Β© 2006 John Bird. All rights reserved. Published by Elsevier.
HIGHER
ENGINEERING
MATHEMATICS
5TH EDITION
JOHN BIRD
SAMPLE OF WORKED SOLUTIONS
TO EXERCISES
Β© 2006 John Bird. All rights reserved. Published by Elsevier. ii
INTRODUCTION
In βHigher Engineering Mathematics 5th Editionβ are some 1750 further problems arranged
regularly throughout the text within 250 Exercises. A sample of solutions for over 1000 of these
further problems has been prepared in this document. The reader should be able to cope with
the remainder by referring to similar worked problems contained in the text.
CONTENTS
Page Chapter 1 Algebra 1
Chapter 2 Inequalities 13
Chapter 3 Partial fractions 19
Chapter 4 Logarithms and exponential functions 25
Chapter 5 Hyperbolic functions 41
Chapter 6 Arithmetic and geometric progressions 48
Chapter 7 The binomial series 55
Chapter 8 Maclaurinβs 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
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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 Moivreβs 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 ΞΈ/2 substitution 372
Chapter 43 Integration by parts 376
Chapter 44 Reduction formulae 384
Chapter 45 Numerical integration 390
Chapter 46 Solution of first order differential equations by separation of variables 398
Chapter 47 Homogeneous first order differential equations 410
Chapter 48 Linear first order differential equations 417
Chapter 49 Numerical methods for first order differential equations 424
Chapter 50 Second order differential equations of the form 2
2
d y dya b cy 0dxdx
+ + = 435
Chapter 51 Second order differential equations of the form 2
2
d y dya b cy f (x)dxdx
+ + = 441
Chapter 52 Power series methods of solving ordinary differential equations 458
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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 2Ο 595
Chapter 70 Fourier series for a non-periodic functions over period 2Ο 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
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1
CHAPTER 1 ALGEBRA
EXERCISE 1 Page 2
2. Find the value of 2 35pq r when p = 25
, q = -2 and r = -1
( ) ( )2 32 3 2 25 5 2 1 5 4 15 5
pq r β β= β β = Γ Γ Γβ =β ββ β
-8
5. Simplify ( )( )2 3 3 2x y z x yz and evaluate when x = 12
, y = 2 and z = 3
( )( )2 3 3 2x y z x yz = 2 3 3 1 1 2 5 4 3x y z x y z+ + + =
When x = 12
, y = 2 and z = 3, 5 4 3x y z = ( ) ( )5 4 3 3 3
4 35 5 4
1 2 3 3 3 272 32 2 2 2 2β
Γβ β = = = = =β ββ β
13 12
6. Evaluate 3 1 1 12 2 2 2a bc a b c
β ββ ββ ββ ββ ββ β β β
when a = 3, b = 4 and c = 2
1 1 13 1 1 1 21
3 3 1 2 22 3 22 2 2 22
a ba bc a b c a b c a b cc
β β β β+ ββ β β β ββ β + ββ β β β β ββ β
= = =β ββ ββ β β β
When a = 3, b = 4 and c = 2, ( )2 2
2 2
9 23 42 4
a bc
Β±= = = Β± 4 1
2
8. Simplify ( )
( )
1 1 13 2 2 3
3
a b c ab
a b c
ββ ββ ββ β
( )
( )
1 1 13 2 2 1 11 13
1 3 1 1 1 1 18 2 9 1 33 3 32 2 3 13 2 2 3 2 2 6 3 2
3 132 2
a b c aba b c a b a b c a b c
a b c a b c
ββ + ββ β β β β β β β+ β + β β β β ββ β β β β β β β
β β β β β β β β
β ββ ββ β = = =
= 11 1 36 3 2a b c
β or
6 11 3
3
a bc
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2
EXERCISE 2 Page 3
3. Remove the brackets and simplify: ( ) ( ) 24 2 3 5 2 2 3p p q p q qβ‘ β€β β β + +β£ β¦
( ) ( ) 24 2 3 5 2 2 3p p q p q qβ‘ β€β β β + +β£ β¦ = 24 2 15 3 2 4 3p p q p q qβ β β β +β‘ β€β£ β¦
= [ ]24 30 6 4 8 3p p q p q qβ β β β +
= [ ]24 26 11p p qβ β = 24p β 26p + 11q = 11q β 2p
6. Simplify 2 4 6 3 4 5y y y+ Γ· + Γ β
2 4 6 3 4 5y y y+ Γ· + Γ β = 4 22 3 4 5 2 12 56 3
y y y yy y
+ + Γ β = + + β = 2 3 123
yyβ +
8. Simplify 2 3 2 6a ab a b abβ Γ Γ· +
2 3 2 6a ab a b abβ Γ Γ· + =
22 2 2 22 63
6 6a a ba ab ab a ab a a abb b
β Γ + = β + = β + = ab
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3
EXERCISE 3 Page 4
3. Solve the equation: 1 1 03 2 5 3a a
+ =β +
1 1
3 2 5 3a a= β
β + from which, (5a + 3) = -(3a β 2)
i.e. 5a + 3 = -3a + 2
and 5a + 3a = 2 β 3
Thus, 8a = -1 and a = - 18
4. Solve the equation: 3 61
tt= β
β
If 3 61
tt= β
β then ( )3 6 1t t= β β
i.e. 3 6 6t t= β + from which, 6 6 3 3t t t= β =
Hence, if 6 = 3 t then 6 23
t = = and t = 22 = 4
6. Make l the subject of 2 ltg
Ο=
If 2 ltg
Ο= then 2t l
gΟ= and
2
2t l
gΟβ β =β ββ β
from which, 2
2tl gΟ
β β= β ββ β
or 2
24gtlΟ
=
7. Transpose LmL rCRΒ΅
=+
for L
If LmL rCRΒ΅
=+
then ( )m L rCR LΒ΅+ = i.e. mL mrCR LΒ΅+ =
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4
from which, ( )mrCR L mL L mΒ΅ Β΅= β = β and mrCRLmΒ΅
=β
8. Make r the subject of the formula 2
2
11
x ry r
+=
β
If 2
2
11
x ry r
+=
β then ( ) ( )2 21 1x r y rβ = +
from which, 2 2x xr y yrβ = +
and ( )2 2 2x y yr xr r y xβ = + = +
Thus, 2 x yrx yβ
=+
and x yrx y
β ββ= β β+β β
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5
EXERCISE 4 Page 5
2. Solve the simultaneous equations 5 1 32 4 0a bb a= β+ + =
5a + 3b = 1 (1)
a + 2b = -4 (2)
5 Γ (2) gives: 5a + 10b = -20 (3)
(1) β (3) gives: - 7b = 21 from which, b = 217β
= -3
Substituting in (1) gives: 5a + 3(-3) = 1
from which, 5a = 1 + 9 = 10 and a = 102
= 2
3. Solve the simultaneous equations
2 49 (1)5 3 153 5 0 (2)7 2 7
x y
x y
+ =
β + =
15 Γ (1) gives: 3x + 10y = 49 (3)
14 Γ (2) gives: 6x β 7y = -10 (4)
2 Γ (3) gives: 6x + 20y = 98 (5)
(5) β (4) gives: 27y = 108 from which, y = 10827
= 4
Substituting in (3) gives: 3x + 40 = 49 and 3x = 49 β 40 = 9
from which, x = 3
4.(b) Solve the quadratic equation by factorisation: 28 2 15 0x x+ β =
If 28 2 15 0x x+ β = then (4x β 5)(2x + 3) = 0
hence, 4x β 5 = 0 i.e. 4x = 5 i.e. x = 54
and 2x + 3 = 0 i.e. 2x = -3 i.e. x = 32
β
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6
5. Determine the quadratic equation in x whose roots are 2 and -5
If roots are x = 2 and x = -5 then (x β 2)(x + 5) = 0 i.e. 2 2 5 10 0x x xβ + β =
i.e. 2 3 10 0x x+ β =
6.(a) Solve the quadratic equation, correct to 3 decimal places: 22 5 4 0x x+ β =
If 22 5 4 0x x+ β = then 25 5 4(2)( 4) 5 (25 32) 5 572(2) 4 4
xβ‘ β€β Β± β β β Β± + β Β±β£ β¦= = =
Hence, x = 5 574
β + = 0.637 or x = 5 574
β β = -3.137
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7
EXERCISE 5 Page 8
3. Determine ( ) ( )210 11 6 2 3x x x+ β Γ· +
5x - 2 22 3 10 11 6x x x+ + β
210 15x x+ - 4x - 6 - 4x - 6
Hence, 210 11 62 3
x xx+ β+
= 5x - 2
5. Divide ( )3 2 2 33 3x x y xy y+ + + by (x + y)
2 22x xy y+ +
3 2 2 33 3x y x x y xy y+ + + +
3 2x x y+
2 22 3x y xy+ 2 22 2x y xy+
2 3xy y+ 2 3xy y+
Hence, 3 2 2 33 3x x y xy y
x y+ + +
+ = 2 22x xy y+ +
6. Find ( )25 4 ( 1)x x xβ + Γ· β
5x + 4 21 5 4x x xβ β +
25 5x xβ 4x + 4 4x - 4 8
Hence, 25 4
1x x
xβ +β
= 5x + 4 + 81x β
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8
8. Determine ( )4 35 3 2 1 ( 3)x x x x+ β + Γ· β
3 25 18 54 160x x x+ + + 4 33 5 3 2 1x x x xβ + β +
4 35 15x xβ
318x 3 218 54x xβ 254 2x xβ 254 162x xβ 160x + 1 160x - 480 481
Hence, 4 35 3 2 1
3x x x
x+ β +
β = 3 2 4815 18 54 160
3x x x
x+ + + +
β
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9
EXERCISE 6 Page 9
2. Use the factor theorem to factorise 3 2 4 4x x x+ β β
Let f(x) = 3 2 4 4x x x+ β β
If x = 1, f(x) = 1 + 1 β 4 β 4 = -6
x = 2, f(x) = 8 + 4 β 8 β 4 = 0 hence, (x β 2) is a factor
x = 3, f(x) = 27 + 9 β 12 β 4 = 20
x = -1, f(x) = -1 + 1 + 4 β 4 = 0 hence, (x + 1) is a factor
x = -2, f(x) = -8 + 4 + 8 β 4 = 0 hence, (x + 2) is a factor
Thus, 3 2 4 4x x x+ β β = (x + 1)(x + 2)(x β 2)
4. Use the factor theorem to factorise 3 22 16 15x x xβ β +
Let f(x) = 3 22 16 15x x xβ β +
If x = 1, f(x) = 2 β 1 β 16 + 15 = 0 hence, (x β 1) is a factor
x = 2, f(x) = 16 β 4 β 32 +15 = -5
x = 3, f(x) = 54 β 9 β 48 + 15 = 12
x = -1, f(x) = β 1 β 1 + 16 + 15 = 29
x = -2, f(x) = -16 β 4 + 32 + 15 = 27
x = -3, f(x) = -54 β 9 + 48 + 15 = 0 hence, (x + 3) is a factor
3 2 3 2
2
2 16 15 2 16 15( 1)( 3) 2 3
x x x x x xx x x xβ β + β β +
=β + + β
2x - 5 2 3 22 3 2 16 15x x x x x+ β β β +
3 22 4 6x x x+ β 25 10 15x xβ β + 25 10 15x xβ β +
Hence, 3 22 16 15x x xβ β + = (x β 1)(x + 3)(2x β 5)
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10
6. Solve the equation 3 22 2 0x x xβ β + =
Let f(x) = 3 22 2x x xβ β +
If x = 1, f(x) = 1 β 2 β 1 + 2 = 0 hence, (x β 1) is a factor
x = 2, f(x) = 8 β 8 β 2 + 2 = 0 hence, (x β 2) is a factor
x = 3, f(x) = 27 β 18 β 3 + 2 = 8
x = -1, f(x) = -1 β 2 + 1 + 2 = 0 hence, (x + 1) is a factor
Hence, 3 22 2x x xβ β + = (x β 1)(x β 2)(x + 1)
If 3 22 2 0x x xβ β + = then (x β 1)(x β 2)(x + 1) = 0
from which, x = 1, x = 2, or x = -1
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11
EXERCISE 7 Page 11
2. Determine the remainder when 3 26 5x x xβ + β is divided by (a) (x + 2) (b) (x β 3)
(a) Remainder is 3 2ap bp cp d+ + + where a = 1, b = -6, c = 1, d = -5 and p = -2
Hence, remainder = 3 21( 2) 6( 2) 1( 2) 5β β β + β β = -8 β 24 β 2 β 5 = -39
(b) When p = 3, remainder = 3 21(3) 6(3) 1(3) 5β + β = 27 β 54 + 3 β 5 = -29
4. Determine the factors of 3 27 14 8x x x+ + + and hence solve the cubic equation
3 27 14 8 0x x x+ + + =
Remainder is 3 2ap bp cp d+ + + where a = 1, b = 7, c = 14, d = 8
Let p = 1, then remainder = 3 21(1) 7(1) 14(1) 8+ + + = 30
Let p = -1, then remainder = 3 21( 1) 7( 1) 14( 1) 8β + β + β + = -1 + 7 β 14 + 8 = 0, hence (x + 1) is a
factor
Let p = -2, then remainder = 3 21( 2) 7( 2) 14( 2) 8β + β + β + = -8 + 28 β 28 + 8 = 0, hence (x + 2) is a
factor
Let p = -3, then remainder = 3 21( 3) 7( 3) 14( 3) 8β + β + β + = -27 + 63 β 42 + 8 = 2
Let p = -4, then remainder = 3 21( 4) 7( 4) 14( 4) 8β + β + β + = -64 + 112 β 56 + 8 = 0, hence (x + 4) is
a factor
Hence, 3 27 14 8x x x+ + + = (x + 1)(x + 2)(x + 4)
If 3 27 14 8 0x x x+ + + = then (x + 1)(x + 2)(x + 4) = 0
from which, x = -1, x = -2 or x = -4
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12
6. Using the remainder theorem, solve the equation 3 22 7 6 0x x xβ β + =
Remainder is 3 2ap bp cp d+ + + where a = 2, b = -1, c = -7, d = 6
Let p = 1, then remainder = 3 22(1) ( 1)(1) ( 7)(1) 6+ β + β + = 2 - 1 β 7 + 6 = 0, hence (x - 1) is a
factor
Let p = 2, then remainder = 3 22(2) ( 1)(2) ( 7)(2) 6+ β + β + = 16 - 4 β 14 + 6 = 4
Let p = -1, then remainder = 3 22( 1) ( 1)( 1) ( 7)( 1) 6β + β β + β β + = -2 - 1 + 7 + 6 = 10
Let p = -2, then remainder = 3 22( 2) ( 1)( 2) ( 7)( 2) 6β + β β + β β + = -16 - 4 + 14 + 6 = 0, hence (x + 2)
is a factor
Let p = -3, then remainder = 3 22( 3) ( 1)( 3) ( 7)( 3) 6β + β β + β β + = -54 - 9 + 21 + 6 = -36
The third root can be found by division, i.e. ( )
3 2 3 2
2
2 7 6 2 7 61)( 2 2
x x x x x xx x x xβ β + β β +
=β + + β
2x - 3 2 3 22 2 7 6x x x x x+ β β β +
3 22 2 4x x x+ β 23 3 6x xβ β + 23 3 6x xβ β +
Hence, 3 22 7 6x x xβ β + = (x β 1)(x + 2)(2x β 3)
If 3 22 7 6 0x x xβ β + = then (x β 1)(x + 2)(2x β 3) = 0
from which, x = 1, x = -2 or x = 1.5
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13
CHAPTER 2 INEQUALITIES EXERCISE 8 Page 13
2. Solve the following inequalities: (a) 2x > 1.5 (b) x + 2 β₯ 5
(a) 2x > 1.5 i.e. x > 2(1.5) i.e. x > 3
(b) x + 2 β₯ 5 i.e. x β₯ 5 β 2 i.e. x β₯ 3
4. Solve the following inequalities: (a) 7 24
kββ€ 1 (b) 3z + 2 > z + 3
(a) 7 24
kββ€ 1 i.e. 7 β 2k β€ 4 i.e. 7 β 4 β€ 2k i.e. 3 β€ 2k and k β₯ 3
2
(b) 3z + 2 > z + 3 i.e. 3z β z > 3 β 2 i.e. 2z >1 and z > 12
5. Solve the following inequalities: (a) 5 β 2y β€ 9 + y (b) 1 - 6x β€ 5 + 2x
(a) 5 β 2y β€ 9 + y i.e. 5 β 9 β€ y + 2y i.e. -4 β€ 3y i.e. 43
β β€ y or y β₯ 43
β
(b) 1 - 6x β€ 5 + 2x i.e. 1 β 5 β€ 2x + 6x i.e. -4 β€ 8x i.e. 48
β β€ x or x β₯ - 12
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14
EXERCISE 9 Page 14 1. Solve the inequality: 1t + < 4 If 1t + < 4 then -4 < t + 1 < 4 -4 < t + 1 becomes -5 < t i.e. t > -5 t + 1 < 4 becomes t < 3 Hence, -5 < t < 3 3. Solve the inequality: 2 1x β < 4 If 2 1x β < 4 then -4 < 2x β 1 < 4
-4 < 2x β 1 becomes -3 < 2x and 32
β < x
2x β 1 < 4 becomes 2x < 5 and x < 52
Hence, 32
β < x < 52
5. Solve the inequality: 1 kβ β₯ 3 1 kβ β₯ 3 means 1 β k β₯ 3 and 1 β k β€ -3 i.e. 1 β 3 β₯ k and 1 + 3 β€ k i.e. k β€ -2 and k β₯ 4
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15
EXERCISE 10 Page 15
2. Solve the inequality: 2 45
tt+β
> 1
If 2 45
tt+β
> 1 then 2 45
tt+β
- 1 > 0 i.e. 2 4 55 5
t tt t+ β
ββ β
> 0 and ( ) ( )2 4 55
t tt
+ β ββ
> 0
i.e. 95
tt+β
> 0 Hence, either (i) t + 9 > 0 and t β 5 > 0
or (ii) t + 9 < 0 and t β 5 < 0 (i) t > -9 and t > 5 and both inequalities are true when t > 5 (ii) t < -9 and t < 5 and both inequalities are true when t < -9
Hence, 2 45
tt+β
> 1 is true when t > 5 or t < -9
3. Solve the inequality: 3 45
zzβ+
β€ 2
If 3 45
zzβ+
β€ 2 then 3 45
zzβ+
- 2 β€ 0 i.e. 3 4 2( 5)5 ( 5)
z zz zβ +
β+ +
β€ 0 and (3 4) 2( 5)5
z zz
β β ++
β€ 0
i.e. 3 4 2 105
z zz
β β β+
β€ 0 i.e. 145
zzβ+
β€ 0
Hence, either (i) z - 14 β€ 0 and z + 5 > 0 or (ii) z β 14 β₯ 0 and z + 5 < 0 (i) z β€ 14 and z > -5 i.e. -5 < z β€ 14 (ii) z β₯ 14 and z β₯ -5 Both of these inequalities are not possible to satisfy.
Hence, 3 45
zzβ+
β€ 2 is true when -5 < z β€ 14
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16
EXERCISE 11 Page 16 3. Solve the inequality: 22x β₯ 6
22x β₯ 6 i.e. 2x β₯ 3 hence, x β₯ 3 or x β€ - 3 4. Solve the inequality: 23 2k β β€ 10
23 2k β β€ 10 i.e. 23k β€ 12 and 2k β€ 4 Hence, - 4 β€ k β€ 4 i.e. -2 β€ k β€ 2 6. Solve the inequality: ( )21t β β₯ 36 ( )21t β β₯ 36 then (t β 1) β₯ 36 or (t β 1) β€ - 36 i.e. (t β 1) β₯ 6 or (t β 1) β€ -6 i.e. t β₯ 7 or t β€ -5 8. Solve the inequality: ( )24 5k + > 9 ( )24 5k + > 9 then 4k + 5 > 9 or 4k + 5 < - 9 i.e. 4k + 5 > 3 or 4k + 5 < -3 i.e. 4k > -2 or 4k < -8
and k > 12
β or k < -2
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17
EXERCISE 12 Page 17 1. Solve the inequality: 2 6x xβ β > 0
2 6x xβ β > 0 thus (x β 3)(x + 2) > 0 Either (i) x β 3 > 0 and x + 2 > 0 or (ii) x β 3 < 0 and x + 2 < 0 (i) x > 3 and x > -2 i.e. x > 3 (ii) x < 3 and x < -2 i.e. x < -2 3. Solve the inequality: 22 3 2x x+ β < 0
22 3 2x x+ β < 0 thus (2x β 1)(x + 2) < 0 Either (i) 2x β 1 > 0 and x + 2 < 0 or (ii) 2x β 1 < 0 and x + 2 > 0
(i) 2x > 1 i.e. x > 12
and x < -2 both of which are not possible
(ii) 2x < 1 i.e. x < 12
and x > -2 thus -2 < x < 12
5. Solve the inequality: 2 4 4z z+ + β€ 4
2 4 4z z+ + β€ 4 i.e. 2 4z z+ β€ 0 i.e. z(z + 4) β€ 0 Either (i) z β€ 0 and z β₯ -4 i.e. -4 β€ z β€ 0 or (ii) z β₯ 0 and z β€ -4 both of which are not possible 7. Solve the inequality: 2 4 7t tβ β β₯ 0 Since 2 4 7t tβ β β₯ 0 then ( )22t β - 7 β 4 β₯ 0
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18
i.e. ( )22t β β₯ 11 and t β 2 β₯ 11 or t β 2 β€ - 11 thus, t β₯ ( )2 11+ or t β€ ( )2 11β
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19
CHAPTER 3 PARTIAL FRACTIONS
EXERCISE 13 Page 20
2. Resolve 2
4( 4)2 3
xx x
ββ β
into partial fractions.
Let 2
4( 4) 4 16 ( 3) ( 1)2 3 ( 1)( 3) ( 1) ( 3) ( 1)( 3)
x x A B A x B xx x x x x x x x
β β β + +β‘ = + =
β β + β + β + β
Hence, 4x β 16 = A(x β 3) + B(x + 1)
If x = -1, -20 = -4A from which, A = 5
If x = 3, 12 β 16 = 4B from which, B = -1
Hence, 2
4( 4) 5 12 3 ( 1) ( 3)
xx x x x
β= β
β β + β
4. Resolve 23(2 8 1)
( 4)( 1)(2 1)x x
x x xβ β
+ + β into partial fractions.
Let
23(2 8 1) ( 1)(2 1) ( 4)(2 1) ( 4)( 1)( 4)( 1)(2 1) ( 4) ( 1) (2 1) ( 4)( 1)(2 1)
x x A B C A x x B x x C x xx x x x x x x x x
β β + β + + β + + +β‘ + + =
+ + β + + β + + β Hence, 26 24 3x xβ β = A(x + 1)(2x β 1) + B(x + 4)(2x β 1) + C(x + 4)(x + 1)
If x = -4, 96 + 96 -3 = A(-3)(-9) from which, 189 = 27A and A = 7
If x = -1, 6 + 24 -3 = B(3)(-3) from which, 27 = -9B and B = -3
If x = 0.5, 1.5 - 12 -3 = C(4.5)(1.5) from which, -13.5 = 6.75C and C = -2
Hence, 23(2 8 1) 7 3 2
( 4)( 1)(2 1) ( 4) ( 1) (2 1)x x
x x x x x xβ β
= β β+ + β + + β
5. Resolve 2
2
9 86
x xx x+ ++ β
into partial fractions.
Since the numerator is of the same degree as the denominator, division is firstly required.
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20
1 2 26 9 8x x x x+ β + +
2 6x x+ β 8x + 14
Hence, 2
2 2
9 8 8 1416 6
x x xx x x x+ + +
= ++ β + β
Let 2
8 14 8 14 ( 2) ( 3)6 ( 3)( 2) ( 3) ( 2) ( 3)( 2)
x x A B A x B xx x x x x x x x
+ + β + += β‘ + =
+ β + β + β + β
Hence, 8x + 14 = A(x β 2) + B(x + 3)
If x = -3, -24 + 14 = -5A form which, -10 = -5A and A = 2
If x = 2, 16 + 14 = 5B from which, 30 = 5B and B = 6
Hence, 2
2
9 8 2 616 ( 3) ( 2)
x xx x x x+ +
= + ++ β + β
7. Resolve 3 23 2 16 20( 2)( 2)
x x xx xβ β +β +
into partial fractions.
3x - 2 2 3 24 3 2 16 20x x x xβ β β +
33 12x xβ
22 4 20x xβ β + 22 8xβ +
- 4x + 12
Hence, 3 2
2
3 2 16 20 12 43 2( 2)( 2) 4
x x x xxx x xβ β + β
β‘ β +β + β
Let 2
12 4 12 4 ( 2) ( 2)4 ( 2)( 2) ( 2) ( 2) ( 2)( 2)x x A B A x B x
x x x x x x xβ β + + β
= β‘ + =β β + β + β +
Hence, 12 β 4x = A(x + 2) + B(x - 2)
If x = 2, 4 = 4A from which, A = 1
If x = -2 20 = -4B from which, B = -5
Hence, 3 23 2 16 20 1 53 2( 2)( 2) ( 2) ( 2)
x x x xx x x xβ β +
β‘ β + ββ + β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
21
EXERCISE 14 PAGE 22
2. Resolve 2
2
7 3( 3)
x xx x+ +
+ into partial fractions.
Let 2 2
2 2 2
7 3 ( )( 3) ( 3)( 3) ( 3) ( 3)
x x A B C A x x B x Cxx x x x x x x+ + + + + +
β‘ + + =+ + +
Hence, 2x + 7x + 3 = A(x)(x + 3) + B (x + 3) + C 2x
If x = 0 3 = 3B from which, B = 1
If x = -3 9 β 21 + 3 = 9C i.e. -9 = 9C f rom which, C = -1
Equating 2x coefficients: 1 = A + C from which, A = 2
Hence, 2
2 2
7 3 2 1 1( 3) ( 3)
x xx x x x x+ +
= + β+ +
4. Resolve 2
2
18 21( 5)( 2)
x xx x+ ββ +
into partial fractions.
Let 2 2
2 2 2
18 21 ( 2) ( 5)( 2) ( 5)( 5)( 2) ( 5) ( 2) ( 2) ( 5)( 2)
x x A B C A x B x x C xx x x x x x x+ β + + β + + β
β‘ + + =β + β + + β +
Hence, 2 218 21 ( 2) ( 5)( 2) ( 5)x x A x B x x C x+ β = + + β + + β
If x = 5 18 + 105 β 25 = 49A i.e. 98 = 49A from which, A = 2
If x = -2 18 β 42 β 4 = -7C i.e. -28 = -7C from which, C = 4
Equating 2x coefficients: -1 = A + B from which, B = -3
Hence, 2
2 2
18 21 2 3 4( 5)( 2) ( 5) ( 2) ( 2)
x xx x x x x+ β
= β +β + β + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
22
EXERCISE 15 PAGE 23
1. Resolve ( )( )
2
2
137 2
x xx x
β β+ β
into partial fractions.
Let ( )( ) ( )
( )( )
22
2 2 2
( )( 2) 713( 2)7 2 7 7 ( 2)
Ax B x C xx x Ax B Cxx x x x x
+ β + +β β +β‘ + =
β+ β + + β
Hence, ( )2 213 ( )( 2) 7x x Ax B x C xβ β = + β + +
If x = 2, 4 β 2 β13 = 11C i.e. -11 = 11C from which, C = -1
Equating 2x coefficients: 1 = A + C from which, A = 2
Equating constant terms: -13 = -2B + 7C = -2B β 7 i.e. 2B = 13 β 7 = 6 from which, B = 3
Hence, ( ) ( ) ( )2
2 2
13 2 3 1( 2)7 2 7
x x xxx x x
β β += β
β+ β +
4. Resolve ( )
3 2
2 2
4 20 7( 1) 8
x x xx x+ + ββ +
into partial fractions.
Let ( ) ( )
( ) ( )( )
2 2 23 2
22 2 2 2 2
( 1) 8 8 ( )( 1)4 20 7( 1) ( 1)( 1) 8 8 ( 1) 8
A x x B x Cx D xx x x A B Cx Dx xx x x x x
β + + + + + β+ + β +β‘ + + =
β ββ + + β +
Hence, ( ) ( )3 2 2 2 24 20 7 ( 1) 8 8 ( )( 1)x x x A x x B x Cx D x+ + β = β + + + + + β
( ) ( )2 2 2( 1) 8 8 ( )( 2 1)A x x B x Cx D x x= β + + + + + β +
If x = 1, 1 + 4 + 20 β 7 = 9B i.e. 18 = 9B from which, B = 2
Equating 3x coefficients: 1 = A + C (1β)
Equating 2x coefficients: 4 = -A + B β 2C + D (2β)
Equating x coefficients: 20 = 8A + C β 2D (3β)
Since B = 2, A + C = 1 (1)
-A β 2C + D = 2 (2)
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
23
8A + C β 2D = 20 (3)
2 Γ (2) gives: -2A β 4C + 2D = 4 (4)
(3) + (4) gives: 6A β 3C = 24 (5)
3 Γ (1) gives: 3A + 3C = 3 (6)
(5) + (6) gives: 9A = 27 from which, A = 3
From (1): 3 + C = 1 from which, C = -2
From (2): -3 + 4 + D = 2 from which, D = 1
Hence, ( ) ( )3 2
22 2 2
4 20 7 3 2 1 2( 1) ( 1)( 1) 8 8
x x x xx xx x x
+ + β β= + +
β ββ + +
5. When solving the differential equation 2
22 6 10 20 td d e
dt dtΞΈ ΞΈ ΞΈβ β = β by Laplace transforms, for
given boundary conditions, the following expression for Ξ ΞΈ results:
Ξ ΞΈ = ( )( )
3 2
2
394 42 4022 6 10
s s s
s s s s
β + β
β β +
Show that the expression can be resolved into partial fractions to give:
Ξ ( ) ( )2
2 1 5 32 2 2 6 10
ss s s s
ΞΈ β= β +
β β +
Let ( )( ) ( )( ) ( )
( )
3 2
2 2
2 2
2
394 42 402
( 2)2 6 10 6 10
( 2) 6 10 ( ) 6 10 ( )( )( 2)
( 2) 6 10
s s s A B Cs Ds ss s s s s s
A s s s B s s s Cs D s s
s s s s
β + β +β‘ + +
ββ β + β +
β β + + β + + + β=
β β +
Hence, ( ) ( )3 2 2 2394 42 40 ( 2) 6 10 ( ) 6 10 ( )( )( 2)2
s s s A s s s B s s s Cs D s sβ + β = β β + + β + + + β
( ) ( )3 2 3 2 28 2 20 6 10 ( )( 2 )A s s s B s s s Cs D s s= β β β + β + + + β
If s = 0, -40 = A(-20) from which, A = 2
If s = 2, 32 β 78 + 84 β 40 = B (8 β 24 + 20) i.e. -2 = 4B from which, B = 12
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
24
Equating 3s coefficients: 4 = A + B + C i.e. 4 = 2 - 12
+ C from which, C = 52
Equating 2s coefficients: 392
β = -8A β 6B β 2C + D i.e. 392
β = -16 + 3 β 5 + D
from which, D = 32
β
Hence, ( )( ) ( )
3 2
2 2
39 1 5 34 42 40 22 2 2 2( 2)2 6 10 6 10
s s s s
s ss s s s s s
β + β β ββ‘ + +
ββ β + β +
i.e. Ξ ( ) ( )2
2 1 5 32 2 2 6 10
ss s s s
ΞΈ β= β +
β β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
25
CHAPTER 4 LOGARITHMS AND EXPONENTIAL FUNCTIONS
EXERCISE 16 Page 26 2. Evaluate: 2log 16 Let x = 2log 16 then 42 16 2x = = from which, x = 4 Hence, 2log 16 = 4
4. Evaluate: 21log8
Let x = 21log8
then 33
1 12 28 2
x β= = = from which, x = -3
Hence, 21log8
= -3
7. Evaluate: 4log 8 Let x = 4log 8 then 4 8x = i.e. ( )2 32 2
x= i.e. 2 32 2x =
from which, 2x = 3 and x = 32
Hence, 4log 8 = 1 12
11. Solve the equation: 41log 22
x = β
If 41log 22
x = β then x = 52
5 552
1 1 14244
β= = = Β± =
132
Β±
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
26
12. Solve the equation: lg 2x = β
If lg x = -2 then 10log 2x = β and x = 22
11010
β = = 1100
or 0.01
16. Write in terms of log 2, log 3 and log 5 to any base: 416 5log
27β βΓβ ββ ββ β
1
1 144 44 3 4 34 4
3
16 5 2 5log log log 2 5 3 log 2 log5 log327 3
β β
β ββ β β βΓ Γβ β= = Γ Γ = + +β β β ββ β β β β β β ββ β β β
= 4 log 2 + 14
log 5 β 3 log 3
17. Write in terms of log 2, log 3 and log 5 to any base: 4
34
125 16log81
β βΓβ ββ ββ β
4
34
125 16log81
β βΓβ ββ ββ β
= ( )3
3 3 3 33
5 2log log 5 2 3 log 5 log 2 log 33
β ββ βΓ= Γ Γ = + +β β
β β
= 3 log 5 + log 2 β 3 log 3 19. Simplify: log 64 + log 32 β log 128
log 64 + log 32 β log 128 = ( )6 5
6 5 7 6 5 7 47
2 2log 2 log 2 log 2 log log 2 log 22
+ ββ βΓ+ β = = =β β
β β = 4 log 2
20. Evaluate:
1 1log16 log82 3
log 4
β
1 1log16 log82 3
log 4
β =
( ) ( )2
1 1114 32 3 232
2 2 2 2
2loglog 2 log 2 2log16 log8 log 2 log 2 log 2log 2 log 2 log 2 log 2 2log 2
β ββ βββ β β β = = = = = 1
2
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
27
22. Solve the equation: 3log 2 log log16 logt t tβ = +
3log 2 log log16 logt t tβ = +
i.e. ( )32log log 16t t
tβ β
=β ββ β
i.e. ( ) ( )2log 2 log 16t t=
from which, 22 16t t= i.e. 22 16 0t tβ = i.e. 2t(t β 8) = 0
Hence, t = 8 (note that t = 0 is not a valid solution to the equation)
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
28
EXERCISE 17 Page 27 1. Solve the equation 3 6.4x = correct to 4 significant figures If 3 6.4x = then 10 10log 3 log 6.4x = and
x = 10
10
log 6.4 0.80617997... 1.689675...log 3 0.47712125...
= = = 1.690, correct to 4 significant figures.
3. Solve the equation 1 2 12 3x xβ β= correct to 4 significant figures If 1 2 12 3x xβ β= then 10 10( 1) log 2 (2 1) log 3x xβ = β i.e. 10 10 10 10log 2 log 2 2 log 3 log 3x xβ = β i.e. ( )10 10 10 10 10 10log 3 log 2 2 log 3 log 2 2log 3 log 2x x xβ = β = β
Hence, x = 10 10
10 10
log 3 log 22log 3 log 2
ββ
= 0.2696, correct to 4 significant figures.
5. Solve the equation 25.28 4.2x= correct to 4 significant figures If 25.28 4.2x= then 10 10log 25.28 log 4.2x=
from which, x = 10
10
log 25.28log 4.2
= 2.251, correct to 4 significant figures.
6. Solve the equation 2 1 24 5x xβ += correct to 4 significant figures If 2 1 24 5x xβ += then 10 10(2 1) log 4 ( 2) log 5β = +x x i.e. 10 10 10 102 log 4 log 4 log 5 2log 5x xβ = + i.e. 10 10 10 102 log 4 log 5 2log 5 log 4x xβ = + i.e. 10 10 10 10(2 log 4 log 5) 2log 5 log 4x β = +
from which, x = 10 10
10 10
2 log 5 log 42log 4 log 5
+β
= 3.959, correct to 4 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
29
8. Solve the equation 0.027 3.26x = correct to 4 significant figures If 0.027 3.26x = then 10 10log 0.027 log 3.26x =
from which, x = 10
10
log 3.26log 0.027
= -0.3272, correct to 4 significant figures.
9. The decibel gain n of an amplifier is given by: n = 210
1
P10logP
β ββ ββ β
where 1P is the power input
and 2P is the power output. Find the power gain 2
1
PP
when n = 25 decibels.
When n = 25 then: 25 = 210
1
P10logP
β ββ ββ β
from which, 210
1
P25 log10 P
β β= β β
β β i.e. 2.5 = 2
101
PlogP
β ββ ββ β
Thus, 2.52
1
P 10P
= i.e. power gain, 2
1
PP
= 316.2
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
30
EXERCISE 18 Page 29
3. Evaluate, correct to 5 significant figures: ( )1.72952.1127 2.1127
2.1347 3.6817
4 15.6823( ) ( ) ( )2
ee ea b ce e
ββ
β
ββ
Using a calculator:
2.1347
5.6823( )aeβ = 48.04106, correct to 5 decimal places.
2.1127 2.1127
( )2
e ebββ = 4.07482, correct to 5 decimal places.
( )1.7295
3.6817
4 1( )
ec
e
β β = -0.08286, correct to 5 decimal places.
4. The length of a bar, l , at a temperature ΞΈ is given by 0l l eΞ±ΞΈ= , where l and Ξ± are constants.
Evaluate l , correct to 4 significant figures, when 0l = 2.587, ΞΈ = 321.7 and Ξ± = 41.771 10βΓ
Using a calculator, ( )4321.7 1.771 100 (2.587)l l e eΞ±ΞΈ
βΓ Γ= = = 2.739, correct to 4 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
31
EXERCISE 19 Page 31 2. Use the power series for xe to determine, correct to 4 significant figures, (a) 2e (b) 0.3eβ and
check your result by using a calculator.
(a) 2 3 4
1 ....2! 3! 4!
x x x xe x= + + + + +
When x = 2, 2e = 2 3 4 5 62 2 2 2 21 2 ...
2! 3! 4! 5! 6!+ + + + + + +
= 1 + 2 + 2 + 1.33333 + 0.66666 + 0.26666 + 0.08888 + 0.02540 + 0.00635
+ 0.00141 + 0.00028 + 0.00005 + β¦
= 7.389, correct to 4 significant figures, which may be checked with a calculator.
(b) When x = -0.3, 0.3eβ = 2 3 4 5( 0.3) ( 0.3) ( 0.3) ( 0.3)1 0.3 ...
2! 3! 4! 5!β β β β
β + + + + +
= 1 β 0.3 + 0.04500 β 0.00450 + 0.00034 β 0.00002 + β¦
= 0.7408, correct to 4 significant figures.
3. Expand (1 β 2x) 2xe to six terms.
(1 β 2x) 2xe = (1 β 2x)2 3 4(2 ) (2 ) (2 )1 2 ...
2! 3! 4!x x xx
β β+ + + + +β β
β β
= (1 β 2x) 2 3 44 21 2 23 3
x x x xβ β+ + + +β ββ β
= 2 3 4 2 3 44 2 81 2 2 2 4 4 ...3 3 3
x x x x x x x x+ + + + β β β β β
= 2 3 481 2 23
x x xβ β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
32
EXERCISE 20 Page 32 1. Plot a graph of y = 3 0.2xe over the range x = -3 to x = 3. Hence determine the value of y when
x = 1.4 and the value of x when y = 4.5
Figure 1
From Figure 1, when x = 1.4, y = 3.95 and when y = 4.5, x = 2.05 4. The rate at which a body cools is given by 0.05250 teΞΈ β= where the excess temperature of a body
above its surroundings at time t minutes is CΞΈΒ° . 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 CΒ°
From Figure 2 on page 33,
(a) after t = 25 minutes, temperature ΞΈ = 70 CΒ°
(b) when the temperature is 195 CΒ° , time t = 5 minutes
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
33
Figure 2
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
34
EXERCISE 21 Page 34 2. Evaluate, correct to 5 significant figures:
1.76 0.1629
1.41
2.946ln 5 ln 4.8629 ln 2.4711( ) ( ) ( )lg10 2ln 0.00165 5.173
e ea b cβ β
Using a calculator,
1.76
1.41
2.946ln( )lg10
ea = (2.946)(1.76)(1.41)
= 3.6773, correct to 5 significant figures.
0.16295( )2 ln 0.00165
ebβ
= -0.33154, correct to 5 significant figures.
ln 4.8629 ln 2.4711( )5.173
c β = 0.13087, correct to 5 significant figures.
4. Solve, correct to 4 significant figures: 1.77.83 2.91 β= xe
If 1.77.83 2.91 β= xe then 1.7 7.832.91
eβ = and 1.7 7.83ln ln2.91
xeβ β β= β ββ β
i.e. -1.7x = 7.83ln2.91
β ββ ββ β
and x = 1 7.83ln1.7 2.91
β ββ β ββ β
= -0.5822, correct to 4 significant figures.
5. Solve, correct to 4 significant figures: 216 24 1t
eββ β
= ββ ββ β
If 216 24 1t
eββ β
= ββ ββ β
then 216 124
t
eβ
= β
from which, 2 16124
t
eβ= β
and 16ln 12 24t β ββ = ββ β
β β
and t = 162ln 124
β ββ ββ ββ β
= 2.197, correct to 4 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
35
7. Solve, correct to 4 significant figures: 1.593.72ln 2.43x
β β =β ββ β
If 1.593.72ln 2.43x
β β =β ββ β
then 1.59 2.43ln3.72
β β =β ββ β x
from which, 2.433.721.59 e
x
β ββ ββ β =
and x = 2.433.72
2.433.72
1.59 1.59ee
β βββ ββ β
β ββ ββ β
= = 0.8274, correct to 4 significant figures.
8. The work done in an isothermal expansion of a gas from pressure 1p to 2p is given by:
10
2
pw w lnp
β β= β β
β β
If the initial pressure 1p = 7.0 kPa, calculate the final pressure 2p if 0w 3w=
If 0w 3w= then 03w = 10
2
pw lnp
β ββ ββ β
i.e. 3 = 1
2
plnp
β ββ ββ β
and 3 1
2 2
p 7000ep p
= =
from which, final pressure, 32 3
7000p 7000ee
β= = = 348.5 Pa
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
36
EXERCISE 22 Page 37
2. The voltage drop, v volts, across an inductor L henrys at time t seconds is given by 200RtLv e
β
= ,
where R = 150 Ξ© and L = 12.5 310βΓ H. Determine (a) the voltage when t = 6160 10βΓ s, and
(b) the time for the voltage to reach 85 V.
(a) Voltage 200RtLv e
β
= = ( )( )6
3
150 160 101.9212.5 10200 200
β
β
Γβ
βΓ =e e = 29.32 volts
(b) When v = 85 V, 3150
12.5 1085 200 ββΓ=
t
e from which, 3150
12.5 1085200
ββΓ=
t
e
and 3
150 85ln12.5 10 200
tβ
β ββ = β βΓ β β
Thus, time t = 312.5 10 85ln
150 200
βΓ β ββ β ββ β
= 671.31 10 sβΓ
4. A belt is in contact with a pulley for a sector ΞΈ = 1.12 radians and the coefficient of friction
between the two surfaces is Β΅ = 0.26. Determine the tension on the taut side of the belt, T
newtons, when tension on the slack side 0T = 22.7 newtons, given that these quantities are
related by the law 0T T e¡θ= . Determine also the value of θ when T = 28.0 newtons.
Tension 0T T e¡θ= = ( )( )0.26 1.12 0.291222.7 22.7e e= = 30.4 N
When T = 28.0 N, 28.0 = 22.7 0.26e ΞΈ from which, 0.2628.022.7
e ΞΈ=
Thus, 28.00.26 ln22.7
ΞΈ β β= β ββ β
and ΞΈ 1 28.0ln0.26 22.7
β β= β ββ β
= 0.807 rad
5. The instantaneous current i at time t is given by: 10t
CRi eβ
= when a capacitor is being charged.
The capacitance C is 67 10βΓ F and the resistance R is 60.3 10Γ ohms. Determine:
(a) the instantaneous current when t is 2.5 seconds, and
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
37
(b) the time for the instantaneous current to fall to 5 amperes.
Sketch a curve of current against time from t = 0 to t = 6 seconds.
(a) Current, i = 6 62.5
7 10 0.3 1010ββ
Γ Γ Γe = 3.04 A
(b) When i = 5, 2.15 10β
=t
e from which, 2.1510
β=
t
e
Thus, 5ln10 2.1
tβ β = ββ ββ β
and time, t = (2.1) ln 0.5β = 1.46 s
Time t 0 1 2 3 4 5 6
Current i 10 6.21 3.86 2.40 1.49 0.92 0.57 A graph of current against time is shown in Figure 3.
Figure 3
7. The current i flowing in a capacitor at time t is given by: 12.5 1t
CRi eββ β
= ββ ββ β
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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
38
(a) Current, 12.5 1ββ β
= ββ ββ β
tCRi e = 6 3
0.520 10 30 1012.5 1 ββΓ Γ Γ
β βββ ββ β
β β e = 7.07 A
(b) When i = 10 A, 0.610 12.5 1ββ β
= ββ ββ β
t
e from which, 0.610 112.5
β= β
t
e
Thus, 0.6 10112.5
β= β
t
e and 10ln 10.6 12.5t β ββ = ββ β
β β
i.e. time, t = 100.6 ln 112.5
β ββ ββ ββ β
= 0.966 s
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
39
EXERCISE 23 Page 40 2. At particular times, t minutes, measurements are made of the temperature, CΞΈΒ° , of a cooling
liquid and the following results are obtained:
Temperature CΞΈΒ° 92.2 55.9 33.9 20.6 12.5
Time t minutes 10 20 30 40 50
Prove that the quantities follow a law of the form 0kteΞΈ ΞΈ= , where 0ΞΈ and k are constants, and
determine the approximate value of 0ΞΈ and k.
Since 0
kteΞΈ ΞΈ= then ( )0 0ln ln ln lnk t k te eΞΈ ΞΈ ΞΈ= = +
i.e. 0ln lnktΞΈ ΞΈ= +
where gradient = k and intercept on vertical axis = 0lnΞΈ
Using log-linear graph paper, a graph of lnΞΈ against t is shown in Figure 4
Figure 4 Since the graph is a straight line the quantities do follow a law of the form 0
kteΞΈ ΞΈ=
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
40
Gradient, k = ln 92.2 ln12.510 50
ABBC
β=
β = -0.05
At point A in Figure 4, 92.2 CΞΈ = Β° , t = 10 min Substituting in 0
kteΞΈ ΞΈ= gives: ( )( )0.05 10092.2 eΞΈ β=
from which, 0ΞΈ0.5
0.5
92.2 92.2eeβ= = = 152
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
41
CHAPTER 5 HYPERBOLIC FUNCTIONS
EXERCISE 24 Page 42 1.(a) Evaluate sh 0.64 correct to 4 significant figures.
(a) sh 0.64 = ( )0.64 0.6412
e eββ = 0.6846, correct to 4 significant figures.
Alternatively, using a scientific calculator, using hyp, sin 0.64 = 0.6846 2.(b) Evaluate ch 2.4625 correct to 4 significant figures.
(b) ch 2.4625 = ( )2.4625 2.462512
e eβ+ = 5.910, correct to 4 significant figures.
Alternatively, using a scientific calculator, using hyp, cos 0.72 = 5.910 3.(a) Evaluate th 0.65 correct to 4 significant figures.
(a) th 0.65 = 0.65 0.65
0.65 0.65
1.39349505...2.4375866...
e ee e
β
β
β=
+ = 0.5717, correct to 4 significant figures.
Alternatively, using a scientific calculator, using hyp, tan 0.65 = 0.5717 4.(b) Evaluate cosech 3.12 correct to 4 significant figures.
(b) cosech 3.12 = 13.12sh
= 0.08849, correct to 4 significant figures, using a calculator.
5.(a) Evaluate sech 0.39 correct to 4 significant figures.
(a) sech 0.39 = 10.39ch
= 0.9285, correct to 4 significant figures, using a calculator.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
42
6.(b) Evaluate coth 1.843 correct to 4 significant figures.
(b) coth 1.843 = 11.843th
= 1.051, correct to 4 significant figures, using a calculator.
7. A telegraph wire hangs so that its shape is described by 5050yy ch= . Evaluate, correct to 4
significant figures, the value of y when x = 25.
When x = 0.25, 5050yy ch= = 2550
50ch = 50 ch 0.50 = 56.38, correct to 4 significant figures.
8. The length l of a heavy cable hanging under gravity is given by l = 2 ( / 2 )c sh L c . Find the value
of l when c = 40 and L = 30.
l = 2 ( / 2 )c sh L c = 2(40) sh 30 3802(40) 8
shβ β β β=β β β ββ β β β
= 30.71
9. 2 0.55 tanh(6.3 / )V L d L= is a formula for velocity V of waves over the bottom of shallow water,
where d is the depth, and L is the wavelength. If d = 8.0 and L = 96, calculate the value of V.
2 0.55 tanh(6.3 / )V L d L= = (6.3)(8.0)0.55(96) tanh
96β‘ β€β’ β₯β£ β¦
= 52.8 tanh 0.525 = 25.425829β¦
Hence, V = 25.425829... = 5.042
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
43
EXERCISE 25 Page 46 2. Prove the following identities: (a) coth x β‘ 2 cosech 2x + th x (b) ch 2ΞΈ - 1 β‘ 2 sh 2 ΞΈ
(a) R.H.S. = 2 cosech 2x + th x = 22 2 1 1
2 2sh x sh x sh x sh x
sh x ch x sh xch x ch x sh x ch x ch x sh xch x+
+ = + = + =
= 2
cothch x ch x xsh xch x sh x
= = = L.H.S
(b) R.H.S. = 2 sh 2 ΞΈ = ( )( )2
2 22 122 4 2
e e e e e e e e e e e eΞΈ ΞΈ
ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈβ
β β β β ββ ββ β‘ β€= β β = β β +β β β£ β¦β β
= 2 2
2 0 0 2 2 21 1 222 2 2 2 2
e ee e e e e eΞΈ ΞΈ
ΞΈ ΞΈ ΞΈ ΞΈβ
β ββ‘ β€ β‘ β€β β + = β + = + ββ£ β¦ β£ β¦
= 2 2
1 2 12
e e chΞΈ ΞΈ
ΞΈβ+
β = β = L.H.S.
4. Prove the following identities: (a) sh(A + B) β‘ sh A ch B + ch A sh B
(b) 2 2
42 2
1 tanh2 cothsh x ch x x
ch x x+ β
β‘
(a) R.H.S. = sh A ch B + ch A sh B = 2 2 2 2
A A B B A A B Be e e e e e e eβ β β ββ ββ β β ββ ββ + + β+β ββ β β ββ β
β β β β β β β β
= 14
A B A B A B A B A B A B A B A Be e e e e e e e+ β β + β β + β β + β ββ‘ β€+ β β + β + ββ£ β¦
= ( )( ) ( )1 2 2 ( )
4 2
+ β +β ++ ββ‘ β€β = = +β£ β¦
A B A BA BA B e ee e sh A B = L.H.S
(b) L.H.S. = 2 2 2 2
2 2 22
2
12 coth
2
sh x ch x sh x sh xch x x ch xch x
sh x
+ β +=
β ββ ββ β
since 2 21ch x sh xβ =
= 2 2 4
42 2 4
2 tanh2
sh x sh x sh x xch x ch x ch x
β β= =β β
β β = R.H.S.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
44
5. Given 6 2x xP e Qe ch x sh xββ β‘ β , find P and Q
6 2x xP e Qe ch x sh xββ β‘ β = ( ) ( )6 2 32 2
x x x xx x x xe e e e e e e e
β ββ ββ β β β+ β
β = + β ββ β β ββ β β β
= 3 3x x x xe e e eβ β+ β + i.e. 2 4x x x xP e Qe e eβ ββ = + from which, P = 2 and Q = -4 6. If 5 4x xe e Ash x Bch xββ β‘ + , find A and B
5 42 2 2 2 2 2
x x x xx x x x x xe e e e A A B Be e Ash x B ch x A B e e e e
β ββ β ββ β β ββ +
β β‘ + = + = β + +β β β ββ β β β
= 2 2 2 2
x xA B A Be eββ β β β+ β ββ β β ββ β β β
i.e. 5 42 2
x x x xA B A Be e e eβ β+ ββ β β ββ = ββ β β ββ β β β
Hence, 52
A B+= i.e. A + B = 10 (1)
and 42
A Bβ= i.e. A β B = 8 (2)
(1) + (2) gives: 2A = 18 from which, A = 9 From (1) B = 1
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
45
EXERCISE 26 Page 48 2. Solve 2 ch x = 3, correct to 4 decimal places.
2 ch x = 3 from which, 32
ch x =
i.e. 32 2
x xe eβ+= or 3 0x xe eβ+ β =
i.e. ( )2
3 0x x x xe e e eβ+ β = or ( )23 1 0x xe eβ + =
Hence, ( ) ( ) ( )( )
( )
23 3 4 1 1 3 52 1 2
xeβ‘ β€β β Β± β β Β±β£ β¦= = = 2.61803 or 0.381966
Thus, x = ln 2.61803 or x = ln 0.381966 i.e. x = 0.9624 or x = -0.9624 i.e. x = Β± 0.9624 3. Solve 3.5 sh x + 2.5 ch x = 0, correct to 4 decimal places. 3.5 sh x + 2.5 ch x = 0
i.e. 3.5 2.5 02 2
x x x xe e e eβ ββ β β ββ ++ =β β β β
β β β β
i.e. 1.75 1.75 1.25 1.25 0β ββ + + =x x x xe e e e and 3 0.5 0x xe eββ = or 3 0.5x xe eβ=
i.e. 0.53
x
x
eeβ = i.e. 2 0.5
3xe =
Hence, 2x = ln 0.53
from which, x = 1 0.5ln2 3
= -0.8959
5. Solve 4 th x - 1 = 0, correct to 4 decimal places.
4 th x - 1 = 0 i.e. 4 1x x
x x
e ee e
β
β
β ββ=β β+β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
46
i.e. ( )4 x x x xe e e eβ ββ = + and 4 4 0x x x xe e e eβ ββ β β = Hence, 3 5 0x xe eββ = and 3 5x xe eβ=
Thus, 53
x
x
eeβ = from which, 2 5
3xe =
i.e. 2x = ln 53
and x = 1 5ln2 3
= 0.2554
6. A chain hangs so that its shape is of the form 56 ( / 56)y ch x= . Determine, correct to 4
significant figures, (a) the value of y when x is 35, and (b) the value of x when y is 62.35
(a) When x = 35, 3556 ( / 56) 5656
y ch x chβ β= = β ββ β
= 67.30, using a calculator.
(b) When, y = 62.35, then 62.35 = 56 ( / 56)ch x
Thus, 62.3556 56
xch= or 56 56 62.35
2 56
x x
e eβ
+=
i.e. 56 56 62.35256
x x
e eβ β β+ = β β
β β = 2.22679
Thus, 2
56 56 56 562.22679 0x x x x
e e e eββ β β ββ β
+ β =β β β ββ ββ β β β β β
i.e. 2
56 562.22679 1 0x x
e eβ β
β + =β ββ β
from which, ( ) ( ) ( )( )
( )
2
562.22679 2.22679 4 1 1 2.22679 0.95859...
2 1 2
β‘ β€β β Β± β β Β±β£ β¦= =x
e
= 1.60293 or 0.623857
Hence, ln1.6029356x= or ln 0.623857
56x=
i.e. x = 56 ln 1.60293 = 26.42 or x = 56 ln 0.623857 = -26.42, which is not possible.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
47
EXERCISE 27 Page 49 3. Expand the following as a power series as far as the term in 5x : (a) sh 3x (b) ch 2x
(a) sh 3x = ( ) ( ) ( )3 53 33 ...
3! 5!x x
x + + + = 3 527 3 3 3 3 336 5 4 3 2 1
x x xΓ Γ Γ Γ+ +
Γ Γ Γ Γ
= 3 59 8132 40
x x x+ + as far as the term in 5x
(b) ch 2x = ( ) ( )2 42 21 ...
2! 4!x x
+ + + = 2 4161 2 ...24
x x+ + +
= 2 421 23
x x+ + as far as the term in 4x
4. Prove the identity: 3 57 3126 120
sh shΞΈ ΞΈ ΞΈ ΞΈ ΞΈβ β‘ + + as far as the term in 5ΞΈ only.
L.H.S. = ( ) ( )3 5 3 52 22 2 ... ...
3! 5! 3! 5!sh sh
ΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈβ β β β
β β‘ + + + β + + +β β β ββ β β β β β
= 3 5 3 58 32 1 12 ... ...6 120 6 120
ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈβ β β β+ + + β + + +β β β ββ β β β
= ( ) 3 58 1 32 126 6 120 120
ΞΈ ΞΈ ΞΈ ΞΈβ β β ββ + β + ββ β β ββ β β β
as far as the term in 5ΞΈ only
= 3 57 316 120
ΞΈ ΞΈ ΞΈ+ + = R.H.S.
5. Prove the identity: 2 3 4 5
2 12 2 8 24 384 1920β β‘ β + β + β +sh chΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈΞΈ as far as the term in 5ΞΈ only.
L.H.S. = ( ) ( ) ( ) ( )3 5 2 4/ 2 / 2 / 2 / 22 2 ... 1 ...
2 2 2 3! 5! 2! 4!sh ch
ΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈβ β β ββ β‘ + + + β + + +β β β β
β β β ββ β β β
= 3 5 2 45
2 2 1 1148 2 (120) 8 (16)(24)
β β β β+ + β + +β β β β
β β β β ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ
= 2 3 4 5
18 24 384 1920ΞΈ ΞΈ ΞΈ ΞΈΞΈβ + β + + + as far as the term in 5ΞΈ only
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
48
CHAPTER 6 ARITHMETIC AND GEOMETRIC PROGRESSIONS
EXERCISE 28 Page 52 1. Find the 11th term of the series 8, 14, 20, 26, β¦ The 11th term of the series 8, 14, 20, 26,β¦ is given by: a + (n β 1)d where a = 8, n = 11 and d = 6 Hence, the 11th term is: 8 + (11 β 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 nβth term of an arithmetic progression is: a + (n β 1)d The 7th term is: a + 6d = 29 (1) The 11th term is: a + 10d = 54 (2)
(2) β (1) gives: 4d = 25 from which, d = 254
Substituting in (1) gives: a + 2564
β ββ ββ β
= 29
i.e. a + 37.5 = 29 from which, a = 29 β 37.5 = -8.5
Hence, the 16th term is: -8.5 + (16 β 1) 254
β ββ ββ β
= -8.5 + 93.75 = 85.25
5. Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13.6, β¦ 29 = 7 + (n β 1)(2.2) from which, 29 β 7 = 2.2(n β 1)
i.e. 222.2
= n β 1 i.e. 10 = n β 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, β¦, 32
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
49
In the series: 6.5, 8.0, 9.5, 11.0, β¦, 32, a = 6.5 and d = 1.5 The nβth term is 32, hence, a + (n β 1)d = 32 i.e. 6.5 + (n - 1)(1.5) = 32
from which, 32 β 6.5 = (n β 1)(1.5) and 25.51.5
= n β 1 i.e. 17 = n - 1 and n = 18
Sum of series, [ ] [ ] [ ]nn 18S 2a (n 1)d 2(6.5) (18 1)(1.5) 9 13 25.52 2
= + β = + β = + = 346.5
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
50
EXERCISE 29 Page 53
2. Three numbers are in arithmetic progression. Their sum is 9 and their product is 20 14
.
Determine the three numbers. Let the three numbers be (a β d), a and (a + d) Thus, (a β d) + a + ( a + d) = 9 i.e. 3a = 9 and a = 3 Also, a(a β d)(a + d) = 20.25 Since, a = 3, then 3 ( )29 dβ = 20.25
i.e. 2 20.259 d 6.753
β = =
and 9 β 6.75 = 2d from which, 2d = 2.25 and d = 2.25 = 1.5 Hence, the three numbers are: (a β d) = 3 β 1.5 = 1.5, a = 3 and (a + d) = 3 + 1.5 = 4.5 4. Find the number of terms of the series 5, 8, 11,β¦ of which the sum is 1025
Sum of n terms is given by: [ ]nnS 2a (n 1)d2
= + β
i.e. 1025 = [ ]n 2(5) (n 1)(3)2
+ β
i.e. [ ]2 1025 n 10 3(n 1)Γ = + β Hence, 2050 = [ ] [ ] 2n 10 3n 3 n 7 3n 7n 3n+ β = + = + i.e. 23n 7n 2050 0+ β =
This is a quadratic equation, hence n = 27 7 4(3)( 2050) 7 24649 7 157
2(3) 6 6β Β± β β β Β± β Β±
= =
i.e. number of terms, n = 25 (the negative answer having no meaning)
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
51
6. The first, tenth and last terms of an arithmetic progression are 9, 40.5 and 425.5 respectively. Find (a) the number of terms, (b) the sum of all the terms, and (c) the 70th term. (a) a = 9 and the 10th term is: a + (10 β 1)d = 40.5 i.e. 9 + 9d = 40.5 and 9d = 40.5 β 9 = 31.5
hence 31.5d 3.59
= =
Last term is given by: a + (n β 1)d i.e. 9 + (n β 1)(3.5) = 425.5 i.e. (n β 1)(3.5) = 425.5 β 9 = 416.5
and n β 1 = 416.5 1193.5
=
Hence, the number of terms, n = 120
(b) Sum of all the terms, [ ] [ ] [ ]nn 120S 2a (n 1)d 2(9) (120 1)(3.5) 60 18 416.52 2
= + β = + β = +
= 26070 (c) The 70th term is: a + (n β 1)d = 9 + (70 β 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 Β£30 for drilling the first metre with an increase in cost of Β£2 per metre for each succeeding metre. The series is: 30, 32, 34, β¦ to 80 terms, i.e. a = 30, d = 2 and n = 80
Thus, total cost, [ ] [ ] [ ]nn 80S 2a (n 1)d 2(30) (80 1)(2) 40 60 158 40(218)2 2
= + β = + β = + = = Β£8720
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
52
EXERCISE 30 Page 55 1. Find the 10th term of the series 5, 10, 20, 40, β¦ The 10th term of the series 5, 10, 20, 40, β¦ is given by: n 1a r β where a = 5, r = 2 and n = 10 i.e. the 10th term = n 1a r β = ( )( )10 1 95 2 (5)(2) 5(512)β = = = 2560 3. The first term of a geometric progression is 4 and the 6th term is 128. Determine the 8th and 11th terms.
The 6th term is given by: 5a r 128= i.e. 54r 128= and 5 128r 324
= =
Thus, r = 5 32 = 2 Hence, the 8th term is: n 1a r β = 8 1 74(2) 4(2) 4(128)β = = = 512 and the 11th term is: 11 1 104(2) 4(2) 4(1024)β = = = 4096
4. Find the sum of the first 7 terms of the series 2, 5, 12 12
,β¦ (correct to 4 significant figures).
Common ratio, r = ar 5a 2= = 2.5 (Also,
2ar 12.5ar 5
= = 2.5)
Sum of 7 terms, ( )( )
( )( )
( )n 7
n
a r 1 2 2.5 1 2 610.35 1S
r 1 2.5 1 1.5β β β
= = =β β
= 812.5, correct to 4 significant
figures.
6. Find the sum to infinity of the series 1 1 52 , 1 , ,.....2 4 8
β
The series is a G.P. where r = 1.252.5
β = -0.5 and a = 2.5
Hence, sum to infinity, a 2.5 2.5 5S1 r 1 ( 0.5) 1.5 3β = = = =β β β
= 1 23
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
53
EXERCISE 31 Page 57 1. In a geometric progression the 5th term is 9 times the 3rd term, and the sum of the 6th and 7th terms is 1944. Determine (a) the common ratio, (b) the first term, and (c) the sum of the 4th to 10th terms inclusive. (a) The 5th term of a geometric progression is: 4ar and the 3rd term is: 2ar
Hence, 4ar = 9 2ar from which, 4
2 9rr
= i.e. 2 9r =
from which, the common ratio, r = 3 (b) The 6th term is 5ar and the 7th term is 6ar Hence, 5ar + 6ar = 1944 Since r = 3, 243a + 729a = 1944
i.e. 972a = 1944 and first term, a = 1944972
= 2
(c) Sum of the 4th to 10th terms inclusive is given by:
( )( )
( )( )
( ) ( )( )
10 3 10 3
10 3
1 1 2 3 1 2 3 11 1 (3 1) 3 1
a r a rS S
r rβ β β β
β = β = ββ β β β
= ( ) ( )10 3 10 33 1 3 1 3 3 59049 27β β β = β = β = 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 Γ 55 million, population after 2 years = ( )20.976 55Γ million Hence, population after 5 years = ( )50.976 55Γ = 48.71 million 6. If Β£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 Β£750.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
54
(a) First term, a = Β£250, common ratio, r = 1.06
Hence, value after 15 years = ( )1515 (250) 1.06ar = = Β£599.14
(b) When Β£750 is reached, 750 = nar
i.e. 750 = 250 ( )1.06 n
and 750250
= 1.06n i.e. 3 = 1.06n
Taking logarithms gives: lg 3 = ( )lg 1.06 lg1.06n n=
from which, n = lg3lg1.06
= 18.85
Hence, it will take 19 years to reach more than Β£750 7. A drilling machine is to have 8 speeds ranging from 100 rev/min to 1000 rev/min. If the speeds form a geometric progression determine their values, each correct to the nearest whole number. First term, a = 100 rev/min
The 8th term is given by: 8 1 1000ar β = from which, 7 1000 10100
r = = and 7 10r = = 1.3895
Hence, 1st term is 100 rev/min 2nd term is ar = (100)(1.3895) = 138.95 3rd term is 2 2(100)(1.3895) 193.07ar = = 4th term is 3 3(100)(1.3895) 268.27ar = = 5th term is 4 4(100)(1.3895) 372.76ar = = 6th term is 5 5(100)(1.3895) 517.96ar = = 7th term is 6 6(100)(1.3895) 719.70ar = = 8th term is 7 7(100)(1.3895) 1000ar = = Hence, correct to the nearest whole numbers, the eight speeds are: 100, 139, 193, 268, 373, 518, 720 and 1000 rev/min
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
55
CHAPTER 7 THE BINOMIAL SERIES
EXERCISE 32 Page 59 1. Expand ( )52 3+a b using Pascalβs triangle From page 58 of textbook, ( )5 5 4 3 2 2 3 4 5a x a 5a x 10a x 10a x 5a x x+ = + + + + + Thus, replacing a with 2a and x with 3b gives: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )5 5 4 3 2 2 3 4 52 3 2 5 2 3 10 2 3 10 2 3 5 2 3 3+ = + + + + +a b a a b a b a b a b b = + + + + +5 4 3 2 2 3 4 532a 240a b 720a b 1080a b 810ab 243b
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
56
EXERCISE 33 Page 61 1. Use the binomial theorem to expand ( )42+a x
( )42+a x = ( ) ( )( ) ( ) ( )( )( ) ( ) ( )2 3 44 3 24 3 4 3 2a 4a 2x a 2x a 2x 2x
2! 3!+ + + +
= 4 3 2 2 3 4a 8a x 24a x 32a x 16x+ + + + 3. Expand ( )42 3βx y
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( ) ( )4 4 3 2 2 3 44 3 4 3 22 3 2 4 2 3 2 3 2 3 3
2! 3!β = + β + β + β + βx y x x y x y x y y
= 4 3 2 2 3 416 96 216 216 81β + β +x x y x y xy y
4. Determine the expansion of 522β β+β β
β β x
x
( ) ( ) ( )( ) ( ) ( )( )( ) ( )
( )( )( )( ) ( )
5 2 35 4 3 2
4 5
5 4 5 4 32 2 2 22 2 5 2 2 22! 3!
5 4 3 2 2 224!
β β β β β β β β+ = + + +β β β β β β β ββ β β β β β β β
β β β β+ +β β β ββ β β β
x x x x xx x x x
xx x
= 5 33 5
320 160 3232 160 320+ + + + +x x xx x x
6. Determine the sixth term of 13
33
β β+β ββ β
qp
The 6th term of ( )+ na x is ( )( )( )
( 1) 11 2 ..... ( 1)1 !
β β ββ β ββ
n r rn n n to r termsa x
r
Hence, the 6th term of 13
33
β β+β ββ β
qp is: ( )( )( )( )( ) ( )5
13 513 12 11 10 93
5! 3β β ββ ββ β
qp = ( )5
81287 33
β ββ ββ β
qp
= 8 534749 p q
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
57
9. Use the binomial theorem to determine, correct to 5 significant figures: (a) ( )70.98 (b) ( )92.01 (a)
( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( ) ( )( )( )( ) ( )7 7 2 3 47 6 7 6 5 7 6 5 40.98 1 0.02 1 7 0.02 0.02 0.02 0.02 ..
2! 3! 4!= β = + β + β + β + β +
= 1 β 0.14 + 0.0084 β 0.00028 + 0.0000056 - β¦ = 0.86813, correct to 5 significant figures.
(b) ( ) ( ) ( )9
9 9 99 90.012.01 2 0.01 2 1 2 1 0.0052
β β= + = + = +β ββ β
= ( )( ) ( )( )( )9 2 39 8 9 8 72 1 9(0.005) (0.005) (0.005) ...
2! 3!β‘ β€+ + + +β’ β₯
β£ β¦
= ( )9 92 1 0.045 0.0009 0.0000105 ... 2 (1.0459105)+ + + + = = 535.51, correct to 5 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
58
EXERCISE 34 Page 63
2. Expand ( )2
11+ x
in ascending powers of x as far as the term in 3x , using the binomial theorem.
Sate the limits of x for which the series is valid.
( )( ) ( )( ) ( ) ( )( )( ) ( )2 2 3
2
2 3 2 3 41 1 1 2 ...2! 3!1
β β β β β β= + = β + + +
+x x x x
x
= 2 21 2 3 4 ...β + β +x x x and x < 1
3. Expand ( )3
12+ x
in ascending powers of x as far as the term in 3x , using the binomial theorem.
Sate the limits of x for which the series is valid.
( )3
12+ x
= ( ) ( )( ) ( )( )( )3 2 33 3 3 3 4 3 4 5
2 2 1 2 1 3 ...2 2 2! 2 3! 2
ββ β β
β‘ β€β β β β ββ β β β β β β β+ = + = β + + +β’ β₯β β β β β β β ββ β β β β β β β β’ β₯β£ β¦
x x x xx
= 2 33
1 3 3 51 ...2 2 2 4
β‘ β€β + β +β’ β₯β£ β¦x x x
= 2 31 3 3 51 ...8 2 2 4β‘ β€β + β +β’ β₯β£ β¦
x x x
The series is true provided 2x < 1 i.e. x < 2
5. Expand 11 3+ x
in ascending powers of x as far as the term in 3x , using the binomial theorem.
Sate the limits of x for which the series is valid.
11 3+ x
= ( ) ( ) ( )1 2 32
1 3 1 3 51 2 2 2 2 21 3 1 (3 ) 3 3 ...2 2! 3!
β
β ββ β β ββ ββ ββ β β β ββ ββ β β ββ ββ ββ β β β β β β β β β β β + = + β + +β ββ β
x x x x
= 2 33 27 13512 8 16
β + βx x x as far as the term in 3x
The series is true provided 3x < 1 i.e. x < 13
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
59
7. When x is very small show that : (a) ( ) ( )2
1 5121 1
β +β β
xx x
(b) ( )( )4
1 21 10
1 3β
β +β
xx
x (c)
( )3
1 5 19161 2
+β +
β
x xx
(a) ( ) ( )
( ) ( ) ( )122
2
1 1 1 1 2 121 1
β β β β= β β β + +β ββ β β β
xx x xx x
β 1 22
+ +x x ignoring the 2x term and above
β 512
+ x
(b) ( )( )
( )( ) ( )( )44
1 21 2 1 3 1 2 1 12
1 3ββ
= β β β β +β
xx x x x
x
β 1 + 12x β 2x ignoring the 2x term and above β 1 + 10x
(c) ( )
( ) ( )1 12 3
3
1 5 5 21 5 1 2 1 12 31 2
β+ β ββ β= + β β + +β ββ ββ β β β β
x x x x xx
β 2 513 2
+ +x x ignoring the 2x term and above
β 1916
+ x 5 2 15 4 192 3 6 6
+β β+ = =β ββ β
9. Express the following as power series in ascending powers of x as far as the term in 2x . State in each case the range of x for which the series is valid.
(a) 11ββ β
β β+β β xx
(b) ( ) ( )
( )
23
2
1 1 3
1
β β
+
x x
x
(a) ( ) ( ) ( )1 1 22 2
1 1 1 31 2 2 2 21 1 1 11 2 2! 2 2!
β
β‘ β€ β‘ β€β ββ β β ββ ββ β ββ ββ β β ββ ββ’ β₯ β’ β₯ββ β β β β β β β β β β’ β₯ β’ β₯= β + β β + β β +β β+β β β’ β₯ β’ β₯β’ β₯ β’ β₯β£ β¦ β£ β¦
x x xx x xx
as far as the term
in 2x
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
60
= 2 2 2 2 23 31 1 1
2 8 2 8 2 8 2 4 8β ββ ββ β β + β β + β + ββ ββ β
β β β β
x x x x x x x x x
β 2
12
β +xx as far as the term in 2x
The series is valid if x < 1
(b) ( ) ( )
( )( )( ) ( )
23 122 23
2
1 1 31 1 3 1
1
ββ β= + β +
+
x xx x x
x
β ( ) ( )2
2
2 13 31 1 2 3 .. 1 ..
2! 2
β‘ β€β ββ βββ ββ ββ’ β₯ β ββ β β β β’ β₯+ β + β + β +β ββ’ β₯ β β β’ β₯β£ β¦
xx x x as far as the term in 2x
β ( )( )2
21 1 2 12
β β+ β β ββ β
β β
xx x x
β ( )2
2 21 2 2 12
β ββ β + β ββ β
β β
xx x x x as far as the term in 2x
β ( )2 2
2 21 3 1 1 32 2
β ββ β β β β β ββ β
β β
x xx x x x neglecting 3x terms and above
β 2712
β βx x
The series is valid provided 3x < 1 i.e. x < 13
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
61
EXERCISE 35 Page 65
2. Kinetic energy is given by 212
mv . Determine the approximate change in the kinetic energy
when mass m is increased by 2.5% and the velocity v is reduced by 3% New mass = 1.025 m = (1 + 0.025)m and new velocity = 0.97 v = (1 β 0.03)v
New kinetic energy = 2 2 21 1(1 0.025) (1 0.03) (1 0.025)(1 0.06)2 2
+ β β + βm v mv
β 2 21 1(1 0.025 0.06) (0.965)2 2
+ β =mv mv
i.e. 96.5% of the original kinetic energy. Thus, the approximate change in kinetic energy is a reduction of 3.5% 3. An error of +1.5% was made when measuring the radius of a sphere. Ignoring the products of small quantities determine the approximate error in calculating (a) the volume, and (b) the surface area.
(a) Volume of sphere, V = 343
rΟ .
New volume = 3 3 3 3 34 4 4 4(1.015 ) (1 0.015) [1 3(0.015)] (1 0.045)3 3 3 3
= + β + = +r r r rΟ Ο Ο Ο
= 1.045 V i.e. the volume has increased by 4.5% (b) Surface area of sphere, A = 24 rΟ New surface area = ( )2 2 2 24 1 0.015 4 [1 2(0.015)] 4 (1 0.03)+ β + = +r r rΟ Ο Ο = 1.03 A i.e. the surface area has increased by 3%
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
62
6. The electric field strength H due to a magnet of length 2 l and moment M at a point on its axis
distance x from the centre is given by: ( ) ( )2 2
1 12
β§ β«βͺ βͺ= ββ¨ β¬β +βͺ βͺβ© β
MHl x l x l
Show that if l is very small compared with x, then H β 3
2Mx
( ) ( )
2 2
2 2 2 2 22 2
1 1 1 1 1 12 2 2
1 1
M M M l lHl l x l x xx l x l l lx x
x x
β β
β§ β«βͺ βͺβ§ β« β§ β«βͺ βͺ βͺ βͺ βͺ βͺβ β β β= β = β = β β +β¨ β¬ β¨ β¬ β¨ β¬β β β β
β β β β β + β β β β βͺ βͺβͺ βͺ βͺ βͺ β© ββ© β β +β β β ββͺ βͺβ β β β β© β
2 2
2 2 41 12 2M l l M lx l x x x l x
β§ β«β β β β β§ β«β + β β ββ¨ β¬ β¨ β¬β β β ββ β β β β© ββ© β
β 3
2Mx
7. The shear stress Ο in a shaft of diameter D under a torque T is given by: Ο = 3
k TDΟ
. Determine
the approximate percentage error in calculating Ο if T is measured 3% too small and D is 1.5% too large. New value of T = (1 β 0.03)T and new value of D = (1 + 0.015)D
Hence, new shear stress = ( )( ) 33 3 3
(1 0.03) 1 0.03 1 0.015(1 0.015)k T kT
D DΟ Οββ β‘ β€= β +β£ β¦+
( )( ) [ ]3 31 0.03 1 0.045 1 0.03 0.045kT kTD DΟ Ο
β β β β β ββ‘ β€β£ β¦
( )3 1 0.075 (1 0.075)kTD
ΟΟ
β β = β
i.e. the new torque has decreased by 7.5% 9. In a series electrical circuit containing inductance L and capacitance C the resonant frequency is
given by: 12
=rf LCΟ. 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
63
New value of inductance = (1 + 0.026)L and new value of capacitance = (1 β 0.008)C
Hence, new resonant frequency = 1 12 (1 0.026) (1 0.008) 2 (1 0.026) (1 0.008)L C L CΟ Ο
=+ β + β
= ( )1 1 112 2 22
1 1 0.026 (1 0.008)2
L CΟ
β β ββ+ β
( )( ) ( )1 12 2
1 11 0.013 1 0.004 1 0.013 0.00422 LCL C ΟΟ
β β + β β +β‘ β€β£ β¦
( )1 0.009rfβ β i.e. the new resonant frequency is 0.9% smaller
10. The viscosity Ξ· of a liquid is given by: Ξ· =4k rlΞ½
, where k is a constant. If there is an error in r
of +2%, in Ξ½ of +4% and l of -3%, what is the resultant error in Ξ·? New value of r = (1 + 0.02)r, new value of Ξ½ = (1 + 0.04) Ξ½ and new value of l = (1 β 0.03) l
Hence, new value of viscosity = ( ) ( ) ( )4 4 4
4 1 1(1 0.02) 1 0.02 1 0.04 1 0.03(1 0.04) (1 0.03)
k r k rl lΞ½ Ξ½
β β+ β‘ β€= + + ββ£ β¦+ β
( )( )( )4
1 0.08 1 0.04 1 0.03k rlΞ½
β + β +β‘ β€β£ β¦
( ) ( )4
1 0.08 0.04 0.03 1 0.07β + β + β +k r
lΞ·
Ξ½
i.e. the viscosity increases by 7%
12. The flow of water through a pipe is given by: G = ( )53d HL
. If d decreases by 2% and H by
1%, use the binomial theorem to estimate the decrease in G. New value of d = (1 - 0.02)d and new value of H = (1 β 0.01)H
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
64
Hence, new value of G = ( )5 1 5 5 1 1
5 5 52 2 2 2 2 2
1 12 2
3 3 3 (1 0.02) (1 0.01)β β= =
d H d H d HL L L
5 15 2 2
12
3 5 11 (0.02) 1 (0.01)2 2
β‘ β€β ββ ββ β ββ ββ ββ’ β₯β β β β β£ β¦
d H
L
( )( ) ( )5 5(3 ) (3 )1 0.05 1 0.005 1 0.05 0.005β β β β β ββ‘ β€β£ β¦
d H d HL L
(1 0.055)Gβ β i.e. the flow G has decreased by 5.5%
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
65
CHAPTER 8 MACLAURINβS SERIES
EXERCISE 36 Page 70 1. Determine the first four terms of the power series for sin 2x using Maclaurinβs series. Let f(x) = sin 2x f(0) = sin 0 = 0 f β²(x) = 2 cos 2x f β²(0) = 2 cos 0 = 2 f β²β²(x) = -4 sin 2x f β²β²(0) = -4 sin 0 = 0 f β²β²β²(x) = -8 cos 2x f β²β²β²(0) = -8 cos 0 = - 8 ( )ivf x = 16 sin 2x (0)ivf = 16 sin 0 = 0 ( )vf x = 32 cos 2x (0)vf = 32 cos 0 = 32 ( )vif x = -64 sin 2x (0)vif = -64 sin 0 = 0 ( )viif x = -128 cos 2x (0)viif = -128 cos 0 = - 128
Maclaurinβs series states: f(x) = f(0) + x f β²(0) + 2
2!x f β²β²(0) +
3
3!x f β²β²β²(0) + β¦.
=2 3 4 5 6 7
0 (2) (0) ( 8) (0) (32) (0) ( 128)2! 3! 4! 5! 6! 7!x x x x x xx+ + + β + + + + β
i.e. f(x) = 3 5 78 32 1282
6 120 5040x x xx β + β
i.e. sin 2x = 3 5 74 4 823 15 315
x x x xβ + β
3. Use Maclaurinβs series to determine the first three terms of the power series for ( )ln 1 xe+ Let f(x) = ( )ln 1 xe+ f(0) = ( )0ln 1 e+ = ln 2
f β²(x) = 1
x
x
ee+
f β²(0) = 0
0
11 2
ee
=+
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
66
f β²β²(x) = ( ) ( )
( )2
1
1
x x x x
x
e e e e
e
+ β
+ f β²β²(0) =
( ) ( )( )
0 0 0 0
2 20
1 2 1 12 41
e e e e
e
+ β β= =
+
Maclaurinβs series states: f(x) = f(0) + x f β²(0) + 2
2!x f β²β²(0) +
3
3!x f β²β²β²(0) + β¦.
=21 1ln 2 ...
2 2! 4xx β β β β+ + +β β β β
β β β β
i.e. ( )ln 1 xe+ = 2
ln 22 8x x
+ +
6. Develop, as far as the term in 4x , the power series for sec 2x. Let f(x) = sec 2x f(0) = sec 0 = 1 f β²(x) = 2 sec 2x tan 2x f β²(0) = 0 f β²β²(x) = ( )( ) ( )( )22sec 2 2sec 2 tan 2 4sec 2 tan 2x x x x x+ = 2 2 2 24sec 2 sec 2 tan 2 4sec 2 sec 2 sec 2 1β‘ β€ β‘ β€+ = + ββ£ β¦ β£ β¦x x x x x x = 2 34sec 2 2sec 2 1 8sec 2 4sec 2x x x xβ‘ β€β = ββ£ β¦ f β²β²(0) = 8 β 4 = 4 f β²β²β²(x) = ( )224sec 2 2sec 2 tan 2 8sec 2 tan 2x x x x xβ = 348sec 2 tan 2 8sec 2 tan 2x x x xβ f β²β²β²(0) = 0 ( )ivf x = ( )( ) ( )( )3 2 248sec 2 2sec 2 tan 2 144sec 2 (2sec 2 tan 2 )x x x x x xβ‘ β€+β£ β¦
( )( ) ( )( )28sec 2 2sec 2 tan 2 16sec 2 tan 2x x x x xβ‘ β€β +β£ β¦
(0)ivf = 96 + 0 β 16 β 0 = 80
Maclaurinβs series states: f(x) = f(0) + x f β²(0) + 2
2!x f β²β²(0) +
3
3!x f β²β²β²(0) + β¦.
= 2 3 4
1 (0) (4) (0) (80)2! 3! 4!x x xx+ + + +
= 2 4801 224
x x+ +
i.e. sec 2x = 2 4101 23
x x+ + as far as the term in 4x
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
67
7. Expand 2 cos3e ΞΈ ΞΈ as far as the term in 2ΞΈ using Maclaurinβs series. Let f(ΞΈ) = 2 cos3e ΞΈ ΞΈ f(0) = 0 cos 0e = 1 f β²(ΞΈ) = ( )( ) ( )( )2 23sin 3 cos3 2e eΞΈ ΞΈΞΈ ΞΈβ +
= ( )2 2cos3 3sin 3e ΞΈ ΞΈ ΞΈβ f β²(0) = ( )0 2cos 0 3sin 0e β = 2 f β²β²(ΞΈ) = ( )( ) ( )( )2 26sin 3 9cos3 2cos3 3sin 3 2e eΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈβ β + β f β²β²(0) = - 9 + 4 = -5
Maclaurinβs series states: f(ΞΈ) = f(0) + ΞΈ f β²(0) + 2
2!ΞΈ f β²β²(0) +
3
3!ΞΈ f β²β²β²(0) + β¦.
= 2
1 (2) ( 5)2!ΞΈΞΈ+ + β
i.e. 2 25cos 3 1 22
e ΞΈ ΞΈ ΞΈ ΞΈ= + β as far as the term in 2ΞΈ
8. Determine the first three terms of the series for 2sin x by applying Maclaurinβs theorem. Let f(x) = 2sin x f(0) = sin 0 = 0 f β²(x) = 2 sin x cos x f β²(0) = 2 sin 0 cos 0 = 0 f β²β²(x) = (2 sin x)(-sin x) + (cos x)(2 cos x) = ( )2 2 2 22sin 2cos 2 cos sin 2cos 2x x x x xβ + = β = f β²β²(0) = 2 cos 0 = 2 f β²β²β²(x) = -4 sin 2x f β²β²β²(0) = -4 sin 0 = 0 ( )ivf x = -8 cos 2x (0)ivf = -8 cos 0 = -8 ( )vf x = 16 sin 2x (0)vf = 16 sin 0 = 0 ( )vif x = 32 cos 2x (0)vif = 32 cos 0 = 32
Maclaurinβs series states: f(x) = f(0) + x f β²(0) + 2
2!x f β²β²(0) +
3
3!x f β²β²β²(0) + β¦.
= 2 3 4 5 6
0 (0) (2) (0) ( 8) (0) (32) ...2! 3! 4! 5! 6!x x x x xx+ + + + β + + +
i.e. 2 2 4 61 2sin3 45
x x x x= β + to three terms
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
68
EXERCISE 37 Page 72
1. Evaluate 0.6 sin
0.23e dΞΈ ΞΈβ« , correct to 3 decimal places, using Maclaurinβs series.
Let f(ΞΈ) = sin3e ΞΈ f(0) = sin 03e = 3 f β²(ΞΈ) = ( )( )sin3 cose ΞΈ ΞΈ f β²(0) = ( )( )sin 03 cos 0e = 3 f β²β²(ΞΈ) = ( )( ) ( )( )sin sin3 sin cos 3 cose eΞΈ ΞΈΞΈ ΞΈ ΞΈβ + = ( )sin 23 cos sine ΞΈ ΞΈ ΞΈβ f β²β²(0) = ( )sin 0 23 cos 0 sin 0e β = 3 f β²β²β²(ΞΈ) = ( )( ) ( )( )sin 2 sin3 2cos sin cos cos sin 3 cose eΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈβ β + β f β²β²β²(0) = ( )( ) ( )( )sin 0 2 sin 03 2cos 0sin 0 cos 0 cos 0 sin 0 3 cos 0e eβ β + β = (3)(-1) + (1)(3) = 0
Maclaurinβs series states: f(ΞΈ) = f(0) + ΞΈ f β²(0) + 2
2!ΞΈ f β²β²(0) +
3
3!ΞΈ f β²β²β²(0) + β¦.
= 233 3 02
ΞΈ ΞΈ+ + +
Hence, 0.6 sin
0.23e dΞΈ ΞΈβ« =
0.630.6 2 2
0.20.2
3 33 3 32 2 2
d ΞΈΞΈ ΞΈ ΞΈ ΞΈ ΞΈβ‘ β€β β+ + = + +β β β’ β₯β β β£ β¦
β«
= ( ) ( ) ( ) ( )3 32 20.6 0.23 33(0.6) 0.6 3(0.2) 0.2
2 2 2 2
β‘ β€ β‘ β€+ + β + +β’ β₯ β’ β₯
β’ β₯ β’ β₯β£ β¦ β£ β¦
= (1.8 + 0.54 + 0.108) β (0.6 + 0.06 + 0.004) = 2.448 β 0.664 = 1.784, correct to 3 decimal places 4. Use Maclaurinβs theorem to expand ln( 1)x x + as a power series. Hence evaluate, correct to 3
decimal places, 0.5
0ln(1 )x x dx+β«
From page XX, 2 3 4
ln( 1) ....2 3 4x x xx x+ = β + β +
Hence, 1 2 3 4 50.5 0.52
0 0ln(1 ) ...
2 3 4 5x x x xx x dx x x dx
β β+ = β + β + +β β
β β β« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
69
5 7 9 11 133 2 2 2 2 20.52
0...
2 3 4 5 6x x x x xx
β ββ β= β + β + ββ ββ ββ β
β«
=
0.55 7 9 11 13 152 2 2 2 2 2
0
...5 7 9 11 13 152 3 4 5 62 2 2 2 2 2
x x x x x xβ‘ β€β’ β₯β’ β₯β + β + β +
β β β β β β β β β ββ’ β₯β β β β β β β β β ββ’ β₯β β β β β β β β β β β£ β¦
= ( ) ( ) ( ) [ ]11 135 7 92 22 2 2
2 1 2 1 20.5 0.5 0.5 (0.5) (0.5) .. 05 7 27 22 65β‘ β€
β + β + + ββ’ β₯β£ β¦
= 0.07071 β 0.0126 + 0.00327 β 0.00100 + 0.00034 - β¦ = 0.06072β¦ = 0.061, correct to 3 decimal places.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
70
EXERCISE 38 Page 74
1. Determine 3
31
2 1lim2 3 5x
x xx xβ
β§ β«β +β¨ β¬+ ββ© β
3
31
2 1lim2 3 5x
x xx xβ
β§ β«β +β¨ β¬+ ββ© β
= 2
21
3 2 3 2lim6 3 6 3x
xxβ
β§ β«β β=β¨ β¬+ +β© β
= 19
4. Determine 2
20
sin 3lim3x
x xx xβ
β§ β«ββ¨ β¬+β© β
2
20
sin 3lim3x
x xx xβ
β§ β«ββ¨ β¬+β© β
= 0
2 3cos3 3lim3 2 3x
x xxβ
β ββ§ β« =β¨ β¬+β© β = -1
5. Determine 30
sin coslimΞΈ
ΞΈ ΞΈ ΞΈΞΈβ
ββ§ β«β¨ β¬β© β
30
sin coslimΞΈ
ΞΈ ΞΈ ΞΈΞΈβ
ββ§ β«β¨ β¬β© β
= ( )( )
2 20 0
cos sin cos sinlim lim3 3ΞΈ ΞΈ
ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈβ β
β§ β«β β +β‘ β€βͺ βͺ β§ β«β£ β¦ =β¨ β¬ β¨ β¬β© ββͺ βͺβ© β
= 0 0
cos sin ( sin ) cos (1) cos 1 1 2lim lim6 6 6 6ΞΈ ΞΈ
ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ ΞΈΞΈβ β
+ β + + +β§ β« β§ β«= = =β¨ β¬ β¨ β¬β© β β© β
= 13
7. Determine 30
sinh sinlimx
x xxβ
ββ§ β«β¨ β¬β© β
30
sinh sinlimx
x xxβ
ββ§ β«β¨ β¬β© β
= 20 0 0
cosh cos sinh sin cosh coslim lim lim3 6 6x x x
x x x x x xx xβ β β
β + +β§ β« β§ β« β§ β«= =β¨ β¬ β¨ β¬ β¨ β¬β© β β© β β© β
= 1 1 26 6+
= = 13
8. Determine 2
sin 1limln sinΟΞΈ
ΞΈΞΈβ
ββ§ β«β¨ β¬β© β
2
sin 1limln sinΟΞΈ
ΞΈΞΈβ
ββ§ β«β¨ β¬β© β
= 2 2
coslim lim sin sin1 2cos
sinΟ ΟΞΈ ΞΈ
ΞΈ ΟΞΈΞΈ
ΞΈβ β
β§ β«βͺ βͺβͺ βͺ = =β¨ β¬β ββͺ βͺβ ββͺ βͺβ β β© β
= 1
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
71
CHAPTER 9 SOLVING EQUATIONS BY ITERATIVE METHODS
EXERCISE 39 Page 80 1. Find the positive root of the equation 2 3 5 0x x+ β = , correct to 3 significant figures, using the
method of bisection, Let f(x) = 2 3 5x x+ β then, using functional notation:
f(0) = - 5
f(1) = 1 + 3 - 5 = - 1
f(2) = 4 + 6 - 5 = + 5
Since there is a change of sign from negative to positive there must be a root of the equation between
x = 1 and x = 2.
The method of bisection suggests that the root is at 1 22+ = 1.5, i.e. the interval between 1 and 2 has been
bisected.
Hence f(1.5) = (1.5) 2 + 3(1.5) - 5 = 1.75
Since f(1) is negative, f(1.5) is positive, and f(2) is also positive, a root of the equation must lie between
x = 1 and x = 1.5, since a sign change has occurred between f(1) and f(1.5)
Bisecting this interval gives 1 1.52+ i.e. 1.25 as the next root.
Hence f(1.25) = (1.25) 2 + 3(1.25) - 5 = 0.3125
Since f(1) is negative and f(1.25) is positive, a root lies between x = 1 and x = 1.25
Bisecting this interval gives 1 1.252
+ i.e. 1.125
Hence f(1.125) = (1.125) 2 + 3(1.125) - 5 = - 0.359375
Since f(1.125) is negative and f(1.25) is positive, a root lies between x = 1.125 and x = 1.25
Bisecting this interval gives 1.125 1.252+ i.e. 1.1875
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
72
Hence f(1.1875) = (1.1875) 2 + 3(1.1875) - 5 = - 0.02734375
Since f(1.1875) is negative and f(1.25) is positive, a root lies between x = 1.1875 and x = 1.25
Bisecting this interval gives 1.1875 1.252+ i.e. 1.21875
Hence f(1.21875) = (1.21875) 2 + 3(1.21875) - 5 = 0.1416016
Since f(1.21875) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.21875
Bisecting this interval gives 1.1875 1.218752+ = 1.203125
Hence f(1.203125) = 0.056885
Since f(1.203125) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.203125
Bisecting this interval gives 1.1875 1.2031252+ = 1.1953125
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 1.18752+ = 1.191406
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 1.19531252+ = 1.193359
The last two values obtained for the root are 1.1914β¦ and 1.1934β¦.
The last two values are both 1.19, correct to 3 significant figures. We therefore stop the iterations here.
Thus, correct to 3 significant figures, the positive root of x 2 + 3x - 5 = 0 is 1.19
2. Using the bisection method solve 2xe xβ = , correct to 4 significant figures
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Let f(x) = 2xe xβ β then f(0) = 1 β 0 β 2 = -1 f(1) = 1 1 2 0.28e β β = β f(2) = 2 2 2 3.38e β β = Hence, a root lies between x = 1 and x = 2. Let the root be x = 1.5 f(1.5) = 1.5 1.5 2 0.98169e β β = Hence, a root lies between x = 1 and x = 1.5 due to a sign change.
Bisecting this interval gives 1 1.52+ = 1.25
f(1.25) = 1.25 1.25 2 0.240343e β β = Hence, a root lies between x = 1 and x = 1.25 due to a sign change.
Bisecting this interval gives 1 1.252
+ = 1.125
f(1.125) = 1.125 1.125 2 0.04478e β β = β Hence, a root lies between x = 1.125 and x = 1.25 due to a sign change.
Bisecting this interval gives 1.125 1.252+ = 1.1875
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 1.1252+ = 1.15625
f(1.15625) = 0.021743 Hence, a root lies between x = 1.15625 and x = 1.125 due to a sign change.
Bisecting this interval gives 1.15625 1.1252+ = 1.140625
f(1.140625) = -0.011902 Hence, a root lies between x = 1.140625 and x = 1.15625 due to a sign change.
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Bisecting this interval gives 1.140625 1.156252+ = 1.1484375
f(1.1484375) = 0.004824586 Hence, a root lies between x = 1.1484375 and x = 1.140625 due to a sign change.
Bisecting this interval gives 1.1484375 1.1406252+ = 1.14453125
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 1.14843752+ = 1.14648
f(1.14648) = 0.0006294 Hence, a root lies between x = 1.14648 and x = 1.14453125 due to a sign change.
Bisecting this interval gives 1.14648 1.144531252
+ = 1.14551
The last two values are both 1.146, correct to 4 significant figures. We therefore stop the iterations here.
Thus, correct to 4 significant figures, the positive root of 2xe xβ = is 1.146
4. Solve x β 2 β ln x = 0 for the root nearest to 3, correct to 3 decimal places using the bisection
method Let f(x) = x β 2 β ln x f(2) = 2 β 2 β ln 2 = - 0.693 f(3) = 3 β 2 β ln 3 = - 0.0986 f(4) = 4 β 2 β 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.
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Table 1 1x 2x 1 2
3 2x xx +β β= β β
β β ( )3f x
3 3 3 3.125 3.125 3.15625 3.140625 3.14844 3.1445325 3.146486
4 3.5 3.25 3.25 3.1875 3.125 3.15625 3.140625 3.14844 3.1445325
3.5 3.25 3.125 3.1875 3.15625 3.140625 3.14844 3.145325 3.146486 3.14550925
0.24724 0.071345 -0.014434 0.028263 0.0068654 -0.003797 0.0015312 -0.0011327 0.0001999
Correct to 3 decimal places, the solution of x β 2 β ln x = 0 is 3.146
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EXERCISE 40 Page 83 2. Use an algebraic method of successive approximation to solve 3 2 14 0x xβ + = , correct to 3
decimal places. Let f(x) = 3 2 14x xβ +
f(0) = 14
f(1) = 1 β 2 + 14 = 13
f(2) = 8 β 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 + Ξ΄1) Let f(x1 + Ξ΄1) = 0, then since x1 = -2.6, (-2.6 + Ξ΄1)3 - 2(-2.6 + Ξ΄1) + 14 = 0 Neglecting terms containing products of Ξ΄1 and using the binomial series gives: [(-2.6)3 + 3(-2.6)2 Ξ΄1] + 5.2 - 2Ξ΄1 + 14 β 0 -17.576 + 20.28Ξ΄1 + 5.2 - 2 Ξ΄1 + 14 β 0 18.28Ξ΄1 β 17.576 β 5.2 - 14
Ξ΄1 β 17.576 5.2 14 0.0888418.28β β
β β
Thus x2 β -2.6 - 0.08884 = -2.6888 Third approximation Let the true value of the root, x3 , be (x2 + Ξ΄2) Let f(x2 + Ξ΄2) = 0, then since x2 = -2.6888,
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(-2.6888 + Ξ΄2)3 - 2(-2.6888 + Ξ΄2) + 14 = 0 Neglecting terms containing products of Ξ΄2 gives: ( ) ( )3 2
2 22.6888 3 2.6888 5.3776 2 14 0β + β Ξ΄ + β Ξ΄ + β -19.439 + 21.6888Ξ΄2 + 5.3776 - 2Ξ΄2 + 14 β 0 19.6888Ξ΄2 β 19.439 β 5.3776 - 14
Ξ΄2 β 19.439 5.3776 1419.6888β β β 0.003119
Thus x3 β (x2 + Ξ΄2) = -2.6888 + 0.003119 β -2.6857 Fourth approximation Let the true value of the root, x4 , be (x3 + Ξ΄3) Let f(x3 + Ξ΄3) = 0, then since x3 = -2.6857, (-2.6857 + Ξ΄3)3 - 2(-2.6857 + Ξ΄3) + 14 = 0 Neglecting terms containing products of Ξ΄3 gives: ( ) ( )3 2
3 32.6857 3 2.6857 5.3714 2 14 0β + β Ξ΄ + β Ξ΄ + β -19.3719 + 21.63895Ξ΄3 + 5.3714 - 2Ξ΄3 + 14 β 0 19.63895Ξ΄2 β 19.3719 β 5.3714 - 14
Ξ΄2 β 19.3719 5.3714 1419.63895β β β 0.00002546
Thus x4 β (x3 + Ξ΄3) = -2.6857 + 0.000025 β -2.6857 Since x3 and x4 are the same when expressed to the required degree of accuracy, then the required root is -2.686, correct to 3 decimal places. 3. Use an algebraic method of successive approximation to solve 4 33 7 5.5 0x x xβ + β = , correct to
3 significant figures. Let f(x) = 4 33 7 5.5x x xβ + β
f(0) = -5.5
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f(1) = 1 β 3 + 7 β 5.5 = -0.5
f(2) = 16 β 24 + 14 β 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 + Ξ΄1)
Let f(x1 + Ξ΄1) = 0, then since x1 = 1.5,
(1.5 + Ξ΄1) 4 - 3(1.5 + Ξ΄1) 3 + 7(1.5 + Ξ΄1) - 5.5 = 0
Neglecting terms containing products of Ξ΄1 and using the binomial series gives:
[(1.5) 4 + 4(1.5) 3 Ξ΄1] β 3[(1.5) 3 + 3(1.5) 2 Ξ΄1] + 10.5 + 7Ξ΄1 - 5.5 β 0
5.0625 + 13.5Ξ΄1 β 10.125 β 20.25Ξ΄1 + 10.5 + 7Ξ΄1 β 5.5 β 0
0.25Ξ΄1 β 0.0625
Ξ΄1 β 0.0625 0.250.25
β
Thus x2 β 1.5 + 0.25 = 1.75 Third approximation Let the true value of the root, x3 , be (x2 + Ξ΄2) Let f(x2 + Ξ΄2) = 0, then since x2 = 1.75,
(1.75 + Ξ΄2) 4 - 3(1.75 + Ξ΄2)3 + 7(1.75 + Ξ΄2) β 5.5 = 0
Neglecting terms containing products of Ξ΄2 gives:
( ) ( ) ( ) ( )4 3 3 22 2 21.75 4 1.75 3[ 1.75 3 1.75 ] 7(1.75 ) 5.5 0+ Ξ΄ β + Ξ΄ + + Ξ΄ β β
9.3789 + 21.4375Ξ΄2 β 16.078125 β 27.5625Ξ΄2 + 12.25 + 7Ξ΄2 β 5.5 β 0
0.875Ξ΄2 β -0.050775
Ξ΄2 β 0.050775 0.058030.875
ββ β
Thus x3 β (x2 + Ξ΄2) = 1.75 β 0.05803 β 1.692 Fourth approximation Let the true value of the root, x4 , be (x3 + Ξ΄3) Let f(x3 + Ξ΄3) = 0, then since x3 = 1.692,
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(1.692 + Ξ΄3) 4 - 3(1.692 + Ξ΄3)3 + 7(1.692 + Ξ΄3) β 5.5 = 0 Neglecting terms containing products of Ξ΄3 gives: ( ) ( ) ( ) ( )4 3 3 2
3 3 31.692 4 1.692 3[ 1.692 3 1.692 ] 7(1.692 ) 5.5 0+ Ξ΄ β + Ξ΄ + + Ξ΄ β β 8.19599 + 19.37586Ξ΄3 β 14.5318977 β 25.76578Ξ΄3 + 11.844 + 7Ξ΄3 β 5.5 β 0 0.61008Ξ΄3 β -0.0080923
Ξ΄3 β 0.0080923 0.01326430.61008
ββ β
Thus x4 β (x3 + Ξ΄3) = 1.692 - 0.0132643 β 1.6787 Fifth approximation Let the true value of the root, x5 , be (x4 + Ξ΄4) Let f(x4 + Ξ΄4) = 0, then since x4 = 1.6787, (1.6787 + Ξ΄4) 4 - 3(1.6787 + Ξ΄4)3 + 7(1.6787 + Ξ΄4) β 5.5 = 0 Neglecting terms containing products of Ξ΄4 gives: ( ) ( ) ( ) ( )4 3 3 2
4 4 41.6787 4 1.6787 3[ 1.6787 3 1.6787 ] 7(1.6787 ) 5.5 0+ Ξ΄ β + Ξ΄ + + Ξ΄ β β 7.941314 + 18.9225Ξ΄4 β 14.1919 β 25.3623Ξ΄4 + 11.7509 + 7Ξ΄4 β 5.5 β 0 0.5602Ξ΄4 β -0.000314
Ξ΄4 β 0.000314 0.000560.5602
ββ β
Thus x5 β (x4 + Ξ΄4) = 1.6787 - 0.00056 β 1.6781 Since x4 and x5 are the same when expressed to the required degree of accuracy, then the required root is 1.68, correct to 3 decimal places. From earlier, f(x) = 4 33 7 5.5x x xβ + β f(0) = -5.5 f(-1) = 1 + 3 β 7 β 5.5 = -8.5
f(-2) = 16 + 24 β 14 β 5.5 = 20.5 Hence, a root lies between x = -1 and x = -2.
This root may be found in exactly the same way as for the positive root above.
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EXERCISE 41 Page 85 2. Use Newtonβs method to solve 33 10 14x xβ = , correct to 4 significant figures. Let f(x) = 33 10 14x xβ β f(0) = -14 f(1) = 3 β 10 β 14 = -21 f(2) = 24 β 20 β 14 = -10 f(3) = 81 β 30 β 14 = 36 Hence, a root lies between x = 2 and x = 3. Let 1r 2.2=
A better approximation is given by: ( )12 1
1
f rr r
f '(r )= β 2f '(x) 9x 10x= β
Hence, 3
2 2
3(2.2) 10(2.2) 14 4.056r 2.2 2.2 2.38819(2.2) 10(2.2) 21.56
β β β= β = β =
β
3f (2.3881) 2.9771577r 2.3881 2.3881 2.2796f '(2.3881) 27.44619
= β = β =
4f (2.2796) 1.257655r 2.2796 2.2796 2.3321f '(2.2796) 23.973184
β= β = β =
5f (2.3321) 0.7297097r 2.3321 2.3321 2.3036f '(2.3321) 25.627214
= β = β =
60.3633356r 2.3036 2.3183
24.7231566β
= β =
70.1962136r 2.3183 2.310525.187634
= β =
80.10180935r 2.3105 2.314624.94069
β= β =
90.05452775r 2.3146 2.312425.070358
= β =
100.02944745r 2.3124 2.3136
25.0007438β
= β =
110.0163322r 2.3136 2.312925.0387046
= β =
120.01037987r 2.3129 2.313
25.0165577β
= β =
Hence, correct to 4 significant figures, x = 2.313
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4. Use Newtonβs method to solve 4 33 4 7 12 0x x xβ + β = , correct to 3 decimal places. Let f(x) = 4 33 4 7 12x x xβ + β f(0) = -12 f(1) = 3 β 4 + 7 β 12 = -6 f(2) = 48 β 32 + 14 β 12 = 18 Hence, a root lies between x = 1 and x = 2 There are no further positive roots since the 4x term predominates. f(-1) = 3 + 4 β 7 β 12 = -12 f(-2) = 48 + 32 β 14 β 12 = 54 Hence, a root lies between x = -1 and x = -2 There are no further negative roots since, once again, the 4x term predominates. For the root between x = 1 and x = 2: Let 1r 1.25= 3 2f '(x) 12x 12x 7= β +
( ) ( )( ) ( )
4 31
2 1 3 21
3 1.25 4 1.25 7(1.25) 12f (r ) 3.73828r r 1.25 1.25 1.56985f '(r ) 11.687512 1.25 12 1.25 7
β + β β= β = β = β =
β +
31.734046r 1.56985 1.49715
23.8522585= β =
40.12925743r 1.49715 1.4908120.3720909
= β =
50.000994417r 1.49081 1.49076
20.089988= β =
Hence, correct to 3 decimal places, the positive root is: x = 1.491 For the root between x = -1 and x = -2: Let 1r 1.2= β
( ) ( )( ) ( )
4 31
2 1 3 21
3 1.2 4 1.2 7( 1.2) 12f (r ) 7.2672r r 1.2 1.2 1.4343f '(r ) 31.01612 1.2 12 1.2 7
β β β + β β β= β = β β = β β = β
ββ β β +
32.4589815r 1.4343 1.388053.094585
= β β = ββ
40.11488764r 1.3880 1.3856
48.207045= β β = β
β
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50.00051387r 1.3856 1.3856
47.960999= β β = β
β
Hence, correct to 3 decimal places, the negative root is: x = -1.386
7. Use Newtonβs method to solve 2300 62
e ΞΈ ΞΈβ + = , correct to 3 significant figures.
Let f(ΞΈ) = 2300 62
e ΞΈ ΞΈβ + β f(0) = 300 β 6 = 294 f(1) = 35.1 f(2) = 0.495
f(3) = -3.756 Hence, a root lies between x = 2 and x = 3, very close to x = 2. There are no further positive roots since the 2300e ΞΈβ term predominates. There are no negative roots since f(x) will always be positive. Let 1r 2= 2f '(x) 600e 0.5β ΞΈ= β +
4
12 1 4
1
f (r ) 300e 1 6 0.494692r r 2 2 2.0492f '(r ) 600e 0.5 10.489383
β
β
+ β= β = β = β =
β + β
30.00436387r 2.0492 2.04979.45952774
β= β =
β
Hence, correct to 3 significant figures, x = 2.05 9. A Fourier analysis of the instantaneous value of a waveform can be represented by:
1sin sin 34 8
y t t tΟβ β= + + +β ββ β
Use Newtonβs method to determine the value of t near to 0.04, correct to 4 decimal places, when the amplitude, y, is 0.880.
When y = 0.88, then 10.88 sin sin 34 8
t t tΟβ β= + + +β ββ β
or 1sin sin 3 0.88 04 8
t t tΟβ β+ + + β =β ββ β
Let f(t) = 1sin sin 3 0.884 8
t t tΟβ β+ + + ββ ββ β
Let 1r = 0.04 3f '(t) 1 cos t cos3t8
= + +
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12 1
1
10.04 sin 0.04 sin 3(0.04) 0.88f (r ) 0.00035154 8r r 0.04 0.043f '(r ) 2.3715033451 cos0.04 cos3(0.04)8
Ο+ + + β
= β = β = β+ +
= 0.03985
30.000004203r 0.03985 0.03985
2.371529496β
= β =
Hence, correct to 4 decimal places, t = 0.0399 10. A damped oscillation of a system is given by the equation: 0.57.4 sin 3ty e t= β Determine the value of t near to 4.2, correct to 3 significant figures, when the magnitude y of the oscillation is zero. Let f(t) = 0.57.4 sin 3β te t ( )( ) ( )( )0.5t 0.5t 0.5t 0.5tf ' (t) 7.4e 3cos3t sin 3t 3.7e 22.2e cos3t 3.7e sin 3t= β + β = β β = ( )0.5te 22.2cos3t 3.7sin 3tβ + Let 1r 4.2=
( )
2.11
2 1 2.11
f (r ) 7.4e sin12.6 2.03182922r r 4.2 4.2 4.189f '(r ) e 22.2cos12.6 3.7sin12.6 182.2023833
β β= β = β = β =
β + β
(Note, sin 12.6 is sin 12.6 rad)
3f (4.189) 0.037825r 4.189 4.189 4.189f '(4.189) 180.3134965
β= β = β =
β
Hence, correct to 3 significant figures, t = 4.19 11. The critical speeds of oscillation, Ξ», of a loaded beam are given by the equation: 3 23.250 0.063 0Ξ» Ξ» Ξ»β + β = Determine the value of Ξ» which is approximately equal to 3.0 by Newtonβs method, correct to 4 decimal places.
