Higher Engineering Mathematics

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Higher Engineering Mathematics

  1. 1. Higher Engineering Mathematics
  2. 2. In memory of Elizabeth
  3. 3. 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
  4. 4. Newnes is an imprint of Elsevier The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, UK 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA 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. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher. Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: permissions@elsevier.com. Alternatively you can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material. Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Because of rapid advances in the medical sciences, in particular, independent verification of diagnoses and drug dosages should be made. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. Library of Congress Cataloging-in-Publication Data A catalogue record for this book is available from the Library of Congress. ISBN: 978-1-85-617767-2 For information on all Newnes publications visit our Web site at www.elsevierdirect.com Typeset by: diacriTech, India Printed and bound in China 10 11 12 13 14 15 10 9 8 7 6 5 4 3 2 1
  5. 5. Contents Preface xiii Syllabus guidance xv 1 Algebra 1 1.1 Introduction 1 1.2 Revision of basic laws 1 1.3 Revision of equations 3 1.4 Polynomial division 6 1.5 The factor theorem 8 1.6 The remainder theorem 10 2 Partial fractions 13 2.1 Introduction to partial fractions 13 2.2 Worked problems on partial fractions with linear factors 13 2.3 Worked problems on partial fractions with repeated linear factors 16 2.4 Worked problems on partial fractions with quadratic factors 17 3 Logarithms 20 3.1 Introduction to logarithms 20 3.2 Laws of logarithms 22 3.3 Indicial equations 24 3.4 Graphs of logarithmic functions 25 4 Exponential functions 27 4.1 Introduction to exponential functions 27 4.2 The power series for ex 28 4.3 Graphs of exponential functions 29 4.4 Napierian logarithms 31 4.5 Laws of growth and decay 34 4.6 Reduction of exponential laws to linear form 37 Revision Test 1 40 5 Hyperbolic functions 41 5.1 Introduction to hyperbolic functions 41 5.2 Graphs of hyperbolic functions 43 5.3 Hyperbolic identities 45 5.4 Solving equations involving hyperbolic functions 47 5.5 Series expansions for cosh x and sinh x 49 6 Arithmetic and geometric progressions 51 6.1 Arithmetic progressions 51 6.2 Worked problems on arithmetic progressions 51 6.3 Further worked problems on arithmetic progressions 52 6.4 Geometric progressions 54 6.5 Worked problems on geometric progressions 55 6.6 Further worked problems on geometric progressions 56 7 The binomial series 58 7.1 Pascal’s triangle 58 7.2 The binomial series 59 7.3 Worked problems on the binomial series 59 7.4 Further worked problems on the binomial series 62 7.5 Practical problems involving the binomial theorem 64 Revision Test 2 67 8 Maclaurin’s series 68 8.1 Introduction 68 8.2 Derivation of Maclaurin’s theorem 68 8.3 Conditions of Maclaurin’s series 69 8.4 Worked problems on Maclaurin’s series 69 8.5 Numerical integration using Maclaurin’s series 73 8.6 Limiting values 74 9 Solving equations by iterative methods 77 9.1 Introduction to iterative methods 77 9.2 The bisection method 77 9.3 An algebraic method of successive approximations 81 9.4 The Newton-Raphson method 84 10 Binary, octal and hexadecimal 87 10.1 Introduction 87 10.2 Binary numbers 87 10.3 Octal numbers 90 10.4 Hexadecimal numbers 92 Revision Test 3 96
  6. 6. vi Contents 11 Introduction to trigonometry 97 11.1 Trigonometry 97 11.2 The theorem of Pythagoras 97 11.3 Trigonometric ratios of acute angles 98 11.4 Evaluating trigonometric ratios 100 11.5 Solution of right-angled triangles 105 11.6 Angles of elevation and depression 106 11.7 Sine and cosine rules 108 11.8 Area of any triangle 108 11.9 Worked problems on the solution of triangles and finding their areas 109 11.10 Further worked problems on solving triangles and finding their areas 110 11.11 Practical situations involving trigonometry 111 11.12 Further practical situations involving trigonometry 113 12 Cartesian and polar co-ordinates 117 12.1 Introduction 117 12.2 Changing from Cartesian into polar co-ordinates 117 12.3 Changing from polar into Cartesian co-ordinates 119 12.4 Use of Pol/Rec functions on calculators 120 13 The circle and its properties 122 13.1 Introduction 122 13.2 Properties of circles 122 13.3 Radians and degrees 123 13.4 Arc length and area of circles and sectors 124 13.5 The equation of a circle 127 13.6 Linear and angular velocity 129 13.7 Centripetal force 130 Revision Test 4 133 14 Trigonometric waveforms 134 14.1 Graphs of trigonometric functions 134 14.2 Angles of any magnitude 135 14.3 The production of a sine and cosine wave 137 14.4 Sine and cosine curves 138 14.5 Sinusoidal form Asin(ωt ± α) 143 14.6 Harmonic synthesis with complex waveforms 146 15 Trigonometric identities and equations 152 15.1 Trigonometric identities 152 15.2 Worked problems on trigonometric identities 152 15.3 Trigonometric equations 154 15.4 Worked problems (i) on trigonometric equations 154 15.5 Worked problems (ii) on trigonometric equations 156 15.6 Worked problems (iii) on trigonometric equations 157 15.7 Worked problems (iv) on trigonometric equations 157 16 The relationship between trigonometric and hyperbolic functions 159 16.1 The relationship between trigonometric and hyperbolic functions 159 16.2 Hyperbolic identities 160 17 Compound angles 163 17.1 Compound angle formulae 163 17.2 Conversion of a sinωt +bcosωt into R sin(ωt +α) 165 17.3 Double angles 169 17.4 Changing products of sines and cosines into sums or differences 170 17.5 Changing sums or differences of sines and cosines into products 171 17.6 Power waveforms in a.c. circuits 173 Revision Test 5 177 18 Functions and their curves 178 18.1 Standard curves 178 18.2 Simple transformations 181 18.3 Periodic functions 186 18.4 Continuous and discontinuous functions 186 18.5 Even and odd functions 186 18.6 Inverse functions 188 18.7 Asymptotes 190 18.8 Brief guide to curve sketching 196 18.9 Worked problems on curve sketching 197 19 Irregular areas, volumes and mean values of waveforms 203 19.1 Areas of irregular figures 203 19.2 Volumes of irregular solids 205 19.3 The mean or average value of a waveform 206 Revision Test 6 212 20 Complex numbers 213 20.1 Cartesian complex numbers 213 20.2 The Argand diagram 214 20.3 Addition and subtraction of complex numbers 214 20.4 Multiplication and division of complex numbers 216
  7. 7. Contents vii 20.5 Complex equations 217 20.6 The polar form of a complex number 218 20.7 Multiplication and division in polar form 220 20.8 Applications of complex numbers 221 21 De Moivre’s theorem 225 21.1 Introduction 225 21.2 Powers of complex numbers 225 21.3 Roots of complex numbers 226 21.4 The exponential form of a complex number 228 22 The theory of matrices and determinants 231 22.1 Matrix notation 231 22.2 Addition, subtraction and multiplication of matrices 231 22.3 The unit matrix 235 22.4 The determinant of a 2 by 2 matrix 235 22.5 The inverse or reciprocal of a 2 by 2 matrix 236 22.6 The determinant of a 3 by 3 matrix 237 22.7 The inverse or reciprocal of a 3 by 3 matrix 239 23 The solution of simultaneous equations by matrices and determinants 241 23.1 Solution of simultaneous equations by matrices 241 23.2 Solution of simultaneous equations by determinants 243 23.3 Solution of simultaneous equations using Cramers rule 247 23.4 Solution of simultaneous equations using the Gaussian elimination method 248 Revision Test 7 250 24 Vectors 251 24.1 Introduction 251 24.2 Scalars and vectors 251 24.3 Drawing a vector 251 24.4 Addition of vectors by drawing 252 24.5 Resolving vectors into horizontal and vertical components 254 24.6 Addition of vectors by calculation 255 24.7 Vector subtraction 260 24.8 Relative velocity 262 24.9 i, j and k notation 263 25 Methods of adding alternating waveforms 265 25.1 Combination of two periodic functions 265 25.2 Plotting periodic functions 265 25.3 Determining resultant phasors by drawing 267 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 275 26.1 The unit triad 275 26.2 The scalar product of two vectors 276 26.3 Vector products 280 26.4 Vector equation of a line 283 Revision Test 8 286 27 Methods of differentiation 287 27.1 Introduction to calculus 287 27.2 The gradient of a curve 287 27.3 Differentiation from first principles 288 27.4 Differentiation of common functions 289 27.5 Differentiation of a product 292 27.6 Differentiation of a quotient 293 27.7 Function of a function 295 27.8 Successive differentiation 296 28 Some applications of differentiation 299 28.1 Rates of change 299 28.2 Velocity and acceleration 300 28.3 Turning points 303 28.4 Practical problems involving maximum and minimum values 307 28.5 Tangents and normals 311 28.6 Small changes 312 29 Differentiation of parametric equations 315 29.1 Introduction to parametric equations 315 29.2 Some common parametric equations 315 29.3 Differentiation in parameters 315 29.4 Further worked problems on differentiation of parametric equations 318 30 Differentiation of implicit functions 320 30.1 Implicit functions 320 30.2 Differentiating implicit functions 320 30.3 Differentiating implicit functions containing products and quotients 321 30.4 Further implicit differentiation 322 31 Logarithmic differentiation 325 31.1 Introduction to logarithmic differentiation 325 31.2 Laws of logarithms 325 31.3 Differentiation of logarithmic functions 325
  8. 8. viii Contents 31.4 Differentiation of further logarithmic functions 326 31.5 Differentiation of [ f (x)]x 328 Revision Test 9 330 32 Differentiation of hyperbolic functions 331 32.1 Standard differential coefficients of hyperbolic functions 331 32.2 Further worked problems on differentiation of hyperbolic functions 332 33 Differentiation of inverse trigonometric and hyperbolic functions 334 33.1 Inverse functions 334 33.2 Differentiation of inverse trigonometric functions 334 33.3 Logarithmic forms of the inverse hyperbolic functions 339 33.4 Differentiation of inverse hyperbolic functions 341 34 Partial differentiation 345 34.1 Introduction to partial derivatives 345 34.2 First order partial derivatives 345 34.3 Second order partial derivatives 348 35 Total differential, rates of change and small changes 351 35.1 Total differential 351 35.2 Rates of change 352 35.3 Small changes 354 36 Maxima, minima and saddle points for functions of two variables 357 36.1 Functions of two independent variables 357 36.2 Maxima, minima and saddle points 358 36.3 Procedure to determine maxima, minima and saddle points for functions of two variables 359 36.4 Worked problems on maxima, minima and saddle points for functions of two variables 359 36.5 Further worked problems on maxima, minima and saddle points for functions of two variables 361 Revision Test 10 367 37 Standard integration 368 37.1 The process of integration 368 37.2 The general solution of integrals of the form axn 368 37.3 Standard integrals 369 37.4 Definite integrals 372 38 Some applications of integration 375 38.1 Introduction 375 38.2 Areas under and between curves 375 38.3 Mean and r.m.s. values 377 38.4 Volumes of solids of revolution 378 38.5 Centroids 380 38.6 Theorem of Pappus 381 38.7 Second moments of area of regular sections 383 39 Integration using algebraic substitutions 392 39.1 Introduction 392 39.2 Algebraic substitutions 392 39.3 Worked problems on integration using algebraic substitutions 392 39.4 Further worked problems on integration using algebraic substitutions 394 39.5 Change of limits 395 Revision Test 11 397 40 Integration using trigonometric and hyperbolic substitutions 398 40.1 Introduction 398 40.2 Worked problems on integration of sin2 x, cos2 x, tan2 x and cot2 x 398 40.3 Worked problems on powers of sines and cosines 400 40.4 Worked problems on integration of products of sines and cosines 401 40.5 Worked problems on integration using the sin θ substitution 402 40.6 Worked problems on integration using tan θ substitution 404 40.7 Worked problems on integration using the sinh θ substitution 404 40.8 Worked problems on integration using the cosh θ substitution 406 41 Integration using partial fractions 409 41.1 Introduction 409 41.2 Worked problems on integration using partial fractions with linear factors 409 41.3 Worked problems on integration using partial fractions with repeated linear factors 411 41.4 Worked problems on integration using partial fractions with quadratic factors 412 42 The t =tan θ 2 substitution 414 42.1 Introduction 414 42.2 Worked problems on the t =tan θ 2 substitution 415
