Bird - Higher Engineering Mathematics - 5e - Solutions Manual 4y2n4y

HIGHER ENGINEERING MATHEMATICS TH

5

EDITION

JOHN BIRD SAMPLE OF WORKED SOLUTIONS TO EXERCISES

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

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 Chapter 1 Algebra

Page 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


ii

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 a

d2 y dy +b + cy = 0 2 dx dx

435

Chapter 51 Second order differential equations of the form a

d2 y dy +b + cy = f (x) dx dx 2

441

Chapter 52 Power series methods of solving ordinary differential equations


458

iii

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


iv

CHAPTER 1 ALGEBRA EXERCISE 1 Page 2 2. Find the value of 5 pq 2 r 3 when p =

2 , q = -2 and r = -1 5

2 2 3 ⎛2⎞ 5 pq 2 r 3 = 5 ⎜ ⎟ ( −2 ) ( −1) = 5 × × 4 × −1 = -8 5 ⎝5⎠ 5. Simplify ( x 2 y 3 z )( x3 yz 2 ) and evaluate when x =

(x

2

1 , y = 2 and z = 3 2

y 3 z )( x3 yz 2 ) = x 2+3 y 3+1 z1+ 2 = x5 y 4 z 3

1 24 × 33 33 33 27 1 4 3 ⎛1⎞ = = = = 13 When x = , y = 2 and z = 3, x5 y 4 z 3 = ⎜ ⎟ ( 2 ) ( 3) = 5 5− 4 2 2 2 2 2 2 ⎝2⎠ 5

⎛ 32 − 12 ⎞⎛ 12 − 12 ⎞ 6. Evaluate ⎜ a bc ⎟⎜ a b c ⎟ when a = 3, b = 4 and c = 2 ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ 1 ⎛ 32 −3 ⎞ ⎛ 12 − 12 ⎞ ⎜ + ⎟ ⎜1− ⎟ −3+1 a2 b 2 2 −2 ⎝ 2 3⎠ ⎝ 2⎠ =a b c = 2 b c ⎜ a bc ⎟ ⎜ a b c ⎟ = a c ⎝ ⎠⎝ ⎠ 1 1

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

1

1 a 2 b 32 4 9 ( ±2 ) = 2 = =±4 2 2 c 2 4

1 ⎛ 3 12 − 12 ⎞ ⎜ a b c ⎟ ( ab ) 3 ⎠ 8. Simplify ⎝ 3 a bc

(

)

1 ⎛ 3 12 − 12 ⎞ 1 1 1 1 − ⎜ a b c ⎟ ( ab ) 3 ⎛ 1 3⎞ ⎛1 1 1⎞ ⎛ 1 ⎞ ⎛ 18 + 2 − 9 ⎞ 1 3 3 2 2 3 3 − ⎜ 3+ − ⎟ ⎜ + − ⎟ − ⎜ − −1⎟ ⎜ ⎟ a b c a b ⎝ ⎠ 6 ⎝ 3 2⎠ ⎝ 2 3 2⎠ ⎝ 2 ⎠ ⎝ ⎠ 3 2 a b c a b c = = = 3 1 a3 b c a 2b 2c

(

)

11

1

= a 6 b3c

−

3 2

6

or

a 11 3 b c3


1

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

24 p − ⎡⎣ 2 {3 ( 5 p − q ) − 2 ( p + 2q )} + 3q ⎤⎦ = 24 p − ⎡⎣ 2 {15 p − 3q − 2 p − 4q} + 3q ⎤⎦ = 24 p − [30 p − 6q − 4 p − 8q + 3q ] = 24 p − [ 26 p − 11q ] = 24p – 26p + 11q = 11q – 2p 6. Simplify 2 y + 4 ÷ 6 y + 3 × 4 − 5 y

2 y + 4 ÷ 6 y + 3× 4 − 5 y = 2 y +

2 4 2 + 3× 4 − 5 y = 2 y + + 12 − 5 y = − 3 y + 12 3y 6y 3y

8. Simplify a 2 − 3ab × 2a ÷ 6b + ab

a 2 − 3ab × 2a ÷ 6b + ab = a 2 − 3ab ×

2a 6a 2b + ab = a 2 − + ab = a 2 − a 2 + ab = ab 6b 6b


2

EXERCISE 3 Page 4 3. Solve the equation:

1 1 + =0 3a − 2 5a + 3

1 1 from which, (5a + 3) = -(3a – 2) =− 3a − 2 5a + 3 i.e.

5a + 3 = -3a + 2

and

5a + 3a = 2 – 3

Thus,

8a = -1

4. Solve the equation:

If

(

6= 3 t

if

If t = 2π

l g

then

t = 2π

µL L + rCR

µL L + rCR

then

t=

then

l g

6 =2 3

and t = 22 = 4

2

l ⎛ t ⎞ and ⎜ ⎟ = g ⎝ 2π ⎠

⎛ t ⎞ l = g⎜ ⎟ ⎝ 2π ⎠

from which,

7. Transpose m =

from which, 6 = 6 t − 3 t = 3 t

l g

6. Make l the subject of t = 2π

If m =

)

3 t = −6 + 6 t

Hence,

1 8

3 t = −6 1− t

3 t = −6 then 3 t = −6 1 − t 1− t

i.e.

and a = -

2

or

l=

gt 2 4π 2

for L

m ( L + rCR ) = µ L

i.e.

mL + mrCR = µ L


3

from which,

mrCR = µ L − mL = L ( µ − m )

8. Make r the subject of the formula

If

x 1+ r2 = y 1− r2

from which, and

Thus,

then

L=

and

mrCR µ−m

x 1+ r2 = y 1− r2

x (1 − r 2 ) = y (1 + r 2 ) x − xr 2 = y + yr 2 x − y = yr 2 + xr 2 = r 2 ( y + x )

r2 =

x− y x+ y

and

⎛ x− r= ⎜ ⎝ x+

y⎞ ⎟ y⎠


4

EXERCISE 4 Page 5

2. Solve the simultaneous equations

5a = 1 − 3b 2b + a + 4 = 0

5a + 3b = 1

(1)

a + 2b = -4

(2)

5 × (2) gives:

5a + 10b = -20

(3)

(1) – (3) gives:

- 7b = 21

from which,

21 = -3 −7

b=

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

5a = 1 + 9 = 10

3. Solve the simultaneous equations

a=

and

10 =2 2

x 2 y 49 + = 5 3 15 3x y 5 − + =0 7 2 7

(1) (2)

15 × (1) gives:

3x + 10y = 49

(3)

14 × (2) gives:

6x – 7y = -10

(4)

6x + 20y = 98

(5)

27y = 108

from which,

2 × (3) gives: (5) – (4) gives: Substituting in (3) gives:

3x + 40 = 49

and

y=

108 =4 27

3x = 49 – 40 = 9

from which,

x=3

4.(b) Solve the quadratic equation by factorisation: 8 x 2 + 2 x − 15 = 0

If

8 x 2 + 2 x − 15 = 0

then

(4x – 5)(2x + 3) = 0 5 4

hence,

4x – 5 = 0

i.e. 4x = 5

i.e.

x=

and

2x + 3 = 0

i.e. 2x = -3

i.e.

x= −


3 2

5

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. x 2 − 2 x + 5 x − 10 = 0

x 2 + 3 x − 10 = 0

i.e.

6.(a) Solve the quadratic equation, correct to 3 decimal places: 2 x 2 + 5 x − 4 = 0

−5 ± ⎡⎣52 − 4(2)(−4) ⎤⎦ −5 ± (25 + 32) −5 ± 57 = = If 2 x + 5 x − 4 = 0 then x = 2(2) 4 4 2

Hence,

x=

−5 + 57 = 0.637 4

or

x=

−5 − 57 = -3.137 4


6

EXERCISE 5 Page 8

(10 x

3. Determine

2

+ 11x − 6 ) ÷ ( 2 x + 3)

5x - 2 2 x + 3 10 x 2 + 11x − 6

10 x 2 + 15 x - 4x - 6 - 4x - 6 Hence,

10 x 2 + 11x − 6 = 5x - 2 2x + 3

5. Divide ( x3 + 3x 2 y + 3xy 2 + y 3 ) by (x + y)

x 2 + 2 xy + y 2 x + y x3 + 3 x 2 y + 3 xy 2 + y 3

x3 + x 2 y

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

Hence,

x 3 + 3 x 2 y + 3 xy 2 + y 3 = x 2 + 2 xy + y 2 x+ y

Hence,

8 5x2 − x + 4 = 5x + 4 + x −1 x −1

6. Find ( 5 x 2 − x + 4 ) ÷ ( x − 1)

5x + 4 x −1 5x2 − x + 4

5x2 − 5x 4x + 4 4x - 4 8


7

8. Determine ( 5 x 4 + 3x 3 − 2 x + 1) ÷ ( x − 3)

5 x3 + 18 x 2 + 54 x + 160 x − 3 5 x 4 + 3 x3

− 2x + 1

5 x 4 − 15 x 3 18 x 3 18 x 3 − 54 x 2 54 x 2 − 2 x 54 x 2 − 162 x 160x + 1 160x - 480 481 481 5 x 4 + 3x3 − 2 x + 1 Hence, = 5 x 3 + 18 x 2 + 54 x + 160 + x−3 x −3


8

EXERCISE 6 Page 9

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

Let f(x) = x3 + x 2 − 4 x − 4 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, x 3 + x 2 − 4 x − 4 = (x + 1)(x + 2)(x – 2)

4. Use the factor theorem to factorise 2 x 3 − x 2 − 16 x + 15

Let f(x) = 2 x 3 − x 2 − 16 x + 15 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 2 x 3 − x 2 − 16 x + 15 2 x 3 − x 2 − 16 x + 15 = ( x − 1)( x + 3) x2 + 2x − 3

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

2 x3 + 4 x 2 − 6 x −5 x 2 − 10 x + 15 −5 x 2 − 10 x + 15

Hence, 2 x 3 − x 2 − 16 x + 15 = (x – 1)(x + 3)(2x – 5) © 2006 John Bird. All rights reserved. Published by Elsevier.

9

6. Solve the equation x3 − 2 x 2 − x + 2 = 0

Let f(x) = x3 − 2 x 2 − x + 2 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, x3 − 2 x 2 − x + 2 = (x – 1)(x – 2)(x + 1) If x3 − 2 x 2 − x + 2 = 0 then (x – 1)(x – 2)(x + 1) = 0 from which,

x = 1, x = 2, or x = -1


10

EXERCISE 7 Page 11

2. Determine the remainder when x3 − 6 x 2 + x − 5 is divided by (a) (x + 2) (b) (x – 3)

(a) Remainder is ap 3 + bp 2 + + d where a = 1, b = -6, c = 1, d = -5 and p = -2 Hence, remainder = 1(−2)3 − 6(−2) 2 + 1(−2) − 5 = -8 – 24 – 2 – 5 = -39 (b) When p = 3, remainder = 1(3)3 − 6(3) 2 + 1(3) − 5 = 27 – 54 + 3 – 5 = -29

4. Determine the factors of x3 + 7 x 2 + 14 x + 8 and hence solve the cubic equation x3 + 7 x 2 + 14 x + 8 = 0

Remainder is ap 3 + bp 2 + + d where a = 1, b = 7, c = 14, d = 8 Let p = 1, then remainder = 1(1)3 + 7(1) 2 + 14(1) + 8 = 30 Let p = -1, then remainder = 1(−1)3 + 7(−1) 2 + 14(−1) + 8 = -1 + 7 – 14 + 8 = 0, hence (x + 1) is a factor Let p = -2, then remainder = 1(−2)3 + 7(−2) 2 + 14(−2) + 8 = -8 + 28 – 28 + 8 = 0, hence (x + 2) is a factor Let p = -3, then remainder = 1(−3)3 + 7(−3) 2 + 14(−3) + 8 = -27 + 63 – 42 + 8 = 2 Let p = -4, then remainder = 1(−4)3 + 7(−4) 2 + 14(−4) + 8 = -64 + 112 – 56 + 8 = 0, hence (x + 4) is a factor Hence, x3 + 7 x 2 + 14 x + 8 = (x + 1)(x + 2)(x + 4) If x3 + 7 x 2 + 14 x + 8 = 0 then (x + 1)(x + 2)(x + 4) = 0 from which,

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


11

6. Using the remainder theorem, solve the equation 2 x 3 − x 2 − 7 x + 6 = 0

Remainder is ap 3 + bp 2 + + d where a = 2, b = -1, c = -7, d = 6 Let p = 1, then remainder = 2(1)3 + (−1)(1) 2 + (−7)(1) + 6 = 2 - 1 – 7 + 6 = 0, hence (x - 1) is a factor Let p = 2, then remainder = 2(2)3 + (−1)(2) 2 + (−7)(2) + 6 = 16 - 4 – 14 + 6 = 4 Let p = -1, then remainder = 2(−1)3 + (−1)(−1) 2 + (−7)(−1) + 6 = -2 - 1 + 7 + 6 = 10 Let p = -2, then remainder = 2(−2)3 + (−1)(−2) 2 + (−7)(−2) + 6 = -16 - 4 + 14 + 6 = 0, hence (x + 2) is a factor Let p = -3, then remainder = 2(−3)3 + (−1)(−3) 2 + (−7)(−3) + 6 = -54 - 9 + 21 + 6 = -36 The third root can be found by division, i.e.

2 x3 − x 2 − 7 x + 6 2 x3 − x 2 − 7 x + 6 = x2 + x − 2 ( x − 1)( x + 2 ) 2x - 3 x 2 + x − 2 2 x3 − x 2 − 7 x + 6

2 x3 + 2 x 2 − 4 x

−3x 2 − 3x + 6 −3 x 2 − 3 x + 6 Hence, 2 x3 − x 2 − 7 x + 6 = (x – 1)(x + 2)(2x – 3) If 2 x 3 − x 2 − 7 x + 6 = 0 then (x – 1)(x + 2)(2x – 3) = 0 from which,

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


12

CHAPTER 2 INEQUALITIES EXERCISE 8 Page 13

2. Solve the following inequalities: (a)

(a)

x > 1.5 2

(b) x + 2 ≥ 5

i.e. x > 2(1.5)

i.e. x > 3

i.e. x ≥ 5 – 2

i.e. x ≥ 3

4. Solve the following inequalities: (a)

(a)

x > 1.5 (b) x + 2 ≥ 5 2

7 − 2k ≤1 4

(b) 3z + 2 > z + 3

i.e. 7 – 2k ≤ 4

7 − 2k ≤ 1 (b) 3z + 2 > z + 3 4

i.e. 7 – 4 ≤ 2k

i.e. 3z – z > 3 – 2

i.e. 2z >1

i.e. 3 ≤ 2k and and

z>

k≥

3 2

1 2

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. −

4 ≤y 3

or

y≥ −

(b) 1 - 6x ≤ 5 + 2x

i.e. 1 – 5 ≤ 2x + 6x

i.e. -4 ≤ 8x

i.e. −

4 ≤x 8

or

x≥-


4 3

1 2

13

EXERCISE 9 Page 14 1. Solve the inequality: t + 1 < 4

If t + 1 < 4 then -4 < t + 1 < 4 -4 < t + 1 becomes -5 < t t + 1 < 4 becomes Hence,

i.e. t > -5

t<3

-5 < t < 3

3. Solve the inequality: 2 x − 1 < 4

If 2 x − 1 < 4

then -4 < 2x – 1 < 4 3 <x 2

-4 < 2x – 1 becomes -3 < 2x

and

−

2x – 1 < 4

and

x<

Hence,

becomes

−

2x < 5

5 2

3 5 < x< 2 2

5. Solve the inequality: 1 − k ≥ 3

1− k ≥ 3

1–k≥3

and 1 – k ≤ -3

i.e.

1–3≥k

and

i.e.

k ≤ -2

and

means

1+3≤k k≥4


14

EXERCISE 10 Page 15

2. Solve the inequality:

If

2t + 4 >1 t −5

2t + 4 >1 t −5

2t + 4 -1>0 t −5

then

t +9 >0 t −5

i.e.

2t + 4 t − 5 − > 0 and t −5 t −5

i.e.

( 2t + 4 ) − ( t − 5) > 0 t −5

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 2t + 4 > 1 is true when t > 5 or t < -9 t −5

Hence,


If

3z − 4 ≤2 z+5

i.e.

then

3z − 4 ≤2 z+5

3z − 4 -2≤0 z+5

3z − 4 − 2 z − 10 ≤0 z +5

Hence,

either (i) z - 14 ≤ 0 or

(ii) z – 14 ≥ 0

i.e. and

i.e.

3z − 4 2( z + 5) − ≤ 0 and ( z + 5) z +5

(3 z − 4) − 2( z + 5) ≤0 z +5

z − 14 ≤0 z +5

z+5>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,

3z − 4 ≤ 2 is true when -5 < z ≤ 14 z+5


15

EXERCISE 11 Page 16

3. Solve the inequality: 2 x 2 ≥ 6

2 x2 ≥ 6

i.e.

x2 ≥ 3

x≥ 3

hence,

or x ≤ - 3

4. Solve the inequality: 3k 2 − 2 ≤ 10

3k 2 − 2 ≤ 10

i.e.

3k 2 ≤ 12

- 4 ≤k≤

Hence,

and

k2 ≤ 4

i.e. -2 ≤ k ≤ 2

4

6. Solve the inequality: ( t − 1) ≥ 36 2

( t − 1)

36

or

(t – 1) ≤ - 36

i.e.

(t – 1) ≥ 6

or

(t – 1) ≤ -6

i.e.

t ≥7

or

t ≤ -5

( 4k + 5 )

>9

2

≥ 36 then (t – 1) ≥


( 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> −

1 2

or

k < -2

2

>9

2

then

4k + 5 >


16

EXERCISE 12 Page 17

1. Solve the inequality: x 2 − x − 6 > 0

x2 − x − 6 > 0

(x – 3)(x + 2) > 0

thus

Either (i) x – 3 > 0

and

x+2>0

(ii) x – 3 < 0

and

x+2<0

or

(i) x > 3

and x > -2

i.e.

x>3

(ii) x < 3

and

x < -2

i.e.

x < -2

3. Solve the inequality: 2 x 2 + 3x − 2 < 0

2 x 2 + 3x − 2 < 0

thus

(2x – 1)(x + 2) < 0

Either (i) 2x – 1 > 0

and

x+2<0

(ii) 2x – 1 < 0

and

x+2>0

or

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

1 2

and

x < -2

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

1 2

and

x > -2

both of which are not possible

thus

-2 < x <

1 2

5. Solve the inequality: z 2 + 4 z + 4 ≤ 4 z2 + 4z + 4 ≤ 4

i.e.

Either (i) z ≤ 0

and

z ≥ -4

and

z ≤ -4

or

(ii) z ≥ 0

z2 + 4z ≤ 0

z(z + 4) ≤ 0

i.e. i.e.

-4 ≤ z ≤ 0

both of which are not possible

7. Solve the inequality: t 2 − 4t − 7 ≥ 0

Since t 2 − 4t − 7 ≥ 0

then

(t − 2)

2

-7–4≥0


17

(t − 2)

i.e. and thus,

t – 2 ≥ 11

(

t ≥ 2 + 11

)

2

≥ 11

or

t – 2 ≤ - 11

or

t ≤ 2 − 11

(

)


18

CHAPTER 3 PARTIAL FRACTIONS EXERCISE 13 Page 20

2. Resolve

Let

4( x − 4) into partial fractions. x2 − 2 x − 3

4( x − 4) 4 x − 16 A B A( x − 3) + B( x + 1) ≡ = + = 2 ( x + 1)( x − 3) x − 2 x − 3 ( x + 1)( x − 3) ( x + 1) ( x − 3)

Hence,

4x – 16 = A(x – 3) + B(x + 1)

If x = -1,

-20 = -4A from which, A = 5

If x = 3,

12 – 16 = 4B

Hence,

4( x − 4) 5 1 = − 2 x − 2 x − 3 ( x + 1) ( x − 3)

4. Resolve

from which, B = -1

3(2 x 2 − 8 x − 1) into partial fractions. ( x + 4)( x + 1)(2 x − 1)

Let 3(2 x 2 − 8 x − 1) A B C A( x + 1)(2 x − 1) + B( x + 4)(2 x − 1) + C ( x + 4)( x + 1) ≡ + + = ( x + 4)( x + 1)(2 x − 1) ( x + 4) ( x + 1) (2 x − 1) ( x + 4)( x + 1)(2 x − 1)

Hence, 6 x 2 − 24 x − 3 = 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, If x = -1,

6 + 24 -3 = B(3)(-3)

from which,

189 = 27A and A = 7 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 3(2 x 2 − 8 x − 1) 7 3 2 = − − ( x + 4)( x + 1)(2 x − 1) ( x + 4) ( x + 1) (2 x − 1)

Hence,

5. Resolve

x2 + 9 x + 8 into partial fractions. x2 + x − 6

Since the numerator is of the same degree as the denominator, division is firstly required. © 2006 John Bird. All rights reserved. Published by Elsevier.

19

1 x2 + x − 6 x2 + 9 x + 8

x2 + x − 6

8x + 14 x + 9x + 8 8 x + 14 = 1+ 2 2 x + x−6 x + x−6 2

Hence, Let

8 x + 14 8 x + 14 A B A( x − 2) + B( x + 3) = ≡ + = 2 ( x + 3)( x − 2) x + x − 6 ( x + 3)( x − 2) ( x + 3) ( x − 2)

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

x2 + 9x + 8 2 6 = 1+ + 2 x + x−6 ( x + 3) ( x − 2)

Hence,

7. Resolve

3 x 3 − 2 x 2 − 16 x + 20 into partial fractions. ( x − 2)( x + 2)

3x - 2 x − 4 3 x3 − 2 x 2 − 16 x + 20 2

3x3

− 12 x −2 x 2 − 4 x + 20 −2 x 2 +8

- 4x + 12 Hence, Let

3 x 3 − 2 x 2 − 16 x + 20 12 − 4 x ≡ 3x − 2 + 2 ( x − 2)( x + 2) x −4

12 − 4 x 12 − 4 x A B A( x + 2) + B( x − 2) = ≡ + = 2 ( x − 2)( x + 2) x − 4 ( x − 2)( x + 2) ( x − 2) ( x + 2)

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 x 3 − 2 x 2 − 16 x + 20 1 5 ≡ 3x − 2 + − ( x − 2)( x + 2) ( x − 2) ( x + 2)


20

EXERCISE 14 PAGE 22

x2 + 7 x + 3 2. Resolve 2 into partial fractions. x ( x + 3)

x2 + 7 x + 3 A B C A( x)( x + 3) + B ( x + 3) + Cx 2 ≡ + 2+ = Let 2 x ( x + 3) x x ( x + 3) x 2 ( x + 3)

Hence, If x = 0 If x = -3

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

from which,

9 – 21 + 3 = 9C

i.e. -9 = 9C f rom which, C = -1

Equating x 2 coefficients: 1 = A + C Hence,

B=1

from which, A = 2

x2 + 7 x + 3 2 1 1 = + 2− 2 x ( x + 3) x x ( x + 3)

18 + 21x − x 2 4. Resolve into partial fractions. ( x − 5)( x + 2) 2

18 + 21x − x 2 A B C A( x + 2) 2 + B( x − 5)( x + 2) + C ( x − 5) ≡ + + = Let ( x − 5)( x + 2) 2 ( x − 5) ( x + 2) ( x + 2) 2 ( x − 5)( x + 2) 2

Hence, If x = 5 If x = -2

18 + 21x − x 2 = A( x + 2) 2 + B( x − 5)( x + 2) + C ( x − 5) 18 + 105 – 25 = 49A

i.e. 98 = 49A

from which, A = 2

18 – 42 – 4 = -7C

i.e. -28 = -7C

from which, C = 4

Equating x 2 coefficients: -1 = A + B Hence,

from which,

B = -3

18 + 21 x − x 2 2 3 4 = − + 2 ( x − 5)( x + 2) ( x − 5) ( x + 2) ( x + 2)2


21

EXERCISE 15 PAGE 23

x 2 − x − 13 1. Resolve 2 into partial fractions. ( x + 7) ( x − 2)

Let

( Ax + B)( x − 2) + C ( x 2 + 7 ) x 2 − x − 13 Ax + B C ≡ + = ( x 2 + 7 ) ( x − 2 ) ( x2 + 7 ) ( x − 2) ( x 2 + 7 ) ( x − 2)

Hence,

x 2 − x − 13 = ( Ax + B )( x − 2) + C ( x 2 + 7 )

If x = 2,

4 – 2 –13 = 11C

Equating x 2 coefficients:

1=A+C

i.e. -11 = 11C from which,

from which, C = -1 A=2

Equating constant : -13 = -2B + 7C = -2B – 7 i.e. 2B = 13 – 7 = 6 from which, B = 3 x 2 − x − 13 2x + 3 1 = 2 − 2 ( x + 7 ) ( x − 2 ) ( x + 7 ) ( x − 2)

Hence,

4. Resolve

x 3 + 4 x 2 + 20 x − 7 into partial fractions. ( x − 1) 2 ( x 2 + 8 )

2 2 2 x 3 + 4 x 2 + 20 x − 7 A B Cx + D A( x − 1) ( x + 8 ) + B ( x + 8 ) + (Cx + D)( x − 1) Let ≡ + + = ( x − 1) ( x − 1) 2 ( x 2 + 8 ) ( x − 1) 2 ( x 2 + 8 ) ( x − 1) 2 ( x 2 + 8 )

x3 + 4 x 2 + 20 x − 7 = A( x − 1) ( x 2 + 8 ) + B ( x 2 + 8 ) + (Cx + D)( x − 1) 2

Hence,

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

If x = 1,

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


1=A+C

(1’)


4 = -A + B – 2C + D

(2’)

Equating x coefficients: Since B = 2,

20 = 8A + C – 2D

(3’)

A+C=1

(1)

-A – 2C + D = 2

(2)


22

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

x 3 + 4 x 2 + 20 x − 7 3 2 1− 2x = + + 2 2 2 2 ( x − 1) ( x − 1) ( x + 8 ) ( x − 1) ( x + 8 )

Hence,

5. When solving the differential equation

d 2θ dθ −6 − 10θ = 20 − e 2t by Laplace transforms, for 2 dt dt

given boundary conditions, the following expression for Λ {θ } results: 39 2 s + 42s − 40 2 Λ {θ } = s ( s − 2 ) ( s 2 − 6s + 10 ) 4s 3 −

Show that the expression can be resolved into partial fractions to give: Λ {θ } =

2 1 5s − 3 − + s 2 ( s − 2 ) 2 ( s 2 − 6s + 10 )

39 2 s + 42s − 40 A B Cs + D 2 ≡ + + 2 2 s ( s − 2 ) ( s − 6s + 10 ) s ( s − 2) ( s − 6s + 10 )

4s 3 −

Let

= Hence, 4 s 3 −

A( s − 2) ( s 2 − 6s + 10 ) + B( s ) ( s 2 − 6 s + 10 ) + (Cs + D)( s )( s − 2) s ( s − 2) ( s 2 − 6s + 10 )

39 2 s + 42 s − 40 = A( s − 2) ( s 2 − 6s + 10 ) + B ( s ) ( s 2 − 6s + 10 ) + (Cs + D)( s )( s − 2) 2 = A ( s 3 − 8s 2 − 2s − 20 ) + B ( s 3 − 6 s 2 + 10s ) + (Cs + D)( s 2 − 2s )

If s = 0, If s = 2,

-40 = A(-20)

from which,

32 – 78 + 84 – 40 = B (8 – 24 + 20)

A=2

i.e. -2 = 4B

from which, B = −


1 2 23

Equating s 3 coefficients:

Equating s 2 coefficients: −

4=A+B+C

i.e.

39 = -8A – 6B – 2C + D 2

4=2-

1 +C 2

i.e.

−

from which,

C=

5 2

39 = -16 + 3 – 5 + D 2 from which, D = −

3 2

39 2 1 5 3 s + 42 s − 40 − s− 2 2 2 + 2 2 ≡ + 2 2 s ( s − 2 ) ( s − 6 s + 10 ) s ( s − 2) ( s − 6s + 10 )

4s 3 −

Hence,

i.e.

Λ {θ } =

2 1 5s − 3 − + 2 s 2 ( s − 2 ) 2 ( s − 6 s + 10 )


24

CHAPTER 4 LOGARITHMS AND EXPONENTIAL FUNCTIONS EXERCISE 16 Page 26

2. Evaluate: log 2 16

then 2 x = 16 = 24

Let x = log 2 16 Hence,

from which,

x=4

log 2 16 = 4

4. Evaluate: log 2

Let x = log 2

1 8

Hence,

1 8

then log 2

2x =

1 1 = = 2−3 8 23

from which, x = -3

1 = -3 8

7. Evaluate: log 4 8

Let x = log 4 8

then

4x = 8

from which, Hence,

2 x

2x = 3 log 4 8 = 1

1 2

then

= 23

and x =

i.e.

2 2 x = 23

3 2

1 2

11. Solve the equation: log 4 x = −2

If log 4 x = −2

(2 )

i.e.

x= 4

−

5 2

=

1 2

1 4

5 2

=

1 4

5

=±

1 1 =± 5 32 2


25

12. Solve the equation: lg x = −2

If lg x = -2 then log10 x = −2 and x = 10−2 =

1 1 = or 0.01 2 100 10

⎛ 16 × 4 5 ⎞ 16. Write in of log 2, log 3 and log 5 to any base: log ⎜⎜ ⎟⎟ ⎝ 27 ⎠ ⎛ 4 14 ⎛ 16 × 4 5 ⎞ ⎜ 2 ×5 log ⎜⎜ ⎟⎟ = log ⎜ 3 ⎜ 3 ⎝ 27 ⎠ ⎝

⎞ 1 1 ⎟ = log ⎛ 24 × 5 4 × 3−3 ⎞ = log 24 + log 5 4 + log 3−3 ⎜ ⎟ ⎟ ⎟ ⎝ ⎠ ⎠ 1 = 4 log 2 + log 5 – 3 log 3 4

⎛ 125 × 4 16 ⎞ 17. Write in of log 2, log 3 and log 5 to any base: log ⎜⎜ ⎟ 4 813 ⎟⎠ ⎝ ⎛ 125 × 4 16 ⎞ ⎛ 53 × 2 ⎞ −3 −3 3 3 log ⎜⎜ = log ⎟⎟ ⎜ 3 ⎟ = log ( 5 × 2 × 3 ) = log 5 + log 2 + log 3 3 4 81 ⎠ ⎝ 3 ⎠ ⎝

= 3 log 5 + log 2 – 3 log 3

19. Simplify: log 64 + log 32 – log 128

⎛ 2 6 × 25 ⎞ 6+5−7 log 64 + log 32 – log 128 = log 26 + log 25 − log 27 = log ⎜ = log 24 = 4 log 2 ) ⎟ = log ( 2 7 ⎝ 2 ⎠

1 1 log16 − log 8 3 20. Evaluate: 2 log 4

⎛ 22 ⎞ 1 1 1 1 1 1 log 4 3 ⎜ ⎟ log16 − log 8 log16 2 − log 83 log ( 2 ) 2 − log ( 2 ) 3 log 22 − log 2 ⎝ 2 ⎠ = log 2 = 1 2 3 = = = = 2 2 2 log 2 log 2 log 2 log 22 2 log 2 2 log 4


26

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

log 2t 3 − log t = log16 + log t ⎛ 2t 3 ⎞ i.e. log ⎜ ⎟ = log (16t ) ⎝ t ⎠

from which, Hence,

2t 2 = 16t t=8

i.e.

i.e.

log ( 2t 2 ) = log (16t )

2t 2 − 16t = 0

i.e.

2t(t – 8) = 0

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


27

EXERCISE 17 Page 27

1. Solve the equation 3x = 6.4 correct to 4 significant figures

If 3x = 6.4

then x log10 3 = log10 6.4 x=

and

log10 6.4 0.80617997... = = 1.689675... = 1.690, correct to 4 significant figures. log10 3 0.47712125...

3. Solve the equation 2 x −1 = 32 x −1 correct to 4 significant figures

If 2 x −1 = 32 x −1

( x − 1) log10 2 = (2 x − 1) log10 3

then

x log10 2 − log10 2 = 2 x log10 3 − log10 3

i.e.

log10 3 − log10 2 = 2 x log10 3 − x log10 2 = x ( 2 log10 3 − log10 2 )

i.e. Hence,

x=

log10 3 − log10 2 = 0.2696, correct to 4 significant figures. 2 log10 3 − log10 2

5. Solve the equation 25.28 = 4.2 x correct to 4 significant figures

If 25.28 = 4.2 x

then

log10 25.28 = x log10 4.2 x=

from which,

log10 25.28 = 2.251, correct to 4 significant figures. log10 4.2

6. Solve the equation 42 x −1 = 5 x + 2 correct to 4 significant figures

If 42 x −1 = 5 x + 2

then

(2 x − 1) log10 4 = ( x + 2) log10 5

2 x log10 4 − log10 4 = x log10 5 + 2 log10 5

i.e. i.e.

2 x log10 4 − x log10 5 = 2 log10 5 + log10 4

i.e.

x(2 log10 4 − log10 5) = 2 log10 5 + log10 4

from which,

x=

2 log10 5 + log10 4 = 3.959, correct to 4 significant figures. 2 log10 4 − log10 5


28

8. Solve the equation 0.027 x = 3.26 correct to 4 significant figures

If 0.027 x = 3.26

then

x log10 0.027 = log10 3.26 x=

from which,

log10 3.26 = -0.3272, correct to 4 significant figures. log10 0.027

⎛P ⎞ 9. The decibel gain n of an amplifier is given by: n = 10 log10 ⎜ 2 ⎟ where P1 is the power input ⎝ P1 ⎠

and P2 is the power output. Find the power gain

When n = 25 then: from which,

Thus,

P2 when n = 25 decibels. P1

⎛P ⎞ 25 = 10 log10 ⎜ 2 ⎟ ⎝ P1 ⎠ ⎛P ⎞ ⎛P ⎞ 25 i.e. 2.5 = log10 ⎜ 2 ⎟ = log10 ⎜ 2 ⎟ 10 ⎝ P1 ⎠ ⎝ P1 ⎠

P2 = 102.5 P1

i.e. power gain,

P2 = 316.2 P1


29

EXERCISE 18 Page 29

5.6823 3. Evaluate, correct to 5 significant figures: (a) −2.1347 e

e 2.1127 − e −2.1127 (b) 2

(c )

4 ( e −1.7295 − 1) e3.6817

Using a calculator: 5.6823 = 48.04106, correct to 5 decimal places. e −2.1347 e 2.1127 − e −2.1127 = 4.07482, correct to 5 decimal places. (b) 2 4 ( e −1.7295 − 1) (c ) = -0.08286, correct to 5 decimal places. e3.6817

(a)

4. The length of a bar, l , at a temperature θ is given by l = l0 eα θ , where l and α are constants.

Evaluate l , correct to 4 significant figures, when l0 = 2.587, θ = 321.7 and α = 1.771×10−4

Using a calculator, l = l0 eα θ = (2.587) e

(321.7×1.771×10 ) = 2.739, correct to 4 significant figures. −4


30

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

check your result by using a calculator.

(a)

x 2 x3 x 4 + + + .... 2! 3! 4!

ex = 1 + x +

When x = 2, e 2 = 1 + 2 +

2 2 23 2 4 25 2 6 + + + + + ... 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, e −0.3 = 1 − 0.3 +

(−0.3) 2 (−0.3)3 (−0.3) 4 (−0.3)5 + + + + ... 2! 3! 4! 5!

= 1 – 0.3 + 0.04500 – 0.00450 + 0.00034 – 0.00002 + … = 0.7408, correct to 4 significant figures.

3. Expand (1 – 2x) e2 x to six .

⎛ ⎞ (2 x) 2 (2 x)3 (2 x) 4 (1 – 2x) e2 x = (1 – 2x) ⎜1 + 2 x + + + + ... ⎟ 2! 3! 4! ⎝ ⎠

4 2 ⎞ ⎛ = (1 – 2x) ⎜1 + 2 x + 2 x 2 + x3 + x 4 ⎟ 3 3 ⎠ ⎝ =1 + 2 x + 2 x2 +

= 1 − 2 x2 −

4 3 2 4 8 x + x − 2 x − 4 x 2 − 4 x3 − x 4 − ... 3 3 3

8 3 x − 2 x4 3


31

EXERCISE 20 Page 32

1. Plot a graph of y = 3 e0.2 x 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 θ = 250e−0.05 t where the excess temperature of a body

above its surroundings at time t minutes is θ °C . 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 © 2006 John Bird. All rights reserved. Published by Elsevier.

32

Figure 2


33

EXERCISE 21 Page 34

2. Evaluate, correct to 5 significant figures: 2.946 ln e1.76 (a) lg101.41

5e −0.1629 (b) 2 ln 0.00165

(c )

ln 4.8629 − ln 2.4711 5.173

Using a calculator, (a)

2.946 ln e1.76 (2.946)(1.76) = 3.6773, correct to 5 significant figures. = 1.41 (1.41) lg10

5e−0.1629 (b) = -0.33154, correct to 5 significant figures. 2 ln 0.00165 (c )

ln 4.8629 − ln 2.4711 = 0.13087, correct to 5 significant figures. 5.173

4. Solve, correct to 4 significant figures: 7.83 = 2.91e−1.7 x

If 7.83 = 2.91e−1.7 x

i.e.

then

e−1.7 =

⎛ 7.83 ⎞ -1.7x = ln ⎜ ⎟ ⎝ 2.91 ⎠

7.83 2.91

and x = −

and

⎛ 7.83 ⎞ ln e −1.7 x = ln ⎜ ⎟ ⎝ 2.91 ⎠

1 ⎛ 7.83 ⎞ ln ⎜ ⎟ = -0.5822, correct to 4 significant figures. 1.7 ⎝ 2.91 ⎠

t − ⎞ ⎛ 2 5. Solve, correct to 4 significant figures: 16 = 24 ⎜1 − e ⎟ ⎝ ⎠

t − ⎞ ⎛ If 16 = 24 ⎜ 1 − e 2 ⎟ ⎝ ⎠

from which, and

and

then

t − 16 = 1− e 2 24

e

−

t 2

= 1−

16 24

−

t ⎛ 16 ⎞ = ln ⎜ 1 − ⎟ 2 ⎝ 24 ⎠

⎛ 16 ⎞ t = −2 ln ⎜1 − ⎟ = 2.197, correct to 4 significant figures. ⎝ 24 ⎠ © 2006 John Bird. All rights reserved. Published by Elsevier.

34

⎛ 1.59 ⎞ 7. Solve, correct to 4 significant figures: 3.72 ln ⎜ ⎟ = 2.43 ⎝ x ⎠

⎛ 1.59 ⎞ If 3.72 ln ⎜ ⎟ = 2.43 ⎝ x ⎠

then

⎛ 1.59 ⎞ 2.43 ln ⎜ ⎟= ⎝ x ⎠ 3.72 ⎛ 2.43 ⎞

⎜ ⎟ 1.59 = e⎝ 3.72 ⎠ x

from which,

and

x=

1.59 e

⎛ 2.43 ⎞ ⎜ ⎟ ⎝ 3.72 ⎠

= 1.59e

⎛ 2.43 ⎞ −⎜ ⎟ ⎝ 3.72 ⎠

= 0.8274, correct to 4 significant figures.

8. The work done in an isothermal expansion of a gas from pressure p1 to p 2 is given by: ⎛p ⎞ w = w 0 ln ⎜ 1 ⎟ ⎝ p2 ⎠

If the initial pressure p1 = 7.0 kPa, calculate the final pressure p 2 if w = 3 w 0

If w = 3 w 0 then

⎛p ⎞ 3 w 0 = w 0 ln ⎜ 1 ⎟ ⎝ p2 ⎠

i.e.

⎛p ⎞ 3 = ln ⎜ 1 ⎟ ⎝ p2 ⎠

and

e3 =

from which,

p1 7000 = p2 p2

final pressure, p 2 =

7000 = 7000 e −3 = 348.5 Pa 3 e


35

EXERCISE 22 Page 37 2. The voltage drop, v volts, across an inductor L henrys at time t seconds is given by v = 200e

− Rt L

,

where R = 150 Ω and L = 12.5 ×10−3 H. Determine (a) the voltage when t = 160 ×10−6 s, and (b) the time for the voltage to reach 85 V.

(a) Voltage v = 200e

− Rt L

−

= 200 e

(b) When v = 85 V, 85 = 200 e

and

Thus,

−

−

(150 )(160×10−6 ) 12.5×10−3

= 200 e −1.92 = 29.32 volts

150 t 12.5×10−3

150 t

from which,

− 85 −3 = e 12.5×10 200

150 t ⎛ 85 ⎞ = ln ⎜ ⎟ −3 12.5 ×10 ⎝ 200 ⎠ time t = −

12.5 × 10−3 ⎛ 85 ⎞ −6 ln ⎜ ⎟ = 71.31 × 10 s 150 ⎝ 200 ⎠

4. A belt is in 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 T0 = 22.7 newtons, given that these quantities are related by the law T = T0 e µθ . Determine also the value of θ when T = 28.0 newtons. Tension T = T0 e µθ = 22.7 e( 0.26)(1.12) = 22.7 e0.2912 = 30.4 N When T = 28.0 N,

Thus,

28.0 = 22.7 e0.26θ ⎛ 28.0 ⎞ 0.26θ = ln ⎜ ⎟ ⎝ 22.7 ⎠

from which,

and

θ =

28.0 = e0.26θ 22.7 1 ⎛ 28.0 ⎞ ln ⎜ ⎟ = 0.807 rad 0.26 ⎝ 22.7 ⎠

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

−

t CR

when a capacitor is being charged.

The capacitance C is 7 ×10−6 F and the resistance R is 0.3 ×106 ohms. Determine: (a) the instantaneous current when t is 2.5 seconds, and © 2006 John Bird. All rights reserved. Published by Elsevier.

36

(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 = 10 e (b) When i = 5,

2.5 7×10−6 × 0.3×106

5 = 10 e

−

= 3.04 A

t 2.1

from which,

t ⎛ 5⎞ ln ⎜ ⎟ = − 2.1 ⎝ 10 ⎠

Thus, Time t

0

1

and 2

3

t − 5 = e 2.1 10

time, t = −(2.1) ln 0.5 = 1.46 s 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

t − ⎛ ⎞ CR 7. The current i flowing in a capacitor at time t is given by: i = 12.5 ⎜ 1 − 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.


37

0.5 t − ⎛ − ⎛ ⎞ 20×10−6 ×30×103 CR (a) Current, i = 12.5 ⎜1 − e ⎟ = 12.5 ⎜⎜1 − e ⎝ ⎠ ⎝ t − ⎛ ⎞ 10 = 12.5 ⎜1 − e 0.6 ⎟ ⎝ ⎠

(b) When i = 10 A,

Thus,

i.e.

e

−

t 0.6

= 1−

10 12.5

and

⎞ ⎟⎟ = 7.07 A ⎠

from which,

−

t − 10 = 1 − e 0.6 12.5

t 10 ⎞ ⎛ = ln ⎜1 − ⎟ 0.6 ⎝ 12.5 ⎠

10 ⎞ ⎛ time, t = −0.6 ln ⎜1 − ⎟ = 0.966 s ⎝ 12.5 ⎠


38

EXERCISE 23 Page 40 2. At particular times, t minutes, measurements are made of the temperature, θ °C , of a cooling liquid and the following results are obtained: Temperature θ °C Time t minutes

92.2 55.9 33.9 20.6 12.5 10

20

30

40

50

Prove that the quantities follow a law of the form θ = θ 0 e kt , where θ 0 and k are constants, and determine the approximate value of θ 0 and k.

Since θ = θ 0 e kt i.e.

then

ln θ = ln (θ 0 e k t ) = ln θ 0 + ln e k t ln θ = kt + ln θ 0

where gradient = k and intercept on vertical axis = ln θ 0 Using log-linear graph paper, a graph of ln θ 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 θ = θ 0 e kt © 2006 John Bird. All rights reserved. Published by Elsevier.

39

Gradient, k =

AB ln 92.2 − ln12.5 = = -0.05 BC 10 − 50

At point A in Figure 4, θ = 92.2°C , t = 10 min Substituting in θ = θ 0 e kt gives: from which,

92.2 = θ 0 e( −0.05)(10)

θ0 =

92.2 = 92.2 e0.5 = 152 −0.5 e


40

CHAPTER 5 HYPERBOLIC FUNCTIONS EXERCISE 24 Page 42

1.(a) Evaluate sh 0.64 correct to 4 significant figures.

(a) sh 0.64 =

1 0.64 −0.64 ( e − e ) = 0.6846, correct to 4 significant figures. 2

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 =

1 2.4625 −2.4625 ( e + e ) = 5.910, correct to 4 significant figures. 2

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 =

e0.65 − e −0.65 1.39349505... = = 0.5717, correct to 4 significant figures. e0.65 + e −0.65 2.4375866...

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 =

1 = 0.08849, correct to 4 significant figures, using a calculator. sh 3.12

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

(a) sech 0.39 =

1 = 0.9285, correct to 4 significant figures, using a calculator. ch 0.39


41

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

(b) coth 1.843 =

1 = 1.051, correct to 4 significant figures, using a calculator. th1.843

7. A telegraph wire hangs so that its shape is described by y = 50 ch

y . Evaluate, correct to 4 50

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

When x = 0.25, y = 50 ch

y 25 = 50 ch = 50 ch 0.50 = 56.38, correct to 4 significant figures. 50 50

8. The length l of a heavy cable hanging under gravity is given by l = 2c sh( L / 2c) . Find the value of l when c = 40 and L = 30.

⎛ 30 ⎞ ⎛3⎞ l = 2c sh( L / 2c) = 2(40) sh ⎜ ⎟ = 80 sh ⎜ ⎟ = 30.71 ⎝8⎠ ⎝ 2(40) ⎠

9. V 2 = 0.55 L tanh(6.3d / 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.

⎡ (6.3)(8.0) ⎤ V 2 = 0.55 L tanh(6.3d / L) = 0.55(96) tanh ⎢ ⎥⎦ = 52.8 tanh 0.525 = 25.425829… ⎣ 96 Hence,

V=

25.425829... = 5.042


42

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 =

2 sh x 2 sh x 1 sh x 1 + sh 2 x + = + = + = sh 2 x ch x 2 sh x ch x ch x sh x ch x ch x sh x ch x =

ch 2 x ch x = = coth x = L.H.S sh x ch x sh x

2

⎛ eθ − e −θ ⎞ 2 θ 1 2θ θ −θ θ −θ −θ −θ −2θ θ (b) R.H.S. = 2 sh θ = 2 ⎜ ⎟ = ( e − e )( e − e ) = ⎡⎣e − e e − e e + e ⎤⎦ 4 2 ⎝ 2 ⎠ 2

=

1 2θ 1 e 2θ e −2θ 2 ⎡⎣e − e0 − e0 + e −2θ ⎦⎤ = ⎣⎡e 2θ − 2 + e−2θ ⎦⎤ = + − 2 2 2 2 2

=

e 2θ + e −2θ − 1 = ch 2θ − 1 = L.H.S. 2

4. Prove the following identities: (a) sh(A + B) ≡ sh A ch B + ch A sh B sh 2 x + ch 2 x − 1 (b) ≡ tanh 4 x 2 2 2ch x coth x

⎛ e A − e− A ⎞ ⎛ e B + e− B ⎞ ⎛ e A + e− A ⎞ ⎛ e B − e− B ⎞ (a) R.H.S. = sh A ch B + ch A sh B = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ 2 2 2 2 ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

(b) L.H.S. =

=

1 A+ B ⎡⎣e + e A− B − e− A+ B − e− A− B + e A+ B − e A− B + e − A+ B − e− A− B ⎤⎦ 4

=

( A+ B ) 1 ⎡ A+ B − e −( A+ B ) −( A+ B ) ⎤=e − = sh ( A + B) = L.H.S 2 e 2 e ⎦ 4⎣ 2

sh 2 x + ch 2 x − 1 sh 2 x + sh 2 x = 2ch 2 x coth 2 x ⎛ ch 2 x ⎞ 2ch 2 x ⎜ 2 ⎟ ⎝ sh x ⎠ =

since ch 2 x − 1 = sh 2 x

2 sh 2 x ⎛ sh 2 x ⎞ sh 4 x = tanh 4 x = R.H.S. ⎜ ⎟= 2 ch 2 x ⎝ ch 2 x ⎠ ch 4 x


43

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

⎛ e x + e− x P e x − Q e − x ≡ 6 ch x − 2 sh x = 6 ⎜ ⎝ 2

⎞ ⎛ e x − e− x ⎟ − 2⎜ ⎠ ⎝ 2

⎞ x −x x −x ⎟ = 3(e + e ) − (e − e ) ⎠

= 3e x + 3e − x − e x + e − x P e x − Q e − x = 2e x + 4e − x

i.e. from which,

P = 2 and Q = -4

6. If 5 e x − 4 e − x ≡ A sh x + B ch x , find A and B

⎛ e x − e− x 5 e x − 4 e − x ≡ A sh x + B ch x = A ⎜ ⎝ 2

⎞ ⎛ e x + e− x B + ⎟ ⎜ ⎠ ⎝ 2

⎞ A x A −x B x B −x ⎟= e − e + e + e 2 2 2 ⎠ 2

⎛ A B⎞ ⎛ A B⎞ = ⎜ + ⎟ e x − ⎜ − ⎟ e− x ⎝2 2⎠ ⎝2 2⎠ i.e.

⎛ A + B ⎞ x ⎛ A − B ⎞ −x 5 e x − 4 e− x = ⎜ ⎟e −⎜ ⎟e ⎝ 2 ⎠ ⎝ 2 ⎠

Hence,

5=

A+ B 2

i.e.

A + B = 10

(1)

and

4=

A− B 2

i.e.

A–B=8

(2)

(1) + (2) gives: From (1)

2A = 18 from which, A = 9

B=1


44

EXERCISE 26 Page 48

2. Solve 2 ch x = 3, correct to 4 decimal places.

2 ch x = 3 from which, ch x = i.e.

i.e.

Hence,

e x + e− x 3 = 2 2

(e )

x 2

3 2 e x + e− x − 3 = 0

or

+ e − x e x − 3e x = 0

(e )

x 2

or

− 3e x + 1 = 0

2 − ( −3) ± ⎡( −3) − 4 (1)(1) ⎤ 3 ± 5 ⎣ ⎦ = e = = 2.61803 or 0.381966 2 (1) 2 x

Thus,

x = ln 2.61803 or

x = ln 0.381966

i.e.

x = 0.9624

x = -0.9624

or

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.

⎛ e x − e− x 3.5 ⎜ ⎝ 2

i.e.

1.75 e x − 1.75 e − x + 1.25 e x + 1.25 e− x = 0

and

⎞ ⎛ e x + e− x 2.5 + ⎟ ⎜ ⎠ ⎝ 2

3 e x − 0.5 e − x = 0

i.e.

e x 0.5 = e− x 3

Hence,

2x = ln

0.5 3

⎞ ⎟=0 ⎠

3 e x = 0.5 e − x

or i.e.

e2 x =

from which,

0.5 3

x=

1 0.5 ln = -0.8959 2 3

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

4 th x - 1 = 0

i.e.

⎛ e x − e− x 4 ⎜ x −x ⎝e +e

⎞ ⎟ =1 ⎠


45

4 ( e x − e− x ) = e x + e− x

i.e.

3e x − 5e− x = 0

Hence, Thus,

ex 5 = e− x 3

i.e.

2x = ln

4e x − 4e− x − e x − e− x = 0

and

3e x = 5e− x

and

e2 x =

from which, 5 3

and

x=

5 3

1 5 ln = 0.2554 2 3

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

⎛ 35 ⎞ (a) When x = 35, y = 56 ch ( x / 56) = 56 ch ⎜ ⎟ = 67.30, using a calculator. ⎝ 56 ⎠ (b) When, y = 62.35, then

62.35 = 56 ch ( x / 56) x

62.35 x = ch 56 56

Thus, x 56

e +e

i.e.

−

x 56

e 56 + e 2

or

−

x 56

=

62.35 56

⎛ 62.35 ⎞ = 2⎜ ⎟ = 2.22679 ⎝ 56 ⎠

2

Thus,

x ⎛ 56x ⎞ ⎛ 56x ⎞ ⎛ − 56x ⎞ 56 ⎜ e ⎟ + ⎜ e ⎟ ⎜ e ⎟ − 2.22679e = 0 ⎝ ⎠ ⎝ ⎠⎝ ⎠

i.e.

x ⎛ 56x ⎞ 56 ⎜ e ⎟ − 2.22679e + 1 = 0 ⎝ ⎠

2

from which,

e

x 56

2 − ( −2.22679 ) ± ⎡( −2.22679 ) − 4 (1)(1) ⎤ 2.22679 ± 0.95859... ⎣ ⎦ = = 2 (1) 2

= 1.60293 or 0.623857 Hence, i.e.

x = ln1.60293 56

x = ln 0.623857 56

or

x = 56 ln 1.60293 = 26.42

or

x = 56 ln 0.623857 = -26.42, which is not possible.


46

EXERCISE 27 Page 49

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

(a) sh 3x = ( 3 x )

( 3x ) + 3!

3

( 3x ) +

5

5!

27 3 3 × 3 × 3 × 3 × 3 5 x + x 6 5 × 4 × 3 × 2 ×1

+ ... = 3x + = 3x +

(b)

( 2x ) ch 2x = 1 +

2

2!

(2x) + 4!

9 3 81 5 x + x as far as the term in x5 2 40

4

+ ... = 1 + 2 x 2 +

16 4 x + ... 24 2 4 x as far as the term in x 4 3

= 1 + 2 x2 +

7 31 5 4. Prove the identity: sh 2θ − sh θ ≡ θ + θ 3 + θ as far as the term in θ 5 only. 6 120 3 5 ⎛ ⎞ ⎛ 2θ ) ( 2θ ) ⎞ ( θ3 θ5 L.H.S. = sh 2θ − sh θ ≡ ⎜ 2θ + + + ... ⎟ − ⎜ θ + + + ... ⎟ ⎜ ⎟ ⎝ 3! 5! 3! 5! ⎠ ⎝ ⎠

8 32 5 1 1 5 ⎛ ⎞ ⎛ ⎞ θ + ... ⎟ − ⎜ θ + θ 3 + θ + ... ⎟ = ⎜ 2θ + θ 3 + 6 120 6 120 ⎝ ⎠ ⎝ ⎠ 1 ⎞ 5 ⎛8 1⎞ ⎛ 32 5 = ( 2θ − θ ) + ⎜ − ⎟ θ 3 + ⎜ − ⎟ θ as far as the term in θ only ⎝6 6⎠ ⎝ 120 120 ⎠ 7 31 5 = θ + θ3 + θ = R.H.S. 6 120

5. Prove the identity: 2 sh

θ 2

− ch

θ 2

≡ −1 + θ −

θ2 8

+

θ3

−

θ4

+

θ5

24 384 1920

as far as the term in θ 5 only.

⎛ θ (θ / 2 )3 (θ / 2 )5 ⎞ ⎛ (θ / 2 )2 (θ / 2 )4 ⎞ L.H.S. = 2 sh − ch ≡ 2 ⎜ + + + ... ⎟ − ⎜1 + + + ... ⎟ ⎜2 ⎟ ⎜ ⎟ 2 2 3! 5! 2! 4! ⎝ ⎠ ⎝ ⎠

θ

θ

⎛ ⎞ ⎛ 1 ⎞ 2 2 1 θ 5 ⎟ − ⎜1 + θ 2 + θ4⎟ = ⎜θ + θ 3 + 5 48 2 (120) ⎠ ⎝ 8 (16)(24) ⎠ ⎝ = −1 + θ −

θ2 8

+

θ3 24

+

θ4 384

+

θ5 1920

as far as the term in θ 5 only


47

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

25 4

from which,

d=

from which,

a = 29 – 37.5 = -8.5

⎛ 25 ⎞ Substituting in (1) gives: a + 6 ⎜ ⎟ = 29 ⎝ 4 ⎠ i.e.

a + 37.5 = 29

Hence, the 16th term is:

⎛ 25 ⎞ -8.5 + (16 – 1) ⎜ ⎟ = -8.5 + 93.75 = 85.25 ⎝ 4 ⎠

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) i.e.

22 =n–1 2.2

i.e.

from which, 10 = n – 1

29 – 7 = 2.2(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


48

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.

from which,

25.5 =n–1 1.5

32 – 6.5 = (n – 1)(1.5)

Sum of series, Sn =

and

6.5 + (n - 1)(1.5) = 32 i.e. 17 = n - 1

and

n = 18

n 18 [ 2a + (n − 1)d ] = [ 2(6.5) + (18 − 1)(1.5)] = 9 [13 + 25.5] = 346.5 2 2


49

EXERCISE 29 Page 53 2. Three numbers are in arithmetic progression. Their sum is 9 and their product is 20

1 . 4

Determine the three numbers.

Let the three numbers be (a – d), a and (a + d) Thus,

(a – d) + a + ( a + d) = 9

i.e. Also,

3a = 9

and

a=3

a(a – d)(a + d) = 20.25

Since, a = 3, then 3 ( 9 − d 2 ) = 20.25

20.25 = 6.75 3

i.e.

9 − d2 =

and

9 – 6.75 = d 2

from which,

d 2 = 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 of the series 5, 8, 11,… of which the sum is 1025

Sum of n is given by:

i.e. i.e. Hence, i.e.

Sn =

1025 =

n [ 2(5) + (n − 1)(3)] 2

2 ×1025 = n [10 + 3(n − 1) ]

2050 = n [10 + 3n − 3] = n [ 7 + 3n ] = 7n + 3n 2 3n 2 + 7n − 2050 = 0

This is a quadratic equation, hence i.e.

n [ 2a + (n − 1)d ] 2

n=

−7 ± 7 2 − 4(3)(−2050) −7 ± 24649 −7 ± 157 = = 2(3) 6 6

number of , n = 25 (the negative answer having no meaning)


50

6. The first, tenth and last of an arithmetic progression are 9, 40.5 and 425.5 respectively.

Find (a) the number of , (b) the sum of all the , 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 d=

hence

and

9d = 40.5 – 9 = 31.5

31.5 = 3.5 9

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 Hence,

n–1=

416.5 = 119 3.5

the number of , n = 120

(b) Sum of all the , Sn =

n 120 [ 2a + (n − 1)d ] = [ 2(9) + (120 − 1)(3.5)] = 60 [18 + 416.5] 2 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 , Thus, total cost, Sn =

i.e. a = 30, d = 2 and n = 80

n 80 [ 2a + (n − 1)d ] = [ 2(30) + (80 − 1)(2)] = 40 [ 60 + 158] = 40(218) = £8720 2 2


51

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: a r n −1 where a = 5, r = 2 and n = 10 i.e. the 10th term = a r n −1 = ( 5 )( 2 )

10 −1

= (5)(2)9 = 5(512) = 2560

3. The first term of a geometric progression is 4 and the 6th term is 128. Determine the 8th and 11th

.

The 6th term is given by:

a r 5 = 128

Thus,

r=

Hence, the 8th term is: and the 11th term is:

5

i.e. 4r 5 = 128

and

r5 =

128 = 32 4

32 = 2

a r n −1 = 4(2)8−1 = 4(2)7 = 4(128) = 512 4(2)11−1 = 4(2)10 = 4(1024) = 4096

1 4. Find the sum of the first 7 of the series 2, 5, 12 ,… (correct to 4 significant figures). 2

Common ratio, r =

ar 5 = = 2.5 a 2

Sum of 7 , Sn =

a ( r n − 1)

( r − 1)

(Also,

=

ar 2 12.5 = = 2.5) ar 5

2 ( 2.57 − 1)

( 2.5 − 1)

=

2 ( 610.35 − 1) = 812.5, correct to 4 significant 1.5 figures.

1 1 5 6. Find the sum to infinity of the series 2 , − 1 , ,..... 2 4 8

The series is a G.P. where r = − Hence, sum to infinity, S∞ =

1.25 = -0.5 and a = 2.5 2.5

2 a 2.5 2.5 5 = = = =1 3 1 − r 1 − (−0.5) 1.5 3


52

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

is 1944. Determine (a) the common ratio, (b) the first term, and (c) the sum of the 4th to 10th inclusive.

(a) The 5th term of a geometric progression is: ar 4 and the 3rd term is: ar 2 r4 Hence, ar 4 = 9 ar 2 from which, = 9 i.e. r2 = 9 r2 from which, the common ratio, r = 3 (b) The 6th term is ar 5 and the 7th term is ar 6 ar 5 + ar 6 = 1944

Hence, Since r = 3,

243a + 729a = 1944

i.e.

972a = 1944

and first term, a =

1944 =2 972

(c) Sum of the 4th to 10th inclusive is given by: S10 − S3 =

a ( r10 − 1)

( r − 1)

−

a ( r 3 − 1)

( r − 1)

=

2 ( 310 − 1) (3 − 1)

−

2 ( 33 − 1)

( 3 − 1)

= ( 310 − 1) − ( 33 − 1) = 310 − 33 = 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 = ( 0.976 ) × 55 million 2

Hence, population after 5 years = ( 0.976 ) × 55 = 48.71 million 5

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

53

(a) First term, a = £250,

common ratio, r = 1.06

Hence, value after 15 years = ar15 = (250) (1.06 ) = £599.14 15

(b) When £750 is reached, 750 = ar n i.e.

750 = 250 (1.06 )

and

750 = 1.06n 250

Taking logarithms gives: from which,

n

i.e.

3 = 1.06n

lg 3 = lg (1.06 ) = n lg1.06 n

n=

lg 3 = 18.85 lg1.06

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:

ar 8−1 = 1000

from which,

r7 =

1000 = 10 100

and r = 7 10 = 1.3895

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


54

CHAPTER 7 THE BINOMIAL SERIES EXERCISE 32 Page 59 1. Expand ( 2a + 3b ) using Pascal’s triangle 5

From page 58 of textbook,

(a + x )

5

= a 5 + 5a 4 x + 10a 3 x 2 + 10a 2 x 3 + 5a x 4 + x 5

Thus, replacing a with 2a and x with 3b gives:

( 2a + 3b )

5

= ( 2a ) + 5 ( 2a ) ( 3b ) + 10 ( 2a ) ( 3b ) + 10 ( 2a ) ( 3b ) + 5 ( 2a )( 3b ) + ( 3b ) 5

4

3

2

2

3

4

5

= 32a5 + 240a4b + 720a 3b 2 + 1080a 2b 3 + 810ab 4 + 243b 5


55

EXERCISE 33 Page 61 1. Use the binomial theorem to expand ( a + 2 x )

( a + 2x)

= a 4 + 4a 3 ( 2x ) +

4

( 4 )( 3) a 2 2!

( 2x )

2

+

4

( 4 )( 3)( 2 ) a 3!

( 2x ) + ( 2x ) 3

4

= a4 + 8a 3 x + 24a 2 x 2 + 32a x 3 + 16x 4

3. Expand ( 2 x − 3 y )

( 2x − 3y )

4

4

= ( 2 x ) + 4 ( 2 x ) ( −3 y ) + 4

3

( 4 )( 3) 2!

( 2 x ) ( −3 y ) 2

2

+

( 4 )( 3)( 2 ) 3!

( 2 x )( −3 y ) + ( −3 y ) 3

4

= 16 x 4 − 96 x 3 y + 216 x 2 y 2 − 216 xy 3 + 81 y 4

2⎞ ⎛ 4. Determine the expansion of ⎜ 2 x + ⎟ x⎠ ⎝

5

( 5)( 4 ) 2 x 3 ⎛ 2 ⎞ + ( 5)( 4 )( 3) 2 x 2 ⎛ 2 ⎞ 2⎞ 5 4⎛2⎞ ⎛ ( )⎜ ⎟ ( )⎜ ⎟ ⎜ 2x + ⎟ = ( 2x) + 5 ( 2x ) ⎜ ⎟ + 2! 3! x⎠ ⎝ ⎝ x⎠ ⎝ x⎠ ⎝ x⎠ 5

2

3

( 5)( 4 )( 3)( 2 ) + 4!

= 32 x 5 + 160 x 3 + 320 x +

4

2 2 ( 2 x ) ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ ⎝ x⎠ ⎝ x⎠

5

320 160 32 + 3 + 5 x x x 13

q⎞ ⎛ 6. Determine the sixth term of ⎜ 3 p + ⎟ 3⎠ ⎝ The 6th term of

(a + x)

n

is

n ( n − 1)( n − 2 ) ..... to (r − 1) n −( r −1) r −1 a x ( r − 1)! 13

q⎞ ⎛ Hence, the 6 term of ⎜ 3 p + ⎟ is: 3⎠ ⎝ th

(13)(12 )(11)(10 )( 9 ) 5!

(3 p )

13− 5

5

8⎛q⎞ ⎛q⎞ ⎜ ⎟ = 1287 ( 3 p ) ⎜ ⎟ ⎝3⎠ ⎝3⎠

5

= 34749 p8 q 5


56

9. Use the binomial theorem to determine, correct to 5 significant figures:

( 0.98)

(a)

7

(b)

( 2.01)

9

(a)

( 0.98)

7

= (1 − 0.02 ) = 1 + 7 ( −0.02 ) + 7

( 7 )( 6 ) 2!

( −0.02 )

2

+

( 7 )( 6 )( 5) 3!

( −0.02 )

3

+

( 7 )( 6 )( 5)( 4 ) 4!

( −0.02 )

4

= 1 – 0.14 + 0.0084 – 0.00028 + 0.0000056 - … = 0.86813, correct to 5 significant figures. 9

9 ⎛ 0.01 ⎞ 9 (b) ( 2.01) = ( 2 + 0.01) = 2 ⎜1 + ⎟ = 2 (1 + 0.005 ) 2 ⎠ ⎝ 9

9

9

⎡ ( 9 )( 8) (0.005)2 + ( 9 )( 8)( 7 ) (0.005)3 + ...⎤ = 29 ⎢1 + 9(0.005) + ⎥ 2! 3! ⎣ ⎦

= 29 (1 + 0.045 + 0.0009 + 0.0000105 + ...) = 29 (1.0459105) = 535.51, correct to 5 significant figures.


57

+ ..

EXERCISE 34 Page 63

2. Expand

1

(1+ x )

in ascending powers of x as far as the term in x3 , using the binomial theorem.

2

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

1

(1 + x )

= (1 + x ) = 1 − 2 x + −2

2

( −2 )( −3) 2!

( x)

2

+

( −2 )( −3)( −4 ) 3!

= 1 − 2 x + 3 x 2 − 4 x 2 + ...

3. Expand

1

(2 + x)

3

and

( x)

3

+ ...

x <1

in ascending powers of x as far as the term in x3 , using the binomial theorem.


1

(2 + x)

3

−3 2 3 ⎡ ⎤ −3 ⎛ x⎞ ⎛ x ⎞ ( −3)( −4 ) ⎛ x ⎞ ( −3)( −4 )( −5 ) ⎛ x ⎞ + = ( 2 + x ) = 2−3 ⎜1 + ⎟ = 2−3 ⎢1 − 3 ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ...⎥ 2! 3! ⎝ 2⎠ ⎝2⎠ ⎝2⎠ ⎝ 2⎠ ⎢⎣ ⎥⎦

The series is true provided

5. Expand

3 2 5 3 ⎡ 3 ⎤ ⎢⎣1 − 2 x + 2 x − 4 x + ...⎥⎦

=

1 23

=

1⎡ 3 3 5 ⎤ 1 − x + x 2 − x 3 + ...⎥ ⎢ 8⎣ 2 2 4 ⎦

x <1 2

i.e.

x <2

1 in ascending powers of x as far as the term in x 3 , using the binomial theorem. 1 + 3x


1 = (1 + 3 x ) 1 + 3x =

1 − 2

⎛ 1 ⎞⎛ 3 ⎞ ⎛ 1 ⎞⎛ 3 ⎞ ⎛ 5 ⎞ − ⎟⎜ − ⎟ ⎜ ⎜ − ⎟⎜ − ⎟ ⎜ − ⎟ 2 3 2 ⎠⎝ 2 ⎠ ⎛ 1⎞ ⎝ = 1 + ⎜ − ⎟ (3 x) + ( 3x ) + ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ( 3x ) ... 2! 3! ⎝ 2⎠

1−

3 27 2 135 3 x+ x − x as far as the term in x3 2 8 16

The series is true provided 3 x < 1

i.e.

x <

1 3


58

1

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

(b)

(a)

(b)

(1 − 2 x ) 4 (1 − 3x )

≈ 1 + 10 x

1

(1 − x ) (1 − x ) 2

(1 − x ) (1 − x ) 2

1 + 5x

(c) 3

= (1 − x )

−2

(1 − x )

−

1 2

(1 − 2 x )

≈ 1+

≈ 1+

5 x 2

19 x 6

⎛ x⎞ ≈ (1 + 2 x ) ⎜1 + ⎟ ⎝ 2⎠ ≈ 1+

x + 2 x ignoring the x 2 term and above 2

≈ 1+

5 x 2

(1 − 2 x ) = 1 − 2 x 1 − 3x −4 ≈ 1 − 2 x 1 + 12 x ( )( ) ( )( ) 4 (1 − 3x ) ≈ 1 + 12x – 2x ignoring the x 2 term and above ≈ 1 + 10x 1 + 5x

(c) 3

(1 − 2 x )

1

= (1 + 5 x ) 2 (1 − 2 x )

−

1 3

⎛ 5 ⎞⎛ 2 ⎞ ≈ ⎜1 + x ⎟⎜ 1 + x ⎟ ⎝ 2 ⎠⎝ 3 ⎠

2 5 x + x ignoring the x 2 term and above 3 2 19 ⎛ 5 2 15 + 4 19 ⎞ ≈ 1+ x = ⎟ ⎜ + = 6 6 6⎠ ⎝2 3 ≈ 1+

9. Express the following as power series in ascending powers of x as far as the term in x 2 . State in

each case the range of x for which the series is valid.

(a)

(a)

⎛ 1− x ⎞ ⎜ ⎟ ⎝ 1+ x ⎠

(b)

(1 − x ) 3 (1 − 3x )

2

(1 + x ) 2

⎛ 1− x ⎞ ⎜ ⎟ = (1 − x ) (1 + x ) ⎝ 1+ x ⎠ 1 2

1 − 2

⎡ ⎤⎡ ⎛ 1 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 3 ⎞ ⎤ ⎜ − ⎟⎜ − ⎟ ⎥ ⎢ x ⎜ 2 ⎟⎜ − 2 ⎟ ⎥ ⎢ ⎠ − x 2 ⎥ ⎢1 − x + ⎝ 2 ⎠⎝ 2 ⎠ ⎥ as far as the term ≈ ⎢1 − + ⎝ ⎠ ⎝ ( ) 2! 2! ⎢ 2 ⎥⎢ 2 ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ in x 2


59

⎛ x x 2 ⎞ ⎛ x 3x 2 ⎞ x 3x 2 x x 2 x 2 = ⎜1 − − ⎟ ⎜1 − + ≈ 1 − + − + − ⎟ 2 8 2 4 8 ⎝ 2 8 ⎠⎝ 2 8 ⎠

≈ 1− x + The series is valid if

(b)

(1 − x ) 3 (1 − 3x )

(1 + x ) 2

2

x2 2

as far as the term in x 2

x <1

= (1 + x )(1 − 3x ) 3 (1 + x 2 ) 2

−

1 2

⎡ ⎤ ⎛ 2 ⎞⎛ 1 ⎞ ⎜ ⎟⎜ − ⎟ ⎢ ⎥ ⎛ x2 ⎞ 2 3 3⎠ ≈ (1 + x ) ⎢1 − 2 x + ⎝ ⎠ ⎝ ( −3x ) + ..⎥ ⎜1 − + .. ⎟ as far as the term in x 2 2! 2 ⎢ ⎥⎝ ⎠ ⎢⎣ ⎥⎦ ⎛ x2 ⎞ ≈ (1 + x ) (1 − 2 x − x 2 ) ⎜ 1 − ⎟ 2⎠ ⎝ ⎛ x2 ⎞ ≈ (1 − 2 x − x 2 + x − 2 x 2 ) ⎜ 1 − ⎟ as far as the term in x 2 2⎠ ⎝ ⎛ x2 ⎞ x2 ≈ (1 − x − 3 x ) ⎜ 1 − ⎟ ≈ 1 − − x − 3 x 2 neglecting x3 and above 2⎠ 2 ⎝ 2

≈ 1− x −

The series is valid provided

7 2 x 2

3x < 1

i.e.

x <

1 3


60

EXERCISE 35 Page 65

1 2 mv . Determine the approximate change in the kinetic energy 2 when mass m is increased by 2.5% and the velocity v is reduced by 3%

2. Kinetic energy is given by

New mass = 1.025 m = (1 + 0.025)m New kinetic energy =

and new velocity = 0.97 v = (1 – 0.03)v

1 1 (1 + 0.025)m (1 − 0.03) 2 v 2 ≈ mv 2 (1 + 0.025)(1 − 0.06) 2 2 ≈

i.e.

1 2 1 mv (1 + 0.025 − 0.06) = mv 2 (0.965) 2 2

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.

4 (a) Volume of sphere, V = π r 3 . 3 4 4 4 4 New volume = π (1.015r )3 = π r 3 (1 + 0.015)3 ≈ π r 3[1 + 3(0.015)] = π r 3 (1 + 0.045) 3 3 3 3 = 1.045 V i.e.

the volume has increased by 4.5%

(b) Surface area of sphere, A = 4π r 2 New surface area = 4π (1 + 0.015 ) r 2 ≈ 4π r 2 [1 + 2(0.015)] = 4π r 2 (1 + 0.03) 2

= 1.03 A i.e.

the surface area has increased by 3%


61

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: H =

M 2l

⎧⎪ 1 1 ⎫⎪ − ⎨ 2 2⎬ ⎪⎩ ( x − l ) ( x + l ) ⎭⎪

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

H=

M 2l

⎪⎧ 1 ⎨ 2 ⎪⎩ ( x − l )

⎧ ⎪ 1 ⎪⎫ M ⎪ 1 1 − = − ⎨ 2⎬ 2 ( x + l ) ⎪⎭ 2l ⎪ x 2 ⎛1 − l ⎞ x 2 ⎛1 + ⎜ ⎪⎩ ⎜⎝ x ⎟⎠ ⎝ ≈

2M x3

⎫ −2 −2 ⎪ ⎪ M ⎪⎧⎛ l ⎞ ⎛ l ⎞ ⎪⎫ = 2 ⎨⎜1 − ⎟ − ⎜1 + ⎟ ⎬ 2⎬ ⎝ x ⎠ ⎭⎪ l ⎞ ⎪ 2 x l ⎪⎩⎝ x ⎠ ⎟ ⎪ x⎠ ⎭

M ⎧⎛ 2l ⎞ ⎛ 2l ⎞ ⎫ M ⎧ 4l ⎫ 2M ⎨⎜1 + ⎟ − ⎜ 1 − ⎟ ⎬ ≈ 2 ⎨ ⎬ ≈ 3 x x⎠ ⎝ x ⎠⎭ 2 x l ⎩ x ⎭ 2 x 2l ⎩⎝

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

kT . Determine π D3

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 Hence, new shear stress =

≈

and

new value of D = (1 + 0.015)D

k (1 − 0.03)T kT ⎡ −3 = 1 − 0.03)(1 + 0.015 ) ⎤ 3 3 3 ⎣( ⎦ π (1 + 0.015) D π D kT kT ⎡ 1 − 0.03)(1 − 0.045 ) ⎤⎦ ≈ [1 − 0.03 − 0.045] 3 ⎣( πD π D3 ≈

i.e.

kT (1 − 0.075 ) = τ (1 − 0.075) π D3

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:

fr =

1 2π LC

. 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.


62

New value of inductance = (1 + 0.026)L Hence, new resonant frequency =

=

and new value of capacitance = (1 – 0.008)C

1 1 = 2π (1 + 0.026) L(1 − 0.008)C 2π (1 + 0.026) L (1 − 0.008)C 1 1 1 −1 − − 1 − (1 + 0.026 ) 2 L 2 (1 − 0.008) 2 C 2 2π

1

≈

1 2

2π L C

1 2

⎡⎣(1 − 0.013)(1 + 0.004 ) ⎤⎦ ≈

1 2π LC

(1 − 0.013 + 0.004 )

≈ f r (1 − 0.009 )

i.e.

the new resonant frequency is 0.9% smaller

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

k r4 , where k is a constant. If there is an error in r νl

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 =

k (1 + 0.02) 4 r 4 k r4 ⎡ 4 −1 −1 = 1 + 0.02 ) (1 + 0.04 ) (1 − 0.03) ⎤ ( ⎦ (1 + 0.04)ν (1 − 0.03) l ν l ⎣ k r4 ≈ ⎡(1 + 0.08 )(1 − 0.04 )(1 + 0.03) ⎤⎦ νl ⎣ ≈

i.e.

k r4 (1 + 0.08 − 0.04 + 0.03) ≈ η (1 + 0.07 ) νl

the viscosity increases by 7%

12. The flow of water through a pipe is given by: G =

( 3d )

5

H

L

. 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


63

Hence, new value of G =

( 3d )

5

5

H

=

L 5 2

1

35 d 2 H 2 1 2

5

=

5

1

1

35 (1 − 0.02) 2 d 2 (1 − 0.01) 2 H 2

L

1

L2

1 2

≈

3 d H ⎡⎛ 5 ⎞⎛ 1 ⎞⎤ 1 ⎢⎜ 1 − 2 (0.02) ⎟ ⎜ 1 − 2 (0.01) ⎟ ⎥ ⎠⎝ ⎠⎦ ⎣⎝ L2

≈

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

5

≈ G (1 − 0.055) i.e.

the flow G has decreased by 5.5%


64

CHAPTER 8 MACLAURIN’S SERIES EXERCISE 36 Page 70

1. Determine the first four 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

f iv ( x) = 16 sin 2x

f iv (0) = 16 sin 0 = 0

f v ( x) = 32 cos 2x

f v (0) = 32 cos 0 = 32

f vi ( x) = -64 sin 2x

f vi (0) = -64 sin 0 = 0

f

vii

( x) = -128 cos 2x

f

vii

(0) = -128 cos 0 = - 128

x2 x3 Maclaurin’s series states: f(x) = f(0) + x f ′(0) + f ′′(0) + f ′′′(0) + …. 2! 3! = 0 + x(2) +

x2 x3 x4 x5 x6 x7 (0) + (−8) + (0) + (32) + (0) + (−128) 2! 3! 4! 5! 6! 7!

8 x 3 32 x 5 128 x 7 + − 6 120 5040

i.e.

f(x) = 2 x −

i.e.

sin 2x = 2 x −

4 3 4 5 8 7 x + x − x 3 15 315

3. Use Maclaurin’s series to determine the first three of the power series for ln (1 + e x )

Let f(x) = ln (1 + e x ) ex f ′(x) = 1 + ex

f(0) = ln (1 + e0 ) = ln 2 e0 1 = f ′(0) = 0 1+ e 2


65

(1 + e ) e − e ( e ) f ′′(x) = (1 + e ) x

x

x

x

x 2

(1 + e ) e − e ( e ) = 2 − 1 = 1 f ′′(0) = 2 4 (1 + e ) 0

0

0

0

0 2

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

2

x2 x3 f ′′(0) + f ′′′(0) + …. 2! 3!

2 ⎛1⎞ x ⎛1⎞ = ln 2 + x ⎜ ⎟ + ⎜ ⎟ + ... ⎝ 2 ⎠ 2! ⎝ 4 ⎠

ln (1 + e

i.e.

x

)

x x2 = ln 2 + + 2 8

6. Develop, as far as the term in x 4 , the power series for sec 2x.

Let f(x) = sec 2x f ′(x) = 2 sec 2x tan 2x

f(0) = sec 0 = 1 f ′(0) = 0

f ′′(x) = ( 2sec 2 x ) ( 2sec 2 2 x ) + ( tan 2 x )( 4sec 2 x tan 2 x ) = 4sec 2 x ⎡⎣sec 2 2 x + tan 2 2 x ⎤⎦ = 4sec 2 x ⎡⎣sec 2 2 x + sec 2 2 x − 1⎤⎦ = 4sec 2 x ⎡⎣ 2sec 2 2 x − 1⎤⎦ = 8sec3 2 x − 4sec 2 x

f ′′(0) = 8 – 4 = 4

f ′′′(x) = 24sec 2 2 x ( 2sec 2 x tan 2 x ) − 8sec 2 x tan 2 x = 48sec3 2 x tan 2 x − 8sec 2 x tan 2 x

f ′′′(0) = 0

f iv ( x) = ⎡⎣( 48sec3 2 x )( 2sec 2 2 x ) + ( tan 2 x ) (144sec2 2 x(2sec 2 x tan 2 x) ) ⎤⎦ − ⎡⎣( 8sec 2 x ) ( 2sec 2 2 x ) + ( tan 2 x )(16sec 2 x tan 2 x ) ⎤⎦ f iv (0) = 96 + 0 – 16 – 0 = 80 x2 x3 Maclaurin’s series states: f(x) = f(0) + x f ′(0) + f ′′(0) + f ′′′(0) + …. 2! 3! x2 x3 x4 = 1 + x(0) + (4) + (0) + (80) 2! 3! 4! = 1 + 2x2 +

i.e.

sec 2x = 1 + 2 x 2 +

80 4 x 24 10 4 x as far as the term in x 4 3


66

7. Expand e 2θ cos 3θ as far as the term in θ 2 using Maclaurin’s series.

Let f(θ) = e2θ cos 3θ

f(0) = e0 cos 0 = 1

f ′(θ) = ( e 2θ ) ( −3sin 3θ ) + ( cos 3θ ) ( 2e 2θ ) = e 2θ ( 2 cos 3θ − 3sin 3θ )

f ′(0) = e0 ( 2 cos 0 − 3sin 0 ) = 2

f ′′(θ) = ( e 2θ ) ( −6sin 3θ − 9 cos 3θ ) + ( 2 cos 3θ − 3sin 3θ ) ( 2e2θ ) Maclaurin’s series states: f(θ) = f(0) + θ f ′(0) +

θ2

= 1 + θ (2) +

2!

2!

f ′′(0) +

θ3 3!

f ′′′(0) + ….

(−5)

5 e 2θ cos 3θ = 1 + 2θ − θ 2 2

i.e.

θ2

f ′′(0) = - 9 + 4 = -5

as far as the term in θ2

8. Determine the first three of the series for sin 2 x by applying Maclaurin’s theorem.

Let f(x) = sin 2 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) = −2sin 2 x + 2 cos 2 x = 2 ( cos 2 x − sin 2 x ) = 2 cos 2 x f ′′′(x) = -4 sin 2x

f ′′′(0) = -4 sin 0 = 0

f iv ( x) = -8 cos 2x

f iv (0) = -8 cos 0 = -8

f v ( x) = 16 sin 2x

f v (0) = 16 sin 0 = 0

f vi ( x) = 32 cos 2x

f vi (0) = 32 cos 0 = 32

Maclaurin’s series states: f(x) = f(0) + x f ′(0) + = 0 + x(0) + i.e.

sin 2 x = x 2 −

f ′′(0) = 2 cos 0 = 2

x2 x3 f ′′(0) + f ′′′(0) + …. 2! 3!

x2 x3 x4 x5 x6 (2) + (0) + (−8) + (0) + (32) + ... 2! 3! 4! 5! 6!

1 4 2 6 x + x to three 3 45


67

EXERCISE 37 Page 72

∫

1. Evaluate

0.6 0.2

3esin θ dθ , correct to 3 decimal places, using Maclaurin’s series.

Let f(θ) = 3esinθ

f(0) = 3esin 0 = 3

f ′(θ) = ( 3esinθ ) ( cos θ )

f ′(0) = ( 3esin 0 ) ( cos 0 ) = 3

f ′′(θ) = ( 3esin θ ) ( − sin θ ) + ( cos θ ) ( 3esin θ cos θ ) = 3esin θ ( cos 2 θ − sin θ )

f ′′(0) = 3esin 0 ( cos 2 0 − sin 0 ) = 3

f ′′′(θ) = ( 3esinθ ) ( −2 cos θ sin θ − cos θ ) + ( cos 2 θ − sin θ )( 3esinθ cos θ ) f ′′′(0) = ( 3esin 0 ) ( −2 cos 0sin 0 − cos 0 ) + ( cos 2 0 − sin 0 )( 3esin 0 cos 0 ) = (3)(-1) + (1)(3) = 0 Maclaurin’s series states: f(θ) = f(0) + θ f ′(0) +

θ2 2!

f ′′(0) +

θ3 3!

f ′′′(0) + ….

3 = 3 + 3θ + θ 2 + 0 2 0.6

Hence,

∫

0.6 0.2

3e

sin θ

⎡ 3 ⎞ 3 θ3 ⎤ ⎛ dθ = ∫ ⎜ 3 + 3θ + θ 2 ⎟ dθ = ⎢3θ + θ 2 + ⎥ 0.2 2 ⎠ 2 2 ⎦ 0.2 ⎝ ⎣ 0.6

3 3 ⎡ 0.6 ) ⎤ ⎡ 0.2 ) ⎤ ( ( 3 3 2 2 ⎢3(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 decimal places,

From page XX,

Hence,

∫

0.5 0

∫

0.5 0

x ln( x + 1) as a power series. Hence evaluate, correct to 3

x ln(1 + x) dx

ln( x + 1) = x − x ln(1 + x) dx = ∫

0.5 0

x 2 x3 x 4 + − + .... 2 3 4 1 ⎛ ⎞ x 2 x3 x 4 x5 x 2 ⎜ x − + − + + ... ⎟ dx 2 3 4 5 ⎝ ⎠


68

=∫

0.5 0

5 7 9 11 13 ⎛ 3 ⎞ 2 2 2 2 2 x x x x x ⎜ x2 − + − + − ... ⎟ ⎜ ⎟ 2 3 4 5 6 ⎜ ⎟ ⎝ ⎠ 0.5

⎡ 5 ⎤ 7 9 11 13 15 ⎢ x2 ⎥ 2 2 2 2 2 x x x x x = ⎢ − + − + − + ...⎥ ⎢ 5 2 ⎛ 7 ⎞ 3 ⎛ 9 ⎞ 4 ⎛ 11 ⎞ 5 ⎛ 13 ⎞ 6 ⎛ 15 ⎞ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢⎣ 2 ⎥⎦ ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎝ 2⎠ 0

11 13 5 7 9 ⎡2 ⎤ 1 2 1 2 = ⎢ ( 0.5 ) 2 − ( 0.5 ) 2 + ( 0.5 ) 2 − (0.5) 2 + (0.5) 2 + ..⎥ − [ 0] 7 27 22 65 ⎣5 ⎦

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


69

EXERCISE 38 Page 74

⎧ x3 − 2 x + 1 ⎫ 1. Determine lim ⎨ 3 ⎬ x →1 2 x + 3 x − 5 ⎩ ⎭ ⎧ x3 − 2 x + 1 ⎫ ⎧ 3x 2 − 2 ⎫ 3 − 2 1 = = lim ⎨ 3 lim ⎬ ⎨ ⎬= x →1 2 x + 3 x − 5 x →1 6 x 2 + 3 ⎩ ⎭ ⎩ ⎭ 6+3 9 ⎧ x 2 − sin 3 x ⎫ 4. Determine lim ⎨ ⎬ 2 x →0 ⎩ 3x + x ⎭ ⎧ x 2 − sin 3 x ⎫ ⎧ 2 x − 3cos 3x ⎫ −3 = lim ⎨ = -1 lim ⎨ ⎬= ⎬ 2 x →0 ⎩ 3 x + x ⎭ x →0 ⎩ 3 + 2 x ⎭ 3

⎧ sin θ − θ cos θ ⎫ 5. Determine lim ⎨ ⎬ θ →0 θ3 ⎩ ⎭ ⎧⎪ cos θ − ⎣⎡(θ )( − sin θ ) + cos θ ⎦⎤ ⎫⎪ ⎧ sin θ − θ cos θ ⎫ ⎧θ sin θ ⎫ lim = lim ⎨ lim ⎨ = ⎨ ⎬ ⎬ 3 2 2 ⎬ θ →0 3θ θ ⎩ ⎭ θ →0 ⎪⎩ ⎪⎭ θ →0 ⎩ 3θ ⎭ 1 ⎧θ cos θ + sin θ ⎫ ⎧θ (− sin θ ) + cos θ (1) + cos θ ⎫ 1 + 1 2 = lim ⎨ = lim ⎨ = = = ⎬ ⎬ θ →0 3 6θ 6 6 6 ⎩ ⎭ θ →0 ⎩ ⎭

⎧ sinh x − sin x ⎫ 7. Determine lim ⎨ ⎬ x →0 x3 ⎩ ⎭ ⎧ sinh x − sin x ⎫ ⎧ cosh x − cos x ⎫ ⎧ sinh x + sin x ⎫ ⎧ cosh x + cos x ⎫ lim ⎨ ⎬ = lim ⎨ ⎬ = lim ⎨ ⎬ = lim ⎨ ⎬ 3 2 0 0 0 x →0 x → x → x → x 3x 6x 6 ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ =

1 1+1 2 = = 3 6 6

⎧ sin θ − 1 ⎫ 8. Determine lim ⎨ ⎬ π θ → ⎩ ln sin θ ⎭ 2 ⎧ ⎫ ⎪ ⎪⎪ cos θ π ⎪ ⎧ sin θ − 1 ⎫ = lim ⎨ lim = lim {sin θ } = sin = 1 ⎬ ⎨ ⎬ π π π 2 θ → ⎩ ln sin θ ⎭ θ→ ⎪⎛ 1 ⎞ θ→ 2 2 ⎜ 2 ⎟ cos θ ⎪ ⎪⎩ ⎝ sin θ ⎠ ⎪⎭


70

CHAPTER 9 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 39 Page 80 1. Find the positive root of the equation x 2 + 3x − 5 = 0 , correct to 3 significant figures, using the

method of bisection,

Let f(x) = x 2 + 3x − 5 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+ 2 = 1.5, i.e. the interval between 1 and 2 has been 2

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 + 1.5 i.e. 1.25 as the next root. 2

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 + 1.25 i.e. 1.125 2

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 + 1.25 i.e. 1.1875 2


71

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 + 1.25 i.e. 1.21875 2

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 + 1.21875 = 1.203125 2

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 Bisecting this interval gives

1.1875 + 1.203125 = 1.1953125 2

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 Bisecting this interval gives

1.1953125 + 1.1875 = 1.191406 2

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 Bisecting this interval gives

1.191406 + 1.1953125 = 1.193359 2

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 e x − x = 2 , correct to 4 significant figures


72

Let f(x) = e x − x − 2 then f(0) = 1 – 0 – 2 = -1 f(1) = e1 − 1 − 2 = −0.28 f(2) = e2 − 2 − 2 = 3.38 Hence, a root lies between x = 1 and x = 2. Let the root be x = 1.5 f(1.5) = e1.5 − 1.5 − 2 = 0.98169 Hence, a root lies between x = 1 and x = 1.5 due to a sign change. Bisecting this interval gives

1 + 1.5 = 1.25 2

f(1.25) = e1.25 − 1.25 − 2 = 0.240343 Hence, a root lies between x = 1 and x = 1.25 due to a sign change. Bisecting this interval gives

1 + 1.25 = 1.125 2

f(1.125) = e1.125 − 1.125 − 2 = −0.04478 Hence, a root lies between x = 1.125 and x = 1.25 due to a sign change. Bisecting this interval gives

1.125 + 1.25 = 1.1875 2

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

1.1875 + 1.125 = 1.15625 2


1.15625 + 1.125 = 1.140625 2

f(1.140625) = -0.011902 Hence, a root lies between x = 1.140625 and x = 1.15625 due to a sign change. © 2006 John Bird. All rights reserved. Published by Elsevier.

73

Bisecting this interval gives

1.140625 + 1.15625 = 1.1484375 2


1.1484375 + 1.140625 = 1.14453125 2

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

1.14453125 + 1.1484375 = 1.14648 2


1.14648 + 1.14453125 = 1.14551 2

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 e x − x = 2 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.


74

Table 1 x1

x2

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

⎛x +x ⎞ x3 = ⎜ 1 2 ⎟ ⎝ 2 ⎠ 3.5 3.25 3.125 3.1875 3.15625 3.140625 3.14844 3.145325 3.146486 3.14550925

f ( x3 )

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


75

EXERCISE 40 Page 83 2. Use an algebraic method of successive approximation to solve x3 − 2 x + 14 = 0 , correct to 3

decimal places.

Let

f(x) = x3 − 2 x + 14 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 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 ≈ −0.08884 18.28

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

76

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

( −2.6888)

3

+ 3 ( −2.6888 ) δ 2 + 5.3776 − 2δ2 + 14 ≈ 0 2

-19.439 + 21.6888δ2 + 5.3776 - 2δ2 + 14 ≈ 0 19.6888δ2 ≈ 19.439 – 5.3776 - 14 δ2 ≈

19.439 − 5.3776 − 14 ≈ 0.003119 19.6888

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 containing products of δ3 gives:

( −2.6857 )

3

+ 3 ( −2.6857 ) δ3 + 5.3714 − 2δ3 + 14 ≈ 0 2

-19.3719 + 21.63895δ3 + 5.3714 - 2δ3 + 14 ≈ 0 19.63895δ2 ≈ 19.3719 – 5.3714 - 14 δ2 ≈

19.3719 − 5.3714 − 14 ≈ 0.00002546 19.63895

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 x 4 − 3x3 + 7 x − 5.5 = 0 , correct to

3 significant figures.

Let

f(x) = x 4 − 3x3 + 7 x − 5.5 f(0) = -5.5 © 2006 John Bird. All rights reserved. Published by Elsevier.

77

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 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 ≈ 0.25 0.25

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 containing products of δ2 gives:

(1.75)

4

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

3

2

9.3789 + 21.4375δ2 – 16.078125 – 27.5625δ2 + 12.25 + 7δ2 – 5.5 ≈ 0 0.875δ2 ≈ -0.050775 δ2 ≈

−0.050775 ≈ −0.05803 0.875

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

78

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

(1.692 )

+ 4 (1.692 ) δ3 − 3[(1.692 ) + 3 (1.692 ) δ3 ] + 7(1.692 + δ3 ) − 5.5 ≈ 0

4

3

3

2

8.19599 + 19.37586δ3 – 14.5318977 – 25.76578δ3 + 11.844 + 7δ3 – 5.5 ≈ 0 0.61008δ3 ≈ -0.0080923 δ3 ≈

−0.0080923 ≈ −0.0132643 0.61008

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 containing products of δ4 gives:

(1.6787 )

4

+ 4 (1.6787 ) δ 4 − 3[(1.6787 ) + 3 (1.6787 ) δ 4 ] + 7(1.6787 + δ 4 ) − 5.5 ≈ 0 3

3

2

7.941314 + 18.9225δ4 – 14.1919 – 25.3623δ4 + 11.7509 + 7δ4 – 5.5 ≈ 0 0.5602δ4 ≈ -0.000314 δ4 ≈

−0.000314 ≈ −0.00056 0.5602

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) = x 4 − 3x3 + 7 x − 5.5 f(-2) = 16 + 24 – 14 – 5.5 = 20.5

f(0) = -5.5

f(-1) = 1 + 3 – 7 – 5.5 = -8.5

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.


79

EXERCISE 41 Page 85

2. Use Newton’s method to solve 3x3 − 10 x = 14 , correct to 4 significant figures.

Let f(x) = 3x3 − 10 x − 14

f(0) = -14

f(2) = 24 – 20 – 14 = -10

f(3) = 81 – 30 – 14 = 36

f(1) = 3 – 10 – 14 = -21

Let r1 = 2.2

Hence, a root lies between x = 2 and x = 3. A better approximation is given by: r2 = r1 −

Hence,

r2 = 2.2 −

f ( r1 ) f '(r1 )

f '(x) = 9x 2 − 10x

3(2.2)3 − 10(2.2) − 14 −4.056 = 2.2 − = 2.3881 2 9(2.2) − 10(2.2) 21.56

r3 = 2.3881 −

f (2.3881) 2.9771577 = 2.3881 − = 2.2796 f '(2.3881) 27.44619

r4 = 2.2796 −

f (2.2796) −1.257655 = 2.2796 − = 2.3321 f '(2.2796) 23.973184

r5 = 2.3321 −

f (2.3321) 0.7297097 = 2.3321 − = 2.3036 f '(2.3321) 25.627214

r6 = 2.3036 −

−0.3633356 = 2.3183 24.7231566

r7 = 2.3183 −

0.1962136 = 2.3105 25.187634

r8 = 2.3105 −

−0.10180935 = 2.3146 24.94069

r9 = 2.3146 −

0.05452775 = 2.3124 25.070358

r10 = 2.3124 −

−0.02944745 = 2.3136 25.0007438

r11 = 2.3136 −

0.0163322 = 2.3129 25.0387046

r12 = 2.3129 −

−0.01037987 = 2.313 25.0165577

Hence, correct to 4 significant figures, x = 2.313


80

4. Use Newton’s method to solve 3x 4 − 4 x3 + 7 x − 12 = 0 , correct to 3 decimal places.

Let f(x) = 3x 4 − 4 x3 + 7 x − 12

f(0) = -12

f(2) = 48 – 32 + 14 – 12 = 18

f(1) = 3 – 4 + 7 – 12 = -6

Hence, a root lies between x = 1 and x = 2

There are no further positive roots since the x 4 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 x 4 term predominates. For the root between x = 1 and x = 2:

Let r1 = 1.25

f '(x) = 12x 3 − 12x 2 + 7

3 (1.25 ) − 4 (1.25 ) + 7(1.25) − 12 f (r ) −3.73828 r2 = r1 − 1 = 1.25 − = 1.25 − = 1.56985 3 2 f '(r1 ) 11.6875 12 (1.25 ) − 12 (1.25 ) + 7 4

3

r3 = 1.56985 −

1.734046 = 1.49715 23.8522585

r4 = 1.49715 −

0.12925743 = 1.49081 20.3720909

r5 = 1.49081 −

0.000994417 = 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 r1 = −1.2 3 ( −1.2 ) − 4 ( −1.2 ) + 7(−1.2) − 12 f (r ) −7.2672 r2 = r1 − 1 = −1.2 − = −1.2 − = −1.4343 3 2 f '(r1 ) −31.016 12 ( −1.2 ) − 12 ( −1.2 ) + 7 4

r3 = −1.4343 −

2.4589815 = −1.3880 −53.094585

r4 = −1.3880 −

0.11488764 = −1.3856 −48.207045

3


81

r5 = −1.3856 −

0.00051387 = −1.3856 −47.960999

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

7. Use Newton’s method to solve 300e−2θ +

Let f(θ) = 300e−2θ + f(3) = -3.756

θ 2

−6

θ 2

= 6 , correct to 3 significant figures.

f(0) = 300 – 6 = 294

f(1) = 35.1

f(2) = 0.495

Hence, a root lies between x = 2 and x = 3, very close to x = 2.

There are no further positive roots since the 300e−2θ term predominates. There are no negative roots since f(x) will always be positive. f '(x) = −600e−2 θ + 0.5

Let r1 = 2

f (r1 ) 300e−4 + 1 − 6 0.494692 = 2− = 2− = 2.0492 r2 = r1 − −4 −600e + 0.5 −10.489383 f '(r1 ) r3 = 2.0492 −

−0.00436387 = 2.0497 −9.45952774

Hence, correct to 3 significant figures, x = 2.05

9. A Fourier analysis of the instantaneous value of a waveform can be represented by:

1 ⎛ π⎞ y = ⎜ t + ⎟ + sin t + sin 3t 8 ⎝ 4⎠ 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

1 ⎛ π⎞ 0.88 = ⎜ t + ⎟ + sin t + sin 3t 8 ⎝ 4⎠

or

1 ⎛ π⎞ ⎜ t + ⎟ + sin t + sin 3t − 0.88 = 0 8 ⎝ 4⎠

1 ⎛ π⎞ Let f(t) = ⎜ t + ⎟ + sin t + sin 3t − 0.88 8 ⎝ 4⎠ Let r1 = 0.04

3 f '(t) = 1 + cos t + cos 3t 8


82

f (r ) r2 = r1 − 1 = 0.04 − f '(r1 )

r3 = 0.03985 −

0.04 +

1 π + sin 0.04 + sin 3(0.04) − 0.88 0.0003515 4 8 = 0.04 − 3 2.371503345 1 + cos 0.04 + cos 3(0.04) 8 = 0.03985

−0.000004203 = 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:

y = −7.4e0.5t sin 3t

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) = −7.4e0.5t sin 3t f '(t) = ( −7.4e0.5t ) ( 3cos 3t ) + ( sin 3t ) ( −3.7e0.5t ) = −22.2e0.5t cos 3t − 3.7e0.5t sin 3t = −e0.5t ( 22.2 cos 3t + 3.7 sin 3t ) Let r1 = 4.2 r2 = r1 −

f (r1 ) −7.4e2.1 sin12.6 −2.03182922 = 4.2 − 2.1 = 4.2 − = 4.189 f '(r1 ) −e ( 22.2 cos12.6 + 3.7 sin12.6 ) −182.2023833 (Note, sin 12.6 is sin 12.6 rad)

r3 = 4.189 −

f (4.189) −0.037825 = 4.189 − = 4.189 f '(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:

λ 3 − 3.250λ 2 + λ − 0.063 = 0 Determine the value of λ which is approximately equal to 3.0 by Newton’s method, correct to 4 decimal places.


83

Let f(λ) = λ 3 − 3.250λ 2 + λ − 0.063 f '(λ) = 3λ 2 − 6.5λ + 1 Let

r1 = 3.0

( 3.0 ) − 3.250 ( 3.0 ) + 3.0 − 0.063 = 3.0 − 0.687 = 2.91918 f (3.0) r2 = 3.0 − = 3.0 − 2 f '(3.0) 8.5 3 ( 3.0 ) − 6.5 ( 3.0 ) + 1 3

r3 = 2.91918 −

0.0370604 = 2.91430 7.5901656

r4 = 2.91430 −

0.00015139 = 2.91428 7.5364835

2

Hence, correct to 4 decimal places, λ = 2.914


84

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) 101012 = 1× 24 + 0 × 23 + 1× 22 + 0 × 21 + 1× 20

= 16 + 0 + 4 + 0 + 1 = 2110 (b) 110012 = 1× 24 + 1× 23 + 0 × 22 + 0 × 21 + 1× 20 = 16 + 8 + 0 + 0 + 1 = 2510 (c) 1011012 = 1× 25 + 0 × 24 + 1× 23 + 1× 22 + 0 × 21 + 1× 20 = 32 + 0 + 8 + 4 + 0 + 1 = 4510 (d) 1100112 = 1× 25 + 1× 24 + 0 × 23 + 0 × 22 + 1× 21 + 1× 20 = 32 + 16 + 0 + 0 + 2 + 1 = 5110

4. Convert the following binary numbers to denary numbers:

(a) 11010.11 (b) 10111.011 (c) 110101.0111 (d) 11010101.10111 (a) 11010.112 = 1× 24 + 1× 23 + 0 × 22 + 1× 21 + 0 × 20 + 1× 2−1 + 1× 2−2 = 16 + 8 + 0 + 2 + 0 +

1 1 + = 26.7510 2 4

(b) 10111.0112 = 1× 24 + 0 × 23 + 1× 22 + 1× 21 + 1× 20 + 0 × 2−1 + 1× 2−2 + 1× 2−3 = 16 + 0 + 4 + 2 + 1 +

1 1 + = 23.37510 4 8

(c) 110101.01112 = 1× 25 + 1× 24 + 0 × 23 + 1× 22 + 0 × 21 + 1× 20 + 0 × 2−1 + 1× 2−2 + 1× 2−3 + 1× 2−4 = 32 + 16 + 0 + 4 + 0 + 1 +

1 1 1 + + =3 4 8 16


85

(d) 11010101.101112 = 1× 27 + 1× 26 + 0 × 25 + 1× 24 + 0 × 23 + 1× 22 + 0 × 21 + 1× 20 + 1× 2−1 + 0 × 2−2 + 1× 2−3 + 1× 2−4 + 1× 2−5 = 128 + 64 + 16 + 4 + 1 +

1 1 1 1 = 213.7187510 + + + 2 8 16 32


86

EXERCISE 43 Page 88

2. Convert the following denary numbers into binary numbers:

(a) 31 (b) 42 (c) 57 (d) 63

(a)

2 2 2 2 2

31 Remainder 15 1 7 1 3 1 1 1 0 1

Hence, (b)

2 2 2 2 2 2

42 Remainder 21 0 10 1 5 0 2 1 1 0 0 1

Hence, (c)

2 2 2 2 2 2

2 2 2 2 2 2

4210 = 101 0102

57 Remainder 28 1 14 0 7 0 3 1 1 1 0 1

Hence, (d)

3110 = 111112

5710 = 111 0012

63 Remainder 31 1 15 1 7 1 3 1 1 1 0 1

Hence,

6310 = 111 1112


87

4. Convert the following denary numbers into binary numbers:

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

2 2 2 2 2 2

47 Remainder 23 1 11 1 5 1 2 1 1 0 0 1

0.40625 × 2 0.8125 × 2 0.625 × 2 0.25 × 2 0.50 × 2

= = = = =

0.8125 1.625 1.25 0.50 1.00

Hence,

4710 = 101 1112

Hence,

0.4062510 = 0.011012

Thus, 47.4062510 = 101 111.011 012 (b)

2 2 2 2 2

30 Remainder 15 0 7 1 3 1 1 1 0 1

0.8125 × 2 0.625 × 2 0.25 × 2 0.50 × 2

= = = =

1.625 1.25 0.50 1.00

Hence,

3010 = 11 1102

Hence,

0.812510 = 0.11012

Hence,

5310 = 110 1012

Thus, 30.812510 = 11 110.110 12 (c)

2 2 2 2 2 2

53 Remainder 26 1 13 0 6 1 3 0 1 1 0 1

0.90625 × 2 0.8125 × 2 0.625 × 2 0.25 × 2 0.50 × 2

= = = = =

1.8125 1.625 1.25 0.50 1.00

Hence,

0.9062510 = 0.111 012

Thus, 53.9062510 = 110 101.111 012 © 2006 John Bird. All rights reserved. Published by Elsevier.

