

Thus, ( ) ( )

1

1

5 4 4 40 2 2

0

3x 3 1 3 1 1dx32 32 3 ( 1)4x 1 4x 1

⎡ ⎤ ⎡ ⎤⎢ ⎥= − = − −⎢ ⎥⎢ ⎥ −⎣ ⎦− −⎣ ⎦∫ = 0.09259

11. The electrostatic potential on all parts of a conducting circular disc of radius r is given by the

equation: V = 9

2 20

R2 dRR r

πσ+

∫

Solve the equation by determining the integral.

Let u = 2 2R r+ then du 2RdR

= i.e. dudR2R

=

( )1

1 22 22

2 2

R R du 1 1 udR u du c u c R r12R 2 2uR r2

−⎛ ⎞= = = + = + = +⎜ ⎟⎝ ⎠+

∫ ∫ ∫ + c


Hence, V = ( ) ( )99 2 2 2 2 2

2 20 0

R2 dR 2 R r 2 9 r rR r

⎡ ⎤ ⎡ ⎤πσ = πσ + = πσ + −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦+∫

= ( ) 2 22 9 r rπσ + −

12. In the study of a rigid motor the following integration occurs: ( )2

2J(J 1)h8 IkT

r 0Z 2J 1 e dJ

− +∞ π= +∫

Determine rZ for constant temperature T assuming h, I and k are constants.

Let u = J(J + 1) = 2J J+ then du 2J 1dJ

= + i.e. dudJ2J 1

=+

( )2 2 2 2

2 2 2 2J(J 1)h u h u h u h8 IkT 8 Ik T 8 Ik T 8 Ik T

2

2

du 12J 1 e dJ (2J 1)e e du e ch(2J 1)

8 I k T

− + −− −

π π π π+ = + = = ++ −

π

∫ ∫ ∫

= 2

2J(J 1)h28 Ik T

2

8 I k T eh

− +ππ

− + c

Thus, ( )2 2

2 2J(J 1)h J(J 1)h2 2

08 IkT 8 Ik Tr 2 20

0

8 I k T 8 I k TZ 2J 1 e dJ e e eh h

∞− + − +

∞ −∞π π⎡ ⎤π π⎢ ⎥ ⎡ ⎤= + = − = − −⎣ ⎦⎢ ⎥⎣ ⎦

∫

= [ ]2

2

8 I k T 0 1h

π− − =

2

2

8 I k Th

π

13. In electrostatics, ( )

2

0 2 2

a sinE d2 a x 2ax cos

π⎧ ⎫

σ θ⎪ ⎪= θ⎨ ⎬ε − − θ⎪ ⎪⎩ ⎭

∫ where a, σ and ε are constants, x is

greater than a, and x is independent of θ. Show that 2aExσ

=ε

Let u = 2 2a x 2ax cos+ − θ then du 2ax sind

= θθ

and dud2ax sin

θ =θ

Hence,

112 2 2 2 22a sin du a du a 1 a 1 uE u du c12 2ax sin 2 2 2ax 2 2axu 2ax u

2

−

⎛ ⎞⎜ ⎟σ θ σ σ σ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = = +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ε θ ε ε ε⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟⎝ ⎠

∫ ∫ ∫

= ( )2 2a a x 2ax cos c2 xσ

+ − θ +ε


Hence, E = 2 2 2 2 2 2

0

a aa x 2ax cos a x 2ax cos a x 2ax cos 02 x 2 x

πσ σ⎡ ⎤ ⎡ ⎤+ − θ = + − π − + −⎣ ⎦ ⎣ ⎦ε ε

= 2 2 2 2a (a x 2ax) (a x 2ax)2 xσ ⎡ ⎤+ + − + −⎣ ⎦ε

= ( ) ( )2 2a x a x a2 xσ ⎡ ⎤+ − −⎢ ⎥⎣ ⎦ε

= [ ] [ ]a a(x a) (x a) 2a2 x 2 xσ σ

+ − − =ε ε

i.e. 2aExσ

=ε


CHAPTER 40 INTEGRATION USING TRIGONOMETRIC AND

HYPERBOLIC SUBSTITUTIONS EXERCISE 156 Page 399 2. Integrate 23cos t with respect to t.

2cos 2t 2cos t 1= − from which, ( )2 1cos t 1 cos 2t2

= +

Hence, ( )2 13cos t dt 3 1 cos 2t dt2

= +∫ ∫ = 3 sin2tt c2 2⎛ ⎞+ +⎜ ⎟⎝ ⎠

4. Integrate 22cot 2t with respect to t.

( )2 2 12cot 2t dt 2 cos ec 2t 1 dt 2 cot 2t t c2

⎡ ⎤= − = − − +⎢ ⎥⎣ ⎦∫ ∫ = - (cot 2t + 2t) + c

5. Evaluate /3 2

03sin 3x dx

π


2cos 2x 1 2sin x= − and 2cos 6x 1 2sin 3x= − from which, 2 1sin 3x (1 cos 6x)

2= −

Hence, /3

/3 /32

0 00

3 3 sin 6x3sin 3x dx (1 cos 6x)dx x2 2 6

ππ π ⎡ ⎤= − = −⎢ ⎥⎣ ⎦∫ ∫

=

6sin3 33 (0 sin0)2 3 6 2 3

⎡ π ⎤⎛ ⎞⎢ ⎥⎜ ⎟π π⎡ ⎤− − − =⎢ ⎥⎜ ⎟ ⎢ ⎥⎣ ⎦⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= 2π or 1.571

7. Evaluate 1 2

02 tan 2t dt∫ correct to 4 significant figures.

( )1

1 12 2

0 00

tan 2t2 tan 2t dt 2 sec 2t 1 dt 2 t2

⎡ ⎤= − = −⎢ ⎥⎣ ⎦∫ ∫

= tan 2 tan 02 1 02 2

⎡ ⎤⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ = tan 2 – 2 = -4.185

(note that ‘tan 2’ means ‘tan 2 radians’)


EXERCISE 157 Page 400 2. Integrate 32cos 2x with respect to x.

( ) ( )3 2 2 22cos 2x dx 2 cos 2x cos 2x dx 2 cos 2x 1 sin 2x dx 2 cos 2x cos2x sin 2x dx= = − = −∫ ∫ ∫ ∫

= 3sin 2x sin 2x2 c

2 6⎛ ⎞

− +⎜ ⎟⎝ ⎠

using the algebraic substitution u = sin 2x

= 3sin 2xsin 2x c3

− +

3. Integrate 3 22sin t cos t with respect to t.

( )3 2 2 2 2 22sin t cos t dt 2sin t sin t cos t dt 2sin t 1 cos t cos t dt= = −∫ ∫ ∫

= ( )3 5

2 4 cos t cos t2 sin t cos t sin t cos t dt 2 c3 5

⎡ ⎤− = − + +⎢ ⎥

⎣ ⎦∫

using the algebraic substitution u = cos t

= 3 52 2cos t cos t c3 5

− + +

5. Integrate 42sin 2θ with respect to θ.

( )2

24 2 21 cos 4 12sin 2 d 2 sin 2 d 2 d (1 2cos 4 cos 4 )d2 2

− θ⎛ ⎞θ θ = θ θ = θ = − θ+ θ θ⎜ ⎟⎝ ⎠∫ ∫ ∫ ∫

= 1 1 cos81 2cos 4 d2 2

⎡ + θ ⎤⎛ ⎞− θ+ θ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫

= 1 sin 4 sin8 c2 2 2 16

θ θ θ⎡ ⎤θ− + + +⎢ ⎥⎣ ⎦ = 3 sin 4 sin 8 c

4 4 32θ θ θ− + +

6. Integrate 2 2sin t cos t with respect to t.

2 2 21 cos 2t 1 cos 2t 1 1 1 cos 4tsin t cos t dt dt (1 cos 2t)dt 1 dt2 2 4 4 2

− + ⎡ + ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞= = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ∫ ∫ ∫


= 1 1 cos 4t 1 t sin 4tdt c4 2 2 4 2 8

⎛ ⎞ ⎡ ⎤− = − +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦∫

= t 1 sin 4t c8 32− +


EXERCISE 158 Page 401 2. Integrate 2 sin 3x sin x with respect to x.

( )12sin 3x sin x dx 2 cos 4x cos 2x dx2

= − −∫ ∫ from 9, page 398 of textbook

= sin 4x sin 2x c4 2

⎡ ⎤− − +⎢ ⎥⎣ ⎦ = sin 2x sin 4x c

2 4− +

4. Integrate 1 cos 4 sin 22

θ θ with respect to θ.

1 1 1cos 4 sin 2 d (sin 6 sin 2 )d2 2 2

θ θ θ = θ− θ θ∫ ∫ from 7, page 398 of textbook

= 1 cos 6 cos 2 c4 6 2

θ θ⎡ ⎤− + +⎢ ⎥⎣ ⎦ = 1 cos 2 cos 6 c

4 2 6θ θ⎛ ⎞− +⎜ ⎟

⎝ ⎠

6. Evaluate 1

02sin 7t cos3t dt∫

1

1

00

1 cos10t cos 4t2sin 7t cos3t dt 2 (sin10t sin 4t)dt2 10 4

⎡ ⎤= + = − −⎢ ⎥⎣ ⎦∫ ∫ from 6, page 398 of textbook

= cos10 cos 4 cos 0 cos 010 4 10 4

⎛ ⎞ ⎛ ⎞− − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= (0.24732) – (-0.35)

= 0.5973

7. Evaluate /3

04 sin 5 sin 2 d

π− θ θ θ∫

[ ]/ 3 /3

0 0

14 sin 5 sin 2 d 4 cos 7 cos3 d2

π π− θ θ θ = − − θ− θ θ∫ ∫ from 9, page 398 of textbook

= /3

0

7 3sin sinsin 7 sin 3 3 32 2 (0 0)7 3 7 3

π⎡ π π ⎤⎛ ⎞⎢ ⎥⎜ ⎟θ θ⎡ ⎤− = − − −⎢ ⎥⎜ ⎟⎢ ⎥⎣ ⎦ ⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= 0.2474



3. Determine ( )24 x dx−∫

( ) ( ) ( )2

2 2 2 1 2 22 x x4 x dx 2 x sin 2 x2 2 2

−− = − = + −∫ ∫ from 11, page 398 of textbook

= ( )1 2x x2sin 4 x c2 2

− + − +

4. Determine ( )216 9t dt−∫

( )2

2 2 216 416 9t dt 9 t dt 9 t dt9 3

⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞− = − = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎣ ⎦∫ ∫ ∫

=

2

21 2

4t t 433 sin t c42 2 33

−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎡ ⎤⎛ ⎞⎝ ⎠⎢ ⎥+ − +⎢ ⎥⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫ from 11, page 398 of textbook

= 2

1 28 3t 3t 4sin t c3 4 2 3

−⎡ ⎤⎛ ⎞+ − +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

= 2

1 2 2 1 2 28 3t t 4 8 3t tsin 3 t c sin (4 9t ) c3 4 2 3 3 4 2

− −⎡ ⎤⎛ ⎞+ − + = + − +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

= ( )1 28 3t tsin 16 9t c3 4 2

− + − +

6. Evaluate ( )1 2

09 4x dx−∫

( )2

1 1 12 2 2

0 0 0

9 39 4x dx 4 x dx 2 x dx4 2

⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞− = − = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎣ ⎦∫ ∫ ∫

=

12

21 2

0

3x x 322 sin x32 2 22

−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎡ ⎤⎛ ⎞⎝ ⎠⎢ ⎥+ −⎢ ⎥⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

from 11, page 398 of textbook

= 19 2 12 sin 1.25 (0 0)8 3 2

−⎡ ⎤⎛ ⎞+ − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ = 2.760



2. Determine 2

5 d16 9

θ+ θ∫

222 2

5 5 5 1d d d1616 9 9 499 3

θ = θ = θ+ θ ⎛ ⎞ ⎛ ⎞+ θ + θ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫ ∫

= 15 1 tan c4 493 3

−

⎛ ⎞⎜ ⎟θ⎜ ⎟ +⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠


= 15 3tan c12 4

− θ+

4. Evaluate 3

20

5 dx4 x+∫

3

3 3

2 2 20 00

5 1 1 xdx 5 dx 5 tan4 x 2 x 2 2

⎡ ⎤= = ⎢ ⎥+ + ⎣ ⎦∫ ∫ from 12, page 398 of textbook

= 1 15 3tan tan 02 2

− −⎡ ⎤−⎢ ⎥⎣ ⎦

= 2.457



2. Find ( )2

3 dx9 5x+

∫

( ) 222 2

3 3 3dx dx dx99 5x 35 x 5 x5 5

= =⎡ ⎤ ⎡ ⎤+ ⎛ ⎞ ⎛ ⎞+⎜ ⎟ +⎢ ⎥ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦ ⎝ ⎠⎢ ⎥⎣ ⎦

∫ ∫ ∫

= 13 xsinh c355

− +⎛ ⎞⎜ ⎟⎝ ⎠


= 13 5sinh x c35

− +

4. Find ( )24t 25 dt+∫

( )2

2 2 225 54t 25 dt 4 t dt 4 t dt4 2

⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎣ ⎦∫ ∫ ∫

=

2

21 2

5t t 522 sinh t c52 2 22

−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎡ ⎤⎛ ⎞⎝ ⎠⎢ ⎥+ + +⎢ ⎥⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


= 2

1 225 2t t 5sinh 4 t c4 5 2 2

−⎡ ⎤⎛ ⎞+ + +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

= 1 2 225 2t tsinh 4t 5 c4 5 2

− ⎡ ⎤+ + +⎣ ⎦

= 1 225 2t tsinh 4t 25 c4 5 2

− ⎡ ⎤+ + +⎣ ⎦

6. Evaluate ( )1 2

016 9 d+ θ θ∫

( )2

1 1 12 2 2

0 0 0

16 416 9 d 9 d 9 d9 3

⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞+ θ θ = + θ θ = + θ θ⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎣ ⎦∫ ∫ ∫


=

12

21 2

0

4433 sinh

42 2 33

−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎡ ⎤θ θ ⎛ ⎞⎝ ⎠⎢ ⎥+ + θ⎢ ⎥⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


= 116 3 1 163 sinh 2.777777 sinh 0 018 4 2 18

−⎡ ⎤⎛ ⎞ ⎛ ⎞+ − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= 4.348



2. Find ( )2

3 dx4x 9−

∫

( ) 222 2

3 3 3dx dx dx94x 9 34 x 2 x4 2

= =⎡ ⎤ ⎡ ⎤− ⎛ ⎞ ⎛ ⎞− −⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦ ⎝ ⎠⎢ ⎥⎣ ⎦

∫ ∫ ∫

= 13 xcosh c322

− +⎛ ⎞⎜ ⎟⎝ ⎠


= 13 2xcosh c2 3

− +

4. Find ( )24 25 dθ − θ∫

( )2

2 2 225 54 25 d 4 d 2 d4 2

⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞θ − θ = θ − θ = θ − θ⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎣ ⎦∫ ∫ ∫

=

2

22 1

55 22 cosh c

52 2 22

−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎡ ⎤θ θ⎛ ⎞ ⎝ ⎠⎢ ⎥θ − − +⎢ ⎥⎜ ⎟⎢ ⎥⎛ ⎞⎝ ⎠⎢ ⎥⎣ ⎦ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


= 2 125 25 2cosh c4 4 5

− θ⎛ ⎞θ θ − − +⎜ ⎟⎝ ⎠

6. Evaluate ( )3 2

2t 4 dt−∫

( ) ( ) ( )323 32 2 2 2 2 1

2 22

t 2 tt 4 dt t 2 dt t 2 cosh2 2 2

−⎡ ⎤− = − = − −⎢ ⎥

⎣ ⎦∫ ∫

= ( )1 13 35 2cosh 0 2cosh 12 2

− −⎛ ⎞− − −⎜ ⎟⎝ ⎠

= 1.429


CHAPTER 41 INTEGRATION USING PARTIAL FRACTIONS EXERCISE 163 Page 409

2. Determine ( )2

4(x 4) dxx 2x 3

−− −∫

2

4( 4) 5 12 3 ( 1) ( 3)−

= −− − + −x

x x x x from question 2, Exercise 13, page 19.

Hence, ( )2

4(x 4) 5 1dx dx(x 1) (x 3)x 2x 3

⎛ ⎞−= −⎜ ⎟+ −− − ⎝ ⎠

∫ ∫ = 5ln(x 1) ln(x 3) c+ − − +

or 5ln(x 1) ln(x 3) c+ − − + or 5(x 1)ln c

(x 3)⎧ ⎫+

+⎨ ⎬−⎩ ⎭ by the laws of logarithms

4. Determine 2

2

x 9x 8 dxx x 6+ ++ −∫

2

2

9 8 2 616 ( 3) ( 2)

x xx x x x+ +

= + ++ − + −

from question 5, Exercise 13, page 19.

2

2

x 9x 8 2 6dx 1 dxx x 6 (x 3) (x 2)

⎛ ⎞+ += + +⎜ ⎟+ − + −⎝ ⎠

∫ ∫ = x + 2 ln(x + 3) + 6 ln(x - 2) + c

or ( ) ( )2 6x ln x 3 ln x 2 c+ + + − + or ( ) 62x ln (x 3) x 2 c+ + − +

6. Evaluate 24

3

x 3x 6 dxx(x 2)(x 1)

− +− −∫ correct to 4 significant figures.

Let 2x 3x 6 A B C A(x 2)(x 1) Bx(x 1) Cx(x 2)

x(x 2)(x 1) x (x 2) (x 1) x(x 2)(x 1)− + − − + − + −

≡ + + =− − − − − −

then 2x 3x 6 A(x 2)(x 1) Bx(x 1) Cx(x 2)− + = − − + − + −

Let x = 0, 6 = 2A i.e. A = 3

Let x = 2, 4 = 2B i.e. B = 2

Let x = 1, 4 = -C i.e. C = -4

Hence, 2x 3x 6 3 2 4

x(x 2)(x 1) x (x 2) (x 1)− +

≡ + −− − − −

Thus, [ ]24 4 4

33 3

x 3x 6 3 2 4dx dx 3ln x 2ln(x 2) 4ln(x 1)x(x 2)(x 1) x (x 2) (x 1)

⎛ ⎞− += + − = + − − −⎜ ⎟− − − −⎝ ⎠

∫ ∫


= (3 ln 4 + 2 ln 2 – 4 ln 3) – (3 ln 3 + 2 ln 1 – 4 ln 2)

= 0.6275

8. Determine the value of k, given that: 1

0

(x k) dx 0(3x 1)(x 1)

−=

+ +∫

Let (x k) A B A(x 1) B(3x 1)(3x 1)(x 1) (3x 1) (x 1) (3x 1)(x 1)

− + + +≡ + =

+ + + + + +

then x – k = A(x + 1) + B(3x + 1)

Let x = -1, -1 – k = -2B i.e. B = 1 k2+

Let x = 13

− , 13

− - k = 2 A3

i.e. A = 3 1 1 3k k2 3 2 2⎛ ⎞− − = − −⎜ ⎟⎝ ⎠

1 1

0 0

1 3 1 kk(x k) 2 2 2dx dx 0(3x 1)(x 1) (3x 1) (x 1)

+⎡ ⎤− −⎢ ⎥−= + =⎢ ⎥+ + + +⎢ ⎥

⎣ ⎦

∫ ∫

i.e. 1

0

1 1 3 1 kk ln(3x 1) ln(x 1) 03 2 2 2⎡ + ⎤⎛ ⎞ ⎛ ⎞− − + + + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

[ ]1 1 3 1 kk ln 4 ln 2 0 0 03 2 2 2⎡ + ⎤⎛ ⎞ ⎛ ⎞− − + − + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

i.e. 1 k 1 1 3ln 2 k (2ln 2)2 3 2 2+⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ since ln 4 = 2ln 2 2ln 2=

i.e. 1 k 1 k2 2 3+ = + from which, 1 1 kk

2 3 2− = −

and 1 k6 2= from which, 1k

3=

9. The velocity constant k of a chemical reaction is given by: 1kt dx(3 0.4x)(2 0.6x)

⎛ ⎞= ⎜ ⎟− −⎝ ⎠∫

where x = 0 when t = 0. Show that: kt = 2(3 0.4x)ln3(2 0.6x)

⎧ ⎫−⎨ ⎬−⎩ ⎭

Let 1 A B A(2 0.6x) B(3 0.4x)(3 0.4x)(2 0.6x) (3 0.4x) (2 0.6x) (3 0.4x)(2 0.6x)

− + −≡ + =

− − − − − −

Hence, 1 = A(2 – 0.6x) + B(3 – 0.4x)


Let x = 3 7.50.4

= , 1 = A(2 – 4.5) i.e. A = 1 0.42.5

= −−

Let x = 2 1 1030.6 3 3

= = , 1 = 4B 33

⎛ ⎞−⎜ ⎟⎝ ⎠

i.e. B = 1 3 0.65 53

= =

Thus, 1 0.4 0.6kt dx dx(3 0.4x)(2 0.6x) (3 0.4x) (2 0.6x)

⎛ ⎞ ⎛ ⎞−= = +⎜ ⎟ ⎜ ⎟− − − −⎝ ⎠ ⎝ ⎠∫ ∫

= 0.4 0.6ln(3 0.4x) ln(2 0.6x)0.4 0.6−

− + −− −

+ c

i.e. kt = ln(3 – 0.4x) – ln(2 – 0.6x) + c

t = 0 when x = 0, hence, 0 = ln 3 – ln 2 + c

i.e. c = ln 2 – ln 3

Hence, kt = ln(3 – 0.4x) – ln(2 – 0.6x) + ln 2 – ln 3

i.e. kt = 2(3 0.4x)ln3(2 0.6x)

⎧ ⎫−⎨ ⎬−⎩ ⎭



2. Determine 2

3

5x 30x 44 dx(x 2)− +−∫

Let 2 2

3 2 3 3

5x 30x 44 A B C A(x 2) B(x 2) C(x 2) (x 2) (x 2) (x 2) (x 2)− + − + − +

≡ + + =− − − − −

Hence, 2 25x 30x 44 A(x 2) B(x 2) C− + = − + − +

Let x = 2, 4 = C

Equating 2x coefficients, 5 = A

Equating x coefficients, -30 = -4A + B i.e. B = -30 + 20 = -10

Thus, 2

3 2 3

5x 30x 44 5 10 4dx dx(x 2) (x 2) (x 2) (x 2)

⎛ ⎞− += − +⎜ ⎟− − − −⎝ ⎠

∫ ∫

= 5 ln(x – 2) + 2

10 2 c(x 2) (x 2)

− +− −

using the algebraic substitution

u = x – 2 in the latter two integrals

4. Evaluate 27

26

18 21x x dx(x 5)(x 2)

+ −− +∫ correct to 4 significant figures.

Let ( ) ( )22

2 2 2

A(x 2) B(x 5) x 2 C x 518 21x x A B C(x 5)(x 2) (x 5) (x 2) (x 2) (x 5)(x 2)

+ + − + + −+ −≡ + + =

− + − + + − +

Hence, 2 218 21x x A(x 2) B(x 5)(x 2) C(x 5)+ − = + + − + + −

Let x = 5, 98 = 49A i.e. A = 2

Let x = -2, -28 = -7C i.e. C = 4

Equating 2x coefficients, -1 = A + B i.e. B = -3

Thus, 27 7

2 26 6

18 21x x 2 3 4dx dx(x 5)(x 2) (x 5) (x 2) (x 2)

⎛ ⎞+ −= − +⎜ ⎟− + − + +⎝ ⎠

∫ ∫

= 7

6

42 ln(x 5) 3ln(x 2)(x 2)

⎡ ⎤− − + −⎢ ⎥+⎣ ⎦

= 4 42ln 2 3ln 9 2ln1 3ln89 8

⎛ ⎞ ⎛ ⎞− − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 1.089

5. Show that 21

20

4t 9t 8 dt(t 2)(t 1)

⎛ ⎞+ +⎜ ⎟+ +⎝ ⎠

∫ = 2.527, correct to 4 significant figures.


Let ( ) ( )22

2 2 2

A(t 1) B(t 2) t 1 C t 24t 9t 8 A B C(t 2)(t 1) (t 2) (t 1) (t 1) (t 2)(t 1)

+ + + + + ++ +≡ + + =

+ + + + + + +

Hence, 2 24t 9t 8 A(t 1) B(t 2)(t 1) C(t 2)+ + = + + + + + +

Let x = -2, 6 = A

Let x = -1, 3 = C

Equating 2x coefficients, 4 = A + B i.e. B = -2

Thus, 21 1

2 20 0

4t 9t 8 6 2 3dt dt(t 2)(t 1) (t 2) (t 1) (t 1)

⎛ ⎞ ⎛ ⎞+ += − +⎜ ⎟ ⎜ ⎟+ + + + +⎝ ⎠⎝ ⎠

∫ ∫

= 1

0

36ln(t 2) 2ln(t 1)(t 1)

⎡ ⎤+ − + −⎢ ⎥+⎣ ⎦

= 3 36ln 3 2ln 2 6ln 2 2ln12 1

⎛ ⎞ ⎛ ⎞− − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= (3.70538) – (1.15888) = 2.546



1. Determine ( )( )

2

2

x x 13 dxx 7 x 2

− −+ −∫

Let ( )( ) ( ) ( )

( )( )( )

22

2 2 2

(Ax B)(x 2) C x 7x x 13 Ax B Cx 2x 7 x 2 x 7 x 7 x 2

+ − + +− − +≡ + =

−+ − + + −

Hence, ( )2 2x x 13 (Ax B)(x 2) C x 7− − = + − + +

Let x = 2, -11 = 11C i.e. C = -1

Equating 2x coefficients, 1 = A + C i.e. A = 2

Equating x coefficients, -1 = -2A + B i.e. B = 3

Hence, ( )( ) ( ) ( ) ( )

2

2 2 2 2

x x 13 2x 3 1 2x 3 1dx dx dx(x 2) (x 2)x 7 x 2 x 7 x 7 x 7

⎛ ⎞ ⎛ ⎞− − +⎜ ⎟ ⎜ ⎟= − = + −⎜ ⎟ ⎜ ⎟− −+ − + + +⎝ ⎠ ⎝ ⎠

∫ ∫ ∫

= ( ) ( )22 2

2x 3 1dx dx dx(x 2)x 7 x 7

+ −−+ +

∫ ∫ ∫

= ( )2 13 xln x 7 tan ln(x 2) c7 7

−+ + − − +

2. Evaluate ( )

6

25

6x 5 dx(x 4) x 3

−− +∫ correct to 4 significant figures.

Let ( )( ) ( )

( )( )( )

2

2 2 2

A x 3 (Bx C)(x 4)6x 5 A Bx C(x 4)x 4 x 3 x 3 x 4 x 3

+ + + −− +≡ + =

−− + + − +

Hence, ( )26x 5 A x 3 (Bx C)(x 4)− = + + + −

Let x = 4, 19 = 19A i.e. A = 1

Equating 2x coefficients, 0 = A + B i.e. B = -1

Equating x coefficients, 6 = -4B + C i.e. C = 2

Thus, ( ) ( ) ( ) ( )

6 6 6

22 2 25 5 5 2

6x 5 1 2 x 1 2 xdx dx dx(x 4) (x 4)(x 4) x 3 x 3 x 3x 3

⎛ ⎞⎛ ⎞− − ⎜ ⎟⎜ ⎟= + = + −⎜ ⎟⎜ ⎟− −− + + +⎜ ⎟+⎝ ⎠ ⎝ ⎠∫ ∫ ∫


= ( )6

1 2

5

2 x 1ln(x 4) tan ln x 323 3

−⎡ ⎤− + − +⎢ ⎥⎣ ⎦

= 1 12 6 1 2 5 1ln 2 tan ln 39 ln1 tan ln 282 23 3 3 3

− −⎛ ⎞ ⎛ ⎞+ − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= (0.35065) – (-0.23736)

= 0.5880

5. Show that ( )

2 32

2 21

2 6 2 d1

⎛ ⎞+ θ+ θ − θ⎜ ⎟ θ⎜ ⎟θ θ +⎝ ⎠

∫ = 1.606, correct to 4 significant figures.

Let ( ) ( )

( ) ( ) ( )( )

2 2 22 3

22 2 2 2 2

A 1 B 1 C D2 6 2 A B C D1 1 1

θ θ + + θ + + θ+ θ+ θ+ θ − θ θ+≡ + + =θ θθ θ + θ + θ θ +

Hence, ( ) ( ) ( )2 3 2 2 22 6 2 A 1 B 1 C D+ θ+ θ − θ = θ θ + + θ + + θ+ θ

Let θ = 0, 2 = B

Equating 3θ coefficients, -2 = A + C (1)

Equating 2θ coefficients, 6 = B + D i.e. D = 4

Equating θ coefficients, 1 = A

From equation (1), C = -3

Hence, ( ) ( )

2 32 2 2

2 2 2 22 2 21 1 1

2 6 2 1 2 4 3 1 2 4 3d d d1 11 1

⎛ ⎞ ⎛ ⎞+ θ+ θ − θ − θ θ⎛ ⎞⎜ ⎟ ⎜ ⎟θ = + + θ = + + − θ⎜ ⎟⎜ ⎟ ⎜ ⎟θ θ θ θ θ + θ +θ θ + θ + ⎝ ⎠⎝ ⎠ ⎝ ⎠∫ ∫ ∫

= ( )2

1 2

1

2 3ln 4 tan ln 12

−⎡ ⎤θ− + θ− θ +⎢ ⎥θ⎣ ⎦

= ( ) ( )1 1ln 2 1 4 tan 2 1.5ln 5 ln1 2 4 tan 1 1.5ln 2− −− + − − − + −

= (1.70759) – (0.10187)

= 1.606


CHAPTER 42 THE t = tan θ/2 SUBSTITUTION EXERCISE 166 Page 415

2. Determine dx1 cos x sin x− +∫

2 2

2 2 2 2

2 2 2

2dt 2dtdx 2dt 11 t 1 t dx

1 t 2t (1 t ) (1 t ) 2t1 cos x sin x 2t 2t t(t 1)11 t 1 t 1 t

+ += = = =− + − − +− + + +− ++ + +

∫ ∫ ∫ ∫ ∫

Let 1 A B A(t 1) Btt(t 1) t t 1 t(t 1)

+ += + =

+ + +

Hence, 1 = A(t + 1) + Bt

Let t = 0, 1 = A

Let t = -1, 1 = -B i.e. B = -1

Thus, dx 1 1 1dx dx1 cos x sin x t(t 1) t t 1

⎛ ⎞= = −⎜ ⎟− + + +⎝ ⎠∫ ∫ ∫

= ln t – ln(t + 1) + c = tln c1 t⎧ ⎫+⎨ ⎬+⎩ ⎭

=

xtan2ln cx1 tan

2

⎧ ⎫⎪ ⎪

+⎨ ⎬⎪ ⎪+⎩ ⎭

4. Determine dx3sin x 4cos x−∫

2 2

2 22

22 2

2dt 2dtdx 2dt1 t 1 t

6t 4 4t3sin x 4cos x 4t 6t 42t 1 t3 41 t1 t 1 t

+ += = =− +− + −⎛ ⎞−⎛ ⎞ −⎜ ⎟ ⎜ ⎟ ++ +⎝ ⎠ ⎝ ⎠

∫ ∫ ∫ ∫

= 2

dt dt2t 3t 2 (2t 1)(t 2)

=+ − − +∫ ∫

Let 1 A B A(t 2) B(2t 1)(2t 1)(t 2) (2t 1) (t 2) (2t 1)(t 2)

+ + −= + =

− + − + − +

Hence, 1 = A(t + 2) + B(2t - 1)

Let t = 0.5, 1 = 2.5A i.e. A = 1 22.5 5

=


Let t = -2, 1 = -5B i.e. B = 15

−

Thus, dx3sin x 4cos x−∫ =

2 1dt 1 15 5 dt ln(2t 1) ln(t 1) c

(2t 1)(t 2) (2t 1) (t 2) 5 5= − = − − + +

− + − +∫ ∫

= 1 2t 1ln c5 t 2

−⎧ ⎫+⎨ ⎬+⎩ ⎭ =

x2tan 11 2ln cx5 tan 22

⎧ ⎫−⎪ ⎪+⎨ ⎬

⎪ ⎪+⎩ ⎭



1. Determine d5 4sin

θ+ θ∫

( )2 2

222

2 2

2dt 2dtd 2dt 2dt1 t 1 t

2t 85 4sin 5t 8t 55 1 t 4(2t)5 4 5 t t 11 t 51 t

θ + += = = =+ θ + +⎛ ⎞ ⎛ ⎞+ ++ + +⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠+

∫ ∫ ∫ ∫ ∫

= 1 12 2

2 4t2 dt 2 5t 45 5tan c tan c3 35 3 34 3t 5 55 5

− −

⎛ ⎞+⎜ ⎟ +⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

∫

= 15 tan 42 2tan c

3 3−

θ⎛ ⎞+⎜ ⎟+⎜ ⎟

⎜ ⎟⎝ ⎠

3. Determine dp3 4sin p 2cos p− +∫

( ) ( )2 2

2 2 2

2 2 2

2dt 2dtdp 1 t 1 t

3 4sin p 2cos p 2t 1 t 3 1 t 4(2t) 2 1 t3 4 21 t 1 t 1 t

+ += =− + ⎛ ⎞− + − + −⎛ ⎞− +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ +⎝ ⎠

∫ ∫ ∫

= ( ) ( )22 2 2 2 2

2dt 2dt 2dt 2dt3 3t 8t 2 2t t 8t 5 (t 4) 11 t 4 11

= = =+ − + − − + − − − −

∫ ∫ ∫ ∫

= 2 t 4 11 1 t 4 11ln c ln2 11 t 4 11 11 t 4 11

⎧ ⎫ ⎧ ⎫− − − −⎪ ⎪ ⎪ ⎪+ =⎨ ⎬ ⎨ ⎬− + − +⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

using partial fractions (see

Problem 9, page 411 of textbook)

=

ptan 4 111 2ln cp11 tan 4 112

⎧ ⎫− −⎪ ⎪+⎨ ⎬

⎪ ⎪− +⎩ ⎭

5. Show that

t2 tandt 1 2ln t1 3cos t 2 2 2 tan2

⎧ ⎫+⎪ ⎪= ⎨ ⎬+ ⎪ ⎪−

⎩ ⎭

∫


2 2

2 2 2 2 22

22

2dt 2dtdt 2dt 2dt1 t 1 t

1(1 t ) 3(1 t )1 3cos t 1 t 3 3t 4 2t1 t1 31 t1 t

+ += = = =+ + −+ + + − −⎛ ⎞−

+ ⎜ ⎟ ++⎝ ⎠

∫ ∫ ∫ ∫ ∫

= ( )22 2

dt dt2 t 2 t

=− −

∫ ∫ = 1 2 tln c2 2 2 t

⎧ ⎫+⎪ ⎪+⎨ ⎬−⎪ ⎪⎩ ⎭

using partial

fractions (see problem 11, page 411 of textbook)

=

t2 tan1 2ln ct2 2 2 tan2

⎧ ⎫+⎪ ⎪+⎨ ⎬

⎪ ⎪−⎩ ⎭

7. Show that / 2

0

d2 cos 3 3

π θ π=

+ θ∫

2 2

2 2 2 2 22

22

2dt 2dtd 2dt 2dt1 t 1 t

2(1 t ) (1 t )2 cos 2 2t 1 t t 31 t21 t1 t

θ + += = = =+ + −+ θ + + − +⎛ ⎞−

+ ⎜ ⎟ ++⎝ ⎠

∫ ∫ ∫ ∫ ∫

= ( )

1 12

2

tandt 2 t 2 22 tan tan3 3 3 3t 3

− −

θ⎛ ⎞⎜ ⎟

= = ⎜ ⎟+ ⎜ ⎟

⎝ ⎠

∫

Hence,

/ 2

/ 2 1 1 1

0

0

tan tand 2 2 tan 02 4tan tan tan2 cos 3 3 3 3 3

π

π − − −

θ ⎡ π ⎤⎡ ⎤ ⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥θ ⎛ ⎞= = −⎢ ⎥⎜ ⎟⎢ ⎥ ⎜ ⎟+ θ ⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

∫

= 12 1 2tan 063 3 3

− π⎡ ⎤ ⎛ ⎞− = =⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦ 3 3π

(Using a calculator on degrees, 1 1tan 30 rad63

− π= ° = )

Note that tan 30° i.e. rad6π⎛ ⎞

⎜ ⎟⎝ ⎠

= 13

(since AD = 3 by Pythagoras)


CHAPTER 43 INTEGRATION BY PARTS EXERCISE 168 Page 420

2. Determine 3x

4x dxe∫

3x

3x

4x dx 4xe dxe

−=∫ ∫ Let u = 4x then dudx

= 4 and du = 4 dx

and dv = 3xe dx− from which, v = 3x 3x1e dx e3

− −= −∫

Hence, 3x 3x 3x1 14xe dx (4x) e e 4dx3 3

− − −⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

= 3x 3x 3x 3x4 4 4 4 1xe e dx xe e c3 3 3 3 3

− − − −⎛ ⎞− + = − + − +⎜ ⎟⎝ ⎠∫

= 3x 3x4 4xe e c3 9

− −− − + = 3x4 1e x c3 3

− ⎛ ⎞− + +⎜ ⎟⎝ ⎠

4. Determine 5 cos 2 dθ θ θ∫

Let u = 5θ then dudθ

= 5 and du = 5 dθ

and dv = cos 2θ dθ from which, v = 1cos 2 d sin 22

θ θ = θ∫

( ) 1 15 cos 2 d 5 sin 2 sin 2 5d2 2

⎛ ⎞ ⎛ ⎞θ θ θ = θ θ − θ θ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

= 5 5 5 5 1sin 2 sin 2 d sin 2 cos 2 c2 2 2 2 2

⎛ ⎞θ θ− θ θ = θ θ− − θ +⎜ ⎟⎝ ⎠∫

= 5 5sin 2 cos 2 c2 4θ θ+ θ+ = 5 1sin 2 cos 2 c

2 2⎛ ⎞θ θ + θ +⎜ ⎟⎝ ⎠

6. Evaluate 2 x

02x e dx∫ correct to 4 significant figures.

Let u = 2x then dudx

= 2 and du = 2 dx

and dv = xe dx from which, v = x xe dx e=∫

( ) ( )x x x x x2x e dx (2x) e e 2dx 2xe 2e c= − = − +∫ ∫

Hence, ( ) ( )2 2x x x 2 2 0

002x e dx 2xe 2e 4e 2e 0 2e⎡ ⎤= − = − − −⎣ ⎦∫ = 22e 2+ = 16.78


8. Evaluate / 2 2

0t cos t dt

π


Let u = 2t then dudt

= 2t and du = 2t dt

and dv = cos t dt from which, v = cos t dt sin t=∫

Hence, ( )( ) ( )2 2t cos t dt t sin t sin t 2t dt= −∫ ∫

= 2t sin t 2 t sin t dt⎡ ⎤− ⎣ ⎦∫ (1)

For t sin t dt∫ , let u = t then dudt

= 1 and du = dt

and dv = sin t dt from which, v = sin t dt cos t= −∫

Hence, t sin t dt∫ = (t)(- cos t) - ( cos t)dt−∫ = -t cos t + sin t

Substituting in (1) gives: [ ]2 2t cos t dt t sin t 2 t cos t sin t c= − − + +∫

= 2t sin t 2t cos t 2sin t c+ − +

Thus, / 2 / 22 2

00t cos t dt t sin t 2t cos t 2sin t

π π⎡ ⎤= + −⎣ ⎦∫

= [ ]2 2sin cos 2sin 0 0 0

2 2 2 2 2⎡ ⎤π π π π π⎛ ⎞ + − − + −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

= 2 2

0 2 22 4π π⎛ ⎞ + − = −⎜ ⎟

⎝ ⎠ = 0.4674

9. Evaluate x2 2 2

13x e dx∫ correct to 4 significant figures.


= 6x and du = 6x dx

and dv = x2e dx from which, v =

xx x22 2ee dx 2e1

2

= =∫

Hence, ( )x x x

2 22 2 23x e dx 3x 2e 2e 6x dx⎛ ⎞ ⎛ ⎞

= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫

= x x

2 2 26x e 12 x e dx⎡ ⎤

− ⎢ ⎥⎣ ⎦∫ (1)

Forx2x e dx∫ , let u = x then du

dx = 1 and du = dx


and dv = x2e dx from which, v =

x x2 2e dx 2e=∫

Hence, x2x e dx∫ = ( )

x x x x2 2 2 2x 2e 2e dx 2x e 4e

⎛ ⎞− = −⎜ ⎟

⎝ ⎠∫

Substituting in (1) gives: x x x x

2 22 2 2 23x e dx 6x e 12 2xe 4e c⎡ ⎤

= − − +⎢ ⎥⎣ ⎦

∫

= x x x

2 2 2 26x e 24x e 48e c− + +

Thus, 2x x x x2 2 22 2 2 2

11

3x e dx 6x e 24x e 48e⎡ ⎤

= − +⎢ ⎥⎣ ⎦

∫

= ( )1 1 1

1 1 1 2 2 224e 48e 48e 6e 24e 48e⎛ ⎞

− + − − +⎜ ⎟⎝ ⎠

= ( )1

1 224e 30e⎛ ⎞

− ⎜ ⎟⎝ ⎠

= 15.78


EXERCISE 169 Page 422 1. Determine 22x ln x dx∫

Let u = ln x then du 1dx x

= and du = 1 dxx

and dv = 22x dx from which, v = 2 322x dx x3

=∫

Hence, ( )2 3 32 2 12x ln x dx ln x x x dx3 3 x

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

= 3

3 2 3 3 32 2 2 2 x 2 2x ln x x dx x ln x c x ln x x c3 3 3 3 3 3 9

⎛ ⎞− = − + = − +⎜ ⎟

⎝ ⎠∫

= 32 1x ln x c3 3

⎛ ⎞− +⎜ ⎟⎝ ⎠

3. Determine 2x sin 3x dx∫


= 2x and du = 2x dx

and dv = sin 3x dx from which, v = 1sin 3x dx cos3x3

= −∫

Hence, ( )2 2 1 1x sin 3x dx x cos3x cos3x 2x dx3 3

⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

= 21 2x cos3x x cos3x dx3 3

⎡ ⎤− + ⎣ ⎦∫ (1)

For x cos3x dx∫ , let u = x then dudx

= 1 and du = dx

and dv = cos 3x dx from which, v = 1cos3x dx sin 3x3

=∫

Hence, x cos3x dx∫ = ( ) 1 1x sin 3x sin 3x dx3 3

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ = 1 1x sin 3x cos3x

3 9+

Substituting in (1) gives: 2 21 2 1 1x sin 3x dx x cos3x x sin 3x cos3x c3 3 3 9

⎡ ⎤= − + + +⎢ ⎥⎣ ⎦∫

= 21 2 2x cos 3x xsin 3x cos 3x c3 9 27

− + + +

or ( )2cos 3x 22 9x xsin 3x c27 9

− + +


5. Determine 22 sec dθ θ θ∫

Let u = 2θ then du 2d

=θ

and du = 2 dθ

and dv = 2sec dθ θ from which, v = 2sec d tanθ θ = θ∫

Hence, ( )( ) ( )22 sec d 2 tan tan 2dθ θ θ = θ θ − θ θ∫ ∫

= 2 tan 2ln(sec ) cθ θ− θ + from Problem 9, chapter 39, page 393 of textbook

= [ ]2 tan ln(sec ) cθ θ − θ +

7. Evaluate 1 3x

02e sin 2x dx∫ correct to 4 significant figures.

Let u = 3x2e then 3xdu 6edx

= and du = 3x6e dx

and dv = sin 2x dx from which, v = 1sin 2x dx cos 2x2

= −∫

Hence, ( )3x 3x 3x1 12e sin 2x dx 2e cos 2x cos 2x 6e dx2 2

⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

= 3x 3xe cos 2x 3 e cos 2x dx⎡ ⎤− + ⎣ ⎦∫ (1)

For 3xe cos 2x dx∫ , let u = 3xe then 3xdu 3edx

= and du = 3x3e dx

and dv = cos 2x dx from which, v = 1cos 2x dx sin 2x2

=∫

Hence, ( )3x 3x 3x1 1e cos 2x dx e sin 2x sin 2x 3e dx2 2

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

= 3x 3x1 3e sin 2x e sin 2x dx2 2

⎡ ⎤− ⎣ ⎦∫

Substituting in (1) gives: 3x 3x 3x 3x1 32e sin 2x dx e cos 2x 3 e sin 2x e sin 2x dx2 2⎡ ⎤= − + −⎢ ⎥⎣ ⎦∫ ∫

= 3x 3x 3x3 9e cos 2x e sin 2x e sin 2x dx2 2

− + − ∫

Hence, 3x 3x 3x9 32 e sin 2x dx e cos 2x e sin 2x2 2

⎛ ⎞+ = − +⎜ ⎟⎝ ⎠ ∫ by combining the far left and far

right-hand integrals

i.e. 3x 3x 3x13 3e sin 2x dx e cos 2x e sin 2x2 2

⎛ ⎞ = − +⎜ ⎟⎝ ⎠ ∫


and 3x 3x 3x2 3e sin 2x dx e cos 2x e sin 2x13 2

⎛ ⎞= − +⎜ ⎟⎝ ⎠∫

Thus, 3x 3x 3x4 32e sin 2x dx e cos 2x e sin 2x13 2

⎛ ⎞= − +⎜ ⎟⎝ ⎠∫

and 1

1 3x 3x 3x

00

4 32e sin 2x dx e cos 2x e sin 2x13 2⎡ ⎤⎛ ⎞= − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫

= 3 34 3 4 3e cos 2 e sin 2 cos 0 sin 013 2 13 2⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− + − − +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

= ( )4 48.3585 27.3956 ( 1)13 13⎡ ⎤ ⎡ ⎤+ − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= (11.0013) + (0.30769) = 11.31

9. Evaluate 4 3

1x ln x dx∫ correct to 4 significant figures.

Let u = ln x then du 1dx x

= and du = 1 dxx

and dv = 3x dx from which, v = 3 52 22x dx x

5=∫

Hence, ( )5 5

3 2 22 2 1x ln x dx ln x x x dx5 5 x

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ ∫

= 3

5 22 2x ln x x dx5 5

− ∫

= 5

5 22 2 2x ln x x c5 5 5

⎛ ⎞− +⎜ ⎟

⎝ ⎠

Thus, 4

4 3 5 5

11

2 4x ln x dx x ln x x5 25⎡ ⎤= −⎢ ⎥⎣ ⎦∫

= 5 5 5 52 4 2 44 ln 4 4 1 ln1 15 25 5 25

⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ( ) ( )2 4 432 ln 4 32 05 25 25

⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 12.78


10. In determining a Fourier series to represent f(x) = x in the range -π to π, Fourier coefficients are

given by: n1a x cos nx dx

π

−π=π ∫ and nb x sin nx dx

π

−π= ∫

where n is a positive integer.

Show by using integration by parts that na = 0 and n2b cos nn

= − π

For n1a x cos nx dx

π

−π=π ∫ , let u = x then du 1

dx= and du = dx

and dv = cos nx dx from which, v = 1cos nx dx sin nxn

=∫

Hence, ( ) 2

1 1 x 1x cos nx dx x sin nx sin nx dx sin nx cos nxn n n n

⎛ ⎞ ⎛ ⎞= − = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

Then n 2

1 1 x 1a x cos nx dx sin nx cos nxn n

ππ

−π−π

⎡ ⎤= = +⎢ ⎥π π ⎣ ⎦∫

= 2 2

1 1 1sin n cos n sin( n ) cos( n )n n n n

⎡ π π ⎤⎛ ⎞ ⎛ ⎞π + π − − − π + − π⎜ ⎟ ⎜ ⎟⎢ ⎥π ⎝ ⎠ ⎝ ⎠⎣ ⎦

sin nπ = sin(-nπ) = 0 for all values of n.

Hence, n 2 2

1 1 1a 0 cos n 0 cos( n )n n

⎡ ⎤⎛ ⎞ ⎛ ⎞= + π − + − π⎜ ⎟ ⎜ ⎟⎢ ⎥π ⎝ ⎠ ⎝ ⎠⎣ ⎦

= [ ]2

1 cos n cos( n )n

π− − ππ

= 0 since cos nπ = cos(-nπ) for all values of n.

For nb x sin nx dxπ

−π= ∫ , let u = x then du 1

dx= and du = dx

and dv = sin nx dx from which, v = 1sin nx dx cos nxn

= −∫

Hence, ( ) 2

1 1 x 1x sin nx dx x cos nx cos nx dx cos nx sin nxn n n n

⎛ ⎞ ⎛ ⎞= − − − = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

Then n 2

1 1 x 1b x sin nx dx cos nx sin nxn n

ππ

−π−π

⎡ ⎤= = − +⎢ ⎥π π ⎣ ⎦∫

= 2 2

1 1 1cos n sin n cos( n ) sin( n )n n n n

⎡ π π ⎤⎛ ⎞ ⎛ ⎞− π+ π − − π + − π⎜ ⎟ ⎜ ⎟⎢ ⎥π ⎝ ⎠ ⎝ ⎠⎣ ⎦

sin nπ = sin(-nπ) = 0 for all values of n.


Hence, n1b cos n 0 cos( n ) 0

n n⎡ π π ⎤⎛ ⎞ ⎛ ⎞= − π+ − − π +⎜ ⎟ ⎜ ⎟⎢ ⎥π ⎝ ⎠ ⎝ ⎠⎣ ⎦

= [ ]cos n cos( n )n

−ππ+ − π

π

cos nπ = cos(-nπ) hence cos nπ + cos(-nπ) ≡ 2 cos nπ

Thus, ( )n1b 2cos nn

= − π = 2 cosnn

− π

11. The equation C = 1 0.4

0e cos1.2 d− θ θ θ∫ and S =

1 0.4

0e sin1.2 d− θ θ θ∫

are involved in the study of damped oscillations. Determine the value of C and S. From problem 9, page 421 of textbook,

( )ax

ax2 2

ee cos bx dx bsin bx a cos bx ca b

= + ++∫

Thus, ( )0.4

0.42 2

ee cos1.2 d 1.2sin1.2 0.4cos1.2( 0.4) (1.2)

− θ− θ θ θ = θ− θ

− +∫

and C = ( )10.41 0.4

00

ee cos1.2 d 1.2sin1.2 0.4cos1.21.6

− θ− θ ⎡ ⎤

θ θ = θ− θ⎢ ⎥⎣ ⎦

∫

= ( )0.4e 11.2sin1.2 0.4cos1.2 (1.2sin 0 0.4cos 0

1.6 1.6

−⎡ ⎤ ⎡ ⎤− − −⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

= (0.40785) – (-0.25)

i.e. C = 0.66

From equation (2) on page 422 (obtained in a similar way to that for question 7 above),

( )ax

ax2 2

ee sin bx dx a sin bx bcos bx ca b

= − ++∫

Thus, 0.4

0.4 ee sin1.2 d ( 0.4sin1.2 1.2cos1.2 )1.6

− θ− θ θ θ = − θ− θ∫

and S = 1 0.4

0e sin1.2 d− θ θ θ∫ ( )

10.4

0

e 0.4sin1.2 1.2cos1.21.6

− θ⎡ ⎤= − θ− θ⎢ ⎥⎣ ⎦

= ( )0.4e 10.4sin1.2 1.2cos1.2 ( 0.4sin 0 1.2cos 0

1.6 1.6

−⎡ ⎤ ⎡ ⎤− − − − −⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

= (-0.33836) – (-0.75)

i.e. S = 0.41


CHAPTER 44 REDUCTION FORMULAE EXERCISE 170 Page 425 2. Determine 3 2tt e dt∫ using a reduction formula.

n 2tt e dt∫ Let u = nt then n 1du nt

dt−= and du = n 1nt dt−

and dv = 2te dt from which, v = 2t 2t1e dt e2

=∫

Thus, ( )n 2t n 2t 2t n 11 1t e dt t e e nt dt2 2

−⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

= n 2t n 1 2t1 nt e t e dt2 2

−− ∫

i.e. n 2tn n 1

1 nI t e I2 2 −= −

Hence, 3 2tt e dt∫ = 3 2t3 2

1 3I t e I2 2

= −

2 2t2 1

1 2I t e I2 2

= −

2t1 0

1 1I t e I2 2

= −

2t 2t0

1I e dt e2

= =∫

Thus, 3 2tt e dt∫ = 3 2t 2 2t1

1 3 1t e t e I2 2 2

⎛ ⎞− −⎜ ⎟⎝ ⎠

= 3 2t 2 2t 2t 2t1 3 3 1 1 1t e t e t e e c2 4 2 2 2 2

⎛ ⎞⎛ ⎞− + − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 3 2t 2 2t 2t 2t1 3 3 3t e t e t e e c2 4 4 8

− + − +

= 2t 3 21 3 3 3e t t t c2 4 4 8

⎛ ⎞− + − +⎜ ⎟⎝ ⎠

3. Use the result of Problem 2 to evaluate 1 3 2t

05t e dt∫ , correct to 3 decimal places.

1

1 3 2t 2t 3 2 2 0

00

1 3 3 3 1 3 3 3 35t e dt 5e t t t 5e 5e2 4 4 8 2 4 4 8 8

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − = − + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∫ = 6.493


EXERCISE 171 Page 427 3. Use a reduction formula to determine 5x sin x dx∫

5 5 4

5 3x sin x dx I x cos x 5x sin x 5(4)I= = − + −∫ from equation (3), page 426 of textbook

3 23 1I x cos x 3x sin x 3(2)I= − + −

1 01I x cos x 1x sin x (1)(0)= − + −

i.e. 5x sin x dx∫ = ( )5 4 3 21x cos x 5x sin x 20 x cos x 3x sin x 6I− + − − + −

= ( )5 4 3 2x cos x 5x sin x 20x cos x 60x sin x 120 x cos x sin x c− + + − + − + +

= 5 4 3 2x cos x 5x sin x 20x cos x 60x sin x 120xcos x 120sin x c− + + − − + +

4. Evaluate 5

0x sin x dx

π

∫ , correct to 2 decimal places.