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Let f(Ξ») = 3 23.250 0.063β + βΞ» Ξ» Ξ» 2f '( ) 3 6.5 1Ξ» = Ξ» β Ξ» + Let 1r 3.0=
( ) ( )( ) ( )
3 2
2 2
3.0 3.250 3.0 3.0 0.063f (3.0) 0.687r 3.0 3.0 3.0 2.91918f '(3.0) 8.53 3.0 6.5 3.0 1
β + β= β = β = β =
β +
30.0370604r 2.91918 2.914307.5901656
= β =
40.00015139r 2.91430 2.914287.5364835
= β =
Hence, correct to 4 decimal places, Ξ» = 2.914
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CHAPTER 10 COMPUTER NUMBERING SYSTEMS
EXERCISE 42 Page 86 2. Convert the following binary numbers to denary numbers:
(a) 10101 (b) 11001 (c) 101101 (d) 110011 (a) 4 3 2 1 0
210101 1 2 0 2 1 2 0 2 1 2= Γ + Γ + Γ + Γ + Γ
= 16 + 0 + 4 + 0 + 1 = 1021 (b) 4 3 2 1 0
211001 1 2 1 2 0 2 0 2 1 2= Γ + Γ + Γ + Γ + Γ
= 16 + 8 + 0 + 0 + 1 = 1025 (c) 5 4 3 2 1 0
2101101 1 2 0 2 1 2 1 2 0 2 1 2= Γ + Γ + Γ + Γ + Γ + Γ
= 32 + 0 + 8 + 4 + 0 + 1 = 1045 (d) 5 4 3 2 1 0
2110011 1 2 1 2 0 2 0 2 1 2 1 2= Γ + Γ + Γ + Γ + Γ + Γ
= 32 + 16 + 0 + 0 + 2 + 1 = 1051 4. Convert the following binary numbers to denary numbers:
(a) 11010.11 (b) 10111.011 (c) 110101.0111 (d) 11010101.10111 (a) 4 3 2 1 0 1 2
211010.11 1 2 1 2 0 2 1 2 0 2 1 2 1 2β β= Γ + Γ + Γ + Γ + Γ + Γ + Γ
= 16 + 8 + 0 + 2 + 0 + 1 12 4+ = 1026.75
(b) 4 3 2 1 0 1 2 3
210111.011 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2β β β= Γ + Γ + Γ + Γ + Γ + Γ + Γ + Γ
= 16 + 0 + 4 + 2 + 1 + 1 14 8+ = 1023.375
(c) 5 4 3 2 1 0 1 2 3 4
2110101.0111 1 2 1 2 0 2 1 2 0 2 1 2 0 2 1 2 1 2 1 2β β β β= Γ + Γ + Γ + Γ + Γ + Γ + Γ + Γ + Γ + Γ
= 32 + 16 + 0 + 4 + 0 + 1 + 1 1 14 8 16+ + = 3
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(d) 7 6 5 4 3 2 1 0 1 2211010101.10111 1 2 1 2 0 2 1 2 0 2 1 2 0 2 1 2 1 2 0 2β β= Γ + Γ + Γ + Γ + Γ + Γ + Γ + Γ + Γ + Γ
+ 3 4 51 2 1 2 1 2β β βΓ + Γ + Γ
= 128 + 64 + 16 + 4 + 1 + 1 1 1 12 8 16 32+ + + = 10213.71875
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EXERCISE 43 Page 88 2. Convert the following denary numbers into binary numbers: (a) 31 (b) 42 (c) 57 (d) 63 (a) 2 31 Remainder 2 15 1 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 231 11111= (b) 2 42 Remainder 2 21 0 2 10 1 2 5 0 2 2 1 2 1 0 0 1 Hence, 10 242 101 010= (c) 2 57 Remainder 2 28 1 2 14 0 2 7 0 2 3 1 2 1 1 0 1 Hence, 10 257 111 001= (d) 2 63 Remainder 2 31 1 2 15 1 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 263 111 111=
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4. Convert the following denary numbers into binary numbers:
(a) 47.40625 (b) 30.8125 (c) 53.90625 (d) 661.65625
(a) 2 47 Remainder 2 23 1 2 11 1 2 5 1 2 2 1 2 1 0 0 1 Hence, 10 247 101111= 0.40625 Γ 2 = 0.8125 0.8125 Γ 2 = 1.625 0.625 Γ 2 = 1.25 0.25 Γ 2 = 0.50 0.50 Γ 2 = 1.00 Hence, 10 20.40625 0.01101= Thus, 10 247.40625 101 111.011 01= (b) 2 30 Remainder 2 15 0 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 230 11110= 0.8125 Γ 2 = 1.625 0.625 Γ 2 = 1.25 0.25 Γ 2 = 0.50 0.50 Γ 2 = 1.00 Hence, 10 20.8125 0.1101= Thus, 10 230.8125 11 110.110 1= (c) 2 53 Remainder 2 26 1 2 13 0 2 6 1 2 3 0 2 1 1 0 1 Hence, 10 253 110 101= 0.90625 Γ 2 = 1.8125 0.8125 Γ 2 = 1.625 0.625 Γ 2 = 1.25 0.25 Γ 2 = 0.50 0.50 Γ 2 = 1.00 Hence, 10 20.90625 0.111 01= Thus, 10 253.90625 110 101.111 01=
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(d) 2 61 Remainder 2 30 1 2 15 0 2 7 1 2 3 1 2 1 1 0 1 Hence, 10 261 111101= 0.65625 Γ 2 = 1.3125 0.3125 Γ 2 = 0.625 0.625 Γ 2 = 1.25 0.25 Γ 2 = 0.50 0.50 Γ 2 = 1.00 Hence, 10 20.65625 0.101 01= Thus, 10 261.65625 111 101.101 01=
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EXERCISE 44 Page 89 3. Convert the following denary numbers to binary numbers, via octal: (a) 247.09375 (b) 514.4375 (c) 1716.78125 (a) 8 247 Remainder 8 30 7 8 3 6 0 3 Hence, 10 8 2247 367 011110 111= = from Table 10.1, page 88 0.09375 Γ 8 = 0.75 0.75 Γ 8 = 6.00 Hence, 10 8 20.09375 0.06 0.000110= = from Table 10.1, page 88 Hence, 10 8 2247.09375 763.06 11 110 111.000 11= = (b) 8 514 Remainder 8 64 2 8 8 0 8 1 0 Hence, 10 8 2514 1002 001 000 000 010= = from Table 10.1, page 88 0 1 0.4375 Γ 8 = 3.50 0.50 Γ 8 = 4.00 Hence, 10 8 20.4375 0.34 0.011100= = from Table 10.1, page 88 Hence, = =10 8 2514.4375 1002.34 1000000 010.0111 (c) 8 1716 Remainder 8 214 4 8 26 6 8 3 2 Hence, 10 8 21716 3264 011 010 110 100= = from Table 10.1, page 88 0 3 0.78175 Γ 8 = 6.25 0.25 Γ 8 = 2.00 Hence, 10 8 20.78125 0.62 0.110 010= = from Table 10.1, page 88 Hence, 10 8 21716.79125 3264.62 11010110 100.11001= = 4. Convert the following binary numbers to denary numbers, via octal: (a) 111.0111 (b) 101001.01 (c) 1110011011010.0011
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(a) 111.0111 = 111.011 100 2
= 7 . 3 84 = 0 1 22
3 47 8 3 8 4 8 78 8
β βΓ + Γ + Γ = + + = 107.4375
(b) 101001.01 = 101 001.010 2
= 5 1 . 82 = 1 0 1 25 8 1 8 2 8 40 18
βΓ + Γ + Γ = + + = 1041.25
(c) 1110011011010.0011 = 001 110 011 011 010.001 100 2 = 1 6 3 3 2 . 1 84 = 4 3 2 1 0 1 21 8 6 8 3 8 3 8 2 8 1 8 4 8β βΓ + Γ + Γ + Γ + Γ + Γ + Γ
= 4096 + 3072 + 192 + 24 + 2 + 1 48 64+ = 107386.1875
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EXERCISE 45 Page 92 2. Convert the hexadecimal number 162C into decimal.
1 0162C 2 16 C 16 32 12= Γ + Γ = + = 1044
4. Convert the hexadecimal number 162F1 into decimal
2 1 0 2 1 0162F1 2 16 F 16 1 16 2 16 15 16 1 16= Γ + Γ + Γ = Γ + Γ + Γ
= 512 + 240 + 1 = 10753 6. Convert the decimal number 10200 into hexadecimal 16 200 Remainder 16 12 8 0 12 (i.e. C) Hence, 10 16200 C8= 8. Convert the decimal number 10238 into hexadecimal 16 238 Remainder 16 14 14 (i.e. E) 0 14 (i.e. E) Hence, 10 16238 EE= 10. Convert the binary number 211101010 into hexadecimal
211101010 = 1110 1010 grouping in 4βs = E A16 from Table 10.2, page 90.
i.e. 2 1611101010 EA=
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12. Convert the binary number 210100101 into hexadecimal
210100101 = 1010 0101 grouping in 4βs = A 516 from Table 10.2, page 90.
i.e. 2 1610100101 A5= 14. Convert the hexadecimal number 16ED into binary
16ED = 1110 1101 2 from Table 10.2, page 90. 16. Convert the hexadecimal number 16A21 into binary
16A21 = 1010 0010 0001 2 from Table 10.2, page 90.
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CHAPTER 11 BOOLEAN ALGEBRA AND LOGIC CIRCUITS
EXERCISE 46 Page 97 1. Determine the Boolean expression and construct a truth table for the following switching circuit:
For the circuit to function, Z = C AND [(A AND B) OR ( A AND B)]
i.e. Z = ( )C. A.B A.B+
The truth table is shown below:
A B C A.B A A .B A.B + A .B Z = C.( A.B + A .B) 0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
0 0 0 0 0 0 1 1
1 1 1 1 0 0 0 0
0 0 1 1 0 0 0 0
0 0 1 1 0 0 1 1
0 0 0 1 0 0 0 1
2. Determine the Boolean expression and construct a truth table for the following switching circuit:
For the circuit to function, Z = C AND [(A AND B ) OR ( A )]
i.e. Z = ( )C. A.B A+
The truth table is shown below:
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A B C A B A. B A. B + A Z = C.( A. B + A ) 0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
1 1 1 1 0 0 0 0
1 1 0 0 1 1 0 0
0 0 0 0 1 1 0 0
1 1 1 1 1 1 0 0
0 1 0 1 0 1 0 0
4. Determine the Boolean expression and construct a truth table for the following switching circuit:
For the circuit to function, Z = C AND [(B AND C AND A ) OR (A AND (B OR C )]
i.e. Z = ( )C. B.C.A A.(B C)+ +
The truth table is shown below:
A B C A
C B.C. A B + C A.( B + C ) B.C. A + A.( B + C ) Z
0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
1 1 1 1 0 0 0 0
1 0 1 0 1 0 1 0
0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1
0 0 0 0 1 0 1 1
0 0 0 1 1 0 1 1
0 0 0 1 0 0 0 1
The remaining problems in this exercise have solutions given.
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EXERCISE 47 Page 100 3. Simplify: ( )F.G F.G G. F F+ + +
( )F.G F.G G. F F+ + + = F.G F.G G(1) F.G F.G G+ + = + + from 10, Table 11.8,
= ( )G. F F G+ + = G.1 G G G+ = + from 10, Table 11.8,
= G from 9, Table 11.8
4. Simplify: ( )F.G F. G G F.G+ + +
( )F.G F. G G F.G+ + + = F.G F.1 F.G F.G F F.G+ + = + + from 10, Table 11.8,
= ( )F. G G F F.1 F+ + = +
= F + F = F from 9 and 10, Table 11.8 6. Simplify: F.G.H F.G.H F.G.H+ +
F.G.H F.G.H F.G.H+ + = ( )F.H. G G F.G.H+ +
= F.H F.G.H+ from 10, Table 11.8,
= ( )H. F F.G+
8. Simplify: P.Q.R P.Q.R P.Q.R+ +
P.Q.R P.Q.R P.Q.R+ + = ( )Q.R. P P P.Q.R+ +
= Q.R P.Q.R+ from 10, Table 11.8 9. Simplify: F.G.H F.G.H F.G.H F.G.H+ + +
F.G.H F.G.H F.G.H F.G.H+ + + = ( ) ( )F.G. H H F.G H H+ + +
= F.G F.G+ from 10, Table 11.8,
= ( )G. F F+
= G from 10, Table 11.8
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12. Simplify: ( ) ( )R. P.Q P.Q P.Q P. Q.R Q.R+ + + +
( ) ( )R. P.Q P.Q P.Q P. Q.R Q.R+ + + + = ( ) ( )R.P.Q R.P. Q Q P.R. Q Q+ + + +
= R.P.Q R.P P.R+ + from 10, Table 11.8,
= ( )R.P.Q P. R R+ +
= R.P.Q P+ from 10, Table 11.8,
= P R.Q+ from 17, Table 11.8
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EXERCISE 48 Page 101 2. Simplify: (A B.C ) (A.B C)+ + +
(A B.C ) (A.B C)+ + + = ( ) ( )A .B.C A B C+ + + by de Morganβs law,
= A.B.C A B C+ + +
= C.(A.B 1) A B+ + +
= C A B+ + or A B C+ + from 8, Table 11.8
3. Simplify: (A.B B.C). A.B+
(A.B B.C). A.B+ = ( ) ( ) ( )A.B . B.C . A B+ by de Morganβs law,
= ( ) ( ) ( )A B . B C . A B+ + +
= ( ) ( ) ( )A B . B C . A B+ + +
= ( ) ( )A.B A.C B.B B.C . A B+ + + +
= ( ) ( )A.B A.C B B.C . A B+ + + + from 13, Table 11.8,
= ( ) ( )A.B A.C B . A B+ + + from 15, Table 11.8,
= ( ) ( )B. A 1 A.C . A Bβ‘ β€+ + +β£ β¦
= ( ) ( )B A.C . A B+ + from 8, Table 11.8,
= A.B B.B A.C.A A.B.C+ + +
= A.B 0 0 A.B.C+ + + from 14, Table 11.8,
= A.B A.B.C+ 5. Simplify: (P.Q P.R).(P.Q.R)+
(P.Q P.R).(P.Q.R)+ = ( ) ( ) ( )P Q P R .P. Q Rβ‘ β€+ + + +β’ β₯β£ β¦ by de Morganβs law,
= ( )( )P Q P R P.Q P.R+ + + +
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= ( ) ( )1 Q R . P.Q P.R+ + +
= P.Q P.R Q.P.Q Q.P.R R .P.Q R .P.R+ + + + +
= P.Q P.R 0 Q.P.R R .P.Q 0+ + + + + from 14 and 9, Table 11.8,
= ( )P. Q R Q.R R .Q+ + +
= ( )( )P. Q R R Q Q+ + +
= ( )P. Q R R+ + from 10, Table 11.8,
= ( )P . Q R+ from 9, Table 11.8
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EXERCISE 49 Page 105 3. Use Karnaugh map techniques to simplify: (P.Q).(P.Q)
P.Q corresponds to P = 0, Q = 0, i.e. the top left hand cell of the Karnaugh map, shown as a 1.
P.Q corresponds to P = 0, Q = 1, i.e. the bottom left hand cell, hence P.Q corresponds to each of
the other three cells, shown as 2βs.
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 1βs in the two right hand cells in the top row
A.B corresponds to A = 0, B = 1, shown as 1βs in the two cells in the second column
A.C corresponds to A = 0, C = 1, shown as 1βs 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
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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.
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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
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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+ + +
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EXERCISE 50 Page 109 5. Simplify the expression given in column 4 of the truth table below and devise a logic circuit to meet the requirements of the simplified expression.
From column 4, 1Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +
The Karnaugh map is shown below.
The vertical 2-cell couple corresponds to: A.B
The horizontal 4-cell couple corresponds to: C
Hence, 1Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +
simplifies to: 1Z A.B C= +
A logic circuit to meet these requirements is shown below.
7. Simplify the expression given in column 6 of the truth table of question 5 above and devise a logic circuit to meet the requirements of the simplified expression. From column 6, 3Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +
The Karnaugh map is shown below.
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The horizontal 2-cell couple corresponds to: A.C
The 4-cell couple corresponds to: B
Hence, 3Z A.B.C A.B.C A.B.C A.B.C A.B.C= + + + +
simplifies to: 3Z B A.C= +
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.
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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+ +
A logic circuit to meet these requirements is shown below.
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 2βs
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P + Q.R is shown with 3βs
P Q.R+ is shown with 4βs
Hence, ( ) ( )P.Q.R . P Q.R+ is represented by the cells containing both 2βs and 4βs
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+
A logic circuit to meet these requirements is shown below.
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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)
A logic circuit to satisfy this Boolean expression is shown below:
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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
A logic circuit to satisfy this Boolean expression is shown below:
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 210Β°C, or
(c) the temperature of the oven is less than 190Β°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)
A logic circuit to satisfy this Boolean expression is shown below:
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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 β QPR
(a) Since triangle PQR in the diagram below is isosceles, PQ = QR From Pythagoras, 2 2 2 2(38.47) (PQ) (QR) 2(PQ)= + =
from which, ( )2
2 38.47PQ2
= and PQ = 238.47 38.47
2 2= = 27.20 cm
Hence, PQ = QR = 27.20 cm
(b) Since triangle PQR is isosceles, β P = β R and since β Q = 90Β°, then β P + β R = 90Β° Hence, β QPR = 45Β° (=β 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 β 20 = 12 km and BC = 24 β 7 = 17 km
Hence, distance between the two men, AC = ( )2 212 17+ = 20.81 km by Pythagoras.
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4. A ladder 3.5 m long is placed against a perpendicular wall with its foot 1.0 m from the wall.
How far up the wall (to the nearest centimetre) does the ladder reach? If the foot of the ladder is
now moved 30 cm further away from the wall, how far does the top of the ladder fall?
Distance up the wall, AB = ( )2 23.5 1.0β = 3.35 m by Pythagoras.
( ) ( ) ( )2 2 2 2A 'B A 'C ' BC' 3.5 1.30 3.25mβ‘ β€= β = β =β£ β¦
Hence, the amount the top of the ladder has moved down the wall, given by AAβ² = 3.35 β 3.25 = 0.10 m or 10 cm
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EXERCISE 53 Page 117
2. If cos A = 1517
find sin A and tan A, in fraction form.
Since adjacentcos inehypotenuse
= then the sides 15 and 17 are as shown in the diagram.
By Pythagoras, BC = ( )2 217 15β = 8
Hence, sin A = opposite BChypotenuse AC
= = 817
and tan A = opposite BCadjacent AB
= = 815
4. Point P lies at co-ordinates (-3, 1) and point Q at (5, -4). Determine (a) the distance PQ, (b) the
gradient of the straight line PQ, and (c) the angle PQ makes with the horizontal.
(a) From the diagram below, PQ = ( )2 25 8+ = 9.434 by Pythagoras
(b) Gradient of PQ = 1 4 5
3 5 8β β
=β β β
= - 0.625
(c) 5Tan8
ΞΈ = from which, the angle PQ makes with the horizontal, ΞΈ = 1 5tan8
β β ββ ββ β
= 32Β°
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EXERCISE 54 Page 119 2. Solve triangle DEF
By Pythagoras, FE = ( )2 24 3+ = 5 cm
Tan E = 43
from which, β E = 1 4tan3
β = 53.13Β° or 53Β°8β²
Hence, β F = 180Β° - 90Β° - 53.13Β° = 36.87Β° or 36Β°52β² 4. Solve triangle JKL and find its area
Sin 51Β° = 6.7JL
from which, JL = 6.7sin 51Β°
= 8.62 cm
Tan 51Β° = 6.7KL
from which, KL = 6.7tan 51Β°
= 5.43 cm
β J = 180Β° - 90Β° - 51Β° = 39Β° (Checking: JL = ( )2 26.7 5.43 8.62cm+ = )
Area of triangle JKL = 12
(KL)(JK) = 12
(5.43)(6.7) = 18.19 2cm
6. Solve triangle PQR and find its area
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By Pythagoras, 2 2 2PR 3.69 8.75+ = from which, PR = ( )2 28.75 3.69β = 7.934 m
Sin R = 3.698.75
from which, β R = 1 3.69sin8.75
β β ββ ββ β
= 24.94Β° or 24Β°57β²
β Q = 180Β° - 90Β° - 24.94Β° = 65.06Β° or 65Β°3β²
Area of triangle PQR = 12
(PQ)(PR) = 12
(3.69)(7.934) = 14.64 2m
7. A ladder rest against the top of the perpendicular wall of a building and makes an angle of 73Β°
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 73Β° = BC2
from which, height of building, BC = 2 tan 73Β° = 6.54 m
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EXERCISE 55 Page 121 2. From the top of a vertical cliff 80.0 m high the angle of depression of two buoys lying due west
of the cliff are 23Β° and 15Β°, respectively. How far are the buoys apart?
In the diagram below, the two buoys are shown as A and B.
Tan 15Β° = 80
AC from which, AC = 80
tan15Β° = 298.56 m
Tan 23Β° = 80BC
from which, BC = 80tan 23Β°
= 188.47 m
Hence, distance apart, AB = AC β BC = 298.56 β 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 32Β° and 30Β° respectively. Determine the
height of the flagpole.
In the diagram below, the flagpole is shown as AB.
Tan 32Β° = AC
200 from which, AC = 200 tan 32Β° = 124.97 m
Tan 30Β° = BC200
from which, BC = 200 tan 30Β° = 115.47 m
Hence, height of flagpole, AB = AC β BC = 124.97 β 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 24Β° and the angle of elevation of the top of the building is 34Β°. Determine,
correct to the nearest centimetre, the width of the road and the height of the building.
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116
In the diagram below, D is the window, the width of the road is AB and the height of the building across the road is BC.
In the triangle ABD, β D = 90Β° - 24Β° = 66Β°
Tan 66Β° = AB4.2
hence, width of road, AB = 4.2 tan 66Β° = 9.43 m
From triangle DEC, tan 34Β° = CE CE CEDE AB 9.43
= = from which, CE = 9.43 tan 34Β° = 6.36 m
Hence, height of building, BC = CE + EB = CE + AD = 6.36 + 4.2 = 10.56 m 7. The elevation of a tower from two points, one due east of the tower and the other due west of it
are 20Β° and 24Β°, 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 20Β° = AB
BC from which, AB = BC tan 20Β°
Tan 24Β° = AB300 BCβ
from which, AB = (300 - BC) tan 24Β°
i.e. BC tan 20Β° = (300 - BC) tan 24Β° = 300 tan 24Β° - BC tan 24Β° i.e. 0.36397 BC = 133.57 β 0.44523 BC i.e. 0.8092 BC = 133.57
and BC = 133.570.8092
= 165.06 m
Tan 20Β° = AB165.06
from which, height of tower, AB = 165.06 tan 20Β° = 60 m, to the nearest metre
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117
EXERCISE 56 Page 123 4.(a) Evaluate correct to 4 decimal places: secant 73Β°
secant 73Β° = 1cos 73Β°
= 3.4203
5.(b) Evaluate correct to 4 decimal places: cosecant 15.62Β°
cosecant 15.62Β° = 1sin15.62Β°
= 3.7139
6.(c) Evaluate correct to 4 decimal places: cotangent 321Β°23β²
cotangent 321Β°23β² = 1 1 123tan 321 23' tan 321.38333tan 32160
= =° °°
= -1.2519
7.(a) Evaluate correct to 4 decimal places: sine 23Ο
Note that sine 23Ο means sine 2
3Ο radians = 0.8660
2 2 180 2rad 120 hence, sin sin1203 3 3Ο Ο Β° Οβ‘ β€= Γ = Β° β‘ Β°β’ β₯Οβ£ β¦
8.(c) Evaluate correct to 4 decimal places: cot 2.612
cot 2.612 = 1tan 2.612 rad
= -1.7083
12. Determine the acute angle 1sec 1.6214β in degrees (correct to 2 decimal places), degrees and
minutes, and in radians (correct to 3 decimal places).
Using a calculator, 1sec 1.6214β = 1 1cos1.6214
β β ββ ββ β
= 51.92Β° or 55Β°55β² or 51.92 rad180Ο
Γ = 0.906 rad
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13. Determine the acute angle 1cosec 2.4891β in degrees (correct to 2 decimal places), degrees and
minutes, and in radians (correct to 3 decimal places).
Using a calculator, 1cosec 2.4891β = 1 1sin2.4891
β β ββ ββ β
= 23.69Β° or 23Β°41β² or 23.69 rad180Ο
Γ = 0.413 rad
16. Evaluate, correct to 4 significant figures: 11.5 tan 49 11' sin 903cos 45
Β° β °°
11.5 tan 49 11' sin 90 11.5 tan 49.18333 sin 90 13.315 1
3cos 45 3cos 45 2.2132Β° β Β° Β°β Β° β
= =Β° Β°
= 5.805
18. Evaluate, correct to 4 significant figures: 6.4cosec 29 5' sec812cot12
Β° β °°
1 16.4
6.4cosec 29 5' sec81 13.1665 6.39245sin 29.08333 cos8112cot12 9.409262
tan12
β β β βββ β β βΒ° β Β° βΒ° Β°β β β β = =Β° β β
β βΒ°β β
= 0.7199
20. If tan x = 1.5276, determine sec x, cosec x and cot x. (Assume x is an acute angle) If tan x = 1.5276, then x = 1tan 1.5276β = 56.79Β°
sec x = sec 56.79Β° = 1cos56.79Β°
= 1.8258
cosec x = cosec 56.79Β° = 1sin 56.79Β°
= 1.1952
cot x = cot 56.79Β° = 1tan 56.79Β°
= 0.6546
21. Evaluate, correct to 4 significant figures: ( )( )( )
sin 34 27 ' cos69 2 '2 tan 53 39 'Β° Β°
Β°
( )( )
( )( )( )sin 34 27 ' cos69 2 ' sin 34.45 cos69.03333
2 tan 53 39 ' 2 tan 53.65Β° Β° Β° Β°
=Β° Β°
= 0.07448
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119
23. Evaluate, correct to 4 significant figures: cos ec 27 19 ' sec 45 29 '1 cosec 27 19 'sec 45 29 '
Β° + Β°β Β° Β°
1 1
cos ec 27 19 ' sec 45 29 ' 2.179086 1.426296sin 27.31666 cos 45.483331 11 cosec 27 19 'sec 45 29 ' 1 (2.179086)(1.426296)1
sin 27.31666 cos 45.48333
β β β β+β β β βΒ° + Β° +Β° Β°β β β β = =β Β° Β° ββ ββ ββ β ββ βΒ° Β°β β β β
= 3.605382.10802β
= -1.710
25. Evaluate, correct to 5 significant figures: (a) cosec(-143Β°) (b) cot(-252Β°) (c) sec(-67Β°22β²)
Using a calculator: (a) cosec(-143Β°) = 1sin( 143 )β Β°
= -1.6616
(b) cot(-252Β°) = 1tan( 252 )β Β°
= -0.32492
(c) sec(-67Β°22β²) = ( )
1cos 67.36666β Β°
= 2.5985
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EXERCISE 57 Page 126 2. Use the sine rule to solve triangle ABC, given B = 71Β°26β², C = 56Β°32β² and b = 8.60 cm, and find
its area.
Triangle ABC is shown below.
β A = 180Β° - 71Β°26β² - 56Β°32β² = 52Β°2β²
From the sine rule, 8.60 csin 71 26 ' sin 56 32 '
=Β° Β°
from which, c = 8.60sin 56 32 'sin 71 26 '
°°
= 7.568 cm
Also from the sine rule, a 8.60sin 52 2 ' sin 71 26 '
=Β° Β°
from which, a = 8.60sin 52 2 'sin 71 26 '
°°
= 7.152 cm
Area = 1 1a csin B (7.152)(7.568)sin 71 26 '2 2
= Β° = 25.65 2cm
4. Use the sine rule to solve triangle DEF, given d = 32.6 mm, e = 25.4 mm and D = 104Β°22β², and
find its area.
Triangle DEF is shown below.
From the sine rule, 32.6 25.4
sin104 22 ' sin E=
Β° from which, sin E = 25.4sin104 22 '
32.6Β° = 0.75477555
and E = 1sin 0.75477555β = 49.0Β° or 49Β°0β² Hence, β F = 180Β° - 104Β°22β² - 49Β°0β² = 26Β°38β²
From the sine rule, 32.6 fsin104 22 ' sin 26 38'
=Β° Β°
from which, f = 32.6sin 26 38'sin104 22 '
°°
= 15.09 mm
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121
Area = 1 1d esin F (32.6)(25.4)sin 26 38'2 2
= Β° = 185.6 2mm
5. Use the sine rule to solve triangle JKL, given j = 3.85 cm, k = 3.23 cm and K = 36Β° and find its
area.
Triangle JKL is shown below.
From the sine rule, 3.23 3.85
sin 36 sin J=
Β° from which, sin J = 3.85sin 36
3.23Β° = 0.7006109
and J = 1sin 0.7006109β = 44.476Β° = 44Β°29β² or J = 180Β° - 44Β°29β² = 135Β°31β² Case 1: When J = 44Β°29β², β L = 180Β° - 36Β° - 44Β°29β² = 99Β°31β²
From the sine rule, l 3.23sin 99 31' sin 36
=Β° Β°
from which, l = 3.23sin 99 31'sin 36
°°
= 5.420 cm
Area = 1 1l jsin K (5.420)(3.85)sin 362 2
= Β° = 6.132 2cm
Case 2: When J = 135Β°31β², β L = 180Β° - 135Β°31β² - 36Β° = 8Β°29β²
From the sine rule, l 3.23sin8 29 ' sin 36
=Β° Β°
from which, l = 3.23sin8 29 'sin 36
°°
= 0.811 cm
Area = 1 1j k sin L (3.85)(3.23)sin8 29 '2 2
= Β° = 0.917 2cm
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122
EXERCISE 58 Page 127 2. Use the cosine and sine rules to solve triangle PQR, given q = 3.25 m, r = 4.42 m and P = 105Β°,
and find its area
Triangle PQR is shown below.
By the cosine rule, 2 2 2p 4.42 3.25 2(4.42)(3.25)cos105= + β Β° = 19.5364 + 10.5625 β (-7.4359) = 37.5348 and p = 37.5348 = 6.127 m
From the sine rule, 6.127 4.42sin105 sin R
=Β°
from which, sin R = 4.42sin1056.127
Β° = 0.696816
and R = 1sin 0.696816β = 44.172Β° or 44Β°10β² β Q = 180Β° - 105Β° - 44Β°10β² = 30Β°50β²
Area = 1 (4.42)(3.25)sin1052
Β° = 6.938 2m
4. Use the cosine and sine rules to solve triangle XYZ, given x = 21 mm, y = 34 mm and
z = 42 mm, and find its area
Triangle XYZ is shown below.
By the cosine rule, 2 2 221 42 34 2(42)(34)cos X= + β
from which, cos X = 2 2 242 34 21 0.8679972(42)(34)+ β
=
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123
and β X = 1cos (0.867997)β = 29.77Β° or 29Β°46β²
From the sine rule, 21 34sin 29 46 ' sin Y
=Β°
from which, sin Y = 34sin 29 46 ' 0.803807021
Β°=
and β Y = 1sin 0.8038070β = 53.495Β° or 53Β°30β² Hence, β Z = 180Β° - 29Β°46β² - 53Β°30β² = 96Β°44β²
Area = 1 (21)(34)sin 96 44 '2
Β° = 355 2mm
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
124
EXERCISE 59 Page 129 2. Two sides of a triangular plot of land are 52.0 m and 34.0 m, respectively. If the area of the plot
is 620 2m find (a) the length of fencing required to enclose the plot, and (b) the angles of the
triangular plot.
The triangular plot of land ABC is shown below.
(a) Area = 620 = 1
2(52.0)(34.0)sin A from which, sin A = 620 0.7013571 (52.0)(34.0)
2
=
and β A = 1sin 0.701357β = 44.54Β° or 44Β°32β² By the cosine rule, 2 2 2BC (52.0) (34.0) 2(52.0)(34.0)cos 44.54= + β Β° = 1339.677 and BC = 1339.677 = 36.6 m Hence, length of fencing required = AB + BC + CA = 52.0 + 36.6 + 34.0 = 122.6 m
(b) Area = 620 = 12
(52.0)(36.6)sin B from which, sin B = 2(620)(52.0)(36.6)
and β B = 1 2(620)sin(52.0)(36.6)
β β ββ ββ β
= 40Β°39β²
β A = 44Β°32β² hence, β C = 180Β° - 44Β°32β² - 40Β°39β² = 94Β°49β² 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
125
(a) Using the cosine rule on triangle PQR shown below gives: 2 2 2RQ 8.0 4.5 2(8.0)(4.5)cos130= + β Β° = 130.53 and jib, RQ = 130.53 = 11.43 m = 11.4 m, correct to 3 significant figures
(b) From the sine rule, 4.5 11.43
sin R sin130=
Β° from which, sin R = 4.5sin130 0.3015923
11.43Β°=
and the angle between the jib and the tie rod, β R = 1sin 0.3015923β = 17.553Β° or 17Β°33β² 4. A building site is in the form of a quadrilateral as shown below, and its area is 1510 2m .
Determine the length of the perimeter of the site.
The quadrilateral is split into two triangles as shown in the diagram below.
Area = 1510 = 12
(52.4)(28.5)sin 72Β° + 12
(34.6)(x)sin 75Β°
i.e. 1510 = 710.15 + 16.71 x
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126
from which, x = 1510 710.1516.71β = 47.87 m
Hence, perimeter of quadrilateral = 52.4 + 28.5 + 34.6 + 47.9 = 163.4 m 5. Determine the length of members BF and EB in the roof truss shown below.
Using the cosine rule on triangle ABF gives: 2 2 2BF 2.5 5 2(2.5)(5)cos50= + β Β°= 15.18 from which, BF = 15.18 = 3.9 m Using the sine rule on triangle ABF gives:
3.9 2.5sin 50 sin B
=Β°
from which, sin B = 2.5sin 50 0.4910543.9
Β°=
and β ABF = 1sin 0.491054β = 29.41Β° Assuming β ABE = 90Β°, then β FBE = 90Β° - 29.41Β° = 60.59Β° Using the sine rule on triangle BEF gives:
4 3.9sin 60.59 sin E
=Β°
from which, sin E = 3.9sin 60.59 0.84934994
Β°=
and β E = 1sin 0.8493499β = 58.14Β° Thus, β EFB =180Β° - 58.14Β° - 60.59Β° = 61.27Β° Using the sine rule on triangle BEF again gives:
BE 4sin 61.27 sin 60.59
=Β°
from which, BE = 4sin 61.27sin 60.59
°°
= 4.0 m
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
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EXERCISE 60 Page 131 1. PQ and QR are the phasors representing the alternating currents in two branches of a circuit.
Phasor PQ is 20.0 A, and is horizontal. Phasor QR (which is joined to the end of PQ to form
triangle PQR) is 14.0 A and is at an angle of 35Β° to the horizontal. Determine the resultant
phasor PR and the angle it makes with phasor PQ.
Phasors PQ and QR are shown in the phasor diagram below.
Using the cosine rule, 2 2 2PR (20.0) (14.0) 2(20.0)(14.0)cos145= + β Β° = 1054.725 from which, resultant phasor, PR = 1054.725 = 32.48 A
Using the sine rule, 14.0 32.48sin P sin145
=Β°
from which, sin P = 14.0sin145 0.24723132.48
Β°=
and β P = 1sin 0.247231β = 14.31Β° or 14Β°19β² 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 ΞΈ between the centre
lines.
The triangle involving angle ΞΈ is shown below, where AB = 45 mm radius + 15 mm radius = 60 mm and BC = 35 mm radius + 15 mm radius = 50 mm
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
128
Using the cosine rule gives: 2 2 299.78 60 50 2(60)(50)cos= + β ΞΈ
from which, 2 2 260 50 99.78cos 0.642675
2(60)(50)+ β
ΞΈ = = β
and angle, ΞΈ = 1cos ( 0.642675)β β = 130Β° 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=
Β° from which, sin C = 12.0sin 40 0.241045
32.0Β°=
and β C = 1sin 0.241045β = 13.95Β° The angle between the crank and the connecting rod, β ABC =180Β° - 40Β° - 13.95Β° = 126.05Β° or 126Β°3β²
Using the sine rule gives:
AC 32.0sin126.05 sin 40
=Β° Β°
from which, AC = 32.0sin126.05sin 40
°°
= 40.25 cm
Alternatively, using the cosine rule, 2 2 2AC 12.0 32.0 2(12.0)(32.0)cos126.05 1619.9611= + β Β° = from which, AC = 1619.9611 = 40.25 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
129
6. In the diagram in question 5 above, determine how far C moves, correct to the nearest millimetre
when angle CAB changes from 40Β° to 160Β°, B moving in an anticlockwise direction.
A diagram showing the position of the crank and connecting rod when angle CAB is 160Β° is shown below.
Using the sine rule, 32.0 12.0
sin160 sin C=
Β° from which, sin C = 12.0sin160 0.1282576
32.0Β°=
and β C = 1sin 0.1282576β = 7.37Β° Hence, β ABβ²Cβ² =180Β° - 7.37Β° - 160Β° = 12.63Β° Using the sine rule again gives:
AC' 32.0sin12.63 sin160
=Β° Β°
from which, ACβ² = 32.0sin12.63sin160
°°
= 20.46 cm
Hence, the distance that C moves, i.e. CCβ² = AC - ACβ² = 40.25 β 20.46 = 19.8 cm
8. An aeroplane is sighted due east from a radar station at an elevation of 40Β° and a height of
8000 m, and later at an elevation of 35Β° and height 5500 m in a direction E 70Β° 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.
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130
Tan 40Β° = 8000OA
from which, OA = 8000tan 40Β°
= 9534.03 m
Tan 35Β° = 5500OB
from which, OB = 5500tan 35Β°
= 7854.81 m
From the cosine rule, 2 2 2AB (9534.03) (7854.81) 2(9534.03)(7854.81)cos 70= + β Β° from which, AB = 10068.24 m From the view shown below, XY = 8000 β 5500 = 2500 m
Hence, angle of descent, ΞΈ = 1 2500tan10068.24
β β ββ ββ β
= 13.95Β° or 13Β°57β²
Flight distance of aeroplane between observations, XZ = ( )2 210068.24 (2500)β‘ β€+β£ β¦
= 10373.98 m or 10.374 km
Hence, speed of descent = dis tan ce 10.374 km 10.374 60 6045time 45h
60 60
Γ Γ= =
Γ
km/h = 829.9 km/h
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131
CHAPTER 13 CARTESIAN AND POLAR CO-ORDINATES
EXERCISE 61 Page 134 2. Express (6.18, 2.35) as polar co-ordinates, correct to 2 decimal places, in both degrees and
radians. From the diagram, 2 2r 6.18 2.35= + = 6.61
and 1 1y 2.35tan tanx 6.18
β βΞΈ = = = 20.82Β° or 20.82180Ο
Γ rad = 0.36 rad
Hence, (6.18, 2.35) in Cartesian co-ordinates corresponds to (6.61, 20.82Β°) 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, 2 2r 5.4 3.7= + = 6.55
and 1 1y 3.7tan tanx 5.4
β βΞ± = = = 34.42Β°
Thus, ΞΈ = 180Β° - Ξ± = 145.58Β° or 145.58180Ο
Γ rad = 2.54 rad
Hence, (-5.4, 3.7) in Cartesian co-ordinates corresponds to (6.55, 145.58Β°) or (6.55, 2.54 rad) in polar co-ordinates.
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132
5. Express (-7, -3) as polar co-ordinates, correct to 2 decimal places, in both degrees and
radians. From the diagram, 2 2r 7 3= + = 7.62
and 1 3tan7
βΞ± = = 23.20Β°
Thus, ΞΈ = 180Β° + 23.20Β° = 203.20Β° or 203.20180Ο
Γ rad = 3.55 rad
Hence, (-7, -3) in Cartesian co-ordinates corresponds to (7.62, 203.20Β°) 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, 2 2r 9.6 12.4= + = 15.68
and 1 12.4tan9.6
βΞ± = = 52.25Β°
Thus, ΞΈ = 360Β° - 52.25Β° = 307.75Β° or 307.75180Ο
Γ rad = 5.37 rad
Hence, (9.6, -12.4) in Cartesian co-ordinates corresponds to (15.68, 307.75Β°) or (15.68, 5.37 rad) in polar co-ordinates.
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133
EXERCISE 62 Page 136 1. Express (5, 75Β°) as Cartesian co-ordinates, correct to 3 decimal places.
In the diagram, x = 5 cos 75Β° = 1.294 and y = 5 sin 75Β° = 4.830 Hence, (5, 75Β°) 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, 210Β°) as Cartesian co-ordinates, correct to 3 decimal places. x = 10.8 cos 210Β° = -9.353 y = 10.8 sin 210Β° = -5.400 Hence, (10.8, 210Β°) 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.
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134
(a) In the diagram below, hole A is at an angle of 90Β° Hence, in polar form, hole A is 40β 90Β°.
The holes will be equally displaced, 3605Β° i.e. 72Β° apart.
Thus, in polar form the holes are at 40β 90Β°, 40β (90Β° + 72Β°), i.e. 40β 162Β°, 40β (162Β° + 72Β°), i.e. 40β 234Β°, 40β (234Β° + 72Β°), i.e. 40β 306Β°, and 40β (306Β° + 72Β°), i.e. 40β 378Β° or 40β 18Β°. Summarising, the holes are at 40β 18Β°, 40β 90Β°, 40β 162Β°, 40β 234Β°, 40β 306Β°
(b) 40β 18Β° = (40 cos 18Β°, 40 sin 18Β°) = (38.04 + j12.36) in Cartesian form 40β 90Β° = (40 cos 90Β°, 40 sin 90Β°) = (0 + j40) in Cartesian form 40β 162Β° = (40 cos 162Β°, 40 sin 162Β°) = (-38.04 + j12.36) in Cartesian form 40β 234Β° = (40 cos 234Β°, 40 sin 234Β°) = (-23.51 β j32.36) in Cartesian form 40β 306Β° = (40 cos 306Β°, 40 sin 306Β°) = (23.51 - j32.36) in Cartesian form In triangle ABC in the above diagram, AC = 40 β 12.36 = 27.64, and BC = 38.04
Thus, by Pythagorasβ theorem, AB = ( )2 227.64 38.04+ = 47.02 mm
i.e. the shortest distance between the centres of two adjacent holes is 47.02 mm.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
135
CHAPTER 14 THE CIRCLE AND ITS PROPERTIES
EXERCISE 63 Page 138 1. If the radius of a circle is 41.3 mm, calculate the circumference of the circle. Circumference, c = 2Οr = 2Ο(41.3) = 259.5 mm 2. Find the diameter of a circle whose perimeter is 149.8 cm. If perimeter, or circumference, c = Οd, then 149.8 = Οd
and diameter, d = 149.8Ο
= 47.68 cm
3. A crank mechanism is shown below, where XY is a tangent to the circle at point X. If the circle
radius 0X is 10 cm and length 0Y is 40 cm, determine the length of the connecting rod XY.
If XY is a tangent to the circle, then β 0XY = 90Β° Thus, by Pythagoras, 2 2 20Y 0X XY= +
from which, 0Y = ( )2 2 2 20Y 0X 40 10 1500β = β = = 38.73 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
136
EXERCISE 64 Page 139 1. Convert to radians in terms of Ο: (a) 30Β° (b) 75Β° (c) 225Β°
(a) 30Β° = 30180Ο
Γ rad = 6Ο rad
(b) 75Β° = 75180Ο
Γ rad = 512Ο rad
(c) 225Β° = 225180Ο
Γ rad = 45 15rad rad36 12Ο Ο
= = 54Ο rad
2. Convert to radians: (a) 48Β° (b) 84Β°51β² (c) 232Β°15β²
(a) 48Β° = 48180Ο
Γ rad = 0.838 rad
(b) 84Β°51β² = 5184 84.9560 180 180
Ο Οβ βΓ = Γβ ββ β
rad = 1.481 rad
(c) 232Β°15β² = 232.25180Ο
Γ rad = 4.054 rad
3. Convert to degrees: (a) 5 rad6Ο (b) 4 rad
9Ο (c) 7 rad
12Ο
(a) 5 rad6Ο = 5 180
6Ο Β°Γ
Ο = 5 Γ 30 = 150Β°
(b) 4 rad9Ο = 4 180
9Ο Β°Γ
Ο = 4 Γ 20 = 80Β°
(c) 7 rad12Ο = 7 180
12Ο Β°Γ
Ο = 7 Γ 15 = 105Β°
4. Convert to degrees and minutes: (a) 0.0125 rad (b) 2.69 rad (c) 7.241 rad
(a) 0.0125 rad = 1800.0125 Β°Γ
Ο = 0.716Β° or 0Β°43β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
137
(b) 2.69 rad = 1802.69 Β°Γ
Ο = 154.126Β° or 154Β°8β²
(c) 7.241 rad = 1807.241 Β°Γ
Ο = 414.879Β° or 414Β°53β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
138
EXERCISE 65 Page 140 2. If the angle subtended at the centre of a circle of diameter 82 mm is 1.46 rad, find the lengths of
the (a) minor arc, (b) major arc
If diameter d = 82 mm, radius r = 822
= 41 mm
(a) Minor arc length, s = rΞΈ = (41)(1.46) = 59.86 mm (b) Major arc length = circumference β minor arc = 2Ο(41) β 59.86 = 257.61 β 59.86 = 197.8 mm 3. A pendulum of length 1.5 m swings through an angle of 10Β° in a single swing. Find, in
centimetres, the length of the arc traced by the pendulum bob.
Arc length of pendulum bob, s = rΞΈ = (1.5) 10180Οβ βΓβ β
β β = 0.262 m or 26.2 cm
5. Determine the angle of lap, in degrees and minutes, if 180 mm of a belt drive are in contact with
a pulley of diameter 250 mm.
Arc length, s = 180 mm, radius, r = 2502
= 125 mm
Since s = rΞΈ, the angle of lap, ΞΈ = s 180r 125= = 1.44 rad = 1801.44Γ
Ο = 82.506Β° or 82Β°30β²
6. Determine the number of complete revolutions a motorcycle wheel will make in travelling 2 km,
if the wheelβs diameter is 85.1 cm If wheel diameter = 85.1 cm, then circumference, c = Οd = Ο(85.1) cm = 267.35 cm = 2.6735 m
Hence, number of revolutions of wheel in travelling 2000 m = 20002.6735
= 748.08
Thus, number of complete revolutions = 748
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
139
8. Determine (a) the shaded area in the diagram below, (b) the percentage of the whole sector that
the area of the shaded portion represents.
(a) Shaded area = 2 21 1(50) (0.75) (38) (0.75)2 2
β = 2 21 (0.75) 50 382
β‘ β€ββ£ β¦ = 396 2mm
(b) Percentage of whole sector = 2
396 100%1 (50) (0.75)2
Γ = 42.24%
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
140
EXERCISE 66 Page 142 1. Determine the radius, and the co-ordinates of the centre of the circle given by the equation
2 2x y 6x 2y 26 0+ + β β =
Method 1: The general equation of a circle 2 2 2(x a) (y b) rβ + β = is 2 2x y 2ex 2fy c 0+ + + + =
where co-ordinate, a = - 2e2
, co-ordinate, b = - 2f2
and radius, r = 2 2a b c+ β
Hence, if 2 2x y 6x 2y 26 0+ + β β = then a = - 2e 62 2= β = -3, b = - 2f 2
2 2β
= β = 1
and radius, r = 2 2( 3) (1) ( 26) (9 1 26) 36β‘ β€β + β β = + + =β£ β¦ = 6
i.e. the circle 2 2x y 6x 2y 26 0+ + β β = has centre at (-3, 1) and radius 6, as shown below.
Method 2: 2 2x y 6x 2y 26 0+ + β β = may be rearranged as: 2 2(x 3) (y 1) 36 0+ + β β = i.e. 2 2 2(x 3) (y 1) 6+ + β = which has a radius of 6 and centre at (-3, 1)
2. Sketch the circle given by the equation 2 2x y 6x 4y 3 0+ β + β = Method 1: The general equation of a circle 2 2 2(x a) (y b) rβ + β = is 2 2x y 2ex 2fy c 0+ + + + =
where co-ordinate, a = - 2e2
, co-ordinate, b = - 2f2
and radius, r = 2 2a b c+ β
Hence, if 2 2x y 6x 4y 3 0+ β + β = then a = - 2e 62 2
β= β = 3, b = - 2f 4
2 2= β = -2
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
141
and radius, r = 2 2(3) ( 2) ( 3) (9 4 3) 16β‘ β€+ β β β = + + =β£ β¦ = 4
i.e. the circle 2 2x y 6x 4y 3 0+ β + β = has centre at (3, -2) and radius 4, as shown below.
Method 2: 2 2x y 6x 4y 3 0+ β + β = may be rearranged as: 2 2(x 3) (y 2) 16 0β + + β = i.e. 2 2 2(x 3) (y 2) 4β + + = which has a radius of 4 and centre at (3, -2)
4. Sketch the curve 2yx 6 1
6β‘ β€β β= ββ’ β₯β β
β β β’ β₯β£ β¦
If 2yx 6 1
6β‘ β€β β= ββ’ β₯β β
β β β’ β₯β£ β¦ then
2x y16 6
β‘ β€β β= ββ’ β₯β ββ β β’ β₯β£ β¦
and 2 2x y1
6 6β β β β= ββ β β ββ β β β
i.e. 2 2
2 2
x y 16 6
+ = and 2 2 2x y 6+ =
which is a circle of radius 6 and co-ordinates of centre at (0, 0), as shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
142
EXERCISE 67 Page 143 2. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm.
Determine the linear velocity of a point on the rim of one of the wheels of the bicycle, and the
angular velocity of the wheels.
Linear velocity, v = 36 km/h = 36 10003600Γ m/s = 10 m/s
(Note that changing from km/h to m/s involves dividing by 3.6)
Radius of wheel, r = 5002
= 250 mm = 0.25 m
Since, v = Οr, then angular velocity, Ο = v 10r 0.25= = 40 rad/s
3. A train is travelling at 108 km/h and has wheels of diameter 800mm.
(a) Determine the angular velocity of the wheels in both rad/s and rev/min.
(b) If the speed remains constant for 2.70 km, determine the number of revolutions made by a
wheel, assuming no slipping occurs.
(a) Linear velocity, v = 108 km/h = 1083.6
m/s = 30 m/s
Radius of wheel = 8002
= 400 mm = 0.4 m
Since, v = Οr, then angular velocity, Ο = v 30r 0.4= = 75 rad/s
rev 60s75rad / s 75rad / s2 rad min
= Γ ΓΟ
= 716.2 rev/min
(b) Linear velocity, v = st
hence, time, t = s 2700mv 30m / s= = 90 s = 90
60 = 1.5 minutes
Since a wheel is rotating at 716.2 rev/min, then in 1.5 minutes it makes 716.2 rev/min Γ 1.5 min = 1074 rev/min
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
143
EXERCISE 68 Page 145 2. Calculate the centripetal force acting on a vehicle of mass 1 tonne when travelling around a bend
of radius 125 m at 40 km/h. If this force should not exceed 750 N, determine the reduction in
speed of the vehicle to meet this requirement.
Centripetal force = 2mv
r where mass, m = 1 tonne = 1000 kg, radius, r = 125 m and
velocity, v = 40 km/h = 403.6
m/s
Hence, centrifugal force =
240(1000)3.6
125
β ββ ββ β = 988 N
If centrifugal force is limited to 750 N, then 2(1000)v750
125=
from which, velocity, v = (750)(125)1000
β ββ ββ β
= 9.6825 m/s
= 9.6825 Γ 3.6 = 34.89 km/h
Hence, reduction in speed is 40 β 34.89 = 5.14 km/h
3. A speed-boat negotiates an S-bend consisting of two circular arcs of radii 100 m and 150 m. If
the speed of the boat is constant at 34 km/h, determine the change in acceleration when leaving
one arc and entering the other.
Speed of the boat, v = 34 km/h = 343.6
m/s
For the first bend of radius 100 m, acceleration =
2
22
1
34v 3.6 0.892 m / sr 100
β ββ ββ β = =
For the second bend of radius 150 m, acceleration =
2
2
343.6 0.595m / s150
β ββ ββ β = β = β , the negative sign
indicating a change in direction Hence, change in acceleration is 0.892 β (-0.595) = 1.49 2m / s
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
144
CHAPTER 15 TRIGONOMETRIC WAVEFORMS
EXERCISE 69 Page 151 2. Determine the angles between 0Β° and 360Β° whose cosecant is 2.5317 Cosecant, and thus sine, is positive in the 1st and 2nd quadrants.
If cosec ΞΈ = 2.5317, then 1 1 1cos ec (2.5317) sin 23.265 or 23 16 '2.5317
β β β βΞΈ = = = Β° Β°β ββ β
.
From the diagram, the two values of ΞΈ between 0Β° and 360Β° are: 23Β°16β² and 180Β° - 23Β°16β² = 156Β°44β² 4. Solve the equation: 1cos ( 0.5316) tβ β = Cosine is negative in the 2nd and 3rd quadrants.
1cos (0.5316)β = 57.886Β° or 57Β°53β² as shown in the diagram below.
From the diagram, t = 180Β° - 57Β°53β² = 122Β°7β² and t = 180Β° + 57Β°53β² = 237Β°53β² 6. Solve the equation: 1tan 0.8314β = ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
145
Tangent is positive in the 1st and 3rd quadrants.
1tan 0.8314βΞΈ = = 39.74Β° or 39Β°44β²
From the diagram, the two values of ΞΈ between 0Β° and 360Β° are: 39Β°44β² and 180Β° + 39Β°44β² = 219Β°44β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
146
EXERCISE 70 Page 157
2. State the amplitude and period of the waveform y = 2 5xsin2
and sketch the curve between 0Β°
and 360Β°.
If y = 2 5xsin2
, amplitude = 2 and period = 36052
Β° = 144Β°
A sketch y = 2 5xsin2
is shown below.
4. State the amplitude and period of the waveform y = 3 cos2ΞΈ and sketch the curve between 0Β°
and 360Β°.
If y = 3 cos2ΞΈ , amplitude = 3 and period = 360
12
Β° = 720Β°
A sketch y = 3 cos2ΞΈ is shown below.
6. State the amplitude and period of the waveform y = 6 sin(t - 45Β°) and sketch the curve between 0Β° and 360Β°.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
147
If y = 6 sin(t - 45Β°), amplitude = 6 and period = 3601Β° = 360Β°
A sketch y = 6 sin(t - 45Β°) is shown below.
7. State the amplitude and period of the waveform y = 4 cos(2ΞΈ + 30Β°) and sketch the curve between 0Β° and 360Β°.
If y = 4 cos(2ΞΈ + 30Β°), amplitude = 4 and period = 3602Β° = 180Β°
A sketch y = 4 cos(2ΞΈ + 30Β°) is shown below.
(Note that y = 4 cos(2ΞΈ + 30Β°) leads y = 4 cos 2ΞΈ by 302Β° = 15Β°)
9. State the amplitude and period of the waveform y = 5 2 3cos2ΞΈ and sketch the curve between 0Β°
and 360Β°.
If y = 5 2 3cos2ΞΈ , amplitude = 5 and period = 180
32
Β° = 120Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
148
A sketch y = 5 2 3cos2ΞΈ is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
149
EXERCISE 71 Page 159 1. Find the amplitude, periodic time, frequency and phase angle (stating whether it is leading or
lagging A sin Οt) of the alternating quantity: i = 40 sin(50Οt + 0.29) mA If i = 40 sin(50Οt + 0.29) mA, then amplitude = 40 mA,
Ο = 50Ο rad/s = 2Οf from which, frequency, f = 502ΟΟ
= 25 Hz,
periodic time, T = 1 1f 25= = 0.040 s or 40 ms,
and phase angle = 0.29 rad leading or 1800.29 Β°Γ
Ο = 16.62Β° leading or 16Β°37β² leading.
4. A sinusoidal voltage has a maximum value of 120 V and a frequency of 50 Hz. At time t = 0, the
voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form
v = A sin(Οt Β± Ξ±). Let v = A sin(Οt Β± Ξ±) = 120 sin(2Οft + Ο) = 120 sin(100Οt + Ο) volts, since f = 50 Hz. (a) When t = 0, v = 0 hence, 0 = 120 sin(0 + Ο), i.e. 0 = 120 sin Ο from which, sin Ο = 0 and Ο = 0 Hence, if v = 0 when t = 0, then v = 120 sin 100Οt volts (b) When t = 0, v = 50 V hence, 50 = 120 sin(0 + Ο)
from which, 50 sin120
= Ο and 1 50sin 24.624 24.624 0.43rad120 180
β Οβ βΟ = = Β°= Γ =β ββ β
Hence, if v = 50 when t = 0, then v = 120 sin(100Οt + 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(Οt Β± Ξ±).
If periodic time T = 25 ms, then frequency, 3
1 1fT 25 10β= =
Γ = 40 Hz
Angular velocity, Ο = 2Οt =2Ο(40) = 80Ο rad/s
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
150
Hence, current i = 20 sin(80Οt + Ο) When t = 0, i = -10, hence -10 = 20 sin Ο
from which, sin Ο = 10 0.520
β = β and 1sin ( 0.5) 30 or rad6
β ΟΟ = β = β Β° β
Thus, i 20sin 80 t A6Οβ β= Ο ββ β
β β
7. The current in a.c. circuit at any time t seconds is given by: i = 5 sin(100Οt β 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(100Οt - 0.432) mA, then amplitude = 5 A,
Ο = 100Ο rad/s = 2Οf from which, frequency, f = 1002ΟΟ
= 50 Hz,
periodic time, T = 1 1f 50= = 0.020 s or 20 ms,
and phase angle = 0.432 rad lagging or 1800.432 Β°Γ
Ο = 24.75Β° lagging or 24Β°45β² lagging.
(b) When t = 0, i = 5 sin(- 0.432) = -2.093 A (note that -0.432 is radians) (c) When t = 8 ms, i = 5 ( )3sin 100 8 10 0.432ββ‘ β€Ο Γ ββ£ β¦ = 5 sin (2.081274) = 4.363 A
(d) When the current is first a maximum, 5 = 5 sin(100Οt β 0.432) i.e. 1 = sin(100Οt β 0.432) and 100Οt β 0.432 = 1sin 1 1.5708β = (again, be sure your calculator is on radians)
from which, time t = 1.5708 0.432100+Ο
= 0.006375 s or 6.375 ms
(e) When i = 3 A, 3 = 5 sin(100Οt β 0.432)
i.e. 35
= sin(100Οt β 0.432)
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
151
and 100Οt β 0.432 = 1 3sin 0.64355
β =
from which, time t = 0.6435 0.432100+Ο
= 0.003423 s or 3.423 ms
A sketch of one cycle of the waveform is shown below.
Note that since phase angle Ο = 24.75Β°, in terms of time tΟ then
t24.75360 20
Οβ‘ from which, tΟ = 1.375 ms
Alternatively, tΟ = 0.432100
Ο=
Ο Ο = 1.375 ms, as shown in the sketch.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
152
EXERCISE 72 Page 165 1. A complex current waveform i comprises a fundamental current of 50 A r.m.s. and frequency
100 Hz, together with a 24% third harmonic, both being in phase with each other at zero time.
(a) Write down an expression to represent current i. (b) Sketch the complex waveform of current
using harmonic synthesis over one cycle of the fundamental.
(a) Fundamental current: r.m.s. = 12
Γ maximum value
from which, maximum value = 2 r.m.s. 2 50Γ = Γ = 70.71 A Hence, fundamental current is: 1i = 70.71 sin 2Ο(100)t = 70.71 sin 628.3t A Third harmonic: amplitude = 24% of 70.71 = 16.97 A Hence, third harmonic current is: 3i = 16.97 sin 3(628.3)t = 16.97 sin 1885t A Thus, current i = 1i + 3i = 70.71 sin 628.3t + 16.97 sin 1885t amperes. (b) The complex waveform for current i is shown sketched below:
2. A complex voltage waveform v is comprised of a 212.1 V r.m.s. fundamental voltage at a
frequency of 50 Hz, a 30% second harmonic component lagging by Ο/2 rad, and a 10% fourth
harmonic leading by Ο/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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
153
(a) Voltage,
v = 212.1 212.1 212.1sin 2 (50)t (0.30) sin 2(2 )(50)t (0.1) sin 4(2 )(50)t0.707 0.707 2 0.707 3
Ο Οβ‘ β€ β‘ β€Ο + Ο β + Ο +β’ β₯ β’ β₯β£ β¦ β£ β¦ volts
i.e. v = 300 sin 314.2 t + 90sin 628.3t 30sin 1256.6t2 3Ο Οβ β β ββ + +β β β β
β β β β volts
(b) The complex waveform representing v is shown sketched below:
5. A voltage waveform is described by:
v = 200 sin 377t + 80sin 1131t 20sin 1885t4 3Ο Οβ β β β+ + ββ β β β
β β β β volts
Determine (a) the fundamental and harmonic frequencies of the waveform, (b) the percentage
third harmonic, and (c) the percentage fifth harmonic. Sketch the voltage waveform using
harmonic synthesis over one cycle of the fundamental.
Β© 2006 John Bird. All rights reserved. Published by Elsevier.
154
(a) From the fundamental voltage, 377 = 1 12 fΟ = Ο i.e. fundamental frequency, 1377f2
=Ο
= 60 Hz
From the 3rd harmonic voltage, 1131 = 3 32 fΟ = Ο
i.e. 3rd harmonic frequency, 31131f2
=Ο
= 180Hz
From the 5th harmonic voltage, 1885 = 5 52 fΟ = Ο
i.e. 5th harmonic frequency, 51885f
2=
Ο = 300Hz
(b) Percentage 3rd harmonic = 80 100%200
Γ = 40%
(c) Percentage 5th harmonic = 20 100%200
Γ = 10%
The complex waveform representing v is shown sketched below:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 155
CHAPTER 16 TRIGONOMETRIC IDENTITIES AND EQUATIONS
EXERCISE 73 Page 167
2. Prove the trigonometric identity: ( )2
1 cos ec1 cos
= ΞΈβ ΞΈ
L.H.S. = ( ) 22
1 1sin1 cos
=ΞΈβ ΞΈ
(since 2 2sin cos 1ΞΈ+ ΞΈ = )
= 1sinΞΈ
= cosec ΞΈ = R.H.S.
4. Prove the trigonometric identity: 3cos x cos x sin x cos x
sin xβ
=
L.H.S. = 3 2 2cos x cos x cos x(1 cos x) cos x sin x
sin x sin x sin xβ β
= = = cos x sin x = sin x cos x = R.H.S.
5. Prove the trigonometric identity: ( ) ( )2 2 21 cot 1 cot 2cos ec+ ΞΈ + β ΞΈ = ΞΈ L.H.S. = ( ) ( )2 2 2 21 cot 1 cot 1 2cot cot 1 2cot cot+ ΞΈ + β ΞΈ = + ΞΈ+ ΞΈ+ β ΞΈ+ ΞΈ
= ( )2 22 2cot 2 2 cos ec 1+ ΞΈ = + ΞΈβ
= 2 22 2cos ec 2 2cos ec+ ΞΈβ = ΞΈ = R.H.S.
6. Prove the trigonometric identity: ( )2sin x sec x cos ec x1 tan x
cos x tan x+
= +
L.H.S. = ( )2 2
21 1 sin x cos xsin x sin xsin x sec x cos ec x cos x sin x cos x sin x
sin xcos x tan x sin xcos xcos x
+β β β β+β β β β+ β β β β = =β ββ ββ β
= sin x cos x sin x cos xsin xcos x sin x cos x
+ +β β =β ββ β
= sin x cos x tan x 1 1 tan xcos x cos x
+ = + = + = R.H.S.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 156
EXERCISE 74 Page 169 2. Solve: 3 cosec A + 5.5 = 0 for angles between 0Β° and 360Β°
Since 3 cosec A + 5.5 = 0 then 3 cosec A = -5.5 and cosec A = 5.53
β
i.e. 1 5.5sin A 3
= β or sin A = 35.5
β
from which, A = 1 3sin5.5
β β βββ ββ β
= -33.056Β° or -33Β°3β²
Since sine is negative, the angle 33Β°3β² occurs in the 3rd and 4th quadrants as shown in the diagram
below.
Hence, the two angles for A between 0Β° and 360Β° whose sine is 3
5.5β are:
180Β° + 33Β°3β² = 213Β°3β² and 360Β° - 33Β°3β² = 326Β°57β² 3. Solve: 4(2.32 β 5.4 cot t) = 0 for angles between 0Β° and 360Β° Since 4(2.32 β 5.4 cot t) = 0 then 2.32 β 5.4 cot t = 0 and 2.32 = 5.4 cot t
i.e. cot t = 2.325.4
from which, tan t = 5.42.32
Hence, t = 1 5.4tan 66.75 or 66 45'2.32
β β β = Β° Β°β ββ β
Since tan is positive, the angle 66Β°45β² occurs in the 1st and 3rd quadrants as shown in the diagram
below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 157
Hence, the two angles for t between 0Β° and 360Β° whose tan is 5.4
2.32 are:
66Β°45β² and 180Β° + 66Β°45β² = 246Β°45β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 158
EXERCISE 75 Page 170 1. Solve: 25sin y 3= for angles between 0Β° and 360Β°
Since 25sin y 3= then 2 3sin y 0.605
= = and sin y 0.60 0.774596...= = Β±
and y = 1sin (0.774596...) 50 46 'β = Β° 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 0Β° and 360Β° are: 50Β°46β², 180Β° - 50Β°46β² = 129Β°14β², 180Β° + 50Β°46β² = 230Β°46β² and 360Β° - 50Β°46β² = 309Β°14β² 2. Solve: 25 3cos ec D 8+ = for angles between 0Β° and 360Β° Since 25 3cosec D 8+ = then 23cosec D 8 5 3= β = i.e. 2cos ec D 1=
Hence, 2
1 1sin D
= and 2sin D 1= from which, sin D = 1 1= Β±
and D = ( )1sin 1β Β±
There are two values of D between 0Β° and 360Β° which satisfy this equation, as shown in the
sinusoidal waveform below
Hence, D = 90Β° and 270Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 159
EXERCISE 76 Page 170 2. Solve: 28 tan 2 tan 15ΞΈ+ ΞΈ = for angles between 0Β° and 360Β° Since 28 tan 2 tan 15ΞΈ+ ΞΈ = then 28 tan 2 tan 15 0ΞΈ+ ΞΈβ = i.e. ( )( )4 tan 5 2 tan 3 0ΞΈβ ΞΈ+ =
i.e. 4 tan ΞΈ - 5 = 0 from which, tan ΞΈ = 5 1.254= and 1tan 1.25βΞΈ = = 51Β°20β²
and 2 tan ΞΈ + 3 = 0 from which, tan ΞΈ = 3 1.52
β = β and 1tan 1.5βΞΈ = β = -56Β°19β²
From the diagram below, the four values of ΞΈ between 0Β° and 360Β° are: 51Β°20β², 180Β° - 56Β°19β² = 123Β°41β², 180Β° + 51Β°20β² = 231Β°20β² and 360Β° - 56Β°19β² = 303Β°41β²
3. Solve: 22cosec t 5cosec t 12β = for angles between 0Β° and 360Β° Since 22cosec t 5cosec t 12β = then 22cosec t 5cosec t 12 0β β = and (2 cosec t + 3)(cosec t β 4) = 0
i.e. 2 cosec t + 3 = 0 from which, cosec t = 32
β and sin t = 23
β from which, t = -41Β°49β²
and cosec t - 4 = 0 from which, cosec t = 4 and sin t = 14
from which, t = 14Β°29β²
From the diagram below, the four values of ΞΈ between 0Β° and 360Β° are:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 160
14Β°29β², 180Β° - 14Β°29β² = 165Β°31β², 180Β° + 41Β°49β² = 221Β°49β² and 360Β° - 41Β°49β² = 318Β°11β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 161
EXERCISE 77 Page 171 1. Solve: 212sin 6 cosΞΈβ = ΞΈ for angles between 0Β° and 360Β° Since 212sin 6 cosΞΈβ = ΞΈ then ( )212 1 cos 6 cosβ ΞΈ β = ΞΈ
i.e. 212 12cos 6 cosβ ΞΈβ = ΞΈ
i.e. 212cos cos 6 0ΞΈ+ ΞΈβ =
Factorising gives: (4 cos ΞΈ + 3)(3 cos ΞΈ - 2) = 0
i.e. 4 cos ΞΈ + 3 = 0 from which, cos ΞΈ = 3 0.754
β = β and ΞΈ = 1cos ( 0.75)β β = -41Β°25β²
and 3 cos ΞΈ - 2 = 0 from which, cos ΞΈ = 23
and ΞΈ = 1 2cos3
β β ββ ββ β
= 48Β°11β²
From the diagram below, the four values of ΞΈ between 0Β° and 360Β° are: 48Β°11β², 180Β° - 41Β°25β² = 138Β°35β², 180Β° + 41Β°25β² = 221Β°25β² and 360Β° - 48Β°11β² = 311Β°49β²
3. Solve: 24cot A 6cosecA 6 0β + = for angles between 0Β° and 360Β° Since 24cot A 6cosecA 6 0β + = then ( )24 cosec A 1 6cosecA 6 0β β + =
and 24cosec A 6cosecA 2 0β + =
Factorising gives: (2 cosec A β 1) (2 cosec A β 2) = 0
i.e. 2 cosec A - 1 = 0 from which, cosec A = 12= and sin A = 2 which has no solutions
and 2 cosec A - 2 = 0 from which, cosec A = 1 and sin A = 1, which has only one solution
between 0Β° and 360Β°, i.e. A = 90Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 162
5. Solve: 22.9cos a 7sin a 1 0β + = for angles between 0Β° and 360Β° Since 22.9cos a 7sin a 1 0β + = then 22.9(1 sin a) 7sin a 1 0β β + =
i.e. 22.9 2.9sin a 7sin A 1 0β β + =
and 22.9sin a 7sin a 3.9 0+ β =
Hence, sin a = 27 7 4(2.9)( 3.9) 7 94.242(2.9) 5.8
β‘ β€β Β± β β β Β±β£ β¦ = = 0.46685 or -2.88064, which has
no solution
Thus, a = 1sin (0.46685)β = 27Β°50β²
and, from the diagram below, 180Β° - 27Β°50β² = 152Β°10β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 163
CHAPTER 17 THE RELATIONSHIP BETWEEN
TRIGONOMETRIC AND HYPERBOLIC FINCTIONS EXERCISE 78 Page 174 2. Verify the following identity by expressing in exponential form:
cos j(A β B) = cos jA cos jB + sin jA sin jB
L.H.S. = cos j(A β B) = ch(A β B) from (5), page 173
R.H.S. = cos jA cos jB + sin jA sin jB
= ch A ch B + j sh A j sh B from (5) and (6), page 173
= ch A ch B + 2j sh A sh B
= ch A ch B β sh A ah B
= ch(A β B)
Hence, L.H.S. = R.H.S. i.e. cos j(A β B) = cos jA cos jB + sin jA sin jB 3. Verify the following identity by expressing in exponential form:
cos j2A = 1 β 2 2sin jA
L.H.S. = cos j2A = ch 2A from (5), page 173
R.H.S. = 1 β 2 2sin jA = 1 β 2(sin jA)(sin jA) = 1 β 2(j sh A)(j sh A)
= 1 - 2 2 2 2j sh A 1 2sh A= +
= ch 2A from Table 5.1, page 45
Hence, L.H.S. = R.H.S. i.e. cos j2A = 1 β 2 2sin jA 5. Verify the following identity by expressing in exponential form:
sin jA β sin jB = A B A B2cos j sin j2 2+ ββ β β β
β β β ββ β β β
L.H.S. = sin jA β sin jB = j sh A β j sh B = j(sh A β sh B)
R.H.S. = A B A B2cos j sin j2 2+ ββ β β β
β β β ββ β β β
= A B A B2c h j s h2 2+ ββ β β β
β β β ββ β β β
= A B A B2 j c h s h2 2+ ββ β β β
β β β ββ β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 164
= A B A B A B A B2j ch ch sh sh sh ch ch sh2 2 2 2 2 2 2 2
β‘ β€ β‘ β€+ ββ’ β₯ β’ β₯β£ β¦ β£ β¦
=
A B A B A B A B A B A Bch ch sh ch ch ch ch sh sh sh sh ch2 2 2 2 2 2 2 2 2 2 2 22j
A B A Bsh sh ch sh2 2 2 2
β‘ β€β +β’ β₯β’ β₯β’ β₯ββ’ β₯β£ β¦
= 2 2 2 2A A B A B B A B B B A A2j ch sh ch ch ch sh sh sh ch sh sh ch2 2 2 2 2 2 2 2 2 2 2 2
β‘ β€β + ββ’ β₯β£ β¦
= 2 2 2 2A A B B B B A A2j sh ch ch sh sh ch sh ch2 2 2 2 2 2 2 2
β‘ β€β β β ββ + ββ β β ββ’ β₯β β β β β£ β¦
= 2 2 2 2A A B B B B A A2j sh ch ch sh sh ch ch sh2 2 2 2 2 2 2 2
β‘ β€β β β ββ β ββ β β ββ’ β₯β β β β β£ β¦
= ( ) ( )A A B B2j sh ch 1 sh ch 12 2 2 2
β‘ β€ββ’ β₯β£ β¦ from Table 5.1, page 45
= A A B Bj 2sh ch 2sh ch2 2 2 2
β‘ β€ββ’ β₯β£ β¦
= A Bj sh 2 sh 22 2
β‘ β€β β β βββ β β ββ’ β₯β β β β β£ β¦ from Table 5.1, page 45
= j [ sh A β sh B ] = L.H.S
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 165
EXERCISE 79 Page 175 1. Use the substitution A = jΞΈ to obtain the hyperbolic identity corresponding to the trigonometric identity: 2 21 tan A sec A+ = Let A = jΞΈ then 2 21 tan A sec A+ =
becomes: 1 + ( ) ( )2 2tan j sec jΞΈ = ΞΈ
i.e. ( ) ( )2 21 tan j sec j+ ΞΈ = ΞΈ
i.e. ( )2
2 11 j tanhcos j
β β+ ΞΈ = β βΞΈβ β
i.e. 2
2 2 11 j tanhch
β β+ ΞΈ = β βΞΈβ β
i.e. 2 21 tanh sec hβ ΞΈ = ΞΈ 3. Use the substitution A = jΞΈ and B = jΟ to obtain the hyperbolic identity corresponding to the trigonometric identity: sin(A β B) = sin A cos B β cos A sin B Substituting A = jΞΈ and B = jΟ in sin(A β B) = sin A cos B β cos A sin B gives: sin(jΞΈ β jΟ) = sin jΞΈ cos jΟ β cos jΞΈ sin jΟ i.e. j sinh (ΞΈ β Ο) = j sinh ΞΈ cosh Ο β j cosh ΞΈ sinh Ο i.e. j sinh (ΞΈ β Ο) = j (sinh ΞΈ cosh Ο β cosh ΞΈ sinh Ο) i.e. sinh (ΞΈ β Ο) = sinh ΞΈ cosh Ο β cosh ΞΈ sinh Ο 4. Use the substitution A = jΞΈ to obtain the hyperbolic identity corresponding to the
trigonometric identity: 2
2 tan Atan 2A1 tan A
=β
Substituting A = jΞΈ in 2
2 tan Atan 2A1 tan A
=β
gives: 2
2 tan jtan 2j1 tan j
ΞΈΞΈ =
β ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 166
i.e. ( )2 2 2
2 j tanh 2jtanhj tanh 21 j tanh1 jtanh
ΞΈ ΞΈΞΈ = =
β ΞΈβ ΞΈ
i.e. 2
2 j tanhj tanh 21 tanh
ΞΈΞΈ =
+ ΞΈ
i.e. 2
2 tanhtanh 21 tanh
ΞΈΞΈ =
+ ΞΈ
6. Use the substitution A = jΞΈ to obtain the hyperbolic identity corresponding to the
trigonometric identity: 3 3 1sin A sin A sin 3A4 4
= β
Substituting A = jΞΈ in 3 3 1sin A sin A sin 3A4 4
= β
gives: 3 3 1sin j sin j sin 3j4 4
ΞΈ = ΞΈβ ΞΈ
i.e. ( )3 3 3 1j sin h j sin h j sin h 34 4
ΞΈ = ΞΈβ ΞΈ
Dividing by j gives: ( )2 3 3 1j sin h sin h sin h 34 4
ΞΈ = ΞΈβ ΞΈ
i.e. ( )3 3 1sin h sin h sin h 34 4
β ΞΈ = ΞΈβ ΞΈ
i.e. 3 1 3sinh sinh 3 sinh4 4
ΞΈ = ΞΈ β ΞΈ
7. Use the substitution A = jΞΈ to obtain the hyperbolic identity corresponding to the trigonometric identity: ( )2 2cot A sec A 1 1β = Substituting A = jΞΈ in ( )2 2cot A sec A 1 1β =
gives: ( )2 2cot j sec j 1 1ΞΈ ΞΈβ =
i.e. 2 2
1 1 1 1tan j cos j
β ββ =β βΞΈ ΞΈβ β
i.e. ( ) ( )2 2
1 1 1 1tan j cos j
β ββ =β β
β βΞΈ ΞΈβ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 167
i.e. ( )2 2
1 1 1 1chjth
β ββ =β βΞΈβ β ΞΈ
and ( )22 2
1 sec h 1 1j th
ΞΈβ =ΞΈ
i.e. ( )2 2coth sec h 1 1β ΞΈ ΞΈβ =
or ( )2 2coth 1 sech 1ΞΈ β ΞΈ =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 168
CHAPTER 18 COMPOUND ANGLES
EXERCISE 80 Page 177
3. Show that (a) 2sin x sin x 3 cos x3 3Ο Οβ β β β+ + + =β β β β
β β β β
(b) 3sin cos2Οβ ββ βΟ = Οβ β
β β
(a) L.H.S. = 2 2 2sin x sin x sin x cos cos x sin sin x cos cos x sin3 3 3 3 3 3Ο Ο Ο Ο Ο Οβ β β β+ + + = + + +β β β β
β β β β
= 1 3 1 3sin x cos x sin x cos x2 2 2 2
β β β ββ β β β+ + β +β β β ββ β β ββ β β ββ β β β β β β β
= 32 cos x2
β ββ ββ ββ β
= 3 cos x = R.H.S.
The diagram below shows an equilateral triangle ABC of side 2 with each angle 60Β°. Angle A is
bisected. By Pythagoras, AD = 2 22 1 3β = . Hence, sin3Ο = sin 60Β° = 3
2 and cos 60Β° = 1
2
(b) L.H.S. = 3 3 3sin sin cos cos sin2 2 2Ο Ο Οβ β β‘ β€β β Ο = β Οβ Οβ β β’ β₯β β β£ β¦
= [ ]( 1)cos (0)sinβ β Οβ Ο = cos Ο = R.H.S.
4. Prove that: (a) 3sin sin 2 (sin cos )4 4Ο Οβ β β βΞΈ + β ΞΈβ = ΞΈ+ ΞΈβ β β β
β β β β
(b) ( )( )
cos 270tan
cos 360Β° + ΞΈ
= ΞΈΒ°β ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 169
(a) L.H.S. = 3 3 3sin sin sin cos cos sin sin cos cos sin4 4 4 4 4 4Ο Ο Ο Ο Ο Οβ β β β β β β βΞΈ+ β ΞΈβ = ΞΈ + ΞΈ β ΞΈ β ΞΈβ β β β β β β β
β β β β β β β β
= 1 1 1 1sin cos sin cos2 2 2 2
β‘ β€ β‘ β€β β β β β β β βΞΈ + ΞΈ β ΞΈ β β ΞΈβ’ β₯ β’ β₯β β β β β β β ββ β β β β β β β β£ β¦ β£ β¦
= [ ] ( )1 2sin cos sin cos sin cos2 2
ΞΈ+ ΞΈ+ ΞΈ+ ΞΈ = ΞΈ+ ΞΈ
= 2 (sin cos )ΞΈ+ ΞΈ = R.H.S. The diagram below shows an isosceles triangle where AB = BC = 1 and angles A and C are both
45Β°. By Pythagoras, AC = 2 21 1 2+ = . Hence, sin 4Ο = sin 45Β° = cos 45Β°= 1
2
(b) L.H.S. = ( )( )
cos 270 cos 270 cos sin 270 sin 0 ( 1)sincos 360 cos360 cos sin 360 sin (1)cos 0
Β° + ΞΈ Β° ΞΈβ Β° ΞΈ β β ΞΈ= =
Β°βΞΈ Β° ΞΈ+ Β° ΞΈ ΞΈ+
= sin tancos
ΞΈ= ΞΈ
ΞΈ = R.H.S.
7. Solve the equation: 4 sin(ΞΈ - 40Β°) = 2 sin ΞΈ for values of ΞΈ between 0Β° and 360Β° 4 sin(ΞΈ - 40Β°) = 2 sin ΞΈ
i.e. 4[sin ΞΈ cos 40Β° - cos ΞΈ sin 40Β°] = 2 sin ΞΈ
i.e. 3.064178 sin ΞΈ - 2.57115 cos ΞΈ = 2 sin ΞΈ
Hence, 1.064178 sin ΞΈ = 2.57115 cos ΞΈ
sin 2.57115 2.4160901cos 1.064178
ΞΈ= =
ΞΈ
i.e. tan ΞΈ = 2.4160901 and ΞΈ = 1tan (2.4160901)β = 67Β°31β² and 247Β°31β² (see diagram below)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 170
EXERCISE 81 Page 181 2. Change the function: 4 sin Οt β 3 cos Οt into the form R sin(Οt Β± Ξ±) Let 4 sin Οt β 3 cos Οt = R sin(Οt + Ξ±)
= R[sin Οt cos Ξ± + cos Οt sin Ξ±]
= (R cos Ξ±) sin Οt + (R sin Ξ±) cos Οt
Hence, 4 = R cos Ξ± from which, cos Ξ± = 4R
and -3 = R sin Ξ± from which, sin Ξ± = 3R
β
There is only one quadrant where both cosine is positive and sine is negative, i.e. the 4th quadrant,
as shown in the diagram below.
R = ( )2 24 3+ = 5 and Ξ± = 1 3tan4
β = 0.644 rad (make sure your calculator is on radians)
Hence, 4 sin Οt β 3 cos Οt = 5 sin(Οt β 0.644)
3. Change the function: -7 sin Οt + 4 cos Οt into the form R sin(Οt Β± Ξ±) Let -7 sin Οt + 4 cos Οt = R sin(Οt + Ξ±)
= R[sin Οt cos Ξ± + cos Οt sin Ξ±]
= (R cos Ξ±) sin Οt + (R sin Ξ±) cos Οt
Hence, -7 = R cos Ξ± from which, cos Ξ± = 7R
β
and 4 = R sin Ξ± from which, sin Ξ± = 4R
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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 171
R = ( )2 27 4+ = 8.062 and Ο = 1 4tan7
β = 0.519 rad
Thus, in the diagram, Ξ± = Ο - 0.519 = 2.622
Hence, -7 sin Οt + 4 cos Οt = 8.062 sin(Οt + 2.622)
5. Solve the following equations for values of ΞΈ between 0Β° and 360Β°:
(a) 2 sin ΞΈ + 4 cos ΞΈ = 3 (b) 12 sin ΞΈ - 9 cos ΞΈ = 7 (a) Let 2 sin ΞΈ + 4 cos ΞΈ = R sin(ΞΈ + Ξ±)
= R[sin ΞΈ cos Ξ± + cos ΞΈ sin Ξ±]
= (R cos Ξ±) sin ΞΈ + (R sin Ξ±) cos ΞΈ
Hence, 2 = R cos Ξ± from which, cos Ξ± = 2R
and 4 = R sin Ξ± from which, sin Ξ± = 4R
There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st
quadrant, as shown in the diagram below.
R = ( )2 24 2+ = 4.472 and Ξ± = 1 4tan
2β = 63.43Β° or 63Β°26β²
Hence, 2 sin ΞΈ + 4 cos ΞΈ = 4.472 sin(ΞΈ + 63Β°26β²)
Thus, since 2 sin ΞΈ + 4 cos ΞΈ = 3 then 4.472 sin(ΞΈ + 63Β°26β²) = 3
i.e. sin(ΞΈ + 63Β°26β²) = 3 0.670844.472
=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 172
and ΞΈ + 63Β°26β² = 1sin 0.67084β = 42Β°8β² or 180Β° - 42Β°8β² = 137Β°52β²
Thus, ΞΈ = 42Β°8β² - 63Β°26β² = -21Β°18β² β‘ 360Β° -21Β°18β² = 338Β°42β²
or ΞΈ = 137Β°52β² - 63Β°26β² = 74Β°26β²
i.e. ΞΈ = 74Β°26β² and 338Β°42β² satisfies the equation 2 sin ΞΈ + 4 cos ΞΈ = 3
(b) Let 12 sin ΞΈ - 9 cos ΞΈ = R sin(ΞΈ + Ξ±)
= R[sin ΞΈ cos Ξ± + cos ΞΈ sin Ξ±]
= (R cos Ξ±) sin ΞΈ + (R sin Ξ±) cos ΞΈ
Hence, 12 = R cos Ξ± from which, cos Ξ± = 12R
and -9 = R sin Ξ± from which, sin Ξ± = 9R
β
There is only one quadrant where both cosine is positive and sine is negative, i.e. the 4th
quadrant, as shown in the diagram below.
R = ( )2 212 9+ = 15 and Ξ± = 1 9tan
12β = 36Β°52β²
Hence, 12 sin ΞΈ - 9 cos ΞΈ = 15 sin(ΞΈ - 36Β°52β²)
Thus, since 12 sin ΞΈ - 9 cos ΞΈ = 7 then 15 sin(ΞΈ - 36Β°52β²) = 7
i.e. sin(ΞΈ - 36Β°52β²) = 715
and ΞΈ - 36Β°52β² = 1 7sin15
β = 27Β°49β² or 180Β° - 27Β°49β² = 152Β°11β²
Thus, ΞΈ = 27Β°49β² + 36Β°52β² = 64Β°41β² or ΞΈ = 152Β°11β² + 36Β°52β² = 189Β°3β²
i.e. ΞΈ = 64Β°41β² and 189Β°3β² satisfies the equation 12 sin ΞΈ - 9 cos ΞΈ = 7 7. The third harmonic of a wave motion is given by: 4.3 cos 3ΞΈ - 6.9 sin 3ΞΈ.
Express this in the form R sin(3ΞΈ Β± Ξ±)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 173
Let 4.3 cos 3ΞΈ - 6.9 sin 3ΞΈ = R sin(3ΞΈ + Ξ±)
= R[sin 3ΞΈ cos Ξ± + cos 3ΞΈ sin Ξ±]
= (R cos Ξ±) sin 3ΞΈ + (R sin Ξ±) cos 3ΞΈ
Hence, -6.9 = R cos Ξ± from which, cos Ξ± = 6.9R
β
and 4.3 = R sin Ξ± from which, sin Ξ± = 4.3R
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 = ( )2 26.9 4.3+ = 8.13 and Ο = 1 4.3tan6.9
β = 31Β°56β²
and Ξ± = 180Β° - 31Β°56β² = 148Β°4β² = 2.584 rad
Hence, 4.3 cos 3ΞΈ - 6.9 sin 3ΞΈ = 8.13 sin(3ΞΈ + 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 Οt + 3.2 cos Οt, where t is the time in seconds. Express x in the form R sin(Οt + Ξ±). Let x = 2.4 sin Οt + 3.2 cos Οt = R sin(Οt + Ξ±)
= R[sin Οt cos Ξ± + cos Οt sin Ξ±]
= (R cos Ξ±) sin Οt + (R sin Ξ±) cos Οt
Hence, 2.4 = R cos Ξ± from which, cos Ξ± = 2.4R
and 3.2 = R sin Ξ± from which, sin Ξ± = 3.2R
There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st quadrant, as
shown in the diagram below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 174
R = ( )2 22.4 3.2+ = 4 and Ξ± = 1 3.2tan2.4
β = 53.13Β° or 0.927 rad
Hence, x = 2.4 sin Οt + 3.2 cos Οt = 4 sin(Οt + 0.927) m
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 175
EXERCISE 82 Page 183
2. Prove the following identities: (a) 22
cos 21 tancos
Οβ = Ο
Ο (b) 2
2
1 cos 2t 2cot tsin t+
=
(c) ( )( )tan 2x 1 tan x 2tan x 1 tan x
+=
β
(d) 2 cosec 2ΞΈ cos 2ΞΈ = cot ΞΈ - tan ΞΈ
(a) L.H.S. = 2 2 2 2
2 2 2 2
cos 2 cos sin cos sin1 1 1cos cos cos cos
β β β βΟ Οβ Ο Ο Οβ = β = β ββ β β βΟ Ο Ο Οβ β β β
= 1- ( )21 tanβ Ο = 2tan Ο = R.H.S.
(b) L.H.S. = ( )2 2
22 2 2
1 2cos t 11 cos 2t 2cos t 2cot tsin t sin t sin t
+ β+= = = = R.H.S.
(c) L.H.S. = ( )( ) ( )( )( )( )( )2
2 tan x 1 tan x2 tan x 1 tan xtan 2x 1 tan x 1 tan x 1 tan x1 tan xtan x tan x tan x
+β β +β β+ β +ββ β = =
= ( )( )
2 tan x1 tan x 2 tan x 2
tan x tan x 1 tan x 1 tan xβ
= =β β
= R.H.S.
(d) L.H.S. = 2 cosec 2ΞΈ cos 2ΞΈ = 2
2 2 2(cos 2 ) 2cot 2 2 tansin 2 tan 21 tan
β β ΞΈ = ΞΈ = =β β ΞΈΞΈ ΞΈβ β β ΞΈ
= ( )2 22 1 tan 1 tan 1 tan cot tan
2 tan tan tanβ ΞΈ β ΞΈ
= = β ΞΈ = ΞΈβ ΞΈΞΈ ΞΈ ΞΈ
= R.H.S.
3. If the third harmonic of a waveform is given by 3V cos3ΞΈ , express the third harmonic in terms of
the first harmonic cos ΞΈ, when 3V = 1. When 3V = 1, 3V cos3ΞΈ = cos 3ΞΈ = cos(2ΞΈ + ΞΈ) = cos 2ΞΈ cos ΞΈ - sin 2ΞΈ sin ΞΈ = ( ) ( )22cos 1 (cos ) 2sin cos (sin )ΞΈβ ΞΈ β ΞΈ ΞΈ ΞΈ = 3 22cos cos 2cos sinΞΈβ ΞΈβ ΞΈ ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 176
= ( )3 22cos cos 2cos 1 cosΞΈβ ΞΈβ ΞΈ β ΞΈ = 3 32cos cos 2cos 2cosΞΈβ ΞΈβ ΞΈ+ ΞΈ = 34cos 3cosΞΈβ ΞΈ = R.H.S.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 177
EXERCISE 83 Page 184 2. Express as a sum or difference: cos 8x sin 2x
cos 8x sin 2x = [ ]1 sin(8x 2x) sin(8x 2x)2
+ β β from (2), page 183
= [ ]1 sin10x sin 6x2
β
4. Express as a sum or difference: 4 cos 3ΞΈ cos ΞΈ
4 cos 3ΞΈ cos ΞΈ = [ ]14 cos(3 ) cos(3 )2
β§ β«ΞΈ+ ΞΈ + ΞΈβΞΈβ¨ β¬β© β
from (3), page 183
= 2[cos 4ΞΈ + cos 2ΞΈ]
6. Determine: 2sin 3t cos t dtβ«
2 sin 3t cost = [ ]12 sin(3t t) sin(3t t)2
β§ β«+ + ββ¨ β¬β© β
from (1), page 183
= sin 4t + sin 2t
Hence, ( )2sin 3t cos t dt sin 4t sin 2t dt= +β« β« = 1 1cos 4t cos 2t c4 2
β β +
8. Solve the equation: 2 sin 2Ο sin Ο = cos Ο in the range Ο = 0Β° to Ο = 180Β°. 2 sin 2Ο sin Ο = cos Ο i.e. 2(2 sin Ο cos Ο) sin Ο = cos Ο i.e. 24sin cos cosΟ Ο = Ο i.e. 24sin cos cos 0Ο Οβ Ο = and ( )2cos 4sin 1 0Ο Οβ = Hence, cos Ο = 0 from which, Ο = 1cos 0β = 90Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 178
and 24sin 1Ο = from which, 2 1sin4
Ο = and sin Ο = 1 0.54= Β±
Hence, Ο = 1sin 0.5β = 30Β° and 150Β° (see diagram below) and Ο = 1sin ( 0.5)β β = 210Β° and 330Β°
Since the range is from Ο = 0Β° to Ο = 180Β°, then the only values of Ο to satisfy: 2 sin 2Ο sin Ο = cos Ο are: Ο = 30Β°, 90Β° and 150Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 179
EXERCISE 84 Page 185 1. Express as a product: sin 3x + sin x
sin 3x + sin x = 3x x 3x x2sin cos2 2+ ββ β β β
β β β ββ β β β
from (5), page 184
= 2 sin 2x cos x
3. Express as a product: cos 5t + cos 3t
cos 5t + cos 3t = 5t 3t 5t 3t2cos cos2 2+ ββ β β β
β β β ββ β β β
from (7), page 184
= 2 cos 4t cos t
5. Express as a product: 1 cos cos2 3 4
Ο Οβ β+β ββ β
1 1 3 4 3 4cos cos 2cos cos2 3 4 2 2 2
β§ Ο Ο Ο Ο β«β β β β+ ββͺ βͺβ β β βΟ Ο βͺ βͺβ β+ = β¨ β¬β β β ββ ββ β βͺ βͺβ β β β
βͺ βͺβ β β β β© β
from (7), page 184
=
712 12cos cos2 2
Ο Οβ β β ββ β β ββ β β ββ β β ββ β β β
= 7cos cos24 24Ο Ο
6. Show that: (a) sin 4x sin 2x tan xcos 4x cos 2x
β=
+ (b) 1 sin(5x ) sin(x ) cos3x sin(2x )
2βΞ± β +Ξ± = βΞ±
(a) L.H.S. =
4x 2x 4x 2x2cos sinsin 4x sin 2x 2 2
4x 2x 4x 2xcos 4x cos 2x 2cos cos2 2
+ ββ β β ββ β β ββ β β β β =
+ β+ β β β ββ β β ββ β β β
= 2cos3x sin x sin x tan x2cos3x cos x cos x
= = = R.H.S.
(b) L.H.S. = 1 sin(5x ) sin(x )2
βΞ± β +Ξ±
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 180
= [ ]1 (sin 5x cos cos5x sin ) (sin x cos cos x sin )2
Ξ± β Ξ± β Ξ± + Ξ±
= [ ]1 cos (sin 5x sin x) sin (cos5x cos x)2
Ξ± β β Ξ± +
= 1 5x x 5x x 5x x 5x xcos 2cos sin sin 2cos cos2 2 2 2 2β‘ β€β§ + β β« β§ + β β«β β β β β β β βΞ± β Ξ±β¨ β¬ β¨ β¬β’ β₯β β β β β β β β
β β β β β β β β β© β β© ββ£ β¦
= [ ]1 2cos (cos3x sin 2x) 2sin (cos3x cos 2x)2
Ξ± β Ξ±
= cos 3x (cos Ξ± sin 2x β sin Ξ± cos 2x)
= cos 3x (sin 2x cos Ξ± β cos 2x sin Ξ±)
= cos 3x sin(2x - Ξ±) = R.H.S.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 181
CHAPTER 19 FUNCTIONS AND THEIR CURVES
EXERCISE 85 Page 199 1. Sketch y = 3x - 5
2. Sketch y = -3x + 4
3. Sketch y = 2x 3+
4. Sketch y = ( )2x 3β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 182
5. Sketch y = ( )2x 4 2β +
6. Sketch y = x - 2x
7. Sketch y = 3x 2+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 183
8. Sketch y = 1 + cos 3x
9. Sketch y = 3 - 2sin x4Οβ β+β β
β β
10. Sketch y = 2 ln x
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 184
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 185
EXERCISE 86 Page 201 1. Determine whether the following functions are even, odd or neither even nor odd: (a) 4x (b) tan 3x (c) 3t2e (d) 2sin x (a) Let f(x) = 4x . Since f(-x) = f(x) then 4x is an even function and is symmetrical about the f(x)
axis as shown below:
(b) Let f(x) = tan 3x. Since f(-x) = - f(x) then tan 3x is an odd function and is symmetrical about
the origin as shown below:
(c) Let f(t) = 3t2e . The function is neither even not odd, and is as shown below:
(d) Let f(x) = 2sin x . Since f(-x) = f(x) then 2sin x is an even function and is symmetrical about
the f(x) axis as shown below:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 186
3. State whether the following functions, which are periodic of period 2Ο, are even or odd:
(a) , when 0
f ( ), when 0ΞΈ β Ο β€ ΞΈ β€β§
ΞΈ = β¨βΞΈ β€ ΞΈ β€ Οβ© (b)
x, when x2 2f (x)
30, when x2 2
Ο Οβ§ β β€ β€βͺβͺ= β¨ Ο Οβͺ β€ β€βͺβ©
(a) A sketch of f(ΞΈ) against ΞΈ is shown below. Since the function is symmetrical about the f(ΞΈ) 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 187
EXERCISE 87 Page 203 2. Determine the inverse function of f(x) = 5x - 1 If y = f(x), then y = 5x - 1
Transposing for x gives: x = y 15+
Interchanging x and y gives: y = x 15+
Hence, if f(x) = 5x β 1, then 1 1f (x) (x 1)5
β = +
4. Determine the inverse function of 1f (x) 2x
= +
If y = f(x), then y = 1 2x+
Transposing for x gives: x = 1y 2β
Interchanging x and y gives: y = 1x 2β
Hence, if 1f (x) 2x
= + , then 1 1f (x)x 2
β =β
6. Determine the principal value of the inverse function 1cos 0.5β
Using a calculator (set on radians), 1cos 0.5β = 1.0472 rad or 3Ο rad
8. Determine the principal value of the inverse function 1cot 2β
Using a calculator (set on radians), 1cot 2β = 1 1tan2
β = 0.4636 rad
10. Determine the principal value of the inverse function 1sec 1.5β
Using a calculator (set on radians), 1sec 1.5β = 1 1cos1.5
β β ββ ββ β
= 0.8411 rad
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 188
12. Evaluate x, correct to 3 decimal places: x = 1 1 11 4 8sin cos tan3 5 9
β β β+ β
Using a calculator (set on radians), x = 1 1 11 4 8sin cos tan3 5 9
β β β+ β
= 0.3398 + 0.6435 β 0.7266 = 0.257 13. Evaluate y, correct to 4 significant figures: y = 1 1 13sec 2 4cos ec 2 5cot 2β β ββ + Using a calculator (set on radians), y = 1 1 13sec 2 4cos ec 2 5cot 2β β ββ +
= 1 1 11 1 13cos 4sin 5 tan22 2
β β ββ β β ββ + β ββ β β β β β
= 2.3562 β 3.1416 + 2.3182 = 1.533
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 189
EXERCISE 88 Page 208
2. Determine the asymptotes parallel to the x- and y-axes for 2 xyx 3
=β
2 xy
x 3=
β hence, 2y (x 3) xβ =
and 2y (x 3) x 0β β = (1) i.e. 2 2y x 3y x 0β β = i.e. ( )2 2x y 1 3y 0β β = (2) From equation (1), equating highest power of y to zero gives: x β 3 = 0, i.e. x = 3 From equation (2), equating highest power of x to zero gives: 2y β 1 = 0, i.e. y = Β± 1 Hence, asymptotes parallel to the x- and y-axes occur at x = 3, y = 1 and y = -1
3. Determine the asymptotes parallel to the x- and y-axes for x(x 3)y(x 2)(x 1)
+=
+ +
x(x 3)y
(x 2)(x 1)+
=+ +
hence, y(x + 2)(x + 1) = x(x + 3) (1)
i.e. ( )2 2y x 3x 2 x 3x 0+ + β β = and 2 2yx 3yx 2y x 3x 0+ + β β = i.e. 2x (y 1) 3xy 2y 3x 0β + β β = (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 β 1 = 0, i.e. y = 1 Hence, asymptotes parallel to the x- and y-axes occur at x = -2, x = -1 and at y = 1 5. Determine all the asymptotes for ( )2 2x y 16 yβ = Equating highest power of x to zero gives: 2y 16β = 0, i.e. y = Β± 4
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 190
Since ( )2 2x y 16 yβ = then 2 2 2x y 16x y 0β β = Equating highest power of y to zero gives: 2x = 0, i.e. x = 0 Let y = mx + c, then ( )22x mx c 16 mx cβ‘ β€+ β = +β£ β¦
i.e. 2 2 2 2x m x 2mxc c 16 mx cβ‘ β€+ + β = +β£ β¦ i.e. 2 4 3 2 2 2m x 2mcx c x 16x mx 1 0+ + β β β = Equating coefficient of highest power of x to zero gives: 2m 0= , i.e. m = 0 Equating next coefficient of highest power of x to zero gives: 2mc = 0, i.e. c = 0 Hence, the only asymptotes occur at y = 4, y = -4 and at x = 0 7. Determine the asymptotes and sketch the curve for 2 2xy x y 2x y 5β + β = 2 2xy x y 2x y 5β + β = (1) Equating the highest power of y to zero gives: x = 0, which is an asymptote. Equating the highest power of x to zero gives: -y = 0, i.e. y = 0, which is an asymptote. Letting y = mx + c in equation (1) gives: ( ) ( )2 2x mx c x mx c 2x (mx c) 5+ β + + β + = i.e. ( )2 2 2 3 2x m x 2mcx c mx cx 2x mx c 5 0+ + β β + β β β = and 2 3 2 2 3 2m x 2mcx c x mx cx 2x mx c 5 0+ + β β + β β β = i.e. ( ) ( ) ( )2 3 2 2m m x 2mc c x x c 2 m c 5 0β + β + + β β β = Equating the coefficient of the highest power of x to zero gives: 2m m 0β = , i.e. m(m β 1) = 0 i.e. m = 0 or m = 1 Equating the coefficient of the next highest power of x to zero gives: 2mc β c = 0 When m = 0, c = 0 and when m = 1, 2c β 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
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 191
A sketch of the curve 2 2xy x y 2x y 5β + β = , together with its asymptotes is shown below:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 192
EXERCISE 89 Page 212
1. Sketch the graphs of (a) 2 7y 3x 9x4
= + + (b) 2y 5x 20x 50= β + +
(a) 2 7y 3x 9x4
= + + dy 6x 9 0dx
= + = for a turning point
from which, x = 96
β = -1.5
When x = -1.5, y = 23( 1.5) 9( 1.5) 1.75β + β + = -5 Hence, a turning point occurs at (-1.5, -5)
2
2
d y 6dx
= , which is positive, hence, (-1.5, -5) is a minimum point.
A sketch of the graph 2 7y 3x 9x4
= + + is shown below.
(b) 2y 5x 20x 50= β + + dy 10x 20 0dx
= β + = for a turning point
from which, 20 = 10x and x = 2 When x = 2, 2y 5(2) 20(2) 50= β + + = 70 Hence, a turning point occurs at (2, 70)
2
2
d y 10dx
= β , which is negative, hence, (2, 70) is a maximum point.
A sketch of the graph 2y 5x 20x 50= β + + is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 193
4. Sketch the curve depicting: 2
2 x 16y4β
=
Since 2
2 x 16y4β
= then 2 24y x 16= β
i.e. 2 216 x 4y= β
i.e. 2 2x 4y 1
16 16β =
and 2 2
2 2
x y 14 2
β = which is a hyperbola, symmetrical about the x- and y-
axes, distance between vertices being 2(4), i.e. 8 units along the x-axis.
A sketch of 2
2 x 16y4β
= , i.e. 2 2
2 2
x y 14 2
β = is shown below.
5. Sketch the curve depicting: 2 2y x5
5 2= β
Since 2 2y x5
5 2= β then
2 2x y 52 5+ =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 194
and 2 2x y 1
10 25+ =
i.e. ( )
2 2
2 2
x y 1510
+ =
which is an ellipse, centre (0, 0), major axis, AB = 2(5) = 10 units along y-axis, and minor axis,
CD = 2 10 along the x-axis.
A sketch of the curve 2 2y x5
5 2= β , i.e.
( )2 2
2 2
x y 1510
+ = is shown below.
7. Sketch the curve depicting: 2 2x y 9=
Since 2 2x y 9= then 22
9yx
= and 3yx
=
which is a rectangular hyperbola, lying in the 1st and 3rd quadrants only, as shown in the sketch below.
9. Sketch the circle given by the equation 2 2x y 4x 10y 25 0+ β + + =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 195
Since 2 2x y 4x 10y 25 0+ β + + = then ( ) ( )2 2x 2 y 5 4 0β + + β = i.e. ( ) ( )2 2 2x 2 y 5 2β + + = which is a circle of centre (2, -5) and radius 2, as shown in the sketch below.
11. Describe the shape of the curve represented by the equation ( )2y 3 x 1β‘ β€= ββ£ β¦
Since ( )2y 3 x 1β‘ β€= ββ£ β¦ then ( )2 2y 3 x 1= β
and 2 2y 3x 3= β
i.e. 2 23 3x y= β
i.e. 2
2 y1 x3
= β
i.e. ( ) ( )
2 2
2 2x y 11 3
β =
which is a hyperbola, symmetrical about the x- and y-axes, with vertices 2(1) = 2 units apart
along the x-axis.
14. Describe the shape of the curve represented by the equation ( )12y 3x=
Since ( )12y 3x= then ( )y 3x= or 2y 3x=
which is a parabola, vertex at (0, 0) and symmetrical about the x-axis.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 196
15. Describe the shape of the curve represented by the equation 2 2y 8 2xβ = β Since 2 2y 8 2xβ = β then 2 2y 2x 8+ =
and 2 2y 2x 1
8 8+ =
i.e. ( )
2 2
2 2
y x 1(2)8
+ =
which is an ellipse, centre (0, 0), with major axis 2 8 units along the y-axis, and minor axis
2(2) = 4 units along the x-axis.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 197
CHAPTER 20 IRREGULAR AREAS, VOLUMES AND MEAN
VALUES OF WAVEFORMS EXERCISE 90 Page 218 1. Plot a graph of 2y 3x x= β 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 Simpsonβs rule. A table of values is shown below.
x 0 0.5 1.0 1.5 2.0 2.5 3.0 2y 3x x= β 0 1.25 2.0 2.25 2.0 1.25 0
A graph of 2y 3x x= β is shown below.
(a) Using the trapezoidal rule, with 6 intervals each of width 0.5 gives:
area ( ) 0 00.5 1.25 2.0 2.25 2.0 1.252
β‘ + β€β ββ + + + + +β ββ’ β₯β β β£ β¦ = (0.5)(8.75) = 4.375 square units
(b) Using the mid-ordinate rule, with 6 intervals, with mid-ordinates occurring at 0.25 0.75 1.25 1.75 2.25 2.75
where the y-values are: 0.6875 1.6875 2.1875 2.1875 1.6875 0.6875 area β (0.5)[0.6875 + 1.6875 + 2.1875 + 1.6875 + 0.6875] = (0.5)(9.125) = 4.563 square units (c) Using Simpsonβs rule, with 6 intervals each of width 0.5 gives:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 198
area ( ) ( ) ( ) [ ]1 1(0.5) 0 0 4 1.25 2.25 1.25 2 2.0 2.0 (0.5) 0 19 83 3
β + + + + + + = + +β‘ β€β£ β¦
= 1 (0.5)(27)3
= 4.5 square units
Simpsonβs 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 Simpsonβs
rule. Using Simpsonβs rule with 6 intervals each of width 1 s gives:
area ( ) ( ) ( ) [ ]1 1(1) 0 29.0 4 2.0 8.0 21.0 2 4.5 14.0 29.0 124 373 3
β + + + + + + = + +β‘ β€β£ β¦
= 1 (190)3
= 63.33 square units
5. The deck of a ship is 35 m long. At equal intervals of 5 m the width is given by the following
table: Width (m) 0 2.8 5.2 6.5 5.8 4.1 3.0 2.3
Estimate the area of the deck. Using the trapezoidal rule with 7 intervals each of width 5 m gives:
area ( ) [ ]0 2.35 2.8 5.2 6.5 5.8 4.1 3.0 (5) 1.15 27.42
β‘ + β€β ββ + + + + + + = +β ββ’ β₯β β β£ β¦
= (5)(28.55) = 143 2m (To use Simpsonβs rule needs an even number of intervals, so could not be used in this question).
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 199
EXERCISE 91 Page 219 1. The areas of equidistantly spaced sections of the underwater form of a small boat are as follows:
1.76 2.78 3.10 3.12 2.61 1.24 0.85 2m
Determine the underwater volume if the sections are 3 m apart.
Underwater volume = [ ]3 (1.76 0.85) 4(2.78 3.12 1.24) 2(3.10 2.61)3
+ + + + + +
= 2.61 + 28.56 + 11.42 = 42.59 3m 3. The circumference of a 12 m long log of timber of varying circular cross-section is measured at
intervals of 2 m along its length and the results are:
Distance from one end (m) 0 2 4 6 8 10 12
Circumference (m) 2.80 3.25 3.94 4.32 5.16 5.82 6.36
Estimate the volume of the timber in cubic metres.
