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Higher Engineering Mathematics

In memory of Elizabeth

Higher Engineering Mathematics
Sixth Edition
John Bird, BSc (Hons), CMath, CEng, CSci, FIMA, FIET, MIEE, FIIE, FCollT

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD
PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO
Newnes is an imprint of Elsevier

Newnes is an imprint of Elsevier
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First edition 2010
Copyright © 2010, John Bird, Published by Elsevier Ltd. All rights reserved.
The right of John Bird to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988.
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A catalogue record for this book is available from the British Library.
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A catalogue record for this book is available from the Library of Congress.
ISBN: 978-1-85-617767-2

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Printed and bound in China
10 11 12 13 14 15 10 9 8 7 6 5 4 3 2 1

Contents
Preface

xiii

Syllabus guidance

xv

1

6

1
1
1
3
6
8
10

2

3

4

Algebra
1.1 Introduction
1.2 Revision of basic laws
1.3 Revision of equations
1.4 Polynomial division
1.5 The factor theorem
1.6 The remainder theorem
Partial fractions
2.1 Introduction to partial fractions
2.2 Worked problems on partial fractions with linear factors
2.3 Worked problems on partial fractions with repeated linear factors
2.4 Worked problems on partial fractions with quadratic factors

17

Logarithms
3.1 Introduction to logarithms
3.2 Laws of logarithms
3.3 Indicial equations
3.4 Graphs of logarithmic functions

20
20
22
24
25

Exponential functions
4.1 Introduction to exponential functions
4.2 The power series for e x
4.3 Graphs of exponential functions
4.4 Napierian logarithms
4.5 Laws of growth and decay
4.6 Reduction of exponential laws to linear form

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34

Revision Test 1

5

Hyperbolic functions
5.1 Introduction to hyperbolic functions
5.2 Graphs of hyperbolic functions
5.3 Hyperbolic identities
5.4 Solving equations involving hyperbolic functions 5.5 Series expansions for cosh x and sinh x

13
13

7

13
16

37
40

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49

Arithmetic and geometric progressions
6.1 Arithmetic progressions
6.2 Worked problems on arithmetic progressions 6.3 Further worked problems on arithmetic progressions 6.4 Geometric progressions
6.5 Worked problems on geometric progressions 6.6 Further worked problems on geometric progressions The binomial series
7.1 Pascal’s triangle
7.2 The binomial series
7.3 Worked problems on the binomial series
7.4 Further worked problems on the binomial series 7.5 Practical problems involving the binomial theorem Revision Test 2

8

Maclaurin’s series
8.1 Introduction
8.2 Derivation of Maclaurin’s theorem
8.3 Conditions of Maclaurin’s series
8.4 Worked problems on Maclaurin’s series
8.5 Numerical integration using Maclaurin’s series 8.6 Limiting values
9 Solving equations by iterative methods
9.1 Introduction to iterative methods
9.2 The bisection method
9.3 An algebraic method of successive approximations 9.4 The Newton-Raphson method

10 Binary, octal and hexadecimal
10.1 Introduction
10.2 Binary numbers
10.3 Octal numbers
10.4 Hexadecimal numbers
Revision Test 3

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vi Contents
11 Introduction to trigonometry
11.1 Trigonometry
11.2 The theorem of Pythagoras
11.3 Trigonometric ratios of acute angles
11.4 Evaluating trigonometric ratios
11.5 Solution of right-angled triangles
11.6 Angles of elevation and depression
11.7 Sine and cosine rules
11.8 Area of any triangle
11.9 Worked problems on the solution of triangles and finding their areas
11.10 Further worked problems on solving triangles and finding their areas
11.11 Practical situations involving trigonometry 11.12 Further practical situations involving trigonometry 97
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12 Cartesian and polar co-ordinates
12.1 Introduction
12.2 Changing from Cartesian into polar co-ordinates 12.3 Changing from polar into Cartesian co-ordinates 12.4 Use of Pol/Rec functions on calculators

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117

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13 The circle and its properties
13.1 Introduction
13.2 Properties of circles
13.3 Radians and degrees
13.4 Arc length and area of circles and sectors
13.5 The equation of a circle
13.6 Linear and angular velocity
13.7 Centripetal force

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Revision Test 4

109

15.5 Worked problems (ii) on trigonometric equations 15.6 Worked problems (iii) on trigonometric equations 15.7 Worked problems (iv) on trigonometric equations 16 The relationship between trigonometric and hyperbolic functions
16.1 The relationship between trigonometric and hyperbolic functions
16.2 Hyperbolic identities

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117

133

14 Trigonometric waveforms
14.1 Graphs of trigonometric functions
14.2 Angles of any magnitude
14.3 The production of a sine and cosine wave
14.4 Sine and cosine curves
14.5 Sinusoidal form A sin(ωt ± α)
14.6 Harmonic synthesis with complex waveforms 134
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15 Trigonometric identities and equations
15.1 Trigonometric identities
15.2 Worked problems on trigonometric identities 15.3 Trigonometric equations
15.4 Worked problems (i) on trigonometric equations 152
152

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17 Compound angles
17.1 Compound angle formulae
17.2 Conversion of a sinωt + b cosωt into
R sin(ωt + α)
17.3 Double angles
17.4 Changing products of sines and cosines into sums or differences
17.5 Changing sums or differences of sines and cosines into products
17.6 Power waveforms in a.c. circuits
Revision Test 5

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18 Functions and their curves
18.1 Standard curves
18.2 Simple transformations
18.3 Periodic functions
18.4 Continuous and discontinuous functions
18.5 Even and odd functions
18.6 Inverse functions
18.7 Asymptotes
18.8 Brief guide to curve sketching
18.9 Worked problems on curve sketching

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19 Irregular areas, volumes and mean values of waveforms 19.1 Areas of irregular figures
19.2 Volumes of irregular solids
19.3 The mean or average value of a waveform

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Revision Test 6

20 Complex numbers
20.1 Cartesian complex numbers
20.2 The Argand diagram
20.3 Addition and subtraction of complex numbers 20.4 Multiplication and division of complex numbers 212

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vii

Contents
20.5
20.6
20.7
20.8

Complex equations
The polar form of a complex number
Multiplication and division in polar form
Applications of complex numbers

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21 De Moivre’s theorem
21.1 Introduction
21.2 Powers of complex numbers
21.3 Roots of complex numbers
21.4 The exponential form of a complex number 225
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226

22 The theory of matrices and determinants
22.1 Matrix notation
22.2 Addition, subtraction and multiplication of matrices
22.3 The unit matrix
22.4 The determinant of a 2 by 2 matrix
22.5 The inverse or reciprocal of a 2 by 2 matrix
22.6 The determinant of a 3 by 3 matrix
22.7 The inverse or reciprocal of a 3 by 3 matrix

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23 The solution of simultaneous equations by matrices and determinants
23.1 Solution of simultaneous equations by matrices 23.2 Solution of simultaneous equations by determinants 23.3 Solution of simultaneous equations using
Cramers rule
23.4 Solution of simultaneous equations using the Gaussian elimination method
Revision Test 7

24 Vectors
24.1
24.2
24.3
24.4
24.5
24.6
24.7
24.8
24.9

Introduction
Scalars and vectors
Drawing a vector
Addition of vectors by drawing
Resolving vectors into horizontal and vertical components
Addition of vectors by calculation
Vector subtraction
Relative velocity i, j and k notation

25 Methods of adding alternating waveforms
25.1 Combination of two periodic functions
25.2 Plotting periodic functions
25.3 Determining resultant phasors by drawing

228

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25.4 Determining resultant phasors by the sine and cosine rules
268
25.5 Determining resultant phasors by horizontal and vertical components
270
25.6 Determining resultant phasors by complex numbers 272
26 Scalar and vector products
26.1 The unit triad
26.2 The scalar product of two vectors
26.3 Vector products
26.4 Vector equation of a line
Revision Test 8

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280
283
286

27 Methods of differentiation
27.1 Introduction to calculus
27.2 The gradient of a curve
27.3 Differentiation from first principles
27.4 Differentiation of common functions
27.5 Differentiation of a product
27.6 Differentiation of a quotient
27.7 Function of a function
27.8 Successive differentiation

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28 Some applications of differentiation
28.1 Rates of change
28.2 Velocity and acceleration
28.3 Turning points
28.4 Practical problems involving maximum and minimum values
28.5 Tangents and normals
28.6 Small changes

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29 Differentiation of parametric equations
29.1 Introduction to parametric equations
29.2 Some common parametric equations
29.3 Differentiation in parameters
29.4 Further worked problems on differentiation of parametric equations

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315

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30 Differentiation of implicit functions
30.1 Implicit functions
30.2 Differentiating implicit functions
30.3 Differentiating implicit functions containing products and quotients
30.4 Further implicit differentiation

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31 Logarithmic differentiation
31.1 Introduction to logarithmic differentiation
31.2 Laws of logarithms
31.3 Differentiation of logarithmic functions

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318

viii Contents
31.4 Differentiation of further logarithmic functions 31.5 Differentiation of [ f (x)]x
Revision Test 9
32 Differentiation of hyperbolic functions
32.1 Standard differential coefficients of hyperbolic functions
32.2 Further worked problems on differentiation of hyperbolic functions
33 Differentiation of inverse trigonometric and hyperbolic functions
33.1 Inverse functions
33.2 Differentiation of inverse trigonometric functions 33.3 Logarithmic forms of the inverse hyperbolic functions
33.4 Differentiation of inverse hyperbolic functions 34 Partial differentiation
34.1 Introduction to partial derivatives
34.2 First order partial derivatives
34.3 Second order partial derivatives
35 Total differential, rates of change and small changes 35.1 Total differential
35.2 Rates of change
35.3 Small changes
36 Maxima, minima and saddle points for functions of two variables
36.1 Functions of two independent variables
36.2 Maxima, minima and saddle points
36.3 Procedure to determine maxima, minima and saddle points for functions of two variables 36.4 Worked problems on maxima, minima and saddle points for functions of two variables 36.5 Further worked problems on maxima, minima and saddle points for functions of two variables
Revision Test 10
37 Standard integration
37.1 The process of integration
37.2 The general solution of integrals of the form ax n
37.3 Standard integrals
37.4 Definite integrals

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38 Some applications of integration
38.1 Introduction
38.2 Areas under and between curves
38.3 Mean and r.m.s. values
38.4 Volumes of solids of revolution
38.5 Centroids
38.6 Theorem of Pappus
38.7 Second moments of area of regular sections 375
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39 Integration using algebraic substitutions
39.1 Introduction
39.2 Algebraic substitutions
39.3 Worked problems on integration using algebraic substitutions
39.4 Further worked problems on integration using algebraic substitutions
39.5 Change of limits

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Revision Test 11
40 Integration using trigonometric and hyperbolic substitutions 40.1 Introduction
40.2 Worked problems on integration of sin2 x, cos2 x, tan2 x and cot2 x
40.3 Worked problems on powers of sines and cosines 40.4 Worked problems on integration of products of sines and cosines
40.5 Worked problems on integration using the sin θ substitution
40.6 Worked problems on integration using tan θ substitution
40.7 Worked problems on integration using the sinh θ substitution
40.8 Worked problems on integration using the cosh θ substitution

397

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41 Integration using partial fractions
41.1 Introduction
41.2 Worked problems on integration using partial fractions with linear factors
41.3 Worked problems on integration using partial fractions with repeated linear factors 41.4 Worked problems on integration using partial fractions with quadratic factors

412

θ
42 The t = tan substitution
2
42.1 Introduction

414
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θ
42.2 Worked problems on the t = tan
2
substitution

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411

415

Contents θ 42.3 Further worked problems on the t = tan
2
substitution
Revision Test 12

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43 Integration by parts
43.1 Introduction
43.2 Worked problems on integration by parts
43.3 Further worked problems on integration by parts

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44 Reduction formulae
44.1 Introduction
44.2 Using reduction formulae for integrals of the form x n e x dx
44.3 Using reduction formulae for integrals of the form x n cos x dx and x n sin x dx
44.4 Using reduction formulae for integrals of the form sinn x dx and cosn x dx
44.5 Further reduction formulae

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45 Numerical integration
45.1 Introduction
45.2 The trapezoidal rule
45.3 The mid-ordinate rule
45.4 Simpson’s rule

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Revision Test 13

46 Solution of first order differential equations by separation of variables
46.1 Family of curves
46.2 Differential equations
46.3 The solution of equations of the form dy = f (x) dx 46.4 The solution of equations of the form dy = f (y) dx 46.5 The solution of equations of the form dy = f (x) · f (y) dx 47 Homogeneous first order differential equations
47.1 Introduction
47.2 Procedure to solve differential equations dy =Q of the form P dx 47.3 Worked problems on homogeneous first order differential equations
47.4 Further worked problems on homogeneous first order differential equations

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48 Linear first order differential equations
48.1 Introduction
48.2 Procedure to solve differential equations dy + Py = Q of the form dx 48.3 Worked problems on linear first order differential equations
48.4 Further worked problems on linear first order differential equations
49 Numerical methods for first order differential equations 49.1 Introduction
49.2 Euler’s method
49.3 Worked problems on Euler’s method
49.4 An improved Euler method
49.5 The Runge-Kutta method
Revision Test 14
50 Second order differential equations of the form dy d2 y a 2 + b + cy= 0 dx dx
50.1 Introduction
50.2 Procedure to solve differential equations dy d2 y of the form a 2 + b + cy = 0 dx dx
50.3 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = 0 dx dx
50.4 Further worked problems on practical differential equations of the form dy d2 y a 2 + b + cy = 0 dx dx
51 Second order differential equations of the form dy d2 y a 2 + b + cy= f (x) dx dx
51.1 Complementary function and particular integral 51.2 Procedure to solve differential equations d2 y dy of the form a 2 + b + cy = f (x) dx dx
51.3 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is a constant or polynomial
51.4 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is an exponential function
51.5 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is a sine or cosine function

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ix

x Contents
51.6 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is a sum or a product
490
52 Power series methods of solving ordinary differential equations
52.1 Introduction
52.2 Higher order differential coefficients as series 52.3 Leibniz’s theorem
52.4 Power series solution by the
Leibniz–Maclaurin method
52.5 Power series solution by the Frobenius method 52.6 Bessel’s equation and Bessel’s functions
52.7 Legendre’s equation and Legendre polynomials 53 An introduction to partial differential equations
53.1 Introduction
53.2 Partial integration
53.3 Solution of partial differential equations by direct partial integration
53.4 Some important engineering partial differential equations
53.5 Separating the variables
53.6 The wave equation
53.7 The heat conduction equation
53.8 Laplace’s equation
Revision Test 15

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58 The normal distribution
58.1 Introduction to the normal distribution
58.2 Testing for a normal distribution

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59 Linear correlation
59.1 Introduction to linear correlation
59.2 The product-moment formula for determining the linear correlation coefficient 59.3 The significance of a coefficient of correlation 59.4 Worked problems on linear correlation

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60 Linear regression
60.1 Introduction to linear regression
60.2 The least-squares regression lines
60.3 Worked problems on linear regression

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576

Revision Test 17

570
571
571

581

516
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54 Presentation of statistical data
54.1 Some statistical terminology
54.2 Presentation of ungrouped data
54.3 Presentation of grouped data

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55 Measures of central tendency and dispersion
55.1 Measures of central tendency
55.2 Mean, median and mode for discrete data
55.3 Mean, median and mode for grouped data
55.4 Standard deviation
55.5 Quartiles, deciles and percentiles

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56 Probability
56.1 Introduction to probability
56.2 Laws of probability
56.3 Worked problems on probability
56.4 Further worked problems on probability

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551

Revision Test 16

57 The binomial and Poisson distributions
57.1 The binomial distribution
57.2 The Poisson distribution

554

61 Introduction to Laplace transforms
61.1 Introduction
61.2 Definition of a Laplace transform
61.3 Linearity property of the Laplace transform 61.4 Laplace transforms of elementary functions 61.5 Worked problems on standard Laplace transforms 582
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62 Properties of Laplace transforms
62.1 The Laplace transform of eat f (t)
62.2 Laplace transforms of the form eat f (t)
62.3 The Laplace transforms of derivatives
62.4 The initial and final value theorems

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63 Inverse Laplace transforms
63.1 Definition of the inverse Laplace transform
63.2 Inverse Laplace transforms of simple functions 63.3 Inverse Laplace transforms using partial fractions 63.4 Poles and zeros

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64 The solution of differential equations using
Laplace transforms
64.1 Introduction
64.2 Procedure to solve differential equations by using Laplace transforms
64.3 Worked problems on solving differential equations using Laplace transforms

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Contents
65 The solution of simultaneous differential equations using Laplace transforms
65.1 Introduction
65.2 Procedure to solve simultaneous differential equations using Laplace transforms 65.3 Worked problems on solving simultaneous differential equations by using Laplace transforms Revision Test 18
66 Fourier series for periodic functions of period 2π
66.1 Introduction
66.2 Periodic functions
66.3 Fourier series
66.4 Worked problems on Fourier series of periodic functions of period 2π
67 Fourier series for a non-periodic function over range 2π
67.1 Expansion of non-periodic functions
67.2 Worked problems on Fourier series of non-periodic functions over a range of 2π
68 Even and odd functions and half-range
Fourier series
68.1 Even and odd functions

68.2 Fourier cosine and Fourier sine series
68.3 Half-range Fourier series

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69 Fourier series over any range
69.1 Expansion of a periodic function of period L
69.2 Half-range Fourier series for functions defined over range L

630

70 A numerical method of harmonic analysis
70.1 Introduction
70.2 Harmonic analysis on data given in tabular or graphical form
70.3 Complex waveform considerations

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71 The complex or exponential form of a
Fourier series
71.1 Introduction
71.2 Exponential or complex notation
71.3 The complex coefficients
71.4 Symmetry relationships
71.5 The frequency spectrum
71.6 Phasors

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Revision Test 19

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658

Essential formulae

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Index

675

xi

xii Contents

Website Chapters

72 Inequalities
72.1 Introduction to inequalities
72.2 Simple inequalities
72.3 Inequalities involving a modulus
72.4 Inequalities involving quotients
72.5 Inequalities involving square functions
72.6 Quadratic inequalities

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73 Boolean algebra and logic circuits
73.1 Boolean algebra and switching circuits
73.2 Simplifying Boolean expressions
73.3 Laws and rules of Boolean algebra
73.4 De Morgan’s laws
73.5 Karnaugh maps
73.6 Logic circuits
73.7 Universal logic gates

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Revision Test 20
74 Sampling and estimation theories
74.1 Introduction
74.2 Sampling distributions

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74.3 The sampling distribution of the means
74.4 The estimation of population parameters based on a large sample size
74.5 Estimating the mean of a population based on a small sample size

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75 Significance testing
75.1 Hypotheses
75.2 Type I and Type II errors
75.3 Significance tests for population means
75.4 Comparing two sample means

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76 Chi-square and distribution-free tests
76.1 Chi-square values
76.2 Fitting data to theoretical distributions
76.3 Introduction to distribution-free tests
76.4 The sign test
76.5 Wilcoxon signed-rank test
76.6 The Mann-Whitney test

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Revision Test 21

82

Preface
This sixth edition of ‘Higher Engineering Mathematics’ covers essential mathematical material suitable for students studying Degrees, Foundation Degrees,
Higher National Certificate and Diploma courses in
Engineering disciplines.
In this edition the material has been ordered into the following twelve convenient categories: number and algebra, geometry and trigonometry, graphs, complex numbers, matrices and determinants, vector geometry, differential calculus, integral calculus, differential equations, statistics and probability, Laplace transforms and
Fourier series. New material has been added on logarithms and exponential functions, binary, octal and hexadecimal, vectors and methods of adding alternating waveforms. Another feature is that a free Internet download is available of a sample (over 1100) of the further problems contained in the book.
The primary aim of the material in this text is to provide the fundamental analytical and underpinning knowledge and techniques needed to successfully complete scientific and engineering principles modules of
Degree, Foundation Degree and Higher National Engineering programmes. The material has been designed to enable students to use techniques learned for the analysis, modelling and solution of realistic engineering problems at Degree and Higher National level. It also aims to provide some of the more advanced knowledge required for those wishing to pursue careers in mechanical engineering, aeronautical engineering, electronics, communications engineering, systems engineering and all variants of control engineering.
In Higher Engineering Mathematics 6th Edition, theory is introduced in each chapter by a full outline of essential definitions, formulae, laws, procedures etc.
The theory is kept to a minimum, for problem solving is extensively used to establish and exemplify the theory.
It is intended that readers will gain real understanding through seeing problems solved and then through solving similar problems themselves.
Access to software packages such as Maple, Mathematica and Derive, or a graphics calculator, will enhance understanding of some of the topics in this text.

Each topic considered in the text is presented in a way that assumes in the reader only knowledge attained in
BTEC National Certificate/Diploma, or similar, in an
Engineering discipline.
‘Higher Engineering Mathematics 6th Edition’ provides a follow-up to ‘Engineering Mathematics 6th
Edition’.
This textbook contains some 900 worked problems, followed by over 1760 further problems (with answers), arranged within 238 Exercises. Some 432 line diagrams further enhance understanding.
A sample of worked solutions to over 1100 of the further problems has been prepared and can be accessed free via the Internet (see next page).
At the end of the text, a list of Essential Formulae is included for convenience of reference.
At intervals throughout the text are some 19 Revision
Tests (plus two more in the website chapters) to check understanding. For example, Revision Test 1 covers the material in Chapters 1 to 4, Revision Test 2 covers the material in Chapters 5 to 7, Revision Test 3 covers the material in Chapters 8 to 10, and so on. An
Instructor’s Manual, containing full solutions to the
Revision Tests, is available free to lecturers adopting this text (see next page).
Due to restriction of extent, five chapters that appeared in the fifth edition have been removed from the text and placed on the website. For chapters on Inequalities, Boolean algebra and logic circuits, Sampling and estimation theories, Significance testing and Chi-square and distribution-free tests (see next page).
‘Learning by example’ is at the heart of ‘Higher
Engineering Mathematics 6th Edition’.

JOHN BIRD
Royal Naval School of Marine Engineering,
HMS Sultan, formerly University of Portsmouth and Highbury College, Portsmouth

xiv Preface
Free web downloads
Extra material available on the Internet at: www.booksite.elsevier.com/newnes/bird. It is recognised that the level of understanding of algebra on entry to higher courses is often inadequate. Since algebra provides the basis of so much of higher engineering studies, it is a situation that often needs urgent attention. Lack of space has prevented the inclusion of more basic algebra topics in this textbook; it is for this reason that some algebra topics – solution of simple, simultaneous and quadratic equations and transposition of formulae – have been made available to all via the Internet. Also included is a Remedial Algebra
Revision Test to test understanding. To access the
Algebra material visit the website.
Five extra chapters
Chapters on Inequalities, Boolean Algebra and logic circuits, Sampling and Estimation theories, Significance testing, and Chi-square and distribution-free tests are available to download at the website.

Sample of worked Solutions to Exercises
Within the text (plus the website chapters) are some 1900 further problems arranged within
260 Exercises. A sample of over 1100 worked solutions has been prepared and can be accessed free via the Internet. To access these worked solutions visit the website.
Instructor’s manual
This provides fully worked solutions and mark scheme for all the Revision Tests in this book
(plus 2 from the website chapters), together with solutions to the Remedial Algebra Revision Test mentioned above. The material is available to lecturers only. To obtain a password please visit the website with the following details: course title, number of students, your job title and work postal address. To download the Instructor’s Manual visit the website and enter the book title in the search box.

Syllabus Guidance
This textbook is written for undergraduate engineering degree and foundation degree courses; however, it is also most appropriate for HNC/D studies and three syllabuses are covered.

The appropriate chapters for these three syllabuses are shown in the table below.
Chapter

Analytical
Methods
for
Engineers

Further
Analytical
Methods for
Engineers

1.

Algebra

×

2.

Partial fractions

×

3.

Logarithms

×

4.

Exponential functions

×

5.

Hyperbolic functions

×

6.

Arithmetic and geometric progressions

×

7.

The binomial series

×

8.

Maclaurin’s series

×

9.

Solving equations by iterative methods

×

10.

Binary, octal and hexadecimal

×

11.

Introduction to trigonometry

×

12.

Cartesian and polar co-ordinates

×

13.

The circle and its properties

×

14.

Trigonometric waveforms

×

15.

Trigonometric identities and equations

×

16.

The relationship between trigonometric and hyperbolic functions ×

17.

Compound angles

×

18.

Functions and their curves

×

19.

Irregular areas, volumes and mean values of waveforms

×

20.

Complex numbers

×

21.

De Moivre’s theorem

×

22.

The theory of matrices and determinants

×

23.

The solution of simultaneous equations by matrices and determinants ×

24.

Vectors

×

25.

Methods of adding alternating waveforms

Engineering
Mathematics

×
(Continued )

xvi Syllabus Guidance
Chapter

Analytical
Methods
for
Engineers

Further
Analytical
Methods for
Engineers

Engineering
Mathematics

×

26.

Scalar and vector products

27.

Methods of differentiation

×

28.

Some applications of differentiation

×

29.

Differentiation of parametric equations

30.

Differentiation of implicit functions

×

31.

Logarithmic differentiation

×

32.

Differentiation of hyperbolic functions

×

33.

Differentiation of inverse trigonometric and hyperbolic functions ×

34.

Partial differentiation

×

35.

Total differential, rates of change and small changes

×

36.

Maxima, minima and saddle points for functions of two variables ×

37.

Standard integration

×

38.

Some applications of integration

×

39.

Integration using algebraic substitutions

×

40.

Integration using trigonometric and hyperbolic substitutions ×

41.

Integration using partial fractions

×

42.

The t = tan θ/2 substitution

43.

Integration by parts

×

44.

Reduction formulae

×

45.

Numerical integration

×

46.

Solution of first order differential equations by separation of variables ×

47.

Homogeneous first order differential equations

48.

Linear first order differential equations

×

49.

Numerical methods for first order differential equations

×

50.

Second order differential equations of the form d2 y dy + cy = 0 a 2 +b dx dx

×

51.

Second order differential equations of the form d2 y dy a 2 +b
+ cy = f (x) dx dx

×

52.

Power series methods of solving ordinary differential equations

×

53.

An introduction to partial differential equations

×

54.

Presentation of statistical data

×

×
(Continued )

Syllabus Guidance
Chapter

Analytical
Methods
for
Engineers

Further
Analytical
Methods for
Engineers

Engineering
Mathematics

55.

Measures of central tendency and dispersion

×

56.

Probability

×

57.

The binomial and Poisson distributions

×

58.

The normal distribution

×

59.

Linear correlation

×

60.

Linear regression

×

61.

Introduction to Laplace transforms

×

62.

Properties of Laplace transforms

×

63.

Inverse Laplace transforms

×

64.

Solution of differential equations using Laplace transforms

×

65.

The solution of simultaneous differential equations using
Laplace transforms

×

66.

Fourier series for periodic functions of period 2π

×

67.

Fourier series for non-periodic functions over range 2π

×

68.

Even and odd functions and half-range Fourier series

×

69.

Fourier series over any range

×

70.

A numerical method of harmonic analysis

×

71.

The complex or exponential form of a Fourier series

×

Website Chapters
72.

Inequalities

73.

Boolean algebra and logic circuits

74.

Sampling and estimation theories

×

75.

Significance testing

×

76.

Chi-square and distribution-free tests

×

×

xvii

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Chapter 1

Algebra
1.1

3x + 2y x−y Introduction

In this chapter, polynomial division and the factor and remainder theorems are explained (in Sections 1.4 to 1.6). However, before this, some essential algebra revision on basic laws and equations is included.
For further Algebra revision, go to website: http://books.elsevier.com/companions/0750681527 1.2

Multiply by x → 3x 2 + 2x y
Multiply by −y →

3x 2 − xy − 2y 2

Adding gives:
Alternatively,

(3x + 2y)(x − y) = 3x 2 − 3x y + 2x y − 2y 2

Revision of basic laws

= 3x 2 − xy − 2y 2

(a) Basic operations and laws of indices
The laws of indices are: am (ii)
= a m−n an √ m (iv) a n = n a m

(i) a m × a n = a m+n
(iii)

(a m )n

(v)

a −n

=

a m×n

1
= n a (vi)

a0

Problem 3. Simplify a = 3, b =

1
8

and c = 2.

=1

4a 2 bc3−2ac

4a bc − 2ac = 4(2)
2

3

2

=

1
2

3
2

3

a 3 b 2 c4 and evaluate when abc−2 a 3 b 2 c4
= a 3−1b2−1c4−(−2) = a 2 bc6 abc−2 When a = 3, b =

Problem 1. Evaluate b = 1 and c = 1 1
2
2

− 3x y − 2y 2

when a = 2,

3
− 2(2)
2

4 × 2 × 2 × 3 × 3 × 3 12

2×2×2×2
2

= 27 − 6 = 21
Problem 2. Multiply 3x + 2y by x − y.

1
8

a 2 bc6 = (3)2

and c = 2,
1
8

(2)6 = (9)

Problem 4. Simplify

1
8

(64) = 72

x 2 y3 + x y2 xy x 2 y3 + x y2 x 2 y3 x y2
=
+ xy xy xy = x 2−1 y 3−1 + x 1−1 y 2−1
= xy 2 + y or y(xy + 1)

2 Higher Engineering Mathematics

Problem 5.

Simplify

√ √
(x 2 y)( x
(x 5 y 3 )

3

y2 )

1
2

√ √
(x 2 y)( x

3

(b) Brackets, factorization and precedence

y2 )

1

(x 5 y 3 ) 2
1

1

5
2

Problem 6.

2

3
2

x2 y 2 x 2 y 3

=

x y

a 2 − (2a − ab) − a(3b + a)
= a 2 − 2a + ab − 3ab − a 2
= −2a − 2ab or −2a(1 + b)

= x 2+ 2 − 2 y 2 + 3 − 2
1

5

1

2

3

= x 0 y− 3
1

Problem 7. expression: 1

= y − 3 or

Simplify a 2 − (2a − ab) − a(3b + a).

1 y 1
3

1 or √
3 y

Remove the brackets and simplify the

2a − [3{2(4a − b) − 5(a + 2b)} + 4a].
Removing the innermost brackets gives:
2a − [3{8a − 2b − 5a − 10b} + 4a]

Now try the following exercise

Collecting together similar terms gives:
Exercise 1 Revision of basic operations and laws of indices
1. Evaluate 2ab + 3bc − abc when a = 2, b = −2 and c = 4.
[−16]
2. Find the value of 5 pq 2r 3 when p = 2 ,
5
q = −2 and r = −1.
[−8]
3. From 4x − 3y + 2z subtract x + 2y − 3z.
[3x − 5y + 5z]

2a − [3{3a − 12b} + 4a]
Removing the ‘curly’ brackets gives:
2a − [9a − 36b + 4a]
Collecting together similar terms gives:
2a − [13a − 36b]
Removing the square brackets gives:

4. Multiply 2a − 5b + c by 3a + b.
[6a 2 − 13ab + 3ac − 5b 2 + bc]
5. Simplify (x y z)(x yz ) and evaluate when
[x 5 y 4 z 3 , 13 1 ] x = 1 , y = 2 and z = 3.
2
2
2 3

3

3
2

2a − 13a + 36b = −11a + 36b or
36b − 11a

2

1
2

6. Evaluate (a bc−3)(a b b = 4 and c = 2.

1
−2

c) when a = 3,
[±4 1 ]
2

Problem 8. Factorize (a) x y − 3x z
(b) 4a 2 + 16ab3 (c) 3a 2 b − 6ab 2 + 15ab.
(a)

7. Simplify

8. Simplify

1+a b a2b + a3b a 2 b2
1
2

−1
2

(a 3 b c )(ab)
√ √
( a 3 b c)
11

1

x y − 3x z = x( y − 3z)

(b) 4a 2 + 16ab3 = 4a(a + 4b3 )
(c) 3a 2 b − 6ab 2 + 15ab = 3ab(a − 2b + 5)

1
3

Problem 9.
3

a 6 b 3 c− 2

√ 11 √
6
a 3b or √ c3 Simplify 3c + 2c × 4c + c ÷ 5c − 8c.

The order of precedence is division, multiplication, addition and subtraction (sometimes remembered by BODMAS). Hence

Algebra
3c + 2c × 4c + c ÷ 5c − 8c c = 3c + 2c × 4c +
− 8c
5c
1
= 3c + 8c2 + − 8c
5
1
1
= 8c2 − 5c + or c(8c − 5) +
5
5

8. Simplify a 2 − 3ab × 2a ÷ 6b + ab.

1.3
Problem 10. Simplify
(2a − 3) ÷ 4a + 5 × 6 −3a.

2a − 3
+ 30 − 3a
4a

Since 4 − 3x = 2x − 11 then 4 + 11 = 2x + 3x
15
i.e. 15 = 5x from which, x =
=3
5
Problem 12. Solve

3
2a

+ 30 − 3a
=
4a 4a
=

Revision of equations

Problem 11. Solve 4 − 3x = 2x − 11.

2a − 3
+ 5 × 6 − 3a
4a

=

[ab]

(a) Simple equations

(2a − 3) ÷ 4a + 5 × 6 − 3a
=

5
−1
y

7. Simplify 3 ÷ y + 2 ÷ y − 1.

4(2a − 3) − 2(a − 4) = 3(a − 3) − 1.

1
3
1
3

+ 30 − 3a = 30 −
− 3a
2 4a
2
4a

Removing the brackets gives:
8a − 12 − 2a + 8 = 3a − 9 − 1
Rearranging gives:
8a − 2a − 3a = −9 − 1 + 12 − 8

Now try the following exercise
Exercise 2 Further problems on brackets, factorization and precedence
1. Simplify 2( p + 3q − r) − 4(r − q + 2 p) + p.
[−5 p + 10q − 6r]
2. Expand and simplify (x + y)(x − 2y).
[x 2 − x y − 2y 2 ]
3. Remove the brackets and simplify:
24 p − [2{3(5 p − q) − 2( p + 2q)} + 3q].
[11q − 2 p]

i.e. and 3a = −6 a= −6
= −2
3

Problem 13. Solve

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

By ‘cross-multiplying’:
Removing brackets gives:

9x + 12 = 4x − 8

Rearranging gives:

9x − 4x = −8 − 12

[7ab(3ab − 4)]

5. Factorize 2x y 2 + 6x 2 y + 8x 3 y.
[2x y(y + 3x + 4x 2 )]

5x = −20

i.e. and 4. Factorize 21a 2b2 − 28ab.

6. Simplify 2y + 4 ÷ 6y + 3 × 4 − 5y.
2
− 3y + 12
3y

Problem 14. Solve

−20
5
= −4

x=

√ t +3

t

= 2.

3

4 Higher Engineering Mathematics

√ t +3

=2 t t √

t +3= 2 t
√ √
3= 2 t − t

3= t

√ t i.e. and i.e.

Rearranging gives: d 2 p + D 2 p = D 2 f − d2 f
Factorizing gives:

Exercise 3 Further problems on simple equations and transposition of formulae

Transpose the formula v = u +

to make f the subject.

ft m m

In problems 1 to 4 solve the equations
1. 3x − 2 − 5x = 2x − 4.

ft ft u+
= v from which,
= v−u m m and f (D 2 − d2 )
(d2 + D2 )

Now try the following exercise

(b) Transposition of formulae
Problem 15.

p=

and

9= t

and

p(d 2 + D 2 ) = f (D 2 − d2 )

1
2

2. 8 + 4(x − 1) − 5(x − 3) = 2(5 − 2x).
[−3]

ft
= m(v − u) m i.e.

f t = m(v − u)

and

3.

m f = (v − u) t 4.

1
1
+
= 0.
3a − 2 5a + 3

3 t
√ = −6.
1− t

−1
8
[4]

3(F − f )
. for f .
L
yL
3F − yL or f = F − f =
3
3

5. Transpose y =

Problem 16. √ impedance of an a.c. circuit is
The
given by Z = R 2 + X 2 . Make the reactance X the subject. 6. Make l the subject of t = 2π
R 2 + X 2 = Z and squaring both sides gives

1
.
g l= R + X = Z , from which,
2

2

2

X 2 = Z 2 − R 2 and reactance X =
D
= d express p in terms of D, d and f .

Problem 17.

Given that

Rearranging gives:
Squaring both sides gives:

Z2 − R2

f +p
,
f −p

D f +p
=
f −p d f +p
D2
= 2 f −p d ‘Cross-multiplying’ gives: d2 ( f + p) = D 2 ( f − p)
Removing brackets gives: d2 f + d2 p = D 2 f − D 2 p

7. Transpose m =

t 2g
4π 2

μL for L.
L + rC R
L=

8. Make r the subject of the formula x 1 + r2
.
r=
=
y
1 − r2

mrC R μ−m x−y x+y (c) Simultaneous equations
Problem 18.

Solve the simultaneous equations:
7x − 2y = 26

(1)

6x + 5y = 29.

(2)

Algebra
5 × equation (1) gives:
35x − 10y = 130

The factors of −4 are +1 and −4 or −1 and
+4, or −2 and +2. Remembering that the product of the two inner terms added to the product of the two outer terms must equal −11x, the only combination to give this is +1 and −4, i.e.,

(3)

2 × equation (2) gives:
12x + 10y = 58

(4)

equation (3) +equation (4) gives:

3x 2 − 11x − 4 = (3x + 1)(x − 4)

47x + 0 = 188
188
from which, x= =4
47
Substituting x = 4 in equation (1) gives:

(3x + 1)(x − 4) = 0 hence

Thus

(3x + 1) = 0 i.e. x = − 1
3

either

(x − 4) = 0 i.e. x = 4

or

28 − 2y = 26

(b) 4x 2 + 8x + 3 = (2x + 3)(2x + 1)

from which, 28 − 26 = 2y and y = 1

(2x + 3)(2x + 1) = 0 hence

Thus

Problem 19. Solve x 5
+ =y
8 2 y 11 + = 3x.
3

either

(2)

(2x + 3) = 0 i.e. x = − 3
2

or

(1)

(2x + 1) = 0 i.e. x = − 1
2

Problem 21. The roots of a quadratic equation are 1 and −2. Determine the equation in x.
3

8 × equation (1) gives:

x + 20 = 8y

(3)

3 × equation (2) gives:

33 + y = 9x

(4)

i.e.

x − 8y = −20

(5)

and

9x − y = 33

(6)

i.e. x 2 + 2x − 1 x − 2 = 0
3
3

(7)

i.e.

x2 + 5 x − 2 = 0
3
3

or

3x2 + 5x −2 = 0

8 × equation (6) gives: 72x − 8y = 264
Equation (7) − equation (5) gives:
71x = 284
284
=4
71
Substituting x = 4 in equation (5) gives: x= from which,

4 − 8y = −20
4 + 20 = 8y and y = 3

from which,

(d) Quadratic equations

If

1
3

Problem 22. Solve 4x 2 + 7x + 2 = 0 giving the answer correct to 2 decimal places.
From the quadratic formula if ax 2 + bx + c = 0 then,

−b ± b2 − 4ac x= 2a
Hence if 4x 2 + 7x + 2 = 0

(b) 4x 2 + 8x + 3 = 0.
The factors of 3x 2 are 3x and x and these are placed in brackets thus:
(3x

)(x

)

72 − 4(4)(2)
2(4)

−7 ± 17
=
8
−7 ± 4.123
=
8
−7 + 4.123
−7 − 4.123
=
or
8
8 x = −0.36 or −1.39

then x =

Problem 20. Solve the following equations by factorization: (a) 3x 2 − 11x − 4 = 0

(a)

and −2 are the roots of a quadratic equation then,
(x − 1 )(x + 2) = 0
3

i.e.

−7 ±

5

6 Higher Engineering Mathematics
Now try the following exercise

For example,

Exercise 4 Further problems on simultaneous and quadratic equations
In problems 1 to 3, solve the simultaneous equations

13
——–
16 208
16
48
48

··


1. 8x − 3y = 51
3x + 4y = 14.

208 is achieved as follows:
16

[x = 6, y = −1]

(1) 16 divided into 2 won’t go
2. 5a = 1 − 3b
2b + a + 4 = 0.
3.

[a = 2, b = −3]

x 2y
49
+
=
5
3
15

(2) 16 divided into 20 goes 1
(3) Put 1 above the zero
(4) Multiply 16 by 1 giving 16
(5) Subtract 16 from 20 giving 4

3x y 5
− + = 0.
7
2 7

[x = 3, y = 4]

(6) Bring down the 8
(7) 16 divided into 48 goes 3 times

4. Solve the following quadratic equations by factorization: (a) x 2 + 4x − 32 = 0

[(a) 4, −8 (b) 5 , − 3 ]
4
2
5. Determine the quadratic equation in x whose roots are 2 and −5.
[x 2 + 3x − 10 = 0]
6. Solve the following quadratic equations, correct to 3 decimal places:
(a)

−4 = 0

(b) 4t 2 − 11t + 3 = 0.
(a) 0.637, −3.137
(b) 2.443, 0.307

1.4

(9) 3 × 16 = 48
(10) 48 − 48 = 0

(b) 8x 2 + 2x − 15 = 0.

2x 2 + 5x

(8) Put the 3 above the 8

Hence
Similarly,

208
= 13 exactly
16

172 is laid out as follows:
15

11
——–
15 172
15
22
15

7

7
7
172
= 11 remainder 7 or 11 +
= 11
Hence
15
15
15
Below are some examples of division in algebra, which in some respects, is similar to long division with numbers. (Note that a polynomial is an expression of the form Polynomial division f (x) = a + bx + cx 2 + d x 3 + · · ·

Before looking at long division in algebra let us revise long division with numbers (we may have forgotten, since calculators do the job for us!)

and polynomial division is sometimes required when resolving into partial fractions—see Chapter 2.)

Algebra
Problem 23. Divide 2x 2 + x − 3 by x − 1.

(3) Subtract
(4)

2x 2 + x − 3 is called the dividend and x − 1 the divisor. The usual layout is shown below with the dividend and divisor both arranged in descending powers of the symbols. 2x + 3
——————–
x − 1 2x 2 + x − 3
2x 2 − 2x
3x − 3
3x − 3
———
· ·
———
Dividing the first term of the dividend by the first term
2x 2 gives 2x, which is put above of the divisor, i.e. x the first term of the dividend as shown. The divisor is then multiplied by 2x, i.e. 2x(x − 1) = 2x 2 − 2x, which is placed under the dividend as shown. Subtracting gives 3x − 3. The process is then repeated, i.e. the first term of the divisor, x, is divided into 3x, giving
+3, which is placed above the dividend as shown. Then
3(x − 1) = 3x − 3 which is placed under the 3x − 3. The remainder, on subtraction, is zero, which completes the process. x into −2x 2 goes −2x. Put −2x above the dividend (5) −2x(x + 1) = −2x 2 − 2x
(6) Subtract
(7)

x into 5x goes 5. Put 5 above the dividend

(8) 5(x + 1) = 5x + 5
(9) Subtract
Thus

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

Problem 25. Simplify

x 3 + y3
.
x+y

(1) (4) (7) x 2 − x y + y2
—————————–
x + y x 3 + 0 + 0 + y3 x3 + x2 y
− x2 y
+ y3
− x 2 y − x y2
———————
x y2 + y3 x y2 + y3
———–
· ·
———–

Thus (2x 2 + x − 3) ÷ (x − 1) = (2x + 3)
[A check can be made on this answer by multiplying
(2x + 3) by (x − 1) which equals 2x 2 + x − 3]
(1)
Problem 24. Divide 3x 3 + x 2 + 3x + 5 by x + 1.
(1) (4) (7)
3x 2 − 2x + 5
—————————
x + 1 3x 3 + x 2 + 3x + 5
3x 3 + 3x 2
− 2x 2 + 3x + 5
− 2x 2 − 2x
————–
5x + 5
5x + 5
———
· ·
———
(1)

x into 3x 3 goes 3x 2 . Put 3x 2 above 3x 3

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

x into x 3 goes x 2 . Put x 2 above x 3 of dividend

(2)

x 2 (x + y) = x 3 + x 2 y

(3) Subtract
(4)

x into −x 2 y goes −x y. Put −x y above dividend

(5) −x y(x + y) = −x 2 y − x y 2
(6) Subtract
(7)

x into x y 2 goes y 2 . Put y 2 above dividend

(8)

y 2 (x + y) = x y 2 + y 3

(9) Subtract
Thus
x 3 + y3
= x 2 − xy + y 2 x+y 7

8 Higher Engineering Mathematics
The zero’s shown in the dividend are not normally shown, but are included to clarify the subtraction process and to keep similar terms in their respective columns.
Problem 26.

Divide (x 2 + 3x − 2) by (x − 2).

x +5
——————–
x − 2 x 2 + 3x − 2 x 2 − 2x

14x 2 − 19x − 3
.
2x − 3

[7x + 1]

6. Find (5x 2 − x + 4) ÷ (x − 1).
5x + 4 +

8 x 2 + 3x − 2
=x +5+ x −2 x−2 Divide 4a 3 − 6a 2 b + 5b 3 by

2a 2 − 2ab − b 2
———————————
2a − b 4a 3 − 6a 2 b
+ 5b 3
3 − 2a 2 b
4a

The factor theorem

There is a simple relationship between the factors of a quadratic expression and the roots of the equation obtained by equating the expression to zero.
For example, consider the quadratic equation x 2 + 2x − 8 = 0.
To solve this we may factorize the quadratic expression x 2 + 2x − 8 giving (x − 2)(x + 4).
Hence (x − 2)(x + 4) = 0.
Then, if the product of two numbers is zero, one or both of those numbers must equal zero. Therefore,

Thus
4a 3 − 6a 2 b + 5b 3
2a − b

either (x − 2) = 0, from which, x = 2 or (x + 4) = 0, from which, x = −4

4b3
2a − b

It is clear then that a factor of (x − 2) indicates a root of +2, while a factor of (x + 4) indicates a root of −4.
In general, we can therefore say that:

Now try the following exercise
Exercise 5 Further problems on polynomial division 2. Divide (3x 2 + 5x − 2) by (x + 2).

8. Determine (5x 4 + 3x 3 − 2x + 1)/(x − 3).
481
5x 3 + 18x 2 + 54x + 160 + x −3

1.5

−4a 2 b
+ 5b3
2 b + 2ab 2
−4a
———— 2
−2ab + 5b 3
−2ab2 + b 3
—————–
4b 3
—————–

1. Divide (2x 2 + x y − y 2 ) by (x + y).

8 x −1

7. Divide (3x 3 + 2x 2 − 5x + 4) by (x + 2).
2
3x 2 − 4x + 3 − x +2

Hence

= 2a 2 − 2ab − b2 +

4. Find

5. Divide (x 3 + 3x 2 y + 3x y 2 + y 3 ) by (x + y).
[x 2 + 2x y + y 2 ]

5x − 2
5x − 10
———
8
———

Problem 27.
2a − b.

3. Determine (10x 2 + 11x − 6) ÷ (2x + 3).
[5x − 2]

[2x − y]
[3x − 1]

a factor of (x − a) corresponds to a root of x = a
In practice, we always deduce the roots of a simple quadratic equation from the factors of the quadratic expression, as in the above example. However, we could reverse this process. If, by trial and error, we could determine that x = 2 is a root of the equation x 2 + 2x − 8 = 0 we could deduce at once that (x − 2) is a factor of the

Algebra expression x 2 + 2x − 8. We wouldn’t normally solve quadratic equations this way — but suppose we have to factorize a cubic expression (i.e. one in which the highest power of the variable is 3). A cubic equation might have three simple linear factors and the difficulty of discovering all these factors by trial and error would be considerable. It is to deal with this kind of case that we use the factor theorem. This is just a generalized version of what we established above for the quadratic expression. The factor theorem provides a method of factorizing any polynomial, f (x), which has simple factors. A statement of the factor theorem says:
‘if x = a is a root of the equation f (x) = 0, then (x − a) is a factor of f (x)’
The following worked problems show the use of the factor theorem.
Problem 28. Factorize x 3 − 7x − 6 and use it to solve the cubic equation x 3 − 7x − 6 = 0.
Let

f (x) = x 3 − 7x − 6

If x = 1, then f (1) = 13 − 7(1) − 6 = −12
If x = 2, then f (2) = 23 − 7(2) − 6 = −12
If x = 3, then f (3)

= 33 − 7(3) − 6

=0

If f (3) = 0, then (x − 3) is a factor — from the factor theorem. We have a choice now. We can divide x 3 − 7x − 6 by
(x − 3) or we could continue our ‘trial and error’ by substituting further values for x in the given expression — and hope to arrive at f (x) = 0.
Let us do both ways. Firstly, dividing out gives: x 2 + 3x + 2
—————————
x − 3 x 3 − 0 − 7x − 6 x 3 − 3x 2
3x 2 − 7x − 6
3x 2 − 9x
————
2x − 6
2x − 6
———
· ·
———
x 3 − 7x − 6
= x 2 + 3x + 2
Hence
x −3
i.e.

x 3 − 7x − 6 = (x − 3)(x 2 + 3x + 2)

x 2 + 3x + 2 factorizes ‘on sight’ as (x + 1)(x + 2).
Therefore
x 3 − 7x − 6 = (x − 3)(x + 1)(x + 2)
A second method is to continue to substitute values of x into f (x).
Our expression for f (3) was 33 − 7(3) − 6. We can see that if we continue with positive values of x the first term will predominate such that f (x) will not be zero.
Therefore let us try some negative values for x.
Therefore f (−1) = (−1)3 − 7(−1) − 6 = 0; hence
(x + 1) is a factor (as shown above). Also f (−2) = (−2)3 − 7(−2) − 6 = 0; hence (x + 2) is a factor (also as shown above).
To solve x 3 − 7x − 6 = 0, we substitute the factors, i.e.,
(x − 3)(x + 1)(x + 2) = 0 from which, x = 3, x = −1 and x = −2.
Note that the values of x, i.e. 3, −1 and −2, are all factors of the constant term, i.e. the 6. This can give us a clue as to what values of x we should consider. Problem 29. Solve the cubic equation x 3 − 2x 2 − 5x + 6 = 0 by using the factor theorem.
Let f (x) = x 3 − 2x 2 − 5x + 6 and let us substitute simple values of x like 1, 2, 3, −1, −2, and so on. f (1) = 13 − 2(1)2 − 5(1) + 6 = 0, hence (x − 1) is a factor f (2) = 23 − 2(2)2 − 5(2) + 6 = 0 f (3) = 33 − 2(3)2 − 5(3) + 6 = 0, hence (x − 3) is a factor f (−1) = (−1)3 − 2(−1)2 − 5(−1) + 6 = 0 f (−2) = (−2)3 − 2(−2)2 − 5(−2) + 6 = 0, hence (x + 2) is a factor x 3 − 2x 2

− 5x + 6 = (x − 1)(x − 3)(x + 2)
Hence
Therefore if x 3 − 2x 2 − 5x + 6 = 0 then (x − 1)(x − 3)(x + 2) = 0 from which, x = 1, x = 3 and x = −2
Alternatively, having obtained one factor, i.e.
(x − 1) we could divide this into (x 3 − 2x 2 − 5x + 6) as follows:

9

10 Higher Engineering Mathematics x2 − x − 6
————————–
x − 1 x 3 − 2x 2 − 5x + 6 x3 − x2

1.6

Dividing a general quadratic expression
(ax 2 + bx + c) by (x − p), where p is any whole number, by long division (see section 1.3) gives: ax + (b + ap)
————————————–
x − p ax 2 + bx
+c
ax 2 − apx

− x 2 − 5x + 6
− x2 + x
————–
− 6x + 6
− 6x + 6
———–
· ·
———–
Hence x 3 − 2x 2 − 5x + 6
= (x − 1)(x 2 − x − 6)
= (x − 1)(x − 3)(x + 2)
Summarizing, the factor theorem provides us with a method of factorizing simple expressions, and an alternative, in certain circumstances, to polynomial division.

Now try the following exercise

Exercise 6 Further problems on the factor theorem Use the factor theorem to factorize the expressions given in problems 1 to 4.
1.

x 2 + 2x − 3

2.

x 3 + x 2 − 4x − 4

[(x − 1)(x + 3)]

3. 2x 3 + 5x 2 − 4x − 7

The remainder theorem

[(x + 1)(x + 2)(x − 2)]
[(x + 1)(2x 2 + 3x − 7)]

4. 2x 3 − x 2 − 16x + 15
[(x − 1)(x + 3)(2x − 5)]
5. Use the factor theorem to factorize x 3 + 4x 2 + x − 6 and hence solve the cubic equation x 3 + 4x 2 + x − 6 = 0.

⎡ 3 x + 4x 2 + x − 6


= (x − 1)(x + 3)(x + 2) ⎦

x = 1, x = −3 and x = −2
6. Solve the equation x 3 − 2x 2 − x + 2 = 0.
[x = 1, x = 2 and x = −1]

(b + ap)x + c
(b + ap)x − (b + ap) p
—————————–
c + (b + ap) p
—————————–
The remainder, c + (b + ap) p = c + bp + ap 2 or ap2 + bp + c. This is, in fact, what the remainder theorem states, i.e.,
‘if (ax 2 + bx + c) is divided by (x − p), the remainder will be ap 2 + bp + c’
If, in the dividend (ax 2 + bx + c), we substitute p for x we get the remainder ap2 + bp + c.
For example, when (3x 2 − 4x + 5) is divided by
(x − 2) the remainder is ap2 + bp + c (where a = 3, b = −4, c = 5 and p = 2),
i.e. the remainder is
3(2)2 + (−4)(2) + 5 = 12 − 8 + 5 = 9
We can check this by dividing (3x 2 − 4x + 5) by
(x − 2) by long division:
3x + 2
——————–
x − 2 3x 2 − 4x + 5
3x 2 − 6x
2x + 5
2x − 4
———
9
———
Similarly, when (4x 2 − 7x + 9) is divided by (x + 3), the remainder is ap 2 + bp + c, (where a = 4, b = −7, c = 9 and p = −3) i.e. the remainder is
4(−3)2 + (−7)(−3) + 9 = 36 + 21 + 9 = 66.
Also, when (x 2 + 3x − 2) is divided by (x − 1), the remainder is 1(1)2 + 3(1) − 2 = 2.
It is not particularly useful, on its own, to know the remainder of an algebraic division. However, if the remainder should be zero then (x − p) is a factor. This is very useful therefore when factorizing expressions.
For example, when (2x 2 + x − 3) is divided by
(x − 1), the remainder is 2(1)2 + 1(1) − 3 = 0, which means that (x − 1) is a factor of (2x 2 + x − 3).

Algebra
In this case the other factor is (2x + 3), i.e.,

i.e. the remainder = (1)(1)3 + (−2)(1)2
+ (−5)(1) + 6

(2x 2 + x − 3) = (x − 1)(2x − 3)
The remainder theorem may also be stated for a cubic equation as:
‘if (ax 3 + bx 2 + cx + d) is divided by
(x − p), the remainder will be ap 3 + bp 2 + cp + d’
As before, the remainder may be obtained by substituting p for x in the dividend.
For example, when (3x 3 + 2x 2 − x + 4) is divided by (x − 1), the remainder is ap 3 + bp2 + cp + d
(where a = 3, b = 2, c = −1, d = 4 and p = 1),
i.e. the remainder is 3(1)3 + 2(1)2 + (−1)(1) + 4 =
3 + 2 − 1 + 4 = 8.
Similarly, when (x 3 − 7x − 6) is divided by (x − 3), the remainder is 1(3)3 + 0(3)2 − 7(3) − 6 = 0, which means that (x − 3) is a factor of (x 3 − 7x − 6).
Here are some more examples on the remainder theorem. Problem 30. Without dividing out, find the remainder when 2x 2 − 3x + 4 is divided by (x − 2).
By the remainder theorem, the remainder is given by ap 2 + bp + c, where a = 2, b = −3, c = 4 and p = 2.
Hence the remainder is:
2(2)2 + (−3)(2) + 4 = 8 − 6 + 4 = 6
Problem 31. Use the remainder theorem to determine the remainder when
(3x 3 − 2x 2 + x − 5) is divided by (x + 2).
By the remainder theorem, the remainder is given by ap 3 + bp2 + cp + d, where a = 3, b = −2, c = 1, d =
−5 and p = −2.
Hence the remainder is:
3(−2)3 + (−2)(−2)2 + (1)(−2) + (−5)
= −24 − 8 − 2 − 5
= −39
Problem 32. Determine the remainder when
(x 3 − 2x 2 − 5x + 6) is divided by (a) (x − 1) and
(b) (x + 2). Hence factorize the cubic expression.
(a)

When (x 3 − 2x 2 − 5x + 6) is divided by (x − 1), the remainder is given by ap 3 + bp2 + cp + d, where a = 1, b = −2, c = −5, d = 6 and p = 1,

11

= 1−2−5+6 = 0
Hence (x − 1) is a factor of (x 3 − 2x 2 − 5x + 6).
(b) When (x 3 − 2x 2 − 5x + 6) is divided by (x + 2), the remainder is given by
(1)(−2)3 + (−2)(−2)2 + (−5)(−2) + 6
= −8 − 8 + 10 + 6 = 0
Hence (x + 2) is also a factor of (x 3 − 2x 2 −
5x + 6). Therefore (x − 1)(x + 2)(x ) = x 3 −
2x 2 − 5x + 6. To determine the third factor (shown blank) we could
(i) divide (x 3 − 2x 2 − 5x + 6) by
(x − 1)(x + 2). or (ii) use the factor theorem where f (x) = x 3 − 2x 2 − 5x + 6 and hoping to choose a value of x which makes f (x) = 0. or (iii) use the remainder theorem, again hoping to choose a factor (x − p) which makes the remainder zero.
(i) Dividing (x 3 − 2x 2 − 5x + 6) by
(x 2 + x − 2) gives: x −3
————————–
x 2 + x − 2 x 3 − 2x 2 − 5x + 6 x 3 + x 2 − 2x
——————
−3x 2 − 3x + 6
−3x 2 − 3x + 6
——————–
·
·
·
——————–
Thus (x 3 − 2x 2 − 5x + 6)
= (x − 1)(x + 2)(x − 3)
(ii) Using the factor theorem, we let f (x) = x 3 − 2x 2 − 5x + 6
Then f (3) = 33 − 2(3)2 − 5(3) + 6
= 27 − 18 − 15 + 6 = 0
Hence (x − 3) is a factor.
(iii) Using the remainder theorem, when
(x 3 − 2x 2 − 5x + 6) is divided by
(x − 3), the remainder is given by

12 Higher Engineering Mathematics ap3 + bp2 + cp + d, where a = 1, b = −2, c = −5, d = 6 and p = 3.
Hence the remainder is:
1(3)3 + (−2)(3)2 + (−5)(3) + 6
= 27 − 18 − 15 + 6 = 0
Hence (x − 3) is a factor.
Thus (x 3 − 2x 2 − 5x + 6)
= (x − 1)(x + 2)(x − 3)

Now try the following exercise
Exercise 7 Further problems on the remainder theorem
1. Find the remainder when 3x 2 − 4x + 2 is divided by
(a) (x − 2) (b) (x + 1).

[(a) 6 (b) 9]

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

[(a) −39 (b) −29]

3. Use the remainder theorem to find the factors of x 3 − 6x 2 + 11x − 6.
[(x − 1)(x − 2)(x − 3)]
4. Determine the factors of x 3 + 7x 2 + 14x + 8 and hence solve the cubic equation x 3 + 7x 2 + 14x + 8 = 0.
[x = −1, x = −2 and x = −4]
5. Determine the value of ‘a’ if (x + 2) is a factor of (x 3 − ax 2 + 7x + 10).
[a = −3]
6. Using the remainder theorem, solve the equation 2x 3 − x 2 − 7x + 6 = 0.
[x = 1, x = −2 and x = 1.5]

Chapter 2

Partial fractions
2.1

When the degree of the numerator is equal to or higher than the degree of the denominator, the numerator must be divided by the denominator until the remainder is of less degree than the denominator (see Problems 3 and 4).
There are basically three types of partial fraction and the form of partial fraction used is summarized in
Table 2.1, where f (x) is assumed to be of less degree than the relevant denominator and A, B and C are constants to be determined.
(In the latter type in Table 2.1, ax 2 + bx + c is a quadratic expression which does not factorize without containing surds or imaginary terms.)
Resolving an algebraic expression into partial fractions is used as a preliminary to integrating certain functions (see Chapter 41) and in determining inverse
Laplace transforms (see Chapter 63).

Introduction to partial fractions

By algebraic addition,
1
3
(x + 1) + 3(x − 2)
+
= x −2 x +1
(x − 2)(x + 1)
=

4x − 5 x2 − x − 2

The reverse process of moving from

4x − 5
−2

x2 − x

1
3
+ is called resolving into partial x −2 x +1 fractions. In order to resolve an algebraic expression into partial fractions: to

(i) the denominator must factorize (in the above example, x 2 − x − 2 factorizes as (x − 2) (x + 1)), and 2.2 Worked problems on partial fractions with linear factors

(ii) the numerator must be at least one degree less than the denominator (in the above example (4x − 5) is of degree 1 since the highest powered x term is x 1 and (x 2 − x − 2) is of degree 2).

Problem 1. Resolve fractions. 11 − 3x into partial x 2 + 2x − 3

Table 2.1
Type

Denominator containing

1

Linear factors
(see Problems 1 to 4)

2

Repeated linear factors
(see Problems 5 to 7)

3

Quadratic factors
(see Problems 8 and 9)

Expression

Form of partial fraction

f (x)
(x + a)(x − b)(x + c)

A
B
C
+
+
(x + a) (x − b) (x + c)

f (x)
(x + a)3

A
C
B
+
+
2
(x + a) (x + a)
(x + a)3

f (x)
+ c)(x + d)

(ax 2 + bx

Ax + B
C
+
+ bx + c) (x + d)

(ax 2

14 Higher Engineering Mathematics
The denominator factorizes as (x − 1) (x + 3) and the numerator is of less degree than the denominator. Thus
11 − 3x may be resolved into partial fractions. x 2 + 2x − 3
Let

Let

2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)


11 − 3x
11 − 3x

− 3 (x − 1)(x + 3)

x 2 + 2x

A
B

+
(x − 1) (x + 3) where A and B are constants to be determined,
11 − 3x
A(x + 3) + B(x − 1)
i.e.

,
(x − 1)(x + 3)
(x − 1)(x + 3) by algebraic addition.
Since the denominators are the same on each side of the identity then the numerators are equal to each other. ≡

When x = 1, then
11 −3(1) ≡ A(1 + 3) + B(0)
8 = 4A
A =2

i.e.
i.e.

2x 2 − 9x − 35 ≡ A(x − 2)(x + 3)
+ B(x + 1)(x + 3) + C(x + 1)(x − 2)
Let x = − 1. Then

20 = −4B
B = −5

Thus

+ B(0)(2) +C(0)(−3)

B=

−45
= −3
15

Let x = − 3. Then

2
5
2(x + 3) − 5(x − 1)

=
(x − 1) (x + 3)
(x − 1)(x + 3)
=

11 − 3x x 2 + 2x − 3

2x 2 − 9x − 35 into (x + 1)(x − 2)(x + 3) the sum of three partial fractions.
Convert

−24
=4
−6

−45 = 15B

i.e.
i.e.

2
5


(x − 1) (x + 3)

Problem 2.

A=

i.e.

11 − 3x
−5
2
+
≡ x2 + 2x − 3 (x − 1) (x + 3)

Check:

−24 = −6 A

i.e.

2(2)2 − 9(2) − 35 ≡ A(0)(5) + B(3)(5) + C(3)(0)

11 −3(−3) ≡ A(0) + B(−3 −1)
i.e.

2(−1)2 − 9(−1) − 35 ≡ A(−3)(2)

Let x = 2. Then

When x = −3, then

i.e.

A(x − 2)(x + 3) + B(x + 1)(x + 3)
+ C(x + 1)(x − 2)
(x + 1)(x − 2)(x + 3)

by algebraic addition.
Equating the numerators gives:

Thus, 11 −3x ≡ A(x + 3) + B(x − 1)
To determine constants A and B, values of x are chosen to make the term in A or B equal to zero.

A
B
C
+
+
(x + 1) (x − 2) (x + 3)

2(−3)2 − 9(−3) − 35 ≡ A(−5)(0) + B(−2)(0)
+ C(−2)(−5)
i.e.

10 = 10C

i.e.

C =1

Thus

2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)


3
1
4

+
(x + 1) (x − 2)
(x + 3)

Problem 3. fractions. Resolve

x2

x2 + 1 into partial
− 3x + 2

Partial fractions
The denominator is of the same degree as the numerator.
Thus dividing out gives:

Thus

x − 10 x 3 − 2x 2 − 4x − 4
≡ x −3+ 2
2 +x −2 x x +x −2

1 x 2 − 3x + 2

+1 x 2 − 3x + 2
—————
3x − 1
———

≡ x −3+

x2

Let

x − 10
A
B

+
(x + 2)(x − 1) (x + 2) (x − 1)


For more on polynomial division, see Section 1.4, page 6.
Hence

3x − 1 x2 + 1
≡1 + 2
2 − 3x + 2 x x − 3x + 2
3x − 1
≡1 +
(x − 1)(x − 2)

A
B
3x − 1

+
Let
(x − 1)(x − 2) (x − 1) (x − 2)


A(x − 2) + B(x − 1)
(x − 1)(x − 2)

Equating numerators gives:

x − 10 ≡ A(x − 1) + B(x + 2)
Let x = −2. Then

−12 = −3 A
A= 4

i.e.
Let x = 1. Then

−9 = 3B
B = −3

i.e.
Hence

x − 10
4
3


(x + 2)(x − 1) (x + 2) (x − 1)

Thus

x3 − 2 x2 − 4x − 4 x2 + x − 2
≡x−3+

Let x = 1. Then 2 = −A
A = −2

A(x − 1) + B(x + 2)
(x + 2)(x − 1)

Equating the numerators gives:

3x − 1 ≡ A(x − 2) + B(x − 1)

i.e.

x − 10
(x + 2)(x − 1)

4
3

(x + 2) (x − 1)

Now try the following exercise

Let x = 2. Then 5 = B
−2
5
3x − 1

+
Hence
(x − 1)(x − 2) (x − 1) (x − 2)
Thus

2
5
x2 + 1
≡ 1−
+
2 − 3x + 2 x (x−1) (x−2)

Problem 4. Express fractions. x 3 − 2x 2 − 4x − 4 in partial x2 + x − 2

The numerator is of higher degree than the denominator.
Thus dividing out gives:

Exercise 8 Further problems on partial fractions with linear factors
Resolve the following into partial fractions.
2
2

(x − 3) (x + 3)

12

1.

x2 − 9

2.

x 2 − 2x

4(x − 4)
−3

5
1

(x + 1) (x − 3)

3.

x 2 − 3x + 6 x(x − 2)(x − 1)

4.

3(2x 2 − 8x − 1)
(x + 4)(x + 1)(2x − 1)

3
2
4
+
− x (x − 2) (x − 1)

x −3 x2 + x − 2

x 3 − 2x 2 − 4x − 4 x 3 + x 2 − 2x
——————
− 3x 2 − 2x − 4
− 3x 2 − 3x + 6
———————
x − 10

7
3
2


(x + 4) (x + 1) (2x − 1)

15

16 Higher Engineering Mathematics
When A = 2 and B = 7,

5.

x 2 + 9x + 8 x2 + x − 6

2
6
1+
+
(x + 3) (x − 2)

6.

x 2 − x − 14 x 2 − 2x − 3

2
3
1−
+
(x − 3) (x + 1)

7.

3x 3 − 2x 2 − 16x + 20
(x − 2)(x + 2)
3x − 2 +

R.H.S. = −2(2) + 7 = 3 = L.H.S.]
Hence

5x 2 − 2x − 19 as the sum
(x + 3)(x − 1)2 of three partial fractions.

Problem 6.
5
1

(x − 2) (x + 2)

Worked problems on partial fractions with repeated linear factors Problem 5.

Resolve

fractions.

2x + 3 into partial
(x − 2)2

The denominator contains a repeated linear factor,
(x − 2)2 .
A
2x + 3
B

Let
+
(x − 2)2 (x − 2) (x − 2)2
A(x − 2) + B
(x − 2)2



Equating the numerators gives:
2x + 3 ≡ A(x − 2) + B
Let x = 2. Then

7 = A(0) + B

i.e.

B =7

2x + 3 ≡ A(x − 2) + B ≡ Ax − 2 A + B
Since an identity is true for all values of the unknown, the coefficients of similar terms may be equated. Hence, equating the coefficients of x gives: 2 = A.
[Also, as a check, equating the constant terms gives:

Express

The denominator is a combination of a linear factor and a repeated linear factor.
Let

2.3

2
7
2x + 3

+
(x − 2)2 (x − 2) (x − 2)2

5x 2 − 2x − 19
(x + 3)(x − 1)2


A
B
C
+
+
(x + 3) (x − 1) (x − 1)2



A(x − 1)2 + B(x + 3)(x − 1) + C(x + 3)
(x + 3)(x − 1)2

by algebraic addition.
Equating the numerators gives:
5x 2 − 2x − 19 ≡ A(x − 1)2 + B(x + 3)(x − 1)
+ C(x + 3)
Let x = −3. Then
5(−3)2 − 2(−3) − 19 ≡ A(−4)2 + B(0)(−4)
+ C(0)
i.e.
32 = 16 A
i.e.
A= 2
Let x = 1. Then
5(1)2 − 2(1) − 19 ≡ A(0)2 + B(4)(0) + C(4)
i.e.
−16 = 4C
i.e.
C = −4
Without expanding the RHS of equation (1) it can be seen that equating the coefficients of x 2 gives:
5 = A + B, and since A = 2, B = 3.
[Check: Identity (1) may be expressed as:
5x 2 − 2x − 19 ≡ A(x 2 − 2x + 1)
+ B(x 2 + 2x − 3) + C(x + 3)
i.e. 5x 2 − 2x − 19 ≡ Ax 2 − 2 Ax + A + Bx 2 + 2Bx

3 = −2 A + B

(1)

− 3B + Cx + 3C

Partial fractions
Equating the x term coefficients gives:

Equating the coefficients of x terms gives:
16 = 6 A + B

−2 ≡ −2 A + 2B + C

Since A = 3, B = −2

When A = 2, B = 3 and C = −4 then

[Check: equating the constant terms gives:

−2 A + 2B + C = −2(2) + 2(3) − 4

15 = 9 A + 3B + C

= −2 = LHS

When A = 3, B = −2 and C = −6,

Equating the constant term gives:

9 A + 3B + C = 9(3) + 3(−2) + (−6)

−19 ≡ A − 3B + 3C

= 27 − 6 − 6 = 15 = LHS]

RHS = 2 − 3(3) + 3(−4) = 2 − 9 − 12
= −19 = LHS]

Hence

Thus

5x2 − 2x − 19
(x + 3)(x − 1)2


2
3
4
+

(x + 3) (x − 1) (x − 1)2

Now try the following exercise

3x 2 + 16x + 15
Problem 7. Resolve into partial
(x + 3)3 fractions. Let

3x2 + 16x + 15
(x + 3)3
3
6
2



2
(x + 3) (x + 3)
(x + 3)3

Exercise 9 Further problems on partial fractions with linear factors




x 2 + 7x + 3 x 2 (x + 3)

3.

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

4x − 3
(x + 1)2

2.

3x 2 + 16x + 15
(x + 3)3

1.

5x 2 − 30x + 44
(x − 2)3

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

1
2
1
+ − x 2 x (x + 3)

5
4
10
+

(x − 2) (x − 2)2 (x − 2)3

Equating the numerators gives:
3x 2 + 16x + 15 ≡ A(x + 3)2 + B(x + 3) + C

4
7

(x + 1) (x + 1)2

(1)

4.

18 + 21x − x 2
(x − 5)(x + 2)2
2
3
4

+
(x − 5) (x + 2) (x + 2)2

Let x = −3. Then
3(−3)2 + 16(−3) + 15 ≡ A(0)2 + B(0) + C
i.e.
−6 = C
Identity (1) may be expanded as:
3x 2 + 16x + 15 ≡ A(x 2 + 6x + 9)
+ B(x + 3) + C

2.4 Worked problems on partial fractions with quadratic factors

i.e. 3x 2 + 16x + 15 ≡ Ax 2 + 6 Ax + 9 A
+ Bx + 3B + C
Equating the coefficients of x 2 terms gives: 3 = A

Problem 8. Express fractions. 7x 2 + 5x + 13 in partial
(x 2 + 2)(x + 1)

17

18 Higher Engineering Mathematics
The denominator is a combination of a quadratic factor,
(x 2 + 2), which does not factorize without introducing imaginary surd terms, and a linear factor, (x + 1).
Let,

Equating the numerators gives:
3 + 6x + 4x 2 − 2x 3 ≡ Ax(x 2 + 3) + B(x 2 + 3)
+ (Cx + D)x 2

7x 2 + 5x + 13
Ax + B
C
≡ 2
+
2 + 2)(x + 1)
(x
(x + 2) (x + 1)


(Ax + B)(x + 1) + C(x 2 + 2)
(x 2 + 2)(x + 1)

≡ Ax 3 + 3 Ax + Bx 2 + 3B
+ Cx 3 + Dx 2
Let x = 0. Then 3 = 3B
i.e.

Equating numerators gives:
7x 2 + 5x + 13 ≡ (Ax + B)(x + 1) + C(x 2 + 2) (1)

Equating the coefficients of x 3 terms gives:

Let x = −1. Then

−2 = A + C

7(−1)2 + 5(−1) + 13 ≡ (Ax

+ B)(0) + C(1 + 2)

15 = 3C
C= 5

i.e.
i.e.

B=1

Equating the coefficients of x 2 terms gives:
4= B+D
Since B = 1, D = 3

Identity (1) may be expanded as:
7x 2 + 5x + 13 ≡ Ax 2 + Ax + Bx + B + Cx 2 + 2C

Equating the coefficients of x terms gives:

Equating the coefficients of x 2 terms gives:
7 = A + C, and since C = 5, A = 2
Equating the coefficients of x terms gives:
5 = A + B, and since A = 2, B = 3

6 = 3A
A=2

i.e.

From equation (1), since A = 2, C = −4
Hence

[Check: equating the constant terms gives:

3 + 6 x + 4x2 − 2 x3
−4x + 3
2
1
≡ + 2+ 2 x2 (x2 + 3) x x x +3

13 = B + 2C



When B = 3 and C = 5,
B + 2C = 3 + 10 = 13 = LHS]
Hence

7x2 + 5x + 13
(x2 + 2)(x + 1)

Problem 9.



Resolve

partial fractions.

2x + 3
5
+
( x2 + 2) (x + 1)

3 + 6x + 4x 2 − 2x 3 into x 2 (x 2 + 3)

Terms such as x 2 may be treated as (x + 0)2 , i.e. they are repeated linear factors.
Let

Now try the following exercise
Exercise 10 Further problems on partial fractions with quadratic factors
1.

x 2 − x − 13
(x 2 + 7)(x − 2)

2x + 3
1

(x 2 + 7) (x − 2)

2.

6x − 5
(x − 4)(x 2 + 3)

1
2−x
+
(x − 4) (x 2 + 3)

3.

15 + 5x + 5x 2 − 4x 3 x 2 (x 2 + 5)

1
2 − 5x
3
+
+
x x 2 (x 2 + 5)

Cx + D
A
B
3 + 6x + 4x 2 − 2x 3
≡ + 2+ 2
2 (x 2 + 3) x x x (x + 3)
Ax(x 2 + 3) + B(x 2 + 3) + (Cx + D)x 2

x 2 (x 2 + 3)

2
3 − 4x
1
+ 2+ 2 x x x +3

(1)

Partial fractions

4.

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

5. When solving the differential equation d2θ dθ
− 6 − 10θ = 20 − e2t by Laplace dt 2 dt transforms, for given boundary conditions, the

following expression for L{θ} results:
39 2 s + 42s − 40
2
L{θ} = s(s − 2)(s 2 − 6s + 10)
4s 3 −

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

1
5s − 3
2

+
2 − 6s + 10) s 2(s − 2) 2(s

19

Chapter 3

Logarithms
3.1

In another example, if we write down that 64 = 82 then the equivalent statement using logarithms is:

Introduction to logarithms

With the use of calculators firmly established, logarithmic tables are now rarely used for calculation. However, the theory of logarithms is important, for there are several scientific and engineering laws that involve the rules of logarithms.
From the laws of indices:

16 = 2

4

The number 4 is called the power or the exponent or the index. In the expression 24 , the number 2 is called the base.
In another example:

64 = 82

In this example, 2 is the power, or exponent, or index.
The number 8 is the base.
What is a logarithm?
Consider the expression 16 = 24.
An alternative, yet equivalent, way of writing this expression is: log2 16 = 4.
This is stated as ‘log to the base 2 of 16 equals 4’.
We see that the logarithm is the same as the power or index in the original expression. It is the base in the original expression which becomes the base of the logarithm. The two statements: 16 = 24 and log2 16 = 4 are equivalent. If we write either of them, we are automatically implying the other.
In general, if a number y can be written in the form a x , then the index ‘x’ is called the ‘logarithm of y to the base of a’,
i.e.

if y = a x then x = loga y

log8 64 = 2
In another example, if we write down that: log3 81 =4 then the equivalent statement using powers is:
34 = 81
So the two sets of statements, one involving powers and one involving logarithms, are equivalent.
Common logarithms
From above, if we write down that: 1000 = 103 , then
3 = log10 1000
This may be checked using the ‘log’ button on your calculator. Logarithms having a base of 10 are called common logarithms and log10 is often abbreviated to lg.
The following values may be checked by using a calculator: lg 27.5 = 1.4393 . . ., lg 378.1 = 2.5776 . . . and lg 0.0204 = −1.6903 . . .
Napierian logarithms
Logarithms having a base of e (where ‘e’ is a mathematical constant approximately equal to 2.7183) are called hyperbolic, Napierian or natural logarithms, and loge is usually abbreviated to ln.
The following values may be checked by using a calculator: ln 3.65 = 1.2947 . . ., ln 417.3 = 6.0338 . . . and ln 0.182 = −1.7037 . . .
More on Napierian logarithms is explained in Chapter 4 following. Here are some worked problems to help understanding of logarithms.

Logarithms
Problem 1. Evaluate log3 9.

Problem 6. Evaluate log3

Let x = log3 9 then 3 x = 9

from the definition of a logarithm,

3 x = 32

i.e.

1
.
81

Let x = log3

from which, x = 2

1
1
1 then 3 x =
= 4 = 3−4
81
81 3 from which, x = −4

log3 9 = 2

Hence,

Hence,

log3

1
= −4
81

Problem 2. Evaluate log10 10.
Problem 7. Solve the equation: lg x = 3.
Let x = log10 10 then 10 x = 10

from the

definition of a logarithm,
10 = 10 x i.e.
Hence,

from which, x = 1

1

log10 10 = 1

(which may be checked by a calculator)

Problem 3. Evaluate log16 8.

If lg x = 3 then log10 x = 3 and x = 103

i.e. x = 1000

Problem 8. Solve the equation: log2 x = 5.
If log2 x = 5 then x = 25 = 32

Let x = log16 8 then 16 x = 8

from the definition

Problem 9. Solve the equation: log5 x = −2.

of a logarithm,
i.e. (24 )x = 23 i.e. 24x = 23 from the laws of indices, from which,
Hence,

4x = 3 and x = log16 8 =

If log5 x = −2 then x = 5−2 =

3
4

3
4

1
1
=
52 25

Now try the following exercise

Problem 4. Evaluate lg 0.001. then 10x = 0.001

Let x = lg 0.001 = log10 0.001
i.e.
Hence,

10 x = 10−3

from which, x = −3

lg 0.001 = −3 (which may be checked

Exercise 11 logarithms Further problems on laws of

In Problems 1 to 11, evaluate the given expressions: 1. log10 10000

[4]

2. log2 16

3. log5 125

[3]

4. log2 1
8

by a calculator)
Problem 5. Evaluate ln e.
5. log8 2
7. lg 100

Let x = ln e = loge e then ex = e
i.e.
Hence,

ex = e1 from which, x = 1 ln e = 1 (which may be checked by a calculator)

1
3
[2]

9. log4 8

1

11. ln e2

6. log7 343

1
2

[2]

[4]
[−3]
[3]

8. lg 0.01

[−2]

10. log27 3

1
3

21

22 Higher Engineering Mathematics
The following may be checked using a calculator:
In Problems 12 to 18 solve the equations:
12.

log10 x = 4

13.

lg x = 5

14.

log3 x = 2

15.

1 log4 x = −2
2

16.

lg x = −2

17.

log8 x = −

18.

ln x = 3

lg 52 = lg 25 = 1.39794. . .

[10000]
[100000]
[9]
1
32
[0.01]
1
16

4
3

[e3 ]

Also, 2 lg 5 = 2 × 0.69897. . . = 1.39794. . . lg 52 = 2 lg 5

Hence,

Here are some worked problems to help understanding of the laws of logarithms.
Problem 10. Write log 4 + log 7 as the logarithm of a single number. log 4 + log 7 = log (7 × 4) by the first law of logarithms
= log 28

3.2

Laws of logarithms

Problem 11. Write log 16 − log 2 as the logarithm of a single number.

There are three laws of logarithms, which apply to any base: (i) To multiply two numbers:

log 16 − log 2 = log

by the second law of logarithms

log (A × B) = log A + log B
The following may be checked by using a calculator: lg 10 = 1
Also, lg 5 + lg 2 = 0.69897. . .
+ 0.301029. . . = 1
Hence,
lg (5 × 2) = lg 10 = lg 5 + lg 2

16
2

= log 8
Problem 12. Write 2 log 3 as the logarithm of a single number.
2 log 3 = log 32

by the third law of logarithms

= log 9

(ii) To divide two numbers:
A
log
B

= log A − log B

1
Problem 13. Write log 25 as the logarithm of a
2
single number.

The following may be checked using a calculator:
5
ln
2
Also,
Hence,

= ln 2.5 = 0.91629. . .

ln 5 − ln 2 = 1.60943. . . − 0.69314. . .
= 0.91629. . .
5
= ln 5 − ln 2 ln 2

(iii) To raise a number to a power: log An = n log A

1
1
log 25 = log 25 2 by the third law of logarithms
2

= log 25 = log 5

Problem 14.

Simplify: log 64 − log 128 + log32.

64 = 26, 128 = 27 and 32 = 25
Hence, log 64 − log 128 + log32
= log 26 − log 27 + log 25

Logarithms
= 6 log2 − 7 log 2 + 5 log2 by the third law of logarithms
= 4 log 2
1
1
Problem 15. Write log16 + log27 − 2 log5
2
3 as the logarithm of a single number.


4
8× 5
i.e. log
81

Problem 18. Evaluate: log 25 − log125 + 1 log 625
2
.
3 log5

1
1
log 16 + log 27 − 2 log5
2
3
1

1

1

= log 16 2 + log 27 3 − log 52 by the third law of logarithms


3
= log 16 + log 27 − log 25 by the laws of indices

1 log 25 − log125 + 2 log 625
3 log5

=

4×3
= log
25
by the first and second laws of logarithms
= log 0.48

1 log 52 − log 53 + 2 log 54
3 log5

=

= log4 + log 3 − log 25

12
= log
25

= log 23 + log 5 4 − log 34 by the laws of indices
1
= 3 log 2 + log 5 − 4 log 3
4
by the third law of logarithms

2 log5 − 3 log 5 + 4 log 5 1 log5 1
2
=
=
3 log5
3 log5 3

Problem 19. Solve the equation: log(x − 1) + log(x + 8) = 2 log(x + 2).
LHS = log (x − 1) + log(x + 8)

Problem 16. Write (a) log30 (b) log 450 in terms of log 2, log3 and log 5 to any base.

= log (x − 1)(x + 8) from the first law of logarithms

(a) log 30 = log(2 × 15) = log(2 × 3 × 5)

= log (x 2 + 7x − 8)

= log 2 + log 3 + log 5 by the first law of logarithms

RHS = 2 log(x + 2) = log (x + 2)2

(b) log 450 = log(2 × 225) = log(2 × 3 × 75)

from the third law of logarithms

= log(2 × 3 × 3 × 25)

= log(x 2 + 4x + 4)

= log(2 × 32 × 52)
= log2 + log 32 + log 52 by the first law of logarithms

Hence,

log(x 2 + 7x − 8) = log (x 2 + 4x + 4) x 2 + 7x − 8 = x 2 + 4x + 4

i.e. log 450 = log 2 + 2 log 3 + 2 log 5 by the third law of logarithms

from which,
i.e.

7x − 8 = 4x + 4


4
8× 5
Problem 17. Write log
81
log 2, log3 and log 5 to any base.

i.e.

3x = 12

and

x=4


4
8× 5 log 81

in terms of

Problem 20. Solve the equation:

4
= log 8 + log 5 − log 81 by the first and second laws of logarithms

1 log 4 = log x.
2

1
1
log 4 = log4 2 from the third law of logarithms

2
= log 4 from the laws of indices

23

24 Higher Engineering Mathematics
1
log4 = log x
2
√ log 4 = log x

Hence, becomes 1
1
log 8 − log81 + log 27
3
2
1
log 4 − 2 log 3 + log45
9.
2
1
10. log 16 + 2 log3 − log 18
4
11. 2 log2 + log 5 − log 10
8.

log2 = log x

i.e.

2=x

from which,

i.e. the solution of the equation is: x = 2
Problem 21. Solve the equation: log x 2 − 3 − log x = log2.

log

x

= log 2

from which,

x2 − 3
=2
x

Rearranging gives:

x 2 − 3 = 2x

log 27 − log9 + log 81
[log 243 or log 35 or 5 log3]

13.

log 64 + log 32 − log 128
[log16 or log24 or 4 log2] log 8 − log4 + log 32
[log64 or log 26 or 6 log2]

15.
16.

x = 3 or x = −1

from which,

x = −1 is not a valid solution since the logarithm of a negative number has no real root.

1
1
2 log 16 − 3 log 8

log 4 log 9 − log3 + 1 log 81
2
2 log3

Now try the following exercise
Further problems on laws of

In Problems 1 to 11, write as the logarithm of a single number:

[0.5]
[1.5]

Solve the equations given in Problems 17 to 22:
17.

log x 4 − log x 3 = log5x − log 2x

18.

log 2t 3 − log t = log 16 + logt

19.

Hence, the solution of the equation is: x = 3

Exercise 12 logarithms 12.

(x − 3)(x + 1) = 0

Factorizing gives:

[log 2]

Evaluate the expressions given in Problems 15 and 16:

x 2 − 2x − 3 = 0

and

[log 1 = 0]

14.

x2 − 3 x from the second law of logarithms

Hence,

[log 10]

Simplify the expressions given in Problems 12 to 14:

log x 2 − 3 − log x = log

x2 − 3

[log 6]

2 logb 2 − 3 logb

20.

log (x + 1) + log(x − 1) = log 3

21.
22.

= log8b − log 4b

1 log 27 = log(0.5a)
3
log x 2 − 5 − log x = log 4

[x = 2.5]
[t = 8]
[b = 2]
[x = 2]
[a = 6]
[x = 5]

1.

log 2 + log 3

[log 6]

2.

log 3 + log 5

[log 15]

3.

log 3 + log 4 − log 6

[log 2]

4.

log 7 + log 21 − log49

[log 3]

5.

2 log 2 + log 3

6.

2 log 2 + 3 log5

[log 500]

The laws of logarithms may be used to solve certain equations involving powers—called indicial equations. For example, to solve, say, 3 x = 27, logarithms to a base of 10 are taken of both sides,

7.

1
2 log 5 − log 81 + log 36
2

[log 100]

i.e. log10 3x = log10 27

[log 12]

3.3

Indicial equations

and x log10 3 = log10 27, by the third law of logarithms

Logarithms
Rearranging gives x= log10 27 1.43136 . . .
=
=3 log10 3
0.4771 . . .

25

log10 41.15
= 0.50449
3.2
Thus x = antilog 0.50449 =100.50449 = 3.195 correct to
4 significant figures.
Hence log10 x =

which may be readily checked
8
log8 is not equal to lg log2 2

Note,

Now try the following exercise
Exercise 13

Problem 22. Solve the equation 2 x = 3, correct to
4 significant figures.
Taking logarithms to base 10 of both sides of 2 x = 3 gives: log10 2x = log10 3
i.e.

x log10 2 = log10 3

Solve the following indicial equations for x, each correct to 4 significant figures:
1. 3x = 6.4

[1.690]

2. 2 x = 9

[3.170]

3. 2 x−1 = 32x−1

log10 3 0.47712125 . . .
=
x= log10 2 0.30102999 . . .
= 1.585, correct to 4 significant figures

[0.2696]

x 1.5 = 14.91

[6.058]

5. 25.28 =4.2x

[2.251]

6. 42x−1 = 5x+2

[3.959]

4.

Rearranging gives:

Indicial equations

7.

x −0.25 = 0.792

[2.542]

8. 0.027x = 3.26 equation 2 x+1 = 32x−5

Problem 23. Solve the correct to 2 decimal places.

9. The decibel gain n of an amplifier is given by: n = 10 log10

Taking logarithms to base 10 of both sides gives: log10 2x+1 = log10 32x−5

x log10 2 + log10 2 = 2x log10 3 − 5 log10 3 x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771)
i.e.

P2
P1

where P1 is the power input and P2 is the
P2
power output. Find the power gain when P1 n =25 decibels.
[316.2]

(x + 1) log10 2 = (2x − 5) log10 3

i.e.

[−0.3272]

0.3010x + 0.3010 = 0.9542x − 2.3855

Hence

3.4
2.3855 + 0.3010 = 0.9542x − 0.3010x
2.6865 = 0.6532x

from which x =

2.6865
= 4.11, correct to
0.6532
2 decimal places

Problem 24. Solve the equation x 3.2 = 41.15, correct to 4 significant figures.
Taking logarithms to base 10 of both sides gives: log10 x 3.2 = log10 41.15
3.2 log10 x = log10 41.15

Graphs of logarithmic functions

A graph of y = log10 x is shown in Fig. 3.1 and a graph of y = loge x is shown in Fig. 3.2. Both are seen to be of similar shape; in fact, the same general shape occurs for a logarithm to any base.
In general, with a logarithm to any base a, it is noted that: (i) loga1 = 0
Let loga = x, then a x = 1 from the definition of the logarithm.
If a x = 1 then x = 0 from the laws of indices.
Hence loga 1 =0. In the above graphs it is seen that log10 1 = 0 and loge 1 = 0

26 Higher Engineering Mathematics y 2

y

1.0
1

0.5
0

0

1 x y 5 log10x

20.5

2
3

3
2

1

21

x
0.5

0.2

0.1

0.48 0.30 0 2 0.30 2 0.70 2 1.0

1

2

3

4

5

6

x

x
6
5
4
3
2 1 0.5
0.2
0.1 y 5 loge x 1.79 1.61 1.39 1.10 0.69 0 20.69 21.61 22.30

22

Figure 3.2

21.0

Figure 3.1

(ii) logaa = 1
Let loga a = x then a x = a from the definition of a logarithm.
If a x = a then x = 1.
Hence loga a = 1. (Check with a calculator that log10 10 = 1 and loge e = 1)

(iii) loga0 → −∞
Let loga 0 = x then a x = 0 from the definition of a logarithm.
If a x = 0, and a is a positive real number, then x must approach minus infinity. (For example, check with a calculator, 2−2 = 0.25,
2−20 = 9.54 × 10−7, 2−200 = 6.22 × 10−61, and so on)
Hence loga 0 → −∞

Chapter 4

Exponential functions
4.1 Introduction to exponential functions An exponential function is one which contains ex , e being a constant called the exponent and having an approximate value of 2.7183. The exponent arises from the natural laws of growth and decay and is used as a base for natural or Napierian logarithms.
The most common method of evaluating an exponential function is by using a scientific notation calculator. Use your calculator to check the following values: e1 = 2.7182818, correct to 8 significant figures, e−1.618 = 0.1982949, each correct to 7 significant figures, 0.0256 e5.21 − e2.49 = 0.0256 (183.094058 . . .
− 12.0612761 . . .)
= 4.3784, correct to 4 decimal places.
Problem 2. Evaluate the following correct to 4 decimal places, using a calculator:
5

5

e0.25 − e−0.25 e0.25 + e−0.25
=5

1.28402541 . . . − 0.77880078 . . .
1.28402541 . . . + 0.77880078 . . .

=5

0.5052246 . . .
2.0628262 . . .

e0.12 = 1.1275, correct to 5 significant figures, e−1.47 = 0.22993, correct to 5 decimal places, e−0.431 = 0.6499, correct to 4 decimal places, e 9.32

e0.25 − e−0.25 e0.25 + e−0.25

= 1.2246, correct to 4 decimal places.

= 11159, correct to 5 significant figures,

e−2.785 = 0.0617291, correct to 7 decimal places.

Problem 1. Evaluate the following correct to 4 decimal places, using a calculator:
0.0256 e5.21 − e2.49

Problem 3. The instantaneous voltage v in a capacitive circuit is related to time t by the equation: v = V e−t /CR where V , C and R are constants. Determine v, correct to 4 significant figures, when t = 50 ms, C = 10 μF, R = 47 k and V = 300 volts. v = V e−t /CR = 300e(−50×10

−3)/(10×10−6 ×47×103)

28 Higher Engineering Mathematics
Using a calculator, v = 300e−0.1063829 ... = 300(0.89908025 . . .)
= 269.7 volts
Now try the following exercise
Exercise 14 Further problems on evaluating exponential functions

(where 3! = 3 ×2 × 1 and is called ‘factorial 3’)
The series is valid for all values of x.
The series is said to converge, i.e. if all the terms are added, an actual value for e x (where x is a real number) is obtained. The more terms that are taken, the closer will be the value of ex to its actual value. The value of the exponent e, correct to say 4 decimal places, may be determined by substituting x = 1 in the power series of equation (1). Thus, e1 = 1 + 1 +

1. Evaluate the following, correct to 4 significant figures: (a) e−1.8 (b) e−0.78 (c) e10
[(a) 0.1653 (b) 0.4584 (c) 22030]

+

2. Evaluate the following, correct to 5 significant figures: (a) e1.629 (b) e−2.7483 (c) 0.62e4.178
[(a) 5.0988 (b) 0.064037 (c) 40.446]
In Problems 3 and 4, evaluate correct to 5 decimal places: 5e2.6921
1
3. (a) e3.4629 (b) 8.52e−1.2651 (c) 1.1171
7
3e
[(a) 4.55848 (b) 2.40444 (c) 8.05124]
5.6823
e2.1127 − e−2.1127
(b)
e−2.1347
2
−1.7295 − 1)
4(e
(c) e3.6817 [(a) 48.04106 (b) 4.07482 (c) −0.08286]

4. (a)

5. The length of a bar, l, at a temperature θ is given by l = l0 eαθ , where l0 and α are constants. Evaluate 1, correct to 4 significant figures, where l0 = 2.587, θ = 321.7 and
[2.739]
α = 1.771 × 10−4.
6. When a chain of length 2L is suspended from two points, 2D metres apart, √ the same horon
L+ L 2 +k 2
. Evalizontal level: D = k ln k uate D when k = 75 m and L = 180 m.
[120.7m]

4.2

The power series for ex

The value of e x can be calculated to any required degree of accuracy since it is defined in terms of the following power series: ex = 1 + x +

x2 x3 x4
+
+
+···
2! 3!
4!

(1)2 (1)3 (1)4 (1)5
+
+
+
2!
3!
4!
5!

(1)6 (1)7 (1)8
+
+
+···
6!
7!
8!

= 1 + 1 + 0.5 + 0.16667 + 0.04167
+ 0.00833 + 0.00139 + 0.00020
+ 0.00002 + · · ·
i.e.

e = 2.71828 = 2.7183, correct to 4 decimal places

The value of e0.05, correct to say 8 significant figures, is found by substituting x = 0.05 in the power series for e x . Thus e0.05 = 1 + 0.05 +

(0.05)2 (0.05)3
+
2!
3!

(0.05)4 (0.05)5
+
+···
4!
5!
= 1 + 0.05 + 0.00125 + 0.000020833
+

+ 0.000000260 + 0.000000003 and by adding, e0.05 = 1.0512711, correct to 8 significant figures
In this example, successive terms in the series grow smaller very rapidly and it is relatively easy to determine the value of e0.05 to a high degree of accuracy.
However, when x is nearer to unity or larger than unity, a very large number of terms are required for an accurate result. If in the series of equation (1), x is replaced by −x, then, e−x = 1 + (−x) +
i.e. e−x = 1 − x +

(−x)2 (−x)3
+
+···
2!
3!

x2 x3

+···
2! 3!

In a similar manner the power series for e x may be used to evaluate any exponential function of the form a ekx ,

Exponential functions where a and k are constants. In the series of equation (1), let x be replaced by kx. Then, a ekx = a 1 + (kx) +

(kx)2 (kx)3
+
+···
2!
3!

Thus 5 e2x = 5 1 + (2x) +

(2x)2 (2x)3
+
+···
2!
3!

= 5 1 + 2x +
i.e.

4x 2
2

+

8x 3
6

+···

ex = 1 + x +

x2
2!

+

x3
3!

+

x4
4!

+···

(0.5)2
(0.5)3
e0.5 = 1 + 0.5 +
+
(2)(1) (3)(2)(1)
(0.5)4

(0.5)5

+
(4)(3)(2)(1) (5)(4)(3)(2)(1)
(0.5)6
+
(6)(5)(4)(3)(2)(1)

+

= 1 + 0.5 + 0.125 + 0.020833
+ 0.0026042 + 0.0002604
+ 0.0000217
i.e.

− 1+x +

e0.5

= 1.64872, correct to 6 significant figures

Hence 5e0.5 = 5(1.64872) = 8.2436, correct to 5 significant figures

= −1 − x + x 2 −
+

Hence e x (x 2 − 1) x2 x3 x4 x5
= 1+x +
+
+
+
+ · · · (x 2 − 1)
2!
3! 4!
5!

x4 x4

2! 4!

+

x2
2!

+ x3 −

x5 x5

3!
5!

x3
3!

+···

1
5
11
19 5
= − 1 −x + x2 + x3 + x4 + x 2
6
24
120
when expanded as far as the term in x 5 .
Now try the following exercise
Exercise 15 series for ex

Further problems on the power

1. Evaluate 5.6 e−1 , correct to 4 decimal places,
[2.0601]
using the power series for e x .
2. Use the power series for ex to determine, correct to 4 significant figures, (a) e2 (b) e−0.3 and check your result by using a calculator.
[(a) 7.389 (b) 0.7408]
3. Expand (1 − 2x) e2x as far as the term in x 4 .
8x 3
1 − 2x 2 −
− 2x 4
3
4. Expand 2 ex

The power series for ex is, x2 x3 x4 x5
+
+
+
+···
2! 3!
4! 5!

x2 x3 x4 x5
+
+
+
+···
2!
3! 4!
5!

ex (x 2 − 1)

Problem 5. Expand ex (x 2 − 1) as far as the term in x 5 .

ex = 1 + x +

x4 x5
+
+···
2!
3!

Grouping like terms gives:

4
5 e2x = 5 1 + 2x + 2x 2 + x 3 + · · ·
3

Problem 4. Determine the value of 5 e0.5 , correct to 5 significant figures by using the power series for ex .

Hence

= x2 + x3 +

29

4.3

2

1

x 2 to six terms.

⎡ 1
5
9
1 13
2x 2 + 2x 2 + x 2 + x 2


3




17
21
1
1
+ x2 + x2
12
60

Graphs of exponential functions

Values of ex and e−x obtained from a calculator, correct to 2 decimal places, over a range x = −3 to x = 3, are shown in the following table.

30 Higher Engineering Mathematics x ex

y

−3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0
0.05

0.08

0.14

0.22

0.37

e−x 20.09 12.18

7.39

4.48

2.72

5

0.61 1.00
1.65 1.00

3.87

y5 2e0.3x

4
3

x

0.5

1.0

1.5

2.0

2.5

3.0

2

ex

1.65

2.72

4.48

7.39

12.18

20.09

1

e−x

0.61

0.37

0.22

0.14

0.08

0.05

1.6

Figure 4.1 shows graphs of y = ex and y = e−x

y 5 ex
16

−1.5 −1.0 −0.5

x
8

−2x e−2x 4

22

21

0

1

2
2.2

3

x

A table of values is drawn up as shown below.

12

23

21
0
20.74

Problem 7. Plot a graph of y = 1 e−2x over the
3
range x = −1.5 to x = 1.5. Determine from the graph the value of y when x = −1.2 and the value of x when y = 1.4.

20 y 22

Figure 4.2

y

5 e2x

23

1

2

3

3

2

0.5

1.0

1.5

0

−1

−2

−3

20.086 7.389 2.718 1.00 0.368 0.135 0.050

1 −2x
6.70
e
3

x

1

0

2.46 0.91 0.33 0.12 0.05 0.02

A graph of 1 e−2x is shown in Fig. 4.3.
3
Figure 4.1

y
1 e22x

y 53

7
6

Problem 6. Plot a graph of y = 2 e0.3x over a range of x = − 2 to x = 3. Hence determine the value of y when x = 2.2 and the value of x when y = 1.6.

5
4

3.67

3
2

A table of values is drawn up as shown below.

1.4

1

x

−3

−2

−1

0

1

2

3

21.5 21.0 20.5

0.5

1.0

1.5

x

21.2 20.72

0.3x

−0.9 −0.6 −0.3

e0.3x

0.407 0.549 0.741 1.000 1.350 1.822 2.460

0

0.3

0.6

0.9

2 e0.3x 0.81 1.10 1.48 2.00 2.70 3.64 4.92
A graph of y = 2 e0.3x is shown plotted in Fig. 4.2.
From the graph, when x = 2.2, y = 3.87 and when y = 1.6, x = −0.74.

Figure 4.3

From the graph, when x = −1.2, y = 3.67 and when y = 1.4, x = −0.72.
Problem 8. The decay of voltage, v volts, across a capacitor at time t seconds is given by v = 250 e

−t
3 .

Draw a graph showing the natural

Exponential functions decay curve over the first 6 seconds. From the graph, find (a) the voltage after 3.4 s, and (b) the time when the voltage is 150 V.

of y when x = 1.4 and the value of x when y = 4.5.
[3.95, 2.05]
2. Plot a graph of y = 1 e−1.5x over a range
2
x = −1.5 to x = 1.5 and hence determine the value of y when x = −0.8 and the value of x when y = 3.5.
[1.65, −1.30]

A table of values is drawn up as shown below. t 0

e

−t
3

2

3

1.00

−t v = 250 e 3

0.7165 0.5134 0.3679

250.0

179.1

t e 1

128.4

6

0.1889

0.1353

65.90

−t v = 250 e 3

5

0.2636

−t
3

3. In a chemical reaction the amount of starting material C cm3 left after t minutes is given by
C = 40 e−0.006t . Plot a graph of C against t and determine (a) the concentration C after 1 hour, and (b) the time taken for the concentration to decrease by half.
[(a) 28 cm3 (b) 116 min]

91.97

4

47.22

33.83

The natural decay curve of v = 250 e
Fig. 4.4.

−t
3

is shown in

250 t y 5 250e2 3

Voltage v (volts)

200

150

4. The rate at which a body cools is given by θ = 250 e−0.05t where the excess of temperature of a body above its surroundings at time t minutes is θ ◦ C. Plot a graph showing the natural decay curve for the first hour of cooling. Hence determine (a) the temperature after 25 minutes, and (b) the time when the temperature is 195◦C.
[(a) 70◦C (b) 5 min]

4.4

100
80
50

0

1 1.5 2
3 3.4 4
Time t(seconds)

5

6

Figure 4.4

Napierian logarithms

Logarithms having a base of ‘e’ are called hyperbolic,
Napierian or natural logarithms and the Napierian logarithm of x is written as loge x, or more commonly as ln x. Logarithms were invented by John Napier, a
Scotsman (1550–1617).
The most common method of evaluating a Napierian logarithm is by a scientific notation calculator. Use your calculator to check the following values:

From the graph:
(a)

31

when time t = 3.4 s, voltage v = 80 V and

(b) when voltage v = 150 V, time t = 1.5 s.
Now try the following exercise

ln 4.328 = 1.46510554 . . .
= 1.4651, correct to 4 decimal places ln 1.812 = 0.59443, correct to 5 significant figures ln 1 = 0 ln 527 = 6.2672, correct to 5 significant figures ln 0.17 = −1.772, correct to 4 significant figures

Exercise 16 Further problems on exponential graphs

ln 0.00042 = −7.77526, correct to 6 significant figures 1. Plot a graph of y = 3 e0.2x over the range x = −3 to x = 3. Hence determine the value

ln e3 = 3 ln e1 = 1

32 Higher Engineering Mathematics
From the last two examples we can conclude that: loge ex = x
This is useful when solving equations involving exponential functions. For example, to solve e3x = 7, take
Napierian logarithms of both sides, which gives: ln e3x = ln 7
i.e.
from which

3x = ln 7
1
x = ln 7 = 0.6486, correct to 4
3
decimal places.

Taking the reciprocal of both sides gives:
4
1
= e3x
=
9 e−3x
Taking Napierian logarithms of both sides gives:
4
= ln(e3x ) ln 9
4
= 3x
Since loge eα = α, then ln
9
1
4
1
Hence, x = ln
= (−0.81093) = −0.2703,
3
9
3
correct to 4 significant figures. t Problem 9. Evaluate the following, each correct to 5 significant figures:
(a)
(a)

(b)

(c)

1 ln 7.8693
3.17 ln 24.07
.
ln 4.7291 (b)
(c)
2
7.8693
e−0.1762

1
1
ln 4.7291 = (1.5537349 . . .) = 0.77687,
2
2 correct to 5 significant figures ln 7.8693 2.06296911 . . .
=
= 0.26215,
7.8693
7.8693 correct to 5 significant figures
3.17 ln 24.07 3.17(3.18096625 . . .)
=
e−0.1762
0.83845027 . . .
= 12.027, correct to 5 significant figures.

Problem 10.

(b)

t

Rearranging 32 = 70(1 − e− 2 ) gives: t 32
= 1 − e− 2
70
t
32 38 and e− 2 = 1 −
=
70 70
Taking the reciprocal of both sides gives: t 70 e2 =
38
Taking Napierian logarithms of both sides gives: t 70 ln e 2 = ln
38

i.e.

t
70
= ln
2
38

Evaluate the following:

ln e2.5
5e2.23 lg 2.23
(a)
(b)
(correct to 3
0.5
ln 2.23 lg 10 decimal places).
(a)

Problem 12. Given 32 = 70 1 − e− 2 determine the value of t , correct to 3 significant figures.

70
38

= 1.22, correct to 3 signifi-

cant figures.
Problem 13.

ln e2.5
2.5
=
=5
lg 100.5 0.5

Solve the equation: 2.68 = ln

4.87 x to find x.

5e2.23 lg 2.23 ln 2.23
5(9.29986607 . . .)(0.34830486 . . .)
=
0.80200158 . . .
= 20.194, correct to 3 decimal places.

Problem 11. Solve the equation: 9 = 4e−3x to find x, correct to 4 significant figures.
Rearranging 9 = 4e−3x gives:

from which, t = 2 ln

9
= e−3x
4

From the definition of a logarithm, since
4.87
4.87 then e2.68 =
2.68 = ln x x
4.87
Rearranging gives: x = 2.68 = 4.87e−2.68 e i.e. x = 0.3339, correct to 4 significant figures.
7
Problem 14. Solve = e3x correct to 4 signi4 ficant figures.

Exponential functions
Taking natural logs of both sides gives: ln 7
= 3x ln e
4

ln

Since ln e = 1

Using the quadratic formula,

7
= ln e3x
4

ln

7
= 3x
4

x = 2.847 or −3.8471

i.e.

x = 0.1865, correct to 4 significant figures.

i.e.

−1 ±

12 − 4(1)(−10.953)
2

−1 ± 44.812 −1 ± 6.6942
=
=
2
2

x=

0.55962 = 3x

i.e.

Problem 15. Solve: e x−1 = 2e3x−4 correct to 4 significant figures.

x = −3.8471 is not valid since the logarithm of a negative number has no real root.
Hence, the solution of the equation is: x = 2.847
Now try the following exercise

Taking natural logarithms of both sides gives:
Exercise 17 Further problems on evaluating Napierian logarithms

ln e x−1 = ln 2e3x−4 and by the first law of logarithms, ln e x−1

= ln 2 + ln

In Problems 1 and 2, evaluate correct to 5 significant figures:

e3x−4

1 ln 82.473 ln 5.2932 (b)
3
4.829
5.62 ln 321.62
(c)
e1.2942
[(a) 0.55547 (b) 0.91374 (c) 8.8941]

x − 1 = ln 2 + 3x − 4

i.e.

1. (a)

Rearranging gives: 4 − 1 − ln 2 = 3x − x
3 − ln 2 = 2x

i.e.

3 − ln 2
2
= 1.153

x=

from which,

In Problems 3 to 7 solve the given equations, each correct to 4 significant figures.

Rearranging gives: ln(x − 2)2 − ln(x − 2) + ln(x + 3) = 1.6

and
i.e.

3. ln x = 2.10

[8.166]

4. 24 + e2x = 45

[1.522]

5. 5 =

and by the laws of logarithms,

Cancelling gives:

5e−0.1629
1.786 ln e1.76
(b)
lg 101.41
2 ln 0.00165 ln 4.8629 − ln 2.4711
(c)
5.173
[(a) 2.2293 (b) −0.33154 (c) 0.13087]

2. (a)

Problem 16. Solve, correct to 4 significant figures: ln(x − 2)2 = ln(x − 2) − ln(x + 3) + 1.6

ln

33

(x − 2)2 (x + 3)
= 1.6
(x − 2) ln {(x − 2)(x + 3)} = 1.6
(x − 2)(x + 3) = e1.6 x2 + x

−6 =

e1.6

or

x 2 + x − 6 − e1.6 = 0

i.e.

x 2 + x − 10.953 = 0

e x+1 − 7

[1.485]

6. 1.5 = 4e2t
7. 7.83 =

[−0.4904]

2.91e−1.7x

[−0.5822]

t

8. 16 = 24 1 − e− 2 x 4.64
1.59
= 2.43
10. 3.72 ln x 9. 5.17 = ln

[2.197]
[816.2]
[0.8274]

34 Higher Engineering Mathematics y 11.
12.

5=8

−x
1−e 2

[1.962]

ln(x + 3) − ln x = ln(x − 1)
− 1)2

− ln 3 = ln(x − 1)

A y 5 Ae2kx

[3]
[4]

13.

ln(x

14.

ln(x + 3) + 2 = 12 − ln(x − 2)

[147.9]

15.

e(x+1)

[4.901]

16.

ln(x + 1)2 = 1.5 − ln(x − 2)
+ ln(x + 1)

17.

18.

19.

=

3e(2x−5)

[3.095] y A

y 5 A(12e2kx )

R1
P
= 10 log10 find the value of R1
If
Q
R2
when P = 160, Q = 8 and R2 = 5.
[500]
If U2 = U1 e

W
PV

x

(a)

Transpose: b = ln t − a ln D to make t the subject. a
[t = eb+a ln D = eb ea ln D = eb eln D
i.e. t = eb D a ]

formula.
20.

0

0

make W the subject of the
U2
W = PV ln
U1

The work done in an isothermal expansion of a gas from pressure p1 to p2 is given by: w = w0 ln

p1 p2 If the initial pressure p1 = 7.0 kPa, calculate the final pressure p2 if w = 3 w0 .
[ p2 = 348.5 Pa]

x

(b)

Figure 4.5

(v) Biological growth

y = y0 ekt

(vi) Discharge of a capacitor q = Q e−t/CR
(vii) Atmospheric pressure

p = p0 e−h/c

(viii) Radioactive decay

N = N0 e−λt

Laws of growth and decay

The laws of exponential growth and decay are of the form y = A e−kx and y = A(1 − e−kx ), where A and k are constants. When plotted, the form of each of these equations is as shown in Fig. 4.5. The laws occur frequently in engineering and science and examples of quantities related by a natural law include.
(i) Linear expansion

l = l0 eαθ

(ii) Change in electrical resistance with temperature
Rθ = R0 eαθ
(iii) Tension in belts

i = I e− Rt /L

(x) Growth of current in a capacitive circuit

4.5

(ix) Decay of current in an inductive circuit

i = I (1 − e−t/CR )

Problem 17. The resistance R of an electrical conductor at temperature θ ◦ C is given by
R = R0 eαθ , where α is a constant and
R0 = 5 × 103 ohms. Determine the value of α, correct to 4 significant figures, when
R = 6 ×103 ohms and θ = 1500◦C. Also, find the temperature, correct to the nearest degree, when the resistance R is 5.4 ×103 ohms.
R
= eαθ .
R0
Taking Napierian logarithms of both sides gives:

Transposing R = R0 eαθ gives

T1 = T0 eμθ

(iv) Newton’s law of cooling θ = θ0 e−kt

ln

R
= ln eαθ = αθ
R0

Exponential functions
Hence α =
=

6 × 103
1
1
R
= ln ln θ R0
1500
5 × 103
1
(0.1823215 . . .)
1500

= 1.215477 · · ·× 10−4
Hence α = 1.215 × 10−4 , correct to 4 significant figures.
R
= αθ
From above, ln
R0
θ=

hence

1
R
ln α R0

When R = 5.4 × 103, α = 1.215477 . . . × 10−4 and
R0 = 5 ×103 θ= =

5.4 × 103
1
ln
−4
1.215477 . . . × 10
5 × 103
104
(7.696104 . . . × 10−2)
1.215477 . . .


= 633 C, correct to the nearest degree.