  9. 9. Contents ix 42.3 Further worked problems on the t = tan θ 2 substitution 416 Revision Test 12 419 43 Integration by parts 420 43.1 Introduction 420 43.2 Worked problems on integration by parts 420 43.3 Further worked problems on integration by parts 422 44 Reduction formulae 426 44.1 Introduction 426 44.2 Using reduction formulae for integrals of the form xnex dx 426 44.3 Using reduction formulae for integrals of the form xncosx dx and xn sinx dx 427 44.4 Using reduction formulae for integrals of the form sinn x dx and cosn x dx 429 44.5 Further reduction formulae 432 45 Numerical integration 435 45.1 Introduction 435 45.2 The trapezoidal rule 435 45.3 The mid-ordinate rule 437 45.4 Simpson’s rule 439 Revision Test 13 443 46 Solution of first order differential equations by separation of variables 444 46.1 Family of curves 444 46.2 Differential equations 445 46.3 The solution of equations of the form dy dx = f (x) 445 46.4 The solution of equations of the form dy dx = f (y) 447 46.5 The solution of equations of the form dy dx = f (x) · f (y) 449 47 Homogeneous first order differential equations 452 47.1 Introduction 452 47.2 Procedure to solve differential equations of the form P dy dx = Q 452 47.3 Worked problems on homogeneous first order differential equations 452 47.4 Further worked problems on homogeneous first order differential equations 454 48 Linear first order differential equations 456 48.1 Introduction 456 48.2 Procedure to solve differential equations of the form dy dx + Py = Q 457 48.3 Worked problems on linear first order differential equations 457 48.4 Further worked problems on linear first order differential equations 458 49 Numerical methods for first order differential equations 461 49.1 Introduction 461 49.2 Euler’s method 461 49.3 Worked problems on Euler’s method 462 49.4 An improved Euler method 466 49.5 The Runge-Kutta method 471 Revision Test 14 476 50 Second order differential equations of the form a d2y dx2 +b dy dx +cy=0 477 50.1 Introduction 477 50.2 Procedure to solve differential equations of the form a d2y dx2 +b dy dx +cy =0 478 50.3 Worked problems on differential equations of the form a d2y dx2 +b dy dx +cy =0 478 50.4 Further worked problems on practical differential equations of the form a d2 y dx2 +b dy dx +cy =0 480 51 Second order differential equations of the form a d2y dx2 +b dy dx +cy=f(x) 483 51.1 Complementary function and particular integral 483 51.2 Procedure to solve differential equations of the form a d2y dx2 +b dy dx +cy = f (x) 483 51.3 Worked problems on differential equations of the form a d2y dx2 +b dy dx + cy = f (x) where f (x) is a constant or polynomial 484 51.4 Worked problems on differential equations of the form a d2y dx2 +b dy dx + cy = f (x) where f (x) is an exponential function 486 51.5 Worked problems on differential equations of the form a d2y dx2 +b dy dx + cy = f (x) where f (x) is a sine or cosine function 488
  10. 10. x Contents 51.6 Worked problems on differential equations of the form a d2 y dx2 +b dy dx + cy = f (x) where f (x) is a sum or a product 490 52 Power series methods of solving ordinary differential equations 493 52.1 Introduction 493 52.2 Higher order differential coefficients as series 493 52.3 Leibniz’s theorem 495 52.4 Power series solution by the Leibniz–Maclaurin method 497 52.5 Power series solution by the Frobenius method 500 52.6 Bessel’s equation and Bessel’s functions 506 52.7 Legendre’s equation and Legendre polynomials 511 53 An introduction to partial differential equations 515 53.1 Introduction 515 53.2 Partial integration 515 53.3 Solution of partial differential equations by direct partial integration 516 53.4 Some important engineering partial differential equations 518 53.5 Separating the variables 518 53.6 The wave equation 519 53.7 The heat conduction equation 523 53.8 Laplace’s equation 525 Revision Test 15 528 54 Presentation of statistical data 529 54.1 Some statistical terminology 529 54.2 Presentation of ungrouped data 530 54.3 Presentation of grouped data 534 55 Measures of central tendency and dispersion 541 55.1 Measures of central tendency 541 55.2 Mean, median and mode for discrete data 541 55.3 Mean, median and mode for grouped data 542 55.4 Standard deviation 544 55.5 Quartiles, deciles and percentiles 546 56 Probability 548 56.1 Introduction to probability 548 56.2 Laws of probability 549 56.3 Worked problems on probability 549 56.4 Further worked problems on probability 551 Revision Test 16 554 57 The binomial and Poisson distributions 556 57.1 The binomial distribution 556 57.2 The Poisson distribution 559 58 The normal distribution 562 58.1 Introduction to the normal distribution 562 58.2 Testing for a normal distribution 566 59 Linear correlation 570 59.1 Introduction to linear correlation 570 59.2 The product-moment formula for determining the linear correlation coefficient 570 59.3 The significance of a coefficient of correlation 571 59.4 Worked problems on linear correlation 571 60 Linear regression 575 60.1 Introduction to linear regression 575 60.2 The least-squares regression lines 575 60.3 Worked problems on linear regression 576 Revision Test 17 581 61 Introduction to Laplace transforms 582 61.1 Introduction 582 61.2 Definition of a Laplace transform 582 61.3 Linearity property of the Laplace transform 582 61.4 Laplace transforms of elementary functions 582 61.5 Worked problems on standard Laplace transforms 583 62 Properties of Laplace transforms 587 62.1 The Laplace transform of eat f (t) 587 62.2 Laplace transforms of the form eat f (t) 587 62.3 The Laplace transforms of derivatives 589 62.4 The initial and final value theorems 591 63 Inverse Laplace transforms 593 63.1 Definition of the inverse Laplace transform 593 63.2 Inverse Laplace transforms of simple functions 593 63.3 Inverse Laplace transforms using partial fractions 596 63.4 Poles and zeros 598 64 The solution of differential equations using Laplace transforms 600 64.1 Introduction 600 64.2 Procedure to solve differential equations by using Laplace transforms 600 64.3 Worked problems on solving differential equations using Laplace transforms 600
  11. 11. Contents xi 65 The solution of simultaneous differential equations using Laplace transforms 605 65.1 Introduction 605 65.2 Procedure to solve simultaneous differential equations using Laplace transforms 605 65.3 Worked problems on solving simultaneous differential equations by using Laplace transforms 605 Revision Test 18 610 66 Fourier series for periodic functions of period 2π 611 66.1 Introduction 611 66.2 Periodic functions 611 66.3 Fourier series 611 66.4 Worked problems on Fourier series of periodic functions of period 2π 612 67 Fourier series for a non-periodic function over range 2π 617 67.1 Expansion of non-periodic functions 617 67.2 Worked problems on Fourier series of non-periodic functions over a range of 2π 617 68 Even and odd functions and half-range Fourier series 623 68.1 Even and odd functions 623 68.2 Fourier cosine and Fourier sine series 623 68.3 Half-range Fourier series 626 69 Fourier series over any range 630 69.1 Expansion of a periodic function of period L 630 69.2 Half-range Fourier series for functions defined over range L 634 70 A numerical method of harmonic analysis 637 70.1 Introduction 637 70.2 Harmonic analysis on data given in tabular or graphical form 637 70.3 Complex waveform considerations 641 71 The complex or exponential form of a Fourier series 644 71.1 Introduction 644 71.2 Exponential or complex notation 644 71.3 The complex coefficients 645 71.4 Symmetry relationships 649 71.5 The frequency spectrum 652 71.6 Phasors 653 Revision Test 19 658 Essential formulae 659 Index 675
  12. 12. xii Contents Website Chapters 72 Inequalities 1 72.1 Introduction to inequalities 1 72.2 Simple inequalities 1 72.3 Inequalities involving a modulus 2 72.4 Inequalities involving quotients 3 72.5 Inequalities involving square functions 4 72.6 Quadratic inequalities 5 73 Boolean algebra and logic circuits 7 73.1 Boolean algebra and switching circuits 7 73.2 Simplifying Boolean expressions 12 73.3 Laws and rules of Boolean algebra 12 73.4 De Morgan’s laws 14 73.5 Karnaugh maps 15 73.6 Logic circuits 19 73.7 Universal logic gates 23 Revision Test 20 28 74 Sampling and estimation theories 29 74.1 Introduction 29 74.2 Sampling distributions 29 74.3 The sampling distribution of the means 29 74.4 The estimation of population parameters based on a large sample size 33 74.5 Estimating the mean of a population based on a small sample size 38 75 Significance testing 42 75.1 Hypotheses 42 75.2 Type I and Type II errors 42 75.3 Significance tests for population means 49 75.4 Comparing two sample means 54 76 Chi-square and distribution-free tests 59 76.1 Chi-square values 59 76.2 Fittingdatatotheoreticaldistributions 60 76.3 Introduction to distribution-free tests 67 76.4 The sign test 68 76.5 Wilcoxon signed-rank test 71 76.6 The Mann-Whitney test 75 Revision Test 21 82
  13. 13. Preface This sixth edition of ‘Higher Engineering Mathe- matics’ 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 equa- tions, statistics and probability, Laplace transforms and Fourier series. New material has been added on log- arithms and exponential functions, binary, octal and hexadecimal, vectors and methods of adding alternat- ing 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 com- plete scientific and engineering principles modules of Degree, Foundation Degree and Higher National Engi- neering programmes. The material has been designed to enable students to use techniques learned for the analysis, modellingand solutionof 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 mechan- ical engineering, aeronautical engineering, electronics, communications engineering, systems engineering and all variants of control engineering. In Higher Engineering Mathematics 6th Edition, the- ory is introduced in each chapter by a full outline of essential definitions, formulae, laws, procedures etc. The theoryiskept to a minimum, forproblem solving is extensively used to establish and exemplify the theory. It is intended that readers will gain real understand- ing through seeing problems solved and then through solving similar problems themselves. Access to software packages such as Maple, Mathemat- ica 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’ pro- vides a follow-up to ‘Engineering Mathematics 6th Edition’. This textbook contains some 900 worked prob- lems, 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 fur- ther 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 cov- ers 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 Inequali- ties, 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
  14. 14. 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, simul- taneous 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 theo- ries, 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 lec- turers 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.