5 5 4 3 2

00x sin x dx x cos x 5x sin x 20x cos x 60x sin x 120x cos x 120sin x

π π⎡ ⎤= − + + − − +⎣ ⎦∫

= ( )5 4 3 2cos 5 sin 20 cos 60 sin 120 cos 120sin−π π+ π π+ π π− π π− π π+ π

( )0 0 0 0 0 120sin 0− + + − − +

= ( ) ( )5 320 120 0π − π + π −

= 62.89



2. Evaluate 3

03sin x dx

π

∫ , using a reduction formula.

3 2

3 11 2sin x dx I sin x cos x I3 3

= = − +∫ from equation (4), page 428 of textbook

01

1I sin x cos x 0 cos x1

= − + = −

Hence, ( )3 21 2sin x dx sin x cos x cos x3 3

= − + −∫

3 2 21 23sin x dx 3 sin x cos x cos x sin x cos x 2cos x3 3

⎛ ⎞= − − = − −⎜ ⎟⎝ ⎠∫

Thus, 3

03sin x dx

π

∫ = 2

0sin x cos x 2cos x

π⎡ ⎤− −⎣ ⎦

= ( ) ( )2 2sin cos 2cos sin 0cos 0 2cos 0− π π− π − − −

= (- 0 – 2(-1)) – (0 – 2)

= 2 - - 2 = 4

4. Determine 6cos x dx∫ , using a reduction formula,

6 5

6 41 5cos x dx I cos x sin x I6 6

= = +∫ from equation (5), page 429 of textbook

34 2

1 3I cos x sin x I4 4

= +

12 0

1 1I cos x sin x I2 2

= +

00I cos x dx x= =∫

Thus, 6cos x dx∫ = 5 32

1 5 1 3cos x sin x cos x sin x I6 6 4 4

⎛ ⎞+ +⎜ ⎟⎝ ⎠

= 5 31 5 15 1 1cos x sin x cos x sin x cos x sin x x6 24 24 2 2

⎛ ⎞+ + +⎜ ⎟⎝ ⎠

= 5 31 5 15 15cos x sin x cos x sin x cos x sin x x6 24 48 48

+ + +

= 5 31 5 5 5cos xsin x cos xsin x cos xsin x x6 24 16 16

+ + +


5. Evaluate / 2 7

0cos x dx

π

∫

Using Wallis’s formula, equation (6), page 430 of textbook

/ 2 7

7 50

6cos x dx I I7

π= =∫

5 34I I5

=

3 12I I3

=

[ ]/ 2 / 21

1 00I cos x dx sin x sin sin 0 1

2π π π

= = = − =∫

Hence, ( )/ 2 7

0

6 4 2cos x dx 17 5 3

π ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫ = 16

35



1. Evaluate / 2 2 5

0cos x sin x dx

π

∫

( )/ 2 / 22 5 2 5

0 0cos x sin x dx 1 sin x sin x dx

π π= −∫ ∫

= ( )/ 2 5 75 70

sin x sin x dx I Iπ

− = −∫

Wallis’s formula states: / 2 n

n n 20

n 1sin x dx I In

π

−

−= =∫ from problem 10, page 429 of textbook

[ ]/ 2 / 21

5 3 3 1 1 00

4 2I I I I I sin x dx cos x [0 1] 15 3

π π= = = = − = − − =∫

Hence, ( )54 2I 15 3

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

7 5 5 3 3 1 16 4 2I I I I I I I 17 5 3

= = = =

Hence, ( )76 4 2I 17 5 3

⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

Thus, / 2 2 5

5 70

4 2 6 4 2cos x sin x dx I I5 3 7 5 3

π ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠∫

= 4 2 6 4 2 115 3 7 5 3 7

⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎛ ⎞⎛ ⎞⎛ ⎞− =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎣ ⎦ ⎝ ⎠⎝ ⎠⎝ ⎠

= 8105

3. Evaluate / 2 5 4

0cos x sin x dx

π

∫

( ) ( )/ 2 / 2 / 225 4 5 2 5 2 4

0 0 0cos x sin x dx cos x 1 cos x dx cos x 1 2cos x cos x dx

π π π= − = − +∫ ∫ ∫

= ( )/ 2 5 7 9

0cos x 2cos x cos x dx

π− +∫

= 5 7 9I 2I I− +

5 34I I5

= from equation (6), page 430, [ ]/ 2 / 2

3 1 1 00

2I I I cos x dx sin x 13

π π= = = =∫

i.e. ( )54 2I 15 3

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

, 7 56 6 4 2I I7 7 5 3

⎛ ⎞⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

and 9 78 8 6 4 2I I9 9 7 5 3

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠


Hence, / 2 5 4

0cos x sin x dx

π

∫ = 5 7 9I 2I I− +

= 4 25 3

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

6 4 227 5 3

⎛ ⎞⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

+ 8 6 4 29 7 5 3

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

= 4 2 6 8 61 25 3 7 9 7

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞− +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦

= 8 12 16 8 21 36 16115 7 21 15 21

− +⎡ ⎤ ⎡ ⎤− + =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= 8 115 21

⎛ ⎞⎜ ⎟⎝ ⎠

= 8315

5. Show that / 2 3 4

0

2sin cos d35

πθ θ θ =∫

( ) ( )/ 2 / 2 / 223 4 3 2 3 2 4

0 0 0sin cos d sin 1 sin d sin 1 2sin sin d

π π πθ θ θ = θ − θ θ = θ − θ+ θ θ∫ ∫ ∫

= / 2 3 5 7

3 5 70(sin 2sin sin )d I 2I I

πθ − θ+ θ θ = − +∫

3 12I I3

= from Problem 10, page 429, and

[ ]/ 2 / 2

1 00I sin d cos cos cos0 (0 1) 1

2π π π⎛ ⎞= θ θ = − θ = − − = − − =⎜ ⎟

⎝ ⎠∫

( )5 3 7 54 4 2 6 6 4 2I I (1), I I 15 5 3 7 7 5 3

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

Hence, / 2 3 4

3 5 70sin cos d I 2I I

πθ θ θ = − +∫

= ( ) ( )2 4 2 6 4 22 1 13 5 3 7 5 3

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞− +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

= 2 8 24 2 35 56 2413 5 35 3 35

− +⎡ ⎤ ⎡ ⎤− + =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= 2 33 35⎛ ⎞⎜ ⎟⎝ ⎠

= 235


CHAPTER 45 NUMERICAL INTEGRATION EXERCISE 174 Page 435 2. Evaluate

3

12 ln 3x dx∫ using the trapezoidal rule with 8 intervals, giving the answer correct to 3

decimal places.

Since 3

12 ln 3x dx∫ , width of interval = 3 1 0.25

8−

=

x 1 1.25 1.50 1.75 2.0 2.25 2.50 2.75 3.0 2 ln 3x 2.1972 2.6435 3.0082 3.3165 3.5835 3.8191 4.0298 4.2204 4.3944

Hence, using the trapezoidal rule,

3

12 ln 3x dx∫

( )1(0.25) 2.1972 4.3944 2.6435 3.0082 3.3165 3.5835 3.8191 4.0298 4.22042⎡ ⎤≈ + + + + + + + +⎢ ⎥⎣ ⎦

= 0.25[27.9168]

= 6.979

4. Evaluate

21.4 x

0e dx−∫ using the trapezoidal rule with 7 intervals, giving the answer correct to 3

decimal places.

Since21.4 x

0e dx−∫ , width of interval = 1.4 0 0.2

7−

=

x 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

2xe− 1.0 0.9608 0.8521 0.6977 0.5273 0.3679 0.2369 0.1409 Hence, using the trapezoidal rule,

21.4 x

0e dx−∫ ( )1(0.2) 1.0 0.1409 0.9608 0.8521 0.6977 0.5273 0.3679 0.2369

2⎡ ⎤≈ + + + + + + +⎢ ⎥⎣ ⎦

= (0.2)[4.21315]

= 0.843



2. Evaluate / 2

0

1 d1 sin

πθ

+ θ∫ using the mid-ordinate rule with 6 intervals, giving the answer correct

to 3 decimal places.

Since/ 2

0

1 d1 sin

πθ

+ θ∫ , width of interval = 0

2 rad or 156 12

π− π

= °

Hence, ordinates occur at 0°, 15°, 30°, 45°, 60°, 75° and 90°,

and mid-ordinates occur at 7.5°, 22.5°, 37.5°, 52.5°, 67.5° and 82.5°.

θ 7.5° 22.5° 37.5° 52.5° 67.5° 82.5° 1

1 sin+ θ

0.8845 0.7232 0.6216 0.5576 0.5198 0.5021

Hence, using the mid-ordinate rule,

/ 2

0

1 d1 sin

πθ

+ θ∫ ≈ [ ]0.8845 0.7232 0.6216 0.5576 0.5198 0.502112π⎛ ⎞ + + + + +⎜ ⎟

⎝ ⎠

= [ ]3.808812π⎛ ⎞

⎜ ⎟⎝ ⎠

= 0.977

3. Evaluate 3

1

ln x dxx∫ using the mid-ordinate rule with 10 intervals, giving the answer correct to 3

decimal places.

Since3

1

ln x dxx∫ , width of interval = 3 1 0.2

10−

=

Hence, ordinates occur at 1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2, 2.4, 2.6, 2.8, 3.0,

and mid-ordinates occur at 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7 and 2.9.

x 1.1 1.3 1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9

ln xx

0.0866 0.2018 0.2703 0.3121 0.3378 0.3533 0.3621 0.3665 0.3679 0.3671

Hence, using the mid-ordinate rule,


3

1

ln x dxx∫ ≈ (0.2)[ 0.0866 + 0.2018 + 0.2703 + 0.3121 + 0.3378 + 0.3533 + 0.3621 + 0.3665

+ 0.3679 + 0.3671]

= (0.2)[3.0255]

= 0.605



2. Evaluate 1.6

40

1 d1

θ+ θ∫ using Simpson’s rule with 8 intervals, correct to 3 decimal places.

Since1.6

40

1 d1

θ+ θ∫ , width of interval = 1.6 0 0.2

8−

=

θ 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

4

11+ θ

1.0 0.9984 0.9750 0.8853 0.7094 0.5000 0.3254 0.2065 0.1324

Hence, using Simpson’s rule,

1.6

40

1 d1

θ+ θ∫

( ) ( ) ( )1 (0.2) 1.0 0.1324 4 0.9984 0.8853 0.5000 0.2065 2 0.9750 0.7094 0.32543

≈ + + + + + + + +⎡ ⎤⎣ ⎦

= [ ]1 (0.2) 1.1324 10.3608 4.01963

+ +

= [ ]1 (0.2) 15.51283

= 1.034

3. Evaluate 1.0

0.2

sin dθ θθ∫ using Simpson’s rule with 8 intervals, correct to 3 decimal places.

Since1.0

0.2

sin dθ θθ∫ , width of interval = 1.0 0.2 0.1

8−

= (note that values of θ are in radians)

θ 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 sinθθ

0.9933 0.9851 0.9735 0.9589 0.9411 0.9203 0.8967 0.8704 0.8415


1.0

0.2

sin dθ θθ∫

( ) ( )( )

0.9933 0.8415 4 0.9851 0.9589 0.9203 0.87041 (0.1)3 2 0.9735 0.9411 0.8967

+ + + + +⎡ ⎤≈ ⎢ ⎥

+ + +⎢ ⎥⎣ ⎦

= [ ]1 (0.1) 1.8348 14.9388 5.62263

+ +

= [ ]1 (0.1) 22.39623

= 0.747


5. Evaluate 2/3 x

0e sin 2x dx

π

∫ using Simpson’s rule with 10 intervals, correct to 3 decimal places.

Since2/3 x

0e sin 2x dx

π

∫ , width of interval = 0

3 rad10 30

π− π

=

x

0 30π 2

30π 3

30π 4

30π 5

30π 6

30π 7

30π 8

30π 9

30π 10

30π

2xe sin 2x

0 0.2102 0.4250 0.6488 0.8857 1.1392 1.4114 1.7021 2.0064 2.3119 2.5929


2/3 x

0e sin 2x dx

π

∫( ) ( )

( )0 2.5929 4 0.2102 0.6488 1.1392 1.7021 2.31191

3 30 2 0.4250 0.8857 1.4114 2.0064

+ + + + + +⎡ ⎤π⎛ ⎞≈ ⎢ ⎥⎜ ⎟ + + + +⎝ ⎠ ⎢ ⎥⎣ ⎦

= [ ]2.5929 24.0488 9.457090π⎛ ⎞ + +⎜ ⎟

⎝ ⎠

= [ ]36.098790π⎛ ⎞

⎜ ⎟⎝ ⎠

= 1.260

7. Evaluate 6

2

1 dx(2x 1)−∫ using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate

rule, (d) Simpson’s rule. Give answers correct to 3 decimal places and use 6 intervals

(a) 6

2

1 dx(2x 1)−∫ Let u = 2x – 1, then du 2

dx= and dx = du

2

Thus, ( )1

1 221 1 du 1 1 udx u du u 2x 112 2 2(2x 1) u

2

−= = = = = −

−∫ ∫ ∫

Hence, ( )66

2 2

1 dx 2x 1 11 3(2x 1)

⎡ ⎤ ⎡ ⎤= − = −⎣ ⎦⎣ ⎦−∫ = 1.585

(b) Width of interval = 6 2 0.58−

=

x 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

( )1

2x 1−

0.5774 0.5000 0.4472 0.4082 0.3780 0.3536 0.3333 0.3162 0.3015



6

2

1 dx(2x 1)−∫

( )1(0.5) 0.5774 0.3015 0.5000 0.4472 0.4082 0.3780 0.3536 0.3333 0.31622⎡ ⎤≈ + + + + + + + +⎢ ⎥⎣ ⎦

= (0.5)[3.17595] = 1.588

(c) Mid-ordinates occur at 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25 and 5.75

x 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75

( )1

2x 1−

0.5345 0.4714 0.4264 0.3922 0.3651 0.3430 0.3244 0.3086

Using the mid-ordinate rule,

6

2

1 dx(2x 1)−∫ ≈ (0.5)[ 0.5345 + 0.4714 + 0.4264 + 0.3922 + 0.3651 + 0.3430

+ 0.3244 + 0.3086] = (0.5)[3.1656] = 1.583 (d) Using the table of values from part (b), using Simpson’s rule,

6

2

1 dx(2x 1)−∫

( ) ( )( )

0.5774 0.3015 4 0.5000 0.4082 0.3536 0.31621 (0.5)3 2 0.4472 0.3780 0.3333

+ + + + +⎡ ⎤≈ ⎢ ⎥

+ + +⎢ ⎥⎣ ⎦

= [ ]1 (0.5) 0.8789 6.312 2.3173

+ +

= [ ]1 (0.5) 9.50793

= 1.585

9. Evaluate ( )

0.7

0.1 2

1 dy1 y−

∫ using (a) the trapezoidal rule, (b) the mid-ordinate rule,

(c) Simpson’s rule. Use 6 intervals and give answers correct to 3 decimal places and use 6

intervals

(a) Since ( )

0.7

0.1 2

1 dy1 y−

∫ then width of interval = 0.7 0.1 0.16−

=

y 0.1 0.2 0.3 0.4 0.5 0.6 0.7

( )2

1

1 y−

1.0050 1.0206 1.0483 1.0911 1.1547 1.2500 1.4003



( )

0.7

0.1 2

1 dy1 y−

∫ ( )1(0.1) 1.0050 1.4003 1.0206 1.0483 1.0911 1.1547 1.25002⎡ ⎤≈ + + + + + +⎢ ⎥⎣ ⎦

= (0.1)[6.76735] = 0.677

(b) Mid-ordinates occur at 0.15, 0.25, 0.35, 0.45, 0.55 and 0.65

y 0.15 0.25 0.35 0.45 0.55 0.65

( )2

1

1 y−

1.0114 1.0328 1.0675 1.1198 1.1974 1.3159

Using the mid-ordinate rule,

( )

0.7

0.1 2

1 dy1 y−

∫ ≈ (0.1)[1.0114 + 1.0328 + 1.0675 + 1.1198 + 1.1974 + 1.3159]

= (0.1)[6.7448] = 0.674 (c) Using the table of values from part (a), using Simpson’s rule,

( )

0.7

0.1 2

1 dy1 y−

∫ ( ) ( ) ( )1 (0.1) 1.0050 1.4003 4 1.0206 1.0911 1.2500 2 1.0483 1.15473

≈ + + + + + +⎡ ⎤⎣ ⎦

= [ ]1 (0.1) 2.4053 13.4468 4.4063

+ +

= [ ]1 (0.1) 20.2583

= 0.675

10. A vehicle starts from rest and its velocity is measured every second for 8 s, with values as

follows: time t (s) 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 velocity v (ms-1) 0 0.4 1.0 1.7 2.9 4.1 6.2 8.0 9.4

The distance travelled in 8.0 s is given by 8.0

0vdt∫

Estimate the distance using Simpson’s rule, giving the answer correct to 3 significant figures.

( )[ ]8.0

0

1vdt 1.0 (0 9.4) 4(0.4 1.7 4.1 8.0) 2(1.0 2.9 6.2)3

≈ + + + + + + + +∫

= [ ]1 19.4 56.8 20.2 (86.4)3 3

+ + = = 28.8 m


11. A pin moves along a straight guide so that its velocity v (m/s) when it is a distance x (m) from

the beginning of the guide at time t (s) is given by the table below.

t (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

v (m/s) 0 0.052 0.082 0.125 0.162 0.175 0.186 0.160 0

Use Simpson’s rule with 8 intervals to determine the approximate total distance travelled by the

pin in the 4.0 s period. Distance travelled by pin

[ ]1 (0.5) (0 0) 4(0.052 0.125 0.175 0.160) 2(0.082 0.162 0.186)3

≈ + + + + + + + +

[ ]1 (0.5) 0 2.048 0.863

= + +

[ ]1 (0.5) 2.9083

=

= 0.485 m


CHAPTER 46 SOLUTION OF FIRST ORDER DIFFERENTIAL

EQUATIONS BY SEPARATION OF VARIABLES EXERCISE 177 Page 444 1. Sketch a family of curves represented by each of the following differential equations:

(a) dy 6dx

= (b) dy 3xdx

= (c) dy x 2dx

= +

(a) If dy 6dx

= , then y = 6dx∫ = 6x + c

There are an infinite number of graphs of y = 6x + c; three curves are shown below.

(b) If dy 3xdx

= , then y = 233x dx x c2

= +∫

A family of three typical curves is shown below.

(c) If dy x 2dx

= + , then y = 2x(x 2)dx 2x c

2+ = + +∫


A family of three typical curves is shown below.

2. Sketch the family of curves given by the equation dy 2x 3dx

= + and determine the equation of

one of the curves which passes through the point (1, 3).

If dy 2x 3dx

= + , then y = ( ) 22x 3 dx x 3x c+ = + +∫

If the curve passes through the point (1, 3) then x = 1 and y = 3,

Hence, 3 = ( ) ( )21 3 1 c+ + i.e. c = -1

and 2y x 3x 1= + −

A family of three curves is shown below, including 2y x 3x 1= + − which passes through the point

(1, 3).



3. Solve the differential equation: dy x 3dx

+ = , given y = 2 when x = 1.

If dy x 3dx

+ = , then dy 3 xdx

= − and y = ( )2x3 x dx 3x c

2− = − +∫

If y = 2 when x = 1, then 2 = 3 - 1 c2+ from which, c = 1

2−

Hence, 2x 1y 3x

2 2= − −

5. Solve the differential equation: x

1 dy2 x 3e dx

+ = − , given y = 1 when x = 0.

If x

1 dy2 x 3e dx

+ = − then xdy3 x e 2dx

−= − − and ( )xdy 1 x e 2dx 3

−= − −

Hence, ( )2

x x1 1 xy x e 2 dx e 2x c3 3 2

− −⎛ ⎞= − − = + − +⎜ ⎟

⎝ ⎠∫

If y = 1 when x = 0, then 1 = 1 (0 1 0) c3

+ + + i.e. c = 23

Thus, y = 2

x1 x 2e 2x3 2 3

−⎛ ⎞+ − +⎜ ⎟

⎝ ⎠ or 2

x

1 2x 4x 46 e⎛ ⎞− + +⎜ ⎟⎝ ⎠

6. The gradient of a curve is given by: 2dy x 3x

dx 2+ = . Find the equation of the curve if it passes

through the point 11,3

⎛ ⎞⎜ ⎟⎝ ⎠

If 2dy x 3x

dx 2+ = , then

2dy x3xdx 2

= −

Hence, 2 2 3x 3x xy 3x dx c

2 2 6⎛ ⎞

= − = − +⎜ ⎟⎝ ⎠∫

If it passes through 11,3

⎛ ⎞⎜ ⎟⎝ ⎠

, x = 1 and y = 13

Thus, 1 3 1 c3 2 6= − + from which, c = 1 3 1 1

3 2 6− + = −

Hence, 3

23 xy x 12 6

= − −


8. An object is thrown vertically upwards with an initial velocity, u, of 20 m/s. The motion of the

object follows the differential equation dsdt

= u – gt, where s is the height of the object in metres

at time t seconds and g = 9.8 2m / s . Determine the height of the object after 3 seconds if s = 0

when t = 0.

If dsdt

= u – gt, then s = ( )2g tu gt dt ut c

2− = − +∫

Since s = 0 when t = 0, then c = 0

Hence, 2g ts ut

2= − and if u = 20 and s = 9.8, then s = 20t -

29.8t2

i.e. u = 20t – 4.9 2t

The height when t = 3, s = 3(20) – 4.9 ( )23

i.e. height = 60 – 44.1 = 15.9 m



2. Solve the differential equation: 2dy 2cos ydx

=

If 2dy 2cos ydx

= then 2

dy 2dxcos y

=

and 2sec ydy 2dx=∫ ∫

i.e. tan y = 2x + c

3. Solve the differential equation: ( )2 dyy 2 5ydx

+ = , given y = 1 when x = 12

If ( )2 dyy 2 5ydx

+ = then 2y 2 dy 5dxy

⎛ ⎞+=⎜ ⎟

⎝ ⎠

and 2y dy 5dxy

⎛ ⎞+ =⎜ ⎟

⎝ ⎠∫ ∫

i.e. 2y 2 ln y 5x c

2+ = +

y = 1 when x = 12

, hence, 1 52ln1 c2 2+ = + from which, c = 1 5 2

2 2− = −

and 2y 2 ln y 5x 2

2+ = −

4. The current in an electric circuit is given by the equation Ri + diLdt

= 0, where L and R are

constants. Show that i =R tLI e

−, given that i = I when t = 0.

If Ri + diLdt

= 0, then diLdt

= - Ri

and di Ridt L

= −

from which, di R dti L= − and di R dt

i L= −∫ ∫

Thus, ln i = Rt cL

− +

i = I when t = 0, thus ln I = c

Hence, ln i = Rt ln IL

− +


i.e. ln i – ln I = RtL

−

i.e. i RtlnI L= −

Taking anti-logarithms gives: RtLi e

I−

= and R tLi I e

−=

5. The velocity of a chemical reaction is given by dxdt

= k(a – x), where x is the amount transferred

in time t, k is a constant and a is the concentration at time t = 0 when x = 0. Solve the equation

and determine x in terms of t.

If dxdt

= k(a – x), then dx k dta x

=−

and dx k dta x

=−∫ ∫

i.e. – ln(a – x) = kt + c

t = 0 when x = 0, hence - ln a = c

Thus, – ln(a – x) = kt – ln a

i.e. ln a – ln(a – x) = kt

i.e. aln kta x

⎛ ⎞ =⎜ ⎟−⎝ ⎠

and kta ea x

=−

i.e. k t

a a xe

= − i.e. k ta e a x− = −

and x = a - k ta e− i.e. ( )k tx a 1 e−= −

6.(a) Charge Q coulombs at time t seconds is given by the differential equation dQ QRdt C

+ = 0,

where C is the capacitance in farads and R the resistance in ohms. Solve the equation for Q

given that Q = Q0 when t = 0.

(b) A circuit possesses a resistance of 250 × 103 Ω and a capacitance of 8.5 × 10-6 F, and after

0.32 seconds the charge falls to 8.0 C. Determine the initial charge and the charge after 1

second, each correct to 3 significant figures.


(a) If dQ QRdt C

+ = 0 then dQ Qdt RC

= −

i.e. dQ 1 dtQ RC

= −∫ ∫

i.e. ln Q = t kRC

− +

Q = Q0 when t = 0, hence, ln Q0 = k

Hence, ln Q = 0t ln Q

RC− +

i.e. ln Q - ln Q0 = tRC

−

i.e. 0

Q tlnQ RC

= −

and t

R C

0

Q eQ

−

= and t

CR0Q Q e

−=

(b) R = 250 × 103 Ω , C = 8.5 × 10-6 F, t = 0.32 s and Q = 8.0 C

Hence, ( )6 30.32

8.5 10 250 100 08.0 Q e Q 0.8602−−

× × ×= =

from which, initial charge, Q08.0

0.8602= = 9.30 C

When t = 1 s, charge, Q = 6 3t 1

CR 8.5 10 250 100Q e 9.30e −

− −× × ×= = 5.81 C

8. The rate of cooling of a body is given by d kdtθ= θ , where k is a constant. If θ = 60°C when t = 2

minutes and θ = 60°C when t = 5 minutes, determine the time taken for θ to fall to 40°C, correct

to the nearest second.

If d kdtθ= θ then d k dtθ

=θ

and d k dtθ=

θ∫ ∫

i.e. ln θ = kt + c

When θ = 60°C, t = 2, i.e. ln 60 = 2k + c (1)

When θ = 50°C, t = 5, i.e. ln 50 = 5k + c (2)


(1) – (2) gives: ln 60 – ln 50 = -3k

from which, k = 1 60ln 0.060773 50

− = −

Substituting in (1): ln 60 = 2(-0.06077) + c

from which, c = ln 60 + 2(0.06077) = 4.2159

Hence, ln θ = kt + c = -0.06077t + 4.2159

When θ = 40°C, ln 40 = -0.06077t + 4.2159

and time, t = 4.2159 ln 40 8.672min0.06077

−= = 8 min 40 s



2. Solve the differential equation (2y – 1) ( )2dy 3x 1dx

= + , given x =1 when y = 2.

If (2y – 1) ( )2dy 3x 1dx

= + , then ( ) ( )22y 1 dy 3x 1 dx− = +∫ ∫

i.e. 2 3y y x x c− = + +

x =1 when y = 2, hence, 4 – 2 = 1 + 1 + c from which, c = 0

Thus, 2 3y y x x− = +

4. Solve the differential equation 2y(1 – x) + x(1 + y) dydx

= 0, given x = 1 when y = 1.

If 2y(1 – x) + x(1 + y) dydx

= 0 then x(1 + y) dydx

= -2y(1 – x) = 2y(x – 1)

Thus, 1 y 2(x 1)dy dxy x

⎛ ⎞+ −⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

∫ ∫

i.e. 1 21 dy 2 dxy x

⎛ ⎞ ⎛ ⎞+ = −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

∫ ∫

ln y + y = 2x – 2 ln x + c

x = 1 when y = 1, hence, ln 1 + 1 = 2 – 2 ln 1 + c from which, c = -1

Thus, ln y + y = 2x – 2 ln x – 1

or ln y + 2 ln x = 2x – y – 1

i.e. ln y + ln 2x = 2x – y – 1

and ( )2ln x y 2x y 1= − −

5. Show that the solution of the equation 2

2

y 1 y dyx 1 x dx

+=

+ is of the form

2

2

y 1x 1

⎛ ⎞+⎜ ⎟+⎝ ⎠

= constant.

Since 2

2

y 1 y dyx 1 x dx

+=

+ then 2 2

y xdy dxy 1 x 1

=+ +∫ ∫

i.e. ( ) ( )2 21 1ln y 1 ln x 1 c2 2

+ = + +


i.e. ( ) ( )1 1

2 22 2ln y 1 ln x 1 c⎡ ⎤

+ − + =⎢ ⎥⎣ ⎦

i.e.

12 2

2

y 1ln cx 1

⎛ ⎞+=⎜ ⎟+⎝ ⎠

or 2

2

y 1ln cx 1

⎛ ⎞+=⎜ ⎟+⎝ ⎠

and 2

c2

y 1 ex 1

⎛ ⎞+=⎜ ⎟+⎝ ⎠

= a constant

7. Determine the equation of the curve which satisfies the equation 2dyxy x 1dx

= − , and which

passes through the point (1, 2).

Since 2dyxy x 1dx

= − then 2x 1 1y dy dx x dxx x− ⎛ ⎞= = −⎜ ⎟

⎝ ⎠∫ ∫ ∫

i.e. 2 2y x ln x c

2 2= − +

If the curve passes through (1, 2) then x = 1 and y = 2,

hence, 2 22 1 ln1 c

2 2= − + from which, c = 3

2

Thus, 2 2y x 3ln x

2 2 2= − +

or 2 2y x 2ln x 3= − +

8. The p.d., V, between the plates of a capacitor C charged by a steady voltage E through a resistor

R is given by the equation dVCR V Edt

+ =

(a) Solve the equation for V given that at t = 0, V = 0.

(b) Calculate V, correct to 3 significant figures, when E = 25 V, C = 20 × 10-6 F,

R = 200 × 103 Ω and t = 3.0 s.

(a) Since dVCR V Edt

+ = then dV E Vdt CR

−=


i.e. dV dtE V CR

=−∫ ∫

from which, ( ) tln E V kCR

− − = +

At t = 0, V = 0, hence, - ln E = k

Thus, ( ) tln E V ln ECR

− − = −

( ) tln E ln E VCR

− − =

and E tlnE V CR

⎛ ⎞ =⎜ ⎟−⎝ ⎠

i.e. t

CRE eE V

=−

i.e. tCR

E E Ve

= −

and t

CRt

CR

EV E E Eee

−

= − = −

i.e. t

CRV E 1 e−⎛ ⎞

= −⎜ ⎟⎜ ⎟⎝ ⎠

volts

(b) Voltage, ( )6 3t 3.0

0.75CR 20 10 200 10V E 1 e 25 1 e 25 1 e−− −

−× × ×⎛ ⎞ ⎛ ⎞

= − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = 13.2 V

9. Determine the value of p, given that 3 dyxdx

= p – x, and that y = 0 when x = 2 and when x = 6.

Since 3 dyxdx

= p – x then 3 3 2

p x p 1dy dx dxx x x− ⎛ ⎞= = −⎜ ⎟

⎝ ⎠∫ ∫ ∫

i.e. ( )3 2dy px x dx− −= −∫ ∫

i.e. 2 1px xy c

2 1

− −

= − +− −

i.e. 2

p 1y c2x x

= − + +


y = 0 when x = 2, hence, p 10 c8 2

= − + + (1)

y = 0 when x = 6, hence, p 10 c72 6

= − + + (2)

(1) – (2) gives: 1 1 1 10 p8 72 2 6

⎛ ⎞ ⎛ ⎞= − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

i.e. p 109 3

= − +

i.e. p 19 3= from which, p = 3


CHAPTER 47 HOMOGENEOUS FIRST ORDER DIFFERENTIAL

EQUATIONS EXERCISE 181 Page 452

1. Find the general solution of: 2 2 dyx ydx

=

(i) Rearranging 2 2 dyx ydx

= gives: 2

2

dy xdx y

= , which is homogeneous in x and y.

(ii) Let y = vx, then dy dvv xdx dx

= +

(iii) Substituting for y and dydx

gives: 2

2 2 2

dv x 1v xdx v x v

+ = =

(iv) Separating the variables gives: 3

2 2

dv 1 1 vx vdx v v

−= − = i.e.

2

3

v1 v−

1dv dxx

=

Integrating both sides gives: 2

3

v 1dv dx1 v x

=−∫ ∫

Hence, ( )31 ln 1 v3

− − = ln x + c (using u = 31 v− substitution)

(v) Replacing v by yx

gives: 3

3

1 yln 13 x

⎛ ⎞− −⎜ ⎟

⎝ ⎠= ln x + c, which is the general solution.

i.e. 3 3

3

1 x yln ln x c3 x

⎛ ⎞−− = +⎜ ⎟

⎝ ⎠

3. Find the particular solution of the differential equation: ( )2 2x y dy+ = x y dx, given that x = 1

when y = 1.

(i) Rearranging ( )2 2x y dy+ = xy dx gives: 2 2

dy xydx x y

=+

, which is homogeneous in x and y.


= +


gives: 2

2 2 2 2 2 2

dv x(vx) vx vv xdx x v x x (1 v ) 1 v

+ = = =+ + +

(iv) Separating the variables gives: ( )2 3 3

2 2 2 2

v v 1 vdv v v v v vx vdx 1 v 1 v 1 v 1 v

− + − − −= − = = =

+ + + +


i.e. 2

3

1 vv+

−1dv dxx

=


3

1 v 1dv dxv x+

− =∫ ∫

i.e. 3

1 1 1dv dv dxv v x

− − =∫ ∫ ∫

Hence, 2v ln v

2

−

− = ln x + c

or 2

1 ln v2v

− = ln x + c


gives: 21 yln

xy2x

−⎛ ⎞⎜ ⎟⎝ ⎠

= ln x + c

or 2

2

x yln2y x

− = ln x + c which is the general solution.

When x = 1, y = 1, thus: 1 ln12− = ln 1 + c from which, c = 1

2

Thus, the particular solution is: 2

2

x yln2y x

− = ln x + 12

i.e. 2

2

x y 1ln (x)2y x 2

⎛ ⎞= +⎜ ⎟⎝ ⎠

i.e. 2

2

x 1ln y2y 2

= +

or 2 2 1x 2y ln y2

⎛ ⎞= +⎜ ⎟⎝ ⎠

5. Find the particular solution of the differential equation: 2y x dy 1y 2x dx

⎛ ⎞−=⎜ ⎟+⎝ ⎠

, given that y = 3 when

x = 2.

(i) Rearranging 2y x dy 1y 2x dx

⎛ ⎞−=⎜ ⎟+⎝ ⎠

gives: dy 2x ydx 2y x

+=

−, which is homogeneous in x and y.


= +


gives: dv 2x vx 2 vv xdx 2vx x 2v 1

+ ++ = =

− −


(iv) Separating the variables gives: 2dv 2 v (2 v) v(2v 1) 2 2v 2vx v

dx 2v 1 2v 1 2v 1+ + − − − +

= − = =− − −

i.e. 2 2dv 2 2v 2v 2v 2v 2x

dx 2v 1 2v 1− + − + +

= =− −

and 2

2v 12v 2v 2

−− + +

1dv dxx

=


2v 1 1dv dx2v 2v 2 x

−=

− + +∫ ∫

i.e. 2

1 2v 1 1dv dx2 1 v v x

−=

+ −∫ ∫

Hence, ( )21 ln 1 v v2

− + − = ln x + c


gives: 21 y yln 1

2 x x⎛ ⎞⎛ ⎞− + −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= ln x + c, which is the general solution.

When x = 2, y = 3, thus: 1 3 9ln 1 ln 2 c2 2 4

⎛ ⎞− + − = +⎜ ⎟⎝ ⎠

i.e. 1 1ln ln 2 c2 4

⎛ ⎞− = +⎜ ⎟⎝ ⎠

and 121ln ln 2 c

4

−⎛ ⎞ = +⎜ ⎟⎝ ⎠

i.e. ( )12ln 4 ln 2 c= + and ln 2 = ln 2 + c, from which, c = 0

Thus, the particular solution is: 21 y yln 1

2 x x⎛ ⎞⎛ ⎞− + −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= ln x

i.e. 2 2

2

1 x xy yln ln x2 x

⎛ ⎞+ −− =⎜ ⎟

⎝ ⎠ i.e.

12 2 2

2

x xy yln ln xx

−⎛ ⎞+ −

=⎜ ⎟⎝ ⎠

or

12 2

2 2

x xx xy y

⎛ ⎞=⎜ ⎟+ −⎝ ⎠

i.e. 2

2 2

x xx xy y

=+ −

i.e. 2 2

x xx xy y

=+ −

from which, 2 2x xy y 1+ − = or 2 2x xy y 1+ − =


EXERCISE 182 Page 454 1. Solve the differential equation: ( )3 4 4xy dy x y dx= +

(i) Rearranging ( )3 4 4xy dy x y dx= + gives: 4 4

3

dy x ydx xy

+= , which is homogeneous in x and y.


= +


gives: ( )

4 4 4 4 4 4

3 4 33 3

dv x v x x (1 v ) 1 vv xdx v x vx v x

+ + ++ = = =

(iv) Separating the variables gives: 4 4 4

3 3 3

dv 1 v 1 v v 1x vdx v v v

+ + −= − = =

i.e. 3 1v dv dxx

=

Integrating both sides gives: 3 1v dv dxx

=∫ ∫

Hence, 4v

4 = ln x + c


gives:

4yx4

⎛ ⎞⎜ ⎟⎝ ⎠ = ln x + c

i.e. 4

4

y ln x c4x

= +

or 4 4y 4x (ln x c)= +

3. Solve the differential equation: dy2x x 3ydx

= + , given that when x = 1, y = 1.

(i) Rearranging dy2x x 3ydx

= + gives: dy x 3ydx 2x

+= , which is homogeneous in x and y.


= +


gives: dv x 3vx x(1 3v) 1 3vv xdx 2x 2x 2

+ + ++ = = =

(iv) Separating the variables gives: dv 1 3v 1 3v 2v 1 vx vdx 2 2 2

+ + − += − = =

i.e. 2 1dv dx1 v x

=+


Integrating both sides gives: 2 1dv dx1 v x

=+∫ ∫

Hence, 2 ln(1+v) = ln x + c


gives: y2 ln 1x

⎛ ⎞+⎜ ⎟⎝ ⎠


When x = 1, y = 1, thus: 2 ln 2 ln1 c= + from which, c = 2 ln 2

Thus, the particular solution is: x y2lnx+⎛ ⎞

⎜ ⎟⎝ ⎠

= ln x + 2 ln 2

i.e. 2

2x yln ln x ln 2 ln 4x2+⎛ ⎞ = + =⎜ ⎟

⎝ ⎠

from which, 2x y 4x

x+⎛ ⎞ =⎜ ⎟

⎝ ⎠ i.e.

2

2

(x y) 4xx+

=

i.e. 2 3(x y) 4x+ =

5. Determine the particular solution of 3 3

2

dy x ydx xy

+= , given that x = 1 when y = 4.

(i) 3 3

2

dy x ydx xy

+= is homogeneous in x and y.


= +


gives: ( )

3 3 3 3 3 3

3 2 22 2

dv x v x x (1 v ) 1 vv xdx x v vx v x

+ + ++ = = =

(iv) Separating the variables gives: 3 3 3

2 2 2

dv 1 v 1 v v 1x vdx v v v

+ + −= − = = , i.e. 2 1v dv dx

x=

Integrating both sides gives 2 1v dv dxx

=∫ ∫

Hence, 3v

3 = ln x + c


gives: 3

3

y3x


When x = 1, y = 4, thus: 64 ln1 c3= + from which, c = 64

3



3

y3x

= ln x + 643

i.e. 3 3 64y 3x ln x3

⎛ ⎞= +⎜ ⎟⎝ ⎠

or ( )3 3y x 3ln x 64= +

6. Show that the solution of the differential equation: 3 2 2 3

2 2 3

dy y xy x y 5xdx xy x y 2x

− − −=

− − is of the form:

2

2

y 4y y 5x18ln ln x 422x x x

−⎛ ⎞+ + = +⎜ ⎟⎝ ⎠

, when x = 1 and y = 6.

(i) 3 2 2 3

2 2 3

dy y xy x y 5xdx xy x y 2x

− − −=

− − is homogeneous in x and y.


= +


gives:

( ) ( )

3 3 2 2 2 3 3 3 2 3 2

22 2 2 3 3 2

dv v x xv x x vx 5x x (v v v 5) v v v 5v xdx v v 2x v x x (vx) 2x x v v 2

− − − − − − − − −+ = = =

− −− − − −

(iv) Separating the variables gives: ( )3 2 23 2

2 2

v v v 5 v v v 2dv v v v 5x vdx v v 2 v v 2

− − − − − −− − −= − =

− − − −

= 2

v 5v v 2

−− −

i.e. 2v v 2v 5− −−

1dv dxx

=

Integrating both sides gives: 2v v 2 1dv dxv 5 x− −

=−∫ ∫

v + 4 2v 5 v v 2− − −

2v 5v− 4v – 2 4v – 20 18

Thus, 18 1v 4 dv dxv 5 x

+ + =−∫ ∫


Hence, 2v 4v 18ln(v 5)

2+ + − = ln x + c


gives: 2

2

y 4y y18ln 52x x x

⎛ ⎞+ + −⎜ ⎟⎝ ⎠


When x = 1, y = 6, thus: 36 24 618ln 52 1 1

⎛ ⎞+ + −⎜ ⎟⎝ ⎠

= ln 1 + c from which, c = 42


2

y 4y y 5x18ln2x x x

−⎛ ⎞+ + ⎜ ⎟⎝ ⎠

= ln x + 42


CHAPTER 48 LINEAR FIRST ORDER DIFFERENTIAL

EQUATIONS EXERCISE 183 Page 457

2. Solve the differential equation: dydx

= x( 1 – 2y)

dydx

= x( 1 – 2y) = x – 2xy i.e. dydx

+ 2xy = x from which, P = 2x and Q = x

Hence, 2x dx 2x dx

y e e x dx∫ ∫= ∫

i.e. 2 2x xye e x dx= ∫

i.e. 2 2x x1ye e c

2= +

or 2x1y ce

2−= +

4. Solve the differential equation: 3dyx 1 x 2ydx

⎛ ⎞+ = −⎜ ⎟⎝ ⎠

, given x = 1 when y = 3.

Since 3dyx 1 x 2ydx

⎛ ⎞+ = −⎜ ⎟⎝ ⎠

then 2dy 2y1 xdx x

+ = −

from which, 2dy 2y x 1dx x

+ = − i.e. P = 2x

and Q = 2x 1−

Hence, ( )2 2dx dx 2x xy e e x 1 dx∫ ∫= −∫

i.e. ( )2 ln x 2 ln x 2ye e x 1 dx= −∫

i.e. ( )2 2ln x ln x 2ye e x 1 dx= −∫

i.e. ( )2 2 2y x x x 1 dx= −∫ since ln Ae A=

i.e. ( )2 4 2y x x x dx= −∫

i.e. 5 3

2 x xyx c5 3

= − +

x = 1 when y = 3, hence, 1 13 c5 3

= − +

from which, c = 1 1 45 5 3 4733 5 15 15

+ −+ − = =


Hence, 5 3

2 x x 47yx5 3 15

= − +

i.e. 3

2

x x 47y5 3 15x

= − +

6. Solve the differential equation: dydx

+ x = 2y

Since dydx

+ x = 2y then dydx

- 2y = - x from which, P = -2 and Q = - x

Hence, 2dx 2dx

y e e ( x)dx− −∫ ∫= −∫

i.e. 2x 2xye e ( x)dx− −= −∫ (1)

Using integration by parts on 2xx e dx−∫ : Let u = x, then du 1dx

= and du = dx

and dv = 2xe dx− and v = 2x 2x1e dx e2

− −= −∫

Thus, 2x 2x 2x1 1x e dx (x) e e dx2 2

− − −⎛ ⎞= − − −⎜ ⎟⎝ ⎠∫ ∫

= 2x 2x 2x 2x1 1 1 1xe e dx xe e2 2 2 4

− − − −− + = − −∫

Substituting in (1) gives: 2x 2x 2x1 1ye xe e c2 4

− − −⎛ ⎞= − − − +⎜ ⎟⎝ ⎠

i.e. 2x 2x 2x1 1ye x e e c2 4

− − −= + +

or 2x

1 1 cy x2 4 e−= + +

i.e. 2x1 1y x ce2 4

= + +



2. Solve the differential equation: ( )dt sec t t sin t cos t sec tdtθ+ + θ = , given t = π when θ = 1.

Since ( )dt sec t t sin t cos t sec tdtθ+ + θ =

then d sec t cos t sec tsec t sin tdt t tθ ⎛ ⎞+ + θ =⎜ ⎟

⎝ ⎠

from which, P =

1 cos tsec t cos t 1 1cos tsec t sin t sin t tan t

t cos t t t

⎛ ⎞⎜ ⎟⎛ ⎞ ⎝ ⎠+ = + = +⎜ ⎟

⎝ ⎠ and Q = sec t

t

Hence, 1 1tan t dt tan t dtt t sec te e dt

t

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫θ = ∫

i.e. (lnsec t ln t ) (ln sec t ln t ) sec te e dtt

+ +θ = ∫

i.e. ln(t sec t ) ln(t sec t ) sec te e dtt

θ = ∫

i.e. 2sec tt sec t t sec t dt sec t dtt

θ = =∫ ∫

and t sec t tan t cθ = +

t = π when θ = 1, hence, π sec π = tan π + c

i.e. - π = c

Thus, t sec t tan tθ = − π

or

sin ttan t sin tcos t cos t1 1t sec t t sec t t tt t

cos t cos t

π π πθ = − = − = −

i.e. ( )1 sin t cos tt

θ = − π

4. Show that the solution of the differential equation ( )3dy 42 x 1 ydx (x 1)

− + =+

is: y = ( ) ( )4 2x 1 ln x 1+ + , given that x = 0 when y = 0

Since ( )3dy 42 x 1 ydx (x 1)

− + =+

from which, ( )3dy 4 y 2 x 1dx (x 1)

− = ++


i.e. P = 4x 1

−+

and Q = 32(x 1)+

Hence, 4 4dx dx 3x 1 x 1y e e 2(x 1) dx

− −+ +∫ ∫= +∫

i.e. 4ln(x 1) 4ln(x 1) 3y e e 2(x 1) dx− + − += +∫

i.e. 4 4ln(x 1) ln(x 1) 3y e e 2(x 1) dx

− −+ += +∫

i.e. 34 4

y 1 2(x 1) dx(x 1) (x 1)

= ++ +∫

and 4

y 2 dx 2ln(x 1) c(x 1) (x 1)

= = + ++ +∫

x = 0 when y = 0, hence, 0 = 2 ln 1 + c from which, c = 0

Thus, 4

y 2 ln(x 1)(x 1)

= ++

or y = ( ) ( )4 2x 1 ln x 1+ +

6. The equation dvdt

= -(av + bt), where a and b are constants, represents an equation of motion

when a particle moves in a resisting medium. Solve the equation for v given that v = u when

t = 0.

Since dvdt

= -(av + bt) then dvdt

+ av = -bt from which, P = a and Q = -bt

Hence, a dt a dt

v e e ( bt)dt∫ ∫= −∫

i.e. at atv e b t e= − ∫ (1)

Using integration by parts on att e dt∫ : Let u = t, then du 1dt

= and du = dt

and dv = ate dt and v = at a t1e dt ea

=∫

Thus, a t a t a t1 1t e dt (t) e e dta a

⎛ ⎞= −⎜ ⎟⎝ ⎠∫ ∫ = a t a t

2

t 1e ea a

−

Substituting in (1) gives: at a t a t2

t 1ve b e e ca a

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

v = u when t = 0, hence, u = 2

b ca

+ from which, c = u - 2

ba


Thus, at a t a t a t at2 2 2 2

t 1 b bt b bve b e e u e e ua a a a a a

⎛ ⎞= − − + − = − + + −⎜ ⎟⎝ ⎠

or at at2 2 2 2

bt b b b bt bv u e or u ea a a a a a

− −⎛ ⎞ ⎛ ⎞= − + + − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

7. In an alternating current circuit containing resistance R and inductance L the current i is given

by: Ri + 0diL E sin tdt

= ω . Given i = 0 when t = 0, show that the solution of the equation is given

by: i = ( )R t

0 0 L2 2 2 2 2 2

E E LR sin t L cos t eR L R L

−ω⎛ ⎞ ⎛ ⎞ω −ω ω +⎜ ⎟ ⎜ ⎟+ω +ω⎝ ⎠ ⎝ ⎠

Since Ri + 0diL E sin tdt

= ω then 0diL E sin t Ridt

= ω − and 0Edi Risin tdt L L

= ω −

i.e. 0Edi Ri sin tdt L L+ = ω from which, P = R

L and Q = 0E sin t

Lω

Hence, R Rdt dt 0L L Ei e e sin t dt

L⎛ ⎞ ⎛ ⎞∫ ∫= ω⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠∫

i.e. R t R t

0L L Ei e e sin t dtL

⎛ ⎞= ω⎜ ⎟⎝ ⎠∫

i.e. R t R t

0L LEi e e sin t dtL⎡ ⎤

= ω⎢ ⎥⎣ ⎦∫ (1)

From page XX of textbook, ( )ax

ax2 2

ee sin bx dx a sin bx bcos bx ca b

= − ++∫

Hence,

R tRt LL

22

e Re sin t dt sin t cos t cLR

L

⎛ ⎞ω = ω −ω ω +⎜ ⎟⎝ ⎠⎛ ⎞ +ω⎜ ⎟

⎝ ⎠

∫

Substituting into (1) gives:

R tRt L

0L2

22

E e Ri e sin t cos t cL LR

L

⎛ ⎞= ω −ω ω +⎜ ⎟⎛ ⎞ ⎝ ⎠+ω⎜ ⎟⎝ ⎠

R tL

02 2 2

2

E e R sin t cos t cL LR L

L

⎛ ⎞= ω −ω ω +⎜ ⎟⎛ ⎞+ω ⎝ ⎠⎜ ⎟⎝ ⎠

i.e. ( )

R tRt L

0L2 2 2

E Le Ri e sin t cos t cLR L

⎛ ⎞= ω −ω ω +⎜ ⎟+ω ⎝ ⎠


i = 0 when t = 0, hence, 0 = 02 2 2

E L ( ) cR L

−ω ++ω

from which, c = 02 2 2

E LR L

ω+ω

Thus, ( ) ( )

R tRt L

0 0L2 2 2 2 2 2

E Le E LRi e sin t cos tLR L R L

ω⎛ ⎞= ω −ω ω +⎜ ⎟+ω +ω⎝ ⎠

i.e. ( ) ( )

R t0 0 L

2 2 2 2 2 2

E L E LRi sin t cos t eLR L R L

−ω⎛ ⎞= ω −ω ω +⎜ ⎟+ω +ω⎝ ⎠

or ( ) ( )R t

0 0 L2 2 22 2 2

E E Li R sin t Lcos t e

R LR L−ω⎛ ⎞= ω −ω ω + ⎜ ⎟+ ω+ ω ⎝ ⎠

8. The concentration, C, of impurities of an oil purifier varies with time t and is described by the

equation dCa b dm Cmdt

= + − , where a, b, d and m are constants. Given C = c0 when t = 0, solve

the equation and show that: m t m ta a

0bC d 1 e c em

− −⎛ ⎞⎛ ⎞= + − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Since dCa b dm Cmdt

= + − then dC b d m Cmdt a a a

= + −

and dC m b d mCdt a a a

+ = + from which, P = ma

and Q = b d ma a+

Hence, m mdt dta a b d mCe e dt

a a⎛ ⎞∫ ∫= +⎜ ⎟⎝ ⎠∫

i.e. m t mta a b d mCe e dt

a a⎛ ⎞= +⎜ ⎟⎝ ⎠∫

i.e.

mtm t aa a e b d mCe k

m a a⎛ ⎞= + +⎜ ⎟⎝ ⎠

C = 0c when t = 0, hence, 0c = a b d m bk d km a a m⎛ ⎞+ + = + +⎜ ⎟⎝ ⎠

from which, k = 0bc dm

− −

Thus,

mtm t aa

0a e b d m bCe c d

m a a m⎛ ⎞= + + − −⎜ ⎟⎝ ⎠

i.e. m ta

0a b d m bC c d em a a m

−⎛ ⎞ ⎛ ⎞= + + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

i.e. m ta

0b bC d c d em m

−⎛ ⎞ ⎛ ⎞= + + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


i.e. m t m ta a

0b bC d d e c em m

− −⎛ ⎞ ⎛ ⎞= + − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or mt mta a

0bC d 1 e c em

− −⎛ ⎞⎛ ⎞= + − +⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

9. The equation of motion of a train is given by: ( )tdvm mk 1 e mcvdt

−= − − , where v is the speed, t

is the time and m, k and c are constants. Determine the speed, v, given v = 0 at t = 0.

Since ( )tdvm mk 1 e mcvdt

−= − − then ( )tdv k 1 e cvdt

−= − −

and ( )tdv cv k 1 edt

−+ = − from which, P = c and Q = ( )tk 1 e−−

Hence, cdt cdt tv e e k(1 e )dt−∫ ∫= −∫

i.e. ( )c t c t t c t c t tv e e k(1 e )dt k e e dt− −= − = −∫ ∫

i.e. c t t (c 1)

c t e ev e k zc c 1

−⎡ ⎤= − +⎢ ⎥−⎣ ⎦

where z is the constant of integration

v = 0 when t = 0, hence, 0 = k k zc c 1− +

−

from which, z = k k ck k(c 1) kc 1 c c(c 1) c(c 1)

− −− = =

− − −

Thus, c t t (c 1)

c t e e kv e kc c 1 c(c 1)

−⎡ ⎤= − +⎢ ⎥− −⎣ ⎦

i.e. v = t

ctk k e k ec c 1 c(c 1)

−−⎛ ⎞

− + ⎜ ⎟− −⎝ ⎠

or t ct1 e ev k

c c 1 c(c 1)

− −⎧ ⎫= − +⎨ ⎬− −⎩ ⎭


CHAPTER 49 NUMERICAL METHODS FOR FIRST ORDER

DIFFERENTIAL EQUATIONS EXERCISE 185 Page 464

1. Use Euler’s method to obtain a numerical solution of the differential equation dy y3dx x

= − , with

the initial conditions that x = 1 when y = 2, for the range x = 1.0 to x = 1.5 with intervals of 0.1.

