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Higher Engineering Mathematics

Why is knowledge of mathematics important in engineering?

A career in any engineering or scientific field willrequire both basic and advanced mathematics. Withoutmathematics to determine principles, calculate dimen-sions and limits, explore variations, prove concepts, andso on, there would be no mobile telephones, televisions,stereo systems, video games, microwave ovens, comput-ers, or virtually anything electronic. There would be nobridges, tunnels, roads, skyscrapers, automobiles, ships,planes, rockets or most things mechanical. There wouldbe no metals beyond the common ones, such as ironand copper, no plastics, no synthetics. In fact, societywould most certainly be less advanced without the useof mathematics throughout the centuries and into thefuture.

Electrical engineers require mathematics to design,develop, test, or supervise the manufacturing and instal-lation of electrical equipment, components, or systemsfor commercial, industrial, military, or scientific use.

Mechanical engineers require mathematics to performengineering duties in planning and designing tools,engines, machines, and other mechanically functioningequipment; they oversee installation, operation, mainte-nance, and repair of such equipment as centralised heat,gas, water, and steam systems.

Aerospace engineers require mathematics to performa variety of engineering work in designing, construct-ing, and testing aircraft, missiles, and spacecraft; theyconduct basic and applied research to evaluate adapt-ability of materials and equipment to aircraft design andmanufacture and recommend improvements in testingequipment and techniques.

Nuclear engineers require mathematics to conductresearch on nuclear engineering problems or apply prin-

ciples and theory of nuclear science to problems con-cerned with release, control, and utilisation of nuclearenergy and nuclear waste disposal.

Petroleum engineers require mathematics to devisemethods to improve oil and gas well production anddetermine the need for new or modified tool designs;they oversee drilling and offer technical advice toachieve economical and satisfactory progress.

Industrial engineers require mathematics to design,develop, test, and evaluate integrated systems for man-aging industrial production processes, including humanwork factors, quality control, inventory control, logis-tics and material flow, cost analysis, and productionco-ordination.

Environmental engineers require mathematics todesign, plan, or perform engineering duties in theprevention, control, and remediation of environmen-tal health hazards, using various engineering disci-plines; their work may include waste treatment, siteremediation, or pollution control technology.

Civil engineers require mathematics in all levels incivil engineering structural engineering, hydraulicsand geotechnical engineering are all fields that employmathematical tools such as differential equations, tensoranalysis, field theory, numerical methods and operationsresearch.

Knowledge of mathematics is therefore needed by eachof the engineering disciplines listed above.

It is intended that this text Higher Engineering Mathe-matics will provide a step-by-step approach to learningfundamental mathematics needed for your engineeringstudies.

To Sue

Higher Engineering Mathematics

Seventh Edition

John Bird, BSc (Hons), CMath, CEng, CSci, FITE, FIMA, FCollT

RoutledgeTaylor & Francis Group

LONDON AND NEW YORK

Seventh edition published 2014by Routledge2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN

and by Routledge711 Third Avenue, New York, NY 10017

Routledge is an imprint of the Taylor & Francis Group, an informa business

2014 John Bird

The right of John Bird to be identified as author of this work has been asserted by him in accordance with sections 77 and 78of the Copyright, Designs and Patents Act 1988.

All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical,or other means, now known or hereafter invented, including photocopying and recording, or in any information storage orretrieval system, without permission in writing from the publishers.

Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identificationand explanation without intent to infringe.

First edition published by Elsevier 1993Sixth edition published by Newnes 2010

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Cataloging in Publication DataBird, J. O.Higher engineering mathematics / John Bird, BSc (Hons), CMath, CEng, CSci, FITE, FIMA, FCoIIT. Seventh edition.pages cmIncludes index.1. Engineering mathematics. I. Title.TA330.B52 2014620.00151dc232013027617

ISBN: 978-0-415-66282-6 (pbk)ISBN: 978-0-315-85882-1 (ebk)

Typeset in Times byServis Filmsetting Ltd, Stockport, Cheshire

Contents

Preface xiii

Syllabus guidance xv

Section A Number and algebra 1

1 Algebra 31.1 Introduction 31.2 Revision of basic laws 31.3 Revision of equations 51.4 Polynomial division 81.5 The factor theorem 101.6 The remainder theorem 12

2 Partial fractions 152.1 Introduction to partial fractions 152.2 Worked problems on partial fractions with

linear factors 162.3 Worked problems on partial fractions with

repeated linear factors 182.4 Worked problems on partial fractions with

quadratic factors 20

3 Logarithms 223.1 Introduction to logarithms 223.2 Laws of logarithms 243.3 Indicial equations 273.4 Graphs of logarithmic functions 28

4 Exponential functions 294.1 Introduction to exponential functions 294.2 The power series for ex 304.3 Graphs of exponential functions 324.4 Napierian logarithms 334.5 Laws of growth and decay 364.6 Reduction of exponential laws to

linear form 40

5 Inequalities 435.1 Introduction to inequalities 435.2 Simple inequalities 445.3 Inequalities involving a modulus 445.4 Inequalities involving quotients 455.5 Inequalities involving square functions 465.6 Quadratic inequalities 47

Revision Test 1 49

6 Arithmetic and geometric progressions 506.1 Arithmetic progressions 506.2 Worked problems on arithmetic

progressions 516.3 Further worked problems on arithmetic

progressions 526.4 Geometric progressions 536.5 Worked problems on geometric

progressions 546.6 Further worked problems on geometric

progressions 55

7 The binomial series 587.1 Pascals triangle 587.2 The binomial series 607.3 Worked problems on the binomial series 607.4 Further worked problems on the binomial

series 627.5 Practical problems involving the binomial

theorem 64

8 Maclaurins series 688.1 Introduction 698.2 Derivation of Maclaurins theorem 698.3 Conditions of Maclaurins series 708.4 Worked problems on Maclaurins series 708.5 Numerical integration using Maclaurins

series 738.6 Limiting values 75

Revision Test 2 78

9 Solving equations by iterative methods 799.1 Introduction to iterative methods 799.2 The bisection method 809.3 An algebraic method of successive

approximations 839.4 The NewtonRaphson method 86

10 Binary, octal and hexadecimal numbers 9010.1 Introduction 9010.2 Binary numbers 9110.3 Octal numbers 9310.4 Hexadecimal numbers 95

vi Contents

11 Boolean algebra and logic circuits 9911.1 Boolean algebra and switching circuits 10011.2 Simplifying Boolean expressions 10411.3 Laws and rules of Boolean algebra 10411.4 De Morgans laws 10611.5 Karnaugh maps 10711.6 Logic circuits 11111.7 Universal logic gates 115

Revision Test 3 118

Section B Geometry and trigonometry 119

12 Introduction to trigonometry 12112.1 Trigonometry 12212.2 The theorem of Pythagoras 12212.3 Trigonometric ratios of acute angles 12312.4 Evaluating trigonometric ratios 12512.5 Solution of right-angled triangles 12912.6 Angles of elevation and depression 13112.7 Sine and cosine rules 13212.8 Area of any triangle 13312.9 Worked problems on the solution of

triangles and finding their areas 13312.10 Further worked problems on solving

triangles and finding their areas 13412.11 Practical situations involving

trigonometry 13612.12 Further practical situations involving

trigonometry 138

13 Cartesian and polar co-ordinates 14113.1 Introduction 14213.2 Changing from Cartesian into polar

co-ordinates 14213.3 Changing from polar into Cartesian

co-ordinates 14413.4 Use of Pol/Rec functions on calculators 145

14 The circle and its properties 14714.1 Introduction 14714.2 Properties of circles 14714.3 Radians and degrees 14914.4 Arc length and area of circles and sectors 15014.5 The equation of a circle 15314.6 Linear and angular velocity 15414.7 Centripetal force 156

Revision Test 4 158

15 Trigonometric waveforms 16015.1 Graphs of trigonometric functions 16015.2 Angles of any magnitude 16115.3 The production of a sine and cosine wave 16415.4 Sine and cosine curves 16515.5 Sinusoidal form A sin (t ) 16915.6 Harmonic synthesis with complex

waveforms 172

16 Hyperbolic functions 17816.1 Introduction to hyperbolic functions 17816.2 Graphs of hyperbolic functions 18016.3 Hyperbolic identities 18216.4 Solving equations involving hyperbolic

functions 18416.5 Series expansions for cosh x and sinh x 186

17 Trigonometric identities and equations 18817.1 Trigonometric identities 18817.2 Worked problems on trigonometric

identities 18917.3 Trigonometric equations 19017.4 Worked problems (i) on trigonometric

equations 19117.5 Worked problems (ii) on trigonometric

equations 19217.6 Worked problems (iii) on trigonometric

equations 19317.7 Worked problems (iv) on trigonometric

equations 193

18 The relationship between trigonometric andhyperbolic functions 196

18.1 The relationship between trigonometricand hyperbolic functions 196

18.2 Hyperbolic identities 197

19 Compound angles 20019.1 Compound angle formulae 20019.2 Conversion of a sint +b cost into

R sin(t +) 20219.3 Double angles 20619.4 Changing products of sines and cosines

into sums or differences 20819.5 Changing sums or differences of sines and

cosines into products 20919.6 Power waveforms in a.c. circuits 210

Revision Test 5 214

Contents vii

Section C Graphs 215

20 Functions and their curves 21720.1 Standard curves 21720.2 Simple transformations 22020.3 Periodic functions 22520.4 Continuous and discontinuous functions 22520.5 Even and odd functions 22620.6 Inverse functions 22720.7 Asymptotes 22920.8 Brief guide to curve sketching 23520.9 Worked problems on curve sketching 236

21 Irregular areas, volumes and mean values ofwaveforms 239

21.1 Areas of irregular figures 23921.2 Volumes of irregular solids 24221.3 The mean or average value of a waveform 243

Revision Test 6 248

Section D Complex numbers 249

22 Complex numbers 25122.1 Cartesian complex numbers 25222.2 The Argand diagram 25322.3 Addition and subtraction of complex

numbers 25322.4 Multiplication and division of complex

numbers 25422.5 Complex equations 25622.6 The polar form of a complex number 25722.7 Multiplication and division in polar form 25922.8 Applications of complex numbers 260

23 De Moivres theorem 26423.1 Introduction 26523.2 Powers of complex numbers 26523.3 Roots of complex numbers 26623.4 The exponential form of a complex

number 26823.5 Introduction to locus problems 269

Section E Matrices and determinants 273

24 The theory of matrices and determinants 27524.1 Matrix notation 27524.2 Addition, subtraction and multiplication

of matrices 27624.3 The unit matrix 27924.4 The determinant of a 2 by 2 matrix 279