88

(d)

2 2 2 2 2 2

61 Remainder 30 1 15 0 7 1 3 1 1 1 0 1

0.65625 × 2 0.3125 × 2 0.625 × 2 0.25 × 2 0.50 × 2

= = = = =

1.3125 0.625 1.25 0.50 1.00

Hence,

6110 = 111 1012

Hence,

0.6562510 = 0.101 012

Thus, 61.6562510 = 111 101.101 012


89

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 8 30 8 3 0

Remainder 7 6 3

Hence, 24710 = 3678 = 011 110 1112 from Table 10.1, page 88

0.09375 × 8 = 0.75 0.75 × 8 = 6.00

Hence, 0.0937510 = 0.068 = 0.0001 102 from Table 10.1, page 88 Hence, 247.0937510 = 763.068 = 11 110 111.000 112

(b)

8 514 8 64 8 8 8 1 0

Remainder 2 0 0 Hence, 51410 = 10028 = 001 000 000 0102 from Table 10.1, page 88 1

0.4375 × 8 = 3.50 0.50 × 8 = 4.00

Hence, 0.437510 = 0.348 = 0.011 1002 from Table 10.1, page 88

Hence, 514.437510 = 1002.348 = 1000 000 010.01112

(c)

8 1716 8 214 8 26 8 3 0

Remainder 4 6 2 Hence, 171610 = 32648 = 011 010 110 1002 from Table 10.1, page 88 3

0.78175 × 8 = 6.25 0.25 × 8 = 2.00

Hence, 0.7812510 = 0.628 = 0.110 0102 from Table 10.1, page 88

Hence, 1716.7912510 = 3264.628 = 11010110 100.110 012

4. Convert the following binary numbers to denary numbers, via octal:

(a) 111.0111 (b) 101001.01 (c) 1110011011010.0011


90

(a) 111.0111 = 111.011 100 2 = 7 . 3

3 4 48 = 7 × 80 + 3 × 8−1 + 4 × 8−2 = 7 + + 2 = 7.437510 8 8

(b) 101001.01 = 101 001.010 2 =

5 1 . 28 = 5 × 81 + 1× 80 + 2 × 8−1 = 40 + 1 +

2 = 41.2510 8

(c) 1110011011010.0011 = 001 110 011 011 010.001 100 2 = 1 6 3 3 2 . 1 48 = 1× 84 + 6 × 83 + 3 × 82 + 3 × 81 + 2 × 80 + 1× 8−1 + 4 × 8−2 = 4096 + 3072 + 192 + 24 + 2 +

1 4 + = 7386.187510 8 64


91

EXERCISE 45 Page 92

2. Convert the hexadecimal number 2C16 into decimal.

2C16 = 2 ×161 + C ×160 = 32 + 12 = 4410

4. Convert the hexadecimal number 2F116 into decimal

2F116 = 2 × 162 + F × 161 + 1×160 = 2 × 162 + 15 × 161 + 1× 160 = 512 + 240 + 1 = 75310

6. Convert the decimal number 20010 into hexadecimal

16 200 Remainder 16 12 8 0 12 (i.e. C) Hence,

20010 = C816

8. Convert the decimal number 23810 into hexadecimal

16 238 Remainder 16 14 14 (i.e. E) 0 14 (i.e. E) Hence,

23810 = EE16

10. Convert the binary number 111010102 into hexadecimal

111010102 = 1110 1010 grouping in 4’s = E A 16 from Table 10.2, page 90. i.e. 111010102 = EA16


92

12. Convert the binary number 101001012 into hexadecimal

101001012 = 1010 0101 grouping in 4’s from Table 10.2, page 90. = A 5 16 i.e. 101001012 = A516

14. Convert the hexadecimal number ED16 into binary

ED16 = 1110 1101 2

from Table 10.2, page 90.

16. Convert the hexadecimal number A2116 into binary

A2116 = 1010 0010 0001 2

from Table 10.2, page 90.


93

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)]

(

Z = C. A.B + A.B

i.e.

)

The truth table is shown below: A 0 0 0 0 1 1 1 1

B 0 0 1 1 0 0 1 1

C 0 1 0 1 0 1 0 1

A.B 0 0 0 0 0 0 1 1

A .B 0 0 1 1 0 0 0 0

A 1 1 1 1 0 0 0 0

A.B + A .B 0 0 1 1 0 0 1 1

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

)

The truth table is shown below:


94

A 0 0 0 0 1 1 1 1

B 0 0 1 1 0 0 1 1

C 0 1 0 1 0 1 0 1

A 1 1 1 1 0 0 0 0

A. B 0 0 0 0 1 1 0 0

B 1 1 0 0 1 1 0 0

A. B + A 1 1 1 1 1 1 0 0

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

(

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

i.e.

)

The truth table is shown below: A B C A C 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

B.C. A

B+ C

A.( B + C )

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

Z

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.


95

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, from 9, Table 11.8

=G

(

)

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

(


)

= F. G + G + F = F.1 + F from 9 and 10, Table 11.8

=F+F=F 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

(

= 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

(

= G. F + F =G

)


) from 10, Table 11.8


96

(

) (

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

(

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

)


)

= R.P.Q + P


= P + R.Q

from 17, Table 11.8


97


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 = 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, from 15, Table 11.8, = ( A.B + A.C + B ) . ( A + B ) = ⎡⎣ B. ( A + 1) + A.C ⎤⎦ . ( A + B ) from 8, Table 11.8, = ( B + A.C ) . ( A + B ) = A+B . B+C . A+B

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


= 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 )

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


)


98

(

)(

= 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

(

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

(

(

= P. Q + R + R Q + Q

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

from 14 and 9, Table 11.8,

)

)) from 10, Table 11.8, from 9, Table 11.8


99


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


100

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 P.Q.R , i.e. P = 1, Q = 1 and R = 0, is shown on the Karnaugh map as a 2

P.Q.R , i.e. P = 1, Q = 1 and R = 1, is shown on the Karnaugh map as a 3 P.Q.R , i.e. P = 1, Q = 0 and R = 1, is shown on the Karnaugh map as a 4

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 A.B.C.D , i.e. A = 1, B = 0, C = 1 and D = 0, is shown as a 3

Two 2-cell couples are formed as shown.


101

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.B.C.D , i.e. A = 1, B = 1, C = 0 and D = 0, is shown as a 2 A.B.C.D , i.e. A = 1, B = 0, C = 0 and D = 0, is shown as a 3 A.B.C.D , i.e. A = 1, B = 1, C = 1 and D = 0, is shown as a 4 A.B.C.D , i.e. A = 1, B = 0, C = 1 and D = 0, is shown as a 5

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


102

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

(

)


103


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, Z1 = 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,

Z1 = A.B.C + A.B.C + A.B.C + A.B.C + A.B.C

simplifies to: Z1 = 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, Z3 = A.B.C + A.B.C + A.B.C + A.B.C + A.B.C The Karnaugh map is shown below.


104

The horizontal 2-cell couple corresponds to: A.C The 4-cell couple corresponds to: B Hence,

Z3 = A.B.C + A.B.C + A.B.C + A.B.C + A.B.C

simplifies to: Z 3 = B + A.C A logic circuit to meet these requirements is shown below.

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

)

A logic circuit to meet these requirements is shown below.


105

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

106

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

)



107

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. The Boolean expression to meet the requirements is: Z = A.(B + C + D)



108

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. The Boolean expression to meet the requirements is: 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. The Boolean expression to meet the requirements is: Z = A.B +A.C i.e.

Z = A.(B + C)



109

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, (38.47) 2 = (PQ) 2 + (QR) 2 = 2(PQ) 2

from which, Hence,

( PQ )

2

38.47 2 = 2

and

PQ =

38.47 2 38.47 = = 27.20 cm 2 2

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 =

(12

2

+ 17 2 ) = 20.81 km by Pythagoras.


110

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 2 A 'B = ⎡( A 'C ' ) − ( BC ' ) ⎤ = ⎣ ⎦

( 3.5

2

− 1.02 ) = 3.35 m by Pythagoras.

( 3.5

2

− 1.302 ) = 3.25 m

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


111

EXERCISE 53 Page 117 2. If cos A =

15 find sin A and tan A, in fraction form. 17

Since cos ine =

adjacent then the sides 15 and 17 are as shown in the diagram. hypotenuse

By Pythagoras, BC =

Hence,

sin A =

(17

2

− 152 ) = 8

8 opposite BC = = hypotenuse AC 17

and

tan A =

8 opposite BC = = adjacent AB 15

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 =

(b) Gradient of PQ =

(c) Tan θ =

(5

2

+ 82 ) = 9.434 by Pythagoras

1 − −4 5 = = - 0.625 −3 − 5 −8

5 ⎛5⎞ from which, the angle PQ makes with the horizontal, θ = tan −1 ⎜ ⎟ = 32° 8 ⎝8⎠


112


2. Solve triangle DEF

By Pythagoras, FE =

Tan E =

(4

2

+ 32 ) = 5 cm

4 4 from which, ∠E = tan −1 = 53.13° or 53°8′ 3 3

Hence, ∠F = 180° - 90° - 53.13° = 36.87° or 36°52′

4. Solve triangle JKL and find its area

Sin 51° =

6.7 6.7 = 8.62 cm from which, JL = JL sin 51°

Tan 51° =

6.7 6.7 = 5.43 cm from which, KL = KL tan 51°

∠J = 180° - 90° - 51° = 39° (Checking: JL =

( 6.7

2

Area of triangle JKL =

+ 5.432 ) = 8.62cm )

1 1 (KL)(JK) = (5.43)(6.7) = 18.19 cm 2 2 2

6. Solve triangle PQR and find its area


113

By Pythagoras, PR 2 + 3.692 = 8.752 from which, PR =

Sin R =

(8.75

2

− 3.692 ) = 7.934 m

3.69 ⎛ 3.69 ⎞ from which, ∠R = sin −1 ⎜ ⎟ = 24.94° or 24°57′ 8.75 ⎝ 8.75 ⎠

∠Q = 180° - 90° - 24.94° = 65.06° or 65°3′

Area of triangle PQR =

1 1 (PQ)(PR) = (3.69)(7.934) = 14.64 m 2 2 2

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° =

BC from which, height of building, BC = 2 tan 73° = 6.54 m 2


114


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 80 = 298.56 m from which, AC = AC tan15°

Tan 23° =

80 BC

from which, BC =

80 = 188.47 m tan 23°

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 from which, AC = 200 tan 32° = 124.97 m 200

Tan 30° =

BC from which, BC = 200 tan 30° = 115.47 m 200

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.


115

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° =

AB hence, width of road, AB = 4.2 tan 66° = 9.43 m 4.2

From triangle DEC, tan 34° =

CE CE CE = = from which, CE = 9.43 tan 34° = 6.36 m DE AB 9.43

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 from which, AB = BC tan 20° BC

Tan 24° =

AB 300 − BC

i.e.

from which, AB = (300 - BC) tan 24°

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 =

Tan 20° =

133.57 = 165.06 m 0.8092

AB from which, height of tower, AB = 165.06 tan 20° = 60 m, to the nearest metre 165.06


116

EXERCISE 56 Page 123 4.(a) Evaluate correct to 4 decimal places: secant 73°

secant 73° =

1 = 3.4203 cos 73°

5.(b) Evaluate correct to 4 decimal places: cosecant 15.62°

cosecant 15.62° =

1 = 3.7139 sin15.62°

6.(c) Evaluate correct to 4 decimal places: cotangent 321°23′

cotangent 321°23′ =

1 1 1 = = = -1.2519 tan 321°23' tan 321 23 ° tan 321.38333° 60

7.(a) Evaluate correct to 4 decimal places: sine

2π 3

2π 2π means sine radians = 0.8660 3 3 2π 180° 2π ⎡ 2π ⎤ ⎢⎣ 3 rad = 3 × π = 120° hence, sin 3 ≡ sin120° ⎥⎦

Note that sine

8.(c) Evaluate correct to 4 decimal places: cot 2.612

cot 2.612 =

1 = -1.7083 tan 2.612 rad

12. Determine the acute angle sec−1 1.6214 in degrees (correct to 2 decimal places), degrees and

minutes, and in radians (correct to 3 decimal places). ⎛ 1 ⎞ Using a calculator, sec−1 1.6214 = cos −1 ⎜ ⎟ ⎝ 1.6214 ⎠ = 51.92° or 55°55′ or 51.92 ×

π rad = 0.906 rad 180


117

13. Determine the acute angle cos ec −1 2.4891 in degrees (correct to 2 decimal places), degrees and

minutes, and in radians (correct to 3 decimal places). ⎛ 1 ⎞ Using a calculator, cos ec −1 2.4891 = sin −1 ⎜ ⎟ ⎝ 2.4891 ⎠ = 23.69° or 23°41′ or 23.69 ×

16. Evaluate, correct to 4 significant figures:

π rad = 0.413 rad 180

11.5 tan 49°11'− sin 90° 3cos 45°

11.5 tan 49°11'− sin 90° 11.5 tan 49.18333° − sin 90° 13.315 − 1 = 5.805 = = 3cos 45° 3cos 45° 2.2132 18. Evaluate, correct to 4 significant figures:

6.4 cos ec 29°5'− sec81° 2 cot12°

1 ⎛ ⎞ ⎛ 1 ⎞ 6.4 ⎜ ⎟−⎜ ⎟ 6.4 cos ec 29°5'− sec81° sin 29.08333° ⎠ ⎝ cos81° ⎠ 13.1665 − 6.39245 ⎝ = = = 0.7199 2 cot12° 9.40926 ⎛ 1 ⎞ 2⎜ ⎟ ⎝ tan12° ⎠

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 = tan −1 1.5276 = 56.79° sec x = sec 56.79° =

1 = 1.8258 cos 56.79°

cosec x = cosec 56.79° =

cot x = cot 56.79° =

1 = 1.1952 sin 56.79°

1 = 0.6546 tan 56.79°


( sin 34°27 ')( cos 69°2 ') ( 2 tan 53°39 ')

( sin 34°27 ')( cos 69°2 ') = ( sin 34.45° )( cos 69.03333° ) 2 tan 53.65° ( 2 tan 53°39 ')

= 0.07448


118


cos ec 27°19 '+ sec 45°29 ' 1 − cos ec 27°19 'sec 45°29 '

1 1 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟+⎜ ⎟ cos ec 27°19 '+ sec 45°29 ' 2.179086 + 1.426296 sin 27.31666° ⎠ ⎝ cos 45.48333° ⎠ = ⎝ = 1 1 1 − cos ec 27°19 'sec 45°29 ' ⎛ ⎞⎛ ⎞ 1 − (2.179086)(1.426296) 1− ⎜ ⎟⎜ ⎟ ⎝ sin 27.31666° ⎠ ⎝ cos 45.48333° ⎠ 3.60538 = -1.710 = −2.10802

25. Evaluate, correct to 5 significant figures: (a) cosec(-143°) (b) cot(-252°) (c) sec(-67°22′)

Using a calculator: (a) cosec(-143°) =

(b) cot(-252°) =

1 = -1.6616 sin(−143°)

1 = -0.32492 tan(−252°)

(c) sec(-67°22′) =

1 = 2.5985 cos ( −67.36666° )


119


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 c 8.60sin 56°32 ' = from which, c = = 7.568 cm sin 71°26 ' sin 56°32 ' sin 71°26 '

Also from the sine rule,

Area =

a 8.60 8.60sin 52°2 ' = from which, a = = 7.152 cm sin 52°2 ' sin 71°26 ' sin 71°26 '

1 1 a c sin B = (7.152)(7.568) sin 71°26 ' = 25.65 cm 2 2 2

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 25.4sin104°22 ' = 0.75477555 = from which, sin E = sin104°22 ' sin E 32.6 E = sin −1 0.75477555 = 49.0° or 49°0′

and Hence, ∠F = 180° - 104°22′ - 49°0′ = 26°38′ From the sine rule,

32.6 f 32.6sin 26°38' = from which, f = = 15.09 mm sin104°22 ' sin 26°38' sin104°22 '


120

Area =

1 1 d esin F = (32.6)(25.4) sin 26°38' = 185.6 mm 2 2 2

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 3.85sin 36° = 0.7006109 = from which, sin J = sin 36° sin J 3.23 J = sin −1 0.7006109 = 44.476° = 44°29′

and or

J = 180° - 44°29′ = 135°31′

Case 1: When J = 44°29′, ∠L = 180° - 36° - 44°29′ = 99°31′

From the sine rule,

Area =

l 3.23 3.23sin 99°31' = from which, l = = 5.420 cm sin 99°31' sin 36° sin 36°

1 1 l jsin K = (5.420)(3.85) sin 36° = 6.132 cm 2 2 2

Case 2: When J = 135°31′, ∠L = 180° - 135°31′ - 36° = 8°29′

From the sine rule,

Area =

l 3.23 3.23sin 8°29 ' = from which, l = = 0.811 cm sin 8°29 ' sin 36° sin 36°

1 1 j k sin L = (3.85)(3.23) sin 8°29 ' = 0.917 cm 2 2 2


121


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, p 2 = 4.422 + 3.252 − 2(4.42)(3.25) cos105° = 19.5364 + 10.5625 – (-7.4359) = 37.5348 and

p=

From the sine rule,

37.5348 = 6.127 m

6.127 4.42 4.42sin105° = 0.696816 = from which, sin R = sin105° sin R 6.127 R = sin −1 0.696816 = 44.172° or 44°10′

and ∠Q = 180° - 105° - 44°10′ = 30°50′ Area =

1 (4.42)(3.25) sin105° = 6.938 m 2 2

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, from which,

212 = 422 + 342 − 2(42)(34) cos X cos X =

422 + 342 − 212 = 0.867997 2(42)(34)


122

∠X = cos −1 (0.867997) = 29.77° or 29°46′

and From the sine rule,

21 34 34sin 29°46 ' = 0.8038070 = from which, sin Y = 21 sin 29°46 ' sin Y ∠Y = sin −1 0.8038070 = 53.495° or 53°30′

and Hence, ∠Z = 180° - 29°46′ - 53°30′ = 96°44′ Area =

1 (21)(34) sin 96°44 ' = 355 mm 2 2


123


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 m 2 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 (52.0)(34.0)sin A 2

from which, sin A =

620 1 (52.0)(34.0) 2

= 0.701357

∠A = sin −1 0.701357 = 44.54° or 44°32′

and By the cosine rule,

BC 2 = (52.0) 2 + (34.0) 2 − 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 =

and

1 (52.0)(36.6)sin B 2

from which, sin B =

2(620) (52.0)(36.6)

⎛ 2(620) ⎞ ∠B = sin −1 ⎜ ⎟ = 40°39′ ⎝ (52.0)(36.6) ⎠

∠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.


124

(a) Using the cosine rule on triangle PQR shown below gives: RQ 2 = 8.02 + 4.52 − 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 4.5sin130° = from which, sin R = = 0.3015923 sin R sin130° 11.43

and the angle between the jib and the tie rod, ∠R = sin −1 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 m 2 . Determine the length of the perimeter of the site.

The quadrilateral is split into two triangles as shown in the diagram below.

Area = 1510 = i.e.

1 1 (52.4)(28.5)sin 72° + (34.6)(x)sin 75° 2 2

1510 = 710.15 + 16.71 x


125

from which,

x=

1510 − 710.15 = 47.87 m 16.71

Hence, perimeter of quadrilateral = 52.4 + 28.5 + 34.6 + 47.9 = 163.4 m

5. Determine the length of BF and EB in the roof truss shown below.

Using the cosine rule on triangle ABF gives: BF2 = 2.52 + 52 − 2(2.5)(5) cos 50° = 15.18

BF = 15.18 = 3.9 m

from which, Using the sine rule on triangle ABF gives: 3.9 2.5 = sin 50° sin B

from which, sin B =

2.5sin 50° = 0.491054 3.9

∠ABF = sin −1 0.491054 = 29.41°

and

Assuming ∠ABE = 90°, then ∠FBE = 90° - 29.41° = 60.59° Using the sine rule on triangle BEF gives: 4 3.9 = sin 60.59° sin E

from which, sin E =

3.9sin 60.59° = 0.8493499 4

∠E = sin −1 0.8493499 = 58.14°

and

Thus, ∠EFB =180° - 58.14° - 60.59° = 61.27° Using the sine rule on triangle BEF again gives: BE 4 = sin 61.27° sin 60.59

from which, BE =

4sin 61.27° = 4.0 m sin 60.59°


126


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 ed 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, PR 2 = (20.0) 2 + (14.0) 2 − 2(20.0)(14.0) cos145° = 1054.725 from which, resultant phasor, PR = 1054.725 = 32.48 A Using the sine rule,

14.0 32.48 14.0sin145° = from which, sin P = = 0.247231 sin P sin145° 32.48 ∠P = sin −1 0.247231 = 14.31° or 14°19′

and

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

127

Using the cosine rule gives: 99.782 = 602 + 502 − 2(60)(50) cos θ cos θ =

from which,

602 + 502 − 99.782 = −0.642675 2(60)(50)

angle, θ = cos −1 (−0.642675) = 130°

and

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

∠C = sin −1 0.241045 = 13.95°

and

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 32.0 = sin126.05° sin 40°

from which, AC =

32.0sin126.05° = 40.25 cm sin 40°

Alternatively, using the cosine rule, AC2 = 12.02 + 32.02 − 2(12.0)(32.0) cos126.05° = 1619.9611 from which,

AC = 1619.9611 = 40.25 cm


128

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 12.0sin160° = from which, sin C = = 0.1282576 sin160° sin C 32.0 ∠C = sin −1 0.1282576 = 7.37°

and Hence, ∠AB′C′ =180° - 7.37° - 160° = 12.63° Using the sine rule again gives: AC ' 32.0 = sin12.63° sin160°

from which, AC′ =

32.0sin12.63° = 20.46 cm sin160°

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.


129

Tan 40° =

8000 8000 from which, OA = = 9534.03 m OA tan 40°

Tan 35° =

5500 5500 from which, OB = = 7854.81 m OB tan 35°

From the cosine rule, AB2 = (9534.03) 2 + (7854.81) 2 − 2(9534.03)(7854.81) cos 70° from which,

AB = 10068.24 m

From the view shown below, XY = 8000 – 5500 = 2500 m

⎛ 2500 ⎞ Hence, angle of descent, θ = tan −1 ⎜ ⎟ = 13.95° or 13°57′ ⎝ 10068.24 ⎠ Flight distance of aeroplane between observations, XZ =

⎡(10068.24 )2 + (2500) 2 ⎤ ⎣ ⎦

= 10373.98 m or 10.374 km Hence, speed of descent =

dis tan ce 10.374 km 10.374 × 60 × 60 = = km/h = 829.9 km/h 45 time 45 h 60 × 60


130

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, r = 6.182 + 2.352 = 6.61 θ = tan −1

and

y 2.35 π = 20.82° or 20.82 × rad = 0.36 rad = tan −1 x 6.18 180

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 radians.

From the diagram, r = 5.42 + 3.7 2 = 6.55 y 3.7 = tan −1 = 34.42° x 5.4

and

α = tan −1

Thus,

θ = 180° - α = 145.58°

or 145.58 ×

π rad = 2.54 rad 180

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

131

5. Express (-7, -3) as polar co-ordinates, correct to 2 decimal places, in both degrees and radians.

From the diagram, r = 7 2 + 32 = 7.62 3 = 23.20° 7

and

α = tan −1

Thus,

θ = 180° + 23.20° = 203.20°

or 203.20 ×

π rad = 3.55 rad 180

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 radians.

From the diagram, r = 9.62 + 12.42 = 15.68 12.4 = 52.25° 9.6

and

α = tan −1

Thus,

θ = 360° - 52.25° = 307.75°

or 307.75 ×

π rad = 5.37 rad 180

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.


132


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.


133

(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,

360° i.e. 72° apart. 5

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 =

( 27.64

2

+ 38.042 ) = 47.02 mm

i.e. the shortest distance between the centres of two adjacent holes is 47.02 mm.


134

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, 149.8 = πd

then

diameter, d =

and

149.8 = 47.68 cm π

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, from which,

0Y 2 = 0X 2 + XY 2

0Y =

( 0Y

2

− 0X 2 ) = 402 − 102 = 1500 = 38.73 cm


135

EXERCISE 64 Page 139 1. Convert to radians in of π: (a) 30° (b) 75° (c) 225°

(a) 30° = 30 ×

π π rad = rad 6 180

(b) 75° = 75 ×

5π π rad = rad 12 180

(c) 225° = 225 ×

5π π 45π 15π rad = rad = rad = rad 4 36 12 180

2. Convert to radians: (a) 48° (b) 84°51′ (c) 232°15′

(a) 48° = 48 ×

π rad = 0.838 rad 180

π ⎛ 51 ⎞ π = 84.95 × rad = 1.481 rad (b) 84°51′ = ⎜ 84 ⎟ × 180 ⎝ 60 ⎠ 180 (c) 232°15′ = 232.25 ×

π rad = 4.054 rad 180

3. Convert to degrees: (a)

5π 4π 7π rad (b) rad (c) rad 6 9 12

(a)

5π 5π 180° rad = × = 5 × 30 = 150° π 6 6

(b)

4π 4π 180° rad = × = 4 × 20 = 80° π 9 9

(c)

7π 7π 180° rad = × = 7 × 15 = 105° π 12 12

4. Convert to degrees and minutes: (a) 0.0125 rad (b) 2.69 rad (c) 7.241 rad 180° = 0.716° or 0°43′ π © 2006 John Bird. All rights reserved. Published by Elsevier.

(a) 0.0125 rad = 0.0125 ×

136

(b) 2.69 rad = 2.69 ×

180° = 154.126° or 154°8′ π

(c) 7.241 rad = 7.241×

180° = 414.879° or 414°53′ π


137


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 =

82 = 41 mm 2

(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.

π ⎞ ⎛ Arc length of pendulum bob, s = rθ = (1.5) ⎜10 × ⎟ = 0.262 m or 26.2 cm 180 ⎠ ⎝

5. Determine the angle of lap, in degrees and minutes, if 180 mm of a belt drive are in with a pulley of diameter 250 mm.

Arc length, s = 180 mm, radius, r =

Since s = rθ, the angle of lap, θ =

250 = 125 mm 2

s 180 180 = = 1.44 rad = 1.44 × = 82.506° or 82°30′ π r 125

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 =

2000 = 748.08 2.6735

Thus, number of complete revolutions = 748


138

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 =

1 1 1 (50) 2 (0.75) − (38) 2 (0.75) = (0.75) ⎡⎣502 − 382 ⎤⎦ = 396 mm 2 2 2 2

(b) Percentage of whole sector =

396 1 (50) 2 (0.75) 2

× 100% = 42.24%


139


1. Determine the radius, and the co-ordinates of the centre of the circle given by the equation x 2 + y 2 + 6x − 2y − 26 = 0 Method 1: The general equation of a circle (x − a) 2 + (y − b) 2 = r 2 is x 2 + y 2 + 2ex + 2fy + c = 0 where co-ordinate, a = -

2e 2f , co-ordinate, b = 2 2

Hence, if x 2 + y 2 + 6x − 2y − 26 = 0 then a = -

and radius, r =

and radius, r =

a 2 + b2 − c

2e 6 2f −2 =1 = − = -3, b = =− 2 2 2 2

⎡⎣(−3) 2 + (1) 2 − (−26) ⎤⎦ = (9 + 1 + 26) = 36 = 6

i.e. the circle x 2 + y 2 + 6x − 2y − 26 = 0 has centre at (-3, 1) and radius 6, as shown below.

Method 2: x 2 + y 2 + 6x − 2y − 26 = 0 may be rearranged as: (x + 3) 2 + (y − 1) 2 − 36 = 0 i.e. (x + 3) 2 + (y − 1) 2 = 62 which has a radius of 6 and centre at (-3, 1)

2. Sketch the circle given by the equation x 2 + y 2 − 6x + 4y − 3 = 0

Method 1: The general equation of a circle (x − a) 2 + (y − b) 2 = r 2 is x 2 + y 2 + 2ex + 2fy + c = 0 where co-ordinate, a = -

2e 2f , co-ordinate, b = 2 2

Hence, if x 2 + y 2 − 6x + 4y − 3 = 0 then a = -

and radius, r =

a 2 + b2 − c

2e −6 2f 4 =− = 3, b = = − = -2 2 2 2 2


140

and radius, r =

⎡⎣(3) 2 + (−2) 2 − (−3) ⎤⎦ = (9 + 4 + 3) = 16 = 4

i.e. the circle x 2 + y 2 − 6x + 4y − 3 = 0 has centre at (3, -2) and radius 4, as shown below.

Method 2: x 2 + y 2 − 6x + 4y − 3 = 0 may be rearranged as: (x − 3) 2 + (y + 2) 2 − 16 = 0 i.e. (x − 3) 2 + (y + 2) 2 = 42 which has a radius of 4 and centre at (3, -2)

⎡ ⎛ y ⎞2 ⎤ 4. Sketch the curve x = 6 ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 6 ⎠ ⎥⎦

⎡ ⎛ y ⎞2 ⎤ If x = 6 ⎢1 − ⎜ ⎟ ⎥ ⎣⎢ ⎝ 6 ⎠ ⎦⎥

i.e.

then

x 2 y2 + =1 62 62

⎡ ⎛ y ⎞2 ⎤ x = ⎢1 − ⎜ ⎟ ⎥ 6 ⎢⎣ ⎝ 6 ⎠ ⎦⎥

and

2

and

⎛x⎞ ⎛ y⎞ ⎜ ⎟ = 1− ⎜ ⎟ ⎝6⎠ ⎝6⎠

2

x 2 + y 2 = 62

which is a circle of radius 6 and co-ordinates of centre at (0, 0), as shown below.


141


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 × 1000 m/s = 10 m/s 3600

(Note that changing from km/h to m/s involves dividing by 3.6) Radius of wheel, r =

500 = 250 mm = 0.25 m 2

Since, v = ωr, then angular velocity, ω =

v 10 = = 40 rad/s r 0.25

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 = Radius of wheel =

108 m/s = 30 m/s 3.6

800 = 400 mm = 0.4 m 2

Since, v = ωr, then angular velocity, ω = 75 rad / s = 75 rad / s ×

(b) Linear velocity, v =

v 30 = = 75 rad/s r 0.4

rev 60s × = 716.2 rev/min 2π rad min

s s 2700 m 90 hence, time, t = = = 1.5 minutes = 90 s = t v 30 m / s 60

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


142


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.

mv 2 where mass, m = 1 tonne = 1000 kg, radius, r = 125 m and Centripetal force = r velocity, v = 40 km/h =

40 m/s 3.6 2

⎛ 40 ⎞ (1000) ⎜ ⎟ ⎝ 3.6 ⎠ = 988 N Hence, centrifugal force = 125

If centrifugal force is limited to 750 N, then 750 =

from which,

velocity, v =

(1000)v 2 125

⎛ (750)(125) ⎞ ⎜ ⎟ = 9.6825 m/s ⎝ 1000 ⎠

= 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 =

34 m/s 3.6 2

⎛ 34 ⎞ 2 ⎜ ⎟ v 3.6 ⎠ =⎝ = 0.892 m / s 2 For the first bend of radius 100 m, acceleration = r1 100 2

⎛ 34 ⎞ ⎜ ⎟ 3.6 ⎠ For the second bend of radius 150 m, acceleration = = − ⎝ = −0.595 m / s 2 , the negative sign 150 indicating a change in direction

Hence, change in acceleration is 0.892 – (-0.595) = 1.49 m / s 2


143

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.

⎛ 1 ⎞ If cosec θ = 2.5317, then θ = cos ec −1 (2.5317) = sin −1 ⎜ ⎟ = 23.265° or 23°16 ' . ⎝ 2.5317 ⎠

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: cos −1 (−0.5316) = t

Cosine is negative in the 2nd and 3rd quadrants. cos −1 (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: tan −1 0.8314 = θ


144

Tangent is positive in the 1st and 3rd quadrants.

θ = tan −1 0.8314 = 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′


145

EXERCISE 70 Page 157 2. State the amplitude and period of the waveform y = 2 sin

5x and sketch the curve between 0° 2

and 360°. 5x 360° , amplitude = 2 and period = = 144° 5 2 2 5x is shown below. A sketch y = 2 sin 2 If y = 2 sin

θ 4. State the amplitude and period of the waveform y = 3 cos and sketch the curve between 0° 2 and 360°. θ 360° If y = 3 cos , amplitude = 3 and period = = 720° 1 2 2 θ A sketch y = 3 cos is shown below. 2

6. State the amplitude and period of the waveform y = 6 sin(t - 45°) and sketch the curve between 0° and 360°. © 2006 John Bird. All rights reserved. Published by Elsevier.

146

If y = 6 sin(t - 45°), amplitude = 6 and period =

360° = 360° 1

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 =

360° = 180° 2

A sketch y = 4 cos(2θ + 30°) is shown below. (Note that y = 4 cos(2θ + 30°) leads y = 4 cos 2θ by

30° = 15°) 2

9. State the amplitude and period of the waveform y = 5 cos 2 and 360°.

If y = 5 cos 2

3 θ and sketch the curve between 0° 2

3 180° θ , amplitude = 5 and period = = 120° 3 2 2


147

A sketch y = 5 cos 2

3 θ is shown below. 2


148


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 = periodic time, T = and phase angle = 0.29 rad leading or 0.29 ×

50π = 25 Hz, 2π 1 1 = 0.040 s or 40 ms, = f 25

180° = 16.62° leading or 16°37′ leading. π

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, from which,

0 = 120 sin(0 + φ), i.e. 0 = 120 sin φ

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 = sin φ 120

π ⎛ 50 ⎞ = 0.43rad and φ = sin −1 ⎜ ⎟ = 24.624° = 24.624 × 180 ⎝ 120 ⎠

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, f =

1 1 = = 40 Hz T 25 × 10−3

Angular velocity, ω = 2πt =2π(40) = 80π rad/s


149

Hence, current i = 20 sin(80πt + φ) -10 = 20 sin φ

When t = 0, i = -10, hence from which,

sin φ = −

10 = −0.5 20

and

φ = sin −1 (−0.5) = −30° or −

π rad 6

π⎞ ⎛ i = 20sin ⎜ 80πt − ⎟ A 6⎠ ⎝

Thus,

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 = periodic time, T = and phase angle = 0.432 rad lagging or 0.432 × (b) When t = 0, i = 5 sin(- 0.432) = -2.093 A

100π = 50 Hz, 2π 1 1 = = 0.020 s or 20 ms, f 50

180° = 24.75° lagging or 24°45′ lagging. π (note that -0.432 is radians)

(c) When t = 8 ms, i = 5 sin ⎡⎣100π ( 8 × 10−3 ) − 0.432 ⎤⎦ = 5 sin (2.081274) = 4.363 A (d) When the current is first a maximum, 5 = 5 sin(100πt – 0.432) 1 = sin(100πt – 0.432)

i.e.

100πt – 0.432 = sin −1 1 = 1.5708

and

from which, (e) When i = 3 A, i.e.

time t =

(again, be sure your calculator is on radians)

1.5708 + 0.432 = 0.006375 s or 6.375 ms 100π

3 = 5 sin(100πt – 0.432) 3 = sin(100πt – 0.432) 5


150

100πt – 0.432 = sin −1

and

time t =

from which,

3 = 0.6435 5

0.6435 + 0.432 = 0.003423 s or 3.423 ms 100π

A sketch of one cycle of the waveform is shown below.

Note that since phase angle φ = 24.75°, in of time φt then 24.75 φt ≡ from which, φt = 1.375 ms 360 20 Alternatively,

φt =

φ 0.432 = = 1.375 ms, as shown in the sketch. ω 100π


151


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. =

from which, maximum value =

1 × maximum value 2 2 × r.m.s. = 2 × 50 = 70.71 A

Hence, fundamental current is: i1 = 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: i3 = 16.97 sin 3(628.3)t = 16.97 sin 1885t A Thus, current i = i1 + i3 = 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.


152

(a) Voltage, v=

i.e.

212.1 212.1 ⎡ π⎤ 212.1 ⎡ π⎤ sin 2π(50)t + (0.30) sin ⎢ 2(2π)(50)t − ⎥ + (0.1) sin ⎢ 4(2π)(50)t + ⎥ volts 0.707 0.707 2⎦ 0.707 3⎦ ⎣ ⎣ π⎞ π⎞ ⎛ ⎛ v = 300 sin 314.2 t + 90 sin ⎜ 628.3t − ⎟ + 30sin ⎜ 1256.6t + ⎟ volts 2⎠ 3⎠ ⎝ ⎝

(b) The complex waveform representing v is shown sketched below:

5. A voltage waveform is described by: π⎞ π⎞ ⎛ ⎛ v = 200 sin 377t + 80sin ⎜1131t + ⎟ + 20sin ⎜1885t − ⎟ volts 4⎠ 3⎠ ⎝ ⎝ 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.


153

(a) From the fundamental voltage, 377 = ω1 = 2πf1 i.e. fundamental frequency, f1 =

377 = 60 Hz 2π

From the 3rd harmonic voltage, 1131 = ω3 = 2πf 3 i.e.

3rd harmonic frequency, f 3 =

1131 = 180Hz 2π

From the 5th harmonic voltage, 1885 = ω5 = 2πf 5 i.e.

5th harmonic frequency, f 5 =

1885 = 300Hz 2π

(b) Percentage 3rd harmonic =

80 × 100% = 40% 200

(c) Percentage 5th harmonic =

20 × 100% = 10% 200

The complex waveform representing v is shown sketched below:


154

CHAPTER 16 TRIGONOMETRIC IDENTITIES AND EQUATIONS EXERCISE 73 Page 167 2. Prove the trigonometric identity:

1

(1 − cos θ )

= cos ec θ

2

L.H.S. =

1

(1 − cos θ )

=

2

=

1

(since sin 2 θ + cos 2 θ = 1 )

sin θ 2

1 = cosec θ = R.H.S. sin θ

4. Prove the trigonometric identity:

L.H.S. =

cos x − cos3 x = sin x cos x sin x

cos x − cos3 x cos x(1 − cos 2 x) cos x sin 2 x = = = cos x sin x = sin x cos x = R.H.S. sin x sin x sin x

5. Prove the trigonometric identity: (1 + cot θ ) + (1 − cot θ ) = 2 cos ec2θ 2

2

L.H.S. = (1 + cot θ ) + (1 − cot θ ) = 1 + 2 cot θ + cot 2 θ + 1 − 2 cot θ + cot 2 θ 2

2

= 2 + 2 cot 2 θ = 2 + 2 ( cos ec 2θ − 1) = 2 + 2 cos ec 2 θ − 2 = 2 cos ec 2 θ = R.H.S. 6. Prove the trigonometric identity:

sin 2 x ( sec x + cos ec x ) = 1 + tan x cos x tan x

1 ⎞ ⎛ 1 ⎛ sin x + cos x ⎞ sin 2 x ⎜ sin 2 x ⎜ + ⎟ ⎟ sin x ( sec x + cos ec x ) ⎝ cos x sin x ⎠ = ⎝ cos x sin x ⎠ L.H.S. = = cos x tan x sin x ⎛ sin x ⎞ cos x ⎜ ⎟ ⎝ cos x ⎠ 2

⎛ sin x + cos x ⎞ sin x + cos x = sin x ⎜ ⎟= cos x ⎝ cos x sin x ⎠ =

sin x cos x + = tan x + 1 = 1 + tan x = R.H.S. cos x cos x


155


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 = − 1 5.5 =− sin A 3

i.e.

from which,

or

sin A = −

5.5 3

3 5.5

⎛ 3 ⎞ A = sin −1 ⎜ − ⎟ = -33.056° or -33°3′ ⎝ 5.5 ⎠

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 − 180° + 33°3′ = 213°3′

3 are: 5.5

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 i.e.

Hence,

cot t =

then 2.32 5.4

2.32 – 5.4 cot t = 0 from which,

tan t =

and

2.32 = 5.4 cot t

5.4 2.32

⎛ 5.4 ⎞ t = tan −1 ⎜ ⎟ = 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. © 2006 John Bird. All rights reserved. Published by Elsevier.

156

Hence, the two angles for t between 0° and 360° whose tan is 66°45′

5.4 are: 2.32

and 180° + 66°45′ = 246°45′


157


1. Solve: 5sin 2 y = 3 for angles between 0° and 360°

Since 5sin 2 y = 3

sin 2 y =

then

3 = 0.60 5

and

sin y = 0.60 = ± 0.774596...

y = sin −1 (0.774596...) = 50°46 '

and

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: 5 + 3cos ec 2 D = 8 for angles between 0° and 360°

Since 5 + 3cos ec2 D = 8 Hence, and

1 =1 sin 2 D

then and

3cos ec2 D = 8 − 5 = 3 sin 2 D = 1

i.e.

from which,

cos ec2 D = 1 sin D = 1 = ± 1

D = sin −1 ( ± 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°


158


2. Solve: 8 tan 2 θ + 2 tan θ = 15 for angles between 0° and 360°

Since 8 tan 2 θ + 2 tan θ = 15

8 tan 2 θ + 2 tan θ − 15 = 0

then

( 4 tan θ − 5 )( 2 tan θ + 3) = 0

i.e.

5 = 1.25 4

i.e.

4 tan θ - 5 = 0

from which,

tan θ =

and

2 tan θ + 3 = 0

from which,

tan θ = −

3 = −1.5 2

and θ = tan −1 1.25 = 51°20′ and θ = tan −1 − 1.5 = -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: 2 cos ec2 t − 5cos ec t = 12 for angles between 0° and 360°

Since 2 cos ec 2 t − 5cos ec t = 12 and

then

2 cos ec 2 t − 5cos ec t − 12 = 0 (2 cosec t + 3)(cosec t – 4) = 0

i.e.

2 cosec t + 3 = 0 from which, cosec t = −

and

cosec t - 4 = 0 from which, cosec t = 4

3 2

and sin t = −

and sin t =

2 from which, t = -41°49′ 3

1 from which, t = 14°29′ 4

From the diagram below, the four values of θ between 0° and 360° are:


159

14°29′, 180° - 14°29′ = 165°31′, 180° + 41°49′ = 221°49′ and

360° - 41°49′ = 318°11′


160


1. Solve: 12sin 2 θ − 6 = cos θ for angles between 0° and 360°

Since 12sin 2 θ − 6 = cos θ

then

12 (1 − cos 2 θ ) − 6 = cos θ

12 − 12 cos 2 θ − 6 = cos θ

i.e.

12 cos 2 θ + cos θ − 6 = 0

i.e. Factorising gives: i.e.

4 cos θ + 3 = 0

and

3 cos θ - 2 = 0

(4 cos θ + 3)(3 cos θ - 2) = 0 from which, cos θ = − from which, cos θ =

2 3

3 = −0.75 and θ = cos −1 (−0.75) = -41°25′ 4 ⎛2⎞ and θ = cos −1 ⎜ ⎟ = 48°11′ ⎝3⎠

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: 4 cot 2 A − 6 cos ecA + 6 = 0 for angles between 0° and 360°

Since 4 cot 2 A − 6 cos ecA + 6 = 0

then

4 ( cos ec 2 A − 1) − 6 cos ecA + 6 = 0 4 cos ec 2 A − 6 cos ecA + 2 = 0

and Factorising gives:

(2 cosec A – 1) (2 cosec A – 2) = 0 1 = and sin A = 2 which has no solutions 2

i.e.

2 cosec A - 1 = 0

from which, cosec A =

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°


161

5. Solve: 2.9 cos 2 a − 7 sin a + 1 = 0 for angles between 0° and 360°

Since 2.9 cos 2 a − 7 sin a + 1 = 0

2.9(1 − sin 2 a) − 7 sin a + 1 = 0

then

i.e.

2.9 − 2.9sin 2 a − 7 sin A + 1 = 0

and

2.9sin 2 a + 7 sin a − 3.9 = 0

Hence,

Thus,

−7 ± ⎡⎣7 2 − 4(2.9)(−3.9) ⎤⎦ −7 ± 94.24 sin a = = = 0.46685 or -2.88064, which has 2(2.9) 5.8 no solution a = sin −1 (0.46685) = 27°50′

and, from the diagram below, 180° - 27°50′ = 152°10′


162

CHAPTER 17 THE RELATIONSHIP BETWEEN TRIGONOMETRIC AND HYPERBOLIC FINCTIONS EXERCISE 78 Page 174

2. 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 + j2 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


cos j2A = 1 – 2 sin 2 jA L.H.S. = cos j2A = ch 2A

from (5), page 173

R.H.S. = 1 – 2 sin 2 jA = 1 – 2(sin jA)(sin jA) = 1 – 2(j sh A)(j sh A) = 1 - 2 j2sh 2 A = 1 + 2sh 2 A = ch 2A from Table 5.1, page 45 Hence, L.H.S. = R.H.S. i.e. cos j2A = 1 – 2 sin 2 jA


⎛ A+B⎞ ⎛ A−B⎞ sin jA – sin jB = 2 cos j ⎜ ⎟ sin j ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ L.H.S. = sin jA – sin jB = j sh A – j sh B = j(sh A – sh B) ⎛ A+B⎞ ⎛ A−B⎞ ⎛ A+B⎞ ⎛A−B⎞ R.H.S. = 2 cos j ⎜ ⎟ sin j ⎜ ⎟ = 2ch⎜ ⎟ j sh⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛A+B⎞ ⎛ A−B⎞ =2 j ch⎜ ⎟s h ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠


163

A B⎤ ⎡ A B A B⎤ ⎡ A B = 2 j ⎢ch ch + sh sh ⎥ ⎢sh ch − ch sh ⎥ 2 2 2⎦⎣ 2 2 2 2⎦ ⎣ 2 A B A B A B A B ⎡ A B A B ⎤ ⎢ch 2 ch 2 sh 2 ch 2 − ch 2 ch 2 ch 2 sh 2 + sh 2 sh 2 sh 2 ch 2 ⎥ = 2j⎢ ⎥ A B A B⎥ ⎢ − sh sh ch sh ⎣⎢ 2 2 2 2 ⎦⎥ B A B B A B B B A A⎤ ⎡ A A = 2 j ⎢ch sh ch 2 − ch 2 ch sh + sh 2 sh ch − sh 2 sh ch ⎥ 2 2 2 2 2 2 2 2 2 2 2⎦ ⎣ 2 ⎡ A A⎛ B B⎞ B B⎛ A A ⎞⎤ = 2 j ⎢sh ch ⎜ ch 2 − sh 2 ⎟ + sh ch ⎜ sh 2 − ch 2 ⎟ ⎥ 2⎝ 2 2⎠ 2 2⎝ 2 2 ⎠⎦ ⎣ 2 ⎡ A A⎛ B B⎞ B B⎛ A A ⎞⎤ = 2 j ⎢sh ch ⎜ ch 2 − sh 2 ⎟ − sh ch ⎜ ch 2 − sh 2 ⎟ ⎥ 2⎝ 2 2⎠ 2 2⎝ 2 2 ⎠⎦ ⎣ 2 B B ⎤ ⎡ A A = 2 j ⎢sh ch (1) − sh ch (1) ⎥ 2 2 2 ⎦ ⎣ 2

from Table 5.1, page 45

A A B B⎤ ⎡ = j ⎢ 2sh ch − 2sh ch ⎥ 2 2 2 2⎦ ⎣ ⎡ ⎛A⎞ ⎛ B ⎞⎤ = j ⎢sh 2 ⎜ ⎟ − sh 2 ⎜ ⎟ ⎥ ⎝2⎠ ⎝ 2 ⎠⎦ ⎣


= j [ sh A – sh B ] = L.H.S


164


1. Use the substitution A = jθ to obtain the hyperbolic identity corresponding to the

trigonometric identity: 1 + tan 2 A = sec2 A

Let A = jθ then becomes:

1 + tan 2 A = sec2 A

1 + tan 2 ( jθ ) = sec 2 ( jθ ) 1 + ( tan jθ ) = ( sec jθ ) 2

i.e.

1 + ( j tanh θ )

i.e.

2

⎛ 1 ⎞ =⎜ ⎟ ⎝ cos jθ ⎠

i.e.

⎛ 1 ⎞ 1 + j tanh θ = ⎜ ⎟ ⎝ ch θ ⎠

i.e.

1 − tanh 2 θ = sec h 2 θ

2

2

2

2

2

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


trigonometric identity:

tan 2A =

2 tan A 1 − tan 2 A

Substituting A = jθ in tan 2A =

2 tan A 1 − tan 2 A

tan 2 jθ =

2 tan jθ 1 − tan 2 jθ

gives:


165

2 j tanh θ

i.e.

j tanh 2θ =

i.e.

j tanh 2θ =

2 jtanh θ 1 + tanh 2 θ

i.e.

tanh 2θ =

2 tanh θ 1 + tanh 2 θ

1 − ( jtanh θ )

2

=

2 jtanh θ 1 − j2 tanh 2 θ


3 1 trigonometric identity: sin 3 A = sin A − sin 3A 4 4 Substituting A = jθ

3 1 sin 3 A = sin A − sin 3A 4 4

in

3 1 sin 3 jθ = sin jθ − sin 3jθ 4 4

gives: i.e.

j3 ( sin h 3θ ) =

3 1 j sin h θ − j sin h 3θ 4 4

Dividing by j gives:

j2 ( sin h 3θ ) =

3 1 sin h θ − sin h 3θ 4 4

i.e.

− ( sin h 3θ ) =

3 1 sin h θ − sin h 3θ 4 4

i.e.

sin h 3 θ =

1 3 sin h 3θ − sin h θ 4 4


trigonometric identity: cot 2 A ( sec 2 A − 1) = 1

Substituting A = jθ in

cot 2 A ( sec 2 A − 1) = 1

gives:

cot 2 jθ ( sec 2 jθ − 1) = 1

i.e.

⎞ 1 ⎛ 1 − 1⎟ = 1 ⎜ 2 2 tan jθ ⎝ cos jθ ⎠

i.e.

⎛ ⎞ 1 − 1 ⎜ ⎟ =1 2 2 ( tan jθ ) ⎜⎝ ( cos jθ ) ⎟⎠ 1


166

i.e.

⎛ 1 ⎞ ⎜ 2 − 1⎟ = 1 ( jth θ ) ⎝ ch θ ⎠

and

1 sec h 2 θ − 1) = 1 2 ( j th θ

i.e.

− coth 2 θ ( sec h 2θ − 1) = 1

or

coth 2 θ ( 1 − sec h 2θ ) = 1

1

2

2


167

CHAPTER 18 COMPOUND ANGLES EXERCISE 80 Page 177 π⎞ 2π ⎞ ⎛ ⎛ 3. Show that (a) sin ⎜ x + ⎟ + sin ⎜ x + ⎟ = 3 cos x 3⎠ 3 ⎠ ⎝ ⎝ ⎛ 3π ⎞ (b) − sin ⎜ − φ ⎟ = cos φ ⎝ 2 ⎠

π⎞ 2π ⎞ π π 2π 2π ⎛ ⎛ (a) L.H.S. = sin ⎜ x + ⎟ + sin ⎜ x + + cos x sin ⎟ = sin x cos + cos x sin + sin x cos 3⎠ 3 ⎠ 3 3 3 3 ⎝ ⎝ ⎛ 3⎞ ⎛ 3⎞ ⎛1⎞ ⎛ 1⎞ = sin x ⎜ ⎟ + cos x ⎜⎜ ⎟⎟ + sin x ⎜ − ⎟ + cos x ⎜⎜ ⎟⎟ ⎝2⎠ ⎝ 2⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 3⎞ = 2 ⎜⎜ ⎟⎟ cos x = 2 ⎝ ⎠

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 = 22 − 12 = 3 . Hence, sin

π 3 1 and cos 60° = = sin 60° = 2 2 3

3π ⎛ 3π ⎞ ⎡ 3π ⎤ (b) L.H.S. = − sin ⎜ − φ ⎟ = − ⎢sin cos φ − cos sin φ ⎥ 2 2 ⎝ 2 ⎠ ⎣ ⎦ = − [ (−1) cos φ − (0) sin φ] = cos φ = R.H.S.

π⎞ 3π ⎞ ⎛ ⎛ 4. Prove that: (a) sin ⎜ θ + ⎟ − sin ⎜ θ − ⎟ = 2 (sin θ + cos θ) 4⎠ 4 ⎠ ⎝ ⎝ cos ( 270° + θ ) = tan θ (b) cos ( 360° − θ )


168

π⎞ 3π ⎞ ⎛ π π⎞ ⎛ 3π 3π ⎞ ⎛ ⎛ (a) L.H.S. = sin ⎜ θ + ⎟ − sin ⎜ θ − ⎟ = ⎜ sin θ cos + cos θ sin ⎟ − ⎜ sin θ cos − cos θ sin ⎟ 4⎠ 4 ⎠ ⎝ 4 4⎠ ⎝ 4 4 ⎠ ⎝ ⎝ ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞⎤ ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞⎤ = ⎢sin θ ⎜ ⎟ + cos θ ⎜ ⎟ ⎥ − ⎢sin θ ⎜ − ⎟ − cos θ ⎜ ⎟⎥ 2⎠ ⎝ 2⎠ ⎝ 2 ⎠⎦ ⎣ ⎝ ⎝ 2 ⎠⎦ ⎣

=

1 2 [sin θ + cos θ + sin θ + cos θ] = ( sin θ + cos θ ) 2 2

=

2 (sin θ + cos 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 = 12 + 12 = 2 . Hence, sin

(b) L.H.S. =

π 1 = sin 45° = cos 45°= 4 2

cos ( 270° + θ ) cos 270° cos θ − sin 270° sin θ 0 − (−1) sin θ = = cos ( 360° − θ ) cos 360° cos θ + sin 360° sin θ (1) cos θ + 0 =

sin θ = tan θ = R.H.S. cos θ

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 θ 2.57115 = = 2.4160901 cos θ 1.064178

i.e. tan θ = 2.4160901 and θ = tan −1 (2.4160901) = 67°31′ and 247°31′ (see diagram below)


169

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 4 R

Hence,

4 = R cos α from which, cos α =

and

-3 = R sin α from which, sin α = −

3 R

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= Hence,

(4

2

+ 32 ) = 5

and α = tan −1

3 = 0.644 rad 4

4 sin ωt – 3 cos ωt = 5 sin(ωt – 0.644)

3. Change the function: -7 sin ωt + 4 cos ωt

Let

(make sure your calculator is on radians)

into the form R sin(ωt ± α)

-7 sin ωt + 4 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, and

-7 = R cos α from which, cos α = − 4 = R sin α from which, sin α =

7 R

4 R

There is only one quadrant where both cosine is negative and sine is positive, i.e. the 2nd quadrant, as shown in the diagram below.


170

R=

(7

2

+ 42 ) = 8.062

Thus, in the diagram,

and φ = tan −1

4 = 0.519 rad 7

α = π - 0.519 = 2.622 -7 sin ωt + 4 cos ωt = 8.062 sin(ωt + 2.622)

Hence,

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

2 sin θ + 4 cos θ = R sin(θ + α)

(a) Let

= R[sin θ cos α + cos θ sin α] = (R cos α) sin θ + (R sin α) cos θ Hence,


2 R

and

4 = R sin α from which, sin α =

4 R

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= Hence,

(4

2

+ 22 ) = 4.472

and α = tan −1

4 = 63.43° or 63°26′ 2

2 sin θ + 4 cos θ = 4.472 sin(θ + 63°26′)

Thus, since 2 sin θ + 4 cos θ = 3

then

i.e.

3 = 0.67084 4.472

sin(θ + 63°26′) =

4.472 sin(θ + 63°26′) = 3


171

and

θ + 63°26′ = sin −1 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′ θ = 74°26′ and 338°42′ satisfies the equation 2 sin θ + 4 cos θ = 3

i.e.

12 sin θ - 9 cos θ = R sin(θ + α)

(b) Let

= R[sin θ cos α + cos θ sin α] = (R cos α) sin θ + (R sin α) cos θ 12 R

Hence,


and

-9 = R sin α from which, sin α = −

9 R

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= Hence,

(12

2

+ 92 ) = 15

and α = tan −1

12 sin θ - 9 cos θ = 15 sin(θ - 36°52′)

Thus, since 12 sin θ - 9 cos θ = 7 i.e.

sin(θ - 36°52′) =

then

15 sin(θ - 36°52′) = 7

7 15

7 = 27°49′ 15

and

θ - 36°52′ = sin −1

Thus,

θ = 27°49′ + 36°52′ = 64°41′

i.e.

9 = 36°52′ 12

or 180° - 27°49′ = 152°11′ or

θ = 152°11′ + 36°52′ = 189°3′

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)


172

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, and

-6.9 = R cos α from which, cos α = − 4.3 = R sin α

from which, sin α =

6.9 R

4.3 R

There is only one quadrant where both cosine is negative and sine is positive, i.e. the 2nd quadrant, as shown in the diagram below.

R=

( 6.9

2

+ 4.32 ) = 8.13

4.3 = 31°56′ 6.9

α = 180° - 31°56′ = 148°4′ = 2.584 rad

and Hence,

and φ = tan −1

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) = R[sin ωt cos α + cos ωt sin α] = (R cos α) sin ωt + (R sin α) cos ωt

Hence,

2.4 = R cos α from which, cos α =

2.4 R

and

3.2 = R sin α

3.2 R

from which, sin α =

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.


173

R= Hence,

( 2.4

2

+ 3.22 ) = 4

and α = tan −1

3.2 = 53.13° or 0.927 rad 2.4

x = 2.4 sin ωt + 3.2 cos ωt = 4 sin(ωt + 0.927) m


174


2. Prove the following identities: (a) 1 −

(c)

cos 2φ = tan 2 φ cos 2 φ

( tan 2x )(1 + tan x ) = tan x

(b)

1 + cos 2t = 2 cot 2 t sin 2 t

2 1 − tan x

(d) 2 cosec 2θ cos 2θ = cot θ - tan θ

(a) L.H.S. = 1 −

⎛ cos 2 φ − sin 2 φ ⎞ ⎛ cos 2 φ sin 2 φ ⎞ cos 2φ = 1 − = 1 − − ⎜ ⎟ ⎜ ⎟ 2 2 cos 2 φ cos 2 φ ⎝ ⎠ ⎝ cos φ cos φ ⎠

= 1- (1 − tan 2 φ ) = tan 2 φ = R.H.S. 1 + cos 2t 1 + ( 2 cos t − 1) 2 cos 2 t (b) L.H.S. = = = = 2 cot 2 t = R.H.S. 2 2 2 sin t sin t sin t 2

( 2 tan x )(1 + tan x ) ⎛ 2 tan x ⎞ 1 + tan x ) ( ⎜ ⎟ 2 (1 − tan x )(1 + tan x ) ( tan 2x )(1 + tan x ) = ⎝ 1 − tan x ⎠ (c) L.H.S. = = tan x tan x tan x

=

2 tan x (1 − tan x ) tan x

=

2 tan x 2 = = R.H.S. tan x (1 − tan x ) 1 − tan x

2 2 ⎛ 2 ⎞ (d) L.H.S. = 2 cosec 2θ cos 2θ = ⎜ = ⎟ (cos 2θ) = 2 cot 2θ = θ 2 tan tan 2θ ⎝ sin 2θ ⎠ 2 1 − tan θ

=

2 (1 − tan 2 θ ) 2 tan θ

1 − tan 2 θ 1 = = − tan θ = cot θ − tan θ = R.H.S. tan θ tan θ

3. If the third harmonic of a waveform is given by V3 cos 3θ , express the third harmonic in of

the first harmonic cos θ, when V3 = 1.

When V3 = 1, V3 cos 3θ = cos 3θ = cos(2θ + θ) = cos 2θ cos θ - sin 2θ sin θ = ( 2 cos 2 θ − 1) (cos θ) − ( 2sin θ cos θ ) (sin θ) = 2 cos3 θ − cos θ − 2 cos θ sin 2 θ © 2006 John Bird. All rights reserved. Published by Elsevier.

175

= 2 cos3 θ − cos θ − 2 cos θ (1 − cos 2 θ ) = 2 cos3 θ − cos θ − 2 cos θ + 2 cos3 θ = 4 cos3 θ − 3cos θ = R.H.S.


176


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 6x] 2

4. Express as a sum or difference: 4 cos 3θ cos θ

⎧1 ⎫ 4 cos 3θ cos θ = 4 ⎨ [ cos(3θ + θ) + cos(3θ − θ) ]⎬ ⎩2 ⎭

from (3), page 183

= 2[cos 4θ + cos 2θ] 6. Determine: ∫ 2sin 3t cos t dt

⎧1 ⎫ 2 sin 3t cost = 2 ⎨ [sin(3t + t) + sin(3t − t) ]⎬ ⎩2 ⎭

from (1), page 183

= sin 4t + sin 2t Hence,

1

1

∫ 2sin 3t cos t dt = ∫ ( sin 4t + sin 2t ) dt = − 4 cos 4t − 2 cos 2t + c

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.

4sin 2 φ cos φ = cos φ

i.e.

4sin 2 φ cos φ − cos φ = 0

and

cos φ ( 4sin 2 φ − 1) = 0

Hence,

cos φ = 0

from which, φ = cos −1 0 = 90°


177

from which, sin 2 φ =

1 4

and

4sin 2 φ = 1

and sin φ =

Hence,

φ = sin −1 0.5 = 30° and 150° (see diagram below)

and

φ = sin −1 (−0.5) = 210° and 330°

1 = ±0.5 4

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°


178


1. Express as a product: sin 3x + sin x

⎛ 3x + x ⎞ ⎛ 3x − x ⎞ sin 3x + sin x = 2sin ⎜ ⎟ cos ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

from (5), page 184

= 2 sin 2x cos x 3. Express as a product: cos 5t + cos 3t

⎛ 5t + 3t ⎞ ⎛ 5t − 3t ⎞ cos 5t + cos 3t = 2 cos ⎜ ⎟ cos ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

from (7), page 184

= 2 cos 4t cos t

5. Express as a product:

π π⎞ 1⎛ ⎜ cos + cos ⎟ 2⎝ 3 4⎠

⎧ ⎛π π⎞ ⎛ π π ⎞⎫ + ⎟ ⎪ ⎜ ⎜ 3 − 4 ⎟ ⎪⎪ 1⎛ π π⎞ 1⎪ 3 4 ⎟ cos ⎜ ⎟⎬ ⎜ cos + cos ⎟ = ⎨2 cos ⎜ 2⎝ 3 4⎠ 2⎪ ⎜ 2 ⎟ ⎜ 2 ⎟⎪ ⎝ ⎠ ⎝ ⎠ ⎭⎪ ⎩⎪

from (7), page 184

⎛ 7π ⎞ ⎛ π ⎞ ⎜ 12 ⎟ ⎜ 12 ⎟ 7π π = cos ⎜ ⎟ cos ⎜ ⎟ = cos cos 24 24 ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠

6. Show that: (a)

sin 4x − sin 2x = tan x cos 4x + cos 2x

(b)

1 { sin(5x − α) − sin(x + α) } = cos 3x sin(2x − α) 2

⎛ 4x + 2x ⎞ ⎛ 4x − 2x ⎞ 2 cos ⎜ ⎟ sin ⎜ ⎟ sin 4x − sin 2x 2 ⎠ ⎝ 2 ⎠ ⎝ (a) L.H.S. = = cos 4x + cos 2x ⎛ 4x + 2x ⎞ ⎛ 4x − 2x ⎞ 2 cos ⎜ ⎟ cos ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ = (b) L.H.S. =

2 cos 3x sin x sin x = = tan x = R.H.S. 2 cos 3x cos x cos x

1 { sin(5x − α) − sin(x + α) } 2


179

=

1 [(sin 5x cos α − cos 5x sin α) − (sin x cos α + cos x sin α)] 2

=

1 [cos α (sin 5x − sin x) − sin α (cos 5x + cos x)] 2

=

1⎡ ⎧ ⎧ ⎛ 5x + x ⎞ ⎛ 5x − x ⎞ ⎫ ⎛ 5x + x ⎞ ⎛ 5x − x ⎞ ⎫⎤ ⎢cos α ⎨2 cos ⎜ ⎟ sin ⎜ ⎟ ⎬ − sin α ⎨2 cos ⎜ ⎟ cos ⎜ ⎟ ⎬⎥ 2⎣ ⎝ 2 ⎠ ⎝ 2 ⎠⎭ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎭⎦ ⎩ ⎩

=

1 [ 2 cos α (cos 3x sin 2x) − 2sin α(cos 3x cos 2x)] 2

= 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.


180

CHAPTER 19 FUNCTIONS AND THEIR CURVES EXERCISE 85 Page 199

1. Sketch y = 3x - 5

2. Sketch y = -3x + 4

3. Sketch y = x 2 + 3

4. Sketch y = ( x − 3)

2


181

5. Sketch y = ( x − 4 ) + 2 2

6. Sketch y = x - x 2

7. Sketch y = x 3 + 2


182

8. Sketch y = 1 + cos 3x

π⎞ ⎛ 9. Sketch y = 3 - 2sin ⎜ x + ⎟ 4⎠ ⎝

10. Sketch y = 2 ln x


183


184


1. Determine whether the following functions are even, odd or neither even nor odd: (a) x 4

(b) tan 3x

(c) 2e3t

(d) sin 2 x

(a) Let f(x) = x 4 . Since f(-x) = f(x) then x 4 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) = 2e3t . The function is neither even not odd, and is as shown below:

(d) Let f(x) = sin 2 x . Since f(-x) = f(x) then sin 2 x is an even function and is symmetrical about the f(x) axis as shown below:


185

3. State whether the following functions, which are periodic of period 2π, are even or odd: ⎧ θ, when − π ≤ θ ≤ 0 (a) f (θ) = ⎨ ⎩−θ, when 0 ≤ θ ≤ π

π π ⎧ ⎪⎪ x, when − 2 ≤ x ≤ 2 (b) f (x) = ⎨ ⎪ 0, when π ≤ x ≤ 3π ⎪⎩ 2 2

(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.


186


2. Determine the inverse function of f(x) = 5x - 1

If y = f(x), then

y = 5x - 1

Transposing for x gives:

x=

y +1 5

Interchanging x and y gives:

y=

x +1 5

1 Hence, if f(x) = 5x – 1, then f −1 (x) = (x + 1) 5

4. Determine the inverse function of f (x) = If y = f(x), then

y=

1 +2 x

Transposing for x gives:

x=

1 y−2

Interchanging x and y gives:

y=

1 x−2

Hence, if f (x) =

1 +2 x

1 1 + 2 , then f −1 (x) = x−2 x

6. Determine the principal value of the inverse function cos −1 0.5

Using a calculator (set on radians), cos −1 0.5 = 1.0472 rad or

π rad 3

8. Determine the principal value of the inverse function cot −1 2

Using a calculator (set on radians), cot −1 2 = tan −1

1 = 0.4636 rad 2

10. Determine the principal value of the inverse function sec−1 1.5

⎛ 1 ⎞ Using a calculator (set on radians), sec−1 1.5 = cos −1 ⎜ ⎟ = 0.8411 rad ⎝ 1.5 ⎠


187

1 4 8 12. Evaluate x, correct to 3 decimal places: x = sin −1 + cos −1 − tan −1 3 5 9 1 4 8 Using a calculator (set on radians), x = sin −1 + cos −1 − tan −1 3 5 9 = 0.3398 + 0.6435 – 0.7266 = 0.257

13. Evaluate y, correct to 4 significant figures: y = 3sec −1 2 − 4 cos ec −1 2 + 5cot −1 2

Using a calculator (set on radians), y = 3sec −1 2 − 4 cos ec −1 2 + 5cot −1 2 ⎛ 1 ⎞ ⎛1⎞ −1 1 + 5 tan −1 ⎜ ⎟ = 3cos −1 ⎜ ⎟ − 4sin 2 ⎝2⎠ ⎝ 2⎠ = 2.3562 – 3.1416 + 2.3182 = 1.533


188

EXERCISE 88 Page 208 2. Determine the asymptotes parallel to the x- and y-axes for y 2 =

y2 =

x hence, x −3

x x −3

y 2 (x − 3) = x

and

y 2 (x − 3) − x = 0

i.e.

y 2 x − 3y 2 − x = 0

i.e.

x ( y 2 − 1) − 3y 2 = 0

(1)

(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: y 2 – 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 y =

y=

x(x + 3) hence, y(x + 2)(x + 1) = x(x + 3) (x + 2)(x + 1)

i.e.

y ( x 2 + 3x + 2 ) − x 2 − 3x = 0

and

yx 2 + 3yx + 2y − x 2 − 3x = 0

i.e.

x 2 (y − 1) + 3xy − 2y − 3x = 0

x(x + 3) (x + 2)(x + 1)

(1)

(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 x 2 ( y 2 − 16 ) = y

Equating highest power of x to zero gives: y 2 − 16 = 0, i.e. y = ± 4


189

Since x 2 ( y 2 − 16 ) = y then

x 2 y 2 − 16x 2 − y = 0

Equating highest power of y to zero gives: x 2 = 0, i.e. x = 0 2 x 2 ⎡( mx + c ) − 16 ⎤ = mx + c ⎣ ⎦

Let y = mx + c, then

x 2 ⎡⎣ m 2 x 2 + 2mxc + c 2 − 16 ⎤⎦ = mx + c

i.e. i.e.

m 2 x 4 + 2mcx 3 + c 2 x 2 − 16x 2 − mx − 1 = 0

Equating coefficient of highest power of x to zero gives: m 2 = 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 xy 2 − x 2 y + 2x − y = 5

xy 2 − x 2 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: x ( mx + c ) − x 2 ( mx + c ) + 2x − (mx + c) = 5 2

i.e.

x ( m 2 x 2 + 2mcx + c 2 ) − mx 3 − cx 2 + 2x − mx − c − 5 = 0

and

m 2 x 3 + 2mcx 2 + c2 x − mx 3 − cx 2 + 2x − mx − c − 5 = 0

(m

i.e.

2

− m ) x 3 + ( 2mc − c ) x 2 + x ( c 2 + 2 − m ) − c − 5 = 0

Equating the coefficient of the highest power of x to zero gives: m 2 − 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


190

A sketch of the curve xy 2 − x 2 y + 2x − y = 5 , together with its asymptotes is shown below:


191

EXERCISE 89 Page 212 1. Sketch the graphs of (a) y = 3x 2 + 9x +

(a) y = 3x 2 + 9x +

7 4

from which,

7 4

(b) y = −5x 2 + 20x + 50

dy = 6x + 9 = 0 for a turning point dx x= −

9 = -1.5 6

When x = -1.5, y = 3(−1.5) 2 + 9(−1.5) + 1.75 = -5 Hence, a turning point occurs at (-1.5, -5) d2 y = 6 , which is positive, hence, (-1.5, -5) is a minimum point. dx 2 A sketch of the graph y = 3x 2 + 9x +

(b) y = −5x 2 + 20x + 50 from which,

7 is shown below. 4

dy = −10x + 20 = 0 for a turning point dx 20 = 10x and x = 2

When x = 2, y = −5(2) 2 + 20(2) + 50 = 70 Hence, a turning point occurs at (2, 70) d2 y = −10 , which is negative, hence, (2, 70) is a maximum point. dx 2 A sketch of the graph y = −5x 2 + 20x + 50 is shown below.


192

4. Sketch the curve depicting: y 2 =

Since y 2 =

x 2 − 16 then 4

x 2 − 16 4

4y 2 = x 2 − 16

i.e.

16 = x 2 − 4y 2

i.e.

x 2 4y 2 − =1 16 16

and

x 2 y2 − = 1 which is a hyperbola, symmetrical about the x- and y42 22

axes, distance between vertices being 2(4), i.e. 8 units along the x-axis. A sketch of y 2 =

x 2 − 16 x 2 y2 , i.e. 2 − 2 = 1 is shown below. 4 4 2

5. Sketch the curve depicting:

Since

y2 x2 = 5− then 5 2

y2 x2 = 5− 5 2

x 2 y2 + =5 2 5


193

x 2 y2 + =1 10 25

and

x2

i.e.

( 10 )

2

+

y2 =1 52

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

y2 x2 = 5 − , i.e. 5 2

x2

( 10 )

2

+

y2 = 1 is shown below. 52

7. Sketch the curve depicting: x 2 y 2 = 9

Since x 2 y 2 = 9 then y 2 =

9 x2

and y =

3 x

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 x 2 + y 2 − 4x + 10y + 25 = 0


194

Since x 2 + y 2 − 4x + 10y + 25 = 0 then

( x − 2 ) + ( y + 5) 2

2

−4=0

i.e.

( x − 2 ) + ( y + 5) 2

2

= 22

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 y = ⎡⎣3 ( x 2 − 1) ⎤⎦

Since y = ⎡⎣3 ( x 2 − 1) ⎤⎦

then y 2 = 3 ( x 2 − 1) y 2 = 3x 2 − 3

and i.e.

3 = 3x 2 − y 2

i.e.

y2 1= x − 3 2

x2

i.e.

(1)

2

−

y2

( 3)

2

=1

which is a hyperbola, symmetrical about the x- and y-axes, with vertices 2(1) = 2 units apart

along the x-axis. 1

14. Describe the shape of the curve represented by the equation y = ( 3x ) 2

1

Since y = ( 3x ) 2 then y =

( 3x )

or y 2 = 3x

which is a parabola, vertex at (0, 0) and symmetrical about the x-axis.


195

15. Describe the shape of the curve represented by the equation y 2 − 8 = −2x 2

Since y 2 − 8 = −2x 2 then

y 2 + 2x 2 = 8

and

y 2 2x 2 + =1 8 8

i.e.

y2

( 8)

2

x2 + 2 =1 (2)

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.


196

CHAPTER 20 IRREGULAR AREAS, VOLUMES AND MEAN VALUES OF WAVEFORMS EXERCISE 90 Page 218 1. Plot a graph of y = 3x − x 2 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 y = 3x − x 2

0 0.5 1.0 1.5 2.0 2.5 3.0 0 1.25 2.0 2.25 2.0 1.25 0

A graph of y = 3x − x 2 is shown below.

(a) Using the trapezoidal rule, with 6 intervals each of width 0.5 gives: ⎡⎛ 0 + 0 ⎞ ⎤ area ≈ ( 0.5 ) ⎢⎜ ⎟ + 1.25 + 2.0 + 2.25 + 2.0 + 1.25⎥ = (0.5)(8.75) = 4.375 square units ⎣⎝ 2 ⎠ ⎦ (b) Using the mid-ordinate rule, with 6 intervals, with mid-ordinates occurring at 0.25 where the y-values are:

0.75

1.25

1.75

2.25

2.75

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:


197

1 1 area ≈ (0.5) ⎡⎣( 0 + 0 ) + 4 (1.25 + 2.25 + 1.25 ) + 2 ( 2.0 + 2.0 ) ⎤⎦ = (0.5) [ 0 + 19 + 8] 3 3 =

1 (0.5)(27) = 4.5 square units 3

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: 1 1 area ≈ (1) ⎡⎣( 0 + 29.0 ) + 4 ( 2.0 + 8.0 + 21.0 ) + 2 ( 4.5 + 14.0 ) ⎤⎦ = [ 29.0 + 124 + 37 ] 3 3 =

1 (190) = 63.33 square units 3

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: ⎡⎛ 0 + 2.3 ⎞ ⎤ area ≈ ( 5 ) ⎢⎜ ⎟ + 2.8 + 5.2 + 6.5 + 5.8 + 4.1 + 3.0 ⎥ = (5) [1.15 + 27.4] ⎣⎝ 2 ⎠ ⎦ = (5)(28.55) = 143 m 2 (To use Simpson’s rule needs an even number of intervals, so could not be used in this question).


198


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 m 2 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 m 3

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 Circumference (m)

2

4

6

8

10

12

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 =

c 2π

2

c2 ⎛ c ⎞ Cross-sectional area = πr = π ⎜ ⎟ = 4π ⎝ 2π ⎠ 2

Hence, the cross-sectional areas are:

2.802 3.252 = 0.6239 m 2 , = 0.8405 m 2 , 1.2353 m 2 , 4π 4π 1.4851 m 2 , 2.1188 m 2 , 2.6955 m 2 , 3.2189 m 2

Hence, volume of timber ≈

2 ⎡( 0.6239 + 3.2189 ) + 4 ( 0.8405 + 1.4851 + 2.6955 ) + 2 (1.2353 + 2.1188 ) ⎤⎦ 3⎣

=

2 2 ( 3.8428 + 20.0844 + 6.7082 ) = ( 30.6354 ) 3 3

= 20.42 m 3


199


1. Determine the mean value of the periodic waveforms shown over half a cycle.

(a) Over half a cycle, mean value =

area under curve (2 × 10 ×10−3 )As = =2A length of base 10 ×10−3 s

1 ( 5 ×10−3 ) (100)Vs 2 = 50 V (b) Over half a cycle, mean value = 5 × 10−3 s 1 (15 ×10−3 ) ( 5) As 2 = 2.5 A (c) Over half a cycle, mean value = 15 ×10−3 s

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


200

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 ≈ (5 × 10−3 ) [ 0.4 + 1.6 + 3.8 + 5.7 + 4.9 + 2.2] = (5 × 10−3 ) [18.6] = 93 ×10−3 As = 0.093 As area 93 × 10−3 As Mean value = = = 3.1 A length of base 30 × 10−3 s

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 + 3.32 ) ⎤⎦ 3 1 1 (2) [9.06 + 49.56 + 15.08] = (2)(73.7) = 49.13 cm 2 3 3

49.13cm 2 Pa Mean value = × 90 ×103 = 368.5 kPa 12 cm cm


201

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.


202

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 =

(6

2

+ 1.7322 ) = 6.24 N

⎛ 1.732 ⎞ and θ = tan −1 ⎜ ⎟ = 16.10° ⎝ 6 ⎠

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 m / s 2

due north, 3 m / s 2 due east, 4 m / s 2 to the south-west and 5 m / s 2 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 = and

( 3.707

2

+ 4.3642 ) = 5.7 m / s 2

⎛ 4.364 ⎞ θ = tan −1 ⎜ ⎟ = 50° correct to 2 significant figures. ⎝ 3.707 ⎠

Hence, the resultant acceleration is 5.7 m / s 2 at 310° (i.e. at E 50° S)


203

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 = and

( 20.794

2

+ 3.3762 ) = 21.07 knots

⎛ 3.376 ⎞ θ = tan −1 ⎜ ⎟ = 9.22° ⎝ 20.794 ⎠

Hence, the speed of the ship is 21.07 knots and its actual direction is E 9.22° S


204

EXERCISE 94 Page 231 2. Calculate the resultant of (a) v1 + v 2 - v 3 (b) v 3 - v 2 + v1 when v1 = 15 m/s at 85°, v 2 = 25 m/s at 175° and v 3 = 12 m/s at 235°

(a) Total horizontal component of v1 + v 2 - v 3 = 15 cos 85° + 25 cos 175° - 12 cos 235° = -16.715 Total vertical component of v1 + v 2 - v 3 = 15 sin 85° + 25 sin 175° - 12 sin 235° = 26.952 From the diagram below, R =

(16.715

2

+ 26.9522 ) = 31.71 m/s,

⎛ 26.952 ⎞ α = tan −1 ⎜ ⎟ = 58.19° and thus θ = 180° - 58.19° = 121.81° ⎝ 16.715 ⎠

i.e. the resultant of v1 + v 2 - v 3 is 31.71 m/s at angle of 121.81° (b) Total horizontal component of v 3 - v 2 + v1 = 12 cos 235° - 25 cos 175° + 15 cos 85° = 19.329 Total vertical component of v 3 - v 2 + v1 = 12 sin 235° - 25 sin 175° + 15 sin 85° = 2.934 From the diagram below, R = and

(19.329

2

+ 2.9342 ) = 19.55 m/s,

⎛ 2.934 ⎞ θ = tan −1 ⎜ ⎟ = 8.63° ⎝ 19.329 ⎠

i.e. the resultant of v 3 - v 2 + v1 is 19.55 m/s at angle of 8.63°


205


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 =

( 79.2

2

+ 26.42 ) = 83.5 km/h

⎛ 79.2 ⎞ and θ = tan −1 ⎜ ⎟ = 71.6° ⎝ 26.4 ⎠

(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=

22 − 12 = 3 km/h =

⎛ 1000 ⎞ 3⎜ ⎟ m/min = 28.87 m/min ⎝ 60 ⎠

Hence, if the width of the river is 142 m, the swimmer will take

142 = 4.919 minutes 28.87

= 4 min 55 s © 2006 John Bird. All rights reserved. Published by Elsevier.

206

In the above diagram, sin θ =

1 2

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

( 22.46

2

+ 3.87 2 ) = 22.79 km/h,

Hence,

v=

and

⎛ 3.87 ⎞ θ = tan −1 ⎜ ⎟ = 9.78° ⎝ 22.46 ⎠

Hence, the ships actual speed is 22.79 km/h in a direction E 9.78° N


207

EXERCISE 96 Page 236 π⎞ ⎛ 2. Two alternating voltages are given by v1 = 10sin ωt volts and v 2 = 14sin ⎜ ωt + ⎟ volts . By 3⎠ ⎝ plotting v1 and v 2 on the same axes over one cycle obtain a sinusoidal expression for (a) v1 + v 2 (b) v1 - v 2

π⎞ ⎛ (a) v1 = 10sin ωt , v 2 = 14sin ⎜ ωt + ⎟ volts and v1 + v 2 are shown sketched below: 3⎠ ⎝

v1 + v 2 leads v1 by 36° = 36 ×

π = 0.63 rad 180

Hence, by measurement, v1 + v 2 = 20.9 sin(ωt + 0.63) volts


208

π⎞ ⎛ (b) v1 = 10sin ωt , v 2 = 14sin ⎜ ωt + ⎟ volts and v1 - v 2 are shown sketched below: 3⎠ ⎝

v1 - v 2 lags v1 by 78° = 78 ×

π = 1.36 rad 180

Hence, by measurement, v1 - v 2 = 12.5 sin(ωt – 1.36) volts π⎞ ⎛ 4. Express 7 sin ωt + 5sin ⎜ ωt + ⎟ in the form A sin ( ωt ± α ) using phasors. 4⎠ ⎝ The space diagram is shown in (a) below and the phasor diagram is shown in (b).

(a) Using the cosine rule: from which,

(b)

R 2 = 7 2 + 52 − 2(7)(5) cos135° = 123.497 R = 123.497 = 11.11


209

Using the sine rule:

5 11.11 = sin θ sin135°

from which,

sin θ =

5sin135° = 0.31823 11.11

θ = sin −1 0.31823 = 18.56° or 0.324 rad

and

π⎞ ⎛ Hence, in sinusoidal form, 7 sin ωt + 5sin ⎜ ωt + ⎟ = 11.11sin(ωt + 0.324) 4⎠ ⎝ π⎞ ⎛ 6. Express i = 25sin ωt − 15sin ⎜ ωt + ⎟ in the form A sin ( ωt ± α ) using phasors. 3⎠ ⎝

The relative positions of currents i1 and i 2 are shown in diagram (a) below. Phasor i 2 is shown reversed in diagram (b) to give - i 2 . The phasor diagram for i = i1 - i 2 is shown in diagram (c)

(a)

(b)

Using the cosine rule: from which, Using the sine rule: and

(c)

i 2 = 252 + 152 − 2(25)(15) cos 60° = 475 i = 475 = 21.79 15 21.79 = sin θ sin 60°

from which,

sin θ =

15sin 60° = 0.5962 21.79

θ = sin −1 0.5962 = 36.60° or 0.639 rad

π⎞ ⎛ Hence, in sinusoidal form, i = 25sin ωt − 15sin ⎜ ωt + ⎟ = 21.79sin(ωt − 0.639) 3⎠ ⎝ π⎞ 3π ⎞ ⎛ ⎛ 8. Express x = 9sin ⎜ ωt + ⎟ − 7 sin ⎜ ωt − ⎟ in the form A sin ( ωt ± α ) using phasors. 3⎠ 8 ⎠ ⎝ ⎝ π π 180° 3π 3π 180° = 60° and = 67.5° rad = × rad = × π π 3 3 8 8

The relative positions of currents x1 and x 2 are shown in diagram (a) below. Phasor x 2 is shown


210

reversed in diagram (b) to give - x 2 . The phasor diagram for x = x1 - x 2 is shown in diagram (c)

(a) Using the cosine rule: from which, Using the sine rule: and

(b)

(c)

x 2 = 92 + 7 2 − 2(9)(7) cos(60° + 67.5°) = 206.704 x = 206.704 = 14.377 7 14.377 = sin α sin127.5°

from which,

sin α =

7 sin127.5° = 0.3863 14.377

α = sin −1 0.3863 = 22.72°

Measured to the horizontal, θ = 60° + 22.72° = 82.72° or 1.444 rad π⎞ 3π ⎞ ⎛ ⎛ Hence, in sinusoidal form, x = 9sin ⎜ ωt + ⎟ − 7 sin ⎜ ωt − ⎟ = 14.38sin(ωt + 1.444) 3⎠ 8 ⎠ ⎝ ⎝


211

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

(a) p =

( (2)

2

+ (−3) 2 + (0) 2 ) = 13

(b) r =

( (1)

2

+ (2) 2 + (−3) 2 ) = 14

(b) r

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, cos θ =

=

from which,

a1b1 + a 2 b 2 + a 3 b3 a + a 2 2 + a 32 b12 + b 2 2 + b32 2 1

(2)(0) + (−3)(4) + (0)(−1) 2 + (−3) + 0 2

2

2

0 + 4 + (−1) 2

2

2

=

−12 = -0.8072 13 17

θ = cos −1 (−0.8072) = 143.82°


212

(0)(1) + (4)(2) + (−1)(−3)

(b) cos θ =

0 + 4 + (−1) 2

2

1 + 2 + (−3)

2

2

2

2

=

11 = 0.713024 17 14

θ = cos −1 (0.713024) = 44.52°

from which,

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, cos α =

(b) For q, cos α =

(c) For r, cos α =

2 2 + (−3) 2

2

0 4 + (−1) 2

2

=

2 −3 0 = 0.555, cos β = =0 = -0.832 and cos γ = 13 13 13

=

0 4 −1 = 0, cos β = = 0.970 and cos γ = = -0.243 17 17 17

1 12 + 22 + (−3) 2

=

1 2 −3 = 0.267, cos β = = 0.535 and cos γ = = -0.802 14 14 14

10. Find the angle between the velocity vector v1 = 5i + 2j + 7k and v 2 = 4i + j - k

cos θ =

(5)(4) + (2)(1) + (7)(−1) 5 +2 +7 2

from which,

2

2

4 + 1 + (−1) 2

2

2

=

15 = 0.40032 78 18

θ = cos −1 (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


213

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

i

j

k

i

j

k

0 2 3 2 3 0 (a) p × q = 3 0 2 = i -j +k = 4i – 7j – 6k −2 3 1 3 1 −2 1 −2 3

(b) q × p = 1 −2 3 = i 3 0 2

−2 3 1 3 1 −2 -j +k = -4i + 7j + 6k 0 2 3 2 3 0

2. When p = 3i + 2k, q = i – 2j + 3k and r = -4i + 3j – k, determine (a) p × r

(a) p × r =

( p ⋅ p )( r ⋅ r ) − ( p ⋅ r )

2

(b) r × q

where p ⋅ p = (3)(3) + (2)(2) = 13, r ⋅ r = (-4)(-4) +(3)(3) + (-1)(-1) = 26 and

Hence, p × r =

⎡⎣(13)(26) − (−14) 2 ⎤⎦ = (338 − 196) = 142 = 11.92

( r ⋅ r )( q ⋅ q ) − ( r ⋅ q )

(b) r × q = =

p ⋅ r = (3)(-4) + (0)(3) + (2)(-1) = -14

2

where r ⋅ r = (-4)(-4) + (3)(3) + (-1)(-1) = 26, q ⋅ q = (1)(1) +(-2)(-2) + (3)(3) = 14 and

Hence, r × q =

r ⋅ q = (-4)(1) + (3)(-2) + (-1)(3) = -13

⎡⎣(26)(14) − (−13) 2 ⎤⎦ = (364 − 169) = 195 = 13.96

4. When p = 3i + 2k, q = i – 2j + 3k and r = -4i + 3j – k, determine

(a) p × (r × q)

i

(b) (3p × 2r) × q

j

k

(a) r × q = − 4 3 −1 = 7i + 11j + 5k 1 −2 3


214

i

Hence,

j

k

p × (r × q) = 3 0 2 = (-22)i – (1)j + (33)k = -22i - j + 33k 7 11 5

(b) 3p × 2r =

i

j

k

9

0

6

= (-36)i – (-18 + 48)j + (54)k = -36i - 30j + 54k

− 8 6 −2 i

j

k

(3p × 2r) × q = − 36 −30 54 = (-90 + 108)i – (-108 - 54)j + (72 + 30)k 1 −2 3 = 18i + 162j + 102k 1 k and b = 6i – 5j – k find (i) a ⋅ b (ii) a × b (iii) a × b 2 (iv) b × a and (v) the angle between the vectors.

6. For vectors a = -7i + 4j +

(i)

1 1 a ⋅ b = (-7)(6) + (4)(-5) +( )(-1) = - 62 2 2

i

j

k 1 5⎞ 1 ⎛ (ii) a × b = − 7 4 = i ⎜ −4 + ⎟ - j(7 – 3) + k(35 – 24) = −1 i − 4j + 11k 2 2⎠ 2 ⎝ 6 −5 − 1 ⎛ 1 ⎞⎛ 1 ⎞ (iii) a ⋅ a = (-7)(-7) + (4)(4) + ⎜ ⎟ ⎜ ⎟ = 65.25 ⎝ 2 ⎠⎝ 2 ⎠ b ⋅ b = (6)(6) + (-5)(-5) + (-1)(-1) = 62 ⎛1⎞ a ⋅ b = (-7)(6) + (4)(-5) + ⎜ ⎟ ( −1) = - 62.5 ⎝2⎠

a×b =

⎡( 65.25 )( 62 ) − ( −62.5 )2 ⎤ = ⎣ ⎦

( 4045.5 − 3906.25) =

139.25 = 11.80

i 6

j k 1 ⎛ 5 ⎞ −5 −1 = i ⎜ − + 4 ⎟ - j(3 – 7) + k(24 – 35) = 1 i + 4j − 11k (iv) b × a = 2 ⎝ 2 ⎠ 1 −7 4 2

(v) cos θ =

⎛1⎞ (−7)(6) + (4)(−5) + ⎜ ⎟ (−1) ⎝2⎠ ⎛1⎞ (−7) + (4) + ⎜ ⎟ ⎝2⎠ 2

from which,

2

2

(6) 2 + (−5) 2 + (−1) 2

=

−62.5 = −0.9826388 65.25 62

θ = cos −1 (−0.9826388) = 169.31°


215

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 i

Moment, M = r × F

j

k

where M = − 4 3 2 = (3 + 2)i – (-4 - 4)j + (4 - 6)k 2 −1 1 = (5i + 8j - 2k) Nm

Magnitude of M, M = r × F = where

( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )

2

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 M =

i.e.

( 29 )( 6 ) − ( −9 )

2

= 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.

i

j

k

Velocity vector, v = ω × r = (3i – j + 2k) × (i – 5j + 4k) = 3 1 2 1 −5 4 = (-4 + 10)i – (12 - 2)j + (-15 + 1)k = 6i - 10j - 14k

Magnitude of v, v =

( ω ⋅ ω )( r ⋅ r ) − ( r ⋅ ω )

2

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 =

(14 )( 42 ) − (16 )

2

= 18.22 m/s


216


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: x − a1 y − a 2 z − a 3 = = =λ b1 b2 b3 Since a = 5i – 2j + 3k,

a1 = 5, a 2 = −2 and a 3 = 3

and b = 2i + 7j – 4k,

b1 = 2, b 2 = 7 and b3 = −4

then the Cartesian equations are:

x −5 y+ 2 z −3 = = =λ −4 2 7

4. Express the straight line equation 2x + 1 =

2x + 1 1 − 4y 3z − 1 = = 1 5 4

or

x−5 y+ 2 3− z = = =λ 2 7 4

1 − 4y 3z − 1 in vector form. = 5 4

1 1 1 z− y− 3 2= 4= 1 5 4 − 2 4 3

x+ i.e.

i.e.

1 1 1 a1 = − , a 2 = and a 3 = 2 4 3

and

1 5 4 b1 = , b 2 = − and b3 = 2 4 3

Hence, in vector form the equation is:

r = ( a1 + λb1 ) i + ( a 2 + λb 2 ) j + ( a 3 + λb3 ) k


217

⎛ 1 1 ⎞ ⎛1 5 ⎞ ⎛1 4 ⎞ = ⎜− + λ⎟i + ⎜ − λ⎟ j + ⎜ + λ⎟k ⎝3 3 ⎠ ⎝ 2 2 ⎠ ⎝4 4 ⎠ or

r=

1 1 1 ( λ − 1) i + ( 1 − 5λ ) j + ( 1 + 4λ ) k 2 4 3


218

CHAPTER 23 COMPLEX NUMBERS EXERCISE 100 Page 250 1. Solve the quadratic equation x 2 + 25 = 0

Since x 2 + 25 = 0

x 2 = −25

then

i.e.

x = −25 = (−1)(25) = −1 25 = j 25

from which,

x=±j5

2. Solve the quadratic equation 2x 2 + 3x + 4 = 0

Since 2x 2 + 3x + 4 = 0 then −3 ± ⎡⎣32 − 4(2)(4) ⎤⎦ −3 ± −23 −3 ± (−1)(23) −3 ± (−1) (23) −3 ± j (23) x= = = = = 2(2) 4 4 4 4

3 23 =- ±j 4 4 4. Evaluate (a) j8 (b) −

1 j7

(c)

or (-0.750 ± j1.199)

4 2 j13

(a) j8 = ( j2 ) = ( −1) = 1 4

4

(b) j7 = j × j6 = j × ( j2 ) = j × ( −1) = − j 3

Hence,

−

3

1 1 1 −j −j −j −j =− = = = 2 = = = -j 7 j − j j j(− j) − j −(−1) 1

(c) j13 = j × j12 = j × ( j2 ) = j × (−1)6 = j 6

Hence,

4 2 2(− j) − j2 − j2 = = = = = -j2 13 2j j j(− j) − j2 1


219


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 Z1 = 1 + j2, Z2 = 4 – j3, Z3 = -2 + j3 and Z4 = -5 – j, evaluate in a + jb form: (a) Z1 + Z2 − Z3

(b) Z2 − Z1 + Z4

(a) Z1 + Z2 − Z3 = 1 + j2 + 4 – j3 – (-2 + j3) = 1 + j2 + 4 – j3 + 2 – j3 = (1 + 4 + 2) + j(2 – 3 – 3) = 7 – j4 (b) Z2 − Z1 + Z4 = (4 – j3) – (1 + j2) + (-5 – j) = 4 – j3 – 1 – j2 – 5 – j = (4 – 1 – 5) + j(-3 – 2 – 1) = -2 - j6

5. Given Z1 = 1 + j2, Z2 = 4 – j3, Z3 = -2 + j3 and Z4 = -5 – j, evaluate in a + jb form: (a) Z1Z3 + Z4

(b) Z1Z2 Z3

(a) Z1Z3 + Z4 = (1 + j2)(-2 + j3) + (-5 – j) = -2 + j3 – j4 + j2 6 - 5 – j = -2 + j3 – j4 - 6 – 5 - j = -13 – j2 (b) Z1Z2 Z3 = (1 + j2)(4 – j3)(-2 + j3) = (4 – j3 + j8 - j2 6)(-2 + j3) = ( 10 + j5)(-2 + j3) = -20 + j30 – j10 + j2 15 = -20 + j30 –j10 – 15 = -35 + j20


220

7. Given Z1 = 1 + j2, Z2 = 4 – j3, Z3 = -2 + j3 and Z4 = -5 – j, evaluate in a + jb form: (a)

(a)

Z1Z3 Z1 + Z3

(b) Z2 +

Z1 + Z3 Z4

Z1Z3 (1 + j2)(−2 + j3) −2 + j3 − j4 + j2 6 −8 − j = = = Z1 + Z3 (1 + j2) + (−2 + j3) −1 + j5 −1 + j5 =

(b) Z2 +

3 41 (−8 − j)(−1 − j5) 8 + j40 + j + j2 5 3 + j41 = +j = = 2 2 26 26 (−1 + j5)(−1 − j5) 1 +5 26

Z1 1 + j2 (1 + j2)(−5 + j) + Z3 = (4 – j3) + + (-2 + j3) = 4 – j3 + - 2 + j3 Z4 −5 − j 52 + 12 = 4 – j3 +

−5 + j − j10 + j2 2 - 2 + j3 26

= 4 – j3 +

−7 − j9 7 9 - 2 + j3 = 2 −j 26 26 26

=

9. Show that

45 9 52 7 9 − −j = −j 26 26 26 26 26

−25 ⎛ 1 + j2 2 − j5 ⎞ − ⎜ ⎟ = 57 + j24 2 ⎝ 3 + j4 −j ⎠

1 + j2 (1 + j2)(3 − j4) 3 − j4 + j6 − j2 8 11 + j2 11 2 = = = = +j 2 2 3 + j4 3 +4 25 25 25 25 2 − j5 (2 − j5)( j) j2 − j2 5 5 + j2 = = = = 5 + j2 −j − j( j) − j2 1

L.H.S. =

−25 ⎛ 1 + j2 2 − j5 ⎞ 25 ⎡⎛ 11 2 ⎞ ⎤ 25 ⎡⎛ 11 ⎞ ⎛ 2 ⎞⎤ − ⎜ ⎟ = − ⎢⎜ + j ⎟ − (5 + j2) ⎥ = − ⎢⎜ − 5 ⎟ + j ⎜ − 2 ⎟ ⎥ 2 ⎝ 3 + j4 −j ⎠ 2 ⎣⎝ 25 25 ⎠ 2 ⎣⎝ 25 ⎠ ⎝ 25 ⎠⎦ ⎦ =−

25 ⎡⎛ 11 − 125 ⎞ ⎛ 2 − 50 ⎞ ⎤ 25 ⎡ 114 48 ⎤ −j ⎥ ⎜ ⎟ + j⎜ ⎟⎥ = − ⎢− ⎢ 2 ⎣⎝ 25 ⎠ ⎝ 25 ⎠ ⎦ 2 ⎣ 25 25 ⎦

= −

25 ⎛ 114 ⎞ 25 ⎛ 48 ⎞ ⎜− ⎟ + j ⎜ ⎟ = 57 + j24 = R.H.S. 2 ⎝ 25 ⎠ 2 ⎝ 25 ⎠


221

EXERCISE 102 Page 254 2. Solve the complex equation:

2+ j = j(x + jy) 1− j

2+ j (2 + j)(1 + j) = j(x + jy) hence, = j(x + jy) 1− j (1 − j)(1 + j) 2 + j2 + j + j2 = jx + j2 y 2 2 1 +1

i.e. i.e.

1 + j3 = jx − y 2

i.e.

1 3 + j = -y + jx 2 2

x=

Hence,

3 2

and y = −

1 2

3. Solve the complex equation: (2 − j3) = (a + jb)

(2 − j3) = (a + jb)

Squaring both sides gives:

( 2 − j3)

2

= a + jb

(2 – j3)(2 – j3) = a + jb i.e.

4 – j6 – j6 + j2 9 = a + jb

i.e.

-5 – j12 = a + jb a = -5 and b = -12

Hence, 5. If Z = R + jωL +

Z = R + jωL +

1 , express Z in (a + jb) form when R = 10, L = 5, C = 0.04 and ω = 4 jω C

1 1 6.25 6.25(− j) = 10 + j(4)(5) + = 10 + j20 + = 10 + j20 + jω C j(4)(0.04) j j(− j) = 10 + j20 -

6.25 = 10 + j20 – j6.25 − j2 = 10 + j13.75


222


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 =

22 + 32 = 13

⎛3⎞ θ = tan −1 ⎜ ⎟ = 56.31° or 56°19 ' ⎝2⎠

and

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, (c) -6 + j and

Thus,

-4 = 4∠180° in polar form

From the diagram below, r = ⎛1⎞ α = tan −1 ⎜ ⎟ = 9.46° or 9°28' ⎝6⎠

-6 + j =

62 + 12 = 37

thus

θ = 180° - 9°28′ = 170°32′

37∠170°32'


223

3. Express the following Cartesian complex numbers in polar form, leaving answers in surd form:

(a) - j3

(a) - j3

(b)

(c) j3 (1 − j)

3

From the diagram below, r = 3 and θ = -90°

- j3 = 3∠-90° in polar form

Hence, (b)

( −2 + j)

( −2 + j)

3

= (-2 + j)(-2 + j)(-2 + j) = (4 – j2 – j2 + j2 )(-2 + j) = (3 – j4)(-2 + j) = -6 + j3 + j8 - j2 4 = -2 + j11

From the diagram below, r =

22 + 112 = 125

and

⎛ 11 ⎞ α = tan −1 ⎜ ⎟ = 79.70° or 79°42 ' ⎝2⎠

θ = 180° - 79°42′ = 100°18′

and

Hence,

( −2 + j )

3

= -2 + j11 =

125∠100°18' in polar form

(c) j3 (1 − j) = (j)( j2 )(1 – j) = -j(1 – j) = -j + j2 = -1 – j From the diagram below, r = 12 + 12 = 2

and

⎛1⎞ α = tan −1 ⎜ ⎟ = 45° ⎝1⎠

θ = 180° - 45° = 135°

and

Hence,

j3 (1 − j) = -1 – j =

2∠ − 135°


224

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°

(a) 6.4∠27° ÷ 2∠ − 15° =

(b) 5∠30°× 4∠80° ÷ 10∠ − 40°

6.4∠27° 6.4 = ∠27° − −15° = 3.2∠42° 2∠ − 15° 2

(b) 5∠30°× 4∠80° ÷ 10∠ − 40° =

5∠30°× 4∠80° 5 × 4 = ∠(30° + 80° − −40°) = 2∠150° 10∠ − 40° 10

π π 8. Evaluate in polar form: (a) 4∠ + 3∠ 6 8

(b) 2∠120° + 5.2∠58° − 1.6∠ − 40°

π π π π⎞ ⎛ π π⎞ ⎛ (a) 4∠ + 3∠ = ⎜ 4 cos + j4sin ⎟ + ⎜ 3cos + j3sin ⎟ = (3.464 + j2) + (2.772 + j1.148) 6 8 6 6⎠ ⎝ 8 8⎠ ⎝ = 6.236 + j3.148 From the diagram below, r = and

6.2362 + 3.1482 = 6.986

⎛ 3.148 ⎞ θ = tan −1 ⎜ ⎟ = 26.79° or 26°47 ' or 0.467rad ⎝ 6.236 ⎠


225

4∠

Hence,

π π + 3∠ = 6.986∠26°47′ 6 8

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 =

0.5302 + 7.1702 = 7.190

⎛ 7.170 ⎞ θ = tan −1 ⎜ ⎟ = 85.77° or 85°46 ' ⎝ 0.530 ⎠

and

Hence,

2∠120° + 5.2∠ 58° − 1.6∠ − 40° = 7.190∠85°46′


226


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, X L = 8 Ω (since the j term is positive) X L = 2πfL = 8 hence, inductance, L =

8 8 = = 0.0255 H or 25.5 mH 2πf 2π(50)

(b) If Z = (2 – j3) Ω then resistance, R = 2 Ω

and capacitive reactance, X C = 3 Ω (since the j term is negative)

XC =

1 1 1 = 3 hence, capacitance, C = = = 1.061× 10−3 or 1061× 10−6 2πfC 2πf (3) 2π(50)(3) = 1061 µF

(c) If Z = j14 Ω i.e. Z = (0 + j14) Ω then resistance, R = 0 Ω i.e.

2πfL = 14 hence, inductance, L =

and X L = 14 Ω

14 = 0.04456 H or 44.56 mH 2π(50)

(d) If Z = 8∠ − 60°Ω = 8 cos(-60°) + j 8 sin(-60°) = (4 – j6.928) Ω Hence, i.e.

resistance, R = 4 Ω

1 = 6.928 2πfC

and X C = 6.928 Ω

and capacitance, C =

1 = 459.4 × 10−6 = 459.4 µF 2π(50)(6.928)

2. Two impedances, Z1 = (3 + j6) Ω and Z2 = (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, ZTOTAL = Z1 + Z2 = (3 + j6) + (4 – j3) = (7 + j3) Ω

⎛3⎞ 7 2 + 32 ∠ tan −1 ⎜ ⎟ ⎝7⎠ = 7.616∠23.20° Ω or 7.616∠23°12′ Ω

=

Since voltage V = 120∠0° V, then current, I =

V 120∠0° = = 15.76∠-23°12′ A Z 7.616∠23°12 '

i.e. the current is 15.76 A and is lagging the voltage by 23°12′


227

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, ZT is given by: 1 1 1 1 1 3 − j6 4 + j3 3 6 4 3 = + = + = 2 + 2 2= −j + +j 2 ZT Z1 Z2 3 + j6 4 − j3 3 + 6 4 + 3 45 45 25 25 1 = it tan ce, YT = 0.22667 – j0.01333 = 0.2271∠3°22′ siemen ZT

i.e.

Current, I =

V = VYT = (120∠0°)(0.2271∠3°22 ') = 27.25∠3°22 ' A ZT

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.

1 1 1 1 1 1 1 30 + j20 40 − j50 1 = + + = + + = 2 + 2 + 2 2 ZT Z1 Z2 Z3 30 − j20 40 + j50 25 30 + 20 40 + 50 25 = i.e.

30 20 40 50 1 +j + −j + 1300 1300 4100 4100 25

1 = it tan ce, YT = 0.07283 + j0.00319 = 0.0729∠2°30′ S ZT


228

Current, I =

V = VYT = (200∠0°)(0.0729∠2°30 ') = 14.6∠2°30 ' A ZT

i.e. the current is 14.6 A and is leading the voltage by 2°30′

7. A delta-connected impedance ZA is given by:

ZA =

Z1Z2 + Z2 Z3 + Z3 Z1 Z2

Determine ZA in both Cartesian and polar form given Z1 = (10 + j0) Ω , Z2 = (0 − j10) Ω and Z3 = (10 + j10) Ω .

ZA =

Z1Z2 + Z2 Z3 + Z3 Z1 (10 + j0)(0 − j10) + (0 − j10)(10 + j10) + (10 + j10)(10 + j0) = Z2 (0 − j10) =

− j100 − j100 − j2100 + 100 + j100 200 − j100 200 j100 = = − − j10 − j10 − j10 − j10

=

j200 + 10 = (10 + j20) Ω 10

From the diagram below, r = 102 + 202 = 22.36

Hence,

and

⎛ 20 ⎞ θ = tan −1 ⎜ ⎟ = 63.43° ⎝ 10 ⎠

Z A = (10 + j20) Ω = 22.36∠63.43° Ω

8. In the hydrogen atom, the angular momentum, p, of the de Broglie wave is given by: ⎛ jh ⎞ pΨ = − ⎜ ⎟ ( ± jm Ψ ) ⎝ 2π ⎠ Determine an expression for p. h 2 ⎛ jh ⎞⎛ ± jm Ψ ⎞ ⎛ jh ⎞ ⎛ jh ⎞ If pΨ = − ⎜ ⎟ ( ± jm Ψ ) then p = − ⎜ ⎟⎜ ⎟ = − ⎜ ⎟ ( ± jm ) = − ( j ) ( ± m ) 2π ⎝ 2π ⎠⎝ Ψ ⎠ ⎝ 2π ⎠ ⎝ 2π ⎠ mh h = ( ±m ) = ± 2π 2π


229

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))

⎛ 1.464 ⎞ α = tan −1 ⎜ ⎟ = 79.63° ⎝ 0.268 ⎠

From the diagram below, r = 1.488 and

θ = 180° − 79.63° = 100.37°

and

Hence,

P – Q – R = 1.488∠100.37°

11. In a Schering bridge circuit, ZX = (R X − jX CX ) , Z2 = − jX C2 , Z3 =

where X C =

1 . At balance: 2πfC

Show that at balance R X =

Since then

Thus,

(R

) )

3

3

− jX C3

and Z4 = R 4 ,

( ZX )( Z3 ) = ( Z2 )( Z4 ) .

C3 R 4 C R and CX = 2 3 C2 R4

( ZX )( Z3 ) = ( Z2 )( Z4 )

(

⎧⎪ ( R 3 ) − jX C 3 (R X − jX CX ) ⎨ ⎪⎩ R 3 − jX C3

) ⎫⎪ = ⎬ ⎪⎭

(R X − jX CX ) =

( − jX ) ( R ) C2

(R

3

4

) ( − jX ) ( R ) ( R ) ( − jX )

− jX C3

C2

3

i.e.

( R 3 ) ( − jX C

(R X − jX CX ) =

C3

− jR 3 X C2 R 4

( R 3 ) ( − jX C

4

3

+

j2 X C3 X C2 R 4

) ( R ) ( − jX ) 3

C3


230

(R X − jX CX ) =

i.e.

=

(R X − jX CX ) =

i.e.

X C2 R 4 X C3 X C2 R 4 X C3 X C2 R 4 X C3

−

+

X C2 R 4

( R 3 ) (− j) X C2 R 4

−j

jR 3 X C2 R 4 R3

Equating the real parts gives:

1 R X C2 R 4 2πfC2 4 2πfC3 = = RX = R4 1 X C3 2πfC2 2πfC3

i.e.