If circumference c = 2Οr then radius, r = c2Ο
Cross-sectional area = 2 2
2 c cr2 4
β βΟ = Ο =β βΟ Οβ β
Hence, the cross-sectional areas are: 2
22.80 0.6239m4
=Ο
, 2
23.25 0.8405m4
=Ο
, 1.2353 2m ,
1.4851 2m , 2.1188 2m , 2.6955 2m , 3.2189 2m Hence, volume of timber
( ) ( ) ( )2 0.6239 3.2189 4 0.8405 1.4851 2.6955 2 1.2353 2.11883
β + + + + + +β‘ β€β£ β¦
= ( ) ( )2 23.8428 20.0844 6.7082 30.63543 3
+ + =
= 20.42 3m
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 200
EXERCISE 92 Page 222 1. Determine the mean value of the periodic waveforms shown over half a cycle.
(a) Over half a cycle, mean value = 3
3
area under curve (2 10 10 )Aslength of base 10 10 s
β
β
Γ Γ=
Γ = 2 A
(b) Over half a cycle, mean value = ( )3
3
1 5 10 (100)Vs2
5 10 s
β
β
Γ
Γ = 50 V
(c) Over half a cycle, mean value = ( )( )3
3
1 15 10 5 As2
15 10 s
β
β
Γ
Γ = 2.5 A
3. An alternating current has the following values at equal intervals of 5 ms.
Time (ms) 0 5 10 15 20 25 30
Current (A) 0 0.9 2.6 4.9 5.8 3.5 0
Plot a graph of current against time and estimate the area under the curve over the 30 ms period
using the mid-ordinate rule and determine its mean value. A graph of current against time is shown plotted below
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 201
Mid-ordinates are shown by the broken lines in the above diagram. The mid-ordinate values are: 0.4, 1.6, 3.8, 5.7, 4.9 and 2.2 area [ ]3(5 10 ) 0.4 1.6 3.8 5.7 4.9 2.2ββ Γ + + + + + = [ ]3(5 10 ) 18.6βΓ = 393 10βΓ As = 0.093 As
Mean value = 3
3
area 93 10 Aslength of base 30 10 s
β
β
Γ=
Γ = 3.1 A
5. An indicator diagram of a steam engine is 12 cm long. Seven evenly spaced ordinates, including
the end ordinates, are measured as follows:
5.90 5.52 4.22 3.63 3.32 3.24 3.16 cm
Determine the area of the diagram and the mean pressure in the cylinder if 1 cm represents 90 kPa
Area ( ) ( ) ( ) ( )1 2 5.90 3.16 4 5.52 3.63 3.24 2 4.22 3.323
β + + + + + +β‘ β€β£ β¦
= [ ]1 1(2) 9.06 49.56 15.08 (2)(73.7)3 3
+ + = = 49.13 2cm
Mean value = 2
349.13cm Pa90 1012cm cm
Γ Γ = 368.5 kPa
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 202
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 90Β°, B is 5 kN at 180Β° and
C is 13 kN at 293Β°. Determine graphically the resultant force. The forces are shown in diagram (a) below. Using the βnose-to-tailβ method, the vector diagram is
shown in Figure (b). The 12 kN force is drawn first, then the 5 kN force is βaddedβ to the end of the
12 kN force. Finally, the 13 kN force is βaddedβ 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 203
Total horizontal component = 3 cos 0Β° + 4 cos 60Β° + 2 cos 300Β° = 6
Total vertical component = 3 sin 0Β° + 4 sin 60Β° + 2 sin 300Β° = 1.732
From the diagram below, R = ( )2 26 1.732+ = 6.24 N and 1 1.732tan6
β β βΞΈ = β ββ β
= 16.10Β°
Hence the direction of the resultant relative to the 2 N force is 16.10Β° + 60Β° = 76.10Β°. Thus, the resultant is 6.24 N at an angle of 76.10Β° to the 2 N force. 6. The acceleration of a body is due to four component, coplanar accelerations. These are 2 2m / s
due north, 3 2m / s due east, 4 2m / s to the south-west and 5 2m / s to the south-east. Calculate the
resultant acceleration and its direction. The space diagram is shown below.
Total horizontal component = 3 cos 0Β° + 2 cos 90Β° + 4 cos 225Β° + 5 cos 315Β° = 3.707
Total vertical component = 3 sin 0Β° + 2 sin 90Β° + 4 sin 225Β° + 5 sin 315Β° = -4.364
From the diagram below, R = ( )2 23.707 4.364+ = 5.7 2m / s
and 1 4.364tan 503.707
β β βΞΈ = = Β°β ββ β
correct to 2 significant figures.
Hence, the resultant acceleration is 5.7 2m / s at 310Β° (i.e. at E 50Β° S)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 204
8. A ship heads in a direction of E 20Β° S at a speed of 20 knots while the current is 4 knots in a
direction of N 30Β° E. Determine the speed and actual direction of the ship. The vector diagram is shown below.
Total horizontal component = 4 cos 60Β° + 20 cos 340Β° = 20.794 knots
Total vertical component = 4 sin 60Β° + 20 sin 340Β° = -3.376 knots
From the diagram below, R = ( )2 220.794 3.376+ = 21.07 knots
and 1 3.376tan 9.2220.794
β β βΞΈ = = Β°β ββ β
Hence, the speed of the ship is 21.07 knots and its actual direction is E 9.22Β° S
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 205
EXERCISE 94 Page 231 2. Calculate the resultant of (a) 1v + 2v - 3v (b) 3v - 2v + 1v when 1v = 15 m/s at 85Β°,
2v = 25 m/s at 175Β° and 3v = 12 m/s at 235Β° (a) Total horizontal component of 1v + 2v - 3v = 15 cos 85Β° + 25 cos 175Β° - 12 cos 235Β° = -16.715
Total vertical component of 1v + 2v - 3v = 15 sin 85Β° + 25 sin 175Β° - 12 sin 235Β° = 26.952
From the diagram below, R = ( )2 216.715 26.952+ = 31.71 m/s,
1 26.952tan 58.1916.715
β β βΞ± = = Β°β ββ β
and thus ΞΈ = 180Β° - 58.19Β° = 121.81Β°
i.e. the resultant of 1v + 2v - 3v is 31.71 m/s at angle of 121.81Β° (b) Total horizontal component of 3v - 2v + 1v = 12 cos 235Β° - 25 cos 175Β° + 15 cos 85Β° = 19.329
Total vertical component of 3v - 2v + 1v = 12 sin 235Β° - 25 sin 175Β° + 15 sin 85Β° = 2.934
From the diagram below, R = ( )2 219.329 2.934+ = 19.55 m/s,
and 1 2.934tan 8.6319.329
β β βΞΈ = = Β°β ββ β
i.e. the resultant of 3v - 2v + 1v is 19.55 m/s at angle of 8.63Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 206
EXERCISE 95 Page 232 1. A car is moving along a straight horizontal road at 79.2 km/h and rain is falling vertically
downwards at 26.4 km/h. Find the velocity of the rain relative to the driver of the car. The space diagram is shown in diagram (a). The velocity diagram is shown in diagram (b) and the velocity of the rain relative to the driver is given by vector rc where rc = re + ec
rc = ( )2 279.2 26.4+ = 83.5 km/h and 1 79.2tan 71.626.4
β β βΞΈ = = Β°β ββ β
(a) (b)
i.e. the velocity of the rain relative to the driver is 83.5 km/h at 71.6Β° 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 β at an angle ΞΈ shown
in the triangle of velocities shown below where v is the swimmers true speed.
v = 2 22 1 3β = km/h = 1000360
β ββ ββ β
m/min = 28.87 m/min
Hence, if the width of the river is 142 m, the swimmer will take 14228.87
= 4.919 minutes
= 4 min 55 s
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 207
In the above diagram, sin ΞΈ = 12
from which, ΞΈ = 30Β°
Hence, the swimmer needs to swim at an angle of 60Β° to the bank (shown as angle Ξ± in the
diagram.
3. A ship is heading in a direction N 60Β° 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 50Β° S. Calculate the shipβs 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 30Β° + 8 cos 310Β° = 22.46
Total vertical component of v = 20 sin 30Β° + 8 sin 310Β° = 3.87
Hence, v = ( )2 222.46 3.87+ = 22.79 km/h,
and 1 3.87tan 9.7822.46
β β βΞΈ = = Β°β ββ β
Hence, the ships actual speed is 22.79 km/h in a direction E 9.78Β° N
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 208
EXERCISE 96 Page 236
2. Two alternating voltages are given by 1v 10sin t= Ο volts and 2v 14sin t volts3Οβ β= Ο +β β
β β . By
plotting 1v and 2v on the same axes over one cycle obtain a sinusoidal expression for
(a) 1v + 2v (b) 1v - 2v
(a) 1v 10sin t= Ο , 2v 14sin t volts3Οβ β= Ο +β β
β β and 1v + 2v are shown sketched below:
1v + 2v leads 1v by 36Β° = 36180Ο
Γ = 0.63 rad
Hence, by measurement, 1v + 2v = 20.9 sin(Οt + 0.63) volts
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 209
(b) 1v 10sin t= Ο , 2v 14sin t volts3Οβ β= Ο +β β
β β and 1v - 2v are shown sketched below:
1v - 2v lags 1v by 78Β° = 78180Ο
Γ = 1.36 rad
Hence, by measurement, 1v - 2v = 12.5 sin(Οt β 1.36) volts
4. Express 7sin t 5sin t4Οβ βΟ + Ο +β β
β β in the form ( )Asin tΟ Β±Ξ± using phasors.
The space diagram is shown in (a) below and the phasor diagram is shown in (b).
(a) (b)
Using the cosine rule: 2 2 2R 7 5 2(7)(5)cos135 123.497= + β Β° = from which, R 123.497= = 11.11
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 210
Using the sine rule: 5 11.11sin sin135
=ΞΈ Β°
from which, 5sin135sin 0.3182311.11
°θ = =
and 1sin 0.31823βΞΈ = = 18.56Β° or 0.324 rad
Hence, in sinusoidal form, 7sin t 5sin t4Οβ βΟ + Ο +β β
β β = 11.11sin( t 0.324)Ο +
6. Express i 25sin t 15sin t3Οβ β= Ο β Ο +β β
β β in the form ( )A sin tΟ Β±Ξ± using phasors.
The relative positions of currents 1i and 2i are shown in diagram (a) below. Phasor 2i is shown
reversed in diagram (b) to give - 2i . The phasor diagram for i = 1i - 2i is shown in diagram (c)
(a) (b) (c)
Using the cosine rule: 2 2 2i 25 15 2(25)(15)cos60 475= + β Β° = from which, i 475= = 21.79
Using the sine rule: 15 21.79sin sin 60
=ΞΈ Β°
from which, 15sin 60sin 0.596221.79
°θ = =
and 1sin 0.5962βΞΈ = = 36.60Β° or 0.639 rad
Hence, in sinusoidal form, i 25sin t 15sin t3Οβ β= Ο β Ο +β β
β β = 21.79sin( t 0.639)Ο β
8. Express 3x 9sin t 7sin t3 8Ο Οβ β β β= Ο + β Ο ββ β β β
β β β β in the form ( )A sin tΟ Β±Ξ± using phasors.
180rad 60
3 3Ο Ο Β°
= Γ = Β°Ο
and 3 3 180rad 67.58 8Ο Ο Β°
= Γ = Β°Ο
The relative positions of currents 1x and 2x are shown in diagram (a) below. Phasor 2x is shown
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 211
reversed in diagram (b) to give - 2x . The phasor diagram for x = 1x - 2x is shown in diagram (c)
(a) (b) (c) Using the cosine rule: 2 2 2x 9 7 2(9)(7)cos(60 67.5 ) 206.704= + β Β°+ Β° = from which, x 206.704 14.377= =
Using the sine rule: 7 14.377sin sin127.5
=Ξ± Β°
from which, 7sin127.5sin 0.386314.377
°α = =
and 1sin 0.3863βΞ± = = 22.72Β° Measured to the horizontal, ΞΈ = 60Β° + 22.72Β° = 82.72Β° or 1.444 rad
Hence, in sinusoidal form, 3x 9sin t 7sin t3 8Ο Οβ β β β= Ο + β Ο ββ β β β
β β β β = 14.38sin( t 1.444)Ο +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 212
CHAPTER 22 SCALAR AND VECTOR PRODUCTS EXERCISE 97 Page 241 2. Given p = 2i β 3j, q = 4j β k and r = i + 2j β 3k, determine the quantities (a) p qβ (b) p rβ (a) p qβ = (2)(0) + (-3)(4) + (0)(-1) = -12 (b) p rβ = (2)(1) + (-3)(2) + (0)(-3) = 2 β 6 = -4 4. Given p = 2i β 3j, q = 4j β k and r = i + 2j β 3k, determine the quantities (a) p (b) r (a) p = ( )2 2 2(2) ( 3) (0)+ β + = 13
(b) r = ( )2 2 2(1) (2) ( 3)+ + β = 14
5. Given p = 2i β 3j, q = 4j β k and r = i + 2j β 3k, determine the quantities (a) ( )p q rβ +
(b) 2r β (q β 2p) (a) ( )p q rβ + = (2i β 3j)( 4j β k + i + 2j β 3k) = (2i β 3j)( i + 6j β 4k)
= (2)(1) + (-3)(6) + (0)(-4) = 2 β 18 + 0 = -16 (b) 2r β (q β 2p) = (2i + 4j β 6k) β (4j β k β 2(2i β 3j)) = (2i + 4j β 6k) β (β 4i + 10j β k)
= (2)(-4) + (4)(10) + (-6)(-1)
= -8 + 40 + 6 = 38 7. If p = 2i β 3j, q = 4j β k and r = i + 2j β 3k find the angle between (a) p and q (b) q and r
(a) From equation (4), page 239, 1 1 2 2 3 32 2 2 2 2 2
1 2 3 1 2 3
a b a b a bcosa a a b b b
+ +ΞΈ =
+ + + +
= 2 2 2 2 2 2
(2)(0) ( 3)(4) (0)( 1) 1213 172 ( 3) 0 0 4 ( 1)
+ β + β β=
+ β + + + β = -0.8072
from which, ΞΈ = 1cos ( 0.8072)β β = 143.82Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 213
(b) cos ΞΈ = 2 2 2 2 2 2
(0)(1) (4)(2) ( 1)( 3) 11 0.71302417 140 4 ( 1) 1 2 ( 3)
+ + β β= =
+ + β + + β
from which, ΞΈ = 1cos (0.713024)β = 44.52Β° 8. If p = 2i β 3j, q = 4j β k and r = i + 2j β 3k, determine the direction cosines of (a) p (b) q
(c) r
(a) For p, 2 2
2 2cos132 ( 3)
Ξ± = =+ β
= 0.555, 3cos13β
Ξ² = = -0.832 and 0cos13
Ξ³ = = 0
(b) For q, 2 2
0 0cos174 ( 1)
Ξ± = =+ β
= 0, 4cos17
Ξ² = = 0.970 and 1cos17β
Ξ³ = = -0.243
(c) For r, 2 2 2
1 1cos141 2 ( 3)
Ξ± = =+ + β
= 0.267, 2cos14
Ξ² = = 0.535 and 3cos14β
Ξ³ = = -0.802
10. Find the angle between the velocity vector 1v = 5i + 2j + 7k and 2v = 4i + j - k
cos ΞΈ = 2 2 2 2 2 2
(5)(4) (2)(1) (7)( 1) 15 0.4003278 185 2 7 4 1 ( 1)
+ + β= =
+ + + + β
from which, ΞΈ = 1cos (0.40032)β = 66.40Β° 11. Calculate the work done by a force F = (-5i + j + 7k) when its point of application moves from
point (-2i β 6j + k) m to the point (i β j + 10k) m. Work done = F β d where d = (i β j + 10k) - (-2i β 6j + k) = 3i + 5j + 9k Hence, work done = (-5i + j + 7k) β (3i + 5j + 9k) = -15 + 5 + 63 = 53 Nm
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 214
EXERCISE 98 Page 244 1. When p = 3i + 2k, q = i β 2j + 3k and r = -4i + 3j β k, determine (a) p qΓ (b) q pΓ
(a) p qΓ = i j k3 0 21 2 3β
= i 0 22 3β
- j 3 21 3
+ k 3 01 2β
= 4i β 7j β 6k
(b) q pΓ = i j k1 2 33 0 2
β = i 2 3
0 2β
- j 1 33 2
+ k 1 23 0
β = -4i + 7j + 6k
2. When p = 3i + 2k, q = i β 2j + 3k and r = -4i + 3j β k, determine (a) p rΓ (b) r qΓ
(a) p rΓ = ( )( ) ( )2p p r r p rβ β β β where p β p = (3)(3) + (2)(2) = 13,
r β r = (-4)(-4) +(3)(3) + (-1)(-1) = 26
and p β r = (3)(-4) + (0)(3) + (2)(-1) = -14
Hence, p rΓ = 2(13)(26) ( 14) (338 196) 142β‘ β€β β = β =β£ β¦ = 11.92
(b) r qΓ = = ( )( ) ( )2r r q q r qβ β β β where r β r = (-4)(-4) + (3)(3) + (-1)(-1) = 26, q β q = (1)(1) +(-2)(-2) + (3)(3) = 14 and r β q = (-4)(1) + (3)(-2) + (-1)(3) = -13
Hence, r qΓ = 2(26)(14) ( 13) (364 169) 195β‘ β€β β = β =β£ β¦ = 13.96
4. When p = 3i + 2k, q = i β 2j + 3k and r = -4i + 3j β k, determine (a) p Γ (r Γ q) (b) (3p Γ 2r) Γ q
(a) r Γ q = i j k4 3 1
1 2 3β β
β = 7i + 11j + 5k
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 215
Hence, p Γ (r Γ q) = i j k3 0 27 11 5
= (-22)i β (1)j + (33)k = -22i - j + 33k
(b) 3p Γ 2r = i j k9 0 68 6 2β β
= (-36)i β (-18 + 48)j + (54)k = -36i - 30j + 54k
(3p Γ 2r) Γ q = i j k36 30 541 2 3
β ββ
= (-90 + 108)i β (-108 - 54)j + (72 + 30)k
= 18i + 162j + 102k
6. For vectors a = -7i + 4j + 12
k and b = 6i β 5j β k find (i) a bβ (ii) a bΓ (iii) a bΓ
(iv) b aΓ and (v) the angle between the vectors.
(i) a bβ = (-7)(6) + (4)(-5) +( 12
)(-1) = - 62 12
(ii) a bΓ =
i j k17 42
6 5 1
β
β β
= i 542
β ββ +β ββ β
- j(7 β 3) + k(35 β 24) = 11 i 4j 11k2
β β +
(iii) a β a = (-7)(-7) + (4)(4) + 1 12 2
β ββ ββ ββ ββ β β β
= 65.25
b β b = (6)(6) + (-5)(-5) + (-1)(-1) = 62
a β b = (-7)(6) + (4)(-5) + ( )1 12
β β ββ ββ β
= - 62.5
a bΓ = ( )( ) ( ) ( )265.25 62 62.5 4045.5 3906.25 139.25β‘ β€β β = β =β£ β¦ = 11.80
(iv) b aΓ = i j k6 5 1
17 42
β β
β
= i 5 42
β ββ +β ββ β
- j(3 β 7) + k(24 β 35) = 11 i 4j 11k2
+ β
(v) cos ΞΈ = 2
2 2 2 2 2
1( 7)(6) (4)( 5) ( 1)62.52 0.9826388
65.25 621( 7) (4) (6) ( 5) ( 1)2
β ββ + β + ββ β ββ β = = ββ ββ + + + β + ββ ββ β
from which, ΞΈ = 1cos ( 0.9826388)β β = 169.31Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 216
8. A force of (2i β j + k) newtonβs 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) β (4i + 0j β k) = -4i + 3j + 2k
Moment, M = r Γ F where M = i j k4 3 2
2 1 1β
β = (3 + 2)i β (-4 - 4)j + (4 - 6)k
= (5i + 8j - 2k) Nm
Magnitude of M, M = ( ) ( ) ( )2r F r r F F r FΓ = β β β β where r β r = (-4)(-4) + (3)(3) +(2)(2) = 29
F β F = (2)(2) + (-1)(-1) +(1)(1) = 6
r β F = (-4)(2) + (3)(-1) +(2)(1) = -9
i.e. M = ( )( ) ( )229 6 9β β = 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 β j + 2k) rad/s when the position vector of the particle is at
(i β 5j + 4k) m.
Velocity vector, v = Ο Γ r = (3i β j + 2k) Γ (i β 5j + 4k) = i j k3 1 21 5 4β
= (-4 + 10)i β (12 - 2)j + (-15 + 1)k
= 6i - 10j - 14k
Magnitude of v, v = ( ) ( ) ( )2r r rΟβ Ο β β β Ο
where Οβ Ο = (3)(3) + (-1)(-1) + (2)(2) = 14
rβ r = (1)(1) + (-5)(-5) + (4)(4) = 42
rβ Ο = (1)(3) + (-5)(-1) + (4)(2) = 16
Hence, v = ( )( ) ( )214 42 16β = 18.22 m/s
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 217
EXERCISE 99 Page 246 1. Find the vector equation of the line through the point with position vector 5i β 2j + 3k which is
parallel to the vector 2i + 7j β 4k. Determine the point on the line corresponding to Ξ» = 2 in the
resulting equation. Vector equation of the line, r = a + Ξ»b = (5i β 2j + 3k) + Ξ»(2i + 7j β 4k) i.e. r = (5 + 2Ξ»)i + (7Ξ» - 2)j + (3 - 4Ξ»)k When Ξ» = 2, r = 9i + 12j β 5k 2. Express the vector equation of the line in problem 1 in standard Cartesian form. The vector equation of a straight line in standard Cartesian form is:
31 2
1 2 3
z ax a y ab b b
ββ β= = = Ξ»
Since a = 5i β 2j + 3k, 1 2 3a 5, a 2 and a 3= = β = and b = 2i + 7j β 4k, 1 2 3b 2, b 7 and b 4= = = β
then the Cartesian equations are: x 5 y 2 z 32 7 4β + β
= = = Ξ»β
or x 5 y 2 3 z2 7 4β + β
= = = Ξ»
4. Express the straight line equation 1 4y 3z 12x 15 4β β
+ = = in vector form.
2x 1 1 4y 3z 11 5 4+ β β
= = i.e.
11 1 zx y32 4
1 5 42 4 3
β+ β= =
β
i.e. 1 2 31 1 1a , a and a2 4 3
= β = =
and 1 2 31 5 4b , b and b2 4 3
= = β =
Hence, in vector form the equation is:
r = ( )1 1a b+ Ξ» i + ( )2 2a b+ Ξ» j + ( )3 3a b+ Ξ» k
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 218
= 1 12 2
β ββ + Ξ»β ββ β
i + 1 54 4
β ββ Ξ»β ββ β
j + 1 43 3
β β+ Ξ»β ββ β
k
or r = ( )1 12
Ξ» β i + ( )1 1 54
β Ξ» j + ( )1 1 43
+ Ξ» k
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 219
CHAPTER 23 COMPLEX NUMBERS EXERCISE 100 Page 250 1. Solve the quadratic equation 2x 25 0+ = Since 2x 25 0+ = then 2x 25= β i.e. x 25 ( 1)(25) 1 25 j 25= β = β = β = from which, x = Β± j 5 2. Solve the quadratic equation 22x 3x 4 0+ + = Since 22x 3x 4 0+ + = then
23 3 4(2)(4) 3 ( 1)(23) 3 ( 1) (23) 3 j (23)3 23x2(2) 4 4 4 4
β‘ β€β Β± β β Β± β β Β± β β Β±β Β± ββ£ β¦= = = = =
= - 3 23j4 4Β± or (-0.750 Β± j1.199)
4. Evaluate (a) 8j (b) 7
1j
β (c) 13
42j
(a) 8j = ( )42j = ( )41β = 1
(b) ( ) ( )3 37 6 2j j j j j j 1 j= Γ = Γ = Γ β = β
Hence, 7 2
1 1 1 j j j jj j j j( j) j ( 1) 1
β β β ββ = β = = = = =
β β β β β = -j
(c) ( )613 12 2 6j j j j j j ( 1) j= Γ = Γ = Γ β =
Hence, 13
42 j
= 2
2 2( j) j2 j2j j( j) j 1
β β β= = =
β β = -j2
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 220
EXERCISE 101 Page 253 1. Evaluate (a) (3 + j2) + (5 β j) and (b) (-2 + j6) β (3 β j2) and show the results on an Argand
diagram. (a) (3 + j2) + (5 β j) = (3 + 5) + j(2 β 1) = 8 + j (b) (-2 + j6) β (3 β j2) = -2 + j6 β 3 + j2 = (-2 β 3) + j(6 + 2) = -5 + j8 (8 + j) and (-5 + j8) are shown on the Argand diagram below.
3. Given 1Z = 1 + j2, 2Z = 4 β j3, 3Z = -2 + j3 and 4Z = -5 β j, evaluate in a + jb form:
(a) 1 2 3Z Z Z+ β (b) 2 1 4Z Z Zβ + (a) 1 2 3Z Z Z+ β = 1 + j2 + 4 β j3 β (-2 + j3) = 1 + j2 + 4 β j3 + 2 β j3 = (1 + 4 + 2) + j(2 β 3 β 3) = 7 β j4 (b) 2 1 4Z Z Zβ + = (4 β j3) β (1 + j2) + (-5 β j) = 4 β j3 β 1 β j2 β 5 β j = (4 β 1 β 5) + j(-3 β 2 β 1) = -2 - j6 5. Given 1Z = 1 + j2, 2Z = 4 β j3, 3Z = -2 + j3 and 4Z = -5 β j, evaluate in a + jb form:
(a) 1 3 4Z Z Z+ (b) 1 2 3Z Z Z (a) 1 3 4Z Z Z+ = (1 + j2)(-2 + j3) + (-5 β j) = -2 + j3 β j4 + 2j 6 - 5 β j
= -2 + j3 β j4 - 6 β 5 - j = -13 β j2 (b) 1 2 3Z Z Z = (1 + j2)(4 β j3)(-2 + j3) = (4 β j3 + j8 - 2j 6)(-2 + j3) = ( 10 + j5)(-2 + j3)
= -20 + j30 β j10 + 2j 15 = -20 + j30 βj10 β 15 = -35 + j20
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 221
7. Given 1Z = 1 + j2, 2Z = 4 β j3, 3Z = -2 + j3 and 4Z = -5 β j, evaluate in a + jb form:
(a) 1 3
1 3
Z ZZ Z+
(b) 12 3
4
ZZ ZZ
+ +
(a) 1 3
1 3
Z ZZ Z+
= 2(1 j2)( 2 j3) 2 j3 j4 j 6 8 j
(1 j2) ( 2 j3) 1 j5 1 j5+ β + β + β + β β
= =+ + β + β + β +
= 2
2 2
( 8 j)( 1 j5) 8 j40 j j 5 3 j41( 1 j5)( 1 j5) 1 5 26β β β β + + + +
= =β + β β +
= 3 41j26 26
+
(b) 12 3
4
ZZ ZZ
+ + = (4 β j3) + 1 j25 j+
β β + (-2 + j3) = 4 β j3 + 2 2
(1 j2)( 5 j)5 1
+ β ++
- 2 + j3
= 4 β j3 + 25 j j10 j 2
26β + β + - 2 + j3
= 4 β j3 + 7 j926
β β - 2 + j3 = 2 - 7 9j26 26
β
= 52 7 9j26 26 26
β β = 45 9j26 26
β
9. Show that 25 1 j2 2 j5 57 j242 3 j4 j
β ββ + ββ = +β β+ ββ β
2
2 2
1 j2 (1 j2)(3 j4) 3 j4 j6 j 8 11 j2 11 2j3 j4 3 4 25 25 25 25+ + β β + β +
= = = = ++ +
2
2
2 j5 (2 j5)( j) j2 j 5 5 j2 5 j2j j( j) j 1
β β β += = = = +
β β β
L.H.S. = 25 1 j2 2 j5 25 11 2 25 11 2j (5 j2) 5 j 22 3 j4 j 2 25 25 2 25 25
β ββ + β β‘ β€ β‘ β€β β β β β ββ = β + β + = β β + ββ β β β β β β ββ’ β₯ β’ β₯+ β β β β β β β β£ β¦ β£ β¦β β
= 25 11 125 2 50 25 114 48j j2 25 25 2 25 25β‘ β β β€β β β β β‘ β€β + = β β ββ β β ββ’ β₯ β’ β₯β β β β β£ β¦β£ β¦
= 25 114 25 48j2 25 2 25β β β ββ β +β β β ββ β β β
= 57 + j24 = R.H.S.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 222
EXERCISE 102 Page 254
2. Solve the complex equation: 2 j j(x jy)1 j+
= +β
2 j j(x jy)1 j+
= +β
hence, (2 j)(1 j) j(x jy)(1 j)(1 j)+ +
= +β +
i.e. 2
22 2
2 j2 j j jx j y1 1
+ + += +
+
i.e. 1 j3 jx y2+
= β
i.e. 1 3j2 2+ = -y + jx
Hence, x = 32
and y = 12
β
3. Solve the complex equation: (2 j3) (a jb)β = + (2 j3) (a jb)β = +
Squaring both sides gives: ( )22 j3 a jbβ = +
(2 β j3)(2 β j3) = a + jb
i.e. 4 β j6 β j6 + 2j 9 = a + jb
i.e. -5 β j12 = a + jb
Hence, a = -5 and b = -12
5. If Z = R + jΟL + 1j CΟ
, express Z in (a + jb) form when R = 10, L = 5, C = 0.04 and Ο = 4
Z = R + jΟL + 1j CΟ
= 10 + j(4)(5) + 1j(4)(0.04)
= 10 + j20 + 6.25j
= 10 + j20 + 6.25( j)j( j)
ββ
= 10 + j20 - 2
6.25jβ
= 10 + j20 β j6.25
= 10 + j13.75
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 223
EXERCISE 103 Page 256 2. Express the following Cartesian complex numbers in polar form, leaving answers in surd form:
(a) 2 + j3 (b) -4 (c) -6 + j (a) 2 + j3 From the diagram below, r = 2 22 3 13+ =
and 1 3tan 56.31 or 56 19 '2
β β βΞΈ = = Β° Β°β ββ β
Hence, 2 + j3 = 13 56 19'β Β° in polar form (b) -4 = -4 + j0 and is shown in the diagram below, where r = 4 and ΞΈ = 180Β°
Hence, -4 = 4 180β Β° in polar form (c) -6 + j From the diagram below, r = 2 26 1 37+ =
and 1 1tan 9.46 or 9 28'6
β β βΞ± = = Β° Β°β ββ β
thus ΞΈ = 180Β° - 9Β°28β² = 170Β°32β²
Thus, -6 + j = 37 170 32'β Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 224
3. Express the following Cartesian complex numbers in polar form, leaving answers in surd form:
(a) - j3 (b) ( )32 jβ + (c) 3j (1 j)β
(a) - j3 From the diagram below, r = 3 and ΞΈ = -90Β°
Hence, - j3 = 3β -90Β° in polar form
(b) ( )32 jβ + = (-2 + j)(-2 + j)(-2 + j) = (4 β j2 β j2 + 2j )(-2 + j)
= (3 β j4)(-2 + j) = -6 + j3 + j8 - 2j 4 = -2 + j11
From the diagram below, r = 2 22 11 125+ = and 1 11tan 79.70 or 79 42 '2
β β βΞ± = = Β° Β°β ββ β
and ΞΈ = 180Β° - 79Β°42β² = 100Β°18β²
Hence, ( )32 jβ + = -2 + j11 = 125 100 18'β Β° in polar form
(c) 3j (1 j)β = (j)( 2j )(1 β j) = -j(1 β j) = -j + 2j = -1 β j
From the diagram below, r = 2 21 1 2+ = and 1 1tan 451
β β βΞ± = = Β°β ββ β
and ΞΈ = 180Β° - 45Β° = 135Β°
Hence, 3j (1 j)β = -1 β j = 2 135β β Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 225
5. Convert the following polar complex numbers into (a + jb) form giving answers correct to 4
significant figures: (a) 6β 125Β° (b) 4β Ο (c) 3.5β -120Β° (a) 6β 125Β° = 6 cos 125Β° + j 6 sin 125Β° = -3.441 + j4.915 (b) 4β Ο = 4 cos Ο + j sin Ο = - 4.000 + j0 (Note that Ο is radians) (c) 3.5β -120Β° = 3.5 cos(-120Β°) + j 3.5 sin(-120Β°) = -1.750 β j3.031 6. Evaluate in polar form: (a) 3 20 15 45β °à β Β° (b) 2.4 65 4.4 21β °à β β Β° (a) 3 20 15 45β °à β Β° = 3 15 (20 45 )Γ β Β°+ Β° = 45β 65Β° (b) 2.4 65 4.4 21β °à β β Β° = 2.4 4.4 (65 21 )Γ β Β°+ β Β° = 10.56β 44Β° 7. Evaluate in polar form: (a) 6.4 27 2 15β °÷ β β Β° (b) 5 30 4 80 10 40β °à β °÷ β β Β°
(a) 6.4 27 2 15β °÷ β β Β° = 6.4 27 6.4 27 152 15 2
β Β°= β Β°β β Β°
β β Β° = 3.2β 42Β°
(b) 5 30 4 80 10 40β °à β °÷ β β Β° = 5 30 4 80 5 4 (30 80 40 )10 40 10β °à β Β° Γ
= β Β°+ Β°β β Β°β β Β°
= 2β 150Β°
8. Evaluate in polar form: (a) 4 36 8Ο Ο
β + β (b) 2 120 5.2 58 1.6 40β Β°+ β Β°β β β Β°
(a) 4 36 8Ο Ο
β + β = 4cos j4sin 3cos j3sin6 6 8 8Ο Ο Ο Οβ β β β+ + +β β β β
β β β β = (3.464 + j2) + (2.772 + j1.148)
= 6.236 + j3.148
From the diagram below, r = 2 26.236 3.148 6.986+ =
and 1 3.148tan 26.79 or 26 47 ' or 0.467rad6.236
β β βΞΈ = = Β° Β°β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 226
Hence, 4 36 8Ο Ο
β + β = 6.986β 26Β°47β² or 6.986β 0.467 rad
(b) 2 120 5.2 58 1.6 40β Β°+ β Β°β β β Β°
= (2 cos 120Β° + j 2 sin 120Β°) + (5.2 cos 58Β° + j 5.2 sin 58Β°) β (1.6 cos(-40Β°) + j 1.6 sin(-40Β°))
= (-1 + j 1.732) + (2.756 + j4.410) β (1.226 β j1.028)
= -1 + j1.732 + 2.756 + j4.410 β 1.226 + j1.028
= 0.530 + j7.170
From the diagram below, r = 2 20.530 7.170 7.190+ =
and 1 7.170tan 85.77 or 85 46 '0.530
β β βΞΈ = = Β° Β°β ββ β
Hence, 2 120 5.2 58 1.6 40β Β° + β Β° β β β Β° = 7.190β 85Β°46β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 227
EXERCISE 104 Page 259 1. Determine the resistance R and series inductance L (or capacitance C) for each of the following
impedances assuming the frequency to be 50 Hz:
(a) (3 + j8) Ξ© (b) (2 β j3) Ξ© (c) j14 Ξ© (d) 8 60β β °Ω (a) If Z = (3 + j8) Ξ© then resistance, R = 3 Ξ© and inductive reactance, LX = 8 Ξ© (since the j
term is positive)
LX = 2ΟfL = 8 hence, inductance, L = 8 82 f 2 (50)
=Ο Ο
= 0.0255 H or 25.5 mH
(b) If Z = (2 β j3) Ξ© then resistance, R = 2 Ξ© and capacitive reactance, CX = 3 Ξ© (since the j
term is negative)
C1X
2 fC=
Ο = 3 hence, capacitance, C = 3 61 1 1.061 10 or 1061 10
2 f (3) 2 (50)(3)β β= = Γ Γ
Ο Ο
= 1061 Β΅F
(c) If Z = j14 Ξ© i.e. Z = (0 + j14) Ξ© then resistance, R = 0 Ξ© and LX = 14 Ξ©
i.e. 2ΟfL = 14 hence, inductance, L = 142 (50)Ο
= 0.04456 H or 44.56 mH
(d) If Z = 8 60β β °Ω = 8 cos(-60Β°) + j 8 sin(-60Β°) = (4 β j6.928) Ξ©
Hence, resistance, R = 4 Ξ© and CX = 6.928 Ξ©
i.e. 1 6.9282 fC
=Ο
and capacitance, C = 61 459.4 102 (50)(6.928)
β= ΓΟ
= 459.4 Β΅F
2. Two impedances, 1Z (3 j6)= + Ξ© and 2Z (4 j3)= β Ξ© are connected in series to a supply
voltage of 120 V. Determine the magnitude of the current and its phase angle relative to the
voltage. In a series circuit, total impedance, TOTAL 1 2Z Z Z= + = (3 + j6) + (4 β j3) = (7 + j3) Ξ©
= 2 2 1 37 3 tan7
β β β+ β β ββ β
= 7.616β 23.20Β° Ξ© or 7.616β 23Β°12β² Ξ©
Since voltage V = 120β 0Β° V, then current, I = V 120 0Z 7.616 23 12 '
β Β°=
β Β° = 15.76β -23Β°12β² A
i.e. the current is 15.76 A and is lagging the voltage by 23Β°12β²
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 228
3. If the two impedances in Problem 2 are connected in parallel determine the current flowing and
its phase relative to the 120 V supply voltage. In a parallel circuit shown below, the total impedance, TZ is given by:
2 2 2 2T 1 2
1 1 1 1 1 3 j6 4 j3 3 6 4 3j jZ Z Z 3 j6 4 j3 3 6 4 3 45 45 25 25
β += + = + = + = β + +
+ β + +
i.e. TT
1 admit tan ce,YZ
= = 0.22667 β j0.01333 = 0.2271β 3Β°22β² siemen
Current, I = TT
V VY (120 0 )(0.2271 3 22 ') 27.25 3 22 'Z
= = β Β° β Β° = β Β° A
i.e. the current is 27.25 A and is leading the voltage by 3Β°22β²
5. For the circuit shown, determine the current I flowing and its phase relative to the applied
voltage.
2 2 2 2T 1 2 3
1 1 1 1 1 1 1 30 j20 40 j50 1Z Z Z Z 30 j20 40 j50 25 30 20 40 50 25
+ β= + + = + + = + +
β + + +
= 30 20 40 50 1j j1300 1300 4100 4100 25
+ + β +
i.e. TT
1 admit tan ce,YZ
= = 0.07283 + j0.00319 = 0.0729β 2Β°30β² S
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 229
Current, I = TT
V VY (200 0 )(0.0729 2 30 ') 14.6 2 30 'Z
= = β Β° β Β° = β Β° A
i.e. the current is 14.6 A and is leading the voltage by 2Β°30β²
7. A delta-connected impedance AZ is given by: 1 2 2 3 3 1A
2
Z Z Z Z Z ZZZ
+ +=
Determine AZ in both Cartesian and polar form given 1Z (10 j0)= + Ξ© , 2Z (0 j10)= β Ξ© and
3Z (10 j10)= + Ξ© .
1 2 2 3 3 1A
2
Z Z Z Z Z Z (10 j0)(0 j10) (0 j10)(10 j10) (10 j10)(10 j0)ZZ (0 j10)
+ + + β + β + + + += =
β
= 2j100 j100 j 100 100 j100 200 j100 200 j100j10 j10 j10 j10
β β β + + β= = β
β β β β
= j200 10 (10 j20)10
+ = + Ξ©
From the diagram below, r = 2 210 20+ = 22.36 and 1 20tan 63.4310
β β βΞΈ = = Β°β ββ β
Hence, AZ (10 j20) 22.36 63.43= + Ξ© = β °Ω 8. In the hydrogen atom, the angular momentum, p, of the de Broglie wave is given by:
( )jhp jm2
β βΞ¨ = β Β± Ξ¨β βΟβ β
Determine an expression for p.
If ( )jhp jm2
β βΞ¨ = β Β± Ξ¨β βΟβ β then p = ( ) ( )( )2jh jm jh hjm j m
2 2 2Β± Ξ¨β ββ β β ββ = β Β± = β Β±β ββ β β βΟ Ξ¨ Ο Οβ β β β β β
= ( )h m2
Β±Ο
= mh2
Β±Ο
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 230
10. Three vectors are represented by P, 2β 30Β°, Q, 3β 90Β° and R, 4β -60Β°. Determine in polar form
the vectors represented by (a) P + Q + R (b) P β Q β R (a) P + Q + R = 2β 30Β° + 3β 90Β° + 4β -60Β° = (1.732 + j1) + (0 + j3) + (2 β j3.464)
= (3.732 + j0.536) = 3.770β 8.17Β° (b) P β Q β R = 2β 30Β° - 3β 90Β° - 4β -60Β° = (1.732 + j1) - (0 + j3) - (2 β j3.464)
= (-0.268 + j1.464))
From the diagram below, r = 1.488 and 1 1.464tan 79.630.268
β β βΞ± = = Β°β ββ β
and 180 79.63 100.37ΞΈ = Β° β Β° = Β°
Hence, P β Q β R = 1.488β 100.37Β°
11. In a Schering bridge circuit, XX X CZ (R jX )= β ,
22 CZ jX= β , ( )( )( )
3
3
3 C3
3 C
R jXZ
R jX
β=
β and 4 4Z R= ,
where C1X
2 fC=
Ο. At balance: ( )( ) ( )( )X 3 2 4Z Z Z Z= .
Show that at balance 3 4X
2
C RRC
= and 2 3X
4
C RCR
=
Since ( )( ) ( )( )X 3 2 4Z Z Z Z=
then ( )( ) ( )( )3
X 2
3
3 CX C C 4
3 C
R jX(R jX ) jX R
R jX
β§ β«ββͺ βͺβ = ββ¨ β¬ββͺ βͺβ© β
Thus, ( )( )( )
( )( )3 2
X
3
3 C C 4X C
3 C
R jX jX R(R jX )
R jX
β ββ =
β
i.e. ( )( ) ( )( )
3 22
X
3 3
2C C 43 C 4
X C3 C 3 C
j X X RjR X R(R jX )
R jX R jX
ββ = +
β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 231
i.e. ( )
2 2
X
3
C 4 C 4X C
C 3
X R X R(R jX )
X R ( j)β = β
β
= 2 2
3
C 4 C 4
C 3
X R X RX jR
+
i.e. XX C(R jX )β = 2 2
3
C 4 C 4
C 3
X R X Rj
X Rβ
Equating the real parts gives: 2
3
4C 4 32
X 4C 2
3
1 RX R 2 fC2 fCR R1X 2 fC2 fC
ΟΟ= = =
ΟΟ
i.e. 3 4X
2
C RR
C=
Equating the imaginary parts gives: 2
X
C 4C
3
X RX
Rβ = β
i.e. 4
2 4
X 3 2 3
1 R2 fC R1
2 fC R 2 fC RΟ
= =Ο Ο
from which, 2 3X
4
C RC
R=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 232
CHAPTER 24 DE MOIVREβS THEOREM EXERCISE 105 Page 261 2. Determine in polar and Cartesian forms (a) [ ] 43 41β Β° (b) ( )52 jβ β (a) [ ] 4 43 41 3 4 41β Β° = β Γ Β° = 81β 164Β° = 8 cos 164Β° + j 8 sin 164Β° = -77.86 + j22.33
(b) ( ) ( )5552 j 5 153.435 5 5 153.435β‘ β€β β = β β Β° = β Γβ Β°β£ β¦
= 55.90β -767.175Β° = 55.90β -47Β°10β²
= 55.90 cos -47Β°10β² + j 55.90 sin -47Β°10β²
= 38 β j41 3. Convert (3 β j) into polar form and hence evaluate ( )73 jβ , giving the answer in polar form.
(3 β j) = 2 2 1 13 1 tan3
β β β+ β ββ ββ β
= 10 18 26'β β Β°
Hence, ( ) ( )7773 j 10 18 26 ' 10 7 18 26 'β‘ β€β = β β Β° = β Γβ Β°β£ β¦ = 3162β -129Β°2β²
5. Express in both polar and rectangular forms: ( )53 j8β
( ) ( )5553 j8 73 69.444 73 5 69.444β‘ β€β = β β Β° = β Γβ Β°β£ β¦ = 45530β -347.22Β° = 3162β 12Β°47β²
= 45530 cos 12Β°47β² + j45530 sin 12Β°47β²
= 44400 + j10070
7. Express in both polar and rectangular forms: ( )616 j9β β
From the diagram below, r = 2 216 9 337+ = and 1 9tan 29.35816
β β βΞ± = = Β°β ββ β
and 180 29.358 209.358ΞΈ = Β°+ Β° = Β°
Hence, ( ) ( )66616 j9 337 209.358 337 6 209.358β‘ β€β β = β Β° = β Γ Β°β£ β¦
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 233
= 638.27 10 1256.148Γ β Β° = 6(38.27 10 ) 176 9'Γ β Β°
( )6 6(38.27 10 ) 176 9 ' 10 38.27 cos176 9 ' j38.27sin176 9 'Γ β Β° = Β° + Β°
= 610 ( 38.18 j2.570)β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 234
EXERCISE 106 Page 263 2. Determine the two square roots of the following complex numbers in Cartesian form and show
the results on an Argand diagram: (a) 3 β j4 (b) -1 β j2
(a) [ ] [ ]123 j4 5 53.13 5 53.13β = β β Β° = β β Β°
The first root is: 12 15 53.13 2.236 26.57 (2 j1)
2β Γβ Β° = β β Β° = β
and the second root is: 52.236 ( 26.57 180 ) ( 2 j1)β β Β°+ Β° = β +
Hence, (3 j4) (2 j)β = Β± β as shown in the Argand diagram of Figure (a) below.
(a) (b)
(b) 121 j2 5 243.435 5 243.435β‘ β€ β‘ β€β β = β Β° = β Β°β£ β¦ β£ β¦
The first root is: ( )12 15 243.435 1.495 121.72 ( 0.786 j1.272)
2β Γ Β° = β Β° = β +
and the second root is: 1.495 (121.72 180 ) 1.495 58.28 (0.786 j1.272)β Β°β Β° = β β = β
Hence, ( 1 j2) (0.786 j1.272)β β = Β± β as shown in the Argand diagram if Figure (b) above.
4. Determine the modulus and argument of the complex number: ( )133 j4+
( ) ( )11 1
333 313 j4 5 53.13 5 53.13 5 17.71 1.710 17 43'3
+ = β Β° = β Γ Β° = β Β° = β Β°
Hence, the modulus is 1.710, and the arguments are 17Β°43β², 17Β°43β² + 3603Β° = 137Β°43β²,
and 137Β°43β² + 3603Β° = 257Β°43β² since the three roots are equally displaced by 120Β°.
5. Determine the modulus and argument of the complex number: ( )142 jβ +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 235
( ) ( )111444
12 j 5 153.435 5 153.435 1.223 38 22 '4
β‘ β€β + = β Β° = β Γ Β° = β Β°β£ β¦
There are 4 roots equally displaced 3604Β° , i.e. 90Β° apart.
Hence, the modulus is 1.223, and the arguments are: 38Β°22β², 38Β°22β² + 90Β° = 128Β°22β²,
128Β°22β² + 90Β° = 218Β°22β² and 218Β°22β² + 90Β° = 308Β°22β²
7. Determine the modulus and argument of the complex number: ( )234 j3 ββ
( ) [ ] ( )22 233 3
24 j3 5 36.87 5 36.87 0.3420 24 35'3
ββ ββ = β β Β° = β β Γβ Β° = β Β°
There are 3 roots equally displaced 3603Β° , i.e. 120Β° apart.
Hence, the modulus is 0.3420, and the arguments are: 24Β°35β², 24Β°35β² + 120Β° = 144Β°35β²,
and 144Β°35β² + 120Β° = 264Β°35β² 8. For a transmission line, the characteristic impedance 0Z and the propagation coefficient Ξ³ are
given by:
0R j LZG j C
β β+ Ο= β β+ Οβ β
and ( )( )R j L G j CΞ³ = + Ο + Οβ‘ β€β£ β¦
Given R = 25 Ξ©, L = 35 10βΓ H, G = 680 10βΓ siemens, C = 60.04 10βΓ F and Ο = 2000Ο rad/s,
determine, in polar form, 0Z and Ξ³.
( )3R j L 25 j(2000 ) 5 10 25 j31.416 40.15 51.49β+ Ο = + Ο Γ = + = β Β°
( )6 6 6 6G j C 80 10 j(2000 ) 0.04 10 10 (80 j251.33) 263.755 10 72.34β β β β+ Ο = Γ + Ο Γ = + = Γ β Β°
Hence, ( )0 6
R j L 40.15 51.49Z 152224.6 20.85G j C 263.755 10 72.34β
β β+ Ο β Β°β β= = = β β Β°β β β β+ Ο Γ β Β°β β β β
= ( )1152224.6 20.852
β β Β° = 390.2β -10.43Β° Ξ©
( )( ) ( )( )6R j L G j C 40.15 51.49 263.755 10 72.34ββ‘ β€Ξ³ = + Ο + Ο = β Β° Γ β Β°β‘ β€β£ β¦ β£ β¦
= ( ) 10.01058976 123.83 0.01058976 123.832
β Β° = β Γ Β°
= 0.1029β 61.92Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 236
EXERCISE 107 Page 265 2. Convert (-2.5 + j4.2) into exponential form. (-2.5 + j4.2) = 4.89 120.76 or 4.89 2.11radβ Β° β
and 4.89β 2.11 β‘ 4.89 j 2.11e
4. Convert 1.2 j 2.51.7e β into rectangular form.
( )( )1.2 j 2.5 1.2 j 2.5 1.21.7e 1.7e e 1.7e 2.5radβ β= = β β = 1.2 1.21.7e cos( 2.5) j1.7e sin( 2.5)β + β
= -4.52 β j3.38
6. If j 2.1z 7e= , determine ln z (a) in Cartesian form, and (b) in polar form. (a) If j 2.1z 7e= then ln z = ( )j 2.1 j 2.1ln 7e ln 7 ln e= + = ln 7 + j2.1 in Cartesian form (b) ln 7 + j2.1 = 2.86β 47.18Β° or 2.86β 0.82 rad 8. Determine in polar form (a) ln(2 + j5) (b) ln(-4 β j3) (a) ln(2 + j5) = ( ) ( )j1.19 j1.19ln 29 1.19 ln 29 e ln 29 ln eβ = = +
= ln 29 j1.19 1.6836 j1.19+ = +
= 2.06β 35.25Β° or 2.06β 0.615 rad (b) ln(-4 β j3) = ( ) ( ) ( )j3.785ln 5 216.87 ln 5 3.785 ln 5eβ Β° = β =
= ln 5 j3.785 1.6094 j3.785+ = +
= 4.11β 66.96Β° or 4.11β 1.17 rad 9. When displaced electrons oscillate about an equilibrium position the displacement x is given by
the equation:
( )24mf hht j t2m 2m a
x A e
β§ β«ββͺ βͺβ +β¨ β¬ββͺ βͺβ© β=
Determine the real part of x in terms of t, assuming ( )24mf hβ is positive.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 237
( )2
24mf hht j t 4mf hh t h t 22m 2m a j t
2m a2m 2m 4mf hx A e Ae e Ae t2m a
β§ β«ββͺ βͺβ +β¨ β¬ βββͺ βͺ β ββ© β β
β β β ββ β β β ββ β= = = β β ββ β β β β ββ β ββ β β β β β β β
= ht h t2 2
2m 2m4mf h 4mf hAe cos t jAe sin t2m a 2m a
β ββ β β ββ β+β β β ββ β β ββ ββ β β β
Hence, the real part is: ht 22m 4mf hAe cos t
2m aβ β ββ
β ββ βββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 238
CHAPTER 25 THE THEORY OF MATRICES AND
DETERMINANTS EXERCISE 108 Page 270
2. Determine
13 624 7 6
22 4 0 5 73
5 7 4 31 05
β ββ β
ββ β β ββ β β ββ + ββ β β ββ ββ β ββ β
β βββ ββ β
1 13 6 (4 3) ( 7 6) (6 )2 24 7 6
2 22 4 0 5 7 ( 2 5) (4 ) (0 7)3 3
5 7 4 3 31 0 (5 1) (7 0) ( 4 )5 5
β β β β+ β + +β β β βββ β β β β β
β β β β β ββ + β = β + + β +β β β β β ββ ββ β β β ββ β β β β ββ + β + β +β β β ββ β β β
=
17 1 62
13 3 73
24 7 35
β βββ ββ ββ ββ ββ ββ βββ ββ β
4. Determine
1 23 1 1.3 7.42 34 7 1 3 2.5 3.9
3 5
β ββ ββ ββ β β β
+ ββ ββ β β ββ ββ ββ β β β β ββ ββ β
1 2 1 2(3 1.3) ( 1 7.43 1 1.3 7.42 3 2 3
4 7 1 3 2.5 3.9 1 3( 4 2.5) (7 3.9)3 5 3 5
β β β β+ β β β + ββ β β ββ ββ β β β+ β =β β β ββ β β ββ ββ β β ββ β β β β β β + β β + β β ββ β β ββ β β β
= (3 0.5 1.3) ( 1 0.666 7.4
( 4 0.333 2.5) (7 0.6 3.9)+ + β + ββ β
β ββ β β β +β β
= 4.8 7.7 3
6.83 10.3
β βββ ββ ββ βββ β
6. Determine
13 624 7 6 3.1 2.4 6.4
22 2 4 0 3 5 7 4 1.6 3.8 1.93
5 7 4 5.3 3.4 4.831 05
β ββ β
ββ β β ββ ββ β β ββ ββ + β β β ββ β β ββ ββ β β ββ ββ ββ β β β
β βββ ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 239
13 624 7 6 3.1 2.4 6.4
22 2 4 0 3 5 7 4 1.6 3.8 1.93
5 7 4 5.3 3.4 4.831 05
β ββ β
ββ β β ββ ββ β β ββ ββ + β β β ββ β β ββ ββ β β ββ ββ ββ β β β
β βββ ββ β
= (8 9 12.4) ( 14 18 9.6) (12 1.5 25.6)
( 4 15 6.4) (8 2 15.2) (0 21 7.6)(10 3 21.2) (14 0 13.6) ( 8 1.8 19.2
+ β β + β + ββ ββ ββ + + β β + +β ββ ββ β + β β + +β β
= 4.6 5.6 12.1
17.4 9.2 28.614.2 0.4 13.0
β ββ ββ βββ ββ βββ β
8. Determine
1 23 1 2 34 7 1 3
3 5
β ββ βββ β
Γ β ββ ββ β ββ β β ββ ββ β
1 2 3 1 3( ) (2 )3 1 2 3 2 3 5
4 7 1 3 7 8 21( 2 ) ( )3 5 3 3 5
β β β β+ +β β β βββ βΓ =β β β ββ ββ β β β ββ β β β β β β ββ β β ββ β β β
=
5 31 26 51 134 63 15
β ββ ββ ββ ββ ββ ββ β
10. Determine 4 7 6 42 4 0 11
5 7 4 7
ββ β β ββ β β ββ Γ ββ β β ββ β β βββ β β β
4 7 6 4 (16 77 42)2 4 0 11 ( 8 44 0)
5 7 4 7 (20 77 28)
β + +β β β β β ββ β β β β ββ Γ β = β β +β β β β β ββ β β β β ββ β ββ β β β β β
= 135
5285
β ββ βββ ββ βββ β
12. Determine 4 7 6 3.1 2.4 6.42 4 0 1.6 3.8 1.9
5 7 4 5.3 3.4 4.8
ββ β β ββ β β ββ Γ β ββ β β ββ β β ββ ββ β β β
4 7 6 3.1 2.4 6.42 4 0 1.6 3.8 1.9
5 7 4 5.3 3.4 4.8
ββ β β ββ β β ββ Γ β ββ β β ββ β β ββ ββ β β β
= (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)
+ + β + + ββ ββ ββ β + β + + β β +β ββ ββ β + β β +β β
= 55.4 3.4 10.112.6 10.4 20.416.9 25.0 37.9
β ββ ββ ββ ββ βββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 240
EXERCISE 109 Page 271
2. Calculate the determinant of
1 22 31 33 5
β ββ ββ ββ ββ ββ ββ β
1 2
1 3 2 1 3 2 27 202 31 3 2 5 3 3 10 9 903 5
β +β ββ β β ββ β= β β β = β + =β ββ β β ββ ββ β β β β β β β β β
= 790
β
4. Evaluate j2 j3
(1 j) jβ
+
j2 j3
(1 j) jβ
+ = (j2)(j) β (-j3)(1 + j) = 2j 2 + j3(1 + j) = -2 + j3 + 2j 3 = -2 + j3 β 3 = -5 + j3
5. Evaluate 2 40 5 207 32 4 117β Β° β β Β°β β Β° β β Β°
( )( ) ( )( )2 40 5 20
2 40 4 117 5 20 7 327 32 4 117β Β° β β Β°
= β Β° β β Β° β β β Β° β β Β°β β Β° β β Β°
= 8β -77Β° - 35β -52Β° = (1.800 β j7.795) β (21.548 β j27.580)
= (-19.75 + j19.79)
From the diagram below, r = 2 2(19.75 19.79 )+ = 27.96 and 1 19.79tan 45.0619.75
β β βΞ± = = Β°β ββ β
and 180 45.06 134.94ΞΈ = Β°β Β° = Β°
Hence, 2 40 5 207 32 4 117β Β° β β Β°β β Β° β β Β°
= (-19.75 + j19.79) or 27.96β 134.94Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 241
EXERCISE 110 Page 272
2. Determine the inverse of
1 22 31 33 5
β ββ ββ ββ ββ ββ ββ β
1 2
3 2 2 3 20 27 72 31 3 10 9 9 10 90 903 5
β= β β β = β = = β
β β
Hence, the inverse of
1 22 31 33 5
β ββ ββ ββ ββ ββ ββ β
is:
90 3 90 23 27 5 7 31 5 3
7 1 1 90 1 90 190 3 2 7 3 7 2
β ββ β β ββ β β β β ββ β β β β ββ ββ β β β β β β β=β ββ ββ β β ββ ββ β ββ β β ββ β β ββ β β β β β β β
=
54 607 730 457 7
β ββ ββ ββ ββ ββ ββ β
=
5 47 87 72 34 67 7
β ββ ββ ββ ββ ββ ββ β
3. Determine the inverse of 1.3 7.4
2.5 3.9ββ ββ βββ β
The inverse of 1.3 7.4
2.5 3.9ββ ββ βββ β
is: 3.9 7.412.5 1.3( 1.3)( 3.9) (7.4)(2.5)
β ββ ββ ββ ββ β β β β
=
3.9 7.43.9 7.41 13.43 13.432.5 1.3 2.5 1.313.43
13.43 13.43
β ββ ββ ββ β
= β ββ ββ ββ β ββ β β ββ β
= 0.290 0.5510.186 0.097β ββ ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 242
EXERCISE 111 Page 274
3. Calculate the determinant of 4 7 62 4 0
5 7 4
ββ ββ βββ ββ βββ β
4 7 62 4 0 4( 16) ( 7)(8) 6( 34)5 7 4
ββ = β β β + β
β using the top row
= -64 + 56 β 204 = -212
5. Calculate the determinant of 3.1 2.4 6.41.6 3.8 1.9
5.3 3.4 4.8
β ββ ββ ββ ββ βββ β
3.1 2.4 6.41.6 3.8 1.9 3.1( 18.24 6.46) 2.4(7.68 10.07) 6.4( 5.44 20.14)
5.3 3.4 4.8β β = β + β + + β β
β using the first row
= 3.1(-11.78) β 2.4(17.75) + 6.4(-25.58) = -36.518 β 42.6 β 163.712 = -242.83
7. Evaluate 3 60 j2 1
0 (1 j) 2 300 2 j5
β Β°+ β Β°
[ ]3 60 j2 1
0 (1 j) 2 30 3 60 j5(1 j) 4 30 j2(0 0) 1(0 0)0 2 j5
β Β°+ β Β° = β Β° + β β Β° β β + + using the top row
= ( ) ( )23 60 j5 j 5 4 30 3 60 j5 5 (3.464 j2)β Β° + β β Β° = β Β° β β +
= ( ) ( )3 60 j5 5 3.464 j2 3 60 8.464 j3β Β° β β β = β Β° β +
= ( )3 60 8.98 160.48 26.94 220.48β Β° β Β° = β
= 26.94β -139.52Β° or (-20.49 β j17.49) 8. Find the eigenvalues Ξ» that satisfy the following equations:
(a) (2 ) 2
01 (5 )βΞ»
=β βΞ»
(b) ( )
(5 ) 7 50 (4 ) 1 02 8 3
βΞ» ββΞ» β =
β βΞ»
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 243
(a) (2 ) 2
01 (5 )βΞ»
=β βΞ»
hence, (2 - Ξ»)(5 - Ξ») β (-2) = 0
i.e. 10 - 7Ξ» + 2Ξ» + 2 = 0
i.e. 2Ξ» - 7Ξ» + 12 = 0
and (Ξ» - 4)( Ξ» - 3) = 0
from which, Ξ» - 4 = 0 or Ξ» - 3 = 0
Thus, eignevalues, Ξ» = 3 or 4
(b) ( )
(5 ) 7 50 (4 ) 1 02 8 3
βΞ» ββΞ» β =
β βΞ»
hence, ( )( ) [ ](5 ) 4 3 8 7(0 2) 5 0 2(4 ) 0βΞ» βΞ» β βΞ» + β + β β βΞ» =β‘ β€β£ β¦
i.e. 2(5 ) 12 8 14 40 10 0β‘ β€β Ξ» β βΞ» + Ξ» + β + β Ξ» =β£ β¦
and ( )( )25 4 26 10 0βΞ» Ξ» βΞ» β + β Ξ» =
i.e. 2 3 25 5 20 4 26 10 0Ξ» β Ξ» β βΞ» + Ξ» + Ξ» + β Ξ» =
and 3 26 11 6 0βΞ» + Ξ» β Ξ» + =
or 3 26 11 6 0Ξ» β Ξ» + Ξ» β =
Let f(Ξ») = 3 26 11 6Ξ» β Ξ» + Ξ» β
f(0) = -6
f(1) = 1 β 6 + 11 β 6 = 0 hence, (Ξ» - 1) is a factor
f(2) = 8 β 24 + 22 β 6 = 0 hence, (Ξ» - 2) is a factor
f(3) = 27 β 54 + 33 β 6 = 0 hence, (Ξ» - 3) is a factor
Thus, (Ξ» - 1)(Ξ» - 2)(Ξ» - 3) = 0
from which, eigenvalues, Ξ» = 1 or 2 or 3
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 244
EXERCISE 112 Page 275
3. Determine the adjoint of 4 7 62 4 0
5 7 4
ββ ββ βββ ββ βββ β
Matrix of cofactors of 4 7 62 4 0
5 7 4
ββ ββ βββ ββ βββ β
is: 16 8 34
14 46 6324 12 2
β β ββ ββ ββ ββ ββ ββ ββ β
Adjoint = transpose of cofactors = 16 14 248 46 1234 63 2
β ββ ββ ββ β ββ ββ ββ ββ β
5. Find the inverse of 4 7 62 4 0
5 7 4
ββ ββ βββ ββ βββ β
From question 3 above, adjoint = 16 14 248 46 63
34 63 2
β ββ ββ ββ β ββ ββ ββ ββ β
4 7 62 4 0 4( 16) 7(8) 6( 34) 2125 7 4
ββ = β + + β = β
β
Hence, the inverse of 4 7 62 4 0
5 7 4
ββ ββ βββ ββ βββ β
is: 16 14 24
1 8 46 12212
34 63 2
β ββ ββ ββ β β ββ ββ ββ ββ β
6. Find the inverse of
13 62
25 73
31 05
β ββ ββ ββ βββ ββ ββ βββ ββ β
Matrix of cofactors of
13 62
25 73
31 05
β ββ ββ ββ βββ ββ ββ βββ ββ β
is:
2 2105 33 33 2 65 101 142 18 323 2
β ββ β ββ ββ ββ ββ ββ ββ ββ ββ ββ ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 245
Transpose of cofactors = adjoint =
2 3 13 425 5 3
3 110 2 1810 2
2 6 323
β ββ ββ ββ ββ ββ ββ ββ ββ ββ β ββ ββ β
13 62
25 73
31 05
β
β
= ( )2 1 2 6 60 1 18 900 5 9233 6 3 75 2 3 5 1 3 15 15
β β ββ β β ββ β + + β = β β β = = ββ β β ββ β β β
Hence, the inverse of
13 62
25 73
31 05
β ββ ββ ββ βββ ββ ββ βββ ββ β
is:
2 3 13 425 5 3
1 3 110 2 18923 10 215 2 6 32
3
β ββ ββ ββ ββ ββ ββ β
β β ββ ββ β ββ ββ β
=
2 3 13 425 5 3
15 3 110 2 18923 10 2
2 6 323
β ββ ββ ββ ββ ββ β ββ ββ ββ ββ β ββ ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 246
CHAPTER 26 THE SOLUTION OF SIMULTANEOUS
EQUATIONS BY MATRICES AND DETERMINANTS EXERCISE 113 Page 279 2. Use matrices to solve: 2p + 5q + 14.6 = 0
3.1p + 1.7 q + 2.06 = 0 2p + 5q = -14.6
3.1p + 1.7 q = -2.06
Hence, 2 5 p 14.6
3.1 1.7 q 2.06ββ ββ β β β
=β ββ β β βββ β β β β β
The inverse of 2 5
3.1 1.7β ββ ββ β
is: 1.7 5 1.7 51 13.1 2 3.1 23.4 15.5 12.1
β ββ β β β=β β β ββ ββ ββ β β β
Thus, p 1.7 5 14.6 14.521 1q 3.1 2 2.06 41.1412.1 12.1
β β ββ β β ββ β β β= =β β β ββ β β ββ ββ ββ β β β β β β β
= 1.23.4
β ββ βββ β
i.e. p = 1.2 and q = -3.4
3. Use matrices to solve: x + 2y + 3z = 5
2x β 3y β z = 3
-3x + 4y + 5z = 3 Since x + 2y + 3z = 5
2x β 3y β z = 3
-3x + 4y + 5z = 3
then, 1 2 3 x 52 3 1 y 33 4 5 z 3
β ββ β β ββ ββ β β ββ β =β ββ β β ββ ββ β β βββ β β β β β
Matrix of cofactors is: 11 7 12 14 107 7 7
β β ββ ββ βββ ββ βββ β
and the transpose of cofactors is: 11 2 77 14 71 10 7
ββ ββ βββ ββ ββ β ββ β
1 2 32 3 13 4 5
β ββ
= 1(-11) β 2 (7) + 3(-1) = -28
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 247
The inverse of 1 2 32 3 13 4 5
β ββ ββ ββ ββ βββ β
is: 11 2 7
1 7 14 728
1 10 7
ββ ββ βββ ββ β ββ β ββ β
Thus, x 11 2 7 5 28 1
1 1y 7 14 7 3 28 128 28
z 1 10 7 3 56 2
β ββ β β ββ β β β β ββ β β ββ β β β β β= β β = β = ββ β β ββ β β β β ββ β β ββ β β β β ββ β β ββ β β β β β β β β β
i.e. x = 1, y = -1 and z = 2
5. Use matrices to solve: p + 2q + 3r + 7.8 = 0
2p + 5q β r β 1.4 = 0
5p β q + 7r β 3.5 = 0 Since p + 2q + 3r = - 7.8
2p + 5q β r = 1.4
5p β q + 7r = 3.5
from which, 1 2 3 p 7.82 5 1 q 1.45 1 7 r 3.5
ββ ββ β β ββ ββ β β ββ =β ββ β β ββ ββ β β βββ β β β β β
Matrix of cofactors is: 34 19 2717 8 1117 7 1
β ββ ββ ββ ββ ββ βββ β
and the transpose of cofactors is: 34 17 1719 8 727 11 1
β ββ ββ ββ ββ ββ βββ β
1 2 32 5 15 1 7
ββ
= 1(34) β 2 (19) + 3(-27) = -85
The inverse of 1 2 32 5 15 1 7
β ββ βββ ββ βββ β
is: 34 17 17
1 19 8 785
27 11 1
β ββ ββ ββ ββ ββ β βββ β
Thus, p 34 17 17 7.8 348.5 4.1
1 1q 19 8 7 1.4 161.5 1.985 85
r 27 11 1 3.5 229.5 2.7
β β β ββ β β ββ β β β β ββ β β ββ β β β β β= β β β = β = ββ β β ββ β β β β ββ β β ββ β β β β ββ ββ β β β β β β β β β
i.e. p = 4.1, q = -1.9 and r = -2.7
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 248
8. In a mechanical system, acceleration x , velocity x and distance x are related by the
simultaneous equations: 3.4 x 7.0 x 13.2x 11.39+ β = β
6.0 x 4.0 x 3.5x 4.98β + + =
2.7 x 6.0 x 7.1x 15.91+ + =
Use matrices to find the values of x , x and x.
Since 3.4 x 7.0 x 13.2x 11.39+ β = β
6.0 x 4.0 x 3.5x 4.98β + + =
2.7 x 6.0 x 7.1x 15.91+ + =
then,
x3.4 7.0 13.2 11.396.0 4.0 3.5 x 4.98
2.7 6.0 7.1 15.91x
β ββ ββ β β ββ ββ β β ββ ββ =β β β ββ ββ β β ββ ββ β β β β β
Matrix of cofactors is: 7.4 52.05 46.8128.9 59.78 1.577.3 67.3 55.6
ββ ββ ββ ββ ββ ββ β
and the transpose of cofactors is: 7.4 128.9 77.3
52.05 59.78 67.346.8 1.5 55.6
ββ ββ ββ ββ ββ ββ β
3.4 7.0 13.26.0 4.0 3.5
2.7 6.0 7.1
ββ = 3.4(7.4) β 7.0 (-52.05) + (-13.2)(-46.8) = 1007.27
The inverse of 3.4 7.0 13.26.0 4.0 3.5
2.7 6.0 7.1
ββ ββ βββ ββ ββ β
is: 7.4 128.9 77.2
1 52.05 59.78 67.31007.27
46.8 1.5 55.6
ββ ββ ββ ββ ββ ββ β
Thus,
x 7.4 128.9 77.3 11.39 503.635 0.51 1x 52.05 59.78 67.3 4.98 775.5979 0.77
1007.27 1007.2746.8 1.5 55.6 15.91 1410.178 1.4x
β β β ββ ββ β β β β ββ β β ββ β β β β ββ β = = =β ββ β β β β ββ β β ββ β β β β ββ ββ β β β β β β β β β β β
i.e. x = 0.5, x = 0.77 and x = 1.4
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 249
EXERCISE 114 Page 282 1. Use determinants to solve the simultaneous equations: 3x β 5y = -17.6
7y β 2x β 22 = 0 Since 3x β 5y + 17.6 = 0
β 2x + 7y β 22 = 0
then x y 15 17.6 3 17.6 3 5
7 22 2 22 2 7
β= =
β ββ β β β
i.e. x y 113.2 30.8 11
β= =
β β
from which, x = 13.211
β = -1.2 and y = 30.811
= 2.8
3. Use determinants to solve the simultaneous equations: 3x + 4y + z = 10
2x β 3y + 5z + 9 = 0
x + 2y β z = 6 Since 3x + 4y + z β 10 = 0
2x β 3y + 5z + 9 = 0
x + 2y β z β 6 = 0
then x y z 14 1 10 3 1 10 3 4 10 3 4 13 5 9 2 5 9 2 3 9 2 3 5
2 1 6 1 1 6 1 2 6 1 2 1
β β= = =
β β ββ β β
β β β β β β
i.e. x y z 14( 21) 1(0) 10( 7) 3( 21) 1( 21) 10( 7) 3(0) 4( 21) 10(7) 3( 7) 4( 7) 1(7)
β β= = =
β β β β β β β β β β β β β β β +
i.e. x y z 114 28 14 14
β β= = =
β
Hence, x = 1414
= 1, y = 2814
= 2 and z = 1414β = -1
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 250
7. Applying mesh-current analysis to an a.c. circuit results in the following equations:
(5 β j4) 1I - (-j4) 2I = 100β 0Β°
(4 + j3 β j4) 2I - (-j4) 1I = 0
Solve the equations for 1I and 2I .
(5 β j4) 1I - (-j4) 2I - 100β 0Β° = 0
- (-j4) 1I + (4 + j3 β j4) 2I + 0 = 0
i.e. (5 β j4) 1I + j4 2I - 100 = 0
j4 1I + (4 β j) 2I + 0 = 0
Hence, 1 2I I 1j4 100 (5 j4) 100 (5 j4) j4
(4 j) 0 j4 0 j4 (4 j)
β= =
β β β ββ β
i.e. ( )
1 22
I I 1100(4 j) j400 (5 j4)(4 j) j4
β= =
β β β β
i.e. 1 2I I 1400 j100 j400 32 j21
= =β β β
Thus, 1I = 400 j100 412.31 14.0432 j21 38.275 33.27
β β β Β°=
β β β Β° = 10.77 19.23 Aβ Β°
and 2I = j400 400 9032 j21 38.275 33.27β β β Β°
=β β β Β°
= 10.45 56.73 Aβ β Β°
9. The forces in three members of a framework are 1F , 2F and 3F . They are related by the
simultaneous equations shown below.
1.4 1F + 2.8 2F + 2.8 3F = 5.6
4.2 1F β 1.4 2F + 5.6 3F = 35.0
4.2 1F + 2.8 2F β 1.4 3F = -5.6
Find the values of 1F , 2F and 3F using determinants.
1.4 1F + 2.8 2F + 2.8 3F - 5.6 = 0
4.2 1F β 1.4 2F + 5.6 3F - 35.0 = 0
4.2 1F + 2.8 2F β 1.4 3F + 5.6 = 0
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 251
Hence, 31 2 FF F 12.8 2.8 5.6 1.4 2.8 5.6 1.4 2.8 5.6 1.4 2.8 2.81.4 5.6 35.0 4.2 5.6 35.0 4.2 1.4 35.0 4.2 1.4 5.62.8 1.4 5.6 4.2 1.4 5.6 4.2 2.8 5.6 4.2 2.8 1.4
β β= = =
β β ββ β β β β β
β β β
i.e. 1 2F F2.8( 17.64) 2.8(90.16) 5.6( 13.72) 1.4( 17.64) 2.8(170.52) 5.6( 29.4)
β=
β β β β β β β β
= 3F 11.4(90.16) 2.8(170.52) 5.6(17.64) 1.4( 13.72) 2.8( 29.4) 2.8(17.64)
β=
β β β β β +
i.e. 31 2 FF F 1225.008 337.512 450.016 112.504
β β= = =
β β β
Thus, 1F = 225.008112.504
= 2 2F = 337.512112.504β = -3 and 3F = 450.016
112.504 = 4
10. Mesh-current analysis produces the following three equations: 1 220 0 (5 3 j4)I (3 j4)Iβ Β° = + β β β 2 1 310 90 (3 j4 2)I (3 j4)I 2Iβ Β° = β + β β β 3 215 0 10 90 (12 2)I 2Iβ β Β°β β Β° = + β Solve the equations for the loop currents 1 2 3I , I and I Rearranging gives: (8 β j4) 1I - (3 β j4) 2I + 0 3I - 20 = 0
-(3 β j4) 1I + (5 β j4) 2I - 2 3I - j10 = 0
0 1I - 2 2I + 14 3I + (15 + j10) = 0 Hence,
31 2 II I(3 j4) 0 20 (8 j4) 0 20 (8 j4) (3 j4) 20
(5 j4) 2 j10 (3 j4) 2 j10 (3 j4) (5 j4) j102 14 (15 j10) 0 14 (15 j10) 0 2 (15 j10)
β= =
β β β β β β β β ββ β β β β β β β β β ββ + + β +
= 1(8 j4) (3 j4) 0(3 j4) (5 j4) 2
0 2 14
ββ β β
β β β ββ
i.e.
[ ] [ ] [ ] [ ]
1 2I I(3 j4) 2(15 j10) j140 20 14(5 j4) 4 (8 j4) 2(15 j10) j140 20 14(3 j4)
β=
β β β + + β β β β β + + β β β
= [ ] [ ] [ ]
3I(8 j4) (5 j4)(15 j10) j20) (3 j4) (3 j4)(15 j10) 20 2(3 j4)β β + β + β β β + β β
= [ ] [ ]
1(8 j4) 14(5 j4) 4 (3 j4) 14(3 j4)
ββ β β + β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 252
i.e. [ ] [ ] [ ] [ ]
1 2I I(3 j4) 30 j120 20 66 j56 (8 j4) 30 j120 20 42 j56)
β=
β β β + β β β β + β β +
= [ ] [ ]
3I(8 j4) 115 j30 (3 j4) 85 j30 40(3 j4)β β + β β + β β
= [ ] [ ]
1(8 j4) 66 j56 (3 j4) 42 j56
ββ β + β β +
i.e. 1 2I I( 90 j360 j120 480) 1320 j1120 240 j960 j120 480 840 j1120
β=
β β + + + β + β + + + + β
= 3I920 j240 j460 120 255 j430 120 120 j160β β β β + + β +
= 1528 j448 j264 224 126 j168 j168 224
ββ β β β + + +
i.e. 31 2 II I 1( 1710 j640) (1080 j40) (545 j110) (402 j376)
β β= = =
β + β β β
Hence, 1I = ( 1710 j640) 1825.84 20.52(402 j376) 550.44 43.09
β β + β β Β°=
β β β Β° = 3.317 22.57 Aβ Β°
2I = (1080 j40) 1080.74 2.12(402 j376) 550.44 43.09
β β β Β°=
β β β Β° = 1.963 40.97 Aβ Β°
3I = (545 j110) 555.99 11.41 555.99 191.41(402 j376) 550.44 43.09 550.44 43.09β β β β β Β° β β Β°
= =β β β Β° β β Β°
= 1.010 148.32 Aβ β Β°
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 253
EXERCISE 115 Page 283 1. Q(3), Exercise 113
Use Cramers rule to solve: x + 2y + 3z = 5
2x β 3y β z = 3
-3x + 4y + 5z = 3
x =
5 2 33 3 13 4 5 5( 11) 2(18) 3(21) 281 2 3 1( 11) 2(7) 3( 1) 282 3 13 4 5
β β
β β + β= =
β β + β ββ β
β
= 1
y =
1 5 32 3 13 3 5 1(18) 5(7) 3(15) 28
28 28 28
ββ β +
= =β β β
= -1
z =
1 2 52 3 33 4 3 1( 21) 2(15) 5( 1) 56
28 28 28
ββ β β + β β
= =β β β
= 2
1. Q(7), Exercise 113
Use Cramers rule to solve: s + 2 v + 2a = 4
3s β v + 4a = 25
3s + 2v β a = -4
s =
4 2 225 1 4
4 2 1 4( 7) 2( 9) 2(46) 821 2 2 1( 7) 2( 15) 2(9) 413 1 43 2 1
ββ β β β β +
= =β β β +
ββ
= 2
v =
1 4 23 25 43 4 1 1( 9) 4( 15) 2( 87) 123
41 41 41β β β β β + β β
= = = -3
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 254
a =
1 2 43 1 253 2 4 1( 46) 2( 87) 4(9) 164
41 41 41
ββ β β β +
= = = 4
2. Q(8), Exercise 114 Use Cramers rule to solve: 1 2 3i 8i 3i 31+ + = β 1 2 33i 2i i 5β + = β 1 2 32i 3i 2i 6β + =
1i =
31 8 35 2 1
6 3 2 31( 1) 8( 16) 3(27) 2401 8 3 1( 1) 8(4) 3( 5) 483 2 12 3 2
ββ β
β β β β β += =
β β + β βββ
= -5
2i =
1 31 33 5 12 6 2 1( 16) 31(4) 3(28) 192
48 48 48
ββ
β + += =
β β β = -4
3i =
1 8 313 2 52 3 6 1( 27) 8(28) 31( 5) 96
48 48 48
ββ ββ β β β β β
= =β β β
= 2
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 255
EXERCISE 116 Page 285 1. In a mass-spring-damper system, the acceleration , velocity and displacement x m are related by
the following simultaneous equations:
6.2 x 7.9 x 12.6x 18.0+ + =
7.5x 4.8x 4.8x 6.39+ + =
13.0 x 3.5x 13.0x 17.4+ β = β
By using Gaussian elimination, determine the acceleration, velocity and displacement for the
system, correct to 2 decimal places.
6.2 x 7.9 x 12.6x 18.0+ + = (1)
7.5x 4.8x 4.8x 6.39+ + = (2)
13.0 x 3.5x 13.0x 17.4+ β = β (3)
(2) - 7.56.2
Γ (1) gives: 0 β 4.7565 x - 10.442 x = -15.384 (2β²)
(3) - 13.06.2
Γ (1) gives: 0 β 13.065 x - 39.419 x = -55.142 (3β²)
(3β²) - 13.0654.7565ββ
Γ (2β²) gives: 0 + 0 β 10.737 x = -12.886
from which, x = 12.88610.737ββ
= 1.2
From (3β²), -13.065 x - 39.419(1.2) = -55.142
i.e. -13.065 x = -55.142 + 39.419(1.2)
and x = 55.142 39.419(1.2)13.065
β +β
= 0.60
From (1), 6.2 x 7.9(0.60) 12.6(1.2) 18.0+ + =
6.2 x 18.0 4.74 15.2 1.86= β β = β
and x = 1.866.2β = -0.30
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 256
2. The tensions, 1 2 3T , T and T in a simple framework are given by the equations:
1 2 35T 5T 5T 7.0+ + =
1 2 3T 2T 4T 2.4+ + =
1 24T 2T 4.0+ =
Determine 1 2 3T , T and T using Gaussian elimination.
1 2 35T 5T 5T 7.0+ + = (1)
1 2 3T 2T 4T 2.4+ + = (2)
1 2 34T 2T 0T 4.0+ + = (3)
(2) - 15
Γ (1) gives: 2 30 T 3T 1.0+ + = (2β²)
(3) - 45
Γ (1) gives: 2 30 2T 4T 1.6β β = β (3β²)
(3β²) - 21ββ β
β ββ β
Γ (2β²) gives: 32T 0.4=
from which, 3T 0.2=
In (3β²) 22T 4(0.2) 1.6β β = β
22T 1.6 0.8 0.8β = β + = β
and 2T 0.4=
In (1) 15T 5(0.4) 5(0.2) 7.0+ + =
i.e. 15T 7.0 2.0 1.0 4.0= β β =
and 1T 0.8=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 257
CHAPTER 27 METHODS OF DIFFERENTIATION EXERCISE 117 Page 291
1.(c) Find the differential coefficient with respect to x of 1x
If y = 11 xx
β= then 2dy 1xdx
β= β = 2
1x
β
2.(a) Find the differential coefficient with respect to x of 2
4xβ
If y = 22
4 4xx
ββ= β then ( )3 3dy ( 4) 2x 8x
dxβ β= β β = or 3
8x
3. Find the differential coefficient with respect to x of (a) 2 x (b) 3 53 x (c) 4x
(a) If y = 122 x 2x= then
1 12 2
12
dy 1 1(2) x xdx 2 x
β ββ β= = =β β
β β = 1
x
(b) If y = 5
3 5 33 x 3x= then 2 23 3dy 5(3) x 5x
dx 3β β
= =β ββ β
= 3 25 x
(c) If y = 12
12
4 4 4xx x
β= = then
3 32 2
32
dy 1 2(4) x 2xdx 2 x
β ββ β= β = β = ββ β
β β =
3
2x
β
4.(a) Find the differential coefficient with respect to x of 3
3xβ
If y = 13
133
3 3 3xx x
ββ = β = β then
4 43 3
43
dy 1 1( 3) x xdx 3 x
β ββ β= β β = =β β
β β =
3 4
1x
5.(c) Find the differential coefficient with respect to x of 5x
3e
If y = 5x5x
3 3ee
β= then ( )5x 5xdy (3) 5e 15edx
β β= β = β = 5x
15e
β
7. Find the gradient of the curve 4 3y 2t 3t t 4= + β + at the points (0, 4) and (1, 8).
If 4 3y 2t 3t t 4= + β + , then gradient, 3 2dy 8t 9t 1dt
= + β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 258
At (0, 4), t = 0, hence gradient = 3 28(0) 9(0) 1+ β = -1
At (1, 8), t = 1, hence gradient = 3 28(1) 9(1) 1+ β = 16
8. Find the co-ordinates of the point on the graph 2y 5x 3x 1= β + where the gradient is 2.
If 2y 5x 3x 1= β + , then gradient = dy 10x 3dx
= β
When the gradient is 2, 10x β 3 = 2 i.e. 10x = 5 and 1x2
=
When 1x2
= , 21 1 5 3 3y 5 3 1 1
2 2 4 2 4β β β β= β + = β + =β β β ββ β β β
Hence, the co-ordinates of the point where the gradient is 2 is 1 3,2 4
β ββ ββ β
9. (a) Differentiate 2 3
2 2y 2ln 2 2(cos5 3sin 2 )e ΞΈ= + ΞΈβ ΞΈ+ ΞΈ β
ΞΈ
(b) Evaluate dydΞΈ
in part (a) when 2Ο
ΞΈ = , correct to 4 significant figures.
(a) 2 3
2 2y 2ln 2 2(cos5 3sin 2 )e ΞΈ= + ΞΈβ ΞΈ+ ΞΈ β
ΞΈ
= 2 32 2ln 2 2cos5 6sin 2 2eβ β ΞΈΞΈ + ΞΈβ ΞΈβ ΞΈβ
Hence, ( )3 3dy 24 2( 5sin 5 ) 6(2cos 2 ) 2 3ed
β β ΞΈ= β ΞΈ + β β ΞΈ β ΞΈ β βΞΈ ΞΈ
= 3 3
4 2 610sin 5 12cos 2e ΞΈβ + + ΞΈ β ΞΈ +
ΞΈ ΞΈ
(b) When 2Ο
ΞΈ = , dydΞΈ
= 33
2
4 2 5 2 610sin 12cos2 2
e22
Οβ ββ ββ β
Ο Οβ + + β +
Οβ βΟβ β β ββ β β β β β
= -1.032049 + 1.2732395 + 10 + 12 + 0.0538997
= 22.30, correct to 4 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 259
EXERCISE 118 Page 293 1. Differentiate 32x cos3x with respect to x.
If y = 32x cos3x , then ( )( ) ( )( )3 2dy 2x 3sin 3x cos3x 6xdx
= β +
= 3 26x sin 3x 6x cos3xβ + = ( )26x cos 3x xsin 3xβ
3. Differentiate 3te sin 4t with respect to x.
If y = 3te sin 4t , then ( )( ) ( )( )3t 3tdy e 4cos 4t sin 4t 3edt
= +
= ( )3te 4cos 4t 3sin 4t+
5. Differentiate te ln t cos t with respect to x.
If y = te ln t cos t , then ( )( ) ( ) ( ) ( )( )t t tdy 1e ln t sin t cos t e ln t edt t
β‘ β€β β= β + +β ββ’ β₯β β β£ β¦
= t cos te ln t sin t cos t ln tt
β§ β«β + +β¨ β¬β© β
= t 1e ln t cos t ln t sin tt
β§ β«β β+ ββ¨ β¬β ββ β β© β
6. Evaluate didt
, correct to 4 significant figures, when t = 0.1, and i = 15t sin3t.
Since i = 15t sin3t, then di (15t)(3cos3t) (sin 3t)(15)dt
= +
= 45t cos 3t + 15 sin 3t
When t = 0.1, didt
= 45(0.1) cos 0.3 + 15 sin 0.3 (note 0.3 is radians)
= 4.2990 + 4.4328
= 8.732, correct to 4 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 260
EXERCISE 119 Page 294
3. Differentiate the quotient 33
2sin 2ΞΈΞΈ
with respect to x.
If y =
33 23 3
2sin 2 2sin 2ΞΈ ΞΈ
=ΞΈ ΞΈ
, then ( ) ( )
( )
1 32 2
2
92sin 2 3 4cos 22dy
d 2sin 2
β β β βΞΈ ΞΈ β ΞΈ ΞΈβ β β ββ β β β =
ΞΈ ΞΈ
=
1 32 2
2
9 sin 2 12 cos 24sin 2
ΞΈ ΞΈβ ΞΈ ΞΈΞΈ
=
2
3 3sin 2 4 cos 24sin 2
ΞΈ ΞΈ β ΞΈ ΞΈΞΈ
4. Differentiate the quotient ln 2tt
with respect to x.
If y = ln 2tt
, ( ) ( )
( )
1 1 12 12 2
2
2
1 1 1t ln 2t t t t ln 2tt 2dy t 12 1 ln 2tdt t t 2t
ββ β
ββ ββ β β β ββ β ββ β β ββ β = = = ββ β
β β
= 32
1 11 ln 2t2t
β βββ ββ β
= 3
1 11 ln 2t2t
β βββ ββ β
6. Find the gradient of the curve 2
2xyx 5
=β
at the point (2, - 4)
If 2
2xyx 5
=β
then gradient,
( )( ) ( ) ( )
2 2 2 2
2 2 22 2 2
x 5 (2) (2x)(2x)dy 2x 10 4x 10 2xdx x 5 x 5 x 5
β β β β β β= = =
β β β
At the point (2, - 4), x = 2, hence gradient = ( )
2
2 22
10 2(2) 10 8 18(4 5) 12 5
β β β β β= =
ββ = -18
7. Evaluate dydx
at x = 2.5, correct to 3 significant figures, given 22x 3y
ln 2x+
=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 261
If 22x 3y
ln 2x+
= then ( )
( )
2
2
1ln 2x (4x) (2x 3)dy xdx ln 2x
β ββ + β ββ β =
When x = 2.5, ( )
( )
2
2
1ln 5 (10) [2(2.5) 3]dy 16.09438 6.22.5dx 2.59029ln 5
β ββ + β β ββ β = = = 3.82, correct to 3
significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 262
EXERCISE 120 Page 296 1. Find the differential coefficient of ( )532x 5xβ with respect to x.
If y = ( )532x 5xβ then ( ) ( )43 2dy 5 2x 5x 6x 5dx
= β β
4. Find the differential coefficient of ( )53
1
x 2x 1β + with respect to x.
If y = ( )
( ) 5353
1 x 2x 1x 2x 1
β= β +
β + then ( ) ( ) ( )
( )
263 2
63
5 3x 2dy 5 x 2x 1 3x 2dx x 2x 1
β β β= β β + β =
β +
= ( )
( )
2
63
5 2 3x
x 2x 1
β
β +
6. Find the differential coefficient of ( )22cot 5t 3+ with respect to x.
If y = cot x = cos xsin x
then
( )2 2
22 2 2
sin x cos xdy (sin x)( sin x) (cos x)(cos x) 1 cosec xdx sin x sin x sin x
β +β β β= = = = β
Thus, if y = ( )22cot 5t 3+ , then ( ) ( ) ( )2 2dy 2 cosec 5t 3 10tdt
β‘ β€= β +β£ β¦ = ( )2 220t cosec 5t 3β +
8. Find the differential coefficient of tan2e ΞΈ with respect to ΞΈ.
If y = tan2e ΞΈ , then ( )tan 2dy 2e secd
ΞΈ= ΞΈΞΈ
= 2 tan2sec e ΞΈΞΈ
9. Differentiate sin3Οβ βΞΈ ΞΈββ β
β β with respect to ΞΈ, and evaluate, correct to 3 significant figures, when
2Ο
ΞΈ =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 263
If y = sin3Οβ βΞΈ ΞΈββ β
β β then ( ) ( )dy cos sin 1
d 3 3Ο Οβ β β β= ΞΈ ΞΈβ + ΞΈββ β β βΞΈ β β β β
When 2Ο
ΞΈ = , dy cos sin cos sind 2 2 3 2 3 2 6 6
Ο Ο Ο Ο Ο Ο Ο Οβ β β β β β= β + β = +β β β β β βΞΈ β β β β β β
= 1.360 + 0.5 = 1.86, correct to 3
significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 264
EXERCISE 121 Page 297
2. (a) Given f(t) = 23
2 1 3t t 15 t t
β + β + determine f β²β²(t).
(b) Evaluate f β²β²(t) when t = 1.
(a) f(t) = 1
2 2 3 1 23
2 1 3 2t t 1 t t 3t t 15 t t 5
β ββ + β + = β + β +
f β²(t) = 1
4 2 24 1t 3t 3t t5 2
ββ β+ β β
f β²β²(t) = 3
5 3 24 112t 6t t5 4
ββ ββ + + = 5 3 3
4 12 6 15 t t 4 tβ + +
(b) When t = 1, f β²β²(t) = 5 3 3
4 12 6 1 4 112 65 (1) (1) 5 44 1β + + = β + + = -4.95
4. Find the second differential coefficient with respect to x of (a) 22cos x (b) ( )42x 3β
(a) If y = 22cos x , dy 4cos x( sin x) 4sin x cos xdx
= β = β
and 2
2 22
d y ( 4sin x)( sin x) (cos x)( 4cos x) 4sin x 4cos xdx
= β β + β = β
= ( )2 24 sin x cos xβ
(b) If y = ( )42x 3β , 3 3dy 4(2x 3) (2) 8(2x 3)dx
= β = β
and 2
22
d y 24(2x 3) (2)dx
= β = ( )248 2x 3β
5. Evaluate f β²β²(ΞΈ) when ΞΈ = 0 given f(ΞΈ) = 2 sec 3ΞΈ If f(ΞΈ) = 2 sec 3ΞΈ, then f β²(ΞΈ) = 6 sec 3ΞΈ tan 3ΞΈ
and f β²β²(ΞΈ) = ( )( ) ( )( )26sec3 3sec 3 tan 3 18sec3 tan 3ΞΈ ΞΈ + ΞΈ ΞΈ ΞΈ
= 3 218sec 3 18sec3 tan 3ΞΈ+ ΞΈ ΞΈ
When ΞΈ = 0, f β²β²(0) = 2
3
18 18 tan 0 18 18(0)cos 0 cos0 1 1
+ = + = 18
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 265
7. Show that, if P and Q are constants and y = P cos(ln t) + Q sin(ln t), then
2
22
d y dyt t y 0dt dt
+ + =
y = P cos(ln t) + Q sin(ln t)
dy P Q Psin(ln t) Qcos(ln t)sin(ln t) cos(ln t)dt t t t
β += β + =
[ ]2
2 2
P cos(ln t) Qsin(ln t)(t) Psin(ln t) Qcos(ln t) (1)d y t tdt t
ββ‘ β€β β β +β’ β₯β£ β¦=
= 2
P cos(ln t) Qsin(ln t) Psin(ln t) Qcos(ln t)t
β β + β
Hence, 2
22
d y dyt t ydt dt
+ + = ( )22
P cos(ln t) Qsin(ln t) Psin(ln t) Qcos(ln t)tt
β β + β
+ (t) Psin(ln t) Qcos(ln t)t
β + + P cos(ln t) + Q sin(ln t)
= P cos(ln t) Qsin(ln t) Psin(ln t) Qcos(ln t) Psin(ln t) Qcos(ln t)β β + β β +
+ P cos(ln t) + Q sin(ln t)
= 0
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 266
CHAPTER 28 SOME APPLICATIONS OF DIFFERENTIATION EXERCISE 122 Page 299 1. An alternating current, i amperes, is given by i = 10 sin 2Οft, 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 2Οft
Rate of change of current, 3di (10)(2 f )cos 2 ft (10)(2 150)cos 2 150 20 10dt
ββ‘ β€= Ο Ο = ΟΓ ΟΓ Γ Γβ£ β¦
= 3000Ο cos 6Ο = 3000Ο A/s
3. The voltage across the plates of a capacitor at any time t seconds is given by v = Vt
CReβ
, where
V, C and R are constants. Given V = 300 volts, C = 60.12 10βΓ F and 6R 4 10= Γ Ξ© find (a) the
initial rate of change of voltage and (b) the rate of change of voltage after 0.5 s.
(a) If v = Vt
CReβ
, then t
CRdv 1V edt CR
ββ β= ββ ββ β
Initial rate of change of voltage, (i.e. when t = 0), 06 6
dv 1(300) edt 0.12 10 4 10β
β β= ββ βΓ Γ Γβ β
= - 3000.48
= -625 V/s
(b) When t = 0.5 s, t 0.5
CR 0.48dv V 300e edt CR 0.48
β ββ β= β = ββ ββ β
= -220.5 V/s
4. The pressure p of the atmosphere at height h above ground level is given by p =hc
0p eβ
, where 0p
is the pressure at ground level and c is a constant. Determine the rate of change of pressure with
height when 0p = 51.013 10Γ pascals and c = 46.05 10Γ at 1450 metres.
Pressure, p =hc
0p eβ
Rate of change of pressure with height, ( ) 41450h
5c 6.05 100 4
dp 1 1(p ) e 1.013 10 edh c 6.05 10
ββΓ
β ββ β= β = Γ ββ ββ β β βΓβ β β β
= -1.635 Pa/m
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 267
EXERCISE 123 Page 301 1. A missile fired from ground level rises x metres vertically upwards in t seconds and
225x 100t t2
= β . Find (a) the initial velocity of the missile, (b) the time when the height of the
missile is a maximum, (c) the maximum height reached, (d) the velocity with which the missile
strikes the ground.
(a) Distance, 225x 100t t2
= β
Initial velocity, (i.e. when t = 0), dx 100 25t 100 25(0)dt
= β = β = 100 m/s
(b) When height is a maximum, velocity = 0, i.e. dx 100 25t 0dt
= β =
from which, 100 = 25t and time t = 4 s
(c) When t = 4 s, maximum height, x = 225100(4) (4) 400 2002
β = β = 200 m
(d) When x = 0 (i.e. on the ground), 2250 100t t2
= β
i.e. 25t 100 t 02
β ββ =β ββ β
Hence, either t = 0 (at the start) or 25 25100 t 0 i.e. 100 t2 2
β = =
and 200t25
= = 8 s
Velocity, i.e. dxdt
, when t = 8 s is given by dxdt
= 100 β 25t = 100 β 25(8)
= 100 β 200 = -100 m/s (negative indicating
reverse direction to the starting velocity)
3. The equation 210 24t 3tΞΈ = Ο+ β gives the angle ΞΈ, in radians, through which a wheel turns in t
seconds. Determine (a) the time the wheel takes to come to rest, (b) the angle turned through in
the last second of movement.
Angle, 210 24t 3tΞΈ = Ο+ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 268
(a) When the wheel comes to rest, angular velocity = 0, i.e. d 0dtΞΈ=
Hence, d 24 6t 0dtΞΈ= β = from which, time, t = 4 s
(b) Distance moved in the last second of movement = (distance after 4 s) β (distance after 3 s)
= 2[10 24(4) 3(4) ]Ο+ β - 2[10 24(3) 3(3) ]Ο+ β
= [10 96 48]Ο+ β - [10 72 27]Ο+ β
= 96 β 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 β 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 β t)
Velocity, v = ( ) 1dx 14 ( 1) 4 1 tdt 1 t
β= + β = β ββ
Initial velocity, i.e. when t = 0, v = 141
β = 3 m/s
Acceleration, a = 2
22 2
d x 1(1 t) ( 1)dt (1 t)
β= β β = ββ
Initial acceleration, i.e. when t = 0, a = 11
β = -1 2m / s
(b) After 1.5 s, velocity, v = 1 1 14 4 4 4 2(1 t) (1 1.5) ( 0.5)
β = β = β = +β β β
= 6 m/s
and acceleration, a = 2 2 2
1 1 1 1(1 t) (1 1.5) ( 0.5) 0.25
β = β = β = ββ β β
= -4 2m / s
(c) When the velocity is zero, 14 0(1 t)
β =β
i.e. 14(1 t)
=β
i.e. 4 β 4t = 1 and 4 β 1 = 4t
from which, time, t = 3 s4
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 269
EXERCISE 124 Page 305 2. Find the turning point on x (6 )= ΞΈ βΞΈ and determine its nature.
2x (6 ) 6= ΞΈ βΞΈ = ΞΈβΞΈ
dx 6 2 0d
= β ΞΈ =ΞΈ
for a turning point, from which, ΞΈ = 3
When ΞΈ = 3, 2 2x 6 6(3) 3 18 9 9= ΞΈβΞΈ = β = β =
Hence, (3, 9) are the co-ordinates of the turning point 2
2
d x 2d
= βΞΈ
, which is negative, hence a maximum occurs at (3, 9)
3. Find the turning point on 3 2y 4x 3x 60x 12= + β β and distinguish between them.
3 2y 4x 3x 60x 12= + β β
2dy 12x 6x 60 0dx
= + β = for a turning point
i.e. 22x x 10 0+ β = i.e. (2x + 5)(x β 2) = 0
from which, 2x + 5 = 0 i.e. x = -2.5
and x β 2 = 0 i.e. x = 2
When x = -2.5, 3 2y 4( 2.5) 3( 2.5) 60( 2.5) 12 94.25= β + β β β β =
When x = 2, 3 2y 4(2) 3(2) 60(2) 12 88= + β β = β
Hence, (-2.5, 94.25) and (2, -88) are the co-ordinates of the turning points
2
2
d x 24x 6d
= +ΞΈ
When x = -2.5, 2
2
d xdΞΈ
is negative, hence (-2.5, 94.25) is a maximum point.
When x = 2, 2
2
d xdΞΈ
is positive, hence (2, -88) is a minimum point.
5. Find the turning point on xy 2x e= β and determine its nature.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 270
xy 2x e= β hence, xdy 2 e 0dx
= β = for a turning point.
i.e. x2 e= and x = ln 2 = 0.6931
When x = 0.6931, 0.6931y 2(0.6931) e 0.6136= β = β
Hence, (0.6931, -0.6136) are the co-ordinates of the turning point.
2x
2
d y edx
= β When x = 0.6931, 2
2
d ydx
is negative, hence (0.6931, -0.6136) is a maximum point.
8. Determine the maximum and minimum values on the graph y = 12 cos ΞΈ - 5 sin ΞΈ in the range
ΞΈ = 0 to ΞΈ = 360Β°. Sketch the graph over one cycle showing relevant points.
y = 12 cos ΞΈ - 5 sin ΞΈ
dy 12sin 5cos 0d
= β ΞΈβ ΞΈ =ΞΈ
for a maximum or minimum value
i.e. -12 sin ΞΈ = 5 cos ΞΈ from which, sin 5cos 12
ΞΈ= β
ΞΈ i.e. tan ΞΈ = 5
12β
Hence, 1 5tan12
β β βΞΈ = ββ ββ β
= - 22Β°37β²
Tangent is negative in the 2nd and 4th quadrants as shown in the diagram below.
Hence, ΞΈ = 180Β° - 22Β°37β² = 157Β°23β² and 360Β° - 22Β°37β² = 337Β°23β²
When ΞΈ = 157Β°23β², y = 12 cos 157Β°23β² - 5 sin 157Β°23β² = -13
When ΞΈ = 337Β°23β², y = 12 cos 337Β°23β² - 5 sin 337Β°23β² = 13
Hence, (157Β°23β², -13) and (337Β°23β², 13) are the co-ordinates of the turning points.
2
2
d y 12cos 5sindx
= β ΞΈ+ ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 271
When ΞΈ = 157Β°23β², 2
2
d ydx
is positive, hence (157Β°23β², -13) is a minimum point
When ΞΈ = 337Β°23β², 2
2
d ydx
is negative, hence (337Β°23β², 13) is a maximum point
A sketch of y = 12 cos ΞΈ - 5 sin ΞΈ is shown below. (When y = 0, 12 cos ΞΈ - 5 sin ΞΈ = 0 and
12 cos ΞΈ = 5 sin ΞΈ; hence, sin 12cos 5
ΞΈ=
ΞΈ from which, tan ΞΈ = 2.4 and ΞΈ = 1tan 2.4 67.4β = Β° and
247.4Β°; also, at ΞΈ = 0, y = 12 cos 0 β 5 sin 0 = 12).
9. Show that the curve 32y (t 1) 2t(t 2)3
= β + β has a maximum value of 23
and a minimum value
of -2.
3 3 22 2y (t 1) 2t(t 2) (t 1) 2t 4t3 3
= β + β = β + β
2dy 2(t 1) 4t 4 0dx
= β + β = for a turning point.
i.e. ( )22 t 2t 1 4t 4 0β + + β =
i.e. 22t 4t 2 4t 4 0β + + β =
i.e. 22t 2 0β = from which, 2t 1= and t = Β±1
When t = 1, 32y (1 1) 2(1 2) 23
= β + β = β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 272
When t = -1, 32 16 2y ( 1 1) 2( 1)( 1 2) 63 3 3
= β β + β β β = β + =
2
2
d y 4(t 1) 4dt
= β + When t = 1, 2
2
d ydt
is positive, hence (1, -2) is a minimum point.
When t = -1, 2
2
d ydt
is negative, hence 21,3
β βββ ββ β
is a maximum point.
Hence, the maximum value of 32y (t 1) 2t(t 2)3
= β + β is 23
and the minimum value is -2
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 273
EXERCISE 125 Page 309 1. The speed, v, of a car (in m/s) is related to time t s by the equation: 2v 3 12t 3t= + β . Determine
the maximum speed of the car in km/h.
Speed, 2v 3 12t 3t= + β
dv 12 6t 0dt
= β = for a maximum value, from which, 12 = 6t and t = 2 s
When t = 2, 2v 3 12(2) 3(2) 3 24 12= + β = + β = 15 m/s
2
2
d v 6dt
= β , which is negative, hence indicating that v = 15 m/s is the maximum speed.
15 m/s = 60 60 s / h15m / s 15 3.61000 m / km
ΓΓ = Γ = 54 km/h = maximum speed.
3. A shell is fired vertically upwards and its vertical height, x metres, is given by x = 24t - 23t ,
where t is the time in seconds. Determine the maximum height reached.
Height, x = 24t - 23t
dx 24 6t 0dt
= β = for a maximum value, from which, 24 = 6t and t = 4 s
2
2
d x 6dt
= β , which is negative β hence a maximum value.
Maximum height = 24(4) - 23(4) = 96 β 48 = 48 m
4. A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area
of the metal for which the volume of the box is 3.5 3m
A lidless box with square ends is shown in the diagram below, having dimensions x by x by y.
Volume of box, V = 2 3x y 3.5m= (1)
Area of metal, A = 2 22
3.52x 3xy 2x 3xx
β β+ = + β ββ β
from equation (1)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 274
i.e. A = 2 12x 10.5xβ+
2dA 4x 10.5x 0
dxβ= β = for a maximum or minimum value.
i.e. 32
10.5 10.54x i.e. x 2.625x 4
= = = from which, 3x 2.625= = 1.3795
23
2
d A 4 21xdx
β= + When x = 1.3795, 2
2
d Adx
is positive β hence a minimum value.
Minimum or least area of metal = 2 210.5 10.52x 2(1.3795)x 1.3795
+ = + = 11.42 2m
6. Calculate the height of a cylinder of maximum volume which can be cut from a cone of height
20 cm and base radius 80 cm.
A sketch of a cylinder within a cone is shown below. Cylinder volume, V = 2r hΟ
A section is shown below.
By similar triangles, 20 h
80 80 r=
β from which, h = 20(80 r) 80 r r20
80 4 4β β
= = β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 275
Hence, cylinder volume, V = 3
2 2r rr 20 20 r4 4
Οβ βΟ β = Ο ββ ββ β
2dV 3 r40 r 0dr 4
Ο= Ο β = for a maximum or minimum value.
i.e. 23 r 3r 16040 r i.e. 40 and r
4 4 3Ο
Ο = = =
2
2
d V 6 r40dr 4
Ο= Οβ When r = 160
3,
2
2
d Vdr
is negative, hence a maximum value.
Hence, height of cylinder, h = 20 -
160r 40320 204 4 3= β = β = 6.67 cm
7. The power developed in a resistor R by a battery of emf E and internal resistance r is given by:
( )
2
2E RP
R r=
+. Differentiate P with respect to R and show that the power is a maximum when
R = r.
( )
2
2E RP
R r=
+ hence
2 2 2
4
dP (R r) (E) E R(2)(R r) 0dR (R r)
+ β += =
+for a maximum value
Thus, 2 2E (R r) 2R(R r) 0β‘ β€+ β + =β£ β¦
i.e. 2 2 2R 2Rr r 2R 2Rr 0+ + β β =
i.e. 2 2r R 0β =
and R = r
2 2 2
4
E r RdPdR (R r)
β‘ β€ββ£ β¦=+
and [ ]4 2 2 2 2 32
2 8
(R r) E 2R E r R 4(R r)d PdR (R r)
β‘ β€+ β β β +β£ β¦=+
When R = r, 2
2
d PdR
is negative, hence power is a maximum when R = r.
9. Resistance to motion, F, of a moving vehicle, is given by F = 5 100xx+ . Determine the
minimum value of resistance.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 276
F = 5 100xx+ = 15x 100xβ +
2
2
dF 55x 100 100 0dx x
β= β + = β + = for a maximum or minimum value.
i.e. 100 = 2
5x
and 2 5x 0.05 and x 0.05 0.2236100
= = = =
2
32 3
d F 1010xdx x
β= = which is positive when x = 0.2236, hence x = 0.2236 gives a minimum value of
resistance.
Maximum resistance to motion, F = 5 100xx+ = 5 100(0.2236)
0.2236+ = 44.72
11. The fuel economy E of a car, in miles per gallon, is given by:
E = 21 + 2 2 6 42.10 10 v 3.80 10 vβ βΓ β Γ
where v is the speed of the car in miles per hour. Determine, correct to 3 significant figures,
the most economical fuel consumption, and the speed at which it is achieved.