24.5 The inverse or reciprocal of a 2 by 2 matrix 28024.6 The determinant of a 3 by 3 matrix 28124.7 The inverse or reciprocal of a 3 by 3 matrix 283

25 Applications of matrices and determinants 28525.1 Solution of simultaneous equations by

matrices 28625.2 Solution of simultaneous equations by

determinants 28825.3 Solution of simultaneous equations using

Cramers rule 29125.4 Solution of simultaneous equations using

the Gaussian elimination method 29225.5 Eigenvalues and eigenvectors 294

Revision Test 7 300

Section F Vector geometry 301

26 Vectors 30326.1 Introduction 30326.2 Scalars and vectors 30326.3 Drawing a vector 30426.4 Addition of vectors by drawing 30526.5 Resolving vectors into horizontal and

vertical components 30726.6 Addition of vectors by calculation 30826.7 Vector subtraction 31226.8 Relative velocity 31426.9 i, j and k notation 315

27 Methods of adding alternating waveforms 31727.1 Combination of two periodic functions 31727.2 Plotting periodic functions 31827.3 Determining resultant phasors by drawing 31927.4 Determining resultant phasors by the sine

and cosine rules 32127.5 Determining resultant phasors by

horizontal and vertical components 32227.6 Determining resultant phasors by complex

numbers 324

28 Scalar and vector products 32828.1 The unit triad 32828.2 The scalar product of two vectors 32928.3 Vector products 33328.4 Vector equation of a line 337

Revision Test 8 339

viii Contents

Section G Differential calculus 341

29 Methods of differentiation 34329.1 Introduction to calculus 34329.2 The gradient of a curve 34329.3 Differentiation from first principles 34429.4 Differentiation of common functions 34529.5 Differentiation of a product 34829.6 Differentiation of a quotient 35029.7 Function of a function 35129.8 Successive differentiation 353

30 Some applications of differentiation 35530.1 Rates of change 35530.2 Velocity and acceleration 35730.3 Turning points 36030.4 Practical problems involving maximum

and minimum values 36330.5 Points of inflexion 36730.6 Tangents and normals 36930.7 Small changes 370

31 Differentiation of parametric equations 37331.1 Introduction to parametric equations 37331.2 Some common parametric equations 37431.3 Differentiation in parameters 37431.4 Further worked problems on

differentiation of parametric equations 376

32 Differentiation of implicit functions 37932.1 Implicit functions 37932.2 Differentiating implicit functions 37932.3 Differentiating implicit functions

containing products and quotients 38032.4 Further implicit differentiation 381

33 Logarithmic differentiation 38533.1 Introduction to logarithmic differentiation 38533.2 Laws of logarithms 38533.3 Differentiation of logarithmic functions 38633.4 Differentiation of further logarithmic

functions 38633.5 Differentiation of [ f (x)]x 388

Revision Test 9 391

34 Differentiation of hyperbolic functions 39234.1 Standard differential coefficients of

hyperbolic functions 39234.2 Further worked problems on

differentiation of hyperbolic functions 393

35 Differentiation of inverse trigonometric andhyperbolic functions 395

35.1 Inverse functions 395

35.2 Differentiation of inverse trigonometricfunctions 397

35.3 Logarithmic forms of the inversehyperbolic functions 400

35.4 Differentiation of inverse hyperbolicfunctions 402

36 Partial differentiation 40636.1 Introduction to partial derivatives 40636.2 First-order partial derivatives 40636.3 Second-order partial derivatives 409

37 Total differential, rates of change and smallchanges 412

37.1 Total differential 41237.2 Rates of change 41337.3 Small changes 416

38 Maxima, minima and saddle points for functionsof two variables 419

38.1 Functions of two independent variables 41938.2 Maxima, minima and saddle points 42038.3 Procedure to determine maxima, minima

and saddle points for functions of twovariables 421

38.4 Worked problems on maxima, minimaand saddle points for functions of twovariables 421

38.5 Further worked problems on maxima,minima and saddle points for functions oftwo variables 424

Revision Test 10 429

Section H Integral calculus 431

39 Standard integration 43339.1 The process of integration 43339.2 The general solution of integrals of the

form axn 43439.3 Standard integrals 43439.4 Definite integrals 437

40 Some applications of integration 44040.1 Introduction 44140.2 Areas under and between curves 44140.3 Mean and rms values 44240.4 Volumes of solids of revolution 44340.5 Centroids 44540.6 Theorem of Pappus 44740.7 Second moments of area of regular

sections 449

41 Integration using algebraic substitutions 45741.1 Introduction 457

Contents ix

41.2 Algebraic substitutions 45741.3 Worked problems on integration using

algebraic substitutions 45841.4 Further worked problems on integration

using algebraic substitutions 45941.5 Change of limits 460


42 Integration using trigonometric and hyperbolicsubstitutions 463

42.1 Introduction 46342.2 Worked problems on integration of sin2 x ,

cos2 x , tan2 x and cot2 x 46342.3 Worked problems on integration of powers

of sines and cosines 46642.4 Worked problems on integration of

products of sines and cosines 46742.5 Worked problems on integration using the

sin substitution 46842.6 Worked problems on integration using

tan substitution 46942.7 Worked problems on integration using the

sinh substitution 47042.8 Worked problems on integration using the

cosh substitution 472

43 Integration using partial fractions 47443.1 Introduction 47443.2 Worked problems on integration using

partial fractions with linear factors 47443.3 Worked problems on integration using

partial fractions with repeated linearfactors 476

43.4 Worked problems on integration usingpartial fractions with quadratic factors 477

44 The t = tan 2

substitution 479

44.1 Introduction 479

44.2 Worked problems on the t = tan 2

substitution 480

44.3 Further worked problems on the t = tan 2

substitution 481


45 Integration by parts 48545.1 Introduction 48545.2 Worked problems on integration by parts 48545.3 Further worked problems on integration

by parts 487

46 Reduction formulae 49146.1 Introduction 491

46.2 Using reduction formulae for integrals ofthe form

xn ex dx 491


xn cos x dx and

xn sin x dx 492


sinn x dx and

cosn x dx 495

46.5 Further reduction formulae 497

47 Double and triple integrals 50047.1 Double integrals 50047.2 Triple integrals 502

48 Numerical integration 50548.1 Introduction 50548.2 The trapezoidal rule 50548.3 The mid-ordinate rule 50848.4 Simpsons rule 509


Section I Differential equations 515

49 Solution of first-order differential equations byseparation of variables 517

49.1 Family of curves 51749.2 Differential equations 51849.3 The solution of equations of the form

dy

dx= f (x) 519

49.4 The solution of equations of the formdy

dx= f (y) 520

49.5 The solution of equations of the formdy

dx= f (x) f (y) 522

50 Homogeneous first-order differential equations 52650.1 Introduction 52650.2 Procedure to solve differential equations

of the form Pdy

dx= Q 526

50.3 Worked problems on homogeneousfirst-order differential equations 527

50.4 Further worked problems on homogeneousfirst-order differential equations 528

51 Linear first-order differential equations 53051.1 Introduction 53051.2 Procedure to solve differential equations

of the formdy

dx+ Py = Q 531

51.3 Worked problems on linear first-orderdifferential equations 531

51.4 Further worked problems on linearfirst-order differential equations 532

x Contents

52 Numerical methods for first-order differentialequations 535

52.1 Introduction 53552.2 Eulers method 53652.3 Worked problems on Eulers method 53752.4 The EulerCauchy method 54152.5 The RungeKutta method 546


53 Second-order differential equations of the form

ad2ydx2

+b dydx

+cy=0 55353.1 Introduction 55353.2 Procedure to solve differential equations

of the form ad2 y

dx2+b dy

dx+cy =0 554

53.3 Worked problems on differential equations

of the form ad2 y

dx2+b dy

dx+cy =0 554

53.4 Further worked problems on practicaldifferential equations of the form

ad2 y

dx2+b dy

dx+cy =0 556

54 Second-order differential equations of the form

ad2ydx2

+b dydx

+cy= f (x) 56054.1 Complementary function and particular

integral 56154.2 Procedure to solve differential equations

of the form ad2 y

dx2+b dy

dx+cy = f (x) 562

54.3 Worked problems on differential equations

of the form ad2 y

dx2+b dy

dx+ cy = f (x)

where f (x) is a constant or polynomial 56254.4 Worked problems on differential equations

of the form ad2 y

dx2+b dy

dx+ cy = f (x)

where f (x) is an exponential function 56354.5 Worked problems on differential equations

of the form ad2 y

dx2+b dy

dx+ cy = f (x)

where f (x) is a sine or cosine function 56554.6 Worked problems on differential equations

of the form ad2 y

dx2+b dy

dx+ cy = f (x)

where f (x) is a sum or a product 567

55 Power series methods of solving ordinarydifferential equations 570

55.1 Introduction 57055.2 Higher order differential coefficients as

series 571

55.3 Leibnizs theorem 57255.4 Power series solution by the

LeibnizMaclaurin method 57555.5 Power series solution by the Frobenius

method 57755.6 Bessels equation and Bessels functions 58455.7 Legendres equation and Legendre

polynomials 589

56 An introduction to partial differential equations 59456.1 Introduction 59556.2 Partial integration 59556.3 Solution of partial differential equations

by direct partial integration 59556.4 Some important engineering partial

differential equations 59756.5 Separating the variables 59856.6 The wave equation 59856.7 The heat conduction equation 60256.8 Laplaces equation 604


Section J Statistics and probability 609

57 Presentation of statistical data 61157.1 Some statistical terminology 61257.2 Presentation of ungrouped data 61357.3 Presentation of grouped data 616

58 Mean, median, mode and standard deviation 62358.1 Measures of central tendency 62358.2 Mean, median and mode for discrete data 62458.3 Mean, median and mode for grouped data 62558.4 Standard deviation 62658.5 Quartiles, deciles and percentiles 628

59 Probability 63159.1 Introduction to probability 63259.2 Laws of probability 63259.3 Worked problems on probability 63359.4 Further worked problems on probability 63459.5 Permutations and combinations 637