35

Problem 19. The current i amperes flowing in a capacitor at time t seconds is given by
−t

i = 8.0(1 − e CR ), where the circuit resistance R is
25 ×103 ohms and capacitance C is
16 ×10−6 farads. Determine (a) the current i after
0.5 seconds and (b) the time, to the nearest millisecond, for the current to reach 6.0 A. Sketch the graph of current against time.
(a)

−t

Current i = 8.0(1 − e CR )
−0.5

= 8.0[1 − e (16 ×10−6 )(25 ×103 ) ] =8.0(1 − e−1.25)
= 8.0(1 − 0.2865047 . . .) = 8.0(0.7134952 . . .)
= 5.71 amperes
−t

(b) Transposing i = 8.0(1 − e CR )

gives

−t i = 1 −e CR
8.0
−t

from which, e CR = 1 −

i
8.0 − i
=
8.0
8.0

Taking the reciprocal of both sides gives:
Problem 18. In an experiment involving
Newton’s law of cooling, the temperature θ(◦ C) is given by θ = θ0 e−kt . Find the value of constant k when θ0 = 56.6◦ C, θ = 16.5◦ C and t = 83.0 seconds.
Transposing

θ0
1
= −kt = ekt θ e

t
8.0
= ln
CR
8.0 − i
Hence
t = CRln

Taking Napierian logarithms of both sides gives: θ0 ln = kt θ from which,
56.6
1
1 θ0 ln k = ln = t θ
83.0
16.5
1
=
(1.2326486 . . .)
83.0
Hence k = 1.485 × 10−2

8.0
8.0 − i

Taking Napierian logarithms of both sides gives:

θ = θ0 e−kt gives θ = e−kt θ0 from which

t

e CR =

8.0
8.0 − i

= (16 × 10−6)(25 × 103 ) ln

8.0
8.0 − 6.0

when i = 6.0 amperes,
i.e.

t=

8.0
400
ln
103
2.0

= 0.4 ln 4.0

= 0.4(1.3862943 . . .) = 0.5545 s
= 555 ms, to the nearest millisecond.
A graph of current against time is shown in Fig. 4.6.

36 Higher Engineering Mathematics i (A)

Hence the time for the temperature θ 2 to be one half of the value of θ 1 is 41.6 s, correct to 1 decimal place.

8
6
5.71

i 5 8.0 (12e2t/CR)

4

Now try the following exercise

2

Exercise 18 Further problems on the laws of growth and decay

0

0.5
0.555

1.0

1.5

t(s)

Figure 4.6

Problem 20. The temperature θ2 of a winding which is being heated electrically at time t is given
−t
θ2 = θ1 (1 − e τ )

by: where θ1 is the temperature (in degrees Celsius) at time t = 0 and τ is a constant.
Calculate,
(a)

θ1 , correct to the nearest degree, when θ2 is
50◦ C, t is 30 s and τ is 60 s

(b) the time t , correct to 1 decimal place, for θ2 to be half the value of θ1 .
(a) Transposing the formula to make θ1 the subject gives: θ1 =

θ2

−t
(1 − e T )

=

50
1−e

−30
60

50
50
=
=
1 − e−0.5 0.393469 . . .
i.e. θ 1 = 127◦ C, correct to the nearest degree.
(b) Transposing to make t the subject of the formula gives: −t θ2 =1−e τ θ1 −t θ2 from which, e τ = 1 − θ1 t θ2 Hence
− = ln 1 − τ θ1 θ2 i.e. t = −τ ln 1 − θ1 1
Since
θ2 = θ1
2
1 t = −60 ln 1 −
2
= −60 ln 0.5 = 41.59 s

1. The temperature, T ◦C, of a cooling object varies with time, t minutes, according to the equation: T = 150e−0.04t . Determine the temperature when (a) t = 0, (b) t = 10 minutes.
[(a) 150◦ C (b) 100.5◦ C ]
2. The pressure p pascals at height h metres
−h

above ground level is given by p = p0 e C , where p0 is the pressure at ground level and C is a constant. Find pressure p when p0 = 1.012 × 105 Pa, height h = 1420 m, and
C = 71500.
[99210]
3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given
− Rt

by v = 200 e L , where R = 150 and L =12.5 × 10−3 H. Determine (a) the voltage when t = 160 ×10−6 s, and (b) the time for the voltage to reach 85 V.
[(a) 29.32 volts (b) 71.31 × 10−6 s]
4. The length l metres of a metal bar at temperature t ◦ C is given by l = l0 eαt , where l0 and α are constants. Determine (a) the value of α when l = 1.993 m, l0 = 1.894 m and t = 250◦C, and (b) the value of l0 when l = 2.416, t = 310◦C and α = 1.682 ×10−4.
[(a) 2.038 × 10−4 (b) 2.293 m]

5. The temperature θ2 C of an electrical conductor at time t seconds is given by: θ2 = θ1 (1 − e−t / T ), where θ1 is the initial temperature and T seconds is a constant.
Determine:
(a) θ2 when θ1 = 159.9◦C, t = 30 s and
T = 80 s, and

(b) the time t for θ2 to fall to half the value of θ1 if T remains at 80 s.
[(a) 50◦ C (b) 55.45 s ]
6. A belt is in contact with a pulley for a sector of θ = 1.12 radians and the coefficient

Exponential functions

of friction between these two surfaces is μ = 0.26. Determine the tension on the taut side of the belt, T newtons, when tension on the slack side T0 = 22.7 newtons, given that these quantities are related by the law
T = T0 eμθ .Determine also the value of θ when
T = 28.0 newtons.
[30.4 N, 0.807 rad]
7. The instantaneous current i at time t is given
−t

by: i = 10 e CR when a capacitor is being charged. The capacitance C is 7 ×10−6 farads and the resistance R is 0.3 × 106 ohms. Determine:
(a) the instantaneous current when t is
2.5 seconds, and
(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) 3.04 A (b) 1.46 s]
8. The amount of product x (in mol/cm3) found in a chemical reaction starting with 2.5 mol/cm3 of reactant is given by x = 2.5(1 − e−4t ) where t is the time, in minutes, to form product x. Plot a graph at 30 second intervals up to 2.5 minutes and determine x after 1 minute.
[2.45 mol/cm3]
9. The current i flowing in a capacitor at time t is given by:
−t

37

be determined. This technique is called ‘determination of law’.
Graph paper is available where the scale markings along the horizontal and vertical axes are proportional to the logarithms of the numbers. Such graph paper is called log-log graph paper.
A logarithmic scale is shown in Fig. 4.7 where the distance between, say 1 and 2, is proportional to lg 2 − lg 1, i.e. 0.3010 of the total distance from 1 to 10.
Similarly, the distance between 7 and 8 is proportional to lg 8 − lg 7, i.e. 0.05799 of the total distance from 1 to
10. Thus the distance between markings progressively decreases as the numbers increase from 1 to 10.
1

2

3

4

5

6 7 8 910

Figure 4.7

With log-log graph paper the scale markings are from
1 to 9, and this pattern can be repeated several times. The number of times the pattern of markings is repeated on an axis signifies the number of cycles. When the vertical axis has, say, 3 sets of values from 1 to 9, and the horizontal axis has, say, 2 sets of values from 1 to 9, then this log-log graph paper is called ‘log 3 cycle × 2 cycle’. Many different arrangements are available ranging from ‘log 1 cycle × 1 cycle’ through to ‘log 5 cycle × 5 cycle’.
To depict a set of values, say, from 0.4 to 161, on an axis of log-log graph paper, 4 cycles are required, from
0.1 to 1, 1 to 10, 10 to 100 and 100 to 1000.
Graphs of the form y = a ekx

i = 12.5(1 − e CR )

Taking logarithms to a base of e of both sides of y = a ekx gives: where resistance R is 30 kilohms and the capacitance C is 20 micro-farads. Determine:

ln y = ln(a ekx ) = ln a + ln ekx = ln a + kx ln e

(a)

the current flowing after 0.5 seconds, and

(b) the time for the current to reach
10 amperes.
[(a) 7.07 A (b) 0.966 s]

4.6 Reduction of exponential laws to linear form
Frequently, the relationship between two variables, say x and y, is not a linear one, i.e. when x is plotted against y a curve results. In such cases the non-linear equation may be modified to the linear form, y = mx + c, so that the constants, and thus the law relating the variables can

i.e. ln y = kx + ln a

(since ln e = 1)

which compares with Y = m X + c
Thus, by plotting ln y vertically against x horizontally, a straight line results, i.e. the equation y = a ekx is reduced to linear form. In this case, graph paper having a linear horizontal scale and a logarithmic vertical scale may be used. This type of graph paper is called log-linear graph paper, and is specified by the number of cycles on the logarithmic scale.
Problem 21. The data given below is believed to be related by a law of the form y = a ekx , where a and b are constants. Verify that the law is true and

38 Higher Engineering Mathematics
The law of the graph is thus y = 18 e0.55x

determine approximate values of a and b. Also determine the value of y when x is 3.8 and the value of x when y is 85. x −1.2 0.38 y 9.3

1.2

2.5

3.4

4.2

When x is 3.8, y = 18 e0.55(3.8) = 18 e2.09
= 18(8.0849) = 146

5.3

When y is 85, 85 = 18 e0.55x

22.2 34.8 71.2 117 181 332

85
= 4.7222
18

Hence,
Since y = a ekx then ln y = kx + ln a (from above), which is of the form Y = m X + c, showing that to produce a straight line graph ln y is plotted vertically against x horizontally. The value of y ranges from 9.3 to 332 hence ‘log 3 cycle × linear’ graph paper is used. The plotted co-ordinates are shown in Fig. 4.8 and since a straight line passes through the points the law y = a ekx is verified.
Gradient of straight line, k= AB ln 100 − ln 10
2.3026
=
=
BC
3.12 − (−1.08)
4.20

e0.55x =

and

0.55x = ln 4.7222 = 1.5523 x= Hence

Problem 22. The voltage, v volts, across an inductor is believed to be related to time, t ms, by t the law v = V e T , where V and T are constants.
Experimental results obtained are: v volts 883

= 0.55, correct to 2 significant figures.
Since ln y = kx + ln a, when x = 0, ln y = ln a, i.e. y = a
The vertical axis intercept value at x = 0 is 18, hence a = 18
1000
y

1.5523
= 2.82
0.55

t ms

347

90

55.5 18.6

5.2

10.4 21.6 37.8 43.6 56.7 72.0

Show that the law relating voltage and time is as stated and determine the approximate values of V and T . Find also the value of voltage after 25 ms and the time when the voltage is 30.0 V. t 1
Since v = V e T then ln v = T t + ln V which is of the form Y = m X + c.
Using ‘log3 cycle × linear’ graph paper, the points are plotted as shown in Fig. 4.9.
Since the points are joined by a straight line the law

y 5a e kx

100

t

v = Ve T is verified.
Gradient of straight line,
1
AB
=
T
BC
ln 100 − ln 10
=
36.5 − 64.2

A

=

B

C

2.3026
−27.7

Hence T =

10

−27.7
2.3026

= −12.0, correct to 3 significant figures.
1
22

21

Figure 4.8

0

1

2

3

4

5

6

x

Since the straight line does not cross the vertical axis at t = 0 in Fig. 4.9, the value of V is determined by selecting any point, say A, having co-ordinates t (36.5,100) and substituting these values into v = V e T .

Exponential functions
Now try the following exercise

1000

v 5Ve

Exercise 19 Further problems on reducing exponential laws to linear form

t
T

1. Atmospheric pressure p is measured at varying altitudes h and the results are as shown below: (36.5, 100)

100

A

Voltage, v volts

Altitude, h m

pressure, p cm

500
1500
3000

C

1
0

10

20

30

40
50
Time, t ms

60

70

80

Figure 4.9

−36.5

correct to 3 significant figures.

−t

Hence the law of the graph is v = 2090 e 12.0 .
When time t = 25 ms,
−t

−t

e 12.0 =

30.0
2090

t

2090
= 69.67
30.0
Taking Napierian logarithms gives: and −5 h

, 37.74 cm

e 12.0 =

t
= ln 69.67 = 4.2438
12.0
from which, time t = (12.0)(4.2438) = 50.9 ms

Temperature θ ◦ C

Time t minutes

92.2

−25

v = 2090 e 12.0 = 260 V

When the voltage is 30.0 volts, 30.0 = 2090 e 12.0 , hence Show that the quantities are related by the law p =a ekh , where a and k are constants.
Determine the values of a and k and state the law. Find also the atmospheric pressure at
10 000 m.

2. At particular times, t minutes, measurements are made of the temperature, θ ◦ C, of a cooling liquid and the following results are obtained: e 12.0
= 2090 volts,

voltage

43.41

p = 76 e−7×10

100

V =

53.56

a = 76, k = −7 × 10−5,

36.5

Thus 100 = V e −12.0
i.e.

90

61.60

8000

B

68.42

5000

10

73.39

10

55.9

20

33.9

30

20.6

40

12.5

50

Prove that the quantities follow a law of the form θ = θ0 ekt , where θ0 and k are constants, and determine the approximate value of θ0 and k.
[θ0 = 152, k = − 0.05]

39

Revision Test 1
This Revision Test covers the material contained in Chapters 1 to 4. The marks for each question are shown in brackets at the end of each question.
1.

Factorise x 3 + 4x 2 + x − 6 using the factor theorem. Hence solve the equation x 3 + 4x 2 + x − 6 =0

2.

(6)

Use the remainder theorem to find the remainder when 2x 3 + x 2 − 7x − 6 is divided by
(a) (x − 2) (b) (x + 1)
Hence factorise the cubic expression
6x 2 + 7x − 5 by dividing out
2x − 1

3.

Simplify

4.

(7)
(4)

Resolve the following into partial fractions
(a)

x − 11
−2

x2 − x

(b)

(x 2

3−x
+ 3)(x + 3)

x 3 − 6x + 9
(c)
x2 + x − 2
5.

8.
(24)

(b) ln x + ln(x – 3) = ln 6x – ln(x – 2)

Evaluate, correct to 3 decimal places,
5e
3 ln0.0173

9.
(2)

Solve the following equations, each correct to 4 significant figures: x 10.

(13)

R
U2
ln find the value of U2
J
U1 given that θ f = 3.5, θi = 2.5, R = 0.315, J = 0.4,
(6)
U1 = 50
If θ f − θi =

Solve, correct to 4 significant figures:
(a) 13e2x−1 = 7ex

(a) ln x = 2.40 (b) 3x−1 = 5x−2
(c) 5 = 8(1 − e− 2 )

Solve the following equations:
(a) log x 2 + 8 − log(2x) = log 3

−0.982

6.

7. (a) The pressure p at height h above ground level is given by: p = p0 e−kh where p0 is the pressure at ground level and k is a constant. When p0 is 101 kilopascals and the pressure at a height of 1500 m is 100 kilopascals, determine the value of k.
(b) Sketch a graph of p against h ( p the vertical axis and h the horizontal axis) for values of height from zero to 12 000 m when p0 is 101 kilopascals. (c) If pressure p = 95 kPa, ground level pressure p0 = 101 kPa, constant k = 5 × 10−6, determine the height above ground level, h, in kilometres correct to 2 decimal places.
(13)

(10)

(b) ln (x + 1)2 = ln(x + 1) – ln(x + 2) + 2

(15)

Chapter 5

Hyperbolic functions
(v) Hyperbolic secant of x,

5.1 Introduction to hyperbolic functions sech x =

Functions which are associated with the geometry of the conic section called a hyperbola are called hyperbolic functions. Applications of hyperbolic functions include transmission line theory and catenary problems.
By definition:
(i) Hyperbolic sine of x, ex − e−x sinh x =
2

1
2
= cosh x e x + e−x

(5)

‘sech x’ is pronounced as ‘shec x’
(vi) Hyperbolic cotangent of x, coth x =

e x + e−x
1
= x −x tanh x e − e

(6)

‘coth x’ is pronounced as ‘koth x’
(1)

Some properties of hyperbolic functions
Replacing x by 0 in equation (1) gives:

‘sinh x’ is often abbreviated to ‘sh x’ and is pronounced as ‘shine x’

sinh 0 =

(ii) Hyperbolic cosine of x, e x + e−x cosh x =
2

Replacing x by 0 in equation (2) gives:
(2)

‘cosh x’ is often abbreviated to ‘ch x’ and is pronounced as ‘kosh x’
(iii) Hyperbolic tangent of x, sinh x e x − e−x
=
tanh x = cosh x e x + e−x

(3)

‘tanh x’ is often abbreviated to ‘th x’ and is pronounced as ‘than x’
(iv) Hyperbolic cosecant of x, cosech x =

1
2
= sinh x e x − e−x

‘cosech x’ is pronounced as ‘coshec x’

e0 − e−0
1−1
=
=0
2
2

(4)

cosh 0 =

e0 + e−0 1 + 1
=
=1
2
2

If a function of x, f (−x) = − f (x), then f (x) is called an odd function of x. Replacing x by −x in equation (1) gives: e−x − e x e−x − e−(−x)
=
2
2
x − e−x e =−
= −sinh x
2

sinh(−x) =

Replacing x by −x in equation (3) gives: e−x − e−(−x) e−x − e x
= −x e−x + e−(−x) e +ex e x − e−x
=− x
= −tanh x e + e−x

tanh(−x) =

42 Higher Engineering Mathematics
Hence sinh x and tanh x are both odd functions
1
and
(see Section 5.1), as also are cosech x = sinh x
1
coth x = tanh x
If a function of x, f (−x) = f (x), then f (x) is called an even function of x. Replacing x by −x in equation (2) gives: e−x + e−(−x) e−x + e x
=
2
2
= cosh x

cosh(−x) =

Hence cosh x is an even function (see Section 5.2), as
1
also is sech x = cosh x
Hyperbolic functions may be evaluated easiest using a calculator. Many scientific notation calculators actually possess sinh and cosh functions; however, if a calculator does not contain these functions, then the definitions given above may be used.
Problem 1. Evaluate sinh 5.4, correct to 4 significant figures.

Problem 3. Evaluate th 0.52, correct to 4 significant figures.
Using a calculator with the procedure similar to that used in Worked Problem 1, th 0.52 = 0.4777, correct to 4 significant figures.
Problem 4. Evaluate cosech 1.4, correct to 4 significant figures. cosech 1.4 =

1 sinh 1.4

Using a calculator,
(i) press hyp
(ii) press 1 and sinh( appears
(iii) type in 1.4
(iv) press ) to close the brackets
(v) press = and 1.904301501 appears
(vi) press x −1

Using a calculator,
(i) press hyp
(ii) press 1 and sinh( appears

(vii) press = and 0.5251269293 appears
Hence, cosech 1.4 = 0.5251, correct to 4 significant figures. (iii) type in 5.4
(iv) press ) to close the brackets
(v) press = and 110.7009498 appears
Hence, sinh 5.4 = 110.7, correct to 4 significant figures.
1
Alternatively, sinh 5.4 = (e5.4 − e−5.4 )
2
1
= (221.406416 . . . − 0.00451658 . . .)
2
1
= (221.401899 . . .)
2
= 110.7, correct to 4 significant figures.
Problem 2. Evaluate cosh 1.86, correct to 3 decimal places.
Using a calculator with the procedure similar to that used in Worked Problem 1, cosh 1.86 = 3.290, correct to 3 decimal places.

Problem 5. Evaluate sech 0.86, correct to 4 significant figures. sech 0.86 =

1 cosh 0.86

Using a calculator with the procedure similar to that used in Worked Problem 4, sech 0.86 = 0.7178, correct to 4 significant figures.
Problem 6. Evaluate coth 0.38, correct to 3 decimal places. coth 0.38 =

1 tanh 0.38

Using a calculator with the procedure similar to that used in Worked Problem 4, coth 0.38 = 2.757, correct to 3 decimal places.

Hyperbolic functions

43

y

Now try the following exercise

10
8
6

Exercise 20 Further problems on evaluating hyperbolic functions

y 5sinh x

4
2

In Problems 1 to 6, evaluate correct to 4 significant figures. 23 22 21 0 1 2
22

1. (a) sh 0.64 (b) sh 2.182

3 x

24
26

[(a) 0.6846 (b) 4.376]

28

2. (a) ch 0.72 (b) ch 2.4625

210

[(a) 1.271 (b) 5.910]
Figure 5.1

3. (a) th 0.65 (b) th 1.81
[(a) 0.5717 (b) 0.9478]
4. (a) cosech 0.543 (b) cosech 3.12
[(a) 1.754 (b) 0.08849]
5. (a) sech 0.39 (b) sech 2.367
[(a) 0.9285 (b) 0.1859]

cosh x is an even function (as stated in Section 5.1).
The shape of y = cosh x is that of a heavy rope or chain hanging freely under gravity and is called a catenary.
Examples include transmission lines, a telegraph wire or a fisherman’s line, and is used in the design of roofs and arches. Graphs of y = tanh x, y = cosech x, y = sech x and y = coth x are deduced in Problems 7 and 8. y 6. (a) coth 0.444 (b) coth 1.843
[(a) 2.398 (b) 1.051]

10

7. A telegraph wire hangs so that its shape is x described by y = 50 ch . Evaluate, correct
50
to 4 significant figures, the value of y when x = 25.
[56.38]
8. The length l of a heavy cable hanging under gravity is given by l = 2c sh (L/2c). Find the value of l when c = 40 and L =30.
[30.71]
9.

V 2 = 0.55L tanh (6.3 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 .
[5.042]

6
4
2
23 22 21 0

Graphs of hyperbolic functions

A graph of y = sinhx may be plotted using calculator values of hyperbolic functions. The curve is shown in
Fig. 5.1. Since the graph is symmetrical about the origin, sinh x is an odd function (as stated in Section 5.1).
A graph of y = cosh x may be plotted using calculator values of hyperbolic functions. The curve is shown in
Fig. 5.2. Since the graph is symmetrical about the y-axis,

1 2

3

x

Figure 5.2

Problem 7. Sketch graphs of (a) y = tanh x and (b) y = coth x for values of x between
−3 and 3.
A table of values is drawn up as shown below
−3

x

5.2

y 5cosh x

8

sh x

−10.02

ch x

10.07

y = th x =

sh x ch x

y = coth x =

ch x sh x

−2

−1

−3.63 −1.18
3.76

1.54

−0.995 −0.97 −0.77
−1.005 −1.04 −1.31

44 Higher Engineering Mathematics x 0

1

2

3

sh x

0

1.18 3.63 10.02

ch x

1

1.54 3.76 10.07

0

0.77 0.97

A table of values is drawn up as shown below
−4

x

sh x ch x

0.995

cosech x =

1 sh x

±∞ 1.31 1.04

1.005

A graph of y = tanh x is shown in Fig. 5.3(a)

sech x =

−0.04

−0.10 −0.28 −0.85
10.07

3.76

1.54

0.04

0.10

0.27

0.65

3

4

1 ch x

x

0

(b) A graph of y = coth x is shown in Fig. 5.3(b)

sh x

0

Both graphs are symmetrical about the origin thus tanh x and coth x are odd functions.

cosech x =

(a)

1 sh x

ch x
Problem 8. Sketch graphs of (a) y = cosech x and (b) y = sech x from x = −4 to x = 4, and, from the graphs, determine whether they are odd or even functions.

y 5 tanh x

y
1
23 22 21

sech x =

1

2

1.18 3.63 10.02 27.29

±∞ 0.85 0.28

0.10

2 3

1 ch x

1.54 3.76 10.07 27.31

1

0.65 0.27

0.10

3

x

2

y 5 cosech x

1

(a)
232221

y

01 2 3
21

y 5 cosech x

3

x

22 y 5coth x

23

1
(a)
23 22 21 0

1

2 3

x

y

21 y 5 coth x

1

22
23

232221 0
(b)

(b)

Figure 5.4
Figure 5.3

0.04

A graph of y = cosech x is shown in Fig. 5.4(a).
The graph is symmetrical about the origin and is thus an odd function.
(b) A graph of y = sech x is shown in Fig. 5.4(b). The graph is symmetrical about the y-axis and is thus an even function.
(a)

21

2

0.04

1

y
0 1

−1

27.31

ch x ch x y = coth x = sh x

−2

−22.29 −10.02 −3.63 −1.18

sh x

y = th x =

−3

y 5 sech x
1 2 3

x

45

Hyperbolic functions
5.3

Hyperbolic identities

For every trigonometric identity there is a corresponding hyperbolic identity. Hyperbolic identities may be proved by either
(i) replacing sh x e x + e−x

by

e x − e−x
2

Problem 9. Prove the hyperbolic identities
(a) ch 2 x − sh2 x = 1 (b) 1 − th2 x = sech2 x
(c) coth 2 x − 1 =cosech2 x.

(a) and ch x

e x + e−x
2

ch x + sh x =

by

e x + e−x
2

ch x − sh x =

, or

2
(ii) by using Osborne’s rule, which states: ‘the six trigonometric ratios used in trigonometrical identities relating general angles may be replaced by their corresponding hyperbolic functions, but the sign of any direct or implied product of two sines must be changed’.
For example, since cos2 x + sin2 x = 1 then, by
Osborne’s rule, ch2 x − sh2 x = 1, i.e. the trigonometric functions have been changed to their corresponding hyperbolic functions and since sin2 x is a product of two sines the sign is changed from + to −. Table 5.1 shows some trigonometric identities and their corresponding hyperbolic identities.

+


i.e. ch2 x − sh2 x = 1

ch2 x sh2 x
1

= 2 ch2 x ch2 x ch x
i.e. 1 −th2 x = sech2 x

cos2 x + sin2 x = 1

ch2 x − sh2 x = 1

1 + tan2 x = sec2 x

1 −th2 x = sech2 x

cot 2 x + 1 =cosec 2 x

coth2 x − 1 = cosech2 x
Compound angle formulae

sin (A ± B) = sin A cos B ± cos A sin B

sh (A ± B) = sh A ch B ± ch A sh B

cos (A ± B) = cos A cos B ∓ sin A sin B

ch (A ± B) = ch A ch B ± sh A sh B th (A ± B) =

th A ± th B
1 ±th A th B

Double angles sin 2x = 2 sin x cos x

sh 2x = 2 sh x ch x

cos 2x = cos2 x − sin2 x

ch 2x =ch2 x + sh2 x

= 2 cos2 x − 1

= 2 ch2 x − 1

= 1 − 2 sin2 x

= 1 + 2sh2 x

2 tan x
1 − tan2 x

(1)

(b) Dividing each term in equation (1) by ch2 x gives: Corresponding hyperbolic identity

tan 2x =

e x − e−x
2

(ch x + sh x)(ch x − sh x) = (e x )(e−x ) = e0 = 1

Trigonometric identity

tan A ± tan B
1 ∓ tan A tan B

= ex

= e+−x

Table 5.1

tan (A ± B) =

ex − e−x
2

th 2x =

2 th x
1 + th2 x

46 Higher Engineering Mathematics
(c)

Dividing each term in equation (1) by sh2 x gives: ch2 x sh2 x
1

= 2 sh2 x sh2 x sh x

Show that th2 x + sech2 x = 1.

Problem 12.

i.e. coth2 x − 1 =cosech2 x

Since ch2 x − sh2 x = 1 then 1 + sh2 x = ch2 x
Thus

From trigonometric ratios, cos 2 A = cos2 A − sin 2 A

(1)

Osborne’s rule states that trigonometric ratios may be replaced by their corresponding hyperbolic functions but the sign of any product of two sines has to be changed. In this case, sin2 A = (sin A)(sin A), i.e. a product of two sines, thus the sign of the corresponding hyperbolic function, sh2 A, is changed from + to −. Hence, from
(1), ch 2A = ch2 A + sh2 A
(b) From trigonometric ratios,
1 + tan2 x

= sec2 x

and tan2 x =

sin2 x cos2 x

(2)
=

(sin x)(sin x) cos2 x

Ae x + Be−x ≡ 4 ch x − 5 sh x
=4

= 1+2

e x − e−x
2

e2x − 2e x e−x + e−2x
4

e2x − 2 + e−2x
2
2x + e−2x e 2
=1+

2
2
=1+

+ e−2x
2

= ch 2x = R.H.S.

e x + e−x
2

−5

e x − e−x
2

5
5
= 2e x + 2e−x − e x + e−x
2
2
1
9
= − e x + e−x
2
2
1
1 and B = 4
2
2

Problem 14. If 4e x − 3e−x ≡ Psh x + Qch x, determine the values of P and Q.
4e x − 3e−x ≡ P sh x + Q ch x
=P
=

= 1 + 2 sh2 x = 1 + 2

=

Problem 13. Given Ae x + Be−x ≡ 4ch x−5 sh x, determine the values of A and B.

Prove that 1 + 2 sh2 x = ch 2x.

Left hand side (L.H.S.)

e2x

sh2 x + 1 ch2 x
= 2 = 1 = R.H.S. ch2 x ch x

Equating coefficients gives: A = −

i.e. a product of two sines.
Hence, in equation (2), the trigonometric ratios are changed to their equivalent hyperbolic function and the sign of th2 x changed + to −, i.e.
1 −th2 x = sech2 x
Problem 11.

sh2 x + 1 ch2 x

=

Problem 10. Prove, using Osborne’s rule
(a) ch 2 A = ch2 A + sh2 A
(b) 1 −th2 x = sech2 x.
(a)

1 sh2 x
+ 2
2
ch x ch x

L.H.S. = th2 x + sech2 x =

e x − e−x
2

+Q

e x + e−x
2

P x P −x Q x Q −x e − e + e + e
2
2
2
2

2

P+Q x e +
2

=

Q − P −x e 2

Equating coefficients gives:
4=

P+Q
Q−P
and −3 =
2
2

i.e. P + Q = 8
−P + Q = −6

(1)
(2)

Adding equations (1) and (2) gives: 2Q = 2, i.e. Q = 1
Substituting in equation (1) gives: P = 7.

Hyperbolic functions
Now try the following exercise
Exercise 21 Further problems on hyperbolic identities
In Problems 1 to 4, prove the given identities.
1. (a) ch (P − Q) ≡ ch P ch Q − sh P sh Q
(b) ch 2x ≡ ch2 x + sh2 x
2. (a) coth x ≡ 2 cosech 2x + th x
(b) ch 2θ − 1 ≡2 sh2 θ th A − th B
1 −th A th B
(b) sh 2 A ≡ 2 sh A ch A

3. (a) th (A − B) ≡

4. (a) sh (A + B) ≡ sh A ch B + ch A sh B
(b)

sh2 x + ch2 x − 1
≡ tanh4 x
2ch2 x coth2 x

5. Given Pe x − Qe−x ≡ 6 ch x − 2 sh x, find P and Q
[P = 2, Q =−4]
6. If 5e x − 4e−x ≡ A sh x + B ch x, find A and B.
[A = 9, B = 1]

5.4 Solving equations involving hyperbolic functions

47

Using a calculator with a similar procedure as in Worked
Problem 15, check that: x = 0.980, correct to 3 decimal places.
With reference to Fig. 5.2, it can be seen that there will be two values corresponding to y = cosh x =
1.52. Hence, x = ±0.980
Problem 17. Solve the equation tanh θ = 0.256, correct to 4 significant figures.
Using a calculator with a similar procedure as in Worked
Problem 15, check that gives θ = 0.2618, correct to 4 significant figures.
Problem 18. Solve the equation sech x = 0.4562, correct to 3 decimal places. sech x = 0.4562, then x = sech −10.4562 =
1
1 cosh−1 since cosh =
0.4562
sech

If

i.e. x = 1.421, correct to 3 decimal places.
With reference to the graph of y = sech x in Fig. 5.4, it can be seen that there will be two values corresponding to y = sech x = 0.4562
Hence, x = ±1.421

Equations such as sinh x = 3.25 or coth x = 3.478 may be determined using a calculator. This is demonstrated in Worked Problems 15 to 21.

Problem 19. Solve the equation cosech y = −0.4458, correct to 4 significant figures.

Problem 15. Solve the equation sh x = 3, correct to 4 significant figures.

If cosech y = − 0.4458, then y = cosech−1 (−0.4458)
1
1 since sinh =
= sinh−1
− 0.4458 cosech i.e. y = −1.547, correct to 4 significant figures.

If sinh x = 3, then x = sinh−1 3
This can be determined by calculator.
(i) Press hyp
(ii) Choose 4, which is sinh−1
(iii) Type in 3
(iv) Close bracket )
(v) Press = and the answer is 1.818448459
i.e. the solution of sh x = 3 is: x = 1.818, correct to 4 significant figures.
Problem 16. Solve the equation ch x = 1.52, correct to 3 decimal places.

Problem 20. Solve the equation coth A = 2.431, correct to 3 decimal places. coth A = 2.431, then A = coth−1 2.431 =
1
1 tanh−1 since tanh =
2.431
coth
i.e. A= 0.437, correct to 3 decimal places.
If

Problem 21. A chain hangs in the form given by x y = 40 ch
. Determine, correct to 4 significant
40
figures, (a) the value of y when x is 25, and (b) the value of x when y = 54.30

48 Higher Engineering Mathematics
(a)

x
, and when x = 25,
40
25 y = 40 ch
= 40 ch 0.625
40

y = 40 ch

= 40(1.2017536 . . .) = 48.07 x (b) When y = 54.30, 54.30 =40 ch , from which
40
x
54.30
ch =
= 1.3575
40
40 x Hence,
= cosh−1 1.3575 =±0.822219 . . ..
40
(see Fig. 5.2 for the reason as to why the answer is ±) from which, x = 40(±0.822219 . . ..) = ±32.89

Following the above procedure:
(i) 2.6 ch x + 5.1 sh x = 8.73
i.e. 2.6

plotting graphs of y = a ch x + b sh x and y = c and noting the points of intersection, or more accurately, (b) by adopting the following procedure:
(i) Change sh x to e x + e−x
2

e x − e−x
2

+ 5.1

e x − e−x
2

= 8.73

(ii) 1.3e x + 1.3e−x + 2.55e x − 2.55e−x = 8.73
i.e. 3.85e x − 1.25e−x − 8.73 =0
(iii) 3.85(e x )2 − 8.73e x − 1.25 =0
(iv) e x
−(−8.73) ± [(−8.73)2 − 4(3.85)(−1.25)]
2(3.85)

8.73 ± 95.463 8.73 ±9.7705
=
=
7.70
7.70
Hence e x = 2.4027 or e x = −0.1351

=

Equations of the form a ch x + b sh x = c, where a, b and c are constants may be solved either by:
(a)

e x + e−x
2

(v)

x = ln 2.4027 or x = ln(−0.1351) which has no real solution.
Hence x = 0.8766, correct to 4 decimal places.

Now try the following exercise and ch x to

(ii) Rearrange the equation into the form pe x + qe−x +r = 0, where p, q and r are constants. Exercise 22 Further problems on hyperbolic equations
In Problems 1 to 8, solve the given equations correct to 4 decimal places.
1.

2.

(iv) Solve the quadratic equation p(e x )2 +re x + q = 0 for e x by factorising or by using the quadratic formula.

3.

(a) cosh B = 1.87 (b) 2 ch x = 3
[(a) ±1.2384 (b) ±0.9624]
(a) tanh y = −0.76 (b) 3 th x = 2.4
[(a) −0.9962 (b) 1.0986]

4.

(a) sech B = 0.235 (b) sech Z = 0.889
[(a) ±2.1272 (b) ±0.4947]

5.

This procedure is demonstrated in Problem 22.

(a) cosech θ = 1.45 (b) 5 cosech x = 4.35
[(a) 0.6442 (b) 0.5401]

6.
Problem 22. Solve the equation
2.6 ch x + 5.1 sh x = 8.73, correct to 4 decimal places. (b) sh A = −2.43
[(a) 0.8814 (b) −1.6209]

(iii) Multiply each term by e x , which produces an equation of the form p(e x )2 +re x + q = 0 (since (e−x )(e x ) = e0 = 1)

(v) Given e x = a constant (obtained by solving the equation in (iv)), take Napierian logarithms of both sides to give x = ln (constant)

(a) sinh x = 1

(a) coth x = 2.54 (b) 2 coth y = −3.64
[(a) 0.4162 (b) −0.6176]

7.

3.5 sh x + 2.5 ch x = 0

[−0.8959]

Hyperbolic functions
8. 2 sh x + 3 ch x = 5

[0.6389 or −2.2484]

9. 4 th x − 1 = 0

[0.2554]

10. A chain hangs so that its shape is of the x . Determine, correct to form y = 56 cosh
56
4 significant figures, (a) the value of y when x is 35, and (b) the value of x when y is 62.35
[(a) 67.30 (b) ±26.42]

x3 x5
i.e. sinh x = x + + + · · · (which is valid for all
3! 5! values of x). sinh x is an odd function and contains only odd powers of x in its series expansion.
Problem 23. Using the series expansion for ch x evaluate ch 1 correct to 4 decimal places. ch x = 1 +
Let

5.5 Series expansions for cosh x and sinh x

x2 x4
+
+ · · ·from above
2! 4!

x = 1,

then ch 1 = 1 +

By definition, x2 x3 x4 x5
+
+
+
+···
2! 3! 4!
5!

from Chapter 4.
Replacing x by −x gives: e−x = 1 − x +

x2 x3 x4 x5

+

+··· .
2!
3! 4! 5!

1 cosh x = (e x + e−x )
2
=

1
2

1+x +

x2 x3 x4 x5

+

+···
2! 3!
4! 5!

+ 1−x +
=

1
2

2+

x2 x3 x4 x5
+
+
+
+···
2!
3! 4! 5!

2x 2 2x 4
+
+···
2!
4!

x2 x4
i.e. cosh x = 1 + + + · · · (which is valid for all
2! 4! values of x). cosh x is an even function and contains only even powers of x in its expansion.
1
sinh x = (e x − e−x )
2
=

1
2

1+x +

− 1−x +
=

x2 x3 x4 x5
+
+
+
+···
2! 3!
4! 5!

x2 x3 x4 x5

+

+···
2! 3!
4! 5!

2x 3 2x 5
1
2x +
+
+ ···
2
3!
5!

14
12
+
2 × 1 4 ×3 × 2 × 1

16
+ ···
6 ×5 × 4 × 3 ×2 × 1

+

ex = 1 + x +

49

= 1 + 0.5 + 0.04167 + 0.001389 + · · ·
i.e. ch 1 = 1.5431, correct to 4 decimal places, which may be checked by using a calculator.
Problem 24. Determine, correct to 3 decimal places, the value of sh 3 using the series expansion for sh x. sh x = x +

x3 x5
+
+ · · · from above
3! 5!

Let x = 3, then
33 35 37 39 311
+
+
+ +
+···
3! 5! 7! 9! 11!
= 3 + 4.5 + 2.025 + 0.43393 + 0.05424

sh 3 = 3 +

+ 0.00444 + · · ·
i.e. sh 3 = 10.018, correct to 3 decimal places.
Problem 25. Determine the power series for θ − sh 2θ as far as the term in θ 5 .
2 ch
2
In the series expansion for ch x, let x =
2 ch

θ
2

=2 1+
=2+

θ then: 2

(θ/2)2 (θ/2)4
+
+···
2!
4!

θ2 θ4 +
+···
4
192

50 Higher Engineering Mathematics
In the series expansion for sh x, let x = 2θ, then:
(2θ)3 (2θ)5
+
+···
3!
5!
4
4
= 2θ + θ 3 + θ 5 + · · ·
3
15

sh 2θ = 2θ +

2. Use the series expansion for sh x to evaluate, correct to 4 decimal places: (a) sh 0.5
(b) sh 2
[(a) 0.5211 (b) 3.6269]

Hence θ θ2 θ4 ch
− sh 2θ = 2 +
+
+···
2
4
192
4
4
− 2θ + θ 3 + θ 5 + · · ·
3
15
= 2 −2θ +


θ2 4 3 θ4
− θ +
4 3
192

4 5 θ + · · · as far the term in θ 5
15

Now try the following exercise
Exercise 23 Further problems on series expansions for cosh x and sinh x
1. Use the series expansion for ch x to evaluate, correct to 4 decimal places: (a) ch 1.5 (b) ch 0.8
[(a) 2.3524 (b) 1.3374]

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


9
81
(a) 3x + x 3 + x 5⎥

2
40 ⎦

2
(b) 1 + 2x 2 + x 4
3
In Problems 4 and 5, prove the given identities, the series being taken as far as the term in θ 5 only. 4. sh 2θ − sh θ ≡ θ +

5. 2 sh

31 5
7 3 θ + θ 6
120

θ θ θ2 θ3 θ4 − ch ≡ − 1 + θ −
+

2
2
8
24 384
+

θ5
1920

Chapter 6

Arithmetic and geometric progressions 6.1

Arithmetic progressions

When a sequence has a constant difference between successive terms it is called an arithmetic progression
(often abbreviated to AP).
Examples include:

i.e.

For example, the sum of the first 7 terms of the series 1,
4, 7, 10, 13, . . . is given by
7
S7 = [2(1) + (7 − 1)3], since a = 1 and d = 3
2

(i) 1, 4, 7, 10, 13, . . . where the common difference is 3 and

7
7
= [2 + 18] = [20] = 70
2
2

(ii) a, a + d, a + 2d, a + 3d,. . .where the common difference is d.
General expression for the n’th term of an AP
If the first term of an AP is ‘a’ and the common difference is ‘d’ then the n’th term is: a + (n − 1)d
In example (i) above, the 7th term is given by 1 +
(7 − 1)3 = 19, which may be readily checked.
Sum of n terms of an AP
The sum S of an AP can be obtained by multiplying the average of all the terms by the number of terms. a +l
, where ‘a’ is the
The average of all the terms =
2
first term and l is the last term, i.e. l = a + (n − 1)d, for n terms.
Hence the sum of n terms,
Sn = n
=

a +l
2

n
{a + [a + (n − 1)d]}
2

n
S n = [2a + (n − 1)d]
2

6.2 Worked problems on arithmetic progressions Problem 1. Determine (a) the ninth, and (b) the sixteenth term of the series 2, 7, 12, 17, . . .
2, 7, 12, 17, . . . is an arithmetic progression with a common difference, d, of 5.
(a)

The n’th term of an AP is given by a + (n −1)d
Since the first term a = 2, d = 5 and n =9 then the
9th term is:
2 + (9 −1)5 = 2 + (8)(5) = 2 + 40 =42

(b) The 16th term is:
2 + (16 −1)5 = 2 +(15)(5) = 2 + 75 =77.
Problem 2. The 6th term of an AP is 17 and the
13th term is 38. Determine the 19th term.

52 Higher Engineering Mathematics
The n’th term of an AP is a + (n −1)d

The sum of the first 21 terms,

a + 5d = 17

(1)

The 13th term is: a + 12d= 38

(2)

The 6th term is:

Equation (2) −equation (1) gives: 7d = 21, from which,
21
d = = 3.
7
Substituting in equation (1) gives: a + 15 =17, from which, a = 2.
Hence the 19th term is: a + (n − 1)d = 2 + (19 − 1)3 = 2 + (18)(3) =
2 + 54 = 56.

is

an

AP

where

a=21
2

1. Find the 11th term of the series 8, 14, 20,
26, . . .
[68]
2. Find the 17th term of the series 11, 10.7, 10.4,
10.1, . . .
[6.2]
and

Hence if the n’th term is 22 then: a + (n − 1)d = 22
i.e. 2 1 + (n − 1) 1 1 = 22
2
2
(n − 1)

11
2

= 22 − 2 1 = 19 1 .
2
2

n −1 =

19 1
2
11
2

= 13 and n = 13 + 1 = 14

i.e. the 14th term of the AP is 22.
Problem 4. Find the sum of the first 12 terms of the series 5, 9, 13, 17, . . .
5, 9, 13, 17, . . . is an AP where a = 5 and d = 4. The sum of n terms of an AP, n Sn = [2a + (n − 1)d]
2
Hence the sum of the first 12 terms,
S12 =

Now try the following exercise
Exercise 24 Further problems on arithmetic progressions Problem 3. Determine the number of the term whose value is 22 in the series 2 1 , 4, 5 1 , 7, . . .
2
2
2 1 , 4, 5 1 , 7, . . .
2
2 d =11.
2

21
[2a + (n − 1)d]
2
21
21
= [2(3.5) + (21 − 1)0.6] = [7 + 12]
2
2
399
21
= 199.5
= (19) =
2
2

S21 =

12
[2(5) + (12 − 1)4]
2

= 6[10 + 44] = 6(54) = 324
Problem 5. Find the sum of the first 21 terms of the series 3.5, 4.1, 4.7, 5.3, . . .
3.5, 4.1, 4.7, 5.3, . . . is an AP where a = 3.5 and d = 0.6

3. The seventh term of a series is 29 and the eleventh term is 54. Determine the sixteenth term. [85.25]
4. Find the 15th term of an arithmetic progression of which the first term is 2.5 and the tenth term is 16.
[23.5]
5. Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13.6, . . .
[11th ]
6. Find the sum of the first 11 terms of the series
4, 7, 10, 13, . . .
[209]
7. Determine the sum of the series 6.5, 8.0, 9.5,
11.0, . . . , 32
[346.5]

6.3 Further worked problems on arithmetic progressions
Problem 6. The sum of 7 terms of an AP is 35 and the common difference is 1.2. Determine the first term of the series. n = 7, d = 1.2 and S7 = 35
Since the sum of n terms of an AP is given by
Sn =

n
[2a + (n − 1)d], then
2

7
7
35 = [2a + (7 − 1)1.2] = [2a + 7.2]
2
2

Arithmetic and geometric progressions
35 × 2
= 2a + 7.2
7
10 = 2a + 7.2
2a = 10 − 7.2 = 2.8,
2.8
a=
= 1.4
2

Hence

Thus from which

i.e. the first term, a = 1.4

53

Problem 9. The first, twelfth and last term of an arithmetic progression are 4, 31 1 , and 376 1
2
2 respectively. Determine (a) the number of terms in the series, (b) the sum of all the terms and (c) the
‘80’th term.
(a)

Problem 7. Three numbers are in arithmetic progression. Their sum is 15 and their product is 80.
Determine the three numbers.

Let the AP be a, a +d, a +2d, . . . , a + (n − 1)d, where a = 4
The 12th term is: a + (12 −1)d = 31 1
2
4 + 11d = 31 1 ,
2

i.e.

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

from which, 11d = 31 1 − 4 = 27 1
2
2

Then (a − d) + a + (a + d) = 15, i.e. 3a = 15, from which, a = 5

Hence d =

27 1
2
=21
2
11
The last term is a + (n − 1)d

Also, a(a − d)(a + d) = 80, i.e. a(a 2 − d 2 ) = 80
Since a = 5, 5(52 − d 2 ) = 80

i.e. 4 + (n − 1) 2 1 = 376 1
2
2

125 − 5d 2 = 80
125 − 80 = 5d 2

(n − 1) =

376 1 − 4
2
21
2

45 = 5d 2

45 from which, d 2 = = 9. Hence d = 9 = ±3.
5
The three numbers are thus (5 − 3), 5 and (5 + 3), i.e.
2, 5 and 8.
Problem 8. Find the sum of all the numbers between 0 and 207 which are exactly divisible by 3.

=

S150 =
=

3 + (n − 1)3 = 207,

Hence

n
[2a + (n − 1)d]
2
69
= [2(3) + (69 − 1)3]
2
69
69
= [6 + 204] = (210) = 7245
2
2

150
1
2(4) + (150 − 1) 2
2
2
1
2

= 85[8 + 372.5]
= 75(380.5) = 28537

The sum of all 69 terms is given by
S69 =

n
[2a + (n − 1)d]
2

= 75 8 + (149) 2

207 − 3
(n − 1) =
= 68
3
n = 68 + 1 = 69

from which

= 149

(b) Sum of all the terms,

a + (n − 1)d = 207

i.e.

21
2

Hence the number of terms in the series, n = 149 +1 =150

The series 3, 6, 9, 12, . . ., 207 is an AP whose first term a = 3 and common difference d = 3
The last term is

372 1
2

(c)

1
2

The 80th term is: a + (n − 1)d = 4 + (80 − 1) 2 1
2
= 4 + (79) 2 1
2
= 4 + 197.5 = 201 1
2

54 Higher Engineering Mathematics
Problem 10. An oil company bores a hole 80 m deep. Estimate the cost of boring if the cost is £30 for drilling the first metre with an increase in cost of
£2 per metre for each succeeding metre.

8. An oil company bores a hole 120 m deep. Estimate the cost of boring if the cost is £70 for drilling the first metre with an increase in cost of £3 per metre for each succeeding metre.
[£29820]

The series is: 30, 32, 34, . . . to 80 terms, i.e. a = 30, d = 2 and n = 80
Thus, total cost, n Sn = 2a + (n − 1)d
2
=

80
[2(30) + (80 − 1)(2)]
2

= 40[60 + 158] = 40(218) = £8720

6.4

Geometric progressions

When a sequence has a constant ratio between successive terms it is called a geometric progression (often abbreviated to GP). The constant is called the common ratio, r.
Examples include
(i) 1, 2, 4, 8, . . . where the common ratio is 2 and

Now try the following exercise

(ii) a, ar, ar 2 , ar 3 , . . . where the common ratio is r.
General expression for the n’th term of a GP

Exercise 25 Further problems on arithmetic progressions If the first term of a GP is ‘a’ and the common ratio is r, then

1. The sum of 15 terms of an arithmetic progression is 202.5 and the common difference is 2.
Find the first term of the series.
[−0.5]

the n’th term is: ar n−1

2. Three numbers are in arithmetic progression.
Their sum is 9 and their product is 20.25.
Determine the three numbers.
[1.5, 3, 4.5]
3. Find the sum of all the numbers between 5 and
250 which are exactly divisible by 4. [7808]
4. Find the number of terms of the series 5, 8,
11, . . . of which the sum is 1025.
[25]
5. Insert four terms between 5 and 22.5 to form an arithmetic progression. [8.5, 12, 15.5, 19]
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) 120 (b) 26070 (c) 250.5]
7. On commencing employment a man is paid a salary of £16000 per annum and receives annual increments of £480. Determine his salary in the 9th year and calculate the total he will have received in the first 12 years.
[£19840, £223,680]

which can be readily checked from the above examples.
For example, the 8th term of the GP 1, 2, 4, 8, . . . is
(1)(2)7 = 128, since a = 1 and r = 2.
Sum of n terms of a GP
Let a GP be a, ar, ar 2 , ar 3 , . . . , ar n−1 then the sum of n terms,
Sn = a + ar + ar 2 + ar 3 + · · · + ar n−1 · · ·

(1)

Multiplying throughout by r gives: r Sn = ar + ar 2 + ar 3 + ar 4
+ · · · + ar n−1 + ar n + · · ·

(2)

Subtracting equation (2) from equation (1) gives:
Sn − r Sn = a − ar n
i.e. Sn (1 − r) = a(1 − r n ) n −r
Thus the sum of n terms, S n = a(1− r ) ) which is valid
(1
when r < 1.

Arithmetic and geometric progressions
Subtracting equation (1) from equation (2) gives a(r n − 1)
Sn = which is valid when r > 1.
(r − 1)
For example, the sum of the first 8 terms of the GP 1, 2,
1(28 − 1)
4, 8, 16, . . . is given by S8 =
, since a = 1 and
(2 − 1) r =2
i.e. S8 =

1(256 − 1)
= 255
1

Sum to infinity of a GP
When the common ratio r of a GP is less than unity, the a(1 −r n )
, which may be written sum of n terms, Sn =
(1 −r) a ar n

as Sn =
(1 −r) (1 −r)
Since r < 1, r n becomes less as n increases, i.e. r n → 0 as n →∞. n a ar
Hence
→ 0 as n →∞. Thus Sn → as (1 −r)
(1 −r) n →∞. a is called the sum to infinity, S∞,
The quantity
(1 −r) and is the limiting value of the sum of an infinite number of terms, a i.e. S ∞ = which is valid when −1 0 is shown in Fig. 18.31 and, again, f −1 (x) is seen to be a reflection of f (x) in the line y = x.
It is noted from the latter example, that not all functions have an inverse. An inverse, however, can be determined if the range is restricted.
Problem 5. Determine the inverse for each of the following functions:
(a) f (x) = x − 1 (b) f (x) = x 2 − 4 (x > 0)
(c) f (x) = x 2 + 1
(a)

If y = f (x), then y = x 2 + 1 √

Transposing for x gives x = y√ 1
Interchanging x and y gives y = x − 1, which has two values.
Hence there is no inverse of f(x) = x2 + 1, since the domain of f (x) is not restricted.

Inverse trigonometric functions

y

0

(c)

If y = f (x), then y = x − 1
Transposing for x gives x = y + 1
Interchanging x and y gives y = x + 1

If y = sin x, then x is the angle whose sine is y.
Inverse trigonometrical functions are denoted by prefixing the function with ‘arc’ or, more commonly,−1 .
Hence transposing y = sin x for x gives x = sin−1 y.
Interchanging x and y gives the inverse y = sin−1 x.
Similarly, y = cos−1 x, y = tan−1 x, y = sec−1 x, y =cosec−1 x and y =cot −1 x are all inverse trigonometric functions. The angle is always expressed in radians. Inverse trigonometric functions are periodic so it is necessary to specify the smallest or principal value of the angle. For sin−1 x, tan−1 x, cosec−1 x and cot −1 x, the π π principal value is in the range − < y < . For cos−1 x
2
2 and sec−1 x the principal value is in the range 0 < y < π.
Graphs of the six inverse trigonometric functions are shown in Fig. 33.1, page 335.
Problem 6. Determine the principal values of
(a) arcsin 0.5



3
2

(c) arccos −

(b) arctan(−1)

(d) arccosec( 2)

Using a calculator,
(a) arcsin 0.5 ≡ sin−1 0.5 = 30◦
=

π rad or 0.5236 rad
6

(b) arctan(−1) ≡ tan−1 (−1) = −45◦
=−

π rad or −0.7854 rad
4

190 Higher Engineering Mathematics
(c) arccos −




3
2

3
2

≡ cos−1 −

=

= 150◦

8. cot −1 2

5π rad or 2.6180 rad
6

9.

[0.4636 rad]

cosec−1 2.5

[0.4115 rad]

10. sec−1 1.5


1
(d) arccosec( 2) = arcsin √
2
1
≡ sin−1 √
2

π or 0.7854 rad
4

7. tan −1 1

[0.8411 rad]

1 π or 0.7854 rad
11. sin−1 √
4
2
12. Evaluate x, correct to 3 decimal places:

= 45◦

x = sin−1

π
= rad or 0.7854 rad
4

1
4
8
+ cos−1 − tan−1
3
5
9
[0.257]

Problem 7. Evaluate (in radians), correct to
3 decimal places: sin−1 0.30 + cos−1 0.65.

13. Evaluate y, correct to 4 significant figures:

√ y = 3 sec−1 2 − 4 cosec−1 2
+ 5 cot−1 2

sin−1 0.30 = 17.4576◦ = 0.3047 rad

[1.533]

cos−1 0.65 = 49.4584◦ = 0.8632 rad
Hence sin−1 0.30 + cos−1 0.65
= 0.3047 +0.8632 = 1.168, correct to 3 decimal places.
Now try the following exercise

Determine the inverse of the functions given in
Problems 1 to 4.
[ f −1(x) = x − 1]

1.

f (x) = x + 1

2.

f (x) = 5x − 1

f −1 (x) = 1 (x + 1)
5

3.

f (x) = x 3 + 1

[ f −1(x) =

4.

f (x) =

1
+2
x


3

x +2 is drawn x +1 up for various values of x and then y plotted against x, the graph would be as shown in Fig. 18.32. The straight lines AB, i.e. x = −1, and CD, i.e. y = 1, are known as asymptotes. An asymptote to a curve is defined as a straight line to which the curve approaches as the distance from the origin increases. Alternatively, an asymptote can be considered as a tangent to the curve at infinity. f −1(x) =

x − 1]

Asymptotes parallel to the x- and y-axes

1 x −2

There is a simple rule which enables asymptotes parallel to the x- and y-axis to be determined. For a curve y = f (x):

Determine the principal value of the inverse functions in Problems 5 to 11.

6. cos−1 0.5

Asymptotes

If a table of values for the function y =

Exercise 79 Further problems on inverse functions 5. sin−1 (−1)

18.7



π or −1.5708 rad
2
π or 1.0472 rad
3

(i) the asymptotes parallel to the x-axis are found by equating the coefficient of the highest power of x to zero.
(ii) the asymptotes parallel to the y-axis are found by equating the coefficient of the highest power of y to zero.

Functions and their curves

191

y

A

5

4

3 y5 2
C

x 12 x 11

D

1

24

23

22

21

0

1

2

3

4

x

21
22

y5

x 12 x 11

23
24
25
B

Figure 18.32

With the above example y =

x +2
, rearranging gives: x +1

y(x + 1) = x + 2
i.e.

yx + y − x − 2 = 0

and

x(y − 1) + y − 2 = 0

(1)

The coefficient of the highest power of x (in this case x 1) is (y − 1). Equating to zero gives: y − 1 = 0
From which, y = 1, which is an asymptote of y = as shown in Fig. 18.32.
Returning to equation (1): from which,

x +2 x +1

yx + y − x − 2 = 0 y(x + 1) − x − 2 = 0.

The coefficient of the highest power of y (in this case y 1 ) is (x + 1). Equating to zero gives: x + 1 = 0 from x +2 which, x = −1, which is another asymptote of y = x +1 as shown in Fig. 18.32.

Problem 8. Determine the asymptotes for the x −3 function y = and hence sketch the curve.
2x + 1
Rearranging y =

x −3 gives: y(2x + 1) = x − 3
2x + 1

i.e.
2x y + y = x − 3 or 2x y + y − x + 3 = 0 and x(2y − 1) + y + 3 = 0
Equating the coefficient of the highest power of x to zero gives: 2y − 1 = 0 from which, y = 1 which is an
2
asymptote.
Since y(2x + 1) = x − 3 then equating the coefficient of the highest power of y to zero gives: 2x + 1 = 0 from which, x = − 1 which is also an asymptote.
2
x − 3 −3
When x = 0, y =
=
= −3 and when y = 0,
2x + 1
1
x −3
0=
from which, x − 3 = 0 and x = 3.
2x + 1 x −3
A sketch of y = is shown in Fig. 18.33.
2x + 1

192 Higher Engineering Mathematics

y

6

4 y5 x 23
2x 11

x 52

1
2

2 y5 1
2

2
28

26

24

22

21

0

1

4

y5

24

26

Figure 18.33

x 23
2x 11

6

8

x

Functions and their curves
Problem 9. Determine the asymptotes parallel to the x- and y-axes for the function x 2 y 2 = 9(x 2 + y 2 ).
Asymptotes parallel to the x-axis:
Rearranging x 2 y 2 = 9(x 2 + y 2 ) gives

(iii) Equating the coefficient of the highest power of x to zero gives m − 1 = 0 from which, m = 1.
Equating the coefficient of the next highest power of x to zero gives m + c + 1 =0. and since m = 1, 1 + c + 1 = 0 from which, c = −2.
Hence y = mx + c = 1x − 2.

x 2 y 2 − 9x 2 − 9y 2 = 0 hence x 2 (y 2

− 9) − 9y 2

193

i.e. y = x − 2 is an asymptote.

=0

To determine any asymptotes parallel to the x-axis:
Equating the coefficient of the highest power of x to zero gives y 2 − 9 = 0 from which, y 2 = 9 and y = ±3.
Asymptotes parallel to the y-axis:
Since x 2 y 2 − 9x 2 − 9y 2 = 0 then y 2 (x 2 − 9) − 9x 2 = 0

Equating the coefficient of the highest power of y to zero gives x 2 − 9 = 0 from which, x 2 = 9 and x = ±3.
Hence asymptotes occur at y = ±3 and x = ±3.

Other asymptotes
To determine asymptotes other than those parallel to x- and y-axes a simple procedure is:
(i) substitute y = mx + c in the given equation
(ii) simplify the expression
(iii) equate the coefficients of the two highest powers of x to zero and determine the values of m and c. y = mx + c gives the asymptote.
Problem 10. Determine the asymptotes for the function: y(x + 1) = (x − 3)(x + 2) and sketch the curve. Rearranging y(x + 1) = (x − 3)(x + 2) yx + y = x 2 − x − 6

gives

The coefficient of the highest power of x (i.e. x 2 ) is 1.
Equating this to zero gives 1 =0 which is not an equation of a line. Hence there is no asymptote parallel to the x-axis. To determine any asymptotes parallel to the y-axis:
Since y(x + 1) = (x − 3)(x + 2) the coefficient of the highest power of y is x + 1. Equating this to zero gives x + 1 = 0, from which, x = −1. Hence x = −1 is an asymptote.
When x = 0, y(1) = (−3)(2), i.e. y = −6.
When y = 0, 0 =(x − 3)(x + 2), i.e. x = 3 and x = −2.
A sketch of the function y(x + 1) = (x − 3)(x + 2) is shown in Fig. 18.34.
Problem 11. Determine the asymptotes for the function x 3 − x y 2 + 2x − 9 =0.
Following the procedure:
(i) Substituting y = mx + c gives x 3 − x(mx + c)2 + 2x − 9 =0.
(ii) Simplifying gives

Following the above procedure:
(i) Substituting y = mx + c into y(x + 1) = (x − 3) (x + 2) gives:
(mx + c)(x + 1) = (x − 3)(x + 2)
(ii) Simplifying gives mx 2 + mx + cx + c = x 2 − x − 6 and (m − 1)x 2 + (m + c + 1)x + c + 6 =0

x 3 − x[m 2 x 2 + 2mcx + c2 ] + 2x − 9 = 0
i.e.

x 3 − m 2 x 3 − 2mcx 2 − c2 x + 2x − 9 = 0

and x 3 (1 − m 2 ) − 2mcx 2 − c2 x + 2x − 9 = 0
(iii) Equating the coefficient of the highest power of x
(i.e. x 3 in this case) to zero gives 1 −m 2 = 0, from which, m = ±1.
Equating the coefficient of the next highest power of x (i.e. x 2 in this case) to zero gives −2mc = 0, from which, c = 0.

194 Higher Engineering Mathematics y 6

x2

2

x 521

y5

4

2

26

24

22

0

2

4

y (x 11) 5 (x 23)(x 12)
22

y (x 11) 5 (x 23)(x 12)

24

26

28

210

Figure 18.34

6

x

Functions and their curves
Hence y = mx + c = ±1x + 0, i.e. y = x and y = −x are asymptotes.
To determine any asymptotes parallel to the x- and y-axes for the function x 3 − x y 2 + 2x − 9 =0:
Equating the coefficient of the highest power of x term to zero gives 1 = 0 which is not an equation of a line.
Hence there is no asymptote parallel with the x-axis.
Equating the coefficient of the highest power of y term to zero gives −x = 0 from which, x = 0.
Hence x = 0, y = x and y = − x are asymptotes for the function x3 − xy2 + 2x − 9 =0.
Problem 12. Find the asymptotes for the function x2 + 1 y= and sketch a graph of the function. x x2 + 1 gives yx = x 2 + 1.
Rearranging y = x Equating the coefficient of the highest power x term to zero gives 1 =0, hence there is no asymptote parallel to the x-axis.

1
Hence 1 = 2 and x 2 = 1, from which, x = ±1. x When x = 1, y= x2 + 1 1 + 1
=
=2 x 1

and when x = −1, y= (−1)2 + 1
= −2
−1

i.e. (1, 2) and (−1, −2) are the co-ordinates of the turning d2 y
2
d2 y points. 2 = 2x −3 = 3 ; when x = 1, 2 is positive, dx x dx which indicates a minimum point and when x = −1, d2 y is negative, which indicates a maximum point, as dx 2 shown in Fig. 18.35.
Now try the following exercise

Exercise 80 Further problems on asymptotes Equating the coefficient of the highest power y term to zero gives x = 0.

In Problems 1 to 3, determine the asymptotes parallel to the x- and y-axes.

Hence there is an asymptote at x = 0 (i.e. the y-axis). 1.

To determine any other asymptotes we substitute y = mx + c into yx = x 2 + 1 which gives

2.

(mx + c)x = x 2 + 1
i.e.

3.

x −2 x +1 x y2 = x −3 y= y=

mx 2 + cx = x 2 + 1

and (m − 1)x 2 + cx − 1 = 0

[y = 1, x = −1]
[x = 3, y = 1 and y = −1]

x(x + 3)
(x + 2)(x + 1)
[x = −1, x = −2 and y = 1]

In Problems 4 and 5, determine all the asymptotes.

Equating the coefficient of the highest power x term to zero gives m − 1 = 0, from which m = 1.
Equating the coefficient of the next highest power x term to zero gives c = 0. Hence y = mx + c = 1x + 0, i.e. y = x is an asymptote.

4. 8x − 10 + x 3 − x y 2 = 0
[x = 0, y = x and y = −x]

x2 + 1 is shown in Fig. 18.35.
A sketch of y = x It is possible to determine maximum/minimum points on the graph (see Chapter 28).

In Problems 6 and 7, determine the asymptotes and sketch the curves.

Since then x2 + 1 x2 1
=
+ = x + x −1 y= x x x dy 1
= 1 − x −2 = 1 − 2 = 0 dx x

for a turning point.

195

5.

6.

7.

x 2 (y 2 − 16) = y

y=

x2 − x − 4 x +1

[y = 4, y = −4 and x = 0]

x = −1, y = x − 2, see Fig 18.40, page 202

x y 2 − x 2 y + 2x − y = 5 x = 0, y = 0, y = x, see Fig. 18.41, page 202

196 Higher Engineering Mathematics y 5

x

6

y5
4

y

x 211 x 2

24

22

2

0

4

x

22

y5

x 211 x 24

26

Figure 18.35

18.8

Brief guide to curve sketching

The following steps will give information from which the graphs of many types of functions y = f (x) can be sketched. (i) Use calculus to determine the location and nature of maximum and minimum points (see Chapter 28)
(ii) Determine where the curve cuts the x- and y-axes
(iii) Inspect the equation for symmetry.

(a)

If the equation is unchanged when −x is substituted for x, the graph will be symmetrical about the y-axis (i.e. it is an even function).

(b) If the equation is unchanged when −y is substituted for y, the graph will be symmetrical about the x-axis.
(c)

If f (−x) = − f (x), the graph is symmetrical about the origin (i.e. it is an odd function). (iv) Check for any asymptotes.

197

Functions and their curves y 18.9 Worked problems on curve sketching 20
15

Problem 13. Sketch the graphs of
(a) y = 2x 2 + 12x + 20

10

y 5 2x 2 1 12x 1 20

(b) y = −3x 2 + 12x − 15

5
2

(a)

y = 2x 2 + 12x + 20 is a parabola since the equation is a quadratic. To determine the turning point: Gradient =

24

23

22

21

0
23
25

1

2

3

x

210

dy
= 4x + 12 = 0 for a turning point. dx y 5 23x 2 1 12x 2 15
215

Hence 4x = −12 and x = −3.

220

When x = −3, y = 2(−3)2 + 12(−3) + 20 =2.

225

Hence (−3, 2) are the co-ordinates of the turning point Figure 18.36

d2 y
= 4, which is positive, hence (−3, 2) is a dx 2 minimum point.
When x = 0, y = 20, hence the curve cuts the y-axis at y = 20.
Thus knowing the curve passes through (−3, 2) and (0, 20) and appreciating the general shape of a parabola results in the sketch given in
Fig. 18.36.
(b)

Problem 14. Sketch the curves depicting the following equations:
(a) x = 9 − y 2 (b) y 2 = 16x
(c) x y = 5
(a)

y = −3x 2 + 12x − 15 is also a parabola (but
‘upside down’ due to the minus sign in front of the x 2 term).
Gradient =

dy
= −6x + 12 = 0 for a turning point. dx Hence 6x = 12 and x = 2.
When x = 2, y = −3(2)2 + 12(2) − 15 =−3.
Hence (2, −3) are the co-ordinates of the turning point d2 y
= −6, which is negative, hence (2, −3) is a dx 2 maximum point.
When x = 0, y = −15, hence the curve cuts the axis at y = −15.
The curve is shown sketched in Fig. 18.36.

Squaring both sides of the equation and transposing gives x 2 + y 2 = 9. Comparing this with the standard equation of a circle, centre origin and radius a, i.e. x 2 + y 2 = a 2, shows that x 2 + y 2 = 9 represents a circle, centre origin and radius 3. A sketch of this circle is shown in
Fig. 18.37(a).

(b) The equation y 2 = 16x is symmetrical about the x-axis and having its vertex at the origin (0, 0).
Also, when x = 1, y = ±4. A sketch of this parabola is shown in Fig. 18.37(b).
(c)

a represents a rectangular
The equation y = x hyperbola lying entirely within the first and third
5
quadrants. Transposing x y = 5 gives y = , and x therefore represents the rectangular hyperbola shown in Fig. 18.37(c).

198 Higher Engineering Mathematics y with the x- and y-axes of a rectangular co-ordinate system, the major axis being 2(3), i.e. 6 units long and the minor axis 2(2), i.e. 4 units long, as shown in Fig. 18.38(a).

3 x y

4
(a)

x

x 5 !(92y 2)
6

y
14

(a) 4x 2 5 36 29y 2 y 1

x x 24

2Œ„
3
(b) y 2 516x

(b) 3y 2 11555x 2

y

Figure 18.38

x

(c) xy 5 5

Figure 18.37

Problem 15. Sketch the curves depicting the following equations:
(a) 4x 2 = 36 −9y 2 (b) 3y 2 + 15 =5x 2
(a) By dividing throughout by 36 and transposing, the equation 4x 2 = 36 − 9y 2 can be written as x 2 y2
+ = 1. The equation of an ellipse is of the
9
4 x 2 y2 form 2 + 2 = 1, where 2a and 2b represent the a b x 2 y2 length of the axes of the ellipse. Thus 2 + 2 = 1
3
2 represents an ellipse, having its axes coinciding

(b) Dividing 3y 2 + 15 = 5x 2 throughout by 15 and x 2 y2 transposing gives
− = 1. The equation
3
5
2
2 y x
− = 1 represents a hyperbola which is syma 2 b2 metrical about both the x- and y-axes, the distance between the vertices being given by 2a. x 2 y2
− = 1 is as shown in
Thus a sketch of
3
5

Fig. 18.38(b), having a distance of 2 3 between its vertices.
Problem 16. Describe the shape of the curves represented by the following equations:
(a) x = 2

1−

(c) y = 6 1 −

x2
16

y
2

2

(b)

y2
= 2x
8

1/2

(a) Squaring the equation gives x 2 = 4 1 − and transposing gives x 2 = 4 − y 2 , i.e.

y
2

2

199

Functions and their curves x 2 + y 2 = 4. Comparing this equation with x 2 + y 2 = a 2 shows that x 2 + y 2 = 4 is the equation of a circle having centre at the origin (0, 0) and of radius 2 units.
(b) Transposing y2 y2
= 2x
8

gives

√ y = 4 x. Thus

Now try the following exercise
Exercise 81 sketching 1. Sketch the graphs of (a) y = 3x 2 + 9x +

= 2x is the equation of a parabola having its
8
axis of symmetry coinciding with the x-axis and its vertex at the origin of a rectangular co-ordinate system. (c)

y =6 1 −
1−

x2
16

1/2

x2
16

can be transposed to

1/2

and squaring both sides gives

This is the equation of an ellipse, centre at the origin of a rectangular co-ordinate system, the major

axis coinciding with the y-axis and being 2 36,
i.e. 12 units long. √ minor axis coincides with
The
the x-axis and is 2 16, i.e. 8 units long.

In Problems 2 to 8, sketch the curves depicting the equations given.
2.

Problem 17. Describe the shape of the curves represented by the following equations:

(a)

x
=
5
Since

1+ x =
5

y
2

2

1+

(b) y 2

15 y =
4 2x

(With reference to Section 18.1 (vii), a is equal to ±5) y 15 a (b) The equation = is of the form y = , a =
4 2x x 60
= 30.
2
This represents a rectangular hyperbola, symmetrical about both the x- and y-axis, and lying entirely in the first and third quadrants, similar in shape to the curves shown in Fig. 18.9.

1−

y
4

2



y x= 9 parabola, symmetrical about x-axis, vertex at (0, 0)

4.

2

y 2 x2 =1+
25
2 x 2 y2
i.e.

=1
25
4
This is a hyperbola which is symmetrical about

both the x- and y-axes, the vertices being 2 25,
i.e. 10 units apart.

x =4

[circle, centre (0, 0), radius 4 units]
3.

(a)

7
4

(b) y = −5x 2 + 20x + 50.


(a) Parabola with minimum

⎢ value at − 3 , −5 and


2
3

passing through 0, 1 4 . ⎥




⎢(b) Parabola with maximum ⎥



value at (2, 70) and passing⎦ through (0, 50).

y
=
6

y2 x2 x 2 y2
= 1 − , i.e.
+
= 1.
36
16
16 36

Further problems on curve

5.

y2 =

x 2 − 16
4


hyperbola, symmetrical about
⎢x- and y-axes, distance



⎣between vertices 8 units along ⎦ x-axis x2 y2 = 5−
5
2

⎡ ellipse, centre (0, 0), major axis
⎣10 units along y-axis, minor axis⎦

2 10 units along x-axis

6.

x = 3 1 + y2

⎤ hyperbola, symmetrical about
⎢x- and y-axes, distance



⎣between vertices 6 units along ⎦ x-axis 7.

x 2 y2 = 9 rectangular hyperbola, lying in first and third quadrants only

200 Higher Engineering Mathematics
8.

9.



hyperbola, symmetrical about x⎣and y-axes, vertices 2 units

apart along x-axis

x = 1 (36 − 18y 2 )
3


ellipse, centre (0, 0),
⎢major axis 4 units along x-axis,⎥



⎣minor axis 2 2 units

along y-axis

12.
13.

y = 7x −1

14.

y = (3x)1/2 parabola, vertex at (0, 0), symmetrical about the x-axis

15.

Sketch the circle given by the equation x 2 + y 2 − 4x + 10y + 25 =0.

√ y = 9 − x2
[circle, centre (0, 0), radius 3 units]

y 2 − 8 =−2x 2

⎤ ellipse, centre (0, 0), major

⎢axis 2 8 units along the ⎥


⎣ y-axis, minor axis 4 units ⎦ along the x-axis

[Centre at (2, −5), radius 2]
In Problems 10 to 15 describe the shape of the curves represented by the equations given.
10.

11.

y = [3(1 − x 2 )]

⎤ ellipse, centre (0, 0), major axis

⎣2 3 units along y-axis, minor ⎦ axis 2 units along x-axis y = [3(x 2 − 1)]



rectangular hyperbola, lying
⎢in first and third quadrants, ⎥


⎣symmetrical about x- and ⎦ y-axes Graphical solutions to Exercise 77, page 186
1.

2.

y
10

y
4

5

2

y 5 3x 25

0

1

2

3

0

x

1

2

22

25
3.

4.

y

3

x

y 5 23x 14

y

8

8

y 5(x 23)2

y 5 x 213

6

4

4
2
22

Figure 18.39

21

0

1

2

x

0

2

4

6

x

201

Functions and their curves
5.

6. y y
0.50

15

y 5x 2x 2

0.25

10

y 5(x24) 212

0

1

x

5

2

0

4

6

8

x

7.

8. y y
10

y 5 11 2 cos 3x

3
2

y 5x 312

5

1
22

21

0

2

1

x



2

0

25

21

3␲
2

210
10.
y
9.

3

y
6

y 5 3 2 2 sin(x 1


)
4

2

y 5 2 ln x

4
1

2

0

p
2

p

3p
2

2p

x

0
21
22

Figure 18.39 (Continued)

1

2

3

4 x

2␲

x

202 Higher Engineering Mathematics
Graphical solutions to Problems 6 and 7, Exercise 80, page 195 y 6

2
2

x 521

y5 x 4

2

26

24

22

x 2 2x2 4 y5 x 11

0

6 x

4

2

x 2 2x 24 y5 x 11

22
24
26

Figure 18.40 y xy 2 2 x 2y 1 2x 2y 5 5
6

y5

x

4

2

26

24

xy 2 2 x 2y 1 2x 2y 5 5

22

0

22

24

26

Figure 18.41

2

4

6

xy 2 2 x 2y 1 2x 2y 5 5

x

Chapter 19

Irregular areas, volumes and mean values of waveforms
19.1

(iii) Areas PQRS

Areas of irregular figures

Areas of irregular plane surfaces may be approximately determined by using (a) a planimeter, (b) the trapezoidal rule, (c) the mid-ordinate rule, and (d) Simpson’s rule.
Such methods may be used, for example, by engineers estimating areas of indicator diagrams of steam engines, surveyors estimating areas of plots of land or naval architects estimating areas of water planes or transverse sections of ships.
(a)

A planimeter is an instrument for directly measuring small areas bounded by an irregular curve.

=d

y1 + y7
+ y2 + y3 + y4 + y5 + y6
2

In general, the trapezoidal rule states:
Area =

⎡ ⎛

⎤ first + sum of width of ⎣ 1 ⎝
⎠ + remaining⎦ last interval
2 ordinate ordinates (c) Mid-ordinate rule
To determine the area ABCD of Fig. 19.2:

(b) Trapezoidal rule
To determine the areas PQRS in Fig. 19.1:

B

C

y1

y2

y3

y4

y5

y6

R y7 y2

y3

y4

y5

y6

d

Q

y1

d

d

d

d

d

D

A
S

P d d

d

d

d

d

Figure 19.2

Figure 19.1

(i) Divide base PSinto any number of equal intervals, each of width d (the greater the number of intervals, the greater the accuracy).
(ii) Accurately measure ordinates y1 , y2 , y3 , etc.

(i) Divide base AD into any number of equal intervals, each of width d (the greater the number of intervals, the greater the accuracy).
(ii) Erect ordinates in the middle of each interval
(shown by broken lines in Fig. 19.2).

204 Higher Engineering Mathematics
(iii) Accurately measure ordinates y1 , y2 , y3 , etc.
(iv) Area ABCD = d(y1 + y2+ y3 + y4 + y5+ y6 )

Area =

width of

sum of

interval

25
Speed (m/s)

In general, the mid-ordinate rule states:

mid-ordinates

Graph of speed/time

30

20
15
10

0

1

2
3
4
Time (seconds)

5

24.0

20.25

17.5

15.0

12.5

8.75

7.0

5.5

2.5

4.0

(i) Divide base PS into an even number of intervals, each of width d (the greater the number of intervals, the greater the accuracy).

1.25

5

To determine the area PQRS of Fig. 19.1:

10.75

(d) Simpson’s rule

6

Figure 19.3

(ii) Accurately measure ordinates y1 , y2 , y3, etc.
(iii) Area PQRS =

d
[(y1 + y7 ) + 4(y2 + y4 +
3
y6 ) + 2(y3 + y5 )]

area = (1)

In general, Simpson’s rule states:

Area =

+ 8.75 + 12.5 + 17.5

ordinate

+2

= 58.75 m sum of even ordinates sum of remaining

2

3

4

5

(b) Mid-ordinate rule (see para. (c) above)
The time base is divided into 6 strips each of width
1 second.
Mid-ordinates are erected as shown in Fig. 19.3 by the broken lines. The length of each mid-ordinate is measured. Thus

odd ordinates

Problem 1. A car starts from rest and its speed is measured every second for 6 s:
0 1

0 + 24.0
+ 2.5 + 5.5
2

first + last

1 width of
3 interval

+4

Time t (s)

Thus

area = (1)[1.25 + 4.0 + 7.0 + 10.75
+ 15.0 + 20.25]

6

Speed v (m/s) 0 2.5 5.5 8.75 12.5 17.5 24.0

= 58.25 m

Determine the distance travelled in 6 seconds (i.e. the area under the v/t graph), by (a) the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson’s rule. (c) Simpson’s rule (see para. (d) above)

A graph of speed/time is shown in Fig. 19.3.

The time base is divided into 6 strips each of width 1 s, and the length of the ordinates measured.
Thus
area = 1 (1)[(0 + 24.0) + 4(2.5 + 8.75
3
+ 17.5) + 2(5.5 + 12.5)]

(a) Trapezoidal rule (see para. (b) above)
The time base is divided into 6 strips each of width 1 s, and the length of the ordinates measured.

= 58.33 m

Irregular areas, volumes and mean values of waveforms

205

Problem 2. A river is 15 m wide. Soundings of the depth are made at equal intervals of 3 m across the river and are as shown below.
Depth (m) 0

2.2 3.3

4.5 4.2 2.4

0
140 160 200 190 180 130

Calculate the cross-sectional area of the flow of water at this point using Simpson’s rule.

50

From para. (d) above,

= (1)[0 + 36.4 + 15] = 51.4 m2

Width (m)

50

50

0 2.8 5.2 6.5 5.8 4.1 3.0 2.3

[143 m2 ]

Estimate the area of the deck.

Exercise 82 Further problems on areas of irregular figures

2. Plot the graph of y = 2x 2 + 3 between x = 0 and x = 4. Estimate the area enclosed by the curve, the ordinates x = 0 and x = 4, and the x-axis by an approximate method.
[54.7 square units]

50

5. The deck of a ship is 35 m long. At equal intervals of 5 m the width is given by the following table:

Now try the following exercise

1. Plot a graph of y = 3x − x 2 by completing a table of values of y from x = 0 to x = 3.
Determine the area enclosed by the curve, the x-axis and ordinate x = 0 and x = 3 by (a) the trapezoidal rule, (b) the mid-ordinate rule and
(c) by Simpson’s rule.
[4.5 square units]

50

Figure 19.4

Area = 1 (3)[(0 + 0) + 4(2.2 + 4.5 + 2.4)
3
+ 2(3.3 + 4.2)]

50

19.2

Volumes of irregular solids

If the cross-sectional areas A1 , A2 , A3 , . . . of an irregular solid bounded by two parallel planes are known at equal intervals of width d (as shown in Fig. 19.5), then by
Simpson’s rule: volume, V =

d
[(A1 + A7 ) + 4(A2 + A4
3
+ A6) + 2 (A3 + A5)]

3. The velocity of a car at one second intervals is given in the following table: time t (s) 0 1 velocity v (m/s)

2

3

4

5

6

A1

A2

A3

A4

A5

A6

A7

0 2.0 4.5 8.0 14.0 21.0 29.0

Determine the distance travelled in 6 seconds
(i.e. the area under the v/t graph) using
Simpson’s rule.
[63.33 m]
4. The shape of a piece of land is shown in
Fig. 19.4. To estimate the area of the land, a surveyor takes measurements at intervals of 50 m, perpendicular to the straight portion with the results shown (the dimensions being in metres). Estimate the area of the land in
[4.70 ha] hectares (1 ha = 104 m2 ).

d

d

d

d

d

d

Figure 19.5

Problem 3. A tree trunk is 12 m in length and has a varying cross-section. The cross-sectional areas at intervals of 2 m measured from one end are:
0.52, 0.55, 0.59, 0.63, 0.72, 0.84, 0.97 m2
Estimate the volume of the tree trunk.

206 Higher Engineering Mathematics
A sketch of the tree trunk is similar to that shown in Fig. 19.5 above, where d = 2 m, A1 = 0.52 m2 ,
A2 = 0.55 m2 , and so on.
Using Simpson’s rule for volumes gives:
Volume =

2
3 [(0.52 + 0.97) + 4(0.55 + 0.63

+ 0.84) + 2(0.59 + 0.72)]
= 2 [1.49 + 8.08 + 2.62] = 8.13 m3
3

1.76, 2.78, 3.10, 3.12, 2.61, 1.24, 0.85 m2
Determine the underwater volume if the sections are 3 m apart.
[42.59 m3 ]
2. To estimate the amount of earth to be removed when constructing a cutting the crosssectional area at intervals of 8 m were estimated as follows:
0, 2.8,

Problem 4. The areas of seven horizontal cross-sections of a water reservoir at intervals of
10 m are:
210, 250, 320, 350, 290, 230, 170 m2
Calculate the capacity of the reservoir in litres.
Using Simpson’s rule for volumes gives:

3.7,

4.5,

4.1,

2.6,

0 m3

Estimate the volume of earth to be excavated.
[147 m3]
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)

Circumference
(m)

0

2.80

2

3.25

4

3.94

6

4.32

= 16400 m3

8

5.16

16400 m3 = 16400 × 106 cm3 and since
1 litre = 1000 cm3 ,

10

5.82

12

6.36

Volume =

10
[(210 + 170) + 4(250 + 350
3
+ 230) + 2(320 + 290)]

=

10
[380 + 3320 + 1220]
3

capacity of reservoir =

16400 × 106 litres 1000

Estimate the volume of the timber in cubic metres. [20.42 m3 ]

= 1 6400000
= 1.64 × 107 litres
Now try the following exercise
Exercise 83
Further problems on volumes of irregular solids
1. The areas of equidistantly spaced sections of the underwater form of a small boat are as follows: 19.3 The mean or average value of a waveform The mean or average value, y, of the waveform shown in Fig. 19.6 is given by:

y=

area under curve length of base, b

Irregular areas, volumes and mean values of waveforms

207

(iv) of a half-wave rectified waveform (see
Fig. 19.7(c)) is 0.318 × maximum value, or
(1/π) maximum value.
Problem 5. Determine the average values over half a cycle of the periodic waveforms shown in
Fig. 19.8.

y

d

d

y3

y4

y5

y6

y7

d

d

d

d

d

Voltage (V)

y1 y2

b

Figure 19.6

20

0

If the mid-ordinate rule is used to find the area under the curve, then:

1

2

3

t (ms)

4

210

y=

Current (A)

(a)

sum of mid-ordinates number of mid-ordinates
=

y1 + y2 + y3 + y4 + y5 + y6 + y7
7

3
2
1
0
21
22
23

for Fig. 19.6

1

2

3

4

5 6

t (s)

(b)
Voltage (V)

For a sine wave, the mean or average value:
(i) over one complete cycle is zero (see Fig. 19.7(a)),

V
Vm

10

0

V
Vm

2

4

6

8

t (ms)

210 t 0

(c)

t

0

Figure 19.8
(a)

(b)

(a)

V
Vm

Area under triangular waveform (a) for a half cycle is given by:
Area =

t

0

(c)

Figure 19.7

(ii) over half a cycle is 0.637 × maximum value, or
(2/π ) × maximum value,
(iii) of a full-wave rectified waveform (see Fig.
19.7(b)) is 0.637 × maximum value,

1
2

(base) (perpendicular height)

= 1 (2 × 10−3)(20)
2
= 20 × 10−3 Vs
Average value of waveform
=

area under curve length of base

=

20 × 10−3 Vs
2 × 10−3 s

= 10 V

208 Higher Engineering Mathematics
(b) Area under waveform (b) for a half cycle = (1 × 1) + (3 × 2) = 7 As.

(a) One cycle of the trapezoidal waveform (a) is completed in 10 ms (i.e. the periodic time is
10 ms).

Average value of waveform

Area under curve = area of trapezium

area under curve
=
length of base

=

1
2

(sum of parallel sides) (perpendicular

distance between parallel sides)
7 As
=
3s

= 1 {(4 + 8) × 10−3}(5 × 10−3 )
2
= 30 × 10−6 As

= 2.33 A

Mean value over one cycle

(c) A half cycle of the voltage waveform (c) is completed in 4 ms.

=

area under curve 30 × 10−6 As
=
length of base
10 × 10−3 s

= 3 mA

Area under curve = 1 {(3 − 1)10−3 }(10)
2
= 10 × 10−3 Vs

(b) One cycle of the sawtooth waveform (b) is completed in 5 ms.

Average value of waveform
=
=

Area under curve = 1 (3 × 10−3)(2)
2

area under curve length of base
10 × 10−3 Vs
4 × 10−3 s

= 3 × 10−3 As
Mean value over one cycle
=

= 2.5 V

area under curve 3 × 10−3 As
=
length of base
5 × 10−3 s

= 0.6 A

Current (mA)

Problem 6. Determine the mean value of current over one complete cycle of the periodic waveforms shown in Fig. 19.9.

5

Time (h)

4

8

12

16

20

24

28 t (ms)

(a)

0

1

2

3

4

5

6

Power (kW)
0

Current (mA)

Problem 7. The power used in a manufacturing process during a 6 hour period is recorded at intervals of 1 hour as shown below.

0

14

29

51

45

23

0

Plot a graph of power against time and, by using the mid-ordinate rule, determine (a) the area under the curve and (b) the average value of the power.

2

The graph of power/time is shown in Fig. 19.10.
(a)
0

2

4

6

8
(b)

Figure 19.9

10

12

t (ms)

The time base is divided into 6 equal intervals, each of width 1 hour. Mid-ordinates are erected (shown by broken lines in Fig. 19.10) and measured. The values are shown in
Fig. 19.10.

Irregular areas, volumes and mean values of waveforms

One cycle of the output voltage is completed in π radians or 180◦ . The base is divided into 6 intervals, each of width 30◦ . The mid-ordinate of each interval will lie at
15◦, 45◦ , 75◦ , etc.
At 15◦ the height of the mid-ordinate is
10 sin 15◦ = 2.588 V.
At 45◦ the height of the mid-ordinate is
10 sin 45◦ = 7.071 V, and so on.
The results are tabulated below:

Graph of power/time
50
40

Power (kW)

209

30
20
10

0

21.5
1

42.0
2

49.5

Mid-ordinate

3
4
Time (hours)

5

= (1)[7.0 + 21.5 + 42.0
+ 49.5 + 37.0 + 10.0]
= 167 kWh (i.e. a measure of electrical energy)

10 sin 105◦ = 9.659 V
10 sin 135◦ = 7.071 V

165◦

Area under curve = (width of interval)
× (sum of mid-ordinates)

10 sin 75◦ = 9.659 V

135◦

Figure 19.10

10 sin 45◦ = 7.071 V

105◦

6

10 sin 15◦ = 2.588 V

45◦

37.0 10.0

Height of mid-ordinate

15◦
75◦

7.0

10 sin 165◦ = 2.588 V sum of mid-ordinates =38.636 V

Mean or average value of output voltage

(b) Average value of waveform

sum of mid-ordinates number of mid-ordinates
38.636
=
6
= 6.439 V
=

=

area under curve length of base

=

167 kWh
= 27.83 kW
6h

(With a larger number of intervals a more accurate answer may be obtained.) For a sine wave the actual mean value is 0.637 ×maximum value, which in this problem gives 6.37 V.

Alternatively, average value
=

sum of mid-ordinates number of mid-ordinates

Voltage (V)

Problem 8. Fig. 19.11 shows a sinusoidal output voltage of a full-wave rectifier. Determine, using the mid-ordinate rule with 6 intervals, the mean output voltage.

10

0

308608908

2

Figure 19.11

1808


2708
3␲
2

3608
2␲



Problem 9. An indicator diagram for a steam engine is shown in Fig. 19.12. The base line has been divided into 6 equally spaced intervals and the lengths of the 7 ordinates measured with the results shown in centimetres. Determine (a) the area of the indicator diagram using Simpson’s rule, and (b) the mean pressure in the cylinder given that 1 cm represents 100 kPa.

3.6

4.0

3.5

2.9

12.0 cm

Figure 19.12

2.2

1.7

1.6

210 Higher Engineering Mathematics

area =

12.0 cm. Using
6

Current (A)

(a) The width of each interval is
Simpson’s rule,

1
3 (2.0)[(3.6 + 1.6) + 4(4.0

0

+ 2.9 + 1.7) + 2(3.5 + 2.2)]

Figure 19.13 (Continued )

(b) Mean height of ordinates
=

30 t (ms)

(c)

area of diagram 34
=
length of base
12

= 2.83 cm
Since 1 cm represents 100 kPa, the mean pressure in the cylinder
= 2.83 cm × 100 kPa/cm = 283 kPa.

2. Find the average value of the periodic waveforms shown in Fig. 19.14 over one complete cycle. [(a) 2.5 V (b) 3 A]
Voltage (mV)

= 34 cm

15

25

= 2 [5.2 + 34.4 + 11.4]
3
2

5

10

0

2

4

6

8

10

t (ms)

8

10

t (ms)

Now try the following exercise
Exercise 84 Further problems on mean or average values of waveforms

Current (A)

(a)

5

0

Current (A)

1. Determine the mean value of the periodic waveforms shown in Fig. 19.13 over a half cycle. [(a) 2 A (b) 50 V (c) 2.5 A]

10

20

t (ms)

Voltage (V)

(a)

(b)

Figure 19.14

Time (ms)

0 5

10

15

20

25

30

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.
[0.093 As, 3.1 A]

100

5

10 t (ms)

2100
(b)

Figure 19.13

6

Current (A) 0 0.9 2.6 4.9 5.8 3.5 0

22

0

4

3. An alternating current has the following values at equal intervals of 5 ms

2

0

2

4. Determine, using an approximate method, the average value of a sine wave of maximum value 50 V for (a) a half cycle and (b) a complete cycle.
[(a) 31.83 V (b) 0]

Irregular areas, volumes and mean values of waveforms

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.
[49.13 cm2 , 368.5 kPa]

211

Revision Test 6
This Revision Test covers the material contained in Chapters 18 and 19. The marks for each question are shown in brackets at the end of each question.
1.

(a)

y = (x − 2)2

(b)

y = 3 −cos 2x (d) 9x 2 − 4y 2 = 36

π
⎪ −1 −π ≤ x ≤ −


2


⎨ π π x − ≤x ≤ f (x) =

2
2


⎪ π ⎪
⎩ 1
≤x ≤π
2

(e)

(c)

x 2 + y 2 − 2x + 4y − 4 = 0

2.

Determine the inverse of f (x) = 3x + 1

3.

Evaluate, correct to 3 decimal places:
2 tan−1 1.64 + sec−1 2.43 − 3 cosec−1 3.85

4.

6.

(x − 1)(x + 4)
(x − 2)(x − 5)

A circular cooling tower is 20 m high. The inside diameter of the tower at different heights is given in the following table:
Height (m)

0

5.0 10.0 15.0 20.0

Diameter (m) 16.0 13.3 10.7

(3)
7.
(3)

(8)

Plot a graph of y = 3x 2 + 5 from x = 1 to x = 4.
Estimate, correct to 2 decimal places, using 6 intervals, the area enclosed by the curve, the ordinates

8.6

8.0

Determine the area corresponding to each diameter and hence estimate the capacity of the tower in cubic metres. (5)

(15)

Determine the asymptotes for the following function and hence sketch the curve: y= 5.

x = 1 and x = 4, and the x-axis by (a) the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson’s rule. (11)

Sketch the following graphs, showing the relevant points: A vehicle starts from rest and its velocity is measured every second for 6 seconds, with the following results:
Time t (s) 0
Velocity
v (m/s)

1

2

0 1.2 2.4

3

4

3.7 5.2

5

6

6.0 9.2

Using Simpson’s rule, calculate (a) the distance travelled in 6 s (i.e. the area under the v/t graph) and (b) the average speed over this period.
(5)

Chapter 20

Complex numbers
20.1

Cartesian complex numbers

There are several applications of complex numbers in science and engineering, in particular in electrical alternating current theory and in mechanical vector analysis. There are two main forms of complex number –
Cartesian form and polar form – and both are explained in this chapter.
If we can add, subtract, multiply and divide complex numbers in both forms and represent the numbers on an Argand diagram then a.c. theory and vector analysis become considerably easier.
(i) If the quadratic equation x 2 + 2x + 5 = 0 is solved using the quadratic formula then, x= −2 ±

− (4)(1)(5)]
2(1)

(iii) In pure √ mathematics the symbol i is used to indicate −1 (i being the first letter of the word imaginary). However i is the symbol of electric current in engineering, and to avoid possible confusion the√ letter in the alphabet, j , is used to next represent −1.
Problem 1. Solve the quadratic equation x 2 + 4 = 0.

Since x 2 + 4 =0 then x 2 = −4 and x = −4.

(−1) 4 = j (±2)

= ± j2, (since j = −1)

i.e., x =

[(−1)(4)] =

(Note that ± j 2 may also be written ±2 j).

[(2)2



−2 ± [−16] −2 ± [(16)(−1)]
=
=
2
2
√ √

−2 ± 16 −1 −2 ± 4 −1
=
=
2
2

= −1 ± 2 −1

It is not possible to evaluate −1 in real terms. However, if an operator j is defined as

j = −1 then the solution may be expressed as x = −1 ± j 2.
(ii) −1 + j 2 and −1 − j 2 are known as complex numbers. Both solutions are of the form a + jb,
‘a’ being termed the real part and jb the imaginary part. A complex number of the form a + jb is called Cartesian complex number.

Problem 2. Solve the quadratic equation
2x 2 + 3x + 5 = 0.
Using the quadratic formula,
−3 ± [(3)2 − 4(2)(5)]
2(2)



−3 ± −31 −3 ± (−1) 31
=
=
4
4

−3 ± j 31
=
4

3
31
Hence x = − ± j or −0.750 ± j1.392,
4
4 correct to 3 decimal places. x= (Note, a graph of y = 2x 2 + 3x + 5 does not cross the x-axis and hence 2x 2 + 3x + 5 = 0 has no real roots.) 214 Higher Engineering Mathematics
Problem 3.
(a)

j3

(b)

Evaluate j4 j 23

(c)

20.2

−4
(d) 9 j (a)

j 3 = j 2 × j = (−1) × j = − j, since j 2 = −1

(b)

j 4 = j 2 × j 2 = (−1) × (−1) = 1

(c)

j 23 = j × j 22 = j × ( j 2)11 = j × (−1)11

(d)

j9=

= j × (−1) = − j j × j 8 = j × ( j 2)4 = j × (−1)4
= j ×1 = j
Hence

The Argand diagram

A complex number may be represented pictorially on rectangular or cartesian axes. The horizontal (or x) axis is used to represent the real axis and the vertical (or y) axis is used to represent the imaginary axis. Such a diagram is called an Argand diagram. In Fig. 20.1, the point A represents the complex number (3 + j 2) and is obtained by plotting the co-ordinates (3, j 2) as in graphical work.
Figure20.1 also showstheArgand points B, C and D representing the complex numbers (−2 + j 4), (−3 − j 5) and (1 − j 3) respectively.

4j
−4 −4 −4 − j
=
=
×
= j9 j j −j
−j2
4j
=
= 4 j or j4
−(−1)

Imaginary axis B

j4 j3 A

j2

Now try the following exercise

j

Exercise 85 Further problems on the introduction to cartesian complex numbers

23

22 21 0
2j

1

2

3

Real axis

2j 2

In Problems 1 to 9, solve the quadratic equations.
[± j 5]

2j 3

x − 2x + 2 = 0

[x = 1 ± j ]

2j 4

3.

x 2 − 4x + 5 =0

[x = 2 ± j ]

4.

x 2 − 6x + 10 =0

[x = 3 ± j ]

5.

2x 2 − 2x + 1 =0

[x = 0.5 ± j 0.5]

6.

x 2 − 4x + 8 =0

7.

25x 2 − 10x + 2 = 0

1.

x 2 + 25 =0

2.

2

2j 5

Figure 20.1

[x = 2 ± j 2]
[x = 0.2 ± j 0.2]

8. 2x 2 + 3x + 4 =0


23
3
− ±j or − 0.750 ± j 1.199
4
4

9. 4t 2 − 5t + 7 =0


87
5
±j
or 0.625 ± j 1.166
8
8

10. Evaluate (a) j 8

C

D

1
4
(b) − 7 (c) 13 j 2j
[(a) 1 (b) − j (c) − j 2]

20.3 Addition and subtraction of complex numbers
Two complex numbers are added/subtracted by adding/ subtracting separately the two real parts and the two imaginary parts.
For example, if Z 1 = a + jb and Z 2 = c + jd, then Z 1 + Z 2 = (a + jb) + (c + j d)
= (a + c) + j (b +d)

and

Z 1 − Z 2 = (a + jb) − (c + j d)
= (a − c) + j (b −d)

Complex numbers
Thus, for example,
(2 + j 3) +(3 − j 4)= 2 + j 3 +3 − j 4
= 5 − j1 and (2 + j 3) −(3 − j 4)= 2 + j 3 −3 + j 4
= −1 + j7
The addition and subtraction of complex numbers may be achieved graphically as shown in the Argand diagram of Fig. 20.2. (2 + j 3) is represented by vector OP and
Imaginary
axis

(3 − j 4) by vector OQ. In Fig. 20.2(a) by vector addition
(i.e. the diagonal of the parallelogram) OP + OQ = OR.
R is the point (5, − j 1).
Hence (2 + j 3) +(3 − j 4) =5 − j1.
In Fig. 20.2(b), vector OQ is reversed (shown as OQ ) since it is being subtracted. (Note OQ = 3 − j 4 and
OQ = −(3 − j 4) =−3 + j 4).
OP − OQ = OP + OQ = OS is found to be the Argand point (−1, j 7).
Hence (2 + j 3) −(3 − j 4) =−1 + j 7
Problem 4. Given Z 1 = 2 + j 4 and Z 2 = 3 − j determine (a) Z 1 + Z 2 , (b) Z 1 − Z 2 , (c) Z 2 − Z 1 and show the results on an Argand diagram.

P (21j3)

j3

215

j2

(a) Z 1 + Z 2 = (2 + j 4) +(3 − j )

j
0
2j

1

3

2

5 Real axis
R (5 2j )

4

= (2 + 3) + j (4 −1) = 5 + j 3
(b) Z 1 − Z 2 = (2 + j 4) −(3 − j )
= (2 − 3) + j (4 −(−1)) = −1 + j 5

2j2

(c) Z 2 − Z 1 = (3 − j ) −(2 + j 4)

2j3
2j4

= (3 − 2) + j (−1 − 4) = 1 − j 5

Q (3 2j 4)

Each result is shown in the Argand diagram of
Fig. 20.3.

(a)
Imaginary
axis
S (211j7)

Imaginary axis j7
(211 j 5)

j6

j4

j5
Q9

j3

j2

P (21j3)

j

j2 j 21 0
2j
1

2

3

Real axis

1

2

2j 2
2j 3

2j2

2j 4

2j3

2j 5

Q (32j4)

2j4
(b)

Figure 20.3
Figure 20.2

( 5 1j 3)

j3

j4

23 22 21 0
2j

j5

( 12 j 5)

3

4

5

Real axis

216 Higher Engineering Mathematics
20.4 Multiplication and division of complex numbers

Problem 5. If Z 1 = 1 − j 3, Z 2 = −2 + j 5 and
Z 3 = −3 − j 4, determine in a + j b form:

(i) Multiplication of complex numbers is achieved by assuming all quantities involved are real and then using j 2 = −1 to simplify.

(a) Z 1 Z 2
Z1 Z2
Z1 + Z2

(c)

Hence (a + j b)(c + j d)
= ac + a( j d) +( j b)c + ( j b)( j d)

Z1
Z3

(d) Z 1 Z 2 Z 3

(a) Z 1 Z 2 = (1 − j 3)(−2 + j 5)
= −2 + j 5 + j 6 − j 215

= ac + j ad + j bc + j 2bd

= (−2 + 15) + j (5 + 6), since j 2 = −1,

= (ac − bd) + j (ad + bc),

= 13 + j11

since j 2 = −1
Thus (3 + j 2)(4 − j 5)

(b)

(b)

= 12 − j 15 + j 8 − j 210

Z1
1 − j3
1 − j3
−3 + j 4
=
=
×
Z 3 −3 − j 4 −3 − j 4 −3 + j 4
=

= 22 − j 7
(ii) The complex conjugate of a complex number is obtained by changing the sign of the imaginary part. Hence the complex conjugate of a + j b is a − j b. The product of a complex number and its complex conjugate is always a real number.

−3 + j 4 + j 9 − j 212
32 + 42

=

= (12 − (−10)) + j (−15 +8)

9 + j 13
9
13
=
+ j
25
25
25
or 0.36 + j0.52

(c)

(1 − j 3)(−2 + j 5)
Z1 Z2
=
Z 1 + Z 2 (1 − j 3) + (−2 + j 5)
=

13 + j 11
, from part (a),
−1 + j 2

=

13 + j 11 −1 − j 2
×
−1 + j 2 −1 − j 2

[(a + j b)(a − j b) may be evaluated ‘on sight’ as a 2 + b2 ].

=

−13 − j 26 − j 11 − j 222
12 + 22

(iii) Division of complex numbers is achieved by multiplying both numerator and denominator by the complex conjugate of the denominator.

=

9 − j 37 9
37
= −j or 1.8 − j 7.4
5
5
5

For example,
(3 + j 4)(3 − j 4)= 9 − j 12 + j 12 − j 216
= 9 + 16 = 25

(d) Z 1 Z 2 Z 3 = (13 + j 11)(−3 − j 4), since

For example,

Z 1 Z 2 = 13 + j 11, from part (a)
= −39 − j 52 − j 33 − j 244

2 − j 5 2 − j 5 (3 − j 4)
=
×
3 + j 4 3 + j 4 (3 − j 4)
=

6 − j 8 − j 15 + j 220
32 + 42

−14 − j 23 −14
23
=
=
−j
25
25
25
or −0.56 − j0.92

= (−39 + 44) − j (52 + 33)
= 5 − j85
Problem 6.
(a)

Evaluate:

1+ j3
2
(b) j
(1 + j )4
1− j2

2

Complex numbers
(a) (1 + j )2 = (1 + j )(1 + j ) =1 + j + j + j 2

4. (a) Z 1 + Z 2 − Z 3 (b) Z 2 − Z 1 + Z 4

=1+ j + j −1= j2
(1 +

= [(1 +

j )4

j )2]2 = (

[(a) 7 − j 4 (b) −2 − j 6]

j 2)2 =

j 24 = −4

5. (a) Z 1 Z 2 (b) Z 3 Z 4
[(a) 10 + j 5 (b) 13 − j 13]

2
2
1
=
Hence
=−
4
(1 + j )
−4
2
(b)

6. (a) Z 1 Z 3 + Z 4 (b) Z 1 Z 2 Z 3
[(a) −13 − j 2 (b) −35 + j 20]

1 + j3 1 + j3 1 + j2
=
×
1 − j2 1 − j2 1 + j2
=

1 + j2+ j3 +
12 + 22

j 26

=

7. (a)

−5 + j 5
5

Z1
Z1 + Z3
(b)
Z2
Z2 − Z4
(a)

= −1 + j 1 = −1 + j
1+ j3
1− j2

2

8. (a)

= (−1 + j )2 = (−1 + j )(−1 + j )

j

1+ j3
1− j2

11
−19
43
−2
+j
(b)
+j
25
25
85
85

Z1 Z3
Z1
(b) Z 2 +
+ Z3
Z1 + Z3
Z4

= 1− j − j + j2 =− j2
Hence

(a)

2

= j (− j 2) =− j 22 =2,

9. Evaluate (a)

since j 2 = −1

41
45
9
3
+ j
(b)
− j
26
26
26
26

1− j
1
(b)
1+ j
1+ j
(a) − j (b)

Now try the following exercise
10. Show that
Exercise 86 Further problems on operations involving Cartesian complex numbers 1. Evaluate
(a)
(3 + j 2) +(5 − j ) and
(b) (−2 + j 6) −(3 − j 2) and show the results on an Argand diagram.
[(a) 8 + j (b) −5 + j 8]
2. Write down the complex conjugates of
(a) 3 + j 4, (b) 2 − j .
[(a) 3 − j 4 (b) 2 + j ]
3. If z = 2 + j and w = 3 − j evaluate
(a) z + w (b) w − z (c) 3z − 2w (d)
5z + 2w (e) j (2w − 3z) (f ) 2 j w − j z
[(a) 5 (b) 1 − j 2 (c)
(e) 5 (f ) 3 + j 4]

217

j 5 (d) 16 + j 3

In Problems 4 to 8 evaluate in a + j b form given Z 1 = 1 + j 2, Z 2 = 4 − j 3, Z 3 = −2 + j 3 and Z 4 = −5 − j .

−25
2

1
1
−j
2
2

1+ j2 2 − j5

3+ j4
−j

= 57 + j 24

20.5

Complex equations

If two complex numbers are equal, then their real parts are equal and their imaginary parts are equal. Hence if a + j b =c + j d, then a = c and b = d.
Problem 7. Solve the complex equations:
(a) 2(x + j y) =6 − j 3
(b) (1 + j 2)(−2 − j 3) =a + j b
(a)

2(x + j y) =6 − j 3 hence 2x + j 2y = 6 − j 3
Equating the real parts gives:
2x = 6, i.e. x = 3
Equating the imaginary parts gives:
2y = −3, i.e. y = − 3
2

218 Higher Engineering Mathematics
(b) (1 + j 2)(−2 − j 3) =a + j b
−2 − j 3 − j 4 − j 26 = a + j b

2.

Hence 4 − j 7 =a + j b

2+ j
= j (x + j y)
1− j

3
1
x = , y =−
2
2


3. (2 − j 3) = (a + j b)

Equating real and imaginary terms gives:

[a = −5, b = −12]

a = 4 and b = −7
4. (x − j 2y) −( y − j x) =2 + j
(a)

Solve the equations:

(2 − j 3) = (a + j b)

[x = 3, y = 1]

Problem 8.

5. If Z = R + j ωL + 1/j ωC, express Z in
(a + j b) form when R = 10, L =5, C = 0.04 and ω = 4.
[Z = 10 + j 13.75]

(b) (x − j 2y) +( y − j 3x) =2 + j 3
(a)


(2 − j 3) = (a + j b)
(2 − j 3)2 = a + j b,

Hence
i.e.

20.6 The polar form of a complex number (2 − j 3)(2 − j 3)= a + j b

Hence 4 − j 6 − j 6 + j 29 = a + j b

Thus a = −5 and b = −12

(i) Let a complex number z be x + j y as shown in the Argand diagram of Fig. 20.4. Let distance
OZ be r and the angle OZ makes with the positive real axis be θ.

(b) (x − j 2y) +( y − j 3x) =2 + j 3

From trigonometry, x = r cos θ and

−5 − j 12= a + j b

and

Hence (x + y) + j (−2y − 3x) = 2 + j 3

y = r sin θ

Equating real and imaginary parts gives: x+y=2 Hence Z = x + j y = r cos θ + j r sin θ
(1)

and −3x − 2y = 3

(2)

i.e. two simultaneous equations to solve.
Multiplying equation (1) by 2 gives:
2x + 2y = 4

= r(cos θ + j sin θ)
Z =r(cos θ + j sin θ) is usually abbreviated to
Z =r∠θ which is known as the polar form of a complex number.

(3)

Imaginary axis Adding equations (2) and (3) gives:

Z

−x = 7, i.e., x = −7 r From equation (1), y = 9, which may be checked in equation (2).

Now try the following exercise
Exercise 87 equations ␪
O

x

A Real axis

Figure 20.4

Further problems on complex

In Problems 1 to 4 solve the complex equations.
1. (2 + j )(3 − j 2) =a + j b

jy

[a = 8, b =−1]

(ii) r is called the modulus (or magnitude) of Z and is written as mod Z or |Z |. r is determined using Pythagoras’ theorem on triangle OAZ in Fig. 20.4,
i.e.

r = (x 2 + y 2 )

Complex numbers
(iii) θ is called the argument (or amplitude) of Z and is written as arg Z .

Imaginary axis (23 1j4)

By trigonometry on triangle OAZ, arg Z = θ = tan−1

j3 r (23 2 j4)

3

Real axis

r

2j4

(3 2 j4)

(b) −3 + j 4 is shown in Fig. 20.6 and lies in the second quadrant.
Modulus, r = 5 and angle α = 53.13◦, from part (a).
Argument =180◦ − 53.13◦ = 126.87◦ (i.e. the argument must be measured from the positive real axis). ␪
2

Hence −3 + j4 = 5∠126.87◦

Real axis

(c)
Figure 20.5

Hence the argument = 180◦ + 53.13◦ = 233.13◦, which is the same as −126.87◦.

Argument, arg Z = θ = tan −1

3
2
= 56.31◦ or

Hence (−3 − j4) = 5∠233.13◦ or 5∠−126.87◦

56◦19

(By convention the principal value is normally used, i.e. the numerically least value, such that
−π < θ < π).

In polar form, 2 + j 3 is written as 3.606∠56.31◦ .
Problem 10. Express the following complex numbers in polar form:

(d) 3 − j 4 is shown in Fig. 20.6 and lies in the fourth quadrant. (b) −3 + j 4

Modulus, r = 5 and angle α = 53.13◦ , as above.
Hence (3 − j4) = 5∠−53.13◦

(c) −3 − j 4 (d) 3 − j 4
3 + j 4 is shown in Fig. 20.6 and lies in the first quadrant. Modulus, r = (32 + 42 ) = 5 and argument θ = tan −1 4 = 53.13◦.
3
= 5∠53.13◦

−3 − j 4 is shown in Fig. 20.6 and lies in the third quadrant. Modulus, r = 5 and α = 53.13◦, as above.


Modulus, |Z | =r = (22 + 32) = 13 or 3.606, correct to 3 decimal places.

Hence 3 + j4

2j2

2

Figure 20.6

r

(a)


␣1

2j3

j3

(a) 3 + j 4

r

r

Problem 9. Determine the modulus and argument of the complex number Z = 2 + j 3, and express Z in polar form.

0

j2

j

23 22 21 ␣
2j

(iv) Whenever changing from cartesian form to polar form, or vice-versa, a sketch is invaluable for determining the quadrant in which the complex number occurs.

Imaginary axis (3 1j4)

j4

y x Z = 2 + j 3 lies in the first quadrant as shown in
Fig. 20.5.