E = 21 + 2 2 6 42.10 10 v 3.80 10 vβ βΓ β Γ
2 6 3dE 4.20 10 v 4(3.80) 10 v 0dv
β β= Γ β Γ = for a maximum (most economical fuel consumption).
i.e. 2 6 34.20 10 v 4(3.80) 10 vβ βΓ = Γ
i.e. 2
26
4.20 10v 2763.1584(3.80) 10
β
β
Γ= =
Γ
from which, speed, v = 2763.158 = 52.6 m.p.h
22 6 2
2
d E 4.20 10 12(3.80) 10 vdv
β β= Γ β Γ and when v = 52.6, 2
2
d Edv
is negative, hence v is the
maximum speed. Maximum fuel economy, E = 21 + 2 2 6 42.10 10 v 3.80 10 vβ βΓ β Γ = 21 + 2 2 6 42.10 10 (52.6) 3.80 10 (52.6)β βΓ β Γ = 50.0 miles/gallon
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 277
EXERCISE 126 Page 311 2. For the curve y = 23x 2xβ find (a) the equation of the tangent, and (b) the equation of the
normal at the point (2, 8)
(a) y = 23x 2xβ Gradient, m = dy 6x 2dx
= β
At the point (2, 8), x = 2, hence, m = 6(2) β 2 = 10
Equation of tangent is: 1 1y y m(x x )β = β
i.e. y β 8 = 10(x β 2)
i.e. y β 8 = 10x β 20
and y = 10x - 12
(b) Equation of normal is: 1 11y y (x x )m
β = β β
i.e. 1y 8 (x 2)10
β = β β
i.e. 10(y β 8) = -x + 2
i.e. 10y β 80 = -x + 2
and 10y + x = 82
4. For the curve y = 21 x x+ β find (a) the equation of the tangent, and (b) the equation of the
normal at the point (-2, -5)
(a) y = 21 x x+ β Gradient, m = dy 1 2xdx
= β
At the point (-2, -5), x = -2, hence, m = 1 β 2(-2) = 5
Equation of tangent is: 1 1y y m(x x )β = β
i.e. y β (-5) = 5(x β -2)
i.e. y + 5 = 5x + 10
and y = 5x + 5
(b) Equation of normal is: 1 11y y (x x )m
β = β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 278
i.e. 1y 5 (x 2)5
β β = β β β
i.e. 5(y + 5) = -x - 2
i.e. 5y + 25 = -x - 2
and 5y + x + 27 = 0
5. For the curve 1t
ΞΈ = find (a) the equation of the tangent, and (b) the equation of the normal at
the point 13,3
β ββ ββ β
(a) 11 tt
βΞΈ = = Gradient, m = 22
d 1tdt t
βΞΈ= β = β
At the point 13,3
β ββ ββ β
, t = 3, hence, m = 2
1 13 9
β = β
Equation of tangent is: 1 1m(t t )ΞΈβΞΈ = β
i.e. 1 1 (t 3)3 9
ΞΈβ = β β
i.e. 9ΞΈ - 3 = -t + 3
and 9ΞΈ + t = 6
(b) Equation of normal is: 1 11 (t t )m
ΞΈβΞΈ = β β
i.e. 1 1 (t 3)139
ΞΈβ = β ββ βββ ββ β
i.e. 1 9(t 3)3
ΞΈβ = β
i.e. 1 9t 273
ΞΈβ = β
and 29t 263
ΞΈ = β or 3ΞΈ = 27t - 80
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 279
EXERCISE 127 Page 312 2. The pressure p and volume v of a mass of gas are related by the equation pv = 50. If the pressure
increases from 25.0 to 25.4, determine the approximate change in the volume of the gas. Find
also the percentage change in the volume of the gas.
pv = 50 i.e. v = 150 50pp
β= and 22
dv 5050pdp p
β= β = β
Approximate change in volume, 2 2
dv 50 50v p (25.4 25.0) (0.4)dp p 25.0
β β β βΞ΄ β β Ξ΄ = β β = β =β β β ββ β β β
-0.032
Percentage change in volume = 0.032 3.2 3.2100%50 50 2p 25.0
β β βΓ = = = -1.6%
4. The radius of a sphere decreases from 6.0 cm to 5.96 cm. Determine the approximate change in
(a) the surface area, and (b) the volume.
(a) Surface area of sphere, A = 24 rΟ and dA 8 rdr
= Ο
Approximate change in surface area, ( )dAA r 8 r (6.0 5.96) 8 (6.0)( 0.04)dr
Ξ΄ β β Ξ΄ = Ο β = Ο β
= -6.03 2cm
(b) Volume of sphere, V = 34 r3Ο and 2dV 4 r
dr= Ο
Approximate change in volume, ( )2 2dVV r 4 r ( 0.04) 4 (6.0) ( 0.04)dr
Ξ΄ β β Ξ΄ = Ο β = Ο β = -18.10 3cm
5. The rate of flow of a liquid through a tube is given by Poiseuillesβs equation as: 4p rQ
8 LΟ
=Ξ·
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 Ξ· is the coefficient of viscosity of the liquid. Ξ·
is obtained by measuring Q, p, r and L. If Q can be measured accurate to Β±0.5%, p accurate to
Β±3%, r accurate to Β±2% and L accurate to Β±1%, calculate the maximum possible percentage error
in the value of Ξ·.
4p rQ
8 LΟ
=Ξ·
from which, 4p r
8 LQΟ
Ξ· =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 280
4 4
2
d p r p rQ ( 0.005Q) ( 0.005)dQ 8LQ 8LQ
β β β βΞ· Ο Οδη β β Ξ΄ = β Β± = Β±β β β β
β β β β
4 4d r p rp ( 0.03p) ( 0.03)dp 8LQ 8LQ
β β β βΞ· Ο Οδη β β Ξ΄ = Β± = Β±β β β β
β β β β
3 4d 4p r p rr ( 0.02r) ( 0.08)dr 8LQ 8LQ
β β β βΞ· Ο Οδη β β Ξ΄ = Β± = Β±β β β β
β β β β
4 4
2
d p r p rL ( 0.01L) ( 0.01)dL 8L Q 8LQ
β β β βΞ· β Ο Οδη β β Ξ΄ = Β± = Β±β β β β
β β β β
Maximum possible percentage error β 0.5% + 3% + 8% + 1% = 12.5%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 281
CHAPTER 29 DIFFERENTIATION OF PARAMETRIC EQUATIONS
EXERCISE 128 Page 316
2. A parabola has parametric equations: x = 2t , y = 2t. Evaluate dydx
when t = 0.5
If x = 2t , then dx 2tdt
=
If y = 2t, then dy 2dt
=
Hence,
dydy 2 1dt
dxdx 2t tdt
= = =
When t = 0.5, dy 1dx 0.5
= = 2
3. The parametric equations for an ellipse are x = 4 cos ΞΈ, y = sin ΞΈ. Determine (a) dydx
(b) 2
2
d ydx
(a) If x = 4 cos ΞΈ, then dx 4sind
= β ΞΈΞΈ
If y = sin ΞΈ, then dy cosd
= ΞΈΞΈ
Hence,
dydy cosd
dxdx 4sind
ΞΈΞΈ= =β ΞΈ
ΞΈ
= 1 cot4
β ΞΈ
(b) ( )2
2
2 2 3
d dy d 1 1cot cosecd y 1 1 1 1d dx d 4 4dxdx 4sin 4sin 16 sin sin 16 sind
β β β ββ ΞΈ β β ΞΈβ β β βΞΈ ΞΈ β β β ββ β β β = = = = β = ββ β β ββ ΞΈ β ΞΈ ΞΈ ΞΈ ΞΈβ β β β ΞΈ
= 31 cosec16
β ΞΈ
4. Evaluate dydx
at 6Ο
ΞΈ = radians for the hyperbola whose parametric equations are x = 3 sec ΞΈ,
y = 6 tan ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 282
If x = 3 sec ΞΈ, then dx 3sec tand
= ΞΈ ΞΈΞΈ
If y = 6 tan ΞΈ, then 2dy 6secd
= ΞΈΞΈ
Hence, 2
dy 1dy 6sec 2sec 2d cos2dx sindx 3sec tan tan sin
d cos
β ββ βΞΈ ΞΈΞΈ ΞΈ= = = = =β βΞΈΞΈ ΞΈ ΞΈ ΞΈβ β
ΞΈ ΞΈβ β
When 6Ο
ΞΈ = , dy 2 2dx 0.5sin
6
= =Ο
= 4
6. The equation of a tangent drawn to a curve at point ( )1 1x , y is given by: ( )11 1
1
dyy y x xdx
β = β .
Determine the equation of the tangent drawn to the ellipse x = 3 cos ΞΈ, y = 2 sin ΞΈ at 6Ο
ΞΈ =
( )11 1
1
dyy y x xdx
β = β
At point ΞΈ, 1x 3cos= ΞΈ and 1dx 3sind
= β ΞΈΞΈ
1y 2sin= ΞΈ and 1dy 2cosd
= ΞΈΞΈ
Hence, 1
1
11
dydy 2cos 2d cotdxdx 3sin 3
d
ΞΈΞΈ= = = β ΞΈβ ΞΈ
ΞΈ
The equation of a tangent is: y β 2 sin ΞΈ = ( )2 cot x 3cos3
β ΞΈ β ΞΈ
At 6Ο
ΞΈ = , y β 2 sin 6Ο = 2 cot x 3cos
3 6 6Ο Οβ ββ ββ ββ β
i.e. y β 1 = ( )( )2 1.732 x 2.5983
β β
i.e. y β 1 = -1.155(x β 2.598)
i.e. y β 1 = -1.155x + 3
and y = -1.155x + 4
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 283
EXERCISE 129 Page 318 1. A cycloid has parametric equations x = 2(ΞΈ - sin ΞΈ), y β 2(1 β cos ΞΈ). Evaluate, at ΞΈ = 0.62 rad,
correct to 4 significant figures, (a) dydx
(b) 2
2
d ydx
(a) If x = 2(ΞΈ - sin ΞΈ), then dx 2 2cosd
= β ΞΈΞΈ
If y = 2(1 β cos ΞΈ), then dy 2sind
= ΞΈΞΈ
Hence,
dydy 2sin sind
dxdx 2(1 cos ) 1 cosd
ΞΈ ΞΈΞΈ= = =β ΞΈ β ΞΈ
ΞΈ
When ΞΈ = 0.62 rad, dy sin 0.62dx 1 cos0.62
=β
= 3.122, correct to 4 significant figures.
(b)
2 2
2 2 2
2
(1 cos )(cos ) (sin )(sin ) cos cos sind dy d sind y (1 cos ) (1 cos )d dx d 1 cos
dxdx 2(1 cos ) 2(1 cos ) 2(1 cos )d
β ΞΈ ΞΈ β ΞΈ ΞΈ ΞΈβ ΞΈβ ΞΈΞΈβ β β ββ β β β β ΞΈ β ΞΈΞΈ ΞΈ β ΞΈβ β β β = = = =
β ΞΈ β ΞΈ β ΞΈΞΈ
= 2 2
3 3
cos (cos sin ) cos 12(1 cos ) 2(1 cos )
ΞΈβ ΞΈ+ ΞΈ ΞΈβ=
β ΞΈ β ΞΈ
When ΞΈ = 0.62 rad, 2
2 3
d y (cos 0.62) 1 0.1861215dx 2(1 cos 0.62) 2(0.00644748)
β β= =
β = -14.43, correct to 4 significant
figures.
2. The equation of a normal drawn to a curve at point ( )1 1x , y is given by: ( )1 11
1
1y y x xdydx
β = β β .
Determine the equation of the normal drawn to the parabola 21 1x t , y t4 2
= = at t = 2.
If 21
1x t ,4
= dx 1 tdt 2
=
If 11y t ,2
= dy 1dt 2
=
Hence, 1
1
11
dy 1dy 1dt 2
dx 1dx tt2dt
= = =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 284
Equation of a normal is: ( )1 11
1
1y y x xdydx
β = β β
i.e. 21 1 1y t x t12 4t
β ββ = β ββ ββ β
At t = 2, equation of normal is: y β 1 = -2(x β 1)
i.e. y β 1 = -2x + 2 or y = -2x + 3
4. Determine the value of 2
2
d ydx
, correct to 4 significant figures, at 6Ο
ΞΈ = rad for the cardioid
x = 5(2ΞΈ - cos 2ΞΈ), y = 5(2 sin ΞΈ β sin 2ΞΈ).
If x = 5(2ΞΈ - cos 2ΞΈ), then ( )dx 10 10sin 2 10 1 sin 2d
= + ΞΈ = + ΞΈΞΈ
If y = 5(2 sin ΞΈ β sin 2ΞΈ), hence dy 10cos 10cos 2 10(cos cos 2 )d
= ΞΈβ ΞΈ = ΞΈβ ΞΈΞΈ
dydy 10(cos cos 2 ) cos cos 2d
dxdx 10(1 sin 2 ) 1 sin 2d
ΞΈβ ΞΈ ΞΈβ ΞΈΞΈ= = =+ ΞΈ + ΞΈ
ΞΈ
( )22
2
(1 sin 2 )( sin 2sin 2 ) (cos cos 2 )(2cos 2 )d dy d cos cos 21 sin 2d y d dx d 1 sin 2
dxdx 10(1 sin 2 ) 10(1 sin 2 )d
+ ΞΈ β ΞΈ+ ΞΈ β ΞΈβ ΞΈ ΞΈΞΈβ ΞΈβ β β ββ β β β + ΞΈΞΈ ΞΈ + ΞΈβ β β β = = =
+ ΞΈ + ΞΈΞΈ
When 6Ο
ΞΈ = rad,
2
2
2
2 2 2 21 sin sin 2sin cos cos 2cos6 6 6 6 6 6
21 sind y 6
2dx 10 1 sin6
Ο Ο Ο Ο Ο Οβ ββ β β ββ β+ β + β ββ ββ β β ββ ββ β β β β β β β
Οβ β+β ββ β =
Οβ β+β ββ β
=
(1.86603)(1.23205) (0.366025)(1)(3.48205)18.660254
β
= 0.02975, correct to 4 significant
figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 285
5. The radius of curvature, Ο, of part of a surface when determining the surface tension of a liquid
is given by:
3/ 22
2
2
dy1dxd ydx
β‘ β€β β+β’ β₯β ββ β β’ β₯β£ β¦Ο =
Find the radius of curvature (correct to 4 significant figures) of the part of the surface having
parametric equations (a) x = 3t, 3yt
= at the point 1t2
=
(b) 3 3x 4cos t, y 4sin t= = at t = rad6Ο
(a) x 3t= , hence dx 3dt
=
13y 3tt
β= = , hence 22
dy 33tdt t
β= β = β
2
2
dy 3dy 1dt t
dxdx 3 tdt
β= = = β and at 1t
2= , 2
dy 1 4dx 1
2
= β = ββ ββ ββ β
( )2
2 32
2 3
d dy d 1 d td y 2t 2dt dx dt t dtdxdx 3 3 3 3tdt
ββ
β β β ββ ββ β β ββ β β β = = = = = and at 1t
2= ,
2
32 3
d y 2 2 16dx 3t 313
2
= = =β ββ ββ β
Hence, radius of curvature, ( )
3/ 2232
3
2
2
dy1 1 4dx 1716 16d y3 3dx
β‘ β€β β+β’ β₯β β β‘ β€+ ββ β β’ β₯ β£ β¦β£ β¦Ο = = = = 13.14
(b) 3x 4cos t= , hence 2 2dx 12cos t( sin t) 12cos t sin tdt
= β = β
3y 4sin t= , hence 2dy 12sin t cos tdt
=
2
2
dydy 12sin t cos t sin tdt tan tdxdx 12cos t sin t cos t
dt
= = = β = ββ
and at t = rad6Ο , dy tan 0.57735
dx 6Ο
= β = β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 286
( )2 2
2 2 2 4
d dy d tan td y sec t 1dt dx dtdxdx 12cos t sin t 12cos t sin t 12cos t sin tdt
β βββ β ββ β = = = =
β β
and at t rad6Ο ,
2
42
d y 1 0.29630dx
12 cos sin6 6
= =Ο Οβ β
β ββ β
Hence, radius of curvature, ( )
3/ 2232
3
2
2
dy1 1 0.57735dx 1.333333d y 0.29630 0.29630dx
β‘ β€β β+β’ β₯β β β‘ β€+ ββ β β’ β₯ β£ β¦β£ β¦Ο = = = = 5.196
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 287
CHAPTER 30 DIFFERENTIATION OF IMPLICIT FUNCTIONS EXERCISE 130 Page 320 2. Differentiate the given functions with respect to x:
(a) 5 ln 3t2
(b) 2y 13 e4
+ (c) 2 tan 3y
(a) d 5 5 1 dtln 3tdx 2 2 t dx
β β β β=β β β ββ β β β
= 5 dt2t dx
(b) ( )2y 1 2y 1d 3 3 dye 2edx 4 4 dx
+ +β β =β ββ β
= 2y 13 dye2 dx
+
(c) ( ) ( )2d dy2 tan 3y (2) 3sec 3ydx dx
= = 2 dy6sec 3ydx
4. Differentiate the following with respect to u:
(a) 2(3x 1)+
(b) 3 sec 2ΞΈ (c) 2y
(a) 1 2d 2 d dx2(3x 1) 2(3x 1) (3)du 3x 1 du du
β ββ β β‘ β€ β‘ β€= + = β +β β β£ β¦ β£ β¦+β β = 2
6 dx(3x 1) du
β+
(b) ( ) ( )d d3sec 2 3 2sec 2 tan 2du du
ΞΈΞΈ = ΞΈ ΞΈ = d6sec 2 tan 2
duΞΈ
ΞΈ ΞΈ
(c) 1 32 2
32
d 2 d 1 dy 1 dy2y (2) ydu du 2 du duy y
β ββ β β β β β= = β = ββ β β β β ββ β β β β β β β
= 3
1 dyduy
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 288
EXERCISE 131 Page 321
1. Determine ( )2 3d 3x ydx
( )2 3d 3x ydx
= ( ) ( )2 3 3 2d d3x y y 3xdx dx
+ using the product rule
= ( ) ( )( )2 2 3dy3x 3y y 6xdx
β β +β ββ β
= 2 2 3dy9x y 6xydx
+
= 2 dy3xy 3x 2ydx
β β+β ββ β
3. Determine d 3udu 4v
β ββ ββ β
d 3udu 4v
β ββ ββ β
= ( )( ) ( )
( )2 2 2
dv dv4v 3 3u 4 12v 12u 12 dvdu du v u16v 16v du4v
β ββ ββ β β ββ β = = ββ ββ β
= 2
3 dvv u4v du
β βββ ββ β
5. Determine dzdy
given z = 32x ln y
z = 32x ln y hence, ( ) ( )3 2dz 1 dx2x ln y 6xdy y dy
β β β β= +β β β β
β β β β
= 3
22x dx6x ln yy dy
+ or 2 x dx2x 3ln yy dy
β β+β β
β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 289
EXERCISE 132 Page 323
1. Determine dydx
given 2 2x y 4x 3y 1 0+ + β + =
( ) ( ) ( ) ( ) ( )2 2d d d d dx y 4x 3y 1 0dx dx dx dx dx
+ + β + =
dy dy2x 2y 4 3 0 0dx dx
+ + β + =
2x + 4 = ( )dy dy dy3 2y 3 2ydx dx dx
β = β
Hence, dy 2x 4dx 3 2y
+=
β
3. Given 2 2x y 9+ = evaluate dydx
when x = 5 and y = 2
2 2d d d(x ) (y ) (9)dx dx dx
+ =
i.e. 2x + 2y dydx
= 0
i.e. 2y dydx
= -2x and dy xdx y
= β
When x = 5 and y = 2, dy 5dx 2
= β
5. Determine dydx
given 2 23y 2xy 4x 0+ β =
Given 2 23y 2xy 4x 0+ β = then ( ) ( )( )dy dy6y 2x 1 y 2 8x 0dx dx
β‘ β€β β+ + β =β ββ’ β₯β β β£ β¦
( )dy 6y 2x 8x 2ydx
+ = β
and dy 8x 2ydx 6y 2x
β=
+ = 4x y
3y xβ+
7. Determine dydx
given 43y 2x ln y y x+ = +
Given 43y 2x ln y y x+ = + then ( ) ( )( ) 3dy 1 dy dy3 2x ln y 2 4y 1dx y dx dx
β‘ β€β β+ + = +β’ β₯β β
β β β£ β¦
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 290
i.e. 3dy 2x3 4y 1 2ln ydx y
β β+ β = ββ β
β β
and 3
dy 1 2ln y2xdx 3 4yy
β=
+ β
8. If 2 2 3 253x 2x y y 04
+ β = evaluate dydx
when x = 12
and y = 1.
Given 2 2 3 253x 2x y y 04
+ β = then ( ) ( )( )2 2 3dy 5 dy6x 2x 3y y 4x y 0dx 2 dx
β‘ β€β β+ + β =β ββ’ β₯β β β£ β¦
i.e. 3 2 25 dy6x 4xy y 6x y2 dx
β β+ = ββ ββ β
and 3
2 2
dy 6x 4xy5dx y 6x y2
+=
β
When x = 12
and y = 1, ( )
( ) ( )
3
22
1 16 4 1dy 3 2 52 2dx 2.5 1.5 15 11 6 1
2 2
β β β β+β β β β +β β β β = = =ββ ββ β β
β β
= 5
10. Find the gradients of the tangents drawn to the ellipse 2 2x y 2
4 9+ = at the point where x = 2.
Given 2 2x y 2
4 9+ = then 2x 2y dy 0
4 9 dx+ = and 2y dy 2x
9 dx 4= β
from which,
2xdy 2x 9 9x4
2ydx 4 2y 4y9
= β = β β = β
If 2 2x y 2
4 9+ = and x = 2, then
24 y 24 9+ = from which,
2y 19=
Hence, 2y 9= and y = Β± 3
Thus, dy 9x (9)(2) 3dx 4y 4( 3) 2
= β = β = Β±Β±
or Β±1.5
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 291
CHAPTER 31 LOGARITHMIC DIFFERENTIATION EXERCISE 133 Page 326
2. Use logarithmic differentiation to differentiate y = 3
2 4
(x 1)(2x 1)(x 3) (x 2)
+ +β +
with respect to x.
If y = 3
2 4
(x 1)(2x 1)(x 3) (x 2)
+ +β +
then ln y = 3
2 4
(x 1)(2x 1)ln(x 3) (x 2)
β§ β«+ +β¨ β¬β +β© β
i.e. = 3 2 4ln(x 1) ln(2x 1) ln(x 3) ln(x 2)+ + + β β β +
i.e. = ln(x 1) 3ln(2x 1) 2ln(x 3) 4ln(x 2)+ + + β β β +
Differentiating w.r.t. x gives: 1 dy 1 3 2 4(2)y dx (x 1) (2x 1) (x 3) (x 2)
= + β β+ + β +
i.e. dy 1 6 2 4ydx (x 1) (2x 1) (x 3) (x 2)
β§ β«= + β ββ¨ β¬+ + β +β© β
i.e. 3
2 4
dy (x 1)(2x 1) 1 6 2 4dx (x 3) (x 2) (x 1) (2x 1) (x 3) (x 2)
β§ β«+ += + β ββ¨ β¬β + + + β +β© β
4. Use logarithmic differentiation to differentiate y = 2xe cos3x(x 4)β
with respect to x.
If y = 2xe cos3x(x 4)β
then ln y = ( )
2xe cos3xlnx 4
β§ β«βͺ βͺβ¨ β¬
ββͺ βͺβ© β =
12x 2ln e ln(cos3x) ln(x 4)+ β β
= 2x + ln(cos 3x) - 12
ln(x - 4)
Differentiating w.r.t. x gives:
11 dy ( 3sin 3x) 22y dx cos3x (x 4)
β= + β
β
i.e. dy 1y 2 3tan 3xdx 2(x 4)
β§ β«= β ββ¨ β¬ββ© β
i.e. 2xdy e cos 3x 12 3tan 3x
dx 2(x 4)(x 4)β§ β«
= β ββ¨ β¬ββ β© β
6. Use logarithmic differentiation to differentiate y = 4
2x
2x tan xe ln 2x
with respect to x.
If y = 4
2x
2x tan xe ln 2x
then ln y = 4
4 2x2x
2x tan xln ln 2x ln(tan x) ln e ln(ln 2x)e ln 2x
β§ β«= + β ββ¨ β¬
β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 292
= ln 2 + ln 4x + ln(tan x) β 2x ln e β ln(ln 2x)
= ln 2 + 4 ln x + ln(tan x) β 2x(1) β ln(ln 2x)
Differentiating w.r.t. x gives: 21 dy 4 1 1 10 (sec x) 2y dx x tan x ln 2x x
β β= + + β β β ββ β
i.e. 2
dy 4 cos x 1 1y 2dx x sin x cos x x ln 2x
β§ β«β β= + β ββ¨ β¬β ββ β β© β
i.e. 4
2x
dy 2x tan x 4 1 12dx e ln 2x x sin xcos x x ln 2x
β§ β«= + β ββ¨ β¬β© β
8. Evaluate dydΞΈ
, correct to 3 significant figures, when 4Ο
ΞΈ = given y = 5
2e sinΞΈ ΞΈ
ΞΈ
If y = 5
2e sinΞΈ ΞΈ
ΞΈ then ln y =
52
5
2e sin ln 2 ln e ln(sin ) lnΞΈ
ΞΈβ§ β«ΞΈ= + + ΞΈ β ΞΈβ¨ β¬
ΞΈβ© β
= 5ln 2 ln(sin ) ln2
+ ΞΈ+ ΞΈ β ΞΈ
Differentiating w.r.t. ΞΈ gives: 1 dy 1 5 10 1 (cos )y d sin 2
β β= + + ΞΈ β β βΞΈ ΞΈ ΞΈβ β
i.e. 5
dy 5 2e sin 5y 1 cot 1 cotd 2 2
ΞΈ ΞΈβ§ β« β§ β«= + ΞΈβ = + ΞΈββ¨ β¬ β¨ β¬ΞΈ ΞΈ ΞΈβ© β β© βΞΈ
When 4Ο
ΞΈ = , 4
5
2e sindy 54 1 cotd 4 2
44
Ο β§ β«Οβͺ βͺΟβͺ βͺ= + ββ¨ β¬ΟΞΈ β ββͺ βͺΟβ β β ββ β βͺ βͺβ β β© ββ β
= 5.673935[1 + 1 β 3.18309886]
= - 6.71, correct to 3 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 293
EXERCISE 134 Page 328 2. Differentiate y = ( )x2x 1β with respect to x. Since y = ( )x2x 1β then ln y = ln ( )x2x 1β = x ln(2x β 1)
Differentiating w.r.t. x gives: 1 dy 2(x) [ln(2x 1)](1)y dx 2x 1
β β= + ββ β+β β by the product rule
i.e. dy 2xy ln(2x 1)dx 2x 1
β§ β«= + ββ¨ β¬ββ© β
i.e. ( )xdy 2x2x 1 ln(2x 1)dx 2x 1
β§ β«= β + ββ¨ β¬ββ© β
3. Differentiate y = x (x 3)+ with respect to x.
y = x (x 3)+ = 1x(x 3)+ and ln y =
1x 1ln(x 3) ln(x 3)
x+ = +
Differentiating w.r.t. x gives: ( )21 dy 1 1 [ln(x 3)] xy dx x x 3
ββ ββ β= + + ββ ββ β+β β β β by the product rule
and 2
dy 1 ln(x 3)ydx x(x 3) x
β§ β«+= ββ¨ β¬+β© β
i.e. x2
dy 1 ln(x 3)(x 3)dx x(x 3) x
β§ β«+= + ββ¨ β¬+β© β
5. Show that when xy 2x= and x = 1, dy 2dx
=
xy 2x= and ln y = x xln 2x ln 2 ln x ln 2 x ln x= + = +
Differentiating w.r.t. x gives: 1 dy 10 (x) (ln x)(1) 1 ln xy dx x
β β= + + = +β ββ β
by the product rule
i.e. xdy y(1 ln x) 2x (1 ln x)dx
= + = +
When x = 1, 1dy 2(1) (1 ln1)dx
= + = 2
6. Evaluate xd (x 2)dx
β when x = 3
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 294
Let y = 1xx (x 2) (x 2)β = β then ln y =
1x 1 ln(x 2)ln(x 2) ln(x 2)
x xβ
β = β =
Differentiating w.r.t. x gives: 2
1(x) [ln(x 2)](1)1 dy x 2y dx x
β β β ββ βββ β = by the quotient rule
and 2
x ln(x 2)dy (x 2)ydx x
β§ β«β ββͺ βͺββͺ βͺ= β¨ β¬βͺ βͺβͺ βͺβ© β
i.e. x
2
(x 2)dy x ln(x 2)dx x x 2
β β§ β«= β ββ¨ β¬ββ© β
When x = 3, 3
2
(3 2)dy 3 1 3ln(3 2) 0dx 3 3 2 9 1
β β§ β« β ββ§ β«= β β = ββ¨ β¬ β¨ β¬β βββ© β β β β© β = 1
3
7. Show that if y ΞΈ= ΞΈ and 2ΞΈ = , dydΞΈ
= 6.77, correct to 3 significant figures.
If y ΞΈ= ΞΈ then ln y = ln lnΞΈΞΈ = ΞΈ ΞΈ
Differentiating w.r.t. ΞΈ gives: ( )1 dy 1 (ln )(1)y d
β β= ΞΈ + ΞΈβ βΞΈ ΞΈβ β = 1 + ln ΞΈ
and dy y(1 ln ) (1 ln )d
ΞΈ= + ΞΈ = ΞΈ + ΞΈΞΈ
When 2ΞΈ = , 2dy 2 (1 ln 2)d
= +ΞΈ
= 6.77, correct to 3 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 295
CHAPTER 32 DIFFERENTIATION OF HYPERBOLIC
FUNCTIONS
EXERCISE 135 Page 331 2. Differentiate the following with respect to the variable:
(a) 2 sec h 5x3
(b) 5 tcosech8 2
(c) 2 coth 7ΞΈ
(a) ( )d 2 2sec h 5x 5sec h 5x tanh 5xdx 3 3
β β β β= ββ β β ββ β β β
= 10 sech 5x tanh 5x3
β
(b) d 5 t 5 1 t tcosec h cosec h cothdt 8 2 8 2 2 2β β β ββ β= ββ β β ββ ββ β β β β β
= 5 t tcosech coth16 2 2
β
(c) ( ) ( )( )2d 2cot h 7 2 7cosec h 7d
ΞΈ = β ΞΈΞΈ
= 214cosech 7β ΞΈ
3. Differentiate the following with respect to the variable:
(a) 2 ln (sh x) (b) 3 ln th4 2
β ΞΈ ββ ββ ββ ββ β β β
(a) ( ) ( ) ( )d 12ln(shx) 2 chxdx shx
β β= β ββ β
= 2cot h x
(b) 2
2
2
1
ch chd 3 3 1 1 3 3 12 2ln th sec hdx 4 2 4 2 2 8 8th sh ch sh
2 2 2 2ch
2
ΞΈ ΞΈβ β β ββ ββ β β ββ ββ‘ β€β ΞΈ β ΞΈβ β β β β β= = =β β β ββ ββ’ β₯β β β β β ββ β ΞΈ ΞΈ ΞΈ ΞΈβ β β β β β β β β£ β¦ β β β ββ ββ β β β β β
ΞΈ
= 3 18 ch sh
2 2
β ββ ββ βΞΈ ΞΈβ ββ β
= 3 sech cosech8 2 2
ΞΈ ΞΈ
5. Differentiate the following with respect to the variable:
(a) 3
3sh 4x2x
(b) ch 2tcos 2t
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 296
(a) ( )( ) ( )( )
( )
3 2
23 3
2x 12ch 4x 3sh 4x 6xd 3sh 4xdx 2x 2x
ββ β =β ββ β
by the quotient rule
= 3 2
6
24x ch 4x 18x sh 4x4xβ = 4
12xch 4x 9sh 4x2xβ
(b) ( )( ) ( )( )2
cos 2t 2sh 2t ch 2t 2sin 2td ch 2tdx cos 2t cos 2t
β ββ β =β ββ β
= ( )
2
2 cos 2t sh 2t ch 2t sin 2tcos 2t
+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 297
CHAPTER 33 DIFFERENTIATION OF INVERSE
TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS
EXERCISE 136 Page 336
1.(a) Differentiate 1sin 4xβ with respect to x.
If y = 1sin 4xβ , then 2
dy 4dx 1 (4x)
=β‘ β€ββ£ β¦
= ( )2
4
1 16xβ
2.(b) Differentiate 12 xcos3 3
β with respect to x.
If y = 12 xcos3 3
β , then ( )2 2
dy 2 1dx 3 3 x
β‘ β€ββ’ β₯=
β’ β₯ββ£ β¦
= ( )2
2
3 9 x
β
β
3.(a) Differentiate 13 tan 2xβ with respect to x.
If y = 13 tan 2xβ , then 2
dy 23dx 1 (2x)
β‘ β€= β’ β₯+β£ β¦
= 2
61 4x+
4.(b) Differentiate 1 3sec x4
β with respect to x.
If y = 1 3sec x4
β , then ( )2 2 2 2
3dy 1 1 14dx 9x 9x 16 x 9x 163 3x x 1 xx 1 16 164 4 4
= = = =β‘ β€ β β β ββ ββ β ββ β β β ββ’ β₯β β β β β β β β β’ β₯β£ β¦
= ( )2
4
x 9x 16β
5.(a) Differentiate 15 cos ec2 2
β ΞΈ with respect to ΞΈ.
If y = 15 cos ec2 2
β ΞΈ , then ( )2 2
dy 5 2d 2 2
β‘ β€ββ’ β₯=
β’ β₯ΞΈ ΞΈ ΞΈ ββ£ β¦
= ( )2
5
4
β
ΞΈ ΞΈ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 298
6.(b) Differentiate 1 2cot 1β ΞΈ β with respect to ΞΈ.
If y = 1 2cot 1β ΞΈ β , then ( )
( ) ( )
12 2
2 2 22
1 1 2dy 2d 1 1 11 1
ββ ΞΈ β ΞΈ β ΞΈ
= =ΞΈ β‘ β€ β‘ β€ΞΈ β + ΞΈ β+ ΞΈ β β£ β¦β’ β₯β£ β¦
= ( )2
1
1
β
ΞΈ ΞΈ β
7. Show that the differential coefficient of 12
xtan1 x
β β ββ βββ β
is 2
2 4
1 x1 x x
+β +
If y = 12
xtan1 x
β β ββ βββ β
then
( )( ) ( )( )( ) ( )
( )( )
2 2 2
2 22 2 2
2 2 2 4 22 2
2 22
1 x 1 x 2x 1 x 2x1 x 1 xdy 1 x
dx 1 2x x xx 1 x x11 x 1 x
β β β β +
β β += = =
β + +β β β ++ β βββ β β
= ( )2
2 4
1 x1 x x
+β +
8.(b) Differentiate 2 1t sec 2tβ with respect to t.
If y = 2 1t sec 2tβ then ( )( )
( )( )2 1
2
dy 2t sec 2t 2tdt 2t 2t 1
β
β ββ β
= +β ββ‘ β€ββ ββ£ β¦β β
= ( )
1
2
t 2t sec 2t4t 1
β+β
9.(a) Differentiate ( )2 1 2cos 1βΞΈ ΞΈ β with respect to ΞΈ.
If y = ( )2 1 2cos 1βΞΈ ΞΈ β then ( )( )
( ) ( )2 1 2
22
dy 2 cos 1 2d 1 1
β
β ββ ββ ΞΈ β‘ β€= ΞΈ + ΞΈ β ΞΈβ β β£ β¦ΞΈ β β‘ β€ ββ ΞΈ ββ ββ’ β₯β£ β¦β β
= ( )( )
31 2
4 2
22 cos 11 2 1
β ΞΈΞΈ ΞΈ β β
β‘ β€β ΞΈ β ΞΈ +β£ β¦
= ( )( )
31 2
2 4
22 cos 12
β ΞΈΞΈ ΞΈ β β
ΞΈ βΞΈ = ( )
( )3
1 2
2 2
22 cos 12
β ΞΈΞΈ ΞΈ β β
ΞΈ βΞΈ
= ( )( )
31 2
2
22 cos 12
β ΞΈΞΈ ΞΈ β β
ΞΈ βΞΈ = ( )
( )2
1 2
2
22 cos 12
β ΞΈΞΈ ΞΈ β β
β ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 299
10.(b) Differentiate 1x cos ec xβ with respect to x.
If y = 1x cos ec xβ then ( )( )
( )
12
1
2
1 xdy 2x cos ec x 1dx
x x 1
β
β
β βββ β
β β β‘ β€= + β£ β¦β ββ‘ β€ββ ββ’ β₯β£ β¦β β
= ( )
1x cos ec x2 x x x 1
ββ ββ β‘ β€β β + β£ β¦β βββ β
= 1 1cosec x2 (x 1)
β ββ
11.(a) Differentiate 1
2
sin 3xx
β
with respect to x.
If y = 1
2
sin 3xx
β
then
( )( )
( )( )
( )
2 1
2
22
3x sin 3x 2x1 3xdy
dx x
β
β ββ β
ββ ββ‘ β€ββ ββ£ β¦β β =
= ( )
( )
( )
21
2
14 4 2
3x 2x sin 3x1 3x x 3x 2sin 3x
x x 1 9x
β
β
β ββ β
ββ ββ‘ β€ β‘ β€ββ ββ£ β¦β β β’ β₯= β
β’ β₯ββ£ β¦
= ( )
13 2
1 3x 2sin 3xx 1 9x
ββ§ β«βͺ βͺββ¨ β¬
ββͺ βͺβ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 300
EXERCISE 137 Page 338 1. Use logarithmic equivalents of inverse hyperbolic functions to evaluate the following correct to
4 decimal places: (a) 1 1sinh2
β (b) 1sinh 4β (c) 1sinh 0.9β
2 2
1 x x a xsinh lna a
ββ§ β«+ +βͺ βͺ= β¨ β¬βͺ βͺβ© β
(a) 1 1sinh2
β = 2 21 2 1 1 5ln ln ln1.618034
2 2
β§ β« β β+ + +βͺ βͺ = =β ββ¨ β¬ β ββͺ βͺ β β β© β = 0.4812
(b) 1 1 4sinh 4 sinh1
β β= = ( )2 24 1 4ln ln 4 17 ln8.123106
1
β§ β«+ +βͺ βͺ = + =β¨ β¬βͺ βͺβ© β
= 2.0947
(c) 1sinh 0.9β = ( )2 20.9 1 0.9ln ln 0.9 1.81 ln 2.245362
1
β§ β«+ +βͺ βͺ = + =β¨ β¬βͺ βͺβ© β
= 0.8089
2. Use logarithmic equivalents of inverse hyperbolic functions to evaluate the following correct to
4 decimal places: (a) 1 5cosh4
β (b) 1cosh 3β (c) 1cosh 4.3β
2 2
1 x x x acosh lna a
ββ§ β«+ ββͺ βͺ= β¨ β¬βͺ βͺβ© β
(a) 1 5cosh4
β = 2 25 5 4 5 3ln ln ln 2
4 4
β§ β«+ β +βͺ βͺ β β= =β¨ β¬ β ββ β βͺ βͺβ© β
= 0.6931
(b) 1cosh 3β = ( )2 23 3 1ln ln 3 8 ln 5.828427
1
β§ β«+ ββͺ βͺ = + =β¨ β¬βͺ βͺβ© β
= 1.7627
(c) 1cosh 4.3β = ( )2 24.3 4.3 1ln ln 4.3 17.49 ln8.482105
1
β§ β«+ ββͺ βͺ = + =β¨ β¬βͺ βͺβ© β
= 2.1380
3. Use logarithmic equivalents of inverse hyperbolic functions to evaluate the following correct to
4 decimal places: (a) 1 1tanh4
β (b) 1 5tanh8
β (c) 1tanh 0.7β
1 x 1 a xtanh ln
a 2 a xβ +β β= β βββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 301
(a) 1 1tanh4
β = 1 4 1 1 5ln ln2 4 1 2 3
+β β =β βββ β = 0.2554
(b) 1 5tanh8
β = 1 8 5 1 13ln ln2 8 5 2 3
+β β =β βββ β = 0.7332
(c) 1tanh 0.7β = 1 1 0.7 1 1.7ln ln2 1 0.7 2 0.3
+β β =β βββ β = 0.8673
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 302
EXERCISE 138 Page 341 1.(b) Differentiate 1sinh 4xβ with respect to x.
If y = 1sinh 4xβ then ( )2
dy 4dx 4x 1
=β‘ β€+β£ β¦
= ( )2
4
16x 1+
2.(a) Differentiate 1 t2cosh3
β with respect to t.
If y = 1 t2cosh3
β then 2 2
dy 12dt t 3
β‘ β€= β’ β₯
ββ£ β¦ =
( )2
2
t 9β
3.(b) Differentiate 13 tanh 3xβ with respect to x.
If y = 13 tanh 3xβ then ( )2
dy 33dx 1 3x
β‘ β€= β’ β₯
ββ’ β₯β£ β¦ = 2
91 9xβ
4.(a) Differentiate 1 3xsech4
β with respect to x.
If y = 1 3xsech4
β then ( )2 2 2 2
3dy 1 1 14
xdx 9x 16 9x3x 3x 16 9xx 1 x1 416 164 4
β β β β= = = =
β‘ β€ β β β ββ ββ β ββ β β β ββ’ β₯β β β β β β β β β’ β₯β£ β¦
= ( )2
4
x 16 9x
β
β
5.(b) Differentiate 11 cos ec h 4x2
β with respect to x.
If y = 11 cos ec h 4x2
β then ( )2
dy 1 4dx 2 4x 4x 1
β‘ β€ββ’ β₯= β’ β₯
β‘ β€+β’ β₯β£ β¦β£ β¦
= ( )2
1
2x 16x 1
β
+
6.(a) Differentiate 1 2xcoth7
β with respect to x.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 303
If y = 1 2xcoth7
β then 2 2 2 2
2 2 2 2 (49)dy 7 7 7 74x 49 4xdx 49 4x2x 11 49 497
= = = =β ββ β ββ β β
β β
= 2
1449 4xβ
7.(b) Differentiate ( )1 21 cosh x 12
β + with respect to x.
If y = ( )1 21 cosh x 12
β + then ( )
( ) ( ) ( )
12 2
2 2 22
1 x 1 (2x)dy 1 x2dx 2 2 x 1 x 1 1x 1 1
ββ‘ β€β’ β₯+β’ β₯= =β’ β₯β‘ β€ + + β+ ββ’ β₯β’ β₯β’ β₯β£ β¦β£ β¦
= ( ) ( ) ( )22 2
x x
2x x 12 x 1 x=
++ =
( )2
1
2 x 1+
8.(a) Differentiate 1sech (x 1)β β with respect to x.
If y = 1sech (x 1)β β then 22
dy 1 1dx (x 1) 1 (x 2x 1)(x 1) 1 (x 1)
β β= =
β β β +β‘ β€β β ββ£ β¦
2
1(x 1) (2x x )
β=
β β =
[ ]1
(x 1) x(2 x)β
β β
9.(b) Differentiate 1coth (cos x)β with respect to x.
If y = 1coth (cos x)β then ( )2 2 2
dy sin x sin x sin x 1dx 1 cos x sin x sin x1 cos x
β β β β= = = = =
ββ = cosec xβ
10.(a) Differentiate 1sinhβΞΈ ΞΈ with respect to ΞΈ.
If y = 1sinhβΞΈ ΞΈ then ( )
( )( )1
2
dy 1 sinh 1d 1
ββ ββ β= ΞΈ + ΞΈβ βΞΈ β βΞΈ +β β
= ( )
1
2sinh
1βΞΈ
+ ΞΈΞΈ +
11.(b) Differentiate ( )
1
2
tanh x1 x
β
β with respect to x.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 304
If y = ( )
1
2
tanh x1 x
β
β then
( ) ( )( )
( )
22
22
11 x tanh x 2xdy 1 xd 1 x
β ββ β ββ βββ β =ΞΈ β
= ( )
1
22
1 2x tanh x
1 x
β+
β
12. Show that 1d x cosh (cosh x) 2xdx
ββ‘ β€ =β£ β¦
( )( )
( )1 1
2
d sinh xx cosh (cosh x) x cosh (cosh x) (1)dx cosh x 1
β ββ‘ β€β’ β₯β‘ β€ = +β£ β¦ β’ β₯ββ£ β¦
= ( )2
x sinh x xsinh x
+ since 1cosh (cosh x) xβ =
= x sinh x xsinh x
+
= x + x = 2x
13.(b) Determine ( )2
3 dx4x 25+
β«
( ) ( )2 22 222
3 1 1 3 1dx 3 dx 3 dx54x 25 2x 5 2x2x 15 1
55
= = =β‘ β€ β‘ β€ β‘ β€+ β β+ β ββ ββ£ β¦ ++β’ β₯ β’ β₯β β β ββ ββ β β β β β β’ β₯β’ β₯ β£ β¦β β β£ β¦
β« β« β« β«
= 2 2
2 23 5 35 5dx dx5 2 22x 2x1 1
5 5
β β =β ββ β β‘ β€ β‘ β€β β β β+ +β’ β₯ β’ β₯β β β β
β β β β β’ β₯ β’ β₯β£ β¦ β£ β¦
β« β«
= 13 2xsinh c2 5
β +
14.(b) Determine ( )2
1 dtt 5β
β«
( ) 2 22 2
11 1 1 1 1 5 5dt dt dt dt
15 5t 5 t t t5 1 1 15 5 5
β β= = = β ββ ββ‘ β€ β‘ β€ β‘ β€β β β β β β β β ββ β ββ’ β₯ β’ β₯ β’ β₯β β β β β ββ β β β β β β£ β¦ β’ β₯ β’ β₯β£ β¦ β£ β¦
β« β« β« β«
= 2
15 dt
t 15
β‘ β€β β ββ’ β₯β ββ β β’ β₯β£ β¦
β« = 1 tcosh c5
β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 305
15.(b) Determine ( )2
3 dx16 2xββ«
( ) ( ) ( ) ( )2 22 2 2 2
3 3 3 1 3 1 8dx dx dx dx2 216 2x 2 8 x 88 x 8 x
β β= = = β ββ β β β β ββ« β« β« β« = 13 xtanh c
2 8 8β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 306
CHAPTER 34 PARTIAL DIFFERENTIATION EXERCISE 139 Page 345
2. Find zxββ
and xyββ
given z = x3 β 2xy + y2
If z = 3 2x 2xy yβ + then ( ) ( ) ( )3 2 2 2z d d dx (2y) (x) y (1) 3x 2y(1) y (0)x dx dx dxβ
= β + = β +β
23x 2y= β
and ( )3 2 3x d d d(x ) (1) (2x) (y) y (x )(0) 2x(1) 2yy dy dy dyβ
= β + = β +β
2x 2y= β +
5. Find zxββ
and xyββ
given z = 3 22
y 1x yx y
β +
If z = 3 2 3 2 2 12
y 1x y x y yx yx y
β ββ + = β + then ( )( ) ( )( )2 2 3z y 3x y 2x 0x
ββ= β β +
β2 2
3
2y3x yx
= +
and ( )( ) ( )3 2 2x x 2y 1 x yy
β ββ= β β
β 3
2 2
1 12x yx y
= β β
6. Find zxββ
and xyββ
given z = cos 3x sin 4y
If z = cos 3x sin 4y then ( ) ( ) ( )( )z dsin 4y cos3x sin 4y 3sin 3xx dxβ
= = ββ
= 3sin 3xsin 4yβ
and ( ) ( ) ( )( )x dcos3x sin 4y cos3x 4cos 4yy dyβ
= =β
= 4cos 3xcos4y
8. The resonant frequency fr in a series electrical circuit is given by fr =1
2 LCΟ. Show that
r3
f 1L 4 CLβ β
=β Ο
fr =1 12 21 1 L C
22 LCβ β
=ΟΟ
and 1 3
r 2 21 3 32 2
f 1 1 1 1C LL 2 2 4 C L4 C L
β ββ ββ ββ= β = β = ββ ββ ββ Ο Οβ β β β Ο
= 3
14 CL
β
Ο
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 307
9. An equation resulting from plucking a string is: n n b n by sin x k cos t csin tL L LΟ β§ Ο Ο β«β β β β β β= +β¨ β¬β β β β β β
β β β β β β β© β
Determine yt
ββ
and yxββ
n n b n b n n b n n by sin x k cos t csin t k sin x cos t csin x sin tL L L L L L LΟ β§ Ο Ο β« Ο Ο Ο Οβ β β β β β β β β β β β β β= + = +β¨ β¬β β β β β β β β β β β β β β
β β β β β β β β β β β β β β β© β
Hence, y n n b n b n n b n bk sin x sin t csin x cos tt L L L L L L
β β‘ Ο β€ β‘ Ο β€ Ο β‘ Ο β€ β‘ Ο β€ Οβ β β β β β β β β β β β= β +β β β β β β β β β β β ββ’ β₯ β’ β₯ β’ β₯ β’ β₯β β β β β β β β β β β β β β£ β¦ β£ β¦ β£ β¦ β£ β¦
= n b n n b n bsin x ccos t k sin tL L L L
β§ β«Ο Ο Ο Οβ β β β β β β βββ¨ β¬β β β β β β β ββ β β β β β β β β© β
and y n n n b n n n bk cos x cos t c cos x sin tx L L L L L Lβ Ο Ο Ο Ο Ο Οβ β β β β β β β β β β β= +β β β β β β β β β β β ββ β β β β β β β β β β β β
= n n n b n bcos x k cos t csin tL L L L
β§ β«Ο Ο Ο Οβ β β β β β β β+β¨ β¬β β β β β β β ββ β β β β β β β β© β
10. In a thermodynamic system, k =T S H
R TAeβ ββ
where R, k and A are constants. Find
(a) kTββ
(b) ATββ
(c) ( S)T
β ββ
(d) ( H)T
β ββ
(a) ( )( ) ( )( )T S HR T
2 2
RT S T S H Rk AeT R T
β βββ‘ β€ β β β βββ‘ β€β= β’ β₯ β’ β₯β β’ β₯ β£ β¦β£ β¦
= T S H
R T2 2
RT S RT S R HAeR T
β βββ‘ β€ β‘ β€β β β + ββ’ β₯ β’ β₯
β£ β¦β’ β₯β£ β¦
= T S H
R T2 2
R HAeR T
β βββ‘ β€ β‘ β€ββ’ β₯ β’ β₯
β£ β¦β’ β₯β£ β¦ =
T S HR T
2
HAeR T
β βββ‘ β€ β‘ β€ββ’ β₯ β’ β₯
β£ β¦β’ β₯β£ β¦
= T S H
RT2
A H eR T
β βββ
(b) Since k = T S H
R TAeβ ββ
then T S H
R TH T S
RTA k e k eβ ββ
β β β β
= =
Thus, ( )( ) ( )( )S T SR T
2 2
RT S H T S RA k eT R T
β β ββ β ββ β β β ββ ββ= β ββ ββ ββ β β β β
= S T SR T
2 2
R T S R H RT Sk eR T
β β ββ ββ ββ β β β + ββ ββ ββ ββ β β β
= S T SR T
2 2
R Hk eR T
β β ββ ββ ββ ββ ββ ββ ββ β β β
= H T S
RT2
k H eR T
β β βββ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 308
(c) If k = T S H
R TAeβ ββ
then T S H
R Tk eA
β ββ
= and k T S HlnA R T
β βββ β =β ββ β
i.e. kR T ln T S HA
β β = β βββ ββ β
(1)
i.e. kT S H R T lnA
β ββ = β + β ββ β
and 1H k kS R ln H T R lnT A A
ββ β β β ββ = + = β +β β β ββ β β β
Hence, ( )2( S) H T 0T
ββ β= β β +
β = 2
HTβ
β
(d) From equation (1), k kH T S R T ln T S R lnA A
β§ β«β β β ββ = β β = β ββ¨ β¬β β β ββ β β β β© β
Hence, ( H)T
β ββ
= kS R lnA
β ββ β β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 309
EXERCISE 140 Page 347
2. If z 2 ln xy= find (a) 2
2
zxββ
(b) 2
2
zyββ
(c) 2z
x yββ β
(d) 2z
y xββ β
(a) 1z 2 2(y) 2xx xy x
ββ ββ= = =β ββ β β
and 2
2
zxββ
= ( )1 22x 2xx
β ββ= β
β= 2
2x
β
(b) 1z 2 2(x) 2yy xy y
ββ ββ= = =β ββ β β
and 2
2
zyββ
= ( )1 22y 2yy
β ββ= β
β= 2
2y
β
(c) ( )2
1z 2yx y x
ββ β=
β β β = 0
(d) ( )2
1z 2xy x y
ββ β=
β β β = 0
3. If ( )( )x y
zx yβ
=+
find (a) 2
2
zxββ
(b) 2
2
zyββ
(c) 2z
x yββ β
(d) 2z
y xββ β
(a) 22 2 2
z (x y)(1) (x y)(1) x y x y 2y 2y(x y)x (x y) (x y) (x y)
ββ + β β + β += = = = +
β + + +
2
32
z 4y(x y) (1)x
ββ= β +
β = 3
4y(x 3)
β+
(b) 22 2 2
z (x y)( 1) (x y)(1) x y x y 2x 2x(x y)y (x y) (x y) (x y)
ββ + β β β β β β + β= = = = β +
β + + +
2
32
z ( 2x) 2(x y) (1)y
ββ β‘ β€= β β +β£ β¦β = 3
4x(x 3)+
(c) 2
2 3 23 2
z 4x 22x(x y) ( 2x) 2(x y) (x y) ( 2)x y x (x y) (x y)
β β ββ β β‘ β€ β‘ β€= β + = β β + + + β = ββ£ β¦ β£ β¦β β β + +
= 3 3
4x 2(x y) 2x 2y(x y) (x y)β + β
=+ +
= 3
2(x y)(x y)
β+
(d) 2
2 3 23 2
z 4y 22y(x y) (2y) 2(x y) (x y) (2)y x y (x y) (x y)
β β ββ β ββ‘ β€ β‘ β€= + = β + + + = +β£ β¦ β£ β¦β β β + +
= 3 3
4y 2(x y) 4y 2x 2y(x y) (x y)
β + + β + +=
+ +
= 3
2x 2y(x y)
β+
= 3
2(x y)(x y)
β+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 310
5. If 2z x sin(x 2y)= β find (a) 2
2
zxββ
(b) 2
2
zyββ
Show also that 2z
x yββ β
= 2z
y xββ β
= 22x sin(x 2y) 4x cos(x 2y)β β β
(a) ( ) [ ]2 2z x cos(x 2y) sin(x 2y) (2x) x cos(x 2y) 2x sin(x 2y)xβ
= β + β = β + ββ
( )[ ] [ ] [ ]2
22
z x sin(x 2y) cos(x 2y) (2x) (2x)cos(x 2y) sin(x 2y) (2)xβ
= β β + β + β + ββ
= 2x sin(x 2y) 2x cos(x 2y) 2x cos(x 2y) 2sin(x 2y)β β + β + β + β
= ( )22 x sin(x 2y) 4xcos(x 2y)β β + β
(b) [ ]2 2z x cos(x 2y) ( 2) 2x cos(x 2y)yβ
= β β = β ββ
2
2
zyββ
= ( )[ ]22x sin(x 2y) ( 2)β β β β
= 24x sin(x 2y)β β
( )[ ] [ ]2
2 2z 2x cos(x 2y) 2x sin(x 2y) cos(x 2y) ( 4x)x y xβ β β‘ β€= β β = β β β + β ββ£ β¦β β β
= 22x sin(x 2y) 4xcos(x 2y)β β β
( )[ ] [ ]2
2 2z x cos(x 2y) 2x sin(x 2y) x sin(x 2y) ( 2) (2x) cos(x 2y) ( 2)y x yβ β β‘ β€= β + β = β β β + β ββ£ β¦β β β
= 22x sin(x 2y) 4xcos(x 2y)β β β
7. Given 3xzy
β β= β β
β β show that
2zx yββ β
= 2z
y xββ β
and evaluate 2
2
zxββ
when 1x2
= and y = 3
1 12 23xz 3x y
yββ β
= =β ββ β
1 1 1 12 2 2 2z 1 33 y x y x
x 2 2β β β ββ ββ
= =β ββ β β
1 3 1 32 2 2 2z 1 33 x y x y
y 2 2β ββ ββ ββ
= β = ββ ββ ββ β β β β
1 3 3 122 2 2 2
3 12 2
z 3 3 1 3x y y xx y x 2 2 2 4y x
β β ββ β β ββ ββ β β= β = β =β β β ββ ββ β β ββ β β β β β β β β
= ( )3
3
4 xy
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 311
1 1 1 322 2 2 2
1 32 2
z 3 3 1 3y x x yy x y 2 2 2 4x y
β β β ββ β β ββ ββ β β= = β =β β β ββ ββ β β ββ β β β β β β β β
= ( )3
3
4 xy
β
Thus, 2z
x yββ β
= 2z
y xββ β
( )1 1 1 322 2 2 2
2 3
z 3 3 1 3y x y xx x 2 2 2 4 yx
β β β ββ β β ββ ββ β β= = β =β β β ββ ββ β β ββ β β β β β β β
When 1x2
= and y = 3,
( )( )2
2 3
3
2 4z 3 3 1 8 2 4 2(2) 2x 4 4 4 4 21 11 4 3 44 (3) 2 82
ββ β β β β β β= = = = = = = = β
β β β β β β ββ β β β β ββ ββ ββ β β β β β β β β β
= 1 or 0.70712
β β
8. An equation used in thermodynamics is the Benedict-Webb-Rubine equation of state for the
expansion of a gas. The equation is:
22
0 V0 0 2 2 3 6 2 3
C 1CRT 1 1 A 1Vp B RT A (bRT a) e
V T V V V T V
Ξ³β
Ξ³β β+β βΞ±β β β ββ β = + β β + β + + β ββ ββ β β β
Show that 22
V02 2 4 2
p 6 C 1 e CT V T V V
Ξ³ββ§ β«β Ξ³βͺ βͺβ β= + ββ¨ β¬β ββ β β βͺ βͺβ© β
( ) 22 2 V0 0 0 2 3 6 2 3
RT 1 1 A 1p B RT A C T (bRT a) T C 1 eV V V V V V
Ξ³β
β βΞ± Ξ³β ββ β= + β β + β + + +β ββ ββ β β β
23
30 0 V2 2 3 2 3
B R 2C Tp R bR 12T C 1 eT V V V V V V
Ξ³β βββ Ξ³β ββ β= + + β β +β ββ ββ β β β β
242
40 V2 2 2 3
6C Tp 16T C 1 eT V V V
Ξ³β ββββ Ξ³β ββ β= + +β ββ ββ β β β β
= 2V02 4 2
6 C 1 e CV T V V
Ξ³ββ§ β«Ξ³βͺ βͺβ β+ ββ¨ β¬β β
β β βͺ βͺβ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 312
CHAPTER 35 TOTAL DIFFERENTIAL, RATES OF CHANGE
AND SMALL CHANGES EXERCISE 141 Page 350 2. Find the total differential, dz, given z = 2xy β cos x
Since z = 2xy β cos x then dz = z zdx dyx yβ β
+β β
= (2y + sin x)dx + 2x dy
3. Find the total differential, dz given z = x yx yβ+
Since z = x yx yβ+
then 2 2
z (x y)(1) (x y)(1) 2yx (x y) (x y)β + β β
= =β + +
and 2 2
z (x y)( 1) (x y)(1) 2xy (x y) (x y)β + β β β β
= =β + +
Thus, dz = z zdx dyx yβ β
+β β
= 2 2
2y 2xdx dy(x y) (x y)
β β β βββ β β β+ +β β β β
5. Find the total differential, dz, given z = xy + xy
- 4
Since z = xy + xy
- 4 then
121 xz 12y y
x y 2y x
β
β= + = +
β
and ( )22
z xx x y xy y
ββ= β = β
β
Thus, dz = z zdx dyx yβ β
+β β
= 2
1 xy dx x dyy2y x
β ββ β+ + ββ ββ β β ββ β β β
7. Given u = ln sin(xy) show that du = cot(xy)(y dx + x dy) Since u = ln sin(xy) then
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 313
du = u udx dyx yβ β
+β β
= 1 1y cos xy dx x cos xy dysin(xy) sin(xy)
β‘ β€ β‘ β€β β β β+β’ β₯ β’ β₯β β β β
β β β β β£ β¦ β£ β¦
= y cot(xy) dx + x cot(xy) dy
i.e. du = cot(xy)[ y dx + x dy ]
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 314
EXERCISE 142 Page 352 1. The radius of a right cylinder is increasing at a rate of 8mm/s and the height is decreasing at a
rate of 15 mm/s. Find the rate at which the volume is changing in cm3/s when the radius is 40 mm
and the height is 150 mm.
Volume of cylinder, V = 2r hΟ
Rate at which volume is changing, dV V dr V dhdt r dt h dt
β β= +β β
= ( ) ( )2dr dh2 rh rdt dt
Ο + Ο
= [ ] 22 (40)(150) (8) (40) ( 15)β‘ β€Ο + Ο ββ£ β¦
= 96 000Ο - 24 000Ο
= 72 000Ο 3mm / s
= 72Ο 3cm / s = 226.2 3cm / s
3. Find the rate of change of k, correct to 4 significant figures, given the following data:
k = f(a, b, c); k = 2b ln a + c2 ea; a is increasing at 2 cm/s; b is decreasing at 3 cm/s; c is
decreasing at 1 cm/s; a = 1.5 cm, b = 6 cm and c = 8 cm.
Since k = 2b ln a + c2 ea
then rate of change of k, dk k da k db k dcdt a dt b dt c dt
β β β= + +β β β
= ( ) ( )2 a a2b da db dcc e 2ln a 2cea dt dt dt
β β+ + +β ββ β
= ( ) ( )2 1.5 1.5(2)(6) 8 e (2) 2 ln1.5 ( 3) 2(8)e ( 1)1.5
β β+ + β + ββ ββ β
= 589.656 β 2.433 β 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 315
Total surface area of a cone, A = Οrl + 2rΟ = ( )2 2 2r r h rΟ + +Ο (see diagram of cone below)
Rate of change of surface area,
dA A dr A dhdt r dt h dt
β β= +β β
= ( ) ( ) ( ) ( ) ( )1 1
2 2 2 2 2 22 21 dr 1 dhr r h (2r) r h ( ) 2 r r r h (2h)2 dt 2 dt
β ββ§ β« β§ β«β‘ β€ β ββͺ βͺ βͺ βͺΟ + + + Ο + Ο + Ο +β¨ β¬ β¨ β¬β ββ’ β₯βͺ βͺ βͺ βͺβ£ β¦ β β β© β β© β
= ( )
( )( )
22 2
2 2 2 2
r dr rh dhr h 2 rdt dtr h r h
β§ β« β§ β«Ο Οβͺ βͺ βͺ βͺ+ Ο + + Ο +β¨ β¬ β¨ β¬+ +βͺ βͺ βͺ βͺβ© β β© β
= ( )
( )( )
22 2
2 2 2 2
(5) (5)(12)5 12 2 (5) (0.5) ( 1.5)5 12 5 12
β§ β« β§ β«Ο Οβͺ βͺ βͺ βͺ+ Ο + + Ο + ββ¨ β¬ β¨ β¬+ +βͺ βͺ βͺ βͺβ© β β© β
in centimetre units
= 25 6013 10 (0.5) ( 1.5)13 13Ο Οβ β β β+ Ο+ Ο + ββ β β β
β β β β
= (78.298)(0.5) + (14.50)(-1.5)
= 39.149 β 21.75
= 17.4 2cm / s
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 316
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 β3% and the error
in measuring time t is +1%
Since H = i2Rt then H H HH i R ti R t
β β βΞ΄ β Ξ΄ + Ξ΄ + Ξ΄
β β β
( ) ( ) ( )2 22iRt (0.02i) i t ( 0.03R) i R (0.01t)β + β +
2 2 20.04i Rt 0.03i Rt 0.01i Rtβ β +
β (0.04 β 0.03 + 0.01) H
β 0.02 H
i.e. the error in H is + 2%
3. fr = 12 LCΟ
represents the resonant frequency of a series connected circuit containing
inductance L and capacitance C. Determine the approximate percentage change in fr when L is
decreased by 3% and C is increased by 5%
Since fr = 1 12 21 1 L C
22 LCβ β
=ΟΟ
then r rr
f ff L CL Cβ β
Ξ΄ β Ξ΄ + Ξ΄β β
3 1 3 12 2 2 21 1L C C L
2 2L C2 2
β β β ββ β β ββ ββ β β β
β Ξ΄ + Ξ΄β β β βΟ Οβ β β ββ β β β
β β β β
3 1 3 12 2 2 2L C C L( 0.03L) (0.05C)
4 4
β β β ββ β β βββ β β ββ β ββ β β βΟ Οβ β β β
β β β β
1 1 1 12 2 2 20.03L C 0.05C L
4 4
β β β β
β βΟ Ο
1 10.015 0.0252 LC 2 LC
β β β ββ ββ β β βΟ Οβ β β β
i.e. r r rf (0.015 0.025)f 0.01fΞ΄ β β β β
i.e. the approximate percentage change in fr is β 1%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 317
4. The second moment of area of a rectangle about its centroid parallel to side b is given by
I =3bd
12 . If b and d are measured as 15 cm and 6 cm respectively, and the measurement errors
are +12 mm in b and β1.5 mm in d, find the error in the calculated value of I.
Since I =3bd
12 then I II b d
b dβ β
Ξ΄ β Ξ΄ + Ξ΄β β
3 2d 3bdb d
12 12β β β β
β β + ββ β β ββ β β β
3 26 3(15)(6)(1.2) ( 0.15)12 12β β β β
β + ββ β β ββ β β β
β 21.6 β 20.25 = 1.35
i.e. error in I = + 1.35 4cm
5. The side b of a triangle is calculated using b2 = a2 + c2 β 2ac cos B. If a, c and B are measured as
3 cm, 4 cm and Ο/4 radians respectively and the measurement errors which occur are +0.8 cm,
- 0.5 cm and +Ο/9 radians respectively, determine the error in the calculated value of b.
b2 = a2 + c2 β 2ac cos B from which, ( )1
2 2 2b a c 2accos B= + β
Approximate error in b, b b bb a c Ba c Bβ β β
Ξ΄ β Ξ΄ + Ξ΄ + Ξ΄β β β
( )1
2 2 21 a c 2accos B (2a 2ccos B a2
ββ‘ β€β + β β ββ’ β₯β£ β¦
+ ( )1
2 2 21 a c 2accos B (2c 2a cos B c2
ββ‘ β€+ β β ββ’ β₯
β£ β¦
+ ( )1
2 2 21 a c 2accos B (2acsin B B2
ββ‘ β€+ β ββ’ β₯
β£ β¦
122 21 3 4 2(3)(4)cos (6 8cos (0.8)
2 4 4
ββ‘ β€Ο Οβ ββ’ β₯β + β ββ ββ’ β₯β β β£ β¦
+ 122 21 3 4 2(3)(4)cos (8 6cos ( 0.5)
2 4 4
ββ‘ β€Ο Οβ ββ’ β₯+ β β ββ ββ’ β₯β β β£ β¦
+ 122 21 3 4 2(3)(4)cos (24sin
2 4 4 90
ββ‘ β€Ο Ο Οβ β β ββ’ β₯+ ββ β β ββ’ β₯β β β β β£ β¦
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 318
β (0.17645)(0.343146)(0.8) + (0.17645)(3.75736)(-0.5)
+ (0.17645(16.97056)90Οβ β
β ββ β
i.e. approximate error in b β 0.04844 β 0.3315 + 0.10453 = - 0.179 cm
7. The rate of flow of gas in a pipe is given by: v =6 5
C dT
, where C is a constant, d is the diameter
of the pipe and T is the thermodynamic temperature of the gas. When determining the rate of
flow experimentally, d is measured and subsequently found to be in error by +1.4%, and T has
an error of β1.8%. Determine the percentage error in the rate of flow based on the measured
values of d and T.
Flow rate, v =5162
6 5
C d Cd TT
β=
Approximate error in flow rate, v vv d Td Tβ β
Ξ΄ β Ξ΄ + Ξ΄β β
12 11
66 5
1(C) d 52 (0.014d) C d T ( 0.018T)6T
β
β
β ββ β β ββ β
β + β ββ β β ββ ββ ββ β β β β β β ββ β
6 65 5
C d C d 5(0.007) (0.018)6T T
β β β ββ ββ +β β β ββ ββ β β ββ β β β β β
β (0.007 + 0.015) v
β 0.022 v
i.e. the percentage error in the rate of flow is + 2.2%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 319
CHAPTER 36 MAXIMA, MINIMA AND SADDLE POINTS FOR
FUNCTIONS OF TWO VARIABLES EXERCISE 144 Page 359 2. Find the maxima, minima and saddle points for the following functions:
(a) f(x, y) = 2 2x y 2x 4y 8+ β + +
(b) f(x, y) = 2 2x y 2x 4y 8β β + +
(c) f(x, y) = 2 22x 2y 2xy 2x y 4+ β β β +
(a) Let f(x, y) = z = 2 2x y 2x 4y 8+ β + +
(i) z 2x 2xβ
= ββ
and z 2y 4yβ
= +β
(ii) For stationary points, 2x - 2 = 0 from which, x = 1
and 2y + 4 = 0 from which, y = -2
(iii) The co-ordinates of the stationary point is (1, -2)
(iv) 2
2
z 2xβ
=β
2
2
z 2yβ
=β
( )2z 2y 4 0
x y xβ β
= + =β β β
(v) When x = 1, y = -2, 2
2
z 2xβ
=β
2
2
z 2yβ
=β
and 2z 0
x yβ
=β β
(vi) 22z 0
x yβ ββ
=β ββ ββ β
(vii) 22 2 2
(1, 2) 2 2
z z zx y x yβ
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )20 2 2 4= β = β which is negative
(viii) Since β < 0 and 2
2
zxββ
> 0 then (1, -2) is a minimum point.
(b) Let f(x, y) = z = 2 2x y 2x 4y 8β β + +
(i) z 2x 2xβ
= ββ
and z 2y 4yβ
= β +β
(ii) For stationary points, 2x - 2 = 0 from which, x = 1
and - 2y + 4 = 0 from which, y = 2
(iii) The co-ordinates of the stationary point is (1, 2)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 320
(iv) 2
2
z 2xβ
=β
2
2
z 2yβ
= ββ
( )2z 2y 4 0
x y xβ β
= β + =β β β
(v) When x = 1, y = 2, 2
2
z 2xβ
=β
2
2
z 2yβ
= ββ
and 2z 0
x yβ
=β β
(vi) 22z 0
x yβ ββ
=β ββ ββ β
(vii) 22 2 2
(1,2) 2 2
z z zx y x y
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )20 2 2 4= β β = which is positive
(viii) Since β > 0 then (1, 2) is a saddle point.
(c) Let f(x, y) = z = 2 22x 2y 2xy 2x y 4+ β β β +
(i) z 2 2y 4xxβ
= β ββ
and z 2 2x 2yyβ
= β ββ
(ii) For stationary points, 2 β 2y β 4x = 0
i.e. 1 β y β 2x = 0 (1)
and 2 β 2x β 2y = 0
i.e. 1 β x β y = 0 (2)
From (1), y = 1 β 2x
Substituting in (2) gives: 1 β x β(1 β 2x) = 0
i.e. 1 β x β 1 + 2x = 0 from which, x = 0
When x = 0 in equations (1) and (2), y = 1
(iii) The co-ordinates of the stationary point is (0, 1)
(iv) 2
2
z 4xβ
= ββ
2
2
z 2yβ
= ββ
( )2z 2 2x 2y 2
x y xβ β
= β β = ββ β β
(v) When x = 0, y = 1, 2
2
z 4xβ
= ββ
2
2
z 2yβ
= ββ
and 2z 2
x yβ
= ββ β
(vi) 22
2z ( 2) 4x y
β ββ= β =β ββ ββ β
(vii) 22 2 2
(0,1) 2 2
z z zx y x y
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( )( )4 4 2 4= β β β = β which is negative
(viii) Since β < 0 and 2
2
zxββ
< 0 then (0, 1) is a maximum point.
4. Locate the stationary points of the function z = 2 212x 6xy 15y+ +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 321
(i) z 24x 6yxβ
= +β
and z 6x 30yyβ
= +β
(ii) For stationary points, 24x + 6y = 0 (1)
and 6x + 30y = 0 (2)
(iii) From (1), 6y = -24x i.e. y = -4x
Substituting in (2) gives: 6x + 30(-4x) = 0
i.e. 6x = 120x i.e. x = 0
When x = 0, y = 0, hence the co-ordinates of the stationary point is (0, 0)
(iv) 2
2
z 24xβ
=β
2
2
z 30yβ
=β
( )2z 6x 30y 6
x y xβ β
= + =β β β
(v) When x = 0, y = 0, 2
2
z 24xβ
=β
2
2
z 30yβ
=β
and 2z 6
x yβ
=β β
(vi) 22z 36
x yβ ββ
=β ββ ββ β
(vii) 22 2 2
(0,0) 2 2
z z zx y x y
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )26 24 30= β which is negative
(viii) Since β < 0 and 2
2
zxββ
> 0 then (0, 0) is a minimum point
5. Find the stationary points of the surface z = 3 3x xy yβ + and distinguish between them.
(i) 2z 3x yxβ
= ββ
and 2z x 3yyβ
= β +β
(ii) For stationary points, 23x - y = 0 (1)
and 2x 3yβ + = 0 (2)
(iii) From (1), y = 23x
Substituting in (2) gives: - x + ( )223 3x = 0
- x + 427x = 0
and ( )3x 27x 1 0β =
i.e. x = 0 or 327x 1 0β = i.e. 3 1x27
= and 31 1x27 3
β β= =β ββ β
Hence, x = 0 or 1x3
=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 322
From (1), when x = 0, y = 0
and when 1x3
= , y = 23x = 21 13
3 3β β =β ββ β
Hence, the stationary points occur at (0, 0) and 1 1,3 3
β ββ ββ β
(iv) 2
2
z 6xxβ
=β
2
2
z 6yyβ
=β
( )2
2z x 3y 1x y xβ β
= β + = ββ β β
(v) For (0, 0), 2
2
z 0xβ
=β
2
2
z 0yβ
=β
and 2z 1
x yβ
= ββ β
For 1 1,3 3
β ββ ββ β
, 2
2
z 2xβ
=β
2
2
z 2yβ
=β
and 2z 1
x yβ
= ββ β
(vi) For (0, 0), 22z 1
x yβ ββ
=β ββ ββ β
For 1 1,3 3
β ββ ββ β
, 22z 1
x yβ ββ
=β ββ ββ β
(vii) ( )( )22 2 2
(0,0) 2 2
z z z 1 0 0 1x y x y
β β β ββ ββ β ββ = β = β =β β β ββ ββ β β ββ β β β β β
which is positive
( )( )1 1,3 3
1 2 2 3β ββ ββ β
β = β = β which is negative
(viii) Since ( )0,0β > 0 then (0, 0) is a saddle point
1 1,3 3
β ββ ββ β
β < 0 and 2
2
zxββ
> 0 then 1 1,3 3
β ββ ββ β
is a minimum point
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 323
EXERCISE 145 Page 363 1. The function z = 2 2x y xy 4x 4y 3+ + + β + has one stationary value. Determine its co-ordinates and its nature.
(i) z 2x y 4xβ
= + +β
and z 2y x 4yβ
= + ββ
(ii) For stationary points, 2x + y + 4 = 0 (1)
and 2y + x β 4 = 0 (2)
(iii) (1) + (2) gives: 3x + 3y = 0 from which, y = -x
Substituting in (1), 2x β x + 4 = 0 i.e. x = -4, thus y = +4
Hence, the stationary point occurs at (-4, 4)
(iv) 2
2
z 2xβ
=β
2
2
z 2yβ
=β
( )2z 2y x 4 1
x y xβ β
= + β =β β β
(v) When x = -4, y = 4, 2
2
z 2xβ
=β
2
2
z 2yβ
=β
and 2z 1
x yβ
=β β
(vi) 22z 1
x yβ ββ
=β ββ ββ β
(vii) 22 2 2
( 4,4) 2 2
z z zx y x yβ
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )21 2 2 3= β = β which is negative
(viii) Since β < 0 and 2
2
zxββ
> 0 then (-4, 4) is a minimum point
3. Determine the stationary values of the function f(x, y) = 4 2 2 2 2x 4x y 2x 2y 1+ β + β and
distinguish between them.
Let f(x, y) = z = 4 2 2 2 2x 4x y 2x 2y 1+ β + β
(i) 3 2z 4x 8xy 4xxβ
= + ββ
and 2z 8x y 4yyβ
= +β
(ii) For stationary points, 3 24x 8xy 4x 0+ β = (1)
and 28x y 4y 0+ = (2)
(iii) From (2), ( )24y 2x 1 0β = from which, y = 0
From (1), if y = 0, 34x 4x 0β = i.e. ( )24x x 1 0β =
from which, x = 0 or x = Β± 1
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 324
Hence, the stationary points occur at (0, 0) and (1, 0) and (-1, 0)
(iv) 2
2 22
z 12x 8y 4xβ
= + ββ
2
22
z 8x 4yβ
= +β
( )2
2z 8x y 4y 16xyx y xβ β
= + =β β β
(v) For (0, 0), 2
2
z 4xβ
= ββ
2
2
z 4yβ
=β
and 2z 0
x yβ
=β β
For (1, 0), 2
2
z 8xβ
=β
2
2
z 12yβ
=β
and 2z 0
x yβ
=β β
For (-1, 0), 2
2
z 8xβ
=β
2
2
z 12yβ
=β
and 2z 0
x yβ
=β β
(vi) For all three stationary points, 22z 0
x yβ ββ
=β ββ ββ β
(vii) 22 2 2
(0,0) 2 2
z z zx y x y
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )20 4 4 32= β β = which is positive
22 2 2
(1,0) 2 2
z z zx y x y
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )20 8 12 96= β = β which is negative
22 2 2
( 1,0) 2 2
z z zx y x yβ
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )20 8 12 96= β = β which is negative
(viii) Since (0,0)β > 0, then (0, 0) is a saddle point.
Since (1,0)β < 0 and 2
2
zxββ
> 0 then (1, 0) is a minimum point.
Since ( 1,0)ββ < 0 and 2
2
zxββ
> 0 then (-1, 0) is a minimum point.
4. Determine the stationary points of the surface f(x, y) = 3 2 2x 6x yβ β Let f(x, y) = z = 3 2 2x 6x yβ β
(i) 2z 3x 12xxβ
= ββ
and z 2yyβ
= ββ
(ii) For stationary points, 23x 12xβ = 0 (1)
and -2y = 0 (2)
(iii) From (1) 3x(x 4)β = 0 from which, x = 0 or x = 4
From (2), y = 0
Hence, the stationary points occurs at (0, 0) and (4, 0)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 325
(iv) 2
2
z 6x 12xβ
= ββ
2
2
z 2yβ
= ββ
( )2z 2y 0
x y xβ β
= β =β β β
(v) At (0, 0), 2
2
z 12xβ
= ββ
2
2
z 2yβ
= ββ
and 2z 0
x yβ
=β β
At (4, 0), 2
2
z 12xβ
=β
2
2
z 2yβ
= ββ
and 2z 0
x yβ
=β β
(vi) For both points, 22z 0
x yβ ββ
=β ββ ββ β
(vii) 22 2 2
(0,0) 2 2
z z zx y x y
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )20 12 2 24= β β β = β which is negative
(4,0)β ( ) ( )( )20 12 2 24= β β = which is positive
(viii) Since (0,0)β < 0 and 2
2
zxββ
< 0 then (0, 0) is a maximum point.
Since (4,0)β >0 then (4, 0) is a saddle point. 6. A large marquee is to be made in the form of a rectangular box-like shape with canvas covering
on the top, back and sides. Determine the minimum surface area of canvas necessary if the
volume of the marquee is to be 250 m3.
A sketch of the marquee is shown above.
Volume of marquee, V = xyz = 250 (1)
Surface area, S = xy + yz + 2xz (2)
From (1), z = 250xy
Substituting in (2) gives: S = 1 1250 250 250 500xy y 2x xy xy 250x 500yxy xy x y
β ββ β β β+ + = + + = + +β β β β
β β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 326
2
S 250yx xβ
= ββ
and 2
S 500xy yβ
= ββ
For a stationary point, 2
S 250y 0x xβ
= β =β
from which, 2
250yx
= or 2yx 250= (3)
and 2
S 500x 0y yβ
= β =β
from which, 2
500xy
= or 2xy 500= (4)
Dividing equation (3) by equation (40 gives:
2
2
yx 250xy 500
= i.e. x 1y 2= and y = 2x
Substituting y = 2x in equation (3) gives: 32x 250= and x = 3 125 = 5 m
and y = 2x = 10 m
From equation (1), xyz = 250 i.e. (5)(10)z = 250 from which, z = 5 m
2
2 3
S 750x xβ
=β
2
2 3
S 1000y yβ
=β
and 2S 1
x yβ
=β β
When x = 5 and y = 10, 2
2
S 6xβ
=β
2
2
S 1yβ
=β
and 2S 1
x yβ
=β β
22 2 2
2 2
S S Sx y x y
β β β ββ ββ β ββ = ββ β β ββ ββ β β ββ β β β β β
( ) ( )( )21 6 1 6= β = β which is negative
Since β < 0 and 2
2
zxββ
> 0 then the surface area is a minimum.
Minimum surface area, S = xy + yz + 2xz
= (5)(10) + (10)(5) + (2)(5)(5)
= 50 + 50 + 50 = 150 2m
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 327
CHAPTER 37 STANDARD INTEGRATION
EXERCISE 146 Page 370
3. Determine (a) 23x 5x dxx
β βββ ββ β β« (b) 2(2 ) d+ ΞΈ ΞΈβ«
(a) ( )2 23x 5x 3x 5xdx dx 3x 5 dxx x x
β β β ββ= β = ββ β β β
β β β β β« β« β« =
23x 5x c2
β +
(b) ( )2 2(2 ) d 4 4 d+ ΞΈ ΞΈ = + ΞΈ+ ΞΈ ΞΈβ« β« = 3
24 2 c3ΞΈ
ΞΈ + ΞΈ + +
4.(b) Determine 4
3 dx4xβ«
4 1 3
4 34
3 3 3 x 3 x 1dx x dx c c x c4x 4 4 4 1 4 3 4
β + ββ ββ β β ββ β β β= = + = + = β +β β β ββ β β ββ + ββ β β β β β β β
β« β« = 3
1 c4x
β +
5.(b) Determine 4 51 x dx4β«
5 915 94 4
4 5 4 41 1 1 x 1 x 1 4x dx x dx c c x c5 94 4 4 4 4 914 4
+β β β ββ β β β β ββ β β β= = + = + = +β β β β β ββ β β β
β β β β β β β β β β+β β β β
β« β« = 4 91 x c9
+
6.(b) Determine 5 4
3 dx7 xβ«
4 114 15 55 5
45 45
3 3 1 3 3 x 3 x 3 5dx dx x dx c c x c4 17 7 7 7 7 17 x 1x5 5
β +β β ββ β= = = + = + = +β ββ ββ β β β β β β β β +β β β β
β β β β
β« β« β« = 515 x c7
+
8. Determine (a) 23 sec 3x dx4β« (b) 22cosec 4 dΞΈ ΞΈβ«
(a) 23 3 1sec 3x dx tan 3x c4 4 3
β ββ β= +β ββ ββ β β β β« = 1 tan 3x c
4+
(b) ( )2 12cosec 4 d 2 cot 4 c4
β βΞΈ ΞΈ = β ΞΈ +β ββ β β« = 1 cot 4 c
2β ΞΈ +
9. Determine (a) 5 cot 2t cosec2t dtβ« (b) 4 sec 4t tan 4t dt3β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 328
(a) ( ) 15 cot 2t cosec2t dt 5 cosec2t c2
β β= β +β ββ β β« = 5 cosec2t c
2β +
(b) 4 4 1sec 4t tan 4t dt sec 4t c3 3 4
β ββ β= +β ββ ββ β β β β« = 1 sec4t c
3+
10.(b) Determine 5x
2 dx3 eβ«
5x 5x 5x
5x
2 dx 2 2 1 2e dx e c e c3 e 3 3 5 15
β β ββ ββ β= = + = β +β ββ βββ β β β β« β« = 5x
2 c15eβ
+
11.(b) Determine 2u 1 duu
β βββ ββ β β«
2 2u 1 u 1 1du du u duu u u u
β β β ββ β β= β = ββ β β β β ββ β β β β β
β« β« β« = 2u lnu c
2β +
12. Determine (a) 2(2 3x) dx
x+
β« (b) 21 2t dt
tβ β+β ββ β β«
(a) 1 1 32 2 22 2 2
1 1 12 2 2
(2 3x) 4 12x 9x 4 12x 9xdx dx dx 4x 12x 9x dxx x x x x
ββ β β β+ + + β β= = + + = + +β ββ β β β β β β« β« β« β«
=
1 3 52 2 24x 12x 9x c1 3 5
2 2 2
+ + + = 3 5188 x 8 x x c5
+ + +
(b) ( )2
2 2 22
1 1 1 12t dt 2t 2t dt 4 4t dt t 4 4t dtt t t t
ββ β β ββ β β β+ = + + = + + = + +β β β ββ β β ββ β β β β β β β β« β« β« β«
= 1 3t 4t4t c1 3
β
+ + +β
= 31 4t4t c
t 3β + + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 329
EXERCISE 147 Page 372
2. Evaluate (a) ( )2 2
13 x dx
βββ« (b) ( )3 2
1x 4x 3 dxβ +β«
(a) ( ) ( )2 33 32 2
11
1x 2 8 13 x dx 3x 3(2) 3( 1) 6 33 3 3 3 3β
β
β‘ β€ββ‘ β€ β‘ β€ β β β ββ = β = β β β β = β β β β ββ’ β₯ β β β ββ’ β₯ β’ β₯ β β β β β’ β₯β£ β¦ β£ β¦ β£ β¦β«
= 1 23 23 3
β ββ ββ ββ β
= 6
(b) ( ) ( )33 23 2
11
x 4x 1x 4x 3 dx 3x 9 18 9 2 33 2 3
β‘ β€ β ββ + = β + = β + β β +β ββ’ β₯ β β β£ β¦β« = ( ) 10 1
3β ββ β ββ β
= 113
β
4. Evaluate (a) /3
/ 62sin 2 d
Ο
ΟΞΈ ΞΈβ« (b)
2
03sin t dtβ«
(a) [ ]/3 /3
/ 6/ 6
2 2 22sin 2 d cos 2 cos cos2 3 6
Ο Ο
ΟΟ
Ο Οβ‘ β€ΞΈ ΞΈ = β ΞΈ = β ββ’ β₯β£ β¦β« (note that 2 2and3 6Ο Ο are in radians)
= -[-0.5 β 0.5] = -[-1] = 1
(b) [ ]2 2
003sin t dt 3 cos t 3[cos 2 cos0]= β = β ββ« (note that 2 is 2 radians)
= -3[-0.41615 β 1] = 4.248
6. Evaluate, correct to 4 significant figures, (a) 2 2
1cosec 4tdtβ« (b) ( )
/ 2
/ 43sin 2x 2cos3x dx
Ο
Οββ«
(a) [ ] [ ]2 22
11
1 1 1 1 1cosec 4tdt cot 4t cot 8 cot 44 4 4 tan8 tan 4
β‘ β€= β = β β = β ββ’ β₯β£ β¦β«
= ( )1 0.147065 0.8636914
β β β = 0.2527
(b) ( )/ 2
/ 2
/ 4/ 4
3 23sin 2x 2cos3x dx cos 2x sin 3x2 3
ΟΟ
ΟΟ
β‘ β€β = β ββ’ β₯β£ β¦β«
= 3 2 2 3 3 2 2 3cos sin cos sin2 2 3 2 2 4 3 4
Ο Ο Ο Οβ β β ββ β β β ββ β β ββ β β β
= 3 2 20 (0.707107)2 3 3
β β β β+ β ββ β β ββ β β β
= 2.638
8. Evaluate, correct to 4 significant figures, (a) 3
2
2 dx3xβ« (b)
23
1
2x 1dxx+
β«
(a) [ ] ( )3 3 3
22 2
2 2 1 2 2dx dx ln x ln 3 ln 23x 3 x 3 3
= = = ββ« β« = 0.2703
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 330
(b) 2 23 3 3 32
11 1 1
2x 1 2x 1 1dx dx 2x dx x ln xx x x x
β β+ β β β‘ β€= + = + = +β β β β β£ β¦β β β β β« β« β« = (9 + ln 3) β (1 + ln 1)
= 9 + ln 3 β 1 = 8 + ln 3 = 9.099
9. The entropy change βS, for an ideal gas is given by: βS = 2 2
1 1
T V
vT V
dT dVC RT V
ββ« β«
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 Γ 10-3 T + 8 Γ 10-6 T2
βS = ( )400 33 6 2
100 1
dT dV45 6 10 T 8 10 T 8.314T V
β β+ Γ + Γ ββ« β«
= 400 33 6
100 1
45 dV6 10 8 10 T dT 8.314T V
β ββ β+ Γ + Γ ββ ββ β β« β«
= [ ]4006 2
331
100
8 10 T45ln T 6 10 T 8.314 ln V2
βββ‘ β€Γ
+ Γ + ββ’ β₯β£ β¦
= 6 2 6 2
3 38 10 (400) 8 10 (100)45ln 400 6 10 (400) 45ln100 6 10 (100)2 2
β ββ ββ‘ β€β β β βΓ Γ
+ Γ + β + Γ +β’ β₯β β β ββ β β β β£ β¦
- 8.314[ln3 β ln 1]
= (269.616 + 2.4 + 0.64) β (207.233 + 0.6 + 0.04) β 9.134
= 272.656 β 207.873 β 9.134
= 55.65
10. The p.d. between boundaries a and b of an electric field is given by: V = b
a0 r
Q dr2 rΟ Ξ΅ Ξ΅β«
If a = 10, b = 20, Q = 2 Γ 10-6 coulombs, Ξ΅0 = 8.85 Γ 10-12 and Ξ΅r = 2.77, show that V = 9 kV.
V = b
a0 r
Q dr2 rΟ Ξ΅ Ξ΅β« = [ ]
b b
aa0 r 0 r 0 r
Q 1 Q Qdr ln r (ln b ln a)2 r 2 2
= = βΟΞ΅ Ξ΅ ΟΞ΅ Ξ΅ ΟΞ΅ Ξ΅β«
= ( )( )
6
12
2 10 (ln 20 ln10)2 8.85 10 2.77
β
β
Γβ
Ο Γ = 9000 V or 9 kV
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 331
11. The average value of a complex voltage waveform is given by:
( )AV 0
1V 10sin t 3sin 3 t 2sin 5 t d( t)Ο
= Ο + Ο + Ο ΟΟ β«
Evaluate VAV correct to 2 decimal places.
( )AV 0
1V 10sin t 3sin 3 t 2sin 5 t d( t)Ο
= Ο + Ο + Ο ΟΟ β«
= 0
1 3 210cos t cos3 t cos5 t3 5
Οβ‘ β€β Ο β Ο β Οβ’ β₯Ο β£ β¦
= ( ) ( )1 10cos cos3 0.4cos5 10cos 0 cos0 0.4cos0β Οβ Οβ Ο β β β ββ‘ β€β£ β¦Ο
= ( ) ( )1 10 1 0.4 10 1 0.4+ + β β β ββ‘ β€β£ β¦Ο
= ( ) ( )1 22.811.4 11.14β β =β‘ β€β£ β¦Ο Ο = 7.26
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 332
CHAPTER 38 SOME APPLICATIONS OF INTEGRATION
EXERCISE 148 Page 375 2. Sketch the curves y = x2 + 3 and y = 7 β 3x and determine the area enclosed by them.
The two curves intersect when x2 + 3 = 7 β 3x
i.e. x2 + 3x β 4 = 0
i.e. (x + 4)(x β 1) = 0
i.e. when x = -4 and x = 1
The two curves are shown below.
Area enclosed by curves =
( ) ( ) ( )1 1 1 12 2 2
4 4 4 4(7 3x)dx x 3 dx (7 3x) x 3 dx 4 3x x dx
β β β ββ β + = β β + = β ββ« β« β« β«
= 12 3
4
3x x4x2 3
β
β‘ β€β ββ’ β₯
β£ β¦
= 3 1 644 16 242 3 3
β β β ββ β β β β +β β β ββ β β β
= 1 2 1 22 18 2 186 3 6 3
β β β ββ β = +β β β ββ β β β
= 5206
square units
3. Determine the area enclosed by the three straight lines y = 3x, 2y = x and y + 2x = 5.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 333
y = -2x + 5 and y = 3x intersect when -2x + 5 = 3x i.e. when 5 = 5x i.e. when x = 1
y = -2x + 5 and y = x2
intersect when -2x + 5 = x2
i.e. when 5 = 2.5x i.e. when x = 2
The three straight lines are shown below.
Shaded area = 1 2
0 1
x x3x dx ( 2x 5) dx2 2
β ββ + β + ββ ββ β β« β«
= ( )1 22 2 2
2
0 1
3x x x 3 1 1x 5x (0) 4 10 1 1 52 4 4 2 4 4
β‘ β€ β‘ β€ β‘ β€ β‘ β€β β β ββ + β + β = β β + β + β β β + ββ β β ββ’ β₯ β’ β₯ β’ β₯ β’ β₯β β β β β£ β¦ β£ β¦β£ β¦ β£ β¦
= ( )1 3 1 11 5 3 1 14 4 4 4
β‘ β€β β β β+ β = +β β β ββ’ β₯β β β β β£ β¦
= 122
square units
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 334
EXERCISE 149 Page 377 2. The distance of points y from the mean value of a frequency distribution are related to the variate
x by the equation y = x + 1x
. Determine the standard deviation (i.e. the r.m.s. value), correct to
4 significant figures for values of x from 1 to 2.
Standard deviation = the r.m.s. value = 2
2 2
1 1
1 1 1 1x dx x x dx2 1 x x x
β β β ββ β+ = + +β β β ββ ββ β β β β β β β« β«
= ( )2 22 2 221 1
1x 2 dx x 2 x dxx
ββ β+ + = + +β ββ β β« β«
= 2 23 1 3
1 1
x x x 12x 2x3 1 3 x
ββ‘ β€ β‘ β€+ + = + ββ’ β₯ β’ β₯ββ£ β¦ β£ β¦
= 8 1 14 2 1 4.83333 2 3
β‘ β€β β β β+ β β + β =β β β ββ’ β₯β β β β β£ β¦ = 2.198
3. The current i = 25 sin 100Οt mA flows in an electrical circuit. Determine, using integral calculus,
its mean and r.m.s. value each correct to 2 decimal places over the range t = 0 to t = 10 ms.
Mean value = ( )3
3 10 1010 10
3 00
1 2525sin100 t dt 100 cos100 t10 10 0 100
ββ Γ
Γ
β
β‘ β€Ο = β Οβ’ β₯Γ β Οβ£ β¦β«
= 3100(25) cos(100 10 10 ) cos 0100
ββ‘ β€β ΟΓ Γ ββ£ β¦Ο
= [ ] [ ]25 25 50cos cos 0 1 1β Οβ = β β β =Ο Ο Ο
= 15.92 mA
r.m.s. value = 3 310 10 10 102 2 2
3 0 0
1 125 sin 100 t dt (100)(25) (1 cos 200 t)dt10 10 0 2
β βΓ Γ
β Ο = β ΟΓ β β« β«
since cos 2A = 1 β 22sin A from which, 2 1sin A (1 cos 2A)2
= β ,
= 310 102 2
3
0
(100)(25) sin 200 t (100)(25) sin 2t 10 10 (0 sin 0)2 200 2 200
βΓβΟ β‘ Ο β€β‘ β€ β ββ = Γ β β ββ ββ’ β₯β’ β₯Ο Οβ£ β¦ β β β£ β¦
= 2 2
3(100)(25) 25 2510 102 2 2
ββ‘ β€Γ = =β£ β¦ = 17.68 mA
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 335
4. A wave is defined by the equation: v = 1 3E sin t E sin 3 tΟ + Ο
where 1E , 3E and Ο are constants. Determine the r.m.s. value of v over the interval 0 t Οβ€ β€
Ο.
r.m.s. value = ( )21 30
1 E sin t E sin 3 t d( t)0
ΟΟ Ο + Ο Ο
Οβ
Ο
β«
= ( )2 2 2 21 1 3 30
E sin t 2E E sin t sin 3 t E sin 3 t d( t)ΟΟΟ
Ο + Ο Ο + Ο ΟΟ β«
2
0 00
1 cos 2 t t sin 2 tsin t dt dt 0 (0 0)2 2 4 2 2
ΟΟ Ο ΟΟ Ο β Ο Ο β‘ Ο β€ Οβ‘ β€ β βΟ = = β = β β β =β ββ’ β₯β’ β₯Ο Ο Οβ£ β¦ β β β£ β¦β« β«
2
0 00
1 cos 6 t t sin 6 tsin 3 t dt dt 0 (0 0)2 2 12 2 2
ΟΟ Ο ΟΟ Ο β Ο Ο β‘ Ο β€ Οβ‘ β€ β βΟ = = β = β β β =β ββ’ β₯β’ β₯Ο Ο Οβ£ β¦ β β β£ β¦β« β«
( )
[ ]
0 0 0
0
1sin t sin 3 t dt sin 3 t sin t dt cos 4 t cos 2 t dt (see page183of textbook)2
1 sin 4 t sin 2 t 1 (0 0) (0 0) 02 4 2 2
Ο Ο ΟΟ Ο Ο
ΟΟ
Ο Ο = Ο Ο = β Ο β Ο
Ο Οβ‘ β€= β β = β β β β =β’ β₯Ο Οβ£ β¦
β« β« β«
Hence, r.m.s. value = 22
2 2 311 3
EEE E2 2 2 2
β βΟ Ο Οβ‘ β€+ = +β ββ’ β₯Ο Ο Οβ£ β¦ β β
= 2 2
1 3E E2
β β+β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 336
EXERCISE 150 Page 378
2. The area between 2
yx
= 1 and y + x2 = 8 is rotated 360Β° about the x-axis. Find the volume
produced.
2y x= and y = 8 - 2x intersect when 2x = 8 - 2x i.e. 2 2x = 8 and 2x = 4
i.e. when x = 4 2= Β± . A sketch of the two curves is shown below.
Volume of solid of revolution = ( ) ( )2 22 22 2
2 28 x dx x dx
β βΟ β β Οβ« β«
= ( ) ( )2 2 22 4 4 2
2 2 264 16x x dx x dx 64 16x dx
β β βΟ β + β Ο = Ο ββ« β« β«
= 23
2
16x 128 12864x 128 1283 3 3
β
β‘ β€ β‘ β€β β β βΟ β = Ο β β β +β β β ββ’ β₯ β’ β₯β β β β β£ β¦β£ β¦
= 256 768 256 5122563 3 3
ββ‘ β€ β‘ β€ β βΟ β = Ο = Οβ ββ’ β₯ β’ β₯β£ β¦ β£ β¦ β β = 2170
3Ο cubic units
3. The curve y = 2x2 + 3 is rotated about (a) the x-axis between the limits x = 0 and x = 3, and
(b) the y-axis, between the same limits. Determine the volume generated in each case.
(a) The curve is shown below.
( ) ( )353 322 4 2 3
x axis 0 00
4xVolume 2x 3 dx 4x 12x 9 dx 4x 9x5β
β‘ β€= Ο + = Ο + + = Ο + +β’ β₯
β£ β¦β« β«
= Ο[(329.4) β (0)] = 329.4Ο cubic units
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 337
(b)
21 2y axis 3
Volume x dyβ = Οβ« Since y = 2x2 + 3, then 2x2 = y β 3 and 2 y 3x2β
=
Hence, 212 221
y axis 33
y 3 y 21 9volume dy 3y 63 92 2 2 2 2 2β
β‘ β€β‘ β€ β ββ Ο Οβ β β β= Ο = β = β β ββ’ β₯β ββ β β ββ’ β₯β β β β β£ β¦ β β β£ β¦β«
= ( ) ( ) 162157.5 4.52 2Ο Ο
β β =β‘ β€β£ β¦
= 81Ο cubic units
4. The profile of a rotor blade is bounded by the lines x = 0.2, y = 2x, xy eβ= , x = 1 and the x-axis.
The blade thickness t varies linearly with x and is given by: t = (1.1 β 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 = xeβ
(c) Calculate the cross-sectional area of the blade, correct to 3 decimal places.
(d) Calculate the volume of the blade in terms of K, correct to 3 decimal places.
(a) A sketch of the rotor blade is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 338
(b) Using the Newton-Raphson method (from chapter 9),
let f(x) = 2x - xeβ
f(0) = 0 - 0e = -1
f(1) = 2 - 1eβ = 1.632 hence a root lies between x = 0 and x = 1
Let 1r = 0.4
xf '(x) 2 eβ= +
12 1
1
f (r ) 0.12967995r r 0.4 0.35144f '(r ) 2.670320046
= β = β =
23 2
2
f (r ) ( 0.00079407)r r 0.35144 0.35173f '(r ) 2.70367407
β= β = β =
34 3
3
f (r ) ( 0.000010033)r r 0.35173 0.35173f '(r ) 2.703470033
β= β = β =
Hence, correct to 3 decimal places, x = 0.352
(c) Cross-sectional area of blade = 0.352 1 x
0.2 0.3522x dx e dxβ+β« β«
= 0.352 12 x
0.2 0.352x e (0.0839) ( 0.3354)ββ‘ β€ β‘ β€β = β ββ£ β¦ β£ β¦
= 0.419 square units
(d) Volume of the blade = 0.352 1 x
0.2 0.3522x(1.1 x)K dx e (1.1 x)K dxββ + ββ« β«
= ( ) ( )0.352 12 x x
0.2 0.352K 2.2x 2x dx K 1.1e xe dxβ ββ + ββ« β«
= 0.3522 3 1x x x
0.3520.2
2.2x 2xK K 1.1e xe e2 3
β β ββ‘ β€ β‘ β€β + β β β ββ’ β₯ β£ β¦β£ β¦ (the latter
integral using integration by parts β see page 418 of textbook)
= 0.3522 3 1x x x
0.3520.2
2.2x 2xK K 1.1e xe e2 3
β β ββ‘ β€β‘ β€β + β + +β’ β₯ β£ β¦
β£ β¦
= [ ] [ ]K 0.107218 0.038667 K 0.331091 0.177227β + β
= 0.068551K + 0.153864K
= 0.222K, correct to 3 decimal places.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 339
EXERCISE 151 Page 380 2. Find the position of the centroid of the area bounded by y = 5x2, the x-axis and ordinates x = 1
and x = 4. A sketch of the area is shown below.
( )44
4 4 4 4 42 3
1 1 1 14 4 4 432 2 3 3
1 1 1
1
5x 5 54 1 (255)xy dx x 5x dx 5x dx 4 318.754 4x 5 5 1055xy dx 5x dx 5x dx 4 1 (63)3 33
β‘ β€β’ β₯ β‘ β€ββ£ β¦β£ β¦
= = = = = = =β‘ β€ β‘ β€ββ£ β¦β’ β₯β£ β¦
β« β« β«β« β« β«
= 3.036
( )4 4 22 2 4541 1 4 5 54 1
11
1 1y dx 5x dx 1 1 25x 5 52 2y 25x dx 4 1 (1023)105 210 210 5 210 210ydx
β‘ β€β‘ β€= = = = = β =β’ β₯ β£ β¦
β£ β¦
β« β«β«
β«
= 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 β x2 which
lies above the x-axis. y = 4x β x2 = x(4 β 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=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 340
( )( )
( )( )
43 4 54 4 422 2 2 3 4
0 0 0 04 4 4 432 2
20 0 0
0
1 16x 8x x1 1 1y dx 4x x dx 16x 8x x dx 2 3 4 52 2 2yxy dx 4x x dx 4x x dx 2x3
β‘ β€β +β’ β₯β β + β£ β¦
= = = =β‘ β€β β ββ’ β₯β£ β¦
β« β« β«β« β« β«
= ( )1 341.333 512 204.8 (0) 17.06652
64 10.66632 (0)3
β + ββ‘ β€β£ β¦=
β ββ ββ ββ β
= 1.6
Hence, the co-ordinates of the centroid are (2, 1.6)
5. Sketch the curve y2 = 9x between the limits x = 0 and x = 4. Determine the position of the
centroid of this area. The curve y2 = 9x is shown sketched below.
By symmetry, y 0=
( )
452
34 4 4 5 520 0 0 0
4 4 1 4 3 34 320 0 20
0
3x5 6 6 6x 4 0 (32)xydx x 3 x dx 3x dx 2 5 5 5x
2(8)2 x 2 4 0ydx 3 x dx 3x dx 3x32
β‘ β€β’ β₯β’ β₯β’ β₯ β‘ β€ β‘ β€ββ£ β¦ β£ β¦ β£ β¦= = = = = = =
β‘ β€ β‘ β€β‘ β€ ββ£ β¦ β£ β¦β’ β₯β’ β₯β’ β₯β£ β¦
β« β« β«β« β« β«
= 38.416
= 2.4
Hence, the centroid is at (2.4, 0)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 341
EXERCISE 152 Page 382 2. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid of a
metal template in the form of a quadrant of a circle of radius 4 cm. (The equation of a circle,
centre 0, radius r is x2 + y2 = r2).
(a) A sketch of the template is shown below.
Using Pappus, volume, V = ( )( )area 2 yΟ
i.e. ( )( )3 21 4 1r r 2 y2 3 4β βΟ = Ο Οβ ββ β
from which, ( )
3
2
2 r 4r 4(4)3y3 3r 2
4
Ο= = =
Ο Οβ βΟΟβ β
β β
= 1.70 cm
By symmetry, x 1.79cm=
Hence, the centroid of the template is at (1.70, 1.70)
(b) ( )( )( )
( )( )
( )
( )
432 2
144 42 2 200 0 0
4 4 4 422 2 2 21 2 2
0 0 0
0
1 16 x2
3x 16 x dxxydx x 16 x dx 2
x4 x xydx 4 x dx 4 x dx sin 4 x2 4 2
β
β‘ β€β ββ’ β₯β’ β₯β’ β₯β β β’ β₯β£ β¦
= = = =β‘ β€β β + ββ’ β₯β£ β¦
β«β« β«β« β« β«
(the numerator being an algebraic substitution - see chapter 39), and the denominator
being a sin ΞΈ substitution - see chapter 40)
i.e. ( )
( ) ( )
32
1 1
1 640 163 3x
12.5668sin 1 2(0) 8sin 0 0(4)β β
β‘ β€β‘ β€β ββ ββ’ β₯β’ β₯β β
β’ β₯β’ β₯β β β£ β¦= =β’ β₯β‘ β€+ β +β’ β₯β£ β¦β’ β₯β£ β¦
= 1.70
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 342
By symmetry, y 1.70cm=
Hence, the centroid lies on the centre line OC (see diagram), at co-ordinates (1.70, 1.70).
The distance from 0 is given by ( )2 21.70 1.70+ = 2.40 cm
3. (a) Determine the area bounded by the curve y = 5x2, the x-axis and the ordinates x = 0 and
x = 3.
(b) If this area is revolved 360Β° about (i) the x-axis, and (ii) the y-axis, find the volume of the
solids of revolution produced in each case.
(c) Determine the co-ordinates of the centroid of the area using (i) integral calculus, and (ii) the
theorem of Pappus. (a) The area is shown in the sketch below.
Shaded area = ( )333 2 3
00
5x 55x dx 3 03 3
β‘ β€= = ββ’ β₯β£ β¦
β« = 45 square units
(b)(i) ( )353 3 322 2 4 5
x axis 0 0 00
xVolume y dx 5x dx 25 x dx 25 5 3 05β
β‘ β€β‘ β€= Ο = Ο = Ο = Ο = Ο ββ’ β₯ β£ β¦
β£ β¦β« β« β«
= 1215Ο cubic units
(ii) y axisVolume β = (volume generated by x = 3) β (volume generated by y = 5x2)
= 45245 45 452
0 0 00
y y y(3) dy dy 9 dy 9y5 5 10
β‘ β€β β β βΟ β Ο = Ο β = Ο ββ β β β β’ β₯β β β β β£ β¦β« β« β«
= 2459(45) (0) [405 202.5]
10β‘ β€β β
Ο β β = Ο ββ’ β₯β ββ β β£ β¦
= 202.5Ο cubic units
(c)(i) ( ) ( )
343 3 3 42 3
0 0 0 03
0
5x 5 3 0xydx x 5x dx 5x dx 4 4x45 45 45 45ydx
β‘ β€β’ β₯ ββ£ β¦
= = = = =β« β« β«β«
= 2.25
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 343
353 32 4 5
0 0 03
0
1 25x1 1 5y dx 25x dx 3 02 52 2 2y45 45 45ydx
β‘ β€β’ β₯ β‘ β€ββ£ β¦β£ β¦
= = = =β« β«β«
= 13.5
Hence, (2.25, 13.5) are the co-ordinates of the centroid.
(ii) Using Pappus, volume generated when the shaded area is revolved about 0y = (area) ( )2 xΟ
i.e. 202.5Ο = (45) ( )2 xΟ
from which, 202.5x(45)(2 )
Ο=
Ο = 2.25
Similarly, volume generated when the shaded area is revolved about 0x = (area) ( )2 yΟ
i.e. 1215Ο = (45) ( )2 yΟ
from which, 1215y(40)(2 )
Ο=
Ο = 13.5
Hence, (2.25, 13.5) are the co-ordinates of the centroid.
4. A metal disc has a radius of 7.0 cm and is of thickness 2.5 cm. A semicircular groove of
diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine the volume of
metal removed using Pappusβ 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 4r3Ο
from its diameter, from problem 2 above,
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 344
i.e. 4(1.0)3Ο
= 0.424 cm from PQ.
Distance of centroid from XX = 7.0 β 0.424 = 6.576 cm.
Distance moved in 1 revolution by the centroid = 2Ο(6.576) cm
Area of semicircle = 2 2
2r (1.0) cm2 2 2Ο Ο Ο
= =
By Pappus, volume generated = area Γ distance moved by the centroid
i.e. volume of metal removed = ( )2 (6.576)2Οβ β Οβ β
β β = 364.90cm
Volume of disc = ( ) ( )22 3r h 7.0 2.5 384.845cmΟ = Ο =
Thus, percentage of metal removed = 64.90 100%384.845
Γ = 16.86%
Mass of metal removed = density Γ volume
= 7800 3
kgm
6 364.90 10 mβΓ Γ = 0.5062 kg or 506.2 g
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 345
EXERCISE 153 Page 389 2. Determine the second moment of area and radius of gyration for the triangle shown below about
(a) axis DD (b) axis EE and (c) an axis through the centroid of the triangle parallel to axis DD.
