Bird's Higher Engineering Mathematics [9 ed.] 0367643731, 9780367643737

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Bird’s Higher Engineering Mathematics

Why is knowledge of mathematics important in engineering? A career in any engineering or scientific field will require both basic and advanced mathematics. Without mathematics to determine principles, calculate dimensions and limits, explore variations, prove concepts and so on, there would be no mobile telephones, televisions, stereo systems, video games, microwave ovens, computers or virtually anything electronic. There would be no bridges, tunnels, roads, skyscrapers, automobiles, ships, planes, rockets or most things mechanical. There would be no metals beyond the common ones, such as iron and copper, no plastics, no synthetics. In fact, society would most certainly be less advanced without the use of mathematics throughout the centuries and into the future. Electrical engineers require mathematics to design, develop, test or supervise the manufacturing and installation of electrical equipment, components or systems for commercial, industrial, military or scientific use. Mechanical engineers require mathematics to perform engineering duties in planning and designing tools, engines, machines and other mechanically functioning equipment; they oversee installation, operation, maintenance and repair of such equipment as centralised heat, gas, water and steam systems. Aerospace engineers require mathematics to perform a variety of engineering work in designing, constructing and testing aircraft, missiles and spacecraft; they conduct basic and applied research to evaluate adaptability of materials and equipment to aircraft design and manufacture and recommend improvements in testing equipment and techniques. Nuclear engineers require mathematics to conduct research on nuclear engineering problems or apply

principles and theory of nuclear science to problems concerned with release, control and utilisation of nuclear energy and nuclear waste disposal. Petroleum engineers require mathematics to devise methods to improve oil and gas well production and determine the need for new or modified tool designs; they oversee drilling and offer technical advice to achieve economical and satisfactory progress. Industrial engineers require mathematics to design, develop, test and evaluate integrated systems for managing industrial production processes, including human work factors, quality control, inventory control, logistics and material flow, cost analysis and production co-ordination. Environmental engineers require mathematics to design, plan or perform engineering duties in the prevention, control and remediation of environmental health hazards, using various engineering disciplines; their work may include waste treatment, site remediation or pollution control technology. Civil engineers require mathematics in all levels in civil engineering – structural engineering, hydraulics and geotechnical engineering are all fields that employ mathematical tools such as differential equations, tensor analysis, field theory, numerical methods and operations research. Knowledge of mathematics is therefore needed by each of the engineering disciplines listed above. It is intended that this text – Bird’s Higher Engineering Mathematics – will provide a step-by-step approach to learning the essential mathematics needed for your engineering studies.

Now in its ninth edition, Bird’s Higher Engineering Mathematics has helped thousands of students to succeed in their exams. Mathematical theories are explained in a straightforward manner, supported by practical engineering examples and applications to ensure that readers can relate theory to practice. Some 1,200 engineering situations/problems have been ‘flagged-up’ to help demonstrate that engineering cannot be fully understood without a good knowledge of mathematics. The extensive and thorough topic coverage makes this an ideal text for undergraduate degree courses, foundation degrees, and for higher-level vocational courses such as Higher National Certificate and Diploma courses in engineering disciplines. Its companion website at www.routledge.com/cw/bird provides resources for both students and lecturers, including full solutions for all 2,100 further questions, lists of essential formulae, multiple-choice tests, and illustrations, as well as full solutions to revision tests for course instructors.

John Bird, BSc (Hons), CEng, CMath, CSci, FIMA, FIET, FCollT, is the former Head of Applied Electronics in the Faculty of Technology at Highbury College, Portsmouth, UK. More recently, he has combined freelance lecturing at the University of Portsmouth, with Examiner responsibilities for Advanced Mathematics with City and Guilds and examining for the International Baccalaureate Organisation. He has over 45 years’ experience of successfully teaching, lecturing, instructing, training, educating and planning trainee engineers study programmes. He is the author of 146 textbooks on engineering, science and mathematical subjects, with worldwide sales of over one million copies. He is a chartered engineer, a chartered mathematician, a chartered scientist and a Fellow of three professional institutions. He has recently retired from lecturing at the Royal Navy’s Defence College of Marine Engineering in the Defence College of Technical Training at H.M.S. Sultan, Gosport, Hampshire, UK, one of the largest engineering training establishments in Europe.

Bird’s Higher Engineering Mathematics Ninth Edition John Bird

Ninth edition published 2021 by Routledge 2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN and by Routledge 52 Vanderbilt Avenue, New York, NY 10017 Routledge is an imprint of the Taylor & Francis Group, an informa business © 2021 John Bird The right of John Bird to be identified as author of this work has been asserted by him in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. First edition published by Elsevier 1993 Eighth edition published by Routledge 2017 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 Names: Bird, J. O., author. Title: Bird’s higher engineering mathematics / John Bird. Other titles: Higher engineering mathematics Description: Ninth edition. | Abingdon, Oxon ; New York : Routledge, 2021. | Includes index. Identifiers: LCCN 2021000158 (print) | LCCN 2021000159 (ebook) | ISBN 9780367643737 (paperback) | ISBN 9780367643751 (hardback) | ISBN 9781003124221 (ebook) Subjects: LCSH: Engineering mathematics. Classification: LCC TA330 .B52 2021 (print) | LCC TA330 (ebook) | DDC 620.001/51–dc23 LC record available at https://lccn.loc.gov/2021000158 LC ebook record available at https://lccn.loc.gov/2021000159 ISBN: 978-0-367-64375-1 (hbk) ISBN: 978-0-367-64373-7 (pbk) ISBN: 978-1-003-12422-1 (ebk) Typeset in Times by KnowledgeWorks Global Ltd. Visit the companion website: www.routledge.com/cw/bird

To Sue

Contents Preface

xiv

5.4

Syllabus guidance

xvi

5.5

Section A 1

Number and algebra

1

Algebra 1.1 Introduction 1.2 Revision of basic laws 1.3 Revision of equations 1.4 Polynomial division 1.5 The factor theorem 1.6 The remainder theorem

3 3 3 5 9 11 13

2

Partial fractions 2.1 Introduction to partial fractions 2.2 Partial fractions with linear factors 2.3 Partial fractions with repeated linear factors 2.4 Partial fractions with quadratic factors

17 17 17 20 22

3

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

24 24 26 28 30

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

32 32 33 35 36 40 43

6

7

Revision Test 1

5

The binomial series 5.1 Pascal’s triangle 5.2 The binomial series 5.3 Worked problems on the binomial series

47

48 48 50 50

51 54

Solving equations by iterative methods 6.1 Introduction to iterative methods 6.2 The bisection method 6.3 An algebraic method of successive approximations

57 57 58

Boolean algebra and logic circuits 7.1 Boolean algebra and switching circuits 7.2 Simplifying Boolean expressions 7.3 Laws and rules of Boolean algebra 7.4 De Morgan’s laws 7.5 Karnaugh maps 7.6 Logic circuits 7.7 Universal logic gates

65 66 70 70 72 73 77 81

Revision Test 2

Section B 8

4

Further worked problems on the binomial series Practical problems involving the binomial theorem

61

85

Geometry and trigonometry 87

Introduction to trigonometry 8.1 Trigonometry 8.2 The theorem of Pythagoras 8.3 Trigonometric ratios of acute angles 8.4 Evaluating trigonometric ratios 8.5 Solution of right-angled triangles 8.6 Angles of elevation and depression 8.7 Sine and cosine rules 8.8 Area of any triangle 8.9 Worked problems on the solution of triangles and finding their areas 8.10 Further worked problems on solving triangles and finding their areas 8.11 Practical situations involving trigonometry 8.12 Further practical situations involving trigonometry

89 90 90 91 93 97 99 100 101 101 102 104 106

Contents vii 9

Cartesian and polar co-ordinates 9.1 Introduction 9.2 Changing from Cartesian into polar co-ordinates 9.3 Changing from polar into Cartesian co-ordinates 9.4 Use of Pol/Rec functions on calculators

10 The circle and its properties 10.1 Introduction 10.2 Properties of circles 10.3 Radians and degrees 10.4 Arc length and area of circles and sectors 10.5 The equation of a circle 10.6 Linear and angular velocity 10.7 Centripetal force Revision Test 3

111 112 112 114 115 117 117 117 119 120 123 125 126 130

14 The relationship between trigonometric and hyperbolic functions 14.1 The relationship between trigonometric and hyperbolic functions 14.2 Hyperbolic identities 15 Compound angles 15.1 Compound angle formulae 15.2 Conversion of a sin ωt + b cos ωt into R sin(ωt + α) 15.3 Double angles 15.4 Changing products of sines and cosines into sums or differences 15.5 Changing sums or differences of sines and cosines into products 15.6 Power waveforms in a.c. circuits Revision Test 4

Section C 11 Trigonometric waveforms 11.1 Graphs of trigonometric functions 11.2 Angles of any magnitude 11.3 The production of a sine and cosine wave 11.4 Sine and cosine curves 11.5 Sinusoidal form A sin(ωt ±α) 11.6 Harmonic synthesis with complex waveforms

132 132 133 136 137 141

12 Hyperbolic functions 12.1 Introduction to hyperbolic functions 12.2 Graphs of hyperbolic functions 12.3 Hyperbolic identities 12.4 Solving equations involving hyperbolic functions 12.5 Series expansions for cosh x and sinh x

151 151 153 155

13 Trigonometric identities and equations 13.1 Trigonometric identities 13.2 Worked problems on trigonometric identities 13.3 Trigonometric equations 13.4 Worked problems (i) on trigonometric equations 13.5 Worked problems (ii) on trigonometric equations 13.6 Worked problems (iii) on trigonometric equations 13.7 Worked problems (iv) on trigonometric equations

161 161

144

157 159

Graphs

169 169 170 173 173 175 179 181 182 183 187

189

16 Functions and their curves 16.1 Standard curves 16.2 Simple transformations 16.3 Periodic functions 16.4 Continuous and discontinuous functions 16.5 Even and odd functions 16.6 Inverse functions 16.7 Asymptotes 16.8 Brief guide to curve sketching 16.9 Worked problems on curve sketching

191 191 194 199 199 200 201 203 208 208

17 Irregular areas, volumes and mean values of waveforms 17.1 Areas of irregular figures 17.2 Volumes of irregular solids 17.3 The mean or average value of a waveform

213 213 216 217

Revision Test 5

223

162 163

Section D

Complex numbers

225

164 165 166 166

18 Complex numbers 18.1 Cartesian complex numbers 18.2 The Argand diagram 18.3 Addition and subtraction of complex numbers

227 228 229 229

viii Contents 18.4 Multiplication and division of complex numbers 18.5 Complex equations 18.6 The polar form of a complex number 18.7 Multiplication and division in polar form 18.8 Applications of complex numbers

230 232 233 235 236

19 De Moivre’s theorem 19.1 Introduction 19.2 Powers of complex numbers 19.3 Roots of complex numbers 19.4 The exponential form of a complex number 19.5 Introduction to locus problems

Section E

241 242 242 243 245 246

Matrices and determinants 251

20 The theory of matrices and determinants 20.1 Matrix notation 20.2 Addition, subtraction and multiplication of matrices 20.3 The unit matrix 20.4 The determinant of a 2 by 2 matrix 20.5 The inverse or reciprocal of a 2 by 2 matrix 20.6 The determinant of a 3 by 3 matrix 20.7 The inverse or reciprocal of a 3 by 3 matrix

253 253 254 257 257 258 259 261

22.5 Resolving vectors into horizontal and vertical components 22.6 Addition of vectors by calculation 22.7 Vector subtraction 22.8 Relative velocity 22.9 i, j and k notation 23 Methods of adding alternating waveforms 23.1 Combination of two periodic functions 23.2 Plotting periodic functions 23.3 Determining resultant phasors by drawing 23.4 Determining resultant phasors by the sine and cosine rules 23.5 Determining resultant phasors by horizontal and vertical components 23.6 Determining resultant phasors by using complex numbers

301 301 302

24 Scalar and vector products 24.1 The unit triad 24.2 The scalar product of two vectors 24.3 Vector products 24.4 Vector equation of a line

312 312 313 317 321

Revision Test 7

Section G 21 Applications of matrices and determinants 21.1 Solution of simultaneous equations by matrices 21.2 Solution of simultaneous equations by determinants 21.3 Solution of simultaneous equations using Cramer’s rule 21.4 Solution of simultaneous equations using the Gaussian elimination method 21.5 Stiffness matrix 21.6 Eigenvalues and eigenvectors Revision Test 6

Section F 22 Vectors 22.1 22.2 22.3 22.4

Vector geometry

Introduction Scalars and vectors Drawing a vector Addition of vectors by drawing

289 290 295 297 298

Differential calculus

303 305 306 308

323

325

265 266 268 271 272 274 275 281

283 285 285 285 286 287

25 Methods of differentiation 25.1 Introduction to calculus 25.2 The gradient of a curve 25.3 Differentiation from first principles 25.4 Differentiation of common functions 25.5 Differentiation of a product 25.6 Differentiation of a quotient 25.7 Function of a function 25.8 Successive differentiation

327 327 327 328 329 332 334 335 337

26 Some applications of differentiation 26.1 Rates of change 26.2 Velocity and acceleration 26.3 The Newton–Raphson method 26.4 Turning points 26.5 Practical problems involving maximum and minimum values 26.6 Points of inflexion 26.7 Tangents and normals 26.8 Small changes

340 340 342 345 348

Revision Test 8

351 355 357 358 362

Contents ix 27 Differentiation of parametric equations 27.1 Introduction to parametric equations 27.2 Some common parametric equations 27.3 Differentiation in parameters 27.4 Further worked problems on differentiation of parametric equations

363 363 364 364 366

28 Differentiation of implicit functions 28.1 Implicit functions 28.2 Differentiating implicit functions 28.3 Differentiating implicit functions containing products and quotients 28.4 Further implicit differentiation

369 369 369

29 Logarithmic differentiation 29.1 Introduction to logarithmic differentiation 29.2 Laws of logarithms 29.3 Differentiation of logarithmic functions 29.4 Differentiation of further logarithmic functions 29.5 Differentiation of [f (x)] x

375

Revision Test 9

30 Differentiation of hyperbolic functions 30.1 Standard differential coefficients of hyperbolic functions 30.2 Further worked problems on differentiation of hyperbolic functions 31 Differentiation of inverse trigonometric and hyperbolic functions 31.1 Inverse functions 31.2 Differentiation of inverse trigonometric functions 31.3 Logarithmic forms of the inverse hyperbolic functions 31.4 Differentiation of inverse hyperbolic functions

370 371

375 375 376 376 378 381

382 382 383 386 386

Revision Test 10

Section H

Integral calculus

412

413

413

415

420

421

35 Standard integration 35.1 The process of integration 35.2 The general solution of integrals of the form axn 35.3 Standard integrals 35.4 Definite integrals

423 423

36 Some applications of integration 36.1 Introduction 36.2 Areas under and between curves 36.3 Mean and rms values 36.4 Volumes of solids of revolution 36.5 Centroids 36.6 Theorem of Pappus 36.7 Second moments of area of regular sections

431 432 432 433 435 436 438 440

Revision Test 11

424 424 427

449

388 391 393

32 Partial differentiation 32.1 Introduction to partial derivatives 32.2 First-order partial derivatives 32.3 Second-order partial derivatives

397 397 397 400

33 Total differential, rates of change and small changes 33.1 Total differential 33.2 Rates of change 33.3 Small changes

404 404 405 408

34 Maxima, minima and saddle points for functions of two variables 34.1 Functions of two independent variables

34.2 Maxima, minima and saddle points 34.3 Procedure to determine maxima, minima and saddle points for functions of two variables 34.4 Worked problems on maxima, minima and saddle points for functions of two variables 34.5 Further worked problems on maxima, minima and saddle points for functions of two variables

411 411

37 Maclaurin’s series and limiting values 37.1 Introduction 37.2 Derivation of Maclaurin’s theorem 37.3 Conditions of Maclaurin’s series 37.4 Worked problems on Maclaurin’s series 37.5 Numerical integration using Maclaurin’s series 37.6 Limiting values

450 451 451 452 452

38 Integration using algebraic substitutions 38.1 Introduction 38.2 Algebraic substitutions 38.3 Worked problems on integration using algebraic substitutions 38.4 Further worked problems on integration using algebraic substitutions 38.5 Change of limits

461 461 461

455 457

462 463 464

x Contents 39 Integration using trigonometric and hyperbolic substitutions 39.1 Introduction 39.2 Worked problems on integration of sin2 x, cos2 x, tan2 x and cot2 x 39.3 Worked problems on integration of powers of sines and cosines 39.4 Worked problems on integration of products of sines and cosines 39.5 Worked problems on integration using the sin θ substitution 39.6 Worked problems on integration using the tan θ substitution 39.7 Worked problems on integration using the sinh θ substitution 39.8 Worked problems on integration using the cosh θ substitution 40 Integration using partial fractions 40.1 Introduction 40.2 Integration using partial fractions with linear factors 40.3 Integration using partial fractions with repeated linear factors 40.4 Integration using partial fractions with quadratic factors θ 41 The t = tan substitution 2 41.1 Introduction

θ 41.2 Worked problems on the t = tan 2 substitution 41.3 Further worked problems on the θ t = tan substitution 2

Revision Test 12

467 467 467 470 471 472 474 474

44 Double and triple integrals 44.1 Double integrals 44.2 Triple integrals

506 506 508

45 Numerical integration 45.1 Introduction 45.2 The trapezoidal rule 45.3 The mid-ordinate rule 45.4 Simpson’s rule 45.5 Accuracy of numerical integration

512 512 512 515 516 520

Revision Test 13

Section I

Differential equations

521

523

476 479 479 479 481 482 485 485 486 487 490

42 Integration by parts 42.1 Introduction 42.2 Worked problems on integration by parts 42.3 Further worked problems on integration by parts

491 491

43 Reduction formulae 43.1 Introduction 43.2 Using reduction formulae for integrals of ∫ the form xn e x dx 43.3 Using reduction formulae ∫for integrals of ∫ the form xn cos x dx and xn sin x dx 43.4 Using reduction formulae ∫ ∫ for integrals of the form sinn x dx and cosn x dx 43.5 Further reduction formulae

497 497

491 493

46 Introduction to differential equations 46.1 Family of curves 46.2 Differential equations 46.3 The solution of equations of the form dy = f (x) dx 46.4 The solution of equations of the form dy = f (y) dx 46.5 The solution of equations of the form dy = f (x).f (y) dx

525 525 526

47 Homogeneous first-order differential equations 47.1 Introduction 47.2 Procedure to solve differential equations dy of the form P = Q dx 47.3 Worked problems on homogeneous first-order differential equations 47.4 Further worked problems on homogeneous first-order differential equations

534 534

48 Linear first-order differential equations 48.1 Introduction 48.2 Procedure to solve differential equations dy of the form + Py = Q dx 48.3 Worked problems on linear first-order differential equations 48.4 Further worked problems on linear first-order differential equations

538 538

527 528 530

534 535

536

539 539 540

497 498 501 503

49 Numerical methods for first-order differential equations 49.1 Introduction 49.2 Euler’s method 49.3 Worked problems on Euler’s method

543 543 544 545

Contents xi 49.4 The Euler–Cauchy method 49.5 The Runge–Kutta method Revision Test 14

50 Second-order differential equations of the d2 y dy form a 2 + b + cy = 0 dy dx 50.1 Introduction 50.2 Procedure to solve differential equations dy d2 y of the form a 2 + b + cy = 0 dx dx 50.3 Worked problems on differential equations of the form d2 y dy a 2 + b + cy = 0 dx dx 50.4 Further worked problems on practical differential equations of the form d2 y dy a 2 + b + cy = 0 dx dx 51 Second-order differential equations of the d2 y dy form a 2 + b + cy = f (x) dx dx 51.1 Complementary function and particular integral 51.2 Procedure to solve differential equations d2 y dy of the form a 2 + b + cy = f (x) dx dx 51.3 Differential equations of the form d2 y dy a 2 + b + cy = f (x) dx dx where f(x) is a constant or polynomial 51.4 Differential equations of the form d2 y dy a 2 + b + cy = f (x) dx dx where f(x) is an exponential function 51.5 Differential equations of the form d2 y dy a 2 + b + cy = f (x) where f(x) is a dx dx sine or cosine function 51.6 Differential equations of the form d2 y dy a 2 + b + cy = f (x) dx dx where f(x) is a sum or a product 52 Power series methods of solving ordinary differential equations 52.1 Introduction 52.2 Higher order differential coefficients as series 52.3 Leibniz’s theorem 52.4 Power series solution by the Leibniz-Maclaurin method

549 554 560

561 561 562

562

52.5 Power series solution by the Frobenius method 52.6 Bessel’s equation and Bessel’s functions 52.7 Legendre’s equation and Legendre polynomials

585 592 597

53 An introduction to partial differential equations 53.1 Introduction 53.2 Partial integration 53.3 Solution of partial differential equations by direct partial integration 53.4 Some important engineering partial differential equations 53.5 Separating the variables 53.6 The wave equation 53.7 The heat conduction equation 53.8 Laplace’s equation

601 602 602 602 604 605 606 610 612

Revision Test 15

615

564

Section J

Laplace transforms

617

568 569 570

570

571

573

575

578 578 579 580 583

54 Introduction to Laplace transforms 54.1 Introduction 54.2 Definition of a Laplace transform 54.3 Linearity property of the Laplace transform 54.4 Laplace transforms of elementary functions 54.5 Worked problems on standard Laplace transforms

619 619 619 620 620

55 Properties of Laplace transforms 55.1 The Laplace transform of e at f (t) 55.2 Laplace transforms of the form e at f (t) 55.3 The Laplace transforms of derivatives 55.4 The initial and final value theorems

625 625 625 627 629

56 Inverse Laplace transforms 56.1 Definition of the inverse Laplace transform 56.2 Inverse Laplace transforms of simple functions 56.3 Inverse Laplace transforms using partial fractions 56.4 Poles and zeros

632

57 The Laplace transform of the Heaviside function 57.1 Heaviside unit step function 57.2 Laplace transforms of H(t – c) 57.3 Laplace transforms of H(t – c).f(t – c) 57.4 Inverse Laplace transforms of Heaviside functions

640 640 644 644

621

632 632 635 637

645

xii Contents 58 The solution of differential equations using Laplace transforms 58.1 Introduction 58.2 Procedure to solve differential equations using Laplace transforms 58.3 Worked problems on solving differential equations using Laplace transforms

648 648 648 649

59 The solution of simultaneous differential equations using Laplace transforms 59.1 Introduction 59.2 Procedure to solve simultaneous differential equations using Laplace transforms 59.3 Worked problems on solving simultaneous differential equations using Laplace transforms

653 653

653

654

Revision Test 16

Section K

Fourier series

659

661

60 Fourier series for periodic functions of period 2π 60.1 Introduction 60.2 Periodic functions 60.3 Fourier series 60.4 Worked problems on Fourier series of periodic functions of period 2π 61 Fourier series for a non-periodic function over period 2π 61.1 Expansion of non-periodic functions 61.2 Worked problems on Fourier series of non-periodic functions over a range of 2π 62 Even and odd functions and half-range Fourier series 62.1 Even and odd functions 62.2 Fourier cosine and Fourier sine series 62.3 Half-range Fourier series

663 664 664 664 665

670 670 671

676 676 676 680

63 Fourier series over any range 63.1 Expansion of a periodic function of period L 63.2 Half-range Fourier series for functions defined over range L

684

64 A numerical method of harmonic analysis 64.1 Introduction 64.2 Harmonic analysis on data given in tabular or graphical form 64.3 Complex waveform considerations

690 690

684

65 The complex or exponential form of a Fourier series 65.1 Introduction 65.2 Exponential or complex notation 65.3 Complex coefficients 65.4 Symmetry relationships 65.5 The frequency spectrum 65.6 Phasors

Section L

Z-transforms

66 An introduction to z-transforms 66.1 Sequences 66.2 Some properties of z-transforms 66.3 Inverse z-transforms 66.4 Using z-transforms to solve difference equations Revision Test 17

Section M

Statistics and probability

698 698 698 699 703 706 707

711

713 714 717 720 722 727

729

67 Presentation of statistical data 67.1 Some statistical terminology 67.2 Presentation of ungrouped data 67.3 Presentation of grouped data

731 732 733 736

68 Mean, median, mode and standard deviation 68.1 Measures of central tendency 68.2 Mean, median and mode for discrete data 68.3 Mean, median and mode for grouped data 68.4 Standard deviation 68.5 Quartiles, deciles and percentiles

743 743 744 745 746 748

69 Probability 69.1 Introduction to probability 69.2 Laws of probability 69.3 Worked problems on probability 69.4 Further worked problems on probability 69.5 Permutations and combinations 69.6 Bayes’ theorem

751 752 752 753 755 758 759

688

690 694

Revision Test 18 70 The binomial and Poisson distributions 70.1 The binomial distribution 70.2 The Poisson distribution

762 764 764 767

Contents xiii 71 The normal distribution 71.1 Introduction to the normal distribution 71.2 Testing for a normal distribution

771 771 776

72 Linear correlation 72.1 Introduction to linear correlation 72.2 The Pearson product-moment formula for determining the linear correlation coefficient 72.3 The significance of a coefficient of correlation 72.4 Worked problems on linear correlation

780 780

781 781

73 Linear regression 73.1 Introduction to linear regression 73.2 The least-squares regression lines 73.3 Worked problems on linear regression

785 785 785 786

Revision Test 19

74 Sampling and estimation theories 74.1 Introduction 74.2 Sampling distributions 74.3 The sampling distribution of the means 74.4 The estimation of population parameters based on a large sample size 74.5 Estimating the mean of a population based on a small sample size

780

791

792 792 792 793 796 801

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

805 805 806 812 817

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

822 822 824 830 830 833 837

Revision Test 20

844

Essential formulae

846

Answers to Practice Exercises

863

Index

910

Preface This ninth edition of ‘Bird’s Higher Engineering Mathematics’ covers essential mathematical material suitable for students studying Degrees, Foundation Degrees and Higher National Certificate and Diploma courses in Engineering disciplines.

required for those wishing to pursue careers in mechanical engineering, aeronautical engineering, electrical and electronic engineering, communications engineering, systems engineering and all variants of control engineering.

The text has been conveniently divided into the following thirteen convenient categories: number and algebra, geometry and trigonometry, graphs, complex numbers, matrices and determinants, vector geometry, differential calculus, integral calculus, differential equations, Laplace transforms, Fourier series, Ztransforms and statistics and probability.

In Bird’s Higher Engineering Mathematics 9th Edition, theory is introduced in each chapter by a full outline of essential definitions, formulae, laws, procedures etc; problem solving is extensively used to establish and exemplify the theory. It is intended that readers will gain real understanding through seeing problems solved and then through solving similar problems themselves.

Increasingly, difficulty in understanding algebra is proving a problem for many students as they commence studying engineering courses. Inevitably there are a lot of formulae and calculations involved with engineering studies that require a sound grasp of algebra. On the website, www.routledge.com/bird is a document which offers a quick revision of the main areas of algebra essential for further study, i.e. basic algebra, simple equations, transposition of formulae, simultaneous equations and quadratic equations.

Access to the plethora of software packages, or a graphics calculator, will enhance understanding of some of the topics in this text.

In this new edition, all but three of the chapters of the previous edition are included (those excluded can be found in Bird’s Engineering Mathematics 8th Edition or on the website), but the order of presenting some of the chapters has been changed. Problems where engineering applications occur have been ‘flagged up’ and some multiple-choice questions have been added to many of the chapters. The primary aim of the material in this text is to provide the fundamental analytical and underpinning knowledge and techniques needed to successfully complete scientific and engineering principles modules of Degree, Foundation Degree and Higher National Engineering programmes. The material has been designed to enable students to use techniques learned for the analysis, modelling and solution of realistic engineering problems at Degree and Higher National level. It also aims to provide some of the more advanced knowledge

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. ‘Birds Higher Engineering Mathematics 9th Edition’ provides a follow-up to ‘Bird’s Engineering Mathematics 9th Edition’. This textbook contains over 1100 worked problems, followed by some 2100 further problems (with answers), arranged within 317 Practice Exercises. Some 450 multiple-choice questions are also included, together with 573 line diagrams to further enhance understanding. Worked solutions to all 2100 of the further problems have been prepared and can be accessed free by students and staff via the website www.routledge.com/bird Where at all possible, the problems mirror practical situations found in engineering and science. In fact, some 1200 engineering situations/problems have been ‘flagged-up’ to help demonstrate that engineering cannot be fully understood without a good knowledge of mathematics. Look out for the symbol

Preface xv

At the end of the text, a list of Essential Formulae is included for convenience of reference. At intervals throughout the text are some 20 Revision Tests to check understanding. For example, Revision Test 1 covers the material in chapters 1 to 4, Revision Test 2 covers the material in chapters 5 to 7, Revision Test 3 covers the material in chapters 8 to 10, and so on. Full solutions to the 20 Revision Tests are available free to lecturers on the website www.routledge.com/cw/bird ‘Learning by example’ is at the heart of ‘Bird’s Higher Engineering Mathematics 9th Edition’. JOHN BIRD Formerly Royal Naval Defence College of Marine and Air Engineering, HMS Sultan, University of Portsmouth and Highbury College, Portsmouth

Free Web downloads are available at www.routledge.com/cw/bird For Students 1.

Full solutions to the 2100 questions contained in the 317 Practice Exercises

2.

Revision of some important algebra topics

3.

List of Essential Formulae

4.

Famous Engineers/Scientists – 32 are mentioned in the text.

5.

Copies of chapters from the previous edition that have been excluded from this text (these being: ‘Inequalities’, ‘Arithmetic and geometric progressions’, and ‘Binary, octal and hexadecimal numbers’)

For instructors/lecturers 1. Full solutions to the 2100 questions contained in the 317 Practice Exercises 2.

Full solutions and marking scheme to each of the 20 Revision Tests

3.

Revision Tests – available to run off to be given to students

4.

List of Essential Formulae

5.

Illustrations – all 573 available on PowerPoint

6.

Famous Engineers/Scientists – 32 are mentioned in the text

Syllabus guidance This textbook is written for undergraduate engineering degree and foundation degree courses; however, it is also most appropriate for HNC/D studies and three syllabuses are covered. The appropriate chapters for these three syllabuses are shown in the table below. Chapter

Analytical Methods for Engineers

Further Analytical Methods for Engineers

1.

Algebra

×

2.

Partial fractions

×

3.

Logarithms

×

4.

Exponential functions

×

5.

The binomial series

×

6.

Solving equations by iterative methods

×

7.

Boolean algebra and logic circuits

×

8.

Introduction to trigonometry

×

9.

Cartesian and polar co-ordinates

×

10.

The circle and its properties

×

11.

Trigonometric waveforms

×

12.

Hyperbolic functions

×

13.

Trigonometric identities and equations

×

14.

×

15.

The relationship between trigonometric and hyperbolic functions Compound angles

16.

Functions and their curves

×

17.

Irregular areas, volumes and mean value of waveforms

×

18.

Complex numbers

×

19.

De Moivre’s theorem

×

20.

The theory of matrices and determinants

×

21.

The solution of simultaneous equations by matrices and determinants

×

22.

Vectors

×

23.

Methods of adding alternating waveforms

×

24.

Scalar and vector products

×

25.

Methods of differentiation

Advanced Mathematics

×

× (Continued )

Syllabus guidance xvii Chapter

Analytical Methods for Engineers

Further Analytical Methods for Engineers

Advanced Mathematics for Engineering

×

26.

Some applications of differentiation

27.

Differentiation of parametric equations

28.

Differentiation of implicit functions

×

29.

Logarithmic differentiation

×

30.

Differentiation of hyperbolic functions

×

31.

Differentiation of inverse trigonometric and hyperbolic functions

×

32.

Partial differentiation

×

33.

Total differential, rates of change and small changes

×

34.

Maxima, minima and saddle points for functions of two variables

×

35.

Standard integration

×

36.

Some applications of integration

×

37.

Maclaurin’s series and limiting values

×

38.

Integration using algebraic substitutions

×

39.

Integration using trigonometric and hyperbolic substitutions

×

40.

Integration using partial fractions

×

41.

The t = tan θ/2 substitution

42.

Integration by parts

×

43.

Reduction formulae

×

44.

Double and triple integrals

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 2 a ddx2y + b dy dx + cy = 0

×

51.

Second-order differential equations of the form 2 a ddx2y + b dy dx + cy = f (x)

×

52.

Power series methods of solving ordinary differential equations

×

53.

An introduction to partial differential equations

×

×

(Continued )

xviii Syllabus guidance Chapter

Analytical Methods for Engineers

Further Analytical Methods for Engineers

Advanced Mathematics for Engineering

54.

Introduction to Laplace transforms

×

55.

Properties of Laplace transforms

×

56.

Inverse Laplace transforms

×

57.

The Laplace transform of the Heaviside function

58.

Solution of differential equations using Laplace transforms

×

59.

The solution of simultaneous differential equations using Laplace transforms

×

60.

Fourier series for periodic functions of period 2π

×

61.

Fourier series for non-periodic functions over range 2π

×

62.

Even and odd functions and half-range Fourier series

×

63.

Fourier series over any range

×

64.

A numerical method of harmonic analysis

×

65.

The complex or exponential form of a Fourier series

×

66.

An introduction to z-transforms

67.

Presentation of statistical data

×

68.

Mean, median, mode and standard deviation

×

69.

Probability

×

70.

The binomial and Poisson distributions

×

71.

The normal distribution

×

72.

Linear correlation

×

73.

Linear regression

×

74.

Sampling and estimation theories

×

75.

Significance testing

×

76.

Chi-square and distribution-free tests

×

Section A

Number and algebra

Chapter 1

Algebra Why it is important to understand: Algebra, polynomial division and the factor and remainder theorems It is probably true to say that there is no branch of engineering, physics, economics, chemistry or computer science which does not require the understanding of the basic laws of algebra, the laws of indices, the manipulation of brackets, the ability to factorise and the laws of precedence. This then leads to the ability to solve simple, simultaneous and quadratic equations which occur so often. The study of algebra also revolves around using and manipulating polynomials. Polynomials are used in engineering, computer programming, software engineering, in management and in business. Mathematicians, statisticians and engineers of all sciences employ the use of polynomials to solve problems; among them are aerospace engineers, chemical engineers, civil engineers, electrical engineers, environmental engineers, industrial engineers, materials engineers, mechanical engineers and nuclear engineers. The factor and remainder theorems are also employed in engineering software and electronic mathematical applications, through which polynomials of higher degrees and longer arithmetic structures are divided without any complexity. The study of algebra, equations, polynomial division and the factor and remainder theorems is therefore of some considerable importance in engineering.

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

1.1

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

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 the website: www.routledge.com/cw/bird

1.2

Revision of basic laws

(a) Basic operations and laws of indices The laws of indices are: am (i) am × an = am+n (ii) = am−n an √ m (iii) (am )n = am×n (iv) a n = n am 1 (v) a−n = n (vi) a0 = 1 a

4 Section A Problem 1. Evaluate 4a2 bc3 −2ac when a = 2, b = 12 and c = 1 12

Problem 5.

( )( )3 ( ) 1 3 3 4a bc − 2ac = 4(2) − 2(2) 2 2 2 2

3

√ √ √ (x2 y)( x 3 y2 )

2

=

Simplify

1

(x5 y3 ) 2

4 × 2 × 2 × 3 × 3 × 3 12 − 2×2×2×2 2

1

1

5

3

2

x2 y 2 x 2 y 3 x2 y2 1

5

1

− 3xy − 2y2 3x2 − xy − 2y2

Adding gives: Alternatively,

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

2

a3 b2 c4 Problem 3. Simplify and evaluate when abc−2 a = 3, b = 18 and c = 2 a3 b2 c4 = a3−1 b2−1 c4−(−2) = a2 bc6 abc−2 When a = 3, b =

1 8

or

a2 bc6 = (3)2

1 8

( ) (2)6 = (9) 18 (64) = 72

Simplify

x2 y3 + xy2 xy

x2 y3 + xy2 x2 y3 xy2 = + xy xy xy = x2−1 y3−1 + x1−1 y2−1 = xy2 + y

1 1

y3

or

1 √ 3 y

Practice Exercise 1 Basic algebraic operations and laws of indices (Answers on page 863) 1.

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

2.

Find the value of 5pq2 r 3 when p = 25 , q = −2 and r = −1

3.

From 4x − 3y + 2z subtract x + 2y − 3z

4.

Multiply 2a − 5b + c by 3a + b

5.

Simplify (x2 y3 z)(x3 yz2 ) and evaluate when x = 12 , y = 2 and z = 3

6.

Evaluate (a 2 bc−3 )(a 2 b− 2 c) when a = 3, b = 4 and c = 2

7.

Simplify

8.

(a3 b 2 c− 2 )(ab) 3 √ √ Simplify ( a3 b c)

3

1

or y(xy + 1)

1

a2 b + a3 b a2 b2 1

and c = 2, ( )

3

Now try the following Practice Exercise

3x2 + 2xy

Multiply by −y →

2

1

Multiply 3x + 2y by x − y

Multiply by x →

1

= x0 y− 3 = y− 3

3x + 2y x−y

Problem 4.

1

(x5 y3 ) 2

= x2+ 2 − 2 y 2 + 3 − 2

= 27 − 6 = 21 Problem 2.

=

√ √ √ (x2 y)( x 3 y2 )

1

1

(b) Brackets, factorisation and precedence Problem 6.

Simplify a2 − (2a − ab) − a(3b + a)

a2 − (2a − ab) − a(3b + a) = a2 − 2a + ab − 3ab − a2 = −2a − 2ab or −2a(1 + b)

Algebra 5 Problem 7. Remove the brackets and simplify the expression:

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

2a − [3{2(4a − b) − 5(a + 2b)} + 4a] (2a − 3) ÷ 4a + 5 × 6 − 3a Removing the innermost brackets gives: 2a − [3{8a − 2b − 5a − 10b} + 4a] Collecting together similar terms gives: 2a − [3{3a − 12b} + 4a]

=

2a − 3 + 5 × 6 − 3a 4a

=

2a − 3 + 30 − 3a 4a

=

2a 3 − + 30 − 3a 4a 4a

=

1 3 1 3 − + 30 − 3a = 30 − − 3a 2 4a 2 4a

Removing the ‘curly’ brackets gives: 2a − [9a − 36b + 4a] Collecting together similar terms gives:

Now try the following Practice Exercise

2a − [13a − 36b]

Practice Exercise 2 Brackets, factorisation and precedence (Answers on page 863)

Removing the square brackets gives: 2a − 13a + 36b = −11a + 36b

or

36b − 11a Problem 8. Factorise (a) xy − 3xz (b) 4a2 + 16ab3 (c) 3a2 b − 6ab2 + 15ab (a) xy − 3xz = x(y − 3z) (b)

4a2 + 16ab3 = 4a(a + 4b3 )

(c) 3a2 b − 6ab2 + 15ab = 3ab(a − 2b + 5)

Problem 9.

1.

Simplify 2(p + 3q − r) − 4(r − q + 2p) + p

2.

Expand and simplify (x + y)(x − 2y)

3.

Remove the brackets and simplify: 24p − [2{3(5p − q) − 2(p + 2q)} + 3q]

4.

Factorise 21a2 b2 − 28ab

5.

Factorise 2xy2 + 6x2 y + 8x3 y

6.

Simplify 2y + 4 ÷ 6y + 3 × 4 − 5y

7.

Simplify 3 ÷ y + 2 ÷ y − 1

8.

Simplify a2 − 3ab × 2a ÷ 6b + ab

Simplify 3c + 2c × 4c + c ÷ 5c − 8c

The order of precedence is division, multiplication, addition, and subtraction (sometimes remembered by BODMAS). Hence 3c + 2c × 4c + c ÷ 5c − 8c (c) = 3c + 2c × 4c + − 8c 5c 1 = 3c + 8c2 + − 8c 5 1 1 = 8c2 − 5c + or c(8c − 5) + 5 5

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 15 i.e. 15 = 5x from which, x = =3 5 Problem 12. Solve 4(2a − 3) − 2(a − 4) = 3(a − 3) − 1

6 Section A Removing the brackets gives: 8a − 12 − 2a + 8 = 3a − 9 − 1 Rearranging gives: 8a − 2a − 3a = −9 − 1 + 12 − 8 i.e. and

3a = −6 −6 a= = −2 3

ft Problem 16. Transpose the formula v = u + m to make f the subject. ft ft = v from which, = v − u m m ( ) ft and m = m(v − u) m i.e. f t = m(v − u) u+

Problem 13. The reaction moment M of a cantilever carrying three, point loads is given by: M = 3.5x + 2.0(x − 1.8) + 4.2(x − 2.6) where x is the length of the cantilever in metres. If M = 55.32 kN m, calculate the value of x.

then

M = 3.5x + 2.0(x − 1.8) + 4.2(x − 2.6) and M = 55.32 55.32 = 3.5x + 2.0(x − 1.8) + 4.2(x − 2.6)

i.e. and

55.32 = 3.5x + 2.0x − 3.6 + 4.2x − 10.92 55.32 + 3.6 + 10.92 = 9.7x

If

(c) Transposition of formulae

3 4 = x − 2 3x + 4

By ‘cross-multiplying’:

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

9x + 12 = 4x − 8

Rearranging gives:

9x − 4x = −8 − 12

i.e.

5x = −20

and

−20 x= 5 = −4

(√ ) t+3 √ Problem 15. Solve =2 t

and i.e. and

(√ ) √ √ t+3 √ t =2 t t √ √ t+3= 2 t √ √ 3= 2 t− t √ 3= t 9= t

√

X2 = Z2 − R2 and reactance X =

D Problem 18. Given that = d express p in terms of D, d and f.

√(

3(3x + 4) = 4(x − 2)

Removing brackets gives:

i.e.

m (v − u) t

Problem 17. √The impedance of an a.c. circuit is given by Z = R2 + X2 . Make the reactance X the subject.

i.e. 69.84 = 9.7x from which, the length of the cantilever, 69.84 x= = 7.2 m 9.7 Problem 14. Solve

f=

and

f +p f −p

√( Rearranging gives: Squaring both sides gives:

Z2 − R2 )

) f +p D = f −p d f + p D2 = 2 f −p d

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

d2 f + d2 p = D2 f − D2 p d p + D2 p = D2 f − d2 f

Factorising gives:

p(d2 + D2 ) = f (D2 − d2 )

and

2

p=

f(D2 − d2 ) (d2 + D2 )

Problem 19. Bernoulli’s equation relates the flow velocity, v, the pressure, p, of the liquid and the height h of the liquid above some reference level.

Algebra 7 Given two locations 1 and 2, the equation states: v21

v22

p1 p2 + + h1 = + + h2 ρ g 2g ρ g 2g

stress, σ and strain, ε length increase, δ strain, ε = length, ℓ For a 50 mm length of a steel bolt under tensile stress, its length increases by 1.25 × 10−4 m. If E = 2 × 1011 N/m2 , calculate the stress.

9. Young’s modulus, E =

where ρ is the density of the liquid. Rearrange the equation to make the velocity of the liquid at location 2, i.e. v2 , the subject. p1 v2 p2 v2 + 1 + h1 = + 2 + h2 ρ g 2g ρ g 2g p1 v21 p2 v2 then + + h1 − − h2 = 2 ρg ( ρ g 2g ) 2g p1 v21 p2 and 2g + + h1 − − h2 = v22 ρ g 2g ρg √ ( ) p1 v2 p2 from which, v2 = 2g + 1 + h1 − − h2 ρ g 2g ρg i.e. the velocity at location 2, v ( ) u 2 u p p v 1 2 1 v2 = t2g − + + h1 − h2 ρg ρg 2g Since

10.

The mass moment of inertia through the centre of gravity for a compound pendulum, IG , is given by: IG = mkG2 . The value of the radius of gyration about G, kG2 , may be determined from the frequency, f ,√ of a compound 1 gh pendulum, given by: f = 2π (kG2 + h2 ) Given that the distance h = 50 mm, g = 9.81 m/s2 and the frequency of oscillation, f = 1.26 Hz, calculate the mass moment of inertia IG when m = 10.5 kg.

(d) Simultaneous equations Now try the following Practice Exercise Practice Exercise 3 Simple equations and transposition of formulae (Answers on page 863) In problems 1 to 4 solve the equations 1. 3x − 2 − 5x = 2x − 4 2. 8 + 4(x − 1) − 5(x − 3) = 2(5 − 2x) 1 1 3. + =0 3a − 2 5a + 3 √ 3 t √ = −6 4. 1− t 5. Transpose y =

3(F − f ) for f L √

6. Make l the subject of t = 2π

7x − 2y = 26

(1)

6x + 5y = 29

(2)

5 × equation (1) gives: 35x − 10y = 130 2 × equation (2) gives: 12x + 10y = 58 Equation (3) + equation (4) gives:

(3) (4)

47x + 0 = 188 188 from which, x= =4 47 Substituting x = 4 in equation (1) gives: 28 − 2y = 26

l g

µL for L L + rCR 8. Make r the subject of the formula x 1 + r2 = y 1 − r2 7. Transpose m =

Problem 20. Solve the simultaneous equations:

from which, 28 − 26 = 2y and y = 1

Problem 21. Solve x 5 + =y 8 2 y 11 + = 3x 3

(1) (2)

8 Section A 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=

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

284 =4 71

1 and −2 are the roots of a quadratic equation then, 3 1 (x − )(x + 2) = 0 3 1 2 2 i.e. x + 2x − x − = 0 3 3 5 2 2 i.e. x + x− = 0 3 3 If

3x2 + 5x − 2 = 0

or

Problem 24. The stress, ρ, set up in a bar of length,ℓ, cross-sectional area, A, by a mass W, falling a distance h is given by the formula:

4 + 20 = 8y and y = 3

from which,

Problem 23. The roots of a quadratic equation are 13 and −2. Determine the equation in x.

(e) Quadratic equations

ρ2 −

Problem 22. Solve the following equations by factorisation: (a) 3x2 − 11x − 4 = 0 (b)

4x2 + 8x + 3 = 0

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

)(x

either or (b)

2W 2WEh ρ− = 0 and E = 12500, ℓ = 110, A Aℓ h = 0.5, W = 2.25 and A = 3.20 then Since ρ2 −

ρ2 −

3x2 − 11x − 4 = (3x + 1)(x − 4) Thus

Given that E = 12500, ℓ = 110, h = 0.5, W = 2.25 and A = 3.20, calculate the positive value of ρ correct to 3 significant figures.

)

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

(3x + 1)(x − 4) = 0 hence (3x + 1) = 0 i.e. x = −

1 3

(x − 4) = 0 i.e. x = 4

2W 2WEh ρ− =0 A Aℓ

2(2.25) 2(2.25)(12500)(0.5) ρ− =0 3.20 (3.20)(110)

i.e. ρ2 − 1.40625ρ − 79.90057 = 0 Solving using the quadratic formula gives: √ − − 1.40625 ± (−1.40625)2 − 4(1)(−79.90057) ρ= 2(1) = 9.669 or −8.263 Hence, correct to 3 significant figures, the positive value of ρ is 9.67 Now try the following Practice Exercise

4x2 + 8x + 3 = (2x + 3)(2x + 1) Thus either or

(2x + 3)(2x + 1) = 0 hence 3 2 1 (2x + 1) = 0 i.e. x = − 2 (2x + 3) = 0 i.e. x = −

Practice Exercise 4 Simultaneous and quadratic equations (Answers on page 863) In problems 1 to 3, solve the simultaneous equations 1.

8x − 3y = 51 3x + 4y = 14

Algebra 9 1.4

2. 5a = 1 − 3b 2b + a + 4 = 0 3.

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!). 208 For example, is achieved as follows: 16

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

4. In an engineering scenario involving reciprocal motion, the following simultaneous √( ) 2 equations resulted: ω r − 0.072 = 8 and √( ) ω r 2 − 0.252 = 2 Calculate the value of radius r and angular velocity ω, each correct to 3 significant figures.

13 ) ——– 16 208 16 48 48 — ·· — (1)

16 divided into 2 won’t go

(2)

16 divided into 20 goes 1

(a) x2 + 4x − 32 = 0

(3)

Put 1 above the zero

(b) 8x2 + 2x − 15 = 0

(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

5. Solve the following quadratic equations by factorisation:

6. Determine the quadratic equation in x whose roots are 2 and −5 7. Solve the following quadratic equations, correct to 3 decimal places: (a) 2x + 5x − 4 = 0 2

(b) 4t − 11t + 3 = 0

(10) 48 − 48 = 0

2

8. A point of contraflexure from the left-hand end of a 6 m beam is given by the value of x in the following equation: −x + 11.25x − 22.5 = 0 2

Determine the point of contraflexure. 9. The vertical height, h, and the horizontal distance travelled, x, of a projectile fired at an angle of 45◦ at an initial velocity,v0 , are related by the equation: h = x−

g x2 v20

If the projectile has an initial velocity of 120 m/s, calculate the values of x when the projectile is at a height of 200 m, assuming that g = 9.81 m/s2 . Give the answers correct to 3 significant figures.

Hence

208 = 13 exactly 16

172 is laid out as follows: 15 11 ) ——– 15 172 15

Similarly,

22 15 — 7 — 172 7 7 Hence = 11 remainder 7 or 11 + = 11 15 15 15 Below are some examples of division in algebra, which in some respects is similar to long division with numbers. (Note that a polynomial is an expression of the form f (x) = a + bx + cx2 + dx3 + · · ·

10 Section A and polynomial division is sometimes required when resolving into partial fractions – see Chapter 2.) Problem 25. Divide 2x2 + x − 3 by x − 1 2x2 + x − 3 is called the dividend and x − 1 the divisor. The usual layout is shown below with the dividend and divisor both arranged in descending powers of the symbols. 2x + 3 ) ——————– x − 1 2x2 + x − 3 2x2 − 2x 3x − 3 3x − 3 ——— · · ———

(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 = 3x2 − 2x + 5 x+1

Problem 27. Simplify

Dividing the first term of the dividend by the first term 2x2 gives 2x, which is put above of the divisor, i.e. x the first term of the dividend as shown. The divisor is then multiplied by 2x, i.e. 2x(x − 1) = 2x2 − 2x, which is placed under the dividend as shown. Subtracting gives 3x − 3. The process is then repeated, i.e. the first term of the divisor, x, is divided into 3x, giving +3, which is placed above the dividend as shown. Then 3(x − 1) = 3x − 3, which is placed under the 3x − 3. The remainder, on subtraction, is zero, which completes the process.

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 ———– · · ———–

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

(1) (4) (7) 3x2 − 2x + 5 ) ————————— x + 1 3x3 + x2 + 3x + 5 3x3 + 3x2

(5)

−xy(x + y) = −x2 y − xy2

(6)

Subtract

(7)

x into xy2 goes y2 . Put y2 above dividend

− 2x2 + 3x + 5 − 2x2 − 2x ————– 5x + 5 5x + 5 ——— · · ———

(8)

y2 (x + y) = xy2 + y3

(9)

Subtract

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

Problem 26. Divide 3x3 + x2 + 3x + 5 by x + 1

Thus x3 + y3 = x2 − xy + y2 x+y

Algebra 11 The zeros shown in the dividend are not normally shown, but are included to clarify the subtraction process and to keep similar terms in their respective columns. Problem 28. Divide (x2 + 3x − 2) by (x − 2) x +5 ) ——————– x − 2 x2 + 3x − 2 x2 − 2x 5x − 2 5x − 10 ——— 8 ——— x2 + 3x − 2 8 =x+5+ x−2 x−2 Problem 29. Divide 4a3 − 6a2 b + 5b3 by 2a − b 2a − 2ab − b ) ——————————— 2a − b 4a3 − 6a2 b + 5b3 3 2 4a − 2a b 2

Find

5.

Divide (x3 + 3x2 y + 3xy2 + y3 ) by (x + y)

6.

Find (5x2 − x + 4) ÷ (x − 1)

7.

Divide (3x3 + 2x2 − 5x + 4) by (x + 2)

8.

Determine (5x4 + 3x3 − 2x + 1)/(x − 3)

1.5

Hence

2

−4a2 b + 5b3 2 2 −4a b + 2ab ———— 2 −2ab + 5b3 −2ab2 + b3 —————– 4b3 —————– Thus 4a3 − 6a2 b + 5b3 2a − b = 2a2 − 2ab − b2 +

4b3 2a − b

Now try the following Practice Exercise Practice Exercise 5 Polynomial division (Answers on page 863) 1. Divide (2x2 + xy − y2 ) by (x + y) 2. Divide (3x2 + 5x − 2) by (x + 2) 3. Determine (10x2 + 11x − 6) ÷ (2x + 3)

14x2 − 19x − 3 2x − 3

4.

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 factorise 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 determine that x = 2 is a root of the equation x2 + 2x − 8 = 0 we could deduce at once that (x − 2) is a factor of the expression x2 + 2x − 8. We wouldn’t normally solve quadratic equations this way – but suppose we have to factorise 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 generalised version of what we established above for the quadratic expression. The factor theorem provides a method of factorising any polynomial, f (x), which has simple factors. A statement of the factor theorem says:

12 Section A ‘if x = a is a root of the equation f(x) = 0, then (x − a) is a factor of f(x)’

To solve x3 − 7x − 6 = 0, we substitute the factors, i.e.

The following worked problems show the use of the factor theorem.

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. 6. This can give us a clue as to what values of x we should consider.

Problem 30. Factorise x3 − 7x − 6 and use it to solve the cubic equation x3 − 7x − 6 = 0.

(x − 3)(x + 1)(x + 2) = 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 and error’ by substituting further values for x in the given expression – and hope to arrive at f (x) = 0 Let us do both ways. Firstly, dividing out gives: 2

x + 3x + 2 ) ————————— x − 3 x3 − 0 − 7x − 6 x3 − 3x2

Hence

3x2 − 7x − 6 3x2 − 9x ———— 2x − 6 2x − 6 ——— · · ——— x3 − 7x − 6 = x2 + 3x + 2 x−3

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

Problem 31. 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 x − 2x − 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: x2 − x − 6 ) ————————– x − 1 x3 − 2x2 − 5x + 6 x3 − x2 3

2

− x2 − 5x + 6 − x2 + x ————– − 6x + 6 − 6x + 6 ———– · · ———–

3

Hence

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

Algebra 13 Summarising, the factor theorem provides us with a method of factorising simple expressions, and an alternative, in certain circumstances, to polynomial division.

Now try the following Practice Exercise Practice Exercise 6 The factor theorem (Answers on page 863) Use the factor theorem to factorise the expressions given in problems 1 to 4. 1. x2 + 2x − 3 2. x3 + x2 − 4x − 4 3. 2x3 + 5x2 − 4x − 7 4. 2x3 − x2 − 16x + 15 5. Use the factor theorem to factorise x3 + 4x2 + x − 6 and hence solve the cubic equation x3 + 4x2 + x − 6 = 0 6. Solve the equation x3 − 2x2 − x + 2 = 0

1.6

The remainder theorem

Dividing a general quadratic expression (ax2 + bx + c) by (x − p), where p is any whole number, by long division (see Section 1.4) 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 + ap 2 or ap 2 + 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 ap 2 + bp + c For example, when (3x2 − 4x + 5) is divided by (x − 2) the remainder is ap 2 + 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 ap 2 + bp + c (where a = 4, b = −7, c = 9 and p = −3), i.e. the remainder is 4(−3)2 + (−7)(−3) + 9 = 36 + 21 + 9 = 66 Also, when (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 factorising expressions. For example, when (2x2 + x − 3) is divided by (x − 1), the remainder is 2(1)2 + 1(1) − 3 = 0, which means that (x − 1) is a factor of (2x2 + x − 3). In this case the other factor is (2x + 3), i.e. (2x2 + x − 3) = (x − 1)(2x − 3) The remainder theorem may also be stated for a 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 substituting p for x in the dividend. For example, when (3x3 + 2x2 − x + 4) is divided by (x − 1), the remainder is ap 3 + bp 2 + 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 32. Without dividing out, find the remainder when 2x2 − 3x + 4 is divided by (x − 2)

14 Section A By the remainder theorem, the remainder is given by ap 2 + bp + c, where a = 2, b = −3, c = 4 and p = 2. Hence the remainder is:

(i)

2(2)2 + (−3)(2) + 4 = 8 − 6 + 4 = 6 Problem 33. 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 ap 3 + bp 2 + cp + d, where a = 3, b = −2, c = 1, d = −5 and p = −2 Hence the remainder is:

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

3(−2)3 + (−2)(−2)2 + (1)(−2) + (−5) = −39

Then f(3) = 33 − 2(3)2 − 5(3) + 6 = 27 − 18 − 15 + 6 = 0

Problem 34. Determine the remainder when (x3 − 2x2 − 5x + 6) is divided by (a) (x − 1) and (b) (x + 2). Hence factorise the cubic expression.

Hence (x − 3) is a factor. (iii)

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

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

= 27 − 18 − 15 + 6 = 0

Hence (x − 1) is a factor of (x − 2x − 5x + 6). 3

Using the remainder theorem, when (x3 − 2x2 − 5x + 6) is divided by (x − 3), the remainder is given by ap 3 + bp 2 + 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

= 1−2−5+6 = 0

(b)

Using the factor theorem, we let f (x) = x3 − 2x2 − 5x + 6

= −24 − 8 − 2 − 5

3

Dividing (x3 − 2x2 − 5x + 6) by (x2 + x − 2) gives: x −3 ) ————————– 2 x + x − 2 x3 − 2x2 − 5x + 6 x3 + x2 − 2x —————— −3x2 − 3x + 6 −3x2 − 3x + 6 ——————– · · · ——————–

2

Hence (x − 3) is a factor.

When (x3 − 2x2 − 5x + 6) is divided by (x + 2), the remainder is given by

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

(1)(−2)3 + (−2)(−2)2 + (−5)(−2) + 6 = −8 − 8 + 10 + 6 = 0 Hence (x + 2) is also a factor of (x3 − 2x2 − 5x + 6). Therefore (x − 1)(x + 2)(x) = x3 − 2x2 − 5x + 6. To determine the third factor (shown blank) we could (i) or (ii)

or (iii)

divide (x3 − 2x2 − 5x + 6) by (x − 1)(x + 2) use the factor theorem where f (x) = x3 − 2x2 − 5x + 6 and hoping to choose a value of x which makes f (x) = 0 use the remainder theorem, again hoping to choose a factor (x − p) which makes the remainder zero.

Now try the following Practice Exercise Practice Exercise 7 The remainder theorem (Answers on page 863) 1.

Find the remainder when 3x2 − 4x + 2 is divided by (a) (x − 2) (b) (x + 1)

2.

Determine the remainder when x3 − 6x2 + x − 5 is divided by (a) (x + 2) (b) (x − 3)

Algebra 15

3. Use the remainder theorem to find the factors of x3 − 6x2 + 11x − 6 4. Determine the factors of x3 + 7x2 + 14x + 8 and hence solve the cubic equation x3 + 7x2 + 14x + 8 = 0

25d 4 (c) 1 + 2d (a)

8.

(2e − 3f )(e + f) is equal to: (a) 2e 2 − 3f 2 (b) 2e 2 − 5ef − 3f 2 (c) 2e 2 + 3f 2 (d) 2e 2 − ef − 3f 2

9.

Factorising 2xy2 + 6x3 y − 8x3 y2 gives: (a) 2x(y2 + 3x2 − 4x2 y) (b) 2xy(y + 3x2 y − 4x2 y2 ) (c) 96x7 y5 (d) 2xy(y + 3x2 − 4x2 y)

5. Determine the value of ‘a’ if (x + 2) is a factor of (x3 − ax2 + 7x + 10) 6. Using the remainder theorem, solve the equation 2x3 − x2 − 7x + 6 = 0

10.

The value of x in the equation 5x − 27 + 3x = 4 + 9 − 2x is: (a) −4 (b) 5 (c) 4 (d) 6

11.

Solving the equation 17 + 19(x + y) = 19(y − x) − 21 for x gives: (a) −1 (b) −2 (c) −3 (d) −4

12.

Solving the equation 1 + 3x = 2(x − 1) gives: (a) x = −1 (b) x = −2 (c) x = 1 (d) x = −3

13.

The current I in an a.c. circuit is given by:

Practice Exercise 8 Multiple-choice questions on algebra (Answers on page 864) Each question has only one correct answer 2

1

1. (16− 4 − 27− 3 ) is equal to: (a) −7 (b)

7 18

(c) 1

8 9

(d) −8

√ 2. ( x)(y3/2 )(x2 y) is equal to: √ √ (a) (xy)5 (b) x√2 y5/2 (c) xy5/2 (d) x y3

1 2

3. Given that 7x + 8 − 2y = 12y − 4 − x then: 4x + 6 7y + 2 (a) y = (b) x = 7 4 3x − 6 5y − 6 (c) y = (d) x = 5 4 4.

5.

6.

V I= √ R2 + X2 When R = 4.8, X = 10.5 and I = 15, the value of voltage V is: (a) 173.18 (b) 1.30 (c) 0.98 (d) 229.50 14.

Transposing the formula R = R0 (1 + αt) for t gives: R − R0 R − R0 − 1 (a) (b) (1 + α) α R − R0 R (c) (d) α R0 R0 α

15.

The height s of a mass projected vertically 1 upwards at time t is given by: s = ut − gt2 . 2 When g = 10, t = 1.5 and s = 3.75, the value of u is: (a) 10 (b) – 5 (c) +5 (d) −10

16.

The quantity of heat Q is given by the formula Q = mc(t2 − t1 ). When m = 5, t1 = 20, c = 8 and Q = 1200, the value of t2 is: (a) 10 (b) 1.5 (c) 21.5 (d) 50

(3x2 y)2 simplifies to: 6xy2 (a)

x y

(b)

3x2 2

(c)

3x3 2

(d)

x 2y

p+q is equivalent to: q p p p (a) p (b) + q (c) + 1 (d) + p q q q (pq2 r)3 simplifies to: (p −2 q r 2 )2 p 3 q4 p3 1 (a) (b) (c) r pr pqr

(d)

p 7 q4 r

7. 3d + 2d × 4d + d ÷ (6d − 2d) simplifies to:

1 + d(3 + 8d) 4 1 (d) 20d2 + 4 (b)

16 Section A 17. Current I in an electrical circuit is given by E−e . Transposing for R gives: I= R+r E − e − Ir E−e (a) (b) I I+r E−e (c) (E − e)(I + r) (d) Ir 18. The solution of the simultaneous equations 3x − 2y = 13 and 2x + 5y = −4 is: (a) x = −2, y = 3 (b) x = 1, y = −5 (c) x = 3, y = −2 (d) x = −7, y = 2

24.

25.

The roots of the quadratic equation 3x2 − 7x − 6 = 0 are: (a) 2 and −5 (b) 2 and −9 2 (d) 3 and −6 (c) 3 and − 3 If one root of the quadratic equation 2x2 + cx − 6 = 0 is 2, the value of c is: (a) 2 (b) −1 (c) −2 (d) 1

26.

The degree of the polynomial equation 6x5 + 7x2 − 4x + 2 = 0 is: (a) 1 (b) 2 (c) 3 (d) 5

19. If x + 2y = 1 and 3x − y = −11 then the value of x is: (a) −3 (b) 4.2 (c) 2 (d) −1.6

27.

6x2 − 5x − 6 divided by 2x − 3 gives: (a) 2x − 1 (b) 3x + 2 (c) 3x − 2 (d) 6x + 1

20. If 4x − 3y = 7 and 2x + 5y = −1 then the value of y is: 59 9 (c) 2 (d) (a) −1 (b) − 13 26

28.

(x3 − x2 − x + 1) divided by (x − 1) gives: (a) x2 − x − 1 (b) x2 + 1 (c) x2 − 1 (d) x2 + x − 1

29.

The remainder when (x2 − 2x + 5) is divided by (x − 1) is: (a) 2 (b) 4 (c) 6 (d) 8

30.

The remainder when (3x3 − x2 + 2x − 5) is divided by (x + 2) is: (a) −19 (b) 21 (c) −29 (d) −37

21. If 5x − 3y = 17.5 and 2x + y = 4.25 then the value of 4x + 3y is: (a) 7.25 (b) 8.5 (c) 12 (d) 14.75 22. 8x + 13x − 6 = (x + p)(qx − 3). The values of p and q are: (a) p = −2, q = 4 (b) p = 3, q = 2 (c) p = 2, q = 8 (d) p = 1, q = 8 2

23. The height S metres of a mass thrown vertically upwards at time t seconds is given by S = 80t − 16t2 . To reach a height of 50 metres on the descent will take the mass: (a) 0.73 s (b) 4.27 s (c) 5.56 s (d) 81.77 s

For more help on basic algebra, simple equations, transposition of formulae, simultaneous equations and quadratic equations, go to the website: www.routledge.com/cw/bird

For fully worked solutions to each of the problems in Practice Exercises 1 to 7 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 2

Partial fractions Why it is important to understand: Partial fractions The algebraic technique of resolving a complicated fraction into partial fractions is often needed by electrical and mechanical engineers for not only determining certain integrals in calculus, but for determining inverse Laplace transforms and for analysing linear differential equations with resonant circuits and feedback control systems.

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

2.1

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

Introduction to partial fractions

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

4x − 5 −x−2

x2

The reverse process of moving from

4x − 5 −x−2

x2

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

(i)

the denominator must factorise (in the above example, x2 − x − 2 factorises 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).

When the degree of the numerator is equal to or higher than the degree of the denominator, the numerator must be divided by the denominator until the remainder is of less degree than the denominator (see Problems 3 and 4). There are basically three types of partial fraction and the form of partial fraction used is summarised 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 factorise without containing surds or imaginary terms.) Resolving an algebraic expression into partial fractions is used as a preliminary to integrating certain functions (see Chapter 40) and in determining inverse Laplace transforms (see Chapter 56).

18 Section A Table 2.1 Type

Denominator containing

1

Linear factors (see Problems 1 to 4)

2

Repeated linear factors (see Problems 5 to 7)

3

Quadratic factors (see Problems 8 and 9)

Expression

Form of partial fraction

f (x) (x + a)(x − b)(x + c)

A B C + + (x + a) (x − b) (x + c)

f (x) (x + a)3

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

(ax2

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

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

2.2 Partial fractions with linear factors

Thus

(ax2

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

Problem 1.

Resolve

fractions.

11 − 3x into partial + 2x − 3

x2

[ Check:

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

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

x2

≡

A B + (x − 1) (x + 3)

where A and B are constants to be determined, i.e.

2x2 − 9x − 35 into (x + 1)(x − 2)(x + 3) the sum of three partial fractions. Let

Convert

2x2 − 9x − 35 (x + 1)(x − 2)(x + 3) ≡

A B C + + (x + 1) (x − 2) (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. 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)

When x = 1, then

2 5 2(x + 3) − 5(x − 1) − = (x − 1) (x + 3) (x − 1)(x + 3) ] 11 − 3x = 2 x + 2x − 3

Problem 2.

11 − 3x A(x + 3) + B(x − 1) ≡ (x − 1)(x + 3) (x − 1)(x + 3)

To determine constants A and B, values of x are chosen to make the term in A or B equal to zero.

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)

11 − 3(1) ≡ A(1 + 3) + B(0) i.e. i.e.

8 = 4A A =2

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

When x = −3, then 11 − 3(−3) ≡ A(0) + B(−3 − 1) i.e.

20 = −4B

i.e.

B = −5

2 5 − (x − 1) (x + 3)

+ B(0)(2) + C(0)(−3) i.e. i.e.

−24 = −6A A=

−24 =4 −6

Partial fractions 19 Let x = 2. Then 2(2) − 9(2) − 35 ≡ A(0)(5) + B(3)(5) + C(3)(0)

Hence

2

−45 = 15B

i.e.

B=

i.e.

Thus

x2

−45 = −3 15

Let x = − 3. Then 2(−3)2 − 9(−3) − 35 ≡ A(−5)(0) + B(−2)(0)

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

Problem 4.

Express

fractions.

x3 − 2x2 − 4x − 4 in partial x2 + x − 2

+ C(−2)(−5) i.e.

10 = 10C

i.e.

C =1

Thus

The numerator is of higher 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

2x2 − 9x − 35 (x + 1)(x − 2)(x + 3) ≡

4 3 1 − + (x + 1) (x − 2) (x + 3)

Problem 3.

Resolve

fractions.

x2 + 1 into partial x2 − 3x + 2 Thus

x − 10 x3 − 2x2 − 4x − 4 ≡ x−3+ 2 x2 + x − 2 x +x−2

The denominator is of the same degree as the numerator. Thus dividing out gives: )1 x − 3x + 2 x2 +1 x2 − 3x + 2 ————— 3x − 1 ——— For more on polynomial division, see Section 1.4, page 9. 2

Hence

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

Let

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

i.e. Let x = 2. Then

5=B

≡ x−3+ Let

x − 10 (x + 2)(x − 1)

x − 10 A B ≡ + (x + 2)(x − 1) (x + 2) (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 A= 4

i.e. Let x = 1. Then

−9 = 3B B = −3

i.e. Hence

x − 10 4 3 ≡ − (x + 2)(x − 1) (x + 2) (x − 1)

Thus

x3 − 2x2 − 4x − 4 x2 + x − 2 ≡x−3+

4 3 − (x + 2) (x − 1)

20 Section A Now try the following Practice Exercise

Equating the numerators gives: 2x + 3 ≡ A(x − 2) + B

Practice Exercise 9 Partial fractions with linear factors (Answers on page 864)

Let x = 2. Then

7 = A(0) + B

Resolve the following into partial fractions.

i.e.

B =7

1.

12 −9

2x + 3 ≡ A(x − 2) + B ≡ Ax − 2A + B

x2

2.

4(x − 4) x2 − 2x − 3

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

3.

x2 − 3x + 6 x(x − 2)(x − 1)

[Also, as a check, equating the constant terms gives:

4.

3(2x2 − 8x − 1) (x + 4)(x + 1)(2x − 1)

3 = −2A + B When A = 2 and B = 7,

2

5.

x + 9x + 8 x2 + x − 6

6.

x2 − x − 14 x2 − 2x − 3

7.

RHS = −2(2) + 7 = 3 = LHS]

Hence

2x + 3 2 7 ≡ + (x − 2)2 (x − 2) (x − 2)2

3x3 − 2x2 − 16x + 20 (x − 2)(x + 2) 5x2 − 2x − 19 as the sum (x + 3)(x − 1)2 of three partial fractions. Problem 6.

2.3

Partial fractions with repeated linear factors

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

Let Problem 5. fractions.

Resolve

2x + 3 into partial (x − 2)2

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

Let

2x + 3 A B ≡ + (x − 2)2 (x − 2) (x − 2)2 A(x − 2) + B ≡ (x − 2)2

Express

5x2 − 2x − 19 (x + 3)(x − 1)2

≡

A B C + + (x + 3) (x − 1) (x − 1)2

≡

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

by algebraic addition. Equating the numerators gives: 5x2 − 2x − 19 ≡ A(x − 1)2 + B(x + 3)(x − 1) + C(x + 3)

(1)

Partial fractions 21 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

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

3x2 + 16x + 15 (x + 3)3 ≡

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

≡

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

Equating the numerators gives:

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

3x2 + 16x + 15 ≡ A(x + 3)2 + B(x + 3) + C Let x = −3. Then

[Check: Identity (1) may be expressed as: 5x2 − 2x − 19 ≡ A(x2 − 2x + 1)

3(−3)2 + 16(−3) + 15 ≡ A(0)2 + B(0) + C −6 = C

i.e.

+ B(x + 2x − 3) + C(x + 3) 2

i.e. 5x2 − 2x − 19 ≡ Ax2 − 2Ax + A + Bx2 + 2Bx

(1)

Identity (1) may be expanded as: 3x2 + 16x + 15 ≡ A(x2 + 6x + 9) + B(x + 3) + C

− 3B + Cx + 3C Equating the x term coefficients gives:

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

−2 ≡ −2A + 2B + C When A = 2, B = 3 and C = −4 then

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

−2A + 2B + C = −2(2) + 2(3) − 4 16 = 6A + B

= −2 = LHS

Since A = 3, B = −2 Equating the constant term gives: −19 ≡ A − 3B + 3C

[Check: equating the constant terms gives: 15 = 9A + 3B + C

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

Hence

When A = 3, B = −2 and C = −6,

5x2 − 2x − 19 (x + 3)(x − 1)2

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

2 3 4 ≡ + − (x + 3) (x − 1) (x − 1)2 Thus Problem 7. fractions.

Resolve

3x2 + 16x + 15 into partial (x + 3)3

3x2 + 16x + 15 (x + 3)3 ≡

3 2 6 − − (x + 3) (x + 3)2 (x + 3)3

22 Section A Equating the coefficients of x2 terms gives:

Now try the following Practice Exercise Practice Exercise 10 Partial fractions with repeated linear factors (Answers on page 864) 1.

4x − 3 (x + 1)2

2.

x2 + 7x + 3 x2 (x + 3)

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,

5x2 − 30x + 44 (x − 2)3

3.

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

18 + 21x − x2 (x − 5)(x + 2)2

4.

7x2 + 5x + 13 2x + 3 5 ≡ 2 + (x2 + 2)(x + 1) ( x + 2) (x + 1)

Problem 9.

2.4 Partial fractions with quadratic factors Problem 8.

Express

fractions.

partial fractions.

Let

3 + 6x + 4x2 − 2x3 A B Cx + D ≡ + 2+ 2 x2 (x2 + 3) x x (x + 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)

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

+ (Cx + D)x2 ≡ Ax3 + 3Ax + Bx2 + 3B

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

Equating numerators gives: 2

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: 2

Let x = 0. Then 3 = 3B B=1

Equating the coefficients of x3 terms gives: −2 = A + C Equating the coefficients of x2 terms gives: 4 = B+D Since B = 1, D = 3

7x + 5x + 13 ≡ Ax + Ax + Bx + B + Cx + 2C 2

+ Cx3 + Dx2

i.e.

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

3 + 6x + 4x2 − 2x3 into x2 (x2 + 3)

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

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

The denominator is a combination of a quadratic factor, (x2 + 2), which does not factorise without introducing imaginary surd terms, and a linear factor, (x + 1). Let,

Resolve

2

Equating the coefficients of x terms gives: 6 = 3A

(1)

Partial fractions 23 A=2

i.e.

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

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

2 1 3 − 4x + + 2 x x2 x +3

Now try the following Practice Exercise Practice Exercise 11 Partial fractions with quadratic factors (Answers on page 864) 1.

x2 − x − 13 (x2 + 7)(x − 2)

2.

6x − 5 (x − 4)(x2 + 3)

3.

15 + 5x + 5x2 − 4x3 x2 (x2 + 5)

4.

x3 + 4x2 + 20x − 7 (x − 1)2 (x2 + 8)

5.

When solving the differential equation d2 θ dθ − 6 − 10θ = 20 − e2t by Laplace transdt2 dt forms, for given boundary conditions, the following expression for L{θ} results: 39 2 s + 42s − 40 2 L{θ} = s(s − 2)(s2 − 6s + 10) 4s3 −

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

2 1 5s − 3 − + s 2(s − 2) 2(s2 − 6s + 10)

For fully worked solutions to each of the problems in Practice Exercises 9 to 11 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 3

Logarithms Why it is important to understand: Logarithms All types of engineers use natural and common logarithms. Chemical engineers use them to measure radioactive decay and pH solutions, both of which are measured on a logarithmic scale. The Richter scale, which measures earthquake intensity, is a logarithmic scale. Biomedical engineers use logarithms to measure cell decay and growth, and also to measure light intensity for bone mineral density measurements. In electrical engineering, a dB (decibel) scale is very useful for expressing attenuations in radio propagation and circuit gains, and logarithms are used for implementing arithmetic operations in digital circuits. Logarithms are especially useful when dealing with the graphical analysis of non-linear relationships and logarithmic scales are used to linearise data to make data analysis simpler. Understanding and using logarithms is clearly important in all branches of engineering.

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

3.1

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

Introduction to logarithms

With the use of calculators firmly established, logarithmic tables are no longer used for calculations. However, the theory of logarithms is important, for there are several scientific and engineering laws that involve the rules of logarithms. From the laws of indices:

16 = 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.

Logarithms 25 The two statements: are equivalent.

16 = 24

and log2 16 = 4

if y = ax

then

x = loga y

In another example, if we write that 64 = 82 then the equivalent statement using logarithms is: log8 64 = 2

from which, x = 2

log3 9 = 2

Hence,

If we write either of them, we are automatically implying 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.

3x = 32

i.e.

Problem 2.

Evaluate log10 10 10 x = 10

Let x = log10 10 then

from the

definition of a logarithm, 10 x = 101

i.e. Hence,

from which, x = 1

log10 10 = 1

(which may be checked

In another example, if we write that log3 81 = 4 then the equivalent statement using powers is: 34 = 81 So the two sets of statements, one involving powers and one involving logarithms, are equivalent. Common logarithms From above, if we write that 1000 = 103 , then 3 = log10 1000 This may be checked using the ‘log’ button on your calculator. Logarithms having a base of 10 are called common logarithms and log10 is often abbreviated to lg. The following values may be checked by using a calculator: lg 27.5 = 1.4393 . . . , lg 378.1 = 2.5776 . . . and

using 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, Hence,

4x = 3 and log16 8 =

x=

3 4

lg 0.0204 = −1.6903 . . .

Napierian logarithms Logarithms having a base of e (where ‘e’ is a mathematical constant approximately equal to 2.7183) are called hyperbolic, Napierian or natural logarithms, and loge is usually abbreviated to ln. The following values may be checked by using a calculator:

Problem 4.

Evaluate lg 0.001 then 10x = 0.001

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

10x = 10−3

from which, x = −3

lg 0.001 = −3 (which may be checked using a calculator)

ln 3.65 = 1.2947 . . . , ln 417.3 = 6.0338 . . . and

ln 0.182 = −1.7037 . . .

More on Napierian logarithms is explained in Chapter 4. Here are some worked problems to help understanding of logarithms. Problem 1. Let x = log3 9

Problem 5.

3x = 9

Evaluate ln e

Let x = ln e = loge e then

ex = e

i.e.

ex = e1

Evaluate log3 9 then

3 4

from which, x = 1 Hence, from the definition of a logarithm,

ln e = 1 (which may be checked using a calculator)

26 Section A

Problem 6.

Let x = log3

Evaluate log3

1 81

then 3x =

Hence,

log3

Problem 7.

In Problems 12 to 18 solve the equations: 12.

log10 x = 4

13. lg x = 5

14.

log3 x = 2

15. log4 x = −2

from which, x = −4

16.

lg x = −2

17.

= −4

18.

ln x = 3

1 1 = = 3−4 81 34

1 81

Solve the equation: lg x = 3

If lg x = 3 then and

1 81

i.e.

4 3

Laws of logarithms

There are three laws of logarithms, which apply to any base:

log10 x = 3

x = 103

3.2

log8 x = −

1 2

x = 1000

(i)

To multiply two numbers: log (A × B) = log A + log B

Problem 8.

Solve the equation: log2 x = 5

The following may be checked by using a calculator:

If log2 x = 5 then x = 25 = 32 Problem 9.

lg 10 = 1

Solve the equation: log5 x = −2

If log5 x = −2 then

x = 5−2 =

1 1 = 2 5 25

Also, lg 5 + lg 2 = 0.69897. . . + 0.301029. . . = 1 Hence, lg (5 × 2) = lg 10 = lg 5 + lg 2 (ii)

Now try the following Practice Exercise

The following may be checked using a calculator: ( ) 5 ln = ln 2.5 = 0.91629. . . 2

Practice Exercise 12 Introduction to logarithms (Answers on page 864) In Problems 1 to 11, evaluate the given expressions: 1. log10 10000

2.

log2 16

3. log5 125

4.

log2 18

5. log8 2

6.

log7 343

7. lg 100

8.

lg 0.01

9. log4 8

10.

log27 3

11. ln e

2

To divide two numbers: ( ) A log = log A − log B B

Also, Hence, (iii)

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

To raise a number to a power: log An = nlog A The following may be checked using a calculator: lg 52 = lg 25 = 1.39794. . . Also, 2 lg 5 = 2 × 0.69897. . . = 1.39794. . . Hence,

lg 52 = 2 lg 5

Logarithms 27 Here are some worked problems to help understanding of the laws of logarithms.

1 1 log 16 + log 27 − 2 log 5 2 3 1

log 4 + log 7 = log (7 × 4)

= log 4 + log 3 − log 25 ( ) 4×3 = log 25

by the first law of logarithms = log 28 Problem 11. Write log 16 − log 2 as the logarithm of a single number. (

16 log 16 − log 2 = log 2

by the first and second laws of logarithms ( ) 12 = log = log0.48 25

)

by the second law of logarithms = log 8

Problem 16. Solve the equation: log(x − 1) + log(x + 8) = 2 log(x + 2) LHS = log (x − 1) + log(x + 8)

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

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

by the third law of logarithms

= log (x2 + 7x − 8)

= log 9

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

1 Problem 13. Write log 25 as the logarithm of a 2 single number. 1 1 log 25 = log 25 2 by the third law of logarithms 2 √ = log 25 = log 5 Problem 14. Simplify: log 64 − log 128 + log 32. 64 = 26 , 128 = 27 and 32 = 25

1

= log 16 2 + log 27 3 − log 52 by the third law of logarithms √ √ 3 = log 16 + log 27 − log 25 by the laws of indices

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

from the third law of logarithms = log (x2 + 4x + 4) Hence, from which,

log (x2 + 7x − 8) = log (x2 + 4x + 4) x2 + 7x − 8 = x2 + 4x + 4

i.e.

7x − 8 = 4x + 4

i.e.

3x = 12

and

x=4

Hence, log 64 − log 128 + log 32 = log 2 − log 2 + log 2 6

7

5

= 6 log 2 − 7 log 2 + 5 log 2 by the third law of logarithms = 4log 2 1 1 Problem 15. Write log 16 + log 27 − 2 log 5 2 3 as the logarithm of a single number.

Problem 17. Solve the equation:

1 log 4 = log x 2

1 1 log 4 = log 4 2 from the third law of logarithms √ 2 = log 4 from the laws of indices

Hence, becomes i.e.

1 log 4 = log x 2 √ log 4 = log x log 2 = log x

28 Section A from which,

2=x

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

10. 11.

Problem 18. Solve ( the equation: ) log x2 − 3 − log x = log 2 ( 2 ) ( 2 ) x −3 log x − 3 − log x = log x from the second law of logarithms ( 2 ) x −3 log = log 2 Hence, x from which,

x2 − 3 =2 x

Rearranging gives:

x2 − 3 = 2x x2 − 2x − 3 = 0

and Factorising gives:

(x − 3)(x + 1) = 0

from which,

x=3

or

x = −1

1 log 16 + 2 log 3 − log 18 4 2 log 2 + log 5 − log 10

Simplify the expressions given in Problems 12 to 14: 12.

log 27 − log 9 + log 81

13.

log 64 + log 32 − log 128

14.

log 8 − log 4 + log 32

Evaluate the expressions given in Problems 15 and 16: 15. 16.

1 2

log 16 − 13 log 8 log 4

log 9 − log 3 + 12 log 81 2 log 3

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

log x4 − log x3 = log 5x − log 2x

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

18.

log 2t3 − log t = log 16 + log t

19.

2 log b2 − 3 log b = log 8b − log 4b

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

20.

log (x + 1) + log(x − 1) = log 3

Now try the following Practice Exercise

21. 22.

1 log 27 = log(0.5a) 3 ( ) log x2 − 5 − log x = log 4

Practice Exercise 13 Laws of logarithms (Answers on page 864) In Problems 1 to 11, write as the logarithm of a single number:

3.3

Indicial equations

3. log 3 + log 4 − log 6

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

4. log 7 + log 21 − log 49

i.e. log10 3x = log10 27

5. 2 log 2 + log 3

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

1. log 2 + log 3 2. log 3 + log 5

6. 2 log 2 + 3 log 5 1 log 81 + log 36 2 1 1 log 8 − log 81 + log 27 3 2 1 log 4 − 2 log 3 + log 45 2

7. 2 log 5 − 8. 9.

x=

log10 27 1.43136 . . . = =3 log10 3 0.4771 . . .

which may be readily checked ( ( ) ( )) log 8 8 Note, is not equal to log log 2 2

Logarithms 29 Problem 19. 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

log10 2x+1 = log10 32x−5

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

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

= 1.585, correct to 4 significant figures.

The velocity v of) a machine is ( 40 1.25 b

given by the formula: v = 15 (a) Transpose the formula to make b the subject. (b)

0.3010x + 0.3010 = 0.9542x − 2.3855

Hence

log10 3 0.47712125 . . . = log10 2 0.30102999 . . .

Problem 20.

(x + 1) log10 2 = (2x − 5) log10 3

i.e.

i.e.

Rearranging gives: x=

Taking logarithms to base 10 of both sides gives:

Evaluate b, correct to 3 significant figures, when v = 12

(a) Taking logarithms to the base of 10 of both sides of the equation gives: (

log10 v = log10 15

40 1.25 b

)

from which x =

2.6865 = 4.11, correct to 0.6532 2 decimal places.

Problem 22. 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 3.2 log10 x = log10 41.15 log10 41.15 Hence log10 x = = 0.50449 3.2 Thus x = antilog 0.50449 = 100.50449 = 3.195 correct to 4 significant figures.

From the third law of logarithms: (

) 40 log10 v = log10 15 1.25 b ( ) ( ) 40 = 1.1761 1.25 b correct to 4 decimal places 47.044 = 1.25 b from which,1.25 b log10 v = 47.044 and b =

(b) When v = 12, i.e.

47.044 1.25 log10 v

47.044 = 34.874 1.25 log10 12 b = 34.9 correct to 3 significant figures. b=

Problem 21. Solve the equation 2x+1 = 32x−5 correct to 2 decimal places.

Problem 23. A gas follows the polytropic law PV1.25 = C. Determine the new volume of the gas, given that its original pressure and volume are 101 kPa and 0.35 m3 , respectively, and its final pressure is 1.18 MPa. If PV1.25 = C then P1 V1.25 = P2 V1.25 1 2 P1 = 101 kPa, P2 = 1.18 MPa and V1 = 0.35 m3 P1 V1.25 = P2 V1.25 1 2 ( ) ( ) ( ) 1.25 i.e. 101 × 103 0.35 = 1.18 × 106 V1.25 2 ( ) ( ) 1.25 101 × 103 0.35 ( ) from which, V1.25 = = 0.02304 2 1.18 × 106 Taking logarithms of both sides of the equation gives: log10 V1.25 = log10 0.02304 2 i.e.

1.25 log10 V2 = log10 0.02304 from the third law of logarithms

30 Section A and log10 V2 =

log10 0.02304 = −1.3100 1.25

from which, volume, V2 = 10− 1.3100 = 0.049 m3

y

1.0

0.5

Now try the following Practice Exercise 0

Practice Exercise 14 Indicial equations (Answers on page 864)

1 x y 5 log10x

20.5

3

2 3

2

1

x 0.5

0.2

0.1

0.48 0.30 0 2 0.30 2 0.70 2 1.0

Solve the following indicial equations for x, each correct to 4 significant figures: 1. 3x = 6.4 2. 2x = 9 3. 2x−1 = 32x−1 1.5

4. x

= 14.91

5. 25.28 = 4.2x

21.0

Figure 3.1 y 2

1

6. 42x−1 = 5x+2 7. x−0.25 = 0.792 8. 0.027x = 3.26 9. The decibel gain n of an amplifier is given by: ( ) P2 n = 10 log10 P1 where P1 is the power input and P2 is the P2 power output. Find the power gain when P1 n = 25 decibels. 10. A gas follows the polytropic law PV1.26 = C. Determine the new volume of the gas, given that its original pressure and volume are 101 kPa and 0.42 m3 , respectively, and its final pressure is 1.25 MPa.

1

0 21

2

3

4

5

6

x

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

22

Figure 3.2

Let loga = x, then ax = 1 from the definition of the logarithm. If ax = 1 then x = 0 from the laws of indices. Hence loga 1 = 0. In the above graphs it is seen that log10 1 = 0 and loge 1 = 0

(ii) loga a = 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)

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)

loga 1 = 0

(iii) loga 0 → −∞ 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 → −∞

Logarithms 31 Now try the following Practice Exercise Practice Exercise 15 Multiple-choice questions on logarithms (Answers on page 864) Each question has only one correct answer 1. log16 8 is equal to: 3 1 (b) 144 (c) (a) 2 4 (b) log 5

(c) log 6

(b) 2

5 3

(b) 5

(d) log

2 3

1 is: 100 (c) 5 (d) −2

log10 4 + log10 25 is equal to: (a) 5 (b) 4 (c) 3 (d) 2

10.

Simplifying (2 log 9) gives:

11.

12.

1 1 6. log 16 − 2 log 5 + log 27 is equivalent to: 4 3 (a) − log 20 (b) log 0.24 (c) − log 5 (d) log 3 7. Solving log10 (3x + 1) = 5 for x gives: 4 (a) 300 (b) 8 (c) 33333 (d) 3

(b) log 18

(c) log 81 (d) log

2 9

Solving the equation log(x + 2) − log(x − 2) = log 2 for x gives: (a) 2

4. Solving (log 30 − log 5) = log x for x gives: (a) 150 (b) 6 (c) 35 (d) 25 5. Solving logx 81 = 4 for x gives: (a) 3 (b) 9 (c) 20.25 (d) 324

1 for x gives: 125 7 (c) −1 (d) 3

9.

(a) log 11

3. The value of log10 (a) −5

Solving 52−x = (a)

(d) 2

2. Simplifying (log 2 + log 3) gives: (a) log 1

8.

(b) 3

(c) 4

(d) 6

1 Writing (3 log 1 + log 16) as a single loga2 rithm gives: (a) log 24 (b) log 18 (c) log 5 (d) log 4

13.

Solving (3 log 10) = log x for x gives: (a) 1000 (b) 30 (c) 13 (d) 0.3

14.

Solving 2x = 6 for x, correct to 1 decimal place, gives: (a) 2.6 (b) 1.6 (c) 12.0 (d) 5.2

15.

The information I contained in a message is given by ( ) 1 I = log2 x 1 If x = then I is equal to: 8 (a) 4 (b) 3 (c) 8 (d) 16

For fully worked solutions to each of the problems in Practice Exercises 12 to 14 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 4

Exponential functions Why it is important to understand: Exponential functions Exponential functions are used in engineering, physics, biology and economics. There are many quantities that grow exponentially; some examples are population, compound interest and charge in a capacitor. With exponential growth, the rate of growth increases as time increases. We also have exponential decay; some examples are radioactive decay, atmospheric pressure, Newton’s law of cooling and linear expansion. Understanding and using exponential functions is important in many branches of engineering.

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

evaluate exponential functions using a calculator state the exponential series for ex plot graphs of exponential functions evaluate Napierian logarithms using a calculator solve equations involving Napierian logarithms appreciate the many examples of laws of growth and decay in engineering and science perform calculations involving the laws of growth and decay reduce exponential laws to linear form using log-linear graph paper

4.1 Introduction to exponential functions An exponential function is one which contains ex , e being a constant called the exponent and having an approximate value of 2.7183. The exponent arises from the natural laws of growth and decay and is used as a base for natural or Napierian logarithms. The most common method of evaluating an exponential function is by using a scientific notation calculator. Use your calculator to check the following values: e1 = 2.7182818, correct to 8 significant figures, e−1.618 = 0.1982949, each correct to 7 significant figures,

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 = 11 159, 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

Exponential functions 33 ( ) 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: ( 0.25 ) e − e−0.25 5 0.25 e + e−0.25 (

) e0.25 − e−0.25 5 e0.25 + e−0.25 ( ) 1.28402541 . . . − 0.77880078 . . . =5 1.28402541 . . . + 0.77880078 . . . ( ) 0.5052246 . . . =5 2.0628262 . . .

In Problems 3 and 4, evaluate correct to 5 decimal places: 1 5e2.6921 3. (a) e3.4629 (b) 8.52e−1.2651 (c) 1.1171 7 3e (a)

5.

The length of a bar l at a temperature θ is given by l = l0 eαθ , where l0 and α are constants. Evaluate l, correct to 4 significant figures, where l0 = 2.587, θ = 321.7 and α = 1.771 × 10−4

6.

When a chain of length 2L is suspended from two points, 2D metres same hor{ apart, ( √on the)} L+ L2 +k 2 izontal level: D = k ln . Evaluk ate D when k = 75 m and L = 180 m.

= 1.2246, correct to 4 decimal places.

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

v = Ve−t/CR = 300e(−50×10

)/(10×10−6 ×47×103 )

Using a calculator, v = 300e−0.1063829 . . . = 300(0.89908025 . . .) = 269.7 volts Now try the following Practice Exercise

5.6823 e2.1127 − e−2.1127 (b) e−2.1347 2 −1.7295 4(e − 1) (c) e3.6817

4.

The power series for e x

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 +

(1)

(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 +

Practice Exercise 16 Evaluating exponential functions (Answers on page 864)

+ 1. Evaluate the following, correct to 4 significant figures: (a) e−1.8 (b) e−0.78 (c) e10 2. Evaluate the following, correct to 5 significant figures: (a) e1.629 (b) e−2.7483 (c) 0.62e4.178

x2 x3 x4 + + + ··· 2! 3! 4!

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

(1)6 (1)7 (1)8 + + + ··· 6! 7! 8!

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

e = 2.71828 = 2.7183, correct to 4 decimal places.

34 Section A 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 +

Hence

e0.5 = 1 + 0.5 + +

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

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

(0.5)2 (0.5)3 + (2)(1) (3)(2)(1)

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

= 1 + 0.5 + 0.125 + 0.020833 + 0.0026042 + 0.0002604

+ 0.000000260 + 0.000000003

+ 0.0000217 i.e.

and by adding,

e0.5 = 1.64872, correct to 6 significant figures.

0.05

e

= 1.0512711, correct to 8 significant figures.

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

(−x)2 (−x)3 + + ··· 2! 3!

x2 x3 − + ··· 2! 3!

In a similar manner the power series for ex may be used to evaluate any exponential function of the form a ekx , where a and k are constants. In the series of equation (1), let x be replaced by kx. Then, { } (kx)2 (kx)3 a ekx = a 1 + (kx) + + + ··· 2! 3! { } (2x)2 (2x)3 Thus 5 e2x = 5 1 + (2x) + + + ··· 2! 3! { } 4x2 8x3 = 5 1 + 2x + + + ··· 2 6 { } 4 i.e. 5 e2x = 5 1 + 2x + 2x2 + x3 + · · · 3 Problem 4. Determine the value of 5 e0.5 , correct to 5 significant figures, by using the power series for ex

Hence 5e0.5 = 5(1.64872) = 8.2436, correct to 5 significant figures. Problem 5. in x5

Expand ex (x2 − 1) as far as the term

The power series for ex is,

ex = 1 + x +

x2 x3 x4 x5 + + + + ··· 2! 3! 4! 5!