RX =

C3 R 4 C2

Equating the imaginary parts gives: − X CX = −

X C2 R 4 R3

1 R4 2πfC2 R4 1 = = 2πfCX R3 2πfC2 R 3

i.e.

from which,

CX =

C2 R 3 R4


231

CHAPTER 24 DE MOIVRE’S THEOREM EXERCISE 105 Page 261 2. Determine in polar and Cartesian forms (a) [3∠41°]

4

(b)

( −2 − j)

5

(a) [3∠41°] = 34 ∠4 × 41° = 81∠164° = 8 cos 164° + j 8 sin 164° = -77.86 + j22.33 4

(b)

( −2 − j)

5

5

= ⎡⎣ 5∠ − 153.435°⎤⎦ =

( 5 ) ∠5 × −153.435° 5

= 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 ( 3 − j) , giving the answer in polar form. 7

(3 – j) =

⎛ 1⎞ 32 + 12 ∠ tan −1 ⎜ − ⎟ = ⎝ 3⎠

10∠ − 18°26'

7 Hence, ( 3 − j) = ⎡⎣ 10∠ − 18°26 '⎤⎦ = 7

( 10 ) ∠7 × −18°26 ' = 3162∠-129°2′ 7

5. Express in both polar and rectangular forms: ( 3 − j8 )

( 3 − j8)

5

5

= ⎡⎣ 73∠ − 69.444°⎤⎦ =

(

)

5

5

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:

( −16 − j9 )

From the diagram below, r = 162 + 92 = 337

and

⎛9⎞ α = tan −1 ⎜ ⎟ = 29.358° ⎝ 16 ⎠

θ = 180° + 29.358° = 209.358°

and Hence,

6

( −16 − j9 )

6

6

= ⎡⎣ 337∠209.358° ⎤⎦ =

(

)

6

337 ∠6 × 209.358°


232

= 38.27 ×106 ∠1256.148° = (38.27 × 106 )∠176°9'

(38.27 × 106 )∠176°9 ' = 106 ( 38.27 cos176°9 '+ j38.27 sin176°9 ')

= 106 ( −38.18 + j2.570)


233


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

(a)

3 − j4 =

(b) -1 – j2

1

[5∠ − 53.13°] = [5∠ − 53.13°] 2

The first root is:

1 1 5 2 ∠ × −53.13° = 2.236∠ − 26.57° = (2 − j1) 2

52.236∠(−26.57° + 180°) = (−2 + j1)

and the second root is:

(3 − j4) = ± (2 − j) as shown in the Argand diagram of Figure (a) below.

Hence,

(b)

(a) 1

(b)

−1 − j2 = ⎡⎣ 5 ∠243.435°⎤⎦ = ⎡⎣ 5 ∠243.435°⎤⎦ 2

The first root is:

( 5)

and the second root is: Hence,

1 2

1 ∠ × 243.435° = 1.495∠121.72° = (−0.786 + j1.272) 2 1.495∠(121.72° − 180°) = 1.495∠ − 58.28 = (0.786 − j1.272)

( −1 − j2) = ± (0.786 − j1.272) as shown in the Argand diagram if Figure (b) above. 1

4. Determine the modulus and argument of the complex number: ( 3 + j4 ) 3

1 1 1 1 3 3 3 3 + j4 = 5 ∠ 53.13 ° = 5 ∠ × 53.13° = 3 5∠17.71° = 1.710∠17°43' ( ) ( ) 3

Hence, the modulus is 1.710, and the arguments are 17°43′, 17°43′ + and

137°43′ +

360° = 137°43′, 3

360° = 257°43′ since the three roots are equally displaced by 120°. 3 1

5. Determine the modulus and argument of the complex number:

( −2 + j) 4


234

1

( −2 + j) 4 = ⎡⎣

1

5∠153.435°⎤⎦ 4 =

( ) 5

1 4

1 ∠ × 153.435° = 1.223∠38°22 ' 4

360° , i.e. 90° apart. 4

There are 4 roots equally displaced

Hence, the modulus is 1.223, and the arguments are: 38°22′, 128°22′ + 90° = 218°22′

and

218°22′ + 90° = 308°22′

7. Determine the modulus and argument of the complex number:

( 4 − j3)

−

2 3

= [5∠ − 36.87°]

−

2 3

= ( 5)

−

There are 3 roots equally displaced

2 3

38°22′ + 90° = 128°22′,

( 4 − j3)

−

2 3

2 ∠ − × −36.87° = 0.3420∠24°35' 3 360° , i.e. 120° apart. 3

Hence, the modulus is 0.3420, and the arguments are: 24°35′,

24°35′ + 120° = 144°35′,

144°35′ + 120° = 264°35′

and

8. For a transmission line, the characteristic impedance Z0 and the propagation coefficient γ are given by:

⎛ R + jω L ⎞ Z0 = ⎜ ⎟ ⎝ G + jωC ⎠

and

γ = ⎡⎣( R + jωL )( G + jωC ) ⎤⎦

Given R = 25 Ω, L = 5 ×10−3 H, G = 80 ×10−6 siemens, C = 0.04 × 10−6 F and ω = 2000π rad/s, determine, in polar form, Z0 and γ. R + jωL = 25 + j(2000π) ( 5 × 10−3 ) = 25 + j31.416 = 40.15∠51.49° G + jωC = 80 ×10−6 + j(2000π) ( 0.04 × 10−6 ) = 10−6 (80 + j251.33) = 263.755 × 10−6 ∠72.34° Hence,

⎛ R + jωL ⎞ 40.15∠51.49° ⎛ ⎞ Z0 = ⎜ ⎟= ⎜ ⎟= −6 ⎝ 263.755 × 10 ∠72.34° ⎠ ⎝ G + jωC ⎠ = 152224.6 ∠

(152224.6∠ − 20.85° )

1 ( −20.85° ) = 390.2∠-10.43° Ω 2

γ = ⎡⎣( R + jωL )( G + jωC ) ⎤⎦ = ⎡⎣( 40.15∠51.49° ) ( 263.755 × 10−6 ∠72.34° ) ⎤⎦ =

( 0.01058976∠123.83° ) =

1 0.01058976 ∠ × 123.83° 2

= 0.1029∠61.92°


235


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 e j 2.11

4. Convert 1.7e1.2 − j 2.5 into rectangular form. 1.7e1.2 − j 2.5 = (1.7e1.2 )( e− j 2.5 ) = 1.7e1.2 ∠ − 2.5 rad = 1.7e1.2 cos(−2.5) + j1.7e1.2 sin(−2.5) = -4.52 – j3.38

6. If z = 7e j 2.1 , determine ln z (a) in Cartesian form, and (b) in polar form. (a) If z = 7e j 2.1 then ln z = ln ( 7e j 2.1 ) = ln 7 + ln e j 2.1 = 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)

(a) ln(2 + j5) = ln

(

)

29∠1.19 = ln

(

(b) ln(-4 – j3)

)

29 e j 1.19 = ln 29 + ln e j 1.19 = ln 29 + j1.19 = 1.6836 + j1.19 = 2.06∠35.25° or 2.06∠0.615 rad

(b) ln(-4 – j3) = ln ( 5∠216.87° ) = ln ( 5∠3.785 ) = ln ( 5e j3.785 ) = 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: x = Ae

⎧ ⎪ ht +j ⎨− ⎪ 2m ⎩

( 4mf − h ) t ⎫⎪ 2

2m − a

⎬ ⎪ ⎭

Determine the real part of x in of t, assuming ( 4mf − h 2 ) is positive.


236

x = Ae

⎧ ⎪ ht +j ⎨− ⎪ 2m ⎩

( 4mf − h ) t ⎫⎪ 2

2m − a

⎬ ⎪ ⎭

ht ⎛ − 2m ⎞⎛ j = ⎜ Ae ⎟ ⎜⎜ e ⎝ ⎠⎝

= Ae

Hence, the real part is:

−

Ae

ht 2m

−

4mf − h 2 t 2m − a

⎞ ⎛ − h t ⎞ ⎛ 4mf − h 2 ⎟ = ⎜ Ae 2m ⎟ ∠⎜ ⎟ ⎠ ⎝⎜ 2m − a ⎠ ⎝

⎛ 4mf − h 2 cos ⎜ ⎜ 2m − a ⎝

ht 2m

⎞ ⎟t ⎟ ⎠

ht ⎞ ⎛ 4mf − h 2 − ⎟ t + jAe 2m sin ⎜ ⎟ ⎜ 2m − a ⎠ ⎝

⎛ 4mf − h 2 cos ⎜ ⎜ 2m − a ⎝

⎞ ⎟t ⎟ ⎠

⎞ ⎟t ⎟ ⎠


237

CHAPTER 25 THE THEORY OF MATRICES AND DETERMINANTS EXERCISE 108 Page 270 ⎛ 6 ⎜3 ⎛ 4 −7 6 ⎞ ⎜ 2 ⎜ ⎟ 2. Determine ⎜ −2 4 0 ⎟ + ⎜ 5 − ⎜ 3 ⎜ 5 7 −4 ⎟ ⎜ ⎝ ⎠ ⎜⎜ −1 0 ⎝ ⎛ 6 ⎜3 4 7 6 − ⎛ ⎞ ⎜ 2 ⎜ ⎟ ⎜ ⎜ −2 4 0 ⎟ + ⎜ 5 − 3 ⎜ 5 7 −4 ⎟ ⎜ ⎝ ⎠ ⎜⎜ −1 0 ⎝

1⎞ 2⎟ ⎟ 7⎟ ⎟ ⎟ 3⎟ ⎟ 5⎠

1⎞ ⎛ 1 ⎞ ⎛ (4 + 3) (−7 + 6) (6 + ) ⎟ ⎜ 7 −1 ⎟ ⎜ 2 2 ⎟ ⎜ ⎟ ⎜ 1 2 ⎟ ⎜ 7 = (−2 + 5) (4 + − ) (0 + 7) ⎟ = ⎜ 3 3 ⎜ ⎟ ⎜ ⎟ 3 3 ⎟ ⎜ ⎟ ⎜ 3⎟ ⎜ 3 ⎜ ⎟ ⎜ (5 + −1) (7 + 0) (−4 + ) ⎟⎟ ⎜ 4 7 5⎠ ⎝ 5 ⎠ ⎝

1 ⎞ 2 ⎟ ⎟ 7 ⎟ ⎟ ⎟ 2⎟ −3 ⎟ 5⎠ 6

2 ⎞ ⎛ 1 ⎜ ⎛ 3 −1⎞ 2 3 ⎟ ⎛ −1.3 7.4 ⎞ 4. Determine ⎜ + ⎜ ⎟−⎜ ⎟ ⎟ ⎝ −4 7 ⎠ ⎜ − 1 − 3 ⎟ ⎝ 2.5 −3.9 ⎠ ⎜ ⎟ 5⎠ ⎝ 3 2 ⎞ 1 2 ⎛ 1 ⎛ ⎞ (3 + − −1.3) (−1 + − 7.4 ⎟ ⎟ ⎜ − 1.3 7.4 ⎛ 3 −1⎞ ⎜ 2 ⎛ ⎞ 3 2 3 ⎟−⎜ ⎟ ⎜ ⎟+⎜ ⎟=⎜ ⎝ −4 7 ⎠ ⎜ − 1 − 3 ⎟ ⎝ 2.5 −3.9 ⎠ ⎜ (−4 + − 1 − 2.5) (7 + − 3 − −3.9) ⎟ ⎜ ⎟ ⎜ ⎟ 5⎠ 3 5 ⎝ 3 ⎝ ⎠ (−1 + 0.666 − 7.4 ⎞ ⎛ (3 + 0.5 + 1.3) =⎜ ⎟ (7 − 0.6 + 3.9) ⎠ ⎝ (−4 − 0.333 − 2.5) ⎛ ⎞ − 4.8 7.7 3 ⎜ ⎟ = ⎜⎜ ⎟⎟ ⎝ −6.8 3 10.3 ⎠

⎛ 6 ⎜3 4 − 7 6 ⎛ ⎞ ⎜ 2 ⎜ ⎟ 6. Determine 2 ⎜ −2 4 0 ⎟ + 3 ⎜ 5 − ⎜ 3 ⎜ 5 7 −4 ⎟ ⎜ ⎝ ⎠ ⎜⎜ −1 0 ⎝

1⎞ 2 ⎟ ⎛ 3.1 2.4 6.4 ⎞ ⎟ ⎜ ⎟ 7 ⎟ − 4 ⎜ −1.6 3.8 −1.9 ⎟ ⎟ ⎟ ⎜ 5.3 3.4 −4.8 ⎟⎠ 3⎟ ⎝ ⎟ 5⎠


238

⎛ 6 ⎜3 ⎛ 4 −7 6 ⎞ ⎜ 2 ⎜ ⎟ 2 ⎜ −2 4 0 ⎟ + 3 ⎜ 5 − ⎜ 3 ⎜ 5 7 −4 ⎟ ⎜ ⎝ ⎠ ⎜ −1 0 ⎜ ⎝

1⎞ 2 ⎟ ⎛ 3.1 2.4 6.4 ⎞ ⎟ ⎜ ⎟ 7 ⎟ − 4 ⎜ −1.6 3.8 −1.9 ⎟ = ⎟ ⎟ ⎜ 5.3 3.4 −4.8 ⎟⎠ 3⎟ ⎝ ⎟ 5⎠

⎛ (8 + 9 − 12.4) (−14 + 18 − 9.6) (12 + 1.5 − 25.6) ⎞ ⎜ ⎟ (0 + 21 + 7.6) ⎟ ⎜ (−4 + 15 + 6.4) (8 − 2 − 15.2) ⎜ (10 − 3 − 21.2) (14 + 0 − 13.6) (−8 + 1.8 + 19.2 ⎟ ⎝ ⎠

−5.6 −12.1 ⎞ ⎛ 4.6 ⎜ ⎟ = ⎜ 17.4 −9.2 28.6 ⎟ ⎜ − 14.2 0.4 13.0 ⎟ ⎝ ⎠ 2 ⎞ ⎛ 1 ⎛ 3 −1⎞ ⎜ 2 3 ⎟ 8. Determine ⎜ × ⎜ ⎟ ⎟ ⎝ −4 7 ⎠ ⎜ − 1 − 3 ⎟ ⎜ ⎟ 5⎠ ⎝ 3 2 ⎞ ⎛ 3 1 3 ⎞ ⎛ 5 ⎛ 1 ( + ) (2 + ) ⎟ ⎜ 1 ⎜ ⎟ ⎜ − 3 1 ⎛ ⎞ 6 2 3 2 3 5 ⎟=⎜ ⎟=⎜ ⎜ ⎟×⎜ ⎝ −4 7 ⎠ ⎜ − 1 − 3 ⎟ ⎜ (−2 − 7 ) (− 8 − 21) ⎟ ⎜ −4 1 ⎜ ⎟ ⎜ ⎟ ⎜ 5⎠ ⎝ 3 3 5 ⎠ ⎝ 3 ⎝ 3

3 ⎞ 5 ⎟ ⎟ 13 ⎟ −6 ⎟ 15 ⎠ 2

⎛ 4 −7 6 ⎞ ⎛ 4 ⎞ 10. Determine ⎜⎜ −2 4 0 ⎟⎟ × ⎜⎜ −11⎟⎟ ⎜ 5 7 −4 ⎟ ⎜ 7 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 4 −7 6 ⎞ ⎛ 4 ⎞ ⎛ (16 + 77 + 42) ⎞ ⎛ 135 ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ −2 4 0 ⎟ × ⎜ −11⎟ = ⎜ (−8 − 44 + 0) ⎟ = ⎜ −52 ⎟ ⎜ 5 7 −4 ⎟ ⎜ 7 ⎟ ⎜ (20 − 77 − 28) ⎟ ⎜ −85 ⎟ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎛ 4 −7 6 ⎞ ⎛ 3.1 2.4 6.4 ⎞ 12. Determine ⎜⎜ −2 4 0 ⎟⎟ × ⎜⎜ −1.6 3.8 −1.9 ⎟⎟ ⎜ 5 7 −4 ⎟ ⎜ 5.3 3.4 −4.8 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 4 −7 6 ⎞ ⎛ 3.1 2.4 6.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −2 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 10.1 ⎞ ⎜ ⎟ = ⎜ −12.6 10.4 −20.4 ⎟ ⎜ −16.9 25.0 37.9 ⎟ ⎝ ⎠


239

EXERCISE 109 Page 271 2 ⎞ ⎛ 1 ⎜ 2 3 ⎟ 2. Calculate the determinant of ⎜ ⎟ ⎜−1 − 3⎟ ⎜ ⎟ 5⎠ ⎝ 3 1 2 7 3 2 −27 + 20 ⎛ 1 ⎞⎛ 3 ⎞ ⎛ 2 ⎞⎛ 1 ⎞ 2 3 = − = ⎜ ⎟⎜ − ⎟ − ⎜ ⎟⎜ − ⎟ = − + = 90 1 3 10 9 90 ⎝ 2 ⎠⎝ 5 ⎠ ⎝ 3 ⎠⎝ 3 ⎠ − − 3 5

4. Evaluate

j2 − j3 (1 + j) j

j2 − j3 = (j2)(j) – (-j3)(1 + j) = j2 2 + j3(1 + j) = -2 + j3 + j2 3 = -2 + j3 – 3 = -5 + j3 (1 + j) j

5. Evaluate

2∠40° 5∠ − 20° 7∠ − 32° 4∠ − 117°

2∠40° 5∠ − 20° = ( 2∠40° )( 4∠ − 117° ) − ( 5∠ − 20° )( 7∠ − 32° ) 7∠ − 32° 4∠ − 117° = 8∠-77° - 35∠-52° = (1.800 – j7.795) – (21.548 – j27.580) = (-19.75 + j19.79) From the diagram below, r = and

Hence,

(19.752 + 19.792 ) = 27.96

and

⎛ 19.79 ⎞ α = tan −1 ⎜ ⎟ = 45.06° ⎝ 19.75 ⎠

θ = 180° − 45.06° = 134.94°

2∠40° 5∠ − 20° = (-19.75 + j19.79) 7∠ − 32° 4∠ − 117°

or 27.96∠134.94°


240

EXERCISE 110 Page 272 2 ⎞ ⎛ 1 ⎜ 2 3 ⎟ 2. Determine the inverse of ⎜ ⎟ ⎜−1 − 3⎟ ⎜ ⎟ 5⎠ ⎝ 3 1 2 3 2 2 3 20 − 27 7 2 3 = − −− = − = =− 1 3 10 9 9 10 90 90 − − 3 5 90 ⎛ 2 ⎞ ⎞ 2 ⎞ ⎛ 90 ⎛ 3 ⎞ 2 ⎞ ⎛ 1 ⎛ 3 − − ⎟ ⎜− ⎜− ⎟ − ⎜− ⎟⎟ ⎜ 2 ⎟ ⎜ 7 ⎝ 5⎠ 7 ⎝ 3 ⎠⎟ 1 3 5 3 Hence, the inverse of ⎜ ⎟ is: ⎜ ⎟=⎜ 7 1 ⎟ ⎜ 90 ⎛ 1 ⎞ 90 ⎛ 1 ⎞ ⎟ ⎜−1 − 3⎟ − ⎜⎜ 1 − ⎜ ⎟ − ⎜ ⎟ ⎟ ⎜ ⎟ ⎟ ⎜ 90 ⎝ 3 5⎠ 2 ⎠ ⎝ 7 ⎝3⎠ ⎝ 3 7 ⎝2⎠ ⎠

⎛ 54 ⎜ 7 =⎜ ⎜ − 30 ⎜ ⎝ 7

60 ⎞ ⎛ 5 7 7 ⎟ ⎜ 7 ⎟=⎜ 45 ⎟ ⎜ 2 − ⎟ ⎜ −4 7 ⎠ ⎝ 7

4 ⎞ 7 ⎟ ⎟ 3⎟ −6 ⎟ 7⎠ 8

⎛ −1.3 7.4 ⎞ 3. Determine the inverse of ⎜ ⎟ ⎝ 2.5 −3.9 ⎠ ⎛ −1.3 7.4 ⎞ ⎛ −3.9 −7.4 ⎞ 1 The inverse of ⎜ ⎟ is: ⎜ ⎟ (−1.3)(−3.9) − (7.4)(2.5) ⎝ −2.5 −1.3 ⎠ ⎝ 2.5 −3.9 ⎠ 7.4 ⎞ ⎛ 3.9 ⎜ 1 ⎛ −3.9 −7.4 ⎞ 13.43 13.43 ⎟ = ⎟ ⎜ ⎟=⎜ 1.3 ⎟ −13.43 ⎝ −2.5 −1.3 ⎠ ⎜ 2.5 ⎜ ⎟ ⎝ 13.43 13.43 ⎠

⎛ 0.290 0.551 ⎞ =⎜ ⎟ ⎝ 0.186 0.097 ⎠


241

EXERCISE 111 Page 274 ⎛ 4 −7 6 ⎞ 3. Calculate the determinant of ⎜⎜ −2 4 0 ⎟⎟ ⎜ 5 7 −4 ⎟ ⎝ ⎠ 4

−7

6

−2

4

0 = 4(−16) − (−7)(8) + 6(−34) using the top row

5

7

−4

= -64 + 56 – 204 = -212 ⎛ 3.1 2.4 6.4 ⎞ 5. Calculate the determinant of ⎜⎜ −1.6 3.8 −1.9 ⎟⎟ ⎜ 5.3 3.4 −4.8 ⎟ ⎝ ⎠ 3.1

2.4

6.4

−1.6 3.8 −1.9 = 3.1(−18.24 + 6.46) − 2.4(7.68 + 10.07) + 6.4(−5.44 − 20.14) using the first row 5.3 3.4 −4.8

= 3.1(-11.78) – 2.4(17.75) + 6.4(-25.58) = -36.518 – 42.6 – 163.712 = -242.83 3∠60°

7. Evaluate

0 0

j2

1

(1 + j) 2∠30°

0 0

3∠60°

j2 2

j5

1

(1 + j) 2∠30° = 3∠60° [ j5(1 + j) − 4∠30°] − j2(0 − 0) + 1(0 + 0) using the top row 2 j5

= 3∠60° ( j5 + j2 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: (2 − λ) 2 (a) =0 (5 − λ) −1

(b)

(5 − λ )

7

0 2

(4 − λ) 8

−5 −1 =0 ( −3 − λ )


242

(a)

(2 − λ) 2 =0 (5 − λ) −1

hence, (2 - λ)(5 - λ) – (-2) = 0

i.e.

10 - 7λ + λ 2 + 2 = 0

i.e.

λ 2 - 7λ + 12 = 0

and

(λ - 4)( λ - 3) = 0 λ-4=0

from which,

eignevalues, λ = 3 or 4

Thus,

(5 − λ ) 7 0 (4 − λ) (b) 2 8 hence,

or λ - 3 = 0

−5 −1

=0

( −3 − λ )

(5 − λ) ⎡⎣( 4 − λ )( −3 − λ ) + 8⎤⎦ − 7(0 + 2) − 5 [ 0 − 2(4 − λ ) ] = 0

i.e.

(5 − λ) ⎡⎣ −12 − λ + λ 2 + 8⎤⎦ − 14 + 40 − 10λ = 0

and

( 5 − λ ) ( λ 2 − λ − 4 ) + 26 − 10λ = 0

i.e.

5λ 2 − 5λ − 20 − λ 3 + λ 2 + 4λ + 26 − 10λ = 0

and

−λ3 + 6λ 2 − 11λ + 6 = 0 λ 3 − 6λ 2 + 11λ − 6 = 0

or Let f(λ) = λ 3 − 6λ 2 + 11λ − 6 f(0) = -6 f(1) = 1 – 6 + 11 – 6 = 0

hence, (λ - 1) is a factor

f(2) = 8 – 24 + 22 – 6 = 0

hence, (λ - 2) is a factor

f(3) = 27 – 54 + 33 – 6 = 0 hence, (λ - 3) is a factor Thus,

(λ - 1)(λ - 2)(λ - 3) = 0

from which,

eigenvalues, λ = 1 or 2 or 3


243

EXERCISE 112 Page 275 ⎛ 4 −7 6 ⎞ 3. Determine the adt of ⎜⎜ −2 4 0 ⎟⎟ ⎜ 5 7 −4 ⎟ ⎝ ⎠ ⎛ 4 −7 6 ⎞ Matrix of cofactors of ⎜⎜ −2 4 0 ⎟⎟ is: ⎜ 5 7 −4 ⎟ ⎝ ⎠

⎛ −16 −8 −34 ⎞ ⎜ ⎟ ⎜ 14 −46 −63 ⎟ ⎜ −24 −12 2 ⎟⎠ ⎝

⎛ −16 14 −24 ⎞ ⎜ ⎟ Adt = transpose of cofactors = ⎜ −8 −46 −12 ⎟ ⎜ −34 −63 2 ⎟ ⎝ ⎠ ⎛ 4 −7 6 ⎞ ⎜ ⎟ 5. Find the inverse of ⎜ −2 4 0 ⎟ ⎜ 5 7 −4 ⎟ ⎝ ⎠ ⎛ −16 14 −24 ⎞ From question 3 above, adt = ⎜⎜ −8 −46 −63 ⎟⎟ ⎜ −34 −63 2 ⎟⎠ ⎝ 4

−7

6

−2 5

4 7

0 = 4(−16) + 7(8) + 6(−34) = −212 −4

⎛ −16 14 −24 ⎞ ⎛ 4 −7 6 ⎞ 1 ⎜ ⎟ ⎜ ⎟ Hence, the inverse of ⎜ −2 4 0 ⎟ is: − ⎜ −8 −46 −12 ⎟ 212 ⎜ −34 −63 2 ⎟ ⎜ 5 7 −4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 6 ⎜3 ⎜ 2 6. Find the inverse of ⎜ 5 − ⎜ 3 ⎜ ⎜⎜ −1 0 ⎝

1⎞ 2⎟ ⎟ 7⎟ ⎟ ⎟ 3⎟ ⎟ 5⎠

⎛ 6 ⎜3 ⎜ 2 Matrix of cofactors of ⎜ 5 − ⎜ 3 ⎜ ⎜⎜ −1 0 ⎝

1⎞ 2⎟ ⎟ 7 ⎟ is: ⎟ ⎟ 3⎟ ⎟ 5⎠

⎛ 2 −10 ⎜ −5 ⎜ ⎜ −3 3 2 3 ⎜ 5 10 ⎜ ⎜ 42 1 −18 1 ⎜ 2 ⎝ 3

2⎞ 3⎟ ⎟ −6 ⎟ ⎟ ⎟ −32 ⎟⎟ ⎠ −


244

3 1 ⎞ ⎛ 2 ⎜ − 5 −3 5 42 3 ⎟ ⎜ ⎟ 3 1⎟ ⎜ Transpose of cofactors = adt = −10 2 −18 ⎜ 10 2⎟ ⎜ ⎟ ⎜ − 2 −6 −32 ⎟⎟ ⎜ ⎝ 3 ⎠ 3 5 −1

6 −

2 3

0

1 2

1⎛ 2⎞ 6 60 1 −18 − 900 − 5 923 ⎛ 2⎞ =− 7 = 3⎜ − ⎟ − 6 (3 + 7 ) + ⎜ − ⎟ = − − − = 2⎝ 3⎠ 5 1 3 15 15 ⎝ 5⎠ 3 5

⎛ 6 ⎜3 ⎜ 2 Hence, the inverse of ⎜ 5 − ⎜ 3 ⎜ ⎜⎜ −1 0 ⎝

3 1 ⎞ 1⎞ ⎛ 2 − −3 42 ⎟ ⎜ ⎟ 5 5 3 2 ⎜ ⎟ ⎟ 1 ⎜ 3 1⎟ ⎟ −10 2 −18 7 is: ⎟ 923 ⎜ 10 2⎟ − ⎜ ⎟ ⎟ 15 ⎜ 2 3⎟ ⎟ − − 6 − 32 ⎟ ⎜ ⎟ 5⎠ ⎝ 3 ⎠

3 ⎛ 2 ⎜ − 5 −3 5 ⎜ 15 ⎜ 3 −10 2 = − ⎜ 923 10 ⎜ ⎜ − 2 −6 ⎜ 3 ⎝


⎞ ⎟ ⎟ ⎟ ⎟ ⎟ −32 ⎟⎟ ⎠

1 3 1 −18 2 42

245

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 5 ⎞ ⎛ p ⎞ ⎛ −14.6 ⎞ ⎛ 2 ⎜ ⎟⎜ ⎟ = ⎜ ⎟ ⎝ 3.1 1.7 ⎠ ⎝ q ⎠ ⎝ −2.06 ⎠

Hence,

5 ⎞ ⎛ 2 ⎛ 1.7 −5 ⎞ 1 1 ⎛ 1.7 −5 ⎞ The inverse of ⎜ ⎜ ⎟= ⎜ ⎟ ⎟ is: 3.4 − 15.5 ⎝ −3.1 2 ⎠ −12.1 ⎝ −3.1 2 ⎠ ⎝ 3.1 1.7 ⎠ ⎛p⎞ 1 ⎛ 1.7 −5 ⎞ ⎛ −14.6 ⎞ 1 ⎛ −14.52 ⎞ ⎛ 1.2 ⎞ ⎜ ⎟= ⎜ ⎟⎜ ⎟= ⎜ ⎟ =⎜ ⎟ ⎝ q ⎠ −12.1 ⎝ −3.1 2 ⎠ ⎝ −2.06 ⎠ −12.1 ⎝ 41.14 ⎠ ⎝ −3.4 ⎠

Thus, i.e.

p = 1.2 and q = -3.4


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 ⎛ 1 2 3 ⎞⎛ x ⎞ ⎛ 5⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ 2 −3 −1⎟ ⎜ y ⎟ = ⎜ 3 ⎟ ⎜ −3 4 5 ⎟ ⎜ z ⎟ ⎜ 3 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

then,

⎛ −11 −7 −1 ⎞ ⎜ ⎟ Matrix of cofactors is: ⎜ 2 14 −10 ⎟ and the transpose of cofactors is: ⎜ 7 7 −7 ⎟⎠ ⎝

1

2

7 ⎞ ⎛ −11 2 ⎜ ⎟ 7 ⎟ ⎜ −7 14 ⎜ −1 −10 −7 ⎟ ⎝ ⎠

3

2 −3 −1 = 1(-11) – 2 (7) + 3(-1) = -28 −3 4 5


246

7⎞ ⎛ −11 2 ⎛1 2 3⎞ 1 ⎜ ⎟ ⎜ ⎟ The inverse of ⎜ 2 −3 −1⎟ is: −7 14 7⎟ ⎜ −28 ⎜ ⎟ ⎜ −3 4 5 ⎟ ⎝ −1 −10 −7 ⎠ ⎝ ⎠ 7 ⎞⎛ 5 ⎞ ⎛x⎞ ⎛ −11 2 ⎛ −28 ⎞ ⎛ 1 ⎞ 1 ⎜ 1 ⎜ ⎜ ⎟ ⎟⎜ ⎟ ⎟ ⎜ ⎟ 7 ⎟⎜ 3 ⎟ = − ⎜ 28 ⎟ = ⎜ −1⎟ ⎜ y ⎟ = − 28 ⎜ −7 14 28 ⎜z⎟ ⎜ −1 −10 −7 ⎟⎜ 3 ⎟ ⎜ −56 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Thus,

i.e.

x = 1, y = -1 and z = 2


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 ⎞ ⎛ −7.8 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ 2 5 −1⎟ ⎜ q ⎟ = ⎜ 1.4 ⎟ ⎜ 5 −1 7 ⎟ ⎜ r ⎟ ⎜ 3.5 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ 34 −19 −27 ⎞ Matrix of cofactors is: ⎜⎜ −17 −8 11 ⎟⎟ and the transpose of cofactors is: ⎜ −17 7 1 ⎟⎠ ⎝ 1

2

3

2

5

−1 = 1(34) – 2 (19) + 3(-27) = -85

5 −1

⎛ 34 −17 −17 ⎞ ⎜ ⎟ 7 ⎟ ⎜ −19 −8 ⎜ −27 11 1 ⎟⎠ ⎝

7

⎛ 34 −17 −17 ⎞ ⎛1 2 3 ⎞ 1 ⎜ ⎟ ⎜ ⎟ The inverse of ⎜ 2 5 −1⎟ is: −19 −8 7 ⎟ ⎜ −85 ⎜ ⎜ 5 −1 7 ⎟ 1 ⎟⎠ ⎝ −27 11 ⎝ ⎠

Thus,

i.e.

⎛p⎞ ⎛ 34 −17 −17 ⎞ ⎛ −7.8 ⎞ ⎛ −348.5 ⎞ ⎛ 4.1 ⎞ 1 ⎜ 1 ⎜ ⎜ ⎟ ⎟⎜ ⎟ ⎟ ⎜ ⎟ 7 ⎟ ⎜ 1.4 ⎟ = − ⎜ 161.5 ⎟ = ⎜ −1.9 ⎟ ⎜ q ⎟ = − 85 ⎜ −19 −8 85 ⎜ ⎜r⎟ ⎜ −27 11 ⎟ ⎜ ⎟ 1 ⎟⎠ ⎜⎝ 3.5 ⎟⎠ ⎝ ⎠ ⎝ ⎝ 229.5 ⎠ ⎝ −2.7 ⎠

p = 4.1, q = -1.9 and r = -2.7


247

8. In a mechanical system, acceleration x , velocity x and distance x are related by the

3.4 x + 7.0 x − 13.2x = −11.39

simultaneous equations:

−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.

3.4 x + 7.0 x − 13.2x = −11.39

Since

−6.0 x + 4.0 x + 3.5x = 4.98 2.7 x + 6.0 x + 7.1x = 15.91 ⎛x⎞ ⎛ 3.4 7.0 −13.2 ⎞ ⎜ ⎟ ⎛ −11.39 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ − 6.0 4.0 3.5 ⎜ ⎟ ⎜ x ⎟ = ⎜ 4.98 ⎟ ⎜ 2.7 6.0 7.1 ⎟ ⎜ 15.91 ⎟ ⎝ ⎠⎜ x ⎟ ⎝ ⎠ ⎝ ⎠

then,

52.05 −46.8 ⎞ ⎛ 7.4 ⎜ Matrix of cofactors is: ⎜ −128.9 59.78 −1.5 ⎟⎟ ⎜ 77.3 67.3 55.6 ⎟⎠ ⎝ ⎛ 7.4 −128.9 77.3 ⎞ and the transpose of cofactors is: ⎜⎜ 52.05 59.78 67.3 ⎟⎟ ⎜ −46.8 −1.5 55.6 ⎟ ⎝ ⎠ 3.4

7.0 −13.2

−6.0 4.0

3.5

2.7

7.1

6.0

= 3.4(7.4) – 7.0 (-52.05) + (-13.2)(-46.8) = 1007.27

⎛ 7.4 −128.9 77.2 ⎞ ⎛ 3.4 7.0 −13.2 ⎞ 1 ⎜ ⎟ ⎜ ⎟ The inverse of ⎜ −6.0 4.0 3.5 ⎟ is: 52.05 59.78 67.3 ⎟ ⎜ 1007.27 ⎜ ⎟ ⎜ 2.7 6.0 7.1 ⎟⎠ ⎝ −46.8 −1.5 55.6 ⎠ ⎝

Thus,

i.e.

⎛x⎞ ⎛ 7.4 −128.9 77.3 ⎞ ⎛ −11.39 ⎞ ⎛ 503.635 ⎞ ⎛ 0.5 ⎞ ⎜ ⎟ 1 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜x⎟ = 52.05 59.78 67.3 ⎟ ⎜ 4.98 ⎟ = 775.5979 ⎟ = ⎜ 0.77 ⎟ ⎜ ⎜ ⎜ ⎟ 1007.27 ⎜ ⎟⎜ ⎟ 1007.27 ⎜ 1410.178 ⎟ ⎜ 1.4 ⎟ ⎜x⎟ ⎝ −46.8 −1.5 55.6 ⎠ ⎝ 15.91 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

x = 0.5, x = 0.77 and x = 1.4


248


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

x −y 1 = = −5 17.6 3 17.6 3 −5 7 −22 −2 −22 −2 7

then

x −y 1 = = −13.2 −30.8 11

i.e.

from which,

x=

−13.2 = -1.2 11

and

y=

30.8 = 2.8 11

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 x −y z −1 = = = 4 1 −10 3 1 −10 3 4 −10 3 4 1 −3 5 9 2 5 9 2 −3 9 2 −3 5 2 −1 −6 1 −1 −6 1 2 −6 1 2 −1

then

i.e.

x −y z −1 = = = 4(−21) − 1(0) − 10(−7) 3(−21) − 1(−21) − 10(−7) 3(0) − 4(−21) − 10(7) 3(−7) − 4(−7) + 1(7) x − y z −1 = = = −14 28 14 14

i.e.

Hence,

x=

14 28 −14 = -1 = 1, y = = 2 and z = 14 14 14


249

7. Applying mesh-current analysis to an a.c. circuit results in the following equations:

(5 – j4) I1 - (-j4) I 2 = 100∠0° (4 + j3 – j4) I 2 - (-j4) I1 = 0 Solve the equations for I1 and I 2 . (5 – j4) I1 - (-j4) I 2 - 100∠0° = 0 - (-j4) I1 + (4 + j3 – j4) I 2 + 0 = 0 (5 – j4) I1 + j4 I 2 - 100 = 0

i.e.

j4 I1 + (4 – j) I 2 + 0 = 0 Hence,

I1 j4

−100

(4 − j)

0

=

−I2 1 = (5 − j4) −100 (5 − j4) j4

0

j4

i.e.

I1 −I 2 1 = = 100(4 − j) j400 (5 − j4)(4 − j) − ( j4 )2

i.e.

I1 I2 1 = = 400 − j100 − j400 32 − j21

j4 (4 − j)

Thus,

I1 =

400 − j100 412.31∠ − 14.04° = = 10.77∠19.23° A 32 − j21 38.275∠ − 33.27°

and

I2 =

− j400 400∠ − 90° = = 10.45∠ − 56.73° A 32 − j21 38.275∠ − 33.27°

9. The forces in three of a framework are F1 , F2 and F3 . They are related by the

simultaneous equations shown below. 1.4 F1 + 2.8 F2 + 2.8 F3 = 5.6 4.2 F1 – 1.4 F2 + 5.6 F3 = 35.0 4.2 F1 + 2.8 F2 – 1.4 F3 = -5.6 Find the values of F1 , F2 and F3 using determinants. 1.4 F1 + 2.8 F2 + 2.8 F3 - 5.6 = 0 4.2 F1 – 1.4 F2 + 5.6 F3 - 35.0 = 0 4.2 F1 + 2.8 F2 – 1.4 F3 + 5.6 = 0


250

Hence,

i.e.

F3 −F2 F1 −1 = = = 2.8 2.8 −5.6 1.4 2.8 −5.6 1.4 2.8 −5.6 1.4 2.8 2.8 −1.4 5.6 −35.0 4.2 5.6 −35.0 4.2 −1.4 −35.0 4.2 −1.4 5.6 2.8 −1.4 5.6 4.2 −1.4 5.6 4.2 2.8 5.6 4.2 2.8 −1.4

F1 −F2 = 2.8(−17.64) − 2.8(90.16) − 5.6(−13.72) 1.4(−17.64) − 2.8(170.52) − 5.6(−29.4) =

F3 −1 = 1.4(90.16) − 2.8(170.52) − 5.6(17.64) 1.4(−13.72) − 2.8(−29.4) + 2.8(17.64) F3 − F2 −1 F1 = = = −225.008 −337.512 −450.016 112.504

i.e. Thus,

F1 =

225.008 =2 112.504

F2 =

−337.512 = -3 112.504

and

F3 =

450.016 =4 112.504

10. Mesh-current analysis produces the following three equations: 20∠0° = (5 + 3 − j4)I1 − (3 − j4)I 2 10∠90° = (3 − j4 + 2)I 2 − (3 − j4)I1 − 2I3 −15∠0° − 10∠90° = (12 + 2)I3 − 2I 2 Solve the equations for the loop currents I1 , I 2 and I3

(8 – j4) I1 - (3 – j4) I 2 + 0 I3 - 20 = 0

Rearranging gives:

-(3 – j4) I1 + (5 – j4) I 2 - 2 I3 - j10 = 0 0 I1 - 2 I 2 + 14 I3 + (15 + j10) = 0 Hence,

I1 − (3 − j4) 0 (5 − j4) −2 −2

14

−20 − j10 (15 + j10)

=

−I 2 (8 − j4) 0 −(3 − j4) −2 0

=

−20 − j10

14 (15 + j10)

I3 (8 − j4) −(3 − j4) −(3 − j4) (5 − j4) −2

0 =

−20 − j10 (15 + j10)

−1 (8 − j4) −(3 − j4) 0 −(3 − j4) (5 − j4) −2 −2 0 14

i.e.

I1 −I2 = −(3 − j4) [ −2(15 + j10) + j140] − 20 [14(5 − j4) − 4] (8 − j4) [ −2(15 + j10) + j140] − 20 [ −14(3 − j4) ]

=

I3 (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) ]


251

i.e.

I1 −I2 = −(3 − j4) [ −30 + j120] − 20 [ 66 − j56] (8 − j4) [ −30 + j120] − 20 [ −42 + j56)]

=

I3 (8 − j4) [115 − j30] + (3 − j4) [ −85 + j30] − 40(3 − j4)

=

i.e.

−1 (8 − j4) [ 66 − j56] + (3 − j4) [ −42 + j56]

I1 −I2 = −(−90 + j360 + j120 + 480) − 1320 + j1120 −240 + j960 + j120 + 480 + 840 − j1120 =

I3 920 − j240 − j460 − 120 − 255 + j430 + 120 − 120 + j160 =

I3 −I2 I1 −1 = = = (−1710 + j640) (1080 − j40) (545 − j110) (402 − j376)

i.e. Hence,

−1 528 − j448 − j264 − 224 − 126 + j168 + j168 + 224

I1 =

−(−1710 + j640) 1825.84∠ − 20.52° = = 3.317∠ 22.57° A (402 − j376) 550.44∠ − 43.09°

I2 =

(1080 − j40) 1080.74∠ − 2.12° = = 1.963∠40.97° A (402 − j376) 550.44∠ − 43.09°

I3 =

−(545 − j110) −555.99∠ − 11.41° 555.99∠ − 191.41° = = = 1.010∠ − 148.32° A (402 − j376) 550.44∠ − 43.09° 550.44∠ − 43.09°


252

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

5 3 3 x= 1 2 −3

2 −3 4 2 −3 4

3 −1 5 5(−11) − 2(18) + 3(21) −28 =1 = = 3 1(−11) − 2(7) + 3(−1) −28 −1 5

1 5 3 2 3 −1 −3 3 5 1(18) − 5(7) + 3(15) 28 y= = -1 = = −28 −28 −28 1 2 5 2 −3 3 − 3 4 3 1(−21) − 2(15) + 5(−1) −56 z= =2 = = −28 −28 −28

1. Q(7), Exercise 113

Use Cramers rule to solve:

s + 2 v + 2a = 4 3s – v + 4a = 25 3s + 2v – a = -4

4 25 −4 s= 1 3 3

2 −1 2 2 −1 2

2 4 −1 4(−7) − 2(−9) + 2(46) 82 =2 = = 2 1(−7) − 2(−15) + 2(9) 41 4 −1

1 4 2 3 25 4 3 −4 −1 1(−9) − 4(−15) + 2(−87) −123 = -3 v= = = 41 41 41


253

1 2 4 3 −1 25 3 2 −4 1(−46) − 2(−87) + 4(9) 164 a= =4 = = 41 41 41

2. Q(8), Exercise 114 Use Cramers rule to solve:

− 31 −5 6 i1 = 1 3 2

8 −2 −3 8 −2 −3

3 1 2 3 1 2

=

i1 + 8i 2 + 3i3 = −31 3i1 − 2i 2 + i3 = −5 2i1 − 3i 2 + 2i3 = 6

−31(−1) − 8(−16) + 3(27) 240 = -5 = 1(−1) − 8(4) + 3(−5) −48

1 −31 3 3 −5 1 2 6 2 1(−16) + 31(4) + 3(28) 192 i2 = = -4 = = −48 −48 −48 1 8 −31 3 −2 −5 2 −3 6 1(−27) − 8(28) − 31(−5) −96 i3 = =2 = = −48 −48 −48


254


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 7.5 x + 4.8 x + 4.8x = 6.39

13.0 x + 3.5 x − 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

7.5 x + 4.8 x + 4.8x = 6.39

13.0 x + 3.5 x − 13.0x = −17.4

(2) -

7.5 × (1) gives: 6.2

0 – 4.7565 x - 10.442 x = -15.384

(3) -

13.0 × (1) gives: 6.2

0 – 13.065 x - 39.419 x = -55.142

(3′) -

−13.065 × (2′) gives: −4.7565

x=

(3) (2′) (3′)

−12.886 = 1.2 −10.737

-13.065 x - 39.419(1.2) = -55.142

-13.065 x = -55.142 + 39.419(1.2)

i.e.

and From (1),

(2)

0 + 0 – 10.737 x = -12.886

from which, From (3′),

(1)

x =

−55.142 + 39.419(1.2) = 0.60 −13.065

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 =

−1.86 = -0.30 6.2


255

2. The tensions, T1 , T2 and T3 in a simple framework are given by the equations:

5T1 + 5T2 + 5T3 = 7.0 T1 + 2T2 + 4T3 = 2.4 4T1 + 2T2

= 4.0

Determine T1 , T2 and T3 using Gaussian elimination. 5T1 + 5T2 + 5T3 = 7.0

(1)

T1 + 2T2 + 4T3 = 2.4

(2)

4T1 + 2T2 + 0T3 = 4.0

(3)

(2) -

1 × (1) gives: 5

0 + T2 + 3T3 = 1.0

(2′)

(3) -

4 × (1) gives: 5

0 − 2T2 − 4T3 = −1.6

(3′)

⎛ −2 ⎞ (3′) - ⎜ ⎟ × (2′) gives: ⎝ 1 ⎠ from which, In (3′)

2T3 = 0.4 T3 = 0.2

−2T2 − 4(0.2) = −1.6 −2T2 = −1.6 + 0.8 = −0.8

and In (1)

T2 = 0.4

5T1 + 5(0.4) + 5(0.2) = 7.0

i.e.

5T1 = 7.0 − 2.0 − 1.0 = 4.0

and

T1 = 0.8


256

CHAPTER 27 METHODS OF DIFFERENTIATION EXERCISE 117 Page 291 1.(c) Find the differential coefficient with respect to x of

If y =

1 1 dy = x −1 then = −1x −2 = − 2 x x dx

2.(a) Find the differential coefficient with respect to x of

If y =

1 x

−4 x2

8 −4 dy = −4x −2 then = (−4) ( −2x −3 ) = 8x −3 or 3 2 x x dx

3. Find the differential coefficient with respect to x of (a) 2 x

1

(a) If y = 2 x = 2x 2 then 5

(b) If y = 3 3 x 5 = 3x 3 then

(c) If y =

4 x

2 ⎛5 2 ⎞ dy = (3) ⎜ x 3 ⎟ = 5x 3 = 5 3 x 2 dx ⎝3 ⎠

1 3 − − ⎛ 1 −3⎞ 2 4 4 dy 2 = 1 = 4x 2 then = (4) ⎜ − x 2 ⎟ = −2x 2 = − 3 = − dx x x3 ⎝ 2 ⎠ x2 x2

−3 3 x

1 4 − − ⎛ 1 − 43 ⎞ 3 3 dy 1 3 3 ( 3) x x 3x = − − = = 4 = = − = − then ⎜ ⎟ 1 3 dx x ⎝ 3 ⎠ x3 x3

5.(c) Find the differential coefficient with respect to x of

If y =

(c)

1 − ⎛1 −1 ⎞ 1 dy 1 = (2) ⎜ x 2 ⎟ = x 2 = 1 = dx x ⎝2 ⎠ x2

4.(a) Find the differential coefficient with respect to x of

If y = −

(b) 3 3 x 5

1 3

x4

3 e5x

15 3 dy = 3e −5x then = (3) ( −5e−5x ) = −15e−5x = − 5x 5x e e dx

7. Find the gradient of the curve y = 2t 4 + 3t 3 − t + 4 at the points (0, 4) and (1, 8).

If y = 2t 4 + 3t 3 − t + 4 , then gradient,

dy = 8t 3 + 9t 2 − 1 dt


257

At (0, 4), t = 0, hence gradient = 8(0)3 + 9(0) 2 − 1 = -1 At (1, 8), t = 1, hence gradient = 8(1)3 + 9(1) 2 − 1 = 16 8. Find the co-ordinates of the point on the graph y = 5x 2 − 3x + 1 where the gradient is 2.

If y = 5x 2 − 3x + 1 , then gradient =

dy = 10x − 3 dx

When the gradient is 2, 10x – 3 = 2

i.e. 10x = 5 and x =

1 2

2

When x =

5 3 3 1 ⎛1⎞ ⎛1⎞ , y = 5⎜ ⎟ − 3⎜ ⎟ +1 = − +1 = 4 2 4 2 ⎝2⎠ ⎝2⎠

⎛ 1 3⎞ Hence, the co-ordinates of the point where the gradient is 2 is ⎜ , ⎟ ⎝ 2 4⎠

9. (a) Differentiate y = (b) Evaluate

(a)

y=

2 2 + 2 ln 2θ − 2(cos 5θ + 3sin 2θ) − 3 θ 2 e θ

dy π in part (a) when θ = , correct to 4 significant figures. 2 dθ

2 2 + 2 ln 2θ − 2(cos 5θ + 3sin 2θ) − 3 θ 2 e θ

= 2θ−2 + 2 ln 2θ − 2 cos 5θ − 6sin 2θ − 2e−3θ Hence,

dy 2 = −4θ−3 + − 2(−5sin 5θ) − 6(2 cos 2θ) − 2 ( −3e−3θ ) θ dθ =−

(b) When θ =

4 2 6 + + 10sin 5θ − 12cos 2θ + 3 θ 3 e θ θ

π dy 4 2 5π 2π 6 , + + 10sin − 12 cos + ⎛π⎞ =− 3 3⎜ ⎟ 2 2 2 dθ ⎛π⎞ ⎛π⎞ e ⎝2⎠ ⎜ ⎟ ⎜⎝ 2 ⎟⎠ ⎝2⎠ = -1.032049 + 1.2732395 + 10 + 12 + 0.0538997 = 22.30, correct to 4 significant figures.


258


1. Differentiate 2x 3 cos 3x with respect to x.

dy = ( 2x 3 ) ( −3sin 3x ) + ( cos 3x ) ( 6x 2 ) dx

If y = 2x 3 cos 3x , then

= −6x 3 sin 3x + 6x 2 cos 3x = 6x 2 ( cos 3x − x sin 3x )

3. Differentiate e3t sin 4t with respect to x.

If y = e3t sin 4t , then

dy = ( e3t ) ( 4 cos 4t ) + ( sin 4t ) ( 3e3t ) dt = e 3t ( 4cos 4t + 3sin 4t )

5. Differentiate e t ln t cos t with respect to x.

If y = e t ln t cos t , then

dy ⎡ ⎤ ⎛1⎞ = ( e t ln t ) ( − sin t ) + ( cos t ) ⎢( e t ) ⎜ ⎟ + ( ln t ) ( e t ) ⎥ dt ⎝t⎠ ⎣ ⎦ cos t ⎧ ⎫ = e t ⎨− ln t sin t + + cos t ln t ⎬ t ⎩ ⎭ ⎧⎛ 1 ⎫ ⎞ = et ⎨⎜ + ln t ⎟ cos t − ln t sin t ⎬ ⎠ ⎩⎝ t ⎭

6. Evaluate

di , correct to 4 significant figures, when t = 0.1, and i = 15t sin3t. dt

Since i = 15t sin3t, then

di = (15t)(3cos 3t) + (sin 3t)(15) dt = 45t cos 3t + 15 sin 3t

When t = 0.1,

di = 45(0.1) cos 0.3 + 15 sin 0.3 dt

(note 0.3 is radians)

= 4.2990 + 4.4328 = 8.732, correct to 4 significant figures.


259


3. Differentiate the quotient

3 θ3 with respect to x. 2sin 2θ

3 2

If y =

dy 3 θ3 3θ = = , then 2sin 2θ 2sin 2θ dθ

⎛ 9 12 ⎞ ⎛ 32 ⎞ θ ⎟ − ⎜ 3θ ⎟ ( 4 cos 2θ ) ⎝2 ⎠ ⎝ ⎠

( 2sin 2θ ) ⎜

( 2sin 2θ )

1

2

3

3 θ { 3sin 2θ − 4θ cos 2θ } 9 θ 2 sin 2θ − 12 θ 2 cos 2θ = = 2 4sin 2 2θ 4sin 2θ

4. Differentiate the quotient

If y =

ln 2t dy = , dt t

ln 2t with respect to x. t

⎛1 −1 ⎞ ⎛1⎞ t ⎜ ⎟ − ( ln 2t ) ⎜ t 2 ⎟ ⎝t⎠ ⎝2 ⎠

( )

( t)

2

=

=

6. Find the gradient of the curve y =

If y =

t

−

1 2

1 − 12 1 − − t ln 2t 2 t ⎛ 1 ⎞ 2 = ⎜1 − ln 2t ⎟ t t ⎝ 2 ⎠

1 ⎛ 1 1⎛ 1 ⎞ ⎞ 1 − ln 2t ⎟ = ⎜ 1 − 2 ln 2t ⎟ 3 ⎜ 3 ⎠ ⎝ 2 ⎠ t ⎝ t2

2x at the point (2, - 4) x −5 2

2x then gradient, x −5 2

2 dy ( x − 5 ) (2) − (2x)(2x) 2x 2 − 10 − 4x 2 −10 − 2x 2 = = = 2 2 2 dx ( x 2 − 5) ( x 2 − 5) ( x 2 − 5)

At the point (2, - 4), x = 2, hence gradient =

7. Evaluate

−10 − 2(2) 2

(2

2

− 5)

2

=

−10 − 8 −18 = = -18 (4 − 5) 2 1

dy 2x 2 + 3 at x = 2.5, correct to 3 significant figures, given y = dx ln 2x


260

If y =

2x + 3 dy then = ln 2x dx 2

When x = 2.5,

dy = dx

( ln 2x ) (4x) − (2x 2 + 3) ⎛⎜ ( ln 2x )

1⎞ ⎟ ⎝x⎠

2

( ln 5) (10) − [2(2.5)2 + 3] ⎛⎜ ( ln 5)

1 ⎞ ⎟ ⎝ 2.5 ⎠

2

=

16.09438 − 6.2 = 3.82, correct to 3 2.59029 significant figures.


261

EXERCISE 120 Page 296 1. Find the differential coefficient of ( 2x 3 − 5x ) with respect to x. 5

If y = ( 2x 3 − 5x ) then 5

4 dy = 5 ( 2x 3 − 5x ) ( 6x 2 − 5 ) dx

4. Find the differential coefficient of

If y =

(x

1

3

− 2x + 1)

= ( x − 2x + 1) 3

5

−5

(x

1

3

− 2x + 1)

5

with respect to x.

−5 ( 3x 2 − 2 ) −6 dy 3 2 = −5 ( x − 2x + 1) ( 3x − 2 ) = then 6 dx ( x 3 − 2x + 1) =

5 ( 2 − 3x 2 )

(x

3

− 2x + 1)

6

6. Find the differential coefficient of 2 cot ( 5t 2 + 3) with respect to x.

If y = cot x =

cos x then sin x

2 2 −1 dy (sin x)(− sin x) − (cos x)(cos x) − ( sin x + cos x ) = = = = − cos ec 2 x 2 2 2 dx sin x sin x sin x

Thus, if y = 2 cot ( 5t 2 + 3) , then

dy = ( 2 ) ⎡⎣ − cos ec2 ( 5t 2 + 3) ⎤⎦ (10t ) = −20t cos ec 2 ( 5t 2 + 3 ) dt

8. Find the differential coefficient of 2e tan θ with respect to θ.

If y = 2e tan θ , then

dy = ( 2e tan θ ) sec2 θ = 2sec2 θ etan θ dθ

π⎞ ⎛ 9. Differentiate θ sin ⎜ θ − ⎟ with respect to θ, and evaluate, correct to 3 significant figures, when 3⎠ ⎝ π θ= 2


262

π⎞ dy π⎞ π⎞ ⎛ ⎛ ⎛ If y = θ sin ⎜ θ − ⎟ then = ( θ ) cos ⎜ θ − ⎟ + sin ⎜ θ − ⎟ (1) dθ 3⎠ 3⎠ 3⎠ ⎝ ⎝ ⎝

When θ =

π , 2

dy ⎛ π ⎞ π π ⎛π π⎞ ⎛π π⎞ π = ⎜ ⎟ cos ⎜ − ⎟ + sin ⎜ − ⎟ = cos + sin dθ ⎝ 2 ⎠ 6 6 ⎝2 3⎠ ⎝2 3⎠ 2 = 1.360 + 0.5 = 1.86, correct to 3 significant figures.


263


2 2 1 3 t − 3 + − t + 1 determine f ′′(t). 5 t t

2. (a) Given f(t) =

(b) Evaluate f ′′(t) when t = 1.

(a) f(t) =

1 2 2 1 3 2 t − 3 + − t + 1 = t 2 − t −3 + 3t −1 − t 2 + 1 5 t t 5

4 1 −1 t + 3t −4 − 3t −2 − t 2 5 2

f ′(t) =

f ′′(t) =

1 4 1 − 3 4 12 6 − 12t −5 + 6t −3 + t 2 = − 5 + 3 + 5 t t 5 4 4 t3

(b) When t = 1, f ′′(t) =

4 12 6 1 4 1 − 5+ 3+ = − 12 + 6 + = -4.95 3 5 (1) (1) 4 1 5 4

4. Find the second differential coefficient with respect to x of (a) 2 cos 2 x (b) ( 2x − 3)

(a) If y = 2 cos 2 x ,

4

dy = 4 cos x(− sin x) = −4sin x cos x dx d2 y = (−4sin x)(− sin x) + (cos x)(−4 cos x) = 4sin 2 x − 4 cos 2 x 2 dx

and

= 4 ( sin 2 x − cos 2 x ) (b) If y = ( 2x − 3) , 4

dy = 4(2x − 3)3 (2) = 8(2x − 3)3 dx d2 y 2 = 24(2x − 3) 2 (2) = 48 ( 2x − 3 ) 2 dx

and

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 f ′′(θ) = ( 6sec 3θ ) ( 3sec2 3θ ) + ( tan 3θ )(18sec 3θ tan 3θ )

and

= 18sec3 3θ + 18sec3θ tan 2 3θ When θ = 0,

18 18 tan 2 0 18 18(0) f ′′(0) = + = + = 18 cos3 0 cos 0 1 1


264

7. Show that, if P and Q are constants and y = P cos(ln t) + Q sin(ln t), then

t2

d2 y dy +t +y=0 2 dt dt

y = P cos(ln t) + Q sin(ln t) dy P Q −P sin(ln t) + Q cos(ln t) = − sin(ln t) + cos(ln t) = dt t t t ⎡ − P cos(ln t) Q sin(ln t) ⎤ (t) ⎢ − ⎥⎦ − [ − P sin(ln t) + Q cos(ln t) ] (1) d y t t ⎣ = dt 2 t2 2

=

−P cos(ln t) − Q sin(ln t) + P sin(ln t) − Q cos(ln t) t2

Hence, t 2

d2 y dy − P cos(ln t) − Q sin(ln t) + P sin(ln t) − Q cos(ln t) + t + y = (t2 ) 2 dt dt t2 + (t)

− P sin(ln t) + Q cos(ln t) + P cos(ln t) + Q sin(ln t) t

= − P cos(ln t) − Q sin(ln t) + P sin(ln t) − Q cos(ln t) − P sin(ln t) + Q cos(ln t) + P cos(ln t) + Q sin(ln t) =0


265

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,

di = (10)(2πf ) cos 2πft = (10)(2π× 150) cos ⎡⎣ 2π× 150 × 20 × 10−3 ⎤⎦ dt = 3000π cos 6π = 3000π A/s

3. The voltage across the plates of a capacitor at any time t seconds is given by v = V e

−

t CR

, where

V, C and R are constants. Given V = 300 volts, C = 0.12 ×10−6 F and R = 4 ×106 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 = V e

−

t CR

, then

dv 1 ⎞ − CRt ⎛ = V⎜− ⎟e dt ⎝ CR ⎠

Initial rate of change of voltage, (i.e. when t = 0),

dv 1 ⎛ ⎞ 0 e = (300) ⎜ − 6 ⎟ −6 dt ⎝ 0.12 ×10 × 4 × 10 ⎠ =-

(b) When t = 0.5 s,

300 = -625 V/s 0.48

0.5 dv ⎛ V ⎞ − CRt 300 − 0.48 e e = -220.5 V/s = ⎜− = − ⎟ dt ⎝ CR ⎠ 0.48

−

h

4. The pressure p of the atmosphere at height h above ground level is given by p = p0 e c , where p0

is the pressure at ground level and c is a constant. Determine the rate of change of pressure with height when p0 = 1.013 ×105 pascals and c = 6.05 ×104 at 1450 metres. Pressure, p = p0 e

−

h c

Rate of change of pressure with height,

1450 − ⎛ ⎛ 1 −h ⎞ dp 1 6.05×104 e = (p 0 ) ⎜ − e c ⎟ = (1.013 × 105 ) ⎜⎜ − 4 dh ⎝ c ⎠ ⎝ 6.05 × 10

⎞ ⎟⎟ ⎠

= -1.635 Pa/m


266


1. A missile fired from ground level rises x metres vertically upwards in t seconds and

x = 100t −

25 2 t . Find (a) the initial velocity of the missile, (b) the time when the height of the 2

missile is a maximum, (c) the maximum height reached, (d) the velocity with which the missile strikes the ground. (a) Distance, x = 100t −

25 2 t 2

Initial velocity, (i.e. when t = 0),

dx = 100 − 25t = 100 − 25(0) = 100 m/s dt

(b) When height is a maximum, velocity = 0, i.e. from which,

dx = 100 − 25t = 0 dt 100 = 25t and time t = 4 s

(c) When t = 4 s, maximum height, x = 100(4) − (d) When x = 0 (i.e. on the ground),

25 2 (4) = 400 − 200 = 200 m 2

0 = 100t −

25 2 t 2

25 ⎞ ⎛ t ⎜100 − t ⎟ = 0 2 ⎠ ⎝

i.e.

Hence, either t = 0 (at the start) or 100 −

25 t=0 2 t=

and Velocity, i.e.

i.e.

100 =

25 t 2

200 =8s 25

dx dx , when t = 8 s is given by = 100 – 25t = 100 – 25(8) dt dt = 100 – 200 = -100 m/s (negative indicating reverse direction to the starting velocity)

3. The equation θ = 10π + 24t − 3t 2 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, θ = 10π + 24t − 3t 2 © 2006 John Bird. All rights reserved. Published by Elsevier.

267

(a) When the wheel comes to rest, angular velocity = 0, i.e. dθ = 24 − 6t = 0 dt

Hence,

from which,

dθ =0 dt

time, t = 4 s

(b) Distance moved in the last second of movement = (distance after 4 s) – (distance after 3 s) = [10π + 24(4) − 3(4) 2 ] - [10π + 24(3) − 3(3) 2 ] = [10π + 96 − 48] - [10π + 72 − 27] = 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 =

dx 1 −1 (−1) = 4 − (1 − t ) = 4+ dt 1− t

Initial velocity, i.e. when t = 0, v = 4 −

1 = 3 m/s 1

d2x 1 Acceleration, a = 2 = (1 − t) −2 (−1) = − dt (1 − t) 2

Initial acceleration, i.e. when t = 0, a = −

(b) After 1.5 s, velocity, v = 4 −

and

acceleration, a = −

1 = -1 m / s 2 1

1 1 1 = 4− = 4− = 4 + 2 = 6 m/s (1 − t) (1 − 1.5) (−0.5)

1 1 1 1 =− =− =− = -4 m / s 2 2 2 2 (1 − t) (1 − 1.5) (−0.5) 0.25 1 =0 (1 − t)

i.e.

i.e.

4 – 4t = 1

and 4 – 1 = 4t

from which,

time, t =

(c) When the velocity is zero, 4 −

4=

1 (1 − t)

3 s 4


268

EXERCISE 124 Page 305 2. Find the turning point on x = θ(6 − θ) and determine its nature.

x = θ(6 − θ) = 6θ − θ2 dx = 6 − 2θ = 0 for a turning point, from which, θ = 3 dθ When θ = 3, x = 6θ − θ2 = 6(3) − 32 = 18 − 9 = 9 Hence, (3, 9) are the co-ordinates of the turning point d2x = −2 , which is negative, hence a maximum occurs at (3, 9) dθ2 3. Find the turning point on y = 4x 3 + 3x 2 − 60x − 12 and distinguish between them.

y = 4x 3 + 3x 2 − 60x − 12 dy = 12x 2 + 6x − 60 = 0 for a turning point dx i.e.

2x 2 + x − 10 = 0

i.e.

from which,

(2x + 5)(x – 2) = 0

2x + 5 = 0 i.e. x = -2.5

and

x–2=0

i.e. x = 2

When x = -2.5, y = 4(−2.5)3 + 3(−2.5) 2 − 60(−2.5) − 12 = 94.25 When x = 2, Hence,

y = 4(2)3 + 3(2) 2 − 60(2) − 12 = −88

(-2.5, 94.25) and (2, -88) are the co-ordinates of the turning points

d2x = 24x + 6 dθ2

When x = -2.5,

When x = 2,

d2 x is negative, hence (-2.5, 94.25) is a maximum point. dθ2

d2x is positive, hence (2, -88) is a minimum point. dθ2

5. Find the turning point on y = 2x − e x and determine its nature.


269

y = 2x − e x

hence,

dy = 2 − e x = 0 for a turning point. dx

2 = ex

i.e.

and

x = ln 2 = 0.6931

When x = 0.6931, y = 2(0.6931) − e0.6931 = −0.6136 Hence,

(0.6931, -0.6136) are the co-ordinates of the turning point.

d2 y = −e x 2 dx

When x = 0.6931,

d2 y is negative, hence (0.6931, -0.6136) is a maximum point. dx 2

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 θ − 5cos θ = 0 for a maximum or minimum value dθ i.e.

Hence,

-12 sin θ = 5 cos θ

from which,

sin θ 5 =− cos θ 12

i.e.

tan θ = −

5 12

⎛ 5⎞ θ = tan −1 ⎜ − ⎟ = - 22°37′ ⎝ 12 ⎠

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. d2 y = −12 cos θ + 5sin θ dx 2


270

When θ = 157°23′,

d2 y is positive, hence (157°23′, -13) is a minimum point dx 2

When θ = 337°23′,

d2 y is negative, hence (337°23′, 13) is a maximum point dx 2

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 θ 12 = cos θ 5

from which, tan θ = 2.4 and θ = tan −1 2.4 = 67.4° and

247.4°; also, at θ = 0, y = 12 cos 0 – 5 sin 0 = 12).

2 2 9. Show that the curve y = (t − 1)3 + 2t(t − 2) has a maximum value of and a minimum value 3 3 of -2. 2 2 y = (t − 1)3 + 2t(t − 2) = (t − 1)3 + 2t 2 − 4t 3 3 dy = 2(t − 1) 2 + 4t − 4 = 0 for a turning point. dx i.e.

2 ( t 2 − 2t + 1) + 4t − 4 = 0

i.e.

2t 2 − 4t + 2 + 4t − 4 = 0

i.e.

2t 2 − 2 = 0

from which,

t2 = 1

and t = ±1

2 When t = 1, y = (1 − 1)3 + 2(1 − 2) = −2 3 © 2006 John Bird. All rights reserved. Published by Elsevier.

271

2 16 2 When t = -1, y = (−1 − 1)3 + 2(−1)(−1 − 2) = − + 6 = 3 3 3 d2 y = 4(t − 1) + 4 dt 2

When t = 1,

When t = -1,

d2 y is positive, hence (1, -2) is a minimum point. dt 2 d2 y is negative, hence dt 2

2⎞ ⎛ ⎜ −1, ⎟ is a maximum point. 3⎠ ⎝

2 2 and the minimum value is -2 Hence, the maximum value of y = (t − 1)3 + 2t(t − 2) is 3 3


272

EXERCISE 125 Page 309 1. The speed, v, of a car (in m/s) is related to time t s by the equation: v = 3 + 12t − 3t 2 . Determine

the maximum speed of the car in km/h. Speed, v = 3 + 12t − 3t 2 dv = 12 − 6t = 0 for a maximum value, from which, 12 = 6t and t = 2 s dt When t = 2, v = 3 + 12(2) − 3(2) 2 = 3 + 24 − 12 = 15 m/s d2v = −6 , which is negative, hence indicating that v = 15 m/s is the maximum speed. dt 2 15 m/s = 15m / s ×

60 × 60 s / h = 15 × 3.6 = 54 km/h = maximum speed. 1000 m / km

3. A shell is fired vertically upwards and its vertical height, x metres, is given by x = 24t - 3t 2 ,

where t is the time in seconds. Determine the maximum height reached. Height, x = 24t - 3t 2 dx = 24 − 6t = 0 for a maximum value, from which, 24 = 6t and t = 4 s dt d2x = −6 , which is negative – hence a maximum value. dt 2 Maximum height = 24(4) - 3(4) 2 = 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 m3 A lidless box with square ends is shown in the diagram below, having dimensions x by x by y. Volume of box, V = x 2 y = 3.5 m3 ⎛ 3.5 ⎞ Area of metal, A = 2x 2 + 3xy = 2x 2 + 3x ⎜ 2 ⎟ ⎝x ⎠

(1) from equation (1)


273

A = 2x 2 + 10.5 x −1

i.e.

dA = 4x − 10.5 x −2 = 0 for a maximum or minimum value. dx i.e.

4x =

10.5 x2

d2A = 4 + 21x −3 2 dx

i.e. x 3 =

10.5 = 2.625 4

When x = 1.3795,

from which,

x = 3 2.625 = 1.3795

d2A is positive – hence a minimum value. dx 2

Minimum or least area of metal = 2x 2 +

10.5 10.5 = 11.42 m 2 = 2(1.3795) 2 + x 1.3795

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 = πr 2 h

A section is shown below.

By similar triangles,

20 h = 80 80 − r

from which, h =

20(80 − r) 80 − r r = = 20 − 80 4 4


274

r⎞ πr 3 ⎛ cylinder volume, V = πr 2 ⎜ 20 − ⎟ = 20πr 2 − 4⎠ 4 ⎝

Hence,

dV 3πr 2 = 40πr − = 0 for a maximum or minimum value. dr 4 40πr =

i.e.

d2V 6πr = 40π − 2 dr 4

3πr 2 4

i.e.

When r =

40 =

3r 4

and

r=

160 3

160 d 2 V , is negative, hence a maximum value. 3 dr 2

160 r 40 Hence, height of cylinder, h = 20 - = 20 − 3 = 20 − = 6.67 cm 4 4 3

7. The power developed in a resistor R by a battery of emf E and internal resistance r is given by:

P=

E2R

(R + r)

2

. Differentiate P with respect to R and show that the power is a maximum when

R = r.

P=

E2R

(R + r)

hence

2

dP (R + r) 2 (E) 2 − E 2 R(2)(R + r) = = 0 for a maximum value dR (R + r) 4

Thus,

E 2 ⎣⎡ (R + r) 2 − 2R(R + r) ⎦⎤ = 0

i.e.

R 2 + 2Rr + r 2 − 2R 2 − 2Rr = 0

i.e.

r2 − R 2 = 0

and

R=r

2 2 2 dP E ⎡⎣ r − R ⎤⎦ = dR (R + r) 4

When R = r,

and

4 2 2 2 2 3 d 2 P (R + r) E [ −2R ] − E ⎡⎣ r − R ⎤⎦ 4(R + r) = dR 2 (R + r)8

d2P is negative, hence power is a maximum when R = r. dR 2

9. Resistance to motion, F, of a moving vehicle, is given by F =

5 + 100x . Determine the x

minimum value of resistance.


275

F=

5 + 100x = 5x −1 + 100x x

dF 5 = −5x −2 + 100 = − 2 + 100 = 0 for a maximum or minimum value. dx x i.e.

100 =

d2F 10 = 10x −3 = 3 2 dx x

5 x2

and

x2 =

5 = 0.05 100

and

x = 0.05 = 0.2236

which is positive when x = 0.2236, hence x = 0.2236 gives a minimum value of resistance.

Maximum resistance to motion, F =

5 5 + 100x = + 100(0.2236) = 44.72 x 0.2236

11. The fuel economy E of a car, in miles per gallon, is given by:

E = 21 + 2.10 × 10−2 v 2 − 3.80 ×10−6 v 4 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 + 2.10 × 10−2 v 2 − 3.80 ×10−6 v 4 dE = 4.20 × 10−2 v − 4(3.80) × 10−6 v3 = 0 for a maximum (most economical fuel consumption). dv 4.20 ×10−2 v = 4(3.80) × 10−6 v3

i.e.

v2 =

i.e.

from which, speed, v =

4.20 × 10−2 = 2763.158 4(3.80) × 10−6

2763.158 = 52.6 m.p.h

d2E d2E −2 −6 2 = 4.20 × 10 − 12(3.80) × 10 v and when v = 52.6, is negative, hence v is the dv 2 dv 2 maximum speed. Maximum fuel economy, E = 21 + 2.10 × 10−2 v 2 − 3.80 ×10−6 v 4

= 21 + 2.10 × 10−2 (52.6) 2 − 3.80 × 10−6 (52.6)4 = 50.0 miles/gallon


276

EXERCISE 126 Page 311 2. For the curve y = 3x 2 − 2x find (a) the equation of the tangent, and (b) the equation of the

normal at the point (2, 8) (a) y = 3x 2 − 2x

Gradient, m =

dy = 6x − 2 dx

At the point (2, 8), x = 2, hence, m = 6(2) – 2 = 10 Equation of tangent is:

y − y1 = m(x − x1 )

i.e.

y – 8 = 10(x – 2)

i.e.

y – 8 = 10x – 20

and

y = 10x - 12

y − y1 = −

(b) Equation of normal is:

1 (x − x1 ) m

1 (x − 2) 10

i.e.

y −8 = −

i.e.

10(y – 8) = -x + 2

i.e.

10y – 80 = -x + 2

and

10y + x = 82

4. For the curve y = 1 + x − x 2 find (a) the equation of the tangent, and (b) the equation of the

normal at the point (-2, -5) (a) y = 1 + x − x 2

Gradient, m =

dy = 1 − 2x dx

At the point (-2, -5), x = -2, hence, m = 1 – 2(-2) = 5 Equation of tangent is:

y − y1 = m(x − x1 )

i.e.

y – (-5) = 5(x – -2)

i.e.

y + 5 = 5x + 10

and

y = 5x + 5

(b) Equation of normal is:

y − y1 = −

1 (x − x1 ) m


277

1 y − −5 = − (x − −2) 5

i.e. i.e.

5(y + 5) = -x - 2

i.e.

5y + 25 = -x - 2

and

5y + x + 27 = 0

1 5. For the curve θ = find (a) the equation of the tangent, and (b) the equation of the normal at t ⎛ 1⎞ the point ⎜ 3, ⎟ ⎝ 3⎠

1 (a) θ = = t −1 t

Gradient, m =

dθ 1 = − t −2 = − 2 dt t

1 1 ⎛ 1⎞ At the point ⎜ 3, ⎟ , t = 3, hence, m = − 2 = − 3 9 ⎝ 3⎠ Equation of tangent is:

θ − θ1 = m(t − t1 )

i.e.

1 1 θ − = − (t − 3) 3 9

i.e.

9θ - 3 = -t + 3

and

9θ + t = 6

(b) Equation of normal is: i.e.

θ − θ1 = −

1 (t − t1 ) m

1 1 θ− = − (t − 3) 3 ⎛ 1⎞ ⎜− ⎟ ⎝ 9⎠

i.e.

1 θ − = 9(t − 3) 3

i.e.

1 θ − = 9t − 27 3

and

θ = 9t − 26

2 3

or 3θ = 27t - 80


278


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 =

50 = 50p −1 p

dv 50 = −50p −2 = − 2 dp p

and

Approximate change in volume, δv ≈

Percentage change in volume =

⎛ 50 ⎞ dv ⎛ 50 ⎞ ⋅δp = ⎜ − 2 ⎟ (25.4 − 25.0) = ⎜ − (0.4) = -0.032 2 ⎟ dp ⎝ 25.0 ⎠ ⎝ p ⎠

−0.032 −3.2 −3.2 × 100% = = = -1.6% 50 50 2 p 25.0

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 = 4πr 2

and

dA = 8πr dr

Approximate change in surface area, δA ≈

dA ⋅δr = ( 8πr ) (6.0 − 5.96) = 8π(6.0)(−0.04) dr = -6.03 cm 2

(b) Volume of sphere, V =

4 3 πr 3

and

Approximate change in volume, δV ≈

dV = 4πr 2 dr dV ⋅δr = ( 4πr 2 ) (−0.04) = 4π(6.0) 2 (−0.04) = -18.10 cm 3 dr

5. The rate of flow of a liquid through a tube is given by Poiseuilles’s equation as: Q =

p π r4 8ηL

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.

Q=

p π r4 8ηL

from which,

η=

p π r4 8 LQ


279

δη ≈

⎛ pπr 4 ⎞ ⎛ pπr 4 ⎞ dη ⋅δQ = ⎜ − ± = ( 0.005Q) ⎜ ⎟ (±0.005) 2 ⎟ dQ ⎝ 8LQ ⎠ ⎝ 8LQ ⎠

δη ≈

⎛ πr 4 ⎞ ⎛ pπr 4 ⎞ dη ⋅δp = ⎜ ( ± 0.03p) = ⎟ ⎜ ⎟ (±0.03) dp ⎝ 8LQ ⎠ ⎝ 8LQ ⎠

⎛ 4pπr 3 ⎞ ⎛ pπr 4 ⎞ dη δη ≈ ⋅δr = ⎜ ⎟ (±0.02r) = ⎜ ⎟ (±0.08) dr ⎝ 8LQ ⎠ ⎝ 8LQ ⎠ δη ≈

⎛ − pπr 4 ⎞ ⎛ pπr 4 ⎞ dη ⋅δL = ⎜ 2 ⎟ (±0.01L) = ⎜ ⎟ (±0.01) dL ⎝ 8L Q ⎠ ⎝ 8LQ ⎠

Maximum possible percentage error ≈ 0.5% + 3% + 8% + 1% = 12.5%


280

CHAPTER 29 DIFFERENTIATION OF PARAMETRIC EQUATIONS EXERCISE 128 Page 316 2. A parabola has parametric equations: x = t 2 , y = 2t. Evaluate

If x = t 2 , then

dx = 2t dt

If y = 2t, then

dy =2 dt

Hence,

dy when t = 0.5 dx

dy dy dt 2 1 = = = dx dx 2t t dt

When t = 0.5,

dy 1 = =2 dx 0.5

dy d2 y (b) 3. The parametric equations for an ellipse are x = 4 cos θ, y = sin θ. Determine (a) dx dx 2

(a) If x = 4 cos θ, then If y = sin θ,

then

dx = −4sin θ dθ dy = cos θ dθ

dy 1 dy dθ cos θ Hence, = = = − cot θ dx 4 dx −4sin θ dθ d ⎛ dy ⎞ d ⎛ 1 ⎞ − cot θ ⎟ − 1 ( − cos ec 2 θ ) ⎜ ⎟ ⎜ d y dθ ⎝ dx ⎠ dθ ⎝ 4 1⎛ 1 1⎛ 1 ⎞ ⎞ ⎠= 4 (b) = = =− ⎜ 2 =− ⎜ 3 ⎟ ⎟ 2 dx −4sin θ −4sin θ dx 16 ⎝ sin θ sin θ ⎠ 16 ⎝ sin θ ⎠ dθ 2

= −

4. Evaluate

1 cos ec 3θ 16

dy π at θ = radians for the hyperbola whose parametric equations are x = 3 sec θ, dx 6

y = 6 tan θ


281

If x = 3 sec θ, then

dx = 3sec θ tan θ dθ

If y = 6 tan θ, then

dy = 6sec 2 θ dθ

Hence,

dy ⎛ 1 ⎞ ⎜ ⎟ dy dθ 6sec 2 θ 2sec θ 2 = = = = 2 ⎜ cos θ ⎟ = sin θ dx dx 3sec θ tan θ tan θ ⎜ ⎟ sin θ dθ ⎝ cos θ ⎠

When θ =

π , 6

dy 2 2 = = =4 dx sin π 0.5 6

6. The equation of a tangent drawn to a curve at point ( x1 , y1 ) is given by: y − y1 =

dy1 ( x − x1 ) . dx1

Determine the equation of the tangent drawn to the ellipse x = 3 cos θ, y = 2 sin θ at θ =

y − y1 =

π 6

dy1 ( x − x1 ) dx1

At point θ, x1 = 3cos θ

and

dx1 = −3sin θ dθ

y1 = 2sin θ

and

dy1 = 2 cos θ dθ

dy1 dy1 dθ 2 cos θ 2 = = = − cot θ dx1 dx1 −3sin θ 3 dθ

Hence,

2 The equation of a tangent is: y – 2 sin θ = − cot θ ( x − 3cos θ ) 3 At θ =

π , 6

y – 2 sin

π 2 π⎛ π⎞ = − cot ⎜ x − 3cos ⎟ 3 6⎝ 6⎠ 6 2 (1.732 )( x − 2.598 ) 3

i.e.

y–1= −

i.e.

y – 1 = -1.155(x – 2.598)

i.e.

y – 1 = -1.155x + 3

and

y = -1.155x + 4


282

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)

(a) If x = 2(θ - sin θ),

then

If y = 2(1 – cos θ), then

dy dx

(b)

d2 y dx 2

dx = 2 − 2 cos θ dθ dy = 2sin θ dθ

dy dy dθ 2sin θ sin θ = = = Hence, dx dx 2(1 − cos θ) 1 − cos θ dθ When θ = 0.62 rad,

dy sin 0.62 = = 3.122, correct to 4 significant figures. dx 1 − cos 0.62

d ⎛ dy ⎞ d ⎛ sin θ ⎞ (1 − cos θ)(cos θ) − (sin θ)(sin θ) cos θ − cos 2 θ − sin 2 θ d 2 y dθ ⎜⎝ dx ⎟⎠ dθ ⎜⎝ 1 − cos θ ⎟⎠ (1 − cos θ) 2 (1 − cos θ) 2 (b) = = = = dx dx 2 2(1 − cos θ) 2(1 − cos θ) 2(1 − cos θ) dθ =

cos θ − (cos 2 θ + sin 2 θ) cos θ − 1 = 3 2(1 − cos θ) 2(1 − cos θ)3

d2 y (cos 0.62) − 1 −0.1861215 = = = -14.43, correct to 4 significant When θ = 0.62 rad, 2 3 dx 2(1 − cos 0.62) 2(0.00644748) figures.

2. The equation of a normal drawn to a curve at point ( x1 , y1 ) is given by: y − y1 = −

Determine the equation of the normal drawn to the parabola x =

If x1 =

1 2 dx 1 t , = t 4 dt 2

If y1 =

1 t, 2

Hence,

1 ( x − x1 ) . dy1 dx1

1 2 1 t , y = t at t = 2. 4 2

dy 1 = dt 2 dy1 1 dy1 1 = dt = 2 = dx1 dx1 1 t t 2 dt


283

y − y1 = −

Equation of a normal is:

1 ( x − x1 ) dy1 dx1

1 1⎛ 1 ⎞ y − t = − ⎜ x − t2 ⎟ 1⎝ 2 4 ⎠ t

i.e.

At t = 2, equation of normal is: y – 1 = -2(x – 1) i.e.

y – 1 = -2x + 2

or

y = -2x + 3

π d2 y , correct to 4 significant figures, at θ = rad for the cardioid 2 dx 6

4. Determine the value of

x = 5(2θ - cos 2θ), y = 5(2 sin θ – sin 2θ).

If x = 5(2θ - cos 2θ), then

dx = 10 + 10sin 2θ = 10 (1 + sin 2θ ) dθ

If y = 5(2 sin θ – sin 2θ), hence

dy = 10 cos θ − 10 cos 2θ = 10(cos θ − cos 2θ) dθ

dy dy dθ 10(cos θ − cos 2θ) cos θ − cos 2θ = = = dx dx 10(1 + sin 2θ) 1 + sin 2θ dθ d ⎛ dy ⎞ d ⎛ cos θ − cos 2θ ⎞ (1 + sin 2θ)(− sin θ + 2sin 2θ) − (cos θ − cos 2θ)(2 cos 2θ) 2 1 + sin 2θ ) ( d 2 y dθ ⎝⎜ dx ⎠⎟ dθ ⎝⎜ 1 + sin 2θ ⎠⎟ = = = dx dx 2 10(1 + sin 2θ) 10(1 + sin 2θ) dθ

When θ =

π rad, 6

2π ⎞ ⎛ π 2π ⎞ ⎛ π 2π ⎞ ⎛ 2π ⎞ ⎛ ⎜1 + sin ⎟ ⎜ − sin + 2sin ⎟ − ⎜ cos − cos ⎟ ⎜ 2 cos ⎟ 6 ⎠⎝ 6 6 ⎠ ⎝ 6 6 ⎠⎝ 6 ⎠ ⎝ 2 2π ⎞ ⎛ 1 + sin ⎟ ⎜ 2 d y 6 ⎠ ⎝ = 2 2π ⎞ dx ⎛ 10 ⎜ 1 + sin ⎟ 6 ⎠ ⎝ (1.86603)(1.23205) − (0.366025)(1) (3.48205) = = 0.02975, correct to 4 significant 18.660254 figures.


284

5. The radius of curvature, ρ, of part of a surface when determining the surface tension of a liquid ⎡ ⎛ dy ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎢ ⎝ dx ⎠ ⎥⎦ ρ= ⎣ d2 y dx 2

is given by:

3/ 2

Find the radius of curvature (correct to 4 significant figures) of the part of the surface having parametric equations (a) x = 3t, y =

3 1 at the point t = t 2

(b) x = 4 cos3 t, y = 4sin 3 t at t =

(a) x = 3t , y=

hence

π rad 6

dx =3 dt

3 dy 3 = 3t −1 , hence = −3t −2 = − 2 t dt t

dy 3 − 2 dy dt 1 = = t =− 2 dx dx 3 t dt

and at t =

1 , 2

dy 1 =− = −4 2 dx ⎛1⎞ ⎜ ⎟ ⎝2⎠

d ⎛ dy ⎞ d ⎛ 1 ⎞ d − − t −2 ) d y dt ⎜⎝ dx ⎟⎠ dt ⎜⎝ t 2 ⎟⎠ dt ( 2t −3 2 = = = = = 3 2 dx dx 3 3 3 3t dt 2

⎡ ⎛ dy ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎢ ⎝ dx ⎠ ⎦⎥ Hence, radius of curvature, ρ = ⎣ d2 y dx 2 (b) x = 4 cos3 t , hence y = 4sin 3 t , hence

and at t =

1 d2 y 2 2 16 , = 3= = 3 2 2 dx 3t 3 ⎛1⎞ 3⎜ ⎟ ⎝2⎠

3/ 2 3

=

⎡1 + ( −4 )2 ⎤ 173 ⎣ ⎦ = 13.14 = 16 16 3 3

dx = 12 cos 2 t(− sin t) = −12 cos 2 t sin t dt dy = 12sin 2 t cos t dt

dy dy dt 12sin 2 t cos t sin t = = =− = − tan t 2 dx dx −12 cos t sin t cos t dt

and at t =

π π dy = − tan = −0.57735 rad , dx 6 6


285

d ⎛ dy ⎞ d ( − tan t ) ⎜ ⎟ − sec 2 t d y dt ⎝ dx ⎠ 1 dt = = = = 2 2 2 dx dx −12 cos t sin t −12 cos t sin t 12 cos 4 t sin t dt 2

and at t

π d2 y rad , = 6 dx 2

1 π⎞ π ⎛ 12 ⎜ cos ⎟ sin 6⎠ 6 ⎝ 4

= 0.29630

⎡ ⎛ dy ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎢ ⎝ dx ⎠ ⎦⎥ ⎣ Hence, radius of curvature, ρ = d2 y dx 2

3/ 2 3

=

⎡1 + ( −0.57735 )2 ⎤ 1.3333333 ⎣ ⎦ = = 5.196 0.29630 0.29630


286

CHAPTER 30 DIFFERENTIATION OF IMPLICIT FUNCTIONS EXERCISE 130 Page 320

2. Differentiate the given functions with respect to x: (a)

5 ln 3t 2

(b)

3 2y +1 e 4

(c) 2 tan 3y

(a)

d ⎛5 ⎞ ⎜ ln 3t ⎟ = dx ⎝ 2 ⎠

5 dt 5 ⎛ 1 ⎞ dt = ⎜ ⎟ 2t dx 2 ⎝ t ⎠ dx

(b)

3 dy d ⎛ 3 2y +1 ⎞ 3 2y +1 dy = e 2y +1 ⎜ e ⎟ = ( 2e ) 2 dx dx ⎝ 4 dx ⎠ 4

(c)

dy d dy ( 2 tan 3y ) = (2) ( 3sec2 3y ) = 6sec2 3y dx dx dx

4. Differentiate the following with respect to u: (a)

2 (3x + 1)

(b) 3 sec 2θ

(c)

2 y

(a)

6 dx d ⎛ 2 ⎞ d dx ⎡ 2(3x + 1) −1 ⎦⎤ = ⎣⎡ −2(3x + 1) −2 (3) ⎦⎤ =− ⎜ ⎟= 2 ⎣ (3x + 1) du du ⎝ 3x + 1 ⎠ du du

(b)

dθ d dθ ( 3sec 2θ ) = 3 ( 2sec 2θ tan 2θ ) = 6sec 2θ tan 2θ du du du

(c)

⎛ 1 − 32 ⎞ dy 1 dy d ⎛ 2 ⎞ d ⎛ − 12 ⎞ 1 dy = =− 3 = − 2y (2) ⎜⎜ ⎟⎟ = ⎜ ⎟ ⎜− y ⎟ du ⎝ y ⎠ du ⎝ du y 3 du ⎠ ⎝ 2 ⎠ du y2


287

EXERCISE 131 Page 321 1. Determine

d 3x 2 y3 ) ( dx

d d 3 d 3x 2 y3 ) = 3x 2 y ) + y3 ( 3x 2 ) ( ( dx dx dx

using the product rule

dy ⎞ ⎛ 3 2 2 dy + 6xy3 = ( 3x 2 ) ⎜ 3y 2 ⎟ + ( y ) ( 6x ) = 9x y dx dx ⎝ ⎠ ⎛ dy ⎞ + 2y ⎟ = 3xy 2 ⎜ 3x ⎝ dx ⎠

3. Determine

d ⎛ 3u ⎞ ⎜ ⎟ = du ⎝ 4v ⎠

d ⎛ 3u ⎞ ⎜ ⎟ du ⎝ 4v ⎠

( 4v )( 3) − ( 3u ) ⎛⎜ 4 ( 4v )

2

dv ⎞ dv ⎟ 12v − 12u du ⎝ ⎠= du = 12 ⎛ v − u dv ⎞ = 3 ⎛ v − u dv ⎞ ⎜ ⎟ ⎜ ⎟ 2 du ⎠ 16v 16v 2 ⎝ du ⎠ 4v 2 ⎝

5. Determine

dz given z = 2x 3 ln y dy

z = 2x 3 ln y

hence,

⎛1⎞ ⎛ dz dx ⎞ = ( 2x 3 ) ⎜ ⎟ + ( ln y ) ⎜ 6x 2 ⎟ dy dy ⎠ ⎝ y⎠ ⎝ ⎛x 2x 3 dx dx ⎞ + 6x 2 ln y or 2x 2 ⎜ + 3 ln y = ⎟ y dy dy ⎠ ⎝y


288


dy given x 2 + y 2 + 4x − 3y + 1 = 0 dx

d 2 d d d d x ) + ( y 2 ) + ( 4x ) − ( 3y ) + (1) = 0 ( dx dx dx dx dx 2x + 2y

dy dy + 4−3 +0 = 0 dx dx dy dy dy − 2y = ( 3 − 2y ) dx dx dx

2x + 4 = 3

dy 2x + 4 = dx 3 − 2y

Hence,

3. Given x 2 + y 2 = 9 evaluate

dy when x = dx

5 and y = 2

d 2 d d (x ) + (y 2 ) = (9) dx dx dx i.e.

2x + 2y

i.e.

When x =

2y

dy =0 dx

dy = -2x dx

5 and y = 2,

5. Determine

and

dy x =− dx y

dy 5 =− dx 2

dy given 3y 2 + 2xy − 4x 2 = 0 dx

Given 3y 2 + 2xy − 4x 2 = 0

then

6y

⎤ dy ⎡ ⎛ dy ⎞ + ⎢( 2x ) ⎜ 1 ⎟ + ( y )( 2 ) ⎥ − 8x = 0 dx ⎣ ⎝ dx ⎠ ⎦ dy ( 6y + 2x ) = 8x − 2y dx 4x − y dy 8x − 2y = = 3y + x dx 6y + 2x

and

7. Determine

dy given 3y + 2x ln y = y 4 + x dx

Given 3y + 2x ln y = y 4 + x then

3

⎤ ⎛ 1 dy ⎞ dy ⎡ 3 dy + ⎢( 2x ) ⎜ +1 ⎟ + ( ln y )( 2 ) ⎥ = 4y dx ⎣ dx ⎝ y dx ⎠ ⎦


289

i.e.

⎞ dy ⎛ 2x − 4y3 ⎟ = 1 − 2 ln y ⎜3+ dx ⎝ y ⎠

and

dy = dx

1 − 2 ln y 2x − 4y 3 3+ y

5 dy 1 8. If 3x 2 + 2x 2 y3 − y 2 = 0 evaluate when x = and y = 1. 4 dx 2 5 Given 3x 2 + 2x 2 y3 − y 2 = 0 then 4

⎡ ⎤ 5 dy dy ⎞ ⎛ 3 6x + ⎢( 2x 2 ) ⎜ 3y 2 =0 ⎟ + ( y ) ( 4x ) ⎥ − y dx ⎠ ⎝ ⎣ ⎦ 2 dx

i.e.

⎛5 ⎞ dy 6x + 4xy3 = ⎜ y − 6x 2 y 2 ⎟ ⎝2 ⎠ dx dy 6x + 4xy3 = dx 5 y − 6x 2 y 2 2

and

When x =

⎛1⎞ ⎛1⎞ 3 6 ⎜ ⎟ + 4 ⎜ ⎟ (1) dy 3+ 2 5 2 2 = ⎝ ⎠ ⎝ 2⎠ = = =5 dx 5 2.5 − 1.5 1 1 2 (1) − 6 ⎛⎜ ⎞⎟ (1) 2 ⎝2⎠

1 and y = 1, 2

10. Find the gradients of the tangents drawn to the ellipse

Given

x 2 y2 + =2 4 9

then

2x 2y dy + =0 4 9 dx

x 2 y2 + = 2 and x = 2, then If 4 9

Thus,

2y dy 2x =− 9 dx 4

2x dy 2x 9 9x =− 4 =− ⋅ =− 2y dx 4 2y 4y 9

from which,

Hence,

and

x 2 y2 + = 2 at the point where x = 2. 4 9

y2 = 9

4 y2 + =2 4 9 and

from which,

y2 =1 9

y=±3

dy 9x (9)(2) 3 =− =− =± or ±1.5 dx 4y 4(±3) 2


290

CHAPTER 31 LOGARITHMIC DIFFERENTIATION EXERCISE 133 Page 326 2. Use logarithmic differentiation to differentiate y = (x + 1)(2x + 1)3 If y = (x − 3) 2 (x + 2) 4

then

(x + 1)(2x + 1)3 with respect to x. (x − 3) 2 (x + 2) 4

⎧ (x + 1)(2x + 1)3 ⎫ ln y = ln ⎨ 2 4⎬ ⎩ (x − 3) (x + 2) ⎭

i.e.

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

i.e.

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

⎧ 1 dy 6 2 4 ⎫ = y⎨ + − − ⎬ dx ⎩ (x + 1) (2x + 1) (x − 3) (x + 2) ⎭

i.e.

dy (x + 1)(2x + 1)3 ⎧ 1 6 2 4 ⎫ = + − − ⎬ 2 4 ⎨ dx (x − 3) (x + 2) ⎩ (x + 1) (2x + 1) (x − 3) (x + 2) ⎭

4. Use logarithmic differentiation to differentiate y =

If y =

e 2x cos3x (x − 4)

then

e 2x cos3x with respect to x. (x − 4)

1 ⎧⎪ e 2x cos 3x ⎫⎪ 2x 2 ln y = ln ⎨ = ln e + ln(cos 3x) − ln(x − 4) ⎬ x − 4 ( ) ⎪⎩ ⎪⎭

= 2x + ln(cos 3x) -


1 ln(x - 4) 2

1 1 dy (−3sin 3x) = 2+ − 2 y dx cos 3x (x − 4)

i.e.

⎧ dy 1 ⎫ = y ⎨2 − 3 tan 3x − ⎬ dx 2(x − 4) ⎭ ⎩

i.e.

dy e 2x cos 3x ⎧ 1 ⎫ = ⎨ 2 − 3 tan 3x − ⎬ dx 2(x − 4) ⎭ (x − 4) ⎩

6. Use logarithmic differentiation to differentiate y =

2x 4 tan x If y = 2x e ln 2x

then

2x 4 tan x with respect to x. e2x ln 2x

⎧ 2x 4 tan x ⎫ 4 2x ln y = ln ⎨ 2x ⎬ = ln 2x + ln(tan x) − ln e − ln(ln 2x) ⎩ e ln 2x ⎭


291

= ln 2 + ln x 4 + 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:

1 dy 4 1 1 ⎛1⎞ = 0+ + (sec 2 x) − 2 − ⎜ ⎟ y dx x tan x ln 2x ⎝ x ⎠

i.e.

dy ⎧ 4 cos x ⎛ 1 ⎞ 1 ⎫ = y⎨ + ⎬ ⎜ ⎟−2− 2 dx x ln 2x ⎭ ⎩ x sin x ⎝ cos x ⎠

i.e.

dy 2x 4 tan x ⎧ 4 1 1 ⎫ = 2x − 2− ⎨ + ⎬ dx e ln 2x ⎩ x sin x cos x x ln 2x ⎭

dy π 2eθ sin θ 8. Evaluate , correct to 3 significant figures, when θ = given y = 4 dθ θ5 If y =

2eθ sin θ θ5

then

5 ⎧ 2eθ sin θ ⎫ θ ln y = ⎨ ⎬ = ln 2 + ln e + ln(sin θ) − ln θ 2 5 θ ⎭ ⎩

5 = ln 2 + θ + ln(sin θ) − ln θ 2 Differentiating w.r.t. θ gives:

1 dy 1 5⎛1⎞ (cos θ) − ⎜ ⎟ = 0 +1+ y dθ sin θ 2⎝θ⎠ dy 5 ⎫ 2eθ sin θ ⎧ 5⎫ ⎧ = y ⎨1 + cot θ − ⎬ = ⎨1 + cot θ − ⎬ 5 dθ 2θ ⎭ 2θ ⎭ ⎩ ⎩ θ

i.e.

⎫ π⎧ ⎪⎪ ⎪ 4 1 + cot π − 5 ⎪ ⎨ ⎬ 5 4 ⎛π⎞ ⎛π⎞ ⎪ 2⎜ ⎟⎪ ⎜ ⎟ ⎪⎩ ⎝ 4 ⎠ ⎭⎪ ⎝4⎠

π

When θ =

π , 4

dy = dθ

2e 4 sin

= 5.673935[1 + 1 – 3.18309886] = - 6.71, correct to 3 significant figures.


292

EXERCISE 134 Page 328 2. Differentiate y = ( 2x − 1) with respect to x. x

Since y = ( 2x − 1)

x

then ln y = ln ( 2x − 1) = x ln(2x – 1) x


1 dy ⎛ 2 ⎞ = (x) ⎜ ⎟ + [ln(2x − 1)](1) y dx ⎝ 2x + 1 ⎠

i.e.

dy ⎧ 2x ⎫ = y⎨ + ln(2x − 1) ⎬ dx ⎩ 2x − 1 ⎭

i.e.

dy x ⎧ 2x ⎫ = ( 2x − 1) ⎨ + ln(2x − 1) ⎬ dx ⎩ 2x − 1 ⎭

3. Differentiate y =

x

(x + 3) with respect to x.

1

y=

x

1

(x + 3) = (x + 3) x

and


ln y = ln(x + 3) x =

1 ln(x + 3) x

1 dy ⎛ 1 ⎞ ⎛ 1 ⎞ −2 = ⎜ ⎟⎜ ⎟ + [ln(x + 3)] ( − x ) by the product rule y dx ⎝ x ⎠ ⎝ x + 3 ⎠ ⎧ 1 dy ln(x + 3) ⎫ = y⎨ − ⎬ dx x2 ⎭ ⎩ x(x + 3)

and

⎧ 1 dy x ln(x + 3) ⎫ = (x + 3) ⎨ − ⎬ dx x2 ⎭ ⎩ x(x + 3)

i.e.

5. Show that when y = 2x x and x = 1,

y = 2x x

by the product rule

dy =2 dx

and ln y = ln 2x x = ln 2 + ln x x = ln 2 + x ln x


1 dy ⎛1⎞ = 0 + (x) ⎜ ⎟ + (ln x)(1) = 1 + ln x y dx ⎝x⎠

i.e.

dy = y(1 + ln x) = 2x x (1 + ln x) dx

When x = 1,

dy = 2(1)1 (1 + ln1) = 2 dx

6. Evaluate

d dx

{

x

by the product rule

}

(x − 2) when x = 3


293

1

Let y =

x

(x − 2) = (x − 2) x

1

then ln y = ln(x − 2) x =

1 ln(x − 2) ln(x − 2) = x x

⎛ 1 ⎞ (x) ⎜ ⎟ − [ln(x − 2)](1) 1 dy x−2⎠ ⎝ = y dx x2


by the quotient rule

and

⎧ x ⎫ − ln(x − 2) ⎪ ⎪ dy ⎪ (x − 2) ⎪ = y⎨ ⎬ 2 dx x ⎪ ⎪ ⎪⎩ ⎭⎪

i.e.

dy x (x − 2) ⎧ x ⎫ = − ln(x − 2) ⎬ ⎨ 2 dx x ⎩x − 2 ⎭

When x = 3,

dy 3 (3 − 2) ⎧ 3 ⎫ ⎛ 1 ⎞ ⎧3 ⎫ 1 = − ln(3 − 2) ⎬ = ⎜ ⎟ ⎨ − 0 ⎬ = ⎨ 2 dx 3 ⎩3 − 2 ⎭ ⎝ 9 ⎠ ⎩1 ⎭ 3

7. Show that if y = θθ and θ = 2 ,

If y = θθ

dy = 6.77, correct to 3 significant figures. dθ

then ln y = ln θθ = θ ln θ

Differentiating w.r.t. θ gives: and When θ = 2 ,

1 dy ⎛1⎞ = ( θ ) ⎜ ⎟ + (ln θ)(1) = 1 + ln θ y dθ ⎝θ⎠ dy = y(1 + ln θ) = θθ (1 + ln θ) dθ dy = 22 (1 + ln 2) = 6.77, correct to 3 significant figures. dθ


294

CHAPTER 32 DIFFERENTIATION OF HYPERBOLIC FUNCTIONS EXERCISE 135 Page 331

2. Differentiate the following with respect to the variable:

(a)

2 sec h 5x 3

(b)

5 t cos ech 8 2

(c) 2 coth 7θ

(a)

10 d ⎛2 ⎞ ⎛2⎞ ⎜ sec h 5x ⎟ = ⎜ ⎟ ( −5sec h 5x tanh 5x ) = − 3 sec h 5x tanh 5x dx ⎝ 3 ⎠ ⎝3⎠

(b)

5 t t d ⎛5 t ⎞ ⎛ 5 ⎞⎛ 1 t t⎞ ⎜ co sec h ⎟ = ⎜ ⎟ ⎜ − co sec h coth ⎟ = − 16 co sec h 2 coth 2 dt ⎝ 8 2 ⎠ ⎝ 8 ⎠⎝ 2 2 2⎠

(c)

d ( 2 cot h 7θ ) = ( 2 ) ( −7 co sec h 2 7θ ) = −14co sec h 2 7θ dθ

3. Differentiate the following with respect to the variable:

(a) 2 ln (sh x)

(a)

(b)

3 ⎛ ⎛ θ ⎞⎞ ln th ⎜ ⎟ 4 ⎜⎝ ⎝ 2 ⎠ ⎟⎠

d 1 ( 2 ln(shx) ) = ( 2 ) ⎛⎜ ⎞⎟ ( chx ) = 2cot h x dx ⎝ shx ⎠ 1

θ ⎛ ⎞ ⎛ ⎞⎛ θ ⎞ ch 2 ⎜ ⎟ ⎜ ch 2 ⎟ θ d ⎡ 3 ⎛ ⎛ θ ⎞ ⎞⎤ ⎛ 3 ⎞ ⎜ 1 ⎟ ⎛ 1 3 3 1 ⎞ 2 2 = = = (b) ln th sec h ⎟⎜ ⎜ ⎟⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟⎜ ⎟ dx ⎣ 4 ⎜⎝ ⎝ 2 ⎠ ⎟⎠ ⎦ ⎝ 4 ⎠ ⎜ th θ ⎟ ⎝ 2 2 ⎠ 8 sh θ 8 ⎜ ch 2 θ ⎟ ⎜ sh θ ⎟ ⎝ 2⎠ ⎝ 2 2 ⎠⎝ 2 ⎠ θ ch 2 ⎛ ⎞ ⎟ 3 θ θ 3⎜ 1 = ⎜ = sec h cos ec h ⎟ 2 2 8 ⎜ ch θ sh θ ⎟ 8 ⎝ 2 2⎠

5. Differentiate the following with respect to the variable: (a)

3sh 4x 2x 3

(b)

ch 2t cos 2t


295

d ⎛ 3sh 4x ⎞ ( 2x (a) ⎜ ⎟= dx ⎝ 2x 3 ⎠ = (b)

3

) (12 ch 4x ) − ( 3sh 4x ) ( 6x ) ( 2x ) 2

3 2

by the quotient rule

12x ch 4x − 9sh 4x 24x 3ch 4x − 18x 2sh 4x = 6 2x 4 4x

d ⎛ ch 2t ⎞ ( cos 2t )( 2sh 2t ) − ( ch 2t )( −2sin 2t ) ⎜ ⎟= dx ⎝ cos 2t ⎠ cos 2 2t

=

2 ( cos 2t sh 2t + ch 2t sin 2t ) cos 2 2t


296

CHAPTER 33 DIFFERENTIATION OF INVERSE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS EXERCISE 136 Page 336 1.(a) Differentiate sin −1 4x with respect to x.

If y = sin −1 4x , then

2.(b) Differentiate

dy = dx

4 ⎡⎣1 − (4x) ⎤⎦

=

2

4

( 1 − 16x ) 2

2 x cos −1 with respect to x. 3 3

⎡ 2 dy 2 ⎢ −1 x = If y = cos , then 3 3 dx 3 ⎢ ⎣

⎤ ⎥ = 2 2 ⎥ (3 − x ) ⎦ 3 −1

−2

(9 − x ) 2

3.(a) Differentiate 3 tan −1 2x with respect to x.