Draw the graph of the solution in this range.

dy yy ' 3dx x

= = −

If initially x0 = 1 and y0 = 2, (and h = 0.1), then (y')0 = 231

− = 1

Line 1 in the table below is completed with x = 1.0 and y = 2

For line 2, where x0 = 1.1 and h = 0.1:

y1 = y0 + h(y')0 = 2.0 + (0.1)(1) = 2.1

and (y')0 = 3 - 0

0

yx

= 3 - 2.11.1

= 1.0909

For line 3, where x0 = 1.2: y1 = y0 + h(y')0 = 2.1 + (0.1)(1.0909) = 2.209091

and (y')0 = 3 - 0

0

yx

= 3 - 2.2090911.2

= 1.159091


The remaining lines of the table are completed in a similar way.

A graph of the solution of dy y3dx x

= − , with initial conditions x = 1 and y = 2 is shown below.


3. (a) The differential equation dy y1dx x

+ = − has the initial conditions that y = 1 at x = 2. Produce a

numerical solution of the differential equation in the range x = 2.0(0.1)2.5.

(b) If the solution of the differential equation by an analytical method is given by y = 4 xx 2− ,

determine the percentage error at x = 2.2.

(a) dy y1dx x

+ = − i.e. dy yy ' 1dx x

= = − −

If initially x0 = 2.0 and y0 = 1, (and h = 0.1), then (y')0 = 112.0

− − = -1.5



y1 = y0 + h(y')0 = 1 + (0.1)(-1.5) = 0.85

and (y')0 = -1 - 0

0

yx

= -1 - 0.852.1

= -1.40476

For line 3, where x0 = 2.2: y1 = y0 + h(y')0 = 0.85 + (0.1)(-1.40476) = 0.709524


and (y')0 = -1 - 0

0

yx

= -1 - 0.7095242.2

= -1.322511

For line 4, where x0 = 2.3: y1 = y0 + h(y')0 = 0.709524 + (0.1)(-1.322511) = 0.577273


(b) If y = 4 xx 2− then when x = 2.2, y = 0.718182

From the Euler method, when x = 2.2, y = 0.709524

Hence, percentage error = 0.718182 0.709524 0.008658100% 100%0.718182 0.718182

−× = ×

= 1.206%

4. Use Euler’s method to obtain a numerical solution of the differential equation dy 2yxdx x

= − ,

given the initial conditions that y = 1 when x = 2, in the range x = 2.0(0.2)3.0.

If the solution of the differential equation is given by y =2x

4, determine the percentage error by

using Euler’s method when x = 2.8.

(a) dy 2yy ' xdx x

= = −

If initially x0 = 2.0 and y0 = 1, (and h = 0.2), then (y')0 = 2(1)2.02.0

− = 1.0



y1 = y0 + h(y')0 = 1 + (0.2)(1.0) = 1.2

and (y')0 = 2.2 - 2(1.2)2.2

= 1.109091



and (y')0 = 2.4 - 2(1.4218182)2.4

= 1.2151515



(b) If y =2x

4, then when x = 2.8, y = 1.96

From the Euler method, when x = 2.8, y = 1.928718

Hence, percentage error = 1.96 1.928718 0.031282100% 100%1.96 1.96−

× = ×

= 1.596%



1. Apply the Euler-Cauchy method to solve the differential equation dy y3dx x

= − for the range

1.0(0.1)1.5, given the initial conditions that x = 1 when y = 2.

dy y 'dx

=y3x

= −

x 0 = 1, y 0 = 2 and h = 0.1

(y ′) 0 0

0

y3x

= −23 11

= − =

x1 = 1.1 and from equation (3), page 465, y1P = y 0 + h(y ′) 0 = 2 + 0.1(1) = 2.1

y1C = y 0 + 1

2h [ (y ′) 0 + f(x1 , y

1P ) ] = y 0 + 12

h [ (y ′) 0 + 1P

1

y3

x− ]

= 2 + 12

(0.1) [ 1 + 2.131.1

− ]

= 2.10454546

(y ′)1 1C

1

y 2.104545463 3x 1.1

= − = − = 1.08677686

Thus the first two lines of the Table below has been completed

x y y ′ 1.0 2 1 1.1 2.10454546 1.08677686 1.2 2.216666672 1.152777773 1.3 2.33461539 1.204142008 1.4 2.457142859 1.244897958 1.5 2.5883333335

For line 3, x1 = 1.2

y1P = y 0 + h(y ′) 0 = 2.10454546 + 0.1(1.08677686) = 2.213223146

y1C = y 0 + 1

2h [ (y ′) 0 + 1P

1

y3

x− ]


= 2.10454546 + 12

(0.1) [ 1.08677686 + 2.21322314631.2

− ] = 2.216666672

(y ′)11C

1

y 2.2166666723 3x 1.2

= − = − = 1.152777773


2. Solving the differential equation in Problem 1 by the integrating method gives y = 3 1x2 2x

+ .

Determine the percentage error, correct to 3 significant figures, when x = 1.3 using (a) Euler’s

method (see Table 49.4, page 464) and (b) the Euler-Cauchy method.

If y = 3 1x2 2x

+ then when x = 1.3, y = 2.334615385

(a) By Euler’s method, when x = 1.3, y = 2.325000

Percentage error = 2.334615385 2.3250 100%2.334615385

−× = 0.412%

(b) By the Euler-Cauchy method, when x = 1.3, y = 2.33461539

Percentage error = 2.334615385 2.33461539 100%2.334615385

−× = 0.000000214%

3. (a) Apply the Euler-Cauchy method to solve the differential equation dy x ydx

− = for the range

x = 0 to x = 0.5 in increments of 0.1, given the initial conditions that when x = 0, y = 1.

(b) The solution of the differential equation in part (a) is given by y = x2e x 1− − . Determine the

percentage error, correct to 3 decimal places, when x = 0.4.

(a) dy y 'dx

= = y + x

x 0 = 0, y 0 = 1 and h = 0.1

(y ′) 0 = 1 + 0 = 1

x1 = 0.1 and from equation (3), page 465, y1P = y 0 + h(y ′) 0 = 1 + 0.1(1) = 1.1


y1C = y 0 + 1

2h [ (y ′) 0 + f(x1 , y

1P ) ] = y 0 + 12

h [ (y ′) 0 + y1P + x1 ]

= 1 + 12

(0.1) [ 1 + 1.1 + 0.1 ]

= 1.11

(y ′)1 = y1C + 0.1 = 1.11 + 0.1 = 1.21

Thus the first two lines of the Table below has been completed

For line 3, x1 = 0.2

y1P = y 0 + h(y ′) 0 = 1.11 + 0.1(1.21) = 1.231

y1C = y 0 + 1

2h [ (y ′) 0 + y

1P + x1 ]

= 1.11 + 12

(0.1) [ 1.21 + 1.231 + 0.2 ] = 1.24205

(y ′)1 = y1C + 0.2 = 1.24205+ 0.2 = 1.44205


(b) If y = x2e x 1− − then when x = 0.4, y = 1.583649395

By the Euler-Cauchy method, when x = 0.4, y = 1.581804

Hence, the percentage error = 1.583649395 1.581804 100%1.583649395

−× = 0.117%



2. Obtain a numerical solution of the differential equation: 1 dy 2y 1x dx

+ = using the Runge-Kutta

method in the range x = 0(0.2)1.0, given the initial conditions that x = 0 when y = 1.

If 1 dy 2y 1x dx

+ = then 1 dy 1 2yx dx

= − and dy x(1 2y)dx

= −

1. 0x = 0, 0y = 1 and since h = 0.2, and the range is from x = 0 to x = 1.0, then

1 2 3 4 5x 0.2, x 0.4, x 0.6, x 0.8, and x 1.0= = = = =

Let n = 0 to determine 1y :

2. ( )1 0 0k f x , y= = f (0, 1); since dy x(1 2y)dx

= − , f (0, 1) = 0(1 - 2) = 0

3. ( )2 0 0 1h h 0.2 0.2k f x , y k f 0 , 1 (0) f 0.1, 12 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0.1(1 - 2) = -0.1

4. 3 0 0 2h h 0.2 0.2k f x , y k f 0 , 1 ( 0.1)2 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ( )f 0.1, 0.99

= 0.1(1 – 1.98) = -0.098

5. ( ) ( )4 0 0 3k f x h, y hk f 0 0.2, 1 0.2( 0.098)= + + = + + − = f (0.2, 0.9804)

= 0.2(1 - 2(0.9804)) = -0.19216

6. n 1y + = ny + 1 2 3 4h k 2k 2k k6

+ + + and when n = 0:

1y = 0y + 1 2 3 4h k 2k 2k k6

+ + + = 1 + 0.2 0 2( 0.1) 2( 0.098) 0.192166

+ − + − + −

= 1 + 0.2 0.588166

− = 0.980395


2. ( )1 1 1k f x , y= = f (0.2, 0.980395); since dydx

= x(1 – 2y),

f (0.2, 0.980395) = 0.2(1 – 2(0.980395)) = -0.192158

3. ( )2 1 1 1h h 0.2 0.2k f x , y k f 0.2 ,0.980395 ( 0.192158) f 0.3, 0.96117922 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0.3(1 – 2(0.9611792) = -0.27670752


4. 3 1 1 2h h 0.2 0.2k f x , y k f 0.2 , 0.980395 ( 0.27670752)2 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ( )f 0.3, 0.952724248 = 0.3(1 – 2(0.952724248)) = -0.271634548

5. ( ) ( )4 1 1 3k f x h, y hk f 0.2 0.2, 0.980395 0.2( 0.271634548)= + + = + + −

= f (0.4, 0.92606809) = 0.4(1 – 2(0.92606809)) = -0.340854472

6. n 1y + = ny + 1 2 3 4h k 2k 2k k6

+ + + and when n = 1:

2y = 1y + 1 2 3 4h k 2k 2k k6

+ + +

= 0.980395 + 0.2 0.192158 2( 0.27670752) 2( 0.271634548) 0.3408544726

− + − + − + −

= 0.980395 + 0.2 1.6296966086

− = 0.926071779

This completes the third row of the table below. In a similar manner 3 4y , y and 5y can be calculated.

3. (a) The differential equation dy y1dx x

+ = − has the initial conditions that y = 1 at x = 2. Produce a

numerical solution of the differential equation, correct to 6 decimal places, using the Runge-

Kutta method in the range x = 2.0(0.1)2.5.

(b) If the solution of the differential equation by an analytical method is given by: y = 4 xx 2−

determine the percentage error at x = 2.2

(a) If dy y1dx x

+ = − then dy y y1 1dx x x

⎛ ⎞= − − = − +⎜ ⎟⎝ ⎠

1. 0x = 2, 0y = 1 and since h = 0.1, and the range is from x = 2.0 to x = 2.5, then

1 2 3 4 5x 2.1, x 2.2, x 2.3, x 2.4, and x 2.5= = = = =



2. ( )1 0 0k f x , y= = f (2, 1); since, dy y 1dx x

⎛ ⎞= − +⎜ ⎟⎝ ⎠

, f (2, 1) = 1 12

⎛ ⎞− +⎜ ⎟⎝ ⎠

= -1.5

3. ( )2 0 0 1h h 0.1 0.1k f x , y k f 2.0 ,1.0 ( 1.5) f 2.05, 0.9252 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0.925 12.05

⎛ ⎞− +⎜ ⎟⎝ ⎠

= -1.451219512

4. 3 0 0 2h h 0.1 0.1k f x , y k f 2.0 , 1.0 ( 1.451219512)2 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ( )f 2.05, 0.927439024

= 0.927439024 12.05

⎛ ⎞− +⎜ ⎟⎝ ⎠

= -1.45240928

5. ( ) ( )4 0 0 3k f x h, y hk f 2.0 0.1, 1.0 0.1( 1.45240928)= + + = + + − = f (2.1, 0.854759072)

= 0.854759072 12.1

⎛ ⎞− +⎜ ⎟⎝ ⎠

= -1.40702813

6. n 1y + = ny + 1 2 3 4h k 2k 2k k6

+ + + and when n = 0:

1y = 0y + 1 2 3 4h k 2k 2k k6

+ + +

= 1.0 + 0.1 1.5 2( 1.451219512) 2( 1.45240928) ( 1.40702813)6

− + − + − + −

= 1.0 + 0.1 8.7142857146

− = 0.854762


2. ( )1 1 1k f x , y= = f (2.1, 0.854762); since dy y 1dx x

⎛ ⎞= − +⎜ ⎟⎝ ⎠

,

f (2.1, 0.854762) = 0.854762 12.1

⎛ ⎞− +⎜ ⎟⎝ ⎠

= - 1.407029524

3. 2 1 1 1h h 0.1 0.1k f x , y k f 2.1 ,0.854762 ( 1.407029524)2 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= f(2.15, 0.784410523) = 0.784410523 12.15

⎛ ⎞− +⎜ ⎟⎝ ⎠

= - 1.364842104

4. 3 1 1 2h h 0.1 0.1k f x , y k f 2.1 , 0.854762 ( 1.364842104)2 2 2 2

⎛ ⎞ ⎛ ⎞= + + = + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ( )f 2.15, 0.786519894 = 0.786519894 12.15

⎛ ⎞− +⎜ ⎟⎝ ⎠

= - 1.365823207


5. ( ) ( )4 1 1 3k f x h, y hk f 2.2, 0.854762 0.1( 1.365823207)= + + = + −

= f (2.2, 0.718179679) = 0.718179679 12.2

⎛ ⎞− +⎜ ⎟⎝ ⎠

= - 1.326445309

6. n 1y + = ny + 1 2 3 4h k 2k 2k k6

+ + + and when n = 1:

2y = 1y + 1 2 3 4h k 2k 2k k6

+ + +

= 0.854762 + 0.1 1.407029524 2( 1.364842104) 2( 1.365823207) 1.3264453096

− + − + − −

= 0.854762 + 0.1 8.1948054556

− = 0.718182

This completes the third row of the Table below. In a similar manner 3 4y , y and 5y can be calculated and the results are as shown.

(b) If y = 4 xx 2− when x = 2.2, y = 0.718182

By the Runge-Kutta method, when x = 2.2, y = 0.718182 also. Hence, there is no error.


CHAPTER 50 SECOND ORDER DIFFERENTIAL EQUATIONS OF

THE FORM 2

2

d y dya b cy 0dx dx

+ + = EXERCISE 188 Page 477

3. Determine the general solution of 2

2

d y dy2 5y 0dx dx

+ + =

2

2

d y dy2 5y 0dx dx

+ + = in D-operator form is: ( )2D 2D 5 y 0+ + =

The auxiliary equation is: 2m 2m 5 0+ + =

and 22 2 4(1)(5) 2 16 2 j4m 1 j2

2(1) 2 2− ± − − ± − − ±

= = = = − ±

Hence, the general solution is: xy e Acos 2x Bsin 2x−= +

5. Find the particular solution of 2

2

d y dy4 5 y 0dt dt

− + = when at t = 0, y = 1 and dydt

= -2.

2

2

d y dy4 5 y 0dx dt

− + = in D-operator form is: ( )24D 5D 1 y 0− + =

The auxiliary equation is: 24m 5m 1 0− + =

i.e. (4m – 1)(m – 1) = 0

from which, 1m4

= and m = 1

Hence, the general solution is: 1 t t4y Ae Be= +

At t = 0, y = 1, hence, 1 = A + B (1)

1 t t4dy 1 Ae Be

dt 4= +

At t = 0, dydt

= -2, hence, - 2 = 1 A B4

+ (2)

(1) – (2) gives: 3 = 3 A4

from which, A = 4

From equation (1), when A = 4, 1 = 4 + B from which, B = -3

Hence, the particular solution is: 1t t4y 4e 3e= −



2

d x dx6 9x 0dt dt

− + = when at t = 0, x = 2 and dxdt

= 0.

2

2

d x dx6 9x 0dt dt

− + = in D-operator form is: ( )2D 6D 9 x 0− + =

The auxiliary equation is: 2m 6m 9 0− + =

i.e. (m – 3)(m – 3) = 0

from which, m = 3 twice

Hence, the general solution is: ( ) 3tx At B e= +

At t = 0, x = 2, hence, 2 = B

( )( ) ( )3t 3tdx At B 3e e Adt

= + +

At t = 0, dxdt

= 0, hence, 0 = 3B + A and since B = 2, A = -6

Hence, the particular solution is: ( ) 3tx 6t 2 e= − + or ( ) 3tx 2 1 3t e= −


2

d y dy6 13y 0dx dx

+ + = when at x = 0, y = 4 and dydx

= 0.

2

2

d y dy6 13y 0dx dx



and 26 6 4(1)(13) 6 16 6 j4m 3 j22(1) 2 2

− ± − − ± − − ±= = = = − ±

Hence, the general solution is: 3xy e A cos 2x Bsin 2x−= +

At x = 0, y = 4, hence, 4 = A

( )( ) ( )( )3x 3xdy e 2Asin 2x 2Bcos 2x A cos 2x Bsin 2x 3edx

− −= − + + + −

At x = 0, dydx

= 0, hence, 0 = 2B – 3A and since A = 4, B = 6

Hence, the particular solution is: 3xy e 4cos 2x 6sin 2x−= +

or 3xy 2e 2cos 2x 3sin 2x−= +


EXERCISE 189 Page 480 1. The charge, q, on a capacitor in a certain electrical circuit satisfies the differential equation

2

2

d q dq4 5q 0dt dt

+ + = . Initially (i.e. when t = 0), q = Q and dqdt

= 0. Show that the charge in the

circuit can be expressed as: q = 2t5 Qe sin(t 0.464)− +

2

2

d q dq4 5q 0dt dt



and 24 4 4(1)(5) 4 4 4 j2m 2 j

2(1) 2 2− ± − − ± − − ±

= = = = − ±

Hence, the general solution is: 2tq e A cos t Bsin t−= +

At t = 0, q = Q, hence, Q = A

( )( ) ( )( )2t 2tdq e Asin t Bcos t A cos t Bsin t 2edt

− −= − + + + −

At t = 0, dqdt

= 0, hence, 0 = B – 2A and since A = Q, B = 2Q

Hence, the particular solution is: 2tq e Q cos t 2Qsin t−= +

i.e. 2tq Qe cos t 2sin t−= +

Let cos t + 2 sin t = R sin(t + α)

= R[sin t cos α + cos t sin α]

= R cos α (sin t) + R sin α (cos t)



There is only one quadrant where both sine and cosine are positive, i.e. the first quadrant.

From the diagram below, R = 2 12 1 5+ =


and α = 1 1tan 26.565 or 0.464 rad2

− = °

Hence, cos t + 2 sin t = 5 sin(t + 0.464)

Since, 2tq Q e cos t 2sin t−= + then 2tq 5 Qe sin(t 0.464)−= +

3. The motion of the pointer of a galvanometer about its position of equilibrium is represented

by the equation 2

2

d dI K F 0dt dtθ θ+ + θ =

If I, the moment of inertia of the pointer about its pivot, is 5 ×10-3, K, the resistance due to

friction at unit angular velocity, is 2 ×10-2 and F, the force on the spring necessary to produce

unit displacement, is 0.20, solve the equation for θ in terms of t given that when t = 0, θ = 0.3

and ddtθ = 0.

2

2

d dI K F 0dt dtθ θ+ + θ = in D-operator form is: ( )2I D K D F 0+ + θ =

The auxiliary equation is: 2I m K m F 0+ + =

and ( ) ( ) ( )( )

( )

22 2 32

3

2 10 2 10 4 5 10 0.2K K 4I F 0.02 0.0036m2I 0.012 5 10

− − −

−

− × ± × − ×− ± − − ±= = =

×

= 0.02 j0.06 2 j60.01

− ±= − ±

Hence, the general solution is: 2te A cos 6t Bsin 6t−θ = +

At t = 0, θ = 0.3, hence, 0.3 = A

( )( ) ( )( )2t 2td e 6Asin 6t 6Bcos 6t A cos t Bsin t 2edt

− −θ= − + + + −

At t = 0, ddtθ = 0, hence, 0 = 6B – 2A and since A = 0.3, B = 0.1


Hence, the particular solution is: 2te 0.3cos 6t 0.1sin 6t−θ = +

4. Determine an expression for x for a differential equation 2

22

d x dx2n n x 0dt dt

+ + = which represents

a critically damped oscillator, given that at time t = 0, x = s and dxdt

= u.

2

22

d x dx2n n x 0dt dt

+ + = in D-operator form is: ( )2 2D 2nD n x 0+ + =

The auxiliary equation is: 2 2m 2n m n 0+ + =

i.e. (m + n)(m + n) = 0

from which, m = -n twice

Hence, the general solution is: ( ) n tx At B e−= +

At t = 0, x = s, hence, s = B

( )( ) ( )n t n tdx At B ne e Adt

− −= + − +

At t = 0, dxdt

= u, hence, u = -nB + A hence A = u + nB = u + ns

Hence, the particular solution is: ( ) ntx u ns t s e−= + + or ( ) ntx s u ns t e−= + +

5. 2

2

d i di 1L R i 0dt dt C

+ + = is an equation representing current i in an electric circuit. If inductance L

is 0.25 henry, capacitance C is 29.76 × 10-6 farads and R is 250 ohms, solve the equation for i

given the boundary conditions that when t = 0, i = 0 and didt

= 34.

2

2

d i di 1L R i 0dt dt C

+ + = in D-operator form is: 2 1L D R D i 0C

⎛ ⎞+ + =⎜ ⎟⎝ ⎠

The auxiliary equation is: 2 1L m R m 0C

+ + =

and ( )

2 26

4L 4(0.25)R R 250 250 250 170C 29.76 10m2L 2 0.25 0.5

−− ± − − ± − − ±×= = = = -160 or -840

Hence, the general solution is: 160t 840ti A e Be− −= +


When t =0, i = 0, hence, 0 = A + B (1)

160t 840tdi 160Ae 840Bedt

− −= − −

When t = 0, didt

= 34, hence, 34 = -160A – 840B (2)

160 × (1) gives: 0 = 160A + 160B (3)

(2) + (3) gives: 34 = -680B from which, B = - 120

From equation (1), A = 120

Hence, 160t 840t1 1i e e20 20

− −= − or ( )160t 840t1i e e20

− −= −

6. The displacement s of a body in a damped mechanical system, with no external forces, satisfies

the following differential equation: 2

2

d s ds2 6 4.5s 0dt dt

+ + = where t represents time. If initially,

when t = 0, s = 0 and dsdt

= 4, solve the differential equation for s in terms of t.

2

2

d s ds2 6 4.5s 0dt dt

+ + = in D-operator form is: ( )22D 6D 4.5 s 0+ + =

The auxiliary equation is: 22m 6m 4.5 0+ + =

or 24m 12m 9 0+ + =

i.e. (2m + 3)(2m + 3) = 0

from which, 3m2

= − twice

Hence, the general solution is: ( )3 t2s At B e

−= +

At t = 0, s = 0, hence, 0 = B

( )3 3t t2 2ds 3At B e e (A)

dt 2− −⎛ ⎞ ⎛ ⎞

= + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

At t = 0, dsdt

= 4, hence, 4 = 3 B A2

− + and since B = 0, A = 4

Hence, the particular solution is: 3t2s 4t e

−=


CHAPTER 51 SECOND ORDER DIFFERENTIAL EQUATIONS OF

THE FORM 2

2

d y dya b cy f (x)dx dx

+ + = EXERCISE 190 Page 483

2. Find the general solution of: 2

2

d y dy6 4 2y 3x 2dx dx

+ − = −

2

2


+ − = − in D-operator form is: ( )26D 4D 2 y 3x 2+ − = −

Auxiliary equation is: 2 26m 4m 2 0 i.e. 3m 2m 1 0+ − = + − =

i.e. (3m – 1)(m + 1) = 0

from which, 1m3

= and m = -1

Hence, the complementary function, C.F., 1x x3u Ae Be−= +

Let the particular integral, P.I., v = ax + b

then ( )( )26D 4D 2 ax b 3x 2+ − + = −

D(ax + b) = a and 2D (ax b) D(a) 0+ = =

Hence, 0 + 4a – 2ax – 2b = 3x – 2

from which, - 2a = 3 and a = 32

−

and 4a – 2b = - 2 and b = 2a + 1 = - 2

Hence, P.I., v = 32

− x – 2

and the general solution, y = u + v = 1 x x3 3Ae Be 2 x

2−+ − −


2


− + = − given that when x = 0, y = 0 and

dy 4dx 3

= − .

2

2


− + = − in D-operator form is: ( )29D 12D 4 y 3x 1− + = −


Auxiliary equation is: 29m 12m 4 0− + =

i.e. (3m – 2)(3m - 2) = 0

from which, 2m3

= twice

Hence, C.F., ( )2 x3u Ax B e= +

Let the particular integral, P.I., v = ax + b

then ( )( )29D 12D 4 ax b 3x 1− + + = −

D(ax + b) = a and 2D (ax b) D(a) 0+ = =

Hence, 0 - 12a + 4ax + 4b = 3x – 1

from which, 4a = 3 and a = 34

and -12a + 4b = - 1 i.e. -9 + 4b = - 1 and b = 2

Hence, P.I., v = 3 x 24

+

and the general solution, y = u + v = ( )2x3 3Ax B e x 2

4+ + +

x = 0 and y = 0, hence, 0 = B + 2 from which, B = - 2

2 2x3 3dy 2 3(Ax B) e e (A)

dx 3 4⎛ ⎞ ⎛ ⎞

= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x = 0 and dy 4dx 3

= − , hence, 4 2 3B A3 3 4

− = + +

and since B = - 2, 4 4 3A3 3 4

− = − + +

from which, A = 34

−

Hence, y = 2 x33 3x 2 e x 2

4 4⎛ ⎞− − + +⎜ ⎟⎝ ⎠

i.e. y = 2x33 32 1 x e 2 x

4 4⎛ ⎞− + + +⎜ ⎟⎝ ⎠


5. The charge q in an electric circuit at time t satisfies the equation 2

2

d q dq 1L R q Edt dt C

+ + = , where

L, R, C and E are constants. Solve the equation given L = 2 H, C = 200 × 10-6 F and E = 250 V,

when (a) R = 200 Ω and (b) R is negligible. Assume that when t = 0, q = 0 and dqdt

= 0

(a)2

2

d q dq 1L R q Edt dt C

+ + = in D-operator form is: 2 1L D R D q EC

⎛ ⎞+ + =⎜ ⎟⎝ ⎠


+ + =

and

2 26

4L 4(2)R R 200 200 200 0C 200 10m 502L 4 4

−− ± − − ± − − ±×= = = = −

Hence, C.F., ( ) 50tu At B e−= +

Let the particular integral, P.I., v = k

then ( )2 1LD RD k EC

⎛ ⎞+ + =⎜ ⎟⎝ ⎠

D(k) = 0 and 2D (k) D(0) 0= =

Hence, 1 (k) EC

= and k = CE = ( )( )6 1200 10 250 0.05 or20

−× =

Hence, P.I., v = 120

and the general solution, y = u + v = ( ) 50t 1At B e20

−+ +

t = 0 and q = 0, hence, 0 = B + 120

from which, B = - 120

( ) ( )50t 50 tdq (At B) 50e e (A)dt

− −= + − +

t = 0 and dq 0dt

= , hence, 0 50B A= − + i.e. 0 = 150 A20

⎛ ⎞− − +⎜ ⎟⎝ ⎠

i.e. A = 52

−

Hence, q = 50 t5 1 1t e2 20 20

−⎛ ⎞− − +⎜ ⎟⎝ ⎠

i.e. q = 50t1 5 1t e20 2 20

−⎛ ⎞− +⎜ ⎟⎝ ⎠


(b) When R = 0,

2 26

4L 4(2)R R 0 0 0 j200C 200 10m 0 j502L 4 4

−− ± − − ± − ±×= = = = ±

and q = (A cos 50 t + B sin 50t) + 120

q = 0 and t = 0, hence, 0 = A + 120

i.e. A = - 120

( )dq 50Asin 50t 50Bcos50tdt

= − +

t = 0 and dq 0dt

= , hence, 0 = 50B i.e. B = 0

Thus, q = - 120

cos 50t + 120

i.e. ( )1q 1 cos 50t20

= −

6. In a galvanometer the deflection θ satisfies the differential equation 2

2

d d4 4 8dt dtθ θ+ + θ = . Solve

the equation for θ given that when t = 0, θ = ddtθ = 2

2

2

d d4 4 8dt dtθ θ+ + θ = in D-operator form is: ( )2D 4D 4 8+ + θ =

Auxiliary equation is: 2m 4m 4 0+ + =

i.e. (m + 2)(m + 2) = 0

from which, m = - 2 twice

Hence, C.F., ( ) 2tu At B e−= +

Let the particular integral, P.I., v = k

then ( )2D 4D 4 k 8+ + =

D(k) = 0 and 2D (k) D(0) 0= =

Hence, 4k = 8 from which, k = 2

Hence, P.I., v = 2


and the general solution, θ = u + v = ( ) 2tAt B e 2−+ +

t = 0 and θ = 2, hence, 2 = B + 2 from which, B = 0

( ) ( )2t 2td (At B) 2e e (A)dt

− −θ= + − +

x = 0 and d 2dtθ= , hence, 2 2B A= − + from which, A = 2

Hence, θ = 2t2 t e 2− +

i.e. θ = ( )2t2 t e 1− +




x2

d y dy3 4y 3edx dx

−− − =

2

x2

d y dy3 4y 3edx dx

−− − = in D-operator form is: ( )2 xD 3D 4 y 3e−− − =

Auxiliary equation is: 2m 3m 4 0− − =

i.e. (m - 4)(m + 1) = 0

from which, m = 4 and m = -1

Hence, C.F., 4x xu Ae Be−= +

As xe− appears in the C.F. and in the right hand side of the differential equation, let the particular

integral, P.I., v = xk x e−

then ( )( )2 x xD 3D 4 kxe 3e− −− − =

D( xkxe− ) = ( )( ) ( )( )x x x xkx e e k kxe ke− − − −− + = − +

and ( )( ) ( )( )2 x x x x x xD (kxe ) kx e e k ke kxe 2ke− − − − − −= − − + − − = −

Hence, ( ) ( )x x x x x xkxe 2ke 3 kxe ke 4kxe 3e− − − − − −− − − + − =

i.e. x x x x x xkxe 2ke 3kxe 3ke 4kxe 3e− − − − − −− + − − =

i.e. x x5ke 3e− −− = i.e. – 5k = 3 and k = 35

−

hence, the particular integral, v = x3 x e5

−−

and the general solution, y = u + v = 4x x x3Ae Be xe5

− −+ −

4. Find the general solution of: t23

2

d y dy9 6 y 12edt dt

− + =

t23

2

d y dy9 6 y 12edt dt

− + = in D-operator form is: ( )t

2 39D 6D 1 y 12e− + =



i.e. (3m - 1)(3m - 1) = 0

from which, m = 13

twice

Hence, C.F., ( )t3u At B e= +

As t3e and

t3t e appears in the C.F. and in the right hand side of the differential equation, let the

particular integral, P.I., v =t

2 3k t e

then ( )t t

2 2 3 39D 6D 1 k t e 12e⎛ ⎞

− + =⎜ ⎟⎝ ⎠

D(t

2 3k t e ) = ( ) ( )t t t t t

2 23 3 3 3 31 k tkt e e 2kt t e 2kte kte 23 3 3

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

and ( ) ( )t t t t

2 2 3 3 3 31 t 1D kt e kte 2 kt e e k3 3 3

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= t t t t t

23 3 3 3 3k k k 2te t e te kte 2ke3 9 3 3

+ + + +

Hence, t t t t t t t

2 2 23 3 3 3 3 3 34 k k9 te t e 2ke 6 t e 2kte kt e 12e3 9 3

⎛ ⎞ ⎛ ⎞+ + − + + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

i.e. t t t t t t t

2 2 23 3 3 3 3 3 312te kt e 18ke 2kt e 12kte kt e 12e+ + − − + =

i.e. t t3 318ke 12e= i.e. and k = 12 2

18 3=

hence, the particular integral, v = t t

2 23 32kt e t e3

=

and the general solution, y = u + v = ( )t t

23 32At B e t e3

+ +


x2

d y dy5 9 2y 3edx dx

+ − = given that when x = 0, y = 14

and dydx

= 0

2

x2

d y dy5 9 2y 3edx dx

+ − = in D-operator form is: ( )2 x5D 9D 2 y 3e+ − =

Auxiliary equation is: 25m 9m 2 0+ − =

i.e. (5m - 1)(m + 2) = 0

from which, m = 15

and m = -2


Hence, C.F., 1 x 2x5u Ae Be−= +

Let P.I., v = xk e

then ( )( )2 x x5D 9D 2 ke 3e+ − =

D( xke ) = xke

and ( )2 x x xD (ke ) D ke ke= =

Hence, x x x x5ke 9ke 2ke 3e+ − =

i.e. x x12ke 3e= i.e. k = 14

hence, the particular integral, v = x1 e4

and the general solution, y = u + v = 1 x 2x x5 1Ae Be e

4−+ +

x = 0 and y = 14

, hence, 14

= A + B + 14

i.e. 0 = A + B (1)

1 x 2x x5dy 1 1Ae 2Be e

dx 5 4−= − +

x = 0 and dydx

= 0, hence, 1 10 A 2B5 4

= − +

i.e. 1 1 A 2B4 5

− = − (2)

2 × (1) gives: 0 = 2A + 2B (3)

(2) + (3) gives: 1 11A4 5

− = i.e. A = 544

− and from (1), B = 544

Hence, y = 1 x 2x x55 5 1e e e

44 44 4−− + +

i.e. 1 x2x x55 1y e e e

44 4−⎛ ⎞

= − +⎜ ⎟⎝ ⎠




2

d y dy2 3y 25sin 2xdx dx

− − =

2

2

d y dy2 3y 25sin 2xdx dx

− − = in D-operator form is: ( )22D D 3 y 25sin 2x− − =

Auxiliary equation is: 22m m 3 0− − =

i.e. (2m - 3)(m + 1) = 0

from which, m = 32

and m = -1

Hence, C.F., 3 x x2u Ae Be−= +

Let P.I., v = A sin 2x + B cos 2x

Hence, ( )( )22D D 3 Asin 2x Bcos 2x 25sin 2x− − + =

D(A sin 2x + B cos 2x) = 2A cos 2x – 2B sin 2x

( ) ( )2D Asin 2x Bcos 2x D 2A cos 2x 2Bsin 2x 4Asin 2x 4Bcos 2x+ = − = − −

Hence, ( ) ( ) ( )2 4Asin 2x 4Bcos 2x 2A cos 2x 2Bsin 2x 3 Asin 2x Bcos 2x 25sin 2x− − − − − + =

i.e. - 8A + 2B - 3A = 25

and - 8B – 2A – 3B = 0

i.e. – 11A + 2B = 25 (1)

and - 2A – 11B = 0 (2)

2 × (1) gives: - 22A + 4B = 50 (3)

11 × (2) gives: - 22A – 121B = 0 (4)

(3) – (4) gives: 125B = 50 from which, B = 50 2125 5

=

Substituting in (1) gives: - 11A + 45

= 25

from which, - 11A = 25 - 45

= 125 4 1215 5−

= and A = 121 115( 11) 5

= −−

Thus, P.I., v = ( )11 2 1sin 2x cos 2x 11sin 2x 2cos 2x5 5 5

− + = − −

and y = u + v = ( )3x x2 1Ae Be 11sin 2x 2cos 2x

5−+ − −



2

d y y 4cos xdx

+ =

2

2

d y y 4cos xdx

+ = in D-operator form is: ( )2D 1 y 4cos x+ =

Auxiliary equation is: 2m 1 0+ =

i.e. 2m 1= − and m = 1 0 j1− = ±

Hence, C.F., u A cos x Bsin x= +

Since cos x occurs in the C.F. and in the right hand side of the differential equation,

let P.I., v = x(C sin x + D cos x)

Hence, ( ) ( )2D 1 x Csin x Dcos x 4cos x+ + =⎡ ⎤⎣ ⎦

( ) ( )( ) ( )( )D x Csin x Dcos x x Ccos x Dsin x 1 Csin x Dcos x+ = − + +⎡ ⎤⎣ ⎦

[ ] ( )( ) ( ) ( )2D v x Csin x Dcos x Ccos x Dsin x Ccos x Dsin x= − − + − + − Hence, since ( )2D 1 v 4cos x+ =

then ( )( ) ( ) ( )x Csin x Dcos x Ccos x Dsin x Ccos x Dsin x− − + − + − + x(C sin x + D cos x)

= 4 cos x

from which, 2C = 4 from which, C = 2

and - 2D = 0 from which, D = 0

Hence, P.I., v = 2x sin x

and y = u + v = A cos x Bsin x 2xsin x+ +

4. Find the particular solution of the differential equation 2

2

d y dy3 4y 3sin xdx dx

− − = given that

when x = 0, y = 0 and dydx

= 0.

2

2

d y dy3 4y 3sin xdx dx

− − = in D-operator form is: ( )2D 3D 4 y 3sin x− − =

Auxiliary equation is: 2m 3m 4 0− − =

i.e. (m - 4)(m + 1) = 0

from which, m = 4 and m = -1


Hence, C.F., 4x xu Ae Be−= +

Let P.I., v = A sin x + B cos x

Hence, ( )( )2D 3D 4 Asin x Bcos x 3sin x− − + =

D(A sin x + B cos x) = A cos x – B sin x

( ) ( )2D Asin x Bcos x D A cos x Bsin x Asin x Bcos x+ = − = − −

Hence, (-A sin x – B cos x) – 3(A cos x – B sin x) – 4(A sin x + B cos x) = 3 sin x

i.e. - A + 3B - 4A = 3

and - B – 3A – 4B = 0

i.e. – 5A + 3B = 3 (1)

and - 3A – 5B = 0 (2)

3 × (1) gives: - 15A + 9B = 9 (3)

5 × (2) gives: - 15A – 25B = 0 (4)

(3) – (4) gives: 34B = 9 from which, B = 934

Substituting in (1) gives: - 5A + 9334

⎛ ⎞⎜ ⎟⎝ ⎠

= 3

from which, 5A = 2734

- 3 = 27 102 7534 34− −

= and A = 75 1534(5) 34−

= −

Thus, P.I., v = 15 9sin x cos x34 34

− +

and y = u + v = 4x x 15 9Ae Be sin x cos x34 34

−+ − +

x = 0 when y = 0, hence, 0 = A + B + 934

i.e. A + B = - 934

(5)

4x xdy 15 94Ae Be cos x sin xdx 34 34

−= − − −

x = 0, when dydx

= 0, hence, 0 = 4A – B 1534

− i.e. 4A – B = 1534

(6)

4 × (5) gives: 4A + 4B = - 3634

(7)

(7) – (6) gives: 5B = - 3634

- 1534

= 5134

−


and B = 51 5134(5) 170

− = −

Substituting in (5) gives: 51 9A170 34

− = − from which, 51 9 51 45 6A170 34 170 170

−= − = =

Hence, y = 4x x6 51 15 9e e sin x cos x170 170 34 34

−− − +

i.e. ( ) ( )4x x1 1y 6e 51e 15sin x 9cos x170 34

−= − − −

7. 2

02

d q dq 1L R q V sin tdt dt C

+ + = ω represents the variation of capacitor charge in an electric circuit.

Determine an expression for q at time t seconds given that R = 40 Ω, L = 0.02 H,

C = 50 × 10-6 F, V0 = 540.8 V and ω = 200 rad/s and given the boundary conditions that when

t = 0, q = 0 and dqdt

= 4.8

2

02

d q dq 1L R q V sin tdt dt C

+ + = ω in D-operator form is: 20

1L D R D q V sin tC

⎛ ⎞+ + = ω⎜ ⎟⎝ ⎠


+ + =

and

2 26

4L 4(0.02)R R 40 40 40 0C 50 10m 10002L 2(0.02) 0.04

−− ± − − ± − − ±×= = = = −

Hence, C.F., ( ) 1000tu At B e−= +

Let P.I., v = A sin ωt + B cos ωt

[ ]20

1L D R D Asin t Bcos t V sin tC

⎛ ⎞+ + ω + ω = ω⎜ ⎟⎝ ⎠

D(v) = Aω cos ωt - Bω sin ωt and 2 2 2D (v) A sin t B cos t= − ω ω − ω ω

Thus, ( ) ( )2 2 21L D R D v 0.02 A sin t B cos t 40 A cos t B sin tC

⎛ ⎞+ + = − ω ω − ω ω + ω ω − ω ω⎜ ⎟⎝ ⎠

( ) 06

1 Asin t Bcos t V sin t50 10−+ ω + ω = ω×

i.e. -800A sin 200t – 800B cos 200t + 8000A cos 200t – 8000B sin 200t + 20000A sin 200t

+ 20000B cos 200t = 540.8 sin 200t


Hence, - 800A – 8000B + 20000A = 540.8

and - 800B + 8000A + 20000B = 0

i.e. 19200A – 8000B = 540.8 (1)

and 8000A + 19200B = 0 (2)

8 × (1) gives: 153600A – 64000B = 4326.4 (3)

19.2 × (2) gives: 153600A + 368640B = 0 (4)

(3) – (4) gives: - 432640B = 4326.4

from which, B = 4326.4 0.01432640

= −

Substituting in (1) gives: 19200A – 8000(-0.01) = 540.8

i.e. 19200A + 80 = 540.8

and A = 540.8 80 460.8 0.02419200 19200

−= =

Hence, P.I., v = 0.024 sin 200t – 0.01 cos 200t

Thus, q = u + v = ( ) 1000tAt B e−+ + 0.024 sin 200t – 0.01 cos 200t

When t = 0, q = 0, hence, 0 = B – 0.01 from which, B = 0.01

( )( )1000t 1000tdq At B 1000e Ae (0.024)(200)cos 200t (0.01)(200)sin 200tdt

− −= + − + + +

When t = 0, dqdt

= 4.8, hence, 4.8 = - 1000B + A + 4.8

i.e. A = 1000B = 1000(0.01) = 10

Thus, q = ( ) 1000t10t 0.01 e−+ + 0.024 sin 200t – 0.010 cos 200t


EXERCISE 193 (Page 490)


2

d y dy8 6 y 2x 40sin xdx dx

− + = +

2

2

d y dy8 6 y 2x 40sin xdx dx

− + = + in D-operator form is: ( )28D 6D 1 y 2x 40sin x− + = +


i.e. (4m - 1)(2m - 1) = 0

from which, m = 14

and m = 12

Hence, C.F., 1 1x x4 2u Ae Be= +

Let P.I., v = ax + b + c sin x + d cos x

Hence, ( )[ ]28D 6D 1 ax b csin x d cos x 2x 40sin x− + + + + = +

D(v) = a + c cos x – d sin x and 2D (v) = - c sin x – d cos x

Hence, ( )28D 6D 1 v− + = 8(-c sin x – d cos x) – 6(a + c cos x – d sin x)

+ (ax + b + c sin x + d cos x) = 2x + 40 sin x

i.e. ax = 2x from which, a = 2

and - 6a + b = 0, from which, b = 12

- 8c + 6d + c = 40 i.e. – 7c + 6d = 40 (1)

- 8d – 6c + d = 0 i.e. – 6c – 7d = 0 (2)

6 × (1) gives: - 42c + 36d = 240 (3)

7 × (2) gives: - 42c – 49d = 0 (4)

(3) – (4) gives: 85d = 240 from which, 240 48d85 17

= =

Substituting in (2) gives: - 6c - 48717⎛ ⎞⎜ ⎟⎝ ⎠

= 0 from which, 7(48) 56c6(17) 17

= − = −

Hence, P.I., v = 2x + 12 5617

− sin x + 4817

cos x


and y = u + v = 1 1x x4 2Ae Be+ + 2x + 12 56

17− sin x + 48

17cos x

or y = 1 1x x4 2Ae Be+ + 2x + 12 + ( )8 6cos x 7sin x

17−


t2

d y dy2 2y e sin tdt dt

− + =

2

t2

d y dy2 2y e sin tdt dt

− + = in D-operator form is: ( )2 tD 2D 2 y e sin t− + =


i.e. 2( 2) ( 2) 4(1)(2) 2 4 2 j2m 1 j

2(1) 2 2− − ± − − ± − ±

= = = = ±

Hence, C.F., ( )tu e A cos t Bsin t= +

Since te sin t occurs in the C.F. and the right hand side of the differential equation,

let P.I., v = ( )tt e Csin t Dcos t+

then ( ) ( )2 t tD 2D 2 t e Csin t Dcos t e sin t⎡ ⎤− + + =⎣ ⎦

D(v) = ( )( ) ( )t t tt e Ccos t Dsin t Csin t Dcos t t e e⎡ ⎤− + + +⎣ ⎦

( )( ) ( ) ( )2 t t t t t tD (v) t e Csin t Dcos t Ccos t Dsin t t e e Csin t Dcos t t e e e⎡ ⎤ ⎡ ⎤= − − + − + + + + +⎣ ⎦ ⎣ ⎦

( )( )t tt e e Ccos t Dsin t+ + −

Hence, ( )( ) ( ) ( )t t t t t tt e Csin t Dcos t Ccos t Dsin t t e e Csin t Dcos t t e e e⎡ ⎤ ⎡ ⎤− − + − + + + + +⎣ ⎦ ⎣ ⎦

( )( )t tt e e Ccos t Dsin t+ + − - 2 ( )( ) ( )t t tt e Ccos t Dsin t 2 Csin t Dcos t t e e⎡ ⎤− − + +⎣ ⎦

+ 2 ( )tt e Csin t Dcos t+ = te sin t

i.e. – D + 2C – D – 2C = 1 i.e. -2D = 1 and D = 12

−

and C + 2D + C – 2D = 0 i.e. C = 0

Hence, P.I., v = t t1 tt e cos t e cos t2 2

⎛ ⎞− = −⎜ ⎟⎝ ⎠

and y = u + v = ( )te Acos t Bsin t+ tt e cos t2

−



2x2

d y dy7 10y e 20dx dx

− + = + given that when x = 0, y = 0 and

dy 1dx 3

= −

2

2x2

d y dy7 10y e 20dx dx

− + = + in D-operator form is: ( )2 2xD 7D 10 y e 20− + = +


i.e. (m - 5)(m - 2) = 0

from which, m = 5 and m = 2

Hence, C.F., 5x 2xu Ae Be= +

Let P.I., v = 2xk x e a+

Thus, ( )2 2x 2xD 7D 10 k x e a e 20⎡ ⎤− + + = +⎣ ⎦

D(v) = ( )( ) ( )2x 2xkx 2e ke+ and ( )( ) ( )( )2 2x 2x 2xD (v) kx 4e 2e k 2ke= + +

Hence, ( )( ) ( )( )2x 2x 2xkx 4e 2e k 2ke+ + ( )( ) ( )2x 2x7 kx 2e 7 ke− − 2x 2x10k x e 10a e 20+ + = +

from which, 10a = 20 and a = 2

Also, 2k + 2k – 7k = 1 i.e. -3k = 1 from which, k = 13

−

Hence, P.I., v = 2x1 x e 23

− +

and y = u + v = 5x 2xAe Be+ 2x1 x e 23

− +

When x = 0, y = 0, hence, 0 = A + B + 2 i.e. A + B = -2 (1)

( ) ( )5x 2x 2x 2xdy 1 15Ae 2Be x 2e edx 3 3

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

When x = 0, dy 1dx 3

= − , hence, 13

− = 5A + 2B 13

− i.e. 5A + 2B = 0 (2)

2 × (1) gives: 2A + 2B = -4 (3)

(2) – (3) gives: 3A = 4 from which, A = 43


Substituting in (1) gives: 43

+ B = - 2 from which, B = - 2 - 43

= 103

−

Hence, y = u + v = 5x 2x4 10e e3 3

− 2x1 xe 23

− +


CHAPTER 52 POWER SERIES METHODS OF SOLVING

ORDINARY DIFFERENTIAL EQUATIONS EXERCISE 194 Page 492

1. (b) Determine the derivative y(5) when y = t28e

If axy e= , then (n) n axy a e= . Hence, if y =1 t28e , then

5 1 1t t(5) 2 21 8y (8) e e2 32

⎛ ⎞= =⎜ ⎟⎝ ⎠

= 1t21 e

4

2. (a) Determine the derivative y(4) when y = sin 3t

If y = sin ax, then (n) n ny a sin ax2π⎛ ⎞= +⎜ ⎟

⎝ ⎠

Hence, if y = sin 3t, then ( )(4) 4 4y 3 sin 3t 81sin 3t 22π⎛ ⎞= + = + π⎜ ⎟

⎝ ⎠ = 81 sin 3t

3. (b) Determine the derivative y(9) when y = 23cos t3

If y = cos ax, then (n) n ny a cos ax2π⎛ ⎞= +⎜ ⎟

⎝ ⎠

Hence, if y = 23cos t3

, then 9 9

(9)8

2 2 9 2 2y (3) cos t cos t3 3 2 3 3 2

π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 9

8

2 2sin t3 3

−

4. (a) Determine the derivative y(7) when y = 92x

If y = ax , then ( )

(n) a na!y xa n !

−=−

Hence, if y = 92x , then ( )

(7) 9 79!y (2) x9 7 !

−=−

= ( ) 29! x

5. (b) Determine the derivative y(6) when y = 2 sinh 3x

If y = sinh ax, then n

(n) n nay 1 ( 1) sinh ax 1 ( 1) cosh ax2

⎡ ⎤ ⎡ ⎤= + − + − −⎣ ⎦ ⎣ ⎦


Hence, if y = 2 sinh 3x, then 6

(6) 6 63y (2) 1 ( 1) sinh 3x 1 ( 1) cosh 3x2

⎡ ⎤ ⎡ ⎤= + − + − −⎣ ⎦ ⎣ ⎦

= 63 2sinh 3x 0+ = 1458 sinh 3x

6. (a) Determine the derivative y(7) when y = cosh 2x

If y = cosh ax, then n

(n) n nay 1 ( 1) sinh ax 1 ( 1) cosh ax2

⎡ ⎤ ⎡ ⎤= − − + + −⎣ ⎦ ⎣ ⎦

Hence, if y = cosh 2x, then 7

(7) 7 72y 1 ( 1) sinh 2x 1 ( 1) cosh 2x2

⎡ ⎤ ⎡ ⎤= − − + + −⎣ ⎦ ⎣ ⎦

= 6 72 2sinh 2x 0 2 sinh 2x+ = = 128 sinh 2x

7. (b) Determine the derivative y(7) when y = 1 ln 2t3

If y = ln ax, then ( ) ( )n 1(n)n

n 1 !y 1

x− −

= −

Hence, if y = 1 ln 2t3

, then ( ) ( )7 1(7)7 7

7 1 !1 6!y 13 t 3 t

− −⎛ ⎞= − =⎜ ⎟⎝ ⎠

= 7

240t


EXERCISE 195 Page 494 2. If y = 3 2xx e find y(n) and hence y(3) by using the theorem of Leibniz. Since y = 3 2xx e then let v = 3x and u = 2xe and the n’th derivative of 2xe is n 2x2 e

Thus, (n) (n) (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!

− −−= + + +

= ( ) ( )( ) ( )( ) ( )( )n 2x 3 n 1 2x 2 n 2 2x (n 3) 2xn n 1 n 2n(n 1)2 e x n 2 e 3x 2 e 6x 2 e (6)2! 3!

− − −− −−+ + +

= ( )( )n 3 2x 2 n 1 2x n 2 2x n 3 2x2 x e 3nx 2 e 3n(n 1)2 e x n n 1 n 2 2 e− − −+ + − + − −

or (n) 2x n 3 3 3 2 2y e 2 2 x 3nx (2) 3n(n 1)x(2) n(n 1)(n 2)−= + + − + − −

= 2x n 3 3 2e 2 8x 12nx n(n 1)(6x) n(n 1)(n 2)− + + − + − −

Hence, (3) 2x 0 3 2y e 2 8x 36x 3(2)6x 3(2)(1)= + + +

= 2x 3 2e 8x 36x 36x 6+ + +

4. Use the theorem of Leibniz to determine the 5th derivative of: y = 3x cos x

Since y = 3x cos x then let u = cos x and v = 3x and (n) n nu 1 cos x2π⎛ ⎞= +⎜ ⎟

⎝ ⎠

(n) n (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!

− −−= + + +

Hence, ( ) ( ) ( )(n) 3 2n (n 1) n(n 1) (n 2)y cos x x n cos x 3x cos x 6x2 2 2! 2π − π − − π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )n(n 1(n 2) (n 3)cos x 63! 2

− − − π⎛ ⎞+ +⎜ ⎟⎝ ⎠

and ( ) ( ) ( )(5) 3 25 4 5(4) 3 5(4)(3) 2y x cos x 5 3x cos x 6x cos x 6 cos x2 2 2! 2 3! 2π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 3 2x sin x 15x cos x 60x sin x 60( cos x)− + + + −

= ( ) ( )3 260x x sin x 15x 60 cos x− + −


6. If y = 5x ln 2x find y(3) by using the theorem of Leibniz.

Since y = 5x ln 2x then let u = 5x and v = ln 2x and ( )

n a n 5 na! 5!u x xa n ! (5 n)!

− −= =− −

(n) n (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!

− −−= + + +

(n) 5 n 6 n 7 n2

8 n3

5! 5! 1 n(n 1) 5! 1y x ln 2x n x x(5 n)! (6 n)! x 2! (7 n)! x

n(n 1)(n 2) 5! 2x3! (8 n)! x

− − −

−

⎡ ⎤ −⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟⎢ ⎥− − −⎝ ⎠ ⎝ ⎠⎣ ⎦− − ⎛ ⎞+ ⎜ ⎟− ⎝ ⎠

Hence, 3 2 3 4 52 3

5! 5! 1 3(2) 5! 1 3(2)(1) 5! 2y x ln 2x (3) x x x2! 3! x 2! (4)! x 3! 5! x

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 2 2 2 260x ln 2x 60x 15x 2x+ − +

= 2 260x ln 2x 47x+

i.e. ( )(3) 2y x 47 60ln 2x= +

8. If y = ( )3 2 2xx 2x e+ determine an expression for y(5) by using the theorem of Leibniz. Since y = ( )3 2 2xx 2x e+ then let u = 2xe and v = ( )3 2x 2x+ and n n 2xu 2 e=

(n) n (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!

− −−= + + +

Hence, ( )( ) ( ) ( )(n) n 2x 3 2 n 1 2x 2 n 2 2xn(n 1)y 2 e x 2x n2 e 3x 4x 2 e 6x 42!