  15. 15. 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 Further Engineering Methods Analytical Mathematics for Methods for Engineers 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 × (Continued)
  16. 16. xvi Syllabus Guidance Chapter Analytical Further Engineering Methods Analytical Mathematics for Methods for Engineers Engineers 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 = tanθ/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 × a d2y dx2 + b dy dx + cy = 0 51. Second order differential equations of the form × a d2y dx2 + b dy dx + cy = f (x) 52. Power series methods of solving ordinary differential equations × 53. An introduction to partial differential equations × 54. Presentation of statistical data × (Continued)
  17. 17. Syllabus Guidance xvii Chapter Analytical Further Engineering Methods Analytical Mathematics for Methods for Engineers Engineers 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 ×
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  19. 19. Chapter 1 Algebra 1.1 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 Revision of basic laws (a) Basic operations and laws of indices The laws of indices are: (i) am × an = am+n (ii) am an = am−n (iii) (am)n = am×n (iv) a m n = n √ am (v) a−n = 1 an (vi) a0 = 1 Problem 1. Evaluate 4a2bc3−2ac when a=2, b = 1 2 and c = 11 2 4a2 bc3 − 2ac = 4(2)2 1 2 3 2 3 − 2(2) 3 2 = 4 × 2 × 2 × 3 × 3 × 3 2 × 2× 2 × 2 − 12 2 = 27 − 6 = 21 Problem 2. Multiply 3x + 2y by x − y. 3x + 2y x − y Multiply by x → 3x2 + 2xy Multiply by −y → −3xy − 2y2 Adding gives: 3x2 − xy − 2y2 Alternatively, (3x + 2y)(x − y) = 3x2 − 3xy + 2xy − 2y2 = 3x2 − xy − 2y2 Problem 3. Simplify a3 b2 c4 abc−2 and evaluate when a = 3, b = 1 8 and c = 2. a3b2c4 abc−2 = a3−1 b2−1 c4−(−2) = a2 bc6 When a = 3, b = 1 8 and c = 2, a2 bc6 = (3)2 1 8 (2)6 = (9) 1 8 (64) = 72 Problem 4. Simplify x2 y3 + xy2 xy x2 y3 + xy2 xy = x2 y3 xy + xy2 xy = x2−1 y3−1 + x1−1 y2−1 = xy2 + y or y(xy + 1)
  20. 20. 2 Higher Engineering Mathematics Problem 5. Simplify (x2√ y)( √ x 3 y2) (x5 y3) 1 2 (x2√ y)( √ x 3 y2) (x5 y3) 1 2 = x2 y 1 2 x 1 2 y 2 3 x 5 2 y 3 2 = x2+ 1 2 − 5 2 y 1 2 + 2 3 − 3 2 = x0 y− 1 3 = y− 1 3 or 1 y 1 3 or 1 3 √ y Now try the following exercise 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 5pq2r3 when p = 2 5 , q = −2 and r = −1. [−8] 3. From 4x − 3y + 2z subtract x + 2y − 3z. [3x − 5y + 5z] 4. Multiply 2a − 5b + c by 3a + b. [6a2 − 13ab + 3ac − 5b2 + bc] 5. Simplify (x2 y3 z)(x3 yz2 ) and evaluate when x = 1 2 , y = 2 and z = 3. [x5 y4z3,131 2] 6. Evaluate (a 3 2 bc−3)(a 1 2 b− 1 2 c) when a=3, b = 4 and c = 2. [±41 2 ] 7. Simplify a2b + a3b a2b2 1 + a b 8. Simplify (a3b 1 2 c− 1 2 )(ab) 1 3 ( √ a3 √ bc) a 11 6 b 1 3 c− 3 2 or 6 √ a 11 3 √ b √ c3 (b) Brackets, factorization 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) Problem 7. Remove the brackets and simplify the expression: 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 or 36b − 11a Problem 8. Factorize (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, multiplica- tion, addition and subtraction (sometimes remembered by BODMAS). Hence
  21. 21. Algebra 3 3c + 2c × 4c + c ÷ 5c − 8c = 3c + 2c × 4c + c 5c − 8c = 3c + 8c2 + 1 5 − 8c = 8c2 − 5c + 1 5 or c(8c − 5)+ 1 5 Problem 10. Simplify (2a − 3)÷4a +5 × 6−3a. (2a − 3)÷4a + 5 × 6 − 3a = 2a − 3 4a + 5 × 6 − 3a = 2a − 3 4a + 30 − 3a = 2a 4a − 3 4a + 30 − 3a = 1 2 − 3 4a + 30 − 3a = 30 1 2 − 3 4a − 3a Now try the following exercise Exercise 2 Further problems on brackets, factorization and precedence 1. Simplify2( p + 3q −r) − 4(r − q + 2p) + p. [−5p + 10q − 6r] 2. Expand and simplify (x + y)(x − 2y). [x2 − xy − 2y2] 3. Remove the brackets and simplify: 24p − [2{3(5p − q) − 2( p + 2q)} + 3q]. [11q − 2p] 4. Factorize 21a2b2 − 28ab. [7ab(3ab − 4)] 5. Factorize 2xy2 + 6x2 y + 8x3 y. [2xy(y + 3x + 4x2)] 6. Simplify 2y + 4 ÷ 6y + 3 × 4− 5y. 2 3y − 3y + 12 7. Simplify 3 ÷ y + 2 ÷ y − 1. 5 y − 1 8. Simplify a2 − 3ab × 2a ÷ 6b + ab. [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 + 3x i.e. 15 = 5x from which, x = 15 5 = 3 Problem 12. Solve 4(2a − 3) − 2(a − 4) = 3(a − 3) − 1. Removing the brackets gives: 8a − 12 − 2a + 8 = 3a − 9 − 1 Rearranging gives: 8a − 2a − 3a = −9 − 1 + 12 − 8 i.e. 3a = −6 and a = −6 3 = −2 Problem 13. Solve 3 x − 2 = 4 3x + 4 . By ‘cross-multiplying’: 3(3x + 4)= 4(x − 2) Removing brackets gives: 9x + 12= 4x − 8 Rearranging gives: 9x − 4x = −8 − 12 i.e. 5x = −20 and x = −20 5 = −4 Problem 14. Solve √ t + 3 √ t = 2.