If y = 3 tan −1 2x , then

⎡ ⎤ 6 dy 2 = 3⎢ = 2 2⎥ dx ⎣1 + (2x) ⎦ 1 + 4x

3 4.(b) Differentiate sec −1 x with respect to x. 4 3 3 dy 1 1 4 If y = sec −1 x , then = = = = 2 2 4 dx 3 ⎡ 3x 2 ⎤ ⎛ ⎞ ⎛ ⎞ − 9x 9x 16 x ⎛ ⎞ − 1⎟ x ⎜ x ⎢⎜ ⎟ − 1⎥ x ⎜ ⎟ 4 ⎣⎢⎝ 4 ⎠ ⎝ 16 ⎠ ⎝ 16 ⎠ ⎦⎥

( 9x

2

− 16 )

4

4

= x

5.(a) Differentiate

1

( 9x

2

− 16 )

5 θ cos ec −1 with respect to θ. 2 2

⎡ dy 5 ⎢ 5 −1 θ , then If y = cos ec = dθ 2 ⎢ θ 2 2 ⎣

⎤ ⎥ = 2 2 ⎥ (θ − 2 ) ⎦ θ −2

−5

(θ

2

− 4)


297

6.(b) Differentiate cot −1 θ2 − 1 with respect to θ.

1 − 1 2 θ − 1) 2 2θ ( dy = 2 = θ2 − 1 , then 2⎤ dθ ⎡ 2 1 + ⎢ ( θ − 1) ⎥ ⎣ ⎦

−

If y = cot −1

−θ

(θ

2

−1

=

− 1) ⎡⎣1 + θ2 − 1⎤⎦

(θ

θ

2

− 1)

1+ x2 ⎛ x ⎞ is 7. Show that the differential coefficient of tan −1 ⎜ 2 ⎟ 2 4 ⎝ 1− x ⎠ 1− x + x

(1 − x ) (1) − ( x )( −2x ) 1 − x + 2x (1 − x ) (1 − x ) dy = = dx ⎛ x ⎞ (1 − x ) + x 1+ ⎜ ⎟ ⎝ 1− x ⎠ (1 − x ) 2

⎛ x ⎞ then If y = tan ⎜ 2 ⎟ ⎝ 1− x ⎠ −1

2

2 2

2

2 2

2

2 2

2

2

1+ x2 = 1 − 2x 2 + x 4 + x 2

2 2

1 + x2 = ( 1 − x 2 + x4 )

8.(b) Differentiate t 2 sec−1 2t with respect to t. ⎛ ⎞ ⎟ dy 2 2 ⎜ If y = t sec 2t then = (t )⎜ + ( sec −1 2t ) ( 2t ) = ⎟ 2 dt ⎜ 2t ⎡⎣( 2t ) − 1⎤⎦ ⎟ ⎝ ⎠ 2

−1

t

( 4t

2

− 1)

+ 2 t sec −1 2t

9.(a) Differentiate θ2 cos −1 ( θ2 − 1) with respect to θ. ⎛ ⎞ ⎜ ⎟ − θ dy 2 −1 2 If y = θ2 cos −1 ( θ2 − 1) then = ( θ2 ) ⎜ ⎟ + ⎡⎣cos ( θ − 1) ⎤⎦ ( 2θ ) 2 dθ ⎜ ⎡1 − ( θ2 − 1) ⎤ ⎟ ⎜ ⎣⎢ ⎦⎥ ⎟⎠ ⎝ 2θ3 −1 2 = 2θ cos ( θ − 1) − ⎡1 − ( θ4 − 2θ2 + 1) ⎤ ⎣ ⎦ 2θ3

= 2θ cos −1 ( θ2 − 1) − = 2θ cos

−1

(θ

2

− 1) −

( 2θ

2

− θ4 )

2θ3 θ

(2 − θ )

= 2θ cos −1 ( θ2 − 1) − = 2θ cos

−1

2


(θ

2

− 1) −

2 θ3 θ2 ( 2 − θ2 ) 2θ 2

(2− θ ) 2

298

10.(b) Differentiate x cos ec −1 x with respect to x.

If y = x cos ec −1 x

⎛ ⎞ 1 −1 − x 2 ⎜ ⎟ dy 2 ⎜ ⎟ + ⎡cos ec −1 x ⎤ (1) = x then ( ) ⎣ ⎦ 2 dx ⎜ ⎡ x − 1⎤ ⎟ x ⎜ ⎢⎣ ⎥⎦ ⎟⎠ ⎝

( )

⎛ ⎞ −x ⎟ + ⎡cos ec −1 x ⎤ =⎜ ⎦ ⎜ 2 x x ( x − 1) ⎟ ⎣ ⎝ ⎠ = cos ec −1 x −

1 2 (x − 1)

sin −1 3x with respect to x. 11.(a) Differentiate x2 ⎛ ⎞ ⎜ ⎟ 3 − ( sin −1 3x ) ( 2x ) (x )⎜ 2 ⎟ ⎜ ⎡⎣1 − ( 3x ) ⎤⎦ ⎟ dy ⎝ ⎠ then = 2 2 dx (x ) 2

If y =

sin −1 3x x2

⎛ ⎞ ⎜ ⎟ 3x 2 − 2x ( sin −1 3x ) ⎜ 2 ⎟ ⎡ ⎜ ⎡⎣1 − ( 3x ) ⎤⎦ ⎟ x ⎢ ⎠ = ⎝ = x4 x4 ⎢ ⎣

⎧ 1 ⎪ = 3⎨ x ⎪ ⎩

3x

(1 − 9x ) 2

⎤ − 2sin −1 3x ⎥ ⎥ ⎦

3x

( 1 − 9x )


2

⎫ ⎪ − 2sin 3x ⎬ ⎭⎪ −1

299


1. Use logarithmic equivalents of inverse hyperbolic functions to evaluate the following correct to

1 2

4 decimal places: (a) sinh −1

sinh −1

⎧⎪ x + a 2 + x 2 x = ln ⎨ a a ⎪⎩

(a) sinh −1

(c) sinh −1 0.9

⎫⎪ ⎬ ⎪⎭

⎧⎪1 + 22 + 12 1 = ln ⎨ 2 2 ⎪⎩

(b) sinh −1 4 = sinh −1

(b) sinh −1 4

⎫⎪ ⎛ 1+ 5 ⎞ ⎟⎟ = ln1.618034 = 0.4812 ⎬ = ln ⎜⎜ ⎪⎭ ⎝ 2 ⎠

⎧⎪ 4 + 12 + 42 4 = ln ⎨ 1 1 ⎩⎪

⎧⎪ 0.9 + 12 + 0.92 (c) sinh 0.9 = ln ⎨ 1 ⎩⎪ −1

⎫⎪ ⎬ = ln 4 + 17 = ln 8.123106 = 2.0947 ⎭⎪

(

)

⎫⎪ ⎬ = ln 0.9 + 1.81 = ln 2.245362 = 0.8089 ⎭⎪

(

)


4 decimal places: (a) cosh −1

cosh −1

⎧⎪ x + x 2 − a 2 x = ln ⎨ a a ⎩⎪

(a) cosh −1

5 4

(b) cosh −1 3

(c) cosh −1 4.3

⎫⎪ ⎬ ⎭⎪

⎧⎪ 5 + 52 − 42 5 = ln ⎨ 4 4 ⎪⎩

⎪⎧ 3 + 32 − 12 (b) cosh −1 3 = ln ⎨ 1 ⎪⎩

⎫⎪ ⎛ 5+3⎞ ⎬ = ln ⎜ ⎟ = ln 2 = 0.6931 ⎝ 4 ⎠ ⎪⎭

⎪⎫ ⎬ = ln 3 + 8 = ln 5.828427 = 1.7627 ⎪⎭

(

⎧⎪ 4.3 + 4.32 − 12 (c) cosh −1 4.3 = ln ⎨ 1 ⎩⎪

)

⎫⎪ ⎬ = ln 4.3 + 17.49 = ln 8.482105 = 2.1380 ⎭⎪

(

)


4 decimal places: (a) tanh −1

tanh −1

1 4

(b) tanh −1

5 8

(c) tanh −1 0.7

x 1 ⎛a+x⎞ = ln ⎜ ⎟ a 2 ⎝a−x⎠ © 2006 John Bird. All rights reserved. Published by Elsevier.

300

(a) tanh −1

1 ⎛ 4 +1 ⎞ 1 5 1 = ln ⎜ ⎟ = ln = 0.2554 2 ⎝ 4 −1 ⎠ 2 3 4

(b) tanh −1

5 1 ⎛ 8 + 5 ⎞ 1 13 = ln ⎜ = 0.7332 ⎟ = ln 8 2 ⎝8−5⎠ 2 3

(c) tanh −1 0.7 =

1 ⎛ 1 + 0.7 ⎞ 1 1.7 = 0.8673 ln ⎜ ⎟ = ln 2 ⎝ 1 − 0.7 ⎠ 2 0.3


301

EXERCISE 138 Page 341 1.(b) Differentiate sinh −1 4x with respect to x.

If y = sinh −1 4x then

dy = dx

2.(a) Differentiate 2 cosh −1

If y = 2 cosh −1

4 ⎡( 4x )2 + 1⎤ ⎣ ⎦

=

4

( 16x

2

+ 1)

t with respect to t. 3

⎡ 1 ⎤ t dy then = 2⎢ ⎥ = 2 2 3 dt ⎣ t −3 ⎦

2

(t

2

− 9)

3.(b) Differentiate 3 tanh −1 3x with respect to x.

If y = 3 tanh −1 3x then

⎡ ⎤ 9 dy 3 = = 3⎢ 2⎥ 2 dx ⎢⎣1 − ( 3x ) ⎥⎦ 1 − 9x

4.(a) Differentiate sech −1

3x with respect to x. 4

3x dy = If y = sech −1 then dx 3x 4 4

−

3 4

⎡ ⎛ 3x ⎞ 2 ⎤ ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 4 ⎠ ⎥⎦

=

−1 ⎛ 9x 2 ⎞ x ⎜1 − ⎟ 16 ⎠ ⎝

=

−1 ⎛ 16 − 9x 2 ⎞ x ⎜ ⎟ ⎝ 16 ⎠

=

−1 x 4

2

−4

= x

5.(b) Differentiate

(16 − 9x )

( 16 − 9x ) 2

1 cos ec h −1 4x with respect to x. 2

⎡ ⎤ − dy 1 4 1 ⎢ ⎥ = ⎢ If y = cos ec h −1 4x then = 2 dx 2 4x ⎡( 4x )2 + 1⎤ ⎥ 2x ⎢ ⎣ ⎦ ⎥⎦ ⎣ 6.(a) Differentiate coth −1

−1

( 16x

2

+ 1)

2x with respect to x. 7


302

2x dy = then If y = coth −1 7 dx

1 cosh −1 2

7.(b) Differentiate

1 If y = cosh −1 2

(x

2

+ 1)

2 2 2 2 (49) 14 7 7 7 7 = = = = 2 2 2 2 49 − 4x 49 − 4x 2 49 − 4x ⎛ 2x ⎞ 1 − 4x 1− ⎜ ⎟ 49 49 ⎝ 7 ⎠

(x

2

+ 1) with respect to x.

⎡ ⎢ dy 1 ⎢ then = dx 2 ⎢ ⎢ ⎣⎢

⎤ 1 − 1 2 2 (2x) ⎥ x 1 + ( ) 2 ⎥= 2 ⎡ ⎤⎥ 2 2 x 1 1 + − ) ⎥⎦ ⎥⎥ ⎢⎣ ( ⎦ x

= 2

(x

2

+ 1)

(x ) 2

x

(x

2

+ 1)

x

=

(x

2x

2

+ 1)

(x

2

+ 1 − 1)

1

=

2

(x

2

+ 1)

8.(a) Differentiate sech −1 (x − 1) with respect to x.

If y = sech −1 (x − 1) then

dy −1 −1 = = dx (x − 1) ⎡1 − (x − 1) 2 ⎤ (x − 1) 1 − (x 2 − 2x + 1) ⎣ ⎦

=

−1

(x − 1) (2x − x 2 )

=

−1

(x − 1)

[ x(2 − x)]

9.(b) Differentiate coth −1 (cos x) with respect to x.

If y = coth −1 (cos x) then

dy − sin x − sin x − sin x −1 = = = = = = − cos ec x 2 2 2 dx 1 − ( cos x ) 1 − cos x sin x sin x

10.(a) Differentiate θ sinh −1 θ with respect to θ.

⎛ dy If y = θ sinh θ then = θ⎜ ⎜⎜ dθ ⎝ −1

⎞ ⎟ + ( sinh −1 θ ) (1) = ( θ2 + 1) ⎟⎟⎠

1

θ

(θ

2

+ 1)

+ sinh −1 θ

tanh −1 x 11.(b) Differentiate with respect to x. (1 − x 2 ) © 2006 John Bird. All rights reserved. Published by Elsevier.

303

tanh x (1 − x 2 )

If y =

12. Show that

(1 − x ) ⎛⎜⎝ 1 −1x

⎞ ⎟ − ( tanh x )( −2x ) 1 + 2x tanh −1 x ⎠ = 2 2 (1 − x 2 ) ( 1 − x2 )

2

−1

dy = dθ

then

2

d ⎡⎣ x cosh −1 (cosh x) ⎤⎦ = 2x dx

⎡ ⎤ d sinh x ⎥ + ( cosh −1 (cosh x) ) (1) ⎡⎣ x cosh −1 (cosh x) ⎤⎦ = ( x ) ⎢ 2 ⎢ cosh x − 1 ⎥ dx )⎦ ⎣ ( x sinh x = since cosh −1 (cosh x) = x +x 2 ( sinh x )

x sinh x +x sinh x

=

= x + x = 2x 13.(b) Determine

∫

3

( 4x

2

+ 25 )

dx = 3∫

14.(b) Determine

∫

1

(t

2

− 5)

∫

dt = ∫

∫

3

( 4x 2 + 25)

dx

1 ⎡( 2x )2 + 52 ⎤ ⎣ ⎦

1

(t

2

− 5)

1

dx = 3∫

dx =

3 1 ∫ 5 ⎡⎛ 2x ⎞ 2 ⎤ ⎢⎜ ⎟ + 1⎥ ⎣⎢⎝ 5 ⎠ ⎦⎥

⎡ 2 ⎛ ⎛ 2x ⎞2 ⎞ ⎤ ⎢5 ⎜⎜ ⎜ ⎟ + 1⎟⎟ ⎥ ⎢⎣ ⎝ ⎝ 5 ⎠ ⎠ ⎥⎦ 2 2 3 ⎛3⎞ 5 5 5 =⎜ ⎟ ∫ dx = ∫ dx 2 5 2 2 ⎝ ⎠ ⎡⎛ 2x ⎞ ⎤ ⎡⎛ 2x ⎞ 2 ⎤ ⎢⎜ ⎟ + 1⎥ ⎢⎜ ⎟ + 1⎥ ⎢⎣⎝ 5 ⎠ ⎥⎦ ⎢⎣⎝ 5 ⎠ ⎥⎦ 3 2x +c = sinh −1 2 5

dt

1 ⎡ ⎛ t 2 ⎞⎤ ⎢5 ⎜ − 1 ⎟ ⎥ ⎠⎦ ⎣ ⎝5

dt =

=

1 5

1 1 1 ⎛ 5⎞ dt = dt ⎜ ⎟∫ ∫ 5 5 ⎜⎝ 1 ⎟⎠ ⎡⎛ t ⎞2 ⎤ ⎡⎛ t ⎞ 2 ⎤ ⎢⎜ ⎢⎜ ⎟ − 1⎥ ⎟ − 1⎥ ⎣⎢⎝ 5 ⎠ ⎦⎥ ⎣⎢⎝ 5 ⎠ ⎦⎥

∫

1 5 ⎡⎛ t ⎞ 2 ⎤ ⎢⎜ ⎟ − 1⎥ ⎢⎣⎝ 5 ⎠ ⎥⎦

dt = cosh −1

t 5

+c


304

15.(b) Determine

3

3

∫ (16 − 2x ) dx 2

3

3

1

∫ (16 − 2x ) dx = ∫ 2 (8 − x ) dx = 2 ∫ ( 8) 2

2

2

⎛3⎞ 1 dx = ⎜ ⎟ ∫ ⎝2⎠ 8 − x2

( ) 8

8 2

− x2

dx =


3 2 8

tanh −1

x 8

+c

305

CHAPTER 34 PARTIAL DIFFERENTIATION EXERCISE 139 Page 345

2. Find

∂z ∂x and given z = x3 – 2xy + y2 ∂x ∂y

If z = x 3 − 2xy + y 2 then

∂z d 3 d d x ) − (2y) (x) + ( y 2 ) (1) = 3x 2 − 2y(1) + ( y 2 ) (0) = ( dx dx ∂x dx = 3x 2 − 2y ∂x d d d = (x 3 ) (1) − (2x) (y) + ( y 2 ) = (x 3 )(0) − 2x(1) + 2y dy dy dy ∂y

and

= −2x + 2y

5. Find

∂z ∂x y 1 and given z = x 3 y 2 − 2 + x y ∂y ∂x

If z = x 3 y 2 −

2y y 1 ∂z + = x 3 y 2 − yx −2 + y −1 then = ( y 2 )( 3x 2 ) − ( y ) ( −2x −3 ) + 0 = 3x 2 y 2 + 3 2 x ∂x x y

1 1 ∂x = ( x 3 ) ( 2y ) − 1( x −2 ) − y −2 = 2x 3 y − 2 − 2 x y ∂y

and

6. Find

∂z ∂x and given z = cos 3x sin 4y ∂x ∂y

If z = cos 3x sin 4y then

∂z d = ( sin 4y ) ( cos 3x ) = ( sin 4y )( −3sin 3x ) = −3sin 3x sin 4y dx ∂x

and

∂x d = ( cos 3x ) ( sin 4y ) = ( cos 3x )( 4 cos 4y ) = 4cos 3x cos 4y dy ∂y

8. The resonant frequency fr in a series electrical circuit is given by fr =

1 . Show that 2π LC

∂f r −1 = ∂L 4π CL3

1 1 − 12 − 12 fr = = L C 2π LC 2π

−1 ∂f r ⎛ 1 − 12 ⎞ ⎛ 1 − 32 ⎞ 1 1 = ⎜ C ⎟⎜ − L ⎟ = − =− and = 1 3 3 ∂L ⎝ 2π 4π CL3 4π C L ⎠⎝ 2 ⎠ 4πC 2 L2


306

⎛ nπ ⎞ ⎧ ⎛ nπb ⎞ ⎛ nπb ⎞ ⎫ 9. An equation resulting from plucking a string is: y = sin ⎜ ⎟ x ⎨ k cos ⎜ ⎟ t + c sin ⎜ ⎟t ⎬ ⎝ L⎠ ⎩ ⎝ L ⎠ ⎝ L ⎠ ⎭ ∂y ∂y Determine and ∂t ∂x

⎛ nπ ⎞ ⎧ ⎛ nπb ⎞ ⎛ nπb ⎞ ⎫ ⎛ nπ ⎞ ⎛ nπb ⎞ ⎛ nπ ⎞ ⎛ nπb ⎞ y = sin ⎜ ⎟ x ⎨ k cos ⎜ ⎟ t + c sin ⎜ ⎟ t ⎬ = k sin ⎜ ⎟ x cos ⎜ ⎟ t + c sin ⎜ ⎟ x sin ⎜ ⎟t ⎝ L⎠ ⎩ ⎝ L ⎠ ⎝ L ⎠ ⎭ ⎝ L⎠ ⎝ L ⎠ ⎝ L⎠ ⎝ L ⎠

Hence,

∂y ⎡ ⎛ nπ ⎞ ⎤ ⎡ ⎛ nπb ⎞ ⎤ ⎛ nπb ⎞ ⎡ ⎛ nπ ⎞ ⎤ ⎡ ⎛ nπb ⎞ ⎤ ⎛ nπb ⎞ = ⎢ k sin ⎜ ⎟ x ⎥ ⎢ − sin ⎜ ⎟ t⎥ ⎜ ⎟ + ⎢ c sin ⎜ ⎟ x ⎥ ⎢ cos ⎜ ⎟ t⎥ ⎜ ⎟ ∂t ⎣ ⎝ L ⎠ ⎦⎣ ⎝ L ⎠ ⎦⎝ L ⎠ ⎣ ⎝ L ⎠ ⎦ ⎣ ⎝ L ⎠ ⎦⎝ L ⎠ ⎛ n πb ⎞ ⎛ n π ⎞ ⎧ ⎛ nπb ⎞ ⎛ nπ b ⎞ ⎫ =⎜ sin ⎜ x ⎨ c cos ⎜ t − k sin ⎜ ⎟ ⎟ ⎟ ⎟t ⎬ ⎝ L ⎠ ⎝ L ⎠ ⎩ ⎝ L ⎠ ⎝ L ⎠ ⎭

and

∂y ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπb ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπb ⎞ = k ⎜ ⎟ cos ⎜ ⎟ x cos ⎜ ⎟ t + c ⎜ ⎟ cos ⎜ ⎟ x sin ⎜ ⎟t ∂x ⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠ ⎝ L⎠ ⎝ L⎠ ⎝ L ⎠ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎧ ⎛ nπb ⎞ ⎛ nπb ⎞ ⎫ =⎜ cos ⎜ x ⎨ k cos ⎜ t + c sin ⎜ ⎟ ⎟ ⎟ ⎟t ⎬ ⎝ L ⎠ ⎝ L ⎠ ⎩ ⎝ L ⎠ ⎝ L ⎠ ⎭

T ∆S−∆H RT

10. In a thermodynamic system, k = Ae where R, k and A are constants. Find ∂k ∂A ∂ (∆S) ∂ (∆H) (a) (b) (c) (d) ∂T ∂T ∂T ∂T T ∆S−∆H H ⎤ ⎡ ( RT )( ∆S) − ( T ∆S − ∆H )( R ) ⎤ ⎡ T ∆RS−∆ ⎤ ⎡ RT∆S − RT∆S + R∆H ⎤ ∂k ⎡ RT T Ae (a) = = ⎢ Ae ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ R 2 T2 R 2 T2 ∂T ⎣⎢ ⎦ ⎦ ⎢⎣ ⎦⎥ ⎣ ⎦⎥ ⎣ H H ⎡ T ∆RS−∆ ⎤ ⎡ R∆H ⎤ ⎡ T ∆RS−∆ ⎤ ⎡ ∆H ⎤ T T = ⎢ Ae ⎥ ⎢ 2 2 ⎥ = ⎢ Ae ⎥⎢ 2⎥ ⎢⎣ ⎥⎦ ⎣ R T ⎦ ⎢⎣ ⎥⎦ ⎣ R T ⎦

A ∆H e = R T2

(b) Since k = Ae

T ∆S−∆H RT

∂A ⎛ Thus, = ⎜ke ∂T ⎜⎝

then

∆S−T ∆S RT

⎛ = ⎜⎜ k e ⎝

∆S−T ∆S RT

A = ke

−

T ∆S−∆H RT

= ke

T ∆S −∆H RT

∆H − T ∆S RT

⎞ ⎛ ( RT )( −∆S) − ( ∆H − T ∆S)( R ) ⎞ ⎟⎟ ⎜ ⎟ R 2 T2 ⎠ ⎠⎝

⎞ ⎛ −R T ∆S − R ∆H + RT ∆S ⎞ ⎛ ⎟⎟ ⎜ ⎟ = ⎜⎜ k e 2 2 R T ⎝ ⎠ ⎝ ⎠

∆S−T ∆S RT

− T ∆S ⎞ ⎛ − R ∆H ⎞ k ∆H ∆HRT = e − ⎟⎟ ⎜ 2 2 ⎟ R T2 ⎠⎝ R T ⎠


307

(c) If k = Ae

T ∆S−∆H RT

then

k =e A

T ∆S−∆H RT

and

⎛k⎞ R T ln ⎜ ⎟ = T ∆S − ∆H ⎝A⎠

i.e.

(1)

⎛k⎞ T ∆S = ∆H + R T ln ⎜ ⎟ ⎝A⎠

i.e.

∆S =

and

∆H ⎛k⎞ ⎛k⎞ + R ln ⎜ ⎟ = ∆H T −1 + R ln ⎜ ⎟ T ⎝A⎠ ⎝A⎠

∆H ∂ (∆S) = ∆H ( −T −2 ) + 0 = − 2 T ∂T

Hence, (d) From equation (1),

Hence,

⎛ k ⎞ T ∆S − ∆H ln ⎜ ⎟ = RT ⎝A⎠

⎧ ⎛k⎞ ⎛ k ⎞⎫ ∆H = T ∆S − R T ln ⎜ ⎟ = T ⎨∆S − R ln ⎜ ⎟ ⎬ ⎝A⎠ ⎝ A ⎠⎭ ⎩

∂ (∆H) ⎛k⎞ = ∆S − R ln ⎜ ⎟ ∂T ⎝A⎠


308

EXERCISE 140 Page 347 2. If z = 2 ln xy find (a)

∂2z ∂x 2

(b)

∂ 2z ∂y 2

(c)

∂2z ∂2z (d) ∂x∂y ∂y∂x

(a)

2 ∂z ⎛ 2 ⎞ = ⎜ ⎟ (y) = = 2x −1 x ∂x ⎝ xy ⎠

and

2 ∂2z ∂ = 2x −1 ) = −2x −2 = − 2 ( 2 x ∂x ∂x

(b)

2 ∂z ⎛ 2 ⎞ = ⎜ ⎟ (x) = = 2y −1 y ∂y ⎝ xy ⎠

and

2 ∂2z ∂ = 2y −1 ) = −2y −2 = − 2 ( 2 y ∂y ∂y

(c)

∂2z ∂ = ( 2y −1 ) = 0 ∂x∂y ∂x

∂2z ∂ (d) = ( 2x −1 ) = 0 ∂y∂x ∂y

3. If z =

(a)

( x − y) ( x + y)

find (a)

∂2z ∂x 2

(b)

∂ 2z ∂y 2

(c)

∂2z ∂ 2z (d) ∂x∂y ∂y∂x

∂z (x + y)(1) − (x − y)(1) x + y − x + y 2y = = = = 2y(x + y) −2 2 2 2 ∂x (x + y) (x + y) (x + y) 4y ∂2z = −4y(x + y) −3 (1) = − 2 (x + 3)3 ∂x

(b)

∂z (x + y)(−1) − (x − y)(1) − x − y − x + y −2x = = = = −2x(x + y) −2 2 2 2 ∂y (x + y) (x + y) (x + y) 4x ∂2z = (−2x) ⎡⎣ −2(x + y) −3 ⎤⎦ (1) = 2 (x + 3)3 ∂y

(c)

(d)

∂2z ∂ 4x 2 ⎡⎣ −2x(x + y) −2 ⎤⎦ = (−2x) ⎡⎣ −2(x + y) −3 ⎤⎦ + (x + y) −2 (−2) = = − 3 ∂x∂y ∂x (x + y) (x + y) 2

=

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

=

2(x − y) (x + y)3

∂2z ∂ −4y 2 ⎡⎣ 2y(x + y) −2 ⎤⎦ = (2y) ⎡⎣ −2(x + y) −3 ⎤⎦ + (x + y) −2 (2) = = + 3 ∂y∂x ∂y (x + y) (x + y) 2

=

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

=

2(x − y) 2x − 2y = 3 (x + y)3 (x + y)


309

5. If z = x 2 sin(x − 2y) find (a)

Show also that

(a)

∂2z ∂x 2

(b)

∂2z ∂y 2

∂ 2z ∂ 2z = = 2x 2 sin(x − 2y) − 4x cos(x − 2y) ∂x∂y ∂y∂x

∂z = ( x 2 ) cos(x − 2y) + [sin(x − 2y) ] (2x) = x 2 cos(x − 2y) + 2x sin(x − 2y) ∂x ∂2z = ( x 2 ) [ − sin(x − 2y) ] + [ cos(x − 2y)] (2x) + (2x) cos(x − 2y) + [sin(x − 2y) ] (2) 2 ∂x = − x 2 sin(x − 2y) + 2x cos(x − 2y) + 2x cos(x − 2y) + 2sin(x − 2y) = ( 2 − x 2 ) sin(x − 2y) + 4x cos(x − 2y)

(b)

∂z = x 2 [ cos(x − 2y) ] (−2) = −2x 2 cos(x − 2y) ∂y ∂ 2z = ( −2x 2 ) [ − sin(x − 2y) ] (−2) ∂y 2

= −4x 2 sin(x − 2y) ∂2z ∂ ⎡⎣ −2x 2 cos(x − 2y) ⎤⎦ = ( −2x 2 ) [ − sin(x − 2y) ] + [ cos(x − 2y) ] (−4x) = ∂x∂y ∂x

= 2x 2 sin(x − 2y) − 4x cos(x − 2y) ∂2z ∂ ⎡⎣ x 2 cos(x − 2y) + 2x sin(x − 2y) ⎤⎦ = ( x 2 ) [ − sin(x − 2y) ] (−2) + (2x) [ cos(x − 2y) ] (−2) = ∂y∂x ∂y

= 2x 2 sin(x − 2y) − 4x cos(x − 2y) ⎛ 3x ⎞ ∂ 2z ∂2z ∂2z 1 7. Given z = ⎜ ⎟ show that = and evaluate when x = and y = 3 2 2 ∂y∂x ∂x ∂x∂y ⎝ y ⎠ 1 1 − ⎛ 3x ⎞ 2 z = ⎜ ⎟ = 3x y 2 ⎝ y ⎠ 1 1 − ⎞1 − 3 − 12 − 12 ∂z ⎛ y x =⎜ 3y 2⎟ x 2 = 2 ∂x ⎝ ⎠2 1 ⎞⎛ 1 − 3 ⎞ 3 12 − 32 ∂z ⎛ x y = ⎜ 3 x 2 ⎟⎜ − y 2 ⎟ = − 2 ∂y ⎝ ⎠⎝ 2 ⎠

− 3 ∂2z ∂ ⎛ 3 12 − 32 ⎞ ⎛ 3 − 32 ⎞ ⎛ 1 − 12 ⎞ − 3 = ⎜⎜ − x y ⎟⎟ = ⎜⎜ − y ⎟⎟ ⎜ x ⎟ = 3 1 = ∂x∂y ∂x ⎝ 2 ⎠ 4y 2 x 2 4 ( xy 3 ) ⎠ ⎝ 2 ⎠⎝ 2


310

− 3 ∂2z ∂ ⎛ 3 − 12 − 12 ⎞ ⎛ 3 − 12 ⎞ ⎛ 1 − 32 ⎞ − 3 = ⎜⎜ y x ⎟⎟ = ⎜⎜ x ⎟⎟ ⎜ − y ⎟ = 1 3 = ∂y∂x ∂y ⎝ 2 ⎠ 4x 2 y 2 4 ( xy 3 ) ⎠ ⎝ 2 ⎠⎝ 2 ∂2z ∂2z = ∂x∂y ∂y∂x

Thus,

∂ 2 z ∂ ⎛ 3 − 12 − 12 ⎞ ⎛ 3 − 12 ⎞ ⎛ 1 − 32 ⎞ − 3 = ⎜⎜ y x ⎟⎟ = ⎜⎜ y ⎟⎟ ⎜ − x ⎟ = 2 ∂x ∂x ⎝ 2 ⎠ 4 ( yx 3 ) ⎠ ⎝ 2 ⎠⎝ 2 When x =

1 and y = 3, 2

∂2z = ∂x 2

− 3 ⎛ ⎛ 1 ⎞3 ⎞ 4 ⎜ (3) ⎜ ⎟ ⎟ ⎜ ⎝2⎠ ⎟ ⎝ ⎠

=

− 3 ⎛1⎞ 4 3 ⎜ 3⎟ ⎝2 ⎠

=

−1 ⎛1⎞ 4 ⎜ ⎟ ⎝8⎠

=

− 8 − = 4

( 2 )( 4 ) 4

=

− 2 4 − 2(2) 2 = =− 4 4 2

=−

1 or − 0.7071 2

8. An equation used in thermodynamics is the Benedict-Webb-Rubine equation of state for the expansion of a gas. The equation is: γ ⎞ ⎛ C ⎜1 + 2 ⎟ γ C0 ⎞ 1 RT ⎛ 1 Aα V ⎠ ⎛ 1 ⎞ − V2 ⎝ + ⎜ B0 RT − A 0 − 2 ⎟ 2 + (bRT − a) 3 + 6 + p= ⎜ 3 ⎟e V ⎝ T ⎠V V V T2 ⎝V ⎠

∂ 2p 6 = 2 4 Show that 2 VT ∂T

γ γ ⎞ − V2 ⎪⎧ C ⎛ ⎪⎫ − C0 ⎬ ⎨ ⎜1 + 2 ⎟ e ⎩⎪ V ⎝ V ⎠ ⎭⎪

γ

p=

RT 1 1 Aα γ ⎞⎛ 1 ⎞ − 2 ⎛ + ( B0 RT − A 0 − C0 T −2 ) 2 + (bRT − a) 3 + 6 + T −2 C ⎜ 1 + 2 ⎟ ⎜ 3 ⎟ e V V V V V ⎝ V ⎠⎝ V ⎠ γ

∂p R B0 R 2C0 T −3 bR γ ⎞⎛ 1 ⎞ − V2 −3 ⎛ 2T C 1 e = + 2 + − − + ⎜ 2 ⎟⎜ 3 ⎟ V2 V3 ∂T V V ⎝ V ⎠⎝ V ⎠ γ

6 ∂ 2 p −6C0 T −4 γ ⎞⎛ 1 ⎞ − 2 ⎛ = + 6T −4 C ⎜1 + 2 ⎟⎜ 3 ⎟ e V = 2 4 2 2 VT V ∂T ⎝ V ⎠⎝ V ⎠

γ ⎧⎪ C ⎛ ⎫⎪ γ ⎞ − V2 + − C0 ⎬ 1 e ⎨ ⎜ 2 ⎟ V ⎠ ⎩⎪ V ⎝ ⎭⎪


311

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 ∂z dx + dy = (2y + sin x)dx + 2x dy ∂x ∂y

3. Find the total differential, dz given z =

Since z =

x−y x+y

∂z (x + y)(1) − (x − y)(1) 2y = = 2 ∂x (x + y) (x + y) 2 ∂z (x + y)(−1) − (x − y)(1) −2x = = 2 ∂y (x + y) (x + y) 2

and

Thus,

then

x−y x+y

dz =

⎛ 2y ⎞ ⎛ 2x ⎞ ∂z ∂z dx + dy = ⎜ dx − ⎜ dy 2 ⎟ 2 ⎟ ∂x ∂y ⎝ (x + y) ⎠ ⎝ (x + y) ⎠

5. Find the total differential, dz, given z = xy +

x -4 y

1

Since z = xy + and Thus,

x - 4 then y

1 −2 x ∂z 1 2 = y+ = y+ ∂x y 2y x

∂z x = x − x ( y −2 ) = x − 2 ∂y y ⎛ ⎛ 1 ⎞ x⎞ ∂z ∂z dx + dy = ⎜ y + dz = ⎟ dx + ⎜⎜ x − 2 ⎟⎟ dy y ⎠ ∂x ∂y 2y x ⎠ ⎝ ⎝

7. Given u = ln sin(xy) show that du = cot(xy)(y dx + x dy)

Since u = ln sin(xy) then


312

du =

⎡⎛ 1 ⎞ ⎤ ⎡⎛ 1 ⎞ ⎤ ∂u ∂u dx + dy = ⎢⎜ ⎟ y cos xy ⎥ dx + ⎢⎜ ⎟ x cos xy ⎥ dy ∂x ∂y ⎣⎝ sin(xy) ⎠ ⎦ ⎣⎝ sin(xy) ⎠ ⎦

= y cot(xy) dx + x cot(xy) dy i.e.

du = cot(xy)[ y dx + x dy ]


313


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 = πr 2 h Rate at which volume is changing,

dV ∂V dr ∂V dh = + dt ∂r dt ∂h dt = ( 2πrh )

dr dh + ( πr 2 ) dt dt

= [ 2π(40)(150) ] (8) + ⎡⎣ π(40) 2 ⎤⎦ (−15) = 96 000π - 24 000π = 72 000π mm3 / s = 72π cm3 / s = 226.2 cm 3 / 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 dc = + + dt ∂a dt ∂b dt ∂c dt db dc ⎛ 2b ⎞ da = ⎜ + c 2 ea ⎟ + ( 2 ln a ) + ( 2cea ) dt dt ⎝ a ⎠ dt ⎛ (2)(6) 2 1.5 ⎞ + 8 e ⎟ (2) + ( 2 ln1.5 ) (−3) + ( 2(8)e1.5 ) (−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. © 2006 John Bird. All rights reserved. Published by Elsevier.

314

Total surface area of a cone, A = πrl + πr 2 = πr

(r

2

+ h 2 ) + πr 2 (see diagram of cone below)

Rate of change of surface area, dA ∂A dr ∂A dh = + dt ∂r dt ∂h dt 1 ⎧⎪ ⎡1 2 2 −2 = ⎨( πr ) ⎢ ( r + h ) (2r) + ⎪⎩ ⎣2

⎧ ⎪ =⎨ ⎪⎩ ⎧ ⎪ =⎨ ⎪⎩

πr 2

(r

2

+ h2 )

2

2

⎫ ⎧ ⎪ dr ⎪ + π ( r + h ) + 2πr ⎬ + ⎨ ⎪⎭ dt ⎪⎩

π(5) 2

(5

1 ⎫⎪ dr ⎧⎪ ⎤ ⎛1 2 ⎞⎫ 2 −2 ( r + h ) (π)⎥ + 2πr ⎬ dt + ⎨( πr ) ⎜ 2 ( r + h ) (2h) ⎟⎪⎬ dh ⎪⎭ ⎪⎩ ⎦ ⎝ ⎠ ⎪⎭ dt 2

+ 122 )

2

+π

2

⎫ ⎪ dh ⎬ ( r 2 + h 2 ) ⎪⎭ dt πrh

⎫ ⎧ ⎫ ⎪ ⎪ π(5)(12) ⎪ 2 2 + + π + 5 12 2 (5) (0.5) ( ) ⎬ ⎨ ⎬ (−1.5) in centimetre units 2 2 ⎪⎭ ⎪⎩ ( 5 + 12 ) ⎪⎭

⎛ 25π ⎞ ⎛ 60π ⎞ + 13π + 10π ⎟ (0.5) + ⎜ =⎜ ⎟ (−1.5) ⎝ 13 ⎠ ⎝ 13 ⎠ = (78.298)(0.5) + (14.50)(-1.5) = 39.149 – 21.75 = 17.4 cm 2 / s


315

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 δH ≈

∂H ∂H ∂H δi + δR + δt ∂i ∂R ∂t

≈ ( 2iRt ) (0.02i) + ( i 2 t ) (−0.03R) + ( i 2 R ) (0.01t) ≈ 0.04i 2 Rt − 0.03i 2 Rt + 0.01i 2 Rt ≈ (0.04 – 0.03 + 0.01) H ≈ 0.02 H i.e. the error in H is + 2% 3. fr =

1 represents the resonant frequency of a series connected circuit containing 2π LC

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 1 − 12 − 12 = L C 2π LC 2π

then δf r ≈

∂f r ∂f δL + r δC ∂L ∂C

⎛ 1 − 32 − 12 ⎜− L C ≈⎜ 2 2π ⎜ ⎜ ⎝ ⎛ − 32 − 12 −L C ≈⎜ ⎜ 4π ⎜ ⎝ −

1

0.03L 2 C ≈ 4π

⎞ ⎛ 1 − 23 − 12 ⎟ ⎜− C L ⎟ δL + ⎜ 2 2π ⎟ ⎜ ⎟ ⎜ ⎠ ⎝

⎞ ⎟ ⎟ δC ⎟ ⎟ ⎠

⎞ ⎛ − 32 − 12 ⎟ (−0.03L) − ⎜ C L ⎟ ⎜ 4π ⎟ ⎜ ⎠ ⎝ −

1 2

−

1

−

0.05C 2 L − 4π

⎞ ⎟ (0.05C) ⎟ ⎟ ⎠

1 2

⎛ 1 ⎞ ⎛ 1 ⎞ ≈ 0.015 ⎜ ⎟ − 0.025 ⎜ ⎟ ⎝ 2π LC ⎠ ⎝ 2π LC ⎠ i.e.

δf r ≈ (0.015 − 0.025)f r ≈ −0.01f r

i.e. the approximate percentage change in fr is – 1% © 2006 John Bird. All rights reserved. Published by Elsevier.

316

4. The second moment of area of a rectangle about its centroid parallel to side b is given by

I=

b d3 . If b and d are measured as 15 cm and 6 cm respectively, and the measurement errors 12

are +12 mm in b and –1.5 mm in d, find the error in the calculated value of I.

Since I =

b d3 12

then

δI ≈

∂I ∂I δb + δd ∂b ∂d

⎛ d3 ⎞ ⎛ 3bd 2 ⎞ ⎛ 63 ⎞ ⎛ 3(15)(6) 2 ⎞ ≈ ⎜ ⎟ ∂b + ⎜ ∂ d ≈ (1.2) + ⎟ ⎜ ⎟ ⎜ ⎟ (−0.15) ⎝ 12 ⎠ ⎝ 12 ⎠ ⎝ 12 ⎠ ⎝ 12 ⎠

≈ 21.6 – 20.25 = 1.35 i.e. error in I = + 1.35 cm 4 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. 1

b2 = a2 + c2 – 2ac cos B from which, b = ( a 2 + c 2 − 2ac cos B ) 2 Approximate error in b, δb ≈

∂b ∂b ∂b δa + δc + δB ∂a ∂c ∂B

1 − ⎡1 2 2 ⎤ ≈ ⎢ ( a + c − 2ac cos B ) 2 (2a − 2c cos B⎥ ∂a ⎣2 ⎦ 1 − ⎡1 ⎤ + ⎢ ( a 2 + c 2 − 2ac cos B ) 2 (2c − 2a cos B⎥ ∂c ⎣2 ⎦ 1 − ⎡1 ⎤ + ⎢ ( a 2 + c 2 − 2ac cos B ) 2 (2ac sin B⎥ ∂B ⎣2 ⎦ 1 − ⎡ ⎤ 2 1 π π⎥ ⎛ ⎞ 2 2 ⎢ ≈ ⎜ 3 + 4 − 2(3)(4) cos ⎟ (6 − 8cos (0.8) ⎢2 ⎝ 4⎠ 4⎥ ⎣ ⎦ 1 − ⎡ ⎤ 2 1 π π⎥ ⎛ ⎞ 2 2 ⎢ 3 + 4 − 2(3)(4) cos ⎟ (8 − 6 cos (−0.5) + ⎢ 2 ⎜⎝ 4⎠ 4⎥ ⎣ ⎦ 1 − ⎡ ⎤ 1 π π ⎛ π⎞ ⎛ ⎞ 2 + ⎢ ⎜ 32 + 42 − 2(3)(4) cos ⎟ (24sin ⎥ ⎜ ⎟ ⎢2 ⎝ 4⎠ 4 ⎥ ⎝ 90 ⎠ ⎣ ⎦


317

≈ (0.17645)(0.343146)(0.8) + (0.17645)(3.75736)(-0.5) ⎛ π⎞ + (0.17645(16.97056) ⎜ ⎟ ⎝ 90 ⎠ 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 =

C d 6

T5

, 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 =

C d 6

T5

1 2

= Cd T

−

5 6

Approximate error in flow rate, δv ≈

∂v ∂v δd + δT ∂d ∂T

⎛ 1 − 12 (C) d ⎜ 2 ≈⎜ ⎜ 6 T5 ⎜ ⎝

⎞ ⎟ ⎛ ⎛ 5 − 116 ⎞ ⎞ ⎟ (0.014d) + ⎜⎜ C d ⎜ − T ⎟ ⎟⎟ (−0.018T) ⎟ ⎝ 6 ⎠⎠ ⎝ ⎟ ⎠

⎛C d ⎞ ⎛ C d ⎞⎛ 5 ⎞ ≈ ⎜⎜ (0.007) + ⎟ ⎜⎜ 6 5 ⎟⎟ ⎜ (0.018) ⎟ 6 5 ⎟ ⎠ ⎝ T ⎠ ⎝ T ⎠⎝ 6 ≈ (0.007 + 0.015) v ≈ 0.022 v i.e. the percentage error in the rate of flow is + 2.2%


318

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) = x 2 + y 2 − 2x + 4y + 8 (b) f(x, y) = x 2 − y 2 − 2x + 4y + 8 (c) f(x, y) = 2x + 2y − 2xy − 2x 2 − y 2 + 4 (a) Let f(x, y) = z = x 2 + y 2 − 2x + 4y + 8 (i)

∂z = 2x − 2 ∂x

∂z = 2y + 4 ∂y

and

(ii) For stationary points, and

2x - 2 = 0 from which, x = 1 2y + 4 = 0 from which, y = -2

(iii) The co-ordinates of the stationary point is (1, -2) (iv)

∂2z =2 ∂x 2

∂2z =2 ∂y 2

(v) When x = 1, y = -2,

∂2z ∂ = ( 2y + 4 ) = 0 ∂x∂y ∂x

∂2z =2 ∂x 2

∂2z = 2 and ∂y 2

∂ 2z =0 ∂x∂y

2

⎛ ∂2z ⎞ (vi) ⎜ ⎟ =0 ⎝ ∂x∂y ⎠ 2

(vii) ∆ (1,−2)

⎛ ∂ 2z ⎞ ⎛ ∂ 2z ⎞ ⎛ ∂2z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = ( 0 ) − ( 2 )( 2 ) = −4 which is negative ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠

(viii) Since ∆ < 0 and

∂2z > 0 then (1, -2) is a minimum point. ∂x 2

(b) Let f(x, y) = z = x 2 − y 2 − 2x + 4y + 8 (i)

∂z = 2x − 2 ∂x

and


∂z = −2y + 4 ∂y 2x - 2 = 0 from which, x = 1 - 2y + 4 = 0 from which, y = 2

(iii) The co-ordinates of the stationary point is (1, 2)


319

(iv)

∂ 2z =2 ∂x 2

∂2z = −2 ∂y 2

(v) When x = 1, y = 2,

∂2z ∂ = ( −2y + 4 ) = 0 ∂x∂y ∂x

∂2z =2 ∂x 2

∂2z = −2 and ∂y 2

∂2z =0 ∂x∂y

2

⎛ ∂2z ⎞ (vi) ⎜ ⎟ =0 ⎝ ∂x∂y ⎠ 2

(vii) ∆ (1,2)

⎛ ∂ 2z ⎞ ⎛ ∂ 2z ⎞ ⎛ ∂ 2z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = ( 0 ) − ( 2 )( −2 ) = 4 which is positive ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠

(viii) Since ∆ > 0 then (1, 2) is a saddle point. (c) Let f(x, y) = z = 2x + 2y − 2xy − 2x 2 − y 2 + 4 (i)

∂z = 2 − 2y − 4x ∂x

∂z = 2 − 2x − 2y ∂y

and

(ii) For stationary points, 2 – 2y – 4x = 0 i.e.

1 – y – 2x = 0

and

2 – 2x – 2y = 0

i.e.

1–x–y=0

(1)

(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) ∂ 2z (iv) = −4 ∂x 2

∂2z = −2 ∂y 2

(v) When x = 0, y = 1,

∂2z ∂ = ( 2 − 2x − 2y ) = −2 ∂x∂y ∂x

∂2z = −4 ∂x 2

∂2z = −2 and ∂y 2

∂2z = −2 ∂x∂y

2

⎛ ∂2z ⎞ 2 (vi) ⎜ ⎟ = (−2) = 4 ⎝ ∂x∂y ⎠ 2

(vii) ∆ (0,1)

⎛ ∂ 2 z ⎞ ⎛ ∂ 2 z ⎞⎛ ∂ 2 z ⎞ =⎜ ⎟ − ⎜ 2 ⎟⎜ 2 ⎟ = 4 − ( −4 )( −2 ) = −4 which is negative ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠⎝ ∂y ⎠


∂2z < 0 then (0, 1) is a maximum point. ∂x 2

4. Locate the stationary points of the function z = 12x 2 + 6xy + 15y 2


320

(i)

∂z = 24x + 6y ∂x

∂z = 6x + 30y ∂y

and

(ii) For stationary points, and (iii) From (1),

6y = -24x

Substituting in (2) gives: i.e.

24x + 6y = 0

(1)

6x + 30y = 0

(2)

i.e. y = -4x

6x + 30(-4x) = 0 6x = 120x

i.e. x = 0

When x = 0, y = 0, hence the co-ordinates of the stationary point is (0, 0) (iv)

∂ 2z = 24 ∂x 2

∂2z = 30 ∂y 2

(v) When x = 0, y = 0,

∂2z ∂ = ( 6x + 30y ) = 6 ∂x∂y ∂x

∂2z = 24 ∂x 2

∂2z = 30 and ∂y 2

∂2z =6 ∂x∂y

2

⎛ ∂2z ⎞ (vi) ⎜ ⎟ = 36 ⎝ ∂x∂y ⎠ 2

(vii) ∆ (0,0)

⎛ ∂ 2 z ⎞ ⎛ ∂ 2 z ⎞⎛ ∂ 2 z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟⎜ 2 ⎟ = ( 6 ) − ( 24 )( 30 ) which is negative ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠⎝ ∂y ⎠


∂2z > 0 then (0, 0) is a minimum point ∂x 2

5. Find the stationary points of the surface z = x 3 − xy + y3 and distinguish between them.

(i)

∂z = 3x 2 − y ∂x

∂z = − x + 3y 2 ∂y

and


3x 2 - y = 0

(1)

− x + 3y 2 = 0

(2)

y = 3x 2

(iii) From (1),

Substituting in (2) gives: - x + 3 ( 3x 2 ) = 0 2

- x + 27x 4 = 0 x ( 27x 3 − 1) = 0

and i.e.

x=0

27x 3 − 1 = 0

or

Hence, x = 0 or x =

i.e.

x3 =

1 27

and

⎛ 1 ⎞ 1 x=3⎜ ⎟= ⎝ 27 ⎠ 3

1 3


321

From (1), when x = 0, y = 0 2

1 ⎛1⎞ 1 and when x = , y = 3x 2 = 3 ⎜ ⎟ = 3 ⎝3⎠ 3 ⎛1 1⎞ Hence, the stationary points occur at (0, 0) and ⎜ , ⎟ ⎝ 3 3⎠

∂2z = 6x (iv) ∂x 2

∂ 2z = 6y ∂y 2

∂2z ∂ = − x + 3y 2 ) = −1 ( ∂x∂y ∂x

(v) For (0, 0),

∂2z =0 ∂x 2

∂2z = 0 and ∂y 2

∂ 2z = −1 ∂x∂y

∂2z =2 ∂x 2

∂ 2z = 2 and ∂y 2

∂2z = −1 ∂x∂y

⎛1 1⎞ For ⎜ , ⎟ , ⎝3 3⎠

2

⎛ ∂2z ⎞ (vi) For (0, 0), ⎜ ⎟ =1 ⎝ ∂x∂y ⎠ 2

2 ⎛1 1⎞ ⎛ ∂ z ⎞ For ⎜ , ⎟ , ⎜ ⎟ =1 ⎝ 3 3 ⎠ ⎝ ∂x∂y ⎠ 2

(vii) ∆ (0,0)

⎛ ∂2z ⎞ ⎛ ∂2z ⎞ ⎛ ∂2z ⎞ =⎜ ⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = 1 − ( 0 )( 0 ) = 1 which is positive ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠

∆ ⎛ 1 1 ⎞ = 1 − ( 2 )( 2 ) = −3 which is negative ⎜ , ⎟ ⎝ 3 3⎠

(viii) Since ∆ ( 0,0) > 0 then (0, 0) is a saddle point ∆ ⎛ 1 1 ⎞ < 0 and ⎜ , ⎟ ⎝ 3 3⎠

∂2z ⎛1 1⎞ > 0 then ⎜ , ⎟ is a minimum point 2 ∂x ⎝ 3 3⎠


322


1. The function z = x 2 + y 2 + xy + 4x − 4y + 3 has one stationary value. Determine its co-ordinates and its nature.

(i)

∂z = 2x + y + 4 ∂x

∂z = 2y + x − 4 ∂y

and

(ii) For stationary points, and (iii) (1) + (2) gives:

2x + y + 4 = 0

(1)

2y + x – 4 = 0

(2)

3x + 3y = 0

Substituting in (1),

2x – x + 4 = 0

from which, y = -x i.e. x = -4, thus y = +4

Hence, the stationary point occurs at (-4, 4) ∂2z (iv) =2 ∂x 2

∂2z =2 ∂y 2

(v) When x = -4, y = 4,

∂2z ∂ = ( 2y + x − 4 ) = 1 ∂x∂y ∂x

∂2z =2 ∂x 2

∂2z = 2 and ∂y 2

∂2z =1 ∂x∂y

2

⎛ ∂2z ⎞ (vi) ⎜ ⎟ =1 ⎝ ∂x∂y ⎠ 2

(vii) ∆ ( −4,4)

⎛ ∂2z ⎞ ⎛ ∂2z ⎞ ⎛ ∂ 2z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = (1) − ( 2 )( 2 ) = −3 which is negative ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠


∂2z > 0 then (-4, 4) is a minimum point ∂x 2

3. Determine the stationary values of the function f(x, y) = x 4 + 4x 2 y 2 − 2x 2 + 2y 2 − 1 and

distinguish between them. Let f(x, y) = z = x 4 + 4x 2 y 2 − 2x 2 + 2y 2 − 1 (i)

∂z = 4x 3 + 8xy 2 − 4x ∂x

(ii) For stationary points,

and

∂z = 8x 2 y + 4y ∂y

4x 3 + 8xy 2 − 4x = 0

(1)

8x 2 y + 4y = 0

and (iii) From (2), From (1), if y = 0, from which,

4y ( 2x 2 − 1) = 0

(2) from which, y = 0

4x 3 − 4x = 0

i.e. 4x ( x 2 − 1) = 0

x = 0 or x = ± 1


323

Hence, the stationary points occur at (0, 0) and (1, 0) and (-1, 0) (iv)

∂2z = 12x 2 + 8y 2 − 4 ∂x 2

∂ 2z = 8x 2 + 4 ∂y 2

∂2z (v) For (0, 0), = −4 ∂x 2 For (1, 0),

∂2z ∂ = (8x 2 y + 4y ) = 16xy ∂x∂y ∂x

∂ 2z = 4 and ∂y 2

∂2z =8 ∂x 2

∂2z = 12 ∂y 2

∂2z For (-1, 0), =8 ∂x 2

∂2z =0 ∂x∂y

and

∂2z = 12 ∂y 2

and

∂2z =0 ∂x∂y ∂ 2z =0 ∂x∂y

2

⎛ ∂2z ⎞ (vi) For all three stationary points, ⎜ ⎟ =0 ⎝ ∂x∂y ⎠ 2

⎛ ∂2z ⎞ ⎛ ∂2z ⎞ ⎛ ∂ 2z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = ( 0 ) − ( −4 )( 4 ) = 32 which is positive ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠

(vii) ∆ (0,0)

2

∆ (1,0)

⎛ ∂2z ⎞ ⎛ ∂2z ⎞ ⎛ ∂ 2z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = ( 0 ) − ( 8 )(12 ) = −96 which is negative ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠ 2

∆ ( −1,0)

⎛ ∂2z ⎞ ⎛ ∂2z ⎞ ⎛ ∂2z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = ( 0 ) − ( 8 )(12 ) = −96 which is negative ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠

(viii) Since ∆ (0,0) > 0, then (0, 0) is a saddle point. Since ∆ (1,0) < 0 and Since ∆ ( −1,0)

∂2z > 0 then (1, 0) is a minimum point. ∂x 2

∂2z < 0 and > 0 then (-1, 0) is a minimum point. ∂x 2

4. Determine the stationary points of the surface f(x, y) = x 3 − 6x 2 − y 2

Let f(x, y) = z = x 3 − 6x 2 − y 2 (i)

∂z = 3x 2 − 12x ∂x

and


∂z = −2y ∂y 3x 2 − 12x = 0

(1)

-2y = 0

(2)

(iii) From (1)

3x(x − 4) = 0

From (2),

y=0

from which, x = 0 or x = 4

Hence, the stationary points occurs at (0, 0) and (4, 0) © 2006 John Bird. All rights reserved. Published by Elsevier.

324

(iv)

∂2z = 6x − 12 ∂x 2

(v) At (0, 0),

At (4, 0),

∂ 2z = −2 ∂y 2

∂2z ∂ = ( −2y ) = 0 ∂x∂y ∂x

∂2z = −12 ∂x 2

∂2z ∂2z 2 and =0 = − ∂x∂y ∂y 2

∂2z = 12 ∂x 2

∂2z = −2 and ∂y 2

∂ 2z =0 ∂x∂y

2

⎛ ∂2z ⎞ (vi) For both points, ⎜ ⎟ =0 ⎝ ∂x∂y ⎠ 2

(vii) ∆ (0,0)

⎛ ∂ 2 z ⎞ ⎛ ∂ 2 z ⎞⎛ ∂ 2 z ⎞ 2 =⎜ ⎟ − ⎜ 2 ⎟⎜ 2 ⎟ = ( 0 ) − ( −12 )( −2 ) = −24 which is negative ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠⎝ ∂y ⎠

∆ (4,0) = ( 0 ) − (12 )( −2 ) = 24 which is positive 2

(viii) Since ∆ (0,0) < 0 and

∂2z < 0 then (0, 0) is a maximum point. ∂x 2

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=

Substituting in (2) gives:

⎛ 250 ⎞ ⎛ 250 ⎞ 250 500 S = xy + y ⎜ + = xy + 250x −1 + 500y −1 ⎟ + 2x ⎜ ⎟ = xy + xy xy x y ⎝ ⎠ ⎝ ⎠

250 xy


325

∂S 250 = y− 2 x ∂x

and

∂S 250 = y− 2 =0 ∂x x

For a stationary point, from which,

y=

∂S 500 =x− 2 y ∂y

250 x2

yx 2 = 250

or

(3)

∂S 500 =x− 2 =0 ∂y y

and from which,

x=

500 y2

xy 2 = 500

or

(4)

Dividing equation (3) by equation (40 gives: yx 2 250 = xy 2 500

i.e.

x 1 = y 2

and

y = 2x

2x 3 = 250

Substituting y = 2x in equation (3) gives:

and

x = 3 125 = 5 m

y = 2x = 10 m

and From equation (1),

xyz = 250

∂ 2S 750 = 3 x ∂x 2 When x = 5 and y = 10,

∂ 2S =6 ∂x 2

i.e.

(5)(10)z = 250 from which, z = 5 m

∂ 2S 1000 = 3 ∂y 2 y ∂ 2S =1 ∂y 2

and

and

∂ 2S =1 ∂x∂y

∂ 2S =1 ∂x∂y

2

⎛ ∂ 2S ⎞ ⎛ ∂ 2S ⎞⎛ ∂ 2S ⎞ 2 ∆ =⎜ ⎟ − ⎜ 2 ⎟⎜ 2 ⎟ = (1) − ( 6 )(1) = −6 ⎝ ∂x∂y ⎠ ⎝ ∂x ⎠⎝ ∂y ⎠ Since ∆ < 0 and

which is negative

∂2z > 0 then the surface area is a minimum. ∂x 2

Minimum surface area, S = xy + yz + 2xz

= (5)(10) + (10)(5) + (2)(5)(5) = 50 + 50 + 50 = 150 m 2


326

CHAPTER 37 STANDARD INTEGRATION EXERCISE 146 Page 370 3. Determine (a)

(a)

(b)

⎛ 3x 2 − 5x ⎞ ⎜ ⎟ dx = ∫ x ⎝ ⎠

∫

∫

2

⎛ 3x 2 5x ⎞ 3x 2 = − 5x + c − = − dx 3x 5 dx ) ⎜ ⎟ ∫( 2 x ⎠ ⎝ x

3

∫ 4x

θ3 +c 3

dx

4

1

∫4

4

x 5 dx

⎛ 5 +1 ⎞ ⎛ 9 4 ⎜ ⎟ 14 5 1 1 ⎜ x4 ⎛1⎞ x x dx = ∫ x dx = ⎜ ⎟ ⎜ +c = ⎜ ⎟ 4 4 4⎜ 9 ⎝ 4 ⎠ ⎜ 5 +1⎟ ⎝4 ⎠ ⎝ 4

⎞ 9 ⎟ 14 9 ⎛ 1 ⎞⎛ 4 4 ⎞ x +c c x + = ⎟ ⎟+c = ⎜ ⎟⎜ 9 4 9 ⎝ ⎠ ⎝ ⎠ ⎟ ⎠

5 4

6.(b) Determine

∫

∫ (2 + θ) dθ

−4 +1 −3 ⎞ 1 3 3 1 −3 ⎛ 3 ⎞⎛ x ⎛ 3 ⎞⎛ x ⎞ −4 dx = ∫ x dx = ⎜ ⎟ ⎜ ⎟ + c = ⎜ ⎟⎜ ⎟+c = − x +c = − 3 +c 4 4x 4x 4 4 ⎝ 4 ⎠ ⎝ −4 + 1 ⎠ ⎝ 4 ⎠ ⎝ −3 ⎠

5.(b) Determine

∫

(b)

(2 + θ) 2 dθ = ∫ ( 4 + 4θ + θ2 ) dθ = 4θ + 2θ 2 +

4.(b) Determine

∫

∫

⎛ 3x 2 − 5x ⎞ ⎜ ⎟ dx x ⎝ ⎠

∫7

3 5

x4

dx 4 − +1

1

4 1 − 15 5 3 1 3 3 x 5 3 x5 ⎛ 3 ⎞⎛ 5 ⎞ dx = ∫ 4 dx = ∫ x 5 dx = x +c +c= + c = ⎜ ⎟⎜ ⎟ x5 + c = 5 4 7 7 7 7⎛ 4 ⎞ 7 ⎛1⎞ 7 ⎠⎝ 1 ⎠ ⎝ 7 x 5 x ⎜ − + 1⎟ ⎜ ⎟ ⎝ 5 ⎠ ⎝5⎠

3

8. Determine (a)

3

3

∫ 4 sec

2

3x dx

(b)

∫ 2 cos ec 4θ dθ 2

1 ⎛ 3 ⎞⎛ 1 ⎞ 3x dx = ⎜ ⎟⎜ tan 3x ⎟ + c = tan 3x + c 4 ⎝ 4 ⎠⎝ 3 ⎠

(a)

∫ 4 sec

(b)

∫ 2 cos ec 4θ dθ = ( 2 ) ⎜⎝ − 4 cot 4θ ⎟⎠ + c = − 2 cot 4θ + c

2

2

⎛ 1

⎞

9. Determine (a) 5∫ cot 2t cos ec2t dt

1

(b)

4

∫ 3 sec 4t tan 4t dt


327

5 ⎛ 1 ⎞ (a) 5∫ cot 2t cos ec2t dt = ( 5 ) ⎜ − cos ec2t ⎟ + c = − cos ec2t + c 2 ⎝ 2 ⎠

(b)

⎛ 4 ⎞⎛ 1

4

⎞

sec 4t ⎟ + c = ∫ 3 sec 4t tan 4t dt = ⎜⎝ 3 ⎟⎜ ⎠⎝ 4 ⎠

10.(b) Determine

1 sec 4t + c 3

2 dx 3 ∫ e5x

−2 2 dx 2 −5x 2 ⎛ 2 ⎞⎛ 1 ⎞ +c = ∫ e dx = ⎜ ⎟ ⎜ e −5x ⎟ + c = − e−5x + c = 5x ∫ 15e5x 3 e 3 15 ⎝ 3 ⎠ ⎝ −5 ⎠

11.(b) Determine

∫

∫

⎛ u2 −1 ⎞ ⎜ ⎟ du ⎝ u ⎠

⎛ u2 −1 ⎞ ⎛ u2 1 ⎞ du = ⎜ ⎟ ∫ ⎜⎝ u − u ⎟⎠ du = ∫ ⎝ u ⎠

12. Determine (a)

(a)

∫

∫

(2 + 3x) 2 dx x

u2 1⎞ ⎛ = − ln u + c u du − ⎜ ⎟ 2 u⎠ ⎝

(b)

∫

2

⎛1 ⎞ ⎜ + 2t ⎟ dt ⎝t ⎠

1 3 ⎛ 4 12x 9x 2 ⎞ ⎛ −1 ⎞ (2 + 3x) 2 4 + 12x + 9x 2 dx = ∫ dx = ∫ ⎜ 1 + 1 + 1 ⎟ dx = ∫ ⎜ 4x 2 + 12x 2 + 9x 2 ⎟ dx ⎜ 2 ⎟ x x ⎝ ⎠ ⎝x x2 x2 ⎠ 1

3

5

18 5 4x 2 12x 2 9x 2 x +c + + + c = 8 x + 8 x3 + = 1 3 5 5 2 2 2

(b)

∫

2

⎛1 ⎞ ⎜ + 2t ⎟ dt = ∫ ⎝t ⎠

⎛1 ⎞⎛ 1 ⎞ ⎛1 −2 2⎞ 2 ⎜ + 2t ⎟⎜ + 2t ⎟ dt = ∫ ⎜ 2 + 4 + 4t ⎟ dt = ∫ ( t + 4 + 4t ) dt t t t ⎝ ⎠⎝ ⎠ ⎝ ⎠ −1 3 1 4t 3 t 4t + 4t + + c = − + 4t + +c = t 3 −1 3


328

EXERCISE 147 Page 372 2. Evaluate (a)

∫ ( 3 − x ) dx 2

2

−1

(b)

∫ (x 3

2

1

− 4x + 3) dx

2 3 −1) ⎤ ⎛ ⎡ ⎡ ( x3 ⎤ 23 ⎤ ⎡ 8⎞ ⎛ 1⎞ (a) ∫ ( 3 − x ) dx = ⎢3x − ⎥ = ⎢3(2) − ⎥ − ⎢3(−1) − ⎥ = ⎜ 6 − ⎟ − ⎜ −3 − − ⎟ −1 3 ⎦ −1 ⎣ 3 ⎦ ⎢⎣ 3 ⎥⎦ ⎝ 3⎠ ⎝ 3⎠ ⎣ 2

2

1 ⎛ 2⎞ = 3 − ⎜ −2 ⎟ = 6 3 ⎝ 3⎠ 3

⎡ x 3 4x 2 ⎤ 1 ⎛ 1⎞ ⎛1 ⎞ + 3x ⎥ = ( 9 − 18 + 9 ) − ⎜ − 2 + 3 ⎟ = ( 0 ) − ⎜1 ⎟ = −1 (b) ∫ ( x − 4x + 3) dx = ⎢ − 1 3 2 ⎝3 ⎠ ⎝ 3⎠ ⎣3 ⎦1 3

2

4. Evaluate (a)

(a)

∫

π/3 π/6

∫

π/3 π/6

2sin 2θ dθ

2sin 2θ dθ = −

(b)

∫

2 0

3sin t dt

2 2π 2π π/3 [cos 2θ] π / 6 = − ⎡⎢cos − cos ⎤⎥ 2 3 6 ⎦ ⎣

(note that

2π 2π are in radians) and 3 6

= -[-0.5 – 0.5] = -[-1] = 1 (b)

∫

2 0

3sin t dt = −3 [ cos t ] 0 = −3[cos 2 − cos 0] 2

(note that 2 is 2 radians)

= -3[-0.41615 – 1] = 4.248 6. Evaluate, correct to 4 significant figures, (a)

(a)

∫

2

cos ec2 4tdt = −

1

∫

2

1

cos ec2 4tdt

π/2

∫ ( 3sin 2x − 2 cos 3x )dx π/ 4

1 1 1 1 1 ⎤ 2 − [cot 4t ]1 = − [cot 8 − cot 4] = − ⎡⎢ 4 4 4 ⎣ tan 8 tan 4 ⎥⎦ =−

(b)

(b)

1 ( −0.147065 − 0.863691) = 0.2527 4 π/2

2 ⎡ 3 ⎤ ∫ π / 4 ( 3sin 2x − 2 cos 3x ) dx = ⎢⎣− 2 cos 2x − 3 sin 3x ⎥⎦ π / 4 π/2

2π 2 3π ⎞ ⎛ 3 2π 2 3π ⎞ ⎛ 3 − sin ⎟ − ⎜ − cos − sin ⎟ = ⎜ − cos 2 3 2 ⎠ ⎝ 2 4 3 4 ⎠ ⎝ 2 2 ⎛3 2⎞ ⎛ ⎞ = ⎜ + ⎟ − ⎜ 0 − (0.707107) ⎟ = 2.638 3 ⎝2 3⎠ ⎝ ⎠ 8. Evaluate, correct to 4 significant figures, (a)

(a)

∫

3 2

∫

3 2

2 dx 3x

(b)

∫

3

1

2x 2 + 1 dx x

2 2 3 1 2 2 3 dx = ∫ dx = [ ln x ] 2 = ( ln 3 − ln 2 ) = 0.2703 2 3x 3 x 3 3 © 2006 John Bird. All rights reserved. Published by Elsevier.

329

∫

(b)

2 3 ⎛ 2x 3 ⎛ 3 2x 2 + 1 1⎞ 1⎞ dx = ∫ ⎜ + ⎟ dx = ∫ ⎜ 2x + ⎟ dx = ⎡⎣ x 2 + ln x ⎤⎦ = (9 + ln 3) – (1 + ln 1) 1 1 1 x x⎠ x⎠ ⎝ ⎝ x

3

1

= 9 + ln 3 – 1 = 8 + ln 3 = 9.099

9. The entropy change ∆S, for an ideal gas is given by: ∆S =

∫

T2 T1

Cv

V2 dV dT − R∫ V1 V T

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

∫ ( 45 + 6 ×10 400

∆S =

=

100

∫

400

100

−3

T + 8 ×10−6 T 2 )

3 dV dT − 8.314∫ 1 V T

3 dV ⎛ 45 −3 −6 ⎞ ⎜ + 6 ×10 + 8 × 10 T ⎟ dT − 8.314 ∫1 V ⎝T ⎠

400

⎡ 8 ×10−6 T 2 ⎤ 3 −3 = ⎢ 45ln T + 6 ×10 T + − 8.314 [ ln V ]1 ⎥ 2 ⎣ ⎦ 100 ⎡⎛ 8 × 10−6 (400) 2 ⎞ ⎛ 8 ×10−6 (100) 2 ⎞ ⎤ −3 45ln100 6 10 (100) = ⎢⎜ 45ln 400 + 6 ×10−3 (400) + − + × + ⎟ ⎜ ⎟⎥ 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 a

Q dr 2πrε 0ε r

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 a

Q Q dr = 2πrε0 ε r 2π ε 0 ε r

∫

b a

1 Q Q b dr = (ln b − ln a) [ln r ] a = r 2π ε 0 ε r 2π ε 0 ε r =

2 ×10−6 (ln 20 − ln10) = 9000 V or 9 kV 2π ( 8.85 ×10−12 ) ( 2.77 )


330

11. The average value of a complex voltage waveform is given by:

VAV =

1 π (10sin ωt + 3sin 3ωt + 2sin 5ωt ) d(ωt) π ∫0

Evaluate VAV correct to 2 decimal places.

VAV =

1 π (10sin ωt + 3sin 3ωt + 2sin 5ωt ) d(ωt) π ∫0 π

1⎡ 3 2 ⎤ = ⎢ −10 cos ωt − cos 3ωt − cos 5ωt ⎥ π⎣ 3 5 ⎦0 =

1 ⎡( −10 cos π − cos 3π − 0.4 cos 5π ) − ( −10 cos 0 − cos 0 − 0.4 cos 0 ) ⎤⎦ π⎣

=

1 ⎡(10 + 1 + 0.4 ) − ( −10 − 1 − 0.4 ) ⎤⎦ π⎣

=

1 22.8 = 7.26 ⎡⎣(11.4 ) − ( −11.14 ) ⎤⎦ = π π


331

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 −4

(7 − 3x) dx − ∫

1 −4

(x

2

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

1

−4

−4

( 4 − 3x − x ) dx 2

1

⎡ 3x 2 x 3 ⎤ = ⎢ 4x − − ⎥ 2 3 ⎦ −4 ⎣ 3 1⎞ ⎛ 64 ⎞ ⎛ = ⎜ 4 − − ⎟ − ⎜ −16 − 24 + ⎟ 2 3⎠ ⎝ 3 ⎠ ⎝ 2⎞ 1 2 ⎛ 1⎞ ⎛ = ⎜ 2 ⎟ − ⎜ −18 ⎟ = 2 + 18 3⎠ 6 3 ⎝ 6⎠ ⎝ 5 = 20 square units 6

3. Determine the area enclosed by the three straight lines y = 3x, 2y = x and y + 2x = 5.


332

y = -2x + 5 and y = 3x intersect when -2x + 5 = 3x i.e. when y = -2x + 5 and y =

x x intersect when -2x + 5 = 2 2

i.e. when

5 = 5x

i.e. when x = 1

5 = 2.5x i.e. when x = 2

The three straight lines are shown below.

Shaded area =

∫

1 0

2 x⎞ x ⎛ ⎜ 3x − ⎟ dx + ∫ 1 (−2x + 5) − dx 2⎠ 2 ⎝ 1

2

⎡ 3x 2 x 2 ⎤ ⎡ 2 x2 ⎤ ⎡⎛ 3 1 ⎞ ⎤ ⎡ 1 ⎞⎤ ⎛ − ⎥ + ⎢ − x + 5x − ⎥ = ⎢⎜ − ⎟ − (0) ⎥ + ⎢( −4 + 10 − 1) − ⎜ −1 + 5 − ⎟ ⎥ = ⎢ 4 ⎦0 ⎣ 4 ⎦ 1 ⎣⎝ 2 4 ⎠ 4 ⎠⎦ ⎝ ⎦ ⎣ ⎣ 2 1 1 ⎛ 1⎞ ⎡ ⎛ 3 ⎞⎤ = ⎜ 1 ⎟ + ⎢( 5 ) − ⎜ 3 ⎟ ⎥ = 1 + 1 4 4 ⎝ 4⎠ ⎣ ⎝ 4 ⎠⎦ 1 = 2 square units 2


333


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 +

1 . Determine the standard deviation (i.e. the r.m.s. value), correct to x

4 significant figures for values of x from 1 to 2. 2

Standard deviation = the r.m.s. value =

2 ⎛ 1 1⎞ x + ⎟ dx = ⎜ ∫ 2 −1 1 ⎝ x⎠

2

1 ⎞ ⎛ 2 ⎜ x + 2 + 2 ⎟ dx = x ⎠ ⎝

∫

2 1

1 ⎞⎛ 1⎞ ⎛ ⎜ x + ⎟⎜ x + ⎟ dx x ⎠⎝ x⎠ ⎝

∫ (x 2

+ 2 + x −2 ) dx

=

∫

=

⎡ x3 ⎡ x3 x −1 ⎤ 1⎤ ⎢ 3 + 2x + −1 ⎥ = ⎢ 3 + 2x − x ⎥ ⎣ ⎦1 ⎣ ⎦1

=

⎡⎛ 8 1⎞ ⎛1 ⎞⎤ ⎢⎜ 3 + 4 − 2 ⎟ − ⎜ 3 + 2 − 1⎟ ⎥ = 4.8333 = 2.198 ⎠ ⎝ ⎠⎦ ⎣⎝

1

1

2

2

2

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.

10×10−3

10×10−3 1 ⎡ 25 ⎤ Mean value = 25sin100πt ) dt = 100 ⎢ − cos100πt ⎥ ( −3 ∫ 10 ×10 − 0 0 ⎣ 100π ⎦0

=−

100(25) ⎡⎣cos(100π× 10 × 10−3 ) − cos 0 ⎤⎦ 100π

= −

25 25 50 [ cos π − cos 0] = − [ −1 − 1] = π π π = 15.92 mA

r.m.s. value =

10×10−3 10×10−3 1 1 2 2 2 25 sin 100 t dt (100)(25) π = ∫ 0 2 (1 − cos 200πt) dt 10 × 10−3 − 0 ∫ 0

since cos 2A = 1 – 2sin 2 A from which, sin 2 A = 10×10−3

⎡ sin 200πt ⎤ ⎢⎣ t − 200π ⎥⎦ 0

(100)(25) 2 = 2

1 (1 − cos 2A) , 2

⎡⎛ sin 2π ⎞ ⎤ −3 ⎢⎜10 × 10 − 200π ⎟ − (0 − sin 0) ⎥ ⎠ ⎣⎝ ⎦

=

(100)(25)2 2

=

(100)(25) 2 252 25 ⎡⎣10 ×10−3 ⎤⎦ = = = 17.68 mA 2 2 2


334

v = E1 sin ωt + E 3 sin 3ωt

4. A wave is defined by the equation:

where E1 , E 3 and ω are constants. Determine the r.m.s. value of v over the interval 0 ≤ t ≤

r.m.s. value =

=

∫

π ω 0

∫

π ω 0

∫

π ω 0

sin 2 ωt dt = ∫

1 π −0 ω

π ω 0

∫ ( E sin ωt + E 1

π . ω

sin 3ωt ) d(ωt) 2

3

ω ωπ 2 2 E1 sin ωt + 2E1E 3 sin ωt sin 3ωt + E 32 sin 2 3ωt ) d(ωt) ( ∫ 0 π π ω 0

sin 2 3ωt dt = ∫

π

1 − cos 2ωt ⎤ π ⎡ t sin 2ωt ⎤ ω ⎡⎛ π ⎞ dt = ⎢ − = ⎢⎜ − 0 ⎟ − (0 − 0) ⎥ = ⎥ 2 4ω ⎦ 0 ⎣ ⎝ 2ω ⎣2 ⎠ ⎦ 2ω

π ω 0

π

1 − cos 6ωt ⎤ π ⎡ t sin 6ωt ⎤ ω ⎡⎛ π ⎞ dt = ⎢ − = ⎢⎜ − 0 ⎟ − (0 − 0) ⎥ = ⎥ 2 12ω ⎦ 0 ⎣⎝ 2ω ⎣2 ⎠ ⎦ 2ω

sin ωt sin 3ωt dt = ∫

π ω 0

π

sin 3ωt sin ωt dt = ∫ ω − 0

1 ( cos 4ωt − cos 2ωt ) dt 2

(see page183of textbook)

π

1 ⎡ sin 4ωt sin 2ωt ⎤ ω 1 =− ⎢ − = − [ (0 − 0) − (0 − 0)] = 0 ⎥ 2 ⎣ 4ω 2ω ⎦ 0 2

Hence, r.m.s. value =

=

⎛ E12 E 32 ⎞ ω ⎡ π 2 π 2⎤ E + E = + ⎜ ⎟ 1 3 ⎥ π ⎢⎣ 2ω 2ω 2 ⎠ ⎦ ⎝ 2

⎛ E12 + E3 2 ⎞ ⎜ ⎟ 2 ⎝ ⎠


335


2. The area between

y = 1 and y + x2 = 8 is rotated 360° about the x-axis. Find the volume x2

produced.

y = x 2 and y = 8 - x 2 intersect when x 2 = 8 - x 2 i.e. when x =

i.e.

2 x2 = 8

and

x2 = 4

4 = ±2 . A sketch of the two curves is shown below.

Volume of solid of revolution =

∫

2

2

−2

= π∫

π ( 8 − x 2 ) dx − ∫ 2

−2

( 64 − 16x

2

2

−2

π ( x 2 ) dx 2

+ x 4 ) dx − π ∫

2

−2

x 4 dx = π ∫

2

−2

( 64 − 16x ) dx 2

2

⎡ 16x 3 ⎤ ⎡⎛ 128 ⎞ ⎛ 128 ⎞ ⎤ = π ⎢64x − = π ⎢⎜128 − ⎟ − ⎜ −128 + ⎟ ⎥ 3 ⎦ −2 3 ⎠ ⎝ 3 ⎠ ⎥⎦ ⎣⎝ ⎣ 2 256 ⎤ ⎡ ⎡ 768 − 256 ⎤ ⎛ 512 ⎞ = π⎢ = π⎜ = π ⎢ 256 − ⎟ = 170 3 π cubic units ⎥ ⎥ 3 ⎦ 3 ⎣ ⎣ ⎦ ⎝ 3 ⎠

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. 3

Volume x −axis

⎡ 4x 5 ⎤ = ∫ π ( 2x + 3) dx = π ∫ ( 4x + 12x + 9 ) dx = π ⎢ + 4x 3 + 9x ⎥ 0 0 ⎣ 5 ⎦0 3

2

2

3

4

2

= π[(329.4) – (0)] = 329.4π cubic units


336

21

(b) Volume y −axis = ∫ πx 2 dy

Since y = 2x2 + 3, then 2x2 = y – 3 and x 2 =

3

y−3 2

⎤ ⎞ ⎛9 π ⎡ y2 π ⎡⎛ 212 ⎛ y−3⎞ ⎞⎤ dy 3y = − = − 63 ⎟ − ⎜ − 9 ⎟ ⎥ ⎢⎜ ⎜ ⎟ ⎢ ⎥ 2⎣ 2 ⎝ 2 ⎠ ⎠⎦ ⎦ 3 2 ⎣⎝ 2 ⎠ ⎝2 21

Hence, volume y −axis = π∫

21 3

=

π 162π ⎡⎣(157.5 ) − ( −4.5 ) ⎤⎦ = 2 2 = 81π cubic units

4. The profile of a rotor blade is bounded by the lines x = 0.2, y = 2x, y = e − x , 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 = e − x (c) Calculate the cross-sectional area of the blade, correct to 3 decimal places. (d) Calculate the volume of the blade in of K, correct to 3 decimal places. (a) A sketch of the rotor blade is shown below.


337

(b) Using the Newton-Raphson method (from chapter 9), let f(x) = 2x - e − x f(0) = 0 - e0 = -1 f(1) = 2 - e −1 = 1.632 hence a root lies between x = 0 and x = 1 Let r1 = 0.4 f '(x) = 2 + e − x r2 = r1 −

f (r1 ) 0.12967995 = 0.4 − = 0.35144 f '(r1 ) 2.670320046

r3 = r2 −

f (r2 ) (−0.00079407) = 0.35144 − = 0.35173 f '(r2 ) 2.70367407

r4 = r3 −

f (r3 ) (−0.000010033) = 0.35173 − = 0.35173 f '(r3 ) 2.703470033

Hence, correct to 3 decimal places, x = 0.352 (c) Cross-sectional area of blade =

∫

0.352 0.2

2x dx + ∫

1 0.352

0.352

e − x dx

1

= ⎡⎣ x 2 ⎤⎦ − ⎡⎣ e − x ⎤⎦ = (0.0839) − (−0.3354) 0.2 0.352 = 0.419 square units (d) Volume of the blade =

∫

0.352 0.2

= K∫

2x(1.1 − x)K dx + ∫

0.352 0.2

1 0.352

( 2.2x − 2x ) dx + K ∫ 2

e − x (1.1 − x)K dx 1 0.352

(1.1e

−x

− xe − x ) dx

0.352

1 ⎡ 2.2x 2 2x 3 ⎤ − + K ⎡⎣ −1.1e − x − {− xe − x − e− x }⎤⎦ = K⎢ (the latter ⎥ 0.352 3 ⎦ 0.2 ⎣ 2 integral using integration by parts – see page 418 of textbook) 0.352

1 ⎡ 2.2x 2 2x 3 ⎤ − + K ⎡⎣ −1.1e − x + xe − x + e − x ⎤⎦ = K⎢ ⎥ 0.352 3 ⎦ 0.2 ⎣ 2

= K [ 0.107218 − 0.038667 ] + K [ 0.331091 − 0.177227 ] = 0.068551K + 0.153864K = 0.222K, correct to 3 decimal places. © 2006 John Bird. All rights reserved. Published by Elsevier.

338

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.

4

∫ xy dx = ∫ x ( 5x ) dx = ∫ x= ∫ y dx ∫ 5x dx ∫ 4

4

1

2

1

4

4

1

2

1

⎡ 5x 4 ⎤ ⎢ 4 ⎥ ⎣ ⎦1

5 4 4 5 ⎡⎣ 4 − 1 ⎤⎦ (255) 318.75 1 4 4 = = = = = 3.036 4 3 4 2 5 5 105 3 3 ⎡ ⎤ 5x 5x dx ⎡ 4 − 1 ⎤⎦ (63) 1 ⎢ 3 ⎥ 3⎣ 3 ⎣ ⎦1 4

5x 3dx

2 1 4 2 1 4 4 y dx 5x 2 ) dx ( ∫ ∫ 1 4 1 ⎡ 25x 5 ⎤ 5 5 1 1 4 2 2 ⎡⎣ 45 − 15 ⎤⎦ = y= 4 25x dx = (1023) = = = ⎢ ⎥ ∫ 1 105 210 210 ⎣ 5 ⎦ 1 210 210 y dx ∫ 1

= 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


339

4

1 ⎡16x 3 8x 4 x 5 ⎤ 4 2 1 4 2 1 4 1 − + 2 2 3 4 ∫0 y dx 2 ∫0 ( 4x − x ) dx 2 ∫0 (16x − 8x + x ) dx 2 ⎢⎣ 3 4 5 ⎥⎦ 0 2 y= 4 = = = 4 4 3 4 2 2 ⎡ ⎤ x y dx 4x x dx 4x x dx − − ( ) ( ) 2 ∫0 ∫0 ∫0 ⎢ 2x − 3 ⎥ ⎣ ⎦0

1 ⎡⎣( 341.333 − 512 + 204.8 ) − (0) ⎤⎦ 17.0665 = 2 = = 1.6 64 ⎞ 10.666 ⎛ ⎜ 32 − ⎟ − (0) 3 ⎠ ⎝ 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 4

∫ xy dx = ∫ x ( 3 x ) dx = ∫ x= ∫ y dx ∫ 3 x dx ∫ 4

4

0

0

4

4

0

0

3 2

⎡ 5⎤ ⎢ 3x 2 ⎥ ⎢ 5 ⎥ ⎢ ⎥ ⎣ 2 ⎦0

6⎡ 5⎤ 6⎡ 5 6 x 4 − 0⎤ (32) ⎣ ⎦ ⎣ ⎦ 5 5 5 = = = = 1 4 4 2(8) ⎡ x 3 ⎤ 2 ⎡ 43 − 0 ⎤ 3 ⎤ ⎡ 2 2 3x dx ⎣ ⎦ ⎣ ⎦ 2 ⎢ 3x ⎥ 0 ⎢ 3 ⎥ ⎢ ⎥ ⎣ 2 ⎦0 38.4 = 2.4 = 16 4

0

3x dx

Hence, the centroid is at (2.4, 0)


340


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 ) 2π y

)

(

1⎛4 3⎞ 1 2 ⎜ πr ⎟ = ( πr ) 2π y 2⎝3 ⎠ 4

i.e.

)

2 3 πr 4r 4(4) 3 y= = = = 1.70 cm 2 3π 3π ⎛ πr ⎞ ⎜ ⎟ ( 2π ) ⎝ 4 ⎠

from which,

By symmetry, x = 1.79 cm

Hence, the centroid of the template is at (1.70, 1.70) 4

4

(b) x =

∫ ∫

0

xy dx

4 0

y dx

=

∫

4 0

x

(

∫ (4 4

0

)

(16 − x 2 ) dx 2

−x

2

) dx

=

1

x (16 − x 2 ) 2 dx

4

∫ ∫ (4 0

4

0

2

− x 2 )dx

=

3 ⎡ 1 2 2 ⎤ 16 x − − ( ) ⎢ 2 ⎥ ⎢ ⎥ 3 ⎢ ⎥ ⎢⎣ ⎥⎦ 0 2

⎡ 42 x −1 x ⎢ 2 sin 4 + 2 ⎣

4

( 4 − x ) ⎤⎥ ⎦0 2

2

(the numerator being an algebraic substitution - see chapter 39), and the denominator being a sin θ substitution - see chapter 40) ⎡ ⎤ ⎛ 3 ⎞⎤ 1⎡ 64 − ⎢( 0 ) − ⎜ 16 2 ⎟ ⎥ ⎢ ⎥ 3 ⎢⎣ ⎢ ⎥ ⎝ ⎠ ⎥⎦ = 3 = 1.70 i.e. x = ⎢ ⎡( 8sin −1 1 + 2(0) ) − ( 8sin −1 0 + 0(4) ) ⎤ ⎥ 12.566 ⎢⎣ ⎦⎥ ⎢⎣ ⎥⎦


341

By symmetry, y = 1.70 cm 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

(1.70

2

+ 1.702 ) = 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.

3

Shaded area =

∫

3 0

⎡ 5x 3 ⎤ 5 = ( 33 − 0 ) = 45 square units 5x dx = ⎢ ⎥ ⎣ 3 ⎦0 3 2

(b)(i) Volume x −axis = ∫ πy dx = ∫ π ( 5x 3

3

2

0

0

)

2 2

3

⎡ x5 ⎤ dx = 25π∫ x dx = 25π ⎢ ⎥ = 5π ⎣⎡35 − 0 ⎦⎤ 0 ⎣ 5 ⎦0 3

4

= 1215π cubic units (ii) Volume y −axis = (volume generated by x = 3) – (volume generated by y = 5x2) 45

=

∫

45 0

π(3) dy − ∫ 2

45 0

45 ⎛ ⎡ y⎞ y2 ⎤ ⎛y⎞ π ⎜ ⎟ dy = π∫ ⎜ 9 − ⎟ dy = π ⎢9y − ⎥ 0 5⎠ 10 ⎦ 0 ⎝5⎠ ⎝ ⎣

⎡⎛ ⎤ 452 ⎞ = π ⎢⎜ 9(45) − ⎟ − (0) ⎥ = π[405 − 202.5] = 202.5π cubic units 10 ⎠ ⎣⎝ ⎦ 3

⎡ 5x 4 ⎤ 5 4 2 3 xy dx ∫ x ( 5x ) dx ∫ 5x dx ⎢⎣ 4 ⎥⎦ (3 − 0) ∫ 0 0 0 0 4 (c)(i) x = 3 = 2.25 = = = = 45 45 45 45 y dx ∫ 3

3

3

0


342

3

1 3 2 1 3 4 y dx ∫ ∫ 0 25x dx 0 2 2 = = y= 3 45 ∫ y dx

1 ⎡ 25x 5 ⎤ 2 ⎢⎣ 5 ⎥⎦ 0 45

5 5 ⎡⎣3 − 0 ⎤⎦ = 13.5 =2 45

0

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) 2π x

(

202.5π = (45) 2π x

i.e. from which,

x=

)

202.5π = 2.25 (45)(2π)

(

Similarly, volume generated when the shaded area is revolved about 0x = (area) 2π y

(

1215π = (45) 2π y

i.e. from which,

y=

)

)

)

1215π = 13.5 (40)(2π)

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

4r from its diameter, from problem 2 above, 3π


343

i.e.

4(1.0) = 0.424 cm from PQ. 3π

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 =

πr 2 π(1.0) 2 π = = cm 2 2 2 2

By Pappus, volume generated = area × distance moved by the centroid ⎛π⎞ i.e. volume of metal removed = ⎜ ⎟ ( 2π(6.576) ) = 64.90cm 3 ⎝2⎠ Volume of disc = πr 2 h = π ( 7.0 ) ( 2.5 ) = 384.845cm3 2

Thus, percentage of metal removed =

64.90 × 100% = 16.86% 384.845

Mass of metal removed = density × volume

= 7800

kg × 64.90 × 10−6 m3 = 0.5062 kg or 506.2 g 3 m


344


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: b h 3 (12.0 )( 9.0 ) = = = 729 cm 4 12 12 3

(a) Second moment of area about DD, I DD Radius of gyration about DD, k DD =

h 9.0 = = 3.67 cm 6 6 b h 3 (12.0 )( 9.0 ) = = = 2187 cm 4 4 4 3

(b) Second moment of area about EE, I EE Radius of gyration about EE, k EE =

h 9.0 = = 6.36 cm 2 2

b h 3 (12.0 )( 9.0 ) (c) Second moment of area about axis through centoid, = = = 243 cm 4 36 36 3

Radius of gyration about axis through centoid, =

h 9.0 = = 2.12 cm 18 18

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.


345

(a) Second moment of area, I LL = IGG + Ad 2 =

bl3 (3.0)(5.0)3 + Ad 2 = + (3.0)(5.0)(2.5 + 2.0) 2 12 12 = 31.25 + 303.75 = 335 cm 4

I LL = Ak LL 2 from which, radius of gyration, k LL =

I LL ⎛ 335 ⎞ = ⎜ ⎟ = 4.73 cm area ⎝ 15.0 ⎠

(b) Second moment of area, I LL = IGG + Ad 2 IGG =

bh 3 (18)(12)3 = = 864 cm 4 where h = 152 − 92 = 12 cm, as shown in the diagram below. 36 36

Hence, area of triangle, A =

1 (18)(12) = 108 cm 2 2 2

Thus, I LL

12 ⎞ ⎛ = IGG + Ad = 864 + 108 ⎜10 + ⎟ = 864 + 108(14) 2 = 22032 cm 4 = 22032 cm 4 3⎠ ⎝ correct to 4 significant figures. 2

I LL = Ak LL 2 from which, radius of gyration, k LL =

I LL ⎛ 22032 ⎞ = ⎜ ⎟ = 14.3 cm area ⎝ 108 ⎠

πr 4 4⎞ π(2.0) 4 2 ⎛ = IGG + Ad = + ( πr ) ⎜ 5 + ⎟ = + π(2.0) 2 (7) 2 4 2⎠ 4 ⎝ 2

(c) Second moment of area, I LL

2

= 12.57 + 615.75 = 628 cm 4 I LL = Ak LL 2 from which, radius of gyration, k LL =

⎛ 628 ⎞ I LL = 7.07 cm = ⎜ 2 ⎟ area ⎝ π(2.0) ⎠

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 ICC

πr 4 πr 4 = − I DD = − ⎡⎣ IGG + Ad 2 ⎤⎦ 4 4


346

=

⎤ π(12) 4 ⎡ π(4.0) 4 −⎢ + π(4.0) 2 (6.0) 2 ⎥ = 16286 – [201 + 1810] = 14275 = 14280 cm 4 , correct 4 ⎣ 4 ⎦ to 4 significant figures.

ICC = Ak CC 2 from which, radius of gyration, k CC =

⎛ ⎞ ICC 14275 14275 = 5.96 cm = ⎜ = 2 2 ⎟ area 128π ⎝ π(12.0) − π(4.0) ⎠

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 CA =

bl3 (18.0)(3.0)3 = = 40.5 m 4 12 12


347

Hence, I XX A = 40.5 + Ad 2 = 40.5 + (3.0)(18.0)(12.0 + 1.5) 2 = 40.5 + 9841.5 = 9882 mm 4 For rectangle B, second moment of area about CB =

bl3 (4.0)(12.0)3 = = 576 m 4 12 12

Hence, I XX B = 576 + (4.0)(12.0)(6.0) 2 = 576 + 1728 = 2304 mm 4 Thus, total second moment of area about XX, I XXT = 9882 + 2304 = 12186 = 12190 mm 4 , correct to 4 significant figures.

Radius of gyration about XX, k XX =

I XXT area

=

12186 = 10.9 mm 54 + 48

(b) Rectangle A (see diagram below): bl3 (6.0)(2.0)3 Second moment of area about CA = = = 4 cm 4 12 12 I XX A = 4 + (2.0)(6.0)(6.0) 2 = 4 + 432 = 436 cm 4

Rectangle B: Second moment of area about CB =

bl3 (2.5)(3.0)3 = = 5.625cm 4 12 12

I XX B = 5.625 + (2.5)(3.0)(3.5) 2 = 97.5 cm 4

Rectangle C: Second moment of area about CC =

bl3 (6.0)(2.0)3 = = 4 cm 4 12 12

I XXC = 4 + (6.0)(2.0)(1.0) 2 = 16 cm 4

Hence, total second moment of area about axis XX, I XX I = 436 + 97.5 + 16 = 549.5 cm 4


I XXI area

=

549.5 = 4.18 cm 12 + 7.5 + 12


348

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, I XX = I XXA + I XXB + I XXC ⎡ ( 6.0 )(1.0 )3 ⎤ ⎡ ( 2.0 )( 8.0 )3 ⎤ ⎡ (10.0 )( 2.0 )3 ⎤ 2 2 =⎢ + (6.0)(10.5) ⎥ + ⎢ + (16)(6.0) ⎥ + ⎢ + (20)(1.0) 2 ⎥ 12 12 12 ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ = 662 + 661.33 + 26.66 = 1350 cm 4


I XX 1350 = 5.67 cm = area 6 + 16 + 20


349

CHAPTER 39 INTEGRATION USING ALGEBRAIC SUBSTITUTIONS EXERCISE 154 Page 392 4. Integrate

1 6 ( 5x − 3) with respect to x. 2

Let u = 5x – 3 then

∫

Hence,

du du = 5 and dx = dx 5

1 6 du 1 1 ⎛ u7 u = ∫ u 6 du = ⎜ 2 5 10 10 ⎝ 7

5. Integrate

−3 with respect to x. (2x − 1)

Let u = 2x – 1 then


−3

∫ (2x − 1) dx = ∫

Hence,

7. Evaluate

∫ ( 3x + 1) 1

0

Let u = 3x + 1 then

∫

Hence,

Thus,

u5

∫ 0 ( 3x + 1) ∫

2 0

Let u = 2x 2 + 1

Hence,

Thus,

∫x ∫

2 0

5

3 −3 du 3 1 3 = − ∫ du = − ln u + c = − ln(2x − 1) + c 2 u 2 2 u 2

dx correct to 4 significant figures.


du 1 1 ⎛ u6 ⎞ 1 = ∫ u 5du = ⎜ ⎟ + c = (3x + 1)6 + c 3 3 3⎝ 6 ⎠ 18

1

8. Evaluate

⎞ 1 (5x − 3)7 + c ⎟+c = 70 ⎠

5

x

dx =

1 ⎡ 1 6 1 3x + 1) ⎤ = ⎡⎣ 46 − 16 ⎤⎦ = 227.5 ( ⎣ ⎦ 0 18 18

( 2x

+ 1) dx correct to 4 significant figures.

then

2

du = 4x dx

(2x 2 + 1) dx = ∫

dx =

du 4x

⎛ 3 1 ⎜ u2 du 1 1 x u = ∫ u 2 du = ⎜ 4x 4 4⎜ 3 ⎝ 2

1⎡ x ( 2x + 1) dx = ⎢ 6⎣ 2

i.e.

⎞ 3 ⎟ 1 1 2 2 +c= + c = 2x + 1 ( ) ⎟ 6 6 ⎟ ⎠

( 2x

2

+ 1) + c 3

2

( 2x + 1) ⎤⎥⎦ = 16 [ 27 − 1] = 4.333 0 2

3


350

9. Evaluate

∫

π/3 0

π⎞ ⎛ 2sin ⎜ 3t + ⎟ dt correct to 4 significant figures. 4⎠ ⎝ π

∫

π/3 0

π⎞ 2⎡ ⎛ π ⎞⎤ 3 2 ⎡ ⎛ 3π π ⎞ π⎤ ⎛ 2sin ⎜ 3t + ⎟ dt = − ⎢cos ⎜ 3t + ⎟ ⎥ = − ⎢cos ⎜ + ⎟ − cos ⎥ (note angles are in radians) 4⎠ 3⎣ ⎝ 4 ⎠⎦ 0 3⎣ ⎝ 3 4⎠ 4⎦ ⎝ = −

2 [ −0.70711 − 0.70711] 3

= 0.9428


351


2. Integrate 5cos5 t sin t with respect to t.

Let u = cos t then

∫

du = − sin t dt

5cos5 t sin t dt = ∫ 5u 5 sin t

dt =

and

du − sin t

⎛ u6 ⎞ 5 du = −5∫ u 5du = −5 ⎜ ⎟ + c = - cos 6 t + c 6 − sin t ⎝ 6 ⎠

3. Integrate 3sec 2 3x tan 3x with respect to x.

du = 3sec 2 3x dx

Let u = tan 3x then

Hence,

∫

3sec 2 3x tan 3x dx = ∫ 3sec 2 3x (u)

Alternatively, let u = sec 3x then

Hence,

∫

i.e.

dx =

du 3sec 2 3x

1 du u2 = u du = + c = tan 2 3x + c 2 ∫ 2 3sec 3x 2

du = 3sec 3x tan 3x dx

3sec2 3x tan 3x dx = ∫ 3u 2 tan 3x

i.e.

dx =

du 3sec 3x tan 3x

du u2 u2 =∫ du = ∫ du = ∫ u du 3sec 3x tan 3x sec 3x u =

5. Integrate

ln θ with respect to θ. θ

Let u = ln θ then Hence,

∫

1 u2 + c = sec 2 3x + c 2 2

du 1 = dθ θ

and dθ = θdu

1 ln θ u u2 2 dθ = ∫ ( θ du ) = ∫ u du = + c = ( ln θ ) + c 2 θ θ 2

6. Integrate 3 tan t with respect to t.

sin 2t

∫ 3 tan 2t dt = 3∫ cos 2t dt Hence, 3∫

Let u = cos 2t then

du = −2sin 2t dt

i.e.

dt =

− du 2sin 2t

sin 2t 3 1 3 ⎛ sin 2t ⎞ ⎛ −du ⎞ dt = 3∫ ⎜ ⎟⎜ ⎟ = − ∫ du = − ln u + c cos 2t 2 u 2 ⎝ u ⎠ ⎝ 2sin 2t ⎠


352

3 3 −1 = − ln cos 2t + c = ln ( cos 2t ) + c 2 2 =

∫

8. Evaluate

1 0

2

3x e

Let u = 2x 2 − 1

∫

Hence,

∫

1 0

( 2x −1)

( 2x −1) 2

3x e

dx =

∫

10. Evaluate

∫

dx = ∫ 3xe u

3x

1 0

( 4x

2

2

− 1)

5

dx =

i.e.

du 4x

du 3 u 3 3 ( 2x 2 −1) = ∫ e du = e u + c = e +c 4x 4 4 4

− 1)

5


du = 8x dx

then

3x

( 4x

du = 4x dx

3 ⎡ ( 2x 2 −1) ⎤ 1 3 1 −1 e = ⎡e − e ⎤⎦ = 1.763 ⎥⎦ 0 4 ⎣ 4 ⎢⎣

Let u = 4x 2 − 1

Hence,


then 2

3xe

( 2x −1)

3 ln sec 2t + c 2

dx = ∫

du 8x

dx =

i.e.

3x ⎛ du ⎞ 3 3 ⎛ u −4 ⎞ 3 3 −5 +c= − +c ⎟+c = − ⎟ = ∫ u du = ⎜ 4 5 ⎜ 4 u ⎝ 8x ⎠ 8 8 ⎝ −4 ⎠ 32u 32 ( 4x 2 − 1) 1

∫

Thus,

1 0

⎡ ⎤ 3x 3 ⎢ 1 ⎥ = − 3 ⎡ 1 − 1 ⎤ = 0.09259 dx = − 5 4 2 2 32 ⎢ ( 4x − 1) ⎥ 32 ⎢⎣ 34 (−1) 4 ⎥⎦ ( 4x − 1) ⎣ ⎦0

11. The electrostatic potential on all parts of a conducting circular disc of radius r is given by the V = 2πσ ∫

equation:

R

9 0

R 2 + r2

dR

Solve the equation by determining the integral.

Let u = R 2 + r 2

∫

R R +r 2

2

dR = ∫

then

du = 2R dR

i.e.

dR = 1 2

du 2R

R ⎛ du ⎞ 1 1u ⎜ ⎟ = ∫ u du = 1 + c = u + c = 2 u ⎝ 2R ⎠ 2 2 −

1 2

(R

2

+ r2 ) + c


353

Hence, V = 2πσ ∫

9 0

dR = 2πσ ⎡ ⎢⎣ 2 2 R +r R

(R

+ r 2 ) ⎤ = 2πσ ⎡ ⎥⎦ 0 ⎢⎣

(9

9

2

= 2πσ

+ r2 ) − r2 ⎤ ⎥⎦

2

{ ( 9 + r ) − r} 2

12. In the study of a rigid motor the following integration occurs:

2

Zr = ∫

∞ 0

( 2J + 1) e

− J(J +1) h 2 8 π2 I kT

dJ

Determine Zr for constant temperature T assuming h, I and k are constants.

Let u = J(J + 1) = J 2 + J

∫ ( 2J + 1) e

− J(J +1) h 2 8 π I kT 2

du = 2J + 1 dJ

then

−u h2

dJ = ∫ (2J + 1)e 8 π

2

IkT

dJ =

i.e. u h2

− 2 du = ∫ e 8 π I k T du = (2J + 1)

du 2J + 1 u h2

− 2 1 8π I k T +c e h2 − 2 8π I k T

8π2 I k T e = − h2 Thus, Zr = ∫

∞ 0

( 2J + 1) e

− J(J +1) h 2 8 π I kT 2

8π 2 I k T ⎡ ⎢e dJ = − h2 ⎢ ⎣

− J(J +1) h 2 8π I k T 2

π 0

⎧ ⎪ ⎨ ⎪⎩ 2 ε

then

⎫ ⎪ dθ ⎬ 2 2 ( a − x − 2ax cos θ ) ⎪⎭

a σ sin θ ⎛ du ⎞ a σ Hence, E = ⎜ ⎟= ∫ 2ε ∫ u ⎝ 2ax sin θ ⎠ 2ε 2

8π 2 I k T 8π2 I k T 0 1 − = [ ] h2 h2

a 2 σ sin θ

du = 2ax sin θ dθ 2

+c

∞

greater than a, and x is independent of θ. Show that E =

Let u = a 2 + x 2 − 2ax cos θ

8 π2 I k T

⎤ 8π2 I k T −∞ 0 ⎥ =− ⎡ e − e ⎤⎦ h2 ⎣ ⎥ ⎦0

= −

13. In electrostatics, E = ∫

− J(J +1) h 2

and

where a, σ and ε are constants, x is a 2σ εx

dθ =

du 2ax sin θ

⎛ 1 du a σ⎛ 1 ⎞ a σ ⎛ 1 ⎞⎜ u2 = ⎜ ⎟ ∫ u du = ⎜ ⎟⎜ 2ε ⎝ 2ax ⎠ 2ε ⎝ 2ax ⎠ ⎜ 1 2ax u ⎝ 2 2

=

aσ 2ε x

−

(a

2

1 2

2

⎞ ⎟ ⎟+c ⎟ ⎠

+ x 2 − 2ax cos θ ) + c


354

Hence, E =

i.e.

π aσ ⎡ 2 aσ ⎡ 2 a + x 2 − 2ax cos θ ⎤ = a + x 2 − 2ax cos π − a 2 + x 2 − 2ax cos 0 ⎤ ⎣ ⎦ ⎣ ⎦ 0 2ε x 2ε x

=

aσ ⎡ (a 2 + x 2 + 2ax) − (a 2 + x 2 − 2ax) ⎤ ⎣ ⎦ 2ε x

=

aσ ⎡ 2 ε x ⎢⎣

=

aσ aσ [(x + a) − (x − a)] = [ 2a ] 2ε x 2ε x

(x + a)

2

−

(x − a)

2

⎤ ⎥⎦

a2σ E= εx


355

CHAPTER 40 INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS EXERCISE 156 Page 399

2. Integrate 3cos 2 t with respect to t.

cos 2t = 2 cos 2 t − 1

∫ 3cos

Hence,

2

from which,

t dt =3∫

cos 2 t =

1 (1 + cos 2t ) 2

3 sin2t ⎞ 1 +c (1 + cos 2t ) dt = ⎜⎛ t + 2⎝ 2 ⎟⎠ 2

4. Integrate 2 cot 2 2t with respect to t.

∫ 2 cot

2

⎡ 1 ⎤ 2t dt = 2∫ ( cos ec2 2t − 1) dt = 2 ⎢ − cot 2t − t ⎥ + c = - (cot 2t + 2t) + c ⎣ 2 ⎦

5. Evaluate

∫

π/3 0

3sin 2 3x dx correct to 4 significant figures.

cos 2x = 1 − 2sin 2 x

Hence,

∫

π/3 0

cos 6x = 1 − 2sin 2 3x

and

3sin 3x dx = ∫ 2

π/3 0

from which,

1 sin 2 3x = (1 − cos 6x) 2

π/3

3 3⎡ sin 6x ⎤ (1 − cos 6x) dx = ⎢ x − 2 2⎣ 6 ⎥⎦ 0

⎡⎛ 6π ⎞ ⎤ sin ⎢ ⎜ ⎟ 3 π 3 − (0 − sin0) ⎥ = 3 ⎡ π ⎤ = π or 1.571 = ⎢⎜ − ⎥ ⎟ ⎢ ⎥ 2 2 ⎢⎜ 3 6 ⎟ ⎥ 2 ⎣3⎦ ⎢⎣⎝ ⎥ ⎠ ⎦

7. Evaluate

∫

1 0

∫

1 0

2 tan 2 2t dt correct to 4 significant figures.