From Table 38.1, page 385:
(a) Second moment of area about DD, ( )( )33
DD
12.0 9.0b hI12 12
= = = 729 4cm
Radius of gyration about DD, DDh 9.0k6 6
= = = 3.67 cm
(b) Second moment of area about EE, ( )( )33
EE
12.0 9.0b hI4 4
= = = 2187 4cm
Radius of gyration about EE, EEh 9.0k2 2
= = = 6.36 cm
(c) Second moment of area about axis through centoid, ( )( )33 12.0 9.0b h36 36
= = = 243 4cm
Radius of gyration about axis through centoid, h 9.018 18
= = = 2.12 cm
5. For each of the areas shown below determine the second moment of area and radius of gyration
about axis LL, by using the parallel axis theorem.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 346
(a) Second moment of area,3 3
2 2 2LL GG
bl (3.0)(5.0)I I Ad Ad (3.0)(5.0)(2.5 2.0)12 12
= + = + = + +
= 31.25 + 303.75 = 335 4cm
2LL LLI Ak= from which, radius of gyration, LL
LLI 335k
area 15.0β β= = β ββ β
= 4.73 cm
(b) Second moment of area, 2LL GGI I Ad= +
3 3
4GG
bh (18)(12)I 864cm36 36
= = = where h = 2 215 9β = 12 cm, as shown in the diagram below.
Hence, area of triangle, A = 1 (18)(12)
2 = 108 2cm
Thus, 2
2 2LL GG
12I I Ad 864 108 10 864 108(14)3
β β= + = + + = +β ββ β
= 22032 4cm = 22032 4cm
correct to 4 significant figures.
2LL LLI Ak= from which, radius of gyration, LL
LLI 22032k
area 108β β= = β ββ β
= 14.3 cm
(c) Second moment of area, ( )24 4
2 2 2 2LL GG
r 4 (2.0)I I Ad r 5 (2.0) (7)4 2 4Ο Οβ β= + = + Ο + = + Οβ β
β β
= 12.57 + 615.75 = 628 4cm
2LL LLI Ak= from which, radius of gyration, LL
LL 2
I 628karea (2.0)
β β= = β βΟβ β
= 7.07 cm
8. A circular cover, centre 0, has a radius of 12.0 cm. A hole of radius 4.0 cm and centre X, where
0X = 6.0 cm, is cut in the cover. Determine the second moment of area and the radius of gyration
of the remainder about a diameter through 0 perpendicular to 0x. Second moment of area about diameter, i.e. axis CC in the diagram below
4 4
2CC DD GG
r rI I I Ad4 4Ο Ο β‘ β€= β = β +β£ β¦
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 347
= 4 4
2 2(12) (4.0) (4.0) (6.0)4 4
β‘ β€Ο Οβ + Οβ’ β₯β£ β¦
= 16286 β [201 + 1810] = 14275 = 14280 4cm , correct
to 4 significant figures.
2
CC CCI Ak= from which, radius of gyration,
CCCC 2 2
I 14275 14275karea (12.0) (4.0) 128
β β= = =β βΟ β Ο Οβ β
= 5.96 cm
9. For the sections shown below, find the second moment of area and the radius of gyration about
axis XX.
(a) (b) (a) For rectangle A in the diagram below,
second moment of area about 3 3
4A
bl (18.0)(3.0)C 40.5m12 12
= = =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 348
Hence, A
2 2XXI 40.5 Ad 40.5 (3.0)(18.0)(12.0 1.5) 40.5 9841.5= + = + + = + = 9882 4mm
For rectangle B, second moment of area about 3 3
4B
bl (4.0)(12.0)C 576m12 12
= = =
Hence, B
2XXI 576 (4.0)(12.0)(6.0) 576 1728= + = + = 2304 4mm
Thus, total second moment of area about XX, TXXI 9882 2304 12186= + = = 12190 4mm ,
correct to 4 significant figures.
Radius of gyration about XX, TXXXX
I 12186karea 54 48
= =+
= 10.9 mm
(b) Rectangle A (see diagram below):
Second moment of area about 3 3
4A
bl (6.0)(2.0)C 4cm12 12
= = =
A
2XXI 4 (2.0)(6.0)(6.0) 4 432= + = + = 436 4cm
Rectangle B:
Second moment of area about 3 3
4B
bl (2.5)(3.0)C 5.625cm12 12
= = =
B
2XXI 5.625 (2.5)(3.0)(3.5)= + = 97.5 4cm
Rectangle C:
Second moment of area about 3 3
4C
bl (6.0)(2.0)C 4cm12 12
= = =
C
2XXI 4 (6.0)(2.0)(1.0)= + = 16 4cm
Hence, total second moment of area about axis XX, IXXI = 436 + 97.5 + 16 = 549.5 4cm
Radius of gyration about XX, IXXXX
I 549.5karea 12 7.5 12
= =+ +
= 4.18 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 349
11. Find the second moment of area and radius of gyration about the axis XX for the beam section
shown below.
Second moment of area about axis XX,
A B CXX XX XX XXI I I I= + +
= ( )( ) ( )( ) ( )( )3 3 32 2 26.0 1.0 2.0 8.0 10.0 2.0
(6.0)(10.5) (16)(6.0) (20)(1.0)12 12 12
β‘ β€ β‘ β€ β‘ β€+ + + + +β’ β₯ β’ β₯ β’ β₯
β’ β₯ β’ β₯ β’ β₯β£ β¦ β£ β¦ β£ β¦
= 662 + 661.33 + 26.66 = 1350 4cm
Radius of gyration about XX, XXXX
I 1350karea 6 16 20
= =+ +
= 5.67 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 350
CHAPTER 39 INTEGRATION USING ALGEBRAIC SUBSTITUTIONS
EXERCISE 154 Page 392
4. Integrate ( )61 5x 32
β with respect to x.
Let u = 5x β 3 then du 5dx
= and dx = du5
Hence, 7
6 61 du 1 1 uu u du c2 5 10 10 7
β β= = +β β
β β β« β« = 71 (5x 3) c
70β +
5. Integrate 3(2x 1)ββ
with respect to x.
Let u = 2x β 1 then du 2dx
= and dx = du2
Hence, 3 3 du 3 1 3dx du ln u c(2x 1) u 2 2 u 2β β
= = β = β +ββ« β« β« = 3 ln(2x 1) c
2β β +
7. Evaluate ( )1 5
03x 1 dx+β« correct to 4 significant figures.
Let u = 3x + 1 then du 3dx
= and dx = du3
Hence, 6
5 5 6du 1 1 u 1u u du c (3x 1) c3 3 3 6 18
β β= = + = + +β β
β β β« β«
Thus, ( ) ( )11 5 6 6 6
0 0
1 13x 1 dx 3x 1 4 118 18
β‘ β€ β‘ β€+ = + = ββ£ β¦β£ β¦β« = 227.5
8. Evaluate ( )2 2
0x 2x 1 dx+β« correct to 4 significant figures.
Let u = 22x 1+ then du 4xdx
= i.e. dudx4x
=
Hence, ( ) ( )3
1 32 32 2 22 2du 1 1 u 1 1x (2x 1) dx x u u du c 2x 1 c 2x 1 c34x 4 4 6 62
β ββ β
+ = = = + = + + = + +β ββ ββ β
β« β« β«
Thus, ( ) ( ) [ ]2
2 32 2
00
1 1x 2x 1 dx 2x 1 27 16 6β‘ β€+ = + = ββ’ β₯β£ β¦β« = 4.333
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 351
9. Evaluate /3
02sin 3t dt
4Ο Οβ β+β β
β β β« correct to 4 significant figures.
3/3
00
2 2 32sin 3t dt cos 3t cos cos4 3 4 3 3 4 4
Ο
Ο Ο β‘ Ο β€ β‘ Ο Ο Οβ€β β β β β β+ = β + = β + ββ β β β β ββ’ β₯ β’ β₯β β β β β β β£ β¦ β£ β¦β« (note angles are in radians)
= [ ]2 0.70711 0.707113
β β β
= 0.9428
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 352
EXERCISE 155 Page 394 2. Integrate 55cos t sin t with respect to t.
Let u = cos t then du sin tdt
= β and dudtsin t
=β
65 5 5du u5cos t sin t dt 5u sin t 5 u du 5 c
sin t 6β β
= = β = β +β ββ β β β« β« β« = - 65 cos t c
6+
3. Integrate 23sec 3x tan 3x with respect to x.
Let u = tan 3x then 2du 3sec 3xdx
= i.e. 2
dudx3sec 3x
=
Hence, 2
2 22
du u3sec 3x tan 3x dx 3sec 3x (u) u du c3sec 3x 2
= = = +β« β« β« = 21 tan 3x c2
+
Alternatively, let u = sec 3x then du 3sec3x tan 3xdx
= i.e. dudx3sec3x tan 3x
=
Hence, 2 2
2 2 du u u3sec 3x tan 3x dx 3u tan 3x du du u du3sec3x tan 3x sec3x u
= = = =β« β« β« β« β«
= 2u c
2+ = 21 sec 3x c
2+
5. Integrate lnΞΈΞΈ
with respect to ΞΈ.
Let u = ln ΞΈ then du 1d
=ΞΈ ΞΈ
and dΞΈ = ΞΈdu
Hence, ( )2ln u ud du u du c
2ΞΈ
ΞΈ = ΞΈ = = +ΞΈ ΞΈβ« β« β« = ( )21 ln c
2ΞΈ +
6. Integrate 3 tan t with respect to t.
sin 2t3 tan 2t dt 3 dtcos 2t
=β« β« Let u = cos 2t then du 2sin 2tdt
= β i.e. dudt2sin 2tβ
=
Hence, sin 2t sin 2t du 3 1 33 dt 3 du ln u ccos 2t u 2sin 2t 2 u 2
ββ ββ β= = β = β +β ββ ββ β β β β« β« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 353
= ( ) 13 3ln cos 2t c ln cos 2t c2 2
ββ + = +
= 3 lnsec 2t c2
+
8. Evaluate ( )22x 11
03x e dx
β
β« correct to 4 significant figures.
Let u = 22x 1β then du 4xdx
= i.e. dudx4x
=
Hence, ( ) ( )2 22x 1 2x 1u u udu 3 3 33xe dx 3xe e du e c e c4x 4 4 4
β β= = = + = +β« β« β«
( ) ( )2
2 12x 11 2x 1 1 1
0 0
3 33x e dx e e e4 4
β β ββ‘ β€ β‘ β€= = ββ£ β¦β’ β₯β£ β¦β« = 1.763
10. Evaluate ( )
1
50 2
3x dx4x 1β
β« correct to 4 significant figures.
Let u = 24x 1β then du 8xdx
= i.e. dudx8x
=
Hence, ( ) ( )
45
5 45 42 2
3x 3x du 3 3 u 3 3dx u du c c cu 8x 8 8 4 32u4x 1 32 4x 1
ββ β ββ β= = = + = β + = β +β ββ β ββ β β β β β
β« β« β«
Thus, ( ) ( )
1
1
5 4 4 40 2 2
0
3x 3 1 3 1 1dx32 32 3 ( 1)4x 1 4x 1
β‘ β€ β‘ β€β’ β₯= β = β ββ’ β₯β’ β₯ ββ£ β¦β ββ£ β¦β« = 0.09259
11. The electrostatic potential on all parts of a conducting circular disc of radius r is given by the
equation: V = 9
2 20
R2 dRR r
ΟΟ+
β«
Solve the equation by determining the integral.
Let u = 2 2R r+ then du 2RdR
= i.e. dudR2R
=
( )1
1 22 22
2 2
R R du 1 1 udR u du c u c R r12R 2 2uR r2
ββ β= = = + = + = +β ββ β +
β« β« β« + c
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 354
Hence, V = ( ) ( )99 2 2 2 2 2
2 20 0
R2 dR 2 R r 2 9 r rR r
β‘ β€ β‘ β€ΟΟ = ΟΟ + = ΟΟ + ββ’ β₯ β’ β₯β£ β¦ β£ β¦+β«
= ( ) 2 22 9 r rΟΟ + β
12. In the study of a rigid motor the following integration occurs: ( )2
2J(J 1)h8 IkT
r 0Z 2J 1 e dJ
β +β Ο= +β«
Determine rZ for constant temperature T assuming h, I and k are constants.
Let u = J(J + 1) = 2J J+ then du 2J 1dJ
= + i.e. dudJ2J 1
=+
( )2 2 2 2
2 2 2 2J(J 1)h u h u h u h8 IkT 8 Ik T 8 Ik T 8 Ik T
2
2
du 12J 1 e dJ (2J 1)e e du e ch(2J 1)
8 I k T
β + ββ β
Ο Ο Ο Ο+ = + = = ++ β
Ο
β« β« β«
= 2
2J(J 1)h28 Ik T
2
8 I k T eh
β +ΟΟ
β + c
Thus, ( )2 2
2 2J(J 1)h J(J 1)h2 2
08 IkT 8 Ik Tr 2 20
0
8 I k T 8 I k TZ 2J 1 e dJ e e eh h
ββ + β +
β ββΟ Οβ‘ β€Ο Οβ’ β₯ β‘ β€= + = β = β ββ£ β¦β’ β₯β£ β¦
β«
= [ ]2
2
8 I k T 0 1h
Οβ β =
2
2
8 I k Th
Ο
13. In electrostatics, ( )
2
0 2 2
a sinE d2 a x 2ax cos
Οβ§ β«
Ο ΞΈβͺ βͺ= ΞΈβ¨ β¬Ξ΅ β β ΞΈβͺ βͺβ© β
β« where a, Ο and Ξ΅ are constants, x is
greater than a, and x is independent of ΞΈ. Show that 2aExΟ
=Ξ΅
Let u = 2 2a x 2ax cos+ β ΞΈ then du 2ax sind
= ΞΈΞΈ
and dud2ax sin
ΞΈ =ΞΈ
Hence,
112 2 2 2 22a sin du a du a 1 a 1 uE u du c12 2ax sin 2 2 2ax 2 2axu 2ax u
2
β
β ββ βΟ ΞΈ Ο Ο Οβ β β β β β= = = = +β ββ β β β β βΞ΅ ΞΈ Ξ΅ Ξ΅ Ξ΅β β β β β β β ββ β
β« β« β«
= ( )2 2a a x 2ax cos c2 xΟ
+ β ΞΈ +Ξ΅
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 355
Hence, E = 2 2 2 2 2 2
0
a aa x 2ax cos a x 2ax cos a x 2ax cos 02 x 2 x
ΟΟ Οβ‘ β€ β‘ β€+ β ΞΈ = + β Ο β + ββ£ β¦ β£ β¦Ξ΅ Ξ΅
= 2 2 2 2a (a x 2ax) (a x 2ax)2 xΟ β‘ β€+ + β + ββ£ β¦Ξ΅
= ( ) ( )2 2a x a x a2 xΟ β‘ β€+ β ββ’ β₯β£ β¦Ξ΅
= [ ] [ ]a a(x a) (x a) 2a2 x 2 xΟ Ο
+ β β =Ξ΅ Ξ΅
i.e. 2aExΟ
=Ξ΅
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 356
CHAPTER 40 INTEGRATION USING TRIGONOMETRIC AND
HYPERBOLIC SUBSTITUTIONS EXERCISE 156 Page 399 2. Integrate 23cos t with respect to t.
2cos 2t 2cos t 1= β from which, ( )2 1cos t 1 cos 2t2
= +
Hence, ( )2 13cos t dt 3 1 cos 2t dt2
= +β« β« = 3 sin2tt c2 2β β+ +β ββ β
4. Integrate 22cot 2t with respect to t.
( )2 2 12cot 2t dt 2 cos ec 2t 1 dt 2 cot 2t t c2
β‘ β€= β = β β +β’ β₯β£ β¦β« β« = - (cot 2t + 2t) + c
5. Evaluate /3 2
03sin 3x dx
Ο
β« correct to 4 significant figures.
2cos 2x 1 2sin x= β and 2cos 6x 1 2sin 3x= β from which, 2 1sin 3x (1 cos 6x)
2= β
Hence, /3
/3 /32
0 00
3 3 sin 6x3sin 3x dx (1 cos 6x)dx x2 2 6
ΟΟ Ο β‘ β€= β = ββ’ β₯β£ β¦β« β«
=
6sin3 33 (0 sin0)2 3 6 2 3
β‘ Ο β€β ββ’ β₯β βΟ Οβ‘ β€β β β =β’ β₯β β β’ β₯β£ β¦β’ β₯β ββ’ β₯β β β£ β¦
= 2Ο or 1.571
7. Evaluate 1 2
02 tan 2t dtβ« correct to 4 significant figures.
( )1
1 12 2
0 00
tan 2t2 tan 2t dt 2 sec 2t 1 dt 2 t2
β‘ β€= β = ββ’ β₯β£ β¦β« β«
= tan 2 tan 02 1 02 2
β‘ β€β β β ββ β ββ β β ββ’ β₯β β β β β£ β¦ = tan 2 β 2 = -4.185
(note that βtan 2β means βtan 2 radiansβ)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 357
EXERCISE 157 Page 400 2. Integrate 32cos 2x with respect to x.
( ) ( )3 2 2 22cos 2x dx 2 cos 2x cos 2x dx 2 cos 2x 1 sin 2x dx 2 cos 2x cos2x sin 2x dx= = β = ββ« β« β« β«
= 3sin 2x sin 2x2 c
2 6β β
β +β ββ β
using the algebraic substitution u = sin 2x
= 3sin 2xsin 2x c3
β +
3. Integrate 3 22sin t cos t with respect to t.
( )3 2 2 2 2 22sin t cos t dt 2sin t sin t cos t dt 2sin t 1 cos t cos t dt= = ββ« β« β«
= ( )3 5
2 4 cos t cos t2 sin t cos t sin t cos t dt 2 c3 5
β‘ β€β = β + +β’ β₯
β£ β¦β«
using the algebraic substitution u = cos t
= 3 52 2cos t cos t c3 5
β + +
5. Integrate 42sin 2ΞΈ with respect to ΞΈ.
( )2
24 2 21 cos 4 12sin 2 d 2 sin 2 d 2 d (1 2cos 4 cos 4 )d2 2
β ΞΈβ βΞΈ ΞΈ = ΞΈ ΞΈ = ΞΈ = β ΞΈ+ ΞΈ ΞΈβ ββ β β« β« β« β«
= 1 1 cos81 2cos 4 d2 2
β‘ + ΞΈ β€β ββ ΞΈ+ ΞΈβ ββ’ β₯β β β£ β¦β«
= 1 sin 4 sin8 c2 2 2 16
ΞΈ ΞΈ ΞΈβ‘ β€ΞΈβ + + +β’ β₯β£ β¦ = 3 sin 4 sin 8 c
4 4 32ΞΈ ΞΈ ΞΈβ + +
6. Integrate 2 2sin t cos t with respect to t.
2 2 21 cos 2t 1 cos 2t 1 1 1 cos 4tsin t cos t dt dt (1 cos 2t)dt 1 dt2 2 4 4 2
β + β‘ + β€β ββ β β β= = β = ββ ββ β β ββ’ β₯β β β β β β β£ β¦β« β« β« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 358
= 1 1 cos 4t 1 t sin 4tdt c4 2 2 4 2 8
β β β‘ β€β = β +β β β’ β₯β β β£ β¦β«
= t 1 sin 4t c8 32β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 359
EXERCISE 158 Page 401 2. Integrate 2 sin 3x sin x with respect to x.
( )12sin 3x sin x dx 2 cos 4x cos 2x dx2
= β ββ« β« from 9, page 398 of textbook
= sin 4x sin 2x c4 2
β‘ β€β β +β’ β₯β£ β¦ = sin 2x sin 4x c
2 4β +
4. Integrate 1 cos 4 sin 22
ΞΈ ΞΈ with respect to ΞΈ.
1 1 1cos 4 sin 2 d (sin 6 sin 2 )d2 2 2
ΞΈ ΞΈ ΞΈ = ΞΈβ ΞΈ ΞΈβ« β« from 7, page 398 of textbook
= 1 cos 6 cos 2 c4 6 2
ΞΈ ΞΈβ‘ β€β + +β’ β₯β£ β¦ = 1 cos 2 cos 6 c
4 2 6ΞΈ ΞΈβ ββ +β β
β β
6. Evaluate 1
02sin 7t cos3t dtβ«
1
1
00
1 cos10t cos 4t2sin 7t cos3t dt 2 (sin10t sin 4t)dt2 10 4
β‘ β€= + = β ββ’ β₯β£ β¦β« β« from 6, page 398 of textbook
= cos10 cos 4 cos 0 cos 010 4 10 4
β β β ββ β β β ββ β β ββ β β β
= (0.24732) β (-0.35)
= 0.5973
7. Evaluate /3
04 sin 5 sin 2 d
Οβ ΞΈ ΞΈ ΞΈβ«
[ ]/ 3 /3
0 0
14 sin 5 sin 2 d 4 cos 7 cos3 d2
Ο Οβ ΞΈ ΞΈ ΞΈ = β β ΞΈβ ΞΈ ΞΈβ« β« from 9, page 398 of textbook
= /3
0
7 3sin sinsin 7 sin 3 3 32 2 (0 0)7 3 7 3
Οβ‘ Ο Ο β€β ββ’ β₯β βΞΈ ΞΈβ‘ β€β = β β ββ’ β₯β ββ’ β₯β£ β¦ β’ β₯β ββ’ β₯β β β£ β¦
= 0.2474
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 360
EXERCISE 159 Page 402
3. Determine ( )24 x dxββ«
( ) ( ) ( )2
2 2 2 1 2 22 x x4 x dx 2 x sin 2 x2 2 2
ββ = β = + ββ« β« from 11, page 398 of textbook
= ( )1 2x x2sin 4 x c2 2
β + β +
4. Determine ( )216 9t dtββ«
( )2
2 2 216 416 9t dt 9 t dt 9 t dt9 3
β‘ β€β‘ β€β β β ββ = β = ββ’ β₯β β β ββ’ β₯β β β β β£ β¦ β’ β₯β£ β¦β« β« β«
=
2
21 2
4t t 433 sin t c42 2 33
β
β‘ β€β ββ’ β₯β β β‘ β€β ββ β β’ β₯+ β +β’ β₯β ββ’ β₯β β β β β’ β₯β£ β¦β ββ’ β₯β β β£ β¦
β« from 11, page 398 of textbook
= 2
1 28 3t 3t 4sin t c3 4 2 3
ββ‘ β€β β+ β +β’ β₯β ββ β β’ β₯β£ β¦
= 2
1 2 2 1 2 28 3t t 4 8 3t tsin 3 t c sin (4 9t ) c3 4 2 3 3 4 2
β ββ‘ β€β β+ β + = + β +β’ β₯β ββ β β’ β₯β£ β¦
= ( )1 28 3t tsin 16 9t c3 4 2
β + β +
6. Evaluate ( )1 2
09 4x dxββ«
( )2
1 1 12 2 2
0 0 0
9 39 4x dx 4 x dx 2 x dx4 2
β‘ β€β‘ β€β β β ββ = β = ββ’ β₯β β β ββ’ β₯β β β β β£ β¦ β’ β₯β£ β¦β« β« β«
=
12
21 2
0
3x x 322 sin x32 2 22
β
β‘ β€β ββ’ β₯β β β‘ β€β ββ β β’ β₯+ ββ’ β₯β ββ’ β₯β β β β β’ β₯β£ β¦β ββ’ β₯β β β£ β¦
from 11, page 398 of textbook
= 19 2 12 sin 1.25 (0 0)8 3 2
ββ‘ β€β β+ β +β ββ’ β₯β β β£ β¦ = 2.760
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 361
EXERCISE 160 Page 403
2. Determine 2
5 d16 9
ΞΈ+ ΞΈβ«
222 2
5 5 5 1d d d1616 9 9 499 3
ΞΈ = ΞΈ = ΞΈ+ ΞΈ β β β β+ ΞΈ + ΞΈβ β β ββ β β β
β« β« β«
= 15 1 tan c4 493 3
β
β ββ βΞΈβ β +β β β ββ ββ β β ββ ββ β β β β β
from 12, page 398 of textbook
= 15 3tan c12 4
β ΞΈ+
4. Evaluate 3
20
5 dx4 x+β«
3
3 3
2 2 20 00
5 1 1 xdx 5 dx 5 tan4 x 2 x 2 2
β‘ β€= = β’ β₯+ + β£ β¦β« β« from 12, page 398 of textbook
= 1 15 3tan tan 02 2
β ββ‘ β€ββ’ β₯β£ β¦
= 2.457
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 362
EXERCISE 161 Page 405
2. Find ( )2
3 dx9 5x+
β«
( ) 222 2
3 3 3dx dx dx99 5x 35 x 5 x5 5
= =β‘ β€ β‘ β€+ β β β β+β β +β’ β₯ β’ β₯β ββ β β£ β¦ β β β’ β₯β£ β¦
β« β« β«
= 13 xsinh c355
β +β ββ ββ β
from 13, page 398 of textbook
= 13 5sinh x c35
β +
4. Find ( )24t 25 dt+β«
( )2
2 2 225 54t 25 dt 4 t dt 4 t dt4 2
β‘ β€β‘ β€β β β β+ = + = +β’ β₯β β β ββ’ β₯β β β β β£ β¦ β’ β₯β£ β¦β« β« β«
=
2
21 2
5t t 522 sinh t c52 2 22
β
β‘ β€β ββ’ β₯β β β‘ β€β ββ β β’ β₯+ + +β’ β₯β ββ’ β₯β β β β β’ β₯β£ β¦β ββ’ β₯β β β£ β¦
from 14, page 398 of textbook
= 2
1 225 2t t 5sinh 4 t c4 5 2 2
ββ‘ β€β β+ + +β’ β₯β β
β β β’ β₯β£ β¦
= 1 2 225 2t tsinh 4t 5 c4 5 2
β β‘ β€+ + +β£ β¦
= 1 225 2t tsinh 4t 25 c4 5 2
β β‘ β€+ + +β£ β¦
6. Evaluate ( )1 2
016 9 d+ ΞΈ ΞΈβ«
( )2
1 1 12 2 2
0 0 0
16 416 9 d 9 d 9 d9 3
β‘ β€β‘ β€β β β β+ ΞΈ ΞΈ = + ΞΈ ΞΈ = + ΞΈ ΞΈβ’ β₯β β β ββ’ β₯β β β β β£ β¦ β’ β₯β£ β¦β« β« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 363
=
12
21 2
0
4433 sinh
42 2 33
β
β‘ β€β ββ’ β₯β β β‘ β€ΞΈ ΞΈ β ββ β β’ β₯+ + ΞΈβ’ β₯β ββ’ β₯β β β β β’ β₯β£ β¦β ββ’ β₯β β β£ β¦
from 14, page 398 of textbook
= 116 3 1 163 sinh 2.777777 sinh 0 018 4 2 18
ββ‘ β€β β β β+ β +β β β ββ’ β₯β β β β β£ β¦
= 4.348
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 364
EXERCISE 162 Page 407
2. Find ( )2
3 dx4x 9β
β«
( ) 222 2
3 3 3dx dx dx94x 9 34 x 2 x4 2
= =β‘ β€ β‘ β€β β β β ββ ββ ββ’ β₯ β’ β₯β ββ β β£ β¦ β β β’ β₯β£ β¦
β« β« β«
= 13 xcosh c322
β +β ββ ββ β
from 15, page 398 of textbook
= 13 2xcosh c2 3
β +
4. Find ( )24 25 dΞΈ β ΞΈβ«
( )2
2 2 225 54 25 d 4 d 2 d4 2
β‘ β€β‘ β€β β β βΞΈ β ΞΈ = ΞΈ β ΞΈ = ΞΈ β ΞΈβ’ β₯β β β ββ’ β₯β β β β β£ β¦ β’ β₯β£ β¦β« β« β«
=
2
22 1
55 22 cosh c
52 2 22
β
β‘ β€β ββ’ β₯β ββ‘ β€ΞΈ ΞΈβ β β β β’ β₯ΞΈ β β +β’ β₯β ββ’ β₯β ββ β β’ β₯β£ β¦ β ββ’ β₯β β β£ β¦
from 16, page 398 of textbook
= 2 125 25 2cosh c4 4 5
β ΞΈβ βΞΈ ΞΈ β β +β ββ β
6. Evaluate ( )3 2
2t 4 dtββ«
( ) ( ) ( )323 32 2 2 2 2 1
2 22
t 2 tt 4 dt t 2 dt t 2 cosh2 2 2
ββ‘ β€β = β = β ββ’ β₯
β£ β¦β« β«
= ( )1 13 35 2cosh 0 2cosh 12 2
β ββ ββ β ββ ββ β
= 1.429
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 365
CHAPTER 41 INTEGRATION USING PARTIAL FRACTIONS EXERCISE 163 Page 409
2. Determine ( )2
4(x 4) dxx 2x 3
ββ ββ«
2
4( 4) 5 12 3 ( 1) ( 3)β
= ββ β + βx
x x x x from question 2, Exercise 13, page 19.
Hence, ( )2
4(x 4) 5 1dx dx(x 1) (x 3)x 2x 3
β ββ= ββ β+ ββ β β β
β« β« = 5ln(x 1) ln(x 3) c+ β β +
or 5ln(x 1) ln(x 3) c+ β β + or 5(x 1)ln c
(x 3)β§ β«+
+β¨ β¬ββ© β by the laws of logarithms
4. Determine 2
2
x 9x 8 dxx x 6+ ++ ββ«
2
2
9 8 2 616 ( 3) ( 2)
x xx x x x+ +
= + ++ β + β
from question 5, Exercise 13, page 19.
2
2
x 9x 8 2 6dx 1 dxx x 6 (x 3) (x 2)
β β+ += + +β β+ β + ββ β
β« β« = x + 2 ln(x + 3) + 6 ln(x - 2) + c
or ( ) ( )2 6x ln x 3 ln x 2 c+ + + β + or ( ) 62x ln (x 3) x 2 c+ + β +
6. Evaluate 24
3
x 3x 6 dxx(x 2)(x 1)
β +β ββ« correct to 4 significant figures.
Let 2x 3x 6 A B C A(x 2)(x 1) Bx(x 1) Cx(x 2)
x(x 2)(x 1) x (x 2) (x 1) x(x 2)(x 1)β + β β + β + β
β‘ + + =β β β β β β
then 2x 3x 6 A(x 2)(x 1) Bx(x 1) Cx(x 2)β + = β β + β + β
Let x = 0, 6 = 2A i.e. A = 3
Let x = 2, 4 = 2B i.e. B = 2
Let x = 1, 4 = -C i.e. C = -4
Hence, 2x 3x 6 3 2 4
x(x 2)(x 1) x (x 2) (x 1)β +
β‘ + ββ β β β
Thus, [ ]24 4 4
33 3
x 3x 6 3 2 4dx dx 3ln x 2ln(x 2) 4ln(x 1)x(x 2)(x 1) x (x 2) (x 1)
β ββ += + β = + β β ββ ββ β β ββ β
β« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 366
= (3 ln 4 + 2 ln 2 β 4 ln 3) β (3 ln 3 + 2 ln 1 β 4 ln 2)
= 0.6275
8. Determine the value of k, given that: 1
0
(x k) dx 0(3x 1)(x 1)
β=
+ +β«
Let (x k) A B A(x 1) B(3x 1)(3x 1)(x 1) (3x 1) (x 1) (3x 1)(x 1)
β + + +β‘ + =
+ + + + + +
then x β k = A(x + 1) + B(3x + 1)
Let x = -1, -1 β k = -2B i.e. B = 1 k2+
Let x = 13
β , 13
β - k = 2 A3
i.e. A = 3 1 1 3k k2 3 2 2β ββ β = β ββ ββ β
1 1
0 0
1 3 1 kk(x k) 2 2 2dx dx 0(3x 1)(x 1) (3x 1) (x 1)
+β‘ β€β ββ’ β₯β= + =β’ β₯+ + + +β’ β₯
β£ β¦
β« β«
i.e. 1
0
1 1 3 1 kk ln(3x 1) ln(x 1) 03 2 2 2β‘ + β€β β β ββ β + + + =β β β ββ’ β₯β β β β β£ β¦
[ ]1 1 3 1 kk ln 4 ln 2 0 0 03 2 2 2β‘ + β€β β β ββ β + β + =β β β ββ’ β₯β β β β β£ β¦
i.e. 1 k 1 1 3ln 2 k (2ln 2)2 3 2 2+β β β β= +β β β β
β β β β since ln 4 = 2ln 2 2ln 2=
i.e. 1 k 1 k2 2 3+ = + from which, 1 1 kk
2 3 2β = β
and 1 k6 2= from which, 1k
3=
9. The velocity constant k of a chemical reaction is given by: 1kt dx(3 0.4x)(2 0.6x)
β β= β ββ ββ β β«
where x = 0 when t = 0. Show that: kt = 2(3 0.4x)ln3(2 0.6x)
β§ β«ββ¨ β¬ββ© β
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)
β + ββ‘ + =
β β β β β β
Hence, 1 = A(2 β 0.6x) + B(3 β 0.4x)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 367
Let x = 3 7.50.4
= , 1 = A(2 β 4.5) i.e. A = 1 0.42.5
= ββ
Let x = 2 1 1030.6 3 3
= = , 1 = 4B 33
β βββ ββ β
i.e. B = 1 3 0.65 53
= =
Thus, 1 0.4 0.6kt dx dx(3 0.4x)(2 0.6x) (3 0.4x) (2 0.6x)
β β β ββ= = +β β β ββ β β ββ β β β β« β«
= 0.4 0.6ln(3 0.4x) ln(2 0.6x)0.4 0.6β
β + ββ β
+ c
i.e. kt = ln(3 β 0.4x) β ln(2 β 0.6x) + c
t = 0 when x = 0, hence, 0 = ln 3 β ln 2 + c
i.e. c = ln 2 β ln 3
Hence, kt = ln(3 β 0.4x) β ln(2 β 0.6x) + ln 2 β ln 3
i.e. kt = 2(3 0.4x)ln3(2 0.6x)
β§ β«ββ¨ β¬ββ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 368
EXERCISE 164 Page 410
2. Determine 2
3
5x 30x 44 dx(x 2)β +ββ«
Let 2 2
3 2 3 3
5x 30x 44 A B C A(x 2) B(x 2) C(x 2) (x 2) (x 2) (x 2) (x 2)β + β + β +
β‘ + + =β β β β β
Hence, 2 25x 30x 44 A(x 2) B(x 2) Cβ + = β + β +
Let x = 2, 4 = C
Equating 2x coefficients, 5 = A
Equating x coefficients, -30 = -4A + B i.e. B = -30 + 20 = -10
Thus, 2
3 2 3
5x 30x 44 5 10 4dx dx(x 2) (x 2) (x 2) (x 2)
β ββ += β +β ββ β β ββ β
β« β«
= 5 ln(x β 2) + 2
10 2 c(x 2) (x 2)
β +β β
using the algebraic substitution
u = x β 2 in the latter two integrals
4. Evaluate 27
26
18 21x x dx(x 5)(x 2)
+ ββ +β« correct to 4 significant figures.
Let ( ) ( )22
2 2 2
A(x 2) B(x 5) x 2 C x 518 21x x A B C(x 5)(x 2) (x 5) (x 2) (x 2) (x 5)(x 2)
+ + β + + β+ ββ‘ + + =
β + β + + β +
Hence, 2 218 21x x A(x 2) B(x 5)(x 2) C(x 5)+ β = + + β + + β
Let x = 5, 98 = 49A i.e. A = 2
Let x = -2, -28 = -7C i.e. C = 4
Equating 2x coefficients, -1 = A + B i.e. B = -3
Thus, 27 7
2 26 6
18 21x x 2 3 4dx dx(x 5)(x 2) (x 5) (x 2) (x 2)
β β+ β= β +β ββ + β + +β β
β« β«
= 7
6
42 ln(x 5) 3ln(x 2)(x 2)
β‘ β€β β + ββ’ β₯+β£ β¦
= 4 42ln 2 3ln 9 2ln1 3ln89 8
β β β ββ β β β ββ β β ββ β β β
= 1.089
5. Show that 21
20
4t 9t 8 dt(t 2)(t 1)
β β+ +β β+ +β β
β« = 2.527, correct to 4 significant figures.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 369
Let ( ) ( )22
2 2 2
A(t 1) B(t 2) t 1 C t 24t 9t 8 A B C(t 2)(t 1) (t 2) (t 1) (t 1) (t 2)(t 1)
+ + + + + ++ +β‘ + + =
+ + + + + + +
Hence, 2 24t 9t 8 A(t 1) B(t 2)(t 1) C(t 2)+ + = + + + + + +
Let x = -2, 6 = A
Let x = -1, 3 = C
Equating 2x coefficients, 4 = A + B i.e. B = -2
Thus, 21 1
2 20 0
4t 9t 8 6 2 3dt dt(t 2)(t 1) (t 2) (t 1) (t 1)
β β β β+ += β +β β β β+ + + + +β β β β
β« β«
= 1
0
36ln(t 2) 2ln(t 1)(t 1)
β‘ β€+ β + ββ’ β₯+β£ β¦
= 3 36ln 3 2ln 2 6ln 2 2ln12 1
β β β ββ β β β ββ β β ββ β β β
= (3.70538) β (1.15888) = 2.546
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 370
EXERCISE 165 Page 412
1. Determine ( )( )
2
2
x x 13 dxx 7 x 2
β β+ ββ«
Let ( )( ) ( ) ( )
( )( )( )
22
2 2 2
(Ax B)(x 2) C x 7x x 13 Ax B Cx 2x 7 x 2 x 7 x 7 x 2
+ β + +β β +β‘ + =
β+ β + + β
Hence, ( )2 2x x 13 (Ax B)(x 2) C x 7β β = + β + +
Let x = 2, -11 = 11C i.e. C = -1
Equating 2x coefficients, 1 = A + C i.e. A = 2
Equating x coefficients, -1 = -2A + B i.e. B = 3
Hence, ( )( ) ( ) ( ) ( )
2
2 2 2 2
x x 13 2x 3 1 2x 3 1dx dx dx(x 2) (x 2)x 7 x 2 x 7 x 7 x 7
β β β ββ β +β β β β= β = + ββ β β ββ β+ β + + +β β β β
β« β« β«
= ( ) ( )22 2
2x 3 1dx dx dx(x 2)x 7 x 7
+ ββ+ +
β« β« β«
= ( )2 13 xln x 7 tan ln(x 2) c7 7
β+ + β β +
2. Evaluate ( )
6
25
6x 5 dx(x 4) x 3
ββ +β« correct to 4 significant figures.
Let ( )( ) ( )
( )( )( )
2
2 2 2
A x 3 (Bx C)(x 4)6x 5 A Bx C(x 4)x 4 x 3 x 3 x 4 x 3
+ + + ββ +β‘ + =
ββ + + β +
Hence, ( )26x 5 A x 3 (Bx C)(x 4)β = + + + β
Let x = 4, 19 = 19A i.e. A = 1
Equating 2x coefficients, 0 = A + B i.e. B = -1
Equating x coefficients, 6 = -4B + C i.e. C = 2
Thus, ( ) ( ) ( ) ( )
6 6 6
22 2 25 5 5 2
6x 5 1 2 x 1 2 xdx dx dx(x 4) (x 4)(x 4) x 3 x 3 x 3x 3
β ββ ββ β β ββ β= + = + ββ ββ ββ ββ + + +β β+β β β β β« β« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 371
= ( )6
1 2
5
2 x 1ln(x 4) tan ln x 323 3
ββ‘ β€β + β +β’ β₯β£ β¦
= 1 12 6 1 2 5 1ln 2 tan ln 39 ln1 tan ln 282 23 3 3 3
β ββ β β β+ β β + ββ β β ββ β β β
= (0.35065) β (-0.23736)
= 0.5880
5. Show that ( )
2 32
2 21
2 6 2 d1
β β+ ΞΈ+ ΞΈ β ΞΈβ β ΞΈβ βΞΈ ΞΈ +β β
β« = 1.606, correct to 4 significant figures.
Let ( ) ( )
( ) ( ) ( )( )
2 2 22 3
22 2 2 2 2
A 1 B 1 C D2 6 2 A B C D1 1 1
ΞΈ ΞΈ + + ΞΈ + + ΞΈ+ ΞΈ+ ΞΈ+ ΞΈ β ΞΈ ΞΈ+β‘ + + =ΞΈ ΞΈΞΈ ΞΈ + ΞΈ + ΞΈ ΞΈ +
Hence, ( ) ( ) ( )2 3 2 2 22 6 2 A 1 B 1 C D+ ΞΈ+ ΞΈ β ΞΈ = ΞΈ ΞΈ + + ΞΈ + + ΞΈ+ ΞΈ
Let ΞΈ = 0, 2 = B
Equating 3ΞΈ coefficients, -2 = A + C (1)
Equating 2ΞΈ coefficients, 6 = B + D i.e. D = 4
Equating ΞΈ coefficients, 1 = A
From equation (1), C = -3
Hence, ( ) ( )
2 32 2 2
2 2 2 22 2 21 1 1
2 6 2 1 2 4 3 1 2 4 3d d d1 11 1
β β β β+ ΞΈ+ ΞΈ β ΞΈ β ΞΈ ΞΈβ ββ β β βΞΈ = + + ΞΈ = + + β ΞΈβ ββ β β βΞΈ ΞΈ ΞΈ ΞΈ ΞΈ + ΞΈ +ΞΈ ΞΈ + ΞΈ + β β β β β β β« β« β«
= ( )2
1 2
1
2 3ln 4 tan ln 12
ββ‘ β€ΞΈβ + ΞΈβ ΞΈ +β’ β₯ΞΈβ£ β¦
= ( ) ( )1 1ln 2 1 4 tan 2 1.5ln 5 ln1 2 4 tan 1 1.5ln 2β ββ + β β β + β
= (1.70759) β (0.10187)
= 1.606
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 372
CHAPTER 42 THE t = tan ΞΈ/2 SUBSTITUTION EXERCISE 166 Page 415
2. Determine dx1 cos x sin xβ +β«
2 2
2 2 2 2
2 2 2
2dt 2dtdx 2dt 11 t 1 t dx
1 t 2t (1 t ) (1 t ) 2t1 cos x sin x 2t 2t t(t 1)11 t 1 t 1 t
+ += = = =β + β β +β + + +β ++ + +
β« β« β« β« β«
Let 1 A B A(t 1) Btt(t 1) t t 1 t(t 1)
+ += + =
+ + +
Hence, 1 = A(t + 1) + Bt
Let t = 0, 1 = A
Let t = -1, 1 = -B i.e. B = -1
Thus, dx 1 1 1dx dx1 cos x sin x t(t 1) t t 1
β β= = ββ ββ + + +β β β« β« β«
= ln t β ln(t + 1) + c = tln c1 tβ§ β«+β¨ β¬+β© β
=
xtan2ln cx1 tan
2
β§ β«βͺ βͺ
+β¨ β¬βͺ βͺ+β© β
4. Determine dx3sin x 4cos xββ«
2 2
2 22
22 2
2dt 2dtdx 2dt1 t 1 t
6t 4 4t3sin x 4cos x 4t 6t 42t 1 t3 41 t1 t 1 t
+ += = =β +β + ββ βββ β ββ β β β ++ +β β β β
β« β« β« β«
= 2
dt dt2t 3t 2 (2t 1)(t 2)
=+ β β +β« β«
Let 1 A B A(t 2) B(2t 1)(2t 1)(t 2) (2t 1) (t 2) (2t 1)(t 2)
+ + β= + =
β + β + β +
Hence, 1 = A(t + 2) + B(2t - 1)
Let t = 0.5, 1 = 2.5A i.e. A = 1 22.5 5
=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 373
Let t = -2, 1 = -5B i.e. B = 15
β
Thus, dx3sin x 4cos xββ« =
2 1dt 1 15 5 dt ln(2t 1) ln(t 1) c
(2t 1)(t 2) (2t 1) (t 2) 5 5= β = β β + +
β + β +β« β«
= 1 2t 1ln c5 t 2
ββ§ β«+β¨ β¬+β© β =
x2tan 11 2ln cx5 tan 22
β§ β«ββͺ βͺ+β¨ β¬
βͺ βͺ+β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 374
EXERCISE 167 Page 416
1. Determine d5 4sin
ΞΈ+ ΞΈβ«
( )2 2
222
2 2
2dt 2dtd 2dt 2dt1 t 1 t
2t 85 4sin 5t 8t 55 1 t 4(2t)5 4 5 t t 11 t 51 t
ΞΈ + += = = =+ ΞΈ + +β β β β+ ++ + +β β β β+β β β β +
β« β« β« β« β«
= 1 12 2
2 4t2 dt 2 5t 45 5tan c tan c3 35 3 34 3t 5 55 5
β β
β β+β β +β β= + = +β β β ββ β β β β β β β+ +β β β β β β β β β β
β«
= 15 tan 42 2tan c
3 3β
ΞΈβ β+β β+β β
β ββ β
3. Determine dp3 4sin p 2cos pβ +β«
( ) ( )2 2
2 2 2
2 2 2
2dt 2dtdp 1 t 1 t
3 4sin p 2cos p 2t 1 t 3 1 t 4(2t) 2 1 t3 4 21 t 1 t 1 t
+ += =β + β ββ + β + ββ ββ +β β β β+ +β β +β β
β« β« β«
= ( ) ( )22 2 2 2 2
2dt 2dt 2dt 2dt3 3t 8t 2 2t t 8t 5 (t 4) 11 t 4 11
= = =+ β + β β + β β β β
β« β« β« β«
= 2 t 4 11 1 t 4 11ln c ln2 11 t 4 11 11 t 4 11
β§ β« β§ β«β β β ββͺ βͺ βͺ βͺ+ =β¨ β¬ β¨ β¬β + β +βͺ βͺ βͺ βͺβ© β β© β
using partial fractions (see
Problem 9, page 411 of textbook)
=
ptan 4 111 2ln cp11 tan 4 112
β§ β«β ββͺ βͺ+β¨ β¬
βͺ βͺβ +β© β
5. Show that
t2 tandt 1 2ln t1 3cos t 2 2 2 tan2
β§ β«+βͺ βͺ= β¨ β¬+ βͺ βͺβ
β© β
β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 375
2 2
2 2 2 2 22
22
2dt 2dtdt 2dt 2dt1 t 1 t
1(1 t ) 3(1 t )1 3cos t 1 t 3 3t 4 2t1 t1 31 t1 t
+ += = = =+ + β+ + + β ββ ββ
+ β β ++β β
β« β« β« β« β«
= ( )22 2
dt dt2 t 2 t
=β β
β« β« = 1 2 tln c2 2 2 t
β§ β«+βͺ βͺ+β¨ β¬ββͺ βͺβ© β
using partial
fractions (see problem 11, page 411 of textbook)
=
t2 tan1 2ln ct2 2 2 tan2
β§ β«+βͺ βͺ+β¨ β¬
βͺ βͺββ© β
7. Show that / 2
0
d2 cos 3 3
Ο ΞΈ Ο=
+ ΞΈβ«
2 2
2 2 2 2 22
22
2dt 2dtd 2dt 2dt1 t 1 t
2(1 t ) (1 t )2 cos 2 2t 1 t t 31 t21 t1 t
ΞΈ + += = = =+ + β+ ΞΈ + + β +β ββ
+ β β ++β β
β« β« β« β« β«
= ( )
1 12
2
tandt 2 t 2 22 tan tan3 3 3 3t 3
β β
ΞΈβ ββ β
= = β β+ β β
β β
β«
Hence,
/ 2
/ 2 1 1 1
0
0
tan tand 2 2 tan 02 4tan tan tan2 cos 3 3 3 3 3
Ο
Ο β β β
ΞΈ β‘ Ο β€β‘ β€ β ββ’ β₯β ββ’ β₯ΞΈ β β= = ββ’ β₯β ββ’ β₯ β β+ ΞΈ β β β’ β₯β ββ’ β₯β’ β₯β£ β¦ β β β£ β¦
β«
= 12 1 2tan 063 3 3
β Οβ‘ β€ β ββ = =β ββ’ β₯ β β β£ β¦ 3 3Ο
(Using a calculator on degrees, 1 1tan 30 rad63
β Ο= Β° = )
Note that tan 30Β° i.e. rad6Οβ β
β ββ β
= 13
(since AD = 3 by Pythagoras)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 376
CHAPTER 43 INTEGRATION BY PARTS EXERCISE 168 Page 420
2. Determine 3x
4x dxeβ«
3x
3x
4x dx 4xe dxe
β=β« β« Let u = 4x then dudx
= 4 and du = 4 dx
and dv = 3xe dxβ from which, v = 3x 3x1e dx e3
β β= ββ«
Hence, 3x 3x 3x1 14xe dx (4x) e e 4dx3 3
β β ββ β β β= β β ββ β β ββ β β β β« β«
= 3x 3x 3x 3x4 4 4 4 1xe e dx xe e c3 3 3 3 3
β β β ββ ββ + = β + β +β ββ β β«
= 3x 3x4 4xe e c3 9
β ββ β + = 3x4 1e x c3 3
β β ββ + +β ββ β
4. Determine 5 cos 2 dΞΈ ΞΈ ΞΈβ«
Let u = 5ΞΈ then dudΞΈ
= 5 and du = 5 dΞΈ
and dv = cos 2ΞΈ dΞΈ from which, v = 1cos 2 d sin 22
ΞΈ ΞΈ = ΞΈβ«
( ) 1 15 cos 2 d 5 sin 2 sin 2 5d2 2
β β β βΞΈ ΞΈ ΞΈ = ΞΈ ΞΈ β ΞΈ ΞΈβ β β ββ β β β β« β«
= 5 5 5 5 1sin 2 sin 2 d sin 2 cos 2 c2 2 2 2 2
β βΞΈ ΞΈβ ΞΈ ΞΈ = ΞΈ ΞΈβ β ΞΈ +β ββ β β«
= 5 5sin 2 cos 2 c2 4ΞΈ ΞΈ+ ΞΈ+ = 5 1sin 2 cos 2 c
2 2β βΞΈ ΞΈ + ΞΈ +β ββ β
6. Evaluate 2 x
02x e dxβ« correct to 4 significant figures.
Let u = 2x then dudx
= 2 and du = 2 dx
and dv = xe dx from which, v = x xe dx e=β«
( ) ( )x x x x x2x e dx (2x) e e 2dx 2xe 2e c= β = β +β« β«
Hence, ( ) ( )2 2x x x 2 2 0
002x e dx 2xe 2e 4e 2e 0 2eβ‘ β€= β = β β ββ£ β¦β« = 22e 2+ = 16.78
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 377
8. Evaluate / 2 2
0t cos t dt
Ο
β« correct to 4 significant figures.
Let u = 2t then dudt
= 2t and du = 2t dt
and dv = cos t dt from which, v = cos t dt sin t=β«
Hence, ( )( ) ( )2 2t cos t dt t sin t sin t 2t dt= ββ« β«
= 2t sin t 2 t sin t dtβ‘ β€β β£ β¦β« (1)
For t sin t dtβ« , let u = t then dudt
= 1 and du = dt
and dv = sin t dt from which, v = sin t dt cos t= ββ«
Hence, t sin t dtβ« = (t)(- cos t) - ( cos t)dtββ« = -t cos t + sin t
Substituting in (1) gives: [ ]2 2t cos t dt t sin t 2 t cos t sin t c= β β + +β«
= 2t sin t 2t cos t 2sin t c+ β +
Thus, / 2 / 22 2
00t cos t dt t sin t 2t cos t 2sin t
Ο Οβ‘ β€= + ββ£ β¦β«
= [ ]2 2sin cos 2sin 0 0 0
2 2 2 2 2β‘ β€Ο Ο Ο Ο Οβ β + β β + ββ’ β₯β ββ β β’ β₯β£ β¦
= 2 2
0 2 22 4Ο Οβ β + β = ββ β
β β = 0.4674
9. Evaluate x2 2 2
13x e dxβ« correct to 4 significant figures.
Let u = 23x then dudx
= 6x and du = 6x dx
and dv = x2e dx from which, v =
xx x22 2ee dx 2e1
2
= =β«
Hence, ( )x x x
2 22 2 23x e dx 3x 2e 2e 6x dxβ β β β
= ββ β β ββ β β β
β« β«
= x x
2 2 26x e 12 x e dxβ‘ β€
β β’ β₯β£ β¦β« (1)
Forx2x e dxβ« , let u = x then du
dx = 1 and du = dx
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 378
and dv = x2e dx from which, v =
x x2 2e dx 2e=β«
Hence, x2x e dxβ« = ( )
x x x x2 2 2 2x 2e 2e dx 2x e 4e
β ββ = ββ β
β β β«
Substituting in (1) gives: x x x x
2 22 2 2 23x e dx 6x e 12 2xe 4e cβ‘ β€
= β β +β’ β₯β£ β¦
β«
= x x x
2 2 2 26x e 24x e 48e cβ + +
Thus, 2x x x x2 2 22 2 2 2
11
3x e dx 6x e 24x e 48eβ‘ β€
= β +β’ β₯β£ β¦
β«
= ( )1 1 1
1 1 1 2 2 224e 48e 48e 6e 24e 48eβ β
β + β β +β ββ β
= ( )1
1 224e 30eβ β
β β ββ β
= 15.78
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 379
EXERCISE 169 Page 422 1. Determine 22x ln x dxβ«
Let u = ln x then du 1dx x
= and du = 1 dxx
and dv = 22x dx from which, v = 2 322x dx x3
=β«
Hence, ( )2 3 32 2 12x ln x dx ln x x x dx3 3 x
β β β β= ββ β β ββ β β β β« β«
= 3
3 2 3 3 32 2 2 2 x 2 2x ln x x dx x ln x c x ln x x c3 3 3 3 3 3 9
β ββ = β + = β +β β
β β β«
= 32 1x ln x c3 3
β ββ +β ββ β
3. Determine 2x sin 3x dxβ«
Let u = 2x then dudx
= 2x and du = 2x dx
and dv = sin 3x dx from which, v = 1sin 3x dx cos3x3
= ββ«
Hence, ( )2 2 1 1x sin 3x dx x cos3x cos3x 2x dx3 3
β β β β= β β ββ β β ββ β β β β« β«
= 21 2x cos3x x cos3x dx3 3
β‘ β€β + β£ β¦β« (1)
For x cos3x dxβ« , let u = x then dudx
= 1 and du = dx
and dv = cos 3x dx from which, v = 1cos3x dx sin 3x3
=β«
Hence, x cos3x dxβ« = ( ) 1 1x sin 3x sin 3x dx3 3
β β β βββ β β ββ β β β β« = 1 1x sin 3x cos3x
3 9+
Substituting in (1) gives: 2 21 2 1 1x sin 3x dx x cos3x x sin 3x cos3x c3 3 3 9
β‘ β€= β + + +β’ β₯β£ β¦β«
= 21 2 2x cos 3x xsin 3x cos 3x c3 9 27
β + + +
or ( )2cos 3x 22 9x xsin 3x c27 9
β + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 380
5. Determine 22 sec dΞΈ ΞΈ ΞΈβ«
Let u = 2ΞΈ then du 2d
=ΞΈ
and du = 2 dΞΈ
and dv = 2sec dΞΈ ΞΈ from which, v = 2sec d tanΞΈ ΞΈ = ΞΈβ«
Hence, ( )( ) ( )22 sec d 2 tan tan 2dΞΈ ΞΈ ΞΈ = ΞΈ ΞΈ β ΞΈ ΞΈβ« β«
= 2 tan 2ln(sec ) cΞΈ ΞΈβ ΞΈ + from Problem 9, chapter 39, page 393 of textbook
= [ ]2 tan ln(sec ) cΞΈ ΞΈ β ΞΈ +
7. Evaluate 1 3x
02e sin 2x dxβ« correct to 4 significant figures.
Let u = 3x2e then 3xdu 6edx
= and du = 3x6e dx
and dv = sin 2x dx from which, v = 1sin 2x dx cos 2x2
= ββ«
Hence, ( )3x 3x 3x1 12e sin 2x dx 2e cos 2x cos 2x 6e dx2 2
β β β β= β β ββ β β ββ β β β β« β«
= 3x 3xe cos 2x 3 e cos 2x dxβ‘ β€β + β£ β¦β« (1)
For 3xe cos 2x dxβ« , let u = 3xe then 3xdu 3edx
= and du = 3x3e dx
and dv = cos 2x dx from which, v = 1cos 2x dx sin 2x2
=β«
Hence, ( )3x 3x 3x1 1e cos 2x dx e sin 2x sin 2x 3e dx2 2
β β β β= ββ β β ββ β β β β« β«
= 3x 3x1 3e sin 2x e sin 2x dx2 2
β‘ β€β β£ β¦β«
Substituting in (1) gives: 3x 3x 3x 3x1 32e sin 2x dx e cos 2x 3 e sin 2x e sin 2x dx2 2β‘ β€= β + ββ’ β₯β£ β¦β« β«
= 3x 3x 3x3 9e cos 2x e sin 2x e sin 2x dx2 2
β + β β«
Hence, 3x 3x 3x9 32 e sin 2x dx e cos 2x e sin 2x2 2
β β+ = β +β ββ β β« by combining the far left and far
right-hand integrals
i.e. 3x 3x 3x13 3e sin 2x dx e cos 2x e sin 2x2 2
β β = β +β ββ β β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 381
and 3x 3x 3x2 3e sin 2x dx e cos 2x e sin 2x13 2
β β= β +β ββ β β«
Thus, 3x 3x 3x4 32e sin 2x dx e cos 2x e sin 2x13 2
β β= β +β ββ β β«
and 1
1 3x 3x 3x
00
4 32e sin 2x dx e cos 2x e sin 2x13 2β‘ β€β β= β +β ββ’ β₯β β β£ β¦
β«
= 3 34 3 4 3e cos 2 e sin 2 cos 0 sin 013 2 13 2β‘ β€ β‘ β€β β β ββ + β β +β β β ββ’ β₯ β’ β₯β β β β β£ β¦ β£ β¦
= ( )4 48.3585 27.3956 ( 1)13 13β‘ β€ β‘ β€+ β ββ’ β₯ β’ β₯β£ β¦ β£ β¦
= (11.0013) + (0.30769) = 11.31
9. Evaluate 4 3
1x ln x dxβ« correct to 4 significant figures.
Let u = ln x then du 1dx x
= and du = 1 dxx
and dv = 3x dx from which, v = 3 52 22x dx x
5=β«
Hence, ( )5 5
3 2 22 2 1x ln x dx ln x x x dx5 5 x
β β β β= ββ β β β
β β β β β« β«
= 3
5 22 2x ln x x dx5 5
β β«
= 5
5 22 2 2x ln x x c5 5 5
β ββ +β β
β β
Thus, 4
4 3 5 5
11
2 4x ln x dx x ln x x5 25β‘ β€= ββ’ β₯β£ β¦β«
= 5 5 5 52 4 2 44 ln 4 4 1 ln1 15 25 5 25
β β β ββ β ββ β β ββ β β β
= ( ) ( )2 4 432 ln 4 32 05 25 25
β β β ββ β ββ β β ββ β β β
= 12.78
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 382
10. In determining a Fourier series to represent f(x) = x in the range -Ο to Ο, Fourier coefficients are
given by: n1a x cos nx dx
Ο
βΟ=Ο β« and nb x sin nx dx
Ο
βΟ= β«
where n is a positive integer.
Show by using integration by parts that na = 0 and n2b cos nn
= β Ο
For n1a x cos nx dx
Ο
βΟ=Ο β« , let u = x then du 1
dx= and du = dx
and dv = cos nx dx from which, v = 1cos nx dx sin nxn
=β«
Hence, ( ) 2
1 1 x 1x cos nx dx x sin nx sin nx dx sin nx cos nxn n n n
β β β β= β = +β β β ββ β β β β« β«
Then n 2
1 1 x 1a x cos nx dx sin nx cos nxn n
ΟΟ
βΟβΟ
β‘ β€= = +β’ β₯Ο Ο β£ β¦β«
= 2 2
1 1 1sin n cos n sin( n ) cos( n )n n n n
β‘ Ο Ο β€β β β βΟ + Ο β β β Ο + β Οβ β β ββ’ β₯Ο β β β β β£ β¦
sin nΟ = sin(-nΟ) = 0 for all values of n.
Hence, n 2 2
1 1 1a 0 cos n 0 cos( n )n n
β‘ β€β β β β= + Ο β + β Οβ β β ββ’ β₯Ο β β β β β£ β¦
= [ ]2
1 cos n cos( n )n
Οβ β ΟΟ
= 0 since cos nΟ = cos(-nΟ) for all values of n.
For nb x sin nx dxΟ
βΟ= β« , let u = x then du 1
dx= and du = dx
and dv = sin nx dx from which, v = 1sin nx dx cos nxn
= ββ«
Hence, ( ) 2
1 1 x 1x sin nx dx x cos nx cos nx dx cos nx sin nxn n n n
β β β β= β β β = β +β β β ββ β β β β« β«
Then n 2
1 1 x 1b x sin nx dx cos nx sin nxn n
ΟΟ
βΟβΟ
β‘ β€= = β +β’ β₯Ο Ο β£ β¦β«
= 2 2
1 1 1cos n sin n cos( n ) sin( n )n n n n
β‘ Ο Ο β€β β β ββ Ο+ Ο β β Ο + β Οβ β β ββ’ β₯Ο β β β β β£ β¦
sin nΟ = sin(-nΟ) = 0 for all values of n.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 383
Hence, n1b cos n 0 cos( n ) 0
n nβ‘ Ο Ο β€β β β β= β Ο+ β β Ο +β β β ββ’ β₯Ο β β β β β£ β¦
= [ ]cos n cos( n )n
βΟΟ+ β Ο
Ο
cos nΟ = cos(-nΟ) hence cos nΟ + cos(-nΟ) β‘ 2 cos nΟ
Thus, ( )n1b 2cos nn
= β Ο = 2 cosnn
β Ο
11. The equation C = 1 0.4
0e cos1.2 dβ ΞΈ ΞΈ ΞΈβ« and S =
1 0.4
0e sin1.2 dβ ΞΈ ΞΈ ΞΈβ«
are involved in the study of damped oscillations. Determine the value of C and S. From problem 9, page 421 of textbook,
( )ax
ax2 2
ee cos bx dx bsin bx a cos bx ca b
= + ++β«
Thus, ( )0.4
0.42 2
ee cos1.2 d 1.2sin1.2 0.4cos1.2( 0.4) (1.2)
β ΞΈβ ΞΈ ΞΈ ΞΈ = ΞΈβ ΞΈ
β +β«
and C = ( )10.41 0.4
00
ee cos1.2 d 1.2sin1.2 0.4cos1.21.6
β ΞΈβ ΞΈ β‘ β€
ΞΈ ΞΈ = ΞΈβ ΞΈβ’ β₯β£ β¦
β«
= ( )0.4e 11.2sin1.2 0.4cos1.2 (1.2sin 0 0.4cos 0
1.6 1.6
ββ‘ β€ β‘ β€β β ββ’ β₯ β’ β₯β£ β¦β£ β¦
= (0.40785) β (-0.25)
i.e. C = 0.66
From equation (2) on page 422 (obtained in a similar way to that for question 7 above),
( )ax
ax2 2
ee sin bx dx a sin bx bcos bx ca b
= β ++β«
Thus, 0.4
0.4 ee sin1.2 d ( 0.4sin1.2 1.2cos1.2 )1.6
β ΞΈβ ΞΈ ΞΈ ΞΈ = β ΞΈβ ΞΈβ«
and S = 1 0.4
0e sin1.2 dβ ΞΈ ΞΈ ΞΈβ« ( )
10.4
0
e 0.4sin1.2 1.2cos1.21.6
β ΞΈβ‘ β€= β ΞΈβ ΞΈβ’ β₯β£ β¦
= ( )0.4e 10.4sin1.2 1.2cos1.2 ( 0.4sin 0 1.2cos 0
1.6 1.6
ββ‘ β€ β‘ β€β β β β ββ’ β₯ β’ β₯β£ β¦β£ β¦
= (-0.33836) β (-0.75)
i.e. S = 0.41
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 384
CHAPTER 44 REDUCTION FORMULAE EXERCISE 170 Page 425 2. Determine 3 2tt e dtβ« using a reduction formula.
n 2tt e dtβ« Let u = nt then n 1du nt
dtβ= and du = n 1nt dtβ
and dv = 2te dt from which, v = 2t 2t1e dt e2
=β«
Thus, ( )n 2t n 2t 2t n 11 1t e dt t e e nt dt2 2
ββ β β β= ββ β β ββ β β β β« β«
= n 2t n 1 2t1 nt e t e dt2 2
ββ β«
i.e. n 2tn n 1
1 nI t e I2 2 β= β
Hence, 3 2tt e dtβ« = 3 2t3 2
1 3I t e I2 2
= β
2 2t2 1
1 2I t e I2 2
= β
2t1 0
1 1I t e I2 2
= β
2t 2t0
1I e dt e2
= =β«
Thus, 3 2tt e dtβ« = 3 2t 2 2t1
1 3 1t e t e I2 2 2
β ββ ββ ββ β
= 3 2t 2 2t 2t 2t1 3 3 1 1 1t e t e t e e c2 4 2 2 2 2
β ββ ββ + β +β ββ ββ β β β
= 3 2t 2 2t 2t 2t1 3 3 3t e t e t e e c2 4 4 8
β + β +
= 2t 3 21 3 3 3e t t t c2 4 4 8
β ββ + β +β ββ β
3. Use the result of Problem 2 to evaluate 1 3 2t
05t e dtβ« , correct to 3 decimal places.
1
1 3 2t 2t 3 2 2 0
00
1 3 3 3 1 3 3 3 35t e dt 5e t t t 5e 5e2 4 4 8 2 4 4 8 8
β‘ β€β β β β β β= β + β = β + β β ββ β β β β ββ’ β₯β β β β β β β£ β¦β« = 6.493
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 385
EXERCISE 171 Page 427 3. Use a reduction formula to determine 5x sin x dxβ«
5 5 4
5 3x sin x dx I x cos x 5x sin x 5(4)I= = β + ββ« from equation (3), page 426 of textbook
3 23 1I x cos x 3x sin x 3(2)I= β + β
1 01I x cos x 1x sin x (1)(0)= β + β
i.e. 5x sin x dxβ« = ( )5 4 3 21x cos x 5x sin x 20 x cos x 3x sin x 6Iβ + β β + β
= ( )5 4 3 2x cos x 5x sin x 20x cos x 60x sin x 120 x cos x sin x cβ + + β + β + +
= 5 4 3 2x cos x 5x sin x 20x cos x 60x sin x 120xcos x 120sin x cβ + + β β + +
4. Evaluate 5
0x sin x dx
Ο
β« , correct to 2 decimal places.
5 5 4 3 2
00x sin x dx x cos x 5x sin x 20x cos x 60x sin x 120x cos x 120sin x
Ο Οβ‘ β€= β + + β β +β£ β¦β«
= ( )5 4 3 2cos 5 sin 20 cos 60 sin 120 cos 120sinβΟ Ο+ Ο Ο+ Ο Οβ Ο Οβ Ο Ο+ Ο
( )0 0 0 0 0 120sin 0β + + β β +
= ( ) ( )5 320 120 0Ο β Ο + Ο β
= 62.89
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 386
EXERCISE 172 Page 430
2. Evaluate 3
03sin x dx
Ο
β« , using a reduction formula.
3 2
3 11 2sin x dx I sin x cos x I3 3
= = β +β« from equation (4), page 428 of textbook
01
1I sin x cos x 0 cos x1
= β + = β
Hence, ( )3 21 2sin x dx sin x cos x cos x3 3
= β + ββ«
3 2 21 23sin x dx 3 sin x cos x cos x sin x cos x 2cos x3 3
β β= β β = β ββ ββ β β«
Thus, 3
03sin x dx
Ο
β« = 2
0sin x cos x 2cos x
Οβ‘ β€β ββ£ β¦
= ( ) ( )2 2sin cos 2cos sin 0cos 0 2cos 0β Ο Οβ Ο β β β
= (- 0 β 2(-1)) β (0 β 2)
= 2 - - 2 = 4
4. Determine 6cos x dxβ« , using a reduction formula,
6 5
6 41 5cos x dx I cos x sin x I6 6
= = +β« from equation (5), page 429 of textbook
34 2
1 3I cos x sin x I4 4
= +
12 0
1 1I cos x sin x I2 2
= +
00I cos x dx x= =β«
Thus, 6cos x dxβ« = 5 32
1 5 1 3cos x sin x cos x sin x I6 6 4 4
β β+ +β ββ β
= 5 31 5 15 1 1cos x sin x cos x sin x cos x sin x x6 24 24 2 2
β β+ + +β ββ β
= 5 31 5 15 15cos x sin x cos x sin x cos x sin x x6 24 48 48
+ + +
= 5 31 5 5 5cos xsin x cos xsin x cos xsin x x6 24 16 16
+ + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 387
5. Evaluate / 2 7
0cos x dx
Ο
β«
Using Wallisβs formula, equation (6), page 430 of textbook
/ 2 7
7 50
6cos x dx I I7
Ο= =β«
5 34I I5
=
3 12I I3
=
[ ]/ 2 / 21
1 00I cos x dx sin x sin sin 0 1
2Ο Ο Ο
= = = β =β«
Hence, ( )/ 2 7
0
6 4 2cos x dx 17 5 3
Ο β ββ β= β ββ ββ β β β β« = 16
35
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 388
EXERCISE 173 Page 432
1. Evaluate / 2 2 5
0cos x sin x dx
Ο
β«
( )/ 2 / 22 5 2 5
0 0cos x sin x dx 1 sin x sin x dx
Ο Ο= ββ« β«
= ( )/ 2 5 75 70
sin x sin x dx I IΟ
β = ββ«
Wallisβs formula states: / 2 n
n n 20
n 1sin x dx I In
Ο
β
β= =β« from problem 10, page 429 of textbook
[ ]/ 2 / 21
5 3 3 1 1 00
4 2I I I I I sin x dx cos x [0 1] 15 3
Ο Ο= = = = β = β β =β«
Hence, ( )54 2I 15 3
β ββ β= β ββ ββ β β β
7 5 5 3 3 1 16 4 2I I I I I I I 17 5 3
= = = =
Hence, ( )76 4 2I 17 5 3
β ββ ββ β= β ββ ββ ββ β β β β β
Thus, / 2 2 5
5 70
4 2 6 4 2cos x sin x dx I I5 3 7 5 3
Ο β ββ β β ββ ββ β= β = ββ ββ β β ββ ββ ββ β β β β β β β β β β«
= 4 2 6 4 2 115 3 7 5 3 7
β ββ β β‘ β€ β ββ ββ ββ =β ββ β β ββ ββ ββ’ β₯β β β β β£ β¦ β β β β β β
= 8105
3. Evaluate / 2 5 4
0cos x sin x dx
Ο
β«
( ) ( )/ 2 / 2 / 225 4 5 2 5 2 4
0 0 0cos x sin x dx cos x 1 cos x dx cos x 1 2cos x cos x dx
Ο Ο Ο= β = β +β« β« β«
= ( )/ 2 5 7 9
0cos x 2cos x cos x dx
Οβ +β«
= 5 7 9I 2I Iβ +
5 34I I5
= from equation (6), page 430, [ ]/ 2 / 2
3 1 1 00
2I I I cos x dx sin x 13
Ο Ο= = = =β«
i.e. ( )54 2I 15 3
β ββ β= β ββ ββ β β β
, 7 56 6 4 2I I7 7 5 3
β ββ ββ β= = β ββ ββ ββ β β β β β
and 9 78 8 6 4 2I I9 9 7 5 3
β ββ ββ ββ β= = β ββ ββ ββ ββ β β β β β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 389
Hence, / 2 5 4
0cos x sin x dx
Ο
β« = 5 7 9I 2I Iβ +
= 4 25 3
β ββ ββ ββ ββ β β β
6 4 227 5 3
β ββ ββ ββ β ββ ββ ββ β β β β β
+ 8 6 4 29 7 5 3
β ββ ββ ββ ββ ββ ββ ββ ββ β β β β β β β
= 4 2 6 8 61 25 3 7 9 7
β‘ β€β ββ β β β β ββ ββ +β ββ β β β β ββ ββ’ β₯β β β β β β β β β β β£ β¦
= 8 12 16 8 21 36 16115 7 21 15 21
β +β‘ β€ β‘ β€β + =β’ β₯ β’ β₯β£ β¦ β£ β¦
= 8 115 21
β ββ ββ β
= 8315
5. Show that / 2 3 4
0
2sin cos d35
ΟΞΈ ΞΈ ΞΈ =β«
( ) ( )/ 2 / 2 / 223 4 3 2 3 2 4
0 0 0sin cos d sin 1 sin d sin 1 2sin sin d
Ο Ο ΟΞΈ ΞΈ ΞΈ = ΞΈ β ΞΈ ΞΈ = ΞΈ β ΞΈ+ ΞΈ ΞΈβ« β« β«
= / 2 3 5 7
3 5 70(sin 2sin sin )d I 2I I
ΟΞΈ β ΞΈ+ ΞΈ ΞΈ = β +β«
3 12I I3
= from Problem 10, page 429, and
[ ]/ 2 / 2
1 00I sin d cos cos cos0 (0 1) 1
2Ο Ο Οβ β= ΞΈ ΞΈ = β ΞΈ = β β = β β =β β
β β β«
( )5 3 7 54 4 2 6 6 4 2I I (1), I I 15 5 3 7 7 5 3
β ββ β β ββ ββ β= = = =β ββ β β ββ ββ ββ β β β β β β β β β
Hence, / 2 3 4
3 5 70sin cos d I 2I I
ΟΞΈ ΞΈ ΞΈ = β +β«
= ( ) ( )2 4 2 6 4 22 1 13 5 3 7 5 3
β β β ββ β β ββ ββ ββ +β β β ββ β β ββ ββ ββ β β β β β β β β β β β
= 2 8 24 2 35 56 2413 5 35 3 35
β +β‘ β€ β‘ β€β + =β’ β₯ β’ β₯β£ β¦ β£ β¦
= 2 33 35β ββ ββ β
= 235
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 390
CHAPTER 45 NUMERICAL INTEGRATION EXERCISE 174 Page 435 2. Evaluate
3
12 ln 3x dxβ« using the trapezoidal rule with 8 intervals, giving the answer correct to 3
decimal places.
Since 3
12 ln 3x dxβ« , width of interval = 3 1 0.25
8β
=
x 1 1.25 1.50 1.75 2.0 2.25 2.50 2.75 3.0 2 ln 3x 2.1972 2.6435 3.0082 3.3165 3.5835 3.8191 4.0298 4.2204 4.3944
Hence, using the trapezoidal rule,
3
12 ln 3x dxβ«
( )1(0.25) 2.1972 4.3944 2.6435 3.0082 3.3165 3.5835 3.8191 4.0298 4.22042β‘ β€β + + + + + + + +β’ β₯β£ β¦
= 0.25[27.9168]
= 6.979
4. Evaluate
21.4 x
0e dxββ« using the trapezoidal rule with 7 intervals, giving the answer correct to 3
decimal places.
Since21.4 x
0e dxββ« , width of interval = 1.4 0 0.2
7β
=
x 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
2xeβ 1.0 0.9608 0.8521 0.6977 0.5273 0.3679 0.2369 0.1409 Hence, using the trapezoidal rule,
21.4 x
0e dxββ« ( )1(0.2) 1.0 0.1409 0.9608 0.8521 0.6977 0.5273 0.3679 0.2369
2β‘ β€β + + + + + + +β’ β₯β£ β¦
= (0.2)[4.21315]
= 0.843
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 391
EXERCISE 175 Page 437
2. Evaluate / 2
0
1 d1 sin
ΟΞΈ
+ ΞΈβ« using the mid-ordinate rule with 6 intervals, giving the answer correct
to 3 decimal places.
Since/ 2
0
1 d1 sin
ΟΞΈ
+ ΞΈβ« , width of interval = 0
2 rad or 156 12
Οβ Ο
= Β°
Hence, ordinates occur at 0Β°, 15Β°, 30Β°, 45Β°, 60Β°, 75Β° and 90Β°,
and mid-ordinates occur at 7.5Β°, 22.5Β°, 37.5Β°, 52.5Β°, 67.5Β° and 82.5Β°.
ΞΈ 7.5Β° 22.5Β° 37.5Β° 52.5Β° 67.5Β° 82.5Β° 1
1 sin+ ΞΈ
0.8845 0.7232 0.6216 0.5576 0.5198 0.5021
Hence, using the mid-ordinate rule,
/ 2
0
1 d1 sin
ΟΞΈ
+ ΞΈβ« β [ ]0.8845 0.7232 0.6216 0.5576 0.5198 0.502112Οβ β + + + + +β β
β β
= [ ]3.808812Οβ β
β ββ β
= 0.977
3. Evaluate 3
1
ln x dxxβ« using the mid-ordinate rule with 10 intervals, giving the answer correct to 3
decimal places.
Since3
1
ln x dxxβ« , width of interval = 3 1 0.2
10β
=
Hence, ordinates occur at 1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2, 2.4, 2.6, 2.8, 3.0,
and mid-ordinates occur at 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7 and 2.9.
x 1.1 1.3 1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9
ln xx
0.0866 0.2018 0.2703 0.3121 0.3378 0.3533 0.3621 0.3665 0.3679 0.3671
Hence, using the mid-ordinate rule,
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 392
3
1
ln x dxxβ« β (0.2)[ 0.0866 + 0.2018 + 0.2703 + 0.3121 + 0.3378 + 0.3533 + 0.3621 + 0.3665
+ 0.3679 + 0.3671]
= (0.2)[3.0255]
= 0.605
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 393
EXERCISE 176 Page 439
2. Evaluate 1.6
40
1 d1
ΞΈ+ ΞΈβ« using Simpsonβs rule with 8 intervals, correct to 3 decimal places.
Since1.6
40
1 d1
ΞΈ+ ΞΈβ« , width of interval = 1.6 0 0.2
8β
=
ΞΈ 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
4
11+ ΞΈ
1.0 0.9984 0.9750 0.8853 0.7094 0.5000 0.3254 0.2065 0.1324
Hence, using Simpsonβs rule,
1.6
40
1 d1
ΞΈ+ ΞΈβ«
( ) ( ) ( )1 (0.2) 1.0 0.1324 4 0.9984 0.8853 0.5000 0.2065 2 0.9750 0.7094 0.32543
β + + + + + + + +β‘ β€β£ β¦
= [ ]1 (0.2) 1.1324 10.3608 4.01963
+ +
= [ ]1 (0.2) 15.51283
= 1.034
3. Evaluate 1.0
0.2
sin dΞΈ ΞΈΞΈβ« using Simpsonβs rule with 8 intervals, correct to 3 decimal places.
Since1.0
0.2
sin dΞΈ ΞΈΞΈβ« , width of interval = 1.0 0.2 0.1
8β
= (note that values of ΞΈ are in radians)
ΞΈ 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 sinΞΈΞΈ
0.9933 0.9851 0.9735 0.9589 0.9411 0.9203 0.8967 0.8704 0.8415
Hence, using Simpsonβs rule,
1.0
0.2
sin dΞΈ ΞΈΞΈβ«
( ) ( )( )
0.9933 0.8415 4 0.9851 0.9589 0.9203 0.87041 (0.1)3 2 0.9735 0.9411 0.8967
+ + + + +β‘ β€β β’ β₯
+ + +β’ β₯β£ β¦
= [ ]1 (0.1) 1.8348 14.9388 5.62263
+ +
= [ ]1 (0.1) 22.39623
= 0.747
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 394
5. Evaluate 2/3 x
0e sin 2x dx
Ο
β« using Simpsonβs rule with 10 intervals, correct to 3 decimal places.
Since2/3 x
0e sin 2x dx
Ο
β« , width of interval = 0
3 rad10 30
Οβ Ο
=
x
0 30Ο 2
30Ο 3
30Ο 4
30Ο 5
30Ο 6
30Ο 7
30Ο 8
30Ο 9
30Ο 10
30Ο
2xe sin 2x
0 0.2102 0.4250 0.6488 0.8857 1.1392 1.4114 1.7021 2.0064 2.3119 2.5929
Hence, using Simpsonβs rule,
2/3 x
0e sin 2x dx
Ο
β«( ) ( )
( )0 2.5929 4 0.2102 0.6488 1.1392 1.7021 2.31191
3 30 2 0.4250 0.8857 1.4114 2.0064
+ + + + + +β‘ β€Οβ ββ β’ β₯β β + + + +β β β’ β₯β£ β¦
= [ ]2.5929 24.0488 9.457090Οβ β + +β β
β β
= [ ]36.098790Οβ β
β ββ β
= 1.260
7. Evaluate 6
2
1 dx(2x 1)ββ« using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate
rule, (d) Simpsonβs rule. Give answers correct to 3 decimal places and use 6 intervals
(a) 6
2
1 dx(2x 1)ββ« Let u = 2x β 1, then du 2
dx= and dx = du
2
Thus, ( )1
1 221 1 du 1 1 udx u du u 2x 112 2 2(2x 1) u
2
β= = = = = β
ββ« β« β«
Hence, ( )66
2 2
1 dx 2x 1 11 3(2x 1)
β‘ β€ β‘ β€= β = ββ£ β¦β£ β¦ββ« = 1.585
(b) Width of interval = 6 2 0.58β
=
x 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
( )1
2x 1β
0.5774 0.5000 0.4472 0.4082 0.3780 0.3536 0.3333 0.3162 0.3015
Hence, using the trapezoidal rule,
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 395
6
2
1 dx(2x 1)ββ«
( )1(0.5) 0.5774 0.3015 0.5000 0.4472 0.4082 0.3780 0.3536 0.3333 0.31622β‘ β€β + + + + + + + +β’ β₯β£ β¦
= (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 1β
0.5345 0.4714 0.4264 0.3922 0.3651 0.3430 0.3244 0.3086
Using the mid-ordinate rule,
6
2
1 dx(2x 1)ββ« β (0.5)[ 0.5345 + 0.4714 + 0.4264 + 0.3922 + 0.3651 + 0.3430
+ 0.3244 + 0.3086] = (0.5)[3.1656] = 1.583 (d) Using the table of values from part (b), using Simpsonβs rule,
6
2
1 dx(2x 1)ββ«
( ) ( )( )
0.5774 0.3015 4 0.5000 0.4082 0.3536 0.31621 (0.5)3 2 0.4472 0.3780 0.3333
+ + + + +β‘ β€β β’ β₯
+ + +β’ β₯β£ β¦
= [ ]1 (0.5) 0.8789 6.312 2.3173
+ +
= [ ]1 (0.5) 9.50793
= 1.585
9. Evaluate ( )
0.7
0.1 2
1 dy1 yβ
β« using (a) the trapezoidal rule, (b) the mid-ordinate rule,
(c) Simpsonβs rule. Use 6 intervals and give answers correct to 3 decimal places and use 6
intervals
(a) Since ( )
0.7
0.1 2
1 dy1 yβ
β« then width of interval = 0.7 0.1 0.16β
=
y 0.1 0.2 0.3 0.4 0.5 0.6 0.7
( )2
1
1 yβ
1.0050 1.0206 1.0483 1.0911 1.1547 1.2500 1.4003
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 396
Hence, using the trapezoidal rule,
( )
0.7
0.1 2
1 dy1 yβ
β« ( )1(0.1) 1.0050 1.4003 1.0206 1.0483 1.0911 1.1547 1.25002β‘ β€β + + + + + +β’ β₯β£ β¦
= (0.1)[6.76735] = 0.677
(b) Mid-ordinates occur at 0.15, 0.25, 0.35, 0.45, 0.55 and 0.65
y 0.15 0.25 0.35 0.45 0.55 0.65
( )2
1
1 yβ
1.0114 1.0328 1.0675 1.1198 1.1974 1.3159
Using the mid-ordinate rule,
( )
0.7
0.1 2
1 dy1 yβ
β« β (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 Simpsonβs rule,
( )
0.7
0.1 2
1 dy1 yβ
β« ( ) ( ) ( )1 (0.1) 1.0050 1.4003 4 1.0206 1.0911 1.2500 2 1.0483 1.15473
β + + + + + +β‘ β€β£ β¦
= [ ]1 (0.1) 2.4053 13.4468 4.4063
+ +
= [ ]1 (0.1) 20.2583
= 0.675
10. A vehicle starts from rest and its velocity is measured every second for 8 s, with values as
follows: time t (s) 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 velocity v (ms-1) 0 0.4 1.0 1.7 2.9 4.1 6.2 8.0 9.4
The distance travelled in 8.0 s is given by 8.0
0vdtβ«
Estimate the distance using Simpsonβs rule, giving the answer correct to 3 significant figures.
( )[ ]8.0
0
1vdt 1.0 (0 9.4) 4(0.4 1.7 4.1 8.0) 2(1.0 2.9 6.2)3
β + + + + + + + +β«
= [ ]1 19.4 56.8 20.2 (86.4)3 3
+ + = = 28.8 m
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 397
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 Simpsonβs 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
β + + + + + + + +
[ ]1 (0.5) 0 2.048 0.863
= + +
[ ]1 (0.5) 2.9083
=
= 0.485 m
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 398
CHAPTER 46 SOLUTION OF FIRST ORDER DIFFERENTIAL
EQUATIONS BY SEPARATION OF VARIABLES EXERCISE 177 Page 444 1. Sketch a family of curves represented by each of the following differential equations:
(a) dy 6dx
= (b) dy 3xdx
= (c) dy x 2dx
= +
(a) If dy 6dx
= , then y = 6dxβ« = 6x + c
There are an infinite number of graphs of y = 6x + c; three curves are shown below.
(b) If dy 3xdx
= , then y = 233x dx x c2
= +β«
A family of three typical curves is shown below.
(c) If dy x 2dx
= + , then y = 2x(x 2)dx 2x c
2+ = + +β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 399
A family of three typical curves is shown below.
2. Sketch the family of curves given by the equation dy 2x 3dx
= + and determine the equation of
one of the curves which passes through the point (1, 3).
If dy 2x 3dx
= + , then y = ( ) 22x 3 dx x 3x c+ = + +β«
If the curve passes through the point (1, 3) then x = 1 and y = 3,
Hence, 3 = ( ) ( )21 3 1 c+ + i.e. c = -1
and 2y x 3x 1= + β
A family of three curves is shown below, including 2y x 3x 1= + β which passes through the point
(1, 3).
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 400
EXERCISE 178 Page 445
3. Solve the differential equation: dy x 3dx
+ = , given y = 2 when x = 1.
If dy x 3dx
+ = , then dy 3 xdx
= β and y = ( )2x3 x dx 3x c
2β = β +β«
If y = 2 when x = 1, then 2 = 3 - 1 c2+ from which, c = 1
2β
Hence, 2x 1y 3x
2 2= β β
5. Solve the differential equation: x
1 dy2 x 3e dx
+ = β , given y = 1 when x = 0.
If x
1 dy2 x 3e dx
+ = β then xdy3 x e 2dx
β= β β and ( )xdy 1 x e 2dx 3
β= β β
Hence, ( )2
x x1 1 xy x e 2 dx e 2x c3 3 2
β ββ β= β β = + β +β β
β β β«
If y = 1 when x = 0, then 1 = 1 (0 1 0) c3
+ + + i.e. c = 23
Thus, y = 2
x1 x 2e 2x3 2 3
ββ β+ β +β β
β β or 2
x
1 2x 4x 46 eβ ββ + +β ββ β
6. The gradient of a curve is given by: 2dy x 3x
dx 2+ = . Find the equation of the curve if it passes
through the point 11,3
β ββ ββ β
If 2dy x 3x
dx 2+ = , then
2dy x3xdx 2
= β
Hence, 2 2 3x 3x xy 3x dx c
2 2 6β β
= β = β +β ββ β β«
If it passes through 11,3
β ββ ββ β
, x = 1 and y = 13
Thus, 1 3 1 c3 2 6= β + from which, c = 1 3 1 1
3 2 6β + = β
Hence, 3
23 xy x 12 6
= β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 401
8. An object is thrown vertically upwards with an initial velocity, u, of 20 m/s. The motion of the
object follows the differential equation dsdt
= u β gt, where s is the height of the object in metres
at time t seconds and g = 9.8 2m / s . Determine the height of the object after 3 seconds if s = 0
when t = 0.
If dsdt
= u β gt, then s = ( )2g tu gt dt ut c
2β = β +β«
Since s = 0 when t = 0, then c = 0
Hence, 2g ts ut
2= β and if u = 20 and s = 9.8, then s = 20t -
29.8t2
i.e. u = 20t β 4.9 2t
The height when t = 3, s = 3(20) β 4.9 ( )23
i.e. height = 60 β 44.1 = 15.9 m
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 402
EXERCISE 179 Page 447
2. Solve the differential equation: 2dy 2cos ydx
=
If 2dy 2cos ydx
= then 2
dy 2dxcos y
=
and 2sec ydy 2dx=β« β«
i.e. tan y = 2x + c
3. Solve the differential equation: ( )2 dyy 2 5ydx
+ = , given y = 1 when x = 12
If ( )2 dyy 2 5ydx
+ = then 2y 2 dy 5dxy
β β+=β β
β β
and 2y dy 5dxy
β β+ =β β
β β β« β«
i.e. 2y 2 ln y 5x c
2+ = +
y = 1 when x = 12
, hence, 1 52ln1 c2 2+ = + from which, c = 1 5 2
2 2β = β
and 2y 2 ln y 5x 2
2+ = β
4. The current in an electric circuit is given by the equation Ri + diLdt
= 0, where L and R are
constants. Show that i =R tLI e
β, given that i = I when t = 0.
If Ri + diLdt
= 0, then diLdt
= - Ri
and di Ridt L
= β
from which, di R dti L= β and di R dt
i L= ββ« β«
Thus, ln i = Rt cL
β +
i = I when t = 0, thus ln I = c
Hence, ln i = Rt ln IL
β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 403
i.e. ln i β ln I = RtL
β
i.e. i RtlnI L= β
Taking anti-logarithms gives: RtLi e
Iβ
= and R tLi I e
β=
5. The velocity of a chemical reaction is given by dxdt
= k(a β x), where x is the amount transferred
in time t, k is a constant and a is the concentration at time t = 0 when x = 0. Solve the equation
and determine x in terms of t.
If dxdt
= k(a β x), then dx k dta x
=β
and dx k dta x
=ββ« β«
i.e. β ln(a β x) = kt + c
t = 0 when x = 0, hence - ln a = c
Thus, β ln(a β x) = kt β ln a
i.e. ln a β ln(a β x) = kt
i.e. aln kta x
β β =β βββ β
and kta ea x
=β
i.e. k t
a a xe
= β i.e. k ta e a xβ = β
and x = a - k ta eβ i.e. ( )k tx a 1 eβ= β
6.(a) Charge Q coulombs at time t seconds is given by the differential equation dQ QRdt C
+ = 0,
where C is the capacitance in farads and R the resistance in ohms. Solve the equation for Q
given that Q = Q0 when t = 0.
(b) A circuit possesses a resistance of 250 Γ 103 Ξ© and a capacitance of 8.5 Γ 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 404
(a) If dQ QRdt C
+ = 0 then dQ Qdt RC
= β
i.e. dQ 1 dtQ RC
= ββ« β«
i.e. ln Q = t kRC
β +
Q = Q0 when t = 0, hence, ln Q0 = k
Hence, ln Q = 0t ln Q
RCβ +
i.e. ln Q - ln Q0 = tRC
β
i.e. 0
Q tlnQ RC
= β
and t
R C
0
Q eQ
β
= and t
CR0Q Q e
β=
(b) R = 250 Γ 103 Ξ© , C = 8.5 Γ 10-6 F, t = 0.32 s and Q = 8.0 C
Hence, ( )6 30.32
8.5 10 250 100 08.0 Q e Q 0.8602ββ
Γ Γ Γ= =
from which, initial charge, Q08.0
0.8602= = 9.30 C
When t = 1 s, charge, Q = 6 3t 1
CR 8.5 10 250 100Q e 9.30e β
β βΓ Γ Γ= = 5.81 C
8. The rate of cooling of a body is given by d kdtΞΈ= ΞΈ , where k is a constant. If ΞΈ = 60Β°C when t = 2
minutes and ΞΈ = 60Β°C when t = 5 minutes, determine the time taken for ΞΈ to fall to 40Β°C, correct
to the nearest second.
If d kdtΞΈ= ΞΈ then d k dtΞΈ
=ΞΈ
and d k dtΞΈ=
ΞΈβ« β«
i.e. ln ΞΈ = kt + c
When ΞΈ = 60Β°C, t = 2, i.e. ln 60 = 2k + c (1)
When ΞΈ = 50Β°C, t = 5, i.e. ln 50 = 5k + c (2)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 405
(1) β (2) gives: ln 60 β ln 50 = -3k
from which, k = 1 60ln 0.060773 50
β = β
Substituting in (1): ln 60 = 2(-0.06077) + c
from which, c = ln 60 + 2(0.06077) = 4.2159
Hence, ln ΞΈ = kt + c = -0.06077t + 4.2159
When ΞΈ = 40Β°C, ln 40 = -0.06077t + 4.2159
and time, t = 4.2159 ln 40 8.672min0.06077
β= = 8 min 40 s
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 406
EXERCISE 180 Page 450
2. Solve the differential equation (2y β 1) ( )2dy 3x 1dx
= + , given x =1 when y = 2.
If (2y β 1) ( )2dy 3x 1dx
= + , then ( ) ( )22y 1 dy 3x 1 dxβ = +β« β«
i.e. 2 3y y x x cβ = + +
x =1 when y = 2, hence, 4 β 2 = 1 + 1 + c from which, c = 0
Thus, 2 3y y x xβ = +
4. Solve the differential equation 2y(1 β x) + x(1 + y) dydx
= 0, given x = 1 when y = 1.
If 2y(1 β x) + x(1 + y) dydx
= 0 then x(1 + y) dydx
= -2y(1 β x) = 2y(x β 1)
Thus, 1 y 2(x 1)dy dxy x
β β+ ββ β=β β β ββ β β β
β« β«
i.e. 1 21 dy 2 dxy x
β β β β+ = ββ β β ββ β β β
β« β«
ln y + y = 2x β 2 ln x + c
x = 1 when y = 1, hence, ln 1 + 1 = 2 β 2 ln 1 + c from which, c = -1
Thus, ln y + y = 2x β 2 ln x β 1
or ln y + 2 ln x = 2x β y β 1
i.e. ln y + ln 2x = 2x β y β 1
and ( )2ln x y 2x y 1= β β
5. Show that the solution of the equation 2
2
y 1 y dyx 1 x dx
+=
+ is of the form
2
2
y 1x 1
β β+β β+β β
= constant.
Since 2
2
y 1 y dyx 1 x dx
+=
+ then 2 2
y xdy dxy 1 x 1
=+ +β« β«
i.e. ( ) ( )2 21 1ln y 1 ln x 1 c2 2
+ = + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 407
i.e. ( ) ( )1 1
2 22 2ln y 1 ln x 1 cβ‘ β€
+ β + =β’ β₯β£ β¦
i.e.
12 2
2
y 1ln cx 1
β β+=β β+β β
or 2
2
y 1ln cx 1
β β+=β β+β β
and 2
c2
y 1 ex 1
β β+=β β+β β
= a constant
7. Determine the equation of the curve which satisfies the equation 2dyxy x 1dx
= β , and which
passes through the point (1, 2).
Since 2dyxy x 1dx
= β then 2x 1 1y dy dx x dxx xβ β β= = ββ β
β β β« β« β«
i.e. 2 2y x ln x c
2 2= β +
If the curve passes through (1, 2) then x = 1 and y = 2,
hence, 2 22 1 ln1 c
2 2= β + from which, c = 3
2
Thus, 2 2y x 3ln x
2 2 2= β +
or 2 2y x 2ln x 3= β +
8. The p.d., V, between the plates of a capacitor C charged by a steady voltage E through a resistor
R is given by the equation dVCR V Edt
+ =
(a) Solve the equation for V given that at t = 0, V = 0.
(b) Calculate V, correct to 3 significant figures, when E = 25 V, C = 20 Γ 10-6 F,
R = 200 Γ 103 Ξ© and t = 3.0 s.
(a) Since dVCR V Edt
+ = then dV E Vdt CR
β=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 408
i.e. dV dtE V CR
=ββ« β«
from which, ( ) tln E V kCR
β β = +
At t = 0, V = 0, hence, - ln E = k
Thus, ( ) tln E V ln ECR
β β = β
( ) tln E ln E VCR
β β =
and E tlnE V CR
β β =β βββ β
i.e. t
CRE eE V
=β
i.e. tCR
E E Ve
= β
and t
CRt
CR
EV E E Eee
β
= β = β
i.e. t
CRV E 1 eββ β
= ββ ββ ββ β
volts
(b) Voltage, ( )6 3t 3.0
0.75CR 20 10 200 10V E 1 e 25 1 e 25 1 eββ β
βΓ Γ Γβ β β β
= β = β = ββ β β ββ ββ β β β β β = 13.2 V
9. Determine the value of p, given that 3 dyxdx
= p β x, and that y = 0 when x = 2 and when x = 6.
Since 3 dyxdx
= p β x then 3 3 2
p x p 1dy dx dxx x xβ β β= = ββ β
β β β« β« β«
i.e. ( )3 2dy px x dxβ β= ββ« β«
i.e. 2 1px xy c
2 1
β β
= β +β β
i.e. 2
p 1y c2x x
= β + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 409
y = 0 when x = 2, hence, p 10 c8 2
= β + + (1)
y = 0 when x = 6, hence, p 10 c72 6
= β + + (2)
(1) β (2) gives: 1 1 1 10 p8 72 2 6
β β β β= β β + ββ β β ββ β β β
i.e. p 109 3
= β +
i.e. p 19 3= from which, p = 3
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 410
CHAPTER 47 HOMOGENEOUS FIRST ORDER DIFFERENTIAL
EQUATIONS EXERCISE 181 Page 452
1. Find the general solution of: 2 2 dyx ydx
=
(i) Rearranging 2 2 dyx ydx
= gives: 2
2
dy xdx y
= , which is homogeneous in x and y.
(ii) Let y = vx, then dy dvv xdx dx
= +
(iii) Substituting for y and dydx
gives: 2
2 2 2
dv x 1v xdx v x v
+ = =
(iv) Separating the variables gives: 3
2 2
dv 1 1 vx vdx v v
β= β = i.e.
2
3
v1 vβ
1dv dxx
=
Integrating both sides gives: 2
3
v 1dv dx1 v x
=ββ« β«
Hence, ( )31 ln 1 v3
β β = ln x + c (using u = 31 vβ substitution)
(v) Replacing v by yx
gives: 3
3
1 yln 13 x
β ββ ββ β
β β = ln x + c, which is the general solution.
i.e. 3 3
3
1 x yln ln x c3 x
β βββ = +β β
β β
3. Find the particular solution of the differential equation: ( )2 2x y dy+ = x y dx, given that x = 1
when y = 1.
(i) Rearranging ( )2 2x y dy+ = xy dx gives: 2 2
dy xydx x y
=+
, which is homogeneous in x and y.
(ii) Let y = vx, then dy dvv xdx dx
= +
(iii) Substituting for y and dydx
gives: 2
2 2 2 2 2 2
dv x(vx) vx vv xdx x v x x (1 v ) 1 v
+ = = =+ + +
(iv) Separating the variables gives: ( )2 3 3
2 2 2 2
v v 1 vdv v v v v vx vdx 1 v 1 v 1 v 1 v
β + β β β= β = = =
+ + + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 411
i.e. 2
3
1 vv+
β1dv dxx
=
Integrating both sides gives: 2
3
1 v 1dv dxv x+
β =β« β«
i.e. 3
1 1 1dv dv dxv v x
β β =β« β« β«
Hence, 2v ln v
2
β
β = ln x + c
or 2
1 ln v2v
β = ln x + c
(v) Replacing v by yx
gives: 21 yln
xy2x
ββ ββ ββ β
= ln x + c
or 2
2
x yln2y x
β = ln x + c which is the general solution.
When x = 1, y = 1, thus: 1 ln12β = ln 1 + c from which, c = 1
2
Thus, the particular solution is: 2
2
x yln2y x
β = ln x + 12
i.e. 2
2
x y 1ln (x)2y x 2
β β= +β ββ β
i.e. 2
2
x 1ln y2y 2
= +
or 2 2 1x 2y ln y2
β β= +β ββ β
5. Find the particular solution of the differential equation: 2y x dy 1y 2x dx
β ββ=β β+β β
, given that y = 3 when
x = 2.
(i) Rearranging 2y x dy 1y 2x dx
β ββ=β β+β β
gives: dy 2x ydx 2y x
+=
β, which is homogeneous in x and y.
(ii) Let y = vx, then dy dvv xdx dx
= +
(iii) Substituting for y and dydx
gives: dv 2x vx 2 vv xdx 2vx x 2v 1
+ ++ = =
β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 412
(iv) Separating the variables gives: 2dv 2 v (2 v) v(2v 1) 2 2v 2vx v
dx 2v 1 2v 1 2v 1+ + β β β +
= β = =β β β
i.e. 2 2dv 2 2v 2v 2v 2v 2x
dx 2v 1 2v 1β + β + +
= =β β
and 2
2v 12v 2v 2
ββ + +
1dv dxx
=
Integrating both sides gives: 2
2v 1 1dv dx2v 2v 2 x
β=
β + +β« β«
i.e. 2
1 2v 1 1dv dx2 1 v v x
β=
+ ββ« β«
Hence, ( )21 ln 1 v v2
β + β = ln x + c
(v) Replacing v by yx
gives: 21 y yln 1
2 x xβ ββ ββ + ββ ββ ββ ββ β β β
= ln x + c, which is the general solution.
When x = 2, y = 3, thus: 1 3 9ln 1 ln 2 c2 2 4
β ββ + β = +β ββ β
i.e. 1 1ln ln 2 c2 4
β ββ = +β ββ β
and 121ln ln 2 c
4
ββ β = +β ββ β
i.e. ( )12ln 4 ln 2 c= + and ln 2 = ln 2 + c, from which, c = 0
Thus, the particular solution is: 21 y yln 1
2 x xβ ββ ββ + ββ ββ ββ ββ β β β
= ln x
i.e. 2 2
2
1 x xy yln ln x2 x
β β+ ββ =β β
β β i.e.
12 2 2
2
x xy yln ln xx
ββ β+ β
=β ββ β
or
12 2
2 2
x xx xy y
β β=β β+ ββ β
i.e. 2
2 2
x xx xy y
=+ β
i.e. 2 2
x xx xy y
=+ β
from which, 2 2x xy y 1+ β = or 2 2x xy y 1+ β =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 413
EXERCISE 182 Page 454 1. Solve the differential equation: ( )3 4 4xy dy x y dx= +
(i) Rearranging ( )3 4 4xy dy x y dx= + gives: 4 4
3
dy x ydx xy
+= , which is homogeneous in x and y.
(ii) Let y = vx, then dy dvv xdx dx
= +
(iii) Substituting for y and dydx
gives: ( )
4 4 4 4 4 4
3 4 33 3
dv x v x x (1 v ) 1 vv xdx v x vx v x
+ + ++ = = =
(iv) Separating the variables gives: 4 4 4
3 3 3
dv 1 v 1 v v 1x vdx v v v
+ + β= β = =
i.e. 3 1v dv dxx
=
Integrating both sides gives: 3 1v dv dxx
=β« β«
Hence, 4v
4 = ln x + c
(v) Replacing v by yx
gives:
4yx4
β ββ ββ β = ln x + c
i.e. 4
4
y ln x c4x
= +
or 4 4y 4x (ln x c)= +
3. Solve the differential equation: dy2x x 3ydx
= + , given that when x = 1, y = 1.
(i) Rearranging dy2x x 3ydx
= + gives: dy x 3ydx 2x
+= , which is homogeneous in x and y.
(ii) Let y = vx, then dy dvv xdx dx
= +
(iii) Substituting for y and dydx
gives: dv x 3vx x(1 3v) 1 3vv xdx 2x 2x 2
+ + ++ = = =
(iv) Separating the variables gives: dv 1 3v 1 3v 2v 1 vx vdx 2 2 2
+ + β += β = =
i.e. 2 1dv dx1 v x
=+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 414
Integrating both sides gives: 2 1dv dx1 v x
=+β« β«
Hence, 2 ln(1+v) = ln x + c
(v) Replacing v by yx
gives: y2 ln 1x
β β+β ββ β
= ln x + c, which is the general solution.
When x = 1, y = 1, thus: 2 ln 2 ln1 c= + from which, c = 2 ln 2
Thus, the particular solution is: x y2lnx+β β
β ββ β
= ln x + 2 ln 2
i.e. 2
2x yln ln x ln 2 ln 4x2+β β = + =β β
β β
from which, 2x y 4x
x+β β =β β
β β i.e.
2
2
(x y) 4xx+
=
i.e. 2 3(x y) 4x+ =
5. Determine the particular solution of 3 3
2
dy x ydx xy
+= , given that x = 1 when y = 4.
(i) 3 3
2
dy x ydx xy
+= is homogeneous in x and y.
(ii) Let y = vx, then dy dvv xdx dx
= +
(iii) Substituting for y and dydx
gives: ( )
3 3 3 3 3 3
3 2 22 2
dv x v x x (1 v ) 1 vv xdx x v vx v x
+ + ++ = = =
(iv) Separating the variables gives: 3 3 3
2 2 2
dv 1 v 1 v v 1x vdx v v v
+ + β= β = = , i.e. 2 1v dv dx
x=
Integrating both sides gives 2 1v dv dxx
=β« β«
Hence, 3v
3 = ln x + c
(v) Replacing v by yx
gives: 3
3
y3x
= ln x + c, which is the general solution.
When x = 1, y = 4, thus: 64 ln1 c3= + from which, c = 64
3
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 415
Thus, the particular solution is: 3
3
y3x
= ln x + 643
i.e. 3 3 64y 3x ln x3
β β= +β ββ β
or ( )3 3y x 3ln x 64= +
6. Show that the solution of the differential equation: 3 2 2 3
2 2 3
dy y xy x y 5xdx xy x y 2x
β β β=
β β is of the form:
2
2
y 4y y 5x18ln ln x 422x x x
ββ β+ + = +β ββ β
, when x = 1 and y = 6.
(i) 3 2 2 3
2 2 3
dy y xy x y 5xdx xy x y 2x
β β β=
β β is homogeneous in x and y.
(ii) Let y = vx, then dy dvv xdx dx
= +
(iii) Substituting for y and dydx
gives:
( ) ( )
3 3 2 2 2 3 3 3 2 3 2
22 2 2 3 3 2
dv v x xv x x vx 5x x (v v v 5) v v v 5v xdx v v 2x v x x (vx) 2x x v v 2
β β β β β β β β β+ = = =
β ββ β β β
(iv) Separating the variables gives: ( )3 2 23 2
2 2
v v v 5 v v v 2dv v v v 5x vdx v v 2 v v 2
β β β β β ββ β β= β =
β β β β
= 2
v 5v v 2
ββ β
i.e. 2v v 2v 5β ββ
1dv dxx
=
Integrating both sides gives: 2v v 2 1dv dxv 5 xβ β
=ββ« β«
v + 4 2v 5 v v 2β β β
2v 5vβ 4v β 2 4v β 20 18
Thus, 18 1v 4 dv dxv 5 x
+ + =ββ« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 416
Hence, 2v 4v 18ln(v 5)
2+ + β = ln x + c
(v) Replacing v by yx
gives: 2
2
y 4y y18ln 52x x x
β β+ + ββ ββ β
= ln x + c, which is the general solution.
When x = 1, y = 6, thus: 36 24 618ln 52 1 1
β β+ + ββ ββ β
= ln 1 + c from which, c = 42
Thus, the particular solution is: 2
2
y 4y y 5x18ln2x x x
ββ β+ + β ββ β
= ln x + 42
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 417
CHAPTER 48 LINEAR FIRST ORDER DIFFERENTIAL
EQUATIONS EXERCISE 183 Page 457
2. Solve the differential equation: dydx
= x( 1 β 2y)
dydx
= x( 1 β 2y) = x β 2xy i.e. dydx
+ 2xy = x from which, P = 2x and Q = x
Hence, 2x dx 2x dx
y e e x dxβ« β«= β«
i.e. 2 2x xye e x dx= β«
i.e. 2 2x x1ye e c
2= +
or 2x1y ce
2β= +
4. Solve the differential equation: 3dyx 1 x 2ydx
β β+ = ββ ββ β
, given x = 1 when y = 3.