  22. 22. 4 Higher Engineering Mathematics √ t √ t + 3 √ t = 2 √ t i.e. √ t + 3= 2 √ t and 3= 2 √ t − √ t i.e. 3= √ t and 9= t (b) Transposition of formulae Problem 15. Transpose the formula v = u + f t m to make f the subject. u + f t m = v from which, f t m = v − u and m f t m = m(v − u) i.e. f t = m(v − u) and f = m t (v − u) Problem 16. The impedance of an a.c. circuit is given by Z = √ R2 + X2. Make the reactance X the subject. R2 + X2 = Z and squaring both sides gives R2 + X2 = Z2 ,from which, X2 = Z2 − R2 and reactance X = Z2 −R2 Problem 17. Given that D d = f + p f − p , express p in terms of D, d and f . Rearranging gives: f + p f − p = D d Squaring both sides gives: f + p f − p = D2 d2 ‘Cross-multiplying’ gives: d2( f + p)= D2( f − p) Removing brackets gives: d2 f + d2 p= D2 f − D2 p Rearranging gives: d2 p + D2 p = D2 f − d2 f Factorizing gives: p(d2 + D2)= f (D2 − d2) and p = f (D2 −d2) (d2 +D2 ) Now try the following exercise Exercise 3 Further problems on simple equations and transposition of formulae In problems 1 to 4 solve the equations 1. 3x − 2 − 5x = 2x − 4. 1 2 2. 8 + 4(x − 1) − 5(x − 3) = 2(5 − 2x). [−3] 3. 1 3a − 2 + 1 5a + 3 = 0. −1 8 4. 3 √ t 1 − √ t = −6. [4] 5. Transpose y = 3(F − f ) L . for f . f = 3F − yL 3 or f = F − yL 3 6. Make l the subject of t = 2π 1 g . l = t2g 4π2 7. Transpose m = μL L +rCR for L. L = mrCR μ − m 8. Make r the subject of the formula x y = 1 +r2 1 −r2 . r = x − y x + y (c) Simultaneous equations Problem 18. Solve the simultaneous equations: 7x − 2y = 26 (1) 6x + 5y = 29. (2)
  23. 23. Algebra 5 5×equation (1) gives: 35x − 10y = 130 (3) 2×equation (2) gives: 12x + 10y = 58 (4) equation (3)+equation (4) gives: 47x + 0 = 188 from which, x = 188 47 = 4 Substituting x = 4 in equation (1) gives: 28 − 2y = 26 from which, 28 − 26 = 2y and y =1 Problem 19. Solve x 8 + 5 2 = y (1) 11 + y 3 = 3x. (2) 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) 8×equation (6) gives: 72x − 8y = 264 (7) Equation (7) − equation (5) gives: 71x = 284 from which, x = 284 71 = 4 Substituting x = 4 in equation (5) gives: 4− 8y = −20 from which, 4 + 20= 8y and y = 3 (d) Quadratic equations Problem 20. Solve the following equations by factorization: (a) 3x2 − 11x − 4 = 0 (b) 4x2 + 8x + 3 = 0. (a) Thefactorsof3x2 are3x and x and theseareplaced in 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 product of the two outer terms must equal −11x, the only combination to give this is +1 and −4, i.e., 3x2 − 11x − 4= (3x + 1)(x − 4) Thus (3x + 1)(x − 4)= 0 hence either (3x + 1)= 0 i.e. x = −1 3 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 =−3 2 or (2x + 1)= 0 i.e. x = −1 2 Problem 21. The roots of a quadratic equation are 1 3 and −2. Determine the equation in x. If 1 3 and −2 are the roots of a quadratic equation then, (x − 1 3 )(x + 2) = 0 i.e. x2 + 2x − 1 3 x − 2 3 = 0 i.e. x2 + 5 3 x − 2 3 = 0 or 3x2 + 5x−2 = 0 Problem 22. Solve 4x2 + 7x + 2 = 0 giving the answer correct to 2 decimal places. From the quadratic formula if ax2 + bx + c = 0 then, x = −b ± √ b2 − 4ac 2a Hence if 4x2 + 7x + 2 = 0 then x = −7 ± 72 − 4(4)(2) 2(4) = −7 ± √ 17 8 = −7 ± 4.123 8 = −7 + 4.123 8 or −7 − 4.123 8 i.e. x= −0.36 or −1.39
  24. 24. 6 Higher Engineering Mathematics Now try the following exercise Exercise 4 Further problems on simultaneous and quadratic equations In problems 1 to 3, solve the simultaneous equa- tions 1. 8x − 3y = 51 3x + 4y = 14. [x = 6, y = −1] 2. 5a = 1 − 3b 2b + a + 4 = 0. [a = 2, b = −3] 3. x 5 + 2y 3 = 49 15 3x 7 − y 2 + 5 7 = 0. [x = 3, y = 4] 4. Solve the following quadratic equations by factorization: (a) x2 + 4x − 32 = 0 (b) 8x2 + 2x − 15 = 0. [(a) 4, −8 (b) 5 4 , −3 2 ] 5. Determine the quadratic equation in x whose roots are 2 and −5. [x2 + 3x − 10 = 0] 6. Solve the following quadratic equations, cor- rect to 3 decimal places: (a) 2x2 + 5x − 4 = 0 (b) 4t2 − 11t + 3 = 0. (a) 0.637,−3.137 (b) 2.443,0.307 1.4 Polynomial division 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!) For example, 208 16 is achieved as follows: 13 ——– 16 208 16 48 48 — · · — (1) 16 divided into 2 won’t 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 Hence 208 16 = 13 exactly Similarly, 172 15 is laid out as follows: 11 ——– 15 172 15 22 15 — 7 — Hence 172 15 = 11 remainder 7 or 11 + 7 15 = 11 7 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 f (x) = a + bx + cx2 + dx3 + ··· and polynomial division is sometimes required when resolving into partial fractions—see Chapter 2.)
  25. 25. Algebra 7 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 dividend and divisor both arranged in descending powers of the symbols. 2x + 3 ——————– x − 1 2x2 + x − 3 2x2 − 2x 3x − 3 3x − 3 ——— · · ——— Dividing the first term of the dividend by the first term of the divisor, i.e. 2x2 x gives 2x, which is put above the first term of the dividend as shown. The divisor is 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. 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. 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 + 5 3x3 + 3x2 −2x2 + 3x + 5 −2x2 − 2x ————– 5x + 5 5x + 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 the dividend (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. Simplify x3 + y3 x + y . (1) (4) (7) x2 − xy + y2 —————————– x + y x3 + 0 + 0 + y3 x3 + x2 y −x2 y + y3 −x2 y − xy2 ——————— xy2 + y3 xy2 + y3 ———– · · ———– (1) x into x3 goes x2. Put x2 above x3 of dividend (2) x2(x + y) = x3 + x2 y (3) Subtract (4) x into −x2 y goes −xy. Put −xy above dividend (5) −xy(x + y) = −x2 y − xy2 (6) Subtract (7) x into xy2 goes y2. Put y2 above dividend (8) y2(x + y) = xy2 + y3 (9) Subtract Thus x3 + y3 x + y = x2 − xy + y2
  26. 26. 8 Higher Engineering Mathematics The zero’s shown in the dividend are not normally shown,but areincludedtoclarifythesubtractionprocess and to keep similar terms in their respective columns. Problem 26. Divide (x2 + 3x − 2) by (x − 2). x + 5 ——————– x − 2 x2 + 3x − 2 x2 − 2x 5x − 2 5x − 10 ——— 8 ——— Hence x2 + 3x − 2 x − 2 = x + 5 + 8 x − 2 Problem 27. Divide 4a3 − 6a2b + 5b3 by 2a − b. 2a2 − 2ab − b2 ——————————— 2a − b 4a3 − 6a2b + 5b3 4a3 − 2a2b −4a2b + 5b3 −4a2b + 2ab2 ———— −2ab2 + 5b3 −2ab2 + b3 —————– 4b3 —————– Thus 4a3 − 6a2b + 5b3 2a − b =2a2 − 2ab − b2 + 4b3 2a − b Now try the following exercise Exercise 5 Further problems on polynomial division 1. Divide (2x2 + xy − y2) by (x + y). [2x − y] 2. Divide (3x2 + 5x − 2) by (x + 2). [3x − 1] 3. Determine (10x2 + 11x − 6) ÷ (2x + 3). [5x − 2] 4. Find 14x2 − 19x − 3 2x − 3 . [7x + 1] 5. Divide (x3 + 3x2 y + 3xy2 + y3) by (x + y). [x2 + 2xy + y2] 6. Find (5x2 − x + 4) ÷ (x − 1). 5x + 4 + 8 x − 1 7. Divide (3x3 + 2x2 − 5x + 4) by (x + 2). 3x2 − 4x + 3 − 2 x + 2 8. Determine (5x4 + 3x3 − 2x + 1)/(x − 3). 5x3 + 18x2 + 54x + 160 + 481 x − 3 1.5 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 x2 + 2x − 8 = 0. To solve this we may factorize the quadratic expression x2 + 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, either (x − 2) = 0, from which, x = 2 or (x + 4) = 0, from which, x = −4 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: 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 deter- mine that x = 2 is a root ofthe equation x2 + 2x − 8 = 0 we could deduce at once that (x − 2) is a factor of the
  27. 27. Algebra 9 expression x2 + 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 x3 − 7x − 6 and use it to solve the cubic equation x3 − 7x − 6 = 0. Let f (x) = x3 − 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 x3 −7x −6 by (x − 3) or we could continue our ‘trial anderror’ by sub- stituting 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: x2 + 3x + 2————————— x − 3 x3 − 0 − 7x − 6 x3 − 3x2 3x2 − 7x − 6 3x2 − 9x ———— 2x − 6 2x − 6 ——— · · ——— Hence x3 − 7x − 6 x − 3 = x2 + 3x + 2 i.e. x3 − 7x − 6 = (x − 3)(x2 + 3x + 2) x2 + 3x + 2 factorizes ‘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 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 x3 − 7x − 6 = 0, we substitute the fac- tors, 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 x3 −2x2 − 5x + 6=0 by using the factor theorem. Let f (x) = x3 − 2x2 − 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 Hence x3 − 2x2 − 5x + 6 = (x − 1)(x − 3)(x + 2) Therefore if x3 − 2x2 − 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 (x3 − 2x2 − 5x + 6) as follows:
  28. 28. 10 Higher Engineering Mathematics x2 − x − 6 ————————– x − 1 x3 − 2x2 − 5x + 6 x3 − 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) Summarizing, the factor theorem provides us with a method of factorizing simple expressions, and an alter- native, 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. x2 + 2x − 3 [(x − 1)(x + 3)] 2. x3 + x2 − 4x − 4 [(x + 1)(x + 2)(x − 2)] 3. 2x3 + 5x2 − 4x − 7 [(x + 1)(2x2 + 3x − 7)] 4. 2x3 − x2 − 16x + 15 [(x − 1)(x + 3)(2x − 5)] 5. Use the factor theorem to factorize x3 + 4x2 + x − 6 and hence solve the cubic equation x3 + 4x2 + x − 6 = 0. ⎡ ⎢ ⎣ x3 + 4x2 + x − 6 = (x − 1)(x + 3)(x + 2) x = 1, x = −3 and x = −2 ⎤ ⎥ ⎦ 6. Solve the equation x3 − 2x2 − x + 2 = 0. [x = 1, x = 2 and x = −1] 1.6 The remainder theorem Dividing a general quadratic expression (ax2 + bx + c) by (x − p), where p is any whole number, by long division (see section 1.3) gives: ax + (b + ap) ————————————– x − p ax2 + bx + c ax2 − apx (b + ap)x + c (b + ap)x − (b + ap)p —————————– c + (b + ap)p —————————– The remainder, c + (b + ap)p = c + bp + ap2 or ap2 + bp + c. This is, in fact, what the remainder theorem 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 x we get the remainder ap2 + bp + c. For 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 = 9 We can check this by dividing (3x2 − 4x + 5) by (x − 2) by long division: 3x + 2 ——————– x − 2 3x2 − 4x + 5 3x2 − 6x 2x + 5 2x − 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 is 4(−3)2 + (−7)(−3) + 9 = 36 + 21 + 9 = 66. Also, when (x2 + 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 (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).