Hence ex (x2 − 1) ( ) x2 x3 x4 x5 = 1 + x + + + + + · · · (x2 − 1) 2! 3! 4! )5! ( x4 x5 = x2 + x3 + + + · · · 2! 3! ( ) x2 x3 x4 x5 − 1 + x + + + + + ··· 2! 3! 4! 5! Grouping like terms gives: ex (x2 − 1)

(

( ) x3 3 + x − 3! ( 4 ) ( 5 ) x x4 x x5 + − + − + ··· 2! 4! 3! 5!

x2 = −1 − x + x − 2! 2

)

1 5 11 19 5 = − 1 − x + x2 + x3 + x4 + x 2 6 24 120 when expanded as far as the term in x5

ex = 1 + x +

x2 x3 x4 + + + ··· 2! 3! 4!

Exponential functions 35 y

Now try the following Practice Exercise

20

Practice Exercise 17 Power series for ex (Answers on page 864)

y 5 e2x

y 5 ex 16

1. Evaluate 5.6 e−1 , correct to 4 decimal places, using the power series for ex

12

2. Use the power series for ex to determine, correct to 4 significant figures, (a) e2 (b) e−0.3 , and check your result by using a calculator.

8

4

3. Expand (1 − 2x) e2x as far as the term in x4 ( 2 )( 1 ) x 2 to six terms. 4. Expand 2 ex

23

22

21

0

1

2

3

x

Figure 4.1

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. x ex

A graph of y = 2 e0.3x is shown plotted in Fig. 4.2. y 5

−3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0 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

3.87

y5 2e0.3x

4 3 2 1.6

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

0.61

0.37

0.22

0.14

0.08

0.05

−x

e

1

23

21 0 20.74

22

1

2 2.2

3

x

Figure 4.2

From the graph, when x = 2.2, y = 3.87 and when y = 1.6, x = −0.74

Fig. 4.1 shows graphs of y = ex and y = e−x Problem 6. Plot a graph of y = 2 e0.3x over a range of x = − 2 to x = 3. Hence determine the value of y when x = 2.2 and the value of x when y = 1.6

Problem 7. Plot a graph of y = 13 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. A table of values is drawn up as shown below. x

−3

−2

−1

0

1

2

3

x

0

0.3

0.6

0.9

−2x

0.3x

−0.9 −0.6 −0.3

e0.3x

0.407 0.549 0.741 1.000 1.350 1.822 2.460

2 e0.3x 0.81 1.10 1.48 2.00 2.70 3.64 4.92

e−2x

−1.5 −1.0 −0.5 3

2

1

0

0.5

1.0

1.5

0

−1

−2

−3

20.086 7.389 2.718 1.00 0.368 0.135 0.050

1 −2x e 6.70 3

2.46 0.91 0.33 0.12 0.05 0.02

36 Section A A graph of 13 e−2x is shown in Fig. 4.3.

250 t

y 5 250e2 3

y 1 y 5 3 e22x

200 Voltage v (volts)

7 6 5 4

3.67

3

150

100 80

2 1.4

50

1 1.0

0.5

21.5 21.0 20.5

1.5

x

3 3.4 4 1 1.5 2 Time t(seconds)

0

21.2 20.72

Figure 4.3

5

6

Figure 4.4

From the graph, when x = −1.2, y = 3.67 and when y = 1.4, x = −0.72

Now try the following Practice Exercise

Problem 8. The decay of voltage, v volts, across a capacitor at time t seconds is given by

Practice Exercise 18 Exponential graphs (Answers on page 865)

v = 250 e 3 . Draw a graph showing the natural decay curve over the first six seconds. From the graph, find (a) the voltage after 3.4 s, and (b) the time when the voltage is 150 V.

1.

Plot a graph of y = 3 e0.2x 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

2.

Plot a graph of y = 12 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

3.

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

4.

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

−t

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

0

e

−t 3

−t v = 250 e 3

t e

−t 3

v = 250 e

−t 3

1

2

3

1.00

0.7165 0.5134 0.3679

250.0

179.1

128.4

91.97

4

5

6

0.2636

0.1889

0.1353

65.90

47.22

33.83

The natural decay curve of v = 250 e Fig. 4.4. From the graph:

−t 3

is shown in

(a) when time t = 3.4 s, voltage v = 80 V and (b)

when voltage v = 150 V, time t = 1.5 s

4.4

Napierian logarithms

Logarithms having a base of ‘e’ are called hyperbolic, Napierian or natural logarithms and the Napierian

Exponential functions 37 logarithm of x is written as loge x, or more commonly as ln x. Logarithms were invented by John Napier∗, a Scotsman (1550–1617). The most common method of evaluating a Napierian logarithm is by a scientific notation calculator. Use your calculator to check the following values: ln 4.328 = 1.46510554 . . . = 1.4651, correct to 4 decimal places ln 1.812 = 0.59443, correct to 5 significant figures ln 1 = 0

This is useful when solving equations involving exponential functions. For example, to solve e3x = 7, take Napierian logarithms of both sides, which gives: ln e3x = ln 7 i.e. from which

Problem 9. Evaluate the following, each correct to 5 significant figures:

ln 527 = 6.2672, correct to 5 significant figures ln 0.17 = −1.772, correct to 4 significant figures ln 0.00042 = −7.77526, correct to 6 significant figures

(a) (a)

ln e3 = 3 ln e1 = 1

(b)

From the last two examples we can conclude that: loge ex = x

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

(c)

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

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

Problem 10. Evaluate the following: (a)

(a)

(b)

ln e2.5 5e2.23 lg 2.23 (b) correct to 3 lg 100.5 ln 2.23 decimal places. ln e2.5 2.5 = =5 0.5 lg 10 0.5 5e2.23 lg 2.23 ln 2.23 5(9.29986607 . . .)(0.34830486 . . .) = 0.80200158 . . . = 20.194, correct to 3 decimal places.

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

Who was Napier? John Napier of Merchiston (1550– 4 April 1617) is best known as the discoverer of logarithms. The inventor of the so-called ‘Napier’s bones’, Napier also made common the use of the decimal point in arithmetic and mathematics. To find out more go to www.routledge.com/cw/bird

9 = e−3x 4

Taking the reciprocal of both sides gives: 4 1 = −3x = e3x 9 e

38 Section A Taking Napierian logarithms of both sides gives: ( ) 4 ln = ln(e3x ) 9 ( ) 4 α Since loge e = α, then ln = 3x 9 ( ) 1 4 1 Hence, x = ln = (−0.81093) = −0.2703, 3 9 3 correct to 4 significant figures. (

t) −2

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

t −2

) gives:

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

(

70 from which, t = 2 ln 38 cant figures.

Taking natural logarithms of both sides gives:

Since ln e = 1

ln

7 = ln e3x 4

ln

7 = 3x ln e 4

ln

7 = 3x 4

0.55962 = 3x

i.e.

x = 0.1865, correct to 4 significant figures.

i.e.

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 = ln 2 + ln e3x−4 x − 1 = ln 2 + 3x − 4

i.e.

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

i.e.

x=

from which,

)

= 1.153

= 1.22, correct to 3 signifi-

( ) 4.87 Problem 13. Solve the equation: 2.68 = ln x to find x.

3 − ln 2 2

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

From the(definition ) of a logarithm, since 4.87 4.87 2.68 = ln then e2.68 = x x Rearranging gives: i.e.

4.87 = 4.87e−2.68 e2.68 x = 0.3339, correct to 4 significant figures.

and by the laws of logarithms, {

(x − 2)2 (x + 3) ln (x − 2)

x=

7 Problem 14. Solve = e3x correct to 4 signi4 ficant figures.

Cancelling gives:

} = 1.6

{ } 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

Exponential functions 39 Using the quadratic formula, √ −1 ± 12 − 4(1)(−10.953) x= 2 √ −1 ± 44.812 −1 ± 6.6942 = = 2 2 x = 2.847 or

i.e.

−3.8471

2. (a) (c)

1.786 ln e1.76 lg 101.41

In Problems 3 to 16 solve the given equations, each correct to 4 significant figures. 3. ln x = 2.10

Problem 17. A steel bar is cooled with running water. Its temperature, θ, in degrees Celsius, is given by: θ = 17 + 1250e −0.17t where t is the time in minutes. Determine the time taken, correct to the nearest minute, for the temperature to fall to 35◦ C.

6. 1.5 = 4e2t

and when θ = 35◦ C, then

35 = 17 + 1250e −0.17t

from which,

35 − 17 = 1250e −0.17t 18 = 1250e −0.17t

i.e.

18 = e −0.17t 1250

and

Taking natural logarithms of both sides gives: ( ) ( ) 18 ln = ln e − 0.17t 1250 ( ) 18 Hence, ln = − 0.17t 1250 ( ) 1 18 from which, t= ln = 24.944 minutes − 0.17 1250 Hence, the time taken, correct to the nearest minute, for the temperature to fall to 35◦ C is 25 minutes

4. 24 + e2x = 45 5. 5 = ex+1 − 7 7. 7.83 = 2.91e−1.7x ( t) 8. 16 = 24 1 − e− 2 ( x ) 9. 5.17 = ln 4.64 ( ) 1.59 10. 3.72 ln = 2.43 x ( ) −x 11. 5 = 8 1 − e 2 12. ln(x + 3) − ln x = ln(x − 1) 13. ln(x − 1)2 − ln 3 = ln(x − 1) 14. ln(x + 3) + 2 = 12 − ln(x − 2) 15. e(x+1) = 3e(2x−5) 16. ln(x + 1)2 = 1.5 − ln(x − 2) + ln(x + 1) 17. Transpose: b = ln t − a ln D to make t the subject. ( ) P R1 18. If = 10 log10 find the value of R1 Q R2 when P = 160, Q = 8 and R2 = 5 (

Now try the following Practice Exercise Practice Exercise 19 Napierian logarithms (Answers on page 865) In Problems 1 and 2, evaluate correct to 5 significant figures: 1.

1 ln 82.473 ln 5.2932 (b) 3 4.829 5.62 ln 321.62 (c) e1.2942

5e−0.1629 2 ln 0.00165

ln 4.8629 − ln 2.4711 5.173

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

θ = 17 + 1250e −0.17 t

(b)

19. If U2 = U1 e formula.

W PV

)

make W the subject of the

20. The work done in an isothermal expansion of a gas from pressure p 1 to p 2 is given by: ( w = w0 ln

p1 p2

)

(a)

If the initial pressure p 1 = 7.0 kPa, calculate the final pressure p 2 if w = 3 w0

40 Section A y = y0 ekt

(v) biological growth 21. The velocity (v2 of ) a rocket is given by: m1 v2 = v1 + C ln m2 where v1 is the initial rocket velocity, C is the velocity of the jet exhaust gases, m1 is the mass of the rocket before the jet engine is fired, and m2 is the mass of the rocket after the jet engine is switched off. Calculate the velocity of the rocket given v1 = 600 m/s, C = 3500 m/s, m1 = 8.50 × 104 kg and m2 = 7.60 × 104 kg.

4.5

Laws of growth and decay

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

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

p = p 0 e−h/c

(viii) radioactive decay (ix) decay of current in an inductive circuit

N = N0 e−λt

(x) growth of current in a capacitive circuit

Transposing R = R0 eαθ gives

R = ln eαθ = αθ R0 ( ) 1 R 1 6 × 103 Hence α = ln = ln θ R0 1500 5 × 103 ln

T1 = T0 eµθ

(iv) newton’s law of cooling

R = eαθ R0

Taking Napierian logarithms of both sides gives:

(ii) change in electrical resistance with temperature Rθ = R0 eαθ (iii) tension in belts

i = I(1 − e−t/CR )

Problem 18. 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.

l = l0 e

(i) linear expansion

i = I e−Rt/L

θ = θ0 e−kt

y

=

A y 5 Ae2kx

1 (0.1823215 . . .) 1500

= 1.215477 · · · × 10−4 Hence α = 1.215 × 10−4 , correct to 4 significant figures.

0

x

(a)

From above, ln

R = αθ R0 θ=

y

hence

A

When R = 5.4 × 103 , R0 = 5 × 103

= (b)

Figure 4.5

α = 1.215477 . . . × 10−4 and

( ) 1 5.4 × 103 θ= ln 1.215477 . . . × 10−4 5 × 103

y 5 A(12e2kx )

0

R 1 ln α R0

x

104 (7.696104 . . . × 10−2 ) 1.215477 . . .

= 633◦ C, correct to the nearest degree.

Exponential functions 41

Problem 19. In an experiment involving Newton’s law of cooling, the temperature θ(◦ C) is given by θ = θ0 e−kt . Find the value of constant k when θ0 = 56.6◦ C, θ = 16.5◦ C and t = 83.0 seconds. θ = θ0 e−kt gives

Transposing

θ = e−kt θ0 θ0 1 from which = −kt = ekt θ e Taking Napierian logarithms of both sides gives:

t

e CR =

Taking Napierian logarithms of both sides gives: ( ) 8.0 t = ln CR 8.0 − i Hence.

( ) 1 θ0 1 56.6 k = ln = ln t θ 83.0 16.5 1 = (1.2326486 . . .) 83.0

(

t = CR ln

8.0 8.0 − i −6

= (16 × 10

θ0 ln = kt θ from which,

8.0 8.0 − i

) (

8.0 )(25 × 10 ) ln 8.0 − 6.0

)

3

when i = 6.0 amperes, ( ) 400 8.0 i.e. t = 3 ln = 0.4 ln 4.0 10 2.0 = 0.4(1.3862943 . . .) = 0.5545 s = 555 ms, to the nearest millisecond.

Hence k = 1.485 × 10−2 Problem 20. The current i amperes flowing in a capacitor at time t seconds is given by

A graph of current against time is shown in Fig. 4.6.

−t

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

i (A) 8

6 5.71 i 5 8.0 (12e2t/CR)

4

−t

(a) Current i = 8.0(1 − e CR )

2

−0.5

= 8.0[1 − e (16 × 10−6 )(25 × 103 ) ] = 8.0(1 − e−1.25 )

0

= 8.0(1 − 0.2865047 . . .) = 8.0(0.7134952 . . .)

0.5 0.555

1.0

1.5

t(s)

Figure 4.6

= 5.71 amperes −t

(b) Transposing i = 8.0(1 − e CR ) gives

i 8.0

Problem 21. The temperature θ2 of a winding which is being heated electrically at time t is given

−t = 1 − e CR

−t

from which, e CR = 1 −

−t

i 8.0 − i = 8.0 8.0

Taking the reciprocal of both sides gives:

by: θ2 = θ1 (1 − e τ ) where θ1 is the temperature (in degrees Celsius) at time t = 0 and τ is a constant. Calculate, (a) θ1 , correct to the nearest degree, when θ2 is 50◦ C, t is 30 s and τ is 60 s

42 Section A (b)

the time t, correct to 1 decimal place, for θ2 to be half the value of θ1

(a) Transposing the formula to make θ1 the subject gives: θ1 =

=

i.e. (b)

θ2 (1 − e

−t τ )

=

3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given −Rt

by v = 200 e L , where R = 150 Ω and L = 12.5 × 10−3 H. Determine (a) the voltage when t = 160 × 10−6 s, and (b) the time for the voltage to reach 85 V.

50 1−e

p 0 = 1.012 × 105 Pa, height h = 1420 m, and C = 71 500

−30 60

50 50 = −0.5 1−e 0.393469 . . .

θ 1 = 127◦ C, correct to the nearest degree.

Transposing to make t the subject of the formula gives: −t θ2 =1−e τ θ1 −t θ2 from which, e τ = 1 − θ ( 1 ) t θ2 Hence − = ln 1 − τ θ ( 1 ) θ2 i.e. t = −τ ln 1 − θ1 1 Since θ2 = θ1 2 ( ) 1 t = −60 ln 1 − 2 = −60 ln 0.5 = 41.59 s Hence the time for the temperature θ 2 to be one half of the value of θ 1 is 41.6 s, correct to 1 decimal place.

Now try the following Practice Exercise Practice Exercise 20 The laws of growth and decay (Answers on page 865) 1. The temperature, T ◦ C, of a cooling object varies with time, t minutes, according to the equation: T = 150e−0.04t . Determine the temperature when (a) t = 0, (b) t = 10 minutes. 2. The pressure p pascals at height h metres −h

above ground level is given by p = p 0 e C , where p 0 is the pressure at ground level and C is a constant. Find pressure p when

4. The length l metres of a metal bar at temperature t ◦ C is given by l = l0 eαt , where l0 and α are constants. Determine (a) the value of α when l = 1.993 m, l0 = 1.894 m and t = 250◦ C, and (b) the value of l0 when l = 2.416, t = 310◦ C and α = 1.682 × 10−4 5. The temperature θ2◦ C of an electrical conductor at time t seconds is given by: θ2 = θ1 (1 − e−t/T ), where θ1 is the initial temperature and T seconds is a constant. Determine: (a) θ2 when θ1 = 159.9◦ C, t = 30 s and T = 80 s, and (b) the time t for θ2 to fall to half the value of θ1 if T remains at 80 s. 6. A belt is in contact with a pulley for a sector of θ = 1.12 radians and the coefficient of friction between these two surfaces is µ = 0.26. Determine the tension on the taut side of the belt, T newtons, when tension on the slack side T0 = 22.7 newtons, given that these quantities are related by the law T = T0 eµθ . Determine also the value of θ when T = 28.0 newtons. 7. The instantaneous current i at time t is −t given by: i = 10 e CR when a capacitor is being charged. The capacitance C is 7 × 10−6 farads and the resistance R is 0.3 × 106 ohms. Determine: (a) the instantaneous current when t is 2.5 seconds, and (b)

the time for the instantaneous current to fall to 5 amperes.

Sketch a curve of current against time from t = 0 to t = 6 seconds.

Exponential functions 43

8. The amount of product x (in mol/cm3 ) found in a chemical reaction starting with 2.5 mol/cm3 of reactant is given by x = 2.5(1 − e−4t ) where t is the time, in minutes, to form product x. Plot a graph at 30-second intervals up to 2.5 minutes and determine x after 1 minute. 9. The current i flowing in a capacitor at time t is given by: −t i = 12.5(1 − e CR ) where resistance R is 30 kilohms and the capacitance C is 20 micro-farads. Determine: (a) the current flowing after 0.5 seconds, and (b)

the time for the current to reach 10 amperes.

10. The percentage concentration C of the starting material in a chemical reaction varies with time t according to the equation C = 100 e− 0.004t . Determine the concentration when (a) t = 0, (b) t = 100 s, (c) t = 1000 s 11. The current i flowing through a diode ( at room) temperature is given by: i = iS e40V − 1 amperes. Calculate the current flowing in a silicon diode when the reverse saturation current iS = 50 nA and the forward voltage V = 0.27 V 12. A formula for chemical decomposition is ( t ) given by: C = A 1 − e− 10 where t is the time in seconds. Calculate the time, in milliseconds, for a compound to decompose to a value of C = 0.12 given A = 8.5 13. The mass, m, of pollutant in a water reservoir decreases according to the law m = m0 e− 0.1 t where t is the time in days and m0 is the initial mass. Calculate the percentage decrease in the mass after 60 days, correct to 3 decimal places. 14. A metal bar is cooled with water. Its temperature, in ◦ C, is given by: θ = 15 + 1300e− 0.2 t where t is the time in minutes. Calculate how long it will take for the temperature, θ, to decrease to 36◦ C, correct to the nearest second.

4.6 Reduction of exponential laws to linear form Frequently, the relationship between two variables, say x and y, is not a linear one, i.e. when x is plotted against y a curve results. In such cases the non-linear equation may be modified to the linear form, y = mx + c, so that the constants, and thus the law relating the variables can be determined. This technique is called ‘determination of law’. Graph paper is available where the scale markings along the horizontal and vertical axes are proportional to the logarithms of the numbers. Such graph paper is called log-log graph paper. A logarithmic scale is shown in Fig. 4.7 where the distance between, say, 1 and 2, is proportional to lg 2 − lg 1, i.e. 0.3010 of the total distance from 1 to 10. Similarly, the distance between 7 and 8 is proportional to lg 8 − lg 7, i.e. 0.05799 of the total distance from 1 to 10. Thus the distance between markings progressively decreases as the numbers increase from 1 to 10. 1

2

3

4

5

6 7 8 910

Figure 4.7

With log-log graph paper the scale markings are from 1 to 9, and this pattern can be repeated several times. The number of times the pattern of markings is repeated on an axis signifies the number of cycles. When the vertical axis has, say, three sets of values from 1 to 9, and the horizontal axis has, say, two sets of values from 1 to 9, then this log-log graph paper is called ‘log 3 cycle × 2 cycle’. Many different arrangements are available ranging from ‘log 1 cycle × 1 cycle’ through to ‘log 5 cycle × 5 cycle’. To depict a set of values, say, from 0.4 to 161, on an axis of log-log graph paper, four cycles are required, from 0.1 to 1, 1 to 10, 10 to 100 and 100 to 1000. Graphs of the form y = a ekx Taking logarithms to a base of e of both sides of y = a ekx gives: ln y = ln(a ekx ) = ln a + ln ekx = ln a + kx ln e i.e. ln y = kx + ln a

(since ln e = 1)

which compares with Y = mX + c Thus, by plotting ln y vertically against x horizontally, a straight line results, i.e. the equation y = a ekx is reduced to linear form. In this case, graph paper having

44 Section A a linear horizontal scale and a logarithmic vertical scale may be used. This type of graph paper is called loglinear graph paper, and is specified by the number of cycles on the logarithmic scale.

Gradient of straight line, k=

AB ln 100 − ln 10 2.3026 = = BC 3.12 − (−1.08) 4.20

= 0.55, correct to 2 significant figures. Since ln y = kx + ln a, when x = 0, ln y = ln a, i.e. y = a The vertical axis intercept value at x = 0 is 18, hence a = 18 The law of the graph is thus y = 18 e0.55x

Problem 22. The data given below are believed to be related by a scientific law of the form y = a ekx , where a and b are constants. Verify that the law is true and determine approximate values of a and b. Also determine the value of y when x is 3.8 and the value of x when y is 85 x −1.2 0.38 y

9.3

1.2

2.5

3.4

4.2

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

5.3

= 18(8.0849) = 146 85 = 18 e0.55x

When y is 85,

22.2 34.8 71.2 117 181 332

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

85 = 4.7222 18

Hence,

e0.55x =

and

0.55x = ln 4.7222 = 1.5523 x=

Hence

1.5523 = 2.82 0.55

Problem 23. The voltage, v volts, across an inductor is believed to be related to time, t ms, by t

the law v = V e T , where V and T are constants. Experimental results obtained are:

1000 y

v volts 883

347

90

55.5

18.6

5.2

t ms

21.6

37.8

43.6

56.7

72.0

10.4

y 5a e kx

100

Show that the law relating voltage and time is as stated and determine the approximate values of V and T. Find also the value of voltage after 25 ms and the time when the voltage is 30.0 V.

A

t

10

Since v = V e T then ln v = T1 t + ln V which is of the form Y = mX + c. Using ‘log 3 cycle × linear’ graph paper, the points are plotted as shown in Fig. 4.9. Since the points are joined by a straight line the law

B

C

t

v = Ve T is verified. Gradient of straight line, 1 AB = T BC

1 22

21

Figure 4.8

0

1

2

3

4

5

6

=

ln 100 − ln 10 36.5 − 64.2

=

2.3026 −27.7

x

Exponential functions 45 1000

e 12.0 =

and

e 12.0 =

t

v 5VeT

t

30.0 2090 2090 = 69.67 30.0

Taking Napierian logarithms gives:

(36.5, 100)

100

−t

hence

A

Voltage, v volts

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

10

B

Now try the following Practice Exercise

C

Practice Exercise 21 Reducing exponential laws to linear form (Answers on page 865) 1.

Atmospheric pressure p is measured at varying altitudes h and the results are as shown:

1 0

10

20

30

40 50 Time, t ms

60

70

80

90

Altitude, h m

Figure 4.9

Hence T =

−27.7 2.3026

= −12.0, correct to 3 significant figures.

t

(36.5,100) and substituting these values into v = V e T . 36.5

i.e.

V=

2.

100 −36.5

e 12.0 = 2090 volts, correct to 3 significant figures. −t

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

−25

v = 2090 e 12.0 = 260 V −t

When the voltage is 30.0 volts, 30.0 = 2090 e 12.0 ,

500

73.39

1500

68.42

3000

61.60

5000

53.56

8000

43.41

Show that the quantities are related by the law p = a ekh , where a and k are constants. Determine the values of a and k and state the law. Find also the atmospheric pressure at 10 000 m.

Since the straight line does not cross the vertical axis at t = 0 in Fig. 4.9, the value of V is determined by selecting any point, say A, having co-ordinates Thus 100 = V e −12.0

pressure, p kPa

At particular times, t minutes, measurements are made of the temperature, θ◦ C, of a cooling liquid and the following results are obtained: Temperature θ◦ C

Time t minutes

92.2

10

55.9

20

33.9

30

20.6

40

12.5

50

46 Section A Prove that the quantities follow a law of the form θ = θ0 ekt , where θ0 and k are constants, and determine the approximate value of θ0 and k.

(

5.

) 2.9 In the equation 5.0 = 3.0 ln , x has a x value correct to 3 significant figures of: (a) 1.59 (b) 0.392 (c) 0.548 (d) 0.0625

6. Practice Exercise 22 Multiple-choice questions on exponential functions (Answers on page 865)

Solving the equation 7 + e 2k−4 = 8 for k gives: (a) 2 (b) 3 (c) 4 (d) 5

7.

Given that 2e x+1 = e 2x−5 , the value of x is: (a) 1 (b) 6.693 (c) 3 (d) 2.231

Each question has only one correct answer

8.

The current i amperes flowing in a capacitor at time t seconds is given by i = 10(1 − e −t/CR ), where resistance R is 25 × 103 ohms and capacitance C is 16 × 10−6 farads. When current i reaches 7 amperes, the time t is: (a) −0.48 s (b) 0.14 s (c) 0.21 s (d) 0.48 s

9.

Solving 13 + 2e 3x = 31 for x, correct to 4 significant figures, gives: (a) 0.4817 (b) 0.09060 (c) 0.3181 (d) 0.7324

10.

Chemical C, is given by ( decomposition, t ) − 15 where t is the time in secC = A 1−e onds. If A = 9, the time, correct to 3 decimal places, for C to reach 0.10 is: (a) 34.388 s (b) 0.007 s (c) 0.168 s (d) 15.168 s

1. The value of figures is: (a) 0.0588

ln 2 , correct to 3 significant e 2 lg 2

(b) 0.312

(c) 17.0

(d) 3.209

3.67 ln 21.28 2. The value of , correct to 4 signife −0.189 icant figures, is: (a) 9.289 (b) 13.56 (c) 13.5566 (d) −3.844 × 109 3. A voltage, v, is given by v = Ve − CR volts. When t = 30 ms, C = 150 nF, R = 50 MΩ and V = 250 V, the value of v, correct to 3 significant figures, is: (a) 249 V (b) 250 V (c) 4.58 V (d) 0 V t

4. The value of logx x2k is: (a) k (b) k 2 (c) 2k (d) 2kx

For fully worked solutions to each of the problems in Practice Exercises 16 to 21 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 1

Algebra, partial fractions, logarithms, and exponentials

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

Factorise x3 + 4x2 + x − 6 using the factor theorem. Hence solve the equation x3 + 4x2 + x − 6 = 0

2.

Use the remainder theorem to find the remainder when 2x3 + x2 − 7x − 6 is divided by (a) (x − 2)

(b) (x + 1)

Hence factorise the cubic expression 6x2 + 7x − 5 by dividing out 2x − 1

3.

Simplify

4.

Resolve the following into partial fractions (a) (c)

5.

x − 11 −x−2

(b)

x2

(x2

(7) (4)

3−x + 3)(x + 3)

x3 − 6x + 9 x2 + x − 2

8. (24)

Evaluate, correct to 3 decimal places, 5 e−0.982 3 ln 0.0173

6.

(6)

(2)

Solve the following equations, each correct to 4 significant figures: (a) ln x = 2.40 (b) 3 (c) 5 = 8(1 − e− 2 ) x

x−1

7. (a) The pressure p at height h above ground level is given by: p = p 0 e−kh where p 0 is the pressure at ground level and k is a constant. When p 0 is 101 kilopascals and the pressure at a height of 1500 m is 100 kilopascals, determine the value of k. (b) Sketch a graph of p against h (p the vertical axis and h the horizontal axis) for values of height from zero to 12 000 m when p 0 is 101 kilopascals. (c) If pressure p = 95 kPa, ground-level pressure p 0 = 101 kPa, constant k = 5 × 10−6 , determine the height above ground level, h, in kilometres correct to 2 decimal places. (13)

x−2

=5

(10)

Solve the following equations: ( ) (a) log x2 + 8 − log(2x) = log 3

(b) ln x + ln(x 3) = ln 6x ln(x 2) (13) ( ) U2 R find the value of U2 9. If θf − θi = ln J U1 given that θf = 3.5, θi = 2.5, R = 0.315, J = 0.4, U1 = 50 (6) 10.

Solve, correct to 4 significant figures: (a) 13e2x−1 = 7ex ( )2 (b) ln x + 1 = ln(x + 1) ln(x + 2) + 2

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 1, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

(15)

Chapter 5

The binomial series Why it is important to understand: The binomial series There are many applications of the binomial theorem in every part of algebra, and in general with permutations, combinations and probability. It is also used in atomic physics, where it is used to count s, p, d and f orbitals. There are applications of the binomial series in financial mathematics to determine the number of stock price paths that leads to a particular stock price at maturity.

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

define a binomial expression use Pascal’s triangle to expand a binomial ( expression )n ( )n state the general binomial expansion of a + x and 1 + x ( )n use the binomial series to expand expressions of the form a + x for positive, negative and fractional values of n • determine the rth term of a binomial expansion • use the binomial expansion with practical applications

5.1

Pascal’s triangle

A binomial expression is one which contains two terms connected by a plus or minus sign. Thus (p + q), (a + x)2 , (2x + y)3 are examples of binomial expressions. Expanding (a + x)n for integer values of n from 0 to 6 gives the results as shown at the top of page 49.

From these results the following patterns emerge: (i)

‘a’ decreases in power moving from left to right.

(ii)

‘x’ increases in power moving from left to right.

(iii)

The coefficients of each term of the expansions are symmetrical about the middle coefficient when n is even and symmetrical about the two middle coefficients when n is odd.

The binomial series 49 (a + x)0 = 1 1 (a + x) = a + x a+x (a + x)2 = (a + x)(a + x) = a2 + 2ax + x2 (a + x)3 = (a + x)2 (a + x) = a3 + 3a2 x + 3ax2 + x3 4 3 4 (a + x) = (a + x) (a + x) = a + 4a3 x + 6a2 x2 + 4ax3 + x4 (a + x)5 = (a + x)4 (a + x) = a5 + 5a4 x + 10a3 x2 + 10a2 x3 + 5ax4 + x5 6 5 6 (a + x) = (a + x) (a + x) = a + 6a5 x + 15a4 x2 + 20a3 x3 + 15a2 x4 + 6ax5 + x6

(iv) The coefficients are shown separately in Table 5.1 and this arrangement is known as Pascal’s triangle.∗ A coefficient of a term may be obtained by adding the two adjacent coefficients immediately above in the previous row. This is shown by the triangles in Table 5.1, where, for example, 1 + 3 = 4, 10 + 5 = 15, and so on. (v)

Table 5.1 (a 1 x)0

1

(a 1 x)1

2

1

(a 1 x)3

3

1

(a 1 x)4

Pascal’s triangle method is used for expansions of the form (a + x)n for integer values of n less than about 8.

1

1

(a 1 x)2

6

1

6

5

1

(a 1 x)6

1

3

4

1

(a 1 x)5

1

10 15

1

4 10

20

1

5 15

6

1

Problem 1. Use Pascal’s triangle method to determine the expansion of (a + x)7 From Table 5.1, the row of Pascal’s triangle corresponding to (a + x)6 is as shown in (1) below. Adding adjacent coefficients gives the coefficients of (a + x)7 as shown in (2) below. 1 1

6 7

15 21

20 35

15 35

6 21

1 7

(1) 1

(2)

The first and last terms of the expansion of (a + x)7 are a7 and x7 , respectively. The powers of a decrease and the powers of x increase moving from left to right. Hence (a + x)7 = a7 + 7a6 x + 21a5 x2 + 35a4 x3 + 35a3 x4 + 21a2 x5 + 7ax6 + x7

∗

Who was Pascal? Blaise Pascal (19 June 1623–19 August 1662) was a French polymath. A child prodigy, he wrote a significant treatise on the subject of projective geometry at the age of 16, and later corresponded with Pierre de Fermat on probability theory, strongly influencing the development of modern economics and social science. To find out more go to www.routledge.com/cw/bird

Problem 2. Determine, using Pascal’s triangle method, the expansion of (2p − 3q)5 Comparing (2p − 3q)5 with (a + x)5 shows that a = 2p and x = −3q Using Pascal’s triangle method: (a + x)5 = a5 + 5a4 x + 10a3 x2 + 10a2 x3 + · · ·

50 Section A If a = 1 in the binomial expansion of (a + x)n then:

Hence (2p − 3q)5 = (2p)5 + 5(2p)4 (−3q)

(1 + x)n = 1 + nx +

+ 10(2p)3 (−3q)2 + 10(2p)2 (−3q)3 + 5(2p)(−3q)4 + (−3q)5

which is valid for −1 < x < 1 When x is small compared with 1 then: (1 + x)n ≈ 1 + nx

i.e. (2p − 3q)5 = 32p5 − 240p4 q + 720p3 q2 − 1080p2 q3 + 810pq4 − 243q5 Now try the following Practice Exercise Practice Exercise 23 Pascal’s triangle (Answers on page 865) 1. Use Pascal’s triangle to expand (x − y)7 2. Expand (2a + 3b)5 using Pascal’s triangle.

5.2

5.3 Worked problems on the binomial series Problem 3. Use the binomial series to determine the expansion of (2 + x)7 The binomial expansion is given by: n(n − 1) n−2 2 a x 2! n(n − 1)(n − 2) n−3 3 + a x + ··· 3!

(a + x)n = an + nan−1 x +

The binomial series

The binomial series or binomial theorem is a formula for raising a binomial expression to any power without lengthy multiplication. The general binomial expansion of (a + x)n is given by: n(n − 1) n−2 2 a x 2! n(n − 1)(n − 2) n−3 3 + a x 3!

(a + x)n = an + nan−1 x +

+ ··· where 3! denotes 3 × 2 × 1 and is termed ‘factorial 3’. With the binomial theorem, n may be a fraction, a decimal fraction, or a positive or negative integer. When n is a positive integer, the series is finite, i.e., it comes to an end; when n is a negative integer, or a fraction, the series is infinite. In the general expansion of (a + x)n it is noted that the n(n − 1)(n − 2) n−3 3 fourth term is: a x . The number 3 is 3! very evident in this expression. For any term in a binomial expansion, say the r th term, (r − 1) is very evident. It may therefore be reasoned that the rth term of the expansion (a + x)n is: n(n − 1)(n − 2). . . to (r − 1) terms n−(r−1) r−1 a x (r − 1)!

n(n − 1) 2 x 2! n(n − 1)(n − 2) 3 + x +··· 3!

When a = 2 and n = 7: (7)(6) 5 2 (2) x (2)(1) (7)(6)(5) 4 3 (7)(6)(5)(4) 3 4 + (2) x + (2) x (3)(2)(1) (4)(3)(2)(1)

(2 + x)7 = 27 + 7(2)6 x +

+

(7)(6)(5)(4)(3) 2 5 (2) x (5)(4)(3)(2)(1)

+

(7)(6)(5)(4)(3)(2) (2)x6 (6)(5)(4)(3)(2)(1)

+

(7)(6)(5)(4)(3)(2)(1) 7 x (7)(6)(5)(4)(3)(2)(1)

i.e. (2 + x)7 = 128 + 448x + 672x2 + 560x3 + 280x4 + 84x5 + 14x6 + x7 Problem 4. Use the binomial series to determine the expansion of (2a − 3b)5 From above, the binomial expansion is given by: n(n − 1) n−2 2 a x 2! n(n − 1)(n − 2) n−3 3 + a x + ··· 3!

(a + x)n = an + nan−1 x +

The binomial series 51 When a = 2a, x = −3b and n = 5:

Problem 7. Find the middle term of ( )10 1 2p − 2q

(2a − 3b) = (2a) + 5(2a) (−3b) 5

5

+

4

(5)(4) (2a)3 (−3b)2 (2)(1)

+

(5)(4)(3) (2a)2 (−3b)3 (3)(2)(1)

+

(5)(4)(3)(2) (2a)(−3b)4 (4)(3)(2)(1)

(5)(4)(3)(2)(1) + (−3b)5 (5)(4)(3)(2)(1) i.e. (2a − 3b)5 = 32a5 −240a4 b + 720a3 b2 −1080a2 b3 + 810ab4 −243b5 ( )5 1 Problem 5. Expand c − using the c binomial series. ( c−

1 c

)5

( ) 1 = c5 + 5c4 − c ( )2 (5)(4) 3 1 + c − (2)(1) c ( )3 (5)(4)(3) 2 1 + c − (3)(2)(1) c ( )4 (5)(4)(3)(2) 1 + c − (4)(3)(2)(1) c ( )5 (5)(4)(3)(2)(1) 1 + − (5)(4)(3)(2)(1) c

( )5 1 10 5 1 i.e. c − = c5 − 5c3 + 10c − + 3 − 5 c c c c Problem 6. Without fully expanding (3 + x)7, determine the fifth term. n

The rth term of the expansion (a + x) is given by: n(n − 1)(n − 2) . . . to (r − 1) terms n−(r−1) r−1 a x (r − 1)! Substituting n = 7, a = 3 and r − 1 = 5 − 1 = 4 gives: (7)(6)(5)(4) 7−4 4 (3) x (4)(3)(2)(1) i.e. the fifth term of (3 + x)7 = 35(3)3 x4 = 945x4

In the expansion of (a + x)10 there are 10 + 1, i.e. 11 terms. Hence the middle term is the sixth. Using the general expression for the rth term where a = 2p, 1 x = − , n = 10 and r − 1 = 5 gives: 2q ( )5 (10)(9)(8)(7)(6) 1 10–5 (2p) − (5)(4)(3)(2)(1) 2q ( ) 1 = 252(32p 5 ) − 32q5 (

1 Hence the middle term of 2p − 2q

)10 is −252

p5 q5

Now try the following Practice Exercise Practice Exercise 24 The binomial series (Answers on page 865) 1. Use the binomial theorem to expand (a + 2x)4 2. Use the binomial theorem to expand (2 − x)6 3. Expand (2x − 3y)4

( )5 2 4. Determine the expansion of 2x + x 5. Expand (p + 2q)11 as far as the fifth term. ( q )13 6. Determine the sixth term of 3p + 3 7. Determine the middle term of (2a − 5b)8

5.4 Further worked problems on the binomial series Problem 8. (a) Expand

1 in ascending powers of x as (1 + 2x)3 far as the term in x3, using the binomial series.

(b) State the limits of x for which the expansion is valid.

52 Section A (a) Using the binomial expansion of (1 + x)n , where n = −3 and x is replaced by 2x gives: 1 = (1 + 2x)−3 (1 + 2x)3 = 1 + (−3)(2x) + +

(−3)(−4) (2x)2 2!

(−3)(−4)(−5) (2x)3 + · · · 3!

= 1 − 6x + 24x2 − 80x3 + · · · (b)

The expansion is valid provided |2x| < 1, i.e.

|x|
0. Thus the inverse is then y√ = + x. A graph of f (x) = x2 and its inverse f −1 (x) = x for x > 0 is shown in Fig. 16.31 and, again, f −1 (x) is seen to be a reflection of f (x) in the line y = x. It is noted from the latter example, that not all functions have an inverse. An inverse, however, can be determined if the range is restricted. y y 5 2x1 1 y5x

4

0

1

2

3

x

Figure 16.31

Problem 5. Determine the inverse for each of the following functions: (a) f (x) = x − 1 (b) f (x) = x2 − 4 (x > 0) (c) f (x) = x2 + 1 (a) If y = f(x), then y = x − 1 Transposing for x gives x = y + 1 Interchanging x and y gives y = x + 1 Hence if f (x) = x − 1, then f−1 (x) = x + 1 (b)

If y = f(x), then y = x2 − 4 √ (x > 0) Transposing for x gives x = y + √4 Interchanging x and y gives y = x + 4 Hence if f√ (x) = x2 − 4 (x > 0) then −1 f (x) = x + 4 if x > −4

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

Inverse trigonometric functions 2 1

21

0 21

Figure 16.30

1

2

3

y5

x 1 2 2 2

4

x

If y = sin x, then x is the angle whose sine is y. Inverse trigonometrical functions are denoted by prefixing the function with ‘arc’ or, more commonly,−1 . Hence transposing y = sin x for x gives x = sin−1 y. Interchanging x and y gives the inverse y = sin−1 x. Similarly, y = cos−1 x, y = tan−1 x, y = sec−1 x, y =cosec−1x and y=cot−1 x are all inverse trigonometric functions. The angle is always expressed in radians.

Functions and their curves 203 Inverse trigonometric functions are periodic so it is necessary to specify the smallest or principal value of the angle. For sin−1 x, tan−1 x, cosec−1 x and cot−1 x, the π π principal value is in the range − < y < . For cos−1 x 2 2 and sec−1 x the principal value is in the range 0 < y < π. Graphs of the six inverse trigonometric functions are shown in Fig. 31.1, page 387.

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

Practice Exercise 83 Inverse functions (Answers on page 875) Problem 6.

Determine the principal values of

(a) arcsin 0.5 ( √ ) 3 (c) arccos − 2

(b) arctan(−1) √ (d) arccosec( 2)

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

π rad or 0.5236 rad 6

Determine the inverse of the functions given in Problems 1 to 4. 1.

f(x) = x + 1

2.

f(x) = 5x − 1

3.

f(x) = x3 + 1

1 f(x) = + 2 x Determine the principal value of the inverse functions in Problems 5 to 11. 4.

(b) arctan(−1) ≡ tan−1 (−1) = −45◦

5.

sin−1 (−1)

π rad or −0.7854 rad 4 ( √ ) ( √ ) 3 3 −1 ≡ cos − = 150◦ (c) arccos − 2 2

6.

cos−1 0.5

7.

tan−1 1

8.

cot−1 2

9.

cosec−1 2.5

=−

=

5π rad or 2.6180 rad 6

( ) √ 1 (d) arccosec( 2) = arcsin √ 2 ( ) 1 −1 √ ≡ sin = 45◦ 2 =

π rad or 0.7854 rad 4

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

10. 11. 12.

sec−1 1.5 ( ) 1 sin−1 √ 2 Evaluate x, correct to 3 decimal places: x = sin−1

13.

4 8 1 + cos−1 − tan−1 3 5 9

Evaluate y, correct to 4 significant figures: √ √ y = 3 sec−1 2 − 4 cosec−1 2 + 5 cot−1 2

16.7

Asymptotes

x+2 is drawn x+1 up for various values of x and then y plotted against x, the graph would be as shown in Fig. 16.32. The straight lines AB, i.e. x = −1, and CD, i.e. y = 1, are known as asymptotes. If a table of values for the function y =

sin−1 0.30 = 17.4576◦ = 0.3047 rad cos−1 0.65 = 49.4584◦ = 0.8632 rad

204 Section C y

A

5

4

3 y5 2 C

x 12 x 11

D

1

24

23

22

0

21

1

2

3

4

x

21

22

y5

x 12 x 11

23

24

25 B

Figure 16.32

An asymptote to a curve is defined as a straight line to which the curve approaches as the distance from the origin increases. Alternatively, an asymptote can be considered as a tangent to the curve at infinity.

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

The asymptotes parallel to the x-axis are found by equating the coefficient of the highest power of x to zero.

(ii)

The asymptotes parallel to the y-axis are found by equating the coefficient of the highest power of y to zero. x+2 With the above example y = , rearranging gives: x+1 y(x + 1) = x + 2

i.e.

yx + y − x − 2 = 0

and

x(y − 1) + y − 2 = 0

(1)

The coefficient of the highest power of x (in this case x1 ) is (y − 1). Equating to zero gives: y − 1 = 0 x+2 From which, y = 1, which is an asymptote of y = x+1 as shown in Fig. 16.32. Returning to equation (1): from which,

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

The coefficient of the highest power of y (in this case y1 ) is (x + 1). Equating to zero gives: x + 1 = 0 from which, x+2 x = −1, which is another asymptote of y = as x+1 shown in Fig. 16.32. Problem 8.

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

Functions and their curves 205 y

6

4

y5

x 23 2x 11

x 52

1 2

2

y5

1 2

2 28

26

24

22

21

0

1

4

y5

6

8

x

x 23 2x 11

24

26

Figure 16.33

Rearranging y = i.e. or and

x−3 gives: y(2x + 1) = x − 3 2x + 1

2xy + y = x − 3 2xy + y − x + 3 = 0 x(2y − 1) + y + 3 = 0

Equating the coefficient of the highest power of x to zero gives: 2y − 1 = 0 from which, y = 21 which is an asymptote. Since y(2x + 1) = x − 3 then equating the coefficient of the highest power of y to zero gives: 2x + 1 = 0 from which, x = − 21 which is also an asymptote. x − 3 −3 When x = 0, y = = = −3 and when y = 0, 2x + 1 1 x−3 0= from which, x − 3 = 0 and x = 3. 2x + 1 x−3 A sketch of y = is shown in Fig. 16.33. 2x + 1 Problem 9. Determine the asymptotes parallel to the x- and y-axes for the function x2 y2 = 9(x2 + y2 ) Asymptotes parallel to the x-axis: Rearranging x2 y2 = 9(x2 + y2 ) gives

x2 y2 − 9x2 − 9y2 = 0 hence

x2 (y2 − 9) − 9y2 = 0

Equating the coefficient of the highest power of x to zero gives y2 − 9 = 0 from which, y2 = 9 and y = ±3 Asymptotes parallel to the y-axis: Since x2 y2 − 9x2 − 9y2 = 0 then

y2 (x2 − 9) − 9x2 = 0

Equating the coefficient of the highest power of y to zero gives x2 − 9 = 0 from which, x2 = 9 and x = ±3. Hence asymptotes occur at y = ±3 and x = ±3

Other asymptotes To determine asymptotes other than those parallel to x- and y-axes a simple procedure is: (i)

substitute y = mx + c in the given equation

(ii)

simplify the expression

(iii)

equate the coefficients of the two highest powers of x to zero and determine the values of m and c. y = mx + c gives the asymptote.

206 Section C y

6

y5

x2

2

x 521

4

2

26

24

22

0

2

4

6

x

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

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

24

26

28

210

Figure 16.34

Problem 10. Determine the asymptotes for the function: y(x + 1) = (x − 3)(x + 2) and sketch the curve. Following the above procedure: (i)

Substituting y = mx + c into y(x + 1) = (x − 3) (x + 2) gives: (mx + c)(x + 1) = (x − 3)(x + 2)

(ii)

Simplifying gives mx2 + mx + cx + c = x2 − x − 6

(iii)

and (m − 1)x2 + (m + c + 1)x + c + 6 = 0 Equating the coefficient of the highest power of x to zero gives m − 1 = 0 from which, m = 1 Equating the coefficient of the next highest power of x to zero gives m + c + 1 = 0 and since m = 1, 1 + c + 1 = 0 from which, c = −2 Hence y = mx + c = 1x − 2. i.e. y = x − 2 is an asymptote.

To determine any asymptotes parallel to the x-axis: Rearranging y(x + 1) = (x − 3)(x + 2) gives

yx + y = x2 − x − 6

The coefficient of the highest power of x (i.e. x2 ) is 1. Equating this to zero gives 1 = 0 which is not an equation of a line. Hence there is no asymptote parallel to the x-axis. To determine any asymptotes parallel to the y-axis: Since y(x + 1) = (x − 3)(x + 2) the coefficient of the highest power of y is x + 1. Equating this to zero gives x + 1 = 0, from which, x = −1. Hence x = −1 is an asymptote. When x = 0, y(1) = (−3)(2), i.e. y = −6 When y = 0, 0 = (x − 3)(x + 2), i.e. x = 3 and x = −2 A sketch of the function y(x + 1) = (x − 3)(x + 2) is shown in Fig. 16.34. Problem 11. Determine the asymptotes for the function x3 − xy2 + 2x − 9 = 0 Following the procedure: (i)

Substituting y = mx + c gives x3 − x(mx + c)2 + 2x − 9 = 0

Functions and their curves 207 (ii)

Simplifying gives x3 − x[m 2 x2 + 2mcx + c2 ] + 2x − 9 = 0 i.e.

x3 − m 2 x3 − 2mcx2 − c2 x + 2x − 9 = 0

and x3 (1 − m 2 ) − 2mcx2 − c2 x + 2x − 9 = 0 (iii)

Equating the coefficient of the highest power of x (i.e. x3 in this case) to zero gives 1 − m 2 = 0, from which, m = ±1 Equating the coefficient of the next highest power of x (i.e. x2 in this case) to zero gives −2mc = 0, from which, c = 0

Hence y = mx + c = ±1x + 0, i.e. y = x and y = −x are asymptotes. To determine any asymptotes parallel to the x- and y-axes for the function x3 − xy2 + 2x − 9 = 0: Equating the coefficient of the highest power of x term to zero gives 1 = 0 which is not an equation of a line. Hence there is no asymptote parallel with the x-axis.

Equating the coefficient of the next highest power x term to zero gives c = 0. Hence y = mx + c = 1x + 0, i.e. y = x is an asymptote. x2 + 1 A sketch of y = is shown in Fig. 16.35. x It is possible to determine maximum/minimum points on the graph (see Chapter 26).

Problem 12. Find the asymptotes for the function x2 + 1 y= and sketch a graph of the function. x x2 + 1 Rearranging y = gives yx = x2 + 1 x Equating the coefficient of the highest power x term to zero gives 1 = 0, hence there is no asymptote parallel to the x-axis. Equating the coefficient of the highest power y term to zero gives x = 0 Hence there is an asymptote at x = 0 (i.e. the y-axis) To determine any other asymptotes we substitute y = mx + c into yx = x2 + 1 which gives

for a turning point. 1 Hence 1 = 2 and x2 = 1, from which, x = ±1 x When x = 1, y=

y=

and (m − 1)x2 + cx − 1 = 0 Equating the coefficient of the highest power x term to zero gives m − 1 = 0, from which m = 1

(−1)2 + 1 = −2 −1

i.e. (1, 2) and (−1, −2) are the co-ordinates of the turning points. d2 y 2 d2 y = 2x−3 = 3 ; when x = 1, is positive, 2 dx x dx2 which indicates a minimum point and when x = −1, d2 y is negative, which indicates a maximum point, as dx2 shown in Fig. 16.35. Now try the following Practice Exercise Practice Exercise 84 on page 875)

Asymptotes (Answers

In Problems 1 to 3, determine the asymptotes parallel to the x- and y-axes. 1. 2.

mx2 + cx = x2 + 1

x2 + 1 1 + 1 = =2 x 1

and when x = −1,

(mx + c)x = x2 + 1 i.e.

x2 + 1 x2 1 = + = x + x−1 x x x

1 dy = 1 − x−2 = 1 − 2 = 0 dx x

then

Equating the coefficient of the highest power of y term to zero gives −x = 0 from which, x = 0 Hence x = 0, y = x and y = − x are asymptotes for the function x3 − xy2 + 2x − 9 = 0

y=

Since

3.

x−2 x+1 x y2 = x−3

y=

y=

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

In Problems 4 and 5, determine all the asymptotes. 4.

8x − 10 + x3 − xy2 = 0

208 Section C y

5

x

6

y5 4

y

x 211 x

2

24

22

2

0

4

x

22

y5

x 211 x

24

26

Figure 16.35

5. x2 (y2 − 16) = y In Problems 6 and 7, determine the asymptotes and sketch the curves. 6. y =

x2 − x − 4 x+1

7. xy2 − x2 y + 2x − y = 5

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

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

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

16.8

Brief guide to curve sketching

The following steps will give information from which the graphs of many types of functions y = f (x) can be sketched. (i)

Use calculus to determine the location and nature of maximum and minimum points (see Chapter 26).

(ii)

Determine where the curve cuts the x- and y-axes.

(iii)

Inspect the equation for symmetry.

16.9 Worked problems on curve sketching Problem 13. Sketch the graphs of (a) y = 2x2 + 12x + 20 (b) y = −3x2 + 12x − 15 (a) y = 2x2 + 12x + 20 is a parabola since the equation is a quadratic. To determine the turning point: Gradient =

dy = 4x + 12 = 0 for a turning point. dx

Functions and their curves 209 Hence 4x = −12 and x = −3

y

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

20

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

15

d2 y = 4, which is positive, hence (−3, 2) is a dx2 minimum point.

y 5 2x 2 1 12x 1 20

When x = 0, y = 20, hence the curve cuts the y-axis at y = 20

24

5 2 23

22

Thus knowing the curve passes through (−3, 2) and (0, 20) and appreciating the general shape of a parabola results in the sketch given in Fig. 16.36. (b)

dy = −6x + 12 = 0 for a turning point. dx

21

0 23 25

1

2

3

x

210 y 5 23x 2 1 12x 2 15 215 220

y = −3x2 + 12x − 15 is also a parabola (but ‘upside down’ due to the minus sign in front of the x2 term). Gradient =

10

225

Figure 16.36

Hence 6x = 12 and x = 2 When x = 2, y = −3(2)2 + 12(2) − 15 = −3 Hence (2, −3) are the co-ordinates of the turning point d2 y = −6, which is negative, hence (2, −3) is a dx2 maximum point. When x = 0, y = −15, hence the curve cuts the axis at y = −15 The curve is shown sketched in Fig. 16.36. Problem 14. Sketch the curves depicting the following equations: √ (a) x = 9 − y2 (b) y2 = 16x (c) xy = 5 (a) Squaring both sides of the equation and transposing gives x2 + y2 = 9. Comparing this with the standard equation of a circle, centre origin and radius a, i.e. x2 + y2 = a2 , shows that x2 + y2 = 9 represents a circle, centre origin and radius 3. A sketch of this circle is shown in Fig. 16.37(a).

(b)

The equation y2 = 16x is symmetrical about the x-axis and having its vertex at the origin (0, 0). Also, when x = 1, y = ±4. A sketch of this parabola is shown in Fig. 16.37(b).

a (c) The equation y = represents a rectangular x hyperbola lying entirely within the first and third 5 quadrants. Transposing xy = 5 gives y = , and x therefore represents the rectangular hyperbola shown in Fig. 16.37(c). Problem 15. Sketch the curves depicting the following equations: (a) 4x2 = 36 − 9y2

(b) 3y2 + 15 = 5x2

(a) By dividing throughout by 36 and transposing, the equation 4x2 = 36 − 9y2 can be written as x2 y2 + = 1. The equation of an ellipse is of the 9 4 x2 y2 form 2 + 2 = 1, where 2a and 2b represent the a b x2 y2 length of the axes of the ellipse. Thus 2 + 2 = 1 3 2 represents an ellipse, having its axes coinciding with the x- and y-axes of a rectangular co-ordinate system, the major axis being 2(3), i.e. 6 units long

210 Section C y

y

3

4 x x 6 (a) 4x 2 5 36 29y 2

(a) x 5 !(92y 2)

y

y 14 x 2Œ„3 1

x

(b) 3y 2 11555x 2

Figure 16.38 24 (b) y 2 516x y

x

(c) xy 5 5

Figure 16.37

and the minor axis 2(2), i.e. 4 units long, as shown in Fig. 16.38(a). (b)

Dividing 3y2 + 15 = 5x2 throughout by 15 and x2 y2 transposing gives − = 1. The equation 3 5 2 2 x y − = 1 represents a hyperbola which is syma2 b2 metrical about both the x- and y-axes, the distance between the vertices being given by 2a x2 y2 Thus a sketch of − = 1 is as shown in 3 5 √ Fig. 16.38(b), having a distance of 2 3 between its vertices.

Problem 16. Describe the shape of the curves represented by the following equations: √[ ( y )2 ] y2 (a) x = 2 1− (b) = 2x 2 8 ( )1/2 x2 (c) y = 6 1 − 16 [ ( y )2 ] (a) Squaring the equation gives x2 = 4 1 − 2 and transposing gives x2 = 4 − y2 , i.e. x2 + y2 = 4. Comparing this equation with x2 + y2 = a2 shows that x2 + y2 = 4 is the equation of a circle having centre at the origin (0, 0) and of radius 2 units. √ y2 y2 Transposing = 2x gives y = 4 x. Thus = 2x 8 8 is the equation of a parabola having its axis of symmetry coinciding with the x-axis and its vertex at the origin of a rectangular co-ordinate system. ( )1/2 x2 y (c) y = 6 1 − can be transposed to = 16 6 ( ) 1/2 x2 1− and squaring both sides gives 16 y2 x2 x2 y2 = 1 − , i.e. + =1 36 16 16 36

(b)

Functions and their curves 211 This is the equation of an ellipse, centre at the origin of a rectangular co-ordinate system, the major √ axis coinciding with the y-axis and being 2 36, i.e. 12 units long. √ The minor axis coincides with the x-axis and is 2 16, i.e. 8 units long. Problem 17. Describe the shape of the curves represented by the following equations: √[ ( y )2 ] x y 15 (a) = 1+ (b) = 5 2 4 2x √[ ( y )2 ] x 1+ (a) Since = 5 2 ( ) 2 2 x y = 1+ 25 2 x2 y2 i.e. − =1 25 4 This is a hyperbola which is symmetrical about √ both the x- and y-axes, the vertices being 2 25, i.e. 10 units apart. (With reference to Section 16.1 (vii), a is equal to ±5) y 15 a The equation = is of the form y = , 4 2x x 60 a= = 30 2 This represents a rectangular hyperbola, symmetrical about both the x- and y-axes, and lying entirely in the first and third quadrants, similar in shape to the curves shown in Fig. 16.9.

(b)

4. y2 =

x2 − 16 4

y2 x2 = 5− 5 2 √ 6. x = 3 1 + y2

5.

7. x2 y2 = 9 √ 8. x = 31 (36 − 18y2 ) 9. Sketch the circle given by the equation x2 + y2 − 4x + 10y + 25 = 0 In Problems 10 to 15 describe the shape of the curves represented by the equations given. √ 10. y = [3(1 − x2 )] √ 11. y = [3(x2 − 1)] √ 12. y = 9 − x2 13. y = 7x−1 14. y = (3x)1/2 15. y2 − 8 = −2x2

Practice Exercise 86 Multiple-choice questions on functions and their curves (Answers on page 876) Each question has only one correct answer

Now try the following Practice Exercise Practice Exercise 85 Curve sketching (Answers on page 875) 1.

Sketch the graphs of (a) y = 3x2 + 9x + (b) y = −5x2 + 20x + 50

7 4

In Problems 2 to 8, sketch the curves depicting the equations given. √[ ( y )2 ] 1− 2. x = 4 4 3.

√ y x= 9

1. The graph of y = tan 3θ is: (a) a continuous, periodic, even function (b) a discontinuous, non-periodic, odd function (c) a discontinuous, periodic, odd function (d) a continuous, non-periodic, even function 2.

3.

x2 y2 + = 1 is the equation of: a2 b2 (a) a hyperbola (b) a circle (c) an ellipse (d) a rectangular hyperbola x2 y2 − = 1 is the equation of: a2 b2 (a) a hyperbola (b) a circle (c) an ellipse (d) a rectangular hyperbola

212 Section C

4.

( )2 ( )2 x − a + y − b = r 2 is the equation of: (a) a hyperbola (b) a circle (c) an ellipse (d) a rectangular hyperbola

5. xy = c is the equation of: (a) a hyperbola (b) a circle (c) an ellipse (d) a rectangular hyperbola

For fully worked solutions to each of the problems in Practice Exercises 81 to 85 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 17

Irregular areas, volumes and mean values of waveforms Why it is important to understand: Irregular areas, volumes and mean values of waveforms Surveyors, farmers and landscapers often need to determine the area of irregularly shaped pieces of land to work with the land properly. There are many applications in business, economics and the sciences, including all aspects of engineering, where finding the areas of irregular shapes, the volumes of solids and the lengths of irregular shaped curves are important applications. Typical earthworks include roads, railway beds, causeways, dams and canals. Other common earthworks are land grading to reconfigure the topography of a site or to stabilise slopes. Engineers need to concern themselves with issues of geotechnical engineering (such as soil density and strength) and with quantity estimation to ensure that soil volumes in the cuts match those of the fills, while minimizing the distance of movement. Simpson’s rule is a staple of scientific data analysis and engineering; it is widely used, for example, by naval architects to numerically determine hull offsets and cross-sectional areas to determine volumes and centroids of ships or lifeboats. There are therefore plenty of examples where irregular areas and volumes need to be determined by engineers.

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

use the trapezoidal rule to determine irregular areas use the mid-ordinate rule to determine irregular areas use Simpson’s rule to determine irregular areas estimate the volume of irregular solids determine the mean values of waveforms

17.1

Areas of irregular figures

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

(a) A planimeter is an instrument for directly measuring small areas bounded by an irregular curve. (b)

Trapezoidal rule To determine the areas PQRS in Fig. 17.1: (i) Divide base PS into any number of equal intervals, each of width d (the greater the number of intervals, the greater the accuracy). (ii) Accurately measure ordinates y1 , y2 , y3 , etc.

214 Section C (d)

Simpson’s rule∗ To determine the area PQRS of Fig. 17.1:

Q y1

y2

y3

y4

y5

y6

P d

d

d

d

d

R y7

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

S

(ii) Accurately measure ordinates y1 , y2 , y3 , etc.

d

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

Figure 17.1

(iii) Areas PQRS ] [ y1 + y7 + y2 + y3 + y4 + y5 + y6 ≈d 2

In general, Simpson’s rule states: 1 Area ≈ 3

In general, the trapezoidal rule states:

( ) [( ) width of first + last interval

Area ≈     ( ) sum of first + width of  1   + remaining last interval 2 ordinates ordinate +2

(c) Mid-ordinate rule

ordinate ( ) sum of even +4 ordinates ( )] sum of remaining odd ordinates

To determine the area ABCD of Fig. 17.2:

B

C y1

y2

y3

y4

y5

y6

d

d

d

d

d

d

D

A

Figure 17.2

(i) Divide base AD into any number of equal intervals, each of width d (the greater the number of intervals, the greater the accuracy). (ii) Erect ordinates in the middle of each interval (shown by broken lines in Fig. 17.2). (iii) Accurately measure ordinates y1 , y2 , y3 , etc. (iv) Area ABCD = d(y1 + y2 + y3 + y4 + y5 + y6 ) In general, the mid-ordinate rule states: ( )( ) width of sum of Area ≈ interval mid-ordinates

∗

Who was Simpson? Thomas Simpson FRS (20 August 1710–14 May 1761) was the British mathematician who invented Simpson’s rule to approximate definite integrals. To find out more go to www.routledge.com/cw/bird

Irregular areas, volumes and mean values of waveforms 215 area ≈ (1)[1.25 + 4.0 + 7.0 + 10.75 + 15.0 + 20.25] = 58.25 m

Problem 1. A car starts from rest and its speed is measured every second for 6 s: Time t(s)

0 1

2

3

4

5

6

(c) Simpson’s rule (see para. (d) on page 214)

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

The time base is divided into six strips each of width 1 s, and the length of the ordinates measured. Thus

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

area ≈ 13 (1)[(0 + 24.0) + 4(2.5 + 8.75 + 17.5) + 2(5.5 + 12.5)]

A graph of speed/time is shown in Fig. 17.3.

= 58.33 m

Graph of speed/time

30

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

Speed (m/s)

25 20

Depth (m)

15

3.3 4.5

4.2 2.4

0

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

1

2 3 4 Time (seconds)

5

20.25

24.0

17.5

15.0

12.5

8.75

5.5

7.0

4.0

2.5

1.25

5

10.75

10

0

0 2.2

From para. (d) on page 214, Area ≈ 13 (3)[(0 + 0) + 4(2.2 + 4.5 + 2.4)

6

+ 2(3.3 + 4.2)] = (1)[0 + 36.4 + 15] = 51.4 m2

Figure 17.3

(a) Trapezoidal rule (see para. (b) on page 213) The time base is divided into six strips each of width 1 s, and the length of the ordinates measured. Thus [( ) 0 + 24.0 area ≈ (1) + 2.5 + 5.5 2 ] + 8.75 + 12.5 + 17.5

Now try the following Practice Exercise Practice Exercise 87 Areas of irregular figures (Answers on page 876) 1.

Plot a graph of y = 3x − x2 by completing a table of values of y from x = 0 to x = 3. Determine the area enclosed by the curve, the x-axis and ordinate x = 0 and x = 3 by (a) the trapezoidal rule, (b) the mid-ordinate rule and (c) by Simpson’s rule.

2.

Plot the graph of y = 2x2 + 3 between x = 0 and x = 4. Estimate the area enclosed by the curve, the ordinates x = 0 and x = 4, and the x-axis by an approximate method.

= 58.75 m (b)

Mid-ordinate rule (see para. (c) on page 214) The time base is divided into six strips each of width 1 s. Mid-ordinates are erected as shown in Fig. 17.3 by the broken lines. The length of each mid-ordinate is measured. Thus

216 Section C

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

2

3

4

5

A1

A2

A4

A5

A6

A7

6 d

velocity v (m/s)

A3

0 2.0 4.5 8.0 14.0 21.0 29.0

Determine the distance travelled in six seconds (i.e. the area under the v/t graph) using Simpson’s rule. 4. The shape of a piece of land is shown in Fig. 17.4. To estimate the area of the land, a surveyor takes measurements at intervals of 50 m, perpendicular to the straight portion with the results shown (the dimensions being in metres). Estimate the area of the land in hectares (1 ha = 104 m2 ).

d

d

d

d

d

Figure 17.5

Problem 3. A tree trunk is 12 m in length and has a varying cross-section. The cross-sectional areas at intervals of 2 m measured from one end are: 0.52, 0.55, 0.59, 0.63, 0.72, 0.84, 0.97 m2 Estimate the volume of the tree trunk. A sketch of the tree trunk is similar to that shown in Fig. 17.5 above, where d = 2 m, A1 = 0.52 m2 , A2 = 0.55 m2 , and so on. Using Simpson’s rule for volumes gives: Volume ≈ 23 [(0.52 + 0.97) + 4(0.55 + 0.63

140 160 200 190 180 130

+ 0.84) + 2(0.59 + 0.72)] 50

50

50

50

50

50

Figure 17.4

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

0 2.8 5.2 6.5 5.8 4.1 3.0 2.3

= 32 [1.49 + 8.08 + 2.62] = 8.13 m3 Problem 4. The areas of seven horizontal cross-sections of a water reservoir at intervals of 10 m are: 210, 250, 320, 350, 290, 230, 170 m2 Calculate the capacity of the reservoir in litres.

Estimate the area of the deck. Using Simpson’s rule for volumes gives:

17.2

Volumes of irregular solids

If the cross-sectional areas A1 , A2 , A3 , . . . of an irregular solid bounded by two parallel planes are known at equal intervals of width d (as shown in Fig. 17.5), then by Simpson’s rule: d volume, V≈ [(A1 + A7 ) + 4(A2 + A4 + A6 ) 3 + 2 (A3 + A5 )]

Volume ≈

10 [(210 + 170) + 4(250 + 350 3 + 230) + 2(320 + 290)]

=

10 [380 + 3320 + 1220] 3

= 16 400 m3 16 400 m3 = 16 400 × 106 cm3 and since 1 litre = 1000 cm3 ,

Irregular areas, volumes and mean values of waveforms 217 16 400 × 106 litres 1000 = 1 6400 000

capacity of reservoir =

= 1.64 × 107 litres

17.3 The mean or average value of a waveform The mean or average value, y, of the waveform shown in Fig. 17.6 is given by:

Now try the following Practice Exercise y= Practice Exercise 88 Volumes of irregular solids (Answers on page 876)

area under curve length of base, b

1. The areas of equidistantly spaced sections of the underwater form of a small boat are as follows: 1.76, 2.78, 3.10, 3.12, 2.61, 1.24, 0.85 m2 y

Determine the underwater volume if the sections are 3 m apart.

y1 y2 d

2. To estimate the amount of earth to be removed when constructing a cutting, the crosssectional area at intervals of 8 m were estimated as follows: 0,

2.8, 3.7,

4.5,

4.1, 2.6,

0 m2

Estimate the volume of earth to be excavated. 3. The circumference of a 12 m long log of timber of varying circular cross-section is measured at intervals of 2 m along its length and the results are:

d

y3

y4

y5

y6

y7

d

d

d

d

d

b

Figure 17.6

If the mid-ordinate rule is used to find the area under the curve, then: y≈

sum of mid-ordinates number of mid-ordinates ( y1 + y2 + y3 + y4 + y5 + y6 + y7 = 7

)

Distance from one end (m)

Circumference (m)

0

2.80

2

3.25

4

3.94

(i)

over one complete cycle is zero (see Fig. 17.7(a)),

6

4.32

(ii)

8

5.16

over half a cycle is 0.637 × maximum value, or (2/π) × maximum value,

10

5.82

(iii)

of a full-wave rectified waveform (see Fig. 17.7(b)) is 0.637 × maximum value,

12

6.36

Estimate the volume of the timber in cubic metres.

for Fig. 17.6

For a sine wave, the mean or average value:

(iv) of a half-wave rectified waveform (see Fig. 17.7(c)) is 0.318 × maximum value, or (1/π) maximum value.

218 Section C V Vm

(a) Area under triangular waveform (a) for a halfcycle is given by:

V Vm

0

0

t

Area =

t

(a)

1 2

(base) (perpendicular height)

= 12 (2 × 10−3 )(20)

(b)

= 20 × 10−3 Vs

V Vm

Average value of waveform

0

=

area under curve length of base

=

20 × 10−3 Vs 2 × 10−3 s

t

(c)

Figure 17.7

= 10 V (b)

Voltage (V)

Problem 5. Determine the average values over half a cycle of the periodic waveforms shown in Fig. 17.8. 20

0

1

2

3

t (ms)

4

Area under waveform (b) for a half-cycle = (1 × 1) + (3 × 2) = 7 As. Average value of waveform =

area under curve length of base

=

7 As 3s

= 2.33 A

210

Current (A)

(a)

(c) A half-cycle of the voltage waveform (c) is completed in 4 ms.

3 2 1

Area under curve = 12 {(3 − 1)10−3 }(10)

0 21 22 23

1

2

4

3

5 6

t (s)

= 10 × 10−3 Vs Average value of waveform

Voltage (V)

(b)

10

area under curve length of base

=

10 × 10−3 Vs 4 × 10−3 s

= 2.5 V 0

2

6

4

210 (c)

Figure 17.8

=

8

t (ms)

Problem 6. Determine the mean value of current over one complete cycle of the periodic waveforms shown in Fig. 17.9.

Current (mA)

Irregular areas, volumes and mean values of waveforms 219

Problem 7. The power used in a manufacturing process during a six hour period is recorded at intervals of one hour as shown below.

5

4

0

8

12

16

20

24

28 t (ms)

Current (mA)

(a)

Time (h)

0

1

2

3

4

5

6

Power (kW)

0

14

29

51

45

23

0

Plot a graph of power against time and, by using the mid-ordinate rule, determine (a) the area under the curve and (b) the average value of the power.

2

0

2

4

6

8

10

12

t (ms)

The graph of power/time is shown in Fig. 17.10.

(b) Graph of power/time

Figure 17.9

Power (kW)

50

(a) One cycle of the trapezoidal waveform (a) is completed in 10 ms (i.e. the periodic time is 10 ms). Area under curve = area of trapezium =

1 2

40 30 20

(sum of parallel sides) (perpendicular

10

distance between parallel sides)

7.0

= 12 {(4 + 8) × 10−3 }(5 × 10−3 ) = 30 × 10−6 As

0

1

area under curve 30 × 10−6 As = = length of base 10 × 10−3 s

2

49.5

37.0 10.0

3 4 Time (hours)

5

6

(a) The time base is divided into six equal intervals, each of width one hour. Mid-ordinates are erected (shown by broken lines in Fig. 17.10) and measured. The values are shown in Fig. 17.10. Area under curve ≈ (width of interval) × (sum of mid-ordinates)

= 3 mA One cycle of the sawtooth waveform (b) is completed in 5 ms.

= (1)[7.0 + 21.5 + 42.0 + 49.5 + 37.0 + 10.0]

Area under curve = 21 (3 × 10−3 )(2)

= 167 kWh (i.e. a measure of electrical energy)

= 3 × 10−3 As (b)

Mean value over one cycle =

42.0

Figure 17.10

Mean value over one cycle

(b)

21.5

area under curve 3 × 10−3 As = length of base 5 × 10−3 s

Average value of waveform area under curve = length of base =

= 0.6 A

167 kWh = 27.83 kW 6h

220 Section C (With a larger number of intervals a more accurate answer may be obtained.) For a sine wave the actual mean value is 0.637 × maximum value, which in this problem gives 6.37 V.

Alternatively, average value =

sum of mid-ordinates number of mid-ordinates

Problem 9. An indicator diagram for a steam engine is shown in Fig. 17.12. The base line has been divided into six equally spaced intervals and the lengths of the seven ordinates measured with the results shown in centimetres. Determine (a) the area of the indicator diagram using Simpson’s rule, and (b) the mean pressure in the cylinder given that 1 cm represents 100 kPa.

Voltage (V)

Problem 8. Fig. 17.11 shows a sinusoidal output voltage of a full-wave rectifier. Determine, using the mid-ordinate rule with six intervals, the mean output voltage.

10

0

308608908 p 2

1808 p

2708 3p 2

3608 2p

u 4.0

3.6

Figure 17.11

2.9

2.2

1.7

1.6

12.0 cm

One cycle of the output voltage is completed in π radians or 180◦ . The base is divided into six intervals, each of width 30◦ . The mid-ordinate of each interval will lie at 15◦ , 45◦ , 75◦ , etc. At 15◦ the height of the mid-ordinate is 10 sin 15◦ = 2.588 V. At 45◦ the height of the mid-ordinate is 10 sin 45◦ = 7.071 V, and so on. The results are tabulated below: Mid-ordinate

Height of mid-ordinate

15◦

10 sin 15◦ = 2.588 V

45◦

10 sin 45◦ = 7.071 V

◦

10 sin 75 = 9.659 V

105◦

10 sin 105◦ = 9.659 V

135◦

10 sin 135◦ = 7.071 V

165◦

10 sin 165◦ = 2.588 V

75

3.5

Figure 17.12

(a) The width of each interval is Simpson’s rule,

12.0 cm. Using 6

area ≈ 31 (2.0)[(3.6 + 1.6) + 4(4.0 + 2.9 + 1.7) + 2(3.5 + 2.2)] = 32 [5.2 + 34.4 + 11.4] = 34 cm2 (b)

Mean height of ordinates =

◦

area of diagram 34 = length of base 12 = 2.83 cm

Since 1 cm represents 100 kPa, the mean pressure in the cylinder = 2.83 cm × 100 kPa/cm = 283 kPa

Sum of mid-ordinates = 38.636 V Now try the following Practice Exercise Mean or average value of output voltage sum of mid-ordinates number of mid-ordinates 38.636 = 6 = 6.439 V =

Practice Exercise 89 Mean or average values of waveforms (Answers on page 876) 1.

Determine the mean value of the periodic waveforms shown in Fig. 17.13 over a halfcycle.

Current (A)

Irregular areas, volumes and mean values of waveforms 221

3. 2

10

0

t (ms)

20

An alternating current has the following values at equal intervals of 5 ms Time (ms)

0 5

10

15

20

25

30

Current (A) 0 0.9 2.6 4.9 5.8 3.5 0

22

Voltage (V)

(a)

Plot a graph of current against time and estimate the area under the curve over the 30 ms period using the mid-ordinate rule and determine its mean value.

100

0

5

10 t (ms)

4.

Determine, using an approximate method, the average value of a sine wave of maximum value 50 V for (a) a half-cycle and (b) a complete cycle.

5.

An indicator diagram of a steam engine is 12 cm long. Seven evenly spaced ordinates, including the end ordinates, are measured as follows:

2100 Current (A)

(b) 5

0

15

30 t (ms)

25

5.90, 5.52, 4.22, 3.63, 3.32, 3.24, 3.16 cm

(c)

Figure 17.13

Determine the area of the diagram and the mean pressure in the cylinder if 1 cm represents 90 kPa.

Voltage (mV)

2. Find the average value of the periodic waveforms shown in Fig. 17.14 over one complete cycle.

Practice Exercise 90 Multiple-choice questions on irregular areas and volumes and mean values of waveforms (Answers on page 876)

10

0

2

4

6

8

10

t (ms)

Each question has only one correct answer

(a) Current (A)

1.

The speed of a car at 1 second intervals is given in the following table:

5

Time t (s) 0

2

4

6 (b)

Figure 17.14

8

10

t (ms)

0 1

2

3

4

5

6

Speed v(m/s) 0 2.5 5.0 9.0 15.0 22.0 30.0 The distance travelled in 6 s (i.e. the area under the v/t graph) using the trapezoidal rule is: (a) 83.5 m (b) 68 m (c) 68.5 m (d) 204 m

222 Section C 2. The mean value of a sine wave over half a cycle is: (a) 0.318 × maximum value (b) 0.707 × maximum value (c) the peak value (d) 0.637 × maximum value 3. An indicator diagram for a steam engine is as shown in Figure 17.15. The base has been divided into 6 equally spaced intervals and the lengths of the 7 ordinates measured, with the results shown in centimetres. Using Simpson’s rule the area of the indicator diagram is: (a) 17.9 cm2 (b) 32 cm2 2 (c) 16 cm (d) 96 cm2

3.1

3.9

3.5

2.8

2.0

1.5

4.

The mean value over half a cycle of a sine wave is 7.5 volts. The peak value of the sine wave is: (a) 15 V (b) 23.58 V (c) 11.77 V (d) 10.61 V

5.

The areas of seven horizontal cross-sections of a water reservoir at intervals of 9 m are 70, 160, 210, 280, 250, 220 and 170 m2 . Given that 1 litre = 1000 cm3 , the capacity of the reservoir is: (a) 11.4 × 106 litres (b) 34.2 × 106 litres (c) 11.4 × 103 litres (d) 32.4 × 103 litres

1.2

12.0 cm

Figure 17.15

For fully worked solutions to each of the problems in Practice Exercises 87 to 89 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 5

Functions and their curves and irregular areas and volumes

This Revision Test covers the material contained in Chapters 16 and 17. The marks for each question are shown in brackets at the end of each question. 1.

ordinates x = 1 and x = 4, and the x-axis by (a) the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson’s rule. (16)

Sketch the following graphs, showing the relevant points: (a) y = (x − 2)2

(c)

x2 + y2 − 2x + 4y − 4 = 0

y = 3 − cos 2x (d) 9x2 − 4y2 = 36  π  −1 −π ≤ x ≤ −   2    π π x − ≤x≤ (e) f (x) = 2 2     π   1 ≤x≤π 2

6.

(b)

2.

Determine the inverse of f (x) = 3x + 1

3.

Evaluate, correct to 3 decimal places: 2 tan−1 1.64 + sec−1 2.43 − 3 cosec−1 3.85

4.

5.

Height (m)

(x − 1)(x + 4) (x − 2)(x − 5)

0

5.0 10.0 15.0 20.0

Diameter (m) 16.0 13.3 10.7

(3) 7. (4)

(9)

Plot a graph of y = 3x2 + 5 from x = 1 to x = 4. Estimate, correct to 2 decimal places, using six intervals, the area enclosed by the curve, the

8.6

8.0

Determine the area corresponding to each diameter and hence estimate the capacity of the tower in cubic metres. (7)

(15)

Determine the asymptotes for the following function and hence sketch the curve: y=

A circular cooling tower is 20 m high. The inside diameter of the tower at different heights is given in the following table:

A vehicle starts from rest and its velocity is measured every second for 6 seconds, with the following results: Time t (s)

0

1

Velocity v (m/s)

0 1.2

2

3

2.4 3.7

4 5.2

5

6

6.0 9.2

Using Simpson’s rule, calculate (a) the distance travelled in 6 s (i.e. the area under the v/t graph) and (b) the average speed over this period. (6)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 5, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Section D

Complex numbers

Chapter 18

Complex numbers Why it is important to understand: Complex numbers Complex numbers are used in many scientific fields, including engineering, electromagnetism, quantum physics and applied mathematics, such as chaos theory. Any physical motion which is periodic, such as an oscillating beam, string, wire, pendulum, electronic signal or electromagnetic wave can be represented by a complex number function. This can make calculations with the various components simpler than with real numbers and sines and cosines. In control theory, systems are often transformed from the time domain to the frequency domain using the Laplace transform. In fluid dynamics, complex functions are used to describe potential flow in two dimensions. In electrical engineering, the Fourier transform is used to analyse varying voltages and currents. Complex numbers are used in signal analysis and other fields for a convenient description for periodically varying signals. This use is also extended into digital signal processing and digital image processing, which utilise digital versions of Fourier analysis (and wavelet analysis) to transmit, compress, restore and otherwise process digital audio signals, still images and video signals. Knowledge of complex numbers is clearly absolutely essential for further studies in so many engineering disciplines.

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

define a complex number solve quadratic equations with imaginary roots use an Argand diagram to represent a complex number pictorially add, subtract, multiply and divide Cartesian complex numbers solve complex equations convert a Cartesian complex number into polar form, and vice-versa multiply and divide polar form complex numbers apply complex numbers to practical applications

228 Section D 18.1

Cartesian complex numbers

There are several applications of complex numbers in science and engineering, in particular in electrical alternating current theory and in mechanical vector analysis. There are two main forms of complex number – Cartesian form (named after Descartes∗) and polar form – and both are explained in this chapter. If we can add, subtract, multiply and divide complex numbers in both forms and represent the numbers on an Argand diagram then a.c. theory and vector analysis become considerably easier. (i)

(iii)

−1 + j2 and −1 − j2 are known as complex numbers. Both solutions are of the form a + jb, ‘a’ being termed the real part and jb the imaginary part. A complex number of the form a + jb is called Cartesian complex number. In pure √ mathematics the symbol i is used to indicate −1 (i being the first letter of the word imaginary). However i is the symbol of electric current in engineering, and to avoid possible confusion the√ next letter in the alphabet, j, is used to represent −1

Solve the quadratic equation x2 + 4 = 0

√ Since x2 + 4 = 0 then x2 = −4 and x = −4 √ √ √ i.e., x = [(−1)(4)] = (−1) 4 = j(±2) √ = ± j2, (since j = −1) (Note that ± j2 may also be written ±2 j) Problem 2.

If the quadratic equation x2 + 2x + 5 = 0 is solved using the quadratic formula then, √ −2 ± [(2)2 − (4)(1)(5)] x = 2(1) √ √ −2 ± [−16] −2 ± [(16)(−1)] = = 2 2 √ √ √ −2 ± 16 −1 −2 ± 4 −1 = = 2 2 √ = −1 ± 2 −1 √ It is not possible to evaluate −1 in real terms. √ However, if an operator j is defined as j = −1 then the solution may be expressed as x = −1 ± j2.

(ii)

Problem 1.

Solve the quadratic equation 2x2 + 3x + 5 = 0

Using the quadratic formula, √ −3 ± [(3)2 − 4(2)(5)] x = 2(2) √ √ √ −3 ± −31 −3 ± (−1) 31 = = 4 4 √ −3 ± j 31 = 4 √ 3 31 or −0.750 ± j1.392, Hence x = − ± j 4 4 correct to 3 decimal places. (Note, a graph of y = 2x2 + 3x + 5 does not cross the x-axis and hence 2x2 + 3x + 5 = 0 has no real roots.) Problem 3. (a) j3

Evaluate

(b) j4

(c) j23

(d)

−4 j9

(a) j 3 = j2 × j = (−1) × j = − j, since j2 = −1 (b) j 4 = j2 × j2 = (−1) × (−1) = 1 (c) j 23 = j × j22 = j × ( j2 )11 = j × (−1)11 = j × (−1) = − j (d) j 9 = j × j8 = j × ( j2 )4 = j × (−1)4 = j×1 = j Hence

∗

Who was Descartes? For image and resume of Descartes, see page 112. To find out more go to www.routledge.com/cw/bird

−4 j9

= =

4j −4 −4 −j = × = 2 j j −j −j 4j = 4 j or j4 −(−1)

Complex numbers 229 Now try the following Practice Exercise Practice Exercise 91 Introduction to Cartesian complex numbers (Answers on page 876) In Problems 1 to 9, solve the quadratic equations. 1. x2 + 25 = 0

axis is used to represent the imaginary axis. Such a diagram is called an Argand diagram∗ . In Fig. 18.1, the point A represents the complex number (3 + j2) and is obtained by plotting the co-ordinates (3, j2) as in graphical work. Fig. 18.1 also shows the Argand points B, C and D representing the complex numbers (−2 + j4), (−3 − j5) and (1 − j3) respectively.

18.3 Addition and subtraction of complex numbers

2. x2 − 2x + 2 = 0 3. x2 − 4x + 5 = 0 5. 2x2 − 2x + 1 = 0

Two complex numbers are added/subtracted by adding/ subtracting separately the two real parts and the two imaginary parts.

6. x2 − 4x + 8 = 0

For example, if Z1 = a + jb and Z2 = c + jd,

7. 25x2 − 10x + 2 = 0

then

4. x2 − 6x + 10 = 0

= (a + c) + j(b + d)

8. 2x2 + 3x + 4 = 0 and

9. 4t2 − 5t + 7 = 0 10.

Z1 + Z2 = (a + jb) + (c + jd)

Evaluate (a) j8 (b) −

Z1 − Z2 = (a + jb) − (c + jd) = (a − c) + j(b − d)

1 4 (c) 13 j7 2j

Thus, for example, (2 + j3) + (3 − j4) = 2 + j3 + 3 − j4 = 5 − j1

18.2

The Argand diagram

A complex number may be represented pictorially on rectangular or Cartesian axes. The horizontal (or x) axis is used to represent the real axis and the vertical (or y) Imaginary axis B

j4 j3 A

j2 j 23

22 21 0 2j

1

2

3

Real axis

2j2 2j3

D

2j4 C

Figure 18.1

2j5

∗

Who was Argand? Jean-Robert Argand (18 July 1768– 13 August 1822) was a highly influential mathematician. He privately published a landmark essay on the representation of imaginary quantities which became known as the Argand diagram. To find out more go to www.routledge.com/cw/bird

230 Section D and (2 + j3) − (3 − j4) = 2 + j3 − 3 + j4 = −1 + j7 The addition and subtraction of complex numbers may be achieved graphically as shown in the Argand diagram of Fig. 18.2. (2 + j3) is represented by vector OP and (3 − j4) by vector OQ. In Fig. 18.2(a) by vector addition (i.e. the diagonal of the parallelogram-see Chapter 22) OP + OQ = OR. R is the point (5, −j1) Hence (2 + j3) + (3 − j4) = 5 − j1

Hence (2 + j3) − (3 − j4) = −1 + j7 Problem 4. Given Z1 = 2 + j4 and Z2 = 3 − j determine (a) Z1 + Z2 , (b) Z1 − Z2 , (c) Z2 − Z1 and show the results on an Argand diagram. (a) Z1 + Z2 = (2 + j4) + (3 − j) = (2 + 3) + j(4 − 1) = 5 + j3 (b) Z1 − Z2 = (2 + j4) − (3 − j)

Imaginary axis P (21j3)

j3

In Fig. 18.2(b), vector OQ is reversed (shown as OQ′ ) since it is being subtracted. (Note OQ = 3 − j4 and OQ′ = −(3 − j4) = −3 + j4) OP − OQ = OP + OQ′ = OS is found to be the Argand point (−1, j7)

= (2 − 3) + j(4 −(−1)) = −1 + j5 (c) Z2 − Z1 = (3 − j) − (2 + j4)

j2

= (3 − 2) + j(−1 − 4) = 1 − j5

j 0 2j

1

2

3

5 Real axis R (5 2j )

4

Each result is shown in the Argand diagram of Fig. 18.3.

2j2

Imaginary axis

2j3 2j4

(211j 5)

Q (3 2j 4)

j5 j4

(a)

( 5 1 j3)

j3 Imaginary axis S (211j7)

j2 j

j7 21 0 2j

j6

1

2

3

4

5

Real axis

j5 2j2 Q9

j4 2j3

P (21j 3)

j3

2j4

j2

2j5

( 12 j5)

j 23 22 21 0 2j

1

2

3

Real axis

Figure 18.3

2j 2 2j 3 Q (32j 4)

2j 4 (b)

Figure 18.2

18.4 Multiplication and division of complex numbers (i)

Multiplication of complex numbers is achieved by assuming all quantities involved are real and then using j2 = −1 to simplify.

Complex numbers 231 (a) Z1 Z2 = (1 − j3)(−2 + j5)

Hence (a + jb)(c + jd) = ac + a(jd) + (jb)c + ( jb)(jd)

= −2 + j5 + j6 − j2 15

= ac + jad + jbc + j2 bd

= (−2 + 15) + j(5 + 6), since j2 = −1,

= (ac − bd) + j(ad + bc),

= 13 + j11

since j2 = −1 Thus (3 + j2)(4 − j5)

(b)

Z1 1 − j3 1 − j3 −3 + j4 = = × Z3 −3 − j4 −3 − j4 −3 + j4

= 12 − j15 + j8 − j2 10

=

−3 + j4 + j9 − j2 12 32 + 42

=

9 + j13 9 13 = + j 25 25 25

= (12 − (−10)) + j(−15 + 8) = 22 − j 7 (ii)

The complex conjugate of a complex number is obtained by changing the sign of the imaginary part. Hence the complex conjugate of a + jb is a − jb. The product of a complex number and its complex conjugate is always a real number.

or 0.36 + j0.52 (c)

Z1 Z2 (1 − j3)(−2 + j5) = Z1 + Z2 (1 − j3) + (−2 + j5) =

13 + j11 , from part (a), −1 + j2

=

13 + j11 −1 − j2 × −1 + j2 −1 − j2

=

−13 − j26 − j11 − j2 22 12 + 22

=

9 − j37 9 37 = − j or 1.8 − j 7.4 5 5 5

For example, (3 + j4)(3 − j4) = 9 − j12 + j12 − j2 16 = 9 + 16 = 25 [(a + jb)(a − jb) may be evaluated ‘on sight’ as a2 + b2 ] (iii)

Division of complex numbers is achieved by multiplying both numerator and denominator by the complex conjugate of the denominator.

(d) Z1 Z2 Z3 = (13 + j11)(−3 − j4), since Z1 Z2 = 13 + j11, from part (a)

For example, 2 − j5 3 + j4

=

2 − j5 (3 − j4) × 3 + j4 (3 − j4)

=

6 − j8 − j15 + j2 20 32 + 42

=

−14 − j23 −14 23 = −j 25 25 25 or −0.56 − j0.92

Problem 5. If Z1 = 1 − j3, Z2 = −2 + j5 and Z3 = −3 − j4, determine in a + jb form: Z1 (a) Z1 Z2 (b) Z3 Z1 Z2 (c) (d) Z1 Z2 Z3 Z1 + Z2

= −39 − j52 − j33 − j2 44 = (−39 + 44) − j(52 + 33) = 5 − j85 Problem 6. (a)

2 (1 + j)4

Evaluate: ( )2 1 + j3 (b) j 1 − j2

(a) (1 + j)2 = (1 + j)(1 + j) = 1 + j + j + j2 = 1 + j + j − 1 = j2 (1 + j)4 = [(1 + j)2 ]2 = ( j2)2 = j2 4 = −4 Hence

2 2 1 = =− 4 (1 + j) −4 2

232 Section D (b)

1 + j3 1 + j3 1 + j2 = × 1 − j2 1 − j2 1 + j2 =

(

1 + j3 1 − j2

)2

Hence

1 + j2 + j3 + j2 6 −5 + j5 = 12 + 22 5

= −1 + j1 = −1 + j = (−1 + j)2 = (−1 + j)(−1 + j) = 1 − j − j + j2 = −j2 ( )2 1 + j3 j = j(− j2) = − j2 2 = 2, 1 − j2 since j2 = −1

18.5

Complex equations

If two complex numbers are equal, then their real parts are equal and their imaginary parts are equal. Hence if a + jb = c + jd, then a = c and b = d Problem 7. Solve the complex equations: (a) 2(x + jy) = 6 − j3 (b)

(1 + j2)(−2 − j3) = a + jb

(a) 2(x + jy) = 6 − j3 hence 2x + j2y = 6 − j3 Equating the real parts gives: 2x = 6, i.e. x = 3

Now try the following Practice Exercise Practice Exercise 92 Operations involving Cartesian complex numbers (Answers on page 876)

Equating the imaginary parts gives: 2y = −3, i.e. y = − 32 (b)

−2 − j3 − j4 − j2 6 = a + jb

1. Evaluate (a) (3 + j2) + (5 − j) and (b) (−2 + j6) − (3 − j2) and show the results on an Argand diagram. 2. Write down the complex conjugates of (a) 3 + j4, (b) 2 − j 3. If z = 2 + j and w = 3 − j evaluate (a) z + w (b) w − z (c) 3z − 2w (d) 5z + 2w (e) j(2w − 3z) (f ) 2jw − jz In Problems 4 to 8 evaluate in a + jb form given Z1 = 1 + j2, Z2 = 4 − j3, Z3 = −2 + j3 and Z4 = −5 − j.

(1 + j2)(−2 − j3) = a + jb Hence 4 − j7 = a + jb Equating real and imaginary terms gives: a = 4 and b = −7

Problem 8.

Solve the equations: √ (a) (2 − j3) = (a + jb) (b)

(x − j2y) + ( y − j3x) = 2 + j3

√ (a) (2 − j3) = (a + jb) Hence

(2 − j3)2 = a + jb,

4. (a) Z1 + Z2 − Z3 (b) Z2 − Z1 + Z4

i.e.

5. (a) Z1 Z2 (b) Z3 Z4

Hence

6. (a) Z1 Z3 + Z4 (b) Z1 Z2 Z3

and

Z1 Z1 + Z3 7. (a) (b) Z2 Z2 − Z4

Thus a = −5 and b = −12

8. (a)

Z1 Z3 Z1 (b) Z2 + + Z3 Z1 + Z3 Z4

1− j 1 9. Evaluate (a) (b) 1+ j 1+ j ( ) −25 1 + j2 2 − j5 10. Show that − 2 3 + j4 −j = 57 + j24

(2 − j3)(2 − j3) = a + jb 4 − j6 − j6 + j2 9 = a + jb −5 − j12 = a + jb

(b) (x − j2y) + ( y − j3x) = 2 + j3 Hence (x + y) + j(−2y − 3x) = 2 + j3 Equating real and imaginary parts gives: x+y = 2

(1)

and −3x − 2y = 3

(2)

i.e. two simultaneous equations to solve.

Complex numbers 233 Multiplying equation (1) by 2 gives: 2x + 2y = 4

Imaginary axis

(3)

Z

Adding equations (2) and (3) gives: −x = 7, i.e., x = −7

r

jy

u

From equation (1), y = 9, which may be checked in equation (2).

O

A Real axis

x

Figure 18.4

Now try the following Practice Exercise

r=

i.e.

√ (x2 + y2 )

Practice Exercise 93 Complex equations (Answers on page 877)

(iii) θ is called the argument (or amplitude) of Z and is written as arg Z

In Problems 1 to 4 solve the complex equations.