219

Problem 11. Convert (a) 4∠30◦ (b) 7∠−145◦ into a + j b form, correct to 4 significant figures.
(a)

4∠30◦ is shown in Fig. 20.7(a) and lies in the first quadrant. 220 Higher Engineering Mathematics
Imaginary
axis

Problem 12.
(a)

4
308
0

(b) 3∠16◦ × 5∠−44◦ × 2∠80◦

jy
Real axis

x

(a) 8∠25◦ ×4∠60◦ = (8 × 4)∠(25◦ +60◦) = 32∠85◦
(a)

(b) 3∠16◦ × 5∠ −44◦ × 2∠80◦
= (3 × 5 × 2)∠[16◦ + (−44◦ )+ 80◦ ] = 30∠52◦

x

jy

Real axis
1458

7

Problem 13.

Figure 20.7

Using trigonometric ratios, x = 4 cos 30◦ = 3.464 and y = 4 sin 30◦ = 2.000.

(a)

Hence 4∠30◦ = 3.464 + j2.000
(b) 7∠145◦ is shown in Fig. 20.7(b) and lies in the third quadrant.


Evaluate in polar form

π π 10∠ × 12∠
16∠75◦
4
2
(b)
(a)
π
2∠15◦
6∠−
3

(b)



Angle α = 180 − 145 = 35



Hence x = 7 cos 35◦ = 5.734 and Determine, in polar form:

8∠25◦ × 4∠60◦

y = 7 sin 35◦ = 4.015

Hence 7∠−145◦ = −5.734 − j4.015

(b)

16∠75◦ 16
= ∠(75◦ − 15◦) = 8∠60◦
2∠15◦
2 π π
× 12∠
4
2 = 10 × 12 ∠ π + π − − π π 6
4 2
3
6∠−
3
13π
11π
= 20∠ or 20∠− or 12
12

10∠

20∠195◦ or 20∠−165◦

Alternatively
7∠−145◦ = 7 cos(−145◦) + j 7 sin(−145◦)
= −5.734 − j4.015

Calculator
Using the ‘Pol’ and ‘Rec’ functions on a calculator enables changing from Cartesian to polar and vice-versa to be achieved more quickly.
Since complex numbers are used with vectors and with electrical engineering a.c. theory, it is essential that the calculator can be used quickly and accurately.

20.7 Multiplication and division in polar form
If Z 1 =r1 ∠θ1 and Z 2 =r2 ∠θ2 then:
(i) Z1 Z2 = r1 r2 ∠(θ1 + θ2 ) and
(ii)

Z1 r1
= ∠(θ1 − θ2 )
Z2 r2

Problem 14. Evaluate, in polar form
2∠30◦ +5∠−45◦ − 4∠120◦.
Addition and subtraction in polar form is not possible directly. Each complex number has to be converted into cartesian form first.
2∠30◦ = 2(cos 30◦ + j sin 30◦ )
= 2 cos 30◦ + j 2 sin30◦ = 1.732 + j 1.000
5∠−45◦ = 5(cos(−45◦) + j sin(−45◦))
= 5 cos(−45◦) + j 5 sin(−45◦)
= 3.536 − j 3.536
4∠120◦ = 4( cos 120◦ + j sin 120◦ )
= 4 cos 120◦ + j 4 sin 120◦
= −2.000 + j 3.464
Hence 2∠30◦ + 5∠−45◦ − 4∠120◦

Complex numbers
= (1.732 + j 1.000) +(3.536 − j 3.536)

6. (a) 3∠20◦ × 15∠45◦
(b) 2.4∠65◦ × 4.4∠−21◦
[(a) 45∠65◦ (b) 10.56∠44◦]

− (−2.000 + j 3.464)
= 7.268 − j 6.000, which lies in the fourth quadrant
= [(7.268)2 + (6.000)2 ]∠ tan−1

7. (a) 6.4∠27◦ ÷ 2∠−15◦

−6.000
7.268

(b) 5∠30◦ × 4∠80◦ ÷ 10∠−40◦
[(a) 3.2∠42◦ (b) 2∠150◦] π π
8. (a) 4∠ + 3∠
6
8
(b) 2∠120◦ + 5.2∠58◦ − 1.6∠−40◦
[(a) 6.986∠26.79◦ (b) 7.190∠85.77◦]

= 9.425∠−39.54◦

Now try the following exercise
Exercise 88 form 221

Further problems on polar

1. Determine the modulus and argument of
(a) 2 + j 4 (b) −5 − j 2 (c) j (2 − j ).


(a) 4.472, 63.43◦


⎣(b)5.385, −158.20◦⎦
(c) 2.236, 63.43◦
In Problems 2 and 3 express the given Cartesian complex numbers in polar form, leaving answers in surd form.
2. (a) 2 + j 3 (b) −4 (c) −6 + j

(a) 13∠56.31◦ (b)4∠180◦

(c) 37∠170.54◦
3. (a) − j 3 (b) (−2 + j )3 (c) j 3(1 − j )

(a) 3∠−90◦ (b) 125∠100.30◦

(c) 2∠−135◦
In Problems 4 and 5 convert the given polar complex numbers into (a + j b) form giving answers correct to 4 significant figures.
4. (a) 5∠30◦ (b) 3∠60◦ (c) 7∠45◦


(a) 4.330 + j 2.500


⎣(b)1.500 + j 2.598⎦
(c) 4.950 + j 4.950
5. (a) 6∠125◦ (b) 4∠π (c) 3.5∠−120◦


(a) −3.441 + j 4.915


⎣(b) −4.000 + j 0


20.8 Applications of complex numbers There are several applications of complex numbers in science and engineering, in particular in electrical alternating current theory and in mechanical vector analysis. The effect of multiplying a phasor by j is to rotate it in a positive direction (i.e. anticlockwise) on an
Argand diagram through 90◦ without altering its length.
Similarly, multiplying a phasor by − j rotates the phasor through −90◦ . These facts are used in a.c. theory since certain quantities in the phasor diagrams lie at 90◦ to each other. For example, in the R−L series circuit shown in Fig. 20.8(a), V L leads I by
90◦ (i.e. I lags V L by 90◦ ) and may be written as j V L , the vertical axis being regarded as the imaginary axis of an Argand diagram. Thus V R + j V L = V and since V R = IR, V = I X L (where X L is the inductive reactance, 2π f L ohms) and V = IZ (where Z is the impedance) then R + j X L = Z .

I

VR

VL

I

V
Phasor diagram
VL

VR I
(a)

In Problems 6 to 8, evaluate in polar form.
Figure 20.8

VR

VC

V
Phasor diagram
VR

V

(c) −1.750 − j 3.031

C

R

L

R



VC
V
(b)

I

222 Higher Engineering Mathematics
Similarly, for the R−C circuit shown in Fig. 20.8(b),
VC lags I by 90◦ (i.e. I leads VC by 90◦) and
V R − j VC = V , from which R − j X C = Z (where X C
1
is the capacitive reactance ohms). 2π fC
Problem 15. Determine the resistance and series inductance (or capacitance) for each of the following impedances, assuming a frequency of
50 Hz:
(a) (4.0 + j 7.0)

Problem 16. An alternating voltage of 240 V,
50 Hz is connected across an impedance of
(60 − j 100) . Determine (a) the resistance (b) the capacitance (c) the magnitude of the impedance and its phase angle and (d) the current flowing.
(a)

Impedance Z = (60 − j 100) .
Hence resistance = 60

(b) Capacitive reactance X C = 100
1
XC = then 2πf C

(b) − j 20

(c) 15∠−60◦
(a) Impedance, Z = (4.0 + j 7.0) hence, resistance = 4.0 and reactance = 7.00 .
Since the imaginary part is positive, the reactance is inductive,

capacitance, C =
=

i.e. X L = 7.0

7.0
XL
=
= 0.0223 H or 22.3 mH
2π f
2π(50)

(b) Impedance, Z = j 20, i.e. Z = (0 − j 20) hence resistance = 0 and reactance = 20 . Since the imaginary part is negative, the reactance is cap1 acitive, i.e., X C = 20 and since X C =
2πf C then: 1
1
= capacitance, C =
F
2πf XC
2π(50)(20)
=

106 μF = 159.2 μF
2π(50)(20)

(c) Impedance, Z
= 15∠−60◦ = 15[ cos (−60◦ ) + j sin (−60◦ )]

(c)

106 μF 2π(50)(100)

Magnitude of impedance,
|Z | =

Hence resistance = 7.50 tance, X C = 12.99

(d) Current flowing, I =

V
240∠0◦
=
Z 116.6∠−59.04◦

Problem 17. For the parallel circuit shown in
Fig. 20.9, determine the value of current I and its phase relative to the 240 V supply, using complex numbers. and capacitive reac-

1
=
μF
C=
2πf XC
2π(50)(12.99)

= −59.04◦

The circuit and phasor diagrams are as shown in
Fig. 20.8(b).

R1 5 4 V

106

−100
60

= 2.058 ∠59.04◦ A

XL 5 3 V

R2 5 10 V

1 then capacitance,
2πf C

= 245 μF

[(60)2 + (−100)2 ] = 116.6

Phase angle, arg Z = tan −1

= 7.50 − j 12.99

Since X C =

1
1
=
2π f X C
2π(50)(100)

= 31.83 μF

Since X L = 2πf L then inductance,
L=

and since

R3 5 12 V

I

XC 5 5 V

240 V, 50 Hz

Figure 20.9

Complex numbers
V
Current I = . Impedance Z for the three-branch
Z
parallel circuit is given by:

10 N

8N
210Њ
120Њ

1
1
1
1
+
+
,
=
Z
Z1 Z2 Z3

45Њ

where Z 1 = 4 + j 3, Z 2 = 10 and Z 3 = 12 − j 5
1
1
=
Z1
4+ j3
1
4 − j3
4− j3
=
×
=
4 + j 3 4 − j 3 42 + 32

Admittance, Y1 =

= 0.160 − j 0.120 siemens

15 N

Figure 20.10

The resultant force

Admittance, Y2 =

1
1
=
= 0.10 siemens
Z2
10

= f A + f B + fC

Admittance, Y3 =

1
1
=
Z3
12 − j 5

= 10(cos 45◦ + j sin 45◦) + 8(cos 120◦

1
12 + j 5
12 + j 5
=
×
=
12 − j 5 12 + j 5 122 + 52

= 10∠45◦ + 8∠120◦ + 15∠210◦
+ j sin 120◦) + 15(cos 210◦ + j sin 210◦ )
= (7.071 + j 7.071) + (−4.00 + j 6.928)

= 0.0710 + j 0.0296 siemens
Total admittance, Y = Y1 + Y2 + Y3
= (0.160 − j 0.120) + (0.10)
+ (0.0710 + j 0.0296)
= 0.331 − j 0.0904
= 0.343∠−15.28◦ siemens
Current I =

V
= VY
Z

+ (−12.99 − j 7.50)
= −9.919 + j 6.499
Magnitude of resultant force
=

[(−9.919)2 + (6.499)2 ] = 11.86 N

Direction of resultant force
= tan −1

6.499
−9.919

= 146.77◦

(since −9.919 + j 6.499 lies in the second quadrant).

= (240∠0◦ )(0.343∠−15.28◦ )
= 82.32 ∠−15.28◦ A
Problem 18. Determine the magnitude and direction of the resultant of the three coplanar forces given below, when they act at a point.
Force A, 10 N acting at 45◦ from the positive horizontal axis.
Force B, 87 N acting at 120◦ from the positive horizontal axis.
Force C, 15 N acting at 210◦ from the positive horizontal axis.
The space diagram is shown in Fig. 20.10. The forces may be written as complex numbers.
Thus force A, f A = 10∠45◦, force B, f B = 8∠120◦ and force C, fC = 15∠210◦.

Now try the following exercise
Exercise 89 Further problems on applications of complex numbers
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 + j 8)
(c) j 14



(b) (2 − j 3)
(d) 8∠−60◦


(a) R = 3 , L = 25.5 mH
⎢ (b) R = 2 , C = 1061 μF ⎥


⎣ (c) R = 0, L = 44.56 mH ⎦
(d) R = 4 , C = 459.4 μF

223

224 Higher Engineering Mathematics
2. Two impedances, Z 1 = (3 + j 6) and Z 2 = (4 − j 3) 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.
[15.76 A, 23.20◦ lagging]
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. [27.25 A, 3.37◦ lagging]
4. A series circuit consists of a 12 resistor, a coil of inductance 0.10 H and a capacitance of
160 μF. Calculate the current flowing and its phase relative to the supply voltage of 240 V,
50 Hz. Determine also the power factor of the circuit. [14.42 A, 43.85◦ lagging, 0.721]
5. For the circuit shown in Fig. 20.11, determine the current I flowing and its phase relative to the applied voltage. [14.6 A, 2.51◦ leading]
6. Determine, using complex numbers, the magnitude and direction of the resultant of the coplanar forces given below, which are acting at a point. Force A, 5 N acting horizontally,
Force B, 9 N acting at an angle of 135◦ to force
A, Force C, 12 N acting at an angle of 240◦ to force A.
[8.394 N, 208.68◦ from force A]
XC 5 20 V

R2 5 40 V

R1 5 30 V

XL 5 50 V

R3 5 25 V

I
V 5 200 V

Figure 20.11

7. A delta-connected impedance Z A is given by: Z1 Z2 + Z2 Z3 + Z3 Z1
ZA =
Z2
Determine Z A in both Cartesian and polar form given Z 1 = (10 + j 0) ,
Z 2 = (0 − j 10) and Z 3 = (10 + j 10) .
[(10 + j 20) , 22.36∠63.43◦ ]
8. In the hydrogen atom, the angular momentum, p, of the de Broglie wave is given jh (±jmψ). Determine an by: pψ = −

mh expression for p.
±

9. An aircraft P flying at a constant height has a velocity of (400 + j 300) km/h. Another aircraft Q at the same height has a velocity of
(200 − j 600) km/h. Determine (a) the velocity of P relative to Q, and (b) the velocity of
Q relative to P. Express the answers in polar form, correct to the nearest km/h.
(a) 922 km/h at 77.47◦
(b) 922 km/h at −102.53◦
10. Three vectors are represented by P, 2∠30◦ ,
Q, 3∠90◦ and R, 4∠−60◦ . Determine in polar form the vectors represented by (a)
P + Q + R, (b) P − Q − R.
(a) 3.770∠8.17◦
(b) 1.488∠100.37◦
11. In a Schering bridge circuit,
Z X = (R X − j X C X ), Z 2 = − j X C2 ,
(R3 )(− j X C3 ) and Z 4 = R4
Z3 =
(R3 − j X C3 )
1
where X C =
2πf C
At balance: (Z X )(Z 3 ) = (Z 2 )(Z 4 ).
C3 R4
Show that at balance R X =
C2
C2 R3
CX =
R4

and

Chapter 21

De Moivre’s theorem
21.1

= 2197∠382.14◦(since 742.14

Introduction

≡ 742.14◦ − 360◦ = 382.14◦)
= 2197∠22.14◦ (since 382.14◦

From multiplication of complex numbers in polar form,

≡ 382.14◦ − 360◦ = 22.14◦)

(r∠θ) × (r ∠θ) = r 2 ∠2θ

or 2197∠22◦8
Similarly, (r∠θ)× (r∠θ)× (r∠θ) = r 3∠3θ, and so on.
In general, De Moivre’s theorem states:
[r∠θ]

n

= r n∠nθ

Problem 2. Determine the value of (−7 + j 5)4, expressing the result in polar and rectangular forms.

The theorem is true for all positive, negative and fractional values of n. The theorem is used to determine powers and roots of complex numbers.

21.2

Powers of complex numbers
◦ 4



For example [3∠20 ] = 3 ∠(4 × 20 ) = 81∠80
De Moivre’s theorem.
4



by

Problem 1. Determine, in polar form
(a) [2∠35◦ ]5 (b) (−2 + j 3)6.
(a)

[2∠35◦]5 = 25 ∠(5 × 35◦), from De Moivre’s theorem

5
[(−7)2 + 52 ]∠ tan−1
−7

= 74∠144.46◦

(−7 + j 5) =

(Note, by considering the Argand diagram, −7 + j 5 must represent an angle in the second quadrant and not in the fourth quadrant.)
Applying De Moivre’s theorem:

(−7 + j 5)4 = [ 74∠144.46◦]4

= 744 ∠4 ×144.46◦
= 5476∠577.84◦
= 5476∠217.84◦

= 32∠175◦
(b)

3
(−2 + j 3)= [(−2)2 + (3)2 ]∠ tan−1
−2

= 13∠123.69◦ , since −2 + j 3 lies in the second quadrant

(−2 + j 3)6 = [ 13∠123.69◦]6

= ( 13)6 ∠(6 × 123.69◦), by De Moivre’s theorem
= 2197∠742.14◦

or 5476∠217◦50 in polar form
Since r∠θ = r cos θ + j r sin θ,
5476∠217.84◦ = 5476 cos217.84◦
+ j 5476 sin217.84◦
= −4325 − j 3359
i.e.

(−7 + j5)4 = −4325 −j3359 in rectangular form

226 Higher Engineering Mathematics
Now try the following exercise
Exercise 90 Further problems on powers of complex numbers

13∠427.38◦. When the angle is divided by 2 an angle less than 360◦ is obtained.
Hence
(5 + j 12) =

1. Determine in polar form (a) [1.5∠15◦]5
(b) (1 + j 2)6.
[(a) 7.594∠75◦ (b) 125∠20.61◦]

1

5. (3 − j 8)5

[45530∠12.78◦, 44400 + j 10070]

6. (−2 + j 7)4

[2809∠63.78◦, 1241 + j 2520]

7. (−16 − j 9)6

1

21.3

=

Roots of complex numbers

= 3.0 + j 2.0


13∠213.69◦ = 13(cos 213.69◦ + j sin 213.69◦)
= −3.0 − j 2.0
Thus, in cartesian form the two roots are
±(3.0 + j2.0).
From the Argand diagram shown in Fig. 21.1 the two roots are seen to be 180◦ apart, which is always true when finding square roots of complex numbers.
Imaginary axis

1
1 1
√ θ
[r∠θ] = [r∠θ] 2 = r 2 ∠ θ = r ∠
2
2

There are two square roots of a real number, equal in size but opposite in sign.



13∠33.69◦ and 13∠213.69◦

Thus, in polar form, the two roots are
3.61∠33.69◦ and 3.61∠−146.31◦.


13∠33.69◦ = 13(cos 33.69◦ + j sin 33.69◦ )

The square root of a complex number is determined by letting n =1/2 in De Moivre’s theorem,
i.e.

1
× 427.38◦
2

= 3.61∠33.69◦ and 3.61∠213.69◦

(38.27 × 106)∠176.15◦ ,
106(−38.18 + j 2.570)

1
× 67.38◦ and
2

13 2 ∠

3. Convert (3 − j ) into polar form and hence evaluate (3 − j√ , giving the answer in polar
)7
form.
[ 10∠−18.43◦ , 3162∠−129◦ ]

[476.4∠119.42◦, −234 + j 415]

1

1

= 13 2 ∠

(a) 81∠164◦, −77.86 + j 22.33
(b) 55.90∠−47.18◦ , 38 − j 41

4. (6 + j 5)3

[13∠427.38◦]

= [13∠67.38◦] 2 and [13∠427.38◦] 2

2. Determine in polar and cartesian forms
(a) [3∠41◦]4 (b) (−2 − j )5.

In problems 4 to 7, express in both polar and rectangular forms.

[13∠67.38◦] and

j2
3.61
213.698

33. 698

23

3

Real axis

3.61

Problem 3. Determine the two square roots of the complex number (5 + j 12) in polar and cartesian forms and show the roots on an Argand diagram.

2j 2

Figure 21.1

(5 + j 12) =

[52 + 122 ]∠ tan−1

12
5

= 13∠67.38◦
When determining square roots two solutions result.
To obtain the second solution one way is to express 13∠67.38◦ also as 13∠(67.38◦ + 360◦ ), i.e.

In general, when finding the nth root of a complex number, there are n solutions. For example, there are three solutions to a cube root, five solutions to a fifth root, and so on. In the solutions to the roots of a complex number, the modulus, r, is always the same, but the

De Moivre’s theorem arguments, θ, are different. It is shown in Problem 3 that arguments are symmetrically spaced on an Argand diagram and are (360/n)◦ apart, where n is the number of the roots required. Thus if one of the solutions to the cube root of a complex number is, say, 5∠20◦, the other two roots are symmetrically spaced (360/3)◦ , i.e. 120◦ from this root and the three roots are 5∠20◦, 5∠140◦ and 5∠260◦ .
1

Problem 4. Find the roots of [(5 + j 3)] 2 in rectangular form, correct to 4 significant figures.

(−14 + j 3) =
(−14 + j 3)

−2
5


205∠167.905◦

=

205

1

(5 + j 3) 2 =

1

34 2 ∠ 1 × 30.96◦
2

= 2.415∠15.48◦or 2.415∠15◦ 29



2
× 167.905◦
5

or 0.3449∠−67◦ 10
There are five roots to this complex number,
−2
5


34∠30.96◦

Applying De Moivre’s theorem:

−2
5 ∠

= 0.3449∠−67.164◦

x
(5 + j 3) =

=

1
2
x5

1
=√
5 2 x The roots are symmetrically displaced from one another (360/5)◦ , i.e. 72◦ apart round an Argand diagram. Thus the required roots are 0.3449∠−67◦ 10 ,
0.3449∠4◦ 50 , 0.3449∠76◦ 50 , 0.3449∠148◦ 50 and
0.3449∠220◦50 .
Now try the following exercise

The second root may be obtained as shown above, i.e. having the same modulus but displaced (360/2)◦ from the first root.
1

Thus, (5 + j 3) 2 = 2.415∠(15.48◦ + 180◦ )
= 2.415∠195.48◦

Exercise 91 Further problems on the roots of complex numbers
In Problems 1 to 3 determine the two square roots of the given complex numbers in Cartesian form and show the results on an Argand diagram.
1. (a) 1 + j (b) j

In rectangular form:

(a) ±(1.099 + j 0.455)
(b) ±(0.707 + j 0.707)

2.415∠15.48◦ = 2.415 cos 15.48◦
+ j 2.415 sin15.48◦
= 2.327 + j0.6446 and 2. (a) 3 − j 4 (b) −1 − j 2
(a) ±(2 − j )
(b) ±(0.786 − j 1.272)

2.415∠195.48◦ = 2.415 cos 195.48◦
+ j 2.415 sin195.48◦

3. (a) 7∠60◦ (b) 12∠

= −2.327 − j0.6446
[(5 + j 3)] 2 = 2.415∠15.48◦and
2.415∠195.48◦or
± (2.327 + j0.6446).
Problem 5. Express the roots of
(−14 + j 3)

−2
5

in polar form.


2
(a) ±(2.291 + j 1.323)
(b) ±(−2.449 + j 2.449)

1

Hence

227

In Problems 4 to 7, determine the moduli and arguments of the complex roots.
1

4. (3 + j 4) 3
Moduli 1.710, arguments 17.71◦ ,
137.71◦ and 257.71◦

228 Higher Engineering Mathematics
1

5. (−2 + j ) 4




Modulus 1.223, arguments
⎣ 38.36◦, 128.36◦,

218.36◦ and 308.36◦


By definition, j = (−1), hence j 2 = −1, j 3 = − j , j 4 = 1, j 5 = j , and so on. θ2 θ3 θ4 θ5 Thus e j θ = 1 + j θ − − j + + j − · · ·
2!
3! 4!
5!
Grouping real and imaginary terms gives:

1

e jθ = 1 −

6. (−6 − j 5) 2
Modulus 2.795, arguments
109.90◦, 289.90◦

+ j θ−

−2

7. (4 − j 3) 3
Modulus 0.3420, arguments 24.58◦,
144.58◦ and 264.58◦

Z0 = γ= θ3 θ5
+
−···
3! 5!

However, from equations (2) and (3):
1−

and

and

θ2 θ4
+
− · · · = cos θ
2! 4!

θ−

8. For a transmission line, the characteristic impedance Z 0 and the propagation coefficient γ are given by:
R + j ωL
G + j ωC

θ2 θ4
+
−···
2! 4!

θ3 θ5
+
− · · · = sin θ
3! 5!

[(R + j ωL)(G + j ωC)]

Given R = 25 , L =5 × 10−3 H,
G = 80 × 10−6 siemens, C = 0.04 × 10−6 F and ω = 2000 π rad/s, determine, in polar
Z 0 = 390.2∠ −10.43◦ , form, Z 0 and γ . γ = 0.1029∠61.92◦

e jθ = cos θ + j sin θ

Thus

(4)

Writing −θ for θ in equation (4), gives: e j (−θ) = cos(−θ) + j sin(−θ)
However, cos(−θ) = cos θ and sin(−θ) = −sin θ

21.4 The exponential form of a complex number

Thus

Certain mathematical functions may be expressed as power series (for example, by Maclaurin’s series—see
Chapter 8), three examples being:
(i) ex = 1 + x +

x2
2!

+

x3
3!

+

x4
4!

+

x5
5!

x3 x5 x7
+

+···
3!
5! 7! x2 x4 x6
+

+···
(iii) cos x = 1 −
2! 4!
6!
(ii) sin x = x −

+···

(1)
(2)
(3)

Replacing x in equation (1) by the imaginary number j θ gives:
( j θ)2 ( j θ)3 ( j θ)4 ( j θ)5
+
+
+
+· · · e j θ = 1+ j θ +
2!
3!
4!
5! j 2θ 2 j 3θ 3 j 4θ 4 j 5θ 5
= 1 + jθ +
+
+
+
+···
2!
3!
4!
5!

e −jθ = cos θ − j sin θ

(5)

The polar form of a complex number z is: z =r(cos θ + j sin θ). But, from equation (4), cos θ + j sin θ = e jθ .
Therefore

z = re jθ

When a complex number is written in this way, it is said to be expressed in exponential form.
There are therefore three ways of expressing a complex number:
1.

z =(a + j b), called Cartesian or rectangular form,

2.

z =r(cos θ + j sin θ) or r∠θ, called polar form, and

3.

z =re j θ called exponential form.

The exponential form is obtained from the polar form. π For example, 4∠30◦ becomes 4e j 6 in exponential form. (Note that in re j θ , θ must be in radians.)

De Moivre’s theorem
Problem 6. Change (3 − j 4) into (a) polar form,
(b) exponential form.
(a)

(3 − j 4) = 5∠−53.13◦ or 5∠−0.927 in polar form

(b) (3 − j 4) = 5∠−0.927 = 5e−j0.927 in exponential form
Problem 7. Convert 7.2e j 1.5 into rectangular form. (a)

Thus if z =4e j 1.3 then ln z = ln(4e j1.3 )
= ln 4 + j1.3
(or 1.386 + j1.300) in Cartesian form.

(b) (1.386 + j 1.300) =1.90∠43.17◦ or 1.90∠0.753 in polar form.
Problem 11. Given z = 3e1− j , find ln z in polar form. If

z = 3e1− j , then

ln
7.2e j 1.5 = 7.2∠1.5 rad(= 7.2∠85.94◦) in polar form

z = ln(3e1− j )
= ln 3 + ln e1− j

= 7.2 cos 1.5 + j 7.2 sin1.5

= ln 3 + 1 − j

= (0.509 + j 7.182) in rectangular form

= (1 + ln 3) − j

Problem 8. Express form. π

z = (2e1 ) e j 3

π z = 2e1+ j 3

= 2.0986 − j 1.0000 in Cartesian

= 2.325∠−25.48◦ or 2.325∠−0.445
Problem 12. Determine, in polar form, ln (3 + j 4).

by the laws of indices

π
(or 2e∠60◦ )in polar form
3
π π = 2e cos + j sin
3
3
= (2e1 )∠

= (2.718 + j4.708) in Cartesian form

ln(3 + j 4) = ln[5∠0.927] = ln[5e j 0.927]
= ln 5 + ln(e j 0.927 )
= ln 5 + j 0.927
= 1.609 + j 0.927
= 1.857∠29.95◦ or 1.857∠0.523

Problem 9. Change 6e2− j 3 into (a + j b) form.
6e2− j 3 = (6e2 )(e− j 3 ) by the laws of indices

Now try the following exercise

= 6e2 ∠−3 rad (or 6e2 ∠−171.890 ) in polar form

Exercise 92 Further problems on the exponential form of complex numbers

= 6e2 [cos (−3) + j sin (−3)]

1. Change (5 + j 3) into exponential form.
[5.83e j 0.54]

= (−43.89 − j6.26) in (a + jb) form
Problem 10. If z = 4e j 1.3 , determine ln z (a) in
Cartesian form, and (b) in polar form.
If
i.e.

229

z = re j θ then ln z = ln(re j θ )
= lnr + ln e j θ ln z = lnr + j θ,

by the laws of logarithms

2. Convert (−2.5 + j 4.2) into exponential form.
[4.89e j 2.11]
3. Change 3.6e j 2 into cartesian form.
[−1.50 + j 3.27] π 4. Express 2e3+ j 6 in (a + j b) form.
[34.79 + j 20.09]
5. Convert 1.7e1.2− j 2.5 into rectangular form.
[−4.52 − j 3.38]

230 Higher Engineering Mathematics
6. If z = 7e j 2.1 , determine ln z (a) in Cartesian form, and (b) in polar form.


(a) ln 7 + j 2.1
⎣(b) 2.86∠47.18◦or⎦
2.86∠0.82
7. Given z =4e1.5− j 2 , determine ln z in polar form. [3.51∠−34.72◦ or 3.51∠−0.61]
8. Determine in polar form (a) ln (2 + j 5)
(b) ln (−4 − j 3)


(a) 2.06∠35.26◦or
⎢ 2.06∠0.615 ⎥


⎣(b) 4.11∠66.96◦or⎦
4.11∠1.17

9. When displaced electrons oscillate about an equilibrium position the displacement x is given by the equation:

x = Ae

ht
− 2m + j

(4m f −h 2 ) t 2m−a

Determine the real part of x in terms of t , assuming (4m f − h 2 ) is positive.

ht
(4m f − h 2 ) t Ae− 2m cos
2m −a

Chapter 22

The theory of matrices and determinants 22.1

Matrix notation

Matrices and determinants are mainly used for the solution of linear simultaneous equations. The theory of matrices and determinants is dealt with in this chapter and this theory is then used in Chapter 23 to solve simultaneous equations.
The coefficients of the variables for linear simultaneous equations may be shown in matrix form.
The coefficients of x and y in the simultaneous equations x + 2y = 3

of the matrix. The number of rows in a matrix is usually specified by m and the number of columns by n and a matrix referred to as an ‘m by n’ matrix. Thus,
2 3 6 is a ‘2 by 3’ matrix. Matrices cannot be
4 5 7 expressed as a single numerical value, but they can often be simplified or combined, and unknown element values can be determined by comparison methods. Just as there are rules for addition, subtraction, multiplication and division of numbers in arithmetic, rules for these operations can be applied to matrices and the rules of matrices are such that they obey most of those governing the algebra of numbers.

4x − 5y = 6 become 1
2
in matrix notation.
4 −5

Similarly, the coefficients of p, q and r in the equations
1.3 p − 2.0q + r = 7
3.7 p + 4.8q − 7r = 3
4.1 p + 3.8q + 12r = −6

1.3 −2.0 become ⎝3.7
4.8
4.1
3.8


1
−7⎠ in matrix form.
12

The numbers within a matrix are called an array and the coefficients forming the array are called the elements

22.2 Addition, subtraction and multiplication of matrices
(i) Addition of matrices
Corresponding elements in two matrices may be added to form a single matrix.
Problem 1. Add the matrices
2 −1
−3
0 and and
−7
4
7 −4




3 1 −4
2 7 −5
(b) ⎝4 3
1⎠ and ⎝−2 1
0⎠
1 4 −3
6 3
4
(a)

232 Higher Engineering Mathematics
(a)


3−2
= ⎝4 − (−2)
1−6

1 −6
=⎝ 6
2
−5
1

Adding the corresponding elements gives:
2 −1
−3
0
+
−7
4
7 −4
=

2 + (−3) −1 + 0
−7 + 7
4 + (−4)

=

−1 −1
0
0

(b) Adding the corresponding elements gives:

⎞ ⎛

3 1 −4
2 7 −5
⎝4 3
1⎠ + ⎝−2 1
0⎠
1 4 −3
6 3
4


3+2
1 + 7 −4 + (−5)

= ⎝4 + (−2) 3 + 1
1+0
1+6
4 + 3 −3 + 4


5 8 −9
1⎠
= ⎝2 4
7 7
1

Problem 3. If
−3
0
2 −1
A=
,B= and 7 −4
−7
4
1
0
C=
find A + B − C.
−2 −4
A+ B =

Problem 2.
−3
7

2
(b) ⎝−2
6
(a)

Subtract

0
2 −1 from and
−4
−7
4



7 −5
3 1 −4
1
0⎠ from ⎝4 3
1⎠
3
4
1 4 −3

To find matrix A minus matrix B, the elements of B are taken from the corresponding elements of A. Thus:
(a)

2 −1
−3
0

−7
4
7 −4
=

2 − (−3) −1 − 0
−7 − 7
4 − (−4)

=

5 −1
−14
8

⎞ ⎛


2 7 −5
3 1 −4
(b) ⎝
1⎠ − ⎝−2 1
0⎠
4 3
1 4 −3
6 3
4

−1 −1
0
0
(from Problem 1)

Hence, A + B − C =

−1 −1
1
0

0
0
−2 −4

=

−1 − 1
−1 − 0
0 − (−2)
0 − (−4)

=

(ii) Subtraction of matrices
If A is a matrix and B is another matrix, then (A − B) is a single matrix formed by subtracting the elements of
B from the corresponding elements of A.


1 − 7 −4 − (−5)

3−1
1−0
4 − 3 −3 − 4

1
1⎠
−7

−2 −1
2
4

Alternatively A + B − C
=

−3
0
2 −1
1
0
+

7 −4
−7
4
−2 −4

=

−3 + 2 − 1
0 + (−1) − 0
7 + (−7) − (−2) −4 + 4 − (−4)

=

−2 −1 as obtained previously
2
4

(iii) Multiplication
When a matrix is multiplied by a number, called scalar multiplication, a single matrix results in which each element of the original matrix has been multiplied by the number.
−3
0
,
7 −4


1
0
2 −1
⎠ find
B=
and C = ⎝
−7
4
−2 −4
2 A − 3B + 4C.

Problem 4.

If A =

The theory of matrices and determinants
For scalar multiplication, each element is multiplied by the scalar quantity, hence
2A = 2

−3
0
−6
0
=
7 −4
14 −8

2 −1
6 −3
3B = 3
=
−7
4
−21 12 and 4C = 4

1
0
4
0
=
−2 −4
−8 −16

=

−6 0
6 −3
4
0

+
14 −8
−21 12
−8 −16

=

−6 − 6 + 4
0 − (−3) + 0
14 − (−21) + (−8) −8 − 12 + (−16)

=

−8
3
27 −36

When a matrix A is multiplied by another matrix B, a single matrix results in which elements are obtained from the sum of the products of the corresponding rows of A and the corresponding columns of B.
Two matrices A and B may be multiplied together, provided the number of elements in the rows of matrix
A are equal to the number of elements in the columns of matrix B. In general terms, when multiplying a matrix of dimensions (m by n) by a matrix of dimensions (n by
r), the resulting matrix has dimensions (m by r). Thus a 2 by 3 matrix multiplied by a 3 by 1 matrix gives a matrix of dimensions 2 by 1.

find A × B.

C21 is the sum of the products of the second row elements of A and the first column elements of B, taken one at a time,
i.e. C21 = (1 × (−5)) + (−4 × (−3)) = 7
Finally, C22 is the sum of the products of the second row elements of A and the second column elements of
B, taken one at a time,
i.e. C22 = (1 × 7) + ((−4) × 4) = −9

Hence 2 A − 3B + 4C

Problem 5. If A =

233

2 3
−5 7 and B =
1 −4
−3 4

Let A × B = C where C =

C11 C12
C21 C22

C11 is the sum of the products of the first row elements of A and the first column elements of B taken one at a time, i.e. C11 = (2 × (−5)) + (3 × (−3)) = −19
C12 is the sum of the products of the first row elements of A and the second column elements of B, taken one at a time,
i.e. C12 = (2 × 7) + (3 × 4) = 26

Thus, A × B =

−19 26
7 −9

Problem 6. Simplify

⎞ ⎛ ⎞
3
4
0
2
⎝−2
6 −3⎠ × ⎝ 5⎠
7 −4
1
−1
The sum of the products of the elements of each row of the first matrix and the elements of the second matrix,
(called a column matrix), are taken one at a time. Thus:


⎞ ⎛ ⎞
3
4
0
2
⎝−2
6 −3⎠ × ⎝ 5⎠
7 −4
1
−1


(3 × 2) + (4 × 5) + (0 × (−1))
= ⎝(−2 × 2) + (6 × 5) + (−3 × (−1))⎠
(7 × 2) + (−4 × 5) + (1 × (−1))
⎛ ⎞
26
= ⎝ 29⎠
−7



3
4
0
Problem 7. If A = ⎝−2
6 −3⎠ and
7 −4
1


2 −5
B = ⎝ 5 −6⎠, find A × B.
−1 −7
The sum of the products of the elements of each row of the first matrix and the elements of each column of the second matrix are taken one at a time. Thus:


⎞ ⎛

3
4
0
2 −5
⎝−2
6 −3⎠ × ⎝ 5 −6⎠
7 −4
1
−1 −7

234 Higher Engineering Mathematics



[(3 × 2)
[(3 × (−5))
⎜ + (4 × 5)
+(4 × (−6)) ⎟


⎜ + (0 × (−1))]
+(0 × (−7))] ⎟


⎜[(−2 × 2)

[(−2 × (−5))


= ⎜ + (6 × 5)
+(6 × (−6)) ⎟


⎜ + (−3 × (−1))]
+(−3 × (−7))]⎟


⎜[(7 × 2)

[(7 × (−5))


⎝ + (−4 × 5)
+(−4 × (−6)) ⎠
+ (1 × (−1))]
+(1 × (−7))]


26 −39
= ⎝ 29 −5⎠
−7 −18

A× B =

2 3
2 3
×
1 0
0 1

=

[(2 × 2) + (3 × 0)] [(2 × 3) + (3 × 1)]
[(1 × 2) + (0 × 0)] [(1 × 3) + (0 × 1)]

=

4 9
2 3

B×A=

2 3
2 3
×
0 1
1 0

[(2 × 2) + (3 × 1)] [(2 × 3) + (3 × 0)]
[(0 × 2) + (1 × 1)] [(0 × 3) + (1 × 0)]

=
Problem 8. Determine

⎞ ⎛

1 0 3
2 2 0
⎝2 1 2⎠ × ⎝1 3 2⎠
1 3 1
3 2 0

=

7 6
1 0

Since

4 9
7 6
=
, then A × B = B × A
2 3
1 0

Now try the following exercise
The sum of the products of the elements of each row of the first matrix and the elements of each column of the second matrix are taken one at a time. Thus:

⎞ ⎛

1 0 3
2 2 0
⎝2 1 2⎠ × ⎝1 3 2⎠
1 3 1
3 2 0


[(1 × 2)
[(1 × 2)
[(1 × 0)
⎜ + (0 × 1)
+ (0 × 3)
+ (0 × 2) ⎟


⎜ + (3 × 3)] + (3 × 2)] + (3 × 0)]⎟


⎜[(2 × 2)

[(2 × 2)
[(2 × 0)


+ (1 × 3)
+ (1 × 2) ⎟
= ⎜ + (1 × 1)


⎜ + (2 × 3)] + (2 × 2)] + (2 × 0)]⎟


⎜[(1 × 2)

[(1 × 2)
[(1 × 0)


⎝ + (3 × 1)
+ (3 × 3)
+ (3 × 2) ⎠
+ (1 × 3)]

11 8 0
= ⎝ 11 11 2⎠
8 13 6

+ (1 × 2)]

+ (1 × 0)]



In algebra, the commutative law of multiplication states that a × b = b × a. For matrices, this law is only true in a few special cases, and in general A × B is not equal to B × A.
Problem 9.

2 3
If A = and 1 0

2 3
B=
show that A × B = B × A.
0 1

Exercise 93 Further problems on addition, subtraction and multiplication of matrices
In Problems 1 to 13, the matrices A to K are:
A=

3 −1
−4
7

B=

5 2
−1 6

−1.3
7.4
2.5 −3.9


4 −7
6
4
0⎠
D = ⎝−2
5
7 −4


3
6 2
E = ⎝ 5 −3 7⎠
−1
0 2


3.1 2.4
6.4
6
F = ⎝−1.6 3.8 −1.9⎠ G =
−2
5.3 3.4 −4.8




4
1 0
−2
H=
J = ⎝−11⎠ K = ⎝0 1⎠
5
7
1 0

C=

Addition, subtraction and multiplication
In Problems 1 to 12, perform the matrix operation stated. 1.

A+ B

8

1

−5 13

The theory of matrices and determinants
⎡⎛
2.

⎢⎜
⎣⎝3
4

D+E

3.

5 A + 6B

6.

2D + 3E −4F

−3

1

9.3 −6.4
−7.5
45

A× B

9.

A×C

10.

D× J

E×K

7

4.6

−5.6 −7.6

The unit matrix

A unit matrix, I, is one in which all elements of the leading diagonal (\) have a value of 1 and all other elements have a value of 0. Multiplication of a matrix by
I is the equivalent of multiplying by 1 in arithmetic.

The determinant of a 2 by 2 matrix,

−26 71
⎡⎛

22.3

22.4 The determinant of a 2 by 2 matrix 16.9

⎞⎤

⎟⎥
28.6⎠⎦

17.2
−11

A× H

8.

11.

⎟⎥
7⎠⎦

7 −2

⎢⎜
⎣⎝ 17.4 −16.2
−14.2
0.4
7.

⎞⎤

−2 −3

A+ B −C

5.

8

1

A− B

4.

7 −1

a b is defined c d

as (ad − bc).
The elements of the determinant of a matrix are written between vertical lines. Thus, the determinant
3 −4
3 −4 of is written as and is equal to
1
6
1
6
(3 × 6) − (−4 × 1), i.e. 18 −(−4) or 22. Hence the determinant of a matrix can be expressed as a single
3 −4 numerical value, i.e.
= 22.
1
6

43
16

Problem 10. Determine the value of

0

3 −2
7
4

−27 34
−6.4

26.1

22.7 −56.9
⎡⎛
⎞⎤
135
⎟⎥
⎢⎜
⎣⎝−52⎠⎦
−85
⎞⎤
⎡⎛
5
6
⎟⎥
⎢⎜
⎣⎝12 −3⎠⎦
1
0
⎞⎤
⎡⎛
55.4 3.4
10.1
⎟⎥
⎢⎜
⎣⎝−12.6 10.4 −20.4⎠⎦
−16.9 25.0
37.9

12.

D× F

13.

Show that A × C = C × A

−6.4
26.1
⎢A × C =

22.7 −56.9



−33.5 −53.1

⎢C × A =
23.1 −29.8

Hence they are not equal

235











3 −2
= (3 × 4) − (−2 × 7)
7
4
= 12 − (−14) = 26

Problem 11. Evaluate

(1 + j ) j2 − j 3 (1 − j 4)

(1 + j ) j2 = (1 + j )(1 − j 4) − ( j 2)(− j 3)
− j 3 (1 − j 4)
= 1 − j 4 + j − j 24 + j 26
= 1 − j 4 + j − (−4) + (−6) since from Chapter 20, j 2 = −1
= 1− j4+ j +4 −6
= −1 − j 3
Problem 12. Evaluate

5∠30◦ 2∠−60◦
3∠60◦ 4∠−90◦

236 Higher Engineering Mathematics
5∠30◦ 2∠−60◦
= (5∠30◦ )(4∠−90◦ )
3∠60◦ 4∠−90◦
− (2∠−60◦ )(3∠60◦ )
= (20∠−60◦ ) − (6∠0◦ )
= (10 − j 17.32) − (6 + j 0)
= (4 − j 17.32) or 17.78∠−77◦
Now try the following exercise
Exercise 94 Further problems on 2 by 2 determinants 1. Calculate the determinant of

3 −1
−4
7
[17]

2. Calculate the determinant of
−2
5
3 −6
3. Calculate the determinant of
−1.3
7.4
2.5 −3.9
4. Evaluate

5. Evaluate

j2
−j3
(1 + j ) j [−3]

[−13.43]

[−5 + j 3]

2∠40◦

5∠−20◦

7∠−32◦

4∠−117◦
(−19.75 + j 19.79) or 27.96∠134.94◦

Multiplying the matrices on the left hand side, gives a + 2c b + 2d
1 0
=
3a + 4c 3b + 4d
0 1
Equating corresponding elements gives: b + 2d = 0, i.e. b = −2d
4
and 3a + 4c = 0, i.e. a = − c
3
Substituting for a and b gives:


4
−2d + 2d ⎟
− c + 2c

3
1 0



⎟=
0 1


4
3 − c + 4c 3(−2d) + 4d
3

⎛2
c 0
⎠= 1 0
⎝3
i.e.
0 1
0 −2d
2
3
1
showing that c = 1, i.e. c = and −2d = 1, i.e. d = −
3
2
2
4
Since b = −2d, b = 1 and since a = − c, a = −2.
3
1 2 a b
Thus the inverse of matrix is that is,
3 4 c d


−2
1
⎝ 3
1⎠

2
2
There is, however, a quicker method of obtaining the inverse of a 2 by 2 matrix. p q
For any matrix the inverse may be r s obtained by:
(i) interchanging the positions of p and s,
(ii) changing the signs of q and r, and
(iii) multiplying this new matrix by the reciprocal of p q the determinant of r s

22.5 The inverse or reciprocal of a
2 by 2 matrix
The inverse of matrix A is A−1 such that A × A−1 = I , the unit matrix.
1 2
Let matrix A be and let the inverse matrix, A−1
3 4 a b be . c d
Then, since A × A−1 = I ,
1 2 a b
1 0
×
=
3 4 c d
0 1

1 2 is 3 4


−2
1
4 −2
1
=⎝ 3
1⎠
4 − 6 −3 1

2
2

Thus the inverse of matrix

as obtained previously.
Problem 13.

Determine the inverse of
3 −2
7
4

The theory of matrices and determinants p q is obtained by interr s changing the positions of p and s, changing the signs of q and r and multiplying by the reciprocal of the p q determinant . Thus, the inverse of r s

The inverse of matrix

1
4 2
3 −2
=
−7 3
7
4
(3 × 4) − (−2 × 7)


1
2
⎜ 13 13 ⎟
1
4 2

=
=⎜
⎝ −7 3 ⎠
26 −7 3
26 26
Now try the following exercise
Exercise 95 Further problems on the inverse of 2 by 2 matrices
3 −1
−4
7
⎞⎤
⎡⎛
7 1
⎢⎜ 17 17 ⎟⎥
⎟⎥
⎢⎜
⎣⎝ 4 3 ⎠⎦
17 17


1
2
⎜ 2
3⎟

2. Determine the inverse of ⎜
⎝ 1
3⎠


5
⎞⎤
⎡⎛3
4
5
8 ⎟⎥
⎢⎜ 7 7
7 ⎟⎥
⎢⎜
⎣⎝ 2
3 ⎠⎦
−6
−4
7
7

237



1 2 3
Thus for the matrix ⎝4 5 6⎠ the minor of
7 8 9 element 4 is obtained ⎛ ⎞covering the row by 1
(4 5 6) and the column ⎝4⎠, leaving the 2 by
7
2 3
, i.e. the minor of element 4
2 determinant
8 9 is (2 × 9) −(3 × 8) = −6.
(ii) The sign of a minor depends on its position within


+ − + the matrix, the sign pattern being ⎝− + −⎠.
+ − +
Thus the signed-minor of element 4 in the matrix


1 2 3
⎝4 5 6⎠ is − 2 3 = −(−6) = 6.
8 9
7 8 9
The signed-minor of an element is called the cofactor of the element.

1. Determine the inverse of

−1.3
7.4
3. Determine the inverse of
2.5 −3.9


0.290 0.551
⎣ 0.186 0.097

correct to 3 dec. places

22.6 The determinant of a 3 by 3 matrix (i) The minor of an element of a 3 by 3 matrix is the value of the 2 by 2 determinant obtained by covering up the row and column containing that element. (iii) The value of a 3 by 3 determinant is the sum of the products of the elements and their cofactors of any row or any column of the corresponding 3 by 3 matrix.
There are thus six different ways of evaluating a 3 × 3 determinant—and all should give the same value.
Problem 14. Find the value of
3
4 −1
2
0
7
1 −3 −2
The value of this determinant is the sum of the products of the elements and their cofactors, of any row or of any column. If the second row or second column is selected, the element 0 will make the product of the element and its cofactor zero and reduce the amount of arithmetic to be done to a minimum.
Supposing a second row expansion is selected.
The minor of 2 is the value of the determinant remaining when the row and column containing the 2 (i.e. the second row and the first column), is covered up.
4 −1
Thus the cofactor of element 2 is
i.e. −11.
−3 −2
The sign of element 2 is minus, (see (ii) above), hence the cofactor of element 2, (the signed-minor) is +11.
3
4
Similarly the minor of element 7 is
i.e. −13,
1 −3 and its cofactor is +13. Hence the value of the sum of

238 Higher Engineering Mathematics the products of the elements and their cofactors is
2 × 11 +7 × 13, i.e.,

= j 2(9) − (1 − j )(5 − j 7)

3
4 −1
2
0
7 = 2(11) + 0 + 7(13) = 113
1 −3 −2

= j 18 − [5 − j 7 − j 5 + j 27]

The same result will be obtained whichever row or column is selected. For example, the third column expansion is
(−1)

= j 2(5 − j 24) − (1 − j )(5 + j 5 − j 12) + 0

2
0
3
4
3 4
−7
+ (−2)
1 −3
1 −3
2 0

= j 18 − [−2 − j 12]
= j 18 + 2 + j 12 = 2 + j 30 or 30.07∠86.19◦

Now try the following exercise

= 6 + 91 + 16 = 113, as obtained previously.

Problem 15.

Exercise 96 Further problems on 3 by 3 determinants 1
4 −3
2
6
Evaluate −5
−1 −4
2

1. Find the matrix of minors of


4 −7
6
⎝−2
4
0⎠
5
7 −4
⎡⎛
⎞⎤
−16
8 −34
⎣⎝−14 −46
63⎠⎦
−24
12
2

1
4 −3
−5
2
6
Using the first row:
−1 −4
2
=1

2 6
−5 6
−5
2
−4
+ (−3)
−4 2
−1 2
−1 −4

2. Find the matrix of cofactors of


4 −7
6
⎝−2
4
0⎠
5
7 −4
⎡⎛
⎞⎤
−16 −8 −34
⎣⎝ 14 −46 −63⎠⎦
−24 −12
2

= (4 + 24) − 4(−10 + 6) − 3(20 + 2)
= 28 + 16 − 66 = −22
1
4 −3
2
6
Using the second column: −5
−1 −4
2
= −4

−5 6
1 −3
1 −3
+2
−(−4)
−1 2
−1
2
−5
6

3. Calculate the determinant of


4 −7
6
⎝−2
4
0⎠
5
7 −4

= −4(−10 + 6) + 2(2 − 3) + 4(6 − 15)
= 16 − 2 − 36 = −22
Problem 16.

Determine the value of

8 −2 −10
4. Evaluate 2 −3 −2
6
3
8

j2
(1 + j ) 3
(1 − j )
1
j
0
j4
5
Using the first column, the value of the determinant is:
( j 2)

1

j

j4 5

− (1 − j )

(1 + j ) 3 j4 5. Calculate the determinant of


3.1 2.4
6.4
⎝−1.6 3.8 −1.9⎠
5.3 3.4 −4.8

[−212]

[−328]

[−242.83]

5
+ (0)

(1 + j ) 3
1

j

j2
2
j
(1 + j ) 1 −3
6. Evaluate
5
−j4 0

[−2 − j ]

The theory of matrices and determinants
3∠60◦
j2
1
7. Evaluate
0
(1 + j ) 2∠30◦
0
2 j5 26.94∠−139.52◦ or
(−20.49 − j 17.49)
8. Find the eigenvalues λ that satisfy the following equations:
(a)

(2 − λ)
2
=0
−1
(5 − λ)

(b)

(5 − λ)
7
−5
0
(4 − λ)
−1
=0
2
8
(−3 − λ)

(You may need to refer to chapter 1, pages
8–12, for the solution of cubic equations).
[(a) λ =3 or 4 (b) λ =1 or 2 or 3]

22.7 The inverse or reciprocal of a
3 by 3 matrix
The adjoint of a matrix A is obtained by:
(i) forming a matrix B of the cofactors of A, and
(ii) transposing matrix B to give B T , where B T is the matrix obtained by writing the rows of B as the columns of B T . Then adj A = BT .
The inverse of matrix A, A−1 is given by
A−1 =

adj A
|A|

where adj A is the adjoint of matrix A and |A| is the determinant of matrix A.
Problem 17. Determine the inverse of the matrix


3
4 −1


0
7⎠
⎝2
1 −3 −2
The inverse of matrix A, A−1 =

adj A
|A|

239

The adjoint of A is found by:
(i) obtaining the matrix of the cofactors of the elements, and
(ii) transposing this matrix.
The cofactor of element 3 is +

0
7
= 21.
−3 −2

2
7
= 11, and so on.
1 −2


21
11 −6
The matrix of cofactors is ⎝11 −5 13⎠
28 −23 −8

The cofactor of element 4 is −

The transpose of the matrix of cofactors, i.e. the adjoint of the matrix, is obtained by writing the rows as columns,


21 11
28
and is ⎝ 11 −5 −23⎠
−6 13 −8
3
4 −1
0
7
From Problem 14, the determinant of 2
1 −3 −2 is 113.


3
4 −1
0
7⎠ is
Hence the inverse of ⎝2
1 −3 −2


28
21 11
⎝ 11 −5 −23⎠


21 11
28
−6 13 −8
1 ⎝
11 −5 −23⎠ or 113
113
−6 13 −8
Problem 18. Find the inverse of


1
5 −2


4⎠
⎝ 3 −1
−3
6 −7
Inverse =

adjoint determinant ⎛

−17
9
15
The matrix of cofactors is ⎝ 23 −13 −21⎠
18 −10 −16
The transpose of the matrix of cofactors (i.e. the


−17
23
18
adjoint) is ⎝ 9 −13 −10⎠
15 −21 −16

240 Higher Engineering Mathematics

1
5 −2
The determinant of ⎝ 3 −1
4⎠
−3
6 −7


= 1(7 − 24) − 5(−21 + 12) − 2(18 − 3)
= −17 + 45 − 30 = −2


1
5 −2
4⎠
Hence the inverse of ⎝ 3 −1
−3
6 −7


−17
23
18
⎝ 9 −13 −10⎠
15 −21 −16
=
−2


8.5 −11.5 −9
6.5
5⎠
= ⎝−4.5
−7.5
10.5
8
Now try the following exercise
Exercise 97 Further problems on the inverse of a 3 by 3 matrix
1. Write down the transpose of


4 −7
6
⎝−2
4
0⎠
5
7 −4
⎡⎛

⎞⎤
4 −2
5
⎣⎝−7
4
7⎠⎦
6
0 −4

2. Write down the transpose of

1⎞
3
6 2
⎝ 5 − 2 7⎠
3
−1
0 3
5
⎡⎛

⎞⎤
3
5 −1
⎣⎝ 6 − 2
0⎠⎦
3
1
3
7
2
5

3. Determine the adjoint of


4 −7
6
⎝−2
4
0⎠
5
7 −4
⎞⎤
⎡⎛
−16
14 −24
⎣⎝ −8 −46 −12⎠⎦
−34 −63
2
4. Determine the adjoint of


1
3
6 2


⎝ 5 − 2 7⎠
3
−1
0 3
5
⎡⎛ 2
⎞⎤
42 1
− 5 −3 3
5
3
⎢⎜
⎟⎥
⎢⎜−10 2 3 −18 1 ⎟⎥
10
2 ⎠⎦
⎣⎝
−2
−6 −32
3
5. Find the inverse of


4 −7
6
⎝−2
4
0⎠
5
7 −4
⎞⎤

−16
14 −24
1
⎝ −8 −46 −12⎠⎦
⎣−
212
−34 −63
2





3
6 1
2


6. Find the inverse of ⎝ 5 − 2 7⎠
3
3
−1
0 5
⎛ 2
⎞⎤

− 5 −3 3
42 1
5
3
⎟⎥
⎢ 15 ⎜
3
⎝−10 2 10 −18 1 ⎠⎦
⎣−
2
923
−2
−6 −32
3

Chapter 23

The solution of simultaneous equations by matrices and determinants (i) Writing the equations in the a1 x + b1 y = c form gives: 23.1 Solution of simultaneous equations by matrices
(a)

The procedure for solving linear simultaneous equations in two unknowns using matrices is:
(i) write the equations in the form a1 x + b1 y = c1 a2 x + b2 y = c2
(ii) write the matrix equation corresponding to these equations, a b x c
i.e. 1 1 ×
= 1 a2 b2 c2 y a b
(iii) determine the inverse matrix of 1 1 a2 b2
1
b2 −b1
i.e.
a1 a1 b2 − b1 a2 −a2
(from Chapter 22)
(iv) multiply each side of (ii) by the inverse matrix, and
(v) solve for x and y by equating corresponding elements. Problem 1. Use matrices to solve the simultaneous equations:
3x + 5y − 7 = 0
4x − 3y − 19 = 0

(1)
(2)

3x + 5y = 7
4x − 3y = 19
(ii) The matrix equation is
3
5 x 7
×
=
4 −3 y 19
(iii) The inverse of matrix

3
5
is
4 −3

1
−3 −5
3
3 × (−3) − 5 × 4 −4
⎛3
5 ⎞


i.e. ⎝ 29 29 ⎠
4 −3
29 29
(iv) Multiplying each side of (ii) by (iii) and remembering that A × A−1 = I , the unit matrix, gives: ⎞

5
3
⎜ 29 29 ⎟
1 0 x
7

=⎜
⎝ 4 −3 ⎠ × 19
0 1 y 29

29

242 Higher Engineering Mathematics

21 95
+ ⎟
⎜ 29 29 x ⎟
Thus
=⎜
⎝ 28 57 ⎠ y −
29 29 x 4
i.e.
= y −1


(i) Writing the equations in the a1 x + b1 y + c1 z = d1 form gives: x +y+z =4
2x − 3y + 4z = 33
3x − 2y − 2z = 2

(v) By comparing corresponding elements:

(ii) The matrix equation is

⎞ ⎛ ⎞ ⎛ ⎞
1
1
1
x
4
⎝2 −3
4⎠ × ⎝ y ⎠ = ⎝33⎠
3 −2 −2 z 2

x = 4 and y = −1
Checking:
equation (1),

(iii) The inverse matrix of


1
1
1
4⎠
A = ⎝2 −3
3 −2 −2

3 × 4 + 5 × (−1) − 7 = 0 = RHS equation (2),
4 × 4 − 3 × (−1) − 19 = 0 = RHS

is given by
(b) The procedure for solving linear simultaneous equations in three unknowns using matrices is:
(i) write the equations in the form a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3
(ii) write the matrix equation corresponding to these equations, i.e.

⎞ ⎛ ⎞ ⎛ ⎞ a1 b1 c1 x d1
⎝a2 b2 c2 ⎠ × ⎝ y ⎠ = ⎝d2 ⎠ a3 b3 c3 d3 z
(iii) determine the inverse matrix of

⎛ a1 b1 c1
⎝a2 b2 c2 ⎠ (see Chapter 22) a3 b3 c3

A−1 =

adj A
|A|

The adjoint of A is the transpose of the matrix of the cofactors of the elements (see Chapter 22). The matrix of cofactors is


14 16 5
⎝ 0 −5 5⎠
7 −2 −5 and the transpose of this matrix gives


14
0
7
adj A = ⎝16 −5 −2⎠
5
5 −5
The determinant of A, i.e. the sum of the products of elements and their cofactors, using a first row expansion is

(iv) multiply each side of (ii) by the inverse matrix, and
(v) solve for x, y and z by equating the corresponding elements.
Problem 2. Use matrices to solve the simultaneous equations: x + y +z −4 = 0

(1)

2x − 3y + 4z − 33 = 0
3x − 2y − 2z − 2 = 0

(2)
(3)

1

−3
4
2
4
2 −3
−1
+1
−2 −2
3 −2
3 −2
= (1 × 14) − (1 × (−16)) + (1 × 5) = 35

Hence the inverse of A,


14
0
7
1 ⎝
16 −5 −2⎠
A−1 =
35
5
5 −5
(iv) Multiplying each side of (ii) by (iii), and remembering that A × A−1 = I , the unit matrix, gives

The solution of simultaneous equations by matrices and determinants

⎞ ⎛ ⎞
1 00 x 1
⎝0 1 0⎠ × ⎝ y ⎠ =
35
0 01 z ⎛

(14 × 4) + (0 × 33) + (7 × 2)
× ⎝(16 × 4) + ((−5) × 33) + ((−2) × 2)⎠
(5 × 4) + (5 × 33) + ((−5) × 2)


⎛ ⎞
(14 × 4) + (0 × 33) + (7 × 2) x 1
⎝(16 × 4) + ((−5) × 33) + ((−2) × 2)

⎝y⎠ =
35 (5 × 4) + (5 × 33) + ((−5) × 2) z ⎛

=



70
1
⎝−105⎠
35
175



2
= ⎝−3⎠
5
(v) By comparing corresponding elements, x = 2, y = −3, z = 5, which can be checked in the original equations.
Now try the following exercise
Exercise 98 Further problems on solving simultaneous equations using matrices

243

6. In two closed loops of an electrical circuit, the currents flowing are given by the simultaneous equations: I1 + 2I2 + 4 = 0
5I1 + 3I2 − 1 = 0
Use matrices to solve for I1 and I2 .
[I1 = 2, I2 = −3]
7. The relationship between the displacement, s, velocity, v, and acceleration, a, of a piston is given by the equations: s + 2v + 2a = 4
3s − v + 4a = 25
3s + 2v − a = −4
Use matrices to determine the values of s, v and a.
[s = 2, v = −3, a = 4]
8. In a mechanical system, acceleration x ,
¨
velocity x and distance x are related by the
˙
simultaneous equations:
3.4 x + 7.0 x − 13.2x = −11.39
¨
˙
−6.0 x + 4.0 x + 3.5x = 4.98
¨
˙
2.7 x + 6.0 x + 7.1x = 15.91
¨
˙
¨ ˙
Use matrices to find the values of x, x and x.
[x = 0.5, x = 0.77, x = 1.4]
¨
˙

In Problems 1 to 5 use matrices to solve the simultaneous equations given.
1. 3x + 4y = 0
2x + 5y + 7 = 0

[x = 4, y = −3]

2. 2 p +5q + 14.6 = 0
3.1 p +1.7q + 2.06 =0

(a)
[ p =1.2, q = −3.4]

3.

x + 2y + 3z =5
2x − 3y − z = 3
−3x + 4y + 5z = 3

When solving linear simultaneous equations in two unknowns using determinants:
(i) write the equations in the form a1 x + b1 y + c1 = 0 a2 x + b2 y + c2 = 0

[x = 1, y = −1, z = 2]
4. 3a + 4b − 3c = 2
−2a + 2b + 2c = 15
7a − 5b + 4c = 26
[a = 2.5, b = 3.5, c = 6.5]
5.

23.2 Solution of simultaneous equations by determinants

p + 2q + 3r + 7.8 = 0
2 p + 5q − r − 1.4 = 0
5 p − q + 7r − 3.5 = 0
[ p = 4.1, q = −1.9, r = −2.7]

and then
(ii) the solution is given by x −y
1
=
=
Dx
Dy
D where Dx =

b1 c1 b2 c2

i.e. the determinant of the coefficients left when the x-column is covered up,

244 Higher Engineering Mathematics a1 c1

Dy =

a2 c2

i.e. the determinant of the coefficients left when the y-column is covered up,
D=

and

find the values of u and a, each correct to 4 significant figures.
Substituting the given values in v = u +at gives:

a1 b1

21 = u + 3.5a

(1)

a2 b2

33 = u + 6.1a

(2)

i.e. the determinant of the coefficients left when the constants-column is covered up.
Problem 3. Solve the following simultaneous equations using determinants:
3x − 4y = 12

(i) The equations are written in the form
i.e.
and

(ii) The solution is given by u −a
1
=
=
Du
Da
D

7x + 5y = 6.5
Following the above procedure:
(i) 3x − 4y − 12 = 0
7x + 5y − 6.5 = 0
(ii)

x
−y
1
=
=
−4 −12
3 −12
3 −4
5 −6.5
7 −6.5
7
5
i.e.

x
(−4)(−6.5) − (−12)(5)
=
=

i.e.

where Du is the determinant of coefficients left when the u column is covered up,
i.e.

Similarly, Da =

i.e.
Since

x
1
86
=
then x =
=2
86 43
43

and since
−y
1
64.5
= then y = −
= −1.5
64.5 43
43
Problem 4. The velocity of a car, accelerating at uniform acceleration a between two points, is given by v = u +at , where u is its velocity when passing the first point and t is the time taken to pass between the two points. If v = 21 m/s when t = 3.5 s and v = 33 m/s when t = 6.1 s, use determinants to

3.5 −21
6.1 −33

1 −21
1 −33

= (1)(−33) − (−21)(1)
= −12 and x
−y
1
=
=
26 + 60 −19.5 + 84 15 + 28 x −y
1
=
=
86 64.5 43

Du =

= (3.5)(−33) − (−21)(6.1)
= 12.6

−y
(3)(−6.5) − (−12)(7)
1
(3)(5) − (−4)(7)

a1 x + b1 y + c1 = 0, u + 3.5a − 21 = 0 u + 6.1a − 33 = 0

D=

1 3.5
1 6.1

= (1)(6.1) − (3.5)(1) = 2.6
Thus
i.e. and u
−a
1
=
=
12.6 −12 26
12.6
= 4.846 m/s
2.6
12 a= = 4.615 m/s2 ,
2.6
each correct to 4 significant figures. u=

Problem 5. Applying Kirchhoff’s laws to an electric circuit results in the following equations:
(9 + j 12)I1 − (6 + j 8)I2 = 5
−(6 + j 8)I1 + (8 + j 3)I2 = (2 + j 4)
Solve the equations for I1 and I2

The solution of simultaneous equations by matrices and determinants
Following the procedure:
(i) (9 + j 12)I1 − (6 + j 8)I2 − 5 = 0
−(6 + j 8)I1 + (8 + j 3)I2 − (2 + j 4) =0
(ii)

I1
−(6 + j 8)
−5
(8 + j 3) −(2 + j 4)
=

−I2
(9 + j 12)
−5
−(6 + j 8) −(2 + j 4)
1

=

(9 + j 12) −(6 + j 8)
−(6 + j 8) (8 + j 3)
I1
(−20 + j 40) + (40 + j 15)
−I2
=
(30 − j 60) − (30 + j 40)
=

1
(36 + j 123) − (−28 + j 96)

I1
−I2
=
20 + j 55 − j 100
=

1
64 + j 27

20 + j 55
Hence I 1 =
64 + j 27
=
and

I2 =

58.52∠70.02◦
= 0.84∠47.15◦A
69.46∠22.87◦
100∠90◦
69.46∠22.87◦

= 1.44∠67.13◦ A

(b) When solving simultaneous equations in three unknowns using determinants:
(i) Write the equations in the form

245

b1 c1 d1 where Dx is b2 c2 d2 b3 c3 d3
i.e. the determinant of the coefficients obtained by covering up the x column. a1 c1 d1
D y is a2 c2 d2 a3 c3 d3
i.e., the determinant of the coefficients obtained by covering up the y column. a1 b1 d1
Dz is a2 b2 d2 a3 b3 d3
i.e. the determinant of the coefficients obtained by covering up the z column. a1 b1 c1 and D is a2 b2 c2 a3 b3 c3
i.e. the determinant of the coefficients obtained by covering up the constants column. Problem 6. A d.c. circuit comprises three closed loops. Applying Kirchhoff’s laws to the closed loops gives the following equations for current flow in milliamperes:
2I1 + 3I2 − 4I3 = 26
I1 − 5I2 − 3I3 = −87
−7I1 + 2I2 + 6I3 = 12
Use determinants to solve for I1 , I2 and I3 .
(i) Writing the equations in the a1 x + b1 y + c1 z +d1 = 0 form gives:
2I1 + 3I2 − 4I3 − 26 = 0

a1 x + b1 y + c1 z + d1 = 0

I1 − 5I2 − 3I3 + 87 = 0

a2 x + b2 y + c2 z + d2 = 0

−7I1 + 2I2 + 6I3 − 12 = 0

a3 x + b3 y + c3 z + d3 = 0 and then
(ii) the solution is given by x −y z −1
=
=
=
Dx
Dy
Dz
D

(ii) the solution is given by
I1
−I2
I3
−1
=
=
=
D I1
D I2
D I3
D
where D I1 is the determinant of coefficients obtained by covering up the I1 column, i.e.

246 Higher Engineering Mathematics
Now try the following exercise

3 −4 −26
87
D I1 = −5 −3
2
6 −12
= (3)

−3
87
−5
87
− (−4)
6 −12
2 −12
+ (−26)

−5 −3
2
6

= 3(−486) + 4(−114) − 26(−24)

Exercise 99 Further problems on solving simultaneous equations using determinants
In Problems 1 to 5 use determinants to solve the simultaneous equations given.
1. 3x − 5y = −17.6
7y − 2x − 22 = 0
[x = −1.2, y = 2.8]

= −1290

2. 2.3m − 4.4n = 6.84
8.5n − 6.7m = 1.23

2 −4 −26
1 −3
87
D I2 =
−7
6 −12

[m = −6.4, n = −4.9]

= (2)(36 − 522) − (−4)(−12 + 609)
+ (−26)(6 − 21)
= −972 + 2388 + 390

3. 3x + 4y + z = 10
2x − 3y + 5z + 9 = 0 x + 2y − z = 6
[x = 1, y = 2, z = −1]

= 1806

4. 1.2 p − 2.3q − 3.1r + 10.1 = 0

2
3 −26
1 −5
87
D I3 =
−7
2 −12

4.7 p + 3.8q − 5.3r − 21.5 = 0
3.7 p − 8.3q + 7.4r + 28.1 = 0
[ p = 1.5, q = 4.5, r = 0.5]

= (2)(60 − 174) − (3)(−12 + 609)
+ (−26)(2 − 35)
= −228 − 1791 + 858 = −1161
D=

and

2
3 −4
1 −5 −3
−7
2
6

= (2)(−30 + 6) − (3)(6 − 21)
+ (−4)(2 − 35)
= −48 + 45 + 132 = 129
Thus
I1
−I2
I3
−1
=
=
=
−1290 1806 −1161 129 giving −1290
= 10 mA,
I1 =
−129
1806
= 14 mA
129
1161 and I 3 =
= 9 mA
129
I2 =

5.

x y 2z
1
− +
=−
2 3
5
20 x 2y z
19
+
− =
4
3
2 40
59
x +y−z =
60
x=

17
5
7
, y = ,z = −
20
40
24

6. In a system of forces, the relationship between two forces F1 and F2 is given by:
5F1 + 3F2 + 6 = 0
3F1 + 5F2 + 18 = 0
Use determinants to solve for F1 and F2 .
[F1 = 1.5, F2 = −4.5]
7. Applying mesh-current analysis to an a.c. circuit results in the following equations:
(5 − j 4)I1 − (− j 4)I2 = 100∠0◦
(4 + j 3 − j 4)I2 − (− j 4)I1 = 0
Solve the equations for I1 and I2.
I1 = 10.77∠19.23◦ A,
I2 = 10.45∠−56.73◦ A

The solution of simultaneous equations by matrices and determinants

8. Kirchhoff’s laws are used to determine the current equations in an electrical network and show that i1 + 8i2 + 3i3 = −31
3i1 − 2i2 + i3 = −5
2i1 − 3i2 + 2i3 = 6
Use determinants to find the values of i1 , i2 and i3 .
[i1 = −5, i2 = −4, i3 = 2]
9. The forces in three members of a framework are F1 , F2 and F3 . They are related by the simultaneous equations shown below.
1.4F1 + 2.8F2 + 2.8F3 = 5.6
4.2F1 − 1.4F2 + 5.6F3 = 35.0
4.2F1 + 2.8F2 − 1.4F3 = −5.6
Find the values of F1 , F2 and F3 using determinants. [F1 = 2, F2 = −3, F3 = 4]
10. Mesh-current analysis produces the following three equations:
20∠0◦ = (5 + 3 − j 4)I1 − (3 − j 4)I2
10∠90◦ = (3 − j 4 + 2)I2 − (3 − j 4)I1 − 2I3
−15∠0◦ − 10∠90◦ = (12 + 2)I3 − 2I2
Solve the equations for the loop currents I1 , I2 and I3 .


I1 = 3.317∠22.57◦ A
⎣ I2 = 1.963∠40.97◦ A ⎦
I3 = 1.010∠−148.32◦ A

23.3 Solution of simultaneous equations using Cramers rule
Cramers rule states that if a11 x + a12 y + a13 z = b1 a21 x + a22 y + a23 z = b2 a31 x + a32 y + a33 z = b3 then x =

where

Dy
Dx
Dz
, y= and z =
D
D
D

a11 a12 a13
D = a21 a22 a23 a31 a32 a33 b1 a12 a13
Dx = b2 a22 a23 b3 a32 a33

247

i.e. the x-column has been replaced by the R.H.S. b column, a11 b1 a13
D y = a21 b2 a23 a31 b3 a33
i.e. the y-column has been replaced by the R.H.S. b column, a11 a12 b1
Dz = a21 a22 b2 a31 a32 b3
i.e. the z-column has been replaced by the R.H.S. b column.
Problem 7. Solve the following simultaneous equations using Cramers rule. x +y+z =4
2x − 3y + 4z = 33
3x − 2y − 2z = 2
(This is the same as Problem 2 and a comparison of methods may be made). Following the above method:
1
1
1
4
D = 2 −3
3 −2 −2
= 1(6 − (−8)) − 1((−4) − 12)
+ 1((−4) − (−9)) = 14 + 16 + 5 = 35
4
1
1
Dx = 33 −3
4
2 −2 −2
= 4(6 − (−8)) − 1((−66) − 8)
+ 1((−66) − (−6)) = 56 + 74 − 60 = 70
1 4
1
Dy = 2 33
4
3 2 −2
= 1((−66) − 8) − 4((−4) − 12) + 1(4 − 99)
= −74 + 64 − 95 = −105
1
1 4
Dz = 2 −3 33
3 −2 2
= 1((−6) − (−66)) − 1(4 − 99)
+ 4((−4) − (−9)) = 60 + 95 + 20 = 175

248 Higher Engineering Mathematics
Hence

Working backwards, from equation (3 ),

x=

Dx
70
Dy
−105
=
= 2, y =
=
= −3
D
35
D
35

z=

−35
= 5,
−7

175
Dz
and z = D = 35 = 5

from equation (2 ),

Now try the following exercise

from which,

−5y + 2(5) = 25,

y=

Exercise 100 Further problems on solving simultaneous equations using Cramers rule

and from equation (1),

1. Repeat problems 3, 4, 5, 7 and 8 of Exercise
98 on page 241, using Cramers rule.
2. Repeat problems 3, 4, 8 and 9 of Exercise 99 on page 244, using Cramers rule.

23.4

x + (−3) + 5 = 4, from which, x = 4+3−5 = 2
(This is the same example as Problems 2 and 7, and a comparison of methods can be made). The above method is known as the Gaussian elimination method.

Solution of simultaneous equations using the Gaussian elimination method

We conclude from the above example that if a11 x + a12 y + a13 z = b1

Consider the following simultaneous equations: x +y+z =4

a21 x + a22 y + a23 z = b2

(1)

2x − 3y + 4z = 33
3x − 2y − 2z = 2

(3)

a31 x + a32 y + a33 z = b3

(2)

Leaving equation (1) as it is gives: x +y+z =4

(1)

Equation (2) − 2 × equation (1) gives:
0 − 5y + 2z = 25

(2 )

and equation (3) − 3 × equation (1) gives:
0 − 5y − 5z = −10

(3 )

the three-step procedure to solve simultaneous equations in three unknowns using the Gaussian elimination method is: a21 × equation (1) to form equa1. Equation (2) − a11 a31
× equation (1) to tion (2 ) and equation (3) − a11 form equation (3 ). a32 × equation (2 ) to form equa2. Equation (3 ) − a22 tion (3 ).
3.

Leaving equations (1) and (2 ) as they are gives: x +y+z =4

(1)

0 − 5y + 2z = 25

(2 )

Equation (3 ) − equation (2 ) gives:
0 + 0 − 7z = −35

25 − 10
= −3
−5

(3 )

By appropriately manipulating the three original equations we have deliberately obtained zeros in the positions shown in equations (2 ) and (3 ).

Determine z from equation (3 ), then y from equation (2 ) and finally, x from equation (1).

Problem 8. A d.c. circuit comprises three closed loops. Applying Kirchhoff’s laws to the closed loops gives the following equations for current flow in milliamperes:
2I1 + 3I2 − 4I3 = 26
I1 − 5I2 − 3I3 = −87
−7I1 + 2I2 + 6I3 = 12

(1)
(2)
(3)

The solution of simultaneous equations by matrices and determinants
Now try the following exercise

Use the Gaussian elimination method to solve for
I1 , I2 and I3 .
(This is the same example as Problem 6 on page 243, and a comparison of methods may be made)
Following the above procedure:
1. 2I1 + 3I2 − 4I3 = 26
1
Equation (2) − × equation (1) gives:
2
0 − 6.5I2 − I3 = −100
−7
× equation (1) gives:
Equation (3) −
2
0 + 12.5I2 − 8I3 = 103

(1)
(2 )

(3 )
(1)
(2 )

12.5
× equation (2 ) gives:
−6.5
0 + 0 − 9.923I3 = −89.308
(3 )
Equation (3 ) −

3. From equation (3 ),
−89.308
I3 =
= 9 mA,
−9.923
from equation (2 ), −6.5I2 − 9 =−100,
−100 +9 from which, I 2 =
= 14 mA
−6.5
and from equation (1), 2I1 + 3(14) − 4(9) = 26,

from which, I 1 =

26 − 42 + 36 20
=
2
2

= 10 mA

1. In a mass-spring-damper system, the acceler˙ ation x m/s2 , velocity x m/s and displacement
¨
x m are related by the following simultaneous equations: 6.2 x + 7.9 x + 12.6x = 18.0
¨
˙
7.5 x + 4.8 x + 4.8x = 6.39
¨
˙

2. 2I1 + 3I2 − 4I3 = 26
0 − 6.5I2 − I3 = −100

Exercise 101 Further problems on solving simultaneous equations using Gaussian elimination 13.0 x + 3.5 x − 13.0x = −17.4
¨
˙
By using Gaussian elimination, determine the acceleration, velocity and displacement for the system, correct to 2 decimal places.
[x = −0.30, x = 0.60, x = 1.20]
¨
˙
2. The tensions, T1 , T2 and T3 in a simple framework are given by the equations:
5T1 + 5T2 + 5T3 = 7.0
T1 + 2T2 + 4T3 = 2.4
4T1 + 2T2

= 4.0

Determine T1 , T2 and T3 using Gaussian elimination.
[T1 = 0.8, T2 = 0.4, T3 = 0.2]
3. Repeat problems 3, 4, 5, 7 and 8 of Exercise 98 on page 241, using the Gaussian elimination method. 4. Repeat problems 3, 4, 8 and 9 of Exercise 99 on page 244, using the Gaussian elimination method. 249

Revision Test 7
This Revision Test covers the material contained in Chapters 20 to 23. The marks for each question are shown in brackets at the end of each question.
1. Solve the quadratic equation x 2 − 2x + 5 =0 and show the roots on an Argand diagram.
(9)
2. If Z 1 = 2 + j 5, Z 2 = 1 − j 3 and Z 3 = 4 − j determine, in both Cartesian and polar forms, the value
Z1 Z2 of + Z 3 , correct to 2 decimal places.
Z1 + Z2
(9)
3. Three vectors are represented by A, 4.2∠45◦ , B,
5.5∠−32◦ and C, 2.8∠75◦. Determine in polar form the resultant D, where D =B + C − A. (8)
4. Two impedances, Z 1 = (2 + j 7) ohms and
Z 2 = (3 − j 4) ohms, are connected in series to a supply voltage V of 150∠0◦ V. Determine the magnitude of the current I and its phase angle relative to the voltage.
(6)

6.

Determine A × B.

(4)

7.

Calculate the determinant of matrix C.

(4)

8.

Determine the inverse of matrix A.

(4)

9.

Determine E × D.

(9)

10.

Calculate the determinant of matrix D.

(6)

11.

Solve the following simultaneous equations:
4x − 3y = 17 x + y+1 = 0 using matrices.

12.

In questions 6 to 10, the matrices stated are:
A=

−5
2
7 −8

B=

1
6
−3 −4

j3
(1 + j 2)
C=
(−1 − j 4) − j 2




2 −1 3
−1
3 0
D = ⎝−5
1 0 ⎠ E = ⎝ 4 −9 2 ⎠
4 −6 2
−5
7 1

Use determinants to solve the following simultaneous equations:
4x + 9y + 2z = 21

5. Determine in both polar and rectangular forms:
(a) [2.37∠35◦]4 (b) [3.2 − j 4.8]5

(c) [−1 − j 3]

(6)

−8x + 6y − 3z = 41
3x + y − 5z = −73

(15)
13.

(10)

The simultaneous equations representing the currents flowing in an unbalanced, three-phase, starconnected, electrical network are as follows:
2.4I1 + 3.6I2 + 4.8I3 = 1.2
−3.9I1 + 1.3I2 − 6.5I3 = 2.6
1.7I1 + 11.9I2 + 8.5I3 = 0
Using matrices, solve the equations for I1 , I2 and I3 .
(10)

Chapter 24

Vectors
24.1

Introduction

This chapter initially explains the difference between scalar and vector quantities and shows how a vector is drawn and represented.
Any object that is acted upon by an external force will respond to that force by moving in the line of the force.
However, if two or more forces act simultaneously, the result is more difficult to predict; the ability to add two or more vectors then becomes important.
This chapter thus shows how vectors are added and subtracted, both by drawing and by calculation, and finding the resultant of two or more vectors has many uses in engineering. (Resultant means the single vector which would have the same effect as the individual vectors.)
Relative velocities and vector i, j , k notation are also briefly explained.

Now try the following exercise
Exercise 102 Further problems on scalar and vector quantities
1. State the difference between scalar and vector quantities. In problems 2 to 9, state whether the quantities given are scalar (S) or vector (V) – answers below. 2. A temperature of 70◦ C
3. 5 m3 volume
4. A downward force of 20 N
5. 500 J of work
6. 30 cm2 area
7. A south-westerly wind of 10 knots

24.2

Scalars and vectors

The time taken to fill a water tank may be measured as, say, 50 s. Similarly, the temperature in a room may be measured as, say, 16◦C, or the mass of a bearing may be measured as, say, 3 kg.
Quantities such as time, temperature and mass are entirely defined by a numerical value and are called scalars or scalar quantities.
Not all quantities are like this. Some are defined by more than just size; some also have direction. For example, the velocity of a car is 90 km/h due west, or a force of 20 N acts vertically downwards, or an acceleration of
10 m/s2 acts at 50◦ to the horizontal.
Quantities such as velocity, force and acceleration, which have both a magnitude and a direction, are called vectors.

8. 50 m distance
9. An acceleration of 15 m/s2 at 60◦ to the horizontal [Answers: 2. S 3. S 4. V 5. S 6. S 7. V
8. S 9. V]

24.3

Drawing a vector

A vector quantity can be represented graphically by a line, drawn so that:
(a)

the length of the line denotes the magnitude of the quantity, and

(b) the direction of the line denotes the direction in which the vector quantity acts.

252 Higher Engineering Mathematics
An arrow is used to denote the sense, or direction, of the vector. The arrow end of a vector is called the ‘nose’ and the other end the ‘tail’.
For example, a force of 9 N acting at 45◦ to the horizontal is shown in Fig. 24.1.
Note that an angle of + 45◦ is drawn from the horizontal and moves anticlockwise. a 9N
458

0

Figure 24.1

In this chapter a vector quantity is denoted by bold print. 24.4

Addition of vectors by drawing

Adding two or more vectors by drawing assumes that a ruler, pencil and protractor are available. Results obtained by drawing are naturally not as accurate as those obtained by calculation.
(a) Nose-to-tail method
Two force vectors, F1 and F2 , are shown in Fig. 24.3.
When an object is subjected to more than one force, the resultant of the forces is found by the addition of vectors. A velocity of 20 m/s at −60◦ is shown in Fig. 24.2.
Note that an angle of −60◦ is drawn from the horizontal and moves clockwise.

F2


0

F1

60Њ

Figure 24.3

20 m/s

b

Figure 24.2

Representing a vector
There are a number of ways of representing vector quantities. These include:
1.
2.

Using bold print


AB where an arrow above two capital letters denotes the sense of direction, where A is the starting point and B the end point of the vector

3.

(i) Force F1 is drawn to scale horizontally, shown as
0a in Fig. 24.4.
(ii) From the nose of F1 , force F2 is drawn at angle θ to the horizontal, shown as ab.
(iii) The resultant force is given by length 0b, which may be measured.

AB or a i.e. a line over the top of letters

4.

To add forces F1 and F2 :

a i.e. an underlined letter

The force of 9 N at 45◦ shown in Fig. 24.1 may be represented as:


0a or 0a or 0a
The magnitude of the force is 0a
Similarly, the velocity of 20 m/s at −60◦ shown in
Fig. 24.2 may be represented as:


0b or 0b or 0b
The magnitude of the velocity is 0b

This procedure is called the ‘nose-to-tail’ or ‘triangle’ method. b

0


F1

F2

a

Figure 24.4

(b) Parallelogram method
To add the two force vectors, F1 and F2 , of Fig. 24.3:
(i) A line cb is constructed which is parallel to and equal in length to 0a (see Fig. 24.5).

253

Vectors
(ii) A line ab is constructed which is parallel to and equal in length to 0c.
(iii) The resultant force is given by the diagonal of the parallelogram, i.e. length 0b.
This procedure is called the ‘parallelogram’ method. c ␪ a F1

(i) In Fig. 24.8, a line is constructed which is parallel to and equal in length to the 8 N force
(ii) A line is constructed which is parallel to and equal in length to the 5 N force
(iii) The resultant force is given by the diagonal of the parallelogram, i.e. length 0b, and is measured as 12 N and angle θ is measured as 17◦ .

b

F2
0

(b) ‘Parallelogram’ method

b

Figure 24.5

Problem 1. A force of 5 N is inclined at an angle of 45◦ to a second force of 8 N, both forces acting at a point. Find the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 8 N force by:
(a) the ‘nose-to-tail’ method, and (b) the
‘parallelogram’ method.
The two forces are shown in Fig. 24.6. (Although the
8 N force is shown horizontal, it could have been drawn in any direction.)

5N

5N
458
0


8N

Figure 24.8

Thus, the resultant of the two force vectors in Fig. 24.6 is 12 N at 17◦ to the 8 N force.

Problem 2. Forces of 15 N and 10 N are at an angle of 90◦ to each other as shown in Fig. 24.9.
Find, by drawing, the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 15 N force.

458
8N

Figure 24.6
10 N

(a) ‘Nose-to tail’ method
(i) The 8 N force is drawn horizontally 8 units long, shown as 0a in Fig. 24.7
(ii) From the nose of the 8 N force, the 5 N force is drawn 5 units long at an angle of 45◦ to the horizontal, shown as ab
(iii) The resultant force is given by length 0b and is measured as 12 N and angle θ is measured as 17◦. b 5N
0

Figure 24.7

458


8N

a

15 N

Figure 24.9

Using the ‘nose-to-tail’ method:
(i) The 15 N force is drawn horizontally 15 units long as shown in Fig. 24.10
(ii) From the nose of the 15 N force, the 10 N force is drawn 10 units long at an angle of 90◦ to the horizontal as shown
(iii) The resultant force is shown as R and is measured as 18 N and angle θ is measured as 34◦ .

254 Higher Engineering Mathematics
195Њ
b

Thus, the resultant of the two force vectors is 18 N at
34◦ to the 15 N force. r R

10 N

␪ a 105Њ

15 N

Figure 24.10

30Њ

O

Problem 3. Velocities of 10 m/s, 20 m/s and
15 m/s act as shown in Fig. 24.11. Determine, by drawing, the magnitude of the resultant velocity and its direction relative to the horizontal.
␷2

Figure 24.12

Worked Problems 1 to 3 have demonstrated how vectors are added to determine their resultant and their direction. However, drawing to scale is time-consuming and not highly accurate. The following sections demonstrate how to determine resultant vectors by calculation using horizontal and vertical components and, where possible, by Pythagoras’s theorem.

20 m/s

10 m/s

␷1

24.5 Resolving vectors into horizontal and vertical components

308
158
␷3

15 m/s

Figure 24.11

When more than two vectors are being added the ‘noseto-tail’ method is used.
The order in which the vectors are added does not matter. In this case the order taken is v1 , then v2 , then v3 . However, if a different order is taken the same result will occur.
(i) v1 is drawn 10 units long at an angle of 30◦ to the horizontal, shown as 0a in Fig. 24.12
(ii) From the nose of v1 , v2 is drawn 20 units long at an angle of 90◦ to the horizontal, shown as ab

A force vector F is shown in Fig. 24.13 at angle θ to the horizontal. Such a vector can be resolved into two components such that the vector addition of the components is equal to the original vector.
F



Figure 24.13

The two components usually taken are a horizontal component and a vertical component.
If a right-angled triangle is constructed as shown in
Fig. 24.14, then 0a is called the horizontal component of F and ab is called the vertical component of F.

(iii) From the nose of v2 , v3 is drawn 15 units long at an angle of 195◦ to the horizontal, shown as br

b
F

(iv) The resultant velocity is given by length 0r and is measured as 22 m/s and the angle measured to the horizontal is 105◦.
Thus, the resultant of the three velocities is 22 m/s at
105◦ to the horizontal.

0

Figure 24.14

␪ a Vectors
From trigonometry (see Chapter 11),
0a
cos θ = from which,
0b

0a = 0b cos θ

17.32 m/s

= F cos θ

i.e. and 308

20

the horizontal component of F = F cos θ ab from which, sin θ =
0b

0

ab = 0b sin θ
= F sin θ

the vertical component of F = F sinθ

i.e.

Problem 4. Resolve the force vector of 50 N at an angle of 35◦ to the horizontal into its horizontal and vertical components.
The horizontal component of the 50 N force,
0a = 50 cos 35◦ = 40.96 N
The vertical component of the 50 N force, ab = 50 sin 35◦ = 28.68 N
The horizontal and vertical components are shown in
Fig. 24.15.

s

0

210 m/s b Problem 6. Resolve the displacement vector of
40 m at an angle of 120◦ into horizontal and vertical components. The horizontal component of the 40 m displacement,
0a = 40 cos 120◦ = −20.0 m
The vertical component of the 40 m displacement, ab = 40 sin 120◦ = 34.64 m
The horizontal and vertical components are shown in
Fig. 24.17. b 40 N
34.64 N a 220.0 N 0

28.68 N

358

a

Figure 24.16

b
50 N

m/

255

1208

Figure 24.17
40.96 N

a

Figure 24.15

and


40.962 + 28.682

= 50 N
28.68
θ = tan −1
= 35◦
40.96

Thus, the vector addition of components 40.96 N and
28.68 N is 50 N at 35◦)
Problem 5. Resolve the velocity vector of 20 m/s at an angle of −30◦ to the horizontal into horizontal and vertical components.

The horizontal component of the 20 m/s velocity,
0a = 20 cos(−30◦) = 17.32 m/s
The vertical component of the 20 m/s velocity, ab = 20 sin(−30◦) = −10 m/s
The horizontal and vertical components are shown in
Fig. 24.16.

24.6 Addition of vectors by calculation Two force vectors, F1 and F2 , are shown in Fig. 24.18,
F1 being at an angle of θ1 and F2 being at an angle of θ2 .
V

F1 sin ␪1
F2 sin ␪2

(Checking: by Pythagoras, 0b =

F2

F1
␪1

␪2

F1 cos ␪1
F2 cos ␪2

Figure 24.18

H

256 Higher Engineering Mathematics
A method of adding two vectors together is to use horizontal and vertical components.
The horizontal component of force F1 is F1 cos θ1 and the horizontal component of force F2 is F2 cos θ2
The total horizontal component of the two forces,
H = F1 cos θ1 + F2 cos θ2
The vertical component of force F1 is F1 sin θ1 and the vertical component of force F2 is F2 sin θ2
The total vertical component of the two forces,
V = F1 sin θ1 + F2 sin θ2
Since we have H and V , the resultant of F1 and F2 is obtained by using the theorem of Pythagoras. From
Fig. 24.19,
0b 2 = H 2 + V 2
i.e.
resultant = H 2 + V 2 at an angle
−1 V given by θ = tan
H

The vertical component of the 8 N force is 8 sin 0◦ and the vertical component of the 5 N force is 5 sin 45◦
The total vertical component of the two forces,
V = 8 sin 0◦ + 5 sin 45◦ = 0 + 3.5355
= 3.5355
From Fig. 24.21, magnitude of resultant vector

= H2 + V 2

= 11.53552 + 3.53552 = 12.07 N



H ϭ11.5355 N

nt

lta

R


0

H

V ϭ 3.5355 N

R

b

u es nt

lta

u es V

Figure 24.21

a

The direction of the resultant vector,

Figure 24.19

θ = tan −1

Problem 7. A force of 5 N is inclined at an angle of 45◦ to a second force of 8 N, both forces acting at a point. Calculate the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 8 N force.

The two forces are shown in Fig. 24.20.

V
H

= tan−1

3.5355
11.5355

= tan −1 0.30648866 . . . = 17.04◦
Thus, the resultant of the two forces is a single vector of 12.07 N at 17.04◦ to the 8 N vector.
Perhaps an easier and quicker method of calculating the magnitude and direction of the resultant is to use complex numbers (see Chapter 20).
In this example, the resultant
= 8∠0◦ + 5∠45◦
= (8 cos 0◦ + j 8 sin0◦) + (5 cos 45◦ + j 5 sin45◦ )

5N

= (8 + j 0) + (3.536 + j 3.536)

458

= (11.536 + j 3.536) N or 12.07∠17.04◦ N

8N

Figure 24.20

The horizontal component of the 8 N force is 8 cos 0◦ and the horizontal component of the 5 N force is
5 cos 45◦
The total horizontal component of the two forces,
H = 8 cos 0◦ + 5 cos 45◦ = 8 + 3.5355
= 11.5355

as obtained above using horizontal and vertical components. Problem 8. Forces of 15 N and 10 N are at an angle of 90◦ to each other as shown in Fig. 24.22.
Calculate the magnitude of the resultant of these two forces and its direction with respect to the
15 N force.

Vectors

257

This is, of course, a special case. Pythagoras can only be used when there is an angle of 90◦ between vectors.
This is demonstrated in the next worked problem.
10 N

Problem 9. Calculate the magnitude and direction of the resultant of the two acceleration vectors shown in Fig. 24.24.

15 N

Figure 24.22

The horizontal component of the 15 N force is 15 cos0◦ and the horizontal component of the 10 N force is
10 cos90◦
The total horizontal component of the two velocities,

28 m/s2

H = 15 cos 0◦ + 10 cos 90◦ = 15 + 0 = 15
15 sin 0◦

The vertical component of the 15 N force is and the vertical component of the 10 N force is 10 sin 90◦
The total vertical component of the two velocities,
V = 15 sin 0◦ + 10 sin 90◦ = 0 + 10 = 10
Magnitude of resultant vector


= H 2 + V 2 = 152 + 102 = 18.03 N
The direction of the resultant vector,
V
10 θ = tan−1
= tan −1
= 33.69◦
H
15
Thus, the resultant of the two forces is a single vector of 18.03 N at 33.69◦ to the 15 N vector.

15 m/s2

Figure 24.24

The 15 m/s2 acceleration is drawn horizontally, shown as 0a in Fig. 24.25.
From the nose of the 15 m/s2 acceleration, the 28 m/s2 acceleration is drawn at an angle of 90◦ to the horizontal, shown as ab. b R

There is an alternative method of calculating the resultant vector in this case.
If we used the triangle method, then the diagram would be as shown in Fig. 24.23.

28



␣ a 15

0

Figure 24.25
R

10 N

The resultant acceleration, R, is given by length 0b.
Since a right-angled triangle results, the theorem of
Pythagoras may be used.


15 N

0b =

Figure 24.23

Since a right-angled triangle results then we could use
Pythagoras’s theorem without needing to go through the procedure for horizontal and vertical components.
In fact, the horizontal and vertical components are 15 N and 10 N respectively.

and

152 + 282 = 31.76 m/s2

α = tan −1

28
15

= 61.82◦

Measuring from the horizontal, θ = 180◦ − 61.82◦ = 118.18◦

258 Higher Engineering Mathematics
Thus, the resultant of the two accelerations is a single vector of 31.76 m/s2 at 118.18◦ to the horizontal.

R
21.118

Problem 10. Velocities of 10 m/s, 20 m/s and
15 m/s act as shown in Fig. 24.26. Calculate the magnitude of the resultant velocity and its direction relative to the horizontal.
␷2





5.829

Figure 24.27

Measuring from the horizontal, θ = 180◦ − 74.57◦ = 105.43◦
Thus, the resultant of the three velocities is a single vector of 21.91 m/s at 105.43◦ to the horizontal.

20 m/s
␷1

Using complex numbers, from Fig. 24.26,

10 m/s
308

resultant = 10∠30◦ + 20∠90◦ + 15∠195◦

158
␷3

= (10 cos 30◦ + j 10 sin30◦)

15 m/s

+ (20 cos 90◦ + j 20 sin90◦ )

Figure 24.26

+ (15 cos 195◦ + j 15 sin195◦)
The horizontal component of the 10 m/s velocity =
10 cos 30◦ = 8.660 m/s, the horizontal component of the 20 m/s velocity is
20 cos 90◦ = 0 m/s, and the horizontal component of the 15 m/s velocity is
15 cos195◦ = −14.489 m/s.
The total horizontal component of the three velocities,
H = 8.660 + 0 − 14.489 = −5.829 m/s
The vertical component of the 10 m/s velocity =
10 sin 30◦ = 5 m/s, the vertical component of the 20 m/s velocity is
20 sin 90◦ = 20 m/s, and the vertical component of the 15 m/s velocity is
15 sin 195◦ = −3.882 m/s.
The total vertical component of the three forces,
V = 5 + 20 − 3.882 = 21.118 m/s
From Fig. 24.27, magnitude of resultant vector,


R = H 2 + V 2 = 5.8292 + 21.1182 = 21.91 m/s
The direction of the resultant vector,
V
21.118 α = tan−1
= tan−1
= 74.57◦
H
5.829

= (8.660 + j 5.000) + (0 + j 20.000)
+ (−14.489 − j 3.882)
= (−5.829 + j 21.118) N or
21.91∠105.43◦ N as obtained above using horizontal and vertical components. The method used to add vectors by calculation will not be specified – the choice is yours, but probably the quickest and easiest method is by using complex numbers. Now try the following exercise
Exercise 103 Further problems on addition of vectors by calculation
1.

A force of 7 N is inclined at an angle of 50◦ to a second force of 12 N, both forces acting at a point. Calculate magnitude of the

Vectors

resultant of the two forces, and the direction of the resultant with respect to the 12 N force. [17.35 N at 18.00◦ to the 12 N force]

8N

2. Velocities of 5 m/s and 12 m/s act at a point at 90◦ to each other. Calculate the resultant velocity and its direction relative to the 12 m/s velocity. [13 m/s at 22.62◦ to the 12 m/s velocity]

708
5N
608

3. Calculate the magnitude and direction of the resultant of the two force vectors shown in
Fig. 24.28.
[16.40 N at 37.57◦ to the 13 N force]
13 N

Figure 24.30
10 N

7. If velocity v1 = 25 m/s at 60◦ and v2 = 15 m/s at −30◦ , calculate the magnitude and direction of v1 + v2 .

13 N

[29.15 m/s at 29.04◦ to the horizontal]

Figure 24.28

4. Calculate the magnitude and direction of the resultant of the two force vectors shown in
Fig. 24.29.
[28.43 N at 129.30◦ to the 18 N force]

8. Calculate the magnitude and direction of the resultant vector of the force system shown in
Fig. 24.31.
[9.28 N at 16.70◦]

4 N 158

8N

22 N
308
18 N

608

Figure 24.29

5. A displacement vector s1 is 30 m at 0◦. A second displacement vector s2 is 12 m at 90◦ .
Calculate magnitude and direction of the resultant vector s1 + s2 .
[32.31 m at 21.80◦ to the 30 m displacement] 6. Three forces of 5 N, 8 N and 13 N act as shown in Fig. 24.30. Calculate the magnitude and direction of the resultant force.
[14.72 N at −14.72◦ to the 5 N force]

6N

Figure 24.31

9. Calculate the magnitude and direction of the resultant vector of the system shown in
Fig. 24.32.
[6.89 m/s at 159.56◦]

259

260 Higher Engineering Mathematics
Fig. 24.34(a) shows that the second diagonal of the
‘parallelogram’ method of vector addition gives the magnitude and direction of vector subtraction of oa from ob.

2 m/s
3.5 m/s
158

b

s

d

b

458 o Ϫa

a
(a)

308

o

a

(b)

Figure 24.34
4 m/s

Problem 11. Accelerations of a1 = 1.5 m/s2 at
90◦ and a2 = 2.6 m/s2 at 145◦ act at a point. Find a1 + a2 and a1 − a2 (i) by drawing a scale vector diagram, and (ii) by calculation.

Figure 24.32

10.

An object is acted upon by two forces of magnitude 10 N and 8 N at an angle of 60◦ to each other. Determine the resultant force on the object. [15.62 N at 26.33◦ to the 10 N force]
A ship heads in a direction of E 20◦ S at a speed of 20 knots while the current is 4 knots in a direction of N 30◦ E. Determine the speed and actual direction of the ship.
[21.07 knots, E 9.22◦ S]

11.

(i) The scale vector diagram is shown in Fig. 24.35.
By measurement, a1 + a2 = 3.7 m/s2 at 126◦ a1 − a2 = 2.1 m/s2 at 0◦ a1 ϩ a2
0

1

2

3

Scale in m/s2 a1 a2

24.7

Vector subtraction

In Fig. 24.33, a force vector F is represented by oa.
The vector (−oa) can be obtained by drawing a vector from o in the opposite sense to oa but having the same magnitude, shown as ob in Fig. 24.33, i.e. ob = (−oa)

2.6 m/s2

1.5 m/s2
126Њ
145Њ

a1 Ϫ a2

Ϫa2
F

2F

a

o

b

Figure 24.33

For two vectors acting at a point, as shown in
Fig. 24.34(a), the resultant of vector addition is: os = oa + ob.
Figure 24.33(b) shows vectors ob + (−oa), that is, ob − oa and the vector equation is ob − oa = od. Comparing od in Fig. 24.34(b) with the broken line ab in

Figure 24.35

(ii) Resolving horizontally and vertically gives:
Horizontal component of a1 + a2 ,
H = 1.5 cos90◦ +2.6 cos 145◦ = −2.13
Vertical component of a1 + a2 ,
V = 1.5 sin90◦ + 2.6 sin145◦ = 2.99
From Fig. 24.36, magnitude of a1 + a2 ,
R = (−2.13)2 + 2.992 = 3.67 m/s2
2.99
= 54.53◦ and
In Fig. 24.36, α = tan−1
2.13
θ = 180◦ − 54.53◦ = 125.47◦
Thus,

a1 + a2 = 3.67 m/s2 at 125.47◦

Vectors

261

The horizontal component of v1 − v2 + v3 = (22 cos 140◦) − (40 cos 190◦)
+ (15 cos 290◦)

R

2.99

= (−16.85) − (−39.39) + (5.13)
= 27.67 units




22.13

The vertical component of

0

v1 − v2 + v3 = (22 sin 140◦ ) − (40 sin 190◦ )
+ (15 sin 290◦ )

Figure 24.36

= (14.14) − (−6.95) + (−14.10)
= 6.99 units

Horizontal component of a1 − a2
= 1.5 cos90◦ − 2.6 cos 145◦ = 2.13

The magnitude of the resultant,
R = 27.672 + 6.992 = 28.54 units
6.99
The direction of the resultant R = tan−1
27.67
= 14.18◦
Thus, v1 − v2 + v3 = 28.54 units at 14.18◦
Using complex numbers, v1 − v2 + v3 = 22∠140◦ − 40∠190◦ + 15∠290◦

Vertical component of a1 − a2
= 1.5 sin 90◦ − 2.6 sin 145◦ = 0

Magnitude of a1 − a2 = 2.132 + 02
= 2.13 m/s2
0
Direction of a1 − a2 = tan −1
= 0◦
2.13
a1 − a2 = 2.13 m/s2 at 0◦

Thus,

= (−16.853 + j 14.141)
Problem 12. Calculate the resultant of (i) v1 − v2 + v3 and (ii) v2 − v1 − v3 when v1 = 22 units at 140◦ , v2 = 40 units at 190◦ and v3 = 15 units at 290◦ .

− (−39.392 − j 6.946)
+ (5.130 − j 14.095)
= 27.669 + j 6.992 =28.54∠14.18◦
(ii) The horizontal component of

(i) The vectors are shown in Fig. 24.37.

v2 − v1 − v3 = (40 cos 190◦) − (22 cos 140◦)
− (15 cos 290◦)

1V

= (−39.39) − (−16.85) − (5.13)
= −27.67 units
The vertical component of

22

v2 − v1 − v3 = (40 sin 190◦ ) − (22 sin 140◦)

1408
1908
2H
40

1H

2908
15

2V

Figure 24.37

− (15 sin 290◦ )
= (−6.95) − (14.14) − (−14.10)
= −6.99 units
From Fig. 24.38 the magnitude of the resultant,
R = (−27.67)2 + (−6.99)2 = 28.54 units
6.99
= 14.18◦ , from which, and α = tan −1
27.67
θ = 180◦ + 14.18◦ = 194.18◦

262 Higher Engineering Mathematics
24.8

227.67



0

26.99
R

Relative velocity

For relative velocity problems, some fixed datum point needs to be selected. This is often a fixed point on the earth’s surface. In any vector equation, only the start and finish points affect the resultant vector of a system. Two different systems are shown in Fig. 24.39, but in each of the systems, the resultant vector is ad. b Figure 24.38

b c Thus, v2 − v1 − v3 = 28.54 units at 194.18◦
This result is as expected, since v2 − v1 − v3 =
− (v1 − v2 + v3 ) and the vector 28.54 units at
194.18◦ is minus times (i.e. is 180◦ out of phase with) the vector 28.54 units at 14.18◦
Using complex numbers, v 2 − v 2 − v 3 = 40∠190◦ − 22∠140◦ − 15∠290◦
= (−39.392 − j 6.946)
− (−16.853 + j 14.141)
− (5.130 − j 14.095)
= −27.669 − j 6.992
= 28.54∠ −165.82◦ or
28.54∠194.18◦

Now try the following exercise
Exercise 104 subtraction Further problems on vector

1. Forces of F1 = 40 N at 45◦ and F2 = 30 N at
125◦ act at a point. Determine by drawing and by calculation: (a) F1 + F2 (b) F1 − F2 .
[(a) 54.0 N at 78.16◦ (b) 45.64 N at 4.66◦ ]
2. Calculate the resultant of (a) v1 + v2 − v3
(b) v3 − v2 + v1 when v1 = 15 m/s at 85◦, v2 =
25 m/s at 175◦ and v3 = 12 m/s at 235◦.
[(a) 31.71 m/s at 121.81◦
(b) 19.55 m/s at 8.63◦ ]

a

d

a d (b)

(a)

Figure 24.39

The vector equation of the system shown in Fig. 24.39(a) is: ad = ab + bd and that for the system shown in Fig. 24.39(b) is: ad = ab + bc + cd
Thus in vector equations of this form, only the first and last letters, ‘a’ and ‘d’, respectively, fix the magnitude and direction of the resultant vector. This principle is used in relative velocity problems.
Problem 13. Two cars, P and Q, are travelling towards the junction of two roads which are at right angles to one another. Car P has a velocity of
45 km/h due east and car Q a velocity of 55 km/h due south. Calculate (i) the velocity of car P relative to car Q, and (ii) the velocity of car Q relative to car P.

(i) The directions of the cars are shown in
Fig. 24.40(a), called a space diagram. The velocity diagram is shown in Fig. 24.40(b), in which pe is taken as the velocity of car P relative to point e on the earth’s surface. The velocity of P relative to Q is vector pq and the vector equation is pq = pe + eq. Hence the vector directions are as shown, eq being in the opposite direction to qe.

Vectors
From the geometry√of the vector triangle, the magnitude of pq = 452 + 552 = 71.06 km/h
55
and the direction of pq = tan −1
= 50.71◦
45
i.e. the velocity of car P relative to car Q is
71.06 km/h at 50.71◦

263

3. A ship is heading in a direction N 60◦ E at a speed which in still water would be 20 km/h.
It is carried off course by a current of 8 km/h in a direction of E 50◦ S. Calculate the ship’s actual speed and direction.
[22.79 km/h, E 9.78◦ N]

N
W

q

q

E
S
P

Q

p

e

45 km/h
(a)

i, j and k notation

24.9

55 km/h

(b)

p

e
(c)

Figure 24.40

A method of completely specifying the direction of a vector in space relative to some reference point is to use three unit vectors, i, j and k, mutually at right angles to each other, as shown in Fig. 24.41. z (ii) The velocity of car Q relative to car P is given by the vector equation qp = qe + ep and the vector diagram is as shown in Fig. 24.40(c), having ep opposite in direction to pe.
From the geometry of this vector triangle, the mag√ nitude of qp = 452 + 552 = 71.06 m/s and the
55
= 50.71◦ but must direction of qp = tan −1
45
lie in the third quadrant, i.e. the required angle is:
180◦ + 50.71◦ = 230.71◦
i.e. the velocity of car Q relative to car P is
71.06 m/s at 230.71◦

Now try the following exercise
Exercise 105 velocity k i 0 j

y

x

Figure 24.41

Calculations involving vectors given in i, j k notation are carried out in exactly the same way as standard algebraic calculations, as shown in the worked example below. Further problems on relative

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.
[83.5 km/h at 71.6◦ to the vertical]
2. Calculate the time needed to swim across a river 142 m wide when the swimmer can swim at 2 km/h in still water and the river is flowing at 1 km/h. At what angle to the bank should the swimmer swim?
[4 minutes 55 seconds, 60◦]

Problem 14. Determine:
(3i + 2j + 2k) − (4i − 3j + 2k)

(3i + 2j + 2k) − (4i − 3j + 2k) = 3i + 2j + 2k
− 4i + 3j − 2k
= −i + 5j
Problem 15. Given p = 3i + 2k, q = 4i − 2j + 3k and r = −3i + 5j − 4k determine: 264 Higher Engineering Mathematics
(a) −r (b) 3p (c) 2p + 3q
(e) 0.2p + 0.6q − 3.2r
(a)

(d) −p + 2r

−r = −(−3i + 5j − 4k) = +3i − 5j + 4k

(b) 3p = 3(3i + 2k) = 9i + 6k
(c)

2p + 3q = 2(3i + 2k) + 3(4i − 2j + 3k)
= 6i + 4k + 12i − 6j + 9k
= 18i − 6j + 13k

(d) −p + 2r = −(3i + 2k) + 2(−3i + 5j − 4k)
= −3i − 2k + (−6i + 10j − 8k)
= −3i − 2k − 6i + 10j − 8k
= −9i + 10j − 10k
(e)

0.2p + 0.6q − 3.2r = 0.2(3i + 2k)
+0.6(4i − 2j + 3k) − 3.2(−3i + 5j − 4k)
= 0.6i + 0.4k + 2.4i − 1.2j + 1.8k
+9.6i − 16j + 12.8k
= 12.6i − 17.2j + 15k

Now try the following exercise
Exercise 106 notation Further problems on i, j , k

Given that p = 2i + 0.5j − 3k, q = −i + j + 4k and r = 6j − 5k, evaluate and simplify the following vectors in i, j , k form:
1. −q
[i − j − 4k]
2. 2p
[4i + j − 6k]
3. q + r
[−i + 7j − k]
4. −q + 2p

[5i − 10k]

5. 3q + 4r

[−3i + 27j − 8k]

6. q − 2 p

[−5i + 10k]

7. p + q + r

[i + 7.5j − 4k]

8. p + 2q + 3r

[20.5j − 10k]

9. 2p + 0.4q + 0.5r [3.6i + 4.4j − 6.9k]
10. 7r − 2q

[2i + 40j − 43k]

Chapter 25

Methods of adding alternating waveforms
25.1 Combination of two periodic functions There are a number of instances in engineering and science where waveforms have to be combined and where it is required to determine the single phasor (called the resultant) that could replace two or more separate phasors. Uses are found in electrical alternating current theory, in mechanical vibrations, in the addition of forces and with sound waves.
There are a number of methods of determining the resultant waveform. These include:
(a) by drawing the waveforms and adding graphically
(b) by drawing the phasors and measuring the resultant (c) by using the cosine and sine rules
(d) by using horizontal and vertical components
(e) by using complex numbers

25.2

Plotting periodic functions

yR = 3 sin A + 2 cos A and obtain a sinusoidal expression for this resultant waveform. y1 = 3 sin A and y2 = 2 cos A are shown plotted in Fig. 25.1. Ordinates may be added at, say, 15◦ intervals. For example, at 0◦, y1 + y2 = 0 + 2 = 2 at 15◦, y1 + y2 = 0.78 + 1.93 = 2.71 at 120◦, y1 + y2 = 2.60 + −1 = 1.6 at 210◦, y1 + y2 = −1.50 −1.73 = −3.23, and so on.
The resultant waveform, shown by the broken line, has the same period, i.e. 360◦ , and thus the same frequency as the single phasors. The maximum value, or

y

348

3.6
3

y1 5 3 sin A y R 5 3.6 sin (A 1 34)8

2

y2 5 2 cos A

1

This may be achieved by sketching the separate functions on the same axes and then adding (or subtracting) ordinates at regular intervals. This is demonstrated in the following worked problems.
Problem 1. Plot the graph of y1 = 3 sin A from
A = 0◦ to A = 360◦ . On the same axes plot y2 = 2 cos A. By adding ordinates, plot

0
21
22
23

Figure 25.1

908

1808

2708

3608

A

266 Higher Engineering Mathematics amplitude, of the resultant is 3.6. The resultant waveπ form leads y1 = 3 sin A by 34◦ or 34 × rad = 0.593
180
rad.
The sinusoidal expression for the resultant waveform is: yR = 3.6 sin(A + 34◦ ) or yR = 3.6 sin(A + 0.593)

y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3) are shown plotted in Fig. 25.2.