Since 3dyx 1 x 2ydx
β β+ = ββ ββ β
then 2dy 2y1 xdx x
+ = β
from which, 2dy 2y x 1dx x
+ = β i.e. P = 2x
and Q = 2x 1β
Hence, ( )2 2dx dx 2x xy e e x 1 dxβ« β«= ββ«
i.e. ( )2 ln x 2 ln x 2ye e x 1 dx= ββ«
i.e. ( )2 2ln x ln x 2ye e x 1 dx= ββ«
i.e. ( )2 2 2y x x x 1 dx= ββ« since ln Ae A=
i.e. ( )2 4 2y x x x dx= ββ«
i.e. 5 3
2 x xyx c5 3
= β +
x = 1 when y = 3, hence, 1 13 c5 3
= β +
from which, c = 1 1 45 5 3 4733 5 15 15
+ β+ β = =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 418
Hence, 5 3
2 x x 47yx5 3 15
= β +
i.e. 3
2
x x 47y5 3 15x
= β +
6. Solve the differential equation: dydx
+ x = 2y
Since dydx
+ x = 2y then dydx
- 2y = - x from which, P = -2 and Q = - x
Hence, 2dx 2dx
y e e ( x)dxβ ββ« β«= ββ«
i.e. 2x 2xye e ( x)dxβ β= ββ« (1)
Using integration by parts on 2xx e dxββ« : Let u = x, then du 1dx
= and du = dx
and dv = 2xe dxβ and v = 2x 2x1e dx e2
β β= ββ«
Thus, 2x 2x 2x1 1x e dx (x) e e dx2 2
β β ββ β= β β ββ ββ β β« β«
= 2x 2x 2x 2x1 1 1 1xe e dx xe e2 2 2 4
β β β ββ + = β ββ«
Substituting in (1) gives: 2x 2x 2x1 1ye xe e c2 4
β β ββ β= β β β +β ββ β
i.e. 2x 2x 2x1 1ye x e e c2 4
β β β= + +
or 2x
1 1 cy x2 4 eβ= + +
i.e. 2x1 1y x ce2 4
= + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 419
EXERCISE 184 Page 458
2. Solve the differential equation: ( )dt sec t t sin t cos t sec tdtΞΈ+ + ΞΈ = , given t = Ο when ΞΈ = 1.
Since ( )dt sec t t sin t cos t sec tdtΞΈ+ + ΞΈ =
then d sec t cos t sec tsec t sin tdt t tΞΈ β β+ + ΞΈ =β β
β β
from which, P =
1 cos tsec t cos t 1 1cos tsec t sin t sin t tan t
t cos t t t
β ββ ββ β β β + = + = +β β
β β and Q = sec t
t
Hence, 1 1tan t dt tan t dtt t sec te e dt
t
β β β β+ +β β β ββ β β β β« β«ΞΈ = β«
i.e. (lnsec t ln t ) (ln sec t ln t ) sec te e dtt
+ +ΞΈ = β«
i.e. ln(t sec t ) ln(t sec t ) sec te e dtt
ΞΈ = β«
i.e. 2sec tt sec t t sec t dt sec t dtt
ΞΈ = =β« β«
and t sec t tan t cΞΈ = +
t = Ο when ΞΈ = 1, hence, Ο sec Ο = tan Ο + c
i.e. - Ο = c
Thus, t sec t tan tΞΈ = β Ο
or
sin ttan t sin tcos t cos t1 1t sec t t sec t t tt t
cos t cos t
Ο Ο ΟΞΈ = β = β = β
i.e. ( )1 sin t cos tt
ΞΈ = β Ο
4. Show that the solution of the differential equation ( )3dy 42 x 1 ydx (x 1)
β + =+
is: y = ( ) ( )4 2x 1 ln x 1+ + , given that x = 0 when y = 0
Since ( )3dy 42 x 1 ydx (x 1)
β + =+
from which, ( )3dy 4 y 2 x 1dx (x 1)
β = ++
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 420
i.e. P = 4x 1
β+
and Q = 32(x 1)+
Hence, 4 4dx dx 3x 1 x 1y e e 2(x 1) dx
β β+ +β« β«= +β«
i.e. 4ln(x 1) 4ln(x 1) 3y e e 2(x 1) dxβ + β += +β«
i.e. 4 4ln(x 1) ln(x 1) 3y e e 2(x 1) dx
β β+ += +β«
i.e. 34 4
y 1 2(x 1) dx(x 1) (x 1)
= ++ +β«
and 4
y 2 dx 2ln(x 1) c(x 1) (x 1)
= = + ++ +β«
x = 0 when y = 0, hence, 0 = 2 ln 1 + c from which, c = 0
Thus, 4
y 2 ln(x 1)(x 1)
= ++
or y = ( ) ( )4 2x 1 ln x 1+ +
6. The equation dvdt
= -(av + bt), where a and b are constants, represents an equation of motion
when a particle moves in a resisting medium. Solve the equation for v given that v = u when
t = 0.
Since dvdt
= -(av + bt) then dvdt
+ av = -bt from which, P = a and Q = -bt
Hence, a dt a dt
v e e ( bt)dtβ« β«= ββ«
i.e. at atv e b t e= β β« (1)
Using integration by parts on att e dtβ« : Let u = t, then du 1dt
= and du = dt
and dv = ate dt and v = at a t1e dt ea
=β«
Thus, a t a t a t1 1t e dt (t) e e dta a
β β= ββ ββ β β« β« = a t a t
2
t 1e ea a
β
Substituting in (1) gives: at a t a t2
t 1ve b e e ca a
β β= β β +β ββ β
v = u when t = 0, hence, u = 2
b ca
+ from which, c = u - 2
ba
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 421
Thus, at a t a t a t at2 2 2 2
t 1 b bt b bve b e e u e e ua a a a a a
β β= β β + β = β + + ββ ββ β
or at at2 2 2 2
bt b b b bt bv u e or u ea a a a a a
β ββ β β β= β + + β β + ββ β β ββ β β β
7. In an alternating current circuit containing resistance R and inductance L the current i is given
by: Ri + 0diL E sin tdt
= Ο . Given i = 0 when t = 0, show that the solution of the equation is given
by: i = ( )R t
0 0 L2 2 2 2 2 2
E E LR sin t L cos t eR L R L
βΟβ β β βΟ βΟ Ο +β β β β+Ο +Οβ β β β
Since Ri + 0diL E sin tdt
= Ο then 0diL E sin t Ridt
= Ο β and 0Edi Risin tdt L L
= Ο β
i.e. 0Edi Ri sin tdt L L+ = Ο from which, P = R
L and Q = 0E sin t
LΟ
Hence, R Rdt dt 0L L Ei e e sin t dt
Lβ β β ββ« β«= Οβ β β β
β β β β β«
i.e. R t R t
0L L Ei e e sin t dtL
β β= Οβ ββ β β«
i.e. R t R t
0L LEi e e sin t dtLβ‘ β€
= Οβ’ β₯β£ β¦β« (1)
From page XX of textbook, ( )ax
ax2 2
ee sin bx dx a sin bx bcos bx ca b
= β ++β«
Hence,
R tRt LL
22
e Re sin t dt sin t cos t cLR
L
β βΟ = Ο βΟ Ο +β ββ β β β +Οβ β
β β
β«
Substituting into (1) gives:
R tRt L
0L2
22
E e Ri e sin t cos t cL LR
L
β β= Ο βΟ Ο +β ββ β β β +Οβ ββ β
R tL
02 2 2
2
E e R sin t cos t cL LR L
L
β β= Ο βΟ Ο +β ββ β+Ο β β β ββ β
i.e. ( )
R tRt L
0L2 2 2
E Le Ri e sin t cos t cLR L
β β= Ο βΟ Ο +β β+Ο β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 422
i = 0 when t = 0, hence, 0 = 02 2 2
E L ( ) cR L
βΟ ++Ο
from which, c = 02 2 2
E LR L
Ο+Ο
Thus, ( ) ( )
R tRt L
0 0L2 2 2 2 2 2
E Le E LRi e sin t cos tLR L R L
Οβ β= Ο βΟ Ο +β β+Ο +Οβ β
i.e. ( ) ( )
R t0 0 L
2 2 2 2 2 2
E L E LRi sin t cos t eLR L R L
βΟβ β= Ο βΟ Ο +β β+Ο +Οβ β
or ( ) ( )R t
0 0 L2 2 22 2 2
E E Li R sin t Lcos t e
R LR LβΟβ β= Ο βΟ Ο + β β+ Ο+ Ο β β
8. The concentration, C, of impurities of an oil purifier varies with time t and is described by the
equation dCa b dm Cmdt
= + β , where a, b, d and m are constants. Given C = c0 when t = 0, solve
the equation and show that: m t m ta a
0bC d 1 e c em
β ββ ββ β= + β +β ββ ββ β β β
Since dCa b dm Cmdt
= + β then dC b d m Cmdt a a a
= + β
and dC m b d mCdt a a a
+ = + from which, P = ma
and Q = b d ma a+
Hence, m mdt dta a b d mCe e dt
a aβ ββ« β«= +β ββ β β«
i.e. m t mta a b d mCe e dt
a aβ β= +β ββ β β«
i.e.
mtm t aa a e b d mCe k
m a aβ β= + +β ββ β
C = 0c when t = 0, hence, 0c = a b d m bk d km a a mβ β+ + = + +β ββ β
from which, k = 0bc dm
β β
Thus,
mtm t aa
0a e b d m bCe c d
m a a mβ β= + + β ββ ββ β
i.e. m ta
0a b d m bC c d em a a m
ββ β β β= + + β ββ β β ββ β β β
i.e. m ta
0b bC d c d em m
ββ β β β= + + β ββ β β ββ β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 423
i.e. m t m ta a
0b bC d d e c em m
β ββ β β β= + β + +β β β ββ β β β
or mt mta a
0bC d 1 e c em
β ββ ββ β= + β +β ββ ββ β β β
9. The equation of motion of a train is given by: ( )tdvm mk 1 e mcvdt
β= β β , where v is the speed, t
is the time and m, k and c are constants. Determine the speed, v, given v = 0 at t = 0.
Since ( )tdvm mk 1 e mcvdt
β= β β then ( )tdv k 1 e cvdt
β= β β
and ( )tdv cv k 1 edt
β+ = β from which, P = c and Q = ( )tk 1 eββ
Hence, cdt cdt tv e e k(1 e )dtββ« β«= ββ«
i.e. ( )c t c t t c t c t tv e e k(1 e )dt k e e dtβ β= β = ββ« β«
i.e. c t t (c 1)
c t e ev e k zc c 1
ββ‘ β€= β +β’ β₯ββ£ β¦
where z is the constant of integration
v = 0 when t = 0, hence, 0 = k k zc c 1β +
β
from which, z = k k ck k(c 1) kc 1 c c(c 1) c(c 1)
β ββ = =
β β β
Thus, c t t (c 1)
c t e e kv e kc c 1 c(c 1)
ββ‘ β€= β +β’ β₯β ββ£ β¦
i.e. v = t
ctk k e k ec c 1 c(c 1)
βββ β
β + β ββ ββ β
or t ct1 e ev k
c c 1 c(c 1)
β ββ§ β«= β +β¨ β¬β ββ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 424
CHAPTER 49 NUMERICAL METHODS FOR FIRST ORDER
DIFFERENTIAL EQUATIONS EXERCISE 185 Page 464
1. Use Eulerβs method to obtain a numerical solution of the differential equation dy y3dx x
= β , with
the initial conditions that x = 1 when y = 2, for the range x = 1.0 to x = 1.5 with intervals of 0.1.
Draw the graph of the solution in this range.
dy yy ' 3dx x
= = β
If initially x0 = 1 and y0 = 2, (and h = 0.1), then (y')0 = 231
β = 1
Line 1 in the table below is completed with x = 1.0 and y = 2
For line 2, where x0 = 1.1 and h = 0.1:
y1 = y0 + h(y')0 = 2.0 + (0.1)(1) = 2.1
and (y')0 = 3 - 0
0
yx
= 3 - 2.11.1
= 1.0909
For line 3, where x0 = 1.2: y1 = y0 + h(y')0 = 2.1 + (0.1)(1.0909) = 2.209091
and (y')0 = 3 - 0
0
yx
= 3 - 2.2090911.2
= 1.159091
For line 4, where x0 = 1.3: 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 y3dx x
= β , with initial conditions x = 1 and y = 2 is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 425
3. (a) The differential equation dy y1dx x
+ = β has the initial conditions that y = 1 at x = 2. Produce a
numerical solution of the differential equation in the range x = 2.0(0.1)2.5.
(b) If the solution of the differential equation by an analytical method is given by y = 4 xx 2β ,
determine the percentage error at x = 2.2.
(a) dy y1dx x
+ = β i.e. dy yy ' 1dx x
= = β β
If initially x0 = 2.0 and y0 = 1, (and h = 0.1), then (y')0 = 112.0
β β = -1.5
Line 1 in the table below is completed with x = 2.0 and y = 1
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 - 0
0
yx
= -1 - 0.852.1
= -1.40476
For line 3, where x0 = 2.2: y1 = y0 + h(y')0 = 0.85 + (0.1)(-1.40476) = 0.709524
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 426
and (y')0 = -1 - 0
0
yx
= -1 - 0.7095242.2
= -1.322511
For line 4, where x0 = 2.3: y1 = y0 + h(y')0 = 0.709524 + (0.1)(-1.322511) = 0.577273
The remaining lines of the table are completed in a similar way.
(b) If y = 4 xx 2β then when x = 2.2, y = 0.718182
From the Euler method, when x = 2.2, y = 0.709524
Hence, percentage error = 0.718182 0.709524 0.008658100% 100%0.718182 0.718182
βΓ = Γ
= 1.206%
4. Use Eulerβs method to obtain a numerical solution of the differential equation dy 2yxdx x
= β ,
given the initial conditions that y = 1 when x = 2, in the range x = 2.0(0.2)3.0.
If the solution of the differential equation is given by y =2x
4, determine the percentage error by
using Eulerβs method when x = 2.8.
(a) dy 2yy ' xdx x
= = β
If initially x0 = 2.0 and y0 = 1, (and h = 0.2), then (y')0 = 2(1)2.02.0
β = 1.0
Line 1 in the table below is completed with x = 2.0 and y = 1
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)2.2
= 1.109091
For line 3, where x0 = 2.4: y1 = y0 + h(y')0 = 1.2 + (0.2)(1.109091) = 1.4218182
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 427
and (y')0 = 2.4 - 2(1.4218182)2.4
= 1.2151515
For line 4, where x0 = 2.6: 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 =2x
4, then when x = 2.8, y = 1.96
From the Euler method, when x = 2.8, y = 1.928718
Hence, percentage error = 1.96 1.928718 0.031282100% 100%1.96 1.96β
Γ = Γ
= 1.596%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 428
EXERCISE 186 Page 468
1. Apply the Euler-Cauchy method to solve the differential equation dy y3dx x
= β for the range
1.0(0.1)1.5, given the initial conditions that x = 1 when y = 2.
dy y 'dx
=y3x
= β
x 0 = 1, y 0 = 2 and h = 0.1
(y β²) 0 0
0
y3x
= β23 11
= β =
x1 = 1.1 and from equation (3), page 465, y1P = y 0 + h(y β²) 0 = 2 + 0.1(1) = 2.1
y1C = y 0 + 1
2h [ (y β²) 0 + f(x1 , y
1P ) ] = y 0 + 12
h [ (y β²) 0 + 1P
1
y3
xβ ]
= 2 + 12
(0.1) [ 1 + 2.131.1
β ]
= 2.10454546
(y β²)1 1C
1
y 2.104545463 3x 1.1
= β = β = 1.08677686
Thus the first two lines of the Table below has been completed
x y y β² 1.0 2 1 1.1 2.10454546 1.08677686 1.2 2.216666672 1.152777773 1.3 2.33461539 1.204142008 1.4 2.457142859 1.244897958 1.5 2.5883333335
For line 3, x1 = 1.2
y1P = y 0 + h(y β²) 0 = 2.10454546 + 0.1(1.08677686) = 2.213223146
y1C = y 0 + 1
2h [ (y β²) 0 + 1P
1
y3
xβ ]
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 429
= 2.10454546 + 12
(0.1) [ 1.08677686 + 2.21322314631.2
β ] = 2.216666672
(y β²)11C
1
y 2.2166666723 3x 1.2
= β = β = 1.152777773
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 1x2 2x
+ .
Determine the percentage error, correct to 3 significant figures, when x = 1.3 using (a) Eulerβs
method (see Table 49.4, page 464) and (b) the Euler-Cauchy method.
If y = 3 1x2 2x
+ then when x = 1.3, y = 2.334615385
(a) By Eulerβs method, when x = 1.3, y = 2.325000
Percentage error = 2.334615385 2.3250 100%2.334615385
βΓ = 0.412%
(b) By the Euler-Cauchy method, when x = 1.3, y = 2.33461539
Percentage error = 2.334615385 2.33461539 100%2.334615385
βΓ = 0.000000214%
3. (a) Apply the Euler-Cauchy method to solve the differential equation dy x ydx
β = for the range
x = 0 to x = 0.5 in increments of 0.1, given the initial conditions that when x = 0, y = 1.
(b) The solution of the differential equation in part (a) is given by y = x2e x 1β β . Determine the
percentage error, correct to 3 decimal places, when x = 0.4.
(a) dy y 'dx
= = y + x
x 0 = 0, y 0 = 1 and h = 0.1
(y β²) 0 = 1 + 0 = 1
x1 = 0.1 and from equation (3), page 465, y1P = y 0 + h(y β²) 0 = 1 + 0.1(1) = 1.1
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 430
y1C = y 0 + 1
2h [ (y β²) 0 + f(x1 , y
1P ) ] = y 0 + 12
h [ (y β²) 0 + y1P + x1 ]
= 1 + 12
(0.1) [ 1 + 1.1 + 0.1 ]
= 1.11
(y β²)1 = y1C + 0.1 = 1.11 + 0.1 = 1.21
Thus the first two lines of the Table below has been completed
For line 3, x1 = 0.2
y1P = y 0 + h(y β²) 0 = 1.11 + 0.1(1.21) = 1.231
y1C = y 0 + 1
2h [ (y β²) 0 + y
1P + x1 ]
= 1.11 + 12
(0.1) [ 1.21 + 1.231 + 0.2 ] = 1.24205
(y β²)1 = y1C + 0.2 = 1.24205+ 0.2 = 1.44205
The remaining lines of the table are completed in a similar way.
(b) If y = x2e x 1β β then when x = 0.4, y = 1.583649395
By the Euler-Cauchy method, when x = 0.4, y = 1.581804
Hence, the percentage error = 1.583649395 1.581804 100%1.583649395
βΓ = 0.117%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 431
EXERCISE 187 Page 472
2. Obtain a numerical solution of the differential equation: 1 dy 2y 1x dx
+ = using the Runge-Kutta
method in the range x = 0(0.2)1.0, given the initial conditions that x = 0 when y = 1.
If 1 dy 2y 1x dx
+ = then 1 dy 1 2yx dx
= β and dy x(1 2y)dx
= β
1. 0x = 0, 0y = 1 and since h = 0.2, and the range is from x = 0 to x = 1.0, then
1 2 3 4 5x 0.2, x 0.4, x 0.6, x 0.8, and x 1.0= = = = =
Let n = 0 to determine 1y :
2. ( )1 0 0k f x , y= = f (0, 1); since dy x(1 2y)dx
= β , f (0, 1) = 0(1 - 2) = 0
3. ( )2 0 0 1h h 0.2 0.2k f x , y k f 0 , 1 (0) f 0.1, 12 2 2 2
β β β β= + + = + + =β β β ββ β β β
= 0.1(1 - 2) = -0.1
4. 3 0 0 2h h 0.2 0.2k f x , y k f 0 , 1 ( 0.1)2 2 2 2
β β β β= + + = + + ββ β β ββ β β β
= ( )f 0.1, 0.99
= 0.1(1 β 1.98) = -0.098
5. ( ) ( )4 0 0 3k f x h, y hk f 0 0.2, 1 0.2( 0.098)= + + = + + β = f (0.2, 0.9804)
= 0.2(1 - 2(0.9804)) = -0.19216
6. n 1y + = ny + 1 2 3 4h k 2k 2k k6
+ + + and when n = 0:
1y = 0y + 1 2 3 4h k 2k 2k k6
+ + + = 1 + 0.2 0 2( 0.1) 2( 0.098) 0.192166
+ β + β + β
= 1 + 0.2 0.588166
β = 0.980395
Let n = 1 to determine 2y :
2. ( )1 1 1k f x , y= = f (0.2, 0.980395); since dydx
= x(1 β 2y),
f (0.2, 0.980395) = 0.2(1 β 2(0.980395)) = -0.192158
3. ( )2 1 1 1h h 0.2 0.2k f x , y k f 0.2 ,0.980395 ( 0.192158) f 0.3, 0.96117922 2 2 2
β β β β= + + = + + β =β β β ββ β β β
= 0.3(1 β 2(0.9611792) = -0.27670752
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 432
4. 3 1 1 2h h 0.2 0.2k f x , y k f 0.2 , 0.980395 ( 0.27670752)2 2 2 2
β β β β= + + = + + ββ β β ββ β β β
= ( )f 0.3, 0.952724248 = 0.3(1 β 2(0.952724248)) = -0.271634548
5. ( ) ( )4 1 1 3k f x h, y hk f 0.2 0.2, 0.980395 0.2( 0.271634548)= + + = + + β
= f (0.4, 0.92606809) = 0.4(1 β 2(0.92606809)) = -0.340854472
6. n 1y + = ny + 1 2 3 4h k 2k 2k k6
+ + + and when n = 1:
2y = 1y + 1 2 3 4h k 2k 2k k6
+ + +
= 0.980395 + 0.2 0.192158 2( 0.27670752) 2( 0.271634548) 0.3408544726
β + β + β + β
= 0.980395 + 0.2 1.6296966086
β = 0.926071779
This completes the third row of the table below. In a similar manner 3 4y , y and 5y can be calculated.
3. (a) The differential equation dy y1dx x
+ = β has the initial conditions that y = 1 at x = 2. Produce a
numerical solution of the differential equation, correct to 6 decimal places, using the Runge-
Kutta method in the range x = 2.0(0.1)2.5.
(b) If the solution of the differential equation by an analytical method is given by: y = 4 xx 2β
determine the percentage error at x = 2.2
(a) If dy y1dx x
+ = β then dy y y1 1dx x x
β β= β β = β +β ββ β
1. 0x = 2, 0y = 1 and since h = 0.1, and the range is from x = 2.0 to x = 2.5, then
1 2 3 4 5x 2.1, x 2.2, x 2.3, x 2.4, and x 2.5= = = = =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 433
Let n = 0 to determine 1y :
2. ( )1 0 0k f x , y= = f (2, 1); since, dy y 1dx x
β β= β +β ββ β
, f (2, 1) = 1 12
β ββ +β ββ β
= -1.5
3. ( )2 0 0 1h h 0.1 0.1k f x , y k f 2.0 ,1.0 ( 1.5) f 2.05, 0.9252 2 2 2
β β β β= + + = + + β =β β β ββ β β β
= 0.925 12.05
β ββ +β ββ β
= -1.451219512
4. 3 0 0 2h h 0.1 0.1k f x , y k f 2.0 , 1.0 ( 1.451219512)2 2 2 2
β β β β= + + = + + ββ β β ββ β β β
= ( )f 2.05, 0.927439024
= 0.927439024 12.05
β ββ +β ββ β
= -1.45240928
5. ( ) ( )4 0 0 3k f x h, y hk f 2.0 0.1, 1.0 0.1( 1.45240928)= + + = + + β = f (2.1, 0.854759072)
= 0.854759072 12.1
β ββ +β ββ β
= -1.40702813
6. n 1y + = ny + 1 2 3 4h k 2k 2k k6
+ + + and when n = 0:
1y = 0y + 1 2 3 4h k 2k 2k k6
+ + +
= 1.0 + 0.1 1.5 2( 1.451219512) 2( 1.45240928) ( 1.40702813)6
β + β + β + β
= 1.0 + 0.1 8.7142857146
β = 0.854762
Let n = 1 to determine 2y :
2. ( )1 1 1k f x , y= = f (2.1, 0.854762); since dy y 1dx x
β β= β +β ββ β
,
f (2.1, 0.854762) = 0.854762 12.1
β ββ +β ββ β
= - 1.407029524
3. 2 1 1 1h h 0.1 0.1k f x , y k f 2.1 ,0.854762 ( 1.407029524)2 2 2 2
β β β β= + + = + + ββ β β ββ β β β
= f(2.15, 0.784410523) = 0.784410523 12.15
β ββ +β ββ β
= - 1.364842104
4. 3 1 1 2h h 0.1 0.1k f x , y k f 2.1 , 0.854762 ( 1.364842104)2 2 2 2
β β β β= + + = + + ββ β β ββ β β β
= ( )f 2.15, 0.786519894 = 0.786519894 12.15
β ββ +β ββ β
= - 1.365823207
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 434
5. ( ) ( )4 1 1 3k f x h, y hk f 2.2, 0.854762 0.1( 1.365823207)= + + = + β
= f (2.2, 0.718179679) = 0.718179679 12.2
β ββ +β ββ β
= - 1.326445309
6. n 1y + = ny + 1 2 3 4h k 2k 2k k6
+ + + and when n = 1:
2y = 1y + 1 2 3 4h k 2k 2k k6
+ + +
= 0.854762 + 0.1 1.407029524 2( 1.364842104) 2( 1.365823207) 1.3264453096
β + β + β β
= 0.854762 + 0.1 8.1948054556
β = 0.718182
This completes the third row of the Table below. In a similar manner 3 4y , y and 5y can be calculated and the results are as shown.
(b) If y = 4 xx 2β when x = 2.2, y = 0.718182
By the Runge-Kutta method, when x = 2.2, y = 0.718182 also. Hence, there is no error.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 435
CHAPTER 50 SECOND ORDER DIFFERENTIAL EQUATIONS OF
THE FORM 2
2
d y dya b cy 0dx dx
+ + = EXERCISE 188 Page 477
3. Determine the general solution of 2
2
d y dy2 5y 0dx dx
+ + =
2
2
d y dy2 5y 0dx dx
+ + = in D-operator form is: ( )2D 2D 5 y 0+ + =
The auxiliary equation is: 2m 2m 5 0+ + =
and 22 2 4(1)(5) 2 16 2 j4m 1 j2
2(1) 2 2β Β± β β Β± β β Β±
= = = = β Β±
Hence, the general solution is: xy e Acos 2x Bsin 2xβ= +
5. Find the particular solution of 2
2
d y dy4 5 y 0dt dt
β + = when at t = 0, y = 1 and dydt
= -2.
2
2
d y dy4 5 y 0dx dt
β + = in D-operator form is: ( )24D 5D 1 y 0β + =
The auxiliary equation is: 24m 5m 1 0β + =
i.e. (4m β 1)(m β 1) = 0
from which, 1m4
= and m = 1
Hence, the general solution is: 1 t t4y Ae Be= +
At t = 0, y = 1, hence, 1 = A + B (1)
1 t t4dy 1 Ae Be
dt 4= +
At t = 0, dydt
= -2, hence, - 2 = 1 A B4
+ (2)
(1) β (2) gives: 3 = 3 A4
from which, A = 4
From equation (1), when A = 4, 1 = 4 + B from which, B = -3
Hence, the particular solution is: 1t t4y 4e 3e= β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 436
7. Find the particular solution of 2
2
d x dx6 9x 0dt dt
β + = when at t = 0, x = 2 and dxdt
= 0.
2
2
d x dx6 9x 0dt dt
β + = in D-operator form is: ( )2D 6D 9 x 0β + =
The auxiliary equation is: 2m 6m 9 0β + =
i.e. (m β 3)(m β 3) = 0
from which, m = 3 twice
Hence, the general solution is: ( ) 3tx At B e= +
At t = 0, x = 2, hence, 2 = B
( )( ) ( )3t 3tdx At B 3e e Adt
= + +
At t = 0, dxdt
= 0, hence, 0 = 3B + A and since B = 2, A = -6
Hence, the particular solution is: ( ) 3tx 6t 2 e= β + or ( ) 3tx 2 1 3t e= β
8. Find the particular solution of 2
2
d y dy6 13y 0dx dx
+ + = when at x = 0, y = 4 and dydx
= 0.
2
2
d y dy6 13y 0dx dx
+ + = in D-operator form is: ( )2D 6D 13 y 0+ + =
The auxiliary equation is: 2m 6m 13 0+ + =
and 26 6 4(1)(13) 6 16 6 j4m 3 j22(1) 2 2
β Β± β β Β± β β Β±= = = = β Β±
Hence, the general solution is: 3xy e A cos 2x Bsin 2xβ= +
At x = 0, y = 4, hence, 4 = A
( )( ) ( )( )3x 3xdy e 2Asin 2x 2Bcos 2x A cos 2x Bsin 2x 3edx
β β= β + + + β
At x = 0, dydx
= 0, hence, 0 = 2B β 3A and since A = 4, B = 6
Hence, the particular solution is: 3xy e 4cos 2x 6sin 2xβ= +
or 3xy 2e 2cos 2x 3sin 2xβ= +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 437
EXERCISE 189 Page 480 1. The charge, q, on a capacitor in a certain electrical circuit satisfies the differential equation
2
2
d q dq4 5q 0dt dt
+ + = . Initially (i.e. when t = 0), q = Q and dqdt
= 0. Show that the charge in the
circuit can be expressed as: q = 2t5 Qe sin(t 0.464)β +
2
2
d q dq4 5q 0dt dt
+ + = in D-operator form is: ( )2D 4D 5 y 0+ + =
The auxiliary equation is: 2m 4m 5 0+ + =
and 24 4 4(1)(5) 4 4 4 j2m 2 j
2(1) 2 2β Β± β β Β± β β Β±
= = = = β Β±
Hence, the general solution is: 2tq e A cos t Bsin tβ= +
At t = 0, q = Q, hence, Q = A
( )( ) ( )( )2t 2tdq e Asin t Bcos t A cos t Bsin t 2edt
β β= β + + + β
At t = 0, dqdt
= 0, hence, 0 = B β 2A and since A = Q, B = 2Q
Hence, the particular solution is: 2tq e Q cos t 2Qsin tβ= +
i.e. 2tq Qe cos t 2sin tβ= +
Let cos t + 2 sin t = R sin(t + Ξ±)
= R[sin t cos Ξ± + cos t sin Ξ±]
= R cos Ξ± (sin t) + R sin Ξ± (cos t)
Hence, 2 = R cos Ξ± from which, cos Ξ± = 2R
and 1 = R sin Ξ± from which, sin Ξ± = 1R
There is only one quadrant where both sine and cosine are positive, i.e. the first quadrant.
From the diagram below, R = 2 12 1 5+ =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 438
and Ξ± = 1 1tan 26.565 or 0.464 rad2
β = Β°
Hence, cos t + 2 sin t = 5 sin(t + 0.464)
Since, 2tq Q e cos t 2sin tβ= + then 2tq 5 Qe sin(t 0.464)β= +
3. The motion of the pointer of a galvanometer about its position of equilibrium is represented
by the equation 2
2
d dI K F 0dt dtΞΈ ΞΈ+ + ΞΈ =
If I, the moment of inertia of the pointer about its pivot, is 5 Γ10-3, K, the resistance due to
friction at unit angular velocity, is 2 Γ10-2 and F, the force on the spring necessary to produce
unit displacement, is 0.20, solve the equation for ΞΈ in terms of t given that when t = 0, ΞΈ = 0.3
and ddtΞΈ = 0.
2
2
d dI K F 0dt dtΞΈ ΞΈ+ + ΞΈ = in D-operator form is: ( )2I D K D F 0+ + ΞΈ =
The auxiliary equation is: 2I m K m F 0+ + =
and ( ) ( ) ( )( )
( )
22 2 32
3
2 10 2 10 4 5 10 0.2K K 4I F 0.02 0.0036m2I 0.012 5 10
β β β
β
β Γ Β± Γ β Γβ Β± β β Β±= = =
Γ
= 0.02 j0.06 2 j60.01
β Β±= β Β±
Hence, the general solution is: 2te A cos 6t Bsin 6tβΞΈ = +
At t = 0, ΞΈ = 0.3, hence, 0.3 = A
( )( ) ( )( )2t 2td e 6Asin 6t 6Bcos 6t A cos t Bsin t 2edt
β βΞΈ= β + + + β
At t = 0, ddtΞΈ = 0, hence, 0 = 6B β 2A and since A = 0.3, B = 0.1
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 439
Hence, the particular solution is: 2te 0.3cos 6t 0.1sin 6tβΞΈ = +
4. Determine an expression for x for a differential equation 2
22
d x dx2n n x 0dt dt
+ + = which represents
a critically damped oscillator, given that at time t = 0, x = s and dxdt
= u.
2
22
d x dx2n n x 0dt dt
+ + = in D-operator form is: ( )2 2D 2nD n x 0+ + =
The auxiliary equation is: 2 2m 2n m n 0+ + =
i.e. (m + n)(m + n) = 0
from which, m = -n twice
Hence, the general solution is: ( ) n tx At B eβ= +
At t = 0, x = s, hence, s = B
( )( ) ( )n t n tdx At B ne e Adt
β β= + β +
At t = 0, dxdt
= u, hence, u = -nB + A hence A = u + nB = u + ns
Hence, the particular solution is: ( ) ntx u ns t s eβ= + + or ( ) ntx s u ns t eβ= + +
5. 2
2
d i di 1L R i 0dt dt C
+ + = is an equation representing current i in an electric circuit. If inductance L
is 0.25 henry, capacitance C is 29.76 Γ 10-6 farads and R is 250 ohms, solve the equation for i
given the boundary conditions that when t = 0, i = 0 and didt
= 34.
2
2
d i di 1L R i 0dt dt C
+ + = in D-operator form is: 2 1L D R D i 0C
β β+ + =β ββ β
The auxiliary equation is: 2 1L m R m 0C
+ + =
and ( )
2 26
4L 4(0.25)R R 250 250 250 170C 29.76 10m2L 2 0.25 0.5
ββ Β± β β Β± β β Β±Γ= = = = -160 or -840
Hence, the general solution is: 160t 840ti A e Beβ β= +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 440
When t =0, i = 0, hence, 0 = A + B (1)
160t 840tdi 160Ae 840Bedt
β β= β β
When t = 0, didt
= 34, hence, 34 = -160A β 840B (2)
160 Γ (1) gives: 0 = 160A + 160B (3)
(2) + (3) gives: 34 = -680B from which, B = - 120
From equation (1), A = 120
Hence, 160t 840t1 1i e e20 20
β β= β or ( )160t 840t1i e e20
β β= β
6. The displacement s of a body in a damped mechanical system, with no external forces, satisfies
the following differential equation: 2
2
d s ds2 6 4.5s 0dt dt
+ + = where t represents time. If initially,
when t = 0, s = 0 and dsdt
= 4, solve the differential equation for s in terms of t.
2
2
d s ds2 6 4.5s 0dt dt
+ + = in D-operator form is: ( )22D 6D 4.5 s 0+ + =
The auxiliary equation is: 22m 6m 4.5 0+ + =
or 24m 12m 9 0+ + =
i.e. (2m + 3)(2m + 3) = 0
from which, 3m2
= β twice
Hence, the general solution is: ( )3 t2s At B e
β= +
At t = 0, s = 0, hence, 0 = B
( )3 3t t2 2ds 3At B e e (A)
dt 2β ββ β β β
= + β +β β β ββ β β β
At t = 0, dsdt
= 4, hence, 4 = 3 B A2
β + and since B = 0, A = 4
Hence, the particular solution is: 3t2s 4t e
β=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 441
CHAPTER 51 SECOND ORDER DIFFERENTIAL EQUATIONS OF
THE FORM 2
2
d y dya b cy f (x)dx dx
+ + = EXERCISE 190 Page 483
2. Find the general solution of: 2
2
d y dy6 4 2y 3x 2dx dx
+ β = β
2
2
d y dy6 4 2y 3x 2dx dx
+ β = β in D-operator form is: ( )26D 4D 2 y 3x 2+ β = β
Auxiliary equation is: 2 26m 4m 2 0 i.e. 3m 2m 1 0+ β = + β =
i.e. (3m β 1)(m + 1) = 0
from which, 1m3
= and m = -1
Hence, the complementary function, C.F., 1x x3u Ae Beβ= +
Let the particular integral, P.I., v = ax + b
then ( )( )26D 4D 2 ax b 3x 2+ β + = β
D(ax + b) = a and 2D (ax b) D(a) 0+ = =
Hence, 0 + 4a β 2ax β 2b = 3x β 2
from which, - 2a = 3 and a = 32
β
and 4a β 2b = - 2 and b = 2a + 1 = - 2
Hence, P.I., v = 32
β x β 2
and the general solution, y = u + v = 1 x x3 3Ae Be 2 x
2β+ β β
4. Find the particular solution of 2
2
d y dy9 12 4y 3x 1dx dx
β + = β given that when x = 0, y = 0 and
dy 4dx 3
= β .
2
2
d y dy9 12 4y 3x 1dx dx
β + = β in D-operator form is: ( )29D 12D 4 y 3x 1β + = β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 442
Auxiliary equation is: 29m 12m 4 0β + =
i.e. (3m β 2)(3m - 2) = 0
from which, 2m3
= twice
Hence, C.F., ( )2 x3u Ax B e= +
Let the particular integral, P.I., v = ax + b
then ( )( )29D 12D 4 ax b 3x 1β + + = β
D(ax + b) = a and 2D (ax b) D(a) 0+ = =
Hence, 0 - 12a + 4ax + 4b = 3x β 1
from which, 4a = 3 and a = 34
and -12a + 4b = - 1 i.e. -9 + 4b = - 1 and b = 2
Hence, P.I., v = 3 x 24
+
and the general solution, y = u + v = ( )2x3 3Ax B e x 2
4+ + +
x = 0 and y = 0, hence, 0 = B + 2 from which, B = - 2
2 2x3 3dy 2 3(Ax B) e e (A)
dx 3 4β β β β
= + + +β β β ββ β β β
x = 0 and dy 4dx 3
= β , hence, 4 2 3B A3 3 4
β = + +
and since B = - 2, 4 4 3A3 3 4
β = β + +
from which, A = 34
β
Hence, y = 2 x33 3x 2 e x 2
4 4β ββ β + +β ββ β
i.e. y = 2x33 32 1 x e 2 x
4 4β ββ + + +β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 443
5. The charge q in an electric circuit at time t satisfies the equation 2
2
d q dq 1L R q Edt dt C
+ + = , where
L, R, C and E are constants. Solve the equation given L = 2 H, C = 200 Γ 10-6 F and E = 250 V,
when (a) R = 200 Ξ© and (b) R is negligible. Assume that when t = 0, q = 0 and dqdt
= 0
(a)2
2
d q dq 1L R q Edt dt C
+ + = in D-operator form is: 2 1L D R D q EC
β β+ + =β ββ β
The auxiliary equation is: 2 1L m R m 0C
+ + =
and
2 26
4L 4(2)R R 200 200 200 0C 200 10m 502L 4 4
ββ Β± β β Β± β β Β±Γ= = = = β
Hence, C.F., ( ) 50tu At B eβ= +
Let the particular integral, P.I., v = k
then ( )2 1LD RD k EC
β β+ + =β ββ β
D(k) = 0 and 2D (k) D(0) 0= =
Hence, 1 (k) EC
= and k = CE = ( )( )6 1200 10 250 0.05 or20
βΓ =
Hence, P.I., v = 120
and the general solution, y = u + v = ( ) 50t 1At B e20
β+ +
t = 0 and q = 0, hence, 0 = B + 120
from which, B = - 120
( ) ( )50t 50 tdq (At B) 50e e (A)dt
β β= + β +
t = 0 and dq 0dt
= , hence, 0 50B A= β + i.e. 0 = 150 A20
β ββ β +β ββ β
i.e. A = 52
β
Hence, q = 50 t5 1 1t e2 20 20
ββ ββ β +β ββ β
i.e. q = 50t1 5 1t e20 2 20
ββ ββ +β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 444
(b) When R = 0,
2 26
4L 4(2)R R 0 0 0 j200C 200 10m 0 j502L 4 4
ββ Β± β β Β± β Β±Γ= = = = Β±
and q = (A cos 50 t + B sin 50t) + 120
q = 0 and t = 0, hence, 0 = A + 120
i.e. A = - 120
( )dq 50Asin 50t 50Bcos50tdt
= β +
t = 0 and dq 0dt
= , hence, 0 = 50B i.e. B = 0
Thus, q = - 120
cos 50t + 120
i.e. ( )1q 1 cos 50t20
= β
6. In a galvanometer the deflection ΞΈ satisfies the differential equation 2
2
d d4 4 8dt dtΞΈ ΞΈ+ + ΞΈ = . Solve
the equation for ΞΈ given that when t = 0, ΞΈ = ddtΞΈ = 2
2
2
d d4 4 8dt dtΞΈ ΞΈ+ + ΞΈ = in D-operator form is: ( )2D 4D 4 8+ + ΞΈ =
Auxiliary equation is: 2m 4m 4 0+ + =
i.e. (m + 2)(m + 2) = 0
from which, m = - 2 twice
Hence, C.F., ( ) 2tu At B eβ= +
Let the particular integral, P.I., v = k
then ( )2D 4D 4 k 8+ + =
D(k) = 0 and 2D (k) D(0) 0= =
Hence, 4k = 8 from which, k = 2
Hence, P.I., v = 2
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 445
and the general solution, ΞΈ = u + v = ( ) 2tAt B e 2β+ +
t = 0 and ΞΈ = 2, hence, 2 = B + 2 from which, B = 0
( ) ( )2t 2td (At B) 2e e (A)dt
β βΞΈ= + β +
x = 0 and d 2dtΞΈ= , hence, 2 2B A= β + from which, A = 2
Hence, ΞΈ = 2t2 t e 2β +
i.e. ΞΈ = ( )2t2 t e 1β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 446
EXERCISE 191 Page 485
2. Find the general solution of: 2
x2
d y dy3 4y 3edx dx
ββ β =
2
x2
d y dy3 4y 3edx dx
ββ β = in D-operator form is: ( )2 xD 3D 4 y 3eββ β =
Auxiliary equation is: 2m 3m 4 0β β =
i.e. (m - 4)(m + 1) = 0
from which, m = 4 and m = -1
Hence, C.F., 4x xu Ae Beβ= +
As xeβ appears in the C.F. and in the right hand side of the differential equation, let the particular
integral, P.I., v = xk x eβ
then ( )( )2 x xD 3D 4 kxe 3eβ ββ β =
D( xkxeβ ) = ( )( ) ( )( )x x x xkx e e k kxe keβ β β ββ + = β +
and ( )( ) ( )( )2 x x x x x xD (kxe ) kx e e k ke kxe 2keβ β β β β β= β β + β β = β
Hence, ( ) ( )x x x x x xkxe 2ke 3 kxe ke 4kxe 3eβ β β β β ββ β β + β =
i.e. x x x x x xkxe 2ke 3kxe 3ke 4kxe 3eβ β β β β ββ + β β =
i.e. x x5ke 3eβ ββ = i.e. β 5k = 3 and k = 35
β
hence, the particular integral, v = x3 x e5
ββ
and the general solution, y = u + v = 4x x x3Ae Be xe5
β β+ β
4. Find the general solution of: t23
2
d y dy9 6 y 12edt dt
β + =
t23
2
d y dy9 6 y 12edt dt
β + = in D-operator form is: ( )t
2 39D 6D 1 y 12eβ + =
Auxiliary equation is: 29m 6m 1 0β + =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 447
i.e. (3m - 1)(3m - 1) = 0
from which, m = 13
twice
Hence, C.F., ( )t3u At B e= +
As t3e and
t3t e appears in the C.F. and in the right hand side of the differential equation, let the
particular integral, P.I., v =t
2 3k t e
then ( )t t
2 2 3 39D 6D 1 k t e 12eβ β
β + =β ββ β
D(t
2 3k t e ) = ( ) ( )t t t t t
2 23 3 3 3 31 k tkt e e 2kt t e 2kte kte 23 3 3
β β β β β β+ = + = +β β β β β ββ β β β β β
and ( ) ( )t t t t
2 2 3 3 3 31 t 1D kt e kte 2 kt e e k3 3 3
β‘ β€β β β β β β β ββ β β β= + + +β’ β₯β β β β β β β ββ β β ββ β β β β’ β₯β β β β β β β β β£ β¦
= t t t t t
23 3 3 3 3k k k 2te t e te kte 2ke3 9 3 3
+ + + +
Hence, t t t t t t t
2 2 23 3 3 3 3 3 34 k k9 te t e 2ke 6 t e 2kte kt e 12e3 9 3
β β β β+ + β + + =β β β β
β β β β
i.e. t t t t t t t
2 2 23 3 3 3 3 3 312te kt e 18ke 2kt e 12kte kt e 12e+ + β β + =
i.e. t t3 318ke 12e= i.e. and k = 12 2
18 3=
hence, the particular integral, v = t t
2 23 32kt e t e3
=
and the general solution, y = u + v = ( )t t
23 32At B e t e3
+ +
5. Find the particular solution of 2
x2
d y dy5 9 2y 3edx dx
+ β = given that when x = 0, y = 14
and dydx
= 0
2
x2
d y dy5 9 2y 3edx dx
+ β = in D-operator form is: ( )2 x5D 9D 2 y 3e+ β =
Auxiliary equation is: 25m 9m 2 0+ β =
i.e. (5m - 1)(m + 2) = 0
from which, m = 15
and m = -2
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 448
Hence, C.F., 1 x 2x5u Ae Beβ= +
Let P.I., v = xk e
then ( )( )2 x x5D 9D 2 ke 3e+ β =
D( xke ) = xke
and ( )2 x x xD (ke ) D ke ke= =
Hence, x x x x5ke 9ke 2ke 3e+ β =
i.e. x x12ke 3e= i.e. k = 14
hence, the particular integral, v = x1 e4
and the general solution, y = u + v = 1 x 2x x5 1Ae Be e
4β+ +
x = 0 and y = 14
, hence, 14
= A + B + 14
i.e. 0 = A + B (1)
1 x 2x x5dy 1 1Ae 2Be e
dx 5 4β= β +
x = 0 and dydx
= 0, hence, 1 10 A 2B5 4
= β +
i.e. 1 1 A 2B4 5
β = β (2)
2 Γ (1) gives: 0 = 2A + 2B (3)
(2) + (3) gives: 1 11A4 5
β = i.e. A = 544
β and from (1), B = 544
Hence, y = 1 x 2x x55 5 1e e e
44 44 4ββ + +
i.e. 1 x2x x55 1y e e e
44 4ββ β
= β +β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 449
EXERCISE 192 Page 487
1. Find the general solution of: 2
2
d y dy2 3y 25sin 2xdx dx
β β =
2
2
d y dy2 3y 25sin 2xdx dx
β β = in D-operator form is: ( )22D D 3 y 25sin 2xβ β =
Auxiliary equation is: 22m m 3 0β β =
i.e. (2m - 3)(m + 1) = 0
from which, m = 32
and m = -1
Hence, C.F., 3 x x2u Ae Beβ= +
Let P.I., v = A sin 2x + B cos 2x
Hence, ( )( )22D D 3 Asin 2x Bcos 2x 25sin 2xβ β + =
D(A sin 2x + B cos 2x) = 2A cos 2x β 2B sin 2x
( ) ( )2D Asin 2x Bcos 2x D 2A cos 2x 2Bsin 2x 4Asin 2x 4Bcos 2x+ = β = β β
Hence, ( ) ( ) ( )2 4Asin 2x 4Bcos 2x 2A cos 2x 2Bsin 2x 3 Asin 2x Bcos 2x 25sin 2xβ β β β β + =
i.e. - 8A + 2B - 3A = 25
and - 8B β 2A β 3B = 0
i.e. β 11A + 2B = 25 (1)
and - 2A β 11B = 0 (2)
2 Γ (1) gives: - 22A + 4B = 50 (3)
11 Γ (2) gives: - 22A β 121B = 0 (4)
(3) β (4) gives: 125B = 50 from which, B = 50 2125 5
=
Substituting in (1) gives: - 11A + 45
= 25
from which, - 11A = 25 - 45
= 125 4 1215 5β
= and A = 121 115( 11) 5
= ββ
Thus, P.I., v = ( )11 2 1sin 2x cos 2x 11sin 2x 2cos 2x5 5 5
β + = β β
and y = u + v = ( )3x x2 1Ae Be 11sin 2x 2cos 2x
5β+ β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 450
3. Find the general solution of: 2
2
d y y 4cos xdx
+ =
2
2
d y y 4cos xdx
+ = in D-operator form is: ( )2D 1 y 4cos x+ =
Auxiliary equation is: 2m 1 0+ =
i.e. 2m 1= β and m = 1 0 j1β = Β±
Hence, C.F., u A cos x Bsin x= +
Since cos x occurs in the C.F. and in the right hand side of the differential equation,
let P.I., v = x(C sin x + D cos x)
Hence, ( ) ( )2D 1 x Csin x Dcos x 4cos x+ + =β‘ β€β£ β¦
( ) ( )( ) ( )( )D x Csin x Dcos x x Ccos x Dsin x 1 Csin x Dcos x+ = β + +β‘ β€β£ β¦
[ ] ( )( ) ( ) ( )2D v x Csin x Dcos x Ccos x Dsin x Ccos x Dsin x= β β + β + β Hence, since ( )2D 1 v 4cos x+ =
then ( )( ) ( ) ( )x Csin x Dcos x Ccos x Dsin x Ccos x Dsin xβ β + β + β + x(C sin x + D cos x)
= 4 cos x
from which, 2C = 4 from which, C = 2
and - 2D = 0 from which, D = 0
Hence, P.I., v = 2x sin x
and y = u + v = A cos x Bsin x 2xsin x+ +
4. Find the particular solution of the differential equation 2
2
d y dy3 4y 3sin xdx dx
β β = given that
when x = 0, y = 0 and dydx
= 0.
2
2
d y dy3 4y 3sin xdx dx
β β = in D-operator form is: ( )2D 3D 4 y 3sin xβ β =
Auxiliary equation is: 2m 3m 4 0β β =
i.e. (m - 4)(m + 1) = 0
from which, m = 4 and m = -1
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 451
Hence, C.F., 4x xu Ae Beβ= +
Let P.I., v = A sin x + B cos x
Hence, ( )( )2D 3D 4 Asin x Bcos x 3sin xβ β + =
D(A sin x + B cos x) = A cos x β B sin x
( ) ( )2D Asin x Bcos x D A cos x Bsin x Asin x Bcos x+ = β = β β
Hence, (-A sin x β B cos x) β 3(A cos x β B sin x) β 4(A sin x + B cos x) = 3 sin x
i.e. - A + 3B - 4A = 3
and - B β 3A β 4B = 0
i.e. β 5A + 3B = 3 (1)
and - 3A β 5B = 0 (2)
3 Γ (1) gives: - 15A + 9B = 9 (3)
5 Γ (2) gives: - 15A β 25B = 0 (4)
(3) β (4) gives: 34B = 9 from which, B = 934
Substituting in (1) gives: - 5A + 9334
β ββ ββ β
= 3
from which, 5A = 2734
- 3 = 27 102 7534 34β β
= and A = 75 1534(5) 34β
= β
Thus, P.I., v = 15 9sin x cos x34 34
β +
and y = u + v = 4x x 15 9Ae Be sin x cos x34 34
β+ β +
x = 0 when y = 0, hence, 0 = A + B + 934
i.e. A + B = - 934
(5)
4x xdy 15 94Ae Be cos x sin xdx 34 34
β= β β β
x = 0, when dydx
= 0, hence, 0 = 4A β B 1534
β i.e. 4A β B = 1534
(6)
4 Γ (5) gives: 4A + 4B = - 3634
(7)
(7) β (6) gives: 5B = - 3634
- 1534
= 5134
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 452
and B = 51 5134(5) 170
β = β
Substituting in (5) gives: 51 9A170 34
β = β from which, 51 9 51 45 6A170 34 170 170
β= β = =
Hence, y = 4x x6 51 15 9e e sin x cos x170 170 34 34
ββ β +
i.e. ( ) ( )4x x1 1y 6e 51e 15sin x 9cos x170 34
β= β β β
7. 2
02
d q dq 1L R q V sin tdt dt C
+ + = Ο represents the variation of capacitor charge in an electric circuit.
Determine an expression for q at time t seconds given that R = 40 Ξ©, L = 0.02 H,
C = 50 Γ 10-6 F, V0 = 540.8 V and Ο = 200 rad/s and given the boundary conditions that when
t = 0, q = 0 and dqdt
= 4.8
2
02
d q dq 1L R q V sin tdt dt C
+ + = Ο in D-operator form is: 20
1L D R D q V sin tC
β β+ + = Οβ ββ β
The auxiliary equation is: 2 1L m R m 0C
+ + =
and
2 26
4L 4(0.02)R R 40 40 40 0C 50 10m 10002L 2(0.02) 0.04
ββ Β± β β Β± β β Β±Γ= = = = β
Hence, C.F., ( ) 1000tu At B eβ= +
Let P.I., v = A sin Οt + B cos Οt
[ ]20
1L D R D Asin t Bcos t V sin tC
β β+ + Ο + Ο = Οβ ββ β
D(v) = AΟ cos Οt - BΟ sin Οt and 2 2 2D (v) A sin t B cos t= β Ο Ο β Ο Ο
Thus, ( ) ( )2 2 21L D R D v 0.02 A sin t B cos t 40 A cos t B sin tC
β β+ + = β Ο Ο β Ο Ο + Ο Ο β Ο Οβ ββ β
( ) 06
1 Asin t Bcos t V sin t50 10β+ Ο + Ο = ΟΓ
i.e. -800A sin 200t β 800B cos 200t + 8000A cos 200t β 8000B sin 200t + 20000A sin 200t
+ 20000B cos 200t = 540.8 sin 200t
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 453
Hence, - 800A β 8000B + 20000A = 540.8
and - 800B + 8000A + 20000B = 0
i.e. 19200A β 8000B = 540.8 (1)
and 8000A + 19200B = 0 (2)
8 Γ (1) gives: 153600A β 64000B = 4326.4 (3)
19.2 Γ (2) gives: 153600A + 368640B = 0 (4)
(3) β (4) gives: - 432640B = 4326.4
from which, B = 4326.4 0.01432640
= β
Substituting in (1) gives: 19200A β 8000(-0.01) = 540.8
i.e. 19200A + 80 = 540.8
and A = 540.8 80 460.8 0.02419200 19200
β= =
Hence, P.I., v = 0.024 sin 200t β 0.01 cos 200t
Thus, q = u + v = ( ) 1000tAt B eβ+ + 0.024 sin 200t β 0.01 cos 200t
When t = 0, q = 0, hence, 0 = B β 0.01 from which, B = 0.01
( )( )1000t 1000tdq At B 1000e Ae (0.024)(200)cos 200t (0.01)(200)sin 200tdt
β β= + β + + +
When t = 0, dqdt
= 4.8, hence, 4.8 = - 1000B + A + 4.8
i.e. A = 1000B = 1000(0.01) = 10
Thus, q = ( ) 1000t10t 0.01 eβ+ + 0.024 sin 200t β 0.010 cos 200t
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 454
EXERCISE 193 (Page 490)
1. Find the general solution of: 2
2
d y dy8 6 y 2x 40sin xdx dx
β + = +
2
2
d y dy8 6 y 2x 40sin xdx dx
β + = + in D-operator form is: ( )28D 6D 1 y 2x 40sin xβ + = +
Auxiliary equation is: 28m 6m 1 0β + =
i.e. (4m - 1)(2m - 1) = 0
from which, m = 14
and m = 12
Hence, C.F., 1 1x x4 2u Ae Be= +
Let P.I., v = ax + b + c sin x + d cos x
Hence, ( )[ ]28D 6D 1 ax b csin x d cos x 2x 40sin xβ + + + + = +
D(v) = a + c cos x β d sin x and 2D (v) = - c sin x β d cos x
Hence, ( )28D 6D 1 vβ + = 8(-c sin x β d cos x) β 6(a + c cos x β 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. β 7c + 6d = 40 (1)
- 8d β 6c + d = 0 i.e. β 6c β 7d = 0 (2)
6 Γ (1) gives: - 42c + 36d = 240 (3)
7 Γ (2) gives: - 42c β 49d = 0 (4)
(3) β (4) gives: 85d = 240 from which, 240 48d85 17
= =
Substituting in (2) gives: - 6c - 48717β ββ ββ β
= 0 from which, 7(48) 56c6(17) 17
= β = β
Hence, P.I., v = 2x + 12 5617
β sin x + 4817
cos x
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 455
and y = u + v = 1 1x x4 2Ae Be+ + 2x + 12 56
17β sin x + 48
17cos x
or y = 1 1x x4 2Ae Be+ + 2x + 12 + ( )8 6cos x 7sin x
17β
4. Find the general solution of: 2
t2
d y dy2 2y e sin tdt dt
β + =
2
t2
d y dy2 2y e sin tdt dt
β + = in D-operator form is: ( )2 tD 2D 2 y e sin tβ + =
Auxiliary equation is: 2m 2m 2 0β + =
i.e. 2( 2) ( 2) 4(1)(2) 2 4 2 j2m 1 j
2(1) 2 2β β Β± β β Β± β Β±
= = = = Β±
Hence, C.F., ( )tu e A cos t Bsin t= +
Since te sin t occurs in the C.F. and the right hand side of the differential equation,
let P.I., v = ( )tt e Csin t Dcos t+
then ( ) ( )2 t tD 2D 2 t e Csin t Dcos t e sin tβ‘ β€β + + =β£ β¦
D(v) = ( )( ) ( )t t tt e Ccos t Dsin t Csin t Dcos t t e eβ‘ β€β + + +β£ β¦
( )( ) ( ) ( )2 t t t t t tD (v) t e Csin t Dcos t Ccos t Dsin t t e e Csin t Dcos t t e e eβ‘ β€ β‘ β€= β β + β + + + + +β£ β¦ β£ β¦
( )( )t tt e e Ccos t Dsin t+ + β
Hence, ( )( ) ( ) ( )t t t t t tt e Csin t Dcos t Ccos t Dsin t t e e Csin t Dcos t t e e eβ‘ β€ β‘ β€β β + β + + + + +β£ β¦ β£ β¦
( )( )t tt e e Ccos t Dsin t+ + β - 2 ( )( ) ( )t t tt e Ccos t Dsin t 2 Csin t Dcos t t e eβ‘ β€β β + +β£ β¦
+ 2 ( )tt e Csin t Dcos t+ = te sin t
i.e. β D + 2C β D β 2C = 1 i.e. -2D = 1 and D = 12
β
and C + 2D + C β 2D = 0 i.e. C = 0
Hence, P.I., v = t t1 tt e cos t e cos t2 2
β ββ = ββ ββ β
and y = u + v = ( )te Acos t Bsin t+ tt e cos t2
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 456
5. Find the particular solution of 2
2x2
d y dy7 10y e 20dx dx
β + = + given that when x = 0, y = 0 and
dy 1dx 3
= β
2
2x2
d y dy7 10y e 20dx dx
β + = + in D-operator form is: ( )2 2xD 7D 10 y e 20β + = +
Auxiliary equation is: 2m 7m 10 0β + =
i.e. (m - 5)(m - 2) = 0
from which, m = 5 and m = 2
Hence, C.F., 5x 2xu Ae Be= +
Let P.I., v = 2xk x e a+
Thus, ( )2 2x 2xD 7D 10 k x e a e 20β‘ β€β + + = +β£ β¦
D(v) = ( )( ) ( )2x 2xkx 2e ke+ and ( )( ) ( )( )2 2x 2x 2xD (v) kx 4e 2e k 2ke= + +
Hence, ( )( ) ( )( )2x 2x 2xkx 4e 2e k 2ke+ + ( )( ) ( )2x 2x7 kx 2e 7 keβ β 2x 2x10k x e 10a e 20+ + = +
from which, 10a = 20 and a = 2
Also, 2k + 2k β 7k = 1 i.e. -3k = 1 from which, k = 13
β
Hence, P.I., v = 2x1 x e 23
β +
and y = u + v = 5x 2xAe Be+ 2x1 x e 23
β +
When x = 0, y = 0, hence, 0 = A + B + 2 i.e. A + B = -2 (1)
( ) ( )5x 2x 2x 2xdy 1 15Ae 2Be x 2e edx 3 3
β‘ β€β β β β= + β +β β β ββ’ β₯β β β β β£ β¦
When x = 0, dy 1dx 3
= β , hence, 13
β = 5A + 2B 13
β i.e. 5A + 2B = 0 (2)
2 Γ (1) gives: 2A + 2B = -4 (3)
(2) β (3) gives: 3A = 4 from which, A = 43
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 457
Substituting in (1) gives: 43
+ B = - 2 from which, B = - 2 - 43
= 103
β
Hence, y = u + v = 5x 2x4 10e e3 3
β 2x1 xe 23
β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 458
CHAPTER 52 POWER SERIES METHODS OF SOLVING
ORDINARY DIFFERENTIAL EQUATIONS EXERCISE 194 Page 492
1. (b) Determine the derivative y(5) when y = t28e
If axy e= , then (n) n axy a e= . Hence, if y =1 t28e , then
5 1 1t t(5) 2 21 8y (8) e e2 32
β β= =β ββ β
= 1t21 e
4
2. (a) Determine the derivative y(4) when y = sin 3t
If y = sin ax, then (n) n ny a sin ax2Οβ β= +β β
β β
Hence, if y = sin 3t, then ( )(4) 4 4y 3 sin 3t 81sin 3t 22Οβ β= + = + Οβ β
β β = 81 sin 3t
3. (b) Determine the derivative y(9) when y = 23cos t3
If y = cos ax, then (n) n ny a cos ax2Οβ β= +β β
β β
Hence, if y = 23cos t3
, then 9 9
(9)8
2 2 9 2 2y (3) cos t cos t3 3 2 3 3 2
Ο Οβ β β β β β= + = +β β β β β ββ β β β β β
= 9
8
2 2sin t3 3
β
4. (a) Determine the derivative y(7) when y = 92x
If y = ax , then ( )
(n) a na!y xa n !
β=β
Hence, if y = 92x , then ( )
(7) 9 79!y (2) x9 7 !
β=β
= ( ) 29! x
5. (b) Determine the derivative y(6) when y = 2 sinh 3x
If y = sinh ax, then n
(n) n nay 1 ( 1) sinh ax 1 ( 1) cosh ax2
β‘ β€ β‘ β€= + β + β ββ£ β¦ β£ β¦
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 459
Hence, if y = 2 sinh 3x, then 6
(6) 6 63y (2) 1 ( 1) sinh 3x 1 ( 1) cosh 3x2
β‘ β€ β‘ β€= + β + β ββ£ β¦ β£ β¦
= 63 2sinh 3x 0+ = 1458 sinh 3x
6. (a) Determine the derivative y(7) when y = cosh 2x
If y = cosh ax, then n
(n) n nay 1 ( 1) sinh ax 1 ( 1) cosh ax2
β‘ β€ β‘ β€= β β + + ββ£ β¦ β£ β¦
Hence, if y = cosh 2x, then 7
(7) 7 72y 1 ( 1) sinh 2x 1 ( 1) cosh 2x2
β‘ β€ β‘ β€= β β + + ββ£ β¦ β£ β¦
= 6 72 2sinh 2x 0 2 sinh 2x+ = = 128 sinh 2x
7. (b) Determine the derivative y(7) when y = 1 ln 2t3
If y = ln ax, then ( ) ( )n 1(n)n
n 1 !y 1
xβ β
= β
Hence, if y = 1 ln 2t3
, then ( ) ( )7 1(7)7 7
7 1 !1 6!y 13 t 3 t
β ββ β= β =β ββ β
= 7
240t
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 460
EXERCISE 195 Page 494 2. If y = 3 2xx e find y(n) and hence y(3) by using the theorem of Leibniz. Since y = 3 2xx e then let v = 3x and u = 2xe and the nβth derivative of 2xe is n 2x2 e
Thus, (n) (n) (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!
β ββ= + + +
= ( ) ( )( ) ( )( ) ( )( )n 2x 3 n 1 2x 2 n 2 2x (n 3) 2xn n 1 n 2n(n 1)2 e x n 2 e 3x 2 e 6x 2 e (6)2! 3!
β β ββ ββ+ + +
= ( )( )n 3 2x 2 n 1 2x n 2 2x n 3 2x2 x e 3nx 2 e 3n(n 1)2 e x n n 1 n 2 2 eβ β β+ + β + β β
or (n) 2x n 3 3 3 2 2y e 2 2 x 3nx (2) 3n(n 1)x(2) n(n 1)(n 2)β= + + β + β β
= 2x n 3 3 2e 2 8x 12nx n(n 1)(6x) n(n 1)(n 2)β + + β + β β
Hence, (3) 2x 0 3 2y e 2 8x 36x 3(2)6x 3(2)(1)= + + +
= 2x 3 2e 8x 36x 36x 6+ + +
4. Use the theorem of Leibniz to determine the 5th derivative of: y = 3x cos x
Since y = 3x cos x then let u = cos x and v = 3x and (n) n nu 1 cos x2Οβ β= +β β
β β
(n) n (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!
β ββ= + + +
Hence, ( ) ( ) ( )(n) 3 2n (n 1) n(n 1) (n 2)y cos x x n cos x 3x cos x 6x2 2 2! 2Ο β Ο β β Οβ β β β β β= + + + + +β β β β β β
β β β β β β
( )n(n 1(n 2) (n 3)cos x 63! 2
β β β Οβ β+ +β ββ β
and ( ) ( ) ( )(5) 3 25 4 5(4) 3 5(4)(3) 2y x cos x 5 3x cos x 6x cos x 6 cos x2 2 2! 2 3! 2Ο Ο Ο Οβ β β β β β β β= + + + + + + +β β β β β β β β
β β β β β β β β
= 3 2x sin x 15x cos x 60x sin x 60( cos x)β + + + β
= ( ) ( )3 260x x sin x 15x 60 cos xβ + β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 461
6. If y = 5x ln 2x find y(3) by using the theorem of Leibniz.
Since y = 5x ln 2x then let u = 5x and v = ln 2x and ( )
n a n 5 na! 5!u x xa n ! (5 n)!
β β= =β β
(n) n (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!
β ββ= + + +
(n) 5 n 6 n 7 n2
8 n3
5! 5! 1 n(n 1) 5! 1y x ln 2x n x x(5 n)! (6 n)! x 2! (7 n)! x
n(n 1)(n 2) 5! 2x3! (8 n)! x
β β β
β
β‘ β€ ββ β β β= + + ββ β β ββ’ β₯β β ββ β β β β£ β¦β β β β+ β ββ β β
Hence, 3 2 3 4 52 3
5! 5! 1 3(2) 5! 1 3(2)(1) 5! 2y x ln 2x (3) x x x2! 3! x 2! (4)! x 3! 5! x
β β β β β β= + + β +β β β β β ββ β β β β β
= 2 2 2 260x ln 2x 60x 15x 2x+ β +
= 2 260x ln 2x 47x+
i.e. ( )(3) 2y x 47 60ln 2x= +
8. If y = ( )3 2 2xx 2x e+ determine an expression for y(5) by using the theorem of Leibniz. Since y = ( )3 2 2xx 2x e+ then let u = 2xe and v = ( )3 2x 2x+ and n n 2xu 2 e=
(n) n (n 1) (1) (n 2) (2)n(n 1)y u v nu v u v .....2!
β ββ= + + +
Hence, ( )( ) ( ) ( )(n) n 2x 3 2 n 1 2x 2 n 2 2xn(n 1)y 2 e x 2x n2 e 3x 4x 2 e 6x 42!
β ββ= + + + + +
( )n 3 2xn(n 1)(n 2) 2 e 63!
ββ β+
and ( ) ( ) ( ) ( )(5) 5 2x 3 2 4 2x 2 3 2x 2 2x5(4) 5(4)(3)y 2 e x 2x (5) 2 e 3x 4x 2 e 6x 4 2 e 62 3!
= + + + + + +
= 2x 5 3 6 2 2e 2 x 2 x (16)15x (16)(20x) 60x(8) (8)(40) 240+ + + + + +
= 2x 5 3 2e 2 x 304x 800x 560+ + +
= 2x 5 3 4 2 4 4e 2 x 2 (19x ) 2 (50)(x) 2 (35)+ + +
= 2x 4 3 2e 2 2x 19x 50x 35+ + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 462
EXERCISE 196 Page 497
1. Determine the power series solution of the differential equation: 2
2
d y dy2x y 0dx dx
+ + = using the
Leibniz-Maclaurin method, given that at x = 0, y = 1 and dydx
= 2.
2
2
d y dy2x y 0dx dx
+ + =
(i) The differential equation is rewritten as: yβ²β² + 2xyβ² + y = 0 and from the Leibniz theorem of
equation (13), page 493 of textbook, each term is differentiated n times, which gives:
(n 2) (n 1) (n) (n)y 2 y (x) n y (1) 0 y 0+ ++ + + + =
i.e. (n 2) (n 1) (n)y 2x y (2n 1) y 0+ ++ + + = (1)
(ii) At x = 0, equation (1) becomes:
(n 2) (n)y (2n 1) y 0+ + + =
from which, (n 2) (n)y (2n 1) y+ = β +
This is the recurrence formula.
(iii) For n = 0, ( ) ( )0 0y '' y= β
n = 1, ( ) ( )0 0y ''' 3 y '= β
n = 2, ( ) ( )(4)000
y 5 y '' 5(y)= β =
n = 3, ( ) ( )(5)00
y 7 y '''= β = ( ) ( )0 07 3 y ' 3 7 y 'β β = Γ
n = 4, ( ) ( )(6) (4)
0 0y 9 y= β = ( ) ( )0 0
9 5 y 5 9 yβ = β Γ
n = 5, ( ) ( )(7) (5)
0 0y 11 y= β = ( ) ( )0 0
11 3 7 y ' 3 7 11 y 'β Γ = β Γ Γ
n = 6, ( ) ( )(8) (6)
0 0y 13 y= β = ( ) ( )0 0
13 5 9 y 5 9 13 yβ β Γ = Γ Γ
(iv) Maclaurinβs theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4
(4)0 0 0 0 0
x x xy x y ' y '' y ''' y ....2! 3! 4!
+ + + + +
Thus, y = ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 5
0 0 0 0 0 0
x x x xy x y ' y 3 y ' 5 y 3 7 y '2! 3! 4! 5!
+ + β + β + + Γ
( ) ( ) 6 7
0 0
x x5 9 y 3 7 11 y '6! 7!
+ β Γ + β Γ Γ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 463
(v) Collecting similar terms together gives:
y = ( )2 4 6 8
0
x 5 x 5 9 x 5 9 13xy 1 ...2! 4! 6! 8!
β§ β«Γ Γ Γβ + β + ββ¨ β¬
β© β
( )3 5 7
0
3x 3 7 x 3 7 11xy ' x ...3! 5! 7!
β§ β«Γ Γ Γ+ β + β +β¨ β¬
β© β
At x = 0, y = 1 and dydx
= 2, hence, ( )0y 1= and ( )0
y ' 2= .
Hence, the power series solution of the differential equation: 2
2
d y dy2x y 0dx dx
+ + = is:
y = 2 4 6 8x 5 x 5 9 x 5 9 13x1 ...
2! 4! 6! 8!β§ β«Γ Γ Γ
β + β + ββ¨ β¬β© β
3 5 73 x 3 7 x 3 7 11x2 x ...3! 5! 7!
β§ β«Γ Γ Γ+ β + β +β¨ β¬
β© β
3. Find the particular solution of the differential equation: ( )2
22
d y dyx 1 x 4y 0dx dx
+ + β = using the
Leibniz-Maclaurin method, given the boundary conditions that at x = 0, y = 1 and dydx
= 1.
( )2
22
d y dyx 1 x 4y 0dx dx
+ + β =
i.e. ( )2x 1+ yβ²β² + xyβ² - 4y = 0
i.e. ( ) 2 (n 2) (n 1) (n) (n 1) n (n)n(n 1)x 1 y ny (2x) y (2) y x ny (1) 4y 02!
+ + +ββ§ β«+ + + + + β =β¨ β¬β© β
i.e. ( ) ( )2 (n 2) (n 1) (n)x 1 y 2nx x y (n(n 1) n 4) y 0+ ++ + + + β + β =
At x = 0, ( )(n 2) 2 (n)y n 4 y 0+ + β =
from which, ( )(n 2) 2 (n)y 4 n y+ = β which is the recurrence formula.
For n = 0, ( ) ( )0 0y '' 4 y=
n = 1, ( ) ( )0 0y ''' 3 y '=
n = 2, ( )(4)
0y 0=
n = 3, ( ) ( )(5)00
y 5 y '''= β = ( ) ( )( )0 05 3 y ' 5 3 y 'β β = β
n = 4, ( ) ( )(6) (4)
0 0y 12 y= β = 12(0) 0β =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 464
n = 5, ( ) ( )(7) (5)
0 0y 21 y= β = ( ) ( )0 0
21 5 3 y ' 315 y 'β β Γ =
Maclaurinβs theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4
(4)0 0 0 0 0
x x xy x y ' y '' y ''' y ....2! 3! 4!
+ + + + +
Thus, y = ( ) ( ) ( ) ( ) ( ) 2 3 4 5
0 0 0 0 0
x x x xy x y ' 4 y 3 y ' 0 3 5 y ' 02! 3! 4! 5!
+ + + + + β Γ + ( ) 7
0
x 315 y '7!
+
i.e. y = ( ) 20
y 1 2x+ ( )3 5 7
0
x x xy ' x ...2 8 16
β§ β«+ + β + +β¨ β¬
β© β
At x = 0, y = 1 and dydx
= 1, hence, ( )0y 1= and ( )0
y ' 1= .
Hence, the power series solution of the differential equation: ( )2
22
d y dyx 1 x 4y 0dx dx
+ + β = is:
y = 21 2x+3 5 7x x xx ...
2 8 16β§ β«
+ + β + +β¨ β¬β© β
i.e. y = 3 5 7
2 x x x1 x 2x .....3 8 16
+ + + β + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 465
EXERCISE 197 Page 503 1. Produce, using Frobeniusβ method, a power series solution for the differential equation:
2
2
d y dy2x y 0dx dx
+ β =
2
2
d y dy2x y 0dx dx
+ β = may be rewritten as: 2xyβ²β² + yβ² - y = 0
(i) Let a trial solution be of the form y = xca0 + a1x + a2x2 + a3x3 + β¦ + arxr+β¦
where a0 β 0,
i.e. y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + β¦ + arxc+r +β¦
(ii) Differentiating gives:
yβ² = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + β¦. + ar(c + r)xc+r-1 + β¦
and yβ²β² = a0c(c β 1)xc-2 + a1c(c + 1)xc-1 + a2(c + 1)(c + 2)xc + β¦. + ar(c + r - 1)(c + r)xc+r-2 + β¦
(iii) Substituting y, yβ² and yβ²β² into each term of the given equation 2xyβ²β² + yβ² - y = 0 gives:
2xyβ²β² = 2a0c(c β 1)xc-1 + 2a1c(c + 1)xc + 2a2(c + 1)(c + 2)xc+1 + β¦
+ 2ar(c + r β1)(c + r)xc+r-1 + β¦ (a)
yβ² = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + β¦. + ar(c + r)xc+r-1 + β¦ (b)
-y = -a0xc - a1xc+1 - a2xc+2 - a3xc+3 - β¦ - arxc+r -β¦ (c)
(iv) The sum of these three terms forms the left-hand side of the equation. Since the right-hand side
is zero, the coefficients of each power of x can be equated to zero.
For example, the coefficient of xc-1 is equated to zero giving: 2a0c(c β 1) + a0c = 0
or a0 c [2c β 2 + 1] = a0 c(2c - 1) = 0 (1)
Equation (1) is the indicial equation, from which, c = 0 or c = 12
The coefficient of xc is equated to zero giving: 2a1c(c + 1) + a1(c + 1) - a0 = 0
i.e. a1 (2c2 + 2c + c + 1) - a0 = a1(2c2 + 3c + 1) - a0 = 0
or a1(2c + 1)(c + 1) - a0 = 0 (2)
Replacing r by (r + 1) will give:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 466
in series (a), 2ar+1(c + r + 1)(c + r)xc+r
in series (b), ar+1(c + r + 1)xc+r
in series (c), -arxc+r
Equating the total coefficients of xc+r to zero gives:
2ar+1(c + r + 1)(c + r) + ar+1(c + r + 1) - ar = 0
which simplifies to: ar+1(c + r + 1)(2c + 2r + 1) - ar = 0 (3)
(a) When c = 0:
From equation (2), if c = 0, a1(1 Γ 1) - a0 = 0, i.e. a1 = 0a
From equation (3), if c = 0, ar+1(r + 1)(2r + 1) - ar = 0, i.e. ar+1 = ra(r 1)(2r 1)+ +
r β₯ 0
Thus, when r = 1, 012
aaa(2 3) (2 3)
= =Γ Γ
since 1 0a a=
when r = 2, 023
aaa(3 5) (2 3)(3 5)
= =Γ Γ Γ
when r = 3, 3 0 04
a a aa(4 7) (2 3)(3 5)(4 7) (2 3 4)(3 5 7)
= = =Γ Γ Γ Γ Γ Γ Γ Γ
and so on.
The trial solution is: y = xca0 + a1x + a2x2 + a3x3 + β¦ + arxr +β¦
Substituting c = 0 and the above values of a1, a2, a3, β¦ into the trial solution gives:
y = 0 2 3 40 0 00 0
a a ax a a x x x x ...(2 3) (2 3)(3 5) (2 3 4)(3 5 7)
β§ β«β β β β β β+ + + + +β¨ β¬β β β β β βΓ Γ Γ Γ Γ Γ Γβ β β β β β β© β
i.e. y = ( ) ( )( ) ( )( )
2 3 4
0x x xa 1 x ...
2 3 2 3 3 5 2 3 4 3 5 7β§ β«βͺ βͺ+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γβͺ βͺβ© β
(4)
(b) When c = 12
:
From equation (2), if c = 12
, a1 ( ) 322
β ββ ββ β
- a0 = 0, i.e. a1 = 0a3
From equation (3), if c = 12
, ar+1 ( )1 r 1 1 2r 12
β β+ + + +β ββ β
- ar = 0,
i.e. ar+1 ( )3r 2r 22
β β+ +β ββ β
- ar = ar+1(2 2r + 5r +3) - ar = 0,
i.e. ar+1 = ra(2r 3)(r 1)+ +
r β₯ 0
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 467
Thus, when r = 1, 012
aaa(2 5) (2 3 5)
= =Γ Γ Γ
since a1 = 0a3
when r = 2, 023
aaa(3 7) (2 3 5)(3 7)
= =Γ Γ Γ Γ
when r = 3, 3 04
a aa(4 9) (2 3 4)(3 5 7 9)
= =Γ Γ Γ Γ Γ Γ
and so on.
The trial solution is: y = xca0 + a1x + a2x2 + a3x3 + β¦ + arxr +β¦
Substituting c = 12
and the above values of a1, a2, a3, β¦ into the trial solution gives:
y =1
2 3 40 0 0 020
a a a ax a x x x x ...3 2 3 5 (2 3 5)(3 7) (2 3 4)(3 5 7 9)
β§ β«β β β ββ β β β+ + + + +β¨ β¬β β β ββ β β βΓ Γ Γ Γ Γ Γ Γ Γ Γ Γβ β β β β β β β β© β
i.e. y = 1 2 3 42
0x x x xa x 1 ...
(1 3) (1 2)(3 5) (1 2 3)(3 5 7) (1 2 3 4)(3 5 7 9)β§ β«+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γβ© β
(5)
Let 0a = A in equation (4), and 0a = B in equation (5).
Hence, y = ( ) ( ) ( ) ( ) ( )
2 3 4x x xA 1 x ...2 3 2 3 3 5 2 3 4 3 5 7
β§ β«βͺ βͺ+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γβͺ βͺβ© β
+ 1 2 3 42 x x x xB x 1 ...
(1 3) (1 2)(3 5) (1 2 3)(3 5 7) (1 2 3 4)(3 5 7 9)β§ β«
+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γβ© β
3. Determine the power series solution of the differential equation: 2
2
d y dy3x 4 y 0dx dx
+ β = using the
Frobenius method.
2
2
d y dy3x 4 y 0dx dx
+ β = may be rewritten as: 3xyβ²β² + 4yβ² - y = 0
(i) Let a trial solution be of the form y = xca0 + a1x + a2x2 + a3x3 + β¦ + arxr+β¦
i.e. y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + β¦ + arxc+r +β¦
(ii) Differentiating gives:
yβ² = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + β¦. + ar(c + r)xc+r-1 + β¦
and yβ²β² = a0c(c β 1)xc-2 + a1c(c + 1)xc-1 + a2(c + 1)(c + 2)xc + β¦. + ar(c + r - 1)(c + r)xc+r-2 + β¦
(iii) Substituting y, yβ² and yβ²β² into each term of the given equation 3xyβ²β² + 4yβ² - y = 0 gives:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 468
3xyβ²β² = 3a0c(c β 1)xc-1 + 3a1c(c + 1)xc + 3a2(c + 1)(c + 2)xc+1 + β¦
+ 3ar(c + r β1)(c + r)xc+r-1 + β¦ (a)
4yβ² = 4a0cxc-1 + 4a1(c + 1)xc + 4a2(c + 2)xc+1 + β¦. + 4ar(c + r)xc+r-1 + β¦ (b)
- y = -a0xc - a1xc+1 - a2xc+2 - a3xc+3 - β¦ - arxc+r -β¦ (c)
(iv) The coefficient of xc-1 is equated to zero giving: 3a0c(c β 1) + 4a0c = 0
or a0 c [3c β 3 + 4] = a0 c(3c + 1) = 0
This is the indicial equation, from which, c = 0 or c = 13
β
The coefficient of xc is equated to zero giving: 3a1c(c + 1) + 4a1(c + 1) - a0 = 0
i.e. a1 (3c(c + 1) +4(c+1)) - a0 = a1(c + 1)(3c + 4) - a0 = 0
or a1(c + 1)(3c + 4) - a0 = 0 (1)
Equating the total coefficients of xc+r to zero gives:
3ar+1(c + r)(c + r + 1) + 4ar+1(c + r + 1) - ar = 0
i.e. ar+1(c + r + 1)(3c + 3r + 4) - ar = 0
which simplifies to: rr 1
aa
(c r 1)(3c 3r 4)+ =+ + + +
(2)
(a) When c = 0:
From equation (1), if c = 0, a1(4) - a0 = 0, i.e. a1 = 0a4
From equation (2), if c = 0, rr 1
aa(r 1)(3r 4)+ =+ +
r β₯ 0
Thus, when r = 1, 012
aaa(2 7) (2 4 7)
= =Γ Γ Γ
since 01
aa4
=
when r = 2, 0 023
a aaa(3 10) (3 10)(2 4 7) (1 2 3)(4 7 10)
= = =Γ Γ Γ Γ Γ Γ Γ Γ
when r = 3, 3 0 04
a a aa(4 13) (4 13)(3 10)(2 4 7) (2 3 4)(4 7 10 13)
= = =Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ
and so on.
The trial solution is: y = xca0 + a1x + a2x2 + a3x3 + β¦ + arxr +β¦
Substituting c = 0 and the above values of a1, a2, a3, β¦ into the trial solution gives:
y = 0 2 3 40 0 0 00
a a a ax a x x x x ...4 (1 2)(4 7) (1 2 3)(4 7 10) (2 3 4)(4 7 10 13)
β§ β«β β β β β ββͺ βͺ+ + + + +β¨ β¬β β β β β βΓ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γβͺ βͺβ β β β β β β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 469
i.e. y = ( ) ( )( )
2 3 4
0x x x xa 1 ...
(1 4) 1 2)(4 7 1 2 3 4 7 10 (2 3 4)(4 7 10 13)β§ β«βͺ βͺ+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γβͺ βͺβ© β
(3)
(b) When c = 13
β :
From equation (1), if c = 13
β , a1 ( )2 33
β ββ ββ β
- a0 = 0, i.e. a1 = 0a2
From equation (2), if c = 13
β , ( )r r r
r 1a a aa 12 (3r 2)(r 1)(3r 2)3(r 1)r 3r 3
33
+ = = =+ +β β + ++ +β β
β β
r β₯ 0
Thus, when r = 1, 0 012 2
a aaa(5 2) (2 5) (1 2)(2 5)
= = =Γ Γ Γ Γ
since a1 = 0a2
when r = 2, 023
aaa(8 3) (1 2 3)(2 5 8)
= =Γ Γ Γ Γ Γ
when r = 3, 3 04
a aa(11 4) (1 2 3 4)(2 5 8 11)
= =Γ Γ Γ Γ Γ Γ Γ
and so on.
The trial solution is: y = xca0 + a1x + a2x2 + a3x3 + β¦ + arxr +β¦
Substituting c = 13
β and the above values of a1, a2, a3, β¦ into the trial solution gives:
y =1
2 30 0 0 030
a a a ax a x x x ...2 (1 2)(2 5) (1 2 3)(2 5 8) (1 2 3 4)(2 5 8 11)
β β§ β«+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γβ© β
i.e. y = 1 2 3 43
0x x x xa x 1 ...
(1 2) (1 2)(2 5) (1 2 3)(2 5 8) (1 2 3 4)(2 5 8 11)β β§ β«
+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γ Γβ© β
(4)
Let 0a = A in equation (3), and 0a = B in equation (4).
Hence, y = ( ) ( ) ( ) ( ) ( )
2 3 4x x xA 1 x ...1 4 1 2 4 7 1 2 3 4 7 10
β§ β«βͺ βͺ+ + + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γβͺ βͺβ© β
+ 1 2 33 x x xB x 1 ...
(1 2) (1 2)(2 5) (1 2 3)(2 5 8)β β§ β«
+ + + +β¨ β¬Γ Γ Γ Γ Γ Γ Γβ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 470
EXERCISE 198 Page 508
1. Determine the power series solution of Besselβs equation: ( )2
2 2 22
d y dyx x x v y 0dx dx
+ + β = when
v = 2, up to and including the term in x6.
The complete solution of Besselβs equation: ( )2
2 2 22
d y dyx x x v y 0dx dx
+ + β = is:
y =2 4 6
v2 4 6
x x xA x 1 ...2 (v 1) 2 2!(v 1)(v 2) 2 3!(v 1)(v 2)(v 3)
β§ β«β + β +β¨ β¬+ Γ + + Γ + + +β© β
+ 2 4 6
v2 4 6
x x xB x 1 ...2 (v 1) 2 2!(v 1)(v 2) 2 3!(v 1)(v 2)(v 3)
β β§ β«+ + + +β¨ β¬β Γ β β Γ β β ββ© β
and y =2 4 6
v2 4 6
x x xA x 1 ...2 (v 1) 2 2!(v 1)(v 2) 2 3!(v 1)(v 2)(v 3)
β§ β«β + β +β¨ β¬+ Γ + + Γ + + +β© β
when v is a
positive integer.
Hence, when v = 2, y = 2 4
22 4
x xA x 1 ...2 (2 1) 2 2!(2 1)(2 2)
β§ β«β + +β¨ β¬+ Γ + +β© β
i.e. y = 2 4 4 6
2 2x x x xA x 1 ... or A x ...12 384 12 384
β§ β« β§ β«β + β β + ββ¨ β¬ β¨ β¬
β© β β© β
2. Find the power series solution of the Bessel function: ( )2 2 2x y '' xy ' x v y 0+ + β = in terms of the
Bessel function 3J (x) when v = 3. Give the answer up to and including the term in x7.
vJ (x) = v 2 4
2 4
x 1 x x ...2 (v 1) 2 (1!) (v 2) 2 (2!) (v 3)
β§ β«β β β + ββ¨ β¬β β Ξ + Ξ + Ξ +β β β© β provided v is not a negative integer.
Hence, when v = 3, 3J (x) = 3 2 4
2 4
x 1 x x ...2 (3 1) 2 (1!) (3 2) 2 (2!) (3 3)
β§ β«β β β + ββ¨ β¬β β Ξ + Ξ + Ξ +β β β© β
i.e. 3J (x) = 3 2 4 3 5 7
2 5 5 8
x 1 x x x x x... or ...2 4 2 5 2 6 8 4 2 5 2 6
β§ β«β β β + β β + ββ¨ β¬β β Ξ Ξ Ξ Ξ Ξ Ξβ β β© β
3. Evaluate the Bessel functions 0J (x) and 1J (x) when x = 1, correct to 3 decimal places.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 471
0J (x) = ( )
2 4 6
22 2 6 24
x x x1 ...2 (1!) 2 (3!)2 2!
β + β +
and when x = 1, 0J (x) = ( )
2 4 6
22 2 6 24
1 1 11 ...2 (1!) 2 (3!)2 2!
β + β +
= 1 β 0.25 + 0.015625 β 0.000434 + β¦
= 0.765 correct to 3 decimal places
1J (x) = 3 5 7
3 5 7
x x x x ...2 2 (1!)(2!) 2 (2!)(3!) 2 (3!)(4!)β + β +
and when x = 1, 1J (x) = 3 5 7
3 5 7
1 1 1 1 ...2 2 (1!)(2!) 2 (2!)(3!) 2 (3!)(4!)β + β +
= 0.5 β 0.0625 + 0.002604 β 0.000054
= 0.440 correct to 3 decimal places
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 472
EXERCISE 199 Page 511 1. Determine the power series solution of the Legrandre equation: ( )21 x y '' 2xy ' k(k 1)y 0β β + + =
when (a) k = 0 (b) k = 2, up to and including the term in x5.
The power series solution of the Legrandre equation is:
y = 2 40
k(k 1) k(k 1)(k 2)(k 3)a 1 x x ..2! 4!+ + β +β§ β«β + ββ¨ β¬
β© β
+ 3 51
(k 1)(k 2) (k 1)(k 3)(k 2)(k 4)a x x x ..3! 5!
β + β β + +β§ β«β + ββ¨ β¬β© β
(a) When k = 0, y = 0a 1 0 0 ..β + β + 3 51
( 1)( 2) ( 1)( 3)( 2)( 4)a x x x ..3! 5!
β + β β + +β§ β«β + ββ¨ β¬β© β
i.e. y = 0a + 3 5
1x xa x ....3 5
β β+ + +β β
β β
(b) When k = 2, y = 2 40
2(3) 2(3)(0)(5)a 1 x x ..2! 4!
β§ β«β + ββ¨ β¬β© β
+ 3 51
(1)(4) (1)( 1)(4)(6)a x x x ..3! 5!
ββ§ β«β + ββ¨ β¬β© β
i.e. y = ( )20a 1 3xβ + 3 5
12 1a x x x ....3 5
β ββ β ββ ββ β
2. Find the following Legrendre polynomials: (a) P1(x) (b) P4(x) (c) P5(x) (a) Since in 1P (x) , n = k = 1, then from the second part of equation (47), page 510 of textbook, i.e.
the odd powers of x:
y = 1a x 0β = 1a x
1a is chosen to make y = 1 when x = 1
i.e. 1 = 1a
Hence, 1P (x) x=
(b) Since in 4P (x) , n = k = 4, then from the first part of equation (47), page 510 of textbook, i.e. the
even powers of x:
y = 2 40
4(5) 4(5)(2)(7)a 1 x x 02! 4!
β§ β«β + +β¨ β¬β© β
= 2 40
35a 1 10x x3
β§ β«β +β¨ β¬β© β
0a is chosen to make y = 1 when x = 1
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 473
i.e. 1 = 0 0 035 2 8a 1 10 a 1 10 11 a3 3 3
β β β ββ + = β + =β β β ββ β β β
, from which, 0a = 38
Hence, 4P (x) = 2 43 351 10x x8 3β§ β«β +β¨ β¬β© β
or 4P (x) = ( )4 21 35x 30x 38
β +
(c) Since in 5P (x) , n = k = 5, then from the second part of equation (47), i.e. the odd powers of x:
y = 3 51
(k 1)(k 2) (k 1)(k 3)(k 2)(k 4)a x x x ...3! 5!
β + β β + +β§ β«β + ββ¨ β¬β© β
i.e. y = 3 5 3 51 1
(4)(7) (4)(2)(7)(9) 14 21a x x x 0 a x x x3! 5! 3 5
β§ β« β§ β«β + β = β +β¨ β¬ β¨ β¬β© β β© β
1a is chosen to make y = 1 when x = 1.
i.e. 1 = 1 1 114 21 15 70 63 8a 1 a a3 5 15 15
β +β§ β« β ββ + = =β¨ β¬ β ββ© β β β
from which, 115a8
=
Hence, 5P (x) = 3 515 14 21x x x8 3 5β ββ +β ββ β
or 5P (x) = ( )5 31 63x 70x 15x8
β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 474
CHAPTER 53 AN INTRODUCTION TO PARTIAL
DIFFERENTIAL EQUATIONS EXERCISE 200 Page 514
2. Solve u 2t cost
β= ΞΈ
β given that u = 2t when ΞΈ = 0
Since u 2t cost
β= ΞΈ
β then u = 2t cos dtΞΈβ« = (2 cos ΞΈ) t dtβ«
= ( )2
2t2cos f ( ) t cos f ( )2
ΞΈ + ΞΈ = ΞΈ+ ΞΈ
u = 2t when ΞΈ = 0, hence, 2t = 2t f ( )+ ΞΈ
from which, ( ) 2f 2t tΞΈ = β
Hence, u = 2 2t cos 2t tΞΈ+ β
or u = ( )2t cos 1 2tΞΈ β +
4. Verify that u = ye cos xβ is a solution of 2 2
2 2
u u 0x yβ β
+ =β β
Since u = ye cos xβ then yu e ( sin x)x
ββ= β
β and
2y
2
u e cos xx
ββ= β
β
Also, yu e cos xy
ββ= β
β and
2y
2
u e cos xx
ββ=
β
Hence, 2 2
y y2 2
u u e cos x e cos x 0x y
β ββ β+ = β + =
β β
6. Solve ( )2
22
u y 4x 1xβ
= ββ
given that x = 0, u = sin y and uxββ
= cos 2y
Since ( )2
22
u y 4x 1xβ
= ββ
then ( )3
2u 4xy 4x 1 dx y x f (y)x 3
β ββ= β = β +β ββ β β β«
x = 0 when uxββ
= cos 2y, hence, cos 2y = 0 + f(y)
Hence, 3u 4xy x cos 2y
x 3β ββ
= β +β ββ β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 475
and u = 3 4 24x x xy x cos 2y dx y x cos 2y F(y)
3 3 2β‘ β€β β β β
β + = β + +β’ β₯β β β ββ β β β β£ β¦
β«
x = 0 when u = sin y, hence, sin y = F(y)
Thus, u =4 2x xy xcos 2y sin y
3 2β β
β + +β ββ β
8. Show that u(x, y) = xxyy
+ is a solution of 2 2
2
u u2x y 2xx y yβ β
+ =β β β
Since, u = xxyy
+ then u 1yx yβ
= +β
and 22
u xx x xyy y
ββ= β = β
β
and 2
2 3
u 2xy yβ
=β
Also, 2
2 2
u x 1x 1x y x y y
β ββ β= β = ββ ββ β β β β
Hence, L.H.S. = 2 2
2 2 3
u u 1 2x2x y 2x 1 yx y y y y
β β β ββ β+ = β +β β β ββ β β β β β β
= 2 2
2x 2x2x 2x R.H.S.y y
β + = =
10. Verify that Ο(x, y) = x cos y + xe sin y satisfies the differential equation
2 2
2 2 x cos y 0x yβ Ο β Ο
+ + =β β
Since Ο = x cos y + xe sin y then xcos y e sin yxβΟ
= +β
and 2
x2 e sin y
xβ Ο
=β
and xx sin y e cos yyβΟ
= β +β
and 2
x2 x cos y e sin y
yβ Ο
= β ββ
Hence, L.H.S. = ( )2 2
x x2 2 x cos y e sin y x cos y e sin y x cos y
x yβ Ο β Ο
+ + = + β β +β β
= x xe sin y x cos y e sin y x cos y 0 R.H.S.β β + = =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 476
EXERCISE 201 Page 516 2. Solve T β²β²- c2Β΅T = 0 given c = 3 and Β΅ = -1 Since T β²β²- c2Β΅T = 0 and c = 3 and Β΅ = -1, then T β²β²- (3)2(-1)T = 0 i.e. T β²β² + 9T = 0
If Tβ²β² + 9T = 0 then the auxiliary equation is:
2m 9 0+ = i.e. 2m 9= β from which, m 9 j3= β = Β± or 0 Β± j3
Thus, the general solution is: T = 0e A cos3t Bsin 3t+
= A cos 3t + B sin 3t
3. Solve X β²β² = Β΅X given Β΅ = 1 Since X β²β² = Β΅X and Β΅ = 1, then X β²β² - X = 0
If X β²β² - X = 0 then the auxiliary equation is:
2m 1 0β = i.e. 2m 1= from which, m = 1 or m = -1
Thus, the general solution is: X = x xAe Beβ+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 477
EXERCISE 202 Page 520 1. An elastic string is stretched between two points 40 cm apart. Its centre point is displaced 1.5 cm
from its position of rest at right angles to the original direction of the string and then released
with zero velocity. Determine the subsequent motion u(x, t) by applying the wave equation
2 2
2 2 2
u 1 ux c tβ β
=β β
with 2c = 9.
The elastic string is shown in the diagram below.
1. The boundary and initial conditions given are:
u(0, t) 0u(40, t) 0
= β«β¬= β
u(x, 0) = f (x) = 1.5 x20
0 β€ x β€ 20
= 1.5 x 320
β + = 60 1.5x20β 20 β€ x β€ 40
t 0
u 0t =
ββ‘ β€ =β’ β₯ββ£ β¦ i.e. zero initial velocity
2. Assuming a solution u = XT, where X is a function of x only, and T is a function of t only,
then u X 'Txβ
=β
and 2
2
u X ''Txβ
=β
and u XT 'yβ
=β
and 2
2
u XT ''yβ
=β
Substituting into the partial differential equation, 2 2
2 2 2
u 1 ux c tβ β
=β β
gives: 2
1X ''T XT ''c
= i.e. 1X ''T XT ''9
= since 2c 9=
3. Separating the variables gives: X '' T ''X 9T
=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 478
Let constant, Β΅ = X '' T ''X 9T
= then Β΅ = X ''X
and Β΅ = T ''9T
from which, Xβ²β² - Β΅X = 0 and Tβ²β² - 9Β΅ T = 0
4. Letting Β΅ = - 2p to give an oscillatory solution gives
Xβ²β² + 2p X = 0 and the auxiliary equation is: 2 2m p+ = 0 from which, 2m p jp= β = Β±
and Tβ²β² + 9 2p T = 0 and the auxiliary equation is:
2 2m 9p+ = 0 from which, 2m 9p j3p= β = Β±
5. Solving each equation gives: X = A cos px + B sin px and T = C cos 3pt + D sin 3pt
Thus, u(x, t) = A cos px + B sin pxC cos 3pt + D sin 3pt
6. Applying the boundary conditions to determine constants A and B gives:
(i) u(0, t) = 0, hence 0 = AC cos 3pt + D sin 3pt from which we conclude that A = 0
Therefore, u(x, t) = B sin px C cos 3pt + D sin 3pt (1)
(ii) u(40, t) = 0, hence 0 = B sin 40pC cos 3pt + D sin 3pt
B β 0 hence sin 40p = 0 from which, 40p = nΟ and p = n40Ο
7. Substituting in equation (1) gives: u(x, t) = B sin n x40Ο 3n t 3n tCcos Dsin
40 40Ο Οβ§ β«+β¨ β¬
β© β
or, more generally, n n nn 1
n x 3n t 3n tu (x, t) sin A cos B sin40 40 40
β
=
Ο Ο Οβ§ β«= +β¨ β¬β© β
β (2)
where nA = BC and nB = BD
8. From equation (8), page 517 of textbook,
nA = L
0
2 n xf (x)sin dxL L
Οβ«
= 20 40
0 20
2 1.5 n x 60 1.5x n xx sin dx sin dx40 20 40 20 40
β‘ Ο β Ο β€β β β β+β β β ββ’ β₯β β β β β£ β¦β« β«
Each integral is determined using integration by parts (see chapter 43, page 418) with the result:
nA = 2 2 2 2
(8)(1.5) n 12 nsin sinn 2 n 2
Ο Ο=
Ο Ο
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 479
From equation (9), page 518 of textbook, L
n 0
2 n xB g(x)sin dxcn L
Ο=
Ο β«
t 0
u 0t =
ββ‘ β€ =β’ β₯ββ£ β¦ = g(x) thus, nB 0=
Substituting into equation (2) gives: n n nn 1
n x 3n t 3n tu (x, t) sin A cos B sin40 40 40
β
=
Ο Ο Οβ§ β«= +β¨ β¬β© β
β
= 2 2n 1
n x 12 n 3n t n tsin sin cos (0)sin40 n 2 40 50
β
=
Ο Ο Ο Οβ§ β«+β¨ β¬Οβ© ββ
Hence, u(x, t) = 2 2n 1
12 1 n n x 3n tsin sin cosn 2 40 40
β
=
Ο Ο ΟΟ β
2. The centre point of an elastic string between two points P and Q, 80 cm apart, is deflected a
distance of 1 cm from its position of rest perpendicular to PQ and released initially with zero
velocity. Apply the wave equation 2 2
2 2 2
u 1 ux c tβ β
=β β
where c = 8, to determine the motion of a
point distance x from P at time t.
The elastic string is shown in the diagram below.
The boundary and initial conditions given are:
u(0, t) 0u(80, t) 0
= β«β¬= β
u(x, 0) = f (x) = 1 x40
0 β€ x β€ 40
= 1 x 240
β + 40 β€ x β€ 80
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 480
t 0
u 0t =
ββ‘ β€ =β’ β₯ββ£ β¦ i.e. zero initial velocity
Assuming a solution u = XT, where X is a function of x only, and T is a function of t only,
then u X 'Txβ
=β
and 2
2
u X ''Txβ
=β
and u XT 'yβ
=β
and 2
2
u XT ''yβ
=β
Substituting into the partial differential equation, 2 2
2 2 2
u 1 ux c tβ β
=β β
gives: 2
1X ''T XT ''c
= i.e. 1X ''T XT ''64
= since c = 8
Separating the variables gives: X '' T ''X 64T
=
Let constant, Β΅ = X '' T ''X 64T
= then Β΅ = X ''X
and Β΅ = T ''64T
from which, Xβ²β² - Β΅X = 0 and Tβ²β² - 64Β΅ T = 0
Letting Β΅ = - 2p to give an oscillatory solution gives
Xβ²β² + 2p X = 0 and the auxiliary equation is: 2 2m p+ = 0 from which, 2m p jp= β = Β±
and Tβ²β² + 64 2p T = 0 and the auxiliary equation is:
2 2m 64p+ = 0 from which, 2m 64p j8p= β = Β±
Solving each equation gives: X = A cos px + B sin px and T = C cos 8pt + D sin 8pt
Thus, u(x, t) = A cos px + B sin pxC cos 8pt + D sin 8pt
Applying the boundary conditions to determine constants A and B gives:
(i) u(0, t) = 0, hence 0 = AC cos 8pt + D sin 8pt from which we conclude that A = 0
Therefore, u(x, t) = B sin px C cos 8pt + D sin 8pt (1)
(ii) u(80, t) = 0, hence 0 = B sin 80pC cos 8pt + D sin 8pt
B β 0 hence sin 80p = 0 from which, 80p = nΟ and p = n80Ο
Substituting in equation (1) gives: u(x, t) = B sin n x80Ο 8n t 8n tCcos Dsin
80 80Ο Οβ§ β«+β¨ β¬
β© β
or, more generally, n n nn 1
n x n t n tu (x, t) sin A cos B sin80 10 10
β
=
Ο Ο Οβ§ β«= +β¨ β¬β© β
β (2)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 481
nA = 2 2 2 2
(8)(1) n 8 nsin sinn 2 n 2
Ο Ο=
Ο Ο
L
n 0
2 n xB g(x)sin dxcn L
Ο=
Ο β« t 0
u 0t =
ββ‘ β€ =β’ β₯ββ£ β¦ = g(x) thus, nB 0=
Substituting into equation (2) gives: n n nn 1
n x n t n tu (x, t) sin A cos B sin80 10 10
β
=
Ο Ο Οβ§ β«= +β¨ β¬β© β
β
= 2 2n 1
n x 8 n n t n tsin sin cos (0)sin80 n 2 10 10
β
=
Ο Ο Ο Οβ§ β«+β¨ β¬Οβ© ββ
Hence, u(x, t) = 2 2n 1
8 1 n n x n tsin sin cosn 2 80 10
β
=
Ο Ο ΟΟ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 482
EXERCISE 203 Page 522 1. A metal bar, insulated along its sides, is 4 m long. It is initially at a temperature of 10Β°C and at
time t = 0, the ends are placed into ice at 0Β°C. Find an expression for the temperature at a point P
at a distance x m from one end at any time t seconds after t = 0. The temperature u along the length of bar is shown in the diagram below.
The heat conduction equation is 2
2 2
u 1 ux c tβ β
=β β
and the given boundary conditions are:
u(0, t) = 0, u(4, t) = 0 and u(x, 0) = 10
Assuming a solution of the form u = XT, then, X = A cos px + B sin px
and T = 2 2p c tk eβ
Thus, the general solution is given by: u(x, t) = P cos px + Q sin px2 2p c teβ
u(0, t) = 0 thus 0 = P2 2p c teβ from which, P = 0 and u(x, t) = Q sin px
2 2p c teβ
Also, u(4, t) = 0 thus 0 = Q sin 4p2 2p c teβ
Since Q β 0, sin 4p = 0 from which, 4p = nΟ where n = 1, 2, 3, β¦ and p = n4Ο
Hence, u(x, t) = 2 2p c t
nn 1
n xQ e sin4
ββ
=
Οβ§ β«β¨ β¬β© β
β
The final initial condition given was that at t = 0, u = 10, i.e. u(x, 0) = f(x) = 10
Hence, 10 = nn 1
n xQ sin4
β
=
Οβ§ β«β¨ β¬β© β
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 483
where, from Fourier coefficients, nQ = 2 Γ mean value of 10 sin n x4Ο from x = 0 to x = 4,
i.e. nQ = 4
0
2 n x10sin dx4 4
Οβ« =
4
0
n xcos45 n
4
Οβ‘ β€β’ β₯ββ’ β₯Οβ’ β₯β£ β¦
= 20 4ncos cos 0n 4
Οβ‘ β€β ββ’ β₯Ο β£ β¦ = ( )20 1 cos n
nβ Ο
Ο
= 0 (when n is even) and 40nΟ
(when n is odd)
Hence, the required solution is: u(x, t) = 2 2p c t
nn 1
n xQ e sin4
ββ
=
Οβ§ β«β¨ β¬β© β
β
= 2 2 2n c t
16
n(odd) 1
40 1 n xe sinn 4
Οβ β
=
ΟΟ β
3. The ends of an insulated rod PQ, 20 units long, are maintained at 0Β°C. At time t = 0, the
temperature within the rod rises uniformly from each end reaching 4Β°C at the mid-point of PQ.
Find an expression for the temperature u(x, t) at any point in the rod, distant x from P at any time
after t = 0. Assume the heat conduction equation to be 2
2 2
u 1 ux c tβ β
=β β
and take c2 = 1
The temperature along the length of the rod is shown in the diagram below.