  29. 29. Algebra 11 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 cubic equation as: ‘if (ax3 + bx2 + cx + d) is divided by (x − p), the remainder will be ap3 + bp2 + cp + d’ As before, the remainder may be obtained by substitut- ing p for x in the dividend. For example, when (3x3 + 2x2 − x + 4) is divided by (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 = 8. Similarly, when (x3 − 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 (x3 − 7x − 6). Here are some more examples on the remainder theorem. Problem 30. Without dividing out, find the remainder when 2x2 −3x + 4 is divided by (x −2). By the remainder theorem, the remainder is given by ap2 + 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 (3x3 −2x2 + x − 5) is divided by (x + 2). By the remainder theorem, the remainder is given by ap3 + 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 (x3 − 2x2 − 5x + 6) is divided by (a) (x − 1) and (b) (x + 2). Hence factorize 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.Todeterminethethirdfactor(shown blank) 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 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 (x3 − 2x2 − 5x + 6) by (x2 + x − 2) gives: x − 3 ————————– x2 + x − 2 x3 − 2x2 − 5x + 6 x3 + x2 − 2x —————— −3x2 − 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 by
  30. 30. 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 (x3 − 2x2 − 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 3x2 − 4x + 2 is divided by (a) (x − 2) (b) (x + 1). [(a) 6 (b) 9] 2. Determine the remainder when x3 − 6x2 + 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 x3 − 6x2 + 11x − 6. [(x − 1)(x − 2)(x − 3)] 4. Determine the factors of x3 + 7x2 + 14x + 8 and hence solve the cubic equation x3 +7x2 + 14x + 8 = 0. [x = −1, x = −2 and x = −4] 5. Determine the value of ‘a’ if (x + 2) is a factor of (x3 − ax2 + 7x + 10). [a = −3] 6. Using the remainder theorem, solve the equation 2x3 − x2 − 7x + 6 = 0. [x = 1, x = −2 and x = 1.5]
  31. 31. Chapter 2 Partial fractions 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 − 5 x2 − x − 2 The reverse process of moving from 4x − 5 x2 − x − 2 to 1 x − 2 + 3 x + 1 is called resolving into partial fractions. In order to resolve an algebraic expression intopartial fractions: (i) the denominator must factorize (in the above example, x2 −x −2 factorizes as (x −2) (x +1)), and (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 x1 and (x2 −x −2) is of degree 2). 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)3 A (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) 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 lessdegreethan thedenominator(seeProblems3 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, ax2 +bx +c is a quadratic expression which does not factorize without containing surds or imaginary terms.) Resolving an algebraic expression into partial frac- tions is used as a preliminary to integrating certain functions (see Chapter 41) and in determining inverse Laplace transforms (see Chapter 63). 2.2 Worked problems on partial fractions with linear factors Problem 1. Resolve 11 − 3x x2 + 2x − 3 into partial fractions.
  32. 32. 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 x2 + 2x − 3 may be resolved into partial fractions. Let 11 − 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 the identity then the numerators are equal to each other. 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. When x =1, then 11−3(1) ≡ A(1+3)+ B(0) i.e. 8 =4A i.e. A =2 When x = −3, then 11−3(−3) ≡ A(0)+ B(−3−1) i.e. 20 =−4B i.e. B =−5 Thus 11 − 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 − 3x x2 + 2x − 3 Problem 2. Convert 2x2 − 9x − 35 (x + 1)(x − 2)(x + 3) into the sum of three partial fractions. Let 2x2 − 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. Then 2(−1)2 −9(−1)−35 ≡ A(−3)(2) +B(0)(2)+C(0)(−3) i.e. −24 =−6A i.e. A = −24 −6 =4 Let x =2. Then 2(2)2 −9(2)−35 ≡ A(0)(5)+ B(3)(5)+C(3)(0) i.e. −45 =15B i.e. B = −45 15 =−3 Let x = − 3. Then 2(−3)2 −9(−3)−35 ≡ A(−5)(0)+ B(−2)(0) +C(−2)(−5) i.e. 10 =10C i.e. C =1 Thus 2x2 −9x −35 (x +1)(x −2)(x +3) ≡ 4 (x + 1) − 3 (x − 2) + 1 (x + 3) Problem 3. Resolve x2 + 1 x2 − 3x + 2 into partial fractions.
  33. 33. Partial fractions 15 The denominatoris ofthe same degree as the numerator. Thus dividing out gives: 1 x2 −3x +2 x2 + 1 x2 −3x +2 ————— 3x − 1 ——— For more on polynomial division, see Section 1.4, page 6. Hence x2 + 1 x2 − 3x + 2 ≡1 + 3x − 1 x2 − 3x + 2 ≡1 + 3x − 1 (x − 1)(x − 2) Let 3x − 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 = −A i.e. A = −2 Let x = 2. Then 5 = B Hence 3x −1 (x −1)(x −2) ≡ −2 (x −1) + 5 (x −2) Thus x2 + 1 x2 − 3x+ 2 ≡1− 2 (x−1) + 5 (x−2) Problem 4. Express x3 − 2x2 − 4x − 4 x2 + x − 2 in partial fractions. The numerator is ofhigher degree than the denominator. Thus dividing out gives: x − 3 x2 + x − 2 x3 − 2x2 − 4x − 4 x3 + x2 − 2x —————— − 3x2 − 2x − 4 − 3x2 − 3x + 6 ——————— x − 10 Thus x3 − 2x2 −4x − 4 x2 +x −2 ≡ x −3+ x −10 x2 +x −2 ≡ x −3+ x −10 (x +2)(x −1) Let x −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 = −3A i.e. A= 4 Let x =1. Then −9 = 3B i.e. B= −3 Hence x −10 (x +2)(x −1) ≡ 4 (x +2) − 3 (x −1) Thus x3 −2x2 −4x−4 x2 +x−2 ≡x−3+ 4 (x+2) − 3 (x−1) Now try the following exercise Exercise 8 Further problems on partial fractions with linear factors Resolve the following into partial fractions. 1. 12 x2 − 9 2 (x − 3) − 2 (x + 3) 2. 4(x − 4) x2 − 2x − 3 5 (x + 1) − 1 (x − 3) 3. x2 − 3x + 6 x(x − 2)(x − 1) 3 x + 2 (x − 2) − 4 (x − 1) 4. 3(2x2 − 8x − 1) (x + 4)(x + 1)(2x − 1) 7 (x + 4) − 3 (x + 1) − 2 (2x − 1)
  34. 34. 16 Higher Engineering Mathematics 5. x2 + 9x + 8 x2 + x − 6 1 + 2 (x + 3) + 6 (x − 2) 6. x2 − x − 14 x2 − 2x − 3 1 − 2 (x − 3) + 3 (x + 1) 7. 3x3 − 2x2 − 16x + 20 (x − 2)(x + 2) 3x − 2 + 1 (x − 2) − 5 (x + 2) 2.3 Worked problems on partial fractions with repeated linear factors Problem 5. Resolve 2x + 3 (x − 2)2 into partial fractions. The denominator contains a repeated linear factor, (x − 2)2. Let 2x + 3 (x − 2)2 ≡ A (x − 2) + B (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 − 2A + 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: 3 = −2A + B When A=2 and B =7, R.H.S. = −2(2) + 7 = 3 = L.H.S.] Hence 2x + 3 (x − 2)2 ≡ 2 (x − 2) + 7 (x − 2)2 Problem 6. Express 5x2 − 2x − 19 (x + 3)(x − 1)2 as the sum of three partial fractions. The denominator is a combination of a linear factor and a repeated linear factor. Let 5x2 − 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) Let x =−3. Then 5(−3)2 − 2(−3) − 19 ≡ A(−4)2 + B(0)(−4) + C(0) i.e. 32 = 16A 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 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 + 2Bx −3B + Cx + 3C
  35. 35. Partial fractions 17 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] Hence 5x2 − 2x − 19 (x + 3)(x− 1)2 ≡ 2 (x + 3) + 3 (x − 1) − 4 (x − 1)2 Problem 7. Resolve 3x2 + 16x + 15 (x + 3)3 into partial fractions. Let 3x2 + 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 Equating the numerators gives: 3x2 + 16x + 15 ≡ A(x + 3)2 + B(x + 3) + C (1) 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: 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 = A Equating the coefficients of x terms gives: 16 = 6A + B Since 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] Thus 3x2 + 16x+ 15 (x + 3)3 ≡ 3 (x + 3) − 2 (x + 3)2 − 6 (x + 3)3 Now try the following exercise Exercise 9 Further problems on partial fractions with linear factors 1. 4x − 3 (x + 1)2 4 (x + 1) − 7 (x + 1)2 2. x2 + 7x + 3 x2(x + 3) 1 x2 + 2 x − 1 (x + 3) 3. 5x2 − 30x + 44 (x − 2)3 5 (x − 2) − 10 (x − 2)2 + 4 (x − 2)3 4. 18 + 21x − x2 (x − 5)(x + 2)2 2 (x − 5) − 3 (x + 2) + 4 (x + 2)2 2.4 Worked problems on partial fractions with quadratic factors Problem 8. Express 7x2 + 5x + 13 (x2 + 2)(x + 1) in partial fractions.