By trigonometry on triangle OAZ, y arg Z = θ = tan−1 x (iv) Whenever changing from Cartesian form to polar form, or vice-versa, a sketch is invaluable for determining the quadrant in which the complex number occurs.

1. (2 + j)(3 − j2) = a + jb 2+ j = j(x + jy) 1− j √ 3. (2 − j3) = (a + jb) 2.

4. (x − j2y) − ( y − jx) = 2 + j 5. If Z = R + jωL + 1/jωC, express Z in (a + jb) form when R = 10, L = 5, C = 0.04 and ω = 4

18.6 The polar form of a complex number

Problem 9. Determine the modulus and argument of the complex number Z = 2 + j3, and express Z in polar form. Z = 2 + j3 lies in the first quadrant as shown in Fig. 18.5. Imaginary axis

(i) Let a complex number z be x + jy as shown in the Argand diagram of Fig. 18.4. Let distance OZ be r and the angle OZ makes with the positive real axis be θ

j3 r

From trigonometry, x = r cos θ and

u 0

y = r sin θ Hence Z = x + jy = r cos θ + jr sin θ = r(cos θ + j sin θ) Z = r(cos θ + j sin θ) is usually abbreviated to Z = r∠θ which is known as the polar form of a complex number. (ii) r is called the modulus (or magnitude) of Z and is written as mod Z or |Z| r is determined using Pythagoras’ theorem on triangle OAZ in Fig. 18.4,

2

Real axis

Figure 18.5

√ √ Modulus, |Z| = r = (22 + 32 ) = 13 or 3.606, correct to 3 decimal places. Argument, arg Z = θ = tan−1

3 2

= 56.31◦ or 56◦ 19′ In polar form, 2 + j3 is written as 3.606∠56.31◦

234 Section D (By convention the principal value is normally used, i.e. the numerically least value, such that −π < θ < π)

Problem 10. Express the following complex numbers in polar form: (a) 3 + j4

(b) −3 + j4

(c) −3 − j4

(d) 3 − j4

(d)

Modulus, r = 5 and angle α = 53.13◦ , as above. Hence (3 − j4) = 5∠−53.13◦

(a) 3 + j4 is shown in Fig. 18.6 and lies in the first quadrant.

Problem 11. Convert (a) 4∠30◦ (b) 7∠−145◦ into a + jb form, correct to 4 significant figures.

Imaginary axis (23 1j4)

(3 1 j4)

j4

(a) 4∠30◦ is shown in Fig. 18.7(a) and lies in the first quadrant.

j3 r

j2

r

j a 23 22 21 a 2j

u a1

2j2 r

3 − j4 is shown in Fig. 18.6 and lies in the fourth quadrant.

Imaginary axis

2

3

4 308

Real axis 0

r

jy Real axis

x (a)

2j3 (23 2 j4)

2j4

(3 2 j4)

x a

Figure 18.6

jy

√ Modulus, r = (32 + 42 ) = 5 and argument θ = tan−1 43 = 53.13◦ Hence 3 + j4 = 5∠53.13◦ (b)

7

Real axis 1458

(b)

Figure 18.7

−3 + j4 is shown in Fig. 18.6 and lies in the second quadrant.

Using trigonometric ratios, x = 4 cos 30◦ = 3.464 and y = 4 sin 30◦ = 2.000

Modulus, r = 5 and angle α = 53.13◦ , from part (a).

Hence 4∠30◦ = 3.464 + j2.000

◦

◦

◦

Argument = 180 − 53.13 = 126.87 (i.e. the argument must be measured from the positive real axis).

(b)

7∠145◦ is shown in Fig. 18.7(b) and lies in the third quadrant. Angle α = 180◦ − 145◦ = 35◦ Hence and

Hence −3 + j4 = 5∠126.87◦ (c) −3 − j4 is shown in Fig. 18.6 and lies in the third quadrant.

x = 7 cos 35◦ = 5.734 y = 7 sin 35◦ = 4.015

Hence 7∠−145◦ = −5.734 − j4.015

Modulus, r = 5 and α = 53.13◦ , as above.

Alternatively

Hence the argument = 180◦ + 53.13◦ = 233.13◦, which is the same as −126.87◦

7∠−145◦ = 7 cos(−145◦ ) + j7 sin(−145◦ )

◦

◦

Hence (−3 − j4) = 5∠233.13 or 5∠−126.87

= −5.734 − j4.015

Complex numbers 235 Calculator Using the ‘Pol’ and ‘Rec’ functions on a calculator enables changing from Cartesian to polar and vice-versa to be achieved more quickly. Since complex numbers are used with vectors and with electrical engineering a.c. theory, it is essential that the calculator can be used quickly and accurately.

Addition and subtraction in polar form is not possible directly. Each complex number has to be converted into Cartesian form first. 2∠30◦ = 2(cos 30◦ + j sin 30◦ ) = 2 cos 30◦ + j2 sin 30◦ = 1.732 + j1.000 5∠−45◦ = 5(cos(−45◦ ) + j sin(−45◦ )) = 5 cos(−45◦ ) + j5 sin(−45◦ )

18.7 Multiplication and division in polar form

= 3.536 − j3.536 4∠120◦ = 4( cos 120◦ + j sin 120◦ )

If Z1 = r1 ∠θ1 and Z2 = r2 ∠θ2 then: (i) (ii)

Z1 Z2 = r1 r2 ∠(θ1 + θ2 ) and Z1 Z2

=

r1 r2

∠(θ1 − θ2 )

Problem 12. Determine, in polar form: (a) 8∠25◦ × 4∠60◦ (b) 3∠16◦ × 5∠−44◦ × 2∠80◦ (a) 8∠25◦ ×4∠60◦ = (8 ×4)∠(25◦ +60◦) = 32∠85◦ (b)

= 4 cos 120◦ + j4 sin 120◦

3∠16◦ × 5∠ −44◦ × 2∠80◦

= −2.000 + j3.464 Hence 2∠30◦ + 5∠−45◦ − 4∠120◦ = (1.732 + j1.000) + (3.536 − j3.536) − (−2.000 + j3.464) = 7.268 − j6.000, which lies in the fourth quadrant ( ) √ −1 −6.000 2 2 = [(7.268) + (6.000) ]∠ tan 7.268 = 9.425∠−39.54◦

= (3 ×5 × 2)∠[16◦ + (−44◦ )+ 80◦ ] = 30∠52◦ Problem 13. Evaluate in polar form 16∠75◦ (a) 2∠15◦

(a)

π π × 12∠ 4 2 π 6∠− 3

10∠ (b)

16∠75◦ 16 = ∠(75◦ − 15◦ ) = 8∠60◦ 2∠15◦ 2 π π ( ( )) × 12∠ 4 2 = 10 × 12 ∠ π + π − − π π 6 4 2 3 6∠− 3 13π 11π = 20∠ or 20∠− or 12 12

10∠ (b)

20∠195◦ or 20∠−165◦ Problem 14. Evaluate, in polar form 2∠30◦ +5∠−45◦ − 4∠120◦

Now try the following Practice Exercise Practice Exercise 94 on page 877) 1.

Polar form (Answers

Determine the modulus and argument of (a) 2 + j4 (b) −5 − j2 (c) j(2 − j)

In Problems 2 and 3 express the given Cartesian complex numbers in polar form, leaving answers in surd form. 2.

(a) 2 + j3 (b) −4 (c) −6 + j

3.

(a) −j3 (b) (−2 + j)3 (c) j3 (1 − j)

In Problems 4 and 5 convert the given polar complex numbers into (a + jb) form giving answers correct to 4 significant figures. 4.

(a) 5∠30◦ (b) 3∠60◦ (c) 7∠45◦

5.

(a) 6∠125◦ (b) 4∠π (c) 3.5∠−120◦

In Problems 6 to 8, evaluate in polar form.

236 Section D VR − jVC = V, from which R − jXC = Z (where XC is the 1 capacitive reactance ohms). 2πfC

6. (a) 3∠20◦ × 15∠45◦ (b) 2.4∠65◦ × 4.4∠−21◦ 7. (a) 6.4∠27◦ ÷ 2∠−15◦

Problem 15. Determine the resistance and series inductance (or capacitance) for each of the following impedances, assuming a frequency of 50 Hz:

(b) 5∠30◦ × 4∠80◦ ÷ 10∠−40◦ π π 8. (a) 4∠ + 3∠ 6 8 (b) 2∠120◦ + 5.2∠58◦ − 1.6∠−40◦

(a) (4.0 + j7.0) Ω

(b) −j20 Ω

◦

(c) 15∠−60 Ω (a) Impedance, Z = (4.0 + j7.0) Ω hence, resistance = 4.0 Ω and reactance = 7.00 Ω. Since the imaginary part is positive, the reactance is inductive,

18.8 Applications of complex numbers There are several applications of complex numbers in science and engineering, in particular in electrical alternating current theory and in mechanical vector analysis. The effect of multiplying a phasor by j is to rotate it in a positive direction (i.e. anticlockwise) on an Argand diagram through 90◦ without altering its length. Similarly, multiplying a phasor by −j rotates the phasor through −90◦ . These facts are used in a.c. theory since certain quantities in the phasor diagrams lie at 90◦ to each other. For example, in the R−L series circuit shown in Fig. 18.8(a), VL leads I by 90◦ (i.e. I lags VL by 90◦ ) and may be written as jVL , the vertical axis being regarded as the imaginary axis of an Argand diagram. Thus VR + jVL = V and since VR = IR, V = IXL (where XL is the inductive reactance, 2πfL ohms) and V = IZ (where Z is the impedance) then R + jXL = Z R

I

VR

VL

I

V Phasor diagram VL

L= (b)

7.0 XL = = 0.0223 H or 22.3 mH 2πf 2π(50)

Impedance, Z = j20, i.e. Z = (0 − j20) Ω hence resistance = 0 and reactance = 20 Ω. Since the imaginary part is negative, the reactance is cap1 acitive, i.e., XC = 20 Ω and since XC = then: 2πf C capacitance, C = =

VR I (a)

106 µF = 159.2 µF 2π(50)(20)

(c) Impedance, Z

VR

VC

Hence resistance = 7.50 Ω and capacitive reactance, XC = 12.99 Ω

VR

I

Since XC =

1 then capacitance, 2πf C

f

C u

1 1 = F 2πf XC 2π(50)(20)

= 7.50 − j12.99 Ω

V Phasor diagram

V

Since XL = 2πf L then inductance,

= 15∠−60◦ = 15[ cos (−60◦ ) + j sin (−60◦ )]

C

R

L

i.e. XL = 7.0 Ω

VC V

=

1 106 = µF 2πf XC 2π(50)(12.99)

= 245 µF

(b)

Figure 18.8

Similarly, for the R−C circuit shown in Fig. 18.8(b), VC lags I by 90◦ (i.e. I leads VC by 90◦ ) and

Problem 16. An alternating voltage of 240 V, 50 Hz is connected across an impedance of (60 − j100) Ω. Determine (a) the resistance, (b) the capacitance, (c) the magnitude of the impedance and its phase angle and (d) the current flowing.

Complex numbers 237 (a)

(b)

1 1 1 1 = + + Z Z1 Z2 Z3

Impedance Z = (60 − j100) Ω. Hence resistance = 60 Ω

where Z1 = 4 + j3, Z2 = 10 and Z3 = 12 − j5

Capacitive reactance XC = 100 Ω and since 1 XC = then 2πf C

Admittance, Y1 =

capacitance, C =

1 1 = Z1 4 + j3 1 4 − j3 4 − j3 = × = 4 + j3 4 − j3 42 + 32

1 1 = 2πfXC 2π(50)(100)

= 0.160 − j0.120 siemens

106 = µF 2π(50)(100)

Admittance, Y2 =

1 1 = = 0.10 siemens Z2 10

Admittance, Y3 =

1 1 = Z3 12 − j5

= 31.83 µF (c)

Magnitude of impedance, √ |Z| = [(60)2 + (−100)2 ] = 116.6 Ω ( Phase angle, arg Z = tan−1

−100 60

Current flowing, I =

12 + j5 12 + j5 1 × = 12 − j5 12 + j5 122 + 52

= −59.04◦ = 0.0710 + j0.0296 siemens ◦

(d)

=

)

V 240∠0 = Z 116.6∠−59.04◦

Total admittance,

Y = Y1 + Y2 + Y3 = (0.160 − j0.120) + (0.10)

= 2.058 ∠ 59.04◦ A

+ (0.0710 + j0.0296) = 0.331 − j0.0904

The circuit and phasor diagrams are as shown in Fig. 18.8(b).

Problem 17. For the parallel circuit shown in Fig. 18.9, determine the value of current I and its phase relative to the 240 V supply, using complex numbers. R1 5 4 V

XL 5 3 V

R2 5 10 V

R3 5 12 V

I

XC 5 5 V

240 V, 50 Hz

Figure 18.9

V Current I = . Impedance Z for the three-branch paralZ lel circuit is given by:

= 0.343∠−15.28◦ siemens Current I =

V = VY Z

= (240∠0◦ )(0.343∠−15.28◦ ) = 82.32 ∠ −15.28◦ A Problem 18. Determine the magnitude and direction of the resultant of the three coplanar forces given below, when they act at a point. Force A, 10 N acting at 45◦ from the positive horizontal axis. Force B, 8 N acting at 120◦ from the positive horizontal axis. Force C, 15 N acting at 210◦ from the positive horizontal axis. The space diagram is shown in Fig. 18.10. The forces may be written as complex numbers. Thus force A, fA = 10∠45◦ , force B, fB = 8∠120◦ and force C, fC = 15∠210◦

238 Section D 10 N

8N 2108

3.

If the two impedances in Problem 2 are connected in parallel determine the current flowing and its phase relative to the 120 V supply voltage.

4.

A series circuit consists of a 12 Ω resistor, a coil of inductance 0.10 H and a capacitance of 160 µF. Calculate the current flowing and its phase relative to the supply voltage of 240 V, 50 Hz. Determine also the power factor of the circuit (where power factor = cos ϕ).

5.

For the circuit shown in Fig. 18.11, determine the current I flowing and its phase relative to the applied voltage.

1208

458

15 N

Figure 18.10

The resultant force = fA + fB + fC = 10∠45◦ + 8∠120◦ + 15∠210◦ = 10(cos 45◦ + j sin 45◦ ) + 8(cos 120◦

XC 5 20 V

R1 5 30 V

R2 5 40 V

XL 5 50 V

+ j sin 120◦ ) + 15(cos 210◦ + j sin 210◦ ) = (7.071 + j7.071) + (−4.00 + j6.928) +(−12.99 − j7.50) R3 5 25 V

= −9.919 + j6.499 Magnitude of resultant force √ = [(−9.919)2 + (6.499)2 ] = 11.86 N Direction of resultant force ( ) 6.499 −1 = tan = 146.77◦ −9.919 (since −9.919 + j6.499 lies in the second quadrant).

I

V 5 200 V

Figure 18.11

6.

Determine, using complex numbers, the magnitude and direction of the resultant of the coplanar forces given below, which are acting at a point. Force A, 5 N acting horizontally, Force B, 9 N acting at an angle of 135◦ to force A, Force C, 12 N acting at an angle of 240◦ to force A.

7.

A delta-connected impedance ZA is given by: Z1 Z2 + Z2 Z3 + Z3 Z1 ZA = Z2 Determine ZA in both Cartesian and polar form given Z1 = (10 + j0) Ω, Z2 = (0 − j10) Ω and Z3 = (10 + j10) Ω.

8.

In the hydrogen atom, the angular momentum p of ( the)de Broglie wave is given by: jh pψ = − (±jmψ). Determine an expres2π sion for p.

Now try the following Practice Exercise Practice Exercise 95 Applications of complex numbers (Answers on page 877) 1.

Determine the resistance R and series inductance L (or capacitance C) for each of the following impedances assuming the frequency to be 50 Hz. (a) (3 + j8) Ω (b) (2 − j3) Ω (c) j14 Ω (d) 8∠−60◦ Ω

2.

Two impedances, Z1 = (3 + j6) Ω and Z2 = (4 − j3) Ω are connected in series to a supply voltage of 120 V. Determine the magnitude of the current and its phase angle relative to the voltage.

Complex numbers 239

9. An aircraft P flying at a constant height has a velocity of (400 + j300) km/h. Another aircraft Q at the same height has a velocity of (200 − j600) km/h. Determine (a) the velocity of P relative to Q, and (b) the velocity of Q relative to P. Express the answers in polar form, correct to the nearest km/h. 10. Three vectors are represented by P, 2∠30◦ , Q, 3∠90◦ and R, 4∠−60◦ . Determine in polar form the vectors represented by (a) P + Q + R, (b) P − Q − R 11. In a Schering bridge circuit, ZX = (RX − jXCX ), Z2 = −jXC2 , (R3 )(− jXC3 ) Z3 = and Z4 = R4 (R3 − jXC3 ) 1 where XC = 2πf C

3. If x2 + 1 = 0, then the value of x is: (a) −1 (b) j (c) ±j (d) −j 4. In a + jb form, (6 + j2) − (3 − j2) is equal to: (a) 3 (b) 9 + j4 (c) 9 (d) 3 + j4 5. The product (j2)(j3) is: (a) −6 (b) j6 (c) −j6 (d) 6 ( )3 6. j2 is equivalent to: (a) j6 (b) j8 (c) −j6 (d) −j8 7. (−4 − j3) in polar form is: (a) 5∠ − 143.13◦ (b) 5∠126.87◦ (c) 5∠143.13◦ (d) 5∠ − 126.87◦ 8. (1 + j)4 is equivalent to: (a) 4 (b) −j4 (c) −4

At balance: (ZX )(Z3 ) = (Z2 )(Z4 ). C 3 R4 Show that at balance RX = C2 C 2 R3 CX = R4

2. Which of the following√statements is true? (a) j2 = 1 (b) j3 = −1 4 (c) j = −1 (d) j5 = j

and

12. An amplifier has a transfer function T given 500 by: T = where ω is the 1 + jω(5 × 10−4 ) angular frequency. The gain of the amplifier is given by the modulus of T and the phase is given by the argument of T. If ω = 2000 rad/s, determine the gain and the phase (in degrees).

(d) j4

9. The product of (2 − j3) and (5 + j4) is: (a) −2 + j7 (b) 22 − j7 (c) −2 + j7 (d) 22 + j7 10. j13 is equivalent to: (a) j (b) −1 (c) −j ( )2 11. 3 − j5 is equal to: (a) 34 (b) −16 − j30 (c) 16 (d) 6 + j30

(d) 1

12. The product √ of (4 + j) and (4 − j) is: (a) 16 (b) 17 (c) 17 (d) 15

Practice Exercise 96 Multiple-choice questions on complex numbers (Answers on page 877)

3 − j4 is equivalent to: 4−j 11 − j7 16 − j13 (a) (b) 15 17 8 − j7 8 − j7 (c) (d) 17 15 20 14. In Cartesian form, is equivalent to: 3+j 15 5 (a) −j (b) 6 + j2 2 2 15 5 (c) +j (d) 6 − j2 2 2

Each question has only one correct answer

15.

13. The sending end current of a transmission VS line is given by: IS = tanh PL. Calculate Z0 the value of the sending current, in polar form, given VS = 200V, Z0 = 560 + j420 Ω, P = 0.20 and L = 10

1.

5 is equivalent to: J6 (a) j5 (b) −5 (c) −j5

(d) 5

13. In Cartesian form,

(1 − j)2 ( )3 is equal to: −j (a) −2 (b) −j2 (c) 2 − j2

(d) j2

240 Section D

16.

17.

18.

1 is equivalent to: 2+j 2 1 2 1 (a) − j (b) − j 5 5 3 3 5 2 1 15 +j (d) + j (c) 2 2 3 3 ( )3 The expression −1 + j is equivalent to: (a) −1 − j (b) 2 + j2 (c) −3j (d) −2 − j2 In Cartesian form,

19.

20.

The expression (2 − j3)2 is equivalent to: (a) −5 + j0 (b) 13 − j12 (c) −5 − j12 (d) 13 + j0 π π 2∠ + 3∠ in polar form is: 3 6 π (a) 5∠ (b) 4.84∠0.84 2 (c) 6∠0.55 (d) 4.84∠0.73

The total impedance, ZT ohms, in an elecZ1 × Z2 trical circuit is given by ZT = ( ) ( Z1 +)Z2 When Z1 = 1 + j3 Ω and Z2 = 1 − j3 Ω, the total impedance is: (a) 2 Ω (b) 1.6 Ω (c) 5 Ω (d) 0.625 Ω

With a calculator such as the CASIO 991ES PLUS it is possible, using the complex mode, to achieve many of the calculations in this chapter much more quickly.

For fully worked solutions to each of the problems in Practice Exercises 91 to 95 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 19

De Moivre’s theorem Why it is important to understand: De Moivre’s theorem There are many, many examples of the use of complex numbers in engineering and science. De Moivre’s theorem has several uses, including finding powers and roots of complex numbers, solving polynomial equations, calculating trigonometric identities, and for evaluating the sums of trigonometric series. The theorem is also used to calculate exponential and logarithmic functions of complex numbers. De Moivre’s theorem has applications in electrical engineering and physics.

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

state de Moivre’s theorem calculate powers of complex numbers calculate roots of complex numbers state the exponential form of a complex number convert Cartesian/polar form into exponential form and vice-versa. determine loci in the complex plane

242 Section D 19.1

Introduction

(a) [2∠35◦ ]5 = 25 ∠(5 × 35◦ ), from de Moivre’s theorem

From multiplication of complex numbers in polar form, (r∠θ) × (r∠θ) = r 2 ∠2θ

(b)

Similarly, (r∠θ) ×(r∠θ) ×(r∠θ) = r 3∠3θ, and so on. In general, de Moivre’s theorem∗ [r∠θ]n = r n∠nθ The theorem is true for all positive, negative and fractional values of n. The theorem is used to determine powers and roots of complex numbers.

19.2

= 32∠175◦ √ 3 (−2 + j3) = [(−2)2 + (3)2 ]∠ tan−1 −2 √ ◦ = 13∠123.69 , since −2 + j3 lies in the second quadrant √ (−2 + j3)6 = [ 13∠123.69◦ ]6 √ = ( 13)6 ∠(6 × 123.69◦ ), by de Moivre’s theorem = 2197∠742.14◦ = 2197∠382.14◦ (since 742.14

Powers of complex numbers

≡ 742.14◦ − 360◦ = 382.14◦ )

For example [3∠20◦ ]4 = 34 ∠(4 × 20◦ ) = 81∠80◦ by de Moivre’s theorem.

= 2197∠22.14◦ (since 382.14◦ ≡ 382.14◦ − 360◦ = 22.14◦ )

Problem 1. Determine, in polar form (a) [2∠35◦ ]5 (b) (−2 + j3)6

or 2197∠22◦ 8′ Problem 2. Determine the value of (−7 + j5)4 , expressing the result in polar and rectangular forms.

(−7 + j5) = =

√ √

[(−7)2 + 52 ]∠ tan−1

5 −7

74∠144.46◦

(Note, by considering the Argand diagram, −7 + j5 must represent an angle in the second quadrant and not in the fourth quadrant.) Applying de Moivre’s theorem: √ (−7 + j5)4 = [ 74∠144.46◦ ]4 √ = 744 ∠4 × 144.46◦ = 5476∠577.84◦ = 5476∠217.84◦ or 5476∠217◦ 50′ in polar form Since r∠θ = r cos θ + jr sin θ, 5476∠217.84◦ = 5476 cos 217.84◦ + j5476 sin 217.84◦

∗

Who was de Moivre? Abraham de Moivre (26 May 1667–27 November 1754) was a French mathematician famous for de Moivre’s formula, which links complex numbers and trigonometry, and for his work on the normal distribution and probability theory. To find out more go to www.routledge.com/cw/bird

= −4325 − j3359 i.e.

(−7 + j5) = −4325 − j3359 in rectangular form 4

De Moivre’s theorem 243 Now try the following Practice Exercise Practice Exercise 97 Powers of complex numbers (Answers on page 877)

Hence √ √ √ (5 + j12) = [13∠67.38◦ ] and [13∠427.38◦ ] 1

1

= [13∠67.38◦ ] 2 and [13∠427.38◦ ] 2

1. Determine in polar form (a) [1.5∠15◦ ]5 (b) (1 + j2)6

=

1 13 2 ∠

2. Determine in polar and Cartesian forms (a) [3∠41◦ ]4 (b) (−2 − j)5

(

) 1 ◦ × 67.38 and 2 1

(

13 2 ∠

1 × 427.38◦ 2

)

3. Convert (3 − j) into polar form and hence evaluate (3 − j)7 , giving the answer in polar form.

=

In problems 4 to 7, express in both polar and rectangular forms.

= 3.61∠33.69◦ and 3.61∠213.69◦

4. (6 + j5)3 5. (3 − j8)5 6. (−2 + j7)4 7. (−16 − j9)6

√ √ 13∠33.69◦ and 13∠213.69◦

Thus, in polar form, the two roots are 3.61∠33.69◦ and 3.61∠−146.31◦ √ √ 13∠33.69◦ = 13(cos 33.69◦ + j sin 33.69◦ ) = 3.0 + j2.0 √ √ 13∠213.69◦ = 13(cos 213.69◦ + j sin 213.69◦ ) = −3.0 − j2.0

19.3

Roots of complex numbers

The square root of a complex number is determined by letting n = 1/2 in de Moivre’s theorem, i.e.

√ 1 1 1 √ θ [r∠θ] = [r∠θ] 2 = r 2 ∠ θ = r∠ 2 2

Thus, in cartesian form the two roots are ±(3.0 + j2.0) From the Argand diagram shown in Fig. 19.1 the two roots are seen to be 180◦ apart, which is always true when finding square roots of complex numbers. Imaginary axis j2

There are two square roots of a real number, equal in size but opposite in sign. Problem 3. Determine the two square roots of the complex number (5 + j12) in polar and Cartesian forms and show the roots on an Argand diagram.

(5 + j12) =

( ) √ 12 [52 + 122 ]∠ tan−1 5

= 13∠67.38◦ When determining square roots two solutions result. To obtain the second solution one way is to express 13∠67.38◦ also as 13∠(67.38◦ + 360◦ ), i.e. 13∠427.38◦ . When the angle is divided by 2 an angle less than 360◦ is obtained.

3.61 213.698

33. 698 3

23

Real axis

3.61 2j 2

Figure 19.1

In general, when finding the nth root of a complex number, there are n solutions. For example, there are three solutions to a cube root, five solutions to a fifth root, and so on. In the solutions to the roots of a complex number, the modulus, r, is always the same, but the arguments, θ, are different. It is shown in Problem 3 that arguments are symmetrically spaced on an Argand

244 Section D diagram and are (360/n)◦ apart, where n is the number of the roots required. Thus if one of the solutions to the cube root of a complex number is, say, 5∠20◦ , the other two roots are symmetrically spaced (360/3)◦ , i.e. 120◦ from this root and the three roots are 5∠20◦ , 5∠140◦ and 5∠260◦

(−14 + j3) = −2 (−14 + j3) 5

√ 205∠167.905◦ √

=

205

−2 5 ∠

[( −

2 5

)

× 167.905◦

]

= 0.3449∠−67.162◦

1

or 0.3449∠−67◦ 10′

Problem 4. Find the roots of [(5 + j3)] 2 in rectangular form, correct to 4 significant figures.

There are five roots to this complex number, √ (5 + j3) = 34∠30.96◦

(

−2 x 5

Applying de Moivre’s theorem: √

1 (5 + j3) 2

=

1 34 2 ∠ 12

× 30.96◦

= 2.415∠15.48◦ or 2.415∠15◦ 29′ The second root may be obtained as shown above, i.e. having the same modulus but displaced (360/2)◦ from the first root. 1

Thus, (5 + j3) 2 = 2.415∠(15.48◦ + 180◦ ) = 2.415∠195.48◦ In rectangular form: 2.415∠15.48◦ = 2.415 cos 15.48◦ + j2.415 sin 15.48◦ = 2.327 + j0.6446 and

2.415∠195.48◦ = 2.415 cos 195.48◦ + j2.415 sin 195.48◦ = −2.327 − j0.6446

Hence

2.415∠195.48◦ or ± (2.327 + j0.6446)

Problem 5.

Express the roots of

(−14 +

in polar form.

−2 j3) 5

1 = 2 =√ 5 2 x x5

The roots are symmetrically displaced from one another (360/5)◦ , i.e. 72◦ apart round an Argand diagram. Thus the required roots are 0.3449∠−67◦ 10′ , 0.3449∠4◦ 50′ , 0.3449∠76◦ 50′ , 0.3449∠148◦ 50′ and 0.3449∠220◦ 50′

Now try the following Practice Exercise Practice Exercise 98 The roots of complex numbers (Answers on page 877) In Problems 1 to 3 determine the two square roots of the given complex numbers in Cartesian form and show the results on an Argand diagram. 1.

(a) 1 + j

2.

(a) 3 − j4 (b) −1 − j2

3.

(a) 7∠60◦ (b) 12∠

(b) j

3π 2

In Problems 4 to 7, determine the moduli and arguments of the complex roots. 1

4.

(3 + j4) 3

5.

(−2 + j) 4

6.

(−6 − j5) 2

7.

(4 − j3)

8.

For a transmission line, the characteristic impedance Z0 and the propagation coefficient

1

[(5 + j3)] 2 = 2.415∠15.48◦ and

) 1

1 1

−2 3

De Moivre’s theorem 245

γ are given by: √( Z0 = γ=

R + jωL G + jωC

However, from equations (2) and (3): ( ) θ2 θ4 1 − + − · · · = cos θ 2! 4! ( ) θ3 θ5 and θ − + − · · · = sin θ 3! 5!

) and

√ [(R + jωL)(G + jωC)]

Given R = 25 Ω, L = 5 × 10−3 H, G = 80 × 10−6 siemens, C = 0.04 × 10−6 F and ω = 2000 π rad/s, determine, in polar form, Z0 and γ

e jθ = cos θ + j sin θ

Thus

(4)

This is known as Euler’s formula and is referred to again in Chapter 65. Writing −θ for θ in equation (4), gives: e j(−θ) = cos(−θ) + j sin(−θ) However, cos(−θ) = cos θ and sin(−θ) = −sin θ

19.4 The exponential form of a complex number

e −jθ = cos θ − j sin θ

Thus

Certain mathematical functions may be expressed as power series (for example, by Maclaurin’s series – see Chapter 37), three examples being: x2 x3 x4 x5 + + + + ··· 2! 3! 4! 5!

(i)

ex = 1 + x +

(ii)

sin x = x −

x3 x5 x7 + − + ··· 3! 5! 7!

(2)

(iii)

cos x = 1 −

x2 x4 x6 + − + ··· 2! 4! 6!

(1)

(5)

The polar form of a complex number z is: z = r(cos θ + j sin θ). But, from equation (4), cos θ + j sin θ = e jθ Therefore

z = re jθ

When a complex number is written in this way, it is said to be expressed in exponential form. There are therefore three ways of expressing a complex number: (i)

z = (a + jb), called rectangular form,

(3)

(ii)

z = r(cos θ + j sin θ) or r∠θ, called polar form, and

Replacing x in equation (1) by the imaginary number jθ gives:

(iii)

z = re jθ called exponential form.

e jθ = 1+jθ+

2

3

4

5

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

= 1 + jθ +

j2 θ 2 j3 θ 3 j4 θ 4 j5 θ 5 + + + + ··· 2! 3! 4! 5!

√ By definition, j = (−1), hence j2 = −1, j3 = −j, j4 = 1, j5 = j, and so on. θ2 θ3 θ4 θ5 Thus e jθ = 1 + jθ − − j + + j − · · · 2! 3! 4! 5! Grouping real and imaginary terms gives: e

jθ

( ) θ2 θ4 = 1 − + − ··· 2! 4! ( ) θ3 θ5 + j θ − + − ··· 3! 5!

Cartesian

or

The exponential form is obtained from the polar form. π For example, 4∠30◦ becomes 4e j 6 in exponential form. (Note that in re jθ , θ must be in radians.) Problem 6. Change (3 − j4) into (a) polar form, (b) exponential form. (a) (3 − j4) = 5∠−53.13◦ or 5∠−0.927 in polar form (b)

(3 − j4) = 5∠−0.927 = 5e−j0.927 in exponential form

Problem 7. form.

Convert 7.2e j1.5 into rectangular

7.2e j1.5 = 7.2∠1.5 rad(= 7.2∠85.94◦ ) in polar form = 7.2 cos 1.5 + j7.2 sin 1.5 = (0.509 + j 7.182) in rectangular form

246 Section D = 2.0986 − j1.0000

π

Express z = 2e1+j 3 in Cartesian

Problem 8. form.

( π) z = (2e1 ) e j 3 by the laws of indices π (or 2e∠60◦ )in polar form 3 ( π π) = 2e cos + j sin 3 3 = (2e1 )∠

= 2.325∠−25.48◦ or 2.325∠−0.445 Problem 12. Determine, in polar form, ln (3 + j4) ln(3 + j4) = ln[5∠0.927] = ln[5e j0.927 ] = ln 5 + ln(e j0.927 ) = ln 5 + j0.927 = 1.609 + j0.927

= (2.718 + j4.708) in Cartesian form

= 1.857∠29.95◦ or 1.857∠0.523 Change 6e2−j3 into (a + jb) form.

Problem 9.

6e2−j3 = (6e2 )(e−j3 ) by the laws of indices = 6e2 ∠−3 rad (or 6e2 ∠−171.89◦ ) in polar form

Practice Exercise 99 The exponential form of complex numbers (Answers on page 878)

= 6e2 [cos (−3) + j sin (−3)]

1.

Change (5 + j3) into exponential form.

= (−43.89 − j6.26) in (a + jb) form

2.

Convert (−2.5 + j4.2) into exponential form.

3.

Change 3.6e j2 into Cartesian form.

4.

Express 2e3+j 6 in (a + jb) form.

5.

Convert 1.7e1.2−j2.5 into rectangular form.

6.

If z = 7e j2.1 , determine ln z (a) in Cartesian form, and (b) in polar form.

7.

Given z = 4e1.5−j2 , determine ln z in polar form.

8.

Determine in polar form (a) ln (2 + j5) (b) ln (−4 − j3)

9.

When displaced electrons oscillate about an equilibrium position the displacement x is given by the equation:   √  2 

Problem 10. If z = 4e j1.3 , determine ln z (a) in Cartesian form, and (b) in polar form. If

z = re jθ then ln z = ln(re jθ ) = ln r + ln e jθ ln z = ln r + jθ,

i.e.

by the laws of logarithms (a) Thus if z = 4e

j1.3

(1.386 + j1.300) = 1.90∠43.17◦ or 1.90∠0.753 in polar form.

Problem 11. Given z = 3e1−j , find ln z in polar form. If

1−j

z = 3e

, then

π

j1.3

then ln z = ln(4e ) = ln 4 + j1.3 (or 1.386 + j1.300) in Cartesian form.

(b)

Now try the following Practice Exercise



x = Ae

ht − 2m + j

(4mf −h ) t 2m−a 

Determine the real part of x in terms of t, assuming (4mf − h2 ) is positive.

ln z = ln(3e1−j ) = ln 3 + ln e1−j = ln 3 + 1 − j = (1 + ln 3) − j

19.5

Introduction to locus problems

The locus is a set of points that satisfy a certain condition. For example, the locus of points that are, say, 2 units from point C, refers to the set of all points that are

De Moivre’s theorem 247 y

2 units from C; this would be a circle with centre C as shown in Fig. 19.2.

locus |z| = 4 ie x 2 + y 2 = 42 4

z

2 C

0

This circle is the locus of points 2 units from C

x

Figure 19.2

It is sometimes needed to find the locus of a point which moves in the Argand diagram according to some stated condition. Loci (the plural of locus) are illustrated by the following worked problems. Problem 13. Determine the locus defined by |z| = 4, given that z = x + jy

Figure 19.4

Hence, in this example, (y) π y π = i.e. = tan = tan 45◦ = 1 tan−1 x 4 x 4 y Thus, if = 1, then y = x x

If z = x + jy, then on an Argand diagram as shown in Figure 19.3, the modulus z, √ |z| = x2 + y2

y

p arg z = — 4 ie y = x

Imaginary axis

p — 4 jy

0

z u 0

x

y Real axis

x

Figure 19.5

Hence, the locus (or path) of arg z = Figure 19.3

In this case,

π

is a straight 4 line y = x (with y > 0) as shown in Fig. 19.5. √ x2 + y2 = 4 from which, x2 + y2 = 42 2

2

2

From Chapter 10, x + y = 4 is a circle, with centre at the origin and with radius 4 The locus (or path) of |z| = 4 is shown in Fig. 19.4. Problem 14. Determine the locus defined by arg π z = , given that z = x + jy 4 ( ) y In Fig. 19.3 above, θ = tan−1 x where θ is called ( )the argument and is written as y arg z = tan−1 x

Problem 15. If z = x + jy, determine the locus π defined by arg (z − 1) = 6 If arg (z − 1) = i.e.

π π , then arg (x + jy − 1) = 6 6 arg[(x − 1) + jy] =

π 6

In Fig. 19.3, (y) (y) θ = tan−1 i.e. arg z = tan−1 x x Hence, in this example, ( ) y π y π 1 −1 tan = i.e. = tan = tan 30◦ = √ x−1 6 x−1 6 3

248 Section D y 1 1 = √ , then y = √ (x − 1) x−1 3 3 π Hence, the locus of arg (z −1) = is a straight line 6 1 1 y = √ x− √ 3 3 Thus,

if

Problem 16. Determine the locus defined by | z − 2 | = 3, given that z = x + jy If

z = x + jy,

then

| z − 2 | = | x + jy − 2 | = | (x − 2) + jy | = 3

On the √ Argand diagram shown in Figure 19.3, | z | = x2 + y2 √ Hence, in this case, | z − 2 | = (x − 2)2 + y2 = 3 from which, (x − 2)2 + y2 = 32 From Chapter 10, (x − a)2 + (y − b)2 = r 2 is a circle, with centre (a, b) and radius r. Hence, (x − 2)2 + y2 = 32 is a circle, with centre (2, 0) and radius 3 The locus of | z − 2 | = 3 is shown in Figure 19.6. y 3 locus |z – 2| = 3 ie (x – 2)2 + y 2 = 32 –1

0

2

5

x

–3

Figure 19.6

Problem 17. If z = x + jy, determine the locus z−1 =3 defined by z+1 z − 1 = x + jy − 1 = (x − 1) + jy z + 1 = x + jy + 1 = (x + 1) + jy z − 1 (x − 1) + jy Hence, z + 1 = (x + 1) + jy √ (x − 1)2 + y2 =√ =3 (x + 1)2 + y2

from which, (x − 1)2 + y2 = 9[(x + 1)2 + y2 ] x2 − 2x + 1 + y2 = 9[x2 + 2x + 1 + y2 ] x2 − 2x + 1 + y2 = 9x2 + 18x + 9 + 9y2 0 = 8x2 + 20x + 8 + 8y2 i.e. 8x2 + 20x + 8 + 8y2 = 0 and dividing by 4 gives: 2x2 + 5x + 2 + 2y2 = 0 which is the equation of the locus. 5 x2 + x + y2 = −1 2 Completing the square gives: ( )2 5 25 x+ − + y2 = −1 4 16 )2 ( 25 5 + y2 = −1 + i.e. x+ 4 16 ( )2 5 9 i.e. x+ + y2 = 4 16 ( )2 ( )2 5 3 2 i.e. x+ +y = which is the 4 4 equation of a circle. z−1 = 3 is a circle of Hence the locus defined by z + 1 ) ( 3 5 centre − , 0 and radius 4 4 Rearranging gives:

Problem 18. (If z = ) x + jy, determine the locus z+1 π defined by arg = z 4 (

z+1 z

)

( =

)

=

[(x + 1) + jy](x − jy) (x + jy)(x − jy)

=

x(x + 1) − j(x + 1)y + jxy + y2 x2 + y2

=

x2 + x − jxy − jy + jxy + y2 x2 + y2

and squaring both sides gives: (x − 1)2 + y2 =9 (x + 1)2 + y2

(x + 1) + jy x + jy

De Moivre’s theorem 249 x2 + x − jy + y2 x2 + y2 ( 2 ) x + x + y2 − jy = x2 + y2

Now try the following Practice Exercise

=

=

Practice Exercise 100 Locus problems (Answers on page 878)

x2 + x + y2 jy − 2 x2 + y2 x + y2 

 −y ( )  x2 + y2  π z+1 π  Since arg = then tan−1  x2 + x + y2  = 4 z 4 x2 + y2 i.e. ( ) −y π tan−1 2 = 2 x +x+y 4 from which, x2

−y π = tan = 1 2 +x+y 4

For each of the following, if z = x + jy, (a) determine the equation of the locus, (b) sketch the locus. 1.

|z| = 2

2.

|z| = 5

3.

arg(z − 2) =

4. 5.

|z − 2| = 4

6.

|z + 3| = 5 z+1 z−1 = 3 z−1 √ z = 2 z−1 π = arg z 4 z+2 π = arg z 4

7.

Hence, −y = x2 + x + y2 Hence, the locus defined by arg x2 + x + y + y2 = 0

(

z+1 z

) =

π 4

8. is:

Completing the square gives: ( )2 ( )2 1 1 1 x+ + y+ = 2 2 2 ( ) 1 1 1 which is a circle, centre − , − and radius √ 2 2 2

π 3 π arg(z + 1) = 6

9. 10. 11.

|z + j| = |z + 2|

12.

| z − 4 | = | z − 2j |

13.

|z − 1| = |z|

Problem 19. Determine the locus defined by | z − j | = | z − 3 |, given that z = x + jy Since | z − j | = | z − 3 | then and

| x + j(y − 1) | = | (x − 3) + jy | √ √ x2 + (y − 1)2 = (x − 3)2 + y2

Squaring both sides gives: x2 + (y − 1)2 = (x − 3)2 + y2 2 2 i.e. x + y − 2y + 1 = x2 − 6x + 9 + y2 from which, i.e. or

−2y + 1 = −6x + 9

Practice Exercise 101 Multiple-choice questions on De Moivre’s theorem (Answers on page 878) Each question has only one correct answer 1.

The two square roots of (−3 + j4) are: (a) ±(1 + j2) (b) ±(0.71 + j2.12) (c) ±(1 − j2) (d) ±(0.71 − j2.12)

2.

[ 2∠30◦ ] in Cartesian form is: (a) (0.50 + j0.06) (b) (−8 + j13.86) (c) (−4 + j6.93) (d) (13.86 + j8)

3.

The first square root of the complex number (6 − j8), correct to 2 decimal places, is:

6x − 8 = 2y y = 3x − 4

Hence, the locus defined by | z − j | = | z − 3 | is a straight line: y = 3x − 4

4

250 Section D (a) 3.16∠ − 7.29◦ (c) 3.16∠ − 26.57◦

(b) 5∠26.57◦ (d) 5∠ − 26.57◦

6.

( )− 2 4. The first of the five roots of −4 + j3 5 is: (a) 1.90∠ − 57.25◦ (b) 0.525∠ − 57.25◦ (c) 1.90∠ − 21.25◦ (d) 0.525∠ − 9.75◦

(4 − j3) in exponential form is: (a) 5e j0.644 (b) 5∠0.643 (c) 5e −j0.644 (d) 5∠ − 0.643

7.

( )2 5. The first of the three roots of 5 − j12 3 is: (a) 6.61∠44.92◦ (b) 5.53∠ − 44.92◦ ◦ (c) 6.61∠ − 44.92 (d) 5.53∠44.92◦

Changing 5e 1−j4 into (a + jb) form gives: (a) 13.59 − j54.37 (b) 8.89 + j10.29 (c) 13.59 − j0.948 (d) −8.89 + j10.29

8.

Given z = 4e 2−j3 , then ln z, in polar form in radians, is: (a) 4.524∠ − 0.725 (b) 3.563∠ − 1.001 (c) 14.422∠ − 0.983 (d) 3.305∠ − 1.138

For fully worked solutions to each of the problems in Practice Exercises 97 to 100 in this chapter, go to the website: www.routledge.com/cw/bird

Section E

Matrices and determinants

Chapter 20

The theory of matrices and determinants Why it is important to understand: The theory of matrices and determinants Matrices are used to solve problems in electronics, optics, quantum mechanics, statics, robotics, linear programming, optimisation, genetics and much more. Matrix calculus is a mathematical tool used in connection with linear equations, linear transformations, systems of differential equations and so on, and is vital for calculating forces, vectors, tensions, masses, loads and a lot of other factors that must be accounted for in engineering to ensure safe and resource-efficient structure. Electrical and mechanical engineers, chemists, biologists and scientists all need knowledge of matrices to solve problems. In computer graphics, matrices are used to project a three-dimensional image on to a two-dimensional screen, and to create realistic motion. Matrices are therefore very important in solving engineering problems.

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

understand matrix notation add, subtract and multiply 2 by 2 and 3 by 3 matrices recognise the unit matrix calculate the determinant of a 2 by 2 matrix determine the inverse (or reciprocal) of a 2 by 2 matrix calculate the determinant of a 3 by 3 matrix determine the inverse (or reciprocal) of a 3 by 3 matrix

20.1

Matrix notation

Matrices and determinants are mainly used for the solution of linear simultaneous equations. The theory of matrices and determinants is dealt with in this chapter and this theory is then used in Chapter 21 to solve simultaneous equations. The coefficients of the variables for linear simultaneous equations may be shown in matrix form.

The coefficients of x and y in the simultaneous equations x + 2y = 3 4x − 5y = 6 ( ) 1 2 become in matrix notation. 4 −5 Similarly, the coefficients of p, q and r in the equations

254 Section E 1.3p − 2.0q + r = 7 3.7p + 4.8q − 7r = 3 4.1p + 3.8q + 12r = −6 

1.3 become 3.7 4.1

−2.0 4.8 3.8

 1 −7 in matrix form. 12

The numbers within a matrix are called an array and the coefficients forming the array are called the elements of the matrix. The number of rows in a matrix is usually specified by m and the number of columns by n and ( a matrix ) referred to as an ‘m by n’ matrix. 2 3 6 Thus, is a ‘2 by 3’ matrix. Matrices cannot 4 5 7 be expressed as a single numerical value, but they can often be simplified or combined, and unknown element values can be determined by comparison methods. Just as there are rules for addition, subtraction, multiplication and division of numbers in arithmetic, rules for these operations can be applied to matrices and the rules of matrices are such that they obey most of those governing the algebra of numbers.

20.2 Addition, subtraction and multiplication of matrices (i) Addition of matrices Corresponding elements in two matrices may be added to form a single matrix. Problem 1. Add the matrices ( ) ( ) 2 −1 −3 0 (a) and −7 4 7 −4     3 1 −4 2 7 −5 1 and −2 1 0 (b) 4 3 1 4 −3 6 3 4 (a) Adding the corresponding elements gives: ( ) ( ) 2 −1 −3 0 + −7 4 7 −4 ( ) 2 + (−3) −1 + 0 = −7 + 7 4 + (−4) ( ) −1 −1 = 0 0

(b)

Adding the corresponding elements gives:     3 1 −4 2 7 −5 4 3 1 + −2 1 0 1 4 −3 6 3 4   3+2 1 + 7 −4 + (−5) 1+0  = 4 + (−2) 3 + 1 1+6 4 + 3 −3 + 4   5 8 −9 1 = 2 4 7 7 1

(ii) Subtraction of matrices If A is a matrix and B is another matrix, then (A − B) is a single matrix formed by subtracting the elements of B from the corresponding elements of A. Problem 2. Subtract ( ) ( ) −3 0 2 −1 (a) from 7 −4 −7 4     2 7 −5 3 1 −4 0 from 4 3 1 (b) −2 1 6 3 4 1 4 −3 To find matrix A minus matrix B, the elements of B are taken from the corresponding elements of A. Thus: ( ) ( ) 2 −1 −3 0 − (a) −7 4 7 −4 ( ) 2 − (−3) −1 − 0 = −7 − 7 4 − (−4) ( ) 5 −1 = −14 8     3 1 −4 2 7 −5 1 − −2 1 0 (b) 4 3 1 4 −3 6 3 4   3−2 1 − 7 −4 − (−5) 1−0  = 4 − (−2) 3 − 1 1−6 4 − 3 −3 − 4   1 −6 1 2 1 = 6 −5 1 −7 Problem 3. If ( ) ( ) −3 0 2 −1 A= ,B= and 7 −4 −7 4

The theory of matrices and determinants 255 (

1 C= −2

) 0 find A + B − C −4 A+B =

( ) −1 −1 0 0

(from Problem 1) ( ) ( ) −1 −1 1 0 Hence, A + B − C = − 0 0 −2 −4 ( ) −1 − 1 −1 − 0 = 0 − (−2) 0 − (−4) ( ) −2 −1 = 2 4 Alternatively A + B − C =

( −3 7

) ( 0 2 + −4 −7

) ( −1 1 − 4 −2

(

−3 + 2 − 1 0 + (−1) − 0 = 7 + (−7) − (−2) −4 + 4 − (−4) ( ) −2 −1 = as obtained previously 2 4

0 −4

)

)

Hence 2A − 3B + 4C ( ) ( ) ( ) −6 0 6 −3 4 0 = − + 14 −8 −21 12 −8 −16 ( ) −6 − 6 + 4 0 − (−3) + 0 = 14 − (−21) + (−8) −8 − 12 + (−16) ( ) −8 3 = 27 −36 When a matrix A is multiplied by another matrix B, a single matrix results in which elements are obtained from the sum of the products of the corresponding rows of A and the corresponding columns of B. Two matrices A and B may be multiplied together, provided the number of elements in the rows of matrix A are equal to the number of elements in the columns of matrix B. In general terms, when multiplying a matrix of dimensions (m by n) by a matrix of dimensions (n by r), the resulting matrix has dimensions (m by r). Thus a 2 by 3 matrix multiplied by a 3 by 1 matrix gives a matrix of dimensions 2 by 1. ( Problem 5. find A × B

If A =

) ( ) 2 3 −5 7 and B = 1 −4 −3 4 (

C11 Let A × B = C where C = C21 (iii) Multiplication When a matrix is multiplied by a number, called scalar multiplication, a single matrix results in which each element of the original matrix has been multiplied by the number. ( ) −3 0 Problem 4. If A = , 7 −4   ( ) 1 0 2 −1  find B= and C =  −7 4 −2 −4 2A − 3B + 4C

For scalar multiplication, each element is multiplied by the scalar quantity, hence ( ) ( ) −3 0 −6 0 2A = 2 = 7 −4 14 −8 ( ) ( ) 2 −1 6 −3 3B = 3 = −7 4 −21 12 ( ) ( ) 1 0 4 0 and 4C = 4 = −2 −4 −8 −16

C12 C22

)

C11 is the sum of the products of the first row elements of A and the first column elements of B taken one at a time, i.e.

C11 = (2 × (−5)) + (3 × (−3)) = −19

C12 is the sum of the products of the first row elements of A and the second column elements of B, taken one at a time, i.e.

C12 = (2 × 7) + (3 × 4) = 26

C21 is the sum of the products of the second row elements of A and the first column elements of B, taken one at a time, i.e.

C21 = (1 × (−5)) + (−4 × (−3)) = 7

Finally, C22 is the sum of the products of the second row elements of A and the second column elements of B, taken one at a time, C22 = (1 × 7) + ((−4) × 4) = −9 ( ) −19 26 Thus, A × B = 7 −9 i.e.

256 Section E Problem 6. Simplify     3 4 0 2 −2 6 −3 ×  5 7 −4 1 −1 The sum of the products of the elements of each row of the first matrix and the elements of the second matrix, (called a column matrix), are taken one at a time. Thus:     3 4 0 2 −2 6 −3 ×  5 7 −4 1 −1   (3 × 2) + (4 × 5) + (0 × (−1)) = (−2 × 2) + (6 × 5) + (−3 × (−1)) (7 × 2) + (−4 × 5) + (1 × (−1))   26 =  29 −7 

 3 4 0 6 −3 and Problem 7. If A = −2 7 −4 1   2 −5 B =  5 −6, find A × B −1 −7 The sum of the products of the elements of each row of the first matrix and the elements of each column of the second matrix are taken one at a time. Thus:     3 4 0 2 −5 −2 6 −3 ×  5 −6 7 −4 1 −1 −7   [(3 × 2) [(3 × (−5))  + (4 × 5) +(4 × (−6))     + (0 × (−1))] +(0 × (−7))]    [(−2 × 2)  [(−2 × (−5))    +(6 × (−6))  =  + (6 × 5)   + (−3 × (−1))] +(−3 × (−7))]   [(7 × 2)  [(7 × (−5))    + (−4 × 5) +(−4 × (−6))  + (1 × (−1))] +(1 × (−7))]   26 −39 −5 =  29 −7 −18

Problem 8. Determine     1 0 3 2 2 0 2 1 2 × 1 3 2 1 3 1 3 2 0 The sum of the products of the elements of each row of the first matrix and the elements of each column of the second matrix are taken one at a time. Thus:     1 0 3 2 2 0 2 1 2 × 1 3 2 1 3 1 3 2 0   [(1 × 2) [(1 × 2) [(1 × 0)  + (0 × 1) + (0 × 3) + (0 × 2)     + (3 × 3)] + (3 × 2)] + (3 × 0)]   [(2 × 2)  [(2 × 2) [(2 × 0)    + (1 × 3) + (1 × 2)  =  + (1 × 1)   + (2 × 3)] + (2 × 2)] + (2 × 0)]   [(1 × 2)  [(1 × 2) [(1 × 0)    + (3 × 1) + (3 × 3) + (3 × 2)  + (1 × 3)] + (1 × 2)] + (1 × 0)]   11 8 0 = 11 11 2 8 13 6 In algebra, the commutative law of multiplication states that a × b = b × a. For matrices, this law is only true in a few special cases, and in general A × B is not equal to B×A (

) 2 3 If A = and 1 0

Problem 9. ( ) 2 3 B= show that A × B ̸= B × A 0 1 ( ) ( ) 2 3 2 3 × 1 0 0 1 ( [(2 × 2) + (3 × 0)] = [(1 × 2) + (0 × 0)] ( ) 4 9 = 2 3 ( ) ( ) 2 3 2 3 B×A = × 0 1 1 0 ( [(2 × 2) + (3 × 1)] = [(0 × 2) + (1 × 1)] ( ) 7 6 = 1 0 A×B =

[(2 × 3) + (3 × 1)] [(1 × 3) + (0 × 1)]

[(2 × 3) + (3 × 0)] [(0 × 3) + (1 × 0)]

)

)

The theory of matrices and determinants 257 ( Since

) ( 4 9 7 ̸ = 2 3 1

) 6 , then A × B ̸= B × A 0

Problem 10. A T-network is a cascade connection of a series impedance, a shunt admittance and then another series impedance. The transmission matrix for such a network is given by: ( ) ( )( )( ) A B 1 Z1 1 0 1 Z2 = C D 0 1 Y 1 0 1 Determine expressions for A, B, C and D in terms of Z1, , Z2 and Y.



 ( ) 3.1 2.4 6.4 6 F = −1.6 3.8 −1.9 G = −2 5.3 3.4 −4.8     ( ) 4 1 0 −2 H= J = −11 K = 0 1 5 7 1 0 In Problems 1 to 12, perform the matrix operation stated. 1. A + B 2. D + E 3. A − B

( A C

) ( B 1 = D 0

Z1 1

)(

( 1 + YZ1 = Y

)( 1 0 1 Y 1 0

Z2 1

)(

)

Z1 1

1 Z2 0 1

)

( ) (1+YZ1 ) [(1+YZ1 )(Z2 )+Z1 ] = Y (YZ2 +1) ( ) (1+YZ1 ) (Z1 +Z2 +YZ1 Z2 ) = Y (1+YZ2 )

4. A + B − C 5. 5A + 6B 6. 2D + 3E − 4F 7. A × H 8. A × B 9. A × C 10.

D×J

11.

E×K

12.

D×F

13.

Show that A × C ̸= C × A

20.3

The unit matrix

Now try the following Practice Exercise Practice Exercise 102 Addition, subtraction and multiplication of matrices (Answers on page 878) In Problems 1 to 13, the matrices A to K are: ( ) ( ) 3 −1 5 2 A= B= −4 7 −1 6 ( ) −1.3 7.4 C= 2.5 −3.9 

 4 −7 6 4 0 D = −2 5 7 −4   3 6 2 E =  5 −3 7 −1 0 2

A unit matrix, I, is one in which all elements of the leading diagonal (\) have a value of 1 and all other elements have a value of 0. Multiplication of a matrix by I is the equivalent of multiplying by 1 in arithmetic.

20.4 The determinant of a 2 by 2 matrix

( ) a b The determinant of a 2 by 2 matrix is defined c d as (ad − bc) The elements of the determinant of a matrix are written Thus, the determinant ( between ) vertical lines. 3 −4 3 −4 and is equal to of is written as 1 6 1 6 (3 × 6) − (−4 × 1), i.e. 18 − (−4) or 22. Hence the

258 Section E determinant of a matrix can be expressed as a single 3 −4 = 22 numerical value, i.e. 1 6 Problem 11. Determine the value of 3 −2 7 4 3 −2 = (3 × 4) − (−2 × 7) 7 4 = 12 − (−14) = 26 + j) j2 Problem 12. Evaluate (1 − j3 (1 − j4) (1 + j) j2 − j3 (1 − j4) = (1 + j)(1 − j4) − ( j2)(− j3) = 1 − j4 + j − j2 4 + j2 6 = 1 − j4 + j − (−4) + (−6) since from Chapter18, j2 = −1 = 1 − j4 + j + 4 − 6 = −1 − j 3 5∠30◦ Problem 13. Evaluate 3∠60◦

2∠−60◦ 4∠−90◦

5∠30◦ 2∠−60◦ ◦ ◦ 3∠60◦ 4∠−90◦ = (5∠30 )(4∠−90 ) − (2∠−60◦ )(3∠60◦ ) = (20∠−60◦ ) − (6∠0◦ ) = (10 − j17.32) − (6 + j0) = (4 − j 17.32) or 17.78∠−77◦ Now try the following Practice Exercise Practice Exercise 103 2 by 2 determinants (Answers on page 878) ( ) 3 −1 1. Calculate the determinant of −4 7

2.

Calculate the ( ) determinant of −2 5 3 −6

3.

Calculate the determinant of ( ) −1.3 7.4 2.5 −3.9 j2 −j3 Evaluate (1 + j) j 2∠40◦ 5∠−20◦ Evaluate 7∠−32◦ 4∠−117◦

4. 5.

(

6.

(x − 2) 6 Given matrix A = 2 (x − 3) determine values of x for which | A | = 0

) ,

20.5 The inverse or reciprocal of a 2 by 2 matrix The inverse of matrix A is A−1 such that A × A−1 = I, the unit matrix. ( ) 1 2 Let matrix A be and let the inverse matrix, A−1 3 4 ( ) a b be c d Then, since A × A−1 = I, ( ) ( ) ( ) 1 2 a b 1 0 × = 3 4 c d 0 1 Multiplying the matrices on the left-hand side, gives ( ) ( ) a + 2c b + 2d 1 0 = 3a + 4c 3b + 4d 0 1 Equating corresponding elements gives: b + 2d = 0, i.e. b = −2d 4 and 3a + 4c = 0, i.e. a = − c 3 Substituting for a and b gives:   4 − c + 2c −2d + 2d   ( 3 1   )  ( = 0   4 3 − c + 4c 3(−2d) + 4d 3 2 i.e.

3 0



c

0 1

)

( ) = 1 0 0 1 −2d 0

The theory of matrices and determinants 259 2 3 1 showing that c = 1, i.e. c = and −2d = 1, i.e. d = − 3 2 2 4 Since b = −2d, b = 1 and since a = − c, a = −2 3 ( ) ( ) 1 2 a b Thus the inverse of matrix is that is, 3 4 c d   −2 1  3 1 − 2 2 There is, however, a quicker method of obtaining the inverse of a 2 by 2 matrix. ( ) p q For any matrix the inverse may be r s obtained by: (i)

interchanging the positions of p and s,

(ii)

changing the signs of q and r, and

(iii) multiplying this new ( matrix ) by the reciprocal of p q the determinant of r s ( ) 1 2 Thus the inverse of matrix is 3 4 ( 4 1 4 − 6 −3

 ) −2 −2 = 3 1 2

Now try the following Practice Exercise Practice Exercise 104 The inverse of 2 by 2 matrices (Answers on page 879) ( ) 3 −1 1. Determine the inverse of −4 7 

2.

3.

20.6 The determinant of a 3 by 3 matrix (i)

The minor of an element of a 3 by 3 matrix is the value of the 2 by 2 determinant obtained by covering up the row and column containing that element.   1 2 3 Thus for the matrix 4 5 6 the minor of 7 8 9 element 4 is obtained bycovering the row 1 (4 5 6) and the column 4, leaving the 2 by 7 2 3 , i.e. the minor of element 4 2 determinant 8 9 is (2 × 9) − (3 × 8) = −6

(ii)

The sign of a minor depends on itsposition within  + − + the matrix, the sign pattern being − + −. + − + Thus the signed-minor of element 4 in the matrix   1 2 3 4 5 6 is − 2 3 = −(−6) = 6 8 9 7 8 9

 1 1 − 2

as obtained previously. Problem 14. Determine the inverse of ( ) 3 −2 7 4 ) p q is obtained by interr s changing the positions of p and s, changing the signs of q and r and by the reciprocal of the multiplying p q determinant . Thus, the inverse of r s

(

The inverse of matrix

(

) ( 1 3 −2 4 = 7 4 (3 × 4) − (−2 × 7) −7  2 ( )  13 1 4 2 = =  −7 26 −7 3 26

2 3

)

 1 13   3 26

 1 2  2 3  Determine the inverse of   1 3 − − 3 5 ( ) −1.3 7.4 Determine the inverse of 2.5 −3.9

The signed-minor of an element is called the cofactor of the element. (iii)

The value of a 3 by 3 determinant is the sum of the products of the elements and their cofactors of any row or any column of the corresponding 3 by 3 matrix.

260 Section E There are thus six different ways of evaluating a 3 × 3 determinant – and all should give the same value. Problem 15. Find the value of 3 4 −1 2 0 7 1 −3 −2

1 4 −3 2 6 Using the second column: −5 −1 −4 2 −5 6 1 −3 1 −3 = −4 + 2 −(−4) −1 2 −1 2 −5 6 = −4(−10 + 6) + 2(2 − 3) + 4(6 − 15) = 16 − 2 − 36 = −22

The value of this determinant is the sum of the products of the elements and their cofactors, of any row or of any column. If the second row or second column is selected, the element 0 will make the product of the element and its cofactor zero and reduce the amount of arithmetic to be done to a minimum. Supposing a second row expansion is selected. The minor of 2 is the value of the determinant remaining when the row and column containing the 2 (i.e. the second row and the first column), is covered up. 4 −1 i.e. −11 Thus the cofactor of element 2 is −3 −2 The sign of element 2 is minus (see (ii) above), hence the cofactor of element 2, (the signed-minor) is +11 3 4 i.e. −13, Similarly the minor of element 7 is 1 −3 and its cofactor is +13. Hence the value of the sum of the products of the elements and their cofactors is 2 × 11 + 7 × 13, i.e., 3 4 −1 2 0 7 = 2(11) + 0 + 7(13) = 113 1 −3 −2 The same result will be obtained whichever row or column is selected. For example, the third column expansion is 2 3 3 4 0 4 (−1) − 7 + (−2) 1 −3 2 0 1 −3 = 6 + 91 + 16 = 113, as obtained previously. 1 Problem 16. Evaluate −5 −1

4 −3 2 6 −4 2

Problem 17. Determine the value of j2 (1 + j) 3 1 j (1 − j) 0 j4 5 Using the first column, the value of the determinant is: 1 j (1 + j) 3 (j2) − (1 − j) j4 5 j4 5 (1 + j) 3 + (0) 1 j = j2(5 − j2 4) − (1 − j)(5 + j5 − j12) + 0 = j2(9) − (1 − j)(5 − j7) = j18 − [5 − j7 − j5 + j2 7] = j18 − [−2 − j12] = j18 + 2 + j12 = 2 + j 30 or 30.07∠86.19◦

Use of calculator Evaluating a 3 by 3 determinant can also be achieved using a calculator. Here is the procedure for the determinant 6 2 −3 −4 1 5 using a CASIO 991 ES PLUS −7 0 9 calculator: 1.

Press ‘Mode’ and choose ‘6’

2.

Select Matrix A by entering ‘1’

3.

Select 3×3 by entering ‘1’

4.

Enter in the nine numbers of the above matrix in this order, following each number by pressing ‘ =′ ; i.e. 6 =, 2 =, −3 =, −4 =, 1 =, and so on

= (4 + 24) − 4(−10 + 6) − 3(20 + 2)

5.

Press ‘ON’ and a zero appears

= 28 + 16 − 66 = −22

6.

Press ‘Shift’ and ‘4’ and eight choices appear

1 4 −3 2 6 Using the first row: −5 −1 −4 2 2 6 2 − 4 −5 6 + (−3) −5 = 1 −1 2 −1 −4 −4 2

The theory of matrices and determinants 261 7.

Select choice ‘7’ and ‘det(’ appears

8.

Press ‘Shift’ and ‘4’ and choose Matrix A by entering ‘3’

9.

Press ‘=’ and the answer 35 appears

(5 − λ) 7 −5 (4 − λ) −1 = 0 (b) 0 2 8 (−3 − λ) (You may need to refer to Chapter 1, pages 11–14, for the solution of cubic equations).

Now try the following Practice Exercise Practice Exercise 105 3 by 3 determinants (Answers on page 879) 1. Find the matrix of minors of 

 4 −7 6 −2 4 0 5 7 −4 2. Find the matrix of cofactors of   4 −7 6 −2 4 0 5 7 −4 3. Calculate the determinant of   4 −7 6 −2 4 0 5 7 −4 8 −2 −10 4. Evaluate 2 −3 −2 6 3 8 5. Calculate the determinant of   3.1 2.4 6.4 −1.6 3.8 −1.9 5.3 3.4 −4.8 j2 2 j 1 −3 6. Evaluate (1 + j) 5 −j4 0 3∠60◦ j2 1 (1 + j) 2∠30◦ 7. Evaluate 0 0 2 j5 8. Find the eigenvalues λ that satisfy the following equations: (2 − λ) 2 =0 (a) −1 (5 − λ)

20.7 The inverse or reciprocal of a 3 by 3 matrix The adjoint of a matrix A is obtained by: (i)

forming a matrix B of the cofactors of A, and

(ii)

transposing matrix B to give BT , where BT is the matrix obtained by writing the rows of B as the columns of BT . Then adj A = BT

The inverse of matrix A, A−1 is given by A−1 =

adj A |A|

where adj A is the adjoint of matrix A and |A| is the determinant of matrix A. Problem 18. Determine the inverse of the matrix   3 4 −1   0 7 2 1 −3 −2 The inverse of matrix A, A−1 =

adj A |A|

The adjoint of A is found by: (i)

obtaining the matrix of the cofactors of the elements, and

(ii)

transposing this matrix.

0 7 The cofactor of element 3 is + = 21 −3 −2 2 7 = 11, and so on. The cofactor of element 4 is − 1 −2 

21 11 The matrix of cofactors is 11 −5 28 −23

 −6 13 −8

262 Section E The transpose of the matrix of cofactors, i.e. the adjoint of the matrix,  is obtained by writing the rows as  21 11 28 columns, and is  11 −5 −23 −6 13 −8 3 4 −1 0 7 From Problem 15, the determinant of 2 1 −3 −2 is 113   3 4 −1 0 7 is Hence the inverse of 2 1 −3 −2   21 11 28  11 −5 −23   21 11 28 −6 13 −8 1  11 −5 −23 or 113 113 −6 13 −8 Problem 19. Find the inverse of   1 5 −2   4  3 −1 −3 6 −7 Inverse =

adjoint determinant



−17 9 The matrix of cofactors is  23 −13 18 −10

 15 −21 −16

The transpose  of the matrix of cofactors (i.e. the −17 23 18 adjoint) is  9 −13 −10 15 −21 −16   1 5 −2 4 The determinant of  3 −1 −3 6 −7



8.5 = −4.5 −7.5

Now try the following Practice Exercise Practice Exercise 106 The inverse of a 3 by 3 matrix (Answers on page 879) 1.

Write down the transpose of   4 −7 6 −2 4 0 5 7 −4

2.

Write down the transpose of   3 6 12  5 − 2 7 3 −1 0 35

3.

Determine the adjoint of   4 −7 6 −2 4 0 5 7 −4

4.

Determine the adjoint of   3 6 12  5 − 2 7 3 −1 0 35

5.

Find the inverse of   4 −7 6 −2 4 0 5



 1 5 −2 4 Hence the inverse of  3 −1 −3 6 −7 

 −17 23 18  9 −13 −10 15 −21 −16 = −2

7 −4 

= 1(7 − 24) − 5(−21 + 12) − 2(18 − 3) = −17 + 45 − 30 = −2

 −11.5 −9 6.5 5 10.5 8

6.

3 6 Find the inverse of  5 − 23 −1 0

1 2

7 3 5

The theory of matrices and determinants 263 Practice Exercise 107 Multiple-choice questions on the theory of matrices and determinants (Answers on page 879)



10.

Each question has only one correct answer 1. The vertical lines of elements in a matrix are called: (a) rows (b) a row matrix (c) columns (d) a column matrix 2. If a matrix P has the same number of rows and columns, then matrix P is called a: (a) row matrix (b) square matrix (c) column matrix (d) rectangular matrix 3. A matrix of order m × 1 is called a: (a) row matrix (b) column matrix (c) square matrix (d) rectangular matrix j2 j is equal to: 4. j −j (a) 5 (b) 4 (c) 3 (d) 2 5. The product ( matrix ) ( ) 2 3 1 −5 is equal to: −1 4 −2 6 ( ) ( ) −13 3 −2 (a) (b) 26 −3 10 ( ) ( ) 1 −2 −4 8 (c) (d) −3 −2 −9 29 2k −1 3 0 = then the value of k 6. If 4 2 2 1 is: 2 1 3 (a) (b) 3 (c) − (d) 3 4 2 6 0 −1 1 4 is: 7. The value of 2 1 1 3 (a) −7 (b) 8 (c) 7 (d) 10 (2k + 5) 3 = 0 for k gives: 8. Solving (5k + 2) 9 17 (a) − (b) −17 (c) 17 (d) −13 11 9. If a matrix has only one row, then it is called a: (a) row matrix (b) column matrix (c) square matrix (d) rectangular matrix

11.

12.

13.

14.

15.

16.

17.

18.

 5 3 2 If P =  0 4 1  then | P | is equal to: 0 0 3 (a) 30 (b) 40 (c) 50 (d) 60 If the order of a matrix P is a × b and the order of a matrix Q is b × c then the order of P × Q is: (a) a × c (b) c × a (c) c × b (d) a × b ( ) 1 + j2 j The value of is: j −j (a) −3 − j (b) −1 − j (c) 3 −j (d) 1 − j   1 0 0  0 1 0  is an example of a: 0 0 1 (a) null matrix (b) triangular matrix (c) rectangular matrix (d) unit matrix ( ) 5 −3 The inverse of the matrix is: −2 1 ( ) ( ) −5 −3 −1 −3 (a) (b) 2 −1 −2 −5 ( ) ( ) −1 3 1 3 (c) (d) 2 −5 2 5 2 −1 4 1 5 The value of the determinant 0 6 0 −1 is: (a) −56 (b) 52 (c) 4 (d) 8 j2 −(1 + j) The value of is: (1 − j) 1 (a) −j2 (b) 2(1 + j) (c) 2 (d) −2 + j2 ( ) 2 −3 The inverse of the matrix is: 1 −4 ( ) ( ) 0.8 0.6 −4 3 (a) (b) −0.2 −0.4 −1 2 ( ) ( ) −0.4 −0.6 0.8 −0.6 (c) (d) 0.2 0.8 0.2 −0.4 ( ) 3 −2 Matrix A is given by: A = 4 −1 The determinant of matrix A is: ( ) −1 2 (a) −11 (b) 5 (c) 4 (d) −4 3

264 Section E (

) −1 4 19. The inverse of the matrix is: −3 2   1 3  1 2  − −  10  (b)  7 7  (a)  10    1 2 1 3 − − 5 5 14 14  1  1 2  3  − −  5  (d)  14 14  (c)  5    1 3 1 2 − − 10 10 7 7

20.

The value(s) of λ given (2 − λ) −1 = 0 is: −4 (−1 − λ) (a) −2 and +3 (b) +2 and −3 (c) −2

(d) 3

For fully worked solutions to each of the problems in Practice Exercises 102 to 106 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 21

Applications of matrices and determinants Why it is important to understand: Applications of matrices and determinants As mentioned previously, matrices are used to solve problems, for example, in electrical circuits, optics, quantum mechanics, statics, robotics, genetics and much more, and for calculating forces, vectors, tensions, masses, loads and a lot of other factors that must be accounted for in engineering. In the main, matrices and determinants are used to solve a system of simultaneous linear equations. The simultaneous solution of multiple equations finds its way into many common engineering problems. In fact, modern structural engineering analysis techniques are all about solving systems of equations simultaneously. Eigenvalues and eigenvectors, which are based on matrix theory, are very important in engineering and science. For example, car designers analyse eigenvalues in order to damp out the noise in a car, eigenvalue analysis is used in the design of car stereo systems, eigenvalues can be used to test for cracks and deformities in a solid and oil companies use eigenvalues analysis to explore land for oil.

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

solve simultaneous equations in two and three unknowns using matrices solve simultaneous equations in two and three unknowns using determinants solve simultaneous equations using Cramer’s rule solve simultaneous equations using Gaussian elimination calculate the stiffness matrix of a system of linear equations determine the eigenvalues of a 2 by 2 and 3 by 3 matrix determine the eigenvectors of a 2 by 2 and 3 by 3 matrix

266 Section E   5 3 ( ) )( )  29 29  0 x × 7 =  4 −3  19 1 y 29 29   21 95 ( )  29 + 29  x  Thus =  28 57  y − 29 29 ( ) ( ) x 4 i.e. = y −1 ( 1 0

21.1 Solution of simultaneous equations by matrices (a) The procedure for solving linear simultaneous equations in two unknowns using matrices is: (i) write the equations in the form a1 x + b1 y = c1 a2 x + b2 y = c2 (ii)

write the matrix equation corresponding to these ( equations, ) ( ) ( ) a1 b1 x c i.e. × = 1 a2 b2 y c2 ( ) a b1 (iii) determine the inverse matrix of 1 ( ) a2 b2 1 b2 −b1 i.e. a1 a1 b2 − b1 a2 −a2

(v)

x=4

(i)

3 × 4 + 5 × (−1) − 7 = 0 = RHS

(ii)

(iii)

equation (2), 4 × 4 − 3 × (−1) − 19 = 0 = RHS

The procedure for solving linear simultaneous equations in three unknowns using matrices is: (i) write the equations in the form a1 x + b1 y + c1 z = d1

Writing the equations in the a1 x + b1 y = c form gives: 3x + 5y = 7 4x − 3y = 19 The matrix equation is ( ) ( ) ( ) 3 5 x 7 × = 4 −3 y 19 ( ) 3 5 The inverse of matrix is 4 −3 ( ) 1 −3 −5 3 3 × (−3) − 5 × 4 −4 3 5 

  i.e.  29 29  4 −3 29 29 (iv) Multiplying each side of (ii) by (iii) and remembering that A × A−1 = I, the unit matrix, gives:

y = −1

equation (1),

(b) (1) (2)

and

Checking:

(from Chapter 20) (iv) multiply each side of (ii) by the inverse matrix, and (v) solve for x and y by equating corresponding elements. Problem 1. Use matrices to solve the simultaneous equations: 3x + 5y − 7 = 0 4x − 3y − 19 = 0

By comparing corresponding elements:

a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3 (ii)

write the matrix equation corresponding to these equations, i.e.       a1 b1 c1 x d1 a2 b2 c2  × y = d2  a3 b3 c3 z d3

(iii)

determine the inverse matrix of   a1 b1 c1 a2 b2 c2  (see Chapter 20) a3 b3 c3

(iv) multiply each side of (ii) by the inverse matrix, and (v)

solve for x, y and z by equating the corresponding elements.

Applications of matrices and determinants 267 Hence the inverse of A,  14 0 1 16 −5 A−1 = 35 5 5

Problem 2. Use matrices to solve the simultaneous equations: x+y+z−4 = 0

(1)

2x − 3y + 4z − 33 = 0 3x − 2y − 2z − 2 = 0

(2) (3)

(i) Writing the equations in the a1 x + b1 y + c1 z = d1 form gives: x+y+z = 4 2x − 3y + 4z = 33 3x − 2y − 2z = 2 (ii) The matrix equation is       1 1 1 x 4 2 −3 4 × y = 33 3 −2 −2 z 2 (iii) The inverse matrix of   1 1 1 4 A = 2 −3 3 −2 −2

 7 −2 −5

(iv) Multiplying each side of (ii) by (iii), and remembering that A × A−1 = I, the unit matrix, gives         1 0 0 x 14 0 7 4 0 1 0 × y = 1  16 −5 −2  ×  33  35 0 0 1 z 5 5 −5 2     x (14 × 4) + (0 × 33) + (7 × 2) 1 y = (16 × 4) + ((−5) × 33) + ((−2) × 2) 35 z (5 × 4) + (5 × 33) + ((−5) × 2)   70 1  −105 = 35 175   2 = −3 5 (v)

By comparing corresponding elements, x = 2, y = −3, z = 5, which can be checked in the original equations.

is given by A−1 =

adj A |A|

The adjoint of A is the transpose of the matrix of the cofactors of the elements (see Chapter 20). The matrix of cofactors is   14 16 5  0 −5 5 7 −2 −5 and the transpose of this matrix gives   14 0 7 adj A = 16 −5 −2 5 5 −5 The determinant of A, i.e. the sum of the products of elements and their cofactors, using a first row expansion is −3 2 2 −3 4 4 1 − 1 + 1 3 −2 −2 −2 3 −2 = (1 × 14) − (1 × (−16)) + (1 × 5) = 35

Now try the following Practice Exercise Practice Exercise 108 Solving simultaneous equations using matrices (Answers on page 879) In Problems 1 to 5 use matrices to solve the simultaneous equations given. 1.

3x + 4y = 0 2x + 5y + 7 = 0

2.

2p + 5q + 14.6 = 0 3.1p + 1.7q + 2.06 = 0

3.

x + 2y + 3z = 5 2x − 3y − z = 3 −3x + 4y + 5z = 3

4.

3a + 4b − 3c = 2 −2a + 2b + 2c = 15 7a − 5b + 4c = 26

5.

p + 2q + 3r + 7.8 = 0 2p + 5q − r − 1.4 = 0 5p − q + 7r − 3.5 = 0

268 Section E i.e. the determinant of the coefficients left when the y-column is covered up, a1 b1 and D= a2 b2

6. In two closed loops of an electrical circuit, the currents flowing are given by the simultaneous equations: I1 + 2I2 + 4 = 0 5I1 + 3I2 − 1 = 0 Use matrices to solve for I1 and I2 7. The relationship between the displacement s, velocity v, and acceleration a, of a piston is given by the equations: s + 2v + 2a = 4 3s − v + 4a = 25 3s + 2v − a = −4 Use matrices to determine the values of s, v and a 8. In a mechanical system, acceleration ¨x, velocity x˙ and distance x are related by the simultaneous equations: 3.4¨x + 7.0˙x − 13.2x = −11.39 −6.0¨x + 4.0˙x + 3.5x = 4.98 2.7¨x + 6.0˙x + 7.1x = 15.91 Use matrices to find the values of ¨x, x˙ and x

i.e. the determinant of the coefficients left when the constants-column is covered up. Problem 3. Solve the following simultaneous equations using determinants: 3x − 4y = 12 7x + 5y = 6.5 Following the above procedure: (i)

3x − 4y − 12 = 0 7x + 5y − 6.5 = 0

(ii)

−4 5 i.e.

x −y 1 = = −12 3 −12 3 −4 −6.5 7 −6.5 7 5 x (−4)(−6.5) − (−12)(5) =

21.2 Solution of simultaneous equations by determinants (a) When solving linear simultaneous equations in two unknowns using determinants: (i) write the equations in the form

=

i.e. i.e.

x −y 1 = = 86 64.5 43

Since

x 1 86 = then x = =2 86 43 43

and then (ii) the solution is given by x −y 1 = = Dx Dy D b1 c1 where Dx = b2 c2 i.e. the determinant of the coefficients left when the x-column is covered up, a1 c1 Dy = a2 c2

1 (3)(5) − (−4)(7)

x −y 1 = = 26 + 60 −19.5 + 84 15 + 28

a1 x + b1 y + c1 = 0 a2 x + b2 y + c2 = 0

−y (3)(−6.5) − (−12)(7)

and since −y 1 64.5 = then y = − = −1.5 64.5 43 43 Problem 4. The velocity of a car, accelerating at uniform acceleration a between two points, is given by v = u + at, where u is its velocity when passing the first point and t is the time taken to pass between the two points. If v = 21 m/s when t = 3.5 s and v = 33 m/s when t = 6.1 s, use determinants to find the values of u and a, each correct to 4 significant figures.

Applications of matrices and determinants 269 Substituting the given values in v = u + at gives: 21 = u + 3.5a 33 = u + 6.1a (i)

Following the procedure: (1) (2)

The equations are written in the form

(i)

−(6 + j8)I1 + (8 + j3)I2 − (2 + j4) = 0 (ii)

a1 x + b1 y + c1 = 0

I1 −(6 + j8) −5 (8 + j3) −(2 + j4)

u + 3.5a − 21 = 0 u + 6.1a − 33 = 0

i.e. and (ii)

(9 + j12)I1 − (6 + j8)I2 − 5 = 0

−I2 = −5 (9 + j12) −(6 + j8) −(2 + j4)

The solution is given by

1 = (9 + j12) −(6 + j8) −(6 + j8) (8 + j3)

u −a 1 = = Du Da D

I1 (−20 + j40) + (40 + j15)

where Du is the determinant of coefficients left when the u column is covered up, 3.5 −21 i.e. Du = 6.1 −33

=

= (3.5)(−33) − (−21)(6.1) = 12.6 1 −21 Similarly, Da = 1 −33

=

Hence I1 = =

= (1)(6.1) − (3.5)(1) = 2.6 Thus

and

u −a 1 = = 12.6 −12 26

i.e.

u=

12.6 = 4.846 m/s 2.6

and

a=

12 = 4.615 m/s2 , 2.6

1 (36 + j123) − (−28 + j96)

I1 −I2 1 = = 20 + j55 −j100 64 + j27

= (1)(−33) − (−21)(1) = −12 1 3.5 D= 1 6.1

and

−I2 (30 − j60) − (30 + j40)

I2 =

20 + j55 64 + j27 58.52∠70.02◦ = 0.84∠47.15◦ A 69.46∠22.87◦ 100∠90◦ 69.46∠22.87◦

= 1.44∠67.13◦ A

(b)

When solving simultaneous equations in three unknowns using determinants: (i) Write the equations in the form

each correct to 4 significant figures.

a1 x + b1 y + c1 z + d1 = 0 a2 x + b2 y + c2 z + d2 = 0

Problem 5. Applying Kirchhoff’s laws to an electric circuit results in the following equations: (9 + j12)I1 − (6 + j8)I2 = 5 −(6 + j8)I1 + (8 + j3)I2 = (2 + j4) Solve the equations for I1 and I2

a3 x + b3 y + c3 z + d3 = 0 and then (ii)

the solution is given by x −y z −1 = = = Dx Dy Dz D

270 Section E b1 c1 d1 where Dx is b2 c2 d2 b3 c3 d3 i.e. the determinant of the coefficients obtained by covering up the x column. a1 c1 d1 Dy is a2 c2 d2 a3 c3 d3

3 −4 −26 87 DI1 = −5 −3 2 6 −12 −3 −5 87 87 = (3) − (−4) 2 −12 6 −12 −5 −3 + (−26) 2 6 = 3(−486) + 4(−114) − 26(−24)

i.e., the determinant of the coefficients obtained by covering up the y column. a1 b1 d1 Dz is a2 b2 d2 a3 b3 d3

= −1290 2 −4 −26 87 DI2 = 1 −3 −7 6 −12

i.e. the determinant of the coefficients obtained by covering up the z column. a1 b1 c1 and D is a2 b2 c2 a3 b3 c3

= (2)(36 − 522) − (−4)(−12 + 609) + (−26)(6 − 21) = −972 + 2388 + 390 = 1806 2 3 −26 87 DI3 = 1 −5 −7 2 −12

i.e. the determinant of the coefficients obtained by covering up the constants column.



= (2)(60 − 174) − (3)(−12 + 609) Problem 6. A d.c. circuit comprises three closed loops. Applying Kirchhoff’s laws to the closed loops gives the following equations for current flow in milliamperes: 2I1 + 3I2 − 4I3 = 26

+ (−26)(2 − 35) = −228 − 1791 + 858 = −1161 2 3 −4 D = 1 −5 −3 −7 2 6

and

I1 − 5I2 − 3I3 = −87

= (2)(−30 + 6) − (3)(6 − 21)

−7I1 + 2I2 + 6I3 = 12

+ (−4)(2 − 35)

Use determinants to solve for I1 , I2 and I3

= −48 + 45 + 132 = 129 (i)

Writing the equations in the a1 x + b1 y + c1 z + d1 = 0 form gives: 2I1 + 3I2 − 4I3 − 26 = 0 I1 − 5I2 − 3I3 + 87 = 0

Thus I1 −I2 I3 −1 = = = −1290 1806 −1161 129 giving

−7I1 + 2I2 + 6I3 − 12 = 0 (ii)

I1 =

−1290 = 10 mA, −129

I2 =

1806 = 14 mA 129

I3 =

1161 = 9 mA 129

the solution is given by I1 −I2 I3 −1 = = = DI1 DI2 DI3 D where DI1 is the determinant of coefficients obtained by covering up the I1 column, i.e.

and

Applications of matrices and determinants 271 Now try the following Practice Exercise 9. The forces in three members of a framework are F1 , F2 and F3 . They are related by the simultaneous equations shown below. 1.4F1 + 2.8F2 + 2.8F3 = 5.6 4.2F1 − 1.4F2 + 5.6F3 = 35.0 4.2F1 + 2.8F2 − 1.4F3 = −5.6 Find the values of F1 , F2 and F3 using determinants.

Practice Exercise 109 Solving simultaneous equations using determinants (Answers on page 879) In Problems 1 to 5 use determinants to solve the simultaneous equations given. 1. 3x − 5y = −17.6 7y − 2x − 22 = 0

10.

2. 2.3m − 4.4n = 6.84 8.5n − 6.7m = 1.23 3. 3x + 4y + z = 10 2x − 3y + 5z + 9 = 0 x + 2y − z = 6

Mesh-current analysis produces the following three equations: 20∠0◦ = (5 + 3 − j4)I1 − (3 − j4)I2 10∠90◦ = (3 − j4 + 2)I2 − (3 − j4)I1 − 2I3 −15∠0◦ − 10∠90◦ = (12 + 2)I3 − 2I2 Solve the equations for the loop currents I1 , I2 and I3

4. 1.2p − 2.3q − 3.1r + 10.1 = 0 4.7p + 3.8q − 5.3r − 21.5 = 0 3.7p − 8.3q + 7.4r + 28.1 = 0 5.

1 x y 2z − + =− 2 3 5 20 19 x 2y z + − = 4 3 2 40 59 x+y−z = 60

21.3 Solution of simultaneous equations using Cramer’s rule Cramer’s∗ rule states that if a11 x + a12 y + a13 z = b1 a21 x + a22 y + a23 z = b2 a31 x + a32 y + a33 z = b3 then

6. In a system of forces, the relationship between two forces F1 and F2 is given by:

x=

Dy Dx Dz , y= and z = D D D

5F1 + 3F2 + 6 = 0 3F1 + 5F2 + 18 = 0 Use determinants to solve for F1 and F2 7. Applying mesh-current analysis to an a.c. circuit results in the following equations: (5 − j4)I1 − (−j4)I2 = 100∠0◦ (4 + j3 − j4)I2 − (−j4)I1 = 0 Solve the equations for I1 and I2 8. Kirchhoff’s laws are used to determine the current equations in an electrical network and show that i1 + 8i2 + 3i3 = −31 3i1 − 2i2 + i3 = −5 2i1 − 3i2 + 2i3 = 6 Use determinants to find the values of i1 , i2 and i3

∗

Who was Cramer? Gabriel Cramer (31 July 1704–4 January 1752) was a Swiss mathematician. His articles cover a wide range of subjects including the study of geometric problems, the history of mathematics, philosophy and the date of Easter. Cramer’s most famous book is a work which Cramer modelled on Newton’s memoir on cubic curves. To find out more go to www.routledge.com/cw/bird

272 Section E where D = Dx =

a11 a21 a31

a12 a22 a32

b1 b2 b3



a13 a23 a33

a12 a22 a32

a13 a23 a33



i.e. the x-column has been replaced by the RHS b column, a11 b1 a13 Dy = a21 b2 a23 a31 b3 a33 i.e. the y-column has been replaced by the RHS b column,

a11 Dz = a21 a31

a12 a22 a32

b1 b2 b3



i.e. the z-column has been replaced by the RHS b column.

1 4 Dy = 2 33 3 2

1 4 −2



= 1((−66) − 8) − 4((−4) − 12) + 1(4 − 99) = −74 + 64 − 95 = −105 1 1 4 Dz = 2 −3 33 3 −2 2 = 1((−6) − (−66)) − 1(4 − 99) + 4((−4) − (−9)) = 60 + 95 + 20 = 175 Hence x=

Dy Dx 70 −105 = = 2, y = = = −3 D 35 D 35

Dz 175 and z = D = 35 = 5

Now try the following Practice Exercise Problem 7. Solve the following simultaneous equations using Cramer’s rule. x+y+z = 4 2x − 3y + 4z = 33 3x − 2y − 2z = 2 (This is the same as Problem 2 and a comparison of methods may be made). Following the above method: 1 1 1 4 D = 2 −3 3 −2 −2 = 1(6 − (−8)) − 1((−4) − 12) + 1((−4) − (−9)) = 14 + 16 + 5 = 35 4 1 33 −3 Dx = 2 −2

−2 1 4

= 4(6 − (−8)) − 1((−66) − 8) + 1((−66) − (−6)) = 56 + 74 − 60 = 70

Practice Exercise 110 Solving simultaneous equations using Cramer’s rule (Answers on page 879) 1.

Repeat problems 3, 4, 5, 7 and 8 of Exercise 108 on page 267, using Cramer’s rule.

2.

Repeat problems 3, 4, 8 and 9 of Exercise 109 on page 271, using Cramer’s rule.

21.4

Solution of simultaneous equations using the Gaussian elimination method

Consider the following simultaneous equations: x+y+z = 4

(1)

2x − 3y + 4z = 33

(2)

3x − 2y − 2z = 2

(3)

Leaving equation (1) as it is gives: x+y+z = 4

(1)

Applications of matrices and determinants 273 above method is known as the Gaussian∗ elimination method.

Equation (2) − 2 × equation (1) gives: (2′ )

0 − 5y + 2z = 25

We conclude from the above example that if a11 x + a12 y + a13 z = b1

and equation (3) − 3 × equation (1) gives:

a21 x + a22 y + a23 z = b2 a31 x + a32 y + a33 z = b3

(3′ )

0 − 5y − 5z = −10 ′

Leaving equations (1) and (2 ) as they are gives: x+y+z = 4

(1)

0 − 5y + 2z = 25

(2′ )

Equation (3′ ) − equation (2′ ) gives: 0 + 0 − 7z = −35

(3′′ )

The three-step procedure to solve simultaneous equations in three unknowns using the Gaussian elimination method is: a21 × equation (1) to form equa(i) Equation (2) − a11 a31 tion (2′ ) and equation (3) − × equation (1) to a11 form equation (3′ ). a32 (ii) Equation (3′ ) − × equation (2′ ) to form equaa22 tion (3′′ ). (iii)

By appropriately manipulating the three original equations we have deliberately obtained zeros in the positions shown in equations (2′ ) and (3′′ ).

Determine z from equation (3′′ ), then y from equation (2′ ) and finally, x from equation (1).

Working backwards, from equation (3′′ ), z=

−35 =5 −7

from equation (2′ ), −5y + 2(5) = 25, from which, y=

25 − 10 = −3 −5

and from equation (1), x + (−3) + 5 = 4 from which, ∗

x = 4+3−5 = 2 (This is the same example as Problems 2 and 7, and a comparison of methods can be made). The

Who was Gauss? Johann Carl Friedrich Gauss (30 April 1777–23 February 1855) was a German mathematician and physical scientist who contributed significantly to many fields, including number theory, statistics, electrostatics, astronomy and optics. To find out more go to www.routledge.com/cw/bird

274 Section E 1.

Problem 8. A d.c. circuit comprises three closed loops. Applying Kirchhoff’s laws to the closed loops gives the following equations for current flow in milliamperes: 2I1 + 3I2 − 4I3 = 26 I1 − 5I2 − 3I3 = −87 −7I1 + 2I2 + 6I3 = 12

In a mass−spring−damper system, the acceleration ¨x m/s2 , velocity x˙ m/s and displacement x m are related by the following simultaneous equations:

(1)

6.2¨x + 7.9˙x + 12.6x = 18.0

(2)

7.5¨x + 4.8˙x + 4.8x = 6.39

(3)

13.0¨x + 3.5˙x − 13.0x = −17.4 By using Gaussian elimination, determine the acceleration, velocity and displacement for the system, correct to 2 decimal places.

Use the Gaussian elimination method to solve for I1 , I2 and I3 (This is the same example as Problem 6 on page 270, and a comparison of methods may be made)

2.

The tensions, T1 , T2 and T3 in a simple framework are given by the equations:

Following the above procedure: 1.

2.

5T1 + 5T2 + 5T3 = 7.0

2I1 + 3I2 − 4I3 = 26 1 Equation (2) − × equation (1) gives: 2 0 − 6.5I2 − I3 = −100

(2′ )

−7 Equation (3) − × equation (1) gives: 2 0 + 12.5I2 − 8I3 = 103

(3′ )

2I1 + 3I2 − 4I3 = 26

(1)

0 − 6.5I2 − I3 = −100

(2′ )

12.5 × equation (2′ ) gives: −6.5 0 + 0 − 9.923I3 = −89.308

(3′′ )

(1)

Equation (3′ ) −

3.

From equation (3′′ ), I3 =

−89.308 = 9 mA −9.923

from equation (2′ ), −6.5I2 − 9 = −100, from which, I2 =

−100 + 9 = 14 mA −6.5

and from equation (1), 2I1 + 3(14) − 4(9) = 26, 26 − 42 + 36 20 = 2 2 = 10 mA

from which, I1 =

Now try the following Practice Exercise Practice Exercise 111 Solving simultaneous equations using Gaussian elimination (Answers on page 879)

T1 + 2T2 + 4T3 = 2.4 4T1 + 2T2

= 4.0

Determine T1 , T2 and T3 using Gaussian elimination. 3.

Repeat problems 3, 4, 5, 7 and 8 of Exercise 108 on page 267, using the Gaussian elimination method.

4.