60 The binomial and Poisson distributions 64160.1 The binomial distribution 64160.2 The Poisson distribution 644

61 The normal distribution 64861.1 Introduction to the normal distribution 64861.2 Testing for a normal distribution 653

Contents xi

62 Linear correlation 65662.1 Introduction to linear correlation 65662.2 The product-moment formula for

determining the linear correlationcoefficient 656

62.3 The significance of a coefficient ofcorrelation 657

62.4 Worked problems on linear correlation 657

63 Linear regression 66163.1 Introduction to linear regression 66163.2 The least-squares regression lines 66163.3 Worked problems on linear regression 662


64 Sampling and estimation theories 66864.1 Introduction 66864.2 Sampling distributions 66864.3 The sampling distribution of the means 66964.4 The estimation of population parameters

based on a large sample size 67264.5 Estimating the mean of a population based

on a small sample size 677

65 Significance testing 68165.1 Hypotheses 68165.2 Type I and type II errors 68265.3 Significance tests for population means 68865.4 Comparing two sample means 693

66 Chi-square and distribution-free tests 69866.1 Chi-square values 69866.2 Fitting data to theoretical distributions 70066.3 Introduction to distribution-free tests 70666.4 The sign test 70666.5 Wilcoxon signed-rank test 70966.6 The MannWhitney test 713


Section K Laplace transforms 723

67 Introduction to Laplace transforms 72567.1 Introduction 72667.2 Definition of a Laplace transform 72667.3 Linearity property of the Laplace

transform 72667.4 Laplace transforms of elementary

functions 72667.5 Worked problems on standard Laplace

transforms 727

68 Properties of Laplace transforms 73168.1 The Laplace transform of eat f (t) 73168.2 Laplace transforms of the form eat f (t) 73168.3 The Laplace transforms of derivatives 73368.4 The initial and final value theorems 735

69 Inverse Laplace transforms 73769.1 Definition of the inverse Laplace transform 73769.2 Inverse Laplace transforms of simple

functions 73769.3 Inverse Laplace transforms using partial

fractions 74069.4 Poles and zeros 742

70 The Laplace transform of the Heaviside function 74470.1 Heaviside unit step function 74470.2 Laplace transform of H(t c) 74870.3 Laplace transform of H(t c). f (t c) 74870.4 Inverse Laplace transforms of Heaviside

functions 749

71 The solution of differential equations usingLaplace transforms 751

71.1 Introduction 75171.2 Procedure to solve differential equations

by using Laplace transforms 75171.3 Worked problems on solving differential

equations using Laplace transforms 752

72 The solution of simultaneous differentialequations using Laplace transforms 756

72.1 Introduction 75672.2 Procedure to solve simultaneous

differential equations using Laplacetransforms 756

72.3 Worked problems on solving simultaneousdifferential equations by using Laplacetransforms 757


Section L Fourier series 763

73 Fourier series for periodic functions ofperiod 2 765

73.1 Introduction 76673.2 Periodic functions 76673.3 Fourier series 76673.4 Worked problems on Fourier series of

periodic functions of period 2 767

xii Contents

74 Fourier series for a non-periodic function overrange 2 772

74.1 Expansion of non-periodic functions 77274.2 Worked problems on Fourier series of

non-periodic functions over a range of 2 773

75 Even and odd functions and half-rangeFourier series 778

75.1 Even and odd functions 77875.2 Fourier cosine and Fourier sine series 77875.3 Half-range Fourier series 782

76 Fourier series over any range 78576.1 Expansion of a periodic function of

period L 78576.2 Half-range Fourier series for functions

defined over range L 789

77 A numerical method of harmonic analysis 79177.1 Introduction 791

77.2 Harmonic analysis on data given in tabularor graphical form 791

77.3 Complex waveform considerations 795

78 The complex or exponential form of aFourier series 799

78.1 Introduction 79978.2 Exponential or complex notation 79978.3 The complex coefficients 80078.4 Symmetry relationships 80478.5 The frequency spectrum 80778.6 Phasors 808


Essential formulae 814

Answers to Practise Exercises 830

Index 873

Preface

This seventh edition of Higher Engineering Mathe-matics covers essential mathematical material suitablefor students studying Degrees, Foundation Degrees,and Higher National Certificate and Diplomacourses in Engineering disciplines.

The text has been conveniently divided into thefollowing 12 convenient categories: number andalgebra, geometry and trigonometry, graphs, complexnumbers, matrices and determinants, vector geome-try, differential calculus, integral calculus, differentialequations, statistics and probability, Laplace transformsand Fourier series.

Increasingly, difficulty in understanding algebra isproving a problem for many students as they commencestudying engineering courses. Inevitably there are a lotof formulae and calculations involved with engineeringstudies that require a sound grasp of algebra. On thewebsite, available to all, is a document which offers aquick revision of the main areas of algebra essentialfor further study, i.e. basic algebra, simple equations,transposition of formulae, simultaneous equations andquadratic equations.

For this edition, new material has been added onloci, eigenvalues and eigenvectors, points of inflexion,double and triple integrals, permutations and combina-tions and Laplace transforms of the Heaviside function,together with material that was previously on the web-site, that is, inequalities, Boolean algebra and logiccircuits, sampling and estimation theories, significancetesting, and Chi square and distribution-free tests.

The primary aim of the material in this text is toprovide the fundamental analytical and underpinningknowledge and techniques needed to successfullycomplete scientific and engineering principles modulesof Degree, Foundation Degree and Higher NationalEngineering programmes. The material has beendesigned to enable students to use techniques learnedfor the analysis, modelling and solution of realisticengineering problems at Degree and Higher Nationallevel. It also aims to provide some of the more advancedknowledge required for those wishing to pursue careers

in mechanical engineering, aeronautical engineering,electrical and electronic engineering, communicationsengineering, systems engineering and all variants ofcontrol engineering.

In Higher Engineering Mathematics 7th Edition,theory is introduced in each chapter by a full outlineof essential definitions, formulae, laws, procedures, etc;problem solving is extensively used to establish andexemplify the theory. It is intended that readers will gainreal understanding through seeing problems solved andthen through solving similar problems themselves.

Access to software packages such as Maple, Mathe-matica and Derive, or a graphics calculator, will enhanceunderstanding of some of the topics in this text.

Each topic considered in the text is presented in away that assumes in the reader only knowledge attainedin BTEC National Certificate/Diploma, or similar, in anEngineering discipline.

Higher Engineering Mathematics 7th Edition pro-vides a follow-up to Engineering Mathematics 7thEdition.

This textbook contains some 1020 worked prob-lems, followed by over 1900 further problems(with answers), arranged within 269 Practice Exer-cises. Some 512 line diagrams further enhanceunderstanding.

Worked solutions to all 1900 of the further problemshas been prepared and can be accessed free by studentsand staff via the website (see page xiv).

At the end of the text, a list of Essential Formulaeis included for convenience of reference.

At intervals throughout the text are some 20 Revi-sion Tests to check understanding. For example, Revi-sion Test 1 covers the material in Chapters 1 to 5,Revision Test 2 covers the material in Chapters 6to 8, Revision Test 3 covers the material in Chap-ters 9 to 11, and so on. An Instructors Manual,containing full solutions to the Revision Tests, isavailable free to lecturers/instructors via the website(see page xiv).

xiv Preface

Learning by example is at the heart of HigherEngineering Mathematics 7th Edition.

JOHN BIRDRoyal Naval School of Marine Engineering,

HMS Sultan,formerly University of Portsmouthand Highbury College, Portsmouth

John Bird is the former Head of Applied Electron-ics in the Faculty of Technology at Highbury College,Portsmouth, UK. More recently, he has combined free-lance lecturing at the University of Portsmouth withexaminer responsibilities for Advanced Mathematicswith City and Guilds, and examining for InternationalBaccalaureate Organisation. He is the author of over125 textbooks on engineering and mathematics withworldwide sales of around one million copies. He iscurrently a Senior Training Provider at the Royal NavalSchool of Marine Engineering in the Defence Collegeof Marine and Air Engineering at HMS Sultan, Gosport,Hampshire, UK.

Free Web downloadsThe following support material is available from

http://www.routledge.com/bird/For Students:

1. Full solutions to all 1900 further questionscontained in the 269 Practice Exercises

2. Revision of some important algebra topics

3. A list of Essential Formulae

4. Information on 31 Mathematicians/Engineersmentioned in the text

For Lecturers/Instructors:1. Full solutions to all 1900 further questions

contained in the 269 Practice Exercises

2. Revision of some important algebra topics

3. Full solutions and marking scheme for eachof the 20 Revision Tests; also, each test maybe downloaded for distribution to students.In addition, solutions to the Revision Testgiven in the Revision of Algebra Topics is alsoincluded

4. A list of Essential Formulae

5. Information on 31 Mathematicians/Engineersmentioned in the text

6. All 512 illustrations used in the text maybe downloaded for use in PowerPointpresentations

http://www.routledge.com/bird/

Syllabus guidance

This textbook is written for undergraduate Engineering Degree and Foundation Degree courses;however, it is also most appropriate for BTEC levels 4 and 5 HNC/D studies in engineering and threesyllabuses are covered. The appropriate chapters for these three syllabuses are shown in the table below.