  36. 36. 18 Higher Engineering Mathematics The denominator is a combination of a quadratic factor, (x2 +2), which does not factorize 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. Then 7(−1)2 + 5(−1) + 13≡(Ax + B)(0) + C(1 + 2) i.e. 15= 3C i.e. C = 5 Identity (1) may be expanded as: 7x2 + 5x + 13 ≡ Ax2 + Ax + Bx + B + Cx2 + 2C Equating the coefficients of x2 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 [Check: equating the constant terms gives: 13 = B + 2C When B =3 and C =5, B + 2C = 3 + 10 = 13 = LHS] Hence 7x2 + 5x + 13 (x2 + 2)(x + 1) ≡ 2x + 3 ( x2 + 2) + 5 (x + 1) Problem 9. Resolve 3 + 6x + 4x2 − 2x3 x2(x2 + 3) into partial fractions. Terms such as x2 may be treated as (x +0)2, i.e. they are repeated linear factors. Let 3 + 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=3B i.e. B = 1 Equating the coefficients of x3 terms gives: −2 = A + C (1) Equating the coefficients of x2 terms gives: 4 = B + D Since B = 1, D = 3 Equating the coefficients of x terms gives: 6 = 3A i.e. A = 2 From equation (1), since A=2, C = −4 Hence 3 + 6 x + 4x2 − 2 x3 x2(x2 + 3) ≡ 2 x + 1 x2 + −4x + 3 x2 + 3 ≡ 2 x + 1 x2 + 3 − 4x x2 + 3 Now try the following exercise Exercise 10 Further problems on partial fractions with quadratic factors 1. x2 − x − 13 (x2 + 7)(x − 2) 2x + 3 (x2 + 7) − 1 (x − 2) 2. 6x − 5 (x − 4)(x2 + 3) 1 (x − 4) + 2 − x (x2 + 3) 3. 15 + 5x + 5x2 − 4x3 x2(x2 + 5) 1 x + 3 x2 + 2 − 5x (x2 + 5)
  37. 37. Partial fractions 19 4. x3 + 4x2 + 20x − 7 (x − 1)2(x2 + 8) 3 (x − 1) + 2 (x − 1)2 + 1 − 2x (x2 + 8) 5. When solving the differential equation d2θ dt2 − 6 dθ dt −10θ =20−e2t by Laplace transforms, for given boundaryconditions,the following expression for L{θ} results: L{θ} = 4s3 − 39 2 s2 + 42s − 40 s(s − 2)(s2 − 6s + 10) Show that the expression can be resolved into partial fractions to give: L{θ} = 2 s − 1 2(s − 2) + 5s − 3 2(s2 − 6s + 10)
  38. 38. Chapter 3 Logarithms 3.1 Introduction to logarithms With the use of calculators firmly established, logarith- mic tables are now rarely used for calculation. However, the theory of logarithms is important, for there are sev- eral scientific and engineering laws that involve the rules of logarithms. From the laws of indices: 16 = 24 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 imply- ing the other. In general, if a number y can be written in the form ax , then the index ‘x’ is called the ‘logarithm of y to the base of a’, i.e. if y = ax then x = loga y In another example, if we write down that 64 = 82 then the equivalent statement using logarithms is: log8 64 = 2 Inanother example, ifwe writedownthat: 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: lg27.5 = 1.4393..., lg378.1 = 2.5776... and lg0.0204 = −1.6903... Napierian logarithms Logarithms having a base of e (where ‘e’ is a math- ematical 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: ln3.65 = 1.2947..., ln417.3 = 6.0338... and ln0.182 = −1.7037... More on Napierian logarithms is explained in Chapter 4 following. Here are some worked problems to help understand- ing of logarithms.
  39. 39. Logarithms 21 Problem 1. Evaluate log3 9. Let x = log3 9 then 3x = 9 from the definition of a logarithm, i.e. 3x = 32 from which, x = 2 Hence, log3 9 = 2 Problem 2. Evaluate log10 10. Let x = log10 10 then 10x = 10 from the definition of a logarithm, i.e. 10x = 101 from which, x = 1 Hence, log10 10 = 1 (which may be checked by a calculator) Problem 3. Evaluate log16 8. Let x = log16 8 then 16x = 8 from the definition of a logarithm, i.e. (24 )x = 23 i.e. 24x = 23 from the laws of indices, from which, 4x = 3 and x = 3 4 Hence, log16 8 = 3 4 Problem 4. Evaluate lg 0.001. Let x = lg 0.001 = log10 0.001 then 10x = 0.001 i.e. 10x = 10−3 from which, x = −3 Hence, lg 0.001 = −3 (which may be checked by a calculator) Problem 5. Evaluate lne. Let x = lne = loge e then ex = e i.e. ex = e1 from which, x = 1 Hence, lne = 1 (which may be checked by a calculator) Problem 6. Evaluate log3 1 81 . Let x = log3 1 81 then 3x = 1 81 = 1 34 = 3−4 from which, x = −4 Hence, log3 1 81 = −4 Problem 7. Solve the equation: lgx = 3. 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 Problem 9. Solve the equation: log5 x = −2. If log5 x = −2 then x = 5−2 = 1 52 = 1 25 Now try the following exercise Exercise 11 Further problems on laws of logarithms In Problems 1 to 11, evaluate the given expressions: 1. log10 10000 [4] 2. log2 16 [4] 3. log5 125 [3] 4. log2 1 8 [−3] 5. log8 2 1 3 6. log7 343 [3] 7. lg100 [2] 8. lg 0.01 [−2] 9. log4 8 1 1 2 10. log27 3 1 3 11. lne2 [2]
  40. 40. 22 Higher Engineering Mathematics In Problems 12 to 18 solve the equations: 12. log10 x = 4 [10000] 13. lgx = 5 [100000] 14. log3 x = 2 [9] 15. log4 x = −2 1 2 1 32 16. lgx = −2 [0.01] 17. log8 x = − 4 3 1 16 18. ln x = 3 [e3] 3.2 Laws of logarithms There are three laws of logarithms, which apply to any base: (i) To multiply two numbers: log(A × B) = log A + log B The following may be checked by using a calcu- lator: lg10 = 1 Also, lg 5 + lg 2 = 0.69897... + 0.301029... = 1 Hence, lg(5 × 2) = lg10 = lg 5 + lg 2 (ii) To divide two numbers: log A B = log A − log B Thefollowingmaybecheckedusingacalculator: ln 5 2 = ln2.5 = 0.91629... Also, ln5 − ln2 = 1.60943... − 0.69314... = 0.91629... Hence, ln 5 2 = ln5 − ln2 (iii) To raise a number to a power: log An = nlog A Thefollowingmaybecheckedusingacalculator: lg 52 = lg 25 = 1.39794... Also, 2lg 5 = 2 × 0.69897... = 1.39794... Hence, lg 52 = 2lg 5 Here are some worked problems to help understand- ing 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 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 2log 3 as the logarithm of a single number. 2log 3 = log 32 by the third law of logarithms = log 9 Problem 13. Write 1 2 log 25 as the logarithm of a single number. 1 2 log 25 = log 25 1 2 by the third law of logarithms = log √ 25 = log 5 Problem 14. Simplify: log64− log128 + log32. 64 = 26,128 = 27 and 32 = 25 Hence, log64 − log128 + log32 = log26 − log27 + log25
  41. 41. Logarithms 23 = 6log2 − 7log2 + 5log2 by the third law of logarithms = 4log2 Problem 15. Write 1 2 log16 + 1 3 log27 − 2log5 as the logarithm of a single number. 1 2 log16 + 1 3 log27 − 2log5 = log16 1 2 + log27 1 3 − log52 by the third law of logarithms = log √ 16 + log 3 √ 27 − log25 by the laws of indices = log4 + log3 − log25 = log 4 × 3 25 by the first and second laws of logarithms = log 12 25 = log0.48 Problem 16. Write (a) log30 (b) log450 in terms of 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 + 2log3 + 2log5 by the third law of logarithms Problem 17. Write log 8 × 4 √ 5 81 in terms of log2,log3 and log5 to any base. log 8 × 4 √ 5 81 = log8 + log 4 √ 5 − log81 by the first and second laws of logarithms = log23 + log5 1 4 − log34 by the laws of indices i.e. log 8× 4 √ 5 81 = 3log2 + 1 4 log5 − 4log3 by the third law of logarithms Problem 18. Evaluate: log25 − log125 + 1 2 log625 3log5 . log25 − log125 + 1 2 log625 3log5 = log52 − log53 + 1 2 log54 3log5 = 2log5 − 3log5 + 4 2 log5 3log5 = 1log5 3log5 = 1 3 Problem 19. Solve the equation: log(x − 1) + log(x + 8) = 2log(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 = 2log(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 2 log 4 = log x. 1 2 log4 = log4 1 2 from the third law of logarithms = log √ 4 from the laws of indices