2 tan 2t dt = ∫ 2

1 0

1

⎡ tan 2t ⎤ 2 ( sec 2t − 1) dt = 2 ⎢ − t⎥ ⎣ 2 ⎦0 2

⎡⎛ tan 2 ⎞ ⎛ tan 0 ⎞⎤ = 2 ⎢⎜ − 1⎟ − ⎜ − 0 ⎟ ⎥ = tan 2 – 2 = -4.185 ⎠ ⎝ 2 ⎠⎦ ⎣⎝ 2

(note that ‘tan 2’ means ‘tan 2 radians’) © 2006 John Bird. All rights reserved. Published by Elsevier.

356


2. Integrate 2 cos3 2x with respect to x.

∫ 2 cos

3

2x dx = 2∫ cos 2x cos 2 2x dx = 2 ∫ cos 2x (1 − sin 2 2x ) dx = 2∫ ( cos 2x − cos2x sin 2 2x ) dx ⎛ sin 2x sin 3 2x ⎞ = 2⎜ − ⎟+c 6 ⎠ ⎝ 2 using the algebraic substitution u = sin 2x

sin 3 2x +c = sin 2x − 3 3. Integrate 2sin 3 t cos 2 t with respect to t.

∫ 2sin

3

t cos 2 t dt = ∫ 2sin t sin 2 t cos 2 t dt = ∫ 2sin t (1 − cos 2 t ) cos 2 t dt ⎡ cos3 t cos5 t ⎤ = 2 ∫ ( sin t cos 2 t − sin t cos 4 t ) dt = 2 ⎢ − + ⎥+c 3 5 ⎦ ⎣

using the algebraic substitution u = cos t 2 2 = − cos 3 t + cos 5 t + c 3 5

5. Integrate 2sin 4 2θ with respect to θ.

∫

2 1 ⎛ 1 − cos 4θ ⎞ 2 2sin 4 2θ dθ = 2 ∫ ( sin 2 2θ ) dθ = 2∫ ⎜ ⎟ dθ = ∫ (1 − 2 cos 4θ + cos 4θ) dθ 2 2 ⎝ ⎠ 2

⎡ ⎛ 1 + cos8θ ⎞ ⎤ ⎟ ⎥ dθ ⎢1 − 2 cos 4θ + ⎜ 2 ⎝ ⎠⎦ ⎣

=

1 2∫

=

3θ sin 4θ sin 8θ 1 ⎡ sin 4θ θ sin 8θ ⎤ θ− + + +c = − + +c ⎢ ⎥ 4 4 32 2⎣ 2 2 16 ⎦

6. Integrate sin 2 t cos 2 t with respect to t.

∫ sin

2

1 1 ⎛ 1 − cos 2t ⎞ ⎛ 1 + cos 2t ⎞ 2 t cos 2 t dt = ∫ ⎜ ⎟⎜ ⎟ dt = ∫ (1 − cos 2t) dt = ∫ 2 2 4 4 ⎝ ⎠⎝ ⎠

⎡ ⎛ 1 + cos 4t ⎞ ⎤ ⎟ ⎥ dt ⎢1 − ⎜ 2 ⎠⎦ ⎣ ⎝


357

=

1 ⎛ 1 cos 4t ⎞ 1 ⎡ t sin 4t ⎤ +c ⎜ − ⎟ dt = ⎢ − ∫ 4 ⎝2 2 ⎠ 4 ⎣2 8 ⎦⎥

=

t 1 − sin 4t + c 8 32


358


2. Integrate 2 sin 3x sin x with respect to x.

1

∫ 2sin 3x sin x dx = 2∫ − 2 ( cos 4x − cos 2x ) dx

from 9, page 398 of textbook

sin 2x sin 4x ⎡ sin 4x sin 2x ⎤ − + c = − +c = −⎢ 2 4 2 ⎥⎦ ⎣ 4

4. Integrate

1 cos 4θ sin 2θ with respect to θ. 2

1

1

1

∫ 2 cos 4θ sin 2θ dθ = 2 ∫ 2 (sin 6θ − sin 2θ) dθ =

6. Evaluate

∫

1 0

∫

1 0


1 ⎛ cos 2θ cos 6θ ⎞ 1 ⎡ cos 6θ cos 2θ ⎤ − + +c = ⎜ − ⎟+c ⎢ ⎥ 4⎝ 2 6 ⎠ 4⎣ 6 2 ⎦

2sin 7t cos 3t dt 1

1 ⎡ cos10t cos 4t ⎤ 2sin 7t cos 3t dt = 2∫ (sin10t + sin 4t) dt = ⎢ − − 2 10 4 ⎥⎦ 0 ⎣


⎛ cos10 cos 4 ⎞ ⎛ cos 0 cos 0 ⎞ − − = ⎜− ⎟−⎜− ⎟ 10 4 ⎠ ⎝ 10 4 ⎠ ⎝

= (0.24732) – (-0.35) = 0.5973 7. Evaluate −4∫ −4 ∫

π/3 0

π/3 0

sin 5θ sin 2θ dθ

sin 5θ sin 2θ dθ = −4 ∫

π/3 0

−

1 [cos 7θ − cos 3θ] dθ 2


⎡⎛ 7π 3π ⎞ ⎤ π/3 ⎢⎜ sin 3 sin 3 ⎟ ⎥ ⎡ sin 7θ sin 3θ ⎤ = 2⎢ 2 (0 0) − = − − − ⎢ ⎥ = 0.2474 ⎜ ⎟ 3 ⎥⎦ 0 3 ⎟ ⎣ 7 ⎢⎜ 7 ⎥ ⎠ ⎣⎢⎝ ⎦⎥


359


∫ ( 4 − x ) dx 2

∫ ( 4 − x ) dx = ∫ 2

22 ( 2 − x ) = 2 sin −1 x2 + x2 2

2

= 2sin −1

4. Determine

x x + 2 2

(2

2

− x2 )


(4 − x ) + c 2

∫ (16 − 9t ) dt 2

⎡ ⎛ 16 2 ⎞ ⎤ ⎢9 ⎜ 9 − t ⎟ ⎥ dt = ∫ ⎠⎦ ⎣ ⎝

∫ (16 − 9t ) dt = ∫ 2

⎡⎛ 4 ⎞ 2 2 ⎤ 9 ⎢⎜ ⎟ − t ⎥ dt ⎣⎢⎝ 3 ⎠ ⎦⎥

⎡ ⎛ 4 ⎞2 ⎤ 2 ⎢⎜ ⎟ ⎤⎥ t t ⎡⎛ 4 ⎞ 3⎠ ⎝ −1 2 ⎥ ⎢ + sin = 3∫ ⎢⎜ ⎟ − t ⎥ + c ⎢ 2 ⎛ 4 ⎞ 2 ⎣⎢⎝ 3 ⎠ ⎦⎥ ⎥ ⎜ ⎟ ⎢ ⎥ ⎝3⎠ ⎣ ⎦ 8 3t 3t = sin −1 + 3 4 2


⎡⎛ 4 ⎞ 2 2 ⎤ ⎢⎜ ⎟ − t ⎥ + c ⎢⎣⎝ 3 ⎠ ⎥⎦

2 ⎤ 8 −1 3t t 2 ⎡⎛ 4 ⎞ 8 3t t + = sin 3 ⎢⎜ ⎟ − t 2 ⎥ + c = sin −1 + (42 − 9t 2 ) + c 3 4 2 3 4 2 ⎣⎢⎝ 3 ⎠ ⎦⎥

=

2

2

0

∫ ( 9 − 4x ) dx = ∫ 0

( 16 − 9t ) + c

∫ ( 9 − 4x ) dx 1

6. Evaluate

1

8 −1 3t t + sin 3 4 2

2

1 0

⎡⎛ 3 ⎞ 2 ⎤ 1 ⎡ ⎛9 2 ⎞⎤ 2 − = − 4 x dx 2 x ⎢ ⎥ dx ⎜ ⎟ ⎜ ⎟ ⎢ 4 ⎥ ∫ 0 ⎢⎝ 2 ⎠ ⎠⎦ ⎣ ⎝ ⎣ ⎦⎥

⎡ ⎛ 3 ⎞2 ⎢⎜ ⎟ x x 2 = 2 ⎢ ⎝ ⎠ sin −1 + ⎢ 2 ⎛3⎞ 2 ⎜ ⎟ ⎢ ⎝2⎠ ⎣

1

⎤ 2 ⎡⎛ 3 ⎞ ⎤⎥ 2 ⎥ x − ⎢⎜ ⎟ ⎥ ⎢⎣⎝ 2 ⎠ ⎥⎦ ⎥ ⎥ ⎦0


⎡⎛ 9 2 1 ⎤ ⎞ 1.25 ⎟ − (0 + 0) ⎥ = 2.760 = 2 ⎢⎜ sin −1 + 3 2 ⎠ ⎣⎝ 8 ⎦


360


5

∫ 16 + 9θ

2

5

∫ 16 + 9θ

dθ = ∫

2

dθ

5 5 1 dθ = ∫ dθ 2 9 ⎛4⎞ ⎛ 16 2 ⎞ 2 9⎜ + θ ⎟ ⎜ ⎟ +θ ⎝9 ⎠ ⎝3⎠ ⎛ ⎞ ⎜ θ ⎟ 5 1 ⎟+c = ⎜ tan −1 9⎜⎛4⎞ ⎛4⎞⎟ ⎜ ⎟⎟ ⎜⎜3⎟ ⎝3⎠⎠ ⎝⎝ ⎠

=

4. Evaluate

∫

3 0

∫

3 0


5 3θ +c tan −1 12 4

5 dx 4 + x2 3

3 5 1 x⎤ ⎡1 dx = 5∫ 2 dx = 5 ⎢ tan ⎥ 2 2 0 4+x 2 +x 2⎦0 ⎣2

=


5 ⎡ −1 3 ⎤ tan − tan −1 0 ⎥ ⎢ 2⎣ 2 ⎦

= 2.457


361

EXERCISE 161 Page 405 2. Find

3

∫

( 9 + 5x ) 2

3

∫

( 9 + 5x ) 2

dx

3

dx = ∫

⎡ ⎛9 2 ⎞⎤ ⎢5 ⎜ 5 + x ⎟ ⎥ ⎠⎦ ⎣ ⎝

=

=

4. Find

∫ ( 4t

2

∫ ( 4t

2

3

dx = ∫

⎡⎛ 3 ⎞ 2 ⎤ 2 5 ⎢⎜ ⎟ +x ⎥ ⎣⎢⎝ 5 ⎠ ⎦⎥

dx

3 x sinh −1 +c ⎛ 3 ⎞ 5 ⎜ ⎟ ⎝ 5⎠

3 5

sinh −1

5 x+c 3

+ 25 ) dt ⎡ ⎛ 2 25 ⎞ ⎤ ⎢ 4 ⎜ t + 4 ⎟ ⎥ dt = ∫ ⎠⎦ ⎣ ⎝

+ 25 ) dt = ∫

⎡ 2 ⎛ 5 ⎞2 ⎤ 4 ⎢ t + ⎜ ⎟ ⎥ dt ⎝ 2 ⎠ ⎦⎥ ⎣⎢

⎡ ⎛ 5 ⎞2 ⎤ 2 ⎥ ⎢⎜ ⎟ t t ⎡ 2 ⎛ 5⎞ ⎤⎥ 2 + = 2 ⎢ ⎝ ⎠ sinh −1 ⎢t + ⎜ ⎟ ⎥ + c ⎢ 2 ⎛ 5 ⎞ 2 ⎣⎢ ⎝ 2 ⎠ ⎦⎥ ⎥ ⎜ ⎟ ⎢ ⎥ ⎝2⎠ ⎣ ⎦ =

2 ⎡ 25 2t t ⎛5⎞ ⎤ sinh −1 + 4 ⎢t2 + ⎜ ⎟ ⎥ + c 4 5 2 ⎝ 2 ⎠ ⎥⎦ ⎢⎣

=

25 2t t ⎡⎣ 4t 2 + 52 ⎤⎦ + c sinh −1 + 4 5 2

=

25 2t t ⎡ 4t 2 + 25 ⎤⎦ + c sinh −1 + 4 5 2 ⎣

1

2

0

∫ (16 + 9θ ) dθ = ∫ 0


∫ (16 + 9θ ) dθ

6. Evaluate

1


2

1 0

⎡⎛ 4 ⎞ 2 ⎤ 1 ⎡ ⎛ 16 2 ⎞ ⎤ 2 ⎢9 ⎜ 9 + θ ⎟ ⎥ dθ = ∫ 0 9 ⎢⎜ 3 ⎟ + θ ⎥ dθ ⎠⎦ ⎣ ⎝ ⎢⎣⎝ ⎠ ⎥⎦


362

⎡ ⎛ 4 ⎞2 ⎢⎜ ⎟ θ θ 3 = 3 ⎢ ⎝ ⎠ sinh −1 + ⎢ 2 ⎛4⎞ 2 ⎜ ⎟ ⎢ ⎝3⎠ ⎣

1

⎤ 2 ⎡⎛ 4 ⎞ ⎤⎥ 2 ⎥ + θ ⎢⎜ ⎟ ⎥ ⎢⎣⎝ 3 ⎠ ⎥⎦ ⎥ ⎥ ⎦0


⎡⎛ 16 3 1 ⎞ ⎛ 16 ⎞⎤ = 3 ⎢⎜ sinh −1 + 2.777777 ⎟ − ⎜ sinh 0 + 0 ⎟ ⎥ 4 2 ⎠ ⎝ 18 ⎠⎦ ⎣⎝ 18 = 4.348


363

EXERCISE 162 Page 407 3

∫

2. Find

( 4x

3

∫

( 4x

2

− 9)

2

− 9)

∫ ( 4θ

∫ ( 4θ

2

⎡ ⎛ 2 9 ⎞⎤ ⎢4 ⎜ x − 4 ⎟⎥ ⎠⎦ ⎣ ⎝

dx = ∫

3 x +c cosh −1 2 ⎛3⎞ ⎜ ⎟ ⎝2⎠

3 2 ⎡ ⎛3⎞ ⎤ 2 ⎢x2 − ⎜ ⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ ⎢⎣

dx


3 2x +c cosh −1 2 3

=

2

3

dx = ∫

=

4. Find

dx

− 25 ) dθ

− 25 ) dθ = ∫

⎡ 2 ⎛ 5 ⎞2 ⎤ ⎡ ⎛ 2 25 ⎞ ⎤ ⎢ 4 ⎜ θ − 4 ⎟ ⎥ dθ = ∫ 2 ⎢θ − ⎜ 2 ⎟ ⎥ dθ ⎠⎦ ⎝ ⎠ ⎥⎦ ⎣ ⎝ ⎢⎣

⎡ ⎢ θ = 2⎢ ⎢2 ⎢ ⎣

2 ⎤ ⎛5⎞ 2 ⎥ ⎜ ⎟ ⎡ 2 ⎛5⎞ ⎤ ⎝2⎠ θ ⎥ +c cosh −1 ⎢θ − ⎜ ⎟ ⎥ − 2 ⎛ 5 ⎞⎥ ⎝ 2 ⎠ ⎥⎦ ⎢⎣ ⎜ ⎟⎥ ⎝ 2 ⎠⎦


25 ⎞ 25 2θ ⎛ = θ ⎜ θ 2 − ⎟ − cosh −1 +c 4 ⎠ 4 5 ⎝

6. Evaluate

∫

3 2

(t

2

∫

3 2

(t

− 4 ) dt = ∫

3 2

2

− 4 ) dt ⎡ ( t − 2 ) dt = ⎢ 2t ⎣ 2

2

3

⎤ 22 − − t 2 ( ) 2 cosh −1 2t ⎥ ⎦2 2

2

3⎞ ⎛3 =⎜ 5 − 2 cosh −1 ⎟ − ( 0 − 2 cosh −1 1) 2⎠ ⎝2 = 1.429


364

CHAPTER 41 INTEGRATION USING PARTIAL FRACTIONS EXERCISE 163 Page 409 2. Determine

∫ (x

4(x − 4) dx 2 − 2x − 3)

4( x − 4) 5 1 = − 2 x − 2 x − 3 ( x + 1) ( x − 3) Hence,

∫ (x

from question 2, Exercise 13, page 19.

⎛ 5 4(x − 4) 1 ⎞ − dx = ∫ ⎜ ⎟ dx = 5 ln(x + 1) − ln(x − 3) + c 2 − 2x − 3) ⎝ (x + 1) (x − 3) ⎠

or ln(x + 1) − ln(x − 3) + c 5

4. Determine

∫

x 2 + 9x + 8 dx x2 + x − 6

2 6 x2 + 9x + 8 = 1+ + 2 x + x−6 ( x + 3) ( x − 2)

∫

x 2 + 9x + 8 dx = ∫ x2 + x − 6

⎛ 2 6 ⎞ + ⎜1 + ⎟ dx = x + 2 ln(x + 3) + 6 ln(x - 2) + c ⎝ (x + 3) (x − 2) ⎠ 2

6. Evaluate

Let

from question 5, Exercise 13, page 19.

x + ln ( x + 3 ) + ln ( x − 2 ) + c

or

∫

4 3

⎧ (x + 1)5 ⎫ or ln ⎨ ⎬ + c by the laws of logarithms ⎩ (x − 3) ⎭

6

or

{

x + ln (x + 3)2 ( x − 2 )

6

}+c

x 2 − 3x + 6 dx correct to 4 significant figures. x(x − 2)(x − 1)

x 2 − 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)

x 2 − 3x + 6 = A(x − 2)(x − 1) + Bx(x − 1) + Cx(x − 2)

then 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,

Thus,

x 2 − 3x + 6 3 2 4 ≡ + − x(x − 2)(x − 1) x (x − 2) (x − 1)

∫

4 3

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

⎛3 2 4 ⎞ 4 − ⎜ + ⎟ dx = [3ln x + 2 ln(x − 2) − 4 ln(x − 1)] 3 ⎝ x (x − 2) (x − 1) ⎠


365

= (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:

Let

∫

1 0

(x − k) dx = 0 (3x + 1)(x + 1)

(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 2 -k= A 3 3

i.e.

A=

1 Let x = − , 3

∫

1 0

−

(x − k) dx = ∫ 0 (3x + 1)(x + 1)

1

1+ k 2 3⎛ 1 1 3 ⎞ ⎜− −k⎟ = − − k 2⎝ 3 2 2 ⎠

1+ k ⎤ ⎡ 1 3 − − k ⎢ 2 2 ⎥ + 2 ⎥ dx = 0 ⎢ ⎢ (3x + 1) (x + 1) ⎥ ⎣ ⎦ 1

⎡1 ⎛ 1 3 ⎞ ⎤ ⎛ 1+ k ⎞ ⎢ 3 ⎜ − 2 − 2 k ⎟ ln(3x + 1) + ⎜ 2 ⎟ ln(x + 1) ⎥ = 0 ⎠ ⎝ ⎠ ⎣ ⎝ ⎦0

i.e.

⎡1 ⎛ 1 3 ⎞ ⎤ ⎛ 1+ k ⎞ ⎢ 3 ⎜ − 2 − 2 k ⎟ ln 4 + ⎜ 2 ⎟ ln 2 ⎥ − [ 0 + 0] = 0 ⎠ ⎝ ⎠ ⎣ ⎝ ⎦ i.e.

1⎛1 3 ⎞ ⎛ 1+ k ⎞ ⎜ ⎟ ln 2 = ⎜ + k ⎟ (2 ln 2) 3⎝ 2 2 ⎠ ⎝ 2 ⎠

i.e.

1 k 1 + = +k 2 2 3

from which, 1 k = 6 2

and

since ln 4 = ln 22 = 2 ln 2 1 1 k − =k− 2 3 2

from which,

k=

1 3

⎛ ⎞ 1 9. The velocity constant k of a chemical reaction is given by: kt = ∫ ⎜ ⎟ dx ⎝ (3 − 0.4x)(2 − 0.6x) ⎠ ⎧ 2(3 − 0.4x) ⎫ where x = 0 when t = 0. Show that: kt = ln ⎨ ⎬ ⎩ 3(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) © 2006 John Bird. All rights reserved. Published by Elsevier.

366

Let x =

3 = 7.5 , 0.4

1 = A(2 – 4.5)

i.e.

A=

Let x =

2 1 10 =3 = , 0.6 3 3

4⎞ ⎛ 1 = B⎜ 3 − ⎟ 3⎠ ⎝

i.e.

B=

1 = −0.4 −2.5 1 3 = = 0.6 5 5 3

⎛ ⎞ ⎛ −0.4 1 0.6 ⎞ Thus, kt = ∫ ⎜ + ⎟ dx = ∫ ⎜ ⎟ dx ⎝ (3 − 0.4x)(2 − 0.6x) ⎠ ⎝ (3 − 0.4x) (2 − 0.6x) ⎠ =

−0.4 0.6 ln(3 − 0.4x) + ln(2 − 0.6x) + c −0.4 −0.6

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.

⎧ 2(3 − 0.4x) ⎫ kt = ln ⎨ ⎬ ⎩ 3(2 − 0.6x) ⎭


367


Let

5x 2 − 30x + 44 dx (x − 2)3

∫

5x 2 − 30x + 44 A B C A(x − 2) 2 + B(x − 2) + C ≡ + + = (x − 2)3 (x − 2) (x − 2) 2 (x − 2)3 (x − 2)3

5x 2 − 30x + 44 = A(x − 2) 2 + B(x − 2) + C

Hence, Let x = 2,

4=C

Equating x 2 coefficients,

5=A

Equating x coefficients, Thus,

-30 = -4A + B

i.e. B = -30 + 20 = -10

⎛ 5 5x 2 − 30x + 44 10 4 ⎞ dx = ∫ ⎜ dx − + 3 2 3 ⎟ (x − 2) ⎝ (x − 2) (x − 2) (x − 2) ⎠

∫

= 5 ln(x – 2) +

10 2 − +c (x − 2) (x − 2)2

using the algebraic substitution u = x – 2 in the latter two integrals

4. Evaluate

∫

7 6

18 + 21x − x 2 dx correct to 4 significant figures. (x − 5)(x + 2) 2

A(x + 2) 2 + B(x − 5) ( x + 2 ) + C ( x − 5 ) 18 + 21x − x 2 A B C ≡ + + = Let (x − 5)(x + 2) 2 (x − 5) (x + 2) (x + 2) 2 (x − 5)(x + 2) 2 18 + 21x − x 2 = A(x + 2) 2 + B(x − 5)(x + 2) + C(x − 5)

Hence, Let x = 5,

98 = 49A

i.e.

Let x = -2,

-28 = -7C

i.e. C = 4

Equating x 2 coefficients, Thus,

∫

7 6

-1 = A + B

i.e.

A=2

B = -3

7⎛ 18 + 21x − x 2 2 3 4 ⎞ dx dx = − + ⎜ 2 2 ⎟ ∫ 6 (x − 5)(x + 2) ⎝ (x − 5) (x + 2) (x + 2) ⎠ 7

⎡ 4 ⎤ = ⎢ 2 ln(x − 5) − 3ln(x + 2) − (x + 2) ⎥⎦ 6 ⎣ 4⎞ ⎛ 4⎞ ⎛ = ⎜ 2 ln 2 − 3ln 9 − ⎟ − ⎜ 2 ln1 − 3ln 8 − ⎟ = 1.089 9⎠ ⎝ 8⎠ ⎝

5. Show that

∫

1 0

⎛ 4t 2 + 9t + 8 ⎞ dt = 2.527, correct to 4 significant figures. ⎜ 2 ⎟ ⎝ (t + 2)(t + 1) ⎠


368

Let

A(t + 1) 2 + B(t + 2) ( t + 1) + C ( t + 2 ) 4t 2 + 9t + 8 A B C ≡ + + = (t + 2)(t + 1) 2 (t + 2) (t + 1) (t + 1) 2 (t + 2)(t + 1) 2 4t 2 + 9t + 8 = A(t + 1) 2 + B(t + 2)(t + 1) + C(t + 2)

Hence, Let x = -2,

6=A

Let x = -1,

3=C


4=A+B

Thus,

∫

1 0

i.e.

B = -2

1 ⎛ ⎛ 4t 2 + 9t + 8 ⎞ 6 2 3 ⎞ dt = ∫ ⎜ dt − + ⎜ 2 ⎟ 2 ⎟ 0 ⎝ (t + 2) (t + 1) (t + 1) ⎠ ⎝ (t + 2)(t + 1) ⎠ 1

⎡ 3 ⎤ = ⎢ 6 ln(t + 2) − 2 ln(t + 1) − (t + 1) ⎥⎦ 0 ⎣ 3⎞ ⎛ 3⎞ ⎛ = ⎜ 6 ln 3 − 2 ln 2 − ⎟ − ⎜ 6 ln 2 − 2 ln1 − ⎟ 2⎠ ⎝ 1⎠ ⎝ = (3.70538) – (1.15888) = 2.546


369


∫

x 2 − x − 13 dx ( x2 + 7) ( x − 2)

(Ax + B)(x − 2) + C ( x 2 + 7 ) x 2 − x − 13 Ax + B C ≡ + = Let 2 ( x + 7 ) ( x − 2) ( x 2 + 7 ) ( x − 2) ( x2 + 7) ( x − 2)

x 2 − x − 13 = (Ax + B)(x − 2) + C ( x 2 + 7 )

Hence,

Let x = 2,

-11 = 11C

i.e.

C = -1


1=A+C

i.e.

A=2

Equating x coefficients,

-1 = -2A + B i.e.

B=3

∫

Hence,

x 2 − x − 13 dx = ( x 2 + 7 ) ( x − 2) ∫

=

⎛ 2x + 3 1 ⎞ ⎜ 2 ⎟ dx = ∫ − ⎜ ( x + 7 ) (x − 2) ⎟ ⎝ ⎠

∫ (x

2x 3 dx + ∫ + 7) x2 + 7

= ln ( x 2 + 7 ) +

2. Evaluate

∫

6 5

( )

2

3 7

tan −1

x 7

2

⎛ 2x 3 1 ⎞ ⎜ 2 ⎟ dx + 2 − ⎜ ( x + 7 ) ( x + 7 ) (x − 2) ⎟ ⎝ ⎠

dx − ∫

1 dx (x − 2)

− ln(x − 2) + c

6x − 5 dx correct to 4 significant figures. (x − 4) ( x 2 + 3)

2 6x − 5 A Bx + C A ( x + 3) + (Bx + C)(x − 4) ≡ + = Let ( x − 4 ) ( x 2 + 3) (x − 4) ( x 2 + 3) ( x − 4 ) ( x 2 + 3)

6x − 5 = A ( x 2 + 3) + (Bx + C)(x − 4)

Hence,

Let x = 4,

19 = 19A

i.e.

A=1


0=A+B

i.e.

B = -1

Equating x coefficients,

6 = -4B + C

i.e.

C=2

Thus,

∫

6 5

6 6x − 5 dx = ∫ 2 5 (x − 4) ( x + 3)

⎛ 1 6 2−x ⎞ ⎜ ⎟ dx = ∫ + 2 5 ⎜ (x − 4) ( x + 3) ⎟ ⎝ ⎠

⎛ 2 ⎜ 1 + ⎜ (x − 4) ⎜ x2 + 3 ⎝


( )

2

⎞ x ⎟ − 2 dx ( x + 3) ⎟⎟⎠

370

6

2 x 1 ⎡ ⎤ tan −1 = ⎢ ln(x − 4) + − ln ( x 2 + 3) ⎥ 3 3 2 ⎣ ⎦5

2 6 1 2 5 1 ⎛ ⎞ ⎛ ⎞ tan −1 tan −1 − ln 39 ⎟ − ⎜ ln1 + − ln 28 ⎟ = ⎜ ln 2 + 3 3 2 3 3 2 ⎝ ⎠ ⎝ ⎠ = (0.35065) – (-0.23736) = 0.5880

5. Show that

∫

2 1

⎛ 2 + θ + 6θ2 − 2θ3 ⎞ ⎜ ⎟ dθ = 1.606, correct to 4 significant figures. ⎜ θ2 ( θ2 + 1) ⎟ ⎝ ⎠

2 2 2 2 + θ + 6θ2 − 2θ3 A B Cθ + D Aθ ( θ + 1) + B ( θ + 1) + ( Cθ + D ) θ ≡ + 2+ 2 = Let θ θ ( θ + 1) θ2 ( θ2 + 1) θ2 ( θ2 + 1)

2 + θ + 6θ2 − 2θ3 = Aθ ( θ2 + 1) + B ( θ2 + 1) + ( Cθ + D ) θ 2

Hence, Let θ = 0,

2=B

Equating θ3 coefficients,

-2 = A + C

Equating θ2 coefficients,

6=B+D

Equating θ coefficients,

1=A

From equation (1),

C = -3

Hence,

∫

2 1

⎛ 2 + θ + 6θ2 − 2θ3 ⎞ 2 ⎜ ⎟ θ = d ∫ 1 ⎜ θ2 ( θ2 + 1) ⎟ ⎝ ⎠

(1) i.e.

D=4

⎛1 2 2 ⎛1 4 − 3θ ⎞ 2 4 3θ ⎞ ⎜ + 2+ 2 ⎟ dθ = ∫ ⎜ + 2 + 2 − 2 ⎟ dθ 1 ⎜ θ θ ( θ + 1) ⎟ ⎝ θ θ θ +1 θ +1⎠ ⎝ ⎠ 2

2 3 ⎡ ⎤ = ⎢ ln θ − + 4 tan −1 θ − ln ( θ2 + 1) ⎥ θ 2 ⎣ ⎦1

= ( ln 2 − 1 + 4 tan −1 2 − 1.5ln 5 ) − ( ln1 − 2 + 4 tan −1 1 − 1.5ln 2 ) = (1.70759) – (0.10187) = 1.606


371

CHAPTER 42 THE t = tan θ/2 SUBSTITUTION EXERCISE 166 Page 415 2. Determine

∫

∫

dx 1 − cos x + sin x

dx = 1 − cos x + sin x ∫

Let

2dt 2dt 2 2dt 1 1+ t 1+ t2 dx =∫ =∫ 2 =∫ 2 2 2 1− t 2t (1 + t ) − (1 − t ) + 2t 2t + 2t t(t + 1) 1− + 1+ t2 1+ t2 1+ t2

1 A B A(t + 1) + Bt = + = t(t + 1) t t + 1 t(t + 1)

Hence,

1 = A(t + 1) + Bt

Let t = 0,

1=A

Let t = -1,

1 = -B

Thus,

dx

i.e.

B = -1 ⎛1

1

1 ⎞

∫ 1 − cos x + sin x = ∫ t(t + 1) dx = ∫ ⎜⎝ t − t + 1 ⎟⎠ dx x ⎫ ⎧ ⎪ tan 2 ⎪ ⎧ t ⎫ = ln t – ln(t + 1) + c = ln ⎨ ⎬+c ⎬ + c = ln ⎨ ⎩1 + t ⎭ ⎪ 1 + tan x ⎪ 2⎭ ⎩

4. Determine

dx

∫

dx 3sin x − 4 cos x

∫ 3sin x − 4 cos x = ∫

2dt 2dt 2 2dt 1+ t 1+ t2 = =∫ 2 2 ∫ 2 6t − 4 + 4t 4t + 6t − 4 ⎛ 2t ⎞ ⎛ 1 − t ⎞ − 4⎜ 3⎜ 2 2 ⎟ 2 ⎟ 1+ t ⎝ 1+ t ⎠ ⎝ 1+ t ⎠ =

Let

∫ 2t

2

dt dt =∫ + 3t − 2 (2t − 1)(t + 2)

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 2 = 2.5 5


372

Let t = -2,

Thus,

1 = -5B

dx

∫ 3sin x − 4 cos x = ∫

i.e.

B= −

dt = (2t − 1)(t + 2) ∫

1 5

2 1 5 − 5 dt = 1 ln(2t − 1) − 1 ln(t + 1) + c (2t − 1) (t + 2) 5 5

x ⎧ ⎫ 1 ⎪ 2 tan 2 − 1 ⎪ 1 ⎧ 2t − 1 ⎫ = ln ⎨ ⎬+c ⎬ + c = ln ⎨ x 5 ⎪ 5 ⎩ t+2 ⎭ tan + 2 ⎪ 2 ⎩ ⎭


373


dθ

∫ 5 + 4sin θ

dθ

∫ 5 + 4sin θ = ∫

2dt 2dt 2 2dt 2dt 1+ t 1+ t2 =∫ =∫ 2 =∫ 2 8 5t + 8t + 5 ⎛ 2t ⎞ ⎛ ⎞ 5 (1 + t ) + 4(2t) 5 + 4⎜ 5 ⎜ t 2 + t + 1⎟ 2 ⎟ 2 5 ⎝ 1+ t ⎠ ⎝ ⎠ 1+ t

=

2 5∫

2 4⎞ ⎛ t+ ⎟ ⎜ dt 5 + c = 2 tan −1 ⎛ 5t + 4 ⎞ + c = 5 tan −1 ⎜ ⎜ ⎟ 2 2 3 3 ⎟⎟ 3 ⎝ 3 ⎠ ⎛ 4⎞ ⎛3⎞ ⎜ + + t ⎜ ⎟ ⎜ ⎟ 5 ⎝ 5 ⎠ ⎝ 5⎠ ⎝5⎠ θ ⎛ ⎞ 5 tan + 4 ⎟ ⎜ 2 2 = tan −1 ⎜ ⎟+c 3 3 ⎜ ⎟ ⎝ ⎠

3. Determine

dp

∫ 3 − 4sin p + 2 cos p

dp

∫ 3 − 4sin p + 2 cos p = ∫

2dt 2dt 2 1+ t 1+ t2 = ∫ 3 (1 + t 2 ) − 4(2t) + 2 (1 − t 2 ) 2 ⎛ 2t ⎞ ⎛ 1 − t ⎞ + 3 − 4⎜ 2 ⎜ 2 ⎟ 2 ⎟ ⎝ 1+ t ⎠ ⎝ 1+ t ⎠ 1+ t2 2dt 2dt 2dt 2dt =∫ 2 =∫ =∫ 2 2 2 − 8t + 2 − 2t t − 8t + 5 (t − 4) − 11 ( t − 4 ) − 11

=

∫ 3 + 3t

=

⎧⎪ t − 4 − 11 ⎫⎪ ⎧⎪ t − 4 − 11 ⎫⎪ 2 1 ln ⎨ ln ⎨ ⎬+c = ⎬ using partial fractions (see 2 11 ⎩⎪ t − 4 + 11 ⎭⎪ 11 ⎩⎪ t − 4 + 11 ⎪⎭

2

( )

2

Problem 9, page 411 of textbook)

=

5. Show that

∫

p ⎧ ⎫ ⎪ tan 2 − 4 − 11 ⎪ ln ⎨ ⎬+c 11 ⎪ tan p − 4 + 11 ⎪ ⎩ ⎭ 2

1

t⎫ ⎧ 2 + tan ⎪ dt 1 2⎪ = ln ⎨ ⎬ 1 + 3cos t 2 2 ⎪ 2 − tan t ⎪ ⎩ 2⎭


374

∫

dt = 1 + 3cos t ∫

2dt 2dt 2 2dt 2dt 1+ t 1+ t2 = =∫ =∫ 2 2 2 2 ∫ 2 1(1 + t ) + 3(1 − t ) 1 + t + 3 − 3t 4 − 2t 2 ⎛ 1− t ⎞ 1+ 3⎜ 2 ⎟ 1+ t2 ⎝ 1+ t ⎠ =

∫

dt = 2 − t2 ∫

dt

( 2)

2

− t2

=

⎧⎪ 2 + t ⎫⎪ ln ⎨ ⎬ + c using partial 2 2 ⎩⎪ 2 − t ⎪⎭ 1

fractions (see problem 11, page 411 of textbook) t⎫ ⎧ 2 tan + ⎪ 1 2⎪+c = ln ⎨ ⎬ 2 2 ⎪ 2 − tan t ⎪ ⎩ 2⎭

7. Show that

∫

dθ = 2 + cos θ ∫

∫

π/2 0

dθ π = 2 + cos θ 3 3

2dt 2dt 2 2dt 2dt 1+ t 1+ t2 = =∫ =∫ 2 2 2 2 2 ∫ 2 2(1 + t ) + (1 − t ) 2 + 2t + 1 − t t +3 ⎛ 1− t ⎞ 2+⎜ 2 2 ⎟ 1+ t ⎝ 1+ t ⎠ = 2∫

dt t2 +

( 3)

2

θ⎞ ⎛ tan ⎟ ⎜ 2 t 2 2 = = tan −1 tan −1 ⎜ ⎟ 3 3 3 ⎜ 3 ⎟ ⎝ ⎠

π/ 2

θ⎤ ⎡⎛ π⎞ ⎤ ⎡ tan ⎥ tan ⎟ ⎢ ⎥ ⎜ π/2 dθ 2 ⎢ −1 2 tan 0 ⎛ ⎞ −1 2 4 − tan −1 = = Hence, ∫ tan tan ⎢ ⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎥ 0 2 + cos θ 3⎢ 3 ⎥ 3 ⎢⎜ 3 ⎟ ⎝ 3 ⎠⎥ ⎢⎣⎝ ⎣ ⎦0 ⎠ ⎦⎥ = (Using a calculator on degrees, tan −1

π 2 ⎡ −1 1 ⎤ 2 ⎛π⎞ − 0⎥ = ⎜ ⎟= ⎢ tan 3⎣ 3 3⎝6⎠ 3 3 ⎦

1 π = 30° = rad ) 6 3

π 1 ⎛ ⎞ (since AD = Note that tan 30° ⎜ i.e. rad ⎟ = 6 3 ⎝ ⎠


3 by Pythagoras)

375

CHAPTER 43 INTEGRATION BY PARTS EXERCISE 168 Page 420

4x

∫e

2. Determine

4x

∫e

3x

3x

dx

dx = ∫ 4xe −3x dx

du = 4 and du = 4 dx dx

Let u = 4x then and dv = e −3x dx

Hence,

∫ 4xe

−3x

⎛ 1 ⎞ dx = (4x) ⎜ − e −3x ⎟ − ∫ ⎝ 3 ⎠

from which, v =

∫e

−3x

1 dx = − e−3x 3

⎛ 1 −3x ⎞ ⎜ − e ⎟ 4 dx ⎝ 3 ⎠

4 4 4 4⎛ 1 ⎞ = − xe −3x + ∫ e −3x dx = − xe −3x + ⎜ − e−3x ⎟ + c 3 3 3 3⎝ 3 ⎠ 4 1⎞ 4 4 ⎛ = − xe −3x − e−3x + c = − e −3x ⎜ x + ⎟ + c 3 3⎠ 3 9 ⎝

∫ 5θ cos 2θ dθ

4. Determine

Let u = 5θ then

du = 5 and du = 5 dθ dθ

and dv = cos 2θ dθ from which, v = ⎛1

⎞

1

∫ cos 2θ dθ = 2 sin 2θ

⎛1

⎞

∫ 5θ cos 2θ dθ = ( 5θ ) ⎜⎝ 2 sin 2θ ⎟⎠ − ∫ ⎜⎝ 2 sin 2θ ⎟⎠ 5dθ

6. Evaluate

=

5 5 5 5⎛ 1 ⎞ θ sin 2θ − ∫ sin 2θ dθ = θ sin 2θ − ⎜ − cos 2θ ⎟ + c 2 2 2 2⎝ 2 ⎠

=

5⎛ 1 5 5 ⎞ θ sin 2θ + cos 2θ + c = ⎜ θ sin 2θ + cos 2θ ⎟ + c 2⎝ 2 2 4 ⎠

∫

2 0

Let u = 2x then and dv = e x dx

∫

2x e x dx correct to 4 significant figures. du = 2 and du = 2 dx dx from which, v =

∫ e dx = e x

x

2x e x dx = (2x) ( e x ) − ∫ ( e x ) 2 dx = 2xe x − 2e x + c

Hence,

∫

2 0

2x e x dx = ⎡⎣ 2xe x − 2e x ⎤⎦ = ( 4e2 − 2e2 ) − ( 0 − 2e0 ) = 2e 2 + 2 = 16.78 2

0


376

8. Evaluate

∫

π/2

0

t 2 cos t dt correct to 4 significant figures. du = 2t and du = 2t dt dt

Let u = t 2 then

and dv = cos t dt from which, v =

∫t

Hence,

2

∫ cos t dt = sin t

cos t dt = ( t 2 ) ( sin t ) − ∫ ( sin t ) 2t dt = t 2 sin t − 2 ⎡⎣ ∫ t sin t dt ⎤⎦

(1)

du = 1 and du = dt dt

For ∫ t sin t dt , let u = t then

and dv = sin t dt from which, v =

∫ t sin t dt

Hence,

= (t)(- cos t) -


∫t

2

∫ (− cos t)dt

∫ sin t dt = − cos t

= -t cos t + sin t

cos t dt = t 2 sin t − 2 [ − t cos t + sin t ] + c = t 2 sin t + 2t cos t − 2sin t + c

Thus,

∫

π/2

0

t 2 cos t dt = ⎡⎣ t 2 sin t + 2t cos t − 2sin t ⎤⎦

π/ 2

0

⎡⎛ π ⎞ 2 π 2π π π⎤ cos − 2sin ⎥ − [ 0 + 0 − 0] = ⎢⎜ ⎟ sin + 2 2 2 2 ⎦⎥ ⎣⎢⎝ 2 ⎠ π2 ⎛π⎞ − 2 = 0.4674 = ⎜ ⎟ +0−2 = 4 ⎝2⎠ 2

9. Evaluate

∫

2 1

x

3x 2 e 2 dx correct to 4 significant figures.

Let u = 3x 2 then x 2

and dv = e dx

Hence,

∫

du = 6x and du = 6x dx dx x 2

from which, v = ∫ e dx =

x ⎛ x⎞ 3x 2 e 2 dx = ( 3x 2 ) ⎜ 2e 2 ⎟ − ∫ ⎝ ⎠

x 2

x e = 2e 2 1 2

⎛ x2 ⎞ ⎜ 2e ⎟ 6x dx ⎝ ⎠

x ⎡ ⎤ = 6x e − 12 ⎢ ∫ x e 2 dx ⎥ ⎣ ⎦ x 2 2

x 2

For ∫ x e dx , let u = x then

(1)

du = 1 and du = dx dx


377

x

x

and dv = e 2 dx

∫

Hence,

x

from which, v = ∫ e 2 dx = 2e 2

x x x x ⎛ x⎞ x e 2 dx = ( x ) ⎜ 2e 2 ⎟ − ∫ 2e 2 dx = 2x e 2 − 4e 2 ⎝ ⎠


∫

x x x x ⎡ ⎤ 3x 2 e 2 dx = 6x 2 e 2 − 12 ⎢ 2xe 2 − 4e 2 ⎥ + c ⎣ ⎦ x

x

x

= 6x 2 e 2 − 24x e 2 + 48e 2 + c Thus,

∫

2 1

2

x x x ⎡ ⎤ 3x e dx = ⎢ 6x 2 e 2 − 24x e 2 + 48e 2 ⎥ ⎣ ⎦1 x 2 2

1 1 ⎛ 1 ⎞ = ( 24e1 − 48e1 + 48e1 ) − ⎜ 6e 2 − 24e 2 + 48e 2 ⎟ ⎝ ⎠ 1 ⎛ ⎞ = ( 24e1 ) − ⎜ 30e 2 ⎟ = 15.78 ⎝ ⎠


378


∫ 2x

Let u = ln x then

2

ln x dx du 1 = dx x

and du =

1 dx x

and dv = 2x 2dx from which, v = ∫ 2x 2 dx = Hence,

∫ 2x

2

2 3 x 3

⎛2 ⎞ ⎛2 ⎞1 ln x dx = ( ln x ) ⎜ x 3 ⎟ − ∫ ⎜ x 3 ⎟ dx ⎝3 ⎠ ⎝3 ⎠x

=

2 3 2 2 2 ⎛ x3 ⎞ 2 2 x ln x − ∫ x 2 dx = x 3 ln x − ⎜ ⎟ + c = x 3 ln x − x 3 + c 3 3 3 3⎝ 3 ⎠ 3 9

=

3. Determine

∫x

Let u = x 2 then

2

2 3⎛ 1⎞ x ⎜ ln x − ⎟ + c 3 ⎝ 3⎠

sin 3x dx du = 2x and du = 2x dx dx

1 and dv = sin 3x dx from which, v = ∫ sin 3x dx = − cos 3x 3 Hence,

∫x

2

⎛ 1 ⎞ ⎛ 1 ⎞ sin 3x dx = ( x 2 ) ⎜ − cos 3x ⎟ − ∫ ⎜ − cos 3x ⎟ 2x dx ⎝ 3 ⎠ ⎝ 3 ⎠

1 2 = − x 2 cos 3x + ⎡⎣ ∫ x cos 3x dx ⎤⎦ 3 3 For ∫ x cos 3x dx , let u = x then

(1)

du = 1 and du = dx dx

1 and dv = cos 3x dx from which, v = ∫ cos 3x dx = sin 3x 3 Hence,

⎛1 ⎞ ∫ x cos 3x dx = ( x ) ⎜⎝ 3 sin 3x ⎟⎠ − ∫


∫x

2

1 1 ⎛1 ⎞ ⎜ sin 3x ⎟ dx = x sin 3x + cos 3x 3 9 ⎝3 ⎠

1 2 ⎡1 1 ⎤ sin 3x dx = − x 2 cos 3x + ⎢ x sin 3x + cos 3x ⎥ + c 3 3 ⎣3 9 ⎦ 1 2 2 = − x 2 cos 3x + x sin 3x + cos 3x + c 3 9 27

or

cos 3x 2 2 − 9x 2 ) + x sin 3x + c ( 27 9


379

∫ 2θ sec

5. Determine

Let u = 2θ

then

and dv = sec2 θ dθ Hence,

∫ 2θ sec

2

2

θ dθ

du =2 dθ

du = 2 dθ

and

from which,

v = ∫ sec 2 θ dθ = tan θ

θ dθ = ( 2θ )( tan θ ) − ∫ ( tan θ ) 2dθ

= 2θ tan θ − 2 ln(sec θ) + c

from Problem 9, chapter 39, page 393 of textbook

= 2 [ θ tan θ − ln(sec θ)] + c 7. Evaluate

∫

1 0

Let u = 2e3x

2e3x sin 2x dx correct to 4 significant figures.

then

and dv = sin 2x dx Hence,

∫ 2e

3x

du = 6e3x dx from which,

and

du = 6e3x dx

1 v = ∫ sin 2x dx = − cos 2x 2

⎛ 1 ⎞ ⎛ 1 ⎞ sin 2x dx = ( 2e3x ) ⎜ − cos 2x ⎟ − ∫ ⎜ − cos 2x ⎟ 6e3x dx ⎝ 2 ⎠ ⎝ 2 ⎠

= −e3x cos 2x + 3 ⎡⎣ ∫ e3x cos 2x dx ⎤⎦ For

∫e

3x

cos 2x dx , let u = e3x

then

and dv = cos 2x dx Hence,

∫e

3x

du = 3e3x dx

(1) du = 3e3x dx

and

from which,

1 v = ∫ cos 2x dx = sin 2x 2

⎛1 ⎞ ⎛1 ⎞ cos 2x dx = ( e3x ) ⎜ sin 2x ⎟ − ∫ ⎜ sin 2x ⎟ 3e3x dx ⎝2 ⎠ ⎝2 ⎠

= Substituting in (1) gives:

1 3x 3 e sin 2x − ⎡⎣ ∫ e3x sin 2x dx ⎤⎦ 2 2

∫

3 ⎡1 ⎤ 2e3x sin 2x dx = −e3x cos 2x + 3 ⎢ e3x sin 2x − ∫ e3x sin 2x dx ⎥ 2 ⎣2 ⎦ 3 9 = −e3x cos 2x + e3x sin 2x − ∫ e3x sin 2x dx 2 2

Hence,

9 ⎞ 3x 3 3x ⎛ 3x ⎜ 2 + ⎟ ∫ e sin 2x dx = −e cos 2x + e sin 2x 2⎠ 2 ⎝

by combining the far left and far right-hand integrals

i.e.

3 3x ⎛ 13 ⎞ 3x 3x ⎜ ⎟ ∫ e sin 2x dx = −e cos 2x + e sin 2x 2 ⎝2⎠ © 2006 John Bird. All rights reserved. Published by Elsevier.

380

∫e

and

3x

∫ 2e

Thus,

3x

sin 2x dx =

2 ⎛ 3x 3 3x ⎞ ⎜ −e cos 2x + e sin 2x ⎟ 13 ⎝ 2 ⎠

sin 2x dx =

4 ⎛ 3x 3 3x ⎞ ⎜ −e cos 2x + e sin 2x ⎟ 13 ⎝ 2 ⎠ 1

∫

and

1 0

⎡4⎛ 3 ⎞⎤ 2e sin 2x dx = ⎢ ⎜ −e3x cos 2x + e3x sin 2x ⎟ ⎥ 2 ⎠⎦ 0 ⎣13 ⎝ 3x

⎡4⎛ 3 3 ⎞⎤ ⎡ 4 ⎛ ⎞⎤ = ⎢ ⎜ −e3 cos 2 + e3 sin 2 ⎟ ⎥ − ⎢ ⎜ − cos 0 + sin 0 ⎟ ⎥ 2 2 ⎠ ⎦ ⎣13 ⎝ ⎠⎦ ⎣13 ⎝ ⎡4 ⎤ ⎡4 ⎤ = ⎢ ( 8.3585 + 27.3956 ) ⎥ − ⎢ (−1) ⎥ ⎣13 ⎦ ⎣13 ⎦

= (11.0013) + (0.30769) = 11.31 9. Evaluate

∫

4 1

Let u = ln x and dv =

∫

Hence,

x 3 ln x dx correct to 4 significant figures.

then x 3 dx

du 1 = dx x from which,

and v=

du =

∫

1 dx x

3

x 2 dx =

2 52 x 5

⎛ 2 52 ⎞ ⎛ 2 52 ⎞ 1 x ln x dx = ( ln x ) ⎜ x ⎟ − ∫ ⎜ x ⎟ dx ⎝5 ⎠ ⎝5 ⎠x 3

3 2 5 2 = x ln x − ∫ x 2 dx 5 5

2 5 2 ⎛ 2 52 ⎞ = x ln x − ⎜ x ⎟ + c 5 5⎝5 ⎠

Thus,

∫

4 1

4

4 ⎡2 5 ⎤ x 3 ln x dx = ⎢ x ln x − x5 ⎥ 25 ⎣5 ⎦1

4 4 5⎞ ⎛2 5 ⎞ ⎛2 =⎜ 4 ln 4 − 45 ⎟ − ⎜ 15 ln1 − 1 ⎟ 25 25 ⎝5 ⎠ ⎝5 ⎠ 4 4 ⎞ ⎛2 ⎞ ⎛ = ⎜ ( 32 ) ln 4 − ( 32 ) ⎟ − ⎜ 0 − ⎟ 25 25 ⎠ ⎝5 ⎠ ⎝ = 12.78


381

10. In determining a Fourier series to represent f(x) = x in the range -π to π, Fourier coefficients are

given by:

an =

1 π x cos nx dx π ∫ −π

bn = ∫

and

π −π

x sin nx dx

where n is a positive integer. 2 Show by using integration by parts that a n = 0 and b n = − cos nπ n For a n =

1 π x cos nx dx , let u = x then π ∫ −π

du =1 dx

and dv = cos nx dx Hence,

⎛1

⎞

∫ x cos nx dx = ( x ) ⎜⎝ n sin nx ⎟⎠ − ∫

and

du = dx

from which, v =

1

∫ cos nx dx = n sin nx

x 1 ⎛1 ⎞ ⎜ sin nx ⎟ dx = sin nx + 2 cos nx n n ⎝n ⎠ π

Then a n =

1 π 1 ⎡x 1 ⎤ x cos nx dx = ⎢ sin nx + 2 cos nx ⎥ ∫ −π π π ⎣n n ⎦ −π =

1 ⎡⎛ π 1 1 ⎞ ⎛ π ⎞⎤ ⎜ sin nπ + 2 cos nπ ⎟ − ⎜ − sin(− nπ) + 2 cos(− nπ) ⎟ ⎥ ⎢ n n π ⎣⎝ n ⎠ ⎝ n ⎠⎦

sin nπ = sin(-nπ) = 0 for all values of n. Hence, a n = =

For b n = ∫

π −π

1 ⎡⎛ 1 1 ⎞ ⎛ ⎞⎤ ⎜ 0 + 2 cos nπ ⎟ − ⎜ 0 + 2 cos(− nπ) ⎟ ⎥ ⎢ n n π ⎣⎝ ⎠ ⎝ ⎠⎦ 1 [ cos nπ − cos(−nπ)] = 0 π n2 x sin nx dx ,

let u = x then

since cos nπ = cos(-nπ) for all values of n.

du =1 dx

and dv = sin nx dx Hence,

⎛ 1

⎞

∫ x sin nx dx = ( x ) ⎜⎝ − n cos nx ⎟⎠ − ∫

and

du = dx

from which, v =

1

∫ sin nx dx = − n cos nx

x 1 ⎛ 1 ⎞ ⎜ − cos nx ⎟ dx = − cos nx + 2 sin nx n n ⎝ n ⎠ π

1 π 1⎡ x 1 ⎤ Then b n = ∫ x sin nx dx = ⎢ − cos nx + 2 sin nx ⎥ −π π π⎣ n n ⎦ −π =

1 ⎡⎛ π 1 1 ⎞ ⎛π ⎞⎤ ⎜ − cos nπ + 2 sin nπ ⎟ − ⎜ cos(− nπ) + 2 sin(− nπ) ⎟ ⎥ ⎢ n n π ⎣⎝ n ⎠ ⎝n ⎠⎦

sin nπ = sin(-nπ) = 0 for all values of n.


382

Hence, b n =

1 ⎡⎛ π ⎞ ⎛π ⎞⎤ ⎜ − cos nπ + 0 ⎟ − ⎜ cos(− nπ) + 0 ⎟ ⎥ ⎢ π ⎣⎝ n ⎠ ⎝n ⎠⎦ −π [cos nπ + cos(−nπ)] πn

=

cos nπ = cos(-nπ) hence cos nπ + cos(-nπ) ≡ 2 cos nπ Thus,

bn = −

2 1 ( 2 cos nπ ) = − cos nπ n n

11. The equation C =

∫

1 0

e − 0.4 θ cos1.2θ dθ

and S =

∫

1 0

e − 0.4 θ sin1.2θ dθ

are involved in the study of damped oscillations. Determine the value of C and S.

From problem 9, page 421 of textbook,

∫ ∫

Thus,

eax cos bx dx =

e −0.4 θ cos1.2θ dθ =

eax ( b sin bx + a cos bx ) + c a 2 + b2 e −0.4 θ (1.2sin1.2θ − 0.4 cos1.2θ ) (−0.4) 2 + (1.2) 2 1

and C =

∫

1 0

e

− 0.4 θ

⎡ e − 0.4 θ ⎤ cos1.2θ dθ = ⎢ (1.2sin1.2θ − 0.4 cos1.2θ ) ⎥ ⎣ 1.6 ⎦0 ⎡ e −0.4 ⎤ 1 = ⎢ (1.2sin1.2 − 0.4 cos1.2 )⎥ − ⎡⎢ (1.2sin 0 − 0.4 cos 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),

∫ Thus,

∫

and

∫

eax e sin bx dx = 2 ( a sin bx − b cos bx ) + c a + b2 ax

e − 0.4 θ sin1.2θ dθ =

e − 0.4 θ (−0.4sin1.2θ − 1.2 cos1.2θ) 1.6 1

S=

1 0

e

−0.4 θ

⎡ e − 0.4 θ ⎤ sin1.2θ dθ = ⎢ ( −0.4sin1.2θ − 1.2 cos1.2θ )⎥ ⎣ 1.6 ⎦0 ⎡ e −0.4 ⎤ 1 =⎢ ( −0.4sin1.2 − 1.2 cos1.2 )⎥ − ⎡⎢ (−0.4sin 0 − 1.2 cos 0⎤⎥ ⎦ ⎣ 1.6 ⎦ ⎣1.6

= (-0.33836) – (-0.75) i.e.

S = 0.41 © 2006 John Bird. All rights reserved. Published by Elsevier.

383

CHAPTER 44 REDUCTION FORMULAE EXERCISE 170 Page 425

∫te

3 2t

2. Determine

∫te

n 2t

Let u = t n then

dt

∫te

n 2t

⎛1 ⎞ dt = ( t n ) ⎜ e 2t ⎟ − ∫ ⎝2 ⎠

= In =

i.e. Hence,

∫te

3 2t

du = nt n −1 dt

dv = e2t dt

and Thus,

dt using a reduction formula.

∫te

3 2t

from which, v =

2t

dt =

1 2t e 2

⎛ 1 2t ⎞ n −1 ⎜ e ⎟ nt dt ⎝2 ⎠

1 n 2t n t e − I n −1 2 2

dt = I3 =

1 3 2t 3 t e − I2 2 2

I2 =

1 2 2t 2 t e − I1 2 2

I1 =

1 2t 1 t e − I0 2 2

dt =

∫e

1 n 2t n n −1 2t t e − ∫ t e dt 2 2

I0 = ∫ e 2t dt = Thus,

and du = nt n −1dt

1 2t e 2

1 3 2t 3 ⎛ 1 2 2t ⎞ t e − ⎜ t e − I1 ⎟ 2 2⎝2 ⎠

=

1 3 2t 3 2 2t 3 ⎛ 1 2t 1 ⎛ 1 2t ⎞ ⎞ t e − t e + ⎜ t e − ⎜ e ⎟⎟ + c 2 4 2⎝2 2 ⎝ 2 ⎠⎠

=

1 3 2t 3 2 2t 3 2t 3 2t t e − t e + te − e +c 2 4 4 8

3 3 3⎞ ⎛1 = e 2t ⎜ t 3 − t 2 + t − ⎟ + c 4 4 8⎠ ⎝2

3. Use the result of Problem 2 to evaluate

∫

1 0

5t 3e 2t dt , correct to 3 decimal places.

1

∫

1 0

⎡ 3 3 3 ⎞⎤ ⎛1 ⎛ 1 3 3 3⎞ ⎛ 3⎞ 5t e dt = ⎢5e2t ⎜ t 3 − t 2 + t − ⎟ ⎥ = 5e2 ⎜ − + − ⎟ − 5e0 ⎜ − ⎟ = 6.493 4 4 8 ⎠⎦ 0 ⎝2 ⎝ 2 4 4 8⎠ ⎝ 8⎠ ⎣ 3 2t


384

EXERCISE 171 Page 427 3. Use a reduction formula to determine

∫x

5

∫x

5

sin x dx

sin x dx = I5 = − x 5 cos x + 5x 4 sin x − 5(4)I3 from equation (3), page 426 of textbook

I3 = − x 3 cos x + 3x 2sin x − 3(2)I1 I1 = − x1 cos x + 1x 0 sin x − (1)(0) i.e.

∫x

5

sin x dx = − x 5 cos x + 5x 4 sin x − 20 ( − x 3 cos x + 3x 2 sin x − 6I1 ) = − x 5 cos x + 5x 4 sin x + 20x 3 cos x − 60x 2 sin x + 120 ( − x cos x + sin x ) + c = − x5 cos x + 5x 4 sin x + 20x 3 cos x − 60x 2 sin x − 120x cos x + 120sin x + c

4. Evaluate

∫

π 0

∫

π 0

x 5 sin x dx , correct to 2 decimal places.

x 5 sin x dx = ⎡⎣ − x 5 cos x + 5x 4 sin x + 20x 3 cos x − 60x 2 sin x − 120x cos x + 120sin x ⎤⎦

π 0

= ( −π5 cos π + 5π4 sin π + 20π3 cos π − 60π2 sin π − 120π cos π + 120sin π ) − ( 0 + 0 + 0 − 0 − 0 + 120sin 0 )

= ( π5 − 20π3 + 120π ) − ( 0 ) = 62.89


385


∫

2. Evaluate

∫ sin

3

π 0

3sin 3 x dx , using a reduction formula.

1 2 x dx = I3 = − sin 2 x cos x + I1 3 3

from equation (4), page 428 of textbook

1 I1 = − sin 0 x cos x + 0 = − cos x 1

∫ sin

Hence,

3

∫ 3sin ∫

Thus,

π 0

3

1 2 x dx = − sin 2 x cos x + ( − cos x ) 3 3 2 ⎛ 1 ⎞ x dx = 3 ⎜ − sin 2 x cos x − cos x ⎟ = − sin 2 x cos x − 2 cos x 3 ⎝ 3 ⎠ π

3sin 3 x dx = ⎡⎣ − sin 2 x cos x − 2 cos x ⎤⎦ 0

= ( − sin 2 π cos π − 2 cos π ) − ( − sin 2 0 cos 0 − 2 cos 0 ) = (- 0 – 2(-1)) – (0 – 2) =2--2=4 4. Determine ∫ cos 6 x dx , using a reduction formula,

∫ cos

6

1 5 x dx = I6 = cos5 x sin x + I 4 6 6

from equation (5), page 429 of textbook

1 3 I 4 = cos3 x sin x + I 2 4 4 1 1 I 2 = cos1 x sin x + I0 2 2 I0 = ∫ cos 0 x dx = x Thus,

∫ cos

6

x dx =

1 5⎛1 3 ⎞ cos5 x sin x + ⎜ cos3 x sin x + I2 ⎟ 6 6⎝ 4 4 ⎠

=

1 5 15 ⎛ 1 1 ⎞ cos5 x sin x + cos3 x sin x + ⎜ cos x sin x + x ⎟ 6 24 24 ⎝ 2 2 ⎠

=

1 5 15 15 cos5 x sin x + cos3 x sin x + cos x sin x + x 6 24 48 48

=

1 5 5 5 cos5 x sin x + cos 3 x sin x + cos x sin x + x 6 24 16 16


386

5. Evaluate

∫

π/2 0

cos 7 x dx

Using Wallis’s formula, equation (6), page 430 of textbook

∫

π/2 0

cos 7 x dx = I7 =

6 I5 7

I5 =

4 I3 5

I3 =

2 I1 3

I1 = ∫ Hence,

∫

π/2 0

π/ 2 0

cos1 x dx = [sin x ] 0 = sin π/2

π − sin 0 = 1 2

16 6 ⎛ 4 ⎞⎛ 2 ⎞ cos 7 x dx = ⎜ ⎟ ⎜ ⎟ (1) = 35 7 ⎝ 5 ⎠⎝ 3 ⎠


387

EXERCISE 173 Page 432 1. Evaluate

∫

π/2 0

∫

π/2 0

cos 2 x sin 5 x dx

cos 2 x sin 5 x dx = ∫ =

4 I3 5

I3 =

∫ ( sin π/2

6 I5 7

∫

2 I1 3

I5 =

π/2 0

x dx

x − sin 7 x ) dx = I5 − I7

sin n x dx = I n =

I1 = ∫

π/ 2 0

n −1 In −2 n

from problem 10, page 429 of textbook

sin1 x dx = − [ cos x ] 0 = −[0 − 1] = 1 π/ 2

4 I3 5

I3 =

2 I1 3

I1 = 1

⎛ 6 ⎞⎛ 4 ⎞⎛ 2 ⎞ I7 = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (1) ⎝ 7 ⎠⎝ 5 ⎠⎝ 3 ⎠

Hence, Thus,

5

0

5

⎛ 4 ⎞⎛ 2 ⎞ I5 = ⎜ ⎟ ⎜ ⎟ (1) ⎝ 5 ⎠⎝ 3 ⎠

Hence, I7 =

2

0

Wallis’s formula states:

I5 =

(1 − sin x ) sin

π/ 2

∫

π/2 0

⎛ 4 ⎞⎛ 2 ⎞ ⎛ 6 ⎞⎛ 4 ⎞⎛ 2 ⎞ cos 2 x sin 5 x dx = I5 − I7 = ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 5 ⎠⎝ 3 ⎠ ⎝ 7 ⎠⎝ 5 ⎠⎝ 3 ⎠ ⎛ 4 ⎞⎛ 2 ⎞ ⎡ 6⎤ ⎛ 4 ⎞⎛ 2 ⎞⎛ 1 ⎞ = ⎜ ⎟ ⎜ ⎟ ⎢1 − ⎥ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 5 ⎠⎝ 3 ⎠ ⎣ 7⎦ ⎝ 5 ⎠⎝ 3 ⎠⎝ 7 ⎠ =

3. Evaluate

∫

π/2 0

∫

π/2 0

8 105

cos5 x sin 4 x dx

cos5 x sin 4 x dx = ∫

π/ 2 0

cos5 x (1 − cos 2 x ) dx = ∫ 2

=

π/ 2 0

cos5 x (1 − 2 cos 2 x + cos 4 x ) dx

∫ ( cos π/2

5

0

x − 2 cos 7 x + cos9 x ) dx

= I5 − 2I7 + I9 I5 =

i.e.

4 I3 from equation (6), page 430, 5 ⎛ 4 ⎞⎛ 2 ⎞ I5 = ⎜ ⎟ ⎜ ⎟ (1) , ⎝ 5 ⎠⎝ 3 ⎠

I7 =

I3 =

2 I1 3

6 ⎛ 6 ⎞⎛ 4 ⎞⎛ 2 ⎞ I5 = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 7 ⎝ 7 ⎠⎝ 5 ⎠⎝ 3 ⎠

I1 = ∫ and

π/ 2 0

cos x dx = [sin x ] 0 = 1 π/2

8 ⎛ 8 ⎞⎛ 6 ⎞⎛ 4 ⎞⎛ 2 ⎞ I9 = I 7 = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 9 ⎝ 9 ⎠⎝ 7 ⎠⎝ 5 ⎠⎝ 3 ⎠


388

Hence,

∫

π/2 0

cos5 x sin 4 x dx = I5 − 2I7 + I9 ⎛ 4 ⎞⎛ 2 ⎞ ⎛ 6 ⎞ ⎛ 4 ⎞ ⎛ 2 ⎞ ⎛ 8 ⎞ ⎛ 6 ⎞ ⎛ 4 ⎞ ⎛ 2 ⎞ = ⎜ ⎟⎜ ⎟ −2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 5 ⎠⎝ 3 ⎠ ⎝ 7 ⎠ ⎝ 5 ⎠ ⎝ 3 ⎠ ⎝ 9 ⎠ ⎝ 7 ⎠ ⎝ 5 ⎠ ⎝ 3 ⎠ ⎛ 4 ⎞⎛ 2 ⎞ ⎡ ⎛ 6 ⎞ ⎛ 8 ⎞ ⎛ 6 ⎞⎤ = ⎜ ⎟ ⎜ ⎟ ⎢1 − 2 ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎥ ⎝ 5 ⎠⎝ 3 ⎠ ⎣ ⎝ 7 ⎠ ⎝ 9 ⎠ ⎝ 7 ⎠⎦ =

8 ⎡ 12 16 ⎤ 8 ⎡ 21 − 36 + 16 ⎤ 1− + = ⎥⎦ 15 ⎢⎣ 7 21 ⎥⎦ 15 ⎢⎣ 21 =

5. Show that

∫

π/2 0

∫

π/2 0

sin 3 θ cos 4 θ dθ =

sin 3 θ cos 4 θ dθ = ∫ =

I3 =

2 I1 3

I1 = ∫ I5 =

π/ 2 0

π/2 0

∫

0

2 35

sin 3 θ (1 − sin 2 θ ) dθ = ∫ 2

π/ 2 0

sin 3 θ (1 − 2sin 2 θ + sin 4 θ ) dθ

(sin 3 θ − 2sin 5 θ + sin 7 θ)dθ = I3 − 2I5 + I7

from Problem 10, page 429,

and

π π/ 2 ⎛ ⎞ sin θ dθ = − [ cos θ] 0 = − ⎜ cos − cos 0 ⎟ = −(0 − 1) = 1 2 ⎝ ⎠

4 ⎛ 4 ⎞⎛ 2 ⎞ I3 = ⎜ ⎟ ⎜ ⎟ (1), 5 ⎝ 5 ⎠⎝ 3 ⎠

Hence,

π/2

8 8⎛ 1 ⎞ ⎜ ⎟ = 315 15 ⎝ 21 ⎠

∫

π/2 0

I7 =

6 ⎛ 6 ⎞⎛ 4 ⎞⎛ 2 ⎞ I5 = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (1) 7 ⎝ 7 ⎠⎝ 5 ⎠⎝ 3 ⎠

sin 3 θ cos 4 θ dθ = I3 − 2I5 + I7 ⎛ 2 ⎞ ⎛ 4 ⎞⎛ 2 ⎞ ⎛ 6 ⎞ ⎛ 4 ⎞⎛ 2 ⎞ = ⎜ ⎟ − 2 ⎜ ⎟ ⎜ ⎟ (1) + ⎜ ⎟ ⎜ ⎟⎜ ⎟ (1) ⎝ 3 ⎠ ⎝ 5 ⎠⎝ 3 ⎠ ⎝ 7 ⎠ ⎝ 5 ⎠⎝ 3 ⎠ =

2 ⎡ 8 24 ⎤ 2 ⎡ 35 − 56 + 24 ⎤ 1− + = ⎥⎦ 3 ⎢⎣ 5 35 ⎥⎦ 3 ⎢⎣ 35 =

2 2⎛ 3 ⎞ ⎜ ⎟ = 35 3 ⎝ 35 ⎠


389

CHAPTER 45 NUMERICAL INTEGRATION EXERCISE 174 Page 435

∫

2. Evaluate

3

1

2 ln 3x dx using the trapezoidal rule with 8 intervals, giving the answer correct to 3

decimal places.

∫

Since

3

1

2 ln 3x dx , width of interval =

x 2 ln 3x

3 −1 = 0.25 8

1 1.25 1.50 1.75 2.0 2.25 2.50 2.75 3.0 2.1972 2.6435 3.0082 3.3165 3.5835 3.8191 4.0298 4.2204 4.3944

Hence, using the trapezoidal rule,

∫

3

1

2 ln 3x dx ⎡1 ⎤ ≈ (0.25) ⎢ ( 2.1972 + 4.3944 ) + 2.6435 + 3.0082 + 3.3165 + 3.5835 + 3.8191 + 4.0298 + 4.2204 ⎥ ⎣2 ⎦ = 0.25[27.9168] = 6.979

4. Evaluate

∫

1.4 0

e − x dx using the trapezoidal rule with 7 intervals, giving the answer correct to 3 2

decimal places. Since ∫

1.4 0

e − x dx , width of interval = 2

x e

− x2

1.4 − 0 = 0.2 7

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.0 0.9608 0.8521 0.6977 0.5273 0.3679 0.2369 0.1409


∫

1.4 0

2 ⎡1 ⎤ e − x dx ≈ (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


390


2. Evaluate

∫

π/2 0

1 dθ using the mid-ordinate rule with 6 intervals, giving the answer correct 1 + sin θ

to 3 decimal places. Since ∫

π/2 0

π −0 π 1 2 dθ , width of interval = = rad or 15° 6 12 1 + sin θ

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 1 1 + sin θ

7.5°

22.5°

37.5°

52.5°

67.5°

82.5°

0.8845 0.7232 0.6216 0.5576 0.5198 0.5021

Hence, using the mid-ordinate rule,

∫

π/2 0

1 ⎛π⎞ dθ ≈ ⎜ ⎟ [ 0.8845 + 0.7232 + 0.6216 + 0.5576 + 0.5198 + 0.5021] 1 + sin θ ⎝ 12 ⎠ ⎛π⎞ = ⎜ ⎟ [3.8088] ⎝ 12 ⎠ = 0.977

3. Evaluate

∫

3 1

ln x dx using the mid-ordinate rule with 10 intervals, giving the answer correct to 3 x

decimal places. Since ∫

3 1

ln x 3 −1 = 0.2 dx , width of interval = x 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 ln x x

1.1

1.3

1.5

1.7

1.9

2.1

2.3

2.5

2.7

2.9

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

391

∫

3 1

ln x dx ≈ (0.2)[ 0.0866 + 0.2018 + 0.2703 + 0.3121 + 0.3378 + 0.3533 + 0.3621 + 0.3665 x + 0.3679 + 0.3671] = (0.2)[3.0255] = 0.605


392

EXERCISE 176 Page 439 2. Evaluate

Since ∫

1.6 0

∫

1.6 0

1 dθ using Simpson’s rule with 8 intervals, correct to 3 decimal places. 1 + θ4

1 1.6 − 0 = 0.2 dθ , width of interval = 4 1+ θ 8

θ 1 1+ θ4

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

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 0

1 dθ 1 + θ4 1 ≈ (0.2) ⎡⎣(1.0 + 0.1324 ) + 4 ( 0.9984 + 0.8853 + 0.5000 + 0.2065 ) + 2 ( 0.9750 + 0.7094 + 0.3254 ) ⎤⎦ 3

=

1 (0.2) [1.1324 + 10.3608 + 4.0196] 3

=

1 (0.2) [15.5128] = 1.034 3

3. Evaluate

Since ∫

1.0 0.2

∫

1.0 0.2

sin θ dθ using Simpson’s rule with 8 intervals, correct to 3 decimal places. θ

sin θ 1.0 − 0.2 dθ , width of interval = = 0.1 θ 8

θ sin θ θ

0.2

0.3

0.4

0.5

(note that values of θ are in radians) 0.6

0.7

0.8

0.9

1.0

0.9933 0.9851 0.9735 0.9589 0.9411 0.9203 0.8967 0.8704 0.8415


∫

1.0 0.2

⎡( 0.9933 + 0.8415 ) + 4 ( 0.9851 + 0.9589 + 0.9203 + 0.8704 ) ⎤ 1 sin θ dθ ≈ (0.1) ⎢ ⎥ 3 θ + 2 ( 0.9735 + 0.9411 + 0.8967 ) ⎦⎥ ⎣⎢ =

1 (0.1) [1.8348 + 14.9388 + 5.6226] 3

=

1 (0.1) [ 22.3962] = 0.747 3


393

∫

5. Evaluate

Since ∫

π/3 0

π/3 0

2

e x sin 2x dx using Simpson’s rule with 10 intervals, correct to 3 decimal places.

π −0 π x2 3 e sin 2x dx , width of interval = = rad 10 30

x

π 30

0

2π 30

3π 30

4π 30

5π 30

6π 30

7π 30

8π 30

9π 30

10π 30

2

e x sin 2x

0 0.2102 0.4250 0.6488 0.8857 1.1392 1.4114 1.7021 2.0064 2.3119 2.5929


∫

π/3 0

⎤ 2 1 ⎛ π ⎞ ⎡( 0 + 2.5929 ) + 4 ( 0.2102 + 0.6488 + 1.1392 + 1.7021 + 2.3119 ) e x sin 2x dx ≈ ⎜ ⎟ ⎢ ⎥ 3 ⎝ 30 ⎠ ⎣⎢ + 2 ( 0.4250 + 0.8857 + 1.4114 + 2.0064 ) ⎦⎥ ⎛ π⎞ = ⎜ ⎟ [ 2.5929 + 24.0488 + 9.4570] ⎝ 90 ⎠ ⎛ π⎞ = ⎜ ⎟ [36.0987 ] = 1.260 ⎝ 90 ⎠

7. Evaluate

∫

6 2

1 dx using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate (2x − 1)

rule, (d) Simpson’s rule. Give answers correct to 3 decimal places and use 6 intervals

(a)

∫

6 2

1 dx (2x − 1)

Hence,

∫

du =2 dx −

∫

Thus,

Let u = 2x – 1, then

6 2

dx =

1 2

1 1 du 1 1u dx = ∫ = ∫ u du = = u= 2 1 (2x − 1) u 2 2 2 6 1 dx = ⎡ ( 2x − 1) ⎤ = ⎡⎣ 11 − 3 ⎤⎦ = 1.585 ⎣ ⎦2 (2x − 1)

(b) Width of interval = x

1 2

and

2.0

du 2

( 2x − 1)

6−2 = 0.5 8 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



394

∫

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 + 0.3162 ⎥ ⎣2 ⎦ = (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 ≈ (0.5)[ 0.5345 + 0.4714 + 0.4264 + 0.3922 + 0.3651 + 0.3430 (2x − 1) + 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

⎡( 0.5774 + 0.3015 ) + 4 ( 0.5000 + 0.4082 + 0.3536 + 0.3162 ) ⎤ 1 1 dx ≈ (0.5) ⎢ ⎥ 3 + 2 ( 0.4472 + 0.3780 + 0.3333) ⎦⎥ (2x − 1) ⎣⎢

∫

9. Evaluate

=

1 (0.5) [ 0.8789 + 6.312 + 2.317 ] 3

=

1 (0.5) [9.5079] = 1.585 3

1

0.7

(1 − y )

0.1

2

dy 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

∫

1

0.7

(1 − y )

0.1

dy then width of interval =

2

y 1

0.1

(1 − y ) 2

0.2

0.3

0.7 − 0.1 = 0.1 6

0.4

0.5

0.6

0.7

1.0050 1.0206 1.0483 1.0911 1.1547 1.2500 1.4003


395


∫

⎡1 ⎤ dy ≈ (0.1) ⎢ (1.0050 + 1.4003) + 1.0206 + 1.0483 + 1.0911 + 1.1547 + 1.2500 ⎥ ⎣2 ⎦ (1 − y ) 1

0.7 0.1

2

= (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

1

(1 − y ) 2

1.0114 1.0328 1.0675 1.1198 1.1974 1.3159

Using the mid-ordinate rule,

∫

1

0.7

(1 − y )

0.1

2

dy ≈ (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

1 dy ≈ (0.1) ⎡⎣(1.0050 + 1.4003) + 4 (1.0206 + 1.0911 + 1.2500 ) + 2 (1.0483 + 1.1547 ) ⎤⎦ 3 (1 − y2 ) 1

=

1 (0.1) [ 2.4053 + 13.4468 + 4.406] 3

=

1 (0.1) [ 20.258] = 0.675 3

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 -1 velocity v (ms ) 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 0

v dt

Estimate the distance using Simpson’s rule, giving the answer correct to 3 significant figures.

∫

8.0 0

v dt ≈ =

1 (1.0 ) [(0 + 9.4) + 4(0.4 + 1.7 + 4.1 + 8.0) + 2(1.0 + 2.9 + 6.2)] 3 1 1 [9.4 + 56.8 + 20.2] = (86.4) = 28.8 m 3 3


396

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 + 0.86] 3 1 = (0.5) [ 2.908] 3 = 0.485 m


397

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)

(a) If

dy =6 dx

(b)

dy = 3x dx

dy = 6 , then y = dx

∫ 6 dx

(c)

dy = x+2 dx

= 6x + c

There are an infinite number of graphs of y = 6x + c; three curves are shown below.

(b) If

dy = 3x , then y = dx

3

∫ 3x dx = 2 x

2

+c

A family of three typical curves is shown below.

(c) If

dy = x + 2 , then y = dx

∫

(x + 2) dx =

x2 + 2x + c 2


398

A family of three typical curves is shown below.

2. Sketch the family of curves given by the equation

dy = 2x + 3 and determine the equation of dx

one of the curves which es through the point (1, 3).

If

dy = 2x + 3 , then y = dx

∫ ( 2x + 3) dx = x

2

+ 3x + c

If the curve es through the point (1, 3) then x = 1 and y = 3, Hence,

and

3 = (1) + 3 (1) + c 2

i.e.

c = -1

y = x 2 + 3x − 1

A family of three curves is shown below, including y = x 2 + 3x − 1 which es through the point (1, 3).


399

EXERCISE 178 Page 445 3. Solve the differential equation:

If

dy dy + x = 3 , then = 3− x dx dx

y = 3x −

1 dy + 2 = x −3 x e dx

then 3 y=

Hence,

from which,

c= −

1 2

1 dy + 2 = x − 3 , given y = 1 when x = 0. x e dx

dy = x − e− x − 2 dx

and

dy 1 = ( x − e− x − 2 ) dx 3

⎞ 1 1 ⎛ x2 −x −x x e 2 dx − − = ( ) ⎜ + e − 2x ⎟ + c ∫ 3 3⎝ 2 ⎠

If y = 1 when x = 0, then 1 =

Thus,

1 +c 2

x2 +c 2

x2 1 − 2 2

5. Solve the differential equation:

If

∫ ( 3 − x ) dx = 3x −

and y =

If y = 2 when x = 1, then 2 = 3 Hence,

dy + x = 3 , given y = 2 when x = 1. dx

1 (0 + 1 + 0) + c 3

⎞ 2 1 ⎛ x2 y = ⎜ + e − x − 2x ⎟ + 3⎝ 2 ⎠ 3

or

i.e. c =

2 3

1⎛ 2 2 ⎞ x − 4x + x + 4 ⎟ ⎜ 6⎝ e ⎠

dy x 2 6. The gradient of a curve is given by: + = 3x . Find the equation of the curve if it es dx 2 ⎛ 1⎞ through the point ⎜1, ⎟ ⎝ 3⎠ dy x 2 dy x2 + = 3x , then = 3x − If dx 2 dx 2 Hence,

⎛ x2 ⎞ 3x 2 x 3 y = ∫ ⎜ 3x − ⎟ dx = − +c 2 ⎠ 2 6 ⎝

1 ⎛ 1⎞ If it es through ⎜1, ⎟ , x = 1 and y = 3 ⎝ 3⎠ Thus, Hence,

1 3 1 = − +c 3 2 6

from which,

c=

1 3 1 − + = −1 3 2 6

3 2 x3 y = x − −1 2 6


400

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

ds = u – gt, where s is the height of the object in metres dt

at time t seconds and g = 9.8 m / s 2 . Determine the height of the object after 3 seconds if s = 0 when t = 0.

If

ds = u – gt, then dt

s=

∫ ( u − gt ) dt = ut −

g t2 +c 2

Since s = 0 when t = 0, then c = 0 Hence,

g t2 9.8t 2 s = ut − and if u = 20 and s = 9.8, then s = 20t 2 2

The height when t = 3,

i.e.

s = 3(20) – 4.9 ( 3)

i.e. u = 20t – 4.9 t 2

2

height = 60 – 44.1 = 15.9 m


401


dy = 2 cos 2 y dx


If

dy = 2 cos 2 y then dx

dy = 2dx cos 2 y

∫ sec

and i.e.

2

y dy = ∫ 2dx

tan y = 2x + c

3. Solve the differential equation: ( y 2 + 2 )

If

(y

2

+ 2)

dy = 5y dx

dy 1 = 5y , given y = 1 when x = dx 2

⎛ y2 + 2 ⎞ ⎜ ⎟ dy = 5dx ⎝ y ⎠

then

⎛

2⎞

and

∫ ⎜⎝ y + y ⎟⎠ dy = ∫ 5dx

i.e.

y2 + 2 ln y = 5x + c 2

y = 1 when x =

1 , hence, 2

1 5 + 2 ln1 = + c 2 2

from which, c =

y2 + 2 ln y = 5x − 2 2

and

4. The current in an electric circuit is given by the equation Ri + L

constants. Show that i = I e

If Ri + L

1 5 − = −2 2 2

−

Rt L

di = 0, where L and R are dt

, given that i = I when t = 0.

di di = 0, then L = - Ri dt dt

and

di Ri =− dt L

from which,

di R = − dt i L

Thus,

ln i = −

i = I when t = 0, thus

ln I = c

Hence,

ln i = −

and

∫

di R = ∫ − dt i L

Rt +c L

Rt + ln I L


402

ln i – ln I = −

i.e.

Rt L

i Rt ln = − I L

i.e.

Rt − i =e L I

Taking anti-logarithms gives:

and

5. The velocity of a chemical reaction is given by

i = Ie

−

Rt L

dx = k(a – x), where x is the amount transferred dt

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 of t.

If

dx = k(a – x), then dt

dx = k dt a−x dx

∫ a − x = ∫ k dt

and i.e.

– ln(a – x) = kt + c

t = 0 when x = 0, hence Thus,

- ln a = c – ln(a – x) = kt – ln a

i.e.

ln a – ln(a – x) = kt

i.e.

⎛ a ⎞ ln ⎜ ⎟ = kt ⎝a−x⎠

and

a = ekt a−x

i.e.

a =a−x ek t

i.e.

and

x = a - a e− k t

i.e.

a e− k t = a − x

x = a ( 1 − e− k t )

6.(a) Charge Q coulombs at time t seconds is given by the differential equation R

dQ Q + = 0, dt C

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.


403

(a) If R

dQ Q + =0 dt C

dQ Q =− dt RC

then

∫

i.e.

dQ 1 =∫ − dt Q RC ln Q = −

i.e.

t +k RC

Q = Q0 when t = 0, hence, ln Q0 = k ln Q = −

Hence, i.e.

ln Q - ln Q0 = −

i.e.

ln

t + ln Q0 RC t RC

Q t =− Q0 RC t

− Q = e RC Q0

and

and

Q = Q 0e

−

t CR

(b) R = 250 × 103 Ω , C = 8.5 × 10-6 F, t = 0.32 s and Q = 8.0 C 8.0 = Q0 e

Hence,

−

0.32 8.5×10−6 × 250×103

from which, initial charge, Q0 = −

When t = 1 s, charge, Q = Q0 e

= Q0 ( 0.8602 )

8.0 = 9.30 C 0.8602

t CR

= 9.30 e

8. The rate of cooling of a body is given by

−

1 8.5×10−6 × 250×103

= 5.81 C

dθ = kθ , where k is a constant. If θ = 60°C when t = 2 dt

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

dθ = kθ dt

then

dθ = k dt θ

and

∫

dθ = k dt θ ∫ ln θ = kt + c

i.e. 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)


404

(1) – (2) gives:

ln 60 – ln 50 = -3k

from which,

1 60 = −0.06077 k = − ln 3 50

Substituting in (1):

ln 60 = 2(-0.06077) + c

from which,

ln θ = kt + c = -0.06077t + 4.2159

Hence, When θ = 40°C, and

c = ln 60 + 2(0.06077) = 4.2159

ln 40 = -0.06077t + 4.2159

time, t =

4.2159 − ln 40 = 8.672 min = 8 min 40 s 0.06077


405

EXERCISE 180 Page 450 2. Solve the differential equation (2y – 1)

If (2y – 1)

dy = ( 3x 2 + 1) , then dx

dy = ( 3x 2 + 1) , given x =1 when y = 2. dx

∫ ( 2y − 1) dy = ∫ ( 3x

2

+ 1) dx

i.e.

y2 − y = x 3 + x + c

x =1 when y = 2, hence,

4–2=1+1+c

Thus,

y 2 − y = x3 + x

4. Solve the differential equation 2y(1 – x) + x(1 + y)

If 2y(1 – x) + x(1 + y)

dy =0 dx

then

x(1 + y)

from which, c = 0

dy = 0, given x = 1 when y = 1. dx

dy = -2y(1 – x) = 2y(x – 1) dx

Thus,

∫

⎛ 1+ y ⎞ ⎛ 2(x − 1) ⎞ ⎜ ⎟ dy = ∫ ⎜ ⎟ dx ⎝ x ⎠ ⎝ y ⎠

i.e.

∫

⎛1 ⎞ 2⎞ ⎛ ⎜ + 1⎟ dy = ∫ ⎜ 2 − ⎟ dx x⎠ ⎝ ⎝y ⎠ ln y + y = 2x – 2 ln x + c

x = 1 when y = 1, hence,

ln 1 + 1 = 2 – 2 ln 1 + c

Thus,

ln y + y = 2x – 2 ln x – 1

or

ln y + 2 ln x = 2x – y – 1

i.e.

ln y + ln x 2 = 2x – y – 1

and

ln ( x 2 y ) = 2x − y − 1

5. Show that the solution of the equation

Since i.e.

y 2 + 1 y dy = then x 2 + 1 x dx

∫

y 2 + 1 y dy = is of the form x 2 + 1 x dx

from which, c = -1

⎛ y2 + 1 ⎞ ⎜ 2 ⎟ = constant. ⎝ x +1⎠

y x dy = ∫ 2 dx y +1 x +1 2

1 1 ln ( y 2 + 1) = ln ( x 2 + 1) + c 2 2


406

i.e.

1 1 ⎡ ⎤ 2 2 − ln x 2 + 1 2 = c ln y + 1 ) ( )⎥ ⎢ ( ⎣ ⎦

i.e.

⎛ y2 + 1 ⎞ 2 ln ⎜ 2 ⎟ = c ⎝ x +1⎠

or

⎛ y2 + 1 ⎞ ln ⎜ 2 ⎟ = c ⎝ x +1⎠

1

⎛ y2 + 1 ⎞ c ⎜ 2 ⎟ = e = a constant ⎝ x +1⎠

and

7. Determine the equation of the curve which satisfies the equation xy

dy = x 2 − 1 , and which dx

es through the point (1, 2).

Since xy

dy = x 2 − 1 then dx

∫

y dy = ∫

x2 −1 1⎞ ⎛ dx = ∫ ⎜ x − ⎟ dx x x⎠ ⎝

y2 x 2 = − ln x + c 2 2

i.e.

If the curve es through (1, 2) then x = 1 and y = 2, hence,

22 12 = − ln1 + c 2 2

Thus,

y2 x 2 3 = − ln x + 2 2 2

or

y 2 = x 2 − 2 ln x + 3

from which, c =

3 2

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 CR

dV +V=E dt

(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 CR

dV + V = E then dt

dV E − V = dt CR


407

dV

dt

∫ E − V = ∫ CR

i.e.

− ln ( E − V ) =

from which, At t = 0, V = 0, hence,

- ln E = k − ln ( E − V ) =

Thus,

t +k CR

ln E − ln ( E − V ) =

t − ln E CR t CR

and

t ⎛ E ⎞ ln ⎜ ⎟= ⎝ E − V ⎠ CR

i.e.

E = eCR E−V

t

E

i.e. e

= E−V

t CR

E

V =E−

and

e

t CR

= E − Ee

t 3.0 − ⎛ ⎞ − ⎛ −6 3 (b) Voltage, V = E ⎜1 − e C R ⎟ = 25 ⎜⎜ 1 − e 20×10 ×200×10 ⎜ ⎟ ⎝ ⎝ ⎠

9. Determine the value of p, given that x 3

i.e.

t CR

t − ⎛ ⎞ V = E ⎜ 1 − e CR ⎟ volts ⎜ ⎟ ⎝ ⎠

i.e.

Since x 3

−

dy =p–x dx

then

∫ dy = ∫

⎞ −0.75 ) = 13.2 V ⎟⎟ = 25 (1 − e ⎠

dy = p – x, and that y = 0 when x = 2 and when x = 6. dx

p−x ⎛ p 1 ⎞ dx = ∫ ⎜ 3 − 2 ⎟ dx 3 x ⎝x x ⎠

∫ dy = ∫ ( px

−3

− x −2 ) dx

px −2 x −1 − +c −2 −1

i.e.

y=

i.e.

y=−

p 1 + +c 2 2x x


408

y = 0 when x = 2, hence,

p 1 0 = − + +c 8 2

y = 0 when x = 6, hence,

0=−

(1) – (2) gives:

⎛1 1 ⎞ ⎛1 1⎞ 0 = −p ⎜ − ⎟ + ⎜ − ⎟ ⎝ 8 72 ⎠ ⎝ 2 6 ⎠

i.e.

p 1 0=− + 9 3

i.e.

p 1 = from which, p = 3 9 3

p 1 + +c 72 6

(1) (2)


409

CHAPTER 47 HOMOGENEOUS FIRST ORDER DIFFERENTIAL EQUATIONS EXERCISE 181 Page 452 1. Find the general solution of: x 2 = y 2

dy dx

dy dy x 2 gives: (i) Rearranging x = y = , which is homogeneous in x and y. dx dx y 2 2

(ii) Let y = vx, then

2

dy dv = v+x dx dx

dy dv x2 1 = 2 2 = 2 (iii) Substituting for y and gives: v + x dx v x v dx (iv) Separating the variables gives: x Integrating both sides gives:

i.e.

v2 1 dv = dx 3 1− v x

v2 1 dv = ∫ dx 3 1− v x

1 − ln (1 − v3 ) = ln x + c 3

Hence,

(v) Replacing v by

∫

dv 1 1 − v3 = 2 −v= 2 dx v v

(using u = 1 − v3 substitution)

y 1 ⎛ y3 ⎞ gives: − ln ⎜ 1 − 3 ⎟ = ln x + c, which is the general solution. x 3 ⎝ x ⎠ 1 ⎛ x3 − y 3 ⎞ − ln ⎜ ⎟ = ln x + c 3 ⎝ x3 ⎠

i.e.

3. Find the particular solution of the differential equation: ( x 2 + y 2 ) dy = x y dx, given that x = 1

when y = 1.

(i) Rearranging ( x 2 + y 2 ) dy = xy dx gives: (ii) Let y = vx, then

dy xy = 2 , which is homogeneous in x and y. dx x + y 2

dy dv = v+x dx dx

(iii) Substituting for y and

dy dv x(vx) vx 2 v = 2 = = gives: v + x 2 2 2 2 dx x + v x x (1 + v ) 1 + v 2 dx

(iv) Separating the variables gives:

v − v (1 + v 2 ) v − v − v3 − v3 dv v = − = = = x v dx 1 + v 2 1 + v2 1 + v2 1 + v2


410

1 1 + v2 dv = dx 3 v x

−

i.e.

Integrating both sides gives: − ∫ −∫

i.e.

1 + v2 1 dv = ∫ dx 3 v x

1 1 1 dv − ∫ dv = ∫ dx 3 v v x

Hence,

v −2 − ln v = ln x + c 2

or

1 − ln v = ln x + c 2v 2

(v) Replacing v by

y gives: x

1 ⎛y⎞ 2⎜ ⎟ ⎝x⎠

2

− ln

y = ln x + c x

x2 y − ln = ln x + c 2 2y x

or When x = 1, y = 1, thus: Thus, the particular solution is:

1 − ln1 = ln 1 + c 2

which is the general solution. from which, c =

1 2

x2 y 1 − ln = ln x + 2 2y x 2

i.e.

x2 y⎞ 1 ⎛ = ln ⎜ (x) ⎟ + 2 2y x⎠ 2 ⎝

i.e.

x2 1 = ln y + 2 2y 2

1⎞ ⎛ x 2 = 2y 2 ⎜ ln y + ⎟ 2⎠ ⎝

or

⎛ 2y − x ⎞ dy 5. Find the particular solution of the differential equation: ⎜ = 1 , given that y = 3 when ⎟ ⎝ y + 2x ⎠ dx x = 2.

⎛ 2y − x ⎞ dy dy 2x + y = , which is homogeneous in x and y. = 1 gives: (i) Rearranging ⎜ ⎟ dx 2y − x ⎝ y + 2x ⎠ dx


dy dv = v+x dx dx


dy dv 2x + vx 2 + v = = gives: v + x dx 2vx − x 2v − 1 dx


411

(iv) Separating the variables gives: x

dv 2 + v (2 + v) − v(2v − 1) 2 − 2v 2 + 2v = −v= = dx 2v − 1 2v − 1 2v − 1 dv 2 − 2v 2 + 2v −2v 2 + 2v + 2 = = dx 2v − 1 2v − 1

i.e.

x

and

2v − 1 1 dv = dx 2 −2v + 2v + 2 x

Integrating both sides gives:

∫

2v − 1 1 dv = ∫ dx 2 −2v + 2v + 2 x

i.e.

1 2v − 1 1 dv = ∫ dx 2 ∫ 2 1+ v − v x

Hence,

1 − ln (1 + v − v 2 ) = ln x + c 2

2 y 1 ⎛ y ⎛y⎞ ⎞ (v) Replacing v by gives: − ln ⎜1 + − ⎜ ⎟ ⎟ = ln x + c, which is the general solution. x 2 ⎜⎝ x ⎝ x ⎠ ⎟⎠

When x = 2, y = 3, thus:

i.e.

1 ⎛1⎞ − ln ⎜ ⎟ = ln 2 + c 2 ⎝4⎠

i.e.

ln ( 4 ) 2 = ln 2 + c

1 ⎛ 3 9⎞ − ln ⎜ 1 + − ⎟ = ln 2 + c 2 ⎝ 2 4⎠ ⎛1⎞ ln ⎜ ⎟ ⎝4⎠

and

−

1 2

= ln 2 + c

1

and

ln 2 = ln 2 + c,

from which, c = 0

2 1 ⎛ y ⎛y⎞ ⎞ Thus, the particular solution is: − ln ⎜1 + − ⎜ ⎟ ⎟ = ln x 2 ⎜⎝ x ⎝ x ⎠ ⎟⎠

i.e.

1 ⎛ x 2 + xy − y 2 ⎞ − ln ⎜ ⎟ = ln x 2 ⎝ x2 ⎠

i.e.

⎛ x 2 + xy − y 2 ⎞ ln ⎜ ⎟ x2 ⎝ ⎠

−

1 2

= ln x

1

or

⎛ ⎞2 x2 =x ⎜ 2 2 ⎟ ⎝ x + xy − y ⎠

from which,

i.e.

x 2 + xy − y 2 = 1

x2 =x x 2 + xy − y 2

or

i.e.

x x 2 + xy − y 2

=x

x 2 + xy − y 2 = 1


412

EXERCISE 182 Page 454 1. Solve the differential equation: xy3dy = ( x 4 + y 4 ) dx dy x 4 + y 4 (i) Rearranging xy dy = ( x + y ) dx gives: = , which is homogeneous in x and y. dx xy3 3


4

4

dy dv = v+x dx dx


dy dv x 4 + v 4 x 4 x 4 (1 + v 4 ) 1 + v 4 gives: v + x = = = 3 dx dx x ( v3 x 3 ) v3 x 4 v

(iv) Separating the variables gives: x

dv 1 + v 4 1 + v4 − v4 1 = 3 −v= = 3 dx v v3 v

v3dv =

i.e. Integrating both sides gives:

1 dx x 1

∫ v dv = ∫ x dx 3

v4 = ln x + c 4

Hence,

4

(v) Replacing v by

y gives: x

⎛y⎞ ⎜ ⎟ ⎝ x ⎠ = ln x + c 4

y4 = ln x + c 4x 4

i.e.

y 4 = 4x4 (ln x + c)

or

3. Solve the differential equation: 2x

dy = x + 3y , given that when x = 1, y = 1. dx

(i) Rearranging 2x

dy dy x + 3y = , which is homogeneous in x and y. = x + 3y gives: dx dx 2x


dy dv = v+x dx dx


dy dv x + 3vx x(1 + 3v) 1 + 3v gives: v + x = = = dx dx 2x 2x 2

(iv) Separating the variables gives: i.e.

x

dv 1 + 3v 1 + 3v − 2v 1 + v = −v= = dx 2 2 2

2 1 dv = dx 1+ v x


413

2

1


∫ 1 + v dv = ∫ x dx

Hence,

2 ln(1+v) = ln x + c

(v) Replacing v by

⎛ y⎞ 2 ln ⎜ 1 + ⎟ = ln x + c, which is the general solution. ⎝ x⎠

y gives: x

2 ln 2 = ln1 + c


from which,

c = 2 ln 2

⎛x+y⎞ Thus, the particular solution is: 2 ln ⎜ ⎟ = ln x + 2 ln 2 ⎝ x ⎠

⎛x+y⎞ 2 ln ⎜ ⎟ = ln x + ln 2 = ln 4x ⎝ 2 ⎠ 2

i.e.

⎛x+y⎞ ⎜ ⎟ = 4x ⎝ x ⎠ 2

from which,

(x + y) 2 = 4x x2

(x + y)2 = 4x 3

i.e.

5. Determine the particular solution of

(i)

i.e.

dy x 3 + y3 = , given that x = 1 when y = 4. dx xy 2

dy x 3 + y3 = is homogeneous in x and y. dx xy 2


dy dv = v+x dx dx


dy dv x 3 + v3 x 3 x 3 (1 + v3 ) 1 + v3 = = = 2 gives: v + x dx x ( v 2 x 2 ) x 3v2 v dx

(iv) Separating the variables gives: x Integrating both sides gives

1

∫ v dv = ∫ x dx 2

v3 = ln x + c 3

Hence,

(v) Replacing v by

dv 1 + v3 1 + v3 − v3 1 1 = 2 −v= = 2 , i.e. v 2 dv = dx 2 dx v v v x

y gives: x


y3 = ln x + c, which is the general solution. 3x 3 64 = ln1 + c 3

from which, c =

64 3


414

64 y3 = ln x + 3 3x 3

Thus, the particular solution is:

i.e.

64 ⎞ ⎛ y3 = 3x 3 ⎜ ln x + ⎟ 3 ⎠ ⎝

or

y 3 = x 3 ( 3 ln x + 64 )

6. Show that the solution of the differential equation:

dy y3 − xy 2 − x 2 y − 5x 3 = is of the form: dx xy 2 − x 2 y − 2x 3

y 2 4y ⎛ y − 5x ⎞ + + 18ln ⎜ ⎟ = ln x + 42 , when x = 1 and y = 6. 2 2x x ⎝ x ⎠ dy y3 − xy 2 − x 2 y − 5x 3 = (i) is homogeneous in x and y. dx xy 2 − x 2 y − 2x 3


dy dv = v+x dx dx

(iii) Substituting for y and v+x

dy gives: dx

dv v3 x 3 − xv 2 x 2 − x 2 vx − 5x 3 x 3 (v3 − v 2 − v − 5) v3 − v 2 − v − 5 = = = dx v2 − v − 2 x ( v 2 x 2 ) − x 2 (vx) − 2x 3 x 3 ( v2 − v − 2)

v3 − v 2 − v − 5 − v ( v 2 − v − 2 ) dv v3 − v 2 − v − 5 = −v= (iv) Separating the variables gives: x dx v2 − v − 2 v2 − v − 2 =

v−5 v −v−2 2

1 v2 − v − 2 dv = dx x v−5

i.e.


∫

v2 − v − 2 1 dv = ∫ dx v−5 x

v+4 v − 5 v2 − v − 2

v 2 − 5v 4v – 2 4v – 20 18

Thus,

18

1

∫ v + 4 + v − 5 dv = ∫ x dx


415

Hence,

v2 + 4v + 18ln(v − 5) = ln x + c 2

y y 2 4y ⎛y ⎞ (v) Replacing v by gives: + + 18ln ⎜ − 5 ⎟ = ln x + c, which is the general solution. 2 2x x x ⎝x ⎠ When x = 1, y = 6, thus:

36 24 ⎛6 ⎞ + + 18ln ⎜ − 5 ⎟ = ln 1 + c 2 1 ⎝1 ⎠

Thus, the particular solution is:

from which, c = 42

y 2 4y ⎛ y − 5x ⎞ + + 18 ln ⎜ ⎟ = ln x + 42 2 2x x ⎝ x ⎠


416

CHAPTER 48 LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS EXERCISE 183 Page 457 2. Solve the differential equation:

dy = x( 1 – 2y) = x – 2xy dx y e∫

Hence,

dy = x( 1 – 2y) dx dy + 2xy = x dx

i.e. 2x dx

2x dx = ∫ e ∫ x dx

i.e.

ye x = ∫ e x x dx

i.e.

2 1 2 ye x = e x + c 2

2

2

y=

or

from which, P = 2x and Q = x

2 1 + c e− x 2

⎛ dy ⎞ 4. Solve the differential equation: x ⎜ + 1⎟ = x 3 − 2y , given x = 1 when y = 3. ⎝ dx ⎠ ⎛ dy ⎞ Since x ⎜ + 1⎟ = x 3 − 2y ⎝ dx ⎠ from which,

then

dy 2y +1 = x2 − dx x dy 2y + = x2 −1 dx x 2

y e∫ x

Hence,

dx

i.e. P =

2 and Q = x 2 − 1 x

2

dx = ∫ e ∫ x ( x 2 − 1) dx

i.e.

ye 2 ln x = ∫ e 2 ln x ( x 2 − 1)dx

i.e.

ye ln x = ∫ eln x ( x 2 − 1)dx

i.e.

y x 2 = ∫ x 2 ( x 2 − 1) dx

i.e.

y x 2 = ∫ ( x 4 − x 2 ) dx

i.e.

yx 2 =

2

2

since eln A = A

x5 x3 − +c 5 3


1 1 3= − +c 5 3

from which,

1 1 45 + 5 − 3 47 c = 3+ − = = 3 5 15 15


417

yx 2 =

Hence,

y=

i.e.

dy + x = 2y dx

dy - 2y = - x dx

then

y e∫

Hence,

x3 x 47 − + 5 3 15x 2

dy + x = 2y dx


Since

x 5 x 3 47 − + 5 3 15

− 2dx

from which, P = -2 and Q = - x

− 2dx (− x) dx = ∫ e∫

ye −2x = ∫ e −2x (− x) dx

i.e. Using integration by parts on

∫xe

−2x

dx :

Let u = x, then and dv = e −2x dx

∫ xe

Thus,

−2x

(1) du = 1 and du = dx dx and v =

∫e

−2x

1 dx = − e −2x 2

1 ⎛ 1 ⎞ dx = (x) ⎜ − e−2x ⎟ − ∫ − e −2x dx 2 ⎝ 2 ⎠ 1 1 1 1 = − xe −2x + ∫ e−2x dx = − xe −2x − e−2x 2 2 2 4


1 ⎛ 1 ⎞ ye −2x = − ⎜ − xe−2x − e−2x ⎟ + c 4 ⎝ 2 ⎠

i.e.

y e −2x =

1 −2x 1 −2x xe + e +c 2 4

or

y=

1 1 c x + + −2x 2 4 e

i.e.

y=

1 1 x + + ce 2x 2 4


418

EXERCISE 184 Page 458 2. Solve the differential equation: t

Since t then

dθ + sec t ( t sin t + cos t ) θ = sec t , given t = π when θ = 1. dt

dθ + sec t ( t sin t + cos t ) θ = sec t dt

dθ ⎛ sec t cos t ⎞ sec t + ⎜ sec t sin t + ⎟θ = dt ⎝ t t ⎠

⎛ 1 ⎞ ⎜ ⎟ cos t sec t cos t ⎛ 1 ⎞ 1 cos t ⎠ ⎝ =⎜ = tan t + from which, P = sec t sin t + ⎟ sin t + t t t ⎝ cos t ⎠ ⎛

1⎞

∫ ⎜⎝ tan t + t ⎟⎠dt

=∫ e

⎛

and Q =

sec t t

1⎞

∫ ⎜⎝ tan t + t ⎟⎠ dt sec t dt t

Hence,

θe

i.e.

θ e(ln sec t + ln t ) = ∫ e(ln sec t + ln t )

i.e.

θ eln(t sec t ) = ∫ eln(t sec t )

i.e.

θ t sec t = ∫ t sec t

and

θ t sec t = tan t + c

sec t dt t

sec t dt t

sec t dt = ∫ sec 2 t dt t

t = π when θ = 1, hence, π sec π = tan π + c -π=c

i.e.

θ t sec t = tan t − π

Thus,

or

sin t tan t sin t π π π θ= − = cos t − = − cos t 1 t sec t t sec t t 1 t t t cos t cos t

i.e.

θ=

1 ( sin t − π cos t ) t

4. Show that the solution of the differential equation

dy 4 3 − 2 ( x + 1) = y dx (x + 1)

is: y = ( x + 1) ln ( x + 1) , given that x = 0 when y = 0 4

Since

2

dy 4 3 − 2 ( x + 1) = y dx (x + 1)

from which,

dy 4 3 y = 2 ( x + 1) − dx (x + 1)


419

P= −

i.e.