Repeat problems 3, 4, 8 and 9 of Exercise 109 on page 271, using the Gaussian elimination method.

21.5

Stiffness matrix

In the finite element method for the numerical solution of elliptic partial differential equations, the stiffness matrix represents the system of linear equations that must be solved in order to ascertain an approximate solution to the differential equation. Here is a typical practical problem that involves matrices. Problem 9. Calculate the stiffness matrix (Q), and derive the (A) and (D) matrices for a sheet of aluminium with thickness h = 1 mm, given that:   0 [ ] E νE 1   0 (Q) = νE E  ( ) 1 − ν2 0 0 1 − ν2 G ( ) ( ) A = Q h

and

( ) ( ) h3 D = Q 12

Applications of matrices and determinants 275 The following material properties can be assumed: Young’s modulus, E = 70 GPa, Poisson’s ratio, υ = 0.35, and for an isotropic E ) material, shear strain, G = ( 2 1+υ E 70 )= ( ) Shear strain, G = ( 2 1+υ 2 1 + 0.35) = 25.93 GPa 1 − υ 2 = 1 − 0.352 = 0.8775 Hence, stiffness matrix,   0 [ ] E νE 1   0 (Q) = νE E  ( ) 1 − ν2 0 0 1 − ν2 G   70 0.35(70) 0 1   70 0 i.e. (Q) = 0.35(70)  0.8775 0 0 0.885(25.93) in GPa i.e. stiffness matrix, 



79.77 27.92 0   0  in GPa (Q) = 27.92 79.77 0 0 25.93

( ) ( ) A = Q h where h = 1.0 mm i.e. (A) = (Q)(1.0)

The units of (A) are: (GPa) mm = (103 MPa) mm = (103 N/mm2 ) mm   79.77 27.92 0   0  in kN/mm (A) = 27.92 79.77 0

0

25.93

( ) ( ) ( ) h3 ( ) 1 1 ( ) 3 D = Q = Q × 1.0 = Q 12 12 12 The units of (D) are: (GPa)(mm3 ) = (103 MPa) (mm3 ) = (103 N/mm2 ) (mm3 ) = 103 N mm   6.648 2.327 0   0  in kN mm (D) = 2.327 6.648 0

0

2.161

Now try the following Practice Exercise Practice Exercise 112 Stiffness matrix (Answers on page 880) 1.

A laminate consists of 4 plies of unidirectional high modulus carbon fibres in an epoxy resin, each 0.125 mm thick. The reduced stiffness matrix (Q) for the single plies along the fibres (i.e. at 0◦ ) is given by: ( ) Q 0=   0 [ ] E1 υ21 E1 1   0 υ12 E2 E2  ( ) 1 − υ12 υ21 0 0 1 − υ12 υ21 G12 Given that E1 = 180 GPa, E2 = 8 GPa,

E2 v12 (E1 ) determine the reduced stiffness matrix, Q 0 G12 = 5 GPa, υ12 = 0.3 and v21 =

21.6

Eigenvalues and eigenvectors

In practical applications, such as coupled oscillations and vibrations, equations of the form: Ax = λ x occur, where A is a square matrix and λ (Greek lambda) is a number. Whenever x ̸= 0, the values of λ are called the eigenvalues of the matrix A; the corresponding solutions of the equation A x = λ x are called the eigenvectors of A. Sometimes, instead of the term eigenvalues, characteristic values or latent roots are used. Also, instead of the term eigenvectors, characteristic vectors is used. From above, if A x = λx then A x − λx = 0 i.e. (A − λI) = 0 where I is the unit matrix. If x ̸= 0 then |A − λI| = 0 |A − λI| is called the characteristic determinant of A and |A − λI| = 0 is called the characteristic equation. Solving the characteristic equation will give the value(s) of the eigenvalues, as demonstrated in the following worked problems. Problem(10. Determine the eigenvalues of the ) 3 4 matrix A = 2 1

276 Section E (

2 1

)

The eigenvalue is determined by solving the characteristic equation |A − λI| = 0 ( ) ( ) 3 4 1 0 i.e. −λ =0 2 1 0 1 ( ) ( ) 3 4 λ 0 i.e. − =0 2 1 0 λ 3−λ 4 i.e. =0 2 1−λ

x1 =

(Given a square matrix, we can get used to going straight to this characteristic equation)

from which,

Hence,

(3 − λ)(1 − λ) − (2)(4) = 0

i.e.

3 − 3λ − λ + λ − 8 = 0

and

λ2 − 4λ − 5 = 0

i.e.

(λ − 5)(λ + 1) = 0

Using the equation (A – λI)x = 0 for λ2 = −1 ( )( ) ( ) 3 − −1 4 x1 0 then = 2 1 − −1 x2 0 i.e. ( )( ) ( ) 4 4 x1 0 = 2 2 x2 0 4x1 + 4x2 = 0 and 2x1 + 2x2 = 0

2

from which, λ − 5 = 0 i.e. λ = 5 or λ + 1 = 0 i.e. λ = −1 (Instead of factorising, the quadratic formula could be used; even electronic calculators can solve quadratic equations.) ( ) 3 4 Hence, the eigenvalues of the matrix are 2 1 5 and –1

Problem(12. Determine ) the eigenvalues of the 5 −2 matrix A = −9 2

Problem 11.( Determine ) the eigenvectors of 3 4 the matrix A = 2 1 ( From Problem 10, the eigenvalues of

3 2

4 1

From either of these two equations, x1 = −x2 or x2 = −x1 Hence, whatever value x1 is, the value of x2 will be –1 times Hence the simplest eigenvector is: ( greater. ) 1 x2 = −1 ( ) 2 Summarising, x1 = is an eigenvector corre1 ( ) 1 sponding to λ1 = 5 and x2 = is an eigenvector –1 corresponding to λ2 = −1

) are

λ1 = 5 and λ2 = –1 Using the equation (A – λI)x = 0 for λ1 = 5 ( )( ) ( ) 3−5 4 x1 0 then = 2 1−5 x2 0 i.e. ( )( ) ( ) −2 4 x1 0 = 2 −4 x2 0 from which, −2x1 + 4x2 = 0 and 2x1 − 4x2 = 0 From either of these two equations, x1 = 2x2 Hence, whatever value x2 is, the value of x1 will be two times greater. Hence the simplest eigenvector is:

The eigenvalue is determined by solving the characteristic equation |A − λI| = 0 5 − λ −2 =0 i.e. −9 2 − λ Hence,

(5 − λ)(2 − λ) − (−9)(−2) = 0

i.e.

10 − 5λ − 2λ + λ2 − 18 = 0

and

λ2 − 7λ − 8 = 0

i.e.

(λ − 8)(λ + 1) = 0

from which, λ − 8 = 0 i.e. λ = 8 or λ + 1 = 0 i.e. λ = −1 (or use the quadratic formula). ( ) 5 −2 Hence, the eigenvalues of the matrix −9 2 are 8 and −1

Applications of matrices and determinants 277

Problem 13.( Determine)the eigenvectors of 5 −2 the matrix A = −9 2 ( From Problem 12, the eigenvalues of λ1 = 8 and λ2 = −1

5 −2 −9 2

Problem14. Determine the  eigenvalues of the 1 2 1 0  matrix A =  6 −1 −1 −2 −1

) are

Using the equation (A − λ I )x = 0 for λ1 = 8 ( )( ) ( ) 5 − 8 −2 x1 0 then = −9 2 − 8 x2 0 i.e. ( )( ) ( ) −3 −2 x1 0 = −9 −6 x2 0

The eigenvalue is determined by solving the characteristic equation |A − λI| = 0 1−λ 2 1 =0 −1 − λ 0 i.e. 6 −1 −2 −1 − λ Hence, using the top row: (1 − λ)[(−1 − λ)(−1 − λ) − (−2)(0)] − 2[6(−1 − λ) − (−1)(0)]

from which, −3x1 − 2x2 = 0

+ 1[(6)(−2) − (−1)(−1 − λ) = 0

and −9x1 − 6x2 = 0 From either of these two equations, 3x1 = −2x2 or 2 x1 = − x2 3 Hence, if x2 = ( 3, x1 =)−2. Hence the simplest eigen–2 vector is: x1 = 3 Using the equation (A – λI)x = 0 for λ2 = – 1 ( )( ) ( ) 5 − −1 −2 x1 0 then = −9 2 − −1 x2 0 i.e. ( )( ) ( ) 6 −2 x1 0 = −9 3 x2 0 from which, 6x1 − 2x2 = 0 and

i.e.

(1 − λ)[1 + λ + λ + λ2 ] − 2[−6 − 6λ] + 1[−12 − 1 − λ] = 0

i.e.

(1 − λ)[λ2 + 2λ + 1] + 12 + 12λ − 13 − λ = 0

and

λ2 + 2λ + 1 − λ3 − 2λ2 − λ + 12 + 12λ − 13 − λ = 0

i.e.

−λ3 − λ2 + 12λ = 0

or

λ3 + λ2 − 12λ = 0

i.e.

λ(λ2 + λ − 12) = 0

i.e.

λ(λ − 3)(λ + 4) = 0 by factorising

from which, λ = 0, λ = 3 or λ = −4 Hence, the eigenvalues of the   1 2 1  6 −1 0  are 0, 3 and −4 −1 −2 −1

matrix

−9x1 + 3x2 = 0 From either of these two equations, x2 = 3x1 Hence, if(x1 =)1, x2 = 3. Hence the simplest eigenvector 1 is: x2 = 3 ( ) –2 Summarising, x1 = is an eigenvector corre3 ( ) 1 sponding to λ1 = 8 and x2 = is an eigenvector 3 corresponding to λ2 = –1

Problem 15. Determine the eigenvectors of  1 2 1 0  the matrix A =  6 −1 −1 −2 −1 From Problem  14, the eigenvalues of  1 2 1  6 −1 0  are λ1 = 0, λ2 = 3 and λ3 = −4 −1 −2 −1 Using the equation (A – λI)x = 0 for λ1 = 0

278 Section E 

 1−0 2 1 x1   x2 −1 − 0 0 then  6 −1 −2 −1 − 0 x3 i.e.     1 2 1 x1  6 −1 0   x2  =  −1 −2 −1 x3





 0 = 0  0 

Using the equation (A – λI)x = 0 for λ2 = – 4  then

0 0  0

from which, x1 + 2x2 + x3 = 0

i.e.

6x1 − x2 = 0 −x1 − 2x2 − x3 = 0

  1 x1   x2  0 −1 − −4 x3   0 = 0  0



    5 2 1 x1 0  6 3 0   x2  =  0  −1 −2 3 x3 0

from which,

From the second equation,

5x1 + 2x2 + x3 = 0

6x1 = x2

6x1 + 3x2 = 0

Substituting in the first equation, x1 + 12x1 + x3 = 0

1 − −4 2  6 −1 − −4 −1 −2

i.e. − 13x1 = x3

−x1 − 2x2 + 3x3 = 0 From the second equation,

Hence, when x1 = 1, x2 = 6 and x3 = −13 Hence the simplesteigenvector corresponding to λ1 = 0 1 6  is: x1 =  −13 Using the equation (A – λI)x = 0 for λ2 = 3      1−3 2 1 x1 0   x2  =  0  −1 − 3 0 then  6 −1 −2 −1 − 3 x3 0 i.e.      −2 2 1 x1 0  6 −4 0   x2  =  0  −1 −2 −4 x3 0 from which,

x2 = −2x1 Substituting in the first equation, 5x1 − 4x1 + x3 = 0

i.e. x3 = −x1

Hence, if x1 = −1, then x2 = 2 and x3 = 1 Hence the simplest eigenvector corresponding to   –1 λ2 = −4 is: x3 =  2  1 Problem16. Determinethe eigenvalues of the 1 −4 −2 3 1  matrix A =  0 1 2 4

−2x1 + 2x2 + x3 = 0 6x1 − 4x2 = 0 −x1 − 2x2 − 4x3 = 0 From the second equation, 3x1 = 2x2

The eigenvalue is determined by solving the characteristic equation |A −λ I| = 0 1 − λ −4 −2 3−λ 1 = 0 i.e. 0 1 2 4−λ Hence, using the top row:

Substituting in the first equation, (1λ)[(3λ)(4λ)(2)(1)](−4)[0(1)(1)]

−2x1 + 3x1 + x3 = 0 i.e. x3 = −x1 Hence, if x2 = 3, then x1 = 2 and x3 = −2 Hence the  simplest  eigenvector corresponding to λ2 = 3 2 is: x2 =  3  –2

− 2[01(3λ)] = 0 i.e.

(1 − λ)[12 − 3λ − 4λ + λ2 − 2] + 4[−1] −2[−3 + λ] = 0

i.e.

(1 − λ)[λ2 − 7λ + 10] − 4 + 6 − 2λ = 0

and λ2 − 7λ + 10 − λ3 + 7λ2 − 10λ − 4 + 6 − 2λ = 0

Applications of matrices and determinants 279 −λ3 + 8λ2 − 19λ + 12 = 0 λ3 − 8λ2 + 19λ − 12 = 0

i.e. or

From the last equation,

To solve a cubic equation, the factor theorem (see Chapter 1) may be used. (Alternatively, electronic calculators can solve such cubic equations.) Let f(λ) = λ3 − 8λ2 + 19λ − 12

x1 = −2x2 − 3x3

i.e.

x1 = −2x2 − 3(−2x2 )

i.e. x1 = 4x2

(i.e. if x2 = 1, x1 = 4)

then (λ − 1)(λ − 3)(λ − 4) = 0

Hence the  simplest  eigenvector corresponding to λ1 = 1 4 is: x1 =  1  −2 Using the equation (A − λI)x = 0 for λ2 = 3      1 − 3 −4 −2 x1 0 3−3 1   x2  =  0  then  0 1 2 4−3 x3 0 i.e.      −2 −4 −2 x1 0  0 0 1   x2  =  0  1 2 1 x3 0

from which, λ = 1 or λ = 3 or λ = 4

from which

If λ = 1, then f (1) = 1 − 8(1) + 19(1) − 12 = 0 3

2

Since f (1) = 0 then (λ − 1) is a factor. If λ = 2, then f (2) = 23 − 8(2)2 + 19(2) − 12 ̸= 0 If λ = 3, then f (3) = 33 − 8(3)2 + 19(3) − 12 = 0 Since f (3) = 0 then (λ − 3) is a factor. If λ = 4, then f (4) = 43 − 8(4)2 + 19(4) − 12 = 0 Since f (4) = 0 then (λ − 4) is a factor. Thus, since λ3 − 8λ2 + 19λ − 12 = 0

Hence, the eigenvalues of  1 −4 −2  0 3 1  are 1, 3 and 4 1 2 4

the

−2x1 − 4x2 − 2x3 = 0 x3 = 0

matrix

x1 + 2x2 + x3 = 0 Since x3 = 0, x1 = −2x2

Problem 17. Determine the eigenvectors of 1 −4 −2 3 1  the matrix A =  0 1 2 4 From  Problem 16, the  eigenvalues of 1 −4 −2  0 3 1  are λ1 = 1, λ2 = 3 and λ3 = 4 1 2 4 Using the equation (A – λI)x = 0 for λ1 = 1      1 − 1 −4 −2 x1 0 3−1 1   x2  =  0  then  0 1 2 4−1 x3 0 i.e.      0 −4 −2 x1 0  0 2 1   x2  =  0  1 2 3 x3 0 from which,

−4x2 − 2x3 = 0 2x2 + x3 = 0 x1 + 2x2 + 3x3 = 0

From the first two equations, x3 = −2x2

(i.e. if x2 = 1, x3 = −2)

(i.e. if x2 = 1, x1 = −2)

Hence the (simplest ) eigenvector corresponding to λ2 = 3 −2 is: x2 = 1 0 Using the equation (A − λI)x = 0 for λ3 = 4 ( )( ) ( ) 1 − 4 −4 −2 x1 0 then = 0 3−4 1 x2 0 1 2 4−4 x3 0 i.e. ( )( ) ( ) −3 −4 −2 x1 0 = 0 −1 1 x2 0 1 2 0 x3 0 from which, −3x1 − 4x2 − 2x3 = 0 −x2 + x3 = 0 x1 + 2x2 = 0 from which, x3 = x2 and x1 = − 2x2 (i.e. if x2 = 1, x1 = −2 and x3 = 1) Hence the  simplest  eigenvector corresponding to λ3 = 4 −2 is: x3 =  1  1

280 Section E Now try the following Practice Exercise 2. Practice Exercise 113 Eigenvalues and eigenvectors (Answers on page 880) For each of the following matrices, determine their (a) eigenvalues (b) eigenvectors ( ( ( ) ) ) 2 −4 3 6 3 1 1. 2. 3. −1 −1 1 4 −2 0     −1 −1 1 1 −1 0 2 −1 4. −4 2 4 5. −1 −1 1 5 0 −1 1 

2 2 6. 1 3 1 2

 −2 1 2



 1 1 2 7.  0 2 2 −1 1 3

3x − 4y + 2z = 11 5x + 3y − 2z = −17 2x − 3y + 5z = 19 using Cramer’s rule, the value of y is: (a) −2 (b) 3 (c) 1 (d) −1 3.

If x + 2y = 1 and 3x − y = −11 then, using determinants, the value of x is: (a) −3 (b) 4.2 (c) 2 (d) −1.6

4.

If 4x − 3y = 7 and 2x + 5y = −1 then, using matrices, the value of y is: 59 9 (c) 2 (d) (a) −1 (b) − 13 26 For the following simultaneous equations:

5. Practice Exercise 114 Multiple-choice questions on applications of matrices and determinants (Answers on page 880) Each question has only one correct answer 1. For the following simultaneous equations:

For the following simultaneous equations:

2a − 3b + 6c = −17 7a + 2b − 4c = 28 a − 5b + 2c = 7 using Cramer’s rule, the value of c is: (a) −3 (b) 2 (c) −5 (d) 3

3x − 4y + 10 = 0 5y − 2x = 9 using matrices, the value of x is: (a) −1 (b) 1 (c) 2 (d) −2

For fully worked solutions to each of the problems in Practice Exercises 108 to 113 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 6

Complex numbers, matrices and determinants

This Revision Test covers the material contained in Chapters 18 to 21. The marks for each question are shown in brackets at the end of each question. 1.

Solve the quadratic equation x2 − 2x + 5 = 0 and show the roots on an Argand diagram. (9)

2.

If Z1 = 2 + j5, Z2 = 1 − j3 and Z3 = 4 − j determine, in both Cartesian and polar forms, the value Z1 Z2 + Z3 , correct to 2 decimal places. of Z1 + Z2 (9)

3.

4.

5.

(4)

8. Determine the inverse of matrix A

(4)

9. Determine E × D

(9)

10. Calculate the determinant of matrix D

Two impedances, Z1 = (2 + j7) ohms and Z2 = (3 − j4) ohms, are connected in series to a supply voltage V of 150∠0◦ V. Determine the magnitude of the current I and its phase angle relative to the voltage. (6) Determine in both polar and rectangular forms:

(18)

In questions 6 to 10, the matrices stated are: ( ) ( ) −5 2 1 6 A= B= 7 −8 −3 −4 ( ) j3 (1 + j2) C= (−1 − j4) −j2     2 −1 3 −1 3 0 1 0  E =  4 −9 2  D = −5 4 −6 2 −5 7 1 Determine A × B

4x − 3y = 17 x+y+1 = 0 using matrices.

(7)

12. Use determinants to solve the following simultaneous equations: 4x + 9y + 2z = 21 −8x + 6y − 3z = 41 3x + y − 5z = −73

(b) [3.2 − j4.8]5

(6)

11. Solve the following simultaneous equations:

Three vectors are represented by A, 4.2∠45◦ , B, 5.5∠−32◦ and C, 2.8∠75◦ . Determine in polar form the resultant D, where D = B + C − A (8)

(a) [2.37∠35◦ ]4 √ (c) [−1 − j3]

6.

7. Calculate the determinant of matrix C

(11)

13. The simultaneous equations representing the currents flowing in an unbalanced, three-phase, starconnected, electrical network are as follows: 2.4I1 + 3.6I2 + 4.8I3 = 1.2 −3.9I1 + 1.3I2 − 6.5I3 = 2.6 1.7I1 + 11.9I2 + 8.5I3 = 0

(4)

Using matrices, solve the equations for I1 , I2 and I3 (11) ( ) 2 −4 14. For the matrix find the (a) eigen−1 −1 values (b) eigenvectors. (14)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 6, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Section F

Vector geometry

Chapter 22

Vectors Why it is important to understand: Vectors Vectors are an important part of the language of science, mathematics and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum and angular momentum are all vector quantities, resolving vectors into components is a most important skill required in any engineering studies.

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

distinguish between scalars and vectors recognise how vectors are represented add vectors using the nose-to-tail method add vectors using the parallelogram method resolve vectors into their horizontal and vertical components add vectors by calculation – horizontal and vertical components, complex numbers perform vector subtraction understand relative velocity understand i, j, k notation

22.1

Introduction

This chapter initially explains the difference between scalar and vector quantities and shows how a vector is drawn and represented. Any object that is acted upon by an external force will respond to that force by moving in the line of the force. However, if two or more forces act simultaneously, the result is more difficult to predict; the ability to add two or more vectors then becomes important. This chapter thus shows how vectors are added and subtracted, both by drawing and by calculation, and finding the resultant of two or more vectors has many uses in engineering. (Resultant means the single vector

which would have the same effect as the individual vectors.) Relative velocities and vector i, j, k notation are also briefly explained.

22.2

Scalars and vectors

The time taken to fill a water tank may be measured as, say, 50 s. Similarly, the temperature in a room may be measured as, say, 16◦ C, or the mass of a bearing may be measured as, say, 3 kg. Quantities such as time, temperature and mass are entirely defined by a numerical value and are called scalars or scalar quantities.

286 Section F Not all quantities are like this. Some are defined by more than just size; some also have direction. For example, the velocity of a car is 90 km/h due west, or a force of 20 N acts vertically downwards, or an acceleration of 2 10 m/s acts at 50◦ to the horizontal. Quantities such as velocity, force and acceleration, which have both a magnitude and a direction, are called vectors.

Note that an angle of + 45◦ is drawn from the horizontal and moves anticlockwise. a

9N

458

0

Now try the following Practice Exercise Practice Exercise 115 Scalar and vector quantities (Answers on page 880) 1. State the difference between scalar and vector quantities.

Figure 22.1

A velocity of 20 m/s at −60◦ is shown in Fig. 22.2. Note that an angle of −60◦ is drawn from the horizontal and moves clockwise. 0 608

In Problems 2 to 9, state whether the quantities given are scalar (S) or vector (V)

20 m/s

2. A temperature of 70◦ C 3. 5 m3 volume

b

4. A downward force of 20 N 5. 500 J of work

Figure 22.2

Representing a vector

6. 30 cm2 area 7. A south-westerly wind of 10 knots

There are a number of ways of representing vector quantities. These include:

8. 50 m distance

1.

9. An acceleration of 15 m/s at 60◦ to the horizontal 2

22.3

Drawing a vector

A vector quantity can be represented graphically by a line, drawn so that: (a) the length of the line denotes the magnitude of the quantity, and (b)

the direction of the line denotes the direction in which the vector quantity acts.

An arrow is used to denote the sense, or direction, of the vector. The arrow end of a vector is called the ‘nose’ and the other end the ‘tail’. For example, a force of 9 N acting at 45◦ to the horizontal is shown in Fig. 22.1.

2.

Using bold print − → AB where an arrow above two capital letters denotes the sense of direction, where A is the starting point and B the end point of the vector

3.

AB or a i.e. a line over the top of letters

4.

a i.e. an underlined letter

The force of 9 N at 45◦ shown in Fig. 22.1 may be represented as: − → 0a or 0a or 0a The magnitude of the force is 0a Similarly, the velocity of 20 m/s at −60◦ shown in Fig. 22.2 may be represented as: − → 0b or 0b or 0b The magnitude of the velocity is 0b In this chapter a vector quantity is denoted by bold print.

Vectors 287 22.4

Addition of vectors by drawing

This procedure is called the ‘parallelogram’ method.

Adding two or more vectors by drawing assumes that a ruler, pencil and protractor are available. Results obtained by drawing are naturally not as accurate as those obtained by calculation. (a) Nose-to-tail method

Figure 22.5

Two force vectors, F1 and F2 , are shown in Fig. 22.3. When an object is subjected to more than one force, the resultant of the forces is found by the addition of vectors.

F2 u F1

Figure 22.3

Problem 1. A force of 5 N is inclined at an angle of 45◦ to a second force of 8 N, both forces acting at a point. Find the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 8 N force by: (a) the ‘nose-to-tail’ method, and (b) the ‘parallelogram’ method. The two forces are shown in Fig. 22.6. (Although the 8 N force is shown horizontal, it could have been drawn in any direction.)

To add forces F1 and F2 : (i) Force F1 is drawn to scale horizontally, shown as 0a in Fig. 22.4. (ii) From the nose of F1 , force F2 is drawn at angle θ to the horizontal, shown as ab.

5N 458

(iii) The resultant force is given by length 0b, which may be measured. This procedure is called the ‘nose-to-tail’ or ‘triangle’ method. b

8N

Figure 22.6

(a) ‘Nose-to tail’ method (i)

F2 u

0 F1

a

Figure 22.4

(b) Parallelogram method To add the two force vectors, F1 and F2 , of Fig. 22.3: (i)

A line cb is constructed which is parallel to and equal in length to 0a (see Fig. 22.5).

(ii)

A line ab is constructed which is parallel to and equal in length to 0c.

(iii)

The resultant force is given by the diagonal of the parallelogram, i.e. length 0b.

The 8 N force is drawn horizontally 8 units long, shown as 0a in Fig. 22.7

(ii) From the nose of the 8 N force, the 5 N force is drawn 5 units long at an angle of 45◦ to the horizontal, shown as ab (iii) The resultant force is given by length 0b and is measured as 12 N and angle θ is measured as 17◦ b 5N 0

458

u 8N

Figure 22.7

a

288 Section F (b) ‘Parallelogram’ method (i)

In Fig. 22.8, a line is constructed which is parallel to and equal in length to the 8 N force

(ii)

A line is constructed which is parallel to and equal in length to the 5 N force

(iii)

Thus, the resultant of the two force vectors is 18 N at 34◦ to the 15 N force.

R

The resultant force is given by the diagonal of the parallelogram, i.e. length 0b, and is measured as 12 N and angle θ is measured as 17◦ . b

10 N

u 15 N

Figure 22.10

Problem 3. Velocities of 10 m/s, 20 m/s and 15 m/s act as shown in Fig. 22.11. Determine, by drawing, the magnitude of the resultant velocity and its direction relative to the horizontal.

5N 458 u 0 8N

y2

Figure 22.8

Thus, the resultant of the two force vectors in Fig. 22.6 is 12 N at 17◦ to the 8 N force. 20 m/s

Problem 2. Forces of 15 N and 10 N are at an angle of 90◦ to each other as shown in Fig. 22.9. Find, by drawing, the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 15 N force.

10 m/s

y1

308 158 y3

15 m/s

Figure 22.11

10 N

15 N

Figure 22.9

Using the ‘nose-to-tail’ method:

When more than two vectors are being added the ‘noseto-tail’ method is used. The order in which the vectors are added does not matter. In this case the order taken is v1 , then v2 , then v3 . However, if a different order is taken the same result will occur. (i) v1 is drawn 10 units long at an angle of 30◦ to the horizontal, shown as 0a in Fig. 22.12. (ii) From the nose of v1 , v2 is drawn 20 units long at an angle of 90◦ to the horizontal, shown as ab.

(i)

The 15 N force is drawn horizontally 15 units long as shown in Fig. 22.10

(iii) From the nose of v2 , v3 is drawn 15 units long at an angle of 195◦ to the horizontal, shown as br.

(ii)

From the nose of the 15 N force, the 10 N force is drawn 10 units long at an angle of 90◦ to the horizontal as shown

(iv) The resultant velocity is given by length 0r and is measured as 22 m/s and the angle measured to the horizontal is 105◦ .

(iii)

The resultant force is shown as R and is measured as 18 N and angle θ is measured as 34◦

Thus, the resultant of the three velocities is 22 m/s at 105◦ to the horizontal.

Vectors 289 1958 b

From trigonometry (see Chapter 8), cos θ =

r

from which,

0a = 0b cos θ = F cos θ

i.e.

the horizontal component of F = Fcos θ

and

sin θ =

ab 0b

from which,

ab = 0b sin θ = F sin θ

the vertical component of F = Fsin θ

i.e.

a

1058

0a 0b

308

O

Problem 4. Resolve the force vector of 50 N at an angle of 35◦ to the horizontal into its horizontal and vertical components.

Figure 22.12

Worked Problems 1 to 3 have demonstrated how vectors are added to determine their resultant and their direction. However, drawing to scale is time-consuming and not highly accurate. The following sections demonstrate how to determine resultant vectors by calculation using horizontal and vertical components and, where possible, by Pythagoras’ theorem.

The horizontal component of the 50 N force, 0a = 50 cos 35◦ = 40.96 N The vertical component of the 50 N force, ab = 50 sin 35◦ = 28.68 N The horizontal and vertical components are shown in Fig. 22.15. b

22.5 Resolving vectors into horizontal and vertical components A force vector F is shown in Fig. 22.13 at angle θ to the horizontal. Such a vector can be resolved into two components such that the vector addition of the components is equal to the original vector.

50 N

28.68 N

358 0 40.96 N

a

Figure 22.15

√ (Checking: by Pythagoras, 0b = 40.962 + 28.682 = 50 N ( ) 28.68 and θ = tan−1 = 35◦ 40.96

F

u

Figure 22.13

The two components usually taken are a horizontal component and a vertical component. If a right-angled triangle is constructed as shown in Fig. 22.14, then 0a is called the horizontal component of F and ab is called the vertical component of F. b

Thus, the vector addition of components 40.96 N and 28.68 N is 50 N at 35◦ )

Problem 5. Resolve the velocity vector of 20 m/s at an angle of −30◦ to the horizontal into horizontal and vertical components.

F

0

u a

Figure 22.14

The horizontal component of the 20 m/s velocity, 0a = 20 cos(−30◦ ) = 17.32 m/s The vertical component of the 20 m/s velocity, ab = 20 sin(−30◦ ) = −10 m/s The horizontal and vertical components are shown in Fig. 22.16.

290 Section F

17.32 m/s 0

a

308

20

m/ s

210 m/s b

Figure 22.16

Problem 6. Resolve the displacement vector of 40 m at an angle of 120◦ into horizontal and vertical components. The horizontal component of the 40 m displacement, 0a = 40 cos 120◦ = −20.0 m The vertical component of the 40 m displacement, ab = 40 sin 120◦ = 34.64 m The horizontal and vertical components are shown in Fig. 22.17.

A method of adding two vectors together is to use horizontal and vertical components. The horizontal component of force F1 is F1 cos θ1 and the horizontal component of force F2 is F2 cos θ2 The total horizontal component of the two forces, H = F1 cosθ1 + F2 cosθ2 The vertical component of force F1 is F1 sin θ1 and the vertical component of force F2 is F2 sin θ2 The total vertical component of the two forces, V = F1 sin θ1 + F2 sin θ2 Since we have H and V, the resultant of F1 and F2 is obtained by using the theorem of Pythagoras. From 2 2 Fig. 22.19, 0b2 = H √+ V 2 2 at an angle i.e. resultant = H + (V ) V −1 given by θ = tan H b

a

Re

b

34.64 m a 220.0 m 0

u

0

40 m

lt su

nt V a

H

Figure 22.19

1208

Problem 7. A force of 5 N is inclined at an angle of 45◦ to a second force of 8 N, both forces acting at a point. Calculate the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 8 N force.

Figure 22.17

22.6 Addition of vectors by calculation Two force vectors, F1 and F2 , are shown in Fig. 22.18, F1 being at an angle of θ1 and F2 being at an angle of θ2

The two forces are shown in Fig. 22.20.

5N V 458

F1 sin u1 F2 sin u2

8N

Figure 22.20 F2

F1 u1

u2

F1 cos u1 F2 cos u2

Figure 22.18

H

The horizontal component of the 8 N force is 8 cos 0◦ and the horizontal component of the 5 N force is 5 cos 45◦ The total horizontal component of the two forces, H = 8 cos 0◦ + 5 cos 45◦ = 8 + 3.5355 = 11.5355

Vectors 291 The vertical component of the 8 N force is 8 sin 0◦ and the vertical component of the 5 N force is 5 sin 45◦ The total vertical component of the two forces, V = 8 sin 0◦ + 5 sin 45◦ = 0 + 3.5355

10 N

= 3.5355 From Fig. 22.21, magnitude of resultant vector √ = H2 + V2 √ = 11.53552 + 3.53552 = 12.07 N

Figure 22.22

The horizontal component of the 15 N force is 15 cos 0◦ and the horizontal component of the 10 N force is 10 cos 90◦ The total horizontal component of the two forces,

t

an

lt su Re

15 N

V 5 3.5355 N

H = 15 cos 0◦ + 10 cos 90◦ = 15 + 0 = 15 N

u H 511.5355 N

Figure 22.21

The direction of the resultant vector, ( ) ( ) V 3.5355 θ = tan−1 = tan−1 H 11.5355 = tan−1 0.30648866 . . . = 17.04◦ Thus, the resultant of the two forces is a single vector of 12.07 N at 17.04◦ to the 8 N vector. Perhaps an easier and quicker method of calculating the magnitude and direction of the resultant is to use complex numbers (see Chapter 18). In this example, the resultant = 8∠0◦ + 5∠45◦

The vertical component of the 15 N force is 15 sin 0◦ and the vertical component of the 10 N force is 10 sin 90◦ The total vertical component of the two forces, V = 15 sin 0◦ + 10 sin 90◦ = 0 + 10 = 10 N Magnitude of resultant vector √ √ = H2 + V2 = 152 + 102 = 18.03 N The direction of the resultant vector, ( ) ( ) −1 V −1 10 θ = tan = tan = 33.69◦ H 15 Thus, the resultant of the two forces is a single vector of 18.03 N at 33.69◦ to the 15 N vector. There is an alternative method of calculating the resultant vector in this case. If we used the triangle method, then the diagram would be as shown in Fig. 22.23.

= (8 cos 0◦ + j8 sin 0◦ ) + (5 cos 45◦ + j5 sin 45◦ ) = (8 + j0) + (3.536 + j3.536) = (11.536 + j3.536) N

or

R

10 N

◦

12.07∠17.04 N u

as obtained above using horizontal and vertical components.

15 N

Figure 22.23

Problem 8. Forces of 15 N and 10 N are at an angle of 90◦ to each other as shown in Fig. 22.22. Calculate the magnitude of the resultant of these two forces and its direction with respect to the 15 N force.

Since a right-angled triangle results then we could use Pythagoras’ theorem without needing to go through the procedure for horizontal and vertical components. In fact, the horizontal and vertical components are 15 N and 10 N respectively.

292 Section F This is, of course, a special case. Pythagoras can only be used when there is an angle of 90◦ between vectors. This is demonstrated in the next worked problem.

Problem 9. Calculate the magnitude and direction of the resultant of the two acceleration vectors shown in Fig. 22.24.

Thus, the resultant of the two accelerations is a single vector of 31.76 m/s2 at 118.18◦ to the horizontal. Problem 10. Velocities of 10 m/s, 20 m/s and 15 m/s act as shown in Fig. 22.26. Calculate the magnitude of the resultant velocity and its direction relative to the horizontal. y2

20 m/s 28 m/s2 y1 10 m/s 308 15 m/s2

158

Figure 22.24

The 15 m/s2 acceleration is drawn horizontally, shown as 0a in Fig. 22.25. From the nose of the 15 m/s2 acceleration, the 28 m/s2 acceleration is drawn at an angle of 90◦ to the horizontal, shown as ab. b

R

u

a a

15

Figure 22.26

The horizontal component of the 10 m/s velocity is 10 cos 30◦ = 8.660 m/s, the horizontal component of the 20 m/s velocity is 20 cos 90◦ = 0 m/s, and the horizontal component of the 15 m/s velocity is 15 cos 195◦ = −14.489 m/s. The total horizontal component of the three velocities, H = 8.660 + 0 − 14.489 = −5.829 m/s

28

0

Figure 22.25

The resultant acceleration R is given by length 0b. Since a right-angled triangle results, the theorem of Pythagoras may be used. √ 152 + 282 = 31.76 m/s2 ( ) 28 = 61.82◦ α = tan−1 15

0b = and

15 m/s

y3

Measuring from the horizontal, θ = 180◦ − 61.82◦ = 118.18◦

The vertical component of the 10 m/s velocity is 10 sin 30◦ = 5 m/s, the vertical component of the 20 m/s velocity is 20 sin 90◦ = 20 m/s, and the vertical component of the 15 m/s velocity is 15 sin 195◦ = −3.882 m/s. The total vertical component of the three velocities, V = 5 + 20 − 3.882 = 21.118 m/s From Fig. 22.27, magnitude of resultant vector, √ √ R = H2 + V2 = 5.8292 + 21.1182 = 21.91 m/s The direction of the resultant ( ) ( vector, ) −1 V −1 21.118 α = tan = tan = 74.57◦ H 5.829 Measuring from the horizontal, θ = 180◦ − 74.57◦ = 105.43◦ Thus, the resultant of the three velocities is a single vector of 21.91 m/s at 105.43◦ to the horizontal.

Vectors 293 The three force vectors are shown in the sketch of Figure 22.29.

R 21.118

270° 200 N u

a

20°

160°

180° 500 N

5.829

u 300 N

Figure 22.27

Using complex numbers, from Fig. 22.26, resultant = 10∠30◦ + 20∠90◦ + 15∠195◦

FR

Figure 22.29

= (10 cos 30◦ + j10 sin 30◦ ) + (20 cos 90◦ + j20 sin 90◦ ) + (15 cos 195◦ + j15 sin 195◦ ) = (8.660 + j5.000) + (0 + j20.000) + (−14.489 − j3.882) = (−5.829 + j21.118) N

or

The method used to add vectors by calculation will not be specified – the choice is yours, but probably the quickest and easiest method is by using complex numbers. Problem 11. A belt-driven pulley is attached to a shaft as shown in Figure 22.28. Calculate the magnitude and direction of the resultant force acting on the shaft.

20° F3 = 300 N

Figure 22.28

= (−500 + j0) + (−187.94 + j68.40) + (0 − j300) = 725.9∠ − 161.39◦ N

as obtained above using horizontal and vertical components.

F2 = 200 N

FR = 500∠180◦ + 200∠160◦ + 300∠270◦ = −687.94 − j231.60

21.91∠105.43◦ N

F1 = 500 N

Using complex numbers, the resultant force,

Shaft

Thus, the resultant force FR is 725.9 N acting at – 161.39◦ to the horizontal

Now try the following Practice Exercise Practice Exercise 116 Addition of vectors by calculation (Answers on page 880) 1. A force of 7 N is inclined at an angle of 50◦ to a second force of 12 N, both forces acting at a point. Calculate the magnitude of the resultant of the two forces, and the direction of the resultant with respect to the 12 N force. 2. Velocities of 5 m/s and 12 m/s act at a point at 90◦ to each other. Calculate the resultant velocity and its direction relative to the 12 m/s velocity. 3. Calculate the magnitude and direction of the resultant of the two force vectors shown in Fig. 22.30.

294 Section F 7. If velocity v1 = 25 m/s at 60◦ and v2 = 15 m/s at −30◦ , calculate the magnitude and direction of v1 + v2

10 N

8. Calculate the magnitude and direction of the resultant vector of the force system shown in Fig. 22.33.

13 N

Figure 22.30

4.

Calculate the magnitude and direction of the resultant of the two force vectors shown in Fig. 22.31.

4 N 158

8N

308 608

22 N

6N

18 N

Figure 22.31

5.

6.

A displacement vector s1 is 30 m at 0◦ . A second displacement vector s2 is 12 m at 90◦ . Calculate magnitude and direction of the resultant vector s1 + s2 Three forces of 5 N, 8 N and 13 N act as shown in Fig. 22.32. Calculate the magnitude and direction of the resultant force.

Figure 22.33

9. Calculate the magnitude and direction of the resultant vector of the system shown in Fig. 22.34.

8N

2 m/s 3.5 m/s 158

708 458

5N 608 308

4 m/s

13 N

Figure 22.32

Figure 22.34

10. An object is acted upon by two forces of magnitude 10 N and 8 N at an angle of 60◦ to each other. Determine the resultant force on the object.

Vectors 295 a1 1 a2

A ship heads in a direction of E 20◦ S at a speed of 20 knots while the current is 4 knots in a direction of N 30◦ E. Determine the speed and actual direction of the ship.

11.

a2 1.5 m/s2 1268 1458

a1 2 a2

Resolving horizontally and vertically gives:

Vertical component of a1 + a2 , V = 1.5 sin 90◦ + 2.6 sin 145◦ = 2.99

o

b

Figure 22.35

For two vectors acting at a point, as shown in Fig. 22.36(a), the resultant of vector addition is: os = oa + ob Figure 22.36(b) shows vectors ob + (−oa), that is, ob − oa and the vector equation is ob − oa = od. Comparing od in Fig. 22.36(b) with the broken line ab in Fig. 22.36(a) shows that the second diagonal of the ‘parallelogram’ method of vector addition gives the magnitude and direction of vector subtraction of oa from ob. s

Figure 22.37

Horizontal component of a1 + a2 , H = 1.5 cos 90◦ + 2.6 cos 145◦ = −2.13

2F

b

2a2

(b)

a

F

d

From√ Fig. 22.38, magnitude of a1 + a2 , R = (−2.13)2 + 2.992 = 3.67 m/s2 ( ) −1 2.99 In Fig. 22.38, α = tan = 54.53◦ and 2.13 θ = 180◦ − 54.53◦ = 125.47◦ a1 + a2 = 3.67 m/s2 at 125.47◦

Thus,

2.99

b

R

a (a)

o

2a

u

a 22.13

o

3

a1

Vector subtraction

In Fig. 22.35, a force vector F is represented by oa. The vector (−oa) can be obtained by drawing a vector from o in the opposite sense to oa but having the same magnitude, shown as ob in Fig. 22.35, i.e. ob = (−oa)

2

Scale in m/s2

2.6 m/s2

22.7

1

0

0

a

(b)

Figure 22.38

Figure 22.36 2

Problem 12. Accelerations of a1 = 1.5 m/s 2 at 90◦ and a2 = 2.6 m/s at 145◦ act at a point. Find a1 + a2 and a1 − a2 (a) by drawing a scale vector diagram, and (b) by calculation. (a) The scale vector diagram is shown in Fig. 22.37. By measurement, a1 + a2 = 3.7 m/s at 126◦ 2

a1 − a2 = 2.1 m/s at 0◦ 2

Horizontal component of a1 − a2 =1.5 cos 90◦ − 2.6 cos 145◦ = 2.13 Vertical component of a1 − a2 = 1.5 sin 90◦ − 2.6 sin 145◦ = 0 √ Magnitude of a1 − a2 = 2.132 + 02 = 2.13 m/s(2 ) 0 −1 Direction of a1 − a2 = tan = 0◦ 2.13 Thus,

a1 − a2 = 2.13 m/s2 at 0◦

296 Section F Using complex numbers, Problem 13. Calculate the resultant of (a) v1 − v2 + v3 and (b) v2 − v1 − v3 when v1 = 22 units at 140◦ , v2 = 40 units at 190◦ and v3 = 15 units at 290◦ .

v1 − v2 + v3 = 22∠140◦ − 40∠190◦ + 15∠290◦ = (−16.853 + j14.141) − (−39.392 − j6.946) + (5.130 − j14.095)

(a) The vectors are shown in Fig. 22.39.

= 27.669 + j6.992 = 28.54∠14.18◦ (b) The horizontal component of

1V

v2 − v1 − v3 = (40 cos 190◦ ) − (22 cos 140◦ ) − (15 cos 290◦ ) = (−39.39) − (−16.85) − (5.13)

22 1408

= −27.67 units

1908 2H 40

1H

2908

The vertical component of

15

v2 − v1 − v3 = (40 sin 190◦ ) − (22 sin 140◦ ) − (15 sin 290◦ ) = (−6.95) − (14.14) − (−14.10) = −6.99 units

2V

Figure 22.39

The horizontal component of v1 − v2 + v3 = (22 cos 140◦ ) − (40 cos 190◦ ) + (15 cos 290◦ )

From Fig. 22.40 the magnitude of the resultant, √ R = (−27.67)2 + (−6.99)2 = 28.54 units ( ) 6.99 −1 and α = tan = 14.18◦ , from which, 27.67 θ = 180◦ + 14.18◦ = 194.18◦

= (−16.85) − (−39.39) + (5.13) = 27.67 units u

The vertical component of 227.67

v1 − v2 + v3 = (22 sin 140◦ ) − (40 sin 190◦ )

a

0

◦

+ (15 sin 290 ) = (14.14) − (−6.95) + (−14.10)

26.99 R

= 6.99 units Figure 22.40

The magnitude√of the resultant, R = 27.672 + 6.992 = 28.54 units ( ) 6.99 −1 The direction of the resultant R = tan 27.67 = 14.18◦ Thus, v1 − v2 + v3 = 28.54 units at 14.18◦

Thus,

v2 − v1 − v3 = 28.54 units at 194.18◦

This result is as expected, since v2 − v1 − v3 = − (v1 − v2 + v3 ) and the vector 28.54 units at 194.18◦ is minus times (i.e. is 180◦ out of phase with) the vector 28.54 units at 14.18◦

Vectors 297 Using complex numbers, v2 − v1 − v3 = 40∠190◦ − 22∠140◦ − 15∠290◦ = (−39.392 − j6.946)

Thus in vector equations of this form, only the first and last letters, ‘a’ and ‘d’, respectively, fix the magnitude and direction of the resultant vector. This principle is used in relative velocity problems.

− (−16.853 + j14.141) − (5.130 − j14.095) = −27.669 − j6.992 = 28.54∠ −165.82◦ or 28.54∠194.18◦

Problem 14. Two cars, P and Q, are travelling towards the junction of two roads which are at right angles to one another. Car P has a velocity of 45 km/h due east and car Q a velocity of 55 km/h due south. Calculate (a) the velocity of car P relative to car Q, and (b) the velocity of car Q relative to car P.

Now try the following Practice Exercise Practice Exercise 117 Vector subtraction (Answers on page 880) 1. Forces of F1 = 40 N at 45◦ and F2 = 30 N at 125◦ act at a point. Determine by drawing and by calculation: (a) F1 + F2 (b) F1 − F2 2. Calculate the resultant of (a) v1 + v2 − v3 (b) v3 − v2 + v1 when v1 = 15 m/s at 85◦ , v2 = 25 m/s at 175◦ and v3 = 12 m/s at 235◦

22.8

Relative velocity

(a) The directions of the cars are shown in Fig. 22.42(a), called a space diagram. The velocity diagram is shown in Fig. 22.42(b), in which pe is taken as the velocity of car P relative to point e on the Earth’s surface. The velocity of P relative to Q is vector pq and the vector equation is pq = pe + eq. Hence the vector directions are as shown, eq being in the opposite direction to qe. From the geometry √ of the vector triangle, the mag552 = nitude of pq = 452 + ( )71.06 km/h and the −1 55 direction of pq = tan = 50.71◦ 45 i.e. the velocity of car P relative to car Q is 71.06 km/h at 50.71◦

For relative velocity problems, some fixed datum point needs to be selected. This is often a fixed point on the Earth’s surface. In any vector equation, only the start and finish points affect the resultant vector of a system. Two different systems are shown in Fig. 22.41, but in each of the systems, the resultant vector is ad.

N W

q

q

E S P

Q

55 km/h p

b

e

45 km/h

p

e

b c a

d

(a)

a d

(b)

(c)

Figure 22.42

(b)

(a)

Figure 22.41

(b) The vector equation of the system shown in Fig. 22.41(a) is: ad = ab + bd and that for the system shown in Fig. 22.41(b) is: ad = ab + bc + cd

The velocity of car Q relative to car P is given by the vector equation qp = qe + ep and the vector diagram is as shown in Fig. 22.42(c), having ep opposite in direction to pe. From the geometry√ of this vector triangle, the 2 magnitude of qp = 452 + 55 ( =)71.06 km/h and 55 = 50.71◦ but the direction of qp = tan−1 45

298 Section F must lie in the third quadrant, i.e. the required angle is: 180◦ + 50.71◦ = 230.71◦ i.e. the velocity of car Q relative to car P is 71.06 km/h at 230.71◦

algebraic calculations, as shown in the worked examples below.

Problem 15. Determine: (3i + 2 j + 2k) − (4i − 3j + 2k)

Now try the following Practice Exercise Practice Exercise 118 Relative velocity (Answers on page 880) 1. A car is moving along a straight horizontal road at 79.2 km/h and rain is falling vertically downwards at 26.4 km/h. Find the velocity of the rain relative to the driver of the car. 2. Calculate the time needed to swim across a river 142 m wide when the swimmer can swim at 2 km/h in still water and the river is flowing at 1 km/h. At what angle to the bank should the swimmer swim? 3. A ship is heading in a direction N 60◦ E at a speed which in still water would be 20 km/h. It is carried off course by a current of 8 km/h in a direction of E 50◦ S. Calculate the ship’s actual speed and direction.

(3i + 2 j + 2k) − (4i − 3j + 2k) = 3i + 2 j + 2k − 4i + 3j − 2k = −i + 5j Problem 16. Given p = 3i + 2k, q = 4i − 2 j + 3k and r = −3i + 5j − 4k determine: (a) −r (b) 3p (c) 2 p + 3q (e) 0.2 p + 0.6q − 3.2r

(d) −p + 2r

(a) −r = −(−3i + 5j − 4k) = +3i − 5j + 4k (b) 3p = 3(3i + 2k) = 9i + 6k (c) 2 p + 3q = 2(3i + 2k) + 3(4i − 2 j + 3k) = 6i + 4k + 12i − 6j + 9k = 18i − 6j + 13k

22.9

i, j and k notation

A method of completely specifying the direction of a vector in space relative to some reference point is to use three unit vectors, i, j and k, mutually at right angles to each other, as shown in Fig. 22.43. z

(d) −p + 2r = −(3i + 2k) + 2(−3i + 5j − 4k) = −3i − 2k + (−6i + 10j − 8k) = −3i − 2k − 6i + 10j − 8k = −9i + 10j − 10k (e) 0.2p + 0.6q − 3.2r = 0.2(3i + 2k) +0.6(4i − 2 j + 3k) − 3.2(−3i + 5j − 4k) = 0.6i + 0.4k + 2.4i − 1.2j + 1.8k +9.6i − 16j + 12.8k = 12.6i − 17.2j + 15k Now try the following Practice Exercise Practice Exercise 119 i, j, k notation (Answers on page 880)

k i

0 j

y

x

Figure 22.43

Calculations involving vectors given in i, j k notation are carried out in exactly the same way as standard

Given that p = 2i + 0.5j − 3k, q = −i + j + 4k and r = 6j − 5k, evaluate and simplify the following vectors in i, j, k form: 1. −q 2. 2p 3. q + r

Vectors 299 4. −q + 2 p

5.

5. 3q + 4r 6. q − 2p 7. p + q + r 8. p + 2q + 3r 9. 2 p + 0.4q + 0.5r 10. 7r − 2q

A force of 4 N is inclined at an angle of 45◦ to a second force of 7 N, both forces acting at a point, as shown in Figure 22.45. The magnitude of the resultant of these two forces and the direction of the resultant with respect to the 7 N force is: (a) 3 N at 45◦ (b) 5 N at 146◦ ◦ (c) 11 N at 135 (d) 10.2 N at 16◦ 4N

458

Practice Exercise 120 Multiple-choice questions on vectors (Answers on page 880)

7N

Figure 22.45

Each question has only one correct answer 1. Which of the following quantities is considered a vector? (a) time (b) temperature (c) force (d) mass

Questions 6 and 7 relate to the following information. Two voltage phasors V1 and V2 are shown in Figure 22.26.

2. A force vector of 40 N is at an angle of 72◦ to the horizontal. Its vertical component is: (a) 40 N (b) 12.36 N (c) 38.04 N (d) 123.11 N 3. A displacement vector of 30 m is at an angle of −120◦ to the horizontal. Its horizontal component is: (a) −15 m (b) −25.98 m (c) 15 m (d) 25.98 m 4. Two voltage phasors are shown in Figure 22.44. If V1 = 40 volts and V2 = 100 volts, the resultant (i.e. length OA) is: (a) 131.4 volts at 32.55◦ to V1 (b) 105.0 volts at 32.55◦ to V1 (c) 131.4 volts at 68.30◦ to V1 (d) 105.0 volts at 42.31◦ to V1

V1 5 15 V 308

V2 5 25 V

Figure 22.46

6.

The resultant V1 + V2 is given by: (a) 14.16 V at 62◦ to V1 (b) 38.72 V at −19◦ to V1 (c) 38.72 V at 161◦ to V1 (d) 14.16 V at 118◦ to V1

7.

The resultant V1 − V2 is given by: (a) 38.72 V at −19◦ to V1 (b) 14.16 V at 62◦ to V1 (c) 38.72 V at 161◦ to V1 (d) 14.16 V at 118◦ to V1

8.

The magnitude of the resultant of velocities of 3 m/s at 20◦ and 7 m/s at 120◦ when acting simultaneously at a point is: (a) 7.12 m/s (b) 10 m/s (c) 21 m/s (d) 4 m/s

A V25 100 volts

458 0 B V1 5 40 volts

Figure 22.44

300 Section F 9. Three forces of 2 N, 3 N and 4 N act as shown in Figure 22.47. The magnitude of the resultant force is: (a) 8.08 N (b) 7.17 N (c) 9 N (d) 1 N

10.

The following three forces are acting upon an object in space: p = 4i − 3 j + 2k q = −5k r = −2i + 7 j + 6k

4N

The resultant of the forces p + q − r is: (a) −3i + 4 j + 4k (c) 6i − 10 j − 9k

(b) 2i + 4 j + 3k (d) 2i + 4 j − 3k

608 308

3N 2N

Figure 22.47

For fully worked solutions to each of the problems in Practice Exercises 115 to 119 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 23

Methods of adding alternating waveforms Why it is important to understand: Methods of adding alternating waveforms In electrical engineering, a phasor is a rotating vector representing a quantity such as an alternating current or voltage that varies sinusoidally. Sometimes it is necessary when studying sinusoidal quantities to add together two alternating waveforms, for example in an a.c. series circuit, where they are not inphase with each other. Electrical engineers, electronics engineers, electronic engineering technicians and aircraft engineers all use phasor diagrams to visualise complex constants and variables. So, given oscillations to add and subtract, the required rotating vectors are constructed, called a phasor diagram, and graphically the resulting sum and/or difference oscillation are added or calculated. Phasors may be used to analyse the behaviour of electrical and mechanical systems that have reached a kind of equilibrium called sinusoidal steady state. Hence, discovering different methods of combining sinusoidal waveforms is of some importance in certain areas of engineering.

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

determine the resultant of two phasors by graph plotting determine the resultant of two or more phasors by drawing determine the resultant of two phasors by the sine and cosine rules determine the resultant of two or more phasors by horizontal and vertical components determine the resultant of two or more phasors by complex numbers

23.1 Combination of two periodic functions

There are a number of methods of determining the resultant waveform. These include: (a) by drawing the waveforms and adding graphically

There are a number of instances in engineering and science where waveforms have to be combined and where it is required to determine the single phasor (called the resultant) that could replace two or more separate phasors. Uses are found in electrical alternating current theory, in mechanical vibrations, in the addition of forces and with sound waves.

(b)

by drawing the phasors and measuring the resultant

(c) by using the cosine and sine rules (d)

by using horizontal and vertical components

(e) by using complex numbers

302 Section F 23.2

Plotting periodic functions

This may be achieved by sketching the separate functions on the same axes and then adding (or subtracting) ordinates at regular intervals. This is demonstrated in the following worked Problems. Problem 1. Plot the graph of y1 = 3 sin A from A = 0◦ to A = 360◦ . On the same axes plot y2 = 2 cos A. By adding ordinates, plot yR = 3 sin A + 2 cos A and obtain a sinusoidal expression for this resultant waveform.

Problem 2. Plot the graphs of y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3) on the same axes, over one cycle. By adding ordinates at intervals plot yR = y1 + y2 and obtain a sinusoidal expression for the resultant waveform. y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3) are shown plotted in Fig. 23.2. y 6.1 6

258 y1 5 4 sin vt

4

y1 = 3 sin A and y2 = 2 cos A are shown plotted in Fig. 23.1. Ordinates may be added at, say, 15◦ intervals. For example, at 0◦, y1 + y2 = 0 + 2 = 2 at 15◦, y1 + y2 = 0.78 + 1.93 = 2.71 at 120◦, y1 + y2 = 2.60 + −1 = 1.6 at 210◦, y1 + y2 = −1.50 −1.73 = −3.23, and so on. The resultant waveform, shown by the broken line, has the same period, i.e. 360◦ , and thus the same frequency as the single phasors. The maximum value, or amplitude, of the resultant is 3.6. The resultant π rad = waveform leads y1 = 3 sin A by 34◦ or 34 × 180 0.593 rad. The sinusoidal expression for the resultant waveform is: yR = 3.6 sin(A + 34◦ ) or

y25 3 sin (vt 2 p/3)

2 0 22

y R 5 y1 1 y2 908 p/2

1808 p

2708 3p/2

3608 2p

vt

258 24 26

Figure 23.2

Ordinates are added at 15◦ intervals and the resultant is shown by the broken line. The amplitude of the resultant is 6.1 and it lags y1 by 25◦ or 0.436 rad. Hence, the sinusoidal expression for the resultant waveform is: yR = 6.1 sin(ωt − 0.436)

yR = 3.6 sin(A + 0.593) y 3.6 3

y1 5 3 sin A y R 5 3.6 sin(A 1 34)8

2

y1 and y2 are shown plotted in Fig. 23.3. At 15◦ intervals y2 is subtracted from y1 . For example:

y2 5 2 cos A

1 0

Problem 3. Determine a sinusoidal expression for y1 − y2 when y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3)

348

908

1808

2708

3608

at 0◦ , y1 − y2 = 0 − (−2.6) = +2.6 A

21

at 30◦ , y1 − y2 = 2 − (−1.5) = +3.5

22

at 150◦ , y1 − y2 = 2 − 3 = −1, and so on.

23

Figure 23.1

The amplitude, or peak value of the resultant (shown by the broken line), is 3.6 and it leads y1 by 45◦ or 0.79 rad. Hence, y1 − y2 = 3.6 sin(ωt + 0.79)

Methods of adding alternating waveforms 303 y

Now try the following Practice Exercise

458 y1 2 y2 y2

y1 4

Practice Exercise 121 Plotting periodic functions (Answers on page 881)

3.6 2 0 22

908 p/2

1808 p

2708 3p/2

3608 2p

vt

1.

Plot the graph of y = 2 sin A from A = 0◦ to A = 360◦ . On the same axes plot y = 4 cos A. By adding ordinates at intervals plot y = 2 sin A + 4 cos A and obtain a sinusoidal expression for the waveform.

2.

Two alternating voltages are given by v1 = 10 sin ωt volts and v2 = 14 sin(ωt + π/3) volts. By plotting v1 and v2 on the same axes over one cycle obtain a sinusoidal expression for (a) v1 + v2 (b) v1 − v2

3.

Express 12 sin ωt + 5 cos ωt in the form A sin(ωt ± α) by drawing and measurement.

24

Figure 23.3

Problem 4. Two alternating currents are given by: i1 = 20(sin ωt amperes and π) amperes. i2 = 10 sin ωt + 3 By drawing the waveforms on the same axes and adding, determine the sinusoidal expression for the resultant i1 + i2 i1 and i2 are shown plotted in Fig. 23.4. The resultant waveform for i1 + i2 is shown by the broken line. It has the same period, and hence frequency, as i1 and i2

30 26.5

198

p iR 5 20 sin vt 110 sin (vt 1 ) 3

20

i1 5 20 sin vt i2 5 10 sin(vt 1 p ) 3

10 908 198

p 2

1808

2708

p

3p 2

3608

23.3 Determining resultant phasors by drawing The resultant of two periodic functions may be found from their relative positions when the time is zero. For example, if y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3) then each may be represented as phasors as shown in Fig. 23.5, y1 being 4 units long and drawn horizontally and y2 being 3 units long, lagging y1 by π/3 radians or 60◦ . To determine the resultant of y1 + y2 , y1 is drawn horizontally as shown in Fig. 23.6 and y2 is joined to

2p angle vt

y1 5 4

210

608 or p/3 rads

220 230 y2 5 3

Figure 23.4

y15 4 0

f

608 y 25 3

The amplitude or peak value is 26.5 A The resultant waveform leads the waveform of i1 = 20 sin ωt by 19◦ or 0.33 rad Hence, the sinusoidal expression for the resultant i1 + i2 is given by: ( ) iR = i1 + i2 = 26.5 sin ωt + 0.33 A

Figure 23.5

yR

Figure 23.6

304 Section F y1 5 4

i2 5 10 A

f

608 yR

i1 5 20 A

y2 5 3

Figure 23.9

Figure 23.7

the end of y1 at 60◦ to the horizontal. The resultant is given by yR . This is the same as the diagonal of a parallelogram that is shown completed in Fig. 23.7. Resultant yR , in Figs. 23.6 and 23.7, may be determined by drawing the phasors and their directions to scale and measuring using a ruler and protractor. In this example, yR is measured as 6 units long and angle ϕ is measured as 25◦ 25◦ = 25 ×

π radians = 0.44 rad 180

Hence, summarising, by drawing: yR = y1 + y2 = 4 sin ωt + 3 sin (ωt − π/3) = 6 sin (ωt − 0.44) If the resultant phasor yR = y1 − y2 is required, then y2 is still 3 units long but is drawn in the opposite direction, as shown in Fig. 23.8.

iR i2 5 10 A

608

f i1 5 20 A

Figure 23.10

The resultant iR is shown and is measured as 26 A and angle ϕ as 19◦ or 0.33 rad leading i1 . Hence, by drawing and measuring: iR = i1 + i2 = 26 sin (ωt + 0.33)A Problem 6. For the currents in Problem 5, determine i1 − i2 by drawing phasors. At time t = 0, current i1 is drawn 20 units long horizontally as shown by 0a in Fig. 23.11. Current i2 is shown, drawn 10 units long in broken line and leading by 60◦ . The current −i2 is drawn in the opposite direction to the broken line of i2 , shown as ab in Fig. 23.11. The resultant iR is given by 0b lagging by angle ϕ. By measurement, iR = 17 A and ϕ = 30◦ or 0.52 rad 10 A

Figure 23.8

It may be shown that yR = y1 − y2 = 3.6 sin(ωt + 0.80)

a

f

Problem 5. Two alternating currents are given by: i1(= 20 sin)ωt amperes and π i2 = 10 sin ωt + amperes. Determine i1 + i2 by 3 drawing phasors. The relative positions of i1 and i2 at time t = 0 are shown π as phasors in Fig. 23.9, where rad = 60◦ 3 The phasor diagram in Fig. 23.10 is drawn to scale with a ruler and protractor.

608

20 A

0

10A

iR b

Figure 23.11

Hence, by drawing phasors: iR = i1 −i2 = 17 sin (ωt − 0.52)A

Methods of adding alternating waveforms 305 Now try the following Practice Exercise

from which,

Using the sine rule,

Practice Exercise 122 Determining resultant phasors by drawing (Answers on page 881) 1. Determine a sinusoidal expression 2 sin θ + 4 cos θ by drawing phasors.

from which, for

2. If v1 = 10 sin ωt volts and v2 = 14 sin(ωt + π/3) volts, determine by drawing phasors sinusoidal expressions for (a) v1 + v2 (b) v1 − v2 3. Express 12 sin ωt + 5 cos ωt in the form R sin(ωt ± α) by drawing phasors.

23.4 Determining resultant phasors by the sine and cosine rules As stated earlier, the resultant of two periodic functions may be found from their relative positions when the time is zero. For example, if y1 = 5 sin ωt and y2 = 4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 23.12, y1 being 5 units long and drawn horizontally and y2 being 4 units long, lagging y1 by π/6 radians or 30◦ . To determine the resultant of y1 + y2 , y1 is drawn horizontally as shown in Fig. 23.13 and y2 is joined to the end of y1 at π/6 radians, i.e. 30◦ to the horizontal. The resultant is given by yR Using the cosine rule on triangle 0ab of Fig. 23.13 gives: y2R = 52 + 42 − [2(5)(4) cos 150◦ ] = 25 + 16 − (−34.641)

yR =

√ 75.641 = 8.697 8.697 4 = sin 150◦ sin ϕ 4 sin 150◦ sin ϕ = 8.697 = 0.22996 ϕ = sin−1 0.22996

and

= 13.29◦ or 0.232 rad Hence, yR = y1 + y2 = 5 sin ωt + 4 sin(ωt − π/6) = 8.697 sin (ωt − 0.232) Problem 7. Given y1 = 2 sin ωt and y2 = 3 sin(ωt + π/4), obtain an expression, by calculation, for the resultant, yR = y1 + y2 When time t = 0, the position of phasors y1 and y2 are as shown in Fig. 23.14(a). To obtain the resultant, y1 is drawn horizontally, 2 units long, y2 is drawn 3 units long at an angle of π/4 rads or 45◦ and joined to the end of y1 as shown in Fig. 23.14(b). From Fig. 23.14(b), and using the cosine rule: y2R = 22 + 32 − [2(2)(3) cos 135◦ ] = 4 + 9 − [−8.485] = 21.49 √ yR = 21.49 = 4.6357

Hence,

Using the sine rule:

3 4.6357 = sin ϕ sin 135◦

from which,

sin ϕ =

3 sin 135◦ = 0.45761 4.6357

ϕ = sin−1 0.45761

Hence,

= 75.641

= 27.23◦ or 0.475 rad. y1 5 5

Thus, by calculation, yR = 4.635 sin (ωt + 0.475)

p/6 or 308 yR y25 3

y25 3

y2 5 4

Figure 23.12 0

p/4 or 458

y1 5 5 a f

308

y1 5 2 y2

54

yR

Figure 23.13

b

(a)

Figure 23.14

f

1358 y1 5 2 (b)

458

306 Section F Problem 8. ( Determine π) A using the cosine 20 sin ωt + 10 sin ωt + 3 and sine rules.

In Problems 3 to 5, express the given expressions in the form A sin(ωt ± α) by using the cosine and sine rules. 3.

From the phasor diagram of Fig. 23.15, and using the cosine rule: iR2 = 202 + 102 − [2(20)(10) cos 120◦ ] = 700 √ Hence, iR = 700 = 26.46 A

4. 5. 6.

iR i2 5 10 A

12 sin ωt + 5 cos ωt ( π) 7 sin ωt + 5 sin ωt + 4 ( π) 6 sin ωt + 3 sin ωt − 6 The sinusoidal currents in two parallel branches of an electrical network are 400 sin ωt and 750 sin(ωt − π/3), both measured in milliamperes. Determine the total current flowing into the parallel arrangement. Give the answer in sinusoidal form and in amperes.

608

f i1 5 20 A

Figure 23.15

Using the sine rule gives : from which,

10 26.46 = sin ϕ sin 120◦ 10 sin 120◦ sin ϕ = 26.46 = 0.327296

23.5 Determining resultant phasors by horizontal and vertical components If a right-angled triangle is constructed as shown in Fig. 23.16, then 0a is called the horizontal component of F and ab is called the vertical component of F. b

−1

0.327296 = 19.10◦ π = 19.10 × = 0.333 rad 180 Hence, by cosine and sine rules, ( ) iR = i1 + i2 = 26.46 sin ωt + 0.333 A and

ϕ = sin

Now try the following Practice Exercise

F

F sin u

u 0

F cos u

a

Figure 23.16

From trigonometry (see Chapter 8), Practice Exercise 123 Resultant phasors by the sine and cosine rules (Answers on page 881) 1. Determine, using the cosine and sine rules, a sinusoidal expression for: y = 2 sin A + 4 cos A 2. Given v1 = 10 sin ωt volts and v2 =14 sin(ωt + π/3) volts use the cosine and sine rules to determine sinusoidal expressions for (a) v1 + v2 (b) v1 − v2

0a from which, 0b 0a = 0b cos θ = F cos θ cos θ =

i.e.

the horizontal component of F, H = Fcosθ ab and sin θ = from which ab = 0b sin θ 0b = F sin θ i.e. the vertical component of F, V = Fsin θ Determining resultant phasors by horizontal and vertical components is demonstrated in the following worked Problems.

Methods of adding alternating waveforms 307 from which, ϕ = tan−1 (−1.8797) = 118.01◦ or 2.06 radians. Hence,

Problem 9. Two alternating voltages are given by v1 = 15 sin ωt volts and v2 = 25 sin(ωt − π/6) volts. Determine a sinusoidal expression for the resultant vR = v1 + v2 by finding horizontal and vertical components.

vR = v1 −v2 = 14.16 sin (ωt + 2.06)V The phasor diagram is shown in Fig. 23.18. 2v2 5 25 V

vR

The relative positions of v1 and v2 at time t = 0 are shown in Fig. 23.17(a) and the phasor diagram is shown in Fig. 23.17(b). The horizontal component of vR , H = 15 cos 0◦ + 25 cos(−30◦ ) = 0a + ab = 36.65 V The vertical component of vR , V = 15 sin 0◦ + 25 sin(−30◦ ) = bc = −12.50 V √ vR = 0c = 36.652 + (−12.50)2 Hence, by Pythagoras’ theorem

f

308 308

v1 5 15 V

v2 5 25 V

Figure 23.18

= 38.72 volts tan ϕ =

V −12.50 = = −0.3411 H 36.65

Problem 11. Determine ( π) 20 sin ωt + 10 sin ωt + A using horizontal and 3 vertical components.

from which, ϕ = tan−1 (−0.3411) = −18.83◦ or Hence,

− 0.329 radians.

vR = v1 + v2 = 38.72 sin (ωt − 0.329)V

From the phasors shown in Fig. 23.19: i2 5 10 A

Problem 10. For the voltages in Problem 9, determine the resultant vR = v1 − v2 using horizontal and vertical components.

608 i15 20 A

The horizontal component of vR , H = 15 cos 0◦ − 25 cos(−30◦ ) = −6.65V The vertical component of vR , V = 15 sin 0◦ − 25 sin(−30◦ ) = 12.50V √ Hence, vR = (−6.65)2 + (12.50)2

Figure 23.19

Total horizontal component, H = 20 cos 0◦ + 10 cos 60◦ = 25.0 Total vertical component, V = 20 sin 0◦ + 10 sin 60◦ = 8.66 √ By Pythagoras, the resultant, iR = [25.02 + 8.662 ] ( ) = 26.46 A −1 8.66 Phase angle, ϕ = tan = 19.11◦ 25.0 or 0.333 rad

by Pythagoras’ theorem = 14.16 volts tan ϕ =

V 12.50 = = −1.8797 H −6.65 (i.e. in the 2nd quadrant) v1 5 15 V

0

p/6 or 308

v1 a

b

f

1508

308

v2 v2 5 25 V (a)

Figure 23.17

vR (b)

c

308 Section F Hence, by using horizontal and vertical components, ( π) 20 sin ωt + 10 sin ωt + = 26.46 sin (ωt + 0.333)A 3

supply are:

( π) volts, v1 = 25 sin 300 πt + 6 ( ) π v2 = 40 sin 300 πt − volts, and 4 ( π) v3 = 50 sin 300 πt + volts. 3 Calculate the: (a) supply voltage, in sinusoidal form, in the form A sin(ωt ± α)

Now try the following Practice Exercise Practice Exercise 124 Resultant phasors by horizontal and vertical components (Answers on page 881) In Problems 1 to 5, express the combination of periodic functions in the form A sin(ωt ± α) by horizontal and vertical components: ( π) A 1. 7 sin ωt + 5 sin ωt + 4 ( π) V 2. 6 sin ωt + 3 sin ωt − 6 ( π) 3. i = 25 sin ωt − 15 sin ωt + A 3 ( π) 4. v = 8 sin ωt − 5 sin ωt − V 4 ( ) ( π) 3π 5. x = 9 sin ωt + −7 sin ωt − m 3 8 6. The voltage drops across two components when connected in series across an a.c. supply are: v1 = 200 sin 314.2t and v2 = 120 sin(314.2t − π/5) volts respectively. Determine the: (a) voltage of the supply (given by v1 + v2 ) in the form A sin(ωt ± α) (b)

frequency of the supply.

7. If the supply to a circuit is v = 20 sin 628.3t volts and the voltage drop across one of the components is v1 = 15 sin(628.3t − 0.52) volts, calculate the: (a) voltage drop across the remainder of the circuit, given by v − v1 , in the form A sin(ωt ± α) (b)

(b)

frequency of the supply

(c) periodic time. 9.

In an electrical circuit, two components are connected in series. The voltage across the first component is given by 80 sin(ωt + π/3) volts, and the voltage across the second component is given by 150 sin(ωt − π/4) volts. Determine the total supply voltage to the two components. Give the answer in sinusoidal form.

23.6 Determining resultant phasors by using complex numbers As stated earlier, the resultant of two periodic functions may be found from their relative positions when the time is zero. For example, if y1 = 5 sin ωt and y2 = 4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 23.20, y1 being 5 units long and drawn horizontally and y2 being 4 units long, lagging y1 by π/6 radians or 30◦ . To determine the resultant of y1 + y2 , y1 is drawn horizontally as shown in Fig. 23.21 y1 5 5 p/6 or 308

y2 5 4

Figure 23.20

supply frequency 0

(c) periodic time of the supply. 8. The voltages across three components in a series circuit when connected across an a.c.

y1 5 5 a f

308

y2

54

yR

Figure 23.21

b

Methods of adding alternating waveforms 309 and y2 is joined to the end of y1 at π/6 radians, i.e. 30◦ to the horizontal. Using complex numbers, from Chapter 18, the resultant is given by yR

Problem 13. For the voltages in Problem 12, determine the resultant vR = v1 − v2 using complex numbers.

π 6 = 5∠0◦ + 4∠ − 30◦

In polar form, yR = 5∠0 + 4∠ −

In polar form, yR = v1 − v2 = 15∠0 − 25∠ −

= (5 + j0) + (3.464 − j2.0)

= 15∠0◦ − 25∠ − 30◦

= 8.464 − j2.0 = 8.70∠ − 13.29◦

= (15 + j0) − (21.65 − j12.5)

= 8.70∠−0.23rad

= −6.65 + j12.5 = 14.16∠118.01◦

Hence, by using complex numbers, the resultant in sinusoidal form is:

= 14.16∠2.06 rad Hence, by using complex numbers, the resultant in sinusoidal form is:

y1 + y2 = 5 sin ωt + 4 sin(ωt − π/6) = 8.70 sin (ωt−0.23)

y1 − y2 = 15 sin ωt − 25 sin(ωt − π/6) = 14.16 sin (ωt − 2.06)V

Problem 12. Two alternating voltages are given by v1 = 15 sin ωt volts and v2 = 25 sin(ωt − π/6) volts. Determine a sinusoidal expression for the resultant vR = v1 + v2 by using complex numbers.

Problem 14. ( Determine π) 20 sin ωt + 10 sin ωt + A using complex 3 numbers.

The relative positions of v1 and v2 at time t = 0 are shown in Fig. 23.22(a) and the phasor diagram is shown in Fig. 23.22(b). In polar form, vR = v1 + v2 = 15∠0 + 25∠ −

From the phasors shown in Fig. 23.23, the resultant may be expressed in polar form as:

π 6

i2 5 10 A

= 15∠0◦ + 25∠ − 30◦

608

= (15 + j0) + (21.65 − j12.5)

i1 5 20 A

= 36.65 − j12.5 = 38.72∠ − 18.83◦

Figure 23.23

= 38.72∠ − 0.329 rad

iR = 20∠0◦ + 10∠60◦ iR = (20 + j0) + (5 + j8.66)

i.e.

Hence, by using complex numbers, the resultant in sinusoidal form is:

= (25 + j8.66) = 26.46∠19.11◦ A or

vR = v1 + v2 = 15 sin ωt + 25 sin(ωt − π/6)

26.46∠0.333 rad A

= 38.72 sin (ωt − 0.329)V v1 5 15 V

v1 f

p/6 or 308

1508

v2 vR

v2 5 25 V

(a)

Figure 23.22

π 6

(b)

310 Section F Hence, by using complex numbers, the resultant in sinusoidal form is: iR = i1 + i2 = 26.46 sin (ωt + 0.333) A Problem 15. If the supply to a circuit is v = 30 sin 100 πt volts and the voltage drop across one of the components is v1 = 20 sin(100 πt − 0.59) volts, calculate the: (a) voltage drop across the remainder of the circuit, given by v − v1 , in the form A sin(ωt ± α) (b)

supply frequency

(c) periodic time of the supply (d)

rms value of the supply voltage.

(a) Supply voltage, v = v1 + v2 where v2 is the voltage across the remainder of the circuit. Hence, v2 = v − v1 = 30 sin 100 πt − 20 sin(100 πt − 0.59) = 30∠0 − 20∠ − 0.59 rad = (30 + j0) − (16.619 − j11.127) = 13.381 + j11.127 = 17.40∠0.694 rad Hence, by using complex numbers, the resultant in sinusoidal form is: v − v1 = 30 sin 100 πt − 20 sin(100 πt − 0.59) = 17.40 sin (100πt + 0.694) volts ω 100 π = = 50 Hz 2π 2π 1 1 (c) Periodic time, T = = = 0.02 s or 20 ms f 50

(b)

Supply frequency, f =

(d)

Rms value of supply voltage, = 0.707 × 30 = 21.21 volts

Now try the following Practice Exercise Practice Exercise 125 Resultant phasors by complex numbers (Answers on page 881) In Problems 1 to 4, express the combination of periodic functions in the form A sin(ωt ± α) by using complex numbers:

( π) 1. 8 sin ωt + 5 sin ωt + V 4 ( π) 2. 6 sin ωt + 9 sin ωt − A 6 ( π) 3. v = 12 sin ωt − 5 sin ωt − V 4 ( ) ( π) 3π 4. x = 10 sin ωt + − 8 sin ωt − m 3 8 5. The voltage drops across two components when connected in series across an a.c. supply are: v1 = 240 sin 314.2t and v2 = 150 sin(314.2t − π/5) volts respectively. Determine the: (a) voltage of the supply (given by v1 + v2 ) in the form A sin(ωt ± α) (b)

frequency of the supply.

6. If the supply to a circuit is v = 25 sin 200πt volts and the voltage drop across one of the components is v1 = 18 sin(200πt − 0.43) volts, calculate the: (a) voltage drop across the remainder of the circuit, given by v − v1 , in the form A sin(ωt ± α) (b)

supply frequency

(c) periodic time of the supply. 7. The voltages across three components in a series circuit when connected across an a.c. supply are: ( π) v1 = 20 sin 300πt − volts, 6 ( π) v2 = 30 sin 300πt + volts, and 4 ( π) v3 = 60 sin 300πt − volts. 3 Calculate the: (a) supply voltage, in sinusoidal form, in the form A sin(ωt ± α) (b)

frequency of the supply

(c) periodic time (d)

rms value of the supply voltage.

Methods of adding alternating waveforms 311

8. Measurements made at a substation at peak demand of the current in the red, yellow and blue phases of a transmission system are: Ired =1248∠ − 15◦ A, Iyellow =1120∠−135◦ A and Iblue = 1310∠95◦ A. Determine the current in the neutral cable if the sum of the currents flows through it.

2.

Which of the phasor diagrams shown represents vR = v1 − v2 ? (a) (i) (b) (ii) (c) (iii) (d) (iv)

3.

Two alternating currents are given by ( π) i1 = 4 sin ωt amperes and i2 = 3 sin ωt + 6 amperes. The sinusoidal expression for the resultant i1 + i2 is: (a) 5 sin(ωt ( + 30)A π) (b) 5 sin ωt + A 6 (c) 6.77 sin(ωt + 0.22)A (d) 6.77 sin(ωt + 12.81)A

4.

Two alternating voltages are given by ( π) v1 = 5 sin ωt volts and v2 = 12 sin ωt − 3 volts. The sinusoidal expression for the resultant v1 + v2 is: (a) 13 sin(ωt − 60)V (b) 15.13 sin(ωt − 0.76)V (c) 13 sin(ωt − 1.05)V (d) 15.13 sin(ωt − 43.37)V

5.

Two alternating currents are given by ( π) i1 = 3 sin ωt amperes and i2 = 4 sin ωt + 4 amperes. The sinusoidal expression for the resultant i1 − i2 is: (a) 2.83 sin(ωt − 1.51)A (b) 6.48 sin(ωt + 0.45)A (c) 1.00 sin(ωt − 176.96)A (d) 6.48 sin(ωt − 25.89)A

9. When a mass is attached to a spring and oscillates in a vertical plane, the displacement, s, at time t is given by: s = (75 sin ωt + 40 cos ωt) mm. Express this in the form s = A sin(ωt + ϕ) 10. The single waveform resulting from the addition of the alternating voltages 9 sin 120t and 12 cos 120t volts is shown on an oscilloscope. Calculate the amplitude of the waveform and its phase angle measured with respect to 9 sin 120t and express in both radians and degrees.

Practice Exercise 126 Multiple-choice questions on methods of adding alternating waveforms (Answers on page 881) Each question has only one correct answer Questions 1 and 2 relate to the following information. Two alternating voltages are given by: v1 = 2 sin ωt ( π) and v2 = 3 sin ωt + volts. 4 1. Which of the phasor diagrams shown in Figure 23.24 represents vR = v1 + v2 ? (a) (i) (b) (ii) (c) (iii) (d) (iv) v2

vR vR

v1

v1 (i)

(ii) v1

v1

vR

v2 vR (iii)

v2

v2

(iv)

Figure 23.24

For fully worked solutions to each of the problems in Practice Exercises 121 to 125 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 24

Scalar and vector products Why it is important to understand: Scalar and vector products Common applications of the scalar product in engineering and physics is to test whether two vectors are perpendicular or to find the angle between two vectors when they are expressed in Cartesian form or to find the component of one vector in the direction of another. In mechanical engineering the scalar and vector product is used with forces, displacement, moments, velocities and the determination of work done. In electrical engineering, scalar and vector product calculations are important in electromagnetic calculations; most electromagnetic equations use vector calculus which is based on the scalar and vector product. Such applications include electric motors and generators and power transformers, and so on. Knowledge of scalar and vector products thus have important engineering applications.

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

define a unit triad determine the scalar (or dot) product of two vectors calculate the angle between two vectors determine the direction cosines of a vector apply scalar products to practical situations determine the vector (or cross) product of two vectors apply vector products to practical situations determine the vector equation of a line

24.1

The unit triad

When a vector x of magnitude x units and direction θ◦ is divided by the magnitude of the vector, the result is a vector of unit length at angle θ◦ . The unit vector for a 10 m/s at 50◦ velocity of 10 m/s at 50◦ is , i.e. 1 at 50◦ . 10 m/s oa In general, the unit vector for oa is , the oa being |oa| a vector and having both magnitude and direction and |oa| being the magnitude of the vector only. One method of completely specifying the direction of a vector in space relative to some reference point is

to use three unit vectors, mutually at right angles to each other, as shown in Fig. 24.1. Such a system is called a unit triad. z

i x

Figure 24.1

k o

j

y

Scalar and vector products 313 24.2 The scalar product of two vectors

r k

z

j

x iO a

b y

Figure 24.2

In Fig. 24.2, one way to get from o to r is to move x units along i to point a, then y units in direction j to get to b and finally z units in direction k to get to r. The vector or is specified as

When vector oa is multiplied by a scalar quantity, say k, the magnitude of the resultant vector will be k times the magnitude of oa and its direction will remain the same. Thus 2 × (5 N at 20◦ ) results in a vector of magnitude 10 N at 20◦ . One of the products of two vector quantities is called the scalar or dot product of two vectors and is defined as the product of their magnitudes multiplied by the cosine of the angle between them. The scalar product of oa and ob is shown as oa • ob. For vectors oa = oa at θ1 , and ob = ob at θ2 where θ2 > θ1 , the scalar product is: oa • ob = oa ob cos (θ 2 − θ 1 )

or = xi + yj + zk Problem 1. With reference to three axes drawn mutually at right angles, depict the vectors (a) op = 4i + 3j − 2k and (b) or = 5i − 2j + 2k

For vectors v1 and v2 shown in Fig. 24.4, the scalar product is: v1 • v2 = v1 v2 cos θ

The required vectors are depicted in Fig. 24.3, op being shown in Fig. 24.3(a) and or in Fig. 24.3(b).

v1 u

k O

v2

j

i

Figure 24.4

4

The commutative law of algebra, a × b = b × a applies to scalar products. This is demonstrated in Fig. 24.5. Let oa represent vector v1 and ob represent vector v2 . Then:

3 22

oa • ob = v1 v2 cos θ (by definition of a scalar product)

P (a)

b

v2 k O

O

u

j

r i

a

2

v1 5

Figure 24.5

22 (b)

Figure 24.3

Similarly, ob • oa = v2 v1 cos θ = v1 v2 cos θ by the commutative law of algebra. Thus oa • ob = ob • oa

314 Section F b

Let

a = a1 i + a2 j + a3 k

and b = b1 i + b2 j + b3 k v2

a • b = (a1 i + a2 j + a3 k) • (b1 i + b2 j + b3 k) Multiplying out the brackets gives:

u

a

O

a • b = a1 b1 i • i + a1 b2 i • j + a1 b3 i • k

c v2 cos u

+ a2 b1 j • i + a2 b2 j • j + a2 b3 j • k

v1

+ a3 b1 k • i + a3 b2 k • j + a3 b3 k • k

(a)

However, the unit vectors i, j and k all have a magnitude of 1 and i • i = (1)(1) cos 0◦ = 1, i • j = (1)(1) cos 90◦ = 0, i • k = (1)(1) cos 90◦ = 0 and similarly j • j = 1, j • k = 0 and k • k = 1. Thus, only terms containing i • i, j • j or k • k in the expansion above will not be zero. Thus, the scalar product

v2 su

v1

co

u

a • b = a1 b1 + a2 b2 + a3 b3 v1

(2)

Both a and b in equation (1) can be expressed in terms of a1 , b1 , a2 , b2 , a3 and b3

(b)

Figure 24.6

The projection of ob on oa is shown in Fig. 24.6(a) and by the geometry of triangle obc, it can be seen that the projection is v2 cos θ. Since, by definition

c

P

oa • ob = v1 (v2 cos θ), O

it follows that oa • ob = v1 (the projection of v2 on v1 )

B

A

Similarly the projection of oa on ob is shown in Fig. 24.6(b) and is v1 cos θ. Since by definition

it follows that

From the geometry of Fig. 24.7, the length of diagonal OP in terms of side lengths a, b and c can be obtained from Pythagoras’ theorem as follows:

ob • oa = v2 (the projection of v1 on v2 ) This shows that the scalar product of two vectors is the product of the magnitude of one vector and the magnitude of the projection of the other vector on it. The angle between two vectors can be expressed in terms of the vector constants as follows: Because a • b = a b cos θ, cos θ =

b

Figure 24.7

ob • oa = v2 (v1 cos θ),

then

a

a•b ab

(1)

OP 2 = OB 2 + BP 2 and OB 2 = OA 2 + AB 2 Thus,

OP 2 = OA 2 + AB 2 + BP 2 = a2 + b2 + c2 , in terms of side lengths

Scalar and vector products 315 Thus, the length or modulus or magnitude or norm of vector OP is given by:

(a)

From equation (2), if

p = a1 i + a2 j + a3 k

(3)

and

q = b1 i + b2 j + b3 k

Relating this result to the two vectors a1 i + a2 j + a3 k and b1 i + b2 j + b3 k, gives:

then

p • q = a1 b1 + a2 b2 + a3 b3

√ OP = (a2 + b2 + c2 )

p = 2i + j − k,

When

√

a1 = 2, a2 = 1 and a3 = −1

(a21 + a22 + a23 )

a=

and when q = i − 3j + 2k,

√ and

b=

(b21

+ b22

+ b23 )

b1 = 1, b2 = −3 and b3 = 2 Hence p • q = (2)(1) + (1)(−3) + (−1)(2)

That is, from equation (1),

i.e. a1 b1 + a2 b2 + a3 b3 √ cos θ = √ 2 (a1 + a22 + a23 ) (b21 + b22 + b23 )

(4)

Problem 2. Find vector a joining points P and Q where point P has co-ordinates (4, −1, 3) and point Q has co-ordinates (2, 5, 0). Also, find |a|, the magnitude or norm of a Let O be the origin, i.e. its co-ordinates are (0, 0, 0). The position vector of P and Q are given by: OP = 4i − j + 3k and OQ = 2i + 5 j

(b)

p • q = −3

p + q = (2i + j − k) + (i − 3 j + 2k) = 3i − 2j + k

(c)

|p + q| = |3i − 2j + k| From equation (3), | p + q| =

(d)

√ √ [32 + (−2)2 + 12 ] = 14

From equation (3), | p| = |2i + j − k| √ √ = [22 + 12 + (−1)2 ] = 6

By the addition law of vectors OP + PQ = OQ Hence a = PQ = OQ − OP i.e.

a = PQ = (2i + 5 j) − (4i − j + 3k) = −2i + 6 j − 3k

From equation (3), the magnitude or norm of a, |a| =

√ (a2 + b2 + c2 )

√ √ = [(−2)2 + 62 + (−3)2 ] = 49 = 7

Problem 3. determine:

If p = 2i + j − k and q = i − 3 j + 2k

Similarly, |q| = |i − 3j + 2k| √ √ = [12 + (−3)2 + 22 ] = 14 √ √ Hence |p| + |q| = 6 + 14 = 6.191, correct to 3 decimal places. Problem 4. Determine the angle between vectors oa and ob when oa = i + 2 j − 3k and ob = 2i − j + 4k An equation for cos θ is given in equation (4)

(a) p • q

(b) p + q

(c) | p + q|

(d) | p| + |q|

a1 b1 + a2 b2 + a3 b3 √ cos θ = √ (a21 + a22 + a23 ) (b21 + b22 + b23 )

316 Section F Since oa = i + 2 j − 3k,

The direction cosines are:

a1 = 1, a2 = 2 and a3 = −3

cos α = √

Since ob = 2i − j + 4k, b1 = 2, b2 = −1 and b3 = 4 Thus,

cos β = √ and

(1 × 2) + (2 × −1) + (−3 × 4) √ cos θ = √ (12 + 22 + (−3)2 ) (22 + (−1)2 + 42 ) −12 = √ √ = −0.6999 14 21

cos γ = √

+ z2

3 = √ = 0.802 14

+ z2

2 = √ = 0.535 14

+ z2

1 = √ = 0.267 14

x x2

+ y2 y

x2

+ y2 z

x2

+ y2

(and hence α = cos−1 0.802 = 36.7◦ , β = cos−1 0.535 =57.7◦ and γ = cos−1 0.267 = 74.5◦) Note that cos2 α + cos2 β + cos2 γ = 0.8022 + 0.5352 +0.2672 = 1

i.e. θ = 134.4◦ or 225.6◦ By sketching the position of the two vectors as shown in Problem 1, it will be seen that 225.6◦ is not an acceptable answer. Thus the angle between the vectors oa and ob, θ = 134.4◦ Direction cosines From Fig. 24.2,√ or = xi + yj + zk and from equation (3), |or| = x2 + y2 + z2 If or makes angles of α, β and γ with the co-ordinate axes i, j and k respectively, then: The direction cosines are: cos α = √ cos β = √ and

cos γ = √ 2

x y

One of the applications of scalar products is to the work done by a constant force when moving a body. The work done is the product of the applied force and the distance moved in the direction of the force. i.e. work done = F • d

AB = AO + OB = OB − OA

x2 + y2 + z2 z

that is AB = (2i − j + 3k) − (i + j + k)

+ y 2 + z2 2

Problem 6. A constant force of F = 10i + 2 j − k newtons displaces an object from A = i + j + k to B = 2i − j + 3k (in metres). Find the work done in newton metres.

The principles developed in Problem 14, page 297, apply equally to this problem when determining the displacement. From the sketch shown in Fig. 24.8,

x2 + y2 + z2

x2

Practical application of scalar product

= i − 2 j + 2k 2

such that cos α + cos β + cos γ = 1 The values of cos α, cos β and cos γ are called the direction cosines of or Problem 5.

B (2, 21, 3)

Find the direction cosines of i+2j+k

A (1,1,1)

. √ √ √ x2 + y2 + z2 = 32 + 22 + 12 = 14

O (0, 0, 0)

Figure 24.8

Scalar and vector products 317 The work done is F • d, that is F • AB in this case i.e. work done = (10i + 2 j − k) • (i − 2 j + 2k) But from equation (2), a • b = a1 b1 + a2 b2 + a3 b3 Hence work done = (10 × 1) + (2 × (−2)) + ((−1) × 2) = 4 N m (Theoretically, it is quite possible to get a negative answer to a ‘work done’ problem. This indicates that the force must be in the opposite sense to that given, in order to give the displacement stated.)

24.3

Vector products

A second product of two vectors is called the vector or cross product and is defined in terms of its modulus and the magnitudes of the two vectors and the sine of the angle between them. The vector product of vectors oa and ob is written as oa × ob and is defined by: |oa × ob| = oa ob sin θ where θ is the angle between the two vectors. The direction of oa × ob is perpendicular to both oa and ob, as shown in Fig. 24.9.

Now try the following Practice Exercise Practice Exercise 127 Scalar products (Answers on page 881)

b o

u

b oa 3 ob a

1. Find the scalar product a • b when (a) a = i + 2 j − k and b = 2i + 3 j + k (b)

o

u

a = i − 3 j + k and b = 2i + j + k

Given p = 2i − 3 j, q = 4 j − k and r = i + 2 j − 3k, determine the quantities stated in problems 2 to 8. 2. (a) p • q (b) p • r 3. (a) q • r

(b) r • q

4. (a) | p |

(b) | r |

5. (a) p • (q + r) 6. (a) | p + r |

(b) 2r • (q − 2p) (b) | p | + | r |

7. Find the angle between (a) p and q (b) q and r 8. Determine the direction cosines of (a) p (b) q (c) r 9. Determine the angle between the forces:

a (a)

10. Find the angle between the velocity vectors υ 1 = 5i + 2 j + 7k and υ 2 = 4i + j − k 11. Calculate the work done by a force F = (−5i + j + 7k) N when its point of application moves from point (−2i − 6 j + k) m to the point (i − j + 10k) m.