  42. 42. 24 Higher Engineering Mathematics 1 2 log4 = logxHence, log √ 4 = logxbecomes log2 = logxi.e. 2 = xfrom which, i.e. the solution of the equation is: x = 2 Problem 21. Solve the equation: log x2 − 3 − logx = log2. log x2 − 3 − logx = log x2 − 3 x from the second law of logarithms log x2 − 3 x = log2Hence, x2 − 3 x = 2from which, x2 − 3 = 2xRearranging gives: x2 − 2x − 3 = 0and (x − 3)(x + 1) = 0Factorizing gives: x = 3 or x = −1from which, x = −1 is not a valid solution since the logarithm of a negative number has no real root. Hence, the solution of the equation is: x = 3 Now try the following exercise Exercise 12 Further problems on laws of logarithms In Problems 1 to 11, write as the logarithm of a single number: 1. log 2 + log3 [log 6] 2. log 3 + log5 [log 15] 3. log 3 + log4 − log6 [log 2] 4. log 7 + log21 − log49 [log 3] 5. 2log 2 + log3 [log 12] 6. 2log 2 + 3log5 [log 500] 7. 2log 5 − 1 2 log81 + log36 [log 100] 8. 1 3 log8 − 1 2 log81 + log27 [log 6] 9. 1 2 log4 − 2log3 + log45 [log 10] 10. 1 4 log16 + 2log3 − log18 [log 1 = 0] 11. 2log2 + log5 − log10 [log2] Simplify the expressions given in Problems 12 to 14: 12. log27 − log9 + log81 [log243 or log35 or 5log3] 13. log64 + log32 − log128 [log16 or log24 or 4log2] 14. log8 − log4 + log32 [log64 or log26 or 6log2] Evaluate the expressions given in Problems 15 and 16: 15. 1 2 log16 − 1 3 log8 log4 [0.5] 16. log9 − log3 + 1 2 log81 2log3 [1.5] Solve the equations given in Problems 17 to 22: 17. logx4 − logx3 = log5x − log2x [x = 2.5] 18. log2t3 − logt = log16 + logt [t = 8] 19. 2logb2 − 3logb = log8b − log4b [b = 2] 20. log(x + 1) + log(x − 1) = log3 [x = 2] 21. 1 3 log 27 = log(0.5a) [a = 6] 22. log x2 − 5 − logx = log4 [x = 5] 3.3 Indicial equations The laws of logarithms may be used to solve certain equations 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
  43. 43. Logarithms 25 Rearranging gives x = log10 27 log10 3 = 1.43136... 0.4771... = 3 which may be readily checked Note, log8 log2 is not equal to lg 8 2 Problem 22. Solve the equation 2x =3, correct to 4 significant figures. Taking logarithms to base 10 of both sides of 2x =3 gives: log10 2x =log10 3 i.e. x log10 2 =log10 3 Rearranging gives: x = log10 3 log10 2 = 0.47712125... 0.30102999... = 1.585,correct to 4 significant figures Problem 23. Solve the equation 2x+1 =32x−5 correct to 2 decimal places. Taking logarithms to base 10 of both sides gives: log10 2x+1 =log10 32x−5 i.e. (x + 1)log10 2 =(2x − 5)log10 3 x log10 2 + log10 2 =2x log10 3 − 5log10 3 x(0.3010) + (0.3010) =2x(0.4771) − 5(0.4771) i.e. 0.3010x + 0.3010 =0.9542x − 2.3855 Hence 2.3855 + 0.3010 =0.9542x − 0.3010x 2.6865 =0.6532x from which x = 2.6865 0.6532 =4.11, correct to 2 decimal places Problem 24. Solve the equation x3.2 =41.15, correct to 4 significant figures. Taking logarithms to base 10 of both sides gives: log10 x3.2 =log10 41.15 3.2log10 x =log10 41.15 Hence log10 x = log10 41.15 3.2 = 0.50449 Thus x=antilog0.50449=100.50449 =3.195 correct to 4 significant figures. Now try the following exercise Exercise 13 Indicial equations Solve the following indicial equations for x, each correct to 4 significant figures: 1. 3x =6.4 [1.690] 2. 2x =9 [3.170] 3. 2x−1 =32x−1 [0.2696] 4. x1.5 =14.91 [6.058] 5. 25.28=4.2x [2.251] 6. 42x−1 =5x+2 [3.959] 7. x−0.25 =0.792 [2.542] 8. 0.027x =3.26 [−0.3272] 9. The decibel gain n of an amplifier is given by: n = 10log10 P2 P1 where P1 is the power input and P2 is the power output. Find the power gain P2 P1 when n =25 decibels. [316.2] 3.4 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 ax =1 from the definition of the logarithm. If ax =1 then x =0 from the laws of indices. Hence loga1=0. In the above graphs it is seen that log101=0 and loge 1=0
  44. 44. 26 Higher Engineering Mathematics y 0.5 1.0 0 1 2 3 20.5 21.0 x x 3 0.48 2 0.30 1 0 0.5 20.30 0.2 20.70 0.1 21.0y5log10x Figure 3.1 (ii) logaa=1 Let loga a=x then ax =a from the definition of a logarithm. If ax =a then x =1. Hence loga a=1. (Check with a calculator that log10 10=1 and loge e=1) y 2 1 0 1 2 3 4 5 6 x x 6 5 4 3 2 1 0.5 0.2 0.1 1.79 1.61 1.39 1.10 0.69 0 20.69 21.61 22.30 21 22 y5logex Figure 3.2 (iii) loga0 → −∞ Let loga 0=x then ax =0 from the definition of a logarithm. If ax =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 → −∞
  45. 45. 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 expo- nential function is by using a scientific notation cal- culator. 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, 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, e9.32 = 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 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 e0.25 − e−0.25 e0.25 + e−0.25 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... = 1.2246, correct to 4 decimal places. 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 = 50ms, C = 10 μF, R = 47k and V = 300 volts. v = V e−t/CR = 300e(−50×10−3)/(10×10−6×47×103)
  46. 46. 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 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: 3. (a) 1 7 e3.4629 (b) 8.52e−1.2651 (c) 5e2.6921 3e1.1171 [(a) 4.55848 (b) 2.40444 (c) 8.05124] 4. (a) 5.6823 e−2.1347 (b) e2.1127 − e−2.1127 2 (c) 4(e−1.7295 − 1) e3.6817 [(a) 48.04106 (b) 4.07482 (c) −0.08286] 5. The length of a bar, l, at a temperature θ is given by l = l0eαθ , where l0 and α are constants. Evaluate 1, correct to 4 signifi- cant figures, where l0 = 2.587,θ = 321.7 and α = 1.771 × 10−4. [2.739] 6. When a chain of length 2L is suspended from two points, 2D metres apart, on the same hor- izontal level: D = k ln L+ √ L2+k2 k . Eval- uate D when k = 75m and L = 180m. [120.7m] 4.2 The power series for ex The value of ex can be calculated to any required degree of accuracy since it is defined in terms of the following power series: ex = 1 + x + x2 2! + x3 3! + x4 4! + ··· (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 ex (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)2 2! + (1)3 3! + (1)4 4! + (1)5 5! + (1)6 6! + (1)7 7! + (1)8 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 ex. Thus e0.05 = 1 + 0.05 + (0.05)2 2! + (0.05)3 3! + (0.05)4 4! + (0.05)5 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 deter- mine 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) + (−x)2 2! + (−x)3 3! + ··· i.e. e−x =1 − x + x2 2! − x3 3! + ··· In a similar manner the power series for ex may be used to evaluate any exponential function of the form aekx ,
  47. 47. Exponential functions 29 where a and k are constants. In the series of equation (1), let x be replaced by kx. Then, aekx =a 1 + (kx) + (kx)2 2! + (kx)3 3! + ··· Thus 5e2x =5 1 + (2x) + (2x)2 2! + (2x)3 3! + ··· =5 1 + 2x + 4x2 2 + 8x3 6 + ··· i.e. 5e2x =5 1 + 2x + 2x2 + 4 3 x3 + ··· Problem 4. Determine the value of 5e0.5, correct to 5 significant figures by using the power series for ex . ex = 1 + x + x2 2! + x3 3! + x4 4! + ··· Hence e0.5 =1 + 0.5 + (0.5)2 (2)(1) + (0.5)3 (3)(2)(1) + (0.5)4 (4)(3)(2)(1) + (0.5)5 (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. e0.5 =1.64872, correct to 6 significant figures Hence 5e0.5 =5(1.64872) = 8.2436, correct to 5 significant figures Problem 5. Expand ex (x2 − 1) as far as the term in x5. The power series for ex is, ex = 1 + x + x2 2! + x3 3! + x4 4! + x5 5! + ··· Hence ex (x2 −1) = 1 + x + x2 2! + x3 3! + x4 4! + x5 5! + ··· (x2 − 1) = x2 + x3 + x4 2! + x5 3! + ··· − 1 + x + x2 2! + x3 3! + x4 4! + x5 5! + ··· Grouping like terms gives: ex (x2 − 1) = −1 − x + x2 − x2 2! + x3 − x3 3! + x4 2! − x4 4! + x5 3! − x5 5! + ··· = −1−x+ 1 2 x2 + 5 6 x3 + 11 24 x4 + 19 120 x5 when expanded as far as the term in x5. Now try the following exercise Exercise 15 Further problems on the power series for ex 1. Evaluate 5.6e−1 , correct to 4 decimal places, using the power series for ex. [2.0601] 2. Use the power series for ex to determine, cor- rect 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 x4. 1−2x2 − 8x3 3 − 2x4 4. Expand 2ex2 x 1 2 to six terms. ⎡ ⎢ ⎢ ⎣ 2x 1 2 + 2x 5 2 + x 9 2 + 1 3 x 13 2 + 1 12 x 17 2 + 1 60 x 21 2 ⎤ ⎥ ⎥ ⎦ 4.3 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.