4 x +1

and Q = 2(x + 1)3 4 dx x +1

4

Hence,

y e∫

i.e.

y e −4ln(x +1) = ∫ e −4ln(x +1) 2(x + 1)3 dx

i.e.

y eln( x +1) = ∫ eln(x +1) 2(x + 1)3 dx

i.e.

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

and

y 2 =∫ dx = 2 ln(x + 1) + c 4 (x + 1) (x + 1)

−

− dx = ∫ e ∫ x +1 2(x + 1)3 dx

−4


−4

0 = 2 ln 1 + c

from which, c = 0

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

Thus,

y = ( x + 1) ln ( x + 1) 4

or

6. The equation

2

dv = -(av + bt), where a and b are constants, represents an equation of motion dt

when a particle moves in a resisting medium. Solve the equation for v given that v = u when t = 0.

Since

dv = -(av + bt) dt

v e∫

Hence,

dv + av = -bt dt

then

a dt

v eat = − b ∫ t eat

Using integration by parts on

∫ te

∫ te

at

P = a and Q = -bt

a dt = ∫ e ∫ (− bt) dt

i.e.

Thus,

from which,

at

dt :

(1)

Let u = t, then

du = 1 and du = dt dt

and dv = eat dt

and v =

∫e

at

1 dt = ea t a

1 t 1 ⎛1 ⎞ dt = (t) ⎜ ea t ⎟ − ∫ ea t dt = ea t − 2 ea t a a a ⎝a ⎠


1 ⎛t ⎞ v eat = − b ⎜ ea t − 2 ea t ⎟ + c a ⎝a ⎠

v = u when t = 0, hence,

u=

b +c a2

from which,

c=u-

b a2


420

1 b bt b b ⎛t ⎞ v eat = − b ⎜ ea t − 2 ea t ⎟ + u − 2 = − ea t + 2 eat + u − 2 a a a a a ⎝a ⎠

Thus,

v=−

or

bt b ⎛ b⎞ + 2 + ⎜ u − 2 ⎟ e− a t a a ⎝ a ⎠

or

b bt ⎛ b⎞ − + ⎜ u − 2 ⎟ e− a t 2 a a ⎝ a ⎠

7. In an alternating current circuit containing resistance R and inductance L the current i is given

di = E 0 sin ωt . Given i = 0 when t = 0, show that the solution of the equation is given dt Rt E ⎛ ⎞ ⎛ E ωL ⎞ − i = ⎜ 2 0 2 2 ⎟ ( R sin ωt − ωL cos ωt ) + ⎜ 2 0 2 2 ⎟ e L ⎝R +ω L ⎠ ⎝R +ω L ⎠

by: Ri + L by:

Since Ri + L i.e. Hence,

di = E 0 sin ωt dt

then

di Ri E 0 + = sin ωt dt L L

L

di = E 0 sin ωt − Ri dt

from which,

R L

P=

and

di E 0 Ri sin ωt − = dt L L

and Q =

E0 sin ωt L

R dt ⎛ E ⎛ R dt ⎞ ⎞ i ⎜ e ∫ L ⎟ = ∫ e ∫ L ⎜ 0 sin ωt ⎟ dt ⎝ L ⎠ ⎝ ⎠

i.e.

⎛E ⎞ i e L = ∫ e L ⎜ 0 sin ωt ⎟ dt ⎝ L ⎠

i.e.

ie L =

Rt

Rt

Rt

From page XX of textbook,

∫

⎤ E 0 ⎡ RLt ⎢ ∫ e sin ωt dt ⎥ L ⎣ ⎦

eax sin bx dx =

(1)

eax ( a sin bx − b cos bx ) + c a 2 + b2 Rt

∫e

Hence,

Rt L

sin ωt dt =

⎛R ⎞ ⎜ sin ωt − ω cos ωt ⎟ + c L ⎠ ⎛R⎞ 2 ⎝ ⎜ ⎟ +ω ⎝L⎠ eL 2

Rt

Substituting into (1) gives:

ie

Rt L

E eL ⎛R ⎞ = 0 ⎜ sin ωt − ω cos ωt ⎟ + c 2 L ⎛R ⎠ 2⎞⎝ L ⎜ 2 +ω ⎟ ⎝L ⎠ Rt

E eL ⎛R ⎞ sin ωt − ω cos ωt ⎟ + c = 0 2 2 2 ⎜ L ⎛R +ω L ⎞⎝ L ⎠ ⎜ ⎟ 2 L ⎝ ⎠ Rt

i.e.

ie

Rt L

=

⎛R ⎞ ⎜ sin ωt − ω cos ωt ⎟ + c ⎠ (R + ω L ) ⎝ L E0 L e L 2

2 2


421

i = 0 when t = 0, hence,

0=

E0 L (−ω) + c R + ω2 L2

from which, c =

2

E0 L ω R 2 + ω2 L2

Rt

Thus,

ie

Rt L

=

E0 L ω ⎛R ⎞ ⎜ sin ωt − ω cos ωt ⎟ + 2 2 2 ⎠ (R + ω L ) (R + ω L ) ⎝ L E0 L e L 2

2 2

Rt − E0 L ω ⎛R ⎞ L ω − ω ω + sin t cos t e ⎟ 2 2 2 ⎜ 2 2 2 L ⎠ (R + ω L ) (R + ω L ) ⎝

E0 L

i.e.

i=

or

Rt ⎛ E0 ω L ⎞ − L i= ( R sin ωt − ωL cos ωt ) + ⎜ 2 2 2 ⎟ e ⎝R +ω L ⎠ ( R 2 + ω2L2 )

E0

8. The concentration, C, of impurities of an oil purifier varies with time t and is described by the equation a

dC = b + dm − Cm , where a, b, d and m are constants. Given C = c0 when t = 0, solve dt

mt mt − − ⎞ ⎛b ⎞⎛ a a the equation and show that: C = ⎜ + d ⎟ ⎜ 1 − e ⎟ + c0 e ⎝m ⎠⎝ ⎠

Since a and

dC = b + dm − Cm dt

dC m b dm + C= + dt a a a m

Hence,

i.e.

dC b d m Cm = + − dt a a a

then

Ce

∫ a dt

Ce

mt a

from which,

=∫ e =∫ e

P=

m b dm and Q = + a a a

∫ a dt ⎛ b d m ⎞ ⎜ + ⎟ dt a ⎠ ⎝a m

mt a

⎛ b dm⎞ ⎜ + ⎟ dt a ⎠ ⎝a

mt

i.e.

Ce

mt a

ae a ⎛ b dm⎞ = ⎜ + ⎟+k m ⎝a a ⎠

C = c 0 when t = 0, hence, c 0 =

a ⎛ b dm⎞ b ⎜ + ⎟+k = +d+k m⎝a a ⎠ m

from which, k = c0 −

b −d m

mt

Thus,

i.e.

i.e.

Ce

mt a

ae a ⎛ b dm⎞ b = ⎜ + ⎟ + c0 − − d m ⎝a a ⎠ m

C=

mt a ⎛ b dm⎞ ⎛ b ⎞ −a + + − − c d e ⎜ ⎟ ⎜ 0 ⎟ m⎝a a ⎠ ⎝ m ⎠

mt b ⎛b ⎞ ⎛ ⎞ −a C = ⎜ + d ⎟ + ⎜ c0 − − d ⎟ e m ⎝m ⎠ ⎝ ⎠


422

i.e.

mt mt − ⎛b ⎞ ⎛b ⎞ − C = ⎜ + d ⎟ − ⎜ + d ⎟ e a + c0 e a ⎝m ⎠ ⎝m ⎠

or

mt mt − − ⎞ ⎛b ⎞⎛ a C = ⎜ + d⎟⎜1− e ⎟ + c 0e a ⎝m ⎠⎝ ⎠

9. The equation of motion of a train is given by: m

dv = mk (1 − e − t ) − mcv , where v is the speed, t dt

is the time and m, k and c are constants. Determine the speed, v, given v = 0 at t = 0.

Since m

dv = mk (1 − e − t ) − mcv dt

then

dv + cv = k (1 − e − t ) dt

and

v e∫

Hence,

cdt

dv = k (1 − e − t ) − cv dt

from which,

P=c

and

Q = k (1 − e − t )

c dt = ∫ e ∫ k(1 − e − t ) dt

i.e.

v ec t = ∫ ec t k(1 − e − t ) dt = k ∫ ( ec t − ec t − t ) dt

i.e.

⎡ ec t e t (c −1) ⎤ v ec t = k ⎢ − ⎥+z c −1 ⎦ ⎣ c

where z is the constant of integration

v = 0 when t = 0, hence,

0=

k k − +z c c −1

from which,

z=

k k ck − k(c − 1) k − = = c −1 c c(c − 1) c(c − 1)

Thus,

⎡ ec t e t (c −1) ⎤ k v ec t = k ⎢ − ⎥+ c − 1 ⎦ c(c − 1) ⎣ c

k k e − t ⎛ k ⎞ − ct − +⎜ ⎟e c c − 1 ⎝ c(c − 1) ⎠

i.e.

v=

or

⎧ 1 e− t e − ct ⎫ v=k⎨ − + ⎬ ⎩ c c − 1 c(c − 1) ⎭


423

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 y = 3 − , with dx x

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 y = y' = 3− dx x If initially x0 = 1 and y0 = 2, (and h = 0.1), then (y')0 = 3 −

2 =1 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 -

For line 3, where x0 = 1.2: and

(y')0 = 3 -

For line 4, where x0 = 1.3:

y0 2.1 =3= 1.0909 x0 1.1

y1 = y0 + h(y')0 = 2.1 + (0.1)(1.0909) = 2.209091 y0 2.209091 = 1.159091 =31.2 x0

y1 = y0 + h(y')0 = 2.209091 + (0.1)(1.15909) = 2.325000

The remaining lines of the table are completed in a similar way. A graph of the solution of

dy y = 3 − , with initial conditions x = 1 and y = 2 is shown below. dx x


424

3. (a) The differential equation

dy y + 1 = − has the initial conditions that y = 1 at x = 2. Produce a dx x

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 x − , x 2

determine the percentage error at x = 2.2.

(a)

dy y +1 = − dx x

i.e.

dy y = y ' = −1 − dx x

If initially x0 = 2.0 and y0 = 1, (and h = 0.1), then (y')0 = −1 −

1 = -1.5 2.0


For line 2, where x0 = 2.1 and h = 0.1: y1 = y0 + h(y')0 = 1 + (0.1)(-1.5) = 0.85 and

(y')0 = -1 -


y0 0.85 = -1 = -1.40476 x0 2.1

y1 = y0 + h(y')0 = 0.85 + (0.1)(-1.40476) = 0.709524


425

and

(y')0 = -1 -

y0 0.709524 = -1.322511 = -1 2.2 x0


y1 = y0 + h(y')0 = 0.709524 + (0.1)(-1.322511) = 0.577273

The remaining lines of the table are completed in a similar way. (b) If y =

4 x then when x = 2.2, y = 0.718182 − x 2

From the Euler method, when x = 2.2, y = 0.709524 Hence, percentage error =

0.718182 − 0.709524 0.008658 × 100% = × 100% 0.718182 0.718182 = 1.206%

4. Use Euler’s method to obtain a numerical solution of the differential equation

dy 2y =x− , dx x

given the initial conditions that y = 1 when x = 2, in the range x = 2.0(0.2)3.0. x2 If the solution of the differential equation is given by y = , determine the percentage error by 4 using Euler’s method when x = 2.8.

(a)

dy 2y = y' = x − dx x If initially x0 = 2.0 and y0 = 1, (and h = 0.2), then (y')0 = 2.0 −

2(1) = 1.0 2.0


For line 2, where x0 = 2.2 and h = 0.2: y1 = y0 + h(y')0 = 1 + (0.2)(1.0) = 1.2 and

(y')0 = 2.2 -


2(1.2) = 1.109091 2.2

y1 = y0 + h(y')0 = 1.2 + (0.2)(1.109091) = 1.4218182


426

and

(y')0 = 2.4 -


2(1.4218182) = 1.2151515 2.4

y1 = y0 + h(y')0 = 1.4218182 + (0.2)(1.2151515) = 1.664849

The remaining lines of the table are completed in a similar way. (b) If y =

x2 , then when x = 2.8, y = 1.96 4

From the Euler method, when x = 2.8, y = 1.928718 Hence, percentage error =

1.96 − 1.928718 0.031282 × 100% = ×100% 1.96 1.96 = 1.596%


427


1. Apply the Euler-Cauchy method to solve the differential equation

dy y = 3− for the range dx x

1.0(0.1)1.5, given the initial conditions that x = 1 when y = 2. dy y = y' = 3− dx x x 0 = 1, y 0 = 2 and h = 0.1 (y ′) 0 = 3 −

y0 2 = 3− =1 1 x0

x 1 = 1.1 and from equation (3), page 465, y P1 = y 0 + h(y ′) 0 = 2 + 0.1(1) = 2.1

y C1 = y 0 +

1 h [ (y ′) 0 + f(x 1 , y P1 ) 2

] = y0+

yP 1 h [ (y ′) 0 + 3 − 1 2 x1

=2+

1 2.1 (0.1) [ 1 + 3 − 2 1.1

] ]

= 2.10454546 (y ′) 1 = 3 −

yC1 x1

= 3−

2.10454546 = 1.08677686 1.1

Thus the first two lines of the Table below has been completed x 1.0 1.1 1.2 1.3 1.4 1.5

y 2 2.10454546 2.216666672 2.33461539 2.457142859 2.5883333335

y′ 1 1.08677686 1.152777773 1.204142008 1.244897958

For line 3, x 1 = 1.2 y P1 = y 0 + h(y ′) 0 = 2.10454546 + 0.1(1.08677686) = 2.213223146

y C1 = y 0 +

yP 1 h [ (y ′) 0 + 3 − 1 x1 2

]


428

= 2.10454546 +

(y ′) 1 = 3 −

yC1 x1

= 3−

1 2.213223146 (0.1) [ 1.08677686 + 3 − 2 1.2

] = 2.216666672

2.216666672 = 1.152777773 1.2

The remaining lines of the table are completed in a similar way.

2. Solving the differential equation in Problem 1 by the integrating method gives y =

3 1 . x+ 2 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 1 x+ then when x = 1.3, y = 2.334615385 2 2x

(a) By Euler’s method, when x = 1.3, y = 2.325000 Percentage error =

2.334615385 − 2.3250 × 100% = 0.412% 2.334615385

(b) By the Euler-Cauchy method, when x = 1.3, y = 2.33461539 Percentage error =

2.334615385 − 2.33461539 × 100% = 0.000000214% 2.334615385

3. (a) Apply the Euler-Cauchy method to solve the differential equation

dy − x = y for the range dx

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 = 2e x − x − 1 . Determine the percentage error, correct to 3 decimal places, when x = 0.4.

(a)

dy = y'= y + x dx x 0 = 0, y 0 = 1 and h = 0.1 (y ′) 0 = 1 + 0 = 1 x 1 = 0.1 and from equation (3), page 465, y P1 = y 0 + h(y ′) 0 = 1 + 0.1(1) = 1.1


429

y C1 = y 0 +

1 h [ (y ′) 0 + f(x 1 , y P1 ) 2

] = y0+

1 h [ (y ′) 0 + y P1 + x 1 2

]

=1+

1 (0.1) [ 1 + 1.1 + 0.1 2

]

= 1.11 (y ′) 1 = y C 1 + 0.1 = 1.11 + 0.1 = 1.21 Thus the first two lines of the Table below has been completed

For line 3, x 1 = 0.2 y P1 = y 0 + h(y ′) 0 = 1.11 + 0.1(1.21) = 1.231 y C1 = y 0 +

1 h [ (y ′) 0 + y P1 + x 1 2

= 1.11 +

]

1 (0.1) [ 1.21 + 1.231 + 0.2 2

] = 1.24205

(y ′) 1 = y C 1 + 0.2 = 1.24205+ 0.2 = 1.44205 The remaining lines of the table are completed in a similar way. (b) If y = 2e x − 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% = 0.117% 1.583649395


430


2. Obtain a numerical solution of the differential equation:

1 dy + 2y = 1 using the Runge-Kutta x dx

method in the range x = 0(0.2)1.0, given the initial conditions that x = 0 when y = 1.

If

1 dy 1 dy + 2y = 1 then = 1 − 2y x dx x dx

dy = x(1 − 2y) dx

and

1. x 0 = 0, y 0 = 1 and since h = 0.2, and the range is from x = 0 to x = 1.0, then x1 = 0.2, x 2 = 0.4, x 3 = 0.6, x 4 = 0.8, and x 5 = 1.0

Let n = 0 to determine y 1 : 2. k1 = f ( x 0 , y 0 ) = f (0, 1); since

dy = x(1 − 2y) , f (0, 1) = 0(1 - 2) = 0 dx

h h ⎞ 0.2 0.2 ⎞ ⎛ ⎛ , 1+ (0) ⎟ = f ( 0.1, 1) = 0.1(1 - 2) = -0.1 3. k 2 = f ⎜ x 0 + , y 0 + k1 ⎟ = f ⎜ 0 + 2 2 ⎠ 2 2 ⎝ ⎝ ⎠ h h ⎞ 0.2 0.2 ⎛ ⎛ ⎞ 4. k 3 = f ⎜ x 0 + , y0 + k 2 ⎟ = f ⎜ 0 + , 1+ (−0.1) ⎟ = f ( 0.1, 0.99 ) 2 2 ⎠ 2 2 ⎝ ⎝ ⎠ = 0.1(1 – 1.98) = -0.098 5. k 4 = f ( x 0 + h, y 0 + hk 3 ) = f ( 0 + 0.2, 1 + 0.2(−0.098) ) = f (0.2, 0.9804) = 0.2(1 - 2(0.9804)) = -0.19216 6. y n +1 = y n + y1 = y0 +

h {k1 + 2k 2 + 2k 3 + k 4 } and when n = 0: 6 h 0.2 {k1 + 2k 2 + 2k 3 + k 4 } = 1 + { 0 + 2(−0.1) + 2(−0.098) + −0.19216 } 6 6 =1+

0.2 { − 0.58816 } = 0.980395 6

Let n = 1 to determine y2 : 2. k1 = f ( x1 , y1 ) = f (0.2, 0.980395); since

dy = x(1 – 2y), dx

f (0.2, 0.980395) = 0.2(1 – 2(0.980395)) = -0.192158 h h ⎞ 0.2 0.2 ⎛ ⎛ ⎞ 3. k 2 = f ⎜ x1 + , y1 + k1 ⎟ = f ⎜ 0.2 + , 0.980395 + (−0.192158) ⎟ = f ( 0.3, 0.9611792 ) 2 2 ⎠ 2 2 ⎝ ⎝ ⎠ = 0.3(1 – 2(0.9611792) = -0.27670752


431

h h ⎞ 0.2 0.2 ⎛ ⎛ ⎞ 4. k 3 = f ⎜ x1 + , y1 + k 2 ⎟ = 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. k 4 = f ( x1 + h, y1 + hk 3 ) = 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. y n +1 = y n + y 2 = y1 +

h {k1 + 2k 2 + 2k 3 + k 4 } and when n = 1: 6 h {k1 + 2k 2 + 2k 3 + k 4 } 6

= 0.980395 +

0.2 { − 0.192158 + 2(−0.27670752) + 2(−0.271634548) + −0.340854472 } 6

= 0.980395 +

0.2 { − 1.629696608 } = 0.926071779 6

This completes the third row of the table below. In a similar manner y3 , y 4 and y5 can be calculated.

3. (a) The differential equation

dy y + 1 = − has the initial conditions that y = 1 at x = 2. Produce a dx x

numerical solution of the differential equation, correct to 6 decimal places, using the RungeKutta 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 x − x 2

determine the percentage error at x = 2.2

(a) If

dy y dy y ⎛y ⎞ +1 = − then = − − 1 = − ⎜ + 1⎟ dx x dx x ⎝x ⎠

1. x 0 = 2, y 0 = 1 and since h = 0.1, and the range is from x = 2.0 to x = 2.5, then x1 = 2.1, x 2 = 2.2, x 3 = 2.3, x 4 = 2.4, and x 5 = 2.5


432

Let n = 0 to determine y 1 : 2. k1 = f ( x 0 , y 0 ) = f (2, 1); since,

dy ⎛y ⎞ ⎛1 ⎞ = − ⎜ + 1⎟ , f (2, 1) = − ⎜ + 1⎟ = -1.5 dx ⎝x ⎠ ⎝2 ⎠

h h ⎞ 0.1 0.1 ⎛ ⎛ ⎞ ⎛ 0.925 ⎞ 3. k 2 = f ⎜ x 0 + , y 0 + k1 ⎟ = f ⎜ 2.0 + ,1.0 + (−1.5) ⎟ = f ( 2.05, 0.925 ) = − ⎜ + 1⎟ 2 2 ⎠ 2 2 ⎝ ⎝ ⎠ ⎝ 2.05 ⎠ = -1.451219512 h h ⎞ 0.1 0.1 ⎛ ⎛ ⎞ , 1.0 + (−1.451219512) ⎟ = f ( 2.05, 0.927439024 ) 4. k 3 = f ⎜ x 0 + , y0 + k 2 ⎟ = f ⎜ 2.0 + 2 2 ⎠ 2 2 ⎝ ⎝ ⎠ ⎛ 0.927439024 ⎞ + 1⎟ = -1.45240928 = −⎜ 2.05 ⎝ ⎠ 5. k 4 = f ( x 0 + h, y 0 + hk 3 ) = f ( 2.0 + 0.1, 1.0 + 0.1(−1.45240928) ) = f (2.1, 0.854759072) ⎛ 0.854759072 ⎞ + 1⎟ = -1.40702813 = −⎜ 2.1 ⎝ ⎠ 6. y n +1 = y n + y1 = y0 +


= 1.0 +

0.1 { − 1.5 + 2(−1.451219512) + 2(−1.45240928) + (−1.40702813) } 6

= 1.0 +

0.1 { − 8.714285714 } = 0.854762 6

Let n = 1 to determine y2 : 2. k1 = f ( x1 , y1 ) = f (2.1, 0.854762); since

dy ⎛y ⎞ = − ⎜ + 1⎟ , dx ⎝x ⎠

⎛ 0.854762 ⎞ + 1⎟ = - 1.407029524 f (2.1, 0.854762) = − ⎜ ⎝ 2.1 ⎠ h h ⎞ 0.1 0.1 ⎛ ⎛ ⎞ , 0.854762 + (−1.407029524) ⎟ 3. k 2 = f ⎜ x1 + , y1 + k1 ⎟ = f ⎜ 2.1 + 2 2 ⎠ 2 2 ⎝ ⎝ ⎠ ⎛ 0.784410523 ⎞ = f(2.15, 0.784410523) = − ⎜ + 1⎟ = - 1.364842104 2.15 ⎝ ⎠ h h ⎞ 0.1 0.1 ⎛ ⎛ ⎞ , 0.854762 + (−1.364842104) ⎟ 4. k 3 = f ⎜ x1 + , y1 + k 2 ⎟ = f ⎜ 2.1 + 2 2 ⎠ 2 2 ⎝ ⎝ ⎠ ⎛ 0.786519894 ⎞ = f ( 2.15, 0.786519894 ) = − ⎜ + 1⎟ = - 1.365823207 2.15 ⎝ ⎠


433

5. k 4 = f ( x1 + h, y1 + hk 3 ) = f ( 2.2, 0.854762 + 0.1(−1.365823207) ) ⎛ 0.718179679 ⎞ + 1⎟ = - 1.326445309 = f (2.2, 0.718179679) = − ⎜ 2.2 ⎝ ⎠ 6. y n +1 = y n + y 2 = y1 +


= 0.854762 +

0.1 { − 1.407029524 + 2(−1.364842104) + 2(−1.365823207) − 1.326445309 } 6

= 0.854762 +

0.1 { − 8.194805455 } = 0.718182 6

This completes the third row of the Table below. In a similar manner y3 , y 4 and y5 can be calculated and the results are as shown.

(b) If y =

4 x − when x = 2.2, y = 0.718182 x 2

By the Runge-Kutta method, when x = 2.2, y = 0.718182 also. Hence, there is no error.


434

CHAPTER 50 SECOND ORDER DIFFERENTIAL EQUATIONS OF d2y dy + cy = 0 THE FORM a 2 + b dx dx EXERCISE 188 Page 477 3. Determine the general solution of

d2 y dy + 2 + 5y = 0 2 dx dx

d2 y dy + 2 + 5y = 0 in D-operator form is: 2 dx dx

m=

+ 2D + 5 ) y = 0

−2 ± 22 − 4(1)(5) −2 ± −16 −2 ± j4 = = = −1 ± j2 2(1) 2 2

Hence, the general solution is:

y = e − x { A cos 2x + B sin 2x}

5. Find the particular solution of 4

4

2

m 2 + 2m + 5 = 0

The auxiliary equation is: and

(D

d2 y dy dy = -2. − 5 + y = 0 when at t = 0, y = 1 and 2 dt dt dt

d2 y dy − 5 + y = 0 in D-operator form is: 2 dx dt

The auxiliary equation is:

( 4D

2

− 5D + 1) y = 0

4m 2 − 5m + 1 = 0

i.e.

(4m – 1)(m – 1) = 0

from which,

m=


y = Ae 4 + Be t

At t = 0, y = 1, hence,

1=A+B

1 and m = 1 4 1

t

(1)

dy 1 14 t = Ae + Be t dt 4 At t = 0,

dy = -2, hence, dt

-2=

1 A+B 4 3 A 4

(1) – (2) gives:

3=

From equation (1), when A = 4,

1=4+B

Hence, the particular solution is:

(2)

from which, A = 4 from which, B = -3

1 t 4

y = 4e − 3et


435

7. Find the particular solution of

d2x dx dx − 6 + 9x = 0 when at t = 0, x = 2 and = 0. 2 dt dt dt

d2x dx − 6 + 9x = 0 in D-operator form is: 2 dt dt

(D

2

− 6D + 9 ) x = 0

m 2 − 6m + 9 = 0

The auxiliary equation is: i.e.

(m – 3)(m – 3) = 0

from which,

m = 3 twice


x = ( At + B ) e3t

At t = 0, x = 2, hence,

2=B dx = ( At + B ) ( 3e3t ) + ( e3t ) A dt

At t = 0,

dx = 0, hence, dt


0 = 3B + A

and since B = 2,

x = ( −6t + 2 ) e3t

or

A = -6

x = 2 ( 1 − 3t ) e 3t

d2 y dy dy + 6 + 13y = 0 when at x = 0, y = 4 and = 0. 8. Find the particular solution of 2 dx dx dx d2 y dy + 6 + 13y = 0 in D-operator form is: 2 dx dx The auxiliary equation is:

(D

2

+ 6D + 13) y = 0

m 2 + 6m + 13 = 0

−6 ± 62 − 4(1)(13) −6 ± −16 −6 ± j4 m= = = = −3 ± j2 2(1) 2 2

and


y = e −3x {A cos 2x + Bsin 2x}

At x = 0, y = 4, hence,

4=A

dy = ( e −3x ) ( −2A sin 2x + 2B cos 2x ) + ( A cos 2x + Bsin 2x ) ( −3e −3x ) dx At x = 0,

dy = 0, hence, dx

0 = 2B – 3A and since A = 4,

B=6


y = e −3x {4 cos 2x + 6sin 2x}

or

y = 2e −3x { 2cos 2x + 3sin 2x}


436


1. The charge, q, on a capacitor in a certain electrical circuit satisfies the differential equation dq d 2q dq + 4 + 5q = 0 . Initially (i.e. when t = 0), q = Q and = 0. Show that the charge in the 2 dt dt dt circuit can be expressed as: q =

5 Q e −2t sin(t + 0.464)

d 2q dq + 4 + 5q = 0 in D-operator form is: 2 dt dt

2

+ 4D + 5 ) y = 0

m 2 + 4m + 5 = 0

The auxiliary equation is: and

(D

m=

−4 ± 42 − 4(1)(5) −4 ± −4 −4 ± j2 = = = −2 ± j 2(1) 2 2


q = e −2t {A cos t + Bsin t}

At t = 0, q = Q, hence,

Q=A

dq = ( e −2t ) ( −A sin t + B cos t ) + ( A cos t + Bsin t ) ( −2e−2t ) dt At t = 0,

dq = 0, hence, dt

0 = B – 2A and since A = Q,


q = e −2t {Q cos t + 2Q sin t}

i.e.

q = Q e −2t {cos t + 2sin t}

B = 2Q

cos t + 2 sin t = R sin(t + α)

Let

= R[sin t cos α + cos t sin α] = R cos α (sin t) + R sin α (cos t) Hence,

2 = R cos α

from which,

cos α =

2 R

and

1 = R sin α

from which,

sin α =

1 R

There is only one quadrant where both sine and cosine are positive, i.e. the first quadrant. From the diagram below,

R=

22 + 11 = 5


437

α = tan −1

and

Hence,

cos t + 2 sin t =

Since, q = Q e −2t {cos t + 2sin t}

1 = 26.565° or 0.464 rad 2

5 sin(t + 0.464)

q = 5 Q e−2t sin(t + 0.464)

then

3. The motion of the pointer of a galvanometer about its position of equilibrium is represented by the equation I

d 2θ dθ + K + Fθ = 0 2 dt dt

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 of t given that when t = 0, θ = 0.3 and

I

dθ = 0. dt

d 2θ dθ + K + Fθ = 0 in D-operator form is: 2 dt dt

(I D

−2 −K ± K 2 − 4 I F − ( 2 ×10 ) ± m= = 2I

= Hence, the general solution is: At t = 0, θ = 0.3, hence,

+ K D + F) θ = 0

I m2 + K m + F = 0


and

2

( 2 ×10 ) − 4 ( 5 ×10 ) ( 0.2 ) −0.02 ± 0.0036 = 0.01 2 ( 5 ×10 ) −2 2

−3

−3

−0.02 ± j0.06 = −2 ± j6 0.01 θ = e −2t {A cos 6t + Bsin 6t}

0.3 = A

dθ = ( e −2t ) ( −6A sin 6t + 6Bcos 6t ) + ( A cos t + Bsin t ) ( −2e −2t ) dt At t = 0,

dθ = 0, hence, dt

0 = 6B – 2A and since A = 0.3,

B = 0.1


438


θ = e −2t {0.3cos 6t + 0.1sin 6t}

4. Determine an expression for x for a differential equation

d2x dx + 2n + n 2 x = 0 which represents 2 dt dt

a critically damped oscillator, given that at time t = 0, x = s and

d2x dx + 2n + n 2 x = 0 in D-operator form is: 2 dt dt

(D

dx = u. dt

+ 2nD + n 2 ) x = 0

2

m 2 + 2n m + n 2 = 0

The auxiliary equation is: i.e.

(m + n)(m + n) = 0

from which,

m = -n

twice


x = ( At + B ) e − n t

At t = 0, x = s, hence,

s=B dx = ( At + B ) ( −ne − n t ) + ( e − n t ) A dt

At t = 0,

dx = u, hence, dt


u = -nB + A

hence

A = u + nB = u + ns

x = {( u + ns ) t + s} e − n t

or

x = {s + ( u + ns ) t} e − n t

d 2i di 1 5. L 2 + R + i = 0 is an equation representing current i in an electric circuit. If inductance L dt dt C 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

L

di = 34. dt

1⎞ d 2i di 1 ⎛ + R + i = 0 in D-operator form is: ⎜ L D 2 + R D + ⎟ i = 0 2 dt dt C C⎠ ⎝


−R ± R 2 − and

m=

2L

L m2 + R m +

1 =0 C

4L 4(0.25) −250 ± 2502 − C = 29.76 × 10−6 = −250 ± 170 = -160 or -840 2 ( 0.25 ) 0.5


i = A e −160t + Be −840t


439

When t =0, i = 0, hence,

0=A+B

(1)

di = −160Ae −160t − 840Be−840t dt When t = 0,

di = 34, hence, dt

160 × (1) gives:

34 = -160A – 840B

(2)

0 = 160A + 160B

(3)

(2) + (3) gives:

34 = -680B

From equation (1),

A=

i=

Hence,

1 −160t 1 −840t e − e 20 20

or

from which,

B=-

1 20

1 20 i=

(

1 −160t −840t −e e 20

)

6. The displacement s of a body in a damped mechanical system, with no external forces, satisfies the following differential equation: 2 when t = 0, s = 0 and

d 2s ds + 6 + 4.5s = 0 2 dt dt

where t represents time. If initially,

ds = 4, solve the differential equation for s in of t. dt

d 2s ds 2 2 + 6 + 4.5s = 0 in D-operator form is: dt dt The auxiliary equation is:

( 2D

2

+ 6D + 4.5 ) s = 0

2m 2 + 6m + 4.5 = 0

4m 2 + 12m + 9 = 0

or i.e.

(2m + 3)(2m + 3) = 0 3 twice 2

from which,

m=−


s = ( At + B ) e

At t = 0, s = 0, hence,

0=B

3 − t 2

⎛ 3 − 3t ⎞ ⎛ − 3t ⎞ ds = ( At + B ) ⎜ − e 2 ⎟ + ⎜ e 2 ⎟ (A) dt ⎝ 2 ⎠ ⎝ ⎠

At t = 0,

ds = 4, hence, dt


3 4 = − B+ A 2 s = 4t e

and since B = 0, A = 4

3 − t 2


440

CHAPTER 51 SECOND ORDER DIFFERENTIAL EQUATIONS OF d2y dy THE FORM a 2 + b + cy = f (x) dx dx EXERCISE 190 Page 483 2. Find the general solution of: 6

6

d2 y dy + 4 − 2y = 3x − 2 2 dx dx

d2 y dy + 4 − 2y = 3x − 2 in D-operator form is: 2 dx dx

6m 2 + 4m − 2 = 0

Auxiliary equation is: i.e.

( 6D

2

+ 4D − 2 ) y = 3x − 2

i.e.

3m 2 + 2m − 1 = 0

(3m – 1)(m + 1) = 0 m=

from which,

1 and m = -1 3 1

x

u = Ae 3 + Be − x

Hence, the complementary function, C.F., Let the particular integral, P.I., v = ax + b

( 6D

then D(ax + b) = a

and

2

+ 4D − 2 ) ( ax + b ) = 3x − 2

D 2 (ax + b) = D(a) = 0

Hence,

0 + 4a – 2ax – 2b = 3x – 2

from which, and

- 2a = 3 4a – 2b = - 2

and a = −

3 2

and b = 2a + 1 = - 2

3 Hence, P.I., v = − x – 2 2 1 x 3 and the general solution, y = u + v = Ae 3 + Be − x − 2 − x 2

4. Find the particular solution of 9

d2 y dy − 12 + 4y = 3x − 1 2 dx dx

given that when x = 0, y = 0 and

dy 4 =− . dx 3 9

d2 y dy − 12 + 4y = 3x − 1 in D-operator form is: ( 9D 2 − 12D + 4 ) y = 3x − 1 2 dx dx


441

9m 2 − 12m + 4 = 0

Auxiliary equation is:

i.e.

(3m – 2)(3m - 2) = 0

m=

from which,

Hence, C.F.,

u = ( Ax + B ) e

2 twice 3

2 x 3

Let the particular integral, P.I., v = ax + b

( 9D

then D(ax + b) = a

and

2

− 12D + 4 ) ( ax + b ) = 3x − 1

D 2 (ax + b) = D(a) = 0

Hence,

0 - 12a + 4ax + 4b = 3x – 1

from which,

4a = 3

and

-12a + 4b = - 1

Hence, P.I., v =

and a =

3 4

i.e. -9 + 4b = - 1 and b = 2

3 x+2 4

2 x 3 and the general solution, y = u + v = ( Ax + B ) e 3 + x + 2 4

x = 0 and y = 0, hence,

0=B+2

from which,

B=-2

⎛ 2 2x ⎞ ⎛ 2 ⎞ dy 3 = (Ax + B) ⎜ e 3 ⎟ + ⎜ e 3 ⎟ (A) + dx 4 ⎝3 ⎠ ⎝ ⎠

x = 0 and

dy 4 4 2 3 = − , hence, − = B + A + dx 3 3 3 4

and since B = - 2,

−

4 4 3 =− +A+ 3 3 4 3 4

from which,

A= −

Hence,

2 ⎛ 3 ⎞ x 3 y = ⎜ − x − 2 ⎟ e3 + x + 2 4 ⎝ 4 ⎠

i.e.

3 ⎞ 23 x 3 ⎛ y = −2 ⎜ 1 + x ⎟ e + 2 + x 4 ⎠ 4 ⎝


442

5. The charge q in an electric circuit at time t satisfies the equation L

d 2q dq 1 +R + q = E , where 2 dt dt C

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

(a) L

dq =0 dt

1⎞ d 2q dq 1 ⎛ +R + q = E in D-operator form is: ⎜ L D 2 + R D + ⎟ q = E 2 C⎠ dt dt C ⎝ L m2 + R m +

The auxiliary equation is: −R ± R 2 − and

m=

2L

Hence, C.F.,

1 =0 C

4L 4(2) −200 ± 2002 − C = 200 ×10−6 = −200 ± 0 = −50 4 4

u = ( At + B ) e −50 t

Let the particular integral, P.I., v = k

1⎞ ⎛ 2 ⎜ LD + RD + ⎟ ( k ) = E C⎠ ⎝

then

D(k) = 0

and

D 2 (k) = D(0) = 0

Hence,

1 (k) = E C

Hence, P.I., v =

1 20

and k = CE = ( 200 × 10−6 ) ( 250 ) = 0.05 or

and the general solution, y = u + v = ( At + B ) e−50 t + t = 0 and q = 0, hence,

0=B+

1 20

1 20

1 20

from which,

B=-

1 20

dq = (At + B) ( −50e− 50 t ) + ( e − 50 t ) (A) dt t = 0 and

dq = 0 , hence, dt

0 = −50B + A

⎛ 1 ⎞ i.e. 0 = −50 ⎜ − ⎟ + A ⎝ 20 ⎠

Hence,

1 ⎞ 1 ⎛ 5 q = ⎜ − t − ⎟ e− 50 t + 20 ⎝ 2 20 ⎠

i.e.

q=

i.e. A = −

5 2

1 ⎛5 1 ⎞ − ⎜ t + ⎟ e − 50t 20 ⎝ 2 20 ⎠


443

−R ± R 2 − (b) When R = 0, m =

2L

4L 4(2) −0 ± 02 − C = 200 × 10−6 = 0 ± j200 = 0 ± j50 4 4

and

q = (A cos 50 t + B sin 50t) +

q = 0 and t = 0, hence,

0=A+

1 20

i.e. A = -

1 20

1 20

dq = ( −50A sin 50t + 50Bcos 50t ) dt t = 0 and

dq = 0 , hence, 0 = 50B dt

Thus,

q=-

i.e.

q=

i.e.

B=0

1 1 cos 50t + 20 20

1 ( 1 − cos 50t ) 20

6. In a galvanometer the deflection θ satisfies the differential equation

dθ =2 dt

the equation for θ given that when t = 0, θ =

d 2θ dθ + 4 + 4θ = 8 in D-operator form is: 2 dt dt

(D

d 2θ dθ + 4 + 4θ = 8 . Solve 2 dt dt

2

+ 4D + 4 ) θ = 8

m 2 + 4m + 4 = 0


(m + 2)(m + 2) = 0

from which, Hence, C.F.,

m = - 2 twice u = ( At + B ) e − 2 t

Let the particular integral, P.I., v = k

(D

then D(k) = 0

and

Hence,

2

+ 4D + 4 ) k = 8

D 2 (k) = D(0) = 0 4k = 8

from which, k = 2

Hence, P.I., v = 2


444

and the general solution, θ = u + v = ( At + B ) e − 2 t + 2 t = 0 and θ = 2, hence,

2=B+2

from which,

B=0

dθ = (At + B) ( −2e −2t ) + ( e−2t ) (A) dt x = 0 and

dθ = 2 , hence, dt

2 = −2B + A

Hence,

θ = 2 t e− 2 t + 2

i.e.

θ = 2 ( t e −2t + 1)

from which, A = 2


445

EXERCISE 191 Page 485 2. Find the general solution of:

d2 y dy − 3 − 4y = 3e − x 2 dx dx

d2 y dy − 3 − 4y = 3e − x in D-operator form is: 2 dx dx

(D

2

− 3D − 4 ) y = 3e − x

m 2 − 3m − 4 = 0


(m - 4)(m + 1) = 0

from which, Hence, C.F.,

m = 4 and

m = -1

u = Ae 4x + Be − x

As e− x appears in the C.F. and in the right hand side of the differential equation, let the particular integral, P.I., v = k x e − x

(D

then

2

− 3D − 4 )( kxe − x ) = 3e − x

D( kxe − x ) = ( kx ) ( −e − x ) + ( e − x ) ( k ) = − kxe− x + ke− x and D 2 (kxe − x ) = ( − kx ) ( −e− x ) + ( e− x ) ( −k ) − ke − x = kxe− x − 2ke− x Hence,

( kxe

−x

− 2ke − x ) − 3 ( − kxe− x + ke − x ) − 4kxe− x = 3e− x

kxe − x − 2ke− x + 3kxe− x − 3ke− x − 4kxe− x = 3e− x

i.e.

−5ke− x = 3e − x

i.e.

i.e.

– 5k = 3

and k = −

3 5

3 hence, the particular integral, v = − x e − x 5 3 and the general solution, y = u + v = Ae4x + Be − x − x e − x 5 t d2 y dy 3 4. Find the general solution of: 9 2 − 6 + y = 12 e dt dt

9

t t d2 y dy 2 3 3 6 y 12 e − + = in D-operator form is: 9D − 6D + 1 y = 12e ( ) dt 2 dt

Auxiliary equation is:

9m 2 − 6m + 1 = 0


446

i.e.

(3m - 1)(3m - 1) = 0

from which,

m=

1 twice 3

t

u = ( At + B ) e 3

Hence, C.F., t

t

As e 3 and t e 3 appears in the C.F. and in the right hand side of the differential equation, let the t

particular integral, P.I., v = k t 2 e 3 t ⎛ 2 3t ⎞ 3 ( 9D − 6D + 1) ⎜ k t e ⎟ = 12e ⎝ ⎠ 2

then

t t t ⎛1 t ⎞ ⎛ t ⎞ k t ⎛t ⎞ D( k t 2 e 3 ) = ( kt 2 ) ⎜ e 3 ⎟ + ⎜ e 3 ⎟ ( 2kt ) = t 2 e 3 + 2kte 3 = kte 3 ⎜ + 2 ⎟ 3 ⎝3 ⎠ ⎝3 ⎠ ⎝ ⎠

t ⎛ ⎞ ⎛ t ⎞⎛ 1 ⎞ ⎛ t ⎛1 t ⎞ ⎛ t ⎞ ⎤ ⎞⎡ and D 2 ⎜ kt 2 e 3 ⎟ = ⎜ kte 3 ⎟ ⎜ ⎟ + ⎜ + 2 ⎟ ⎢( kt ) ⎜ e 3 ⎟ + ⎜ e 3 ⎟ ( k ) ⎥ ⎠ ⎢⎣ ⎝ ⎠ ⎝ ⎠⎝ 3 ⎠ ⎝ 3 ⎝ 3 ⎠ ⎝ ⎠ ⎥⎦ t k 3t k 2 3t k 3t 2 3t 3 = te + t e + te + kte + 2ke 3 9 3 3 t t t t t ⎛4 t k ⎞ ⎛k t ⎞ Hence, 9 ⎜ te 3 + t 2 e 3 + 2ke 3 ⎟ − 6 ⎜ t 2 e 3 + 2kte 3 ⎟ + kt 2 e 3 = 12e 3 9 ⎝3 ⎠ ⎝3 ⎠ t 3

t 2 3

t 3

t 2 3

t 3

t 2 3

12te + kt e + 18ke − 2kt e − 12kte + kt e = 12e

i.e.

t 3

18ke = 12e

i.e.

t 3

i.e.

and k =

t 3

12 2 = 18 3

2 2 3t hence, the particular integral, v = kt e = t e 3 2

t 3

t 3

2 2 3t and the general solution, y = u + v = ( At + B ) e + t e 3 d2 y dy 1 dy =0 5. Find the particular solution of 5 2 + 9 − 2y = 3e x given that when x = 0, y = and dx dx 4 dx d2 y dy 5 2 + 9 − 2y = 3e x in D-operator form is: ( 5D 2 + 9D − 2 ) y = 3e x dx dx 5m 2 + 9m − 2 = 0


(5m - 1)(m + 2) = 0

from which,

m=

1 5

and

m = -2


447

1

Hence, C.F.,

x

u = Ae 5 + Be −2x

Let P.I., v = k e x

( 5D

then

+ 9D − 2 )( ke x ) = 3e x

2

D( ke x ) = ke x and D 2 (ke x ) = D ( ke x ) = ke x 5ke x + 9ke x − 2ke x = 3e x

Hence,

12ke x = 3e x

i.e.

hence, the particular integral, v =

i.e.

k=

1 4

1 x e 4 1 x 5

1 and the general solution, y = u + v = Ae + Be −2x + e x 4 1 x = 0 and y = , hence, 4

1 1 =A+B+ 4 4

i.e.

0=A+B

(1)

dy 1 15 x 1 = Ae − 2Be −2x + e x dx 5 4 x = 0 and

dy = 0, hence, dx

1 1 0 = A − 2B + 5 4 −

i.e. 2 × (1) gives: (2) + (3) gives:

−

1 1 = A − 2B 4 5

(2)

0 = 2A + 2B

(3)

1 11 = A 4 5

Hence,

y= −

i.e.

y=

i.e.

A= −

5 5 and from (1), B = 44 44

5 15 x 5 −2x 1 x e + e + e 44 44 4

1 x⎞ 5 ⎛ −2x 1 x 5 ⎜e −e ⎟ + e 44 ⎝ ⎠ 4


448

EXERCISE 192 Page 487 1. Find the general solution of: 2 2

d 2 y dy − − 3y = 25sin 2x dx 2 dx

d 2 y dy − − 3y = 25sin 2x in D-operator form is: dx 2 dx

( 2D

2

− D − 3) y = 25sin 2x

2m 2 − m − 3 = 0


(2m - 3)(m + 1) = 0

from which,

m= 3

3 2

and

m = -1

x

u = Ae 2 + Be − x

Hence, C.F.,

Let P.I., v = A sin 2x + B cos 2x Hence,

( 2D

2

− D − 3) ( A sin 2x + B cos 2x ) = 25sin 2x

D(A sin 2x + B cos 2x) = 2A cos 2x – 2B sin 2x D 2 ( A sin 2x + Bcos 2x ) = D ( 2A cos 2x − 2Bsin 2x ) = −4A sin 2x − 4B cos 2x

Hence, 2 ( −4A sin 2x − 4Bcos 2x ) − ( 2A cos 2x − 2Bsin 2x ) − 3 ( A sin 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

(3) – (4) gives:

125B = 50 - 11A +

from which,

- 11A = 25 -

and

from which,

B=

50 2 = 125 5

4 = 25 5


Thus, P.I., v = −

(4)

4 125 − 4 121 = = 5 5 5

and A =

121 11 =− 5(−11) 5

11 2 1 sin 2x + cos 2x = − (11sin 2x − 2 cos 2x ) 5 5 5 3 x 2

y = u + v = Ae + Be − x −

1 ( 11sin 2x − 2cos 2x ) 5


449

3. Find the general solution of:

d2 y + y = 4 cos x dx 2

d2 y + y = 4 cos x in D-operator form is: dx 2

m 2 = −1

Hence, C.F.,

2

+ 1) y = 4 cos x

m2 + 1 = 0


(D

and

−1 = 0 ± j1

m=

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)

(D

Hence,

2

+ 1) ⎡⎣ x ( Csin x + D cos x ) ⎤⎦ = 4 cos x

D ⎡⎣ x ( Csin x + D cos x ) ⎤⎦ = ( x )( C cos x − D sin x ) + (1)( Csin x + D cos x ) D 2 [ v ] = ( x )( −C sin x − D cos x ) + ( C cos x − D sin x ) + ( C cos x − D sin x )

Hence, since then

(D

2

+ 1) v = 4 cos x

( x )( −Csin x − D cos x ) + ( C cos x − D sin x ) + ( C cos x − D sin 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 y = u + v = A cos x + B sin x + 2x sin x

and

d2 y dy 4. Find the particular solution of the differential equation − 3 − 4y = 3sin x given that 2 dx dx dy when x = 0, y = 0 and = 0. dx d2 y dy − 3 − 4y = 3sin x in D-operator form is: 2 dx dx Auxiliary equation is: i.e.

(D

2

− 3D − 4 ) y = 3sin x

m 2 − 3m − 4 = 0 (m - 4)(m + 1) = 0

from which,

m = 4 and

m = -1


450

u = Ae 4x + Be− x

Hence, C.F.,

Let P.I., v = A sin x + B cos x Hence,

(D

2

− 3D − 4 ) ( A sin x + Bcos x ) = 3sin x

D(A sin x + B cos x) = A cos x – B sin x D 2 ( A sin x + Bcos x ) = D ( A cos x − Bsin x ) = − A sin 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


⎛ 9 ⎞ - 5A + 3 ⎜ ⎟ = 3 ⎝ 34 ⎠

from which,

5A =

Thus, P.I., v = −

27 27 − 102 −75 -3 = = 34 34 34

and A =

−75 15 =− 34(5) 34

15 9 sin x + cos x 34 34 y = u + v = Ae4x + Be − x −

and

9 34

from which, B =


0=A+B+

9 34

15 9 sin x + cos x 34 34

i.e.

A+B=-

9 34

(5)

dy 15 9 = 4Ae 4x − Be− x − cos x − sin x dx 34 34 x = 0, when

dy 15 = 0, hence, 0 = 4A – B − dx 34

4 × (5) gives: (7) – (6) gives:

i.e. 4A – B =

15 34

4A + 4B = 5B = -

36 34

(6)

(7)

36 15 51 =− 34 34 34


451

B= −

and A−


51 9 =− 170 34

from which, A =

51 51 =− 34(5) 170

51 9 51 − 45 6 − = = 170 34 170 170

Hence,

y=

6 4x 51 − x 15 9 e − e − sin x + cos x 170 170 34 34

i.e.

y=

1 1 6e4x − 51e − x ) − ( 15sin x − 9cos x ) ( 170 34

7. L

d 2q dq 1 +R + q = V0 sin ωt represents the variation of capacitor charge in an electric circuit. 2 dt dt C

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

dq = 4.8 dt

d 2q dq 1 1⎞ ⎛ L 2 +R + q = V0 sin ωt in D-operator form is: ⎜ L D 2 + R D + ⎟ q = V0 sin ωt dt dt C C⎠ ⎝ L m2 + R m +

The auxiliary equation is: −R ± R 2 − m=

and

Hence, C.F.,

2L

1 =0 C

4L 4(0.02) −40 ± 402 − C = 50 ×10−6 = −40 ± 0 = −1000 2(0.02) 0.04

u = ( At + B ) e −1000 t

Let P.I., v = A sin ωt + B cos ωt 1⎞ ⎛ 2 ⎜ L D + R D + ⎟ [ A sin ωt + Bcos ωt ] = V0 sin ωt C⎠ ⎝ D(v) = Aω cos ωt - Bω sin ωt

and

D 2 (v) = −Aω2 sin ωt − Bω2 cos ωt

1⎞ ⎛ Thus, ⎜ L D 2 + R D + ⎟ v = 0.02 ( − Aω2 sin ωt − Bω2 cos ωt ) + 40 ( Aω cos ωt − Bω sin ωt ) C⎠ ⎝ 1 + ( A sin ωt + Bcos ωt ) = V0 sin ωt 50 ×10−6 i.e. -800A sin 200t – 800B cos 200t + 8000A cos 200t – 8000B sin 200t + 20000A sin 200t + 20000B cos 200t = 540.8 sin 200t © 2006 John Bird. All rights reserved. Published by Elsevier.

452

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

19.2 × (2) gives:

(3)

153600A + 368640B = 0

(3) – (4) gives:

(4)

- 432640B = 4326.4

from which,

B=


4326.4 = −0.01 432640

19200A – 8000(-0.01) = 540.8

i.e.

19200A + 80 = 540.8

and

A=

540.8 − 80 460.8 = = 0.024 19200 19200

Hence, P.I., v = 0.024 sin 200t – 0.01 cos 200t q = u + v = ( At + B ) e −1000 t + 0.024 sin 200t – 0.01 cos 200t

Thus,

When t = 0, q = 0, hence,

0 = B – 0.01

from which,

B = 0.01

dq = ( At + B ) ( −1000e−1000t ) + Ae −1000t + (0.024)(200) cos 200t + (0.01)(200) sin 200t dt When t = 0, i.e. Thus,

dq = 4.8, hence, dt

4.8 = - 1000B + A + 4.8 A = 1000B = 1000(0.01) = 10

q = ( 10t + 0.01) e −1000t + 0.024 sin 200t – 0.010 cos 200t


453

EXERCISE 193 (Page 490) 1. Find the general solution of: 8

8

d2 y dy − 6 + y = 2x + 40sin x 2 dx dx

d2 y dy − 6 + y = 2x + 40sin x in D-operator form is: ( 8D2 − 6D + 1) y = 2x + 40sin x 2 dx dx 8m 2 − 6m + 1 = 0


(4m - 1)(2m - 1) = 0

from which,

m=

u = Ae

Hence, C.F.,

1 x 4

+ Be

1 4

and

m=

1 2

1 x 2

Let P.I., v = ax + b + c sin x + d cos x Hence, ( 8D2 − 6D + 1) [ ax + b + c sin x + d cos x ] = 2x + 40sin x D(v) = a + c cos x – d sin x Hence,

(8D

2

D 2 (v) = - c sin x – d cos x

and

− 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:


Hence, P.I., v = 2x + 12 −

85d = 240 ⎛ 48 ⎞ - 6c - 7 ⎜ ⎟ = 0 ⎝ 17 ⎠

from which, d =

from which,

240 48 = 85 17

c=−

7(48) 56 =− 6(17) 17

56 48 sin x + cos x 17 17


454

1

x

1

x

and

y = u + v = Ae 4 + Be 2 + 2x + 12 −

or

y = Ae 4 + Be 2 + 2x + 12 +

1

x

1

x

4. Find the general solution of:

56 48 sin x + cos x 17 17

8 ( 6cos x − 7 sin x ) 17

d2 y dy − 2 + 2y = e t sin t 2 dt dt

d2 y dy − 2 + 2y = e t sin t in D-operator form is: 2 dt dt

(D

m=

Hence, C.F.,

− 2D + 2 ) y = e t sin t

m 2 − 2m + 2 = 0


2

−(−2) ± (−2) 2 − 4(1)(2) 2 ± −4 2 ± j2 = = = 1± j 2(1) 2 2

u = e t ( A cos t + Bsin t )

Since e t sin t occurs in the C.F. and the right hand side of the differential equation, let P.I., v = t e t ( Csin t + D cos t ) then

(D

2

− 2D + 2 ) ⎡⎣ t e t ( Csin t + D cos t ) ⎤⎦ = e t sin t

D(v) = ( t e t ) ( C cos t − D sin t ) + ( Csin t + D cos t ) ⎡⎣ t e t + e t ⎤⎦ D 2 (v) = ( t e t ) ( −Csin t − D cos t ) + ( C cos t − Dsin t ) ⎡⎣ t e t + e t ⎤⎦ + ( Csin t + D cos t ) ⎡⎣ t e t + e t + e t ⎤⎦ + ( t e t + e t ) ( C cos t − D sin t ) Hence,

( t e ) ( −Csin t − D cos t ) + ( C cos t − D sin t ) ⎡⎣ t e t

t

+ e t ⎤⎦ + ( Csin t + D cos t ) ⎡⎣ t e t + e t + e t ⎤⎦

+ ( t e t + e t ) ( C cos t − D sin t ) - 2 ( t e t ) ( C cos t − D sin t ) − 2 ( Csin t + D cos t ) ⎡⎣ t e t + e t ⎤⎦ + 2 t e t ( Csin t + D cos t ) = e t sin t i.e.

– D + 2C – D – 2C = 1

i.e.

and

C + 2D + C – 2D = 0

i.e.

-2D = 1 and D = −

1 2

C=0

t ⎛ 1 ⎞ Hence, P.I., v = t e t ⎜ − cos t ⎟ = − e t cos t 2 ⎝ 2 ⎠ and

t y = u + v = et ( A cos t + B sin t ) − et cos t 2


455

5. Find the particular solution of

d2 y dy − 7 + 10y = e 2x + 20 given that when x = 0, y = 0 and 2 dx dx

dy 1 =− dx 3 d2 y dy − 7 + 10y = e 2x + 20 in D-operator form is: 2 dx dx

(D

2

− 7D + 10 ) y = e 2x + 20

m 2 − 7m + 10 = 0


(m - 5)(m - 2) = 0

from which,

m = 5 and m = 2

u = Ae5x + Be 2x

Hence, C.F.,

Let P.I., v = k x e2x + a Thus,

(D

2

− 7D + 10 ) ⎡⎣ k x e2x + a ⎤⎦ = e2x + 20

D(v) = ( kx ) ( 2e2x ) + ( ke2x ) Hence,

D 2 (v) = ( kx ) ( 4e 2x ) + ( 2e2x ) ( k ) + 2ke 2x

and

( kx ) ( 4e2x ) + ( 2e2x ) ( k ) + 2ke2x − 7 ( kx ) ( 2e2x ) − 7 ( ke2x ) +10k x e2x + 10a = e2x + 20

from which,

10a = 20

Also,

2k + 2k – 7k = 1

i.e.

and -3k = 1

a=2 from which,

k= −

1 3

1 Hence, P.I., v = − x e2x + 2 3 1 y = u + v = Ae5x + Be2x − x e2x + 2 3

and

When x = 0, y = 0, hence,

0=A+B+2

i.e.

A + B = -2

(1)

dy ⎡⎛ 1 ⎞ ⎛ 1 ⎞⎤ = 5Ae5x + 2Be 2x − ⎢⎜ x ⎟ ( 2e2x ) + ( e2x ) ⎜ ⎟ ⎥ dx ⎝ 3 ⎠⎦ ⎣⎝ 3 ⎠ dy 1 = − , hence, dx 3

−

1 1 = 5A + 2B − 3 3

i.e. 5A + 2B = 0

(2)

2 × (1) gives:

2A + 2B = -4

(3)

(2) – (3) gives:

3A

When x = 0,

=4


from which, A =

4 3 456


Hence,

4 +B=-2 3

y=u+v=

from which,

B=-2-

4 10 =− 3 3

4 5x 10 2x 1 2x e − e − xe + 2 3 3 3


457

CHAPTER 52 POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS EXERCISE 194 Page 492 t

1. (b) Determine the derivative y(5) when y = 8e 2

If y = e , then y ax

(n )

1 t 2

= a e . Hence, if y = 8e , then y n ax

5

(5)

1 1 1 ⎛ 1 ⎞ 2t 8 2 t 1 2t = (8) ⎜ ⎟ e = e = e 4 32 ⎝2⎠

2. (a) Determine the derivative y(4) when y = sin 3t nπ ⎞ ⎛ If y = sin ax, then y (n ) = a n sin ⎜ ax + ⎟ 2 ⎠ ⎝ 4π ⎞ ⎛ Hence, if y = sin 3t, then y (4) = 34 sin ⎜ 3t + ⎟ = 81sin ( 3t + 2π ) = 81 sin 3t 2 ⎠ ⎝ 2 3. (b) Determine the derivative y(9) when y = 3cos t 3 nπ ⎞ ⎛ If y = cos ax, then y (n ) = a n cos ⎜ ax + ⎟ 2 ⎠ ⎝ 9 29 2 π⎞ 2 ⎛2⎞ ⎛ 2 9π ⎞ 2 ⎛2 Hence, if y = 3cos t , then y (9) = (3) ⎜ ⎟ cos ⎜ t + ⎟ = 8 cos ⎜ t + ⎟ = − 8 sin t 3 3 3 2 ⎠ 3 2⎠ ⎝3⎠ ⎝3 ⎝3 9

4. (a) Determine the derivative y(7) when y = 2x 9

If y = x a , then y (n ) =

a! x a −n − a n ! ( )

Hence, if y = 2x 9 , then y (7) = (2)

9! x 9−7 = ( 9!) x 2 ( 9 − 7 )!

5. (b) Determine the derivative y(6) when y = 2 sinh 3x

If y = sinh ax, then y (n ) =

{

}

an ⎡⎣1 + (−1) n ⎤⎦ sinh ax + ⎡⎣1 − (−1) n ⎤⎦ cosh ax 2


458

Hence, if y = 2 sinh 3x, then y (6) = (2)

{

}

36 ⎡1 + (−1)6 ⎤⎦ sinh 3x + ⎡⎣1 − (−1)6 ⎤⎦ cosh 3x 2 ⎣

= 36 {2sinh 3x + 0} = 1458 sinh 3x

6. (a) Determine the derivative y(7) when y = cosh 2x

If y = cosh ax, then y

(n )

{

}

an ⎡⎣1 − (−1) n ⎤⎦ sinh ax + ⎡⎣1 + (−1) n ⎤⎦ cosh ax = 2

Hence, if y = cosh 2x, then y

(7)

{

}

27 ⎡⎣1 − (−1)7 ⎤⎦ sinh 2x + ⎡⎣1 + (−1)7 ⎤⎦ cosh 2x = 2

= 26 {2sinh 2x + 0} = 27 sinh 2x = 128 sinh 2x

7. (b) Determine the derivative y(7) when y =

If y = ln ax, then y (n ) = ( −1) Hence, if y =

n −1

1 ln 2t 3

( n − 1)! xn

240 1 6! 7 −1 ( 7 − 1) ! ⎛1⎞ = 7 = 7 ln 2t , then y (7) = ⎜ ⎟ ( −1) 7 t t 3t 3 ⎝3⎠


459


2. If y = x 3e 2x find y(n) and hence y(3) by using the theorem of Leibniz.

Since y = x 3e 2x then let v = x 3 and u = e2x and the n’th derivative of e2x is 2n e2x Thus, y (n ) = u (n ) v + nu (n −1) v(1) +

n(n − 1) (n − 2) (2) u v + ..... 2!

= 2n e 2x ( x 3 ) + n ( 2n −1 e 2x )( 3x 2 ) +

n ( n − 1)( n − 2 ) (n −3) 2x n(n − 1) n − 2 2x 2 e ) ( 6x ) + 2 e (6) ( 2! 3!

= 2n x 3e 2x + 3nx 2 2n −1 e 2x + 3n(n − 1)2n − 2 e 2x x + n ( n − 1)( n − 2 ) 2n −3 e 2x or

y (n ) = e 2x 2n −3 {23 x 3 + 3nx 2 (2) 2 + 3n(n − 1)x(2) + n(n − 1)(n − 2)} = e 2x 2n − 3 {8x 3 + 12nx 2 + n(n − 1)(6x) + n(n − 1)(n − 2)}

Hence, y (3) = e 2x 20 {8x 3 + 36x 2 + 3(2)6x + 3(2)(1)}

{

}

= e 2x 8x 3 + 36x 2 + 36x + 6

4. Use the theorem of Leibniz to determine the 5th derivative of: y = x 3 cos x

Since y = x 3 cos x then let u = cos x and v = x 3 y (n ) = u n v + nu (n −1) v(1) +

and

nπ ⎞ ⎛ u (n ) = 1n cos ⎜ x + ⎟ 2 ⎠ ⎝

n(n − 1) (n − 2) (2) u v + ..... 2!

nπ ⎞ (n − 1)π ⎞ n(n − 1) (n − 2)π ⎞ ⎛ ⎛ ⎛ 2 cos ⎜ x + Hence, y (n ) = cos ⎜ x + ⎟ ( x 3 ) + n cos ⎜ x + ⎟ ( 3x ) + ⎟ ( 6x ) 2 ⎠ 2 ⎠ 2! 2 ⎝ ⎝ ⎝ ⎠ n(n − 1(n − 2) (n − 3)π ⎞ ⎛ + cos ⎜ x + ⎟ ( 6) 3! 2 ⎠ ⎝ 5π ⎞ 4π ⎞ 5(4) 3π 5(4)(3) 2π ⎛ ⎛ and y (5) = x 3 cos ⎜ x + ⎟ + 5 ( 3x 2 ) cos ⎜ x + ( 6x ) cos ⎛⎜ x + ⎞⎟ + ( 6 ) cos ⎛⎜ x + ⎞⎟ ⎟+ 2 ⎠ 2 ⎠ 2! 2 ⎠ 3! 2 ⎠ ⎝ ⎝ ⎝ ⎝ = − x 3 sin x + 15x 2 cos x + 60x sin x + 60(− cos x) = ( 60x − x 3 ) sin x + ( 15x 2 − 60 ) cos x


460

6. If y = x 5 ln 2x find y(3) by using the theorem of Leibniz.

Since y = x 5 ln 2x then let u = x 5 and v = ln 2x y (n ) = u n v + nu (n −1) v(1) +

and

un =

a! 5! x a −n = x 5− n (5 − n)! ( a − n )!

n(n − 1) (n − 2) (2) u v + ..... 2!

⎡ 5! ⎤ 5! ⎛ 1 ⎞ n(n − 1) 5! ⎛ 1 ⎞ y (n ) = ⎢ x 5− n ⎥ ln 2x + n x 6− n ⎜ ⎟ + x 7−n ⎜ − 2 ⎟ (6 − n)! 2! (7 − n)! ⎝x⎠ ⎝ x ⎠ ⎣ (5 − n)! ⎦ n(n − 1)(n − 2) 5! ⎛ 2 ⎞ x 8− n ⎜ 3 ⎟ + 3! (8 − n)! ⎝x ⎠ Hence, y3 =

5! 2 5! ⎛ 1 ⎞ 3(2) 5! 4 ⎛ 1 ⎞ 3(2)(1) 5! 5 ⎛ 2 ⎞ x ln 2x + (3) x 3 ⎜ ⎟ + x ⎜− ⎟+ x ⎜ ⎟ 2! 3! ⎝ x ⎠ 2! (4)! ⎝ x 2 ⎠ 3! 5! ⎝ x 3 ⎠

= 60x 2 ln 2x + 60x 2 − 15x 2 + 2x 2 = 60x 2 ln 2x + 47x 2 i.e.

y (3) = x 2 ( 47 + 60 ln 2x )

8. If y = ( x 3 + 2x 2 ) e 2x determine an expression for y(5) by using the theorem of Leibniz. Since y = ( x 3 + 2x 2 ) e 2x then let u = e2x y (n ) = u n v + nu (n −1) v(1) +

and v = ( x 3 + 2x 2 )

n(n − 1) (n − 2) (2) u v + ..... 2!

Hence, y (n ) = ( 2n e 2x )( x 3 + 2x 2 ) + n2n −1 e 2x ( 3x 2 + 4x ) +

and

and u n = 2n e2x

y (5) = 25 e 2x ( x 3 + 2x 2 ) + (5) 24 e 2x ( 3x 2 + 4x ) +

n(n − 1) n − 2 2x 2 e ( 6x + 4 ) 2! n(n − 1)(n − 2) n −3 2x + 2 e (6) 3!

5(4) 3 2x 5(4)(3) 2 2x 2 e ( 6x + 4 ) + 2 e (6) 2 3!

= e 2x {25 x 3 + 26 x 2 + (16)15x 2 + (16)(20x) + 60x(8) + (8)(40) + 240} = e 2x {25 x 3 + 304x 2 + 800x + 560} = e 2x {25 x 3 + 24 (19x 2 ) + 24 (50)(x) + 24 (35)} = e 2x 24 { 2x 3 + 19x 2 + 50x + 35}


461

EXERCISE 196 Page 497 1. Determine the power series solution of the differential equation: Leibniz-Maclaurin method, given that at x = 0, y = 1 and

d2 y dy + 2x + y = 0 using the 2 dx dx

dy = 2. dx

d2 y dy + 2x +y=0 2 dx 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: y (n + 2) + 2 { y(n +1) (x) + n y(n ) (1) + 0} + y(n ) = 0 y (n + 2) + 2x y(n +1) + (2n + 1) y(n ) = 0

i.e.

(1)

(ii) At x = 0, equation (1) becomes: y (n + 2) + (2n + 1) y(n ) = 0 y (n + 2) = −(2n + 1) y (n )

from which,

This is the recurrence formula. (iii) For n = 0,

( y '')0 = − ( y )0

n = 1,

( y ''')0 = −3 ( y ')0

n = 2,

(y )

n = 3,

(y )

n = 4,

(y )

0

= −9 ( y (4) ) = −9 {5 ( y )0 } = −5 × 9 ( y )0

n = 5,

(y )

0

= −11( y (5) ) = −11{3 × 7 ( y ' )0 } = −3 × 7 × 11( y ')0

n = 6,

(y )

0

= −13 ( y (6) ) = −13 −5 × 9 ( y )0 = 5 × 9 × 13 ( y )0

0

= −5 ( y '')0 = 5(y)0

0

= −7 ( y ''' )0 = −7 −3 ( y ')0 = 3 × 7 ( y ' )0

(4)

(5)

(6)

(7)

(8)

{

0

0

(iv) Maclaurin’s theorem is: Thus,

}

y = ( y )0 + x ( y ' )0 +

0

{

y = ( y )0 + x ( y ' )0 +

}

x2 x3 x4 ( y '')0 + ( y ''')0 + ( y(4) )0 + .... 2! 3! 4!

x2 x3 x4 x5 − ( y )0 } + {−3 ( y ')0 } + {5 ( y )0 } + {3 × 7 ( y ' )0 } { 2! 3! 4! 5! +

x6 x7 −5 × 9 ( y )0 } + {−3 × 7 × 11( y ')0 } { 6! 7!


462

(v) Collecting similar together gives: ⎧ ⎫ x 2 5 x 4 5 × 9 x 6 5 × 9 × 13x 8 y = ( y )0 ⎨1 − + − + − ...⎬ 2! 4! 6! 8! ⎩ ⎭ ⎧ ⎫ 3 x 3 3 × 7 x 5 3 × 7 × 11x 7 + ( y ')0 ⎨ x − + − + ...⎬ 3! 5! 7! ⎩ ⎭

At x = 0, y = 1 and

dy = 2, hence, dx

( y )0 = 1

and

( y ' )0 = 2 .

Hence, the power series solution of the differential equation:

d2 y dy + 2x + y = 0 is: 2 dx dx

⎧ ⎫ ⎧ ⎫ x 2 5 x 4 5 × 9 x 6 5 × 9 × 13x 8 3 x 3 3 × 7 x 5 3 × 7 × 11x 7 y = ⎨1 − + − + − ...⎬ + 2 ⎨ x − + − + ...⎬ 2! 4! 6! 8! 3! 5! 7! ⎩ ⎭ ⎩ ⎭

3. Find the particular solution of the differential equation: ( x 2 + 1)

d2 y dy +x − 4y = 0 using the 2 dx dx

Leibniz-Maclaurin method, given the boundary conditions that at x = 0, y = 1 and

dy = 1. dx

d2 y dy − 4y = 0 ( x + 1) dx 2 + x dx 2

i.e.

(x

i.e.

n(n − 1) (n ) ⎫ ⎧ 2 (n + 2) + ny (n +1) (2x) + y (2) ⎬ + { y(n +1) x + ny n (1)} − 4y (n ) = 0 ⎨( x + 1) y 2! ⎩ ⎭

2

+ 1) y′′ + xy′ - 4y = 0

(x

i.e.

2

+ 1) y(n + 2) + ( 2nx + x ) y(n +1) + (n(n − 1) + n − 4) y (n ) = 0 y (n + 2) + ( n 2 − 4 ) y(n ) = 0

At x = 0, from which,

y (n + 2) = ( 4 − n 2 ) y (n) which is the recurrence formula.

For n = 0,

( y '')0 = 4 ( y )0

n = 1,

( y ''')0 = 3 ( y ')0

n = 2,

(y )

n = 3,

(y )

n = 4,

(y )

(4)

0

(5)

0

(6)

0

=0 = −5 ( y ''')0 = −5{−3 ( y ')0 } = −5 ( 3)( y ')0 = −12 ( y (4) ) = −12(0) = 0 0


463

n = 5,

(y ) (7)

0

= −21( y (5) ) = −21{−5 × 3 ( y ')0 } = 315 ( y ' )0 0

y = ( y )0 + x ( y ' )0 +

Maclaurin’s theorem is: Thus, y = ( y )0 + x ( y ')0 + i.e.

x2 x3 x 4 (4) y '' + y ''' + ( )0 ( )0 ( y )0 + .... 2! 3! 4!

x2 x3 x4 x5 x7 4 y + 3 y ' + 0 + − 3 × 5 y ' + 0 + { ( )0 } 3! { ( )0 } 4! { } 5! { ( )0 } {315 ( y ')0 } 2! 7!

⎧ ⎫ x3 x5 x7 y = ( y )0 {1 + 2x } + ( y ')0 ⎨ x + − + + ...⎬ 2 8 16 ⎩ ⎭ 2

At x = 0, y = 1 and

dy = 1, hence, dx

( y )0 = 1

and

( y ')0 = 1 .

Hence, the power series solution of the differential equation:

(x

2

+ 1)

d2 y dy +x − 4y = 0 is: 2 dx dx

⎧ ⎫ x3 x5 x7 y = {1 + 2x 2 } + ⎨ x + − + + ...⎬ 2 8 16 ⎩ ⎭

i.e.

y = 1 + x + 2x 2 +

x 3 x5 x7 − + + ..... 3 8 16


464


1. Produce, using Frobenius’ method, a power series solution for the differential equation: 2x

d 2 y dy + −y=0 dx 2 dx

d 2 y dy 2x 2 + − y = 0 may be rewritten as: 2xy′′ + y′ - y = 0 dx dx (i) Let a trial solution be of the form y = xc{a0 + 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 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 a0 c [2c – 2 + 1] = a0 c(2c - 1) = 0

or

Equation (1) is the indicial equation, from which, c = 0 or c =

(1)

1 2

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

a1(2c + 1)(c + 1) - a0 = 0

or

(2)

Replacing r by (r + 1) will give:


465

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

ar+1{(c + r + 1)(2c + 2r + 1)} - ar = 0

which simplifies to:

(3)

(a) When c = 0: From equation (2), if c = 0, a1(1 × 1) - a0 = 0, i.e. a1 = a 0 From equation (3), if c = 0, ar+1(r + 1)(2r + 1) - ar = 0, i.e. ar+1 =

ar (r + 1)(2r + 1)

r≥0

Thus, when r = 1,

a2 =

a0 a1 = since a1 = a 0 (2 × 3) (2 × 3)

when r = 2,

a3 =

a0 a2 = (3 × 5) (2 × 3)(3 × 5)

when r = 3,

a4 =

a3 a0 a0 = = and so on. (4 × 7) (2 × 3)(3 × 5)(4 × 7) (2 × 3 × 4)(3 × 5 × 7)

The trial solution is:

y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}

Substituting c = 0 and the above values of a1, a2, a3, … into the trial solution gives: ⎧ ⎫ ⎛ a ⎞ ⎛ ⎞ 3 ⎛ ⎞ 4 a0 a0 y = x 0 ⎨a 0 + a 0 x + ⎜ 0 ⎟ x 2 + ⎜ ⎟x +⎜ ⎟ x + ...⎬ ⎝ (2 × 3) ⎠ ⎝ (2 × 3)(3 × 5) ⎠ ⎝ (2 × 3 × 4)(3 × 5 × 7) ⎠ ⎩ ⎭

x2 x3 x4 ⎪⎧ ⎪⎫ i.e. y = a 0 ⎨1 + x + + + + ...⎬ ( 2 × 3) ( 2 × 3)( 3 × 5 ) ( 2 × 3 × 4 )( 3 × 5 × 7 ) ⎭⎪ ⎩⎪

(4)

1 (b) When c = : 2

From equation (2), if c =

a 1 ⎛3⎞ , a1 ( 2 ) ⎜ ⎟ - a0 = 0, i.e. a1 = 0 3 2 ⎝2⎠

From equation (3), if c =

1 ⎛1 ⎞ , ar+1 ⎜ + r + 1⎟ (1 + 2r + 1) - ar = 0, 2 ⎝2 ⎠

i.e.

i.e.

⎛ 3⎞ ar+1 ⎜ r + ⎟ ( 2r + 2 ) - ar = ar+1(2 r 2 + 5r +3) - ar = 0, ⎝ 2⎠

ar+1 =

ar (2r + 3)(r + 1)

r≥0


466

Thus, when r = 1,

a2 =

a0 a1 = (2 × 5) (2 × 3 × 5)

when r = 2,

a3 =

a0 a2 = (3 × 7) (2 × 3 × 5)(3 × 7)

when r = 3,

a4 =

a3 a0 = and so on. (4 × 9) (2 × 3 × 4)(3 × 5 × 7 × 9)

The trial solution is: Substituting c =

since a1 =

a0 3

y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}

1 and the above values of a1, a2, a3, … into the trial solution gives: 2

1 ⎧ ⎫ ⎞ 3 ⎛ ⎞ 4 a0 a0 ⎛a ⎞ ⎛ a0 ⎞ 2 ⎛ y = x 2 ⎨a 0 + ⎜ 0 ⎟ x + ⎜ ⎟x +⎜ ⎟ x + ...⎬ ⎟x +⎜ ⎝ 3⎠ ⎝ 2 × 3× 5 ⎠ ⎝ (2 × 3 × 5)(3 × 7) ⎠ ⎝ (2 × 3 × 4)(3 × 5 × 7 × 9) ⎠ ⎩ ⎭

⎧ ⎫ x x2 x3 x4 i.e. y = a 0 x ⎨1 + + + + + ...⎬ (5) ⎩ (1× 3) (1× 2)(3 × 5) (1× 2 × 3)(3 × 5 × 7) (1× 2 × 3 × 4)(3 × 5 × 7 × 9) ⎭ 1 2

Let a 0 = A in equation (4), and a 0 = B in equation (5).

x2 x3 x4 ⎪⎧ ⎪⎫ + + + ...⎬ Hence, y = A ⎨1 + x + ( 2 × 3 ) ( 2 × 3 )( 3 × 5 ) ( 2 × 3 × 4 )( 3 × 5 × 7 ) ⎭⎪ ⎩⎪ ⎧ ⎫ x x2 x3 x4 + B x ⎨1 + + + + + ...⎬ ⎩ (1 × 3) (1 × 2)(3 × 5) (1 × 2 × 3)(3 × 5 × 7) (1 × 2 × 3 × 4)(3 × 5 × 7 × 9) ⎭ 1 2

3. Determine the power series solution of the differential equation: 3x

d2 y dy + 4 − y = 0 using the 2 dx dx

Frobenius method.

3x

d2 y dy + 4 − y = 0 may be rewritten as: 3xy′′ + 4y′ - y = 0 2 dx dx

(i) Let a trial solution be of the form y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr+…} y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +…

i.e. (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:


467

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 a0 c [3c – 3 + 4] = a0 c(3c + 1) = 0

or

This is the indicial equation, from which,

c = 0 or c = −

1 3

The coefficient of xc is equated to zero giving: 3a1c(c + 1) + 4a1(c + 1) - a0 = 0 a1 (3c(c + 1) +4(c+1)) - a0 = a1(c + 1)(3c + 4) - a0 = 0

i.e.

a1(c + 1)(3c + 4) - a0 = 0

or

(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 ar+1(c + r + 1)(3c + 3r + 4) - ar = 0

i.e.

ar +1 =

which simplifies to:

ar (c + r + 1)(3c + 3r + 4)

(2)

(a) When c = 0: From equation (1), if c = 0, a1(4) - a0 = 0, i.e. a1 = From equation (2), if c = 0, Thus, when r = 1, a 2 =

a r +1 =

ar (r + 1)(3r + 4)

a0 4

r≥0

a0 a a1 = since a1 = 0 (2 × 7) (2 × 4 × 7) 4

when r = 2, a 3 =

a0 a0 a2 = = (3 × 10) (3 × 10)(2 × 4 × 7) (1× 2 × 3)(4 × 7 × 10)

when r = 3, a 4 =

a3 a0 a0 = = and so on. (4 × 13) (4 × 13)(3 ×10)(2 × 4 × 7) (2 × 3 × 4)(4 × 7 × 10 × 13)

The trial solution is:

y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}

Substituting c = 0 and the above values of a1, a2, a3, … into the trial solution gives: ⎪⎧

y = x 0 ⎨a 0 + ⎩⎪

⎛ ⎞ 2 ⎛ ⎞ 3 ⎛ ⎞ 4 a0 a0 a0 a0 ⎪⎫ x+⎜ ⎟x + ⎜ ⎟x +⎜ ⎟ x + ...⎬ 4 ⎝ (1 × 2)(4 × 7) ⎠ ⎝ (1 × 2 × 3)(4 × 7 × 10) ⎠ ⎝ (2 × 3 × 4)(4 × 7 × 10 × 13) ⎠ ⎭⎪


468

⎧⎪ ⎫⎪ x x2 x3 x4 + + + + ...⎬ (3) i.e. y = a 0 ⎨1 + ⎪⎩ (1× 4) (1× 2)(4 × 7 ) (1× 2 × 3)( 4 × 7 × 10 ) (2 × 3 × 4)(4 × 7 × 10 × 13) ⎪⎭ 1 (b) When c = − : 3 a 1 ⎛2⎞ From equation (1), if c = − , a1 ⎜ ⎟ ( 3) - a0 = 0, i.e. a1 = 0 2 3 ⎝3⎠

ar ar ar 1 From equation (2), if c = − , a r +1 = = = 1 3 (3r + 2)(r + 1) ⎛ 2⎞ ⎜ r + ⎟ ( 3r + 3) 3 (3r + 2)3(r + 1) ⎝ 3⎠ Thus, when r = 1,

a2 =

a a0 a1 = 20 = (5 × 2) (2 × 5) (1× 2)(2 × 5)

when r = 2,

a3 =

a0 a2 = (8 × 3) (1× 2 × 3)(2 × 5 × 8)

when r = 3,

a4 =

a3 a0 = and so on. (11× 4) (1× 2 × 3 × 4)(2 × 5 × 8 × 11)

The trial solution is: Substituting c = −

since a1 =

r≥0

a0 2

y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}

1 and the above values of a1, a2, a3, … into the trial solution gives: 3

1 − ⎧ ⎫ a a0 a0 a0 + ...⎬ y = x 3 ⎨a 0 + 0 x + x2 + x3 + 2 (1× 2)(2 × 5) (1× 2 × 3)(2 × 5 × 8) (1× 2 × 3 × 4)(2 × 5 × 8 × 11) ⎩ ⎭

i.e. y = a 0 x

−

1 3

⎧ ⎫ x x2 x3 x4 + + + + ...⎬ ⎨1 + ⎩ (1× 2) (1× 2)(2 × 5) (1× 2 × 3)(2 × 5 × 8) (1× 2 × 3 × 4)(2 × 5 × 8 × 11) ⎭

(4) Let a 0 = A in equation (3), and a 0 = B in equation (4).

Hence,

⎧⎪ ⎫⎪ x2 x3 x4 y = A ⎨1 + x + + + + ...⎬ ( 1 × 4 ) ( 1 × 2 )( 4 × 7 ) ( 1 × 2 × 3 )( 4 × 7 × 10 ) ⎪⎭ ⎪⎩

+ Bx

−

1 3

⎧ ⎫ x x2 x3 + + + ...⎬ ⎨1 + ⎩ (1 × 2) (1 × 2)(2 × 5) (1 × 2 × 3)(2 × 5 × 8) ⎭


469

EXERCISE 198 Page 508 1. Determine the power series solution of Bessel’s equation: x 2

d2 y dy +x + ( x 2 − v 2 ) y = 0 when 2 dx dx

v = 2, up to and including the term in x6.

The complete solution of Bessel’s equation: x 2

d2 y dy +x + ( x 2 − v 2 ) y = 0 is: 2 dx dx

⎧ ⎫ x2 x4 x6 y = A x v ⎨1 − 2 + 4 − 6 + ...⎬ ⎩ 2 (v + 1) 2 × 2!(v + 1)(v + 2) 2 × 3!(v + 1)(v + 2)(v + 3) ⎭ ⎧ ⎫ x2 x4 x6 + B x ⎨1 + 2 + 4 + 6 + ...⎬ ⎩ 2 (v − 1) 2 × 2!(v − 1)(v − 2) 2 × 3!(v − 1)(v − 2)(v − 3) ⎭ −v

⎧ ⎫ x2 x4 x6 and y = A x v ⎨1 − 2 + 4 − 6 + ...⎬ when v is a ⎩ 2 (v + 1) 2 × 2!(v + 1)(v + 2) 2 × 3!(v + 1)(v + 2)(v + 3) ⎭

positive integer. ⎧ ⎫ x2 x4 Hence, when v = 2, y = A x 2 ⎨1 − 2 + 4 + ...⎬ ⎩ 2 (2 + 1) 2 × 2!(2 + 1)(2 + 2) ⎭ ⎧ x2 x4 ⎫ y = A x 2 ⎨1 − + − ...⎬ ⎩ 12 384 ⎭

i.e.

or

⎧ ⎫ x4 x6 A ⎨ x2 − + − ...⎬ 12 384 ⎩ ⎭

2. Find the power series solution of the Bessel function: x 2 y ''+ xy '+ ( x 2 − v 2 ) y = 0 in of the Bessel function J 3 (x) when v = 3. Give the answer up to and including the term in x7. ⎫ x2 x4 ⎛x⎞ ⎧ 1 J v (x) = ⎜ ⎟ ⎨ − 2 + 4 − ...⎬ provided v is not a negative integer. ⎝ 2 ⎠ ⎩ Γ(v + 1) 2 (1!)Γ(v + 2) 2 (2!)Γ(v + 3) ⎭ v

⎫ x2 x4 ⎛x⎞ ⎧ 1 − 2 + 4 − ...⎬ Hence, when v = 3, J 3 (x) = ⎜ ⎟ ⎨ ⎝ 2 ⎠ ⎩ Γ(3 + 1) 2 (1!)Γ(3 + 2) 2 (2!)Γ(3 + 3) ⎭ 3

⎫ x2 x4 ⎛ x⎞ ⎧ 1 J 3 (x) = ⎜ ⎟ ⎨ − 2 + 5 − ...⎬ ⎝ 2 ⎠ ⎩ Γ 4 2 Γ5 2 Γ6 ⎭ 3

i.e.

or

x3 x5 x7 − 5 + 8 − ... 8 Γ4 2 Γ5 2 Γ6

3. Evaluate the Bessel functions J 0 (x) and J1 (x) when x = 1, correct to 3 decimal places.


470

J 0 (x) = 1 −

x2 x4 x6 + − + ... 22 (1!) 2 24 ( 2!)2 26 (3!) 2

and when x = 1, J 0 (x) = 1 −

12 14 16 + − + ... 22 (1!) 2 24 ( 2!)2 26 (3!) 2

= 1 – 0.25 + 0.015625 – 0.000434 + … = 0.765 correct to 3 decimal places J1 (x) =

x x3 x5 x7 − 3 + 5 − 7 + ... 2 2 (1!)(2!) 2 (2!)(3!) 2 (3!)(4!)

and when x = 1, J 1 (x) =

1 13 15 17 − 3 + 5 − 7 + ... 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


471

EXERCISE 199 Page 511 1. Determine the power series solution of the Legrandre equation: (1 − x 2 ) 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: ⎧ k(k + 1) 2 k(k + 1)(k − 2)(k + 3) 4 ⎫ y = a 0 ⎨1 − x + x − ..⎬ 2! 4! ⎩ ⎭

(a) When k = 0,

⎛ ⎞ x 3 x5 y = a0 + a1 ⎜ x + + + .... ⎟ 3 5 ⎝ ⎠

i.e.

(b) When k = 2,

i.e.

(k − 1)(k + 2) 3 (k − 1)(k − 3)(k + 2)(k + 4) 5 ⎫ ⎧ x + x − ..⎬ + a1 ⎨ x − 3! 5! ⎩ ⎭ (−1)(+2) 3 ( −1)(−3)(+2)(+4) 5 ⎫ ⎧ y = a 0 {1 − 0 + 0 − ..} + a1 ⎨ x − x + x − ..⎬ 3! 5! ⎩ ⎭

(1)(4) 3 (1)(−1)(4)(6) 5 ⎫ ⎧ 2(3) 2 2(3)(0)(5) 4 ⎫ ⎧ y = a 0 ⎨1 − x + x − ..⎬ + a1 ⎨ x − x + x − ..⎬ 2! 4! 3! 5! ⎩ ⎭ ⎩ ⎭ 2 1 ⎛ ⎞ y = a0 1 − 3x 2 + a1 ⎜ x − x 3 − x 5 − .... ⎟ 3 5 ⎝ ⎠

(

)

2. Find the following Legrendre polynomials: (a) P1(x) (b) P4(x)

(c) P5(x)

(a) Since in P1 (x) , n = k = 1, then from the second part of equation (47), page 510 of textbook, i.e. the odd powers of x: y = a1 {x − 0} = a1 x a1 is chosen to make y = 1 when x = 1 1 = a1

i.e. Hence,

P1 (x) = x

(b) Since in P4 (x) , n = k = 4, then from the first part of equation (47), page 510 of textbook, i.e. the even powers of x: 35 ⎫ ⎧ ⎧ 4(5) 2 4(5)(2)(7) 4 ⎫ x + x + 0 ⎬ = a 0 ⎨1 − 10x 2 + x 4 ⎬ y = a 0 ⎨1 − 2! 4! 3 ⎭ ⎩ ⎩ ⎭ a 0 is chosen to make y = 1 when x = 1


472

35 ⎞ 2⎞ 8 3 ⎛ ⎛ 1 = a 0 ⎜1 − 10 + ⎟ = a 0 ⎜1 − 10 + 11 ⎟ = a 0 , from which, a 0 = 3 ⎠ 3⎠ 3 8 ⎝ ⎝

i.e.

Hence,

P4 (x) =

3⎧ 35 4 ⎫ 2 ⎨1 − 10x + x ⎬ 8⎩ 3 ⎭

or

P4 (x) =

1 35x 4 − 30x 2 + 3 8

(

)

(c) Since in P5 (x) , n = k = 5, then from the second part of equation (47), i.e. the odd powers of x: (k − 1)(k + 2) 3 (k − 1)(k − 3)(k + 2)(k + 4) 5 ⎧ ⎫ y = a1 ⎨ x − x + x − ...⎬ 3! 5! ⎩ ⎭ (4)(7) 3 (4)(2)(7)(9) 5 21 ⎫ ⎧ ⎫ ⎧ 14 x + x − 0 ⎬ = a1 ⎨ x − x 3 + x 5 ⎬ i.e. y = a1 ⎨ x − 3! 5! 3 5 ⎭ ⎩ ⎭ ⎩ a1 is chosen to make y = 1 when x = 1.

⎧ 14 21 ⎫ ⎛ 15 − 70 + 63 ⎞ 8 i.e. 1 = a1 ⎨1 − + ⎬ = a1 ⎜ ⎟ = a1 15 ⎩ 3 5⎭ ⎝ ⎠ 15 Hence,

P5 (x) =

15 ⎛ 14 3 21 5 ⎞ ⎜x − x + x ⎟ 8⎝ 3 5 ⎠

or

P5 (x) =

1 63x 5 − 70x 3 + 15x 8

(

from which, a1 =

15 8

)


473

CHAPTER 53 AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS EXERCISE 200 Page 514 2. Solve

Since

∂u = 2t cos θ given that u = 2t when θ = 0 ∂t

∂u = 2t cos θ then u = ∫ 2t cos θ dt = (2 cos θ) ∫ t dt ∂t = ( 2 cos θ )

u = 2t when θ = 0, hence,

t2 + f (θ) = t 2 cos θ + f (θ) 2

2t = t 2 + f (θ) f ( θ ) = 2t − t 2

from which, Hence,

u = t 2 cos θ + 2t − t 2

or

u = t 2 ( cos θ − 1) + 2t

4. that u = e − y cos x is a solution of

∂2u ∂2u + =0 ∂x 2 ∂y 2

Since u = e − y cos x then

∂u = e − y (− sin x) ∂x

and

∂2u = −e − y cos x 2 ∂x

Also,

∂u = −e − y cos x ∂y

and

∂2u = e − y cos x 2 ∂x

Hence,

∂2u ∂2u + 2 = − e − y cos x + e − y cos x = 0 2 ∂x ∂y

6. Solve

∂2u ∂u = cos 2y = y ( 4x 2 − 1) given that x = 0, u = sin y and 2 ∂x ∂x

∂2u Since = y ( 4x 2 − 1) 2 ∂x x = 0 when

Hence,

then

⎛ 4x 3 ⎞ ∂u 2 = ∫ y ( 4x − 1) dx = y ⎜ − x ⎟ + f (y) ∂x ⎝ 3 ⎠

∂u = cos 2y, hence, cos 2y = 0 + f(y) ∂x ⎛ 4x 3 ⎞ ∂u = y⎜ − x ⎟ + cos 2y ∂x ⎝ 3 ⎠


474

and

u=

∫

⎡ ⎛ 4x 3 ⎤ ⎞ ⎛ x4 x2 ⎞ y − x + cos 2y dx = y ⎢ ⎜ ⎥ ⎟ ⎜ − ⎟ + x cos 2y + F(y) 3 2 ⎠ ⎝ ⎠ ⎝ 3 ⎣ ⎦

x = 0 when u = sin y, hence, sin y = F(y) ⎛ x4 x2 ⎞ u = y ⎜ − ⎟ + x cos 2y + sin y 2 ⎠ ⎝ 3

Thus,

x ∂2u ∂2u 8. Show that u(x, y) = xy + is a solution of 2x + y 2 = 2x ∂x∂y ∂y y Since, u = xy +

x ∂u 1 = y+ then ∂x y y

and

∂ 2 u 2x = ∂y 2 y3

and

x ⎞ 1 ∂2u ∂ ⎛ = ⎜ x − 2 ⎟ = 1− 2 ∂x∂y ∂x ⎝ y ⎠ y

Also,

Hence,

∂u x = x − 2 = x − xy −2 ∂y y

L.H.S. = 2x

⎛ ⎛ 2x ⎞ 1 ⎞ ∂2u ∂2u + y 2 = 2x ⎜1 − 2 ⎟ + y ⎜ 3 ⎟ ∂x∂y ∂y ⎝ y ⎠ ⎝y ⎠ = 2x −

2x 2x + = 2x = R.H.S. y2 y2

10. that φ(x, y) = x cos y + e x sin y satisfies the differential equation ∂ 2φ ∂ 2φ + + x cos y = 0 ∂x 2 ∂y 2

Since φ = x cos y + e x sin y then

∂φ = cos y + e x sin y ∂x

and

∂ 2φ = e x sin y 2 ∂x

and

∂φ = − x sin y + e x cos y ∂y

and

∂ 2φ = − x cos y − e x sin y 2 ∂y

Hence,

L.H.S. =

∂ 2φ ∂ 2φ + 2 + x cos y = e x sin y + ( − x cos y − e x sin y ) + x cos y 2 ∂x ∂y

= e x sin y − x cos y − e x sin y + x cos y = 0 = R.H.S.


475

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:

m 2 + 9 = 0 i.e. m 2 = −9

from which, m = −9 = ± j3 or 0 ± j3

T = e0 {A cos 3t + Bsin 3t}

Thus, the general solution is:

= 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: m2 − 1 = 0

i.e.

m2 = 1

from which,

m = 1 or m = -1

Thus, the general solution is: X = A e x + B e − x


476


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 ∂2u 1 ∂2u = 2 2 with c 2 = 9. 2 ∂x c ∂t The elastic string is shown in the diagram below.

1. The boundary and initial conditions given are: u(0, t) = 0 ⎫ ⎬ u(40, t) = 0 ⎭ u(x, 0) = f (x) =

1.5 x 20

= −

0 ≤ x ≤ 20

60 − 1.5x 1.5 x+3= 20 20

20 ≤ x ≤ 40

⎡ ∂u ⎤ ⎢⎣ ∂t ⎥⎦ = 0 i.e. zero initial velocity t =0 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 'T ∂x

and

∂2u = X ''T ∂x 2

and

Substituting into the partial differential equation, X ''T =

gives: 3. Separating the variables gives:

∂u = XT ' ∂y

and

∂ 2u = XT '' ∂y 2

∂2u 1 ∂2u = ∂x 2 c 2 ∂t 2

1 XT '' c2

i.e.

1 X ''T = XT '' since c 2 = 9 9

X '' T '' = X 9T


477

Let constant, µ =

X '' T '' X '' = then µ = X 9T X

µ=

and

T '' 9T

X′′ - µX = 0 and T′′ - 9µ T = 0

from which,

4. Letting µ = - p 2 to give an oscillatory solution gives X′′ + p 2 X = 0 and

and the auxiliary equation is: m 2 + p 2 = 0 from which, m = − p 2 = ± jp

T′′ + 9 p 2 T = 0

and the auxiliary equation is:

m 2 + 9p 2 = 0 from which, m = −9p 2 = ± 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 px}{C cos 3pt + D sin 3pt} 6. Applying the boundary conditions to determine constants A and B gives: (i) u(0, t) = 0, hence 0 = A{C 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 40p{C cos 3pt + D sin 3pt} B ≠ 0 hence sin 40p = 0 from which, 40p = nπ and p = 7. Substituting in equation (1) gives: u(x, t) = B sin ∞

nπx 40

u n (x, t) = ∑ sin

or, more generally,

n =1

nπ 40

3nπt 3nπt ⎫ ⎧ + D sin ⎨C cos ⎬ 40 40 ⎭ ⎩

nπx ⎧ 3nπt 3nπt ⎫ + Bn sin ⎨A n cos ⎬ 40 ⎩ 40 40 ⎭

(2)

where A n = BC and Bn = BD 8. From equation (8), page 517 of textbook, An =

=

2 L nπx f (x) sin dx ∫ L 0 L 40 ⎛ 60 − 1.5x ⎞ 2 ⎡ 20 ⎛ 1.5 ⎞ nπx nπx ⎤ x ⎟ sin dx + ∫ ⎜ sin dx ⎥ ⎜ ⎟ ⎢ ∫ 20 40 ⎣ 0 ⎝ 20 ⎠ 40 40 ⎝ 20 ⎠ ⎦

Each integral is determined using integration by parts (see chapter 43, page 418) with the result: An =

(8)(1.5) nπ 12 nπ sin = 2 2 sin 2 2 n π 2 n π 2


478

From equation (9), page 518 of textbook,

2 L nπx g(x) sin dx ∫ 0 cnπ L

Bn =

⎡ ∂u ⎤ ⎢⎣ ∂t ⎥⎦ = 0 = g(x) thus, Bn = 0 t =0 ∞

Substituting into equation (2) gives: u n (x, t) = ∑ sin n =1

∞

=

∑ sin n =1

Hence,

u(x, t) =

nπx ⎧ 3nπt 3nπt ⎫ + Bn sin ⎨A n cos ⎬ 40 ⎩ 40 40 ⎭ nπx ⎧ 12 nπ 3nπt nπt ⎫ + (0) sin ⎨ 2 2 sin cos ⎬ 40 ⎩ n π 2 40 50 ⎭

12 ∞ 1 nπ nπx 3nπt sin sin cos 2 ∑ 2 π n =1 n 2 40 40

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

∂ 2u 1 ∂ 2u = where c = 8, to determine the motion of a ∂x 2 c 2 ∂t 2

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) = 0 ⎫ ⎬ u(80, t) = 0 ⎭ u(x, 0) = f (x) =

1 x 40

= −

1 x+ 2 40

0 ≤ x ≤ 40 40 ≤ x ≤ 80


479

⎡ ∂u ⎤ ⎢⎣ ∂t ⎥⎦ = 0 i.e. zero initial velocity t =0 Assuming a solution u = XT, where X is a function of x only, and T is a function of t only, then

∂u = X 'T ∂x

and

∂2u = X ''T ∂x 2

∂u = XT ' ∂y

and

∂ 2u = XT '' ∂y 2

and

∂2u 1 ∂2u = Substituting into the partial differential equation, ∂x 2 c 2 ∂t 2 X ''T =

gives:

1 XT '' c2

1 XT '' since c = 8 64

X '' T '' = X 64T

Separating the variables gives: Let constant, µ =

X ''T =

i.e.

X '' T '' X '' then µ = = X 64T X

and

µ=

T '' 64T

X′′ - µX = 0 and T′′ - 64µ T = 0

from which,

Letting µ = - p 2 to give an oscillatory solution gives X′′ + p 2 X = 0 and

and the auxiliary equation is: m 2 + p 2 = 0 from which, m = − p 2 = ± jp

T′′ + 64 p 2 T = 0

and the auxiliary equation is:

m 2 + 64p 2 = 0 from which, m = −64p 2 = ± 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 px}{C cos 8pt + D sin 8pt} Applying the boundary conditions to determine constants A and B gives: (i) u(0, t) = 0, hence 0 = A{C 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 80p{C cos 8pt + D sin 8pt} B ≠ 0 hence sin 80p = 0 from which, 80p = nπ and p = Substituting in equation (1) gives: u(x, t) = B sin ∞

or, more generally,

8nπt 8nπt ⎫ nπx ⎧ + D sin ⎨C cos ⎬ 80 ⎩ 80 80 ⎭

u n (x, t) = ∑ sin n =1

nπ 80

nπx ⎧ nπt nπt ⎫ + Bn sin ⎨A n cos ⎬ 80 ⎩ 10 10 ⎭


(2)

480

An =

(8)(1) nπ 8 nπ sin = 2 2 sin 2 2 n π 2 n π 2

Bn =

2 L nπx g(x) sin dx ∫ 0 cnπ L

⎡ ∂u ⎤ ⎢⎣ ∂t ⎥⎦ = 0 = g(x) thus, Bn = 0 t =0 ∞

Substituting into equation (2) gives: u n (x, t) = ∑ sin n =1

∞

=

∑ sin n =1

Hence,

u(x, t) =

8 π2

∞

nπx ⎧ nπt nπt ⎫ + Bn sin ⎨A n cos ⎬ 80 ⎩ 10 10 ⎭ nπx ⎧ 8 nπ nπt nπt ⎫ + (0) sin ⎨ 2 2 sin cos ⎬ 80 ⎩ n π 2 10 10 ⎭ 1

∑n n =1

2

sin

nπ nπx nπt sin cos 2 80 10


481


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.

∂ 2 u 1 ∂u = and the given boundary conditions are: The heat conduction equation is ∂x 2 c 2 ∂t 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 T = k e− p c t 2 2

and

Thus, the general solution is given by: u(x, t) = {P cos px + Q sin px} e − p c t 2 2

u(0, t) = 0 thus 0 = P e − p c t from which, P = 0 and u(x, t) = {Q sin px} e − p c t 2 2

2 2

Also, u(4, t) = 0 thus 0 = {Q sin 4p} e − p c t 2 2

Since Q ≠ 0, sin 4p = 0 from which, 4p = nπ where n = 1, 2, 3, … and p = ∞

Hence,

u(x, t) =

⎧

∑ ⎨⎩Q n =1

e − p c t sin 2 2

n

nπ 4

nπx ⎫ ⎬ 4 ⎭

The final initial condition given was that at t = 0, u = 10, i.e. u(x, 0) = f(x) = 10 Hence,

∞ nπx ⎫ ⎧ 10 = ∑ ⎨Q n sin ⎬ 4 ⎭ n =1 ⎩


482

where, from Fourier coefficients, Q n = 2 × mean value of 10 sin

nπx from x = 0 to x = 4, 4

nπx ⎤ ⎡ ⎢ cos 4 ⎥ 20 ⎡ 4nπ 2 4 nπx 20 ⎤ − cos 0 ⎥ = Q n = ∫ 10sin = − ⎢ cos dx = 5 ⎢ − (1 − cos nπ ) ⎥ nπ n 4 π 4 0 4 n π ⎣ ⎦ ⎢ ⎥ 4 ⎦0 ⎣ 4

i.e.

= 0 (when n is even) and ∞

Hence, the required solution is: u(x, t) =

⎧

∑ ⎨⎩Q n =1

e − p c t sin 2 2

n

40 (when n is odd) nπ

nπx ⎫ ⎬ 4 ⎭

40 ∞ 1 − n π16c t nπx = e sin ∑ π n(odd)=1 n 4 2 2 2

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 u 1 ∂u = ∂x 2 c 2 ∂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 u 1 ∂u = and the given boundary conditions are: ∂x 2 c 2 ∂t

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 = k e− p c t 2 2

Thus, the general solution is given by: u(x, t) = {P cos px + Q sin px} e − p c t 2 2


483

u(0, t) = 0 thus 0 = P e − p c t from which, P = 0 and u(x, t) = {Q sin px} e − p c t 2 2

2 2

Also, u(20, t) = 0 thus 0 = {Q sin 20p} e − p c t 2 2

Since Q ≠ 0, sin 20p = 0 from which, 20p = nπ where n = 1, 2, 3, … and p = ∞

Hence,

u(x, t) =

⎧

∑ ⎨⎩Q n =1

e − p c t sin 2 2

n

nπ 20

nπx ⎫ ⎬ 20 ⎭

where, from Fourier coefficients, 2 × the mean value from x = 0 to x = 20 Qn =

20 ⎛ 2 ⎡ 10 ⎛ 2 ⎞ nπx 2 nπx ⎤ ⎞ dx + ∫ ⎜ − x + 8 ⎟ sin ⎜ x ⎟ sin ⎢ ∫ 0 10 20 ⎣ ⎝ 5 ⎠ 20 20 ⎥⎦ ⎝ 5 ⎠

(see above diagram)

10 20 ⎧⎡ 2 ⎤ ⎡⎛ 2 ⎞ ⎤ ⎫ π π n x n x ⎛ ⎞ π π π 2 n x 2 n x n x ⎪ ⎢ − x cos ⎥ ⎢ ⎜ x ⎟ cos ⎥ ⎪ sin sin 8cos 1 ⎪ ⎢ ⎜⎝ 5 ⎟⎠ 20 5 5 ⎠ 20 5 ⎝ 20 20 20 ⎥ ⎢ ⎥ ⎪⎬ = + − − ⎨ 2 2 10 ⎪ ⎢ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎥ ⎪ ⎛ nπ ⎞ ⎥ ⎢ ⎛ nπ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ ⎢ ⎥ ⎢ ⎜ ⎟ ⎜ ⎟ ⎪⎣ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎦ 10 ⎪ ⎝ 20 ⎠ ⎦ 0 ⎣ ⎝ 20 ⎠ ⎩ ⎭

⎧ ⎡⎛ ⎫ ⎤ ⎡⎛ ⎞ ⎞ ⎛ nπ nπ nπ ⎞ ⎤⎥ ⎪ ⎪ ⎢⎜ −4cos nπ 4sin nπ ⎟ ⎥ ⎢⎜ ⎜ ⎟ 4cos 4sin 8cos ⎟ 1 ⎪ 2 + 2 ⎟ − ( 0 ) ⎥ + ⎢⎜ 8cos nπ + 0 − 8cos nπ ⎟ − ⎜ 2 − 2 − 2 ⎟⎥ ⎪ = ⎨ ⎢⎜ ⎬ 2 2 ⎢ ⎥ ⎢ ⎜ ⎟ ⎜ n n n n n 10 ⎪ ⎛ π⎞ ⎛ π⎞ ⎟ ⎛ π⎞ ⎛ π ⎞ ⎟⎥ ⎪ ⎜ ⎛ π⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎢ ⎥ ⎢⎜ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎟⎥ ⎪ ⎜ ⎟ ⎟ ⎜ ⎟ ⎪ ⎢⎝⎜ ⎝⎜ 20 ⎠⎟ ⎝ 20 ⎠ ⎠ ⎝ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎠ ⎥⎦ ⎭ ⎥⎦ ⎢⎣⎝ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎠ ⎝ 20 ⎠ ⎩⎣

⎧ nπ nπ nπ nπ nπ ⎫ ⎪ −4 cos 4sin 4 cos 4sin 8cos ⎪ 1 ⎪ 2 + 2 + 8cos nπ − 8cos nπ − 2 + 2 + 2 ⎪ = ⎨ ⎬ 2 2 10 ⎪ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎪ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎪⎩ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎪⎭ ⎝ 20 ⎠ ⎝ 20 ⎠ ⎧ ⎧ nπ nπ nπ ⎫ nπ ⎫ ⎪ −8cos ⎪ 8sin ⎪ 8sin 8cos ⎪ 1 ⎪ 2 + 2 + 2 ⎪= 1 ⎪ 2 ⎪ = ⎨ ⎬ ⎨ ⎬ 2 10 ⎪ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎪ 10 ⎪ ⎛ nπ ⎞ 2 ⎪ ⎛ nπ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎪⎩ ⎝ 20 ⎠ ⎪⎩ ⎜⎝ 20 ⎟⎠ ⎪⎭ ⎝ 20 ⎠ ⎪⎭ ⎝ 20 ⎠ = 0 when n is even 2 8 ⎧⎪⎛ 20 ⎞ nπ ⎫⎪ 320 nπ when n is odd = ⎨⎜ ⎟ sin ⎬ = 2 2 sin 10 ⎩⎪⎝ nπ ⎠ 2 ⎪⎭ n π 2 π n (1) t ⎧⎪ 320 nπ − 2 nπx ⎫⎪ Hence, the required solution is: u(x, t) = ∑ ⎨ 2 2 sin e 20 sin ⎬ 2 20 ⎭⎪ n =1 ⎪ n π ⎩ ∞

2 2

⎛ n2π2 t ⎞ ⎟ 400 ⎟⎠

320 ∞ 1 nπ nπx − ⎜⎜⎝ = 2 ∑ sin sin e π n(odd)=1 n 2 2 20


484


1. A rectangular plate is bounded by the lines x = 0, y = 0, x = 1 and y = 3. Apply the Laplace

equation

∂2u ∂2u + = 0 to determine the potential distribution u(x, y) over the plate, subject to ∂x 2 ∂y 2

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

∂2u ∂2u + = 0 gives: ∂x 2 ∂y 2

X′′Y + XY′′ = 0 X '' Y '' =− X Y

Separating the variables gives:

Letting each side equal a constant, - p 2 gives the two equations: X′′ + p 2 X = 0 from which,

and

Y′′ - p 2 Y = 0

X = A cos px + B sin px

and Y = C e py + D e − py 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 u(x, y) = Q sin px sinh p(y + φ)

Hence,

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, Hence,

sinh p(2 + φ) = 0 and

φ = -2

u(x, y) = Q sin px sinh p(y – 2)

Since there are many solutions for integer values of n,


485

∞

u(x, y) =

∑Q n =1

n

sin px sinh p(y − 2) =

∞

∑Q n =1

n

sin nπx sinh nπ(y − 2)

(a)

∞

The fourth boundary condition is: u(x, 3) = 5 = f (x), hence, f (x) =

∑Q n =1

n

sin nπx sinh nπ(3 − 2)

From Fourier series coefficients, Q n sinh nπ = 2 × the mean value of f (x) sin nπx from x = 0 to x = 1

2 1 10 10 ⎡ cos nπx ⎤ = − ( cos nπ − cos 0 ) = = ∫ 5sin nπx dx = 10 ⎢ − (1 − cos nπ ) ⎥ nπ nπ 1 0 nπ ⎦ 0 ⎣ 1

i.e.