Chapter Analytical Further AdvancedMethods Analytical Mathematicsfor Engineers Methods for for

Engineers Engineering

1. Algebra 2. Partial fractions 3. Logarithms 4. Exponential functions 5. Inequalities

6. Arithmetic and geometric progressions 7. The binomial series 8. Maclaurins series 9. Solving equations by iterative methods

10. Binary, octal and hexadecimal numbers 11. Boolean algebra and logic circuits 12. Introduction to trigonometry 13. Cartesian and polar co-ordinates 14. The circle and its properties 15. Trigonometric waveforms 16. Hyperbolic functions 17. Trigonometric identities and equations 18. The relationship between trigonometric and hyperbolic

functions

19. Compound angles 20. Functions and their curves 21. Irregular areas, volumes and mean values of waveforms 22. Complex numbers 23. De Moivres theorem 24. The theory of matrices and determinants 25. Applications of matrices and determinants

(Continued )

xvi Syllabus Guidance



26. Vectors 27. Methods of adding alternating waveforms 28. Scalar and vector products 29. Methods of differentiation 30. Some applications of differentiation 31. Differentiation of parametric equations

32. Differentiation of implicit functions 33. Logarithmic differentiation 34. Differentiation of hyperbolic functions 35. Differentiation of inverse trigonometric and hyperbolic

functions

36. Partial differentiation 37. Total differential, rates of change and small changes 38. Maxima, minima and saddle points for functions of two

variables

39. Standard integration 40. Some applications of integration 41. Integration using algebraic substitutions 42. Integration using trigonometric and hyperbolic

substitutions

43. Integration using partial fractions 44. The t = tan/2 substitution45. Integration by parts 46. Reduction formulae 47. Double and triple integrals

48. Numerical integration 49. Solution of first-order differential equations by separation of

variables

50. Homogeneous first-order differential equations

51. Linear first-order differential equations 52. Numerical methods for first-order differential equations 53. Second-order differential equations of the form

ad2 y

dx2+ b dy

dx+ cy = 0

(Continued )

Syllabus Guidance xvii



54. Second-order differential equations of the form a

d2 y

dx2+ b dy

dx+ cy = f (x)

55. Power series methods of solving ordinary differential equations 56. An introduction to partial differential equations 57. Presentation of statistical data 58. Mean, median, mode and standard deviation 59. Probability 60. The binomial and Poisson distributions 61. The normal distribution 62. Linear correlation 63. Linear regression 64. Sampling and estimation theories 65. Significance testing 66. Chi-square and distribution-free tests 67. Introduction to Laplace transforms 68. Properties of Laplace transforms 69. Inverse Laplace transforms 70. The Laplace transform of the Heaviside function

71. Solution of differential equations using Laplace transforms

72. The solution of simultaneous differential equations using Laplace transforms

73. Fourier series for periodic functions of period 2 74. Fourier series for non-periodic functions over range 2 75. Even and odd functions and half-range Fourier series 76. Fourier series over any range 77. A numerical method of harmonic analysis 78. The complex or exponential form of a Fourier series

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Section A

Number and algebra

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Chapter 1

Algebra

Why it is important to understand: Algebra, polynomial division and the factor and remainder theorems

It is probably true to say that there is no branch of engineering, physics, economics, chemistry or computerscience which does not require the understanding of the basic laws of algebra, the laws of indices, themanipulation of brackets, the ability to factorise and the laws of precedence. This then leads to the abilityto solve simple, simultaneous and quadratic equations which occur so often. The study of algebra alsorevolves around using and manipulating polynomials. Polynomials are used in engineering, computerprogramming, software engineering, in management, and in business. Mathematicians, statisticians andengineers of all sciences employ the use of polynomials to solve problems; among them are aerospaceengineers, chemical engineers, civil engineers, electrical engineers, environmental engineers, industrialengineers, materials engineers, mechanical engineers and nuclear engineers. The factor and remaindertheorems are also employed in engineering software and electronic mathematical applications, throughwhich polynomials of higher degrees and longer arithmetic structures are divided without any complexity.The study of algebra, equations, polynomial division and the factor and remainder theorems is thereforeof some considerable importance in engineering.

At the end of this chapter, you should be able to:

understand and apply the laws of indices understand brackets, factorisation and precedence transpose formulae and solve simple, simultaneous and quadratic equations divide algebraic expressions using polynomial division factorise expressions using the factor theorem use the remainder theorem to factorise algebraic expressions

1.1 Introduction

In this chapter, polynomial division and the factorand remainder theorems are explained (in Sections 1.4to 1.6). However, before this, some essential algebrarevision on basic laws and equations is included.For further algebra revision, go to the website:www.routledge.com/cw/bird

1.2 Revision of basic laws

(a) Basic operations and laws of indicesThe laws of indices are:

(i) am an = am+n (ii) am

an= amn

(iii) (am)n = amn (iv) a mn = nam(v) an = 1

an(vi) a0 = 1

Higher Engineering Mathematics. 978-0-415-66282-6, 2014 John Bird. Published by Taylor & Francis. All rights reserved.

http://www.routledge.com/cw/bird

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4 Higher Engineering Mathematics

Problem 1. Evaluate 4a2bc32ac when a =2,b = 12 and c = 1 12

4a2bc3 2ac = 4(2)2(

12

)(3

2

)3 2(2)

(3

2

)

= 4 2 2 3 3 32 2 2 2

12

2

= 27 6 = 21

Problem 2. Multiply 3x + 2y by x y

3x + 2yx y

Multiply by x 3x2 + 2xyMultiply by y 3xy 2y2

Adding gives: 3x2 x y 2 y2

Alternatively,

(3x + 2y)(x y)= 3x2 3xy + 2xy 2y2

= 3x2 x y 2 y2

Problem 3. Simplifya3b2c4

abc2and evaluate when

a = 3, b = 18 and c = 2

a3b2c4

abc2= a31b21c4(2) = a2bc6

When a = 3, b = 18 and c = 2,

a2bc6 = (3)2(

18

)(2)6 = (9)

(18

)(64)= 72

Problem 4. Simplifyx2y3 + xy2

xy

x2y3 + xy2xy

= x2 y3

xy+ xy

2

xy

= x21 y31 + x11y21

= x y2 + y or y(x y + 1)

Problem 5. Simplify(x2

y)(

x 3

y2)

(x5y3)12

(x2

y)(

x 3

y2)

(x5y3)12

= x2y

12 x

12 y

23

x52 y

32

= x2+ 12 52 y 12 + 23 32= x0y 13

= y 13 or 1y

13

or1

3

y

Now try the following Practice Exercise

Practice Exercise 1 Basic algebraicoperations and laws of indices (Answers onpage 830)

1. Evaluate 2ab + 3bc abc when a = 2,b = 2 and c = 4

2. Find the value of 5 pq2r3 when p = 25 ,q = 2 and r = 1

3. From 4x 3y + 2z subtract x + 2y 3z.4. Multiply 2a 5b + c by 3a + b5. Simplify (x2 y3z)(x3yz2) and evaluate when

x = 12 , y = 2 and z = 3

6. Evaluate (a32 bc3)(a

12 b

12 c) when a =3,

b = 4 and c = 2

7. Simplifya2b + a3b

a2b2

8. Simplify(a3b

12 c

12 )(ab)

13

(

a3

b c)

(b) Brackets, factorisation and precedence

Problem 6. Simplify a2 (2a ab) a(3b + a)

a2 (2a ab) a(3b+ a)= a2 2a + ab 3ab a2

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

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Algebra 5

Problem 7. Remove the brackets and simplify theexpression:

2a [3{2(4a b) 5(a + 2b)} + 4a]

Removing the innermost brackets gives:

2a [3{8a 2b 5a 10b} + 4a]

Collecting together similar terms gives:

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:

2a 13a + 36b = 11a+36b or36b 11a

Problem 8. Factorise (a) xy 3xz(b) 4a2 + 16ab3 (c) 3a2b 6ab2 + 15ab

(a) xy 3xz = x( y 3z)(b) 4a2 + 16ab3 = 4a(a + 4b3)(c) 3a2b 6ab2 + 15ab = 3ab(a 2b + 5)

Problem 9. Simplify 3c + 2c 4c + c 5c 8c

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

3c + 2c 4c + c 5c 8c= 3c + 2c 4c +

( c

5c

) 8c

= 3c + 8c2 + 15

8c

= 8c2 5c + 15

or c(8c 5)+ 15

Problem 10. Simplify(2a 3)4a +5 63a

(2a 3)4a + 5 6 3a

= 2a 34a

+ 5 6 3a

= 2a 34a

+ 30 3a

= 2a4a

34a

+ 30 3a

= 12

34a

+ 30 3a = 30 12

34a

3a


Practice Exercise 2 Brackets, factorisationand precedence (Answers on page 830)

1. Simplify 2( p + 3q r) 4(r q + 2 p)+ p2. Expand and simplify (x + y)(x 2y)3. Remove the brackets and simplify:

24 p [2{3(5 p q) 2( p + 2q)} + 3q]4. Factorise 21a2b2 28ab5. Factorise 2xy2 + 6x2y + 8x3y6. Simplify 2y + 4 6y + 3 4 5y7. Simplify 3 y + 2 y 18. Simplify a2 3ab 2a 6b + ab

1.3 Revision of equations

(a) Simple equations

Problem 11. Solve 4 3x = 2x 11

Since 4 3x = 2x 11 then 4 + 11 = 2x + 3xi.e. 15 = 5x from which, x = 15

5= 3

Problem 12. Solve

4(2a 3) 2(a 4)= 3(a 3) 1

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Removing the brackets gives:

8a 12 2a + 8 = 3a 9 1Rearranging gives:

8a 2a 3a = 9 1 + 12 8i.e. 3a = 6and a = 6

3= 2

Problem 13. Solve3

x 2 =4

3x + 4

By cross-multiplying: 3(3x + 4)= 4(x 2)Removing brackets gives: 9x + 12= 4x 8Rearranging gives: 9x 4x = 8 12i.e. 5x = 20

and x = 205

= 4

Problem 14. Solve(

t + 3t

)

= 2

t

(t + 3

t

)

= 2t

i.e.

t + 3= 2tand 3= 2t ti.e. 3= tand 9= t

(c) Transposition of formulae

Problem 15. Transpose the formula v = u + f tm

to make f the subject.

u + f tm

= v from which, f tm

= v u

and m

(f t

m

)

= m(v u)i.e. f t = m(v u)

and f = mt

(v u)

Problem 16. The impedance of an a.c. circuit isgiven by Z = R2 + X2. Make the reactance X thesubject.

R2 + X2 = Z and squaring both sides givesR2 + X2 = Z2, from which,

X2 = Z2 R2 and reactance X =

Z2 R2

Problem 17. Given thatD

d=

(f + pf p

)

express p in terms of D, d and f .

Rearranging gives:

(f + pf p

)

= Dd

Squaring both sides gives:f + pf p =

D2

d2

Cross-multiplying gives:

d2( f + p)= D2( f p)Removing brackets gives:

d2 f + d2 p = D2 f D2 pRearranging gives: d2 p + D2 p = D2 f d2 fFactorising gives: p(d2 + D2)= f (D2 d2)

and p = f (D2 d2)

(d2 + D2)


Practice Exercise 3 Simple equations andtransposition of formulae (Answers onpage 830)

In problems 1 to 4 solve the equations

1. 3x 2 5x = 2x 42. 8 + 4(x 1) 5(x 3)= 2(5 2x)3.