The heat conduction equation is 2
2 2
u 1 ux c tβ β
=β β
and the given boundary conditions are:
u(0, t) = 0, u(20, t) = 0 and u(x, 0) = 0
Assuming a solution of the form u = XT, then, X = A cos px + B sin px
and T = 2 2p c tk eβ
Thus, the general solution is given by: u(x, t) = P cos px + Q sin px2 2p c teβ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 484
u(0, t) = 0 thus 0 = P2 2p c teβ from which, P = 0 and u(x, t) = Q sin px
2 2p c teβ
Also, u(20, t) = 0 thus 0 = Q sin 20p2 2p c teβ
Since Q β 0, sin 20p = 0 from which, 20p = nΟ where n = 1, 2, 3, β¦ and p = n20Ο
Hence, u(x, t) = 2 2p c t
nn 1
n xQ e sin20
ββ
=
Οβ§ β«β¨ β¬β© β
β
where, from Fourier coefficients, 2 Γ the mean value from x = 0 to x = 20
nQ = 10 20
0 10
2 2 n x 2 n xx sin dx x 8 sin20 5 20 5 20
β‘ Ο Ο β€β β β β+ β +β β β ββ’ β₯β β β β β£ β¦β« β« (see above diagram)
=
10 20
2 2
0 10
2 n x 2 n x2 n x 2 n x n xx cos x cossin sin 8cos1 5 20 5 205 20 5 20 20n n n10 n n20 20 2020 20
β§ β«β‘ β€ β‘ β€Ο Οβ β β βΟ Ο Οβͺ βͺββ’ β₯ β’ β₯β β β ββͺ βͺβ β β β β’ β₯ β’ β₯+ β ββ¨ β¬Ο Ο Οβ’ β₯ β’ β₯β β β β β βΟ Οβ β β ββͺ βͺβ β β β β ββ’ β₯ β’ β₯β β β ββͺ βͺβ β β β β β β β β β β£ β¦ β£ β¦β© β
= ( )2 2
n n n n n4cos 4sin 4cos 4sin 8cos1 8cos n 8cosn2 2 2 2 20 0n n n n n10 n n20 20 20 20 2020 20
β§β‘ β€ β‘ β€β β β ββ βΟ Ο Ο Ο Οβͺβ’ β₯ β’ β₯β β β ββ β βΟ Οβ’ β₯ β’ β₯β β β ββ β+ β + + β β β ββ¨β’ β₯ β’ β₯β β β βΟ Ο Ο Ο Οβ β β β β β β β β ββ βΟ Οβ β β ββ’ β₯ β’ β₯β β β ββ β β β β β β β β ββ ββ β β ββ β β β β β β β β β β β β’ β₯ β’ β₯β β β β β β β β β£ β¦ β£ β¦
β«βͺ
βͺ βͺβ¬
βͺ βͺβͺ βͺβ© β
= 2 2
n n n n n4cos 4sin 4cos 4sin 8cos1 8cos n 8cos n2 2 2 2 2n n n n n10 n n20 20 20 20 2020 20
β§ β«Ο Ο Ο Ο Οβͺ βͺβ Ο Οβͺ βͺ+ + β β + +β¨ β¬Ο Ο Ο Ο Οβ β β β β β β β β βΟ Οβ β β ββͺ βͺβ β β β β β β β β ββ β β ββͺ βͺβ β β β β β β β β β β β β β β© β
= 2 2
n n n n8cos 8sin 8cos 8sin1 12 2 2 2n n10 10n n20 2020 20
β§ β« β§ β«Ο Ο Ο Οβͺ βͺ βͺ βͺββͺ βͺ βͺ βͺ+ + =β¨ β¬ β¨ β¬Ο Οβ β β βΟ Οβ β β ββͺ βͺ βͺ βͺβ β β ββ β β ββͺ βͺ βͺ βͺβ β β β β β β β β© β β© β
= 0 when n is even
= 2
2 2
8 20 n 320 nsin sin10 n 2 n 2
β§ β«Ο Οβͺ βͺβ β =β¨ β¬β βΟ Οβ β βͺ βͺβ© β when n is odd
Hence, the required solution is: u(x, t) = 2 2
2n (1) t20
2 2n 1
320 n n xsin e sinn 2 20
Οβ β
=
β§ β«Ο Οβͺ βͺβ¨ β¬Οβͺ βͺβ© β
β
= 2 2n t400
2 2n(odd) 1
320 1 n n xsin sin en 2 20
β βΟβ β β ββ ββ β
=
Ο ΟΟ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 485
EXERCISE 204 Page 524 1. A rectangular plate is bounded by the lines x = 0, y = 0, x = 1 and y = 3. Apply the Laplace
equation 2 2
2 2
u u 0x yβ β
+ =β β
to determine the potential distribution u(x, y) over the plate, subject to
the following boundary conditions:
u = 0 when x = 0 0 β€ y β€ 2 u = 0 when x = 1 0 β€ y β€ 2
u = 0 when x = 2 0 β€ x β€ 1 u = 5 when x = 3 0 β€ x β€ 1
Initially a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and
Y is a function of y only. Simplifying to u = XY, determining partial derivatives, and substituting
into 2 2
2 2
u u 0x yβ β
+ =β β
gives: Xβ²β²Y + XYβ²β² = 0
Separating the variables gives: X '' Y ''X Y
= β
Letting each side equal a constant, - 2p gives the two equations:
Xβ²β² + 2p X = 0 and Yβ²β² - 2p Y = 0
from which, X = A cos px + B sin px
and Y = py pyCe Deβ+ or Y = C cosh py + D sinh py or Y = E sinh p(y + Ο)
Hence u(x, y) = XY = A cos px + B sin px E sinh p(y + Ο)
or u(x, y) = P cos px + Q sin px sinh p(y + Ο) where P = AE and Q = BE
The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + Ο) from which, P = 0
Hence, u(x, y) = Q sin px sinh p(y + Ο)
The second boundary condition is: u(1, y) = 0, hence 0 = Q sin p(1) sinh p(y + Ο)
from which, sin p = 0, hence, p = nΟ for n = 1, 2, 3, β¦
The third boundary condition is: u(x, 2) = 0, hence, 0 = Q sin px sinh p(2 + Ο)
from which, sinh p(2 + Ο) = 0 and Ο = -2
Hence, u(x, y) = Q sin px sinh p(y β 2)
Since there are many solutions for integer values of n,
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 486
u(x, y) = nn 1
Q sin px sinh p(y 2)β
=
ββ = nn 1
Q sin n x sinh n (y 2)β
=
Ο Ο ββ (a)
The fourth boundary condition is: u(x, 3) = 5 = f (x), hence, f (x) = nn 1
Q sin n x sinh n (3 2)β
=
Ο Ο ββ
From Fourier series coefficients,
nQ sinh nΟ = 2 Γ the mean value of f (x) sin n xΟ from x = 0 to x = 1
i.e. = 1
1
00
2 cos n x5sin n x dx 101 n
Οβ‘ β€Ο = ββ’ β₯Οβ£ β¦β« = ( ) ( )10 10cos n cos 0 1 cos nn n
β Οβ = β ΟΟ Ο
= 0 (for even values of n), = 20nΟ
(for odd values of n)
Hence, n20 20Q
n (sinh n ) n= =
Ο Ο Οcosech nΟ
Hence, from equation (a), u(x, y) = nn 1
Q sin n x sinh n (y 2)β
=
Ο Ο ββ
= ( )n odd) 1
20 1 cosechn sinn x sinhn (y 2)n
β
=
Ο Ο Ο βΟ β
2. A rectangular plate is bounded by the lines x = 0, y = 0, x = 3, y = 2. Determine the potential
distribution u(x, y) over the rectangle using the Laplace equation 2 2
2 2
u u 0x yβ β
+ =β β
subject to the
following boundary conditions:
u(0, y) = 0 0 β€ y β€ 2
u(3, y) = 0 0 β€ y β€ 2
u(x, 2) = 0 0 β€ x β€ 3
u(x, 0) = x(3 β x) 0 β€ x β€ 3 Initially a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and
Y is a function of y only. Simplifying to u = XY, determining partial derivatives, and substituting
into 2 2
2 2
u u 0x yβ β
+ =β β
gives: Xβ²β²Y + XYβ²β² = 0
Separating the variables gives: X '' Y ''X Y
= β
Letting each side equal a constant, - 2p gives the two equations:
Xβ²β² + 2p X = 0 and Yβ²β² - 2p Y = 0
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 487
from which, X = A cos px + B sin px
and Y = py pyCe Deβ+ or Y = C cosh py + D sinh py or Y = E sinh p(y + Ο)
Hence u(x, y) = XY = A cos px + B sin px E sinh p(y + Ο)
or u(x, y) = P cos px + Q sin px sinh p(y + Ο)
The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + Ο) from which, P = 0
Hence, u(x, y) = Q sin px sinh p(y + Ο)
The second boundary condition is: u(3, y) = 0, hence 0 = Q sin 3p sinh p(y + Ο)
from which, sin 3p = 0, hence, 3p = nΟ i.e. p = n3Ο for n = 1, 2, 3, β¦
The third boundary condition is: u(x, 2) = 0, hence, 0 = Q sin px sinh p(2 + Ο)
from which, sinh p(2 + Ο) = 0 and Ο = - 2
Hence, u(x, y) = Q sin px sinh p(y β 2)
Since there are many solutions for integer values of n,
u(x, y) = nn 1
Q sin px sinh p(y 2)β
=
ββ = nn 1
n nQ sin x sinh (y 2)3 3
β
=
Ο Οββ (a)
The fourth boundary condition is: u(x, 0) = x(3 β x) = 3x - 2x = f (x),
hence, f (x) = nn 1
n nQ sin x sinh ( 2)3 3
β
=
Ο Οββ
From Fourier series coefficients,
n2nQ sinh3
β Ο = 2 Γ the mean value of f (x) sin n x3Ο from x = 0 to x = 3
= ( )3 2
0
2 n3x x sin x dx1 3
Οββ«
=
3 3
2
2 2 3
0 0
n x n x n x n x n x3x cos 3sin x cos 2xsin 2cos3 3 3 3 32
n nn n n3 33 3 3
β§ β«β‘ β€ β‘ β€Ο Ο Ο Ο Οβ β β β β β β β β ββͺ βͺβ ββ’ β₯ β’ β₯β β β β β β β β β ββͺ βͺβ β β β β β β β β β β’ β₯ β’ β₯+ β + +β¨ β¬β’ β₯ β’ β₯Ο Οβ β β βΟ Ο Οβ β β β β ββͺ βͺβ β β ββ’ β₯ β’ β₯β β β β β ββͺ βͺβ β β β β β β β β β β£ β¦ β£ β¦β© β
by integration by parts (see chapter 43)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 488
= ( )3 3 3 3
9cos n 9cos n 2cos n 2 272 2 2 2cos nn n nn n3 3 3 3
β§ β«βͺ βͺβ Ο Ο Οβͺ βͺ β§ β«+ β + = β Οβ¨ β¬ β¨ β¬Ο Ο Οβ β β β β© βΟ Οβ β β ββͺ βͺβ β β β β β β ββͺ βͺβ β β β β β β β β© β
= ( )3 3
54 2 2cos nn
β ΟΟ
= 0 (for even values of n), = 3 3
216n Ο
(for odd values of n)
Hence, n 3 33 3
216 216 2nQ cos ech2n n 3n sinh3
β Ο= =
β Ο Οβ βΟ β ββ β
Hence, from equation (a),
u(x, y) = n 3 3n 1 n 1
n n 216 2n n x nQ sin x sinh (y 2) cos ech sin sinh (y 2)3 3 n 3 3 3
β β
= =
Ο Ο β Ο Ο Οβ = β
Οβ β
= ( )
3 3n odd) 1
216 1 2n n x ncosech sin sinh (2 y)n 3 3 3
β
=
Ο Ο Οβ
Ο β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 489
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 β could be any amount of petrol.
(b) Continuous β could be any amount of coal.
(c) Discrete β can only be a whole number of bottles of milk.
(d) Continuous β 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 β can only be a whole number of people.
(b) Continuous β could be any time taken.
(c) Discrete β can only be a whole number of stamps.
(d) Discrete β can only be a whole number of defective items.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 490
EXERCISE 206 Page 530 3. The number of vehicles passing a stationary observer on a road in six ten-minute intervals is as
shown. Draw a horizontal bar chart.
Period of time 1 2 3 4 5 6
Number of vehicles 35 44 62 68 49 41 A horizontal bar chart is shown below.
6. The number of components produced by a factory in a week is as shown below:
Day Mon Tues Wed Thur Fri
Number of components 1580 2190 1840 2385 1280
Depict the data on a vertical bar chart. A vertical bar chart is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 491
8. A company has five distribution centres and the mass of goods in tonnes sent to each centre
during four, one-week periods, is as shown.
Week 1 2 3 4
Centre A 147 160 174 158
Centre B 54 63 77 69
Centre C 283 251 237 211
Centre D 97 104 117 144
Centre E 224 218 203 194
Use a percentage component bar chart to present these data and comment on any trends. Week 1: Total = 147 + 54 + 283 + 97 + 224 = 805
A = 147 100% 18%805
Γ β , B = 54 100% 7%805
Γ β , C β 35%, D β 12%, E β 28%
Week 2: Total = 160 + 63 + 251 + 104 + 218 = 796
A = 160 100% 20%796
Γ β , B = 63 100% 8%796
Γ β , C β 32%, D β 13%, E β 27%
Week 3: Total = 174 + 77 + 237 + 117 + 203 = 808
A = 174 100% 22%808
Γ β , B = 77 100% 10%808
Γ β , C β 29%, D β 14%, E β 25%
Week 4: Total = 158 + 69 + 211 + 144 + 194 = 776
A = 158 100% 20%776
Γ β , B = 69 100% 9%776
Γ β , C β 27%, D β 19%, E β 25%
A percentage component bar chart is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 492
From the above percentage component bar chart, it is seen that there is little change in centres A
and B, there is a reduction of around 8% in centre C, an increase of around 7% in centre D and a
reduction of about 3% in centre E.
10. The way in which an apprentice spent his time over a one-month period is as follows:
drawing office 44 hours, production 64 hours, training 12 hours, at college 28 hours
Use a pie chart to depict this information. Total hours = 44 + 64 + 12 + 28 = 148
Drawing office, D = 44 360 107148
Γ Β° β Β° , Production, P = 64 360 156148
Γ Β° β Β° ,
Training, T = 12 360 29148
Γ Β° β Β° , College, C = 28 360 68148
Γ Β° β Β°
A pie chart to depict this information is shown below.
12. (a) If the company sell 23500 units per annum of the product depicted in Fig 54.5 on page XX
of textbook, determine the cost of their overheads per annum.
(b) If 1% of the dwellings represented in year 1 of Fig 54.4 on page XX of the textbook
corresponds to 2 dwellings, find the number of houses sold in that year.
(a) Overheads = 126 100% 35%360
Γ = of total costs.
Cost per unit = Β£2, hence total income per annum = 23500 Γ 2 = Β£47000
Cost of overheads per annum = 35% of Β£47000 = 35 47000100
Γ = Β£16450
(b) Percentage of houses sold in year 1 = 22 + 32 + 15 = 69%
If 1% corresponds to 2 dwellings then the number of houses sold = 69 Γ 2 = 138 houses
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 493
EXERCISE 207 Page 536 3. The information given below refers to the value of resistance in ohms of a batch of 48 resistors
of similar value. Form a frequency distribution for the data having about 6 classes, and draw a
frequency polygon and histogram to represent these data diagrammatically.
21.0 22.4 22.8 21.5 22.6 21.1 21.6 22.3
22.9 20.5 21.8 22.2 21.0 21.7 22.5 20.7
23.2 22.9 21.7 21.4 22.1 22.2 22.3 21.3
22.1 21.8 22.0 22.7 21.7 21.9 21.1 22.6
21.4 22.4 22.3 20.9 22.8 21.2 22.7 21.6
22.2 21.6 21.3 22.1 21.5 22.0 23.4 21.2 The range is from 20.5 to 23.4, i.e. range = 23.4 β 20.5 = 2.9
2.9 Γ· 6 β 0.5 hence, classes of 20.5 β 20.9, 21.0 β 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 494
The histogram for the above frequency distribution is shown below.
5. Form a cumulative frequency distribution and hence draw the ogive for the frequency
distribution given in the solution to Problem 3 above. A cumulative frequency distribution is shown below.
Class Frequency Upper class boundary Cumulative frequency 20.5 β 20.9 21.0 β 21.4 21.5 β 21.9 22.0 β 22.4 22.5 β 22.9 23.0 β 23.4
3 10 11 13 9 2
Less than 20.95 21.45 21.95 22.45 22.95 23.45
3 13 24 37 46 48
An ogive for the above frequency distribution is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 495
9. The diameter in millimetres of a reel of wire is measured in 48 places and the results are as
shown.
2.10 2.29 2.32 2.21 2.14 2.22 2.28 2.18 2.17 2.20 2.23 2.13 2.26 2.10 2.21 2.17 2.28 2.15 2.16 2.25 2.23 2.11 2.27 2.34 2.24 2.05 2.29 2.18 2.24 2.16 2.15 2.22 2.14 2.27 2.09 2.21 2.11 2.17 2.22 2.19 2.12 2.30 2.23 2.07 2.13 2.26 2.16 2.12
(a) Form a frequency distribution of diameters having about 6 classes
(b) Draw a histogram depicting the data.
(c) Form a cumulative frequency distribution.
(d) Draw an ogive for the data. (a) Range = 2.34 β 2.05 = 0.29
0.29 Γ· 6 β 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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 496
(c) A cumulative frequency distribution is shown below.
Class Frequency Upper class boundary Cumulative frequency 2.05 β 2.09 2.10 β 2.14 2.15 β 2.19 2.20 β 2.24 2.25 β 2.29 2.30 β 2.34
3 10 11 13 9 2
Less than 2.095 2.145 2.195 2.245 2.295 2.345
3 13 24 37 46 48
(d) An ogive for the above data is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 497
CHAPTER 55 MEASURES OF CENTRAL TENDENCY AND
DISPERSION EXERCISE 208 Page 539
2. Determine the mean, median and modal values for the set:
26, 31, 21, 29, 32, 26, 25, 28
Mean = 26 31 21 29 32 26 25 28 2188 8
+ + + + + + += = 27.25
Ranking gives: 21 25 26 26 28 29 31 32
Median = middle value = 26 282+ = 27
Most commonly occurring value, i.e. mode = 26
4. Determine the mean, median and modal values for the set:
73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9
Mean = 73.8 126.4 40.7 141.7 28.5 237.4 157.9 806.47 7
+ + + + + += = 115.2
Ranking gives: 28.5 40.7 73.8 126.4 141.7 157.9 237.4
Middle value = median = 126.4
There is no mode since all the values are different.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 498
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.
150β156 5, 157-163 18, 164-170 20,
171-177 27, 178-184 22, 185-191 8
Mean value = ( ) ( ) ( ) ( ) ( ) ( )5 153 18 160 20 167 27 174 22 181 8 188100
Γ + Γ + Γ + Γ + Γ + Γ
= 17169100
= 171.7 cm
3. The diameters, in centimetres, of 60 holes bored in engine castings are measured and the results
are as shown. Draw a histogram depicting these results and hence determine the mean, median
and modal values of the distribution.
2.011-2.014 7, 2.016-2.019 16, 2.021-2.024 23, 2.026-2.029 9,
2.031-2.034 5
The histogram is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 499
The mean value lies at the centroid of the histogram. With reference to axis YY at 2.010 cm,
A M (a m)=β
where A = area of histogram = 35 + 80 + 115 + 45 + 25 = 300 and M = horizontal distance of
centroid from YY. (Actually, the area of, say, 35 square units is 335 10βΓ square units; however, the
310β will cancel on each side of the equation so has been omitted).
Hence, 300 M = (35 Γ 0.0025) + (80 Γ 0.0075) + (115 Γ 0.0125)
+ (45 Γ 0.0175) + (25 Γ 0.0225)
i.e. 300 M = 3.475
i.e. M = 3.475 0.01158300
= cm
Thus, the mean is at 2.010 + 0.01158 = 2.02158 cm
The median is the diameter where the area on each side of it is the same, i.e. 300/2, i.e. 150 square
units on each side.
The first two rectangles have an area of 35 + 80 = 115; hence, 35 more square units are needed from
the third rectangle. 35 100% 30.43%115
Γ = of the distance from 2.020 to 2.025
i.e. 0.3043 Γ (2.025 β 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
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 500
EXERCISE 210 Page 542 1. Determine the standard deviation from the mean of the set of numbers:
35, 22, 25, 23, 28, 33, 30
Mean, 35 22 25 23 28 33 30 196x 287 7
+ + + + + += = =
Standard deviation,
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2x x 35 28 22 28 25 28 23 28 28 28 33 28 30 28n 7
β§ β«ββͺ βͺ β§ β«β + β + β + β + β + β + ββͺ βͺ βͺ βͺΟ = =β¨ β¬ β¨ β¬βͺ βͺβͺ βͺ β© β
βͺ βͺβ© β
β
= 148 21.1437
= = 4.60, correct to 3 significant figures.
3. The tensile strength in megapascals for 15 samples of tin were determined and found to be:
34.61, 34.57, 34.40, 34.63, 34.63, 34.51, 34.49, 34.61,
34.52, 34.55, 34.58, 34.53, 34.44, 34.48 and 34.40
Calculate the mean and standard deviation from the mean for these 15 values, correct to 4
significant figures.
Mean, 34.61 34.57 34.40 34.63 ...... 517.95x15 15
+ + + += = = 34.53 MPa
Standard deviation,
( ) ( ) ( ) ( )2 2 2x x 34.61 34.53 34.57 34.53 34.40 34.53 .....
n 15
β§ β«ββͺ βͺ β§ β«β + β + β +βͺ βͺ βͺ βͺΟ = =β¨ β¬ β¨ β¬βͺ βͺβͺ βͺ β© β
βͺ βͺβ© β
β
= 0.0838 0.00558666615
= = 0.07474 MPa
5. Calculate the standard deviation from the mean for the data given in Problem 3 of Exercise 209
above, correct to 3 significant figures. From Problem 3, Exercise 209, mean value, x 2.02158 cm=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 501
Standard deviation, Ο = ( ) f x x
f
β§ β«ββͺ βͺβͺ βͺβ¨ β¬βͺ βͺβͺ βͺβ© β
β
β
( ) ( ) ( )( ) ( )
2 2 2
2 2
7 2.0125 2.02158 16 2.0175 2.02158 23 2.0225 2.02158
9 2.0275 2.02158 5 2.0325 2.0215860
β§ β«β + β + ββͺ βͺβͺ βͺ+ β + ββͺ βͺ= β¨ β¬βͺ βͺβͺ βͺβͺ βͺβ© β
= 0.000577124 0.000266342 0.000019467 0.000315417 0.00059623260
+ + + +
= 0.00177458260
= 0.00544 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 502
EXERCISE 211 Page 544 2. The number of faults occurring on a production line in a nine-week period are as shown below.
Determine the median and quartiles values for the data.
30 27 25 24 27 37 31 27 35 Ranking gives: 24 25 27 27 27 30 31 35 37 β β β
Median = middle value = 27 faults
1st quartile value = 25 272+ = 26 faults
3rd quartile value = 31 352+ = 33 faults
3. Determine the quartile values and semi-interquartile range for the frequency distribution given in
Problem 1 of Exercise 209 above. The frequency distribution is shown below.
Upper class boundary values Frequency Cumulative frequency 156.5 163.5 170.5 177.5 184.5 191.5
5 18 20 27 22 8
5 23 43 70 92 100
The ogive is shown below. From the ogive, 1Q 164.5cm= , 2Q 172.5cm= and 3Q 179cm=
and semi-interquartile range = 179 164.5 14.52 2β
= = 7.25 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 503
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
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 504
CHAPTER 56 PROBABILITY EXERCISE 212 Page 547
2. A box of fuses are all of the same shape and size and comprises 23 2 A fuses, 47 5A fuses and
69 13 A fuses. Determine the probability of selecting at random (a) a 2 A fuse, (b) a 5 A fuse,
and (c) a 13 A fuse.
(a) 2Anumber of 2A fuses 23ptotal number of fuses 23 47 69
= =+ +
= 23139
or 0.1655
(b) 5Anumber of 5A fusesptotal number of fuses
= = 47139
or 0.3381
(c) 13Anumber of 13A fusesptotal number of fuses
= = 69139
or 0.4964
4. The probability of event A happening is 35
and the probability of event B happening is 23
.
Calculate the probabilities of (a) both A and B happening, (b) only event A happening, i.e. event
A happening and event B not happening, (c) only event B happening, and (d) either A, or B, or A
and B happening.
Let A3p5
= and B2p3
= and thus the probability of events not happening, A2p5
= and B1p3
=
(a) The probability of both A and B happening = A B3 2p p5 3
Γ = Γ = 25
(b) The probability of event A happening and event B not happening = A B3 1p p5 3
Γ = Γ = 15
(c) The probability of only event B happening = B A2 2p p3 5
Γ = Γ = 415
(d) The probability of either A, or B, or A and B happening = ( ) ( ) ( )A B B A A Bp p p p p pβ‘ β€Γ + Γ + Γβ£ β¦
= 3 1 2 2 3 25 3 3 5 5 3
β‘ β€β β β β β βΓ + Γ + Γβ β β β β ββ’ β₯β β β β β β β£ β¦
= 3 4 6 7 615 15 15 15 15β β+ + = +β ββ β
= 1315
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 505
5. When testing 1000 soldered joints, 4 failed during a vibration test and 5 failed due to having a
high resistance. Determine the probability of a joint failing due to (a) vibration, (b) high
resistance, (c) vibration or high resistance, and (d) vibration and high resistance.
(a) The probability of a joint failing due to vibration, v4p
1000= = 1
250
(b) The probability of a joint failing due to high resistance, R5p
1000= = 1
200
(c) The probability of a joint failing due to vibration or high resistance,
v R1 1 4 5p p
250 200 1000+
+ = + = = 91000
(d) The probability of a joint failing due to vibration and high resistance,
v R1 1p p
250 200Γ = Γ = 1
50000
6. Find the probability that the score is 8 if two like dice are thrown.
A score of 8 is achieved with a (2 + 6), (3 + 5), (4 + 4), (5 + 3) and (6 + 2) - see above diagram,
i.e. 5 possibilities, and there are 36 possible scores when throwing two dice.
Hence, the probability of a score of 8 is 536
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 506
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 Ap = 0.8 and Bp = 0.75, and unsatisfactory operations be Ap = 0.2
and Bp = 0.25
(a) The probability that both components operate satisfactorily, A Bp p 0.8 0.75Γ = Γ = 0.6
(b) The probability that only component A will operate satisfactorily, A Bp p 0.8 0.25Γ = Γ = 0.2
(c) The probability that only component B will operate satisfactorily, B Ap p 0.75 0.2Γ = Γ = 0.15
4. A batch of 1 kW fire elements contain 16 which are within a power tolerance and 4 which are
not. If 3 elements are selected at random from the batch, calculate the probabilities that (a) all
three are within the power tolerance and (b) two are within but one is not within the power
tolerance.
(a) The probability that all three are within the power tolerance = 16 15 1420 19 18
Γ Γ = 0.4912
(b) The probability that two are within but one is not within the power tolerance
= 16 15 4 16 4 15 4 16 1520 19 18 20 19 18 20 19 18
β β β β β βΓ Γ + Γ Γ + Γ Γβ β β β β ββ β β β β β
= 3(0.14035) = 0.4211
5. An amplifier is made up of three transistors A, B and C. The probabilities of A, B and C being
defective are 1 1,20 25
and 150
, respectively. Calculate the percentage of amplifiers produced
(a) which work satisfactorily and (b) which have just one defective transistor
Let the probability of transistors working be: A19p20
= , B24p25
= and C49p50
=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 507
(a) The probability of amplifiers working satisfactorily, A B C19 24 49p p p20 25 50
Γ Γ = Γ Γ
= 0.8938 or 89.38%
(b) The probability of amplifiers having just one defective transistor
= 1 24 49 1 19 49 1 19 2420 25 50 25 20 50 50 20 25
β β β β β βΓ Γ + Γ Γ + Γ Γβ β β β β ββ β β β β β
= 0.04704 + 0.03724 + 0.01824
= 0.10252 = 10.25%
6. A box contains 14 40 W lamps, 28 60 W lamps and 58 25 W lamps, all the lamps being of the
same shape and size. Three lamps are drawn at random from the box, first one, them a second,
then a third. Determine the probabilities of: (a) getting one 25 W, one 40 W and one 60 W lamp,
with replacement, (b) getting one 25 W, one 40 W and one 60 W lamp without replacement, and
(c) getting either one 25 W and two 40 W or one 60 W and two 40 W lamps with replacement.
Let 40W
14 14p 0.1414 28 58 100
= = =+ +
, 60W28p 0.28
100= = and 25W
58p 0.58100
= =
(a) The probability of getting one 25 W, one 40 W and one 60 W lamp, with replacement
= 0.58 Γ 0.14 Γ 0.28 = 0.0227
(b) The probability of getting one 25 W, one 40 W and one 60 W lamp, without replacement
= 58 14 28100 99 98
Γ Γ = 0.0234
(c) The probability of getting either one 25 W and two 40 W or one 60 W and two 40 W lamps,
with replacement = (0.58 Γ 0.14 Γ 0.14) + (0.28 Γ 0.14 Γ 0.14)
= 0.011368 + 0.005488 = 0.0169
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 508
CHAPTER 57 THE BINOMIAL AND POISSON DISTRIBUTION EXERCISE 214 Page 555
1. Concrete blocks are tested and it is found that, on average, 7% fail to meet the required
specification. For a batch of 9 blocks, determine the probabilities that (a) three blocks and (b) less
than four blocks will fail to meet the specification.
Let probability of failure to meet specification, p = 0.07 and probability of success, q = 0.93
By the binomial expansion, ( )9 9 8 7 2 6 3(9)(8) (9)(8)(7)q p q 9q p q p q p .....2! 3!
+ = + + + + hence,
( )9 9 8 7 2 6 3(9)(8) (9)(8)(7)0.93 0.07 (0.93) 9(0.93) (0.07) (0.93) (0.07) (0.93) (0.07) .....2! 3!
+ = + + + +
= 0.5204 + 0.3525 + 0.1061 + 0.0186 +
which corresponds to 0, 1, 2, 3, β¦ failing to meet the specification.
(a) Probability that three blocks fail to meet specification = 0.0186
(b) Probability that less than four blocks fail = 0.5204 + 0.3525 + 0.1061 + 0.0186 = 0.9976
3. The average number of employees absent from a firm each day is 4%. An office within the firm
has seven employees. Determine the probabilities that (a) no employee and (b) three employees
will be absent on a particular day.
Let p = 4% = 0.04 then q = 0.96 (i.e. 96% present)
( ) ( )7 7 7 6 5 2
4 3
(7)(6)q p 0.96 0.04 (0.96) 7(0.96) (0.04) (0.96) (0.04)2!
(7)(6)(5) (0.96) (0.07) .....3!
+ = + = + +
+ +
Which corresponds to 0, 1, 2, 3, β¦. employees being absent.
(a) The probability that no employee will be absent on a particular day = ( )70.96 = 0.7514
(b) The probability that three employees will be absent on a particular day
= ( ) ( )4 3(7)(6)(5) 0.96 0.043!
= 0.0019
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 509
5. Five coins are tossed simultaneously. Determine the probabilities of having 0, 1, 2, 3, 4 and 5
heads upwards, and draw a histogram depicting the results.
Let probability of a head, Hp = 0.5 and the probability of a tail, Tp = 0.5
( )
( )( ) ( )
5 5 4 3 2 2 3
4 5
(5)(4) (5)(4)(3)0.5 0.5 (0.5) 5(0.5) (0.5) (0.5) (0.5) (0.5) (0.5)2! 3!
(5)(4)(3)(2) 0.5 0.5 0.54!
+ = + + +
+ +
= 0.03125 + 0.15625 + 0.3125 + 0.3125 + 0.15625 + 0.03125
which corresponds to 0, 1, 2, 3, 4 and 5 heads landing upwards.
A histogram depicting this data is shown below.
7. An automatic machine produces, on average, 10% of its components outside of the tolerance
required. In a sample of 10 components from this machine, determine the probability of having
three components outside of the tolerance required by assuming a binomial distribution.
Let the probability of a component being outside the tolerance, p = 10% = 0.1 and the probability of
a component being within tolerance, q = 0.9
( ) ( )10 10 10 9 8 2 7 3(10)(9) (10)(9)(8)q p 0.9 0.1 (0.9) 10(0.9) (0.1) (0.9) (0.1) (0.9) (0.1) ....2! 3!
+ = + = + + + +
which corresponds to the probability of 0, 1, 2, 3, β¦. components being outside the tolerance.
The probability of having three components outside of the required tolerance
= ( ) ( )7 3(10)(9)(8) 0.9 0.13!
= 0.0574
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 510
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, Ξ» = np = (2000)(0.0015) = 3
The probability of 0, 1, 2, 3, β¦. employees going to hospital is given by the terms of:
2 3
e 1 ...2 3
βΞ» β βΞ» Ξ»+ Ξ» + + +β β
β β =
2 33 3 3 33 3e 3e e e ...
2! 3!β β β β+ + + +
= 0.0498 + 0.1494 + 0.2240 + β¦
The probability of more than two employees going to hospital = 1 β (0.0498 + 0.1494 + 0.2240)
= 1 β 0.4232 = 0.5768
3. When packaging a product, a manufacturer finds that one packet in twenty is underweight.
Determine the probabilities that in a box of 72 packets (a) two and (b) less than four will be
underweight.
Probability of a packet being underweight, p = 120
= 0.05 and n = 72,
Hence, the average occurrence of event, Ξ» = np = (72)(0.05) = 3.6
The probability of 0, 1, 2, 3, β¦.packets being underweight is given by the terms of:
2 3
e 1 ...2 3
βΞ» β βΞ» Ξ»+ Ξ» + + +β β
β β =
2 33.6 3.6 3.6 3.63.6 3.6e 3.6e e e ...
2! 3!β β β β+ + + +
= 0.0273 + 0.0984 + 0.1771 + 0.2125 + β¦
(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
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 511
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, Ξ» = 5
The probability of 0, 1, 2, 3, β¦ tools being demanded is given by the terms:
eβΞ» , eβλλ , 2
e2!
βλλ , 3
e3!
βλλ , β¦. i.e. 5eβ , 55eβ , 2
55 e2!
β , 3
55 e3!
β , β¦.
i.e. 0.0067, 0.0333, 0.0842, 0.1404, 0.1755, 0.1755, 0.1462, 0.1044, 0.0653, β¦
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, Ξ» = 0.7
The probability of 0, 1, 2, 3, 4 and 5 failures in a week is given by the terms:
eβΞ» , eβλλ , 2
e2!
βλλ , 3
e3!
βλλ , 4
e4!
βλλ and 5
e5!
βλλ
i.e. 0.7eβ , 0.70.7eβ , 2
0.70.7 e2!
β , 3
0.70.7 e3!
β , 4
0.70.7 e4!
β and 5
0.70.7 e5!
β
i.e. 0.4966, 0.3476, 0.1217, 0.0284, 0.0050 and 0.0007
A histogram depicting these results is shown below:
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 512
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 513
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 xβΟ
i.e. 69 752.8β = -2.14 standard deviations
From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -2.14 is 0.4838, i.e. the
shaded area of the diagram below.
Thus the area to the left of the z = -2.14 ordinate is 0.5000 β 0.4838 = 0.0162
The number likely to be classed as defective = 0.0162 Γ 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 xβΟ
i.e. 1025 10108.7β = 1.72 standard
deviations
From Table 58.1 on page 561 of textbook, the area between z = 0 and z = 1.72 is 0.4573, i.e. the
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 Γ 500 = 479, correct to nearest whole
number.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 514
(b) The z-value corresponding to 1000 ml is given by: x xβΟ
i.e. 1000 10108.7β = -1.15 standard
deviations
From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -1.15 is 0.3749, i.e.
the shaded area of the diagram below.
Thus the area to the left of the z = -1.15 ordinate is 0.5000 - 0.3749 = 0.1251
The number likely to have less than 1000 ml = 0.1251 Γ 500 = 63, correct to nearest whole
number.
(c) The z-value corresponding to 995 ml is given by: x xβΟ
i.e. 995 10108.7β = -1.72 standard
deviations
From Table 58.1, the area between z = 0 and z = -1.72 is 0.4573, i.e. the shaded area of the
diagram below.
Thus the area to the left of the z = -1.72 ordinate is 0.5000 - 0.4573 = 0.0427
The number likely to have less than 995 ml = 0.0427 Γ 500 = 21, correct to nearest whole
number.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 515
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 xβΟ
i.e. 81.5 752.8β = 2.32 standard deviations.
From Table 58.1 on page 561 of textbook, the area between z = 0 and z = 2.32 is 0.4898, i.e. the
shaded area of the diagram below.
Thus the area to the right of the z = 2.32 ordinate is 0.5000 β 0.4838 = 0.0102
The number likely to be classed as oversized = 0.0102 Γ 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 xβΟ
i.e. 4.048 4.050.0028
β = -0.71 standard
deviations
From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -0.71 is 02611
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 516
The z-value corresponding to 4.0553 mm is given by: x xβΟ
i.e. 4.0553 4.050.0028
β = 1.89 standard
deviations
From Table 58.1, the area between z = 0 and z = 1.89 is 0.4706
The probability of the diameter being between 4.048 mm and 4.0553 mm is 0.2611 + 0.4706 =
0.7317 (see shaded area in diagram below).
The number likely to have diameter between 4.048 mm and 4.0553 mm = 0.7317 Γ 20 =
14.63 = 15, correct to nearest whole number.
(b) The z-value corresponding to 4.052 mm is given by: x xβΟ
i.e. 4.052 4.050.0028
β = 0.71 standard
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 xβΟ
i.e. 4.056 4.050.0028
β = 2.14 standard
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 Γ 20 =
4.454 = 4, correct to nearest whole number.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 517
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 xβΟ
i.e. 4.88 5.000.036β = -3.33 standard
deviations
From Table 58.1 on page 561 of textbook, the area between z = 0 and z = -3.33 is 0.4996.
The z-value corresponding to 4.92 g is given by: 4.92 5.000.036β = -2.22 standard deviations.
From Table 58.1, the area between z = 0 and z = -2.22 is 0.4868
The probability of having masses being between 4.88 g and 4.92 g is 0.4996 - 0.4868 = 0.0128
(see shaded area in diagram below).
The number of tablets likely to have a mass between 4.88 g and 4.92 g = 0.0128 Γ 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 xβΟ
i.e. 5.04 5.000.036β = 1.11 standard
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 Γ 100 = 85,
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 518
correct to nearest whole number.
(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 Γ 100 = 13,
correct to nearest whole number.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 519
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 verify the results obtained
To test the normality of a distribution, the upper class boundary values are plotted against
percentage cumulative frequency values on normal probability paper.
The table below shows the upper class boundary values for the distribution, together with the
cumulative frequency and percentage cumulative frequency.
Class mid-point
value
Upper class boundary
value
Frequency Cumulative frequency % cumulative
frequency
26.4
26.6
26.8
27.0
27.2
27.4
27.6
26.5
26.7
26.9
27.1
27.3
27.5
27.7
5
12
24
36
36
25
12
5
5+12 = 17
17 + 24 = 41
77
113
138
150
5/150 = 3
17/150 = 11
41/150 = 27
51
75
92
100
The co-ordinates of upper class boundary values/percentage cumulative frequency values are shown
plotted below. Since the points plotted lie very nearly in a straight line, the data is
approximately normally distributed.
From the graph, the mean occurs at 50%, i.e. mean, x = 27.1 at point P.
At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 27.38
At 16% cumulative frequency value, i.e. point R, upper class boundary value = 26.78
Hence, standard deviation, Ο = 27.38 26.78 0.62 2β
= = 0.3
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 520
By calculation, mean,
x = ( ) ( ) ( ) ( ) ( ) ( ) ( )5 26.4 12 26.6 24 26.8 36 27.0 36 27.2 25 27.4 12 27.6150
Γ + Γ + Γ + Γ + Γ + Γ + Γ
= 4061.8150
= 27.079
Standard deviation, Ο = ( ) ( ) ( )2 2 25 26.4 27.079 12 26.6 27.079 24 26.8 27.079 ....150
β§ β«β + β + β +βͺ βͺβ¨ β¬βͺ βͺβ© β
= 13.51175150
β§ β«β¨ β¬β© β
= 0.3001
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 521
2. A frequency distribution of the class mid-point values of the breaking loads for 275 similar
fibres is as shown below:
Load (kN) 17 19 21 23 25 27 29 31
Frequency 9 23 55 78 64 28 14 4
Use normal probability paper to show that this distribution is approximately normally distributed
and determine the mean and standard deviation of the distribution (a) from the graph and (b) by
calculation.
To test the normality of a distribution, the upper class boundary values are plotted against
percentage cumulative frequency values on normal probability paper.
The table below shows the upper class boundary values for the distribution, together with the
cumulative frequency and percentage cumulative frequency.
Class mid-point
value (kN)
Upper class boundary
value
Frequency Cumulative frequency % cumulative
frequency
17
19
21
23
25
27
29
31
18
20
22
24
26
28
30
32
9
23
55
78
64
28
14
4
9
9 +23 = 32
32 + 55 = 87
165
229
257
271
275
9/275 = 3
32/275 = 12
87/275 = 32
60
83
93
99
100
The co-ordinates of upper class boundary values/percentage cumulative frequency values are shown
plotted below. Since the points plotted lie very nearly in a straight line, the data is
approximately normally distributed.
(a) From the graph, the mean occurs at 50%, at point P , i.e. mean, x = 23.5 kN
At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 26.2
At 16% cumulative frequency value, i.e. point R, upper class boundary value = 20.4
Hence, standard deviation, 26.2 20.4 5.82 2β
Ο = = = 2.9 kN
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 522
(b) By calculation, mean, x = ( ) ( ) ( ) ( ) ( )9 17 23 19 55 21 78 23 64 25 ......275
Γ + Γ + Γ + Γ + Γ +
= 6425275
= 23.364 kN
Standard deviation, Ο = ( ) ( ) ( )2 2 29 17 23.364 23 19 23.364 55 21 23.364 ....275
β§ β«β + β + β +βͺ βͺβ¨ β¬βͺ βͺβ© β
= 2339.6364275
β§ β«β¨ β¬β© β
= 2.917 kN
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 523
CHAPTER 59 LINEAR CORRELATION EXERCISE 218 Page 570
2. Determine the coefficient of correlation for the data given below, correct to 3 decimal places.
X 2.7 4.3 1.2 1.4 4.9
Y 11.9 7.10 33.8 25.0 7.50
A tabular method to determine the quantities is shown below.
X Y x =
( )X Xβ
y =
( )Y Yβ
xy 2x 2y
2.7
4.3
1.2
1.4
4.9
11.9
7.10
33.8
25.0
7.50
-0.2
1.4
-1.7
-1.5
2
-5.16
-9.96
16.74
7.94
-9.56
1.032
-13.944
-28.458
-11.91
-19.12
0.04
1.96
2.89
2.25
4
26.6256
99.2016
280.2276
63.0436
91.3936
X 14.5=β
14.5X 2.95
= =
Y 85.3=β
85.3Y 17.065
= =
xyβ
= - 72.4
2xβ
= 11.14
2yβ
= 560.492
Coefficient of correlation, r = ( )( ) ( )( )2 2
xy 72.411.14 560.492x y
β=β
β β = - 0.916
4. In an experiment to determine the relationship between the current flowing in an electrical
circuit and the applied voltage, the results obtained are:
Current (mA) 5 11 15 19 24 28 33
Applied voltage (V) 2 4 6 8 10 12 14
Determine, using the product-moment formula, the coefficient of correlation for these results.
A tabular method to determine the quantities is shown below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 524
I V i = ( )I Iβ v = ( )V Vβ iv 2i 2v
5
11
15
19
24
28
33
2
4
6
8
10
12
14
-14.286
-8.286
-4.286
-0.286
4.714
8.714
13.714
-6
-4
-2
0
2
4
6
85.716
33.144
8.572
0
9.428
34.856
82.284
204.09
68.66
18.37
0.082
22.22
75.93
188.07
36
16
4
0
4
16
36
I 135=β
135I 19.2867
= =
V 56=β
56V 87
= =
ivβ
= 254
2iβ
= 577.43
2vβ
= 112
Coefficient of correlation, r = ( )( ) ( )( )2 2
iv 254 254254.307577.43 112i v
= =ββ β
= 0.999
5. A gas is being compressed in a closed cylinder and the values of pressures and corresponding
volumes at constant temperature are as follows:
Pressure (kPa) 160 180 200 220 240 260 280 300
Volume ( 3m ) 0.034 0.036 0.030 0.027 0.024 0.025 0.020 0.019
Find the coefficient of correlation for these values.
A tabular method to determine the quantities is shown on page 526.
From the table,
coefficient of correlation, r = ( )( ) ( )( )62 2
xy 2.03 2.032.1094816800 264.875 10x y β
β β= =
Γ
ββ β
= - 0.962
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 525
P V x =
( )P Pβ
y =
( )V Vβ
xy 2x 2y
160
180
200
220
240
260
280
300
0.034
0.036
0.030
0.027
0.024
0.025
0.020
0.019
-70
-50
-30
-10
10
30
50
70
0.007125
0.009125
0.003125
0.000125
-0.002875
-0.001875
-0.006875
0.007875
-0.49875
-0.45625
-0.09375
-0.00125
-0.02875
-0.05625
-0.34375
-0.55125
4900
2500
900
100
100
900
2500
4900
50.7656 610βΓ
83.2656 610βΓ
9.7656 610βΓ
0.0156 610βΓ
8.2656 610βΓ
3.5156 610βΓ
47.2656 610βΓ
62.0156 610βΓ
P 1840=β1840P
8230
=
=
V 0.215=β
0.215V8
0.026875
=
=
xyβ
= -2.03
2xβ
= 16800
2vβ
= 264.875 610βΓ
7. The data shown below refers to the number of times machine tools had to be taken out of
service, in equal time periods, due to faults occurring and the number of hours worked by
maintenance teams. Calculate the coefficient of correlation for this data.
Machines out of service: 4 13 2 9 16 8 7
Maintenance hours: 400 515 360 440 570 380 415
A tabular method to determine the quantities is shown below, where X = machines out of service,
and Y = maintenance hours.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 526
X Y x =
( )X Xβ
y =
( )Y Yβ
xy 2x 2y
4
13
2
9
16
8
7
400
515
360
440
570
380
415
-4.4286
4.5714
-6.4286
0.5714
7.5714
-0.4286
-1.4286
-40
75
-80
0
130
-60
-25
177.144
342.855
514.288
0
984.282
25.716
35.715
19.6125
20.8977
41.3269
0.3265
57.3261
0.1837
2.0409
1600
5625
6400
0
16900
3600
625
X 59=β
59X 8.42867
= =
Y 3080=β
3080Y 4407
= =
xyβ
= 2080.04
2xβ =
141.714
2yβ =
34750
Coefficient of correlation, r = ( )( ) ( )( )2 2
xy 2080.04 2080.042219.135141.714 34750x y
= =ββ β
= 0.937
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 527
CHAPTER 60 LINEAR REGRESSION EXERCISE 219 Page 575
2. Determine the equation of the regression line of Y on X, correct to 3 significant figures, for the
following data:
X 6 3 9 15 2 14 21 13
Y 1.3 0.7 2.0 3.7 0.5 2.9 4.5 2.7
X Y 2X XY 2Y
6
3
9
15
2
14
21
13
1.3
0.7
2.0
3.7
0.5
2.9
4.5
2.7
36
9
81
225
4
196
441
169
7.8
2.1
18.0
55.5
1.0
40.6
94.5
35.1
1.69
0.49
4.0
13.69
0.25
8.41
20.25
7.29
X 83=β Y 18.3=β 2X 1161=β XY 254.6=β 2Y 56.07=β
Substituting into 0 1Y a N a X= +β β
and 20 1XY a X a X= +β β β
gives: 18.3 = 8 0a + 83 1a (1)
and 254.6 = 83 0a + 1161 1a (2)
83 Γ (1) gives: 1518.9 = 664 0a + 6889 1a (3)
8 Γ (2) gives: 2036.8 = 664 0a + 9288 1a (4)
(4) β (3) gives: 517.9 = 2399 1a from which, 1517.9a2399
= = 0.216
Substituting in (1) gives: 18.3 = 8 0a + 83(0.216) from which, 018.3 83(0.216)a
8β
= = 0.0477
Hence, the equation of the regression line of Y on X is: 0 1Y a a X= +
i.e. Y = 0.0477 + 0.216X
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 528
4. Determine the equation of the regression line of X on Y, correct to 3 significant figures, for the
data given in Problem 2.
Substituting into 0 1X b N b Y= +β β
and 20 1XY b Y b Y= +β β β
gives: 83 = 8 0b + 18.3 1b (1)
and 254.6 = 18.3 0b + 56.07 1b (2)
18.3 Γ (1) gives: 1518.9 = 146.4 0b + 334.89 1b (3)
8 Γ (2) gives: 2036.8 = 146.4 0b + 448.56 1b (4)
(4) β (3) gives: 517.9 = 113.67 1b from which, 1517.9b
113.67= = 4.56
Substituting in (1) gives: 83 = 8 0b + 18.3(4.56) from which, 083 18.3(4.56)b
8β
= = -0.056
Hence, the equation of the regression line of X on Y is: 0 1X b b Y= +
i.e. X = - 0.056 + 4.56Y
5. The relationship between the voltage applied to an electrical circuit and the current flowing is as
shown:
Current (mA) 2 4 6 8 10 12 14
Applied voltage (V) 5 11 15 19 24 28 33
Assuming a linear relationship, determine the equation of the regression line of applied voltage,
Y, on current, X, correct to 4 significant figures.
A table is produced as shown below, where current I = X and voltage V = Y
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 529
X Y 2X XY 2Y
2
4
6
8
10
12
14
5
11
15
19
24
28
33
4
16
36
64
100
144
196
10
44
90
152
240
336
462
25
121
225
361
576
784
1089
X 56=β Y 135=β 2X 560=β XY 1334=β 2Y 3181=β
Substituting into 0 1Y a N a X= +β β
and 20 1XY a X a X= +β β β
gives: 135 = 7 0a + 56 1a (1)
and 1334 = 56 0a + 560 1a (2)
8 Γ (1) gives: 1080 = 56 0a + 448 1a (3)
(2) β (3) gives: 254 = 112 1a from which, 1254a112
= = 2.268
Substituting in (1) gives: 135 = 7 0a + 56(2.268) from which, 0135 56(2.268)a
7β
= = 1.142
Hence, the equation of the regression line of Y on X is: 0 1Y a a X= +
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 0 1X b N b Y= +β β
and 20 1XY b Y b Y= +β β β
gives: 56 = 7 0b + 135 1b (1)
and 1334 = 135 0b + 3181 1b (2)
135 Γ (1) gives: 7560 = 945 0b + 18225 1b (3)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 530
7 Γ (2) gives: 9338 = 945 0b + 22267 1b (4)
(4) β (3) gives: 1778 = 4042 1b from which, 11778b4042
= = 0.43988
Substituting in (1) gives: 56 = 7 0b + 135(0.43988) from which, 056 135(0.43988)b
7β
= = -0.483
Hence, the equation of the regression line of X on Y is: 0 1X b b Y= +
i.e. X = - 0.483 + 0.440Y, correct to 3 significant figures.
8. In an experiment to determine the relationship between force and momentum, a force X, is
applied to a mass, by placing the mass on an inclined plane, and the time, Y, for the velocity to
change from u m/s to v m/s is measured. The results obtained are as follows:
Force (N) 11.4 18.7 11.7 12.3 14.7 18.8 19.6
Time (s) 0.56 0.35 0.55 0.52 0.43 0.34 0.31
Determine the equation of the regression line of time on force, assuming a linear relationship
between the quantities, correct to 3 significant figures.
Let force F = X and time = Y. A table is produced as shown below.
X Y 2X XY 2Y
11.4
18.7
11.7
12.3
14.7
18.8
19.6
0.56
0.35
0.55
0.52
0.43
0.34
0.31
129.96
349.69
136.89
151.29
216.09
353.44
384.16
6.384
6.545
6.435
6.396
6.321
6.392
6.076
0.3136
0.1225
0.3025
0.2704
0.1849
0.1156
0.0961
Xβ
= 107.2
Yβ
= 3.06
2Xβ
= 1721.52
XYβ
= 44.549
2Yβ
= 1.4056
Substituting into 0 1Y a N a X= +β β
and 20 1XY a X a X= +β β β
gives: 3.06 = 7 0a + 107.2 1a (1)
and 44.549 = 107.2 0a + 1721.52 1a (2)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 531
107.2 Γ (1) gives: 328.032 = 750.4 0a + 11491.84 1a (3)
7 Γ (2) gives: 311.843 = 750.4 0a + 12050.64 1a (4)
(3) β (4) gives: 16.189 = - 558.8 1a from which, 116.189a558.8β
= = - 0.0290
Substituting in (1) gives: 3.06 = 7 0a + 107.2(-0.0290)
from which, 03.06 107.2( 0.0290)a
7β β
= = 0.881
Hence, the equation of the regression line of Y on X is: 0 1Y a a X= +
i.e. Y = 0.881 β 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.
Substituting into 0 1X b N b Y= +β β
and 20 1XY b Y b Y= +β β β
gives: 107.2 = 7 0b + 3.06 1b (1)
and 44.549 = 3.06 0b + 1.4056 1b (2)
3.06 Γ (1) gives: 328.032 = 21.42 0b + 9.3636 1b (3)
7 Γ (2) gives: 311.843 = 21.42 0b + 9.8392 1b (4)
(3) β (4) gives: 16.189 = - 0.4756 1b from which, 116.189b0.4756
=β
= - 34.039
Substituting in (1) gives: 107.2 = 7 0b + 3.06(-34.039)
from which, 0107.2 3.06(34.039)b
7+
= = 30.194
Hence, the equation of the regression line of X on Y is: 0 1X b b Y= +
i.e. X = 30.194 β 34.039Y
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 532
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 β 0.0290X, and
force on time, X = 30.194 β 34.039Y are shown and to the scale drawn are seen to coincide.
(a) When force X = 16 N then time, Y = 0.881 β 0.0290(16) = 0.417 s
(b) When time Y = 0.25 s, force, X = 30.194 β 34.039(0.25) = 21.7 N
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 533
CHAPTER 61 SAMPLING AND ESTIMATION THEORIES EXERCISE 220 Page 580
1. The lengths of 1500 bolts are normally distributed with a mean of 22.4 cm and a standard
deviation of 0.0438 cm. If 30 samples are drawn at random from this population, each sample
being 36 bolts, determine the mean of the sampling distributions and standard error of the means
when sampling is done with replacement.
For the population, number of bolts, pN = 1500, standard deviation, Ο = 0.0438 cm,
mean, Β΅ = 22.4 cm. For the sample, N = 30
The mean of the sampling distributions, xΒ΅ = Β΅ = 22.4 cm
The standard error of the means with replacement, x
0.0438N 30Ο
Ο = = = 0.0080 cm
2. Determine the standard error of the means in problem 1, if sampling is done without
replacement, correct to 4 decimal places.
Standard error of means without replacement, px
p
N N 0.0438 1500 30N 1 1500 1N 30
β ββΟ ββ βΟ = =β β β ββ ββ ββ β β β
= (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 hourβs output from the punch, (a) with replacement, and (b) without
replacement, correct to three significant figures.
For the population, number of bolts, pN = 1800, standard deviation, Ο = 0.013 cm,
mean, Β΅ = 1.70 cm. For the sample, N = 20
(a) With replacement.
The mean of the sampling distributions, xΒ΅ = Β΅ = 1.70 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 534
The standard error of the means, x
0.013N 20Ο
Ο = = = 2.907 310βΓ = 32.91 10βΓ cm, correct
to 3 significant figures.
(b) Without replacement.
The mean of the sampling distributions, xΒ΅ = Β΅ = 1.70 cm
Standard error of means, px
p
N N 0.013 1800 20N 1 1800 1N 20
β ββΟ ββ βΟ = =β β β ββ ββ ββ β β β
= ( )( )32.907 10 0.9947βΓ = 32.89 10βΓ cm
5. A large batch of electric light bulbs have a mean time to failure of 800 hours and the standard
deviation of the batch is 60 hours. Determine the probability that the mean time to failure of a
random sample of 16 light bulbs will be between 790 hours and 810 hours, correct to three
decimal places.
N = 16 and x
60N 16Ο
Ο = = = 15 h
790 h corresponds to a z-value of: 1790 800z
15β
= = - 0.67 standard deviations and the area
between z = 0 and z = -0.67 is 0.2486 (from Table 58.1, page 561 of textbook).
810 h corresponds to a z-value of: 2810 800z
15β
= = 0.67 standard deviations and the area between
z = 0 and z = 0.67 is 0.2486.
Hence, the probability that the mean time to failure will be between 790 hours and 810 hours
(i.e. the shaded area of the diagram below) = 2 Γ 0.2486 = 0.497, correct to 3 decimal places.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 535
7. The contents of a consignment of 1200 tins of a product have a mean mass of 0.504 kg and a
standard deviation of 92 g. Determine the probability that a random sample of 40 tins drawn
from the consignment will have a combined mass of (a) less than 20.13 kg, (b) between 20.13 kg
and 20.17 kg, and (c) more than 20.17 kg, correct to three significant figures.
Number of tins, pN = 1200, mean mass, Β΅ = 0.504 kg and standard deviation, Ο = 92 g = 0.092 kg,
For the sample, N = 40
Standard error of means, px
p
N N 0.092 1200 40N 1 1200 1N 40
β ββΟ ββ βΟ = =β β β ββ ββ ββ β β β
= (0.01455)(0.9836) = 0.0143 kg
The mean mass of 40 tins = 40 Γ 0.504 = 20.16 kg
(a) z-value corresponding to 20.13 kg, z = 20.13 20.160.0143β = - 2.1 standard deviations.
The area between z = 0 and z = - 2.1 is 0.4821 (from table 58.1, page 561 of textbook) and is the
shaded area in the diagram below.
Hence, the probability that the combined mass will be less than 20.13 kg is:
0.5000 β 0.4821 = 0.0179
(b) The z-value corresponding to 20.13 kg, is - 2.1 standard deviations from above, and the area
between z = 0 and z = - 2.1 is 0.4821
The z-value corresponding to 20.17 kg, z = 20.17 20.160.0143β = 0.7 standard deviations and the area
between z = 0 and z = 0.7 is 0.2580 (from table 58.1, page 561 of textbook)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 536
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 β 0.2580 = 0.242
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 537
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, pN 1546= and Ο = 2.93 mm
For the sample, N = 100 and x = 67.45 mm
For a 95% confidence limit, Cz = 1.96 from Table 61.1, page 582 of textbook.
An estimate of the confidence limits of the population mean is:
( )( )pC
p
N N 1.96 2.93z 1546 100x 67.45N 1 1546 1N 100
β ββΟ ββ βΒ± = Β±β β β ββ ββ ββ β β β
= 67.45 Β± (0.57428)(0.96743)
= 67.45 Β± 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, Cz = 2.58 from Table 61.1, page 582 of textbook.
An estimate of the confidence limits of the population mean is:
( )( )pC
p
N N 2.58 2.93z 1546 100x 67.45N 1 1546 1N 100
β ββΟ ββ βΒ± = Β±β β β ββ ββ ββ β β β
= 67.45 Β± (0.75594)(0.96743)
= 67.45 Β± 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
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 538
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 Β± 0.035 Mg?
pN 500= and Ο = 150 kg = 0.15 Mg, N = 40 and x = 2.40 Mg
(a) For a 95% confidence limit, Cz = 1.96 from Table 61.1, page 582 of textbook.
An estimate of the confidence limits of the population mean is:
( )( )pC
p
N N 1.96 0.15z 500 40x 2.40N 1 500 1N 40
β ββΟ ββ βΒ± = Β±β β β ββ ββ ββ β β β
= 2.40 Β± (0.0465)(0.9601)
= 2.40 Β± 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, Cz = 2.58 from Table 61.1, page 582 of textbook.
An estimate of the confidence limits of the population mean is:
( )( )pC
p
N N 2.58 0.15z 500 40x 2.40N 1 500 1N 40
β ββΟ ββ βΒ± = Β±β β β ββ ββ ββ β β β
= 2.40 Β± (0.06119)(0.9601)
= 2.40 Β± 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 Β± 0.035 = CzxNΟ
Β±
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 539
x = 2.40, hence, Cz 0.035NΟ
Β± = Β±
and ( )C
0.035 400.035 Nz0.15
= =Ο
= Β± 1.48
From Table 58.1, page 561 of textbook, 1.48 corresponds to an area of 0.4306.
The area between the mean and Β± 1.48 is 2 Γ 0.4306 = 0.8612 (see diagram below).
Thus, the confidence level corresponding to 2.40 Β± 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.
Ο = 100 h The confidence limits for the mean of a population is Czx
NΟ
Β±
For a 90% confidence level, Cz = 1.645 from Table 61.1, page 582 of textbook.
(a) ( )( )1.645 100x 20 x
NΒ± = Β±
from which, ( )( )1.645 10020
N= and ( )( )1.645 100
N20
= = 8.225
i.e. N = 2(8.225) = 67.65
Hence, at least 68 components are required to be 90% confident that the error will not
exceed 20 hours.
(b) ( )( )1.645 100x 10 x
NΒ± = Β±
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 540
from which, ( )( )1.645 10010
N= and ( )( )1.645 100
N10
= = 16.45
i.e. N = 2(16.45) = 270.6
Hence, at least 271 components are required to be 90% confident that the error will not
exceed 10 hours.
6. The time taken to assemble a servo-mechanism is measured for 40 operatives and the mean time
is 14.63 minutes with a standard deviation of 2.45 minutes. Determine the maximum error in
estimating the true mean time to assemble the servo-mechanism for all operatives, based on a
95% confidence level
N = 40, x = 14.63 minutes, Ο = 2.45 minutes, at 95% confidence level, Cz = 1.96 (from Table
61.1, page 582 of textbook).
Hence, ( )( )1.96 2.4514.63 14.63 0.759
40Β± = Β±
i.e. the maximum error is 0.759 minutes = 0.759 Γ 60 = 45.6 s
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 541
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, Ξ½ = 10 β 1 = 9
From Table 61.2, page 587 of textbook, 0.95t , 9Ξ½ = gives Ct = 1.83
Estimated value of the mean of the U.T.S. is: ( )( )C 1.83 0.06t sx 5.17N 1 10 1
Β± = Β±β β
= 5.17 Β± 0.0366
= 5.17 β 0.0366 or 5.17 + 0.0366
Thus, the 95% confidence interval is: 5.133 MPa to 5.207 MPa.
3. The specific resistance of a reel of German silver wire of nominal diameter 0.5 mm is estimated
by determining the resistance of 7 samples of the wire. These were found to have resistance
values (in ohms per metre) of:
1.12, 1.15, 1.10, 1.14, 1.15, 1.10 and 1.11
Determine the 99% confidence interval for the true specific resistance of the reel of wire.
N = 7, mean, 1.12 1.15 1.10 1.14 1.15 1.10 1.11x7
+ + + + + += = 1.1243 Ξ© 1mβ
and standard deviation, s = ( ) ( ) ( )2 2 21.12 1.1243 1.15 1.1243 1.10 1.1243 ...7
β§ β«β + β + β +βͺ βͺβ¨ β¬βͺ βͺβ© β
= 0.002971437
= 0.0206 Ξ© 1mβ
From Table 61.2, page 587 of textbook, 0.99t , 7 1 6Ξ½ = β = gives Ct = 3.14
Hence, 99% confidence limits is given by: ( )( )C 3.14 0.0206t sx 1.1243N 1 7 1
Β± = Β±β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 542
= 1.1243 Β± 0.0264
= 1.1243 β 0.0264 or 1.1243 + 0.0264
Thus, the 99% confidence interval is: 1.10 1mβΞ© to 1.15 1mβΞ©
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 132.27Β°C and 0.742Β°C. Calculate the
confidence with which the prediction βthe melting point of the metal is between 131.48Β°C and
133.06Β°C β can be made?
N = 5, mean, x = 132.27Β°C and standard deviation, s = 0.742Β°C
133.06 β 131.48 = 0.79Β°C
Hence, Ct s132.27 0.79 xN 1
Β± = Β±β
i.e. ( )Ct 0.7420.79
5 1=
β
from which, ( )C
0.79 4t
0.742= = 2.13
From Table 61.2, page 587 of textbook, at Ξ½ = N β 1 = 5 β 1 = 4, 2.13 corresponds to 0.95t , i.e.
the confidence with which the prediction βthe melting point of the metal is between 131.48Β°C
and 133.06Β°C β can be made is 95%.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 543
CHAPTER 62 SIGNIFICANCE TESTING EXERCISE 223 Page 597
1. An automatic machine produces piston rings for car engines. Random samples of 1000 rings are
drawn from the output of the machine periodically for inspection purposes. A defect rate of 5% is
acceptable to the manufacturer, but if the defect rate is believed to have exceeded this value, the
machine producing the rings is stopped and adjusted. Determine the type 1 errors which occur for
the following decision rule: Stop production and adjust the machine if a sample contains (a) 54
(b) 62 and (c) 70 or more defective rings.
(a) N = 1000, p = 0.05, q = 0.95, mean of normal distribution = Np = 50,
standard deviation of the normal distribution = ( ) ( )( )( )Npq 1000 0.05 0.95 6.892= = .
A type I error is the probability of stopping production when getting more than 54 defective
rings in the sample, even though defect rate is 5%.
z-value = var iate means tan dard deviation
β = 54 506.892β = 0.58 and from Table 58.1, page 561 of textbook, the
area between the mean and a z-value of 0.58 is 0.2190.
Thus, the probability of more than 54 defective rings = 0.5000 β 0.2190 = 0.281
Hence, the type I error is 28.1%
(b) z-value = 62 506.892β = 1.74 and from Table 58.1, the area between the mean and a z-value of 1.74
is 0.4591.
Thus, the probability of more than 62 defective rings = 0.5000 β 0.4591 = 0.0409
Hence, the type I error is 4.09%
(c) z-value = 70 506.892β = 2.90 and from Table 58.1, the area between the mean and a z-value of
2.90 is 0.4981.
Thus, the probability of more than 70 defective rings = 0.5000 β 0.4981 = 0.0019
Hence, the type I error is 0.19%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 544
2. For the data in Problem 1, determine the type II errors which are made if the decision rule is to
stop production if there are more than 60 defective components in the sample when the actual
defect rate has risen to (a) 6% (b) 7.5% and (c) 9%.
(a) N = 1000, p = 0.06, q = 0.94
Mean = Np = 60, standard deviation = ( ) ( )( )( )Npq 1000 0.06 0.94 7.51= =
z-value = 61 607.51β = 0.13 (note that βmore than 60 components defectiveβ means 61 or more)
and from Table 58.1, page 561 of textbook, the area between the mean and a z-value of
0.13 is 0.0517.
Thus, the probability of more than 60 defective components = 0.5000 + 0.0517 = 0.5517
Hence, the type II error is 55.2%
(b) N = 1000, p = 0.075, q = 0.925
Mean = Np = 75, standard deviation = ( ) ( )( )( )Npq 1000 0.075 0.925 8.329= =
z-value = 61 758.329β = -1.68 and from Table 58.1, the area between the mean and a z-value of
-1.68 is 0.4535.
Thus, the probability of more than 75 defective components = 0.5000 - 0.4535 = 0.0465
Hence, the type II error is 4.65%
(c) N = 1000, p = 0.09, q = 0.91
Mean = Np = 90, standard deviation = ( ) ( )( )( )Npq 1000 0.09 0.91 9.05= =
z-value = 61 909.05β = -3.20 and from Table 58.1, the area between the mean and a z-value of
-3.20 is 0.4993.
Thus, the probability of more than 90 defective components = 0.5000 - 0.4993 = 0.0007
Hence, the type II error is 0.07%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 545
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 β₯ 50 and Np β€ 5 the Poisson distribution is used.
Ξ» = Np = 3
The probability of 0, 1, 2, 3, 4, 5, β¦ defective components are given by the terms:
3e e 0.0498βΞ» β= = , 3e 3e 0.1494βΞ» βΞ» = = , 2 2
33e e 0.22402! 2
βΞ» βΞ»= = ,
3 333e e 0.2240
3! 6βΞ» βΞ»= = ,
4 433e e 0.1680
4! 24βΞ» βΞ»= = ,
5 533e e 0.1008
5! 120βΞ» βΞ»= =
(a) The probability of a sample containing 4 or more defective components is:
1 β (0.0498 + 0.1494 + 0.2240 + 0.2240)
= 1 β 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 β (0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680)
= 1 β 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 β (0.8152 + 0.1008)
= 1 β 0.9160 = 0.0840
i.e. the type I error is 8.4%
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 546
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.
Β΅ = 2000 kN, Ο = 100 kN, N = 50, x 2050kN=
The null hypothesis, 0H : x > Β΅
The alternative hypothesis, 1H : x = Β΅
z = x 2050 2000 50 3.54100 14.142N 50
βΒ΅ β= = Β± = Β±
Ο
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 Β± 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.
Β΅ = 3.0000 cm, Ο = 0.015 cm, N = 30, x 3.0078cm=
The null hypothesis, 0H : mean diameter = 3.0000 cm
The alternative hypothesis, 1H : mean diameter β 3.0000 cm
z = x 3.0078 3.0000 0.0078 2.850.015 0.00274N 30
βΒ΅ β= = Β± = Β±
Ο
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 Β± 2.58.
Since the z-value of the sample is outside of this range, the hypothesis is rejected.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 547
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.
Β΅ = 10.25 kN and N = 20
Sample mean, ( ) ( ) ( ) ( )9.8 1 10.0 1 10.1 4 10.2 5 .... 207.6x 10.3820 20
Γ + Γ + Γ + Γ += = =
Sample standard deviation,
s = ( ) ( ) ( ) ( )2 2 2 21 9.8 10.38 1 10.0 10.38 4 10.1 10.38 5 10.2 10.38 ... 2.21220 20
β + β + β + β += = 0.33
The null hypothesis is that the sample breaking strength is better than the mean breaking strength.
N < 30, therefore a t distribution is used.
( ) ( ) ( ) ( )x N 1 10.38 10.25 20 1
ts 0.33
βΒ΅ β β β= = = 1.72
At a level of significance of 0.1, the t value is 0.112
tβ βββ ββ β
i.e. 0.95t and Ξ½ = N β 1 = 20 β 1 = 19, and
from Table 61.1, page 587 of textbook, 0.95t 19Ξ½ = 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.
Β΅ = 0.50 cm, N = 10, x = 0.53 cm and s = 0.03 cm
The null hypothesis is: 0H : Β΅ = 0.50 cm
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 548
( ) ( ) ( ) ( )x N 1 0.53 0.50 10 1
ts 0.03
βΒ΅ β β β= = = 3
(a) From Table 61.1, page 587 of the textbook, 0.9750.0512
t t , 9β βββ ββ β
= Ξ½ = has a value of 2.26
Since 3 is outside of this range, the hypothesis is rejected.
(b) From Table 61.1, 0.9950.0112
t t , 9β βββ ββ β
= Ξ½ = has a value of 3.25
Since 3 is within this range, the hypothesis is accepted.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 549
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: Ax 24= , A 4Ο = and AN 100=
Battery B: Bx 40= , B 6Ο = and BN 80=
The hypothesis is: H: x = Ax + 15
Let x = 24 + 15 = 39
z = B
2 22 2A B
A B
x x 39 40 1 1.280.7810254 6
100 80N N
β β β= = = β
β β β βΟ Ο ++β β β ββ β β β
From Table 62.1, page 594 of textbook, for Ξ± = 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 Β΅F 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.
(Besselβs correction is $22 s N
N 1Ο =
β)
N = 5, 121.4 23.6 24.8 22.4 26.3x 23.7
5+ + + +
= =
( ) ( ) ( ) ( ) ( )2 2 2 2 2
1
21.4 23.7 23.6 23.7 24.8 23.7 22.4 23.7 26.3 23.7s 1.73
5
β ββ + β + β + β + β= =β β
β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 550
1 1N 5s 1.73 1.93
N 1 4β β β βΟ = = =β β β βββ β β β
222.4 27.7 23.5 29.1 25.8x 25.7
5+ + + +
= =
( ) ( ) ( ) ( ) ( )2 2 2 2 2
2
22.4 25.7 27.7 25.7 23.5 25.7 29.1 25.7 25.8 25.7s 2.50
5
β ββ + β + β + β + β= =β β
β ββ β
2 2N 5s 2.50 2.80
N 1 4β β β βΟ = = =β β β βββ β β β
2 1
2 222yx
x y
x x 25.7 23.7 2t 1.321.52081.93 2.80
5 5N N
β β= = = =
β β β βΟΟ ++β β β ββ β β β β β
0.9950.0112
t tβ βββ ββ β
= and 1 2N N 2 5 5 2 8Ξ½ = + β = + β =
From Table 61.2, page 587 of textbook, 0.995t , 8Ξ½ = has a value of 3.36
Since the t value of the difference of the means, i.e. 1.32, is within the range Β± 3.36, the hypothesis
is accepted.
5. A sample of 12 car engines produced by manufacturer A showed that the mean petrol
consumption over a measured distance was 4.8 litres with a standard deviation of 0.40 litres.
Twelve similar engines for manufacturer B were tested over the same distance and the mean
petrol consumption was 5.1 litres with a standard deviation of 0.36 litres. Test the hypothesis
that the engines produced by manufacturer A are more economical than those produced by
manufacturer B at a level of significance of (a) 0.01 and (b) 0.1.
AA AN 12, x 4.8 litre, 0.40 litre= = Ο =
BB BN 12, x 5.1 litre, 0.36 litre= = Ο =
The hypothesis is: H: manufacturer A is more economical than manufacturer B.
( ) ( )2 22 2A A B B
A B
12 0.40 12 0.36N s N s 3.4752 0.397N N 2 12 12 2 22
β β+β β+Ο = = = =β ββ β β β+ β + ββ β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 551
( )( )A B
A B
x x 4.8 5.1 0.3t 1.850.397 0.408251 11 1 0.397
12 12N N
β β= = = β = β
β β β β+Ο + β ββ β β β β β
(a) The t value is 0.0112
tβ βββ ββ β
i.e. 0.995t and Ξ½ = 12 + 12 β 2 = 22, hence, from Table 61.2, page 587 of
textbook, 0.995t , 22Ξ½ = has a value of 2.82
Since the t value of the difference of the means is outside the range Β± 1.85, the hypothesis is
rejected.
(b) From Table 61.2, 0.112
tβ βββ ββ β
i.e. 0.95t , 22Ξ½ = has a value of 1.72
Since the t value of the difference of the means is within the range Β± 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.
4 4N 5, x 21.4 km, s 0.54 km= = =
un unx 22.6 km, s 0.48 km= =
The hypothesis is: H: unx > 4x
( ) ( )2 22 24 4 un un
4 un
5 0.54 5 0.48N s N s 2.61 0.571N N 2 5 5 2 8
β β+β β+Ο = = = =β ββ β β β+ β + ββ β β β
( )( )un 4
un 4
x x 22.6 21.4 1.2t 3.320.571 0.632461 11 1 0.571
5 5N N
β β= = = =
β β β β+Ο + β ββ β β β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 552
The t value is 0.112
tβ βββ ββ β
i.e. 0.95t and Ξ½ = 5 + 5 β 2 = 8, hence, from Table 61.2, page 587 of textbook,
0.95t , 8Ξ½ = 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 Β± 3.32, the
hypothesis is rejected.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 553
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 Ο2-value for this distribution.
Face Observed
frequency, o
Expected
frequency, e
o - e ( )2o eβ ( )2o eeβ
1
2
3
4
5
6
49
35
32
46
49
29
40
40
40
40
40
40
9
-5
-8
6
9
-11
81
25
64
36
81
121
2.025
0.625
1.6
0.9
2.025
3.025
( )22 o e
10.2e
β§ β«ββͺ βͺΟ = =β¨ β¬βͺ βͺβ© β
β
Hence, the Chi-square value, 2 10.2Ο =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 554
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 Ο2-value for this data.
Number
of calls
Observed
frequency, o
Expected
frequency, e
o - e ( )2o eβ
( )2o eeβ
0
1
2
3
4
5
6
7
11
44
53
46
24
12
7
3
16
42
52
42
26
14
6
2
-5
2
1
4
-2
-2
1
1
25
4
1
16
4
4
1
1
1.5625
0.0952
0.0192
0.3810
0.1538
0.2857
0.1667
0.5000
( )22 o e
3.16e
β§ β«ββͺ βͺΟ = =β¨ β¬βͺ βͺβ© β
β
Hence, the Chi-square value, 2 3.16Ο =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 555
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 βtoo goodβ at a level of confidence of 90%?
( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )( ) ( )
6 5 2 4 37
7 3 4 2 5
6 7
(7)(6) (7)(6)(5)0.6 7 0.6 0.4 0.6 0.4 0.6 0.42! 3!
(7)(6)(5)(4) (7)(6)(5)(4)(3)250 0.6 0.4 250 0.6 0.4 0.6 0.44! 5!
(7)(6)(5)(4)(3)(2) 0.6 (0.4 0.46!
β‘ β€+ + +β’ β₯β’ β₯β’ β₯+ = + +β’ β₯β’ β₯β’ β₯+ +β’ β₯β£ β¦
= 250[0.02799 + 0.13064 + 0.26127 + 0.29030 + 0.19354 + 0.07741 + 0.01720
+ 0.00164]
= 7 + 33 + 65 + 73 + 48 + 19 + 4 + 0 correct to the nearest whole number
Observed
frequency, o
Expected
frequency, e
o - e ( )2o eβ ( )2o eeβ
8
27
62
79
45
24
5
0
7
33
65
73
48
19
4
0
1
-6
-3
6
-3
5
1
0
1
36
9
36
9
25
1
0
0.14286
1.09091
0.13846
0.49315
0.18750
1.31579
0.25000
0
( )22 o e
3.62e
β§ β«ββͺ βͺΟ = =β¨ β¬βͺ βͺβ© β
β
Degrees of freedom, Ξ½ = N β 1 = 8 β 1 = 7
For 20.95Ο and Ξ½ = 7 from Table 63.1, page 609 of textbook is 14.1
Hence, the hypothesis is accepted, i.e. the observed data fits.
20.10 7,Ο Ξ½ = 2.83, hence the data is not βtoo goodβ.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 556
3. The resistances of a sample of carbon resistors are as shown below.
Resistance (MΞ©) 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.
(7 1.28) (19 1.29) (41 1.30) (50 1.31) .... 390.5x 1.327 19 41 50 73 52 28 17 9 296
Γ + Γ + Γ + Γ += = =
+ + + + + + + +
( ) ( ) ( ) ( )2 2 2 27 1.28 1.32 19 1.29 1.32 41 1.30 1.32 50 1.31 1.32 ....s
296
0.0958 0.0180296
β§ β«β + β + β + β +βͺ βͺ= β¨ β¬βͺ βͺβ© β
= =
Class
mid-point
Class
boundaries, x
z-value for class
boundary =
x 1.320.0180β
Area from
0 to z
from Table
58.1, page 561
Area for
class
Expected
frequency
1.28
1.29
1.30
1.31
1.32
1.33
1.34
1.35
1.36
1.275
1.285
1.295
1.305
1.315
1.325
1.335
1.345
1.355
1.365
-2.50
-1.94
-1.39
-0.83
-0.28
0.28
0.83
1.39
1.94
2.5
0.4938
0.4738
0.4177
0.2967
0.1103
0.1103
0.2967
0.4177
0.4738
0.4938
0.02
0.0561
0.121
0.1864
0.2206
0.1864
0.121
0.0561
0.02
6
17
36
55
65
55
36
17 6
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 557
Resistance Observed
frequency,
o
Expected
frequency, e
o - e ( )2o eβ
( )2o eeβ
1.28
1.29
1.30
1.31
1.32
1.33
1.34
1.35
1.36
7
19
41
50
73
52
28
17
9
6
17
36
55
65
55
36
17
6
1
2
5
-5
8
-3
-8
0
3
1
4
25
25
64
9
64
0
9
0.1667
0.2353
0.6944
0.4545
0.9846
0.1636
1.7778
0
1.5000
( )22 o e
5.98e
β§ β«ββͺ βͺΟ = =β¨ β¬βͺ βͺβ© β
β
Degrees of freedom, Ξ½ = N β 1 β M = 9 β 1 β 2 = 6
For 20.95Ο and Ξ½ = 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 supported by certain cables produced
by a company follows a normal distribution at a level of significance of 0.05, based on the
experimental data given below. Also, test to see if the data is βtoo goodβ at a level of significance
of 0.05.
Maximum load (MN) 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0
Number of cables 2 5 12 17 14 6 3 1
(2 8.5) (5 9.0) (12 9.5) (17 10.0) .... 605.5x 10.09MN
2 5 12 17 14 6 3 1 60Γ + Γ + Γ + Γ +
= = =+ + + + + + +
( ) ( ) ( ) ( )2 2 2 22 8.5 10.09 5 9.0 10.09 12 9.50 10.09 17 10.0 10.09 ....s
60
32.246 0.733MN60
β§ β«β + β + β + β +βͺ βͺ= β¨ β¬βͺ βͺβ© β
= =
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 558
Class
mid-point
Class
boundaries, x
z-value for class
boundary =
x 10.090.733β
Area from
0 to z
Area for class Expected
frequency
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
8.25
8.75
9.25
9.75
10.25
10.75
11.25
11.75
12.25
-2.51
-1.83
-1.15
-0.46
0.22
0.90
1.58
2.26
2.95
0.4940
0.4664
0.3749
0.1772
0.0871
0.3159
0.4430
0.4881
0.4984
0.0276
0.0915
0.1977
0.2643
0.2288
0.1271
0.0451
0.0103
2 5
12
16
14 8 3 1
Load Observed
frequency,
o
Expected
frequency, e
o - e ( )2o eβ
( )2o eeβ
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
2
5
12
17
14
6
3
1
2
5
12
16
14
8
3
1
0
0
0
1
0
-2
0
0
0
0
0
1
0
4
0
0
0
0
0
0.0625
0
0.5000
0
0
( )22 o e
0.563e
β§ β«ββͺ βͺΟ = =β¨ β¬βͺ βͺβ© β
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 559
Degrees of freedom, Ξ½ = N β 1 β M = 8 β 1 β 2 = 5
For 20.95Ο and Ξ½ = 5 from Table 63.1, page 609 of textbook is 11.1
Hence, the hypothesis is accepted.
20.05 5, 1.15Ο Ξ½ = , hence the results are βtoo good to be trueβ.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 560
EXERCISE 228 Page 616
2. In a laboratory experiment, 18 measurements of the coefficient of friction, Β΅, 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 Β΅ = 0.56 against an
alternative hypothesis Β΅ β 0.56.
Using the procedure for the sign test:
(i) Null hypothesis, 0H : 0.56Β΅ =
Alternative hypothesis, 1H : 0.56Β΅ β
(ii) Significance level, 2 5%Ξ± =
(iii) With + sign β₯ 0.56 and - sign < 0.56:
+ + - - + + - + + - + + + + + - + -
(iv) There are 12 + signs and 6 β signs, hence, S = 6
(v) From Table 63.3, page 614 of textbook, with n = 18 and 2 5%Ξ± = S β€ 4
hence, the null hypothesis is accepted.
3. 18 random samples of two types of 9 V batteries are taken and the mean lifetime (in hours) of
each are:
Type A 8.2 7.0 11.3 13.9 9.0 13.8 16.2 8.6 9.4
3.6 7.5 6.5 18.0 11.5 13.4 6.9 14.2 12.4
Type B 15.3 15.4 11.2 16.1 18.1 17.1 17.7 8.4 13.5
7.8 9.8 10.6 16.4 12.7 16.8 9.9 12.9 14.7
Use the sign test, at a level of significance of 5%, to test the null hypothesis that the two samples
come from the same population.
Using the procedure for the sign test:
(i) Null hypothesis, 0 A BH : Β΅ = Β΅
Alternative hypothesis, 1 A BH : Β΅ β Β΅
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 561
(ii) Significance level, 2 5%Ξ± =
(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 β signs, hence, S = 4
(v) From Table 63.3, page 614 of textbook, with n = 18 and 2 5%Ξ± = S β€ 4
Since from (iv) S is equal to 4, then the result is significant at 2 5%Ξ± = hence, the alternative
hypothesis 1H is accepted.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 562
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) 0H : t = 220 h
1H : t β 220 h
(ii) 2 5%Ξ± =
(iii) Taking the time difference between the time taken for repair and 220 h gives:
-2 +55 +44 -10 -59 +154 -42 +45
-70 +140 -35 -49 -5 -120 +254 + 28
(iv) Ranking gives:
Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Difference -2 -5 -10 +28 -35 -42 +44 +45 -49 +55 -59 -70 -120 +140 +154 +254
(v) T = 4 + 7 + 8 + 10 + 14 + 15 + 16 = 74
(vi) From Table 63.4, page 617 of textbook, for n = 16, 2 5%Ξ± = , T β€ 29
Hence, since 74 > 29, the hypothesis 0H is accepted, i.e. the mean repair time is 220 hours.
3. A paint supplier claims that a new additive will reduce the drying time of their acrylic paint. To
test his claim, 12 pieces of wood are painted, one half of each piece with paint containing the
regular additive and the other half with paint containing the new additive. The drying time (in
hours) were measured as follows:
New additive 4.5 5.5 3.9 3.6 4.1 6.3 5.9 6.7 5.1 3.6 4.0 3.0
Regular additive 4.7 5.9 3.9 3.8 4.4 6.5 6.9 6.5 5.3 3.6 3.9 3.9
Use the Wilcoxon signed-rank test at a significance level of 5% to test the hypothesis that there is
no difference, on average, in the drying times of the new and regular additive paints.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 563
Using the procedure for the Wilcoxon signed-rank test:
(i) 0H : N = R
1H : N β R
(ii) 2 5%Ξ± =
(iii) Taking the time difference between N and R gives:
(N β 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 zeroβs gives:
Rank 1 4 4 4 4 4 7 8 9 10
Difference +0.1 -0.2 -0.2 -0.2 -0.2 +0.2 -0.3 -0.4 -0.9 -1.0
(v) T = 1 + 4 = 5
(vi) From Table 63.4, page 617 of textbook, for n = 10, 2 5%Ξ± = , T β€ 8
Since from (v) T is less than 8, there is a significant difference in the drying times.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 564
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) 0 A BH : T T=
1 A BH : T Tβ
(ii) 2 5%Ξ± =
(iii) Brand P 0.7 1.7 2.0 2.3 3.2 3.9 4.1 5.6 6.1 22.6
Brand Q 2.1 3.4 3.5 3.8 4.7 5.5 6.2 6.3
(iv) P P P Q P P Q Q Q P P Q Q P P Q Q P
0 0 0 1 1 4 4 6 6 8 writing the Qβs that precede the Pβs
(v) U = 1 + 1 + 4 + 4 + 6 + 6 + 8 = 30
(vi) From Table 63.5, page 622 of textbook, for a sample size of 10 and 8 at 2 5%Ξ± = , U β€ 17
Hence, 0H is accepted, i.e. there is no difference between brands P and Q
3. An experiment, designed to compare two preventive methods against corrosion gave the
following results for the maximum depths of pits (in mm) in metal strands:
Method A 143 106 135 147 139 132 153 140
Method B 98 105 137 94 112 103
Use the Mann-Whitney test, at a level of significance of 0.05, to determine whether the two tests
are equally effective.
Using the procedure for the Mann-Whitney test:
(i) 0H : A B=
1H : A Bβ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 565
(ii) 2 5%Ξ± =
(iii) A 106 132 135 139 140 143 147 153
B 94 98 103 105 112 137
(iv) B B B B A B A A B A A A A A
1 3 writing the Aβs that precede the Bβs
(v) U = 1 + 3 = 4
(vi) From Table 63.5, page 621 of textbook, for a sample size of 8 and 6 at 2 5%Ξ± = , U β€ 8
Hence, the null hypothesis 0H is rejected, i.e. the two methods are not equally effective.
4. Repeat problem 3 of Exercise 228, using the Mann-Whitney test.
Using the procedure for the Mann-Whitney test:
(i) Null hypothesis, 0 A BH : Β΅ = Β΅
Alternative hypothesis, 1 A BH : Β΅ β Β΅
(ii) 2 5%Ξ± =
(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 2 5%Ξ± = , U β€ 99
Hence, the null hypothesis 0H is rejected and 1H is accepted, i.e. the two means are not
equal.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 566
CHAPTER 64 INTRODUCTION TO LAPLACE TRANSFORMS EXERCISE 231 Page 630
1. Determine the Laplace transforms of (a) 2t β 3 (b) 25t 4t 3+ β
(a) β 2
1 12t 3 2 3s s
β β β ββ = ββ β β ββ β β β
= 2
2 3s sβ
(b) β 23 2
2! 1 15t 4t 3 5 4 3s s s
β β β β β β+ β = + ββ β β β β ββ β β β β β
= 3 2
10 4 3s s s
+ β
2. Determine the Laplace transforms of (a) 3t
24 - 3t + 2 (b)
5 24t t2t
15 2β +
(a) β3
3 1 2
t 1 3! 1 13t 2 3 224 24 s s s+
β§ β« β ββ β β β β ββ + = β +β¨ β¬ β ββ β β β β ββ β β β β β β β β© β
= 4 2
1 3 24s s s
β +
(b) β5 2
45 1 4 1 3
t t 1 5! 4! 1 2!2t 215 2 15 s s 2 s+ +
β§ β« β ββ β β β β ββ ββ + = β +β¨ β¬ β ββ β β β β ββ ββ β β β β β β β β β β© β
= 6 5 3
8 48 1s s sβ +
3. Determine the Laplace transform of (a) 3t5e (b) 2t2eβ
(a) β 3t 15e 5s 3
β β= β βββ β = 5
s 3β
(b) β 2t 12e 2s 2
β β β= β β+β β = 2
s 2+
4. Determine the Laplace transform of (a) 4 sin 3t (b) 3 cos 2t
(a) β 2 2
34sin 3t 4s 3
β β= β β+β β = 2
12s 9+
(b) β 2 2
s3cos 2t 3s 2
β β= β β+β β = 2
3ss 4+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 567
5. Determine the Laplace transforms of (a) 7 cosh 2x (b) 1 sinh 3t3
(a) β 2 2
s7cosh 2x 7s 2
β β= β βββ β = 2
7ss 4β
(b) β 2 2
1 1 3sinh 3t3 3 s 3
β§ β« β β=β¨ β¬ β βββ© β β β = 3
1s 9β
6.(a) Determine the Laplace transform of 22cos t
cos 2t = 22cos t 1β from which, 2 1 cos 2tcos t2
+=
β 22cos t = β 1 cos 2t22
β§ + β«β ββ¨ β¬β β
β β β© β = β ( )
( )2 2
2 2 2 2
s 4 s1 s 1 s1 cos 2ts s 2 s s 4 s s 4
+ ++ = + = + =
+ + +
= ( )
2
2
2s 4s s 4
++
= ( )( )
2
2
2 s 2
s s 4
+
+
7.(b) Determine the Laplace transform of 22sinh 2ΞΈ
cosh 4ΞΈ = 1 + 22sinh 2ΞΈ from which, 22sinh 2ΞΈ = cosh 4ΞΈ - 1
β 22sinh 2ΞΈ = β ( )( )
2 2
2 2 2
s s 16s 1cosh 4 1s 4 s s s 16
β βΞΈβ = β =
β β = ( )2
16s s 16β
8. Determine the Laplace transform of 4 sin (at + b), where a and b are constants.
β 4sin(at b)+ = β 4 sin at cos b cos at sin b+ = β (4cos b)sin at (4sin b)cos at+
= ( ) ( )2 2 2 2
a s4cos b 4sin bs a s a
β β β β+β β β β+ +β β β β
= ( )2 2
4 acosb ssinbs a
++
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 568
10. Show that β ( )2 22
scos 3t sin 3ts 36
β =+
2cos 6t 2cos 3t 1= β from which, 2 1 cos 6tcos 3t2
+=
and 2cos6t 1 2sin 3t= β from which, 2 1 cos 6tsin 3t2
β=
Hence, β ( )2 2cos 3t sin 3tβ = β 1 cos 6t 1 cos 6t2 2
β§ + β β«β β β βββ¨ β¬β β β ββ β β β β© β
= β 1 cos 6t 1 cos 6t2 2 2 2
β§ β«+ β +β¨ β¬β© β
= βcos 6t = 2
ss 36+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 569
CHAPTER 65 PROPERTIES OF LAPLACE TRANSFORMS EXERCISE 232 Page 634
1. Determine the Laplace transforms of (a) 2t2te (b) 2 tt e
(a) β 2t2te 2= β ( )
1 2t1 1
1!t e (2)s 2 +=β
= ( )2
2s 2β
(b) β ( )
2 t2 1
2!t es 1 +=β
= ( )3
2s 1β
2.(b) Determine the Laplace transform of 4 3t1 t e2
β
β( )
4 3t4 1
1 1 4!t e2 2 s 3
β+
β ββ§ β« = β ββ¨ β¬ β ββ© β +β β =
( )5
12s 3+
3.(b) Determine the Laplace transform of 2t3e sin 2t
β ( )
2t2 22
2 63e sin 2t 3s 4s 4 4s 2 2
β β= =β β
β β β + +β +β β = 2
6s 4s 8β +
4.(a) Determine the Laplace transform of 2t5e cos3tβ
β ( )
( )2t2 22
5 s 2s 25e cos3t 5s 4s 4 9s 2 3
ββ β ++
= =β ββ β + + ++ +β β
= ( )2
5 s 2s 4s 13
++ +
5.(a) Determine the Laplace transform of t 22e sin t
β t 22e sin t = β t 1 cos 2t2e2
β§ β β«β ββ¨ β¬β β
β β β© β = β te - β
( )t
2 2
1 s 1e cos 2ts 1 s 1 2
β= β
β β +
= 2
1 s 1s 1 s 2s 5
ββ
β β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 570
6.(b) Determine the Laplace transform of 2t3e cosh 4t
β ( )
2t2 22
s 2 3(s 2)3e cosh 4t 3s 4s 4 16s 2 4
β ββ β= =β β
β β β + ββ ββ β =
( )2
3 s 2s 4s 12
ββ β
7.(a) Determine the Laplace transform of t2e sinh 3tβ
β ( )
t2 22
3 62e sinh 3t 2s 2s 1 9s 1 3
ββ β
= =β ββ β + + β+ ββ β
= 2
6s 2s 8+ β
8. Determine the Laplace transforms of (a) ( )t2e cos3t 3sin 3tβ (b) ( )2t3e sinh 2t 2cosh 2tβ β
(a) β ( ) t2e cos3t 3sin 3tβ = β t2e cos3t - β ( ) ( )
t2 22 2
s 1 36e sin 3t 2 6s 1 3 s 1 3
β β β ββ= ββ β β β
β β β ββ + β +β β β β
= ( )2 2 2 2
2 s 1 18 2s 2 18 2s 20s 2s 10 s 2s 10 s 2s 10 s 2s 10
β β β ββ = =
β + β + β + β +
= ( )2
2 s 10s 2s 10
ββ +
(b) β ( ) 2t3e sinh 2t 2cosh 2tβ β = β 2t3e sinh 2tβ - β 2t6e cosh 2tβ
= ( ) ( )2 22 2
2 s 23 6s 2 2 s 2 2
β β β β+ββ β β β
β β β β+ β + ββ β β β
= ( ) ( )2 2 2 2
6 s 2 6 s 26 6s 4s 4 4 s 4s 4 4 s 4s s 4s
+ +β = β
+ + β + + β + +
= ( ) ( )
6 6s 12 6s 6s s 4 s s 4β β β β
=+ +
= ( )( )6 s 1
s s 4β +
+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 571
EXERCISE 233 Page 635
2. Use the Laplace transform of the first derivative to derive the transforms:
(a) β ate = 1s aβ
(b) β 23t = 3
6s
(a) Let f(t) = ate then f β²(t) = a ate and f(0) = 1
From equation (3), page 634 of textbook, β f '(t) s= β f (t) f (0)β
Hence, β ata e = sβ ate - 1
i.e. 1 = (s β a)β ate
and β ate = 1s aβ
(b) Let f(t) = 23t then f β²(t) = 6t and f(0) = 0
Since β f '(t) s= β f (t) f (0)β
then, β6t = sβ 23t + 0
i.e. 2
6s
= sβ 23t
and β 23t = 3
6s
3. Derive the Laplace transform of the second derivative from the definition of a Laplace
transform. Hence derive the transform βsin at = 2 2
as a+
For the derivation, see page 634 of textbook.
Let f(t) = sin at, then f β²(t) = a cos at and f β²β²(t) = 2a sin atβ , f(0) = 0 and f β²(0) = a
From equation (4), page 635 of textbook, β 2f ''(t) s= β f (t) sf (0) f '(0)β β
Hence, β 2 2a sin at sβ = β sin at - s(0) - a
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 572
i.e. 2aβ β sin at = 2s β sin at - a
Hence, a = ( )2 2s a+ β sin at
and β sinat = 2 2
as a+
4.(b) Use the Laplace transform of the second derivative to derive the transform:
βcosh at = 2 2
ss aβ
Let f(t) = cosh at then f β²(t) = a sinh at and f β²β²(t) = 2a cosh at , f(0) = 1 and f β²(0) = 0
β 2f ''(t) s= β f (t) sf (0) f '(0)β β
Hence, β 2 2a cosh at s= β cosh at - s(1) - 0
i.e. 2a β cosh at = 2s β cosh at - s
i.e. s = ( )2 2s aβ β cosh at
and β coshat = 2 2
ss aβ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 573
EXERCISE 234 Page 637
1. State the initial value theorem. Verify the theorem for the function (a) 3 β 4 sin t (b) ( )2t 4β
and state their initial values.
The initial value theorem states: [ ] [
t 0 slimit f (t) limit s
β ββ= β ]f (t)
(a) Let f(t) = 3 β 4 sin t then βf(t) = β3 β 4 sin t = 2
3 4s s 1β
+
Hence, [ ] 2t 0 s
3 4limit 3 4sin t limit ss s 1β ββ
β‘ β€β ββ = ββ ββ’ β₯+β β β£ β¦
= 2s
4slimit 3s 1ββ
β‘ β€ββ’ β₯+β£ β¦
i.e. 3 β 4 sin 0 = 3 - 2 1β
β +
i.e. 3 = 3 which verifies the theorem.
The initial value is 3
(b) Let f(t) = ( )2 2t 4 t 8t 16β = β +
then β 23 2
2 8 16t 8t 16s s s
β + = β +
Hence, 23 2t 0 s
2 8 16limit t 8t 16 limit ss s sβ ββ
β‘ β€β ββ‘ β€β + = β +β ββ’ β₯β£ β¦ β β β£ β¦
= 2s
2 8limit 16s sββ
β‘ β€β +β’ β₯β£ β¦
i.e. 16 = 16 which verifies the theorem
The initial value is 16
3. State the final value theorem and state a practical application where it is of use. Verify the
theorem for the function 4 + 2teβ (sin t + cos t) representing a displacement and state its final
value.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 574
The final value theorem states: [ ] [t s 0
limit f (t) limit sββ β
= β ]f (t)
The final value theorem is used in investigating the stability of systems such as in automatic
aircraft-landing systems.
Let f(t) = 4 + 2teβ (sin t + cos t) = 4 + 2t 2te sin t e cos tβ β+
then β ( ) ( )2 2
4 1 s 2f (t)s s 2 1 s 2 1
+= + +
+ + + +
Hence, [( ) ( )
2t 2t2 2t s 0
4 1 s 2limit 4 e sin t e cos t limit ss s 2 1 s 2 1
β β
ββ β
β β+β‘ β€+ + = + +β ββ£ β¦ β β+ + + +β β
= ( )
( )( )2 2s 0
s s 2slimit 4s 2 1 s 2 1β
β‘ β€++ +β’ β₯
+ + + +β’ β₯β£ β¦
i.e. 4 + 0 + 0 = 4 + 0 + 0
i.e. 4 = 4 which verifies the theorem
The final value is 4
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 575
CHAPTER 66 INVERSE LAPLACE TRANSFORMS EXERCISE 235 Page 640
2. (a) Determine the inverse Laplace transform of 32s 1+
β 1 32s 1
β β§ β«β¨ β¬+β© β
= 3β 1 112 s2
β
β§ β«βͺ βͺβͺ βͺβ¨ β¬
β ββͺ βͺ+β ββͺ βͺβ β β© β
= 32β 1 1
1s2
β
β§ β«βͺ βͺβͺ βͺβ¨ β¬β ββͺ βͺ+β ββͺ βͺβ β β© β
= 1t23 e
2β
3. (a) Determine the inverse Laplace transform of 2
1s 25+
β 12
1s 25
β β§ β«β¨ β¬+β© β
= 15β 1
2 2
5s 5
β β§ β«β¨ β¬+β© β
= 1 sin 5t5
4. (a) Determine the inverse Laplace transform of 2
5s2s 18+
β 12
5s2s 18
β β§ β«β¨ β¬+β© β
= 5β( )
12
s2 s 9
ββ§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β = 5
2β 1
2 2
ss 3
β β§ β«β¨ β¬+β© β
= 5 cos 3t2
5. (a) Determine the inverse Laplace transform of 3
5s
β 13
5s
β β§ β«β¨ β¬β© β
= 52!β 1
3
2!s
β β§ β«β¨ β¬β© β
= 25 t2
6. (a) Determine the inverse Laplace transform of 2
3s1 s 82
β
β 1
2
3s1 s 82
β
β§ β«βͺ βͺβ¨ β¬βͺ βͺββ© β
= 3β( )
1
2
s1 s 162
β
β§ β«βͺ βͺβ¨ β¬βͺ βͺββ© β
= 6β 12 2
ss 4
β β§ β«β¨ β¬ββ© β
= 6 cosh 4t
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 576
8. (b) Determine the inverse Laplace transform of ( )5
3s 3β
β( )
15
3s 3
ββ§ β«βͺ βͺβ¨ β¬
ββͺ βͺβ© β = 3β
( )1
4 11
s 3β
+
β§ β«βͺ βͺβ¨ β¬
ββͺ βͺβ© β = 3
4!β
( )1
4 14!
s 3β
+
β§ β«βͺ βͺβ¨ β¬
ββͺ βͺβ© β = 1
8β
( )1
4 14!
s 3β
+
β§ β«βͺ βͺβ¨ β¬
ββͺ βͺβ© β = 3t 41 e t
8
9. (b) Determine the inverse Laplace transform of 2
3s 6s 13+ +
β 12
3s 6s 13
β β§ β«β¨ β¬+ +β© β
= 3β( )
12 2
1s 3 2
ββ§ β«βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β = 3
2β
( )1
2 2
2s 3 2
ββ§ β«βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β = 3t3 e sin 2t
2β
10. (a) Determine the inverse Laplace transform of 2
2(s 3)s 6s 13
ββ +
β ( )12
2 s 3s 6s 13
β ββ§ β«β¨ β¬β +β© β
= 2β( )
12 2
s 3s 3 2
ββ§ β«ββͺ βͺβ¨ β¬
β +βͺ βͺβ© β = 3t2e cos 2t
11. Determine the inverse Laplace transforms of (a) 2
2s 5s 4s 5
++ β
(b) 2
3s 2s 8s 25
+β +
(a) β 12
2s 5s 4s 5
β +β§ β«β¨ β¬+ ββ© β
= β ( )( ) ( )
12 22 2
2 s 2 1s 2 3 s 2 3
ββ§ β«+βͺ βͺ+β¨ β¬
+ β + ββͺ βͺβ© β
= 2β( )
12 2
s 2s 2 3
ββ§ β«+βͺ βͺβ¨ β¬
+ ββͺ βͺβ© β+ 1
3β
( )1
2 2
3s 2 3
ββ§ β«βͺ βͺβ¨ β¬
+ ββͺ βͺβ© β = 2t 2t12e cosh 3t e sinh 3t
3β β+
(b) β 12
3s 2s 8s 25
β +β§ β«β¨ β¬β +β© β
= β( )
12 2
3s 2s 4 3
ββ§ β«+βͺ βͺβ¨ β¬
β +βͺ βͺβ© β = β ( )
( )1
2 2
3 s 4 14s 4 3
ββ§ β«β +βͺ βͺβ¨ β¬
β +βͺ βͺβ© β
= 3β( )
12 2
s 4s 4 3
ββ§ β«ββͺ βͺβ¨ β¬
β +βͺ βͺβ© β+14
3β
( )1
2 2
3s 4 3
ββ§ β«βͺ βͺβ¨ β¬
β +βͺ βͺβ© β = 4t 4t143e cos 3t e sin 3t
3+
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 577
EXERCISE 236 Page 642
2. Use partial fractions to find the inverse Laplace transform of: ( )( )( )
22s 9s 35s 1 s 2 s 3
β β+ β +
( )( )( ) ( ) ( ) ( )22s 9s 35 4 3 1
s 1 s 2 s 3 s 1 s 2 s 3β β
= β ++ β + + β +
from Problem 2, page 19 of textbook
Hence, β( )( )( )
21 2s 9s 35
s 1 s 2 s 3β β§ β«β ββͺ βͺβ¨ β¬+ β +βͺ βͺβ© β
= β( ) ( ) ( )
1 4 3 1s 1 s 2 s 3
β β§ β«βͺ βͺβ +β¨ β¬+ β +βͺ βͺβ© β = t 2t 3t4e 3e eβ ββ +
4. Use partial fractions to find the inverse Laplace transform of: ( )
2
33s 16s 15
s 3+ +
+
( ) ( ) ( ) ( )
2
3 2 33s 16s 15 3 2 6
s 3s 3 s 3 s 3+ +
= β β++ + +
from Problem 7, page 22 of textbook
Hence, β( )
21
33s 16s 15
s 3ββ§ β«+ +βͺ βͺβ¨ β¬
+βͺ βͺβ© β = β
( ) ( ) ( )1
2 33 2 6
s 3 s 3 s 3ββ§ β«βͺ βͺβ ββ¨ β¬+ + +βͺ βͺβ© β
= 3β( )
1 1s 3
β β§ β«βͺ βͺβ¨ β¬+βͺ βͺβ© β
- 2β( )
12
1s 3
ββ§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β - 6β
( )1
31
s 3ββ§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β
= 3β( )
1 1s 3
β β§ β«βͺ βͺβ¨ β¬+βͺ βͺβ© β
- 2β( )
11 1
1s 3
β+
β§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β - 6
2!β
( )1
2 12!
s 3β
+
β§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β
= 3t 3t 3t 23e 2e t 3e tβ β ββ β or 3t 2e (3 2t 3t )β β β
6. Use partial fractions to find the inverse Laplace transform of: ( )
2 3
2 2
3 6s 4s 2ss s 3
+ + β+
( )2 3
2 22 2
3 6s 4s 2s 2 1 3 4ss s s 3s s 3
+ + β β= + +
++ from Problem 9, page 23 of textbook
Hence, β( )
2 31
2 2
3 6s 4s 2ss s 3
ββ§ β«+ + ββͺ βͺβ¨ β¬
+βͺ βͺβ© β = β 1
2 2
2 1 3 4ss s s 3
β ββ§ β«+ +β¨ β¬+β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 578
= 2β 1 1s
β β§ β«β¨ β¬β© β
+ β 12
1s
β β§ β«β¨ β¬β© β
+ β( )
12
2
3 4s
s 3β
β§ β«ββͺ βͺ
β¨ β¬βͺ βͺ+β© β
= 2β 1 1s
β β§ β«β¨ β¬β© β
+ β 12
1s
β β§ β«β¨ β¬β© β
+ β( )
12
2
3
s 3β
β§ β«βͺ βͺβ¨ β¬βͺ βͺ+β© β
- β( )
12
2
4s
s 3β
β§ β«βͺ βͺβ¨ β¬βͺ βͺ+β© β
= 2β 1 1s
β β§ β«β¨ β¬β© β
+ β 12
1s
β β§ β«β¨ β¬β© β
+ 33β
( )1
22
3
s 3β
β§ β«βͺ βͺβ¨ β¬βͺ βͺ+β© β
- 4β( )
12
2
s
s 3β
β§ β«βͺ βͺβ¨ β¬βͺ βͺ+β© β
= 2 t 3 sin 3 t 4cos 3 t+ + β
7. Use partial fractions to find the inverse Laplace transform of: ( )
2
2
26 ss s 4s 13
β+ +
Let ( )
( ) ( )( )
22
22 2
A s 4s 13 Bs C s26 s A Bs Cs s 4s 13s s 4s 13 s s 4s 13
+ + + +β +β‘ + =
+ ++ + + +
Hence, ( )2 2 226 s A s 4s 13 Bs Csβ = + + + +
When s = 0: 26 = 13A + 0 + 0 i.e. A = 2
Equating 2s coefficients: -1 = A + B i.e. B = -3
Equating s coefficients: 0 = 4A + C i.e. C = -8
Thus, ( )
2
22
26 s 2 3s 8s s 4s 13s s 4s 13
β β ββ‘ +
+ ++ +
Hence, β( )
21
2
26 ss s 4s 13
ββ§ β«ββͺ βͺβ¨ β¬
+ +βͺ βͺβ© β = β 1 2
sβ β§ β«β¨ β¬β© β
- β 12
3s 8s 4s 13
β +β§ β«β¨ β¬+ +β© β
= 2β 1 1s
β β§ β«β¨ β¬β© β
- β( )
12 2
3s 8s 2 3
ββ§ β«+βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β
= 2β 1 1s
β β§ β«β¨ β¬β© β
- β ( )( )
12 2
3 s 2s 2 3
ββ§ β«+βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β - β
( )1
2 2
2s 2 3
ββ§ β«βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 579
= 2β 1 1s
β β§ β«β¨ β¬β© β
- 3β ( )( )
12 2
s 2s 2 3
ββ§ β«+βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β - 2
3β
( )1
2 2
3s 2 3
ββ§ β«βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β
= 2t 2t22 3e cos 3t e sin 3t3
β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 580
EXERCISE 237 (Page XX)
1. Determine for the transfer function: R(s) = ( )( )( )2
50 s 4s s 2 s 8s 25
+
+ β +
(a) the zero and (b) the poles. Show the poles and zeros on a pole-zero diagram.
(a) For the numerator to be zero, (s + 4) = 0 hence, s = -4 is a zero of R(s)
(b) For the denominator to be zero, s = 0 or s = -2 or 2s 8s 25 0β + =
i.e. s = ( )28 8 4(1)(25) 8 36 8 j6 4 j3
2(1) 2 2β β Β± β β Β± β β Β±
= = = Β±
Hence, poles occur at s = 0, s = -2, s = 4 + j3 and 4 - j3
A pole-zero diagram is shown below.
3. For the function G(s) = ( )( )2
s 1s 2 s 2s 5
β+ + +
determine the poles and zeros and show them on a
pole-zero diagram.
For the denominator to be zero, s = -2 or 2s 2s 5 0+ + =
i.e. s = ( )22 2 4(1)(5) 2 16 2 j4 1 j2
2(1) 2 2β Β± β β Β± β β Β±
= = = β Β±
Hence, poles occur at s = -2, s = -1 + j2 and -1 β j2
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 581
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.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 582
CHAPTER 67 THE SOLUTION OF DIFFERENTIAL EQUATIONS
USING LAPLACE TRANSFORMS EXERCISE 238 Page 648
1. A first order differential equation involving current i in a series R-L circuit is given by:
di E5idt 2+ = and i = 0 at time t = 0.
Use Laplace transforms to solve for i when (a) E = 20 (b) E = 3t40eβ and (c) E = 50 sin 5t.
Taking the Laplace transform of each term of di E5idt 2+ = gives:
β didt
β§ β«β¨ β¬β© β
+ 5βi = β E2
β§ β«β¨ β¬β© β
i.e. sβi β i(0) + 5βi = E / 2s
i = 0 at t = 0, hence, i(0) = 0
Hence, (s + 5)βi = E / 2s
i.e. βi = E / 2s(s 5)+
and i = β 1 E / 2s(s 5)
β β§ β«β¨ β¬+β© β
= E2β 1 1
s(s 5)β β§ β«β¨ β¬+β© β
Let 1 A B A(s 5) Bss(s 5) s (s 5) s(s 5)
+ +β‘ + =
+ + +
Hence, 1 = A(s + 5) + Bs
When s = 0: 1 = 5A i.e. A = 15
When s = -5: 1 = -5B i.e. B = - 15
Thus, i = E2β 1 1
s(s 5)β β§ β«β¨ β¬+β© β
= E2β 1
1 15 5s (s 5)
β
β§ β«βͺ βͺ
ββ¨ β¬+βͺ βͺβ© β
= E2
5t1 1 e5 5
ββ βββ ββ β
(a) When E = 20, i = 202
5t1 1 e5 5
ββ βββ ββ β
= ( )5t2 1 eββ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 583
(b) When E = 3t40eβ β didt
β§ β«β¨ β¬β© β
+ 5βi = β3t40e
2
ββ§ β«β¨ β¬β© β
i.e. sβi β i(0) + 5βi = 20s 3+
i = 0 at t = 0, hence, i(0) = 0
Hence, (s + 5)βi = 20s 3+
i.e. βi = 20(s 3)(s 5)+ +
and i = β 1 20(s 3)(s 5)
β β§ β«β¨ β¬+ +β© β
Let 20 A B A(s 5) B(s 3)(s 3)(s 5) (s 3) (s 5) (s 3)(s 5)
+ + +β‘ + =
+ + + + + +
Hence, 20 = A(s + 5) + B(s + 3)
When s = -3: 20 = 2A i.e. A = 10
When s = -5: 20 = -2B i.e. B = -10
Thus, i = β 1 20(s 3)(s 5)
β β§ β«β¨ β¬+ +β© β
= β 1 10 10(s 3) (s 5)
β β§ β«ββ¨ β¬+ +β© β
i.e. i = 3t 5t10e 10eβ ββ = ( )3t 5t10 e eβ ββ
(c) When E = 50 sin 5t β didt
β§ β«β¨ β¬β© β
+ 5βi = β 50sin 5t2
β§ β«β¨ β¬β© β
i.e. sβi β i(0) + 5βi = 2 2
25(5)s 5+
i = 0 at t = 0, hence, i(0) = 0
Hence, (s + 5)βi = 2
125s 25+
i.e. βi = 2
125(s 5)(s 25)+ +
and i = β 12
125(s 5)(s 25)
β β§ β«β¨ β¬+ +β© β
Let 2
2 2 2
125 A Bs C A(s 25) (Bs C)(s 5)(s 5)(s 25) (s 5) (s 25) (s 5)(s 25)
+ + + + +β‘ + =
+ + + + + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 584
Hence, 125 = ( ) ( )( )2A s 25 Bs C s 5+ + + +
When s = -5: 125 = 50A i.e. A = 52
Equating 2s coefficients: 0 = A + B i.e. B = - 52
Equating constant terms: 125 = 25A + 5C i.e. 125 = 1252
+ 5C
from which, C =
125252
5 2=
Thus, i = β 12
125(s 5)(s 25)
β β§ β«β¨ β¬+ +β© β
= β( )
12
5 5 25s2 2 2
(s 5) s 25β
β§ β«β +βͺ βͺ+β¨ β¬+ +βͺ βͺ
β© β
= 52β 1 1
(s 5)β β§ β«β¨ β¬+β© β
+ β( )
12
252
s 25β
β§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β
- β( )
12
5 s2
s 25β
β§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β
= 52β 1 1
(s 5)β β§ β«β¨ β¬+β© β
+ 52β
( )1
2 2
5s 5
ββ§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β - 5
2β
( )1
2 2
ss 5
ββ§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β
= 5t5 5 5e sin 5t cos5t2 2 2
β + β
i.e. i = ( )5t5 e sin 5t cos 5t2
β + β
3. Use Laplace transforms to solve the differential equation: 2
2
d x 100x 0dt
+ = given x(0) = 2 and
xβ²(0) = 0.