− −−= + + + + +

( )n 3 2xn(n 1)(n 2) 2 e 63!

−− −+

and ( ) ( ) ( ) ( )(5) 5 2x 3 2 4 2x 2 3 2x 2 2x5(4) 5(4)(3)y 2 e x 2x (5) 2 e 3x 4x 2 e 6x 4 2 e 62 3!

= + + + + + +

= 2x 5 3 6 2 2e 2 x 2 x (16)15x (16)(20x) 60x(8) (8)(40) 240+ + + + + +

= 2x 5 3 2e 2 x 304x 800x 560+ + +

= 2x 5 3 4 2 4 4e 2 x 2 (19x ) 2 (50)(x) 2 (35)+ + +

= 2x 4 3 2e 2 2x 19x 50x 35+ + +



1. Determine the power series solution of the differential equation: 2

2

d y dy2x y 0dx dx

+ + = using the

Leibniz-Maclaurin method, given that at x = 0, y = 1 and dydx

= 2.

2

2

d y dy2x y 0dx dx

+ + =

(i) The differential equation is rewritten as: y′′ + 2xy′ + y = 0 and from the Leibniz theorem of

equation (13), page 493 of textbook, each term is differentiated n times, which gives:

(n 2) (n 1) (n) (n)y 2 y (x) n y (1) 0 y 0+ ++ + + + =

i.e. (n 2) (n 1) (n)y 2x y (2n 1) y 0+ ++ + + = (1)

(ii) At x = 0, equation (1) becomes:

(n 2) (n)y (2n 1) y 0+ + + =

from which, (n 2) (n)y (2n 1) y+ = − +

This is the recurrence formula.

(iii) For n = 0, ( ) ( )0 0y '' y= −

n = 1, ( ) ( )0 0y ''' 3 y '= −

n = 2, ( ) ( )(4)000

y 5 y '' 5(y)= − =

n = 3, ( ) ( )(5)00

y 7 y '''= − = ( ) ( )0 07 3 y ' 3 7 y '− − = ×

n = 4, ( ) ( )(6) (4)

0 0y 9 y= − = ( ) ( )0 0

9 5 y 5 9 y− = − ×

n = 5, ( ) ( )(7) (5)

0 0y 11 y= − = ( ) ( )0 0

11 3 7 y ' 3 7 11 y '− × = − × ×

n = 6, ( ) ( )(8) (6)

0 0y 13 y= − = ( ) ( )0 0

13 5 9 y 5 9 13 y− − × = × ×

(iv) Maclaurin’s theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4

(4)0 0 0 0 0

x x xy x y ' y '' y ''' y ....2! 3! 4!

+ + + + +

Thus, y = ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 5

0 0 0 0 0 0

x x x xy x y ' y 3 y ' 5 y 3 7 y '2! 3! 4! 5!

+ + − + − + + ×

( ) ( ) 6 7

0 0

x x5 9 y 3 7 11 y '6! 7!

+ − × + − × ×


(v) Collecting similar terms together gives:

y = ( )2 4 6 8

0

x 5 x 5 9 x 5 9 13xy 1 ...2! 4! 6! 8!

⎧ ⎫× × ×− + − + −⎨ ⎬

⎩ ⎭

( )3 5 7

0

3x 3 7 x 3 7 11xy ' x ...3! 5! 7!

⎧ ⎫× × ×+ − + − +⎨ ⎬

⎩ ⎭

At x = 0, y = 1 and dydx

= 2, hence, ( )0y 1= and ( )0

y ' 2= .

Hence, the power series solution of the differential equation: 2

2

d y dy2x y 0dx dx

+ + = is:

y = 2 4 6 8x 5 x 5 9 x 5 9 13x1 ...

2! 4! 6! 8!⎧ ⎫× × ×

− + − + −⎨ ⎬⎩ ⎭

3 5 73 x 3 7 x 3 7 11x2 x ...3! 5! 7!

⎧ ⎫× × ×+ − + − +⎨ ⎬

⎩ ⎭

3. Find the particular solution of the differential equation: ( )2

22

d y dyx 1 x 4y 0dx dx

+ + − = using the

Leibniz-Maclaurin method, given the boundary conditions that at x = 0, y = 1 and dydx

= 1.

( )2

22


+ + − =

i.e. ( )2x 1+ y′′ + xy′ - 4y = 0

i.e. ( ) 2 (n 2) (n 1) (n) (n 1) n (n)n(n 1)x 1 y ny (2x) y (2) y x ny (1) 4y 02!

+ + +−⎧ ⎫+ + + + + − =⎨ ⎬⎩ ⎭

i.e. ( ) ( )2 (n 2) (n 1) (n)x 1 y 2nx x y (n(n 1) n 4) y 0+ ++ + + + − + − =

At x = 0, ( )(n 2) 2 (n)y n 4 y 0+ + − =

from which, ( )(n 2) 2 (n)y 4 n y+ = − which is the recurrence formula.

For n = 0, ( ) ( )0 0y '' 4 y=

n = 1, ( ) ( )0 0y ''' 3 y '=

n = 2, ( )(4)

0y 0=

n = 3, ( ) ( )(5)00

y 5 y '''= − = ( ) ( )( )0 05 3 y ' 5 3 y '− − = −

n = 4, ( ) ( )(6) (4)

0 0y 12 y= − = 12(0) 0− =


n = 5, ( ) ( )(7) (5)

0 0y 21 y= − = ( ) ( )0 0

21 5 3 y ' 315 y '− − × =

Maclaurin’s theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4

(4)0 0 0 0 0

x x xy x y ' y '' y ''' y ....2! 3! 4!

+ + + + +

Thus, y = ( ) ( ) ( ) ( ) ( ) 2 3 4 5

0 0 0 0 0

x x x xy x y ' 4 y 3 y ' 0 3 5 y ' 02! 3! 4! 5!

+ + + + + − × + ( ) 7

0

x 315 y '7!

+

i.e. y = ( ) 20

y 1 2x+ ( )3 5 7

0

x x xy ' x ...2 8 16

⎧ ⎫+ + − + +⎨ ⎬

⎩ ⎭

At x = 0, y = 1 and dydx

= 1, hence, ( )0y 1= and ( )0

y ' 1= .

Hence, the power series solution of the differential equation: ( )2

22


+ + − = is:

y = 21 2x+3 5 7x x xx ...

2 8 16⎧ ⎫

+ + − + +⎨ ⎬⎩ ⎭

i.e. y = 3 5 7

2 x x x1 x 2x .....3 8 16

+ + + − + +


EXERCISE 197 Page 503 1. Produce, using Frobenius’ method, a power series solution for the differential equation:

2

2

d y dy2x y 0dx dx

+ − =

2

2

d y dy2x y 0dx dx

+ − = may be rewritten as: 2xy′′ + y′ - y = 0

(i) Let a trial solution be of the form y = xca0 + a1x + a2x2 + a3x3 + … + arxr+…

where a0 ≠ 0,

i.e. y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +…

(ii) Differentiating gives:

y′ = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + …. + ar(c + r)xc+r-1 + …

and y′′ = a0c(c – 1)xc-2 + a1c(c + 1)xc-1 + a2(c + 1)(c + 2)xc + …. + ar(c + r - 1)(c + r)xc+r-2 + …

(iii) Substituting y, y′ and y′′ into each term of the given equation 2xy′′ + y′ - y = 0 gives:

2xy′′ = 2a0c(c – 1)xc-1 + 2a1c(c + 1)xc + 2a2(c + 1)(c + 2)xc+1 + …

+ 2ar(c + r –1)(c + r)xc+r-1 + … (a)

y′ = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + …. + ar(c + r)xc+r-1 + … (b)

-y = -a0xc - a1xc+1 - a2xc+2 - a3xc+3 - … - arxc+r -… (c)

(iv) The sum of these three terms forms the left-hand side of the equation. Since the right-hand side

is zero, the coefficients of each power of x can be equated to zero.

For example, the coefficient of xc-1 is equated to zero giving: 2a0c(c – 1) + a0c = 0

or a0 c [2c – 2 + 1] = a0 c(2c - 1) = 0 (1)

Equation (1) is the indicial equation, from which, c = 0 or c = 12

The coefficient of xc is equated to zero giving: 2a1c(c + 1) + a1(c + 1) - a0 = 0

i.e. a1 (2c2 + 2c + c + 1) - a0 = a1(2c2 + 3c + 1) - a0 = 0

or a1(2c + 1)(c + 1) - a0 = 0 (2)

Replacing r by (r + 1) will give:


in series (a), 2ar+1(c + r + 1)(c + r)xc+r

in series (b), ar+1(c + r + 1)xc+r

in series (c), -arxc+r

Equating the total coefficients of xc+r to zero gives:

2ar+1(c + r + 1)(c + r) + ar+1(c + r + 1) - ar = 0

which simplifies to: ar+1(c + r + 1)(2c + 2r + 1) - ar = 0 (3)

(a) When c = 0:

From equation (2), if c = 0, a1(1 × 1) - a0 = 0, i.e. a1 = 0a

From equation (3), if c = 0, ar+1(r + 1)(2r + 1) - ar = 0, i.e. ar+1 = ra(r 1)(2r 1)+ +

r ≥ 0

Thus, when r = 1, 012

aaa(2 3) (2 3)

= =× ×

since 1 0a a=

when r = 2, 023

aaa(3 5) (2 3)(3 5)

= =× × ×

when r = 3, 3 0 04

a a aa(4 7) (2 3)(3 5)(4 7) (2 3 4)(3 5 7)

= = =× × × × × × × ×

and so on.

The trial solution is: y = xca0 + a1x + a2x2 + a3x3 + … + arxr +…

Substituting c = 0 and the above values of a1, a2, a3, … into the trial solution gives:

y = 0 2 3 40 0 00 0

a a ax a a x x x x ...(2 3) (2 3)(3 5) (2 3 4)(3 5 7)

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟× × × × × × ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

i.e. y = ( ) ( )( ) ( )( )

2 3 4

0x x xa 1 x ...

2 3 2 3 3 5 2 3 4 3 5 7⎧ ⎫⎪ ⎪+ + + + +⎨ ⎬× × × × × × ×⎪ ⎪⎩ ⎭

(4)

(b) When c = 12

:

From equation (2), if c = 12

, a1 ( ) 322

⎛ ⎞⎜ ⎟⎝ ⎠

- a0 = 0, i.e. a1 = 0a3


, ar+1 ( )1 r 1 1 2r 12

⎛ ⎞+ + + +⎜ ⎟⎝ ⎠

- ar = 0,

i.e. ar+1 ( )3r 2r 22

⎛ ⎞+ +⎜ ⎟⎝ ⎠

- ar = ar+1(2 2r + 5r +3) - ar = 0,

i.e. ar+1 = ra(2r 3)(r 1)+ +

r ≥ 0



aaa(2 5) (2 3 5)

= =× × ×

since a1 = 0a3

when r = 2, 023

aaa(3 7) (2 3 5)(3 7)

= =× × × ×

when r = 3, 3 04

a aa(4 9) (2 3 4)(3 5 7 9)

= =× × × × × ×

and so on.


Substituting c = 12

and the above values of a1, a2, a3, … into the trial solution gives:

y =1

2 3 40 0 0 020

a a a ax a x x x x ...3 2 3 5 (2 3 5)(3 7) (2 3 4)(3 5 7 9)

⎧ ⎫⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞+ + + + +⎨ ⎬⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟× × × × × × × × × ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

i.e. y = 1 2 3 42

0x x x xa x 1 ...

(1 3) (1 2)(3 5) (1 2 3)(3 5 7) (1 2 3 4)(3 5 7 9)⎧ ⎫+ + + + +⎨ ⎬× × × × × × × × × × × × ×⎩ ⎭

(5)

Let 0a = A in equation (4), and 0a = B in equation (5).

Hence, y = ( ) ( ) ( ) ( ) ( )

2 3 4x x xA 1 x ...2 3 2 3 3 5 2 3 4 3 5 7

⎧ ⎫⎪ ⎪+ + + + +⎨ ⎬× × × × × × ×⎪ ⎪⎩ ⎭

+ 1 2 3 42 x x x xB x 1 ...

(1 3) (1 2)(3 5) (1 2 3)(3 5 7) (1 2 3 4)(3 5 7 9)⎧ ⎫

+ + + + +⎨ ⎬× × × × × × × × × × × × ×⎩ ⎭

3. Determine the power series solution of the differential equation: 2

2

d y dy3x 4 y 0dx dx

+ − = using the

Frobenius method.

2

2

d y dy3x 4 y 0dx dx

+ − = may be rewritten as: 3xy′′ + 4y′ - y = 0

(i) Let a trial solution be of the form y = xca0 + a1x + a2x2 + a3x3 + … + arxr+…

i.e. y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +…

(ii) Differentiating gives:

y′ = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + …. + ar(c + r)xc+r-1 + …

and y′′ = a0c(c – 1)xc-2 + a1c(c + 1)xc-1 + a2(c + 1)(c + 2)xc + …. + ar(c + r - 1)(c + r)xc+r-2 + …

(iii) Substituting y, y′ and y′′ into each term of the given equation 3xy′′ + 4y′ - y = 0 gives:


3xy′′ = 3a0c(c – 1)xc-1 + 3a1c(c + 1)xc + 3a2(c + 1)(c + 2)xc+1 + …

+ 3ar(c + r –1)(c + r)xc+r-1 + … (a)

4y′ = 4a0cxc-1 + 4a1(c + 1)xc + 4a2(c + 2)xc+1 + …. + 4ar(c + r)xc+r-1 + … (b)

- y = -a0xc - a1xc+1 - a2xc+2 - a3xc+3 - … - arxc+r -… (c)

(iv) The coefficient of xc-1 is equated to zero giving: 3a0c(c – 1) + 4a0c = 0

or a0 c [3c – 3 + 4] = a0 c(3c + 1) = 0

This is the indicial equation, from which, c = 0 or c = 13

−

The coefficient of xc is equated to zero giving: 3a1c(c + 1) + 4a1(c + 1) - a0 = 0

i.e. a1 (3c(c + 1) +4(c+1)) - a0 = a1(c + 1)(3c + 4) - a0 = 0

or a1(c + 1)(3c + 4) - a0 = 0 (1)

Equating the total coefficients of xc+r to zero gives:

3ar+1(c + r)(c + r + 1) + 4ar+1(c + r + 1) - ar = 0

i.e. ar+1(c + r + 1)(3c + 3r + 4) - ar = 0

which simplifies to: rr 1

aa

(c r 1)(3c 3r 4)+ =+ + + +

(2)

(a) When c = 0:

From equation (1), if c = 0, a1(4) - a0 = 0, i.e. a1 = 0a4

From equation (2), if c = 0, rr 1

aa(r 1)(3r 4)+ =+ +

r ≥ 0


aaa(2 7) (2 4 7)

= =× × ×

since 01

aa4

=

when r = 2, 0 023

a aaa(3 10) (3 10)(2 4 7) (1 2 3)(4 7 10)

= = =× × × × × × × ×

when r = 3, 3 0 04

a a aa(4 13) (4 13)(3 10)(2 4 7) (2 3 4)(4 7 10 13)

= = =× × × × × × × × × ×

and so on.


Substituting c = 0 and the above values of a1, a2, a3, … into the trial solution gives:

y = 0 2 3 40 0 0 00

a a a ax a x x x x ...4 (1 2)(4 7) (1 2 3)(4 7 10) (2 3 4)(4 7 10 13)

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪+ + + + +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟× × × × × × × × × × ×⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭


i.e. y = ( ) ( )( )

2 3 4

0x x x xa 1 ...

(1 4) 1 2)(4 7 1 2 3 4 7 10 (2 3 4)(4 7 10 13)⎧ ⎫⎪ ⎪+ + + + +⎨ ⎬× × × × × × × × × × × ×⎪ ⎪⎩ ⎭

(3)

(b) When c = 13

− :


− , a1 ( )2 33

⎛ ⎞⎜ ⎟⎝ ⎠

- a0 = 0, i.e. a1 = 0a2


− , ( )r r r

r 1a a aa 12 (3r 2)(r 1)(3r 2)3(r 1)r 3r 3

33

+ = = =+ +⎛ ⎞ + ++ +⎜ ⎟

⎝ ⎠

r ≥ 0

Thus, when r = 1, 0 012 2

a aaa(5 2) (2 5) (1 2)(2 5)

= = =× × × ×

since a1 = 0a2

when r = 2, 023

aaa(8 3) (1 2 3)(2 5 8)

= =× × × × ×

when r = 3, 3 04

a aa(11 4) (1 2 3 4)(2 5 8 11)

= =× × × × × × ×

and so on.


Substituting c = 13

− and the above values of a1, a2, a3, … into the trial solution gives:

y =1

2 30 0 0 030

a a a ax a x x x ...2 (1 2)(2 5) (1 2 3)(2 5 8) (1 2 3 4)(2 5 8 11)

− ⎧ ⎫+ + + + +⎨ ⎬× × × × × × × × × × × ×⎩ ⎭

i.e. y = 1 2 3 43

0x x x xa x 1 ...

(1 2) (1 2)(2 5) (1 2 3)(2 5 8) (1 2 3 4)(2 5 8 11)− ⎧ ⎫

+ + + + +⎨ ⎬× × × × × × × × × × × × ×⎩ ⎭

(4)

Let 0a = A in equation (3), and 0a = B in equation (4).

Hence, y = ( ) ( ) ( ) ( ) ( )

2 3 4x x xA 1 x ...1 4 1 2 4 7 1 2 3 4 7 10

⎧ ⎫⎪ ⎪+ + + + +⎨ ⎬× × × × × × ×⎪ ⎪⎩ ⎭

+ 1 2 33 x x xB x 1 ...

(1 2) (1 2)(2 5) (1 2 3)(2 5 8)− ⎧ ⎫

+ + + +⎨ ⎬× × × × × × ×⎩ ⎭



1. Determine the power series solution of Bessel’s equation: ( )2

2 2 22

d y dyx x x v y 0dx dx

+ + − = when

v = 2, up to and including the term in x6.

The complete solution of Bessel’s equation: ( )2

2 2 22

d y dyx x x v y 0dx dx

+ + − = is:

y =2 4 6

v2 4 6

x x xA x 1 ...2 (v 1) 2 2!(v 1)(v 2) 2 3!(v 1)(v 2)(v 3)

⎧ ⎫− + − +⎨ ⎬+ × + + × + + +⎩ ⎭

+ 2 4 6

v2 4 6

x x xB x 1 ...2 (v 1) 2 2!(v 1)(v 2) 2 3!(v 1)(v 2)(v 3)

− ⎧ ⎫+ + + +⎨ ⎬− × − − × − − −⎩ ⎭

and y =2 4 6

v2 4 6

x x xA x 1 ...2 (v 1) 2 2!(v 1)(v 2) 2 3!(v 1)(v 2)(v 3)

⎧ ⎫− + − +⎨ ⎬+ × + + × + + +⎩ ⎭

when v is a

positive integer.

Hence, when v = 2, y = 2 4

22 4

x xA x 1 ...2 (2 1) 2 2!(2 1)(2 2)

⎧ ⎫− + +⎨ ⎬+ × + +⎩ ⎭

i.e. y = 2 4 4 6

2 2x x x xA x 1 ... or A x ...12 384 12 384

⎧ ⎫ ⎧ ⎫− + − − + −⎨ ⎬ ⎨ ⎬

⎩ ⎭ ⎩ ⎭

2. Find the power series solution of the Bessel function: ( )2 2 2x y '' xy ' x v y 0+ + − = in terms of the

Bessel function 3J (x) when v = 3. Give the answer up to and including the term in x7.

vJ (x) = v 2 4

2 4

x 1 x x ...2 (v 1) 2 (1!) (v 2) 2 (2!) (v 3)

⎧ ⎫⎛ ⎞ − + −⎨ ⎬⎜ ⎟ Γ + Γ + Γ +⎝ ⎠ ⎩ ⎭ provided v is not a negative integer.

Hence, when v = 3, 3J (x) = 3 2 4

2 4

x 1 x x ...2 (3 1) 2 (1!) (3 2) 2 (2!) (3 3)

⎧ ⎫⎛ ⎞ − + −⎨ ⎬⎜ ⎟ Γ + Γ + Γ +⎝ ⎠ ⎩ ⎭

i.e. 3J (x) = 3 2 4 3 5 7

2 5 5 8

x 1 x x x x x... or ...2 4 2 5 2 6 8 4 2 5 2 6

⎧ ⎫⎛ ⎞ − + − − + −⎨ ⎬⎜ ⎟ Γ Γ Γ Γ Γ Γ⎝ ⎠ ⎩ ⎭

3. Evaluate the Bessel functions 0J (x) and 1J (x) when x = 1, correct to 3 decimal places.


0J (x) = ( )

2 4 6

22 2 6 24

x x x1 ...2 (1!) 2 (3!)2 2!

− + − +

and when x = 1, 0J (x) = ( )

2 4 6

22 2 6 24

1 1 11 ...2 (1!) 2 (3!)2 2!

− + − +

= 1 – 0.25 + 0.015625 – 0.000434 + …

= 0.765 correct to 3 decimal places

1J (x) = 3 5 7

3 5 7

x x x x ...2 2 (1!)(2!) 2 (2!)(3!) 2 (3!)(4!)− + − +

and when x = 1, 1J (x) = 3 5 7

3 5 7

1 1 1 1 ...2 2 (1!)(2!) 2 (2!)(3!) 2 (3!)(4!)− + − +

= 0.5 – 0.0625 + 0.002604 – 0.000054

= 0.440 correct to 3 decimal places


EXERCISE 199 Page 511 1. Determine the power series solution of the Legrandre equation: ( )21 x y '' 2xy ' k(k 1)y 0− − + + =

when (a) k = 0 (b) k = 2, up to and including the term in x5.

The power series solution of the Legrandre equation is:

y = 2 40

k(k 1) k(k 1)(k 2)(k 3)a 1 x x ..2! 4!+ + − +⎧ ⎫− + −⎨ ⎬

⎩ ⎭

+ 3 51

(k 1)(k 2) (k 1)(k 3)(k 2)(k 4)a x x x ..3! 5!

− + − − + +⎧ ⎫− + −⎨ ⎬⎩ ⎭

(a) When k = 0, y = 0a 1 0 0 ..− + − + 3 51

( 1)( 2) ( 1)( 3)( 2)( 4)a x x x ..3! 5!

− + − − + +⎧ ⎫− + −⎨ ⎬⎩ ⎭

i.e. y = 0a + 3 5

1x xa x ....3 5

⎛ ⎞+ + +⎜ ⎟

⎝ ⎠

(b) When k = 2, y = 2 40

2(3) 2(3)(0)(5)a 1 x x ..2! 4!

⎧ ⎫− + −⎨ ⎬⎩ ⎭

+ 3 51

(1)(4) (1)( 1)(4)(6)a x x x ..3! 5!

−⎧ ⎫− + −⎨ ⎬⎩ ⎭

i.e. y = ( )20a 1 3x− + 3 5

12 1a x x x ....3 5

⎛ ⎞− − −⎜ ⎟⎝ ⎠

2. Find the following Legrendre polynomials: (a) P1(x) (b) P4(x) (c) P5(x) (a) Since in 1P (x) , n = k = 1, then from the second part of equation (47), page 510 of textbook, i.e.

the odd powers of x:

y = 1a x 0− = 1a x

1a is chosen to make y = 1 when x = 1

i.e. 1 = 1a

Hence, 1P (x) x=

(b) Since in 4P (x) , n = k = 4, then from the first part of equation (47), page 510 of textbook, i.e. the

even powers of x:

y = 2 40

4(5) 4(5)(2)(7)a 1 x x 02! 4!

⎧ ⎫− + +⎨ ⎬⎩ ⎭

= 2 40

35a 1 10x x3

⎧ ⎫− +⎨ ⎬⎩ ⎭

0a is chosen to make y = 1 when x = 1


i.e. 1 = 0 0 035 2 8a 1 10 a 1 10 11 a3 3 3

⎛ ⎞ ⎛ ⎞− + = − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, from which, 0a = 38

Hence, 4P (x) = 2 43 351 10x x8 3⎧ ⎫− +⎨ ⎬⎩ ⎭

or 4P (x) = ( )4 21 35x 30x 38

− +

(c) Since in 5P (x) , n = k = 5, then from the second part of equation (47), i.e. the odd powers of x:

y = 3 51

(k 1)(k 2) (k 1)(k 3)(k 2)(k 4)a x x x ...3! 5!

− + − − + +⎧ ⎫− + −⎨ ⎬⎩ ⎭

i.e. y = 3 5 3 51 1

(4)(7) (4)(2)(7)(9) 14 21a x x x 0 a x x x3! 5! 3 5

⎧ ⎫ ⎧ ⎫− + − = − +⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭

1a is chosen to make y = 1 when x = 1.

i.e. 1 = 1 1 114 21 15 70 63 8a 1 a a3 5 15 15

− +⎧ ⎫ ⎛ ⎞− + = =⎨ ⎬ ⎜ ⎟⎩ ⎭ ⎝ ⎠

from which, 115a8

=

Hence, 5P (x) = 3 515 14 21x x x8 3 5⎛ ⎞− +⎜ ⎟⎝ ⎠

or 5P (x) = ( )5 31 63x 70x 15x8

− +


CHAPTER 53 AN INTRODUCTION TO PARTIAL

DIFFERENTIAL EQUATIONS EXERCISE 200 Page 514

2. Solve u 2t cost

∂= θ

∂ given that u = 2t when θ = 0

Since u 2t cost

∂= θ

∂ then u = 2t cos dtθ∫ = (2 cos θ) t dt∫

= ( )2

2t2cos f ( ) t cos f ( )2

θ + θ = θ+ θ

u = 2t when θ = 0, hence, 2t = 2t f ( )+ θ

from which, ( ) 2f 2t tθ = −

Hence, u = 2 2t cos 2t tθ+ −

or u = ( )2t cos 1 2tθ − +

4. Verify that u = ye cos x− is a solution of 2 2

2 2

u u 0x y∂ ∂

+ =∂ ∂

Since u = ye cos x− then yu e ( sin x)x

−∂= −

∂ and

2y

2

u e cos xx

−∂= −

∂

Also, yu e cos xy

−∂= −

∂ and

2y

2

u e cos xx

−∂=

∂

Hence, 2 2

y y2 2

u u e cos x e cos x 0x y

− −∂ ∂+ = − + =

∂ ∂

6. Solve ( )2

22

u y 4x 1x∂

= −∂

given that x = 0, u = sin y and ux∂∂

= cos 2y

Since ( )2

22

u y 4x 1x∂

= −∂

then ( )3

2u 4xy 4x 1 dx y x f (y)x 3

⎛ ⎞∂= − = − +⎜ ⎟∂ ⎝ ⎠∫

x = 0 when ux∂∂

= cos 2y, hence, cos 2y = 0 + f(y)

Hence, 3u 4xy x cos 2y

x 3⎛ ⎞∂

= − +⎜ ⎟∂ ⎝ ⎠


and u = 3 4 24x x xy x cos 2y dx y x cos 2y F(y)

3 3 2⎡ ⎤⎛ ⎞ ⎛ ⎞

− + = − + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∫

x = 0 when u = sin y, hence, sin y = F(y)

Thus, u =4 2x xy xcos 2y sin y

3 2⎛ ⎞

− + +⎜ ⎟⎝ ⎠

8. Show that u(x, y) = xxyy

+ is a solution of 2 2

2

u u2x y 2xx y y∂ ∂

+ =∂ ∂ ∂

Since, u = xxyy

+ then u 1yx y∂

= +∂

and 22

u xx x xyy y

−∂= − = −

∂

and 2

2 3

u 2xy y∂

=∂

Also, 2

2 2

u x 1x 1x y x y y

⎛ ⎞∂ ∂= − = −⎜ ⎟∂ ∂ ∂ ⎝ ⎠

Hence, L.H.S. = 2 2

2 2 3

u u 1 2x2x y 2x 1 yx y y y y

⎛ ⎞ ⎛ ⎞∂ ∂+ = − +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠

= 2 2

2x 2x2x 2x R.H.S.y y

− + = =

10. Verify that φ(x, y) = x cos y + xe sin y satisfies the differential equation

2 2

2 2 x cos y 0x y∂ φ ∂ φ

+ + =∂ ∂

Since φ = x cos y + xe sin y then xcos y e sin yx∂φ

= +∂

and 2

x2 e sin y

x∂ φ

=∂

and xx sin y e cos yy∂φ

= − +∂

and 2

x2 x cos y e sin y

y∂ φ

= − −∂

Hence, L.H.S. = ( )2 2

x x2 2 x cos y e sin y x cos y e sin y x cos y

x y∂ φ ∂ φ

+ + = + − − +∂ ∂

= x xe sin y x cos y e sin y x cos y 0 R.H.S.− − + = =


EXERCISE 201 Page 516 2. Solve T ′′- c2µT = 0 given c = 3 and µ = -1 Since T ′′- c2µT = 0 and c = 3 and µ = -1, then T ′′- (3)2(-1)T = 0 i.e. T ′′ + 9T = 0

If T′′ + 9T = 0 then the auxiliary equation is:

2m 9 0+ = i.e. 2m 9= − from which, m 9 j3= − = ± or 0 ± j3

Thus, the general solution is: T = 0e A cos3t Bsin 3t+

= A cos 3t + B sin 3t

3. Solve X ′′ = µX given µ = 1 Since X ′′ = µX and µ = 1, then X ′′ - X = 0

If X ′′ - X = 0 then the auxiliary equation is:

2m 1 0− = i.e. 2m 1= from which, m = 1 or m = -1

Thus, the general solution is: X = x xAe Be−+


EXERCISE 202 Page 520 1. An elastic string is stretched between two points 40 cm apart. Its centre point is displaced 1.5 cm

from its position of rest at right angles to the original direction of the string and then released

with zero velocity. Determine the subsequent motion u(x, t) by applying the wave equation

2 2

2 2 2

u 1 ux c t∂ ∂

=∂ ∂

with 2c = 9.

The elastic string is shown in the diagram below.

1. The boundary and initial conditions given are:

u(0, t) 0u(40, t) 0

= ⎫⎬= ⎭

u(x, 0) = f (x) = 1.5 x20

0 ≤ x ≤ 20

= 1.5 x 320

− + = 60 1.5x20− 20 ≤ x ≤ 40

t 0

u 0t =

∂⎡ ⎤ =⎢ ⎥∂⎣ ⎦ i.e. zero initial velocity

2. Assuming a solution u = XT, where X is a function of x only, and T is a function of t only,

then u X 'Tx∂

=∂

and 2

2

u X ''Tx∂

=∂

and u XT 'y∂

=∂

and 2

2

u XT ''y∂

=∂

Substituting into the partial differential equation, 2 2

2 2 2

u 1 ux c t∂ ∂

=∂ ∂

gives: 2

1X ''T XT ''c

= i.e. 1X ''T XT ''9

= since 2c 9=

3. Separating the variables gives: X '' T ''X 9T

=


Let constant, µ = X '' T ''X 9T

= then µ = X ''X

and µ = T ''9T

from which, X′′ - µX = 0 and T′′ - 9µ T = 0

4. Letting µ = - 2p to give an oscillatory solution gives

X′′ + 2p X = 0 and the auxiliary equation is: 2 2m p+ = 0 from which, 2m p jp= − = ±

and T′′ + 9 2p T = 0 and the auxiliary equation is:

2 2m 9p+ = 0 from which, 2m 9p j3p= − = ±

5. Solving each equation gives: X = A cos px + B sin px and T = C cos 3pt + D sin 3pt

Thus, u(x, t) = A cos px + B sin pxC cos 3pt + D sin 3pt

6. Applying the boundary conditions to determine constants A and B gives:

(i) u(0, t) = 0, hence 0 = AC cos 3pt + D sin 3pt from which we conclude that A = 0

Therefore, u(x, t) = B sin px C cos 3pt + D sin 3pt (1)

(ii) u(40, t) = 0, hence 0 = B sin 40pC cos 3pt + D sin 3pt

B ≠ 0 hence sin 40p = 0 from which, 40p = nπ and p = n40π

7. Substituting in equation (1) gives: u(x, t) = B sin n x40π 3n t 3n tCcos Dsin

40 40π π⎧ ⎫+⎨ ⎬

⎩ ⎭

or, more generally, n n nn 1

n x 3n t 3n tu (x, t) sin A cos B sin40 40 40

∞

=

π π π⎧ ⎫= +⎨ ⎬⎩ ⎭

∑ (2)

where nA = BC and nB = BD

8. From equation (8), page 517 of textbook,

nA = L

0

2 n xf (x)sin dxL L

π∫

= 20 40

0 20

2 1.5 n x 60 1.5x n xx sin dx sin dx40 20 40 20 40

⎡ π − π ⎤⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ∫

Each integral is determined using integration by parts (see chapter 43, page 418) with the result:

nA = 2 2 2 2

(8)(1.5) n 12 nsin sinn 2 n 2

π π=

π π


From equation (9), page 518 of textbook, L

n 0

2 n xB g(x)sin dxcn L

π=

π ∫

t 0

u 0t =

∂⎡ ⎤ =⎢ ⎥∂⎣ ⎦ = g(x) thus, nB 0=

Substituting into equation (2) gives: n n nn 1

n x 3n t 3n tu (x, t) sin A cos B sin40 40 40

∞

=

π π π⎧ ⎫= +⎨ ⎬⎩ ⎭

∑

= 2 2n 1

n x 12 n 3n t n tsin sin cos (0)sin40 n 2 40 50

∞

=

π π π π⎧ ⎫+⎨ ⎬π⎩ ⎭∑

Hence, u(x, t) = 2 2n 1

12 1 n n x 3n tsin sin cosn 2 40 40

∞

=

π π ππ ∑

2. The centre point of an elastic string between two points P and Q, 80 cm apart, is deflected a

distance of 1 cm from its position of rest perpendicular to PQ and released initially with zero

velocity. Apply the wave equation 2 2

2 2 2

u 1 ux c t∂ ∂

=∂ ∂

where c = 8, to determine the motion of a

point distance x from P at time t.

The elastic string is shown in the diagram below.

The boundary and initial conditions given are:

u(0, t) 0u(80, t) 0

= ⎫⎬= ⎭

u(x, 0) = f (x) = 1 x40

0 ≤ x ≤ 40

= 1 x 240

− + 40 ≤ x ≤ 80


t 0

u 0t =

∂⎡ ⎤ =⎢ ⎥∂⎣ ⎦ i.e. zero initial velocity

Assuming a solution u = XT, where X is a function of x only, and T is a function of t only,

then u X 'Tx∂

=∂

and 2

2

u X ''Tx∂

=∂

and u XT 'y∂

=∂

and 2

2

u XT ''y∂

=∂

Substituting into the partial differential equation, 2 2

2 2 2

u 1 ux c t∂ ∂

=∂ ∂

gives: 2

1X ''T XT ''c

= i.e. 1X ''T XT ''64

= since c = 8

Separating the variables gives: X '' T ''X 64T

=

Let constant, µ = X '' T ''X 64T

= then µ = X ''X

and µ = T ''64T

from which, X′′ - µX = 0 and T′′ - 64µ T = 0

Letting µ = - 2p to give an oscillatory solution gives

X′′ + 2p X = 0 and the auxiliary equation is: 2 2m p+ = 0 from which, 2m p jp= − = ±

and T′′ + 64 2p T = 0 and the auxiliary equation is:

2 2m 64p+ = 0 from which, 2m 64p j8p= − = ±

Solving each equation gives: X = A cos px + B sin px and T = C cos 8pt + D sin 8pt

Thus, u(x, t) = A cos px + B sin pxC cos 8pt + D sin 8pt

Applying the boundary conditions to determine constants A and B gives:

(i) u(0, t) = 0, hence 0 = AC cos 8pt + D sin 8pt from which we conclude that A = 0

Therefore, u(x, t) = B sin px C cos 8pt + D sin 8pt (1)

(ii) u(80, t) = 0, hence 0 = B sin 80pC cos 8pt + D sin 8pt

B ≠ 0 hence sin 80p = 0 from which, 80p = nπ and p = n80π

Substituting in equation (1) gives: u(x, t) = B sin n x80π 8n t 8n tCcos Dsin

80 80π π⎧ ⎫+⎨ ⎬

⎩ ⎭

or, more generally, n n nn 1

n x n t n tu (x, t) sin A cos B sin80 10 10

∞

=

π π π⎧ ⎫= +⎨ ⎬⎩ ⎭

∑ (2)


nA = 2 2 2 2

(8)(1) n 8 nsin sinn 2 n 2

π π=

π π

L

n 0

2 n xB g(x)sin dxcn L

π=

π ∫ t 0

u 0t =

∂⎡ ⎤ =⎢ ⎥∂⎣ ⎦ = g(x) thus, nB 0=

Substituting into equation (2) gives: n n nn 1

n x n t n tu (x, t) sin A cos B sin80 10 10

∞

=

π π π⎧ ⎫= +⎨ ⎬⎩ ⎭

∑

= 2 2n 1

n x 8 n n t n tsin sin cos (0)sin80 n 2 10 10

∞

=

π π π π⎧ ⎫+⎨ ⎬π⎩ ⎭∑

Hence, u(x, t) = 2 2n 1

8 1 n n x n tsin sin cosn 2 80 10

∞

=

π π ππ ∑


EXERCISE 203 Page 522 1. A metal bar, insulated along its sides, is 4 m long. It is initially at a temperature of 10°C and at

time t = 0, the ends are placed into ice at 0°C. Find an expression for the temperature at a point P

at a distance x m from one end at any time t seconds after t = 0. The temperature u along the length of bar is shown in the diagram below.

The heat conduction equation is 2

2 2

u 1 ux c t∂ ∂

=∂ ∂

and the given boundary conditions are:

u(0, t) = 0, u(4, t) = 0 and u(x, 0) = 10

Assuming a solution of the form u = XT, then, X = A cos px + B sin px

and T = 2 2p c tk e−

Thus, the general solution is given by: u(x, t) = P cos px + Q sin px2 2p c te−

u(0, t) = 0 thus 0 = P2 2p c te− from which, P = 0 and u(x, t) = Q sin px

2 2p c te−

Also, u(4, t) = 0 thus 0 = Q sin 4p2 2p c te−

Since Q ≠ 0, sin 4p = 0 from which, 4p = nπ where n = 1, 2, 3, … and p = n4π

Hence, u(x, t) = 2 2p c t

nn 1

n xQ e sin4

∞−

=

π⎧ ⎫⎨ ⎬⎩ ⎭

∑

The final initial condition given was that at t = 0, u = 10, i.e. u(x, 0) = f(x) = 10

Hence, 10 = nn 1

n xQ sin4

∞

=

π⎧ ⎫⎨ ⎬⎩ ⎭

∑


where, from Fourier coefficients, nQ = 2 × mean value of 10 sin n x4π from x = 0 to x = 4,

i.e. nQ = 4

0

2 n x10sin dx4 4

π∫ =

4

0

n xcos45 n

4

π⎡ ⎤⎢ ⎥−⎢ ⎥π⎢ ⎥⎣ ⎦

= 20 4ncos cos 0n 4

π⎡ ⎤− −⎢ ⎥π ⎣ ⎦ = ( )20 1 cos n

n− π

π

= 0 (when n is even) and 40nπ

(when n is odd)

Hence, the required solution is: u(x, t) = 2 2p c t

nn 1

n xQ e sin4

∞−

=

π⎧ ⎫⎨ ⎬⎩ ⎭

∑

= 2 2 2n c t

16

n(odd) 1

40 1 n xe sinn 4

π∞ −

=

ππ ∑

3. The ends of an insulated rod PQ, 20 units long, are maintained at 0°C. At time t = 0, the

temperature within the rod rises uniformly from each end reaching 4°C at the mid-point of PQ.

Find an expression for the temperature u(x, t) at any point in the rod, distant x from P at any time

after t = 0. Assume the heat conduction equation to be 2

2 2

u 1 ux c t∂ ∂

=∂ ∂

and take c2 = 1

The temperature along the length of the rod is shown in the diagram below.

The heat conduction equation is 2

2 2

u 1 ux c t∂ ∂

=∂ ∂

and the given boundary conditions are:

u(0, t) = 0, u(20, t) = 0 and u(x, 0) = 0

Assuming a solution of the form u = XT, then, X = A cos px + B sin px

and T = 2 2p c tk e−

Thus, the general solution is given by: u(x, t) = P cos px + Q sin px2 2p c te−


u(0, t) = 0 thus 0 = P2 2p c te− from which, P = 0 and u(x, t) = Q sin px

2 2p c te−

Also, u(20, t) = 0 thus 0 = Q sin 20p2 2p c te−

Since Q ≠ 0, sin 20p = 0 from which, 20p = nπ where n = 1, 2, 3, … and p = n20π

Hence, u(x, t) = 2 2p c t

nn 1

n xQ e sin20

∞−

=

π⎧ ⎫⎨ ⎬⎩ ⎭

∑

where, from Fourier coefficients, 2 × the mean value from x = 0 to x = 20

nQ = 10 20

0 10

2 2 n x 2 n xx sin dx x 8 sin20 5 20 5 20

⎡ π π ⎤⎛ ⎞ ⎛ ⎞+ − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ∫ (see above diagram)

=

10 20

2 2

0 10

2 n x 2 n x2 n x 2 n x n xx cos x cossin sin 8cos1 5 20 5 205 20 5 20 20n n n10 n n20 20 2020 20

⎧ ⎫⎡ ⎤ ⎡ ⎤π π⎛ ⎞ ⎛ ⎞π π π⎪ ⎪−⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥+ − −⎨ ⎬π π π⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞π π⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎩ ⎭

= ( )2 2

n n n n n4cos 4sin 4cos 4sin 8cos1 8cos n 8cosn2 2 2 2 20 0n n n n n10 n n20 20 20 20 2020 20

⎧⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞π π π π π⎪⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟− ⎜ ⎟π π⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟+ − + + − − − −⎨⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟π π⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

⎫⎪

⎪ ⎪⎬

⎪ ⎪⎪ ⎪⎩ ⎭

= 2 2

n n n n n4cos 4sin 4cos 4sin 8cos1 8cos n 8cos n2 2 2 2 2n n n n n10 n n20 20 20 20 2020 20

⎧ ⎫π π π π π⎪ ⎪− π π⎪ ⎪+ + − − + +⎨ ⎬π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞π π⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎩ ⎭

= 2 2

n n n n8cos 8sin 8cos 8sin1 12 2 2 2n n10 10n n20 2020 20

⎧ ⎫ ⎧ ⎫π π π π⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪+ + =⎨ ⎬ ⎨ ⎬π π⎛ ⎞ ⎛ ⎞π π⎛ ⎞ ⎛ ⎞⎪ ⎪ ⎪ ⎪⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎪ ⎪ ⎪ ⎪⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎩ ⎭ ⎩ ⎭

= 0 when n is even

= 2

2 2

8 20 n 320 nsin sin10 n 2 n 2

⎧ ⎫π π⎪ ⎪⎛ ⎞ =⎨ ⎬⎜ ⎟π π⎝ ⎠⎪ ⎪⎩ ⎭ when n is odd

Hence, the required solution is: u(x, t) = 2 2

2n (1) t20

2 2n 1

320 n n xsin e sinn 2 20

π∞ −

=

⎧ ⎫π π⎪ ⎪⎨ ⎬π⎪ ⎪⎩ ⎭

∑

= 2 2n t400

2 2n(odd) 1

320 1 n n xsin sin en 2 20

⎛ ⎞π∞ − ⎜ ⎟⎜ ⎟⎝ ⎠

=

π ππ ∑


EXERCISE 204 Page 524 1. A rectangular plate is bounded by the lines x = 0, y = 0, x = 1 and y = 3. Apply the Laplace

equation 2 2

2 2

u u 0x y∂ ∂

+ =∂ ∂

to determine the potential distribution u(x, y) over the plate, subject to

the following boundary conditions:

u = 0 when x = 0 0 ≤ y ≤ 2 u = 0 when x = 1 0 ≤ y ≤ 2

u = 0 when x = 2 0 ≤ x ≤ 1 u = 5 when x = 3 0 ≤ x ≤ 1

Initially a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and

Y is a function of y only. Simplifying to u = XY, determining partial derivatives, and substituting

into 2 2

2 2

u u 0x y∂ ∂

+ =∂ ∂

gives: X′′Y + XY′′ = 0

Separating the variables gives: X '' Y ''X Y

= −

Letting each side equal a constant, - 2p gives the two equations:

X′′ + 2p X = 0 and Y′′ - 2p Y = 0

from which, X = A cos px + B sin px

and Y = py pyCe De−+ or Y = C cosh py + D sinh py or Y = E sinh p(y + φ)

Hence u(x, y) = XY = A cos px + B sin px E sinh p(y + φ)

or u(x, y) = P cos px + Q sin px sinh p(y + φ) where P = AE and Q = BE

The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + φ) from which, P = 0

Hence, u(x, y) = Q sin px sinh p(y + φ)

The second boundary condition is: u(1, y) = 0, hence 0 = Q sin p(1) sinh p(y + φ)

from which, sin p = 0, hence, p = nπ for n = 1, 2, 3, …

The third boundary condition is: u(x, 2) = 0, hence, 0 = Q sin px sinh p(2 + φ)

from which, sinh p(2 + φ) = 0 and φ = -2

Hence, u(x, y) = Q sin px sinh p(y – 2)

Since there are many solutions for integer values of n,


u(x, y) = nn 1

Q sin px sinh p(y 2)∞

=

−∑ = nn 1

Q sin n x sinh n (y 2)∞

=

π π −∑ (a)

The fourth boundary condition is: u(x, 3) = 5 = f (x), hence, f (x) = nn 1

Q sin n x sinh n (3 2)∞

=

π π −∑

From Fourier series coefficients,

nQ sinh nπ = 2 × the mean value of f (x) sin n xπ from x = 0 to x = 1

i.e. = 1

1

00

2 cos n x5sin n x dx 101 n

π⎡ ⎤π = −⎢ ⎥π⎣ ⎦∫ = ( ) ( )10 10cos n cos 0 1 cos nn n

− π− = − ππ π

= 0 (for even values of n), = 20nπ

(for odd values of n)

Hence, n20 20Q

n (sinh n ) n= =

π π πcosech nπ

Hence, from equation (a), u(x, y) = nn 1

Q sin n x sinh n (y 2)∞

=

π π −∑

= ( )n odd) 1

20 1 cosechn sinn x sinhn (y 2)n

∞

=

π π π −π ∑

2. A rectangular plate is bounded by the lines x = 0, y = 0, x = 3, y = 2. Determine the potential

distribution u(x, y) over the rectangle using the Laplace equation 2 2

2 2

u u 0x y∂ ∂

+ =∂ ∂

subject to the

following boundary conditions:

u(0, y) = 0 0 ≤ y ≤ 2

u(3, y) = 0 0 ≤ y ≤ 2

u(x, 2) = 0 0 ≤ x ≤ 3

u(x, 0) = x(3 – x) 0 ≤ x ≤ 3 Initially a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and

Y is a function of y only. Simplifying to u = XY, determining partial derivatives, and substituting

into 2 2

2 2

u u 0x y∂ ∂

+ =∂ ∂

gives: X′′Y + XY′′ = 0

Separating the variables gives: X '' Y ''X Y

= −

Letting each side equal a constant, - 2p gives the two equations:

X′′ + 2p X = 0 and Y′′ - 2p Y = 0


from which, X = A cos px + B sin px

and Y = py pyCe De−+ or Y = C cosh py + D sinh py or Y = E sinh p(y + φ)

Hence u(x, y) = XY = A cos px + B sin px E sinh p(y + φ)

or u(x, y) = P cos px + Q sin px sinh p(y + φ)

The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + φ) from which, P = 0

Hence, u(x, y) = Q sin px sinh p(y + φ)

The second boundary condition is: u(3, y) = 0, hence 0 = Q sin 3p sinh p(y + φ)

from which, sin 3p = 0, hence, 3p = nπ i.e. p = n3π for n = 1, 2, 3, …

The third boundary condition is: u(x, 2) = 0, hence, 0 = Q sin px sinh p(2 + φ)

from which, sinh p(2 + φ) = 0 and φ = - 2

Hence, u(x, y) = Q sin px sinh p(y – 2)

Since there are many solutions for integer values of n,

u(x, y) = nn 1

Q sin px sinh p(y 2)∞

=

−∑ = nn 1

n nQ sin x sinh (y 2)3 3

∞

=

π π−∑ (a)

The fourth boundary condition is: u(x, 0) = x(3 – x) = 3x - 2x = f (x),

hence, f (x) = nn 1

n nQ sin x sinh ( 2)3 3

∞

=

π π−∑

From Fourier series coefficients,

n2nQ sinh3

− π = 2 × the mean value of f (x) sin n x3π from x = 0 to x = 3

= ( )3 2

0

2 n3x x sin x dx1 3

π−∫

=

3 3

2

2 2 3

0 0

n x n x n x n x n x3x cos 3sin x cos 2xsin 2cos3 3 3 3 32

n nn n n3 33 3 3

⎧ ⎫⎡ ⎤ ⎡ ⎤π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪− −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥+ − + +⎨ ⎬⎢ ⎥ ⎢ ⎥π π⎛ ⎞ ⎛ ⎞π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎩ ⎭

by integration by parts (see chapter 43)


= ( )3 3 3 3

9cos n 9cos n 2cos n 2 272 2 2 2cos nn n nn n3 3 3 3

⎧ ⎫⎪ ⎪− π π π⎪ ⎪ ⎧ ⎫+ − + = − π⎨ ⎬ ⎨ ⎬π π π⎛ ⎞ ⎛ ⎞ ⎩ ⎭π π⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

= ( )3 3

54 2 2cos nn

− ππ

= 0 (for even values of n), = 3 3

216n π

(for odd values of n)

Hence, n 3 33 3

216 216 2nQ cos ech2n n 3n sinh3

− π= =

− π π⎛ ⎞π ⎜ ⎟⎝ ⎠

Hence, from equation (a),

u(x, y) = n 3 3n 1 n 1

n n 216 2n n x nQ sin x sinh (y 2) cos ech sin sinh (y 2)3 3 n 3 3 3

∞ ∞

= =

π π − π π π− = −

π∑ ∑

= ( )

3 3n odd) 1

216 1 2n n x ncosech sin sinh (2 y)n 3 3 3

∞

=

π π π−

π ∑


CHAPTER 54 PRESENTATION OF STATISTICAL DATA EXERCISE 205 Page 527 1. State whether the following data is discrete or continuous.

(a) The amount of petrol produced daily, for each of 31 days, by a refinery.

(b) The amount of coal produced daily by each of 15 miners.

(c) The number of bottles of milk delivered by each of 20 milkmen.

(d) The size of 10 samples of rivets produced by a machine. (a) Continuous – could be any amount of petrol.

(b) Continuous – could be any amount of coal.

(c) Discrete – can only be a whole number of bottles of milk.

(d) Continuous – could be any size of rivet.

2. State whether the following data is discrete or continuous.

(a) The number of people visiting an exhibition on each of 5 days.

(b) The time taken by each of 12 athletes to run 100 metres.

(c) The value of stamps sold in a day by each of 20 post offices.

(d) The number of defective items produced in each of 10 one-hour periods by a machine.

(a) Discrete – can only be a whole number of people.

(b) Continuous – could be any time taken.

(c) Discrete – can only be a whole number of stamps.

(d) Discrete – can only be a whole number of defective items.


EXERCISE 206 Page 530 3. The number of vehicles passing a stationary observer on a road in six ten-minute intervals is as

shown. Draw a horizontal bar chart.

Period of time 1 2 3 4 5 6

Number of vehicles 35 44 62 68 49 41 A horizontal bar chart is shown below.

6. The number of components produced by a factory in a week is as shown below:

Day Mon Tues Wed Thur Fri

Number of components 1580 2190 1840 2385 1280

Depict the data on a vertical bar chart. A vertical bar chart is shown below.


8. A company has five distribution centres and the mass of goods in tonnes sent to each centre

during four, one-week periods, is as shown.

Week 1 2 3 4

Centre A 147 160 174 158

Centre B 54 63 77 69

Centre C 283 251 237 211

Centre D 97 104 117 144

Centre E 224 218 203 194

Use a percentage component bar chart to present these data and comment on any trends. Week 1: Total = 147 + 54 + 283 + 97 + 224 = 805

A = 147 100% 18%805

× ≈ , B = 54 100% 7%805

× ≈ , C ≈ 35%, D ≈ 12%, E ≈ 28%

Week 2: Total = 160 + 63 + 251 + 104 + 218 = 796

A = 160 100% 20%796

× ≈ , B = 63 100% 8%796

× ≈ , C ≈ 32%, D ≈ 13%, E ≈ 27%

Week 3: Total = 174 + 77 + 237 + 117 + 203 = 808

A = 174 100% 22%808

× ≈ , B = 77 100% 10%808

× ≈ , C ≈ 29%, D ≈ 14%, E ≈ 25%

Week 4: Total = 158 + 69 + 211 + 144 + 194 = 776

A = 158 100% 20%776

× ≈ , B = 69 100% 9%776

× ≈ , C ≈ 27%, D ≈ 19%, E ≈ 25%

A percentage component bar chart is shown below.


From the above percentage component bar chart, it is seen that there is little change in centres A

and B, there is a reduction of around 8% in centre C, an increase of around 7% in centre D and a

reduction of about 3% in centre E.

10. The way in which an apprentice spent his time over a one-month period is as follows:

drawing office 44 hours, production 64 hours, training 12 hours, at college 28 hours

Use a pie chart to depict this information. Total hours = 44 + 64 + 12 + 28 = 148

Drawing office, D = 44 360 107148

× ° ≈ ° , Production, P = 64 360 156148

× ° ≈ ° ,

Training, T = 12 360 29148

× ° ≈ ° , College, C = 28 360 68148

× ° ≈ °

A pie chart to depict this information is shown below.

12. (a) If the company sell 23500 units per annum of the product depicted in Fig 54.5 on page XX

of textbook, determine the cost of their overheads per annum.

(b) If 1% of the dwellings represented in year 1 of Fig 54.4 on page XX of the textbook

corresponds to 2 dwellings, find the number of houses sold in that year.

(a) Overheads = 126 100% 35%360

× = of total costs.