(b)

Figure 24.9

The direction is obtained by considering that a righthanded screw is screwed along oa × ob with its head at the origin and if the direction of oa × ob is correct, the head should rotate from oa to ob, as shown in Fig. 24.9(a). It follows that the direction of ob × oa is as shown in Fig. 24.9(b). Thus oa × ob is not equal to ob × oa. The magnitudes of oa ob sin θ are the same but their directions are 180◦ displaced, i.e. oa × ob = −ob × oa The vector product of two vectors may be expressed in terms of the unit vectors. Let two vectors, a and b, be such that: a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k

F1 = 3i + 4 j + 5k and F2 = i + j + k

ob 3 oa

Then, a × b = (a1 i + a2 j + a3 k) × (b1 i + b2 j + b3 k) = a1 b1 i × i + a1 b2 i × j + a1 b3 i × k + a2 b1 j × i + a2 b2 j × j + a2 b3 j × k + a3 b1 k × i + a3 b2 k × j + a3 b3 k × k

318 Section F But θ = 0◦, thus a • a = a2

But by the definition of a vector product, i × j = k, j × k = i and k × i = j

a•b ab

Also i × i = j × j = k × k = (1)(1) sin 0◦ = 0

Also, cos θ =

Remembering that a × b = −b × a gives:

Multiplying both sides of this equation by a2 b2 and squaring gives:

a × b = a1 b2 k − a1 b3 j − a2 b1 k + a2 b3 i + a3 b1 j − a3 b2 i

a2 b2 cos2 θ =

Grouping the i, j and k terms together, gives:

Substituting in equation (6) above for a2 = a • a, b2 = b • b and a2 b2 cos2 θ = (a • b)2 gives:

a × b = (a2 b3 − a3 b2 )i + (a3 b1 − a1 b3 ) j + (a1 b2 − a2 b1 )k The vector product can form as: i a × b = a1 b1 i The 3 × 3 determinant a1 b1 a a3 a1 − j i 2 b2 b3 b1 where



(|a × b|)2 = (a • a)(b • b) − (a • b)2

be written in determinant j a2 b2

k a3 b3

That is, (5) |a × b| =

k a3 is evaluated as: b3 a1 a2 a3 + k b1 b2 b3 j a2 b2

i j k a × b = a1 a2 a3 b1 b2 b3 a2 a3 a1 − j = i b2 b3 b1

a • b = a1 b1 + a2 b2 + a3 b3 Squaring both sides of a vector product equation gives: 2

(|a × b|) = a2 b2 sin2 θ = a2 b2 (1 − cos2 θ) (6)

It is stated in Section 24.2 that a • b = ab cos θ, hence a • a = a cos θ

[(a • a)(b • b) − (a • b)2 ]

(7)

(a) From equation (5),

The magnitude of the vector product of two vectors can be found by expressing it in scalar product form and then using the relationship of equation (2)

2

√

Problem 7. For the vectors a = i + 4 j − 2k and b = 2i − j + 3k find (a) a × b and (b) |a × b|

a2 a3 = a2 b3 − a3 b2 , b2 b3 a1 a3 = a1 b3 − a3 b1 and b1 b3 a1 a2 = a1 b2 − a2 b1 b1 b2

= a2 b2 − a2 b2 cos2 θ

a2 b2 (a • b)2 = (a • b)2 a2 b2

a3 b3

+ k a1 b1

a2 b2

Hence i j k 4 −2 a × b = 1 2 −1 3 4 −2 4 − j 1 −2 + k 1 = i 2 −1 −1 3 2 3 = i(12 − 2) − j(3 + 4) + k(−1 − 8) = 10i − 7j − 9k



Scalar and vector products 319 (b)

From equation (7) |a × b| = Now

Hence √

p × (2r × 3q) = (4i + j − 2k)

[(a • a)(b • b) − (a • b)2 ]

× (−24i − 42 j − 12k) i j k 4 1 −2 = −24 −42 −12

a • a = (1)(1) + (4 × 4) + (−2)(−2) = 21 b • b = (2)(2) + (−1)(−1) + (3)(3) = 14

and Thus

a • b = (1)(2) + (4)(−1) + (−2)(3)

= i(−12 − 84) − j(−48 − 48)

= −8 √ |a × b| = (21 × 14 − 64) √ = 230 = 15.17

+ k(−168 + 24) = −96i + 96 j − 144k or −48(2i − 2 j + 3k)

Problem 8. If p = 4i + j − 2k, q = 3i − 2 j + k and r = i − 2k find (a) ( p − 2q) × r (b) p × (2r × 3q)

Practical applications of vector products

(a) ( p − 2q) × r = [4i + j − 2k − 2(3i − 2 j + k)] × (i − 2k) = (−2i + 5 j − 4k) × (i − 2k) i j k = −2 5 −4 1 0 −2 from equation (5) 5 −4 −2 −4 = i − j 0 −2 1 −2 −2 + k 1

Problem 9. Find the moment and the magnitude of the moment of a force of (i + 2 j − 3k) newtons about point B having co-ordinates (0, 1, 1), when the force acts on a line through A whose co-ordinates are (1, 3, 4) The moment M about point B of a force vector F which has a position vector of r from A is given by: M=r×F

5 0

r is the vector from B to A, i.e. r = BA But BA = BO + OA = OA − OB (see Problem 14, page 297), that is: r = (i + 3 j + 4k) − ( j + k)

= i(−10 − 0) − j(4 + 4)

= i + 2 j + 3k

+ k(0 − 5), i.e. ( p − 2q) × r = −10i − 8 j − 5k (b) (2r × 3q) = (2i − 4k) × (9i − 6 j + 3k) i j k = 2 0 −4 9 −6 3 = i(0 − 24) − j(6 + 36)

Moment, M = r × F = (i + 2 j + 3k) × (i + 2 j − 3k) i j k 3 = 1 2 1 2 −3 = i(−6 − 6) − j(−3 − 3) + k(2 − 2)

+ k(−12 − 0) = −24i − 42 j − 12k

= −12i + 6 j N m

320 Section F The magnitude of M,

Now try the following Practice Exercise

|M| = |r × F| √ = [(r • r)(F • F) − (r • F)2 ]

Practice Exercise 128 Vector products (Answers on page 881)

r • r = (1)(1) + (2)(2) + (3)(3) = 14

In Problems 1 to 4, determine the quantities stated when

F • F = (1)(1) + (2)(2) + (−3)(−3) = 14

p = 3i + 2k, q = i − 2 j + 3k and

r • F = (1)(1) + (2)(2) + (3)(−3) = −4 √ |M| = [14 × 14 − (−4)2 ] √ = 180 Nm = 13.42 N m

r = −4i + 3 j − k 1. (a) p × q (b) q × p 2. (a) |p × r| (b) |r × q| 3. (a) 2p × 3r

Problem 10. The axis of a circular cylinder coincides with the z-axis and it rotates with an angular velocity of (2i − 5 j + 7k) rad/s. Determine the tangential velocity at a point P on the cylinder, whose co-ordinates are ( j + 3k) metres, and also determine the magnitude of the tangential velocity. The velocity v of point P on a body rotating with angular velocity ω about a fixed axis is given by: v=ω ×r where r is the point on vector P Thus

v = (2i − 5 j + 7k) × ( j + 3k) i j k = 2 −5 7 0 1 3 = i(−15 − 7) − j(6 − 0) + k(2 − 0) = (−22i − 6 j + 2k) m/s

The magnitude of v, √ |v| = [(ω • ω)(r • r) − (ω • r)2 ] ω • ω = (2)(2) + (−5)(−5) + (7)(7) = 78 r • r = (0)(0) + (1)(1) + (3)(3) = 10 ω • r = (2)(0) + (−5)(1) + (7)(3) = 16 Hence,

√ (78 × 10 − 162 ) √ = 524 m/s = 22.89 m/s

|v| =

(b) (p + r) × q

4. (a) p × (r × q) (b) (3p × 2r) × q 5. For vectors p = 4i − j + 2k and q = −2i + 3 j − 2k determine: (a) p • q (b) p × q (c) |p × q| (d) q × p and (e) the angle between the vectors. 6. For vectors a = −7i + 4 j + 12 k and b = 6i − 5 j − k find (a) a • b (b) a × b (c) |a × b| (d) b × a and (e) the angle between the vectors. 7. Forces of (i + 3 j), (−2i − j), (−i − 2 j) newtons act at three points having position vectors of (2i + 5 j), 4 j and (−i + j) metres respectively. Calculate the magnitude of the moment. 8. A force of (2i − j + k) newtons acts on a line through point P having co-ordinates (0, 3, 1) metres. Determine the moment vector and its magnitude about point Q having co-ordinates (4, 0, −1) metres. 9. A sphere is rotating with angular velocity ω about the z-axis of a system, the axis coinciding with the axis of the sphere. Determine the velocity vector and its magnitude at position (−5i + 2 j − 7k) m, when the angular velocity is (i + 2 j) rad/s. 10.

Calculate the velocity vector and its magnitude for a particle rotating about the z-axis at an angular velocity of (3i − j + 2k) rad/s when the position vector of the particle is at (i − 5 j + 4k) m.

Scalar and vector products 321 24.4

Vector equation of a line

The equation of a straight line may be determined, given that it passes through point A with position vector a relative to O, and is parallel to vector b. Let r be the position vector of a point P on the line, as shown in Fig. 24.10.

b

Problem 11. (a) Determine the vector equation of the line through the point with position vector 2i + 3 j − k which is parallel to the vector i − 2 j + 3k. (b) Find the point on the line corresponding to λ = 3 in the resulting equation of part (a). (c) Express the vector equation of the line in standard Cartesian form. (a) From equation (8),

P

r = a + λb A

i.e. r = (2i + 3 j − k) + λ(i − 2 j + 3k) or

r a

r = (2 + λ)i + (3 − 2λ) j + (3λ − 1)k

which is the vector equation of the line. (b) O

When λ = 3,

r = 5i − 3 j + 8k

(c) From equation (9), y − a2 z − a3 x − a1 = = =λ b1 b2 b3

Figure 24.10

By vector addition, OP = OA + AP, i.e. r = a + AP However, as the straight line through A is parallel to the free vector b (free vector means one that has the same magnitude, direction and sense), then AP = λ b, where λ is a scalar quantity. Hence, from above, r=a+λb

(8)

Since a = 2i + 3 j − k, then a1 = 2, a2 = 3 and a3 = −1 and b = i − 2 j + 3k, then b1 = 1, b2 = −2 and b3 = 3 Hence, the Cartesian equations are:

If, say, r = xi + yj + zk, a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, then from equation (8),

x − 2 y − 3 z − (−1) = = =λ 1 −2 3

xi + yj + zk = (a1 i + a2 j + a3 k)

i.e.

x−2 =

3−y z+1 = =λ 2 3

+ λ(b1 i + b2 j + b3 k) Problem 12. The equation Hence x = a1 + λb1 , y = a2 + λb2 and z = a3 + λb3 . Solving for λ gives: x − a1 y − a2 z − a3 = = =λ b1 b2 b3

(9)

Equation (9) is the standard Cartesian form for the vector equation of a straight line.

2x − 1 y + 4 −z + 5 = = 3 3 2 represents a straight line. Express this in vector form. Comparing the given equation with equation (9) shows that the coefficients of x, y and z need to be equal to unity.

322 Section F Thus

2x − 1 y + 4 −z + 5 = = becomes: 3 3 2 x − 21 3 2

=

y+4 z−5 = 3 −2

Again, comparing with equation (9), shows that

Practice Exercise 130 Multiple-choice questions on scalar and vector products (Answers on page 882) Each question has only one correct answer 1.

A vector q joins points A and B, where A has co-ordinates (3, −2, 1) and B has co-ordinates (4, 7, −5). The norm of q is: (a) i +9j −6k (b) 10.86 (c) i−9j +6k (d) 118

2.

If a = 3i + 2j − 4k and b = i − 2j + 5k then a • b is equal to: (a) 21 (b) 4i + k (c) −21 (d) −2i −4 j +9k

3.

If a = 4i −5j −3k and b = 6i −2j + 4k then | a + b | is equal to: √ (b) √ 10i −7j + k (a) 150 (c) 14.55 (d) 52

4.

A constant force of F = 9i + 5j −2k N displaces an object from P = 2i + 3j + 4k to Q = 7i −4j +6k (in metres). The work done is: (a) 2.652 kN m (b) −6 N m (c) 14i − 2j N m (d) 6 N m

5.

If p = 2i + 3j − 4k and q = i −4j + 3k then p × q is equal to: (a) −7i −10j −11k (b) −22 (c) 16.43 (d) −25i + 10j + 11k

6.

The moment of a force of (2i + 3j − k) N about a point H having co-ordinates (1, 2, 2) when the force acts on a line through G whose co-ordinates are (3, 2, 5) is: (a) 521 N m (b) −9i + 8j + 6k N m (c) 22.83 N m (d) 9i − 8j −6k N m

1 a1 = , a2 = −4 and a3 = 5 and 2 3 b1 = , b2 = 3 and b3 = −2 2 In vector form the equation is: r = (a1 + λb1 )i + (a2 + λb2 ) j + (a3 + λb3 )k, from equation (8) ( ) 1 3 i.e. r = + λ i + (−4 + 3λ) j + (5 − 2λ)k 2 2 1 or r = (1 + 3λ)i + (3λ − 4) j + (5 − 2λ)k 2 Now try the following Practice Exercise Practice Exercise 129 The vector equation of a line (Answers on page 882) 1. Find the vector equation of the line through the point with position vector 5i − 2j + 3k which is parallel to the vector 2i + 7j − 4k. Determine the point on the line corresponding to λ = 2 in the resulting equation. 2. Express the vector equation of the line in problem 1 in standard Cartesian form. In problems 3 and 4, express the given straight line equations in vector form. 3x − 1 5y + 1 4 − z = = 4 2 3 1 − 4y 3z − 1 4. 2x + 1 = = 5 4 3.

For fully worked solutions to each of the problems in Practice Exercises 127 to 129 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 7

Vectors

This Revision Test covers the material contained in Chapters 22 to 24. The marks for each question are shown in brackets at the end of each question. 1.

State whether the following are scalar or vector quantities: (a) A temperature of 50◦ C (b)

4.

v1 = 150 sin(ωt + π/3) volts and

A downward force of 80 N

v2 = 90 sin(ωt − π/6) volts

(c) 300 J of work (d)

A south-westerly wind of 15 knots

Plot the two voltages on the same axes to scales π of 1 cm = 50 volts and 1 cm = rad. 6 Obtain a sinusoidal expression for the resultant v1 + v2 in the form R sin(ωt + α): (a) by adding ordinates at intervals and (b) by calculation. (13)

(e) 70 m distance (f) 2.

The instantaneous values of two alternating voltages are given by:

An acceleration of 25 m/s2 at 30◦ to the horizontal (6)

Calculate the resultant and direction of the force vectors shown in Fig. RT7.1, correct to 2 decimal places. (7)

5.

If velocity v1 = 26 m/s at 52◦ and v2 = 17 m/s at −28◦ calculate the magnitude and direction of v1 + v2 , correct to 2 decimal places, using complex numbers. (10)

6.

Given a = −3i + 3j + 5k, b = 2i − 5j + 7k and c = 3i + 6j − 4k, determine the following: (a) −4b (b) a + b − c (c) 5b − 3c (8)

7.

If a = 2i + 4 j − 5k and b = 3i − 2j + 6k determine: (a) a · b (b) |a + b| (c) a × b (d) the angle between a and b (14)

8.

Determine the work done by a force of F newtons acting at a point A on a body, when A is displaced to point B, the co-ordinates of A and B being (2, 5, −3) and (1, −3, 0) metres respectively, and when F = 2i − 5j + 4k newtons. (4)

9.

A force of F = 3i − 4 j + k newtons acts on a line passing through a point P. Determine moment M and its magnitude of the force F about a point Q when P has co-ordinates (4, −1, 5) metres and Q has co-ordinates (4, 0, −3) metres. (8)

5N

7N

Figure RT7.1

3.

Four coplanar forces act at a point A as shown in Fig. RT7.2 Determine the value and direction of the resultant force by (a) drawing (b) by calculation using horizontal and vertical components. (10) 4N A 5N

458

458

7N 8N

Figure RT7.2

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 7, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Section G

Differential calculus

Chapter 25

Methods of differentiation Why it is important to understand: Methods of differentiation There are many practical situations engineers have to analyse which involve quantities that are varying. Typical examples include the stress in a loaded beam, the temperature of an industrial chemical, the rate at which the speed of a vehicle is increasing or decreasing, the current in an electrical circuit or the torque on a turbine blade. Further examples include the voltage on a transmission line, the rate of growth of a bacteriological culture and the rate at which the charge on a capacitor is changing. Differential calculus, or differentiation, is a mathematical technique for analysing the way in which functions change. There are many methods and rules of differentiation which are individually covered in the following chapters. A good knowledge of algebra, in particular, laws of indices, is essential. Calculus is one of the most powerful mathematical tools used by engineers. This chapter explains how to differentiate common functions, products, quotients and function of a function – all important methods providing a basis for further study in later chapters.

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

differentiate common functions differentiate a product using the product rule differentiate a quotient using the quotient rule differentiate a function of a function differentiate successively

25.1

Introduction to calculus

Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions – such as velocity and acceleration, rates of change and maximum and minimum values of curves. Calculus has widespread applications in science and engineering and is used to solve complicated problems for which algebra alone is insufficient. Calculus is a subject that falls into two parts:

(i)

differential calculus, or differentiation, which is covered in Chapters 25 to 34, and

(ii)

integral calculus, or integration, which is covered in Chapters 35 to 45.

25.2

The gradient of a curve

If a tangent is drawn at a point P on a curve, then the gradient of this tangent is said to be the gradient of the curve at P. In Fig. 25.1, the gradient of the curve at P is equal to the gradient of the tangent PQ.

328 Section G f(x) f(x) 10

Q

B

f(x) 5 x 2

8 6

P

4 0

C

x 2 A

Figure 25.1 0 f(x)

1

D

1.5

2

3

x

Figure 25.3 B

(iii)

the gradient of chord AD =

f(x2)

A C

(iv) if E is the point on the curve (1.1, f (1.1)) then the gradient of chord AE

f(x1) E x1

0

D x2

x

Figure 25.2

= (v)

The gradient of the chord AB BC BD − CD f (x2 ) − f (x1 ) = = AC ED (x2 − x1 )

For the curve f (x) = x2 shown in Fig. 25.3. (i)

the gradient of chord AB

=

(ii)

f(3) − f (1) 9 − 1 = =4 3−1 2

the gradient of chord AC

=

f(2) − f (1) 4 − 1 = =3 2−1 1

f(1.1) − f (1) 1.21 − 1 = = 2.1 1.1 − 1 0.1

if F is the point on the curve (1.01, f(1.01)) then the gradient of chord AF

For the curve shown in Fig. 25.2, let the points A and B have co-ordinates (x1 , y1 ) and (x2 , y2 ), respectively. In functional notation, y1 = f(x1 ) and y2 = f(x2 ) as shown.

=

f(1.5) − f (1) 2.25 − 1 = = 2.5 1.5 − 1 0.5

=

f(1.01) − f (1) 1.0201 − 1 = = 2.01 1.01 − 1 0.01

Thus as point B moves closer and closer to point A the gradient of the chord approaches nearer and nearer to the value 2. This is called the limiting value of the gradient of the chord AB and when B coincides with A the chord becomes the tangent to the curve.

25.3 Differentiation from first principles In Fig. 25.4, A and B are two points very close together on a curve, δx (delta x) and δy (delta y) representing small increments in the x and y directions, respectively. δy Gradient of chord AB = ; however, δx δy = f(x + δx) − f (x) Hence

δy f(x + δx) − f (x) = δx δx

Methods of differentiation 329 y

B (x 1 dx, y 1 dy)

Substituting (x + δx) for x gives f (x + δx) = (x + δx)2 = x2 + 2xδx + δx2 , hence { 2 } (x + 2xδx + δx2 ) − (x2 ) f ′ (x) = limit δx→0 δx {

dy

= limit δx→0

A(x, y)

(2xδx + δx2 ) δx

}

f(x 1 dx)

= limit [2x + δx]

dx

f(x)

δx→0

0

x

Figure 25.4

δy approaches a limiting value As δx approaches zero, δx and the gradient of the chord approaches the gradient of the tangent at A. When determining the gradient of a tangent to a curve there are two notations used. The gradient of the curve at A in Fig. 25.4 can either be written as } { δy f(x + δx) − f (x) limit or limit δx→0 δx δx→0 δx In Leibniz∗ notation,

dy δy = limit dx δx→0 δx

In functional notation, {

′

f (x) = limit

δx→0

f(x + δx) − f(x) δx

}

dy is the same as f ′ (x) and is called the differential dx coefficient or the derivative. The process of finding the differential coefficient is called differentiation. Problem 1. Differentiate from first principle f (x) = x2 and determine the value of the gradient of the curve at x = 2

As δx → 0, [2x + δx] → [2x + 0]. Thus f ′ (x) = 2x, i.e. the differential coefficient of x2 is 2x. At x = 2, the gradient of the curve, f ′ (x) = 2(2) = 4 Differentiation from first principles can be a lengthy process and it would not be convenient to go through this procedure every time we want to differentiate a function. In reality we do not have to because a set of general rules have evolved from the above procedure, which we consider in the following section.

25.4 Differentiation of common functions From differentiation by first principles of a number of examples such as in Problem 1 above, a general rule for differentiating y = axn emerges, where a and n are constants. The rule is: if y = axn then

dy = anxn−1 dx

(or, if f(x) = axn then f ′ (x) = anxn−1 ) and is true for all real values of a and n. For example, if y = 4x3 then a = 4 and n = 3, and dy = anxn−1 = (4)(3)x3−1 = 12x2 dx If y = axn and n = 0 then y = ax0 and

To ‘differentiate from first principles’ means ‘to find f ′ (x)’ by using the expression { } f (x + δx) − f (x) f ′ (x) = limit δx→0 δx f(x) = x2 ∗

Who was Leibniz? For image and resume go to www.routledge.com/cw/bird

dy = (a)(0)x0−1 = 0, dx i.e. the differential coefficient of a constant is zero. Fig. 25.5(a) shows a graph of y = sin x. The gradient is continually changing as the curve moves from 0 to dy A to B to C to D. The gradient, given by , may be dx plotted in a corresponding position below y = sin x, as shown in Fig. 25.5(b).

330 Section G y

The standard derivatives summarised below may be proved theoretically and are true for all real values of x

A y 5 sin x 1 B (a)

0

D

p

p 2

3p 2

2p

x rad

2 C 09 dy dx 1 (b)

A9 p 2

0

D9

d (sin x) 5 cos x dx

3p 2

2p

axn

anxn−1

sin ax

a cos ax

cos ax

−a sin ax

e

aeax

ln ax

1 x

x rad

2

The differential coefficient of a sum or difference is the sum or difference of the differential coefficients of the separate terms.

B9

Thus, if f (x) = p(x) + q(x) − r(x), (where f, p, q and r are functions),

Figure 25.5

(i)

At 0, the gradient is positive and is at its steepest. Hence 0′ is a maximum positive value.

(ii)

Between 0 and A the gradient is positive but is decreasing in value until at A the gradient is zero, shown as A′ .

(iii)

dy or f ′ (x) dx

ax

C9 p

y or f(x)

Between A and B the gradient is negative but is increasing in value until at B the gradient is at its steepest negative value. Hence B′ is a maximum negative value.

(iv) If the gradient of y = sin x is further investigated dy between B and D then the resulting graph of dx is seen to be a cosine wave. Hence the rate of change of sin x is cos x, i.e. if y = sin x then

dy = cos x dx

By a similar construction to that shown in Fig. 25.5 it may be shown that: if y = sin ax then

dy = a cos ax dx

If graphs of y = cos x, y = ex and y = ln x are plotted and their gradients investigated, their differential coefficients may be determined in a similar manner to that shown for y = sin x. The rate of change of a function is a measure of the derivative.

then

f ′ (x) = p′ (x) + q′ (x) − r′ (x)

Differentiation of common functions is demonstrated in the following worked Problems. Problem 2.

Find the differential coefficients of 12 (a) y = 12x3 (b) y = 3 x If y = axn then

dy = anxn−1 dx

(a) Since y = 12x3 , a = 12 and n = 3 thus dy = (12)(3)x3−1 = 36x2 dx 12 (b) y = 3 is rewritten in the standard axn form as x y = 12x−3 and in the general rule a = 12 and n=−3 dy 36 Thus = (12)(−3)x−3−1 = −36x−4 = − 4 dx x

Problem 3.

Differentiate (a) y = 6 (b) y = 6x

(a) y = 6 may be written as y = 6x0 , i.e. in the general rule a = 6 and n = 0 Hence

dy = (6)(0)x0−1 = 0 dx

Methods of differentiation 331 In general, the differential coefficient of a constant is always zero. (b)

Thus

Since y = 6x, in the general rule a = 6 and n = 1 Hence

dy = (6)(1)x1−1 = 6x0 = 6 dx

In general, the differential coefficient of kx, where k is a constant, is always k.

dy 1 = (5)(4)x4−1 + (4)(1)x1−1 − (−2)x−2−1 dx 2 ( ) 1 − 1 −1 + (1) − x 2 −0 2 1 3 = 20x3 + 4 + x−3 − x− 2 2

i.e.

dy 1 1 = 20x3 + 4 + 3 − √ dx x 2 x3

Problem 6. Find the differential coefficients of (a) y = 3 sin 4x (b) f (t) = 2 cos 3t with respect to the variable.

Problem 4. Find the derivatives of √ 5 (a) y = 3 x (b) y = √ 3 4 x √ (a) y = 3 x is rewritten in the standard differential

(a) When y = 3 sin 4x then

1

form as y = 3x 2 In the general rule, a = 3 and n =

Thus

( ) 1 dy 1 3 1 = (3) x 2 −1 = x− 2 dx 2 2 =

(b)

1 2

3 1 2x 2

3 = √ 2 x

4 5 5 y= √ = 4 = 5x− 3 in the standard differential 3 4 x x3 form. In the general rule, a = 5 and n = − 43

Thus

( ) dy 4 − 4 −1 −20 − 7 = (5) − x 3 = x 3 dx 3 3 =

−20 7 3x 3

−20 = √ 3 3 x7

Problem 5.

Differentiate, with respect to x, 1 1 y = 5x4 + 4x − 2 + √ − 3 2x x y = 5x4 + 4x −

1 1 + √ − 3 is rewritten as 2x2 x

1 1 y = 5x4 + 4x − x−2 + x− 2 −3 2

When differentiating a sum, each term is differentiated in turn.

(b)

dy = (3)(4 cos 4x) dx = 12 cos 4x

When f (t) = 2 cos 3t then f ′ (t) = (2)(−3 sin 3t) = −6 sin 3t

Problem 7. Determine the derivatives of 2 (a) y = 3e5x (b) f (θ) = 3θ (c) y = 6 ln 2x e (a) When y = 3e5x then (b)

f (θ) =

dy = (3)(5)e5x = 15e5x dx

2 = 2e−3θ , thus e3θ

−6 e3θ ( ) dy 1 6 (c) When y = 6 ln 2x then =6 = dx x x f ′ (θ) = (2)(−3)e−30 = −6e−3θ =

Problem 8. Find the gradient of the curve y = 3x4 − 2x2 + 5x − 2 at the points (0, −2) and (1, 4) The gradient of a curve at a given point is given by the corresponding value of the derivative. Thus, since y = 3x4 − 2x2 + 5x − 2 Then the gradient =

dy = 12x3 − 4x + 5 dx

At the point (0, −2), x = 0 Thus the gradient = 12(0)3 − 4(0) + 5 = 5 At the point (1, 4), x = 1 Thus the gradient = 12(1)3 − 4(1) + 5 = 13

332 Section G Problem 9. Determine the co-ordinates of the point on the graph y = 3x2 − 7x + 2 where the gradient is −1 The gradient of the curve is given by the derivative.

Evaluate

11.

A mass, m, is held by a spring with a stiffness constant k. The potential energy, p, of the 1 system is given by: p = kx2 − mgx where x 2 is the displacement and g is acceleration due to gravity. dp The system is in equilibrium if = 0. dx Determine the expression for x for system equilibrium.

12.

The current i flowing in an inductor of inductance 100 mH is given by: i = 5 sin 100t amperes, where t is the time t in seconds. The voltage v across the inductor is given by: di v = L volts. dt Determine the voltage when t = 10 ms.

dy When y = 3x2 − 7x + 2 then = 6x − 7 dx Since the gradient is −1 then 6x − 7 = −1, from which, x=1 When x = 1, y = 3(1)2 − 7(1) + 2 = −2 Hence the gradient is −1 at the point (1, −2) Now try the following Practice Exercise Practice Exercise 131 Differentiating common functions (Answers on page 882) In Problems 1 to 6 find the differential coefficients of the given functions with respect to the variable. 1. (a) 5x5 (b) 2.4x3.5 (c)

1 x

25.5

−4 (b) 6 (c) 2x x2 √ √ 4 3 3. (a) 2 x (b) 3 x5 (c) √ x 2. (a)

3 e5x

7. Find the gradient of the curve y = 2t4 + 3t3 − t + 4 at the points (0, 4) and (1, 8)

This is known as the product rule. Problem 10. Find the differential coefficient of y = 3x2 sin 2x 3x2 sin 2x is a product of two terms 3x2 and sin 2x Let u = 3x2 and v = sin 2x Using the product rule:

8. Find the co-ordinates of the point on the graph y = 5x2 − 3x + 1 where the gradient is 2 (a)

(b)

dy dv du =u +v dx dx dx

then

√ ex − e−x 1− x 6. (a) 4 ln 9x (b) (c) 2 x

9.

Differentiation of a product

When y = uv, and u and v are both functions of x,

−3 4. (a) √ (b) (x − 1)2 (c) 2 sin 3x 3 x 5. (a) −4 cos 2x (b) 2e6x (c)

ds , correct to 3 significant figures, dt √ π when t = given s = 3 sin t − 3 + t 6

10.

2 + 2 ln 2θ − θ2 2 2 (cos 5θ + 3 sin 2θ) − 3θ e dy π Evaluate in part (a) when θ = , dθ 2 correct to 4 significant figures.

dy = dx

Differentiate y =

gives: i.e.

dy dx

u

dv dx

↓

↓

+

v

du dx

↓

↓

2

= (3x )(2 cos 2x) + (sin 2x)(6x)

dy = 6x2 cos 2x + 6x sin 2x dx = 6x(xcos 2x + sin 2x)

Methods of differentiation 333 Note that the differential coefficient of a product is not obtained by merely differentiating each term and multiplying the two answers together. The product rule formula must be used when differentiating products.

Problem 13. Determine the rate of change of voltage, given v = 5t sin 2t volts when t = 0.2 s Rate of change of voltage =

Problem 11. Find the√rate of change of y with respect to x given y = 3 x ln 2x dy The rate of change of y with respect to x is given by dx 1 √ y = 3 x ln 2x = 3x 2 ln 2x, which is a product. 1

Let u = 3x 2 and v = ln 2x dy dv Then = u dx dx ( =

↓ 1 3x 2

+

v

du dx

↓ ↓ ↓ )( ) [ ( ) ] 1 1 1 + (ln 2x) 3 x 2 −1 x 2

( ) 3 −1 + (ln 2x) x 2 2 ( ) 1 1 = 3x− 2 1 + ln 2x 2 ( ) dy 3 1 = √ 1 + ln 2x dx 2 x 1 = 3x 2 −1

i.e.

dv dt

= (5t)(2 cos 2t) + (sin 2t)(5) = 10t cos 2t + 5 sin 2t When t = 0.2,

dv = 10(0.2) cos 2(0.2) + 5 sin 2(0.2) dt

= 2 cos 0.4 + 5 sin 0.4 (where cos 0.4 means the cosine of 0.4 radians) dv Hence = 2(0.92106) + 5(0.38942) dt = 1.8421 + 1.9471 = 3.7892 i.e. the rate of change of voltage when t = 0.2 s is 3.79 volts/s, correct to 3 significant figures. Now try the following Practice Exercise Practice Exercise 132 Differentiating products (Answers on page 882) In Problems 1 to 8 differentiate the given products with respect to the variable. 1. x sin x

Problem 12. Differentiate y = x3 cos 3x ln x 2. x2 e2x 3

Let u = x cos 3x (i.e. a product) and v = ln x Then

dy dv du =u +v dx dx dx

where

du = (x3 )(−3 sin 3x) + (cos 3x)(3x2 ) dx

3. x2 ln x 4. 2x3 cos 3x √ 5. x3 ln 3x

dv 1 and = dx x ( ) dy 1 Hence = (x3 cos 3x) + (ln x)[−3x3 sin 3x dx x + 3x2 cos 3x]

6. e3t sin 4t 7. e4θ ln 3θ 8. et ln t cos t di , correct to 4 significant figures, dt when t = 0.1, and i = 15t sin 3t

9. Evaluate

= x2 cos 3x + 3x2 ln x(cos 3x − x sin 3x) i.e.

dy = x2 {cos 3x + 3 ln x(cos 3x − x sin 3x)} dx

10.

dz , correct to 4 significant figures, dt when t = 0.5, given that z = 2e3t sin 2t Evaluate

334 Section G 25.6

Differentiation of a quotient

u When y = , and u and v are both functions of x v du dv dy v dx − u dx = dx v2

then

This is known as the quotient rule. Problem 14. Find the differential coefficient of 4 sin 5x y= 5x4 4 sin 5x is a quotient. Let u = 4 sin 5x and v = 5x4 5x4 (Note that v is always the denominator and u the numerator.) du dv dy v dx − u dx = dx v2 du where = (4)(5) cos 5x = 20 cos 5x dx dv and = (5)(4)x3 = 20x3 dx dy (5x4 )(20 cos 5x) − (4 sin 5x)(20x3 ) = Hence dx (5x4 )2

i.e.

=

100x4 cos 5x − 80x3 sin 5x 25x8

=

20x3 [5x cos 5x − 4 sin 5x] 25x8

dy 4 = (5x cos 5x − 4 sin 5x) dx 5x5

Note that the differential coefficient is not obtained by merely differentiating each term in turn and then dividing the numerator by the denominator. The quotient formula must be used when differentiating quotients. Problem 15. Determine the differential coefficient of y = tan ax sin ax (from Chapter 13). Differentiation cos ax of tan ax is thus treated as a quotient with u = sin ax and v = cos ax y = tan ax =

du dv dy v dx − u dx = dx v2 (cos ax)(a cos ax) − (sin ax)(−a sin ax) = (cos ax)2 2 a cos ax + a sin2 ax a(cos2 ax + sin2 ax) = = (cos ax)2 cos2 ax a = , since cos2 ax + sin2 ax = 1 cos2 ax (see Chapter 13) Hence

dy 1 = a sec2 ax since sec2 ax = (see dx cos2 ax Chapter 8).

Problem 16. Find the derivative of y = sec ax 1 y = sec ax = (i.e. a quotient). Let u = 1 and cos ax v = cos ax du dv dy v dx − u dx = dx v2 =

(cos ax)(0) − (1)(−a sin ax) (cos ax)2

( )( ) a sin ax 1 sin ax = =a cos2 ax cos ax cos ax i.e.

dy = a sec ax tan ax dx

Problem 17. Differentiate y =

te2t 2 cos t

te2t The function is a quotient, whose numerator is a 2 cos t product. Let u = te2t and v = 2 cos t then du = (t)(2e2t ) + (e2t )(1) from the product rule and dt dv = −2 sin t dt du dv dy v dt − u dt = Hence dt v2 =

(2 cos t)[2te2t + e2t ] − (te2t )(−2 sin t) (2 cos t)2

=

4te2t cos t + 2e2t cos t + 2te2t sin t 4 cos2 t

Methods of differentiation 335 = i.e.

2e2t [2t cos t + cos t + t sin t] 4 cos2 t

dy e2t = (2t cos t + cos t + t sin t) dt 2 cos2 t

Problem 18. Determine the of the ( gradient √ ) √ 5x 3 curve y = 2 at the point 3, 2x + 4 2

2xe4x sin x

7.

8. Find the gradient of the curve y = point (2, −4)

dy at x = 2.5, correct to 3 significant dx 2x2 + 3 figures, given y = ln 2x

9. Evaluate

Let y = 5x and v = 2x2 + 4 du dv dy v dx − u dx (2x2 + 4)(5) − (5x)(4x) = = 2 dx v (2x2 + 4)2 10x2 + 20 − 20x2 20 − 10x2 = 2 2 (2x + 4) (2x2 + 4)2 ( √ ) √ √ 3 At the point 3, , x = 3, 2 =

√ dy 20 − 10( 3)2 hence the gradient = = √ dx [2( 3)2 + 4]2 =

20 − 30 1 =− 100 10

Now try the following Practice Exercise Practice Exercise 133 Differentiating quotients (Answers on page 882) In Problems 1 to 7, differentiate the quotients with respect to the variable. 1. 2. 3. 4. 5. 6.

sin x x 2 cos 3x x3 2x 2 x +1 √ x cos x √ 3 θ3 2 sin 2θ ln 2t √ t

2x at the x2 − 5

25.7

Function of a function

It is often easier to make a substitution before differentiating. dy dy du = × dx du dx

If y is a function of x then

This is known as the ‘function of a function’ rule (or sometimes the chain rule). For example, if y = (3x − 1)9 then, by making the substitution u = (3x − 1), y = u9 , which is of the ‘standard’ form. Hence Then

dy du = 9u8 and =3 du dx

dy dy du = × = (9u8 )(3) = 27u8 dx du dx

Rewriting u as (3x − 1) gives:

dy = 27(3x − 1)8 dx

Since y is a function of u, and u is a function of x, then y is a function of a function of x. Problem 19. Differentiate y = 3 cos(5x2 + 2) Let u = 5x2 + 2 then y = 3 cos u Hence

du dy = 10x and = −3 sin u dx du

Using the function of a function rule, dy dy du = × = (−3 sin u)(10x) = −30x sin u dx du dx Rewriting u as 5x2 + 2 gives: dy = −30x sin(5x2 + 2) dx

336 Section G Problem 20. Find the derivative of y = (4t3 − 3t)6

Problem 23. Find the differential coefficient of 2 y= 3 (2t − 5)4

Let u = 4t3 − 3t, then y = u6 du dy Hence = 12t2 − 3 and = 6u5 dt du Using the function of a function rule, dy dy du = × = (6u5 )(12t2 − 3) dt du dt

y=

2 (2t3 − 5)4 −4

y = 2u

Hence Then

Rewriting u as (4t3 − 3t) gives:

= 2(2t3 − 5)−4 . Let u = (2t3 − 5), then

du dy −8 = 6t2 and = −8u−5 = 5 dt du u ( ) dy dy du −8 = × = (6t2 ) dt du dt u5

dy = 6(4t3 − 3t)5 (12t2 − 3) dt

=

−48t2 (2t3 − 5)5

= 18(4t2 − 1)(4t3 − 3t)5 Now try the following Practice Exercise Problem 21. Determine the differential √ 2 coefficient of y = (3x + 4x − 1) y=

√

1

(3x2 + 4x − 1) = (3x2 + 4x − 1) 2

Let u = 3x2 + 4x − 1 then

1 y=u2

du dy 1 − 1 1 = 6x + 4 and = u 2= √ dx du 2 2 u Using the function of a function rule, ( ) dy dy du 1 3x + 2 √ (6x + 4) = √ = × = dx du dx 2 u u Hence

i.e.

dy 3x + 2 =√ dx (3x2 + 4x − 1)

Problem 22. Differentiate y = 3 tan4 3x 4

Let u = tan 3x then y = 3u Hence

du = 3 sec2 3x, (from Problem 15), and dx dy = 12u3 du

Then

dy dy du = × = (12u3 )(3 sec2 3x) dx du dx = 12(tan 3x)3 (3 sec2 3x)

i.e.

Practice Exercise 134 Function of a function (Answers on page 882)

dy = 36 tan3 3x sec2 3x dx

In Problems 1 to 9, find the differential coefficients with respect to the variable. 1. (2x − 1)6 2. (2x3 − 5x)5 3. 2 sin(3θ − 2) 4. 2 cos5 α 1 5. (x3 − 2x + 1)5 6. 5e2t+1 7. 2 cot(5t2 + 3) 8. 6 tan(3y + 1) 9. 2etan θ 10.

11.

( π) Differentiate θ sin θ − with respect to θ, 3 and evaluate, correct to 3 significant figures, π when θ = 2 The extension, x metres, of an undamped vibrating spring after t seconds is given by: x = 0.54 cos(0.3t − 0.15) + 3.2 Calculate the speed of the spring, given by dx , when (a) t = 0, (b) t = 2s dt

Methods of differentiation 337 25.8

y = 2xe−3x (i.e. a product)

Successive differentiation

When a function y = f(x) is differentiated with respect dy to x the differential coefficient is written as or f ′ (x). dx If the expression is differentiated again, the second difd2 y ferential coefficient is obtained and is written as 2 dx (pronounced dee two y by dee x squared) or f ′′ (x) (pronounced f double-dash x). By successive differentiation further higher derivatives d3 y d4 y such as 3 and 4 may be obtained. dx dx Thus if y = 3x4 , 3

dy d2 y = 12x3 , 2 = 36x2 , dx dx

4

Hence

dy = (2x)(−3e−3x ) + (e−3x )(2) dx = −6xe−3x + 2e−3x d2 y = [(−6x)(−3e−3x ) + (e−3x )(−6)] dx2 + (−6e−3x ) = 18xe−3x − 6e−3x − 6e−3x d2 y = 18xe−3x − 12e−3x dx2

i.e.

Substituting values into

5

d y d y d y = 72x, 4 = 72 and 5 = 0 3 dx dx dx

(18xe−3x − 12e−3x ) + 6(−6xe−3x + 2e−3x )

Problem 24. If f(x) = 2x5 − 4x3 + 3x − 5, find f ′′ (x)

+ 9(2xe−3x ) = 18xe−3x − 12e−3x − 36xe−3x + 12e−3x + 18xe−3x = 0

f (x) = 2x5 − 4x3 + 3x − 5 f ′ (x) = 10x4 − 12x2 + 3 ′′

Thus when y = 2xe−3x ,

f (x) = 40x − 24x = 4x(10x − 6) 3

2

Problem 25. If y = cos x − sin x, evaluate x, in the π d2 y range 0 ≤ x ≤ , when 2 is zero. 2 dx Since

y = cos x − sin x,

d2 y = −cos x + sin x dx2

d2 y dy + 6 + 9y gives: 2 dx dx

dy = −sin x − cos x dx

d2 y dy + 6 + 9y = 0 dx2 dx

d2 y Problem 27. Evaluate 2 when θ = 0 given dθ y = 4 sec 2θ Since y = 4 sec 2θ,

and then

dy = (4)(2) sec 2θ tan 2θ (from Problem 16) dθ = 8 sec 2θ tan 2θ (i.e. a product)

d2 y is zero, −cos x + sin x = 0, dx2 sin x i.e. sin x = cos x or =1 cos x When

d2 y = (8 sec 2θ)(2 sec2 2θ) dθ2

π Hence tan x = 1 and x = arctan1 = 45◦ or rads in 4 π the range 0 ≤ x ≤ 2 Problem 26. Given y = 2xe−3x show that d2 y dy + 6 + 9y = 0 2 dx dx

+ (tan 2θ)[(8)(2) sec 2θ tan 2θ] = 16 sec3 2θ + 16 sec 2θ tan2 2θ When

θ = 0,

d2 y = 16 sec3 0 + 16 sec 0 tan2 0 dθ2

= 16(1) + 16(1)(0) = 16

338 Section G Now try the following Practice Exercise 10. Practice Exercise 135 Successive differentiation (Answers on page 882) 1. If y = 3x4 + 2x3 − 3x + 2 find (a) 2.

d2 y d3 y (b) dx2 dx3 2 1 3 √ f (t) = t2 − 3 + − t + 1 5 t t determine f ′′ (t)

(a) Given (b)

Evaluate f ′′ (t) when t = 1

3. The charge q on the plates of a capacitor is t given by q = CVe− CR , where t is the time, C is the capacitance and R the resistance. Determine (a) the rate of change of charge, which is dq given by , (b) the rate of change of current, dt d2 q which is given by 2 dt In Problems 4 and 5, find the second differential coefficient with respect to the variable.

Practice Exercise 136 Multiple-choice questions on methods of differentiation (Answers on page 883) Each question has only one correct answer 1.

2.

4. (a) 3 sin 2t + cos t (b) 2 ln 4θ 2

5. (a) 2 cos x

(b) (2x − 3)

3.

2 3x 2 (c) 6e x + x 4.

8. Show that, if P and Q are constants and y = P cos(ln t) + Q sin(ln t), then d2 y dy +t +y = 0 dt2 dt

9. The displacement, s, of a (mass in ) a vibrating system is given by: s = 1 + t e−ωt where ω is the natural frequency of vibration. Show that: d2 s ds + 2ω + ω 2 s = 0 2 dt dt

5.

6.

with

(b) 6(sin 2t − cos 3t) (d) 6(cos 2t + sin 3t)

Given that y = 3e x + 2 ln 3x, (a) 6e x +

7. Show that the differential equation d2 y dy − 4 + 4y = 0 is satisfied dx2 dx when y = xe2x

t2

Differentiating y = 4x5 gives: dy 2 6 dy (a) = x (b) = 20x4 dx 3 dx dy dy (c) = 4x6 (d) = 5x4 dx dx Differentiating i = 3 sin 2t − 2 cos 3t respect to t gives: (a) 3 cos 2t + 2 sin 3t 3 2 (c) cos 2t + sin 3t 2 3

4

6. Evaluate f ′′ (θ) when θ = 0 given f (θ) = 2 sec 3θ

An equation for the deflection of a cantilever is: ( ) Px2 ( x ) Wx2 x2 y= L− + 3L2 + 2xL + 2EI 3 12EI 2 where P, W, L, E and I are all constants. If d2 y the bending moment M = −EI 2 deterdx mine an equation for M in terms of P, W, L and x.

dy is equal to: dx

2 x 2 (d) 3e x + 3

(b) 3e x +

Given f (t) = 3t4 − 7, f ′ (t) is equal to: 3 (a) 12t3 − 2 (b) t5 −2t + c 4 (c) 12t3 (d) 3t5 − 2 1 dy If y = 5x + 3 then is equal to: 2x dx 1 3 (a) 5 + 2 (b) 5 − 4 6x 2x 6 1 (c) 5 − 4 (d) 5x2 − 2 x 6x √ dy If y = t5 then is equal to: dx 5 2 5√ 3 5 (a) − √ (b) √ (c) t (d) t2 5 2 2 2 t7 5 t3

Methods of differentiation 339

dy 7. Given that y = cos 2x − e − 3x + ln 4x then dx is equal to: 1 (a) 2 sin 2x + 3e −3x + 4x 1 (b) −2 sin 2x − 3e −3x + 4x 1 (c) 2 sin 2x − 3e −3x + 4x 1 (d) −2 sin 2x + 3e −3x + x ( )2 dy is equal to: 8. Given that y = 2 x − 1 then dx (a) 4x + 4 (b) 4x − 2 (c) 4(x − 1) (d) 4x

14.

If f (x) = e 2x ln 2x, then f ′ (x) is equal to: 2e 2x e 2x (a) (b) + 2e 2x ln 2x x 2x ( ) 1 e 2x (c) (d) e 2x + 2 ln 2x 2x x

15.

The gradient of the curve y = 4x2 − 7x + 3 at the point (1, 0) is (a) 1 (b) 3 (c) 0 (d) −7

16.

Given y = 4x3 − 3 cos x − e − x + ln x − 12 dy is equal to: then dx 1 (a) 12x2 + 3 sin x − e − x + x 1 2 −x (b) 12x − 3 sin x − e + x (c) x4 − 3 sin x − e − x + x(1 + ln x) + 30x + c 1 (d) 12x2 + 3 sin x + e − x + x ds If s = 2t ln t then is equal to: dt 2 (a) 2 + 2 ln t (b) 2t + t (c) 2 − 2 ln t (d) 2 + 2e t

9. The gradient of the curve y = −2x3 + 3x + 8 at x = 2 is: (a) −21 (b) 27 (c) −16 (d) −5 10.

11.

12.

13.

An alternating current is given by i = 4 sin 150t amperes, where t is the time in seconds. The rate of change of current at t = 0.025 s is: (a) 3.99 A/s (b) −492.3 A/s (c) −3.28 A/s (d) 598.7 A/s √ dy If y = 5 x3 − 2, then is equal to: dx √ 15 √ x (b) 2 x5 − 2x + c (a) 2 √ 5√ (c) x − 2 (d) 5 x − 2x 2 1 If f(t) = 5t − √ , then f ′ (t) is equal to: t √ 1 (b) 5 − 2 t (a) 5 + √ 3 2 t √ 5t2 1 (c) − 2 t + c (d) 5 + √ 2 t3 d2 y If y = 3x − ln 5x then 2 is equal to: dx 1 1 (a) 6 + 2 (b) 6x − 5x x 1 1 (d) 6 + 2 (c) 6 − 5x x 2

17.

18.

19.

20.

An alternating voltage is given by v = 10 sin 300t volts, where t is the time in seconds. The rate of change of voltage when t = 0.01 s is: (a) −2996 V/s (b) 157 V/s (c) −2970 V/s (d) 0.523 V/s ( )2 dy Given that y = 2x2 − 3 , then is equal to: dx 3 3 (a) 16x (b) 16x − 24x (c) 4x2 − 6 (d) 16x − 18 √ If a function is given by y = 4x3 − 4 x, then d2 y is given by: dx2 1 1 (a) 24x + x 2 (b) 12x2 − 2x− 2 3

(c) 24x + x− 2

1

(d) 12x2 − x 2

For fully worked solutions to each of the problems in Practice Exercises 131 to 135 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 26

Some applications of differentiation Why it is important to understand: Some applications of differentiation In the previous chapter some basic differentiation techniques were explored, sufficient to allow us to look at some applications of differential calculus. Some practical rates of change problems are initially explained, followed by some practical velocity and acceleration problems. Determining maximum and minimum points and points of inflexion on curves, together with some practical maximum and minimum problems follow. Tangents and normals to curves and errors and approximations complete this initial look at some applications of differentiation. In general, with these applications, the differentiation tends to be straightforward.

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

determine rates of change using differentiation solve velocity and acceleration problems understand turning points determine the turning points on a curve and determine their nature solve practical problems involving maximum and minimum values determine points of inflexion on a curve determine tangents and normals to a curve determine small changes in functions

26.1

Rates of change

If a quantity y depends on and varies with a quantity x dy then the rate of change of y with respect to x is . Thus, dx for example, the rate of change of pressure p with height dp h is dh A rate of change with respect to time is usually just called ‘the rate of change’, the ‘with respect to time’ being assumed. Thus, for example, a rate of change of

di current, i, is and a rate of change of temperature, dt dθ θ, is , and so on. dt Problem 1. The length l metres of a certain metal rod at temperature θ◦ C is given by l = 1 + 0.00005θ + 0.0000004θ2 . Determine the rate of change of length, in mm/◦ C, when the temperature is (a) 100◦ C and (b) 400◦ C.

Some applications of differentiation 341 The rate of change of length means Since length then

dl dθ

l = 1 + 0.00005θ + 0.0000004θ2 , dl = 0.00005 + 0.0000008θ dθ

(a) When θ = 100◦ C, dl = 0.00005 + (0.0000008)(100) dθ = 0.00013 m/◦ C

The rate of change of temperature is Since

dθ = (θ0 )(−k)e−kt = −kθ0 e−kt dt

When

θ0 = 16, k = 0.03

then

dθ = −(0.03)(16)e−(0.03)(40) dt

dl = 0.00005 + (0.0000008)(400) dθ = 0.00037 m/◦ C

and

t = 40

= −0.48e−1.2 = −0.145◦ C/s

◦

When θ = 400◦ C,

θ = θ0 e−kt

then

= 0.13 mm/ C (b)

dθ dt

Problem 4. The displacement s cm of the end of a stiff spring at time t seconds is given by s = ae−kt sin 2πf t. Determine the velocity of the end of the spring after 1 s, if a = 2, k = 0.9 and f =5

◦

= 0.37 mm/ C

Problem 2. The luminous intensity I candelas of a lamp at varying voltage V is given by I = 4 ×10−4 V2 . Determine the voltage at which the light is increasing at a rate of 0.6 candelas per volt. The rate of change of light with respect to voltage is dI given by dV Since

ds Velocity, v = where s = ae−kt sin 2πf t (i.e. a dt product). Using the product rule, ds = (ae−kt )(2πf cos 2πf t) dt + (sin 2πft)(−ake−kt ) When a = 2, k = 0.9, f = 5 and t = 1, velocity, v = (2e−0.9 )(2π5 cos 2π5)

I = 4 × 10−4 V2 ,

+ (sin 2π5)(−2)(0.9)e−0.9

dI = (4 × 10−4 )(2)V = 8 × 10−4 V dV

= 25.5455 cos 10π − 0.7318 sin 10π = 25.5455(1) − 0.7318(0)

When the light is increasing at 0.6 candelas per volt then +0.6 = 8 × 10−4 V, from which, voltage 0.6 V= = 0.075 × 10+4 8 × 10−4 = 750 volts

Problem 3. Newton’s law of cooling is given by θ = θ0 e−kt , where the excess of temperature at zero time is θ0◦ C and at time t seconds is θ◦ C. Determine the rate of change of temperature after 40 s, given that θ0 = 16◦ C and k = 0.03

= 25.55 cm/s (Note that cos10π means ‘the cosine of 10π radians’, not degrees, and cos 10π ≡ cos 2π = 1) Now try the following Practice Exercise Practice Exercise 137 Rates of change (Answers on page 883) 1.

An alternating current, i amperes, is given by i = 10 sin 2πf t, where f is the frequency in hertz and t the time in seconds. Determine the

342 Section G

3. The voltage across the plates of a capacitor at any time t seconds is given by v = Ve−t/CR , where V, C and R are constants. Given V = 300 volts, C = 0.12 × 10−6 F and R = 4 × 106 Ω find (a) the initial rate of change of voltage, and (b) the rate of change of voltage after 0.5 s. 4. The pressure p of the atmosphere at height h above ground level is given by p = p 0 e−h/c , where p 0 is the pressure at ground level and c is a constant. Determine the rate of change of pressure with height when p 0 = 1.013 × 105 pascals and c = 6.05 × 104 at 1450 metres. 5. The volume, v cubic metres, of water in a reservoir varies with time t, in minutes. When a valve is opened the relationship between v and t is given by: v = 2 × 104 − 20t2 − 10t3 . Calculate the rate of change of water volume at the time when t = 3 minutes.

26.2

Velocity and acceleration

When a car moves a distance x metres in a time t seconds along a straight road, if the velocity v is constant then x v = m/s, i.e. the gradient of the distance/time graph t shown in Fig. 26.1 is constant. If, however, the velocity of the car is not constant then the distance/time graph will not be a straight line. It may be as shown in Fig. 26.2. The average velocity over a small time δt and distance δx is given by the gradient of the chord AB, i.e. δx the average velocity over time δt is δt As δt → 0, the chord AB becomes a tangent, such that at point A, the velocity is given by: dx v= dt

x

t

Time

Figure 26.1

Distance

2. The luminous intensity, I candelas, of a lamp is given by I = 6 × 10−4 V2 , where V is the voltage. Find (a) the rate of change of luminous intensity with voltage when V = 200 volts, and (b) the voltage at which the light is increasing at a rate of 0.3 candelas per volt.

Distance

rate of change of current when t = 20 ms, given that f = 150 Hz.

B

dx A dt Time

Figure 26.2

Hence the velocity of the car at any instant is given by the gradient of the distance/time graph. If an expression for the distance x is known in terms of time t then the velocity is obtained by differentiating the expression. The acceleration a of the car is defined as the rate of change of velocity. A velocity/time graph is shown in Fig. 26.3. If δv is the change in v and δt the correspondδv ing change in time, then a = δt As δt → 0, the chord CD becomes a tangent, such that at point C, the acceleration is given by: a=

dv dt

Hence the acceleration of the car at any instant is given by the gradient of the velocity/time graph. If an expression for velocity is known in terms of time t then the acceleration is obtained by differentiating the expression.

Some applications of differentiation 343

Velocity

(b)

D

When time t = 1.5 s, velocity v = 9(1.5)2 − 4(1.5) + 4 = 18.25 m/s and acceleration a = 18(1.5) − 4 = 23 m/s2

Problem 6. Supplies are dropped from a helicoptor and the distance fallen in a time t seconds is given by x = 12 gt2 , where g = 9.8 m/s2 . Determine the velocity and acceleration of the supplies after it has fallen for 2 seconds.

dv C dt

1 1 x = gt2 = (9.8)t2 = 4.9t2 m 2 2 dv v = = 9.8t m/s dt

Distance

Time

Velocity Figure 26.3

and acceleration dv dx Acceleration a = . However, v = . Hence dt dt ( ) d dx d2 x a= = 2 dt dt dt The acceleration is given by the second differential coefficient of distance x with respect to time t. Summarising, if a body moves a distance x metres in a time t seconds then: (i)

distance x = f(t)

(ii)

dx velocity v = f ′ (t) or , which is the gradient of dt the distance/time graph.

(iii)

acceleration a =

dv d2 x = f ′′ (t) or 2 , which is the dt dt gradient of the velocity/time graph.

a=

d2 x = 9.8 m/s2 dt2

When time t = 2 s, velocity, v = (9.8)(2) = 19.6 m/s 2

and acceleration a = 9.8 m/s (which is acceleration due to gravity). Problem 7. The distance x metres travelled by a vehicle in time t seconds after the brakes are applied is given by x = 20t − 53 t2 . Determine (a) the speed of the vehicle (in km/h) at the instant the brakes are applied, and (b) the distance the car travels before it stops. (a) Distance, x = 20t − 53 t2 dx 10 = 20 − t dt 3 At the instant the brakes are applied, time = 0. Hence velocity, v = 20 m/s Hence velocity v =

Problem 5. The distance x metres moved by a car in a time t seconds is given by x = 3t3 − 2t2 + 4t − 1. Determine the velocity and acceleration when (a) t = 0 and (b) t = 1.5 s. Distance

x = 3t3 − 2t2 + 4t − 1 m

Velocity

v=

Acceleration a =

dx = 9t2 − 4t + 4 m/s dt

(b)

d2 x 2 = 18t − 4 m/s dt2

(a) When time t = 0, velocity v = 9(0)2 − 4(0) + 4 = 4 m/s and acceleration a = 18(0) − 4 = −4 m/s2 (i.e. deceleration)

20 × 60 × 60 km/h 1000 = 72 km/h (Note: changing from m/s to km/h merely involves multiplying by 3.6) When the car finally stops, the velocity is zero, i.e. 10 10 v = 20 − t = 0, from which, 20 = t, giving 3 3 t = 6 s. Hence the distance travelled before the car stops is given by: =

a

x = 20t − 53 t2 = 20(6) − 53 (6)2 = 120 − 60 = 60 m

344 Section G Now try the following Practice Exercise Problem 8. The angular displacement θ radians of a flywheel varies with time t seconds and follows the equation θ = 9t2 − 2t3 . Determine (a) the angular velocity and acceleration of the flywheel when time, t = 1 s, and (b) the time when the angular acceleration is zero. (a) Angular displacement θ = 9t2 − 2t3 rad Angular velocity ω =

dθ = 18t − 6t2 rad/s dt

When time t = 1 s, ω = 18(1) − 6(1)2 = 12 rad/s d2 θ Angular acceleration α = 2 = 18 − 12t rad/s2 dt When time t = 1 s, α = 18 − 12(1) = 6 rad/s (b)

2

When the angular acceleration is zero, 18 − 12t = 0, from which, 18 = 12t, giving time, t = 1.5 s

Problem 9. The displacement x cm of the slide valve of an engine is given by x = 2.2 cos 5πt + 3.6 sin 5πt Evaluate the velocity (in m/s) when time t = 30 ms. Displacement x = 2.2 cos 5πt + 3.6 sin 5πt Velocity v =

dx dt

= (2.2)(−5π) sin 5πt + (3.6)(5π) cos 5πt = −11π sin 5πt + 18π cos 5πt cm/s When time t = 30 ms, velocity ( ) ( ) 30 30 = −11π sin 5π · 3 + 18π cos 5π · 3 10 10 = −11π sin 0.4712 + 18π cos 0.4712 = −11π sin 27◦ + 18π cos 27◦ = −15.69 + 50.39 = 34.7 cm/s = 0.347 m/s

Practice Exercise 138 Velocity and acceleration (Answers on page 883) 1. A missile fired from ground level rises x metres vertically upwards in t seconds and 25 x = 100t − t2 . Find (a) the initial velocity 2 of the missile, (b) the time when the height of the missile is a maximum, (c) the maximum height reached, (d) the velocity with which the missile strikes the ground. 2. The distance s metres travelled by a car in t seconds after the brakes are applied is given by s = 25t − 2.5t2 . Find (a) the speed of the car (in km/h) when the brakes are applied, (b) the distance the car travels before it stops. 3. The equation θ = 10π + 24t − 3t2 gives the angle θ, in radians, through which a wheel turns in t seconds. Determine (a) the time the wheel takes to come to rest, (b) the angle turned through in the last second of movement. 4. At any time t seconds the distance x metres of a particle moving in a straight line from a fixed point is given by x = 4t + ln(1 − t). Determine (a) the initial velocity and acceleration (b) the velocity and acceleration after 1.5 s (c) the time when the velocity is zero. 5. The angular displacement θ of a rotating disc t is given by θ = 6 sin , where t is the time 4 in seconds. Determine (a) the angular velocity of the disc when t is 1.5 s, (b) the angular acceleration when t is 5.5 s, and (c) the first time when the angular velocity is zero. 20t3 23t2 6. x = − + 6t + 5 represents the dis3 2 tance, x metres, moved by a body in t seconds. Determine (a) the velocity and acceleration at the start, (b) the velocity and acceleration when t = 3 s, (c) the values of t when the body is at rest, (d) the value of t when the acceleration is 37 m/s2 and (e) the distance travelled in the third second.

Some applications of differentiation 345

7. A particle has a displacement s given by s = 30t + 27t2 − 3t3 metres, where time t is in seconds. Determine the time at which the acceleration will be zero.

(i.e. ones containing trigonometric, exponential, logarithmic, hyperbolic and algebraic functions), and it is usually easier to apply than the algebraic method. Problem 10. Use Newton’s method to determine the positive root of the quadratic equation 5x2 + 11x − 17 = 0, correct to 3 significant figures. Check the value of the root by using the quadratic formula.

8. The angular displacement θ radians of the spoke of a wheel is given by the expression: 1 1 θ = t4 − t3 2 3 where t is the time in seconds. Calculate (a) the angular velocity after 2 seconds, (b) the angular acceleration after 3 seconds, (c) the time when the angular acceleration is zero.

The functional notation method is used to determine the first approximation to the root. f(x) = 5x2 + 11x − 17 f (0) = 5(0)2 + 11(0) − 17 = −17

9. A mass of 4000 kg moves along a straight line so that the distance s metres travelled in a time t seconds is given by: s = 4t2 + 3t + 2 Kinetic energy is given by the formula 21 mv2 joules, where m is the mass in kg and v is the velocity in m/s. Calculate the kinetic energy after 0.75s, giving the answer in kJ.

f (1) = 5(1)2 + 11(1) − 17 = −1 f (2) = 5(2)2 + 11(2) − 17 = 25

10. The height, h metres, of a liquid in(a chemical) t storage tank is given by: h = 100 1 − e − 40 where t seconds is the time from which the tank starts to be filled. Calculate the speed in m/s at which the liquid level rises in the tank after 2 seconds. 11. The position of a missile, neglecting gravitational force, is described by: s = (40t) ln(1 + t) where s is the displacement of the missile, in metres, and t is the time, in seconds. 10 seconds after the missile has been fired, calculate (a) the displacement, (b) the velocity, and (c) the acceleration

26.3

The Newton–Raphson method

The Newton–Raphson formula,∗ often just referred to as Newton’s method, may be stated as follows: If r1 is the approximate value of a real root of the equation f (x) = 0, then a closer approximation to the root r2 is given by: f (r1 ) r2 = r1 − ′ f (r1 ) The advantages of Newton’s method over the algebraic method of successive approximations is that it can be used for any type of mathematical equation

∗

Who were Newton and Raphson? Sir Isaac Newton PRS MP (25 December 1642–20 March 1727) was an English polymath. Newton showed that the motions of objects are governed by the same set of natural laws, by demonstrating the consistency between Kepler’s laws of planetary motion and his theory of gravitation. To find out more go to www.routledge.com/cw/bird Joseph Raphson was an English mathematician known best for the Newton–Raphson method for approximating the roots of an equation. To find out more go to www.routledge.com/cw/bird

346 Section G This shows that the value of the root is close to x = 1 Let the first approximation to the root, r1 , be 1 Newton’s formula states that a closer approximation, f(r1 ) r2 = r1 − ′ f (r1 ) f (x) = 5x2 + 11x − 17 thus, f(r1 ) = 5(r1 )2 + 11(r1 ) − 17 = 5(1)2 + 11(1) − 17 = −1 f ′ (x) is the differential coefficient of f(x), i.e.

Problem 11. Taking the first approximation as 2, determine the root of the equation x2 − 3 sin x + 2 ln(x + 1) = 3.5, correct to 3 significant figures, by using Newton’s method. f (r1 ) , where r1 f ′ (r1 ) is a first approximation to the root and r2 is a better approximation to the root. Newton’s formula states that r2 = r1 −

Since f (x) = x2 − 3 sin x + 2 ln (x + 1) − 3.5

f ′ (x) = 10x + 11

f (r1 ) = f (2) = 22 − 3 sin 2 + 2 ln 3 − 3.5

′

Thus f (r1 ) = 10(r1 ) + 11 = 10(1) + 11 = 21 By Newton’s formula, a better approximation to the root is: −1 r2 = 1 − = 1 − (−0.048) = 1.05, 21 correct to 3 significant figures. A still better approximation to the root, r3 , is given by:

where sin 2 means the sine of 2 radians = 4 − 2.7279 + 2.1972 − 3.5 = −0.0307 f ′ (x) = 2x − 3 cos x +

f ′ (r1 ) = f ′ (2) = 2(2) − 3 cos 2 +

[5(1.05)2 + 11(1.05) − 17] [10(1.05) + 11]

= 1.05 −

0.0625 21.5

= 5.9151 Hence,

r2 = r1 − = 2−

Since the values of r2 and r3 are the same when expressed to the required degree of accuracy, the required root is 1.05, correct to 3 significant figures. Checking, using the quadratic equation formula, √ −11 ± [121 − 4(5)(−17)] x= (2)(5) =

−11 ± 21.47 10

The positive root is 1.047, i.e. 1.05, correct to 3 significant figures (This root was determined in Problem 1, page 58, using the bisection method; Newton’s method is clearly quicker).

f(r1 ) f ′ (r1 ) −0.0307 5.9151

= 2.005 or 2.01, correct to

= 1.05 − 0.003 = 1.047, i.e. 1.05, correct to 3 significant figures.

2 3

= 4 + 1.2484 + 0.6667

f(r2 ) r3 = r2 − ′ f (r2 ) = 1.05 −

2 x+1

3 significant figures. A still better approximation to the root, r3 , is given by: r3 = r2 −

f (r2 ) f ′ (r2 )

[(2.005)2 − 3 sin 2.005 + 2 ln 3.005 − 3.5] ] = 2.005 − [ 2 2(2.005) − 3 cos 2.005 + 2.005 + 1 = 2.005 −

(−0.00104) = 2.005 + 0.000175 5.9376

i.e. r3 = 2.01, correct to 3 significant figures. Since the values of r2 and r3 are the same when expressed to the required degree of accuracy, then the required root is 2.01, correct to 3 significant figures.

Some applications of differentiation 347 Problem 12. Use Newton’s method to find the positive root of: x (x + 4)3 − e1.92x + 5 cos = 9, 3 correct to 3 significant figures. The functional notational method is used to determine the approximate value of the root. x f (x) = (x + 4)3 − e1.92x + 5 cos − 9 3

Now try the following Practice Exercise Practice Exercise 139 Newton’s method (Answers on page 883) In Problems 1 to 7, use Newton’s method to solve the equations given to the accuracy stated. 1. x2 − 2x − 13 = 0, correct to 3 decimal places. 2. 3x3 − 10x = 14, correct to 4 significant figures.

f(0) = (0 + 4)3 − e0 + 5 cos 0 − 9 = 59 1 f(1) = 53 − e1.92 + 5 cos − 9 ≈ 114 3 2 3 3.84 f(2) = 6 − e + 5 cos − 9 ≈ 164 3 f(3) = 73 − e5.76 + 5 cos 1 − 9 ≈ 19 4 f(4) = 83 − e7.68 + 5 cos − 9 ≈ −1660 3

3. x4 − 3x3 + 7x = 12, correct to 3 decimal places.

From these results, let a first approximation to the root be r1 = 3 Newton’s formula states that a better approximation to the root,

θ 7. 300e−2θ + = 6, correct to 3 significant 2 figures.

f(r1 ) r2 = r1 − ′ f (r1 ) f (r1 ) = f (3) = 73 − e5.76 + 5 cos 1 − 9 = 19.35 5 x sin 3 3 5 f ′ (r1 ) = f ′ (3) = 3(7)2 − 1.92e5.76 − sin 1 3 = −463.7 19.35 Thus, r2 = 3 − = 3 + 0.042 −463.7 = 3.042 = 3.04, correct to 3 significant figures. f (3.042) Similarly, r3 = 3.042 − ′ f (3.042) (−1.146) = 3.042 − (−513.1) = 3.042 − 0.0022 = 3.0398 = 3.04, correct to 3 significant figures. f ′ (x) = 3(x + 4)2 − 1.92e1.92x −

Since r2 and r3 are the same when expressed to the required degree of accuracy, then the required root is 3.04, correct to 3 significant figures.

4. 3x4 − 4x3 + 7x − 12 = 0, correct to 3 decimal places. 5. 3 ln x + 4x = 5, correct to 3 decimal places. 6. x3 = 5 cos 2x, correct to 3 significant figures.

8. A Fourier analysis of the instantaneous value of a waveform can be represented by: ( π) 1 y = t+ + sin t + sin 3t 4 8 Use Newton’s method to determine the value of t near to 0.04, correct to 4 decimal places, when the amplitude, y, is 0.880 9. A damped oscillation of a system is given by the equation: y = −7.4e0.5t sin 3t Determine the value of t near to 4.2, correct to 3 significant figures, when the magnitude y of the oscillation is zero. 10. The critical speeds of oscillation, λ, of a loaded beam are given by the equation: λ3 − 3.250λ2 + λ − 0.063 = 0 Determine the value of λ which is approximately equal to 3.0 by Newton’s method, correct to 4 decimal places.

348 Section G 26.4

Procedure for finding and distinguishing between stationary points:

Turning points

In Fig. 26.4, the gradient (or rate of change) of the curve changes from positive between O and P to negative between P and Q, and then positive again between Q and R. At point P, the gradient is zero and, as x increases, the gradient of the curve changes from positive just before P to negative just after. Such a point is called a maximum point and appears as the ‘crest of a wave’. At point Q, the gradient is also zero and, as x increases, the gradient of the curve changes from negative just before Q to positive just after. Such a point is called a minimum point, and appears as the ‘bottom of a valley’. Points such as P and Q are given the general name of turning points. y R P Positive gradient

Negative gradient

Positive gradient

Q

O

x

Given y = f(x), determine

(ii)

Let

(iii)

Substitute the values of x into the original equation, y = f (x), to find the corresponding y-ordinate values. This establishes the co-ordinates of the stationary points.

dy = 0 and solve for the values of x. dx

To determine the nature of the stationary points: Either d2 y (iv) Find 2 and substitute into it the values of x dx found in (ii). If the result is: (a) positive − the point is a minimum one, (b) negative − the point is a maximum one, (c) zero − the point is a point of inflexion, or (v) Determine the sign of the gradient of the curve just before and just after the stationary points. If the sign change for the gradient of the curve is: (a) positive to negative − the point is a maximum one,

Figure 26.4

(b)

It is possible to have a turning point, the gradient on either side of which is the same. Such a point is given the special name of a point of inflexion, and examples are shown in Fig. 26.5.

dy (i.e. f ′ (x)) dx

(i)

negative to positive − the point is a minimum one,

(c) positive to positive or negative to negative − the point is a point of inflexion. For more on points of inflexion, see Section 26.6, page 355.

Maximum point

y Maximum point

Problem 13. Locate the turning point on the curve y = 3x2 − 6x and determine its nature by examining the sign of the gradient on either side.

Points of inflexion

Following the above procedure: (i) 0

Minimum point

x

Figure 26.5

Maximum and minimum points and points of inflexion are given the general term of stationary points.

(ii) (iii)

dy = 6x − 6 dx dy At a turning point, = 0. Hence 6x − 6 = 0, dx from which, x = 1

Since y = 3x2 − 6x,

When x = 1, y = 3(1)2 − 6(1) = −3 Hence the co-ordinates of the turning point are (1, −3)

Some applications of differentiation 349 If x is slightly less than −1, say −1.1, then

(iv) If x is slightly less than 1, say, 0.9, then

dy = 3(−1.1)2 − 3, dx

dy = 6(0.9) − 6 = −0.6, dx

which is positive.

i.e. negative. If x is slightly greater than 1, say, 1.1, then

If x is slightly more than −1, say −0.9, then dy = 3(−0.9)2 − 3, dx

dy = 6(1.1) − 6 = 0.6, dx

which is negative.

i.e. positive. Since the gradient of the curve is negative just before the turning point and positive just after (i.e. − ∨ +), (1, −3) is a minimum point. Problem 14. Find the maximum and minimum values of the curve y = x3 − 3x + 5 by (a) examining the gradient on either side of the turning points, and (b)

determining the sign of the second derivative.

dy = 3x2 − 3 dx dy For a maximum or minimum value =0 dx Since y = x3 − 3x + 5 then

Since the gradient changes from positive to negative, the point (−1, 7) is a maximum point. (b)

dy d2 y = 3x2 − 3, then 2 = 6x dx dx d2 y When x = 1, is positive, hence (1, 3) is a dx2 minimum value. d2 y When x = −1, 2 is negative, hence (−1, 7) is a dx maximum value. Thus the maximum value is 7 and the minimum value is 3

Since

It can be seen that the second differential method of determining the nature of the turning points is, in this case, quicker than investigating the gradient.

Hence 3x2 − 3 = 0, from which, 3x2 = 3 and x = ± 1 When x = 1, y = (1)3 − 3(1) + 5 = 3 When x = −1, y = (−1)3 − 3(−1) + 5 = 7 Hence (1, 3) and (−1, 7) are the co-ordinates of the turning points. (a) Considering the point (1, 3): If x is slightly less than 1, say 0.9, then dy = 3(0.9)2 − 3, dx which is negative. If x is slightly more than 1, say 1.1, then dy = 3(1.1)2 − 3, dx which is positive. Since the gradient changes from negative to positive, the point (1, 3) is a minimum point. Considering the point (−1, 7):

Problem 15. Locate the turning point on the following curve and determine whether it is a maximum or minimum point: y = 4θ + e−θ y = 4θ + e−θ dy then = 4 − e−θ = 0 dθ for a maximum or minimum value.

Since

Hence 4 = e−θ , 41 = eθ, giving θ = ln 41 = −1.3863 (see Chapter 4). When θ = − 1.3863, y = 4(−1.3863) + e−(−1.3863) = 5.5452 + 4.0000 = −1.5452 Thus (−1.3863, −1.5452) are the co-ordinates of the turning point. d2 y = e−θ dθ2 When θ = −1.3863, d2 y = e+1.3863 = 4.0, dθ2

350 Section G which is positive, hence (−1.3863, −1.5452) is a minimum point.

y 12

Problem 16. Determine the co-ordinates of the maximum and minimum values of the graph x3 x2 5 y = − − 6x + and distinguish between 3 2 3 them. Sketch the graph.

8

9 3 2 y5 x3 2 x2 26x 1 5 3

4

22

0

21

1

2

3

x

Following the given procedure: (i)

Since y =

24

5 x3 x2 − − 6x + then 3 2 3

dy = x2 − x − 6 dx (ii)

dy = 0. Hence dx x2 − x − 6 = 0, i.e. (x + 2)(x − 3) = 0,

2115

28

6

212

At a turning point,

Figure 26.6

from which x = −2 or x = 3 (iii)

When x = −2, y=

(−2)3 (−2)2 5 − − 6(−2) + = 9 3 2 3

When x = 3, (3)3 (3)2 5 5 y= − − 6(3) + = −11 3 2 3 6 Thus the co-ordinates ( )of the turning points are (−2, 9) and 3, −11 56 dy d2 y = x2 − x − 6 then 2 = 2x−1 dx dx When x = −2,

(iv) Since

d2 y = 2(−2) − 1 = −5, dx2 which is negative. Hence (−2, 9) is a maximum point. When x = 3, d2 y = 2(3) − 1 = 5, dx2 which is positive. ( ) Hence 3, −11 56 is a minimum point. Knowing (−2, 9)( is a maximum point (i.e. crest ) of a wave), and 3, −11 56 is a minimum point (i.e. bottom of a valley) and that when x = 0, y = 53 , a sketch may be drawn as shown in Fig. 26.6.

Problem 17. Determine the turning points on the curve y = 4 sin x − 3 cos x in the range x = 0 to x = 2π radians, and distinguish between them. Sketch the curve over one cycle. Since y = 4 sin x − 3 cos x dy = 4 cos x + 3 sin x = 0, dx for a turning point, from which,

then

4 cos x = −3 sin x and −4 sin x = = tan x from Chapter 13 3 cos x ( ) −1 −4 Hence x = tan = 126.87◦ or 306.87◦ , since 3 tangent is negative in the second and fourth quadrants. When x = 126.87◦ , y = 4 sin 126.87◦ − 3 cos 126.87◦ = 5 When x = 306.87◦ , y = 4 sin 306.87◦ − 3 cos 306.87◦ = −5 ( π ) 126.87◦ = 126.87◦ × radians 180 = 2.214 rad ( π ) 306.87◦ = 306.87◦ × radians 180 = 5.356 rad

Some applications of differentiation 351 Hence (2.214, 5) and (5.356, −5) are the co-ordinates of the turning points. d2 y = −4 sin x + 3 cos x dx2 When x = 2.214 rad, d2 y = −4 sin 2.214 + 3 cos 2.214, dx2

6.

x = θ(6 − θ)

7.

y = 4x3 + 3x2 − 60x − 12

8.

y = 5x − 2 ln x

9.

y = 2x − ex

10.

which is negative. Hence (2.214, 5) is a maximum point. When x = 5.356 rad,

11.

d2 y = −4 sin 5.356 + 3 cos 5.356, dx2

12.

which is positive. Hence (5.356, −5) is a minimum point. A sketch of y = 4 sin x − 3 cos x is shown in Fig. 26.7.

13.

y 5

y = t3 −

t2 − 2t + 4 2

1 2t2 Determine the maximum and minimum values on the graph y = 12 cos θ − 5 sin θ in the range θ = 0 to θ = 360◦ . Sketch the graph over one cycle showing relevant points. x = 8t +

Show that the curve y = 23 (t − 1)3 + 2t(t − 2) has a maximum value of 32 and a minimum value of −2

y 5 4 sin x 2 3 cos x

0

p/2 2.214

p

5.356 3p/2

x(rads) 2p

23

26.5 Practical problems involving maximum and minimum values There are many practical problems involving maximum and minimum values which occur in science and engineering. Usually, an equation has to be determined from given data, and rearranged where necessary, so that it contains only one variable. Some examples are demonstrated in Problems 18 to 23.

25

Figure 26.7

Now try the following Practice Exercise Practice Exercise 140 Turning points (Answers on page 883) In Problems 1 to 11, find the turning points and distinguish between them.

Problem 18. A rectangular area is formed having a perimeter of 40 cm. Determine the length and breadth of the rectangle if it is to enclose the maximum possible area. Let the dimensions of the rectangle be x and y. Then the perimeter of the rectangle is (2x + 2y). Hence

1. y = x2 − 6x 2. y = 8 + 2x − x2

2x + 2y = 40 or

x + y = 20

(1)

3. y = x − 4x + 3 2

4. y = 3 + 3x2 − x3 5. y = 3x2 − 4x + 2

Since the rectangle is to enclose the maximum possible area, a formula for area A must be obtained in terms of one variable only. Area A = xy. From equation (1), x = 20 − y

352 Section G Hence, area A = (20 − y)y = 20y − y2

Using the quadratic formula, √ 32 ± (−32)2 − 4(3)(60) x= 2(3) = 8.239 cm or 2.427 cm.

dA = 20 − 2y = 0 dy for a turning point, from which, y = 10 cm

Since the breadth is (12 − 2x) cm then x = 8.239 cm is not possible and is neglected. Hence x = 2.427 cm

d2 A = −2, dy2 which is negative, giving a maximum point. When y = 10 cm, x = 10 cm, from equation (1) Hence the length and breadth of the rectangle are each 10 cm, i.e. a square gives the maximum possible area. When the perimeter of a rectangle is 40 cm, the maximum possible area is 10 × 10 = 100 cm2 .

d2 V = −128 + 24x dx2 d2 V When x = 2.427, 2 is negative, giving a maxdx imum value. The dimensions of the box are: length = 20 − 2(2.427) = 15.146 cm,

Problem 19. A rectangular sheet of metal having dimensions 20 cm by 12 cm has squares removed from each of the four corners and the sides bent upwards to form an open box. Determine the maximum possible volume of the box. The squares to be removed from each corner are shown in Fig. 26.8, having sides x cm. When the sides are bent upwards the dimensions of the box will be: length (20 − 2x) cm, breadth (12 − 2x) cm and height x cm. x

x x

x (20 2 2x )

12 cm

(12 2 2x )

x

x x

x 20 cm

Figure 26.8

breadth = 12 − 2(2.427) = 7.146 cm, and height = 2.427 cm Maximum volume = (15.146)(7.146)(2.427) = 262.7 cm3 Problem 20. Determine the height and radius of a cylinder of volume 200 cm3 which has the least surface area. Let the cylinder have radius r and perpendicular height h. Volume of cylinder, V = πr 2 h = 200 Surface area of cylinder, A = 2πrh + 2πr 2 Least surface area means minimum surface area and a formula for the surface area in terms of one variable only is required. From equation (1),

Volume of box, h= V = (20 − 2x)(12 − 2x)(x) = 240x − 64x2 + 4x3 dV = 240 − 128x + 12x2 = 0 dx for a turning point. Hence 4(60 − 32x + 3x2 ) = 0, i.e.

3x2 − 32x + 60 = 0

(1)

200 πr 2

Hence surface area, ( ) 200 A = 2πr + 2πr 2 πr 2 400 + 2πr 2 = 400r −1 + 2πr 2 r dA −400 = + 4πr = 0, dr r2 =

for a turning point.

(2)

Some applications of differentiation 353 Hence 4πr = from which, √( r=

3

400 400 and r 3 = 2 r 4π

100 π

for a turning point, from which, y = 25 m d2 A = −4, dy2

)

which is negative, giving a maximum value. When y = 25 m, x = 50 m from equation (1). Hence the maximum possible area = xy = (50)(25) = 1250 m2

= 3.169 cm

d2 A 800 = 3 + 4π dr 2 r d2 A When r = 3.169 cm, 2 is positive, giving a mindr imum value. From equation (2), when r = 3.169 cm, 200 = 6.339 cm h= π(3.169)2

Problem 22. An open rectangular box with square ends is fitted with an overlapping lid which covers the top and the front face. Determine the maximum volume of the box if 6 m2 of metal are used in its construction. A rectangular box having square ends of side x and length y is shown in Fig. 26.10.

Hence for the least surface area, a cylinder of volume 200 cm3 has a radius of 3.169 cm and height of 6.339 cm. x

Problem 21. Determine the area of the largest piece of rectangular ground that can be enclosed by 100 m of fencing, if part of an existing straight wall is used as one side. Let the dimensions of the rectangle be x and y as shown in Fig. 26.9, where PQ represents the straight wall.

P

Figure 26.9

y=

From Fig. 26.9, (1)

Hence volume

(

V=x y=x (2)

Since the maximum area is required, a formula for area A is needed in terms of one variable only. From equation (1), x = 100 − 2y Hence area A = xy = (100 − 2y)y = 100y − 2y2 dA = 100 − 4y = 0, dy

(1)

6 − 2x2 6 2x = − 5x 5x 5

2

Area of rectangle, A = xy

Surface area of box, A, consists of two ends and five faces (since the lid also covers the front face). Hence

Since it is the maximum volume required, a formula for the volume in terms of one variable only is needed. Volume of box, V = x2 y From equation (1),

y x

x + 2y = 100

Figure 26.10

A = 2x2 + 5xy = 6

Q y

y

x

2

6 2x − 5x 5

(2) ) =

6x 2x3 − 5 5

dV 6 6x2 = − =0 dx 5 5 for a maximum or minimum value. Hence 6 = 6x2 , giving x = 1 m (x = −1 is not possible, and is thus neglected). d2 V −12x = dx2 5

354 Section G d2 V When x = 1, 2 is negative, giving a maximum value. dx From equation (2), when x = 1, y=

2(1) 4 6 − = 5(1) 5 5

Substituting into equation (1) gives: ( ) h2 πh3 V = π 144 − h = 144πh − 4 4 dV 3πh2 = 144π − = 0, dh 4

Hence the maximum volume of the box is given by

for a maximum or minimum value. Hence

( ) V = x2 y = (1)2 45 = 45 m3

3πh2 4 √ (144)(4) = 13.86 cm h= 3

144π = Problem 23. Find the diameter and height of a cylinder of maximum volume which can be cut from a sphere of radius 12 cm. A cylinder of radius r and height h is shown enclosed in a sphere of radius R = 12 cm in Fig. 26.11. Volume of cylinder, V = πr 2 h

(1)

Using the right-angled triangle OPQ shown in Fig. 26.11,

r2 +

by Pythagoras’ theorem,

h2 = 144 4

(2)

h2 4

r 2 = 144 −

h2 13.862 = 144 − 4 4

P

Q

h 2 R

O

5

Hence the cylinder having the maximum volume that can be cut from a sphere of radius 12 cm is one in which the diameter is 19.60 cm and the height is 13.86 cm. Now try the following Practice Exercise Practice Exercise 141 Practical maximum and minimum problems (Answers on page 883)

r

h

d2 V When h = 13.86, 2 is negative, giving a maximum dh value. From equation (2),

Diameter of cylinder = 2r = 2(9.80) = 19.60 cm.

Since the maximum volume is required, a formula for the volume V is needed in terms of one variable only. From equation (2), r 2 = 144 −

d2 V −6πh = dh2 4

from which, radius r = 9.80 cm

( )2 h 2 = R2 r + 2 i.e.

from which,

12

cm

1. The speed, v, of a car (in m/s) is related to time t s by the equation v = 3 + 12t − 3t2 . Determine the maximum speed of the car in km/h. 2. Determine the maximum area of a rectangular piece of land that can be enclosed by 1200 m of fencing. 3. A shell is fired vertically upwards and its vertical height, x metres, is given by x = 24t − 3t2 , where t is the time in seconds. Determine the maximum height reached.

Figure 26.11

Some applications of differentiation 355

4. A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area of the metal for which the volume of the box is 3.5 m3 . 5. A closed cylindrical container has a surface area of 400 cm2 . Determine the dimensions for maximum volume.

13.

The signalling range, x, of a submarine ( ) 1 2 cable is given by the formula: x = r ln r where r is the ratio of the radii of the conductor and cable. Determine the value of r for maximum range.

14.

The equation y = 2e −t sin 2t represents the motion of a particle performing damped vibrations, y being the displacement from its mean position at time t. Calculate, correct to 3 significant figures, the maximum displacement.

6. Calculate the height of a cylinder of maximum volume which can be cut from a cone of height 20 cm and base radius 80 cm. 7. The power developed in a resistor R by a battery of emf E and internal resistance r is E2 R given by P = . Differentiate P with (R + r)2 respect to R and show that the power is a maximum when R = r. 8. Find the height and radius of a closed cylinder of volume 125 cm3 which has the least surface area. 9. Resistance to motion, F, of a moving vehicle is given by F = 5x + 100x. Determine the minimum value of resistance. 10. An electrical voltage E is given by E = (15 sin 50πt + 40 cos 50πt) volts, where t is the time in seconds. Determine the maximum value of voltage. 11. The fuel economy E of a car, in miles per gallon, is given by: E = 21 + 2.10 × 10−2 v2 − 3.80 × 10−6 v4 where v is the speed of the car in miles per hour. Determine, correct to 3 significant figures, the most economical fuel consumption, and the speed at which it is achieved. 12. The horizontal range of a projectile, x, launched with velocity u at an angle θ to 2u2 sin θ cos θ the horizontal is given by: x = g To achieve maximum horizontal range, determine the angle the projectile should be launched at.

26.6

Points of inflexion

As mentioned earlier in the chapter, it is possible to have a turning point, the gradient on either side of which is the same. This is called a point of inflexion. Procedure to determine points of inflexion: (i)

Given y = f(x), determine

(ii)

Solve the equation

(iii)

dy d2 y and 2 dx dx

d2 y =0 dx2 Test whether there is a change of sign occurring d2 y in 2 . This is achieved by substituting into the dx d2 y expression for 2 firstly a value of x just less dx than the solution and then a value just greater than the solution.

d2 y (iv) A point of inflexion has been found if 2 = 0 dx and there is a change of sign. This procedure is demonstrated in the following worked Problems.

Problem 24. Determine the point(s) of inflexion (if any) on the graph of the function y = x3 − 6x2 + 9x + 5 Find also any other turning points. Hence, sketch the graph.

356 Section G Using the above procedure: (i)

(ii)

Given y = x3 − 6x2 + 9x + 5, dy d2 y = 3x2 − 12x + 9 and 2 = 6x − 12 dx dx d2 y Solving the equation 2 = 0 gives: 6x − 12 = 0 dx from which, 6x = 12 and x = 2 Hence, if there is a point of inflexion, it occurs at x=2

Taking a value just less than 2, say, 1.9: d2 y = 6x − 12 = 6(1.9) − 12, which is negative. dx2 Taking a value just greater than 2, say, 2.1: d2 y = 6x − 12 = 6(2.1) − 12, which is positive. dx2 (iv) Since a change of sign has occurred a point of inflexion exists at x = 2

y 10 9 8 7 6 5 4 2

(iii)

0

i.e. x2 − 4x + 3 = 0 Using the quadratic formula or factorising (or by calculator), the solution is: x = 1 or x = 3 Since y = x3 − 6x2 + 9x + 5, y = 13 − 6(1)2 + 9(1) + 5 = 9

then

when

x = 1,

and when x = 3, y = 33 − 6(3)2 + 9(3) + 5 = 5 Hence, there are turning points at (1, 9) and at (3, 5) d2 y d2 y Since = 6x − 12, when x = 1, = 6(1) − 12 dx2 dx2 which is negative – hence a maximum point d2 y and when x = 3, 2 = 6(3) − 12 which is positive – dx hence a minimum point. Thus, (1, 9) is a maximum point and (3, 5) is a minimum point. A sketch of the graph y = x3 − 6x2 + 9x + 5 is shown in Figure 26.12

Problem 25. Determine the point(s) of inflexion (if any) on the graph of the function y = x4 − 24x2 + 5x + 60

1

2

3

4x

Figure 26.12

Using the above procedure: (i)

Given y = x4 − 24x2 + 5x + 60, dy d2 y = 4x3 − 48x + 5 and 2 = 12x2 − 48 dx dx

(ii)

Solving the equation

(iii)

Taking a value just less than 2, say, 1.9: d2 y = 12x2 − 48 = 12(1.9)2 − 48, which is dx2 negative. Taking a value just greater than 2, say, 2.1: d2 y = 12x2 − 48 = 12(2.1)2 − 48, which is dx2 positive.

When x = 2, y = 23 − 6(2)2 + 9(2) + 5 = 7 i.e. a point of inflexion occurs at the co-ordinates (2, 7) dy From above, = 3x2 − 12x + 9 = 0 for a turning dx point

y = x 3 – 6x 2 + 9x + 5

d2 y = 0 gives: dx2 2 2 12x − 48 = 0 from which, 12x √ = 48 2 and x = 4 from which, x = 4 = ±2 Hence, if there are points of inflexion, they occur at x = 2 and at x = −2

Taking a value just less than −2, say, −2.1: d2 y = 12x2 − 48 = 12(−2.1)2 − 48, which is dx2 positive. Taking a value just greater than −2, say, −1.9: d2 y = 12x2 − 48 = 12(−1.9)2 − 48, which is dx2 negative. (iv) Since changes of signs have occurred, points of inflexion exist at x = 2 and x = −2 When x = 2, y = 24 − 24(2)2 + 5(2) + 60 = −10 When x = −2, y = (−2)4 − 24(−2)2 + 5(−2) + 60 = −30 i.e. points of inflexion occur at the co-ordinates (2, − 10) and at ( − 2, − 30)

Some applications of differentiation 357 Now try the following Practice Exercise Practice Exercise 142 Further problems on points of inflexion (Answers on page 883) 1. Find the points of inflexion (if any) on the graph of the function 1 1 1 y = x3 − x2 − 2x + 3 2 12

The graph of y = x2 − x − 2 is shown in Fig. 26.13. The line AB is the tangent to the curve at the point C, i.e. (1, −2), and the equation of this line is y = x − 3

Normals The normal at any point on a curve is the line which passes through the point and is at right angles to the tangent. Hence, in Fig. 26.13, the line CD is the normal. y

2. Find the points of inflexion (if any) on the graph of the function 5 y = 4x3 + 3x2 − 18x − 8

y 5 x 2 2 x2 2

2 1

3. Find the point(s) of inflexion on the graph of the function y = x + sin x for 0 < x < 2π

22

21

4. Find the point(s) of inflexion on the graph of the function y = 3x3 − 27x2 + 15x + 17

0

1

3

2

x

B 21 C 22

5. Find the point(s) of inflexion on the graph of the function y = 2xe−x 6. The displacement, s, of a particle is given by: s = 3t3 − 9t2 + 10. Determine the maximum, minimum and point of inflexion of s.

26.7

Tangents and normals

Tangents The equation of the tangent to a curve y = f(x) at the point (x1 , y1 ) is given by:

23 A

Figure 26.13

It may be shown that if two lines are at right angles then the product of their gradients is −1. Thus if m is the gradient of the tangent, then the gradient of the normal 1 is − m Hence the equation of the normal at the point (x1 , y1 ) is given by: y − y1 = −

y − y1 = m(x − x1 ) where m =

dy = gradient of the curve at (x1 , y1 ) dx

Problem 26. Find the equation of the tangent to the curve y = x2 − x − 2 at the point (1, −2) Gradient, m =

dy = 2x − 1 dx

At the point (1, −2), x = 1 and m = 2(1) − 1 = 1 Hence the equation of the tangent is: i.e. i.e. or

y − y1 = m(x − x1 ) y − (−2) = 1(x − 1) y+2 = x−1 y=x−3

D

1 (x − x1 ) m

Problem 27. Find the equation of the normal to the curve y = x2 − x − 2 at the point (1, −2) m = 1 from Problem 26, hence the equation of the normal is 1 y − y1 = − (x − x1 ) m i.e.

1 y − (−2) = − (x − 1) 1

i.e.

y + 2 = −x + 1

or

y = −x − 1

Thus the line CD in Fig. 26.13 has the equation y = −x − 1

358 Section G Problem 28. Determine the equations of the x3 tangent and normal to the curve y = at the point 5 ( ) 1 −1, − 5 x3 Gradient m of curve y = is given by 5 dy 3x2 m= = dx 5 ( ) 3(−1)2 3 1 At the point −1, − 5 , x = −1 and m = = 5 5 Equation of the tangent is:

i.e.

y − y1 = m(x − x1 ) ( ) 1 3 y− − = (x − (−1)) 5 5 1 3 = (x + 1) 5 5

i.e.

y+

or

5y + 1 = 3x + 3 5y − 3x = 2

or

1.

y = 2x2 at the point (1, 2)

2.

y = 3x2 − 2x at the point (2, 8)

3.