258 y1 5 4 sin ␻t

4

y25 3 sin(␻t 2 ␲/3)

2
0
22
24

y R 5 y1 1 y2
908
␲/2

1808


2708
3␲/2

458

3.6

y1 2 y2 y2 y1

4
2

0
22

908
␲/2

1808


2708
3␲/2

3608
2␲

␻t

24

Problem 2. Plot the graphs of y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3) on the same axes, over one cycle. By adding ordinates at intervals plot yR = y1 + y2 and obtain a sinusoidal expression for the resultant waveform.

y
6.1
6

y

3608
2␲

␻t

258

26

Figure 25.3

The amplitude, or peak value of the resultant (shown by the broken line), is 3.6 and it leads y1 by 45◦ or 0.79 rad. Hence, y1 − y2 = 3.6 sin(ωt + 0.79)
Problem 4. Two alternating currents are given by: i1 = 20 sin ωt amperes and π i2 = 10 sin ωt + amperes. 3
By drawing the waveforms on the same axes and adding, determine the sinusoidal expression for the resultant i1 + i2 . i1 and i2 are shown plotted in Fig. 25.4. The resultant waveform for i1 + i2 is shown by the broken line. It has the same period, and hence frequency, as i1 and i2 .

Figure 25.2

Ordinates are added at 15◦ intervals and the resultant is shown by the broken line. The amplitude of the resultant is 6.1 and it lags y1 by 25◦ or 0.436 rad.
Hence, the sinusoidal expression for the resultant waveform is: yR = 6.1 sin(ωt − 0.436)

30
26.5

y1 and y2 are shown plotted in Fig. 25.3. At 15◦ intervals y2 is subtracted from y1. For example:

␲ iR 5 20 sin ␻t 110 sin (␻t 1 )
3

20

i1 5 20 sin ␻t i2 5 10 sin(␻t 1 ␲ )
3

10
908
198

210

Problem 3. Determine a sinusoidal expression for y1 − y2 when y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3).

198


2

1808

2708



3␲
2

3608
2␲ angle ␻t

220
230

Figure 25.4



at 0 , y1 − y2 = 0 − (−2.6) = +2.6 at 30◦ , y1 − y2 = 2 − (−1.5) = +3.5 at 150◦ , y1 − y2 = 2 − 3 = −1, and so on.

The amplitude or peak value is 26.5 A.
The resultant waveform leads the waveform of i1 = 20 sin ωt by 19◦ or 0.33 rad

Methods of adding alternating waveforms

267

y1 5 4

Hence, the sinusoidal expression for the resultant i1 + i2 is given by:

608 or ␲/3 rads

iR = i1 + i 2 = 26.5 sin(ωt + 0.33) A y2 5 3

Now try the following exercise

Figure 25.5 y15 4

Exercise 107 Further problems on plotting periodic functions

3. Express 12 sin ωt + 5 cos ωt in the form
A sin(ωt ± α) by drawing and measurement.
[13 sin(ωt + 0.395)]

25.3 Determining resultant phasors by drawing
The resultant of two periodic functions may be found from their relative positions when the time is zero.
For example, if y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3) then each may be represented as phasors as shown in
Fig. 25.5, y1 being 4 units long and drawn horizontally and y2 being 3 units long, lagging y1 by π/3 radians or
60◦ . To determine the resultant of y1 + y2 , y1 is drawn horizontally as shown in Fig. 25.6 and y2 is joined to the end of y1 at 60◦ to the horizontal. The resultant is given by yR . This is the same as the diagonal of a parallelogram that is shown completed in Fig. 25.7.
Resultant yR , in Figs. 25.6 and 25.7, may be determined by drawing the phasors and their directions to scale and measuring using a ruler and protractor.

608

3

2. Two alternating voltages are given by v1 = 10 sin ωt volts and v2 = 14 sin(ωt + π/3) volts. By plotting v1 and v2 on the same axes over one cycle obtain a sinusoidal expression for (a) v1 + v2 (b) v1 − v2 .
(a) 20.9 sin(ωt + 0.63) volts
(b) 12.5 sin(ωt − 1.36) volts



y 25

1. Plot the graph of y = 2 sin A from A = 0◦ to A = 360◦ . On the same axes plot y = 4 cos A. By adding ordinates at intervals plot y = 2 sin A + 4 cos A and obtain a sinusoidal expression for the waveform.
[4.5 sin(A + 63.5◦ )]

0

yR

Figure 25.6 y1 5 4


yR y2 5 3

Figure 25.7

In this example, yR is measured as 6 units long and angle φ is measured as 25◦.
25◦ = 25 ×

π radians = 0.44 rad
180

Hence, summarising, by drawing: y R = y 1 + y 2 =
4 sin ωt + 3 sin(ωt − π/3) = 6 sin(ωt − 0.44)
If the resultant phasor yR = y1 − y2 is required, then y2 is still 3 units long but is drawn in the opposite direction, as shown in Fig. 25.8. yR ␾

2y2 5 3

608 y1 5 4

608

y2

Figure 25.8

268 Higher Engineering Mathematics
Problem 5. Two alternating currents are given by: i1 = 20 sin ωt amperes and π amperes. Determine i1 + i2 i2 = 10 sin ωt +
3
by drawing phasors.
The relative positions of i1 and i2 at time t = 0 are shown π as phasors in Fig. 25.9, where rad = 60◦ .
3
The phasor diagram in Fig. 25.10 is drawn to scale with a ruler and protractor.

10 A

0

608

20 A a ␾

iR

210A b Figure 25.11

i2 5 10 A

Now try the following exercise
608

Exercise 108 Further problems on determining resultant phasors by drawing i1 5 20 A

Figure 25.9 iR i2 5 10 A



608 i1 5 20 A

Figure 25.10

The resultant iR is shown and is measured as 26 A and angle φ as 19◦ or 0.33 rad leading i1 . Hence, by drawing and measuring:

1. Determine a sinusoidal expression for
2 sin θ + 4 cos θ by drawing phasors.
[4.5 sin(A + 63.5◦ )]
2. If v1 = 10 sin ωt volts and v2 = 14 sin(ωt + π/3) volts, determine by drawing phasors sinusoidal expressions for (a) v1 + v2
(b) v1 − v2.
(a) 20.9 sin(ωt + 0.62) volts
(b) 12.5 sin(ωt − 1.33) volts
3. Express 12 sin ωt + 5 cos ωt in the form
R sin(ωt ± α) by drawing phasors.
[13 sin(ωt + 0.40)]

i R = i 1 + i 2 = 26 sin(ωt + 0.33)A
Problem 6. For the currents in Problem 5, determine i1 − i2 by drawing phasors.
At time t = 0, current i1 is drawn 20 units long horizontally as shown by 0a in Fig. 25.11. Current i2 is shown, drawn 10 units long in broken line and leading by 60◦ . The current −i2 is drawn in the opposite direction to the broken line of i2 , shown as ab in
Fig. 25.11. The resultant iR is given by 0b lagging by angle φ.
By measurement, iR = 17 A and φ = 30◦ or
0.52 rad
Hence, by drawing phasors: i R = i 1 −i2 = 17 sin(ωt − 0.52)

25.4 Determining resultant phasors by the sine and cosine rules
As stated earlier, the resultant of two periodic functions may be found from their relative positions when the time is zero. For example, if y1 = 5 sin ωt and y2 =
4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 25.12, y1 being 5 units long and drawn horizontally and y2 being 4 units long, lagging y1 by π/6 radians or 30◦ . To determine the resultant of y1 + y2 , y1 is drawn horizontally as shown in Fig. 25.13 and y2 is joined to the end of y1 at π/6 radians, i.e. 30◦ to the horizontal. The resultant is given by yR .

Methods of adding alternating waveforms yR y1 5 5

y25 3

␲/6 or 308

y1 5 5 a


308

458

y1 5 2

(a)

(b)

Figure 25.14 y2 54

3
4.6357
= sin φ sin 135◦ sin φ =

b

Figure 25.13

Using the cosine rule on triangle 0ab of Fig. 25.13 gives:
2
yR = 52 + 42 − [2(5)(4) cos 150◦]

= 25 + 16 − (−34.641)
= 75.641

from which, yR = 75.641 = 8.697
Using the sine rule,

Using the sine rule: from which,

yR

and

1358



y1 5 2

Figure 25.12

from which,

y25 3

␲/4 or 458

y2 5 4

0

8.697
4
=

sin 150 sin φ
4 sin 150◦ sin φ =
8.697
= 0.22996 φ = sin−1 0.22996
= 13.29◦ or 0.232 rad

3 sin 135◦
= 0.45761
4.6357

φ = sin−1 0.45761

Hence,

= 27.23◦ or 0.475 rad.
Thus, by calculation,

y R = 4.635 sin(ωt + 0.475)

Problem 8. Determine π 20 sin ωt + 10 sin ωt +
3
and sine rules.

using the cosine

From the phasor diagram of Fig. 25.15, and using the cosine rule:
2
iR = 202 + 102 − [2(20)(10) cos 120◦]

= 700

Hence, iR = 700 = 26.46 A iR Hence, yR = y1 + y2 = 5 sin ωt + 4 sin(ωt − π/6)

i2 5 10 A

= 8.697 sin(ωt − 0.232)
Problem 7. Given y1 = 2 sin ωt and y2 = 3 sin(ωt + π/4), obtain an expression, by calculation, for the resultant, yR = y1 + y2 .
When time t = 0, the position of phasors y1 and y2 are as shown in Fig. 25.14(a). To obtain the resultant, y1 is drawn horizontally, 2 units long, y2 is drawn
3 units long at an angle of π/4 rads or 45◦ and joined to the end of y1 as shown in Fig. 25.14(b).
From Fig. 25.14(b), and using the cosine rule:
2
yR = 22 + 32 − [2(2)(3) cos 135◦]

Hence,

269

= 4 + 9 − [−8.485] = 21.49

yR = 21.49 = 4.6357

608

␾ i1 5 20 A

Figure 25.15

Using the sine rule gives : from which,

10
26.46
= sin φ sin 120◦
10 sin 120◦ sin φ =
26.46
= 0.327296

and

0.327296 = 19.10◦ π = 19.10 ×
= 0.333 rad
180

φ = sin

−1

270 Higher Engineering Mathematics b Hence, by cosine and sine rules, iR = i1 + i 2 = 26.46 sin(ωt + 0.333) A

F


Now try the following exercise

0

Exercise 109 Resultant phasors by the sine and cosine rules
1. Determine, using the cosine and sine rules, a sinusoidal expression for: y = 2 sin A + 4 cos A.
[4.5 sin(A + 63.5◦ )]
2. Given v1 = 10 sin ωt volts and v2 =14 sin(ωt + π/3) volts use the cosine and sine rules to determine sinusoidal expressions for (a) v1 + v2 (b) v1 − v2 .
(a) 20.88 sin(ωt + 0.62) volts
(b) 12.50 sin(ωt − 1.33)volts
In Problems 3 to 5, express the given expressions in the form A sin(ωt ± α) by using the cosine and sine rules.
3. 12 sin ωt + 5 cos ωt
[13 sin(ωt + 0.395)] π 4. 7 sin ωt + 5 sin ωt +
4
[11.11 sin(ωt + 0.324)]
5. 6 sin ωt + 3 sin ωt −

F sin ␪

π
6
[8.73 sin(ωt − 0.173)]

F cos ␪

a

Figure 25.16

i.e.

the horizontal component of F, H = F cos θ

and sin θ =

i.e.

ab from which ab = 0b sin θ
0b
= F sin θ

the vertical component of F, V = F sin θ

Determining resultant phasors by horizontal and vertical components is demonstrated in the following worked problems. Problem 9. Two alternating voltages are given by v1 = 15 sin ωt volts and v2 = 25 sin(ωt − π/6) volts. Determine a sinusoidal expression for the resultant vR = v1 + v2 by finding horizontal and vertical components.
The relative positions of v1 and v2 at time t = 0 are shown in Fig. 25.17(a) and the phasor diagram is shown in Fig. 25.17(b).
The horizontal component of vR ,
H = 15 cos0◦ + 25 cos(−30◦ ) = 0a + ab = 36.65 V
The vertical component of vR ,
V = 15 sin 0◦ + 25 sin(−30◦ ) = bc = −12.50 V
Hence,

vR = 0c =

36.652 + (−12.50)2 by Pythagoras’ theorem

= 38.72 volts

25.5 Determining resultant phasors by horizontal and vertical components If a right-angled triangle is constructed as shown in
Fig. 25.16, then 0a is called the horizontal component of F and ab is called the vertical component of F.

tan φ =

V
−12.50
=
= −0.3411
H
36.65

from which, φ = tan−1 (−0.3411) = −18.83◦ or − 0.329 radians.
Hence,

v R = v 1 + v2 = 38.72sin(ωt − 0.329)V

From trigonometry (see Chapter 11),
0a
from which,
0b
0a = 0b cos θ = F cos θ cos θ =

Problem 10. For the voltages in Problem 9, determine the resultant vR = v1 − v2 using horizontal and vertical components.

Methods of adding alternating waveforms v1 5 15 V

0

␲/6 or 308

v1 a


271

b

1508

308

v2 v2 5 25 V

vR

(a)

c

(b)

Figure 25.17 i2 5 10 A

The horizontal component of vR ,
H = 15 cos0◦ − 25 cos(−30◦ ) = −6.65V
608

The vertical component of vR ,
V = 15 sin0◦ − 25 sin(−30◦ ) = 12.50V
Hence, vR =

i15 20 A

Figure 25.19

(−6.65)2 + (12.50)2 by Pythagoras’ theorem

= 14.16 volts tan φ =

V
12.50
=
= −1.8797
H
−6.65

from which, φ = tan −1(−1.8797) = 118.01◦ or 2.06 radians.
Hence,

Total vertical component,
V = 20 sin 0◦ + 10 sin 60◦ = 8.66
By Pythagoras, the resultant, iR = 25.02 + 8.662
= 26.46 A
−1 8.66 = 19.11◦
Phase angle, φ = tan
25.0
or 0.333 rad
Hence, by using horizontal and vertical components,
20 sin ωt + 10 sin ωt +

π
= 26.46 sin(ωt + 0.333)
3

vR = v1 −v2 = 14.16 sin(ωt + 2.06)V
The phasor diagram is shown in Fig. 25.18. vR 2v2 5 25 V



Now try the following exercise
Exercise 110 Further problems on resultant phasors by horizontal and vertical components 308 v1 5 15 V

308

v2 5 25 V

Figure 25.18

Problem 11. Determine π 20 sin ωt + 10 sin ωt + using horizontal and
3
vertical components.
From the phasors shown in Fig. 25.19:
Total horizontal component,
H = 20 cos0◦ + 10 cos60◦ = 25.0

In Problems 1 to 4, express the combination of periodic functions in the form A sin(ωt ± α) by horizontal and vertical components: π 1. 7 sin ωt + 5 sin ωt +
4
[11.11 sin(ωt + 0.324)] π 2. 6 sin ωt + 3 sin ωt −
6
[8.73 sin(ωt − 0.173)] π 3. i = 25 sin ωt − 15 sin ωt +
3
[i = 21.79 sin(ωt − 0.639)]
4.

x = 9 sin ωt +

3π π −7 sin ωt −
3
8
[x = 14.38 sin(ωt + 1.444)]

272 Higher Engineering Mathematics
5. The voltage drops across two components when connected in series across an a.c. supply are: v1 = 200 sin314.2t and v2 = 120 sin(314.2t − π/5) volts respectively.
Determine the:
(a) voltage of the supply (given by v1 + v2 ) in the form A sin(ωt ± α).

the time is zero. For example, if y1 = 5 sin ωt and y2 =
4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 25.20, y1 being 5 units long and drawn horizontally and y2 being 4 units long, lagging y1 by π/6 radians or 30◦ . To determine the resultant of y1 + y2 , y1 is drawn horizontally as shown in Fig. 25.21 and y2 is joined to the end of y1 at π/6 radians, i.e. 30◦ to the horizontal. The resultant is given by yR .

(b) frequency of the supply. y1 5 5

[(a) 305.3 sin(314.2t − 0.233)V
(b) 50 Hz]
6. If the supply to a circuit is v = 20 sin 628.3t volts and the voltage drop across one of the components is v1 = 15 sin(628.3t − 0.52) volts, calculate the:
(a) voltage drop across the remainder of the circuit, given by v − v1 , in the form
A sin(ωt ± α).

␲/6 or 308

y2 5 4

Figure 25.20

(b) supply frequency.
(c) periodic time of the supply.
[(a) 10.21 sin(628.3t + 0.818)V
(b) 100 Hz (c) 10 ms]
7. The voltages across three components in a series circuit when connected across an a.c. supply are: π volts,
6
π v2 = 40 sin 300 πt − volts, and
4
π volts. v3 = 50 sin 300 πt +
3
Calculate the:
(a) supply voltage, in sinusoidal form, in the form A sin(ωt ± α). v1 = 25 sin 300 πt +

(b) frequency of the supply.
(c)

periodic time.
[(a) 79.83 sin (300 πt + 0.352)V
(b) 150 Hz (c) 6.667 ms]

25.6 Determining resultant phasors by complex numbers
As stated earlier, the resultant of two periodic functions may be found from their relative positions when

0

y1 5 5 a


308

y2

54

yR

b

Figure 25.21

π
6
= 5∠0◦ + 4∠ − 30◦

In polar form, yR = 5∠0 + 4∠ −

= (5 + j 0) + (4.33 − j 2.0)
= 9.33 − j 2.0 = 9.54∠ − 12.10◦
= 9.54∠−0.21rad
Hence, by using complex numbers, the resultant in sinusoidal form is: y1 + y2 = 5 sin ωt + 4 sin(ωt − π/6)
= 9.54 sin(ωt−0.21)
Problem 12. Two alternating voltages are given by v1 = 15 sin ωt volts and v2 = 25 sin(ωt − π/6) volts. Determine a sinusoidal expression for the resultant vR = v1 + v2 by using complex numbers.
The relative positions of v1 and v2 at time t = 0 are shown in Fig. 25.22(a) and the phasor diagram is shown in Fig. 25.22(b).

Methods of adding alternating waveforms v1 5 15 V

273

v1


␲/6 or 308

1508

v2 5 25 V

vR

(a)

(b)

Figure 25.22

In polar form, vR = v1 + v2 = 15∠0 + 25∠ −

π
6

= 15∠0◦ + 25∠ − 30◦

From the phasors shown in Fig. 25.23, the resultant may be expressed in polar form as: i2 5 10 A

= (15 + j 0) + (21.65 − j 12.5)
= 36.65 − j 12.5 = 38.72∠ − 18.83◦

608

= 38.72∠ − 0.329 rad
Hence, by using complex numbers, the resultant in sinusoidal form is:

i1 5 20 A

Figure 25.23

iR = 20∠0◦ + 10∠60◦

vR = v1 + v2 = 15 sin ωt + 25 sin(ωt − π/6)
= 38.72 sin(ωt − 0.329)

i.e.

= (25 + j 8.66) = 26.46∠19.11◦A or

Problem 13. For the voltages in Problem 12, determine the resultant vR = v1 − v2 using complex numbers. π
In polar form, yR = v1 − v2 = 15∠0 − 25∠ −
6

iR = (20 + j 0) + (5 + j 8.66)

26.46∠0.333 rad A
Hence, by using complex numbers, the resultant in sinusoidal form is: iR = i1 + i2 = 26.46 sin(ωt + 0.333)A

= 15∠0◦ − 25∠ − 30◦
= (15 + j 0) − (21.65 − j 12.5)
= −6.65 + j 12.5 = 14.16∠118.01◦
= 14.16∠2.06 rad
Hence, by using complex numbers, the resultant in sinusoidal form is: y1 − y2 = 15 sin ωt − 25 sin(ωt − π/6)
= 14.16 sin(ωt − 2.06)
Problem 14. Determine π 20 sin ωt + 10 sin ωt + using complex
3
numbers.

Problem 15. If the supply to a circuit is v = 30 sin 100 πt volts and the voltage drop across one of the components is v1 = 20 sin(100 πt − 0.59) volts, calculate the:
(a) voltage drop across the remainder of the circuit, given by v − v1 , in the form
A sin(ωt ± α)
(b) supply frequency
(c)

periodic time of the supply

(d) r.m.s. value of the supply voltage
(a)

Supply voltage, v =v1 + v2 where v2 is the voltage across the remainder of the circuit.

274 Higher Engineering Mathematics
Hence, v2 = v − v1 = 30 sin 100 πt
− 20 sin(100 πt − 0.59)
= 30∠0 − 20∠ − 0.59 rad
= (30 + j 0) − (16.619 − j 11.127)
= 13.381 + j 11.127
= 17.40∠0.694 rad
Hence, by using complex numbers, the resultant in sinusoidal form is: v − v1 = 30 sin 100 πt − 20 sin(100 πt − 0.59)
= 17.40 sin(ωt + 0.694) volts ω 100 π
=
= 50 Hz


1
1
(c) Periodic time, T = =
= 0.02 s or 20 ms f 50

(b) Supply frequency, f =

(d) R.m.s. value of supply voltage, = 0.707 × 30
= 21.21 volts

Now try the following exercise
Exercise 111 Further problems on resultant phasors by complex numbers
In Problems 1 to 4, express the combination of periodic functions in the form A sin(ωt ± α) by using complex numbers: π 1. 8 sin ωt + 5 sin ωt +
4
[12.07 sin(ωt + 0.297)]

5. The voltage drops across two components when connected in series across an a.c. supply are: v1 = 240 sin 314.2t and v2 = 150 sin(314.2t − π/5) volts respectively.
Determine the:
(a) voltage of the supply (given by v1 + v2 ) in the form A sin(ωt ± α).
(b) frequency of the supply.
[(a) 371.95 sin(314.2t − 0.239)V
(b) 50 Hz]
6. If the supply to a circuit is v = 25 sin200πt volts and the voltage drop across one of the components is v1 = 18 sin(200πt − 0.43) volts, calculate the:
(a) voltage drop across the remainder of the circuit, given by v − v1 , in the form
A sin(ωt ± α).
(b) supply frequency.
(c) periodic time of the supply.
[(a) 11.44 sin(200πt + 0.715)V
(b) 100 Hz (c) 10 ms]
7. The voltages across three components in a series circuit when connected across an a.c. supply are: π volts,
6
π volts, and v2 = 30 sin 300πt +
4
π volts. v3 = 60 sin 300πt −
3
v1 = 20 sin 300πt −

π
6
[14.51 sin(ωt − 0.315)]

Calculate the:

π
4
[9.173 sin(ωt + 0.396)]

2. 6 sin ωt + 9 sin ωt −

(b) frequency of the supply.

3. v = 12 sin ωt − 5 sin ωt −

4.

x = 10 sin ωt +

3π π − 8 sin ωt −
3
8
[16.168 sin(ωt + 1.451)]

(a) supply voltage, in sinusoidal form, in the form A sin(ωt ± α).

(c) periodic time.
(d) r.m.s. value of the supply voltage.
[(a) 79.73 sin(300π − 0.536) V
(b) 150 Hz (c) 6.667 ms (d) 56.37 V]

Chapter 26

Scalar and vector products
26.1

The unit triad

When a vector x of magnitude x units and direction θ ◦ is divided by the magnitude of the vector, the result is a vector of unit length at angle θ ◦ . The unit vector for a
10 m/s at 50◦ velocity of 10 m/s at 50◦ is
, i.e. 1 at 50◦.
10 m/s oa In general, the unit vector for oa is
, the oa being
|oa|
a vector and having both magnitude and direction and
|oa| being the magnitude of the vector only.
One method of completely specifying the direction of a vector in space relative to some reference point is to use three unit vectors, mutually at right angles to each other, as shown in Fig. 26.1. Such a system is called a unit triad.

r k z

j

x iO a

b y Figure 26.2

k
O

j

i
4

z
3
22 i k o j

P

y

(a)

x

Figure 26.1

In Fig. 26.2, one way to get from o to r is to move x units along i to point a, then y units in direction j to get to b and finally z units in direction k to get to r. The vector or is specified as or =xi + yj + zk

k
O
r

j i 2
5

Problem 1. With reference to three axes drawn mutually at right angles, depict the vectors
(i) op = 4i +3j −2k and (ii) or= 5i − 2j +2k.
The required vectors are depicted in Fig. 26.3, op being shown in Fig. 26.3(a) and or in Fig. 26.3(b).

22
(b)

Figure 26.3

276 Higher Engineering Mathematics b 26.2 The scalar product of two vectors v2

When vector oa is multiplied by a scalar quantity, say k, the magnitude of the resultant vector will be k times the magnitude of oa and its direction will remain the same.
Thus 2 ×(5 N at 20◦) results in a vector of magnitude
10 N at 20◦ .
One of the products of two vector quantities is called the scalar or dot product of two vectors and is defined as the product of their magnitudes multiplied by the cosine of the angle between them. The scalar product of oa and ob is shown as oa • ob. For vectors oa = oa at θ1 , and ob = ob at θ2 where θ2 > θ1 , the scalar product is:



a

O

c v2 cos ␪ v1 (a)

v2

oa • ob = oa ob cos(θ 2 − θ 1 )

s␪

v1

For vectors v1 and v 2 shown in Fig. 26.4, the scalar product is:

co



v 1 • v2 = v1 v2 cos θ

v1
(b)
v1

Figure 26.6



The projection of ob on oa is shown in Fig. 26.6(a) and by the geometry of triangle obc, it can be seen that the projection is v2 cos θ. Since, by definition

v2

Figure 26.4

oa • ob = v1 (v2 cos θ),
The commutative law of algebra, a × b = b × a applies to scalar products. This is demonstrated in Fig. 26.5. Let oa represent vector v1 and ob represent vector v2 . Then: oa • ob = v1 v2 cos θ (by definition of a scalar product)

it follows that oa • ob = v1 (the projection of v2 on v1 )
Similarly the projection of oa on ob is shown in
Fig. 26.6(b) and is v1 cos θ. Since by definition ob • oa = v2 (v1 cos θ),

b

v2

O

it follows that ob • oa = v2 (the projection of v1 on v2 )



v1

a

Figure 26.5

Similarly, ob • oa = v2 v1 cos θ = v1 v2 cos θ by the commutative law of algebra. Thus oa • ob = ob • oa.

This shows that the scalar product of two vectors is the product of the magnitude of one vector and the magnitude of the projection of the other vector on it.
The angle between two vectors can be expressed in terms of the vector constants as follows:
Because a • b = a b cos θ, then cos θ =

a•b ab (1)

Scalar and vector products
Let

a = a1 i + a2 j + a3 k

and

b = b1 i + b2 j + b3 k

277

Thus, the length or modulus or magnitude or norm of vector OP is given by:

a • b = (a1 i + a2 j + a3 k) • (b1 i + b2 j + b3 k)

(a 2 + b2 + c 2 )

OP =

Multiplying out the brackets gives:

(3)

Relating this result to the two vectors a1 i + a2 j + a3k and b1 i + b2 j + b3k, gives:

a • b = a1 b1 i • i + a1 b2 i • j + a1 b3 i • k
+ a2 b1 j • i + a2 b2 j • j + a2 b3 j • k

a=

2
2
2
(a1 + a2 + a3 )

and b =

2
2
2
(b1 + b2 + b3 ).

+ a3 b1 k • i + a3 b2 k • j + a3 b3 k • k
However, the unit vectors i, j and k all have a magnitude of 1 and i • i = (1)(1) cos 0◦ = 1, i • j = (1)(1) cos 90◦ = 0, i • k = (1)(1) cos 90◦ = 0 and similarly j • j = 1, j • k = 0 and k • k = 1. Thus, only terms containing i • i, j • j or k • k in the expansion above will not be zero.
Thus, the scalar product a • b = a 1 b1 + a 2 b2 + a 3 b3

That is, from equation (1), a 1 b1 + a 2 b2 + a 3 b3

cos θ =

(4)

(a 2 + a 2 + a 2 ) (b2 + b2 + b2 )
1
2
3
1
2
3

(2)

Both a and b in equation (1) can be expressed in terms of a1 , b1 , a2 , b2 , a3 and b3 .

Problem 2. Find vector a joining points P and Q where point P has co-ordinates (4, −1, 3) and point
Q has co-ordinates (2, 5, 0). Also, find |a|, the magnitude or norm of a.
Let O be the origin, i.e. its co-ordinates are (0, 0, 0). The position vector of P and Q are given by:

c

P

OP = 4i − j + 3k and OQ = 2i + 5j
By the addition law of vectors OP + PQ = OQ.

O

Hence a =PQ = OQ − OP
B

A

a

i.e.

a =PQ = (2i + 5j) − (4i − j + 3k)

b

= −2i + 6j − 3k

Figure 26.7

From equation (3), the magnitude or norm of a,

From the geometry of Fig. 26.7, the length of diagonal
OP in terms of side lengths a, b and c can be obtained from Pythagoras’ theorem as follows:

|a| =
=

(a 2 + b2 + c2 )
[( 2 + 62 + ( 2 ] =
−2)
−3)


49 = 7

OP2 = OB2 + BP2 and
OB2 = OA2 + AB2

Problem 3. If p = 2i + j −k and q = i −3j + 2k determine: Thus, OP2 = OA2 + AB2 + BP2
= a +b +c , in terms of side lengths
2

2

2

(i) p • q
(iii) |p + q|

(ii) p +q
(iv) |p| +|q|

278 Higher Engineering Mathematics
Since oa = i + 2j − 3k,

(i) From equation (2), if and

q = b1 i + b2 j + b3 k

then

a1 = 1, a2 = 2 and a3 = −3

p = a1 i + a2 j + a3 k p • q = a1 b1 + a2 b2 + a3 b3

Since ob = 2i − j + 4k, b1 = 2, b2 = −1 and b3 = 4

p = 2i + j − k,

When

a1 = 2, a2 = 1 and a3 = −1 and when q = i − 3j +2k,

cos θ =

b1 = 1, b2 = −3 and b3 = 2
Hence p • q = (2)(1) + (1)(−3) + (−1)(2) p • q = −3

i.e.

= 3i −2j + k
(iii) |p +q| = |3i − 2 j + k|
From equation (3),
[32 + (−2)2 + 12 ] =

(1 × 2) + (2 × −1) + (−3 × 4)
(12 + 22 + (−3)2 )

=√

(22 + (−1)2 + 42 )

−12
√ = −0.6999
14 21

i.e. θ = 134.4◦ or 225.6◦ .

(ii) p +q = (2i + j −k) + (i − 3j +2k)

|p + q| =

Thus,


14

By sketching the position of the two vectors as shown in
Problem 1, it will be seen that 225.6◦ is not an acceptable answer. Thus the angle between the vectors oa and ob, θ = 134.4◦
Direction cosines

(iv) From equation (3),
|p| = |2i + j − k|
=

[22 + 12 + (−1)2 ] =


6

Similarly,
|q| = |i − 3 j + 2k|
=

[12 + (−3)2 + 22 ] =


14

√ √
Hence |p| +|q|= 6 + 14 = 6.191, correct to 3 decimal places.
Problem 4. Determine the angle between vectors oa and ob when oa = i + 2j − 3k

From Fig. 26.2, or= xi + yj + zk and from equation (3), |or| = x 2 + y 2 + z 2 .
If or makes angles of α, β and γ with the co-ordinate axes i, j and k respectively, then:
The direction cosines are: cos α = cos β = and cos γ =

x x2 + y2 + z2 y x2

+ y2 + z2 y x2

+ y2 + z2

such that cos2 α + cos2 β + cos2 γ = 1.
The values of cos α, cos β and cos γ are called the direction cosines of or.

and ob = 2i − j + 4k.
An equation for cos θ is given in equation (4) cos θ =

Problem 5.
3i + 2j +k.

Find the direction cosines of

a1 b1 + a2 b2 + a3 b3
2
2
2
2
2
2
(a1 + a2 + a3 ) (b1 + b2 + b3 )

x 2 + y2 + z2 =

32 + 22 + 12 =


14

Scalar and vector products
The direction cosines are: cos α =

The work done is F • d, that is F • AB in this case

3
= √ = 0.802
2 + y2 + z2
14
x x y

cos β =

2
= √ = 0.535
14
x 2 + y2 + z2

and cos γ =

1
= √ = 0.267
2 + y2 + z2
14
x y (and hence α = cos−1 0.802 = 36.7◦, β = cos−1 0.535 =
57.7◦ and γ = cos−1 0.267 =74.5◦).
Note that cos2 α + cos2 β + cos2 γ = 0.8022 + 0.5352 +
0.2672 = 1.

Practical application of scalar product
Problem 6. A constant force of
F =10i + 2j −k newtons displaces an object from
A =i + j +k to B =2i − j +3k (in metres). Find the work done in newton metres.

i.e. work done = (10i + 2j − k) • (i − 2j + 2k)
But from equation (2), a • b = a1 b1 + a2 b2 + a3 b3
Hence work done =
(10 × 1) + (2 × (−2)) + ((−1) × 2) = 4 Nm.
(Theoretically, it is quite possible to get a negative answer to a ‘work done’ problem. This indicates that the force must be in the opposite sense to that given, in order to give the displacement stated.)
Now try the following exercise
Exercise 112 products 1.

Further problems on scalar

Find the scalar product a • b when
(i) a =i + 2j − k and b =2i + 3j +k
(ii) a =i − 3j +k and b = 2i + j +k
[(i) 7 (ii) 0]

One of the applications of scalar products is to the work done by a constant force when moving a body. The work done is the product of the applied force and the distance moved in the direction of the force.

Given p =2i − 3j, q = 4j −k and r =i + 2j −3k, determine the quantities stated in problems 2 to 8.
2.

The principles developed in Problem 13, page 262, apply equally to this problem when determining the displacement. From the sketch shown in Fig. 26.8,

(a) p • q (b) p • r

[(a) −12 (b) −4]

3.

i.e. work done = F • d

(a) q • r (b) r • q

4.

(a) | p | (b) | r |

[(a) 11 (b) 11]


[(a) 13 (b) 14]

5.

(a) p • (q + r) (b) 2r • (q − 2p)
[(a) −16 (b) 38]

AB = AO+ OB = OB − OA

6.

(a) | p +r | (b) | p | +| r |

that is AB = (2i − j + 3k) − (i + j + k)
= i − 2j + 2k

[(a)


19 (b) 7.347]

7.

A (1,1,1)

Find the angle between (a) p and q
(b) q and r.
[(a) 143.82◦ (b) 44.52◦]

8.

B (2, 21, 3)

Determine the direction cosines of (a) p
(b) q (c) r.


(a) 0.555, −0.832, 0
⎣ (b) 0, 0.970, −0.243

(c) 0.267, 0.535, −0.802

9.

Determine the angle between the forces:
F1 = 3i + 4j + 5k and

O (0, 0, 0)

Figure 26.8

279

F2 = i + j + k

[11.54◦]

280 Higher Engineering Mathematics
Then,
10. Find the angle between the velocity vectors υ 1 = 5i +2j + 7k and υ 2 = 4i +j − k.
[66.40◦ ]

a ×b = (a1 i + a2 j + a3 k) × (b1 i + b2 j + b3 k)
= a1 b1 i × i + a1 b2 i × j

11. Calculate the work done by a force
F =(−5i + j +7k) N when its point of application moves from point (−2i − 6j +k) m to the point (i − j + 10k) m.
[53 Nm]

+ a1 b3 i × k + a2 b1 j × i + a2 b2 j × j
+ a2 b3 j × k + a3 b1 k × i + a3 b2 k × j
+ a3 b3 k × k

26.3

But by the definition of a vector product,

Vector products

i × j = k, j ×k = i and k × i = j
A second product of two vectors is called the vector or cross product and is defined in terms of its modulus and the magnitudes of the two vectors and the sine of the angle between them. The vector product of vectors oa and ob is written as oa ×ob and is defined by:

Also i × i = j ×j = k × k = (1)(1) sin 0◦ = 0.
Remembering that a ×b = −b × a gives: a × b = a1 b2 k − a1 b3 j − a2 b1 k + a2 b3 i
+ a3 b1 j − a3 b2 i

|oa × ob| =oa obsinθ where θ is the angle between the two vectors.
The direction of oa × ob is perpendicular to both oa and ob, as shown in Fig. 26.9.

Grouping the i, j and k terms together, gives: a ×b = (a2 b3 − a3 b2)i + (a3 b1 − a1 b3 ) j
+ (a1 b2 − a2 b1 )k

b


o

The vector product can be written in determinant form as:

b

oa ϫ ob

i j k a ×b = a 1 a 2 a 3 b1 b2 b3

a o ob ϫ oa

␪ a (a)

(b)

i j k
The 3 × 3 determinant a1 a2 a3 is evaluated as: b1 b2 b3

Figure 26.9

i
The direction is obtained by considering that a righthanded screw is screwed along oa ×ob with its head at the origin and if the direction of oa × ob is correct, the head should rotate from oa to ob, as shown in Fig. 26.9(a). It follows that the direction of ob ×oa is as shown in Fig. 26.9(b). Thus oa × ob is not equal to ob × oa. The magnitudes of oa ob sin θ are the same but their directions are 180◦ displaced, i.e. oa ×ob = −ob ×oa
The vector product of two vectors may be expressed in terms of the unit vectors. Let two vectors, a and b, be such that: a = a1i + a2 j + a3 k and b = b 1 i + b2 j + b3 k

(5)

a a a a a2 a3
−j 1 3 +k 1 2 b2 b3 b1 b3 b1 b2

where a2 a3
= a2 b3 − a3 b2 , b2 b3 a1 a3
= a1 b3 − a3 b1 and b1 b3 a1 a2
= a1 b2 − a2 b1 b1 b2
The magnitude of the vector product of two vectors can be found by expressing it in scalar product form and then using the relationship a • b = a1 b1 + a2 b2 + a3 b3

Scalar and vector products
Squaring both sides of a vector product equation gives:

(ii) From equation (7)

(|a × b|)2 = a 2 b2 sin2 θ = a 2 b2(1 − cos2 θ)
= a 2 b2 − a 2 b2 cos2 θ

(6)

Now

a a = a cos θ.
2

b • b = (2)(2) + (−1)(−1) + (3)(3)

But θ = 0◦, thus a • a = a 2
Also, cos θ =

a • a = (1)(1) + (4 × 4) + (−2)(−2)
= 21

It is stated in Section 26.2 that a • b = ab cos θ, hence


[(a • a)(b • b) − (a • b)2 ]

|a × b| =

= 14 and a•b
.
ab

a • b = (1)(2) + (4)(−1) + (−2)(3)
= −8

Multiplying both sides of this equation by squaring gives: a 2b2 cos2 θ =

a 2 b2

and

a 2b2 (a • b)2
= (a • b)2 a 2b2

Substituting in equation (6) above for a 2 = a • a, b2 = b • b

Thus

|a × b| =

(21 × 14 − 64)

= 230 = 15.17

Problem 8. If p = 4i + j −2k, q =3i − 2j + k and r = i −2k find (a) ( p −2q) × r (b) p × (2r × 3q).

and a 2 b2 cos2 θ = (a • b)2 gives:

(a) ( p − 2q) × r = [4i + j − 2k

(|a × b|)2 = (a • a)(b • b) − (a • b)2

− 2(3i − 2j + k)] × (i − 2k)

That is,

= (−2i + 5j − 4k) × (i − 2k)
|a × b| =

[(a • a)(b • b) − (a • b) ]
2

(7)

Problem 7. For the vectors a =i + 4j −2k and b =2i − j +3k find (i) a × b and (ii) |a × b|.
(i) From equation (5), i j k a × b = a1 a2 a3 b1 b2 b3

i

j

k

= −2 5 −4
1 0 −2 from equation (5)
=i

5 −4
−2 −4
−j
0 −2
1 −2
+k

−2 5
1 0

= i(−10 − 0) − j(4 + 4)

a a a a a a
= i 2 3 −j 1 3 +k 1 2 b2 b3 b1 b3 b1 b2

+ k(0 − 5), i.e.
( p − 2q) × r = −10i − 8j −5k

Hence i j k 4 −2 a×b = 1
2 −1
3
=i

4 −2
1 −2
1
4
−j
+k
−1
3
2
3
2 −1

= i(12 − 2) − j(3 + 4) + k(−1 − 8)
= 10i − 7j −9k

(b) (2r × 3q) = (2i − 4k) × (9i − 6j + 3k) i j k
= 2 0 −4
9 −6 3
= i(0 − 24) − j(6 + 36)
+ k(−12 − 0)
= −24i − 42j −12k

281

282 Higher Engineering Mathematics
The magnitude of M,

Hence

|M| = |r × F|

p × (2r × 3q) = (4i + j − 2k)

=

× (−24i − 42j − 12k)

r • r = (1)(1) + (2)(2) + (3)(3) = 14

i j k
=
4
1 −2
−24 −42 −12

F • F = (1)(1) + (2)(2) + (−3)(−3) = 14 r • F = (1)(1) + (2)(2) + (3)(−3) = −4

= i(−12 − 84) − j(−48 − 48)

[14 × 14 − (−4)2 ]

= 180 Nm = 13.42 Nm

|M| =

+ k(−168 + 24)
= −96i +96j − 144k or −48(2i − 2j +3k)
Practical applications of vector products
Problem 9. Find the moment and the magnitude of the moment of a force of (i + 2j −3k) newtons about point B having co-ordinates (0, 1, 1), when the force acts on a line through A whose co-ordinates are (1, 3, 4).
The moment M about point B of a force vector F which has a position vector of r from A is given by:

Problem 10. The axis of a circular cylinder coincides with the z-axis and it rotates with an angular velocity of (2i − 5j + 7k) rad/s. Determine the tangential velocity at a point P on the cylinder, whose co-ordinates are ( j + 3k) metres, and also determine the magnitude of the tangential velocity.
The velocity v of point P on a body rotating with angular velocity ω about a fixed axis is given by: v = ω × r, where r is the point on vector P. v = (2i − 5j + 7k) × ( j + 3k)

Thus
M =r×F

i j k
= 2 −5 7
0
1 3

r is the vector from B to A, i.e. r = BA.
But BA = BO + OA = OA − OB (see Problem 13, page 262), that is:

= i(−15 − 7) − j(6 − 0) + k(2 − 0)
= (−22i − 6j +2k) m/s

r = (i + 3j + 4k) − ( j + k)
= i + 2j + 3k

The magnitude of v,
|v| =

Moment,
M = r × F = (i + 2j + 3k) × (i + 2j − 3k) k i j
3
= 1 2
1 2 −3

[(ω • ω)(r • r) − (r • ω)2 ]

ω • ω = (2)(2) + (−5)(−5) + (7)(7) = 78 r • r = (0)(0) + (1)(1) + (3)(3) = 10 ω • r = (2)(0) + (−5)(1) + (7)(3) = 16
Hence,

= i(−6 − 6) − j(−3 − 3)
+ k(2 − 2)
= −12i + 6j Nm

[(r • r)(F • F) − (r • F)2 ]

|v| =

(78 × 10 − 162 )

= 524 m/s = 22.89 m/s

Scalar and vector products

283

Now try the following exercise
Exercise 113 products Further problems on vector

In problems 1 to 4, determine the quantities stated when p =3i +2k, q =i − 2j +3k and r =−4i +3j − k.
1. (a) p × q (b) q × p
[(a) 4i − 7j −6k (b) −4i + 7j +6k]
2. (a) |p × r| (b) |r × q|
[(a) 11.92 (b) 13.96]
3. (a) 2p × 3r (b) (p +r) × q
(a) −36i −30j −54k
(b) 11i +4j −k

magnitude about point Q having co-ordinates
(4, 0, −1) metres.
M = (5i + 8j − 2k) Nm,
|M| = 9.64 Nm
9. A sphere is rotating with angular velocity ω about the z-axis of a system, the axis coinciding with the axis of the sphere. Determine the velocity vector and its magnitude at position
(−5i +2j − 7k) m, when the angular velocity is (i + 2j) rad/s. υ = −14i +7j +12k,
|υ|= 19.72 m/s
10. Calculate the velocity vector and its magnitude for a particle rotating about the z-axis at an angular velocity of (3i − j +2k) rad/s when the position vector of the particle is at
(i − 5j +4k) m.
[6i −10j −14k, 18.22 m/s]

4. (a) p × (r × q) (b) (3p × 2r) × q
(a) −22i − j +33k
(b) 18i +162j +102k
5. For vectors p =4i − j +2k and q =−2i +3j − 2k determine: (i) p • q
(ii) p × q (iii) |p ×q| (iv) q × p and
(v) the angle between the vectors.


(i) −15 (ii) −4i + 4j +10k


⎣ (iii) 11.49 (iv) 4i −4j − 10k ⎦

26.4

Vector equation of a line

The equation of a straight line may be determined, given that it passes through the point A with position vector a relative to O, and is parallel to vector b. Let r be the position vector of a point P on the line, as shown in
Fig. 26.10.

(v) 142.55◦

b
P

6. For vectors a =−7i + 4j + 1 k and b =6i −
2
5j −k find (i) a • b (ii) a × b (iii) |a ×b|
(iv) b ×a and (v) the angle between the vectors. ⎤

(i) −62 1 (ii) −1 1 i − 4j +11k
2
2


⎣(iii) 11.80 (iv) 1 1 i +4j − 11k ⎦
2
(v) 169.31◦
7. Forces of (i + 3j), (−2i − j), (i − 2j) newtons act at three points having position vectors of
(2i + 5j), 4j and (−i + j) metres respectively.
Calculate the magnitude of the moment.

A

r a O

Figure 26.10

[10 Nm]
8. A force of (2i − j + k) newtons acts on a line through point P having co-ordinates (0, 3, 1) metres. Determine the moment vector and its

By vector addition, OP = OA + AP,
i.e. r = a +AP.
However, as the straight line through A is parallel to the free vector b (free vector means one that has the same

284 Higher Engineering Mathematics magnitude, direction and sense), then AP = λb, where λ is a scalar quantity. Hence, from above, r = a +λ b

Hence, the Cartesian equations are: x −2 y − 3 z − (−1)
=
=

1
−2
3

(8)

If, say, r = xi + yj + zk, a =a1 i +a2 j + a3k and b = b1 i + b2 j + b3 k, then from equation (8),

i.e. x −2 =

xi + yj + zk = (a1 i + a2 j + a3 k)

Problem 12.

+ λ(b1 i + b2 j + b3 k)
Hence x = a1 + λb1, y = a2 + λb2 and z = a3 + λb3 .
Solving for λ gives: x −a1 y − a2 z − a3
=
=

b1 b2 b3

2x − 1 y + 4 −z + 5
=
=
3
3
2
represents a straight line. Express this in vector form. Comparing the given equation with equation (9), shows that the coefficients of x, y and z need to be equal to unity. Thus

Problem 11. (a) Determine the vector equation of the line through the point with position vector
2i + 3j −k which is parallel to the vector i − 2j + 3k.
(b) Find the point on the line corresponding to λ =3 in the resulting equation of part (a).
(c) Express the vector equation of the line in standard Cartesian form.

r = a + λb
i.e. r = (2i + 3j −k) +λ(i − 2j + 3k) or r = (2 + λ)i + (3 − 2λ)j + (3λ − 1)k

which is the vector equation of the line.
(b) When λ =3,

r = 5i −3j + 8k.

(c) From equation (9),

The equation

(9)

Equation (9) is the standard Cartesian form for the vector equation of a straight line.

(a) From equation (8),

3−y z+1
=

2
3

y + 4 −z + 5
2x − 1
=
= becomes: 3
3
2 x− 3
2

1
2

=

y +4 z−5
=
3
−2

Again, comparing with equation (9), shows that
1
a1 = , a2 = −4 and a3 = 5 and
2
3 b1 = , b2 = 3 and b3 = −2
2
In vector form the equation is: r = (a1 + λb1 )i + (a2 + λb2 ) j + (a3 + λb3 )k, from equation (8)
i.e. r =

1 3
+ λ i + (−4 + 3λ) j + (5 − 2λ)k
2 2

1 or r = (1 + 3λ)i + (3λ − 4) j + (5 − 2λ)k
2

y − a2 z − a3 x − a1
=
=

b1 b2 b3
Now try the following exercise
Since a = 2i + 3j − k, then a1 = 2, a2 = 3 and a3 = −1 and b = i − 2j + 3k, then b1 = 1, b2 = −2 and b3 = 3

Exercise 114 Further problems on the vector equation of a line
1. Find the vector equation of the line through the point with position vector 5i −2j + 3k which

Scalar and vector products is parallel to the vector 2i + 7j −4k. Determine the point on the line corresponding to λ =2 in the resulting equation.

⎤ r = (5 + 2λ)i + (7λ − 2)j


+ (3 − 4λ)k; r = 9i + 12j − 5k
2. Express the vector equation of the line in problem 1 in standard Cartesian form. x −5 y +2 3−z
=
=

2
7
4

In problems 3 and 4, express the given straight line equations in vector form.
3.

3x − 1 5y + 1 4 − z
=
=
4
2
3
r = 1 (1 + 4λ)i + 1 (2λ − 1)j
3
5
+ (4 − 3λ)k

4. 2x + 1 =

1 −4y 3z −1
=
5
4
r = 1 (λ − 1)i + 1 (1 − 5λ)j
2
4
+ 1 (1 + 4λ)k
3

285

Revision Test 8
This Revision Test covers the material contained in Chapters 24 to 26. The marks for each question are shown in brackets at the end of each question.
1. State whether the following are scalar or vector quantities: (a) A temperature of 50◦C
(b) A downward force of 80 N
(c)

70 m distance

(f) An acceleration of 25 m/s2 at 30◦ to the horizontal (6)
2. Calculate the resultant and direction of the force vectors shown in Fig. RT8.1, correct to 2 decimal places. (7)
5N

7N

3. Four coplanar forces act at a point A as shown in Fig. RT8.2 Determine the value and direction of the resultant force by (a) drawing (b) by calculation using horizontal and vertical components.
(10)
4N
A
458

Figure RT8.2

5. If velocity v1 = 26 m/s at 52◦ and v2 = 17 m/s at −28◦ calculate the magnitude and direction of v1 + v2 , correct to 2 decimal places, using complex numbers.
(10)

7. If a = 2i + 4j −5k and b =3i − 2j +6k determine:
(i) a ·b (ii) |a +b| (iii) a × b (iv) the angle between a and b.
(14)
8. Determine the work done by a force of F newtons acting at a point A on a body, when A is displaced to point B, the co-ordinates of A and B being (2, 5,
−3) and (1, −3, 0) metres respectively, and when
F = 2i −5j + 4k newtons.
(4)

458

7N
8N

Plot the two voltages on the same axes to scales π of 1 cm = 50 volts and 1 cm = rad.
6
Obtain a sinusoidal expression for the resultant v1 + v2 in the form R sin(ωt + α): (a) by adding ordinates at intervals and (b) by calculation.
(13)

6. Given a = −3i + 3j + 5k, b = 2i − 5j + 7k and c = 3i + 6j − 4k, determine the following:
(i) −4b (ii) a + b − c (iii) 5b − 3c.
(8)

Figure RT8.1

5N

v1 = 150 sin(ωt + π/3) volts and v2 = 90 sin(ωt − π/6) volts

300 J of work

(d) A south-westerly wind of 15 knots
(e)

4. The instantaneous values of two alternating voltages are given by:

9. A force of F =3i −4j + k newtons acts on a line passing through a point P. Determine moment M and its magnitude of the force F about a point Q when P has co-ordinates (4, −1, 5) metres and Q has co-ordinates (4, 0, −3) metres.
(8)

Chapter 27

Methods of differentiation
27.1

Introduction to calculus

Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions – such as velocity and acceleration, rates of change and maximum and minimum values of curves. Calculus has widespread applications in science and engineering and is used to solve complicated problems for which algebra alone is insufficient.
Calculus is a subject that falls into two parts:
(i) differential calculus, or differentiation, which is covered in Chapters 27 to 36, and

f(x)
B

A
C

f(x2)

f(x1)
E
x1

0

D x2 x

Figure 27.2

(ii) integral calculus, or integration, which is covered in Chapters 37 to 44.

27.2

For the curve shown in Fig. 27.2, let the points A and
B have co-ordinates (x 1 , y1) and (x 2 , y2), respectively.
In functional notation, y1 = f (x 1 ) and y2 = f (x 2 ) as shown. The gradient of a curve

If a tangent is drawn at a point P on a curve, then the gradient of this tangent is said to be the gradient of the curve at P. In Fig. 27.1, the gradient of the curve at P is equal to the gradient of the tangent PQ. f (x)

The gradient of the chord AB
=

BC BD − CD f (x 2 ) − f (x 1 )
=
=
AC
ED
(x 2 − x 1 )

For the curve f (x) = x 2 shown in Fig. 27.3.
(i) the gradient of chord AB

Q

=
P

f (3) − f (1) 9 − 1
=
=4
3−1
2

(ii) the gradient of chord AC
0

Figure 27.1

x

=

f (2) − f (1) 4 − 1
=
=3
2−1
1

288 Higher Engineering Mathematics y f(x)
10

B

f(x) 5 x 2

8
B (x 1 ␦x, y 1 ␦y)
6
␦y

4

C

2

A

f(x 1 ␦x)

A(x, y)

D

␦x

f(x)
0

1

1.5

2

3

x x 0

Figure 27.3
Figure 27.4

(iii) the gradient of chord AD f (1.5) − f (1) 2.25 − 1
=
= 2.5
=
1.5 − 1
0.5
(iv) if E is the point on the curve (1.1, f (1.1)) then the gradient of chord AE
=

f (1.1) − f (1) 1.21 − 1
=
= 2.1
1.1 − 1
0.1

(v) if F is the point on the curve (1.01, f (1.01)) then the gradient of chord AF
=

f (1.01) − f (1) 1.0201 − 1
=
= 2.01
1.01 − 1
0.01

Thus as point B moves closer and closer to point A the gradient of the chord approaches nearer and nearer to the value 2. This is called the limiting value of the gradient of the chord AB and when B coincides with A the chord becomes the tangent to the curve.

27.3 Differentiation from first principles In Fig. 27.4, A and B are two points very close together on a curve, δx (delta x) and δy (delta y) representing small increments in the x and y directions, respectively. δy Gradient of chord AB = ; however, δx δy = f (x + δx) − f (x). δy f (x + δx) − f (x)
Hence
=
.
δx δx δy
As δx approaches zero, approaches a limiting value δx and the gradient of the chord approaches the gradient of the tangent at A.
When determining the gradient of a tangent to a curve there are two notations used. The gradient of the curve at A in Fig. 27.4 can either be written as limit δy

δx→0 δx

or limit

δx→0

In Leibniz notation,

f (x + δx) − f (x) δx δy dy = limit dx δx→0 δx

In functional notation, f (x) = limit δx→0 f (x +δx) − f (x) δx dy is the same as f (x) and is called the differential dx coefficient or the derivative. The process of finding the differential coefficient is called differentiation.
Problem 1. Differentiate from first principle f (x) = x 2 and determine the value of the gradient of the curve at x = 2.
To ‘differentiate from first principles’ means ‘to find f (x)’ by using the expression f (x) = limit

δx→0

f (x) = x 2

f (x + δx) − f (x) δx Methods of differentiation
Substituting (x + δx) for x gives f (x + δx) = (x + δx)2 = x 2 + 2xδx + δx 2 , hence

y
A

(x 2 + 2xδx + δx 2 ) − (x 2 ) f (x) = limit δx→0 δx
= limit

δx→0

+ δx 2 )

(2xδx δx (a)

27.4 Differentiation of common functions From differentiation by first principles of a number of examples such as in Problem 1 above, a general rule for differentiating y = ax n emerges, where a and n are constants. The rule is: if y = axn then f (x)= axn then f

dy
= anxn−1 dx (x)= anxn−1 ) and is true for all

(or, if real values of a and n.
For example, if y = 4x 3 then a = 4 and n =3, and dy = anx n−1 = (4)(3)x 3−1 = 12x 2 dx If y = ax n and n =0 then y = ax 0 and

dy
= (a)(0)x 0−1 = 0, dx i.e. the differential coefficient of a constant is zero.
Figure 27.5(a) shows a graph of y = sin x. The gradient is continually changing as the curve moves from dy 0 to A to B to C to D. The gradient, given by
, may dx be plotted in a corresponding position below y = sin x, as shown in Fig. 27.5(b).

B
0
Ϫ

D




2

3␲
2

2␲

x rad

C

δx→0

Differentiation from first principles can be a lengthy process and it would not be convenient to go through this procedure every time we want to differentiate a function.
In reality we do not have to because a set of general rules have evolved from the above procedure, which we consider in the following section.

y ϭ sin x

ϩ

= limit [2x + δx]
As δx → 0, [2x + δx] →[2x + 0]. Thus f (x) = 2x, i.e. the differential coefficient of x 2 is 2x. At x = 2, the gradient of the curve, f (x) = 2(2) = 4.

289

0Ј dy dx ϩ (b)

0
Ϫ



d
(sin x) ϭ cos x dx AЈ

2




3␲
2

2␲

x rad



Figure 27.5

(i) At 0, the gradient is positive and is at its steepest.
Hence 0 is a maximum positive value.
(ii) Between 0 and A the gradient is positive but is decreasing in value until at A the gradient is zero, shown as A .
(iii) Between A and B the gradient is negative but is increasing in value until at B the gradient is at its steepest negative value. Hence B is a maximum negative value.
(iv) If the gradient of y = sin x is further investigated dy between B and D then the resulting graph of dx is seen to be a cosine wave. Hence the rate of change of sin x is cos x,
i.e. if y = sin x then

dy
= cos x dx By a similar construction to that shown in Fig. 27.5 it may be shown that: if y = sin ax then

dy
= a cos ax dx If graphs of y = cos x, y = ex and y = ln x are plotted and their gradients investigated, their differential coefficients may be determined in a similar manner to that shown for y = sin x. The rate of change of a function is a measure of the derivative.

290 Higher Engineering Mathematics
The standard derivatives summarized below may be proved theoretically and are true for all real values of x

In general, the differential coefficient of a constant is always zero.
(b) Since y = 6x, in the general rule a = 6 and n =1.

y or f (x)

dy or f (x) dx ax n

anx n−1

sin ax

a cos ax

cos ax

−a sin ax

eax

aeax

ln ax

1 x The differential coefficient of a sum or difference is the sum or difference of the differential coefficients of the separate terms.

Hence

In general, the differential coefficient of kx, where k is a constant, is always k.
Problem 4. Find the derivatives of

5
(a) y = 3 x (b) y = √
3 4 x (a)

√ y = 3 x is rewritten in the standard differential
1

form as y = 3x 2 .
In the general rule, a = 3 and n =

Thus, if f (x) = p(x) + q(x) − r(x),
(where f, p, q and r are functions), then dy
= (6)(1)x 1−1 = 6x 0 = 6 dx Thus

f (x) = p (x) + q (x) − r (x)

1
1
3 1 dy = (3) x 2 −1 = x − 2 dx 2
2

Differentiation of common functions is demonstrated in the following worked problems.
Problem 2.

Find the differential coefficients of
12
(a) y = 12x 3 (b) y = 3 x If y = ax n then

dy
= anx n−1 dx (a) Since y = a = 12 and n =3 thus dy 3−1 = 36x2
= (12)(3)x dx 12
(b) y = 3 is rewritten in the standard ax n form as x y = 12x −3 and in the general rule a = 12 and n = − 3.
36
dy
Thus
= (12)( 3)x −3−1 = −36x −4 = − 4

dx x 12x 3 ,

Problem 3.
(a)

Differentiate (a) y = 6 (b) y = 6x.

y = 6 may be written as y = 6x 0 , i.e. in the general rule a = 6 and n =0.
Hence

dy
= (6)(0)x 0−1 = 0 dx =

3
2x

(b)

1
2

1
2

3
= √
2 x

4
5
5 y = √ = 4 = 5x − 3 in the standard differen3 4 x x3 tial form.
In the general rule, a = 5 and n =− 4
3

Thus

dy
4 − 4 −1 −20 − 7
= (5) − x 3 = x 3 dx 3
3
=

−20
7
3x 3

−20
= √
3
3 x7

Problem 5.

Differentiate, with respect to x,
1
1 y = 5x 4 + 4x − 2 + √ − 3.
2x
x y = 5x 4 + 4x −

1
1
+ √ − 3 is rewritten as
2x 2 x 1
1
y = 5x 4 + 4x − x −2 + x − 2 −3
2
When differentiating a sum, each term is differentiated in turn.

Methods of differentiation
Thus

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

1 − 1 −1 x 2 −0
2

1 3
= 20x 3 + 4 + x −3 − x − 2
2

dy
= (3)(4 cos 4x) dx = 12 cos 4x

(b) When f (t ) = 2 cos 3t then f (t ) = (2)(−3 sin 3t ) =−6 sin 3t

When y = 3e5x then

(b)

f (θ) =

dy
= (3)(5)e 5x = 15e5x dx Hence the gradient is −1 at the point (1, −2).

Now try the following exercise

In Problems 1 to 6 find the differential coefficients of the given functions with respect to the variable. 1.

(a) 5x 5 (b) 2.4x 3.5 (c)

2
= 2e−3θ , thus e3θ f (θ) = (2)(−3)e−30 = −6e−3θ =
(c)

When x = 1, y = 3(1)2 − 7(1) + 2 = −2

Exercise 115 Further problems on differentiating common functions

Problem 7. Determine the derivatives of
2
(a) y = 3e5x (b) f (θ) = 3θ (c) y = 6 ln 2x. e (a)

dy
= 6x − 7 dx Since the gradient is −1 then 6x − 7 =−1, from which, x =1

When y = 3x 2 − 7x + 2 then

Problem 6. Find the differential coefficients of
(a) y = 3 sin 4x (b) f (t ) = 2 cos3t with respect to the variable.
When y = 3 sin 4x then

Problem 9. Determine the co-ordinates of the point on the graph y = 3x 2 − 7x + 2 where the gradient is −1.
The gradient of the curve is given by the derivative.

dy
1
1
i.e.
= 20x3 + 4 + 3 − √ dx x
2 x3

(a)

When y = 6 ln 2x then

1 x (a) 25x 4 (b) 8.4x 2.5 (c) −
−6
e3θ

1 dy =6 dx x

=

6 x The gradient of a curve at a given point is given by the corresponding value of the derivative. Thus, since y = 3x 4 − 2x 2 + 5x − 2

(a)

3.

At the point (1, 4), x = 1
Thus the gradient =12(1)3 − 4(1) + 5 = 13.

1 x2 8
(b) 0 (c) 2 x3 √

4
3
(a) 2 x (b) 3 x 5 (c) √ x (a)


1
2
3
(a) √ (b) 5 x 2 (c) − √ x x3
4.

dy
= 12x 3 − 4x + 5 dx At the point (0, −2), x = 0
Thus the gradient =12(0)3 − 4(0) + 5 =5

−4
(b) 6 (c) 2x x2 2.

Problem 8. Find the gradient of the curve y = 3x 4 − 2x 2 + 5x − 2 at the points (0, −2) and (1, 4).

Then the gradient =

291

5.

−3
(a) √ (b) (x − 1)2 (c) 2 sin 3x
3
x


1
(a) √ (b) 2(x − 1) ⎥
3 4

x


(c) 6 cos 3x
(a) −4 cos 2x (b) 2e6x (c)

3 e5x (a) 8 sin 2x (b) 12e6x (c)

−15 e5x 292 Higher Engineering Mathematics

6.

7.

8.

9.

√ e x − e−x
1− x
(a) 4 ln 9x (b)
(c)
2 x ⎡
4
ex + e−x
(a)
(b)

x
2


1
−1
(c) 2 + √ x 2 x3

Using the product rule:





Find the gradient of the curve y = 2t 4 +
3t 3 − t + 4 at the points (0, 4) and (1, 8).
[−1, 16]
Find the co-ordinates of the point on the graph y = 5x 2 − 3x + 1 where the gradient
1 3 is 2.
2, 4
2
+ 2 ln 2θ − θ2 2
2 (cos 5θ + 3 sin 2θ) − 3θ e dy π (b) Evaluate in part (a) when θ = , dθ 2 correct to 4 significant figures.


−4 2
(a) 3 + + 10 sin 5θ

⎢ θ θ


6


−12 cos 2θ + 3θ

⎣ e (b) 22.30
(a)

dy
=
dx

Differentiate y =

ds
, correct to 3 significant figures, dt π when t = given
6

[3.29]
s = 3 sin t − 3 + t.

10. Evaluate

gives:
i.e.

u

dv dx ↓

Differentiation of a product

then

du dx ↓

dy
= 6x 2 cos 2x + 6x sin 2x dx = 6x(xcos 2x +sin 2x)

Note that the differential coefficient of a product is not obtained by merely differentiating each term and multiplying the two answers together. The product rule formula must be used when differentiating products.
Problem 11. Find the√ of change of y with rate respect to x given y = 3 x ln 2x.
The rate of change of y with respect to x is given by
1
√ y = 3 x ln 2x = 3x 2 ln 2x, which is a product.
1

Let u = 3x 2 and v = ln 2x dy dv du Then
= u
+ v dx dx dx ↓



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

= 3x 2 −1 + (ln 2x)
1

i.e.

3 −1 x 2
2

1 ln 2x
2

dy
3
1
= √ 1 + ln 2x dx 2 x When y = uv, and u and v are both functions of x, dv du dy =u
+v
dx dx dx

v



dy
= (3x 2 )(2 cos 2x) + (sin 2x)(6x) dx = 3x − 2 1 +

27.5

+

Problem 12.

Differentiate y = x 3 cos 3x ln x.

Let u = x 3 cos 3x (i.e. a product) and v = ln x

This is known as the product rule.
Then
Problem 10. Find the differential coefficient of y = 3x 2 sin 2x.
3x 2 sin 2x is a product of two terms 3x 2 and sin 2x
Let u = 3x 2 and v = sin 2x

dy dv du
=u
+v dx dx dx where

du
= (x 3 )(−3 sin 3x) + (cos 3x)(3x 2 ) dx and

dv 1
=
dx x

dy dx Methods of differentiation
Hence

1 dy = (x 3 cos 3x)
+ (ln x)[−3x 3 sin 3x dx x
+ 3x 2 cos 3x]

8. et ln t cos t et = x 2 cos 3x + 3x 2 ln x(cos 3x − x sin 3x) dy = x2 {cos 3x + 3 lnx(cos 3x −x sin 3x)} dx i.e.

Problem 13. Determine the rate of change of voltage, given v = 5t sin 2t volts when t = 0.2 s. dv dt
= (5t )(2 cos 2t ) + (sin 2t )(5)

Rate of change of voltage =
= 10t cos 2t + 5 sin 2t

= 2 cos 0.4 + 5 sin 0.4 (where cos 0.4 means the cosine of 0.4 radians) dv = 2(0.92106) + 5(0.38942)
Hence
dt
= 1.8421 + 1.9471 = 3.7892
i.e. the rate of change of voltage when t = 0.2 s is
3.79 volts/s, correct to 3 significant figures.

Now try the following exercise
Exercise 116 Further problems on differentiating products
In Problems 1 to 8 differentiate the given products with respect to the variable.
1.

x sin x

[x cos x + sin x]

2.

x 2 e2x

[2x e2x (x + 1)]

3.

x 2 ln x

[x(1 + 2 ln x)]

4. 2x 3 cos 3x

x 3 ln 3x
5.

[6x 2(cos 3x − x sin 3x)]

x 1 + 3 ln 3x
2

6. e3t sin 4t

di
, correct to 4 significant figures, dt when t = 0.1, and i = 15t sin 3t .
[8.732]
dz
10. Evaluate
, correct to 4 significant figures, dt when t = 0.5, given that z =2e3t sin 2t .
[32.31]

e4θ

1
+ 4 ln 3θ θ Differentiation of a quotient

u
When y = , and u and v are both functions of x v du dv dy v dx − u dx then = dx v2
This is known as the quotient rule.
Problem 14. Find the differential coefficient of
4 sin 5x y= 5x 4
4 sin 5x is a quotient. Let u = 4 sin 5x and v = 5x 4
5x 4
(Note that v is always the denominator and u the numerator.) du dv dy v dx − u dx
=
dx v2 du where = (4)(5) cos 5x = 20 cos5x dx dv and = (5)(4)x 3 = 20x 3 dx dy (5x 4 )(20 cos 5x) − (4 sin 5x)(20x 3 )
Hence
= dx (5x 4 )2

[e3t (4 cos 4t + 3 sin 4t )]

7. e4θ ln 3θ

1
+ ln t cos t − ln t sin t t 9. Evaluate

27.6

dv
When t = 0.2,
= 10(0.2) cos 2(0.2) + 5 sin 2(0.2) dt 293

=
=
i.e.

100x 4 cos 5x − 80x 3 sin 5x
25x 8
20x 3 [5x cos 5x − 4 sin 5x]
25x 8

dy
4
= 5 (5x cos 5x − 4 sin 5x) dx 5x

294 Higher Engineering Mathematics
Note that the differential coefficient is not obtained by merely differentiating each term in turn and then dividing the numerator by the denominator. The quotient formula must be used when differentiating quotients.
Problem 15. Determine the differential coefficient of y = tan ax.

Let u = t e2t and v = 2 cos t then du dv
= (t )(2e2t ) + (e2t )(1) and
= −2 sin t dt dt du dv dy v dx − u dx
=
Hence dx v2
=
=

4t e2t cos t + 2e2t cos t + 2t e2t sin t
4 cos2 t

=

sin ax
. Differentiation of tan ax is thus cos ax treated as a quotient with u = sin ax and v = cos ax

(2 cos t )[2t e2t + e2t ] − (t e2t )(−2 sin t )
(2 cos t )2

2e2t [2t cos t + cos t + t sin t ]
4 cos2 t

y = tan ax =

du dv dy v dx − u dx
=
dx v2 (cos ax)(a cos ax) − (sin ax)(−a sin ax)
=
(cos ax)2 a cos2 ax + a sin2 ax a(cos2 ax + sin2 ax)
=
(cos ax)2 cos2 ax a =
, sincecos2 ax + sin2 ax = 1 cos2 ax
(see Chapter 15)
=

dy
1
Hence
= a sec2 ax since sec2 ax =
(see
dx cos2 ax
Chapter 11).
Problem 16.

Find the derivative of y = sec ax.

1 y = sec ax =
(i.e. a quotient). Let u = 1 and cos ax v = cos ax du dv v −u dy = dx 2 dx dx v
=

i.e.

a sin ax
1
=a
2 ax cos cos ax

Problem 18. Determine the gradient of the curve


3
5x
.
3, at the point y= 2
2x + 4
2
Let y = 5x and v = 2x 2 + 4 du dv v −u
2
dy dx dx = (2x + 4)(5) − (5x)(4x)
=
2 dx v
(2x 2 + 4)2
10x 2 + 20 − 20x 2
20 − 10x 2
=
(2x 2 + 4)2
(2x 2 + 4)2



3
, x = 3,
At the point
3,
2

dy
20 − 10( 3)2

hence the gradient =
=
dx
[2( 3)2 + 4]2
=

(cos ax)(0) − (1)(−a sin ax)
(cos ax)2

=

dy e2t =
(2t cos t + cos t +t sin t) dx 2 cos2 t

i.e.

sin ax cos ax

dy
= a sec ax tan ax dx Problem 17.

Differentiate y =

=

20 − 30
1
=−
100
10

Now try the following exercise
Exercise 117
Further problems on differentiating quotients

t e2t
2 cost

t e2t
The function is a quotient, whose numerator is a
2 cost product. In Problems 1 to 7, differentiate the quotients with respect to the variable.
1.

sin x x x cos x − sin x x2 Methods of differentiation

2.

3.

−6
(x sin 3x + cos 3x) x4 2 cos3x x3 2x x2 + 1


cos x

2 x

x cos x

4.

cos2 x


3 θ3
2 sin 2θ

5.


3 θ(3 sin 2θ − 4θ cos 2θ)
4 sin2 2θ


6.

7.




ln 2t

t
2xe4x
sin x

2(1 − x 2 )
(x 2 + 1)2

+ x sin x



2e4x
{(1 + 4x) sin x − x cos x} sin2 x

the point (2, −4).

2x x2 − 5

[−18]

If y is a function of x then

du dy = 9u 8 and
=3
du dx dy dy du
=
×
= (9u 8 )(3) = 27u 8 dx du dx

dy du = 10x and
= −3 sin u. dx du

Using the function of a function rule, dy dy du
=
×
= (−3 sin u)(10x) = −30x sin u dx du dx
Rewriting u as 5x 2 + 2 gives: dy = −30x sin(5x2 + 2) dx Problem 20. Find the derivative of y = (4t 3 − 3t )6 .
Let u =4t 3 − 3t , then y = u 6 du dy
= 12t 2 − 3 and
= 6u 5 dt du
Using the function of a function rule,
Hence

dy dy du
=
×
= (6u 5 )(12t 2 − 3) dx du dx
Rewriting u as (4t 3 − 3t ) gives: dy = 6(4t 3 − 3t )5 (12t 2 − 3) dt dy dy du
=
× dx du dx

This is known as the ‘function of a function’ rule (or sometimes the chain rule).
For example, if y = (3x − 1)9 then, by making the substitution u = (3x − 1), y = u 9 , which is of the ‘standard’ form. Then

Let u =5x 2 + 2 then y = 3 cosu

Function of a function

It is often easier to make a substitution before differentiating.

Hence

Problem 19. Differentiate y = 3 cos(5x 2 + 2).

at

dy at x = 2.5, correct to 3 significant
9. Evaluate dx 2x 2 + 3
.
figures, given y = ln 2x
[3.82]

27.7

dy
= 27(3x −1)8 dx Since y is a function of u, and u is a function of x, then y is a function of a function of x.

Hence

1
1 − ln 2t

2


3
t

8. Find the gradient of the curve y =

Rewriting u as (3x − 1) gives:

295

= 18(4t 2 − 1)(4t 3 − 3t)5
Problem 21. Determine the differential coefficient of y = (3x 2 + 4x − 1).
1

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

Let u =3x 2 + 4x − 1 then y = u 2
Hence

du
1
dy 1 − 1
= 6x + 4 and
= u 2= √ dx du 2
2 u

296 Higher Engineering Mathematics
Using the function of a function rule,
3.

2 sin(3θ − 2)

4.

2 cos5 α

[− cos4 α sin α]
10

5.

1
(x 3 − 2x + 1)5

5(2 − 3x 2 )
(x 3 − 2x + 1)6

6.

5e2t +1

7.

2 cot(5t 2 + 3)

[−20t cosec2 (5t 2 + 3)]

du
= 3 sec2 3x, (from Problem 15), and dx 8.

6 tan(3y + 1)

[18 sec2 (3y + 1)]

dy
= 12u 3 du 9.

2etan θ

dy dy du
=
×
=
dx du dx
i.e.

dy
=
dx

1
3x + 2
√ (6x + 4) = √
2 u u 3x + 2
(3x2 + 4x − 1)

Problem 22.

Differentiate y = 3 tan4 3x.

Let u = tan 3x then y = 3u 4
Hence

Then

dy dy du
=
×
= (12u 3 )(3 sec2 3x) dx du dx
= 12(tan 3x)3 (3 sec2 3x)

[6 cos(3θ − 2)]

[10e2t +1]

[2 sec2 θ etan θ ]

π with respect to θ,
10. Differentiate θ sin θ −
3
and evaluate, correct to 3 significant figures, π [1.86] when θ = .
2

dy
= 36 tan3 3x sec2 3x dx i.e.

Problem 23. Find the differential coefficient of
2
y= 3
(2t − 5)4
2

y=

(2t 3 − 5)4 y = 2u −4

= 2(2t 3 − 5)−4 . Let u = (2t 3 − 5), then

Hence

du dy −8
= 6t 2 and
= −8u −5 = 5 dt du u Then

dy dy du
=
×
=
dt du dt =

−8
(6t 2 ) u5 −48t 2

Exercise 118 Further problems on the function of a function
In Problems 1 to 9, find the differential coefficients with respect to the variable.

2.

(2x 3 − 5x)5

When a function y = f (x) is differentiated with respect dy to x the differential coefficient is written as or f (x). dx If the expression is differentiated again, the second difd2 y ferential coefficient is obtained and is written as dx 2
(pronounced dee two y by dee x squared) or f (x)
(pronounced f double-dash x).
By successive differentiation further higher derivatives d4 y d3 y such as 3 and 4 may be obtained. dx dx dy d2 y
= 12x 3 , 2 = 36x 2 , dx dx

d3 y d4 y d5 y
= 72x, 4 = 72 and 5 = 0. dx 3 dx dx

Now try the following exercise

(2x − 1)6

Successive differentiation

Thus if y = 3x 4 ,

(2t 3 − 5)5

1.

27.8

[12(2x − 1)5 ]
[5(6x 2 − 5)(2x 3 − 5x)4 ]

Problem 24. f (x).

If f (x) = 2x 5 − 4x 3 + 3x − 5, find

f (x) = 2x 5 − 4x 3 + 3x − 5 f (x) = 10x 4 − 12x 2 + 3 f (x) = 40x 3 − 24x = 4x(10x2 − 6)

Methods of differentiation
Problem 25. If y = cos x − sin x, evaluate x, in π d2 y the range 0 ≤ x ≤ , when 2 is zero.
2
dx
Since

y = cos x − sin x,

d2 y
= −cos x + sin x. dx 2

dy
= −sin x − cos x dx d2 y
Problem 27. Evaluate 2 when θ = 0 given dθ y = 4 sec 2θ. and Since y = 4 sec 2θ, then d2 y is zero, −cos x + sin x = 0, dx 2 sin x
i.e. sin x = cos x or
= 1. cos x
When

π
Hence tan x = 1 and x =arctan1 =45◦ or rads in the
4
π range 0 ≤ x ≤
2

dy
= (4)(2) sec 2θ tan 2θ (from Problem 16) dθ = 8 sec 2θ tan 2θ (i.e. a product) d2 y
= (8 sec 2θ)(2 sec 2 2θ) dθ 2
+ (tan 2θ)[(8)(2) sec 2θ tan 2θ]
= 16 sec3 2θ + 16 sec 2θ tan2 2θ

When

θ = 0,

= 16(1) + 16(1)(0) = 16.

Problem 26. Given y = 2xe−3x show that d2 y dy + 6 + 9y = 0. dx 2 dx Now try the following exercise

y = 2xe−3x (i.e. a product)
Hence

Exercise 119 Further problems on successive differentiation

dy
= (2x)(−3e−3x ) + (e −3x )(2) dx 1. If y = 3x 4 + 2x 3 − 3x + 2 find

= −6xe−3x + 2e−3x

(a)

d2 y
= [(−6x)(−3e−3x ) + (e−3x )(−6)] dx 2
+ (−6e−3x )
= 18xe−3x − 6e−3x − 6e−3x
i.e.

d2 y
= 18xe−3x − 12e−3x dx 2

Substituting values into

d2 y
= 16 sec3 0 + 16 sec 0 tan2 0 dθ 2

d2 y dy + 6 + 9y gives: dx 2 dx (18xe−3x − 12e−3x ) + 6(−6xe−3x + 2e−3x )
+ 9(2xe−3x ) = 18xe−3x − 12e−3x − 36xe−3x
+ 12e−3x + 18xe−3x = 0 d2 y dy Thus when y = 2xe−3x , 2 + 6 + 9y = 0 dx dx

2.

d2 y d3 y
(b) 3 . dx 2 dx [(a) 36x 2 + 12x (b) 72x + 12]
1 3 √
2
f (t ) = t 2 − 3 + − t + 1
5
t t determine f (t ).

(a) Given

(b) Evaluate f (t ) when t = 1.


6
1
4 12
(a) − 5 + 3 + √

5 t t 4 t3 ⎥


(b) −4.95
In Problems 3 and 4, find the second differential coefficient with respect to the variable.
3. (a) 3 sin 2t + cos t (b) 2 ln 4θ
(a) −(12 sin 2t + cos t ) (b)

−2 θ2 4. (a) 2 cos2 x (b) (2x − 3)4
[(a) 4(sin2 x − cos2 x) (b) 48(2x − 3)2 ]

297

298 Higher Engineering Mathematics
5. Evaluate f (θ) when θ = 0 given f (θ) = 2 sec 3θ.
6. Show that the differential equation dy d2 y
− 4 + 4y = 0 is satisfied dx 2 dx when y = xe2x .

[18]

7. Show that, if P and Q are constants and y = P cos(ln t ) +Q sin(ln t ), then t2 d2 y dy +t
+y=0
dt 2 dt Chapter 28

Some applications of differentiation 28.1

Rates of change

If a quantity y depends on and varies with a quantity dy x then the rate of change of y with respect to x is
.
dx
Thus, for example, the rate of change of pressure p with dp height h is
.
dh
A rate of change with respect to time is usually just called ‘the rate of change’, the ‘with respect to time’ being assumed. Thus, for example, a rate of change of di current, i, is and a rate of change of temperature, dt dθ θ, is
, and so on. dt Problem 1. The length l metres of a certain metal rod at temperature θ ◦ C is given by l = 1 + 0.00005θ + 0.0000004θ 2. Determine the rate of change of length, in mm/◦ C, when the temperature is (a) 100◦ C and (b) 400◦C. dl The rate of change of length means
.

Since length then (a)

l = 1 +0.00005θ + 0.0000004θ 2, dl = 0.00005 + 0.0000008θ dθ When θ = 100◦C, dl = 0.00005 + (0.0000008)(100) dθ = 0.00013 m/◦C
= 0.13 mm/◦ C

(b) When θ = 400◦C, dl = 0.00005 + (0.0000008)(400) dθ = 0.00037 m/◦C
= 0.37 mm/◦ C
Problem 2. The luminous intensity I candelas of a lamp at varying voltage V is given by
I = 4 ×10−4 V 2 . Determine the voltage at which the light is increasing at a rate of 0.6 candelas per volt.
The rate of change of light with respect to voltage is dI given by
.
dV
Since

I = 4 × 10−4 V 2 , dI = (4 × 10−4)(2)V = 8 × 10−4 V dV When the light is increasing at 0.6 candelas per volt then
+0.6 = 8 × 10−4 V , from which, voltage
V=

0.6
= 0.075 × 10+4
8 × 10−4

= 750 volts
Problem 3. Newtons law of cooling is given by θ = θ0 e−kt , where the excess of temperature at zero

time is θ0 C and at time t seconds is θ ◦ C. Determine the rate of change of temperature after 40 s, given that θ0 = 16◦C and k = −0.03

300 Higher Engineering Mathematics
The rate of change of temperature is
Since

θ = θ0 e−kt dθ = (θ0 )(−k)e−kt = −kθ0 e−kt dt then
When
then

dθ dt θ0 = 16, k = −0.03 and t = 40 dθ = −(−0.03)(16)e−(−0.03)(40) dt = 0.48e1.2 = 1.594◦C/s

Problem 4. The displacement s cm of the end of a stiff spring at time t seconds is given by s = ae−kt sin 2π f t . Determine the velocity of the end of the spring after 1 s, if a = 2, k = 0.9 and f = 5. ds Velocity, v = where s = ae−kt sin 2π f t (i.e. a dt product).
Using the product rule, ds = (ae−kt )(2π f cos 2π f t ) dt + (sin 2π f t )(−ake−kt )

rate of change of current when t = 20 ms, given that f = 150 Hz.
[3000π A/s]
2. The luminous intensity, I candelas, of a lamp is given by I = 6 × 10−4 V 2 , where V is the voltage. Find (a) the rate of change of luminous intensity with voltage when V = 200 volts, and
(b) the voltage at which the light is increasing at a rate of 0.3 candelas per volt.
[(a) 0.24 cd/V (b) 250 V]
3. The voltage across the plates of a capacitor at any time t seconds is given by v = V e−t /C R , where V , C and R are constants.
Given V = 300 volts, C = 0.12 × 10−6 F and
R = 4 ×106 find (a) the initial rate of change of voltage, and (b) the rate of change of voltage after 0.5 s.
[(a) −625 V/s (b) −220.5 V/s]
4. The pressure p of the atmosphere at height h above ground level is given by p = p0e−h/c , where p0 is the pressure at ground level and c is a constant. Determine the rate of change of pressure with height when p0 = 1.013 × 105 pascals and c = 6.05 × 104 at
1450 metres.
[−1.635 Pa/m]

When a = 2, k = 0.9, f = 5 and t = 1, velocity, v = (2e−0.9 )(2π5 cos 2π5)
+ (sin 2π5)(−2)(0.9)e−0.9
= 25.5455 cos10π − 0.7318 sin 10π
= 25.5455(1) − 0.7318(0)
= 25.55 cm/s
(Note that cos10π means ‘the cosine of 10π radians’, not degrees, and cos 10π ≡ cos 2π = 1.)
Now try the following exercise
Exercise 120 change Further problems on rates of

1. An alternating current, i amperes, is given by i = 10 sin 2πf t , where f is the frequency in hertz and t the time in seconds. Determine the

28.2

Velocity and acceleration

When a car moves a distance x metres in a time t seconds along a straight road, if the velocity v is constant then x v = m/s, i.e. the gradient of the distance/time graph t shown in Fig. 28.1 is constant.
If, however, the velocity of the car is not constant then the distance/time graph will not be a straight line. It may be as shown in Fig. 28.2.
The average velocity over a small time δt and distance δx is given by the gradient of the chord AB, i.e. the δx . average velocity over time δt is δt As δt → 0, the chord AB becomes a tangent, such that at point A, the velocity is given by: v= dx dt Hence the velocity of the car at any instant is given by the gradient of the distance/time graph. If an expression

Velocity

Distance

Some applications of differentiation

x

301

D

␦v t C
␦t
Time

Time

Figure 28.1

Distance

Figure 28.3

The acceleration is given by the second differential coefficient of distance x with respect to time t .
Summarizing, if a body moves a distance x metres in a time t seconds then:

B

(i) distance x = f(t).
␦x
A
␦t
Time

dx
, which is the gradient of
(ii) velocity v = f (t) or dt the distance/time graph. d2 x dv (iii) acceleration a = = f (t) or 2 , which is the dt dt gradient of the velocity/time graph.

Figure 28.2

for the distance x is known in terms of time t then the velocity is obtained by differentiating the expression.
The acceleration a of the car is defined as the rate of change of velocity. A velocity/time graph is shown in Fig. 28.3. If δv is the change in v and δt the δv corresponding change in time, then a = . δt As δt → 0, the chord CD becomes a tangent, such that at point C, the acceleration is given by: a= dv dt Hence the acceleration of the car at any instant is given by the gradient of the velocity/time graph. If an expression for velocity is known in terms of time t then the acceleration is obtained by differentiating the expression. dx dv Acceleration a = . However, v = . Hence dt dt d2 x d dx
= 2 a= dt dt dx Problem 5. The distance x metres moved by a car in a time t seconds is given by x = 3t 3 − 2t 2 + 4t − 1. Determine the velocity and acceleration when (a) t = 0 and (b) t = 1.5 s.
Distance

x = 3t 3 − 2t 2 + 4t − 1 m

Velocity

v=

Acceleration a =
(a)

dx
= 9t 2 − 4t + 4 m/s dt d2 x
= 18t − 4 m/s2 dx 2

When time t = 0, velocity v = 9(0)2 − 4(0) + 4 =4 m/s and acceleration a = 18(0) − 4 = −4 m/s2 (i.e. deceleration) (b) When time t = 1.5 s, velocity v = 9(1.5)2 − 4(1.5) + 4 =18.25 m/s and acceleration a = 18(1.5) − 4 =23 m/s2

a

302 Higher Engineering Mathematics
Problem 6. Supplies are dropped from a helicoptor and the distance fallen in a time t seconds is given by x = 1 gt 2, where g = 9.8 m/s2.
2
Determine the velocity and acceleration of the supplies after it has fallen for 2 seconds.
1
1 x = gt 2 = (9.8)t 2 = 4.9t 2 m
2
2 dv v=
= 9.8t m/s dt Distance
Velocity
and acceleration

a=

d2 x
= 9.8 m/s2 dt 2

When time t = 2 s, velocity, v = (9.8)(2) = 19.6 m/s and acceleration a = 9.8 m/s2
(which is acceleration due to gravity).
Problem 7. The distance x metres travelled by a vehicle in time t seconds after the brakes are applied is given by x = 20t − 5 t 2. Determine (a) the
3
speed of the vehicle (in km/h) at the instant the brakes are applied, and (b) the distance the car travels before it stops.
(a) Distance, x = 20t − 5 t 2.
3

Problem 8. The angular displacement θ radians of a flywheel varies with time t seconds and follows the equation θ = 9t 2 − 2t 3 . Determine (a) the angular velocity and acceleration of the flywheel when time, t = 1 s, and (b) the time when the angular acceleration is zero.
(a) Angular displacement θ = 9t 2 − 2t 3 rad
Angular velocity ω =


= 18t − 6t 2 rad/s dt When time t = 1 s, ω = 18(1) − 6(1)2 = 12 rad/s
Angular acceleration α =
When time t = 1 s,

d2θ
= 18 − 12t rad/s2 dt 2

α = 18 − 12(1) = 6 rad/s2
(b) When the angular acceleration is zero,
18 − 12t = 0, from which, 18 =12t , giving time, t = 1.5 s.
Problem 9. The displacement x cm of the slide valve of an engine is given by x = 2.2 cos 5πt + 3.6 sin 5πt . Evaluate the velocity (in m/s) when time t = 30 ms.

10 dx = 20 − t . dt 3
At the instant the brakes are applied, time = 0.

Displacement x = 2.2 cos 5πt + 3.6 sin 5πt

Hence velocity, v = 20 m/s

Velocity v =

Hence velocity v =

=

20 × 60 × 60 km/h 1000

= 72 km/h
(Note: changing from m/s to km/h merely involves multiplying by 3.6.)
(b) When the car finally stops, the velocity is zero, i.e.
10
10 v = 20 − t = 0, from which, 20 = t , giving
3
3 t = 6 s.
Hence the distance travelled before the car stops is given by: x = 20t − 5 t 2 = 20(6) − 5 (6)2
3
3
= 120 − 60 = 60 m

dx dt = (2.2)(−5π) sin 5πt + (3.6)(5π) cos 5πt
= −11π sin 5πt + 18π cos 5πt cm/s
When time t = 30 ms, velocity
= −11π sin 5π ·

30
30
+ 18π cos 5π · 3
3
10
10

= −11π sin 0.4712 + 18π cos 0.4712
= −11π sin 27◦ + 18π cos 27◦
= −15.69 + 50.39 = 34.7 cm/s
= 0.347 m/s

303

Some applications of differentiation
Now try the following exercise



(a) 100 m/s (b) 4 s
(c) 200 m

(d) −100 m/s

2. The distance s metres travelled by a car in t seconds after the brakes are applied is given by s = 25t − 2.5t 2. Find (a) the speed of the car (in km/h) when the brakes are applied, (b) the distance the car travels before it stops. [(a) 90 km/h (b) 62.5 m]
3. The equation θ = 10π + 24t − 3t 2 gives the angle θ, in radians, through which a wheel turns in t seconds. Determine (a) the time the wheel takes to come to rest, (b) the angle turned through in the last second of movement. [(a) 4 s (b) 3 rads]
4. At any time t seconds the distance x metres of a particle moving in a straight line from a fixed point is given by x = 4t + ln(1 − t ).
Determine (a) the initial velocity and acceleration (b) the velocity and acceleration after 1.5 s (c) the time when the velocity is zero. ⎤

(a) 3 m/s; −1 m/s2


⎢(b) 6 m/s; −4 m/s2⎥


(c)

(c) t = 6.28 s
6.

20t 3 23t 2 x= −
+ 6t + 5 represents the dis3
2
tance, x metres, moved by a body in t seconds.
Determine (a) the velocity and acceleration at the start, (b) the velocity and acceleration when t = 3 s, (c) the values of t when the body is at rest, (d) the value of t when the acceleration is 37 m/s2 and (e) the distance travelled in the third second.


(a) 6 m/s; −23 m/s2


⎢(b) 117 m/s; 97 m/s2⎥



⎢(c) 3 s or 2 s


4
5



⎣(d) 1 1 s
2
(e) 75 1 m
6

28.3

Turning points

In Fig. 28.4, the gradient (or rate of change) of the curve changes from positive between O and P to negative between P and Q, and then positive again between Q and R. At point P, the gradient is zero and, as x increases, the gradient of the curve changes from positive just before P to negative just after. Such a point is called a maximum point and appears as the
‘crest of a wave’. At point Q, the gradient is also zero and, as x increases, the gradient of the curve changes from negative just before Q to positive just after. Such a point is called a minimum point, and appears as the
‘bottom of a valley’. Points such as P and Q are given the general name of turning points. y R

3
4s

5. The angular displacement θ of a rotating disc is t given by θ = 6 sin , where t is the time in sec4 onds. Determine (a) the angular velocity of the disc when t is 1.5 s, (b) the angular acceleration when t is 5.5 s, and (c) the first time when the angular velocity is zero.





⎣(b) α = −0.37 rad/s2 ⎦

Exercise 121 Further problems on velocity and acceleration
1. A missile fired from ground level rises x metres vertically upwards in t seconds and
25
x = 100t − t 2. Find (a) the initial velocity
2
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) ω = 1.40 rad/s

P
Positive
gradient

O

Negative gradient Positive gradient Q x Figure 28.4

304 Higher Engineering Mathematics
It is possible to have a turning point, the gradient on either side of which is the same. Such a point is given the special name of a point of inflexion, and examples are shown in Fig. 28.5.
Maximum
point

y
Maximum
point

Problem 10. Locate the turning point on the curve y = 3x 2 − 6x and determine its nature by examining the sign of the gradient on either side.

dy
= 6x − 6. dx dy
= 0. Hence 6x − 6 = 0,
(ii) At a turning point, dx from which, x = 1.
(i) Since y = 3x 2 − 6x,

x

Minimum point

Maximum and minimum points and points of inflexion are given the general term of stationary points. Procedure for finding and distinguishing between stationary points:
(i) Given y = f (x), determine

(iii) When x = 1, y = 3(1)2 − 6(1) = −3.
Hence the co-ordinates of the turning point are (1, −3).

Figure 28.5

(ii) Let

positive to positive or negative to negative— the point is a point of inflexion.

Following the above procedure:
Points of inflexion 0

(c)

dy
(i.e. f (x)) dx dy
= 0 and solve for the values of x. dx (iii) Substitute the values of x into the original equation, y = f (x), to find the corresponding yordinate values. This establishes the co-ordinates of the stationary points.

(iv) If x is slightly less than 1, say, 0.9, then dy = 6(0.9) − 6 = −0.6, dx i.e. negative.
If x is slightly greater than 1, say, 1.1, then dy = 6(1.1) − 6 = 0.6, dx i.e. positive.
Since the gradient of the curve is negative just before the turning point and positive just after
(i.e. − ∨ +), (1, −3) is a minimum point.

To determine the nature of the stationary points:
Either

Problem 11. Find the maximum and minimum values of the curve y = x 3 − 3x + 5 by

d2 y and substitute into it the values of x
(iv) Find dx 2 found in (ii).
If the result is:
(a) positive—the point is a minimum one,
(b) negative—the point is a maximum one,
(c) zero—the point is a point of inflexion, or (a) examining the gradient on either side of the turning points, and

(v) Determine the sign of the gradient of the curve just before and just after the stationary points. If the sign change for the gradient of the curve is:
(a) positive to negative—the point is a maximum one, (b) negative to positive—the point is a minimum one, (b) determining the sign of the second derivative. dy = 3x 2 − 3 dx dy
For a maximum or minimum value
=0
dx
Since y = x 3 − 3x + 5 then

Hence 3x 2 − 3 = 0, from which, 3x 2 = 3 and x = ± 1
When x = 1, y = (1)3 − 3(1) + 5 =3
When x = −1, y = (−1)3 − 3(−1) + 5 =7
Hence (1, 3) and (−1, 7) are the co-ordinates of the turning points.

Some applications of differentiation
(a)

Considering the point (1, 3):
If x is slightly less than 1, say 0.9, then dy = 3(0.9)2 − 3, dx which is negative.
If x is slightly more than 1, say 1.1, then dy = 3(1.1)2 − 3, dx which is positive.
Since the gradient changes from negative to positive, the point (1, 3) is a minimum point.
Considering the point (−1, 7):
If x is slightly less than −1, say −1.1, then dy = 3(−1.1)2 − 3, dx which is positive.
If x is slightly more than −1, say −0.9, then dy = 3(−0.9)2 − 3, dx which is negative.

d2 y

dy
= 3x 2 − 3, then 2 = 6x dx dx d2 y
When x = 1, is positive, hence (1, 3) is a dx 2 minimum value. d2 y is negative, hence (−1, 7) is
When x = −1, dx 2 a maximum value.
Thus the maximum value is 7 and the minimum value is 3.
It can be seen that the second differential method of determining the nature of the turning points is, in this case, quicker than investigating the gradient.

Problem 12. Locate the turning point on the following curve and determine whether it is a maximum or minimum point: y = 4θ + e−θ . y = 4θ + e−θ dy then
= 4 − e−θ = 0 dθ for a maximum or minimum value.

Since

1
4

= eθ, giving θ = ln 1 = −1.3863 (see
4

When θ = − 1.3863, y = 4(−1.3863) + e−(−1.3863)
= 5.5452 +4.0000 = −1.5452
Thus (−1.3863, −1.5452) are the co-ordinates of the turning point. d2 y
= e−θ . dθ 2
When θ = −1.3863, d2 y
= e+1.3863 = 4.0, dθ 2 which is positive, hence (−1.3863, −1.5452) is a minimum point.
Problem 13. Determine the co-ordinates of the maximum and minimum values of the graph x3 x2
5
y = − − 6x + and distinguish between
3
2
3
them. Sketch the graph.
Following the given procedure:

Since the gradient changes from positive to negative, the point (−1, 7) is a maximum point.
(b) Since

Hence 4 = e−θ ,
Chapter 4).

305

(i) Since y =

5 x3 x2
− − 6x + then
3
2
3

dy
= x2 − x −6 dx dy
= 0. Hence dx x 2 − x − 6 = 0, i.e. (x + 2)(x − 3) = 0,

(ii) At a turning point,

from which x = −2 or x = 3.
(iii) When x = −2, y= 5
(−2)3 (−2)2

− 6(−2) + = 9
3
2
3

When x = 3, y= 5
5
(3)3 (3)2

− 6(3) + = −11
3
2
3
6

Thus the co-ordinates of the turning points are
(−2, 9) and 3, −11 5 .
6
d2 y dy = x 2 − x − 6 then 2 = 2x−1. dx dx
When x = −2,

(iv) Since

d2 y
= 2(−2) − 1 = −5, dx 2 which is negative.

306 Higher Engineering Mathematics
When x = 126.87◦,

Hence (−2, 9) is a maximum point.
When x = 3,

y = 4 sin 126.87◦ − 3 cos126.87◦ = 5

d2 y
= 2(3) − 1 = 5, dx 2 which is positive.

When x = 306.87◦, y = 4 sin 306.87◦ − 3 cos 306.87◦ = −5

Hence 3, −11 5 is a minimum point.
6

126.87◦ = 126.87◦ ×

Knowing (−2, 9) is a maximum point (i.e. crest of a wave), and 3, −11 5 is a minimum point (i.e.
6
bottom of a valley) and that when x = 0, y = 5 , a
3
sketch may be drawn as shown in Fig. 28.6.

= 2.214 rad
306.87◦ = 306.87◦ ×
= 5.356 rad

12

21

9

0

3
2
y5 x 2 x 26x 1 5
3
2
3

When x = 2.214 rad, d2 y
= −4 sin 2.214 + 3 cos 2.214, dx 2

1

2

3

x

24

2115

the

d2 y
= −4 sin x + 3 cos x dx 2

4

22

π radians 180

Hence (2.214, 5) and (5.356, −5) are co-ordinates of the turning points.

y

8

π radians 180

which is negative.
Hence (2.214, 5) is a maximum point.
When x = 5.356 rad,

28

d2 y
= −4 sin 5.356 + 3 cos5.356, dx 2

6

212

Figure 28.6

which is positive.
Hence (5.356, −5) is a minimum point.
A sketch of y = 4 sin x − 3 cos x is shown in Fig. 28.7.

Problem 14. Determine the turning points on the curve y = 4 sin x − 3 cos x in the range x = 0 to x = 2π radians, and distinguish between them.
Sketch the curve over one cycle.

y
5

y ϭ 4 sin x Ϫ 3 cos x

Since y = 4 sin x − 3 cos x dy = 4 cos x + 3 sin x = 0, dx for a turning point, from which, then 4 cos x = −3 sin x and
−4
sin x
=
= tan x
3
cos x
−4
= 126.87◦ or 306.87◦, since
3
tangent is negative in the second and fourth quadrants.

Hence x = tan−1

0
Ϫ3
Ϫ5

Figure 28.7

␲/2 2.214



5.356
3␲/2

x (rads)
2␲

307

Some applications of differentiation
Now try the following exercise
Exercise 122 points Further problems on turning

13. Show that the curve y = 2 (t − 1)3 + 2t (t − 2)
3
has a maximum value of 2 and a minimum
3
value of −2.

In Problems 1 to 11, find the turning points and distinguish between them.
1.

y = x 2 − 6x

2.

y = 8 + 2x − x 2

[(1, 9) Maximum]

3.

y = x 2 − 4x + 3

[(2, −1) Minimum]

4.

y = 3 + 3x 2 − x 3

(0, 3) Minimum,
(2, 7) Maximum

5.

y = 3x 2 − 4x + 2

Minimum at

6.

x = θ(6 − θ)

[Maximum at (3, 9)]

7.

y = 4x 3 + 3x 2 − 60x − 12

[(3, −9) Minimum]

2 2
3, 3

Minimum (2, −88);
Maximum(−2.5, 94.25)

28.4 Practical problems involving maximum and minimum values
There are many practical problems involving maximum and minimum values which occur in science and engineering. Usually, an equation has to be determined from given data, and rearranged where necessary, so that it contains only one variable. Some examples are demonstrated in Problems 15 to 20.
Problem 15. A rectangular area is formed having a perimeter of 40 cm. Determine the length and breadth of the rectangle if it is to enclose the maximum possible area.
Let the dimensions of the rectangle be x and y. Then the perimeter of the rectangle is (2x + 2y). Hence
2x + 2y = 40,

8.

y = 5x − 2 ln x

or
[Minimum at (0.4000, 3.8326)]

9.

10.

11.

y = 2x − ex

y =t3−

x = 8t +

[Maximum at (0.6931, −0.6136)]

t2
− 2t + 4
2


Minimum at (1, 2.5);



2 22 ⎦
Maximum at − , 4
3 27
1
2t 2

[Minimum at (0.5, 6)]

12. Determine the maximum and minimum values on the graph y = 12 cosθ − 5 sin θ in the range θ = 0 to θ = 360◦. Sketch the graph over one cycle showing relevant points.
Maximum of 13 at 337.38◦,
Minimum of −13 at 157.38◦

x + y = 20

(1)

Since the rectangle is to enclose the maximum possible area, a formula for area A must be obtained in terms of one variable only.
Area A = x y. From equation (1), x = 20 − y
Hence, area A = (20 − y)y = 20y − y 2 dA = 20 − 2y = 0 dy for a turning point, from which, y = 10 cm d2 A
= −2, d y2 which is negative, giving a maximum point.
When y = 10 cm, x = 10 cm, from equation (1).
Hence the length and breadth of the rectangle are each 10 cm, i.e. a square gives the maximum possible area. When the perimeter of a rectangle is 40 cm, the maximum possible area is 10 × 10 = 100 cm2 .
Problem 16. A rectangular sheet of metal having dimensions 20 cm by 12 cm has squares removed from each of the four corners and the sides bent

308 Higher Engineering Mathematics upwards to form an open box. Determine the maximum possible volume of the box.
The squares to be removed from each corner are shown in Fig. 28.8, having sides x cm. When the sides are bent upwards the dimensions of the box will be: length (20 − 2x) cm, breadth (12 − 2x) cm and height, x cm. x x x x

Maximum volume = (15.146)(7.146)(2.427)
= 262.7 cm 3
Problem 17. Determine the height and radius of a cylinder of volume 200 cm3 which has the least surface area.
Let the cylinder have radius r and perpendicular height h.
Volume of cylinder,
V = πr 2 h = 200

12 cm

Surface area of cylinder,

(12 2 2x )

x

x x x
20 cm

Figure 28.8

A = 2πrh + 2πr 2
Least surface area means minimum surface area and a formula for the surface area in terms of one variable only is required.
From equation (1),

Volume of box,
V = (20 − 2x)(12 − 2x)(x)
= 240x − 64x 2 + 4x 3 dV = 240 − 128x + 12x 2 = 0 dx for a turning point.
Hence 4(60 − 32x + 3x 2 ) = 0,
3x 2 − 32x + 60 = 0

i.e.

Using the quadratic formula,
32 ±

(−32)2 − 4(3)(60)
2(3)
= 8.239 cm or 2.427 cm.

x=

(1)

(20 2 2x )

Since the breadth is (12 − 2x) cm then x = 8.239 cm is not possible and is neglected. Hence x = 2.427 cm d2 V
= −128 + 24x. dx 2 d2 V
When x = 2.427, 2 is negative, giving a maxdx imum value.
The dimensions of the box are: length = 20 − 2(2.427) = 15.146 cm, breadth = 12 − 2(2.427) = 7.146 cm, and height = 2.427 cm

h=

200 πr 2

(2)

Hence surface area,
A = 2πr

200
+ 2πr 2 πr 2

400
+ 2πr 2 = 400r −1 + 2πr 2 r d A −400
= 2 + 4πr = 0, dr r
=

for a turning point.
Hence 4πr =

400
400
and r 3 =
,
r2


from which, r= 3

100 π = 3.169 cm

d 2 A 800
= 3 + 4π. dr 2 r d2 A
When r = 3.169 cm, 2 is positive, giving a mindr imum value.
From equation (2), when r = 3.169 cm,
200
= 6.339 cm h= π(3.169)2

Some applications of differentiation
Hence for the least surface area, a cylinder of volume 200 cm3 has a radius of 3.169 cm and height of
6.339 cm.
Problem 18. Determine the area of the largest piece of rectangular ground that can be enclosed by
100 m of fencing, if part of an existing straight wall is used as one side.
Let the dimensions of the rectangle be x and y as shown in Fig. 28.9, where P Q represents the straight wall.

P y x

Figure 28.10

Surface area of box, A, consists of two ends and five faces (since the lid also covers the front face.)
Hence

x

Figure 28.9

y=
From Fig. 28.9,
(1)

(2)

Since the maximum area is required, a formula for area
A is needed in terms of one variable only.
From equation (1), x = 100 −2y
Hence area A =xy = (100 −2y)y = 100y −2y2 dA = 100 − 4y = 0, dy for a turning point, from which, y = 25 m d2 A d y2

6 − 2x 2
6
2x
=

5x
5x
5

= −4,

which is negative, giving a maximum value.
When y = 25 m, x = 50 m from equation (1).
Hence the maximum possible area = x y = (50)(25) =
1250 m2 .
Problem 19. An open rectangular box with square ends is fitted with an overlapping lid which covers the top and the front face. Determine the maximum volume of the box if 6 m2 of metal are used in its construction.
A rectangular box having square ends of side x and length y is shown in Fig. 28.10.

(2)

Hence volume
6
2x

5x
5

V = x2 y = x2

Area of rectangle,
A = xy

(1)

Since it is the maximum volume required, a formula for the volume in terms of one variable only is needed.
Volume of box, V = x 2 y.
From equation (1),

y

x + 2y = 100

y

x

A = 2x 2 + 5x y = 6

Q

309

=

6x 2x 3

5
5

dV
6 6x 2
= −
=0
dx
5
5 for a maximum or minimum value.
Hence 6 =6x 2 , giving x = 1 m (x = −1 is not possible, and is thus neglected).
−12x
d2 V
=
2 dx 5 d2 V
When x = 1, 2 is negative, giving a maximum value. dx From equation (2), when x = 1, y= 2(1) 4
6

=
5(1)
5
5

Hence the maximum volume of the box is given by
V = x 2 y = (1)2

4
5

= 4 m3
5

Problem 20. Find the diameter and height of a cylinder of maximum volume which can be cut from a sphere of radius 12 cm.
A cylinder of radius r and height h is shown enclosed in a sphere of radius R = 12 cm in Fig. 28.11.

310 Higher Engineering Mathematics d2 V
When h = 13.86, 2 is negative, giving a maximum dh value.
From equation (2),

r

P h 2 h Q

R

5

12

cm

r 2 = 144 −

O

h2
13.862
= 144 −
4
4

from which, radius r = 9.80 cm
Diameter of cylinder = 2r = 2(9.80) = 19.60 cm.
Hence the cylinder having the maximum volume that can be cut from a sphere of radius 12 cm is one in which the diameter is 19.60 cm and the height is
13.86 cm.

Figure 28.11

Volume of cylinder,

Now try the following exercise

V = πr 2 h

(1)

Using the right-angled triangle OPQ shown in
Fig. 28.11, r2 i.e.

h
+
2 r2 +

Exercise 123 Further problems on practical maximum and minimum problems
1.

2

= R2

2.

Determine the maximum area of a rectangular piece of land that can be enclosed by
1200 m of fencing.
[90000 m2]

3.

A shell is fired vertically upwards and its vertical height, x metres, is given by x = 24t − 3t 2, where t is the time in seconds.
Determine the maximum height reached.
[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
[11.42 m2 ] of the box is 3.5 m3.

5.

A closed cylindrical container has a surface area of 400 cm2 . Determine the dimensions for maximum volume. radius = 4.607 cm;

by Pythagoras’ theorem,

h2
= 144
4

(2)

Since the maximum volume is required, a formula for the volume V is needed in terms of one variable only.
From equation (2), r 2 = 144 −

The speed, v, of a car (in m/s) is related to time t s by the equation v = 3 +12t − 3t 2.
Determine the maximum speed of the car in km/h.
[54 km/h]

h2
4

Substituting into equation (1) gives:
V = π 144 −

h2 πh 3 h = 144πh −
4
4

dV
3πh 2
= 144π −
= 0, dh 4 for a maximum or minimum value.
Hence
144π = from which,

h=

3πh 2
4
(144)(4)
= 13.86 cm
3

−6πh d2 V
=
2 dh 4

height = 9.212 cm
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.
[6.67 cm]

Some applications of differentiation

7.

8.

The power developed in a resistor R by a battery of emf E and internal resistance r is
E2 R
. Differentiate P with given by P =
(R + r)2 respect to R and show that the power is a maximum when R = r.
Find the height and radius of a closed cylinder of volume 125 cm3 which has the least surface area. height = 5.42 cm; radius = 2.71 cm

Problem 21. Find the equation of the tangent to the curve y = x 2 − x − 2 at the point (1, −2).
Gradient, m
=

10.

11.

Resistance to motion, F, of a moving vehicle, is given by F = 5 + 100x. Determine the x minimum value of resistance.
[44.72]
An electrical voltage E is given by
E =(15 sin 50πt + 40 cos 50πt ) volts, where t is the time in seconds. Determine the maximum value of voltage.
[42.72 volts]
The fuel economy E of a car, in miles per gallon, is given by:
E = 21 + 2.10 × 10−2v 2
− 3.80 × 10−6v 4 where v is the speed of the car in miles per hour. Determine, correct to 3 significant figures, the most economical fuel consumption, and the speed at which it is achieved.
[50.0 miles/gallon, 52.6 miles/hour]

dy
= 2x − 1 dx At the point (1, −2), x = 1 and m = 2(1) − 1 =1.
Hence the equation of the tangent is: y − y1 = m(x − x 1)
i.e. y − (−2) = 1(x − 1)
i.e.

9.

311

y+2 = x −1 y = x−3

or

The graph of y = x 2 − x − 2 is shown in Fig. 28.12. The line AB is the tangent to the curve at the point C, i.e. (1,
−2), and the equation of this line is y = x − 3.

Normals
The normal at any point on a curve is the line which passes through the point and is at right angles to the tangent. Hence, in Fig. 28.12, the line CD is the normal.
It may be shown that if two lines are at right angles then the product of their gradients is −1. Thus if m is the gradient of the tangent, then the gradient of the normal
1
is − m Hence the equation of the normal at the point (x 1 , y1) is given by: y − y1 = −

1
(x − x1 ) m y y ϭ x 2 Ϫ xϪ 2

2

28.5

1

Tangents and normals
Ϫ2

Tangents

Ϫ1

2

Ϫ3 A

Figure 28.12

3
B

Ϫ2

y − y1 = m(x − x1) dy = gradient of the curve at (x 1, y1). dx 1

Ϫ1

The equation of the tangent to a curve y = f (x) at the point (x 1, y1) is given by:

where m =

0

C
D

x

312 Higher Engineering Mathematics
Multiplying each term by 15 gives:

Problem 22. Find the equation of the normal to the curve y = x 2 − x − 2 at the point (1, −2).