Cost per unit = £2, hence total income per annum = 23500 × 2 = £47000

Cost of overheads per annum = 35% of £47000 = 35 47000100

× = £16450

(b) Percentage of houses sold in year 1 = 22 + 32 + 15 = 69%

If 1% corresponds to 2 dwellings then the number of houses sold = 69 × 2 = 138 houses


EXERCISE 207 Page 536 3. The information given below refers to the value of resistance in ohms of a batch of 48 resistors

of similar value. Form a frequency distribution for the data having about 6 classes, and draw a

frequency polygon and histogram to represent these data diagrammatically.

21.0 22.4 22.8 21.5 22.6 21.1 21.6 22.3

22.9 20.5 21.8 22.2 21.0 21.7 22.5 20.7

23.2 22.9 21.7 21.4 22.1 22.2 22.3 21.3

22.1 21.8 22.0 22.7 21.7 21.9 21.1 22.6

21.4 22.4 22.3 20.9 22.8 21.2 22.7 21.6

22.2 21.6 21.3 22.1 21.5 22.0 23.4 21.2 The range is from 20.5 to 23.4, i.e. range = 23.4 – 20.5 = 2.9

2.9 ÷ 6 ≈ 0.5 hence, classes of 20.5 – 20.9, 21.0 – 21.4, and so on are chosen as shown in the frequency distribution below.

A frequency polygon is shown below where class mid-point values are plotted against frequency

values. Class mid-points occur at 20.7, 21.2, 21.7, and so on.


The histogram for the above frequency distribution is shown below.

5. Form a cumulative frequency distribution and hence draw the ogive for the frequency

distribution given in the solution to Problem 3 above. A cumulative frequency distribution is shown below.

Class Frequency Upper class boundary Cumulative frequency 20.5 – 20.9 21.0 – 21.4 21.5 – 21.9 22.0 – 22.4 22.5 – 22.9 23.0 – 23.4

3 10 11 13 9 2

Less than 20.95 21.45 21.95 22.45 22.95 23.45

3 13 24 37 46 48

An ogive for the above frequency distribution is shown below.


9. The diameter in millimetres of a reel of wire is measured in 48 places and the results are as

shown.

2.10 2.29 2.32 2.21 2.14 2.22 2.28 2.18 2.17 2.20 2.23 2.13 2.26 2.10 2.21 2.17 2.28 2.15 2.16 2.25 2.23 2.11 2.27 2.34 2.24 2.05 2.29 2.18 2.24 2.16 2.15 2.22 2.14 2.27 2.09 2.21 2.11 2.17 2.22 2.19 2.12 2.30 2.23 2.07 2.13 2.26 2.16 2.12

(a) Form a frequency distribution of diameters having about 6 classes

(b) Draw a histogram depicting the data.

(c) Form a cumulative frequency distribution.

(d) Draw an ogive for the data. (a) Range = 2.34 – 2.05 = 0.29

0.29 ÷ 6 ≈ 0.5, hence classes of 2.05 - 2.09, 2.10 -2.14, and so on are chosen, as shown in the

frequency distribution below.

(b) A histogram depicting the data is shown below.


(c) A cumulative frequency distribution is shown below.

Class Frequency Upper class boundary Cumulative frequency 2.05 – 2.09 2.10 – 2.14 2.15 – 2.19 2.20 – 2.24 2.25 – 2.29 2.30 – 2.34

3 10 11 13 9 2

Less than 2.095 2.145 2.195 2.245 2.295 2.345

3 13 24 37 46 48

(d) An ogive for the above data is shown below.


CHAPTER 55 MEASURES OF CENTRAL TENDENCY AND

DISPERSION EXERCISE 208 Page 539

2. Determine the mean, median and modal values for the set:

26, 31, 21, 29, 32, 26, 25, 28

Mean = 26 31 21 29 32 26 25 28 2188 8

+ + + + + + += = 27.25

Ranking gives: 21 25 26 26 28 29 31 32

Median = middle value = 26 282+ = 27

Most commonly occurring value, i.e. mode = 26

4. Determine the mean, median and modal values for the set:

73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9

Mean = 73.8 126.4 40.7 141.7 28.5 237.4 157.9 806.47 7

+ + + + + += = 115.2

Ranking gives: 28.5 40.7 73.8 126.4 141.7 157.9 237.4

Middle value = median = 126.4

There is no mode since all the values are different.



1. The frequency distribution given below refers to the height in centimetres of 100 people.

Determine the mean value of the distribution, correct to the nearest millimetre.

150–156 5, 157-163 18, 164-170 20,

171-177 27, 178-184 22, 185-191 8

Mean value = ( ) ( ) ( ) ( ) ( ) ( )5 153 18 160 20 167 27 174 22 181 8 188100

× + × + × + × + × + ×

= 17169100

= 171.7 cm

3. The diameters, in centimetres, of 60 holes bored in engine castings are measured and the results

are as shown. Draw a histogram depicting these results and hence determine the mean, median

and modal values of the distribution.

2.011-2.014 7, 2.016-2.019 16, 2.021-2.024 23, 2.026-2.029 9,

2.031-2.034 5

The histogram is shown below.


The mean value lies at the centroid of the histogram. With reference to axis YY at 2.010 cm,

A M (a m)=∑

where A = area of histogram = 35 + 80 + 115 + 45 + 25 = 300 and M = horizontal distance of

centroid from YY. (Actually, the area of, say, 35 square units is 335 10−× square units; however, the

310− will cancel on each side of the equation so has been omitted).

Hence, 300 M = (35 × 0.0025) + (80 × 0.0075) + (115 × 0.0125)

+ (45 × 0.0175) + (25 × 0.0225)

i.e. 300 M = 3.475

i.e. M = 3.475 0.01158300

= cm

Thus, the mean is at 2.010 + 0.01158 = 2.02158 cm

The median is the diameter where the area on each side of it is the same, i.e. 300/2, i.e. 150 square

units on each side.

The first two rectangles have an area of 35 + 80 = 115; hence, 35 more square units are needed from

the third rectangle. 35 100% 30.43%115

× = of the distance from 2.020 to 2.025

i.e. 0.3043 × (2.025 – 2.020) = 0.00152

i.e. median occurs at 2.020 + 0.00152 = 2.02152 cm

The mode is at the intersection of AC and BD, i.e. at 2.02167 cm


EXERCISE 210 Page 542 1. Determine the standard deviation from the mean of the set of numbers:

35, 22, 25, 23, 28, 33, 30

Mean, 35 22 25 23 28 33 30 196x 287 7

+ + + + + += = =

Standard deviation,

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2x x 35 28 22 28 25 28 23 28 28 28 33 28 30 28n 7

⎧ ⎫−⎪ ⎪ ⎧ ⎫− + − + − + − + − + − + −⎪ ⎪ ⎪ ⎪σ = =⎨ ⎬ ⎨ ⎬⎪ ⎪⎪ ⎪ ⎩ ⎭

⎪ ⎪⎩ ⎭

∑

= 148 21.1437

= = 4.60, correct to 3 significant figures.

3. The tensile strength in megapascals for 15 samples of tin were determined and found to be:

34.61, 34.57, 34.40, 34.63, 34.63, 34.51, 34.49, 34.61,

34.52, 34.55, 34.58, 34.53, 34.44, 34.48 and 34.40

Calculate the mean and standard deviation from the mean for these 15 values, correct to 4


Mean, 34.61 34.57 34.40 34.63 ...... 517.95x15 15

+ + + += = = 34.53 MPa

Standard deviation,

( ) ( ) ( ) ( )2 2 2x x 34.61 34.53 34.57 34.53 34.40 34.53 .....

n 15

⎧ ⎫−⎪ ⎪ ⎧ ⎫− + − + − +⎪ ⎪ ⎪ ⎪σ = =⎨ ⎬ ⎨ ⎬⎪ ⎪⎪ ⎪ ⎩ ⎭

⎪ ⎪⎩ ⎭

∑

= 0.0838 0.00558666615

= = 0.07474 MPa

5. Calculate the standard deviation from the mean for the data given in Problem 3 of Exercise 209

above, correct to 3 significant figures. From Problem 3, Exercise 209, mean value, x 2.02158 cm=


Standard deviation, σ = ( ) f x x

f

⎧ ⎫−⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

∑

∑

( ) ( ) ( )( ) ( )

2 2 2

2 2

7 2.0125 2.02158 16 2.0175 2.02158 23 2.0225 2.02158

9 2.0275 2.02158 5 2.0325 2.0215860

⎧ ⎫− + − + −⎪ ⎪⎪ ⎪+ − + −⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

= 0.000577124 0.000266342 0.000019467 0.000315417 0.00059623260

+ + + +

= 0.00177458260

= 0.00544 cm


EXERCISE 211 Page 544 2. The number of faults occurring on a production line in a nine-week period are as shown below.

Determine the median and quartiles values for the data.

30 27 25 24 27 37 31 27 35 Ranking gives: 24 25 27 27 27 30 31 35 37 ↑ ↑ ↑

Median = middle value = 27 faults

1st quartile value = 25 272+ = 26 faults

3rd quartile value = 31 352+ = 33 faults

3. Determine the quartile values and semi-interquartile range for the frequency distribution given in

Problem 1 of Exercise 209 above. The frequency distribution is shown below.

Upper class boundary values Frequency Cumulative frequency 156.5 163.5 170.5 177.5 184.5 191.5

5 18 20 27 22 8

5 23 43 70 92 100

The ogive is shown below. From the ogive, 1Q 164.5cm= , 2Q 172.5cm= and 3Q 179cm=

and semi-interquartile range = 179 164.5 14.52 2−

= = 7.25 cm


5. Determine the numbers in the 6th decile group and in the 81st to 90th percentile group for the set

of numbers:

43 47 30 25 15 51 17 21 36 44 33 17 35 58 51

35 37 33 44 56 40 49 22 44 40 31 41 55 50 16 Ranking gives: 15 16 17 17 21 22 25 30 31 33 33 35 35 36 37 40 40 41 43 44 44 44 47 49 50 51 51 55 56 58 The numbers in the 6th decile group are: 40, 40 and 41 The numbers in the 81st to 90th percentile group are: 50, 51 and 51


CHAPTER 56 PROBABILITY EXERCISE 212 Page 547

2. A box of fuses are all of the same shape and size and comprises 23 2 A fuses, 47 5A fuses and

69 13 A fuses. Determine the probability of selecting at random (a) a 2 A fuse, (b) a 5 A fuse,

and (c) a 13 A fuse.

(a) 2Anumber of 2A fuses 23ptotal number of fuses 23 47 69

= =+ +

= 23139

or 0.1655

(b) 5Anumber of 5A fusesptotal number of fuses

= = 47139

or 0.3381

(c) 13Anumber of 13A fusesptotal number of fuses

= = 69139

or 0.4964

4. The probability of event A happening is 35

and the probability of event B happening is 23

.

Calculate the probabilities of (a) both A and B happening, (b) only event A happening, i.e. event

A happening and event B not happening, (c) only event B happening, and (d) either A, or B, or A

and B happening.

Let A3p5

= and B2p3

= and thus the probability of events not happening, A2p5

= and B1p3

=

(a) The probability of both A and B happening = A B3 2p p5 3

× = × = 25

(b) The probability of event A happening and event B not happening = A B3 1p p5 3

× = × = 15

(c) The probability of only event B happening = B A2 2p p3 5

× = × = 415

(d) The probability of either A, or B, or A and B happening = ( ) ( ) ( )A B B A A Bp p p p p p⎡ ⎤× + × + ×⎣ ⎦

= 3 1 2 2 3 25 3 3 5 5 3

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞× + × + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= 3 4 6 7 615 15 15 15 15⎛ ⎞+ + = +⎜ ⎟⎝ ⎠

= 1315


5. When testing 1000 soldered joints, 4 failed during a vibration test and 5 failed due to having a

high resistance. Determine the probability of a joint failing due to (a) vibration, (b) high

resistance, (c) vibration or high resistance, and (d) vibration and high resistance.

(a) The probability of a joint failing due to vibration, v4p

1000= = 1

250

(b) The probability of a joint failing due to high resistance, R5p

1000= = 1

200

(c) The probability of a joint failing due to vibration or high resistance,

v R1 1 4 5p p

250 200 1000+

+ = + = = 91000

(d) The probability of a joint failing due to vibration and high resistance,

v R1 1p p

250 200× = × = 1

50000

6. Find the probability that the score is 8 if two like dice are thrown.

A score of 8 is achieved with a (2 + 6), (3 + 5), (4 + 4), (5 + 3) and (6 + 2) - see above diagram,

i.e. 5 possibilities, and there are 36 possible scores when throwing two dice.

Hence, the probability of a score of 8 is 536



1. The probability that component A will operate satisfactorily for 5 years is 0.8 and that B will

operate satisfactorily over the same period of time is 0.75. Find the probabilities that in a 5 year

period: (a) both components operate satisfactorily, (b) only component A will operate

satisfactorily, and (c) only component B will operate satisfactorily.

Let satisfactory operations be Ap = 0.8 and Bp = 0.75, and unsatisfactory operations be Ap = 0.2

and Bp = 0.25

(a) The probability that both components operate satisfactorily, A Bp p 0.8 0.75× = × = 0.6

(b) The probability that only component A will operate satisfactorily, A Bp p 0.8 0.25× = × = 0.2

(c) The probability that only component B will operate satisfactorily, B Ap p 0.75 0.2× = × = 0.15

4. A batch of 1 kW fire elements contain 16 which are within a power tolerance and 4 which are

not. If 3 elements are selected at random from the batch, calculate the probabilities that (a) all

three are within the power tolerance and (b) two are within but one is not within the power

tolerance.

(a) The probability that all three are within the power tolerance = 16 15 1420 19 18

× × = 0.4912

(b) The probability that two are within but one is not within the power tolerance

= 16 15 4 16 4 15 4 16 1520 19 18 20 19 18 20 19 18

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × + × × + × ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 3(0.14035) = 0.4211

5. An amplifier is made up of three transistors A, B and C. The probabilities of A, B and C being

defective are 1 1,20 25

and 150

, respectively. Calculate the percentage of amplifiers produced

(a) which work satisfactorily and (b) which have just one defective transistor

Let the probability of transistors working be: A19p20

= , B24p25

= and C49p50

=


(a) The probability of amplifiers working satisfactorily, A B C19 24 49p p p20 25 50

× × = × ×

= 0.8938 or 89.38%

(b) The probability of amplifiers having just one defective transistor

= 1 24 49 1 19 49 1 19 2420 25 50 25 20 50 50 20 25

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × + × × + × ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 0.04704 + 0.03724 + 0.01824

= 0.10252 = 10.25%

6. A box contains 14 40 W lamps, 28 60 W lamps and 58 25 W lamps, all the lamps being of the

same shape and size. Three lamps are drawn at random from the box, first one, them a second,

then a third. Determine the probabilities of: (a) getting one 25 W, one 40 W and one 60 W lamp,

with replacement, (b) getting one 25 W, one 40 W and one 60 W lamp without replacement, and

(c) getting either one 25 W and two 40 W or one 60 W and two 40 W lamps with replacement.

Let 40W

14 14p 0.1414 28 58 100

= = =+ +

, 60W28p 0.28

100= = and 25W

58p 0.58100

= =

(a) The probability of getting one 25 W, one 40 W and one 60 W lamp, with replacement

= 0.58 × 0.14 × 0.28 = 0.0227

(b) The probability of getting one 25 W, one 40 W and one 60 W lamp, without replacement

= 58 14 28100 99 98

× × = 0.0234

(c) The probability of getting either one 25 W and two 40 W or one 60 W and two 40 W lamps,

with replacement = (0.58 × 0.14 × 0.14) + (0.28 × 0.14 × 0.14)

= 0.011368 + 0.005488 = 0.0169


CHAPTER 57 THE BINOMIAL AND POISSON DISTRIBUTION EXERCISE 214 Page 555

1. Concrete blocks are tested and it is found that, on average, 7% fail to meet the required

specification. For a batch of 9 blocks, determine the probabilities that (a) three blocks and (b) less

than four blocks will fail to meet the specification.

Let probability of failure to meet specification, p = 0.07 and probability of success, q = 0.93

By the binomial expansion, ( )9 9 8 7 2 6 3(9)(8) (9)(8)(7)q p q 9q p q p q p .....2! 3!

+ = + + + + hence,

( )9 9 8 7 2 6 3(9)(8) (9)(8)(7)0.93 0.07 (0.93) 9(0.93) (0.07) (0.93) (0.07) (0.93) (0.07) .....2! 3!

+ = + + + +

= 0.5204 + 0.3525 + 0.1061 + 0.0186 +

which corresponds to 0, 1, 2, 3, … failing to meet the specification.

(a) Probability that three blocks fail to meet specification = 0.0186

(b) Probability that less than four blocks fail = 0.5204 + 0.3525 + 0.1061 + 0.0186 = 0.9976

3. The average number of employees absent from a firm each day is 4%. An office within the firm

has seven employees. Determine the probabilities that (a) no employee and (b) three employees

will be absent on a particular day.

Let p = 4% = 0.04 then q = 0.96 (i.e. 96% present)

( ) ( )7 7 7 6 5 2

4 3

(7)(6)q p 0.96 0.04 (0.96) 7(0.96) (0.04) (0.96) (0.04)2!

(7)(6)(5) (0.96) (0.07) .....3!

+ = + = + +

+ +

Which corresponds to 0, 1, 2, 3, …. employees being absent.

(a) The probability that no employee will be absent on a particular day = ( )70.96 = 0.7514

(b) The probability that three employees will be absent on a particular day

= ( ) ( )4 3(7)(6)(5) 0.96 0.043!

= 0.0019


5. Five coins are tossed simultaneously. Determine the probabilities of having 0, 1, 2, 3, 4 and 5

heads upwards, and draw a histogram depicting the results.

Let probability of a head, Hp = 0.5 and the probability of a tail, Tp = 0.5

( )

( )( ) ( )

5 5 4 3 2 2 3

4 5

(5)(4) (5)(4)(3)0.5 0.5 (0.5) 5(0.5) (0.5) (0.5) (0.5) (0.5) (0.5)2! 3!

(5)(4)(3)(2) 0.5 0.5 0.54!

+ = + + +

+ +

= 0.03125 + 0.15625 + 0.3125 + 0.3125 + 0.15625 + 0.03125

which corresponds to 0, 1, 2, 3, 4 and 5 heads landing upwards.

A histogram depicting this data is shown below.

7. An automatic machine produces, on average, 10% of its components outside of the tolerance

required. In a sample of 10 components from this machine, determine the probability of having

three components outside of the tolerance required by assuming a binomial distribution.

Let the probability of a component being outside the tolerance, p = 10% = 0.1 and the probability of

a component being within tolerance, q = 0.9

( ) ( )10 10 10 9 8 2 7 3(10)(9) (10)(9)(8)q p 0.9 0.1 (0.9) 10(0.9) (0.1) (0.9) (0.1) (0.9) (0.1) ....2! 3!

+ = + = + + + +

which corresponds to the probability of 0, 1, 2, 3, …. components being outside the tolerance.

The probability of having three components outside of the required tolerance

= ( ) ( )7 3(10)(9)(8) 0.9 0.13!

= 0.0574



2. The probability that an employee will go to hospital in a certain period of time is 0.0015. Use a

Poisson distribution to determine the probability of more than two employees going to hospital

during this period of time if there are 2000 employees on the payroll.

Average occurrence of the event, λ = np = (2000)(0.0015) = 3

The probability of 0, 1, 2, 3, …. employees going to hospital is given by the terms of:

2 3

e 1 ...2 3

−λ ⎛ ⎞λ λ+ λ + + +⎜ ⎟

⎝ ⎠ =

2 33 3 3 33 3e 3e e e ...

2! 3!− − − −+ + + +

= 0.0498 + 0.1494 + 0.2240 + …

The probability of more than two employees going to hospital = 1 – (0.0498 + 0.1494 + 0.2240)

= 1 – 0.4232 = 0.5768

3. When packaging a product, a manufacturer finds that one packet in twenty is underweight.

Determine the probabilities that in a box of 72 packets (a) two and (b) less than four will be

underweight.

Probability of a packet being underweight, p = 120

= 0.05 and n = 72,

Hence, the average occurrence of event, λ = np = (72)(0.05) = 3.6

The probability of 0, 1, 2, 3, ….packets being underweight is given by the terms of:

2 3

e 1 ...2 3

−λ ⎛ ⎞λ λ+ λ + + +⎜ ⎟

⎝ ⎠ =

2 33.6 3.6 3.6 3.63.6 3.6e 3.6e e e ...

2! 3!− − − −+ + + +

= 0.0273 + 0.0984 + 0.1771 + 0.2125 + …

(a) The probability that two will be underweight = 0.1771

(b) The probability that less than four will be underweight = sum of probabilities of 0, 1, 2 and 3

= 0.0273 + 0.0984 + 0.1771 + 0.2125

= 0.5153


5. The demand for a particular tool from a store is, on average, five times a day and the demand

follows a Poisson distribution. How many of these tools should be kept in the stores so that the

probability of there being one available when required is greater than 10%?

Average occurrence of demand, λ = 5

The probability of 0, 1, 2, 3, … tools being demanded is given by the terms:

e−λ , e−λλ , 2

e2!

−λλ , 3

e3!

−λλ , …. i.e. 5e− , 55e− , 2

55 e2!

− , 3

55 e3!

− , ….

i.e. 0.0067, 0.0333, 0.0842, 0.1404, 0.1755, 0.1755, 0.1462, 0.1044, 0.0653, …

Hence, the probability of wanting a tool 8 times a day is 0.0653, i.e. 6.53% which is less than 10%.

Thus, 7 tools should be kept in the store so that the probability of there being one available when

required is greater than 10%.

6. Failure of a group of particular machine tools follows a Poisson distribution with a mean value

of 0.7. Determine the probabilities of 0, 1, 2, 3, 4 and 5 failures in a week and present these

results on a histogram.

Mean value, λ = 0.7

The probability of 0, 1, 2, 3, 4 and 5 failures in a week is given by the terms:

e−λ , e−λλ , 2

e2!

−λλ , 3

e3!

−λλ , 4

e4!

−λλ and 5

e5!

−λλ

i.e. 0.7e− , 0.70.7e− , 2

0.70.7 e2!

− , 3

0.70.7 e3!

− , 4

0.70.7 e4!

− and 5

0.70.7 e5!

−

i.e. 0.4966, 0.3476, 0.1217, 0.0284, 0.0050 and 0.0007

A histogram depicting these results is shown below:



CHAPTER 58 THE NORMAL DISTRIBUTION EXERCISE 216 Page 563

1. A component is classed as defective if it has a diameter of less than 69 mm. In a batch of 350

components, the mean diameter is 75 mm and the standard deviation is 2.8 mm. Assuming the

diameters are normally distributed, determine how many are likely to be classed as defective.

The z-value corresponding to 69 mm is given by: x x−σ

i.e. 69 752.8− = -2.14 standard deviations

From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -2.14 is 0.4838, i.e. the

shaded area of the diagram below.

Thus the area to the left of the z = -2.14 ordinate is 0.5000 – 0.4838 = 0.0162

The number likely to be classed as defective = 0.0162 × 350 = 5.67 or 6, correct to nearest whole

number.

3. 500 tins of paint have a mean content of 1010 ml and the standard deviation of the contents is

8.7 ml. Assuming the volumes of the contents are normally distributed, calculate the number of

tins likely to have contents whose volumes are less than (a) 1025 ml (b) 1000 ml and (c) 995 ml.

(a) The z-value corresponding to 1025 ml is given by: x x−σ

i.e. 1025 10108.7− = 1.72 standard

deviations

From Table 58.1 on page 561 of textbook, the area between z = 0 and z = 1.72 is 0.4573, i.e. the


Thus the area to the left of the z = 1.72 ordinate is 0.5000 + 0.4573 = 0.9573

The number likely to have less than 1025 ml = 0.9573 × 500 = 479, correct to nearest whole

number.


(b) The z-value corresponding to 1000 ml is given by: x x−σ

i.e. 1000 10108.7− = -1.15 standard

deviations

From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -1.15 is 0.3749, i.e.

the shaded area of the diagram below.

Thus the area to the left of the z = -1.15 ordinate is 0.5000 - 0.3749 = 0.1251


number.

(c) The z-value corresponding to 995 ml is given by: x x−σ

i.e. 995 10108.7− = -1.72 standard

deviations

From Table 58.1, the area between z = 0 and z = -1.72 is 0.4573, i.e. the shaded area of the

diagram below.

Thus the area to the left of the z = -1.72 ordinate is 0.5000 - 0.4573 = 0.0427


number.


4. For the 350 components in Problem 1, if those having a diameter of more than 81.5 mm are

rejected, find, correct to the nearest component, the number likely to be rejected due to being

oversized.

The z-value corresponding to 81.5 mm is given by: x x−σ

i.e. 81.5 752.8− = 2.32 standard deviations.

From Table 58.1 on page 561 of textbook, the area between z = 0 and z = 2.32 is 0.4898, i.e. the


Thus the area to the right of the z = 2.32 ordinate is 0.5000 – 0.4838 = 0.0102

The number likely to be classed as oversized = 0.0102 × 350 = 4, correct to nearest whole

number.

6. The mean diameter of holes produced by a drilling machine bit is 4.05 mm and the standard

deviation of the diameters is 0.0028 mm. For twenty holes drilled using this machine, determine,

correct to the nearest whole number, how many are likely to have diameters of between (a) 4.048

and 4.0553 mm and (b) 4.052 and 4.056mm, assuming the diameters are normally distributed.

(a) The z-value corresponding to 4.048 mm is given by: x x−σ

i.e. 4.048 4.050.0028

− = -0.71 standard

deviations

From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -0.71 is 02611



i.e. 4.0553 4.050.0028

− = 1.89 standard

deviations

From Table 58.1, the area between z = 0 and z = 1.89 is 0.4706

The probability of the diameter being between 4.048 mm and 4.0553 mm is 0.2611 + 0.4706 =

0.7317 (see shaded area in diagram below).

The number likely to have diameter between 4.048 mm and 4.0553 mm = 0.7317 × 20 =

14.63 = 15, correct to nearest whole number.

(b) The z-value corresponding to 4.052 mm is given by: x x−σ

i.e. 4.052 4.050.0028

− = 0.71 standard

deviations

From Table 58.1, the area between z = 0 and z = 0.71 is 02611


i.e. 4.056 4.050.0028

− = 2.14 standard

deviations


The probability of the diameter being between 4.052 mm and 4.056 mm is 0.4838 - 0.2611 =

0.2227 (see shaded area in diagram below).

The number likely to have diameter between 4.052 mm and 4.056 mm = 0.2227 × 20 =

4.454 = 4, correct to nearest whole number.


8. The mean mass of active material in tablets produced by a manufacturer is 5.00 g and the

standard deviation of the masses is 0.036 g. In a bottle containing 100 tablets, find how many

tablets are likely to have masses of (a) between 4.88 and 4.92 g, (b) between 4.92 and 5.04 g

and (c) more than 5.04 g.

(a) The z-value corresponding to 4.88 g is given by: x x−σ

i.e. 4.88 5.000.036− = -3.33 standard

deviations

From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -3.33 is 0.4996.

The z-value corresponding to 4.92 g is given by: 4.92 5.000.036− = -2.22 standard deviations.

From Table 58.1, the area between z = 0 and z = -2.22 is 0.4868

The probability of having masses being between 4.88 g and 4.92 g is 0.4996 - 0.4868 = 0.0128

(see shaded area in diagram below).

The number of tablets likely to have a mass between 4.88 g and 4.92 g = 0.0128 × 100 = 1,

correct to nearest whole number.

(b) The z-value corresponding to 4.92 g is -2.22 standard deviations, from above, and the area between z = 0 and z = -2.22 is 0.4868.

The z-value corresponding to 5.04 g is given by: x x−σ

i.e. 5.04 5.000.036− = 1.11 standard

deviations


The probability of having masses being between 4.92 g and 5.04 g is 0.4868 + 0.3665 = 0.8533


The number of tablets likely to have a mass between 4.92 g and 5.04 g = 0.8533 × 100 = 85,



(c) The z-value corresponding to 5.04 g is 1.11 standard deviations, from above, and the area between z = 0 and z = 1.11 is 0.3665 The probability of having a mass greater than 5.04 g is 0.5000 - 0.3665 = 0.1335


The number of tablets likely to have a mass greater than 5.04g = 0.1335 × 100 = 13,




1. A frequency distribution of 150 measurements is as shown:

Class mid-point value 26.4 26.6 26.8 27.0 27.2 27.4 27.6

Frequency 5 12 24 36 36 25 12

Use normal probability paper to show that this data approximates to a normal distribution and

hence determine the approximate values of the mean and standard deviation of the distribution.

Use the formula for mean and standard deviation to verify the results obtained

To test the normality of a distribution, the upper class boundary values are plotted against

percentage cumulative frequency values on normal probability paper.

The table below shows the upper class boundary values for the distribution, together with the

cumulative frequency and percentage cumulative frequency.

Class mid-point

value

Upper class boundary

value

Frequency Cumulative frequency % cumulative

frequency

26.4

26.6

26.8

27.0

27.2

27.4

27.6

26.5

26.7

26.9

27.1

27.3

27.5

27.7

5

12

24

36

36

25

12

5

5+12 = 17

17 + 24 = 41

77

113

138

150

5/150 = 3

17/150 = 11

41/150 = 27

51

75

92

100

The co-ordinates of upper class boundary values/percentage cumulative frequency values are shown

plotted below. Since the points plotted lie very nearly in a straight line, the data is

approximately normally distributed.

From the graph, the mean occurs at 50%, i.e. mean, x = 27.1 at point P.

At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 27.38

At 16% cumulative frequency value, i.e. point R, upper class boundary value = 26.78

Hence, standard deviation, σ = 27.38 26.78 0.62 2−

= = 0.3


By calculation, mean,

x = ( ) ( ) ( ) ( ) ( ) ( ) ( )5 26.4 12 26.6 24 26.8 36 27.0 36 27.2 25 27.4 12 27.6150

× + × + × + × + × + × + ×

= 4061.8150

= 27.079

Standard deviation, σ = ( ) ( ) ( )2 2 25 26.4 27.079 12 26.6 27.079 24 26.8 27.079 ....150

⎧ ⎫− + − + − +⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

= 13.51175150

⎧ ⎫⎨ ⎬⎩ ⎭

= 0.3001


2. A frequency distribution of the class mid-point values of the breaking loads for 275 similar

fibres is as shown below:

Load (kN) 17 19 21 23 25 27 29 31

Frequency 9 23 55 78 64 28 14 4

Use normal probability paper to show that this distribution is approximately normally distributed

and determine the mean and standard deviation of the distribution (a) from the graph and (b) by

calculation.

To test the normality of a distribution, the upper class boundary values are plotted against

percentage cumulative frequency values on normal probability paper.

The table below shows the upper class boundary values for the distribution, together with the

cumulative frequency and percentage cumulative frequency.

Class mid-point

value (kN)

Upper class boundary

value

Frequency Cumulative frequency % cumulative

frequency

17

19

21

23

25

27

29

31

18

20

22

24

26

28

30

32

9

23

55

78

64

28

14

4

9

9 +23 = 32

32 + 55 = 87

165

229

257

271

275

9/275 = 3

32/275 = 12

87/275 = 32

60

83

93

99

100

The co-ordinates of upper class boundary values/percentage cumulative frequency values are shown

plotted below. Since the points plotted lie very nearly in a straight line, the data is

approximately normally distributed.

(a) From the graph, the mean occurs at 50%, at point P , i.e. mean, x = 23.5 kN

At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 26.2

At 16% cumulative frequency value, i.e. point R, upper class boundary value = 20.4

Hence, standard deviation, 26.2 20.4 5.82 2−

σ = = = 2.9 kN


(b) By calculation, mean, x = ( ) ( ) ( ) ( ) ( )9 17 23 19 55 21 78 23 64 25 ......275

× + × + × + × + × +

= 6425275

= 23.364 kN

Standard deviation, σ = ( ) ( ) ( )2 2 29 17 23.364 23 19 23.364 55 21 23.364 ....275

⎧ ⎫− + − + − +⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

= 2339.6364275

⎧ ⎫⎨ ⎬⎩ ⎭

= 2.917 kN


CHAPTER 59 LINEAR CORRELATION EXERCISE 218 Page 570

2. Determine the coefficient of correlation for the data given below, correct to 3 decimal places.

X 2.7 4.3 1.2 1.4 4.9

Y 11.9 7.10 33.8 25.0 7.50

A tabular method to determine the quantities is shown below.

X Y x =

( )X X−

y =

( )Y Y−

xy 2x 2y

2.7

4.3

1.2

1.4

4.9

11.9

7.10

33.8

25.0

7.50

-0.2

1.4

-1.7

-1.5

2

-5.16

-9.96

16.74

7.94

-9.56

1.032

-13.944

-28.458

-11.91

-19.12

0.04

1.96

2.89

2.25

4

26.6256

99.2016

280.2276

63.0436

91.3936

X 14.5=∑

14.5X 2.95

= =

Y 85.3=∑

85.3Y 17.065

= =

xy∑

= - 72.4

2x∑

= 11.14

2y∑

= 560.492

Coefficient of correlation, r = ( )( ) ( )( )2 2

xy 72.411.14 560.492x y

−=∑

∑ ∑ = - 0.916

4. In an experiment to determine the relationship between the current flowing in an electrical

circuit and the applied voltage, the results obtained are:

Current (mA) 5 11 15 19 24 28 33

Applied voltage (V) 2 4 6 8 10 12 14

Determine, using the product-moment formula, the coefficient of correlation for these results.

A tabular method to determine the quantities is shown below.


I V i = ( )I I− v = ( )V V− iv 2i 2v

5

11

15

19

24

28

33

2

4

6

8

10

12

14

-14.286

-8.286

-4.286

-0.286

4.714

8.714

13.714

-6

-4

-2

0

2

4

6

85.716

33.144

8.572

0

9.428

34.856

82.284

204.09

68.66

18.37

0.082

22.22

75.93

188.07

36

16

4

0

4

16

36

I 135=∑

135I 19.2867

= =

V 56=∑

56V 87

= =

iv∑

= 254

2i∑

= 577.43

2v∑

= 112


iv 254 254254.307577.43 112i v

= =∑∑ ∑

= 0.999

5. A gas is being compressed in a closed cylinder and the values of pressures and corresponding

volumes at constant temperature are as follows:

Pressure (kPa) 160 180 200 220 240 260 280 300

Volume ( 3m ) 0.034 0.036 0.030 0.027 0.024 0.025 0.020 0.019

Find the coefficient of correlation for these values.

A tabular method to determine the quantities is shown on page 526.

From the table,

coefficient of correlation, r = ( )( ) ( )( )62 2

xy 2.03 2.032.1094816800 264.875 10x y −

− −= =

×

∑∑ ∑

= - 0.962


P V x =

( )P P−

y =

( )V V−

xy 2x 2y

160

180

200

220

240

260

280

300

0.034

0.036

0.030

0.027

0.024

0.025

0.020

0.019

-70

-50

-30

-10

10

30

50

70

0.007125

0.009125

0.003125

0.000125

-0.002875

-0.001875

-0.006875

0.007875

-0.49875

-0.45625

-0.09375

-0.00125

-0.02875

-0.05625

-0.34375

-0.55125

4900

2500

900

100

100

900

2500

4900

50.7656 610−×

83.2656 610−×

9.7656 610−×

0.0156 610−×

8.2656 610−×

3.5156 610−×

47.2656 610−×

62.0156 610−×

P 1840=∑1840P

8230

=

=

V 0.215=∑

0.215V8

0.026875

=

=

xy∑

= -2.03

2x∑

= 16800

2v∑

= 264.875 610−×

7. The data shown below refers to the number of times machine tools had to be taken out of

service, in equal time periods, due to faults occurring and the number of hours worked by

maintenance teams. Calculate the coefficient of correlation for this data.

Machines out of service: 4 13 2 9 16 8 7

Maintenance hours: 400 515 360 440 570 380 415

A tabular method to determine the quantities is shown below, where X = machines out of service,

and Y = maintenance hours.


X Y x =

( )X X−

y =

( )Y Y−

xy 2x 2y

4

13

2

9

16

8

7

400

515

360

440

570

380

415

-4.4286

4.5714

-6.4286

0.5714

7.5714

-0.4286

-1.4286

-40

75

-80

0

130

-60

-25

177.144

342.855

514.288

0

984.282

25.716

35.715

19.6125

20.8977

41.3269

0.3265

57.3261

0.1837

2.0409

1600

5625

6400

0

16900

3600

625

X 59=∑

59X 8.42867

= =

Y 3080=∑

3080Y 4407

= =

xy∑

= 2080.04

2x∑ =

141.714

2y∑ =

34750


xy 2080.04 2080.042219.135141.714 34750x y

= =∑∑ ∑

= 0.937


CHAPTER 60 LINEAR REGRESSION EXERCISE 219 Page 575

2. Determine the equation of the regression line of Y on X, correct to 3 significant figures, for the

following data:

X 6 3 9 15 2 14 21 13

Y 1.3 0.7 2.0 3.7 0.5 2.9 4.5 2.7

X Y 2X XY 2Y

6

3

9

15

2

14

21

13

1.3

0.7

2.0

3.7

0.5

2.9

4.5

2.7

36

9

81

225

4

196

441

169

7.8

2.1

18.0

55.5

1.0

40.6

94.5

35.1

1.69

0.49

4.0

13.69

0.25

8.41

20.25

7.29

X 83=∑ Y 18.3=∑ 2X 1161=∑ XY 254.6=∑ 2Y 56.07=∑

Substituting into 0 1Y a N a X= +∑ ∑

and 20 1XY a X a X= +∑ ∑ ∑

gives: 18.3 = 8 0a + 83 1a (1)

and 254.6 = 83 0a + 1161 1a (2)

83 × (1) gives: 1518.9 = 664 0a + 6889 1a (3)

8 × (2) gives: 2036.8 = 664 0a + 9288 1a (4)

(4) – (3) gives: 517.9 = 2399 1a from which, 1517.9a2399

= = 0.216

Substituting in (1) gives: 18.3 = 8 0a + 83(0.216) from which, 018.3 83(0.216)a

8−

= = 0.0477

Hence, the equation of the regression line of Y on X is: 0 1Y a a X= +

i.e. Y = 0.0477 + 0.216X


4. Determine the equation of the regression line of X on Y, correct to 3 significant figures, for the

data given in Problem 2.

Substituting into 0 1X b N b Y= +∑ ∑

and 20 1XY b Y b Y= +∑ ∑ ∑

gives: 83 = 8 0b + 18.3 1b (1)

and 254.6 = 18.3 0b + 56.07 1b (2)

18.3 × (1) gives: 1518.9 = 146.4 0b + 334.89 1b (3)

8 × (2) gives: 2036.8 = 146.4 0b + 448.56 1b (4)

(4) – (3) gives: 517.9 = 113.67 1b from which, 1517.9b

113.67= = 4.56

Substituting in (1) gives: 83 = 8 0b + 18.3(4.56) from which, 083 18.3(4.56)b

8−

= = -0.056

Hence, the equation of the regression line of X on Y is: 0 1X b b Y= +

i.e. X = - 0.056 + 4.56Y

5. The relationship between the voltage applied to an electrical circuit and the current flowing is as

shown:

Current (mA) 2 4 6 8 10 12 14

Applied voltage (V) 5 11 15 19 24 28 33

Assuming a linear relationship, determine the equation of the regression line of applied voltage,

Y, on current, X, correct to 4 significant figures.

A table is produced as shown below, where current I = X and voltage V = Y


X Y 2X XY 2Y

2

4

6

8

10

12

14

5

11

15

19

24

28

33

4

16

36

64

100

144

196

10

44

90

152

240

336

462

25

121

225

361

576

784

1089

X 56=∑ Y 135=∑ 2X 560=∑ XY 1334=∑ 2Y 3181=∑


and 20 1XY a X a X= +∑ ∑ ∑

gives: 135 = 7 0a + 56 1a (1)

and 1334 = 56 0a + 560 1a (2)

8 × (1) gives: 1080 = 56 0a + 448 1a (3)

(2) – (3) gives: 254 = 112 1a from which, 1254a112

= = 2.268

Substituting in (1) gives: 135 = 7 0a + 56(2.268) from which, 0135 56(2.268)a

7−

= = 1.142


i.e. Y = 1.142 + 2.268X

6. For the data given in Problem 5, determine the equation of the regression line of current on

applied voltage, correct to 3 significant figures.


and 20 1XY b Y b Y= +∑ ∑ ∑

gives: 56 = 7 0b + 135 1b (1)

and 1334 = 135 0b + 3181 1b (2)

135 × (1) gives: 7560 = 945 0b + 18225 1b (3)


7 × (2) gives: 9338 = 945 0b + 22267 1b (4)

(4) – (3) gives: 1778 = 4042 1b from which, 11778b4042

= = 0.43988

Substituting in (1) gives: 56 = 7 0b + 135(0.43988) from which, 056 135(0.43988)b

7−

= = -0.483


i.e. X = - 0.483 + 0.440Y, correct to 3 significant figures.

8. In an experiment to determine the relationship between force and momentum, a force X, is

applied to a mass, by placing the mass on an inclined plane, and the time, Y, for the velocity to

change from u m/s to v m/s is measured. The results obtained are as follows:

Force (N) 11.4 18.7 11.7 12.3 14.7 18.8 19.6

Time (s) 0.56 0.35 0.55 0.52 0.43 0.34 0.31

Determine the equation of the regression line of time on force, assuming a linear relationship

between the quantities, correct to 3 significant figures.

Let force F = X and time = Y. A table is produced as shown below.

X Y 2X XY 2Y

11.4

18.7

11.7

12.3

14.7

18.8

19.6

0.56

0.35

0.55

0.52

0.43

0.34

0.31

129.96

349.69

136.89

151.29

216.09

353.44

384.16

6.384

6.545

6.435

6.396

6.321

6.392

6.076

0.3136

0.1225

0.3025

0.2704

0.1849

0.1156

0.0961

X∑

= 107.2

Y∑

= 3.06

2X∑

= 1721.52

XY∑

= 44.549

2Y∑

= 1.4056


and 20 1XY a X a X= +∑ ∑ ∑

gives: 3.06 = 7 0a + 107.2 1a (1)

and 44.549 = 107.2 0a + 1721.52 1a (2)


107.2 × (1) gives: 328.032 = 750.4 0a + 11491.84 1a (3)

7 × (2) gives: 311.843 = 750.4 0a + 12050.64 1a (4)

(3) – (4) gives: 16.189 = - 558.8 1a from which, 116.189a558.8−

= = - 0.0290

Substituting in (1) gives: 3.06 = 7 0a + 107.2(-0.0290)

from which, 03.06 107.2( 0.0290)a

7− −

= = 0.881


i.e. Y = 0.881 – 0.0290X

9. Find the equation for the regression line of force on time for the data given in Problem 8, correct

to 3 decimal places.


and 20 1XY b Y b Y= +∑ ∑ ∑

gives: 107.2 = 7 0b + 3.06 1b (1)

and 44.549 = 3.06 0b + 1.4056 1b (2)

3.06 × (1) gives: 328.032 = 21.42 0b + 9.3636 1b (3)

7 × (2) gives: 311.843 = 21.42 0b + 9.8392 1b (4)

(3) – (4) gives: 16.189 = - 0.4756 1b from which, 116.189b0.4756

=−

= - 34.039

Substituting in (1) gives: 107.2 = 7 0b + 3.06(-34.039)

from which, 0107.2 3.06(34.039)b

7+

= = 30.194


i.e. X = 30.194 – 34.039Y


10. Draw a scatter diagram for the data given in Problem 8 and show the regression lines of time on

force and force on time. Hence find (a) the time corresponding to a force of 16 N, and (b) the

force at a time of 0.25 s, assuming the relationship is linear outside of the range of values given.

A scatter diagram is shown below. The regression line of force on time, Y = 0.881 – 0.0290X, and

force on time, X = 30.194 – 34.039Y are shown and to the scale drawn are seen to coincide.

(a) When force X = 16 N then time, Y = 0.881 – 0.0290(16) = 0.417 s

(b) When time Y = 0.25 s, force, X = 30.194 – 34.039(0.25) = 21.7 N


CHAPTER 61 SAMPLING AND ESTIMATION THEORIES EXERCISE 220 Page 580

1. The lengths of 1500 bolts are normally distributed with a mean of 22.4 cm and a standard

deviation of 0.0438 cm. If 30 samples are drawn at random from this population, each sample

being 36 bolts, determine the mean of the sampling distributions and standard error of the means

when sampling is done with replacement.

For the population, number of bolts, pN = 1500, standard deviation, σ = 0.0438 cm,

mean, µ = 22.4 cm. For the sample, N = 30

The mean of the sampling distributions, xµ = µ = 22.4 cm

The standard error of the means with replacement, x

0.0438N 30σ

σ = = = 0.0080 cm

2. Determine the standard error of the means in problem 1, if sampling is done without

replacement, correct to 4 decimal places.

Standard error of means without replacement, px

p

N N 0.0438 1500 30N 1 1500 1N 30

⎛ ⎞−σ −⎛ ⎞σ = =⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

= (0.0080)(0.9903) = 0.0079 cm

3. A power punch produces 1800 washers per hour. The mean inside diameter of the washers is

1.70 cm and the standard deviation is 0.013 mm. Random samples of 20 washers are drawn every

5 minutes. Determine the mean of the sampling distribution of means and the standard error of

the means for the one hour’s output from the punch, (a) with replacement, and (b) without

replacement, correct to three significant figures.

For the population, number of bolts, pN = 1800, standard deviation, σ = 0.013 cm,

mean, µ = 1.70 cm. For the sample, N = 20

(a) With replacement.



The standard error of the means, x

0.013N 20σ

σ = = = 2.907 310−× = 32.91 10−× cm, correct

to 3 significant figures.

(b) Without replacement.


Standard error of means, px

p

N N 0.013 1800 20N 1 1800 1N 20

⎛ ⎞−σ −⎛ ⎞σ = =⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

= ( )( )32.907 10 0.9947−× = 32.89 10−× cm

5. A large batch of electric light bulbs have a mean time to failure of 800 hours and the standard

deviation of the batch is 60 hours. Determine the probability that the mean time to failure of a

random sample of 16 light bulbs will be between 790 hours and 810 hours, correct to three

decimal places.

N = 16 and x

60N 16σ

σ = = = 15 h

790 h corresponds to a z-value of: 1790 800z

15−

= = - 0.67 standard deviations and the area

between z = 0 and z = -0.67 is 0.2486 (from Table 58.1, page 561 of textbook).

810 h corresponds to a z-value of: 2810 800z

15−

= = 0.67 standard deviations and the area between

z = 0 and z = 0.67 is 0.2486.

Hence, the probability that the mean time to failure will be between 790 hours and 810 hours

(i.e. the shaded area of the diagram below) = 2 × 0.2486 = 0.497, correct to 3 decimal places.


7. The contents of a consignment of 1200 tins of a product have a mean mass of 0.504 kg and a

standard deviation of 92 g. Determine the probability that a random sample of 40 tins drawn

from the consignment will have a combined mass of (a) less than 20.13 kg, (b) between 20.13 kg

and 20.17 kg, and (c) more than 20.17 kg, correct to three significant figures.

Number of tins, pN = 1200, mean mass, µ = 0.504 kg and standard deviation, σ = 92 g = 0.092 kg,

For the sample, N = 40

Standard error of means, px

p

N N 0.092 1200 40N 1 1200 1N 40

⎛ ⎞−σ −⎛ ⎞σ = =⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

= (0.01455)(0.9836) = 0.0143 kg

The mean mass of 40 tins = 40 × 0.504 = 20.16 kg

(a) z-value corresponding to 20.13 kg, z = 20.13 20.160.0143− = - 2.1 standard deviations.

The area between z = 0 and z = - 2.1 is 0.4821 (from table 58.1, page 561 of textbook) and is the

shaded area in the diagram below.

Hence, the probability that the combined mass will be less than 20.13 kg is:

0.5000 – 0.4821 = 0.0179

(b) The z-value corresponding to 20.13 kg, is - 2.1 standard deviations from above, and the area

between z = 0 and z = - 2.1 is 0.4821

The z-value corresponding to 20.17 kg, z = 20.17 20.160.0143− = 0.7 standard deviations and the area

between z = 0 and z = 0.7 is 0.2580 (from table 58.1, page 561 of textbook)


Hence, the probability that the combined mass will be between 20.13 kg and 20.17 kg is:

0.4821 + 0.2580 = 0.740 (see diagram below)

(c) z-value corresponding to 20.17 kg, is z = 0.7 standard deviations and the area between z = 0 and

z = 0.7 is 0.2580 from part (b) and is the shaded area in the diagram below.

Hence, the probability that the combined mass will be greater than 20.17 kg is:

0.5000 – 0.2580 = 0.242



1. Measurements are made on a random sample of 100 components drawn from a population of

size 1546 and having a standard deviation of 2.93 mm. The mean measurement of the

components in the sample is 67.45 mm. Determine the 95% and 99% confidence limits for an

estimate of the mean of the population.

For the population, pN 1546= and σ = 2.93 mm

For the sample, N = 100 and x = 67.45 mm

For a 95% confidence limit, Cz = 1.96 from Table 61.1, page 582 of textbook.

An estimate of the confidence limits of the population mean is:

( )( )pC

p

N N 1.96 2.93z 1546 100x 67.45N 1 1546 1N 100

⎛ ⎞−σ −⎛ ⎞± = ±⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

= 67.45 ± (0.57428)(0.96743)

= 67.45 ± 0.556 mm

= 67.45 - 0.556 mm or 67.45 + 0.556 mm

Thus, the 95% confidence limits are 66.89 mm and 68.01 mm



( )( )pC

p

N N 2.58 2.93z 1546 100x 67.45N 1 1546 1N 100

⎛ ⎞−σ −⎛ ⎞± = ±⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

= 67.45 ± (0.75594)(0.96743)

= 67.45 ± 0.731 mm

= 67.45 - 0.731 mm or 67.45 + 0.731 mm

Thus, the 99% confidence limits are 66.72 mm and 68.18 mm


2. The standard deviation of the masses of 500 blocks is 150 kg. A random sample of 40 blocks has

a mean mass of 2.40 Mg.

(a) Determine the 95% and 99% confidence intervals for estimating the mean mass of the

remaining 460 blocks

(b) With what degree of confidence can be said that the mean mass of the remaining 460 blocks

is 2.40 ± 0.035 Mg?

pN 500= and σ = 150 kg = 0.15 Mg, N = 40 and x = 2.40 Mg

(a) For a 95% confidence limit, Cz = 1.96 from Table 61.1, page 582 of textbook.


( )( )pC

p

N N 1.96 0.15z 500 40x 2.40N 1 500 1N 40

⎛ ⎞−σ −⎛ ⎞± = ±⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

= 2.40 ± (0.0465)(0.9601)

= 2.40 ± 0.0446 Mg

= 2.40 - 0.0446 Mg or 2.40 + 0.0446 Mg

Thus, the 95% confidence limits are 2.355 mg and 2.445 Mg



( )( )pC

p

N N 2.58 0.15z 500 40x 2.40N 1 500 1N 40

⎛ ⎞−σ −⎛ ⎞± = ±⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

= 2.40 ± (0.06119)(0.9601)

= 2.40 ± 0.0587 Mg

= 2.40 - 0.0587 Mg or 2.40 + 0.0587 Mg

Thus, the 99% confidence limits are 2.341 Mg and 2.459 Mg

(b) 2.40 ± 0.035 = CzxNσ

±


x = 2.40, hence, Cz 0.035Nσ

± = ±

and ( )C

0.035 400.035 Nz0.15

= =σ

= ± 1.48

From Table 58.1, page 561 of textbook, 1.48 corresponds to an area of 0.4306.

The area between the mean and ± 1.48 is 2 × 0.4306 = 0.8612 (see diagram below).

Thus, the confidence level corresponding to 2.40 ± 0.035 is 86%

4. The standard deviation of the time to failure of an electronic component is estimated as 100

hours. Determine how large a sample of these components must be, in order to be 90% confident

that the error in the estimated time to failure will not exceed (a) 20 hours and (b) 10 hours.

σ = 100 h The confidence limits for the mean of a population is Czx

Nσ

±

For a 90% confidence level, Cz = 1.645 from Table 61.1, page 582 of textbook.

(a) ( )( )1.645 100x 20 x

N± = ±

from which, ( )( )1.645 10020

N= and ( )( )1.645 100

N20

= = 8.225

i.e. N = 2(8.225) = 67.65

Hence, at least 68 components are required to be 90% confident that the error will not

exceed 20 hours.

(b) ( )( )1.645 100x 10 x

N± = ±


from which, ( )( )1.645 10010

N= and ( )( )1.645 100

N10

= = 16.45

i.e. N = 2(16.45) = 270.6

Hence, at least 271 components are required to be 90% confident that the error will not

exceed 10 hours.

6. The time taken to assemble a servo-mechanism is measured for 40 operatives and the mean time

is 14.63 minutes with a standard deviation of 2.45 minutes. Determine the maximum error in

estimating the true mean time to assemble the servo-mechanism for all operatives, based on a

95% confidence level

N = 40, x = 14.63 minutes, σ = 2.45 minutes, at 95% confidence level, Cz = 1.96 (from Table

61.1, page 582 of textbook).

Hence, ( )( )1.96 2.4514.63 14.63 0.759

40± = ±

i.e. the maximum error is 0.759 minutes = 0.759 × 60 = 45.6 s



1. The value of the ultimate tensile strength of a material is determined by instruments on 10

samples of the materials. The mean and standard deviation of the results are found to be 5.17

MPa and 0.06 MPa respectivly. Determine the 95% confidence interval for the mean of the

ultimate tensile strength of the material.

N = 10, x = 5.17 MPa, s = 0.06 MPa.

Since the sample size is less than 30, the degrees of freedom, ν = 10 – 1 = 9

From Table 61.2, page 587 of textbook, 0.95t , 9ν = gives Ct = 1.83

Estimated value of the mean of the U.T.S. is: ( )( )C 1.83 0.06t sx 5.17N 1 10 1

± = ±− −

= 5.17 ± 0.0366

= 5.17 – 0.0366 or 5.17 + 0.0366

Thus, the 95% confidence interval is: 5.133 MPa to 5.207 MPa.