1

3a 2 +1

5a + 3 = 0

4.3

t

1 t = 6

5. Transpose y = 3(F f )L

for f

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Algebra 7

6. Make l the subject of t = 2

l

g

7. Transpose m = LL + rC R for L

8. Make r the subject of the formulax

y= 1 + r

2

1 r2

(d) Simultaneous equations

Problem 18. Solve the simultaneous equations:

7x 2y = 26 (1)6x + 5y = 29 (2)

5equation (1) gives:35x 10y = 130 (3)

2equation (2) gives:12x + 10y = 58 (4)

Equation (3)+equation (4) gives:47x + 0 = 188

from which, x = 18847

= 4Substituting x = 4 in equation (1) gives:

28 2y = 26from which, 28 26 = 2y and y=1

Problem 19. Solvex

8+ 5

2= y (1)

11 + y3

= 3x (2)

8equation (1) gives: x + 20 = 8y (3)3equation (2) gives: 33 + y = 9x (4)i.e. x 8y = 20 (5)and 9x y = 33 (6)8equation (6) gives: 72x 8y = 264 (7)Equation (7) equation (5) gives:

71x = 284

from which, x= 28471

= 4

Substituting x = 4 in equation (5) gives:4 8y = 20

from which, 4 + 20 = 8y and y = 3

(e) Quadratic equations

Problem 20. Solve the following equations byfactorisation:(a) 3x2 11x 4 = 0(b) 4x2 + 8x + 3 = 0

(a) The factors of 3x2 are 3x and x and these are placedin brackets thus:

(3x )(x )

The factors of 4 are +1 and 4 or 1 and+4, or 2 and +2. Remembering that the prod-uct of the two inner terms added to the productof the two outer terms must equal 11x , the onlycombination to give this is +1 and 4, i.e.,

3x 2 11x 4= (3x + 1)(x 4)Thus (3x + 1)(x 4)= 0 hence

either (3x + 1)= 0 i.e. x = 13

or (x 4)= 0 i.e. x = 4

(b) 4x2 + 8x + 3 = (2x + 3)(2x + 1)Thus (2x + 3)(2x + 1)= 0 hence

either (2x + 3)= 0 i.e. x =32

or (2x + 1)= 0 i.e. x = 12

Problem 21. The roots of a quadratic equationare 13 and 2. Determine the equation in x .

If1

3and 2 are the roots of a quadratic equation then,

(x 13)(x + 2)= 0

i.e. x2 + 2x 13

x 23

= 0

i.e. x2 + 53

x 23

= 0

or 3x2 + 5x2= 0

Problem 22. Solve 4x2 + 7x + 2 = 0 giving theanswer correct to 2 decimal places.

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From the quadratic formula if ax2 + bx + c = 0 then,

x = b

b2 4ac2a

Hence if 4x2 + 7x + 2 = 0

then x = 7

72 4(4)(2)2(4)

= 7

17

8

= 7 4.1238

= 7 + 4.1238

or7 4.123

8

i.e. x= 0.36 or 1.39


Practice Exercise 4 Simultaneous andquadratic equations (Answers on page 830)

In problems 1 to 3, solve the simultaneous equa-tions

1. 8x 3y = 513x + 4y = 14

2. 5a = 1 3b2b + a + 4 = 0

3.x

5+ 2y

3= 49

15

3x

7 y

2+ 5

7= 0

4. Solve the following quadratic equations byfactorisation:

(a) x2 + 4x 32 = 0(b) 8x2 + 2x 15 = 0

5. Determine the quadratic equation in x whoseroots are 2 and 5

6. Solve the following quadratic equations, cor-rect to 3 decimal places:

(a) 2x2 + 5x 4 = 0(b) 4t2 11t + 3 = 0

1.4 Polynomial division

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

For example,208

16is achieved as follows:

13

16)

20816

4848

(1) 16 divided into 2 wont go

(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

(6) Bring down the 8

(7) 16 divided into 48 goes 3 times

(8) Put the 3 above the 8

(9) 3 16 = 48(10) 48 48 = 0

Hence208

16= 13 exactly

Similarly,172

15is laid out as follows:

11

15)

17215

2215

7

Hence172

15= 11 remainder 7 or 11 + 7

15= 11 7

15Below are some examples of division in algebra, whichin some respects is similar to long division with num-bers.(Note that a polynomial is an expression of the form

f (x)= a + bx + cx2 + dx3 +

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Algebra 9

and polynomial division is sometimes required whenresolving into partial fractions see Chapter 2.)

Problem 23. Divide 2x2 + x 3 by x 1

2x2 + x 3 is called the dividend and x 1 the divi-sor. The usual layout is shown below with the dividendand divisor both arranged in descending powers of thesymbols.

2x + 3

x 1)

2x2 + x 32x2 2x

3x 33x 3

Dividing the first term of the dividend by the first term

of the divisor, i.e.2x2

xgives 2x , which is put above

the first term of the dividend as shown. The divisoris then multiplied by 2x , i.e. 2x(x 1)= 2x2 2x ,which is placed under the dividend as shown. Subtract-ing gives 3x 3. The process is then repeated, i.e. thefirst term of the divisor, x , is divided into 3x , giving+3, which is placed above the dividend as shown. Then3(x 1)=3x 3, which is placed under the 3x 3. Theremainder, on subtraction, is zero, which completes theprocess.

Thus (2x2 + x 3) (x 1) = (2x + 3)

[A check can be made on this answer by multiplying(2x + 3) by (x 1) which equals 2x2 + x 3.]

Problem 24. Divide 3x3 + x2 + 3x + 5 by x + 1

(1) (4) (7)

3x2 2x + 5

x + 1)

3x3 + x2 + 3x + 53x3 + 3x2

2x2 + 3x + 52x2 2x

5x + 55x + 5

(1) x into 3x3 goes 3x2. Put 3x2 above 3x3

(2) 3x2(x + 1)= 3x3 + 3x2

(3) Subtract

(4) x into 2x2 goes 2x . Put 2x above thedividend

(5) 2x(x + 1)= 2x2 2x(6) Subtract

(7) x into 5x goes 5. Put 5 above the dividend

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

Thus 3x3 + x2 + 3x + 5

x + 1 = 3x2 2x + 5

Problem 25. Simplifyx3 + y3x + y

(1) (4) (7)

x2 xy + y2

x + y)

x3 + 0 + 0 + y3x3 + x2 y

x2 y + y3 x2 y xy2

xy2 + y3xy2 + y3

(1) x into x3 goes x2. Put x2 above x3 of dividend

(2) x2(x + y)= x3 + x2y(3) Subtract

(4) x into x2y goes xy. Put xy above dividend(5) xy(x + y)= x2y xy2

(6) Subtract

(7) x into xy2 goes y2. Put y2 above dividend

(8) y2(x + y)= xy2 + y3

(9) Subtract

Thus

x3 + y3x + y = x

2 x y + y2

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The zeros shown in the dividend are not normally shown,but are included to clarify the subtraction process andto keep similar terms in their respective columns.

Problem 26. Divide (x2 + 3x 2) by (x 2)

x + 5

x 2)

x2 + 3x 2x2 2x

5x 25x 10

8

Hence

x2 + 3x 2x 2 = x + 5 +

8x 2

Problem 27. Divide 4a3 6a2b + 5b3 by2a b

2a2 2ab b2

2a b)

4a3 6a2b + 5b34a3 2a2b

4a2b + 5b34a2b + 2ab2

2ab2 + 5b32ab2 + b3

4b3

Thus

4a3 6a2b + 5b32a b

=2a2 2ab b2 + 4b3

2a b


Practice Exercise 5 Polynomial division(Answers on page 830)

1. Divide (2x2 + xy y2) by (x + y)2. Divide (3x2 + 5x 2) by (x + 2)3. Determine (10x2 + 11x 6) (2x + 3)

4. Find14x2 19x 3

2x 3

5. Divide (x3 + 3x2y + 3xy2 + y3) by (x + y)6. Find (5x2 x + 4) (x 1)7. Divide (3x3 + 2x2 5x + 4) by (x + 2)8. Determine (5x4 + 3x3 2x + 1)/(x 3)

1.5 The factor theorem

There is a simple relationship between the factors ofa quadratic expression and the roots of the equationobtained by equating the expression to zero.For example, consider the quadratic equationx2 + 2x 8 = 0To solve this we may factorise the quadratic expressionx2 + 2x 8 giving (x 2)(x + 4)Hence (x 2)(x + 4)= 0Then, if the product of two numbers is zero, one or bothof those numbers must equal zero. Therefore,

either (x 2)= 0, from which, x = 2or (x + 4)= 0, from which, x = 4It is clear, then, that a factor of (x 2) indicates a rootof +2, while a factor of (x + 4) indicates a root of 4In general, we can therefore say that:

a factor of (x a) corresponds to aroot of x = a

In practice, we always deduce the roots of a simplequadratic equation from the factors of the quadraticexpression, as in the above example. However, we couldreverse this process. If, by trial and error, we could deter-mine that x = 2 is a root of the equation x2 + 2x 8 = 0we could deduce at once that (x 2) is a factor of theexpression x2 + 2x 8. We wouldnt normally solvequadratic equations this way but suppose we haveto factorise a cubic expression (i.e. one in which thehighest power of the variable is 3). A cubic equationmight have three simple linear factors and the difficultyof discovering all these factors by trial and error wouldbe considerable. It is to deal with this kind of case thatwe use the factor theorem. This is just a generalisedversion of what we established above for the quadraticexpression. The factor theorem provides a method offactorising any polynomial, f (x), which has simplefactors.A statement of the factor theorem says:

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Algebra 11

if x = a is a root of the equationf (x) = 0, then (x a) is a factor of f (x)

The following worked problems show the use of thefactor theorem.

Problem 28. Factorise x3 7x 6 and use it tosolve the cubic equation x3 7x 6 = 0.

Let f (x)= x3 7x 6If x = 1, then f (1)= 13 7(1) 6 = 12If x = 2, then f (2)= 23 7(2) 6 = 12If x = 3, then f (3)= 33 7(3) 6 = 0If f (3)= 0, then (x 3) is a factor from the factortheorem.We have a choice now. We can divide x3 7x 6 by(x 3) or we could continue our trial and error bysubstituting further values for x in the given expression and hope to arrive at f (x)=0Let us do both ways. Firstly, dividing out gives:

x2 + 3x + 2x 3

)x3 0 7x 6x3 3x2

3x2 7x 63x2 9x

2x 62x 6

Hencex3 7x 6

x 3 = x2 + 3x + 2

i.e. x3 7x 6 = (x 3)(x2 + 3x + 2)x2 + 3x + 2 factorises on sight as (x + 1)(x + 2).Therefore

x3 7x 6 = (x 3)(x + 1)(x + 2)A second method is to continue to substitute values ofx into f (x).Our expression for f (3) was 33 7(3) 6. We cansee that if we continue with positive values of x thefirst term will predominate such that f (x) will notbe 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). Alsof (2)= (2)3 7(2) 6 = 0; hence (x + 2) isa factor (also as shown above).