15y + 3 = −25x − 25

m = 1 from Problem 21, hence the equation of the normal is
1
y − y1 = − (x − x 1 ) m 1
i.e. y − (−2) = − (x − 1)
1
i.e.

y = −x − 1

Thus the line CD in Fig. 28.12 has the equation y = −x − 1.
Problem 23.

Determine the equations of the x3 tangent and normal to the curve y = at the point
5
1
−1, −
5

Equation of the tangent is: y − y1 = m(x − x 1 )
i.e. y − −
i.e.
or or 1
5

y+

3
= (x − (−1))
5

(a) y = 4x − 2
(b) 4y + x = 9

1.

y = 2x 2 at the point (1, 2)

2.

y = 3x 2 − 2x at the point (2, 8)
(a) y = 10x − 12

3.
3(−1)2
5

=

y=

1 x3 at the point −1, −
2
2
(a) y = 3 x + 1
2

3
5

(b) 6y + 4x + 7 = 0
4.

y = 1 + x − x 2 at the point (−2, −5)
(a) y = 5x + 5
(b) 5y + x + 27 = 0

5. θ =

1
1
at the point 3, t 3
(a) 9θ + t = 6
(b) θ = 9t − 26 2 or 3θ = 27t − 80
3

5y + 1 = 3x + 3
5y − 3x = 2

1 y − y1 = − (x − x 1 ) m 1
−1
i.e. y − −
=
(x − (−1))
5
(3/5)

i.e.

For the curves in problems 1 to 5, at the points given, find (a) the equation of the tangent, and (b) the equation of the normal.

1 3
= (x + 1)
5 5

Equation of the normal is:

i.e.

Now try the following exercise

(b) 10y + x = 82

x3
Gradient m of curve y = is given by
5
d y 3x 2
=
m= dx 5
At the point −1, − 1 , x = − 1 and m =
5

15y + 25x + 28 = 0

Exercise 124 Further problems on tangents and normals

y + 2 = −x + 1

or

Hence equation of the normal is:

1
5
= − (x + 1)
5
3
5
5
1
y+ =− x−
5
3
3
y+

28.6

Small changes

If y is a function of x, i.e. y = f (x), and the approximate change in y corresponding to a small change δx in x is required, then: δy dy

δx dx dy and δy ≈
· δx or δy ≈ f (x) · δx dx Some applications of differentiation
Problem 24. Given y = 4x 2 − x, determine the approximate change in y if x changes from 1 to
1.02.
Since y = 4x 2 − x, then

=

=

approximate change in T original value of T

100%

k
√ (−0.1)
2 l

× 100% k l
−0.1
−0.1
100% =
100%
2l
2(32.1)

= −0.156%

Approximate change in y, δy ≈

Percentage error

=

dy
= 8x − 1 dx 313

dy
· δx ≈ (8x − 1)δx dx When x = 1 and δx = 0.02, δy ≈ [8(1) − 1](0.02)
≈ 0.14
[Obviously, in this case, the exact value of dy may be obtained by evaluating y when x = 1.02, i.e. y = 4(1.02)2 − 1.02 = 3.1416 and then subtracting from it the value of y when x = 1, i.e. y = 4(1)2 − 1 = 3, giving δy = 3.1416 −3 =0.1416. dy Using δy =
· δx above gave 0.14, which shows that dx the formula gives the approximate change in y for a small change in x.]

Hence the change in the time of swing is a decrease of 0.156%.
Problem 26. A circular template has a radius of
10 cm (±0.02). Determine the possible error in calculating the area of the template. Find also the percentage error.
Area of circular template, A = πr 2 , hence dA = 2πr dr Approximate change in area, δA ≈

dA
· δr ≈ (2πr)δr dr When r = 10 cm and δr = 0.02,
Problem 25. The time of swing T of a pendulum

is given by T = k l, where k is a constant.
Determine the percentage change in the time of swing if the length of the pendulum l changes from
32.1 cm to 32.0 cm.



dT δl ≈ dl 0.4π
100%
π(10)2

= 0.40% k = √
2 l

Approximate change in T , δt ≈

i.e. the possible error in calculating the template area is approximately 1.257 cm2.
Percentage error ≈

1

If T = k l = kl 2 , then

dT
1 −1
=k
l 2 dl 2

δ A = (2π10)(0.02) ≈ 0.4π cm 2

k
√ δl
2 l

k
√ (−0.1)
2 l

(negative since l decreases)

Now try the following exercise
Exercise 125 changes Further problems on small

1. Determine the change in y if x changes from
2.50 to 2.51 when
5
(a) y = 2x − x 2 (b) y = x [(a) −0.03 (b) −0.008]

314 Higher Engineering Mathematics
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.
[−0.032, −1.6%]
3. Determine the approximate increase in (a) the volume, and (b) the surface area of a cube of side x cm if x increases from 20.0 cm to
20.05 cm.
[(a) 60 cm3 (b) 12 cm2 ]
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) −6.03 cm2 (b) −18.10 cm3 ]

5. The rate of flow of a liquid through a tube is given by Poiseuilles’s equation as: pπr 4
Q=
where Q is the rate of flow, p
8ηL
is the pressure difference between the ends of the tube, r is the radius of the tube, L is the length of the tube and η is the coefficient of viscosity of the liquid. η is obtained by measuring Q, p, r and L. If Q can be measured accurate to ±0.5%, p accurate to
±3%, r accurate to ±2% and L accurate to
±1%, calculate the maximum possible percentage error in the value of η.
[12.5%]

Chapter 29

Differentiation of parametric equations 29.1 Introduction to parametric equations Certain mathematical functions can be expressed more simply by expressing, say, x and y separately in terms of a third variable. For example, y =r sin θ, x =r cos θ.
Then, any value given to θ will produce a pair of values for x and y, which may be plotted to provide a curve of y = f (x).
The third variable, θ, is called a parameter and the two expressions for y and x are called parametric equations. The above example of y =r sin θ and x =r cos θ are the parametric equations for a circle. The equation of any point on a circle, centre at the origin and of radius r is given by: x 2 + y 2 =r 2 , as shown in Chapter 13.
To show that y =r sin θ and x =r cos θ are suitable parametric equations for such a circle:
Left hand side of equation

29.2 Some common parametric equations The following are some of the most common parametric equations, and Fig. 29.1 shows typical shapes of these curves. (a)

Ellipse

x = a cos θ, y = b sin θ

(b) Parabola

x = a t 2, y = 2a t

(c)

x = a sec θ, y = b tan θ c x = c t, y = t Hyperbola

(d) Rectangular hyperbola (e)

Cardioid

x = a (2 cosθ − cos 2θ), y = a (2 sin θ − sin 2θ )

(f ) Astroid

x = a cos3 θ, y = a sin3 θ

(g) Cycloid

x = a (θ − sin θ ) , y = a (1− cos θ)

= x 2 + y2
= (r cos θ)2 + (r sin θ)2
= r 2 cos2 θ + r 2 sin2 θ
= r 2 cos2 θ + sin2 θ
= r = right hand side
2

(since cos2 θ + sin2 θ = 1, as shown in
Chapter 15)

29.3

Differentiation in parameters

When x and y are given in terms of a parameter, say θ, then by the function of a function rule of differentiation
(from Chapter 27): dy d y dθ
=
× dx dθ dx

316 Higher Engineering Mathematics
Given x = 5θ − 1 and dy y = 2θ (θ − 1), determine in terms of θ. dx Problem 1.

x = 5θ − 1, hence
(a) Ellipse

(b) Parabola

dy
=5


y = 2θ(θ − 1) = 2θ 2 − 2θ, hence dy
= 4θ − 2 =2 (2θ − 1) dθ From equation (1),

(c) Hyperbola

dy dy 2
2(2θ − 1)
= dθ = or (2θ − 1) dx dx
5
5 dθ (d) Rectangular hyperbola

Problem 2. The parametric equations of a function are given by y = 3 cos2t , x = 2 sin t . dy d2 y
Determine expressions for (a)
(b) 2 . dx dx
(e) Cardioid

(f) Astroid

(a)

(g) Cycloid

Figure 29.1

It may be shown that this can be written as: dy dy dθ = dx dx dθ (1)

dy dx =

d dθ (b) From equation (2), d dy d 2 y dt dx
=
dx dx 2 dt For the second differential, d d2 y
=
2 dx dx

dy
= −6 sin 2t dt dx
= 2 cos t x = 2 sin t , hence dt From equation (1), dy dy
−6(2 sin t cos t )
−6 sin 2t
= dt =
=
dx dx 2 cos t
2 cos t dt from double angles, Chapter 17 dy i.e.
= −6 sin t dx y = 3 cos 2t , hence

dy dx ·

dθ dx i.e.

or d dy d2 y dθ dx
=
dx dx2 dθ

d
(−6 sin t) −6 cost dt =
=
2 cost
2 cost

d2 y
= −3 dx2 Problem 3. The equation of a tangent drawn to a curve at point (x 1, y1) is given by:
(2)

y − y1 =

d y1
(x − x 1 ) dx 1

Differentiation of parametric equations

dx 1
At point t , x 1 = 2t , hence
= 4t dt d y1 and y1 = 4t , hence
=4
dt
From equation (1),
2

the

equation

of

the

tangent is:
1
y − 4t = x − 2t 2 t Problem 4. The parametric equations of a cycloid are x = 4(θ − sin θ), y = 4(1 − cosθ). Determine dy d2 y
(a)
(b) 2 dx dx
(a)

x = 4(θ − sin θ), dx = 4 −4 cos θ = 4(1 − cos θ) dθ dy
= 4 sin θ y = 4(1 − cos θ), hence dθ From equation (1),

hence

dy sin θ
4 sin θ dy dθ =
=
= dx dx
4(1 − cos θ) (1 − cos θ) dθ (b) From equation (2),

d2 y dx 2

=

d dθ dy dx dx dθ cos θ − 1
4(1 − cos θ )3
−(1 − cos θ)
−1
=
4(1 − cos θ )3
4(1 − cos θ)2

Now try the following exercise

dy
4 1 dy = dt = = dx dx
4t t dt Hence,

=
=

Determine the equation of the tangent drawn to the parabola x = 2t 2, y = 4t at the point t .

Exercise 126 Further problems on differentiation of parametric equations
1. Given x = 3t − 1 and y = t (t − 1), determine
1
dy in terms of t .
(2t − 1) dx 3
2. A parabola has parametric equations: x = t 2 , dy y = 2t . Evaluate when t = 0.5.
[2]
dx
3. The parametric equations for an ellipse dy are x = 4 cos θ, y = sin θ. Determine (a) dx d2 y
1
1

(b) 2 .
(a) − cot θ (b) − cosec dx 4
16
dy π 4. Evaluate at θ = radians for the dx 6 hyperbola whose parametric equations are x = 3 sec θ, y = 6 tan θ.
[4]
5. The parametric equations for a rectangular dy 2 hyperbola are x = 2t , y = . Evaluate t dx when t = 0.40.
[−6.25]
The equation of a tangent drawn to a curve at point (x 1 , y1) is given by: y − y1 =

sin θ d dθ 1 − cos θ
=
4(1 − cos θ)

(1 − cos θ)(cos θ) − (sin θ)(sin θ)
(1 − cos θ)2
=
4(1 − cos θ)
− cos2 θ

− sin2 θ

=

cos θ
4(1 − cos θ)3

=

cos θ − cos2 θ + sin2 θ
4(1 − cos θ )3

d y1
(x − x 1) dx 1

Use this in Problems 6 and 7.
6. Determine the equation of the tangent drawn π to the ellipse x = 3 cos θ, y = 2 sin θ at θ = .
6
[y = −1.155x + 4]
7. Determine the equation of the tangent drawn
5
to the rectangular hyperbola x = 5t , y = at t t = 2.
1
y =− x +5
4

317

318 Higher Engineering Mathematics
From equation (1),

29.4

Further worked problems on differentiation of parametric equations Problem 5. The equation of the normal drawn to a curve at point (x 1, y1) is given by: y − y1 = −

1
(x − x 1 ) d y1 dx 1

Determine the equation of the normal drawn to the π astroid x = 2 cos3 θ, y = 2 sin3 θ at the point θ =
4
x = 2 cos3 θ, hence

dx
= −6 cos2 θ sin θ dθ y = 2 sin3 θ, hence

dy
= 6 sin2 θ cos θ dθ From equation (1), dy 2 dy dθ = 6 sin θ cos θ =− sin θ = −tanθ
=
dx dx −6 cos2 θ sin θ cos θ dθ π dy π
,
= −tan = −1
4
dx
4
π π x 1 = 2 cos3 = 0.7071 and y1 = 2 sin3 = 0.7071
4
4

When θ =

Hence, the equation of the normal is: y − 0.7071 = −
i.e.
i.e.

1
(x − 0.7071)
−1

y − 0.7071 = x − 0.7071 y =x

Problem 6. The parametric equations for a hyperbola are x = 2 sec θ, y = 4 tan θ. Evaluate dy d2 y
(a)
(b) 2 , correct to 4 significant figures, dx dx when θ = 1 radian.
(a)

x = 2 sec θ, hence

dx
= 2 sec θ tan θ dθ y = 4 tan θ, hence

dy
= 4 sec2 θ dθ dy dy 2 sec θ
4 sec2 θ
= dθ =
=
dx dx 2 sec θ tan θ tan θ dθ 1
2
2 cos θ
=
= or 2 cosec θ sin θ sin θ cos θ
2
dy
=
= 2.377, correct to 4
When θ = 1 rad, dx sin 1 significant figures.
(b) From equation (2), d dy d (2 cosec θ) d2 y dθ dx
=
= dθ dx dx 2
2 sec θ tan θ dθ −2 cosec θ cot θ
=
2 sec θ tan θ

=

=−
=−

cos θ sin θ sin θ cos θ

1 sin θ
1
cos θ cos θ sin2 θ

cos2 θ sin θ

cos3 θ
= − cot 3 θ sin3 θ

d2 y
1
= − cot 3 1 =−
2
dx
(tan 1)3
= −0.2647, correct to 4 significant figures.
When θ = 1 rad,

Problem 7. When determining the surface tension of a liquid, the radius of curvature, ρ, of part of the surface is given by:
1+
ρ=

dy dx 2 3

d2 y dx 2

Find the radius of curvature of the part of the surface having the parametric equations x = 3t 2, y = 6t at the point t = 2.

Differentiation of parametric equations dx = 6t dt dy
=6
y = 6t , hence dt x = 3t 2, hence

The equation of the normal drawn to a curve at point (x 1 , y1) is given by:
1
y − y1 = −
(x − x 1) d y1 dx 1
Use this in Problems 2 and 3.

dy d y dt
6 1
From equation (1),
=
= = dx dx 6t t dt From equation (2),
2

d y
=
dx 2

d dt dy dx dx dt =

1 t d dt 1 t2 = − 1
=
6t
6t 3


6t

dy dx 1+
Hence, radius of curvature, ρ =

1+ t = 2,

ρ=



1
2

1
6 (2)3

2

1 t −

=

2

3

d2 y dx 2
1+

When

2. Determine the equation of the normal drawn
1
1 to the parabola x = t 2, y = t at t = 2.
4
2
[y = −2x + 3]

3

1
6t 3

d2 y
, correct to 4 sigdx 2 π nificant figures, at θ = rad for the cardioid
6
x = 5(2θ − cos 2θ), y = 5(2 sin θ − sin 2θ).

4. Determine the value of

[0.02975]

2 3

=

3. Find the equation of the normal drawn to the cycloid x = 2(θ − sin θ), y = 2(1 − cos θ) π [y = −x + π] at θ = rad.
2

(1.25)3
1

48

= − 48 (1.25)3 = −67.08
Now try the following exercise
Exercise 127 Further problems on differentiation of parametric equations
1. A cycloid has parametric equations x = 2(θ − sin θ), y = 2(1 −cos θ). Evaluate, at θ = 0.62 rad, correct to 4 significant figures, dy d2 y
(a)
(b) 2 . dx dx
[(a) 3.122 (b) −14.43]

5. The radius of curvature, ρ, of part of a surface when determining the surface tension of a liquid is given by: dy dx

1+ ρ= 2 3/2

d2 y dx 2

Find the radius of curvature (correct to 4 significant figures) of the part of the surface having parametric equations
3
1 at the point t = t 2 π (b) x = 4 cos3 t, y = 4 sin3 t at t = rad.
6
(a) x = 3t , y =

[(a) 13.14 (b) 5.196]

319

Chapter 30

Differentiation of implicit functions 30.1

Implicit functions

A simple rule for differentiating an implicit function is summarised as:

When an equation can be written in the form y = f (x) it is said to be an explicit function of x. Examples of explicit functions include y = 2x 3 − 3x + 4, y = 2x ln x
3ex
and y = cos x
In these examples y may be differentiated with respect to x by using standard derivatives, the product rule and the quotient rule of differentiation respectively.
Sometimes with equations involving, say, y and x, it is impossible to make y the subject of the formula.
The equation is then called an implicit function and examples of such functions include y 3 + 2x 2 = y 2 − x and sin y = x 2 + 2x y.

30.2 Differentiating implicit functions It is possible to differentiate an implicit function by using the function of a function rule, which may be stated as du d y du =
×
dx d y dx
Thus, to differentiate y 3 with respect to x, the subdu stitution u = y 3 is made, from which,
= 3y 2 . Hence, dy d 3 dy (y ) = (3y 2 ) × , by the function of a function rule. dx dx

d d dy
[ f ( y)] = [ f ( y)] × dx dy dx (1)

Problem 1. Differentiate the following functions with respect to x:
(a) 2y 4 (b) sin 3t .
(a) Let u =2y 4 , then, by the function of a function rule: du dy du dy d =
×
=
(2y 4 ) × dx dy dx dy dx dy = 8y3 dx (b) Let u = sin 3t , then, by the function of a function rule: du du dt d dt
=
×
= (sin 3t ) × dx dt dx dt dx dt
= 3 cos 3t dx Problem 2. Differentiate the following functions with respect to x:
(a) 4 ln 5y

1
(b) e3θ−2
5

(a) Let u = 4 ln 5y, then, by the function of a function rule: 321

Differentiation of implicit functions du du dy d dy
=
×
=
(4 ln 5y) × dx dy dx dy dx



1
(b) Let u = e3θ−2, then, by the function of a function
5
rule:
1 3θ−2 dθ e
×
5 dx 30.3


3
= e3θ −2
5
dx

Now try the following exercise

Exercise 128 Further problems on differentiating implicit functions

2. (a)

5
3
ln 3t (b) e2y+1 (c) 2 tan 3y
2
4

dy ⎤
3
5 dt
(b) e2y+1
(a)

2t dx
2
dx ⎥

⎦ dy (c) 6 sec2 3y dx 3. Differentiate the following with respect to y:

2
(a) 3 sin 2θ (b) 4 x 3 (c) t e ⎤

√ dx dθ (b) 6 x
(a) 6 cos 2θ

dy dy ⎥




−2 dt
(c) t e dy
4. Differentiate the following with respect to u:
(a)

2
2
(b) 3 sec 2θ (c) √
(3x + 1) y ⎤

Differentiating implicit functions containing products and quotients

The product and quotient rules of differentiation must be applied when differentiating functions containing products and quotients of two variables.
For example,

d d d 2
(x y) = (x 2 ) (y) + (y) (x 2 ), dx dx dx by the product rule

In Problems 1 and 2 differentiate the given functions with respect to x.

1. (a) 3y 5 (b) 2 cos 4θ (c) k

⎤ dy dθ
(b) −8 sin 4θ
(a) 15y 4

dx dx ⎥




1 dk
(c) √ dx 2 k

dx
−6
(3x + 1)2 du





dθ ⎥

⎢ (b) 6 sec 2θ tan 2θ

⎢ du ⎥




−1 dy
(c)
y 3 du

4 dy
=
y dx

du du dθ d =
×
= dx dθ dx dθ (a)

= (x 2 ) 1

= x2

Problem 3. Determine

dy
+ y(2x), dx by using equation (1)

dy
+ 2xy dx d
(2x 3 y 2 ). dx In the product rule of differentiation let u = 2x 3 and v = y2 .
Thus

d d d
(2x 3 y 2 ) = (2x 3 ) (y 2 ) + (y 2 ) (2x 3 ) dx dx dx = (2x 3 ) 2y
= 4x 3 y

d dx + (y 2 )(6x 2 )

dy
+ 6x 2 y 2 dx = 2x2 y 2x

Problem 4. Find

dy dx dy
+ 3y dx 3y
.
2x

In the quotient rule of differentiation let u = 3y and v = 2x.

322 Higher Engineering Mathematics d Thus dx 3y
2x

d d (2x) (3y) − (3y) (2x) dx dx
=
(2x)2 dy − (3y)(2)
(2x) 3 dx =
4x 2 dy 6x
− 6y
3
dy dx =
= 2 x −y
2
4x
2x
dx

Problem 5. Differentiate z = x 2 + 3x cos 3y with respect to y. dz d d 2
=
(x ) + (3x cos 3y) dy dy dy = 2x

dx dx + (3x)(−3 sin 3y) + (cos 3y) 3 dy dy

Exercise 129 Further problems on differentiating implicit functions involving products and quotients d (3x 2 y 3 ). dx 3. Determine

2y
.
5x d du

dy
+ 2y dx 2 dy −y x 5x 2 dx 3u
.
4v

3 dv v −u
4v 2 du dz

4. Given z = 3 y cos 3x find
.
dx cos 3x dy

− 9 y sin 3x
3

2 y dx
5. Determine

d d d d d
(3x 2 ) + (y 2 ) − (5x) + (y) =
(2)
dx dx dx dx dx dy dy
− 5 + 1 = 0, dx dx using equation (1) and standard derivatives. dy An expression for the derivative in terms of x and dx y may be obtained by rearranging this latter equation.
Thus:
dy
(2y + 1)
= 5 − 6x dx 6x + 2y

i.e.

5 −6x dy = dx 2y + 1

Each term in turn is differentiated with respect to x:
Hence

d d d d (2y 2 ) − (5x 4 ) − (2) − (7y 3 ) dx dx dx dx
=

3x y 2 3x d dx

An implicit function such as 3x 2 + y 2 − 5x + y = 2, may be differentiated term by term with respect to x. This gives: Problem 6. Given 2y 2 − 5x 4 − 2 − 7y 3 = 0, dy determine dx Now try the following exercise

2. Find

Further implicit differentiation

from which,

dx dx = 2x − 9x sin 3y +3 cos 3y dy dy

1. Determine

30.4

dz given z = 2x 3 ln y. dy x dx 2x 2
+ 3 ln y y dy

i.e.

4y

d
(0)
dx

dy dy − 20x 3 − 0 − 21y 2
=0
dx dx Rearranging gives:
(4y − 21y 2 )

dy
= 20x 3 dx dy
20x3
= dx (4y − 21y2 )

i.e.

Problem 7.

Determine the values of

x = 4 given that x 2 + y 2 = 25.

dy when dx

Differentiating each term in turn with respect to x gives: d 2 d d
(x ) + (y 2 ) =
(25)
dx dx dx

Differentiation of implicit functions
2x + 2y

i.e.

dy
=0
dx

i.e. 8x + (2x) 3y 2

dy dx + (y 3 )(2)

2x x dy
=−
=− dx 2y y Hence

− 10y

Since x 2 +y 2 = 25, when x = 4, y = (25 − 42 ) = ±3
Thus when x = 4 and y = ±3,

4
4
dy
=−
=± dx ±3
3

dy dx 4x + y3 dy 8x + 2y 3
=
= dx 10y − 6x y 2 y(5 − 3xy)

8x + 2y 3 = (10y − 6x y 2 ) and Gradient
4
52
3

5

dy
4(1) + (2)3
12
=
=
= −6 dx 2[5 − (3)(1)(2)]
−2

3

0

dy dy + 2y 3 − 10y
=0
dx dx (b) When x = 1 and y = 2,

y

25

4

5

x

Problem 9. Find the gradients of the tangents drawn to the circle x 2 + y 2 − 2x − 2y = 3 at x = 2.

23
25

dy
=0
dx

Rearranging gives:

x 2 + y 2 = 25 is the equation of a circle, centre at the origin and radius 5, as shown in Fig. 30.1. At x = 4, the two gradients are shown.

x 2 1 y 2 5 25

8x + 6x y 2

i.e.

323

Gradient
4
5
3

The gradient of the tangent is given by

Differentiating each term in turn with respect to x gives: d d d 2 d d
(x ) + (y 2 ) − (2x) − (2y) =
(3)
dx dx dx dx dx

Figure 30.1

Above, x 2 + y 2 = 25 was differentiated implicitly; actually, the equation could be transposed to y = (25 − x 2 ) and differentiated using the function of a function rule. This gives
−1
dy x 1
= (25 − x 2 ) 2 (−2x) = − dx 2
(25 − x 2 )

4
4
dy
= ± as obtained
=−
and when x = 4,
2)
dx
3
(25 − 4 above. dy dx 2x + 2y

i.e.
Hence

(2y − 2)

from which

dy dy −2−2
=0
dx dx dy
= 2 − 2x, dx dy
2 − 2x
1−x
=
=
dx
2y − 2 y−1 The value of y when x = 2 is determined from the original equation.

Problem 8. dy (a) Find in terms of x and y given dx 4x 2 + 2x y 3 − 5y 2 = 0.

Hence (2)2 + y 2 − 2(2) − 2y = 3

dy
(b) Evaluate when x = 1 and y = 2. dx or

(a)

Differentiating each term in turn with respect to x gives: d d d d (4x 2 ) + (2x y 3 ) − (5y 2 ) =
(0)
dx dx dx dx i.e.

4 + y 2 − 4 − 2y = 3 y 2 − 2y − 3 = 0

Factorizing gives: (y + 1)(y − 3) =0, from which y = −1 or y = 3.
When x = 2 and y = −1, dy 1−x
1−2
−1 1
=
=
=
= dx y − 1 −1 − 1 −2 2

324 Higher Engineering Mathematics
When x = 2 and y = 3,

Now try the following exercise

dy
1 − 2 −1
=
= dx 3−1
2
1
Hence the gradients of the tangents are ±
2
The circle having the given equation has its centre at

(1, 1) and radius 5 (see Chapter 13) and is shown in
Fig. 30.2 with the two gradients of the tangents. y 22y 5 3 3
2

1.

2y 3 − y + 3x − 2 = 0

3.

r5 5

1
0

1

2

21

4

22

dy when Given x 2 + y 2 = 9 evaluate dx √

x = 5 and y = 2.
− 25

In Problems 4 to 7, determine

4.

x 2 + 2x sin 4y = 0

5.

Problem 10. Pressure p and volume v of a gas are related by the law pv γ = k, where γ and k are constants. Show that the rate of change of pressure dp p dv
= −γ dt v dt

6.

2x 2 y + 3x 3 = sin y

k
= kv −γ vγ d p d p dv
=
× dt dv dt 7.

3y + 2x ln y = y 4 + x

8.

by the function of a function rule dp d
=
(kv −γ ) dv dv
= −γ kv

−γ −1

dp
−γ k dv
= γ +1 × dt v dt −γ k
= γ +1 v Since k = pv γ ,

dp p dv
= −γ dt v dt

−(x + sin 4y)
4x cos 4y
4x − y
3y + x x(4y + 9x) cos y − 2x 2
1 − 2 ln y
3 +(2x/y) − 4y 3

5 dy when
If 3x 2 + 2x 2 y 3 − y 2 = 0 evaluate
4
dx
1
x = and y = 1.
[5]
2

9.

Determine the gradients of the tangents drawn to the circle x 2 + y 2 = 16 at the point where x = 2. Give the answer correct to 4 significant figures.
[±0.5774]

10.

Find the gradients of the tangents drawn to x 2 y2 the ellipse
+ = 2 at the point where
4
9 x = 2.
[±1.5]

11.

Determine the gradient of the curve
3x y + y 2 = −2 at the point (1,−2).

d p −γ ( pv γ ) dv
−γ pv γ dv
=
= γ 1 γ +1 dt v dt v v dt
i.e.

dy dx 3y 2 + 2x y − 4x 2 = 0

Figure 30.2

Since pv γ = k, then p =

3
1 − 6y 2

x

Gradient
51
2

2x + 4
3 −2y

x 2 + y 2 + 4x − 3y + 1 = 0

2.

Gradient
52 1
2

4

x 2 1y 2 2 2x

Exercise 130 Further problems on implicit differentiation dy
In Problems 1 and 2 determine dx [−6]

Chapter 31

Logarithmic Differentiation
31.1 Introduction to logarithmic differentiation With certain functions containing more complicated products and quotients, differentiation is often made easier if the logarithm of the function is taken before differentiating. This technique, called ‘logarithmic differentiation’ is achieved with a knowledge of (i) the laws of logarithms, (ii) the differential coefficients of logarithmic functions, and (iii) the differentiation of implicit functions.

31.2

Laws of logarithms

Three laws of logarithms may be expressed as:
(i) log(A × B) = log A + log B
(ii) log

A
= log A − log B
B

(iii) log An = n log A
In calculus, Napierian logarithms (i.e. logarithms to a base of ‘e’) are invariably used. Thus for two functions f (x) and g(x) the laws of logarithms may be expressed as:
(i) ln[ f (x) · g(x)] = ln f (x) + ln g(x)
(ii) ln

f (x)
= ln f (x) − ln g(x) g(x) (iii) ln[ f (x)]n = n ln f (x)
Taking Napierian logarithms of both sides of the equaf (x) · g(x) tion y = gives: h(x) f (x) · g(x) ln y = ln h(x) which may be simplified using the above laws of logarithms, giving: ln y = ln f (x) + ln g(x) − ln h(x)
This latter form of the equation is often easier to differentiate. 31.3 Differentiation of logarithmic functions The differential coefficient of the logarithmic function ln x is given by: d 1
(lnx) = dx x
More generally, it may be shown that: d f (x)
[ln f (x)] = dx f (x)

(1)

For example, if y = ln(3x 2 + 2x − 1) then, dy 6x + 2
= 2 dx 3x + 2x − 1
Similarly, if y = ln(sin 3x) then dy 3 cos 3x
=
= 3 cot 3x. dx sin 3x
Now try the following exercise
Exercise 131 Further problems on differentiating logarithmic functions
Differentiate the following using the laws for logarithms. 1. ln (4x − 10)

4
4x − 10

326 Higher Engineering Mathematics
i.e. ln y = 2 ln(1 + x) +
2. ln(cos 3x)

[−3 tan 3x]

of Section 31.2

10x + 10
5x 2 + 10x − 7

4. ln(5x 2 + 10x − 7)

(iii) Differentiate each term in turn with respect to x using equations (1) and (2).

1 x 5. ln 8x

Thus

2x
−1

6. ln(x 2 − 1)

3 x 8. 2 ln(sin x)

Thus

[2 cot x]
+3
3 − 6x 2 + 3x
4x

+

As explained in Chapter 30, by using the function of a function rule:
(2)

Differentiation of an expression such as

(1 + x)2 (x − 1) y= √ may be achieved by using the x (x + 2) product and quotient rules of differentiation; however the working would be rather complicated. With logarithmic differentiation the following procedure is adopted: (i) Take Napierian logarithms of both sides of the equation. √
(1 + x)2 (x − 1)
Thus ln y = ln

x (x + 2)
= ln



(1 + x) (x − 1)
2

1
2(x + 2)

(v) Substitute for y in terms of x.

dy
2
(1 + x)2 (x − 1)
Thus
=

dx
(1 + x) x (x + 2)

31.4 Differentiation of further logarithmic functions

1 dy y dx

dy the subject. dx 2
1
1 dy =y
+
− dx (1 + x) 2(x − 1) x

12x 2 − 12x

9. ln(4x 3 − 6x 2 + 3x)

d
(ln y) = dx 1
1
1 dy
2
1
2
2
=
+
− − y dx
(1 + x) (x − 1) x (x + 2)

(iv) Rearrange the equation to make

x2

7. 3 ln 4x

ln(x − 1)

− ln x − 1 ln(x + 2), by law (iii)
2

9x 2 + 1
3x 3 + x

3. ln(3x 3 + x)

1
2

1
2

1

x(x + 2) 2

(ii) Apply the laws of logarithms.
1
Thus ln y = ln(1 + x)2 + ln(x − 1) 2
1
2

− ln x − ln(x + 2) , by laws (i) and (ii) of Section 31.2

1
1
1
− −
2(x − 1) x 2(x + 2)

Problem 1.

Use logarithmic differentiation to
(x + 1)(x − 2)3 differentiate y =
(x − 3)

Following the above procedure:
(i) Since

y=

(x + 1)(x − 2)3
(x − 3)

then ln y = ln

(x + 1)(x − 2)3
(x − 3)

(ii) ln y = ln(x + 1) + ln(x − 2)3 − ln(x − 3), by laws (i) and (ii) of Section 31.2,
i.e. ln y = ln(x + 1) + 3 ln(x − 2) − ln(x − 3), by law (iii) of Section 31.2.
(iii) Differentiating with respect to x gives:
1 dy
1
3
1
=
+

,
y dx
(x + 1) (x − 2) (x − 3) by using equations (1) and (2)
(iv) Rearranging gives:
1
3
1
dy
=y
+

dx
(x + 1) (x − 2) (x − 3)

Logarithmic Differentiation
(v) Substituting for y gives:

Using logarithmic differentiation and following the procedure gives:

dy (x + 1)(x − 2)3
=
dx
(x − 3)
+

327

1
(x + 1)

(i) Since

3e2θ sec 2θ y= √
(θ − 2)

1
3

(x − 2) (x − 3) then ln y = ln

(x − 2)3
(x + 1)2 (2x − 1) dy with respect to x and evaluate when x = 3. dx Problem 2. Differentiate y =

= ln

3e2θ sec 2θ

(θ − 2)
3e2θ sec 2θ
1

(θ − 2) 2
1

Using logarithmic differentiation and following the above procedure:
(i) Since

(ii) ln y = ln 3e2θ + ln sec 2θ − ln(θ − 2) 2
i.e. ln y = ln 3 + ln e2θ + ln sec 2θ

(x − 2)3 y= (x + 1)2 (2x − 1)

− 1 ln(θ − 2)
2
i.e. ln y = ln 3 + 2θ + ln sec 2θ − 1 ln(θ − 2)
2

(x − 2)3 then ln y = ln
(x + 1)2 (2x − 1)

(iii) Differentiating with respect to θ gives:

3

(x − 2) 2
= ln
(x + 1)2 (2x − 1)

1
1 dy
2 sec 2θ tan 2θ
2
= 0+2 +

y dθ sec 2θ
(θ − 2)

from equations (1) and (2)

3

(ii) ln y = ln(x − 2) 2 − ln(x + 1)2 − ln(2x − 1)
i.e. ln y =

3
2

ln(x − 2) − 2 ln(x + 1)

(iv) Rearranging gives:

− ln(2x − 1)
(iii)

3
1 dy
2
2
2
=

− y dx
(x − 2) (x + 1) (2x − 1)

(iv)

3
2
2 dy =y

− dx 2(x − 2) (x + 1) (2x − 1)

(v)

3
(x − 2)3 dy = dx (x + 1)2 (2x − 1) 2(x − 2)

dy
1
= y 2 + 2 tan 2θ − dθ 2(θ − 2)



2
2

(x + 1) (2x − 1)

(1)3 dy When x = 3,
=
dx
(4)2 (5)


1
80

3
5



3 2 2
− −
2 4 5

3 or ±0.0075
400

3e2θ sec 2θ dy Problem 3. Given y = √ determine dθ
(θ − 2)

(v) Substituting for y gives: dy 3e2θ sec 2θ
1
= √
2 + 2 tan 2θ − dθ 2(θ − 2)
(θ − 2) x 3 ln 2x
Problem 4. Differentiate y = x with e sin x respect to x.
Using logarithmic differentiation and following the procedure gives:
(i) ln y = ln

x 3 ln 2x ex sin x

(ii) ln y = ln x 3 + ln(ln 2x) − ln(ex ) − ln(sin x)
i.e. ln y = 3 ln x + ln(ln 2x) − x − ln(sin x)
(iii)

1
1 dy
3
cos x
= + x −1− y dx x ln 2x sin x

328 Higher Engineering Mathematics
(iv)
(v)

3 dy 1
=y
+
− 1 − cot x dx x x ln 2x

dy when x = 1 given dx √
(x + 1)2 (2x − 1)

7. Evaluate

dy x3 ln 2x 3
1
= x
+
− 1 − cot x dx e sin x x x ln 2x

y=

(x

+ 3)3

13
16

dy
, correct to 3 significant figures, dθ 2eθ sin θ π when θ = given y = √
4
θ5
[−6.71]

Now try the following exercise

8. Evaluate

Exercise 132 Further problems on differentiating logarithmic functions
In Problems 1 to 6, use logarithmic differentiation to differentiate the given functions with respect to the variable.
(x − 2)(x + 1)
1. y =
(x − 1)(x + 3)


1
(x − 2)(x + 1)
1
+

⎢ (x − 1)(x + 3) (x − 2) (x + 1)




1
1


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

1
(x + 1)(2x + 1)3
6
+

⎢ (x − 3)2 (x + 2)4 (x + 1) (2x + 1)


2
4


(x − 3) (x + 2)

(2x − 1) (x + 2)
3. y =
(x − 3) (x + 1)3


(2x − 1) (x + 2)
2
1
+

3 (2x − 1)
2(x + 2)
⎢ (x − 3) (x + 1)


1
3


(x − 3) 2(x + 1)

31.5

Differentiation of [ f (x)]x

Whenever an expression to be differentiated contains a term raised to a power which is itself a function of the variable, then logarithmic differentiation must be used. For example, the differentiation of expressions

such as x x , (x + 2)x , x (x − 1) and x 3x+2 can only be achieved using logarithmic differentiation.

2.

4.

y = 3θ sin θ cos θ
1
3θ sin θ cos θ
+ cot θ − tan θ θ 6.

y=

2x 4 tan x e2x ln 2x












Problem 5.

Determine

dy given y = x x . dx Taking Napierian logarithms of both sides of y = x x gives: ln y = ln x x = x ln x, by law (iii) of Section 31.2
Differentiating both sides with respect to x gives:
1
1 dy
= (x)
+ (ln x)(1), using the product rule y dx x i.e.

1 dy
= 1 + ln x, y dx

from which,

e2x cos 3x y= √
(x − 4) e2x cos 3x
1
2 − 3 tan 3x −

2(x − 4)
(x − 4)

5.



2x 4 tan x e2x ln 2x

4
1
+ x sin x cos x
1
−2 − x ln 2x

dy
= y(1 + ln x) dx i.e.

dy
= xx (1 + ln x) dx Problem 6. y = (x + 2)x .

Evaluate

dy when x = −1 given dx Taking Napierian logarithms of both sides of y = (x + 2)x gives: ln y = ln(x + 2)x = x ln(x + 2), by law (iii) of Section 31.2

Logarithmic Differentiation
Differentiating both sides with respect to x gives:

Let y = x 3x+2

1 dy
1
+ [ln(x + 2)](1),
= (x) y dx x +2 by the product rule. dy x
Hence
=y
+ ln(x + 2) dx x +2

Taking Napierian logarithms of both sides gives:

= (x + 2)x
When x = −1,

x
+ ln (x + 2) x+2 −1 dy = (1)−1
+ ln 1 dx 1

ln y = ln x 3x+2
i.e. ln y = (3x + 2) ln x, by law (iii) of Section 31.2.
Differentiating each term with respect to x gives:
1 dy
1
+ (ln x)(3),
= (3x + 2) y dx x by the product rule.
Hence

3x + 2 dy =y
+ 3 ln x dx x

= (+1)(−1) = −1

= x 3x+2

Problem 7. Determine (a) the differential

dy coefficient of y = x (x − 1) and (b) evaluate dx when x = 2.
(a)


1
y = x (x − 1) = (x − 1) x , since by the laws of

m indices n a m = a n
Taking Napierian logarithms of both sides gives:
1

ln y = ln(x − 1) x =

1 ln(x − 1), x by law (iii) of Section 31.2.

= x3x+2 3 +

Hence
i.e.

1 x 1
−1
+ [ln(x − 1)]
,
x −1 x2 by the product rule.

dy
1
ln(x − 1)
=y
− dx x(x − 1) x2 1 ln(x − 1) dy √
= x (x − 1)

dx x(x − 1) x2 (b) When x = 2,

1 dy √ ln(1) = 2 (1)

dx
2(1)
4
= ±1

1
1
−0 = ±
2
2

Problem 8. Differentiate x 3x+2 with respect to x.

2
+ 3 ln x x Now try the following exercise
Exercise 133 Further problems on differentiating [ f (x)]x type functions
In Problems 1 to 4, differentiate with respect to x.
[2x 2x (1 + ln x)]

1.

y = x 2x

2.

y = (2x − 1)x
(2x − 1)x

Differentiating each side with respect to x gives:
1 dy
=
y dx

3x + 2
+ 3 ln x x √ x (x + 3)

x
(x + 3)

3.

y=

4.

y = 3x 4x+1

2x
+ ln(2x − 1)
2x − 1
1
ln(x + 3)

x(x + 3) x2 3x 4x+1 4 +

1
+ 4 ln x x 5. Show that when y = 2x x and x = 1,
6. Evaluate

dy
= 2. dx d √ x (x − 2) when x = 3. dx 1
3

dy
7. Show that if y = θ θ and θ = 2,
= 6.77, dθ correct to 3 significant figures.

329

Revision Test 9
This Revision Test covers the material contained in Chapters 27 to 31. The marks for each question are shown in brackets at the end of each question.
1. Differentiate the following with respect to the variable: √
1
(a) y = 5 +2 x 3 − 2 (b) s = 4e2θ sin 3θ x 3 ln 5t
(c) y = cos 2t
2
(d) x =
(13)
2 − 3t + 5)
(t
2. If f (x) = 2.5x 2 − 6x + 2 find the co-ordinates at the point at which the gradient is −1.
(5)
3. The displacement s cm of the end of a stiff spring at time t seconds is given by: s = ae−kt sin 2π f t . Determine the velocity and acceleration of the end of the spring after
2 seconds if a = 3, k = 0.75 and f = 20.
(10)
4. Find the co-ordinates of the turning points on the curve y = 3x 3 + 6x 2 + 3x − 1 and distinguish between them.
(7)
5. The heat capacity C of a gas varies with absolute temperature θ as shown:
C = 26.50 + 7.20 × 10

−3

θ − 1.20 × 10

−6 2

θ

Determine the maximum value of C and the temperature at which it occurs.
(5)
6. Determine for the curve y = 2x 2 − 3x at the point
(2, 2): (a) the equation of the tangent (b) the equation of the normal.
(6)
7. A rectangular block of metal with a square crosssection has a total surface area of 250 cm2 . Find the maximum volume of the block of metal. (7)

8. A cycloid has parametric equations given by: x = 5(θ − sin θ) and y = 5(1 − cos θ). Evaluate d2 y dy when θ = 1.5 radians. Give
(b)
(a) dx dx 2 answers correct to 3 decimal places.
(8)
9. Determine the equation of (a) the tangent, and (b) the normal, drawn to an ellipse x = 4 cos θ, π (8) y = sin θ at θ = .
3
10. Determine expressions for

dz for each of the dy following functions:
(a) z =5y 2 cos x (b) z = x 2 + 4x y − y 2 .

(5)

dy
11. If x 2 + y 2 + 6x + 8y + 1 = 0, find in terms of x dx and y.
(3)
12. Determine the gradient of the tangents drawn to
(3)
the hyperbola x 2 − y 2 = 8 at x = 3.
13. Use logarithmic differentiation to differentiate

(x + 1)2 (x − 2) y= with respect to x.
(6)
(2x − 1) 3 (x − 3)4
3eθ sin 2θ

and hence evaluate θ5 dy π , correct to 2 decimal places, when θ = . dθ 3
(9)

14. Differentiate y =

d √ t (2t + 1) when t = 2, correct to 4
15. Evaluate dt significant figures.
(5)

Chapter 32

Differentiation of hyperbolic functions 32.1 Standard differential coefficients of hyperbolic functions

(a)

=

From Chapter 5, d d ex − e−x
(sinh x) = dx dx
2
=

e x + e−x
2

=

e x − (−e−x )
2

= cosh x

=
(b)

=

e x + e−x
2

ex − e−x
2

=

If y = cosh ax, where ‘a’ is a constant, then dy = a sinh ax dx Using the quotient rule of differentiation the derivatives of tanh x, sech x, cosech x and coth x may be determined using the above results.
Problem 1. Determine the differential coefficient of: (a) th x (b) sech x.

ch2 x − sh2 x
1
= 2 = sech2 x
2x
ch ch x

d d (sech x) = dx dx

1 ch x

(ch x)(0) − (1)(sh x) ch2 x

=

ex + (−e −x )
2

= sinh x

(ch x)(ch x) − (sh x)(sh x) ch2 x using the quotient rule

=

If y = sinh ax, where ‘a’ is a constant, then dy = a cosh ax dx d d (cosh x) = dx dx

sh x ch x

d d (th x) = dx dx

1
−sh x
=−
ch x ch2 x

sh x ch x

= −sech x th x dy Problem 2. Determine given dθ
(a) y = cosech θ (b) y = coth θ.
(a)

d d (cosec θ) = dθ dθ

1 sh θ

=

(sh θ)(0) − (1)(ch θ) sh2 θ

=

1
−ch θ
=−
2θ sh θ sh = −cosech θ coth θ

ch θ sh θ

332 Higher Engineering Mathematics
(b)

d d (coth θ) = dθ dθ
=

=
=

ch θ sh θ

(sh θ)(sh θ) − (ch θ)(ch θ) sh2 θ
−(ch2 θ − sh2 θ) sh2 θ − ch2 θ
=
2 sh θ sh2 θ
−1
= −cosech2 θ sh2 θ

x
− 2 coth 4x
2
x
1
dy
− 2(−4 cosech2 4x)
=5
sech2 dx 2
2

(b) y = 5 th

=

x
5
sech2 + 8 cosech2 4x
2
2

Problem 4. Differentiate the following with respect to the variable: (a) y = 4 sin 3t ch 4t
(b) y = ln (sh 3θ) − 4 ch2 3θ.
(a) y = 4 sin 3t ch 4t (i.e. a product)

Summary of differential coefficients y or f (x)

dy or f (x) dx sinh ax

a cosh ax

cosh ax

a sinh ax

tanh ax

a sech2 ax

sech ax

−a sech ax tanh ax

cosech ax −a cosech ax coth ax coth ax

32.2

−a cosech 2 ax

Further worked problems on differentiation of hyperbolic functions Problem 3. Differentiate the following with respect to x:
3
(a) y = 4 sh 2x − ch 3x
7
x
(b) y = 5 th − 2 coth 4x.
2
(a) y = 4 sh 2x −

3 ch 3x
7

dy
3
= 4(2 cosh 2x) − (3 sinh 3x) dx 7
9
= 8 cosh 2x − sinh 3x
7

dy
= (4 sin 3t )(4 sh 4t ) + (ch 4t )(4)(3 cos 3t ) dx = 16 sin 3t sh 4t + 12 ch 4t cos 3t
= 4(4 sin 3t sh 4t + 3 cos 3t ch 4t)
(b) y = ln (sh 3θ) − 4 ch 2 3θ
(i.e. a function of a function) dy = dθ 1
(3 ch 3θ) − (4)(2 ch 3θ)(3 sh 3θ) sh 3θ

= 3 coth 3θ − 24 ch 3θ sh 3θ
= 3(coth 3θ − 8 ch 3θ sh 3θ)
Problem 5. of Show that the differential coefficient

y=

3x 2 is: 6x sech 4x (1 − 2x th4x). ch 4x

y=

3x 2 ch 4x

(i.e. a quotient)

dy
(ch 4x)(6x) − (3x 2 )(4 sh 4x)
=
dx
(ch 4x)2
=

6x(ch 4x − 2x sh 4x) ch2 4x

= 6x

ch 4x
2x sh 4x

2 ch 4x ch2 4x

= 6x

1
− 2x ch 4x

sh 4x ch 4x

1 ch 4x

= 6x[sech 4x − 2x th 4x sech 4x]
= 6x sech 4x (1 −2x th 4x)

333

Differentiation of hyperbolic functions
Now try the following exercise
3. (a) 2 ln (sh x) (b)
Exercise 134 Further problems on differentiation of hyperbolic functions

(a) 2 coth x (b)

In Problems 1 to 5 differentiate the given functions with respect to the variable:








θ θ 3 sech cosech
8
2
2

4. (a) sh 2x ch 2x (b) 3e2x th 2x

1. (a) 3 sh 2x (b) 2 ch 5θ (c) 4 th9t
(a) 6 ch 2x (b) 10 sh 5θ (c) 36 sech2 9t
2
5 t 2. (a) sech 5x (b) cosech
(c) 2 coth 7θ
3
2
⎡ 8
10
⎢ (a) − 3 sech 5x th 5x

⎢ t t
5
⎢ (b) − cosech coth

16
2
2
(c) −14 cosech2 7θ

θ
3
ln th
4
2

(a) 2(sh2 2x + ch2 2x)
(b) 6e2x (sech 2 2x + th 2x)
5. (a)

3 sh 4x ch 2t
(b)
2x 3 cos 2t

12x ch 4x − 9 sh 4x
⎢ (a)
2x 4


2(cos 2t sh 2t + ch 2t sin 2t )
(b)
cos2 2t






Chapter 33

Differentiation of inverse trigonometric and hyperbolic functions
33.1

Inverse functions

y +2
If y = 3x − 2, then by transposition, x =
. The
3
y +2 function x = is called the inverse function of
3
y = 3x − 2 (see page 188).
Inverse trigonometric functions are denoted by prefixing the function with ‘arc’ or, more commonly, by using the −1 notation. For example, if y = sin x, then x = arcsin y or x = sin−1 y. Similarly, if y = cos x, then x = arccos y or x = cos−1 y, and so on. In this chapter the −1 notation will be used. A sketch of each of the inverse trigonometric functions is shown in Fig. 33.1.
Inverse hyperbolic functions are denoted by prefixing the function with ‘ar’ or, more commonly, by using the −1 notation. For example, if y = sinh x, then x = arsinh y or x = sinh−1 y. Similarly, if y = sech x, then x = arsech y or x = sech−1 y, and so on. In this chapter the −1 notation will be used. A sketch of each of the inverse hyperbolic functions is shown in Fig. 33.2.

33.2 Differentiation of inverse trigonometric functions
(i) If y = sin−1 x, then x = sin y.
Differentiating both sides with respect to y gives: dx = cos y = 1 − sin2 y dy since cos2 y + sin2 y = 1, i.e.
However

dx √
= 1 − x2 dy dy
1
= dx dx dy Hence, when y = sin−1 x then dy 1
=√
dx
1 −x2
(ii) A sketch of part of the curve of y = sin−1 x is shown in Fig. 33.1(a). The principal value of sin−1 x is defined as the value lying between
−π/2 and π/2. The gradient of the curve between points A and B is positive for all values of x and thus only the positive value is taken when
1
evaluating √
1 − x2 x (iii) Given y = sin−1 a x = a sin y
Hence

then

x
= sin y a dx
= a cos y = a 1 − sin2 y dy =a

1−

x a 2

=a

a2 − x 2 a2 √ a a2 − x 2 √ 2
=
= a − x2 a and

Differentiation of inverse trigonometric and hyperbolic functions y y

y
3␲/2

3␲/2

y 5 sin21x


␲/2

y 5cos21x

␲/2

B

21
0
A 2␲/2

y 5 tan21x



D

11 x

21

11 x

0
2␲/2

2␲

23␲/2

(a)

2␲/2

(b)

(c) y y
3␲/2

3␲/2


␲/2

y




y 5 sec21x

y 5 cosec21x

␲/2

␲/2

21 0 11
2␲/2

x

21 0
2␲/2

2␲

11

x

x

2␲/2

23␲/2

(d)

y 5 cot21x

0

2␲

23␲/2

x

0

2␲

23␲/2

␲/2

C



(e)

(f)

Figure 33.1

y
3
2

y 5 sinh21x

y

y
3

y 5cosh21x

2

y 5 tanh21x

1

1
01 2 3x
23 22 21
21

22 21 0
21

22

21

2 3x

11 x

0

22

23

1

23
(b)

(a) y 3

(c)

y y 5 sech21x

y y 5cosech21x

2

y 5 coth21x

1
0
21
22

1

x

0

x

21 0 11

23
(c)

Figure 33.2

(e)

(f)

x

335

336 Higher Engineering Mathematics dy 1
1
=
=√
dx dx a2 − x 2 dy x dy 1
i.e. when y = sin−1 then
=√
2 − x2 a dx a Table 33.1 Differential coefficients of inverse trigonometric functions

Thus

y or f (x)

Since integration is the reverse process of differentiation then:
1
x

dx = sin−1 + c
2 − x2 a a

(i)

(iv) Given y = sin−1 f (x) the function of a function dy rule may be used to find dx Then

x a 1

2 − x2 a cos−1

dy du 1
= f (x) and
=√
dx du 1 − u2

tan −1

x a (iv)

Problem 1.

Find

dy given y = sin−1 5x 2 . dx (v)

dy
=
dx

Hence, if y = sin−1 5x 2 f (x) = 10x.
10x
1 − (5x 2 )2

cosec−1

x a cosec−1 f (x)
(vi)

cot −1

f (x)

f (x)
1 + [ f (x)]2 x x a cot −1 f (x)

y = sin−1 f (x) then

dy
=
dx

x a sec−1 f (x)

From Table 33.1(i), if

Thus

sec−1

1 − [ f (x)]2 a a2 + x 2

tan −1 f (x)

(v) The differential coefficients of the remaining inverse trigonometric functions are obtained in a similar manner to that shown above and a summary of the results is shown in Table 33.1.

− f (x)

cos−1 f (x)
(iii)

dy dy du
1
f (x)
=
×
=√
dx du dx
1 − u2 f (x)
=
1 −[ f (x)]2

1 − [ f (x)]2
−1
√ a2 − x 2

x a (see para. (i))
Thus

f (x)

sin−1 f (x)
(ii)

Let u = f (x) then y = sin−1 u

sin−1

dy or f (x) dx √

a x 2 − a2 f (x)

f (x) [ f (x)]2 − 1


−a

x x 2 − a2
− f (x) f (x) [ f (x)]2 − 1
−a
a2 + x 2
− f (x)
1 + [ f (x)]2

1 − [ f (x)]2

then

f (x) = 5x 2

and

(a) If y = cos−1 x then x = cos y.
Differentiating with respect to y gives:

10x
=√
1 −25x4

Problem 2.
(a) Show that if y = cos−1 x then dy 1
=√
dx
1 − x2
(b) Hence obtain the differential coefficient of y = cos−1 (1 − 2x 2 ).

dx
= −sin y = − 1 − cos2 y dy √
=− 1 − x2 dy 1
1
=
=−√
dx dx 1 −x2 dy The principal value of y = cos−1 x is defined as the angle lying between 0 and π, i.e. between points C and D shown in Fig. 33.1(b). The gradient of the curve
Hence

Differentiation of inverse trigonometric and hyperbolic functions is negative between C and D and thus the differential dy coefficient is negative as shown above. dx (b) If y = cos−1 f (x) then by letting u = f (x), y = cos−1 u dy 1
(from part (a))
=−√
du
1 − u2

Then

du and = f (x) dx From the function of a function rule,

dy between these two values is always positive dx (see Fig. 33.1(c)).
2x
x
3
Comparing tan −1 with tan−1 shows that a =
3
a
2
2x
Hence if y = tan −1 then 3
i.e.

3
2
=
=
9
9 + 4x 2
3 2
2
+x
+ x2
4
4
2
3
(4)
6
= 2 2=
9 + 4x
9 + 4x2

− f (x)
1 − [ f (x)]2

−1

Hence, when y = cos (1 − 2x )

=
=

1 − [1 − 2x 2 ]2

4x
1 − (1 − 4x 2

+ 4x 4 )

4x
[4x 2 (1 − x 2 )]

=

By the function of a function rule,

4x

=

dy dy du
1
d
=
·
= × (cos−1 3x) dx du dx u dx

(4x 2 − 4x 4 )

4x
2
=√

2x 1 − x 2
1 − x2

Problem 3. Determine the differential coefficient x of y = tan −1 and show that the differential a 2x
6
coefficient of tan−1 is 3
9 + 4x 2
If y = tan −1

Problem 4. Find the differential coefficient of y = ln (cos−1 3x).
Let u = cos−1 3x then y = ln u.

−(−4x)

dy
=
dx

then

2

3
2

3
2

dy
=
dx

dy dy du
1
f (x)
=
·
= −√ dx du dx
1 − u2
=

x x then = tan y and x = a tan y a a

=

i.e.

= a 1+

x a 2

=a

−3

1 cos−1 3x

Problem 5. If y = tan−1

a2

a2 + x 2 a dy
1
a
Hence
=
= 2 dx dx a + x2 dy =

The principal value of y = tan−1 x is defined as π π the angle lying between − and and the gradient
2
2

3 dy find t2 dt

Using the general form from Table 33.1(iii), f (t ) =

a2 + x 2

1 − (3x)2

−3 d [ln(cos−1 3x)]= √ dx 1 − 9x2 cos−1 3x

dx
= a sec2 y = a(1 + tan2 y) since dy sec2 y = 1 + tan2 y

337

from which f (t ) =
Hence

3 d tan−1 2 dt t
6
− 3 t =
3
1+ 2 t = −

6 t3 3
= 3t −2, t2 −6 t3 =

2

f (t )
1 + [ f (t )]2
6
− 3 t = 4 t +9 t4 t4 t4 + 9

=−

6t t4 + 9

338 Higher Engineering Mathematics
Differentiate y =

Problem 6.

cot −1 2x
1 + 4x 2

Using the quotient rule:
−2
− (cot −1 2x)(8x)
(1 + 4x 2 ) dy 1 + (2x)2
=
dx
(1 + 4x 2 )2 from Table 33.1(vi)
=

−2(1 +4x cot−1 2x)

−1 cos t − 1
=
(cos t − 1)2 + (sin t )2
(cos t − 1)2
=

=

−1 cos t − 1

(cos t − 1)2 cos2 t − 2 cos t + 1 + sin2 t

−(cos t − 1)
1 − cos t
1
=
=
2 − 2 cos t
2(1 − cos t ) 2

(1 +4x2 )2
Now try the following exercise
Differentiate y = x cosec

Problem 7.

−1

x.
Exercise 135 Further problems on differentiating inverse trigonometric functions Using the product rule:
−1
dy
= (x) √
+ (cosec −1 x) (1)
2 −1 dx x x from Table 33.1(v)
−1
+ cosec −1 x
=√
2 −1 x Problem 8. y = tan−1

If

Show that if dy sin t
1
then
=
cos t − 1 dt 2 sin t cos t − 1

f (t ) =

then f (t ) =
=

In Problems 1 to 6, differentiate with respect to the variable. x
1. (a) sin−1 4x (b) sin−1
2
4
1
(a) √
(b) √
2
1 − 16x
4 − x2
2.

−3
−2
(a) √
(b) √
1 − 9x 2
3 9 − x2
3.

(a) 3 tan−1 2x (b)

(cos t − 1)(cos t ) − (sin t )(−sin t )
(cos t − 1)2 cos2 t − cos t + sin2 t
1 − cos t
=
2
(cos t − 1)
(cos t − 1)2

(a)

4.

−(cos t − 1)
−1
=
(cos t − 1)2 cos t − 1

y

dy then = dt 3 x 4

θ
5
cosec−1 (b) cosec−1 x 2
2
2

5.

(a)

6.

−2
−5
(b) √
(a) √
2 −4 θ θ x x4 − 1

(a) 3 cot −1 2t (b) cot −1 θ 2 − 1

sin t cos t − 1

−1 cos t − 1 sin t
1+
cos t − 1

1
6
(b) √
2
1 + 4x
4 x (1 + x)

2
4
(a) √
(b) √ t 4t 2 − 1 x 9x 2 − 16

Using Table 33.1(iii), when
= tan −1


1
tan−1 x
2

(a) 2 sec−1 2t (b) sec−1

since sin2 t + cos2 t = 1
=

x
2
cos−1
3
3

(a) cos−1 3x (b)

2

(a)

−1
−6
(b) √
2
1 + 4t θ θ2 − 1

339

Differentiation of inverse trigonometric and hyperbolic functions

7.

Show that the differential coefficient of x 1 + x2 tan−1 . is 1 − x2
1 − x2 + x4

Hence e y =
=

In Problems 8 to 11 differentiate with respect to the variable.
8.

9.

10.

11.

(a) 2x sin−1 3x (b) t 2 sec−1 2t


6x
(a) √
+ 2 sin−1 3x


1 − 9x 2



⎣ t + 2t sec−1 2t
(b) √
4t 2 − 1
(a) θ 2 cos−1 (θ 2 − 1) (b) (1 − x 2 ) tan −1 x


2θ 2
(a) 2θ cos−1 (θ 2 − 1) − √



2 − θ2 ⎥




1 − x2
−1 x
(b)
− 2x tan
1 + x2


(a) 2 t cot −1 t (b) x cosec−1 x



1
−2 t
−1 t
⎢ (a) 1 + t 2 + √t cot






1
−1 x− √
(b) cosec
2 (x − 1)
(a)

cos−1 x sin−1 3x
(b) √ x2 1 − x2

3x
1
− 2 sin−1 3x
⎢ (a) x 3 √
1 − 9x 2



x

⎢ cos−1 x
−1 + √

1 − x2
(b)
(1 − x 2 )










33.3 Logarithmic forms of inverse hyperbolic functions
Inverse hyperbolic functions may be evaluated most conveniently when expressed in a logarithmic form. x x For example, if y = sinh−1 then = sinh y. a a
From Chapter 5, e y = cosh y + sinh y and cosh 2 y − sinh2 y = 1, from which, cosh y = 1 + sinh2 y which is positive since cosh y is always positive (see Fig. 5.2, page 43).

1 + sinh2 y + sinh y x a

1+

2

+

a2 + x 2 a2 x
=
a

+

x a √

a2 + x 2 x x + a2 + x 2
=
+ or a a a
Taking Napierian logarithms of both sides gives:

x + a2 + x 2 y = ln a x x + a2 + x2
= ln a a

Hence, sinh−1

(1)

3
Thus to evaluate sinh−1 , let x = 3 and a = 4 in
4
equation (1).

3 + 42 + 32
−1 3
Then sin h
= ln
4
4
= ln

3+5
4

= ln 2 = 0.6931

By similar reasoning to the above it may be shown that:

x + x2 − a2
−1 x cosh = ln a a and tanh−1

a+x x 1
= ln a 2 a−x Problem 9. Evaluate, correct to 4 decimal places, sinh−1 2.

2
2
−1 x = ln x + a + x
From above, sinh a a
With x = 2 and a = 1,

2 + 12 + 22 sinh−1 2 = ln
1

= ln(2 + 5) = ln 4.2361
= 1.4436, correct to 4 decimal places
Using a calculator,
(i) press hyp
(ii) press 4 and sinh−1 ( appears
(iii) type in 2

340 Higher Engineering Mathematics
(iv) press ) to close the brackets
(v) press = and 1.443635475 appears
−1

Hence, sinh

2 = 1.4436, correct to 4 decimal places.

Problem 10. Show that a+x x 1 tanh−1 = ln and evaluate, correct a 2 a−x 3 to 4 decimal places, tanh −1
5
If y = tanh−1

x x then = tanh y. a a

If y = cosh−1

x x then = cos y a a

e y = cosh y + sinh y = cosh y ± cosh2 y − 1 x a

x
= ± a =





2

x
−1 = ± a √

x 2 − a2 a x 2 − a2 a Taking Napierian logarithms of both sides gives:

x ± x 2 − a2 y = ln a From Chapter 5,
1 y
(e − e−y ) e2y − 1 sinh x
= 2y
= 2 tanh y =
1 y
−y
cosh x e +1
2 (e + e )

by dividing each term by e−y x e2y − 1
=
a e2y + 1

Thus,

from which, x(e2y + 1) = a(e2y − 1)
Hence x + a = ae2y − xe2y = e2y (a − x) a+x a−x

from which e2y =

Taking Napierian logarithms of both sides gives:
2y = ln and y=

a+x a−x 1 a+x ln
2
a−x

5+3
3 1
= ln
5 2
5−3

=

1 ln 4
2

= ln 2.3798 = 0.8670, correct to 4 decimal places.

In Problems 1 to 3 use logarithmic equivalents of inverse hyperbolic functions to evaluate correct to
4 decimal places.

Prove that

x + x 2 − a2
−1 x cosh = ln a a correct to

1. (a) sinh−1

1
(b) sinh−1 4 (c) sinh−1 0.9
2
[(a) 0.4812 (b) 2.0947 (c) 0.8089]
5
(b) cosh−1 3 (c) cosh−1 4.3
4
[(a) 0.6931 (b) 1.7627 (c) 2.1380]

3. (a) tanh−1

Problem 11.

cosh −1 1.4

Exercise 136 Further problems on logarithmic forms of the inverse hyperbolic functions 2. (a) cosh−1

= 0.6931, correct to 4 decimal places.

and hence evaluate
4 decimal places.

14
7
= cosh−1
10
5
−1 x , let x = 7 and a = 5
In the equation for cosh a √
7
7 + 72 − 52
Then cosh−1 = ln
5
5 cosh−1 1.4 = cosh−1

Now try the following exercise

a+x x 1
Hence, tanh−1 = ln a 2 a−x Substituting x = 3 and a = 5 gives: tanh−1 Thus, assuming the principal value,

x + x2 − a2
−1 x
= ln cosh a a 1
5
(b) tanh−1 (c) tanh−1 0.7
4
8
[(a) 0.2554 (b) 0.7332 (c) 0.8673]

Differentiation of inverse trigonometric and hyperbolic functions
Table 33.2 Differential coefficients of inverse hyperbolic functions

33.4 Differentiation of inverse hyperbolic functions x x
If y = sinh−1 then = sinh y and x = a sinh y a a dx = a cosh y (from Chapter 32). dy Also cosh2 y − sinh2 y = 1, from which,
1 + sinh2 y =

cosh y =

x a 1+

(i) sinh−1

x a √

(ii) cosh−1

[ f (x)]2 + 1

x a √

cosh−1 f (x)

√ a a2 + x 2 √ 2 dx = a cosh y =
= a + x2 dy a

(iii) tanh−1

dy
1
1
Then
=
=
dx dx a2 + x2 dy x
[An alternative method of differentiating sinh−1 a is to differentiate the logarithmic form

x + a2 + x 2 with respect to x.] ln a

x a x a (v) cosech−1



or

ln

x 2 + a2

x+

dx = sinh

√ a2 + x 2 a −1

(vi) coth−1
+c

− f (x) f (x) 1 − [ f (x)]2

x a √

d
(sinh−1 x)= dx −a

x x 2 + a2

x a − f (x) f (x) [ f (x)]2 + 1 a a2 − x 2 f (x)
1 − [ f (x)]2

coth−1 f (x)

It may be shown that

−a

x a2 − x 2

cosech−1 f (x)

x
+c
a

[ f (x)]2 − 1



It follows from above that
1

f (x)

f (x)
1 − [ f (x)]2

sech−1 f (x)

From the sketch of y = sinh−1 x shown in Fig. 33.2(a) dy is always positive. it is seen that the gradient i.e. dx 1 x 2 − a2

a a2 − x 2

tanh−1 f (x)
(iv) sech−1

1 x 2 + a2 f (x)

sinh−1 f (x)

2

√ a2 + x 2
=
a

Hence

dy or f (x) dx y or f (x)

1
Problem 12. Find the differential coefficient of y = sinh−1 2x.

x2 + 1

or more generally
From Table 33.2(i), d [sinh−1 f (x)] = dx f (x)

d
[sinh−1 f (x)] = dx [ f (x)] + 1
2

by using the function of a function rule as in
Section 33.2(iv).
The remaining inverse hyperbolic functions are differentiated in a similar manner to that shown above and the results are summarized in Table 33.2.

Hence

d
(sinh−1 2x) = dx =

f (x)
[ f (x)]2 + 1
2
[(2x)2 + 1]
2
[4x2 + 1]

341

342 Higher Engineering Mathematics
Problem 13. Determine d cosh−1 (x 2 + 1) dx If y = cosh −1 f (x),

Problem 15.

From Table 33.2(v), f (x)

dy
=
dx

[

f (x)]2

− f (x) d [cosech−1 f (x)] = dx f (x) [ f (x)]2 + 1

−1

If y = cosh−1 (x 2 + 1), then f (x) = (x 2 + 1) and
1
x f (x) = (x + 1)−1/2 (2x) =
2 + 1)
2
(x
Hence,

d cosh−1 dx

(x 2 + 1)
=
x

2

−1

(x 2 + 1 − 1)

(x2 + 1)

x x then = tanh y and x = a tanh y a a

dx
= a sech2 y = a(1 − tanh2 y), since dy 1 − sech2 y = tanh2 y x 2 a2 − x 2 a2 − x 2
=a
=
2
a a a dy 1 a Hence
=
= 2 dx dx a − x2 dy 4x x 3
Comparing tanh−1 with tanh−1 shows that a =
3
a
4
3
3
4x d 4
4
tanh−1
=
=
Hence
9 dx 3
3 2
− x2
− x2
16
4
3
3
16
12
4
=
= ·
=
2
2)
4 (9 − 16x
9 −16x2
9 − 16x
16
=a 1−

=

−cosh θ

since cosh2 θ − sinh 2 θ = 1 sinh θ cosh2 θ
−1
−cosh θ
=
= −cosech θ sinh θ cosh θ sinh θ

1

x d Problem 14. Show that tanh−1 = dx a a and hence determine the differential a2 − x 2
4x
coefficient of tanh−1
3
If y = tanh−1

sinh θ [sinh2 θ + 1]

=

=

(x 2 + 1)

x
=

d
[cosech−1 (sinh θ)] dθ −cosh θ

x

x

(x 2 + 1)

Hence
=

(x 2 + 1)

(x 2 + 1)

=

Differentiate cosech−1 (sinh θ).

Problem 16. Find the differential coefficient of y = sech−1 (2x − 1).
From Table 33.2(iv), d − f (x)
[sech−1 f (x)] = dx f (x) 1 − [ f (x)]2
Hence,
=
=
=
=

d
[sech −1 (2x − 1)] dx −2

(2x − 1) [1 − (2x − 1)2 ]
−2
(2x − 1) [1 − (4x 2 − 4x + 1)]
−2
(2x − 1) (4x

− 4x 2 )

=

−2

(2x −1) [4x(1−x)]

−2
−1
=


(2x − 1)2 [x(1 − x)] (2x − 1) [x(1 −x)]

Problem 17. Show that d [coth−1 (sin x)] = sec x. dx From Table 33.2(vi), f (x) d [coth−1 f (x)] = dx 1 − [ f (x)]2

Differentiation of inverse trigonometric and hyperbolic functions
Hence

d cos x
[coth−1 (sin x)] = dx [1 − (sin x)2 ]

Since

tanh−1

x a =

a a2 − x 2

=

cos x since cos2 x + sin2 x = 1 cos2 x

then

x a dx = tanh−1 + c a2 − x 2 a =

1
= sec x cos x

i.e.

x
1
1 dx = tanh−1 + c a2 − x 2 a a

Problem 18. Differentiate y = (x 2 − 1) tanh−1 x.

Hence

2 dx = 2
(9 − 4x 2 )

Using the product rule,
1
dy
= (x 2 − 1) dx 1 − x2

=
+ (tanh

−1

=

Problem 19. Determine

then

Hence

d x sinh−1
=
dx a dx
(x 2 + a 2 )
1
(x 2 + 4)

dx
(x 2 + 4)
1

x
+c
a
1

(x 2 + 22 )

Problem 20. Determine

Since

then

Hence

x d cosh−1
=
dx a 1
(x 2 − a 2 )
4
(x 2 − 3)

4
(x 2 − 3)

dx

1

(a)
2.

(a) 2 cosh −1

(x 2 − a 2 )

(x 2

+ 9)

(b)

4
(16x 2 + 1)

t
1
(b) cosh −1 2θ
3
2

(a)

x
+c
a

2
(t 2 − 9)

(b)

1
(4θ 2 − 1)

3. dx = 4

(a) tanh −1

2x
(b) 3 tanh−1 3x
5
9
10
(b)
(a)
25 − 4x 2
(1 − 9x 2 )

4.

(a) sech−1

3x
1
(b) − sech −1 2x
4
2

1

dx

[x 2 − ( 3)2 ]

x
= 4 cosh−1 √ + c
3
Problem 21. Find

+c

3
2

In Problems 1 to 11, differentiate with respect to the variable. x 1. (a) sinh−1 (b) sinh−1 4x
3

dx.

1

dx = cosh −1

x

tanh−1

3
2

Exercise 137 Further problems on differentiation of inverse hyperbolic functions x
+c
2

= sinh−1

1

Now try the following exercise

(x 2 + a 2)

= sinh−1

dx =

1
2

dx

3 2
2
2 −x

2
2x
1 dx = tanh−1
+c
2)
(9 − 4x
3
3

i.e.

.

dx

1

1
2

x)(2x)

−(1 − x 2 )
=
+ 2x tanh−1 x = 2x tanh−1 x − 1
(1 − x 2 )

Since

1
− x2

9
4

4

2 dx. (9 − 4x 2 )

(a)

−4 x (16 − 9x 2 )

(b)

1
2x (1 − 4x 2 )

343

344 Higher Engineering Mathematics

5.

(a) cosech−1
(a)

6.

7.

x
1
(b) cosech−1 4x
4
2
−4

x (x 2 + 16)

(a) coth−1

2x (16x 2 + 1)

2x
1
(b) coth−1 3t
7
4
14
3
(a)
(b)
49 − 4x 2
4(1 − 9t 2)

1 cosh −1
2
(a)

(x 2 − 1)

(a) cosh−1

13. (a)
(b)

1
2 (x 2 + 1)

−1

(b) 1
(x − 1) [x(2 − x)]

t t −1

(b)






d
[x cosh−1 (cosh x)] = 2x. dx (b) coth−1 (cos x)

−1

(b) −cosec x
(t − 1) (2t − 1)

(a) θ sinh−1 θ (b) x cosh−1 x

⎡ θ + sinh−1 θ
(a)


(θ 2 + 1)





−1 ⎦

x cosh x
+
(b)

2 x
(x 2 − 1)

1
(x 2

14. (a)
(b)

dx

+ 9)
3

x
2x
3
+ c (b) sinh−1
+c
3
2
5

1
(x 2

− 16)

1
(t 2 − 5)

dx

dt

(a) cosh−1
15. (a)

dx

(4x 2 + 25)
(a) sinh−1

(a)

10.



In Problems 13 to 15, determine the given integrals. (x 2 + 1)
2


2 sec h−1 t tan h −1 x
(b)
2 t (1 − x 2 )


1
−1
(a) 3 √
+ 4 sech −1 t

t
(1 − t )



1 + 2x tanh−1 x
(b)
(1 − x 2 )2

12. Show that

(a) sech−1 (x − 1) (b) tanh−1(tanh x)
(a)

9.

−1

(a) 2 sinh−1 (x 2 − 1)
(b)

8.

(b)

11. (a)

x t + c (b) cosh−1 √ + c
4
5


(36 + θ 2 )

(b)

3 dx (16 − 2x 2 )

⎤ θ 1 tan −1 + c


6
6



⎣ x 3
−1 √ + c
(b) √ tanh
2 8
8


(a)

Chapter 34

Partial differentiation constant’. Thus,

34.1

Introduction to partial derivatives In engineering, it sometimes happens that the variation of one quantity depends on changes taking place in two, or more, other quantities. For example, the volume V of a cylinder is given by V = πr 2 h. The volume will change if either radius r or height h is changed.
The formula for volume may be stated mathematically as V = f (r, h) which means ‘V is some function of r and h’. Some other practical examples include: l i.e. t = f (l, g). g (i) time of oscillation, t = 2π

(ii) torque T = I α, i.e. T = f (I, α).
(iii) pressure of an ideal gas p =
i.e. p = f (T, V ).
(iv) resonant frequency fr =

mRT
V

1


2π LC
i.e. fr = f (L , C), and so on.

When differentiating a function having two variables, one variable is kept constant and the differential coefficient of the other variable is found with respect to that variable. The differential coefficient obtained is called a partial derivative of the function.

34.2

First order partial derivatives

A ‘curly dee’, ∂, is used to denote a differential coefficient in an expression containing more than one variable. ∂V means ‘the partial
Hence if V = πr 2 h then
∂r
derivative of V with respect to r, with h remaining

∂V d = (πh) (r 2 ) = (πh)(2r) = 2πrh.
∂r
dr
∂V
Similarly, means ‘the partial derivative of V with
∂h
respect to h, with r remaining constant’. Thus, d ∂V
= (πr 2 ) (h) = (πr 2 )(1) = πr 2 .
∂h
dh
∂V
∂V and are examples of first order partial
∂r
∂h derivatives, since n =1 when written in the form
∂n V
.
∂r n
First order partial derivatives are used when finding the total differential, rates of change and errors for functions of two or more variables (see Chapter 35), when finding maxima, minima and saddle points for functions of two variables (see Chapter 36), and with partial differential equations (see Chapter 53).
Problem 1. If z = 5x 4 + 2x 3 y 2 − 3y find
∂z
∂z
(a)
and (b)
.
∂x
∂y
(a)

∂z
, y is kept constant.
∂x
Since z = 5x 4 + (2y 2 )x 3 − (3y) then, To find

d d d
∂z
=
(5x 4 ) + (2y 2 ) (x 3 ) − (3y) (1)
∂x
dx dx dx
= 20x 3 + (2y 2 )(3x 2 ) − 0.
Hence

∂z
= 20x3 + 6x2 y2 .
∂x

346 Higher Engineering Mathematics
∂z
, x is kept constant.
∂y

(b) To find

Problem 4.

Since z =(5x 4 ) + (2x 3 )y 2 − 3y

1

z=

then,
∂z
d d d
= (5x 4 ) (1) + (2x 3 ) (y 2 ) − 3 ( y)
∂y
dy dy dy
= 0 + (2x )(2y) − 3

1

z=

(x 2 + y 2 )

∂y
, t is kept constant.
To find
∂x
Hence

i.e.
To find

Hence

i.e.

d
∂y
= (4 cos 2t ) (sin 3x)
∂x
dx
= (4 cos 2t )(3 cos3x)
∂y
= 12 cos 3x cos 2t
∂x
∂y
, x is kept constant.
∂t
d
∂y
= (4 sin 3x) (cos 2t )
∂t
dt
= (4 sin 3x)(−2 sin 2t )
∂y
= −8 sin 3x sin 2t
∂t

Problem 3.

If z =sin x y show that
1 ∂z 1 ∂z
=
y ∂x x ∂ y

∂z
= y cos x y, since y is kept constant.
∂x
∂z
= x cos x y, since x is kept constant.
∂y
1 ∂z
=
y ∂x and Hence

1
( y cos x y) = cos x y y 1 ∂z
1
(x cos x y) = cos x y.
=
x ∂y x 1 ∂z
1 ∂z
=
y ∂x x ∂y

= (x 2 + y 2 )

−1
2

−3
∂z
1
= − (x 2 + y 2 ) 2 (2x), by the function of a
∂x
2 function rule (keeping y constant)

∂z
Hence
= 4x3 y − 3.
∂y
Given y = 4 sin 3x cos 2t , find

∂z
∂z
and when ∂x
∂y

(x 2 + y 2 )

3

Problem 2.
∂y
and
∂t

Determine

∂y
∂x

−x

=

3 y2 ) 2

(x 2 +

=

−x
(x2 + y2 )3

∂z
−3
1
= − (x 2 + y 2 ) 2 (2y), (keeping x constant)
∂y
2
=

−y
(x2 + y2 )3

Problem 5. Pressure p of a mass of gas is given by pV = mRT, where m and R are constants, V is the volume and T the temperature. Find expressions
∂p
∂p for and
.
∂T
∂V
Since pV = mRT then p =

mRT
V

To find

∂p
, V is kept constant.
∂T

Hence

∂p
=
∂T

To find

∂p
, T is kept constant.
∂V

Hence

mR
V

d mR (T ) = dT V

d
∂p
= (mRT)
∂V
dV

1
V

= (m RT )(− −2 ) =
V
Problem 6.

−mRT
V2

The time of oscillation, t , of l where l is the a pendulum is given by t = 2π g length of the pendulum and g the free fall
∂t
∂t acceleration due to gravity. Determine and ∂l
∂g

Partial differentiation
To find

∂t
, g is kept constant.
∂l

To find

∂t
=
∂l



g

=

Hence

2π √ l= √ g l
=
g

t = 2π



g

4.



g

d 1
(l 2 ) = dl 1

2 l

=

z =sin(4x + 3y) ⎡

5.

z = x 3 y2 −


∂z
= 4 cos(4x + 3y) ⎥
⎢ ∂x


⎣ ∂z

= 3 cos(4x + 3y)
∂y


1
√ l2 g 1 −1 l 2
2

347

y
1
+
2
x y ⎡


∂z
2y
= 3x 2 y 2 + 3
⎢ ∂x

x


⎣ ∂z
1
1 ⎦
3y −

= 2x
∂y
x 2 y2

π lg ∂t
, l is kept constant.
∂g
6. t = 2π

z =cos 3x sin 4y

7.

The volume of a cone of height h and base radius r is given by V = 1 πr 2 h. Determine
3
∂V
∂V
and
∂h
∂r
∂V 1 2 ∂V 2
= πr
= πrh
∂h 3
∂r
3

8.

The resonant frequency fr in a series electri1 cal circuit is given by fr = √
. Show
2π LC
∂ fr
−1
that
= √
∂ L 4π C L 3

9.

An equation resulting from plucking a string is: nπb nπ nπb t + c sin t x k cos y = sin
L
L
L
∂y
∂y
Determine and ∂t
∂x


nπ nπb ∂ y nπb t ⎥
⎢ ∂t = L sin L x c cos L





⎢ nπb ⎢ t ⎥
− k sin


L





⎢ ∂ y nπ nπ nπb

t ⎥

⎢ ∂x = L cos L x k cos L





⎣ nπb t
+ c sin
L


1
l
= (2π l) √ g g

√ −1
= (2π l)g 2
Hence


∂t
1 −3
= (2π l) − g 2
∂g
2

= (2π l)

=


−π l g3 −1
2 g3

= −π

l g3 Now try the following exercise
Exercise 138 Further problems on first order partial derivatives
In Problems 1 to 6, find
1.

z =2x y

∂z
∂z
and
∂x
∂y
∂z
∂z
= 2y
= 2x
∂x
∂y


2.

3.

z = x 3 − 2x y + y 2

z=

x y ⎤
∂z
= 3x 2 − 2y ⎥
⎢ ∂x


⎣ ∂z

= −2x + 2y
∂y
⎡ ∂z
1 ⎤
=
⎢ ∂x y ⎥
⎣ ∂z
−x ⎦
= 2
∂y
y


∂z
= −3 sin 3x sin 4y

⎢ ∂x



⎣ ∂z
= 4 cos3x cos 4y
∂y


348 Higher Engineering Mathematics
10. In a thermodynamic system, k = Ae where R, k and A are constants.

T S− H
RT

as

,

∂A
∂( S)
∂( H )
∂k
(b)
(c)
(d)
Find (a)
∂T
∂T
∂T
∂T


A H T S− H
∂k
e RT
=
(a)


∂T
RT 2






∂A
k H H −T S
⎢ (b)

e RT
=−


2
∂T
RT






∂( S)
H


=− 2
⎢ (c)



∂T
T




⎣ k ⎦
∂( H )
(d)
= S − R ln
∂T
A

∂2V
. Thus,
∂r∂h
∂V
∂h

∂2V

=
∂r∂h
∂r

=


(πr 2 ) = 2π r.
∂r

∂V with respect to h, keeping r
∂r
∂ ∂V
, which is written as constant, gives
∂h ∂r
∂2V
. Thus,
∂h∂r

(iv) Differentiating

∂V
∂r

∂2V

=
∂h∂r
∂h
(v)

=


(2πrh) = 2π r.
∂h

∂2V
∂2V ∂2 V ∂2V
,
, and are examples of
2
2 ∂r∂h
∂r
∂h
∂h∂r
second order partial derivatives.

34.3

Second order partial derivatives

∂2V
∂2V
=
∂r∂h ∂h∂r and such a result is always true for continuous functions (i.e. a graph of the function which has no sudden jumps or breaks).

(vi) It is seen from (iii) and (iv) that

As with ordinary differentiation, where a differential coefficient may be differentiated again, a partial derivative may be differentiated partially again to give higher order partial derivatives.
∂V
(i) Differentiating of Section 34.2 with respect
∂r
∂ ∂V to r, keeping h constant, gives which ∂r ∂r
∂2V
is written as
∂r 2
Thus if V = πr 2 h, then ∂2V

= (2πrh) = 2π h.
2
∂r
∂r

∂V with respect to h, keeping
∂h
∂ ∂V r constant, gives which is written
∂h ∂h
∂2V
as
∂h 2

(ii) Differentiating

∂2V

= (πr 2 ) = 0.
2
∂h
∂h
∂V
(iii) Differentiating with respect to r, keeping
∂h
∂ ∂V which is written h constant, gives
∂r ∂h
Thus

Second order partial derivatives are used in the solution of partial differential equations, in waveguide theory, in such areas of thermodynamics covering entropy and the continuity theorem, and when finding maxima, minima and saddle points for functions of two variables (see
Chapter 36).
Problem 7. Given z =4x 2 y 3 − 2x 3 + 7y 2 find
∂2z
∂2 z
∂2 z
∂2z
(a) 2 (b) 2 (c)
(d)
∂x
∂y
∂x∂ y
∂ y∂x
(a)

∂z
= 8x y 3 − 6x 2
∂x
∂2 z

=
∂x 2
∂x

∂z
∂x

=


(8x y 3 − 6x 2 )
∂x

= 8y3 − 12 x
(b)

∂z
= 12x 2 y 2 + 14y
∂y
∂2 z

=
2
∂y
∂y

∂z
∂y

=

= 24x2y + 14


(12x 2 y 2 + 14y)
∂y

Partial differentiation
(c)

∂2 z

=
∂x∂ y
∂x

∂z

= (12x 2 y 2+14y) = 24xy2
∂y
∂x

(d)

∂2 z

=
∂ y∂x
∂y

∂z
∂x

It is noted that

=

=x


(8x y 3 − 6x 2 ) = 24xy2
∂y

∂2z

=
∂ y∂x
∂y

∂2z
∂2z
=
∂x∂ y ∂ y∂x

( y)
=

z = e−t sin θ,
∂2z

Problem 8. Show that when
∂2 z
∂2z
∂2 z
(a) 2 = − 2 , and (b)
=
∂t
∂θ
∂t ∂θ ∂θ∂t
(a)

∂z
∂2 z
= e−t sin θ
= −e−t sin θ and
∂t
∂t 2
∂2z
∂z
= − e−t sin θ
= e−t cos θ and
∂θ
∂θ 2
2

Hence

2

(b)

∂ z
∂ z
=− 2
2
∂t
∂θ

∂2z
∂ ∂z
(b)
=
∂t ∂θ
∂t ∂θ
∂2 z

=
∂θ∂t
∂θ

∂z
∂t


(−e−t sin θ)
∂θ
= −e−t cos θ
=

1
− (ln y)(1) y y2 using the quotient rule

∂z x ∂ 2z
= (1 − ln y)=
∂y∂ x y2
∂y
∂z
∂y

=


∂y

x
(1 − ln y) y2 =

x
[−y − 2y + 2y ln y] y4 =

x ln y, then y ∂z
∂2 z
∂2 z
(a)
=x and (b) evaluate 2 when
∂y
∂ y∂x
∂y
x = −3 and y = 1.

using the quotient rule

ln y y using the quotient rule

Problem 9. Show that if z =

∂z d ln y
= (x)
∂y
dy y ⎧

⎪ ( y) 1 − (ln y)(1) ⎪



⎬ y = (x)

⎪ y2 ⎪





∂y

=

d 1 − ln y dy y2


⎪ ( y 2 ) − 1 − (1 − ln y)(2y) ⎪



⎬ y = (x)

⎪ y4 ⎪




∂ 2z
∂ 2z
Hence
=
∂t∂θ ∂θ∂t

(a)

∂z
∂x

x
(1 − ln y) y2 = (x)


= ( e−t cos θ)
∂t
= −e−t cos θ

∂z
To find
, y is kept constant.
∂x
1 d 1
∂z
= ln y
(x) = ln y
Hence
∂x y dx y ∂z
, x is kept constant.
To find
∂y
Hence

∂2z

=
∂ y2
∂y

=

1
(1 − ln y) y2 =

Hence x

1 − ln y y2 x xy [−3 + 2 ln y] = 3 (2 ln y − 3)
4
y y When x = −3 and y = 1,
∂ 2 z (−3)
=
(2 ln 1− 3) = (−3)(−3) = 9
∂ y 2 (1)3
Now try the following exercise
Exercise 139 Further problems on second order partial derivatives
In Problems 1 to 4, find (a)
(c)

∂2z
∂2z
(d)
∂x∂ y
∂ y∂x

1.

z =(2x − 3y)2

349

∂2 z
∂2 z
(b)
2
∂x
∂ y2

(a)
8 (b) 18
(c) −12 (d) −12

350 Higher Engineering Mathematics

2.



−2
−2
⎢(a) x 2 (b) y 2 ⎥


(c) 0
(d) 0

z = 2 ln x y


3.

z=

(x − y)
(x + y)







4x
−4y
(b)
(x + y)3
(x + y)3 ⎥


2(x − y)
2(x − y) ⎦
(c)
(d)
(x + y)3
(x + y)3
(a)


4.

z = sinh x cosh 2y

(a) sinh x cosh 2y

⎢ (b) 4 sinh x cosh 2y


⎢ (c) 2 cosh x sinh 2y

(d) 2 cosh x sinh 2y

5. Given z = x 2 sin(x − 2y) find (a)
(b)

∂2 z
∂ y2








∂2z and ∂x 2

∂2 z
∂2 z
=
∂x∂ y ∂ y∂x
= 2x 2 sin(x − 2y) − 4x cos(x − 2y).


(a) (2 − x 2 ) sin(x − 2y)


+ 4x cos(x − 2y)




(b) − 4x 2 sin(x − 2y)

Show also that

∂2z

∂2 z

∂2z

∂2z

, and show that
=
∂x 2 ∂ y 2
∂x∂ y ∂ y∂x x when z = cos−1 y 6. Find


−x
∂2 z
=
,
(a)

2 − x 2 )3
∂x 2
(y



1
−x
1
∂2 z

=
+ 2
⎢(b)
2
2 − x 2 ) y2

∂y
( y − x2)
(y



∂2z
∂2z y (c)
=
=
∂x∂ y ∂ y∂x
( y 2 − x 2 )3
3x
y

7. Given z =













show that

∂2 z
∂2 z
∂2 z
=
and evaluate 2 when
∂x∂ y ∂ y∂x
∂x
1
1
x = and y = 3.
−√
2
2
8. An equation used in thermodynamics is the
Benedict-Webb-Rubine equation of state for the expansion of a gas. The equation is: p= Show that
=

RT
C0 1
+ B0 RT − A0 − 2
V
T
V2
1

+ (b RT − a) 3 + 6
V
V γ C 1+ 2
1
− γ
V
+ e V2
T2
V3

∂2 p
∂T 2
6
V 2T 4

γ
C
γ

1 + 2 e V 2 − C0 .
V
V

Chapter 35

Total differential, rates of change and small changes
35.1

Problem 2. If z = f (u, v, w) and z =3u 2 − 2v + 4w 3 v 2 find the total differential, dz.

Total differential

In Chapter 34, partial differentiation is introduced for the case where only one variable changes at a time, the other variables being kept constant. In practice, variables may all be changing at the same time.
If z = f (u, v, w, . . .), then the total differential, dz, is given by the sum of the separate partial differentials of z,
i.e. dz =

∂z
∂z
∂z du + dv + dw + · · ·
∂u
∂v
∂w

Problem 1. If z = f (x, y) and z = x 2 y 3 +

(1)

2x
+ 1, y determine the total differential, dz.

The total differential
∂z
∂z
∂z
dz = du + dv + dw ∂u
∂v
∂w
∂z
= 6u (i.e. v and w are kept constant)
∂u
∂z
= −2 + 8w 3v
∂v
(i.e. u and w are kept constant)
∂z
= 12w 2 v 2 (i.e. u and v are kept constant)
∂w
Hence dz = 6u du + (8vw 3 − 2) dv + (12v2 w2 ) dw

The total differential is the sum of the partial differentials,
i.e.

∂z
∂z
dx + dy dz =
∂x
∂y
2
∂z
= 2x y 3 +
(i.e. y is kept constant)
∂x
y
2x
∂z
= 3x 2 y 2 2
∂y
y

Problem 3. The pressure p, volume V and temperature T of a gas are related by pV = kT , where k is a constant. Determine the total differentials (a) dp and (b) dT in terms of p, V and T .
(a)

(i.e. x is kept constant)

2
2x
dx + 3x2 y2 − 2 dy
Hence dz = 2xy3 + y y

∂p
∂p
dT + dV .
∂T
∂V kT = kT then p =
V
k
∂p
kT
= and
=− 2
V
∂V
V
k kT = dT − 2 dV
V
V

Total differential dp =
Since pV hence ∂p
∂T

Thus

dp

352 Higher Engineering Mathematics
Since

pV = kT, k = pV T
V

Hence d p = dp =

i.e.

hence

pV
T
T dV V2

dT −

p p dT − dV
T
V

(b) Total differential dT =
Since

pV
T

∂T
∂T
dp+ dV ∂p
∂V

pV k ∂T V
∂T
p
= and
=
∂p k
∂V
k pV = kT, T =

V pV T

p pV T

dp +

In Problems 1 to 5, find the total differential dz.

z = 2x y − cos x

3.

z=

[3x 2 dx + 2y dy]

4.

z = x ln y

5.

z=xy+

[(2y + sin x) dx + 2x dy]

x−y x+y 2y
2x
dx − dy (x + y)2
(x + y)2 ln y d x +

x dy y



x
−4
y y+ Sometimes it is necessary to solve problems in which different quantities have different rates of change. From dz equation (1), the rate of change of z, is given by: dt (2)

Problem 4. If z = f (x, y) and z = 2x 3 sin 2y find the rate of change of z, correct to 4 significant figures, when x is 2 units and y is π/6 radians and when x is increasing at 4 units/s and y is decreasing at 0.5 units/s.

Exercise 140 Further problems on the total differential

2.

Rates of change

dz
∂z du ∂z dv
∂z dw
=
+
+
+ ··· dt ∂u dt
∂v dt
∂w dt

dV

Now try the following exercise

z = x 3 + y2

+ b(a − 3b) dc

35.2

T
T
i.e. dT = dp + dV p V

1.

b(2 + c) da + (2a − 6bc + ac) db

7. Given u = ln sin(x y) show that du = cot(x y)(y dx + x dy).

V p pV
Thus dT = d p + dV and substituting k = k k
T
gives: dT =

6. If z = f (a, b, c) and z =2ab − 3b2 c + abc, find the total differential, dz.

√ x 1
√ dx + x − 2 dy
2y x y Using equation (2), the rate of change of z,
∂z dx ∂z dy dz =
+
dt
∂x dt
∂ y dt
Since z =2x 3 sin 2y, then
∂z
∂z
= 6x 2 sin 2y and
= 4x 3 cos 2y
∂x
∂y dx = +4 dt dy and since y is decreasing at 0.5 units/s,
= −0.5 dt dz
Hence
= (6x 2 sin 2y)(+4) + (4x 3 cos 2y)(−0.5) dt = 24x 2 sin 2y − 2x 3 cos 2y π When x = 2 units and y = radians, then
6
Since x is increasing at 4 units/s,

dz
= 24(2)2 sin[2(π/6)] − 2(2)3 cos[2(π/6)] dt = 83.138 − 8.0

Total differential, rates of change and small changes dz Hence the rate of change of z,
= 75.14 units/s, dt correct to 4 significant figures.
Problem 5. The height of a right circular cone is increasing at 3 mm/s and its radius is decreasing at
2 mm/s. Determine, correct to 3 significant figures, the rate at which the volume is changing (in cm3 /s) when the height is 3.2 cm and the radius is 1.5 cm.
1
Volume of a right circular cone, V = πr 2 h
3
Using equation (2), the rate of change of volume, dV ∂V dr ∂V dh
=
+ dt ∂r dt
∂h dt
∂V
2
∂V
1
= πrh and
= πr 2
∂r
3
∂h
3
Since the height is increasing at 3 mm/s, dh i.e. 0.3 cm/s, then
= +0.3 dt and since the radius is decreasing at 2 mm/s, dr i.e. 0.2 cm/s, then
= −0.2 dt 2
1 2 dV Hence
=
πrh (−0.2) + πr (+0.3) dt 3
3
=
However,
Hence

−0.4 πrh + 0.1πr 2
3

Since

and
Hence

∂A
1
1
= c sin B,
A = ac sin B,
2
∂a
2
∂A
1
∂A
1
= a sin B and
= ac cos B
∂c
2
∂B
2 da dc
= 0.4 units/s,
= −0.8 units/s dt dt dB = 0.2 units/s dt 1
1
dA
=
c sin B (0.4) + a sin B (−0.8) dt 2
2
+

π then: 6
1
π
(0.4) +
(3) sin
2
6

1 dA π
=
(4) sin dt 2
6

+

1 π (3)(4) cos
2
6

d A ∂ A da ∂ A dc ∂ A dB
=
+
+
dt
∂a dt
∂c dt ∂ B dt

(0.2)

Problem 7. Determine the rate of increase of diagonal AC of the rectangular solid, shown in
Fig. 35.1, correct to 2 significant figures, if the sides x, y and z increase at 6 mm/s, 5 mm/s and 4 mm/s when these three sides are 5 cm, 4 cm and 3 cm respectively. C b B

z 5 3 cm

= −2.011 + 0.707 = −1.304 cm3 /s

Using equation (2), the rate of change of area,

(−0.8)

= 0.4 − 0.6 + 1.039 = 0.839 units2/s, correct to 3 significant figures.

dV
−0.4
= π(1.5)(3.2) + (0.1)π(1.5)2 dt 3

Problem 6. The area A of a triangle is given by
A = 1 ac sin B, where B is the angle between sides a
2
and c. If a is increasing at 0.4 units/s, c is decreasing at 0.8 units/s and B is increasing at 0.2 units/s, find the rate of change of the area of the triangle, correct to 3 significant figures, when a is 3 units, c is 4 units and B is π/6 radians.

1 ac cos B (0.2)
2

When a = 3, c = 4 and B =

h = 3.2 cm and r = 1.5 cm.

Thus the rate of change of volume is 1.30 cm3/s decreasing. 353

y5

4 cm

x5

5 cm

A

Figure 35.1

Diagonal AB =

(x 2 + y 2 )

Diagonal AC =

(BC 2 + AB 2 )

=

[z 2 + { (x 2 + y 2 )}2

=

(z 2 + x 2 + y 2 )

Let AC = b, then b = (x 2 + y 2 + z 2 )

354 Higher Engineering Mathematics
Using equation (2), the rate of change of diagonal b is given by: db ∂b dx ∂b dy ∂b dz
=
+
+
dt
∂x dt
∂ y dt
∂z dt

Exercise 141 change Since b = (x 2 + y 2 + z 2 )
−1
∂b
1
= (x 2 + y 2 + z 2 ) 2 (2x) =
∂x
2

Similarly,

∂b
=
∂y

and

∂b
=
∂z

x
(x 2 + y 2 + z 2 )

y
(x 2

+ y2 + z2 )

(x 2 + y 2 + z 2 )

dz
= 4 mm/s = 0.4 cm/s dt db
=
dt

x
(x 2

+ y2 + z2 )

(0.6)

y

+

(x 2 + y 2

+ z2)

z

+

(x 2 + y 2

+ z2)

(0.5)

(0.4)

When x = 5 cm, y = 4 cm and z = 3 cm, then: db = dt 5
(52

+ 42 + 32 )
+

+

1. The radius of a right cylinder is increasing at a rate of 8 mm/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.
[+226.2 cm3 /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.
[515.5 cm/s]

dy
= 5 mm/s = 0.5 cm/s, dt Hence

Further problems on rates of

2. If z = f (x, y) and z = 3x 2 y 5 , find the rate of change of z when x is 3 units and y is 2 units when x is decreasing at 5 units/s and y is increasing at 2.5 units/s.
[2520 units/s]

z

dx
= 6 mm/s = 0.6 cm/s, dt and

Now try the following exercise

4. A rectangular box has sides of length x cm, y cm and z cm. Sides x and z are expanding at rates of 3 mm/s and 5 mm/s respectively and side y is contracting at a rate of 2 mm/s. Determine the rate of change of volume when x is
3 cm, y is 1.5 cm and z is 6 cm.
[1.35 cm3 /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.
[17.4 cm2 /s]

(0.6)

35.3
4
(52 + 42 + 32 )
3
(52 + 42 + 32 )

(0.5)

(0.4)

= 0.4243 + 0.2828 + 0.1697 = 0.8768 cm/s
Hence the rate of increase of diagonal AC is
0.88 cm/s or 8.8 mm/s, correct to 2 significant figures.

Small changes

It is often useful to find an approximate value for the change (or error) of a quantity caused by small changes (or errors) in the variables associated with the quantity. If z = f (u, v, w, . . .) and δu, δv, δw, . . . denote small changes in u, v, w, . . . respectively, then the corresponding approximate change δz in z is obtained from equation (1) by replacing the differentials by the small changes. Thus δz ≈

∂z
∂z
∂z δu + δv + δw + · · ·
∂u
∂v
∂w

(3)

Total differential, rates of change and small changes
Problem 8. Pressure p and volume V of a gas are connected by the equation pV 1.4 = k. Determine the approximate percentage error in k when the pressure is increased by 4% and the volume is decreased by 1.5%.

Hence δG ≈

∂k
∂k
δp + δV ∂p
∂V

δk ≈ (V )

1.4

(0.04 p) + (1.4 pV

0.4

)(
−0.015V )

≈ pV 1.4[0.04 − 1.4(0.015)]
≈ pV 1.4[0.019] ≈

1.9
1.9
pV 1.4 ≈ k 100
100

i.e. the approximate error in k is a 1.9% increase.
Problem 9. Modulus of rigidity G = (R 4 θ)/L, where R is the radius, θ the angle of twist and L the length. Determine the approximate percentage error in G when R is increased by 2%, θ is reduced by
5% and L is increased by 4%.

R4
(
−0.05θ)
L
R4 θ
L2

(0.04L)

R4 θ
R4 θ
[0.08 − 0.05 − 0.04] ≈ −0.01
,
L
L
1
G
δG ≈ −
100


i.e.

Let p, V and k refer to the initial values.
∂k
Since k = pV 1.4 then
= V 1.4
∂p
∂k and = 1.4 pV 0.4
∂V
Since the pressure is increased by 4%, the change in
4
pressure δp =
× p = 0.04 p.
100
Since the volume is decreased by 1.5%, the change in
−1.5
× V = −0.015V . volume δV =
100
Hence the approximate error in k,

(0.02R) +
+ −

Using equation (3), the approximate error in k, δk ≈

4R 3 θ
L

355

Hence the approximate percentage error in G is a
1% decrease.
Problem 10. The second moment of area of a rectangle is given by I = (bl 3 )/3. If b and l are measured as 40 mm and 90 mm respectively and the measurement errors are −5 mm in b and +8 mm in l, find the approximate error in the calculated value of I .
Using equation (3), the approximate error in I , δI ≈

∂I
∂I
δb + δl
∂b
∂l

l3
∂I
3bl 2
∂I
= and
=
= bl 2
∂b
3
∂l
3 δb = −5 mm and δl = +8 mm l3 (−5) + (bl 2 )(+8)
3
Since b = 40 mm and l = 90 mm then
Hence δ I ≈

δI ≈

903
(−5) + 40(90)2 (8)
3

≈ −1215000 + 2592000
Using δG ≈

Since

and

G=

∂G
∂G
∂G δR + δθ + δL ∂R
∂θ
∂L
4R 3 θ ∂G
R4
R 4 θ ∂G
,
=
,
=
L ∂R
L
∂θ
L

−R 4 θ
∂G
=
∂L
L2

2
R = 0.02R
100
Similarly, δθ = −0.05θ and δL =0.04L
Since R is increased by 2%, δ R =

≈ 1377000 mm4 ≈ 137.7 cm4
Hence the approximate error in the calculated value of I is a 137.7 cm4 increase.
Problem 11. The time of oscillation t of a l pendulum is given by t = 2π
. Determine the g approximate percentage error in t when l has an error of 0.2% too large and g 0.1% too small.

356 Higher Engineering Mathematics
Using equation (3), the approximate change in t ,

H if the error in measuring current i is +2%, the error in measuring resistance R is −3% and the error in measuring time t is +1%.
[+2%]

∂t
∂t
δt ≈ δl + δg
∂l
∂g
Since t = 2π

and

l ∂t π ,
=√
g ∂l lg ∂t
= −π
∂g
δl =

3.

l
(from Problem 6, Chapter 34) g3 0.2 l = 0.002 l and δg = −0.001g
100

π hence δt ≈ √ (0.002l) + −π lg ≈ 0.002π

l
+ 0.001π g ≈ (0.001) 2π

≈ 0.0015t ≈

l g l
(−0.001 g) g3 l g + 0.0005 2π

l g 0.15 t 100

Hence the approximate error in t is a 0.15% increase.
Now try the following exercise
Exercise 142 changes Further problems on small

1. The power P consumed in a resistor is given by
P = V 2 /R watts. Determine the approximate change in power when V increases by 5% and
R decreases by 0.5% if the original values of V and R are 50 volts and 12.5 ohms respectively.
[+21 watts]
2. An equation for heat generated H is H = i 2 Rt .
Determine the error in the calculated value of

fr =

1


represents the resonant
2π LC 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%.
[−1%]

4. The second moment of area of a rectangle about its centroid parallel to side b is given by
I = bd 3/12. If b and d are measured as 15 cm and 6 cm respectively and the measurement errors are +12 mm in b and −1.5 mm in d, find the error in the calculated value of I .
[+1.35 cm4 ]
5. The side b of a triangle is calculated using b2 = a 2 + c2 − 2ac cos B. If a, c and B are measured as 3 cm, 4 cm and π/4 radians respectively and the measurement errors which occur are +0.8 cm, −0.5 cm and +π/90 radians respectively, determine the error in the calculated value of b.
[−0.179 cm]
6.

Q factor in a resonant electrical circuit is given
1 L
. Find the percentage change in by: Q =
R C
Q when L increases by 4%, R decreases by 3% and C decreases by 2%.
[+6%]

7. The rate of flow of gas in a pipe is given by:

C d v = √ , where C is a constant, d is the diam6
T5
eter of the pipe and T is the thermodynamic temperature of the gas. When determining the rate of flow experimentally, d is measured and subsequently found to be in error by +1.4%, and T has an error of −1.8%. Determine the percentage error in the rate of flow based on the measured values of d and T .
[+2.2%]

Chapter 36

Maxima, minima and saddle points for functions of two variables 36.1 Functions of two independent variables If a relation between two real variables, x and y, is such that when x is given, y is determined, then y is said to be a function of x and is denoted by y = f (x); x is called the independent variable and y the dependent variable. If y = f (u, v), then y is a function of two independent variables u and v. For example, if, say, y = f (u, v) = 3u 2 − 2v then when u = 2 and v = 1, y = 3(2)2 − 2(1) = 10. This may be written as f (2, 1) = 10. Similarly, if u = 1 and v = 4, f (1, 4) = −5.

Consider a function of two variables x and y defined by z = f (x, y) = 3x 2 − 2y. If (x, y) = (0, 0), then f (0, 0) = 0 and if (x , y) =(2, 1), then f (2, 1)=10.
Each pair of numbers, (x, y), may be represented by a point P in the (x, y) plane of a rectangular
Cartesian co-ordinate system as shown in Fig. 36.1.
The corresponding value of z = f (x, y) may be represented by a line PP drawn parallel to the z-axis.
Thus, if, for example, z =3x 2 − 2y, as above, and P is the co-ordinate (2, 3) then the length of PP is
3(2)2 − 2(3) = 6. Figure 36.2 shows that when a large number of (x, y) co-ordinates are taken for a function z z
6
p9

o

3

0

y
2
p

x

Figure 36.1

x

Figure 36.2

y

358 Higher Engineering Mathematics f (x, y), and then f (x, y) calculated for each, a large number of lines such as P P can be constructed, and in the limit when all points in the (x, y) plane are considered, a surface is seen to result as shown in Fig. 36.2.
Thus the function z = f (x, y) represents a surface and not a curve.

z

Minimum point q y 36.2 Maxima, minima and saddle points Partial differentiation is used when determining stationary points for functions of two variables. A function f (x, y) is said to be a maximum at a point (x, y) if the value of the function there is greater than at all points in the immediate vicinity, and is a minimum if less than at all points in the immediate vicinity. Figure 36.3 shows geometrically a maximum value of a function of two variables and it is seen that the surface z = f (x, y) is higher at (x, y) = (a, b) than at any point in the immediate vicinity. Figure 36.4 shows a minimum value of a function of two variables and it is seen that the surface z = f (x, y) is lower at (x, y) = ( p, q) than at any point in the immediate vicinity.

p x Figure 36.4 z t1
Maximum
point t2 b

O

y

z

Maximum point a x Figure 36.5

b y a x With functions of two variables there are three types of stationary points possible, these being a maximum point, a minimum point, and a saddle point. A saddle point Q is shown in Fig. 36.6 and is such that a point Q is a maximum for curve 1 and a minimum for curve 2.

Figure 36.3
Curve 2

If z = f (x, y) and a maximum occurs at (a, b), the curve lying in the two planes x = a and y = b must also have a maximum point (a, b) as shown in Fig. 36.5. Consequently, the tangents (shown as t1 and t2) to the curves at (a, b) must be parallel to Ox and Oy respectively.
∂z
∂z
This requires that
= 0 and
= 0 at all maximum
∂x
∂y and minimum values, and the solution of these equations gives the stationary (or critical) points of z.

Q

Curve 1

Figure 36.6

Maxima, minima and saddle points for functions of two variables

36.3 Procedure to determine maxima, minima and saddle points for functions of two variables Given z = f (x, y):
(i) determine

∂z
∂z
and
∂x
∂y

(ii) for stationary points,

∂z
∂z
= 0 and
= 0,
∂x
∂y

∂z
= 0 and
(iii) solve the simultaneous equations
∂x
∂z
= 0 for x and y, which gives the co-ordinates
∂y
of the stationary points,
(iv) determine

∂2z
∂x∂ y

36.4 Worked problems on maxima, minima and saddle points for functions of two variables
Problem 1. Show that the function z =(x − 1)2 + (y − 2)2 has one stationary point only and determine its nature. Sketch the surface represented by z and produce a contour map in the x-y plane.
Following the above procedure:
(i)

(ii) 2(x − 1) =0

(1)
(2)

(iii) From equations (1) and (2), x = 1 and y = 2, thus the only stationary point exists at (1, 2).
(iv) Since

2

for each stationary point,

into the equation

∂2z

∂2 z

and

∂2 z

, and ∂x 2 ∂ y 2
∂x∂ y
(v)

∂2z
∂x∂ y

2



∂2 z
∂x 2

∂2z
∂ y2

and evaluate,
(viii) (a) if > 0 then the stationary point is a saddle point.
2

(b)

∂ z if < 0 and 2 < 0, then the stationary
∂x
point is a maximum point,

and
(c)

∂2z
> 0, then the stationary
∂x2
point is a minimum point. if < 0 and

∂2z
∂z
= 2(x − 1) = 2x − 2, 2 = 2
∂x
∂x

and since

(vii) substitute the values of

=

∂z
∂z
= 2(x − 1) and
= 2(y − 2)
∂x
∂y

2(y − 2) = 0

∂2 z
∂2z ∂2z
, 2 and
∂x 2 ∂ y
∂x∂ y

(v) for each of the co-ordinates of the stationary
∂2z ∂2z points, substitute values of x and y into 2 , 2
∂x ∂ y
∂2 z and and evaluate each,
∂x∂ y
(vi) evaluate

359

∂z
∂2z
= 2(y − 2) = 2y − 4, 2 = 2
∂y
∂y

∂2z

=
∂x∂ y ∂x

∂z

= (2y − 4) = 0
∂y
∂x

∂2 z ∂2z
∂2z
= 2 = 2 and
=0
∂x 2 ∂ y
∂x∂ y
2

(vi)

∂2 z
=0
∂x∂ y

(vii)

= (0)2 − (2)(2) = −4

∂2z
(viii) Since < 0 and 2 > 0, the stationary point
∂x
(1, 2) is a minimum.
The surface z = (x − 1)2 + (y − 2)2 is shown in three dimensions in Fig. 36.7. Looking down towards the x-y plane from above, it is possible to produce a contour map. A contour is a line on a map which gives places having the same vertical height above a datum line (usually the mean sea-level on a geographical map).