( ) x3 1 y = at the point −1, − 2 2

4. 5.

y = 1 + x − x2 at the point (−2, −5) ( ) 1 1 θ = at the point 3, t 3

26.8

Small changes

If y is a function of x, i.e. y = f (x), and the approximate change in y corresponding to a small change δx in x is required, then: δy dy ≈ δx dx

and δy ≈

dy · δx or δy ≈ f ′ (x) · δx dx

Equation of the normal is:

i.e. i.e. i.e.

1 y − y1 = − (x − x1 ) m ( ) 1 −1 y− − = (x − (−1)) 5 (3/5) 1 5 = − (x + 1) 5 3 1 5 5 y+ = − x− 5 3 3 y+

Multiplying each term by 15 gives: 15y + 3 = −25x − 25 Hence equation of the normal is: 15y + 25x + 28 = 0 Now try the following Practice Exercise Practice Exercise 143 Tangents and normals (Answers on page 883) For the curves in problems 1 to 5, at the points given, find (a) the equation of the tangent, and (b) the equation of the normal.

Problem 29. Given y = 4x2 − x, determine the approximate change in y if x changes from 1 to 1.02 Since y = 4x2 − x, then dy = 8x − 1 dx Approximate change in y, dy · δx ≈ (8x − 1)δx dx When x = 1 and δx = 0.02, δy ≈ [8(1) − 1](0.02) δy ≈

≈ 0.14 [Obviously, in this case, the exact value of dy may be obtained by evaluating y when x = 1.02, i.e. y = 4(1.02)2 − 1.02 = 3.1416 and then subtracting from it the value of y when x = 1, i.e. y = 4(1)2 − 1 = 3, giving δy = 3.1416 − 3 = 0.1416 dy Using δy = · δx above gave 0.14, which shows that dx the formula gives the approximate change in y for a small change in x]

Some applications of differentiation 359

Problem 30. The time√of swing T of a pendulum is given by T = k l, where k is a constant. Determine the percentage change in the time of swing if the length of the pendulum l changes from 32.1 cm to 32.0 cm. 1 √ If T = k l = kl 2 , then ( ) dT 1 −1 k =k l 2 = √ dl 2 2 l

Approximate change in T, ( ) dT k √ δl δt ≈ δl ≈ dl 2 l ( ≈

) k √ (−0.1) 2 l

(negative since l decreases) Percentage change ( ) approximate change in T = 100% original value of T ( ) k √ (−0.1) 2 l√ = × 100% k l ( ) ( ) −0.1 −0.1 = 100% = 100% 2l 2(32.1) = −0.156%

i.e. the possible error in calculating the template area is approximately 1.257 cm2 . ( Percentage error ≈

Area of circular template, A = πr 2 , hence dA = 2πr dr Approximate change in area, δA ≈

dA · δr ≈ (2πr)δr dr

When r = 10 cm and δr = 0.02, δA = (2π10)(0.02) ≈ 0.4π cm2

) 100%

= 0.40%

Now try the following Practice Exercise Practice Exercise 144 Small changes (Answers on page 884) 1.

Determine the change in y if x changes from 2.50 to 2.51 when 5 (a) y = 2x − x2 (b) y = x

2.

The pressure p and volume v of a mass of gas are related by the equation pv = 50. If the pressure increases from 25.0 to 25.4, determine the approximate change in the volume of the gas. Find also the percentage change in the volume of the gas.

3.

Determine the approximate increase in (a) the volume, and (b) the surface area of a cube of side x cm if x increases from 20.0 cm to 20.05 cm.

4.

The radius of a sphere decreases from 6.0 cm to 5.96 cm. Determine the approximate change in (a) the surface area, and (b) the volume.

5.

The rate of flow of a liquid through a tube is pπr 4 given by Poiseuilles’ equation as: Q = 8ηL where Q is the rate of flow, p is the pressure difference between the ends of the tube, r is the radius of the tube, L is the length of the tube and η is the coefficient of viscosity of the liquid. η is obtained by measuring Q, p, r and L. If Q can be measured accurate to ±0.5%, p accurate to ±3%, r accurate to ±2% and L accurate to ±1%, calculate the maximum possible percentage error in the value of η.

Hence the change in the time of swing is a decrease of 0.156% Problem 31. A circular template has a radius of 10 cm (±0.02). Determine the possible error in calculating the area of the template. Find also the percentage error.

0.4π π(10)2

360 Section G Practice Exercise 145 Multiple-choice questions on some applications of differentiation (Answers on page 884) Each question has only one correct answer 1. The length ℓ metres of a certain metal rod at temperature t◦ C is given by: ℓ = 1 + 4 × 10−5 t + 4 × 10−7 t2 The rate of change of length, in mm/◦ C, when the temperature is 400◦ C, is: (a) 3.6 × 10−4 (b) 1.00036 (c) 0.36 (d) 3.2 × 10−4 2. An alternating current, i amperes, is given by i = 100 sin 2π ft amperes, where f is the frequency in hertz and t is the time in seconds. The rate of change of current when t = 12 ms and f = 50 Hz is: (a) 31348 A/s (b) −58.78 A/s (c) 627.0 A/s (d) −25416 A/s 3. For the curve shown in Figure 26.14, which of the following statements is incorrect? (a) P is a turning point (b) Q is a minimum point (c) R is a maximum value (d) Q is a stationary value

6. The vertical displacement, s, of a prototype model in a tank is given by s = 40 sin(0.1t) mm, where t is the time in seconds. The vertical velocity of the model, in mm/s, is: (a) − cos 0.1t (b) 400 cos 0.1t (c) −400 cos 0.1t (d) 4 cos 0.1t 7. The motion of a particle in an electrostatic field is described by the equation y = x3 + 3x2 + 5x − 28 When x = 2, y is approximately zero. Using one iteration of the Newton-Raphson method, a better approximation (correct to 2 decimal places) is: (a) 1.89 (b) 2.07 (c) 2.11 (d) 1.93 8. The turning point on the curve y = x2 − 4x is at: (a) (2, 0) (b) (0, 4) (c) (−2, 12) (d) (2, −4) 9. The current i in a circuit at time t seconds is given by i = 0.20(1 − e −20 t ) A. When time t = 0.1 s, the rate of change of current is: (a) −1.022 A/s (b) 0.541 A/s (c) 0.173 A/s (d) 0.373 A/s 10.

y R P

11. Q 0

x

Figure 26.14

1 1 The equation of a curve is x3 + x2 − 6x. 3 2 The maximum value of the curve is: (a) 13.5 (b) −3 (c) −7.33 (d) 2

12.

A first approximation indicates that a root of the equation x3 − 9x + 1 = 0 lies close to 3. Using the Newton-Raphson method, a second approximation, correct to 4 decimal places, is: (a) 2.9420 (b) 3.0580 (c) 2.9444 (c) 3.0556

13.

The maximum value of the function y = 6x − x2 is: (a) 11 (b) 10 (c) 9 (d) 8

4. The equation of a curve is y = 2x − 6x + 1. The maximum value of the curve is: (a) −3 (b) 5 (c) 1 (d) −6 3

5. The resistance to motion F of a moving vehi5 cle is given by F = + 100x. The minimum x value of resistance is: (a) 44.72 (b) 0.2236 (c) −44.72 (d) −0.2236

The velocity of a car (in m/s) is related to time t seconds by the equation v = 4.5 + 18t − 4.5t2 The maximum speed of the car, in km/h, is: (a) 81 (b) 6.25 (c) 22.5 (d) 77

Some applications of differentiation 361

14. The equation of the tangent to the curve y = x2 − 3x + 4 at the point (2, −1) is: (a) y = x − 1 (b) y = −6x − 12 (c) y = x − 3 (d) y = x + 3

15.

Using Newton’s method to find the real root of the equation 3x3 − 7x = 15, correct to 3 decimal places, and starting at x = 2, gives: (a) 1.987 (b) 2.157 (c) 2.132 (d) 2.172

For fully worked solutions to each of the problems in Practice Exercises 137 to 144 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 8

Introduction to Differentiation

This Revision Test covers the material contained in Chapters 25 and 26. The marks for each question are shown in brackets at the end of each question. 1. Differentiate the following with respect to the variable: √ 1 (a) y = 5 + 2 x3 − 2 (b) s = 4e2θ sin 3θ x 3 ln 5t (c) y = cos 2t 2 (d) x = √ (15) 2 (t − 3t + 5) 2. If f(x) = 2.5x2 − 6x + 2 find the co-ordinates at the point at which the gradient is −1 (5) 3. The displacement s cm of the end of a stiff spring at time t seconds is given by: s = ae−kt sin 2πft. Determine the velocity and acceleration of the end of the spring after two seconds if a = 3, k = 0.75 and f = 20 (10)

4.

Find the co-ordinates of the turning points on the curve y = 3x3 + 6x2 + 3x − 1 and distinguish between them. Find also the point(s) of inflexion. (14)

5.

The heat capacity C of a gas varies with absolute temperature θ as shown: C = 26.50 + 7.20 × 10−3 θ − 1.20 × 10−6 θ2 Determine the maximum value of C and the temperature at which it occurs. (7)

6.

Determine for the curve y = 2x2 − 3x at the point (2, 2): (a) the equation of the tangent (b) the equation of the normal. (7)

7.

A rectangular block of metal with a square crosssection has a total surface area of 250 cm2 . Find the maximum volume of the block of metal. (7)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 8, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Chapter 27

Differentiation of parametric equations Why it is important to understand: Differentiation of parametric equations Rather than using a single equation to define two variables with respect to one another, parametric equations exist as a set that relates the two variables to one another with respect to a third variable. Some curves are easier to describe using a pair of parametric equations. The coordinates x and y of the curve are given using a third variable t, such as x = f(t) and y = g(t), where t is referred to as the parameter. Hence, for a given value of t, a point (x, y) is determined. For example, let t be the time while x and y are the positions of a particle; the parametric equations then describe the path of the particle at different times. Parametric equations are useful in defining three-dimensional curves and surfaces, such as determining the velocity or acceleration of a particle following a three-dimensional path. CAD systems use parametric versions of equations. Sometimes in engineering, differentiation of parametric equations is necessary, for example, when determining the radius of curvature of part of the surface when finding the surface tension of a liquid. Knowledge of standard differentials and the function of a function rule from previous chapters are needed to be able to differentiate parametric equations.

At the end of this chapter, you should be able to: • recognise parametric equations – ellipse, parabola, hyperbola, rectangular hyperbola, cardioids, asteroid and cycloid • differentiate parametric equations

27.1 Introduction to parametric equations Certain mathematical functions can be expressed more simply by expressing, say, x and y separately in terms of a third variable. For example, y = r sin θ, x = r cos θ. Then, any value given to θ will produce a pair of values for x and y, which may be plotted to provide a curve of y = f (x).

The third variable, θ, is called a parameter and the two expressions for y and x are called parametric equations. The above example of y = r sin θ and x = r cos θ are the parametric equations for a circle. The equation of any point on a circle, centre at the origin and of radius r is given by: x2 + y2 = r 2 , as shown in Chapter 10. To show that y = r sin θ and x = r cos θ are suitable parametric equations for such a circle:

364 Section G Left-hand side of equation = x2 + y2

(

= (r cos θ)2 + r sin θ 2

2

2

)2

(a) Ellipse

x = a cos θ, y = b sin θ

(b)

x = a t2 , y = 2a t

Parabola

x = a sec θ, y = b tan θ c x = c t, y = t

(c) Hyperbola

2

= r cos θ + r sin θ ( ) = r 2 cos2 θ + sin2 θ

(d)

= r 2 = right-hand side

(e) Cardioid

Rectangular hyperbola

( ) x = a (2 cos θ − cos 2θ) , y = a 2 sin θ − sin 2θ

(since cos2 θ + sin2 θ = 1, as shown in Chapter 13)

27.2 Some common parametric equations The following are some of the most common parametric equations, and Fig. 27.1 shows typical shapes of these curves.

(f )

Astroid

(g)

Cycloid

27.3

x = a cos3 θ, y = a sin3 θ ( ) ( ) x = a θ− sin θ , y = a 1− cos θ

Differentiation in parameters

When x and y are given in terms of a parameter, say θ, then by the function of a function rule of differentiation (from Chapter 25): dy dy dθ = × dx dθ dx It may be shown that this can be written as: dy dy

(a) Ellipse

dx

(b) Parabola

= dθ dx

(1)

dθ For the second differential, d2 y d = dx2 dx (c) Hyperbola

(d) Rectangular hyperbola

(

dy dx

) =

d dθ

(

dy dx

) ·

dθ dx

or d d2 y dx2

=

dθ

(

dy

)

dx dx dθ

(e) Cardioid

(f) Astroid

Given x = 5θ − 1 and dy y = 2θ (θ − 1), determine in terms of θ. dx Problem 1.

x = 5θ − 1, hence (g) Cycloid

Figure 27.1

dx =5 dθ

y = 2θ(θ − 1) = 2θ2 − 2θ, hence

( ) dy = 4θ − 2 = 2 2θ − 1 dθ

(2)

Differentiation of parametric equations 365 From equation (1),

From equation (1),

dy dy 2(2θ − 1) dθ = = dx dx 5 dθ

or

dy dy 4 1 = dt = = dx dx 4t t dt

2 (2θ − 1) 5

Problem 2. The parametric equations of a function are given by y = 3 cos 2t, x = 2 sin t. dy d2 y Determine expressions for (a) (b) 2 dx dx dy = −6 sin 2t dt dx x = 2 sin t, hence = 2 cos t dt From equation (1),

(a) y = 3 cos 2t, hence

Hence, the equation of the tangent is: ) 1( y − 4t = x − 2t2 t Problem 4. The parametric equations of a cycloid are x = 4(θ − sin θ), y = 4(1 − cos θ). d2 y dy (b) 2 Determine (a) dx dx (a) x = 4(θ − sin θ), dx = 4 − 4 cos θ = 4(1 − cos θ) dθ dy y = 4(1 − cos θ), hence = 4 sin θ dθ From equation (1),

hence

dy −6 sin 2t −6(2 sin t cos t) dy = dt = = dx dx 2 cos t 2 cos t dt

dy sin θ dy 4 sin θ dθ = = = dx dx 4(1 − cos θ) (1 − cos θ) dθ

from double angles, Chapter 15 dy

i.e. (b)

dx

= −6 sin t

From equation (2),

d2 y = dx2

i.e.

d dt

(

dy dx dx dt

)

(b) ) d( −6 sin t −6 cos t dt = = 2 cos t 2 cos t

d2 y = dx2

Problem 3. (The equation of a tangent drawn to a ) curve at point x1 , y1 is given by: dy1 (x − x1 ) dx1

Determine the equation of the tangent drawn to the parabola x = 2t2 , y = 4t at the point t.

(

dy dx dx dθ

)

( ) sin θ d dθ 1 − cos θ = 4(1 − cos θ)

dx1 = 4t dt dy1 y1 = 4t, hence =4 dt

cos θ − cos2 θ − sin2 θ 4(1 − cos θ)3 ( ) cos θ − cos2 θ + sin2 θ = 4(1 − cos θ)3

=

=

cos θ − 1 4(1 − cos θ)3

=

−1 −(1 − cos θ) = 4(1 − cos θ)3 4(1 − cos θ)2

At point t, x1 = 2t2 , hence and

d dθ

(1 − cos θ)(cos θ) − (sin θ)(sin θ) ( )2 1 − cos θ = 4(1 − cos θ)

d2 y = −3 dx2

y − y1 =

From equation (2),

366 Section G Now try the following Practice Exercise

x = 2 cos3 θ, hence

Practice Exercise 146 Differentiation of parametric equations (Answers on page 884) dy 1. Given x = 3t − 1 and y = t(t − 1), determine dx in terms of t

y = 2 sin3 θ, hence

dx = −6 cos2 θ sin θ dθ by the function of a function rule dy = 6 sin2 θ cos θ dθ by the function of a function rule

From equation (1),

2

2. A parabola has parametric equations: x = t , dy y = 2t. Evaluate when t = 0.5 dx 3. The parametric equations for an ellipse are dy d2 y x = 4 cos θ, y = sin θ. Determine (a) (b) 2 dx dx dy π 4. Evaluate at θ = radians for the hyperbola dx 6 whose parametric equations are x = 3 sec θ, y = 6 tan θ 5. The parametric equations for a rectangular 2 dy hyperbola are x = 2t, y = . Evaluate when t dx t = 0.40 The equation of a tangent drawn to a curve at point (x1 , y1 ) is given by: y − y1 =

) dy1 ( x − x1 dx1

Use this in Problems 6 and 7. 1. Determine the equation of the tangent drawn π to the ellipse x = 3 cos θ, y = 2 sin θ at θ = 6 2. Determine the equation of the tangent drawn 5 to the rectangular hyperbola x = 5t, y = at t t=2

27.4

Further worked problems on differentiation of parametric equations

Problem 5. The of the normal drawn to ( equation ) a curve at point x1 , y1 is given by: ) 1 ( y − y1 = − x − x1 dy1 dx1 Determine the equation of the normal drawn to the π astroid x = 2 cos3 θ, y = 2 sin3 θ at the point θ = 4

dy dy 6 sin2 θ cos θ sin θ dθ = = =− = −tanθ 2 dx dx −6 cos θ sin θ cos θ dθ π dy π When θ = , = −tan = −1 4 dx 4 π π x1 = 2 cos3 = 0.7071 and y1 = 2 sin3 = 0.7071 4 4 Hence, the equation of the normal is: y − 0.7071 = − i.e. i.e.

1 (x − 0.7071) −1

y − 0.7071 = x − 0.7071 y =x

Problem 6. The parametric equations for a hyperbola are x = 2 sec θ, y = 4 tan θ. Evaluate dy d2 y (a) (b) 2 , correct to 4 significant figures, dx dx when θ = 1 radian. (a) x = 2 sec θ, hence

dx = 2 sec θ tan θ dθ

dy y = 4 tan θ, hence = 4 sec2 θ dθ From equation (1), dy dy 4 sec2 θ 2 sec θ dθ = = = dx dx 2 sec θ tan θ tan θ dθ ( ) 1 2 2 cos θ ) = = ( or 2 cosec θ sin θ sin θ cos θ dy 2 When θ = 1 rad, = = 2.377, correct to 4 dx sin 1 significant figures.

Differentiation of parametric equations 367 (b)

From equation (2), ( ) d dy ) d ( 2 cosec θ d2 y dθ dx = = dθ dx dx2 2 sec θ tan θ dθ −2 cosec θ cot θ = 2 sec θ tan θ ( )( ) 1 cos θ − sin θ sin θ )( ) = ( 1 sin θ cos θ cos θ ( )( 2 ) cos θ cos θ =− 2 sin θ sin θ =−

v[ u ( )2 ] 3 u t 1 + dy dx Hence, radius of curvature, ρ =

v[ u ( )2 ] 3 u t 1+ 1 t = − v[ u ( )2 ] 3 u t 1+ 1 2 When

cos3 θ = − cot3 θ sin3 θ

d2 y 1 = − cot3 1 = − ( )3 dx2 tan 1 = −0.2647, correct to 4 significant figures.

d2 y dx2

t = 2,

ρ=

1 − ( )3 6 2

1 6t3

√( )3 1.25 = 1 − 48

√( )3 = − 48 1.25 = −67.08

When θ = 1 rad,

Problem 7. When determining the surface tension of a liquid, the radius of curvature, ρ, of part of the surface is given by: v[ u ( )2 ] 3 u t 1 + dy dx ρ= d2 y dx2

Now try the following Practice Exercise Practice Exercise 147 Differentiation of parametric equations (Answers on page 884) 1.

Find the radius of curvature of the part of the surface having the parametric equations x = 3t2 , y = 6t at the point t = 2 dx x = 3t2 , hence = 6t dt dy y = 6t, hence =6 dt

dy dy dt 6 1 From equation (1), = = = dx dx 6t t dt From equation (2), ( ) ( ) d dy d 1 1 −2 d2 y dt dx 1 dt t t = = = =− 3 2 dx dx 6t 6t 6t dt

A cycloid has parametric equations x = 2(θ − sin θ), y = 2(1 − cos θ). Evaluate, at θ = 0.62 rad, correct to 4 significant figures, dy d2 y (a) (b) 2 dx dx The equation of the drawn to a ( normal ) curve at point x1 , y1 is given by: ) 1 ( y − y1 = − x − x1 dy1 dx1 Use this in Problems 2 and 3.

2.

3.

Determine the equation of the normal drawn 1 1 to the parabola x = t2 , y = t at t = 2 4 2 Find the equation of the normal drawn to the cycloid x = 2(θ − sin θ), y = 2(1 − cos θ) at π θ = rad. 2

4.

Determine the value of

d2 y , correct to 4 sigdx2 π nificant figures, at θ = rad for the cardioid 6 x = 5(2θ − cos 2θ), y = 5(2 sin θ − sin 2θ)

368 Section G

5. The radius of curvature, ρ, of part of a surface when determining the surface tension of a liquid is given by: [

(

1+ ρ=

dy dx

)2 ] 3/2 3.

2

d y dx2

Find the radius of curvature (correct to 4 significant figures) of the part of the surface having parametric equations (a) x = 3t, y =

2.

4.

The parametric equations for a hyperbola are d2 y x = 3 sec θ, y = 2 tan θ. The value of 2 dx when θ = 1.2 rad, correct to 3 decimal places is: (a) 0.259 (b) −0.013 (c) −0.086 (d) −0.778

5.

If the parametric equations of a cycloid are ( ) ( ) dy x = 5 θ − sin θ , x = 5 1 − cos θ then is dx equal to: sin θ 1 − cos θ (a) (b) 1 − cos θ sin θ sin θ 1 + cos θ (c) (d) 1 + cos θ sin θ

1 3 at the point t = t 2

π (b) x = 4 cos3 t, y = 4 sin3 t at t = rad. 6

Practice Exercise 148 Multiple-choice questions on differentiation of parametric equations (Answers on page 884) Each question has only one correct answer 1. Given x = 3t − 1 and y = 3t(t − 1) then which of the following is correct? d2 y dy = 2t − 1 (b) 2 = 2 (a) dx dx d2 y 1 dy 1 (c) 2 = (d) = dx 2 dx 2t − 1

Given x = 2 sin t and y = 4 cos 2t then: d2 y dy = −8 sin t (b) 2 = 4 (a) dx dx d2 y dy 4 sin 2t (c) 2 = − tan t (d) =− dx dx cos t The parametric equations for a hyperbola are dy when x = sec θ, y = 3 tan θ. The value of dx θ = 1.5 rad, correct to 2 decimal places is: (a) 0.33 (b) 114.61 (c) 3.01 (d) 42.52

For fully worked solutions to each of the problems in Practice Exercises 146 and 147 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 28

Differentiation of implicit functions Why it is important to understand: Differentiation of implicit functions Differentiation of implicit functions is another special technique, but it occurs often enough to be important. It is needed for more complicated problems involving different rates of change. Up to this chapter we have been finding derivatives of functions of the form y = f(x); unfortunately not all functions fall into this form. However, implicit differentiation is nothing more than a special case of the function of a function (or chain rule) for derivatives. Engineering applications where implicit differentiation is needed are found in optics, electronics, control and even some thermodynamics.

At the end of this chapter, you should be able to: • recognise implicit functions • differentiate simple implicit functions • differentiate implicit functions containing products and quotients

28.1

Implicit functions

When an equation can be written in the form y = f (x) it is said to be an explicit function of x. Examples of explicit functions include y = 2x3 − 3x + 4, and y =

The equation is then called an implicit function and examples of such functions include y3 + 2x2 = y2 − x and sin y = x2 + 2xy

28.2 Differentiating implicit functions

y = 2x ln x

3ex cos x

In these examples y may be differentiated with respect to x by using standard derivatives, the product rule and the quotient rule of differentiation respectively. Sometimes with equations involving, say, y and x, it is impossible to make y the subject of the formula.

It is possible to differentiate an implicit function by using the function of a function rule, which may be stated as du du dy = × dx dy dx Thus, to differentiate y3 with respect to x, the du substitution u = y3 is made, from which, = 3y2 dy

370 Section G d dy Hence, (y3 ) = (3y2 ) × , by the function of a funcdx dx tion rule. A simple rule for differentiating an implicit function is summarised as: d d dy [ f ( y)] = [ f ( y)] × dx dy dx

(1)

Problem 1. Differentiate the following functions with respect to x: 4

(a) 2y

(b) sin 3t

(a) Let u = 2y4 , then, by the function of a function rule: du du dy d dy = × = (2y4 ) × dx dy dx dy dx dy = 8y3 dx (b)

Now try the following Practice Exercise Practice Exercise 149 Differentiating implicit functions (Answers on page 884) In Problems 1 and 2 differentiate the given functions with respect to x. (a) 3y5

2.

(a)

3.

Differentiate the following with respect to y: √ 2 (a) 3 sin 2θ (b) 4 x3 (c) t e

4.

Differentiate the following with respect to u: (a)

(b) 2 cos 4θ

(c)

5 3 ln 3t (b) e2y+1 2 4

2 (3x + 1)

(c) 2 tan 3y

(b) 3 sec 2θ

2 (c) √ y

Let u = sin 3t, then, by the function of a function rule: du du dt d dt = × = (sin 3t) × dx dt dx dt dx dt = 3 cos 3t dx

Problem 2. Differentiate the following functions with respect to x: (a) 4 ln 5y

1 (b) e3θ−2 5

28.3

Differentiating implicit functions containing products and quotients

The product and quotient rules of differentiation must be applied when differentiating functions containing products and quotients of two variables. For example,

(a) Let u = 4 ln 5y, then, by the function of a function rule: du du dy d dy = × = (4 ln 5y) × dx dy dx dy dx

1 Let u = e3θ−2 , then, by the function of a function 5 rule: ( ) du du dθ d 1 3θ−2 dθ = × = e × dx dθ dx dθ 5 dx 3 dθ = e3θ−2 5 dx

d 2 d d (x y) = (x2 ) (y) + (y) (x2 ), dx dx dx by the product rule ( ) dy = (x2 ) 1 + y(2x), dx by using equation (1)

4 dy = y dx (b)

√ k

1.

= x2

Problem 3.

Determine

dy + 2xy dx

d (2x3 y2 ) dx

In the product rule of differentiation let u = 2x3 and v = y2 Thus

d d d (2x3 y2 ) = (2x3 ) (y2 ) + (y2 ) (2x3 ) dx dx dx

Differentiation of implicit functions 371 ( ) dy = (2x3 ) 2y + (y2 )(6x2 ) dx dy + 6x2 y2 dx ( ) dy 2 = 2x y 2x + 3y dx

4.

dz √ Given z = 3 y cos 3x find dx

5.

Determine

= 4x3 y

Problem 4.

( ) d 3y Find dx 2x

In the quotient rule of differentiation let u = 3y and v = 2x ( ) (2x) d (3y) − (3y) d (2x) d 3y dx dx Thus = dx 2x (2x)2 ( ) dy (2x) 3 − (3y)(2) dx = 4x2 dy ( ) 6x − 6y 3 dy = dx 2 = 2 x −y 4x 2x dx Problem 5. Differentiate z = x2 + 3x cos 3y with respect to y. dz d d = (x2 ) + (3x cos 3y) dy dy dy [ ( )] dx dx = 2x + (3x)(−3 sin 3y) + (cos 3y) 3 dy dy = 2x

dx dx − 9x sin 3y + 3 cos 3y dy dy

Now try the following Practice Exercise

28.4

Further implicit differentiation

An implicit function such as 3x2 + y2 − 5x + y = 2, may be differentiated term by term with respect to x. This gives: d d d d d (3x2 ) + (y2 ) − (5x) + (y) = (2) dx dx dx dx dx dy dy − 5 + 1 = 0, dx dx using equation (1) and standard derivatives. dy An expression for the derivative in terms of x and dx y may be obtained by rearranging this latter equation. Thus: dy (2y + 1) = 5 − 6x dx 6x + 2y

i.e.

Problem 6. Given 2y2 − 5x4 − 2 − 7y3 = 0, dy determine dx Each term in turn is differentiated with respect to x: Hence

d d d d (2y2 ) − (5x4 ) − (2) − (7y3 ) dx dx dx dx =

i.e. d (3x2 y3 ) dx ( ) d 2y 2. Find dx 5x ( ) d 3u 3. Determine du 4v

dy 5 − 6x = dx 2y + 1

from which,

Practice Exercise 150 Differentiating implicit functions involving products and quotients (Answers on page 884) 1. Determine

dz given z = 2x3 ln y dy

4y

dy dy − 20x3 − 0 − 21y2 = 0 dx dx

Rearranging gives: (4y − 21y2 ) i.e.

dy = 20x3 dx dy 20x3 = dx (4y − 21y2 )

d (0) dx

372 Section G

Problem 7.

Determine the values of

x = 4 given that x2 + y2 = 25

dy when dx

Differentiating each term in turn with respect to x gives: d d d 2 (x ) + (y2 ) = (25) dx dx dx 2x + 2y

i.e.

(a) Find

(b)

dy when x = 1 and y = 2 dx

Evaluate

(a) Differentiating each term in turn with respect to x gives: d d d d (4x2 ) + (2xy3 ) − (5y2 ) = (0) dx dx dx dx [ ( ) ] 2 dy 3 i.e. 8x + (2x) 3y + (y )(2) dx

dy =0 dx dy 2x x =− =− dx 2y y

Hence

dy in terms of x and y dx 2 3 2 given 4x + 2xy − 5y = 0 Problem 8.

− 10y

√ Since x2 +y2 = 25, when x = 4, y = (25 − 42 ) = ±3 8x + 6xy2

dy dy + 2y3 − 10y = 0 dx dx

dy 4 4 Thus when x = 4 and y = ±3, =− =± dx ±3 3

i.e.

From Chapter 10, x2 + y2 = 25 is the equation of a circle, centre at the origin and radius 5, as shown in Fig. 28.1. At x = 4, the two gradients are shown.

Rearranging gives: 8x + 2y3 = (10y − 6xy2 ) and

y Gradient 4 52 3

5 x 2 1 y 2 5 25

(b)

0

dy 8x + 2y3 4x + y3 = = 2 dx 10y − 6xy y(5 − 3xy)

dy 4(1) + (2)3 12 = = = −6 dx 2[5 − (3)(1)(2)] −2 4

5

x

23 25

dy dx

When x = 1 and y = 2,

3

25

dy =0 dx

Gradient 4 5 3

Problem 9. Find the gradients of the tangents drawn to the circle x2 + y2 − 2x − 2y = 3 at x = 2 The gradient of the tangent is given by

Figure 28.1

dy dx

Differentiating each term in turn with respect to x gives: Above, x2 + y2 = 25 was differentiated implicitly; actually, the equation could be transposed to √ y = (25 − x2 ) and differentiated using the function of a function rule. This gives −1 dy 1 x = (25 − x2 ) 2 (−2x) = − √ dx 2 (25 − x2 )

and when x = 4, above.

dy 4 4 =−√ = ± as obtained dx 3 (25 − 42 )

d 2 d d d d (x ) + (y2 ) − (2x) − (2y) = (3) dx dx dx dx dx

2x + 2y

i.e. Hence

(2y − 2)

from which

dy dy −2−2 = 0 dx dx

dy = 2 − 2x, dx dy 2 − 2x 1 − x = = dx 2y − 2 y − 1

Differentiation of implicit functions 373 The value of y when x = 2 is determined from the original equation.

Since pvγ = k, then p =

dp dp dv = × dt dv dt

Hence (2)2 + y2 − 2(2) − 2y = 3 4 + y2 − 4 − 2y = 3

i.e.

by the function of a function rule

y2 − 2y − 3 = 0

or

dp d = (kv−γ ) dv dv

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

= −γkv−γ−1 =

When x = 2 and y = −1,

−γk vγ+1

−γk dv dp = γ+1 × dt v dt

dy 1 − x 1−2 −1 1 = = = = dx y − 1 −1 − 1 −2 2

Since k = pvγ ,

When x = 2 and y = 3,

dp −γ(pvγ ) dv −γpvγ dv = = γ 1 dt vγ+1 dt v v dt

dy 1 − 2 −1 = = dx 3 − 1 2 Hence the gradients of the tangents are ±

k = kv−γ vγ

1 2

The circle having the given √ equation has its centre at (1, 1) and radius 5 (see Chapter 10) and is shown in Fig. 28.2 with the two gradients of the tangents.

i.e.

dp p dv = −γ dt v dt

Now try the following Practice Exercise Practice Exercise 151 Implicit differentiation (Answers on page 884) In Problems 1 and 2 determine

dy dx

1. x2 + y2 + 4x − 3y + 1 = 0 2. 2y3 − y + 3x − 2 = 0 dy 3. Given x2 + y2 = 9 evaluate when dx √ x = 5 and y = 2 In Problems 4 to 7, determine Figure 28.2

4. x2 + 2x sin 4y = 0 Problem 10. Pressure p and volume v of a gas are related by the law pvγ = k, where γ and k are constants. Show that the rate of change of pressure dp p dv = −γ dt v dt

5. 3y2 + 2xy − 4x2 = 0 6. 2x2 y + 3x3 = sin y 7. 3y + 2x ln y = y4 + x

dy dx

374 Section G 5 dy 8. If 3x2 + 2x2 y3 − y2 = 0 evaluate when 4 dx 1 x = and y = 1 2

2.

9. Determine the gradients of the tangents drawn to the circle x2 + y2 = 16 at the point where x = 2. Give the answer correct to 4 significant figures. 10. Find the gradients of the tangents drawn to x2 y2 the ellipse + = 2 at the point where 4 9 x=2

(c) 20x4 − 6y − 24y2 3.

d ( 2 5) 3x y is equal to: dx ( ) dx 5 4 (a) 6xy (b) 3xy 2y + 5x dy ( ) dy 2 4 5 4 (c) 15x y + 6xy (d) 3xy 5x + 2y dx

(

5y 3x

(d)

) is equal to:

) 5 x−y (a) 3x2 ( ) dy 5 x −y dx (c) 3x2

20x4 − 1 6y − 24y2

( ) dx 15 x − y dy (b) 9x2

(d)

5y 3

4.

Given 5y2 + 3x4 − 6y3 + 4 = 0 then the value dy of when x = 2 and y = 1 is equal to: dx (a) −52 (b) 12 (c) −48 (d) 0.194

5.

The gradient of the curve y2 + 5xy = − 3 at the point (2, −1) is: (a) 7 (b) −2.5 (c) 0.625 (d) −0.125

Practice Exercise 152 Multiple-choice questions on differentiation of implicit functions (Answers on page 884)

1. The function

d dx

(

11. Determine the gradient of the curve 3xy + y2 = −2 at the point (1,−2)

Each question has only one correct answer

dy Given 3y2 + 1 − 4x5 − 8y3 = 0 then is dx equal to: 10x4 3y − 12y2 ) (a) ( (b) 24x4 3y 1 − 4y

For fully worked solutions to each of the problems in Practice Exercises 154 to 156 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 29

Logarithmic differentiation Why it is important to understand: Logarithmic differentiation Logarithmic differentiation is a means of differentiating algebraically complicated functions or functions for which the ordinary rules of differentiation do not apply. The technique is performed in cases where it is easier to differentiate the logarithm of a function rather than the function itself. Logarithmic differentiation relies on the function of a function rule (i.e. chain rule) as well as properties of logarithms (in particular, the natural logarithm, or logarithm to the base e) to transform products into sums and divisions into subtractions, and can also be applied to functions raised to the power of variables of functions. Logarithmic differentiation occurs often enough in engineering calculations to make it an important technique.

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

state the laws of logarithms differentiate simple logarithmic functions differentiate an implicit function involving logarithms differentiate more difficult logarithmic functions involving products and quotients x differentiate functions of the form y = [ f (x)]

29.1 Introduction to logarithmic differentiation With certain functions containing more complicated products and quotients, differentiation is often made easier if the logarithm of the function is taken before differentiating. This technique, called ‘logarithmic differentiation’ is achieved with a knowledge of (i) the laws of logarithms, (ii) the differential coefficients of logarithmic functions, and (iii) the differentiation of implicit functions.

29.2

Laws of logarithms

Three laws of logarithms may be expressed as:

(i) (ii) (iii)

log(A × B) = log A + log B ( ) A log = log A − log B B log An = n log A

In calculus, Napierian logarithms (i.e. logarithms to a base of ‘e’) are invariably used. Thus for two functions f (x) and g(x) the laws of logarithms may be expressed as: (i) (ii) (iii)

ln[ f (x) · g(x)] = ln f (x) + ln g(x) ( ) f(x) ln = ln f(x) − ln g(x) g(x) ln[ f (x)]n = n ln f(x)

376 Section G Taking Napierian logarithms of both sides of the equaf(x) · g(x) tion y = gives: h(x) ) ( f(x) · g(x) ln y = ln h(x)

7.

3 ln 4x

8.

2 ln(sin x)

9.

ln(4x3 − 6x2 + 3x)

which may be simplified using the above laws of logarithms, giving: ln y = ln f(x) + ln g(x) − ln h(x) This latter form of the equation is often easier to differentiate.

As explained in Chapter 28, by using the function of a function rule: ( ) 1 dy d (ln y) = (2) dx y dx

29.3 Differentiation of logarithmic functions The differential coefficient of the logarithmic function ln x is given by: 1 d (ln x) = dx x More generally, it may be shown that: f ′ (x) d [ln f (x)] = dx f (x)

29.4 Differentiation of further logarithmic functions

(1)

Differentiation of an expression such as √ (1 + x)2 (x − 1) √ y= may be achieved by using the x (x + 2) product and quotient rules of differentiation; however the working would be rather complicated. With logarithmic differentiation the following procedure is adopted: (i)

For example, if y = ln(3x2 + 2x − 1) then,

Take Napierian logarithms of both sides of the equation. √ } (1 + x)2 (x − 1) √ Thus ln y = ln x (x + 2) {

dy 6x + 2 = 2 dx 3x + 2x − 1 Similarly, if y = ln(sin 3x) then

{

dy 3 cos 3x = = 3 cot 3x dx sin 3x Now try the following Practice Exercise

= ln

(ii)

}

1

x(x + 2) 2

Apply the laws of logarithms. 1 Thus ln y = ln(1 + x)2 + ln(x − 1) 2

Practice Exercise 153 Differentiating logarithmic functions (Answers on page 884)

1

− ln x − ln(x + 2) 2 , by laws (i) and (ii) of Section 29.2

Differentiate the following:

i.e. ln y = 2 ln(1 + x) + 12 ln(x − 1)

1. ln(4x − 10)

− ln x − 12 ln(x + 2), by law (iii)

2. ln(cos 3x)

of Section 29.2

3. ln(3x3 + x) 4. ln(5x2 + 10x − 7)

1

(1 + x)2 (x − 1) 2

(iii)

5. ln 8x

Differentiate each term in turn with respect to x using equations (1) and (2).

6. ln(x2 − 1)

Thus

1 1 1 dy 2 1 = + 2 − − 2 y dx (1 + x) (x − 1) x (x + 2)

Logarithmic differentiation 377 (iv) Rearrange the equation to make Thus

dy the subject. dx

{ dy 2 1 1 =y + − dx (1 + x) 2(x − 1) x

1 − 2(x + 2)

}

√

(x − 2)3 (x + 1)2 (2x − 1) dy with respect to x and evaluate when x = 3 dx Differentiate y =

Problem 2.

Using logarithmic differentiation and following the above procedure: √ (x − 2)3 (i) Since y = (x + 1)2 (2x − 1)

(v)

Substitute for y in terms of x. √ { dy (1 + x)2 (x − 1) 2 √ Thus = dx (1 + x) x (x + 2) 1 1 1 + − − 2(x − 1) x 2(x + 2)

{

} then

√

(x − 2)3 ln y = ln (x + 1)2 (2x − 1) {

Problem 1.

Use logarithmic differentiation to (x + 1)(x − 2)3 differentiate y = (x − 3) (ii)

(ii)

y=

(iii)

ln y = ln(x + 1) + ln(x − 2)3 − ln(x − 3), by laws (i) and (ii) of Section 29.2, i.e. ln y = ln(x + 1) + 3 ln(x − 2) − ln(x − 3), by law (iii) of Section 29.2.

(iii)

3

(iv) (v)

ln(x − 2) − 2 ln(x + 1)

3 2

1 dy 2 2 = − − y dx (x − 2) (x + 1) (2x − 1) { } dy 3 2 2 =y − − dx 2(x − 2) (x + 1) (2x − 1) √ { dy (x − 2)3 3 = 2 dx (x + 1) (2x − 1) 2(x − 2)

Differentiating with respect to x gives:

−

1 dy 1 3 1 = + − , y dx (x + 1) (x − 2) (x − 3) (iv) Rearranging gives: { } dy 1 3 1 =y + − dx (x + 1) (x − 2) (x − 3)

1 =± 80

Substituting for y gives: dy (x + 1)(x − 2)3 = dx (x − 3) +

{

Problem 3. 1 (x + 1)

3 1 − (x − 2) (x − 3)

}

2 2 − (x + 1) (2x − 1)

}

√ ( ) dy (1)3 3 2 2 When x = 3, = − − dx (4)2 (5) 2 4 5

by using equations (1) and (2)

(v)

3 2

− ln(2x − 1)

(x + 1)(x − 2)3 (x − 3) { } (x + 1)(x − 2)3 then ln y = ln (x − 3)

Since

}

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

Following the above procedure: (i)

3

(x − 2) 2 = ln (x + 1)2 (2x − 1)

}

( ) 3 3 or ±0.0075 =± 5 400

3e2θ sec 2θ dy Given y = √ determine dθ (θ − 2)

Using logarithmic differentiation and following the procedure gives: (i)

Since

3e2θ sec 2θ y= √ (θ − 2)

378 Section G {

then

(ii)

} 3e2θ sec 2θ ln y = ln √ (θ − 2) { 2θ } 3e sec 2θ = ln 1 (θ − 2) 2

Now try the following Practice Exercise Practice Exercise 154 Differentiating logarithmic functions (Answers on page 885) 1

ln y = ln 3e2θ + ln sec 2θ − ln(θ − 2) 2 i.e. ln y = ln 3 + ln e2θ + ln sec 2θ − 12 ln(θ − 2) i.e. ln y = ln 3 + 2θ + ln sec 2θ − 12 ln(θ − 2)

(iii)

2.

Differentiating with respect to θ gives: 1 1 dy 2 sec 2θ tan 2θ = 0+2+ − 2 y dθ sec 2θ (θ − 2) from equations (1) and (2)

(iv) Rearranging gives: { } dy 1 = y 2 + 2 tan 2θ − dθ 2(θ − 2) (v)

Substituting for y gives: dy 3e2θ sec 2θ = √ dθ (θ − 2)

{ 2 + 2 tan 2θ −

1 2(θ − 2)

}

x3 ln 2x Problem 4. Differentiate y = x with e sin x respect to x.

Using logarithmic differentiation and following the procedure gives: { 3 } x ln 2x (i) ln y = ln x e sin x (ii)

ln y = ln x3 + ln(ln 2x) − ln(ex ) − ln(sin x) i.e. ln y = 3 ln x + ln(ln 2x) − x − ln(sin x)

(iii) (iv)

(v)

1 1 dy 3 cos x = + x −1− y dx x ln 2x sin x { } dy 3 1 =y + − 1 − cot x dx x x ln 2x { } dy x3 ln 2x 3 1 = x + − 1 − cot x dx e sin x x x ln 2x

In Problems 1 to 6, use logarithmic differentiation to differentiate the given functions with respect to the variable. (x − 2)(x + 1) 1. y = (x − 1)(x + 3)

3.

(x + 1)(2x + 1)3 (x − 3)2 (x + 2)4 √ (2x − 1) (x + 2) √ y= (x − 3) (x + 1)3 y=

4.

e2x cos 3x y= √ (x − 4)

5.

y = 3θ sin θ cos θ

6.

y=

7.

2x4 tan x e2x ln 2x dy Evaluate when x = 1 given dx √ (x + 1)2 (2x − 1) √ y= (x + 3)3

8.

Evaluate

dy , correct to 3 significant figures, dθ π 2eθ sin θ when θ = given y = √ 4 θ5

29.5

Differentiation of [ f(x)]x

Whenever an expression to be differentiated contains a term raised to a power which is itself a function of the variable, then logarithmic differentiation must be used. For example, the √ differentiation of expressions such as xx , (x + 2)x , x (x − 1) and x3x+2 can only be achieved using logarithmic differentiation. Problem 5.

Determine

dy given y = xx dx

Taking Napierian logarithms of both sides of y = xx gives:

Logarithmic differentiation 379 ln y = ln xx = x ln x, by law (iii) of Section 29.2. Differentiating both sides with respect to x gives: ( ) 1 dy 1 = (x) + (ln x)(1), using the product rule y dx x

Differentiating each side with respect to x gives: ( )( ) ( ) 1 dy 1 1 −1 = + [ln(x − 1)] y dx x x−1 x2 by the product rule. { } dy 1 ln(x − 1) Hence =y − dx x(x − 1) x2 { } dy √ 1 ln(x − 1) i.e. = x (x − 1) − dx x(x − 1) x2

1 dy = 1 + ln x, y dx

i.e. from which,

dy = y(1 + ln x) dx

i.e.

dy = xx (1 + ln x) dx

Problem 6. y = (x + 2)x

Evaluate

dy when x = −1 given dx

Taking Napierian logarithms of both sides of y = (x + 2)x gives: ln y = ln(x + 2)x = x ln(x + 2), by law (iii) of Section 29.2 Differentiating both sides with respect to x gives: ( ) 1 dy 1 = (x) + [ln(x + 2)](1), y dx x+2 (

Hence

by ) the product rule. + ln(x + 2)

x dy =y dx x+2 } { x = (x + 2)x + ln (x + 2) x+2

When x = −1,

When x = 2,

(b)

{ } dy √ 1 ln(1) 2 = (1) − dx 2(1) 4

dy = (1)−1 dx

(

−1 + ln 1 1

)

{ } 1 1 = ±1 −0 = ± 2 2 Problem 8. Let y = x3x+2 Taking Napierian logarithms of both sides gives: ln y = ln x3x+2 i.e. ln y = (3x + 2) ln x, by law (iii) of Section 29.2. Differentiating each term with respect to x gives: ( ) 1 dy 1 = (3x + 2) + (ln x)(3), y dx x by the product rule. { } dy 3x + 2 Hence =y + 3 ln x dx x { = x3x+2

= (+1)(−1) = −1 Determine (a) the differential √ dy coefficient of y = x (x − 1) and (b) evaluate dx when x = 2

Taking Napierian logarithms of both sides gives: 1

ln y = ln(x − 1) x =

1 ln(x − 1), x

by law (iii) of Section 29.2.

} 3x + 2 + 3 ln x x

{ } 2 = x3x+2 3 + + 3 ln x x

Problem 7.

√ 1 (a) y = x (x√ − 1) = (x − 1) x , since by the laws of m indices n am = a n

Differentiate x3x+2 with respect to x

Now try the following Practice Exercise Practice Exercise 155 Differentiating x [ f(x)] type functions (Answers on page 885) In Problems 1 to 4, differentiate with respect to x. 1.

y = x2x

2.

y = (2x − 1)x

380 Section G √ 3. y = x (x + 3)

2.

4. y = 3x4x+1 5. Show that when y = 2xx and x = 1,

dy =2 dx

} d {√ x (x − 2) when x = 3 dx dy = 6.77, 7. Show that if y = θθ and θ = 2, dθ correct to 3 significant figures.

3.

6. Evaluate

Practice Exercise 156 Multiple-choice questions on logarithmic differentiation (Answers on page 885) Each question has only one correct answer

4.

) d ( π 2 ln cos θ when θ = is: dθ 6 (a) −3.464 (b) −7.464 (c) −9.774 (d) 4 The value of

dy is equal to: dx (a) 6x2x (1 + ln x) (b) ln 3 + 2x ln x (c) 2x(3x)x (d) 3x2x (2x ln 3x) If y = 3x2x then

The value of is: (a) 1

5.

d ln(x3 − 2x2 + 1) when x = 2 dx

(b) 12

(c) 5 (d) 4 √ dy Given that y = x (x − 2) , the value of when dx x = 3 is: 1 1 1 1 (a) − (b) − (c) (d) 3 9 3 9

dy is equal 1. Given that y = ln(4x3 − 3x2 ) then dx to: 6x 6(2x − 1) (a) (b) 3 2 4x + 3x x(4x − 3) 12x2 1 (c) 3 (d) 3 2 4x + 3x 4x + 3x2

For fully worked solutions to each of the problems in Practice Exercises 153 to 155 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 9

Further differentiation

This Revision Test covers the material contained in Chapters 27 to 29. The marks for each question are shown in brackets at the end of each question. 1.

2.

3.

A cycloid has parametric equations given by: x = 5(θ − sin θ) and y = 5(1 − cos θ). Evaluate

5.

Determine the gradient of the tangents drawn to the hyperbola x2 − y2 = 8 at x = 3 (4)

d2 y dy (b) 2 when θ = 1.5 radians. Give answers (a) dx dx correct to 3 decimal places. (8)

6.

Use logarithmic differentiation to differentiate √ (x + 1)2 (x − 2) √ with respect to x. (6) y= (2x − 1) 3 (x − 3)4

Determine the equation of (a) the tangent, and (b) the normal, drawn to an ellipse x = 4 cos θ, π (8) y = sin θ at θ = 3 Determine expressions for lowing functions: (a) z = 5y2 cos x

4.

dz for each of the foldy

(b) z = x2 + 4xy − y2

(5)

7.

8.

3eθ sin 2θ dy √ and hence evaluate , dθ θ5 π correct to 2 decimal places, when θ = 3 (9) Differentiate y =

] d [√t (2t + 1) when t = 2, correct to 4 dt significant figures. (6) Evaluate

dy If x2 + y2 + 6x + 8y + 1 = 0, find in terms of x dx and y. (4)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 9, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Chapter 30

Differentiation of hyperbolic functions Why it is important to understand: Differentiation of hyperbolic functions Hyperbolic functions have applications in many areas of engineering. For example, the shape formed by a wire freely hanging between two points (known as a catenary curve) is described by the hyperbolic cosine. Hyperbolic functions are also used in electrical engineering applications and for solving differential equations; other applications of hyperbolic functions are found in fluid dynamics, optics, heat, mechanical engineering and in astronomy when dealing with the curvature of light in the presence of black holes. Differentiation of hyperbolic functions is quite straightforward.

At the end of this chapter, you should be able to: • derive the differential coefficients of hyperbolic functions • differentiate hyperbolic functions

30.1 Standard differential coefficients of hyperbolic functions From Chapter 12, ( ) [ x ] d d ex − e−x e − (−e−x ) (sinh x) = = dx dx 2 2 ( x ) −x e +e = = cosh x 2 If y = sinhax, where a is then dy = a cosh ax dx

a

constant,

d d (cosh x) = dx dx ( =

(

ex + e−x 2

ex − e−x 2

)

[

ex + (−e−x ) = 2

]

) = sinh x

If y = cosh ax, where a is a constant, then dy = a sinh ax dx Using the quotient rule of differentiation the derivatives of tanh x, sech x, cosech x and coth x may be determined using the above results.

Differentiation of hyperbolic functions 383 Summary of differential coefficients

Problem 1. Determine the differential coefficient of: (a) th x (b) sech x

(a)

d d (th x) = dx dx =

(

sh x ch x

)

(ch x)(ch x) − (sh x)(sh x) ch2 x using the quotient rule ch x − sh x 1 = 2 = sech2 x 2 ch x ch x 2

=

(b)

2

d d (sech x) = dx dx

(

1 ch x

30.2

(ch x)(0) − (1)(sh x) ch2 x ( )( ) −sh x 1 sh x = 2 =− ch x ch x ch x = −sech x th x dy Problem 2. Determine given dθ (a) y = cosech θ (b) y = coth θ (a)

d d (cosech θ) = dθ dθ

1 sh θ

)

(sh θ)(0) − (1)(ch θ) sh2 θ ( )( ) −ch θ 1 ch θ = 2 =− sh θ sh θ sh θ =

= −cosech θ coth θ (b)

d d (coth θ) = dθ dθ

(

ch θ sh θ

dy or f ′ (x) dx

sinh ax

a cosh ax

cosh ax

a sinh ax

tanh ax

a sech2 ax

sech ax

−a sech ax tanh ax

cosech ax

−a cosech ax coth ax

coth ax

−a cosech2 ax

)

=

(

y or f(x)

)

(sh θ)(sh θ) − (ch θ)(ch θ) = sh2 θ =

sh2 θ − ch2 θ −(ch2 θ − sh2 θ) = sh2 θ sh2 θ

=

−1 = −cosech2 θ sh2 θ

Further worked problems on differentiation of hyperbolic functions

Problem 3. Differentiate the following with respect to x: 3 (a) y = 4 sh 2x − ch 3x 7 x (b) y = 5 th − 2 coth 4x 2 (a)

y = 4 sh 2x −

3 ch 3x 7

dy 3 = 4(2 cosh 2x) − (3 sinh 3x) dx 7 9 = 8 cosh 2x − sinh 3x 7 x − 2 coth 4x 2 ( ) dy 1 x =5 sech2 − 2(−4 cosech2 4x) dx 2 2

(b) y = 5 th

=

5 x sech2 + 8 cosech2 4x 2 2

Problem 4. Differentiate the following with respect to the variable: (a) y = 4 sin 3t ch 4t (b) y = ln (sh 3θ) − 4 ch2 3θ

384 Section G (a) y = 4 sin 3t ch 4t (i.e. a product) dy = (4 sin 3t)(4 sh 4t) + (ch 4t)(4)(3 cos 3t) dt = 16 sin 3t sh 4t + 12 ch 4t cos 3t

2. 3.

= 4(4 sin 3t sh 4t + 3 cos 3t ch 4t) (b)

y = ln (sh 3θ) − 4 ch2 3θ (i.e. a function of a function) ( ) dy 1 = (3 ch 3θ) − (4)(2 ch 3θ)(3 sh 3θ) dθ sh 3θ

2 (a) sech 5x 3

5 t (b) cosech (c) 2 coth 7θ 8 2 ( ( )) 3 θ (a) 2 ln (sh x) (b) ln th 4 2

4.

(a) sh 2x ch 2x (b) 3e2x th 2x

5.

(a)

3 sh 4x 2x3

(b)

ch 2t cos 2t

= 3 coth 3θ − 24 ch 3θ sh 3θ = 3(coth 3θ − 8 ch 3θ sh 3θ) Problem 5. Show that the differential coefficient 3x2 of y = is: ch 4x 6x sech 4x (1 − 2x th 4x)

Practice Exercise 158 Multiple-choice questions on differentiation of hyperbolic functions (Answers on page 885) Each question has only one correct answer

3x2 y= ch 4x

(i.e. a quotient)

1.

dy (ch 4x)(6x) − (3x2 )(4 sh 4x) = dx (ch 4x)2 6x(ch 4x − 2x sh 4x) (ch2 4x) [ ] ch4x 2x sh 4x = 6x − ch2 4x ch2 4x [ ( )( )] 1 sh 4x 1 = 6x − 2x ch 4x ch 4x ch 4x =

2.

Now try the following Practice Exercise Practice Exercise 157 Differentiation of hyperbolic functions (Answers on page 885) In Problems 1 to 5 differentiate the given functions with respect to the variable: 1. (a) 3 sh 2x

(b) 2 ch 5θ

(c) 4 th 9t

If y = 6 tanh 2x − 5 coth 3x then

dy is equal to: dx

(a) 3(4sech2 2x + 5cosech2 3x) (b) 3(4sech2 2x − 5cosech2 3x) (c) − 3(4sech2 2x + 5cosech2 3x) (d) 3(− 4sech2 2x + 5cosech2 3x)

= 6x[sech 4x − 2x th 4x sech 4x] = 6x sech 4x (1 − 2x th 4x)

dy If y = 3 sinh 2x − 2cosech 3x then is equal dx to: (a) 6(cosh 2x − cosech 3x coth 3x) (b) 6(cosh 2x + cosech 3x coth 3x) (c) 6(− cosh 2x + cosech 3x coth 3x) (d) −6(cosh 2x + cosech 3x coth 3x)

3.

dy Given that y = 3 ln(sinh 2t), then is given dt by: 3 (a) sinh 2t 3 (b) cosh 2t 3 (c) 2 cosh 2t 6 (d) tanh 2t

Differentiation of hyperbolic functions 385

4. If y = sinh 3x cosh 3x, then (a) 3 cosh 3x sinh 3x (b) 3(sinh2 3x + cosh2 3x) (c) −3 cosh 3x sinh 3x (d) 3(sinh2 3x − cosh2 3x)

dy is equal to: dx

5.

If y =

sinh 2x dy then is equal to: 3x2 dx

1 (x cosh 2x − sinh 2x) 3x2 1 (b) 2 (x cosh 2x − sinh 2x) x 2 (c) 3 (x cosh 2x − sinh 2x) 3x 2 (d) 3 (x cosh 2x + sinh 2x) 3x

(a)

For fully worked solutions to each of the problems in Practice Exercise 157 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 31

Differentiation of inverse trigonometric and hyperbolic functions Why it is important to understand: Differentiation of inverse trigonometric hyperbolic functions As has been mentioned earlier, hyperbolic functions have applications in many areas of engineering. For example, the hyperbolic sine arises in the gravitational potential of a cylinder, the hyperbolic cosine function is the shape of a hanging cable, the hyperbolic tangent arises in the calculation of and rapidity of special relativity, the hyperbolic secant arises in the profile of a laminar jet and the hyperbolic cotangent arises in the Langevin function for magnetic polarisation. So there are plenty of applications for inverse functions in engineering and this chapter explains how to differentiate inverse trigonometric and hyperbolic functions.

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

understand inverse functions differentiate inverse trigonometric functions evaluate inverse hyperbolic functions using logarithmic forms differentiate inverse hyperbolic functions

31.1

Inverse functions

y+2 . The x and 3 x+2 y are interchanged and the function y = is called 3 the inverse function of y = 3x − 2 (see page 201). Inverse trigonometric functions are denoted by prefixing the function with ‘arc’ or, more commonly, by using the −1 notation. For example, if y = sin x, then x = arcsin y or x = sin−1 y. Similarly, if y = cos x, then If y = 3x − 2, then by transposition, x =

x = arccos y or x = cos−1 y, and so on. In this chapter the −1 notation will be used. A sketch of each of the inverse trigonometric functions is shown in Fig. 31.1. Inverse hyperbolic functions are denoted by prefixing the function with ‘ar’ or, more commonly, by using the −1 notation. For example, if y = sinh x, then x = arsinh y or x = sinh−1 y. Similarly, if y = sech x, then x = arsech y or x = sech−1 y, and so on. In this chapter the −1 notation will be used. A sketch of each of the inverse hyperbolic functions is shown in Fig. 31.2.

Differentiation of inverse trigonometric and hyperbolic functions 387 y

y

y 3p/2

3p/2

y 5sin21x

y 5 tan21x p

D

p p/2

B

21 0 A 2p/2

y 5 cos21x

p/2

11 x

0 2p/2

21

2p

2p

23p/2

23p/2

p/2

C 0

11 x

x

2p/2

(a)

(b)

(c) y

y 3p/2

3p/2

p p/2

p

y 5sec21x

y p

y 5 cosec21x

p/2

p/2

21 0 11 2p/2

x

21 0 2p/2

2p

2p

23p/2

23p/2

(d)

0

x

11

y 5 cot21x x

2p/2 p

(e)

(f)

Figure 31.1

y 3 2

y 5 sinh21x

1

y 5 cosh21x

2

y 5 tanh21x

1

01 2 3x 23 22 21 21

22 21 0 21

22

22

23

23

1

y y 5 cosech21x

2

11 x

(c)

y y 5sech21x

0

21

2 3x

(b)

(a) y 3

y

y 3

y 5coth21x

1 0 21 22

1

x

0

x

21 0 11

23 (c)

Figure 31.2

(e)

(f)

x

388 Section G 31.2 Differentiation of inverse trigonometric functions (i)

If y = sin

−1

Then

x, then x = sin y.

dy 1 =√ dx 1 − x2

(iii)

A sketch of part of the curve of y = sin−1 x is shown in Fig. 31.1(a). The principal value of sin−1 x is defined as the value lying between −π/2 and π/2. The gradient of the curve between points A and B is positive for all values of x and thus only the positive value is taken 1 when evaluating √ 1 − x2 x Given y = sin−1 a x = a sin y

then

x = sin y a

and

√ dx Hence = a cos y = a 1 − sin2 y dy √( √[ ) ( x )2] a2 − x2 =a 1− =a a a2 √ a a2 − x2 √ 2 = = a − x2 a dy 1 1 Thus = =√ 2 dx dx a − x2 dy x dy 1 i.e. when y = sin−1 then =√ 2 a dx a − x2 Since integration is the reverse process of differentiation then: ∫

Thus

(iv) Given y = sin−1 f (x) the function of a function dy rule may be used to find dx

dy dy du 1 = × =√ f ′ (x) dx du dx 1 − u2 f ′ (x) =√ 1 − [ f(x)]2

(v)

The differential coefficients of the remaining inverse trigonometric functions are obtained in a similar manner to that shown above and a summary of the results is shown in Table 31.1.

Table 31.1 Differential coefficients of inverse trigonometric functions dy or f ′ (x) dx

y or f (x) (i)

sin−1

1 √ 2 a − x2

x a

sin−1 f (x) (ii)

cos−1

(iii)

tan−1

x a

sec−1

x a

cosec−1

x a

cosec−1 f(x) (vi)

cot−1

f ′ (x) 1 + [ f (x)]2 a √ 2 x x − a2

sec−1 f (x) (v)

−f ′ (x) √ 1 − [ f(x)]2 a a2 + x2

tan−1 f(x) (iv)

f ′ (x) √ 1 − [ f(x)]2 −1 √ a2 − x2

x a

cos−1 f(x)

1 x √ dx = sin−1 + c 2 2 a a −x

Let u = f (x) then y = sin−1 u

dy 1 =√ du 1 − u2 (see para. (i))

Differentiating√ both sides with respect to y gives: dx = cos y = 1 − sin2 y dy dx √ = 1 − x2 since cos2 y + sin2 y = 1, i.e. dy dy 1 However = dx dx dy Hence, when y = sin−1 x then

(ii)

du ′ = f (x) and dx

x a

cot−1 f (x)

f ′ (x) √ f (x) [ f (x)]2 − 1 −a √ x x2 − a2 −f ′ (x) √ f (x) [ f (x)]2 − 1 −a + x2

a2

−f ′ (x) 1 + [ f (x)]2

Differentiation of inverse trigonometric and hyperbolic functions 389

Problem 1.

dy Find given y = sin−1 5x2 dx

dy −(−4x) =√ dx 1 − [1 − 2x2 ]2

then

From Table 31.1(i), if y = sin−1 f (x) then

f ′ (x) dy =√ dx 1 − [ f(x)]2

=√

10x dy 10x =√ =√ 2 2 dx 1 − 25x4 1 − (5x )

4x 1 − (1 − 4x2

+ 4x4 )

4x =√ (4x2 − 4x4 )

4x 4x 2 =√ = √ =√ 2 2 2 1 − x2 [4x (1 − x )] 2x 1 − x

Hence, if y = sin−1 5x2 then f(x) = 5x2 and f ′ (x) = 10x Thus

y = cos−1(1 − 2x2 )

Hence, when

Problem 3.

Determine the differential coefficient x of y = tan and show that the differential a 6 2x is coefficient of tan−1 3 9 + 4x2 −1

Problem 2. (a) Show that if y = cos−1 x then 1 dy =√ dx 1 − x2 (b)

Hence obtain the differential coefficient of y = cos−1 (1 − 2x2 )

(a) If y = cos−1 x then x = cos y Differentiating with respect to y gives: √ dx = −sin y = − 1 − cos2 y dy √ = − 1 − x2 dy 1 1 = = −√ dx dx 1 − x2 dy The principal value of y = cos−1 x is defined as the angle lying between 0 and π, i.e. between points C and D shown in Fig. 31.1(b). The gradient of the curve is negative between C and D and thus the differential dy coefficient is negative as shown above. dx

If y = tan−1

x x then = tan y and x = a tan y a a

dx = a sec2 y = a(1 + tan2 y) since dy sec2 y = 1 + tan2 y [ ] ( 2 ) ( x )2 a + x2 = a 1+ =a a a2 =

a2 + x2 a

dy 1 a = = 2 dx dx a + x2 dy

Hence

Hence

(b) If y = cos−1 f(x) then by letting u = f(x), y = cos−1 u

The principal value of y = tan−1 x is defined as the π π angle lying between − and and the gradient 2 2 ) ( dy between these two values is always positive i.e. dx (see Fig. 31.1(c)). 2x x 3 Comparing tan−1 with tan−1 shows that a = 3 a 2 2x Hence if y = tan−1 then 3 3 3 3 dy 2 2 2 = ( )2 = = 9 dx 9 + 4x2 3 2 +x + x2 4 4 2

Then

dy 1 = −√ (from part (a)) du 1 − u2

du = f ′ (x) dx From the function of a function rule, and

dy dy du 1 = · = −√ f ′ (x) dx du dx 1 − u2 −f ′ (x) =√ 1 − [ f(x)]2

3 (4) 6 = 2 2= 9 + 4x 9 + 4x2 Problem 4. Find the differential coefficient of y = ln (cos−1 3x)

390 Section G Let u = cos−1 3x then y = ln u By the function of a function rule,

dy dy du 1 d = · = × (cos−1 3x) dx du dx u dx { } 1 −3 √ = cos−1 3x 1 − (3x)2

i.e.

−3 d [ln(cos−1 3x)] = √ dx 1 − 9x2 cos−1 3x

Problem 5.

If y = tan−1

3 dy find t2 dt

Using the product rule: [ ] dy −1 = (x) √ + (cosec−1 x) (1) dx x x2 − 1 from Table 31.1(v) −1 =√ + cosec−1 x x2 − 1 Problem 8. Show that if ( ) sin t dy 1 y = tan−1 then = cos t − 1 dt 2 ( If

f (t) =

then

f ′ (t) =

Using the general form from Table 31.1(iii), = f (t) =

−6 t3 ( ) d 3 f ′ (t) tan−1 2 = dt t 1 + [ f (t)]2

from which Hence

3 = 3t−2 t2

( )( 4 ) 6 t 6t = −3 =−4 4 t t +9 t +9 Problem 6.

Differentiate y =

cot−1 2x 1 + 4x2

Using the quotient rule: ( ) −2 (1 + 4x2 ) − (cot−1 2x)(8x) dy 1 + (2x)2 = dx (1 + 4x2 )2 from Table 31.1(vi)

)

(cos t − 1)(cos t) − (sin t)(−sin t) (cos t − 1)2 cos2 t − cos t + sin2 t 1 − cos t = (cos t − 1)2 (cos t − 1)2 since sin2 t + cos2 t = 1

=

f ′ (t) =

6 6 −3 −3 t t ={ ( )2 } = t4 + 9 3 1+ 2 t4 t

sin t cos t − 1

−(cos t − 1) −1 = (cos t − 1)2 cos t − 1

Using Table 31.1(iii), ( when ) sin t −1 y = tan cos t − 1 dy then = dt

−1 cos t − 1 ( )2 sin t 1+ cos t − 1

−1 cos t−1 = (cos t − 1)2 + (sin t)2 (cos t − 1)2 ( )( ) −1 (cos t − 1)2 = cos t − 1 cos2 t − 2 cos t + 1 + sin2 t =

−(cos t − 1) 1 − cos t 1 = = 2 − 2 cos t 2(1 − cos t) 2

Now try the following Practice Exercise

−1

=

−2(1 + 4x cot 2x) (1 + 4x2 )2

Problem 7.

Differentiate y = x cosec−1 x

Practice Exercise 159 Differentiating inverse trigonometric functions (Answers on page 885)

Differentiation of inverse trigonometric and hyperbolic functions 391 Taking Napierian logarithms of both sides gives: In Problems 1 to 6, differentiate with respect to the variable. x 1. (a) sin−1 4x (b) sin−1 2 2 x 2. (a) cos−1 3x (b) cos−1 3 3 √ 1 3. (a) 3 tan−1 2x (b) tan−1 x 2 3 −1 4. (a) 2 sec 2t (b) sec−1 x 4 5 −1 θ (b) cosec−1 x2 5. (a) cosec 2 2 √ 6. (a) 3 cot−1 2t (b) cot−1 θ2 − 1 7.

{ y = ln

Hence,

−1

sinh

8.

(a) 2x sin−1 3x (b) t2 sec−1 2t

9. 10.

(a) θ2 cos−1 (θ2 − 1) (b) (1 − x2 ) tan−1 x √ √ (a) 2 t cot−1 t (b) x cosec−1 x

11.

(a)

sin−1 3x cos−1 x (b) √ 2 x 1 − x2

√ { } x x + a2 + x2 = ln a a

(1)

3 Thus to evaluate sinh−1 , let x = 3 and a = 4 in equa4 tion (1). Then sinh

−1

} √ 3 + 42 + 32 4 ( ) 3+5 = ln = ln 2 = 0.6931 4

3 = ln 4

Show(that the coefficient of ) differential x 1 + x2 −1 tan is 1 − x2 1 − x2 + x4

In Problems 8 to 11 differentiate with respect to the variable.

} √ x + a2 + x2 a

{

By similar reasoning to the above it may be shown that: cosh−1 and

tanh−1

Problem 9. sinh−1 2

√ { } x x + x2 − a2 = ln a a ( ) x 1 a+x = ln a 2 a−x

Evaluate, correct to 4 decimal places,

{ } √ x x + a2 + x2 From above, sinh = ln a a With x = 2 and a = 1, { } √ 2 + 12 + 22 −1 sinh 2 = ln 1 √ = ln(2 + 5) = ln 4.2361 −1

31.3 Logarithmic forms of inverse hyperbolic functions Inverse hyperbolic functions may be evaluated most conveniently when expressed in a logarithmic form. x x For example, if y = sinh−1 then = sinh y. a a From Chapter 12, e y = cosh y + sinh y and cosh2 y − sinh2 y = 1, from which, √ cosh y = 1 + sinh2 y which is positive since cosh y is always positive (see Fig. 12.2, page 154). √ Hence e y = 1 + sinh2 y + sinh y √( √[ ) ( x )2 ] x a2 + x2 x = 1+ + = + 2 a a a a =

√ a2 + x2 x + a a

or

√ x + a2 + x2 a

= 1.4436, correct to 4 decimal places Using a calculator, (i)

press hyp

(ii)

press 4 and sinh−1 ( appears

(iii)

type in 2

(iv) press ) to close the brackets (v)

press = and 1.443635475 appears

Hence, sinh−1 2 = 1.4436, correct to 4 decimal places.

392 Section G Problem 10. Show that ( ) 1 a+x −1 x = ln and evaluate, correct tanh a 2 a−x 3 to 4 decimal places, tanh−1 5 If y = tanh−1

x x then = cosh y a a √ e y = cosh y + sinh y = cosh y ± cosh2 y − 1 √[ √ ] ( x )2 x x x2 − a2 = ± −1 = ± a a a a

If y = cosh−1

√ x ± x2 − a2 = a

x x then = tanh y a a

From Chapter 12,

tanh y =

Taking Napierian logarithms of both sides gives: {

1 y (e − e−y ) e2y − 1 sinh y = 21 y = 2y −y cosh y e +1 2 (e + e )

y = ln

} √ x ± x2 − a2 a

by dividing each term by e−y Thus, assuming the principal value, x e2y − 1 = a e2y + 1

Thus, from which,

x(e2y + 1) = a(e2y − 1)

Hence x + a = ae2y − xe2y = e2y (a − x) ( ) a+x 2y from which e = a−x Taking Napierian logarithms of both sides gives: ) a+x 2y = ln a−x ( ) 1 a+x y = ln 2 a−x (

and

cosh−1

x 1 Hence, tanh−1 = ln a 2

(

a+x a−x

√ { } x x + x2 − a2 = ln a a

14 7 = cosh−1 10 5 −1 x In the equation for cosh , let x = 7 and a = 5 a } { √ 7 + 72 − 52 −1 7 Then cosh = ln 5 5 cosh−1 1.4 = cosh−1

= ln 2.3798 = 0.8670 correct to 4 decimal places.

)

Substituting x = 3 and a = 5 gives: ( ) 1 5+3 1 −1 3 tanh = ln = ln 4 5 2 5−3 2 = 0.6931, correct to 4 decimal places. Problem 11. Prove that { } √ x + x2 − a 2 −1 x = ln cosh a a and hence evaluate cosh−1 1.4 correct to 4 decimal places.

Now try the following Practice Exercise

Practice Exercise 160 Logarithmic forms of the inverse hyperbolic functions (Answers on page 886) In Problems 1 to 3 use logarithmic equivalents of inverse hyperbolic functions to evaluate correct to 4 decimal places. 1. 2. 3.

1 2 5 (a) cosh−1 4 1 (a) tanh−1 4 (a) sinh−1

(b) sinh−1 4

(c) sinh−1 0.9

(b) cosh−1 3

(c) cosh−1 4.3

5 8

(c) tanh−1 0.7

(b) tanh−1

Differentiation of inverse trigonometric and hyperbolic functions 393 Table 31.2 Differential coefficients of inverse hyperbolic functions

31.4 Differentiation of inverse hyperbolic functions

dy or f ′ (x) dx

y or f(x) x x If y = sinh−1 then = sinh y and x = a sinh y a a dx = a cosh y (from Chapter 30). dy Also cosh2 y − sinh2 y = 1, from which, √[ √ ( x )2 ] 2 1+ cosh y = 1 + sinh y = a √ a2 + x2 = a √ dx a a2 + x2 √ 2 Hence = a cosh y = = a + x2 dy a

(i) sinh−1

sinh−1 f (x) (ii) cosh−1

(iii) tanh−1

(iv) sech−1

It follows from above that ∫ 1 x √ dx = sinh−1 + c 2 2 a x +a { } √ x + a2 + x2 or ln +c a

coth−1

by using the function of a function rule as in Section 31.2(iv). The remaining inverse hyperbolic functions are differentiated in a similar manner to that shown above and the results are summarised in Table 31.2.

x a

−f ′ (x) √ f(x) 1 − [ f(x)]2 −a √ x x2 + a2 −f ′ (x) √ f(x) [ f(x)]2 + 1 a a2 − x2

coth−1 f (x)

It may be shown that

d f ′ (x) [sinh−1 f (x)] = √ dx [ f(x)]2 + 1

x a

cosech−1 f (x) (vi)

f ′ (x) 1 − [ f (x)]2 −a √ x a2 − x2

x a

cosech−1

f ′ (x) √ [ f (x)]2 − 1 a a2 − x2

x a

sech−1 f(x) (v)

f ′ (x) √ [ f (x)]2 + 1 1 √ 2 x − a2

tanh−1 f(x)

−1 From the sketch of y = sinh in Fig. 31.2(a) ( x shown ) dy it is seen that the gradient i.e. is always positive. dx

or more generally

x a

cosh−1 f (x)

dy 1 1 Then = =√ 2 dx dx a + x2 dy x [An alternative method of differentiating sinh−1 a is to differentiate the logarithmic form { } √ x + a2 + x2 ln with respect to x] a

d 1 (sinh−1 x) = √ 2 dx x +1

1 √ 2 x + a2

x a

f ′ (x) 1 − [ f (x)]2

Problem 12. Find the differential coefficient of y = sinh−1 2x From Table 31.2(i), d f ′ (x) [sinh−1 f(x)] = √ dx [ f(x)]2 + 1 Hence

d 2 (sinh−1 2x) = √ dx [(2x)2 + 1] 2 =√ [4x2 + 1]

394 Section G Problem 13. Determine ] √ d [ cosh−1 (x2 + 1) dx If y = cosh−1 f (x),

dy f ′ (x) =√ dx [ f (x)]2 − 1

√ √ If y = cosh−1 (x2 + 1), then f(x) = (x2 + 1) and 1 x f ′ (x) = (x + 1)−1/2 (2x) = √ 2 2 (x + 1) [ ] √ d cosh−1 (x2 + 1) Hence, dx x x √ √ 2 2 (x + 1) (x + 1) = √[ ]=√ 2 (√ )2 (x + 1 − 1) (x2 + 1) − 1 x √ 2 1 (x + 1) = =√ 2 x (x + 1) Problem 14. Show that d [ a x] tanh−1 = 2 2 and hence determine the dx a a −x 4x differential coefficient of tanh−1 3 If y = tanh−1

x x then = tanh y and x = a tanh y a a

dx = a sech2 y = a(1 − tanh2 y), since dy 1 − sech2 y = tanh2 y [ ( 2 ) ( x )2 ] a − x2 a2 − x2 =a 1− =a = 2 a a a dy 1 a Hence = = 2 dx dx a − x2 dy x 3 4x with tanh−1 shows that a = Comparing tanh−1 3 a 4 3 3 [ ] d 4x 4 Hence tanh−1 = ( )42 = 9 dx 3 3 − x2 − x2 16 4 3 3 16 12 4 = = · = 2 2 4 (9 − 16x ) 9 − 16x2 9 − 16x 16

Problem 15. Differentiate cosech−1 (sinh θ) From Table 31.2(v), −f ′ (x) d √ [cosech−1 f (x)] = dx f(x) [ f(x)]2 + 1 d Hence [cosech−1 (sinh θ)] dθ −cosh θ √ = sinh θ [sinh2 θ + 1] =

−cosh θ √ since cosh2 θ − sinh2 θ = 1 sinh θ cosh2 θ

=

−cosh θ −1 = = −cosech θ sinh θ cosh θ sinh θ

Problem 16. Find the differential coefficient of y = sech−1 (2x − 1) From Table 31.2(iv), d −f ′ (x) √ [sech−1 f(x)] = dx f (x) 1 − [ f (x)]2 d [sech−1 (2x − 1)] dx −2 √ = (2x − 1) [1 − (2x − 1)2 ]

Hence,

=

−2 √ (2x − 1) [1 − (4x2 − 4x + 1)]

−2 −2 √ √ = 2 (2x − 1) (4x − 4x ) (2x−1) [4x(1−x)] −1 −2 √ √ = = (2x − 1)2 [x(1 − x)] (2x − 1) [x(1 − x)] =

Problem 17. Show that d [coth−1 (sin x)] = sec x dx From Table 31.2(vi), f ′ (x) d [coth−1 f (x)] = dx 1 − [ f(x)]2 d cos x Hence [coth−1 (sin x)] = dx [1 − (sin x)2 ] cos x = since cos2 x + sin2 x = 1 cos2 x =

1 = sec x cos x

Differentiation of inverse trigonometric and hyperbolic functions 395 ∫ Hence

Problem 18. Differentiate y = (x2 − 1) tanh−1 x

2 dx = 2 (9 − 4x2 ) =

Using the product rule, ( ) dy 1 = (x2 − 1) + (tanh−1 x)(2x) dx 1 − x2 =

Problem 19. Determine

d ( x) 1 sinh−1 =√ dx a (x2 + a2 )

Since

∫

dx x √ = sinh−1 + c a (x2 + a2 ) ∫ ∫ 1 1 √ Hence dx = √ dx (x2 + 4) (x2 + 22 )

∫ Problem 20. Determine

√

x +c 2

4 (x2 − 3)

1 x √ dx = cosh−1 + c 2 2 a (x − a ) ∫ ∫ 4 1 √ Hence dx = 4 √ dx √ 2 (x − 3) [x2 − ( 3)2 ] x = 4 cosh−1 √ + c 3

Since then i.e.

2 dx (9 − 4x2 )

d x a tanh−1 = 2 dx a a − x2 ∫ a x dx = tanh−1 + c 2 − x2 a a ∫ 1 1 x dx = tanh−1 + c 2 2 a −x a a

x (b) sinh−1 4x 3 t 1 (a) 2 cosh−1 (b) cosh−1 2θ 3 2 2x (a) tanh−1 (b) 3 tanh−1 3x 5 3x 1 (a) sech−1 (b) − sech−1 2x 4 2 x 1 (a) cosech−1 (b) cosech−1 4x 4 2 2x 1 (a) coth−1 (b) coth−1 3t 7 4 √ −1 2 (a) 2 sinh (x − 1)

1. (a) sinh−1

3. 4.

∫

Problem 21. Find

2 1 2x dx = tanh−1 +c 2 (9 − 4x ) 3 3

In Problems 1 to 11, differentiate with respect to the variable.

2.

then

∫

1 [( )2 ] dx 3 2 − x 2

Practice Exercise 161 Differentiation of inverse hyperbolic functions (Answers on page 886)

dx

d ( x) 1 cosh−1 =√ 2 dx a (x − a2 )

Since

∫

1 ) dx 2 4 −x

(9

Now try the following Practice Exercise

then

= sinh−1

∫ i.e.

dx √ (x2 + 4)

1 2

4

[ ] 1 1 −1 x ( ) tanh ( 3 ) + c = 2 32 2

−(1 − x2 ) + 2x tanh−1 x = 2x tanh−1 x − 1 (1 − x2 ) ∫

∫

5. 6. 7.

(b)

√ 1 cosh−1 (x2 + 1) 2

8. (a) sech−1 (x − 1) (b) tanh−1 (tanh x) ( ) t 9. (a) cosh−1 (b) coth−1 (cos x) t−1 √ 10. (a) θ sinh−1 θ (b) x cosh−1 x √ 2 sec h−1 t tanh−1 x 11. (a) (b) t2 (1 − x2 )

396 Section G d [x cosh−1 (cosh x)] = 2x dx In Problems 13 to 15, determine the given integrals. ∫ 1 dx 13. (a) √ 2 (x + 9) ∫ 3 (b) √ dx 2 (4x + 25) ∫ 1 14. (a) √ dx 2 (x − 16) 12. Show that

∫ √

(b) ∫ 15. (a)

1 (t2 − 5) dθ

2.

3.

√ (36 + θ2 )

(b)

3 dx (16 − 2x2 )

Practice Exercise 162 Multiple-choice questions on differentiation of inverse trigonometric and hyperbolic functions (Answers on page 886)

If y = cosec−1 f (x) then f ′ (x) √ 2 f (x) [f(x)] − 1 − f ′ (x) √ (c) 2 f (x) [f(x)] − 1 (a)

dt ∫

dy is equal to: dx ′ f ′ (x) f (x) √ (b) √ (a) 2 2 f (x) [f(x)] − 1 1 − [f (x)] − f ′ (x) − f ′ (x) √ (c) (d) √ 2 2 f (x) [f(x)] − 1 1 − [f (x)] If y = cos−1 f (x) then

4.

5.

dy is equal to: dx ′ f (x) f ′ (x) √ (a) (b) √ 2 2 f (x) [f(x)] − 1 1 − [f (x)] − f ′ (x) − f ′ (x) √ (c) (d) √ 2 2 f (x) [f(x)] − 1 1 − [f (x)] If y = sec−1 f(x) then

(a) √

dy is equal to: dx f ′ (x) f ′ (x) √ (a) (b) √ 2 2 f (x) [ f(x)] − 1 1 − [ f (x)] − f ′ (x) − f ′ (x) √ (c) (d) √ 2 2 f (x) [ f(x)] − 1 1 − [ f (x)]

(c) √

If y = sin−1 f(x) then

f ′ (x) 2

[f (x)] − 1 f ′ (x) 2

[f (x)] + 1 6.

dy is equal to: dx f ′ (x) (b) √ 2 1 − [f(x)] − f ′ (x) (d) √ 2 1 − [f(x)]

If y = sinh−1 f (x) then

Each question has only one correct answer 1.

dy is equal to: dx f ′ (x) (b) √ 2 1 − [f (x)] − f ′ (x) (d) √ 2 1 − [f (x)]

If y = coth−1 f (x) then − f ′ (x) √ 2 f (x) [f(x)] + 1 f ′ (x) (c) √ 2 [f (x)] + 1 (a)

dy is equal to: dx f ′ (x) (b) 2 1 − [f(x)] − f ′ (x) (d) √ 2 1 − [f (x)]

For fully worked solutions to each of the problems in Practice Exercises 159 to 161 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 32

Partial differentiation Why it is important to understand: Partial differentiation First-order partial derivatives can be used for numerous applications, from determining the volume of different shapes to analysing anything from water to heat flow. Second-order partial derivatives are used in many fields of engineering. One of its applications is in solving problems related to dynamics of rigid bodies and in determination of forces and strength of materials. Partial differentiation is used to estimate errors in calculated quantities that depend on more than one uncertain experimental measurement. Thermodynamic energy functions (enthalpy, Gibbs free energy, Helmholtz free energy) are functions of two or more variables. Most thermodynamic quantities (temperature, entropy, heat capacity) can be expressed as derivatives of these functions. Many laws of nature are best expressed as relations between the partial derivatives of one or more quantities. Partial differentiation is hence important in many branches of engineering.

At the end of this chapter, you should be able to: • determine first-order partial derivatives • determine second-order partial derivatives

32.1

Introduction to partial derivatives

In engineering, it sometimes happens that the variation of one quantity depends on changes taking place in two, or more, other quantities. For example, the volume V of a cylinder is given by V = πr 2 h. The volume will change if either radius r or height h is changed. The formula for volume may be stated mathematically as V = f (r, h) which means ‘V is some function of r and h’. Some other practical examples include: √ l (i) time of oscillation, t = 2π i.e. t = f (l, g) g (ii)

torque T = Iα, i.e. T = f (I, α)

(iii)

mRT pressure of an ideal gas p = V i.e. p = f (T, V) 1 √ 2π LC i.e. fr = f(L, C), and so on.

(iv) resonant frequency fr =

When differentiating a function having two variables, one variable is kept constant and the differential coefficient of the other variable is found with respect to that variable. The differential coefficient obtained is called a partial derivative of the function.

32.2

First-order partial derivatives

A ‘curly dee’, ∂, is used to denote a differential coefficient in an expression containing more than one variable.

398 Section G ∂V Hence if V = πr 2 h then means ‘the partial derivative ∂r of V with respect to r, with h remaining constant’. Thus, ∂V d = (πh) (r 2 ) = (πh)(2r) = 2πrh ∂r dr ∂V Similarly, means ‘the partial derivative of V with ∂h respect to h, with r remaining constant’. Thus, ∂V d = (πr 2 ) (h) = (πr 2 )(1) = πr 2 ∂h dh ∂V ∂V and are examples of first-order partial ∂r ∂h derivatives, since n = 1 when written in the form ∂nV ∂r n

Problem 2. ∂y and ∂t To find

Hence

i.e.

To find First-order partial derivatives are used when finding the total differential, rates of change and errors for functions of two or more variables (see Chapter 33), when finding maxima, minima and saddle points for functions of two variables (see Chapter 34), and with partial differential equations (see Chapter 53). Problem 1. If z = 5x4 + 2x3 y2 − 3y find ∂z ∂z (a) and (b) ∂x ∂y ∂z , y is kept constant. ∂x Since z = 5x4 + (2y2 )x3 − (3y) then,

(a) To find

(b)

∂z d d d = (5x4 ) + (2y2 ) (x3 ) − (3y) (1) ∂x dx dx dx 3 2 2 = 20x + (2y )(3x ) − 0 ∂z Hence = 20x3 + 6x2 y2 ∂x ∂z To find , x is kept constant. ∂y Since z = (5x ) + (2x )y − 3y 4

3

2

Hence

i.e.

Given y = 4 sin 3x cos 2t, find

∂y , t is kept constant. ∂x ∂y d = (4 cos 2t) (sin 3x) ∂x dx = (4 cos 2t)(3 cos 3x) ∂y = 12 cos 3x cos 2t ∂x ∂y , x is kept constant. ∂t ∂y d = (4 sin 3x) (cos 2t) ∂t dt = (4 sin 3x)(−2 sin 2t) ∂y = −8 sin 3x sin 2t ∂t

Problem 3.

If z = sin xy show that 1 ∂z 1 ∂z = y ∂x x ∂y

∂z = y cos xy, since y is kept constant. ∂x ∂z = x cos xy, since x is kept constant. ∂y ( ) 1 (y cos xy) = cos xy y ( ) 1 ∂z 1 = (x cos xy) = cos xy x ∂y x 1 ∂z = y ∂x

and Hence

1 ∂z 1 ∂z = y ∂x x ∂y

then, ∂z d d d = (5x4 ) (1) + (2x3 ) (y2 ) − 3 ( y) ∂y dy dy dy = 0 + (2x )(2y) − 3 3

Hence

∂z ∂y

= 4x3 y − 3

Problem 4.

Determine

∂z ∂z and when ∂x ∂y

1 z= √ 2 (x + y2 ) −1 1 z= √ = (x2 + y2 ) 2 (x2 + y2 )

∂y ∂x

Partial differentiation 399 ( −3 ∂z 1 = − (x2 + y2 ) 2 (2x), by the function of a ∂x 2 function rule (keeping y constant)

−x

= (x2

3 + y2 ) 2

= To find

−x =√ 3 2 (x + y2 )

−3 ∂z 1 = − (x2 + y2 ) 2 (2y), (keeping x constant) ∂y 2

−y =√ (x2 + y2 )3

Hence

Problem 5. Pressure p of a mass of gas is given by pV = mRT, where m and R are constants, V is the volume and T the temperature. Find ∂p ∂p expressions for and ∂T ∂V Since pV = mRT then p =

mRT V

∂p , V is kept constant. ∂T ( ) ∂p mR d mR Hence = (T) = ∂T V dT V

−mRT V2

)

d 1 (l 2 ) = dl

∂z ∂z and ∂x ∂y

4. z = sin(4x + 3y) y 1 5. z = x3 y2 − 2 + x y

(

2π √ g

)(

1 −1 l 2 2

6. z = cos 3x sin 4y 7. The volume of a cone of height h and base radius r is given by V = 13 πr 2 h. Determine ∂V ∂V and ∂h ∂r 8. The resonant frequency fr in a series elec1 trical circuit is given by fr = √ . Show 2π LC ∂fr −1 √ that = ∂L 4π CL3

∂t To find , g is kept constant. ∂l √ ( ) ( ) 1 l 2π √ 2π t = 2π = √ l = √ l2 g g g 2π √ g

√ −1 = (2π l)g 2 ( ) √ ∂t 1 −3 = (2π l) − g 2 ∂g 2 ( ) √ −1 √ = (2π l) 2 g3 √ √ −π l l = √ = −π g3 g3

2. z = x3 − 2xy + y2 x 3. z = y

The time of oscillation, t, of √ l a pendulum is given by t = 2π where l is the g length of the pendulum and g the free fall ∂t ∂t acceleration due to gravity. Determine and ∂l ∂g

(

π =√ lg

1. z = 2xy

Problem 6.

∂t = ∂l

)

∂t , l is kept constant. ∂g √ ( ) √ l 1 t = 2π = (2π l) √ g g

In Problems 1 to 6, find

( ) ∂p d 1 Hence = (mRT) ∂V dV V

Hence

1 √ 2 l

Practice Exercise 163 First-order partial derivatives (Answers on page 886)

∂p , T is kept constant. ∂V

= (mRT)(−V−2 ) =

)(

Now try the following Practice Exercise

To find

To find

2π √ g

)

9. An equation resulting from plucking a string is:

400 Section G ( ) ( )} ( nπ ) { nπb nπb x k cos t + c sin t y = sin L L L ∂y ∂y Determine and ∂t ∂x 10. In a thermodynamic system, k = Ae where R, k and A are constants. Find (a)

T∆S−∆H RT

∂V with respect to h, keeping r ∂r ( ) ∂ ∂V constant, gives , which is written as ∂h ∂r ∂2V . Thus, ∂h∂r ( ) ∂ ∂V ∂ ∂2V = = (2πrh) = 2πr ∂h∂r ∂h ∂r ∂h

(iv) Differentiating

,

∂k ∂A ∂(∆S) ∂(∆H) (b) (c) (d) ∂T ∂T ∂T ∂T

(v)

∂2V ∂2V ∂2V ∂2V , , and are examples of ∂r 2 ∂h2 ∂r∂h ∂h∂r second-order partial derivatives.

(vi)

32.3

Second-order partial derivatives

As with ordinary differentiation, where a differential coefficient may be differentiated again, a partial derivative may be differentiated partially again to give higherorder partial derivatives. ∂V of Section 32.2 with respect ∂r ( ) ∂ ∂V to r, keeping h constant, gives which ∂r ∂r ∂2V is written as ∂r 2

(i) Differentiating

Thus if then

V = πr 2 h,

∂2V ∂ = (2πrh) = 2πh ∂r 2 ∂r

∂V with respect to h, keeping ∂h ( ) ∂ ∂V r constant, gives which is written ∂h ∂h ∂2V as ∂h2

Second-order partial derivatives are used in the solution of partial differential equations, in waveguide theory, in such areas of thermodynamics covering entropy and the continuity theorem, and when finding maxima, minima and saddle points for functions of two variables (see Chapter 34). Problem 7. Given z = 4x2 y3 − 2x3 + 7y2 find ∂2z ∂2z ∂2z ∂2z (a) 2 (b) 2 (c) (d) ∂x ∂y ∂x∂y ∂y∂x (a)

(ii) Differentiating

Thus

∂2V ∂ = (πr 2 ) = 0 ∂h2 ∂h

∂V (iii) Differentiating with respect to r, keeping ∂h ( ) ∂ ∂V h constant, gives which is written ∂r ∂h ∂2V as . Thus, ∂r∂h ( ) ∂2V ∂ ∂V ∂ = = (πr 2 ) = 2πr ∂r∂h ∂r ∂h ∂r

∂2V ∂2V = ∂r∂h ∂h∂r and such a result is always true for continuous functions (i.e. a graph of the function which has no sudden jumps or breaks).