To solve x3 7x 6 = 0, we substitute the factors, i.e.(x 3)(x + 1)(x + 2)= 0

from which, x = 3, x = 1 and x = 2Note that the values of x , i.e. 3, 1 and 2, areall factors of the constant term, i.e. 6. This cangive us a clue as to what values of x we shouldconsider.

Problem 29. Solve the cubic equationx3 2x2 5x + 6=0 by using the factor theorem.

Let f (x)= x3 2x2 5x + 6 and let us substitutesimple 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 = 0f (3)= 33 2(3)2 5(3)+ 6 = 0,

hence (x 3) is a factorf (1)= (1)3 2(1)2 5(1)+ 6 = 0f (2)= (2)3 2(2)2 5(2)+ 6 = 0,

hence (x + 2) is a factorHence x3 2x2 5x + 6 = (x 1)(x 3)(x + 2)Therefore if x3 2x2 5x + 6 = 0then (x 1)(x 3)(x + 2)= 0from which, x = 1, x = 3 and x = 2Alternatively, having obtained one factor, i.e.(x 1) we could divide this into (x3 2x2 5x + 6)as follows:

x2 x 6

x 1)

x3 2x2 5x + 6x3 x2

x2 5x + 6 x2 + x 6x + 6

6x + 6

Hence x3 2x2 5x + 6= (x 1)(x2 x 6)= (x 1)(x 3)(x + 2)

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Summarising, the factor theorem provides us with amethod of factorising simple expressions, and an alter-native, in certain circumstances, to polynomial division.


Practice Exercise 6 The factor theorem(Answers on page 830)

Use the factor theorem to factorise the expressionsgiven in problems 1 to 4.

1. x2 + 2x 32. x3 + x2 4x 43. 2x3 + 5x2 4x 74. 2x3 x2 16x + 155. Use the factor theorem to factorise

x3 + 4x2 + x 6 and hence solve the cubicequation x3 + 4x2 + x 6 = 0

6. Solve the equation x3 2x2 x + 2 = 0

1.6 The remainder theorem

Dividing a general quadratic expression(ax2 + bx + c) by (x p), where p is any wholenumber, by long division (see Section 1.4) gives:

ax + (b + ap)

x p)

ax2 + bx + cax2 apx

(b + ap)x + c(b + ap)x (b + ap)p

c + (b + ap)p

The remainder, c + (b + ap)p = c + bp + ap2 orap2 + bp + c. This is, in fact, what the remaindertheorem states, i.e.

if (ax2 + bx + c) is divided by (x p),the remainder will be ap2 + bp + c

If, in the dividend (ax2 + bx + c), we substitute p for xwe get the remainder ap2 + bp + cFor example, when (3x2 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 = 9We can check this by dividing (3x2 4x + 5) by (x 2)by long division:

3x + 2

x 2)

3x2 4x + 53x2 6x

2x + 52x 4

9

Similarly, when (4x2 7x + 9) is divided by (x +3),the remainder is ap2 + bp + c (where a = 4, b = 7,c = 9 and p = 3), i.e. the remainder is4(3)2 + (7)(3)+ 9 = 36 + 21 + 9 = 66Also, when (x2 + 3x 2) is divided by (x 1), theremainder is 1(1)2 + 3(1) 2 = 2It is not particularly useful, on its own, to know theremainder of an algebraic division. However, if theremainder should be zero then (x p) is a factor. Thisis very useful therefore when factorising expressions.For example, when (2x2 + 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 (2x2 + x 3).In this case the other factor is (2x + 3), i.e.

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

The remainder theorem may also be stated for a cubicequation as:

if (ax3 + bx2 + cx + d) is divided by(x p), the remainder will be

ap3 + bp2 + cp + dAs before, the remainder may be obtained by substitut-ing p for x in the dividend.For example, when (3x3 + 2x2 x + 4) is dividedby (x 1), the remainder is ap3 + 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 = 8Similarly, when (x3 7x 6) is divided by (x 3),the remainder is 1(3)3 + 0(3)2 7(3) 6 = 0, whichmeans that (x 3) is a factor of (x3 7x 6)Here are some more examples on the remaindertheorem.

Problem 30. Without dividing out, find theremainder when 2x2 3x + 4 is divided by (x 2)

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Algebra 13

By the remainder theorem, the remainder is given byap2 + 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 todetermine the remainder when(3x3 2x2 + x 5) is divided by (x + 2)

By the remainder theorem, the remainder is given byap3 + bp2 + cp + d , where a = 3, b = 2, c = 1, d =5 and p = 2Hence the remainder is:

3(2)3 + (2)(2)2 + (1)(2)+ (5)= 24 8 2 5= 39

Problem 32. Determine the remainder when(x3 2x2 5x + 6) is divided by (a) (x 1) and(b) (x + 2). Hence factorise the cubic expression.

(a) When (x3 2x2 5x + 6) is divided by (x 1),the remainder is given by ap3 + bp2 + cp + d ,where a = 1, b = 2, c = 5, d = 6 and p = 1,i.e. the remainder = (1)(1)3 + (2)(1)2

+ (5)(1)+ 6= 1 2 5 + 6 = 0

Hence (x 1) is a factor of (x3 2x2 5x + 6).(b) When (x3 2x2 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 (x3 2x2 5x + 6). Therefore (x 1)(x + 2)(x )= x3 2x2 5x + 6. To determine the third factor (shownblank) we could

(i) divide (x3 2x2 5x +6) by(x 1)(x + 2)

or (ii) use the factor theorem where f (x)=x3 2x2 5x +6 and hoping to choosea value of x which makes f (x)= 0

or (iii) use the remainder theorem, again hopingto choose a factor (x p) which makesthe remainder zero.

(i) Dividing (x3 2x2 5x + 6) by(x2 + x 2) gives:

x 3

x2 + x 2)

x3 2x2 5x + 6x3 + x2 2x3x2 3x + 6

3x2 3x + 6

Thus (x3 2x2 5x + 6)= (x 1)(x + 2)(x 3)

(ii) Using the factor theorem, we let

f (x)= x3 2x2 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

(x3 2x2 5x + 6) is divided by(x 3), the remainder is given byap3 + bp2 + cp + d , where a = 1,b = 2, c = 5, d = 6 and p = 3Hence the remainder is:

1(3)3 + (2)(3)2 + (5)(3)+ 6= 27 18 15 + 6 = 0

Hence (x 3) is a factor.Thus (x3 2x2 5x + 6)

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


Practice Exercise 7 The remaindertheorem (Answers on page 830)

1. Find the remainder when 3x2 4x + 2 isdivided by

(a) (x 2) (b) (x + 1)2. Determine the remainder when

x3 6x2 + x 5 is divided by(a) (x + 2) (b) (x 3)

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3. Use the remainder theorem to find the factorsof x3 6x2 + 11x 6

4. Determine the factors of x3 + 7x2 + 14x + 8and hence solve the cubic equationx3 +7x2 + 14x + 8 = 0

5. Determine the value of a if (x + 2) is afactor of (x3 ax2 + 7x + 10)

6. Using the remainder theorem, solve theequation 2x3 x2 7x + 6 = 0

For fully worked solutions to each of the problems in Practice Exercises 1 to 7 in this chapter,go to the website:

www.routledge.com/cw/bird


Chapter 2

Partial fractions

Why it is important to understand: Partial fractions

The algebraic technique of resolving a complicated fraction into partial fractions is often needed byelectrical and mechanical engineers for not only determining certain integrals in calculus, but for deter-mining inverse Laplace transforms, and for analysing linear differential equations with resonant circuitsand feedback control systems.


understand the term partial fraction appreciate the conditions needed to resolve a fraction into partial fractions resolve into partial fractions a fraction containing linear factors in the denominator resolve into partial fractions a fraction containing repeated linear factors in the denominator resolve into partial fractions a fraction containing quadratic factors in the denominator

2.1 Introduction to partial fractions

By algebraic addition,

1

x 2 +3

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

= 4x 5x2 x 2

The reverse process of moving from4x 5

x2 x 2to

1

x 2 +3

x + 1 is called resolving into partialfractions.In order to resolve an algebraic expression into partialfractions:

(i) the denominator must factorise (in the aboveexample, x2 x 2 factorises as (x 2) (x +1)),and

(ii) the numerator must be at least one degree less thanthe denominator (in the above example (4x 5) isof degree 1 since the highest powered x term is x1

and (x2 x 2) is of degree 2).When the degree of the numerator is equal to or higherthan the degree of the denominator, the numerator mustbe divided by the denominator until the remainder is ofless degree than the denominator (see Problems 3 and 4).There are basically three types of partial fraction and theform of partial fraction used is summarised in Table 2.1,where f (x) is assumed to be of less degree than therelevant denominator and A, B and C are constants tobe determined.(In the latter type in Table 2.1, ax2+bx +c is a quadraticexpression which does not factorise without containingsurds or imaginary terms.)Resolving an algebraic expression into partial fractionsis used as a preliminary to integrating certain functions(see Chapter 43) and in determining inverse Laplacetransforms (see Chapter 69).


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Table 2.1

Type Denominator containing Expression Form of partial fraction

1 Linear factors(see Problems 1 to 4)

f (x)

(x + a)(x b)(x + c)A

(x + a) +B

(x b) +C

(x + c)

2 Repeated linear factors(see Problems 5 to 7)

f (x)

(x + a)3A

(x + a) +B

(x + a)2 +C

(x + a)3

3 Quadratic factors(see Problems 8 and 9)

f (x)

(ax2 + bx + c)(x + d)Ax + B

(ax2 + bx + c) +C

(x + d)

2.2 Worked problems on partialfractions with linear factors

Problem 1. Resolve11 3x

x2 + 2x 3 into partialfractions.

The denominator factorises as (x 1) (x +3) and thenumerator is of less degree than the denominator. Thus

11 3xx2 + 2x 3 may be resolved into partial fractions.

Let11 3x

x2 + 2x 3 11 3x

(x 1)(x + 3)

A(x 1) +

B

(x + 3)where A and B are constants to be determined,

i.e.11 3x

(x 1)(x + 3) A(x + 3)+ B(x 1)(x 1)(x + 3)

by algebraic addition.Since the denominators are the same on each side of theidentity then the numerators are equal to each other.

Thus, 113x A(x +3)+ B(x 1)To determine constants A and B, values of x are chosento make the term in A or B equal to zero.