β2
2
d xdx
β§ β«β¨ β¬β© β
+ β100x = β0
i.e. 2s βx β s x(0) - xβ²(0) + 100βx = 0
x(0) = 2 and xβ²(0) = 0, hence 2s βx β 2s + 100βx = 0
i.e. ( 2s + 100)βx = 2s
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 585
from which, βx = 2
2ss 100+
and x = β 12
2ss 100
β β§ β«β¨ β¬+β© β
i.e. x = 2β 12 2
ss 10
β β§ β«β¨ β¬+β© β
= 2 cos 10t
4. Use Laplace transforms to solve the differential equation: 2
2
d i di1000 250000i 0dt dt
+ + =
given i(0) = 0 and iβ²(0) = 100.
β2
2
d idt
β§ β«β¨ β¬β© β
+ 1000β didt
β§ β«β¨ β¬β© β
+ 250000βi = β0
i.e. [ 2s βi β s i(0) - iβ²(0)] + 1000[sβi β i(0)] + 250000βi = 0
i(0) = 0 and iβ²(0) = 100, hence
2s βi β 100 + 1000sβi + 250000βi = 0
i.e. ( 2s +1000s + 250000)βi = 100
from which, βi = 2
100s 1000s 250000+ +
and i = β( )
12
100s 500
ββ§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β = 100β
( )1
1 11
s 500β
+
β§ β«βͺ βͺβ¨ β¬
+βͺ βͺβ© β
i.e. i = 500t100t eβ
6. Use Laplace transforms to solve the differential equation: 2
4x2
d y dy2 y 3edx dx
β + = given
y(0) = 23
β and yβ²(0) = 143
β2
2
d ydx
β§ β«β¨ β¬β© β
- 2β dydx
β§ β«β¨ β¬β© β
+ βy = β 4x3e
i.e. [ 2s βy β s y(0) - yβ²(0)] - 2[sβy β y(0)] + βy = 3(s 4)β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 586
y(0) = 23
β and yβ²(0) = 143
, hence
2s βy β s 23
β βββ ββ β
- 133
- 2sβy + 2 23
β βββ ββ β
+ βy = 3(s 4)β
i.e. ( 2s - 2s + 1)βy + 2 13 4s3 3 3
β β = 3(s 4)β
from which, ( 2s - 2s + 1)βy = 3(s 4)β
- 2 17 9 2s(s 4) 17(s 4)s3 3 3(s 4)
β β + β+ =
β
βy = ( )( ) ( )( )
2 2
22
9 2s 8s 17s 68 59 25s 2s3 s 4 s 2s 1 3 s 4 s 1β + + β β + β
=β β + β β
and y = β( )( )
21
259 25s 2s
3 s 4 s 1ββ§ β«β + ββͺ βͺβ¨ β¬
β ββͺ βͺβ© β
Let ( )( )( ) ( ) ( ) ( )
( ) ( )( ) ( )( )( )
2 2
2 2 2
1 59 25s 2s A s 1 B s 4 s 1 C s 4A B C3s 4 s 1s 4 s 1 s 1 s 4 s 1
β + β β + β β + ββ‘ + + =
β ββ β β β β
Hence, ( ) ( )( ) ( )2259 25 2s s A s 1 B s 4 s 1 C s 43 3 3
β + β = β + β β + β
When s = 4: 59 100 32 9A 0 03 3 3
β + β = + + i.e. 3 = 9A and A = 13
When s = 1: 59 25 2 0 0 3C3 3 3
β + β = + β i.e. -12 = -3C and C = 4
Equating 2s coefficients: 2 A B3
β = + i.e. 2 1 B3 3
β = + and B = -1
Hence, y = β( ) ( ) ( )
12
11 43
s 4 s 1 s 1β
β§ β«βͺ βͺ
β +β¨ β¬β β ββͺ βͺβ© β
i.e. y = 4x x x1 e e 4x e3
β + or y = ( ) x 4x14x 1 e e3
β +
8. Use Laplace transforms to solve the differential equation: 2
2
d y dy 2y 3cos3x 11sin 3xdx dx
+ β = β
given y(0) = 0 and yβ²(0) = 6.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 587
β2
2
d ydx
β§ β«β¨ β¬β© β
+ β dydx
β§ β«β¨ β¬β© β
- 2βy = β 3cos3x 11sin 3xβ
i.e. [ 2s βy β s y(0) - yβ²(0)] + [sβy β y(0)] - 2βy = 2 2
3s 33s 9 s 9
β+ +
y(0) = 0 and yβ²(0) = 6, hence
2s βy - 6 + sβy - 2βy = 2
3s 33s 9β+
i.e. ( 2s + s - 2)βy = 6 + 2
3s 33s 9β+
from which, ( 2s + s - 2)βy = ( )2 2
2 2
6 s 9 3s 33 6s 3s 21s 9 s 9+ + β + +
=+ +
βy = ( )( )
2
2 2
6s 3s 21s 9 s s 2
+ ++ + β
and y = β( )( )( )
21
2
6s 3s 21s 9 s 1 s 2
ββ§ β«+ +βͺ βͺβ¨ β¬
+ β +βͺ βͺβ© β
Let ( )( )( ) ( ) ( ) ( )
2
2 2
6s 3s 21 A B Cs Ds 2 s 1s 9 s 1 s 2 s 9
+ + +β‘ + +
+ β+ β + +
= ( )( ) ( )( ) ( )( )( )
( )( )( )2 2
2
A s 1 s 9 B s 2 s 9 Cs D s 2 s 1
s 2 s 1 s 9
β + + + + + + + β
+ β +
Hence, ( )( ) ( )( ) ( )( )( )2 2 26s 3s 21 A s 1 s 9 B s 2 s 9 Cs D s 2 s 1+ + = β + + + + + + + β
When s = 1: 6 3 21 B(3)(10)+ + = i.e. 30 = 30B and B = 1
When s = -2: 24 6 21 A( 3)(13)β + = β i.e. 39 = -39A and A = -1
Equating 3s coefficients: 0 A B C= + + i.e. C = 0
Equating constant terms: 21 = -9A + 18B β 2D i.e. 21 = 9 + 18 β 2D and D = 3
Hence, y = β( ) ( ) ( )
12 2
1 1 3s 2 s 1 s 3
ββ§ β«ββͺ βͺ+ +β¨ β¬+ β +βͺ βͺβ© β
i.e. y = 2x xe e sin 3xββ + + or y = x 2xe e sin 3xββ +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 588
9. Use Laplace transforms to solve the differential equation: 2
x2
d y dy2 2y 3e cos 2xdx dx
β + = given
y(0) = 2 and yβ²(0) = 5.
β2
2
d ydx
β§ β«β¨ β¬β© β
- 2β dydx
β§ β«β¨ β¬β© β
+ 2βy = β x3e cos 2x
i.e. [ 2s βy β s y(0) - yβ²(0)] - 2[sβy β y(0)] + 2βy = ( )2 2
s 13s 1 2
β βββ ββ ββ +β β
y(0) = 2 and yβ²(0) = 5, hence
2s βy β 2s - 5 - 2sβy + 4 + 2βy = ( )2 2
s 13s 1 2
β βββ ββ ββ +β β
i.e. ( 2s - 2s + 2)βy = 2s + 1 +( )2 2
s 13s 1 2
β βββ ββ ββ +β β
= ( ) ( )( )
( )2
2
3s 3 2s 1 s 2s 5
s 2s 5
β + + β +
β +
= ( )
3 2 2
2
3s 3 2s 4s 10s s 2s 5s 2s 5
β + β + + β +β +
= ( )
3 2
2
2s 3s 11s 2s 2s 5β + +β +
from which, βy = ( )( )
3 2
2 2
2s 3s 11s 2s 2s 5 s 2s 2
β + +β + β +
and y = β( )( )
3 21
2 2
2s 3s 11s 2s 2s 5 s 2s 2
ββ§ β«β + +βͺ βͺβ¨ β¬
β + β +βͺ βͺβ© β
Let ( )( ) ( ) ( )
3 2
2 2 2 2
2s 3s 11s 2 As B Cs Ds 2s 5 s 2s 2 s 2s 5 s 2s 2
β + + + +β‘ +
β + β + β + β +
= ( )( ) ( )( )
( )( )2 2
2 2
As B s 2s 2 Cs D s 2s 5
s 2s 5 s 2s 2
+ β + + + β +
β + β +
Hence, ( )( ) ( )( )3 2 2 22s 3s 11s 2 As B s 2s 2 Cs D s 2s 5β + + = + β + + + β +
Equating 3s coefficients: 2 = A + C (1)
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 589
Equating 2s coefficients: -3 = -2A + B β 2C + D (2)
Equating s coefficients: 11 = 2A β 2B + 5C β 2D (3)
Equating constant terms: 2 = 2B + 5D (4)
From (1), C = 2 β A and
substituting C = 2 β A in (2): -3 = -2A + B β 2(2 β A) + D
i.e. 1 = B + D (5)
2 Γ (5) gives: 2 = 2B + 2D (6)
(4) β (6) gives: 0 = 3D i.e. D = 0
From (4), if D = 0, then B = 1
In (2), if B = 1 and D = 0, then -3 = -2A + 1 β 2C
i.e. 4 = 2A + 2C (7)
From (3), 11 = 2A β 2 + 5C
i.e. 13 = 2A + 5C (8)
(8) β (7) gives: 9 = 3C i.e. C = 3
In (1), if C = 3, then A = -1
Hence, y = β( ) ( )
12 2
1 s 3ss 2s 5 s 2s 2
ββ§ β«ββͺ βͺ+β¨ β¬
β + β +βͺ βͺβ© β
= -β( )
12 2
s 1s 1 2
ββ§ β«ββͺ βͺβ¨ β¬
β +βͺ βͺβ© β+β
( )1
2 2
3ss 1 1
ββ§ β«βͺ βͺβ¨ β¬
β +βͺ βͺβ© β
= -β( )
12 2
s 1s 1 2
ββ§ β«ββͺ βͺβ¨ β¬
β +βͺ βͺβ© β+β ( )
( )1
2 2
3 s 1 3s 1 1
ββ§ β«β +βͺ βͺβ¨ β¬
β +βͺ βͺβ© β
= -β( )
12 2
s 1s 1 2
ββ§ β«ββͺ βͺβ¨ β¬
β +βͺ βͺβ© β+β ( )
( )1
2 2
3 s 1s 1 1
ββ§ β«ββͺ βͺβ¨ β¬
β +βͺ βͺβ© β+ β
( )1
2 2
3s 1 1
ββ§ β«βͺ βͺβ¨ β¬
β +βͺ βͺβ© β
i.e. y = x x xe cos 2x 3e cos x 3e sin xβ + + or y = ( )x x3e cos x sin x e cos 2x+ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 590
CHAPTER 68 THE SOLUTION OF SIMULTANEOUS
DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS EXERCISE 239 Page 654
2. Solve the following pair of simultaneous differential equations:
t
dy dx2 y x 5sin t 0dt dt
dy dx3 x y 2 e 0dt dt
β + + β =
+ β + β =
given that when t = 0, x = 0 and y = 0.
Taking Laplace transforms of each term in each equation gives:
2[sβy β y(0)] - βy + βx + [sβx β x(0)] - 2
5s 1+
= 0
3[sβy β y(0)] + βx - βy + 2[sβx β x(0)] - 1s 1β
= 0
y(0) = 0 and x(0) = 0, hence
(2s β 1)βy + (s + 1)βx = 2
5s 1+
(1)
and (3s β 1)βy +(2s + 1)βx = 1s 1β
(2)
(3s β 1) Γ (1) gives: (3s β 1)(2s β 1)βy + (3s β 1)(s + 1)βx = (3s β 1) 2
5s 1+
(3)
(2s β 1) Γ (2) gives: (2s β 1)(3s β 1)βy + (2s β 1)(2s + 1)βx = (2s β 1) 1s 1β
(4)
(3) β (4) gives: ( ) ( )2 23s 2s 1 4s 1β‘ β€+ β β ββ£ β¦βx = ( )2
5 3s 1 2s 1s 1 s 1
β ββ
+ β (5)
i.e. ( )2s 2sβ + βx = ( )( ) ( )( )
( )( )2
2
5 3s 1 s 1 2s 1 s 1
s 1 s 1
β β β β +
β +
i.e. βx = ( )( )( )
2 3 2
2
15s 20s 5 2s 2s s 1s s 2 s 1 s 1
β§ β«β + β β + +βͺ βͺββ¨ β¬β β +βͺ βͺβ© β
= ( )( )( )
3 2
2
2s 16s 22s 6s s 2 s 1 s 1
β§ β«β + ββͺ βͺβ¨ β¬
β β +βͺ βͺβ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 591
and x = β( )( )( )
3 21
2
2s 16s 22s 6s s 2 s 1 s 1
ββ§ β«β + ββͺ βͺβ¨ β¬
β β +βͺ βͺβ© β
Let ( )( )( ) ( ) ( ) ( )
3 2
2 2
2s 16s 22s 6 A B C Ds Es s 2 s 1s s 2 s 1 s 1 s 1
β + β +β‘ + + +
β ββ β + +
= ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( )
( )( )( )2 2 2
2
A s 2 s 1 s 1 B s s 1 s 1 C s s 2 s 1 Ds E s s 2 s 1
s s 2 s 1 s 1
β β + + β + + β + + + β β
β β +
from which, ( )( )( ) ( )( )( ) ( )( )( )3 2 2 2 22s 16s 22s 6 A s 2 s 1 s 1 B s s 1 s 1 C s s 2 s 1β + β = β β + + β + + β +
+ ( )( )( )( )Ds E s s 2 s 1+ β β
When s = 0: -6 = A(-2)(-1)(1) i.e. A = -3
When s = 1: 2 β 16 + 22 β 6 = C(1)(-1)(2) i.e. C = -1
When s = 2: 16 β 64 + 44 β 6 = B(2)(1)(5) i.e. B = -1
Equating 4s coefficients: 0 = A + B + C + D i.e. D = 5
Equating 3s coefficients: 2 = -3A β B β 2C β 3D + E
i.e. 2 = 9 + 1 + 2 β 15 + E i.e. E = 5
Hence, x = β( ) ( ) ( )
12
3 1 1 5s 5s s 2 s 1 s 1
ββ§ β«+βͺ βͺβ β β +β¨ β¬β β +βͺ βͺβ© β
i.e. x = 2t t3 e e 5cos t 5sin tββ β β + + or x = 2t t5cos t 5sin t e e 3+ β β β
From equations (1) and (2)
(2s + 1) Γ (1) gives: (2s + 1)(2s β 1)βy + (2s + 1)(s + 1)βx = (2s + 1) 2
5s 1+
(6)
(s + 1) Γ (2) gives: (s + 1)(3s β 1)βy + (s + 1)(2s + 1)βx = (s + 1) 1s 1β
(7)
(6) β (7): ( ) ( )2 24s 1 3s 2s 1β‘ β€β β + ββ£ β¦βy = ( ) ( )( )
( )( ) ( )( )( )( )
2
2 2
10s 5 s 1 s 1 s 15 2s 1 s 1s 1 s 1 s 1 s 1
+ β β + ++ +β =
+ β β +
i.e. ( )2s 2sβ βy = ( )( )
2 3 2
2
10s 5s 5 s s s 1s 1 s 1
β β β β β ββ +
and βy = ( )( )( )
3 2
2
s 9s 6s 6s s 2 s 1 s 1β + β ββ β +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 592
and y = β( )( )( )
3 21
2
s 9s 6s 6s s 1 s 2 s 1
ββ§ β«β + β ββͺ βͺβ¨ β¬
β β +βͺ βͺβ© β
Let ( )( )( ) ( ) ( ) ( )
3 2
2 2
s 9s 6s 6 A B C Ds Es s 1 s 2s s 1 s 2 s 1 s 1
β + β β +β‘ + + +
β ββ β + +
= ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( )
( )( )( )2 2 2
2
A s 1 s 2 s 1 B s s 2 s 1 C s s 1 s 1 Ds E s s 1 s 2
s s 1 s 2 s 1
β β + + β + + β + + + β β
β β +
from which, ( )( )( ) ( )( )( ) ( )( )( )3 2 2 2 2s 9s 6s 6 A s 1 s 2 s 1 B s s 2 s 1 C s s 1 s 1β + β β = β β + + β + + β +
+ ( )( )( )( )Ds E s s 1 s 2+ β β
When s = 0: -6 = A(-1)(-2)(1) i.e. A = -3
When s = 1: -1 + 9 - 6 β 6 = B(1)(-1)(2) i.e. B = 2
When s = 2: -8 + 36 - 12 β 6 = C(2)(1)(5) i.e. C = 1
Equating 4s coefficients: 0 = A + B + C + D i.e. D = 0
Equating 3s coefficients: -1 = -3A β 2B β C β 3D + E
i.e. -1 = 9 - 4 - 1 + 0 + E i.e. E = -5
Hence, y = β( ) ( ) ( )
12
3 2 1 5s s 1 s 2 s 1
ββ§ β«βͺ βͺβ + + ββ¨ β¬β β +βͺ βͺβ© β
i.e. y = t 2t3 2e e 5sin tβ + + β or y = 2t te 2e 3 5sin t+ β β
3. Solve the following pair of simultaneous differential equations:
2
2
2
2
d x 2x ydtd y 2y xdt
+ =
+ =
given that when t = 0, x = 4, y = 2, dxdt
= 0 and dydt
= 0.
Taking Laplace transforms of each term in each equation gives:
[ 2s βx β s x(0) - xβ²(0)] + 2βx = βy
and [ 2s βy β s y(0) - yβ²(0)] + 2βy = βx
x(0) = 4 and xβ²(0), hence [ 2s βx β 4s] + 2βx = βy
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 593
y(0) = 2 and yβ²(0) = 0, hence [ 2s βy β 2s] + 2βy = βx
i.e. ( 2s +2)βx - βy = 4s (1)
and - βx + ( 2s +2)βy = 2s (2)
( 2s +2) Γ (2) gives: - ( 2s +2)βx + ( 2s + 2) ( 2s +2)βy = 2s( 2s +2) (3)
(1) + (3) gives: ( )22s 2 1β‘ β€+ ββ’ β₯β£ β¦βy = 2s( 2s +2) + 4s
i.e. ( )4 2s 4s 3+ + βy = 32s 8s+
and βy = ( ) ( )( )
3 3
4 2 2 2
2s 8s 2s 8ss 4s 3 s 3 s 1
+ +=
+ + + +
and y = β( )( )
31
2 2
2s 8ss 3 s 1
ββ§ β«+βͺ βͺβ¨ β¬
+ +βͺ βͺβ© β
Let ( )( ) ( ) ( )
( )( ) ( )( )( )( )
2 23
2 2 2 2 2 2
As B s 1 Cs D s 32s 8s As B Cs Ds 3 s 1 s 3 s 1 s 3 s 1
+ + + + ++ + +β‘ + =
+ + + + + +
from which, ( )( ) ( )( )3 2 22s 8s As B s 1 Cs D s 3+ = + + + + +
Equating 3s coefficients: 2 = A + C (4)
Equating 2s coefficients: 0 = B + D (5)
Equating s coefficients: 8 = A + 3C (6)
(6) β (4) gives: 6 = 2C i.e. C = 3
and from (4), A = 2 β C i.e. A = -1
Equating constant terms: 0 = B + 3D (7)
(7) β (5) gives: 0 = 2D i.e. D = 0 and from (5), B = 0
Hence, y = β( ) ( )
12 2
s 3ss 3 s 1
ββ§ β«ββͺ βͺ+β¨ β¬
+ +βͺ βͺβ© β = β
( )( ) ( )( )1
2 22 2
s 3s
s 3 s 1
β
β§ β«βͺ βͺβ +β¨ β¬βͺ βͺ+ +β© β
i.e. y = ( )cos 3 t 3cos tβ + or y = 3 cos t - ( )cos 3 t
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 594
If y = 3 cos t - ( )cos 3 t then dy 3sin t 3 sin 3 tdt
= β +
and 2
2
d y 3cos t 3cos 3 tdt
= β +
Since from one of the original equations,
2
2
d y 2y xdt
+ =
then ( ) ( )( )3cos t 3cos 3 t 2 3cos t cos 3 t xβ + + β =
i.e. x = ( ) ( )3cos t 3cos 3 t 6cos t 2cos 3 tβ + + β
i.e. x = 3 cos t + ( )cos 3 t
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 595
CHAPTER 69 FOURIER SERIES FOR PERIODIC FUNCTIONS
OF PERIOD 2Ο EXERCISE 240 Page 661
1. Determine the Fourier series for the periodic function:
f(x) =2, when x 02, when 0 x
β β Ο β€ β€β§β¨ + β€ β€ Οβ©
which is periodic outside this range of period 2Ο.
The periodic function is shown in the diagram below.
The Fourier series is given by: f(x) = ( )0 n nn 1
a a cos nx b sin nxβ
=
+ +β (1)
[ ] [ ] [ ] [ ]
0 00 00
1 1 1a f (x)dx 2dx 2dx 2x 2x2 2 2
1 (0) (2 ) (2 ) (0) 02
Ο Ο Ο
βΟβΟ βΟ= = β + = β +
Ο Ο Ο
= β Ο + Ο β =Ο
β« β« β«
0
n 0
1 1a f (x)cos nx dx 2cos nx dx 2cos nx dxΟ Ο
βΟ βΟ= = β +Ο Οβ« β« β«
= 0
0
1 2 2sin nx sin nxn n
Ο
βΟ
β§ β«βͺ βͺβ‘ β€ β‘ β€β +β¨ β¬β’ β₯ β’ β₯Ο β£ β¦ β£ β¦βͺ βͺβ© β = 0
0
n 0
1 1b f (x)sin nx dx 2sin nx dx 2sin nx dxΟ Ο
βΟ βΟ= = β +Ο Οβ« β« β«
= [ ] [ ] 0
0
1 2 2 2cos nx cos nx cos 0 cos n( ) cos n cos 0n n n
Ο
βΟ
β§ β«βͺ βͺβ‘ β€ β‘ β€+ β = β βΟ β Οββ¨ β¬β’ β₯ β’ β₯Ο Οβ£ β¦ β£ β¦βͺ βͺβ© β
When n is even, [ ] [ ] n2b 1 1 1 1 0n
= β β β =Ο
When n is odd, [ ] [ ] ( )n2 2 8b 1 1 1 1 4n n n
= β β β β β = =Ο Ο Ο
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 596
Hence, 18b =Ο
, 38b
3=
Ο, 5
8b5
=Ο
and so on.
Substituting onto equation (1) gives:
f(x) = 0 + 0 + 8 8 8sin x sin 3x sin 5x .....3 5
+ + +Ο Ο Ο
i.e. f(x) = 8 1 1sin x sin 3x sin 5x ......3 5
β β+ + +β βΟ β β
2. For the Fourier series in problem 1, deduce a series for 4Ο at the point where x =
2Ο
When x = 2Ο , f(x) = 2, hence 2 = 8 8 3 8 5sin sin sin .....
2 3 2 5 2Ο Ο Ο+ + +
Ο Ο Ο
i.e. 2 = 8 1 1 11 ....3 5 7
β ββ + β +β βΟ β β
and 2 1 1 11 .....8 3 5 7Ο= β + β +
i.e. 1 1 11 ....4 3 5 7Ο= β + β +
5. Find the term representing the third harmonic for the periodic function of period 2Ο given by:
f(x) =0, when x 01, when 0 x
β Ο β€ β€β§β¨ β€ β€ Οβ©
The periodic function is shown in the diagram below.
n 00
1 1 1 sin nxa f (x)cos nx dx 1cos nx dx 0n
ΟΟ Ο
βΟ
β‘ β€= = = =β’ β₯Ο Ο Ο β£ β¦β« β«
n 0
1 1b f (x)sin nx dx sin nx dxΟ Ο
βΟ= =Ο Οβ« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 597
= 0
1 cos nx 1 1cos n cos 0 1 cos nn n n
Οβ§ β«βͺ βͺβ‘ β€β = β Οβ = β Οβ¨ β¬β’ β₯Ο Ο Οβ£ β¦βͺ βͺβ© β
The third harmonic is when n = 3,
i.e. 31 1 2b 1 cos3 1 1
3 3 3= β Ο = β β =
Ο Ο Ο
Since the Fourier series is given by: f(x) = ( )0 n nn 1
a a cos nx b sin nxβ
=
+ +β ,
the 3rd harmonic term is: 2 sin 3x3Ο
6. Determine the Fourier series for the periodic function of period 2Ο defined by:
f(t) =
0, when t 0
1, when 0 t2
1, when t2
β§βͺ β Ο β€ β€βͺ Οβͺ β€ β€β¨βͺ
Οβͺβ β€ β€ Οβͺβ©
The function has a period of 2Ο.
The periodic function is shown in the diagram below.
[ ] [ ] ( ) ( )
0 / 2 / 20 0 / 20 / 2
1 1 1a f (t)dt 0dt 1dt 1dt t t2 2 2
1 0 02 2 2
Ο Ο Ο Ο Ο
ΟβΟ βΟ Ο= = + + β = + β
Ο Ο Οβ§ β«β‘ Ο β€ β‘ Ο β€β β β β= β + βΟ β β =β¨ β¬β β β ββ’ β₯ β’ β₯Ο β β β β β£ β¦ β£ β¦β© β
β« β« β« β«
/ 2
n 0 / 2
1 1a f (t) cos nt dt cos nt dt cos nx dxΟ Ο Ο
βΟ Ο= = + βΟ Οβ« β« β«
= / 2
0 / 2
1 sin nt sin nt 1 n nsin 0 sin n sinn n n 2 2
Ο Ο
Ο
β§ β« β§ Ο Ο β«βͺ βͺβ‘ β€ β‘ β€ β‘ β€ β‘ β€β = β β Οββ¨ β¬ β¨ β¬β’ β₯ β’ β₯ β’ β₯ β’ β₯Ο Οβ£ β¦ β£ β¦ β£ β¦ β£ β¦β© ββͺ βͺβ© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 598
When n is even, na 0=
When n = 1, [ ] [ ] 11 1 2a sin 0 sin sin 1 0 0 1
2 2β§ Ο Ο β«β‘ β€ β‘ β€= β β Οβ = β β β =β¨ β¬β’ β₯ β’ β₯Ο Ο Οβ£ β¦ β£ β¦β© β
When n = 3, [ ] [ ] 31 3 3 1 2a sin 0 sin 3 sin 1 0 0 1
3 2 2 3 3β§ Ο Ο β«β‘ β€ β‘ β€= β β Οβ = β β β β β = ββ¨ β¬β’ β₯ β’ β₯Ο Ο Οβ£ β¦ β£ β¦β© β
When n = 5, [ ] [ ] 51 5 5 1 2a sin 0 sin 5 sin 1 0 0 1
5 2 2 5 5β§ Ο Ο β«β‘ β€ β‘ β€= β β Οβ = β β β =β¨ β¬β’ β₯ β’ β₯Ο Ο Οβ£ β¦ β£ β¦β© β
It follows that 72a
7= β
Ο , 9
2a9
=Ο
and so on.
/ 2
n 0 / 2
1 1b f (t)sin nt dt sin nt dt sin nt dtΟ Ο Ο
βΟ Ο= = + βΟ Οβ« β« β«
= / 2
0 / 2
1 cos nt cos nt 1 n ncos cos 0 cos n cosn n n 2 2
Ο Ο
Ο
β§ β« β§ Ο Ο β«βͺ βͺβ‘ β€ β‘ β€ β‘ β€ β‘ β€β + = β β + Οββ¨ β¬ β¨ β¬β’ β₯ β’ β₯ β’ β₯ β’ β₯Ο Οβ£ β¦ β£ β¦ β£ β¦ β£ β¦β© ββͺ βͺβ© β
= 1 n n 1 n1 cos cos n cos 1 2cos cos nn 2 2 n 2β§ Ο Ο β« Οβ β β β β§ β«β + Οβ = β + Οβ¨ β¬ β¨ β¬β β β βΟ Οβ β β β β© ββ© β
When n is odd, n1b 1 0 1 0n
= β β =Ο
When n is even, 21 4 2b 1 2( 1) 1
2 2= β β + = =
Ο Ο Ο,
41b 1 2(1) 1 0
4= β + =
Ο,
61 4 2b 1 2( 1) 1
6 6 3= β β + = =
Ο Ο Ο
Similarly, 8b 0= , 102b
5=
Ο, and so on.
Substituting into f(t) = ( )0 n nn 1
a a cos nt b sin ntβ
=
+ +β
gives: f(x) = 0 + 2 2 2 2cos t cos3t cos5t cos 7t .....3 5 7
β + β +Ο Ο Ο Ο
+ 2 2 2sin 2t sin 6t sin10t ....3 5
+ + +Ο Ο Ο
i.e. f(x) = 2 1 1 1 1cos t cos 3t cos 5t ...... sin 2t sin 6t sin10t .....3 5 3 5
β ββ + β + + + +β βΟ β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 599
7. Show that the Fourier series for the periodic function of period 2Ο defined by:
f(ΞΈ) =0, when 0sin , when 0
β Ο β€ΞΈ β€β§β¨ ΞΈ β€ ΞΈ β€ Οβ©
is given by: ( ) 2 1 cos 2 cos 4 cos 6f .....2 (3) (3)(5) (5)(7)
β βΞΈ ΞΈ ΞΈΞΈ = β β β ββ βΟ β β
The periodic function is shown in the diagram below.
[ ] ( ) ( ) 0
0 00
1 1 1 1a f ( )d 0d sin d cos cos cos 02 2 2 2
Ο Ο Ο
βΟ βΟ= ΞΈ ΞΈ = ΞΈ+ ΞΈ ΞΈ = β ΞΈ = β Ο β ββ‘ β€β£ β¦Ο Ο Ο Οβ« β« β«
= ( ) ( )1 1 11 cos 1 12 2
β Ο = β β =Ο Ο Ο
0
n 0
1a 0cos n d sin cos n dΟ
βΟ= ΞΈ ΞΈ+ ΞΈ ΞΈ ΞΈΟ β« β«
= ( ) ( ) ( ) ( ) 0 0
1 1 1sin n sin n sin 1 n sin 1 n d2 2
Ο Οβ§ β«ΞΈ+ ΞΈ + ΞΈβ ΞΈ = ΞΈ + + ΞΈ β ΞΈβ‘ β€β¨ β¬β£ β¦Ο Οβ© ββ« β« from 6, page 398
of textbook
= ( ) ( ) ( ) ( )0
cos 1 n 1 n cos 1 n cos 1 n1 1 1 1cos2 1 n 1 n 2 1 n 1 n 1 n 1 n
Ο β‘ β€ΞΈ + ΞΈ β Ο + Ο ββ‘ β€ β β β ββ β = β β β β ββ’ β₯β ββ’ β₯ β βΟ + β Ο + β + ββ β β’ β₯β£ β¦ β β β£ β¦
When n is odd, n1 1 1 1 1a 0
2 1 n 1 n 1 n 1 nβ‘ β€= β β + + =β’ β₯Ο + β + ββ£ β¦
When n = 2, 21 cos3 cos( ) 1 1 1 1 1 1 4 2a 1 1
2 3 1 3 1 2 3 3 2 3 3Ο βΟβ§ β« β§ β« β§ β«= β β + + = β + β = β = ββ¨ β¬ β¨ β¬ β¨ β¬Ο β β Ο Ο Οβ© β β© β β© β
When n = 4, 41 cos5 cos( 3 ) 1 1 1 1 1 1 1a
2 5 3 5 3 2 5 3 5 3Ο β Οβ§ β« β§ β«= β β + + = β + ββ¨ β¬ β¨ β¬Ο β β Οβ© β β© β
= 1 3 5 3 5 1 4 22 (3)(5) 2 (3)(5) (3)(5)
β§ β« β§ β«β + β= β = ββ¨ β¬ β¨ β¬Ο Ο Οβ© β β© β
When n = 6, 61 cos 7 cos( 5 ) 1 1 1 1 1 1 1a
2 7 5 7 5 2 7 5 7 5Ο β Οβ§ β« β§ β«= β β + + = β + ββ¨ β¬ β¨ β¬Ο β β Οβ© β β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 600
= 1 5 7 5 7 1 4 22 (5)(7) 2 (5)(7) (5)(7)
β§ β« β§ β«β + β= β = ββ¨ β¬ β¨ β¬Ο Ο Οβ© β β© β
0
n 0
1b 0sin n d sin sin n dΟ
βΟ= ΞΈ ΞΈ+ ΞΈ ΞΈ ΞΈΟ β« β«
= ( ) ( )0
0
1 1 1 sin (1 n) sin (1 n)cos n cos n2 2 1 n 1 n
ΟΟ ΞΈ + ΞΈ ββ§ β« β‘ β€β ΞΈ+ ΞΈ β ΞΈβ ΞΈ = β ββ‘ β€β¨ β¬β£ β¦ β’ β₯Ο Ο + ββ© β β£ β¦β« = 0 from 9, page
398, of textbook
Substituting into f(ΞΈ) = ( )0 n nn 1
a a cos n b sin nβ
=
+ ΞΈ+ ΞΈβ
gives: f(ΞΈ) = 1Ο
2 2 2cos 2 cos 4 cos 6 .....3 (3)(5) (5)(7)
β ΞΈβ ΞΈβ ΞΈβΟ Ο Ο
+ 0
i.e. f(ΞΈ) = 2 1 cos 2 cos 4 cos 6 ......2 (3) (3)(5) (5)(7)
β βΞΈ ΞΈ ΞΈβ β β ββ βΟ β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 601
CHAPTER 70 FOURIER SERIES FOR A NON-PERIODIC
FUNCTION OVER RANGE 2Ο EXERCISE 241 Page 667
2. Determine the Fourier series for the function defined by:
f(t) =1 t, when t 01 t, when 0 tβ β Ο β€ β€β§
β¨ + β€ β€ Οβ©
Draw a graph of the function within and outside of the given range.
The periodic function is shown in the diagram below.
( ) ( )
02 20
0 00
2 2 2 2
1 1 1 t ta f (t)dt (1 t)dt (1 t)dt t t2 2 2 2 2
1 1 2 20 0 2 12 2 2 2 2 2 4 2
ΟΟ Ο
βΟ βΟβΟ
β§ β«β‘ β€ β‘ β€βͺ βͺ= = β + + = β + +β¨ β¬β’ β₯ β’ β₯Ο Ο Ο β£ β¦ β£ β¦βͺ βͺβ© ββ§ β«β‘ β€ β‘ β€ β§ β«β β β β β βΟ Ο Ο Ο Ο Οβͺ βͺ βͺ βͺ= β βΟβ + Ο+ β = Ο+ = + = +β¨ β¬ β¨ β¬β’ β₯ β’ β₯β β β β β βΟ Ο Ο Οβͺ βͺβ β β β β β βͺ βͺβ£ β¦ β£ β¦ β© ββ© β
β« β« β«
0
n 0
1 1a f (t) cos nt dt (1 t)cos nt dt (1 t)cos nt dtΟ Ο
βΟ βΟ= = β + +Ο Οβ« β« β«
= ( ) ( ) 0
0
1 cos nt t cos nt dt cos nt t cos nt dtΟ
βΟβ + +
Ο β« β«
= 0
2 20
1 sin nt t sin nt cos nt sin nt t sin nt cos ntn n n n n n
Ο
βΟ
β‘ β€ β‘ β€β β + + +β’ β₯ β’ β₯Ο β£ β¦ β£ β¦ by integration by parts
= 2 2 2 2
1 cos 0 cos( n ) cos n cos 00 0 0 0 0 0 0 0n n n n
β§ β«β‘ β Ο β€ β‘ Ο β€β β β β β β β ββ β β β β + + + β + +β¨ β¬β β β β β β β ββ’ β₯ β’ β₯Ο β β β β β β β β β£ β¦ β£ β¦β© β
= ( )2 2 2 2 2
1 1 cos( n ) cos n 1 1 2cos n 2n n n n n
β Ο Οβ§ β«β + + β = Οββ¨ β¬Ο Οβ© β since cos(-nΟ) = cos nΟ
= ( )2
2 cos n 1n
ΟβΟ
When n is even, na 0=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 602
When n = 1, ( )1 2
2 4a 1 1(1)
== β β = βΟ Ο
When n = 3, ( )3 2 2
2 4a 1 1(3) (3)
= β β = βΟ Ο
When n = 5, ( )5 2 2
2 4a 1 1(5) (5)
= β β = βΟ Ο
and so on.
0
n 0
1 1b f (t)sin nt dt (1 t)sin nt dt (1 t)sin nt dtΟ Ο
βΟ βΟ= = β + +Ο Οβ« β« β«
= ( ) ( ) 0
0
1 sin nt t sin nt dt sin nt t sin nt dtΟ
βΟβ + +
Ο β« β«
= 0
2 20
1 cos nt t cos nt sin nt cos nt t cos nt sin ntn n n n n n
Ο
βΟ
β‘ β€ β‘ β€β + β + β β +β’ β₯ β’ β₯Ο β£ β¦ β£ β¦ by integration by parts
=
cos 0 cos( n ) cos( n )0 0 0n n n1
cos n cos n cos 00 0 0n n n
β§ β«β‘ β Ο Ο β Ο β€β β β ββ + β β β β ββͺ βͺβ β β ββ’ β₯β β β β βͺβ£ β¦ βͺβ¨ β¬Ο β‘ Ο Ο Ο β€β β β ββͺ βͺ+ β β + β β + +β β β ββ’ β₯βͺ βͺβ β β β β£ β¦β© β
= 1 1 cos( n ) cos( n ) cos n cos n 1n n n n n n
β Ο Ο β Ο Ο Ο Οβ§ β«β + + β β +β¨ β¬Ο β© β = 0 since cos nΟ = cos(-nΟ)
Substituting into f(t) = ( )0 n nn 1
a a cos nt b sin ntβ
=
+ +β
gives: f(t) = 12Ο+ 2 2
4 4 4cos t cos3t cos5t .....(3) (5)
β β β βΟ Ο Ο
+ 0
i.e. f(t) = 2 2
4 cos 3t cos 5t1 cos t ......2 3 5Ο β β+ β + + +β βΟ β β
4. Determine the Fourier series up to and including the third harmonic for the function defined by:
f(x) =x, when 0 x2 x, when x 2
β€ β€ Οβ§β¨ Οβ Ο β€ β€ Οβ©
Sketch a graph of the function within and outside of the given range, assuming the period is 2Ο.
The periodic function is shown in the diagram below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 603
( ) ( )
22 22
0 00
2 2 22 2 2
1 1 1 x xa f (x)dx x dx (2 x)dx 2 x2 2 2 2 2
1 4 10 4 22 2 2 2 2 2
Ο ΟΟ Ο Ο
βΟ ΟΟ
β§ β«β‘ β€ β‘ β€βͺ βͺ= = + Οβ = + Ο ββ¨ β¬β’ β₯ β’ β₯Ο Ο Ο β£ β¦ β£ β¦βͺ βͺβ© ββ§ β«β‘ β€ β‘ β€β β β β β βΟ Ο Ο Οβͺ βͺ= β + Ο β β Ο β = Ο =β¨ β¬β’ β₯ β’ β₯β β β β β βΟ Οβ β β β β β βͺ βͺβ£ β¦ β£ β¦β© β
β« β« β«
2 2
n 0 0
1 1a f (x)cos nx dx x cos nx dx (2 x)cos nx dxΟ Ο Ο
Ο= = + ΟβΟ Οβ« β« β«
= 2
2 20
1 x sin nx cos nx 2 sin nx x sin nx cos nxn n n n n
Ο Ο
Ο
β§ β«Οβͺ βͺβ‘ β€ β‘ β€+ + β ββ¨ β¬β’ β₯ β’ β₯Ο β£ β¦ β£ β¦βͺ βͺβ© β by integration by parts
= 2 2 2 2
1 cos n cos 0 cos 2 n cos n0 0 0 0 0 0n n n n
β§ β«β‘ Ο β€ β‘ Ο Ο β€β β β β β β β β+ β + + β β β β ββ¨ β¬β β β β β β β ββ’ β₯ β’ β₯Ο β β β β β β β β β£ β¦ β£ β¦β© β
= ( )2 2
1 2cos n 1 cos 2 n cos n cos n 1n n
Οβ β Ο + Ο = ΟβΟ Ο
When n is even, na 0=
When n = 1, ( )1 2
2 4a 1 1(1)
= β β = βΟ Ο
When n = 3, ( )3 2 2
2 4a 1 1(3) (3)
= β β = βΟ Ο
When n = 5, ( )5 2 2
2 4a 1 1(5) (5)
= β β = βΟ Ο
and so on.
2 0
n 0 0
1 1b f (x)sin nx dx x sin nx dx (2 x)sin nx dxΟ Ο
βΟ= = + ΟβΟ Οβ« β« β«
= 2
2 20
1 x cos nx sin nx 2 cos nx x cos nx sin nxn n n n n
Ο Ο
Ο
Οβ‘ β€ β‘ β€β + + β + ββ’ β₯ β’ β₯Ο β£ β¦ β£ β¦ by integration by parts
= ( )
2 cos 2n 2 cos 2n 0n n1 cos n 0 0 0
n 2 cos n cos n 0n n
β§ β«β‘ Ο Ο Ο Ο β€β ββ + +βͺ βͺβ ββ’ β₯β‘ Ο Ο β€βͺ β β βͺβ β β’ β₯β + β + +β¨ β¬β ββ’ β₯ β’ β₯Ο Ο Ο Ο Οβ β β ββ£ β¦βͺ βͺβ β + ββ’ β₯β ββͺ βͺβ β β£ β¦β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 604
= 1 cos n 2 cos n cos n 0n
βΟ Ο+ Ο Οβ Ο Ο =Ο
Substituting into f(x) = ( )0 n nn 1
a a cos nx b sin nxβ
=
+ +β
gives: f(x) = 2Ο
2 2
4 4 4cos x cos3x cos5x .....(3) (5)
β β β βΟ Ο Ο
+ 0
i.e. f(x) = 2 2
4 cos 3x cos 5xcos x ......2 3 5Ο β ββ + + +β βΟ β β
5. Expand the function ( ) 2f ΞΈ = ΞΈ in a Fourier series in the range - Ο < ΞΈ < Ο
Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below.
3 3 2
2 3 30
1 1 1 1 2a f (x)dx d2 2 2 3 6 6 3
ΟΟ Ο
βΟ βΟβΟ
β§ β«β‘ β€ΞΈ Ο Οβͺ βͺ= = ΞΈ ΞΈ = = Ο β βΟ = =β¨ β¬β’ β₯Ο Ο Ο Ο Οβ£ β¦βͺ βͺβ© ββ« β«
2n
1 1a f ( ) cos n d cos n dΟ Ο
βΟ βΟ= ΞΈ ΞΈ ΞΈ = ΞΈ ΞΈ ΞΈΟ Οβ« β«
= 2
2 3
1 sin n 2 cos n 2sin nn n n
Ο
βΟ
β‘ β€ΞΈ ΞΈ ΞΈ ΞΈ ΞΈ+ ββ’ β₯Ο β£ β¦
by integration by parts
= 2 2 2 2
1 2 cos n 2 cos( n ) 4 cos n 40 0 0 0 cos nn n n n
β§ β«β‘ Ο Ο Ο β Ο β€ Ο Οβ β β β+ β β β β = = Οβ¨ β¬β β β ββ’ β₯Ο Οβ β β β β£ β¦β© β
since cos nΟ = cos(-nΟ)
Hence, ( )1 2 2
4 4a 1(1) 1
= β = β , ( )2 2 2
4 4a 1(2) 2
= = , ( )3 2 2
4 4a 1(3) 3
= β = β , 4 2
4a4
= , and so on.
2n
1 1b f ( )sin n d sin n dΟ Ο
βΟ βΟ= ΞΈ ΞΈ ΞΈ = ΞΈ ΞΈ ΞΈΟ Οβ« β«
= 2
2 3
1 cos n 2 sin n 2cos nn n n
Ο
βΟ
β‘ β€ΞΈ ΞΈ ΞΈ ΞΈ ΞΈβ + +β’ β₯Ο β£ β¦
by integration by parts
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 605
= 2 2
3 3
1 cos n 2cos n cos( n ) 2cos( n )0 0n n n n
β§ β«β‘ β€β β β βΟ Ο Ο Ο β Ο β Οβͺ βͺβ + + β β + +β¨ β¬β’ β₯β β β βΟ β β β β βͺ βͺβ£ β¦β© β = 0
Substituting into f(ΞΈ) = ( )0 n nn 1
a a cos n b sin nβ
=
+ ΞΈ+ ΞΈβ
gives: f(ΞΈ) = 2
3Ο
2 2 2 2
4 4 4 4cos cos 2 cos3 cos 4 .....1 2 3 4
β ΞΈ+ ΞΈβ ΞΈ+ ΞΈβ + 0
i.e. f(ΞΈ) = 2
2 2
1 14 cos cos 2 cos 3 ......3 2 3Ο β ββ ΞΈ β ΞΈ + ΞΈ ββ β
β β
6. For the Fourier series in problem 5, let ΞΈ = Ο and deduce a series for 2n 1
1n
β
=β
When ΞΈ = Ο in Problem 5 above, f(ΞΈ) = 2Ο
Thus, 2Ο = 2
2 2 2
1 1 14 cos cos 2 cos3 cos 4 ......3 2 3 4Ο β ββ Οβ Ο+ Οβ Ο+β β
β β
i.e. 2Ο = 2
2 2 2 2
4 4 4 4 .....3 1 2 3 4Ο
+ + + + +
i.e. 2
22 2 2 2
1 1 1 14 .....3 1 2 3 4Ο β βΟ β = + + + +β β
β β
i.e. 2
2 2 2 2
2 1 1 1 14 .....3 1 2 3 4Ο β β= + + + +β β
β β
and 2
2 2 2 2
2 1 1 1 1 ....3(4) 1 2 3 4Ο
= + + + +
i.e. 2
2 2 2 2
1 1 1 1 ....1 2 3 4 6
Ο+ + + + =
i.e. 2n 1
1n
β
=β =
2
6Ο
8. Sketch the waveform defined by: f(x) =
2x1 , when x 0
2x1 , when 0 x
β§ + β Ο β€ β€βͺβͺ Οβ¨βͺ β β€ β€ Οβͺ Οβ©
Determine the Fourier series in this range.
The periodic function is shown in the diagram below.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 606
( ) ( ) ( ) ( )
02 20
0 00
2 2
1 1 2x 2x 1 x xa f (x)dx 1 dx (1 )dx x x2 2 2
1 10 0 02 2
ΟΟ Ο
βΟ βΟβΟ
β§ β«β‘ β€ β‘ β€β§ β« βͺ βͺβ β= = + + β = + + ββ¨ β¬ β¨ β¬β β β’ β₯ β’ β₯Ο Ο Ο Ο Ο Ο Οβ β β© β β£ β¦ β£ β¦βͺ βͺβ© ββ§ β«β‘ β€ β‘ β€β β β βΟ Οβͺ βͺ= β βΟ+ + Οβ β = Οβ Ο + Οβ Ο =β¨ β¬β’ β₯ β’ β₯β β β βΟ Ο Ο Οβ β β β βͺ βͺβ£ β¦ β£ β¦β© β
β« β« β«
0
n 0
1 1 2x 2xa f (x)cos nx dx 1 cos nx dx (1 )cos nx dxΟ Ο
βΟ βΟ
β§ β«β β= = + + ββ¨ β¬β βΟ Ο Ο Οβ β β© ββ« β« β«
= 0
2 20
1 sin nx 2 x sin nx cos nx sin nx 2 x sin nx cos nxn n n n n n
Ο
βΟ
β‘ β€ β‘ β€β β β β+ + + β +β β β ββ’ β₯ β’ β₯Ο Ο Οβ β β β β£ β¦ β£ β¦ by integration by
parts
= 2 2 2 2
1 2 1 2 cos( n ) 2 cos n 2 10 0 0 0 0 0 0 0n n n n
β§ β«β‘ β€ β‘ β€β β β β Ο β β Ο β β ββͺ βͺβ β β β β β β β+ + β + + + β β β β ββ¨ β¬β’ β₯ β’ β₯β β β β β β β ββ β β β β β β βΟ Ο Ο Ο Οβ β β β β β β β β β β β β β β β βͺ βͺβ£ β¦ β£ β¦β© β
= ( )2 2 2 2
1 42 2cos( n ) 2cos n 2 1 cos nn n
β β Ο β Ο+ = β ΟΟ Ο
since cos nΟ = cos(-nΟ)
When n is even, na 0=
Hence, ( )1 2 2 2
4 8a 1 1(1)
= β β =Ο Ο
, ( )3 2 2 2 2
4 8a 1 1(3) (3)
= β β =Ο Ο
, 5 2 2
8a(5)
=Ο
and so on.
0
n 0
1 1 2x 2xb f (x)sin nx dx 1 sin nx dx 1 sin nx dxΟ Ο
βΟ βΟ
β§ β«β β β β= = + + ββ¨ β¬β β β βΟ Ο Ο Οβ β β β β© ββ« β« β«
= 0
2 20
1 cos nx 2 x cos nx sin nx cos nx 2 x cos nx sin nxn n n n n n
Ο
βΟ
β§ β«β‘ β€ β‘ β€βͺ βͺβ β β ββ + β + + β β β +β¨ β¬β β β ββ’ β₯ β’ β₯Ο Ο Οβ β β β β£ β¦ β£ β¦βͺ βͺβ© β by
integration by parts
=
cos n 2 cos n 0n n1 1 cos( n ) 2 cos( n )0 0 0
n n n 1 0 0n
β§ β«β‘ Ο Ο Ο β€β ββ + +βͺ βͺβ ββ’ β₯β‘ β Ο Ο β Ο β€βͺ β β βͺβ β β β β’ β₯β β + β β + + +β¨ β¬β β β ββ’ β₯ β’ β₯Ο β β β β β ββ£ β¦βͺ βͺβ β + ββ’ β₯β ββͺ βͺβ β β£ β¦β© β
= 1 1 cos( n ) 2 cos( n ) cos n 2 cos n 1 0n n n n n n
β Ο Ο β Ο Ο Ο Οβ§ β«β + β β + + =β¨ β¬Ο β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 607
Substituting into f(x) = ( )0 n n nn 1 n 1
a a cos nx b sin nx a cos nxβ β
= =
+ + =β β
gives: f(x) = 2 2 2 2 2 2 2
8 8 8 8cos x cos3x cos5x cos 7x .....(3) (5) (7)
+ + + +Ο Ο Ο Ο
i.e. f(x) = 2 2 2 2
8 1 1 1cos x cos 3x cos 5x cos 7x ......3 5 7
β β+ + + +β βΟ β β
9. For the Fourier series of Problem 8, deduce a series for 2
8Ο
When f(x) = 1 in the series of problem 8 above, x = 0,
hence, 1 = 2 2 2 2
8 1 1 1cos 0 cos 0 cos 0 cos 0 ......3 5 7
β β+ + + +β βΟ β β
i.e. 2
2 2 2
1 1 11 .....8 3 5 7Ο
= + + + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 608
CHAPTER 71 EVEN AND ODD FUNCTIONS AND HALF-RANGE
FOURIER SERIES EXERCISE 242 Page 672 2. Obtain the Fourier series of the function defined by:
f(t) =t , when t 0t , when 0 t+ Ο β Ο β€ β€β§
β¨ β Ο β€ β€ Οβ©
which is periodic of period 2Ο. Sketch the given function.
The periodic function is shown in the diagram below. Since it is symmetrical about the origin, the
function is odd, and nn 1
f (t) b sin ntβ
=
=β
0
n 0
1 1b f (t)sin nt dt (t )sin nt dt (t ) sin ntdtΟ Ο
βΟ βΟ= = + Ο + β ΟΟ Οβ« β« β«
= 0
2 20
1 t cos nt sin nt cos nt t cos nt sin nt cos ntn n n n n n
Ο
βΟ
Ο Οβ‘ β€ β‘ β€β + β + β + +β’ β₯ β’ β₯Ο β£ β¦ β£ β¦ by integration by parts
=
cos n cos n0n n1 cos( n ) cos( n )0 0 0
n n n0 0
n
β§ β«β‘ Ο Ο Ο Ο β€β ββ + +βͺ βͺβ ββ’ β₯β‘ Ο βΟ β Ο Ο β Ο β€βͺ β β βͺβ β β β β’ β₯+ β β β + β +β¨ β¬β β β ββ’ β₯ β’ β₯Ο Οβ β β β β ββ£ β¦βͺ βͺβ + +β’ β₯β ββͺ βͺβ β β£ β¦β© β
= 1 cos( n ) cos( n ) cos n cos n 1 2 2n n n n n n n nΟ Ο β Ο Ο β Ο Ο Ο Ο Ο Ο Οβ§ β« β ββ β + β + β = β = ββ¨ β¬ β βΟ Οβ© β β β
Hence, 12b1
= β , 22b2
= β , 32b3
= β , 42b4
= β , and so on.
i.e. f(t) = 2 2 2 2sin t sin 2t sin 3t sin 4t ......1 2 3 4
β β β β β
i.e. 1 1 1f (t) 2 sin t sin 2t sin 3t sin 4t .....2 3 4
β β= β + + + +β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 609
3. Determine the Fourier series defined by: f(x) =1 x, when x 01 x, when 0 xβ β Ο β€ β€β§
β¨ + β€ β€ Οβ©
which is periodic of period 2Ο.
The periodic function is shown in the diagram below. Since it is symmetrical about the f(x) axis, the
function is even, and 0 nn 1
f (x) a a cos nxβ
=
= +β
0 0
1 1a f (x)dx f (x)dx2
Ο Ο
βΟ= =
Ο Οβ« β« due to symmetry
= ( )2 2
00
1 1 x 1(1 x)dx x 0 12 2 2
ΟΟ β‘ β€β‘ β€ β βΟ Ο
+ = + = Ο+ β = +β’ β₯β ββ’ β₯Ο Ο Οβ£ β¦ β β β£ β¦β«
0
n 0
1 1a f (x)cos nx dx (1 x)cos nx dx (1 x)cos nx dxΟ Ο
βΟ βΟ= = β + +Ο Οβ« β« β«
= ( ) ( ) 0
0
1 cos nx x cos nx dx cos nx x cos nx dxΟ
βΟβ + +
Ο β« β«
= 0
2 20
1 sin nx x sin nx cos nx sin nx x sin nx cos nxn n n n n n
Ο
βΟ
β‘ β€ β‘ β€β β + + +β’ β₯ β’ β₯Ο β£ β¦ β£ β¦ by integration by parts
= 2 2 2 2
1 1 cos( n ) cos n 10 0 0 0 0 0 0 0n n n n
β§ β«β‘ β Ο β€ β‘ Ο β€β β β β β β β ββ β β β β + + + β + +β¨ β¬β β β β β β β ββ’ β₯ β’ β₯Ο β β β β β β β β β£ β¦ β£ β¦β© β
= ( )2 2 2 2 2
1 1 cos( n ) cos n 1 2 cos n 1n n n n n
β Ο Οβ§ β«β + + β = Οββ¨ β¬Ο Οβ© β since cos(-nΟ) = cos nΟ
When n is even, na 0=
When n = 1, ( )1 2
2 4a 1 1(1)
= β β = βΟ Ο
When n = 3, ( )3 2 2
2 4a 1 1(3) (3)
= β β = βΟ Ο
When n = 5, ( )5 2 2
2 4a 1 1(5) (5)
= β β = βΟ Ο
and so on.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 610
0 nn 1
f (x) a a cos nxβ
=
= +β = 12Ο+ 2 2
4 4 4cos x cos3x cos5x .....(3) (5)
β β β βΟ Ο Ο
i.e. f(x) = 2 2
4 1 11 cos x cos 3x cos 5x ......2 3 5Ο β β+ β + + +β βΟ β β
4. In the Fourier series of Problem 3, let x = 0 and deduce a series for 2
8Ο
When x = 0 in the series of Problem 3 above, f(x) = 1,
hence, 1 = 2 2
4 cos0 cos01 cos0 ......2 3 5Ο β β+ β + + +β βΟ β β
i.e. 1 = 2 2 2
4 1 1 11 1 .....2 3 5 7Ο β β+ β + + + +β βΟ β β
i.e. 2 2 2
4 1 1 11 .....2 3 5 7Ο β ββ = β + + + +β βΟ β β
and 2
2 2 2
1 1 11 ....8 3 5 7Ο
= + + + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 611
EXERCISE 243 Page 675
1. Determine the half-range series for the function defined by: f(x) =x, when 0 x
2
0, when x2
Οβ§ β€ β€βͺβͺβ¨ Οβͺ β€ β€ Οβͺβ©
The periodic function is shown in the diagram below. Since a half-range sine series is required, the
function is symmetrical about the origin and nn 1
f (x) b sin nxβ
=
=β
/ 2
/ 2
n 20 00
2 2 2 x cos nx sin nxb f (x)sin nx dx x sin nx dxn n
ΟΟ Ο β‘ β€= = = β +β’ β₯Ο Ο Ο β£ β¦β« β« by integration by parts
= ( )2
n ncos sin2 2 2 2 0n n
β‘ Ο Ο Ο β€β ββ’ β₯β ββ + ββ’ β₯β βΟ β’ β₯β β
β’ β₯β β β£ β¦
Hence, 1 2 2
cos sin2 2 1 22 2 2b 01 1 1
β‘ Ο Ο Ο β€β ββ’ β₯β β β β= β + = + =β’ β₯β β β βΟ Ο Οβ β β’ β₯β ββ’ β₯β β β£ β¦
,
2 2
cos2 sin 2 22b 02 2 4 4
β‘ Ο β€β βΟβ’ β₯β βΟ Ο Οβ β β β= β + = + =β’ β₯β β β β β βΟ Ο Οβ β β β β’ β₯β ββ’ β₯β β β£ β¦
,
3 2 2 2
3 3cos sin2 2 1 22 2 2b 03 3 3 (3)
β‘ Ο Ο Ο β€β ββ’ β₯β β β β= β + = β = ββ’ β₯β β β βΟ Ο Οβ β β’ β₯β ββ’ β₯β β β£ β¦
,
4 2
cos 22 sin 2 2 22b 04 4 8 8
β‘ Ο β€β βΟβ’ β₯β βΟ Ο Οβ β β β= β + = β + = ββ’ β₯β β β β β βΟ Ο Οβ β β β β’ β₯β ββ’ β₯β β β£ β¦
, and so on
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 612
Hence, nn 1
f (x) b sin nxβ
=
=β = 2
2 2 2 1 2sin x sin 2x sin 3x sin 4x ...4 3 8Ο Οβ β β β β β+ β β +β β β β β βΟ Ο Ο Οβ β β β β β
i.e. 2 1f (x) sin x sin 2x sin 3x sin 4x .....4 9 8Ο Οβ β= + β β +β βΟ β β
2. Obtain (a) the half-range cosine series and (b) the half-range sine series for the function:
f(t) =0, when 0 t
2
1, when t2
Οβ§ β€ β€βͺβͺβ¨ Οβͺ β€ β€ Οβͺβ©
(a) The periodic function is shown in the diagram below. Since a half-range cosine series is
required, the function is symmetrical about the f(t) axis and 0 nn 1
f (t) a a cos ntβ
=
= +β
[ ]0 / 2/ 2
1 1 1 1a 1dt t2 2
Ο Ο
ΟΟ
Οβ‘ β€= = = Οβ =β’ β₯Ο Ο Ο β£ β¦β«
n / 2/ 2
n nsin 2sin2 2 sin nt 2 2 2a 1cos nt dt 0n n n
ΟΟ
ΟΟ
Ο Οβ ββ ββ‘ β€= = = β = ββ ββ’ β₯Ο Ο Ο Οβ£ β¦ β ββ β
β«
When n is even, na 0=
and 1
2sin 22a
Ο
= β = βΟ Ο
, 3
32sin 2 12a3 3
Οβ β= β = β βΟ Ο β β
, 5
52sin 2 12a5 5
Οβ β= β = β β βΟ Ο β β
, and so on.
Thus, 0 nn 1
1 2 2 1 2 1f (t) a a cos nt cos t cos3t cos5t ....2 3 5
β
=
β β β β= + = β + β +β β β βΟ Ο Οβ β β β β
i.e. 1 2 1 1f (t) cos t cos 3t cos 5t ......2 3 5
β β= β β + ββ βΟ β β
(b) The periodic function is shown in the diagram below. Since a half-range sine series is required,
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 613
the function is symmetrical about the origin and nn 1
f (x) b sin nxβ
=
=β
n / 2/ 2
2 2 cos nt 2 nb 1sin nt dt cos n cosn n 2
ΟΟ
ΟΟ
Οβ‘ β€ β‘ β€= = β = β Οββ’ β₯ β’ β₯Ο Ο Οβ£ β¦ β£ β¦β«
Hence, ( )12 2 2b cos cos 1 0
2Οβ β= β Οβ = β β β =β βΟ Ο Οβ β
,
( ) ( )22 2 2b cos 2 cos 1 1
2 2= β Οβ Ο = β β β = β
Ο Ο Ο ,
( )32 3 2 2b cos3 cos 1 0
3 2 3 3Οβ β= β Οβ = β β β =β βΟ Ο Οβ β
,
( ) ( )42 2b cos 4 cos 2 0 0 0
4 4= β Οβ Ο = β β =
Ο Ο ,
( )52 5 2 2b cos5 cos 1 0
5 2 5 5Οβ β= β Οβ = β β β =β βΟ Ο Οβ β
,
( ) ( )62 2 2b cos 6 cos3 1 1
6 6 3= β Οβ Ο = β β β = β
Ο Ο Ο , and so on.
Thus, nn 1
f (t) b sin ntβ
=
=β = 2 2 2 2 2sin t sin 2t sin 3t 0 sin 5t sin 6t ....3 5 3
β + + + β +Ο Ο Ο Ο Ο
i.e. 2 1 1 1f (t) sin t sin 2t sin 3t sin 5t sin 6t .....3 5 3
β β= β + + β +β βΟ β β
4. Determine the half-range Fourier cosine series in the range x = 0 to x = Ο for the function
defined by: f(x) =( )
x, when 0 x2
x , when x2
Οβ§ β€ β€βͺβͺβ¨ Οβͺ Οβ β€ β€ Οβͺβ©
The periodic function is shown in the diagram below. Since a half-range cosine series is required,
the function is symmetrical about the f(x) axis and 0 nn 1
f (x) a a cos nxβ
=
= +β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 614
/ 22 2/ 2
0 0 / 20 / 2
1 1 x xa x dx ( x)dx x2 2
Ο ΟΟ Ο
ΟΟ
β§ β«β‘ β€ β‘ β€βͺ βͺ= + Οβ = + Ο ββ¨ β¬β’ β₯ β’ β₯Ο Ο β£ β¦ β£ β¦βͺ βͺβ© ββ« β«
= 2 2 2 2
218 2 2 8
β§ β«β β β βΟ Ο Ο Οβͺ βͺ+ Ο β β ββ¨ β¬β β β βΟ βͺ βͺβ β β β β© β
= 2 2 2 2 21 1 2
8 2 2 8 8 4β β β βΟ Ο Ο Ο Ο Ο
+ β + = =β β β βΟ Οβ β β β
/ 2
n 0 / 2
2a x cos nx dx ( x)cos nx dxΟ Ο
Ο= + ΟβΟ β« β«
= / 2
2 20 / 2
2 x sin nx cos nx sin n x sin nx cos nxn n n n n
Ο Ο
Ο
β§ β«Ο Οβͺ βͺβ‘ β€ β‘ β€+ + β ββ¨ β¬β’ β₯ β’ β₯Ο β£ β¦ β£ β¦βͺ βͺβ© β
= 2 2 2 2
n n n n nsin cos sin sin cos2 1 cos n2 2 2 2 2 2 20 0 0n n n n n n n
β§ Ο Ο Ο Ο Ο Ο Ο β«β β β βΟβͺ βͺβ β β βΟβͺ βͺβ β β β+ β + + β β β β ββ¨ β¬β β β ββ β β βΟ β β β β βͺ βͺβ β β ββͺ βͺβ β β β β© β
= 2 2 2 2
n n n2cos sin sin2 1 cos n 2 n2 2 2 2cos 1 cos nn n n n n n 2
Ο Ο Οβ§ β«Ο Οβͺ βͺΟ Οβ§ β«+ β β β = β β Οβ¨ β¬ β¨ β¬Ο Ο β© ββͺ βͺβ© β
When n is odd, na 0=
and ( ) ( )2 2
2 2 8 2a 2cos 1 cos 2 2 1 1(2) 4 4
= Οβ β Ο = β β β = β = βΟ Ο Ο Ο
,
( ) ( )4 2
2 2a 2cos 2 1 cos 4 2 1 1 0(4) 16
= Οβ β Ο = β β =Ο Ο
,
( ) ( )6 2 2
2 2 8 2a 2cos3 1 cos 6 2 1 1(6) 36 36 (3)
= Οβ β Ο = β β β = β = βΟ Ο Ο Ο
,
8a 0= ,
( ) ( )10 2 2
2 2 8 2a 2cos5 1 cos10 2 1 1(10) 100 100 (5)
= Οβ β Ο = β β β = β = βΟ Ο Ο Ο
, and so on.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 615
Thus, 0 n 2 2n 1
2 2 2f (t) a a cos nt cos 2t cos 6t cos10t ....4 (3) (5)
β
=
Ο= + = β β β +
Ο Ο Οβ
i.e. 2 2
2 1 1f (t) cos 2t cos 6t cos10t ......4 3 5Ο β β= β + + +β βΟ β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 616
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: f(x) = 0 n nn 1
2 nx 2 nxa a cos b sinL L
β
=
β‘ Ο Ο β€β β β β+ +β β β ββ’ β₯β β β β β£ β¦β where L = 5
52 2L 5
0 0 00
1 1 1 x 1 5 5a f (x)dx x dxL 5 5 2 5 2 2
β‘ β€ β‘ β€= = = = =β’ β₯ β’ β₯
β£ β¦ β£ β¦β« β«
5
L 5
n 20 0
0
2 nx 2 nxx sin cos2 2 nx 2 2 nx 2 5 5a f (x)cos dx x cos dx
2 nL L 5 5 5 2 n5 5
β‘ β€Ο Οβ β β ββ’ β₯β β β βΟ Οβ β β β β β β β β’ β₯= = = +β β β β Οβ’ β₯β ββ β β β Οβ ββ ββ’ β₯β ββ β β β β£ β¦
β« β« by parts
= 2 22 5sin 2 n cos 2 n 10
2 n5 2 n 2 n5 5 5
β‘ β€β β β ββ’ β₯β β β βΟ Οβ’ β₯β β β β+ β +β’ β₯β β β βΟβ β Ο Οβ β β ββ ββ’ β₯β β β ββ β β ββ β β β β β β β β β β£ β¦
= 0
5
L 5
n 20 0
0
2 nx 2 nxx cos sin2 2 nx 2 2 nx 2 5 5b f (x)sin dx x sin dx
2 nL L 5 L 5 2 nx5 5
β‘ β€Ο Οβ β β ββ’ β₯β β β βΟ Οβ β β β β β β β β’ β₯= = = β +β β β β Οβ’ β₯β ββ β β β Οβ ββ ββ’ β₯β ββ β β β β£ β¦
β« β« by parts
= ( )22 5cos 2 n sin 2 n 2 5cos 2 n 5 50 0 cos 2 n
2 n 2 n5 5 n n2 n5 55
β‘ β€β β β‘ β€β’ β₯β β β’ β₯Ο Ο Οβ’ β₯β β β’ β₯β + β + = β = β Ο = ββ’ β₯β βΟ Ο Ο Οβ β β βΟ β’ β₯β ββ β β ββ’ β₯β ββ β β’ β₯β β β β β£ β¦β β β β β£ β¦
Hence, 1 2 3 45 5 5 5b , b , b , b ,
2 3 4= β = β = β = β
Ο Ο Ο Ο and so on.
Thus, f(x) = 0 n nn 1
2 nx 2 nxa a cos b sinL L
β
=
β‘ Ο Ο β€β β β β+ +β β β ββ’ β₯β β β β β£ β¦β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 617
i.e. f(x) = 5 5 2 x 5 4 x 5 6 xsin sin sin .....2 5 2 5 3 5
Ο Ο Οβ β β β β ββ β β ββ β β β β βΟ Ο Οβ β β β β β
i.e. f(x) = 5 5 2 x 1 4 x 1 6 xsin sin sin ......2 5 2 5 3 5
β‘ β€Ο Ο Οβ β β β β ββ + + +β β β β β ββ’ β₯Ο β β β β β β β£ β¦
3. A periodic function of period 4 is defined by: f(x) =3, when 2 x 03, when 0 x 2
β β β€ β€β§β¨ + β€ β€β©
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. na 0=
Thus, f(x) = 0 nn 1
2 nxa b sinL
β
=
β‘ Ο β€β β+ β ββ’ β₯β β β£ β¦β where L = 4
[ ] [ ] ( ) ( ) ( ) ( ) L / 2 0 2 0 20 2 0L / 2 2 0
1 1 1 1a f (x)dx 3dx 3dx 3x 3x 0 6 6 0L 4 4 4ββ β
= = β + = β + = β + ββ« β« β« = 0
L / 2 0 2
n L / 2 2 0
2 2 nx 2 2 nx 2 nxb f (x)sin dx 3sin dx 3sin dxL L 4 4 4β β
Ο β§ Ο Ο β«β β β β β β= = β +β¨ β¬β β β β β ββ β β β β β β© β
β« β« β«
=
0 2
2 0
nx nx3cos 3cos1 1 3cos 0 3cos( n) 3cos n 3cos 02 2
n n n n n n2 22 2 2 2 2 2β
β§ β« β§ β«β‘ Ο β€ β‘ Ο β€ β β β ββ β β ββͺ βͺβ β β β βͺ βͺβ β β ββ’ β₯ β’ β₯ βΟ Οβͺ βͺ βͺ βͺβ β β β β β β ββ’ β₯ β’ β₯+ β = β + β β ββ¨ β¬ β¨ β¬Ο Ο Ο Ο Ο Οβ β β β β β β β β β β ββ β β ββ’ β₯ β’ β₯βͺ βͺ βͺ βͺβ β β β β β β β β β β ββ β β ββ’ β₯ β’ β₯βͺ βͺ βͺ βͺβ β β β β β β β β β β β β£ β¦ β£ β¦ β β β β β© ββ© β
= ( )1 6 6 6cos n 1 cos nn n2 n2 2
β‘ β€β’ β₯β’ β₯β Ο = β ΟΟ Ο Οβ β β ββ’ β₯
β β β ββ’ β₯β β β β β£ β¦
When n is even, nb 0=
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 618
Hence, ( ) ( ) ( )1 3 56 12 6 12 6 12b 1 1 , b 1 1 , b 1 1 ,
3 3 5 5= β β = = β β = = β β =Ο Ο Ο Ο Ο Ο
and so on.
Thus, f(x) = 0 nn 1
2 nxa b sinL
β
=
β‘ Ο β€β β+ β ββ’ β₯β β β£ β¦β = 0 + 12 x 12 3 x 12 5 xsin sin sin ......
2 3 2 5 2Ο Ο Οβ β β β β β+ + +β β β β β βΟ Ο Οβ β β β β β
i.e. f(x) = 12 x 1 3 x 1 5 xsin sin sin .....2 3 2 5 2
β§ β«Ο Ο Οβ β β β β β+ + +β¨ β¬β β β β β βΟ β β β β β β β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 619
EXERCISE 245 Page 681
1. Determine the half-range Fourier cosine series for the function f(x) = x in the range 0 β€ x β€ 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 0 nn 1
n xf (x) a a cosL
β
=
Οβ β= + β ββ β
β
32L 3
0 0 00
1 1 1 x 3a f (x)dx x dxL 3 3 2 2
β§ β«β‘ β€βͺ βͺ= = = =β¨ β¬β’ β₯β£ β¦βͺ βͺβ© β
β« β«
L 3
n 0 0
2 n x 2 n xa f (x)cos dx x cos dxL L 3 3β§ Ο β« Οβ β β β= =β¨ β¬β β β β
β β β β β© ββ« β«
=
3
2 2 2
0
n x n xx sin cos2 2 3sin n cos n 13 3 0
n n3 3n n n3 33 3 3
β§ β« β‘ β€β‘ β€ β β β βΟ Οβ β β ββͺ βͺ β’ β₯β β β ββ’ β₯β β β β Ο Οβͺ βͺβ β β β β’ β₯β β β ββ’ β₯+ = + β +β¨ β¬ β’ β₯β β β βΟ Οβ’ β₯β β β βΟ Ο Οβ β β β β ββͺ βͺβ β β ββ’ β₯β β β ββ’ β₯β β β β β ββͺ βͺβ β β β β β β β β β β£ β¦ β β β β β£ β¦β© β
by parts
= ( )2
2 2 2 2
2 cos n 1 2 3 60 cos n 1 cos n 13 3 n nn n
3 3
β§ β«βͺ βͺΟβͺ βͺ β β+ β = Οβ = Οββ¨ β¬ β βΟ Οβ β Ο Οβ β β ββͺ βͺ
β β β ββͺ βͺβ β β β β© β
When n is even, na 0=
and ( )1 2 2 2
6 12a 2(1)
= β = βΟ Ο
, ( )3 2 2 2 2
6 12a 2(3) (3)
= β = βΟ Ο
, 5 2 2
12a(5)
= βΟ
, and so on.
Thus, 0 n 2 2 2 2 2n 1
n x 3 12 x 12 3 x 12 5 xf (x) a a cos cos cos cos ....L 2 3 (3) 3 (5) 3
β
=
Ο Ο Ο Οβ β β β β β β β= + = β β β ββ β β β β β β βΟ Ο Οβ β β β β β β β β
i.e. 2 2 2
3 12 x 1 3 x 1 5 xf (x) cos cos cos .....2 3 3 3 5 3
β§ β«Ο Ο Οβ β β β β β= β + + +β¨ β¬β β β β β βΟ β β β β β β β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 620
2. Find the half-range Fourier sine series for the function f(x) = x in the range 0 β€ x β€ 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 nn 1
n xf (x) b sinL
β
=
Οβ β= β ββ β
β
3
L 3
n 20 0
0
n x n xx cos sin2 n x 2 n x 2 3 3b f (x)sin dx x sin dx
nL L 3 3 3 n3 3
β‘ β€Ο Οβ β β ββ’ β₯β β β βΟ β§ Ο β«β β β β β β β β β’ β₯= = = β +β¨ β¬β β β β Οβ’ β₯β ββ β β β Οβ© β β ββ ββ’ β₯β ββ β β β β£ β¦
β« β« by parts
= ( )22 3cos n sin n 2cos n 60 0 cos n
n n3 nn3 33
β‘ β€β ββ’ β₯β βΟ Ο Οβ’ β₯β ββ + β + = β = β Οβ’ β₯β βΟ Ο Οβ β β βΟβ ββ β β ββ’ β₯β ββ ββ β β β β β β β β£ β¦
1 2 3 46 6 6 6b , b , b , b ,
2 3 4= = β = = βΟ Ο Ο Ο
and so on.
Thus, nn 1
n xf (x) b sinL
β
=
Οβ β= β ββ β
β = 6 x 6 2 x 6 3 x 6 4 xsin sin sin sin ....3 2 3 3 3 4 3Ο Ο Ο Οβ β β β β β β ββ + β +β β β β β β β βΟ Ο Ο Οβ β β β β β β β
i.e. f(x) = 6 x 1 2 x 1 3 x 1 4 xsin sin sin sin .....3 2 3 3 3 4 3
β§ β«Ο Ο Ο Οβ β β β β β β ββ + β +β¨ β¬β β β β β β β βΟ β β β β β β β β β© β
3. Determine the half-range Fourier sine series for the function defined by:
f(t) =( )
t, when 0 t 12 t , when 1 t 2
β€ β€β§βͺβ¨ β β€ β€βͺβ©
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 621
The periodic function is shown in the diagram below. Since a half-range sine series is required, the
function is symmetrical about the origin and nn 1
n tf (t) b sinL
β
=
Οβ β= β ββ β
β
L 1 2
n 0 0 1
2 n t 2 n t n tb f (t)sin dt t sin dt (2 t)sin dtL L 2 2 2
Ο β§ Ο Ο β«β β β β β β= = + ββ¨ β¬β β β β β ββ β β β β β β© β
β« β« β«
=
1 2
2 2
0 1
n t n t n t n t n tt cos sin 2cos t cos sin2 2 2 2 2
n n nn n2 2 22 2
β§ β«β‘ β€ β‘ β€Ο Ο Ο Ο Οβ β β β β β β β β ββͺ βͺβ’ β₯ β’ β₯β β β β β β β β β ββͺ βͺβ β β β β β β β β β β’ β₯ β’ β₯β + + β + ββ¨ β¬Ο Ο Οβ’ β₯ β’ β₯β β β β β βΟ Οβ β β ββͺ βͺβ β β β β ββ’ β₯ β’ β₯β β β ββͺ βͺβ β β β β β β β β β β£ β¦ β£ β¦β© β
by parts
= ( )
2
2
2
2cos n 2cos n sin nn n nnn sincos 2 2 222 0 0
n n n n n2cos cos sin2 2 2 2 2n n n2 2 2
β ββ βΟ Ο Οβ ββ + ββ βΟ Οβ β β ββ‘ β€ Οβ β β βΟβ βΟ β β β ββ ββ ββ’ β₯β ββ β β β β β β β β β β β β’ β₯β ββ + β + +
β’ β₯β βΟβ β Ο ββ β Ο Ο Οβ β β β β ββ ββ’ β₯β ββ β β β β β β β ββ β β β β β β£ β¦ β β β β β β β β + βΟ Οβ β β β Οβ ββ β β β β ββ β β β β β β
β§ β«β‘ β€βͺ βͺβ’ β₯βͺ βͺβ’ β₯βͺ βͺβ’ β₯βͺ βͺβ’ β₯βͺ βͺβ’ β₯β¨ β¬β’ β₯ββͺ βͺβ’ β₯ββͺ βͺβ’ β₯β ββͺ βͺβ’ β₯β ββͺ βͺβ’ β₯β ββͺ βͺβ’ β₯β β£ β¦β© β
2 2 2
n2sin8 n2 sin
n 2n2
Οβ ββ β Οβ ββ β = = β βΟ β β Οβ β
β ββ β
When n is even, nb 0=
1 3 52 2 2 2 2
8 8 8b , b , b ,(3) (5)
= = β =Ο Ο Ο
and so on.
Thus, nn 1
n tf (t) b sinL
β
=
Οβ β= β ββ β
β = 2 2 2 2 2
8 t 8 3 t 8 5 tsin sin sin ....2 (3) 2 (5) 2Ο Ο Οβ β β β β ββ + ββ β β β β βΟ Ο Οβ β β β β β
i.e. f(t) = 2 2 2
8 t 1 3 t 1 5 tsin sin sin .....2 3 2 5 2
β§ β«Ο Ο Οβ β β β β ββ + ββ¨ β¬β β β β β βΟ β β β β β β β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 622
4. Show that the half-range Fourier cosine series for the function f(ΞΈ) = ΞΈ2 in the range 0 to 4 is
given by: ( ) 2 2 2
16 64 1 2 1 3f cos cos cos .....3 4 2 4 3 4
β§ ΟΞΈ ΟΞΈ ΟΞΈ β«β β β β β βΞΈ = β β + ββ¨ β¬β β β β β βΟ β β β β β β β© β
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(ΞΈ) axis and 0 nn 1
nf ( ) a a cosL
β
=
ΟΞΈβ βΞΈ = + β ββ β
β
43L 4 2
0 0 00
1 1 1 1 64 16a f ( )d dL 4 4 3 4 3 3
β§ β«β‘ β€ΞΈβͺ βͺ β β= ΞΈ ΞΈ = ΞΈ ΞΈ = = =β¨ β¬ β ββ’ β₯ β β β£ β¦βͺ βͺβ© ββ« β«
L 4 2n 0 0
2 n 2 na f ( )cos d cos dL L 4 4β§ ΟΞΈ β« ΟΞΈβ β β β= ΞΈ ΞΈ = ΞΈ ΞΈβ¨ β¬β β β β
β β β β β© ββ« β«
=
4
2
2 3
0
n n nsin 2 cos 2sin1 4 4 4
n2 n n4 4 4
β§ β«β‘ β€ΟΞΈ ΟΞΈ ΟΞΈβ β β β β ββͺ βͺΞΈ ΞΈβ’ β₯β β β β β ββͺ βͺβ β β β β β β’ β₯+ ββ¨ β¬Οβ’ β₯β β Ο Οβ β β ββͺ βͺβ ββ’ β₯β β β ββͺ βͺβ β β β β β β£ β¦β© β
by parts
= ( )2 3 2 2 2
1 16sin n 8cos n 2sin n 1 8cos n 1 8(16)0 cos nn2 2 2 nn n n4 4 4 4
β§ β«β β β§ β«βͺ βͺβ β βͺ βͺΟ Ο Ο Οβͺ βͺ βͺ βͺ β ββ β+ β β = = Οβ¨ β¬ β¨ β¬ β ββ βΟ Οβ β β β Ο Ο Οβ β β β β ββͺ βͺ βͺ βͺβ ββ ββ β β β β ββͺ βͺ βͺ βͺβ β β β β β β β β β β© ββ© β
= 2 2
64 cos nn
ΟΟ
and ( )1 2 2 2
64 64a 1(1)
= β = βΟ Ο
, ( )2 2 2 2 2
64 64a 1(2) (2)
= =Ο Ο
, 3 2 2 2 2
64 64a ( 1)(3) (3)
= β = βΟ Ο
, and so on.
Thus, 0 n 2 2 2 2 2n 1
n 16 64 64 2 64 3f ( ) a a cos cos cos cos ....L 3 4 (2) 4 (3) 4
β
=
ΟΞΈ ΟΞΈ ΟΞΈ ΟΞΈβ β β β β β β βΞΈ = + = β + β +β β β β β β β βΟ Ο Οβ β β β β β β β β
i.e. 2 2 2
16 64 1 2 1 3f ( ) cos cos cos .....3 4 2 4 3 4
β§ β«ΟΞΈ ΟΞΈ ΟΞΈβ β β β β βΞΈ = β β + ββ¨ β¬β β β β β βΟ β β β β β β β© β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 623
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 ΞΈΒ° 30 60 90 120 150 180 210 240 270 300 330 360
Displacement y 40 43 38 30 23 17 11 9 10 13 21 32
ΞΈΒ° y
cosΞΈ y cos ΞΈ sin ΞΈ y sin ΞΈ cos 2ΞΈ y cos 2ΞΈ sin 2ΞΈ y sin 2ΞΈ cos 3ΞΈ ycos3ΞΈ sin3ΞΈ ycos3ΞΈ
30 40
60 43
90 38
120 30
150 23
180 17
210 11
240 9
270 10
300 13
330 21
360 32
0.866
0.5
0
-0.5
-0.866
-1
-0.866
-0.5
0
0.5
0.866
1
34.64
21.50
0
-15.0
-19.92
-17
-9.53
-4.5
0
6.5
18.19
32
0.5
0.866
1
0.866
0.5
0
-0.5
-0.866
-1
-0.866
-0.5
0
20
37.24
38
25.98
11.5
0
-5.5
-7.79
-10
-11.26
-10.5
0
0.5
-0.5
-1
-0.5
0.5
1
0.5
-0.5
-1
-0.5
0.5
1
20
-21.5
-38
-15
11.5
17
5.5
-4.5
-10
-6.5
10.5
32
0.866
0.866
0
-0.866
-0.866
0
0.866
0.866
0
-0.866
-0.866
0
34.64
37.24
0
-25.98
-19.92
0
9.53
7.79
0
-11.26
-18.19
0
0
-1
0
1
0
-1
0
1
0
-1
0
1
0
-43
0
30
0
-17
0
9
0
-13
0
32
1
0
-1
0
1
0
-1
0
1
0
-1
0
40
0
-38
0
23
0
-11
0
10
0
-21
0
12
kk 1
y 287=
=β
12
k kk 1
y cos 46.88=
ΞΈ =β
12
k kk 1
y sin 87.67=
ΞΈ =β
12
k kk 1
y cos 2 1=
ΞΈ =β
12
k kk 1
y sin 2 13.85=
ΞΈ =β
12
k kk 1
y cos3 2=
ΞΈ = ββ
12
k kk 1
y sin 3 3=
ΞΈ =β
p
0 kk 1
1 1a y (287) 23.92p 12=
β = =β
p
n k kk 1
2a y cos nxp =
β β hence, 12a (46.88) 7.81
12β = , 2
2a (1) 0.1712
β = , 32a ( 2) 0.33
12β β = β
p
n k kk 1
2b y sin nxp =
β β hence, 12b (87.67) 14.61
12β = , 2
2b (13.85) 2.3112
β = , 32b (3) 0.50
12β =
Substituting these values into the Fourier series: ( )0 n nn 1
f (x) a a cos nx b sin nxβ
=
= + +β
gives: y = 23.92 + 7.81 cos ΞΈ + 0.17 cos 2ΞΈ - 0.33 cos 3ΞΈ + β¦β¦
+ 14.61 sin ΞΈ + 2.31 sin 2ΞΈ + 0.50 sin 3ΞΈ
or y = 23.92 + 7.81 cos ΞΈ +14.61 sin ΞΈ + 0.17 cos 2ΞΈ + 2.31 sin 2ΞΈ - 0.33 cos 3ΞΈ + 0.50 sin 3ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 624
3. Determine the Fourier series to represent the periodic function given by the table of values
below, up to and including the third harmonic and each coefficient correct to 2 decimal places.
Use 12 ordinates.
Angle ΞΈΒ° 30 60 90 120 150 180 210 240 270 300 330 360
Current i 0 -1.4 -1.8 -1.9 -1.8 -1.3 0 2.2 3.8 3.9 3.5 2.5
ΞΈΒ° i
cosΞΈ i cos ΞΈ sinΞΈ i sin ΞΈ cos 2ΞΈ i cos 2ΞΈ sin 2ΞΈ i sin 2ΞΈ cos 3ΞΈ i cos3ΞΈ sin3ΞΈ i cos3ΞΈ
30 0
60 -1.4
90 -1.8
120 -1.9
150 -1.8
180 -1.3
210 0
240 2.2
270 3.8
300 3.9
330 3.5
360 2.5
0.866
0.5
0
-0.5
-0.866
-1
-0.866
-0.5
0
0.5
0.866
1
0
-0.7
0
0.95
1.56
1.3
0
-1.1
0
1.95
3.03
2.5
0.5
0.866
1
0.866
0.5
0
-0.5
-0.866
-1
-0.866
-0.5
0
0
-1.212
-1.8
-1.645
-0.9
0
0
-1.905
-3.8
-3.377
-1.75
0
0.5
-0.5
-1
-0.5
0.5
1
0.5
-0.5
-1
-0.5
0.5
1
0
0.7
1.8
0.95
-0.90
-1.3
0
-1.1
-3.8
-1.95
1.75
2.5
0.866
0.866
0
-0.866
-0.866
0
0.866
0.866
0
-0.866
-0.866
0
0
-1.212
0
1.645
1.548
0
0
1.905
0
-3.377
-3.031
0
0
-1
0
1
0
-1
0
1
0
-1
0
1
0
1.4
0
-1.9
0
1.3
0
2.2
0
-3.9
0
2.5
1
0
-1
0
1
0
-1
0
1
0
-1
0
0
0
1.8
0
-1.8
0
0
0
3.8
0
-3.5
0
12
kk 1
y 7.7=
=β
12
k kk 1
y cos 9.49=
ΞΈ =β
12
k kk 1
y sin 16.389=
ΞΈ = ββ
12
k kk 1
y cos 2 1.35=
ΞΈ = ββ
12
k kk 1
y sin 2 2.522=
ΞΈ = ββ
12
k kk 1
y cos3 1.6=
ΞΈ =β
12
k kk 1
y sin 3 0.3=
ΞΈ =β
p
0 kk 1
1 1a y (7.7) 0.64p 12=
β = =β
p
n k kk 1
2a y cos nxp =
β β hence, 12a (9.49) 1.58
12β = , 2
2a ( 1.35) 0.2312
β β = β , 32a (1.6) 0.27
12β =
p
n k kk 1
2b y sin nxp =
β β hence, 12b ( 16.389) 2.73
12β β = β , 2
2b ( 2.522) 0.4212
β β = β ,
32b (0.3) 0.05
12β =
Substituting these values into the Fourier series: ( )0 n nn 1
f (x) a a cos nx b sin nxβ
=
= + +β
gives: i = 0.64 + 1.58 cos ΞΈ - 0.23 cos 2ΞΈ + 0.27 cos 3ΞΈ + β¦β¦
- 2.73 sin ΞΈ - 0.42 sin 2ΞΈ + 0.05 sin 3ΞΈ
or i = 0.64 + 1.58 cos ΞΈ - 2.73 sin ΞΈ - 0.23 cos 2ΞΈ - 0.42 sin 2ΞΈ + 0.27 cos 3ΞΈ + 0.05 sin 3ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 625
EXERCISE 247 Page 688
2. Analyse the periodic waveform of displacement y against angle ΞΈ in the diagram below into its
constituent harmonics as far as and including the third harmonic, by taking 30Β° intervals.
ΞΈΒ° y
cosΞΈ y cos ΞΈ sinΞΈ y sin ΞΈ cos 2ΞΈ y cos 2ΞΈ sin 2ΞΈ y sin 2ΞΈ cos 3ΞΈ y cos3ΞΈ sin3ΞΈ y cos3ΞΈ
30 10
60 -6
90 -17
120 -17
150 -13
180 -4
210 10
240 24
270 33
300 36
330 33
360 24
0.866
0.5
0
-0.5
-0.866
-1
-0.866
-0.5
0
0.5
0.866
1
8.66
-3.0
0
8.5
11.26
4
-8.66-
12
0
18
28.58
24
0.5
0.866
1
0.866
0.5
0
-0.5
-0.866
-1
-0.866
-0.5
0
5
-5.196
-17
-14.72
-6.5
0
-5.0
-20.78
-33
-31.18
-16.5
0
0.5
-0.5
-1
-0.5
0.5
1
0.5
-0.5
-1
-0.5
0.5
1
5
3
17
8.5
-6.5
-4
5
-12
-33
-18
16.5
24
0.866
0.866
0
-0.866
-0.866
0
0.866
0.866
0
-0.866
-0.866
0
8.66
-5.196
0
14.722
11.258
0
8.66
20.784
0
-31.176
-28.578
0
0
-1
0
1
0
-1
0
1
0
-1
0
1
0
6
0
-17
0
4
0
24
0
-36
0
24
1
0
-1
0
1
0
-1
0
1
0
-1
0
10
0
17
0
-13
0
-10
0
33
0
-33
0
12
kk 1
y 113=
=β
12
k kk 1
y cos 79.34=
ΞΈ =β
12
k kk 1
y sin
144.88=
ΞΈ
= β
β
12
k kk 1
y cos 2 5.5=
ΞΈ =β
12
k kk 1
y sin 2 0.866=
ΞΈ = ββ
12
k kk 1
y cos3 5=
ΞΈ =β
12
k kk 1
y sin 3 4=
ΞΈ =β
p
0 kk 1
1 1a y (113) 9.4p 12=
β = =β
p
n k kk 1
2a y cos nxp =
β β hence, 12a (79.34) 13.2
12β = , 2
2a (5.5) 0.9212
β = , 32a (5) 0.83
12β =
p
n k kk 1
2b y sin nxp =
β β hence, 12b ( 144.88) 24.1
12β β = β , 2
2b ( 0.866) 0.1412
β β = β ,
32b (4) 0.67
12β =
Substituting these values into the Fourier series: ( )0 n nn 1
f (x) a a cos nx b sin nxβ
=
= + +β
gives: y = 9.4 + 13.2 cos ΞΈ + 0.92 cos 2ΞΈ + 0.83 cos 3ΞΈ + β¦β¦
- 24.1 sin ΞΈ - 0.14 sin 2ΞΈ + 0.67 sin 3ΞΈ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 626
or y = 9.4 + 13.2 cos ΞΈ - 24.1 sin ΞΈ + 0.92 cos 2ΞΈ - 0.14 sin 2ΞΈ + 0.83 cos 3ΞΈ + 0.67 sin 3ΞΈ
3. For the waveform of current shown below state why only a d.c. component and even cosine
terms will appear in the Fourier series and determine the series, using Ο/6 intervals, up to and
including the sixth harmonic.
The function is even, thus no sine terms will be present.
The function repeats itself every half cycle, hence only even terms will be present.
Hence, the Fourier series will contains a d.c. component and even cosine terms only.
ΞΈΒ° i
cos 2ΞΈ i cos 2ΞΈ cos 4ΞΈ i cos 4ΞΈ cos 6ΞΈ i cos 6ΞΈ
30 1.5
60 5.5
90 10
120 5.5
150 1.5
180 0
210 1.5
240 5.5
270 10
300 5.5
330 1.5
360 0
0.5
-0.5
-1
-0.5
0.5
1
0.5
-0.5
-1
-0.5
0.5
1
0.75
-2.75
-10
-2.75
0.75
0
0.75
-2.75
-10
-2.75
0.75
0
-0.5
-0.5
1
-0.5
-0.5
1
-0.5
-0.5
1
-0.5
-0.5
1
-0.75
-2.75
10
-2.75
-0.75
0
-0.75
-2.75
10
-2.75
-0.75
0
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1.5
5.5
-10
5.5
-1.5
0
-1.5
5.5
-10
5.5
-1.5
0
12
kk 1
y 48=
=β
12
k kk 1
y cos 2 28=
ΞΈ = ββ
12
k kk 1
y cos 4 6=
ΞΈ =β
12
k kk 1
y cos 6 4=
ΞΈ = ββ
p
0 kk 1
1 1a y (48) 4.00p 12=
β = =β
p
n k kk 1
2a y cos nxp =
β β hence, 22a ( 28) 4.67
12β β = β , 4
2a (6) 1.0012
β = , 32a ( 4) 0.66
12β β = β
Substituting these values into the Fourier series: ( )0 nn 1
f (x) a a cos nxβ
=
= +β
gives: i = 4.00 β 4.67 cos 2ΞΈ + 1.00 cos 4ΞΈ - 0.66 cos 6ΞΈ + β¦β¦
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 627
CHAPTER 74 THE COMPLEX OR EXPONENTIAL FORM OF A
FOURIER SERIES EXERCISE 248 Page 694
1. Determine the complex Fourier series for the function defined by:
f(t) =0, when t 02, when 0 t
β Ο β€ β€β§β¨ β€ β€ Οβ©
The function is periodic outside of this range of period 2Ο.
The periodic function is shown in the diagram below.
The complex Fourier series is given by: 2 n tj
Ln
nf (t) c e
Οβ
=ββ
= β
where 2 n tL / 2 j
Ln L / 2
1c f (t) e dtL
Οβ
β= β«
i.e. 2 n t jn tj0 jn t jn 02
n 0 00
1 1 1 e 1c 0dt 2e dt e e e2 jn j n
ΟΟ ββΟ Ο β β ΟΟβΟ
β§ β« β‘ β€β‘ β€= + = = = β ββ¨ β¬ β’ β₯ β£ β¦Ο Ο Ο β Οβ£ β¦β© β
β« β« β«
= ( )jn jn2
j je 1 e 1j n n
β Ο β Οβ‘ β€β β = ββ£ β¦Ο Ο
= [ ] [ ]j jcos n jsin n 1 cos n 1n n
Οβ Οβ = ΟβΟ Ο
for all
integer values of n
Hence, ( )2 ntj jntL
nn n
jf (t) c e cosn 1 en
Οβ β
=β β =β β
= = Ο βΟβ β
0 02c a mean value 12ΓΟ
= = = =Ο
1j 2c ( 1 1) j= β β = βΟ Ο
, 2jc (1 1) 0
2= β =
Ο and all even terms will be zero
32c j
3= β
Ο, 5
2c j5
= βΟ
, and so on.
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 628
1j 2c ( 2) jβ = β =βΟ Ο
, 3j 2c ( 2) j3 3β = β =β Ο Ο
, 52c j
5=
Ο , and so on.
Thus, f(t) = 1 jt j3t j5t jt j3t j5t2 2 2 2 2 2j e j e j e .... j e j e j e3 5 3 5
β β ββ β β β + + +Ο Ο Ο Ο Ο Ο
i.e. jt j3t j5t jt j3t j5t2 1 1 2 1 1f (t) 1 j e e e .... j e e e ....3 5 3 5
β β ββ β β β= β + + + + + + +β β β βΟ Οβ β β β
2. Show that the complex Fourier series for the waveform shown below, that has period 2, may be
represented by: ( ) j n t
n(n 0)
j2f (t) 2 cos n 1 en
βΟ
=β ββ
= + ΟβΟβ
The complex Fourier series is given by: 2 n tj
Ln
n
f (t) c eΟβ
=β β
= β
where 2 n tL / 2 j
Ln L / 2
1c f (t) e dtL
Οβ
β= β«
i.e. 12 n t j n t1 1j j n t j n 02
n 0 00
1 e 2c 4e 2 e 2 e e2 j n j n
Ο β Οβ β Ο β Οβ§ β« β‘ β€β‘ β€= = = = β ββ¨ β¬ β’ β₯ β£ β¦β Ο Οβ£ β¦β© β
β« β«
= ( )jn jn2 j2e 1 e 1j n n
β Ο β Οβ‘ β€β β = ββ£ β¦Ο Ο
= [ ] [ ]j2 j2cos n jsin n 1 cos n 1n n
Οβ Οβ = ΟβΟ Ο
for all
integer values of n
[ ]1 1
0 0 00
1 1c a mean value 4dt 4t 22 2
= = = = =β«
Hence, ( ) j nt
n(n 0)
j2f (t) 2 cosn 1 en
βΟ
=βββ
= + Ο βΟβ
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 629
3. Show that the complex Fourier series of Problem 2 is equivalent to:
8 1 1f (t) 2 sin t sin 3 t sin 5 t ....3 5
β β= + Ο + Ο + Ο +β βΟ β β
1j2 4c ( 1 1) j(1)
= β β = βΟ Ο
, 2j2c (1 1) 02
= β =Ο
and all even terms will be zero
32 4c j ( 2) j
3 3= β = β
Ο Ο, 5
4c j5
= βΟ
, and so on.
1j2 4c ( 2) jβ = β =βΟ Ο
, 3j2 4c ( 2) j3 3β = β =β Ο Ο
, 54c j
5=
Ο , and so on.
Thus, ( ) j n t
n(n 0)
j2f (t) 2 cos n 1 en
βΟ
=β ββ
= + ΟβΟβ
i.e. f(t) = 2 j t j3 t j5 t j t j3 t j5 t4 4 4 4 4 4j e j e j e .... j e j e j e3 5 3 5
Ο Ο Ο β Ο β Ο β Οβ β β β + + +Ο Ο Ο Ο Ο Ο
i.e. j t j3 t j5 t j t j3 t j5 t4 1 1 4 1 1f (t) 2 j e e e .... j e e e ....3 5 3 5
Ο Ο Ο β Ο β Ο β Οβ β β β= β + + + + + + +β β β βΟ Οβ β β β
= ( ) ( ) ( )j t j t j3 t j3 t j5 t j5 t4 1 12 j e e e e e e ....3 5
Ο β Ο Ο β Ο Ο β Οβ‘ β€β β + β + β +β’ β₯Ο β£ β¦
= j t j t j3 t j3 t j5 t j5 t
2 8 e e 1 e e 1 e e2 j ....2 j 3 2j 5 2j
Ο β Ο Ο β Ο Ο β Οβ‘ β€β β β β β ββ β ββ + + +β’ β₯β β β β β βΟ β β β β β β β£ β¦
i.e. 8 1 1f (t) 2 sin t sin 3 t sin 5 t ....3 5
β‘ β€= + Ο + Ο + Ο +β’ β₯Ο β£ β¦
4. Determine the exponential form of the Fourier series for the function defined by:
f(t) = e2t when β 1 < t < 1 and has period 2.
The function is shown in the diagram below.
The complex Fourier series is given by: 2 n tj
Ln
nf (t) c e
Οβ
=ββ
= β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 630
where 2 n tL / 2 j
Ln L / 2
1c f (t) e dtL
Οβ
β= β«
i.e. ( ) ( ) ( )1t 2 j n 2 j n 2 j n2 n t1 1j2t 2t j n t2
n 1 11
1 1 1 e 1 e ec e e dt e dt2 2 2 2 j n 2 2 j n
β Ο β Ο β β ΟΟβ β Ο
β ββ
β‘ β€ β‘ β€β§ β« β= = = =β¨ β¬ β’ β₯ β’ β₯β Ο β Οβ© β β£ β¦ β£ β¦
β« β«
Thus, f(t) =2 n tj
Ln
n
c eΟβ
=β ββ =
(2 j n) (2 j n)j nt
n
1 e e e2 2 j n
β Ο β β ΟβΟ
=β β
β βββ ββ Οβ β
β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 631
EXERCISE 249 Page 698
1. Determine the exponential form of the Fourier series for the periodic function defined by:
f(x) =
2, when x2
2, when x2 2
2, when x2
Οβ§ β β Ο β€ β€ ββͺβͺ
Ο Οβͺ β β€ β€ +β¨βͺ
Οβͺ β + β€ β€ +Οβͺβ©
and which has a period of 2Ο.
The periodic waveform is shown below. It is an even function and contains no sine terms, hence
nb 0= and between -Ο and +Ο, the mean value is zero, hence 0a 0= .
L / 2
n 0 0
2 2 nx 2 2 nxc f (x)cos dx f (x)cos dxL L 2 2
ΟΟ Οβ β β β= =β β β βΟ Οβ β β β β« β« since L = 2Ο
= / 2
0 / 2
1 2cos nx dx 2cos nx dxΟ Ο
Ο+ β
Ο β« β«
= / 2
0 / 2
1 2sin nx 2sin nxn n
Ο Ο
Ο
β§ β«βͺ βͺβ‘ β€ β‘ β€ββ¨ β¬β’ β₯ β’ β₯Ο β£ β¦ β£ β¦βͺ βͺβ© β
= 1 n n 4 n2sin 0 2sin n 2sin sinn 2 2 n 2β‘ Ο Ο β€ Οβ β β ββ β Οβ =β β β ββ’ β₯Ο Οβ β β β β£ β¦
Hence, f(x) = 2 n xj
Ln
n
c eΟβ
=β ββ = jnx
n
4 nsin en 2
β
=β β
β§ β«Οβ ββ¨ β¬β βΟ β β β© β
β
2. Show that the exponential form of the Fourier series in Problem 1 above is equivalent to:
8 1 1 1f (x) cos x cos3x cos5x cos 7x ....3 5 7
β β= β + β +β βΟ β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 632
Since from Problem 1, n4 nc sinn 2
Ο=Ο
, then
0c 0= , 14 4c sin
2Ο
= =Ο Ο
, 2 4 6 2 44 2c sin 0 c c c c andsoon
2 2 β β
Ο= = = = = =
Ο,
34 3 4c sin
3 2 3Ο
= = βΟ Ο
, 54c
5=
Ο, 7
4c7
= βΟ
and so on.
14 4c sin
2β
βΟ= β =
Ο Ο, 3
4 3 4c sin3 2 3β
β Ο= β = β
Ο Ο, 5
4c5β =Ο
and so on.
f(x) = 2 n xj
Ln
n
c eΟβ
=β ββ = jn x
n
4 nsin en 2
β
=β β
β§ Ο β«β ββ¨ β¬β βΟ β β β© β
β
= jx j3x j5x jx j3x j5x4 4 4 4 4 4e e e ..... e e e .....3 5 3 5
β β ββ + + + β + βΟ Ο Ο Ο Ο Ο
= jx jx j3x j3x j5x j5x4 4 4 4 4 4e e e e e e .....3 3 5 5
β β ββ β β β β β+ β + + + ββ β β β β βΟ Ο Ο Ο Ο Οβ β β β β β
= jx jx j3x j3x j5x j5x8 e e 8 e e 8 e e .....
2 3 2 5 2
β β ββ β β β β β+ + +β + ββ β β β β βΟ Ο Οβ β β β β β
= 8 8 8cos x cos3x cos5x .....3 5
β + βΟ Ο Ο
i.e. 8 1 1 1f (x) cos x cos 3x cos 5x cos 7x .....3 5 7
β β= β + β +β βΟ β β
3. Determine the complex Fourier series to represent the function f(t) = 2t in the range - Ο to + Ο.
The triangular waveform shown below is an odd function since it is symmetrical about the origin.
The period of the waveform, L = 2Ο.
Thus, nc = L2
0
2 2 ntj f (t) sin dtL L
Οβ ββ β ββ β β«
= 0 0
2 2 nt 2j 2t sin dt j t sin nt dt2 2
Ο ΟΟβ ββ = ββ βΟ Ο Οβ β β« β«
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 633
= ( )2 20
2 t cos nt sin nt 2 cos n sin nj j 0 0n n n n
Οβ β‘ βΟ Ο Ο β€β‘ β€ β ββ + = β + β +β ββ’ β₯β’ β₯Ο Οβ£ β¦ β β β£ β¦ by parts
i.e. n2c j cos nn
= Ο
Hence, the complex Fourier series is given by:
f(t) = 2 ntj
Ln
n
c eΟβ
=β ββ = jnt
n
j2 cosn en
β
=β β
β βΟβ ββ β
β
4. Show that the complex Fourier series is Problem 3 above is equivalent to:
f(t) = 1 1 14 sin t sin 2t sin 3t sin 4t ....2 3 4
β ββ + β +β ββ β
From Problem 3 above, n2c j cos nn
= Ο
When n = 1, ( )12 2 j2c j cos j 1
(1) (1) 1= Ο = β = β
When n = 2, 22 2c j cos 2 j2 2
= Ο =
When n = 3, ( )32 2 j2c j cos3 j 13 3 3
= Ο = β = βΟ
By similar reasoning, 4 5j2 j2c , c4 5
= = β , and so on.
When n = -1, ( )12 2 j2c j cos( ) j 1
( 1) ( 1) 1β = βΟ = + β =β β
When n = -2, ( )22 2 j2c j cos( 2 ) j 1
( 2) ( 2) 2β = β Ο = = ββ β
By similar reasoning, 3 42 j2c j , c3 4β β= = β , and so on.
Since the waveform is odd, 0 0c a 0= =
From Problem 3, f(t) = 2 ntj
Ln
nc e
Οβ
=βββ = jn t
n
j2 cos n en
β
=ββ
β βΟβ ββ β
β
Hence, f(t) = jt j2t j3t j4 tj2 j2 j2 j2e e e e ...1 2 3 4
β + β + β + jt j2 t j3t j4tj2 j2 j2 j2e e e e ...1 2 3 4
β β β ββ + β +
= jt jt j2 t j2 t j3t j3tj2 j2 j2 j2 j2 j2e e e e e e ....1 1 2 2 3 3
β β ββ β β β β ββ + + β + β + +β β β β β ββ β β β β β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 634
= jt jt j2t j2 t j3t j3te e j4 e e j4 e ej4 ....
2 2 2 3 2
β β ββ β β β β ββ β ββ + β +β β β β β β
β β β β β β
= jt jt 2 j2t j2 t 2 j3t j3t
2 e e j 4 e e j 4 e ej 4 ....2 j 2 2 j 3 2j
β β ββ β β β β ββ β ββ + β +β β β β β β
β β β β β β by multiplying top and
bottom by j
= 4 44sin t sin 2t sin 3t ....2 3
β + +
i.e. f(t) = 1 1 14 sin t sin 2t sin 3t sin 4t ...2 3 4
β ββ + β +β ββ β
Hence, f(t) = jnt
n
j2 cosn en
β
=ββ
β βΟβ ββ β
β β‘ 1 1 14 sin t sin 2t sin 3t sin 4t ...2 3 4
β ββ + β +β ββ β
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 635
EXERCISE 250 Page 703
2. Determine the pair of phasors that can represent the harmonic given by:
v = 10 cos 2t β 12 sin 2t
v = 10 cos 2t β 12 sin 2t
= ( ) ( )j2 t j2 t j2 t j2t1 110 e e 12 e e2 2j
β ββ‘ β€β‘ β€+ β ββ’ β₯β’ β₯β£ β¦ β£ β¦
= j2 t j2 t j2 t j2 t6 65e 5e e ej j
β β+ β +
= j2 t j2 t j2 t j2 t5e 5e 6je 6jeβ β+ + β (note: 1 j 1 jorj 1 j 1
β= =
β)
i.e. v = ( ) ( )j2t j2t5 j6 e 5 j6 eβ+ + β
Hence, v = 7.81β 0.88 rad, rotating anticlockwise with an angular velocity, Ο = 2 rad/s
and v = 7.81β -0.88 rad, rotating clockwise with an angular velocity, Ο = 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
= ( ) ( ) ( ) ( )j t j t j t jt jt jt jt jt1 1 36 e e 4 e e e e 2 e e2j 2 j
β β β ββ‘ β€ β‘ β€β + + = β + +β’ β₯ β’ β₯β£ β¦β£ β¦
= ( ) ( )jt jt jt jt3j e e 2 e eβ ββ β + +
i.e. i = ( ) ( )j t j t2 j3 e 2 j3 eββ + +
Β© 2006 John Bird. All rights reserved. Published by Elsevier. 636
Hence, i = 3.61β -0.98 rad, rotating anticlockwise with an angular velocity, Ο = 1 rad/s
and i = 3.61β 0.98 rad, rotating clockwise with an angular velocity, Ο = 1 rad/s, as
shown in the diagram below.