It is seen from (iii) and (iv) that

∂z = 8xy3 − 6x2 ∂x ( ) ∂2z ∂ ∂z ∂ = = (8xy3 − 6x2 ) ∂x2 ∂x ∂x ∂x = 8y3 − 12 x

(b)

∂z = 12x2 y2 + 14y ∂y ( ) ∂2z ∂ ∂z ∂ = = (12x2 y2 + 14y) 2 ∂y ∂y ∂y ∂y

= 24x2 y + 14 ( ) ∂2z ∂ ∂z ∂ (c) = = (12x2 y2+14y) = 24xy2 ∂x∂y ∂x ∂y ∂x ( ) 2 ∂ z ∂ ∂z ∂ (d) = = (8xy3 − 6x2 ) = 24xy2 ∂y∂x ∂y ∂x ∂y [ ] 2 ∂ z ∂2z It is noted that = ∂x∂y ∂y∂x

Partial differentiation 401 ( ) 1 (y) − (ln y)(1) y = y2 using the quotient rule

Problem 8. Show that when z = e−t sin θ, ∂2z ∂2z ∂2z ∂2z (a) 2 = − 2 , and (b) = ∂t ∂θ ∂t∂θ ∂θ∂t (a)

∂z ∂2z = −e−t sin θ and = e−t sin θ ∂t ∂t2

=

∂z ∂2z = e−t cos θ and = − e−t sin θ ∂θ ∂θ2

Hence x

∂2z ∂2z Hence = − ∂t2 ∂θ 2 ( ) ∂2z ∂ ∂z ∂ (b) = = ( e−t cos θ) ∂t∂θ ∂t ∂θ ∂t

∂2z ∂ = ∂θ∂t ∂θ

(

∂z ∂t

)

(b)

∂ (−e−t sin θ) ∂θ

= −e−t cos θ ∂2z

Hence

∂t∂θ

=

∂2z ∂θ∂t

∂z To find , x is kept constant. ∂y Hence ( ) ∂z d ln y = (x) ∂y dy y  ( )  1     (y) y − (ln y)(1)   = (x)   y2     using the quotient rule

∂2z ∂ = ∂y∂x ∂y

(

∂z ∂x

) =

= ∂ ∂y

x (1 − ln y) y2 (

y2

(1 − ln y) =

∂z

=

x [−y − 2y + 2y ln y] y4

=

xy x [−3 + 2 ln y] = 3 (2 ln y − 3) y4 y

When x = −3 and y = 1, ∂ 2 z (−3) = (2 ln 1− 3) = (−3)(−3) = 9 ∂y2 (1)3

∂z (a) To find , y is kept constant. ∂x ( ) ∂z 1 d 1 Hence = ln y (x) = ln y ∂x y dx y

)

x

using the quotient rule

x Show that if z = ln y, then y ∂z ∂2z ∂2z (a) =x and (b) evaluate 2 when ∂y ∂y∂x ∂y x = −3 and y = 1

1 − ln y =x y2

=

∂y∂x ∂y ) { } ∂ z ∂ ∂z ∂ x = = (1 − ln y) 2 ∂y ∂y ∂y ∂y y2 ( ) d 1 − ln y = (x) dy y2 ( )   1   2   (y ) − y − (1 − ln y)(2y)   = (x)   y4    

Problem 9.

(

∂2z

(

2

= −e−t cos θ =

1 (1 − ln y) y2

ln y y

)

10. Van der Waal’s law states: ( Problem ) a )( p + 2 V − b = RT where p is the pressure, V V is the volume and T the thermodynamic temperature. R, a and b are constants. The critical ∂p ∂2p point of a pure substance occurs when and ∂V ∂V2 are both zero. Determine expressions for p and V in terms of a, b, R and T where the substance has its critical point. (

( ) a )( a) RT ) V − b = RT i.e. p + =( 2 2 V V V−b RT a )− 2 and p= ( (1) V V−b ( )−1 i.e. p = RT V − b − aV−2 ( )−2 ∂p Hence, = −RT V − b + 2aV−3 (2) ∂V RT 2a = −( )2 + 3 V V−b p+

402 Section G Since

Hence, From equation (2),

Since

Hence,

∂p = 0 at its critical point, ∂V RT 2a −( )2 + 3 = 0 V V−b RT 2a (3) ( )2 = 3 V V−b ( )−3 ∂2p = 2RT V − b − 6aV−4 2 ∂V 2RT 6a =( )3 − 4 V V−b ∂2p = 0 at its critical point, ∂V2 2RT 6a ( )3 − 4 = 0 V V−b 2RT 6a (4) ( )3 = 4 V V−b

1.

z = (2x − 3y)2

2.

z = 2 ln xy

3.

z=

4.

z = sinh x cosh 2y

5.

Given z = x2 sin(x − 2y) find (a) (b)

6.

RT

2a )2 V−b 3 V = 2RT 6a ( )3 V4 V−b Dividing fractions gives: ( )3 V−b RT 2a V4 = 3× ( )2 × 2RT V 6a V−b ( ) V−b V Cancelling gives: = 2 3 ‘Cross-multiplying’ gives: 3(V − b) = 2V i.e.

3V − 3b = 2V

from which,

V = 3b RT a )− 2 p= ( V V−b RT a )− p= ( (3b)2 3b − b RT a p= − 2 2b 9b

From equation (1), If V = 3b, then i.e.

∂2z ∂y2

∂2z ∂2z = ∂x∂y ∂y∂x = 2x2 sin(x − 2y) − 4x cos (x − 2y)

7.

8.

∂2z ∂2z ∂2z ∂2z , and show that = ∂x2 ∂y2 ∂x∂y ∂y∂x −1 x when z = cos y √( ) 3x Given z = show that y ∂2z ∂2z ∂2z = and evaluate 2 when ∂x∂y ∂y∂x ∂x 1 x = and y = 3 2 Find

An equation used in thermodynamics is the Benedict–Webb–Rubine equation of state for the expansion of a gas. The equation is: ( ) RT C0 1 p= + B0 RT − A0 − 2 V T V2 1 Aα + (bRT − a) 3 + 6 V V ( γ )( ) C 1+ 2 γ 1 V + e− V2 2 3 T V Show that

∂2p ∂T2

=

6 V2 T4

Now try the following Practice Exercise Practice Exercise 164 Second-order partial derivatives (Answers on page 887) In Problems 1 to 4, find (a) (c)

∂2z ∂2z (d) ∂x∂y ∂y∂x

∂2z ∂2z (b) ∂x2 ∂y2

∂2z and ∂x2

Show also that

Dividing equation (3) by equation (4) gives: (

(x − y) (x + y)

{

C( γ) γ 1 + 2 e− V2 − C0 V V

}

Partial differentiation 403 Practice Exercise 165 Multiple-choice questions on partial differentiation (Answers on page 887)

3.

Each question has only one correct answer ∂z is 1. Given that z = 4x3 − 3x2 y3 + 6y then ∂x equal to: (a) 12x2 − 9x2 y2 (b) 12x2 − 6xy3 + 6 (c) 12x2 − 6xy3 dy dy (d) 12x2 − 9x2 y2 − 6xy3 + 6 dx dx 2y 1 ∂z 2. Given that z = 2x3 y + 2 − then is 3x x ∂x equal to: y 1 4y 1 (a) 6x2 y − 3 + 2 (b) 6x2 y − 3 + 2 3x x 3x x y 1 y 1 (c) 6x2 y + 3 + 2 (d) 6x2 y − 3 − 2 3x x 3x x

4.

5.

2y 1 ∂z Given that z = 2x3 y + 2 − then is 3x x ∂x equal to: 2 2 1 (a) 6x2 + 2 (b) 2x3 + 2 + 2 3x 3x x 2 1 2 (d) 2x3 + 2 − 2 (c) 2x3 + 2 3x 3x x ∂2z Given that z = 5x3 y2 + 3x2 − 4y3 then 2 is ∂x equal to: (a) 30xy + 6 (b) 30xy2 + 6 − 24y 2 (c) 30xy + 6 + 24y (d) 30xy2 + 6 Given that z = 5x3 y2 + 3x2 − 4y3 then equal to: (a) 30x2 y (c) 30xy2

∂2z is ∂y∂x

(b) 30xy2 + 24y (d) 30x2 y + 6

For fully worked solutions to each of the problems in Practice Exercises 163 and 164 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 33

Total differential, rates of change and small changes Why it is important to understand: Total differential, rates of change and small changes This chapter looks at some applications of partial differentiation, i.e. the total differential for variables which may all be changing at the same time, rates of change, where different quantities have different rates of change, and small changes, where approximate errors may be calculated in situations where small changes in the variables associated with the quantity occur.

At the end of this chapter, you should be able to: • determine the total differential of a function having more than one variable • determine the rates of change of functions having more than one variable • determine an approximate error in a function having more than one variable

33.1

Total differential

In Chapter 32, partial differentiation is introduced for the case where only one variable changes at a time, the other variables being kept constant. In practice, variables may all be changing at the same time. If z = f (u, v, w, . . .), then the total differential, dz, is given by the sum of the separate partial differentials of z, i.e. dz =

∂z ∂z ∂z du + dv + dw + · · · ∂u ∂v ∂w

Problem 1.

If z = f(x, y) and z = x2 y3 +

determine the total differential, dz

(1)

2x + 1, y

The total differential is the sum of the partial differentials, ∂z ∂z i.e. dz = dx + dy ∂x ∂y ∂z 2 = 2xy3 + ∂x y

(i.e. y is kept constant)

∂z 2x = 3x2 y2 − 2 (i.e. x is kept constant) ∂y y ( ) ( ) 2 2x 3 2 2 Hence dz = 2xy + dx + 3x y − 2 dy y y

Problem 2. If z = f (u, v, w) and z = 3u2 − 2v + 4w3 v2 find the total differential, dz

Total differential, rates of change and small changes 405 The total differential

dz =

Thus dT =

∂z ∂z ∂z du + dv + dw ∂u ∂v ∂w

∂z = 6u (i.e. v and w are kept constant) ∂u ∂z = −2 + 8w3 v ∂v (i.e. u and w are kept constant) ∂z = 12w2 v2 (i.e. u and v are kept constant) ∂w

Hence

V p pV dp + dV and substituting k = gives: k k T

dT = (

i.e.

dT =

V p ) dp + ( ) dV pV pV T T

T T dp + dV p V

Now try the following Practice Exercise Practice Exercise 166 Total differential (Answers on page 887) In Problems 1 to 5, find the total differential dz. 1.

z = x3 + y2

2.

z = 2xy − cos x

Problem 3. The pressure p, volume V and temperature T of a gas are related by pV = kT, where k is a constant. Determine the total differentials (a) dp and (b) dT in terms of p, V and T.

3.

z=

4.

z = x ln y

∂p ∂p dT + dV ∂T ∂V kT Since pV = kT then p = V ∂p k ∂p kT hence = and =− 2 ∂T V ∂V V k kT Thus dp = dT − 2 dV V V pV Since pV = kT, k = T ( ) ( ) pV pV T T T Hence dp = dT − dV V V2 p p i.e. dp= dT − dV T V

5.

√ x z = xy + −4 y

6.

If z = f(a, b, c) and z = 2ab − 3b2 c + abc, find the total differential, dz

7.

Given u = ln sin(xy) show that du = cot(xy)(y dx + x dy)

dz = 6u du + (8vw3 − 2) dv + (12v2 w2 ) dw

(a) Total differential dp =

(b)

Total differential dT = Since hence

∂T ∂T dp + dV ∂p ∂V

pV k ∂T V ∂T p = and = ∂p k ∂V k pV = kT, T =

33.2

x−y x+y

Rates of change

Sometimes it is necessary to solve problems in which different quantities have different rates of change. From dz equation (1), the rate of change of z, is given by: dt dz ∂z du ∂z dv ∂z dw = + + + ··· dt ∂u dt ∂v dt ∂w dt

(2)

Problem 4. If z = f (x, y) and z = 2x3 sin 2y find the rate of change of z, correct to 4 significant figures, when x is 2 units and y is π/6 radians and when x is increasing at 4 units/s and y is decreasing at 0.5 units/s.

406 Section G Using equation (2), the rate of change of z,

Hence

dz ∂z dx ∂z dy = + dt ∂x dt ∂y dt Since z = 2x3 sin 2y, then ∂z ∂z = 6x2 sin 2y and = 4x3 cos 2y ∂x ∂y dx = +4 dt dy and since y is decreasing at 0.5 units/s, = −0.5 dt dz Hence = (6x2 sin 2y)(+4) + (4x3 cos 2y)(−0.5) dt = 24x2 sin 2y − 2x3 cos 2y π When x = 2 units and y = radians, then 6 Since x is increasing at 4 units/s,

dz = 24(2)2 sin[2(π/6)] − 2(2)3 cos[2(π/6)] dt = 83.138 − 8.0

dV = dt =

(

) ( ) 2 1 2 πrh (−0.2) + πr (+0.3) 3 3

−0.4 πrh + 0.1πr 2 3

However, h = 3.2 cm and r = 1.5 cm. Hence

dV −0.4 = π(1.5)(3.2) + (0.1)π(1.5)2 dt 3 = −2.011 + 0.707 = −1.304 cm3 /s

Thus the rate of change of volume is 1.30 cm3 /s decreasing. Problem 6. The area A of a triangle is given by A = 12 ac sin B, where B is the angle between sides a and c. If a is increasing at 0.4 units/s, c is decreasing at 0.8 units/s and B is increasing at 0.2 units/s, find the rate of change of the area of the triangle, correct to 3 significant figures, when a is 3 units, c is 4 units and B is π/6 radians. Using equation (2), the rate of change of area, dA ∂A da ∂A dc ∂A dB = + + dt ∂a dt ∂c dt ∂B dt

dz Hence the rate of change of z, = 75.14 units/s, dt correct to 4 significant figures.

1 ∂A 1 A = ac sin B, = c sin B, 2 ∂a 2 ∂A 1 ∂A 1 = a sin B and = ac cos B ∂c 2 ∂B 2 da dc = 0.4 units/s, = −0.8 units/s dt dt dB and = 0.2 units/s dt ( ) ( ) dA 1 1 Hence = c sin B (0.4) + a sin B (−0.8) dt 2 2 ( ) 1 + ac cos B (0.2) 2 π When a = 3, c = 4 and B = then: 6 ( ) ( ) dA 1 π 1 π = (4) sin (0.4) + (3) sin (−0.8) dt 2 6 2 6 ( ) 1 π + (3)(4) cos (0.2) 2 6 Since

Problem 5. The height of a right circular cone is increasing at 3 mm/s and its radius is decreasing at 2 mm/s. Determine, correct to 3 significant figures, the rate at which the volume is changing (in cm3 /s) when the height is 3.2 cm and the radius is 1.5 cm. 1 Volume of a right circular cone, V = πr 2 h 3 Using equation (2), the rate of change of volume, dV ∂V dr ∂V dh = + dt ∂r dt ∂h dt ∂V 2 ∂V 1 2 = πrh and = πr ∂r 3 ∂h 3 Since the height is increasing at 3 mm/s, dh i.e. 0.3 cm/s, then = +0.3 dt and since the radius is decreasing at 2 mm/s, dr i.e. 0.2 cm/s, then = −0.2 dt

= 0.4 − 0.6 + 1.039 = 0.839 units2 /s, correct to 3 significant figures.

Total differential, rates of change and small changes 407

Problem 7. Determine the rate of increase of diagonal AC of the rectangular solid, shown in Fig. 33.1, correct to 2 significant figures, if the sides x, y and z increase at 6 mm/s, 5 mm/s and 4 mm/s when these three sides are 5 cm, 4 cm and 3 cm respectively.

[ ] db x Hence = √ (0.6) dt (x2 + y2 + z2 ) [

(0.5) + √ (x2 + y2 + z2 ) [

C

]

+ √

b B

]

y

z (x2 + y2 + z2 )

(0.4)

z 5 3 cm

When x = 5 cm, y = 4 cm and z = 3 cm, then: y5

5 x5

4 cm

cm

A

[ ] 5 db = √ (0.6) dt (52 + 42 + 32 ) [

Figure 33.1

+ √ (0.5) (52 + 42 + 32 )

√ Diagonal AB = (x2 + y2 ) √ Diagonal AC = (BC2 + AB2 ) √ √ = [z2 + { (x2 + y2 )}2 √ = (z2 + x2 + y2 ) √ Let AC = b, then b = (x2 + y2 + z2 ) Using equation (2), the rate of change of diagonal b is given by: db ∂b dx ∂b dy ∂b dz = + + dt ∂x dt ∂y dt ∂z dt Since b =

√ (x2 + y2 + z2 )

[

and

y ∂b =√ 2 ∂y (x + y2 + z2 )

dy = 5 mm/s = 0.5 cm/s, dt and

dz = 4 mm/s = 0.4 cm/s dt

]

= 0.4243 + 0.2828 + 0.1697 = 0.8768 cm/s Hence the rate of increase of diagonal AC is 0.88 cm/s or 8.8 mm/s, correct to 2 significant figures. Now try the following Practice Exercise Practice Exercise 167 Rates of change (Answers on page 887) 1.

The radius of a right cylinder is increasing at a rate of 8 mm/s and the height is decreasing at a rate of 15 mm/s. Find the rate at which the volume is changing in cm3 /s when the radius is 40 mm and the height is 150 mm.

2.

If z = f(x, y) and z = 3x2 y5 , find the rate of change of z when x is 3 units and y is 2 units when x is decreasing at 5 units/s and y is increasing at 2.5 units/s.

3.

Find the rate of change of k, correct to 4 significant figures, given the following data: k = f (a, b, c); k = 2b ln a + c2 e a ; a is increasing at 2 cm/s; b is decreasing at 3 cm/s; c is decreasing at 1 cm/s; a = 1.5 cm, b = 6 cm and c = 8 cm.

∂b z =√ 2 ∂z (x + y2 + z2 ) dx = 6 mm/s = 0.6 cm/s, dt

3

+ √ (0.4) (52 + 42 + 32 )

∂b 1 2 x −1 = (x + y2 + z2 ) 2 (2x) = √ 2 ∂x 2 (x + y2 + z2 ) Similarly,

] 4

408 Section G Hence the approximate error in k, 4. A rectangular box has sides of length x cm, y cm and z cm. Sides x and z are expanding at rates of 3 mm/s and 5 mm/s respectively and side y is contracting at a rate of 2 mm/s. Determine the rate of change of volume when x is 3 cm, y is 1.5 cm and z is 6 cm. 5. Find the rate of change of the total surface area of a right circular cone at the instant when the base radius is 5 cm and the height is 12 cm if the radius is increasing at 5 mm/s and the height is decreasing at 15 mm/s.

33.3

Small changes

It is often useful to find an approximate value for the change (or error) of a quantity caused by small changes (or errors) in the variables associated with the quantity. If z = f (u, v, w, . . .) and δu, δv, δw, . . . denote small changes in u, v, w, . . . respectively, then the corresponding approximate change δz in z is obtained from equation (1) by replacing the differentials by the small changes.

δk ≈ (V)1.4 (0.04p) + (1.4pV 0.4 )(−0.015V) ≈ pV1.4 [0.04 − 1.4(0.015)] ≈ pV1.4 [0.019] ≈

i.e. the approximate error in k is a 1.9% increase. Problem 9. Modulus of rigidity G = (R 4 θ)/L, where R is the radius, θ the angle of twist and L the length. Determine the approximate percentage error in G when R is increased by 2%, θ is reduced by 5% and L is increased by 4%

Using

δG ≈

Since

G=

and Thus δz ≈

∂z ∂z ∂z δu + δv + δw + · · · ∂u ∂v ∂w

(3)

Problem 8. Pressure p and volume V of a gas are connected by the equation pV1.4 = k. Determine the approximate percentage error in k when the pressure is increased by 4% and the volume is decreased by 1.5%

∂k ∂k δp + δV ∂p ∂V

Let p, V and k refer to the initial values. ∂k Since k = pV1.4 then = V1.4 ∂p ∂k and = 1.4pV 0.4 ∂V Since the pressure is increased by 4%, the change in 4 pressure δp = × p = 0.04p 100 Since the volume is decreased by 1.5%, the change in −1.5 volume δV = × V = −0.015V 100

∂G ∂G ∂G δR + δθ + δL ∂R ∂θ ∂L R 4 θ ∂G 4R3 θ ∂G R 4 , = , = L ∂R L ∂θ L

∂G −R 4 θ = ∂L L2

2 R = 0.02R 100 Similarly, δθ = −0.05θ and δL = 0.04L ( 3 ) ( 4) 4R θ R Hence δG ≈ (0.02R) + (−0.05θ) L L ( ) R 4θ + − 2 (0.04L) L Since R is increased by 2%, δR =

R 4θ R 4θ [0.08 − 0.05 − 0.04] ≈ −0.01 L L 1 δG ≈ − G 100 ≈

Using equation (3), the approximate error in k, δk ≈

1.9 1.4 1.9 pV ≈ k 100 100

i.e.

Hence the approximate percentage error in G is a 1% decrease. Problem 10. The second moment of area of a rectangle is given by I = (bl 3 )/3. If b and l are measured as 40 mm and 90 mm respectively and the measurement errors are −5 mm in b and +8 mm in l, find the approximate error in the calculated value of I.

Total differential, rates of change and small changes 409 [

[ √ ] √ ] l l ≈ (0.001) 2π + 0.0005 2π g g

Using equation (3), the approximate error in I, δI ≈

∂I ∂I δb + δl ∂b ∂l

≈ 0.0015t ≈

∂I l3 ∂I 3bl 2 = and = = bl 2 ∂b 3 ∂l 3

Hence the approximate error in t is a 0.15% increase.

δb = −5 mm and δl = +8 mm ( 3) l Hence δI ≈ (−5) + (bl 2 )(+8) 3 Since b = 40 mm and l = 90 mm then ( 3) 90 δI ≈ (−5) + 40(90)2 (8) 3

Now try the following Practice Exercise Practice Exercise 168 Small changes (Answers on page 887) 1.

The power P consumed in a resistor is given by P = V2 /R watts. Determine the approximate change in power when V increases by 5% and R decreases by 0.5% if the original values of V and R are 50 volts and 12.5 ohms respectively.

2.

An equation for heat generated H is H = i 2 Rt. Determine the error in the calculated value of H if the error in measuring current i is +2%, the error in measuring resistance R is −3% and the error in measuring time t is +1% 1 fr = √ represents the resonant 2π LC frequency of a series-connected circuit containing inductance L and capacitance C. Determine the approximate percentage change in fr when L is decreased by 3% and C is increased by 5%

≈ −1 215 000 + 2 592 000 ≈ 1 377 000 mm4 ≈ 137.7 cm4 Hence the approximate error in the calculated value of I is a 137.7 cm4 increase.

Problem 11.

The time √ of oscillation t of a l pendulum is given by t = 2π . Determine the g approximate percentage error in t when l has an error of 0.2% too large and g 0.1% too small.

3.

Using equation (3), the approximate change in t, ∂t ∂t δl + δg ∂l ∂g √ l ∂t π Since t = 2π , =√ g ∂l lg δt ≈

4.

The second moment of area of a rectangle about its centroid parallel to side b is given by I = bd3/12. If b and d are measured as 15 cm and 6 cm respectively and the measurement errors are +12 mm in b and −1.5 mm in d, find the error in the calculated value of I.

5.

The side b of a metal triangular template is calculated using b2 = a2 + c2 − 2ac cos B. If a, c and B are measured as 3 cm, 4 cm and π/4 radians respectively and the measurement errors which occur are +0.8 cm, −0.5 cm and +π/90 radians respectively, determine the error in the calculated value of b.

6.

Q factor in√ a resonant electrical circuit is given 1 L by: Q = . Find the percentage change in R C Q when L increases by 4%, R decreases by 3% and C decreases by 2%

√ and

∂t = −π ∂g

l (from Problem 6, Chapter 32) g3

0.2 l = 0.002 l and δg = −0.001g 100 √ π l hence δt ≈ √ (0.002l) + −π (−0.001 g) g3 lg δl =

√ ≈ 0.002π

√ l + 0.001π g

0.15 t 100

l g

410 Section G

7. The rate √ of flow of gas in a pipe is given by: C d , where C is a constant, d is the diamv= √ 6 T5 eter of the pipe and T is the thermodynamic temperature of the gas. When determining the rate of flow experimentally, d is measured and subsequently found to be in error by +1.4%, and T has an error of −1.8%. Determine the percentage error in the rate of flow based on the measured values of d and T.

3.

If z = f(x, y) and z = 4x3 y4 , the rate of change of z when x = 2 units and y = 3 units when x is increasing at 4 units/s and y is decreasing at 5 units/s, is: (a) 1728 units/s (b) 32832 units/s (c) 5616 units/s (d) 33264 units/s

4.

The second moment of area, I, of a rectangle about its centroid parallel to side b is given by bd3 . If b = 12 cm and d = 8 cm, and the I= 12 measurement errors are +9 mm in b and −1.5 mm in d, the error in the calculated value of I is: (a) 166.4 cm4 decrease (b) 9.6 cm4 increase (c) 67.2 cm4 increase (d) 9.6 cm4 decrease

5.

The power, P, consumed in a resistor is given V2 by P = . Originally, V = 60V and R = 15Ω. R If V increases by 4% and R decreases by 1%, the approximate change in power is: (a) 16.8 W increase (b) 37.2% decrease (c) 559.2 W increase (d) 21.6 W increase

Practice Exercise 169 Multiple-choice questions on total differential, rates of change and small changes (Answers on page 887) Each question has only one correct answer 1. Given that z = f (x, y) and z = 3x ln 2y, the total differential is equal to: ( ) ( ) 3 dy (a) 3 ln 2y dx + y ( ) ( ) 3x (b) 3 ln 2y dx + dy y ( ) ( ) 3x (c) 3 ln 2y dx − dy y ( ) 3x (d) + 3 ln 2y dx y 2. Given that z = f(x, y) and z = 3x2 cos 2y the rate of change of z, correct to 4 significant π figures, when x = 3 units and y = radi6 ans and x is increasing by 3 units/s and y is decreasing at 0.8 units/s, is: (a) 147.50 units/s (b) −64.41 units/s (c) 64.41 units/s (d) −64.60 units/s

For fully worked solutions to each of the problems in Practice Exercises 166 to 168 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 34

Maxima, minima and saddle points for functions of two variables Why it is important to understand: Maxima, minima and saddle points for functions of two variables The problem of finding the maximum and minimum values of functions is encountered in mechanics, physics, geometry, and in many other fields. Finding maxima, minima and saddle points for functions of two variables requires the application of partial differentiation (which was explained in the previous chapters). This is demonstrated in this chapter.

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

understand functions of two independent variables explain a saddle point determine the maxima, minima and saddle points for a function of two variables sketch a contour map for functions of two variables

34.1 Functions of two independent variables If a relation between two real variables, x and y, is such that when x is given, y is determined, then y is said to be a function of x and is denoted by y = f (x); x is called the independent variable and y the dependent variable. If y = f (u, v), then y is a function of two independent variables u and v. For example, if, say, y = f(u, v) = 3u2 − 2v then when u = 2 and v = 1, y = 3(2)2 − 2(1) = 10. This may be written as f (2, 1) = 10. Similarly, if u = 1 and v = 4, f(1, 4) = −5

Consider a function of two variables x and y defined by z = f(x, y) = 3x2 − 2y. If (x, y) = (0, 0), then f (0, 0) = 0 and if (x, y) =(2, 1), then f (2, 1)=10. Each pair of numbers, (x, y), may be represented by a point P in the (x, y) plane of a rectangular Cartesian co-ordinate system as shown in Fig. 34.1. The corresponding value of z = f (x, y) may be represented by a line PP′ drawn parallel to the z-axis. Thus, if, for example, z = 3x2 − 2y, as above, and P is the co-ordinate (2, 3) then the length of PP′ is 3(2)2 − 2(3) = 6. Fig. 34.2 shows that when a large number of (x, y) co-ordinates are taken for a

412 Section G Maximum point

z z 6 p9

b 3

0

y y

2

a p x

x

Figure 34.3

Figure 34.1

function f(x, y), and then f (x, y) calculated for each, a large number of lines such as PP′ can be constructed, and in the limit when all points in the (x, y) plane are considered, a surface is seen to result as shown in Fig. 34.2. Thus the function z = f (x, y) represents a surface and not a curve.

shows geometrically a maximum value of a function of two variables and it is seen that the surface z = f (x, y) is higher at (x, y) = (a, b) than at any point in the immediate vicinity. Fig. 34.4 shows a minimum value of a function of two variables and it is seen that the surface z = f (x, y) is lower at (x, y) = (p, q) than at any point in the immediate vicinity.

z

z

Minimum point q o

y

y

p

x x

Figure 34.4

Figure 34.2

34.2 Maxima, minima and saddle points Partial differentiation is used when determining stationary points for functions of two variables. A function f(x, y) is said to be a maximum at a point (x, y) if the value of the function there is greater than at all points in the immediate vicinity, and is a minimum if less than at all points in the immediate vicinity. Fig. 34.3

If z = f(x, y) and a maximum occurs at (a, b), the curve lying in the two planes x = a and y = b must also have a maximum point (a, b) as shown in Fig. 34.5. Consequently, the tangents (shown as t1 and t2 ) to the curves at (a, b) must be parallel to Ox and Oy respectively. This ∂z ∂z requires that = 0 and = 0 at all maximum and ∂x ∂y minimum values, and the solution of these equations gives the stationary (or critical) points of z With functions of two variables there are three types of stationary points possible, these being a maximum

Maxima, minima and saddle points for functions of two variables 413 z

(iv) determine

t1 Maximum point

(v) for each of the co-ordinates of the stationary ∂2z ∂2z points, substitute values of x and y into 2 , 2 ∂x ∂y ∂2z and and evaluate each, ∂x∂y

t2

b

O

∂2z ∂2z ∂2z , and ∂x2 ∂y2 ∂x∂y

y

( (vi) evaluate

∂2z ∂x∂y

)2 for each stationary point,

(vii) substitute the values of

a x

∂2z ∂2z ∂2z , 2 and 2 ∂x ∂y ∂x∂y

into the equation (

Figure 34.5

∆= point, a minimum point, and a saddle point. A saddle point Q is shown in Fig. 34.6 and is such that a point Q is a maximum for curve 1 and a minimum for curve 2.

∂2z ∂x∂y

)2

( −

∂2z ∂x2

)(

∂2z ∂y2

)

and evaluate, (viii) (a) (b)

Curve 2 Q

if ∆ > 0 then the stationary point is a saddle point. ∂2z if ∆ < 0 and < 0, then the stationary ∂x2 point is a maximum point,

and (c)

∂2z if ∆ < 0 and > 0, then the stationary ∂x2 point is a minimum point.

Curve 1

34.4 Worked problems on maxima, minima and saddle points for functions of two variables

Figure 34.6

34.3 Procedure to determine maxima, minima and saddle points for functions of two variables Given z = f(x, y): ∂z ∂z and ∂x ∂y

(i)

determine

(ii)

for stationary points,

(iii)

∂z ∂z = 0 and = 0, ∂x ∂y

∂z solve the simultaneous equations = 0 and ∂x ∂z = 0 for x and y, which gives the co-ordinates ∂y of the stationary points,

Problem 1. Show that the function z = (x − 1)2 + (y − 2)2 has one stationary point only and determine its nature. Sketch the surface represented by z and produce a contour map in the x–y plane. Following the above procedure: (i)

∂z ∂z = 2(x − 1) and = 2(y − 2) ∂x ∂y

(ii)

2(x − 1) = 0

(1)

2(y − 2) = 0

(2)

414 Section G (iii) From equations (1) and (2), x = 1 and y = 2, thus the only stationary point exists at (1, 2) (iv) Since

y

∂z ∂2z = 2(x − 1) = 2x − 2, 2 = 2 ∂x ∂x

and since

∂z ∂2z = 2(y − 2) = 2y − 4, 2 = 2 ∂y ∂y

∂ ∂2z = and ∂x∂y ∂x

(

2

) ∂z ∂ = (2y − 4) = 0 ∂y ∂x

z51

z 5 16

2

x

∂2z ∂2z ∂2z = = 2 and =0 ∂x2 ∂y2 ∂x∂y (

(vi)

z59

1

1

(v)

z54

)2 ∂2z =0 ∂x∂y Figure 34.8

(vii) ∆ = (0)2 − (2)(2) = −4 ∂2z (viii) Since ∆ < 0 and 2 > 0, the stationary point ∂x (1, 2) is a minimum. The surface z = (x − 1)2 + (y − 2)2 is shown in three dimensions in Fig. 34.7. Looking down towards the x–y plane from above, it is possible to produce a contour map. A contour is a line on a map which gives places having the same vertical height above a datum line (usually the mean sea-level on a geographical map). A contour map for z = (x − 1)2 + (y − 2)2 is shown in Fig. 34.8. The values of z are shown on the map and these give an indication of the rise and fall to a stationary point.

Problem 2. Find the stationary points of the surface f(x, y) = x3 − 6xy + y3 and determine their nature. Let z = f (x, y) = x3 − 6xy + y3 Following the procedure: (i)

∂z ∂z = 3x2 − 6y and = −6x + 3y2 ∂x ∂y

(ii)

for stationary points, 3x2 − 6y = 0

(1)

−6x + 3y2 = 0

(2)

and

(iii) from equation (1), 3x2 = 6y

z

and y 1

y=

3x2 1 2 = x 6 2

and substituting in equation (2) gives:

2

(

1 2 −6x + 3 x 2

o 1

x

Figure 34.7

)2 =0

3 −6x + x4 = 0 4 ( 3 ) x 3x −2 = 0 4 from which, x = 0 or

x3 −2=0 4

Maxima, minima and saddle points for functions of two variables 415 i.e. x3 = 8 and x = 2 2.

Find the maxima, minima and saddle points for the following functions: (a) f(x, y) = x2 + y2 − 2x + 4y + 8 (b) f(x, y) = x2 − y2 − 2x + 4y + 8 (c) f(x, y) = 2x + 2y − 2xy − 2x2 − y2 + 4

3.

Determine the stationary values of the function f (x, y) = x3 − 6x2 − 8y2 and distinguish between them.

4.

Locate the stationary point of the function z = 12x2 + 6xy + 15y2

5.

Find the stationary points of the surface z = x3 − xy + y3 and distinguish between them.

When x = 0, y = 0 and when x = 2, y = 2 from equations (1) and (2). Thus stationary points occur at (0, 0) and (2, 2) ( ) ∂2z ∂2z ∂2z ∂ ∂z (iv) = 6x, 2 = 6y and = ∂x2 ∂y ∂x∂y ∂x ∂y = (v)

(vi)

∂ (−6x + 3y2 ) = −6 ∂x

∂2z ∂2z = 0, =0 ∂x2 ∂y2 ∂2z and = −6 ∂x∂y ∂2z ∂2z for (2, 2), = 12, = 12 ∂x2 ∂y2 ∂2z and = −6 ∂x∂y ( 2 )2 ∂ z for (0, 0), = (−6)2 = 36 ∂x∂y ( 2 )2 ∂ z for (2, 2), = (−6)2 = 36 ∂x∂y for (0, 0)

( (vii)

∆(0, 0) =

∂2z ∂x∂y

)2

( −

∂2z ∂x2

)(

∂2z ∂y2

34.5 Further worked problems on maxima, minima and saddle points for functions of two variables Problem 3. Find the co-ordinates of the stationary points on the surface )

= 36 − (0)(0) = 36 ∆(2, 2) = 36 − (12)(12) = −108

z = (x2 + y2 )2 − 8(x2 − y2 ) and distinguish between them. Sketch the approximate contour map associated with z Following the procedure: (i)

(viii) Since ∆(0, 0) > 0 then (0, 0) is a saddle point. ∂2z Since ∆(2, 2) < 0 and > 0, then (2, 2) is a ∂x2 minimum point.

(ii)

∂z = 2(x2 + y2 )2x − 16x ∂x ∂z = 2(x2 + y2 )2y + 16y ∂y

and

for stationary points, 2(x2 + y2 )2x − 16x = 0 i.e.

Now try the following Practice Exercise Practice Exercise 170 Maxima, minima and saddle points for functions of two variables (Answers on page 887) 1. Find the stationary point of the surface f (x, y) = x2 + y2 and determine its nature. Sketch the contour map represented by z

(iii)

4x3 + 4xy2 − 16x = 0 2

(1)

2

and

2(x + y )2y + 16y = 0

i.e.

4y(x2 + y2 + 4) = 0

From equation (1), y2 =

(2)

16x − 4x3 = 4 − x2 4x

Substituting y2 = 4 − x2 in equation (2) gives 4y(x2 + 4 − x2 + 4) = 0 i.e. 32y = 0 and y = 0

416 Section G When y = 0 in equation (1),

4x3 − 16x = 0

i.e.

4x(x2 − 4) = 0

y 4

from which, x = 0 or x = ±2

i

The co-ordinates of the stationary points are (0, 0), (2, 0) and (−2, 0)

z5

128

2

z59 c

g

∂2z (iv) = 12x2 + 4y2 − 16, ∂x2

0

S f

b

z5

3 22

3 2

a

e

x

d

2

2

∂ z ∂ z = 4x2 + 12y2 + 16 and = 8xy 2 ∂y ∂x∂y (v)

h

22

For the point (0, 0), j

24

∂2z ∂2z ∂2z = −16, = 16 and =0 ∂x2 ∂y2 ∂x∂y

For the point (2, 0),

∂2z ∂2z ∂2z = 32, 2 = 32 and =0 2 ∂x ∂y ∂x∂y

For the point (−2, 0),

∂2z ∂2z ∂2z = 32, = 32 and =0 ∂x2 ∂y2 ∂x∂y (

)2

∂2z = 0 for each stationary point ∂x∂y (vii) ∆(0, 0) = (0)2 − (−16)(16) = 256 (vi)

∆(2, 0) = (0)2 − (32)(32) = −1024 ∆(−2, 0) = (0)2 − (32)(32) = −1024 (viii) Since ∆(0, 0) > 0, the point (0, 0) is a saddle point. ( 2 ) ∂ z Since ∆(0, 0) < 0 and > 0, the point ∂x2 (2, 0) (2, 0) is a minimum point. ( 2 ) ∂ z Since ∆(−2, 0) < 0 and > 0, the ∂x2 (−2, 0) point (−2, 0) is a minimum point.

Figure 34.9

Looking down towards the x–y plane from above, an approximate contour map can be constructed to represent the value of z. Such a map is shown in Fig. 34.9. To produce a contour map requires a large number of x–y co-ordinates to be chosen and the values of z at each co-ordinate calculated. Here are a few examples of points used to construct the contour map. When z = 0, 0 = (x2 + y2 )2 − 8(x2 − y2 ) In addition, when, say, y = 0 (i.e. on the x-axis) 0 = x4 − 8x2 , i.e. x2 (x2 − 8) = 0 from which,

√ x = 0 or x = ± 8

√ Hence the contour z = 0 crosses the x-axis at 0 and ± 8, i.e. at co-ordinates (0, 0), (2.83, 0) and (−2.83, 0) shown as points S, a and b respectively. When z = 0 and x = 2 then 0 = (4 + y2 )2 − 8(4 − y2 ) i.e.

0 = 16 + 8y2 + y4 − 32 + 8y2

i.e.

0 = y4 + 16y2 − 16

Let

y2 = p, then p 2 + 16p − 16 = 0 and √ −16 ± 162 − 4(1)(−16) p= 2 −16 ± 17.89 = 2 = 0.945 or −16.945

Maxima, minima and saddle points for functions of two variables 417

Hence y =

√

p=

√

(0.945) or

√ (−16.945)

= ± 0.97 or complex roots. Hence the z = 0 contour passes through the co-ordinates (2, 0.97) and (2, −0.97) shown as a c and d in Fig. 34.9. Similarly, for the z = 9 contour, when y = 0,

Let z = f(x, y) = x3 − 3x2 − 4y2 + 2 Following the procedure: (i)

∂z ∂z = 3x2 − 6x and = − 8y ∂x ∂y

(ii)

for stationary points,

9 = x4 − 8x2

i.e.

x − 8x − 9 = 0 4

(1)

−8y = 0

(2)

and (iii)

9 = (x2 + 02 )2 − 8(x2 − 02 ) i.e.

3x2 −6x = 0

From equation (1), 3x(x − 2) = 0 from which, x = 0 and x = 2 From equation (2), y = 0 Hence the stationary points are (0, 0) and (2, 0)

2

Hence (x2 − 9)(x2 + 1) = 0 from which, x = ±3 or complex roots.

(iv)

∂2z ∂2z ∂2z = 6x − 6, = −8 and =0 ∂x2 ∂y2 ∂x∂y

Thus the z = 9 contour passes through (3, 0) and (−3, 0), shown as e and f in Fig. 34.9.

(v)

For the point (0, 0),

If z = 9 and x = 0, 9 = y4 + 8y2 i.e.

y4 + 8y2 − 9 = 0

i.e.

(y2 + 9)(y2 − 1) = 0

∂2z ∂2z ∂2z = −6, = −8 and =0 ∂x2 ∂y2 ∂x∂y For the point (2, 0),

from which, y = ±1 or complex roots. ∂2z ∂2z ∂2z = 6, 2 = −8 and =0 2 ∂x ∂y ∂x∂y

Thus the z = 9 contour also passes through (0, 1) and (0, −1), shown as g and h in Fig. 34.9. When, say, x = 4 and y = 0, z = (42 )2 − 8(42 ) = 128 when z = 128 and x = 0, 128 = y4 + 8y2 i.e.

y4 + 8y2 − 128 = 0

i.e. (y2 + 16)(y2 − 8) = 0 √ from which, y = ± 8 or complex roots. Thus the z = 128 contour passes through (0, 2.83) and (0, −2.83), shown as i and j in Fig. 34.9. In a similar manner many other points may be calculated with the resulting approximate contour map shown in Fig. 34.9. It is seen that two ‘hollows’ occur at the minimum points, and a ‘cross-over’ occurs at the saddle point S, which is typical of such contour maps. Problem 4.

Show that the function

f(x, y) = x3 − 3x2 − 4y2 + 2 has one saddle point and one maximum point. Determine the maximum value.

( (vi)

∂2z ∂x∂y

)2 = (0)2 = 0 for both stationary points

(vii) ∆(0, 0) = 0 − (−6)(−8) = −48 ∆(2, 0) = 0 −(6)(−8) = 48 (

(viii)

) ∂2z < 0, the point ∂x2 (0, 0) (0, 0) is a maximum point and hence the maximum value is 0 Since ∆(0, 0) < 0 and

Since ∆(2, 0) > 0, the point (2, 0) is a saddle point. The value of z at the saddle point is 23 − 3(2)2 − 4(0)2 + 2 = −2 An approximate contour map representing the surface f (x, y) is shown in Fig. 34.10 where a ‘hollow effect’ is seen surrounding the maximum point and a ‘cross-over’ occurs at the saddle point S.

418 Section G ∂s 125 = y − 2 = 0 for a stationary point, ∂x x

y 2

hence x2 y =125 ∂s 125 = x − 2 = 0 for a stationary point, ∂y y

2 z5

MAX

S

3 z

2 22

z5

z5

4

x

21

1 52

z5

0

22

24

hence xy2 = 125 Dividing equation (3) by (4) gives:

Figure 34.10

Hence y = 5 m also From equation (1),

Problem 5. An open rectangular container is to have a volume of 62.5 m3 . Determine the least surface area of material required.

V = xyz = 62.5

(1)

S = xy + 2yz + 2xz

From equation (1), z =

62.5 = 2.5 m 25

∂ 2 S 250 ∂ 2 S 250 ∂2S = , = and =1 ∂x2 x3 ∂y2 y3 ∂x∂y When x = y = 5,

∂2S ∂2S ∂2S = 2, = 2 and =1 ∂x2 ∂y2 ∂x∂y

∆ = (1)2 − (2)(2) = −3

(2)

∂2S Since ∆ < 0 and > 0, then the surface area S is a ∂x2 minimum.

62.5 xy

Substituting in equation (2) gives: ( ) ( ) 62.5 62.5 S = xy + 2y + 2x xy xy

Hence the minimum dimensions of the container to have a volume of 62.5 m3 are 5 m by 5 m by 2.5 m.

125 125 S = xy + + x y which is a function of two variables

i.e.

(5) (5) z = 62.5 z=

from which,

Let the dimensions of the container be x, y and z as shown in Fig. 34.11.

Surface area,

(4)

x2 y x = 1, i.e. = 1, i.e. x = y 2 xy y Substituting y = x in equation (3) gives x3 = 125, from which, x = 5 m.

22

Volume

(3)

From equation (2), minimum surface area, S = (5)(5) + 2(5)(2.5) + 2(5)(2.5) = 75 m2 Now try the following Practice Exercise Practice Exercise 171 Maxima, minima and saddle points for functions of two variables (Answers on page 887) 1.

The function z = x2 + y2 + xy + 4x − 4y + 3 has one stationary value. Determine its co-ordinates and its nature.

2.

An open rectangular container is to have a volume of 32 m3 . Determine the dimensions and the total surface area such that the total surface area is a minimum.

y z

x

Figure 34.11

Maxima, minima and saddle points for functions of two variables 419

3. Determine the stationary values of the function

5.

f (x, y) = 2x3 + 2y3 − 6x − 24y + 16

f(x, y) = x4 + 4x2 y2 − 2x2 + 2y2 − 1

and determine their nature.

and distinguish between them. 4. Determine the stationary points of the surface f (x, y) = x3 − 6x2 − y2

Locate the stationary points on the surface

6.

A large marquee is to be made in the form of a rectangular box-like shape with canvas covering on the top, back and sides. Determine the minimum surface area of canvas necessary if the volume of the marquee is to be 250 m3 .

For fully worked solutions to each of the problems in Practice Exercises 170 and 171 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 10

Further differentiation

This Revision Test covers the material contained in Chapters 30 to 34. The marks for each question are shown in brackets at the end of each question. 1. Differentiate the following functions with respect to x: (b) 3 ch3 2x

(a) 5 ln (shx)

6.

2x

(c) e sech 2x

(7)

2. Differentiate the following functions with respect to the variable: x 1 (a) y = cos−1 5 2 (b) y = 3esin

−1

Show that

2 sec−1 5x x √ (d) y = 3 sinh−1 (2x2 − 1)

8.

The volume V of a liquid of viscosity coefficient η delivered after time t when passed through a tube of length L and diameter d by a pressure p is given pd4 t by V = . If the errors in V, p and L are 1%, 128ηL 2% and 3% respectively, determine the error in η. (8)

9.

Determine and distinguish between the stationary values of the function

3. Evaluate the following, each correct to 3 decimal places: (6)

4. If z = f(x, y) and z = x cos(x + y) determine 2

2

2

2

∂z ∂z ∂ z ∂ z ∂ z ∂ z , , 2, 2, and ∂x ∂y ∂x ∂y ∂x∂y ∂y∂x

(12)

5. The magnetic field vector H due to a steady current I flowing around a circular wire of radius r and at a distance x from its centre is given by ( ) I ∂ x √ H=± 2 ∂x r 2 + x2

(6)

An engineering function z = f(x, y) and y z = e 2 ln(2x + 3y). Determine the rate of increase of z, correct to 4 significant figures, when x = 2 cm, y = 3 cm, x is increasing at 5 cm/s and y is increasing at 4 cm/s. (8)

(14)

(a) sinh−1 3 (b) cosh−1 2.5 (c) tanh−1 0.8

If xyz = c, where c is constant, show that ( ) dx dy dz = −z + x y

(7)

7.

t

(c) y =

r 2I H=± √ 2 (r 2 + x2 )3

f (x, y) = x3 − 6x2 − 8y2 and sketch an approximate contour map to represent the surface f(x, y) (20) 10.

An open, rectangular fish tank is to have a volume of 13.5 m3 . Determine the least surface area of glass required. (12)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 10, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Section H

Integral calculus

Chapter 35

Standard integration Why it is important to understand: Standard integration Engineering is all about problem solving and many problems in engineering can be solved using calculus. Physicists, chemists, engineers and many other scientific and technical specialists use calculus in their everyday work; it is a technique of fundamental importance. Both integration and differentiation have numerous applications in engineering and science and some typical examples include determining areas, mean and rms values, volumes of solids of revolution, centroids, second moments of area, differential equations and Fourier series. Besides the standard integrals covered in this chapter, there are a number of other methods of integration covered in later chapters. For any further studies in engineering, differential and integral calculus are unavoidable.

At the end of this chapter, you should be able to: • understand that integration is the reverse process of differentiation • determine integrals of the form axn where n is fractional, zero, or a positive or negative integer

1 • integrate standard functions: cos ax, sin ax, sec2 ax, cosec 2 ax, cosec ax, cot ax, sec ax, tan ax, eax , x • evaluate definite integrals ∫

35.1

The process of integration

The process of integration reverses the process of differentiation. In differentiation, if f (x) = 2x2 then f ′ (x) = 4x. Thus the integral of 4x is 2x2 , i.e. integration is the process of moving from f ′ (x) to f (x). By similar reasoning, the integral of 2t is t2 Integration is a process of summation or∫adding parts together and an elongated S, shown as , is used to replace the words ∫ ∫ ‘the integral of’. Hence, from above, 4x = 2x2 and 2t is t2 dy In differentiation, the differential coefficient indidx cates that a function of x is being differentiated with respect to x, the dx indicating that it is ‘with respect to x’. In integration the variable of integration is shown by adding d (the variable) after the function to be integrated.

Thus ∫ and

4x dx means ‘the integral of 4x with respect to x’, 2t dt means ‘the integral of 2t with respect to t’.

As stated ∫above, the differential coefficient of 2x2 is 4x, hence 4x dx = 2x2 . However, the ∫ differential coefficient of 2x2 + 7 is also 4x. Hence 4x dx is also equal to 2x2 + 7. To allow for the possible presence of a constant, whenever the process of integration is performed, a constant ‘c’ is added to the result. ∫ ∫ Thus 4x dx = 2x2 + c and 2t dt = t2 + c ‘c’ is called the arbitrary constant of integration.

424 Section H 35.2 The general solution of integrals of the form axn

Table 35.1 Standard integrals ∫

∫ The general solution of integrals of the form axn dx, where a and n are constants is given by: ∫

axn dx =

(i)

(except when n = −1) ∫

axn+1 ax dx = +c n+1

(ii)

n

∫ This rule is true when n is fractional, zero, or a positive or negative integer, with the exception of n = −1

∫ (iii)

√ x dx =

∫

1 x2

1

(iii) ∫

Using this rule gives: ∫ 3 3x4+1 + c = x5 + c (i) 3x4 dx = 4+1 5 ∫ ∫ 2x−2+1 2 −2 dx = 2x dx = +c (ii) x2 −2 + 1 −2 2x−1 = +c= + c, and −1 x

(iv) ∫ (v) ∫ (vi) 3

x 2 +1 x2 dx = +c= +c 1 3 +1 2 2

2√ 3 x +c 3 Each of these three results may be checked by differentiation.

∫ (vii) ∫

=

(a)

(b)

The integral of a constant k is kx + c. For example, ∫ 8 dx = 8x + c When a sum of several terms is integrated the result is the sum of the integrals of the separate terms. For example, ∫ (3x + 2x2 − 5) dx ∫ ∫ ∫ = 3x dx + 2x2 dx − 5 dx =

35.3

3x 2 2

+

2x 3 3

− 5x + c

Standard integrals

Since integration is the reverse process of differentiation the standard integrals listed in Table 35.1 may be deduced and readily checked by differentiation.

axn+1 +c n+1

(viii) ∫ (ix)

1 cos ax dx = sin ax + c a 1 sin ax dx = − cos ax + c a 1 sec2 ax dx = tan ax + c a 1 cosec 2 ax dx = − cot ax + c a 1 cosec ax cot ax dx = − cosec ax + c a 1 sec ax tan ax dx = sec ax + c a 1 eax dx = eax + c a 1 dx = ln x + c x

Problem 1.

∫ ∫ Determine (a) 5x2 dx (b) 2t3 dt

∫ axn+1 The standard integral, axn dx = +c n+1 (a) When a = 5 and n = 2 then ∫ 5x 3 5x2+1 +c= +c 5x2 dx = 2+1 3 (b)

When a = 2 and n = 3 then ∫ 2t3 dt =

1 2t3+1 2t4 +c= + c = t4 + c 3+1 4 2

Each of these results may be checked by differentiating them.

Standard integration 425 ∫ Problem 2.

(b)

Determine ) ∫( 3 2 4 + x − 6x dx 7

(1 − t)2 dt gives:

Rearranging ∫

(1 − 2t + t2 ) dt = t −

∫ (4 + 37 x − 6x2 ) dx may be written as ∫ ∫ ∫ 4 dx + 37 x dx − 6x2 dx, i.e. each term is integrated separately. (This splitting up of terms only applies, however, for addition and subtraction.)

Hence

) ∫( 3 4 + x − 6x2 dx 7

2t2 t3 + +c 2 3

1 = t−t2 + t3 +c 3 This problem shows that functions often have to be ∫ rearranged into the standard form of axn dx before it is possible to integrate them.

( ) 1+1 3 x x2+1 = 4x + − (6) +c 7 1+1 2+1 ( ) 2 3 x x3 = 4x + − (6) + c 7 2 3 = 4x +

= t−

2t1+1 t2+1 + +c 1+1 2+1

∫ Problem 4.

Determine

3 dx x2

∫

∫ 3 dx = 3x−2 dx. Using the standard integral, 2 x ∫ axn dx when a = 3 and n = −2 gives: ∫

3 2 x − 2x 3 + c 14

Note that when an integral contains more than one term there is no need to have an arbitrary constant for each; just a single constant at the end is sufficient.

3x−2+1 3x−1 +c = +c −2 + 1 −1 −3 = −3x−1 + c = +c x

3x−2 dx =

Problem 5.

∫ √ Determine 3 x dx

For fractional powers it is necessary to appreciate √ m n am = a n

Problem 3. Determine ∫ ∫ 2x3 − 3x (a) dx (b) (1 − t)2 dt 4x

∫

√ 3 x dx =

∫

1

1

3x 2 dx =

(a) Rearranging into standard integral form gives:

3x 2 +1 +c 1 +1 2

3

∫ (

2x3 − 3x 4x ∫ ( =

=

√ 3 3x 2 = + c = 2x 2 + c = 2 x 3 + c 3 2

) dx

∫ 3

2x 3x − 4x 4x

)

∫ ( dx =

2

x 3 − 2 4

)

( ) 2+1 1 x 3 − x+c 2 2+1 4

( ) 3 1 x 3 1 3 = − x + c = x3 − x + c 2 3 4 6 4

Problem 6.

Determine

dx ∫

−5 √ dt = 4 9 t3

∫

−5 √ dt 4 9 t3

) ∫ ( 5 −3 − t 4 dt 3 dt = 9 9t 4 −5

( = −

5 9

)

3 +1 t 4 +c 3 − +1 4 −

426 Section H ( ) 1 ( )( ) 5 t4 5 4 1 = − +c = − t4 + c 1 9 4 9 1 20 √ 4 =− t+c 9 ∫ Problem 7. ∫

Determine

(1 + θ)2 √ dθ = θ

(a) From Table 35.1(iv), ∫

2

(1 + θ) √ dθ θ

7 sec2 4t dt = (7)

∫

(1 + 2θ + θ2 ) √ dθ θ ) ∫ ( 1 2θ θ2 = dθ 1 + 1 + 1 θ2 θ2 θ2 ∫( (1 ) ( 1 )) −1 = θ 2 + 2θ1− 2 + θ2− 2 dθ ∫ (

=

θ (

−1

−1 2

1 2

+ 2θ + θ

)

3 2

1

θ2 1 2

3

+

2θ 2 3 2

(b)

∫ 3

(

) 1 cosec 2θ dθ = (3) − cot 2θ + c 2 2

3 = − cot 2θ + c 2 (3)

Problem 10. Determine ∫ ∫ 2 3x (a) 5 e dx (b) dt 3 e4t

5

+

θ2 5 2

+c (a) From Table 35.1(viii), ∫

1 4 3 2 5 = 2θ 2 + θ 2 + θ 2 + c 3 5 √ √ 4 2√ 5 = 2 θ+ θ3 + θ +c 3 5

Problem 8. Determine ∫ ∫ (a) 4 cos 3x dx (b) 5 sin 2θ dθ (a) From Table 35.1(ii), ( ) ∫ 1 4 cos 3x dx = (4) sin 3x + c 3 4 = sin 3x + c 3 (b)

From Table 35.1(v),

dθ

(1)

+1

( ) 1 tan 4t + c 4

7 = tan 4t + c 4

)

θ 2 2θ 2 +1 θ 2 +1 = 1 + 1 + 3 +c −2 + 1 2 +1 2 +1 =

Problem 9. Determine ∫ ∫ (a) 7 sec2 4t dt (b) 3 cosec 2 2θ dθ

From Table 35.1(iii), ( ) ∫ 1 5 sin 2θ dθ = (5) − cos 2θ + c 2 5 = − cos 2θ + c 2

( ) 1 3x 5 5 e dx = (5) e + c = e3x + c 3 3 3x

∫ (b)

2 dt = 3 e4t

∫

( )( ) 2 −4t 2 1 −4t e dt = − e +c 3 3 4

1 1 = − e−4t + c = − 4t + c 6 6e Problem 11. Determine ) ∫ ∫( 2 3 2m + 1 (a) dx (b) dm 5x m ∫ (a)

(b)

3 dx = 5x

∫ ( )( ) 3 1 3 dx = ln x + c 5 x 5

(from Table 35.1(ix)) ) ) ∫( 2 ∫( 2 2m + 1 1 2m + dm = dm m m m =

) ∫ ( 1 2m + dm m

Standard integration 427

=

2m 2 + ln m + c 2

35.4

= m 2 + ln m + c

Now try the following Practice Exercise Practice Exercise 172 Standard integrals (Answers on page 888) In Problems 1 to 12, determine the indefinite integrals. ∫ ∫ 1. (a) 4 dx (b) 7x dx

Definite integrals

Integrals containing an arbitrary constant c in their results are called indefinite integrals since their precise value cannot be determined without further information. Definite integrals are those in which limits are applied. If an expression is written as [x]ba , ‘b’ is called the upper limit and ‘a’ the lower limit. The operation of applying the limits is defined as [x]ba = (b) − (a) The increase in the value of the integral x2 as x increases ∫3 from 1 to 3 is written as 1 x2 dx. Applying the limits gives: ∫

[

3

x2 dx = 1

∫ 2. (a)

2 2 x dx 5

∫( 3. (a)

∫ (b)

) 3x2 − 5x dx x

5 3 x dx 6 ∫ (b)

(2 + θ)2 dθ

∫

∫ 4 3 dx (b) dx 3x2 4x4 ∫√ ∫ √ 14 5 5. (a) 2 x3 dx (b) x dx 4 ∫ ∫ −5 3 √ 6. (a) √ dt (b) dx 5 4 3 t 7 x ∫ ∫ 7. (a) 3 cos 2x dx (b) 7 sin 3θ dθ 4. (a)

∫

3 8. (a) sec2 3x dx (b) 4 ∫ 9. (a) 5 cot 2t cosec 2t dt ∫ (b) ∫

4 sec 4t tan 4t dt 3

2 cosec 4θ dθ

(

) ( 3 ) 33 1 +c − +c 3 3 1 ( ) 1 2 = (9 + c) − + c =8 3 3 =

Problem 12. Evaluate ∫2 ∫3 (a) 1 3x dx (b) −2 (4 − x2 ) dx ∫

2

(a) 1

[

3x2 3x dx = 2

]2

{ =

1

3 2 (2) 2

}

{ −

3 2 (1) 2

}

1 1 =6 − 1 =4 2 2 [ ]3 x3 (4 − x2 ) dx = 4x − 3 −2 −2

∫ 2

]3

Note that the ‘c’ term always cancels out when limits are applied and it need not be shown with definite integrals.

∫

∫ 3 2x 2 dx e dx (b) 4 3 e5x ) ∫ ∫( 2 2 u −1 11. (a) dx (b) du 3x u )2 ∫ ∫( (2+3x)2 1 √ 12. (a) dx (b) + 2t dt t x 10. (a)

x3 +c 3

(b)

3

{ } { } (3)3 (−2)3 = 4(3) − − 4(−2) − 3 3 { } −8 = {12 − 9} − −8 − 3 { } 1 1 = {3} − −5 =8 3 3 ∫ 4( Problem 13. Evaluate positive square roots only.

1

) θ+2 √ dθ, taking θ

428 Section H ∫ 4( 1

) ) ∫ 4( θ+2 θ 2 √ dθ = dθ 1 + 1 θ θ2 θ2 1 =

Note that limits of trigonometric functions are always expressed in radians – thus, for example, sin 6 means the sine of 6 radians = −0.279415 . . . ∫ 2 Hence 4 cos 3t dt

) ∫ 4( 1 −1 θ 2 + 2θ 2 dθ 1

1



( ) 1 +1 θ  2

=

 =

3 θ2 3 2

+

1 2θ 2 1 2

1 +1 2

(

+

2θ

4

) −1 2 +1 

 1 − +1 2 1

4

[ √ ] √ 4  = 2 θ3 + 4 θ 3 1

{

=

} { } 4 4 (−0.279415 . . .) − (0.141120 . . .) 3 3

= (−0.37255) − (0.18816) = −0.5607 Problem 16. Evaluate ∫ 2 ∫ (a) 4 e2x dx (b) 1

1

{ √ } { √ } √ √ 2 2 (4)3 + 4 4 − (1)3 + 4 (1) 3 3 { } { } 16 2 = +8 − +4 3 3

∫

2

[

4 2x 4 e dx = e 2

(a) 1

∫ (b)

4

[ ]4 3 3 3 du = ln u = [ln 4 − ln 1] 4u 4 4 1 3 = [1.3863 − 0] = 1.040 4

3 sin 2x dx 0

∫

= 2[ e2x ]21 = 2[ e4 − e2 ] 1

= 2[54.5982 − 7.3891] = 94.42

π 2

Problem 14. Evaluate

]2

2x

1

∫

3 du, 4u

each correct to 4 significant figures.

=

1 2 2 = 5 +8− −4 = 8 3 3 3

1

4

π 2

3 sin 2x dx

Now try the following exercise

0

[ ( ) ] π2 [ ] π2 1 3 = (3) − cos 2x = − cos 2x 2 2 0 0 { } { } (π) 3 3 = − cos 2 − − cos 2(0) 2 2 2 { } { } 3 3 = − cos π − − cos 0 2 2 { } { } 3 3 3 3 = − (−1) − − (1) = + = 3 2 2 2 2 ∫

In problems 1 to 8, evaluate the definite integrals (where necessary, correct to 4 significant figures). ∫ 4 ∫ 1 3 2 1. (a) 5x dx (b) − t2 dt 4 1 −1 ∫ 2 ∫ 3 2. (a) (3 − x2 ) dx (b) (x2 − 4x + 3) dx −1

∫

4 cos 3t dt

∫

1

∫ 1

2

[

π

( ) ]2 [ ]2 1 4 4 cos 3t dt = (4) sin 3t = sin 3t 3 3 1 1 { } { } 4 4 = sin 6 − sin 3 3 3

4. (a) ∫

0 π 3 π 6

1

3 cos θ dθ 2

3. (a)

2

Problem 15. Evaluate

Practice Exercise 173 Definite integrals (Answers on page 888)

∫ (b) ∫

2 sin 2θ dθ

2

(b)

3 sin t dt 0

∫ 5 cos 3x dx (b)

0

4 cos θ dθ

0

1

5. (a)

π 2

0

π 6

3 sec2 2x dx

Standard integration 429 ∫

2

Practice Exercise 174 Multiple-choice questions on standard integration (Answers on page 888)

cosec 2 4t dt

6. (a) 1

∫ (b) ∫

π 2

π 4

(3 sin 2x − 2 cos 3x) dx ∫

1 3t

7. (a)

3 e dt (b) ∫

0 3

8. (a) 2

2 dx 3x

∫

2

2 dx 2x −1 3 e 3

(b) 1

2

2x + 1 dx x

9. The entropy change ∆S, for an ideal gas is given by: ∫ T2 ∫ V2 dT dV ∆S = Cv −R Joules/Kelvin T T1 V1 V where T is the thermodynamic temperature, V is the volume and R = 8.314. Determine the entropy change when a gas expands from 1 litre to 3 litres for a temperature rise from 100 K to 400 K given that: Cv = 45 + 6 × 10−3 T + 8 × 10−6 T2 10. The p.d. between boundaries a and b of an ∫ b Q electric field is given by: V = dr 2πrε 0 εr a If a = 10, b = 20, Q = 2 × 10−6 coulombs, ε0 = 8.85 × 10−12 and εr = 2.77, show that V = 9 kV. 11. The average value of a complex voltage waveform is given by: ∫ 1 π VAV = (10 sin ωt + 3 sin 3ωt π 0 + 2 sin 5ωt) d(ωt) Evaluate VAV correct to 2 decimal places. 12. The ∫volume of liquid in a tank is given by: t2 v= q dt. Determine the volume of a t1

chemical, given q = (5 − 0.05t + 0.003t2 )m3 /s, t1 = 0 and t2 = 16s.

Each question has only one correct answer ∫ 1. (5 − 3t2 ) dt is equal to: (a) 5 − t3 + c (b) −3t3 + c (c) −6t + c (d) 5t − t3 + c ∫ 2 ( 2 ) 2. Evaluating 3x + 4x3 dx gives: −1

(a) 54 (b) 22 (c) 24 (d) 26 ) ∫ ( 5x − 1 3. dx is equal to: x (a) 5x − ln x + c

(b)

5x2 − x x2 2

5x2 1 1 (c) + 2 + c (d) 5x + 2 + c 2 x x ∫ 23 4. t dt is equal to: 9 t4 2 (a) + c (b) t2 + c 18 3 24 2 (c) t + c (d) t3 + c 9 9 ∫ 5. (5 sin 3t − 3 cos 5t) dt is equal to: (a) −5 cos 3t + 3 sin 5t + c (b) 15(cos 3t + sin 3t) + c 5 3 (c) − cos 3t − sin 5t + c 3 5 3 5 (d) cos 3t − sin 5t + c 5 3 ∫ √ 6. ( x − 3) dx is equal to: 3√ 3 2√ 3 x − 3x + c (b) x +c 2 3 1 2√ 3 (c) √ + c (d) x − 3x + c 3 2 x ∫ π/3 7. Evaluating 3 sin 3x dx gives: (a)

0

(a) 1.503

(b) 2

(c) −18

(d) 6

430 Section H ) ∫ ( 4 8. 1 + 2x dx is equal to: e 8 2 (a) 2x + c (b) x − 2x + c e e 4 8 (c) x + 2x + c (d) x − 2x + c e e

∫ 9.

2

2e 3t dt, correct to 4 significant

Evaluating 1

10.

figures, gives: (a) 2300 (b) 255.6 (c) 766.7 (d) 282.3 ∫ 3π ) 2 ( 2 sin θ − 3 cos θ dθ is equal to: π 2

(a) 0

(b) 1

(c) 2

(d) 6

For fully worked solutions to each of the problems in Practice Exercises 172 and 173 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 36

Some applications of integration Why it is important to understand: Some applications of integration Engineering is all about problem solving and many problems in engineering can be solved using integral calculus. One important application is to find the area bounded by a curve; often such an area can have a physical significance like the work done by a motor or the distance travelled by a vehicle. Other examples can involve position, velocity, force, charge density, resistivity and current density. Electrical currents and voltages often vary with time and engineers may wish to know the average or mean value of such a current or voltage over some particular time interval. An associated quantity is the root mean square (rms) value of a current which is used, for example, in the calculation of the power dissipated by a resistor. Mean and rms values are required with alternating currents and voltages, pressure of sound waves and much more. Revolving a plane figure about an axis generates a volume, called a solid of revolution, and integration may be used to calculate such a volume. There are many applications in engineering, and particularly in manufacturing. Centroids of basic shapes can be intuitive − such as the centre of a circle; centroids of more complex shapes can be found using integral calculus − as long as the area, volume or line of an object can be described by a mathematical equation. Centroids are of considerable importance in manufacturing, and in mechanical, civil and structural design engineering. The second moment of area is a property of a cross-section that can be used to predict the resistance of a beam to bending and deflection around an axis that lies in the cross-sectional plane. The stress in, and deflection of, a beam under load depends not only on the load but also on the geometry of the beam’s cross-section; larger values of second moment cause smaller values of stress and deflection. This is why beams with larger second moments of area, such as I-beams, are used in building construction in preference to other beams with the same crosssectional area. The second moment of area has applications in many scientific disciplines including fluid mechanics, engineering mechanics and biomechanics − for example to study the structural properties of bone during bending. The static roll stability of a ship depends on the second moment of area of the waterline section − short fat ships are stable, long thin ones are not. It is clear that calculations involving areas, mean and rms values, volumes, centroids and second moment of area are very important in many areas of engineering.

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

calculate areas under and between curves determine the mean and rms value of a function over a given range determine the volume of a solid of revolution between given limits determine the centroid of an area between a curve and given axes define and use the theorem of Pappus to determine the centroid of an area determine the second moment of area and radius of gyration of regular sections and composite areas

432 Section H y

36.1

Introduction

10

There are a number of applications of integral calculus in engineering. The determination of areas, mean and rms values, volumes, centroids and second moments of area and radius of gyration are included in this chapter.

36.2

22

y 5 x 3 2 2x 2 2 8x

0

21

2

1

3

4

x

210

Areas under and between curves 220

In Fig. 36.1, ∫ total shaded area =

b

∫

c

f (x)dx −

a

b

f (x)dx ∫ d + f(x)dx c

Figure 36.2

Problem 2. Determine the area enclosed between the curves y = x2 + 1 and y = 7 − x.

y

At the points of intersection the curves are equal. Thus, equating the y values of each curve gives:

y 5 f (x) G

x2 + 1 = 7 − x

E 0

a

b

c

F

x

d

x2 + x − 6 = 0

from which,

Factorising gives (x − 2)(x + 3) = 0 Figure 36.1

Problem 1. Determine the area between the curve y = x3 − 2x2 − 8x and the x-axis. y = x3 −2x2 − 8x = x(x2 −2x − 8) = x(x + 2)(x − 4) When y = 0, x = 0 or (x + 2) = 0 or (x − 4) = 0, i.e. when y = 0, x = 0 or −2 or 4, which means that the curve crosses the x-axis at 0, −2, and 4. Since the curve is a continuous function, only one other coordinate value needs to be calculated before a sketch of the curve can be produced. When x = 1, y = −9, showing that the part of the curve between x = 0 and x = 4 is negative. A sketch of y = x3 − 2x2 − 8x is shown in Fig. 36.2. (Another method of sketching Fig. 36.2 would have been to draw up a table of values.)

from which x = 2 and x = −3 By firstly determining the points of intersection the range of x-values has been found. Tables of values are produced as shown below. x

−3 −2 −1

0 1 2

y = x2 + 1

10

1 2

−2

] 2 0

0

[

x4 2x3 8x x4 2x3 8x = − − − − − 4 3 2 −2 4 3 2 ( ) ( ) 2 2 1 = 6 − −42 = 49 square units 3 3 3

2

x

−3

0 2

y = 7−x

10

7

5

y 10

y 5 x 2 11

5

] 2 4 0

5

A sketch of the two curves is shown in Fig. 36.3.

Shaded area ∫ 0 ∫ 4 3 2 (x − 2x − 8x)dx − (x3 − 2x2 − 8x)dx = [

5

y572x 23

Figure 36.3

22

21

0

1

2

x

Some applications of integration 433 ∫ Shaded area = ∫

2 −3 2

= ∫

−3 2

= −3

∫ (7 − x)dx −

) ( ) ( 1 1 + 6−3 = 4 square units = 1 3 3

2 2

−3

(x + 1)dx

[(7 − x) − (x2 + 1)]dx

Now try the following Practice Exercise

(6 − x − x2 )dx

Practice Exercise 175 Areas under and between curves (Answers on page 888)

]2 x2 x3 = 6x − − 2 3 −3 ) ( ) ( 9 8 − −18 − + 9 = 12 − 2 − 3 2 ( ) ( ) 1 1 = 7 − −13 3 2 [

5 = 20 square units 6 Problem 3. Determine by integration the area bounded by the three straight lines y = 4 − x, y = 3x and 3y = x Each of the straight lines are shown sketched in Fig. 36.4. y y542x

y 5 3x

Find the area enclosed by the curve y = 4 cos 3x, the x-axis and ordinates x = 0 π and x = 6

2.

Sketch the curves y = x2 + 3 and y = 7 − 3x and determine the area enclosed by them.

3.

Determine the area enclosed by the three straight lines y = 3x, 2y = x and y + 2x = 5

36.3

Mean and rms values

With reference to Fig. 36.5, ∫ b 1 mean value, y = y dx b−a a v{ } u ∫ b u 1 2 and rms value = t y dx b−a a y

4

2

0

1.

y 5 f(x)

3y 5 x (or y 5 x3 )

1

2

3

4

y

x

Figure 36.4 0

Shaded area ∫ 1( ∫ 3[ x) x] = 3x − dx + (4 − x) − dx 3 3 0 1 ]1 [ ]3 3x2 x2 x2 x2 − + 4x − − 2 6 0 2 6 1 [( ) ] [( ) 3 1 9 9 = − − (0) + 12 − − 2 6 2 6 ( )] 1 1 − 4− − 2 6 [

=

x5a

x5b

x

Figure 36.5

Problem 4. A sinusoidal voltage v = 100 sin ωt volts. Use integration to determine over half a cycle (a) the mean value, and (b) the rms value. (a) Half a cycle means the limits are 0 to π radians. ∫ π 1 Mean value, y = v d(ωt) π−0 0 ∫ 1 π = 100 sin ωt d(ωt) π 0

434 Section H √{

100 [− cos ωt]π0 π 100 [(− cos π) − (− cos 0)] = π 100 200 = [(+1) − (−1)] = π π = 63.66 volts =

= √{ =

(b)

=

1 π−0

∫

1 rms value = √ × 100 = 70.71 V] 2

2 × 100 = 63.66 V] π

Now try the following Practice Exercise

}

π

v2 d(ωt)

Practice Exercise 176 Mean and rms values (Answers on page 888)

0

√{ ∫ } 1 π (100 sin ωt)2 d(ωt) = π 0 √{ } ∫ 10 000 π 2 = sin ωt d(ωt) π 0 which is not a ‘standard’ integral.

1.

The vertical height h km of a missile varies with the horizontal distance d km, and is given by h = 4d − d2 . Determine the mean height of the missile from d = 0 to d = 4 km.

2.

The distances of points y from the mean value of a frequency distribution are related 1 to the variate x by the equation y = x + . x Determine the standard deviation (i.e. the rms value), correct to 4 significant figures for values of x from 1 to 2

3.

A current i = 25 sin 100πt mA flows in an electrical circuit. Determine, using integral calculus, its mean and rms values each correct to 2 decimal places over the range t = 0 to t = 10 ms.

4.

A wave is defined by the equation:

It is shown in Chapter 15 that cos 2A = 1 − 2 sin2 A and this formula is used whenever sin2 A needs to be integrated. Rearranging cos 2A = 1 − 2 sin2 A gives 1 sin2 A = (1 − cos 2A) 2 √{ Hence √{ = √{ =

10 000 π

10 000 π

∫

∫

}

π 2

sin ωt d(ωt) 0 π

0

[

100 = √ = 70.71 volts 2

In this case,

rms value √{

}

1 rms value = √ × maximum value. 2

2 × maximum value π

In this case, mean value =

10 000 2

[Note that for a sine wave,

[Note that for a sine wave, mean value =

} 10 000 1 [π] π 2

} 1 (1 − cos 2ωt) d(ωt) 2

10 000 1 sin 2ωt ωt − π 2 2

v = E1 sin ωt + E3 sin 3ωt where E1 , E3 and ω are constants. Determine the rms value of v over the π interval 0 ≤ t ≤ ω

]π } 0

v [( ) u 10 000 1 sin 2π u u π −  u π 2 2( )] =u sin 0 t  − 0−  2

      

5.

A current waveform, i, is given by: i = (0.1 sin 20πt +0.02 sin 30πt) amperes, where t is the time in seconds. Calculate its mean value between the limits of t = 0 to t = 200 ms.

Some applications of integration 435 [

36.4

Volumes of solids of revolution

=π

With reference to Fig. 36.6, the volume of revolution V obtained by rotating area A through one revolution about the x-axis is given by: ∫ b V= πy2 dx

]4 x5 8x3 + + 16x 5 3 1

= π[(204.8 + 170.67 + 64) − (0.2 + 2.67 + 16)] = 420.6π cubic units

a

y

Problem 6. Determine the area enclosed by the two curves y = x2 and y2 = 8x. If this area is rotated 360◦ about the x-axis determine the volume of the solid of revolution produced.

y 5 f (x)

A

0

x5b

x5a

At the points of intersection the co-ordinates of the curves are equal. Since y = x2 then y2 = x4 . Hence equating the y2 values at the points of intersection:

x

Figure 36.6

If a curve x = f( y) is rotated 360◦ about the y-axis between the limits y = c and y = d then the volume generated, V, is given by: ∫ d V= πx2 dy c

Problem 5. The curve y = x2 + 4 is rotated one revolution about the x-axis between the limits x = 1 and x = 4. Determine the volume of solid of revolution produced.

x4 = 8x x(x3 − 8) = 0

and

Hence, at the points of intersection, x = 0 and x = 2 When x = 0, y = 0 and when x = 2, y = 4. The points of intersection of the curves y = x2 and y2 = 8x are therefore at (0,0) and (2,4). √ A sketch is shown in Fig. 36.8. If y2 = 8x then y = 8x

1

y 2 5 8x (or y 5Œ(8x)

4

1

2

4 4

=

y5x2

y

Revolving the shaded area shown in Fig. 36.7 360◦ about the x-axis produces a solid of revolution given by: ∫ 4 ∫ 4 Volume = πy2 dx = π(x2 + 4)2 dx ∫

x4 − 8x = 0

from which,

2

π(x + 8x + 16) dx 1

0

y

2

1

x

30

Figure 36.8 y 5 x21 4

20 A 10 5 D 4 0

B

Shaded area ) ∫ 2 (√ ∫ 2 (√ ) ) 1 = 8x − x2 dx = 8 x 2 − x2 dx 0

C

1

0



2

3

4

5

x

2 { √ √ } (√ ) x 23 3 x 8 8 8 − − {0} = 8 3 −  = 3 3 3 2 2 0

Figure 36.7

436 Section H =

16 8 8 2 − = = 2 square units 3 3 3 3

The volume produced by revolving the shaded area about the x-axis is given by: {(volume produced by revolving y2 = 8x) − (volume produced by revolving y = x2 )} ∫

2

i.e. volume =

∫ π(8x)dx −

0

2

π(x4 )dx 0

∫

2

[

8x2 x5 − (8x − x )dx = π 2 5

]2

4

=π 0

0

) ] [( 32 − (0) = π 16 − 5 = 9.6π cubic units

Now try the following Practice Exercise

(d) Calculate the volume of the blade in terms of K, correct to 3 decimal places.

36.5

Centroids

A lamina is a thin, flat sheet having uniform thickness. The centre of gravity of a lamina is the point where it balances perfectly, i.e. the lamina’s centre of mass. When dealing with an area (i.e. a lamina of negligible thickness and mass) the term centre of area or centroid is used for the point where the centre of gravity of a lamina of that shape would lie. If x and y denote the co-ordinates of the centroid C of area A of Fig. 36.9, then: ∫ b ∫ b 1 xy dx y2 dx 2 a a and y = ∫ b x= ∫ b y dx y dx a

Practice Exercise 177 page 888)

Volumes (Answers on

1. The curve xy = 3 is revolved one revolution about the x-axis between the limits x = 2 and x = 3. Determine the volume of the solid produced. y 2. The area between 2 = 1 and y + x2 = 8 is x rotated 360◦ about the x-axis. Find the volume produced. 3. The curve y = 2x2 + 3 is rotated about (a) the x-axis between the limits x = 0 and x = 3, and (b) the y-axis, between the same limits. Determine the volume generated in each case. 4. The profile of a rotor blade is bounded by the lines x = 0.2, y = 2x, y = e −x , x = 1 and the x-axis. The blade thickness t varies linearly with x and is given by: t = (1.1 − x)K, where K is a constant. (a) Sketch the rotor blade, labelling the limits. (b) Determine, using an iterative method, the value of x, correct to 3 decimal places, where 2x = e −x (c) Calculate the cross-sectional area of the blade, correct to 3 decimal places.

a

y y 5 f(x) Area A C x y 0

x5a

x5b

x

Figure 36.9

Problem 7. Find the position of the centroid of the area bounded by the curve y = 3x2 , the x-axis and the ordinates x = 0 and x = 2 If (x, y) are co-ordinates of the centroid of the given area then: ∫ 2 ∫ 2 xy dx x(3x2 ) dx 0 x = ∫0 2 = ∫ 2 y dx 3x2 dx 0

0

]2 3x4 3x dx 4 0 = ∫0 2 = [x3 ]20 3x2 dx ∫

[

2

3

0

Some applications of integration 437 =

12 = 1.5 8 1 2

∫

2 2

y dx 0 2

y= ∫

=

1 2

∫

625 625 625 − 3 4 = = 12 125 125 125 − 2 3 6

2

(3x2 )2 dx

(

0

8

y dx

=

0

[ ]2 9 x5 9x dx 2 5 0 0 = = 8 8 ( ) 9 32 18 2 5 = = 3.6 = 8 5 1 2

∫

2

1 2

4

∫

5

x = ∫0

= ∫0

5

=

5

0

∫

5

= ∫0 5

=[ (5x − x2 ) dx

0 5

= ∫

4

5x3 3

− x4

5x2 2

−

x3 3

]5 0

]5

1 2

∫

(5x − x2 )2 dx

(5x − x2 ) dx

0 5

(25x2 − 10x3 + x4 ) dx

0

125 6

Now try the following Practice Exercise Practice Exercise 178 on page 888)

y

Centroids (Answers

In Problems 1 and 2, find the position of the centroids of the areas bounded by the given curves, the x-axis and the given ordinates.

8 y 5 5x 2 x 2

4

1.

y = 3x + 2

2.

y = 5x2

3.

Determine the position of the centroid of a sheet of metal formed by the curve y = 4x − x2 which lies above the x-axis.

C

x

0 5

0

0

6

5

(Note from Fig. 36.10 that the curve is symmetrical about x = 2.5 and thus x could have been determined ‘on sight’.)

0

[

y dx

∫

Hence the centroid of the area lies at (2.5, 2.5)

(5x − x2 ) dx

(5x2 − x3 ) dx

1 2

2

5 = 2.5 2

=

[ ]5 1 25x3 10x4 x5 − + 2 3 4 5 0 = 125 6 ( ) 1 25(125) 6250 − + 625 2 3 4 = = 2.5 125 6

x(5x − x2 ) dx

y dx

5

)

y dx

y = 5x − x2 = x(5 − x). When y = 0, x = 0 or x = 5. Hence the curve cuts the x-axis at 0 and 5 as shown in Fig. 36.10. Let the co-ordinates of the centroid be (x, y) then, by integration, xy dx

6 125

0

Problem 8. Determine the co-ordinates of the centroid of the area lying between the curve y = 5x − x2 and the x-axis.

5

∫

y= ∫

Hence the centroid lies at (1.5, 3.6)

∫

)(

625 12

2

x = 0, x = 4

x = 1, x = 4

y 0

Figure 36.10

1

2

3

4

5

x

438 Section H (a) The required area is shown shaded in Fig. 36.12. 4. Find the co-ordinates of the centroid of the area which lies between the curve y/x = x − 2 and the x-axis.

∫

36.6

=

Theorem of Pappus

0

18

‘If a plane area is rotated about an axis in its own plane but not intersecting it, the volume of the solid formed is given by the product of the area and the distance moved by the centroid of the area’.

12

With reference to Fig. 36.11, when the curve y = f (x) is rotated one revolution about the x-axis between the limits x = a and x = b, the volume V generated is given by:

2x3 3

]3

= 18 square units 0

V 2πA

y 5 2x 2

x

6

y 1

0

(b)

∫

∫

3

3

πy2 dx =

= 0

π(2x2 )2 dx 0

[

]3 x5 = 4πx dx = 4π 5 0 0 ( ) 243 = 4π = 194.4πcubic units 5 ∫

C

x

(i) When the shaded area of Fig. 36.12 is revolved 360◦ about the x-axis, the volume generated

y 5 f(x) Area A

3

2

Figure 36.12

y

3

4

y x5a

2x2 dx 0

y

A theorem of Pappus∗ states:

volume V = (A)(2πy), from which, y =

3

y dx = [

5. Sketch the curve y2 = 9x between the limits x = 0 and x = 4. Determine the position of the centroid of this area.

∫

3

Area =

x5b x

Figure 36.11

Problem 9. (a) Calculate the area bounded by the curve y = 2x2 , the x-axis and ordinates x = 0 and x = 3. (b) If this area is revolved (i) about the x-axis and (ii) about the y-axis, find the volumes of the solids produced. (c) Locate the position of the centroid using (i) integration, and (ii) the theorem of Pappus.

∗ Who

was Pappus? Pappus of Alexandria (c. 290–c. 350) was one of the last great Greek mathematicians of Antiquity. Collection, his best-known work, is a compendium of mathematics in eight volumes. It covers a wide range of topics, including geometry, recreational mathematics, doubling the cube, polygons and polyhedra. To find out more go to www.routledge.com/cw/bird

(ii) When the shaded area of Fig. 36.12 is revolved 360◦ about the y-axis, the volume generated = (volume generated by x = 3) − (volume generated by y = 2x2 ) ∫ 18 ∫ 18 ( ) y 2 = π(3) dy − π dy 2 0 0 [ ]18 ∫ 18 ( y) y2 =π 9− dy = π 9y − 2 4 0 0 = 81π cubic units (c) If the co-ordinates of the centroid of the shaded area in Fig. 36.12 are (x, y) then:

Some applications of integration 439 2.0 cm

(i) by integration, ∫

∫

3

x(2x ) dx =

3

0

0

[

3

2x3 dx = =

0

18

5.0 cm

18

y dx ∫

∫

2x 4 = 18

3

y dx 0 3

=

1 2

∫

(2x2 )2 dx 0

18

0

=

1 2

[ ]3 1 4x5 4x dx 2 5 0 0 = = 5.4 18 18 3

4

(ii) using the theorem of Pappus: Volume generated when shaded area is revolved about OY = (area)(2πx) i.e. from which,

81π = (18)(2πx) x=

81π = 2.25 36π

Volume generated when shaded area is revolved about OX = (area)(2πy) i.e. from which,

X

0

3

y dx ∫

R

X

Figure 36.13

2

y= ∫

S

] 4 3

81 = 2.25 36 1 2

Q

2

xy dx x = ∫0

P

3

194.4π = (18)(2πy) y=

194.4π = 5.4 36π

Hence the centroid of the shaded area in Fig. 36.12 is at (2.25, 5.4)

Problem 10. A metal disc has a radius of 5.0 cm and is of thickness 2.0 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine, using Pappus’ theorem, the volume and mass of metal removed and the volume and mass of the pulley if the density of the metal is 8000 kg m−3 A side view of the rim of the disc is shown in Fig. 36.13.

When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid of the semicircular area removed is at a distance of 4r from its diameter (see Bird’s Engineer3π ing Mathematics 9th edition, Chapter 54), i.e. 4(1.0) , i.e. 0.424 cm from PQ. Thus the dis3π tance of the centroid from XX is 5.0 − 0.424, i.e. 4.576 cm. The distance moved through in one revolution by the centroid is 2π(4.576) cm. πr 2 π(1.0)2 π Area of semicircle = = = cm2 2 2 2 By the theorem of Pappus, volume generated = area × distance (π ) centroid = (2π)(4.576) 2

moved

by

i.e. volume of metal removed = 45.16 cm3 Mass of metal removed = density × volume = 8000 kg m−3 ×

45.16 3 m 106

= 0.3613 kg or 361.3 g volume of pulley = volume of cylindrical disc − volume of metal removed = π(5.0)2 (2.0) − 45.16 = 111.9 cm3 Mass of pulley = density × volume = 8000 kg m−3 ×

111.9 3 m 106

= 0.8952 kg or 895.2 g

440 Section H Now try the following Practice Exercise

Second moments of areas are usually denoted by I and have units of mm4 , cm4 , and so on.

Practice Exercise 179 The theorem of Pappus (Answers on page 888)

Radius of gyration

1. A right-angled isosceles triangle having a hypotenuse of 8 cm is revolved one revolution about one of its equal sides as axis. Determine the volume of the solid generated using Pappus’ theorem.

Several areas, a1 , a2 , a3 , . . . at distances y1 , y2 , y3 , . . . from a fixed axis, may be replaced by a single area A, where A = a1 + a2 ∑ + a3 + · · · at distance k from the axis, such that Ak 2 = ay2 . k is called the radius of ∑ gyration of area A about the given axis. Since Ak 2 = ay2 = I then the radius of gyration,

2. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid of a metal template in the form of a quadrant of a circle of radius 4 cm. (The equation of a circle, centre 0, radius r is x2 + y2 = r 2 ) 3.

(a) Determine the area bounded by the curve y = 5x2 , the x-axis and the ordinates x = 0 and x = 3 (b)

If this area is revolved 360◦ about (i) the x-axis, and (ii) the y-axis, find the volumes of the solids of revolution produced in each case.

(c) Determine the co-ordinates of the centroid of the area using (i) integral calculus, and (ii) the theorem of Pappus. 4. A metal disc has a radius of 7.0 cm and is of thickness 2.5 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine the volume of metal removed using Pappus’ theorem and express this as a percentage of the original volume of the disc. Find also the mass of metal removed if the density of the metal is 7800 kg m−3

√ k=

I A

The second moment of area is a quantity much used in the theory of bending of beams, in the torsion of shafts, and in calculations involving water planes and centres of pressure. The procedure to determine the second moment of area of regular sections about a given axis is (i) to find the second moment of area of a typical element and (ii) to sum all such second moments of area by integrating between appropriate limits. For example, the second moment of area of the rectangle shown in Fig. 36.14 about axis PP is found by initially considering an elemental strip of width δx, parallel to and distance x from axis PP. Area of shaded strip = bδx P

l

b x

For more on areas, mean and rms values, volumes, centroids and second moments of area, see Bird’s Engineering Mathematics 9th edition, Chapters 51 to 55. dx

36.7 Second moments of area of regular sections The first moment of area about a fixed axis of a lamina of area A, perpendicular distance y from the centroid of the lamina is defined as Ay cubic units. The second moment of area of the same lamina as above is given by Ay2 , i.e. the perpendicular distance from the centroid of the area to the fixed axis is squared.

P

Figure 36.14

Second moment of area of the shaded strip about PP = (x2 )(b δx) The second moment of area of the whole rectangle about PP is obtained by∑ summing all such strips x=l between x = 0 and x = l, i.e. x=0 x2 bδx

Some applications of integration 441 It is a fundamental theorem of integration that limit δx→0

x=l ∑

∫

G

P

l

l 2

x2 b dx

2

x b δx =

l 2

0

x=0

Thus the second moment of area of the rectangle about PP [ 3 ]l ∫ l x bl 3 2 = b x dx = b = 3 0 3 0

C

b

x

dx

Since the total area of the rectangle, A = lb, then ( 2) Al 2 l Ipp = (lb) = 3 3 l2 2 2 Ipp = Akpp thus kpp = 3 i.e. the radius of gyration about axes PP, √ kpp =

l l2 =√ 3 3

Parallel axis theorem In Fig. 36.15, axis GG passes through the centroid C of area A. Axes DD and GG are in the same plane, are parallel to each other and distance d apart. The parallel axis theorem states: IDD = IGG + Ad2 Using the parallel axis theorem the second moment of area of a rectangle about an axis through the

G

P

Figure 36.16

centroid may be determined. In the rectangle shown in bl 3 (from above). Fig. 36.16, Ipp = 3 From the parallel axis theorem ( )2 1 Ipp = IGG + (bl) 2 bl 3 bl 3 i.e. = IGG + 3 4 3 bl 3 bl3 bl − = from which, IGG = 3 4 12 Perpendicular axis theorem In Fig. 36.17, axes OX, OY and OZ are mutually perpendicular. If OX and OY lie in the plane of area A then the perpendicular axis theorem states: IOZ = IOX + IOY

G Z d

D Y

Area A O

C Area A

X

Figure 36.17 G D

Figure 36.15

A summary of derived standard results for the second moment of area and radius of gyration of regular sections are listed in Table 36.1.

442 Section H Table 36.1 Summary of standard results of the second moments of areas of regular sections Shape

Position of axis

Second moment of area, I

Radius of gyration, k

Rectangle

(1) Coinciding with b

bl 3 3

l √ 3

(2) Coinciding with l

lb3 3

(3) Through centroid, parallel to b

bl 3 12

(4) Through centroid, parallel to l

lb3 12

b √ 3 l √ 12 b √ 12

(1) Coinciding with b

bh3 12

h √ 6

(2) Through centroid, parallel to base

bh3 36

h √ 18

(3) Through vertex, parallel to base

bh3 4

h √ 2

(1) Through centre, perpendicular to

πr 4 2

r √ 2

(2) Coinciding with diameter

πr 4 4

(3) About a tangent

5πr 4 4

r 2 √ 5 r 2

Coinciding with diameter

πr 4 8

r 2

length l, breadth b

Triangle Perpendicular height h, base b

Circle

plane (i.e. polar axis)

radius r

Semicircle radius r

Problem 11. Determine the second moment of area and the radius of gyration about axes AA, BB and CC for the rectangle shown in Fig. 36.18. l 5 12.0 cm C

A

C b 5 4.0 cm

B

B A

Figure 36.18

From Table 36.1, the second moment of area about axis AA, bl 3 (4.0)(12.0)3 IAA = = = 2304 cm4 3 3 l 12.0 Radius of gyration, kAA = √ = √ = 6.93 cm 3 3 lb3 (12.0)(4.0)3 = = 256 cm4 3 3 b 4.0 and kBB = √ = √ = 2.31 cm 3 3 The second moment of area about the centroid of a bl 3 rectangle is when the axis through the centroid 12 Similarly,

IBB =

Some applications of integration 443 is parallel with the breadth b. In this case, the axis CC is parallel with the length l. Hence and

ICC =

(12.0)(4.0)3 lb3 = = 64 cm4 12 12

Using the parallel axis theorem: IQQ = IGG + Ad2 , where IGG is the second moment of area about the centroid of the triangle, i.e.

b 4.0 kCC = √ = √ = 1.15 cm 12 12

(8.0)(12.0)3 bh3 = = 384 cm4 , 36 36

A is the area of the triangle, Problem 12. Find the second moment of area and the radius of gyration about axis PP for the rectangle shown in Fig. 36.19.

= 12 bh = 21 (8.0)(12.0) = 48 cm2 and d is the distance between axes GG and QQ,

40.0 mm G

G 15.0 mm

= 6.0 + 31 (12.0) = 10 cm Hence the second moment of area about axis QQ, IQQ = 384 + (48)(10)2 = 5184 cm4

25.0 mm P

P

Figure 36.19

lb3 where l = 40.0 mm and b = 15.0 mm 12 (40.0)(15.0)3 Hence IGG = = 11 250 mm4 12 From the parallel axis theorem, IPP = IGG + Ad2 , where A = 40.0 × 15.0 = 600 mm2 and d = 25.0 + 7.5 = 32.5 mm, the perpendicular distance between GG and PP. Hence,

Radius of gyration, √( √ ) IQQ 5184 kQQ = = = 10.4 cm A 48

IGG =

Problem 14. Determine the second moment of area and radius of gyration of the circle shown in Fig. 36.21 about axis YY.

IPP = 11 250 + (600)(32.5)2

r 5 2.0 cm

= 645 000 mm4

G

G

2 IPP = AkPP , from which, √( √ ) IPP 645 000 kPP = = = 32.79 mm A 600

3.0 cm

Problem 13. Determine the second moment of area and radius of gyration about axis QQ of the triangle BCD shown in Fig. 36.20. B

12.0 cm

G

G

C

Q

Figure 36.20

8.0 cm

Y

Y

Figure 36.21

In Fig. 36.21, IGG =

πr 4 π = (2.0)4 = 4π cm4 . 4 4

Using the parallel axis theorem, IYY = IGG + Ad2 , where d = 3.0 + 2.0 = 5.0 cm.