= 0 (for even values of n), = Hence,

Qn =

20 (for odd values of n) nπ

20 20 = cosech nπ nπ (sinh nπ) nπ ∞

Hence, from equation (a), u(x, y) =

∑Q n =1

=

20 π

n

sin nπx sinh nπ(y − 2) ∞

∑

n ( odd) ) =1

1 cos ech nπ sin nπx sinh nπ(y − 2) n

2. A rectangular plate is bounded by the lines x = 0, y = 0, x = 3, y = 2. Determine the potential ∂2u ∂2u distribution u(x, y) over the rectangle using the Laplace equation + = 0 subject to the ∂x 2 ∂y 2

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

∂2u ∂2u + = 0 gives: ∂x 2 ∂y 2

X′′Y + XY′′ = 0 X '' Y '' =− X Y

Separating the variables gives:

Letting each side equal a constant, - p 2 gives the two equations: X′′ + p 2 X = 0

and

Y′′ - p 2 Y = 0


486

from which,

X = A cos px + B sin px

and Y = C e py + D e − py 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 u(x, y) = Q sin px sinh p(y + φ)

Hence,

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 =

nπ 3

for n = 1, 2, 3, …

The third boundary condition is: u(x, 2) = 0, hence, 0 = Q sin px sinh p(2 + φ) sinh p(2 + φ) = 0 and

from which, Hence,

φ=-2

u(x, y) = Q sin px sinh p(y – 2)

Since there are many solutions for integer values of n, ∞

u(x, y) =

∑Q n =1

n

sin px sinh p(y − 2) =

∞

∑Q n =1

n

sin

nπ nπ x sinh (y − 2) 3 3

(a)

The fourth boundary condition is: u(x, 0) = x(3 – x) = 3x - x 2 = f (x), ∞

hence,

f (x) =

∑Q n =1

n

sin

nπ nπ x sinh (−2) 3 3

From Fourier series coefficients, Q n sinh

−2nπ nπ = 2 × the mean value of f (x) sin x from x = 0 to x = 3 3 3 =

2 3 nπ 3x − x 2 ) sin x dx ( ∫ 1 0 3

3 3 ⎧⎡ ⎤ ⎡ 2 ⎤ ⎫ n π x n π x n π x n π x n π x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎪ ⎢ −3x cos ⎜ ⎟ 3sin ⎜ ⎟ ⎥ ⎢ − x cos ⎜ ⎟ 2x sin ⎜ ⎟ 2cos ⎜ ⎟⎥ ⎪ ⎪⎢ 3 ⎠ 3 ⎠⎥ ⎢ 3 ⎠ 3 ⎠ 3 ⎠⎥ ⎪ ⎝ ⎝ ⎝ ⎝ ⎝ + − + + = 2⎨ ⎬ 2 2 3 ⎥ ⎢ ⎥ ⎪ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎪⎢ ⎜ ⎟ ⎜ ⎟ ⎥ ⎢ ⎥ ⎪ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎪ ⎢⎣ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎦0 ⎣ ⎦0⎭ ⎩

by integration by parts (see chapter 43)


487

⎧ ⎫ ⎪ ⎪ 2 ⎪ ⎪ −9 cos nπ 9 cos nπ 2 cos nπ ⎧ 27 ⎫ + − + = 2 ⎨ 3 3 ( 2 − 2 cos nπ ) ⎬ = 2⎨ 3 3⎬ ⎛ nπ ⎞ ⎩n π ⎭ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎪ ⎪ ⎜⎛ nπ ⎟⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎭⎪ ⎩⎪ ⎝ 3 ⎠ =

54 ( 2 − 2 cos nπ ) n 3 π3

= 0 (for even values of n), Hence,

Qn =

216 −2nπ ⎞ ⎛ n 3 π3 ⎜ sinh ⎟ 3 ⎠ ⎝

=

=

216 (for odd values of n) n 3 π3

216 −2nπ cos ech 3 3 nπ 3

Hence, from equation (a), ∞

u(x, y) =

∑ Qn sin n =1

∞ nπ nπ 216 −2nπ nπx nπ x sinh (y − 2) = ∑ 3 3 cos ech sin sinh (y − 2) 3 3 3 3 3 n =1 n π

=

216 π3

∞

∑

n ( odd) ) =1

1 2nπ n πx nπ cos ech sin sinh (2 − y) 3 n 3 3 3


488

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.


489


3. The number of vehicles ing 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.


490

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 54 ×100% ≈ 18% , B = ×100% ≈ 7% , C ≈ 35%, D ≈ 12%, E ≈ 28% 805 805

Week 2: Total = 160 + 63 + 251 + 104 + 218 = 796

A=

160 63 × 100% ≈ 20% , B = × 100% ≈ 8% , C ≈ 32%, D ≈ 13%, E ≈ 27% 796 796

Week 3: Total = 174 + 77 + 237 + 117 + 203 = 808

A=

174 77 ×100% ≈ 22% , B = × 100% ≈ 10% , C ≈ 29%, D ≈ 14%, E ≈ 25% 808 808

Week 4: Total = 158 + 69 + 211 + 144 + 194 = 776

A=

158 69 × 100% ≈ 20% , B = × 100% ≈ 9% , C ≈ 27%, D ≈ 19%, E ≈ 25% 776 776

A percentage component bar chart is shown below.


491

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 = Training, T =

44 64 × 360° ≈ 107° , Production, P = × 360° ≈ 156° , 148 148

12 × 360° ≈ 29° , 148

College, C =

28 × 360° ≈ 68° 148

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% of total costs. 360

Cost per unit = £2, hence total income per annum = 23500 × 2 = £47000 Cost of overheads per annum = 35% of £47000 =

35 × 47000 = £16450 100

(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


492


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.


493

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 20.5 – 20.9 21.0 – 21.4 21.5 – 21.9 22.0 – 22.4 22.5 – 22.9 23.0 – 23.4

Frequency 3 10 11 13 9 2

Upper class boundary Less than 20.95 21.45 21.95 22.45 22.95 23.45

Cumulative frequency 3 13 24 37 46 48

An ogive for the above frequency distribution is shown below.


494

9. The diameter in millimetres of a reel of wire is measured in 48 places and the results are as

shown. 2.10 2.28 2.26 2.16 2.24 2.15 2.11 2.23

2.29 2.18 2.10 2.25 2.05 2.22 2.17 2.07

2.32 2.17 2.21 2.23 2.29 2.14 2.22 2.13

2.21 2.20 2.17 2.11 2.18 2.27 2.19 2.26

2.14 2.23 2.28 2.27 2.24 2.09 2.12 2.16

2.22 2.13 2.15 2.34 2.16 2.21 2.30 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.


495

(c) A cumulative frequency distribution is shown below.

Class 2.05 – 2.09 2.10 – 2.14 2.15 – 2.19 2.20 – 2.24 2.25 – 2.29 2.30 – 2.34

Frequency 3 10 11 13 9 2

Upper class boundary Less than 2.095 2.145 2.195 2.245 2.295 2.345


(d) An ogive for the above data is shown below.


496

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 + 28 218 = = 27.25 8 8

Ranking gives: 21 25 26 26 28 29 31 32 Median = middle value =

26 + 28 = 27 2

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 + 157.9 806.4 = = 115.2 7 7

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.


497

EXERCISE 209 Page 540 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,

171-177 27, 178-184

Mean value =

=

22,

164-170 20, 185-191 8

( 5 ×153) + (18 ×160 ) + ( 20 ×167 ) + ( 27 ×174 ) + ( 22 ×181) + (8 ×188 ) 100 17169 = 171.7 cm 100

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.


498

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 35 ×10−3 square units; however, the

10−3 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 = 0.01158 cm 300

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% of the distance from 2.020 to 2.025 115

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


499


1. Determine the standard deviation from the mean of the set of numbers:

{ 35, 22, 25, 23, 28, 33, 30 } 35 + 22 + 25 + 23 + 28 + 33 + 30 196 = = 28 7 7

Mean, x =

Standard deviation, ⎧ ∑ x−x ⎪⎪ σ= ⎨ n ⎪ ⎪⎩

(

) ⎪⎫⎪

2 2 2 2 2 2 2 ⎪⎧ ( 35 − 28 ) + ( 22 − 28 ) + ( 25 − 28 ) + ( 23 − 28 ) + ( 28 − 28 ) + ( 33 − 28 ) + ( 30 − 28 ) ⎫⎪ ⎬= ⎨ ⎬ 7 ⎪ ⎩⎪ ⎭⎪ ⎪⎭

148 = 21.143 = 4.60, correct to 3 significant figures. 7

=

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 significant figures.

Mean, x =

34.61 + 34.57 + 34.40 + 34.63 + ...... 517.95 = = 34.53 MPa 15 15

Standard deviation, ⎧ ⎪⎪ ∑ x − x σ= ⎨ n ⎪ ⎩⎪

(

) ⎫⎪⎪

2 2 2 ⎪⎧ ( 34.61 − 34.53) + ( 34.57 − 34.53) + ( 34.40 − 34.53) + ..... ⎪⎫ = ⎬ ⎨ ⎬ 15 ⎪ ⎩⎪ ⎭⎪ ⎭⎪

=

0.0838 = 0.005586666 = 0.07474 MPa 15

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

500

Standard deviation, σ =

{(

⎧ ⎪⎪ ∑ f x − x ⎨ ⎪ ∑f ⎩⎪

)} ⎫⎪⎪ ⎬ ⎪ ⎭⎪

⎧ 7 ( 2.0125 − 2.02158 )2 + 16 ( 2.0175 − 2.02158 )2 + 23 ( 2.0225 − 2.02158 )2 ⎫ ⎪ ⎪ 2 ⎪⎪ + 9 ( 2.0275 − 2.02158 ) + 5 ( 2.0325 − 2.021582 ) ⎪⎪ = ⎨ ⎬ 60 ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎪⎭

=

0.000577124 + 0.000266342 + 0.000019467 + 0.000315417 + 0.000596232 60

=

0.001774582 = 0.00544 cm 60


501


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 + 27 = 26 faults 2

3rd quartile value =

31 + 35 = 33 faults 2

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 156.5 163.5 170.5 177.5 184.5 191.5

Frequency 5 18 20 27 22 8


The ogive is shown below. From the ogive, Q1 = 164.5cm , Q 2 = 172.5cm and and semi-interquartile range =

Q 3 = 179cm

179 − 164.5 14.5 = = 7.25 cm 2 2


502

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


503

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) p 2A =

23 number of 2 A fuses 23 = = total number of fuses 23 + 47 + 69 139

(b) p5A =

47 number of 5 A fuses = or 0.3381 total number of fuses 139

(c) p13A =

69 number of 13A fuses = or 0.4964 139 total number of fuses

4. The probability of event A happening is

or 0.1655

3 2 and the probability of event B happening is . 5 3

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 p A =

3 2 2 1 and p B = and thus the probability of events not happening, p A = and p B = 5 3 5 3

2 3 2 (a) The probability of both A and B happening = p A × p B = × = 5 5 3

3 1 1 (b) The probability of event A happening and event B not happening = p A × p B = × = 5 3 5 4 2 2 (c) The probability of only event B happening = p B × p A = × = 15 3 5

(

) (

)

(d) The probability of either A, or B, or A and B happening = ⎡ p A × p B + p B × p A ⎤ + ( p A × p B ) ⎣ ⎦ ⎡⎛ 3 1 ⎞ ⎛ 2 2 ⎞ ⎤ ⎛ 3 2 ⎞ = ⎢⎜ × ⎟ + ⎜ × ⎟ ⎥ + ⎜ × ⎟ ⎣⎝ 5 3 ⎠ ⎝ 3 5 ⎠ ⎦ ⎝ 5 3 ⎠ 13 7 6 ⎛ 3 4⎞ 6 = ⎜ + ⎟+ = + = ⎝ 15 15 ⎠ 15 15 15 15


504

5. When testing 1000 soldered ts, 4 failed during a vibration test and 5 failed due to having a

high resistance. Determine the probability of a t failing due to (a) vibration, (b) high resistance, (c) vibration or high resistance, and (d) vibration and high resistance.

(a) The probability of a t failing due to vibration, p v =

1 4 = 250 1000

(b) The probability of a t failing due to high resistance, p R =

1 5 = 200 1000

(c) The probability of a t failing due to vibration or high resistance, pv + pR =

9 1 1 4+5 + = = 250 200 1000 1000

(d) The probability of a t failing due to vibration and high resistance, pv × pR =

1 1 1 × = 50000 250 200

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

5 36


505

EXERCISE 213 Page 550 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 p A = 0.8 and p B = 0.75, and unsatisfactory operations be p A = 0.2 and p B = 0.25 (a) The probability that both components operate satisfactorily, p A × p B = 0.8 × 0.75 = 0.6 (b) The probability that only component A will operate satisfactorily, p A × p B = 0.8 × 0.25 = 0.2 (c) The probability that only component B will operate satisfactorily, p B × p A = 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 14 × × = 0.4912 20 19 18

(b) The probability that two are within but one is not within the power tolerance

⎛ 16 15 4 ⎞ ⎛ 16 4 15 ⎞ ⎛ 4 16 15 ⎞ = ⎜ × × ⎟+⎜ × × ⎟+⎜ × × ⎟ ⎝ 20 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 1 , and , respectively. Calculate the percentage of amplifiers produced 20 25 50

(a) which work satisfactorily and (b) which have just one defective transistor Let the probability of transistors working be: p A =

19 24 49 , pB = and pC = 20 25 50


506

(a) The probability of amplifiers working satisfactorily, p A × p B × p C =

19 24 49 × × 20 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 24 ⎞ = ⎜ × × ⎟+⎜ × × ⎟+⎜ × × ⎟ ⎝ 20 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 p 40 W =

14 14 28 58 = = 0.14 , p60 W = = 0.28 and p 25W = = 0.58 14 + 28 + 58 100 100 100

(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 28 × × = 0.0234 100 99 98

(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


507

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,

( 0.93 + 0.07 )

9

(q + p)

9

= q 9 + 9q8 p +

= (0.93)9 + 9(0.93)8 (0.07) +

(9)(8) 7 2 (9)(8)(7) 6 3 qp + q p + ..... 2! 3!

hence,

(9)(8) (9)(8)(7) (0.93)7 (0.07) 2 + (0.93)6 (0.07)3 + ..... 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)

(q + p)

7

= ( 0.96 + 0.04 ) = (0.96)7 + 7(0.96)6 (0.04) + 7

(7)(6) (0.96)5 (0.04) 2 2! (7)(6)(5) (0.96) 4 (0.07)3 + ..... + 3!

Which corresponds to 0, 1, 2, 3, …. employees being absent. (a) The probability that no employee will be absent on a particular day = ( 0.96 ) = 0.7514 7

(b) The probability that three employees will be absent on a particular day =

(7)(6)(5) 4 3 ( 0.96 ) ( 0.04 ) = 0.0019 3!


508

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, p H = 0.5 and the probability of a tail, pT = 0.5

( 0.5 + 0.5)

5

= (0.5)5 + 5(0.5) 4 (0.5) +

(5)(4) (5)(4)(3) (0.5)3 (0.5) 2 + (0.5) 2 (0.5)3 2! 3! (5)(4)(3)(2) 4 5 + ( 0.5)( 0.5) + ( 0.5) 4!

= 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

(q + p)

10

= ( 0.9 + 0.1) = (0.9)10 + 10(0.9)9 (0.1) + 10

(10)(9) (10)(9)(8) (0.9)8 (0.1) 2 + (0.9)7 (0.1)3 + .... 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 =

(10)(9)(8) 7 3 ( 0.9 ) ( 0.1) = 0.0574 3!


509

EXERCISE 215 Page 558 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 of: ⎛ ⎞ 32 33 λ 2 λ3 e −λ ⎜1 + λ + + + ... ⎟ = e−3 + 3e−3 + e −3 + e−3 + ... 2! 3! 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 =

1 = 0.05 and n = 72, 20

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 of: ⎛ ⎞ 3.62 −3.6 3.63 −3.6 λ 2 λ3 e + e + ... e −λ ⎜1 + λ + + + ... ⎟ = e−3.6 + 3.6e−3.6 + 2! 3! 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


510

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 :

e−λ , λe −λ ,

λ 2 −λ λ 3 −λ e , e , …. 2! 3!

i.e.

e−5 , 5e−5 ,

52 −5 53 −5 e , e , …. 2! 3!

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 :

e−λ , λe −λ ,

i.e.

e −0.7 , 0.7e−0.7 ,

λ 2 −λ λ 3 −λ λ 4 −λ λ 5 −λ e , e , e and e 2! 3! 4! 5!

0.7 2 −0.7 0.73 −0.7 0.7 4 −0.7 0.75 −0.7 e , e , e e and 2! 3! 4! 5!

i.e. 0.4966, 0.3476, 0.1217, 0.0284, 0.0050 and 0.0007 A histogram depicting these results is shown below:


511


512

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 − 75 = -2.14 standard deviations 2.8

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 − 1010 = 1.72 standard 8.7

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 shaded area of the diagram below. 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. © 2006 John Bird. All rights reserved. Published by Elsevier.

513

(b) The z-value corresponding to 1000 ml is given by:

x−x σ

i.e.

1000 − 1010 = -1.15 standard 8.7

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 The number likely to have less than 1000 ml = 0.1251 × 500 = 63, correct to nearest whole

number.

(c) The z-value corresponding to 995 ml is given by:

x−x σ

i.e.

995 − 1010 = -1.72 standard 8.7

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 The number likely to have less than 995 ml = 0.0427 × 500 = 21, correct to nearest whole

number.


514

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 81.5 − 75 i.e. = 2.32 standard deviations. σ 2.8

From Table 58.1 on page 561 of textbook, the area between z = 0 and z = 2.32 is 0.4898, i.e. the shaded area of the diagram below. 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 − 4.05 = -0.71 standard 0.0028

deviations From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -0.71 is 02611 © 2006 John Bird. All rights reserved. Published by Elsevier.

515

The z-value corresponding to 4.0553 mm is given by:

x−x 4.0553 − 4.05 i.e. = 1.89 standard 0.0028 σ

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 − 4.05 = 0.71 standard 0.0028

deviations From Table 58.1, the area between z = 0 and z = 0.71 is 02611 The z-value corresponding to 4.056 mm is given by:

x−x σ

i.e.

4.056 − 4.05 = 2.14 standard 0.0028

deviations From Table 58.1, the area between z = 0 and z = 2.14 is 0.4838 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.


516

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 − 5.00 = -3.33 standard 0.036

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 − 5.00 = -2.22 standard deviations. 0.036

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 − 5.00 = 1.11 standard 0.036

deviations From Table 58.1, the area between z = 0 and z = 1.11 is 0.3665 The probability of having masses being between 4.92 g and 5.04 g is 0.4868 + 0.3665 = 0.8533 (see shaded area in diagram below). The number of tablets likely to have a mass between 4.92 g and 5.04 g = 0.8533 × 100 = 85, © 2006 John Bird. All rights reserved. Published by Elsevier.

517


(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 (see shaded area in diagram below). The number of tablets likely to have a mass greater than 5.04g = 0.1335 × 100 = 13,



518

EXERCISE 217 Page 565 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 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

Upper class boundary Frequency

Cumulative frequency

% cumulative

value

value

frequency

26.4

26.5

5

5

5/150 = 3

26.6

26.7

12

5+12 = 17

17/150 = 11

26.8

26.9

24

17 + 24 = 41

41/150 = 27

27.0

27.1

36

77

51

27.2

27.3

36

113

75

27.4

27.5

25

138

92

27.6

27.7

12

150

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 0.6 = = 0.3 2 2


519

By calculation, mean,

x =

=

( 5 × 26.4 ) + (12 × 26.6 ) + ( 24 × 26.8) + ( 36 × 27.0 ) + ( 36 × 27.2 ) + ( 25 × 27.4 ) + (12 × 27.6 ) 150 4061.8 = 27.079 150


=

⎧⎪ 5 ( 26.4 − 27.079 )2 + 12 ( 26.6 − 27.079 )2 + 24 ( 26.8 − 27.079 )2 + .... ⎫⎪ ⎨ ⎬ 150 ⎩⎪ ⎭⎪ ⎧13.51175 ⎫ ⎨ ⎬ = 0.3001 ⎩ 150 ⎭


520

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

Upper class boundary Frequency

Cumulative frequency

% cumulative

value (kN)

value

frequency

17

18

9

9

9/275 = 3

19

20

23

9 +23 = 32

32/275 = 12

21

22

55

32 + 55 = 87

87/275 = 32

23

24

78

165

60

25

26

64

229

83

27

28

28

257

93

29

30

14

271

99

31

32

4

275

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, σ =

26.2 − 20.4 5.8 = = 2.9 kN 2 2


521

(b) By calculation, mean, x =

=


=

( 9 ×17 ) + ( 23 ×19 ) + ( 55 × 21) + ( 78 × 23) + ( 64 × 25 ) + ...... 275 6425 = 23.364 kN 275 ⎧⎪ 9 (17 − 23.364 )2 + 23 (19 − 23.364 )2 + 55 ( 21 − 23.364 )2 + .... ⎫⎪ ⎨ ⎬ 275 ⎩⎪ ⎭⎪ ⎧ 2339.6364 ⎫ ⎨ ⎬ = 2.917 kN ⎩ 275 ⎭


522

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

y=

(X − X)

(Y − Y)

xy

x2

y2

2.7

11.9

-0.2

-5.16

1.032

0.04

26.6256

4.3

7.10

1.4

-9.96

-13.944

1.96

99.2016

1.2

33.8

-1.7

16.74

-28.458

2.89

280.2276

1.4

25.0

-1.5

7.94

-11.91

2.25

63.0436

4.9

7.50

2

-9.56

-19.12

4

91.3936

∑ X = 14.5 X=

x=

14.5 = 2.9 5

∑ Y = 85.3 Y=

∑ xy = - 72.4

85.3 = 17.06 5

Coefficient of correlation, r =

∑x

−72.4 ∑ xy = {( ∑ x )( ∑ y )} (11.14 )(560.492 ) 2

2

= 11.14

∑y

2

= 560.492

= - 0.916

2

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.


523

(

)

iv

i2

v2

-6

85.716

204.09

36

-8.286

-4

33.144

68.66

16

6

-4.286

-2

8.572

18.37

4

19

8

-0.286

0

0

0.082

0

24

10

4.714

2

9.428

22.22

4

28

12

8.714

4

34.856

75.93

16

33

14

13.714

6

82.284

188.07

36

V

i = I−I

v = V−V

5

2

-14.286

11

4

15

∑ I = 135 I=

( )

I

∑ V = 56

135 = 19.286 7

V=

∑ iv = 254

56 =8 7


∑i

∑v

2

= 577.43

254 254 ∑ iv = = {( ∑ i )( ∑ v )} ( 577.43)(112) 254.307 2

2

= 112

= 0.999

2

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 ( m3 ) 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 =

−2.03 −2.03 ∑ xy = = {( ∑ x )( ∑ y )} (16800) ( 264.875 ×10 ) 2.10948 2

2

−6

= - 0.962


524

P

V

x=

y=

(P − P)

(V − V)

xy

x2

y2

160

0.034

-70

0.007125

-0.49875

4900

50.7656 ×10−6

180

0.036

-50

0.009125

-0.45625

2500

83.2656 ×10−6

200

0.030

-30

0.003125

-0.09375

900

9.7656 ×10−6

220

0.027

-10

0.000125

-0.00125

100

0.0156 ×10−6

240

0.024

10

-0.002875

-0.02875

100

260

0.025

30

-0.001875

-0.05625

900

280

0.020

50

-0.006875

-0.34375

2500

300

0.019

70

0.007875

-0.55125

4900

8.2656 ×10−6 3.5156 ×10−6 47.2656 ×10−6 62.0156 ×10−6

∑ P = 1840 ∑ V = 0.215 1840 8 = 230

P=

∑ xy

∑x

= -2.03

0.215 8 = 0.026875

V=

2

= 16800

∑v

2

= 264.875 ×10−6

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 Maintenance hours:

13

2

9

16

8

7

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.


525

X

Y

x=

y=

xy

x2

y2

(X − X) (Y − Y) 4

400

-4.4286

-40

177.144

19.6125

1600

13

515

4.5714

75

342.855

20.8977

5625

2

360

-6.4286

-80

514.288

41.3269

6400

9

440

0.5714

0

0

0.3265

0

16

570

7.5714

130

984.282

57.3261

16900

8

380

-0.4286

-60

25.716

0.1837

3600

7

415

-1.4286

-25

35.715

2.0409

625

∑ X = 59 X=

∑ Y = 3080

59 = 8.4286 7

Y=

3080 = 440 7


∑ xy

∑x

= 2080.04

141.714

2

=

2080.04 2080.04 ∑ xy = = {( ∑ x )( ∑ y )} (141.714)( 34750) 2219.135 2

∑y

2

=

34750

= 0.937

2


526

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

X2

XY

Y2

6

1.3

36

7.8

1.69

3

0.7

9

2.1

0.49

9

2.0

81

18.0

4.0

15

3.7

225

55.5

13.69

2

0.5

4

1.0

0.25

14

2.9

196

40.6

8.41

21

4.5

441

94.5

20.25

13

2.7

169

35.1

7.29

∑ X = 83 ∑ Y = 18.3 ∑ X Substituting into and

∑Y = a N +a ∑X ∑ XY = a ∑ X + a ∑ X 0

0

∑ XY = 254.6 ∑ Y

= 1161

2

2

1

2

1

18.3 = 8 a 0 + 83 a1

gives:

= 56.07

(1)

254.6 = 83 a 0 + 1161 a1

(2)

83 × (1) gives:

1518.9 = 664 a 0 + 6889 a1

(3)

8 × (2) gives:

2036.8 = 664 a 0 + 9288 a1

(4)

and

(4) – (3) gives:

517.9 =

2399 a1

Substituting in (1) gives: 18.3 = 8 a 0 + 83(0.216)

from which, a1 = from which,

a0 =

517.9 = 0.216 2399

18.3 − 83(0.216) = 0.0477 8

Hence, the equation of the regression line of Y on X is:

Y = a 0 + a 1X

i.e.

Y = 0.0477 + 0.216X © 2006 John Bird. All rights reserved. Published by Elsevier.

527

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 and

∑X = b N + b ∑Y ∑ XY = b ∑ Y + b ∑ Y 0

0

gives:

1

2

1

83 = 8 b0 + 18.3 b1

(1)

254.6 = 18.3 b0 + 56.07 b1

and

(2)

18.3 × (1) gives:

1518.9 = 146.4 b0 + 334.89 b1

(3)

8 × (2) gives:

2036.8 = 146.4 b0 + 448.56 b1

(4)

(4) – (3) gives:

113.67 b1

517.9 =

Substituting in (1) gives: 83 = 8 b0 + 18.3(4.56)

from which, b1 =

from which,

b0 =

517.9 = 4.56 113.67

83 − 18.3(4.56) = -0.056 8

Hence, the equation of the regression line of X on Y is:

X = b0 + b1Y

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


528

X

Y

X2

XY

Y2

2

5

4

10

25

4

11

16

44

121

6

15

36

90

225

8

19

64

152

361

10

24

100

240

576

12

28

144

336

784

14

33

196

462

1089

∑ X = 56 ∑ Y = 135 ∑ X Substituting into and

2

∑Y = a N + a ∑X ∑ XY = a ∑ X + a ∑ X 0

∑Y

2

= 3181

1

0

gives:

∑ XY = 1334

= 560

2

1

135 = 7 a 0 + 56 a1

(1)

and

1334 = 56 a 0 + 560 a1

(2)

8 × (1) gives:

1080 = 56 a 0 + 448 a1

(3)

(2) – (3) gives:

from which, a1 =

112 a1

254 =

Substituting in (1) gives: 135 = 7 a 0 + 56(2.268)

from which,

254 = 2.268 112

a0 =

135 − 56(2.268) = 1.142 7


Y = a 0 + a 1X

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. Substituting into and

∑X = b N + b ∑Y ∑ XY = b ∑ Y + b ∑ Y 0

0

gives:

1

2

1

56 = 7 b0 + 135 b1

(1)

and

1334 = 135 b0 + 3181 b1

(2)

135 × (1) gives:

7560 = 945 b0 + 18225 b1

(3)


529

7 × (2) gives:

9338 = 945 b0 + 22267 b1

(4) – (3) gives:

1778 =


4042 b1

(4) from which, b1 =

56 = 7 b0 + 135(0.43988) from which, b0 =


1778 = 0.43988 4042

56 − 135(0.43988) = -0.483 7

X = b0 + b1Y

X = - 0.483 + 0.440Y, correct to 3 significant figures.

i.e.

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

X2

XY

Y2

11.4

0.56

129.96

6.384

0.3136

18.7

0.35

349.69

6.545

0.1225

11.7

0.55

136.89

6.435

0.3025

12.3

0.52

151.29

6.396

0.2704

14.7

0.43

216.09

6.321

0.1849

18.8

0.34

353.44

6.392

0.1156

19.6

0.31

384.16

6.076

0.0961

∑X = 107.2

Substituting into and gives: and

∑Y = 3.06

∑X

∑ XY

2

= 1721.52

= 44.549

∑Y = a N +a ∑X ∑ XY = a ∑ X + a ∑ X 0

0

∑Y

2

= 1.4056

1

2

1

3.06 = 7 a 0 + 107.2 a1 44.549 = 107.2 a 0 + 1721.52 a1 © 2006 John Bird. All rights reserved. Published by Elsevier.

(1) (2) 530

107.2 × (1) gives:

328.032 = 750.4 a 0 + 11491.84 a1

(3)

7 × (2) gives:

311.843 = 750.4 a 0 + 12050.64 a1

(4)

(3) – (4) gives:

from which, a1 =

- 558.8 a1

16.189 =

−16.189 = - 0.0290 558.8

Substituting in (1) gives: 3.06 = 7 a 0 + 107.2(-0.0290) from which,

a0 =

3.06 − 107.2(−0.0290) = 0.881 7


Y = a 0 + a 1X

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

∑X = b N + b ∑Y ∑ XY = b ∑ Y + b ∑ Y

gives:

107.2 = 7 b0 + 3.06 b1

Substituting into

0

0

1

2

1

(1)

44.549 = 3.06 b0 + 1.4056 b1

(2)

3.06 × (1) gives:

328.032 = 21.42 b0 + 9.3636 b1

(3)

7 × (2) gives:

311.843 = 21.42 b0 + 9.8392 b1

(4)

and

(3) – (4) gives:

16.189 =

- 0.4756 b1

from which, b1 =

16.189 = - 34.039 −0.4756

Substituting in (1) gives: 107.2 = 7 b0 + 3.06(-34.039) from which,

b0 =

107.2 + 3.06(34.039) = 30.194 7


X = b0 + b1Y

i.e.

X = 30.194 – 34.039Y


531

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


532

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, N p = 1500, standard deviation, σ = 0.0438 cm,

mean, µ = 22.4 cm. For the sample, N = 30 The mean of the sampling distributions, µ x = µ = 22.4 cm The standard error of the means with replacement, σ x =

σ 0.0438 = = 0.0080 cm N 30

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, σ x =

σ N

⎛ N p − N ⎞ 0.0438 ⎛ 1500 − 30 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ 30 ⎝ 1500 − 1 ⎠ ⎝ Np −1 ⎠

= (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, N p = 1800, standard deviation, σ = 0.013 cm, mean, µ = 1.70 cm. For the sample, N = 20 (a) With replacement. The mean of the sampling distributions, µ x = µ = 1.70 cm © 2006 John Bird. All rights reserved. Published by Elsevier.

533

The standard error of the means, σ x =

σ 0.013 = = 2.907 ×10−3 = 2.91 × 10−3 cm, correct N 20 to 3 significant figures.

(b) Without replacement. The mean of the sampling distributions, µ x = µ = 1.70 cm

Standard error of means, σ x =

σ N

⎛ N p − N ⎞ 0.013 ⎛ 1800 − 20 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ 20 ⎝ 1800 − 1 ⎠ ⎝ Np −1 ⎠

= ( 2.907 × 10−3 ) ( 0.9947 ) = 2.89 × 10−3 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 σ x =

σ 60 = = 15 h N 16

790 h corresponds to a z-value of: z1 =

790 − 800 = - 0.67 standard deviations and the area 15

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: z 2 =

810 − 800 = 0.67 standard deviations and the area between 15

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.


534

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, N p = 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, σ x =

σ N

⎛ N p − N ⎞ 0.092 ⎛ 1200 − 40 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ 40 ⎝ 1200 − 1 ⎠ ⎝ Np −1 ⎠

= (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 − 20.16 = - 2.1 standard deviations. 0.0143

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 − 20.16 = 0.7 standard deviations and the area 0.0143

between z = 0 and z = 0.7 is 0.2580 (from table 58.1, page 561 of textbook)


535

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


536

EXERCISE 221 Page 585 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, N p = 1546 and σ = 2.93 mm For the sample, N = 100 and x = 67.45 mm For a 95% confidence limit, z C = 1.96 from Table 61.1, page 582 of textbook. An estimate of the confidence limits of the population mean is: x±

zC σ N

⎛ Np − N ⎞ (1.96 )( 2.93) ⎛ 1546 − 100 ⎞ ⎜⎜ ⎟⎟ = 67.45 ± ⎜ ⎟ 100 ⎝ 1546 − 1 ⎠ ⎝ Np −1 ⎠

= 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 For a 99% confidence limit, z C = 2.58 from Table 61.1, page 582 of textbook. An estimate of the confidence limits of the population mean is: x±

zC σ ⎛ N p − N ⎞ ( 2.58)( 2.93) ⎛ 1546 − 100 ⎞ ⎜⎜ ⎟⎟ = 67.45 ± ⎜ ⎟ N ⎝ Np −1 ⎠ 100 ⎝ 1546 − 1 ⎠

= 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


537

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? N p = 500 and σ = 150 kg = 0.15 Mg, N = 40 and x = 2.40 Mg (a) For a 95% confidence limit, z C = 1.96 from Table 61.1, page 582 of textbook. An estimate of the confidence limits of the population mean is: x±

zC σ ⎛ Np − N ⎞ (1.96 )( 0.15) ⎛ 500 − 40 ⎞ ⎜⎜ ⎟⎟ = 2.40 ± ⎜ ⎟ N ⎝ Np −1 ⎠ 40 ⎝ 500 − 1 ⎠

= 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 For a 99% confidence limit, z C = 2.58 from Table 61.1, page 582 of textbook. An estimate of the confidence limits of the population mean is: x±

zC σ ⎛ N p − N ⎞ ( 2.58)( 0.15) ⎛ 500 − 40 ⎞ ⎜⎜ ⎟⎟ = 2.40 ± ⎜ ⎟ N ⎝ Np −1 ⎠ 40 ⎝ 500 − 1 ⎠

= 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 = x ±

zC σ N


538

x = 2.40, hence,

±

zC σ = ±0.035 N zC =

and

0.035 N ( 0.035 ) 40 = = ± 1.48 σ 0.15

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 x ±

zC σ N

For a 90% confidence level, z C = 1.645 from Table 61.1, page 582 of textbook. x ± 20 = x ±

(a)

(1.645)(100 ) N

(1.645 )(100 )

from which,

20 =

i.e.

N = (8.225) 2 = 67.65

N

and

N=

(1.645)(100 ) 20

= 8.225

Hence, at least 68 components are required to be 90% confident that the error will not exceed 20 hours. (b)

x ± 10 = x ±

(1.645)(100 ) N


539

from which, i.e.

10 =

(1.645)(100 ) N

and

N=

(1.645)(100 ) 10

= 16.45

N = (16.45) 2 = 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, z C = 1.96 (from Table 61.1, page 582 of textbook). Hence,

14.63 ±

(1.96 )( 2.45) = 14.63 ± 0.759 40

i.e. the maximum error is 0.759 minutes = 0.759 × 60 = 45.6 s


540

EXERCISE 222 Page 588 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, t 0.95 , ν = 9 gives t C = 1.83 Estimated value of the mean of the U.T.S. is: x ±

tC s N −1

= 5.17 ±

(1.83)( 0.06 ) 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, x =

1.12 + 1.15 + 1.10 + 1.14 + 1.15 + 1.10 + 1.11 = 1.1243 Ω m −1 7

and standard deviation, s =

=

⎧⎪ (1.12 − 1.1243)2 + (1.15 − 1.1243)2 + (1.10 − 1.1243)2 + ... ⎫⎪ ⎨ ⎬ 7 ⎪⎩ ⎪⎭ 0.00297143 = 0.0206 Ω m −1 7

From Table 61.2, page 587 of textbook, t 0.99 , ν = 7 − 1 = 6 gives t C = 3.14 Hence, 99% confidence limits is given by: x ±

tC s N −1

= 1.1243 ±

( 3.14 )( 0.0206 ) 7 −1


541

= 1.1243 ± 0.0264 = 1.1243 – 0.0264 or 1.1243 + 0.0264 Thus, the 99% confidence interval is: 1.10 Ω m −1 to 1.15 Ω m −1

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 tC s N −1

132.27 ± 0.79 = x ±

Hence,

0.79 =

i.e.

from which,

tC =

t C ( 0.742 ) 5 −1

( 0.79 ) 0.742

4

= 2.13

From Table 61.2, page 587 of textbook, at ν = N – 1 = 5 – 1 = 4, 2.13 corresponds to t 0.95 , 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%.


542

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 − mean 54 − 50 = = 0.58 and from Table 58.1, page 561 of textbook, the s tan dard deviation 6.892

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 − 50 = 1.74 and from Table 58.1, the area between the mean and a z-value of 1.74 6.892

is 0.4591. Thus, the probability of more than 62 defective rings = 0.5000 – 0.4591 = 0.0409 Hence, the type I error is 4.09% (c) z-value =

70 − 50 = 2.90 and from Table 58.1, the area between the mean and a z-value of 6.892

2.90 is 0.4981. Thus, the probability of more than 70 defective rings = 0.5000 – 0.4981 = 0.0019 Hence, the type I error is 0.19%


543

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 = z-value =

( Npq ) = (1000 )( 0.06 )( 0.94 ) = 7.51

61 − 60 = 0.13 (note that ‘more than 60 components defective’ means 61 or more) 7.51

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 Mean = Np = 75, standard deviation = z-value =

( Npq ) = (1000 )( 0.075)( 0.925 ) = 8.329

61 − 75 = -1.68 and from Table 58.1, the area between the mean and a z-value of 8.329

-1.68 is 0.4535. Thus, the probability of more than 75 defective components = 0.5000 - 0.4535 = 0.0465 Hence, the type II error is 4.65% (c) N = 1000, p = 0.09, q = 0.91 Mean = Np = 90, standard deviation = z-value =

( Npq ) = (1000 )( 0.09 )( 0.91) = 9.05

61 − 90 = -3.20 and from Table 58.1, the area between the mean and a z-value of 9.05

-3.20 is 0.4993. Thus, the probability of more than 90 defective components = 0.5000 - 0.4993 = 0.0007 Hence, the type II error is 0.07%


544

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 :

e−λ = e−3 = 0.0498 ,

λe−λ = 3e −3 = 0.1494 ,

λ 2 −λ 32 −3 λ 3 −λ 33 −3 e = e = 0.2240 , e = e = 0.2240 , 2! 2 3! 6

λ 4 −λ 34 −3 λ 5 −λ 35 −3 e = e = 0.1680 , e = e = 0.1008 4! 24 5! 120 (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 i.e. the type I error is 18.5% (c) The probability of a sample containing 6 or more defective components is: 1 – (0.8152 + 0.1008) = 1 – 0.9160 = 0.0840 i.e. the type I error is 8.4%


545

EXERCISE 224 Page 601 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 = 2050 kN The null hypothesis,

H0 : x > µ

The alternative hypothesis, H1 : x = µ z=

x − µ 2050 − 2000 50 = =± = ± 3.54 100 σ 14.142 N 50

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,

H 0 : mean diameter = 3.0000 cm

The alternative hypothesis, H1 : mean diameter ≠ 3.0000 cm z=

x − µ 3.0078 − 3.0000 0.0078 = =± = ± 2.85 0.015 σ 0.00274 N 30

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.


546

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 significance of 0.1. µ = 10.25 kN and N = 20 Sample mean, x =

( 9.8 ×1) + (10.0 ×1) + (10.1× 4 ) + (10.2 × 5) + .... = 207.6 = 10.38 20

20

Sample standard deviation, 1( 9.8 − 10.38 ) + 1(10.0 − 10.38 ) + 4 (10.1 − 10.38 ) + 5 (10.2 − 10.38 ) + ... 2.212 = = 0.33 20 20 2

s=

2

2

2

The null hypothesis is that the sample breaking strength is better than the mean breaking strength. N < 30, therefore a t distribution is used. t =

(x − µ)

( N − 1) (10.38 − 10.25) ( 20 − 1) s

=

0.33

At a level of significance of 0.1, the t value is t ⎛

0.1 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

= 1.72

i.e. t 0.95 and ν = N – 1 = 20 – 1 = 19, and

from Table 61.1, page 587 of textbook, t 0.95 ν = 19 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: H 0 : µ = 0.50 cm


547

t =

( x − µ)

( N − 1) ( 0.53 − 0.50 ) (10 − 1) =

s

0.03

(a) From Table 61.1, page 587 of the textbook, t ⎛

=3

0.05 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

= t 0.975 , ν = 9 has a value of 2.26

Since 3 is outside of this range, the hypothesis is rejected. (b) From Table 61.1, t ⎛

0.01 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

= t 0.995 , ν = 9 has a value of 3.25

Since 3 is within this range, the hypothesis is accepted.


548

EXERCISE 225 Page 605 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: x A = 24 , σ A = 4 and N A = 100 Battery B: x B = 40 , σ B = 6 and N B = 80

The hypothesis is: H: x = x A + 15 Let x = 24 + 15 = 39 z=

x − xB ⎛ σA 2 σB2 ⎞ + ⎜ ⎟ ⎝ NA NB ⎠

=

39 − 40 ⎛ 42 62 ⎞ + ⎟ ⎜ ⎝ 100 80 ⎠

=

−1 = −1.28 0.781025

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. 2 $2 = s N ) (Bessel’s correction is σ N −1

N = 5, x1 =

21.4 + 23.6 + 24.8 + 22.4 + 26.3 = 23.7 5

⎛ ( 21.4 − 23.7 )2 + ( 23.6 − 23.7 )2 + ( 24.8 − 23.7 )2 + ( 22.4 − 23.7 )2 + ( 26.3 − 23.7 )2 ⎞ s1 = ⎜ ⎟ = 1.73 ⎜ ⎟ 5 ⎝ ⎠


549

⎛ N ⎞ ⎛5⎞ σ1 = s1 ⎜ ⎟ = 1.73 ⎜ ⎟ = 1.93 ⎝ N −1 ⎠ ⎝4⎠

x2 =

22.4 + 27.7 + 23.5 + 29.1 + 25.8 = 25.7 5

⎛ ( 22.4 − 25.7 )2 + ( 27.7 − 25.7 )2 + ( 23.5 − 25.7 )2 + ( 29.1 − 25.7 )2 + ( 25.8 − 25.7 )2 ⎞ s2 = ⎜ ⎟ = 2.50 ⎜ ⎟ 5 ⎝ ⎠ ⎛ N ⎞ ⎛5⎞ σ2 = s2 ⎜ ⎟ = 2.50 ⎜ ⎟ = 2.80 ⎝ N −1 ⎠ ⎝4⎠

t =

t⎛

x 2 − x1 ⎛ σx 2 σy 2 ⎞ + ⎜⎜ ⎟ N N y ⎟⎠ x ⎝

0.01 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

=

25.7 − 23.7 ⎛ 1.932 2.802 ⎞ + ⎜ ⎟ 5 ⎠ ⎝ 5

=

2 = 1.32 1.5208

= t 0.995 and ν = N1 + N 2 − 2 = 5 + 5 − 2 = 8

From Table 61.2, page 587 of textbook, t 0.995 , ν = 8 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.

N A = 12, x A = 4.8 litre, σ A = 0.40 litre N B = 12, x B = 5.1 litre, σ B = 0.36 litre

The hypothesis is: H: manufacturer A is more economical than manufacturer B. ⎛ 12 ( 0.40 )2 + 12 ( 0.36 )2 ⎞ ⎛ N A sA 2 + N B sB2 ⎞ 3.4752 σ= ⎜ = 0.397 ⎟= ⎟ = ⎜⎜ ⎟ 12 + 12 − 2 22 ⎝ NA + NB − 2 ⎠ ⎝ ⎠ © 2006 John Bird. All rights reserved. Published by Elsevier.

550

t =

xA − xB ⎛ 1 1 ⎞ σ ⎜ + ⎟ ⎝ NA NB ⎠

4.8 − 5.1

=

⎛1 1⎞ 0.397 ⎜ + ⎟ ⎝ 12 12 ⎠

(a) The t value is t ⎛

0.01 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

=−

0.3 = −1.85 ( 0.397 )( 0.40825)

i.e. t 0.995 and ν = 12 + 12 – 2 = 22, hence, from Table 61.2, page 587 of

textbook, t 0.995 , ν = 22 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, t ⎛

0.1 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

i.e. t 0.95 , ν = 22 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 significance of 0.1. N = 5, x 4 = 21.4 km, s 4 = 0.54 km x un = 22.6 km, s un = 0.48 km

The hypothesis is: H: x un > x 4 ⎛ 5 ( 0.54 )2 + 5 ( 0.48 )2 ⎞ ⎛ N 4 s 4 2 + N un s un 2 ⎞ 2.61 σ= ⎜ = 0.571 ⎟= ⎟ = ⎜⎜ ⎟ 5 + 5 − 2 8 ⎝ N 4 + N un − 2 ⎠ ⎝ ⎠ t =

x un − x 4 ⎛ 1 1 ⎞ σ ⎜ + ⎟ ⎝ N un N 4 ⎠

=

22.6 − 21.4 ⎛1 1⎞ 0.571 ⎜ + ⎟ ⎝5 5⎠

=

1.2 = 3.32 ( 0.571)( 0.63246 )


551

The t value is t ⎛

0.1 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

i.e. t 0.95 and ν = 5 + 5 – 2 = 8, hence, from Table 61.2, page 587 of textbook,

t 0.95 , ν = 8 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.


552

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

o-e

(o − e)

(o − e)

Observed

Expected

frequency, o

frequency, e

1

49

40

9

81

2.025

2

35

40

-5

25

0.625

3

32

40

-8

64

1.6

4

46

40

6

36

0.9

5

49

40

9

81

2.025

6

29

40

-11

121

3.025

2

2

e

⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 10.2 ⎩⎪ e ⎭⎪ 2

Hence, the Chi-square value, χ 2 = 10.2


553

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.

o-e

(o − e)

(o − e)

Number

Observed

Expected

of calls

frequency, o

frequency, e

0

11

16

-5

25

1.5625

1

44

42

2

4

0.0952

2

53

52

1

1

0.0192

3

46

42

4

16

0.3810

4

24

26

-2

4

0.1538

5

12

14

-2

4

0.2857

6

7

6

1

1

0.1667

7

3

2

1

1

0.5000

2

2

e

⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 3.16 ⎪⎩ e ⎪⎭ 2

Hence, the Chi-square value, χ 2 = 3.16


554

EXERCISE 227 Page 612 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%?

250 ( 0.6 + 0.4 )

7

(7)(6) (7)(6)(5) 6 5 2 4 3 ⎡ 7 ⎤ ( 0.6 ) ( 0.4 ) ⎢0.6 + 7 ( 0.6 ) ( 0.4 ) + 2! ( 0.6 ) ( 0.4 ) + ⎥ 3! ⎢ ⎥ (7)(6)(5)(4) (7)(6)(5)(4)(3) 3 4 2 5 ⎢ ⎥ = 250 + ( 0.6 ) ( 0.4 ) + ( 0.6 ) ( 0.4 ) ⎢ ⎥ 4! 5! ⎢ ⎥ (7)(6)(5)(4)(3)(2) 6 7 ⎢ + 0.6 )( (0.4 ) + ( 0.4 ) ⎥ ( ⎢⎣ ⎥⎦ 6!

= 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 o-e

(o − e)

(o − e)

Observed

Expected

frequency, o

frequency, e

8

7

1

1

0.14286

27

33

-6

36

1.09091

62

65

-3

9

0.13846

79

73

6

36

0.49315

45

48

-3

9

0.18750

24

19

5

25

1.31579

5

4

1

1

0.25000

0

0

0

0

0

2

2

e

⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 3.62 e ⎩⎪ ⎭⎪ 2

Degrees of freedom, ν = N – 1 = 8 – 1 = 7 2 For χ 0.95 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. 2 χ 0.10 , ν 7 = 2.83, hence the data is not ‘too good’.


555

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 significance of 0.05.

x=

(7 × 1.28) + (19 × 1.29) + (41× 1.30) + (50 × 1.31) + .... 390.5 = = 1.32 7 + 19 + 41 + 50 + 73 + 52 + 28 + 17 + 9 296

2 2 2 2 ⎪⎧ 7 (1.28 − 1.32 ) + 19 (1.29 − 1.32 ) + 41(1.30 − 1.32 ) + 50 (1.31 − 1.32 ) + .... ⎫⎪ s= ⎨ ⎬ 296 ⎩⎪ ⎭⎪

=

0.0958 = 0.0180 296

Class

Class

z-value for class

Area from

Area for

Expected

mid-point

boundaries, x

boundary =

0 to z

class

frequency

x − 1.32 0.0180


-2.50

0.4938 0.02

6

0.0561

17

0.121

36

0.1864

55

0.2206

65

0.1864

55

0.121

36

0.0561

17

0.02

6

1.275 1.28 1.285

-1.94

0.4738

1.29 1.295

-1.39

0.4177

1.30 1.305

-0.83

0.2967

1.315

-0.28

0.1103

1.325

0.28

0.1103

1.335

0.83

0.2967

1.345

1.39

0.4177

1.355

1.94

0.4738

1.31 1.32 1.33 1.34 1.35 1.36 1.365

2.5

0.4938


556

Resistance

Observed

Expected

frequency,

frequency, e

(o − e)

o-e

(o − e)

2

2

e

o 1.28

7

6

1

1

0.1667

1.29

19

17

2

4

0.2353

1.30

41

36

5

25

0.6944

1.31

50

55

-5

25

0.4545

1.32

73

65

8

64

0.9846

1.33

52

55

-3

9

0.1636

1.34

28

36

-8

64

1.7778

1.35

17

17

0

0

0

1.36

9

6

3

9

1.5000 ⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 5.98 ⎪⎩ e ⎪⎭ 2

Degrees of freedom, ν = N – 1 – M = 9 – 1 – 2 = 6 2 For χ 0.95 and ν = 6 from Table 63.1, page 609 of textbook is 12.6

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 ed 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 x=

2

5

12

17

14

6

3

1

(2 × 8.5) + (5 × 9.0) + (12 × 9.5) + (17 × 10.0) + .... 605.5 = = 10.09 MN 2 + 5 + 12 + 17 + 14 + 6 + 3 + 1 60

⎧⎪ 2 ( 8.5 − 10.09 )2 + 5 ( 9.0 − 10.09 )2 + 12 ( 9.50 − 10.09 )2 + 17 (10.0 − 10.09 )2 + .... ⎫⎪ s= ⎨ ⎬ 60 ⎩⎪ ⎭⎪ =

32.246 = 0.733MN 60


557

Class

Class

z-value for class

Area from

mid-point

boundaries, x

boundary =

0 to z

Area for class

Expected frequency

x − 10.09 0.733 8.25

0.4940

-2.51

8.5 8.75

-1.83

0.4664

9.25

-1.15

0.3749

9.75

-0.46

0.1772

10.25

0.22

0.0871

10.75

0.90

0.3159

11.25

1.58

0.4430

11.75

2.26

0.4881

9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.25

Load

Expected

frequency,

frequency, e

2

0.0915

5

0.1977

12

0.2643

16

0.2288

14

0.1271

8

0.0451

3

0.0103

1

0.4984

2.95

Observed

0.0276

o-e

(o − e)

(o − e)

2

2

e

o 8.5

2

2

0

0

0

9.0

5

5

0

0

0

9.5

12

12

0

0

0

10.0

17

16

1

1

0.0625

10.5

14

14

0

0

0

11.0

6

8

-2

4

0.5000

11.5

3

3

0

0

0

12.0

1

1

0

0

0 ⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 0.563 e ⎩⎪ ⎭⎪ 2


558

Degrees of freedom, ν = N – 1 – M = 8 – 1 – 2 = 5 2 and ν = 5 from Table 63.1, page 609 of textbook is 11.1 For χ 0.95

Hence, the hypothesis is accepted. 2 χ 0.05 , ν 5 = 1.15 , hence the results are ‘too good to be true’.


559

EXERCISE 228 Page 616 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, H 0 : µ = 0.56 Alternative hypothesis, H1 : µ ≠ 0.56 (ii) Significance level, α 2 = 5% (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 α 2 = 5% 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

3.6

7.5

9.0 13.8 16.2

6.5 18.0 11.5 13.4

9.8 10.6 16.4 12.7 16.8

9.4

6.9 14.2 12.4

Type B 15.3 15.4 11.2 16.1 18.1 17.1 17.7 7.8

8.6

8.4 13.5

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, H 0 : µ A = µ B Alternative hypothesis, H1 : µ A ≠ µ B


560

(ii) Significance level, α 2 = 5% (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 α 2 = 5% S ≤ 4 Since from (iv) S is equal to 4, then the result is significant at α 2 = 5% hence, the alternative

hypothesis H1 is accepted.


561

EXERCISE 229 Page 619 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) H 0 : t = 220 h H1 : t ≠ 220 h (ii) α 2 = 5% (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, α 2 = 5% , T ≤ 29 Hence, since 74 > 29, the hypothesis H 0 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.


562

Using the procedure for the Wilcoxon signed-rank test: (i) H 0 : N = R H1 : N ≠ R (ii) α 2 = 5% (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, α 2 = 5% , T ≤ 8 Since from (v) T is less than 8, there is a significant difference in the drying times.


563

EXERCISE 230 Page 624 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)

H 0 : TA = TB H1 : TA ≠ TB

(ii) α 2 = 5% (iii) Brand P

0.7 1.7 2.0 2.3 3.2

Brand Q

2.1

3.9 4.1

3.4 3.5 3.8

5.6 6.1

4.7 5.5

22.6

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 α 2 = 5% , U ≤ 17 Hence, H 0 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. Using the procedure for the Mann-Whitney test: (i)

H0 : A = B H1 : A ≠ B


564

(ii) α 2 = 5% (iii) A B

106 94 98 103 105

132 135

112

139 140 143 147 153 137

(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 (vi) From Table 63.5, page 621 of textbook, for a sample size of 8 and 6 at α 2 = 5% , U ≤ 8 Hence, the null hypothesis H 0 is rejected, i.e. the two methods are not equally effective.

4. Repeat problem 3 of Exercise 228, using the Mann-Whitney test.

Using the procedure for the Mann-Whitney test: (i) Null hypothesis, H 0 : µ A = µ B Alternative hypothesis, H1 : µ A ≠ µ B (ii) α 2 = 5% (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 (vi) From Table 63.5, page 622 of textbook, for a sample size of 18 and 18 at α 2 = 5% , U ≤ 99 Hence, the null hypothesis H 0 is rejected and H1 is accepted, i.e. the two means are not

equal.


565

CHAPTER 64 INTRODUCTION TO LAPLACE TRANSFORMS EXERCISE 231 Page 630 1. Determine the Laplace transforms of (a) 2t – 3 (b) 5t 2 + 4t − 3 2 3 ⎛ 1 ⎞ ⎛1⎞ (a) ℒ {2t − 3} = 2 ⎜ 2 ⎟ − 3 ⎜ ⎟ = 2 − s s ⎝s ⎠ ⎝s⎠ 10 4 3 ⎛ 2! ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ (b) ℒ {5t 2 + 4t − 3} = 5 ⎜ 3 ⎟ + 4 ⎜ 2 ⎟ − 3 ⎜ ⎟ = 3 + 2 − s s s ⎝s ⎠ ⎝s ⎠ ⎝s⎠

2. Determine the Laplace transforms of (a)

t3 t5 t2 − 2t 4 + - 3t + 2 (b) 15 2 24

⎧ t3 ⎫ ⎛ 1 ⎞ ⎛ 3! ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 3 2 (a) ℒ ⎨ − 3t + 2 ⎬ = ⎜ ⎟ ⎜ 3+1 ⎟ − 3 ⎜ 2 ⎟ + 2 ⎜ ⎟ = 4 − 2 + 4s s s ⎩ 24 ⎭ ⎝ 24 ⎠ ⎝ s ⎠ ⎝ s ⎠ ⎝ s ⎠ ⎧ t5 8 48 1 t 2 ⎫ ⎛ 1 ⎞⎛ 5! ⎞ ⎛ 4! ⎞ ⎛ 1 ⎞⎛ 2! ⎞ (b) ℒ ⎨ − 2t 4 + ⎬ = ⎜ ⎟⎜ 5+1 ⎟ − 2 ⎜ 4+1 ⎟ + ⎜ ⎟⎜ 3 ⎟ = 6 − 5 + 3 s s s 2 ⎭ ⎝ 15 ⎠⎝ s ⎠ ⎝ s ⎠ ⎝ 2 ⎠⎝ s ⎠ ⎩15

3. Determine the Laplace transform of (a) 5e3t (b) 2e−2t 5 ⎛ 1 ⎞ (a) ℒ {5e3t } = 5 ⎜ ⎟ = s−3 ⎝ s−3⎠ 2 ⎛ 1 ⎞ (b) ℒ {2e −2t } = 2 ⎜ ⎟ = s+2 ⎝s+2⎠

4. Determine the Laplace transform of (a) 4 sin 3t (b) 3 cos 2t 12 ⎛ 3 ⎞ (a) ℒ { 4sin 3t } = 4 ⎜ 2 2 ⎟ = 2 s +9 ⎝s +3 ⎠ 3s ⎛ s ⎞ (b) ℒ { 3cos 2t } = 3 ⎜ 2 = 2 2 ⎟ s +4 ⎝s +2 ⎠


566

5. Determine the Laplace transforms of (a) 7 cosh 2x (b)

1 sinh 3t 3

7s ⎛ s ⎞ (a) ℒ { 7 cosh 2x } = 7 ⎜ 2 = 2 2 ⎟ s −4 ⎝s −2 ⎠ 1 ⎧1 ⎫ 1⎛ 3 ⎞ (b) ℒ ⎨ sinh 3t ⎬ = ⎜ 2 2 ⎟ = 3 s −9 ⎩3 ⎭ 3⎝ s −3 ⎠

6.(a) Determine the Laplace transform of 2 cos 2 t cos 2t = 2 cos 2 t − 1 ℒ { 2 cos t } 2

from which,

cos 2 t =

1 + cos 2t 2

s2 + 4 ) + s2 ( ⎧ ⎛ 1 + cos 2t ⎞ ⎫ 1 s 1 s = + = = ℒ ⎨2 ⎜ ⎟ ⎬ = ℒ {1 + cos 2t} = + 2 2 s s + 22 s s 2 + 4 s ( s2 + 4 ) ⎠⎭ ⎩ ⎝ 2 ( s2 + 2 ) 2s 2 + 4 = = s (s2 + 4) s ( s2 + 4)

7.(b) Determine the Laplace transform of 2sinh 2 2θ cosh 4θ = 1 + 2sinh 2 2θ

from which,

ℒ { 2sinh 2 2θ } = ℒ {cosh 4θ − 1} =

2sinh 2 2θ = cosh 4θ - 1

2 2 16 s 1 s − ( s − 16 ) − = = 2 2 s −4 s s ( s 2 − 16 ) s ( s 2 − 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 { (4 cos b) sin at + (4sin b) cos at }

⎛ a ⎞ ⎛ s ⎞ + 4sin b ) ⎜ 2 = ( 4 cos b ) ⎜ 2 2 ⎟ ( 2 ⎟ ⎝s +a ⎠ ⎝s +a ⎠

=

4 ( a cos b + s sin b ) s + a2 2


567

10. Show that ℒ ( cos 2 3t − sin 2 3t ) =

s s + 36 2

cos 6t = 2cos 2 3t − 1

from which,

cos 2 3t =

1 + cos 6t 2

and cos 6t = 1 − 2sin 2 3t

from which,

sin 2 3t =

1 − cos 6t 2

⎧⎛ 1 + cos 6t ⎞ ⎛ 1 − cos 6t ⎞ ⎫ ⎧ 1 cos 6t 1 cos 6t ⎫ Hence, ℒ ( cos 2 3t − sin 2 3t ) = ℒ ⎨⎜ − + ⎬ ⎟−⎜ ⎟⎬ = ℒ ⎨ + 2 2 2 2 2 ⎭ ⎠ ⎝ ⎠⎭ ⎩2 ⎩⎝ = ℒ{cos 6t} =


s s + 36 2

568

CHAPTER 65 PROPERTIES OF LAPLACE TRANSFORMS EXERCISE 232 Page 634 1. Determine the Laplace transforms of (a) 2te 2t (b) t 2 e t

(a) ℒ {2te 2t } = 2 ℒ {t1e2t } = (2)

(b) ℒ {t 2 e t } =

2!

( s − 1)

2 +1

=

1!

(s − 2)

1+1

=

2

(s − 2)

2

2

( s − 1)

3

2.(b) Determine the Laplace transform of

1 4 −3t te 2

12 4! ⎞ ⎧1 ⎫ 1⎛ ℒ ⎨ t 4 e −3t ⎬ = ⎜ ⎟ = 5 4 +1 ⎩2 ⎭ 2 ⎜⎝ ( s + 3) ⎟⎠ ( s + 3)

3.(b) Determine the Laplace transform of 3e 2t sin 2t ⎛ ⎞ 6 2 6 ℒ {3e 2t sin 2t} = 3 ⎜ = = ⎟ 2 2 2 ⎜ ( s − 2 ) + 22 ⎟ s − 4s + 4 + 4 s − 4s + 8 ⎝ ⎠

4.(a) Determine the Laplace transform of 5e−2t cos 3t ⎛ ⎞ 5 ( s + 2) 5 (s + 2) s+2 ℒ {5e −2t cos 3t} = 5 ⎜ = = ⎟ ⎜ ( s + 2 )2 + 32 ⎟ s 2 + 4s + 4 + 9 s 2 + 4s + 13 ⎝ ⎠

5.(a) Determine the Laplace transform of 2e t sin 2 t

⎧ ⎛ 1 − cos 2t ⎞ ⎫ 1 s −1 t t ℒ {2e t sin 2 t} = ℒ ⎨2e t ⎜ − ⎟ ⎬ = ℒ {e } - ℒ {e cos 2t} = 2 s − 1 ( s − 1)2 + 22 ⎠⎭ ⎩ ⎝ =

1 s−1 − 2 s − 1 s − 2s + 5


569

6.(b) Determine the Laplace transform of 3e 2t cosh 4t

⎛ ⎞ 3 ( s − 2) s−2 3(s − 2) ℒ {3e 2t cosh 4t} = 3 ⎜ = = ⎟ ⎜ ( s − 2 )2 − 42 ⎟ s 2 − 4s + 4 − 16 s 2 − 4s − 12 ⎝ ⎠

7.(a) Determine the Laplace transform of 2e− t sinh 3t

⎛ ⎞ 6 3 6 = 2 ℒ {2e − t sinh 3t} = 2 ⎜ ⎟= 2 ⎜ ( s + 1) − 32 ⎟ s 2 + 2s + 1 − 9 s + 2s − 8 ⎝ ⎠

8. Determine the Laplace transforms of (a) 2e t ( cos 3t − 3sin 3t )

(b) 3e −2t ( sinh 2t − 2 cosh 2t )

⎛ ⎞ ⎛ ⎞ s −1 3 (a) ℒ {2e t ( cos 3t − 3sin 3t )} = ℒ {2e t cos 3t} - ℒ {6e t sin 3t} = 2 ⎜ 6 − ⎟ ⎜ ⎟ ⎜ ( s − 1)2 + 32 ⎟ ⎜ ( s − 1)2 + 32 ⎟ ⎝ ⎠ ⎝ ⎠ =

2 ( s − 1) 18 2s − 2 − 18 2s − 20 − 2 = 2 = 2 s − 2s + 10 s − 2s + 10 s − 2s + 10 s − 2s + 10 2

=

2 ( s − 10 ) s 2 − 2s + 10

(b) ℒ {3e−2t ( sinh 2t − 2 cosh 2t )} = ℒ {3e−2t sinh 2t} - ℒ {6e −2t cosh 2t}

⎛ ⎞ ⎛ ⎞ 2 s+2 − 6 = 3⎜ ⎟ ⎜ ⎟ 2 2 ⎜ ( s + 2 ) − 22 ⎟ ⎜ ( s + 2 ) − 22 ⎟ ⎝ ⎠ ⎝ ⎠ =

6 (s + 2) 6 (s + 2) 6 6 − = − s 2 + 4s + 4 − 4 s 2 + 4s + 4 − 4 s 2 + 4s s 2 + 4s

=

−6 ( s + 1 ) 6 − 6s − 12 −6s − 6 = = s ( s + 4) s (s + 4) s (s + 4)


570

EXERCISE 233 Page 635 2. Use the Laplace transform of the first derivative to derive the transforms:

(a) ℒ{ eat } =

1 s−a

(b) ℒ{ 3t 2 } =

6 s3

(a) Let f(t) = eat then f ′(t) = a eat and f(0) = 1 From equation (3), page 634 of textbook, ℒ {f '(t)} = s ℒ {f (t)} − f (0) ℒ{ a eat } = sℒ {eat } - 1

Hence,

1 = (s – a)ℒ {eat }

i.e.

ℒ {eat } =

and

1 s−a

(b) Let f(t) = 3t 2 then f ′(t) = 6t and f(0) = 0 Since

ℒ {f '(t)} = s ℒ {f (t)} − f (0)

then,

ℒ{ 6t } = sℒ {3t 2 } + 0

i.e.

6 = sℒ {3t 2 } s2

and

ℒ { 3t 2 } =

6 s3

3. Derive the Laplace transform of the second derivative from the definition of a Laplace

transform. Hence derive the transform ℒ{sin at} =

a s + a2 2

For the derivation, see page 634 of textbook. Let f(t) = sin at, then f ′(t) = a cos at and f ′′(t) = −a 2 sin at , f(0) = 0 and f ′(0) = a From equation (4), page 635 of textbook, Hence,

ℒ {f ''(t)} = s 2 ℒ {f (t)} − sf (0) − f '(0)

ℒ {−a 2 sin at} = s 2 ℒ {sin at} - s(0) - a


571

−a 2 ℒ {sin at} = s 2 ℒ {sin at} - a

i.e.

a = ( s 2 + a 2 ) ℒ {sin at}

Hence,

ℒ {sin at} =

and

a s + a2 2

4.(b) Use the Laplace transform of the second derivative to derive the transform: ℒ{cosh at} =

s s − a2 2

Let f(t) = cosh at then f ′(t) = a sinh at and f ′′(t) = a 2 cosh at , f(0) = 1 and f ′(0) = 0 ℒ {f ''(t)} = s 2 ℒ {f (t)} − sf (0) − f '(0)

Hence,

ℒ {a 2 cosh at} = s 2 ℒ {cosh at} - s(1) - 0

i.e.

a 2 ℒ {cosh at} = s 2 ℒ {cosh at} - s

i.e. and

s = ( s 2 − a 2 ) ℒ {cosh at} ℒ {cosh at} =

s s − a2 2


572

EXERCISE 234 Page 637 1. State the initial value theorem. the theorem for the function (a) 3 – 4 sin t (b) ( t − 4 )

2

and state their initial values. The initial value theorem states: lim it [ f (t) ] = lim it [ s ℒ {f (t)} t →0

s →∞

then ℒ{f(t)} = ℒ{3 – 4 sin t} =

(a) Let f(t) = 3 – 4 sin t

]

3 4 − 2 s s +1

⎡ ⎛3 4 ⎞⎤ lim it [3 − 4sin t ] = lim it ⎢s ⎜ − 2 ⎟ ⎥ t →0 s →∞ ⎣ ⎝ s s + 1 ⎠⎦

Hence,

4s ⎤ ⎡ = lim it ⎢3 − 2 ⎥ s →∞ ⎣ s + 1⎦ i.e.

3 – 4 sin 0 = 3 -

i.e.

3=3

∞ ∞ +1 2

which verifies the theorem.

The initial value is 3

(b) Let f(t) = ( t − 4 ) = t 2 − 8t + 16 2

then

Hence,

ℒ {t 2 − 8t + 16} =

2 8 16 − + s3 s 2 s

⎡ ⎛ 2 8 16 ⎞ ⎤ lim it ⎡⎣ t 2 − 8t + 16 ⎤⎦ = lim it ⎢s ⎜ 3 − 2 + ⎟ ⎥ t →0 s →∞ s ⎠⎦ ⎣ ⎝s s ⎡2 8 ⎤ = lim it ⎢ 2 − + 16 ⎥ s →∞ ⎣s 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. the

theorem for the function 4 + e−2t (sin t + cos t) representing a displacement and state its final value.


573

The final value theorem states:

lim it [ f (t) ] = lim it [ s ℒ {f (t)} t →∞

s→0

]

The final value theorem is used in investigating the stability of systems such as in automatic aircraft-landing systems. Let f(t) = 4 + e−2t (sin t + cos t) = 4 + e−2t sin t + e−2t cos t then

Hence,

ℒ {f (t)} =

4 1 s+2 + + 2 s ( s + 2 ) + 1 ( s + 2 )2 + 1

⎛4 1 s+2 ⎞ lim it ⎡⎣ 4 + e−2t sin t + e−2t cos t ⎤⎦ = lim it [ s ⎜ + + ⎟ t →∞ s→0 ⎜ s ( s + 2 )2 + 1 ( s + 2 )2 + 1 ⎟ ⎝ ⎠ ⎡ s (s + 2) ⎤ s + = lim it ⎢ 4 + ⎥ 2 2 s→0 ⎢⎣ ( s + 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


574

CHAPTER 66 INVERSE LAPLACE TRANSFORMS EXERCISE 235 Page 640

2. (a) Determine the inverse Laplace transform of

3 2s + 1

⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ 3 − 12 t 3 −1 ⎪ 1 ⎪⎪ 1 ⎪ ⎧ 3 ⎫ −1 ⎪ ℒ −1 ⎨ = 3 ℒ = ℒ = e ⎬ ⎨ ⎬ ⎨ ⎬ 1 1 2 2 ⎛ ⎞ ⎛ ⎞ ⎩ 2s + 1 ⎭ ⎪2⎜s + ⎟⎪ ⎪⎜ s + ⎟ ⎪ ⎪⎩ ⎝ 2 ⎠ ⎭⎪ ⎩⎪ ⎝ 2 ⎠ ⎭⎪


1 s + 25 2

1 1 −1 ⎧ 5 ⎫ ⎧ 1 ⎫ ℒ −1 ⎨ 2 ⎬ = ℒ ⎨ 2 2 ⎬ = sin 5t 5 5 ⎩ s + 25 ⎭ ⎩s + 5 ⎭


5s 2s + 18 2

⎧ ⎫⎪ 5 −1 ⎧ s ⎫ 5 s ⎧ 5s ⎫ −1 ⎪ ℒ −1 ⎨ 2 ⎬ = 5ℒ ⎨ ⎬ = ℒ ⎨ 2 2 ⎬ = cos 3t 2 2 ⎩ 2s + 18 ⎭ ⎩s + 3 ⎭ 2 ⎪⎩ 2 ( s + 9 ) ⎪⎭


5 s3

5 ⎧5⎫ ⎧ 2!⎫ 5 ℒ −1 ⎨ 3 ⎬ = ℒ −1 ⎨ 3 ⎬ = t 2 2! ⎩s ⎭ ⎩s ⎭ 2


3s 1 2 s −8 2

⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ 3s s ⎧ s ⎫ ℒ −1 ⎨ = 3ℒ −1 ⎨ = 6ℒ −1 ⎨ 2 = 6 cosh 4t ⎬ ⎬ 2⎬ 1 1 − s 4 2 2 ⎩ ⎭ ⎪ s −8⎪ ⎪ ( s − 16 ) ⎪ ⎩2 ⎭ ⎩2 ⎭


575

8. (b) Determine the inverse Laplace transform of

3

( s − 3)

5

⎧⎪ 3 ⎫⎪ ⎧⎪ ⎫⎪ 1 3 −1 ⎧⎪ 4! ⎫⎪ 1 −1 ⎧⎪ 4! ⎫⎪ 1 3t 4 = 3ℒ −1 ⎨ = ℒ ⎨ = ℒ ⎨ = e t ℒ −1 ⎨ 5⎬ 4 +1 ⎬ 4 +1 ⎬ 4 +1 ⎬ 4! 8 − − − − s 3 s 3 s 3 s 3 ( ) ( ) ( ) ( ) ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ 8

9. (b) Determine the inverse Laplace transform of

3 s + 6s + 13 2

⎧ ⎫⎪ ⎫⎪ 3 −3t 3 1 3 −1 ⎧⎪ 2 ⎧ ⎫ −1 ⎪ ℒ −1 ⎨ 2 = ℒ ⎬ ⎨ ⎬ = e sin 2t ⎬ = 3ℒ ⎨ 2 2 2 2 2 ⎩ s + 6s + 13 ⎭ ⎩⎪ ( s + 3) + 2 ⎭⎪ ⎩⎪ ( s + 3) + 2 ⎭⎪ 2


2(s − 3) s − 6s + 13 2

⎧ ⎫⎪ ⎧ 2 ( s − 3) ⎫ s−3 −1 ⎪ 3t ℒ −1 ⎨ 2 ⎬ = 2e cos 2t ⎬ = 2ℒ ⎨ 2 2 ⎩ s − 6s + 13 ⎭ ⎩⎪ ( s − 3) + 2 ⎭⎪

11. Determine the inverse Laplace transforms of (a)

2s + 5 s + 4s − 5 2

(b)

3s + 2 s − 8s + 25 2

⎧ 2 (s + 2) ⎫⎪ 1 ⎧ 2s + 5 ⎫ −1 ⎪ (a) ℒ −1 ⎨ 2 + ⎬ ⎬ =ℒ ⎨ 2 2 2 2 ⎩ s + 4s − 5 ⎭ ⎩⎪ ( s + 2 ) − 3 ( s + 2 ) − 3 ⎭⎪ ⎧⎪ ⎫⎪ 1 −1 ⎧⎪ ⎫⎪ 1 −2t s+2 3 −2t 2e cosh 3t e sinh 3t + = 2ℒ −1 ⎨ ℒ = + ⎬ ⎨ ⎬ 2 2 2 2 3 ⎪⎩ ( s + 2 ) − 3 ⎪⎭ 3 ⎪⎩ ( s + 2 ) − 3 ⎭⎪ ⎧ 3s + 2 ⎫⎪ ⎧ ⎫ ⎧ 3s + 2 ⎫ −1 ⎪ −1 ⎪ 3 ( s − 4 ) + 14 ⎪ = ℒ (b) ℒ −1 ⎨ 2 ⎬ ⎬ =ℒ ⎨ ⎨ ⎬ 2 2 2 2 ⎩ s − 8s + 25 ⎭ ⎪⎩ ( s − 4 ) + 3 ⎪⎭ ⎪⎩ ( s − 4 ) + 3 ⎭⎪ ⎧⎪ ⎫⎪ 14 −1 ⎧⎪ ⎫⎪ 14 4t s−4 3 4t = 3ℒ −1 ⎨ + ℒ ⎬ ⎨ ⎬ = 3e cos 3t + e sin 3t 2 2 2 2 3 ⎪⎩ ( s − 4 ) + 3 ⎪⎭ 3 ⎪⎩ ( s − 4 ) + 3 ⎭⎪


576


2. Use partial fractions to find the inverse Laplace transform of:

2s 2 − 9s − 35 4 3 1 = − + ( s + 1)( s − 2 )( s + 3) ( s + 1) ( s − 2 ) ( s + 3)

2s 2 − 9s − 35 ( s + 1)( s − 2 )( s + 3)

from Problem 2, page 19 of textbook

⎧ 4 3 1 ⎪⎫ ⎪⎧ 2s 2 − 9s − 35 ⎪⎫ −1 ⎪ −t 2t −3t − + Hence, ℒ −1 ⎨ ⎬ =ℒ ⎨ ⎬ = 4e − 3e + e + − + + − + s 1 s 2 s 3 s 1 s 2 s 3 ( )( )( ) ( ) ( ) ( ) ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪


3s 2 + 16s + 15

( s + 3)

3

=

3 2 6 − − 2 ( s + 3 ) ( s + 3 ) ( s + 3 )3

3s 2 + 16s + 15

( s + 3)

3


⎧⎪ 3s 2 + 16s + 15 ⎫⎪ ⎧ 3 2 6 ⎫⎪ −1 ⎪ Hence, ℒ −1 ⎨ = ℒ − − ⎬ ⎨ 2 3⎬ 3 ⎪⎩ ( s + 3) ⎪⎭ ⎪⎩ ( s + 3) ( s + 3) ( s + 3) ⎪⎭ ⎧ 1 ⎫⎪ ⎧ 1 ⎫⎪ ⎪⎧ 1 ⎪⎫ −1 ⎪ −1 ⎪ = 3ℒ −1 ⎨ 6 ℒ ⎬ - 2ℒ ⎨ ⎬ ⎨ 2 3⎬ ⎪⎩ ( s + 3) ⎭⎪ ⎪⎩ ( s + 3) ⎭⎪ ⎩⎪ ( s + 3) ⎭⎪ ⎧ ⎫⎪ 6 −1 ⎧⎪ 2! ⎫⎪ ⎧⎪ 1 ⎫⎪ 1 −1 ⎪ = 3ℒ −1 ⎨ - ℒ ⎨ ⎬ - 2ℒ ⎨ 1+1 ⎬ 2 +1 ⎬ ⎪⎩ ( s + 3) ⎭⎪ 2! ⎩⎪ ( s + 3) ⎭⎪ ⎩⎪ ( s + 3) ⎭⎪

= 3e −3t − 2e −3t t − 3e −3t t 2

or

e −3t (3 − 2t − 3t 2 )

3 + 6s + 4s 2 − 2s3 6. Use partial fractions to find the inverse Laplace transform of: s 2 ( s 2 + 3) 3 + 6s + 4s 2 − 2s3 2 1 3 − 4s = + 2+ 2 s s s +3 s 2 ( s 2 + 3)


⎧⎪ 3 + 6s + 4s 2 − 2s3 ⎫⎪ 1 3 − 4s ⎫ −1 ⎧ 2 Hence, ℒ −1 ⎨ ⎬ ⎬ =ℒ ⎨ + 2 + 2 2 2 ⎩s s s +3⎭ ⎩⎪ s ( s + 3) ⎭⎪


577

⎧ 3 − 4s ⎧1 ⎫ −1 ⎧ 1 ⎫ −1 ⎪ = 2ℒ ⎨ ⎬ + ℒ ⎨ 2 ⎬ + ℒ ⎨ ⎩s ⎭ ⎩s ⎭ ⎪ s2 + 3 ⎩ −1

( )

⎫ ⎪ 2⎬ ⎪ ⎭

⎧ ⎫ ⎧ 3 4s ⎪ ⎧1 ⎫ −1 ⎧ 1 ⎫ −1 ⎪ −1 ⎪ = 2ℒ ⎨ ⎬ + ℒ ⎨ 2 ⎬ + ℒ ⎨ -ℒ ⎨ 2⎬ ⎩s ⎭ ⎩s ⎭ ⎪ s2 + 3 ⎪ ⎪ s2 + 3 ⎩ ⎭ ⎩ −1

( )

( )

⎫ ⎪ 2⎬ ⎪ ⎭

⎧ ⎫ ⎧ ⎫ 3 3 s ⎪ ⎪ ⎧1 ⎫ −1 ⎧ 1 ⎫ −1 ⎪ −1 ⎪ = 2ℒ ⎨ ⎬ + ℒ ⎨ 2 ⎬ + ℒ ⎨ - 4ℒ ⎨ 2⎬ 2⎬ 3 ⎩s ⎭ ⎩s ⎭ ⎪ s2 + 3 ⎪ ⎪ s2 + 3 ⎪ ⎩ ⎭ ⎩ ⎭ −1

( )

( )

= 2 + t + 3 sin 3 t − 4cos 3 t


26 − s 2 s ( s 2 + 4s + 13)

A ( s 2 + 4s + 13) + ( Bs + C ) s 26 − s 2 A Bs + C ≡ + 2 = Let s ( s 2 + 4s + 13) s s + 4s + 13 s ( s 2 + 4s + 13)

26 − s 2 = A ( s 2 + 4s + 13) + Bs 2 + Cs

Hence,

When s = 0:

26 = 13A + 0 + 0

i.e. A = 2


-1 = A + B

i.e. B = -3

Equating s coefficients: Thus,

0 = 4A + C

i.e. C = -8

26 − s 2 2 −3s − 8 ≡ + 2 2 s ( s + 4s + 13) s s + 4s + 13

⎧⎪ ⎫⎪ 3s + 8 ⎫ 26 − s 2 −1 ⎧ 2 ⎫ −1 ⎧ Hence, ℒ −1 ⎨ 2 ⎬ ⎬ =ℒ ⎨ ⎬ -ℒ ⎨ 2 ⎩ s + 4s + 13 ⎭ ⎩s ⎭ ⎩⎪ s ( s + 4s + 13) ⎭⎪

⎧⎪ 3s + 8 ⎫⎪ ⎧1 ⎫ = 2ℒ −1 ⎨ ⎬ - ℒ −1 ⎨ ⎬ 2 2 ⎩s ⎭ ⎩⎪ ( s + 2 ) + 3 ⎭⎪ ⎧⎪ 3 ( s + 2 ) ⎫⎪ ⎧ ⎫⎪ 2 ⎧1 ⎫ −1 ⎪ ℒ = 2ℒ −1 ⎨ ⎬ - ℒ −1 ⎨ ⎬ ⎨ ⎬ 2 2 2 2 ⎩s ⎭ ⎩⎪ ( s + 2 ) + 3 ⎭⎪ ⎩⎪ ( s + 2 ) + 3 ⎭⎪


578

⎧⎪ ( s + 2 ) ⎫⎪ 2 −1 ⎧⎪ ⎫⎪ 3 ⎧1 ⎫ = 2ℒ −1 ⎨ ⎬ - 3ℒ −1 ⎨ ℒ ⎬ ⎨ ⎬ 2 2 2 2 ⎩s ⎭ ⎩⎪ ( s + 2 ) + 3 ⎭⎪ 3 ⎩⎪ ( s + 2 ) + 3 ⎭⎪ 2 = 2 − 3e −2t cos 3t − e −2t sin 3t 3


579

EXERCISE 237 (Page XX)

1. Determine for the transfer function: R(s) =

50 ( s + 4 )

s ( s + 2 ) ( s 2 − 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 s 2 − 8s + 25 = 0

i.e.

s=

( −8 )

− −8±

2

− 4(1)(25)

2(1)

=

8 ± −36 −8 ± j6 = = 4 ± j3 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) =

s −1 determine the poles and zeros and show them on a ( s + 2 ) ( s 2 + 2s + 5 )

pole-zero diagram. For the denominator to be zero, s = -2 or s 2 + 2s + 5 = 0

i.e.

s=

−2 ±

( 2)

2

− 4(1)(5)

2(1)

=

−2 ± −16 −2 ± j4 = = −1 ± j2 2 2

Hence, poles occur at s = -2, s = -1 + j2 and -1 – j2 © 2006 John Bird. All rights reserved. Published by Elsevier.

580

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.


581

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 E + 5i = dt 2

and i = 0 at time t = 0.

Use Laplace transforms to solve for i when (a) E = 20 (b) E = 40e −3t and (c) E = 50 sin 5t.

Taking the Laplace transform of each term of

di E + 5i = gives: dt 2

⎧ di ⎫ ⎧E ⎫ ℒ ⎨ ⎬ + 5ℒ{i} = ℒ ⎨ ⎬ ⎩2⎭ ⎩ dt ⎭ sℒ{i} – i(0) + 5ℒ{i} =

i.e. i = 0 at t = 0, hence,

E/2 s

i(0) = 0 (s + 5)ℒ{i} =

Hence,

ℒ{i} =

i.e.

E/2 s E/2 s(s + 5)

⎧ E / 2 ⎫ E −1 ⎧ 1 ⎫ i = ℒ −1 ⎨ ⎬ ⎬= ℒ ⎨ ⎩ s(s + 5) ⎭ ⎩ s(s + 5) ⎭ 2

and

1 A B A(s + 5) + Bs ≡ + = s(s + 5) s (s + 5) s(s + 5)

Let Hence,

1 = A(s + 5) + Bs

When s = 0:

1 = 5A

i.e.

When s = -5:

1 = -5B

i.e.

Thus,

A=

1 5

B=-

1 5

1 ⎫ ⎧1 E −1 ⎧ 1 ⎫ E −1 ⎪ 5 5 ⎪ = E ⎛ 1 − 1 e −5t ⎞ i= ℒ ⎨ ⎬= ℒ ⎨ − ⎬ ⎜ ⎟ 2 2 2 ⎝5 5 ⎠ ⎩ s(s + 5) ⎭ ⎪ s (s + 5) ⎪ ⎩ ⎭

(a) When E = 20,

i=

20 ⎛ 1 1 −5t ⎞ −5t ⎜ − e ⎟ = 2 1− e 2 ⎝5 5 ⎠

(

)


582

⎧ 40e −3t ⎫ ⎧ di ⎫ ℒ ⎨ ⎬ + 5ℒ{i} = ℒ ⎨ ⎬ ⎩ dt ⎭ ⎩ 2 ⎭

(b) When E = 40e −3t

sℒ{i} – i(0) + 5ℒ{i} =

i.e. i = 0 at t = 0, hence,

20 s+3

i(0) = 0 (s + 5)ℒ{i} =

Hence,

ℒ{i} =

i.e.

20 s+3 20 (s + 3)(s + 5)

and

⎧ ⎫ 20 i = ℒ −1 ⎨ ⎬ ⎩ (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,

⎧ ⎫ 20 10 ⎫ −1 ⎧ 10 − i = ℒ −1 ⎨ ⎬ =ℒ ⎨ ⎬ ⎩ (s + 3)(s + 5) ⎭ ⎩ (s + 3) (s + 5) ⎭

i.e.

i = 10e −3t − 10e −5t = 10 ( e−3t − e−5t ) ⎧ di ⎫ ⎧ 50sin 5t ⎫ ℒ ⎨ ⎬ + 5ℒ{i} = ℒ ⎨ ⎬ ⎩ dt ⎭ ⎩ 2 ⎭

(c) When E = 50 sin 5t

sℒ{i} – i(0) + 5ℒ{i} =

i.e. i = 0 at t = 0, hence, Hence, i.e.

and

Let

25(5) s 2 + 52

i(0) = 0 (s + 5)ℒ{i} = ℒ{i} =

125 s + 25 2

125 (s + 5)(s 2 + 25)

⎧ ⎫ 125 i = ℒ −1 ⎨ ⎬ 2 ⎩ (s + 5)(s + 25) ⎭ 125 A Bs + C A(s 2 + 25) + (Bs + C)(s + 5) ≡ + = (s + 5)(s 2 + 25) (s + 5) (s 2 + 25) (s + 5)(s 2 + 25)


583

Hence,

125 = A ( s 2 + 25 ) + ( Bs + C )( s + 5 )

When s = -5:

125 = 50A


0=A+B

Equating constant :

i.e. i.e.

125 = 25A + 5C

from which,

A=

5 2

B=-

i.e. 125 =

5 2

125 + 5C 2

125 25 C= 2 = 5 2

5 25 ⎫ ⎧ 5 − + s ⎪ ⎧ ⎫ 125 −1 2 + 2 2 ⎪ = ℒ i = ℒ −1 ⎨ ⎬ ⎨ ⎬ 2 2 ⎩ (s + 5)(s + 25) ⎭ ⎪ (s + 5) ( s + 25 ) ⎪ ⎩ ⎭

Thus,

⎧ 25 ⎫ ⎧ 5 ⎫ s ⎪ ⎪ ⎪ ⎪ 5 −1 ⎧ 1 ⎫ −1 −1 2 2 = ℒ ⎨ ⎬ +ℒ ⎨ 2 ⎬ -ℒ ⎨ 2 ⎬ 2 ⎩ (s + 5) ⎭ ⎪ ( s + 25 ) ⎪ ⎪ ( s + 25 ) ⎪ ⎩ ⎭ ⎩ ⎭

i.e.

=

⎫⎪ 5 ⎧ ⎫⎪ 5 −1 ⎧ 1 ⎫ 5 −1 ⎧⎪ 5 s −1 ⎪ ℒ ⎨ ⎬+ ℒ ⎨ 2 2 ⎬- ℒ ⎨ 2 2 ⎬ 2 2 ⎩ (s + 5) ⎭ 2 ⎪⎩ ( s + 5 ) ⎪⎭ ⎩⎪ ( s + 5 ) ⎭⎪

=

5 −5t 5 5 e + sin 5t − cos 5t 2 2 2

i=

5 −5t ( e + sin 5t − cos 5t ) 2

d2x 3. Use Laplace transforms to solve the differential equation: 2 + 100x = 0 given x(0) = 2 and dt x′(0) = 0. ⎧ d2x ⎫ ℒ ⎨ 2 ⎬ + ℒ{100x} = ℒ{0} ⎩ dx ⎭

i.e.

s 2 ℒ{x} – s x(0) - x′(0) + 100ℒ{x} = 0

x(0) = 2 and x′(0) = 0, hence i.e.

s 2 ℒ{x} – 2s + 100ℒ{x} = 0 ( s 2 + 100)ℒ{x} = 2s


584

from which,

ℒ{x} =

2s s + 100 2

and

⎧ 2s ⎫ x = ℒ −1 ⎨ 2 ⎬ ⎩ s + 100 ⎭

⎧ s ⎫ x = 2ℒ −1 ⎨ 2 = 2 cos 10t 2⎬ ⎩ s + 10 ⎭

i.e.

4. Use Laplace transforms to solve the differential equation:

d 2i di + 1000 + 250000i = 0 2 dt dt

given i(0) = 0 and i′(0) = 100.

⎧ d 2i ⎫ ⎧ di ⎫ ℒ ⎨ 2 ⎬ + 1000ℒ ⎨ ⎬ + 250000ℒ{i} = ℒ{0} ⎩ dt ⎭ ⎩ dt ⎭

i.e. [ s 2 ℒ{i} – s i(0) - i′(0)] + 1000[sℒ{i} – i(0)] + 250000ℒ{i} = 0 i(0) = 0 and i′(0) = 100, hence s 2 ℒ{i} – 100 + 1000sℒ{i} + 250000ℒ{i} = 0 ( s 2 +1000s + 250000)ℒ{i} = 100

i.e.

ℒ{i} =

from which,

100 s + 1000s + 250000 2

⎧⎪ 100 ⎫⎪ ⎧ ⎫⎪ 1 −1 ⎪ i = ℒ −1 ⎨ = 100 ℒ ⎨ 2⎬ 1+1 ⎬ ⎪⎩ ( s + 500 ) ⎭⎪ ⎪⎩ ( s + 500 ) ⎭⎪

and

i = 100 t e −500t

i.e.

6. Use Laplace transforms to solve the differential equation: y(0) = −

d2 y dy − 2 + y = 3e 4x given 2 dx dx

2 1 and y′(0) = 4 3 3 ⎧ d2 y ⎫ ⎧ dy ⎫ ℒ ⎨ 2 ⎬ - 2ℒ ⎨ ⎬ + ℒ{y} = ℒ {3e 4x } ⎩ dx ⎭ ⎩ dx ⎭

i.e.

[ s 2 ℒ{y} – s y(0) - y′(0)] - 2[sℒ{y} – y(0)] + ℒ{y} =

3 (s − 4)


585

y(0) = −

2 1 and y′(0) = 4 , hence 3 3

3 ⎛ 2 ⎞ 13 ⎛ 2⎞ - 2sℒ{y} + 2 ⎜ − ⎟ + ℒ{y} = s 2 ℒ{y} – s ⎜ − ⎟ (s − 4) ⎝ 3⎠ 3 ⎝ 3⎠ ( s 2 - 2s + 1)ℒ{y} +

i.e.

( s 2 - 2s + 1)ℒ{y} =

from which,

ℒ{y} =

3 2 17 9 − 2s(s − 4) + 17(s − 4) - s+ = (s − 4) 3 3 3(s − 4) 9 − 2s 2 + 8s + 17s − 68 −59 + 25s − 2s 2 = 2 3 ( s − 4 ) ( s 2 − 2s + 1) 3 ( s − 4 )( s − 1)

⎧⎪ −59 + 25s − 2s 2 ⎫⎪ y = ℒ −1 ⎨ 2 ⎬ ⎩⎪ 3 ( s − 4 )( s − 1) ⎭⎪

and

Let

2 13 4 3 s− − = 3 3 3 (s − 4)

1 2 −59 + 25s − 2s 2 ) ( A ( s − 1) + B ( s − 4 )( s − 1) + C ( s − 4 ) A B C 3 ≡ + + = 2 2 ( s − 4 ) ( s − 1) ( s − 1)2 ( s − 4 )( s − 1) ( s − 4 )( s − 1) −

Hence,

59 25 2 2 + s − s 2 = A ( s − 1) + B ( s − 4 )( s − 1) + C ( s − 4 ) 3 3 3 −

When s = 4:

59 100 32 + − = 9A + 0 + 0 3 3 3 −

When s = 1:

i.e. 3 = 9A

59 25 2 + − = 0 + 0 − 3C 3 3 3


−

2 =A+B 3

i.e. -12 = -3C and i.e.

Hence,

⎧ 1 ⎫ ⎪ ⎪ 1 4 y = ℒ −1 ⎨ 3 − + 2⎬ ⎪ ( s − 4 ) ( s − 1) ( s − 1) ⎪ ⎩ ⎭

i.e.

y=

1 4x x e − e + 4x e x 3

or

and

−

2 1 = + B and 3 3

A=

1 3

C=4 B = -1

1 y = ( 4x − 1) e x + e4x 3

d 2 y dy + − 2y = 3cos 3x − 11sin 3x 8. Use Laplace transforms to solve the differential equation: dx 2 dx given y(0) = 0 and y′(0) = 6.


586

⎧ d2 y ⎫ ⎧ dy ⎫ ℒ ⎨ 2 ⎬ + ℒ ⎨ ⎬ - 2ℒ{y} = ℒ {3cos 3x − 11sin 3x} ⎩ dx ⎭ ⎩ dx ⎭

i.e.

[ s 2 ℒ{y} – s y(0) - y′(0)] + [sℒ{y} – y(0)] - 2ℒ{y} =

3s 33 − 2 s +9 s +9 2

y(0) = 0 and y′(0) = 6, hence s 2 ℒ{y} - 6 + sℒ{y} - 2ℒ{y} =

( s 2 + s - 2)ℒ{y} = 6 +

i.e.

from which,

( s + s - 2)ℒ{y} = 2

6 ( s 2 + 9 ) + 3s − 33

ℒ{y} =

s2 + 9

=

3s − 33 s2 + 9

6s 2 + 3s + 21 s2 + 9

6s 2 + 3s + 21 ( s2 + 9 )( s2 + s − 2 )

⎧⎪ ⎫⎪ 6s 2 + 3s + 21 y = ℒ −1 ⎨ 2 ⎬ ⎪⎩ ( s + 9 ) ( s − 1)( s + 2 ) ⎭⎪

and

Let

3s − 33 s2 + 9

6s 2 + 3s + 21 A B Cs + D ≡ + + 2 2 ( s + 9 ) ( s − 1)( s + 2 ) ( s + 2 ) ( s − 1) ( s + 9 ) =

A ( s − 1) ( s 2 + 9 ) + B ( s + 2 ) ( s 2 + 9 ) + ( Cs + D )( s + 2 )( s − 1)

( s + 2 )( s − 1) ( s 2 + 9 )

6s 2 + 3s + 21 = A ( s − 1) ( s 2 + 9 ) + B ( s + 2 ) ( s 2 + 9 ) + ( Cs + D )( s + 2 )( s − 1)

Hence, When s = 1:

6 + 3 + 21 = B(3)(10)

When s = -2:

24 − 6 + 21 = A(−3)(13)

i.e. 30 = 30B

and

B=1

i.e. 39 = -39A and

A = -1

Equating s3 coefficients:

0 = A+B+C

Equating constant :

21 = -9A + 18B – 2D

i.e.

C=0

Hence,

⎧⎪ −1 ⎫⎪ 1 3 y = ℒ −1 ⎨ + + 2 2 ⎬ s + 2 ) ( s − 1) ( s + 3 ) ⎪ ⎩⎪ ( ⎭

i.e.

y = −e−2x + e x + sin 3x

i.e.

21 = 9 + 18 – 2D

and D = 3

or y = ex − e−2x + sin 3x


587

d2 y dy − 2 + 2y = 3e x cos 2x given 9. Use Laplace transforms to solve the differential equation: 2 dx dx y(0) = 2 and y′(0) = 5. ⎧ d2 y ⎫ ⎧ dy ⎫ ℒ ⎨ 2 ⎬ - 2ℒ ⎨ ⎬ + 2ℒ{y} = ℒ {3e x cos 2x} ⎩ dx ⎭ ⎩ dx ⎭

⎛ ⎞ s −1 i.e. [ s 2 ℒ{y} – s y(0) - y′(0)] - 2[sℒ{y} – y(0)] + 2ℒ{y} = 3 ⎜ ⎟ ⎜ ( s − 1)2 + 22 ⎟ ⎝ ⎠ y(0) = 2 and y′(0) = 5, hence ⎛ ⎞ s −1 s 2 ℒ{y} – 2s - 5 - 2sℒ{y} + 4 + 2ℒ{y} = 3 ⎜ ⎟ ⎜ ( s − 1)2 + 22 ⎟ ⎝ ⎠ ⎛ ⎞ s −1 ( s 2 - 2s + 2)ℒ{y} = 2s + 1 + 3 ⎜ ⎟ ⎜ ( s − 1)2 + 22 ⎟ ⎝ ⎠

i.e.

=

=

( 3s − 3) + ( 2s + 1) ( s 2 − 2s + 5)

(s

2

− 2s + 5 )

3s − 3 + 2s3 − 4s 2 + 10s + s 2 − 2s + 5 ( s2 − 2s + 5)

2s3 − 3s 2 + 11s + 2 = ( s2 − 2s + 5) ℒ{y} =

from which,

⎧⎪ 2s3 − 3s 2 + 11s + 2 ⎫⎪ y = ℒ −1 ⎨ 2 ⎬ 2 ⎩⎪ ( s − 2s + 5 )( s − 2s + 2 ) ⎭⎪

and

Let

2s3 − 3s 2 + 11s + 2 As + B Cs + D ≡ 2 + 2 2 2 ( s − 2s + 5)( s − 2s + 2 ) ( s − 2s + 5) ( s − 2s + 2 ) =

Hence,

2s3 − 3s 2 + 11s + 2 ( s2 − 2s + 5)( s2 − 2s + 2 )

( As + B) ( s 2 − 2s + 2 ) + ( Cs + D ) ( s 2 − 2s + 5 )

(s

2

− 2s + 5 )( s 2 − 2s + 2 )

2s3 − 3s 2 + 11s + 2 = ( As + B ) ( s 2 − 2s + 2 ) + ( Cs + D ) ( s 2 − 2s + 5 )


2=A+C

(1)


588


-3 = -2A + B – 2C + D

(2)

Equating s coefficients:

11 = 2A – 2B + 5C – 2D

(3)

Equating constant :

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 i.e.

-3 = -2A + 1 – 2C 4 = 2A + 2C

From (3),

11 = 2A – 2 + 5C

i.e.

13 = 2A + 5C

(8) – (7) gives:

9 = 3C

i.e.

(7)

(8)

C=3

In (1), if C = 3, then A = -1

Hence,

⎧⎪ ⎫⎪ 1− s 3s y = ℒ −1 ⎨ 2 + 2 ⎬ ⎩⎪ ( s − 2s + 5 ) ( s − 2s + 2 ) ⎭⎪

⎧⎪ ⎫⎪ ⎧ ⎫⎪ s −1 3s −1 ⎪ = -ℒ −1 ⎨ +ℒ ⎬ ⎨ ⎬ 2 2 2 2 ⎪⎩ ( s − 1) + 1 ⎭⎪ ⎩⎪ ( s − 1) + 2 ⎭⎪ ⎧⎪ ⎫⎪ ⎧ ⎫ s −1 −1 ⎪ 3 ( s − 1) + 3 ⎪ = -ℒ −1 ⎨ +ℒ ⎬ ⎨ ⎬ 2 2 2 2 ⎪⎩ ( s − 1) + 1 ⎭⎪ ⎩⎪ ( s − 1) + 2 ⎭⎪ ⎧⎪ ⎫⎪ ⎧ ⎫ ⎧ ⎫⎪ s −1 3 −1 ⎪ 3 ( s − 1) ⎪ −1 ⎪ = -ℒ −1 ⎨ +ℒ + ℒ ⎬ ⎨ ⎬ ⎨ ⎬ 2 2 2 2 2 2 ⎪⎩ ( s − 1) + 2 ⎪⎭ ⎪⎩ ( s − 1) + 1 ⎭⎪ ⎪⎩ ( s − 1) + 1 ⎭⎪ i.e.

y = −e x cos 2x + 3e x cos x + 3e x sin x

or

y = 3e x ( cos x + sin x ) − e x cos 2x


589

CHAPTER 68 THE SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS EXERCISE 239 Page 654 2. Solve the following pair of simultaneous differential equations: 2

dy dx −y+x+ − 5sin t = 0 dt dt dy dx 3 + x − y + 2 − et = 0 dt dt

given that when t = 0, x = 0 and y = 0. Taking Laplace transforms of each term in each equation gives: 2[sℒ{y} – y(0)] - ℒ{y} + ℒ{x} + [sℒ{x} – x(0)] -

5 =0 s +1

3[sℒ{y} – y(0)] + ℒ{x} - ℒ{y} + 2[sℒ{x} – x(0)] -

2

1 =0 s −1

y(0) = 0 and x(0) = 0, hence

and

(2s – 1)ℒ{y} + (s + 1)ℒ{x} =

5 s +1

(1)

(3s – 1)ℒ{y} +(2s + 1)ℒ{x} =

1 s −1

(2)

2

(3s – 1) × (1) gives: (3s – 1)(2s – 1)ℒ{y} + (3s – 1)(s + 1)ℒ{x} = (3s – 1)

5 s +1

(3)

1 s −1

(4)

(2s – 1) × (2) gives: (2s – 1)(3s – 1)ℒ{y} + (2s – 1)(2s + 1)ℒ{x} = (2s – 1) (3) – (4) gives:

i.e.

i.e.

⎡( 3s 2 + 2s − 1) − ( 4s 2 − 1) ⎤ ℒ{x} = ⎣ ⎦

( −s

2

+ 2s ) ℒ{x} =

2

5 ( 3s − 1) 2s − 1 − s2 + 1 s −1

(5)

5 ( 3s − 1)( s − 1) − ( 2s − 1) ( s 2 + 1)

( s − 1) ( s 2 + 1)

⎧⎪15s 2 − 20s + 5 − 2s3 − 2s + s 2 + 1 ⎫⎪ ℒ{x} = − ⎨ ⎬ s ( s − 2 )( s − 1) ( s 2 + 1) ⎪⎩ ⎭⎪ ⎧⎪ 2s3 − 16s 2 + 22s − 6 ⎫⎪ = ⎨ ⎬ 2 ⎪⎩ s ( s − 2 )( s − 1) ( s + 1) ⎭⎪


590

⎧⎪ 2s3 − 16s 2 + 22s − 6 ⎫⎪ x = ℒ −1 ⎨ ⎬ 2 ⎪⎩ s ( s − 2 )( s − 1) ( s + 1) ⎭⎪

and

Let

2s3 − 16s 2 + 22s − 6 A B C Ds + E ≡ + + + 2 2 s ( s − 2 )( s − 1) ( s + 1) s ( s − 2 ) ( s − 1) ( s + 1) =

A ( s − 2 )( s − 1) ( s 2 + 1) + B ( s )( s − 1) ( s 2 + 1) + C ( s )( s − 2 ) ( s 2 + 1) + ( Ds + E )( s )( s − 2 )( s − 1) s ( s − 2 )( s − 1) ( s 2 + 1)

from which, 2s3 − 16s 2 + 22s − 6 = A ( s − 2 )( s − 1) ( s 2 + 1) + B ( s )( s − 1) ( s 2 + 1) + C ( s )( s − 2 ) ( s 2 + 1) + ( Ds + E )( s )( s − 2 )( s − 1) When s = 0: When s = 1: When s = 2:

-6 = A(-2)(-1)(1) 2 – 16 + 22 – 6 = C(1)(-1)(2)

i.e. A = -3 i.e. C = -1

16 – 64 + 44 – 6 = B(2)(1)(5)

i.e. B = -1


0=A+B+C+D


2 = -3A – B – 2C – 3D + E

i.e.