When x =1, then113(1) A(1+3)+ B(0)

i.e. 8 =4Ai.e. A =2When x = 3, then

113(3) A(0)+ B(31)i.e. 20 =4Bi.e. B =5

Thus11 3x

x2 + 2x 3 2

(x 1) +5

(x + 3)

2(x 1)

5(x + 3)

[

Check:2

(x 1) 5

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

= 11 3xx2 + 2x 3

]

Problem 2. Convert2x2 9x 35

(x + 1)(x 2)(x + 3) intothe sum of three partial fractions.

Let2x2 9x 35

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

A(x + 1) +

B

(x 2) +C

(x + 3)

(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:2x2 9x 35 A(x 2)(x + 3)

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

Let x = 1. Then2(1)2 9(1)35 A(3)(2)

+ B(0)(2)+C(0)(3)i.e. 24 =6A

i.e. A = 246 =4

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Partial fractions 17

Let x =2. Then2(2)2 9(2)35 A(0)(5)+ B(3)(5)+C(3)(0)i.e. 45 =15B

i.e. B = 4515

=3

Let x = 3. Then2(3)2 9(3)35 A(5)(0)+ B(2)(0)

+C(2)(5)i.e. 10 =10Ci.e. C =1

Thus2x2 9x 35

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

4(x + 1)

3(x 2) +

1(x + 3)

Problem 3. Resolvex2 + 1

x2 3x + 2 into partialfractions.

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

1

x2 3x +2)

x2 + 1x2 3x +2

3x 1

For more on polynomial division, see Section 1.4,page 8.

Hencex2 + 1

x2 3x + 2 1 +3x 1

x2 3x + 2

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

Let3x 1

(x 1)(x 2)A

(x 1) +B

(x 2)

A(x 2)+ B(x 1)(x 1)(x 2)

Equating numerators gives:3x 1 A(x 2)+ B(x 1)

Let x = 1. Then 2 = Ai.e. A = 2Let x = 2. Then 5 = B

Hence3x 1

(x 1)(x 2) 2

(x 1)+5

(x 2)

Thusx2 + 1

x2 3x + 2 12

(x1)+5

(x2)

Problem 4. Expressx3 2x2 4x 4

x2 + x 2 in partialfractions.

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

x 3x2 + x 2

)x3 2x2 4x 4x3 + x2 2x 3x2 2x 4

3x2 3x + 6

x 10

Thusx3 2x2 4x 4

x2 + x 2 x 3+x 10

x2 + x 2

x 3+ x 10(x +2)(x 1)

Letx 10

(x +2)(x 1) A

(x +2)+B

(x 1)

A(x 1)+ B(x +2)(x +2)(x 1)

Equating the numerators gives:

x 10 A(x 1)+ B(x +2)Let x =2. Then 12 = 3Ai.e. A= 4Let x =1. Then 9 = 3Bi.e. B= 3

Hencex 10

(x +2)(x 1)4

(x +2)3

(x 1)

Thusx3 2x2 4x4

x2 +x2

x3+ 4(x+2)

3(x1)

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Practice Exercise 8 Partial fractions withlinear factors (Answers on page 830)

Resolve the following into partial fractions.

1.12

x2 9

2.4(x 4)

x2 2x 3

3.x2 3x + 6

x(x 2)(x 1)

4.3(2x2 8x 1)

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

5.x2 + 9x + 8x2 + x 6

6.x2 x 14x2 2x 3

7.3x3 2x2 16x + 20

(x 2)(x + 2)

2.3 Worked problems on partialfractions with repeated linearfactors

Problem 5. Resolve2x + 3(x 2)2 into partial

fractions.

The denominator contains a repeated linear factor,(x 2)2.

Let2x + 3(x 2)2

A

(x 2) +B

(x 2)2

A(x 2)+ B(x 2)2


2x + 3 A(x 2)+ BLet x = 2. Then 7 = A(0)+ Bi.e. B =72x + 3 A(x 2)+ B Ax 2A + B

Since an identity is true for all values of theunknown, the coefficients of similar terms may beequated.Hence, equating the coefficients of x gives: 2 = A[Also, as a check, equating the constant terms gives:

3 = 2A + B

When A=2 and B =7,

RHS = 2(2)+ 7 = 3 = LHS]

Hence2x + 3(x 2)2

2(x 2) +

7(x 2)2

Problem 6. Express5x2 2x 19(x + 3)(x 1)2 as the sum

of three partial fractions.

The denominator is a combination of a linear factor anda repeated linear factor.

Let5x2 2x 19(x + 3)(x 1)2

A(x + 3) +

B

(x 1) +C

(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:

5x2 2x 19 A(x 1)2 + B(x + 3)(x 1)+ C(x + 3) (1)

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Let x =3. Then5(3)2 2(3) 19 A(4)2+B(0)(4)+C(0)i.e. 32 = 16Ai.e. A = 2Let x =1. Then

5(1)2 2(1) 19 A(0)2 + B(4)(0)+ C(4)i.e. 16 = 4Ci.e. C = 4Without expanding the RHS of equation (1) it canbe seen that equating the coefficients of x2 gives:5= A+ B, and since A=2, B=3[Check: Identity (1) may be expressed as:

5x2 2x 19 A(x2 2x + 1)+ B(x2 + 2x 3)+ C(x + 3)

i.e. 5x2 2x 19 Ax2 2Ax + A + Bx2 + 2Bx3B + Cx + 3C

Equating the x term coefficients gives:

2 2A + 2B + C

When A=2, B =3 and C =4 then

2A + 2B + C = 2(2)+ 2(3) 4= 2 = LHS

Equating the constant term gives:

19 A 3B + 3C

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

Hence5x2 2x 19(x + 3)(x 1)2

2(x + 3) +

3(x 1)

4

(x 1)2

Problem 7. Resolve3x2 + 16x + 15

(x + 3)3 into partialfractions.

Let3x2 + 16x + 15

(x + 3)3

A(x + 3) +

B

(x + 3)2 +C

(x + 3)3

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


3x2 + 16x + 15 A(x + 3)2 + B(x + 3)+ C (1)

Let x =3. Then3(3)2 + 16(3)+ 15 A(0)2 + B(0)+ C

i.e. 6=CIdentity (1) may be expanded as:

3x2 + 16x + 15 A(x2 + 6x + 9)+ B(x + 3)+ C

i.e. 3x2 + 16x + 15 Ax2 + 6Ax + 9A+Bx + 3B + C

Equating the coefficients of x2 terms gives: 3 = AEquating the coefficients of x terms gives:

16 = 6A + BSince A = 3,B = 2

[Check: equating the constant terms gives:

15 = 9A + 3B + C

When A=3, B =2 and C =6,

9A + 3B + C = 9(3)+ 3(2)+ (6)= 27 6 6 = 15 = LHS]

Thus3x2 + 16x + 15

(x + 3)3

3(x + 3)

2(x + 3)2

6(x + 3)3

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Practice Exercise 9 Partial fractions withrepeated linear factors (Answers on page831)

1.4x 3(x + 1)2

2.x2 + 7x + 3x2(x + 3)

3.5x2 30x + 44

(x 2)3

4.18 + 21x x2(x 5)(x + 2)2

2.4 Worked problems on partialfractions with quadratic factors

Problem 8. Express7x2 + 5x + 13(x2 + 2)(x + 1) in partial

fractions.

The denominator is a combination of a quadratic factor,(x 2 +2), which does not factorise without introduc-ing imaginary surd terms, and a linear factor, (x +1).Let,

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

Ax + B(x2 + 2) +

C

(x + 1)

(Ax + B)(x + 1)+ C(x2 + 2)

(x2 + 2)(x + 1)Equating numerators gives:

7x2 + 5x + 13 (Ax + B)(x + 1)+ C(x2 + 2) (1)

Let x =1. Then7(1)2 + 5(1)+ 13(Ax + B)(0)+ C(1 + 2)i.e. 15= 3Ci.e. C= 5Identity (1) may be expanded as:

7x 2 + 5x + 13 Ax2 + Ax + Bx + B + Cx2 + 2C

Equating the coefficients of x2 terms gives:

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

5 = A + B,and since A = 2,B = 3[Check: equating the constant terms gives:

13 = B + 2CWhen B =3 and C =5,

B + 2C = 3 + 10 = 13 = LHS]

Hence7x2 + 5x + 13(x2 + 2)(x + 1)

2x + 3( x2 + 2) +

5(x + 1)

Problem 9. Resolve3 + 6x + 4x2 2x3

x2(x2 + 3) intopartial fractions.

Terms such as x2 may be treated as (x +0)2, i.e. theyare repeated linear factors.

Let3 + 6x + 4x2 2x3

x2(x2 + 3) A

x+ B

x2+ Cx + D(x2 + 3)

Ax(x2 + 3)+ B(x2 + 3)+ (Cx + D)x2

x2(x2 + 3)Equating the numerators gives:

3 + 6x + 4x2 2x3 Ax(x2 + 3)+ B(x2 + 3)+ (Cx + D)x2

Ax3 + 3Ax + Bx2 + 3B+ Cx3 + Dx2

Let x =0. Then 3=3Bi.e. B = 1Equating the coefficients of x3 terms gives:

2 = A + C (1)Equating the coefficients of x2 terms gives:

4 = B + DSince B = 1, D = 3

Equating the coefficients of x terms gives:

6 = 3A

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i.e. A = 2From equation (1), since A=2, C = 4

Hence3 + 6 x + 4x2 2 x3

x2(x2 + 3) 2

x+ 1

x2+ 4x + 3

x2 + 3

2x

+ 1x2

+ 3 4xx2 + 3


Practice Exercise 10 Partial fractions withquadratic factors (Answers on page 831)

1.x2 x 13

(x2 + 7)(x 2)

2.6x 5

(x 4)(x2 + 3)

3.15 + 5x + 5x2 4x3

x2(x2 + 5)

4.x3 + 4x2 + 20x 7(x 1)2(x2 + 8)

5. When solving the differential equationd2

dt2 6 d

dt10=20e2t by Laplace

transforms, for given boundary conditions,the following expression for L{} results:

L{} =4s3 39

2s2 + 42s 40

s(s 2)(s2 6s + 10)Show that the expression can be resolved intopartial fractions to give:

L{} = 2s

12(s 2) +

5s 32(s2 6s + 10)

For fully worked solutions to each of the problems in Practice Exercises 8 to 10 in this chapter,go to the website:

www.routledge.com/cw/bird


Chapter 3

Logarithms

Why it is important to understand: Logarithms

All types of engineers use natural and common logarithms. Chemical engineers use them to measureradioactive decay and pH solutions, both of which are measured on a logarithmic scale. The Richterscale, which measures earthquake intensity, is a logarithmic scale. Biomedical engineers use logarithmsto measure cell decay and growth, and also to measure light intensity for bone mineral density measure-ments. In electrical engineering, a dB (decibel) scale is very useful for expressing attenuations in radiopropagation and circuit gains, and logarithms are used for implementing arithmetic operations in digitalcircuits. Logarithms are especially useful when dealing with the graphical analysis of non-linear relation-ships and logarithmic scales are used to linearise data to make data analysis simpler. Understanding andusing logarithms is clearly important in all branches of engineering.


define base, power, exponent and index define a logarithm distinguish between common and Napierian (i.e. hyperbolic or natural) logarithms evaluate logarithms to any base state the laws of logarithms simplify logarithmic expressions solve equations involving logarithms solve indicial equations sketch graphs of log10 x and logex

3.1 Introduction to logarithms

With the use of calculators firmly established, logarith-mic tables are no longer used for calculations. However,the theory of logarithms is important, for there are sev-eral scientific and engineering laws that involve the rulesof logarithms.