3. The specific resistance of a reel of German silver wire of nominal diameter 0.5 mm is estimated

by determining the resistance of 7 samples of the wire. These were found to have resistance

values (in ohms per metre) of:

1.12, 1.15, 1.10, 1.14, 1.15, 1.10 and 1.11

Determine the 99% confidence interval for the true specific resistance of the reel of wire.

N = 7, mean, 1.12 1.15 1.10 1.14 1.15 1.10 1.11x7

+ + + + + += = 1.1243 Ω 1m−

and standard deviation, s = ( ) ( ) ( )2 2 21.12 1.1243 1.15 1.1243 1.10 1.1243 ...7

⎧ ⎫− + − + − +⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

= 0.002971437

= 0.0206 Ω 1m−

From Table 61.2, page 587 of textbook, 0.99t , 7 1 6ν = − = gives Ct = 3.14

Hence, 99% confidence limits is given by: ( )( )C 3.14 0.0206t sx 1.1243N 1 7 1

± = ±− −


= 1.1243 ± 0.0264

= 1.1243 – 0.0264 or 1.1243 + 0.0264

Thus, the 99% confidence interval is: 1.10 1m−Ω to 1.15 1m−Ω

4. In determining the melting point of a metal, five determinations of the melting point are made.

The mean and standard deviation of the five results are 132.27°C and 0.742°C. Calculate the

confidence with which the prediction ‘the melting point of the metal is between 131.48°C and

133.06°C ’ can be made?

N = 5, mean, x = 132.27°C and standard deviation, s = 0.742°C

133.06 – 131.48 = 0.79°C

Hence, Ct s132.27 0.79 xN 1

± = ±−

i.e. ( )Ct 0.7420.79

5 1=

−

from which, ( )C

0.79 4t

0.742= = 2.13

From Table 61.2, page 587 of textbook, at ν = N – 1 = 5 – 1 = 4, 2.13 corresponds to 0.95t , i.e.

the confidence with which the prediction ‘the melting point of the metal is between 131.48°C

and 133.06°C ’ can be made is 95%.


CHAPTER 62 SIGNIFICANCE TESTING EXERCISE 223 Page 597

1. An automatic machine produces piston rings for car engines. Random samples of 1000 rings are

drawn from the output of the machine periodically for inspection purposes. A defect rate of 5% is

acceptable to the manufacturer, but if the defect rate is believed to have exceeded this value, the

machine producing the rings is stopped and adjusted. Determine the type 1 errors which occur for

the following decision rule: Stop production and adjust the machine if a sample contains (a) 54

(b) 62 and (c) 70 or more defective rings.

(a) N = 1000, p = 0.05, q = 0.95, mean of normal distribution = Np = 50,

standard deviation of the normal distribution = ( ) ( )( )( )Npq 1000 0.05 0.95 6.892= = .

A type I error is the probability of stopping production when getting more than 54 defective

rings in the sample, even though defect rate is 5%.

z-value = var iate means tan dard deviation

− = 54 506.892− = 0.58 and from Table 58.1, page 561 of textbook, the

area between the mean and a z-value of 0.58 is 0.2190.

Thus, the probability of more than 54 defective rings = 0.5000 – 0.2190 = 0.281

Hence, the type I error is 28.1%

(b) z-value = 62 506.892− = 1.74 and from Table 58.1, the area between the mean and a z-value of 1.74

is 0.4591.



(c) z-value = 70 506.892− = 2.90 and from Table 58.1, the area between the mean and a z-value of

2.90 is 0.4981.




2. For the data in Problem 1, determine the type II errors which are made if the decision rule is to

stop production if there are more than 60 defective components in the sample when the actual

defect rate has risen to (a) 6% (b) 7.5% and (c) 9%.

(a) N = 1000, p = 0.06, q = 0.94

Mean = Np = 60, standard deviation = ( ) ( )( )( )Npq 1000 0.06 0.94 7.51= =

z-value = 61 607.51− = 0.13 (note that ‘more than 60 components defective’ means 61 or more)

and from Table 58.1, page 561 of textbook, the area between the mean and a z-value of

0.13 is 0.0517.

Thus, the probability of more than 60 defective components = 0.5000 + 0.0517 = 0.5517

Hence, the type II error is 55.2%

(b) N = 1000, p = 0.075, q = 0.925


z-value = 61 758.329− = -1.68 and from Table 58.1, the area between the mean and a z-value of

-1.68 is 0.4535.

Thus, the probability of more than 75 defective components = 0.5000 - 0.4535 = 0.0465


(c) N = 1000, p = 0.09, q = 0.91


z-value = 61 909.05− = -3.20 and from Table 58.1, the area between the mean and a z-value of

-3.20 is 0.4993.

Thus, the probability of more than 90 defective components = 0.5000 - 0.4993 = 0.0007



3. A random sample of 100 components is drawn from the output of a machine whose defect rate is

3%. Determine the type 1 error if the decision rule is to stop production when the sample contains

(a) 4 or more defective components, (b) 5 or more defective components, and (c) 6 or more

defective components.

N = 100, p = 0.03 and Np = 3. Since N ≥ 50 and Np ≤ 5 the Poisson distribution is used.

λ = Np = 3

The probability of 0, 1, 2, 3, 4, 5, … defective components are given by the terms:

3e e 0.0498−λ −= = , 3e 3e 0.1494−λ −λ = = , 2 2

33e e 0.22402! 2

−λ −λ= = ,

3 333e e 0.2240

3! 6−λ −λ= = ,

4 433e e 0.1680

4! 24−λ −λ= = ,

5 533e e 0.1008

5! 120−λ −λ= =

(a) The probability of a sample containing 4 or more defective components is:

1 – (0.0498 + 0.1494 + 0.2240 + 0.2240)

= 1 – 0.6472 = 0.3528

i.e. the type I error is 35.3%

(b) The probability of a sample containing 5 or more defective components is:

1 – (0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680)

= 1 – 0.8152 = 0.1848


(c) The probability of a sample containing 6 or more defective components is:

1 – (0.8152 + 0.1008)

= 1 – 0.9160 = 0.0840




1. A batch of cables produced by a manufacturer have a mean breaking strength of 2000 kN and a

standard deviation of 100 kN. A sample of 50 cables is found to have a mean breaking strength of

2050 kN. Test the hypothesis that the breaking strength of the sample is greater than the breaking

strength of the population from which it is drawn at a level of significance of 0.01.

µ = 2000 kN, σ = 100 kN, N = 50, x 2050kN=

The null hypothesis, 0H : x > µ

The alternative hypothesis, 1H : x = µ

z = x 2050 2000 50 3.54100 14.142N 50

−µ −= = ± = ±

σ

The value for a two-tailed test is given in Table 62.1, page 594 of textbook, and at a significance

level of 0.01 is ± 2.58.

Since the z-value of the sample is outside of this range, the hypothesis is rejected.

3. The internal diameter of a pipe has a mean diameter of 3.0000 cm with a standard deviation of

0.015 cm. A random sample of 30 measurements are taken and the mean of the samples is

3.0078 cm. Test the hypothesis that the mean diameter of the pipe is 3.0000 cm at a level of

significance of 0.01.

µ = 3.0000 cm, σ = 0.015 cm, N = 30, x 3.0078cm=

The null hypothesis, 0H : mean diameter = 3.0000 cm

The alternative hypothesis, 1H : mean diameter ≠ 3.0000 cm

z = x 3.0078 3.0000 0.0078 2.850.015 0.00274N 30

−µ −= = ± = ±

σ

The value for a two-tailed test is given in Table 62.1, page 594 of textbook, and at a significance

level of 0.01 is ± 2.58.

Since the z-value of the sample is outside of this range, the hypothesis is rejected.


4. A fishing line has a mean breaking strength of 10.25 kN. Following a special treatment on the

line, the following results are obtained for 20 specimens taken from the line.

Breaking strength (kN) 9.8 10.0 10.1 10.2 10.5 10.7 10.8 10.9 11.0

Frequency 1 1 4 5 3 2 2 1 1

Test the hypothesis that special treatment has improved the breaking strength at a level of


µ = 10.25 kN and N = 20

Sample mean, ( ) ( ) ( ) ( )9.8 1 10.0 1 10.1 4 10.2 5 .... 207.6x 10.3820 20

× + × + × + × += = =

Sample standard deviation,

s = ( ) ( ) ( ) ( )2 2 2 21 9.8 10.38 1 10.0 10.38 4 10.1 10.38 5 10.2 10.38 ... 2.21220 20

− + − + − + − += = 0.33

The null hypothesis is that the sample breaking strength is better than the mean breaking strength.

N < 30, therefore a t distribution is used.

( ) ( ) ( ) ( )x N 1 10.38 10.25 20 1

ts 0.33

−µ − − −= = = 1.72

At a level of significance of 0.1, the t value is 0.112

t⎛ ⎞−⎜ ⎟⎝ ⎠

i.e. 0.95t and ν = N – 1 = 20 – 1 = 19, and

from Table 61.1, page 587 of textbook, 0.95t 19ν = has a value of 1.73

Since 1.72 is within this range, the hypothesis is accepted.

5. A machine produces ball bearings having a mean diameter of 0.50 cm. A sample of 10 ball

bearings is drawn at random and the sample mean is 0.53 cm with a standard deviation of

0.03 cm. Test the hypothesis that the mean diameter is 0.50 cm at a level of significance of

(a) 0.05 and (b) 0.01.

µ = 0.50 cm, N = 10, x = 0.53 cm and s = 0.03 cm

The null hypothesis is: 0H : µ = 0.50 cm


( ) ( ) ( ) ( )x N 1 0.53 0.50 10 1

ts 0.03

−µ − − −= = = 3

(a) From Table 61.1, page 587 of the textbook, 0.9750.0512

t t , 9⎛ ⎞−⎜ ⎟⎝ ⎠

= ν = has a value of 2.26

Since 3 is outside of this range, the hypothesis is rejected.

(b) From Table 61.1, 0.9950.0112

t t , 9⎛ ⎞−⎜ ⎟⎝ ⎠

= ν = has a value of 3.25

Since 3 is within this range, the hypothesis is accepted.



1. A comparison is being made between batteries used in calculators. Batteries of type A have a

mean lifetime of 24 hours with a standard deviation of 4 hours, this data being calculated from a

sample of 100 of the batteries. A sample of 80 of the type B batteries has a mean lifetime of 40

hours with a standard deviation of 6 hours. Test the hypothesis that the type B batteries have a mean

lifetime of at least 15 hours more than those of type A, at a level of significance of 0.05.

Battery A: Ax 24= , A 4σ = and AN 100=

Battery B: Bx 40= , B 6σ = and BN 80=

The hypothesis is: H: x = Ax + 15

Let x = 24 + 15 = 39

z = B

2 22 2A B

A B

x x 39 40 1 1.280.7810254 6

100 80N N

− − −= = = −

⎛ ⎞ ⎛ ⎞σ σ ++⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

From Table 62.1, page 594 of textbook, for α = 0.05, one-tailed test, z = 1.645

Since the z-value is within this range, the hypothesis is accepted.

3. Capacitors having a nominal capacitance of 24 µF but produced by two different companies are

tested. The values of actual capacitances are:

Company 1 21.4 23.6 24.8 22.4 26.3

Company 2 22.4 27.7 23.5 29.1 25.8

Test the hypothesis that the mean capacitance of capacitors produced by company 2 are higher

than those produced by company 1 at a level of significance of 0.01.

(Bessel’s correction is $22 s N

N 1σ =

−)

N = 5, 121.4 23.6 24.8 22.4 26.3x 23.7

5+ + + +

= =

( ) ( ) ( ) ( ) ( )2 2 2 2 2

1

21.4 23.7 23.6 23.7 24.8 23.7 22.4 23.7 26.3 23.7s 1.73

5

⎛ ⎞− + − + − + − + −= =⎜ ⎟

⎜ ⎟⎝ ⎠


1 1N 5s 1.73 1.93

N 1 4⎛ ⎞ ⎛ ⎞σ = = =⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

222.4 27.7 23.5 29.1 25.8x 25.7

5+ + + +

= =

( ) ( ) ( ) ( ) ( )2 2 2 2 2

2

22.4 25.7 27.7 25.7 23.5 25.7 29.1 25.7 25.8 25.7s 2.50

5

⎛ ⎞− + − + − + − + −= =⎜ ⎟

⎜ ⎟⎝ ⎠

2 2N 5s 2.50 2.80

N 1 4⎛ ⎞ ⎛ ⎞σ = = =⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

2 1

2 222yx

x y

x x 25.7 23.7 2t 1.321.52081.93 2.80

5 5N N

− −= = = =

⎛ ⎞ ⎛ ⎞σσ ++⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

0.9950.0112

t t⎛ ⎞−⎜ ⎟⎝ ⎠

= and 1 2N N 2 5 5 2 8ν = + − = + − =

From Table 61.2, page 587 of textbook, 0.995t , 8ν = has a value of 3.36

Since the t value of the difference of the means, i.e. 1.32, is within the range ± 3.36, the hypothesis

is accepted.

5. A sample of 12 car engines produced by manufacturer A showed that the mean petrol

consumption over a measured distance was 4.8 litres with a standard deviation of 0.40 litres.

Twelve similar engines for manufacturer B were tested over the same distance and the mean

petrol consumption was 5.1 litres with a standard deviation of 0.36 litres. Test the hypothesis

that the engines produced by manufacturer A are more economical than those produced by

manufacturer B at a level of significance of (a) 0.01 and (b) 0.1.

AA AN 12, x 4.8 litre, 0.40 litre= = σ =

BB BN 12, x 5.1 litre, 0.36 litre= = σ =

The hypothesis is: H: manufacturer A is more economical than manufacturer B.

( ) ( )2 22 2A A B B

A B

12 0.40 12 0.36N s N s 3.4752 0.397N N 2 12 12 2 22

⎛ ⎞+⎛ ⎞+σ = = = =⎜ ⎟⎜ ⎟ ⎜ ⎟+ − + −⎝ ⎠ ⎝ ⎠


( )( )A B

A B

x x 4.8 5.1 0.3t 1.850.397 0.408251 11 1 0.397

12 12N N

− −= = = − = −

⎛ ⎞ ⎛ ⎞+σ + ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

(a) The t value is 0.0112

t⎛ ⎞−⎜ ⎟⎝ ⎠

i.e. 0.995t and ν = 12 + 12 – 2 = 22, hence, from Table 61.2, page 587 of

textbook, 0.995t , 22ν = has a value of 2.82

Since the t value of the difference of the means is outside the range ± 1.85, the hypothesis is

rejected.

(b) From Table 61.2, 0.112

t⎛ ⎞−⎜ ⎟⎝ ⎠

i.e. 0.95t , 22ν = has a value of 1.72

Since the t value of the difference of the means is within the range ± 1.85, the hypothesis is

accepted.

6. Four-star and unleaded petrol is tested in 5 similar cars under identical conditions. For four-star

petrol, the cars covered a mean distance of 21.4 kilometres with a standard deviation of 0.54

kilometres for a given mass of petrol. For the same mass of unleaded petrol the mean distance

covered was 22.6 kilometres with a standard deviation of 0.48 kilometres. Test the hypothesis

that unleaded petrol gives more kilometres per litre than four-star petrol at a level of


4 4N 5, x 21.4 km, s 0.54 km= = =

un unx 22.6 km, s 0.48 km= =

The hypothesis is: H: unx > 4x

( ) ( )2 22 24 4 un un

4 un

5 0.54 5 0.48N s N s 2.61 0.571N N 2 5 5 2 8

⎛ ⎞+⎛ ⎞+σ = = = =⎜ ⎟⎜ ⎟ ⎜ ⎟+ − + −⎝ ⎠ ⎝ ⎠

( )( )un 4

un 4

x x 22.6 21.4 1.2t 3.320.571 0.632461 11 1 0.571

5 5N N

− −= = = =

⎛ ⎞ ⎛ ⎞+σ + ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠


The t value is 0.112

t⎛ ⎞−⎜ ⎟⎝ ⎠

i.e. 0.95t and ν = 5 + 5 – 2 = 8, hence, from Table 61.2, page 587 of textbook,

0.95t , 8ν = has a value of 1.86.

Since the t value of the difference of the means, i.e. 3.32, is outside of the range ± 3.32, the

hypothesis is rejected.


CHAPTER 63 CHI-SQUARE AND DISTRIBUTION-FREE TESTS EXERCISE 226 Page 607

1. A dice is rolled 240 times and the observed and expected frequencies are as shown.

Face Observed frequency Expected frequency

1 49 40

2 35 40

3 32 40

4 46 40

5 49 40

6 29 40

Determine the χ2-value for this distribution.

Face Observed

frequency, o

Expected

frequency, e

o - e ( )2o e− ( )2o ee−

1

2

3

4

5

6

49

35

32

46

49

29

40

40

40

40

40

40

9

-5

-8

6

9

-11

81

25

64

36

81

121

2.025

0.625

1.6

0.9

2.025

3.025

( )22 o e

10.2e

⎧ ⎫−⎪ ⎪χ = =⎨ ⎬⎪ ⎪⎩ ⎭

∑

Hence, the Chi-square value, 2 10.2χ =


2. The numbers of telephone calls received by the switchboard of a company in 200 five-minute

intervals are shown in the distribution below.

Number of calls Observed frequency Expected frequency

0 11 16

1 44 42

2 53 52

3 46 42

4 24 26

5 12 14

6 7 6

7 3 2

Calculate the χ2-value for this data.

Number

of calls

Observed

frequency, o

Expected

frequency, e

o - e ( )2o e−

( )2o ee−

0

1

2

3

4

5

6

7

11

44

53

46

24

12

7

3

16

42

52

42

26

14

6

2

-5

2

1

4

-2

-2

1

1

25

4

1

16

4

4

1

1

1.5625

0.0952

0.0192

0.3810

0.1538

0.2857

0.1667

0.5000

( )22 o e

3.16e

⎧ ⎫−⎪ ⎪χ = =⎨ ⎬⎪ ⎪⎩ ⎭

∑

Hence, the Chi-square value, 2 3.16χ =



1. Test the null hypothesis that the observed data given below fits a binomial distribution of the

form 250(0.6 + 0.4)7 at a level of significance of 0.05.

Observed frequency 8 27 62 79 45 24 5 0

Is the fit of the data ‘too good’ at a level of confidence of 90%?

( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )( ) ( )

6 5 2 4 37

7 3 4 2 5

6 7

(7)(6) (7)(6)(5)0.6 7 0.6 0.4 0.6 0.4 0.6 0.42! 3!

(7)(6)(5)(4) (7)(6)(5)(4)(3)250 0.6 0.4 250 0.6 0.4 0.6 0.44! 5!

(7)(6)(5)(4)(3)(2) 0.6 (0.4 0.46!

⎡ ⎤+ + +⎢ ⎥⎢ ⎥⎢ ⎥+ = + +⎢ ⎥⎢ ⎥⎢ ⎥+ +⎢ ⎥⎣ ⎦

= 250[0.02799 + 0.13064 + 0.26127 + 0.29030 + 0.19354 + 0.07741 + 0.01720

+ 0.00164]

= 7 + 33 + 65 + 73 + 48 + 19 + 4 + 0 correct to the nearest whole number

Observed

frequency, o

Expected

frequency, e

o - e ( )2o e− ( )2o ee−

8

27

62

79

45

24

5

0

7

33

65

73

48

19

4

0

1

-6

-3

6

-3

5

1

0

1

36

9

36

9

25

1

0

0.14286

1.09091

0.13846

0.49315

0.18750

1.31579

0.25000

0

( )22 o e

3.62e

⎧ ⎫−⎪ ⎪χ = =⎨ ⎬⎪ ⎪⎩ ⎭

∑

Degrees of freedom, ν = N – 1 = 8 – 1 = 7

For 20.95χ and ν = 7 from Table 63.1, page 609 of textbook is 14.1

Hence, the hypothesis is accepted, i.e. the observed data fits.

20.10 7,χ ν = 2.83, hence the data is not ‘too good’.


3. The resistances of a sample of carbon resistors are as shown below.

Resistance (MΩ) 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36

Frequency 7 19 41 50 73 52 28 17 9

Test the null hypothesis that this data corresponds to a normal distribution at a level of


(7 1.28) (19 1.29) (41 1.30) (50 1.31) .... 390.5x 1.327 19 41 50 73 52 28 17 9 296

× + × + × + × += = =

+ + + + + + + +

( ) ( ) ( ) ( )2 2 2 27 1.28 1.32 19 1.29 1.32 41 1.30 1.32 50 1.31 1.32 ....s

296

0.0958 0.0180296

⎧ ⎫− + − + − + − +⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

= =

Class

mid-point

Class

boundaries, x

z-value for class

boundary =

x 1.320.0180−

Area from

0 to z

from Table

58.1, page 561

Area for

class

Expected

frequency

1.28

1.29

1.30

1.31

1.32

1.33

1.34

1.35

1.36

1.275

1.285

1.295

1.305

1.315

1.325

1.335

1.345

1.355

1.365

-2.50

-1.94

-1.39

-0.83

-0.28

0.28

0.83

1.39

1.94

2.5

0.4938

0.4738

0.4177

0.2967

0.1103

0.1103

0.2967

0.4177

0.4738

0.4938

0.02

0.0561

0.121

0.1864

0.2206

0.1864

0.121

0.0561

0.02

6

17

36

55

65

55

36

17 6


Resistance Observed

frequency,

o

Expected

frequency, e

o - e ( )2o e−

( )2o ee−

1.28

1.29

1.30

1.31

1.32

1.33

1.34

1.35

1.36

7

19

41

50

73

52

28

17

9

6

17

36

55

65

55

36

17

6

1

2

5

-5

8

-3

-8

0

3

1

4

25

25

64

9

64

0

9

0.1667

0.2353

0.6944

0.4545

0.9846

0.1636

1.7778

0

1.5000

( )22 o e

5.98e

⎧ ⎫−⎪ ⎪χ = =⎨ ⎬⎪ ⎪⎩ ⎭

∑

Degrees of freedom, ν = N – 1 – M = 9 – 1 – 2 = 6


Hence, the null hypothesis is accepted, i.e. the data does correspond to a normal distribution.

5. Test the hypothesis that the maximum load before breaking supported by certain cables produced

by a company follows a normal distribution at a level of significance of 0.05, based on the

experimental data given below. Also, test to see if the data is ‘too good’ at a level of significance

of 0.05.

Maximum load (MN) 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0

Number of cables 2 5 12 17 14 6 3 1

(2 8.5) (5 9.0) (12 9.5) (17 10.0) .... 605.5x 10.09MN

2 5 12 17 14 6 3 1 60× + × + × + × +

= = =+ + + + + + +

( ) ( ) ( ) ( )2 2 2 22 8.5 10.09 5 9.0 10.09 12 9.50 10.09 17 10.0 10.09 ....s

60

32.246 0.733MN60

⎧ ⎫− + − + − + − +⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

= =


Class

mid-point

Class

boundaries, x

z-value for class

boundary =

x 10.090.733−

Area from

0 to z

Area for class Expected

frequency

8.5

9.0

9.5

10.0

10.5

11.0

11.5

12.0

8.25

8.75

9.25

9.75

10.25

10.75

11.25

11.75

12.25

-2.51

-1.83

-1.15

-0.46

0.22

0.90

1.58

2.26

2.95

0.4940

0.4664

0.3749

0.1772

0.0871

0.3159

0.4430

0.4881

0.4984

0.0276

0.0915

0.1977

0.2643

0.2288

0.1271

0.0451

0.0103

2 5

12

16

14 8 3 1

Load Observed

frequency,

o

Expected

frequency, e

o - e ( )2o e−

( )2o ee−

8.5

9.0

9.5

10.0

10.5

11.0

11.5

12.0

2

5

12

17

14

6

3

1

2

5

12

16

14

8

3

1

0

0

0

1

0

-2

0

0

0

0

0

1

0

4

0

0

0

0

0

0.0625

0

0.5000

0

0

( )22 o e

0.563e

⎧ ⎫−⎪ ⎪χ = =⎨ ⎬⎪ ⎪⎩ ⎭

∑


Degrees of freedom, ν = N – 1 – M = 8 – 1 – 2 = 5


Hence, the hypothesis is accepted.

20.05 5, 1.15χ ν = , hence the results are ‘too good to be true’.



2. In a laboratory experiment, 18 measurements of the coefficient of friction, µ, between metal and

leather gave the following results:

0.60 0.57 0.51 0.55 0.66 0.56 0.52 0.59 0.58

0.48 0.59 0.63 0.61 0.69 0.57 0.51 0.58 0.54

Use the sign test at a level of significance of 5% to test the null hypothesis µ = 0.56 against an

alternative hypothesis µ ≠ 0.56.

Using the procedure for the sign test:

(i) Null hypothesis, 0H : 0.56µ =

Alternative hypothesis, 1H : 0.56µ ≠

(ii) Significance level, 2 5%α =

(iii) With + sign ≥ 0.56 and - sign < 0.56:

+ + - - + + - + + - + + + + + - + -

(iv) There are 12 + signs and 6 – signs, hence, S = 6

(v) From Table 63.3, page 614 of textbook, with n = 18 and 2 5%α = S ≤ 4

hence, the null hypothesis is accepted.

3. 18 random samples of two types of 9 V batteries are taken and the mean lifetime (in hours) of

each are:

Type A 8.2 7.0 11.3 13.9 9.0 13.8 16.2 8.6 9.4

3.6 7.5 6.5 18.0 11.5 13.4 6.9 14.2 12.4

Type B 15.3 15.4 11.2 16.1 18.1 17.1 17.7 8.4 13.5

7.8 9.8 10.6 16.4 12.7 16.8 9.9 12.9 14.7

Use the sign test, at a level of significance of 5%, to test the null hypothesis that the two samples

come from the same population.

Using the procedure for the sign test:

(i) Null hypothesis, 0 A BH : µ = µ

Alternative hypothesis, 1 A BH : µ ≠ µ


(ii) Significance level, 2 5%α =

(iii) A - B -7.1 -8.4 +0.1 -2.2 -9.1 -3.3 -1.5 +0.2 -4.1

-4.2 -2.3 -4.1 +1.6 -1.2 -3.4 -3.0 +1.3 -2.3

(iv) There are 4 + signs and 14 – signs, hence, S = 4

(v) From Table 63.3, page 614 of textbook, with n = 18 and 2 5%α = S ≤ 4

Since from (iv) S is equal to 4, then the result is significant at 2 5%α = hence, the alternative

hypothesis 1H is accepted.



1. The time to repair an electronic instrument is a random variable. The repair times (in hours) for

16 instruments are as follows:

218 275 264 210 161 374 178 265 150 360 185 171 215 100 474 248

Use the Wilcoxon signed-rank test, at a 5% level of significance, to test the hypothesis that the

mean repair time is 220 hours

Using the procedure for the Wilcoxon signed-rank test:

(i) 0H : t = 220 h

1H : t ≠ 220 h

(ii) 2 5%α =

(iii) Taking the time difference between the time taken for repair and 220 h gives:

-2 +55 +44 -10 -59 +154 -42 +45

-70 +140 -35 -49 -5 -120 +254 + 28

(iv) Ranking gives:

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Difference -2 -5 -10 +28 -35 -42 +44 +45 -49 +55 -59 -70 -120 +140 +154 +254

(v) T = 4 + 7 + 8 + 10 + 14 + 15 + 16 = 74

(vi) From Table 63.4, page 617 of textbook, for n = 16, 2 5%α = , T ≤ 29

Hence, since 74 > 29, the hypothesis 0H is accepted, i.e. the mean repair time is 220 hours.

3. A paint supplier claims that a new additive will reduce the drying time of their acrylic paint. To

test his claim, 12 pieces of wood are painted, one half of each piece with paint containing the

regular additive and the other half with paint containing the new additive. The drying time (in

hours) were measured as follows:

New additive 4.5 5.5 3.9 3.6 4.1 6.3 5.9 6.7 5.1 3.6 4.0 3.0

Regular additive 4.7 5.9 3.9 3.8 4.4 6.5 6.9 6.5 5.3 3.6 3.9 3.9

Use the Wilcoxon signed-rank test at a significance level of 5% to test the hypothesis that there is

no difference, on average, in the drying times of the new and regular additive paints.


Using the procedure for the Wilcoxon signed-rank test:

(i) 0H : N = R

1H : N ≠ R

(ii) 2 5%α =

(iii) Taking the time difference between N and R gives:

(N – R) -0.2 -0.4 0 -0.2 -0.3 -0.2 -1.0 +0.2 -0.2 0 +0.1 -0.9

(iv) Ranking and ignoring the zero’s gives:

Rank 1 4 4 4 4 4 7 8 9 10

Difference +0.1 -0.2 -0.2 -0.2 -0.2 +0.2 -0.3 -0.4 -0.9 -1.0

(v) T = 1 + 4 = 5

(vi) From Table 63.4, page 617 of textbook, for n = 10, 2 5%α = , T ≤ 8

Since from (v) T is less than 8, there is a significant difference in the drying times.



1. The tar content of two brands of cigarettes (in mg) was measured as follows:

Brand P 22.6 4.1 3.9 0.7 3.2 6.1 1.7 2.3 5.6 2.0

Brand Q 3.4 6.2 3.5 4.7 6.3 5.5 3.8 2.1

Use the Mann-Whitney test at a 0.05 level of significance to determine if the tar contents of the

two brands are equal.

Using the procedure for the Mann-Whitney test:

(i) 0 A BH : T T=

1 A BH : T T≠

(ii) 2 5%α =

(iii) Brand P 0.7 1.7 2.0 2.3 3.2 3.9 4.1 5.6 6.1 22.6

Brand Q 2.1 3.4 3.5 3.8 4.7 5.5 6.2 6.3

(iv) P P P Q P P Q Q Q P P Q Q P P Q Q P

0 0 0 1 1 4 4 6 6 8 writing the Q’s that precede the P’s

(v) U = 1 + 1 + 4 + 4 + 6 + 6 + 8 = 30

(vi) From Table 63.5, page 622 of textbook, for a sample size of 10 and 8 at 2 5%α = , U ≤ 17

Hence, 0H is accepted, i.e. there is no difference between brands P and Q

3. An experiment, designed to compare two preventive methods against corrosion gave the

following results for the maximum depths of pits (in mm) in metal strands:

Method A 143 106 135 147 139 132 153 140

Method B 98 105 137 94 112 103

Use the Mann-Whitney test, at a level of significance of 0.05, to determine whether the two tests

are equally effective.


(i) 0H : A B=

1H : A B≠


(ii) 2 5%α =

(iii) A 106 132 135 139 140 143 147 153

B 94 98 103 105 112 137

(iv) B B B B A B A A B A A A A A

1 3 writing the A’s that precede the B’s

(v) U = 1 + 3 = 4


Hence, the null hypothesis 0H is rejected, i.e. the two methods are not equally effective.

4. Repeat problem 3 of Exercise 228, using the Mann-Whitney test.


(i) Null hypothesis, 0 A BH : µ = µ

Alternative hypothesis, 1 A BH : µ ≠ µ

(ii) 2 5%α =

(iii) A 3.6 6.5 6.9 7.0 7.5 8.2 8.6 9.0 9.4 11.3 11.5 12.4 13.4 13.8 13.9 14.2 16.2 18.0

B 7.8 8.4 9.8 9.9 10.6 11.2 12.7 12.9 13.5 14.7 15.3 15.4 16.1 16.4 16.8 17.1 17.7 18.1

(iv) A A A A A B A B A A A B B B B A A A B B A B A A A B B B B A B B B B A B

1 2 2 2 6 6 6 8 9 9 9 13 17

(v) U = 1 + 2 + 2 + 2 + 6 + 6 + 6 + 8 + 9 + 9 + 9 + 13 + 17 = 90


Hence, the null hypothesis 0H is rejected and 1H is accepted, i.e. the two means are not

equal.


CHAPTER 64 INTRODUCTION TO LAPLACE TRANSFORMS EXERCISE 231 Page 630

1. Determine the Laplace transforms of (a) 2t – 3 (b) 25t 4t 3+ −

(a) ℒ 2

1 12t 3 2 3s s

⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 2

2 3s s−

(b) ℒ 23 2

2! 1 15t 4t 3 5 4 3s s s

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − = + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 3 2

10 4 3s s s

+ −

2. Determine the Laplace transforms of (a) 3t

24 - 3t + 2 (b)

5 24t t2t

15 2− +

(a) ℒ3

3 1 2

t 1 3! 1 13t 2 3 224 24 s s s+

⎧ ⎫ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + = − +⎨ ⎬ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

= 4 2

1 3 24s s s

− +

(b) ℒ5 2

45 1 4 1 3

t t 1 5! 4! 1 2!2t 215 2 15 s s 2 s+ +

⎧ ⎫ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞− + = − +⎨ ⎬ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎩ ⎭

= 6 5 3

8 48 1s s s− +

3. Determine the Laplace transform of (a) 3t5e (b) 2t2e−

(a) ℒ 3t 15e 5s 3

⎛ ⎞= ⎜ ⎟−⎝ ⎠ = 5

s 3−

(b) ℒ 2t 12e 2s 2

− ⎛ ⎞= ⎜ ⎟+⎝ ⎠ = 2

s 2+

4. Determine the Laplace transform of (a) 4 sin 3t (b) 3 cos 2t

(a) ℒ 2 2

34sin 3t 4s 3

⎛ ⎞= ⎜ ⎟+⎝ ⎠ = 2

12s 9+

(b) ℒ 2 2

s3cos 2t 3s 2

⎛ ⎞= ⎜ ⎟+⎝ ⎠ = 2

3ss 4+


5. Determine the Laplace transforms of (a) 7 cosh 2x (b) 1 sinh 3t3

(a) ℒ 2 2

s7cosh 2x 7s 2

⎛ ⎞= ⎜ ⎟−⎝ ⎠ = 2

7ss 4−

(b) ℒ 2 2

1 1 3sinh 3t3 3 s 3

⎧ ⎫ ⎛ ⎞=⎨ ⎬ ⎜ ⎟−⎩ ⎭ ⎝ ⎠ = 3

1s 9−

6.(a) Determine the Laplace transform of 22cos t

cos 2t = 22cos t 1− from which, 2 1 cos 2tcos t2

+=

ℒ 22cos t = ℒ 1 cos 2t22

⎧ + ⎫⎛ ⎞⎨ ⎬⎜ ⎟

⎝ ⎠⎩ ⎭ = ℒ ( )

( )2 2

2 2 2 2

s 4 s1 s 1 s1 cos 2ts s 2 s s 4 s s 4

+ ++ = + = + =

+ + +

= ( )

2

2

2s 4s s 4

++

= ( )( )

2

2

2 s 2

s s 4

+

+

7.(b) Determine the Laplace transform of 22sinh 2θ

cosh 4θ = 1 + 22sinh 2θ from which, 22sinh 2θ = cosh 4θ - 1

ℒ 22sinh 2θ = ℒ ( )( )

2 2

2 2 2

s s 16s 1cosh 4 1s 4 s s s 16

− −θ− = − =

− − = ( )2

16s s 16−

8. Determine the Laplace transform of 4 sin (at + b), where a and b are constants.

ℒ 4sin(at b)+ = ℒ 4 sin at cos b cos at sin b+ = ℒ (4cos b)sin at (4sin b)cos at+

= ( ) ( )2 2 2 2

a s4cos b 4sin bs a s a

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

= ( )2 2

4 acosb ssinbs a

++


10. Show that ℒ ( )2 22

scos 3t sin 3ts 36

− =+

2cos 6t 2cos 3t 1= − from which, 2 1 cos 6tcos 3t2

+=

and 2cos6t 1 2sin 3t= − from which, 2 1 cos 6tsin 3t2

−=

Hence, ℒ ( )2 2cos 3t sin 3t− = ℒ 1 cos 6t 1 cos 6t2 2

⎧ + − ⎫⎛ ⎞ ⎛ ⎞−⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎩ ⎭

= ℒ 1 cos 6t 1 cos 6t2 2 2 2

⎧ ⎫+ − +⎨ ⎬⎩ ⎭

= ℒcos 6t = 2

ss 36+


CHAPTER 65 PROPERTIES OF LAPLACE TRANSFORMS EXERCISE 232 Page 634

1. Determine the Laplace transforms of (a) 2t2te (b) 2 tt e

(a) ℒ 2t2te 2= ℒ ( )

1 2t1 1

1!t e (2)s 2 +=−

= ( )2

2s 2−

(b) ℒ ( )

2 t2 1

2!t es 1 +=−

= ( )3

2s 1−

2.(b) Determine the Laplace transform of 4 3t1 t e2

−

ℒ( )

4 3t4 1

1 1 4!t e2 2 s 3

−+

⎛ ⎞⎧ ⎫ = ⎜ ⎟⎨ ⎬ ⎜ ⎟⎩ ⎭ +⎝ ⎠ =

( )5

12s 3+

3.(b) Determine the Laplace transform of 2t3e sin 2t

ℒ ( )

2t2 22

2 63e sin 2t 3s 4s 4 4s 2 2

⎛ ⎞= =⎜ ⎟

⎜ ⎟ − + +− +⎝ ⎠ = 2

6s 4s 8− +

4.(a) Determine the Laplace transform of 2t5e cos3t−

ℒ ( )

( )2t2 22

5 s 2s 25e cos3t 5s 4s 4 9s 2 3

−⎛ ⎞ ++

= =⎜ ⎟⎜ ⎟ + + ++ +⎝ ⎠

= ( )2

5 s 2s 4s 13

++ +

5.(a) Determine the Laplace transform of t 22e sin t

ℒ t 22e sin t = ℒ t 1 cos 2t2e2

⎧ − ⎫⎛ ⎞⎨ ⎬⎜ ⎟

⎝ ⎠⎩ ⎭ = ℒ te - ℒ

( )t

2 2

1 s 1e cos 2ts 1 s 1 2

−= −

− − +

= 2

1 s 1s 1 s 2s 5

−−

− − +


6.(b) Determine the Laplace transform of 2t3e cosh 4t

ℒ ( )

2t2 22

s 2 3(s 2)3e cosh 4t 3s 4s 4 16s 2 4

⎛ ⎞− −= =⎜ ⎟

⎜ ⎟ − + −− −⎝ ⎠ =

( )2

3 s 2s 4s 12

−− −

7.(a) Determine the Laplace transform of t2e sinh 3t−

ℒ ( )

t2 22

3 62e sinh 3t 2s 2s 1 9s 1 3

−⎛ ⎞

= =⎜ ⎟⎜ ⎟ + + −+ −⎝ ⎠

= 2

6s 2s 8+ −

8. Determine the Laplace transforms of (a) ( )t2e cos3t 3sin 3t− (b) ( )2t3e sinh 2t 2cosh 2t− −

(a) ℒ ( ) t2e cos3t 3sin 3t− = ℒ t2e cos3t - ℒ ( ) ( )

t2 22 2

s 1 36e sin 3t 2 6s 1 3 s 1 3

⎛ ⎞ ⎛ ⎞−= −⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟− + − +⎝ ⎠ ⎝ ⎠

= ( )2 2 2 2

2 s 1 18 2s 2 18 2s 20s 2s 10 s 2s 10 s 2s 10 s 2s 10

− − − −− = =

− + − + − + − +

= ( )2

2 s 10s 2s 10

−− +

(b) ℒ ( ) 2t3e sinh 2t 2cosh 2t− − = ℒ 2t3e sinh 2t− - ℒ 2t6e cosh 2t−

= ( ) ( )2 22 2

2 s 23 6s 2 2 s 2 2

⎛ ⎞ ⎛ ⎞+−⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟+ − + −⎝ ⎠ ⎝ ⎠

= ( ) ( )2 2 2 2

6 s 2 6 s 26 6s 4s 4 4 s 4s 4 4 s 4s s 4s

+ +− = −

+ + − + + − + +

= ( ) ( )

6 6s 12 6s 6s s 4 s s 4− − − −

=+ +

= ( )( )6 s 1

s s 4− +

+



2. Use the Laplace transform of the first derivative to derive the transforms:

(a) ℒ ate = 1s a−

(b) ℒ 23t = 3

6s

(a) Let f(t) = ate then f ′(t) = a ate and f(0) = 1

From equation (3), page 634 of textbook, ℒ f '(t) s= ℒ f (t) f (0)−

Hence, ℒ ata e = sℒ ate - 1

i.e. 1 = (s – a)ℒ ate

and ℒ ate = 1s a−

(b) Let f(t) = 23t then f ′(t) = 6t and f(0) = 0

Since ℒ f '(t) s= ℒ f (t) f (0)−

then, ℒ6t = sℒ 23t + 0

i.e. 2

6s

= sℒ 23t

and ℒ 23t = 3

6s

3. Derive the Laplace transform of the second derivative from the definition of a Laplace

transform. Hence derive the transform ℒsin at = 2 2

as a+

For the derivation, see page 634 of textbook.

Let f(t) = sin at, then f ′(t) = a cos at and f ′′(t) = 2a sin at− , f(0) = 0 and f ′(0) = a

From equation (4), page 635 of textbook, ℒ 2f ''(t) s= ℒ f (t) sf (0) f '(0)− −

Hence, ℒ 2 2a sin at s− = ℒ sin at - s(0) - a


i.e. 2a− ℒ sin at = 2s ℒ sin at - a

Hence, a = ( )2 2s a+ ℒ sin at

and ℒ sinat = 2 2

as a+

4.(b) Use the Laplace transform of the second derivative to derive the transform:

ℒcosh at = 2 2

ss a−

Let f(t) = cosh at then f ′(t) = a sinh at and f ′′(t) = 2a cosh at , f(0) = 1 and f ′(0) = 0

ℒ 2f ''(t) s= ℒ f (t) sf (0) f '(0)− −

Hence, ℒ 2 2a cosh at s= ℒ cosh at - s(1) - 0

i.e. 2a ℒ cosh at = 2s ℒ cosh at - s

i.e. s = ( )2 2s a− ℒ cosh at

and ℒ coshat = 2 2

ss a−



1. State the initial value theorem. Verify the theorem for the function (a) 3 – 4 sin t (b) ( )2t 4−

and state their initial values.

The initial value theorem states: [ ] [

t 0 slimit f (t) limit s

→ →∞= ℒ ]f (t)

(a) Let f(t) = 3 – 4 sin t then ℒf(t) = ℒ3 – 4 sin t = 2

3 4s s 1−

+

Hence, [ ] 2t 0 s

3 4limit 3 4sin t limit ss s 1→ →∞

⎡ ⎤⎛ ⎞− = −⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦

= 2s

4slimit 3s 1→∞

⎡ ⎤−⎢ ⎥+⎣ ⎦

i.e. 3 – 4 sin 0 = 3 - 2 1∞

∞ +

i.e. 3 = 3 which verifies the theorem.

The initial value is 3

(b) Let f(t) = ( )2 2t 4 t 8t 16− = − +

then ℒ 23 2

2 8 16t 8t 16s s s

− + = − +

Hence, 23 2t 0 s

2 8 16limit t 8t 16 limit ss s s→ →∞

⎡ ⎤⎛ ⎞⎡ ⎤− + = − +⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

= 2s

2 8limit 16s s→∞

⎡ ⎤− +⎢ ⎥⎣ ⎦

i.e. 16 = 16 which verifies the theorem

The initial value is 16

3. State the final value theorem and state a practical application where it is of use. Verify the

theorem for the function 4 + 2te− (sin t + cos t) representing a displacement and state its final

value.


The final value theorem states: [ ] [t s 0

limit f (t) limit s→∞ →

= ℒ ]f (t)

The final value theorem is used in investigating the stability of systems such as in automatic

aircraft-landing systems.

Let f(t) = 4 + 2te− (sin t + cos t) = 4 + 2t 2te sin t e cos t− −+

then ℒ ( ) ( )2 2

4 1 s 2f (t)s s 2 1 s 2 1

+= + +

+ + + +

Hence, [( ) ( )

2t 2t2 2t s 0

4 1 s 2limit 4 e sin t e cos t limit ss s 2 1 s 2 1

− −

→∞ →

⎛ ⎞+⎡ ⎤+ + = + +⎜ ⎟⎣ ⎦ ⎜ ⎟+ + + +⎝ ⎠

= ( )

( )( )2 2s 0

s s 2slimit 4s 2 1 s 2 1→

⎡ ⎤++ +⎢ ⎥

+ + + +⎢ ⎥⎣ ⎦

i.e. 4 + 0 + 0 = 4 + 0 + 0

i.e. 4 = 4 which verifies the theorem

The final value is 4


CHAPTER 66 INVERSE LAPLACE TRANSFORMS EXERCISE 235 Page 640

2. (a) Determine the inverse Laplace transform of 32s 1+

ℒ 1 32s 1

− ⎧ ⎫⎨ ⎬+⎩ ⎭

= 3ℒ 1 112 s2

−

⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬

⎛ ⎞⎪ ⎪+⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

= 32ℒ 1 1

1s2

−

⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬⎛ ⎞⎪ ⎪+⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

= 1t23 e

2−

3. (a) Determine the inverse Laplace transform of 2

1s 25+

ℒ 12

1s 25

− ⎧ ⎫⎨ ⎬+⎩ ⎭

= 15ℒ 1

2 2

5s 5

− ⎧ ⎫⎨ ⎬+⎩ ⎭

= 1 sin 5t5


5s2s 18+

ℒ 12

5s2s 18

− ⎧ ⎫⎨ ⎬+⎩ ⎭

= 5ℒ( )

12

s2 s 9

−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭ = 5

2ℒ 1

2 2

ss 3

− ⎧ ⎫⎨ ⎬+⎩ ⎭

= 5 cos 3t2


5s

ℒ 13

5s

− ⎧ ⎫⎨ ⎬⎩ ⎭

= 52!ℒ 1

3

2!s

− ⎧ ⎫⎨ ⎬⎩ ⎭

= 25 t2


3s1 s 82

−

ℒ 1

2

3s1 s 82

−

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪−⎩ ⎭

= 3ℒ( )

1

2

s1 s 162

−

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪−⎩ ⎭

= 6ℒ 12 2

ss 4

− ⎧ ⎫⎨ ⎬−⎩ ⎭

= 6 cosh 4t


8. (b) Determine the inverse Laplace transform of ( )5

3s 3−

ℒ( )

15

3s 3

−⎧ ⎫⎪ ⎪⎨ ⎬

−⎪ ⎪⎩ ⎭ = 3ℒ

( )1

4 11

s 3−

+

⎧ ⎫⎪ ⎪⎨ ⎬

−⎪ ⎪⎩ ⎭ = 3

4!ℒ

( )1

4 14!

s 3−

+

⎧ ⎫⎪ ⎪⎨ ⎬

−⎪ ⎪⎩ ⎭ = 1

8ℒ

( )1

4 14!

s 3−

+

⎧ ⎫⎪ ⎪⎨ ⎬

−⎪ ⎪⎩ ⎭ = 3t 41 e t

8

9. (b) Determine the inverse Laplace transform of 2

3s 6s 13+ +

ℒ 12

3s 6s 13

− ⎧ ⎫⎨ ⎬+ +⎩ ⎭

= 3ℒ( )

12 2

1s 3 2

−⎧ ⎫⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭ = 3

2ℒ

( )1

2 2

2s 3 2

−⎧ ⎫⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭ = 3t3 e sin 2t

2−


2(s 3)s 6s 13

−− +

ℒ ( )12

2 s 3s 6s 13

− −⎧ ⎫⎨ ⎬− +⎩ ⎭

= 2ℒ( )

12 2

s 3s 3 2

−⎧ ⎫−⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭ = 3t2e cos 2t

11. Determine the inverse Laplace transforms of (a) 2

2s 5s 4s 5

++ −

(b) 2

3s 2s 8s 25

+− +

(a) ℒ 12

2s 5s 4s 5

− +⎧ ⎫⎨ ⎬+ −⎩ ⎭

= ℒ ( )( ) ( )

12 22 2

2 s 2 1s 2 3 s 2 3

−⎧ ⎫+⎪ ⎪+⎨ ⎬

+ − + −⎪ ⎪⎩ ⎭

= 2ℒ( )

12 2

s 2s 2 3

−⎧ ⎫+⎪ ⎪⎨ ⎬

+ −⎪ ⎪⎩ ⎭+ 1

3ℒ

( )1

2 2

3s 2 3

−⎧ ⎫⎪ ⎪⎨ ⎬

+ −⎪ ⎪⎩ ⎭ = 2t 2t12e cosh 3t e sinh 3t

3− −+

(b) ℒ 12

3s 2s 8s 25

− +⎧ ⎫⎨ ⎬− +⎩ ⎭

= ℒ( )

12 2

3s 2s 4 3

−⎧ ⎫+⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭ = ℒ ( )

( )1

2 2

3 s 4 14s 4 3

−⎧ ⎫− +⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭

= 3ℒ( )

12 2

s 4s 4 3

−⎧ ⎫−⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭+14

3ℒ

( )1

2 2

3s 4 3

−⎧ ⎫⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭ = 4t 4t143e cos 3t e sin 3t

3+



2. Use partial fractions to find the inverse Laplace transform of: ( )( )( )

22s 9s 35s 1 s 2 s 3

− −+ − +

( )( )( ) ( ) ( ) ( )22s 9s 35 4 3 1

s 1 s 2 s 3 s 1 s 2 s 3− −

= − ++ − + + − +

from Problem 2, page 19 of textbook

Hence, ℒ( )( )( )

21 2s 9s 35

s 1 s 2 s 3− ⎧ ⎫− −⎪ ⎪⎨ ⎬+ − +⎪ ⎪⎩ ⎭

= ℒ( ) ( ) ( )

1 4 3 1s 1 s 2 s 3

− ⎧ ⎫⎪ ⎪− +⎨ ⎬+ − +⎪ ⎪⎩ ⎭ = t 2t 3t4e 3e e− −− +

4. Use partial fractions to find the inverse Laplace transform of: ( )

2

33s 16s 15

s 3+ +

+

( ) ( ) ( ) ( )

2

3 2 33s 16s 15 3 2 6

s 3s 3 s 3 s 3+ +

= − −++ + +

from Problem 7, page 22 of textbook

Hence, ℒ( )

21

33s 16s 15

s 3−⎧ ⎫+ +⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭ = ℒ

( ) ( ) ( )1

2 33 2 6

s 3 s 3 s 3−⎧ ⎫⎪ ⎪− −⎨ ⎬+ + +⎪ ⎪⎩ ⎭

= 3ℒ( )

1 1s 3

− ⎧ ⎫⎪ ⎪⎨ ⎬+⎪ ⎪⎩ ⎭

- 2ℒ( )

12

1s 3

−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭ - 6ℒ

( )1

31

s 3−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

= 3ℒ( )

1 1s 3

− ⎧ ⎫⎪ ⎪⎨ ⎬+⎪ ⎪⎩ ⎭

- 2ℒ( )

11 1

1s 3

−+

⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭ - 6

2!ℒ

( )1

2 12!

s 3−

+

⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

= 3t 3t 3t 23e 2e t 3e t− − −− − or 3t 2e (3 2t 3t )− − −


2 3

2 2

3 6s 4s 2ss s 3

+ + −+

( )2 3

2 22 2

3 6s 4s 2s 2 1 3 4ss s s 3s s 3

+ + − −= + +

++ from Problem 9, page 23 of textbook

Hence, ℒ( )

2 31

2 2

3 6s 4s 2ss s 3

−⎧ ⎫+ + −⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭ = ℒ 1

2 2

2 1 3 4ss s s 3

− −⎧ ⎫+ +⎨ ⎬+⎩ ⎭


= 2ℒ 1 1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒ 12

1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒ( )

12

2

3 4s

s 3−

⎧ ⎫−⎪ ⎪

⎨ ⎬⎪ ⎪+⎩ ⎭

= 2ℒ 1 1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒ 12

1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒ( )

12

2

3

s 3−

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪+⎩ ⎭

- ℒ( )

12

2

4s

s 3−

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪+⎩ ⎭

= 2ℒ 1 1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒ 12

1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

+ 33ℒ

( )1

22

3

s 3−

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪+⎩ ⎭

- 4ℒ( )

12

2

s

s 3−

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪+⎩ ⎭

= 2 t 3 sin 3 t 4cos 3 t+ + −


2

2

26 ss s 4s 13

−+ +

Let ( )

( ) ( )( )

22

22 2

A s 4s 13 Bs C s26 s A Bs Cs s 4s 13s s 4s 13 s s 4s 13

+ + + +− +≡ + =

+ ++ + + +

Hence, ( )2 2 226 s A s 4s 13 Bs Cs− = + + + +

When s = 0: 26 = 13A + 0 + 0 i.e. A = 2

Equating 2s coefficients: -1 = A + B i.e. B = -3

Equating s coefficients: 0 = 4A + C i.e. C = -8

Thus, ( )

2

22

26 s 2 3s 8s s 4s 13s s 4s 13

− − −≡ +

+ ++ +

Hence, ℒ( )

21

2

26 ss s 4s 13

−⎧ ⎫−⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭ = ℒ 1 2

s− ⎧ ⎫⎨ ⎬⎩ ⎭

- ℒ 12

3s 8s 4s 13

− +⎧ ⎫⎨ ⎬+ +⎩ ⎭

= 2ℒ 1 1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

- ℒ( )

12 2

3s 8s 2 3

−⎧ ⎫+⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭

= 2ℒ 1 1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

- ℒ ( )( )

12 2

3 s 2s 2 3

−⎧ ⎫+⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭ - ℒ

( )1

2 2

2s 2 3

−⎧ ⎫⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭


= 2ℒ 1 1s

− ⎧ ⎫⎨ ⎬⎩ ⎭

- 3ℒ ( )( )

12 2

s 2s 2 3

−⎧ ⎫+⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭ - 2

3ℒ

( )1

2 2

3s 2 3

−⎧ ⎫⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭

= 2t 2t22 3e cos 3t e sin 3t3

− −− −


EXERCISE 237 (Page XX)

1. Determine for the transfer function: R(s) = ( )( )( )2

50 s 4s s 2 s 8s 25

+

+ − +

(a) the zero and (b) the poles. Show the poles and zeros on a pole-zero diagram.