2 = 9 + 1 + 2 – 15 + E

i.e. D = 5

i.e. E = 5

Hence,

⎧⎪ 3 1 1 5s + 5 ⎫⎪ x = ℒ −1 ⎨− − − + 2 ⎬ ⎪⎩ s ( s − 2 ) ( s − 1) ( s + 1) ⎪⎭

i.e.

x = −3 − e−2t − e t + 5cos t + 5sin t

or

x = 5cos t + 5sin t − e 2t − et − 3

From equations (1) and (2) 5 s +1

(6)

1 s −1

(7)

(2s + 1) × (1) gives: (2s + 1)(2s – 1)ℒ{y} + (2s + 1)(s + 1)ℒ{x} = (2s + 1) (s + 1) × (2) gives:

(6) – (7):

i.e.

and

(s + 1)(3s – 1)ℒ{y} + (s + 1)(2s + 1)ℒ{x} = (s + 1)

⎡( 4s 2 − 1) − ( 3s 2 + 2s − 1) ⎤ ℒ{y} = ⎣ ⎦

( s2 − 2s ) ℒ{y} =

2

2 5 ( 2s + 1) ( s + 1) (10s + 5 )( s − 1) − ( s + 1) ( s + 1) − = s2 + 1 ( s − 1) ( s − 1) ( s 2 + 1)

10s 2 − 5s − 5 − s3 − s − s 2 − 1 ( s − 1) ( s 2 + 1)

−s3 + 9s 2 − 6s − 6 ℒ{y} = s ( s − 2 )( s − 1) ( s 2 + 1)


591

⎧⎪ −s3 + 9s 2 − 6s − 6 ⎫⎪ y = ℒ −1 ⎨ ⎬ 2 ⎪⎩ s ( s − 1)( s − 2 ) ( s + 1) ⎭⎪

and

Let

A B C Ds + E −s3 + 9s 2 − 6s − 6 ≡ + + + 2 2 s ( s − 1)( s − 2 ) ( s + 1) s ( s − 1) ( s − 2 ) ( s + 1) =

from which,

A ( s − 1)( s − 2 ) ( s 2 + 1) + B ( s )( s − 2 ) ( s 2 + 1) + C ( s )( s − 1) ( s 2 + 1) + ( Ds + E )( s )( s − 1)( s − 2 ) s ( s − 1)( s − 2 ) ( s 2 + 1)

−s3 + 9s 2 − 6s − 6 = A ( s − 1)( s − 2 ) ( s 2 + 1) + B ( s )( s − 2 ) ( s 2 + 1) + C ( s )( s − 1) ( s 2 + 1) + ( Ds + E )( s )( s − 1)( s − 2 )

When s = 0: When s = 1: When s = 2:

-6 = A(-1)(-2)(1)

i.e. A = -3

-1 + 9 - 6 – 6 = B(1)(-1)(2)

i.e. B = 2

-8 + 36 - 12 – 6 = C(2)(1)(5)

i.e. C = 1


0=A+B+C+D


-1 = -3A – 2B – C – 3D + E

i.e.

-1 = 9 - 4 - 1 + 0 + E

i.e. D = 0

i.e. E = -5

Hence,

⎧⎪ 3 2 1 5 ⎫⎪ y = ℒ −1 ⎨− + + − 2 ⎬ ⎪⎩ s ( s − 1) ( s − 2 ) ( s + 1) ⎪⎭

i.e.

y = −3 + 2e t + e 2t − 5sin t

or

y = e2t + 2et − 3 − 5sin t

3. Solve the following pair of simultaneous differential equations: d2x + 2x = y dt 2 d2 y + 2y = x dt 2 given that when t = 0, x = 4, y = 2,

dx dy = 0 and = 0. dt dt

Taking Laplace transforms of each term in each equation gives: [ s 2 ℒ{x} – s x(0) - x′(0)] + 2ℒ{x} = ℒ{y} and

[ s 2 ℒ{y} – s y(0) - y′(0)] + 2ℒ{y} = ℒ{x}

x(0) = 4 and x′(0), hence

[ s 2 ℒ{x} – 4s] + 2ℒ{x} = ℒ{y}


592

y(0) = 2 and y′(0) = 0, hence [ s 2 ℒ{y} – 2s] + 2ℒ{y} = ℒ{x} i.e.

( s 2 +2)ℒ{x} - ℒ{y} = 4s

(1)

and

- ℒ{x} + ( s 2 +2)ℒ{y} = 2s

(2)

( s 2 +2) × (2) gives: - ( s 2 +2)ℒ{x} + ( s 2 + 2) ( s 2 +2)ℒ{y} = 2s( s 2 +2)

(3)

2 2 ⎡ 2 ⎤ ⎢⎣( s + 2 ) − 1⎥⎦ ℒ{y} = 2s( s +2) + 4s

(1) + (3) gives:

(s

i.e.

4

+ 4s 2 + 3) ℒ{y} = 2s3 + 8s ℒ{y} =

and

2s3 + 8s 2s3 + 8s = ( s4 + 4s2 + 3) ( s2 + 3)( s2 + 1)

⎧⎪ 2s3 + 8s ⎫⎪ y = ℒ −1 ⎨ 2 ⎬ 2 ⎪⎩ ( s + 3)( s + 1) ⎭⎪

and

2 2 2s3 + 8s As + B Cs + D ( As + B ) ( s + 1) + ( Cs + D ) ( s + 3) ≡ + = Let ( s2 + 3)( s2 + 1) ( s2 + 3)( s2 + 1) ( s2 + 3) ( s2 + 1)

2s3 + 8s = ( As + B ) ( s 2 + 1) + ( Cs + D ) ( s 2 + 3)

from which,


2=A+C

(4)


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 :

0 = B + 3D

(7) – (5) gives:

0 = 2D

Hence,

i.e.

(7) i.e.

D=0

⎧ ⎧⎪ −s ⎫⎪ 3s s −1 ⎪ y = ℒ −1 ⎨ 2 + 2 ⎬ = ℒ ⎨− ⎪ s2 + 3 ⎩⎪ ( s + 3) ( s + 1) ⎭⎪ ⎩

(

y = − cos

(

)

3 t + 3cos t

or

and from (5), B = 0

( )) 2

+

y = 3 cos t - cos

(

(

3s s2 + 3t

( )) 2

1

⎫ ⎪ ⎬ ⎪ ⎭

)


593

If y = 3 cos t - cos

(

3t

)

dy = −3sin t + 3 sin 3 t dt

then

d2 y = −3cos t + 3cos 3 t dt 2

and

Since from one of the original equations, d2 y + 2y = x dt 2 then

−3cos t + 3cos

(

) (

3 t + 2 3cos t − cos

i.e.

x = −3cos t + 3cos

i.e.

x = 3 cos t + cos

(

(

)

(

3t

)) = x

3 t + 6 cos t − 2 cos 3t

(

3t

)

)


594

CHAPTER 69 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD 2π EXERCISE 240 Page 661 1. Determine the Fourier series for the periodic function: ⎧ − 2, when − π ≤ x ≤ 0 f(x) = ⎨ ⎩ + 2, when 0 ≤ x ≤ 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) = a 0 + ∑ ( a n cos nx + b n sin nx )

(1)

n =1

a0 =

1 π 1 f (x) dx = ∫ 2π −π 2π

{∫

π

0 −π

} 21π {[−2x]

−2 dx + ∫ 2 dx = 0

= an =

1 π 1 f (x) cos nx dx = ∫ −π π π

{∫

0 −π

+ [ 2x ] 0

π

0 −π

}

1 {[(0) − (2π)] + [(2π) − (0)]} = 0 2π

}

π

−2 cos nx dx + ∫ 2 cos nx dx 0

0 π 1 ⎪⎧ ⎡ 2 ⎤ ⎡2 ⎤ ⎪⎫ = ⎨ ⎢ − sin nx ⎥ + ⎢ sin nx ⎥ ⎬ = 0 π ⎪⎩ ⎣ n ⎦ −π ⎣ n ⎦ 0 ⎪⎭

bn =

1 π 1 f (x) sin nx dx = ∫ −π π π =

0

−π

π

bn =

}

−2sin nx dx + ∫ 2sin nx dx 0

0 π 1 ⎧⎪ ⎡ 2 ⎤ ⎡ 2 ⎤ ⎫⎪ 2 + − cos nx cos nx ⎨ ⎥⎦ ⎢⎣ n ⎥⎦ ⎬ = πn {[ cos 0 − cos n(−π)] − [ cos nπ − cos 0]} π ⎪⎩ ⎢⎣ n −π 0⎪ ⎭

bn =

When n is even, When n is odd,

{∫

2 {[1 − 1] − [1 − 1]} = 0 πn

2 2 8 [1 − −1] − [ −1 − 1]} = ( 4 ) = { πn πn πn


595

Hence,

b1 =

8 , π

b3 =

8 , 3π

8 and so on. 5π

b5 =

Substituting onto equation (1) gives: f(x) = 0 + 0 +

f(x) =

i.e.

8 8 8 sin x + sin 3x + sin 5x + ..... π 3π 5π

8⎛ 1 1 ⎞ sin x + sin 3x + sin 5x + ...... ⎟ ⎜ π⎝ 3 5 ⎠

2. For the Fourier series in problem 1, deduce a series for

When x = i.e.

π , f(x) = 2, hence 2

π π at the point where x = 4 2

2=

8 π 8 3π 8 5π sin + sin + sin + ..... π 2 3π 2 5π 2

2=

8⎛ 1 1 1 ⎞ ⎜1 − + − + .... ⎟ π⎝ 3 5 7 ⎠

and

2π 1 1 1 = 1 − + − + ..... 8 3 5 7

i.e.

π 1 1 1 = 1 − + − + .... 4 3 5 7

5. Find the term representing the third harmonic for the periodic function of period 2π given by: ⎧ 0, when − π ≤ x ≤ 0 f(x) = ⎨ ⎩ 1, when 0 ≤ x ≤ π The periodic function is shown in the diagram below.

π

1 π 1 π 1 ⎡ sin nx ⎤ =0 a n = ∫ f (x) cos nx dx = ∫ 1cos nx dx = ⎢ 0 −π π π π ⎣ n ⎥⎦ 0 bn =

1 π 1 π f (x) sin nx dx = ∫ sin nx dx ∫ π −π π 0


596

π 1 ⎧⎪ ⎡ cos nx ⎤ ⎫⎪ 1 1 − {1 − cos nπ} ⎨⎢ ⎬ = − {cos nπ − cos 0} = ⎥ π ⎪⎩ ⎣ πn πn n ⎦ 0 ⎪⎭

=

The third harmonic is when n = 3, 1 1 2 {1 − cos 3π} = {1 − −1} = 3π 3π 3π

b3 =

i.e.

∞

Since the Fourier series is given by: f(x) = a 0 + ∑ ( a n cos nx + b n sin nx ) , n =1

2 sin 3x 3π

the 3rd harmonic term is:

6. Determine the Fourier series for the periodic function of period 2π defined by: ⎧ ⎪ 0, when − π ≤ t ≤ 0 ⎪ π ⎪ f(t) = ⎨ 1, when 0 ≤ t ≤ 2 ⎪ π ⎪ ⎪⎩−1, when 2 ≤ t ≤ π The function has a period of 2π. The periodic function is shown in the diagram below.

a0 =

1 π 1 f (t) dt = ∫ −π 2π 2π

{∫

0 −π

0 dt + ∫

π/ 2 0

1dt + ∫

π π/ 2

} 21π {[ t ]

−1dt =

= an =

1 π 1 f (t) cos nt dt = ∫ −π π π

{∫

π/ 2 0

cos nt dt + ∫

π π/ 2

π/2 0

+ [−t] π / 2 π

}

⎤ ⎡ 1 ⎧ ⎡⎛ π ⎞ ⎛ π ⎞⎤ ⎫ ⎨ ⎢⎜ ⎟ − ( 0 ) ⎥ + ⎢( −π ) − ⎜ − ⎟ ⎥ ⎬ = 0 2π ⎩ ⎣⎝ 2 ⎠ ⎝ 2 ⎠⎦ ⎭ ⎦ ⎣

}

− cos nx dx

π/ 2 π 1 ⎧⎪ ⎡ sin nt ⎤ nπ ⎤ ⎫ ⎡ sin nt ⎤ ⎫⎪ 1 ⎧ ⎡ nπ ⎤ ⎡ = ⎨⎢ −⎢ − 0 ⎥ − ⎢sin nπ − sin ⎥ ⎬ ⎬= ⎨ ⎢sin ⎥ ⎥ π ⎪⎩ ⎣ n ⎦ 0 2 2 ⎦⎭ ⎣ n ⎦ π / 2 ⎪⎭ nπ ⎩ ⎣ ⎦ ⎣


597

When n is even,

an = 0

When n = 1,

a1 =

1 ⎧⎡ π ⎤ ⎡ π ⎤⎫ 1 2 ⎨ ⎢sin − 0 ⎥ − ⎢sin π − sin ⎥ ⎬ = {[1 − 0] − [ 0 − 1]} = π ⎩⎣ 2 ⎦ ⎣ 2 ⎦⎭ π π

When n = 3,

a3 =

1 ⎧ ⎡ 3π ⎤ ⎡ 3π ⎤ ⎫ 1 2 ⎨ ⎢sin − 0 ⎥ − ⎢sin 3π − sin ⎥ ⎬ = {[ −1 − 0] − [ 0 − −1]} = − 3π ⎩ ⎣ 2 2 ⎦ ⎭ 3π 3π ⎦ ⎣

When n = 5,

a5 =

1 ⎧ ⎡ 5π ⎤ ⎡ 5π ⎤ ⎫ 1 2 ⎨ ⎢sin − 0 ⎥ − ⎢sin 5π − sin ⎥ ⎬ = {[1 − 0] − [ 0 − 1]} = 5π ⎩ ⎣ 2 2 ⎦ ⎭ 5π 5π ⎦ ⎣

It follows that

a7 = −

bn =

2 , 7π

1 π 1 f (t) sin nt dt = ∫ π −π π

{∫

a9 = π/ 2 0

2 9π

and so on.

sin nt dt + ∫

π π/2

}

− sin nt dt

π/ 2 π 1 ⎧⎪ ⎡ cos nt ⎤ nπ nπ ⎤ ⎫ ⎡ cos nt ⎤ ⎫⎪ 1 ⎧ ⎡ ⎤ ⎡ +⎢ = ⎨⎢− ⎬= ⎨− ⎢ cos − cos 0 ⎥ + ⎢ cos nπ − cos ⎥ ⎬ ⎥ ⎥ n ⎦0 2 2 ⎦⎭ π ⎩⎪ ⎣ ⎣ n ⎦ π / 2 ⎭⎪ πn ⎩ ⎣ ⎦ ⎣

=

1 ⎧⎛ nπ ⎞ ⎛ nπ ⎞ ⎫ 1 ⎧ nπ ⎫ ⎨⎜1 − cos ⎟ + ⎜ cos nπ − cos ⎟ ⎬ = ⎨1 − 2 cos + cos nπ ⎬ πn ⎩⎝ 2 ⎠ ⎝ 2 ⎠ ⎭ πn ⎩ 2 ⎭

When n is odd,

bn =

1 {1 − 0 − 1} = 0 πn

When n is even,

b2 =

1 4 2 {1 − 2(−1) + 1} = = , 2π 2π π

b4 =

1 {1 − 2(1) + 1} = 0 , 4π

b6 =

1 4 2 {1 − 2(−1) + 1} = = 6π 6π 3π

Similarly,

b8 = 0 ,

b10 =

2 , and so on. 5π

∞

Substituting into f(t) = a 0 + ∑ ( a n cos nt + b n sin nt ) n =1

gives:

f(x) = 0 +

2 2 2 2 cos t − cos 3t + cos 5t − cos 7t + ..... π 3π 5π 7π +

i.e. f(x) =

2 2 2 sin 2t + sin 6t + sin10t + .... π 3π 5π

2⎛ 1 1 1 1 ⎞ cos t − cos 3t + cos 5t − ...... + sin 2t + sin 6t + sin10t + ..... ⎟ ⎜ π⎝ 3 5 3 5 ⎠


598

7. Show that the Fourier series for the periodic function of period 2π defined by: when − π ≤ θ ≤ 0 ⎧ 0, f(θ) = ⎨ ⎩ sin θ, when 0 ≤ θ ≤ π is given by:

f ( θ) =

⎞ 2 ⎛ 1 cos 2θ cos 4θ cos 6θ − − − ..... ⎟ ⎜ − π⎝ 2 (3) (3)(5) (5)(7) ⎠

The periodic function is shown in the diagram below.

a0 =

1 π 1 f (θ) dθ = ∫ −π 2π 2π

{∫

0 −π

} 21π {[− cos θ] } = 21π {⎡⎣( − cos π) − ( − cos 0)⎤⎦}

π

0

= an = =

1 π

{∫

−π

0

1 1 1 (1 − cos π ) = (1 − −1) = π 2π 2π

}

π

0

π

0 dθ + ∫ sin θ dθ =

0 cos nθ dθ + ∫ sin θ cos nθ dθ 0

1⎧ π1 ⎫ 1 ⎨ ∫ 0 ⎡⎣sin ( θ + nθ ) + sin ( θ − nθ ) ⎤⎦ ⎬ = π⎩ 2 ⎭ 2π

{∫ sin θ (1 + n ) + sin θ (1 − n ) dθ} from 6, page 398 π

0

of textbook π

θ (1 − n ) ⎤ 1 ⎡ cos θ (1 + n ) 1 ⎡⎛ cos π (1 + n ) cos π (1 − n ) ⎞ ⎛ 1 1 ⎞⎤ − cos − − = ⎢⎜ − ⎟−⎜− ⎢− ⎥ = ⎟⎥ 2π ⎣ 1+ n 1 − n ⎦ 0 2π ⎢⎣⎝ 1+ n 1− n ⎠ ⎝ 1 + n 1 − n ⎠ ⎥⎦ When n is odd, a n =

1 ⎡ 1 1 1 1 ⎤ − − + + =0 ⎢ 2π ⎣ 1 + n 1 − n 1 + n 1 − n ⎥⎦

When n = 2,

a2 =

1 ⎧ cos 3π cos(−π) 1 1 ⎫ 1 ⎧ 1 1 ⎫ 1 ⎧ 4⎫ 2 − + + ⎬= ⎨− ⎨ − 1 + − 1⎬ = ⎨− ⎬ = − 2π ⎩ 3 −1 3 −1 ⎭ 2 π ⎩ 3 3 ⎭ 2π ⎩ 3 ⎭ 3π

When n = 4,

a4 =

1 ⎧ cos 5π cos(−3π) 1 1 ⎫ 1 ⎧ 1 1 1 1 ⎫ − + + ⎬= ⎨− ⎨ − + − ⎬ −3 2π ⎩ 5 5 −3 ⎭ 2 π ⎩ 5 3 5 3 ⎭

=

When n = 6,

a6 =

1 ⎧3 − 5 + 3 − 5 ⎫ 1 ⎧ 4 ⎫ 2 ⎨ ⎬= ⎨− ⎬=− 2π ⎩ (3)(5) ⎭ 2π ⎩ (3)(5) ⎭ π(3)(5) 1 ⎧ cos 7π cos(−5π) 1 1 ⎫ 1 ⎧ 1 1 1 1 ⎫ − + + ⎬= ⎨− ⎨ − + − ⎬ −5 2π ⎩ 7 7 −5 ⎭ 2π ⎩ 7 5 7 5 ⎭


599

= bn =

1 π

{∫

1 ⎧5 − 7 + 5 − 7 ⎫ 1 ⎧ 4 ⎫ 2 ⎨ ⎬= ⎨− ⎬=− 2π ⎩ (5)(7) ⎭ 2π ⎩ (5)(7) ⎭ π(5)(7)

}

π

0 −π

0sin nθ dθ + ∫ sin θ sin nθ dθ 0

π

1⎧ π 1 1 ⎡ sin θ(1 + n) sin θ(1 − n) ⎤ ⎫ = ⎨ ∫ − ⎡⎣cos ( θ + nθ ) − cos ( θ − nθ ) ⎤⎦ ⎬ = − =0 − 0 2 2π ⎣⎢ 1 + n 1 − n ⎦⎥ 0 π⎩ ⎭

from 9, page 398, of textbook

∞

Substituting into f(θ) = a 0 + ∑ ( a n cos nθ + b n sin nθ ) n =1

gives:

f(θ) =

1 2 2 2 − cos 2θ − cos 4θ − cos 6θ − ..... + 0 (3)(5)π (5)(7)π π 3π

i.e.

f(θ) =

⎞ 2 ⎛ 1 cos 2θ cos 4θ cos 6θ − − − ...... ⎟ ⎜ − π⎝2 (3) (3)(5) (5)(7) ⎠


600

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:

⎧ 1 − t, when − π ≤ t ≤ 0 f(t) = ⎨ ⎩ 1 + t, when 0 ≤ t ≤ π Draw a graph of the function within and outside of the given range. The periodic function is shown in the diagram below.

1 π 1 a0 = f (t) dt = ∫ 2π −π 2π =

an = =

{∫

0 −π

{

0

}

0 π ⎡ t 2 ⎤ ⎫⎪ 1 ⎧⎪ ⎡ t 2 ⎤ (1 + t) dt = + t+ ⎨ t− ⎬ 2π ⎪ ⎢⎣ 2 ⎥⎦ −π ⎢⎣ 2 ⎥⎦ 0 ⎪ ⎩ ⎭

⎤ ⎫⎪ 1 ⎧⎪ ⎛ ⎛ 1 ⎧⎪ ⎡ π 2 ⎞ ⎤ ⎡⎛ π2 ⎞ π2 ⎞ ⎫⎪ 2π 2π2 π + = 1+ ⎨ ⎢( 0 ) − ⎜ −π − ⎟ ⎥ + ⎢⎜ π + ⎟ − ( 0 ) ⎥ ⎬ = ⎨2 ⎜ π + ⎟ ⎬ = 2π ⎩⎪ ⎣ 2 ⎠ ⎦ ⎣⎝ 2 ⎠ 2 ⎠ ⎭⎪ 2π 4π 2 ⎝ ⎦ ⎭⎪ 2π ⎩⎪ ⎝

1 π 1 f (t) cos nt dt = ∫ π −π π 1 π

(1 − t) dt + ∫

π

{∫

0 −π

}

π

(1 − t) cos nt dt + ∫ (1 + t) cos nt dt 0

}

π

∫ −π ( cos nt − t cos nt ) dt + ∫ 0 ( cos nt + t cos nt ) dt 0

π

0

1 ⎡ sin nt t sin nt cos nt ⎤ ⎡ sin nt t sin nt cos nt ⎤ = ⎢ − − 2 ⎥ +⎢ + + 2 ⎥ by integration by parts n n ⎦ −π ⎣ n n n ⎦0 π⎣ n =

1 ⎧ ⎡⎛ cos 0 ⎞ ⎛ cos(− nπ) ⎞ ⎤ ⎡⎛ cos nπ ⎞ ⎛ cos 0 ⎞ ⎤ ⎫ ⎨ ⎢⎜ 0 − 0 − 2 ⎟ − ⎜ 0 − 0 − ⎟ ⎥ + ⎢⎜ 0 + 0 + ⎟ − ⎜ 0 + 0 + 2 ⎟⎥ ⎬ 2 2 n ⎠ ⎝ n n ⎠ ⎝ n ⎠⎦ ⎭ π ⎩ ⎣⎝ ⎠ ⎦ ⎣⎝

=

1 ⎧ 1 cos(− nπ) cos nπ 1 ⎫ 1 + − 2 ⎬ = 2 ( 2 cos nπ − 2 ) ⎨− 2 + 2 2 π⎩ n n n n ⎭ πn =

When n is even,

since cos(-nπ) = cos nπ

2 ( cos nπ − 1) πn 2

an = 0


601

2 4 −1 − 1) = − 2 ( π(1) π

When n = 1,

a1 ==

When n = 3,

a3 =

2 4 −1 − 1) = − 2 ( π(3) π(3) 2

When n = 5,

a5 =

2 4 −1 − 1) = − 2 ( π(5) π(5) 2

bn =


{∫

{∫

0 −π

and so on.

}

π

(1 − t) sin nt dt + ∫ (1 + t) sin nt dt 0

}

π

=

1 π

=

1 ⎡ cos nt t cos nt sin nt ⎤ ⎡ cos nt t cos nt sin nt ⎤ − + − + − − + 2 ⎥ by integration by parts 2 n n n ⎥⎦ −π ⎢⎣ n n n ⎦0 π ⎢⎣

0 −π

( sin nt − t sin nt ) dt + ∫ 0 ( sin nt + t sin nt ) dt 0

π

⎧ ⎡⎛ cos 0 ⎫ ⎞ ⎛ cos(− nπ) π cos(− nπ) ⎞⎤ + 0 − 0⎟ − ⎜ − − − 0 ⎟⎥ ⎪ ⎢⎜ − ⎪ n n n ⎠ ⎝ ⎠⎦ 1 ⎪⎣⎝ ⎪ = ⎨ ⎬ π⎪ ⎡⎛ cos nπ π cos nπ ⎞ ⎛ cos 0 ⎞⎤ ⎪ + ⎢⎜ − − + 0⎟ − ⎜ − + 0 + 0 ⎟⎥ ⎪ n n n ⎠ ⎝ ⎠ ⎦ ⎪⎭ ⎣⎝ ⎩ =

1 ⎧ 1 cos(−nπ) π cos(−nπ) cos nπ π cos nπ 1 ⎫ + − − + ⎬ =0 ⎨− + n n n n n⎭ π⎩ n

since cos nπ = cos(-nπ)

∞

Substituting into f(t) = a 0 + ∑ ( a n cos nt + b n sin nt ) n =1

gives:

f(t) =

π 4 4 4 cos 3t − cos 5t − ..... + 0 + 1 − cos t − 2 π π(3) π(5) 2 2

i.e.

f(t) =

π 4⎛ cos 3t cos 5t ⎞ + 1 − ⎜ cos t + + + ...... ⎟ 2 2 2 3 5 π⎝ ⎠

4. Determine the Fourier series up to and including the third harmonic for the function defined by:

when 0 ≤ x ≤ π ⎧ x, f(x) = ⎨ ⎩ 2π − x, when π ≤ x ≤ 2π Sketch a graph of the function within and outside of the given range, assuming the period is 2π. The periodic function is shown in the diagram below.


602

1 π 1 a0 = f (x) dx = ∫ 2π −π 2π

0

x dx + ∫

2π π

}

1 ⎧⎪ ⎡ x 2 ⎤ ⎡ x2 ⎤ (2π − x) dx = 2 x + π − ⎨ 2π ⎪ ⎢⎣ 2 ⎥⎦ 0 ⎢⎣ 2 ⎥⎦ π ⎩ π

2π

{ }

1 2π 1 f (x) cos nx dx = ∫ 0 π π

{∫

π 0

}

2π

x cos nx dx + ∫ (2π − x) cos nx dx π

1 ⎪⎧ ⎡ x sin nx cos nx ⎤ ⎡ 2π sin nx x sin nx cos nx ⎤ + + − − = ⎨⎢ n 2 ⎥⎦ 0 ⎢⎣ n n n 2 ⎥⎦ π π ⎪⎩ ⎣ n π

2π

⎪⎫ ⎬ by integration by parts ⎪⎭

=

1 ⎧ ⎡⎛ cos nπ ⎞ ⎛ cos 0 ⎞ ⎤ ⎡⎛ cos 2πn ⎞ ⎛ cos nπ ⎞ ⎤ ⎫ − 0 + + 0 − 0 − − 0 − 0 − ⎨ ⎢⎜ 0 + ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎬ n2 ⎠ ⎝ n 2 ⎠ ⎥⎦ ⎢⎣⎝ n2 ⎠ ⎝ n 2 ⎠ ⎥⎦ ⎭ π ⎩ ⎣⎝

=

1 2 cos nπ − 1 − cos 2πn + cos nπ} = 2 ( cos nπ − 1) 2 { πn πn

When n is even,

an = 0

When n = 1,

a1 =

2 4 −1 − 1) = − 2 ( π(1) π

When n = 3,

a3 =

2 4 −1 − 1) = − 2 ( π(3) π(3) 2

When n = 5,

a5 =

2 4 −1 − 1) = − 2 ( π(5) π(5) 2

bn =

1 2π 1 f (x) sin nx dx = ∫ π 0 π

{∫ π

=

⎫⎪ ⎬ ⎪⎭

⎤ ⎡⎛ 2 4π2 ⎞ ⎛ 2 π2 ⎞ ⎤ ⎫⎪ 1 1 ⎧⎪ ⎡⎛ π2 ⎞ π π2 ) = ( ⎨ ⎢⎜ ⎟ − ( 0 ) ⎥ + ⎢ ⎜ 4π − ⎟ − ⎜ 2π − ⎟ ⎥ ⎬ = 2π ⎩⎪ ⎣⎝ 2 ⎠ 2 ⎠ ⎝ 2 ⎠ ⎦ ⎭⎪ 2π 2 ⎦ ⎣⎝

=

an =

{∫

π

0 −π

and so on. π

}

x sin nx dx + ∫ (2π − x) sin nx dx 0

2π

1 ⎡ x cos nx sin nx ⎤ ⎡ 2π cos nx x cos nx sin nx ⎤ − + + − + − n n 2 ⎥⎦ 0 ⎢⎣ n n n 2 ⎥⎦ π π ⎢⎣

by integration by parts

⎧ ⎡⎛ 2π cos 2nπ 2π cos 2nπ ⎤⎫ ⎞ + + 0⎟ ⎪ ⎢⎜ − ⎥⎪ n n 1 ⎪ ⎡⎛ π cos nπ ⎤ ⎢⎝ ⎠ ⎞ ⎥ ⎪⎬ = ⎨ ⎢⎜ − + 0 ⎟ − ( 0 + 0 )⎥ + π ⎪ ⎣⎝ n ⎠ ⎛ 2π cos nπ π cos nπ ⎞⎥ ⎦ ⎢ −⎜− + − 0 ⎟⎥ ⎪ ⎢ n n ⎝ ⎠ ⎦ ⎭⎪ ⎣ ⎩⎪


603

=

1 {−π cos nπ + 2π cos nπ − π cos nπ} = 0 nπ ∞

Substituting into f(x) = a 0 + ∑ ( a n cos nx + b n sin nx ) n =1

gives:

f(x) =

π 4 4 4 cos 3x − cos 5x − ..... + 0 − cos x − 2 π(3) π(5) 2 2 π

i.e.

f(x) =

π 4⎛ cos 3x cos 5x ⎞ − ⎜ cos x + + + ...... ⎟ 2 2 2 π⎝ 3 5 ⎠

5. Expand the function f ( θ ) = θ2 in a Fourier series in the range - π < θ < π

Sketch the function within and outside of the given range. The periodic function is shown in the diagram below.

1 π 1 a0 = f (x) dx = ∫ 2π −π 2π

an =

{

}

π 1 ⎧⎪ ⎡ θ3 ⎤ ⎫⎪ 1 3 2 π3 π 2 3 θ θ = = π − −π = = d { } ⎨ ⎬ ∫ −π 2π ⎪ ⎢⎣ 3 ⎥⎦ −π ⎪ 6π 6π 3 ⎩ ⎭ π

2

1 π 1 f (θ) cos nθ dθ = ∫ π −π π

{∫

π −π

}

θ2 cos nθ dθ π

1 ⎡ θ2 sin nθ 2θ cos nθ 2sin nθ ⎤ + − = ⎢ by integration by parts π⎣ n n2 n 3 ⎥⎦ −π =

1 ⎧ ⎡⎛ 2π cos nπ 2π cos(− nπ) ⎞ ⎛ ⎞ ⎤ ⎫ 4π cos nπ 4 − 0 − 0 − − 0 = 2 cos nπ ⎨ ⎢⎜ 0 + ⎟ ⎜ ⎟⎥ ⎬ = π ⎩ ⎣⎝ πn 2 n2 n2 n ⎠ ⎝ ⎠⎦ ⎭

since cos nπ = cos(-nπ) Hence, bn =

a1 =

4 4 −1) = − 2 , 2 ( (1) 1

1 π 1 f (θ) sin nθ dθ = ∫ −π π π

{∫

π −π

a2 =

4 4 1 = 2, 2 ( ) (2) 2

a3 =

4 4 −1) = − 2 , 2 ( (3) 3

a4 =

4 , and so on. 42

}

θ2 sin nθ dθ π

1 ⎡ θ2 cos nθ 2θ sin nθ 2 cos nθ ⎤ + + = ⎢− by integration by parts π⎣ n n2 n 3 ⎥⎦ −π © 2006 John Bird. All rights reserved. Published by Elsevier.

604

1 ⎧⎪ ⎡⎛ π2 cos nπ 2 cos nπ ⎞ ⎛ π2 cos(− nπ) 2 cos(− nπ) ⎞ ⎤ ⎫⎪ +0+ − − + + 0 = ⎨ ⎢⎜ − ⎟ ⎜ ⎟⎥ ⎬ = 0 n n3 ⎠ ⎝ n n3 π ⎩⎪ ⎣⎝ ⎠ ⎦ ⎭⎪ ∞

Substituting into f(θ) = a 0 + ∑ ( a n cos nθ + b n sin nθ ) n =1

gives:

f(θ) =

π2 4 4 4 4 − 2 cos θ + 2 cos 2θ − 2 cos 3θ + 2 cos 4θ − ..... + 0 2 3 4 3 1

i.e.

f(θ) =

π2 1 1 ⎛ ⎞ − 4 ⎜ cos θ − 2 cos 2θ + 2 cos 3θ − ...... ⎟ 3 2 3 ⎝ ⎠

6. For the Fourier series in problem 5, let θ = π and deduce a series for

∞

1

∑n n =1

2

When θ = π in Problem 5 above, f(θ) = π2 Thus,

π =

π2 1 1 1 ⎛ ⎞ − 4 ⎜ cos π − 2 cos 2π + 2 cos 3π − 2 cos 4π + ...... ⎟ 3 2 3 4 ⎝ ⎠

i.e.

π2 =

π2 4 4 4 4 + + + + + ..... 3 12 22 32 42

2

π2 −

i.e.

π2 ⎛1 1 1 1 ⎞ = 4 ⎜ 2 + 2 + 2 + 2 + ..... ⎟ 3 ⎝1 2 3 4 ⎠

2π 2 ⎛1 1 1 1 ⎞ = 4 ⎜ 2 + 2 + 2 + 2 + ..... ⎟ 3 ⎝1 2 3 4 ⎠

i.e.

and

2π2 1 1 1 1 = + + + + .... 3(4) 12 22 32 42

i.e.

1 1 1 1 π2 .... + + + + = 12 22 32 42 6

i.e.

1 π2 = ∑ 2 6 n =1 n ∞

⎧ 2x ⎪⎪ 1 + π , when − π ≤ x ≤ 0 8. Sketch the waveform defined by: f(x) = ⎨ ⎪ 1 − 2x , when 0 ≤ x ≤ π π ⎩⎪ Determine the Fourier series in this range. The periodic function is shown in the diagram below. © 2006 John Bird. All rights reserved. Published by Elsevier.

605

0 π π ⎡ 1 π 1 ⎧ 0 ⎛ 2x ⎞ 2x x2 ⎤ x 2 ⎤ ⎫⎪ ⎫ 1 ⎧⎪ ⎡ a0 = f (x) dx = ⎨ ⎜1 + ⎨⎢ x + ⎥ + ⎢ x − ⎥ ⎬ ⎟ dx + ∫ 0 (1 − ) dx ⎬ = 2π ∫ −π 2π ⎩ ∫ −π ⎝ π ⎠ π π ⎦ −π ⎣ π ⎦0⎪ ⎭ 2π ⎩⎪ ⎣ ⎭

=

an =

⎤ ⎪⎫ 1 ⎛ 1 ⎧⎪ ⎡ π 2 ⎞ ⎤ ⎡⎛ π2 ⎞ 0 − −π + + π − ( ) {( π − π ) + ( π − π )} = 0 ⎨⎢ ⎜ ⎟ ⎥ ⎢⎜ ⎟ − ( 0 )⎥ ⎬ = 2π ⎪⎩ ⎣ π ⎠ ⎦ ⎣⎝ π⎠ ⎝ ⎦ ⎪⎭ 2π

π 1 π 1 ⎧ 0 ⎛ 2x ⎞ 2x ⎫ f (x) cos nx dx = ⎨ ∫ ⎜1 + cos nx dx + ∫ (1 − ) cos nx dx ⎬ ⎟ ∫ 0 −π −π π π⎩ ⎝ π ⎠ π ⎭ π

0

1 ⎡ sin nx 2 ⎛ x sin nx cos nx ⎞ ⎤ ⎡ sin nx 2 ⎛ x sin nx cos nx ⎞ ⎤ = ⎢ by integration by + ⎜ + − ⎜ + ⎟⎥ + ⎢ ⎟ 2 n ⎠ ⎦ −π ⎣ n n 2 ⎠ ⎥⎦ 0 π⎣ n π⎝ n π⎝ n parts =

1 ⎧⎪ ⎡⎛ 2 ⎛ 1 ⎞⎞ ⎛ 2 ⎛ cos(− nπ) ⎞ ⎞ ⎤ ⎡⎛ 2 ⎛ cos nπ ⎞ ⎞ ⎛ 2 ⎛ 1 ⎞ ⎞ ⎤ ⎫⎪ 0 0 0 0 + − − − − − ⎨ ⎢⎜ 0 + 0 + ⎜ 2 ⎟ ⎟ − ⎜ 0 + 0 + ⎜ ⎟ ⎟ ⎥ ⎢⎜ ⎜ ⎟ ⎜ ⎟ ⎥⎬ n2 π ⎩⎪ ⎣⎝ π ⎝ n ⎠⎠ ⎝ π⎝ π ⎝ n 2 ⎠ ⎟⎠ ⎜⎝ π ⎝ n 2 ⎠ ⎟⎠ ⎦ ⎭⎪ ⎠ ⎠ ⎦ ⎣⎝

=

1 4 2 − 2 cos(− nπ) − 2 cos nπ + 2} = 2 2 (1 − cos nπ ) 2 { πn πn 2

an = 0

When n is even, Hence, bn =

a1 =

since cos nπ = cos(-nπ)

4 8 1 − −1) = 2 , 2 ( π (1) π 2

a3 =

4 8 1 − −1) = 2 2 , 2 ( π (3) π (3) 2

a5 =

8 π (5) 2 2

and so on.

π⎛ 1 π 1 ⎧ 0 ⎛ 2x ⎞ 2x ⎞ ⎫ f (x) sin nx dx = ⎨ ∫ ⎜1 + ⎟ sin nx dx + ∫ 0 ⎜1 − ⎟ sin nx dx ⎬ ∫ −π −π π π⎩ ⎝ π ⎠ π ⎠ ⎝ ⎭

0 π 1 ⎧⎪ ⎡ cos nx 2 ⎛ x cos nx sin nx ⎞ ⎤ ⎡ cos nx 2 ⎛ x cos nx sin nx ⎞ ⎤ ⎫⎪ = ⎨⎢− + ⎜− + + − − ⎜− + ⎟ ⎟ ⎬ by n n n 2 ⎠ ⎥⎦ −π ⎢⎣ n n n 2 ⎠ ⎥⎦ 0 ⎭⎪ π ⎩⎪ ⎣ π⎝ π⎝

integration by parts ⎧ ⎡⎛ cos nπ 2π cos nπ ⎤⎫ ⎞ + 0⎟ ⎪ ⎢⎜ − n + ⎥⎪ n 1 ⎪ ⎡⎛ 1 ⎠ ⎞ ⎛ cos(− nπ) 2π cos(− nπ) ⎞ ⎤ ⎢⎝ ⎥ ⎪⎬ = ⎨ ⎢⎜ − − 0 + 0 ⎟ − ⎜ − + + 0 ⎟⎥ + π ⎪ ⎣⎝ n n n ⎠ ⎝ ⎠⎦ ⎢ ⎛ 1 ⎞⎥ − ⎜ − + 0 − 0 ⎟⎥ ⎪ ⎢ ⎪⎩ ⎝ n ⎠ ⎦ ⎪⎭ ⎣

=

1 ⎧ 1 cos(− nπ) 2π cos(− nπ) cos nπ 2π cos nπ 1 ⎫ − − + + ⎬=0 ⎨− + n n n n n⎭ π⎩ n


606

∞

∞

n =1

n =1

Substituting into f(x) = a 0 + ∑ ( a n cos nx + b n sin nx ) = ∑ a n cos nx

gives:

f(x) =

i.e.

f(x) =

8 8 8 8 cos x + 2 2 cos 3x + 2 2 cos 5x + 2 2 cos 7x + ..... 2 π π (3) π (5) π (7) 8 ⎛ 1 1 1 ⎞ cos x + 2 cos 3x + 2 cos 5x + 2 cos 7x + ...... ⎟ 2 ⎜ π ⎝ 3 5 7 ⎠

π2 9. For the Fourier series of Problem 8, deduce a series for 8 When f(x) = 1 in the series of problem 8 above, x = 0, hence,

i.e.

1=

8 ⎛ 1 1 1 ⎞ cos 0 + 2 cos 0 + 2 cos 0 + 2 cos 0 + ...... ⎟ 2 ⎜ 3 5 7 π ⎝ ⎠

π2 1 1 1 = 1 + 2 + 2 + 2 + ..... 8 3 5 7


607

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: ⎧ t + π, when − π ≤ t ≤ 0 f(t) = ⎨ ⎩ t − π, when 0 ≤ t ≤ 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 f (t) = ∑ b n sin nt n =1

bn =


{∫

0 −π

}

π

(t + π) sin nt dt + ∫ (t − π) sin ntdt 0

π

1 ⎡ t cos nt sin nt π cos nt ⎤ ⎡ t cos nt sin nt π cos nt ⎤ = ⎢− + 2 − + ⎢− + 2 + by integration by parts ⎥ π⎣ n n n ⎦ −π ⎣ n n n ⎥⎦ 0 0

⎧ ⎡⎛ π cos nπ π cos nπ ⎞ ⎤⎫ +0+ ⎪ ⎟ ⎢⎜ − ⎥⎪ n n π ⎞ ⎛ −π cos(− nπ) π cos(− nπ) ⎞ ⎤ ⎢⎝ 1 ⎪ ⎡⎛ ⎠ ⎥ ⎪⎬ = ⎨ ⎢⎜ 0 + 0 − ⎟ − ⎜ − +0− + ⎟⎥ π ⎪ ⎣⎝ n⎠ ⎝ n n π ⎞⎥ ⎠⎦ ⎢ ⎛ − ⎜ 0 + 0 + ⎟⎥ ⎪ ⎢ n ⎠ ⎦ ⎭⎪ ⎝ ⎣ ⎩⎪

=

1 ⎧ π π cos(− nπ) π cos(− nπ) π cos nπ π cos nπ π ⎫ 1 ⎛ 2π ⎞ 2 + − + − ⎬ = ⎜− ⎟ = − ⎨− − π⎩ n n n n n n⎭ π⎝ n ⎠ n

Hence,

b1 = −

2 , 1

b2 = −

2 , 2

b3 = −

2 , 3

b4 = −

2 , and so on. 4

i.e.

2 2 2 2 f(t) = − sin t − sin 2t − sin 3t − sin 4t − ...... 1 2 3 4

i.e.

1 1 1 ⎛ ⎞ f (t) = −2 ⎜ sin t + sin 2t + sin 3t + sin 4t + ..... ⎟ 2 3 4 ⎝ ⎠


608

⎧ 1 − x, when − π ≤ x ≤ 0 3. Determine the Fourier series defined by: f(x) = ⎨ ⎩ 1 + x, when 0 ≤ x ≤ 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 f (x) = a 0 + ∑ a n cos nx n =1

a0 =

1 π 1 π f (x) dx = ∫ f (x) dx due to symmetry ∫ π 0 2π −π π

⎤ 1 π 1⎡ x2 ⎤ 1 ⎡⎛ π2 ⎞ π = ∫ (1 + x) dx = ⎢ x + ⎥ = ⎢⎜ π + ⎟ − ( 0 ) ⎥ = 1 + π 0 π⎣ 2 ⎦ 0 π ⎣⎝ 2 ⎠ 2 ⎦ an = =

1 π 1 f (x) cos nx dx = ∫ π −π π 1 π

{∫

{∫

0 −π

}

π

(1 − x) cos nx dx + ∫ (1 + x) cos nx dx 0

}

π

( cos nx − x cos nx ) dx + ∫ 0 ( cos nx + x cos nx ) dx −π 0

π

0

1 ⎡ sin nx x sin nx cos nx ⎤ ⎡ sin nx x sin nx cos nx ⎤ by integration by parts = ⎢ − − +⎢ + + 2 ⎥ π⎣ n n n ⎦ −π ⎣ n n n 2 ⎥⎦ 0 =

1 ⎧ ⎡⎛ 1 ⎞ ⎛ cos(−nπ) ⎞ ⎤ ⎡⎛ cos nπ ⎞ ⎛ 1 ⎞⎤ ⎫ ⎨ ⎢⎜ 0 − 0 − 2 ⎟ − ⎜ 0 − 0 − ⎟ ⎥ + ⎢⎜ 0 + 0 + ⎟ − ⎜ 0 + 0 + 2 ⎟⎥ ⎬ 2 2 π ⎩ ⎣⎝ n ⎠ ⎝ n n ⎠ ⎝ n ⎠⎦ ⎭ ⎠ ⎦ ⎣⎝

=

1 ⎧ 1 cos(− nπ) cos nπ 1 ⎫ 2 + − 2 ⎬ = 2 ( cos nπ − 1) ⎨− 2 + 2 2 π⎩ n n n n ⎭ πn

When n is even,

since cos(-nπ) = cos nπ

an = 0

When n = 1,

a1 =

2 4 −1 − 1) = − 2 ( π(1) π

When n = 3,

a3 =

2 4 −1 − 1) = − 2 ( π(3) π(3) 2

When n = 5,

a5 =

2 4 −1 − 1) = − 2 ( π(5) π(5) 2

and so on.


609

∞

f (x) = a 0 + ∑ a n cos nx = n =1

i.e.

f(x) =

π 4 4 4 cos 3x − cos 5x − ..... + 1 − cos x − 2 π π(3) π(5) 2 2

π 4⎛ 1 1 ⎞ + 1 − ⎜ cos x + 2 cos 3x + 2 cos 5x + ...... ⎟ π⎝ 2 3 5 ⎠

4. In the Fourier series of Problem 3, let x = 0 and deduce a series for

π2 8

When x = 0 in the series of Problem 3 above, f(x) = 1, hence,

1=

π 4⎛ cos 0 cos 0 ⎞ + 1 − ⎜ cos 0 + 2 + 2 + ...... ⎟ 2 π⎝ 3 5 ⎠

i.e.

1=

π 4⎛ 1 1 1 ⎞ + 1 − ⎜ 1 + 2 + 2 + 2 + ..... ⎟ 2 π⎝ 3 5 7 ⎠

i.e.

and

−

π 4⎛ 1 1 1 ⎞ = − ⎜1 + 2 + 2 + 2 + ..... ⎟ 2 π⎝ 3 5 7 ⎠

π2 1 1 1 = 1 + 2 + 2 + 2 + .... 8 3 5 7


610


π ⎧ ⎪⎪ x, when 0 ≤ x ≤ 2 1. Determine the half-range series for the function defined by: f(x) = ⎨ ⎪ 0, when π ≤ x ≤ π ⎪⎩ 2 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

f (x) = ∑ b n sin nx n =1

2 π 2 b n = ∫ f (x) sin nx dx = 0 π π

{∫

π/ 2 0

}

π/2

2 ⎡ x cos nx sin nx ⎤ + x sin nx dx = ⎢ − π⎣ n n 2 ⎥⎦ 0

by integration by parts

⎡⎛ π nπ nπ ⎞ ⎤ cos sin ⎟ 2 ⎢⎜ 2 2 + 2 − ( 0)⎥ = ⎢⎜ − ⎥ ⎟ 2 n n ⎟ π ⎢⎜ ⎥ ⎢⎣⎝ ⎥⎦ ⎠

Hence,

⎡⎛ π π π ⎞⎤ cos sin ⎟ ⎥ 2 ⎢⎜ 2 2+ 2 = 2 ⎛0+ 1 ⎞ = 2 , b1 = ⎢⎜ − ⎟⎥ ⎜ ⎟ 2 π ⎢⎜ 1 1 ⎟ ⎥ π ⎝ 12 ⎠ π ⎠ ⎦⎥ ⎣⎢⎝ ⎡⎛ π ⎞⎤ cos π ⎢ ⎜ 2 sin π ⎟ ⎥ 2 ⎛ π ⎞ 2⎛π⎞ b 2 = ⎢⎜ − 2 + 2 ⎟⎥ = ⎜ + 0 ⎟ = ⎜ ⎟ , 2 2 ⎟⎥ π ⎝ 4 π ⎢⎜ ⎠ π⎝ 4⎠ ⎠ ⎦⎥ ⎣⎢⎝ ⎡⎛ π 3π 3π ⎞ ⎤ cos sin ⎢ ⎜ 2 2 + 2 ⎟⎥ = 2 ⎛ 0 − 1 ⎞ = − 2 , b 3 = ⎢⎜ − 2 ⎟⎥ ⎜ ⎟ 2 π ⎢⎜ π(3) 2 3 3 ⎟⎥ π ⎝ 32 ⎠ ⎢⎣⎝ ⎠ ⎥⎦ ⎡⎛ π ⎞⎤ cos 2π 2 ⎢⎜ 2 sin 2π ⎟ ⎥ 2 ⎛ π 2⎛π⎞ ⎞ b 4 = ⎢⎜ − + ⎟ ⎥ = ⎜ − + 0 ⎟ = − ⎜ ⎟ , and so on 2 4 4 ⎟⎥ π ⎝ 8 π ⎢⎜ π⎝8⎠ ⎠ ⎢⎣⎝ ⎠ ⎥⎦ © 2006 John Bird. All rights reserved. Published by Elsevier.

611

∞

Hence,

f (x) = ∑ b n sin nx = n =1

f (x) =

i.e.

2 2⎛π⎞ 2⎛ 1 ⎞ 2⎛π⎞ sin x + ⎜ ⎟ sin 2x − ⎜ 2 ⎟ sin 3x − ⎜ ⎟ sin 4x + ... π π⎝ 4⎠ π⎝3 ⎠ π⎝ 8⎠

2⎛ π 1 π ⎞ ⎜ sin x + sin 2x − sin 3x − sin 4x + ..... ⎟ π⎝ 4 9 8 ⎠

2. Obtain (a) the half-range cosine series and (b) the half-range sine series for the function:

⎧ ⎪⎪ 0, when f(t) = ⎨ ⎪ 1, when ⎪⎩

0≤t ≤

π 2

π ≤t≤π 2

(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

f (t) = a 0 + ∑ a n cos nt n =1

a0 =

1 π 1 π 1⎡ π⎤ 1 1dt = [ t ] π / 2 = ⎢ π − ⎥ = ∫ π π/ 2 π π⎣ 2⎦ 2

an =

2 2 ⎡ sin nt ⎤ 1cos nt dt = ⎢ ∫ π / 2 π π ⎣ n ⎥⎦ π / 2

π

π

nπ ⎞ nπ ⎛ sin 2sin ⎜ ⎟ 2 2 =− 2 = ⎜0− ⎟ n ⎟ π⎜ πn ⎝ ⎠

When n is even, a n = 0

and

a1 = −

π 2 =−2 , π π

2sin

∞

a3 = −

Thus, f (t) = a 0 + ∑ a n cos nt = n =1

i.e.

f (t) =

3π 5π 2sin 2 1 ⎛ ⎞ 2 = 2 = − 2 ⎛ 1 ⎞ , and so on. ⎜ ⎟ , a5 = − ⎜ ⎟ π⎝3⎠ π⎝5⎠ 3π 5π

2sin

1 2 2⎛1⎞ 2⎛1⎞ − cos t + ⎜ ⎟ cos 3t − ⎜ ⎟ cos 5t + .... π⎝3⎠ π⎝5⎠ 2 π

1 2⎛ 1 1 ⎞ − ⎜ cos t − cos 3t + cos 5t − ...... ⎟ 2 π⎝ 3 5 ⎠

(b) The periodic function is shown in the diagram below. Since a half-range sine series is required, © 2006 John Bird. All rights reserved. Published by Elsevier.

612

∞

f (x) = ∑ b n sin nx

the function is symmetrical about the origin and

n =1

π

bn =

2 π 2 ⎡ cos nt ⎤ 2 ⎡ nπ ⎤ = − ⎢cos nπ − cos ⎥ 1sin nt dt = ⎢ − ∫ ⎥ π / 2 n ⎦ π/2 nπ ⎣ 2⎦ π π⎣

Hence,

2⎛ π⎞ 2 2 b1 = − ⎜ cos π − cos ⎟ = − ( −1 − 0 ) = , π⎝ π π 2⎠ b2 = −

2 2 2 ( cos 2π − cos π ) = − (1 − −1) = − , 2π 2π π

b3 = −

2 ⎛ 3π ⎞ 2 2 , ⎜ cos 3π − cos ⎟ = − ( −1 − 0 ) = 3π ⎝ 2 ⎠ 3π 3π

b4 = −

2 2 ( cos 4π − cos 2π ) = − ( 0 − 0 ) = 0 , 4π 4π

b5 = −

2 ⎛ 5π ⎞ 2 2 , ⎜ cos 5π − cos ⎟ = − ( −1 − 0 ) = 5π ⎝ 2 ⎠ 5π 5π

b6 = −

2 2 2 ( cos 6π − cos 3π ) = − (1 − −1) = − , and so on. 6π 6π 3π ∞

Thus,

f (t) = ∑ b n sin nt = n =1

i.e.

f (t) =

2 2 2 2 2 sin t − sin 2t + sin 3t + 0 + sin 5t − sin 6t + .... π π 3π 5π 3π

2⎛ 1 1 1 ⎞ ⎜ 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:

⎧ when ⎪⎪ x, f(x) = ⎨ ⎪ ( π − x ) , when ⎪⎩

0 ≤x ≤

π 2

π ≤x≤π 2

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

f (x) = a 0 + ∑ a n cos nx n =1


613

1 a0 = π

{∫

π/ 2 0

}

π/ 2 π ⎡ 1 ⎧⎪ ⎡ x 2 ⎤ x 2 ⎤ ⎫⎪ x dx + ∫ (π − x) dx = ⎨ ⎢ ⎥ + ⎢ πx − ⎥ ⎬ π/2 π ⎪⎣ 2 ⎦ 0 2 ⎦ π/ 2 ⎪ ⎣ ⎩ ⎭ π

1 ⎧⎪ π2 ⎛ 2 π2 ⎞ ⎛ π2 π2 ⎞ ⎫⎪ = ⎨ + ⎜ π − ⎟ − ⎜ − ⎟⎬ π ⎪⎩ 8 ⎝ 2 ⎠ ⎝ 2 8 ⎠ ⎭⎪ = an =

2 π

{∫

π/ 2 0

x cos nx dx + ∫

π π/2

1 ⎛ π 2 π 2 π 2 π 2 ⎞ 1 ⎛ 2π 2 ⎞ π ⎜ + − + ⎟= ⎜ ⎟= π⎝ 8 2 2 8 ⎠ π⎝ 8 ⎠ 4

}

(π − x) cos nx dx

π/ 2 π 2 ⎪⎧ ⎡ x sin nx cos nx ⎤ ⎡ π sin nπ x sin nx cos nx ⎤ ⎪⎫ = ⎨⎢ + +⎢ − − ⎬ π ⎪⎩ ⎣ n n 2 ⎥⎦ 0 n n 2 ⎥⎦ π / 2 ⎪⎭ ⎣ n

⎧⎛ π nπ nπ ⎞ nπ π nπ nπ ⎞ ⎫ ⎛ sin cos π sin sin cos ⎟ ⎜ π 2 ⎪⎪⎜ 2 1 cos n ⎛ ⎞ ⎛ ⎞ 2 + 2 − 0+ 2 −2 2 − 2 ⎟ ⎪⎪ = ⎨⎜ + − − − 0 0 ⎟ ⎜ ⎟⎬ ⎜ ⎟ ⎜ ⎟ π ⎪⎜ n n2 ⎟ ⎝ n2 ⎠ ⎝ n2 ⎠ ⎜ n n n 2 ⎟⎪ ⎠ ⎝ ⎠ ⎭⎪ ⎩⎪⎝ nπ nπ nπ ⎧ ⎫ 2 cos π sin π sin ⎪ 2⎪ 1 cos n π 2 ⎧ nπ ⎫ 2 + 2 − − 2 − = ⎨ − 1 − cos nπ ⎬ ⎬ = 2 ⎨2 cos 2 2 2 n n n n ⎪ πn ⎩ 2 π⎪ n ⎭ ⎩ ⎭

When n is odd, a n = 0 and

a2 =

2 2 8 2 2 cos π − 1 − cos 2π ) = ( −2 − 1 − 1) = − = − , 2 ( π(2) 4π 4π π

a4 =

2 2 2 cos 2π − 1 − cos 4π ) = ( 2 − 1 − 1) = 0 , 2 ( π(4) 16π

a6 =

2 2 8 2 2 cos 3π − 1 − cos 6π ) = =− 2 , ( −2 − 1 − 1) = − 2 ( π(6) 36π 36π (3) π

a8 = 0 , a10 =

2 2 8 2 2 cos 5π − 1 − cos10π ) = = − 2 , and so on. ( −2 − 1 − 1) = − 2 ( π(10) 100π 100π (5) π


614

∞

Thus, f (t) = a 0 + ∑ a n cos nt = n =1

i.e.

f (t) =

π 2 2 2 − cos 2t − 2 cos 6t − 2 cos10t + .... 4 π (3) π (5) π

π 2⎛ 1 1 ⎞ − ⎜ cos 2t + 2 cos 6t + 2 cos10t + ...... ⎟ 4 π⎝ 3 5 ⎠


615

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 periodic function is shown in the diagram below.

The Fourier series is given by:

∞ ⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ f(x) = a 0 + ∑ ⎢a n cos ⎜ ⎟ + b n sin ⎜ ⎟ ⎥ where L = 5 ⎝ L ⎠ ⎝ L ⎠⎦ n =1 ⎣ 5

1 L 1 5 1 ⎡ x2 ⎤ 1 ⎡ 52 ⎤ 5 a 0 = ∫ f (x) dx = ∫ x dx = ⎢ ⎥ = ⎢ ⎥ = L 0 5 0 5 ⎣ 2 ⎦0 5 ⎣ 2 ⎦ 2 5

⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ x sin ⎜ cos ⎜ ⎢ ⎟ ⎟⎥ 2 L 2 5 2⎢ 5 ⎠ 5 ⎠⎥ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎝ ⎝ + by parts a n = ∫ f (x) cos ⎜ ⎟ dx = ∫ 0 x cos ⎜ ⎟ dx = ⎢ 2 L 0 5 5 ⎛ 2πn ⎞ ⎝ L ⎠ ⎝ 5 ⎠ ⎛ 2πn ⎞ ⎥ ⎢ ⎜ 5 ⎟ ⎜ ⎟ ⎥ ⎠ ⎝ 5 ⎠ ⎦0 ⎣ ⎝

⎡⎛ ⎞ ⎛ ⎞⎤ ⎢⎜ ⎟ ⎜ ⎟⎥ 2 5sin 2πn cos 2πn ⎟ ⎜ 1 ⎟⎥ = 0 = ⎢⎜ + − 0 + 2 5 ⎢⎜ ⎛ 2πn ⎞ ⎛ 2πn ⎞ 2 ⎟ ⎜ ⎛ 2πn ⎞ ⎟ ⎥ ⎢⎜ ⎜ 5 ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎟⎥ ⎠ ⎝ 5 ⎟⎠ ⎠ ⎝ ⎝ 5 ⎠ ⎠⎦ ⎣⎝ ⎝ 5

⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ x cos ⎜ sin ⎜ ⎢ ⎟ ⎟⎥ 2 L 2 5 2⎢ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎝ 5 ⎠+ ⎝ 5 ⎠ ⎥ by parts = = − b n = ∫ f (x) sin ⎜ dx x sin dx ⎟ ⎜ ⎟ 2 L 0 5 ∫0 5⎢ ⎛ 2πn ⎞ ⎝ L ⎠ ⎝ L ⎠ ⎛ 2πnx ⎞ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎦0 ⎣ ⎡⎛ ⎤ ⎞ ⎡ ⎤ ⎢⎜ ⎥ ⎟ ⎢ ⎥ 2 5cos 2πn sin 2πn ⎟ 2 5cos 2πn 5 5 ⎥ = − cos 2πn = − = ⎢⎜ − + − ( 0 + 0 )⎥ = ⎢− 2 ⎥ 5 ⎢ ⎛ 2πn ⎞ ⎥ πn πn 5 ⎢⎜ ⎛ 2πn ⎞ ⎛ 2πn ⎞ ⎟ ⎜ ⎟ ⎥ ⎢⎜ ⎜ 5 ⎟ ⎜ ⎥ ⎟ ⎟ ⎢ ⎠ ⎝ 5 ⎠ ⎠ ⎣ ⎝ 5 ⎠ ⎦ ⎣⎝ ⎝ ⎦

Hence, b1 = − Thus,

5 , π

b2 = −

5 , 2π

b3 = −

5 , 3π

b4 = −

5 , and so on. 4π

∞ ⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ f(x) = a 0 + ∑ ⎢a n cos ⎜ ⎟ + b n sin ⎜ ⎟⎥ ⎝ L ⎠ ⎝ L ⎠⎦ n =1 ⎣


616

i.e.

f(x) =

5 5 ⎛ 2πx ⎞ 5 ⎛ 4πx ⎞ 5 ⎛ 6πx ⎞ − sin ⎜ ⎟ − sin ⎜ ⎟ − sin ⎜ ⎟ − ..... 2 π ⎝ 5 ⎠ 2π ⎝ 5 ⎠ 3π ⎝ 5 ⎠

i.e.

f(x) =

⎤ 5 5 ⎡ ⎛ 2πx ⎞ 1 ⎛ 4 πx ⎞ 1 ⎛ 6πx ⎞ − ⎢sin ⎜ + sin ⎜ + sin ⎜ + ......⎥ ⎟ ⎟ ⎟ 2 π⎣ ⎝ 5 ⎠ 2 ⎝ 5 ⎠ 3 ⎝ 5 ⎠ ⎦

3. A periodic function of period 4 is defined by:

⎧ − 3, when − 2 ≤ x ≤ 0 f(x) = ⎨ ⎩ + 3, when 0 ≤ x ≤ 2

Sketch the function and obtain the Fourier series for the function. The periodic function is shown in the diagram below.

The function is odd since it is symmetrical about the origin, i.e. a n = 0 Thus,

∞ ⎡ ⎛ 2πnx ⎞ ⎤ f(x) = a 0 + ∑ ⎢ b n sin ⎜ ⎟⎥ ⎝ L ⎠⎦ n =1 ⎣

{∫

where L = 4

} 14 {[−3x]

}

1 {( 0 ) − ( 6 ) + ( 6 ) − ( 0 )} = 0 4

a0 =

1 L/2 1 f (x) dx = ∫ L −L / 2 4

bn =

2 2 L/2 2⎧ 0 ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎫ f (x) sin ⎜ dx = ⎨ ∫ −3sin ⎜ dx + ∫ 3sin ⎜ ⎟ ⎟ ⎟ dx ⎬ ∫ 0 L −L / 2 4 ⎩ −2 ⎝ L ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎭

0

−2

2

−3dx + ∫ 3dx = 0

0

−2

+ [3x ] 0 = 2

0 2 ⎧⎡ ⎧⎛ ⎡ ⎞ ⎛ ⎞⎫ ⎛ πnx ⎞ ⎤ ⎛ πnx ⎞ ⎤ ⎫ 3cos ⎜ ⎪ ⎢ 3cos ⎜ ⎪ ⎟ ⎟ ⎪ ⎜ ⎟ ⎜ ⎟⎪ ⎥ ⎢ ⎥ 1⎪ ⎝ 2 ⎠ ⎥ + ⎢− ⎝ 2 ⎠ ⎥ ⎪ = 1 ⎪⎜ 3cos 0 − 3cos(−πn) ⎟ + ⎜ − 3cos πn − − 3cos 0 ⎟ ⎪ = ⎨⎢ ⎬ ⎨ ⎬ 2 ⎪ ⎢ ⎛ πn ⎞ ⎥ ⎛ πn ⎞ ⎥ ⎪ 2 ⎪⎜ ⎛ πn ⎞ ⎛ πn ⎞ ⎟ ⎜ ⎛ πn ⎞ ⎛ πn ⎞ ⎟ ⎪ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟⎟ ⎢ ⎜ ⎟ ⎥ ⎢ ⎪⎩⎝⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ 0 ⎭⎪ ⎝ 2 ⎠ ⎠ ⎝ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎠ ⎪⎭ ⎩⎪ ⎣ ⎝ 2 ⎠ ⎦ − 2 ⎣

⎡ ⎤ ⎢ ⎥ 6 1 6 6 = ⎢ − cos πn ⎥ = (1 − cos πn ) 2 ⎢ ⎛ πn ⎞ ⎛ πn ⎞ ⎥ πn ⎢⎣ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎥⎦

When n is even, b n = 0


617

Hence, b1 = Thus,

i.e.

6 12 (1 − −1) = , π π

b3 =

6 12 (1 − −1) = , 3π 3π

b5 =

6 12 (1 − −1) = , and so on. 5π 5π

∞ 12 ⎛ πx ⎞ 12 ⎡ ⎛ 3πx ⎞ 12 ⎛ 5πx ⎞ ⎛ 2πnx ⎞ ⎤ sin ⎜ ⎟ + sin ⎜ f(x) = a 0 + ∑ ⎢ b n sin ⎜ ⎟ + sin ⎜ ⎟ + ...... ⎟⎥ = 0 + π ⎝ 2 ⎠ 3π ⎝ 2 ⎠ 5π ⎝ 2 ⎠ ⎝ L ⎠⎦ n =1 ⎣

f(x) =

⎫ 12 ⎧ ⎛ πx ⎞ 1 ⎛ 3πx ⎞ 1 ⎛ 5πx ⎞ + sin ⎜ + .....⎬ ⎨sin ⎜ ⎟ + sin ⎜ ⎟ ⎟ π ⎩ ⎝ 2 ⎠ 3 ⎝ 2 ⎠ 5 ⎝ 2 ⎠ ⎭


618

EXERCISE 245 Page 681 1. Determine the half-range Fourier cosine series for the function f(x) = x in the range 0 ≤ x ≤ 3.

Sketch the function within and outside of the given range. 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

1 a0 = L

an =

{

∞ ⎛ nπx ⎞ f (x) = a 0 + ∑ a n cos ⎜ ⎟ ⎝ L ⎠ n =1

}

3 1 3 1 ⎧⎪ ⎡ x 2 ⎤ ⎫⎪ 3 ∫ 0 f (x) dx = 3 ∫ 0 x dx = 3 ⎨⎢⎣ 2 ⎥⎦ ⎬ = 2 ⎪ 0⎭ ⎩⎪ L

2⎧ L ⎛ nπx ⎞ ⎫ 2 3 ⎛ nπx ⎞ ⎨ ∫ 0 f (x) cos ⎜ ⎟ dx ⎬ = ∫ 0 x cos ⎜ ⎟ dx L⎩ ⎝ L ⎠ ⎭ 3 ⎝ 3 ⎠

3 ⎧⎡ ⎡⎛ ⎤ ⎫ ⎞ ⎛ ⎞⎤ n x n x π π ⎛ ⎞ ⎛ ⎞ ⎪ ⎢ x sin ⎜ cos ⎜ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎥ ⎪ ⎟ ⎟ 2 ⎪⎢ 1 ⎟⎥ 3 ⎠ 3 ⎠ ⎥ ⎪ 2 ⎢⎜ 3sin nπ cos nπ ⎟ ⎜ ⎝ ⎝ = ⎨ + + − 0+ ⎬= 2 2 2 3 ⎪ ⎢ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎥ ⎪ 3 ⎢⎜ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎟ ⎜ ⎛ nπ ⎞ ⎟ ⎥ ⎜ ⎟ ⎢⎜ ⎜ 3 ⎟ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎥ ⎪ ⎜ ⎟ ⎟⎥ ⎪⎢ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎦0⎭ ⎝ 3 ⎠ ⎠⎦ ⎣⎝ ⎝ ⎠ ⎝ 3 ⎠ ⎠ ⎝ ⎩⎣

⎧ ⎫ ⎪ ⎪ 2⎪ cos nπ 1 ⎪ − = ⎨0 + ⎬= 3 ⎪ ⎛ nπ ⎞ 2 ⎛ nπ ⎞ 2 ⎪ ⎪⎩ ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎪⎭

by parts

2

2⎛ 3 ⎞ 6 ⎜ ⎟ {cos nπ − 1} = 2 2 ( cos nπ − 1) 3 ⎝ nπ ⎠ n π

When n is even, a n = 0 and

Thus,

i.e.

a1 =

6 12 6 12 −2 ) = − 2 , a 3 = 2 2 ( −2 ) = − 2 2 , 2 ( π (1) π π (3) π (3) 2

a5 = −

12 , and so on. π (5) 2 2

∞ 12 12 ⎛ nπx ⎞ 3 12 ⎛ πx ⎞ ⎛ 3πx ⎞ ⎛ 5πx ⎞ f (x) = a 0 + ∑ a n cos ⎜ ⎟ = − 2 cos ⎜ ⎟ − 2 2 cos ⎜ ⎟ − 2 2 cos ⎜ ⎟ − .... ⎝ L ⎠ 2 π ⎝ 3 ⎠ π (3) ⎝ 3 ⎠ π (5) ⎝ 3 ⎠ n =1

f (x) =

⎫ 3 12 ⎧ ⎛ πx ⎞ 1 ⎛ 3πx ⎞ 1 ⎛ 5πx ⎞ − 2 ⎨cos ⎜ ⎟ + 2 cos ⎜ + 2 cos ⎜ + .....⎬ ⎟ ⎟ 2 π ⎩ ⎝ 3 ⎠ 3 ⎝ 3 ⎠ 5 ⎝ 3 ⎠ ⎭


619

2. Find the half-range Fourier sine series for the function f(x) = x in the range 0 ≤ x ≤ 3.

Sketch the function within and outside of the given range. 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

∞ ⎛ nπx ⎞ f (x) = ∑ b n sin ⎜ ⎟ ⎝ L ⎠ n =1

3

⎡ ⎛ nπx ⎞ ⎛ nπx ⎞ ⎤ x cos ⎜ sin ⎜ ⎢ ⎟ ⎟⎥ 2 L 2⎧ 3 3 ⎠ 3 ⎠⎥ ⎛ nπx ⎞ ⎛ nπx ⎞ ⎫ 2 ⎢ ⎝ ⎝ b n = ∫ f (x) sin ⎜ by parts + ⎟ dx = ⎨ ∫ 0 x sin ⎜ ⎟ dx ⎬ = − 2 L 0 3⎩ ⎛ nπ ⎞ ⎝ L ⎠ ⎝ 3 ⎠ ⎭ 3⎢ ⎛ nπ ⎞ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎦0 ⎣ ⎡⎛ ⎤ ⎞ ⎢⎜ ⎥ ⎟ 2 3cos nπ sin nπ ⎟ ⎥ = − 2 cos nπ = − 6 cos nπ 0 0 = ⎢⎜ − + − + ( ) 2 ⎢ ⎥ ⎜ ⎟ 3 nπ ⎛ nπ ⎞ ⎛ n π ⎞ ⎛ nπ ⎞ ⎜ ⎟ ⎢⎜ ⎜ 3 ⎟ ⎜ ⎟ ⎟ ⎥ ⎝ 3 ⎠ ⎣⎝ ⎝ ⎠ ⎝ 3 ⎠ ⎠ ⎦

b1 =

6 , π

b2 = −

6 , 2π

b3 =

6 , 3π

b4 = −

6 , and so on. 4π

∞ ⎛ nπx ⎞ 6 ⎛ πx ⎞ 6 ⎛ 2πx ⎞ 6 ⎛ 3πx ⎞ 6 ⎛ 4πx ⎞ Thus, f (x) = ∑ b n sin ⎜ ⎟ = sin ⎜ ⎟ − sin ⎜ ⎟ + sin ⎜ ⎟ − sin ⎜ ⎟ + .... ⎝ L ⎠ π ⎝ 3 ⎠ 2π ⎝ 3 ⎠ 3π ⎝ 3 ⎠ 4π ⎝ 3 ⎠ n =1

i.e.

f(x) =

⎫ 6 ⎧ ⎛ πx ⎞ 1 ⎛ 2πx ⎞ 1 ⎛ 3πx ⎞ 1 ⎛ 4 πx ⎞ + sin ⎜ − sin ⎜ + .....⎬ ⎨sin ⎜ ⎟ − sin ⎜ ⎟ ⎟ ⎟ π⎩ ⎝ 3 ⎠ 2 ⎝ 3 ⎠ 3 ⎝ 3 ⎠ 4 ⎝ 3 ⎠ ⎭

3. Determine the half-range Fourier sine series for the function defined by:

t, when ⎧⎪ f(t) = ⎨ ⎪⎩ ( 2 − t ) , when

0 ≤t ≤1 1≤ t ≤ 2


620

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

bn =

∞ ⎛ nπt ⎞ f (t) = ∑ b n sin ⎜ ⎟ ⎝ L ⎠ n =1

2 2 L 2⎧ 1 ⎛ nπt ⎞ ⎛ nπt ⎞ ⎛ nπt ⎞ ⎫ f (t) sin ⎜ dt = ⎨ ∫ t sin ⎜ dt + ∫ (2 − t) sin ⎜ ⎟ ⎟ ⎟ dt ⎬ ∫ 1 L 0 2⎩ 0 ⎝ L ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎭

1 2 ⎧⎡ ⎤ ⎡ ⎤ ⎫ n t n t n t n t n t π π π π π ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎪ ⎢ t cos ⎜ ⎟ sin ⎜ ⎟ ⎥ ⎢ 2 cos ⎜ ⎟ t cos ⎜ ⎟ sin ⎜ ⎟⎥ ⎪ ⎪⎢ 2 ⎠ 2 ⎠⎥ ⎢ 2 ⎠ 2 ⎠ 2 ⎠⎥ ⎪ ⎝ ⎝ ⎝ ⎝ ⎝ = ⎨ − + + − + − ⎬ 2 2 n n n π π π ⎢ ⎥ ⎢ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ n π ⎛ ⎞ ⎛ nπ ⎞ ⎥ ⎪ ⎪ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎥ ⎪ ⎪ ⎢⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 ⎝ ⎠ ⎝ 2 ⎠ ⎦1 ⎭ ⎦ ⎣ 0 ⎩

by parts

⎧ ⎡⎛ ⎤⎫ ⎞ ⎪ ⎢⎜ ⎥⎪ ⎟ π π π 2 cos n 2 cos n sin n ⎪ ⎢⎜ − ⎥⎪ ⎟ + − 2 ⎪ ⎡⎛ ⎢ ⎥⎪ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎟ ⎤ ⎜ ⎛ nπ ⎞ ⎞ ⎪ ⎢⎜ cos nπ sin ⎜⎛ nπ ⎟⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎪ ⎜ ⎟ ⎥ 2 ⎠ 2 ⎠ ⎜⎝ 2 ⎟⎠ ⎠ ⎝ ⎝ ⎪ ⎢⎜ ⎝ 2 ⎢ ⎥⎪ ⎝ ⎠ 2 + ⎟ − ( 0 + 0)⎥ + =⎨ − 2 ⎢ ⎥⎬ ⎥ ⎞ ⎛ ⎛ nπ ⎞ ⎟ n n n π π π ⎪ ⎢⎢⎜ ⎜⎛ nπ ⎟⎞ ⎪ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎥ ⎢ 2 cos ⎜ ⎟ cos ⎜ ⎟ sin ⎜ ⎟ ⎟ ⎥ ⎪ ⎜ ⎟ ⎟ ⎜ ⎪ ⎣⎝⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎠ ⎦ ⎢ ⎝ 2 ⎠+ ⎝ 2 ⎠− ⎝ 2 ⎠ ⎟⎥ ⎪ ⎪ −⎜− ⎢ ⎥ 2 ⎜ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎪ ⎛ nπ ⎞ ⎟ ⎥ ⎪ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎟⎥ ⎪ ⎪ ⎢⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎠⎦ ⎭ ⎝ ⎩ ⎛ nπ ⎞ 2sin ⎜ ⎟ ⎝ 2 ⎠ = 8 sin ⎛ nπ ⎞ = ⎜ ⎟ 2 n 2 π2 ⎝ 2 ⎠ ⎛ nπ ⎞ ⎜ ⎟ ⎝ 2 ⎠

When n is even, b n = 0 b1 =

8 , π2

b3 = −

8 , (3) 2 π2

b5 =

8 , and so on. (5) 2 π2

∞ 8 8 8 ⎛ nπt ⎞ ⎛ πt ⎞ ⎛ 3πt ⎞ ⎛ 5πt ⎞ Thus, f (t) = ∑ b n sin ⎜ ⎟ = 2 sin ⎜ ⎟ − 2 2 sin ⎜ ⎟ + 2 2 sin ⎜ ⎟ − .... ⎝ 2 ⎠ (3) π ⎝ 2 ⎠ (5) π ⎝ 2 ⎠ ⎝ L ⎠ π n =1

i.e.

f(t) =

⎫ 8 ⎧ ⎛ πt ⎞ 1 ⎛ 3πt ⎞ 1 ⎛ 5πt ⎞ sin ⎜ ⎟ − 2 sin ⎜ + 2 sin ⎜ − .....⎬ ⎟ ⎟ 2 ⎨ π ⎩ ⎝ 2⎠ 3 ⎝ 2 ⎠ 5 ⎝ 2 ⎠ ⎭


621

4. Show that the half-range Fourier cosine series for the function f(θ) = θ2 in the range 0 to 4 is

given by: f ( θ ) =

16 64 ⎧ ⎛ πθ ⎞ 1 ⎫ ⎛ 2πθ ⎞ 1 ⎛ 3πθ ⎞ − 2 ⎨cos ⎜ ⎟ − 2 cos ⎜ ⎟ + 2 cos ⎜ ⎟ − .....⎬ 3 π ⎩ ⎝ 4 ⎠ 2 ⎝ 4 ⎠ 3 ⎝ 4 ⎠ ⎭

Sketch the function within and outside of the given range. The periodic function is shown in the diagram below. Since a half-range cosine series is required, the function is symmetrical about the f(θ) axis and

1 a0 = L

an =

{∫

L 0

}

4 1 4 2 1 ⎧⎪ ⎡ θ3 ⎤ ⎫⎪ f (θ) dθ = ∫ θ dθ = ⎨ ⎢ ⎥ ⎬ = 4 0 4 ⎪⎣ 3 ⎦ 0 ⎪ ⎩ ⎭

∞ ⎛ nπθ ⎞ f (θ) = a 0 + ∑ a n cos ⎜ ⎟ ⎝ L ⎠ n =1

1 ⎛ 64 ⎞ 16 ⎜ ⎟= 4⎝ 3 ⎠ 3

2⎧ L ⎛ nπθ ⎞ ⎫ 2 4 2 ⎛ nπθ ⎞ ⎨ ∫ 0 f (θ) cos ⎜ ⎟ dθ ⎬ = ∫ 0 θ cos ⎜ ⎟ dθ L⎩ ⎝ L ⎠ ⎭ 4 ⎝ 4 ⎠

4 ⎧⎡ ⎫ ⎤ n πθ n πθ n πθ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎪ ⎢ θ2 sin ⎜ θ 2 cos 2sin ⎥ ⎟ ⎜ ⎟ ⎜ ⎟ ⎪ 1 ⎪⎢ 4 ⎠ 4 ⎠ 4 ⎠⎥ ⎪ ⎝ ⎝ ⎝ = ⎨ + − ⎬ by parts 2 3 ⎥ ⎪ 2 ⎪⎢ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎜ ⎟ ⎥ ⎪ ⎜ ⎟ ⎜ ⎟ ⎪ ⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎦0⎭ ⎩

⎧⎛ ⎫ ⎞ ⎧ ⎫ ⎪⎜ ⎪ ⎟ ⎪ ⎪ 64 1 ⎪⎜ 16sin nπ 8cos nπ 2sin nπ ⎟ ⎪ 1 ⎪ 8cos nπ ⎪ 1 ⎛ 8(16) ⎞ = ⎨ + − − ( 0 )⎬ = ⎨ = ⎜ 2 2 ⎟ cos nπ = 2 2 cos nπ 2 3 2 ⎬ n π 2 ⎪⎜ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎟ ⎪ 2 ⎪ ⎛ nπ ⎞ ⎪ 2 ⎝ n π ⎠ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎪⎩⎝ ⎝ 4 ⎠ ⎪⎭ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎠ ⎩⎪ ⎝ 4 ⎠ ⎭⎪

and a1 =

64 64 64 64 64 64 −1) = − 2 , a 2 = 2 2 (1) = 2 2 , a 3 = 2 2 (−1) = − 2 2 , and so on. 2 ( π (1) π π (2) π (2) π (3) π (3) 2

Thus,

∞ 64 64 ⎛ nπθ ⎞ 16 64 ⎛ πθ ⎞ ⎛ 2πθ ⎞ ⎛ 3πθ ⎞ f (θ) = a 0 + ∑ a n cos ⎜ ⎟ = − 2 cos ⎜ ⎟ + 2 2 cos ⎜ ⎟ − 2 2 cos ⎜ ⎟ + .... ⎝ L ⎠ 3 π ⎝ 4 ⎠ π (2) ⎝ 4 ⎠ π (3) ⎝ 4 ⎠ n =1

i.e.

f (θ ) =

⎫ 16 64 ⎧ ⎛ πθ ⎞ 1 ⎛ 2πθ ⎞ 1 ⎛ 3πθ ⎞ − 2 ⎨cos ⎜ − 2 cos ⎜ + 2 cos ⎜ − .....⎬ ⎟ ⎟ ⎟ 3 π ⎩ ⎝ 4 ⎠ 2 ⎝ 4 ⎠ 3 ⎝ 4 ⎠ ⎭


622

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 θ°

sin θ

30

y

cosθ

y cos θ

y sin θ

30

40

0.866

34.64

0.5

20

60

43

0.5

21.50

0.866

37.24

23

cos 2θ

17

11

9

10

13

21

y cos 2θ

sin 2θ

y sin 2θ

cos 3θ

0.5

20

0.866

34.64

-0.5

-21.5

0.866

37.24 0

0 1

32

ycos3θ

sin3θ

ycos3θ

0

0

1

40

-1

-43

0

0

0

-1

-38

30

0

0

90

38

0

0

1

38

-1

-38

0

120

30

-0.5

-15.0

0.866

25.98

-0.5

-15

-0.866

-25.98

150

23

-0.866

-19.92

0.5

11.5

0.5

11.5

-0.866

-19.92

0

0

1

23

180

17

-1

-17

0

0

1

17

0

0

-1

-17

0

0

210

11

-0.866

-9.53

-0.5

-5.5

0.5

5.5

0.866

9.53

0

0

-1

-11

240

9

-0.5

-4.5

-0.866

-7.79

-0.5

-4.5

0.866

7.79

1

9

0

0

270

10

0

0

-1

-10

-1

-10

0

0

0

1

10

300

13

0.5

6.5

-0.866

-11.26

-0.5

-6.5

-0.866

-11.26

-1

-13

0

0

330

21

0.866

18.19

-0.5

-10.5

0.5

10.5

-0.866

-18.19

0

0

360

32

1

32

0

0

1

32

0

1

32

12

∑y k =1

k

= 287

12

∑y k =1

k

cos θk = 46.88

12

∑y k =1

k

sin θk = 87.67

12

∑y k =1

k

cos 2θk = 1

0

0

12

∑y k =1

k

sin 2θk = 13.85

12

∑y k =1

k

cos 3θk = −2

-21

-1

0

0

12

∑y k =1

k

sin 3θk = 3

a0 ≈

1 p 1 y k = (287) = 23.92 ∑ p k =1 12

an ≈

2 p ∑ yk cos nx k p k =1

hence, a1 ≈

2 2 2 (46.88) = 7.81 , a 2 ≈ (1) = 0.17 , a 3 ≈ (−2) = −0.33 12 12 12

2 p b n ≈ ∑ y k sin nx k p k =1

hence, b1 ≈

2 2 2 (87.67) = 14.61 , b 2 ≈ (13.85) = 2.31 , b3 ≈ (3) = 0.50 12 12 12 ∞

Substituting these values into the Fourier series: f (x) = a 0 + ∑ ( a n cos nx + b n sin nx ) n =1

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


623

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.

θ°

Angle θ°

30 60

Current i

0 -1.4 -1.8 -1.9 -1.8 -1.3

i

cosθ

30

0

0.866

60

-1.4

0.5

i cos θ

0 -0.7

90

sinθ

120 150 180 210 240 270 300 330 360

i sin θ

cos 2θ

0

i cos 2θ

2.2 3.8 3.9 3.5 2.5 sin 2θ

i sin 2θ

cos 3θ

i cos3θ

sin3θ

i cos3θ

0.5

0

0.866

0

0

0

1

0

0.866

-1.212

-0.5

0.7

0.866

-1.212

-1

1.4

0

0

0

0.5

90

-1.8

0

0

1

-1.8

-1

1.8

0

0

0

0

-1

1.8

120

-1.9

-0.5

0.95

0.866

-1.645

-0.5

0.95

-0.866

1.645

1

-1.9

0

0

150

-1.8

-0.866

1.56

0.5

-0.9

0.5

-0.90

-0.866

1.548

0

0

1

-1.8

180

-1.3

-1

1.3

0

1

-1.3

0

0

-1

1.3

0

0

0

-0.5

0.5

0

0.866

0

0

0

-1

0

-0.866

-1.905

-0.5

-1.1

0.866

1.905

1

2.2

0

0

0

-1

-3.8

-1

-3.8

0

0

0

0

1

3.8

1.95

-0.866

-3.377

-0.5

-1.95

-0.866

-3.377

-1

-3.9

0

0

0.866

3.03

-0.5

-1.75

0.5

1.75

-0.866

-3.031

0

0

-1

-3.5

1

2.5

0

0

1

2.5

0

0

1

2.5

0

0

210

0

-0.866

240

2.2

-0.5

270

3.8

0

300

3.9

0.5

330

3.5

360

2.5

12

∑y k =1

k

= 7.7

0

-1.1

12

∑y k =1

k

cos θk = 9.49

0

12

∑y k =1

k

sin θk = −16.389

12

∑y k =1

k

cos 2θk = −1.35

12

∑y k =1

k

sin 2θk = −2.522

12

∑y k =1

k

cos 3θk = 1.6

12

∑y k =1

k

sin 3θk = 0.3

a0 ≈

1 p 1 y k = (7.7) = 0.64 ∑ p k =1 12

an ≈

2 p 2 2 2 y k cos nx k hence, a1 ≈ (9.49) = 1.58 , a 2 ≈ (−1.35) = −0.23 , a 3 ≈ (1.6) = 0.27 ∑ 12 12 12 p k =1

2 p 2 2 b n ≈ ∑ y k sin nx k hence, b1 ≈ (−16.389) = −2.73 , b 2 ≈ (−2.522) = −0.42 , p k =1 12 12 b3 ≈

2 (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


624

EXERCISE 247 Page 688 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

30

10

0.866

8.66

0.5

60

-6

0.5

-3.0

0.866

-5.196

90

-17

0

0

1

-17

-1

120

-17

-0.5

8.5

0.866

-14.72

150

-13

-0.866

0.5

-6.5

180

-4

-1

0

210

10

-0.866

-8.66-

-0.5

240

24

-0.5

12

-0.866

270

33

0

0

-1

300

36

0.5

18

330

33

0.866

28.58

360

24

1

24

12

∑y k =1

k

= 113

k =1

11.26 4

12

∑y

y cos θ

k

cos θk = 79.34

sinθ

cos 2θ

y cos 2θ

sin 2θ

0.5

5

0.866

-0.5

3

0.866

17

0

-0.5

8.5

0.5

-6.5

0

1

-4

-5.0

0.5

5

-20.78

-0.5

-33

-1

-0.866

-31.18

-0.5

-0.5

-16.5

0.5

16.5

0

0

1

24

0

5

12

∑y k =1

y sin θ

k

12

∑y

sin θk

k =1

k

cos 3θ

y cos3θ

sin3θ

y cos3θ

8.66

0

0

1

10

-5.196

-1

6

0

0

0

0

0

-1

17

-0.866

14.722

1

-17

0

0

-0.866

11.258

0

0

1

-13

0

0

-1

4

0

0

0.866

8.66

0

0

-1

-10

-12

0.866

20.784

1

24

0

0

-33

0

0

0

0

1

33

-18

-0.866

-31.176

-1

-36

0

0

-0.866

-28.578

0

0

-1

-33

1

24

0

0

cos 2θk = 5.5

0

12

∑y k =1

y sin 2θ

k

sin 2θk = −0.866

12

∑y k =1

k

cos 3θk = 5

12

∑y k =1

k

sin 3θk = 4

= −144.88

a0 ≈

1 p 1 y k = (113) = 9.4 ∑ p k =1 12

an ≈

2 p 2 2 2 y k cos nx k hence, a1 ≈ (79.34) = 13.2 , a 2 ≈ (5.5) = 0.92 , a 3 ≈ (5) = 0.83 ∑ p k =1 12 12 12

bn ≈

2 p ∑ yk sin nx k p k =1

hence, b1 ≈ b3 ≈

2 2 (−144.88) = −24.1 , b 2 ≈ (−0.866) = −0.14 , 12 12 2 (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 © 2006 John Bird. All rights reserved. Published by Elsevier.

625

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

or

3. For the waveform of current shown below state why only a d.c. component and even cosine

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 will be present. The function repeats itself every half cycle, hence only even will be present. Hence, the Fourier series will contains a d.c. component and even cosine only. θ°

i

cos 2θ

i cos 2θ

cos 4θ

i cos 4θ

cos 6θ

i cos 6θ

-1

-1.5

30

1.5

0.5

0.75

-0.5

-0.75

60

5.5

-0.5

-2.75

-0.5

-2.75

90

10

-1

-10

1

10

120

5.5

-0.5

-2.75

-0.5

-2.75

1

5.5

150

1.5

0.5

0.75

-0.5

-0.75

-1

-1.5

180

0

1

0

1

0

210

1.5

0.5

0.75

-0.5

-0.75

-1

-1.5

240

5.5

-0.5

-2.75

-0.5

-2.75

1

5.5

270

10

-1

-10

1

10

-1

-10

300

5.5

-0.5

-2.75

-0.5

-2.75

1

5.5

330

1.5

0.5

0.75

-0.5

-0.75

-1

-1.5

360

0

1

0

1

0

12

∑y k =1

k

= 48

12

∑y k =1

k

cos 2θk = −28

12

∑y k =1

k

1

5.5

-1

-10

0

1

0

1

12

∑y

cos 4θk = 6

k =1

k

cos 6θk = −4

a0 ≈

1 p 1 y k = (48) = 4.00 ∑ p k =1 12

an ≈

2 p 2 2 2 y k cos nx k hence, a 2 ≈ (−28) = −4.67 , a 4 ≈ (6) = 1.00 , a 3 ≈ (−4) = −0.66 ∑ 12 12 12 p k =1 ∞

Substituting these values into the Fourier series: f (x) = a 0 + ∑ ( a n cos nx ) n =1

gives:

i = 4.00 – 4.67 cos 2θ + 1.00 cos 4θ - 0.66 cos 6θ + ……


626

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:

⎧ 0, when − π ≤ t ≤ 0 f(t) = ⎨ ⎩ 2, when 0 ≤ t ≤ π The function is periodic outside of this range of period 2π. The periodic function is shown in the diagram below.

f (t) =

The complex Fourier series is given by:

∞

∑

n =−∞

cn e

j

2π n t L

2π n t −j 1 L/2 c n = ∫ f (t) e L dt L −L / 2

where

π

2π n t −j π ⎫ 1 π − jn t 1 ⎡ e − jn t ⎤ 1 ⎧ 0 1 2π ⎡⎣e − jn π − e0 ⎤⎦ cn = 0 dt 2 e dt + = ∫ e = ⎢ =− ⎨ ∫ −π ⎬ ⎥ ∫ 0 0 π ⎣ − jn ⎦ 0 2π ⎩ jπ n ⎭ π

i.e.

= −

=

j j − jn π ⎡⎣e − jnπ − 1⎤⎦ = ( e − 1) πn j πn 2

j j [cos nπ − jsin nπ − 1] = [ cos nπ − 1] πn πn

for all

integer values of n f (t) =

Hence,

∞

∑

n=− ∞

c0 = a 0 = mean value = c1 =

j 2 (−1 − 1) = − j , π π

c3 = − j

2 , 3π

c5 = − j

cn e

j

2πn t L

=

∞

j ( cos nπ − 1) e jn t n= − ∞ n π

∑

2× π =1 2π c2 =

j (1 − 1) = 0 and all even will be zero 2π

2 , and so on. 5π


627

c −1 =

j 2 (−2) = j , −π π

c −3 =

j 2 , (−2) = j −3π 3π

c5 = j

2 , and so on. 5π

2 jt 2 2 2 2 2 e − j e j3t − j e j5t − .... + j e− jt + j e− j3t + j e− j5t π π 3π 5π 3π 5π

Thus,

f(t) = 1 − j

i.e.

2⎛ 1 1 2⎛ 1 1 ⎞ ⎞ f (t) = 1 − j ⎜ e jt + e j3t + e j5t + .... ⎟ + j ⎜ e − jt + e − j3t + e − j5t + .... ⎟ π⎝ 3 5 3 5 ⎠ π⎝ ⎠

2. Show that the complex Fourier series for the waveform shown below, that has period 2, may be

represented by:

f (t) = 2 +

∞

j2 ( cos nπ − 1) e jπ n t n = − ∞ πn

∑

(n ≠ 0)

f (t) =


∞

∑

n=− ∞

cn e

j

2π n t L

2π n t −j 1 L/2 c n = ∫ f (t) e L dt L −L / 2

where

1

i.e.

1 ⎡ e− jπ n t ⎤ 1 ⎧ 1 − j2π n t ⎫ 2 ⎡⎣ e− jπ n − e0 ⎤⎦ c n = ⎨ ∫ 4e 2 ⎬ = 2 ∫ e − jπ n t = 2 ⎢ =− ⎥ 0 0 2⎩ jπ n ⎣ − jπ n ⎦ 0 ⎭

= −

=

2 j2 − jn π ⎡e − jn π − 1⎦⎤ = ( e − 1) ⎣ πn jπ n

j2 j2 [cos nπ − jsin nπ − 1] = [ cos nπ − 1] πn πn

for all

integer values of n c0 = a 0 = mean value =

1 1 1 1 4 dt = [ 4t ] 0 = 2 ∫ 2 0 2

Hence,

f (t) = 2 +

∞

j2

∑ n π ( cos nπ − 1) e

jπ n t

n =−∞ (n ≠ 0)


628

3. Show that the complex Fourier series of Problem 2 is equivalent to:

8⎛ 1 1 ⎞ f (t) = 2 + ⎜ sin πt + sin 3πt + sin 5πt + .... ⎟ π⎝ 3 5 ⎠ j2 4 (−1 − 1) = − j , π(1) π

c2 =

c3 = j

2 4 (−2) = − j , 3π 3π

c5 = − j

c −1 =

j2 4 (−2) = j , −π π

c1 =

Thus,

f (t) = 2 +

c −3 =

j2 (1 − 1) = 0 and all even will be zero 2π 4 , and so on. 5π

j2 4 , (−2) = j −3π 3π

c5 = j

4 , and so on. 5π

∞

j2 ( cos nπ − 1) e jπ n t n=− ∞ n π

∑

(n ≠ 0)

4 j πt 4 4 4 4 4 e − j e j3πt − j e j5 πt − .... + j e− jπ t + j e− j3πt + j e− j5πt π π 3π 5π 3π 5π

i.e.

f(t) = 2 − j

i.e.

4⎛ 1 1 1 1 ⎞ 4⎛ ⎞ f (t) = 2 − j ⎜ e j πt + e j3πt + e j5πt + .... ⎟ + j ⎜ e − j πt + e− j3πt + e− j5 πt + .... ⎟ π⎝ 3 5 3 5 ⎠ π⎝ ⎠ 4⎡ 1 1 ⎤ = 2 − j ⎢( e j π t − e − jπ t ) + ( e j3 π t − e− j3 π t ) + ( e j5 πt − e− j5 π t ) + ....⎥ π⎣ 3 5 ⎦ = 2 − j2

i.e.

f (t) = 2 +

8 ⎡⎛ e j π t − e − j π t ⎢⎜ 2j π ⎣⎝

⎞ 1 ⎛ e j3 π t − e− j3 π t ⎟+ ⎜ 2j ⎠ 3⎝

⎞ 1 ⎛ e j5 π t − e − j5 π t ⎟+ ⎜ 2j ⎠ 5⎝

⎤ ⎞ ⎟ + ....⎥ ⎠ ⎦

8⎡ 1 1 ⎤ sin πt + sin 3πt + sin 5πt + ....⎥ ⎢ π⎣ 3 5 ⎦

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.


f (t) =

∞

∑

n =−∞

cn e

j

2π n t L


629

cn =

where

2π n t −j 1 L/2 L f (t) e dt L ∫ −L / 2

− 2− j π n ) t 2− jπ n ) 2− jπ n ) 2πn t ⎤ ⎤ −j ⎫ 1 1 −e ( 1⎧ 1 1 ⎡e ( 1 ⎡ e( = ⎢ c n = ⎨ ∫ e 2t e 2 dt ⎬ = ∫ e 2t − j π n t dt = ⎢ ⎥ ⎥ −1 2 ⎩ −1 2 ⎣ 2 − jπ n ⎦ −1 2 ⎣ 2 − jπ n ⎭ 2 ⎦ 1

i.e.

∞

Thus,

f(t) =

∑

n =− ∞

cn e

j

2π n t L

=

1 ∞ ⎛ e( 2− j π n) − e − ( 2− j π n) ⎞ j π n t ∑ ⎜ 2 − jπ n ⎟e 2 n=− ∞ ⎝ ⎠


630

EXERCISE 249 Page 698 1. Determine the exponential form of the Fourier series for the periodic function defined by: π ⎧ ⎪ − 2, when − π ≤ x ≤ − 2 ⎪ π π ⎪ f(x) = ⎨ 2, when − ≤ x ≤ + 2 2 ⎪ π ⎪ ⎪ − 2, when + 2 ≤ x ≤ +π ⎩

and which has a period of 2π.

The periodic waveform is shown below. It is an even function and contains no sine , hence b n = 0 and between -π and +π, the mean value is zero, hence a 0 = 0 .

cn =

2 L/2 2 π ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ f (x) cos ⎜ dx = f (x) cos ⎜ ⎟ ⎟ dx ∫ ∫ L 0 2π 0 ⎝ L ⎠ ⎝ 2π ⎠ =

1 π

{∫

π/ 2 0

2 cos nx dx + ∫

π π/2

since L = 2π

}

−2 cos nx dx

π/2 π 1 ⎧⎪ ⎡ 2sin nx ⎤ ⎡ 2sin nx ⎤ ⎫⎪ −⎢ = ⎨⎢ ⎬ π ⎩⎪ ⎣ n ⎥⎦ 0 ⎣ n ⎥⎦ π / 2 ⎭⎪

∞

Hence,

f(x) =

∑ce

n =− ∞

n

j

2π n x L

=

1 ⎡⎛ nπ nπ ⎞ ⎤ 4 nπ ⎞ ⎛ − 0 ⎟ − ⎜ 2sin nπ − 2sin ⎟ ⎥ = sin ⎜ 2sin ⎢ 2 2 ⎠ ⎦ πn 2 πn ⎣⎝ ⎠ ⎝

=

⎧ 4 ⎛ nπ ⎞ ⎫ jn x ⎨ sin ⎜ ⎟⎬ e ⎝ 2 ⎠⎭ n = − ∞ ⎩ πn ∞

∑

2. Show that the exponential form of the Fourier series in Problem 1 above is equivalent to:

f (x) =

8⎛ 1 1 1 ⎞ ⎜ cos x − cos 3x + cos 5x − cos 7x + .... ⎟ π⎝ 3 5 7 ⎠


631

Since from Problem 1, c n = c0 = 0 , c1 = c3 =

4 π 4 4 2π = 0 = c 4 = c6 = c −2 = c −4 and so on , sin = , c 2 = sin π 2 π 2π 2

4 3π 4 sin =− , 3π 2 3π

4 −π 4 c −1 = − sin = , π 2 π ∞

f(x) =

∑

n =− ∞

4 nπ sin , then πn 2

cn e

j

2π n x L

c5 =

c −3 = − ∞

=

⎧4

4 −3π 4 4 and so on. sin = − , c −5 = 3π 2 3π 5π ⎛ nπ ⎞ ⎫ jn x ⎟⎬ e 2 ⎠⎭

∑ ⎨⎩ πn sin ⎜⎝

n =− ∞

=

4 4 and so on. , c7 = − 5π 7π

4 jx 4 j3x 4 j5x 4 4 4 e − e + e + ..... + e − jx − e − j3x + e − j5x − ..... π π 3π 5π 3π 5π

4 4 4 ⎛4 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ = ⎜ e jx + e − jx ⎟ − ⎜ e j3x + e − j3x ⎟ + ⎜ e j5x + e − j5x ⎟ − ..... π 3π 5π ⎝π ⎠ ⎝ 3π ⎠ ⎝ 5π ⎠

i.e.

=

8 ⎛ e jx + e − jx ⎞ 8 ⎛ e j3x + e − j3x ⎞ 8 ⎛ e j5x + e− j5x ⎞ ⎜ ⎟− ⎜ ⎟+ ⎜ ⎟ − ..... π⎝ 2 2 2 ⎠ 3π ⎝ ⎠ 5π ⎝ ⎠

=

8 8 8 cos x − cos 3x + cos 5x − ..... π 3π 5π

f (x) =

8⎛ 1 1 1 ⎞ ⎜ cos x − cos 3x + cos 5x − cos 7x + ..... ⎟ π⎝ 3 5 7 ⎠

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,

2 L2 ⎛ 2πnt ⎞ c n = − j ∫ f (t) sin ⎜ ⎟ dt 0 L ⎝ L ⎠

= −j

2 π 2 π ⎛ 2πnt ⎞ 2t sin ⎜ dt = − j ∫ t sin nt dt ⎟ ∫ 2π 0 π 0 ⎝ 2π ⎠


632

π

2 ⎡ − t cos nt sin nt ⎤ 2 ⎡⎛ −π cos nπ sin nπ ⎞ ⎤ = −j ⎢ + 2 ⎥ = − j ⎢⎜ + ⎟ − ( 0 + 0 )⎥ 2 n n ⎦0 n n ⎠ π⎣ π ⎣⎝ ⎦ i.e.

by parts

2 c n = j cos nπ n

Hence, the complex Fourier series is given by: ∞

f(t) =

∑c

n =− ∞

n e

j

2 πnt L

∞

=

⎛ j2

⎞

∑ ⎜⎝ n cos nπ ⎟⎠ e

jnt

n=− ∞

4. Show that the complex Fourier series is Problem 3 above is equivalent to:

1 1 1 ⎛ ⎞ f(t) = 4 ⎜ sin t − sin 2t + sin 3t − sin 4t + .... ⎟ 2 3 4 ⎝ ⎠ 2 From Problem 3 above, c n = j cos nπ n When n = 1, c1 = j

2 2 j2 cos π = j ( −1) = − (1) (1) 1

2 2 When n = 2, c 2 = j cos 2π = j 2 2 2 2 j2 When n = 3, c3 = j cos 3π = j ( −1) = − 3 3 3π By similar reasoning, c 4 =

j2 , 4

c5 = −

j2 , and so on. 5

When n = -1, c −1 = j

2 2 j2 cos(−π) = + j ( −1) = (−1) (−1) 1

When n = -2, c −2 = j

2 2 j2 cos(−2π) = j (1) = − (−2) (−2) 2

2 j2 By similar reasoning, c −3 = j , c −4 = − , and so on. 3 4 Since the waveform is odd, c0 = a 0 = 0 ∞

From Problem 3,

f(t) =

∑c

n =−∞

Hence,

f(t) = −

n e

j

2 πnt L

∞

=

⎛ j2

⎞

∑ ⎜⎝ n cos nπ ⎟⎠ e

jn t

n =−∞

j2 jt j2 j2t j2 j3t j2 j4 t j2 − jt j2 − j2 t j2 − j3t j2 − j4t + e − e + ... e + e − e + e − ... + e − e 1 2 3 4 1 2 3 4

j2 j2 j2 ⎛ j2 ⎞ ⎛ j2 ⎞ ⎛ j2 ⎞ = ⎜ − e jt + e − jt ⎟ + ⎜ e j2 t − e− j2 t ⎟ + ⎜ − e j3t + e− j3t ⎟ + .... 1 2 3 ⎝ 1 ⎠ ⎝2 ⎠ ⎝ 3 ⎠


633

⎛ e jt − e − jt = − j4 ⎜ 2 ⎝

⎞ j4 ⎛ e j2t − e − j2 t ⎟+ ⎜ 2 ⎠ 2⎝

⎛ e jt − e − jt = − j2 4 ⎜ ⎝ 2j

⎞ j4 ⎛ e j3t − e − j3t ⎟− ⎜ 2 ⎠ 3⎝

⎞ j2 4 ⎛ e j2t − e − j2 t ⎟+ ⎜ 2j ⎠ 2 ⎝

⎞ ⎟ + .... ⎠

⎞ j2 4 ⎛ e j3t − e− j3t ⎟− ⎜ 2j ⎠ 3 ⎝

⎞ ⎟ + .... by multiplying top and ⎠ bottom by j

4 4 = 4sin t − sin 2t + sin 3t + .... 2 3 i.e.

1 1 1 ⎛ ⎞ f(t) = 4 ⎜ sin t − sin 2t + sin 3t − sin 4t + ... ⎟ 2 3 4 ⎝ ⎠

Hence,

f(t) =

∞

⎛ j2

⎞

∑ ⎜⎝ n cos nπ ⎟⎠ e

n =−∞

jnt

1 1 1 ⎛ ⎞ ≡ 4 ⎜ sin t − sin 2t + sin 3t − sin 4t + ... ⎟ 2 3 4 ⎝ ⎠


634

EXERCISE 250 Page 703 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 ⎡1 ⎤ ⎡1 ⎤ = 10 ⎢ ( e j2 t + e − j2 t ) ⎥ − 12 ⎢ ( e j2 t − e− j2t ) ⎥ ⎣2 ⎦ ⎣2j ⎦ 6 6 = 5e j2 t + 5e − j2 t − e j2 t + e− j2 t j j = 5e j2 t + 5e − j2 t + 6 je j2 t − 6 je− j2 t

(note:

1 −j = j 1

or

1 j = ) −j 1

i.e. v = ( 5 + j6 ) e j2t + ( 5 − j6 ) e − j2t 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 shown in the diagram below.

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 ⎡1 ⎤ ⎡1 ⎤ 3 = 6 ⎢ ( e j t − e − j t ) ⎥ + 4 ⎢ ( e j t + e − jt ) ⎥ = ( e jt − e− jt ) + 2 ( e jt + e− jt ) ⎣2 ⎦ j ⎣2j ⎦ = −3j ( e jt − e − jt ) + 2 ( e jt + e− jt )

i.e. i = ( 2 − j3 ) e j t + ( 2 + j3 ) e − j t


635

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 shown in the diagram below.


636

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