From the laws of indices: 16 = 24

The number 4 is called the power or the exponent orthe index. In the expression 24, the number 2 is calledthe base.

In another example: 64 = 82In 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 thisexpression 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 poweror index in the original expression. It is the base inthe original expression which becomes the base of thelogarithm.


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Logarithms 23

The two statements: 16=24 and log2 16=4 areequivalent.

If we write either of them, we are automatically imply-ing the other.

In general, if a number y can be written in the formax , then the index x is called the logarithm of y to thebase of a,

i.e. if y = ax then x = loga y

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

log8 64 = 2In another example, if we write that log3 81=4 then theequivalent statement using powers is:

34 = 81So the two sets of statements, one involving powers andone involving logarithms, are equivalent.

Common logarithmsFrom above, if we write that 1000 = 103, then 3 =log10 1000This may be checked using the log button on yourcalculator.Logarithms having a base of 10 are called commonlogarithms and log10 is often abbreviated to lg.The following values may be checked by using acalculator:

lg27.5 = 1.4393 . . ., lg378.1 = 2.5776 . . .and lg0.0204 = 1.6903 . . .

Napierian logarithmsLogarithms having a base of e (where e is a math-ematical constant approximately equal to 2.7183) arecalled hyperbolic, Napierian or natural logarithms,and loge is usually abbreviated to ln.The following values may be checked by using acalculator:

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.

Here are some worked problems to help understandingof logarithms.

Problem 1. Evaluate log3 9

Let x = log3 9 then 3x = 9 from the definitionof a logarithm,

i.e. 3x = 32 from which, x = 2Hence, log3 9 = 2


Let x = log10 10 then 10 x = 10 from thedefinition of a logarithm,

i.e. 10 x = 101 from which,x = 1Hence, log10 10 = 1 (which may be checked

using a calculator)


Let x = log16 8 then 16x = 8 from the definitionof a logarithm,

i.e. (24)x = 23 i.e. 24x = 23 from the laws of indices,

from which, 4x = 3 and x = 34

Hence, log16 8 =34

Problem 4. Evaluate lg 0.001

Let x = lg 0.001 = log10 0.001 then 10x = 0.001i.e. 10x = 103 from which, x = 3Hence, lg 0.001 = 3 (which may be checked

using a calculator)

Problem 5. Evaluate ln e

Let x = ln e = loge e then ex = ei.e. ex = e1

from which, x = 1Hence, ln e = 1 (which may be checked

using a calculator)

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Problem 6. Evaluate log31

81

Let x = log31

81then 3x = 1

81= 1

34= 34

from which, x = 4

Hence, log31

81= 4

Problem 7. Solve the equation: lg x = 3

If lg x = 3 then log10 x = 3and x = 103 i.e. x = 1000

Problem 8. Solve the equation: log2 x = 5

If log2 x = 5 then x = 25 = 32

Problem 9. Solve the equation: log5 x = 2

If log5 x = 2 then x = 52 =1

52= 1

25


Practice Exercise 11 Introduction tologarithms (Answers on page 831)

In Problems 1 to 11, evaluate the givenexpressions:

1. log10 10000 2. log2 16

3. log5 125 4. log218

5. log8 2 6. log7 343

7. lg100 8. lg 0.01

9. log4 8 10. log27 3

11. ln e2

In Problems 12 to 18 solve the equations:

12. log10 x = 4 13. lg x = 5

14. log3 x = 2 15. log4 x = 21

2

16. lg x = 2 17. log8 x = 4

3

18. ln x = 3

3.2 Laws of logarithms

There are three laws of logarithms, which apply to anybase:

(i) To multiply two numbers:

log (A B) = log A + log BThe following may be checked by using a calcu-lator:

lg10 = 1Also, lg 5 + lg 2 = 0.69897. . .

+ 0.301029. . . = 1Hence, lg (5 2) = lg10 = lg 5 + lg 2

(ii) To divide two numbers:

log(

AB

)

= log A log B

The following may be checked using a calculator:

ln

(5

2

)

= ln 2.5 = 0.91629. . .

Also, ln 5 ln 2 = 1.60943. . . 0.69314. . .= 0.91629. . .

Hence, ln

(5

2

)

= ln 5 ln 2

(iii) To raise a number to a power:

log An = n log AThe following may be checked using a calculator:

lg 52 = lg 25 = 1.39794. . .Also, 2 lg 5 = 2 0.69897. . . = 1.39794. . .Hence, lg 52 = 2 lg 5

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Here are some worked problems to help understandingof the laws of logarithms.

Problem 10. Write log 4 + log 7 as the logarithmof a single number.

log 4 + log 7 = log (7 4)by the first law of logarithms

= log 28

Problem 11. Write log 16 log 2 as the logari-thm of a single number.

log16 log 2 = log(

16

2

)

by the second law of logarithms

= log 8

Problem 12. Write 2 log 3 as the logarithm of asingle number.

2 log 3 = log 32 by the third law of logarithms= log 9

Problem 13. Write1

2log 25 as the logarithm of a

single number.

1

2log 25 = log 25 12 by the third law of logarithms

= log25 = log 5

Problem 14. Simplify: log64 log128 + log32.

64 = 26,128 = 27 and 32 = 25

Hence, log64 log128 + log32= log26 log27 + log25

= 6log2 7log2 + 5log2by the third law of logarithms

= 4 log2

Problem 15. Write1

2log16 + 1

3log27 2 log5

as the logarithm of a single number.

1

2log16 + 1

3log27 2 log5

= log16 12 + log27 13 log52by the third law of logarithms

= log

16 + log 3

27 log25by the laws of indices

= log4 + log3 log25

= log(

4 325

)

by the first and second laws of logarithms

= log(

12

25

)

= log 0.48

Problem 16. Write (a) log30 (b) log450 in termsof log2, log3 and log5 to any base.

(a) log30 = log(2 15) = log(2 3 5)= log2 + log3 + log5

by the first law of logarithms

(b) log450 = log(2 225) = log(2 3 75)= log(2 3 3 25)= log(2 32 52)= log2 + log32 + log52

by the first law of logarithms

i.e. log450 = log2 + 2 log3 + 2 log5by the third law of logarithms

Problem 17. Write log

(8 45

81

)

in terms of

log2, log3 and log5 to any base.

log

(8 45

81

)

= log8 + log 45 log81by the first and second

laws of logarithms

= log23 + log5 14 log34by the laws of indices

i.e. log

(8 45

81

)

= 3log2 + 14

log5 4 log3by the third law of logarithms

Sect

ion

A


Problem 18. Evaluate:

log25 log125 + 12 log6253log5

log25 log125 + 12 log6253log5

= log52 log53 + 12 log54

3 log5

= 2 log5 3log5 +42 log5

3log5= 1log5

3log5= 1

3

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

LHS = log(x 1) + log(x + 8)= log(x 1)(x + 8)

from the first law of logarithms

= log(x2 + 7x 8)RHS = 2 log(x + 2) = log (x + 2)2

from the third law of logarithms

= log (x2 + 4x + 4)log(x2 + 7x 8) = log(x2 + 4x + 4)Hence,

x2 + 7x 8 = x2 + 4x + 4from which,7x 8 = 4x + 4i.e.

3x = 12i.e.x = 4and

Problem 20. Solve the equation:1

2log 4 = log x

1

2log4 = log4 12 from the third law of logarithms

= log

4 from the laws of indices

1

2log4 = log xHence,

log

4 = log xbecomeslog2 = log xi.e.

2 = xfrom which,

i.e. the solution of the equation is: x = 2

Problem 21. Solve the equation:log

(x2 3) log x = log2

log(x2 3) log x = log

(x2 3

x

)

from the second law of logarithms

log

(x2 3

x

)

= log2Hence,

x2 3x

= 2from which,

x2 3 = 2xRearranging gives:x2 2x 3 = 0and

(x 3)(x + 1) = 0Factorising gives:x = 3 or x = 1from which,

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

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


Practice Exercise 12 Laws of logarithms(Answers on page 831)

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

1. log 2 + log32. log 3 + log53. log 3 + log4 log64. log 7 + log21 log495. 2 log 2 + log36. 2 log 2 + 3log5

7. 2 log 5 12

log81 + log36

8.1

3log8 1

2log81 + log27

9.1

2log4 2 log3 + log45

10.1

4log16 + 2 log3 log18

11. 2 log2 + log5 log10

Sect

ion

A

Logarithms 27

Simplify the expressions given in Problems 12to 14:

12. log27 log9 + log8113. log64 + log32 log12814. log8 log4 + log32Evaluate the expressions given in Problems 15and 16:

15.12 log16 13 log8

log4

16.log9 log3 + 12 log81

2 log3

Solve the equations given in Problems 17 to 22:

17. log x4 log x3 = log5x log2x18. log2t3 log t = log16 + log t19. 2 logb2 3logb = log8b log4b20. log(x + 1) + log(x 1) = log3

21.1

3log 27 = log(0.5a)

22. log(x2 5) log x = log4

3.3 Indicial equations

The laws of logarithms may be used to solve certainequations involving powers called indicial equa-tions. For example, to solve, say, 3x =27, logari-thms to a base of 10 are taken of both sides,

i.e. log10 3x = log10 27

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

x = log10 27log10 3

= 1.43136 . . .0.4771 . . .

=

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