(a) For the numerator to be zero, (s + 4) = 0 hence, s = -4 is a zero of R(s)

(b) For the denominator to be zero, s = 0 or s = -2 or 2s 8s 25 0− + =

i.e. s = ( )28 8 4(1)(25) 8 36 8 j6 4 j3

2(1) 2 2− − ± − − ± − − ±

= = = ±

Hence, poles occur at s = 0, s = -2, s = 4 + j3 and 4 - j3

A pole-zero diagram is shown below.

3. For the function G(s) = ( )( )2

s 1s 2 s 2s 5

−+ + +

determine the poles and zeros and show them on a

pole-zero diagram.

For the denominator to be zero, s = -2 or 2s 2s 5 0+ + =

i.e. s = ( )22 2 4(1)(5) 2 16 2 j4 1 j2

2(1) 2 2− ± − − ± − − ±

= = = − ±

Hence, poles occur at s = -2, s = -1 + j2 and -1 – j2


For the numerator to be zero, (s - 1) = 0 hence, s = 1 is a zero of G(s)

A pole-zero diagram is shown below.


CHAPTER 67 THE SOLUTION OF DIFFERENTIAL EQUATIONS

USING LAPLACE TRANSFORMS EXERCISE 238 Page 648

1. A first order differential equation involving current i in a series R-L circuit is given by:

di E5idt 2+ = and i = 0 at time t = 0.

Use Laplace transforms to solve for i when (a) E = 20 (b) E = 3t40e− and (c) E = 50 sin 5t.

Taking the Laplace transform of each term of di E5idt 2+ = gives:

ℒ didt

⎧ ⎫⎨ ⎬⎩ ⎭

+ 5ℒi = ℒ E2

⎧ ⎫⎨ ⎬⎩ ⎭

i.e. sℒi – i(0) + 5ℒi = E / 2s

i = 0 at t = 0, hence, i(0) = 0

Hence, (s + 5)ℒi = E / 2s

i.e. ℒi = E / 2s(s 5)+

and i = ℒ 1 E / 2s(s 5)

− ⎧ ⎫⎨ ⎬+⎩ ⎭

= E2ℒ 1 1

s(s 5)− ⎧ ⎫⎨ ⎬+⎩ ⎭

Let 1 A B A(s 5) Bss(s 5) s (s 5) s(s 5)

+ +≡ + =

+ + +

Hence, 1 = A(s + 5) + Bs

When s = 0: 1 = 5A i.e. A = 15

When s = -5: 1 = -5B i.e. B = - 15

Thus, i = E2ℒ 1 1

s(s 5)− ⎧ ⎫⎨ ⎬+⎩ ⎭

= E2ℒ 1

1 15 5s (s 5)

−

⎧ ⎫⎪ ⎪

−⎨ ⎬+⎪ ⎪⎩ ⎭

= E2

5t1 1 e5 5

−⎛ ⎞−⎜ ⎟⎝ ⎠

(a) When E = 20, i = 202

5t1 1 e5 5

−⎛ ⎞−⎜ ⎟⎝ ⎠

= ( )5t2 1 e−−


(b) When E = 3t40e− ℒ didt

⎧ ⎫⎨ ⎬⎩ ⎭

+ 5ℒi = ℒ3t40e

2

−⎧ ⎫⎨ ⎬⎩ ⎭

i.e. sℒi – i(0) + 5ℒi = 20s 3+

i = 0 at t = 0, hence, i(0) = 0

Hence, (s + 5)ℒi = 20s 3+

i.e. ℒi = 20(s 3)(s 5)+ +

and i = ℒ 1 20(s 3)(s 5)

− ⎧ ⎫⎨ ⎬+ +⎩ ⎭

Let 20 A B A(s 5) B(s 3)(s 3)(s 5) (s 3) (s 5) (s 3)(s 5)

+ + +≡ + =

+ + + + + +

Hence, 20 = A(s + 5) + B(s + 3)

When s = -3: 20 = 2A i.e. A = 10

When s = -5: 20 = -2B i.e. B = -10

Thus, i = ℒ 1 20(s 3)(s 5)

− ⎧ ⎫⎨ ⎬+ +⎩ ⎭

= ℒ 1 10 10(s 3) (s 5)

− ⎧ ⎫−⎨ ⎬+ +⎩ ⎭

i.e. i = 3t 5t10e 10e− −− = ( )3t 5t10 e e− −−

(c) When E = 50 sin 5t ℒ didt

⎧ ⎫⎨ ⎬⎩ ⎭

+ 5ℒi = ℒ 50sin 5t2

⎧ ⎫⎨ ⎬⎩ ⎭

i.e. sℒi – i(0) + 5ℒi = 2 2

25(5)s 5+

i = 0 at t = 0, hence, i(0) = 0

Hence, (s + 5)ℒi = 2

125s 25+

i.e. ℒi = 2

125(s 5)(s 25)+ +

and i = ℒ 12

125(s 5)(s 25)

− ⎧ ⎫⎨ ⎬+ +⎩ ⎭

Let 2

2 2 2

125 A Bs C A(s 25) (Bs C)(s 5)(s 5)(s 25) (s 5) (s 25) (s 5)(s 25)

+ + + + +≡ + =

+ + + + + +


Hence, 125 = ( ) ( )( )2A s 25 Bs C s 5+ + + +

When s = -5: 125 = 50A i.e. A = 52

Equating 2s coefficients: 0 = A + B i.e. B = - 52

Equating constant terms: 125 = 25A + 5C i.e. 125 = 1252

+ 5C

from which, C =

125252

5 2=

Thus, i = ℒ 12

125(s 5)(s 25)

− ⎧ ⎫⎨ ⎬+ +⎩ ⎭

= ℒ( )

12

5 5 25s2 2 2

(s 5) s 25−

⎧ ⎫− +⎪ ⎪+⎨ ⎬+ +⎪ ⎪

⎩ ⎭

= 52ℒ 1 1

(s 5)− ⎧ ⎫⎨ ⎬+⎩ ⎭

+ ℒ( )

12

252

s 25−

⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

- ℒ( )

12

5 s2

s 25−

⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

= 52ℒ 1 1

(s 5)− ⎧ ⎫⎨ ⎬+⎩ ⎭

+ 52ℒ

( )1

2 2

5s 5

−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭ - 5

2ℒ

( )1

2 2

ss 5

−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

= 5t5 5 5e sin 5t cos5t2 2 2

− + −

i.e. i = ( )5t5 e sin 5t cos 5t2

− + −

3. Use Laplace transforms to solve the differential equation: 2

2

d x 100x 0dt

+ = given x(0) = 2 and

x′(0) = 0.

ℒ2

2

d xdx

⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒ100x = ℒ0

i.e. 2s ℒx – s x(0) - x′(0) + 100ℒx = 0

x(0) = 2 and x′(0) = 0, hence 2s ℒx – 2s + 100ℒx = 0

i.e. ( 2s + 100)ℒx = 2s


from which, ℒx = 2

2ss 100+

and x = ℒ 12

2ss 100

− ⎧ ⎫⎨ ⎬+⎩ ⎭

i.e. x = 2ℒ 12 2

ss 10

− ⎧ ⎫⎨ ⎬+⎩ ⎭

= 2 cos 10t


2

d i di1000 250000i 0dt dt

+ + =

given i(0) = 0 and i′(0) = 100.

ℒ2

2

d idt

⎧ ⎫⎨ ⎬⎩ ⎭

+ 1000ℒ didt

⎧ ⎫⎨ ⎬⎩ ⎭

+ 250000ℒi = ℒ0

i.e. [ 2s ℒi – s i(0) - i′(0)] + 1000[sℒi – i(0)] + 250000ℒi = 0

i(0) = 0 and i′(0) = 100, hence

2s ℒi – 100 + 1000sℒi + 250000ℒi = 0

i.e. ( 2s +1000s + 250000)ℒi = 100

from which, ℒi = 2

100s 1000s 250000+ +

and i = ℒ( )

12

100s 500

−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭ = 100ℒ

( )1

1 11

s 500−

+

⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

i.e. i = 500t100t e−


4x2

d y dy2 y 3edx dx

− + = given

y(0) = 23

− and y′(0) = 143

ℒ2

2

d ydx

⎧ ⎫⎨ ⎬⎩ ⎭

- 2ℒ dydx

⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒy = ℒ 4x3e

i.e. [ 2s ℒy – s y(0) - y′(0)] - 2[sℒy – y(0)] + ℒy = 3(s 4)−


y(0) = 23

− and y′(0) = 143

, hence

2s ℒy – s 23

⎛ ⎞−⎜ ⎟⎝ ⎠

- 133

- 2sℒy + 2 23

⎛ ⎞−⎜ ⎟⎝ ⎠

+ ℒy = 3(s 4)−

i.e. ( 2s - 2s + 1)ℒy + 2 13 4s3 3 3

− − = 3(s 4)−

from which, ( 2s - 2s + 1)ℒy = 3(s 4)−

- 2 17 9 2s(s 4) 17(s 4)s3 3 3(s 4)

− − + −+ =

−

ℒy = ( )( ) ( )( )

2 2

22

9 2s 8s 17s 68 59 25s 2s3 s 4 s 2s 1 3 s 4 s 1− + + − − + −

=− − + − −

and y = ℒ( )( )

21

259 25s 2s

3 s 4 s 1−⎧ ⎫− + −⎪ ⎪⎨ ⎬

− −⎪ ⎪⎩ ⎭

Let ( )( )( ) ( ) ( ) ( )

( ) ( )( ) ( )( )( )

2 2

2 2 2

1 59 25s 2s A s 1 B s 4 s 1 C s 4A B C3s 4 s 1s 4 s 1 s 1 s 4 s 1

− + − − + − − + −≡ + + =

− −− − − − −

Hence, ( ) ( )( ) ( )2259 25 2s s A s 1 B s 4 s 1 C s 43 3 3

− + − = − + − − + −

When s = 4: 59 100 32 9A 0 03 3 3

− + − = + + i.e. 3 = 9A and A = 13

When s = 1: 59 25 2 0 0 3C3 3 3

− + − = + − i.e. -12 = -3C and C = 4

Equating 2s coefficients: 2 A B3

− = + i.e. 2 1 B3 3

− = + and B = -1

Hence, y = ℒ( ) ( ) ( )

12

11 43

s 4 s 1 s 1−

⎧ ⎫⎪ ⎪

− +⎨ ⎬− − −⎪ ⎪⎩ ⎭

i.e. y = 4x x x1 e e 4x e3

− + or y = ( ) x 4x14x 1 e e3

− +


2

d y dy 2y 3cos3x 11sin 3xdx dx

+ − = −

given y(0) = 0 and y′(0) = 6.


ℒ2

2

d ydx

⎧ ⎫⎨ ⎬⎩ ⎭

+ ℒ dydx

⎧ ⎫⎨ ⎬⎩ ⎭

- 2ℒy = ℒ 3cos3x 11sin 3x−

i.e. [ 2s ℒy – s y(0) - y′(0)] + [sℒy – y(0)] - 2ℒy = 2 2

3s 33s 9 s 9

−+ +

y(0) = 0 and y′(0) = 6, hence

2s ℒy - 6 + sℒy - 2ℒy = 2

3s 33s 9−+

i.e. ( 2s + s - 2)ℒy = 6 + 2

3s 33s 9−+

from which, ( 2s + s - 2)ℒy = ( )2 2

2 2

6 s 9 3s 33 6s 3s 21s 9 s 9+ + − + +

=+ +

ℒy = ( )( )

2

2 2

6s 3s 21s 9 s s 2

+ ++ + −

and y = ℒ( )( )( )

21

2

6s 3s 21s 9 s 1 s 2

−⎧ ⎫+ +⎪ ⎪⎨ ⎬

+ − +⎪ ⎪⎩ ⎭

Let ( )( )( ) ( ) ( ) ( )

2

2 2

6s 3s 21 A B Cs Ds 2 s 1s 9 s 1 s 2 s 9

+ + +≡ + +

+ −+ − + +

= ( )( ) ( )( ) ( )( )( )

( )( )( )2 2

2

A s 1 s 9 B s 2 s 9 Cs D s 2 s 1

s 2 s 1 s 9

− + + + + + + + −

+ − +

Hence, ( )( ) ( )( ) ( )( )( )2 2 26s 3s 21 A s 1 s 9 B s 2 s 9 Cs D s 2 s 1+ + = − + + + + + + + −

When s = 1: 6 3 21 B(3)(10)+ + = i.e. 30 = 30B and B = 1

When s = -2: 24 6 21 A( 3)(13)− + = − i.e. 39 = -39A and A = -1

Equating 3s coefficients: 0 A B C= + + i.e. C = 0

Equating constant terms: 21 = -9A + 18B – 2D i.e. 21 = 9 + 18 – 2D and D = 3

Hence, y = ℒ( ) ( ) ( )

12 2

1 1 3s 2 s 1 s 3

−⎧ ⎫−⎪ ⎪+ +⎨ ⎬+ − +⎪ ⎪⎩ ⎭

i.e. y = 2x xe e sin 3x−− + + or y = x 2xe e sin 3x−− +



x2

d y dy2 2y 3e cos 2xdx dx

− + = given

y(0) = 2 and y′(0) = 5.

ℒ2

2

d ydx

⎧ ⎫⎨ ⎬⎩ ⎭

- 2ℒ dydx

⎧ ⎫⎨ ⎬⎩ ⎭

+ 2ℒy = ℒ x3e cos 2x

i.e. [ 2s ℒy – s y(0) - y′(0)] - 2[sℒy – y(0)] + 2ℒy = ( )2 2

s 13s 1 2

⎛ ⎞−⎜ ⎟⎜ ⎟− +⎝ ⎠

y(0) = 2 and y′(0) = 5, hence

2s ℒy – 2s - 5 - 2sℒy + 4 + 2ℒy = ( )2 2

s 13s 1 2

⎛ ⎞−⎜ ⎟⎜ ⎟− +⎝ ⎠

i.e. ( 2s - 2s + 2)ℒy = 2s + 1 +( )2 2

s 13s 1 2

⎛ ⎞−⎜ ⎟⎜ ⎟− +⎝ ⎠

= ( ) ( )( )

( )2

2

3s 3 2s 1 s 2s 5

s 2s 5

− + + − +

− +

= ( )

3 2 2

2

3s 3 2s 4s 10s s 2s 5s 2s 5

− + − + + − +− +

= ( )

3 2

2

2s 3s 11s 2s 2s 5− + +− +

from which, ℒy = ( )( )

3 2

2 2

2s 3s 11s 2s 2s 5 s 2s 2

− + +− + − +

and y = ℒ( )( )

3 21

2 2

2s 3s 11s 2s 2s 5 s 2s 2

−⎧ ⎫− + +⎪ ⎪⎨ ⎬

− + − +⎪ ⎪⎩ ⎭

Let ( )( ) ( ) ( )

3 2

2 2 2 2

2s 3s 11s 2 As B Cs Ds 2s 5 s 2s 2 s 2s 5 s 2s 2

− + + + +≡ +

− + − + − + − +

= ( )( ) ( )( )

( )( )2 2

2 2

As B s 2s 2 Cs D s 2s 5

s 2s 5 s 2s 2

+ − + + + − +

− + − +

Hence, ( )( ) ( )( )3 2 2 22s 3s 11s 2 As B s 2s 2 Cs D s 2s 5− + + = + − + + + − +

Equating 3s coefficients: 2 = A + C (1)


Equating 2s coefficients: -3 = -2A + B – 2C + D (2)

Equating s coefficients: 11 = 2A – 2B + 5C – 2D (3)

Equating constant terms: 2 = 2B + 5D (4)

From (1), C = 2 – A and

substituting C = 2 – A in (2): -3 = -2A + B – 2(2 – A) + D

i.e. 1 = B + D (5)

2 × (5) gives: 2 = 2B + 2D (6)

(4) – (6) gives: 0 = 3D i.e. D = 0

From (4), if D = 0, then B = 1

In (2), if B = 1 and D = 0, then -3 = -2A + 1 – 2C

i.e. 4 = 2A + 2C (7)

From (3), 11 = 2A – 2 + 5C

i.e. 13 = 2A + 5C (8)

(8) – (7) gives: 9 = 3C i.e. C = 3

In (1), if C = 3, then A = -1

Hence, y = ℒ( ) ( )

12 2

1 s 3ss 2s 5 s 2s 2

−⎧ ⎫−⎪ ⎪+⎨ ⎬

− + − +⎪ ⎪⎩ ⎭

= -ℒ( )

12 2

s 1s 1 2

−⎧ ⎫−⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭+ℒ

( )1

2 2

3ss 1 1

−⎧ ⎫⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭

= -ℒ( )

12 2

s 1s 1 2

−⎧ ⎫−⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭+ℒ ( )

( )1

2 2

3 s 1 3s 1 1

−⎧ ⎫− +⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭

= -ℒ( )

12 2

s 1s 1 2

−⎧ ⎫−⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭+ℒ ( )

( )1

2 2

3 s 1s 1 1

−⎧ ⎫−⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭+ ℒ

( )1

2 2

3s 1 1

−⎧ ⎫⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭

i.e. y = x x xe cos 2x 3e cos x 3e sin x− + + or y = ( )x x3e cos x sin x e cos 2x+ −


CHAPTER 68 THE SOLUTION OF SIMULTANEOUS

DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS EXERCISE 239 Page 654

2. Solve the following pair of simultaneous differential equations:

t

dy dx2 y x 5sin t 0dt dt

dy dx3 x y 2 e 0dt dt

− + + − =

+ − + − =

given that when t = 0, x = 0 and y = 0.

Taking Laplace transforms of each term in each equation gives:

2[sℒy – y(0)] - ℒy + ℒx + [sℒx – x(0)] - 2

5s 1+

= 0

3[sℒy – y(0)] + ℒx - ℒy + 2[sℒx – x(0)] - 1s 1−

= 0

y(0) = 0 and x(0) = 0, hence

(2s – 1)ℒy + (s + 1)ℒx = 2

5s 1+

(1)

and (3s – 1)ℒy +(2s + 1)ℒx = 1s 1−

(2)

(3s – 1) × (1) gives: (3s – 1)(2s – 1)ℒy + (3s – 1)(s + 1)ℒx = (3s – 1) 2

5s 1+

(3)

(2s – 1) × (2) gives: (2s – 1)(3s – 1)ℒy + (2s – 1)(2s + 1)ℒx = (2s – 1) 1s 1−

(4)

(3) – (4) gives: ( ) ( )2 23s 2s 1 4s 1⎡ ⎤+ − − −⎣ ⎦ℒx = ( )2

5 3s 1 2s 1s 1 s 1

− −−

+ − (5)

i.e. ( )2s 2s− + ℒx = ( )( ) ( )( )

( )( )2

2

5 3s 1 s 1 2s 1 s 1

s 1 s 1

− − − − +

− +

i.e. ℒx = ( )( )( )

2 3 2

2

15s 20s 5 2s 2s s 1s s 2 s 1 s 1

⎧ ⎫− + − − + +⎪ ⎪−⎨ ⎬− − +⎪ ⎪⎩ ⎭

= ( )( )( )

3 2

2

2s 16s 22s 6s s 2 s 1 s 1

⎧ ⎫− + −⎪ ⎪⎨ ⎬

− − +⎪ ⎪⎩ ⎭


and x = ℒ( )( )( )

3 21

2

2s 16s 22s 6s s 2 s 1 s 1

−⎧ ⎫− + −⎪ ⎪⎨ ⎬

− − +⎪ ⎪⎩ ⎭

Let ( )( )( ) ( ) ( ) ( )

3 2

2 2

2s 16s 22s 6 A B C Ds Es s 2 s 1s s 2 s 1 s 1 s 1

− + − +≡ + + +

− −− − + +

= ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( )

( )( )( )2 2 2

2

A s 2 s 1 s 1 B s s 1 s 1 C s s 2 s 1 Ds E s s 2 s 1

s s 2 s 1 s 1

− − + + − + + − + + + − −

− − +

from which, ( )( )( ) ( )( )( ) ( )( )( )3 2 2 2 22s 16s 22s 6 A s 2 s 1 s 1 B s s 1 s 1 C s s 2 s 1− + − = − − + + − + + − +

+ ( )( )( )( )Ds E s s 2 s 1+ − −

When s = 0: -6 = A(-2)(-1)(1) i.e. A = -3

When s = 1: 2 – 16 + 22 – 6 = C(1)(-1)(2) i.e. C = -1

When s = 2: 16 – 64 + 44 – 6 = B(2)(1)(5) i.e. B = -1

Equating 4s coefficients: 0 = A + B + C + D i.e. D = 5

Equating 3s coefficients: 2 = -3A – B – 2C – 3D + E

i.e. 2 = 9 + 1 + 2 – 15 + E i.e. E = 5

Hence, x = ℒ( ) ( ) ( )

12

3 1 1 5s 5s s 2 s 1 s 1

−⎧ ⎫+⎪ ⎪− − − +⎨ ⎬− − +⎪ ⎪⎩ ⎭

i.e. x = 2t t3 e e 5cos t 5sin t−− − − + + or x = 2t t5cos t 5sin t e e 3+ − − −

From equations (1) and (2)

(2s + 1) × (1) gives: (2s + 1)(2s – 1)ℒy + (2s + 1)(s + 1)ℒx = (2s + 1) 2

5s 1+

(6)

(s + 1) × (2) gives: (s + 1)(3s – 1)ℒy + (s + 1)(2s + 1)ℒx = (s + 1) 1s 1−

(7)

(6) – (7): ( ) ( )2 24s 1 3s 2s 1⎡ ⎤− − + −⎣ ⎦ℒy = ( ) ( )( )

( )( ) ( )( )( )( )

2

2 2

10s 5 s 1 s 1 s 15 2s 1 s 1s 1 s 1 s 1 s 1

+ − − + ++ +− =

+ − − +

i.e. ( )2s 2s− ℒy = ( )( )

2 3 2

2

10s 5s 5 s s s 1s 1 s 1

− − − − − −− +

and ℒy = ( )( )( )

3 2

2

s 9s 6s 6s s 2 s 1 s 1− + − −− − +


and y = ℒ( )( )( )

3 21

2

s 9s 6s 6s s 1 s 2 s 1

−⎧ ⎫− + − −⎪ ⎪⎨ ⎬

− − +⎪ ⎪⎩ ⎭

Let ( )( )( ) ( ) ( ) ( )

3 2

2 2

s 9s 6s 6 A B C Ds Es s 1 s 2s s 1 s 2 s 1 s 1

− + − − +≡ + + +

− −− − + +

= ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( )

( )( )( )2 2 2

2

A s 1 s 2 s 1 B s s 2 s 1 C s s 1 s 1 Ds E s s 1 s 2

s s 1 s 2 s 1

− − + + − + + − + + + − −

− − +

from which, ( )( )( ) ( )( )( ) ( )( )( )3 2 2 2 2s 9s 6s 6 A s 1 s 2 s 1 B s s 2 s 1 C s s 1 s 1− + − − = − − + + − + + − +

+ ( )( )( )( )Ds E s s 1 s 2+ − −

When s = 0: -6 = A(-1)(-2)(1) i.e. A = -3

When s = 1: -1 + 9 - 6 – 6 = B(1)(-1)(2) i.e. B = 2

When s = 2: -8 + 36 - 12 – 6 = C(2)(1)(5) i.e. C = 1

Equating 4s coefficients: 0 = A + B + C + D i.e. D = 0

Equating 3s coefficients: -1 = -3A – 2B – C – 3D + E

i.e. -1 = 9 - 4 - 1 + 0 + E i.e. E = -5

Hence, y = ℒ( ) ( ) ( )

12

3 2 1 5s s 1 s 2 s 1

−⎧ ⎫⎪ ⎪− + + −⎨ ⎬− − +⎪ ⎪⎩ ⎭

i.e. y = t 2t3 2e e 5sin t− + + − or y = 2t te 2e 3 5sin t+ − −

3. Solve the following pair of simultaneous differential equations:

2

2

2

2

d x 2x ydtd y 2y xdt

+ =

+ =

given that when t = 0, x = 4, y = 2, dxdt

= 0 and dydt

= 0.

Taking Laplace transforms of each term in each equation gives:

[ 2s ℒx – s x(0) - x′(0)] + 2ℒx = ℒy

and [ 2s ℒy – s y(0) - y′(0)] + 2ℒy = ℒx

x(0) = 4 and x′(0), hence [ 2s ℒx – 4s] + 2ℒx = ℒy


y(0) = 2 and y′(0) = 0, hence [ 2s ℒy – 2s] + 2ℒy = ℒx

i.e. ( 2s +2)ℒx - ℒy = 4s (1)

and - ℒx + ( 2s +2)ℒy = 2s (2)

( 2s +2) × (2) gives: - ( 2s +2)ℒx + ( 2s + 2) ( 2s +2)ℒy = 2s( 2s +2) (3)

(1) + (3) gives: ( )22s 2 1⎡ ⎤+ −⎢ ⎥⎣ ⎦ℒy = 2s( 2s +2) + 4s

i.e. ( )4 2s 4s 3+ + ℒy = 32s 8s+

and ℒy = ( ) ( )( )

3 3

4 2 2 2

2s 8s 2s 8ss 4s 3 s 3 s 1

+ +=

+ + + +

and y = ℒ( )( )

31

2 2

2s 8ss 3 s 1

−⎧ ⎫+⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭

Let ( )( ) ( ) ( )

( )( ) ( )( )( )( )

2 23

2 2 2 2 2 2

As B s 1 Cs D s 32s 8s As B Cs Ds 3 s 1 s 3 s 1 s 3 s 1

+ + + + ++ + +≡ + =

+ + + + + +

from which, ( )( ) ( )( )3 2 22s 8s As B s 1 Cs D s 3+ = + + + + +

Equating 3s coefficients: 2 = A + C (4)

Equating 2s coefficients: 0 = B + D (5)

Equating s coefficients: 8 = A + 3C (6)

(6) – (4) gives: 6 = 2C i.e. C = 3

and from (4), A = 2 – C i.e. A = -1

Equating constant terms: 0 = B + 3D (7)

(7) – (5) gives: 0 = 2D i.e. D = 0 and from (5), B = 0

Hence, y = ℒ( ) ( )

12 2

s 3ss 3 s 1

−⎧ ⎫−⎪ ⎪+⎨ ⎬

+ +⎪ ⎪⎩ ⎭ = ℒ

( )( ) ( )( )1

2 22 2

s 3s

s 3 s 1

−

⎧ ⎫⎪ ⎪− +⎨ ⎬⎪ ⎪+ +⎩ ⎭

i.e. y = ( )cos 3 t 3cos t− + or y = 3 cos t - ( )cos 3 t


If y = 3 cos t - ( )cos 3 t then dy 3sin t 3 sin 3 tdt

= − +

and 2

2

d y 3cos t 3cos 3 tdt

= − +

Since from one of the original equations,

2

2

d y 2y xdt

+ =

then ( ) ( )( )3cos t 3cos 3 t 2 3cos t cos 3 t x− + + − =

i.e. x = ( ) ( )3cos t 3cos 3 t 6cos t 2cos 3 t− + + −

i.e. x = 3 cos t + ( )cos 3 t


CHAPTER 69 FOURIER SERIES FOR PERIODIC FUNCTIONS

OF PERIOD 2π EXERCISE 240 Page 661

1. Determine the Fourier series for the periodic function:

f(x) =2, when x 02, when 0 x

− − π ≤ ≤⎧⎨ + ≤ ≤ π⎩

which is periodic outside this range of period 2π.

The periodic function is shown in the diagram below.

The Fourier series is given by: f(x) = ( )0 n nn 1

a a cos nx b sin nx∞

=

+ +∑ (1)

[ ] [ ] [ ] [ ]

0 00 00

1 1 1a f (x)dx 2dx 2dx 2x 2x2 2 2

1 (0) (2 ) (2 ) (0) 02

π π π

−π−π −π= = − + = − +

π π π

= − π + π − =π

∫ ∫ ∫

0

n 0

1 1a f (x)cos nx dx 2cos nx dx 2cos nx dxπ π

−π −π= = − +π π∫ ∫ ∫

= 0

0

1 2 2sin nx sin nxn n

π

−π

⎧ ⎫⎪ ⎪⎡ ⎤ ⎡ ⎤− +⎨ ⎬⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭ = 0

0

n 0

1 1b f (x)sin nx dx 2sin nx dx 2sin nx dxπ π

−π −π= = − +π π∫ ∫ ∫

= [ ] [ ] 0

0

1 2 2 2cos nx cos nx cos 0 cos n( ) cos n cos 0n n n

π

−π

⎧ ⎫⎪ ⎪⎡ ⎤ ⎡ ⎤+ − = − −π − π−⎨ ⎬⎢ ⎥ ⎢ ⎥π π⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭

When n is even, [ ] [ ] n2b 1 1 1 1 0n

= − − − =π

When n is odd, [ ] [ ] ( )n2 2 8b 1 1 1 1 4n n n

= − − − − − = =π π π


Hence, 18b =π

, 38b

3=

π, 5

8b5

=π

and so on.

Substituting onto equation (1) gives:

f(x) = 0 + 0 + 8 8 8sin x sin 3x sin 5x .....3 5

+ + +π π π

i.e. f(x) = 8 1 1sin x sin 3x sin 5x ......3 5

⎛ ⎞+ + +⎜ ⎟π ⎝ ⎠

2. For the Fourier series in problem 1, deduce a series for 4π at the point where x =

2π

When x = 2π , f(x) = 2, hence 2 = 8 8 3 8 5sin sin sin .....

2 3 2 5 2π π π+ + +

π π π

i.e. 2 = 8 1 1 11 ....3 5 7

⎛ ⎞− + − +⎜ ⎟π ⎝ ⎠

and 2 1 1 11 .....8 3 5 7π= − + − +

i.e. 1 1 11 ....4 3 5 7π= − + − +

5. Find the term representing the third harmonic for the periodic function of period 2π given by:

f(x) =0, when x 01, when 0 x

− π ≤ ≤⎧⎨ ≤ ≤ π⎩


n 00

1 1 1 sin nxa f (x)cos nx dx 1cos nx dx 0n

ππ π

−π

⎡ ⎤= = = =⎢ ⎥π π π ⎣ ⎦∫ ∫

n 0

1 1b f (x)sin nx dx sin nx dxπ π

−π= =π π∫ ∫


= 0

1 cos nx 1 1cos n cos 0 1 cos nn n n

π⎧ ⎫⎪ ⎪⎡ ⎤− = − π− = − π⎨ ⎬⎢ ⎥π π π⎣ ⎦⎪ ⎪⎩ ⎭

The third harmonic is when n = 3,

i.e. 31 1 2b 1 cos3 1 1

3 3 3= − π = − − =

π π π

Since the Fourier series is given by: f(x) = ( )0 n nn 1


=

+ +∑ ,

the 3rd harmonic term is: 2 sin 3x3π

6. Determine the Fourier series for the periodic function of period 2π defined by:

f(t) =

0, when t 0

1, when 0 t2

1, when t2

⎧⎪ − π ≤ ≤⎪ π⎪ ≤ ≤⎨⎪

π⎪− ≤ ≤ π⎪⎩

The function has a period of 2π.


[ ] [ ] ( ) ( )

0 / 2 / 20 0 / 20 / 2

1 1 1a f (t)dt 0dt 1dt 1dt t t2 2 2

1 0 02 2 2

π π π π π

π−π −π π= = + + − = + −

π π π⎧ ⎫⎡ π ⎤ ⎡ π ⎤⎛ ⎞ ⎛ ⎞= − + −π − − =⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎩ ⎭

∫ ∫ ∫ ∫

/ 2

n 0 / 2

1 1a f (t) cos nt dt cos nt dt cos nx dxπ π π

−π π= = + −π π∫ ∫ ∫

= / 2

0 / 2

1 sin nt sin nt 1 n nsin 0 sin n sinn n n 2 2

π π

π

⎧ ⎫ ⎧ π π ⎫⎪ ⎪⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤− = − − π−⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥π π⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎩ ⎭⎪ ⎪⎩ ⎭


When n is even, na 0=

When n = 1, [ ] [ ] 11 1 2a sin 0 sin sin 1 0 0 1

2 2⎧ π π ⎫⎡ ⎤ ⎡ ⎤= − − π− = − − − =⎨ ⎬⎢ ⎥ ⎢ ⎥π π π⎣ ⎦ ⎣ ⎦⎩ ⎭

When n = 3, [ ] [ ] 31 3 3 1 2a sin 0 sin 3 sin 1 0 0 1

3 2 2 3 3⎧ π π ⎫⎡ ⎤ ⎡ ⎤= − − π− = − − − − − = −⎨ ⎬⎢ ⎥ ⎢ ⎥π π π⎣ ⎦ ⎣ ⎦⎩ ⎭

When n = 5, [ ] [ ] 51 5 5 1 2a sin 0 sin 5 sin 1 0 0 1

5 2 2 5 5⎧ π π ⎫⎡ ⎤ ⎡ ⎤= − − π− = − − − =⎨ ⎬⎢ ⎥ ⎢ ⎥π π π⎣ ⎦ ⎣ ⎦⎩ ⎭

It follows that 72a

7= −

π , 9

2a9

=π

and so on.

/ 2

n 0 / 2

1 1b f (t)sin nt dt sin nt dt sin nt dtπ π π

−π π= = + −π π∫ ∫ ∫

= / 2

0 / 2

1 cos nt cos nt 1 n ncos cos 0 cos n cosn n n 2 2

π π

π

⎧ ⎫ ⎧ π π ⎫⎪ ⎪⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤− + = − − + π−⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥π π⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎩ ⎭⎪ ⎪⎩ ⎭

= 1 n n 1 n1 cos cos n cos 1 2cos cos nn 2 2 n 2⎧ π π ⎫ π⎛ ⎞ ⎛ ⎞ ⎧ ⎫− + π− = − + π⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎩ ⎭⎩ ⎭

When n is odd, n1b 1 0 1 0n

= − − =π

When n is even, 21 4 2b 1 2( 1) 1

2 2= − − + = =

π π π,

41b 1 2(1) 1 0

4= − + =

π,

61 4 2b 1 2( 1) 1

6 6 3= − − + = =

π π π

Similarly, 8b 0= , 102b

5=

π, and so on.

Substituting into f(t) = ( )0 n nn 1

a a cos nt b sin nt∞

=

+ +∑

gives: f(x) = 0 + 2 2 2 2cos t cos3t cos5t cos 7t .....3 5 7

− + − +π π π π

+ 2 2 2sin 2t sin 6t sin10t ....3 5

+ + +π π π

i.e. f(x) = 2 1 1 1 1cos t cos 3t cos 5t ...... sin 2t sin 6t sin10t .....3 5 3 5

⎛ ⎞− + − + + + +⎜ ⎟π ⎝ ⎠


7. Show that the Fourier series for the periodic function of period 2π defined by:

f(θ) =0, when 0sin , when 0

− π ≤θ ≤⎧⎨ θ ≤ θ ≤ π⎩

is given by: ( ) 2 1 cos 2 cos 4 cos 6f .....2 (3) (3)(5) (5)(7)

⎛ ⎞θ θ θθ = − − − −⎜ ⎟π ⎝ ⎠


[ ] ( ) ( ) 0

0 00

1 1 1 1a f ( )d 0d sin d cos cos cos 02 2 2 2

π π π

−π −π= θ θ = θ+ θ θ = − θ = − π − −⎡ ⎤⎣ ⎦π π π π∫ ∫ ∫

= ( ) ( )1 1 11 cos 1 12 2

− π = − − =π π π

0

n 0

1a 0cos n d sin cos n dπ

−π= θ θ+ θ θ θπ ∫ ∫

= ( ) ( ) ( ) ( ) 0 0

1 1 1sin n sin n sin 1 n sin 1 n d2 2

π π⎧ ⎫θ+ θ + θ− θ = θ + + θ − θ⎡ ⎤⎨ ⎬⎣ ⎦π π⎩ ⎭∫ ∫ from 6, page 398

of textbook

= ( ) ( ) ( ) ( )0

cos 1 n 1 n cos 1 n cos 1 n1 1 1 1cos2 1 n 1 n 2 1 n 1 n 1 n 1 n

π ⎡ ⎤θ + θ − π + π −⎡ ⎤ ⎛ ⎞ ⎛ ⎞− − = − − − − −⎢ ⎥⎜ ⎟⎢ ⎥ ⎜ ⎟π + − π + − + −⎝ ⎠⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

When n is odd, n1 1 1 1 1a 0

2 1 n 1 n 1 n 1 n⎡ ⎤= − − + + =⎢ ⎥π + − + −⎣ ⎦

When n = 2, 21 cos3 cos( ) 1 1 1 1 1 1 4 2a 1 1

2 3 1 3 1 2 3 3 2 3 3π −π⎧ ⎫ ⎧ ⎫ ⎧ ⎫= − − + + = − + − = − = −⎨ ⎬ ⎨ ⎬ ⎨ ⎬π − − π π π⎩ ⎭ ⎩ ⎭ ⎩ ⎭

When n = 4, 41 cos5 cos( 3 ) 1 1 1 1 1 1 1a

2 5 3 5 3 2 5 3 5 3π − π⎧ ⎫ ⎧ ⎫= − − + + = − + −⎨ ⎬ ⎨ ⎬π − − π⎩ ⎭ ⎩ ⎭

= 1 3 5 3 5 1 4 22 (3)(5) 2 (3)(5) (3)(5)

⎧ ⎫ ⎧ ⎫− + −= − = −⎨ ⎬ ⎨ ⎬π π π⎩ ⎭ ⎩ ⎭

When n = 6, 61 cos 7 cos( 5 ) 1 1 1 1 1 1 1a

2 7 5 7 5 2 7 5 7 5π − π⎧ ⎫ ⎧ ⎫= − − + + = − + −⎨ ⎬ ⎨ ⎬π − − π⎩ ⎭ ⎩ ⎭


= 1 5 7 5 7 1 4 22 (5)(7) 2 (5)(7) (5)(7)

⎧ ⎫ ⎧ ⎫− + −= − = −⎨ ⎬ ⎨ ⎬π π π⎩ ⎭ ⎩ ⎭

0

n 0

1b 0sin n d sin sin n dπ

−π= θ θ+ θ θ θπ ∫ ∫

= ( ) ( )0

0

1 1 1 sin (1 n) sin (1 n)cos n cos n2 2 1 n 1 n

ππ θ + θ −⎧ ⎫ ⎡ ⎤− θ+ θ − θ− θ = − −⎡ ⎤⎨ ⎬⎣ ⎦ ⎢ ⎥π π + −⎩ ⎭ ⎣ ⎦∫ = 0 from 9, page

398, of textbook

Substituting into f(θ) = ( )0 n nn 1

a a cos n b sin n∞

=

+ θ+ θ∑

gives: f(θ) = 1π

2 2 2cos 2 cos 4 cos 6 .....3 (3)(5) (5)(7)

− θ− θ− θ−π π π

+ 0

i.e. f(θ) = 2 1 cos 2 cos 4 cos 6 ......2 (3) (3)(5) (5)(7)

⎛ ⎞θ θ θ− − − −⎜ ⎟π ⎝ ⎠


CHAPTER 70 FOURIER SERIES FOR A NON-PERIODIC

FUNCTION OVER RANGE 2π EXERCISE 241 Page 667

2. Determine the Fourier series for the function defined by:

f(t) =1 t, when t 01 t, when 0 t− − π ≤ ≤⎧

⎨ + ≤ ≤ π⎩

Draw a graph of the function within and outside of the given range.


( ) ( )

02 20

0 00

2 2 2 2

1 1 1 t ta f (t)dt (1 t)dt (1 t)dt t t2 2 2 2 2

1 1 2 20 0 2 12 2 2 2 2 2 4 2

ππ π

−π −π−π

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪= = − + + = − + +⎨ ⎬⎢ ⎥ ⎢ ⎥π π π ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞π π π π π π⎪ ⎪ ⎪ ⎪= − −π− + π+ − = π+ = + = +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π π⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦ ⎣ ⎦ ⎩ ⎭⎩ ⎭

∫ ∫ ∫

0

n 0

1 1a f (t) cos nt dt (1 t)cos nt dt (1 t)cos nt dtπ π

−π −π= = − + +π π∫ ∫ ∫

= ( ) ( ) 0

0

1 cos nt t cos nt dt cos nt t cos nt dtπ

−π− + +

π ∫ ∫

= 0

2 20

1 sin nt t sin nt cos nt sin nt t sin nt cos ntn n n n n n

π

−π

⎡ ⎤ ⎡ ⎤− − + + +⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦ by integration by parts

= 2 2 2 2

1 cos 0 cos( n ) cos n cos 00 0 0 0 0 0 0 0n n n n

⎧ ⎫⎡ − π ⎤ ⎡ π ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − − − + + + − + +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎩ ⎭

= ( )2 2 2 2 2

1 1 cos( n ) cos n 1 1 2cos n 2n n n n n

− π π⎧ ⎫− + + − = π−⎨ ⎬π π⎩ ⎭ since cos(-nπ) = cos nπ

= ( )2

2 cos n 1n

π−π



When n = 1, ( )1 2

2 4a 1 1(1)

== − − = −π π

When n = 3, ( )3 2 2

2 4a 1 1(3) (3)

= − − = −π π

When n = 5, ( )5 2 2

2 4a 1 1(5) (5)

= − − = −π π

and so on.

0

n 0

1 1b f (t)sin nt dt (1 t)sin nt dt (1 t)sin nt dtπ π

−π −π= = − + +π π∫ ∫ ∫

= ( ) ( ) 0

0

1 sin nt t sin nt dt sin nt t sin nt dtπ

−π− + +

π ∫ ∫

= 0

2 20

1 cos nt t cos nt sin nt cos nt t cos nt sin ntn n n n n n

π

−π

⎡ ⎤ ⎡ ⎤− + − + − − +⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦ by integration by parts

=

cos 0 cos( n ) cos( n )0 0 0n n n1

cos n cos n cos 00 0 0n n n

⎧ ⎫⎡ − π π − π ⎤⎛ ⎞ ⎛ ⎞− + − − − − −⎪ ⎪⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎪⎣ ⎦ ⎪⎨ ⎬π ⎡ π π π ⎤⎛ ⎞ ⎛ ⎞⎪ ⎪+ − − + − − + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎪ ⎪⎝ ⎠ ⎝ ⎠⎣ ⎦⎩ ⎭

= 1 1 cos( n ) cos( n ) cos n cos n 1n n n n n n

− π π − π π π π⎧ ⎫− + + − − +⎨ ⎬π ⎩ ⎭ = 0 since cos nπ = cos(-nπ)

Substituting into f(t) = ( )0 n nn 1

a a cos nt b sin nt∞

=

+ +∑

gives: f(t) = 12π+ 2 2

4 4 4cos t cos3t cos5t .....(3) (5)

− − − −π π π

+ 0

i.e. f(t) = 2 2

4 cos 3t cos 5t1 cos t ......2 3 5π ⎛ ⎞+ − + + +⎜ ⎟π ⎝ ⎠

4. Determine the Fourier series up to and including the third harmonic for the function defined by:

f(x) =x, when 0 x2 x, when x 2

≤ ≤ π⎧⎨ π− π ≤ ≤ π⎩

Sketch a graph of the function within and outside of the given range, assuming the period is 2π.



( ) ( )

22 22

0 00

2 2 22 2 2

1 1 1 x xa f (x)dx x dx (2 x)dx 2 x2 2 2 2 2

1 4 10 4 22 2 2 2 2 2

π ππ π π

−π ππ

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪= = + π− = + π −⎨ ⎬⎢ ⎥ ⎢ ⎥π π π ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭⎧ ⎫⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞π π π π⎪ ⎪= − + π − − π − = π =⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

∫ ∫ ∫

2 2

n 0 0

1 1a f (x)cos nx dx x cos nx dx (2 x)cos nx dxπ π π

π= = + π−π π∫ ∫ ∫

= 2

2 20

1 x sin nx cos nx 2 sin nx x sin nx cos nxn n n n n

π π

π

⎧ ⎫π⎪ ⎪⎡ ⎤ ⎡ ⎤+ + − −⎨ ⎬⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭ by integration by parts

= 2 2 2 2

1 cos n cos 0 cos 2 n cos n0 0 0 0 0 0n n n n

⎧ ⎫⎡ π ⎤ ⎡ π π ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + + − − − − −⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎩ ⎭

= ( )2 2

1 2cos n 1 cos 2 n cos n cos n 1n n

π− − π + π = π−π π


When n = 1, ( )1 2

2 4a 1 1(1)

= − − = −π π

When n = 3, ( )3 2 2

2 4a 1 1(3) (3)

= − − = −π π

When n = 5, ( )5 2 2

2 4a 1 1(5) (5)

= − − = −π π

and so on.

2 0

n 0 0

1 1b f (x)sin nx dx x sin nx dx (2 x)sin nx dxπ π

−π= = + π−π π∫ ∫ ∫

= 2

2 20

1 x cos nx sin nx 2 cos nx x cos nx sin nxn n n n n

π π

π

π⎡ ⎤ ⎡ ⎤− + + − + −⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦ by integration by parts

= ( )

2 cos 2n 2 cos 2n 0n n1 cos n 0 0 0

n 2 cos n cos n 0n n

⎧ ⎫⎡ π π π π ⎤⎛ ⎞− + +⎪ ⎪⎜ ⎟⎢ ⎥⎡ π π ⎤⎪ ⎝ ⎠ ⎪⎛ ⎞ ⎢ ⎥− + − + +⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥π π π π π⎝ ⎠ ⎛ ⎞⎣ ⎦⎪ ⎪− − + −⎢ ⎥⎜ ⎟⎪ ⎪⎝ ⎠⎣ ⎦⎩ ⎭


= 1 cos n 2 cos n cos n 0n

−π π+ π π− π π =π

Substituting into f(x) = ( )0 n nn 1


=

+ +∑

gives: f(x) = 2π

2 2

4 4 4cos x cos3x cos5x .....(3) (5)

− − − −π π π

+ 0

i.e. f(x) = 2 2

4 cos 3x cos 5xcos x ......2 3 5π ⎛ ⎞− + + +⎜ ⎟π ⎝ ⎠

5. Expand the function ( ) 2f θ = θ in a Fourier series in the range - π < θ < π

Sketch the function within and outside of the given range.


3 3 2

2 3 30

1 1 1 1 2a f (x)dx d2 2 2 3 6 6 3

ππ π

−π −π−π

⎧ ⎫⎡ ⎤θ π π⎪ ⎪= = θ θ = = π − −π = =⎨ ⎬⎢ ⎥π π π π π⎣ ⎦⎪ ⎪⎩ ⎭∫ ∫

2n

1 1a f ( ) cos n d cos n dπ π

−π −π= θ θ θ = θ θ θπ π∫ ∫

= 2

2 3

1 sin n 2 cos n 2sin nn n n

π

−π

⎡ ⎤θ θ θ θ θ+ −⎢ ⎥π ⎣ ⎦

by integration by parts

= 2 2 2 2

1 2 cos n 2 cos( n ) 4 cos n 40 0 0 0 cos nn n n n

⎧ ⎫⎡ π π π − π ⎤ π π⎛ ⎞ ⎛ ⎞+ − − − − = = π⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥π π⎝ ⎠ ⎝ ⎠⎣ ⎦⎩ ⎭

since cos nπ = cos(-nπ)

Hence, ( )1 2 2

4 4a 1(1) 1

= − = − , ( )2 2 2

4 4a 1(2) 2

= = , ( )3 2 2

4 4a 1(3) 3

= − = − , 4 2

4a4

= , and so on.

2n

1 1b f ( )sin n d sin n dπ π

−π −π= θ θ θ = θ θ θπ π∫ ∫

= 2

2 3

1 cos n 2 sin n 2cos nn n n

π

−π

⎡ ⎤θ θ θ θ θ− + +⎢ ⎥π ⎣ ⎦

by integration by parts


= 2 2

3 3

1 cos n 2cos n cos( n ) 2cos( n )0 0n n n n

⎧ ⎫⎡ ⎤⎛ ⎞ ⎛ ⎞π π π π − π − π⎪ ⎪− + + − − + +⎨ ⎬⎢ ⎥⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭ = 0

Substituting into f(θ) = ( )0 n nn 1

a a cos n b sin n∞

=

+ θ+ θ∑

gives: f(θ) = 2

3π

2 2 2 2

4 4 4 4cos cos 2 cos3 cos 4 .....1 2 3 4

− θ+ θ− θ+ θ− + 0

i.e. f(θ) = 2

2 2

1 14 cos cos 2 cos 3 ......3 2 3π ⎛ ⎞− θ − θ + θ −⎜ ⎟

⎝ ⎠

6. For the Fourier series in problem 5, let θ = π and deduce a series for 2n 1

1n

∞

=∑

When θ = π in Problem 5 above, f(θ) = 2π

Thus, 2π = 2

2 2 2

1 1 14 cos cos 2 cos3 cos 4 ......3 2 3 4π ⎛ ⎞− π− π+ π− π+⎜ ⎟

⎝ ⎠

i.e. 2π = 2

2 2 2 2

4 4 4 4 .....3 1 2 3 4π

+ + + + +

i.e. 2

22 2 2 2

1 1 1 14 .....3 1 2 3 4π ⎛ ⎞π − = + + + +⎜ ⎟

⎝ ⎠

i.e. 2

2 2 2 2

2 1 1 1 14 .....3 1 2 3 4π ⎛ ⎞= + + + +⎜ ⎟

⎝ ⎠

and 2

2 2 2 2

2 1 1 1 1 ....3(4) 1 2 3 4π

= + + + +

i.e. 2

2 2 2 2

1 1 1 1 ....1 2 3 4 6

π+ + + + =

i.e. 2n 1

1n

∞

=∑ =

2

6π

8. Sketch the waveform defined by: f(x) =

2x1 , when x 0

2x1 , when 0 x

⎧ + − π ≤ ≤⎪⎪ π⎨⎪ − ≤ ≤ π⎪ π⎩

Determine the Fourier series in this range.



( ) ( ) ( ) ( )

02 20

0 00

2 2

1 1 2x 2x 1 x xa f (x)dx 1 dx (1 )dx x x2 2 2

1 10 0 02 2

ππ π

−π −π−π

⎧ ⎫⎡ ⎤ ⎡ ⎤⎧ ⎫ ⎪ ⎪⎛ ⎞= = + + − = + + −⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎢ ⎥ ⎢ ⎥π π π π π π π⎝ ⎠⎩ ⎭ ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭⎧ ⎫⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞π π⎪ ⎪= − −π+ + π− − = π− π + π− π =⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟π π π π⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

∫ ∫ ∫

0

n 0

1 1 2x 2xa f (x)cos nx dx 1 cos nx dx (1 )cos nx dxπ π

−π −π

⎧ ⎫⎛ ⎞= = + + −⎨ ⎬⎜ ⎟π π π π⎝ ⎠⎩ ⎭∫ ∫ ∫

= 0

2 20

1 sin nx 2 x sin nx cos nx sin nx 2 x sin nx cos nxn n n n n n

π

−π

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞+ + + − +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π π π⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ by integration by

parts

= 2 2 2 2

1 2 1 2 cos( n ) 2 cos n 2 10 0 0 0 0 0 0 0n n n n

⎧ ⎫⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ − π ⎞ ⎛ π ⎞ ⎛ ⎞⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − + + + − − − − −⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

= ( )2 2 2 2

1 42 2cos( n ) 2cos n 2 1 cos nn n

− − π − π+ = − ππ π

since cos nπ = cos(-nπ)


Hence, ( )1 2 2 2

4 8a 1 1(1)

= − − =π π

, ( )3 2 2 2 2

4 8a 1 1(3) (3)

= − − =π π

, 5 2 2

8a(5)

=π

and so on.

0

n 0

1 1 2x 2xb f (x)sin nx dx 1 sin nx dx 1 sin nx dxπ π

−π −π

⎧ ⎫⎛ ⎞ ⎛ ⎞= = + + −⎨ ⎬⎜ ⎟ ⎜ ⎟π π π π⎝ ⎠ ⎝ ⎠⎩ ⎭∫ ∫ ∫

= 0

2 20

1 cos nx 2 x cos nx sin nx cos nx 2 x cos nx sin nxn n n n n n

π

−π

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⎛ ⎞ ⎛ ⎞− + − + + − − − +⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π π π⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭ by

integration by parts

=

cos n 2 cos n 0n n1 1 cos( n ) 2 cos( n )0 0 0

n n n 1 0 0n

⎧ ⎫⎡ π π π ⎤⎛ ⎞− + +⎪ ⎪⎜ ⎟⎢ ⎥⎡ − π π − π ⎤⎪ ⎝ ⎠ ⎪⎛ ⎞ ⎛ ⎞ ⎢ ⎥− − + − − + + +⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π ⎝ ⎠ ⎝ ⎠ ⎛ ⎞⎣ ⎦⎪ ⎪− − + −⎢ ⎥⎜ ⎟⎪ ⎪⎝ ⎠⎣ ⎦⎩ ⎭

= 1 1 cos( n ) 2 cos( n ) cos n 2 cos n 1 0n n n n n n

− π π − π π π π⎧ ⎫− + − − + + =⎨ ⎬π ⎩ ⎭


Substituting into f(x) = ( )0 n n nn 1 n 1

a a cos nx b sin nx a cos nx∞ ∞

= =

+ + =∑ ∑

gives: f(x) = 2 2 2 2 2 2 2

8 8 8 8cos x cos3x cos5x cos 7x .....(3) (5) (7)

+ + + +π π π π

i.e. f(x) = 2 2 2 2

8 1 1 1cos x cos 3x cos 5x cos 7x ......3 5 7

⎛ ⎞+ + + +⎜ ⎟π ⎝ ⎠

9. For the Fourier series of Problem 8, deduce a series for 2

8π

When f(x) = 1 in the series of problem 8 above, x = 0,

hence, 1 = 2 2 2 2

8 1 1 1cos 0 cos 0 cos 0 cos 0 ......3 5 7

⎛ ⎞+ + + +⎜ ⎟π ⎝ ⎠

i.e. 2

2 2 2

1 1 11 .....8 3 5 7π

= + + + +


CHAPTER 71 EVEN AND ODD FUNCTIONS AND HALF-RANGE

FOURIER SERIES EXERCISE 242 Page 672 2. Obtain the Fourier series of the function defined by:

f(t) =t , when t 0t , when 0 t+ π − π ≤ ≤⎧

⎨ − π ≤ ≤ π⎩

which is periodic of period 2π. Sketch the given function.