360 Higher Engineering Mathematics z Problem 2. Find the stationary points of the surface f (x, y) = x 3 − 6x y + y 3 and determine their nature. y
1

Let z = f (x, y) = x 3 − 6x y + y 3

2

Following the procedure:
(i)

o
1

∂z
∂z
= 3x 2 − 6y and
= −6x + 3y 2
∂x
∂y

(ii) for stationary points, 3x 2 − 6y = 0 x −6x + 3y 2 = 0

and

Figure 36.7

(iii) from equation (1), 3x 2 = 6y

A contour map for z =(x − 1)2 + (y − 2)2 is shown in
Fig. 36.8. The values of z are shown on the map and these give an indication of the rise and fall to a stationary point.

and

y=

3x 2 1 2
= x
6
2

y

z51

2

z54

z59

z 5 16

1

1

Figure 36.8

2

x

(1)
(2)

Maxima, minima and saddle points for functions of two variables and substituting in equation (2) gives:

Now try the following exercise

1 2 2
=0
x
2
3
−6x + x 4 = 0
4
3 x 3x
−2 = 0
4

−6x + 3

from which, x = 0 or

Exercise 143 Further problems on maxima, minima and saddle points for functions of two variables
1. Find the stationary point of the surface f (x, y) = x 2 + y 2 and determine its nature.
Sketch the surface represented by z.
[Minimum at (0, 0)]

x3
− 2 =0
4

i.e. x 3 = 8 and x = 2
When x = 0, y = 0 and when x = 2, y = 2 from equations (1) and (2).
Thus stationary points occur at (0, 0) and (2, 2).
(iv)

∂2z
∂2z

∂2z
= 6x, 2 = 6y and
=
2
∂x
∂y
∂x∂ y
∂x
=

(v)


(−6x + 3y 2 ) = −6
∂x

∂2 z
∂x 2
∂2 z and ∂x∂ y
∂2 z for (2, 2),
∂x 2
∂2 z and ∂x∂ y for (0, 0)

= 0,

(vii)

∂2 z
=0
∂ y2

= 12,

[Minimum at (0, 0)]
∂2 z
∂ y2

= 12

= −6

∂2z
∂x∂ y

2

for (0, 0),

∂2z
∂x∂ y

3. Determine the stationary values of the function f (x, y) = x 3 − 6x 2 − 8y 2 and distinguish between them. Sketch an approximate contour map to represent the surface f (x, y).
Maximum point at (0, 0), saddle point at (4, 0)
4. Locate the stationary point of the function z =12x 2 + 6x y + 15y 2 .

= −6

5. Find the stationary points of the surface z = x 3 − x y + y 3 and distinguish between them. saddle point at (0, 0), minimum at 1 , 1
3 3

2

for (2, 2),

(vi)

∂z
∂y

2. Find the maxima, minima and saddle points for the following functions:
(a) f (x, y) = x 2 + y 2 − 2x + 4y + 8
(b) f (x, y) = x 2 − y 2 − 2x + 4y + 8
(c) f (x, y) = 2x + 2y − 2x y − 2x 2 − y 2 + 4.⎤

(a) Minimum at (1, −2)
⎣ (b) Saddle point at (1, 2) ⎦
(c) Maximum at (0, 1)

∂2z
(0, 0) =
∂x∂ y

= (−6)2 = 36
= (−6)2 = 36
2



∂2z
∂x 2

∂2 z
∂ y2

= 36 − (0)(0) = 36
(2, 2) = 36 − (12)(12) = −108

(viii) Since

(0, 0) > 0 then (0, 0) is a saddle point.

∂2 z
> 0, then (2, 2) is a
Since (2, 2) < 0 and
∂x 2 minimum point.

36.5 Further worked problems on maxima, minima and saddle points for functions of two variables Problem 3. Find the co-ordinates of the stationary points on the surface z = (x 2 + y 2 )2 − 8(x 2 − y 2 ) and distinguish between them. Sketch the approximate contour map associated with z.

361

362 Higher Engineering Mathematics

∂z
(i)
= 2(x 2 + y 2 )2x − 16x and
∂x
∂z
= 2(x 2 + y 2 )2y + 16y
∂y

2(x 2 + y 2 )2y + 16y = 0
4y(x 2 + y 2 + 4) = 0

(iii) From equation (1), y 2 =
Substituting

y2 = 4 − x 2

(1)
(2)

16x − 4x 3
=4 − x2
4x
in equation (2) gives

4y(x 2 + 4 − x 2 + 4) = 0
i.e. 32y = 0 and y = 0
When y = 0 in equation (1),

4x 3 − 16x = 0

i.e.

4x(x 2 − 4) = 0

from which, x = 0 or x = ±2
The co-ordinates of the stationary points are
(0, 0), (2, 0) and (−2, 0).
∂2z
= 12x 2 + 4y 2 − 16,
(iv)
∂x 2
∂2 z
∂ y2

= 4x 2 + 12y 2 + 16 and

∂2z
∂x∂ y

= 8x y

(v) For the point (0, 0),
∂2z
∂2z
∂2z
= −16, 2 = 16 and
=0
∂x 2
∂y
∂x∂ y
For the point (2, 0),
∂2z
∂2z
∂2z
= 32, 2 = 32 and
=0
∂x 2
∂y
∂x∂ y
For the point (−2, 0),
∂2z
∂2z
∂2z
= 32, 2 = 32 and
=0
2
∂x
∂y
∂x∂ y

(vi)

∂2 z
∂x∂ y

2

= 0 for each stationary point

(0, 0) > 0,

the point (0, 0) is a saddle

∂2z
∂x 2
(2, 0) is a minimum point.
Since

4x 3 + 4x y 2 − 16x = 0

i.e.

= (0)2 − (32)(32) = −1024

(viii) Since point. 2(x 2 + y 2 )2x − 16x = 0 and = (0)2 − (−16)(16) = 256

2
(−2, 0) = (0) − (32)(32) = −1024

(ii) for stationary points,
i.e.

(0, 0)
(2, 0)

(vii)

Following the procedure:

(0, 0) < 0 and

(2, 0)

> 0, the point

∂2z
> 0, the
∂x 2 (−2, 0) point (−2, 0) is a minimum point.

Since

(−2, 0) < 0 and

Looking down towards the x-y plane from above, an approximate contour map can be constructed to represent the value of z. Such a map is shown in Fig. 36.9.
To produce a contour map requires a large number of x-y co-ordinates to be chosen and the values of z at each co-ordinate calculated. Here are a few examples of points used to construct the contour map.
When z = 0, 0 =(x 2 + y 2 )2 − 8(x 2 − y)2
In addition, when, say, y = 0 (i.e. on the x-axis)
0 = x 4 − 8x 2 , i.e. x 2 (x 2 − 8) = 0

from which, x = 0 or x = ± 8

Hence the contour z = 0 crosses the x-axis at 0 and ± 8,
i.e. at co-ordinates (0, 0), (2.83, 0) and (−2.83, 0) shown as points, S, a and b respectively.
When z = 0 and x =2 then
0 = (4 + y 2 )2 − 8(4 − y 2 )
i.e. 0 = 16 + 8y 2 + y 4 − 32 + 8y 2
i.e. 0 = y 4 + 16y 2 − 16
Let y 2 = p, then p2 + 16 p − 16 = 0 and
−16 ± 162 − 4(1)(−16)
2
−16 ± 17.89
=
2
= 0.945 or −
16.945

p=

Hence y =



p=

(0.945) or

(−16.945)

= ± 0.97 or complex roots.

Maxima, minima and saddle points for functions of two variables

363

y
4

i

z5

128

2

z59 c g

0 z5 S f b

3
22

3
2

a

e

x

d h 22

j
24

Figure 36.9

Hence the z =0 contour passes through the co-ordinates
(2, 0.97) and (2, −0.97) shown as a c and d in Fig. 36.9.
Similarly, for the z = 9 contour, when y = 0,
9=

(x 2

+ 02 )2

i.e.

9 = x 4 − 8x 2

i.e.

− 8(x 2 − 02 )

x 4 − 8x 2 − 9 =0

Hence (x 2 − 9)(x 2 + 1) = 0. from which, x = ±3 or complex roots.
Thus the z = 9 contour passes through (3, 0) and (−3, 0), shown as e and f in Fig. 36.9.
If z = 9 and x = 0, 9 = y 4 + 8y 2
i.e.

y 4 + 8y 2 − 9 = 0

i.e.

(y 2 + 9)(y 2 − 1) = 0

from which, y = ±1 or complex roots.
Thus the z = 9 contour also passes through (0, 1) and
(0, −1), shown as g and h in Fig. 36.9.

When, say, x = 4 and y = 0, z = (42 )2 − 8(42 ) = 128. when z = 128 and x = 0, 128 = y 4 + 8y 2
i.e.

y 4 + 8y 2 − 128 = 0

i.e. (y 2 + 16)(y 2 − 8) = 0

from which, y = ± 8 or complex roots.
Thus the z = 128 contour passes through (0, 2.83) and
(0, −2.83), shown as i and j in Fig. 36.9.
In a similar manner many other points may be calculated with the resulting approximate contour map shown in
Fig. 36.9. It is seen that two ‘hollows’ occur at the minimum points, and a ‘cross-over’ occurs at the saddle point S, which is typical of such contour maps.
Problem 4. Show that the function f (x, y) = x 3 − 3x 2 − 4y 2 + 2 has one saddle point and one maximum point.
Determine the maximum value.

364 Higher Engineering Mathematics
Let z = f (x, y) = x 3 − 3x 2 − 4y 2 + 2.
∂z
∂z
= 3x 2 − 6x and
= − 8y
∂x
∂y

(0, 0) = 0 −(−6)(−8) = −48

(1)

−8y = 0

(2)

∂2z
< 0, the
∂x 2 (0, 0) point (0, 0) is a maximum point and hence the maximum value is 0.

(viii) Since

(iii) From equation (1), 3x(x − 2) = 0 from which, x = 0 and x = 2.

Since point. From equation (2), y = 0.
Hence the stationary points are (0, 0) and (2, 0).
(iv)

∂2z

= 6x − 6,

∂x 2

∂2z
∂ y2

= −8 and

= (0)2 = 0

(2, 0) = 0 −(6)(−8) = 48

(ii) for stationary points, 3x 2 −6x = 0 and ∂2 z
∂x∂ y

(vii)

Following the procedure:
(i)

2

(vi)

∂2z
∂x∂ y

(0, 0) < 0

(2, 0) > 0,

and

the point (2, 0) is a saddle

The value of z at the saddle point is
23 − 3(2)2 − 4(0)2 + 2 =−2.

=0

An approximate contour map representing the surface f (x, y) is shown in Fig. 36.10 where a ‘hollow effect’ is seen surrounding the maximum point and a ‘cross-over’ occurs at the saddle point S.

(v) For the point (0, 0),
∂2 z
∂2 z
∂2 z
= −6, 2 = −8 and
=0
∂x 2
∂y
∂x∂ y
For the point (2, 0),

Problem 5. An open rectangular container is to have a volume of 62.5 m3 . Determine the least surface area of material required.

∂2z
∂2z
∂2 z
= 6, 2 = −8 and
=0
2
∂x
∂y
∂x∂ y

y
2

z5

0

MAX

S
2
2
52

z

z5

22

Figure 36.10

24

21

21

z5

z5

22

3

2

4

x

Maxima, minima and saddle points for functions of two variables
From equation (1),

(5) (5) z =62.5 z= from which,

365

62.5
= 2.5 m
25

∂ 2 S 250 ∂ 2 S 250
∂2 S
= 3 , 2 = 3 and
=1
2
∂x
x
∂y
y
∂x∂ y
When x = y = 5,

y z ∂2 S
∂2 S
∂2 S
= 2, 2 = 2 and
=1
∂x 2
∂y
∂x∂ y

= (1)2 − (2)(2) = −3
∂2 S
> 0, then the surface area S is a
Since < 0 and
∂x 2 minimum. x

Figure 36.11

Hence the minimum dimensions of the container to have a volume of 62.5 m3 are 5 m by 5 m by 2.5 m.
Let the dimensions of the container be x, y and z as shown in Fig. 36.11.

From equation (2), minimum surface area, S
= (5)(5) + 2(5)(2.5) + 2(5)(2.5)

Volume

V = x yz = 62.5

(1)

Surface area,

S = x y + 2yz + 2x z

(2)
Now try the following exercise

62.5
From equation (1), z = xy Exercise 144 Further problems on maxima, minima and saddle points for functions of two variables

Substituting in equation (2) gives:
S = x y + 2y

i.e.

S=xy +

62.5
+ 2x xy 62.5 xy 1. The function z = x 2 + y 2 + x y + 4x − 4y + 3 has one stationary value. Determine its co-ordinates and its nature.
[Minimum at (−4, 4)]

125 125
+
x y which is a function of two variables
∂s
125
= y − 2 = 0 for a stationary point,
∂x
x hence x 2 y =125
∂s
125
= x − 2 = 0 for a stationary point,
∂y
y hence x y 2 = 125

(3)

(4)

Dividing equation (3) by (4) gives: x2 y x = 1, i.e. = 1, i.e. x = y x y2 y Substituting y = x in equation (3) gives x 3 = 125, from which, x = 5 m.
Hence y = 5 m also

= 75 m2

2. An open rectangular container is to have a volume of 32 m3 . Determine the dimensions and the total surface area such that the total surface area is a minimum.
4 m by 4 m by 2 m, surface area = 48m2
3. Determine the stationary values of the function f (x, y) = x 4 + 4x 2 y 2 − 2x 2 + 2y 2 − 1 and distinguish between them.


Minimum at (1, 0),
⎣ minimum at (−1, 0), ⎦ saddle point at (0, 0)

366 Higher Engineering Mathematics
4. Determine the stationary points of the surface f (x, y) = x 3 − 6x 2 − y 2 .
Maximum at (0, 0), saddle point at (4, 0)
5. Locate the stationary points on the surface f (x, y) = 2x 3 + 2y 3 − 6x − 24y + 16 and determine their nature.


Minimum at (1, 2),
⎣ maximum at (−1, −2),

saddle points at (1, −2) and (−1, 2)

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 the 250 m3.
[150 m2 ]

Revision Test 10
This Revision Test covers the material contained in Chapters 32 to 36. The marks for each question are shown in brackets at the end of each question.
1.

Differentiate the following functions with respect to x:
(a) 5 ln (shx) (b) 3 ch3 2x
2x

(c) e
2.

(7)

Differentiate the following functions with respect to the variable: x 1
(a) y = cos−1
5
2

(c) y =

(2x 2 − 1)

dx d y
+
x y (6)

An engineering function z = f (x, y) and y z = e 2 ln(2x + 3y). Determine the rate of increase of z, correct to 4 significant figures, when x = 2 cm, y = 3 cm, x is increasing at 5 cm/s and y is increasing at 4 cm/s.
(8)
The volume V of a liquid of viscosity coefficient η delivered after time t when passed through a tube of length L and diameter d by a pressure p pd 4t
. If the errors in V , p and is given by V =
128ηL
L are 1%, 2% and 3% respectively, determine the error in η.
(8)

9.

Determine and distinugish between the stationary values of the function

(14)

(6)

If z = f (x, y) and z = x cos(x + y) determine
∂z ∂z
,
,
, , and .
∂x ∂ y ∂x 2 ∂ y 2 ∂x∂ y
∂ y∂x

5.

(7)

7.

Evaluate the following, each correct to 3 decimal places: ∂2z

2 (r 2 + x 2 )3

8.

2 sec−1 5x x (a) sinh−1 3 (b) cosh−1 2.5 (c) tanh−1 0.8
4.

r2 I

If x yz = c, where c is constant, show that dz = −z

−1 t

(d) y = 3 sinh−1
3.

6.

sech 2x

(b) y = 3esin

Show that H = ±

∂2z

∂2 z

∂2 z

(12)

The magnetic field vector H due to a steady current I flowing around a circular wire of radius r and at a distance x from its centre is given by
H =±

I ∂
2 ∂x

x

2 + x2 r f (x, y) = x 3 − 6x 2 − 8y 2 and sketch an approximate contour map to represent the surface f (x, y).
(20)
10. An open, rectangular fish tank is to have a volume of 13.5 m3 . Determine the least surface area of glass required.
(12)

Chapter 37

Standard integration
37.1

The process of integration

The process of integration reverses the process of differentiation. In differentiation, if f (x) = 2x 2 then f (x) = 4x. Thus the integral of 4x is 2x 2 , i.e. integration is the process of moving from f (x) to f (x). By similar reasoning, the integral of 2t is t 2.
Integration is a process of summation or adding parts together and an elongated S, shown as , is used to replace the words ‘the integral of’. Hence, from above,
4x = 2x 2 and 2t is t 2. dy In differentiation, the differential coefficient indidx cates that a function of x is being differentiated with respect to x, the dx indicating that it is ‘with respect to x’. In integration the variable of integration is shown by adding d (the variable) after the function to be integrated. Thus

and

The general solution of integrals of the form ax n dx, where a and n are constants is given by: ax n dx =

2t dt means ‘the integral of 2t with respect to t ’.

4x dx = 2x 2 + c and

2t dt = t 2 + c

‘c’ is called the arbitrary constant of integration.

ax n+1
+c
n+1

This rule is true when n is fractional, zero, or a positive or negative integer, with the exception of n = −1.
Using this rule gives:
(i)

3x 4 dx =

(ii)

2 dx = x2 =

4x dx means ‘the integral of 4x with respect to x’,

As stated above, the differential coefficient of 2x 2 is
4x, hence 4x dx = 2x 2 . However, the differential coefficient of 2x 2 + 7 is also 4x. Hence 4x dx is also equal to 2x 2 + 7. To allow for the possible presence of a constant, whenever the process of integration is performed, a constant ‘c’ is added to the result.
Thus

37.2 The general solution of integrals of the form ax n

(iii)



3x 4+1
3
+ c = x5 + c
4+1
5
2x −2 dx =

2x −2+1
+c
−2 +1

2x −1
−2
+c=
+ c, and
−1
x
1

x dx =

1 x2 3

x 2 +1 x2 dx =
+c=
+c
1
3
+1
2
2

2√ 3 x +c
3
Each of these three results may be checked by differentiation.
=

(a)

The integral of a constant k is kx + c. For example, 8 dx = 8x + c

(b) When a sum of several terms is integrated the result is the sum of the integrals of the separate terms.

Standard integration
(b) When a = 2 and n = 3 then

For example,
(3x + 2x 2 − 5) dx
=
=

37.3

3x dx +

2x 2 dx −

5 dx

3x 2
2x 3
+
− 5x + c
2
3

Standard integrals

Table 37.1 Standard integrals ax n+1
+c
n +1
(except when n =−1)

ax n dx =

Each of these results may be checked by differentiating them. Problem 2. Determine
3
4 + x − 6x 2 dx.
7
(4 + 3 x − 6x 2 ) dx may be written as
7
4 dx + 3 x dx − 6x 2 dx, i.e. each term is integrated
7
separately. (This splitting up of terms only applies, however, for addition and subtraction.)
3
4 + x − 6x 2 dx
7

Hence

= 4x +

cos ax dx =

(iii)

1 sin ax dx = − cos ax + c a (iv)

sec 2 ax dx =

(v)

1 cosec 2 ax dx = − cot ax + c a (vi)

1 cosec ax cot ax dx = − cosec ax + c a (vii)

1 sec ax tan ax dx = sec ax + c a (viii)

eax dx =

1 tan ax + c a (a)

3
7

x2 x3 − (6) + c
2
3

3 2 x − 2x 3 + c
14

Note that when an integral contains more than one term there is no need to have an arbitrary constant for each; just a single constant at the end is sufficient.
Problem 3. Determine
(a)

(a)

2x 3 − 3x dx (b)
4x

(1 − t )2 dt.

Rearranging into standard integral form gives:
2x 3 − 3x dx 4x

Problem 1. Determine (a) 5x 2 dx (b) 2t 3 dt .
The standard integral, ax n dx =

x 1+1 x 2+1
− (6)
+c
1+1
2+1

= 4x +

1 ax e +c a 1 dx = ln x + c x 3
7

= 4x +

1 sin ax + c a (ii)

(ix)

2t 3+1
2t 4
1
+c=
+c= t4 +c
3+1
4
2

2t 3 dt =

Since integration is the reverse process of differentiation the standard integrals listed in Table 37.1 may be deduced and readily checked by differentiation.

(i)

369

ax n+1
+c
n +1

When a = 5 and n =2 then
5x 2+1
5x 3
5x 2 dx =
+c=
+c
2+1
3

=

2x 3 3x

dx =
4x
4x

x2 3
− dx
2
4

=

1
2

x 2+1 3
− x +c
2+1 4

=

1
2

x3 3
3
1
− x + c = x3 − x + c
3
4
6
4

370 Higher Engineering Mathematics
(1 − t )2 dt gives:

(b) Rearranging

2t 1+1 t 2+1
+
+c
1+1 2+1

(1 − 2t + t 2) dt = t −

2t 2 t 3
+ +c
2
3

=t−

1
= t −t 2 + t 3 +c
3
This problem shows that functions often have to be rearranged into the standard form of ax n dx before it is possible to integrate them.
Problem 4.
3
dx = x2 Determine

3 dx. x2

3x −2 dx. Using the standard integral,

1

Problem 7.

3x −2+1
3x −1
+c =
+c
−2 + 1
−1
−3
= −3x −1 + c =
+c
x

Determine 3 x dx.

Problem 5.

For fractional powers it is necessary to appreciate

m n m a =a n

(1 + θ)2 dθ. √ θ Determine

(1 + θ)2 dθ =

θ

(1 + 2θ + θ 2 ) dθ √ θ 1

=

θ

=

θ

=

θ

ax n dx when a = 3 and n =−2 gives:
3x −2 dx =

4 1 t4 +c
1

5
5 t4
+c = −
= −
9 1
9
4
20 √
4
t+c
=−
9

=

−1
2

θ2
1
2

θ

θ2



1

θ2

1
2

1−



1

+1

+


1
2

3

+

+

1
2

2−

1
2



3

+ 2θ 2 + θ 2 dθ

−1 + 1
2
1

=



+ 2θ

−1
2

−1
2

θ

+

1
2

2θ 2
3
2

1
2

+1

+1

+

θ
3
2

3
2

+1

+1

5

+

θ2
5
2

+c

1
4 3 2 5
= 2θ 2 + θ 2 + θ 2 + c
3
5

4 3 2 5 θ + θ +c
= 2 θ+
3
5

1



3 x dx =

1
3x 2

3x 2 +1 dx =
+c
1
+1
2

3
3
3x 2
+ c = 2x 2 + c = 2 x 3 + c
=
3
2

Problem 6.
−5
√ dt =
4
9 t3

Determine
−5
3

9t 4

5
= −
9

dt =

−5
√ dt .
4
9 t3


5 −3 t 4 dt
9

3
− +1 t 4
+c
3
− +1
4

Problem 8. Determine
(a) 4 cos3x dx (b) 5 sin 2θ dθ.
(a) From Table 37.1(ii),
4 cos3x dx = (4)

1 sin 3x + c
3

4
= sin 3x + c
3
(b) From Table 37.1(iii),
5 sin 2θ dθ = (5) −

1 cos 2θ + c
2

5
= − cos 2θ + c
2

+c

Standard integration
Problem 9. Determine
(a)

2m 2
+ ln m + c
2

=

7 sec2 4t dt (b) 3 cosec 2 2θ dθ.

= m 2 + ln m + c
(a)

From Table 37.1(iv),
1
tan 4t + c
4

7 sec2 4t dt = (7)

7
= tan 4t + c
4

Exercise 145 integrals Further problems on standard

In Problems 1 to 12, determine the indefinite integrals. (b) From Table 37.1(v),
3

Now try the following exercise

cosec 2 2θ dθ = (3) −

1 cot 2θ + c
2

1.

(a)

4 dx

(b)

7x dx

3
= − cot 2θ + c
2
Problem 10. Determine
(a)

2.

2 dt. 3 e4t

5 e3x dx (b)

(a) 4x + c (b)

(a)

2 2 x dx
5

(b)
(a)

(a)

From Table 37.1(viii),
5 e3x dx = (5)

(b)

2 dt =
3 e4t

1 3x
5
e + c = e3x + c
3
3

2
2 −4t e dt =
3
3



3.

(a)

4.

(a)

1 −4t e +c
4

1
1
= − e−4t + c = − 4t + c
6
6e

3 dx =
5x

(a) 2

1
3
dx = ln x +c x 5

3
5

2m 2 + 1 dm = m =

2m 2 1
+
dm m m
1
dm
2m + m x 3 dx (b)
(a)

(from Table 37.1(ix))
(b)

3 dx 4x 4
(a)

5.
(a)

5 4
2 3 x + c (b) x +c
15
24

4 dx (b)
3x 2

2m 2 + 1 dm. m

3 dx (b)
5x

6.

(a)

5 3 x dx
6

3x 2 − 5x dx (b) (2 + θ)2 dθ x ⎡

3x 2
(a)
− 5x + c


2




3
2+ θ +c
(b) 4θ + 2θ
3

Problem 11. Determine
(a)

7x 2
+c
2

−1
−4
+ c (b) 3 + c
3x
4x
14 5 x dx
4

1√ 9
4√ 5
4
x + c (b) x +c
5
9

−5
√ dt (b) t3 3
√ dx
5 4
7 x

10
15 √
5
(a) √ + c (b) x +c
7
t

371

372 Higher Engineering Mathematics

7.

(a)

3 cos2x dx (b)

7 sin 3θ dθ



3
⎢ (a) 2 sin 2x + c ⎥




7
(b) − cos 3θ + c
3
8.

9.

(a)

3 sec2 3x dx (b) 2 cosec 2 4θ dθ
4
1
1
(a) tan 3x +c (b) − cot 4θ +c
4
2

limit and ‘a’ the lower limit. The operation of applying the limits is defined as [x]b = (b) − (a). a The increase in the value of the integral x 2 as x increases
3
from 1 to 3 is written as 1 x 2 dx.
Applying the limits gives:
3

x 2 dx =

1

4 sec 4t tan 4t dt
3


(a)

(a)

2 dx (b)
3x
(a)

(a)

2

1
2
+ c =8
3
3

u

(b)

3x dx =

3
2
−2 (4 − x ) dx.

3x 2
2

2
1

−2

dt

−8
3

1
1
=8
3
3 θ +2

dθ, taking θ 4

Problem 13.

Evaluate

positive square roots only.
4
1

Definite integrals

−2

(3)3
(−2)3
− 4(−2) −
3
3

= {3} − −5




18 √ 5
(a) 8 x + 8 x 3 + x +c


5




3
1
4t
(b) − + 4t +
+c
t
3

3

x3
3

= {12 − 9} − −8 −

2



3 2
3 2
(2) −
(1)
2
2

=

(4 − x 2 ) dx = 4x −
= 4(3) −

du

1
+ 2t t (b)

1
1
=6 − 1 =4
2
2

2 u2 ln x + c (b)
− ln u + c
3
2

(2+3x)2

dx (b) x Evaluate

1

3

−2
3 2x
+c
e + c (b)
8
15 e5x u2 − 1

2
1 3x dx

(a)

dx e5x 2
3 2x e dx (b)
4
3
(a)

37.4

1

Note that the ‘c’ term always cancels out when limits are applied and it need not be shown with definite integrals.

(a)


5
(a) − cosec 2t + c ⎥

2




1
(b) sec 4t + c
3

12.

33
13
+c −
+c
3
3

=

(a) 5 cot 2t cosec 2t dt
(b)

11.

3

= (9 + c) −

Problem 12.

10.

x3
+c
3

θ +2

dθ = θ 4

θ

1
4

=

θ
1
2

1

+

2
1

θ2

1

θ 2 + 2θ

−1
2





1

Integrals containing an arbitrary constant c in their results are called indefinite integrals since their precise value cannot be determined without further information.
Definite integrals are those in which limits are applied.
If an expression is written as [x]b, ‘b’ is called the upper a ⎡
⎢θ
=⎣

1
2 +1

1
+1
2

+

−1
2 +1

⎤4



1
− +1
2
1



Standard integration

=⎣

3

1

θ2

+

3
2

2θ 2
1
2

⎤4


2 3 θ +4 θ
3

⎦ =
1

Problem 16. Evaluate

4

2
1

(a)


2
2
=
(4)3 + 4 4 −
(1)3 + 4 (1)
3
3
16
2
+8 −
+4
3
3

=

1

Problem 14. Evaluate

2

(a)

4 e2x dx =

1

4

(b)
1

3
3
du = ln u
4u
4

3 sin 2x dx.

= 2[ e2x ]2 = 2[ e4 − e2 ]
1
1

4

3
= [ln 4 − ln 1]
4
1

3
= [1.3863 −0] =1.040
4

3 sin 2x dx

Now try the following exercise

0

= (3) −

2

4 2x e 2

= 2[54.5982 −7.3891] =94.42

0 π 2

1

3 du, 4u

each correct to 4 significant figures.

2
2
1
= 5 +8− −4 = 8
3
3
3
π
2

4

4 e2x dx (b)

1 cos 2x
2

π
3
= − cos 2
2
2

π
2

0

3
= − cos 2x
2

π
2

Exercise 146 integrals 0

3
− − cos 2(0)
2

Further problems on definite

In problems 1 to 8, evaluate the definite integrals
(where necessary, correct to 4 significant figures).

3
3
= − cos π − − cos 0
2
2

4

1. (a)

5x 2 dx (b)

1

1

3
− t 2 dt
−1 4

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

(a) 105 (b) −
2

2. (a)

2

Problem 15. Evaluate

4 cos 3t dt.

3

(3 − x 2 ) dx (b)

−1

(x 2 − 4x + 3) dx

1

1
2

4 cos3t dt = (4)

1

=

1 sin 3t
3

2

=
1

4 sin 3t
3

(a) 6 (b) −1

2
1

4
4
sin 6 − sin 3
3
3

Note that limits of trigonometric functions are always expressed in radians—thus, for example, sin 6 means the sine of 6 radians= −0.279415 . . .

π

3 cos θ dθ
2

3. (a)
0

π
2

(b)

4. (a)

π
3
π
6

2

2 sin 2θ dθ (b)

3 sin t dt
0

[(a) 1 (b) 4.248]

1

= (−0.37255) − (0.18816) = −0.5607

4 cos θ dθ
[(a) 0 (b) 4]

4 cos 3t dt
4
4
= (−0.279415 . . .) −
(0.141120 . . .)
3
3

1
3

0

2

Hence

1
2

π
6

1

5. (a)

5 cos3x dx (b)
0

3 sec2 2x dx

0

[(a) 0.2352 (b) 2.598]

373

374 Higher Engineering Mathematics
2

6. (a)

1 litre to 3 litres for a temperature rise from
100 K to 400 K given that:

cosec 2 4t dt

1

(b)

π
2

π
4

Cv = 45 + 6 × 10−3 T + 8 × 10−6 T 2 .

(3 sin 2x − 2 cos3x) dx

[55.65]

[(a) 0.2527 (b) 2.638]
1

7. (a)

2

2 dx 3 e2x
−1

3 e3t dt (b)

0

[(a) 19.09 (b) 2.457]
3

8. (a)
2

2 dx (b)
3x

3
1

9. The entropy change given by:
S=

T2
T1

2x 2 + 1 dx x
[(a) 0.2703 (b) 9.099]
S, for an ideal gas is

dT
Cv
−R
T

10. The p.d. between boundaries a and b of an b Q electric field is given by: V = dr 2πrε0 εr a If a = 10, b = 20, Q =2 × 10−6 coulombs, ε0 = 8.85 ×10−12 and εr = 2.77, show that
V = 9 kV.
11. The average value of a complex voltage waveform is given by:
V AV =

V2
V1

dV
V

where T is the thermodynamic temperature,
V is the volume and R = 8.314. Determine the entropy change when a gas expands from

1 π π

(10 sin ωt + 3 sin 3ωt

0

+ 2 sin 5ωt) d(ωt)
Evaluate V AV correct to 2 decimal places.
[7.26]

Chapter 38

Some applications of integration 38.1

Introduction

There are a number of applications of integral calculus in engineering. The determination of areas, mean and
r.m.s. values, volumes, centroids and second moments of area and radius of gyration are included in this chapter. 38.2

Areas under and between curves

When y = 0, x = 0 or (x + 2) = 0 or (x − 4) = 0, i.e. when y = 0, x = 0 or −2 or 4, which means that the curve crosses the x-axis at 0, −2, and 4. Since the curve is a continuous function, only one other co-ordinate value needs to be calculated before a sketch of the curve can be produced. When x = 1, y = −9, showing that the part of the curve between x = 0 and x = 4 is negative. A sketch of y = x 3 − 2x 2 − 8x is shown in
Fig. 38.2. (Another method of sketching Fig. 38.2 would have been to draw up a table of values.) y In Fig. 38.1,

10 b total shaded area =

c

f (x)dx −

a

y 5 x 3 2 2x 2 2 8x

f (x)dx b 22

d

+

21

0

1

2

3

4

f (x)dx
210

c

y
220
y 5 f (x)
G

Figure 38.2

E
0

a

b

F

c

d

x

Shaded area
=

Figure 38.1

0
−2

4

(x 3 − 2x 2 − 8x)dx −

(x 3 − 2x 2 − 8x)dx

0
0

x 4 2x 3 8x 2 x 4 2x 3 8x 2





4
3
2 −2
4
3
2
2
2
1
= 6
− −42
= 49 square units
3
3
3

4

=

Problem 1. Determine the area between the curve y = x 3 − 2x 2 − 8x and the x-axis. y = x 3 −2x 2 − 8x = x(x 2 −2x − 8) = x(x + 2)(x − 4)

0

x

376 Higher Engineering Mathematics
= 7

Problem 2. Determine the area enclosed between the curves y = x 2 + 1 and y = 7 − x.
At the points of intersection the curves are equal. Thus, equating the y values of each curve gives: x2 + 1 = 7 − x x2 + x − 6 = 0

from which,

Factorizing gives (x − 2)(x + 3) = 0 from which x = 2 and x = −3
By firstly determining the points of intersection the range of x-values has been found. Tables of values are produced as shown below. x 10

5
= 20 square units
6
Problem 3. Determine by integration the area bounded by the three straight lines y = 4 − x, y = 3x and 3y = x.
Each of the straight lines are shown sketched in
Fig. 38.4. y y542x

y 5 3x
4

−3 −2 −1 0 1 2

y = x2 + 1

5

2 1 2

x

−3
10

7

5

3y 5 x (or y 5 x )
3

2

0 2

y = 7−x

1
1
− −13
3
2

1

0

2

3

4

x

5

A sketch of the two curves is shown in Fig. 38.3.

Figure 38.4

Shaded area y 10

y 5 x 2 11

5

23

22

21

0

=

y572x

0

1

2

1

=

3x −

x dx +
3

3x 2 x 2

2
6

x

=

3
1

1

+ 4x −
0

3 1
− (0) +

2 6

(4 − x) − x2 x2

2
6
12 −

x dx 3

3
1

9 9

2 6

Figure 38.3

− 4−
Shaded area =
=
=

2
−3
2
−3
2
−3

(7 − x)dx −

2
−3

(x 2 + 1)dx

= 1

1 1

2 6

1
1
+ 6−3
= 4 square units
3
3

[(7 − x) − (x 2 + 1)]dx
Now try the following exercise
(6 − x − x 2 )dx

x2 x3
= 6x −

2
3
= 12 − 2 −

2
−3

9
8
− −18 − + 9
3
2

Exercise 147 Further problems on areas under and between curves
1. Find the area enclosed by the curve y = 4 cos 3x, the x-axis and ordinates x = 0 π [1 1 square units] and x =
3
6

Some applications of integration

377

[Note that for a sine wave,
2. Sketch the curves y = x 2 + 3 and y = 7 − 3x and determine the area enclosed by them.
[20 5 square units]
6
3. Determine the area enclosed by the three straight lines y = 3x, 2y = x and y + 2x = 5.
[2 1 square units]
2

In this case, mean value =

Mean and r.m.s. values

With reference to Fig. 38.5,
1
mean value, y = b −a and =

π

1 π −0 π 1 π y dx

=

a b v 2 d(ωt )

0

(100 sin ωt )2 d(ωt )

0

b

1 b−a 2
× 100 = 63.66 V] π (b) r.m.s. value
=

38.3

2
× maximum value π mean value=

10000 π π

sin2 ωt d(ωt ) ,

0

y

y2 dx

which is not a ‘standard’ integral.

y 5 f(x)

r.m.s. value =

It is shown in Chapter 17 that cos 2 A = 1 − 2 sin2 A and this formula is used whenever sin2 A needs to be integrated.

a

Rearranging cos 2 A = 1 − 2 sin2 A gives
1
sin2 A = (1 − cos 2 A)
2
y

Hence
0

x5a

x5b

x

Figure 38.5

Problem 4. A sinusoidal voltage v = 100 sin ωt volts. Use integration to determine over half a cycle
(a) the mean value, and (b) the r.m.s. value.
(a)

Half a cycle means the limits are 0 to π radians. π 1
Mean value, y = v d(ωt ) π −0 0
1 π
=
100 sinωt d(ωt ) π 0
100
=
[−cos ωt ]π
0
π
100
=
[(−cos π) − (−cos 0)] π 200
100
[(+1) − (−1)] =
=
π π = 63.66 volts

=

=

=

π

10000 π 10000 π 0 π 0

sin2 ωt d(ωt )

1
(1 − cos 2ωt ) d(ωt )
2

10000 1 sin 2ωt ωt − π 2
2

⎪ 10000 1

⎨ π 2




π
0






sin 2π π− 2
− 0−

sin 0
2

=

10000 1
[π]
π 2

=





100
10000
= √ = 70.71 volts
2
2

[Note that for a sine wave,
1
r.m.s. value= √ × maximum value.
2

378 Higher Engineering Mathematics y In this case,

y 5 f (x)

1
r.m.s. value = √ × 100 = 70.71 V]
2
A

Now try the following exercise x5a 0

Exercise 148 Further problems on mean and r.m.s. values
1. The vertical height h km of a missile varies with the horizontal distance d km, and is given by h = 4d − d 2 . Determine the mean height of the missile from d = 0 to d = 4 km.
[2 2 km].
3
2. The distances of points y from the mean value of a frequency distribution are related to the
1
variate x by the equation y = x + . Deterx mine the standard deviation (i.e. the r.m.s. value), correct to 4 significant figures for values of x from 1 to 2.
[2.198]
3. A current i = 25 sin 100πt mA flows in an electrical circuit. Determine, using integral calculus, its mean and r.m.s. values each correct to 2 decimal places over the range t = 0 to t = 10 ms.
[15.92 mA, 17.68 mA]

generated, V , is given by: d V=

πx2 dy

c

Problem 5. The curve y = x 2 + 4 is rotated one revolution about the x-axis between the limits x = 1 and x = 4. Determine the volume of solid of revolution produced.
Revolving the shaded area shown in Fig. 38.7, 360◦ about the x-axis produces a solid of revolution given by: π(x 2 + 4)2 dx

1
4

=

π(x 4 + 8x 2 + 16) dx

1

x 5 8x 3

+
+ 16x
5
3

4

1

= π[(204.8 + 170.67 + 64)
− (0.2 + 2.67 + 16)]
= 420.6π cubic units y 30

20

Volumes of solids of revolution

With reference to Fig. 38.6, the volume of revolution,
V , obtained by rotating area A through one revolution about the x-axis is given by:

A

10
5 D
4
0

πy2 dx

a

If a curve x = f ( y) is rotated 360◦ about the y-axis between the limits y = c and y = d then the volume

4

π y 2 dx =

1

where E 1 , E 3 and ω are constants.
Determine the r.m.s. value of v over the π interval 0 ≤ t ≤ . ω ⎤

2
2
E1 + E3


2

b

4

Volume =

v = E 1 sin ωt + E 3 sin 3ωt

V=

x

Figure 38.6

4. A wave is defined by the equation:

38.4

x5b

Figure 38.7

y 5 x21 4

B

C

1

2

3

4

5

x

Some applications of integration
Problem 6. Determine the area enclosed by the two curves y = x 2 and y 2 = 8x. If this area is rotated 360◦ about the x-axis determine the volume of the solid of revolution produced.

{(volume produced by revolving y 2 = 8x)
− (volume produced by revolving y = x 2 )}
2

i.e. volume =

x 4 = 8x

Hence, at the points of intersection, x = 0 and x = 2.
When x = 0, y = 0 and when x = 2, y = 4. The points of intersection of the curves y = x 2 and y 2 = 8x are therefore at (0,0) and (2,4).√A sketch is shown in
Fig. 38.8. If y 2 = 8x then y = 8x.

Shaded area
2

=


8x − x 2 dx =

0


1
8 x 2 − x 2 dx

2
0

⎤2
3
3
√ x2 x =⎣ 8 3 − ⎦ =
3
2


√ √
8 8
3
2

0

=



8
− {0}
3

16 8 8
2
− = = 2 square units
3
3 3
3
y5x2

y

y 2 5 8x
(or y 5Œ(8x)

4

2

0

8x 2 x 5
= π (8x − x )dx = π

2
5
0

2

4

16 −

0

32
− (0)
5

= 9.6π cubic units

x(x 3 − 8) = 0

and

π(x 4 )dx

0
2



x 4 − 8x = 0

from which,

2

π(8x)dx −

0

At the points of intersection the co-ordinates of the curves are equal. Since y = x 2 then y 2 = x 4 . Hence equating the y 2 values at the points of intersection:

Now try the following exercise
Exercise 149

Further problems on volumes

1. The curve x y = 3 is revolved one revolution about the x-axis between the limits x = 2 and x = 3. Determine the volume of the solid produced. [1.5π cubic units] y 2. The area between 2 = 1 and y + x 2 = 8 is x rotated 360◦ about the x-axis. Find the volume produced.
[170 2 π cubic units]
3
3. The curve y = 2x 2 + 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) 329.4π (b) 81π]
4. The profile of a rotor blade is bounded by the lines x = 0.2, y = 2x, y = e−x , x = 1 and the x-axis. The blade thickness t varies linearly with x and is given by: t = (1.1 − x)K, where
K is a constant.
(a) Sketch the rotor blade, labelling the limits.
(b) Determine, using an iterative method, the value of x, correct to 3 decimal places, where 2x = e−x

1

2

x

Figure 38.8

The volume produced by revolving the shaded area about the x-axis is given by:

379

(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.
[(b) 0.352 (c) 0.419 square units
(d) 0.222 K]

380 Higher Engineering Mathematics
38.5

Centroids

A lamina is a thin flat sheet having uniform thickness.
The centre of gravity of a lamina is the point where it balances perfectly, i.e. the lamina’s centre of mass.
When dealing with an area (i.e. a lamina of negligible thickness and mass) the term centre of area or centroid is used for the point where the centre of gravity of a lamina of that shape would lie.
If x and y denote the co-ordinates of the centroid C of area A of Fig. 38.9, then: b xy dx x= and y =

a b b

1
2

y2 dx

=

0
2

2

1
2

y 2 dx

(3x 2 )2 dx

0

8

y dx
0

=

=

2

1
2

9 x5
2 5

4

9x dx
=

0

8
32
5
8

9
2

2

0

8

18
= 3.6
5

=

Hence the centroid lies at (1.5, 3.6)

a b y dx

y=

2

1
2

y dx

a

Problem 8. Determine the co-ordinates of the centroid of the area lying between the curve y = 5x − x 2 and the x-axis.

a

y y 5 f(x)

y = 5x − x 2 = x(5 − x). When y = 0, x = 0 or x = 5.
Hence the curve cuts the x-axis at 0 and 5 as shown in Fig. 38.10. Let the co-ordinates of the centroid be
(x , y) then, by integration,

Area A
C
x

5

y x5a 0

x5b

x

x=

x(5x − x 2 ) dx

5

0
5

0

Problem 7. Find the position of the centroid of the area bounded by the curve y = 3x 2 , the x-axis and the ordinates x = 0 and x = 2.
If (x , y) are co-ordinates of the centroid of the given area then:
2

2

x y dx
=

0
2

x(3x 2 ) dx

0

(5x 2 − x 3 ) dx
=

0
5

(5x − x ) dx
2

6

5x 2
2



5 x4 4 0
5
x3
3 0

2

4

y 5 5x 2 x 2

3

12
= 1.5
8



y

3x dx
0

0

5x 3
3

0

2

3x 4
3x dx
4 0
=
= 02
[x 3 ]2
0
3x 2 dx
2

5

=

(5x − x 2 ) dx

0

8
2

y dx
0

=

=

0

y dx

Figure 38.9

x=

5

x y dx

C

x
2

y
0

Figure 38.10

1

2

3

4

5

x

Some applications of integration
625

= 3
125

2

6
125

625
12

=

y=

625
625
4 = 12
125
125
3
6

1
2

5

y 2 dx
=

0
5

5
0
5

y dx
0

=

=

=

1
2

4. Find the co-ordinates of the centroid of the area which lies between the curve y/x = x − 2 and the x-axis.
[(1, −0.4)]
5. Sketch the curve y 2 = 9x between the limits x = 0 and x = 4. Determine the position of the centroid of this area.
[(2.4, 0)]

5
= = 2.5
2
1
2

(5x − x 2 )2 dx

(5x − x 2 ) dx

0
5

38.6

Theorem of Pappus

A theorem of Pappus states:
(25x − 10x + x ) dx
2

3

4

‘If a plane area is rotated about an axis in its own plane but not intersecting it, the volume of the solid formed is given by the product of the area and the distance moved by the centroid of the area’.
With reference to Fig. 38.11, when the curve y = f (x) is rotated one revolution about the x-axis between the limits x = a and x = b, the volume V generated is given by:

0

125
6
1 25x 3 10x 4 x 5

+
2
3
4
5

5

0

125
6
1
2

381

volume V = (A)(2π y ), from which, y =

25(125) 6250

+ 625
3
4
125
6

V
2π A

y

= 2.5

y 5 f(x)
Area A

Hence the centroid of the area lies at (2.5, 2.5).

C

(Note from Fig. 38.10 that the curve is symmetrical about x = 2.5 and thus x could have been determined
‘on sight’.)

y x5a x5b x

Figure 38.11

Now try the following exercise
Exercise 150 Further problems on centroids
In Problems 1 and 2, find the position of the centroids of the areas bounded by the given curves, the x-axis and the given ordinates.
1.

y = 3x + 2 x = 0, x = 4

2.

y=

5x 2

x = 1, x = 4

[(2.5, 4.75)]
[(3.036, 24.36)]

3. Determine the position of the centroid of a sheet of metal formed by the curve y = 4x − x 2 which lies above the x-axis.
[(2, 1.6)]

Problem 9. (a) Calculate the area bounded by the curve y = 2x 2 , the x-axis and ordinates x = 0 and x = 3. (b) If this area is revolved (i) about the x-axis and (ii) about the y-axis, find the volumes of the solids produced. (c) Locate the position of the centroid using (i) integration, and (ii) the theorem of Pappus.
(a)

The required area is shown shaded in Fig. 38.12.
3

Area =
0

=

3

y dx =

2x 3
3

2x 2 dx

0
3

= 18 square units
0

382 Higher Engineering Mathematics y y 5 2x 2

y=

18

3

1
2

=

0
3

1
2

3

(2x 2 )2 dx

0

18

y dx

12

0

x

6

y
0

1

2

=

x

3

Figure 38.12

(b)

y 2 dx

3

1
2

4x 4 dx
=

0

18

1 4x 5
2 5

3

0

18

= 5.4

(ii) using the theorem of Pappus:

(i) When the shaded area of Fig. 38.12 is revolved 360◦ about the x-axis, the volume generated 3

=
0

π(2x 2 )2 dx

81π = (18)(2π x ),

i.e. from which,

x=

0
3

=

3

π y 2 dx =

Volume generated when shaded area is revolved about OY= (area)(2π x ).

3

x5
5

4π x 4 dx = 4π

0

Volume generated when shaded area is revolved about OX = (area)(2π y).

0

194.4π = (18)(2π y),

i.e.

243
= 194.4πcubic units
5

= 4π

y=

from which,
(ii) When the shaded area of Fig. 38.12 is revolved 360◦ about the y-axis, the volume generated = (volume generated by x = 3)
− (volume generated by y = 2x 2 )
18

=

18

π(3)2 dy −

0

y dy 2

π

0
18



9−

0

y2 y dy = π 9y −
2
4

18
0

= 81π cubic units
(c) If the co-ordinates of the centroid of the shaded area in Fig. 38.12 are (x, y) then:
(i) by integration,
3

3

x y dx x= =

0
3

y dx

Problem 10. A metal disc has a radius of 5.0 cm and is of thickness 2.0 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine, using Pappus’ theorem, the volume and mass of metal removed and the volume and mass of the pulley if the density of the metal is 8000 kg m−3.
A side view of the rim of the disc is shown in Fig. 38.13.
2.0 cm
P

x(2x 2 ) dx

=

18

5.0 cm

=

S

0

18
81
= 2.25
36

=

Q

0

3
2x 4

2x 3 dx

194.4π
= 5.4
36π

Hence the centroid of the shaded area in
Fig. 38.12 is at (2.25, 5.4).

0
3

81π
= 2.25
36π

4
18

X

0

Figure 38.13

R
X

Some applications of integration
When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid of the
4r
semicircular area removed is at a distance of from its

diameter (see ‘Engineering Mathematics 6th edition’,
4(1.0)
Chapter 58), i.e.
, i.e. 0.424 cm from PQ. Thus

the distance of the centroid from XX is 5.0 − 0.424,
i.e. 4.576 cm.
The distance moved through in one revolution by the centroid is 2π(4.576) cm. π(1.0)2 π πr 2
=
= cm2
Area of semicircle =
2
2
2
By the theorem of Pappus, volume generated = area × distance moved by π (2π)(4.576). centroid =
2
i.e. volume of metal removed = 45.16 cm3
Mass of metal removed = density × volume
45.16 3 m 106
= 0.3613 kg or 361.3 g

= 8000 kg m−3×

volume of pulley = volume of cylindrical disc
− volume of metal removed
= π(5.0)2 (2.0) − 45.16
= 111.9 cm3
Mass of pulley = density× volume
= 8000 kg m−3 ×

111.9 3 m 106

= 0.8952 kg or 895.2 g

Now try the following exercise
Exercise 151 Further problems on the theorem of Pappus
1. A right angled isosceles triangle having a hypotenuse of 8 cm is revolved one revolution about one of its equal sides as axis. Determine the volume of the solid generated using
Pappus’ theorem.
[189.6 cm3 ]
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

383

of a circle of radius 4 cm. (The equation of a circle, centre 0, radius r is x 2 + y 2 = r 2 ).


On the centre line, distance
⎢ 2.40 cm from the centre, ⎥




⎣ i.e. at co-ordinates

(1.70, 1.70)
3.

(a) Determine the area bounded by the curve y = 5x 2 , the x-axis and the ordinates x = 0 and x = 3.
(b) If this area is revolved 360◦ about (i) the x-axis, and (ii) the y-axis, find the volumes 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) 45 square units
⎢(b) (i) 1215π cubic units ⎥




⎣ (ii) 202.5π cubic units⎦
(c) (2.25, 13.5)

4. A metal disc has a radius of 7.0 cm and is of thickness 2.5 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine the volume of metal removed using Pappus’ theorem and express this as a percentage of the original volume of the disc. Find also the mass of metal removed if the density of the metal is
7800 kg m−3.
[64.90 cm3 , 16.86%, 506.2 g]
For more on areas, mean and r.m.s. values, volumes and centroids, see ‘Engineering Mathematics 6th edition’,
Chapters 55 to 58.

38.7 Second moments of area of regular sections
The first moment of area about a fixed axis of a lamina of area A, perpendicular distance y from the centroid of the lamina is defined as Ay cubic units. The second moment of area of the same lamina as above is given by Ay 2 , i.e. the perpendicular distance from the centroid of the area to the fixed axis is squared.

384 Higher Engineering Mathematics
Second moments of areas are usually denoted by I and have units of mm4 , cm4 , and so on.

x=l δx→0 Several areas, a1 , a2, a3 , . . . at distances y1 , y2, y3 , . . . from a fixed axis, may be replaced by a single area
A, where A = a1 + a2 + a3 + · · · at distance k from the axis, such that Ak 2 = ay 2 . k is called the radius of gyration of area A about the given axis. Since Ak 2 = ay 2 = I then the radius of gyration, The second moment of area is a quantity much used in the theory of bending of beams, in the torsion of shafts, and in calculations involving water planes and centres of pressure.
The procedure to determine the second moment of area of regular sections about a given axis is (i) to find the second moment of area of a typical element and (ii) to sum all such second moments of area by integrating between appropriate limits.
For example, the second moment of area of the rectangle shown in Fig. 38.14 about axis PP is found by initially considering an elemental strip of width δx, parallel to and distance x from axis PP. Area of shaded strip = bδx.

x 2 b dx

0

x=0

Thus the second moment of area of the rectangle about PP l =b

x 2 dx = b

0

x3
3

l

=
0

bl 3
3

Since the total area of the rectangle, A = lb, then
I pp = (lb)

I
A

l

x 2 b δx =

limit

Radius of gyration

k=

It is a fundamental theorem of integration that

l2
3

=

Al 2
3

l2
3
i.e. the radius of gyration about axes PP,
I pp = Ak 2 thus k 2 = pp pp

kpp =

l2 l =√
3
3

Parallel axis theorem
In Fig. 38.15, axis GG passes through the centroid C of area A. Axes DD and GG are in the same plane, are parallel to each other and distance d apart. The parallel axis theorem states:
IDD = IGG + Ad 2
Using the parallel axis theorem the second moment of area of a rectangle about an axis through the centroid

P

G l d

b

x

Area A
C
␦x
P

Figure 38.14

Second moment of area of the shaded strip about
PP = (x 2 )(b δx).
The second moment of area of the whole rectangle about
PP is obtained by summing all such strips between x =
0 and x = l, i.e. x=l x 2 bδx. x=0 G
D

Figure 38.15

D

Some applications of integration
P

G l 2

A summary of derived standard results for the second moment of area and radius of gyration of regular sections are listed in Table 38.1.

l
2

C

Problem 11. Determine the second moment of area and the radius of gyration about axes AA, BB and CC for the rectangle shown in Fig. 38.18.

b

x

l 5 12.0 cm

␦x
G

P

C

A

C b 5 4.0 cm

Figure 38.16
B

may be determined. In the rectangle shown in Fig. 38.16, bl 3
I pp =
(from above).
3
From the parallel axis theorem
1
I pp = IGG + (bl)
2

from which, IGG =

B
A

Figure 38.18

From Table 38.1, the second moment of area about axis AA,

2

bl 3 bl 3
= IGG +
3
4

i.e.

385

IAA =

bl 3 bl 3 bl 3

=
3
4
12

bl 3
(4.0)(12.0)3
=
= 2304 cm4
3
3

Perpendicular axis theorem

12.0 l Radius of gyration,kAA = √ = √ = 6.93 cm
3
3

In Fig. 38.17, axes OX , OY and OZ are mutually perpendicular. If OX and OY lie in the plane of area A then the perpendicular axis theorem states:

Similarly, IBB =

IOZ = IOX + IOY

and
Z

Y

O

lb3 (12.0)(4.0)3
=
= 256 cm4
3
3

4.0 b kBB = √ = √ = 2.31 cm
3
3

The second moment of area about the centroid of a bl 3 rectangle is when the axis through the centroid is
12
parallel with the breadth b. In this case, the axis CC is parallel with the length l.
Hence ICC =

lb3 (12.0)(4.0)3
=
= 64 cm4
12
12

Area A
X

Figure 38.17

and

4.0 b kCC = √ = √ = 1.15 cm
12
12

386 Higher Engineering Mathematics
Table 38.1 Summary of standard results of the second moments of areas of regular sections
Shape

Position of axis

Radius of

of area, I

gyration, k

bl 3
3

l

3

(2) Coinciding with l

lb3
3

b

3

(3) Through centroid, parallel to b

bl 3
12

l

12

(4) Through centroid, parallel to l

lb3
12

b

12

(1) Coinciding with b

bh 3
12

h

6

(2) Through centroid, parallel to base

bh 3
36

h

18

(3) Through vertex, parallel to base

bh 3
4

h

2

(1) Through centre, perpendicular to

πr 4
2

r

2

(2) Coinciding with diameter

πr 4
4

(3) About a tangent

5πr 4
4

r
2

5
r
2

Coinciding with diameter

Rectangle

Second moment

πr 4
8

r
2

(1) Coinciding with b

length l, breadth b

Triangle
Perpendicular height h, base b

Circle

plane (i.e. polar axis)

radius r

Semicircle radius r

Problem 12. Find the second moment of area and the radius of gyration about axis PP for the rectangle shown in Fig. 38.19.
40.0 mm
G

G
15.0 mm

25.0 mm
P

Figure 38.19

P

IGG =

lb3 where 1 = 40.0 mm and b = 15.0 mm
12

Hence IGG =

(40.0)(15.0)3
= 11250 mm4
12

From the parallel axis theorem, I PP = IGG + Ad 2 , where A = 40.0 × 15.0 = 600 mm2 and d = 25.0 +7.5 = 32.5 mm, the perpendicular distance between GG and PP. Hence,
IPP = 11 250 + (600)(32.5)2
= 645000 mm4

Some applications of integration
2
IPP = AkPP , from which,

kPP =

Problem 14. Determine the second moment of area and radius of gyration of the circle shown in
Fig. 38.21 about axis YY .

645000
= 32.79 mm
600

IPP
=
area

387

Problem 13. Determine the second moment of area and radius of gyration about axis QQ of the triangle BCD shown in Fig. 38.20.

r 5 2.0 cm
G

G

B
3.0 cm
12.0 cm

G

G

Y
C

8.0 cm

Figure 38.21

D
6.0 cm

Q

Y

Q

In Fig. 38.21, IGG =

Figure 38.20

Using the parallel axis theorem: I QQ = IGG + where IGG is the second moment of area about the centroid of the triangle,
Ad 2 ,

bh 3
(8.0)(12.0)3
i.e.
=
= 384 cm4 ,
36
36
A is the area of the triangle,

Using the parallel axis theorem, IYY = IGG + Ad 2 , where d = 3.0 + 2.0 = 5.0 cm.
Hence

= 6.0 + 1 (12.0) = 10 cm.
3

IYY = 4π + [π(2.0)2 ](5.0)2
= 4π + 100π = 104π = 327 cm4

Radius of gyration, kYY =

= 1 bh = 1 (8.0)(12.0) = 48 cm2
2
2 and d is the distance between axes GG and QQ,

πr 4 π = (2.0)4 = 4π cm4 .
4
4

IY Y
=
area

104π π(2.0)2 =


26 = 5.10 cm

Problem 15. Determine the second moment of area and radius of gyration for the semicircle shown in Fig. 38.22 about axis XX .

Hence the second moment of area about axis QQ,
G

IQQ = 384 + (48)(10)2 = 5184 cm4

B

Radius of gyration,

kQQ =

IQ Q
=
area

10.0 mm

G
B

15.0 mm

5184
= 10.4 cm
48

X

Figure 38.22

X

388 Higher Engineering Mathematics
4r
The centroid of a semicircle lies at from its

diameter.
Using the parallel axis theorem:
IBB = IGG + Ad 2 ,
IBB =

where

πr 4
(from Table 38.1)
8

πr 4
The polar second moment of area of a circle=
2
The polar second moment of area of the shaded area is given by the polar second moment of area of the 7.0 cm diameter circle minus the polar second moment of area of the 6.0 cm diameter circle.
Hence the polar second moment of area of the crosssection shown
=

π(10.0)4
=
= 3927 mm4,
8
π(10.0)2 πr 2
=
= 157.1 mm2
2
2
4r
4(10.0) d= =
= 4.244 mm



Hence

3927 = IGG + (157.1)(4.244)2

i.e.

3927 = IGG + 2830,

from which, IGG = 3927 − 2830 = 1097 mm4

7.0
2

4



π
2

6.0
2

4

= 235.7 − 127.2 = 108.5 cm4

A= and π
2

Problem 17. Determine the second moment of area and radius of gyration of a rectangular lamina of length 40 mm and width 15 mm about an axis through one corner, perpendicular to the plane of the lamina.
The lamina is shown in Fig. 38.24.

Using the parallel axis theorem again:
I XX = IGG + A(15.0 + 4.244)2
i.e. IXX =

Y
Z

1097 + (157.1)(19.244)2

m

0m

l54

b 5 15 mm
X

= 1097 + 58 179
= 59276 mm4 or 59280 mm4 , correct to 4 significant figures.

X
Z

Y

Radius of gyration, kXX =

I XX
=
area

59 276
157.1

Figure 38.24

= 19.42 mm

From the perpendicular axis theorem:
I ZZ = I XX + IYY

Problem 16. Determine the polar second moment of area of the propeller shaft cross-section shown in
Fig. 38.23.

IYY =

bl 3
(15)(40)3
=
= 320000 mm4
3
3

Hence
7.0 cm

lb 3
(40)(15)3
=
= 45000 mm4
3
3

and

6.0 cm

I XX =

IZZ = 45 000 + 320 000
= 365000 mm4 or 36.5 cm4

Radius of gyration, kZZ =

Figure 38.23

IZ Z
=
area

365 000
(40)(15)

= 24.7 mm or 2.47 cm

Some applications of integration
Problem 18. Determine correct to 3 significant figures, the second moment of area about axis XX for the composite area shown in Fig. 38.25.

389

Problem 19. Determine the second moment of area and the radius of gyration about axis XX for the
I -section shown in Fig. 38.26.
S
8.0 cm
CD

m
0c
4.

X
1.0 cm

CE

7.0 cm

X
1.0 cm

3.0 cm
8.0 cm

2.0 cm

3.0 cm

2.0 cm
C

C y CT
T

T
6.0 cm

X

CF

4.0 cm

15.0 cm

X

S

Figure 38.26
Figure 38.25

The I -section is divided into three rectangles, D, E and F and their centroids denoted by CD , CE and CF respectively. For the semicircle,
I XX =

πr 4 π(4.0)4 =
= 100.5 cm4
8
8

For the rectangle,
I XX =

bl 3
3

=

(6.0)(8.0)3
3

= 1024 cm4

For the triangle, about axis TT through centroid C T ,
ITT =

bh 3
(10)(6.0)3
=
= 60 cm4
36
36

By the parallel axis theorem, the second moment of area of the triangle about axis XX
2
= 60 + 1 (10)(6.0) 8.0 + 1 (6.0) = 3060 cm4 .
2
3
Total second moment of area about XX
= 100.5 + 1024 + 3060
= 4184.5
= 4180 cm4 , correct to 3 significant figures.

For rectangle D:
The second moment of area about C D (an axis through
CD parallel to XX )
=

bl 3
(8.0)(3.0)3
=
= 18 cm4
12
12

Using the parallel axis theorem:
I XX = 18 + Ad 2 where A = (8.0)(3.0) = 24 cm2 and d = 12.5 cm
Hence I XX = 18 + 24(12.5)2 = 3768 cm4.
For rectangle E:
The second moment of area about CE (an axis through
CE parallel to XX )
=

bl 3
(3.0)(7.0)3
=
= 85.75 cm4
12
12

Using the parallel axis theorem:
I XX = 85.75 + (7.0)(3.0)(7.5)2 = 1267 cm4.

390 Higher Engineering Mathematics
For rectangle F:
I XX

E

bl 3 (15.0)(4.0)3
=
=
= 320 cm4
3
3

E

Total second moment of area for the I-section about axis XX,

9.0 cm

I XX = 3768 + 1267 + 320 = 5355 cm4
D

Total area of I -section
= (8.0)(3.0) + (3.0)(7.0) + (15.0)(4.0)

Figure 38.28

= 105 cm2 .

3. For the circle shown in Fig. 38.29, find the second moment of area and radius of gyration about (a) axis FF and (b) axis HH .

Radius of gyration, k XX =

I XX
=
area

D

12.0 cm

5355
105

= 7.14 cm

(a) 201 cm4 , 2.0 cm
(b) 1005 cm4, 4.47 cm
H

Now try the following exercise
H

Exercise 152 Further problems on second moment of areas of regular sections

m

0c

r5

1. Determine the second moment of area and radius of gyration for the rectangle shown in
Fig. 38.27 about (a) axis AA (b) axis BB and
(c) axis CC.


(a) 72 cm4 , 1.73 cm
⎣(b) 128 cm4, 2.31 cm⎦
(c) 512 cm4 , 4.62 cm
B

C

4.

F

F

Figure 38.29

4. For the semicircle shown in Fig. 38.30, find the second moment of area and radius of gyration about axis J J .
[3927 mm4 , 5.0 mm]

8.0 cm

A

m m A

r5

10

.0

3.0 cm

J
B

C

J

Figure 38.30

Figure 38.27

2. Determine the second moment of area and radius of gyration for the triangle shown in
Fig. 38.28 about (a) axis DD (b) axis EE and
(c) an axis through the centroid of the triangle parallel to axis DD.⎡

(a) 729 cm4 , 3.67 cm
⎣(b) 2187 cm4 , 6.36 cm⎦
(c) 243 cm4, 2.12 cm

5. For each of the areas shown in Fig. 38.31 determine the second moment of area and radius of gyration about axis LL, by using the parallel axis theorem.


(a) 335 cm4, 4.73 cm


⎣(b) 22030 cm4, 14.3 cm⎦
(c) 628 cm4, 7.07 cm

Some applications of integration

3.0 cm
15 cm

m

.0 c

15 cm

4 ia 5

D

5.0 cm
2.0 cm

18 cm 10 cm

5.0 cm

L

L
(a)

(b)

(c)

391

10. Determine the second moments of areas about the given axes for the shapes shown in
Fig. 38.33. (In Fig. 38.33(b), the circular area is removed.)


I AA = 4224 cm4 ,
⎣ I BB = 6718 cm4 , ⎦
ICC = 37300 cm4

Figure 38.31
3.0 cm

6. Calculate the radius of gyration of a rectangular door 2.0 m high by 1.5 m wide about a vertical axis through its hinge.
[0.866 m]

B

4.5 cm
9.0 cm

16.0 cm

m

.0 c

7 ia 5

7. A circular door of a boiler is hinged so that it turns about a tangent. If its diameter is
1.0 m, determine its second moment of area and radius of gyration about the hinge.
[0.245 m4 , 0.559 m]
8. A circular cover, centre 0, has a radius of
12.0 cm. A hole of radius 4.0 cm and centre X , where OX = 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 OX .
[14280 cm4 , 5.96 cm]
9. For the sections shown in Fig. 38.32, find the second moment of area and the radius of gyration about axis XX .
(a) 12190 mm4 , 10.9 mm

D

4.0 cm
15.0 cm
A

9.0 cm
(a)

A
C
B

Figure 38.33

11. Find the second moment of area and radius of gyration about the axis XX for the beam section shown in Fig. 38.34.
1350 cm4 ,
5.67 cm

6.0 cm

(b) 549.5 cm4 , 4.18 cm
18.0 mm

2.0 cm
8.0 cm

2.0 cm

12.0 mm
X

1.0 cm

6.0 cm

3.0 mm
2.5 cm
4.0 mm

3.0 cm
2.0 cm

X

2.0 cm

X
(a)

Figure 38.32

C

10.0 cm
(b)

X

X
(b)

Figure 38.34

10.0 cm

X

Chapter 39

Integration using algebraic substitutions 39.1

39.3 Worked problems on integration using algebraic substitutions

Introduction

Functions which require integrating are not always in the ‘standard form’ shown in Chapter 37. However, it is often possible to change a function into a form which can be integrated by using either:
(i) an algebraic substitution (see Section 39.2),
(ii) a trigonometric or hyperbolic substitution (see
Chapter 40),
(iii) partial fractions (see Chapter 41),
(iv) the t = tan θ/2 substitution (see Chapter 42),
(v) integration by parts (see Chapter 43), or
(vi) reduction formulae (see Chapter 44).

Problem 1.

Determine

cos(3x + 7) dx.

cos(3x + 7) dx is not a standard integral of the form shown in Table 37.1, page 369, thus an algebraic substitution is made. du = 3 and rearranging gives
Let u = 3x + 7 then dx du dx = . Hence,
3
cos(3x + 7) dx =

(cos u)

du
=
3

1 cos u du,
3

which is a standard integral

39.2

Algebraic substitutions

With algebraic substitutions, the substitution usually made is to let u be equal to f (x) such that f (u) du is a standard integral. It is found that integrals of the forms, k

[ f (x)] f (x) dx and k n f (x) dx [ f (x)]n

(where k and n are constants) can both be integrated by substituting u for f (x).

=

1 sin u + c
3

Rewriting u as (3x + 7) gives: cos(3x + 7) dx =

1 sin(3x + 7) + c,
3

which may be checked by differentiating it.
Problem 2.

Find (2x − 5)7 dx.

(2x − 5) may be multiplied by itself 7 times and then each term of the result integrated. However, this would

Integration using algebraicsubstitutions be a lengthy process, and thus an algebraic substitution is made. du du
= 2 and dx =
Let u =(2x − 5) then dx 2
Hence

3x(4x 2 + 3)5 dx.

Problem 5. Determine
Let u =(4x 2 + 3) then

393

du du = 8x and dx = dx 8x

Hence du 1 u =
2
2

(2x − 5) dx =
7

=

7

u8
8

1
2

7

u du
1 8 u +c
16

+c =

Rewriting u as (2x − 5) gives:
(2x − 5)7 dx =

1
(2x −5)8 + c
16

4 dx. (5x − 3)

Problem 3. Find

3x(4x 2 + 3)5 dx =
=

=

4 du 4
=
u 5
5

Hence

3
8

u 5 du =
=

1 du u

1 6x−1 dx, 0 2e

correct to

Let u = sin θ then

du du = 6 and dx = dx 6

2e

dx =

du
1
2e
=
6
3
u

= 24

1 6
1
u + c = (4x2 + 3)6 + c
16
16 π 6

24 sin5 θ cos θ dθ.

du du = cos θ and dθ = dθ cos θ
24u 5 cos θ

π
6

Thus
0

u6
+ c = 4u 6 + c = 4(sin θ)6 + c
6

π

24 sin5 θ cos θ dθ = [4 sin6 θ]06

=4

1
1
2e6x−1 dx = [e6x−1 ]1 = [e5 − e−1 ] = 49.35,
0
3
3
correct to 4 significant figures.

du cos θ

u 5 du, by cancelling

e du

Thus

0

+c

= 4 sin6 θ + c

u

1
1
= eu + c = e6x−1 + c
3
3

1

u6
6

24 sin5 θ cos θ dθ =

Hence

Hence
6x−1

u 5 du, by cancelling

0

= 24
Let u =6x − 1 then

3
8

Problem 6. Evaluate

4
4
ln u + c = ln(5x −3) + c
5
5

Problem 4. Evaluate
4 significant figures.

3
8

du
8x

The original variable ‘x’ has been completely removed and the integral is now only in terms of u and is a standard integral.

du du = 5 and dx =
Let u =(5x − 3) then dx 5
Hence
4 dx =
(5x − 3)

3x(u)5

sin

=4

1
2

π
6

6

− (sin 0)6

6

−0 =

1 or 0.0625
16

394 Higher Engineering Mathematics
Now try the following exercise

Hence

Exercise 153 Further problems on integration using algebraic substitutions

2. 3 cos(2θ − 5)
3. 4 sec2 (3t + 1)

4 tan(3t + 1) +c
3

4.

1
(5x − 3)7 + c
70

1
(5x − 3)6
2

5.

−3
(2x − 1)

6.

3e3θ+5

=

Problem 8.

1

=
=

[e3θ + 5 + c]

=

[227.5]

x (2x 2 + 1) dx

8.

=
[4.333]

0 π 3

9.
0

1

10.

1
4

3 cos(4x − 3) dx

[0.7369]

Problem 7.

Find

Let u = 2 +3x 2 then

x dx. 2 + 3x 2 du du
= 6x and dx = dx 6x

(4x 2 − 1)

dx.

du =



−1
2 +1



1 ⎢u

⎣ 1
⎦+c
4
− +1
2

1√

u +c
⎣ 1 ⎦+c =
4
2
2
1
(4x2 − 1) + c
2
Show that tan θ dθ = ln(sec θ) + c. sin θ dθ. Let u = cos θ cos θ

tan θ dθ = then Further worked problems on integration using algebraic substitutions −1
2

1
1 ⎢u 2

[0.9428]

0

39.4

u

Problem 9. π dt
2 sin 3t +
4

2x

1
√ du, by cancelling u ⎡

1
4



0
2

Determine

du du = 8x and dx = dx 8x
2x du
2x
dx = √ u 8x
(4x 2 − 1)

Hence

3
− ln(2x − 1) +c
2

(3x + 1)5 dx

1
1
ln u + c = ln(2 + 3x2) + c
6
6

Let u = 4x 2 − 1 then

In Problems 7 to 10, evaluate the definite integrals correct to 4 significant figures.
7.

1 du, u by cancelling

In Problems 1 to 6, integrate with respect to the variable. 1
1. 2 sin(4x + 9)
− cos(4x + 9) +c
2
3 sin(2θ − 5) +c
2

x du
1
= u 6x
6

x dx =
2 + 3x 2

−du du = −sin θ and dθ = dθ sin θ

Hence sin θ dθ = cos θ
=−

sin θ u −du sin θ

1 du = − ln u + c u = − ln(cos θ) + c = ln(cos θ)−1 + c, by the laws of logarithms.

Integration using algebraicsubstitutions
Hence

tan θ dθ = ln(sec θ)+ c,
(cos θ)−1 =

since

2

Hence
0

39.5

3x
(2x 2

+ 1)

x=0

=

du du Let u =2x 2 + 7, then
= 4x and dx = dx 4x
It is possible in this case to change the limits of integration. Thus when x = 3, u =2(3)2 + 7 =25 and when x = 1, u = 2(1)2 + 7 = 9.

Thus

3
4

5x (2x 2 + 7) dx =

x=1

u=9

√ du
5x u
4x

x=2

u

−1
2

du

x=0

x=2

u

−1
2

x=0

du =

3
4

u=9

u

−1
2

du,

u=1

i.e. the limits have been changed


=

Hence u=25 3
4

3x du

u 4x

Since u = 2x 2 + 1, when x = 2, u =9 and when x = 0, u =1.

3

Problem 10. Evaluate 1 5x (2x 2 + 7) d x, taking positive values of square roots only.

x=2

dx =

Change of limits

When evaluating definite integrals involving substitutions it is sometimes more convenient to change the limits of the integral as shown in Problems 10 and 11.

x=3

du du = 4x and dx = dx 4x

Let u =2x 2 + 1 then

1
= sec θ cos θ

1
3 ⎢u2

⎤9


3 √

9 − 1 = 3,
⎣ 1 ⎦ =
4
2
2 1

taking positive values of square roots only.

Now try the following exercise
=

=

5
4
5
4

25 √

u du

Exercise 154 Further problems on integration using algebraic substitutions

9
25

1

u 2 du
9

Thus the limits have been changed, and it is unnecessary to change the integral back in terms of x. x=3 Thus x=1 ⎡ 3 ⎤25
5⎣ u2 ⎦
5x (2x 2 + 7) dx =
4 3/2
9

=

5
6

25
5 √ 3 √ 3 u3 =
25 − 9
9
6

5
2
= (125 − 27) = 81
6
3
2
0

In Problems 1 to 7, integrate with respect to the variable. 1
1. 2x(2x 2 − 3)5
(2x 2 − 3)6 + c
12
5
− cos6 t + c
6

2.

5 cos5 t sin t

3.

3 sec2 3x tan 3x
1
1 sec2 3x + c or tan2 3x + c
2
2

4.

2t (3t 2 − 1)

5.

ln θ θ 3x

dx,
(2x 2 + 1) taking positive values of square roots only.

Problem 11. Evaluate

395

2
(3t 2 − 1)3 + c
9
1
(ln θ)2 + c
2

396 Higher Engineering Mathematics

6.

7.

3 ln(sec 2t ) + c
2

3 tan 2t


4 (et + 4) + c

2et

√ t
(e + 4)

In Problems 8 to 10, evaluate the definite integrals correct to 4 significant figures.
1

8.

3x e(2x

π
2

2 −1)

dx

[1.763]

3 sin θ cos θ dθ
4

[0.6000]

0
1

10.
0

11.

V = 2πσ

(92 + r 2 ) − r

12. In the study of a rigid rotor the following integration occurs:


Zr =

(2 J + 1)e

−J (J +1) h 2
8π 2 I k T

dJ

0

0

9.

Solve the equation by determining the integral.

3x dx 2 − 1)5
(4x

[0.09259]

The electrostatic potential on all parts of a conducting circular disc of radius r is given by the equation:
9

V = 2πσ
0



R
R2 + r 2

dR

Determine Z r for constant temperature T assuming h, I and k are constants.
8π 2 I kT h2 13. In electrostatics,



π⎨ a2 σ sin θ
E=


0 ⎩ 2ε a2 − x 2 − 2ax cos θ where a, σ and ε are constants, x is greater than a, and x is independent of θ. Show that a2 σ
E=
εx

Revision Test 11
This Revision Test covers the material contained in Chapters 37 to 39. The marks for each question are shown in brackets at the end of each question.
1.

Determine: (a)

π
3

2
1

1

(c)
0

5.

6.

7.

(9)

3 sin 2t dt (b)

0

4.

theorem of Pappus to determine the volume of material removed, in cm3 , correct to 3 significant figures. (8)

Evaluate the following integrals, each correct to
4 significant figures:
(a)

3.

2
√ dx
3 2 x (2 + θ)2 dθ

(c)
2.

3 t 5 dt (b)

2
1 3
+ + dx x2 x 4

3 dt e2t

Calculate the area between the curve y = x 3 − x 2 − 6x and the x-axis.

400 mm

(15)

50 mm
200 mm

(10)

A voltage v = 25 sin 50πt volts is applied across an electrical circuit. Determine, using integration, its mean and r.m.s. values over the range t = 0 to t = 20 ms, each correct to 4 significant figures.
(12)
Sketch on the same axes the curves x 2 = 2y and y 2 = 16x and determine the co-ordinates of the points of intersection. Determine (a) the area enclosed by the curves, and (b) the volume of the solid produced if the area is rotated one revolution about the x-axis.
(13)

Figure RT11.1

8.

A circular door is hinged so that it turns about a tangent. If its diameter is 1.0 m find its second moment of area and radius of gyration about the hinge. (5)

9.

Determine the following integrals:

Calculate the position of the centroid of the sheet of metal formed by the x-axis and the part of the curve y = 5x − x 2 which lies above the x-axis.
(9)
A cylindrical pillar of diameter 400 mm has a groove cut around its circumference as shown in
Fig. RT11.1. The section of the groove is a semicircle of diameter 50 mm. Given that the centroid
4r
of a semicircle from its base is
, use the


(a)
(c)
10.

5(6t + 5)7 dt

3 ln x dx x

2


(2θ − 1)

(b)

(9)

Evaluate the following definite integrals: π 2

(a)
0

2 sin 2t +

π dt (b)
3

1
0

3x e4x

2 −3

dx
(10)

Chapter 40

Integration using trigonometric and hyperbolic substitutions


40.1

Introduction

Table 40.1 gives a summary of the integrals that require the use of trigonometric and hyperbolic substitutions and their application is demonstrated in Problems 1 to 27.

⎢π
=⎣ +
4
=

π or 0.7854
4

Problem 2.

40.2 Worked problems on integration of sin2 x, cos2 x, tan2 x and cot2 x

Problem 1.

π
4

Evaluate

2 cos 2 4t dt.

0

Since cos 2t = 2 cos 2 t − 1 (from Chapter 17),
1
then cos 2 t = (1 + cos 2t ) and
2
1 cos 2 4t = (1 + cos 8t )
2
π
4

Hence
0

2 cos 2 4t dt π 4

1
(1 + cos 8t ) dt
0 2 π sin 8t 4
= t+
8
0
=2

π ⎤
4 ⎥ − 0 + sin 0

8
8

sin 8

Determine

sin 2 3x dx.

Since cos 2x = 1 − 2 sin 2 x (from Chapter 17),
1
then sin 2 x = (1 − cos 2x) and
2
2 3x = 1 (1 − cos 6x) sin 2
1
2
(1 − cos 6x) dx
Hence
sin 3x dx =
2
sin 6x
1
x−
+c
=
2
6
Problem 3.

Find 3 tan 2 4x dx.

Since 1 + tan2 x = sec2 x, then tan2 x = sec2 x − 1 and tan2 4x = sec2 4x − 1.
Hence 3

tan 2 4x dx = 3

(sec 2 4x − 1) dx

=3

tan 4x
−x +c
4

Integration using trigonometric and hyperbolic substitutions
Table 40.1 Integrals using trigonometric and hyperbolic substitutions f (x)

f (x)dx

Method

See problem

1. cos 2 x

sin 2x
1
x+
2
2

+c

Use cos 2x = 2 cos 2 x − 1

1

2. sin 2 x

sin 2x
1
x−
2
2

+c

Use cos 2x = 1 − 2 sin 2 x

2

3. tan2 x

tan x − x + c

Use 1 + tan2 x = sec2 x

3

4. cot 2 x

− cot x − x + c

Use cot 2 x + 1 = cosec2 x

4

5.

cos m x

sin n x

(a) If either m or n is odd (but not both), use cos 2 x + sin 2 x = 1

5, 6

(b) If both m and n are even, use either cos 2x = 2 cos 2 x − 1 or cos 2x = 1 − 2 sin 2 x
Use 1 [ sin(A + B) + sin(A − B)]
2

6. sin A cos B
7. cos A sin B

Use

8. cos A cos B

Use

9. sin A sin B

Use

10.

11.

12.

13.

1
(a 2 − x 2 )
(a 2 − x 2 )
1
a2 + x 2
1
(x 2 + a 2 )

sin−1

x
+c
a

15.

(x 2 + a 2 )
1
(x 2

− a2)

(x 2 − a 2 )

10
11
12

13, 14

a 2 −1 x x
(a 2 − x 2 ) + c sin +
2
a 2

Use x = a sin θ substitution

15, 16

1 −1 x tan +c a a

Use x = a tan θ substitution

17–19

Use x = a sinh θ substitution

20–22

Use x = a sinh θ substitution

23

Use x = a cosh θ substitution

24, 25

Use x = a cosh θ substitution

26, 27

sinh−1

x
+c
a x+ (x 2 + a 2 )
+c
a

a2 x x
(x 2 + a 2 ) + c sinh−1 +
2
a 2 cosh−1 or ln

16.

9

Use x = a sin θ substitution

or ln

14.

1
2 [ sin(A + B) − sin(A − B)]
1
2 [ cos(A + B) + cos(A − B)]
− 1 [ cos(A + B) − cos(A − B)]
2

7, 8

x
+c
a x+ (x 2 − a 2 )
+c
a

x x a2
(x 2 − a 2 ) − cosh−1 + c
2
2 a 399

400 Higher Engineering Mathematics

Problem 4.

Evaluate

π
3
π
6

40.3 Worked problems on powers of sines and cosines

1 2 cot 2θ dθ.
2

Since cot 2 θ +1 = cosec2 θ, then cot 2 θ = cosec2 θ−1 and cot 2 2θ = cosec 2 2θ − 1. π 3

Hence

π
6

π
3

sin 5 θ dθ

Hence π 2

1
= [(0.2887 − 1.0472) − (−0.2887 − 0.5236)]
2
= 0.0269

Now try the following exercise

=

sin θ(sin 2 θ)2 dθ =

=

sin θ(1 − 2 cos 2 θ + cos 4 θ) dθ

=

(sin θ − 2 sin θ cos 2 θ + sin θ cos 4 θ) dθ

Exercise 155 Further problems on integration of sin2 x, cos2 x, tan2 x and cot2 x
In Problems 1 to 4, integrate with respect to the variable. 1 sin 4x
1. sin 2 2x x− +c
2
4
3
sin 2t t+ 2
2

3 cos 2 t

2 cos3 θ cos5 θ

+c
3
5
Whenever a power of a cosine is multiplied by a sine of power 1, or vice-versa, the integral may be determined by inspection as shown.

5

2 cot 2 2t

π
3

5.

2

3 sin 3x dx

0 π 4

6.

2

cos 4x dx

0
1

7.

2 tan2 2t dt

8.

π
6

π
2

π or 0.3927
8
[−4.185]

cot θ dθ
2

[0.6311]

π
2

Evaluate π 2

sin 2 x cos 3 x dx =

0

π
2

=

π or 1.571
2

=

−cos n+1 θ
+c
(n + 1) sin n+1 θ
+c
(n + 1)

sin 2 x cos 3 x dx.

0

=

0 π 3

Problem 6.

1 tan 3θ − θ + c
3

In Problems 5 to 8, evaluate the definite integrals, correct to 4 significant figures.

cos n θ sin θ dθ =

sin n θ cos θ dθ

and

[−(cot 2t + 2t ) + c]

3. 5 tan2 3θ
4.

+c

sin θ(1 − cos 2 θ)2 dθ

= −cos θ +

In general,

2.

sin 5 θ dθ.

Determine

Since cos 2 θ + sin 2 θ = 1 then sin 2 θ = (1 − cos 2 θ).

1 2 cot 2θ dθ
2

3
1 −cot 2θ
(cosec 2θ − 1) dθ =
−θ
π π 2
2
⎡6

⎞ ⎛
⎞6
⎤ π π
−cot 2
−cot 2
1⎢


3 − π⎟−⎜
6 − π⎟
= ⎣

⎠ ⎝


2
2
3
2
6

1
=
2

Problem 5.

0

π
2

0

(sin 2 x)(1 − sin 2 x)(cos x) dx
(sin 2 x cos x − sin 4 x cos x) dx

0

sin 3 x sin 5 x

3
5

=


π sin ⎢
2
=⎣
3
=

sin 2 x cos 2 x cos x dx

3



π
2
0

π
2
5

sin

5⎤


⎦ − [0 − 0]

2
1 1
− = or 0.1333
3 5 15

Problem 7.

π
4

Evaluate

significant figures.

0

4 cos 4 θ dθ, correct to 4

Integration using trigonometric and hyperbolic substitutions π 4

π
4

4 cos θ dθ = 4
4

0

401

Now try the following exercise

(cos 2 θ)2 dθ

0 π 4

=4
0
π
4

=

Exercise 156 Further problems on integration of powers of sines and cosines

2

1
(1 + cos 2θ)
2



(1 + 2 cos 2θ + cos 2θ) dθ
2

In Problems 1 to 6, integrate with respect to the variable. 0 π 4

=
0

π
4

=
0

=
=
=

1
1 + 2 cos 2θ + (1 + cos 4θ) dθ
2
3
1
+ 2 cos 2θ + cos 4θ dθ
2
2

3θ sin 4θ
+ sin 2θ +
2
8

π
4

sin 3 θ

(a)−cos θ + sin 2x −

2. 2 cos 3 2x
3. 2 sin 3 t cos 2 t

4.

3θ 1
1
− sin 4θ + sin 8θ + c
4
4
32

correct to 4 significant figures.

Problem 8. Find

sin 2 t cos 4 t dt.

sin 2 t cos 4 t dt =

sin 3 2x
+c
3

− cos 5 x cos 7 x
+
+c
5
7

sin 3 x cos 4 x

5. 2 sin 4 2θ

cos 3 θ
+c
3

−2
2
cos 3 t + cos 5 t + c
3
5

0

3 π

sin 4(π/4)
+ sin
+
− [0]
2 4
4
8

+ 1 = 2.178,
8

1.

t
1
− sin 4t + c
8 32

sin 2 t (cos 2 t )2 dt

1 − cos 2t
2

=

1 + cos 2t
2

2

dt

=

1
8

(1 − cos 2t )(1 + 2 cos 2t + cos 2 2t ) dt

=

1
8

(1 + 2 cos 2t + cos 2 2t − cos 2t

1
=
8
=

1
8

− 2 cos 2 2t − cos 3 2t ) dt
(1 + cos 2t − cos 2 2t − cos 3 2t ) dt
1 + cos 2t −

1 + cos 4t
2
− cos 2t (1 − sin 2 2t ) dt

=

1
8

=

1
8

1 cos 4t

+ cos 2t sin 2 2t dt
2
2 t sin 4t sin3 2t

+
2
8
6

6.

sin 2 t cos 2 t

40.4 Worked problems on integration of products of sines and cosines
Problem 9. Determine

sin 3t cos 2t dt.

sin 3t cos 2t dt
=

1
[sin (3t + 2t ) + sin (3t − 2t )] dt,
2

from 6 of Table 40.1, which follows from Section 17.4, page 170,
=

1
2

(sin 5t + sin t ) dt

=

1
2

−cos 5t
− cos t + c
5

+c
Problem 10. Find

1 cos 5x sin 2x dx.
3

402 Higher Engineering Mathematics
1
cos 5x sin 2x dx
3
1 1
[sin (5x + 2x) − sin (5x − 2x)] dx,
=
3 2 from 7 of Table 40.1
1
6
1
=
6

−cos 7x cos 3x
+
7
3

+c
1

Problem 11.

Evaluate

2 cos 6θ cos θ dθ,

3. 3 cos 6x cos x

correct to 4 decimal places.

2 cos 6θ cos θ dθ
1
0

1
[ cos (6θ + θ) + cos (6θ − θ)] dθ,
2
from 8 of Table 40.1

1

=
0

sin 7θ sin 5θ
(cos 7θ + cos 5θ) dθ =
+
7
5

sin 7 sin 5
+

7
5

=

sin 0 sin 0
+
7
5

‘sin 7’ means ‘the sine of 7 radians’ sin 5 ≡286◦29 .
1

Hence

sin 7x sin 5x
+
7
5

+c

cos 2θ cos 6θ

2
6

+c

1 cos 4θ sin 2θ
2

In Problems 5 to 8, evaluate the definite integrals.

1
0

3
2

1
4

4.

0

=2

sin 2x sin 4x

+c
2
4

2. 2 sin 3x sin x

0

1

Exercise 157 Further problems on integration of products of sines and cosines
In Problems 1 to 4, integrate with respect to the variable. 1 cos 7t cos 3t
+c
1. sin 5t cos 2t

+
2
7
3

(sin 7x − sin 3x) dx

=

Now try the following exercise

π
2

5.

(a)

cos 4x cos 3x dx

0

3 or 0.4286
7

1

2 sin 7t cos 3t dt

6.
(≡401◦4

[0.5973]

0

) and

π
3

7. −4

sin 5θ sin 2θ dθ

[0.2474]

0

2 cos 6θ cos θ dθ

2

0

= (0.09386 + (−0.19178)) − (0)

8.

3 cos 8t sin 3t dt

[−0.1999]

1

= −0.0979, correct to 4 decimal places.
Problem 12.

3

Find 3

sin 5x sin 3x dx.

sin 5x sin 3x dx
=3
3
2
3
=−
2
=−

Problem 13.

1
− [ cos (5x + 3x) − cos (5x − 3x)] dx,
2
from 9 of Table 40.1
( cos 8x − cos 2x) dx sin 8 sin 2x

8
2

40.5 Worked problems on integration using the sin θ substitution

+ c or
3
(4 sin 2x −sin 8x) + c
16

Determine

Let x = a sin θ, then
Hence
=
=

1
(a 2 − x 2 )

dx.

dx
= a cos θ and dx = a cos θ dθ. dθ 1
(a 2 − x 2 )

dx

1
(a 2 − a 2 sin 2 θ) a cos θ dθ
[a 2(1 − sin 2 θ)]

a cos θ dθ

Integration using trigonometric and hyperbolic substitutions a cos θ dθ

=

(a 2 cos 2 θ)

, since sin 2 θ + cos 2 θ = 1

a cos θ dθ
=
= dθ = θ + c a cos θ x x
Since x = a sin θ, then sin θ = and θ = sin−1 . a a x 1 dx = sin−1 + c
Hence
a
(a 2 − x 2 )
3

Problem 14. Evaluate

(9 − x 2 )

0
3

0

(1 − sin 2 θ) =

cos θ =

a2 − x 2 a2 =

1−

2

(a 2 − x 2 ) a =

a2
[θ + sin θ cos θ]
2

(a 2 − x 2 ) dx =

Thus

x a (a 2 − x 2 )
+c
a

x x a2 sin−1 +
2
a a , since a = 3

=

a2 x x
(a2 − x2 ) + c sin−1 +
2
a 2

(9 − x 2 )

3

Also, cos 2 θ + sin 2 θ = 1, from which,

=

0

x
3

dx.

x x and θ = sin−1 a a

dx

1

From Problem 13,
= sin−1

1

Since x = a sin θ, then sin θ =

= (sin −1 1 − sin−1 0) =

π or 1.5708
2

4

(16 − x 2 ) dx.

Problem 16. Evaluate
0

(a 2 − x 2 ) dx.

Problem 15. Find

4

From Problem 15,
Let x = a sin θ then
Hence

dx
= a cos θ and dx = a cos θ dθ. dθ (a 2 − x 2 ) dx

=

16 x x
(16 − x 2 ) sin−1 +
2
4 2

= 8 sin

=

(a 2 − a 2 sin 2 θ) (a cos θ dθ)

=

−1

1 + 2 (0) − [8 sin

4
0

−1

0 + 0]

π
= 4π or 12.57
2

[a 2 (1 − sin 2 θ)] (a cos θ dθ)

=

(16 − x 2 ) dx

0

(a 2 cos 2 θ) (a cos θ dθ)

= 8 sin −11 = 8

Now try the following exercise

=

Exercise 158 Further problems on integration using the sine θ substitution

(a cos θ)(a cos θ dθ)

= a2

1 + cos 2θ
2

cos 2 θ dθ = a 2



1. Determine

5
(4 − t 2)

dt .
5 sin−1

=
=

=

sin 2θ a2 θ+
2
2

+c

a2
2 sin θ cos θ θ+ +c
2
2 since from Chapter 17, sin 2θ = 2 sin θ cos θ a2 [θ + sin θ cos θ] + c
2

2. Determine

3. Determine

3
(9 − x 2 )

x
+c
2

3 sin−1

(since cos 2θ = 2 cos 2 θ − 1)

x
+c
3

dx.

(4 − x 2 ) dx. x x
(4 − x 2 ) + c
2 sin−1 +
2 2

403

404 Higher Engineering Mathematics
1

(16 − 9t 2) dt .

4. Determine

Problem 19.

8
3t
t
(16 − 9t 2 ) + c sin−1 +
3
4
2
4

1

5. Evaluate

(16 − x 2 )

0
1

6. Evaluate

dx.

π or 1.571
2

(9 − 4x 2 ) dx.

Problem 17.

1 dx. 2 + x 2)
(a

Determine

Let x = a tan θ then

1
0

dx
= a sec 2 θ and dx = a sec2 θ dθ. dθ 1 dx (a 2 + x 2 )
1
(a sec2 θ dθ)
=
2 + a 2 tan2 θ)
(a
a sec2 θ dθ
=
2 (1 + tan 2 θ) a a sec 2 θ dθ
=
, since 1+tan2 θ = sec 2 θ a 2 sec2 θ
1
1 dθ = (θ) + c
=
a a x
Since x = a tan θ, θ = tan −1 a x
1
1
Hence
dx = tan−1 + c
(a2 + x2 ) a a
Hence

=

=

5
2

1
0

2

Evaluate
0
2

5 dx 2[(3/2) + x 2 ]

1 dx √
[ (3/2)]2 + x 2

2
3

5
2

1
0

2
− tan −1 0
3

tan−1

= (2.0412)[0.6847 − 0]
= 1.3976, correct to 4 decimal places.
Now try the following exercise
Exercise 159 Further problems on integration using the tan θ substitution
3
t tan −1 + c
2
2

1. Determine

3 dt .
4 + t2

2. Determine

5 dθ. 16 + 9θ 2
5
3θ tan −1
+c
12
4
1

3. Evaluate
0
3

4. Evaluate

3 dt .
1 + t2

[2.356]

5 dx. 4 + x2

[2.457]

1 dx. (4 + x 2 )

1 dx 2
0 (4 + x ) x 2
1
since a = 2 tan −1
=
2
2 0
1 π
1
−0
= (tan −1 1 −tan −1 0) =
2
2 4 π = or 0.3927
8

From Problem 17,

0

5 dx, correct
(3 + 2x 2 )

1
5
x tan−1 √

2
(3/2)
(3/2)

0

Problem 18.

1

5 dx =
(3 + 2x 2 )

=

40.6 Worked problems on integration using tan θ substitution

0

to 4 decimal places.

[2.760]

0

Evaluate

40.7 Worked problems on integration using the sinh θ substitution
Problem 20.

Determine

1
(x 2

+ a2)

dx
Let x = a sinh θ, then
= a cosh θ and dθ dx = a cosh θ dθ

dx.

Integration using trigonometric and hyperbolic substitutions
1

Hence

(x 2 + a 2 )
(a 2

sinh2 θ + a 2 )

a cosh θ dθ

=

Since the integral contains a term of the form
(a 2 + x 2 ), then let x = sinh θ, from which dx = cosh θ and dx = cosh θ dθ dθ 2 dx Hence
2 (1 + x 2 ) x dx
1

=

(a 2 cosh2 θ)

(a cosh θ dθ)

,

a cosh θ dθ = dθ = θ + c a cosh θ x = sinh−1 + c, since x = a sinh θ a It is shown on page 339 that
=

x x+ = ln a 2(cosh θ dθ)

=

since cosh2 θ − sinh2 θ = 1

sinh−1

sinh2 θ

=2

=2

(x 2 + a 2 )
,
a coth θ =
Hence

dx

1

2

Problem 21. Evaluate
0

to 4 decimal places.
2
0

1
(x 2

+ 4)

1
(x 2

dx = sinh −1

x
2

x+

ln

+ 4)


=2
sinh2 θ

cosech 2 θ dθ

dx, correct

(1 + sinh2 θ)
=
sinh θ

2 x2 1 + x2)

(1 + x 2 ) x dx
2

(1 + x 2 )
= −[2 coth θ]2 = −2
1
x
1


5
2
= 0.592,
= −2

2
1
correct to 3 significant figures

2
0

or

Problem 23. Find

(x 2 + 4)
2

(x 2 + a 2 ) dx.

2

0

dx
Let x = a sinh θ then
= a cosh θ and dθ dx = a cosh θ dθ

from Problem 20, where a = 2
Hence

Using the logarithmic form,
2

cosh θ dθ
,
sinh2 θ cosh θ since cosh2 θ − sinh2 θ = 1

cosh θ
=
sinh θ
2

(x 2 + a 2 )

(1 + sinh2 θ)

= −2 coth θ + c

which provides an alternative solution to
1

1

(x 2 + a 2 ) dx

=

(a 2 sinh2 θ + a 2 )(a cosh θ dθ)

=

[a 2(sinh2 θ + 1)](a cosh θ dθ)

=

dx
(x 2 + 4)


0+ 4
2+ 8
− ln
= ln
2
2

(a 2 cosh2 θ) (a cosh θ dθ),

0

= ln 2.4142 − ln 1 = 0.8814, correct to 4 decimal places.
2

Problem 22. Evaluate

2

x 2 (1 + x 2 ) correct to 3 significant figures.
1

405

since cosh2 θ − sinh2 θ = 1
=

dx,
= a2

(a cosh θ)(a cosh θ) dθ = a 2
1 + cosh 2θ
2



cosh2 θ dθ

406 Higher Engineering Mathematics
=

a2 sinh 2θ θ+ 2
2

+c

3

4

5. Evaluate

a2
= [θ + sinh θ cosh θ] + c,
2
since sinh 2θ = 2 sinh θ cosh θ x x
Since x = a sinh θ, then sinh θ = and θ = sinh −1 a a

(t 2 + 9)

0
1

dt .

[3.525]

(16 + 9θ 2 ) dθ.

6. Evaluate

[4.348]

0

Also since cosh2 θ − sinh2 θ = 1

40.8 Worked problems on integration using the cosh θ substitution

then cosh θ = (1 + sinh 2 θ)
=
=

=

2

=

a2 + x 2 a2 Problem 24.

(a 2 + x 2 ) a (x 2 + a 2 ) dx

Hence
=

x a 1+

(x 2 + a 2 )
+c
a

x a2 x sinh −1 +
2
a a a2
2

sinh−1

x x
(x2 + a2) + c
+
a 2

=

=

Exercise 160 Further problems on integration using the sinh θ substitution
1. Find

2. Find

3. Find

2
(x 2 + 16)
3
(9 + 5x 2 )

dx.

2 sinh

−1

x
+c
4

dx.

3
−1 5 x + c
√ sinh
3
5

(x 2 + 9) dx.
9
x x
(x 2 + 9) + c sinh −1 +
2
3 2

4. Find

(4t 2 + 25) dt .
25
2t t (4t 2 + 25) + c sinh −1 +
4
5
2

1
(x 2

− a2)

dx.

dx
Let x = a cosh θ then
= a sinh θ and dθ dx = a sinh θ dθ
1
dx
Hence
(x 2 − a 2 )

=

Now try the following exercise

Determine

=

1
(a 2 cosh2 θ

− a2)

(a sinh θ dθ)

a sinh θ dθ
[a 2 (cosh2 θ − 1)] a sinh θ dθ

,
(a 2 sinh2 θ) since cosh2 θ − sinh2 θ = 1

a sinh θ dθ
=
a sinh θ

dθ = θ + c

x
= cosh−1 + c, since x = a cosh θ a It is shown on page 339 that cosh−1 x x + (x2 − a2 )
= ln a a

which provides as alternative solution to
1
(x 2

− a2)

Problem 25.

dx

Determine

2x − 3
(x 2 − 9)

dx.

407

Integration using trigonometric and hyperbolic substitutions
2x − 3
(x 2 − 9)

2x

dx =

(x 2 − 9)

sinh θ =

dx


3
(x 2

− 9)

dx

The first integral is determined using the algebraic substitution u =(x 2 − 9), and the second integral is of the
1
form dx (see Problem 24)
2 − a2)
(x
2x

Hence

(x 2

− 9)

3

dx −

(x 2

= 2 (x2 − 9) − 3 cosh−1

− 9)

2

x a =

(x 2 − a 2 ) x x +c
− cosh −1 a a a =

a2
2

=

x x a2
(x2 − a2 ) − cosh−1 + c
2
2 a 3

Problem 27. Evaluate
Problem 26.

3
2

(a 2 cosh2 θ − a 2 ) (a sinh θ dθ)

=

[a 2 (cosh2 θ − 1)] (a sinh θ dθ)

= a2

sinh2 θ dθ = a 2

since cosh 2θ = 1 + 2 sinh2 θ

3√
3
5 − 2 cosh−1
5
2

x x + (x 2 − a 2 )
Since cosh−1 = ln a a cosh−1 =

a2
[sinh θ cosh θ − θ] + c,
2

(32 − 22 )
2

Similarly, cosh−11 = 0
3

Hence

(x 2 − 4) dx

2

=

a 2 sinh 2θ
−θ +c
2
2

3
3+
= ln
2

then

= ln 2.6180 = 0.9624

from Table 5.1, page 45,
=

2

− (0 − 2 cosh −1 1)

(a 2 sinh2 θ) (a sinh θ dθ) cosh 2θ − 1 dθ 2

3

from Problem 26, when a = 2,
=

=

=

x x 4
(x 2 − 4) − cosh−1
2
2
2

(x 2 − 4) dx =

(x 2 − a 2 ) dx

Hence

(x 2 − 4) dx.

2

(x 2 − a 2 ) dx.

dx
= a sinh θ and
Let x = a cosh θ then dθ dx = a sinh θ dθ

(x 2 − a 2 ) a −1 =

(x 2 − a 2 ) dx

Hence

dx

x
+c
3

(cosh 2 θ − 1)

3√
5 − 2(0.9624) − [0]
2

= 1.429, correct to 4 significant figures.
Now try the following exercise

since sinh 2θ = 2 sinh θ cosh θ
Since x = a cosh θ then cosh θ =

x and a

x a Also, since cosh 2 θ − sinh2 θ = 1, then

θ = cosh−1

Exercise 161 Further problems on integration using the cosh θ substitution
1. Find

1
(t 2 − 16)

dt .

cosh−1

x
+c
4

408 Higher Engineering Mathematics
3

2. Find

(4x 2 − 9)

dx.

3
2x
cosh−1
+c
2
3

(θ 2 − 9) dθ.

3. Find

θ θ 9
(θ 2 − 9) − cosh−1 + c
2
2
3
(4θ 2 − 25) dθ.

4. Find θ θ2 −

25

25

cosh−1
+c
4
4
5

2

5. Evaluate
1
3

6. Evaluate
2

2
(x 2 − 1)

dx.

(t 2 − 4) dt .

[2.634]

[1.429]

Chapter 41

Integration using partial fractions
41.1

(by algebraic substitutions — see Chapter 39)

Introduction

The process of expressing a fraction in terms of simpler fractions—called partial fractions—is discussed in
Chapter 2, with the forms of partial fractions used being summarized in Table 2.1, page 13.
Certain functions have to be resolved into partial fractions before they can be integrated as demonstrated in the following worked problems.

41.2

Worked problems on integration using partial fractions with linear factors

Problem 1. Determine

11 −3x dx. x 2 + 2x − 3

As shown in Problem 1, page 13:
11 − 3x
2
5

− x 2 + 2x − 3 (x − 1) (x + 3)
11 − 3x dx 2 + 2x − 3 x Hence
=

2
5
− dx (x − 1) (x + 3)

= 2 ln(x −1) − 5 ln(x + 3) + c

or ln

(x −1)2
+ c by the laws of logarithms
(x +3)5

Problem 2. Find
2x 2 − 9x − 35 dx. (x + 1)(x − 2)(x + 3)

It was shown in Problem 2, page 14:
2x 2 − 9x − 35
4
3
1


+
(x + 1)(x − 2)(x + 3) (x + 1) (x − 2) (x + 3)
2x 2 − 9x − 35 dx (x + 1)(x − 2)(x + 3)

Hence


3
1
4

+ dx (x + 1) (x − 2) (x + 3)

= 4 ln(x+ 1) − 3 ln(x− 2) + ln(x+ 3) + c or ln

(x + 1)4 (x + 3)
(x −2)3

Problem 3. Determine

+c

x2 + 1 dx. x 2 − 3x + 2

410 Higher Engineering Mathematics
By dividing out (since the numerator and denominator are of the same degree) and resolving into partial fractions it was shown in Problem 3, page 14: x2 + 1 x 2 − 3x + 2

≡ 1−

2
5
+
(x − 1) (x − 2)



Exercise 162 Further problems on integration using partial fractions with linear factors
In Problems 1 to 5, integrate with respect to x.

x2 + 1 dx x 2 − 3x + 2

Hence

Now try the following exercise

12 dx (x 2 − 9)


2 ln(x − 3) − 2 ln(x + 3) + c



⎣ x −3 2
+c
or ln x +3

1.

5
2
dx
+
(x − 1) (x − 2)

1−

= (x −2) ln(x − 1) + 5 ln(x −2) + c
(x −2)
+c
(x −1)2

or x + ln

Problem 4.

4(x − 4) dx (x 2 − 2x − 3)


5 ln(x + 1) − ln(x − 3) + c




5

⎣ or ln (x + 1) + c
(x − 3)

2.

5

Evaluate
3
2

x 3 − 2x 2 − 4x − 4 dx, x2 + x − 2

3(2x 2 − 8x − 1) dx (x + 4)(x + 1)(2x − 1)


7 ln(x + 4) − 3 ln(x + 1)


⎢ − ln(2x − 1) + c or







(x + 4)7 ln +c
(x + 1)3 (2x − 1)

3.

correct to 4 significant figures.
By dividing out and resolving into partial fractions it was shown in Problem 4, page 15: x 3 − 2x 2 − 4x − 4
4
3
≡ x −3+

2 +x −2 x (x + 2) (x − 1)
3

Hence
2

2

=

=

x 3 − 2x 2 − 4x − 4 dx x2 + x − 2
3



x 2 + 9x + 8 dx x2 + x − 6

4.

x −3+

x + 2 ln(x + 3) + 6 ln(x − 2) + c or x + ln{(x + 3)2 (x − 2)6 } + c

3
4
dx

(x + 2) (x − 1)

x2
− 3x + 4 ln(x + 2) − 3 ln(x − 1)
2

3x 3 − 2x 2 − 16x + 20 dx (x − 2)(x + 2)

⎡ 2
3x
⎣ 2 − 2x + ln(x − 2) ⎦
−5 ln(x + 2) + c

5.
3
2

9
− 9 + 4 ln 5 − 3 ln 2
2

In Problems 6 and 7, evaluate the definite integrals correct to 4 significant figures.

− (2 − 6 + 4 ln 4 − 3 ln 1)
= −1.687, correct to 4 significant figures.

4

6.
3

x 2 − 3x + 6 dx x(x − 2)(x − 1)

[0.6275]

Integration usingpartial fractions

6

7.
4

x 2 − x − 14 dx x 2 − 2x − 3

[0.8122]

8. Determine the value of k, given that:
1
0

(x − k) dx = 0
(3x + 1)(x + 1)

1
3

9. The velocity constant k of a given chemical reaction is given by: kt =

1 dx (3 − 0.4x)(2 − 0.6x)

where x = 0 when t = 0. Show that: kt = ln

2(3 − 0.4x)
3(2 − 0.6x)

Problem 6. Find

5x 2 − 2x − 19 dx. (x + 3)(x − 1)2

It was shown in Problem 6, page 16:
5x 2 − 2x − 19
2
3
4

+

2
(x + 3)(x − 1)
(x + 3) (x − 1) (x − 1)2
5x 2 − 2x − 19 dx (x + 3)(x − 1)2

Hence

3
4
2
+

(x + 3) (x − 1) (x − 1)2



= 2 ln (x +3) + 3 ln (x −1) + or ln (x +3)2 (x −1)3 +

dx

4
+c
(x − 1)

4
+c
(x − 1)

Problem 7. Evaluate

41.3

Worked problems on integration using partial fractions with repeated linear factors 3x 2 + 16x + 15 dx, (x + 3)3
−2
1

correct to 4 significant figures.
It was shown in Problem 7, page 17:

Problem 5. Determine

2x + 3 dx. (x − 2)2

3
6
3x 2 + 16x + 15
2



3
2
(x + 3)
(x + 3) (x + 3)
(x + 3)3
3x 2 + 16x + 15 dx (x + 3)3

It was shown in Problem 5, page 16:
Hence
2x + 3
2
7

+
(x − 2)2
(x − 2) (x − 2)2
Thus

2x + 3 dx ≡
(x − 2)2

2
7
+ dx (x − 2) (x − 2)2

7
+c
= 2 ln(x −2) −
(x −2)

7 dx is determined using the algebraic

⎣ (x − 2)2 substitution u = (x − 2) — see Chapter 39.




1
−2

3
6
2


(x + 3) (x + 3)2 (x + 3)3

= 3 ln(x + 3) +

= 3 ln 4 +

2
3
+
(x + 3) (x + 3)2

dx

1
−2

2 3
2
3
− 3 ln 1 + +
+
4 16
1 1

= −0.1536, correct to 4 significant figures.

411

412 Higher Engineering Mathematics
3 + 6x + 4x 2 − 2x 3 dx x 2 (x 2 + 3)

Now try the following exercise
Thus
Exercise 163 Further problems on integration using partial fractions with repeated linear factors
In Problems 1 and 2, integrate with respect to x.
4x − 3 dx 1.
(x + 1)2
7
4 ln(x + 1) +
+c
(x + 1)
5x 2 − 30x

+ 44 dx (x − 2)3


10
5 ln(x − 2) +

(x − 2) ⎥




2
+c

(x − 2)2

2.



=

2
1
7

4.
6

x 2 + 7x + 3 x 2 (x + 3)

4x x2 + 3

1

dx
+ ( 3)2

dx is determined using the algebraic substi-

tution u =(x 2 + 3).
Hence

2
3
1
4x
dx
+
+

x x 2 (x 2 + 3) (x 2 + 3)

1 x 3
+ √ tan−1 √ − 2 ln(x 2 + 3) + c x 3
3
2 x x
1 √
= ln 2
− + 3 tan−1 √ + c x +3 x 3
= 2 ln x −

[1.089]
Problem 9.

1

5. Show that
0

Determine

4t 2 + 9t + 8 dt = 2.546,
(t + 2)(t + 1)2
Let

correct to 4 significant figures.

Worked problems on integration using partial fractions with quadratic factors

Problem 8.

Find

+ 4x 2 − 2x 3

3 + 6x x 2 (x 2 + 3)

dx.

It was shown in Problem 9, page 18:
3 − 4x
2
1
3 + 6x + 4x 2 − 2x 3
≡ + 2+ 2
2 (x 2 + 3) x x x
(x + 3)

(x 2

1 dx. − a2)

A
B
1

+
(x 2 − a 2 ) (x − a) (x + a)


41.4

x2

dx

x
3
= √ tan −1 √ , from 12, Table 40.1, page 399.
3
3

[1.663]

18 + 21x − x 2 dx (x − 5)(x + 2)2

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

3 dx = 3
(x 2 + 3)

In Problems 3 and 4, evaluate the definite integrals correct to 4 significant figures.
3.

2
(3 − 4x)
1
+ 2+ 2 x x
(x + 3)

A(x + a) + B(x − a)
(x + a)(x − a)

Equating the numerators gives:
1 ≡ A(x + a) + B(x − a)
1
Let x = a, then A = , and let x = −a, then
2a
1
B =−
2a
1
Hence
dx
(x 2 − a 2 )


1
1
1

dx
2a (x − a) (x + a)

Integration usingpartial fractions
1
[ln(x − a) − ln(x + a)] + c
2a
x −a
1
ln
+c
=
2a
x +a
=

413

Problem 12. Evaluate
2
0

5 dx, (9 − x 2 )

correct to 4 decimal places.
Problem 10. Evaluate
4
3

3 dx, (x 2 − 4)

From Problem 11,
2

correct to 3 significant figures.
0

1
3+x
5 dx = 5 ln (9 − x 2 )
2(3)
3−x

From Problem 9,
4
3

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

=

4
3

=

3 5 ln = 0.383, correct to 3
4 3 significant figures.