  48. 48. 30 Higher Engineering Mathematics x −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0 ex 0.05 0.08 0.14 0.22 0.37 0.61 1.00 e−x 20.09 12.18 7.39 4.48 2.72 1.65 1.00 x 0.5 1.0 1.5 2.0 2.5 3.0 ex 1.65 2.72 4.48 7.39 12.18 20.09 e−x 0.61 0.37 0.22 0.14 0.08 0.05 Figure 4.1 shows graphs of y =ex and y =e−x y 20 16 y 5exy 5e2x 12 8 4 21 0 1 2 3 x2223 Figure 4.1 Problem 6. Plot a graph of y =2e0.3x 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. A table of values is drawn up as shown below. x −3 −2 −1 0 1 2 3 0.3x −0.9 −0.6 −0.3 0 0.3 0.6 0.9 e0.3x 0.407 0.549 0.741 1.000 1.350 1.822 2.460 2e0.3x 0.81 1.10 1.48 2.00 2.70 3.64 4.92 A graph of y=2e0.3x 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. y 5 y52e0.3x 4 3 1.6 3.87 1 20.74 2.2 21 0 1 2 3 x2223 2 Figure 4.2 Problem 7. Plot a graph of y = 1 3 e−2x over the 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. A table of values is drawn up as shown below. x −1.5 −1.0 −0.5 0 0.5 1.0 1.5 −2x 3 2 1 0 −1 −2 −3 e−2x 20.086 7.389 2.718 1.00 0.368 0.135 0.050 1 3 e−2x 6.70 2.46 0.91 0.33 0.12 0.05 0.02 A graph of 1 3 e−2x is shown in Fig. 4.3. 7 6 5 4 3 3.67 1.4 2 1 0.520.5 20.72 21.0 21.2 21.5 1.0 1.5 y x 1 3 e22x y5 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=250e −t 3 . Draw a graph showing the natural
  49. 49. Exponential functions 31 decay curve over the first 6seconds. From the graph, find (a) the voltage after 3.4s, and (b) the time when the voltage is 150V. A table of values is drawn up as shown below. t 0 1 2 3 e −t 3 1.00 0.7165 0.5134 0.3679 v=250e −t 3 250.0 179.1 128.4 91.97 t 4 5 6 e −t 3 0.2636 0.1889 0.1353 v=250e −t 3 65.90 47.22 33.83 The natural decay curve of v=250e −t 3 is shown in Fig. 4.4. 250 200 Voltagev(volts) Time t(seconds) 150 100 80 50 0 1 1.5 2 3 3.4 4 5 6 y5 250e t 3 2 Figure 4.4 From the graph: (a) when time t =3.4 s, voltage v=80V and (b) when voltage v=150V, time t =1.5s. Now try the following exercise Exercise 16 Further problems on exponential graphs 1. Plot a graph of y=3e0.2x over the range x =−3 to x =3. Hence determine the value of y when x =1.4 and the value of x when y =4.5. [3.95, 2.05] 2. Plot a graph of y = 1 2 e−1.5x over a range 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] 3. In a chemical reaction the amount of starting material C cm3 left after t minutes is given by C =40e−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) 28cm3 (b) 116min] 4. The rate at which a body cools is given by θ =250e−0.05t where the excess of tempera- ture 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) 5min] 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 logarithmisbyascientificnotation calculator.Useyour calculator to check the following values: ln4.328 = 1.46510554... = 1.4651, correct to 4 decimal places ln1.812 = 0.59443, correct to 5 significant figures ln1 = 0 ln527 = 6.2672, correct to 5 significant figures ln0.17 = −1.772, correct to 4 significant figures ln0.00042 = −7.77526, correct to 6 significant figures lne3 = 3 lne1 = 1
  50. 50. 32 Higher Engineering Mathematics From the last two examples we can conclude that: loge ex = x This is useful when solving equations involving expo- nential functions. For example, to solve e3x = 7, take Napierian logarithms of both sides, which gives: lne3x = ln7 i.e. 3x = ln7 from which x = 1 3 ln7 = 0.6486, correct to 4 decimal places. Problem 9. Evaluate the following, each correct to 5 significant figures: (a) 1 2 ln4.7291 (b) ln7.8693 7.8693 (c) 3.17ln24.07 e−0.1762 . (a) 1 2 ln4.7291 = 1 2 (1.5537349...) = 0.77687, correct to 5 significant figures (b) ln7.8693 7.8693 = 2.06296911... 7.8693 = 0.26215, correct to 5 significant figures (c) 3.17ln24.07 e−0.1762 = 3.17(3.18096625...) 0.83845027... = 12.027, correct to 5 significant figures. Problem 10. Evaluate the following: (a) lne2.5 lg100.5 (b) 5e2.23 lg2.23 ln2.23 (correct to 3 decimal places). (a) lne2.5 lg100.5 = 2.5 0.5 = 5 (b) 5e2.23 lg2.23 ln2.23 = 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: 9 4 = e−3x Taking the reciprocal of both sides gives: 4 9 = 1 e−3x = e3x Taking Napierian logarithms of both sides gives: ln 4 9 = ln(e3x) Since loge eα = α, then ln 4 9 = 3x Hence, x = 1 3 ln 4 9 = 1 3 (−0.81093)=−0.2703, correct to 4 significant figures. Problem 12. Given 32 = 70 1 − e− t 2 determine the value of t, correct to 3 significant figures. Rearranging 32 = 70(1 − e− t 2 ) gives: 32 70 = 1 − e− t 2 and e− t 2 = 1 − 32 70 = 38 70 Taking the reciprocal of both sides gives: e t 2 = 70 38 Taking Napierian logarithms of both sides gives: lne t 2 = ln 70 38 i.e. t 2 = ln 70 38 from which, t =2ln 70 38 = 1.22, correct to 3 signifi- cant figures. Problem 13. Solve the equation: 2.68= ln 4.87 x to find x. From the definition of a logarithm, since 2.68=ln 4.87 x then e2.68 = 4.87 x Rearranging gives: x = 4.87 e2.68 = 4.87e−2.68 i.e. x = 0.3339, correct to 4 significant figures. Problem 14. Solve 7 4 = e3x correct to 4 signi- ficant figures.
  51. 51. Exponential functions 33 Taking natural logs of both sides gives: ln 7 4 = lne3x ln 7 4 = 3x lne Since lne = 1 ln 7 4 = 3x i.e. 0.55962 = 3x i.e. x = 0.1865, correct to 4 significant figures. Problem 15. Solve: ex−1 = 2e3x−4 correct to 4 significant figures. Taking natural logarithms of both sides gives: ln ex−1 = ln 2e3x−4 and by the first law of logarithms, ln ex−1 = ln2 + ln e3x−4 i.e. x − 1 = ln2 + 3x − 4 Rearranging gives: 4 − 1 − ln2 = 3x − x i.e. 3 − ln2 = 2x from which, x = 3 − ln2 2 = 1.153 Problem 16. Solve, correct to 4 significant figures: ln(x −2)2 = ln(x −2) − ln(x +3) +1.6 Rearranging gives: ln(x − 2)2 − ln(x − 2) + ln(x + 3) = 1.6 and by the laws of logarithms, ln (x − 2)2(x + 3) (x − 2) = 1.6 Cancelling gives: ln{(x − 2)(x + 3)} = 1.6 and (x − 2)(x + 3) = e1.6 i.e. x2 + x − 6 = e1.6 or x2 + x − 6 − e1.6 = 0 i.e. x2 + x − 10.953 = 0 Using the quadratic formula, x = −1 ± 12 − 4(1)(−10.953) 2 = −1 ± √ 44.812 2 = −1 ± 6.6942 2 i.e. x = 2.847 or −3.8471 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 Exercise 17 Further problems on evaluating Napierian logarithms In Problems 1 and 2, evaluate correct to 5 signifi- cant figures: 1. (a) 1 3 ln5.2932 (b) ln82.473 4.829 (c) 5.62ln321.62 e1.2942 [(a) 0.55547 (b) 0.91374 (c) 8.8941] 2. (a) 1.786lne1.76 lg101.41 (b) 5e−0.1629 2ln0.00165 (c) ln4.8629 − ln2.4711 5.173 [(a) 2.2293 (b) −0.33154 (c) 0.13087] In Problems 3 to 7 solve the given equations, each correct to 4 significant figures. 3. ln x = 2.10 [8.166] 4. 24 + e2x = 45 [1.522] 5. 5 = ex+1 − 7 [1.485] 6. 1.5 = 4e2t [−0.4904] 7. 7.83 = 2.91e−1.7x [−0.5822] 8. 16 = 24 1 − e− t 2 [2.197] 9. 5.17 = ln x 4.64 [816.2] 10. 3.72ln 1.59 x = 2.43 [0.8274]
  52. 52. 34 Higher Engineering Mathematics 11. 5 = 8 1 − e −x 2 [1.962] 12. ln(x + 3) − ln x = ln(x − 1) [3] 13. ln(x − 1)2 − ln3 = ln(x − 1) [4] 14. ln(x + 3) + 2 = 12 − ln(x − 2) [147.9] 15. e(x+1) = 3e(2x−5) [4.901] 16. ln(x + 1)2 = 1.5 − ln(x − 2) + ln(x + 1) [3.095] 17. Transpose: b = lnt − aln D to make t the subject. [t = eb+a ln D = ebea ln D = ebeln Da i.e. t = eb Da] 18. If P Q = 10log10 R1 R2 find the value of R1 when P = 160, Q = 8 and R2 = 5. [500] 19. If U2 = U1e W P V make W the subject of the formula. W = PV ln U2 U1 20. 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 = 3w0. [p2 = 348.5Pa] 4.5 Laws of growth and decay The laws of exponential growth and decay are of the form y = Ae−kx and y= A(1−e−kx ), where A and k are constants. When plotted, the form of each of these equa- tions 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 T1 = T0 eμθ (iv) Newton’s law of cooling θ =θ0 e−kt (a) y x A 0 y5Ae2kx (b) y x A 0 y5A(12e2kx) 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 (ix) Decay of current in an inductive circuit i = I e−Rt/L (x) Growth of current in a capacitive 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. Transposing R= R0 eαθ gives R R0 =eαθ . Taking Napierian logarithms of both sides gives: ln R R0 = lneαθ = αθ
  53. 53. Exponential functions 35 Hence α = 1 θ ln R R0 = 1 1500 ln 6 × 103 5 × 103 = 1 1500 (0.1823215...) =1.215477···× 10−4 Hence α =1.215 × 10−4, correct to 4 significant figures. From above, ln R R0 =αθ hence θ = 1 α ln R R0 When R= 5.4×103, α =1.215477... ×10−4 and R0 =5×103 θ = 1 1.215477...× 10−4 ln 5.4 × 103 5 × 103 = 104 1.215477... (7.696104...× 10−2 ) = 633◦ C,correct to the nearest degree. 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 =83.0seconds. Transposing θ =θ0 e−kt gives θ θ0 =e−kt from which θ0 θ = 1 e−kt = ekt Taking Napierian logarithms of both sides gives: ln θ0 θ = kt from which, k = 1 t ln θ0 θ = 1 83.0 ln 56.6 16.5 = 1 83.0 (1.2326486...) Hence k= 1.485×10−2 Problem 19. The current i amperes flowing in a capacitor at time t seconds is given by i =8.0(1−e −t CR ), where the circuit resistance R is 25×103 ohms and capacitance C is 16×10−6 farads. Determine (a) the current i after 0.5seconds and (b) the time, to the nearest millisecond, for the current to reach 6.0A. Sketch the graph of current against time. (a) Current i =8.0(1−e −t CR ) = 8.0[1−e −0.5 (16×10−6)(25×103) ]=8.0(1−e−1.25) =8.0(1−0.2865047...)=8.0(0.7134952...) = 5.71amperes (b) Transposing i =8.0(1−e −t CR ) gives i 8.0 =1−e −t CR from which, e −t CR =1− i 8.0 = 8.0−i 8.0 Taking the reciprocal of both sides gives: e t CR = 8.0 8.0 − i Taking Napierian logarithms of both sides gives: t CR = ln 8.0 8.0 − i Hence t = CRln 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 = 400 103 ln 8.0 2.0 = 0.4ln4.0 =0.4(1.3862943...) = 0.5545s =555ms, to the nearest millisecond. Agraphofcurrent against timeisshowninFig.4.6.

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