The periodic function is shown in the diagram below. Since it is symmetrical about the origin, the

function is odd, and nn 1

f (t) b sin nt∞

=

=∑

0

n 0

1 1b f (t)sin nt dt (t )sin nt dt (t ) sin ntdtπ π

−π −π= = + π + − ππ π∫ ∫ ∫

= 0

2 20

1 t cos nt sin nt cos nt t cos nt sin nt cos ntn n n n n n

π

−π

π π⎡ ⎤ ⎡ ⎤− + − + − + +⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦ by integration by parts

=

cos n cos n0n n1 cos( n ) cos( n )0 0 0

n n n0 0

n

⎧ ⎫⎡ π π π π ⎤⎛ ⎞− + +⎪ ⎪⎜ ⎟⎢ ⎥⎡ π −π − π π − π ⎤⎪ ⎝ ⎠ ⎪⎛ ⎞ ⎛ ⎞ ⎢ ⎥+ − − − + − +⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π π⎝ ⎠ ⎝ ⎠ ⎛ ⎞⎣ ⎦⎪ ⎪− + +⎢ ⎥⎜ ⎟⎪ ⎪⎝ ⎠⎣ ⎦⎩ ⎭

= 1 cos( n ) cos( n ) cos n cos n 1 2 2n n n n n n n nπ π − π π − π π π π π π π⎧ ⎫ ⎛ ⎞− − + − + − = − = −⎨ ⎬ ⎜ ⎟π π⎩ ⎭ ⎝ ⎠

Hence, 12b1

= − , 22b2

= − , 32b3

= − , 42b4

= − , and so on.

i.e. f(t) = 2 2 2 2sin t sin 2t sin 3t sin 4t ......1 2 3 4

− − − − −

i.e. 1 1 1f (t) 2 sin t sin 2t sin 3t sin 4t .....2 3 4

⎛ ⎞= − + + + +⎜ ⎟⎝ ⎠


3. Determine the Fourier series defined by: f(x) =1 x, when x 01 x, when 0 x− − π ≤ ≤⎧

⎨ + ≤ ≤ π⎩

which is periodic of period 2π.

The periodic function is shown in the diagram below. Since it is symmetrical about the f(x) axis, the

function is even, and 0 nn 1

f (x) a a cos nx∞

=

= +∑

0 0

1 1a f (x)dx f (x)dx2

π π

−π= =

π π∫ ∫ due to symmetry

= ( )2 2

00

1 1 x 1(1 x)dx x 0 12 2 2

ππ ⎡ ⎤⎡ ⎤ ⎛ ⎞π π

+ = + = π+ − = +⎢ ⎥⎜ ⎟⎢ ⎥π π π⎣ ⎦ ⎝ ⎠⎣ ⎦∫

0

n 0

1 1a f (x)cos nx dx (1 x)cos nx dx (1 x)cos nx dxπ π

−π −π= = − + +π π∫ ∫ ∫

= ( ) ( ) 0

0

1 cos nx x cos nx dx cos nx x cos nx dxπ

−π− + +

π ∫ ∫

= 0

2 20

1 sin nx x sin nx cos nx sin nx x sin nx cos nxn n n n n n

π

−π

⎡ ⎤ ⎡ ⎤− − + + +⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦ by integration by parts

= 2 2 2 2

1 1 cos( n ) cos n 10 0 0 0 0 0 0 0n n n n

⎧ ⎫⎡ − π ⎤ ⎡ π ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − − − + + + − + +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎩ ⎭

= ( )2 2 2 2 2

1 1 cos( n ) cos n 1 2 cos n 1n n n n n

− π π⎧ ⎫− + + − = π−⎨ ⎬π π⎩ ⎭ since cos(-nπ) = cos nπ


When n = 1, ( )1 2

2 4a 1 1(1)

= − − = −π π

When n = 3, ( )3 2 2

2 4a 1 1(3) (3)

= − − = −π π

When n = 5, ( )5 2 2

2 4a 1 1(5) (5)

= − − = −π π

and so on.


0 nn 1

f (x) a a cos nx∞

=

= +∑ = 12π+ 2 2

4 4 4cos x cos3x cos5x .....(3) (5)

− − − −π π π

i.e. f(x) = 2 2

4 1 11 cos x cos 3x cos 5x ......2 3 5π ⎛ ⎞+ − + + +⎜ ⎟π ⎝ ⎠

4. In the Fourier series of Problem 3, let x = 0 and deduce a series for 2

8π

When x = 0 in the series of Problem 3 above, f(x) = 1,

hence, 1 = 2 2

4 cos0 cos01 cos0 ......2 3 5π ⎛ ⎞+ − + + +⎜ ⎟π ⎝ ⎠

i.e. 1 = 2 2 2

4 1 1 11 1 .....2 3 5 7π ⎛ ⎞+ − + + + +⎜ ⎟π ⎝ ⎠

i.e. 2 2 2

4 1 1 11 .....2 3 5 7π ⎛ ⎞− = − + + + +⎜ ⎟π ⎝ ⎠

and 2

2 2 2

1 1 11 ....8 3 5 7π

= + + + +



1. Determine the half-range series for the function defined by: f(x) =x, when 0 x

2

0, when x2

π⎧ ≤ ≤⎪⎪⎨ π⎪ ≤ ≤ π⎪⎩

The periodic function is shown in the diagram below. Since a half-range sine series is required, the

function is symmetrical about the origin and nn 1

f (x) b sin nx∞

=

=∑

/ 2

/ 2

n 20 00

2 2 2 x cos nx sin nxb f (x)sin nx dx x sin nx dxn n

ππ π ⎡ ⎤= = = − +⎢ ⎥π π π ⎣ ⎦∫ ∫ by integration by parts

= ( )2

n ncos sin2 2 2 2 0n n

⎡ π π π ⎤⎛ ⎞⎢ ⎥⎜ ⎟− + −⎢ ⎥⎜ ⎟π ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

Hence, 1 2 2

cos sin2 2 1 22 2 2b 01 1 1

⎡ π π π ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎛ ⎞= − + = + =⎢ ⎥⎜ ⎟ ⎜ ⎟π π π⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

,

2 2

cos2 sin 2 22b 02 2 4 4

⎡ π ⎤⎛ ⎞π⎢ ⎥⎜ ⎟π π π⎛ ⎞ ⎛ ⎞= − + = + =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

,

3 2 2 2

3 3cos sin2 2 1 22 2 2b 03 3 3 (3)

⎡ π π π ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎛ ⎞= − + = − = −⎢ ⎥⎜ ⎟ ⎜ ⎟π π π⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

,

4 2

cos 22 sin 2 2 22b 04 4 8 8

⎡ π ⎤⎛ ⎞π⎢ ⎥⎜ ⎟π π π⎛ ⎞ ⎛ ⎞= − + = − + = −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

, and so on


Hence, nn 1

f (x) b sin nx∞

=

=∑ = 2

2 2 2 1 2sin x sin 2x sin 3x sin 4x ...4 3 8π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i.e. 2 1f (x) sin x sin 2x sin 3x sin 4x .....4 9 8π π⎛ ⎞= + − − +⎜ ⎟π ⎝ ⎠

2. Obtain (a) the half-range cosine series and (b) the half-range sine series for the function:

f(t) =0, when 0 t

2

1, when t2

π⎧ ≤ ≤⎪⎪⎨ π⎪ ≤ ≤ π⎪⎩

(a) The periodic function is shown in the diagram below. Since a half-range cosine series is

required, the function is symmetrical about the f(t) axis and 0 nn 1

f (t) a a cos nt∞

=

= +∑

[ ]0 / 2/ 2

1 1 1 1a 1dt t2 2

π π

ππ

π⎡ ⎤= = = π− =⎢ ⎥π π π ⎣ ⎦∫

n / 2/ 2

n nsin 2sin2 2 sin nt 2 2 2a 1cos nt dt 0n n n

ππ

ππ

π π⎛ ⎞⎜ ⎟⎡ ⎤= = = − = −⎜ ⎟⎢ ⎥π π π π⎣ ⎦ ⎜ ⎟⎝ ⎠

∫


and 1

2sin 22a

π

= − = −π π

, 3

32sin 2 12a3 3

π⎛ ⎞= − = ⎜ ⎟π π ⎝ ⎠

, 5

52sin 2 12a5 5

π⎛ ⎞= − = − ⎜ ⎟π π ⎝ ⎠

, and so on.

Thus, 0 nn 1

1 2 2 1 2 1f (t) a a cos nt cos t cos3t cos5t ....2 3 5

∞

=

⎛ ⎞ ⎛ ⎞= + = − + − +⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠∑

i.e. 1 2 1 1f (t) cos t cos 3t cos 5t ......2 3 5

⎛ ⎞= − − + −⎜ ⎟π ⎝ ⎠

(b) The periodic function is shown in the diagram below. Since a half-range sine series is required,


the function is symmetrical about the origin and nn 1

f (x) b sin nx∞

=

=∑

n / 2/ 2

2 2 cos nt 2 nb 1sin nt dt cos n cosn n 2

ππ

ππ

π⎡ ⎤ ⎡ ⎤= = − = − π−⎢ ⎥ ⎢ ⎥π π π⎣ ⎦ ⎣ ⎦∫

Hence, ( )12 2 2b cos cos 1 0

2π⎛ ⎞= − π− = − − − =⎜ ⎟π π π⎝ ⎠

,

( ) ( )22 2 2b cos 2 cos 1 1

2 2= − π− π = − − − = −

π π π ,

( )32 3 2 2b cos3 cos 1 0

3 2 3 3π⎛ ⎞= − π− = − − − =⎜ ⎟π π π⎝ ⎠

,

( ) ( )42 2b cos 4 cos 2 0 0 0

4 4= − π− π = − − =

π π ,

( )52 5 2 2b cos5 cos 1 0

5 2 5 5π⎛ ⎞= − π− = − − − =⎜ ⎟π π π⎝ ⎠

,

( ) ( )62 2 2b cos 6 cos3 1 1

6 6 3= − π− π = − − − = −

π π π , and so on.

Thus, nn 1

f (t) b sin nt∞

=

=∑ = 2 2 2 2 2sin t sin 2t sin 3t 0 sin 5t sin 6t ....3 5 3

− + + + − +π π π π π

i.e. 2 1 1 1f (t) sin t sin 2t sin 3t sin 5t sin 6t .....3 5 3

⎛ ⎞= − + + − +⎜ ⎟π ⎝ ⎠

4. Determine the half-range Fourier cosine series in the range x = 0 to x = π for the function

defined by: f(x) =( )

x, when 0 x2

x , when x2

π⎧ ≤ ≤⎪⎪⎨ π⎪ π− ≤ ≤ π⎪⎩

The periodic function is shown in the diagram below. Since a half-range cosine series is required,

the function is symmetrical about the f(x) axis and 0 nn 1

f (x) a a cos nx∞

=

= +∑


/ 22 2/ 2

0 0 / 20 / 2

1 1 x xa x dx ( x)dx x2 2

π ππ π

ππ

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪= + π− = + π −⎨ ⎬⎢ ⎥ ⎢ ⎥π π ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭∫ ∫

= 2 2 2 2

218 2 2 8

⎧ ⎫⎛ ⎞ ⎛ ⎞π π π π⎪ ⎪+ π − − −⎨ ⎬⎜ ⎟ ⎜ ⎟π ⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

= 2 2 2 2 21 1 2

8 2 2 8 8 4⎛ ⎞ ⎛ ⎞π π π π π π

+ − + = =⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

/ 2

n 0 / 2

2a x cos nx dx ( x)cos nx dxπ π

π= + π−π ∫ ∫

= / 2

2 20 / 2

2 x sin nx cos nx sin n x sin nx cos nxn n n n n

π π

π

⎧ ⎫π π⎪ ⎪⎡ ⎤ ⎡ ⎤+ + − −⎨ ⎬⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭

= 2 2 2 2

n n n n nsin cos sin sin cos2 1 cos n2 2 2 2 2 2 20 0 0n n n n n n n

⎧ π π π π π π π ⎫⎛ ⎞ ⎛ ⎞π⎪ ⎪⎜ ⎟ ⎜ ⎟π⎪ ⎪⎛ ⎞ ⎛ ⎞+ − + + − − − − −⎨ ⎬⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠⎪ ⎪⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

= 2 2 2 2

n n n2cos sin sin2 1 cos n 2 n2 2 2 2cos 1 cos nn n n n n n 2

π π π⎧ ⎫π π⎪ ⎪π π⎧ ⎫+ − − − = − − π⎨ ⎬ ⎨ ⎬π π ⎩ ⎭⎪ ⎪⎩ ⎭

When n is odd, na 0=

and ( ) ( )2 2

2 2 8 2a 2cos 1 cos 2 2 1 1(2) 4 4

= π− − π = − − − = − = −π π π π

,

( ) ( )4 2

2 2a 2cos 2 1 cos 4 2 1 1 0(4) 16

= π− − π = − − =π π

,

( ) ( )6 2 2

2 2 8 2a 2cos3 1 cos 6 2 1 1(6) 36 36 (3)

= π− − π = − − − = − = −π π π π

,

8a 0= ,

( ) ( )10 2 2

2 2 8 2a 2cos5 1 cos10 2 1 1(10) 100 100 (5)

= π− − π = − − − = − = −π π π π

, and so on.


Thus, 0 n 2 2n 1

2 2 2f (t) a a cos nt cos 2t cos 6t cos10t ....4 (3) (5)

∞

=

π= + = − − − +

π π π∑

i.e. 2 2

2 1 1f (t) cos 2t cos 6t cos10t ......4 3 5π ⎛ ⎞= − + + +⎜ ⎟π ⎝ ⎠


CHAPTER 72 FOURIER SERIES OVER ANY RANGE EXERCISE 244 Page 679

2. Find the Fourier series for f(x) = x in the range x = 0 to x = 5.


The Fourier series is given by: f(x) = 0 n nn 1

2 nx 2 nxa a cos b sinL L

∞

=

⎡ π π ⎤⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∑ where L = 5

52 2L 5

0 0 00

1 1 1 x 1 5 5a f (x)dx x dxL 5 5 2 5 2 2

⎡ ⎤ ⎡ ⎤= = = = =⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦∫ ∫

5

L 5

n 20 0

0

2 nx 2 nxx sin cos2 2 nx 2 2 nx 2 5 5a f (x)cos dx x cos dx

2 nL L 5 5 5 2 n5 5

⎡ ⎤π π⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟π π⎛ ⎞ ⎛ ⎞ ⎝ ⎠ ⎝ ⎠⎢ ⎥= = = +⎜ ⎟ ⎜ ⎟ π⎢ ⎥⎛ ⎞⎝ ⎠ ⎝ ⎠ π⎛ ⎞⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∫ ∫ by parts

= 2 22 5sin 2 n cos 2 n 10

2 n5 2 n 2 n5 5 5

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟π π⎢ ⎥⎜ ⎟ ⎜ ⎟+ − +⎢ ⎥⎜ ⎟ ⎜ ⎟π⎛ ⎞ π π⎛ ⎞ ⎛ ⎞⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦

= 0

5

L 5

n 20 0

0

2 nx 2 nxx cos sin2 2 nx 2 2 nx 2 5 5b f (x)sin dx x sin dx

2 nL L 5 L 5 2 nx5 5

⎡ ⎤π π⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟π π⎛ ⎞ ⎛ ⎞ ⎝ ⎠ ⎝ ⎠⎢ ⎥= = = − +⎜ ⎟ ⎜ ⎟ π⎢ ⎥⎛ ⎞⎝ ⎠ ⎝ ⎠ π⎛ ⎞⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∫ ∫ by parts

= ( )22 5cos 2 n sin 2 n 2 5cos 2 n 5 50 0 cos 2 n

2 n 2 n5 5 n n2 n5 55

⎡ ⎤⎛ ⎞ ⎡ ⎤⎢ ⎥⎜ ⎟ ⎢ ⎥π π π⎢ ⎥⎜ ⎟ ⎢ ⎥− + − + = − = − π = −⎢ ⎥⎜ ⎟π π π π⎛ ⎞ ⎛ ⎞π ⎢ ⎥⎛ ⎞⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎝ ⎠⎝ ⎠⎣ ⎦

Hence, 1 2 3 45 5 5 5b , b , b , b ,

2 3 4= − = − = − = −

π π π π and so on.

Thus, f(x) = 0 n nn 1

2 nx 2 nxa a cos b sinL L

∞

=

⎡ π π ⎤⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∑


i.e. f(x) = 5 5 2 x 5 4 x 5 6 xsin sin sin .....2 5 2 5 3 5

π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i.e. f(x) = 5 5 2 x 1 4 x 1 6 xsin sin sin ......2 5 2 5 3 5

⎡ ⎤π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

3. A periodic function of period 4 is defined by: f(x) =3, when 2 x 03, when 0 x 2

− − ≤ ≤⎧⎨ + ≤ ≤⎩

Sketch the function and obtain the Fourier series for the function.


The function is odd since it is symmetrical about the origin, i.e. na 0=

Thus, f(x) = 0 nn 1

2 nxa b sinL

∞

=

⎡ π ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∑ where L = 4

[ ] [ ] ( ) ( ) ( ) ( ) L / 2 0 2 0 20 2 0L / 2 2 0

1 1 1 1a f (x)dx 3dx 3dx 3x 3x 0 6 6 0L 4 4 4−− −

= = − + = − + = − + −∫ ∫ ∫ = 0

L / 2 0 2

n L / 2 2 0

2 2 nx 2 2 nx 2 nxb f (x)sin dx 3sin dx 3sin dxL L 4 4 4− −

π ⎧ π π ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

∫ ∫ ∫

=

0 2

2 0

nx nx3cos 3cos1 1 3cos 0 3cos( n) 3cos n 3cos 02 2

n n n n n n2 22 2 2 2 2 2−

⎧ ⎫ ⎧ ⎫⎡ π ⎤ ⎡ π ⎤ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎪ ⎪⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ −π π⎪ ⎪ ⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥+ − = − + − − −⎨ ⎬ ⎨ ⎬π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠⎩ ⎭⎩ ⎭

= ( )1 6 6 6cos n 1 cos nn n2 n2 2

⎡ ⎤⎢ ⎥⎢ ⎥− π = − ππ π π⎛ ⎞ ⎛ ⎞⎢ ⎥

⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

When n is even, nb 0=


Hence, ( ) ( ) ( )1 3 56 12 6 12 6 12b 1 1 , b 1 1 , b 1 1 ,

3 3 5 5= − − = = − − = = − − =π π π π π π

and so on.

Thus, f(x) = 0 nn 1

2 nxa b sinL

∞

=

⎡ π ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∑ = 0 + 12 x 12 3 x 12 5 xsin sin sin ......

2 3 2 5 2π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i.e. f(x) = 12 x 1 3 x 1 5 xsin sin sin .....2 3 2 5 2

⎧ ⎫π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭



1. Determine the half-range Fourier cosine series for the function f(x) = x in the range 0 ≤ x ≤ 3.



the function is symmetrical about the f(x) axis and 0 nn 1

n xf (x) a a cosL

∞

=

π⎛ ⎞= + ⎜ ⎟⎝ ⎠

∑

32L 3

0 0 00

1 1 1 x 3a f (x)dx x dxL 3 3 2 2

⎧ ⎫⎡ ⎤⎪ ⎪= = = =⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭

∫ ∫

L 3

n 0 0

2 n x 2 n xa f (x)cos dx x cos dxL L 3 3⎧ π ⎫ π⎛ ⎞ ⎛ ⎞= =⎨ ⎬⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎩ ⎭∫ ∫

=

3

2 2 2

0

n x n xx sin cos2 2 3sin n cos n 13 3 0

n n3 3n n n3 33 3 3

⎧ ⎫ ⎡ ⎤⎡ ⎤ ⎛ ⎞ ⎛ ⎞π π⎛ ⎞ ⎛ ⎞⎪ ⎪ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ π π⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥+ = + − +⎨ ⎬ ⎢ ⎥⎜ ⎟ ⎜ ⎟π π⎢ ⎥⎛ ⎞ ⎛ ⎞π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎩ ⎭

by parts

= ( )2

2 2 2 2

2 cos n 1 2 3 60 cos n 1 cos n 13 3 n nn n

3 3

⎧ ⎫⎪ ⎪π⎪ ⎪ ⎛ ⎞+ − = π− = π−⎨ ⎬ ⎜ ⎟π π⎝ ⎠π π⎛ ⎞ ⎛ ⎞⎪ ⎪

⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭


and ( )1 2 2 2

6 12a 2(1)

= − = −π π

, ( )3 2 2 2 2

6 12a 2(3) (3)

= − = −π π

, 5 2 2

12a(5)

= −π

, and so on.

Thus, 0 n 2 2 2 2 2n 1

n x 3 12 x 12 3 x 12 5 xf (x) a a cos cos cos cos ....L 2 3 (3) 3 (5) 3

∞

=

π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑

i.e. 2 2 2

3 12 x 1 3 x 1 5 xf (x) cos cos cos .....2 3 3 3 5 3

⎧ ⎫π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭


2. Find the half-range Fourier sine series for the function f(x) = x in the range 0 ≤ x ≤ 3.




n xf (x) b sinL

∞

=

π⎛ ⎞= ⎜ ⎟⎝ ⎠

∑

3

L 3

n 20 0

0

n x n xx cos sin2 n x 2 n x 2 3 3b f (x)sin dx x sin dx

nL L 3 3 3 n3 3

⎡ ⎤π π⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟π ⎧ π ⎫⎛ ⎞ ⎛ ⎞ ⎝ ⎠ ⎝ ⎠⎢ ⎥= = = − +⎨ ⎬⎜ ⎟ ⎜ ⎟ π⎢ ⎥⎛ ⎞⎝ ⎠ ⎝ ⎠ π⎩ ⎭ ⎛ ⎞⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∫ ∫ by parts

= ( )22 3cos n sin n 2cos n 60 0 cos n

n n3 nn3 33

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟π π π⎢ ⎥⎜ ⎟− + − + = − = − π⎢ ⎥⎜ ⎟π π π⎛ ⎞ ⎛ ⎞π⎛ ⎞⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠⎣ ⎦

1 2 3 46 6 6 6b , b , b , b ,

2 3 4= = − = = −π π π π

and so on.

Thus, nn 1

n xf (x) b sinL

∞

=

π⎛ ⎞= ⎜ ⎟⎝ ⎠

∑ = 6 x 6 2 x 6 3 x 6 4 xsin sin sin sin ....3 2 3 3 3 4 3π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i.e. f(x) = 6 x 1 2 x 1 3 x 1 4 xsin sin sin sin .....3 2 3 3 3 4 3

⎧ ⎫π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

3. Determine the half-range Fourier sine series for the function defined by:

f(t) =( )

t, when 0 t 12 t , when 1 t 2

≤ ≤⎧⎪⎨ − ≤ ≤⎪⎩




n tf (t) b sinL

∞

=

π⎛ ⎞= ⎜ ⎟⎝ ⎠

∑

L 1 2

n 0 0 1

2 n t 2 n t n tb f (t)sin dt t sin dt (2 t)sin dtL L 2 2 2

π ⎧ π π ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + −⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

∫ ∫ ∫

=

1 2

2 2

0 1

n t n t n t n t n tt cos sin 2cos t cos sin2 2 2 2 2

n n nn n2 2 22 2

⎧ ⎫⎡ ⎤ ⎡ ⎤π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥− + + − + −⎨ ⎬π π π⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞π π⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎩ ⎭

by parts

= ( )

2

2

2

2cos n 2cos n sin nn n nnn sincos 2 2 222 0 0

n n n n n2cos cos sin2 2 2 2 2n n n2 2 2

⎛ ⎞⎜ ⎟π π π⎜ ⎟− + −⎜ ⎟π π⎛ ⎞ ⎛ ⎞⎡ ⎤ π⎛ ⎞ ⎛ ⎞π⎛ ⎞π ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠⎢ ⎥⎜ ⎟− + − + +

⎢ ⎥⎜ ⎟π⎛ ⎞ π ⎛⎛ ⎞ π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠− − + −π π⎛ ⎞ ⎛ ⎞ π⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝

⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎨ ⎬⎢ ⎥⎞⎪ ⎪⎢ ⎥⎟⎪ ⎪⎢ ⎥⎜ ⎟⎪ ⎪⎢ ⎥⎜ ⎟⎪ ⎪⎢ ⎥⎜ ⎟⎪ ⎪⎢ ⎥⎠⎣ ⎦⎩ ⎭

2 2 2

n2sin8 n2 sin

n 2n2

π⎛ ⎞⎜ ⎟ π⎛ ⎞⎝ ⎠= = ⎜ ⎟π ⎝ ⎠π⎛ ⎞

⎜ ⎟⎝ ⎠

When n is even, nb 0=

1 3 52 2 2 2 2

8 8 8b , b , b ,(3) (5)

= = − =π π π

and so on.

Thus, nn 1

n tf (t) b sinL

∞

=

π⎛ ⎞= ⎜ ⎟⎝ ⎠

∑ = 2 2 2 2 2

8 t 8 3 t 8 5 tsin sin sin ....2 (3) 2 (5) 2π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i.e. f(t) = 2 2 2

8 t 1 3 t 1 5 tsin sin sin .....2 3 2 5 2

⎧ ⎫π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + −⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭


4. Show that the half-range Fourier cosine series for the function f(θ) = θ2 in the range 0 to 4 is

given by: ( ) 2 2 2

16 64 1 2 1 3f cos cos cos .....3 4 2 4 3 4

⎧ πθ πθ πθ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞θ = − − + −⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭



the function is symmetrical about the f(θ) axis and 0 nn 1

nf ( ) a a cosL

∞

=

πθ⎛ ⎞θ = + ⎜ ⎟⎝ ⎠

∑

43L 4 2

0 0 00

1 1 1 1 64 16a f ( )d dL 4 4 3 4 3 3

⎧ ⎫⎡ ⎤θ⎪ ⎪ ⎛ ⎞= θ θ = θ θ = = =⎨ ⎬ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦⎪ ⎪⎩ ⎭∫ ∫

L 4 2n 0 0

2 n 2 na f ( )cos d cos dL L 4 4⎧ πθ ⎫ πθ⎛ ⎞ ⎛ ⎞= θ θ = θ θ⎨ ⎬⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎩ ⎭∫ ∫

=

4

2

2 3

0

n n nsin 2 cos 2sin1 4 4 4

n2 n n4 4 4

⎧ ⎫⎡ ⎤πθ πθ πθ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪θ θ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥+ −⎨ ⎬π⎢ ⎥⎛ ⎞ π π⎛ ⎞ ⎛ ⎞⎪ ⎪⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎩ ⎭

by parts

= ( )2 3 2 2 2

1 16sin n 8cos n 2sin n 1 8cos n 1 8(16)0 cos nn2 2 2 nn n n4 4 4 4

⎧ ⎫⎛ ⎞ ⎧ ⎫⎪ ⎪⎜ ⎟ ⎪ ⎪π π π π⎪ ⎪ ⎪ ⎪ ⎛ ⎞⎜ ⎟+ − − = = π⎨ ⎬ ⎨ ⎬ ⎜ ⎟⎜ ⎟π π⎛ ⎞ ⎝ ⎠π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪ ⎪ ⎪⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪ ⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎩ ⎭⎩ ⎭

= 2 2

64 cos nn

ππ

and ( )1 2 2 2

64 64a 1(1)

= − = −π π

, ( )2 2 2 2 2

64 64a 1(2) (2)

= =π π

, 3 2 2 2 2

64 64a ( 1)(3) (3)

= − = −π π

, and so on.

Thus, 0 n 2 2 2 2 2n 1

n 16 64 64 2 64 3f ( ) a a cos cos cos cos ....L 3 4 (2) 4 (3) 4

∞

=

πθ πθ πθ πθ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞θ = + = − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑

i.e. 2 2 2

16 64 1 2 1 3f ( ) cos cos cos .....3 4 2 4 3 4

⎧ ⎫πθ πθ πθ⎛ ⎞ ⎛ ⎞ ⎛ ⎞θ = − − + −⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭


CHAPTER 73 A NUMERICAL METHOD OF HARMONIC

ANALYSIS EXERCISE 246 Page 686

1. Determine the Fourier series to represent the periodic function given by the table of values

below, up to and including the third harmonic and each coefficient correct to 2 decimal places.

Use 12 ordinates.

Angle θ° 30 60 90 120 150 180 210 240 270 300 330 360

Displacement y 40 43 38 30 23 17 11 9 10 13 21 32

θ° y

cosθ y cos θ sin θ y sin θ cos 2θ y cos 2θ sin 2θ y sin 2θ cos 3θ ycos3θ sin3θ ycos3θ

30 40

60 43

90 38

120 30

150 23

180 17

210 11

240 9

270 10

300 13

330 21

360 32

0.866

0.5

0

-0.5

-0.866

-1

-0.866

-0.5

0

0.5

0.866

1

34.64

21.50

0

-15.0

-19.92

-17

-9.53

-4.5

0

6.5

18.19

32

0.5

0.866

1

0.866

0.5

0

-0.5

-0.866

-1

-0.866

-0.5

0

20

37.24

38

25.98

11.5

0

-5.5

-7.79

-10

-11.26

-10.5

0

0.5

-0.5

-1

-0.5

0.5

1

0.5

-0.5

-1

-0.5

0.5

1

20

-21.5

-38

-15

11.5

17

5.5

-4.5

-10

-6.5

10.5

32

0.866

0.866

0

-0.866

-0.866

0

0.866

0.866

0

-0.866

-0.866

0

34.64

37.24

0

-25.98

-19.92

0

9.53

7.79

0

-11.26

-18.19

0

0

-1

0

1

0

-1

0

1

0

-1

0

1

0

-43

0

30

0

-17

0

9

0

-13

0

32

1

0

-1

0

1

0

-1

0

1

0

-1

0

40

0

-38

0

23

0

-11

0

10

0

-21

0

12

kk 1

y 287=

=∑

12

k kk 1

y cos 46.88=

θ =∑

12

k kk 1

y sin 87.67=

θ =∑

12

k kk 1

y cos 2 1=

θ =∑

12

k kk 1

y sin 2 13.85=

θ =∑

12

k kk 1

y cos3 2=

θ = −∑

12

k kk 1

y sin 3 3=

θ =∑

p

0 kk 1

1 1a y (287) 23.92p 12=

≈ = =∑

p

n k kk 1

2a y cos nxp =

≈ ∑ hence, 12a (46.88) 7.81

12≈ = , 2

2a (1) 0.1712

≈ = , 32a ( 2) 0.33

12≈ − = −

p

n k kk 1

2b y sin nxp =

≈ ∑ hence, 12b (87.67) 14.61

12≈ = , 2

2b (13.85) 2.3112

≈ = , 32b (3) 0.50

12≈ =

Substituting these values into the Fourier series: ( )0 n nn 1

f (x) a a cos nx b sin nx∞

=

= + +∑

gives: y = 23.92 + 7.81 cos θ + 0.17 cos 2θ - 0.33 cos 3θ + ……

+ 14.61 sin θ + 2.31 sin 2θ + 0.50 sin 3θ

or y = 23.92 + 7.81 cos θ +14.61 sin θ + 0.17 cos 2θ + 2.31 sin 2θ - 0.33 cos 3θ + 0.50 sin 3θ


3. Determine the Fourier series to represent the periodic function given by the table of values

below, up to and including the third harmonic and each coefficient correct to 2 decimal places.

Use 12 ordinates.

Angle θ° 30 60 90 120 150 180 210 240 270 300 330 360

Current i 0 -1.4 -1.8 -1.9 -1.8 -1.3 0 2.2 3.8 3.9 3.5 2.5

θ° i

cosθ i cos θ sinθ i sin θ cos 2θ i cos 2θ sin 2θ i sin 2θ cos 3θ i cos3θ sin3θ i cos3θ

30 0

60 -1.4

90 -1.8

120 -1.9

150 -1.8

180 -1.3

210 0

240 2.2

270 3.8

300 3.9

330 3.5

360 2.5

0.866

0.5

0

-0.5

-0.866

-1

-0.866

-0.5

0

0.5

0.866

1

0

-0.7

0

0.95

1.56

1.3

0

-1.1

0

1.95

3.03

2.5

0.5

0.866

1

0.866

0.5

0

-0.5

-0.866

-1

-0.866

-0.5

0

0

-1.212

-1.8

-1.645

-0.9

0

0

-1.905

-3.8

-3.377

-1.75

0

0.5

-0.5

-1

-0.5

0.5

1

0.5

-0.5

-1

-0.5

0.5

1

0

0.7

1.8

0.95

-0.90

-1.3

0

-1.1

-3.8

-1.95

1.75

2.5

0.866

0.866

0

-0.866

-0.866

0

0.866

0.866

0

-0.866

-0.866

0

0

-1.212

0

1.645

1.548

0

0

1.905

0

-3.377

-3.031

0

0

-1

0

1

0

-1

0

1

0

-1

0

1

0

1.4

0

-1.9

0

1.3

0

2.2

0

-3.9

0

2.5

1

0

-1

0

1

0

-1

0

1

0

-1

0

0

0

1.8

0

-1.8

0

0

0

3.8

0

-3.5

0

12

kk 1

y 7.7=

=∑

12

k kk 1

y cos 9.49=

θ =∑

12

k kk 1

y sin 16.389=

θ = −∑

12

k kk 1

y cos 2 1.35=

θ = −∑

12

k kk 1

y sin 2 2.522=

θ = −∑

12

k kk 1

y cos3 1.6=

θ =∑

12

k kk 1

y sin 3 0.3=

θ =∑

p

0 kk 1

1 1a y (7.7) 0.64p 12=

≈ = =∑

p

n k kk 1

2a y cos nxp =

≈ ∑ hence, 12a (9.49) 1.58

12≈ = , 2

2a ( 1.35) 0.2312

≈ − = − , 32a (1.6) 0.27

12≈ =

p

n k kk 1

2b y sin nxp =

≈ ∑ hence, 12b ( 16.389) 2.73

12≈ − = − , 2

2b ( 2.522) 0.4212

≈ − = − ,

32b (0.3) 0.05

12≈ =



=

= + +∑

gives: i = 0.64 + 1.58 cos θ - 0.23 cos 2θ + 0.27 cos 3θ + ……

- 2.73 sin θ - 0.42 sin 2θ + 0.05 sin 3θ

or i = 0.64 + 1.58 cos θ - 2.73 sin θ - 0.23 cos 2θ - 0.42 sin 2θ + 0.27 cos 3θ + 0.05 sin 3θ



2. Analyse the periodic waveform of displacement y against angle θ in the diagram below into its

constituent harmonics as far as and including the third harmonic, by taking 30° intervals.

θ° y

cosθ y cos θ sinθ y sin θ cos 2θ y cos 2θ sin 2θ y sin 2θ cos 3θ y cos3θ sin3θ y cos3θ

30 10

60 -6

90 -17

120 -17

150 -13

180 -4

210 10

240 24

270 33

300 36

330 33

360 24

0.866

0.5

0

-0.5

-0.866

-1

-0.866

-0.5

0

0.5

0.866

1

8.66

-3.0

0

8.5

11.26

4

-8.66-

12

0

18

28.58

24

0.5

0.866

1

0.866

0.5

0

-0.5

-0.866

-1

-0.866

-0.5

0

5

-5.196

-17

-14.72

-6.5

0

-5.0

-20.78

-33

-31.18

-16.5

0

0.5

-0.5

-1

-0.5

0.5

1

0.5

-0.5

-1

-0.5

0.5

1

5

3

17

8.5

-6.5

-4

5

-12

-33

-18

16.5

24

0.866

0.866

0

-0.866

-0.866

0

0.866

0.866

0

-0.866

-0.866

0

8.66

-5.196

0

14.722

11.258

0

8.66

20.784

0

-31.176

-28.578

0

0

-1

0

1

0

-1

0

1

0

-1

0

1

0

6

0

-17

0

4

0

24

0

-36

0

24

1

0

-1

0

1

0

-1

0

1

0

-1

0

10

0

17

0

-13

0

-10

0

33

0

-33

0

12

kk 1

y 113=

=∑

12

k kk 1

y cos 79.34=

θ =∑

12

k kk 1

y sin

144.88=

θ

= −

∑

12

k kk 1

y cos 2 5.5=

θ =∑

12

k kk 1

y sin 2 0.866=

θ = −∑

12

k kk 1

y cos3 5=

θ =∑

12

k kk 1

y sin 3 4=

θ =∑

p

0 kk 1

1 1a y (113) 9.4p 12=

≈ = =∑

p

n k kk 1

2a y cos nxp =

≈ ∑ hence, 12a (79.34) 13.2

12≈ = , 2

2a (5.5) 0.9212

≈ = , 32a (5) 0.83

12≈ =

p

n k kk 1

2b y sin nxp =

≈ ∑ hence, 12b ( 144.88) 24.1

12≈ − = − , 2

2b ( 0.866) 0.1412

≈ − = − ,

32b (4) 0.67

12≈ =



=

= + +∑

gives: y = 9.4 + 13.2 cos θ + 0.92 cos 2θ + 0.83 cos 3θ + ……

- 24.1 sin θ - 0.14 sin 2θ + 0.67 sin 3θ


or y = 9.4 + 13.2 cos θ - 24.1 sin θ + 0.92 cos 2θ - 0.14 sin 2θ + 0.83 cos 3θ + 0.67 sin 3θ

3. For the waveform of current shown below state why only a d.c. component and even cosine

terms will appear in the Fourier series and determine the series, using π/6 intervals, up to and

including the sixth harmonic.

The function is even, thus no sine terms will be present.

The function repeats itself every half cycle, hence only even terms will be present.

Hence, the Fourier series will contains a d.c. component and even cosine terms only.

θ° i

cos 2θ i cos 2θ cos 4θ i cos 4θ cos 6θ i cos 6θ

30 1.5

60 5.5

90 10

120 5.5

150 1.5

180 0

210 1.5

240 5.5

270 10

300 5.5

330 1.5

360 0

0.5

-0.5

-1

-0.5

0.5

1

0.5

-0.5

-1

-0.5

0.5

1

0.75

-2.75

-10

-2.75

0.75

0

0.75

-2.75

-10

-2.75

0.75

0

-0.5

-0.5

1

-0.5

-0.5

1

-0.5

-0.5

1

-0.5

-0.5

1

-0.75

-2.75

10

-2.75

-0.75

0

-0.75

-2.75

10

-2.75

-0.75

0

-1

1

-1

1

-1

1

-1

1

-1

1

-1

1

-1.5

5.5

-10

5.5

-1.5

0

-1.5

5.5

-10

5.5

-1.5

0

12

kk 1

y 48=

=∑

12

k kk 1

y cos 2 28=

θ = −∑

12

k kk 1

y cos 4 6=

θ =∑

12

k kk 1

y cos 6 4=

θ = −∑

p

0 kk 1

1 1a y (48) 4.00p 12=

≈ = =∑

p

n k kk 1

2a y cos nxp =

≈ ∑ hence, 22a ( 28) 4.67

12≈ − = − , 4

2a (6) 1.0012

≈ = , 32a ( 4) 0.66

12≈ − = −

Substituting these values into the Fourier series: ( )0 nn 1

f (x) a a cos nx∞

=

= +∑

gives: i = 4.00 – 4.67 cos 2θ + 1.00 cos 4θ - 0.66 cos 6θ + ……


CHAPTER 74 THE COMPLEX OR EXPONENTIAL FORM OF A

FOURIER SERIES EXERCISE 248 Page 694

1. Determine the complex Fourier series for the function defined by:

f(t) =0, when t 02, when 0 t

− π ≤ ≤⎧⎨ ≤ ≤ π⎩

The function is periodic outside of this range of period 2π.


The complex Fourier series is given by: 2 n tj

Ln

nf (t) c e

π∞

=−∞

= ∑

where 2 n tL / 2 j

Ln L / 2

1c f (t) e dtL

π−

−= ∫

i.e. 2 n t jn tj0 jn t jn 02

n 0 00

1 1 1 e 1c 0dt 2e dt e e e2 jn j n

ππ −−π π − − ππ−π

⎧ ⎫ ⎡ ⎤⎡ ⎤= + = = = − −⎨ ⎬ ⎢ ⎥ ⎣ ⎦π π π − π⎣ ⎦⎩ ⎭

∫ ∫ ∫

= ( )jn jn2

j je 1 e 1j n n

− π − π⎡ ⎤− − = −⎣ ⎦π π

= [ ] [ ]j jcos n jsin n 1 cos n 1n n

π− π− = π−π π

for all

integer values of n

Hence, ( )2 ntj jntL

nn n

jf (t) c e cosn 1 en

π∞ ∞

=− ∞ =− ∞

= = π −π∑ ∑

0 02c a mean value 12×π

= = = =π

1j 2c ( 1 1) j= − − = −π π

, 2jc (1 1) 0

2= − =

π and all even terms will be zero

32c j

3= −

π, 5

2c j5

= −π

, and so on.


1j 2c ( 2) j− = − =−π π

, 3j 2c ( 2) j3 3− = − =− π π

, 52c j

5=

π , and so on.

Thus, f(t) = 1 jt j3t j5t jt j3t j5t2 2 2 2 2 2j e j e j e .... j e j e j e3 5 3 5

− − −− − − − + + +π π π π π π

i.e. jt j3t j5t jt j3t j5t2 1 1 2 1 1f (t) 1 j e e e .... j e e e ....3 5 3 5

− − −⎛ ⎞ ⎛ ⎞= − + + + + + + +⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

2. Show that the complex Fourier series for the waveform shown below, that has period 2, may be

represented by: ( ) j n t

n(n 0)

j2f (t) 2 cos n 1 en

∞π

=− ∞≠

= + π−π∑


Ln

n

f (t) c eπ∞

=− ∞

= ∑

where 2 n tL / 2 j

Ln L / 2

1c f (t) e dtL

π−

−= ∫

i.e. 12 n t j n t1 1j j n t j n 02

n 0 00

1 e 2c 4e 2 e 2 e e2 j n j n

π − π− − π − π⎧ ⎫ ⎡ ⎤⎡ ⎤= = = = − −⎨ ⎬ ⎢ ⎥ ⎣ ⎦− π π⎣ ⎦⎩ ⎭

∫ ∫

= ( )jn jn2 j2e 1 e 1j n n

− π − π⎡ ⎤− − = −⎣ ⎦π π

= [ ] [ ]j2 j2cos n jsin n 1 cos n 1n n

π− π− = π−π π

for all

integer values of n

[ ]1 1

0 0 00

1 1c a mean value 4dt 4t 22 2

= = = = =∫

Hence, ( ) j nt

n(n 0)

j2f (t) 2 cosn 1 en

∞π

=−∞≠

= + π −π∑


3. Show that the complex Fourier series of Problem 2 is equivalent to:

8 1 1f (t) 2 sin t sin 3 t sin 5 t ....3 5

⎛ ⎞= + π + π + π +⎜ ⎟π ⎝ ⎠

1j2 4c ( 1 1) j(1)

= − − = −π π

, 2j2c (1 1) 02

= − =π

and all even terms will be zero

32 4c j ( 2) j

3 3= − = −

π π, 5

4c j5

= −π

, and so on.

1j2 4c ( 2) j− = − =−π π

, 3j2 4c ( 2) j3 3− = − =− π π

, 54c j

5=

π , and so on.

Thus, ( ) j n t

n(n 0)

j2f (t) 2 cos n 1 en

∞π

=− ∞≠

= + π−π∑

i.e. f(t) = 2 j t j3 t j5 t j t j3 t j5 t4 4 4 4 4 4j e j e j e .... j e j e j e3 5 3 5

π π π − π − π − π− − − − + + +π π π π π π

i.e. j t j3 t j5 t j t j3 t j5 t4 1 1 4 1 1f (t) 2 j e e e .... j e e e ....3 5 3 5

π π π − π − π − π⎛ ⎞ ⎛ ⎞= − + + + + + + +⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

= ( ) ( ) ( )j t j t j3 t j3 t j5 t j5 t4 1 12 j e e e e e e ....3 5

π − π π − π π − π⎡ ⎤− − + − + − +⎢ ⎥π ⎣ ⎦

= j t j t j3 t j3 t j5 t j5 t

2 8 e e 1 e e 1 e e2 j ....2 j 3 2j 5 2j

π − π π − π π − π⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −− + + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

i.e. 8 1 1f (t) 2 sin t sin 3 t sin 5 t ....3 5

⎡ ⎤= + π + π + π +⎢ ⎥π ⎣ ⎦

4. Determine the exponential form of the Fourier series for the function defined by:

f(t) = e2t when – 1 < t < 1 and has period 2.

The function is shown in the diagram below.


Ln

nf (t) c e

π∞

=−∞

= ∑


where 2 n tL / 2 j

Ln L / 2

1c f (t) e dtL

π−

−= ∫

i.e. ( ) ( ) ( )1t 2 j n 2 j n 2 j n2 n t1 1j2t 2t j n t2

n 1 11

1 1 1 e 1 e ec e e dt e dt2 2 2 2 j n 2 2 j n

− π − π − − ππ− − π

− −−

⎡ ⎤ ⎡ ⎤⎧ ⎫ −= = = =⎨ ⎬ ⎢ ⎥ ⎢ ⎥− π − π⎩ ⎭ ⎣ ⎦ ⎣ ⎦

∫ ∫

Thus, f(t) =2 n tj

Ln

n

c eπ∞

=− ∞∑ =

(2 j n) (2 j n)j nt

n

1 e e e2 2 j n

− π − − π∞π

=− ∞

⎛ ⎞−⎜ ⎟− π⎝ ⎠

∑



1. Determine the exponential form of the Fourier series for the periodic function defined by:

f(x) =

2, when x2

2, when x2 2

2, when x2

π⎧ − − π ≤ ≤ −⎪⎪

π π⎪ − ≤ ≤ +⎨⎪

π⎪ − + ≤ ≤ +π⎪⎩

and which has a period of 2π.

The periodic waveform is shown below. It is an even function and contains no sine terms, hence

nb 0= and between -π and +π, the mean value is zero, hence 0a 0= .

L / 2

n 0 0

2 2 nx 2 2 nxc f (x)cos dx f (x)cos dxL L 2 2

ππ π⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠∫ ∫ since L = 2π

= / 2

0 / 2

1 2cos nx dx 2cos nx dxπ π

π+ −

π ∫ ∫

= / 2

0 / 2

1 2sin nx 2sin nxn n

π π

π

⎧ ⎫⎪ ⎪⎡ ⎤ ⎡ ⎤−⎨ ⎬⎢ ⎥ ⎢ ⎥π ⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭

= 1 n n 4 n2sin 0 2sin n 2sin sinn 2 2 n 2⎡ π π ⎤ π⎛ ⎞ ⎛ ⎞− − π− =⎜ ⎟ ⎜ ⎟⎢ ⎥π π⎝ ⎠ ⎝ ⎠⎣ ⎦

Hence, f(x) = 2 n xj

Ln

n

c eπ∞

=− ∞∑ = jnx

n

4 nsin en 2

∞

=− ∞

⎧ ⎫π⎛ ⎞⎨ ⎬⎜ ⎟π ⎝ ⎠⎩ ⎭

∑

2. Show that the exponential form of the Fourier series in Problem 1 above is equivalent to:

8 1 1 1f (x) cos x cos3x cos5x cos 7x ....3 5 7

⎛ ⎞= − + − +⎜ ⎟π ⎝ ⎠


Since from Problem 1, n4 nc sinn 2

π=π

, then

0c 0= , 14 4c sin

2π

= =π π

, 2 4 6 2 44 2c sin 0 c c c c andsoon

2 2 − −

π= = = = = =

π,

34 3 4c sin

3 2 3π

= = −π π

, 54c

5=

π, 7

4c7

= −π

and so on.

14 4c sin

2−

−π= − =

π π, 3

4 3 4c sin3 2 3−

− π= − = −

π π, 5

4c5− =π

and so on.

f(x) = 2 n xj

Ln

n

c eπ∞

=− ∞∑ = jn x

n

4 nsin en 2

∞

=− ∞

⎧ π ⎫⎛ ⎞⎨ ⎬⎜ ⎟π ⎝ ⎠⎩ ⎭

∑

= jx j3x j5x jx j3x j5x4 4 4 4 4 4e e e ..... e e e .....3 5 3 5

− − −− + + + − + −π π π π π π

= jx jx j3x j3x j5x j5x4 4 4 4 4 4e e e e e e .....3 3 5 5

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= jx jx j3x j3x j5x j5x8 e e 8 e e 8 e e .....

2 3 2 5 2

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +− + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟π π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 8 8 8cos x cos3x cos5x .....3 5

− + −π π π

i.e. 8 1 1 1f (x) cos x cos 3x cos 5x cos 7x .....3 5 7

⎛ ⎞= − + − +⎜ ⎟π ⎝ ⎠

3. Determine the complex Fourier series to represent the function f(t) = 2t in the range - π to + π.

The triangular waveform shown below is an odd function since it is symmetrical about the origin.

The period of the waveform, L = 2π.

Thus, nc = L2

0

2 2 ntj f (t) sin dtL L

π⎛ ⎞− ⎜ ⎟⎝ ⎠∫

= 0 0

2 2 nt 2j 2t sin dt j t sin nt dt2 2

π ππ⎛ ⎞− = −⎜ ⎟π π π⎝ ⎠∫ ∫


= ( )2 20

2 t cos nt sin nt 2 cos n sin nj j 0 0n n n n

π− ⎡ −π π π ⎤⎡ ⎤ ⎛ ⎞− + = − + − +⎜ ⎟⎢ ⎥⎢ ⎥π π⎣ ⎦ ⎝ ⎠⎣ ⎦ by parts

i.e. n2c j cos nn

= π

Hence, the complex Fourier series is given by:

f(t) = 2 ntj

Ln

n

c eπ∞

=− ∞∑ = jnt

n

j2 cosn en

∞

=− ∞

⎛ ⎞π⎜ ⎟⎝ ⎠

∑

4. Show that the complex Fourier series is Problem 3 above is equivalent to:

f(t) = 1 1 14 sin t sin 2t sin 3t sin 4t ....2 3 4

⎛ ⎞− + − +⎜ ⎟⎝ ⎠

From Problem 3 above, n2c j cos nn

= π

When n = 1, ( )12 2 j2c j cos j 1

(1) (1) 1= π = − = −

When n = 2, 22 2c j cos 2 j2 2

= π =

When n = 3, ( )32 2 j2c j cos3 j 13 3 3

= π = − = −π

By similar reasoning, 4 5j2 j2c , c4 5

= = − , and so on.

When n = -1, ( )12 2 j2c j cos( ) j 1

( 1) ( 1) 1− = −π = + − =− −

When n = -2, ( )22 2 j2c j cos( 2 ) j 1

( 2) ( 2) 2− = − π = = −− −

By similar reasoning, 3 42 j2c j , c3 4− −= = − , and so on.

Since the waveform is odd, 0 0c a 0= =

From Problem 3, f(t) = 2 ntj

Ln

nc e

π∞

=−∞∑ = jn t

n

j2 cos n en

∞

=−∞

⎛ ⎞π⎜ ⎟⎝ ⎠

∑

Hence, f(t) = jt j2t j3t j4 tj2 j2 j2 j2e e e e ...1 2 3 4

− + − + − + jt j2 t j3t j4tj2 j2 j2 j2e e e e ...1 2 3 4

− − − −− + − +

= jt jt j2 t j2 t j3t j3tj2 j2 j2 j2 j2 j2e e e e e e ....1 1 2 2 3 3

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + − + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠


= jt jt j2t j2 t j3t j3te e j4 e e j4 e ej4 ....

2 2 2 3 2

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −− + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= jt jt 2 j2t j2 t 2 j3t j3t

2 e e j 4 e e j 4 e ej 4 ....2 j 2 2 j 3 2j

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −− + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ by multiplying top and

bottom by j

= 4 44sin t sin 2t sin 3t ....2 3

− + +

i.e. f(t) = 1 1 14 sin t sin 2t sin 3t sin 4t ...2 3 4

⎛ ⎞− + − +⎜ ⎟⎝ ⎠

Hence, f(t) = jnt

n

j2 cosn en

∞

=−∞

⎛ ⎞π⎜ ⎟⎝ ⎠

∑ ≡ 1 1 14 sin t sin 2t sin 3t sin 4t ...2 3 4

⎛ ⎞− + − +⎜ ⎟⎝ ⎠



2. Determine the pair of phasors that can represent the harmonic given by:

v = 10 cos 2t – 12 sin 2t

v = 10 cos 2t – 12 sin 2t

= ( ) ( )j2 t j2 t j2 t j2t1 110 e e 12 e e2 2j

− −⎡ ⎤⎡ ⎤+ − −⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

= j2 t j2 t j2 t j2 t6 65e 5e e ej j

− −+ − +

= j2 t j2 t j2 t j2 t5e 5e 6je 6je− −+ + − (note: 1 j 1 jorj 1 j 1

−= =

−)

i.e. v = ( ) ( )j2t j2t5 j6 e 5 j6 e−+ + −

Hence, v = 7.81∠0.88 rad, rotating anticlockwise with an angular velocity, ω = 2 rad/s

and v = 7.81∠-0.88 rad, rotating clockwise with an angular velocity, ω = 2 rad/s, as


3. Find the pair of phasors that can represent the fundamental current:

i = 6 sin t + 4 cos t

i = 6 sin t + 4 cos t

= ( ) ( ) ( ) ( )j t j t j t jt jt jt jt jt1 1 36 e e 4 e e e e 2 e e2j 2 j

− − − −⎡ ⎤ ⎡ ⎤− + + = − + +⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

= ( ) ( )jt jt jt jt3j e e 2 e e− −− − + +

i.e. i = ( ) ( )j t j t2 j3 e 2 j3 e−− + +


Hence, i = 3.61∠-0.98 rad, rotating anticlockwise with an angular velocity, ω = 1 rad/s

and i = 3.61∠0.98 rad, rotating clockwise with an angular velocity, ω = 1 rad/s, as


Documents

Bird - Higher Engineering Mathematics - 5e - Solutions Manual.pdf