Problem 11. Determine

(a 2

1 dx. − x 2)

Exercise 164 Further problems on integration using partial fractions with quadratic factors

Using partial fractions, let
1
A
B
1


+
(a 2 − x 2 ) (a − x)(a + x) (a − x) (a + x)
A(a + x) + B(a − x)
(a − x)(a + x)

Then 1 ≡ A(a + x) + B(a − x)
1
1
Let x = a then A = . Let x = −a then B =
2a
2a
1

Hence

(a 2 − x 2 )
=
=
=

dx

1
1
1
+
dx
2a (a − x) (a + x)
1
[−ln(a − x) + ln(a + x)] + c
2a
1 a+x +c ln 2a a−x 5
5
ln − ln 1
6
1

Now try the following exercise

1. Determine



0

= 1.3412, correct to 4 decimal places.

3
2
1 ln − ln
4
6
5

=

2

x 2 − x − 13 dx. (x 2 + 7)(x − 2)

⎡ x 3
2 + 7) + √ tan−1 √
⎣ ln(x
7
7⎦
− ln(x − 2) + c

In Problems 2 to 4, evaluate the definite integrals correct to 4 significant figures.
6

2.
5
2

3.
1
5

4.
4

6x − 5 dx (x − 4)(x 2 + 3)

[0.5880]

4 dx (16 − x 2 )

[0.2939]

2 dx (x 2 − 9)

[0.1865]
2

5. Show

that
1

2 +θ + 6θ 2 − 2θ 3 dθ θ 2 (θ 2 + 1)

= 1.606, correct to 4 significant figures.

Chapter 42 θ The t = tan 2 substitution
42.1

Introduction

1 dθ, where a cos θ + b sin θ + c a, b and c are constants, may be determined by using the θ substitution t = tan . The reason is explained below.
2
If angle A in the right-angled triangle ABC shown in θ Fig. 42.1 is made equal to then, since tangent =
2
opposite θ , if BC = t and AB = 1, then tan = t . adjacent 2

By Pythagoras’ theorem, AC = 1 +t 2
Integrals of the form

C
Ί1 1 t 2

A


2
1

sin θ =

i.e.

t

2t
(1 + t2 )

θ θ Since cos 2x = cos2 − sin2
2
2
=

i.e.

1

1 + t2

cos θ =

2

t
− √
1 + t2

1 −t 2
1 +t 2

θ
Also, since t = tan ,
2
θ 1 θ dt 1
1 + tan 2
= sec 2 = dθ 2
2 2
2
identities,

2

(2)

from trigonometric

B

i.e.

Figure 42.1

θ θ t
1
and cos = √
Since
=√
2
2
2
1 +t
1 +t 2 sin 2x = 2 sin x cos x (from double angle formulae,
Chapter 17), then

dt
1
= (1 + t 2 ) dθ 2

Therefore sin

θ θ sin θ = 2 sin cos
2
2 t =2 √
1 + t2

(1)

t

1 + t2

from which,

dθ =

2 dt
1 +t 2

(3)

Equations (1), (2) and (3) are used to determine
1
integrals of the form dθ where a cos θ + b sin θ + c a, b or c may be zero.

The t = tan θ substitution
2
When

42.2

Worked problems on the θ t = tan substitution
2

415

t = −1, 2 = 2B, from which, B = 1
2 dt
=
1 − t2

Hence

1
1
+ dt (1 − t ) (1 + t )

= −ln(1 − t ) + ln(1 + t ) + c
(1 + t )
+c
(1 − t )


⎪ 1 +tan x ⎪

⎬ dx 2
= ln x ⎪+c
⎪ 1 −tan ⎭ cos x

2

dθ sin θ

Problem 1. Determine

θ
2 dt
2t
and dθ = from then sin θ =
2
1+t2
1 +t 2 equations (1) and (3).

= ln

If t = tan


=
sin θ

1 dθ sin θ
1
2 dt
2t
=
1 + t2 1 + t2
1
= dt = ln t + c t dθ θ +c
= ln tan sin θ
2

Thus

Hence

Thus

π
Note that since tan = 1, the above result may be
4
written as:


⎪ tan π + tan x ⎪

⎬ dx 4
2
= ln π x +c

cos x
⎩ 1 − tan tan ⎪

4
2
= ln tan

π x
+
4 2

+c

from compound angles, Chapter 17. dx cos x

Problem 2. Determine

2 dt x 1 − t2 and dx = from then cos x =
2
2
1+t
1 + t2 equations (2) and (3).
If tan

Thus

dx
=
cos x
=

1
1 − t2
1 + t2

2 dt
1 + t2

2 dt x 1 −t 2 and dx = from then cos x =
2
2
1 +t
1 +t 2 equations (2) and (3).
If tan

Thus

2 dt 1 − t2

When

A(1 + t ) + B(1 − t )
(1 − t )(1 + t )

2 = A(1 + t ) + B(1 − t ) t = 1, 2 = 2 A, from which, A = 1

1 dx 1 + cos x

1
1 − t2
1+
1 + t2

2 dt
1 + t2

1
(1 + t 2 ) + (1 − t 2 )

2 dt
1 + t2

1 +t 2
=

B
A
+
(1 − t ) (1 + t )

=
Hence

=

2
2
=
2
1−t
(1 − t )(1 + t )
=

dx
=
1 + cos x
=

2 may be resolved into partial fractions (see
1 − t2
Chapter 2).
Let

dx
1 +cos x

Problem 3. Determine

Hence

dt

dx x = t + c = tan + c
1 +cos x
2

Problem 4. Determine


5 +4 cos θ

416 Higher Engineering Mathematics θ 2 dt
1 −t 2 and dx = then cos θ =
2
1 +t 2
1+t2
from equations (2) and (3).
2 dt dθ 1 + t2
Thus
=
5 + 4 cosθ
1 − t2
5+4
1 + t2
If t = tan

2 dt
1 + t2
=
2 ) + 4(1 − t 2 )
5(1 + t
(1 + t 2) dt dt
=2
=2
2+9
2 + 32 t t
1 −1 t
=2
tan
+ c,
3
3 from 12 of Table 40.1, page 399. Hence
1
2 θ dθ
= tan−1 tan +c
5 +4 cos θ 3
3
2
Now try the following exercise

42.3

Problem 5.





1 + sin θ

2.

dx
1 − cos x + sin x

3.

−2

θ
1 + tan
2

1 − t2 x 2t
, cos x = then sin x =
2
1 + t2
1 + t2
2 dt dx = from equations (1), (2) and (3).
1 + t2
Thus
2 dt dx 1 + t2
=
sin x + cos x
2t
1 − t2
+
2
1+t
1 + t2
2 dt
2 dt
1 + t2 =
=
1 + 2t − t 2
2t + 1 − t 2
1 + t2
−2 dt
−2 dt
=
= t 2 − 2t − 1
(t − 1)2 − 2


3 + 2 cosα
2
1 α √ tan−1 √ tan
+c
2
5
5

4.

dx
3 sin x − 4 cos x


2 dt

2 − (t − 1)2
( 2)

2 + (t − 1)
1
= 2 √ ln √
2 2
2 − (t − 1)




⎪ 2 tan x − 1 ⎪



⎢1
2
+ c⎦
⎣ ln
5 ⎪ tan x + 2 ⎪


2

and

=


+ c⎦


⎡ ⎧

⎪ tan x ⎪




2
+ c⎦
⎣ln
⎪ 1 + tan x ⎪


2

dx sin x + cos x

Determine

If tan

Exercise 165 Further problems on the θ t =tan substitution
2
Integrate the following with respect to the variable:


1.

Further worked problems on the θ t = tan substitution
2

+c

(see Problem 11, Chapter 41, page 413),
i.e.

dx sin x + cos x
⎧√

⎪ 2 − 1 +tan x ⎪


1
2 +c
= √ ln √
2 ⎪ 2 + 1 −tan x ⎪


2

Problem 6. Determine dx 7 − 3 sin x + 6 cos x
From equations (1) and (3), dx 7 − 3 sin x + 6 cos x
2 dt
1 + t2

=
7−3

2t
1 + t2

+6

1 − t2
1 + t2

The t = tan θ substitution
2

=

=



2 dt
1 + t2
2 ) − 3(2t ) + 6(1 − t 2 )
7(1 + t
1 + t2
2 dt
7 + 7t 2 − 6t + 6 − 6t 2
2 dt
=
+ 13

=

1 −1 t − 3 tan 2
2

+c

1
5
2
4


4 cos θ + 3 sin θ


⎪ 1 + tan θ ⎪


1
2 +c
= ln 2
5 ⎪
⎩ 2 − tan θ ⎪

2


⎪ 1 +2 tan θ ⎪


1
2 +c ln 5 ⎪
⎩ 4 − 2 tan θ ⎪

2

Hence

from 12, Table 40.1, page 399. Hence dx 7 − 3 sin x + 6 cos x

⎛ x tan − 3


2
= tan−1 ⎝
⎠+c
2

1⎢

2⎣

⎫⎤


⎬⎥
⎥+c
⎪⎦



from Problem 11, Chapter 41, page 413


⎪ 1 +t ⎪


1
+c
= ln 2
5 ⎪ 2−t ⎪



2 dt
(t − 3)2 + 22

t 2 − 6t

=2

=


⎪5+ t−3

⎨4
4
ln
⎪5
3

⎩ − t−
4
4

or

Now try the following exercise dθ 4 cosθ + 3 sin θ

Problem 7. Determine
From equations (1) to (3), dθ 4 cos θ + 3 sin θ

2 dt
1 + t2

=
4

1 − t2
2t
+3
1 + t2
1 + t2

=−

1
2

=−

1
2

Exercise 166 Further problems on the θ t = tan substitution
2
In Problems 1 to 4, integrate with respect to the variable. dθ
1.
5 + 4 sin θ



⎛ θ 5 tan + 4


⎢ 2 −1 ⎜
2
⎠ + c⎦
⎣ tan ⎝
3
3

2.

dx
1 + 2 sin x




√ ⎫
⎪ tan x + 2 − 3 ⎪


⎢ 1

2
+ c⎦
⎣ √ ln
⎪ tan x + 2 + √3 ⎪
3 ⎩

2

dt
3
t2 − t − 1
2

1
2

=

dt
2 + 3t − 2t 2

2 dt
=
4 − 4t 2 + 6t

=

dt t− 3
4

3.

2



25
16

5
4

− t−

3
4

dp
3 − 4 sin p +2 cos p


√ ⎫
⎪ tan p − 4 − 11 ⎪


⎢ 1

2
+ c⎦
⎣ √ ln


11 ⎩ tan p − 4 + 11 ⎪

2


dt
2

2

417

418 Higher Engineering Mathematics

4.


3 − 4 sin θ




√ ⎫
⎪ 3 tan θ − 4 − 7 ⎪


⎢ 1

2
⎣ √ ln
√ ⎪ + c⎦ θ ⎪
7 ⎩ 3 tan − 4 + 7 ⎭
2
5. Show that
⎧√

⎪ 2 + tan t ⎪


1
dt
2 + c.
= √ ln √
1 + 3 cost 2 2 ⎪ 2 − tan t ⎪


2

π/3

3 dθ
= 3.95, correct to 3 cos θ
0
significant figures.

6. Show that

7. Show that π/2 0

dθ π = √ .
2 + cos θ
3 3

Revision Test 12
This Revision Test covers the material contained in Chapters 40 to 42. The marks for each question are shown in brackets at the end of each question.
1. Determine the following integrals:
(a)

2

cos3 x sin2 x dx (b)
2

(c)

(4x 2 − 9)

(9 − 4x 2 )

dx

dx

(14)

2

2. Evaluate the following definite integrals, correct to
4 significant figures: π 2

(a)

3 sin t dt (b)

0

4. Evaluate figures. 5 dx 4 + x2

3 x 2 (x + 2)

1

3 cos5θ sin 3θ dθ
(15)

6. Evaluate places. π
2
π
3

(21)

dx correct to 4 significant

dx
2 sin x + cos x

5. Determine:

0
2

(c)

π
3

2

0

3. Determine: x − 11 dx (a)
2 −x −2 x 3−x dx (b)
(x 2 + 3)(x + 3)

(12)
(8)

dx correct to 3 decimal
3 − 2 sin x
(10)

Chapter 43

Integration by parts
43.1

43.2 Worked problems on integration by parts

Introduction

From the product rule of differentiation: d du dv (uv) = v
+u , dx dx dx Problem 1.

where u and v are both functions of x. dv d du Rearranging gives: u
=
(uv) − v dx dx dx Integrating both sides with respect to x gives: u i.e.

dv dx = dx u

d
(uv) dx − dx v

du dx dx

dv du dx = uv− v dx dx dx u dv = uv −

u dv = uv −

v du

du
= 1, i.e. du = dx and let dx dv = cos x dx, from which v = cos x dx = sin x.
Expressions for u, du and v are now substituted into the ‘by parts’ formula as shown below.
Let u = x, from which

dv

v du x cos x dx

This is known as the integration by parts formula and provides a method of integrating such prodt sin t dt , ucts of simple functions as xex dx, θ cos θ dθ and x ln x dx. e Given a product of two terms to integrate the initial choice is: ‘which part to make equal to u’ and ‘which part to make equal to v’. The choice must be such that the
‘u part’ becomes a constant after successive differentiation and the ‘dv part’ can be integrated from standard integrals. Invariably, the following rule holds: If a product to be integrated contains an algebraic term (such as x, t 2 or 3θ) then this term is chosen as the u part. The one exception to this rule is when a ‘ln x’ term is involved; in this case ln x is chosen as the ‘u part’.

x cos x dx.

From the integration by parts formula,

u

or

Determine

i.e.

ϭ ϭ u

v

(x) (sin x)

Ϫ
Ϫ

v

du

(sin x) (dx)

x cos x dx = x sin x − (−cos x) + c
= x sin x +cos x + c

[This result may be checked by differentiating the right hand side,
i.e.

d
(x sin x + cos x + c) dx = [(x)(cos x) + (sin x)(1)] − sin x + 0 using the product rule
= x cos x, which is the function being integrated]

Integration by parts
Problem 2. Find

3t e2t dt .

1

Problem 4. Evaluate

du
= 3, i.e. du = 3 dt and dt 1 let dv = e2t dt , from which, v = e2t dt = e2t
2
Substituting into u dv = uv − v du gives:
Let u =3t , from which,

3t e2t dt = (3t )

1 2t e −
2

3
3
= t e2t −
2
2

1 2t e (3 dt )
2

e2t dt

5xe4x dx, correct to

0

3 significant figures.

du
= 5, i.e. du = 5 dx and dx let dv = e4x dx, from which, v = e4x dx = 1 e4x .
4
Substituting into u dv = uv − v du gives:
Let u =5x, from which

5xe4x dx = (5x)

e4x
4



e4x
4

e2t
2

=

+c

Hence
3t e2t dt = 3 e2t t − 1 + c,
2
2
1

Problem 3. Evaluate

Hence
2θ sin θ dθ.

=

du
= 2, i.e. du =2 dθ and let
Let u = 2θ, from which, dθ dv = sin θ dθ, from which,

2θ sin θ dθ = (2θ)(−cos θ) −
= −2θ cos θ + 2

π
2

Hence

0

5 4
1
e 1−
4
4

15 4
5
e − −
16
16



5 0
1
e 0−
4
4

= 51.186 + 0.313 = 51.499 = 51.5, correct to 3 significant figures

(−cos θ)(2 dθ)

= −2θ cos θ + 2 sin θ + c

1

5 4x
1
e x− 4
4

=

v du gives:

cos θ dθ

+c

5xe4x dx

=

sin θ dθ = −cos θ

Substituting into u dv = uv −

e4x
4

0

0

v=

e4x dx

5 4x 5 xe −
4
4

(5 dx)

5
1
+c
= e4x x −
4
4

which may be checked by differentiating. π 2

5 4x 5 xe −
4
4

=

3
3
= t e2t −
2
2

Problem 5. Determine

x 2 sin x dx.

du
= 2x, i.e. du =2x dx, and
Let u = x 2 , from which, dx let dv = sin x dx, from which,

2θ sin θ dθ

0 π 2
= [−2θ cos θ + 2 sin θ]0

= −2

421

π π π cos + 2 sin
− [0 + 2 sin 0]
2
2
2

= (−0 + 2) − (0 + 0) = 2 π π sincecos = 0 and sin = 1
2
2

v=

sin x dx = −cos x

Substituting into u dv = uv − x 2 sin x dx = (x 2 )(−cos x) −
= −x 2 cos x + 2

v du gives:
(−cos x)(2x dx) x cos x dx

422 Higher Engineering Mathematics
The integral, x cos x dx, is not a ‘standard integral’ and it can only be determined by using the integration by parts formula again.
From Problem 1, x cos x dx = x sin x + cos x

π
2 2

8.

t cos t dt

2

9.

2

x sin x dx

Hence

[0.4674]

0 x 3x 2 e 2 dx

[15.78]

1

= −x 2 cos x + 2{x sin x + cos x} + c
= −x 2 cos x + 2x sin x + 2 cos x + c

43.3 Further worked problems on integration by parts

= (2 −x2 )cos x +2x sin x +c
In general, if the algebraic term of a product is of power n, then the integration by parts formula is applied n times. Now try the following exercise
Exercise 167 Further problems on integration by parts

Problem 6.

e2x
2

xe dx

1.

4x dx e3x

2.

1 x− 2

x ln x dx = (ln x)

4.

3t 2e2t dt

5.

θ sin 2θ +
3 2t
2e

1
2 cos 2θ

+c

t2 − t + 1 + c
2

Evaluate the integrals in Problems 6 to 9, correct to 4 significant figures.
2

6.

2xex dx

x sin 2x dx

Determine

ln x dx.

ln x dx = (ln x)(x) −

[16.78]

[0.2500]

+c

x2
1
lnx −
+ c or
2
2

= x ln x − π 4

0

x2
2

ln x dx is the same as (1) ln x dx du 1 dx Let u = ln x, from which,
= , i.e. du = dx x x and let dv = 1dx, from which, v = 1 dx = x
Substituting into u dv = uv − v du gives:

0

7.

x dx

x2
1
ln x −
2
2

dx x x2
(2 ln x −1) + c
4
Problem 7.

5
2

x2
1
ln x −
2
2

x ln x dx =

[−x cos x + sin x + c]

5θ cos 2θ dθ

x2
2



=

+c

4
1
− e−3x x +
+c
3
3

x sin x dx

x2
2

=

Hence
3.

x ln x dx.

The logarithmic function is chosen as the ‘u part’. du 1 dx Thus when u = ln x, then
= , i.e. du = dx x x 2 x Letting dv = x dx gives v = x dx =
2
Substituting into u dv = uv − v du gives:

Determine the integrals in Problems 1 to 5 using integration by parts.
2x

Find

Hence

x

dx x dx = x ln x − x + c

ln x dx = x(ln x −1) + c

Integration by parts
9√
x ln x dx,
1

Problem 8. Evaluate
3 significant figures.

correct to

= (eax )

2 3 x2 3

x ln x dx = (ln x)

2 3
2
x ln x −
3
3
2 3
2
= x ln x −
3
3

Hence
=

=

2 3 x2 3



=

=

dx x 2 3 x2 3

v=

2
3

1 sin bx dx = − cos bx b +c
Substituting into the integration by parts formula gives:
1
eax sin bx dx = (eax ) − cos bx b 9
1





(1)

Let u = eax then du =aeax dx, and let dv = sin bx dx, from which



2√ 3
2
9 ln 9 −
3
3

eax sin bx dx

1

x 2 dx

2 3
2
x ln x −
+c
3
3

2 3
2
x ln x −
3
3

1 sin bx (aeax dx) b eax sin bx dx is now determined separately using integration by parts again:

9√
1 x ln x dx

= 18 ln 9 −

1 sin bx − b 1 a = eax sin bx − b b

1
2 3 x 2 dx = x 2
3

Substituting into u dv = uv − v du gives:


v du gives:

eax cos bx dx

dx
Let u = ln x, from which du = x 1

and let dv = x dx = x 2 dx, from which, v= Substituting into u dv = uv −

423

2√ 3
2
1 ln1 −
3
3

2
2
0−
3
3

1
− cos bx (aeax dx) b 1 a = − eax cos bx + b b

eax cos bx dx

Substituting this result into equation (1) gives:
1
1 a eax cos bx dx = eax sin bx −
− eax cos bx b b b = 27.550 + 0.444 = 27.994 = 28.0,

+

correct to 3 significant figures.

a b eax cos bx dx

1 a = eax sin bx + 2 eax cos bx b b
Problem 9. Find

eax cos bx dx.

When integrating a product of an exponential and a sine or cosine function it is immaterial which part is made equal to ‘u’. du Let u =eax , from which
= aeax , dx i.e. du =aeax dx and let dv = cos bx dx, from which, v= cos bx dx =

1 sin bx b −

a2 b2 eax cos bx dx

The integral on the far right of this equation is the same as the integral on the left hand side and thus they may be combined. eax cos bx dx +

a2 b2 eax cos bx dx
1
a
= eax sin bx + 2 eax cos bx b b

424 Higher Engineering Mathematics
i.e.

1+

π

a2 b2 eax cos bx dx
=

i.e.

b2 + a 2 b2 1 ax a e sin bx + 2 eax cos bx b b

= 0.8387, correct to 4 decimal places.

eax cos bx dx
Now try the following exercise eax = 2 (b sin bx + a cos bx) b Exercise 168 Further problems on integration by parts

eax cos bx dx

Hence

b2

eax

b2 + a 2

=

π

1 e4 e4
2
(1 − 0) − (0 − 2) =
+
5
5
5
5

=

b2

Determine the integrals in Problems 1 to 5 using integration by parts.

(b sin bx + a cos bx)

1.

=

a2 + b2

2x 2 ln x dx

2.

eax

2 ln 3x dx

3.

x 2 sin 3x dx

(b sin bx + a cos bx) + c

Using a similar method to above, that is, integrating by parts twice, the following result may be proved: eax sin bx dx
=

eax a2 + b2

(a sin bx − b cos bx)+ c π 4

Evaluate

4 decimal places.

(2)
2e5x cos 2x dx

et sin 2t dt , correct to

2 5x e (2 sin 2x + 5 cos 2x) + c
29

0

Comparing et sin 2t dt with eax sin bx dx shows that x = t , a = 1 and b = 2.
Hence, substituting into equation (2) gives: π 4

[2x(ln 3x − 1) + c]

cos 3x
2
(2 − 9x 2 ) + x sin 3x + c
27
9
4.

Problem 10.

2 3
1
+c x ln x −
3
3

et sin 2t dt

2θ sec 2 θ dθ

5.

[2[θ tan θ − ln(sec θ)] + c]

Evaluate the integrals in Problems 6 to 9, correct to 4 significant figures.
2

x ln x dx

6.

0

[0.6363]

1

et
= 2
(1 sin 2t − 2 cos 2t )
1 + 22 π e4
=
5

π
4

1

7.

0

2e3x sin 2x dx

[11.31]

0

π π sin 2
− 2 cos2
4
4

π
2

et cos 3t dt

[−1.543]

x 3 ln x dx

8.

[12.78]

0
4



e0
5

(sin 0 − 2 cos 0)

9.
1

Integration by parts

10. In determining a Fourier series to represent f (x) = x in the range −π to π, Fourier coefficients are given by: an = and bn =

1 π 1 π 1

11. The equation C =

π
−π
π
−π

x cos nx dx

e−0.4θ cos 1.2θ dθ

0

and

1

S=

e−0.4θ sin 1.2θ dθ

0

x sin nx dx

where n is a positive integer. Show by using integration by parts that an = 0 and
2
bn = − cos nπ. n are involved in the study of damped oscillations. Determine the values of C and S.
[C = 0.66, S = 0.41]

425

Chapter 44

Reduction formulae
44.1

then

Introduction

When using integration by parts in Chapter 43, an integral such as x 2 e x dx requires integration by parts twice. Similarly, x 3 e x dx requires integration by parts three times. Thus, integrals such as x 5 e x dx, x 6 cos x dx and x 8 sin 2x dx for example, would take a long time to determine using integration by parts. Reduction formulae provide a quicker method for determining such integrals and the method is demonstrated in the following sections.

x n−1 ex dx = In−1

Hence

x n ex dx = x n ex − n

can be written as:
In = xn ex − nIn−1

To determine

x 2 ex dx = I2 = x 2 ex − 2I1
I1 = x 1 ex − 1I0

and

du
= nx n−1 and du =nx n−1 dx dx dv = ex dx from which,

and

v=

I0 =

e x nx n−1 dx

i.e.

x2 ex dx = x 2 ex − 2xex + 2e x + 2c1
= ex (x2 − 2x +2) + c

x n−1 ex dx

The integral on the far right is seen to be of the same form as the integral on the left-hand side, except that n has been replaced by n −1.
Thus, if we let, n x

x e dx = In ,

e x dx = ex + c1

= x 2 ex − 2[xex − 1(e x + c1 )]

using the integration by parts formula,
= x n ex − n

x 0 ex dx =

I2 = x 2 ex − 2[xex − 1I0 ]

Hence

e x dx = ex

x n e x dx = x n e x −

Thus,

x 2 e x dx using a

Using equation (1) with n = 2 gives:

x n e x dx using integration by parts, u = x n from which,

let

(1)

Equation (1) is an example of a reduction formula since it expresses an integral in n in terms of the same integral in n −1.
Problem 1. Determine reduction formula.

44.2 Using reduction formulae for integrals of the form xn exdx

x n−1 e x dx

(where c = 2c1 )
As with integration by parts, in the following examples the constant of integration will be added at the last step with indefinite integrals.
Problem 2. x 3 ex dx.

Use a reduction formula to determine

427

Reduction formulae
From equation (1), In = x n ex − n In−1
Hence

Hence In = x n sin x −

x 3 e x dx = I3 = x 3 ex − 3I2

Using integration by parts again, this time with u = x n−1 : du = (n − 1)x n−2 , and dv = sin x dx, dx I1 = x 1 ex − 1I0
I0 =

Thus

x 0 ex dx =

e x dx = ex

x 3 e x dx = x 3 ex − 3[x 2e x − 2I1 ]
= x 3 ex − 3[x 2e x − 2(xe x − I0 )]

from which, v= = x 3 ex − 3[x 2e x − 2(xe x − ex )]
= x 3 ex − 3x 2 ex + 6(xe x − ex )

sin x dx = −cos x

Hence In = x n sin x − n x n−1 (−cos x)

= x 3 ex − 3x 2 ex + 6xe x − 6e x
i.e.

x n−1 sin x dx

= x n sin x − n

I2 = x 2 e x − 2I1

and

(sin x)nx n−1 dx

x3ex dx = ex (x3 − 3x2 + 6x −6) + c



(−cos x)(n − 1)x n−2 dx

= x n sin x + nx n−1 cos x

Now try the following exercise

x n−2 cos x dx

− n(n − 1)
Exercise 169 Further problems on using reduction formulae for integrals of the form xn ex dx
1. Use a reduction formula to determine x 4 ex dx.
[ex (x 4 − 4x 3 + 12x 2 − 24x + 24) + c]
2. Determine mula. i.e.

I n = xn sin x + nxn−1 cos x

(2)

− n(n −1)In−2

Problem 3. Use a reduction formula to determine x 2 cos x dx.

t 3e2t dt using a reduction forUsing the reduction formula of equation (2): e2t 1 3
3 2
3
3
2t − 4t + 4t − 8

+c

x 2 cos x dx = I2

3. Use the result of Problem 2 to evaluate
1 3 2t
0 5t e dt, correct to 3 decimal places.
[6.493]

= x 2 sin x + 2x 1 cos x − 2(1)I0
I0 =

and

44.3 Using reduction formulae for integrals of the form xn cos x dx and xn sin x dx
(a) xn cos x dx

=

x 0 cos x dx cos x dx = sin x

Hence x2 cos x dx = x2 sin x +2x cos x − 2 sin x +c

Let In =

x n cos x dx then, using integration by parts: du if u = x n then
= nx n−1 dx and if dv = cos x dx then v= cos x dx = sin x

Problem 4. Evaluate significant figures.
Let us firstly t 3 cos t dt .

find

2 3
1 4t cos t dt ,

a

reduction

correct to 4

formula

for

428 Higher Engineering Mathematics
When n =2,

From equation (2),

π

t cos t dt = I3 = t sin t + 3t cos t − 3(2)I1
3

3

2

and

x 2 cos x dx = I2 = −2π 1 − 2(1)I0

0 π I0 =

and

x 0 cos x dx

0

I1 = t 1 sin t + 1t 0 cos t − 1(0)In−2

π

=

= t sin t + cos t

cos x dx
0

= [sin x]π = 0
0

Hence t 3 cos t dt = t 3 sin t + 3t 2 cos t
− 3(2)[t sin t + cos t ]

Hence π x 4 cos x dx = −4π 3 − 4(3)[−2π − 2(1)(0)]

0

= −4π 3 + 24π or −48.63,

= t sin t + 3t cos t − 6t sin t − 6 cost
3

2

correct to 2 decimal places.

Thus
2

4t 3 cos t dt

(b) xn sin x dx

1

= [4(t 3 sin t + 3t 2 cos t − 6t sin t − 6 cost )]2
1
= [4(8 sin 2 +12 cos 2 −12 sin 2 − 6 cos 2)]
− [4(sin 1 +3 cos 1 − 6 sin 1 −6 cos 1)]

Let In = x n sin x dx
Using integration by parts, if u = x n then du = nx n−1 and if dv = sin x dx then dx v = sin x dx = −cos x. Hence

= (−24.53628) −(−23.31305) x n sin x dx

= −1.223
Problem 5. Determine a reduction formula π for 0 x n cos x dx and hence evaluate π 4
0 x cos x dx, correct to 2 decimal places.

= In = x n (−cos x) −
= −x n cos x + n

(−cos x)nx n−1 dx

x n−1 cos x dx

From equation (2),
In = x n sin x + nx n−1 cos x − n(n − 1)In−2 . π hence
0

x n cos x dx = [x n sin x + nx n−1 cos x]π
0

Using integration by parts again, with u = x n−1 , from du which,
= (n − 1)x n−2 and dv = cos x, from which, dx v = cos x dx = sin x. Hence

− n(n − 1)In−2

In = −x n cos x + n x n−1 (sin x)

= [(π n sin π + nπ n−1 cos π)


− (0 + 0)] − n(n − 1)In−2
= − nπ n−1 − n(n − 1)In−2

= −x n cos x + nx n−1 (sin x)

Hence π − n(n − 1)

x cos x dx = I4
4

0

= −4π 3 − 4(3)I2 since n = 4

(sin x)(n − 1)x n−2 dx

i.e.

x n−2 sin x dx

In = −xn cos x + nxn−1 sin x − n(n − 1)In−2 (3)

Reduction formulae
Problem 6. Use a reduction formula to determine x 3 sin x dx.

Hence

π
2

3

429

θ 4 sin θ dθ

0

= 3I4

Using equation (3),

=3 4

I1 = −x 1 cos x + 1x 0 sin x

and

= −x cos x + sin x
Hence

= −x3cos x + 3x2 sin x
+ 6x cos x − 6 sin x + c
3θ 4 sin θ dθ, correct to 2

0

decimal places.
From equation (3),

π

In = [−x n cos x + nx n−1 (sin x)]02 − n(n − 1)In−2
=



π
2

n

cos

π π +n
2
2

n−1

sin

π
2

− (0)

− n(n − 1)In−2 π =n
2

n−1

3

π
2

3

π
2

1

− 4(3) 2

π
2

1

− 24

π
2

1

− 2(1)I0
− 2(1)(1)

+ 24

= 3(1.8039) = 5.41

− 6[−x cos x + sin x]

π
2

π
2

− 4(3) 2

= 3(15.503 − 37.699 + 24)

x 3 sin x dx = −x 3 cos x + 3x 2 sin x

Problem 7. Evaluate

3

=3 4

= −x 3 cos x + 3x 2 sin x − 3(2)I1

π
2

=3 4

x 3 sin x dx = I3

Now try the following exercise
Exercise 170 Further problems on reduction formulae for integrals of the form xn cos x dx and xn sin x dx
1. Use a reduction formula to determine dx. x 5 cos x⎡

x 5 sin x + 5x 4 cos x − 20x 3 sin x


⎣ − 60x 2 cos x + 120x sin x

+ 120 cos x + c
2. Evaluate places. π
0

x 5 cos x dx, correct to 2 decimal
[−134.87]

3. Use a reduction formula to determine x 5 sin x dx.
⎡ 5

−x cos x + 5x 4 sin x + 20x 3 cos x


⎣ − 60x 2 sin x − 120x cos x


− n(n − 1)In−2

Hence

+ 120 sin x + c π 2

π
2

3θ sin θ dθ = 3
4

0

4. Evaluate places. θ sin θ dθ
4

π
0

x 5 sin x dx, correct to 2 decimal
[62.89]

0

= 3I4 π 2

=3 4 π 2

I2 = 2

π
2

I0 =
0

1

3

44.4 Using reduction formulae for integrals of the form sinn x dx and cosn x dx

− 4(3)I2

− 2(1)I0 and π 2

θ sin θ dθ = [−cos x]0
0

= [−0 − (−1)] = 1

(a) sinn x dx
Let In = sin n x dx ≡ sinn−1 x sin x dx from laws of indices. Using integration by parts, let u = sinn−1 x, from which,

430 Higher Engineering Mathematics du = (n − 1) sinn−2 x cos x and dx and

du = (n − 1) sin n−2 x cos x dx

Hence

1 dx = x

1 sin4 x dx = I4 = − sin3 x cos x
4

and let dv = sin x dx, from which, v = sin x dx = −cos x. Hence,
In =

sin0 x dx =

I0 =

+

sinn−1 x sin x dx

= (sinn−1 x)(−cos x)

3
1
1
− sin x cos x + (x)
4
2
2

1
3
= − sin3x cos x − sin x cos x
4
8

(−cos x)(n − 1) sinn−2 x cos x dx



+

= −sinn−1 x cos x
+ (n − 1)

cos2 x sinn−2 x dx

Problem 9. Evaluate significant figures.

1
5
0 4 sin t dt ,

3 x+c 8

correct to 3

= −sinn−1 x cos x
+ (n − 1)
= −sin

n−1

Using equation (4),

(1 − sin2 x) sinn−2 x dx

1
4
sin5 t dt = I5 = − sin4 t cos t + I3
5
5
1 2
2
I3 = − sin t cos t + I1
3
3
1 0 and I1 = − sin t cos t + 0 = −cos t
1
Hence

x cos x

+ (n − 1)

sinn−2 x dx −

sinn x dx

In = −sinn−1 x cos x

i.e.

+ (n − 1)In−2 −(n − 1)In

1 sin5 t dt = − sin4 t cos t
5

In + (n − 1)In

i.e.

= −sinn−1 x cos x + (n − 1)In−2

+

n In = −sinn−1 x cos x + (n − 1)In−2

and

4
2
1
− sin2 t cos t + (−cos t )
5
3
3

4
1
= − sin4 t cos t − sin2 t cos t
5
15

from which,



sin x dx = n 1 n−1 In = − sinn−1 xcos x +
In−2
n n Problem 8. sin4 x dx.

t

(4)

Use a reduction formula to determine

and

4 sin5 t dt

0

1
= 4 − sin4 t cos t
5


Using equation (4),
1
3 sin4 x dx = I4 = − sin3 x cos x + I2
4
4
1
1
I2 = − sin1 x cos x + I0
2
2

8 cos t + c
15

=4

4
8
sin2 t cos t − cos t
15
15

1
0

1
4
− sin4 1 cos1 − sin2 1 cos1
5
15


8
8
cos 1 − −0 − 0 −
15
15

Reduction formulae
= 4[(−0.054178 − 0.1020196
− 0.2881612) − (−0.533333)]
= 4(0.0889745) = 0.356

Using integration by parts, let u = cosn−1 x from which, du
= (n − 1) cosn−2 x(−sin x) dx and

du = (n − 1) cosn−2 x(−sin x) dx

and let

dv = cos x dx

Problem 10. Determine a reduction formula for π 2

π
2

n

sin x dx and hence evaluate

0

sin6 x dx

0

from which, v =

In = (cosn−1 x)(sin x)

sinn x dx
1
n −1
= In = − sinn−1 x cos x +
In−2
n n hence
1
sin x dx = − sinn−1 x cos x n n

0

= [0 − 0] +
In =

i.e.

π
2

+

0

n −1
In−2
n

n −1
In−2
n



0

(sin x)(n − 1) cosn−2 x(−sin x) dx

= (cosn−1 x)(sin x)
+ (n − 1)

+ (n − 1)

n−1
In−2
n

sin2 x cosn−2 x dx

= (cosn−1 x)(sin x)
(1 − cos2 x) cosn−2 x dx

= (cosn−1 x)(sin x) cosn−2 x dx −

+ (n − 1)

Hence π 2

cos x dx = sin x

Then

From equation (4),

π
2

5 sin6 x dx = I6 = I4
6

π
2

I0 =

i.e. In + (n − 1)In = (cos n−1 x)(sin x) + (n − 1)In−2
i.e. n In = (cosn−1 x)(sin x) + (n −1)In−2 π 2

sin0 x dx =

0

cosn x dx

i.e. In = (cosn−1 x)(sin x) + (n − 1)In−2 − (n − 1)In

3
1
I4 = I2 , I2 = I0
4
2 and 1 dx =

0

π
2

Thus

1
In = cosn−1 x sin x + n −1 In−2 n n

Thus π 2

0

431

Problem 11. Use a reduction formula to determine cos4 x dx.

5
5 3 sin6 x dx = I6 = I4 =
I2
6
6 4

Using equation (5),

5 3 1
=
I0
6 4 2
=

5 3 1 π
6 4 2 2

1
3
cos4 x dx = I4 = cos3 x sin x + I2
4
4
=

15 π 96

(b) cosn x dx
Let In = cosn x dx ≡ cosn−1 x cos x dx from laws of indices.

and

1
1
I2 = cos x sin x + I0
2
2

and

I0 =
=

cos0 x dx
1 dx = x

(5)

432 Higher Engineering Mathematics
Now try the following exercise

cos4 x dx

Hence
=

Exercise 171 Further problems on reduction formulae for integrals of the form sinn x dx and cosn x dx

1
1
cos x sin x + x
2
2

3
1
cos3 x sin x +
4
4

1
3
3
= cos3 x sin x + cos x sin x + x + c
4
8
8
Problem 12. π 2

for

1. Use a reduction formula to determine sin7 x dx.


6
1
− sin6 x cos x − sin4 x cos x

⎢ 7
35

⎣ 8
16
− sin2 x cos x − cos x + c
35
35

Determine a reduction formula π 2

cosn x dx and hence evaluate

0

cos5 x dx

0

π
0

2. Evaluate formula. From equation (5), cosn x dx =

1 n −1 cosn−1 x sin x +
In−2
n n π
2

3. Evaluate

1 cosn−1 x sin x n cosn x dx =

0

= [0 −0] +

i.e.

cosn x dx = In =

0

8
15

formula.

π
2

4. Determine, using a reduction formula,

0

cos6 x dx.


n −1
+
In−2 n π
2

sin5 x dx using a reduction

0

and hence π 2

3 sin3 x dx using a reduction
[4]


1
5 x sin x + 5 cos3 x sin x
⎢ 6 cos

24


5
5
+ cos x sin x + x + c
16
16

n −1
In−2
n

n−1
In−2
n

(6)

π
2

5. Evaluate

16
35

cos7 x dx.

0

(Note that this is the same reduction formula as for π 2

sinn x dx (in Problem 10) and the result is usually

0

known as Wallis’s formula).
Thus, from equation (6), π 2

0

4 cos5 x dx = I3 ,
5
π
2

I1 =

and

2
I3 = I1
3
1

cos x dx

44.5

Further reduction formulae

The following worked problems demonstrate further examples where integrals can be determined using reduction formulae.
Problem 13. Determine a reduction formula for tann x dx and hence find tan7 x dx.

0 π 2
= [sin x]0 = (1 − 0) = 1

Let In =

tann x dx ≡

tann−2 x tan 2 x dx by the laws of indices

π
2

Hence
0

4
4 2 cos5 x dx = I3 =
I1
5
5 3
4 2
8
=
(1) =
5 3
15

=

tan n−2 x(sec 2 x − 1) dx since 1 + tan2 x = sec2 x

=

tan n−2 x sec2 x dx −

tann−2 x dx

Reduction formulae

=
i.e. In =

tann−2 x sec2 x dx − In−2

and from equation (6),
5
5 3
I2
I6 = I4 =
6
6 4

tann−1 x
− In−2 n−1 =

5 3
6 4

When n =7,
I7 =

0

cos0 t dt π 2
1 dt = [x]0 =

I6 =

π
2

5 3 1 π
· · ·
6 4 2 2

=

Hence

tan x dx = ln(sec x)

π
2

=

tan4 x tan2 x
I5 =
− I3 and I3 =
− I1
4
2

15π

or
96
32

from Problem 9, Chapter 39, page 394

7
7 5π
Similarly, I8 = I6 = ·
8
8 32
Thus

Thus tan7 x dx =

1
I0
2

0

7

I1 =

π
2

I0 =

and

tan6 x tan x dx =
− I5
6

tan 6 x tan4 x

6
4

π
2

tan2 x

− ln(sec x)
2

sin2 t cos6 t dt = I6 − I8

0

1
1
1
= tan6 x − tan4 x + tan2 x
6
4
2
− ln(sec x) + c
Problem 14. Evaluate, using a reduction formula, sin2 t cos6 t dt .

0 π 2

π
2

sin t cos t dt =
2

6

0

π
2

=

(1 − cos t ) cos t dt
2

Let In = (ln x)n dx.
Using integration by parts, let u =(ln x)n , from which, du = n(ln x)n−1 dx π
2

cos8 t dt

and du = n(ln x)n−1

1 x 1 dx x

0 π 2

In =

and let dv = dx, from which, v = dx = x cosn t dt

0

then

0

6

cos6 t dt −

0

π
2

1 5π

·
=
8 32
256

Problem 15. Use integration by parts to determine a reduction formula for (ln x)n dx.
Hence determine (ln x)3 dx.

0

If

5π 7 5π
− ·
32 8 32

=

π
2

=

tan7 x dx

Hence

sin2 t cos6 t dt = I6 − I8

433

Then In =

(ln x)n dx

= (ln x)n (x) −

(x)n(ln x)n−1

1 dx x

434 Higher Engineering Mathematics
= x(ln x)n − n

Now try the following exercise

(ln x)n−1 dx

i.e. In = x(ln x)n − nIn−1

Exercise 172 Further problems on reduction formulae

When n =3,
(ln x)3 dx = I3 = x(ln x)3 − 3I2
I2 = x(ln x)2 − 2I1 and I1 = ln x dx = x(ln x − 1) from
Problem 7, page 422.

π
2

1. Evaluate
0

tan6 x dx by using reduction for-

2. Determine

π
4

mulae and hence evaluate

Hence

8
105

cos2 x sin5 x dx.

tan 6 x dx.

0

13 π

15 4

(ln x)3 dx = x(ln x)3 − 3[x(ln x)2 − 2I1 ] + c
= x(ln x)3 − 3[x(ln x)2
− 2[x(ln x − 1)]] + c
= x(ln x) − 3[x(ln x)
3

2

− 2x ln x + 2x] + c

3. Evaluate

4. Use a reduction formula to determine
(ln x)4 dx. x(ln x)4 − 4x(ln x)3 + 12x(ln x)2
− 24x ln x + 24x + c

= x(ln x) − 3x(ln x)2
+ 6x ln x − 6x + c
= x[(ln x) − 3(ln x)2
+ 6 ln x − 6] + c

8
315

cos5 x sin4 x dx.

0

3

3

π
2

π
2

5. Show that
0

sin3 θ cos4 θ dθ =

2
35

Chapter 45

Numerical integration y ϭ f(x )

y

45.1

Introduction

Even with advanced methods of integration there are many mathematical functions which cannot be integrated by analytical methods and thus approximate methods have then to be used. Approximate methods of definite integrals may be determined by what is termed numerical integration.
It may be shown that determining the value of a definite integral is, in fact, finding the area between a curve, the horizontal axis and the specified ordinates. Three methods of finding approximate areas under curves are the trapezoidal rule, the mid-ordinate rule and Simpson’s rule, and these rules are used as a basis for numerical integration. y1 y2 y3 y4

xϭa

O

d

45.2

ynϩ1

x ϭb d x

d

Figure 45.1

The trapezoidal rule b Let a required definite integral be denoted by a y dx and be represented by the area under the graph of y = f (x) between the limits x = a and x = b as shown in Fig. 45.1.
Let the range of integration be divided into n equal intervals each of width d, such that nd = b − a, i.e. b−a d= n The ordinates are labelled y1 , y2, y3, . . . , yn+1 as shown. An approximation to the area under the curve may be determined by joining the tops of the ordinates by straight lines. Each interval is thus a trapezium, and since the area of a trapezium is given by:

b a 1
1
y dx ≈ (y1 + y2 )d + (y2 + y3 )d
2
2
1
1
+ (y3 + y4 )d + · · · (yn + yn+1 )d
2
2
1
≈ d y1 + y2 + y3 + y4 + · · · + yn
2
1
+ yn+1
2

i.e. the trapezoidal rule states: b a

1 area = (sum of parallel sides) (perpendicular
2
distance between them) then

y dx ≈

1 first + last
2 ordinate sum of remaining
+
ordinates

width of interval (1)

436 Higher Engineering Mathematics
Problem 1.

(a) Use integration to evaluate,
3 2 correct to 3 decimal places,
√ dx (b) Use the x 1 trapezoidal rule with 4 intervals to evaluate the integral in part (a), correct to 3 decimal places.
3

(a)
1

3

2
√ dx = x 1

2x − 2 dx

1



−1
2 +1

⎤3

1
⎢ 2x

=⎣
⎦ = 4x 2
1
− +1
2
1


√ 3
= 4 x 1 = 4 3− 1

3

Problem 2.

Use the trapezoidal rule with 8
3 2 intervals to evaluate,
√ dx correct to 3 x 1 decimal places.

3−1
With 8 intervals, the width of each is
i.e. 0.25
8
giving ordinates at 1.00, 1.25, 1.50, 1.75, 2.00, 2.25,
2
2.50, 2.75 and 3.00. Corresponding values of √ are x shown in the table below.

1

x
1.00

1.7889

1.50

1.6330

1.75

1.5119

2.00

1.4142

2.25

1.3333

2.50

1.2649

2.75

1.2060

3.00

(b) The range of integration is the difference between the upper and lower limits, i.e. 3 − 1 = 2. Using the trapezoidal rule with 4 intervals gives an inter3−1 val width d =
= 0.5 and ordinates situated
4
at 1.0, 1.5, 2.0, 2.5 and 3.0. Corresponding values
2
of √ are shown in the table below, each correct x to 4 decimal places (which is one more decimal place than required in the problem).

2.0000

1.25

= 2.928, correct to 3 decimal places

2

x

1.1547

x

2

x

1.0

2.0000

1.5

1.6330

2.0

1.4142

+ 1.6330 + 1.5119 + 1.4142

2.5

1.2649

+ 1.3333 + 1.2649 + 1.2060

3.0

1.1547

From equation (1):
3
1

1
2
√ dx ≈ (0.25) (2.000 + 1.1547) + 1.7889
2
x

= 2.932, correct to 3 decimal places.
From equation (1):
3
1

2
1
√ dx ≈ (0.5) (2.0000 + 1.1547)
2
x
+ 1.6330 + 1.4142 + 1.2649
= 2.945, correct to 3 decimal places

This problem demonstrates that even with just 4 intervals a close approximation to the true value of 2.928
(correct to 3 decimal places) is obtained using the trapezoidal rule.

This problem demonstrates that the greater the number of intervals chosen (i.e. the smaller the interval width) the more accurate will be the value of the definite integral. The exact value is found when the number of intervals is infinite, which is, of course, what the process of integration is based upon.
Problem 3. Use the trapezoidal rule to evaluate π 2
1
dx using 6 intervals. Give the answer
0 1 + sin x correct to 4 significant figures.

437

Numerical integration π −0
With 6 intervals, each will have a width of 2
6
π
i.e.
rad (or 15◦) and the ordinates occur at
12
π π π π 5π π 0, , , , , and 12 6 4 3 12
2
1
Corresponding values of are shown in the
1 + sin x table below.

1

2 dx 1 + x2

(Use 8 intervals)

[1.569]

2 ln 3x dx

(Use 8 intervals)

[6.979]

1.
0
3

2.
1

π
3

3.

1.4

x

1
1 + sin x

0

0.79440

π
(or 30◦ )
6

0.66667

π
(or 45◦)
4

0.58579

π
(or 60◦ )
3

2

0.50000

4.

0

(Use 7 intervals)

45.3

The mid-ordinate rule

Let a required definite integral be denoted again b by a y dx and represented by the area under the graph of y = f (x) between the limits x = a and x = b, as shown in Fig. 45.2. y y ϭ f(x)

y1

π
1
dx ≈
1 + sin x
12

[0.843]

0

From equation (1): π 2

[0.672]

0.50867

π
(or 90◦ )
2

e−x dx

0.53590


(or 75◦)
12

(Use 6 intervals)

1.0000

π
(or 15◦)
12

(sin θ) dθ

0

y2

y3

yn

1
(1.00000 + 0.50000)
2
+ 0.79440 + 0.66667

a

O

b x

+ 0.58579 + 0.53590
+ 0.50867
= 1.006, correct to 4 significant figures.

Now try the following exercise
Exercise 173 Further problems on the trapezoidal rule
In Problems 1 to 4, evaluate the definite integrals using the trapezoidal rule, giving the answers correct to 3 decimal places.

d

d

d

Figure 45.2

With the mid-ordinate rule each interval of width d is assumed to be replaced by a rectangle of height equal to the ordinate at the middle point of each interval, shown as y1 , y2, y3 , . . . , yn in Fig. 45.2. b Thus

y dx ≈ d y1 + d y2 + d y3 + · · · + d yn

a

≈ d( y1 + y2 + y3 + · · · + yn )

i.e. the mid-ordinate rule states: b a

y dx ≈ (width of interval) (sum of mid-ordinates)

(2)

438 Higher Engineering Mathematics
From equation (2):

Problem 4.

Use the mid-ordinate rule with (a) 4
3 2 intervals, (b) 8 intervals, to evaluate
√ dx, x 1 correct to 3 decimal places.

3−1
,
(a) With 4 intervals, each will have a width of
4
i.e. 0.5 and the ordinates will occur at 1.0, 1.5, 2.0,
2.5 and 3.0. Hence the mid-ordinates y1 , y2, y3 and y4 occur at 1.25, 1.75, 2.25 and 2.75. Corre2 sponding values of √ are shown in the following x table. x 2

x

1.25

1.7889

1.75

1.5119

2.25

1.3333

2.75

1.2060

From equation (2):
3
1

2
√ dx ≈ (0.5)[1.7889 + 1.5119 x + 1.3333 + 1.2060]
= 2.920, correct to 3 decimal places.

3
1

2
√ dx ≈ (0.25)[1.8856 + 1.7056 x + 1.5689 + 1.4606 + 1.3720
+ 1.2978 + 1.2344 + 1.1795]
= 2.926, correct to 3 decimal places.

As previously, the greater the number of intervals the nearer the result is to the true value (of 2.928, correct to 3 decimal places).
2.4

Problem 5.

Evaluate

e

−x 2
3

dx, correct to 4

0

significant figures, using the mid-ordinate rule with
6 intervals.
2.4 − 0
With 6 intervals each will have a width of
, i.e.
6
0.40 and the ordinates will occur at 0, 0.40, 0.80, 1.20,
1.60, 2.00 and 2.40 and thus mid-ordinates at 0.20, 0.60,
1.00, 1.40, 1.80 and 2.20. Corresponding values of e are shown in the following table.

(b) With 8 intervals, each will have a width of 0.25 and the ordinates will occur at 1.00, 1.25, 1.50,
1.75, . . . and thus mid-ordinates at 1.125, 1.375,
1.625, 1.875 . . .
2
Corresponding values of √ are shown in the x following table.

−x 2
3

e

0.20

0.98676

0.60

0.88692

1.00

0.71653

1.40 x x

0.52031

1.80

0.33960

2.20

2

x

−x 2
3

0.19922

1.125 1.8856
1.375 1.7056

From equation (2):

1.625 1.5689
1.875 1.4606

2.4

e

−x 2
3

dx ≈ (0.40)[0.98676 + 0.88692

0

2.125 1.3720

+ 0.71653 + 0.52031

2.375 1.2978

+ 0.33960 + 0.19922]

2.625 1.2344
2.875 1.1795

= 1.460, correct to 4 significant figures.

Numerical integration

439

y

Now try the following exercise

y ϭ a ϩ bx ϩcx 2

Exercise 174 Further problems on the mid-ordinate rule
In Problems 1 to 4, evaluate the definite integrals using the mid-ordinate rule, giving the answers correct to 3 decimal places. y1 2

3 dt 1 + t2

1.
0
π
2

2.
0

(Use 8 intervals)

y3

[3.323]
Ϫd

1 dθ (Use 6 intervals)
1 + sin θ

y2

O

d

x

[0.997]
Figure 45.3

3 ln x

3.

x

1

(Use 10 intervals) [0.605]
Since

(cos3 x) dx (Use 6 intervals) [0.799]

0

y = a + bx + cx 2 ,

at

x = −d, y1 = a − bd + cd 2

at

π
3

4.

dx

x = 0, y2 = a

and at x = d, y3 = a + bd + cd 2

45.4

Hence y1 + y3 = 2a + 2cd 2

Simpson’s rule

And

The approximation made with the trapezoidal rule is to join the top of two successive ordinates by a straight line, i.e. by using a linear approximation of the form a + bx. With Simpson’s rule, the approximation made is to join the tops of three successive ordinates by a parabola, i.e. by using a quadratic approximation of the form a + bx + cx 2 .
Figure 45.3 shows a parabola y = a + bx + cx 2 with ordinates y1 , y2 and y3 at x = −d, x = 0 and x = d respectively. Thus the width of each of the two intervals is d. The area enclosed by the parabola, the x-axis and ordinates x = −d and x = d is given by: d bx 2 cx 3
(a + bx + cx )dx = ax +
+
2
3
−d

d

2

= ad +

−d

bd 2 cd 3
+
2
3

− −ad +

bd 2 cd 3

2
3

(4)

Thus the area under the parabola between x = −d and x =d in Fig. 45.3 may be expressed as
1
3 d(y1 + 4y2 + y3 ), from equations (3) and (4), and the result is seen to be independent of the position of the origin. b
Let a definite integral be denoted by a y dx and represented by the area under the graph of y = f (x) between the limits x = a and x = b, as shown in Fig. 45.4.
The range of integration, b − a, is divided into an even number of intervals, say 2n, each of width d.
Since an even number of intervals is specified, an odd number of ordinates, 2n + 1, exists. Let an approximation to the curve over the first two intervals be a parabola of the form y = a + bx + cx 2 which passes through the tops of the three ordinates y1, y2 and y3. Similarly, let an approximation to the curve over the next two intervals be the parabola which passes through the tops of the ordinates y3, y4 and y5 , and so on. b Then

y dx a 2
= 2ad + cd 3 or
3
1 d(6a + 2cd 2 )
3

y1 + 4y2 + y3 = 6a + 2cd 2


(3)

1
1
d(y1 + 4y2 + y3 ) + d(y3 + 4y4 + y5 )
3
3
1
+ d(y2n−1 + 4y2n + y2n+1 )
3

440 Higher Engineering Mathematics y Thus, from equation (5):
3

y ϭf(x)

1

2
1
√ dx ≈ (0.5) [(2.0000 + 1.1547)
3
x
+ 4(1.6330 + 1.2649) + 2(1.4142)]
1
= (0.5)[3.1547 + 11.5916
3

y2

y1

y3

y4

+ 2.8284]

y2nϩ1

= 2.929, correct to 3 decimal places.

a

O

b d d

x

d

Figure 45.4



1 d[(y1 + y2n+1 ) + 4(y2 + y4 + · · · + y2n )
3
+ 2(y3 + y5 + · · · + y2n−1 )]

(b) With 8 intervals, each will have a width of
3−1
, i.e. 0.25 and the ordinates occur at 1.00,
8
1.25, 1.50, 1.75, . . . , 3.0. The values of the ordinates are as shown in the table in Problem 2, page 436.
Thus, from equation (5):
3
1

2
1
√ dx ≈ (0.25) [(2.0000 + 1.1547) x 3
+ 4(1.7889 + 1.5119 + 1.3333

i.e. Simpson’s rule states: b y dx ≈

a

1 width of
3 interval
+4
+2

+ 1.2060) + 2(1.6330 + 1.4142

first + last ordinate sum of even ordinates + 1.2649)]
1
= (0.25)[3.1547 + 23.3604
3

(5)

+ 8.6242]

sum of remaining odd ordinates

Note that Simpson’s rule can only be applied when an even number of intervals is chosen, i.e. an odd number of ordinates.
Use Simpson’s rule with (a) 4
3 2 intervals, (b) 8 intervals, to evaluate
√ dx, x 1 correct to 3 decimal places.

= 2.928, correct to 3 decimal places.
It is noted that the latter answer is exactly the same as that obtained by integration. In general, Simpson’s rule is regarded as the most accurate of the three approximate methods used in numerical integration.

Problem 6.

3−1
,
(a) With 4 intervals, each will have a width of
4
i.e. 0.5 and the ordinates will occur at 1.0, 1.5,
2.0, 2.5 and 3.0. The values of the ordinates are as shown in the table of Problem 1(b), page 436.

Problem 7.

Evaluate π 3
0

1−

1 2 sin θ dθ,
3

correct to 3 decimal places, using Simpson’s rule with 6 intervals.

Numerical integration π −0
With 6 intervals, each will have a width of 3
6
π
◦ ), and the ordinates will occur at
i.e.
rad (or 10
18
π π π 2π 5π π 0, , , ,
,
and
18 9 6 9 18
3
1
1 − sin2 θ are shown in
3

Corresponding values of the table below. θ π
18

0

π
9

1.0000 0.9950 0.9803 0.9574

9

θ

(or
1−

40◦)

(or

0



2.0

12.0

0

Charge, q, in millicoulombs, is given by
12.0
q = 0 i dt.
Use Simpson’s rule to determine the approximate charge in the 12 millisecond period.

From equation (5):

(or

12.0

Charge, q =
0

1 i dt ≈ (2.0) [(0 + 0) + 4(3.5
3
+10.0 + 2.0) + 2(8.2 + 7.3)]

= 62 mC

60◦)
Now try the following exercise

1 2 sin θ
3

0.9286

0.8969

0.8660
Exercise 175 Further problems on
Simpson’s rule

From Equation (5) π 3

10.0

π
3


18
50◦)

7.3

π
6

(or 10◦ ) (or 20◦) (or 30◦)
1
1 − sin2 θ
3

8.0

In Problems 1 to 5, evaluate the definite integrals using Simpson’s rule, giving the answers correct to 3 decimal places.

1
1 − sin2 θ dθ
3
1 π
[(1.0000 + 0.8660) + 4(0.9950
3 18
+ 0.9574 + 0.8969)
+ 2(0.9803 + 0.9286)]

1 π
[1.8660 + 11.3972 + 3.8178]
=
3 18
= 0.994, correct to 3 decimal places.

π
2

1.

1.6

[1.187]

1 dθ (Use 8 intervals)
1 + θ4

[1.034]

sin θ dθ θ

(Use 8 intervals)

[0.747]

x cos x dx

(Use 6 intervals)

[0.571]

2.
0
1.0

3.
0.2

Problem 8. An alternating current i has the following values at equal intervals of
2.0 milliseconds:

(sin x) dx (Use 6 intervals)

0

π
2

4.
0

Time (ms)

Current i (A)

0

0

2.0

3.5

4.0

8.2

6.0

10.0

π
3

5.
0

2

ex sin 2x dx (Use 10 intervals)
[1.260]

In Problems 6 and 7 evaluate the definite integrals using (a) integration, (b) the trapezoidal rule,

441

442 Higher Engineering Mathematics
(c) the mid-ordinate rule, (d) Simpson’s rule. Give answers correct to 3 decimal places.
4

6.
1

6

7.
2

4 dx x3

In Problems 8 and 9 evaluate the definite integrals using (a) the trapezoidal rule, (b) the mid-ordinate rule, (c) Simpson’s rule. Use 6 intervals in each case and give answers correct to 3 decimal places.
3

(1 + x 4 ) dx

0

(a) 10.194 (b) 10.007
(c) 10.070
0.7

9.
0.1

1
(1 − y 2 )

7.0

8.0

8.0

9.4

The distance travelled in 8.0 s is given by
8.0
0 v dt
Estimate this distance using Simpson’s rule, giving the answer correct to 3 significant figures. [28.8 m]
11. A pin moves along a straight guide so that its velocity v (m/s) when it is a distance x(m) from the beginning of the guide at time t (s) is given in the table below.

time t (s) velocity v (ms−1 )
0

1.0

1.0

3.0

1.7

v (m/s)
0
0.052

1.0

0.082

1.5

0.125

2.0

0.162

2.5

0.175

3.0

0.186

3.5

0.160

4.0

0

0.4

2.0

t (s)

0.5

A vehicle starts from rest and its velocity is measured every second for 8 s, with values as follows: 0

6.2

0 dy (a) 0.677 (b) 0.674
(c) 0.675

10.

4.1

6.0

(a) 1.875 (b) 2.107
(c) 1.765 (d) 1.916

2.9

5.0

(Use 6 intervals)

1

dx (Use 8 intervals)
(2x − 1)
(a) 1.585 (b) 1.588
(c) 1.583 (d) 1.585

8.

4.0

Use Simpson’s rule with 8 intervals to determine the approximate total distance travelled by the pin in the 4.0 s period.
[0.485 m]

Revision Test 13
This Revision Test covers the material contained in Chapters 43 to 45. The marks for each question are shown in brackets at the end of each question.
3

(13)

5 dx using (a) integration (b) the x2 1 trapezoidal rule (c) the mid-ordinate rule
(d) Simpson’s rule. In each of the approximate methods use 8 intervals and give the answers correct to 3 decimal places.
(19)

(10)

6. An alternating current i has the following values at equal intervals of 5 ms:

1. Determine the following integrals:
5x e2x dx

(a)

5. Evaluate

t 2 sin 2t dt

(b)

2. Evaluate correct to 3 decimal places:
4√

x ln x dx

1

Time t (ms)

3. Use reduction formulae to determine:

0 5

Current i(A) 0 4.8
(a)

3 3x

x e dx π 2

4. Evaluate formula. 0

(b)

4

t sin t dt

(13)

using

a

reduction
(6)

15

20

9.1 12.7

8.8

25
3.5

30
0

Charge q, in coulombs, is given by q= cos6 x dx

10

30×10−3 i dt .
0

Use Simpson’s rule to determine the approximate charge in the 30 ms period.
(4)

Chapter 46

Solution of first order differential equations by separation of variables
46.1

Family of curves

dy
= 3 with dx respect to x gives y = 3 dx, i.e., y = 3x + c, where c is an arbitrary constant.

Integrating both sides of the derivative

y = 3x + c represents a family of curves, each of the curves in the family depending on the value of c.
Examples include y = 3x + 8, y = 3x + 3, y = 3x and y = 3x − 10 and these are shown in Fig. 46.1. y y 5 3x 1 3

12

y 5 3x

8 y 5 3x 2 10

4

28
212
216

Figure 46.1

Problem 1.

Sketch the family of curves given by dy the equation
= 4x and determine the equation of dx one of these curves which passes through the point
(2, 3).

y 5 3x 1 8

16

24 23 22 21 0
24

Each are straight lines of gradient 3. A particular curve of a family may be determined when a point on the curve is specified. Thus, if y = 3x + c passes through the point
(1, 2) then 2 = 3(1) + c, from which, c = −1. The equation of the curve passing through (1, 2) is therefore y = 3x − 1.

1

2

3

4

x

dy
= 4x with respect to x
Integrating both sides of dx gives: dy dx = dx 4x dx, i.e., y = 2x 2 + c

Some members of the family of curves having an equation y = 2x 2 + c include y = 2x 2 + 15, y = 2x 2 + 8, y = 2x 2 and y = 2x 2 − 6, and these are shown in Fig. 46.2. To determine the equation of the curve passing through the point (2, 3), x = 2 and y = 3 are substituted into the equation y = 2x 2 + c.
Thus 3 =2(2)2 + c, from which c = 3 −8 =−5.
Hence the equation of the curve passing through the point (2, 3) is y = 2x2 − 5.

Solution of first order differential equations by separation of variables

Ϫ6



20

2x 2

30



yϭ yϭ 2
2x 2
2x 2 x 2 ϩ ϩ1 8
5

y

10

Ϫ4

Ϫ3

Ϫ2

Ϫ1

0

1

2

3

4

x

Ϫ10

Figure 46.2

Now try the following exercise
Exercise 176 of curves

Differential equations

A differential equation is one that contains differential coefficients. Examples include
(i)

The degree of a differential equation is that of the highest power of the highest differential which the equation contains after simplification.
3

d2 x dx 5
Thus
+2
= 7 is a second order differdt 2 dt ential equation of degree three.
Starting with a differential equation it is possible, by integration and by being given sufficient data to determine unknown constants, to obtain the original function. This process is called ‘solving the differential equation’. A solution to a differential equation which contains one or more arbitrary constants of integration is called the general solution of the differential equation. When additional information is given so that constants may be calculated the particular solution of the differential equation is obtained. The additional information is called boundary conditions. It was shown in
Section 46.1 that y = 3x + c is the general solution of dy = 3. the differential equation dx Given the boundary conditions x = 1 and y = 2, produces the particular solution of y = 3x − 1.
Equations which can be written in the form

Further problems on families

1. Sketch a family of curves represented by each of the following differential equations: dy dy dy = 6 (b)
= 3x (c)
= x +2
(a)
dx dx dx
2. Sketch the family of curves given by the equady tion = 2x + 3 and determine the equation dx of one of these curves which passes through the point (1, 3).
[ y = x 2 + 3x − 1]

46.2

445

dy d2 y dy = 7x and (ii) 2 + 5
+ 2y = 0 dx dx dx Differential equations are classified according to the highest derivative which occurs in them. Thus example (i) above is a first order differential equation, and example (ii) is a second order differential equation.

dy dy dy
= f (x),
= f ( y) and
= f (x) · f ( y) dx dx dx can all be solved by integration. In each case it is possible to separate the y’s to one side of the equation and the x’s to the other. Solving such equations is therefore known as solution by separation of variables.

46.3

The solution of equations of the dy form
= f (x) dx dy
A differential equation of the form
= f (x) is solved dx by direct integration, y= i.e.

f (x) dx

Problem 2. Determine the general solution of dy x
= 2 − 4x 3 dx Rearranging x

dy
= 2 − 4x 3 gives: dx dy
2 − 4x 3
2 4x 3
2
=
= −
= − 4x 2 dx x x x x 446 Higher Engineering Mathematics
Integrating both sides gives:

Problem 5.

2 y= − 4x 2 dx x 4 3
i.e. y = 2 ln x − x + c,
3
which is the general solution.
Find the particular solution of the dy differential equation 5 + 2x = 3, given the dx 2 boundary conditions y = 1 when x = 2.
5

The bending moment M of the beam dM is given by
= −w(l − x), where w and x are dx constants. Determine M in terms of x given:
M = 1 wl 2 when x = 0.
2
dM
= −w(l − x) = −wl + wx dx Problem 3.

Since 5

d y 3 − 2x
3 2x dy + 2x = 3 then
=
= − dx dx
5
5
5

3 2x dx −
5
5
3x x 2
i.e.
y=

+ c,
5
5 which is the general solution.
Substituting the boundary conditions y = 1 2 and x = 2
5
to evaluate c gives:
1 2 = 6 − 4 + c, from which, c = 1
5
5
5
3x x2
Hence the particular solution is y = − + 1.
5
5

Hence y =

Problem 4. Solve the equation dθ = 5, given θ = 2 when t = 1.
2t t − dt M = −wlx +


5

5
= and =t − dt 2t dt 2t
Integrating gives: t− 5 dt θ= t− 2t t2 5
i.e. θ = − ln t + c,
2 2

which is the general solution.
When M = 1 wl 2 , x = 0.
2
w(0)2
1 2
+c
Thus wl = −wl(0) +
2
2
1 2 from which, c = wl .
2
Hence the particular solution is: w(x)2 1 2
M = −wlx +
+ wl
2
2
1 2
2)
i.e. M = w(l − 2lx + x
2
1 or M = w(l − x)2
2
Now try the following exercise

1.

dy
= cos 4x − 2x dx 2. 2x which is the general solution.


When θ = 2, t = 1, thus c= 3.
2
Hence the particular solution is:
2= 1
2

t2 5
3
− ln t +
2
2
2
1 2
i.e. θ = (t − 5 ln t + 3)
2

5
2 ln

wx 2
+c
2

Exercise 177 Further problems on dy equations of the form
= f (x). dx In Problems 1 to 5, solve the differential equations. Rearranging gives:

θ=

Integrating with respect to x gives:

1 + c from which,

3.

dy
=3 − x3 dx y=

sin 4x
− x2 +c
4

x3
3
+c y = ln x −
2
6

dy
+ x = 3, given y = 2 when x = 1. dx y = 3x −

4. 3

x2 1

2
2

dy
2
π
+ sin θ = 0, given y = when θ = dθ 3
3
1
1
y = cos θ +
3
2

Solution of first order differential equations by separation of variables

447

Integrating both sides gives:
5.

1 dy + 2 = x − 3 , given y = 1 when x = 0. ex dx
2
1 2 x − 4x + x + 4 y= 6 e 6. The gradient of a curve is given by: dy x2
+ = 3x dx 2
Find the equation of the curve if it passes through the point 1, 1 .
3
x3
3
y = x2 − −1
2
6
7. The acceleration, a, of a body is equal to its rate dv of change of velocity, . Find an equation for dt v in terms of t , given that when t = 0, velocity v = u.
[v = u +at]
8. An object is thrown vertically upwards with an initial velocity, u, of 20 m/s. The motion of the object follows the differential equation ds = u − gt , where s is the height of the object dt in metres at time t seconds and g = 9.8 m/s2 .
Determine the height of the object after 3 seconds if s = 0 when t = 0.
[15.9 m]

46.4

The solution of equations of the dy form
= f ( y) dx dy
A differential equation of the form
= f ( y) is initially dx dy rearranged to give dx = and then the solution is f ( y) obtained by direct integration, dx =

i.e.

dy f ( y)

Problem 6. Find the general solution of dy = 3 + 2y. dx Rearranging

dx =

dy
= 3 + 2y gives: dx dy
3 + 2y

dx =

dy
3 + 2y

Thus, by using the substitution u = (3 + 2y) — see
Chapter 39, x = 1 ln (3 +2y) + c
2

(1)

It is possible to give the general solution of a differential equation in a different form. For example, if c = ln k, where k is a constant, then: x = 1 ln(3 + 2y) + ln k,
2
1

i.e.

x = ln(3 + 2y) 2 + ln k

or

x = ln[k (3 +2y)]

(2)

by the laws of logarithms, from which, ex = k (3 + 2y)

(3)

Equations (1), (2) and (3) are all acceptable general solutions of the differential equation dy = 3 + 2y dx Problem 7. Determine the particular solution of dy 1
( y 2 − 1) = 3y given that y = 1 when x = 2 dx 6
Rearranging gives: dx =

y2 − 1 dy =
3y

y
1
dy

3 3y

Integrating gives: y 1

dy
3 3y y2 1
− ln y + c, x= 6
3
which is the general solution.

dx =
i.e.

When y = 1, x = 2 1 , thus 2 1 = 1 − 1 ln 1 +c, from
6
6
6
3 which, c = 2.
Hence the particular solution is: x= y2
1
− ln y + 2
6
3

448 Higher Engineering Mathematics
Problem 8. (a) The variation of resistance,
R ohms, of an aluminium conductor with dR temperature θ ◦ C is given by
= α R, where α dθ is the temperature coefficient of resistance of aluminium. If R = R0 when θ = 0◦C, solve the equation for R. (b) If α = 38 ×10−4 /◦C, determine the resistance of an aluminium conductor at 50◦ C, correct to 3 significant figures, when its resistance at 0◦C is 24.0 .
(a)

dy dR = α R is of the form
= f ( y) dθ dx dR Rearranging gives: dθ = αR Integrating both sides gives: dθ = θ= 1 ln R0 + c α 1 from which c = − ln R0 α Hence the particular solution is
0=

1
1
1 ln R − ln R0 = ( ln R − ln R0 ) α α α R
1
R
i.e. θ = ln or αθ = ln α R0
R0
θ=

Hence eαθ =

dy
= 2 +3y dx 2.

dy
= 2 cos2 y dx 3. ( y 2 + 2)

R from which, R = R0eαθ
R0

(b) Substituting α = 38 ×10−4 , R0 = 24.0 and θ = 50 into R = R0 eαθ gives the resistance at 50◦ C, i.e.
−4
R50 = 24.0 e(38×10 ×50) = 29.0 ohms

1 x = ln(2 + 3y) + c
3
[tan y = 2x + c]

dy
1
= 5y, given y = 1 when x = dx 2 y2 + 2 ln y = 5x − 2
2

4. The current in an electric circuit is given by the equation
Ri + L

dR αR 1 ln R + c, α which is the general solution.
Substituting the boundary conditions R = R0 when θ = 0 gives:
i.e.

1.

di
= 0, dt where L and R are constants. Show that
− Rt i = I e L , given that i = I when t = 0.
5. The velocity of a chemical reaction is given by dx = k(a − x), where x is the amount transdt ferred 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 .
[x = a(1 − e−kt )]
6.

(a)

Charge Q coulombs at time t seconds is given by the differential equation dQ Q
R
+ = 0, where C is the capacidt
C
tance in farads and R the resistance in ohms. Solve the equation for Q given that Q = Q 0 when t = 0.

(b) A circuit possesses a resistance of and a capacitance of
250 ×103
8.5 × 10−6 F, and after 0.32 seconds the charge falls to 8.0 C. Determine the initial charge and the charge after
1 second, each correct to 3 significant figures. −t

[(a) Q = Q 0 e CR (b) 9.30 C, 5.81 C]
Now try the following exercise
Exercise 178 Further problems on dy equations of the form
= f ( y) dx In Problems 1 to 3, solve the differential equations. 7. A differential equation relating the difference in tension T , pulley contact angle θ and coefdT
= μT . When θ = 0, ficient of friction μ is dθ T = 150 N, and μ = 0.30 as slipping starts.
Determine the tension at the point of slipping when θ = 2 radians. Determine also the value of θ when T is 300 N.
[273.3 N, 2.31 rads]

449

Solution of first order differential equations by separation of variables
i.e. ( y 2 − 1)2 = Ax
8. The rate of cooling of a body is given by dθ = kθ, where k is a constant. If θ = 60◦C dt when t = 2 minutes and θ = 50◦C when t = 5 minutes, determine the time taken for θ to fall to 40◦C, correct to the nearest second.
[8 m 40 s]

46.5 The solution of equations of the dy form
= f (x) · f ( y) dx dy
A differential equation of the form
= f (x) · f ( y), dx where f (x) is a function of x only and f ( y) is a function dy = f (x) dx, and of y only, may be rearranged as f ( y) then the solution is obtained by direct integration, i.e.

Equations (1) to (3) are thus three valid solutions of the differential equations
4x y


= 2e3t −2θ = 2(e3t )(e−2θ ), dt by the laws of indices.
Separating the variables gives: dθ = 2e3t dt, e−2θ i.e. e2θ dθ = 2e3t dt
Integrating both sides gives: e2θ dθ =

1 2θ 2 3t e = e +c
2
3
When t = 0, θ = 0, thus:

4y
1
d y = dx y2 − 1 x 1 0 2 0 e = e +c
2
3
1
1 2 from which, c = − = −
2 3
6
Hence the particular solution is:

Integrating both sides gives:
4y
dy = y2 − 1

1 dx x

substitution u = y 2 − 1,
2 ln ( y 2 − 1) = lnx + c

or from which, and the

(1)

(2)

If in equation (1), c = ln A, where A is a different constant, then ln( y 2 − 1)2 = ln x + ln A
i.e. ln( y 2 − 1)2 = ln Ax

or

3e2θ = 4e3t − 1

Problem 11. Find the curve which satisfies the dy equation x y = (1 + x 2 ) and passes through the dx point (0, 1).

=c

( y2 − 1)2
= ec x 1 2θ 2 3t 1 e = e −
2
3
6

general

ln( y 2 − 1)2 − ln x = c
( y 2 − 1)2 ln x

2e3t dt

Thus the general solution is:

dy
= y2 − 1 dx Separating the variables gives:

Using the solution is:

dy
= y2 − 1 dx Problem 10. Determine the particular solution of dθ = 2e3t −2θ , given that t = 0 when θ = 0. dt dy
= f (x) dx f ( y)
Problem 9. Solve the equation 4x y

(3)

Separating the variables gives: dy x dx =
(1 + x 2 ) y Integrating both sides gives:
1
2

ln (1 + x 2 ) = ln y + c

450 Higher Engineering Mathematics
When x = 0, y = 1 thus c = 0.

1
2 ln 1 = ln 1 +c,

from which,

Hence the particular solution is 1 ln(1 + x 2 ) = ln y
2
1

1

i.e. ln(1 + x 2 ) 2 = ln y, from which, (1 + x 2 ) 2 = y

(by making
Chapter 39).

In the R − L series circuit shown in Fig. 46.3, the supply
p.d., E, is given by
E = V R + VL
V R = iR and V L = L
Hence
from which

di
E = iR + L dt di
E − L = Ri dt di dt −

VR

u = E − Ri,

see

t
1
1 ln (E − Ri) = − ln E
R
L
R

Transposing gives:


1 t 1 ln (E − Ri) + ln E =
R
R
L
1 t [ln E − ln (E − Ri)] =
R
L ln E
E − Ri

=

Rt
L

E
Rt
=e L
E − Ri
E − Ri
−Rt
Hence and =e L
E
−Rt
Ri = E − Ee L .

from which

E − Ri = Ee

−Rt
L

and

Hence current, i= L

R

substitution

1
When t = 0, i = 0, thus − ln E =c
R
Thus the particular solution is:

Hence the equation of the curve is y = (1 +x2 ).
Problem 12. The current i in an electric circuit containing resistance R and inductance L in series with a constant voltage source E is given by the di = Ri. Solve the differential equation E − L dt equation and find i in terms of time t given that when t = 0, i = 0.

a

VL

−Rt
E
1−e L ,
R

which represents the law of growth of current in an inductive circuit as shown in Fig. 46.4.

i
E

i
E
R

Figure 46.3

Most electrical circuits can be reduced to a differential equation. di E − Ri di Rearranging E − L = Ri gives = dt dt
L

E i ϭ R (1ϪeϪRt/L )

and separating the variables gives: di dt
=
E − Ri
L

0

Integrating both sides gives:

Figure 46.4

di
=
E − Ri

Time t

dt
L

Hence the general solution is:
1
t
− ln (E − Ri) = + c
R
L

Problem 13.
Cv

For an adiabatic expansion of a gas

dp dV +Cp
= 0, p V

Solution of first order differential equations by separation of variables

where C p and Cv are constants. Given n = show that pV n = constant.

Cp
,
Cv

Separating the variables gives:

when y = 1.

dV
V

y2 + 1
= constant. x2 +1

Cv ln p = −C p ln V + k

i.e.

Dividing throughout by constant Cv gives: ln p = −

Since

Cp k ln V +
Cv
Cv

Cp
= n, then ln p +n ln V = K ,
Cv

where K =
i.e.
logarithms.

[ y 2 = x 2 − 2 ln x + 3] or ln

pV n = K ,

by the laws of

Hence pV n = e K , i.e. pV n = constant.

Now try the following exercise
Further problems on dy equations of the form
= f (x) · f ( y) dx In Problems 1 to 4, solve the differential equations. dy = 2y cos x dx 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 dV + V = E.
CR
dt
(a)

Exercise 179

1.

dy
6. Solve x y = (1 − x 2 ) for y, given x = 0 dx when y = 1.
1
y=
(1 − x 2 )
7. Determine the equation of the curve which dy satisfies the equation x y = x 2 − 1, and dx which passes through the point (1, 2).

k
.
Cv

ln p +ln V n = K

dy
= 0, given x = 1 dx [ln (x 2 y) = 2x − y − 1]

5. Show that the solution of the equation y2 + 1 y d y
=
is of the form x 2 + 1 x dx

Integrating both sides gives: dp = −C p p dy
= e2x−y , given x = 0 when y = 0. dx 1
1
e y = e2x +
2
2

4. 2y(1 − x) + x(1 + y)

dp dV Cv = −C p p V

Cv

3.

[ln y = 2 sin x + c]

dy
2. (2y − 1) = (3x 2 + 1), given x = 1 when dx y = 2.
[ y 2 − y = x 3 + x]

Solve the equation for V given that at t = 0, V = 0.

(b) Calculate V , correct to 3 significant figures, when E =25V, C = 20 ×10−6 F,
R = 200 ×103 and t = 3.0 s.


−t
CR

⎣(a) V = E 1 − e
(b) 13.2 V
9. Determine the value of p, given that dy x 3 = p − x, and that y = 0 when x = 2 and dx when x = 6.
[3]

451

Chapter 47

Homogeneous first order differential equations
47.1

Introduction

Certain first order differential equations are not of the
‘variable-separable’ type, but can be made separable by changing the variable. dy An equation of the form P
= Q, where P and Q are dx functions of both x and y of the same degree throughout, is said to be homogeneous in y and x. For example, f (x, y) = x 2 + 3x y + y 2 is a homogeneous function since each of the three terms are of degree 2. However, x2 − y is not homogeneous since the term f (x, y) = 2
2x + y 2 in y in the numerator is of degree 1 and the other three terms are of degree 2.

(iv) Separate the variables and solve using the method shown in Chapter 46. y (v) Substitute v = to solve in terms of the original x variables.

47.3 Worked problems on homogeneous first order differential equations
Problem 1. Solve the differential equation: dy y − x =x
, given x = 1 when y = 2. dx Using the above procedure:

47.2 Procedure to solve differential dy equations of the form P
=Q
dx
(i) Rearrange P

dy dy Q
= Q into the form
= . dx dx P

(ii) Make the substitution y = vx (where v is a funcdy dv tion of x), from which,
= v(1) + x
, by the dx dx product rule. dy in the equation
(iii) Substitute for both y and dx dy Q
= . Simplify, by cancelling, and an equation dx P results in which the variables are separable.

(i) Rearranging y − x = x

dy gives: dx

dy y − x
=
, dx x which is homogeneous in x and y.
(ii) Let y = vx, then

dy dv =v+x dx dx

(iii) Substituting for y and v+x dy gives: dx

dv vx − x x(v − 1)
=
=
=v − 1 dx x x Homogeneous first order differential equations
(iv) Separating the variables gives: x dv
1
= v − 1 − v = −1, i.e. dv = − dx dx x

Integrating both sides gives: dv =

1
− dx x Hence, v = −ln x + c y y
(v) Replacing v by gives: = −ln x + c, which is x x the general solution.
2
= − ln 1 + c from
When x = 1, y = 2, thus:
1
which, c = 2 y Thus, the particular solution is: = − ln x + 2 x or y = −x(ln x −2) or y = x(2 − ln x)
Problem 2. Find the particular solution of the d y x 2 + y2 equation: x
=
, given the boundary dx y conditions that y = 4 when x = 1.

1
Hence, v dv = dx x Integrating both sides gives: v dv =

1 v2 dx i.e.
= ln x + c x 2

y2 y gives: 2 = ln x + c, which is x 2x the general solution.

(v) Replacing v by

When x = 1, y = 4, thus: which, c = 8

16
= ln 1 + c from
2

Hence, the particular solution is: or y2 = 2x2 (8 + lnx)

y2
= ln x + 8
2x 2

Now try the following exercise

Exercise 180 Further problems on homogeneous first order differential equations 1. Find the general solution of: x 2 = y 2

Using the procedure of section 47.2: d y x 2 + y2
(i) Rearranging x
=
gives: dx y d y x 2 + y2
=
which is homogeneous in x and y dx xy since each of the three terms on the right hand side are of the same degree (i.e. degree 2).
(ii) Let y = vx then

dy dv =v+ x dx dx

dy in the equation
(iii) Substituting for y and dx 2 + y2 dy x
=
gives: dx xy v+x dv
1 + v2 x 2 + v2 x 2 x 2 + v2 x 2
=
=
=
dx x(vx) vx 2 v (iv) Separating the variables gives: x dv 1 + v 2
1 + v2 − v2 1
=
−v=
=
dx v v v 453

x 3 − y3
1
− ln
3
x3

dy
.
dx

= ln x + c

2. Find the general solution of: dy = 0.
[y = x(c − ln x)] x − y+x dx 3. Find the particular solution of the differential equation: (x 2 + y 2 )d y = x y dx, given that x = 1 when y = 1. x 2 = 2y 2 ln y +

1
2

x + y dy
= . y − x dx


1
2y y 2
⎣ − 2 ln 1 + x − x 2 = ln x + c ⎦ or x 2 + 2x y − y 2 = k

4. Solve the differential equation:

5. Find the particular solution of the differential
2y − x d y equation: = 1 given that y = 3 y + 2x dx when x = 2.
[x 2 + x y − y 2 = 1]

454 Higher Engineering Mathematics
(v) Replacing v by

47.4 Further worked problems on homogeneous first order differential equations

4
3y
ln 1 +
3
x

Problem 3. Solve the equation:
7x(x − y)d y = 2(x 2 + 6x y − 5y 2 )dx given that x = 1 when y = 0.

d y 2x 2 + 12x y − 10y 2
=
dx
7x 2 − 7x y which is homogeneous in x and y since each of the terms on the right hand side is of degree 2.

dy dv =v+ x dx dx dy (iii) Substituting for y and gives: dx
2x 2 + 12x(vx) − 10 (vx)2 dv = v+x dx
7x 2 − 7x(vx)

x + 3y
4
ln
3
x

i.e.

x +3y x 4
3

+ ln
4
3

2x − y x 2x − y
= ln(2x) x from the laws of logarithms

2x − y
= 2x x Using the procedure of section 47.2:
(i) Rearranging gives:

(2 + 12v − 10v 2 ) − v(7 − 7v)
=
7 − 7v

d y 3y 2 − x 2 dy = 3y 2 − x 2 and
=
dx dx 2x y dy dv
(ii) Let y = vx then
=v+ x dx dx dy (iii) Substituting for y and gives: dx
2x y

2 + 5v − 3v 2
7 − 7v

dx
7 − 7v dv =
2 + 5v − 3v 2 x Integrating both sides gives:

v+x
1
dx x 3 (vx)2 − x 2
3v 2 − 1 dv =
=
dx
2x(vx)
2v

(iv) Separating the variables gives:

7 − 7v into partial fractions
2 + 5v − 3v 2
4
1

(see chapter 2) gives: (1 + 3v) (2 − v)

dv 3v 2 − 1
3v 2 − 1 − 2v 2 v 2 − 1
=
−v=
=
dx
2v
2v
2v

Resolving

x

Hence,

1 dx x
Integrating both sides gives:
1
2v dv = dx 2 −1 v x

i.e.

= ln + ln 2

Show that the solution of the dy differential equation: x 2 − 3y 2 + 2x y = 0 is: dx √ y = x (8x + 1), given that y = 3 when x = 1.

dv 2 + 12v − 10v 2
=
−v dx 7 − 7v

7 − 7v dv =
2 + 5v − 3v 2

= ln + c

Problem 4.

2 + 12v − 10v 2
7 − 7v

(iv) Separating the variables gives:

Hence,

2x − y x Hence, the particular solution is:

x + 3y
i.e. ln x (ii) Let y = vx then

=

+ ln

y
= ln + c x 4
When x = 1, y = 0, thus: ln 1 + ln 2 = ln 1 + c
3
from which, c = ln 2

(i) Rearranging gives:

x

+ ln 2 −

4 x + 3y ln 3 x or

Using the procedure of section 47.2:

=

y gives: x

1
4
dv =

(1 + 3v) (2 − v)

4 ln(1 + 3v) + ln(2 − v) = ln x + c
3

1 dx x

Hence,

2v

v2 − 1

dv =

i.e. ln(v 2 − 1) = ln x + c

Homogeneous first order differential equations
(v) Replacing v by ln y gives: x

2. Solve: (9x y − 11x y)

y2
− 1 = ln x + c, x2 which is the general solution.
When y = 3, x = 1, thus: ln from which, c = ln 8

1 3
13y − 3x ln 5 13 x y2
− 1 = ln x + ln 8 = ln 8x x2 by the laws of logarithms

y2 y2 Hence,
− 1 = 8x i.e. 2 = 8x + 1 and x2 x
2 = x 2 (8x + 1) y √
i.e.
y = x (8x + 1)

Now try the following exercise
Exercise 181 Further problems on homogeneous first order differential equations 1. Solve the differential equation: x y 3 d y = (x 4 + y 4 )dx. y 4 = 4x 4 (ln x + c)

− ln

y−x x = ln x + c

9
− 1 = ln 1 +c
1

Hence, the particular solution is: ln dy
= 11y 2 − 16x y + 3x 2 . dx 3. Solve the differential equation: dy 2x
= x + 3y, given that when x = 1, y = 1. dx (x + y)2 = 4x 3
4. Show that the solution of the differential equady
= x 2 + y 2 can be expressed as: tion: 2x y dx x = K(x 2 − y 2 ), where K is a constant.
5. Determine the particular solution of d y x 3 + y3
, given that x = 1 when y = 4.
=
dx x y2 y 3 = x 3 (3 ln x + 64)
6. Show that the solution of the differential equad y y 3 − x y 2 − x 2 y − 5x 3 tion: is of the
=
dx x y 2 − x 2 y − 2x 3 form: y2
4y
y − 5x
+
+ 18 ln
2x 2 x x when x = 1 and y = 6.

= ln x + 42,

455

Chapter 48

Linear first order differential equations 48.1

Integrating both sides gives:

Introduction

dy
An equation of the form
+ P y = Q, where P and dx Q are functions of x only is called a linear differential equation since y and its derivatives are of the first degree. dy
+ P y = Q is obtained by
(i) The solution of dx multiplying throughout by what is termed an integrating factor. dy (ii) Multiplying
+ P y = Q by say R, a function dx of x only, gives:
R

dy
+ RPy = RQ dx (1)

dR
=
R

R=e
i.e. R = Ae

dR dx dR
= RP, then separating the variables
(iv) If dx dR gives = P dx.
R

=e

P dx+c
P dx ,

Ae

dy dx P dx

i.e. e

P dx

P dx c

e

where A = ec = a constant.

(v) Substituting R = Ae

P dx

in equation (1) gives:

+ Ae

P dx

P y = Ae

+e

P dx

Py = e

dy dx P dx

P dx

Q

Q (2)

(vi) The left hand side of equation (2) is d ye dx P dx

which may be checked by differentiating ye P dx with respect to x, using the product rule.
(vii) From equation (2),

which is the same as the left hand side of equation (1), when R is chosen such that
RP =

P dx + c

from which,

(iii) The differential coefficient of a product Ry is obtained using the product rule, dy dR d (Ry) = R
+y
,
i.e.
dx dx dx

P dx i.e. ln R =

d ye dx

P dx

=e

P dx

Q

Integrating both sides gives: ye (viii) e

P dx

P dx

=

e

P dx

Q dx

is the integrating factor.

(3)

Linear first order differential equations

48.2 Procedure to solve differential equations of the form dy + Py = Q dx (i) Rearrange the differential equation into the form dy + P y = Q, where P and Q are functions of x. dx (ii) Determine

Problem 2. Show that the solution of the equation dy y
3 −x2
+ 1 =− is given by y =
, given dx x
2x
x = 1 when y = 1.
Using the procedure of Section 48.2:
(i) Rearranging gives:

(iv) Substitute e

P dx

P dx .

into equation (3).

(v) Integrate the right hand side of equation (3) to give the general solution of the differential equation. Given boundary conditions, the particular solution may be determined.

dy
+
dx

1 x y = −1, which is

dy
1
of the form
+ P y = Q, where P = and dx x
Q = −1. (Note that Q can be considered to be
−1x 0 , i.e. a function of x).

P dx.

(iii) Determine the integrating factor e

457

(ii)

1 dx = ln x. x P dx =

(iii) Integrating factor e P dx = eln x = x (from the definition of logarithm).
(iv) Substituting into equation (3) gives:

48.3 Worked problems on linear first order differential equations
1 dy
Problem 1. Solve
+ 4y = 2 given the x dx boundary conditions x = 0 when y = 4.

yx =

(v) Hence the general solution is: yx =

dy
(i) Rearranging gives
+ 4x y = 2x, which is of the dx dy form + P y = Q where P = 4x and Q = 2x. dx Pdx =

4xdx = 2x 2 .

yx =

P dx

(iii) Integrating factor e

−x 2
+c
2

−1
+ c, from
When x = 1, y = 1, thus 1 =
2
3 which, c =
2
Hence the particular solution is:

Using the above procedure:

(ii)

x(−1) dx

2

= e2x .

i.e.

−x 2 3
+
2
2

2yx = 3 − x 2 and y =

(iv) Substituting into equation (3) gives:
2

Problem 3. Determine the particular solution of dy − x + y = 0, given that x = 0 when y = 2. dx 2

ye2x =

e2x (2x) dx

(v) Hence the general solution is:
2

2

ye2x = 1 e2x + c,
2

Using the procedure of Section 48.2:

by using the substitution u = 2x 2 When x = 0, y = 4, thus 4e0 = 1 e0 + c, from which, c = 7 .
2
2
Hence the particular solution is
2

2

ye2x = 1 e2x +
2

7
2

or y = 1 + 7 e−2x or y = 1 1 +7e−2x
2
2
2
2

3 − x2
2x

2

dy
(i) Rearranging gives
+ y = x, which is of the dx dy form + P, = Q, where P = 1 and Q = x. dx (In this case P can be considered to be 1x 0 , i.e. a function of x).
(ii)

P dx = 1dx = x.

(iii) Integrating factor e

P dx = e x .

458 Higher Engineering Mathematics
(iv) Substituting in equation (3) gives: ye x =
(v)

Using the procedure of Section 48.2:

e x (x) dx

(4)

e x (x) dx is determined using integration by parts (see Chapter 43).

dy
(i) Rearranging gives
− (tan θ)y = sec θ, which is dθ dy of the form
+ P y = Q where P = −tan θ and dθ Q = sec θ.
P dx = − tan θdθ = − ln(sec θ)

(ii)

xe x dx = xex − e x + c
Hence from equation (4): ye x = xe x − e x + c, which is the general solution.
When x = 0, y = 2 thus 2e0 = 0 − e0 + c, from which, c = 3.
Hence the particular solution is:

= ln(sec θ)−1 = ln(cos θ).
(iii) Integrating factor e P dθ = eln(cosθ) = cos θ
(from the definition of a logarithm).
(iv) Substituting in equation (3) gives:

yex = xex − ex + 3 or y = x − 1 + 3e−x

y cos θ =

Now try the following exercise

i.e.

Solve the following differential equations. dy =3− y dx dy
= x(1 − 2y)
2.
dx dy 3. t
−5t = −y dt 4.

5.
6.



c x y cos θ = θ + 1 or y = (θ + 1) sec θ

y = 1 + ce−x
2

Problem 5.
(a) Find the general solution of the equation

y =3+

x

y=

2

5t c
+
2 t

dy
+ 1 = x 3 − 2y, given x = 1 when dx 47 x3 x y =3 y= − +
5 3 15x 2

x

1 dy
+ y =1 x dx dy + x = 2y dx y cos θ =

(v) Integrating gives: y cos θ = θ + c, which is the general solution. When θ = 0, y = 1, thus
1 cos0 = 0 +c, from which, c = 1.
Hence the particular solution is:

Exercise 182 Further problems on linear first order differential equations

1.

cos θ(sec θ) dθ

y = 1 +ce−x y= 2 /2

x 1
+ + ce2x
2 4

(x − 2)
(b)

(a)

Given the boundary conditions that y = 5 when x = −1, find the particular solution of the equation given in (a).
Using the procedure of Section 48.2:
(i) Rearranging gives: dy 3(x − 1)
1
+ y= dx (x + 1)(x − 2)
(x − 2) which is of the form dy 3(x − 1)
+ P y = Q, where P = dx (x + 1)(x − 2)
1
and Q =
(x − 2)

48.4 Further worked problems on linear first order differential equations Problem 4. Solve the differential equation dy = sec θ + y tan θ given the boundary conditions dθ y = 1 when θ = 0.

d y 3(x − 1)
+
y =1 dx (x + 1)

(ii)

3(x − 1) dx, which
(x + 1)(x − 2) integrated using partial fractions.
P dx =

is

Linear first order differential equations
Let

3x − 3
(x + 1)(x − 2)

Now try the following exercise

A
B
+
(x + 1) (x − 2)
A(x − 2) + B(x + 1)

(x + 1)(x − 2)

Exercise 183 Further problems on linear first order differential equations



from which, 3x − 3 = A(x − 2) + B(x + 1)
When x = −1,
−6 = −3 A, from which, A = 2

3 = 3B, from which, B = 1
3x − 3 dx (x + 1)(x − 2)

= 2 ln (x + 1) + ln (x − 2)
= ln [(x + 1)2 (x − 2)]
(iii) Integrating factor
P dx

= eln[(x+1)

2 (x−2)]

= (x + 1)2 (x − 2)

(iv) Substituting in equation (3) gives:
2

y(x + 1) (x − 2)
=

(x + 1)2 (x − 2)

=

dy
2
=
− y show dx x + 2
2
that the particular solution is y = ln (x + 2), x given the boundary conditions that x = −1 when y = 0.

3. Given the equation x

1
2
dx
+
x +1 x −2

=

e


+ sec t (t sin t + cos t )θ = sec t , given dt 1 t = π when θ = 1. θ = (sin t − π cos t ) t 2. t

When x = 2,

Hence

In problems 1 and 2, solve the differential equations. π dy
= 1 − 2y, given y = 1 when x = .
1. cot x dx 4
[y = 1 + cos2 x]
2

1 dx x −2

(x + 1)2 dx

(v) Hence the general solution is:

4. Show that the solution of the differential equation 4 dy − 2(x + 1)3 = y dx
(x + 1) is y = (x + 1)4 ln (x + 1)2 , given that x = 0 when y = 0.
5. Show that the solution of the differential equation dy
+ ky = a sin bx dx is given by: y= a
(k sin bx − b cos bx) k2 + b2
+

y(x + 1)2 (x − 2) = 1 (x + 1)3 + c
3
(b) When x = −1, y = 5 thus 5(0)(−3) = 0 + c, from which, c = 0.
Hence y(x + 1)2 (x − 2) = 1 (x + 1)3
3
i.e. y =

(x + 1)3
3(x + 1)2 (x − 2)

and hence the particular solution is y= (x + 1)
3(x − 2)

k 2 + b2 + ab −kx e , k2 + b2

given y = 1 when x = 0. dv 6. The equation
= −(av + bt ), where a and dt 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. v= b bt b − + u − 2 e−at a2 a a 459

460 Higher Engineering Mathematics
7. In an alternating current circuit containing resistance R and inductance L the current di i is given by: Ri + L = E 0 sin ωt . Given dt i = 0 when t = 0, show that the solution of the equation is given by: i= R2

E0
(R sin ωt − ωL cosωt )
+ ω2 L 2
+

E 0 ωL e− Rt /L
R 2 + ω2 L 2

8. The concentration, C, of impurities of an oil purifier varies with time t and is described by the equation

dC a = b + dm − Cm, where a, b, d and m are dt constants. Given C = c0 when t = 0, solve the equation and show that:
C=

b
+ d (1 − e−mt /a ) + c0 e−mt /a m 9. The equation of motion of a train is given dv by: m = mk(1 − e−t ) − mcv, where v is the dt speed, t is the time and m, k and c are constants. Determine the speed, v, given v = 0 at t = 0.
1
e−t e−ct −
+
v=k c c − 1 c(c − 1)

Chapter 49

Numerical methods for first order differential equations y

49.1

Introduction

Not all first order differential equations may be solved by separating the variables (as in Chapter 46) or by the integrating factor method (as in Chapter 48). A number of other analytical methods of solving differential equations exist. However the differential equations that can be solved by such analytical methods is fairly restricted.
Where a differential equation and known boundary conditions are given, an approximate solution may be obtained by applying a numerical method. There are a number of such numerical methods available and the simplest of these is called Euler’s method.

49.2

P f (h) f (0)
0

x

h

Figure 49.1 y P f (a 1 x)

f (a)

From Chapter 8, Maclaurin’s series may be stated as: x2 2!

f (0) + · · ·

Hence at some point f (h) in Fig. 49.1: h2 f (0) + · · · f (h) = f (0) + h f (0) +
2!
If the y-axis and origin are moved a units to the left, as shown in Fig. 49.2, the equation of the same curve

y 5 f (a 1 x)

Q

Euler’s method

f (x) = f (0) + x f (0) +

y 5 f (x )

Q

0

x a h

Figure 49.2

relative to the new axis becomes y = f (a + x) and the function value at P is f (a).
At point Q in Fig. 49.2: f (a + h) = f (a) + h f (a) +

h2 f (a) + · · ·
2!

(1)

462 Higher Engineering Mathematics which is a statement called Taylor’s series.
If h is the interval between two new ordinates y0 and y1 , as shown in Fig. 49.3, and if f (a) = y0 and y1 = f (a + h), then Euler’s method states:

y1 = y0 + h(y )0 , from equation (2) y1 = 4 + (0.2)(2) = 4.4, since h = 0.2

Hence

f (a + h) = f (a) + h f (a) y1 = y0 + h ( y )0

i.e.

By Euler’s method:

(2)

At point Q in Fig. 49.4, x 1 = 1.2, y1 = 4.4 and (y )1 = 3(1 + x 1 ) − y1
i.e. ( y )1 = 3(1 + 1.2) − 4.4 = 2.2

y

y 5 f (x)

Q
P

y0

y1 = y0 + h(y )0 from equation (2)

y1

(a 1 h)

a

0

If the values of x, y and y found for point Q are regarded as new starting values of x 0, y0 and (y )0 , the above process can be repeated and values found for the point R shown in Fig. 49.5.
Thus at point R,

= 4.4 + (0.2)(2.2) = 4.84 x h

When x 1 = 1.4 and y1 = 4.84,
( y )1 = 3(1 + 1.4) − 4.84 = 2.36

Figure 49.3 y The approximation used with Euler’s method is to take only the first two terms of Taylor’s series shown in equation (1).
Hence if y0 , h and (y )0 are known, y1 , which is an approximate value for the function at Q in Fig. 49.3, can be calculated.
Euler’s method is demonstrated in the worked problems following. Q

4.4
P

4

y0

y1

x0 5 1

0

49.3 Worked problems on Euler’s method x1 5 1.2

x

h

Figure 49.4

Problem 1. Obtain a numerical solution of the differential equation

y

dy
= 3(1 + x) − y dx R
Q
P

given the initial conditions that x = 1 when y = 4, for the range x = 1.0 to x = 2.0 with intervals of 0.2.
Draw the graph of the solution. dy = y = 3(1 + x) − y dx With x 0 = 1 and y0 = 4, ( y )0 = 3(1 + 1) − 4 = 2.

y0

0

1.0

y1

x0 5 1.2

x1 5 1.4 h Figure 49.5

x

463

Numerical methods for first order differential equations
This step by step Euler’s method can be continued and it is easiest to list the results in a table, as shown in Table 49.1. The results for lines 1 to 3 have been produced above.

y

6.0

Table 49.1 x0 (y )0

y0

1.

1

4

2

2.

1.2

4.4

2.2

3.

1.4

4.84

2.36

4.

1.6

5.312

2.488

5.

1.8

5.8096

2.5904

6.

2.0

6.32768

5.0

4.0
1.0

1.2

1.4

1.6

1.8

2.0

x

Figure 49.6

For line 4, where x 0 = 1.6: y1 = y0 + h( y )0
= 4.84 + (0.2)(2.36) = 5.312 and ( y )0 = 3(1 + 1.6) − 5.312 = 2.488
For line 5, where x 0 = 1.8: y1 = y0 + h(y )0
= 5.312 + (0.2)(2.488) = 5.8096 and ( y )0 = 3(1 + 1.8) − 5.8096 = 2.5904
For line 6, where x 0 = 2.0: y1 = y0 + h(y )0

Problem 2. Use Euler’s method to obtain a numerical solution of the differential equation dy + y = 2x, given the initial conditions that at dx x = 0, y = 1, for the range x = 0(0.2)1.0. Draw the graph of the solution in this range. x = 0(0.2)1.0 means that x ranges from 0 to 1.0 in equal intervals of 0.2 (i.e. h =0.2 in Euler’s method). dy + y = 2x, dx dy
= 2x − y, i.e. y = 2x − y hence dx
If initially x 0 = 0 and y0 = 1, then
( y )0 = 2(0) − 1 = −1.
Hence line 1 in Table 49.2 can be completed with x = 0, y = 1 and y (0) = −1.

= 5.8096 + (0.2)(2.5904)
= 6.32768

Table 49.2 x0 (As the range is 1.0 to 2.0 there is no need to calculate
(y )0 in line 6). The particular solution is given by the value of y against x. dy A graph of the solution of
= 3(1 + x) − y with initial dx conditions x = 1 and y = 4 is shown in Fig. 49.6.
In practice it is probably best to plot the graph as each calculation is made, which checks that there is a smooth progression and that no calculation errors have occurred. y0

(y )0

1.

0

1

−1

2.

0.2

0.8

−0.4

3.

0.4

0.72

0.08

4.

0.6

0.736

0.464

5.

0.8

0.8288

0.7712

6.

1.0

0.98304

464 Higher Engineering Mathematics dy A graph of the solution of
+ y = 2x, with initial dx conditions x = 0 and y = 1 is shown in Fig. 49.7.

For line 2, where x 0 = 0.2 and h = 0.2: y1 = y0 + h(y ), from equation (2)
= 1 + (0.2)(−1) = 0.8

Problem 3.

and ( y )0 = 2x 0 − y0 = 2(0.2) − 0.8 = −0.4

(a)

For line 3, where x 0 = 0.4: y1 = y0 + h(y )0
= 0.8 + (0.2)(−0.4) = 0.72

Obtain a numerical solution, using
Euler’s method, of the differential equation dy = y − x, with the initial conditions that at dx x = 0, y = 2, for the range x = 0(0.1)0.5. Draw the graph of the solution.

(b) By an analytical method (using the integrating factor method of Chapter 48), the solution of the above differential equation is given by y = x + 1 + ex .

and ( y )0 = 2x 0 − y0 = 2(0.4) − 0.72 = 0.08
For line 4, where x 0 = 0.6:

Determine the percentage error at x = 0.3

y1 = y0 + h(y )0
= 0.72 + (0.2)(0.08) = 0.736

(a)

and ( y )0 = 2x 0 − y0 = 2(0.6) − 0.736 = 0.464
For line 5, where x 0 = 0.8: y1 = y0 + h(y )0

dy
= y = y − x. dx If initially x 0 = 0 and y0 = 2, then (y )0 = y0 − x 0 = 2 − 0 =2.
Hence line 1 of Table 49.3 is completed.

For line 2, where x 0 = 0.1:

= 0.736 + (0.2)(0.464) = 0.8288 and ( y )0 = 2x 0 − y0 = 2(0.8) − 0.8288 = 0.7712
For line 6, where x 0 = 1.0:

y1 = y0 + h(y )0 , from equation (2),
= 2 + (0.1)(2) = 2.2 and (y )0 = y0 − x 0

y1 = y0 + h(y )0

= 2.2 − 0.1 = 2.1

= 0.8288 + (0.2)(0.7712) = 0.98304
As the range is 0 to 1.0, ( y )0 in line 6 is not needed.

For line 3, where x 0 = 0.2: y1 = y0 + h(y )0
= 2.2 + (0.1)(2.1) = 2.41

y

and ( y )0 = y0 − x 0 = 2.41 − 0.2 = 2.21
1.0

Table 49.3 x0 y0

( y )0

1.

Figure 49.7

0.4

0.6

0.8

1.0

x

2

0.1

2.2

2.1

0.2

2.41

2.21

4.

0.3

2.631

2.331

5.

0.2

2

3.
0

0

2.

0.5

0.4

2.8641

2.4641

6.

0.5

3.11051

Numerical methods for first order differential equations
For line 4, where x 0 = 0.3:

465

Percentage error
=

actual − estimated actual =

y1 = y0 + h(y )0

2.649859 − 2.631
× 100%
2.649859

= 2.41 + (0.1)(2.21) = 2.631 and ( y )0 = y0 − x 0
= 2.631 − 0.3 = 2.331

× 100%

= 0.712%

For line 5, where x 0 = 0.4:
Euler’s method of numerical solution of differential equations is simple, but approximate. The method is most useful when the interval h is small.

y1 = y0 + h(y )0
= 2.631 + (0.1)(2.331) = 2.8641 and ( y )0 = y0 − x 0

Now try the following exercise

= 2.8641 − 0.4 = 2.4641
For line 6, where x 0 = 0.5:

Exercise 184 method y1 = y0 + h(y )0
= 2.8641 + (0.1)(2.4641) = 3.11051 dy = y − x with x = 0, y = 2
A graph of the solution of dx is shown in Fig. 49.8.
(b) If the solution of the differential equation dy = y − x is given by y = x + 1 +ex , then when dx x = 0.3, y = 0.3 + 1 +e0.3 = 2.649859.

Further problems on Euler’s

1. Use Euler’s method to obtain a numerical solution of the differential equation dy y
= 3 − , with the initial conditions that dx x 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. [see Table 49.4]
Table 49.4 x By Euler’s method, when x = 0.3 (i.e. line 4 in
Table 49.3), y = 2.631.

y

1.0

2

1.1

2.1

1.2

2.209091

1.3

2.325000

1.4

2.446154

1.5

2.571429

y

3.0

2. Obtain a numerical solution of the differen1 dy tial equation
+ 2y = 1, given the initial x dx conditions that x = 0 when y = 1, in the range x = 0(0.2)1.0
[see Table 49.5]

2.5

3. (a)
2.0
0

Figure 49.8

0.1

0.2

0.3

0.4

0.5

x

y dy +1 = −
The differential equation dx x has the initial conditions that y = 1 at x = 2. Produce a numerical solution of the differential equation in the range x = 2.0(0.1)2.5

466 Higher Engineering Mathematics
Table 49.5

49.4

x

An improved Euler method

y

0

1

0.2

1

0.4

0.96

0.6

0.8864

0.8

0.793664

1.0

In Euler’s method of Section 49, the gradient ( y )0 at
P(x0 , y0 ) in Fig. 49.9 across the whole interval h is used to obtain an approximate value of y1 at point Q. QR in
Fig. 49.9 is the resulting error in the result.

0.699692

y

(b) If the solution of the differential equation by an analytical method is given
4 x by y = − , determine the percentage x 2 error at x = 2.2
[(a) see Table 49.6 (b) 1.206%]
Table 49.6 x Q
R
P

y0
0

x1

x0

y

x

h

Figure 49.9

2.0

1

2.1

0.85

2.2

0.709524

2.3

0.577273

2.4

0.452174

2.5

0.333334

4. Use Euler’s method to obtain a numerical soludy
2y
tion of the differential equation
=x − , dx x given the initial conditions that y = 1 when x = 2, in the range x = 2.0(0.2)3.0.
If the solution of the differential equation is x2 given by y = , determine the percentage
4
error by using Euler’s method when x = 2.8
[see Table 49.7, 1.596%]
Table 49.7 x In an improved Euler method, called the Euler-Cauchy method, the gradient at P(x0 , y0 ) across half the interval is used and then continues with a line whose gradient approximates to the gradient of the curve at x 1, shown in Fig. 49.10.
Let y P1 be the predicted value at point R using Euler’s method, i.e. length RZ, where yP1 = y0 + h( y )0

The error shown as QT in Fig. 49.10 is now less than the error QR used in the basic Euler method and the calculated results will be of greater accuracy. The y Q
T

y

2.0

1.421818

2.6

1.664849

2.8

1.928718

3.0

2.213187

S

1.2

2.4

R

P

1

2.2

(3)

Z
0

x0

x0 1 1 h
2
h

Figure 49.10

x1

x

Numerical methods for first order differential equations corrected value, yC1 in the improved Euler method is given by: yC1 = y0 +

1
2 h[( y )0 + f (x1 , yP1 )]

(4)

The following worked problems demonstrate how equations (3) and (4) are used in the Euler-Cauchy method. Problem 4. Apply the Euler-Cauchy method to solve the differential equation dy = y−x dx For line 3, x 1 = 0.2 y P1 = y0 + h(y )0 = 2.205 + (0.1)(2.105)
= 2.4155 yC1 = y0 + 1 h[(y )0 + f (x 1 , y P1 )]
2
= 2.205 + 1 (0.1)[2.105 + (2.4155 − 0.2)]
2
= 2.421025
( y )0 = yC1 − x 1 = 2.421025 − 0.2 = 2.221025
For line 4, x 1 = 0.3

in the range 0(0.1)0.5, given the initial conditions that at x = 0, y = 2. dy = y = y−x dx Since the initial conditions are x 0 = 0 and y0 = 2 then (y )0 = 2 − 0 = 2. Interval h = 0.1, hence x 1 = x 0 + h = 0 + 0.1 = 0.1.
From equation (3), y P1 = y0 + h(y )0 = 2 + (0.1)(2) = 2.2

y P1 = y0 + h(y )0
= 2.421025 + (0.1)(2.221025)
= 2.6431275 yC1 = y0 + 1 h[(y )0 + f (x 1 , y P1 )]
2
= 2.421025 + 1 (0.1)[2.221025
2
+ (2.6431275 − 0.3)]
= 2.649232625
(y )0 = yC1 − x 1 = 2.649232625 − 0.3

From equation (4), yC1 = y0 + 1 h[(y )0 + f (x 1 , y P1 )]
2
= y0 +

467

1
2 h[(y )0 + (y P1

− x 1 )], in this case

1
= 2 + 2 (0.1)[2 + (2.2 − 0.1)] = 2.205

(y )1 = yC1 − x 1 = 2.205 − 0.1 = 2.105
If we produce a table of values, as in Euler’s method, we have so far determined lines 1 and 2 of Table 49.8.
The results in line 2 are now taken as x 0 , y0 and (y )0 for the next interval and the process is repeated.

= 2.349232625
For line 5, x 1 = 0.4 y P1 = y0 + h(y )0
= 2.649232625 + (0.1)(2.349232625)
= 2.884155887 yC1 = y0 + 1 h[(y )0 + f (x 1 , y P1 )]
2
= 2.649232625 + 1 (0.1)[2.349232625
2
+ (2.884155887 − 0.4)]

Table 49.8 x y

y

1.

0

2

2

2.

0.1

2.205

2.105

3.

0.2

2.421025

2.221025

4.

0.3

2.649232625

2.349232625

5.

0.4

2.89090205

2.49090205

6.

0.5

3.147446765

= 2.89090205
(y )0 = yC1 − x 1 = 2.89090205 − 0.4
= 2.49090205
For line 6, x 1 = 0.5 y P1 = y0 + h(y )0
= 2.89090205 + (0.1)(2.49090205)
= 3.139992255

468 Higher Engineering Mathematics
Table 49.10

yC1 = y0 + 1 h[(y )0 + f (x 1 , y P1 )]
2

x

Error in
Euler method

= 2.89090205 + 1 (0.1)[2.49090205
2

Error in
Euler-Cauchy method

0

=

0.472%

0.0156%

0.712%

0.0236%

0.959%

0.0319%

0.5

1.214%

0.0405%

Problem 5. Obtain a numerical solution of the differential equation dy = 3(1 + x) − y dx 2.649858808 − 2.631
× 100%
2.649858808

in the range 1.0(0.2)2.0, using the Euler-Cauchy method, given the initial conditions that x = 1 when y = 4.

= 0.712%

This is the same as Problem 1 on page 462, and a comparison of values may be made.

% error with Euler-Cauchy method
=

0.00775%

0.4

actual − estimated
× 100% actual =

0.234%

0.3

Problem 4 is the same example as Problem 3 and
Table 49.9 shows a comparison of the results, i.e. it compares the results of Tables 49.3 and 49.8. dy = y − x may be solved analytically by the intedx grating factor method of Chapter 48 with the solution y = x + 1 +ex . Substituting values of x of 0, 0.1, 0.2, . . . give the exact values shown in Table 49.9.
The percentage error for each method for each value of x is shown in Table 49.10. For example when x = 0.3,
% error with Euler method

0

0.2

= 3.147446765

0

0.1

+ (3.139992255 − 0.5)]

dy
= y = 3(1 + x) − y i.e. y = 3 + 3x − y dx 2.649858808 − 2.649232625
× 100%
2.649858808

x 0 = 1.0, y0 = 4 and h = 0.2

= 0.0236%

(y )0 = 3 + 3x 0 − y0 = 3 + 3(1.0) − 4 = 2

This calculation and the others listed in Table 49.10 show the Euler-Cauchy method to be more accurate than the Euler method.

x 1 = 1.2 and from equation (3),

Table 49.9 x 1.

0

2.

Euler method y Euler-Cauchy method y Exact value y = x + 1 + ex

2

2

2

0.1

2.2

2.205

2.205170918

3.

0.2

2.41

2.421025

2.421402758

4.

0.3

2.631

2.649232625

2.649858808

5.

0.4

2.8641

2.89090205

2.891824698

6.

0.5

3.11051

3.147446765

3.148721271

Numerical methods for first order differential equations
(y )1 = 3 + 3x 1 − y P1

y P1 = y0 + h(y )0 = 4 + 0.2(2) = 4.4

= 3 + 3(1.6) − 5.351368

yC1 = y0 + 1 h[(y )0 + f (x 1 , y P1 )]
2

= 2.448632

= y0 + 1 h[(y )0 + (3 + 3x 1 − y P1 )]
2
= 4 + 1 (0.2)[2 + (3 + 3(1.2) − 4.4)]
2
= 4.42
(y )1 = 3 + 3x 1 − y P1 = 3 + 3(1.2) − 4.42 = 2.18
Thus the first two lines of Table 49.11 have been completed. For line 3, x 1 = 1.4

For line 5, x 1 = 1.8 y P1 = y0 + h(y )0 = 5.351368 + 0.2(2.448632)
= 5.8410944 yC1 = y0 + 1 h[(y )0 + (3 + 3x 1 − y P1 )]
2
= 5.351368 + 1 (0.2)[2.448632
2
+ (3 + 3(1.8) − 5.8410944)]

y P1 = y0 + h(y )0 = 4.42 + 0.2(2.18) = 4.856 yC1 = y0 + 1 h[(y )0 + (3 + 3x 1 − y P1 )]
2
= 4.42 +

469

= 5.85212176
(y )1 = 3 + 3x 1 − y P1

1
2 (0.2)[2.18

= 3 + 3(1.8) − 5.85212176

+ (3 + 3(1.4) − 4.856)]
= 4.8724

= 2.54787824
For line 6, x 1 = 2.0

(y )1 = 3 + 3x 1 − y P1 = 3 + 3(1.4) − 4.8724
= 2.3276

y P1 = y0 + h(y )0
= 5.85212176 + 0.2(2.54787824)
= 6.361697408

For line 4, x 1 = 1.6 y P1 = y0 + h(y )0 = 4.8724 + 0.2(2.3276)

1 yC1 = y0 + 2 h[(y )0 + (3 + 3x 1 − y P1 )]

= 5.85212176 + 1 (0.2)[2.54787824
2

= 5.33792

+ (3 + 3(2.0) − 6.361697408)]

yC1 = y0 + 1 h[(y )0 + (3 + 3x 1 − y P1 )]
2

= 6.370739843

= 4.8724 + 1 (0.2)[2.3276
2
+ (3 + 3(1.6) − 5.33792)]
= 5.351368
Table 49.11 x0 y0

y0

Problem 6. Using the integrating factor method the solution of the differential equation dy = 3(1 + x) − y of Problem 5 is y = 3x + e1 − x . dx When x = 1.6, compare the accuracy, correct to 3 decimal places, of the Euler and the Euler-Cauchy methods. 1.

1.0

4

2

2.

1.2

4.42

2.18

3.

1.4

4.8724

2.3276

When x = 1.6, y = 3x + e1−x = 3(1.6) + e1−1.6 =
4.8 + e−0.6 = 5.348811636.
From Table 49.1, page 463, by Euler’s method, when x = 1.6, y = 5.312

4.

1.6

5.351368

2.448632

% error in the Euler method

5.

1.8

5.85212176

2.54787824

6.

2.0

6.370739847

=

5.348811636 − 5.312
× 100%
5.348811636

= 0.688%

470 Higher Engineering Mathematics
From Table 49.11 of Problem 5, by the Euler-Cauchy method, when x = 1.6, y = 5.351368

for the range x = 0 to x = 0.5 in increments of 0.1, given the initial conditions that when x = 0, y = 1

% error in the Euler-Cauchy method
=

5.348811636 − 5.351368
× 100%
5.348811636

= −0.048%
The Euler-Cauchy method is seen to be more accurate than the Euler method when x = 1.6.

(b) The solution of the differential equation in part (a) is given by y = 2ex − x − 1.
Determine the percentage error, correct to
3 decimal places, when x = 0.4
[(a) see Table 49.13 (b) 0.117%]

Now try the following exercise
Table 49.13
Exercise 185 Further problems on an improved Euler method

x
0

1

1

1. Apply the Euler-Cauchy method to solve the differential equation

0.1

1.11

1.21

0.2

1.24205

1.44205

0.3

1.398465

1.698465

0.4

1.581804

1.981804

0.5

1.794893

dy y = 3− dx x for the range 1.0(0.1)1.5, given the initial conditions that x = 1 when y = 2.
[see Table 49.12]

y

y

Table 49.12 x y

y

1.0

2

1

1.1

2.10454546

1.08677686

1.2

2.216666672

1.152777773

1.3

2.33461539

1.204142008

1.4

2.457142859

1.2448987958

1.5

4. Obtain a numerical solution of the differential equation 2.583333335

2. Solving the differential equation in Problem 1 by the integrating factor method gives
3
1 y = x + . Determine the percentage error,
2
2x correct to 3 significant figures, when x = 1.3 using (a) Euler’s method (see Table 49.4, page 465), and (b) the Euler-Cauchy method.

1 dy
+ 2y = 1 x dx using the Euler-Cauchy method in the range x = 0(0.2)1.0, given the initial conditions that x = 0 when y = 1.
[see Table 49.14]

Table 49.14 x y

y

0

1

0

0.2

0.99

−0.196

[(a) 0.412% (b) 0.000000214%]

0.4

0.958336

−0.3666688

3. (a) Apply the Euler-Cauchy method to solve the differential equation dy −x = y dx 0.6

0.875468851

−0.450562623

0.8

0.784755575

−0.45560892

1.0

0.700467925

Numerical methods for first order differential equations
49.5

The Runge-Kutta method

The Runge-Kutta method for solving first order differential equations is widely used and provides a high degree of accuracy. Again, as with the two previous methods, the Runge-Kutta method is a step-by-step process where results are tabulated for a range of values of x.
Although several intermediate calculations are needed at each stage, the method is fairly straightforward.
The 7 step procedure for the Runge-Kutta method, without proof, is as follows: dy = f (x, y) given the
To solve the differential equation dx initial condition y = y0 at x = x 0 for a range of values of x = x 0 (h)x n :
1. Identify x 0 , y0 and h, and values of x 1 , x 2, x 3 , . . ..

Using the above procedure:
1.

2. k1 = f (x 0 , y0 ) = f (0, 2); dy since
= y − x, f (0, 2) =2 − 0 = 2 dx 3. k2 = f x 0 + h , y0 + h k1
2
2
= f 0+

4.

= 2.1025 − 0.05 = 2.0525
5.

= f (0.1, 2.20525)
= 2.20525 − 0.1 = 2.10525
6.

in the range 0(0.1)0.5, given the initial conditions that at x = 0, y = 2.

k4 = f (x 0 + h, y0 + hk3 )
= f (0 + 0.1, 2 + 0.1(2.0525))

h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }
6

dy
=y−x
dx

0.1
0.1
,2+
(2.05)
2
2

= f (0.05, 2.1025)

6. Use the values determined from steps 2 to 5 to evaluate: Problem 7. Use the Runge-Kutta method to solve the differential equation:

h h k3 = f x 0 + , y0 + k2
2
2
= f 0+

h h 4. Evaluate k3 = f x n + , yn + k2
2
2

Thus, step 1 is given, and steps 2 to 5 are intermediate steps leading to step 6. It is usually most convenient to construct a table of values.
The Runge-Kutta method is demonstrated in the following worked problems.

0.1
0.1
,2+
(2)
2
2

= f (0.05, 2.1) = 2.1 − 0.05 = 2.05

h h 3. Evaluate k2 = f x n + , yn + k1
2
2

7. Repeat steps 2 to 6 for n = 1, 2, 3, . . .

x 0 = 0, y0 = 2 and since h = 0.1, and the range is from x = 0 to x = 0.5, then x 1 = 0.1, x 2 = 0.2, x 3 =
0.3, x 4 = 0.4, and x 5 = 0.5

Let n =0 to determine y1 :

2. Evaluate k1 = f(x n , yn ) starting with n =0

5. Evaluate k4 = f (x n + h, yn + hk3 )

471

h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 } and when
6
n = 0: h y1 = y0 + {k1 + 2k2 + 2k3 + k4 }
6
= 2+

0.1
{2 + 2(2.05) + 2(2.0525)
6
+ 2.10525}

= 2+

0.1
{12.31025} = 2.205171
6

A table of values may be constructed as shown in
Table 49.15. The working has been shown for the first two rows.

472 Higher Engineering Mathematics
Table 49.15 n xn

k1

k2

0

0

1

0.1

2.0

2.05

2.0525

2.10525

2.205171

2

0.2

2.105171

2.160430

2.163193

2.221490

2.421403

3

0.3

2.221403

2.282473

2.285527

2.349956

2.649859

4

0.4

2.349859

2.417339

2.420726

2.491932

2.891824

5

k4

yn

0.5

2.491824

2.566415

2.570145

2.648838

3.148720

2

Let n =1 to determine y2 :
2.

k3

k1 = f (x 1 , y1 ) = f (0.1, 2.205171); since dy = y − x, f (0.1, 2.205171) dx = 2.205171 − 0.1 = 2.105171

h y2 = y1 + {k1 + 2k2 + 2k3 + k4 }
6
= 2.205171+

+ 2(2.163193) + 2.221490}
= 2.205171 +

3.

h h k2 = f x 1 + , y1 + k1
2
2
0.1
0.1
, 2.205171 +
(2.105171)
= f 0.1 +
2
2
= f (0.15, 2.31042955)
= 2.31042955 − 0.15 = 2.160430

4.

h h k3 = f x 1 + , y1 + k2
2
2
= f 0.1 +

0.1
0.1
, 2.205171 +
(2.160430)
2
2

= f (0.15, 2.3131925) = 2.3131925 − 0.15
= 2.163193
5.

k4 = f (x 1 + h, y1 + hk3 )
= f (0.1 + 0.1, 2.205171 + 0.1(2.163193))

0.1
{2.105171+2(2.160430)
6

0.1
{12.973907} = 2.421403
6

This completes the third row of Table 49.15. In a similar manner y3 , y4 and y5 can be calculated and the results are as shown in Table 49.15. Such a table is best produced by using a spreadsheet, such as Microsoft Excel.
This problem is the same as problem 3, page 459 which used Euler’s method, and problem 4, page 461 which used the improved Euler’s method, and a comparison of results can be made. dy = y − x may be solved
The differential equation dx analytically using the integrating factor method of chapter 48, with the solution: y = x + 1 +ex
Substituting values of x of 0, 0.1, 0.2, . . ., 0.5 will give the exact values. A comparison of the results obtained by Euler’s method, the Euler-Cauchy method and the
Runga-Kutta method, together with the exact values is shown in Table 49.16.
It is seen from Table 49.16 that the Runge-Kutta method is exact, correct to 5 decimal places.

= f (0.2, 2.421490)
= 2.421490 − 0.2 = 2.221490
6.

h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }
6
and when n = 1:

Problem 8. Obtain a numerical solution of the dy differential equation:
= 3(1 + x) − y in the dx range 1.0(0.2)2.0, using the Runge-Kutta method, given the initial conditions that x = 1.0 when y = 4.0.

Numerical methods for first order differential equations

473

Table 49.16

x

Euler’s method y

Euler-Cauchy method y

Runge-Kutta method y

Exact value y = x +1 + e x

0

2

2

2

2

0.1

2.2

2.205

2.205171

2.205170918

0.2

2.41

2.421025

2.421403

2.421402758

0.3

2.631

2.649232625

2.649859

2.649858808

0.4

2.8641

2.89090205

2.891824

2.891824698

0.5

3.11051

3.147446765

3.148720

3.148721271

Using the above procedure:
1.

x 0 = 1.0, y0 = 4.0 and since h = 0.2, and the range is from x = 1.0 to x = 2.0, then x 1 = 1.2, x 2 = 1.4, x 3 = 1.6, x 4 = 1.8, and x 5 = 2.0

Let n = 0 to determine y1 :
2.

k1 = f (x 0 , y0 ) = f (1.0, 4.0); since dy = 3(1 + x) − y, dx f (1.0, 4.0) = 3(1 + 1.0) − 4.0 = 2.0

h h 3. k2 = f x 0 + , y0 + k1
2
2

6.

2.

k1 = f (x 1 , y1 ) = f (1.2, 4.418733); since dy = 3(1 + x) − y, f (1.2, 4.418733) dx = 3(1 + 1.2) − 4.418733 = 2.181267

3.

h h k2 = f x 1 + , y1 + k1
2
2
= f 1.2 +

0.2
0.2
, 4.0 +
(2.1)
2
2

0.2
0.2
, 4.418733 +
(2.181267)
2
2

= f (1.3, 4.636860)

= f (1.1, 4.21)

= 3(1 + 1.3) − 4.636860 = 2.263140

= 3(1 + 1.1) − 4.21 = 2.09
5. k4 = f (x 0 + h, y0 + hk3 )

when

Let n = 1 to determine y2 :

= f (1.1, 4.2) = 3(1 + 1.1) − 4.2 = 2.1

= f 1.0 +

and

h y1 = y0 + {k1 + 2k2 + 2k3 + k4 }
6
0.2
{2.0 + 2(2.1) + 2(2.09) + 2.182}
= 4.0 +
6
0.2
{12.562} = 4.418733
= 4.0 +
6
A table of values is compiled in Table 49.17. The working has been shown for the first two rows.

0.2
0.2
= f 1.0 +
, 4.0 +
(2)
2
2

h h 4. k3 = f x 0 + , y0 + k2
2
2

h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }
6
n = 0:

4.

h h k3 = f x 1 + , y1 + k2
2
2
0.2
0.2
, 4.418733 +
(2.263140)
2
2

= f (1.0 + 0.2, 4.1 + 0.2(2.09))

= f 1.2 +

= f (1.2, 4.418)

= f (1.3, 4.645047) = 3(1 + 1.3) − 4.645047

= 3(1 + 1.2) − 4.418 = 2.182

= 2.254953

474 Higher Engineering Mathematics
Table 49.17 n 0

1.0

1

1.2

2.0

2.1

2.09

2.182

4.418733

2

1.4

2.181267

2.263140

2.254953

2.330276

4.870324

3

1.6

2.329676

2.396708

2.390005

2.451675

5.348817

4

1.8

2.451183

2.506065

2.500577

2.551068

5.849335

5

5.

xn

k1

k2

2.0

2.550665

2.595599

2.591105

2.632444

6.367886

k4 = f (x 1 + h, y1 + hk3 )
= f (1.4, 4.869724) = 3(1 + 1.4) − 4.869724
= 2.330276 h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }
6
n =1:

and

when

h
{k1 + 2k2 + 2k3 + k4 }
6
0.2
= 4.418733 +
{2.181267 + 2(2.263140)
6
+ 2(2.254953) + 2.330276}

y2 = y1 +

= 4.418733 +

k4

yn
4.0

= f (1.2 + 0.2, 4.418733 + 0.2(2.254953))

6.

k3

0.2
{13.547729} = 4.870324
6

This completes the third row of Table 49.17. In a similar manner y3 , y4 and y5 can be calculated and the

results are as shown in Table 49.17. As in the previous problem such a table is best produced by using a spreadsheet. This problem is the same as Problem 1, page 462 which used Euler’s method, and Problem 5, page 468 which used the Euler-Cauchy method, and a comparison of results can be made. dy = 3(1 + x) − y may be
The differential equation dx solved analytically using the integrating factor method of chapter 48, with the solution: y = 3x +e1−x
Substituting values of x of 1.0, 1.2, 1.4, . . ., 2.0 will give the exact values. A comparison of the results obtained by Euler’s method, the Euler-Cauchy method and the
Runga-Kutta method, together with the exact values is shown in Table 49.18.
It is seen from Table 49.18 that the Runge-Kutta method is exact, correct to 4 decimal places.

Table 49.18

x

Euler’s method y

Euler-Cauchy method y

Runge-Kutta method y

Exact value y = 3x + e1−x

1.0

4

4

4

4

1.2

4.4

4.42

4.418733

4.418730753

1.4

4.84

4.8724

4.870324

4.870320046

1.6

5.312

5.351368

5.348817

5.348811636

1.8

5.8096

5.85212176

5.849335

5.849328964

2.0

6.32768

6.370739847

6.367886

6.367879441

Numerical methods for first order differential equations
The percentage error in the Runge-Kutta method when, say, x = 1.6 is:
5.348811636 − 5.348817
×100% = −0.0001%
5.348811636

Table 49.20 n xn

yn

Exercise 186 Further problems on the
Runge-Kutta method
1. Apply the Runge-Kutta method to solve the dy y differential equation:
= 3 − for the range dx x
1.0(0.1)1.5, given that the initial conditions that x = 1 when y = 2.
[see Table 49.19]
Table 49.19 yn 1.0

1

0.2

0.980395

2

0.4

0.926072

3

0.6

0.848838

0.8

0.763649

5

Now try the following exercise

0

4

From Problem 6, page 469, when x = 1.6, the percentage error for the Euler method was 0.688%, and for the Euler-Cauchy method −0.048%. Clearly, the
Runge-Kutta method is the most accurate of the three methods. 0

1.0

0.683952

dy y +1 = −
3. (a) The differential equation: dx x has the initial conditions that y = 1 at x = 2. Produce a numerical solution of the differential equation, correct to 6 decimal places, using the Runge-Kutta method in the range x = 2.0(0.1)2.5
(b) If the solution of the differential equation by an analytical method is given by:
4 x y = − determine the percentage error x 2 at x = 2.2
[(a) see Table 49.21 (b) no error]

n

xn

0

1.0

2.0

1

1.1

2.104545

2

1.2

2.216667

n

xn

3

1.3

2.334615

0

2.0

1.0

4

1.4

2.457143

1

2.1

0.854762

5

1.5

2.533333

2

2.2

0.718182

3

2.3

0.589130

4

2.4

0.466667

5

2.5

0.340000

2. Obtain a numerical solution of the differential
1 dy equation: + 2y = 1 using the Rungex dx
Kutta method in the range x = 0(0.2)1.0, given the initial conditions that x = 0 when y = 1.
[see Table 49.20]

Table 49.21 yn 475

Revision Test 14
This Revision Test covers the material contained in Chapters 46 to 49. The marks for each question are shown in brackets at the end of each question.
1.
2.

3.

Determine the equation of the curve which satisfies dy the differential equation 2x y = x 2 + 1 and which dx passes through the point (1, 2).
(5)

6.

dV
+V = E dt Solve the equation for V given that when time t = 0, V = 0.

given the initial conditions that x = 1 when y = 3, for the range x = 1.0 (0.1) 1.5
(b) Apply the Euler-Cauchy method to the differential equation given in part (a) over the same range. (c)

(b) Evaluate voltage V when E =50 V, C =10 μF,
R = 200 k and t = 1.2 s.
(14)
4.

5.

Show that the solution to the differential equation: d y x 2 + y2
=
is of the form
4x
dx y √

3y 2 = x 1 − x 3 given that y = 0 when x = 1.
(12)
Show that the solution to the differential equation dy + (x sin x + cos x)y = 1 x cos x dx (a) Use Euler’s method to obtain a numerical solution of the differential equation: y dy
= + x2 − 2 dx x

A capacitor C is charged by applying a steady voltage E through a resistance R. The p.d. between the plates, V , is given by the differential equation:
CR

(a)

is given by: x y = sin x + k cos x where k is a constant. (11)

dy
+ x 2 = 5 given
Solve the differential equation: x dx that y = 2.5 when x = 1.
(4)

Apply the integrating factor method to solve the differential equation in part (a) analytically. (d) Determine the percentage error, correct to 3 significant figures, in each of the two numerical methods when x = 1.2
(30)
7.

Use the Runge-Kutta method to solve the difdy y ferential equation:
= + x 2 − 2 in the range dx x
1.0(0.1)1.5, given the initial conditions that at x = 1, y = 3. Work to an accuracy of 6 decimal places. (24)

Chapter 50

Second order differential equations of the form d2 y dy a dx 2 + b dx + cy = 0
50.1

Introduction

d2 y dy An equation of the form a 2 + b + cy = 0, where dx dx a, b and c are constants, is called a linear second order differential equation with constant coefficients. When the right-hand side of the differential equation is zero, it is referred to as a homogeneous differential equation. When the right-hand side is not equal to zero (as in Chapter 51) it is referred to as a non-homogeneous differential equation.
There are numerous engineering examples of second order differential equations. Three examples are:
(i)

dq
1
d2q
+ q = 0, representing an equaL 2 +R dt dt
C
tion for charge q in an electrical circuit containing resistance R, inductance L and capacitance C in series. ds d2 s
(ii) m 2 + a + ks = 0, defining a mechanical sysdt dt tem, where s is the distance from a fixed point after t seconds, m is a mass, a the damping factor and k the spring stiffness.
P
d2 y
+
y = 0, representing an equation for the
(iii)
2 dx EI deflected profile y of a pin-ended uniform strut

of length l subjected to a load P. E is Young’s modulus and I is the second moment of area. d2 d
If D represents and D2 represents 2 then the above dx dx equation may be stated as
(aD2 + bD + c)y = 0. This equation is said to be in
‘D-operator’ form. dy d2 y
= Amem x and 2 = Am 2 em x .
If y = Aem x then dx dx d2 y dy Substituting these values into a 2 + b + cy = 0 dx dx gives: a(Am 2 em x ) + b(Amem x ) + c(Aem x ) = 0
i.e.

Aem x (am 2 + bm + c) = 0

Thus y = Aem x is a solution of the given equation provided that (am 2 + bm +c) = 0. am 2 + bm + c = 0 is called the auxiliary equation, and since the equation is a quadratic, m may be obtained either by factorizing or by using the quadratic formula. Since, in the auxiliary equation, a, b and c are real values, then the equation may have either
(i) two different real roots (when b2 > 4ac) or
(ii) two equal real roots (when b 2 = 4ac) or
(iii) two complex roots (when b2 < 4ac).

478 Higher Engineering Mathematics
Using the above procedure:

50.2

Procedure to solve differential equations of the form d2 y dy a 2 + b + cy = 0 dx dx

(a) Rewrite the differential equation dy d2 y a 2 +b
+ cy = 0 dx dx as (aD2 + bD + c)y = 0

(b) Substitute m for D and solve the auxiliary equation am 2 + bm + c = 0 for m.
(c) If the roots of the auxiliary equation are:
(i) real and different, say m = α and m = β, then the general solution is y = Aeαx + Beβx
(ii) real and equal, say m = α twice, then the general solution is y = (Ax + B)eαx
(iii) complex, say m = α ± jβ, then the general solution is y = eαx {A cosβx + B sinβx}
(d) Given boundary conditions, constants A and B, may be determined and the particular solution of the differential equation obtained.
The particular solutions obtained in the worked problems of Section 50.3 may each be verified by substid2 y dy and 2 into the original tuting expressions for y, dx dx equation. dy d2 y
+ 5 − 3y = 0 in D-operator form is dx 2 dx d
(2D2 + 5D − 3)y = 0, where D ≡ dx (b) Substituting m for D gives the auxiliary equation
(a) 2

2m 2 + 5m − 3 = 0.
Factorising gives: (2m − 1)(m + 3) = 0, from which, m = 1 or m = −3.
2
(c) Since the roots are real and different the general
1
solution is y = Ae 2 x + Be−3x .
(d) When x = 0, y = 4, hence 4= A+ B

Since

y = Ae 2 x + Be−3x

(1)

1

dy 1 1 x
= Ae 2 − 3Be−3x dx 2 dy When x = 0,
=9
dx
1
(2) thus 9 = A − 3B
2
Solving the simultaneous equations (1) and (2) gives A = 6 and B = −2. then Hence the particular solution is y = 6e 2 x − 2e−3x
1

Problem 2. Find the general solution of d2 y dy 9 2 − 24 + 16y = 0 and also the particular dt dt solution given the boundary conditions that when dy t = 0, y =
= 3. dt Using the procedure of Section 50.2:

50.3

Worked problems on differential equations of d2 y dy the form a 2 + b + cy = 0 dx dx

Problem 1. Determine the general solution of d2 y dy 2 2 + 5 − 3y = 0. Find also the particular dx dx dy solution given that when x = 0, y = 4 and
= 9. dx d2 y dy − 24 + 16y = 0 in D-operator form is dt 2 dt 2 − 24D +16)y = 0 where D ≡ d
(9D
dt
(b) Substituting m for D gives the auxiliary equation
9m 2 − 24m + 16 =0.
(a) 9

Factorizing gives: (3m − 4)(3m − 4) = 0, i.e. m = 4 twice.
3
(c) Since the roots are real and equal, the general
4
solution is y = (At +B)e 3 t .

2

Second order differential equations of the form a d y + b dy + cy = 0 dx dx 2
(d) When t = 0, y = 3 hence 3 = (0 + B)e0, i.e. B = 3.

then

4

Since y = (At + B)e 3 t
4
dy
4 4t
= (At + B) e 3 + Ae 3 t , by the dt 3 product rule.

dy
When t = 0,
=3
dt
4
thus 3= (0 + B) e0 + Ae0
3

by the product rule,
−3x

=e

4

y = (−t + 3)e 3 t or y = (3 − t)e 3 t
Problem 3. Solve the differential equation d2 y dy + 6 + 13y = 0, given that when x = 0, y = 3
2
dx dx dy and = 7. dx dy
+ 6 + 13y = 0 in D-operator form is dx 2 dx d
(D2 + 6D + 13)y = 0, where D ≡ dx dy
= 7, dx hence 7 =e0 [(2B − 3 A) cos 0 − (2 A + 3B) sin 0]
i.e. 7 =2B − 3 A, from which, B = 8, since A = 3.
Hence the particular solution is y = e−3x(3 cos 2x + 8 sin 2x)
Since, from Chapter 17, page 165, a cos ωt + b sin ωt = R sin(ωt + α), where a R = (a 2 + b2) and α = tan −1 then b 3 cos2x + 8 sin 2x
=
=

Using the procedure of Section 50.2: d2 y

[(2B − 3 A) cos 2x
− (2 A + 3B) sin 2x]

When x = 0,

4
i.e. 3 = B + A from which, A = −1, since
3
B = 3.
Hence the particular solution is

(a)

dy
= e−3x (−2 A sin 2x + 2B cos 2x) dx − 3e−3x (A cos 2x + B sin 2x),

then

4

=




(32 + 82 ) sin(2x + tan−1 3 )
8
73 sin(2x + 20.56◦ )
73 sin(2x + 0.359)

Thus the particular solution may also be expressed as

y = 73 e−3x sin(2x + 0.359)

(b) Substituting m for D gives the auxiliary equation m 2 + 6m + 13 =0.
Now try the following exercise
Using the quadratic formula:
[(6)2 − 4(1)(13)]
2(1)

−6 ± (−16)
=
2

m=

−6 ±

−6 ± j 4
= −3 ± j 2
i.e. m=
2
(c)

Since the roots are complex, the general solution is

Exercise 187 Further problems on differential equations of the form dy d2 y a 2 + b + cy = 0 dx dx
In Problems 1 to 3, determine the general solution of the given differential equations.
1. 6

Since y = e−3x (A cos 2x + B sin 2x)

d2 y d y

− 2y = 0 dt 2 dt y = Ae 3 t + Be− 2 t
2

y = e−3x (A cos 2x + B sin 2x)
(d) When x = 0, y = 3, hence
3 =e0 (A cos 0 + B sin 0), i.e. A = 3.

479

2. 4

1

d2θ dθ +4 +θ =0
2
dt dt 1

θ = (At + B)e− 2 t

480 Higher Engineering Mathematics

3.

d2 y dy + 2 + 5y = 0
2
dx dx [y = e−x (A cos 2x + B sin 2x)]

In Problems 4 to 9, find the particular solution of the given differential equations for the stated boundary conditions. dy d2 y
4. 6 2 + 5 − 6y = 0; when x = 0, y = 5 and dx dx
2
3 dy = −1. y = 3e 3 x + 2e− 2 x dx dy d2 y
5. 4 2 − 5 + y = 0; when t = 0, y = 1 and dt dt
1
dy
= −2. y = 4e 4 t − 3et dt d
6. (9D2 + 30D +25)y = 0, where D ≡ ; when dx dy x = 0, y = 0 and
= 2. dx 5 y = 2xe− 3 x
7.

8.

d2 x dx − 6 + 9x = 0; when t = 0, x = 2 and dt 2 dt dx
= 0.
[x = 2(1 − 3t )e3t ] dt dy d2 y
+ 6 + 13y = 0; when x = 0, y = 4 and
2
dx dx dy
= 0.
[y = 2e−3x (2 cos 2x + 3 sin 2x)] dx d
9. (4D2 + 20D + 125)θ = 0, where D ≡ ; when dt dθ t = 0, θ = 3 and
= 2.5. dt [θ = e−2.5t (3 cos 5t + 2 sin 5t )]

where x is the displacement in metres of the body from its equilibrium position after time t seconds.
Determine x in terms of t given that at time t = 0, dx x = 2m and
= 0. dt d2 x
+ m 2 x = 0 is a differAn equation of the form dt 2 ential equation representing simple harmonic motion
(S.H.M.). Using the procedure of Section 50.2:
(a)

d2 x
+ 100x = 0 in D-operator form is dt 2
(D2 + 100)x = 0.

(b) The auxiliary equation is m 2 + 100 = 0, i.e.

m 2 = −100 and m = (−100), i.e. m = ± j 10.
(c) Since the roots are complex, the general solution is x = e0 (A cos 10t + B sin 10t ),
i.e. x =(A cos 10t +B sin10t) metres
(d) When t = 0, x = 2, thus 2 = A dx = −10 A sin 10t + 10B cos 10t dt dx
When t = 0,
=0
dt thus 0 = −10 A sin 0 + 10B cos 0, i.e. B = 0
Hence the particular solution is x = 2 cos 10t metres
Problem 5. Given the differential equation d2 V
= ω2 V , where ω is a constant, show that its dt 2 solution may be expressed as:
V = 7 cosh ωt + 3 sinh ωt given the boundary conditions that when

50.4

Further worked problems on practical differential equations d2 y dy of the form a 2 + b + cy = 0 dx dx

Problem 4. The equation of motion of a body oscillating on the end of a spring is d2 x
+ 100x = 0, dt 2

t = 0, V = 7 and

dV
= 3ω. dt Using the procedure of Section 50.2:
(a)

d2 V d2 V
= ω2 V , i.e.
− ω2 V = 0 in D-operator dt 2 dt 2 d form is (D2 − ω2 )v = 0, where D ≡
.
dx

(b) The auxiliary equation is m 2 − ω2 = 0, from which, m 2 = ω2 and m = ±ω.

2

Second order differential equations of the form a d y + b dy + cy = 0 dx dx 2
(c)

Since the roots are real and different, the general solution is

(d) When t = 0, V = 7 hence 7 = A + B

t = 0,

d2 i R di
1
+
+
i = 0 in D-operator form is dt 2 L dt LC

(1)

dV
= Aωeωt − Bωe−ωt dt When

Using the procedure of Section 50.2:
(a)

V = Aeωt + Be−ωt

D2 +

d
R
1 i = 0 where D ≡
D+
L
LC
dt

(b) The auxiliary equation is m 2 +

dV
= 3ω, dt −

3ω = Aω − Bω,

thus

(2)

− 4(1)

Hence the particular solution is



V = 5eωt + 2e−ωt

and

sinh ωt = 1 (eωt − e−ωt )
2
cosh ωt =

+ e−ωt )

then sinh ωt + cosh ωt = eωt and 2

200
±
0.20

200
0.20

m=

1 ωt
2 (e

cosh ωt − sinh ωt = e−ωt from Chapter 5.

Hence the particular solution may also be written as
V = 5(sinh ωt + cosh ωt )
+ 2(cosh ωt − sinh ωt )
i.e. V = (5 + 2) cosh ωt + (5 − 2) sinh ωt
i.e. V = 7 cosh ωt + 3 sinh ωt
Problem 6. The equation d2i R di
1
+
+
i =0 dt 2 L dt LC represents a current i flowing in an electrical circuit containing resistance R, inductance L and capacitance C connected in series. If R = 200 ohms,
L =0.20 henry and C = 20 ×10−6 farads, solve the equation for i given the boundary conditions that di when t = 0, i = 0 and = 100. dt 1
LC

When R = 200, L =0.20 and C = 20 ×10−6, then

From equations (1) and (2), A = 5 and B = 2

Since

R
1
m+
=0
L
LC

2

R
L

R
±
L

Hence m =
3 = A− B

i.e.

481

=

(c)

2



4
(0.20)(20 × 10−6 )

2
−1000 ±
2



0

= −500

Since the two roots are real and equal (i.e. −500 twice, since for a second order differential equation there must be two solutions), the general solution is i = (At +B)e−500t .

(d) When t = 0, i = 0, hence B = 0 di = (At + B)(−500e−500t ) + (e−500t )(A), dt by the product rule di = 100, thus 100 =−500B + A dt i.e. A = 100, since B = 0
When t = 0,

Hence the particular solution is i = 100te−500t
Problem 7. The oscillations of a heavily damped pendulum satisfy the differential equation dx d2 x
+ 6 + 8x = 0, where x cm is the dt 2 dt displacement of the bob at time t seconds.
The initial displacement is equal to +4 cm and the dx is 8 cm/s. Solve the initial velocity i.e. dt equation for x.

482 Higher Engineering Mathematics
Using the procedure of Section 50.2:

from

2. A body moves in a straight line so that its distance s metres from the origin after time d2 s t seconds is given by 2 + a2 s = 0, where a dt is a constant. Solve the equation for s given ds 2π that s = c and
= 0 when t =
.
dt a [s = c cos at ]

(c) Since the roots are real and different, the general solution is x =Ae−2t + Be−4t .

3. The motion of the pointer of a galvanometer about its position of equilibrium is represented by the equation

dx d2 x
+ 6 + 8x = 0 in D-operator form is
(a)
2 dt dt
2 + 6D + 8)x = 0, where D ≡ d .
(D
dt
(b) The auxiliary equation is m 2 + 6m + 8 =0.
Factorising gives: (m + 2)(m + 4) = 0, which, m = −2 or m = −4.

(d) Initial displacement means that time t = 0. At this instant, x = 4.
Thus 4 = A + B

I

If I , the moment of inertia of the pointer about its pivot, is 5 ×10−3, K , the resistance due to friction at unit angular velocity, is 2 × 10−2 and F, the force on the spring necessary to produce unit displacement, is 0.20, solve the equation for θ in terms of t given that when dθ t = 0, θ = 0.3 and
= 0. dt [θ = e−2t (0.3 cos 6t + 0.1 sin 6t )]

(1)

Velocity, dx = −2 Ae−2t − 4Be−4t dt dx
= 8 cm/s when t = 0, dt thus

8 = −2 A − 4B

(2)

From equations (1) and (2),
A = 12 and B = −8
Hence the particular solution is x = 12e−2t − 8e−4t

4. Determine an expression for x for a differential dx d2 x equation 2 + 2n + n 2 x = 0 which repredt dt sents a critically damped oscillator, given that dx at time t = 0, x = s and
= u. dt [x = {s + (u + ns)t }e−nt ]
5.

i.e. displacement, x = 4(3e−2t − 2e−4t ) cm

Now try the following exercise
Exercise 188 Further problems on second order differential equations of the form dy d2 y a 2 + b + cy = 0 dx dx
1. The charge, q, on a capacitor in a certain electrical circuit satisfies the differential equadq d2 q tion 2 + 4 + 5q = 0. Initially (i.e. when dt dt dq t = 0), q = Q and
= 0. Show that the dt charge in the circuit can be expressed as:

q = 5 Qe−2t sin(t + 0.464).

dθ d2θ +K
+ Fθ = 0.
2
dt dt di 1 d2i L 2 + R + i = 0 is an equation repredt dt C senting current i in an electric circuit. If inductance L is 0.25 henry, capacitance C is 29.76 ×10−6 farads and R is 250 ohms, solve the equation for i given the boundary di conditions that when t = 0, i = 0 and = 34. dt 1 −160t
− e−840t e i=
20

6. The displacement s of a body in a damped mechanical system, with no external forces, satisfies the following differential equation:
2

ds d2 s
+ 6 + 4.5s = 0
2
dt dt where t represents time. If initially, when ds t = 0, s = 0 and
= 4, solve the differential dt 3 equation for s in terms of t .
[s = 4t e− 2 t ]

Chapter 51

Second order differential equations of the form d2 y dy a dx 2 + b dx
51.1 Complementary function and particular integral
If in the differential equation a d2 y dy +b
+ cy = f (x)
2
dx dx (1)

the substitution y = u + v is made then: a + cy = f (x)
The general solution, u, of equation (3) will contain two unknown constants, as required for the general solution of equation (1). The method of solution of equation (3) is shown in Chapter 50. The function u is called the complementary function (C.F.).
If the particular solution, v, of equation (2) can be determined without containing any unknown constants then y = u +v will give the general solution of equation (1).
The function v is called the particular integral (P.I.).
Hence the general solution of equation (1) is given by:

d(u + v) d2(u + v)
+b
+ c(u + v) = f (x) dx 2 dx y = C.F. + P.I.

Rearranging gives: a 51.2

du dv d2u d2 v
+b
+ cu + a 2 + b +cv dx 2 dx dx dx = f (x)

If we let a d2 v dx 2

+b

dv
+ cv = f (x) dx (i) Rewrite the given differential equation as
(aD2 + bD+ c)y = f (x).
(2)

then du d2 u a 2 +b
+ cu = 0 dx dx

Procedure to solve differential equations of the form d2 y dy a 2 + b + cy = f (x) dx dx

(3)

(ii) Substitute m for D, and solve the auxiliary equation am 2 + bm +c = 0 for m.
(iii) Obtain the complementary function, u, which is achieved using the same procedure as in
Section 50.2(c), page 478.

484 Higher Engineering Mathematics
Table 51.1 Form of particular integral for different functions
Type

Straightforward cases
‘Snag’ cases
Try as particular integral: Try as particular integral:

(a) f (x) = a constant

v=k

(b) f (x) = polynomial (i.e.

v = a + bx + cx 2 + · · ·

f (x) = L + M x + N x 2 +

v = kx (used when C.F. contains a constant)

See problem 1, 2
3

···

where any of the coefficients may be zero)
(c) f (x) = an exponential function
(i.e. f (x) =

v = keax

(i) v = kxeax (used when eax

Aeax )

4, 5

appears in the C.F.)
(ii) v = kx 2 eax (used when eax and xeax both appear in the C.F.)

(d) f (x) = a sine or cosine function v = A sin px + B cos px

v = x(A sin px + B cos px)

(i.e. f (x) = a sin px + b cos px,

7, 8

(used when sin px and/or

where a or b may be zero)

6

cos px appears in the C.F.)

(e) f (x) = a sum e.g.
(i)

f (x) = 4x 2 − 3 sin 2x

(ii)

f (x) = 2 − x + e3x

9
(i)

v = ax 2 + bx + c
+ d sin 2x + e cos 2x

(f ) f (x) = a product e.g. f (x) = 2ex

(ii) v = ax + b + ce3x v = ex (A sin 2x + B cos 2x)

10

cos 2x

(iv) To determine the particular integral, v, firstly assume a particular integral which is suggested by f (x), but which contains undetermined coefficients. Table 51.1 gives some suggested substitutions for different functions f (x).
(v) Substitute the suggested P.I. into the differential equation (aD2 + bD +c)v = f (x) and equate relevant coefficients to find the constants introduced. (vi) The general solution is given by y = C.F. + P.I., i.e. y = u +v.
(vii) Given boundary conditions, arbitrary constants in the C.F. may be determined and the particular solution of the differential equation obtained.

51.3

Worked problems on differential equations of the d2 y dy form a 2 + b
+ cy = f (x) dx dx where f (x) is a constant or polynomial Problem 1. Solve the differential equation d2 y d y
+
− 2y = 4. dx 2 dx
Using the procedure of Section 51.2:
(i)

d2 y d y
+
− 2y = 4 in D-operator form is dx 2 dx
(D2 + D − 2)y = 4.

2

Second order differential equations of the form a d y + b dy + cy = f (x) dx dx 2
(ii) Substituting m for D gives the auxiliary equation m 2 + m − 2 = 0. Factorizing gives: (m − 1)
(m + 2) = 0, from which m = 1 or m = −2.
(iii) Since the roots are real and different, the C.F., u = Aex + Be−2x .
(iv) Since the term on the right hand side of the given equation is a constant, i.e. f (x) = 4, let the P.I. also be a constant, say v = k (see
Table 51.1(a)).
(v) Substituting v = k into (D2 + D − 2)v = 4 gives (D2 + D − 2)k = 4. Since D(k) = 0 and
D2 (k) = 0 then −2k = 4, from which, k = −2.
Hence the P.I., v = −2.
(vi) The general solution is given by y = u + v, i.e. y = Aex + Be−2x − 2.
Problem 2. Determine the particular solution of d2 y dy the equation 2 − 3 = 9, given the boundary dx dx dy conditions that when x = 0, y = 0 and
= 0. dx d2 y dy − 3 =9
2
dx dx (D2 − 3D)y = 9.

in

D-operator

Hence the particular solution is y = −1 + 1e3x − 3x,
i.e. y = e3x − 3x − 1

Problem 3. Solve the differential equation d2 y dy 2 2 − 11 + 12y = 3x − 2. dx dx
Using the procedure of Section 51.2: dy d2 y
(i) 2 2 − 11 + 12y = 3x − 2 dx dx form is

in

D-operator

(2D2 − 11D + 12)y = 3x − 2.
(ii) Substituting m for D gives the auxiliary equation 2m 2 − 11m + 12 =0. Factorizing gives:
(2m − 3)(m − 4) = 0, from which, m = 3 or
2
m = 4.
(iii) Since the roots are real and different, the C.F.,
3

u =Ae 2 x + Be4x
(iv) Since f (x) = 3x − 2 is a polynomial, let the P.I., v = ax + b (see Table 51.1(b)).

Using the procedure of Section 51.2:
(i)

485

form

is

(ii) Substituting m for D gives the auxiliary equation m 2 − 3m =0. Factorizing gives: m(m − 3) = 0, from which, m = 0 or m = 3.

(v) Substituting v = ax + b into
(2D2 − 11D +12)v = 3x − 2 gives:
(2D2 − 11D + 12)(ax + b) = 3x − 2,
i.e. 2D2 (ax + b) − 11D(ax + b)
+ 12(ax + b) = 3x − 2

(iii) Since the roots are real and different, the C.F., u = Ae0 + Be3x , i.e. u = A +Be3x .

i.e.

(iv) Since the C.F. contains a constant (i.e. A) then let the P.I., v = kx (see Table 51.1(a)).

Equating the coefficients of x gives: 12a = 3, from which, a = 1 .
4

(v) Substituting v = kx into (D2 − 3D)v = 9 gives
(D2 − 3D)kx = 9.
D(kx) = k and D2 (kx) = 0.
Hence (D2 − 3D)kx = 0 −3k = 9, from which, k = −3.
Hence the P.I., v = −3x.
(vi) The general solution is given by y = u + v, i.e. y = A +Be3x −3x.
(vii) When x = 0, y = 0, thus 0 = A + Be0 − 0, i.e.
0= A+ B
(1)
dy dy = 3Be3x − 3;
= 0 when x = 0, thus dx dx
0 = 3Be0 − 3 from which, B = 1. From equation (1), A = −1.

0 − 11a + 12ax + 12b = 3x − 2

Equating the constant terms gives:
−11a + 12b = −2.
i.e. −11

1
4

+ 12b = −2 from which,

1
11 3
= i.e. b =
4 4
16
1
1
Hence the P.I., v = ax + b = x +
4
16
(vi) The general solution is given by y = u + v, i.e.
12b = −2 +

3
1
1 y = Ae 2 x + Be4x + x +
4
16

486 Higher Engineering Mathematics
Now try the following exercise

6. In a galvanometer the deflection θ satisfies d2θ dθ the differential equation 2 + 4 + 4 θ = 8. dt dt
Solve the equation for θ given that when t = 0, dθ θ=
= 2.
[θ = 2(t e−2t + 1)] dt Exercise 189 Further problems on differential equations of the form dy d2 y a 2 +b
+ cy = f (x) where f (x) is a dx dx constant or polynomial.
In Problems 1 and 2, find the general solutions of the given differential equations.
1. 2

51.4

dy d2 y
+ 5 − 3y = 6 dx 2 dx 1

y = Ae 2 x + Be−3x − 2
2. 6

d2 y dy + 4 − 2y = 3x − 2 dx 2 dx 1

y = Ae 3 x + Be−x − 2 − 3 x
2
In Problems 3 and 4 find the particular solutions of the given differential equations. d2 y d y
3. 3 2 +
− 4y = 8; when x = 0, y = 0 and dx dx dy = 0. dx 4

y = 2 (3e− 3 x + 4ex ) − 2
7
d2 y dy − 12 + 4y = 3x − 1; when x = 0,
2
dx dx dy
4
y = 0 and
=−
dx
3

4. 9

y = − 2+

3
4x

2 e3x +2+

3
4x

5. The charge q in an electric circuit at time t satd2 q dq 1 isfies the equation L 2 + R
+ q = E, dt dt C where L, R, C and E are constants. Solve the equation given L = 2H , C = 200 ×10−6 F and
E = 250 V, when (a) R = 200 and (b) R is negligible. Assume that when t = 0, q = 0 and dq =0 dt ⎡

5
1
1 e−50t ⎥

t+
(a) q =

20
2
20




1
(b) q = (1 − cos 50t )
20

Worked problems on differential equations of the form d2 y dy a 2 +b
+ cy = f (x) where dx dx f (x) is an exponential function

Problem 4. Solve the equation d2 y dy − 2 + y = 3e4x given the boundary
2
dx dx dy conditions that when x = 0, y = − 2 and
=41
3
3
dx
Using the procedure of Section 51.2:
(i)

d2 y dy − 2 + y = 3e4x in D-operator form is dx 2 dx (D2 − 2D + 1)y = 3e4x .

(ii) Substituting m for D gives the auxiliary equation m 2 − 2m + 1 =0. Factorizing gives:
(m − 1)(m − 1) = 0, from which, m = 1 twice.
(iii) Since the roots are real and equal the C.F., u = (Ax + B)ex .
(iv) Let the particular integral, v = ke4x
Table 51.1(c)).

(see

(v) Substituting v = ke4x into
(D2 − 2D + 1)v = 3e4x gives:
(D2 − 2D + 1)ke4x = 3e4x
i.e. D2 (ke4x ) − 2D(ke4x ) + 1(ke4x ) = 3e4x
i.e.

16ke4x − 8ke4x + ke4x = 3e4x

Hence 9ke4x = 3e4x , from which, k = 1
3
Hence the P.I., v = ke4x = 1 e4x .
3
(vi) The general solution is given by y = u + v, i.e. y = (Ax + B)ex + 1 e4x .
3
(vii) When x = 0, y = − 2 thus
3

2

487

Second order differential equations of the form a d y + b dy + cy = f (x) dx dx 2
− 2 = (0 + B)e0 + 1 e0 , from which, B = −1.
3
3 dy x + ex (A) + 4 e4x .
= (Ax + B)e
3
dx dy 1
13
4
When x = 0,
= 4 , thus
= B + A+ dx 3
3
3 from which, A = 4, since B = −1.
Hence the particular solution is:

3

= 2 ke 2 x

i.e.

(v) The

3

(see Table 51.1(c), snag case (i)).
3

(iv) Substituting v = kxe 2 x into (2D2 − D − 3)v =
3
3
(2D2 − D − 3)kxe 2 x = 5e 2 x .
3
3 2x
2e

3

+ e 2 x (k), by the product rule,

=
3

D2 kxe 2 x

3 ke 2 x 3 x
2
3

= D ke 2 x
3

= ke 2 x

+1
3
2x

3

general

solution

is

Problem 6. Solve

y = u + v,

i.e.

d2 y dy − 4 + 4y = 3e2x . dx 2 dx Using the procedure of Section 51.2: dy d2 y
− 4 + 4y = 3e2x in D-operator form is dx 2 dx (D2 − 4D +4)y = 3e2x .

(ii) Substituting m for D gives the auxiliary equation m 2 − 4m + 4 = 0. Factorizing gives:
(m − 2)(m − 2) = 0, from which, m = 2 twice.
(iii) Since the roots are real and equal, the C.F., u =(Ax + B)e2x .
(iv) Since e2x and xe2x both appear in the C.F. let the
P.I., v = kx 2 e2x (see Table 51.1(c), snag case (ii)).
(v) Substituting v = kx 2 e2x into (D2 − 4D + 4)v =
3e2x gives: (D2 − 4D + 4)(kx 2 e2x ) = 3e2x

= 2ke2x (x 2 + x)
D2 (kx 2 e2x ) = D[2ke2x (x 2 + x)]

3
2

9
4x

3

D(kx 2 e2x ) = (kx 2 )(2e2x ) + (e2x )(2kx)

+1

= (2ke2x )(2x + 1) + (x 2 + x)(4ke2x )

+
= ke 2 x

3

3

3
3
y = Ae 2 x + Be−x + xe 2 x .

(i)

(iii) Since e 2 x appears in the C.F. and in the right hand side of the differential equation, let the

= (kx)

3

3

3

3

3

+ 6ke 2 x − 3 xke 2 x − ke 2 x
2

Hence the P.I., v = kxe 2 x = xe 2 x .

the C.F., u =Ae 2 x + Be−x .

D kxe 2 x

3

3

3
− 3y = 5e 2 x

gives:

+1

Equating coefficients of e 2 x gives: 5k = 5, from which, k = 1.

(ii) Substituting m for D gives the auxiliary equation 2m 2 − m − 3 = 0. Factorizing gives:
(2m − 3)(m + 1) = 0, from which, m = 3 or
2
m = −1. Since the roots are real and different then

3
5e 2 x

3
9
2x
2 kxe

3

dy

in D-operator form is dx 2 dx
3
(2D2 − D − 3)y = 5e 2 x .

P.I.,

3
2x

− 3kxe 2 x = 5e 2 x

Using the procedure of Section 51.2:

3 v = kxe 2 x

3

− ke 2 x
3

Problem 5. Solve the differential equation
3
d2 y d y
2 2−
− 3y = 5e 2 x . dx dx

(i) 2

+3

− 3 kxe 2 x = 5e 2 x

y = (4x − 1)ex + 1 e4x
3

d2 y

9
4x

+3
3

Hence (2D2 − D − 3) kxe 2 x

3
2x

+1

3
3
2x
2 ke

= 2ke2x (4x + 1 + 2x 2 )
Hence (D2 − 4D + 4)(kx 2 e2x )
= [2ke2x (4x + 1 + 2x 2 )]
− 4[2ke2x (x 2 + x)] + 4[kx 2 e2x ]
= 3e2x

488 Higher Engineering Mathematics from which, 2ke2x = 3e2x and k = 3
2

51.5

Hence the P.I., v = kx2 e2x = 3 x2 e2x .
2
(vi) The general solution, y = u + v, i.e.
3
y = (Ax + B)e2x + 2 x2e2x

Now try the following exercise

Worked problems on differential equations of the d2 y dy form a 2 + b + cy= f (x) dx dx where f (x) is a sine or cosine function Problem 7. Solve the differential equation d2 y dy 2 2 + 3 − 5y = 6 sin 2x. dx dx

Exercise 190 Further problems on differential equations of the form dy d2 y a 2 + b +cy = f (x) where f (x) is an dx dx exponential function

Using the procedure of Section 51.2:
In Problems 1 to 4, find the general solutions of the given differential equations.
1.

d2 y d y

− 6y = 2ex dx 2 dx y = Ae3x + Be−2x − 1 ex
3

2.

d2 y
+ 9y = 26e2x dx 2
[ y = A cos 3x + B sin 3x + 2e2x ]

4. 9

t dy d2 y
− 6 + y = 12e 3 dt 2 dt 1

1

y = (At + B)e 3 t + 2 t 2e 3 t
3
In problems 5 and 6 find the particular solutions of the given differential equations. dy 1 d2 y
+ 9 − 2y = 3ex ; when x = 0, y =
2
dx dx 4 dy and
= 0. dx 1
1
5 e−2x − e 5 x + ex y= 44
4

5. 5

6.

(ii) The auxiliary equation is 2m 2 + 3m −5 = 0, from which, (m − 1)(2m + 5) = 0,

dy d2 y
− 3 − 4y = 3e−x
2
dx dx y = Ae4x + Be−x − 3 xe−x
5

3.

d2 y dy + 3 − 5y = 6 sin 2x in D-operator form dx 2 dx is (2D2 + 3D − 5)y = 6 sin 2x

(i) 2

dy d2 y
− 6 + 9y = 4e3t ; when t = 0, y = 2 dt 2 dt dy and =0
[ y = 2e3t (1 − 3t + t 2)] dt i.e. m = 1 or m = −5
2
(iii) Since the roots are real and different the C.F.,
5
u = Aex + Be− 2 x.
(iv) Let the P.I.,
Table 51.1(d)).

v = A sin 2x + B cos 2x

(see

(v) Substituting v = A sin 2x + B cos 2x into
(2D2 + 3D −5)v = 6 sin 2x gives:
(2D2 + 3D−5)(A sin 2x + B cos 2x) = 6 sin 2x.
D(A sin 2x + B cos 2x)
= 2 A cos 2x − 2B sin 2x
D2 (A sin 2x + B cos 2x)
= D(2 A cos 2x − 2B sin 2x)
= −4 A sin 2x − 4B cos 2x
Hence (2D2 + 3D −5)(A sin 2x + B cos 2x)
= −8 A sin 2x − 8B cos 2x + 6 A cos 2x
− 6B sin 2x − 5 A sin 2x − 5B cos 2x
= 6 sin 2x
Equating coefficient of sin 2x gives:
−13 A − 6B = 6

(1)

2

Second order differential equations of the form a d y + b dy + cy = f (x) dx dx 2
D[x(C sin 4x + D cos 4x)]

Equating coefficients of cos 2x gives:
6 A − 13B = 0

(2)

6 × (1)gives : −78 A − 36B = 36
13 × (2)gives :

by the product rule
D2 [x(C sin 4x + D cos 4x)]

−36 into equation (1) or (2)
Substituting B =
205
−78 gives A =
205
−78
36
Hence the P.I., v = sin 2x − cos 2x.
205
205
(vi) The general solution, y = u +v, i.e.
−5x
2

x

+ (C sin 4x + D cos 4x)(1),

(4)

− 205B = 36
−36
B=
205

from which,

= x(4C cos 4x − 4D sin 4x)

(3)

78 A − 169B = 0

(3) + (4)gives :

489

y = Ae + Be
2

(39 sin 2x + 18 cos 2x)
205

= x(−16C sin 4x − 16D cos 4x)
+ (4C cos 4x − 4D sin 4x)
+ (4C cos 4x − 4D sin 4x)
Hence (D2 + 16)[x(C sin 4x + D cos 4x)]
= −16Cx sin 4x −16Dx cos 4x + 4C cos 4x
− 4D sin 4x + 4C cos 4x − 4D sin 4x
+ 16Cx sin 4x + 16Dx cos 4x
= 10 cos4x,

d2 y
Problem 8. Solve 2 + 16y = 10 cos4x given dx dy y = 3 and
= 4 when x = 0. dx i.e. −8D sin 4x + 8C cos 4x = 10 cos4x
Equating coefficients of cos 4x gives:
10 5
8C = 10, from which, C = =
8 4

Using the procedure of Section 51.2:
(i)

Equating coefficients of sin 4x gives:
−8D = 0, from which, D = 0.

d2 y
+ 16y = 10 cos 4x in D-operator form is dx 2

Hence the P.I., v = x

(D + 16)y = 10 cos4x
2

(ii) The auxiliary equation is

which m = −16 = ± j 4.

m 2 + 16 = 0,

from

(iv) Since sin 4x occurs in the C.F. and in the right hand side of the given differential equation, let the P.I., v = x(C sin 4x + D cos 4x) (see
Table 51.1(d), snag case—constants C and D are used since A and B have already been used in the
C.F.).

= 10 cos 4x

5 y = A cos 4x + B sin 4x + 4 x sin 4x

(vii) When x = 0, y = 3, thus
3 = A cos 0 + B sin 0 + 0, i.e. A = 3.

i.e. u =Acos 4x + B sin4x

(D2 + 16)[x(C sin 4x + D cos 4x)]

sin 4x .

(vi) The general solution, y = u +v, i.e.

(iii) Since the roots are complex the C.F., u =e0 (A cos 4x + B sin 4x)

(v) Substituting v = x(C sin 4x + D cos 4x)
(D2 + 16)v = 10 cos 4x gives:

5
4

into

dy
= −4 A sin 4x + 4B cos 4x dx + 5 x(4 cos 4x) + 5 sin 4x
4
4
When x = 0,

dy
= 4, thus dx 4 = −4 A sin 0 + 4B cos 0 + 0 + 5 sin 0
4
i.e. 4 =4B, from which, B = 1
Hence the particular solution is y = 3 cos 4x + sin 4x + 5 x sin 4x
4

490 Higher Engineering Mathematics
Now try the following exercise given by:
Exercise 191 Further problems on differential equations of the form dy d2 y a 2 + b + cy = f (x) where f (x) is a sine or dx dx cosine function

y = e−4t (A cos 2t + B sin 2t )
15
+
(sin 4t − 8 cos4t )
13
7.

In Problems 1 to 3, find the general solutions of the given differential equations.
1. 2

d2 y d y

− 3y = 25 sin 2x dx 2 dx
3

y = Ae 2 x + Be−x
− 1 (11 sin 2x − 2 cos 2x)
5
2.

dq 1 d2q L 2 + R + q = V0 sin ωt represents the dt dt C variation of capacitor charge in an electric circuit. Determine an expression for q at time t seconds given that R = 40 ,
L =0.02 H, C = 50 × 10−6 F, V0 = 540.8 V and ω = 200 rad/s and given the boundary dq conditions that when t = 0, q = 0 and
= 4.8 dt q = (10t + 0.01)e−1000t
+ 0.024 sin 200t − 0.010 cos 200t

d2 y dy − 4 + 4y = 5 cos x dx 2 dx y = (Ax + B)e2x − 4 sin x + 3 cos x
5
5

3.

d2 y
+ y = 4 cos x dx 2

51.6

[ y = A cos x + B sin x + 2x sin x]
4. Find the particular solution of the differend2 y dy tial equation 2 − 3 − 4y = 3 sin x; when dx dx dy x = 0, y = 0 and
= 0. dx ⎤

1
(6e4x − 51e−x ) ⎥ y= ⎢
170




1
− (15 sin x − 9 cos x)
34
5. A differential equation representing the d2 y
+ n 2 y = k sin pt , motion of a body is dt 2 where k, n and p are constants. Solve the equation (given n =0 and p2 = n 2) given that when dy t = 0, y =
= 0. dt y=

k p sin pt − sin nt n 2 − p2 n 6. The motion of a vibrating mass is given by d2 y dy + 8 + 20y = 300 sin4t . Show that the
2
dt dt general solution of the differential equation is

Worked problems on differential equations of the d2 y dy form a 2 + b
+ cy = f (x) dx dx where f (x) is a sum or a product

Problem 9. Solve d2 y d y
+
− 6y = 12x − 50 sin x. dx 2 dx
Using the procedure of Section 51.2:
(i)

d2 y d y
+
− 6y = 12x − 50 sin x in D-operator dx 2 dx form is
(D2 + D − 6)y = 12x − 50 sin x

(ii) The auxiliary equation is (m 2 + m − 6) = 0, from which, (m − 2)(m + 3) = 0,
i.e. m = 2 or m = −3
(iii) Since the roots are real and different, the C.F., u = Ae2x + Be−3x .
(iv) Since the right hand side of the given differential equation is the sum of a polynomial and a sine function let the P.I. v = ax + b + c sin x + d cos x
(see Table 51.1(e)).

2

Second order differential equations of the form a d y + b dy + cy = f (x) dx dx 2
(v) Substituting v into

Using the procedure of Section 51.2:

(D2 + D −6)v = 12x − 50 sin x gives:

(i)

(D + D − 6)(ax + b + c sin x + d cos x)
2

= 12x − 50 sin x

D2 (ax + b + c sin x + d cos x)
= −c sin x − d cos x
(D2 + D − 6)(v)

= (−c sin x − d cos x) + (a + c cos x
− d sin x) − 6(ax + b + c sin x + d cos x)
= 12x − 50 sin x
Equating constant terms gives: a − 6b = 0

(1)

Equating coefficients of x gives: −6a = 12, from which, a = −2.
Hence, from (1), b = − 1
3

D-operator

(iii) Since the roots are complex, the C.F., u = ex (A cos x + B sin x).
(iv) Since the right hand side of the given differential equation is a product of an exponential and a cosine function, let the P.I., v = ex (C sin 2x + D cos 2x) (see Table 51.1(f) — again, constants C and D are used since A and B have already been used for the C.F.).
(v) Substituting v into (D2 − 2D +2)v = 3ex cos 2x gives: (D2 − 2D + 2)[ex (C sin 2x + D cos 2x)]

Equating the coefficients of cos x gives:

= 3ex cos 2x

−d + c − 6d = 0

(2)

c − 7d = 0

−c − d − 6c = −50

(≡ex {(2C + D) cos 2x

Solving equations (2) and (3) gives: c = 7 and d = 1.
Hence the P.I., υ = −2x − 1 + 7 sin x + cos x
3
(vi) The general solution, y = u +v,
i.e. y = Ae + Be

+ (C − 2D) sin 2x})

(3)

i.e. − 7c − d = −50

−3x

D(v) = ex (2C cos 2x − 2D sin 2x)
+ ex (C sin 2x + D cos 2x)

Equating the coefficients of sin x gives:

2x

in

(ii) The auxiliary equation is m 2 − 2m + 2 = 0
Using the quadratic formula,

2 ± [4 − 4(1)(2)] m= 2

2 ± −4 2 ± j 2
=
=
i.e. m = 1 ± j 1.
2
2

= a + c cos x − d sin x

i.e.

dy d2 y
− 2 + 2y = 3ex cos 2x dx 2 dx form is

(D2 − 2D + 2)y = 3ex cos 2x

D(ax + b + c sin x + d cos x)

Hence

491

− 2x

− 1 + 7 sin x + cos x
3
Problem 10. Solve the differential equation d2 y dy − 2 + 2y = 3ex cos 2x, given that when
2
dx dx dy x = 0, y = 2 and
= 3. dx D2 (v) = ex (−4C sin 2x − 4D cos 2x)
+ ex (2C cos 2x − 2D sin 2x)
+ ex (2C cos 2x − 2D sin 2x)
+ ex (C sin 2x + D cos 2x)
≡ ex {(−3C − 4D) sin 2x
+ (4C − 3D) cos 2x}
Hence (D2 − 2D + 2)v
= ex {(−3C − 4D) sin 2x
+ (4C − 3D) cos 2x}
− 2ex {(2C + D) cos 2x
+ (C − 2D) sin 2x}
+ 2ex (C sin 2x + D cos 2x)
= 3ex cos 2x

492 Higher Engineering Mathematics
Equating coefficients of ex sin 2x gives:
−3C − 4D − 2C + 4D + 2C = 0

1. 8

i.e. −3C = 0, from which, C = 0.
Equating coefficients of ex cos 2x gives:
4C − 3D − 4C − 2D + 2D = 3
i.e. −3D = 3, from which, D = −1.
Hence the P.I., υ = ex (−cos 2x).
(vi) The general solution, y = u + v, i.e.

2.

y = ex (A cos x +B sinx) − ex cos 2x

d2 y dy − 6 + y = 2x + 40 sin x dx 2 dx ⎤

x x y = Ae 4 + Be 2 + 2x + 12


8
+ (6 cos x − 7 sin x)
17

d2 y dy − 3 + 2y = 2 sin 2 θ − 4 cos 2 θ dθ 2 dθ y = Ae2θ + Beθ + 1 (sin 2 θ + cos 2 θ)
2

(vii) When x = 0, y = 2 thus
2 = e0 (A cos 0 + B sin 0)
− e0 cos 0
i.e.

When thus 2 = A − 1, from which, A = 3 dy = ex (− A sin x + B cos x) dx + ex (A cos x + B sin x)
− [ex (−2 sin 2x) + ex cos 2x] dy x = 0,
=3
dx
0 (− A sin 0 + B cos 0)
3= e
+ e0 (A cos 0 + B sin 0)
− e0 (−2 sin 0) − e0 cos 0

i.e.

3.

− 1 x − 1 x 2 + 1 e2x
2
2
4

4.

Hence the particular solution is y = ex (3 cos x + sin x) − ex cos 2x

d2 y dy − 2 + 2y = et sin t
2
dt dt t y = et (A cos t + B sin t ) − 2 et cos t

In Problems 5 to 6 find the particular solutions of the given differential equations.

3 = B + A − 1, from which,
B = 1, since A = 3

d2 y d y
+
− 2y = x 2 + e2x dx 2 dx y = Aex + Be−2x − 3
4

5.

d2 y dy − 7 + 10y = e2x + 20; when x = 0,
2
dx dx dy
1
y = 0 and
=−
dx
3
10
1
4 y = e5x − e2x − xe2x + 2
3
3
3

Now try the following exercise
Exercise 192 Further problems on second order differential equations of the form dy d2 y a 2 +b
+ cy = f (x) where f (x) is a sum or dx dx product In Problems 1 to 4, find the general solutions of the given differential equations.

d2 y d y

− 6y = 6ex cos x; when x = 0, dx 2 dx
21
dy
20
y = − and
= −6
29
dx
29


3 y = 2e− 2 x − 2e2x


3ex
+
(3 sin x − 7 cos x)
29

6. 2

Chapter 52

Power series methods of solving ordinary differential equations
52.1

Introduction

Second order ordinary differential equations that cannot be solved by analytical methods (as shown in
Chapters 50 and 51), i.e. those involving variable coefficients, can often be solved in the form of an infinite series of powers of the variable. This chapter looks at some of the methods that make this possible—by the Leibniz–
Maclaurin and Frobinius methods, involving Bessel’s and Legendre’s equations, Bessel and gamma functions and Legendre’s polynomials. Before introducing
Leibniz’s theorem, some trends with higher differential coefficients are considered. To better understand this chapter it is necessary to be able to:
(i) differentiate standard functions (as explained in
Chapters 27 and 32),

as the differential coefficient of common functions rises. dy d2 y
= a 2 eax , and so
= aeax ,
(i) If y = eax , then dx dx 2 on. If we abbreviate

dy d2 y as y , … and as y , dx dx 2

dn y as y (n) , then y = aeax , y = a 2eax , and the dx n emerging pattern gives:

y(n) = an eax

For example, if y = 3e2x , then d7 y
= y (7) = 3(27 ) e2x = 384e2x dx 7
(ii) If y = sin ax, π 2

(ii) appreciate the binomial theorem (as explained in
Chapters 7), and

y = a cos ax = a sin ax +

(iii) use Maclaurins theorem (as explained in Chapter 8).

y = −a 2 sin ax = a 2 sin(ax + π)
= a 2 sin ax +

52.2 Higher order differential coefficients as series
The following is an extension of successive differentiation (see page 296), but looking for trends, or series,


2

y = −a 3 cos x
= a 3 sin ax +

3π and so on.
2

(1)

494 Higher Engineering Mathematics
In general, y(n) = an sin ax + nπ
2

(2)

For example, if d5 y
= y (5) dx 5

π
= 35 sin 3x +
= 35 sin 3x +
2
2

y = sin 3x, then

y(n) =

y = sinh 2x, then

(iii) If y = cos ax, y = −a sin ax = a cos ax +

π
2

y = −a 2 cos ax = a 2 cos ax + y = a 3 sin ax = a 3 cos ax +

=

then

dx 6


2

y(n) = an cos ax +

and so on. nπ 2

(3)

25
{[1 + (−1)5 ] sinh 2x
2

25
{[0] sinh 2x + [2] cosh 2x}
2
= 32 cosh 2x

y = a 2 cosh ax


2
6 ) cos (2x + 3π)
= 4(2

y = a 3 sinh ax, and so on
Since cosh ax is not periodic (see graph on page
43), again it is more difficult to find a general statement for y (n) . However, this is achieved with the following general series:

= −256 cos 2x
(iv) If y = x a, y = a x a−1 , y = a(a − 1)x a−2 , y = a(a − 1)(a − 2)x a−3 , and y(n) = a(a − 1)(a − 2) . . . . . (a − n + 1) x a−n a! xa−n
(a − n)!

where a is a positive integer. d4 y
For example, if y = 2x6 , then 4 = y (4) dx 6!
= (2) x 6−4
(6 − 4)!
6 × 5 × 4× 3 × 2× 1 2
= (2) x 2×1
= 720x2
(v) If y = sinh ax, y = a cosh ax y = a sinh ax
2

y = a cosh ax, and so on

(vi) If y = cosh ax, y = a sinh ax

= y (6) = 4(26 ) cos 2x +

3

d5 y
= y (5) dx 5

=

= 4(26 ) cos (2x + π)

or y(n) =

(5)

+ [1 − (−1)5 ] cosh 2x}


2

For example, if y = 4 cos 2x, d6 y

an
{[1 +(−1)n ] sinh ax
2
+ [1 −(−1)n ] cosh ax}

For example, if

= 243 cos 3x

In general,

Since sinh ax is not periodic (see graph on page
43), it is more difficult to find a general statement for y (n) . However, this is achieved with the following general series:

(4)

y(n) =

an
{[1 − (−1)n ] sinh ax
2
+ [1 + (−1)n ] cosh ax}

(6)

1 cosh 3x,
9
7y
1 37 d (2 sinh 3x) then 7 = y (7) = dx 9 2
= 243 sinh 3x

For example, if y =

1
1
2
(vii) If y = ln ax, y = , y = − 2 , y = 3 , and so x x x on.
In general, y(n) = (−1)n−1

(n − 1)! xn For example, if y = ln 5x, then d6 y
5!
= y (6) = (−1)6−1 6 dx 6 x =−

120 x6 (7)

Power series methods of solving ordinary differential equations
1
Note that if y = ln x, y = ; if in equation (7), x 0 (0)! n = 1 then y = (−1) 1 x 1
(−1)0 = 1 and if y = then (0)!= 1 (Check that x (−1)0 = 1 and (0)! = 1 on a calculator).

52.3

Leibniz’s theorem y = uv

If

495

(8)

where u and v are each functions of x, then by using the product rule, y = uv + vu

(9)

y = uv + v u + vu + u v

Now try the following exercise

= u v + 2u v + uv

Exercise 193 Further problems on higher order differential coefficients as series

y = u v + vu + 2u v + 2v u + uv + v u
= u v + 3u v + 3u v + uv

1. (a) y (4) when y = e2x (b) y (5) when y
[(a) 16 e2x

t
= 8e2

1 t
(b) e 2 ]
4

+ 4u (1)v (3) + uv (4)

[(a) 81 sin3t (b) −1562.5 cos5θ]

2
29
sin t
38
3 t7 8
(b) 630 t ]

4. (a) y (7) when y = 2x 9 (b) y (6) when y =

1
5. (a) y (7) when y = sinh 2x
4
(b) y (6) when y = 2 sinh 3x
[(a) 32 cosh 2x (b) 1458 sinh 3x]
6. (a) y (7) when y = cosh 2x
1
(b) y (8) when y = cosh 3x
9
[(a) 128 sinh 2x (b) 729 cosh 3x]
7. (a) y (4) when y = 2ln 3θ
1
(b) y (7) when y = ln 2t
3

the n’th derivative of u decreases by 1 moving from left to right,

(b) the n’th derivative of v increases by 1 moving from left to right,
(c)

3. (a) y (8) when y = cos 2x
2
(b) y (9) when y = 3 cos t
3

(12)

From equations (8) to (12) it is seen that
(a)

2. (a) y (4) when y = sin 3t
1
(b) y (7) when y = sin 5θ
50

[(a) (9! )x 2

(11)

y (4) = u (4)v + 4u (3)v (1) + 6u (2)v (2)

Determine the following derivatives:

(a) 256 cos2x (b) −

(10)

the coefficients 1, 4, 6, 4, 1 are the normal binomial coefficients (see page 58).

In fact, (uv)(n) may be obtained by expanding (u + v)(n) using the binomial theorem (see page 59), where the
‘powers’ are interpreted as derivatives. Thus, expanding
(u + v)(n) gives: y(n) = (uv)(n) = u(n) v + nu(n−1) v (1) n(n− 1) (n−2) (2) v u
2!
n(n− 1)(n −2) (n−3) (3)
+
v +··· u 3!
+

(13)

Equation (13) is a statement of Leibniz’s theorem, which can be used to differentiate a product n times.
The theorem is demonstrated in the following worked problems. Problem 1. Determine y (n) when y = x 2 e3x .
For a product y = uv, the function taken as

(a) −

240
6
(b) 7 θ4 t

(i) u is the one whose nth derivative can readily be determined (from equations (1) to (7)),
(ii) v is the one whose derivative reduces to zero after a few stages of differentiation.

496 Higher Engineering Mathematics
Thus, when y = x 2 e3x , v = x 2 , since its third derivative is zero, and u = e3x since the nth derivative is known from equation (1), i.e. 3n eax
Using Leinbiz’s theorem (equation (13), y (n)

=u

(n)

n(n − 1) (n−2) (2) v + nu v + v u
2!
n(n − 1)(n − 2) (n−3) (3)
+
v + ··· u 3!

By Leibniz’s equation, equation (13), y (n+2)(1 + x 2 ) + n y (n+1)(2x)+

n(n − 1) (n) y (2)+ 0
2!

+ 2{y (n+1) (x) + n y (n) (1) + 0} − 3{y (n) } = 0

(n−1) (1)

i.e. (1 + x 2 )y (n+2) + 2n x y (n+1) + n(n − 1)y (n)
+ 2x y (n+1) + 2 ny (n) − 3y (n) = 0
(1 + x 2 )y (n+2) + 2(n + 1)x y (n+1)

where in this case v = x 2 , v (1) = 2x, v (2) = 2 and v (3) = 0

or

Hence, y (n) = (3n e3x )(x 2 ) + n(3n−1 e3x )(2x)

i.e. (1 + x2 )y(n+2) + 2(n + 1)xy(n+1)

n(n − 1) n−2 3x
(3 e )(2)
2!
n(n − 1)(n − 2) n−3 3x
(3 e )(0)
+
3!
= 3n−2 e3x (32 x 2 + n(3)(2x)

+ (n 2 − n + 2n − 3)y (n) = 0

+ (n2 + n − 3)y(n) = 0

+

+ n(n − 1) + 0) y(n) = e3x 3n−2 (9x2 + 6nx + n(n− 1))

i.e.

Problem 2. If x 2 y + 2x y + y = 0 show that: x y (n+2) + 2(n + 1)x y (n+1) + (n 2 + n + 1)y (n) = 0

Problem 4.

If y = x 4 sin x, then using Leibniz’s equation with u = sin x and v = x 4 gives: y (n) = sin x +

y

x +n y

(n+1)

nπ 4 x 2
(n − 1)π
2

+ n sin x +

Differentiating each term of x 2 y + 2x y + y = 0 n times, using Leibniz’s theorem of equation (13), gives: (n+2) 2

Find the 5th derivative of y = x 4 sin x.

+

n(n − 1)(n − 2)
(n − 3)π sin x +
24x
3!
2

+

+ {y (n+1) (2x) + n y (n) (2) + 0} + {y (n) } = 0

(n − 2)π n(n − 1) sin x +
12x 2
2!
2

+

n(n − 1) (n)
(2x) + y (2) + 0
2!

n(n − 1)(n − 2)(n − 3) sin x
4!
+

i.e. x 2 y (n+2) + 2n x y (n+1) + n(n − 1)y (n)
+ 2x y (n+1) + 2n y (n) + y (n) = 0

+ (n − n + 2n + 1)y or 2 (n+2)

x y

(n)

+

=0

+ 2(n + 1) x y

(5)(4)

(12x 2 ) sin x +
2
2

(n+1)

+ (n + n + 1)y
2

(n)

+

=0
+

Problem 3. Differentiate the following differential equation n times:
(1 + x 2 )y + 2x y − 3y = 0.

Since

(n − 4)π
24
2


+ 20x 3 sin(x + 2π)
2

and y (5) = x 4 sin x +

i.e. x 2 y (n+2) + 2(n + 1)x y (n+1)
2

4x 3

(5)(4)(3)
(24x) sin (x + π)
(3)(2)

π
(5)(4)(3)(2)
(24) sin x +
(4)(3)(2)
2

sin x +


2

≡ sin x +

π
≡ cos x,
2

497

Power series methods of solving ordinary differential equations sin(x + 2π) ≡ sin x, sin x + and 3π
2

≡ −cos x,

52.4 Power series solution by the
Leibniz–Maclaurin method

sin (x + π) ≡ −sin x,

then y (5) = x 4 cos x + 20x 3 sin x + 120x 2 (−cos x)
+ 240x(−sin x) + 120 cos x
i.e. y(5) = (x4 − 120x2 + 120)cos x
+ (20x3 − 240x) sin x

(i) Differentiate the given equation n times, using the Leibniz theorem of equation (13),

Now try the following exercise

(ii) rearrange the result to obtain the recurrence relation at x = 0,

Exercise 194 Further problems on
Leibniz’s theorem
Use the theorem of Leibniz in the following problems: 1. Obtain the n’th derivative of: x 2 y. x 2 y (n) + 2n x y (n−1) + n(n − 1)y (n−2)
2. If




y = x 3 e2x

find

y (n)

and hence

y (3) .



y (n) = e2x 2n−3 {8x 3 + 12nx 2


+ n(n − 1)(6x) + n(n − 1)(n − 2)} ⎦

y (3) = e2x (8x 3 + 36x 2 + 36x + 6)
3. Determine the 4th derivative of: y = 2x 3 e−x .
[ y (4) = 2e−x (x 3 − 12x 2 + 36x − 24)]
4. If y = x 3 cos x determine the 5th derivative.
[ y (5) = (60x − x 3 ) sin x +
(15x 2 − 60) cos x]
5. Find an expression for y (4) if y = e−t sin t .
[ y (4)

=

−4 e−t sin t ]

6. If y = x 5 ln 2x find y (3) .
[ y (3) = x 2 (47 + 60 ln 2x)]
7. Given 2x 2 y + x y + 3y = 0 show that
2x 2 y (n+2) + (4n + 1)x y (n+1) + (2n 2 − n +
3)y (n) = 0.
8. If y = (x 3 + 2x 2 )e2x determine an expansion for y (5).
[ y (5)

=

e2x 24 (2x 3

For second order differential equations that cannot be solved by algebraic methods, the Leibniz–Maclaurin method produces a solution in the form of infinite series of powers of the unknown variable. The following simple 5-step procedure may be used in the
Leibniz–Maclaurin method:

+ 19x 2 + 50x

+ 35)]

(iii) determine the values of the derivatives at x = 0,
i.e. find ( y)0 and ( y )0 ,
(iv) substitute in the Maclaurin expansion for y = f (x) (see page 69, equation (5)),
(v) simplify the result where possible and apply boundary condition (if given).
The Leibniz–Maclaurin method is demonstrated, using the above procedure, in the following worked problems.
Problem 5. Determine the power series solution of the differential equation: dy d2 y
+ x + 2y = 0 using Leibniz–Maclaurin’s
2
dx dx method, given the boundary conditions that at dy = 2. x = 0, y = 1 and dx Following the above procedure:
(i) The differential equation is rewritten as: y + x y + 2y = 0 and from the Leibniz theorem of equation (13), each term is differentiated n times, which gives: y (n+2) +{y (n+1) (x)+n y (n) (1)+0}+2 y (n) = 0
i.e.

y (n+2) + x y (n+1) + (n + 2) y (n) = 0
(14)

(ii) At x = 0, equation (14) becomes: y (n+2) + (n + 2) y (n) = 0 from which, y (n+2) = −(n +2) y (n)

498 Higher Engineering Mathematics
This equation is called a recurrence relation or recurrence formula, because each recurring term depends on a previous term.
(iii) Substituting n =0, 1, 2, 3, … will produce a set of relationships between the various coefficients. For n =0,

( y )0 = −2( y)0

(v) Collecting similar terms together gives: y = ( y)0 1 −


2x 2 2 × 4x 4
+
2!
4!

2 × 4 × 6x 6 2 × 4 × 6 × 8x 8
+
6!
8!

− · · · + ( y )0 x −

n =1, ( y )0 = −3( y )0 n =2, ( y (4) )0 = −4( y )0 = −4{−2( y)0 }

3x 3 3 × 5x 5
+
3!
5!


= 2 × 4( y)0 n =3, ( y (5) )0 = −5( y )0 = −5{−3( y )0 }

i.e. y = ( y)0 1 −

= 3 × 5( y )0 n =4,

( y (6) )0 = −6( y (4) )0

x4 x6 x2
+

1
1×3 3×5
+

= −6{2 × 4( y)0 }

= −2 × 4 × 6( y)0 n =5, ( y (7) )0 = −7( y (5) )0 = −7{3×5( y )0 }

+ ( y )0 ×



n =6, ( y (8) )0 = −8( y (6) )0 =

(iv) Maclaurin’s theorem from page 69 may be written as: y = ( y)0 + x( y )0 +

x2 x3 ( y )0 + ( y )0
2!
3! x 4 (4)
+
( y )0 + · · ·
4!

Substituting the above values into Maclaurin’s theorem gives: y = ( y)0 + x( y )0 +

x2
{−2( y)0 }
2!

x4 x3 +
{−3( y )0 } + {2 × 4( y)0 }
3!
4!
+

x6 x5 {3 × 5( y )0 } + {−2 × 4 ×6( y)0 }
5!
6!

+

x7
{−3 × 5 × 7( y )0 }
7!
+

x8
8!

{2 × 4 × 6 × 8( y)0 }

x8
− ···
3×5×7

x x3 x5

+
1 1×2 2×4

= −3 × 5 × 7( y )0

−8{−2 × 4 × 6( y)0}= 2 × 4 × 6×8(y)0

3 × 5 × 7x 7
+ ···
7!

x7
+···
2×4×6

The boundary conditions are that at x = 0, y = 1 dy = 2, i.e. ( y)0 = 1 and ( y )0 = 2. and dx
Hence, the power series solution of the differendy d2 y tial equation: 2 + x
+ 2y = 0 is: dx dx y = 1−
+

x2 x4 x6
+

1 1 ×3 3 ×5

x x8 x3
−··· +2

3 ×5 × 7
1 1×2
+

x5 x7 −
+···
2×4 2×4×6

Problem 6. Determine the power series solution of the differential equation: d2 y d y
+
+ x y = 0 given the boundary conditions dx 2 dx dy that at x = 0, y = 0 and
= 1, using dx Leibniz–Maclaurin’s method.
Following the above procedure:
(i) The differential equation is rewritten as: y + y + x y = 0 and from the Leibniz theorem of

Power series methods of solving ordinary differential equations equation (13), each term is differentiated n times, which gives: y (n+2)

i.e.

+y

(n+1)

+y

(n)

(x) + n y

(n−1)

(1) + 0 = 0

y (n+2) + y (n+1) + x y (n) + n y (n−1) = 0
(15)

(ii) At x = 0, equation (15) becomes: y (n+2) + y (n+1) + n y (n−1) = 0 from which, y (n+2) = −{y (n+1) + n y (n−1) }
This is the recurrence relation and applies for n ≥1
(iii) Substituting n = 1, 2, 3, . . . will produce a set of relationships between the various coefficients.
For n = 1, ( y )0 = −{( y )0 + ( y)0 } n = 2, ( y (4) )0 = −{( y )0 + 2( y )0 } n = 3, ( y (5) )0 = −{( y (4) )0 + 3( y )0 } n = 4, ( y (6) )0 = −{( y (5) )0 + 4( y )0 } n = 5, ( y (7) )0 = −{( y (6) )0 + 5( y (4) )0 } n = 6, ( y (8) )0 = −{( y (7) )0 + 6( y (5) )0 }
From the given boundary conditions, at x = 0, dy y = 0, thus ( y)0 = 0, and at x = 0,
= 1, thus dx ( y )0 = 1
From the given differential equation, y + y + x y = 0, and, at x = 0,
( y )0 + ( y )0 + (0)y = 0 from which,
( y )0 = −( y )0 = −1
Thus, ( y)0 = 0, ( y )0 = 1, ( y )0 = −1,
( y )0 = −{( y )0 + ( y)0 } = −(−1 +0) = 1
( y (4) )0 = −{( y )0 + 2( y )0 }
= −[1 + 2(1)] = −3
( y (5) )0 = −{( y (4) )0 + 3( y )0 }
= −[−3 +3(−1)] =6
( y (6) )0 = −{( y (5) )0 + 4( y )0 }
= −[6 + 4(1)] = −10
( y (7) )0 = −{( y (6) )0 + 5( y (4) )0 }
= −[−10 +5(−3)] =25

499

( y (8) )0 = −{( y (7) )0 + 6( y (5) )0 }
= −[25 +6(6)] = −61
(iv) Maclaurin’s theorem states: x2 x3 y = ( y)0 + x( y )0 + ( y )0 + ( y )0
2!
3! x 4 (4)
( y )0 + · · ·
4!
and substituting the above values into
Maclaurin’s theorem gives:
+

y = 0 + x(1) +

x2 x3 x4
{−1} + {1} + {−3}
2!
3!
4!

+

x6 x7 x5
{6} + {−10} + {25}
5!
6!
7!

x8
{−61} + · · ·
8!
(v) Simplifying, the power series solution of d2 y d y
+
the differential equation:
+ x y = 0 is dx 2 dx given by:
+

y = x−

x2 x3 3x4 6x5 10x6
+ −
+

2! 3! 4!
5!
6!
+

25x7 61x8

+···
7!
8!

Now try the following exercise
Exercise 195 Further problems on power series solutions by the Leibniz–Maclaurin method 1. Determine the power series solution of the difdy d2 y ferential equation: 2 + 2x
+ y = 0 using dx dx the Leibniz–Maclaurin method, given that at dy x = 0, y = 1 and
= 2. dx ⎤

x 2 5x 4 5 × 9x 6

⎢ y = 1 − 2! + 4! − 6!




3⎥
⎢ 5 × 9 × 13x 8
3x ⎥
⎢ +
−··· +2 x −

8!
3! ⎥







⎣ 3 × 7x 5 3 × 7 × 11x 7
+

+···
5!
7!

500 Higher Engineering Mathematics
2. Show that the power series solution of the difd2 y dy ferential equation: (x + 1) 2 + (x − 1) − dx dx
2y = 0, using the Leibniz–Maclaurin method, is given by: y = 1 + x 2 + ex given the boundary dy conditions that at x = 0, y =
= 1. dx 3. Find the particular solution of the differd2 y dy − 4y = 0 ential equation: (x 2 + 1) 2 + x dx dx using the Leibniz–Maclaurin method, given the boundary conditions that at x = 0, y = 1 dy and
= 1. dx x3 x5 x7 y = 1 + x + 2x 2 +

+
+···
2
8
16
4. Use the Leibniz–Maclaurin method to determine the power series solution for the differend2 y d y
+ x y = 1 given that tial equation: x 2 + dx dx dy at x = 0, y = 1 and
= 2. dx ⎤

x4 x6 x2
⎢ y = 1 − 22 + 22 × 42 − 22 × 42 × 62 ⎥



⎢ x3 x5




+ ··· +2 x − 2 + 2
2


3
3 ×5




7
x

− 2
+··· ⎦
3 × 52 × 72

(iv) equate coefficients of corresponding powers of the variable on each side of the equation; this enables index c and coefficients a1 , a2 , a3 , … from the trial solution, to be determined.
This introductory treatment of the Frobenius method covering the simplest cases is demonstrated, using the above procedure, in the following worked problems.
Problem 7. Determine, using the Frobenius method, the general power series solution of the d2 y d y differential equation: 3x 2 +
− y = 0. dx dx
The differential equation may be rewritten as:
3x y + y − y = 0.
(i) Let a trial solution be of the form y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·
+ ar x r + · · ·

(16)

where a0 = 0,
i.e. y = a0 x + a1 x c c+1

+ a2 x

c+2

+ a3 x c+3

+ · · · + ar x c+r + · · ·

(17)

(ii) Differentiating equation (17) gives: y = a0cx c−1 + a1 (c + 1)x c
+ a2(c + 2)x c+1 + · · ·
+ ar (c + r)x c+r−1 + · · · and y = a0c(c − 1)x c−2 + a1 c(c + 1)x c−1
+ a2 (c + 1)(c + 2)x c + · · ·

52.5 Power series solution by the
Frobenius method
A differential equation of the form y + P y + Qy = 0, where P and Q are both functions of x, such that the equation can be represented by a power series, may be solved by the Frobenius method.
The following 4-step procedure may be used in the
Frobenius method:
(i) Assume a trial solution of the form y = xc a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · ·

+ ar (c + r − 1)(c + r)x c+r−2 + · · ·
(iii) Substituting y, y and y into each term of the given equation 3x y + y − y = 0 gives:
3x y = 3a0 c(c − 1)x c−1 + 3a1 c(c + 1)x c
+ 3a2(c + 1)(c + 2)x c+1 + · · ·
+ 3ar (c + r − 1)(c+r)x c+r−1 +· · · (a) y = a0 cx c−1 +a1 (c + 1)x c +a2 (c + 2)x c+1
+ · · · + ar (c + r)x c+r−1 + · · ·

(b)

(ii) differentiate the trial series,
(iii) substitute the results in the given differential equation, −y = −a0 x c − a1 x c+1 − a2 x c+2 − a3 x c+3
− · · · − ar x c+r − · · ·

(c)

501

Power series methods of solving ordinary differential equations
(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 x c−1 is equated to zero giving: 3a0 c(c − 1) + a0 c = 0 or a0 c[3c − 3 + 1] = a0 c(3c − 2) = 0

(18)

The coefficient of x c is equated to zero giving:
3a1c(c + 1) + a1 (c + 1) − a0 = 0
i.e.

a1 (3c2 + 3c + c + 1) − a0
= a1(3c2 + 4c + 1) − a0 = 0

or

a1 (3c + 1)(c + 1) − a0 = 0

a1 a0 =
(2 × 4) (2 × 4) since a1 = a0 a2 a0 when r = 2, a3 =
=
(3 × 7) (2 × 4)(3 × 7) a0 or
(2 × 3)(4 × 7) a3 when r = 3, a4 =
(4 × 10) a0 =
(2 × 3 × 4)(4 × 7 × 10) and so on.

Thus, when r = 1, a2 =

From equation (16), the trial solution was:
(19)

In each of series (a), (b) and (c) an x c term is involved, after which, a general relationship can be obtained for x c+r , where r ≥ 0.
In series (a) and (b), terms in x c+r−1 are present; replacing r by (r + 1) will give the corresponding terms in x c+r , which occurs in all three equations, i.e.

y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·+ ar x r + · · ·}
Substituting c = 0 and the above values of a1 , a2 , a3, … into the trial solution gives: y = x 0 a0 + a0 x +

a0 x2 (2 × 4)

+

a0 x3 (2 × 3)(4 × 7)

+

in series (a), 3ar+1 (c + r)(c + r + 1)x c+r

a0 x4 + · · ·
(2 × 3 × 4)(4 × 7 × 10)

in series (b), ar+1 (c + r + 1)x c+r in series (c), −ar x c+r
Equating the total coefficients of x c+r to zero gives: 3ar+1 (c + r)(c + r + 1) + ar+1 (c + r + 1)
− ar = 0

i.e. y = a0 1 + x +
+

x3 x2 +
(2 × 4) (2 × 3) (4 × 7)

x4
+···
(2 × 3 × 4)(4 × 7 × 10)

(21)

which simplifies to: ar+1 {(c + r + 1)(3c + 3r +1)} − ar = 0

(20)

Equation (18), which was formed from the coefficients of the lowest power of x, i.e. x c−1, is called the indicial equation, from which, the value of c is obtained. From equation (18), since a0 = 0,
2
then c = 0 or c =
3

(a) When c = 0:
From equation (19), if c = 0, a1 (1 × 1) − a0 = 0,
i.e. a1 = a0
From equation (20), if c = 0, ar+1 (r + 1)(3r + 1) − ar = 0, ar i.e. ar+1 = r ≥0
(r + 1)(3r + 1)

2
(b) When c = :
3
5
2
− a0 = 0, i.e.
From equation (19), if c = , a1(3)
3
3 a0 a1 =
5
2
From equation (20), if c =
3
2 ar+1 + r + 1 (2 + 3r + 1) − ar = 0,
3
5
(3r + 3) − ar
3
= ar+1 (3r 2 + 8r + 5) − ar = 0, ar i.e. ar+1 = r ≥0
(r + 1)(3r + 5)
i.e. ar+1 r +

502 Higher Engineering Mathematics a1 a0
=
(2 × 8) (2 × 5 × 8) a0 since a1 =
5
a2 when r = 2, a3 =
(3 × 11) a0 =
(2 × 3)(5 × 8 × 11) a3 when r = 3, a4 =
(4 × 14) a0 =
(2×3×4)(5×8×11×14)
and so on.

Thus, when r = 1, a2 =

y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·+ ar x r + · · ·}
2
Substituting c = and the above values of a1 , a2 ,
3
a3 , … into the trial solution gives: y=x a0 a0 x+ x2 a0 +
5
2×5×8

+

2

x4
+ ···
(2 × 3 × 4)(5 × 8 × 11 × 14)

+

x4
+···
(2 ×3 × 4)(4 × 7 × 10)

2 x2 x
+Bx3 1+ +
5 (2 × 5 ×8)

y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·
+ ar x r + · · ·}

(23)

where a0 = 0,
i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3
(24)

and y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1
+ a2(c + 1)(c + 2)x c + · · ·
+ ar (c + r − 1)(c + r)x c+r−2 + · · ·
(iii) Substituting y, y and y into each term of the given equation 2x 2 y − x y + (1 − x)y = 0 gives: (22)

Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different.
Let a0 = A in equation (21), and a0 = B in equation (22).
Also, if the first solution is denoted by u(x) and the second by v(x), then the general solution of the given differential equation is y = u(x) + v(x). Hence, x3 x2
+
y = A 1 +x +
(2 × 4) (2 × 3)(4 × 7)

The differential equation may be rewritten as:
2x 2 y − x y + (1 − x)y = 0.

+ · · · + ar (c + r)x c+r−1 + · · ·

x3
(2 × 3)(5 × 8 × 11)

+

Problem 8. Use the Frobenius method to determine the general power series solution of the differential equation: d2 y dy 2x 2 2 − x
+ (1 − x)y = 0. dx dx

y = a0 cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1

x2 x +
5 (2 × 5 × 8)

+

x4
+···
(2 × 3 × 4)(5 × 8 × 11 × 14)

(ii) Differentiating equation (24) gives:

a0 x4 + · · ·
(2 × 3 × 4)(5 × 8 × 11 × 14)

i.e. y = a0 x 3 1 +

+

+ · · · + ar x c+r + · · ·

a0 x3 (2 × 3)(5 × 8 × 11)

+

x3
(2 × 3)(5 × 8 × 11)

(i) Let a trial solution be of the form

From equation (16), the trial solution was:

2
3

+

2x 2 y = 2a0 c(c − 1)x c + 2a1 c(c + 1)x c+1
+ 2a2 (c + 1)(c + 2)x c+2 + · · ·
+ 2ar (c + r − 1)(c + r)x c+r + · · ·
(a)
−x y = −a0 cx c − a1 (c + 1)x c+1
− a2 (c + 2)x c+2 − · · ·
− ar (c + r)x c+r − · · ·

(b)

(1 − x)y = (1 − x)(a0 x c + a1 x c+1 + a2 x c+2
+ a3 x c+3 + · · · + ar x c+r + · · ·)
= a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3
+ · · · + ar x c+r + · · ·

503

Power series methods of solving ordinary differential equations
− a0 x c+1 − a1 x c+2 − a2 x c+3
− a3 x

c+4

− · · · − ar x

c+r+1

−···

(c)

(iv) The indicial equation, which is obtained by equating the coefficient of the lowest power of x to zero, gives the value(s) of c. Equating the total coefficients of x c (from equations (a) to (c)) to zero gives:

+ ar x r + · · ·

a0 [2c2 − 2c − c + 1] = 0

i.e.
i.e.

a0 [2c2 − 3c + 1] = 0 a0 [(2c − 1)(c − 1)] = 0

Substituting c = 1 and the above values of a1 , a2 , a3, … into the trial solution gives:

1
2
The coefficient of the general term, i.e. x c+r , gives (from equations (a) to (c)): from which,

From equation (23), the trial solution was: y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·

2a0c(c − 1) − a0 c + a0 = 0
i.e. a0 [2c(c − 1) − c + 1] = 0
i.e.

when r = 4, a3 a3
=
a4 =
4(8 + 1) 4 × 9 a0 =
(1 × 2 × 3 × 4) × (3 × 5 × 7 × 9) and so on.

a0 a0 x+ x2 (1×3)
(1×2)×(3×5)
a0
+
x3
(1 × 2 × 3) × (3 × 5 × 7) a0 + x4 (1×2×3×4)×(3×5×7×9)

y = x 1 a0 +

c = 1 or c =

2ar (c + r − 1)(c + r) − ar (c + r)
+ ar − ar−1 = 0

+ ···

from which, ar [2(c + r − 1)(c + r) − (c + r) + 1] = ar−1 and ar =

ar−1
2(c +r −1)(c +r)−(c +r) +1

i.e. y = a0 x 1 1+
+

ar−1
2(r)(1 + r) − (1 + r ) +1 ar−1 =
2 −1−r +1
2r + 2r ar−1 ar−1
= 2
=
2r + r r (2r + 1)

when r = 2, a1 a1
=
2(4 + 1) (2 × 5) a0 a0
=
or
(1 × 3)(2 × 5)
(1 × 2) × (3 × 5)

a2 =

when r = 3, a2 a2 a3 =
=
3(6 + 1) 3 × 7 a0 =
(1 × 2 × 3) × (3 × 5 × 7)

x3
(1 × 2 × 3) × (3 × 5 × 7)

+

(25)

(a) With c = 1, ar =

Thus, when r = 1, a0 a0 a1 =
=
1(2 + 1) 1 × 3

x2 x +
(1×3) (1 × 2) × (3 × 5)

x4
(1×2×3×4)×(3×5×7×9)
+ ···

(b) With c =

(26)

1
2

ar−1
2(c + r − 1)(c + r) − (c + r) + 1 from equation (25) ar−1 i.e. ar =
1
1
1
2 +r −1
+r −
+ r +1
2
2
2
ar−1
=
1
1
1
2 r− r+ − −r +1
2
2
2
ar−1
=
1
1
2 r2 −
− −r +1
4
2 ar =

504 Higher Engineering Mathematics ar−1 ar−1
= 2
1 1
2r − r
2r 2 − − − r + 1
2 2 ar−1 = r(2r − 1)

solution of the given differential equation is y = u(x) + v(x),

=

i.e. y = A x 1 +

a0 a0 =
1(2 − 1) 1 × 1 a1 a1 when r = 2, a2 =
=
2(4 − 1) (2 × 3) a0 =
(2 × 3) a2 a2 when r = 3, a3 =
=
3(6 − 1) 3 × 5 a0 =
(2 × 3) × (3 × 5) a3 a3 when r = 4, a4 =
=
4(8 − 1) 4 × 7 a0 =
(2×3×4)×(3×5×7)

Thus, when r = 1, a1 =

and so on.
From equation (23), the trial solution was: y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·

1
Substituting c = and the above values of a1 , a2 ,
2
a3 , … into the trial solution gives:
1

+

a0 2 a0 x + x3 (2×3)
(2×3)×(3×5)

a0 x4 + · · ·
(2 × 3 × 4) × (3 × 5 × 7)
1

i.e. y = a0 x 2 1 + x +

x3
(1 × 2 × 3) × (3 × 5 ×7)

+

x4
(1 × 2 × 3×4)×(3×5×7×9)
1

+··· +Bx2 1+x+

x2
(2 × 3)

+

x3
(2 × 3) × (3 × 5)

+

x4
+···
(2 × 3 × 4) ×(3 × 5 ×7)

Problem 9. Use the Frobenius method to determine the general power series solution of the d2 y differential equation: 2 − 2y = 0. dx (i) Let a trial solution be of the form y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·
+ ar x r + · · ·

where a0 = 0,

+ · · · + ar x c+r + · · ·

x2
(2 × 3)

(29)

(ii) Differentiating equation (29) gives: y = a0cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1

x4
(2 × 3 × 4) × (3 × 5 × 7)
+ ···

(28)

i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3

x3
+
(2 × 3) × (3 × 5)
+

+

The differential equation may be rewritten as: y − 2y = 0.

+ ar x r + · · ·

y=x 2 a0 +a0 x +

x2 x +
(1 × 3) (1 × 2) × (3 × 5)

+ · · · + ar (c + r)x c+r−1 + · · ·
(27)

Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different.
Let a0 = A in equation (26), and a0 = B in equation (27). Also, if the first solution is denoted by u(x) and the second by v(x), then the general

and y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1
+ a2(c + 1)(c + 2)x c + · · ·
+ ar (c + r − 1)(c + r)x c+r−2 + · · ·
(iii) Replacing r by (r + 2) in ar (c + r − 1)(c + r) x c+r−2 gives: ar+2 (c + r + 1)(c + r + 2)x c+r

Power series methods of solving ordinary differential equations
Substituting y and y into each term of the given equation y − 2y = 0 gives:

= a0 1 +

2x 2 4x 4
+
+···
2!
4!

+ a1 x +

y − 2y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1
+ [a2(c+1)(c + 2)−2a0 ]x c +· · ·
(30)

(iv) The indicial equation is obtained by equating the coefficient of the lowest power of x to zero.

y=P 1 +

Hence, a0 c(c − 1) =0 from which, c = 0 or c = 1 since a0 = 0

a2 =
Since

For the term in x c , a2 (c + 1)(c + 2) − 2a0 = 0 from which,
2a0
a2 =
(31)
(c + 1)(c + 2)

a3 =

a4 =

from which,

c = 1,

a5 =

2a1
= 0 since a1 = 0
(3 × 4)
2a2
2
2a0 4a0
=
×
=
(4 × 5) (4 × 5)
3!
5!
2a3
=0
(5 × 6)

Hence, when c = 1,

2a0
2a0
a2 =
=
(1 × 2)
2!

y = x 1 a0 +

2ar and (r + 1)(r + 2)
2a1
2a1
2a1
=
=
when r = 1, a3 =
(2 × 3) (1 × 2 × 3)
3!
2a2
4a0
= when r = 2, a4 =
3×4
4!
In general, ar + 2 =

+

4a0 4 x + ···
4!

from equation (28)

2ar
(c + r + 1)(c + r + 2)
2ar
=
(r + 2)(r + 3)

ar+2 =

when r = 3,

(32)

(a) When c = 0: a1 is indeterminate, and from equation (31)

Hence, y = x

2a0
2a0
=
(2 × 3)
3!

when r = 2,

ar+2 (c + r + 1)(c + r + 2) − 2ar = 0

2a0 2 2a1 3 x + x a0 + a1 x +
2!
3!

(33)

from equation (32) and when r = 1,

For the term in x c+r ,

0

2x3 4x5
+
+···
3!
5!

(b) When c =1: a1 = 0, and from equation (31),

With c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined with the zero value of c would make the product zero. 2ar
(c + r + 1)(c + r + 2)

2x2 4x4
+
+···
2!
4!

+Q x+

For the term in x c−1 , i.e. a1c(c + 1) = 0

ar+2 =

2x 3 4x 5
+
+···
3!
5!

Since a0 and a1 are arbitrary constants depending on boundary conditions, let a0 = P and a1 = Q, then:

+ [ar+2 (c + r + 1)(c + r + 2)
− 2ar ] x c+r + · · · = 0

505

2a0 2
4a0 4 x + x +···
3!
5! from equation (28)

i.e. y = a0 x +

2x 3 4x 5
+
+ ...
3!
5!

Again, a0 is an arbitrary constant; let a0 = K , then y=K x+

2x3 4x5
+
+···
3!
5!

However, this latter solution is not a separate solution, for it is the same form as the second series in equation
(33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solution. This is always

506 Higher Engineering Mathematics the case when the two values of c differ by an integer (i.e. whole number). From the above three worked problems, the following can be deduced, and in future assumed:
(i) if two solutions of the indicial equation differ by a quantity not an integer, then two independent solutions y = u(x) + v(x) result, the general solution of which is y = Au + Bv (note: Problem 7
1
2 had c = 0 and and Problem 8 had c = 1 and ;
3
2 in neither case did c differ by an integer)
(ii) if two solutions of the indicial equation do differ by an integer, as in Problem 9 where c = 0 and 1, and if one coefficient is indeterminate, as with when c = 0, then the complete solution is always given by using this value of c. Using the second value of c, i.e. c = 1 in Problem 9, always gives a series which is one of the series in the first solution.
Now try the following exercise
Exercise 196 Further problems on power series solution by the Frobenius method
1. Produce, using Frobenius’ method, a power series solution for the differential equation: d2 y d y
2x 2 +
− y = 0. dx dx

⎡ x2 ⎥
⎢y = A 1 + x +


(2 × 3)





⎢ x3 ⎢
+
+··· ⎥


(2 × 3)(3 × 5)






1
⎥ x2 x


⎢ +Bx2 1+
+

(1 × 3) (1 × 2)(3 × 5) ⎥




3


x

+
+ ··· ⎦
(1 × 2 × 3)(3 × 5 × 7)
2. Use the Frobenius method to determine the general power series solution of the differend2 y tial equation: 2 + y = 0. dx ⎡

x2 x4
⎢ y = A 1 − 2! + 4! − · · ·





3
5


x x ⎢
+ B x − + − ··· ⎥


3! 5!


= P cos x + Q sin x

3. Determine the power series solution of the d2 y dy differential equation: 3x 2 + 4 − y = 0 dx dx using the Frobenius method.

⎤ x x2
⎢y = A 1 + (1 × 4) + (1 × 2)(4 × 7) ⎥



⎥ x3 ⎢


+
+··· ⎥


(1 × 2 × 3)(4 × 7 × 10)




2
1 x x


−3
+
1+
⎢ + Bx


(1 × 2) (1 × 2)(2 × 5)⎥



⎥ x3 ⎣
+
+ ··· ⎦
(1 × 2 × 3)(2 × 5 × 8)
4. Show, using the Frobenius method, that the power series solution of the differential d2 y
− y = 0 may be expressed as equation: dx 2 y = P cosh x + Q sinh x, where P and Q are constants. [Hint: check the series expansions for cosh x and sinh x on page 47]

52.6 Bessel’s equation and Bessel’s functions One of the most important differential equations in applied mathematics is Bessel’s equation and is of the form: d2 y dy + (x 2 − v 2 )y = 0 x2 2 + x dx dx where v is a real constant. The equation, which has applications in electric fields, vibrations and heat conduction, may be solved using Frobenius’ method of the previous section.
Problem 10. Determine the general power series solution of Bessels equation. d2 y dy +x
+ (x 2 − v 2 )y = 0 may
2
dx dx be rewritten as: x 2 y + x y + (x 2 − v 2 )y = 0
Bessel’s equation x 2

Using the Frobenius method from page 500:
(i) Let a trial solution be of the form y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·
+ ar x r + · · ·} where a0 = 0,

(34)

507

Power series methods of solving ordinary differential equations
i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3
+ · · · + ar x

c+r

+···

(35)

Similarly, if c = −va1[1 − 2v] = 0
The terms (2v + 1) and (1 − 2v) cannot both be zero since v is a real constant, hence a1 = 0.

(ii) Differentiating equation (35) gives: y = a0cx c−1 + a1 (c + 1)x c
+ a2 (c + 2)x c+1 + · · ·
+ ar (c + r)x c+r−1 + · · · and y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1

Since a1 = 0, then from a3 = a5 = a7 = . . . = 0

+ ar (c + r − 1)(c + r)x c+r−2 + · · ·
(iii) Substituting y, y and y into each term of the given equation: x 2 y + x y + (x 2 − v 2 )y = 0 gives: + a2(c + 1)(c + 2)x c+2 + · · ·

a0 v 2 − (c + 2)2

a4 =

a0
2 − (c + 2)2 ][v 2 − (c + 4)2 ]
[v

a6 =

a0
[v 2 − (c + 2)2 ][v 2 −(c + 4)2 ][v 2 − (c + 6)2 ] and so on.

When c = +v,

+ ar (c + r − 1)(c + r)x c+r + · · · + a0 cx c

a2 =

+ a1(c + 1)x c+1 + a2 (c + 2)x c+2 + · · ·
+ ar (c + r)x c+r + · · · + a0 x c+2 + a1 x c+3
+ a2 x

− a1 v x

+ · · · + ar x

2 c+1

2 c

+ · · · − a0 v x

− · · · − ar v x

2 c+r

+··· = 0
(36)

(iv) The indicial equation is obtained by equating the coefficient of the lowest power of x to zero.
Hence,

a0 [c2 − c + c − v 2 ] = 0

from which, c = +v or c = −v since a0 = 0
For the term in x c+r ,
− ar v 2 = 0

ar [(c + r)2 − v 2 ] =−ar−2

For the term in x c+1 , a1[c(c + 1) + (c + 1) − v 2 ] = 0
[(c + 1)2

− v2] = 0

i.e.

a1

but if c = v

a1 [(v + 1)2 − v 2 ] = 0

a6 =

=

ar [(c + r − 1)(c + r) + (c + r) − v 2 ] =