D 6.0 cm

Hence Q

IYY = 4π + [π(2.0)2 ](5.0)2 = 4π + 100π = 104π = 327 cm4

444 Section H Radius of gyration, √( √ ) √ IYY 104π kYY = = = 26 = 5.10 cm A π(2.0)2

Problem 16. Determine the polar second moment of area of the propeller shaft cross-section shown in Fig. 36.23.

G B

B 15.0 mm

Figure 36.23

X

X

Figure 36.22

The centroid of a semicircle lies at

4r from its diameter. 3π

Using the parallel axis theorem: IBB = IGG + Ad2

Hence

πr 4 (from Table 36.1) 8 π(10.0)4 = = 3927 mm4 , 8 πr 2 π(10.0)2 A= = = 157.1 mm2 2 2 4(10.0) 4r = = 4.244 mm d= 3π 3π 3927 = IGG + (157.1)(4.244)2

i.e.

3927 = IGG + 2830

where

and

7.0 cm

10.0 mm

G

6.0 cm

Problem 15. Determine the second moment of area and radius of gyration for the semicircle shown in Fig. 36.22 about axis XX.

IBB =

from which, IGG = 3927 − 2830 = 1097 mm4

πr 4 The polar second moment of area of a circle = 2 The polar second moment of area of the shaded area is given by the polar second moment of area of the 7.0 cm diameter circle minus the polar second moment of area of the 6.0 cm diameter circle. Hence the polar second moment of area of the crosssection shown ( )4 ( )4 π 7.0 π 6.0 = − 2 2 2 2 = 235.7 − 127.2 = 108.5 cm4 Problem 17. Determine the second moment of area and radius of gyration of a rectangular lamina of length 40 mm and width 15 mm about an axis through one corner, perpendicular to the plane of the lamina. The lamina is shown in Fig. 36.24. Y

m

0m

l54

Z

Using the parallel axis theorem again:

b 5 15 mm X

2

IXX = IGG + A(15.0 + 4.244)

i.e. IXX = 1097 + (157.1)(19.244)2 = 1097 + 58 179 = 59 276 mm4 or 59 280 mm4 correct to 4 significant figures. √( √ ) IXX 59 276 Radius of gyration, kXX = = A 157.1

X Y

Z

Figure 36.24

From the perpendicular axis theorem: IZZ = IXX + IYY

= 19.42 mm IXX =

lb3 (40)(15)3 = = 45 000 mm4 3 3

Some applications of integration 445 and

IYY =

bl 3 (15)(40)3 = = 320 000 mm4 3 3

Hence IZZ = 45 000 + 320 000

Total second moment of area about XX

= 365 000 mm4 or 36.5 cm4 Radius of gyration, √ kZZ =

By the parallel axis theorem, the second moment of area of the triangle about axis XX [ ][ ]2 = 60 + 12 (10)(6.0) 8.0 + 31 (6.0) = 3060 cm4

= 100.5 + 1024 + 3060

IZZ = A

√(

365 000 (40)(15)

)

= 4184.5 = 4180 cm4 , correct to 3 significant figures.

= 24.7 mm or 2.47 cm Problem 18. Determine correct to 3 significant figures, the second moment of area about axis XX for the composite area shown in Fig. 36.25.

Problem 19. Determine the second moment of area and the radius of gyration about axis XX for the I-section shown in Fig. 36.26. S 8.0 cm

0 4.

X 1.0 cm

CD

3.0 cm

CE

7.0 cm

cm 3.0 cm

X 1.0 cm 8.0 cm

2.0 cm

C

2.0 cm

C y CF

4.0 cm

15.0 cm

X

CT T

T 6.0 cm

X

S

Figure 36.26

Figure 36.25

For the semicircle, IXX =

πr 4 π(4.0)4 = = 100.5 cm4 8 8

For the rectangle, IXX =

bl 3 (6.0)(8.0)3 = = 1024 cm4 3 3

The I-section is divided into three rectangles, D, E and F and their centroids denoted by CD , CE and CF respectively. For rectangle D: The second moment of area about CD (an axis through CD parallel to XX) =

bl 3 (8.0)(3.0)3 = = 18 cm4 12 12

Using the parallel axis theorem: IXX = 18 + Ad2

For the triangle, about axis TT through centroid CT , where A = (8.0)(3.0) = 24 cm2 and d = 12.5 cm ITT =

bh3 (10)(6.0)3 = = 60 cm4 36 36

Hence IXX = 18 + 24(12.5)2 = 3768 cm4 .

446 Section H For rectangle E: The second moment of area about CE (an axis through CE parallel to XX) =

and (c) an axis through the centroid of the triangle parallel to axis DD. E

(3.0)(7.0)3 bl 3 = = 85.75 cm4 12 12

E

Using the parallel axis theorem:

9.0 cm

IXX = 85.75 + (7.0)(3.0)(7.5)2 = 1267 cm4 For rectangle F:

D

D

bl 3 (15.0)(4.0)3 IXX = = = 320 cm4 3 3

12.0 cm

Figure 36.28

Total second moment of area for the I-section about axis XX, IXX = 3768 + 1267 + 320 = 5355 cm4

3. For the circle shown in Fig. 36.29, find the second moment of area and radius of gyration about (a) axis FF and (b) axis HH.

Total area of I-section

H

= (8.0)(3.0) + (3.0)(7.0) + (15.0)(4.0) H

= 105 cm2

m

0c

4.

Radius of gyration, √( √ ) IXX 5355 = = 7.14 cm kXX = A 105

r5 F

F

Figure 36.29

1. Determine the second moment of area and radius of gyration for the rectangle shown in Fig. 36.27 about (a) axis AA (b) axis BB and (c) axis CC. B

C

8.0 cm

A

A 3.0 cm

m m .0 10

Practice Exercise 180 Second moments of areas of regular sections (Answers on page 889)

4. For the semicircle shown in Fig. 36.30, find the second moment of area and radius of gyration about axis JJ.

r5

Now try the following Practice Exercise

J

J

Figure 36.30

5. For each of the areas shown in Fig. 36.31 determine the second moment of area and radius of gyration about axis LL, by using the parallel axis theorem. 3.0 cm 15 cm

B

C

m

5 Dia

c 4.0

5.0 cm

Figure 36.27

2. Determine the second moment of area and radius of gyration for the triangle shown in Fig. 36.28 about (a) axis DD (b) axis EE

15 cm

2.0 cm

18 cm 10 cm

5.0 cm

L

L (a)

Figure 36.31

(b)

(c)

Some applications of integration 447

6. Calculate the radius of gyration of a rectangular door 2.0 m high by 1.5 m wide about a vertical axis through its hinge.

11.

Find the second moment of area and radius of gyration about the axis XX for the beam section shown in Fig. 36.34.

7. A circular door of a boiler is hinged so that it turns about a tangent. If its diameter is 1.0 m, determine its second moment of area and radius of gyration about the hinge.

6.0 cm 1.0 cm

8. A circular cover, centre O, has a radius of 12.0 cm. A hole of radius 4.0 cm and centre X, where OX = 6.0 cm, is cut in the cover. Determine the second moment of area and the radius of gyration of the remainder about a diameter through O perpendicular to OX.

2.0 cm 8.0 cm

2.0 cm X

9. For the sections shown in Fig. 36.32, find the second moment of area and the radius of gyration about axis XX. 18.0 mm

6.0 cm

3.0 mm

2.0 cm

12.0 mm X

2.5 cm 4.0 mm

3.0 cm

X

2.0 cm X

X

Practice Exercise 181 Multiple-choice questions on some applications of integration (Answers on page 889) Each question has only one correct answer 1.

A vehicle has a velocity v = (2 + 3t) m/s after t seconds. The distance travelled is equal to the area under the v/t graph. In the first 3 seconds the vehicle has travelled: (a) 11 m (b) 33 m (c) 13.5 m (d) 19.5 m

2.

The area, in square units, enclosed by the curve y = 2x + 3, the x-axis and ordinates x = 1 and x = 4 is: (a) 24 (b) 2 (c) 28 (d) 39

3.

The area enclosed by the curve y = 3 cos 2θ, π the ordinates θ = 0 and θ = and the θ axis 4 is: (a) −3 (b) 6 (c) 1.5 (d) 3

4.

The area under a force/distance graph gives the work done. The shaded area shown between p and q in Figure 36.35 is: ( ) c 1 1 (a) c(ln p − ln q) (b) − − 2 q2 p 2 c q (c) (ln q − ln p) (d) c ln 2 p

Figure 36.32

10. Determine the second moments of areas about the given axes for the shapes shown in Fig. 36.33. (In Fig. 36.33(b), the circular area is removed.) 3.0 cm B

4.5 cm 9.0 cm

16.0 cm .0 c 57 Dia

m

4.0 cm A

9.0 cm (a)

15.0 cm

A C B

Figure 36.33

10.0 cm (b)

C

X

Figure 36.34

(b)

(a)

10.0 cm

448 Section H

Force F

0

10.

A metal template is bounded by the curve y = x2 , the x-axis and ordinates x = 0 and x = 2. The x-co-ordinate of the centroid of the area is: (a) 1.0 (b) 2.0 (c) 1.5 (d) 2.5

11.

Using the theorem of Pappus, the position of the centroid of a semicircle of radius r lies on the axis of symmetry at a distance from the diameter of: 3r 4r 4π 3π (b) (c) (d) (a) 4r 4π 3π 3r The distance of the centroid from the y-axis of the surface area described by y = 3x between x = 0 and x = 3 is: (a) ¯x = 1 (b) ¯x = 1.5 (c) ¯x = 2 (d) ¯x = 3

c F5 s

p

q

Distance s

Figure 36.35

5. The mean value of y = 4x between x = 1 and x = 4 is: (a) 1.33 (b) 10 (c) 3.75 (d) 30 6. The mean value of y = 2x2 between x = 1 and x = 3 is: 2 1 (d) 8 (a) 2 (b) 4 (c) 4 3 3 7. The r.m.s. value of y = x2 between x = 1 and x = 3, correct to 2 decimal places, is: (a) 2.08 (b) 4.92 (c) 6.96 (d) 24.2 8. The volume of the solid of revolution when the curve y = 2x is rotated one revolution about the x-axis between the limits x = 0 and x = 4 cm is: 1 (a) 85 π cm3 (b) 8 cm3 3 1 (c) 85 cm3 (d) 64π cm3 3 9. A volume of a solid of revolution is formed when the area under the curve y = x2 is rotated once about the x-axis between x = 0 and x = 2. Leaving the answer as a multiple of π, the volume is: (a) 32π cubic units (b) 16π cubic units (c) 4π cubic units (d) 6.4π cubic units

12.

13.

14.

15.

A rectangle has length x and breadth y. The second moment of area of the rectangle about its centroid parallel to y is: yx3 xy3 yx3 xy3 (a) (b) (c) (d) 12 3 3 12 A circle has a diameter d. The second moment of area of the circle coinciding with the diameter is: πd4 πd4 5πd4 πd2 (a) (b) (c) (d) 32 64 64 8 A triangle has perpendicular height h and base b. The second moment of area of the rectangle through the centroid parallel to the base is: bh3 bh3 bh3 bh3 (a) (b) (c) (d) 12 36 24 4

For fully worked solutions to each of the problems in Practice Exercises 175 to 180 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 11

Introduction to integration and its applications

This Revision Test covers the material contained in Chapters 35 and 36. The marks for each question are shown in brackets at the end of each question. ∫ ∫ √ 2 7. A cylindrical pillar of diameter 400 mm has a 5 dx 1. Determine: (a) 3 t dt (b) √ 3 2 groove cut around its circumference as shown in x ∫ Fig. RT11.1. The section of the groove is a semi(c) (2 + θ)2 dθ (9) circle of diameter 50 mm. Given that the cen4r troid of a semicircle from its base is , use the 2. Evaluate the following integrals, each correct to 3π theorem of Pappus to determine the volume of 4 significant figures: 3 material removed, in cm , correct to 3 significant ) ∫ π ∫ 2( 3 2 1 3 figures. (a) 3 sin 2t dt (b) + + dx 2 (8) x x 4 0 1 ∫ 1 3 (c) dt (13) 2t e 0 3. 4.

5.

6.

Calculate the area between the curve y = x3 − x2 − 6x and the x-axis.

(10)

400 mm

A voltage v = 25 sin 50πt volts is applied across an electrical circuit. Determine, using integration, its mean and rms values over the range t = 0 to t = 20 ms, each correct to 4 significant figures. (9)

50 mm 200 mm

Sketch on the same axes the curves x2 = 2y and y2 = 16x and determine the co-ordinates of the points of intersection. Determine (a) the area enclosed by the curves, and (b) the volume of the solid produced if the area is rotated one revolution about the x-axis. (12) Calculate the position of the centroid of the sheet of metal formed by the x-axis and the part of the curve y = 5x − x2 which lies above the x-axis. (9)

Figure RT11.1

8.

A circular door is hinged so that it turns about a tangent. If its diameter is 1.0 m find its second moment of area and radius of gyration about the hinge. (5)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 11, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Chapter 37

Maclaurin’s series and limiting values Why it is important to understand: Maclaurin’s series One of the simplest kinds of function to deal with, in either algebra or calculus, is a polynomial (i.e. an expression of the form a + bx + cx2 + dx3 + . . .). Polynomials are easy to substitute numerical values into, and they are easy to differentiate. One useful application of Maclaurin’s series is to approximate, to a polynomial, functions which are not already in polynomial form. In the simple theory of flexure of beams, the slope, bending moment, shearing force, load and other quantities are functions of a derivative of y with respect to x. The elastic curve of a transversely loaded beam can be represented by the Maclaurin series. Substitution of the values of the derivatives gives a direct solution of beam problems. Another application of Maclaurin series is in relating inter-atomic potential functions. At this stage, not all of the above applications would have been met or understood; however, sufficient to say that Maclaurin’s series has a number of applications in engineering and science.

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

determine simple derivatives derive Maclaurin’s theorem appreciate the conditions of Maclaurin’s series use Maclaurin’s series to determine the power series for simple trigonometric, logarithmic and exponential functions • evaluate a definite integral using Maclaurin’s series • state L’Hˆopital’s rule • determine limiting values of functions

Maclaurin’s series and limiting values 451 37.1

Introduction

Some mathematical functions may be represented as power series, containing terms in ascending powers of the variable. For example, ex = 1 + x + sin x = x −

x2 x3 + + ··· 2! 3!

x3 x5 x7 + − + ··· 3! 5! 7!

x2 x4 and cosh x = 1 + + + · · · 2! 4! (as shown in Chapter 12) Using a series, called Maclaurin’s∗ series, mixed functions containing, say, algebraic, trigonometric and exponential functions, may be expressed solely as

algebraic functions, and differentiation and integration can often be more readily performed. To expand a function using Maclaurin’s theorem, some knowledge of differentiation is needed. Here is a revision of derivatives of the main functions needed in this chapter. or f ′ (x)

y or f (x)

dy dx

axn

anxn−1

sin ax

a cos ax

cos ax

−a sin ax

eax

aeax

ln ax

1 x

sinh ax

a cosh ax

cosh ax

a sinh ax

Given a general function f(x), then f ′ (x) is the first derivative, f ′′ (x) is the second derivative, and so on. Also, f (0) means the value of the function when x = 0, f ′ (0) means the value of the first derivative when x = 0, and so on.

37.2 Derivation of Maclaurin’s theorem Let the power series for f (x) be f(x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + · · ·

(1)

where a0 , a1 , a2 , . . . are constants. When x = 0, f (0) = a0 Differentiating equation (1) with respect to x gives: f ′ (x) = a1 + 2a2 x + 3a3 x2 + 4a4 x3 + 5a5 x4 + · · ·

(2)

When x = 0, f ′ (0) = a1 Differentiating equation (2) with respect to x gives: ∗

Who was Maclaurin? Colin Maclaurin (February 1698– 14 June 1746) was a Scottish mathematician who made important contributions to geometry and algebra. The Maclaurin series are named after him. To find out more go to www.routledge.com/cw/bird

f ′′ (x) = 2a2 + (3)(2)a3 x + (4)(3)a4 x2 + (5)(4)a5 x3 + · · · When x = 0, f ′′ (0) = 2a2 = 2!a2 , i.e. a2 =

f ′′ (0) 2!

(3)

452 Section H and the sum of the terms must reach a limiting value. For example, the series 1 + 12 + 14 + 18 + · · · is convergent since the value of the terms is getting smaller and the sum of the terms is approaching a limiting value of 2.

Differentiating equation (3) with respect to x gives: f ′′′ (x) = (3)(2)a3 + (4)(3)(2)a4 x + (5)(4)(3)a5 x2 + · · ·

(4)

f ′′′ (0) 3! f iv (0) Continuing the same procedure gives a4 = , 4! v f (0) a5 = , and so on. 5! Substituting for a0 , a1 , a2 , . . . in equation (1) gives: When x = 0, f ′′′ (0) = (3)(2)a3 = 3!a3 , i.e. a3 =

f(x) = f(0) + f ′ (0)x +

f ′′ (0) 2 x 2!

+ i.e.

f(x) = f(0) + x f ′ (0) + +

f (x) = cos x

f (0) = cos 0 = 1

′

2

x ′′ f (0) 2! (5)

Equation (5) is a mathematical statement called Maclaurin’s theorem or Maclaurin’s series.

37.3

Problem 1. Determine the first four (non-zero) terms of the power series for cos x The values of f(0), f ′ (0), f ′′ (0), . . . in Maclaurin’s series are obtained as follows:

f ′′′ (0) 3 x + ··· 3!

x3 ′′′ f (0) + · · · 3!

37.4 Worked problems on Maclaurin’s series

Conditions of Maclaurin’s series

f (x) = −sin x

f ′ (0) = −sin 0 = 0

f ′′ (x) = −cos x

f ′′ (0) = −cos 0 = −1

f ′′′ (x) = sin x

f ′′′ (0) = sin 0 = 0

f iv (x) = cos x

f iv (0) = cos 0 = 1

f v (x) = −sin x

f v (0) = −sin 0 = 0

f vi (x) = −cos x

f vi (0) = −cos 0 = −1

Substituting these values into equation (5) gives:

Maclaurin’s series may be used to represent any function, say f(x), as a power series provided that at x = 0 the following three conditions are met:

f(x) = cos x = 1 + x(0) + +

(a) f(0) ̸= ∞ For example, for the function f (x) = cos x, f (0) = cos 0 = 1, thus cos x meets the condition. However, if f (x) = ln x, f (0) = ln 0 = −∞, thus ln x does not meet this condition. (b)

′

′′

f (0), f (0), f

′′′

i.e.

cos x = 1 −

Problem 2.

x2 x3 (−1) + (0) 2! 3!

x4 x5 x6 (1) + (0) + (−1) + · · · 4! 5! 6!

x2 x4 x6 + − + ··· 2! 4! 6!

Determine the power series for cos 2θ

(0), . . . ̸= ∞

For example, for the function f (x) = cos x, f ′ (0) = −sin 0 = 0, f ′′ (0) = −cos 0 = −1, and so on; thus cos x meets this condition. However, if f (x) = ln x, f ′ (0) = 10 = ∞, thus ln x does not meet this condition (c) The resultant Maclaurin’s series must be convergent. In general, this means that the values of the terms, or groups of terms, must get progressively smaller

Replacing x with 2θ in the series obtained in Problem 1 gives: cos 2θ = 1 − = 1−

(2θ)2 (2θ)4 (2θ)6 + − + ··· 2! 4! 6! 4θ2 16θ4 64θ6 + − + ··· 2 24 720

2 4 i.e. cos 2θ = 1 − 2θ2 + θ4 − θ6 + · · · 3 45

Maclaurin’s series and limiting values 453

Problem 3. Using Maclaurin’s series, find the first 4 (non zero) terms for the function f (x) = sin x f (x) = sin x

f (0) = sin 0 = 0

f ′′ (x) = −sin x

f ′′ (0) = −sin 0 = 0

f ′′′ (x) = −cos x

f ′′′ (0) = −cos 0 = −1

iv

i.e.

Problem 5. Determine the power series for tan x as far as the term in x3 f(x) = tan x

iv

f (x) = sin x

f (0) = sin 0 = 0

v

v

f (x) = cos x

f (0) = cos 0 = 1

f vi (x) = −sin x

f vi (0) = −sin 0 = 0

f vii (x) = −cos x

f vii (0) = −cos 0 = −1

f(0) = tan 0 = 0 f ′ (x) = sec2 x f ′ (0) = sec2 0 =

+

x3 x5 x7 + − + ··· 3! 5! 7!

Problem 4. Using Maclaurin’s series, find the first five terms for the expansion of the function f (x) = e3x f (x) = e3x

f (0) = e0 = 1

′

′

3x

f (0) = 3 e = 3

f ′′ (x) = 9 e3x

f ′′ (0) = 9 e0 = 9

′′′

′′′

f (x) = 3 e

f ′′ (0) = 2 sec2 0 tan 0 = 0 f ′′′ (x) = (2 sec2 x)(sec2 x)

= 2 sec4 x + 4 sec2 x tan2 x f ′′′ (0) = 2 sec4 0 + 4 sec2 0 tan2 0 = 2 Substituting these values into equation (5) gives: f (x) = tan x = 0 + (x)(1) +

Problem 6.

f (0) = 27 e = 27

f iv (x) = 81 e3x

f iv (0) = 81 e0 = 81

x2 x3 (0) + (2) 2! 3!

Expand ln(1 + x) to five terms.

0

Substituting the above values into Maclaurin’s series of equation (5) gives: ( ) x2 ( ) x3 ( ) e3x = 1 + x 3 + 9 + 27 2! 3!

f (x) = ln(1 + x) f ′ (x) = f ′′ (x) = f ′′′ (x) =

x4 ( ) 81 + · · · 4!

f iv (x) =

9x2 27x3 81x4 + + + ··· 2! 3! 4!

f v (x) =

+ e3x = 1 + 3x +

+ (tan x)(4 sec x sec x tan x), by the product rule,

1 i.e. tan x = x + x3 3

0

3x

f (x) = 27 e

= 2 sec2 x tan x

x5 ( ) x6 ( ) x7 ( ) 1 + 0 + −1 + · · · 5! 6! 7!

i.e. sin x = x −

1 =1 cos2 0

f ′′ (x) = (2 sec x)(sec x tan x)

Substituting the above values into Maclaurin’s series of equation (5) gives: ( ) x2 ( ) x3 ( ) x4 ( ) sin x = 0 + x 1 + 0 + −1 + 0 2! 3! 4!

9x2 9x3 27x4 + + + ··· 2 2 8

e3x = 1 + 3x +

1 (1 + x) −1 (1 + x)2 2 (1 + x)3 −6 (1 + x)4 24 (1 + x)5

f(0) = ln(1 + 0) = 0 f ′ (0) = f ′′ (0) = f ′′′ (0) = f iv (0) = f v (0) =

1 =1 1+0 −1 = −1 (1 + 0)2 2 =2 (1 + 0)3 −6 = −6 (1 + 0)4 24 = 24 (1 + 0)5

454 Section H Substituting these values into equation (5) gives: x2 f (x) = ln(1 + x) = 0 + x(1) + (−1) 2! 3 x x4 x5 + (2) + (−6) + (24) 3! 4! 5!

f ′′ (x) = 12(2 + x)2

f ′′ (0) = 12(2)2 = 48

f ′′′ (x) = 24(2 + x)1

f ′′′ (0) = 24(2) = 48

f iv (x) = 24

f iv (0) = 24

Substituting in equation (5) gives: (2 + x)4

i.e. ln(1 + x) = x −

2

3

4

5

x x x x + − + −··· 2 3 4 5

Expand ln(1 − x) to five terms.

Problem 7.

Replacing x by −x in the series for ln(1 + x) in Problem 6 gives: (−x)2 (−x)3 ln(1 − x) = (−x) − + 2 3 (−x)4 (−x)5 − + − ··· 4 5 i.e. ln(1 − x) = −x − Problem 8. ( ) 1+x ln 1−x

x2 x3 x4 x5 − − − −··· 2 3 4 5

Determine the power series for

(

) 1+x ln = ln(1 + x) − ln(1 − x) by the laws of 1−x logarithms, and from Problems 6 and 7, ( ) ( ) 1+x x2 x3 x4 x5 ln = x − + − + − ··· 1−x 2 3 4 5 ( ) 2 3 4 x x x x5 − −x − − − − − · · · 2 3 4 5

= f(0) + x f ′ (0) +

x2 ′′ x3 x4 f (0) + f ′′′ (0) + f iv (0) 2! 3! 4!

= 16 + (x)(32) +

x2 x3 x4 (48) + (48) + (24) 2! 3! 4!

= 16 + 32x + 24x2 + 8x3 + x4 (This expression could have been obtained by applying the binomial series.) x

Problem 10. Expand e 2 as far as the term in x4 x

f (x) = e 2 1 x f ′ (x) = e 2 2

1 1 f ′ (0) = e0 = 2 2

1 x f ′′ (x) = e 2 4

1 1 f ′′ (0) = e0 = 4 4

1 x f ′′′ (x) = e 2 8

1 1 f ′′′ (0) = e0 = 8 8

f iv (x) =

(

1+x 1−x

)

x

f ′ (0) = 4(2)3 = 32

x2 ′′ f (0) 2!

x3 ′′′ x4 f (0) + f iv (0) + · · · 3! 4! ( ) ( ) ( ) 1 x2 1 x3 1 = 1 + (x) + + 2 2! 4 3! 8 ( ) x4 1 + + ··· 4! 16 i.e.

f ′ (x) = 4(2 + x)3

1 0 1 e = 16 16

+

( ) x3 x5 =2 x+ + + ··· 3 5

f(0) = 24 = 16

f iv (0) =

e 2 = f(0) + x f ′ (0) +

Problem 9. Use Maclaurin’s series to find the expansion of (2 + x)4 f(x) = (2 + x)4

1 x e2 16

Substituting in equation (5) gives:

2 2 = 2x + x3 + x5 + · · · 3 5 i.e. ln

f (0) = e0 = 1

x 1 1 1 1 4 e 2 = 1 + x + x2 + x3 + x + ··· 2 8 48 384

Maclaurin’s series and limiting values 455 Problem 11. Develop a series for sinh x using Maclaurin’s series. f (x) = sinh x f ′ (x) = cosh x f ′′ (x) = sinh x f ′′′ (x) = cosh x

e0 − e−0 =0 2 0 −0 e +e f ′ (0) = cosh 0 = =1 2 ′′ f (0) = sinh 0 = 0 f ′′′ (0) = cosh 0 = 1 f (0) = sinh 0 =

f iv (x) = sinh x f iv (0) = sinh 0 = 0 f v (x) = cosh x f v (0) = cosh 0 = 1 Substituting in equation (5) gives: x3 x2 sinh x = f (0) + x f ′ (0) + f ′′ (0) + f ′′′ (0) 2! 3! x5 x4 + f iv (0) + f v (0) + · · · 4! 5! x2 x3 x4 = 0 + (x)(1) + (0) + (1) + (0) 2! 3! 4! x5 + (1) + · · · 5! x5 x3 i.e. sinh x = x + + + · · · 3! 5! (as shown in Chapter 12) Problem 12. Produce a power series for cos2 2x as far as the term in x6 From double angle formulae, cos 2 A = 2 cos A − 1 (see Chapter 15). 1 from which, cos2 A = (1 + cos 2A) 2 1 and cos2 2x = (1 + cos 4x) 2 2

From Problem 1, x2 x4 x6 cos x = 1 − + − + · · · 2! 4! 6! (4x)2 (4x)4 (4x)6 + − + ··· hence cos 4x = 1 − 2! 4! 6! 32 256 6 = 1 − 8x2 + x4 − x + ··· 3 45 1 Thus cos2 2x = (1 + cos 4x) 2 ( ) 1 32 4 256 6 2 x + ··· = 1 + 1 − 8x + x − 2 3 45 i.e. cos2 2x = 1− 4x2 +

16 4 128 6 x − x +··· 3 45

Now try the following Practice Exercise Practice Exercise 182 Maclaurin’s series (Answers on page 889) 1.

Determine the first four terms of the power series for sin 2x using Maclaurin’s series.

2.

Use Maclaurin’s series to produce a power series for cosh 3x as far as the term in x6

3.

Use Maclaurin’s theorem to determine the first three terms of the power series for ln(1 + ex )

4.

Determine the power series for cos 4t as far as the term in t6

5.

Expand e 2 x in a power series as far as the term in x3

6.

Develop, as far as the term in x4 , the power series for sec 2x

7.

Expand e2θ cos 3θ as far as the term in θ2 using Maclaurin’s series.

8.

Determine the first three terms of the series for sin2 x by applying Maclaurin’s theorem.

9.

Use Maclaurin’s series to determine the expansion of (3 + 2t)4

3

37.5 Numerical integration using Maclaurin’s series The value of many integrals cannot be determined using the various analytical methods. In Chapter 45, the trapezoidal, mid-ordinate and Simpson’s rules are used to numerically evaluate such integrals. Another method of finding the approximate value of a definite integral is to express the function as a power series using Maclaurin’s series, and then integrating each algebraic term in turn. This is demonstrated in the following worked problems. As a reminder, the general solution of integrals of ∫ the form axn dx, where a and n are constants, is given by: ∫ axn+1 axn dx = +c n+1

456 Section H ∫

∫

0.4

2 esin θ dθ, correct to

Problem 13. Evaluate 0.1

3 significant figures.

A power series for esin θ is firstly obtained using Maclaurin’s series. sin θ

sin 0

f (θ) = e

f (0) = e

f ′ (θ) = cos θ esin θ ′′

sin θ dθ using θ 0 Maclaurin’s series, correct to 3 significant figures. Let f(θ) = sin θ

f (0) = 0

f ′ (θ) = cos θ

f ′ (0) = 1

f ′′ (θ) = −sin θ

f ′′ (0) = 0

f ′′′ (θ) = −cos θ

f ′′′ (0) = −1

f iv (θ) = sin θ

f iv (0) = 0

f v (θ) = cos θ

f v (0) = 1

0

=e =1

f ′ (0) = cos 0 esin 0 = (1)e0 = 1 sin θ

f (θ) = (cos θ)(cos θ e

sin θ

) + (e

)(−sin θ),

by the product rule = esin θ (cos2 θ − sin θ); f ′′ (0) = e0 (cos2 0 − sin 0) = 1 ′′′

sin θ

f (θ) = (e

Hence from equation (5): sin θ = f (0) + θf ′ (0) +

)[(2 cos θ(−sin θ) − cos θ)] + (cos2 θ − sin θ)(cos θ esin θ )

+

= esin θ cos θ[−2 sin θ − 1 + cos2 θ − sin θ]

= 0 + θ(1) +

f ′′′ (0) = e0 cos 0[(0 − 1 + 1 − 0)] = 0

θ2 ′′ θ3 f (0) + f ′′′ (0) + · · · 2! 3!

θ2 +0 2 ∫ 0.4 ∫ sin θ Thus 2 e dθ = = 1+θ+

) 0.4 ( θ2 2 1+θ+ dθ 2 0.1

0.1

∫

0.4

(2 + 2θ + θ2 )dθ

= [

i.e. Hence ∫

sin θ = θ −

= 0.98133 − 0.21033 = 0.771, correct to 3 significant figures.

θ4 iv θ5 f (0) + f v (0) + · · · 4! 5! θ2 θ3 (0) + (−1) 2! 3! θ4 θ5 (0) + (1) + · · · 4! 5!

θ3 θ5 + − ··· 3! 5!

1

sin θ dθ θ 0 ( ) θ3 θ5 θ7 ∫ 1 θ − + − + ··· 3! 5! 7! = dθ θ 0

0.1

]0.4 2θ2 θ3 = 2θ + + 2 3 0.1 ( ) (0.4)3 2 = 0.8 + (0.4) + 3 ( ) (0.1)3 − 0.2 + (0.1)2 + 3

θ2 ′′ θ3 f (0) + f ′′′ (0) 2! 3!

+

Hence from equation (5): esin θ = f(0) + θf ′ (0) +

1

Problem 14. Evaluate

=

) ∫ 1( θ2 θ4 θ6 1− + − + · · · dθ 6 120 5040 0

[ ]1 θ3 θ5 θ7 = θ− + − + ··· 18 600 7(5040) 0 ( ) 1 1 1 = 1− + − + · · · − (0) 18 600 7(5040) = 0.946, correct to 3 significant figures.

Maclaurin’s series and limiting values 457 ∫

0.4

x ln(1 + x) dx using

Problem 15. Evaluate

4.

Use Maclaurin’s theorem to expand √ x ln(x + 1) as a power series. Hence evaluate, correct to 3 decimal places, ∫ 0.5 √ x ln (x + 1) dx.

0

Maclaurin’s theorem, correct to 3 decimal places. From Problem 6, ln(1 + x) = x − ∫ Hence

0 2

3

4

5

x x x x + − + − ··· 2 3 4 5

37.6

0.4

x ln(1 + x)dx 0

( ) x2 x3 x4 x5 = x x − + − + − · · · dx 2 3 4 5 0 ) ∫ 0.4 ( x3 x4 x5 x6 2 = x − + − + − · · · dx 2 3 4 5 0 ∫

0.4

[

]0.4

x3 x4 x5 x6 x7 − + − + − ··· 3 8 15 24 35 0 ( (0.4)3 (0.4)4 (0.4)5 (0.4)6 = − + − 3 8 15 24

For example, { lim

x→1

x2 + 3x − 4 x2 − 7x + 6

} =

1+3−4 0 = 1−7+6 0

For example, {

) − (0)

= 0.02133 − 0.0032 + 0.0006827 − · · · = 0.019, correct to 3 decimal places. Now try the following Practice Exercise Practice Exercise 183 Numerical integration using Maclaurin’s series (Answers on page 889) ∫ 0.6 1. Evaluate 3esin θ dθ, correct to 3 decimal 0.2

places, using Maclaurin’s series. 2. Use Maclaurin’s theorem to expand cos 2θ and hence evaluate, correct to 2 decimal places, ∫ 1 cos 2θ dθ 1 0 θ3 ∫ 1√ 3. Determine the value of θ cos θ dθ, cor0

It is{sometimes necessary to find limits of the form } f (x) lim , where f (a) = 0 and g(a) = 0 x→a g(x)

and 00 is generally referred to as indeterminate. For certain limits a knowledge of series can sometimes help.

=

(0.4)7 + − ··· 35

Limiting values

rect to 2 significant figures, using Maclaurin’s series.

} tan x − x x→0 x3   1   x + x3 + · · · − x   3 ≡ lim from Problem 5 3 x→0   x   lim

  1  { }  x3 + · · ·   1 1 = lim 3 3 = lim = x→0  x→0  x 3 3   Similarly, {

} sinh x lim x→0 x   x3 x5    x + +  + ... 3! 5! ≡ lim from Problem 11 x→0   x     { } x2 x4 = lim 1 + + + · · · = 1 x→0 3! 5! However, a knowledge of series does not help with { 2 } x + 3x − 4 examples such as lim 2 x→1 x − 7x + 6

458 Section H L’Hˆopital’s∗ rule will enable us to determine such limits when the differential coefficients of the numerator and denominator can be found. L’Hˆopital’s rule states:

The first step is to substitute x = 1 into both numerator and denominator. In this case we obtain 00 . It is only when we obtain such a result that we then use L’Hˆopital’s rule. Hence applying L’Hˆopital’s rule, {

x2 + 3x − 4 lim 2 x→1 x − 7x + 6

} { ′ } { f (x) f(x) = lim ′ lim x→a g (x) x→a g(x)

}

{

2x + 3 = lim x→1 2x − 7

i.e. both numerator and denominator have been differentiated

′

provided g (a) ̸= 0 { ′ } f (x) is still 00 ; if so, the It can happen that lim ′ x→a g (x) numerator and denominator are differentiated again (and again) until a non-zero value is obtained for the denominator. The following worked problems demonstrate how L’Hˆopital’s rule is used. Refer to Chapter 25 for methods of differentiation. { Problem 16. Determine lim

x→1

x2 + 3x − 4 x2 − 7x + 6

}

}

=

5 = −1 −5 {

Problem 17. Determine lim

x→0

sin x − x x2

}

Substituting x = 0 gives { } sin x − x sin 0 − 0 0 lim = = 2 x→0 x 0 0 Applying L’Hˆopital’s rule gives { } { } sin x − x cos x − 1 lim = lim x→0 x→0 x2 2x Substituting x = 0 gives cos 0 − 1 1 − 1 0 = = 0 0 0

again

Applying L’Hˆopital’s rule again gives { } { } cos x − 1 −sin x lim = lim =0 x→0 x→0 2x 2 { Problem 18. Determine lim

x→0

x − sin x x − tan x

}

Substituting x = 0 gives { } x − sin x 0 − sin 0 0 lim = = x→0 x − tan x 0 − tan 0 0

∗

Who was L’Hˆ opital? Guillaume François Antoine, Marquis de l’Hˆ opital (1661–2 February 1704, Paris) was a French mathematician. His name is firmly associated with l’Hˆ opital’s rule for calculating limits involving indeterminate forms 0/0 and ∞/∞. To find out more go to www.routledge.com/cw/bird

Applying L’Hˆopital’s rule gives { } { } x − sin x 1 − cos x lim = lim x→0 x − tan x x→0 1 − sec2 x Substituting x = 0 gives } { 1 − cos x 1 − cos 0 1−1 0 lim = = = again x→0 1 − sec2 x 1 − sec2 0 1 − 1 0

Maclaurin’s series and limiting values 459 Applying L’Hˆ opital’s rule gives { } { } 1 − cos x sin x lim = lim x→0 1 − sec2 x x→0 (−2 sec x)(sec x tan x) { } sin x = lim x→0 −2 sec2 x tan x Substituting x = 0 gives

{

6. 7. 8.

sin 0 0 = −2 sec2 0 tan 0 0

again

Applying L’Hˆ opital’s rule gives { } sin x lim x→0 −2 sec2 x tan x         cos x = lim 2 2 x→0  x)     (−2 sec x)(sec  2 + (tan x)(−4 sec x tan x) using the product rule

9.

Practice Exercise 185 Multiple-choice questions on Maclaurin’s series and limiting values (Answers on page 889) Each question has only one correct answer 1.

Substituting x = 0 gives cos 0 −2 sec4 0 − 4 sec2 0 tan2 0

=

1 −2 − 0

=− { Hence lim

x→0

x − sin x x − tan x

} =−

1 2

} ln t lim t→1 t2 − 1 { } sinh x − sin x lim x→0 x3 { } sin θ − 1 limπ θ→ 2 ln sin θ { } sec t − 1 lim t→0 t sin t

Using Maclaurin’s series, the first 4 terms of the power series for cos θ are: θ2 θ4 θ6 θ2 θ4 θ6 (a) 1 + − + (b) 1 − + − 2! 4! 6! 2 4 6 θ2 θ4 θ6 θ3 θ5 θ7 + − (d) θ − + − 2! 4! 6! 3! 5! 7! Using Maclaurin’s series, the first 4 terms of the power series for sin t are: t2 t4 t6 t2 t4 t6 (a) t − + − (b) t + − + 3 5 7 2! 4! 6! (c) 1 −

2.

1 2

t2 t4 t6 t3 t5 t7 + − (d) t − + − 2! 4! 6! 3! 5! 7! Using Maclaurin’s series, the first 4 terms of the power series for e θ are: θ2 θ3 θ2 θ3 θ4 (a) 1 − θ + − (b) θ + − + 2! 3! 2! 3! 4! (c) 1 −

Now try the following Practice Exercise

3.

Practice Exercise 184 Limiting values (Answers on page 889) Determine the following limiting values { 3 } x − 2x + 1 1. lim x→1 2x3 + 3x − 5 { } sin x 2. lim x→0 x { } ln(1 + x) 3. lim x→0 x { 2 } x − sin 3x 4. lim x→0 3x + x2 { } sin θ − θ cos θ 5. lim θ→0 θ3

θ2 θ3 θ2 θ3 + (d) 1 + θ + + 2! 3! 2 3 By using Maclaurin’s series to expand cos 2x, ∫1 the value of 0 3x cos 2x dx, correct to 3 decimal places, is: (a) 1.500 (b) 0.101 (c) 0.455 (d) 0.302 (c) 1 + θ +

4.

5.

By using Maclaurin’s series to expand sin 3x, ∫2 the value of 1 x sin 3x dx, correct to 3 decimal places, is: (a) 0.122 (b) −1.017 (c) 0.078 (d) −0.650

460 Section H { 6. The limiting value lim

x →1

equal to: 1 3 (a) 2 (b) 3 7

3 (c) − 4

2x3 − 3x + 1 x3 + 4x − 5 (d) 0

}

{ is

7.

The limiting value lim

x →0

to: 1 (a) − 2

(b) 1

(c)

1 2

1 − cos x x2

} is equal

(d) 0

For fully worked solutions to each of the problems in Practice Exercises 182 to 184 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 38

Integration using algebraic substitutions Why it is important to understand: Integration using algebraic substitutions As intimated in previous chapters, most complex engineering problems cannot be solved without calculus. Calculus has widespread applications in science, economics and engineering and can solve many problems for which algebra alone is insufficient. For example, calculus is needed to calculate the force exerted on a particle a specific distance from an electrically charged wire, and is needed for computations involving arc length, centre of mass, work and pressure. Sometimes the integral is not a standard one; in these cases it may be possible to replace the variable of integration by a function of a new variable. A change in variable can reduce an integral to a standard form, and this is demonstrated in this chapter.

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

appreciate when an algebraic substitution is required to determine an integral integrate functions which require an algebraic substitution determine definite integrals where an algebraic substitution is required appreciate that it is possible to change the limits when determining a definite integral

38.1

Introduction

Functions which require integrating are not always in the ‘standard form’ shown in Chapter 35. However, it is often possible to change a function into a form which can be integrated by using either: (i) (ii)

an algebraic substitution (see Section 38.2), a trigonometric or hyperbolic substitution (see Chapter 39), (iii) partial fractions (see Chapter 40), (iv) the t = tan θ/2 substitution (see Chapter 41), (v) integration by parts (see Chapter 42), or (vi) reduction formulae (see Chapter 43).

38.2

Algebraic substitutions

With algebraic substitutions, the substitution usually made is to let u be equal to f (x) such that f(u) du is a standard integral. It is found that integrals of the forms, ∫ k

n ′

[ f(x)] f (x) dx and k

∫

f ′ (x) dx [ f (x)]n

(where k and n are constants) can both be integrated by substituting u for f (x).

462 Section H 38.3 Worked problems on integration using algebraic substitutions Problem 1.

Determine

∫

Let u = (5x − 3) then Hence ∫

4 dx = (5x − 3)

cos (3x + 7) dx =

∫

cos(3x + 7) dx is not a standard integral of the form shown in Table 35.1, page 424, thus an algebraic substitution is made. du Let u = (3x + 7) then = 3 and rearranging gives dx du dx = . Hence, 3 ∫ ∫ ∫ du 1 cos (3x + 7) dx = (cos u) = cos u du, 3 3

∫

4 du 4 = u 5 5

Problem 4.

Let u = (6x − 1) then

du du = 6 and dx = dx 6 ∫

2e6x−1 dx =

∫ Find (2x − 5)7 dx

u8 8

) +c =

1 8 u +c 16

Rewriting u as (2x − 5) gives: ∫ (2x − 5)7 dx = ∫ Problem 3.

Find

Determine

Let u = (4x2 + 3) then Hence

∫

eu du

3x(4x2 + 3)5 dx

du du = 8x and dx = dx 8x

∫

∫ 3x(4x2 + 3)5 dx =

=

3x(u)5 3 8

du 8x

∫ u5 du, by cancelling.

The original variable ‘x’ has been completely removed and the integral is now only in terms of u and is a standard integral. ( ) ∫ 3 u6 3 5 u du = +c Hence 8 8 6

1 (2x − 5)8 + c 16

4 dx (5x − 3)

∫

correct to 4 significant figures. Problem 5.

(2x − 5) may be multiplied by itself seven times and then each term of the result integrated. However, this would be a lengthy process, and thus an algebraic substitution is made. du du Let u = (2x − 5) then = 2 and dx = dx 2 Hence ∫ ∫ ∫ du 1 (2x − 5)7 dx = u7 = u7 du 2 2 (

du 1 = 6 3

Thus ∫ 1 1 1 2e6x−1 dx = [e6x−1 ]10 = [e5 − e−1 ] = 49.35, 3 3 0

which may be checked by differentiating it.

1 = 2

2eu

1 1 = eu + c = e6x−1 + c 3 3

Rewriting u as (3x + 7) gives: ∫ 1 cos(3x + 7) dx = sin(3x + 7) + c, 3

Problem 2.

2e6x−1 dx, correct to 0

∫

1 sin u + c 3

1 du u

1

Evaluate

4 significant figures.

Hence

∫

4 4 ln u + c = ln(5x − 3) + c 5 5 ∫

which is a standard integral =

du du = 5 and dx = dx 5

=

1 6 1 u + c = (4x 2 + 3)6 + c 16 16 ∫

Problem 6.

π 6

Evaluate 0

24 sin5 θ cos θ dθ

Integration using algebraic substitutions 463 du du = cos θ and dθ = dθ cos θ ∫ ∫ du Hence 24 sin5 θ cos θ dθ = 24u5 cos θ cos θ ∫ = 24 u5 du, by cancelling Let u = sin θ then

= 24

∫

1

10.

3 cos(4x − 3) dx

0

11.

The mean time to failure, M years, for a set of components is given by: ∫ 4 M= (1 − 0.25t)1.5 dt 0

u6 + c = 4u6 + c = 4(sin θ)6 + c 6

Determine the mean time to failure.

= 4 sin6 θ + c ∫

π 6

Thus 0

π

24 sin5 θ cos θ dθ = [4 sin6 θ]06 [(

] π )6 − (sin 0)6 6 [( ) ] 6 1 1 =4 −0 = or 0.0625 2 16 =4

Practice Exercise 186 Integration using algebraic substitutions (Answers on page 889) In Problems 1 to 6, integrate with respect to the variable. 1. 2 sin(4x + 9) 2. 3 cos(2θ − 5) 3. 4 sec2 (3t + 1)

5.

1 (5x − 3)6 2 −3 (2x − 1)

6. 3e3θ+5 In Problems 7 to 10, evaluate the definite integrals correct to 4 significant figures. ∫ 1 7. (3x + 1)5 dx ∫

0 2

8. ∫

√ x (2x2 + 1) dx

0 π 3

9. 0

Further worked problems on integration using algebraic substitutions

sin

Now try the following Practice Exercise

4.

38.4

( π) 2 sin 3t + dt 4

∫ Problem 7.

Find

x dx 2 + 3x2

du du Let u = (2 + 3x2 ) then = 6x and dx = dx 6x Hence ∫ ∫ ∫ x x du 1 1 dx = = du, 2 + 3x2 u 6x 6 u by cancelling 1 ln u + c = 6 ∫ Determine

= Problem 8.

1 ln(2 + 3x 2 ) + c 6 2x √ dx (4x2 − 1)

du du Let u = (4x2 − 1) then = 8x and dx = dx 8x ∫ ∫ 2x 2x du √ Hence dx = √ u 8x (4x2 − 1) ∫ 1 1 √ du, by cancelling = 4 u  ( )  −1 ∫ +1 −1 1 1 u 2  = u 2 du =  +c 1 4 4 − +1 2   1

1 u2  1√ =  +c = u+c 1 4 2 2 =

1√ (4x 2 − 1) + c 2

464 Section H ∫ 2( Problem 9.

∫

Therefore, distance, d =

Show that ∫ tan θ dθ = ln(sec θ) + c

∫

0

38.5 Hence ∫

(

∫

Change of limits

When evaluating definite integrals involving substitutions it is sometimes more convenient to change the limits of the integral as shown in Problems 11 and 12.

)

sin θ −du u sin θ ∫ 1 =− du = − ln u + c u

sin θ dθ = cos θ

dv

)] 2 1 [ ( ln 9.81 − 0.3v2 0 0.6 ) ( )] 1 [ ( =− ln 9.81 − 0.3(2)2 − ln 9.81 − 0.3(0)2 0.6 ) ( )] 1 [ ( =− ln 8.61 − ln 9.81 = 0.2175 m 0.6 i.e. the distance a body has fallen in the resistive medium is 0.2175 m

du −du = −sin θ and dθ = dθ sin θ

then

)

=−

sin θ dθ. Let u = cos θ cos θ

tan θ dθ =

v 9.81 − 0.3v2

∫

3

Problem 11. Evaluate

= − ln(cos θ) + c = ln(cos θ)−1 + c,

√ 5x (2x2 + 7) dx,

1

taking positive values of square roots only.

by the laws of logarithms. du du = 4x and dx = dx 4x It is possible in this case to change the limits of integration. Thus when x = 3, u = 2(3)2 + 7 = 25 and when x = 1, u = 2(1)2 + 7 = 9

∫ Hence since

Let u = (2x2 + 7), then tan θ dθ = ln(sec θ) + c, (cos θ)−1 =

1 = sec θ cos θ

Hence Problem 10. The distance, d, a body has fallen in a resistive medium when its velocity, v, is 0.3 m/s is given by the integral: ) ∫ 2( v d= dv 9.81 − 0.3v2 0 Evaluate distance d, correct to 4 significant figures. ∫ 2( Distance, d = 0

v 9.81 − 0.3v2

Let u = 9.81 − 0.3v2 then

∫ (

v 9.81 − 0.3v2 =

dv =

1 − 0.6

∫

5x

∫ √ (2x2 + 7) dx =

x=1

∫ ( ) v du u − 0.6v

du −0.6v

1 1 du = − ln u + c u 0.6

Since u = 9.81 − 0.3v2 then ) ∫ ( ) v 1 ( dv = − ln 9.81 − 0.3v2 + c 9.81 − 0.3v2 0.6

u=25 u=9

5 = 4

=

dv

du = −0.6v and dx

)

x=3

)

dx = Hence,

∫

5 4

∫

√ du 5x u 4x

25 √

u du

9

∫

25

1

u 2 du 9

Thus the limits have been changed, and it is unnecessary to change the integral back in terms of x.  3 25 ∫ x=3 √ 5 u2  Thus 5x (2x2 + 7) dx =  4 3/2 x=1 9

5 [√ 3 ]25 5 [√ 3 √ 3 ] = u = 25 − 9 6 6 9 5 2 = (125 − 27) = 81 6 3

Integration using algebraic substitutions 465 ∫

2

3x dx, 2 + 1) (2x 0 taking positive values of square roots only. Problem 12. Evaluate

Let u = (2x2 + 1) then

du du = 4x and dx = dx 4x

∫

∫

2

Hence 0

6. 3 tan 2t 2et 7. √ t (e + 4)

√

3x √ dx = (2x2 + 1)

=

x=2 x=0

3 4

∫

In Problems 8 to 10, evaluate the definite integrals correct to 4 significant figures. ∫ 1 2 8. 3x e(2x −1) dx

3x du √ u 4x

0

∫ 9.

x=2

u

−1 2

π 2

0

∫

du

1

10.

x=0

0

Since u = 2x2 + 1, x = 0, u = 1

Thus

3 4

∫

x=2

u

−1 2

x = 2, u = 9

when

du =

x=0

3 4

∫

u=9

u

−1 2

and

when

11.

=

The electrostatic potential on all parts of a conducting circular disc of radius r is given by the equation: ∫ 0

12.

In the study of a rigid rotor the following integration occurs: ∫

(2J + 1)e

−J(J+1) h2 8π 2 Ik T

dJ

0

Determine Zr for constant temperature T assuming h, I and k are constants.

Now try the following Practice Exercise 13.

In electrostatics,  ∫ π E=

In Problems 1 to 7, integrate with respect to the variable.

0

1. 2x(2x2 − 3)5

 2ε

√(

a2 σ sin θ a2 − x2 − 2ax cos θ

  ) dθ

where a, σ and ε are constants, x is greater than a, and x is independent of θ. Show that a2 σ E= εx

2. 5 cos5 t sin t 3. 3 sec2 3x tan 3x √ 4. 2t (3t2 − 1) ln θ θ

∞

Zr =

taking positive values of square roots only.

5.

R √ dR 2 R + r2

Solve the equation by determining the integral.

9

Practice Exercise 187 Integration using algebraic substitutions (Answers on page 889)

9

V = 2πσ

u=1

√ ] 3 3 [√  9 − 1 = 3,  1  = 4 2 2 1 1 u2

3x dx (4x2 − 1)5

du,

i.e. the limits have been changed 

3 sin4 θ cos θ dθ

14.

The time taken, t hours, for a vehicle to reach a velocity of 130 km/h with an initial speed

466 Section H ∫

130

of 60 km/h is given by: t = 60

dv 650 − 3v

∫

Each question has only one correct answer ∫ π/6 ( π) 1. The value of 2 sin 3t + dt is: 2 0 2 2 (c) −6 (d) (a) 6 (b) − 3 3 ∫ ( 2 )3 2. The indefinite integral x x − 3 dx is equal to: )4 1( 2 (a) x −3 +c 4 )2 3( 2 (c) x −3 +c 2

5 dx, correct to 3 decimal 4x − 3

Evaluating

4.

places, gives: (a) 2.747 (b) 1.758 (c) −0.025 (d) 11.513 ∫ π/3 The value of 16 cos4 θ sin θ dθ is:

1

where v is the velocity in km/h. Determine t, correct to the nearest second.

Practice Exercise 188 Multiple-choice questions on integration using algebraic substitutions (Answers on page 889)

3

3.

0

(a) −0.1 5.

(b) 3.1 (c) 0.1 (d) −3.1 ∫ 2 12x √ dx, correct to 3 The value of 3x2 − 1 1 decimal places, is: (a) −1.249 (b) 0.852 (c) 7.610 (d) 10.228

( )4 (b) 2 x2 − 3 + c (d)

)4 1( 2 x −3 +c 8

For fully worked solutions to each of the problems in Practice Exercises 186 and 187 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 39

Integration using trigonometric and hyperbolic substitutions Why it is important to understand: Integration using trigonometric and hyperbolic substitutions Calculus is the most powerful branch of mathematics. It is capable of computing many quantities accurately, which cannot be calculated using any other branch of mathematics. Many integrals are not ‘standard’ ones that we can determine from a list of results. Some need substitutions to rearrange them into a standard form. There are a number of trigonometric and hyperbolic substitutions that may be used for certain integrals to change them into a form that can be integrated. These are explained in this chapter which provides another piece of the integral calculus jigsaw.

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

integrate functions of the form sin 2 x, cos 2 x, tan2 x and cot2 x integrate functions having powers of sines and cosines integrate functions that are products of sines and cosines integrate using the sin θ substitution integrate using the tan θ substitution integrate using the sinh θ substitution integrate using the cosh θ substitution

39.1

Introduction

Table 39.1 gives a summary of the integrals that require the use of trigonometric and hyperbolic substitutions and their application is demonstrated in Problems 1 to 28.

39.2 Worked problems on integration of sin2 x, cos2 x, tan2 x and cot2 x ∫ Problem 1.

Evaluate 0

π 4

2 cos 2 4t dt

468 Section H Table 39.1 Integrals using trigonometric and hyperbolic substitutions ∫ f(x) f (x)dx Method ( ) 1 sin 2x 1. cos 2 x x+ +c Use cos 2x = 2 cos 2 x − 1 2 2

See problem 1

( ) 1 sin 2x x− +c 2 2

Use cos 2x = 1 − 2 sin 2 x

2

3. tan2 x

tan x − x + c

Use 1 + tan2 x = sec2 x

3

4. cot2 x

− cot x − x + c

Use cot2 x + 1 = cosec2 x

4

2.

5.

2

sin x

m

n

cos x sin x

(a) If either m or n is odd (but not both), use cos 2 x + sin 2 x = 1

5, 6

(b) If both m and n are even, use either cos 2x = 2 cos 2 x − 1 or cos 2x = 1 − 2 sin 2 x 6.

7, 8

Use 12 [ sin(A + B) + sin(A − B)]

sin A cos B

1 2 [ sin(A + B) −

sin(A − B)]

9

7.

cos A sin B

Use

8.

cos A cos B

Use 12 [ cos(A + B) + cos(A − B)]

11

9.

sin A sin B

Use − 12 [ cos(A + B) − cos(A − B)]

12

10.

1 √ (a2 − x2 )

sin−1

√ (a2 − x2 )

11.

12.

13.

14.

15.

16.

a2

1 + x2

x +c a

10

Use x = a sin θ substitution

13, 14

a2 −1 x x √ 2 sin + (a − x2 ) + c 2 a 2

Use x = a sin θ substitution

15, 16

1 −1 x tan +c a a

Use x = a tan θ substitution

17–19

1 √ 2 (x + a2 )

sinh−1

x +c a { } √ x + (x2 + a2 ) or ln +c a

Use x = a sinh θ substitution

20–22

√ (x2 + a2 )

a2 x x√ 2 sinh−1 + (x + a2 ) + c 2 a 2

Use x = a sinh θ substitution

23

1 √ (x2 − a2 )

cosh−1

Use x = a cosh θ substitution

24, 25

√ (x2 − a2 )

x√ 2 a2 x (x − a2 ) − cosh−1 + c 2 2 a

Use x = a cosh θ substitution

26, 27

x +c a { } √ x + (x2 − a2 ) or ln +c a

Integration using trigonometric and hyperbolic substitutions 469 Since cos 2t = 2 cos 2 t − 1 (from Chapter 15), 1 then cos 2 t = (1 + cos 2t) and 2

Hence

π 4

Problem 4.

2 cos 2 4t dt π 4

1 (1 + cos 8t) dt 2 0 [ ]π sin 8t 4 = t+ 8 0 =2

 π = + 4

=

(π )

[ ] 4  − 0 + sin 0 8 8

sin 8

π or 0.7854 4

Problem 2.

Determine

∫

sin 2 3x dx

Since cos 2x = 1 − 2 sin 2 x (from Chapter 15), 1 then sin 2 x = (1 − cos 2x) and 2 1 2 sin 3x = (1 − cos 6x) 2 ∫ ∫ 1 2 Hence sin 3x dx = (1 − cos 6x) dx 2 ( ) 1 sin 6x = x− +c 2 6

1 = 2

∫

π 3

2

Find 3 tan 4x dx

π 6

[ ]π 3 1 −cot 2θ −θ (cosec 2θ − 1) dθ = π 2 2 6

(π)

 1 −cot 2 3 = 2 2

Hence 3

∫ (sec 4x − 1) dx

2

2

tan 4x dx = 3

) tan 4x =3 −x +c 4 (

(π )

  π −cot 2 6 − − 3 2

 π − 6

1 = [(0.2887 − 1.0472) − (−0.2887 − 0.5236)] 2 = 0.0269 Problem 5. In a thin-walled closed tube of uniform thickness, t, the second moment of area about its neutral axis is given by: ∫ 0.206 I = 2× tds1 (s1 sin α)2 + 0.1 × t × 0.052 × 2 0 ∫ π + (t R dϕ)(R cos ϕ)2 0

Given that angle α = 14.04◦ and radius R = 0.05 m, determine I in terms of t ∫

0.206

I = 2×

tds1 (s1 sin α)2 + 0.1 × t × 0.052 × 2

0

∫

π

(t R dϕ)(R cos ϕ)2

+ = 2t ×

0 0.206

(s1 sin 14.04◦ )2 ds1 + 5 × 10−4 t

0

Since 1 + tan2 x = sec2 x, then tan2 x = sec2 x − 1 and tan2 4x = sec2 4x − 1 ∫

1 2 cot 2θ dθ 2

2

∫

∫

Problem 3.

π 6

6

0

∫

Evaluate

π 3

Since cot2 θ +1 = cosec2 θ, then cot2 θ = cosec2 θ−1 and cot2 2θ = cosec2 2θ − 1 ∫ π 3 1 cot2 2θ dθ Hence π 2

1 cos 2 4t = (1 + cos 8t) 2 ∫

∫

∫

π

+ t(0.05)3

cos 2 ϕ dϕ 0

∫

0.206

= 0.1177t × 0

s21 ds1 + 5 × 10−4 t π(

∫ + t(0.05)3 0

1 + cos 2ϕ 2

) dϕ

since from 1 of Table 39.1, cos 2ϕ = 2 cos 2 ϕ − 1 from 1 + cos 2ϕ which, cos 2 ϕ = 2

470 Section H [

s3 = 0.1177t × 1 3

] 0.206 + 5 × 10

−4

t

0

[ ]π t(0.05)3 sin 2ϕ + ϕ+ 2 2 0 [ ] 3 3 (0.206) (0) = 0.1177t × − + 5 × 10−4 t 3 3 [( ) ] ( ) t(0.05)3 sin 2π + π+ − 0 2 2

Since cos 2 θ + sin 2 θ = 1 then sin 2 θ = (1 − cos 2 θ) ∫ Hence sin 5 θ dθ ∫ ∫ = sin θ(sin 2 θ)2 dθ = sin θ(1 − cos 2 θ)2 dθ ∫ = sin θ(1 − 2 cos 2 θ + cos 4 θ) dθ ∫ = (sin θ − 2 sin θ cos 2 θ + sin θ cos 4 θ) dθ

= 3.430 × 10−4 t + 5 × 10−4 t + 1.963 × 10−4 t Hence, the second moment of area, I = 1.039 × 10−3 t Now try the following Practice Exercise Practice Exercise 189 Integration of sin2 x, cos2 x, tan2 x and cot2 x (Answers on page 890) In Problems 1 to 4, integrate with respect to the variable. 1.

= −cos θ +

2 cos3 θ cos5 θ − +c 3 5

Whenever a power of a cosine is multiplied by a sine of power 1, or vice-versa, the integral may be determined by inspection as shown. ∫ −cos n+1 θ In general, cos n θ sin θ dθ = +c (n + 1) ∫ sin n+1 θ and sin n θ cos θ dθ = +c (n + 1) ∫

sin 2 2x Problem 7.

2. 3 cos 2 t

π 2

Evaluate

sin 2 x cos 3 x dx

0

2

3. 5 tan 3θ

∫

4. 2 cot2 2t In Problems 5 to 8, evaluate the definite integrals, correct to 4 significant figures. ∫ π 3 5. 3 sin 2 3x dx ∫

π 4

8.

sin x cos x dx = ∫

(sin 2 x)(1 − sin 2 x)(cos x) dx

0

∫

π 2

(sin 2 x cos x − sin 4 x cos x) dx

0

[ =

sin 3 x sin 5 x − 3 5

(

0

π 6

π 2

=

cos 2 4x dx

2 tan2 2t dt

 =

cot2 θ dθ

=

sin

π )3 2

3

( −

Problem 6.

Determine

sin θ dθ

2

0

 π )5 2   − [0 − 0] 5

sin

∫ Problem 8.

5

]π

1 1 2 − = or 0.1333 3 5 15

39.3 Worked problems on integration of powers of sines and cosines ∫

sin 2 x cos 2 x cos x dx

0

=

1

π 3

π 2

3

0

0

7. ∫

∫ 2

0

6. ∫

π 2

π 4

Evaluate

significant figures.

0

4 cos 4 θ dθ, correct to 4

Integration using trigonometric and hyperbolic substitutions 471 ∫

π 4

∫

π 4

4

4 cos θ dθ = 4

0

∫

π 4

=4 0 π 4

∫ = 0

[

Now try the following Practice Exercise

(cos 2 θ)2 dθ

0

1 (1 + cos 2θ) 2

]2

Practice Exercise 190 Integration of powers of sines and cosines (Answers on page 890)

dθ

In Problems 1 to 6, integrate with respect to the variable.

(1 + 2 cos 2θ + cos 2 2θ) dθ

] 1 1 + 2 cos 2θ + (1 + cos 4θ) dθ 2 0 ) ∫ π( 4 3 1 = + 2 cos 2θ + cos 4θ dθ 2 2 0 ]π [ sin 4θ 4 3θ + sin 2θ + = 2 8 0 [ ( ) ] 3 π 2π sin 4(π/4) = + sin − [0] + 2 4 4 8 ∫

=

=

π 4

[

3π + 1 = 2.178, 8

Problem 9.

∫

Find

∫

correct to 4 significant figures.

∫ ( = = =

1 8 1 8

1 = 8 =

1 8

∫

1 − cos 2t 2

2 cos 3 2x

3.

2 sin 3 t cos 2 t

4.

sin 3 x cos 4 x

5.

2 sin 4 2θ

6.

sin 2 t cos 2 t

Problem 10. Determine sin 2 t(cos 2 t)2 dt

)(

1 + cos 2t 2

dt

from 6 of Table 39.1, which follows from Section 15.4, page 181,

∫ (1 + 2 cos 2t + cos 2 2t − cos 2t − 2 cos 2 2t − cos 3 2t) dt

=

1 2

=

1 2

(1 + cos 2t − cos 2t − cos 2t) dt 2

∫ [

( 1 + cos 2t −

3

1 + cos 4t 2

sin 3t cos 2t dt

sin 3t cos 2t dt ∫ 1 [sin (3t + 2t) + sin (3t − 2t)] dt, = 2

)2

(1 − cos 2t)(1 + 2 cos 2t + cos 2 2t) dt

∫

∫

∫

)

) 1 cos 4t − + cos 2t sin 2 2t dt 2 2 ( ) 1 t sin 4t sin3 2t = − + +c 8 2 8 6

∫ (sin 5t + sin t) dt (

) −cos 5t − cos t + c 5

]

∫

− cos 2t(1 − sin 2t) dt 1 8

2.

sin 2 t cos 4 t dt

2

=

sin 3 θ

39.4 Worked problems on integration of products of sines and cosines

∫ sin 2 t cos 4 t dt =

1.

Problem 11. Find

∫ (

∫

1 cos 5x sin 2x dx 3

1 cos 5x sin 2x dx 3 ∫ 1 1 = [sin (5x + 2x) − sin (5x − 2x)] dx, 3 2 from 7 of Table 39.1

472 Section H ∫ 1 (sin 7x − sin 3x) dx 6 ( ) 1 −cos 7x cos 3x = + +c 6 7 3

Now try the following Practice Exercise

=

∫

Practice Exercise 191 Integration of products of sines and cosines (Answers on page 890)

1

In Problems 1 to 4, integrate with respect to the variable.

2 cos 6θ cos θ dθ,

Problem 12. Evaluate 0

correct to 4 decimal places. ∫

1

2 cos 6θ cos θ dθ 0 ∫ 1 1 =2 [ cos (6θ + θ) + cos (6θ − θ)] dθ, 0 2 from 8 of Table 39.1 ∫

[

1

sin 7θ sin 5θ = + (cos 7θ + cos 5θ) dθ = 7 5 0 ) ( ) ( sin 5 sin 0 sin 0 sin 7 + − + = 7 5 7 5

]1 0

1.

sin 5t cos 2t

2.

2 sin 3x sin x

3.

3 cos 6x cos x

4.

1 cos 4θ sin 2θ 2

In Problems 5 to 8, evaluate the definite integrals. ∫ 5.

π 2

∫

1

6. ‘sin 7’ means ‘the sine of 7 radians’ (≡401◦ 4′ ) and sin 5 ≡ 286◦ 29′ ∫

2 sin 7t cos 3t dt 0

∫ 7.

−4

π 3

sin 5θ sin 2θ dθ

0

1

2 cos 6θ cos θ dθ

Hence

cos 4x cos 3x dx

0

∫

0

= (0.09386 + (−0.19178)) − (0)

8.

2

3 cos 8t sin 3t dt 1

= −0.0979, correct to 4 decimal places. ∫ Problem 13. Find 3

sin 5x sin 3x dx

∫

∫ 3

39.5 Worked problems on integration using the sin θ substitution Problem 14. Determine

sin 5x sin 3x dx

1 √ dx 2 (a − x2 )

∫

1 − [ cos (5x + 3x) − cos (5x − 3x)] dx, 2 from 9 of Table 39.1 ∫ 3 =− ( cos 8x − cos 2x) dx 2 ( ) 3 sin 8x sin 2x =− − + c or 2 8 2 =3

3 (4 sin 2x − sin 8x) + c 16

dx = a cos θ and dx = a cos θ dθ Let x = a sin θ, then dθ ∫ 1 Hence √ dx 2 (a − x2 ) ∫ 1 = √ a cos θ dθ 2 (a − a2 sin 2 θ) ∫ a cos θ dθ = √ [a2 (1 − sin 2 θ)]

Integration using trigonometric and hyperbolic substitutions 473 ∫

a cos θ dθ √ , since sin 2 θ + cos 2 θ = 1 (a2 cos 2 θ) ∫ ∫ a cos θ dθ = dθ = θ + c = a cos θ x x Since x = a sin θ, then sin θ = and θ = sin−1 a a ∫ 1 x Hence √ dx = sin−1 + c a (a2 − x2 ) =

∫

3

Problem 15. Evaluate

√

0

∫

3

√

From Problem 14, 0

[ x ]3 = sin−1 , 3 0

(9 − x2 )

1 (9 − x2 )

∫ √

=

Thus

dx

π or 1.5708 2

(since cos 2θ = 2 cos 2 θ − 1) ( ) sin 2θ θ+ +c 2 ( ) a2 2 sin θ cos θ = θ+ +c 2 2 a2 2

)

√ (a2 − x2 ) = a

a2 x x√ 2 sin−1 + (a − x2 ) + c 2 a 2

Problem 17. Evaluate

∫ 4√ (16 − x2 ) dx 0

∫ 4√ From Problem 16, (16 − x2 ) dx [ =

0

]4 16 x x√ (16 − x2 ) sin−1 + 2 4 2 0

[ √ ] = 8 sin−1 1 + 2 (0) − [8 sin−1 0 + 0] = 8 sin−1 1 = 8

(π) 2

= 4π or 12.57

Now try the following Practice Exercise Practice Exercise 192 Integration using the sin θ substitution (Answers on page 890) ∫ 5 1. Determine √ dt (4 − t2 ) ∫

3 √ dx (9 − x2 )

2.

Determine

3.

∫ √ (4 − x2 ) dx Determine

since from Chapter 15, sin 2θ = 2 sin θ cos θ a2 = [θ + sin θ cos θ] + c 2

a2 − x2 a2

∫ √ a2 (a2 − x2 ) dx = [θ + sin θ cos θ] + c 2 [ ] ( x ) √(a2 − x2 ) a2 −1 x = sin + +c 2 a a a =

dx Let x = a sin θ then = a cos θ and dx = a cos θ dθ dθ ∫ √ Hence (a2 − x2 ) dx ∫ √ = (a2 − a2 sin 2 θ) (a cos θ dθ) ∫ √ = [a2 (1 − sin 2 θ)] (a cos θ dθ) ∫ √ = (a2 cos 2 θ) (a cos θ dθ) ∫ = (a cos θ)(a cos θ dθ) ) ∫ ∫ ( 1 + cos 2θ = a2 cos 2 θ dθ = a2 dθ 2

=

√(

dx

(a2 − x2 ) dx

x x and θ = sin−1 a a

Also, cos 2 θ + sin 2 θ = 1, from which, √[ √ ( x )2 ] 2 cos θ = (1 − sin θ) = 1− a

since a = 3

= (sin−1 1 − sin−1 0) =

Problem 16. Find

1

Since x = a sin θ, then sin θ =

474 Section H

4. Determine

∫ √

∫ (16 − 9t2 ) dt

0

4 decimal places.

∫

4

5. Evaluate

√

0

1 (16 − x2 )

∫

dx

1 0

5 dx = (3 + 2x2 )

∫ 1√ 6. Evaluate (9 − 4x2 ) dx

5 = 2

0

Problem 18. Determine

(a2

∫

2 0

2

1 dx (4 + x2 )

1 dx (4 + x2 ) 0 [ 1 x ]2 = tan−1 since a = 2 2 2 0 ) 1 1 (π = (tan−1 1 − tan−1 0) = −0 2 2 4 π = or 0.3927 8

From Problem 18,

1

0

1 0

5 dx 2[(3/2) + x2 ]

1 √ dx [ (3/2)]2 + x2

√( ) [ √( ) ] 2 2 −1 −1 tan − tan 0 3 3

= (2.0412)[0.6847 − 0] = 1.3976, correct to 4 decimal places.

Now try the following Practice Exercise Practice Exercise 193 Integration using the tan θ substitution (Answers on page 890) ∫ 3 1. Determine dt 4 + t2 ∫ 5 2. Determine dθ 16 + 9θ2 ∫ 3.

1

3 dt 1 + t2

3

5 dx 4 + x2

Evaluate 0

∫ 4.

Problem 19. Evaluate ∫

5 = 2

1 dx + x2 )

dx Letx = a tan θ then = a sec2 θ and dx = a sec2 θ dθ dθ ∫ 1 Hence dx (a2 + x2 ) ∫ 1 = (a sec2 θ dθ) 2 (a + a2 tan2 θ) ∫ a sec2 θ dθ = a2 (1 + tan2 θ) ∫ a sec2 θ dθ = , since 1+tan2 θ = sec2 θ a2 sec2 θ ∫ 1 1 = dθ = (θ) + c a a x Since x = a tan θ, θ = tan−1 a ∫ 1 1 x Hence dx = tan−1 + c 2 2 (a + x ) a a

∫

∫

5 dx, correct to (3 + 2x2 )

[ ]1 5 1 x −1 √ = tan √ 2 (3/2) (3/2) 0

39.6 Worked problems on integration using the tan θ substitution ∫

1

Problem 20. Evaluate

Evaluate 0

39.7 Worked problems on integration using the sinh θ substitution ∫ Problem 21. Determine

1 √ dx 2 (x + a2 )

dx Let x = a sinh θ, then = a cosh θ and dθ dx = a cosh θ dθ

Integration using trigonometric and hyperbolic substitutions 475 ∫ Hence

1 √ dx (x2 + a2 ) ∫ 1 = √ (a cosh θ dθ) 2 (a sinh2 θ + a2 ) ∫ a cosh θ dθ = √ (a2 cosh2 θ)

since cosh2 θ − sinh2 θ = 1 ∫ ∫ a cosh θ = dθ = dθ = θ + c a cosh θ x = sinh−1 + c, since x = a sinh θ a It is shown on page 391 that { sinh

−1

x = ln a

x+

} √ (x2 + a2 ) a

which provides an alternative solution to ∫ √

1 (x2

+ a2 )

dx ∫

2

Problem 22. Evaluate 0

√

1 (x2

+ 4)

dx, correct

to 4 decimal places. ∫ 0

2

[ 1 x ]2 √ dx = sinh−1 or 2 0 (x2 + 4) [ { }]2 √ x + (x2 + 4) ln 2 0

from Problem 21, where a = 2 Using the logarithmic form, ∫ 2 1 √ dx 2 (x + 4) 0 [ ( ( √ )] √ ) 2+ 8 0+ 4 = ln − ln 2 2 = ln 2.4142 − ln 1 = 0.8814, correct to 4 decimal places. ∫

2

2 √ dx, 1 x2 (1 + x2 ) correct to 3 significant figures.

Since the integral contains a term of the form √ (a2 + x2 ), then let x = sinh θ, from which dx = cosh θ and dx = cosh θ dθ dθ ∫ 2 √ Hence dx 2 x (1 + x2 ) ∫ 2(cosh θ dθ) √ = 2 sinh θ (1 + sinh2 θ) ∫ cosh θ dθ =2 sinh2 θ cosh θ since cosh2 θ − sinh2 θ = 1 ∫ ∫ dθ =2 = 2 cosech2 θ dθ sinh2 θ = −2 coth θ + c √ √ cosh θ (1 + sinh2 θ) (1 + x2 ) coth θ = = = sinh θ sinh θ x ∫ 2 2 √ Hence dx 2 1 + x2 ) 1 x [√ ]2 (1 + x2 ) 2 = −[2 coth θ]1 = −2 x 1 [√ √ ] 5 2 = −2 − = 0.592, 2 1 correct to 3 significant figures Problem 24. Find

∫ √ (x2 + a2 ) dx

dx Let x = a sinh θ then = a cosh θ and dθ dx = a cosh θ dθ ∫ √ Hence (x2 + a2 ) dx ∫ √ (a2 sinh2 θ + a2 )(a cosh θ dθ) ∫ √ = [a2 (sinh2 θ + 1)](a cosh θ dθ) ∫ √ = (a2 cosh2 θ) (a cosh θ dθ) =

since cosh2 θ − sinh2 θ = 1 ∫ ∫ = (a cosh θ)(a cosh θ) dθ = a2 cosh2 θ dθ

Problem 23. Evaluate

∫ ( 2

=a

) 1 + cosh 2θ dθ 2

476 Section H (

)

=

a2 2

=

a2 [θ + sinh θ cosh θ] + c, 2

θ+

sinh 2θ 2

39.8 Worked problems on integration using the cosh θ substitution

+c

∫

since sinh 2θ = 2 sinh θ cosh θ Since x = a sinh θ, then sinh θ =

x x and θ = sinh−1 a a

Also since cosh2 θ − sinh2 θ = 1 √ then cosh θ = (1 + sinh2 θ) √[ =

Hence

a

a 2

[ sinh−1

a2 + x2 a2

( x ) √(x2 + a2 )

x + a a

x√

2

=

=

)

√ (a2 + x2 ) = a

∫ √ (x2 + a2 ) dx 2

=

1+

√(

( x )2 ]

a x sinh−1 + 2 a 2

Problem 25. Determine

dx Let x = a cosh θ then = a sinh θ and dθ dx = a sinh θ dθ ∫ 1 Hence √ dx (x2 − a2 ) ∫ 1 = √ (a sinh θ dθ) 2 (a cosh2 θ − a2 ) ∫ a sinh θ dθ = √ [a2 (cosh2 θ − 1)] ∫ a sinh θ dθ = √ (a2 sinh2 θ)

]

a

since cosh2 θ − sinh2 θ = 1 ∫

+c

(x2 + a2 ) + c

a sinh θ dθ = a sinh θ

=

{ x = ln a

4. Find

∫

3

5. Evaluate 0

6. Evaluate

cosh

+ 25) dt

√

4 (t2

+ 9)

dt

∫ 1√ (16 + 9θ2 ) dθ 0

x+

} √ (x2 − a2 ) a

which provides as alternative solution to ∫ √

1 (x2

− a2 )

dx ∫

Problem 26. Determine ∫

(4t2

dθ = θ + c

x + c, since x = a cosh θ a It is shown on page 391 that −1

Practice Exercise 194 Integration using the sinh θ substitution (Answers on page 890) ∫ 2 1. Find √ dx 2 (x + 16) ∫ 3 2. Find √ dx (9 + 5x2 ) ∫ √ 3. Find (x2 + 9) dx

∫

= cosh−1

Now try the following Practice Exercise

∫ √

1 √ dx 2 (x − a2 )

2x − 3 √ dx = (x2 − 9)

∫

2x − 3 √ dx (x2 − 9)

2x √ dx (x2 − 9) ∫ 3 − √ dx 2 (x − 9)

The first integral is determined using the algebraic sub2 stitution ∫ u = (x − 9), and the second integral is of the 1 form √ dx (see Problem 25) (x2 − a2 )

Integration using trigonometric and hyperbolic substitutions 477 ∫ √

Hence

∫

2x (x2 − 9)

dx −

3 √ dx (x2 − 9)

√ x = 2 (x2 − 9) − 3 cosh−1 + c 3 Problem 27.

∫ √ (x2 − a2 ) dx

dx Let x = a cosh θ then = a sinh θ and dθ dx = a sinh θ dθ ∫ √ Hence (x2 − a2 ) dx

Hence

∫ √ (x2 − a2 ) dx

a2 = 2 =

[√ ] (x2 − a2 ) ( x ) −1 x − cosh +c a a a

x x√ 2 a2 cosh−1 + c (x − a2 ) − 2 2 a

Problem 28. Evaluate

2

[ ]3 ∫ 3√ x√ 2 4 −1 x 2 (x − 4) dx = (x − 4) − cosh 2 2 2 2 2

∫ √ (a2 cosh2 θ − a2 ) (a sinh θ dθ) =

=

( =

∫ √ [a2 (cosh2 θ − 1)] (a sinh θ dθ)

∫ √ = (a2 sinh2 θ) (a sinh θ dθ) ∫ (

∫ = a2

sinh2 θ dθ = a2

) cosh 2θ − 1 dθ 2

∫ 3√ (x2 − 4) dx

Since

from Problem 27, when a = 2, ) 3√ −1 3 5 − 2 cosh 5 2

− (0 − 2 cosh−1 1) } { √ 2 − a2 ) (x x x + then cosh−1 = ln a a { } √ 2 − 22 ) (3 3 3 + cosh−1 = ln 2 2 = ln 2.6180 = 0.9624 −1

since cosh 2θ = 1 + 2 sinh2 θ from Table 12.1, page 156, [ ] a2 sinh 2θ = −θ +c 2 2

Similarly, cosh 1 = 0 ∫ 3√ Hence (x2 − 4) dx 2

=

[

] 3√ 5 − 2(0.9624) − [0] 2

= 1.429, correct to 4 significant figures.

2

=

a [sinh θ cosh θ − θ] + c, 2 since sinh 2θ = 2 sinh θ cosh θ

x Since x = a cosh θ then cosh θ = and a −1 x θ = cosh a Also, since cosh2 θ − sinh2 θ = 1, then sinh θ =

√ (cosh2 θ − 1)

√[ ] √ 2 ( x )2 (x − a2 ) = −1 = a a

Now try the following Practice Exercise Practice Exercise 195 Integration using the cosh θ substitution (Answers on page 890) ∫ 1 1. Find √ dt (t2 − 16) ∫ 3 2. Find √ dx (4x2 − 9) ∫ √ 3. Find (θ2 − 9) dθ

478 Section H

4. Find

∫ √

(4θ2 − 25) dθ

∫

2

5. Evaluate 1

6. Evaluate

√

2 (x2

− 1)

3. dx

∫ 3√ (t2 − 4) dt 2

Practice Exercise 196 Multiple-choice questions on integration using trigonometric and hyperbolic substitutions (Answers on page 890) Each question has only one correct answer ∫ π/3 1. The value of 4 cos 2 2θ dθ, correct to 3 0

decimal places, is: (a) 1.661 (b) 3.464 (c) −1.167 (d) −2.417 ∫ π/2 2. 2 sin 3 t dt is equal to: 0

(a) 1.33

(b) −0.25

(c) −1.33

4.

∫ √ (

) 4 − x2 dx is equal to: √( ) 2 4 − x2 −1 x + +c (a) 2 cos 2 x √ ) x x ( (b) 2 sin −1 + 4 − x2 + c 2 2 x −1 x + +c (c) 2 sin 2 2√ ( ) x (d) 4 sin −1 + x 4 − x2 + c 2 ∫ π/2 Evaluating 4 cos 6x sin 3x dx, correct to The integral

0

5.

3 decimal places, gives: (a) −0.888 (b) 4 (c) −0.444 (d) 0.888 ∫ 3 5 dx, correct to 3 deciThe value of 2 0 9+x mal places, is: (a) 2.596 (b) 75.000 (c) 0.536 (d) 1.309

(d) 0.25

For fully worked solutions to each of the problems in Practice Exercises 189 to 195 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 40

Integration using partial fractions Why it is important to understand: Integration using partial fractions Sometimes expressions which at first sight look impossible to integrate using standard techniques may in fact be integrated by first expressing them as simpler partial fractions and then using earlier learned techniques. As explained in Chapter 2, the algebraic technique of resolving a complicated fraction into partial fractions is often needed by electrical and mechanical engineers for not only determining certain integrals in calculus, but for determining inverse Laplace transforms, and for analysing linear differential equations like resonant circuits and feedback control systems.

At the end of this chapter, you should be able to: • integrate functions using partial fractions with linear factors • integrate functions using partial fractions with repeated linear factors • integrate functions using partial fractions with quadratic factors

40.1

Introduction

The process of expressing a fraction in terms of simpler fractions – called partial fractions – is discussed in Chapter 2, with the forms of partial fractions used being summarised in Table 2.1, page 18. Certain functions have to be resolved into partial fractions before they can be integrated as demonstrated in the following worked problems.

40.2

Integration using partial fractions with linear factors ∫

Problem 1.

Determine

11 − 3x x2 + 2x − 3

dx

As shown in Problem 1, page 18: 11 − 3x 2 5 ≡ − x2 + 2x − 3 (x − 1) (x + 3) ∫ 11 − 3x Hence dx x2 + 2x − 3 ∫ { =

5 2 − (x − 1) (x + 3)

} dx

= 2 ln(x − 1) − 5 ln(x + 3) + c (by algebraic substitutions — see Chapter 38) { } (x − 1)2 or ln + c by the laws of logarithms (x + 3)5

480 Section H

Problem 2.

∫

By dividing out and resolving into partial fractions it was shown in Problem 4, page 19:

Find 2x2 − 9x − 35 dx (x + 1)(x − 2)(x + 3)

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

It was shown in Problem 2, page 18:

2

2x2 − 9x − 35 4 3 1 ≡ − + (x + 1)(x − 2)(x + 3) (x + 1) (x − 2) (x + 3) ∫ Hence ∫ { ≡

≡

∫ 3{ x−3+ 2

2x2 − 9x − 35 dx (x + 1)(x − 2)(x + 3) 4 3 1 − + (x + 1) (x − 2) (x + 3)

x3 − 2x2 − 4x − 4 dx x2 + x − 2

3

Hence

[ =

} dx

x2 − 3x + 4 ln(x + 2) − 3 ln(x − 1) 2

( =

= 4 ln(x + 1) − 3 ln(x − 2) + ln(x + 3) + c { } (x + 1)4 (x + 3) or ln +c (x − 2)3

} 4 3 − dx (x + 2) (x − 1)

9 − 9 + 4 ln 5 − 3 ln 2 2

]3 2

)

− (2 − 6 + 4 ln 4 − 3 ln 1) = −1.687, correct to 4 significant figures.

by the laws of logarithms ∫ Problem 3.

Determine

Now try the following Practice Exercise 2

x +1 dx x2 − 3x + 2

By dividing out (since the numerator and denominator are of the same degree) and resolving into partial fractions it was shown in Problem 3, page 19: x2 + 1 2 5 ≡ 1− + 2 x − 3x + 2 (x − 1) (x − 2) ∫

x2 + 1 dx x2 − 3x + 2 } ∫ { 2 5 ≡ 1− + dx (x − 1) (x − 2)

Hence

= x − 2 ln(x − 1) + 5 ln(x − 2) + c { } (x − 2)5 or x + ln +c (x − 1)2

Practice Exercise 197 Integration using partial fractions with linear factors (Answers on page 890) In Problems 1 to 5, integrate with respect to x. ∫ 12 1. dx (x2 − 9) ∫ 4(x − 4) 2. dx 2 (x − 2x − 3) ∫ 3(2x2 − 8x − 1) 3. dx (x + 4)(x + 1)(2x − 1) ∫ 2 x + 9x + 8 4. dx x2 + x − 6 ∫ 5.

In Problems 6 and 7, evaluate the definite integrals correct to 4 significant figures.

by the laws of logarithms Problem 4.

Evaluate ∫ 3 3 x − 2x2 − 4x − 4 dx, x2 + x − 2 2

correct to 4 significant figures.

3x3 − 2x2 − 16x + 20 dx (x − 2)(x + 2)

∫ ∫

x2 − 3x + 6 dx x(x − 2)(x − 1)

4

6. 3

6

7. 4

x2 − x − 14 dx x2 − 2x − 3

Integration using partial fractions 481

8. Determine the value of k, given that: ∫

1

0

(x − k) dx = 0 (3x + 1)(x + 1)

9. The velocity constant k of a given chemical reaction is given by: ) ∫ ( 1 dx kt = (3 − 0.4x)(2 − 0.6x)

∫ Problem 6.

{

2(3 − 0.4x) 3(2 − 0.6x)

2 3 4 5x2 − 2x − 19 ≡ + − 2 (x + 3)(x − 1) (x + 3) (x − 1) (x − 1)2 ∫ Hence

}

10. The velocity v of an object in a medium at time t seconds is given by: ∫ 80 dv t= v(2v − 1) 20 Evaluate t, in milliseconds, correct to 2 decimal places.

5x2 − 2x − 19 dx (x + 3)(x − 1)2 } ∫ { 2 3 4 ≡ + − dx (x + 3) (x − 1) (x − 1)2

= 2 ln (x + 3) + 3 ln (x − 1) +

Integration using partial fractions with repeated linear factors ∫

Problem 5.

Determine

2x + 3 dx (x − 2)2

It was shown in Problem 5, page 20:

∫ 

= 2 ln(x − 2) −

7 +c (x − 2)

 7 dx is determined using the algebraic  (x − 2)2

substitution u = (x − 2) – see Chapter 38.

Evaluate ∫ 1 2 3x + 16x + 15 dx, (x + 3)3 −2

correct to 4 significant figures. It was shown in Problem 7, page 21: 3 2 6 3x2 + 16x + 15 ≡ − − (x + 3)3 (x + 3) (x + 3)2 (x + 3)3 ∫

2x + 3 2 7 ≡ + (x − 2)2 (x − 2) (x − 2)2 } ∫ ∫ { 2x + 3 7 2 Thus dx ≡ + dx (x − 2)2 (x − 2) (x − 2)2

4 +c (x − 1)

using an algebraic substitution, u = (x − 1), for the third integral { } 4 or ln (x + 3)2 (x − 1)3 + +c (x − 1) by the laws of logarithms Problem 7.

40.3

5x2 − 2x − 19 dx. (x + 3)(x − 1)2

It was shown in Problem 6, page 20:

where x = 0 when t = 0. Show that: kt = ln

Find

Hence

3x2 + 16x + 15 dx (x + 3)3 } ∫ 1{ 3 2 6 ≡ − − dx (x + 3)2 (x + 3)3 −2 (x + 3) ]1 [ 2 3 = 3 ln(x + 3) + + (x + 3) (x + 3)2 −2 ( ) ( ) 2 3 2 3 = 3 ln 4 + + − 3 ln 1 + + 4 16 1 1 = −0.1536, correct to 4 significant figures.

482 Section H Now try the following Practice Exercise Practice Exercise 198 Integration using partial fractions with repeated linear factors (Answers on page 891) In Problems 1 and 2, integrate with respect to x. ∫ 4x − 3 dx 1. (x + 1)2 ∫ 5x2 − 30x + 44 2. dx (x − 2)3 In Problems 3 and 4, evaluate the definite integrals correct to 4 significant figures. ∫ 2 2 x + 7x + 3 dx 3. x2 (x + 3) 1 ∫ 7 18 + 21x − x2 4. dx 2 6 (x − 5)(x + 2) ) ∫ 1( 2 4t + 9t + 8 5. Show that dt = 2.546, (t + 2)(t + 1)2 0 correct to 4 significant figures.

3 x = √ tan−1 √ , from 12, Table 39.1, page 468. 3 3 ∫ 4x dx is determined using the algebraic substitux2 + 3 tion u = (x2 + 3) } ∫ { 2 1 3 4x Hence + + − dx x x2 (x2 + 3) (x2 + 3) 1 3 x + √ tan−1 √ − 2 ln(x2 + 3) + c x 3 3 ( )2 √ x 1 x = ln 2 − + 3 tan−1 √ + c x +3 x 3 = 2 ln x −

by the laws of logarithms ∫ Problem 9.

Let

Determine

1 dx (x2 − a2 )

1 A B ≡ + (x2 − a2 ) (x − a) (x + a) ≡

A(x + a) + B(x − a) (x + a)(x − a)

Equating the numerators gives: 1 ≡ A(x + a) + B(x − a)

40.4

Integration using partial fractions with quadratic factors ∫

Problem 8.

Find

3 + 6x + 4x2 − 2x3 dx. x2 (x2 + 3)

It was shown in Problem 9, page 22: 3 + 6x + 4x2 − 2x3 2 1 3 − 4x ≡ + 2+ 2 2 2 x (x + 3) x x (x + 3) ∫ 2 3 3 + 6x + 4x − 2x Thus dx x2 (x2 + 3) ) ∫ ( 2 1 (3 − 4x) ≡ + 2+ 2 dx x x (x + 3) ∫ { = ∫

2 1 3 4x + + − x x2 (x2 + 3) (x2 + 3)

3 dx = 3 (x2 + 3)

∫

1 Let x = a, then A = , and let x = −a, then 2a 1 B=− 2a ∫ 1 Hence dx 2 (x − a2 ) [ ] ∫ 1 1 1 ≡ − dx 2a (x − a) (x + a) 1 [ln(x − a) − ln(x + a)] + c 2a ( ) x−a 1 ln = +c 2a x+a =

by the laws of logarithms

} dx

Problem 10. Evaluate ∫ 4 3

1 √ dx 2 x + ( 3)2

(x2

3 dx, − 4)

correct to 3 significant figures.

Integration using partial fractions 483 From Problem 9, [ ( )]4 ∫ 4 3 1 x−2 dx = 3 ln 2 2(2) x+2 3 3 (x − 4) [ ] 3 2 1 = ln − ln 4 6 5 =

3 5 ln = 0.383, correct to 3 4 3 significant figures. ∫

Problem 11. Determine

1 dx (a2 − x2 )

Now try the following Practice Exercise Practice Exercise 199 Integration using partial fractions with quadratic factors (Answers on page 891) ∫ 1.

Determine

In Problems 2 to 4, evaluate the definite integrals correct to 4 significant figures. ∫

Using partial fractions, let 1 1 A B ≡ ≡ + 2 2 (a − x ) (a − x)(a + x) (a − x) (a + x) ≡

A(a + x) + B(a − x) (a − x)(a + x)

Then 1 ≡ A(a + x) + B(a − x) 1 1 Let x = a then A = . Let x = −a then B = 2a 2a ∫ 1 Hence dx 2 (a − x2 ) [ ] ∫ 1 1 1 ≡ + dx 2a (a − x) (a + x) 1 [−ln(a − x) + ln(a + x)] + c 2a ( ) 1 a+x = ln +c 2a a−x

x2 − x − 13 dx (x2 + 7)(x − 2)

6

6x − 5 dx (x − 4)(x2 + 3)

2

4 dx (16 − x2 )

2. 5

∫ 3. 1

∫ 4. 4

5

(x2

2 dx − 9) ∫ 2(

5.

Show

that 1

2 + θ + 6θ2 − 2θ3 θ2 (θ2 + 1)

) dθ

= 1.606, correct to 4 significant figures.

=

Problem 12. Evaluate ∫ 2 0

5 dx, (9 − x2 )

correct to 4 decimal places. From Problem 11, ∫

2 0

[ ( )]2 5 1 3+x dx = 5 ln (9 − x2 ) 2(3) 3−x 0 [ ] 5 5 = ln − ln 1 6 1 = 1.3412, correct to 4 decimal places.

Practice Exercise 200 Multiple-choice questions on integration using partial fractions (Answers on page 891) Each question has only one correct answer ∫ 4 3x + 4 ( )( ) dx is equal to: 1. x−2 x+3 3 (a) 1.540 (b) 1.001 (c) 1.232 (d) 0.847 ∫ 6 8 2. dx is equal to: 2 − 16 x 5 (a) −0.588 (b) 6.388 (c) 0.588 (d) −0.748 ∫ 3 3 3. dx is equal to: 2+x−2 x 2 1 (a) 3 ln 2.5 (b) lg 1.6 3 (c) ln 40 (d) ln 1.6

484 Section H ( ) 2 x−2 4. ( )2 dx is equal to: 4 x−3 (a) 0.386 (b) 2.386 (c) -2.500 ∫

∫

5

5. (d) 1.000

5

1 dx is equal to: 2 −9 x 4 (a) −0.0933 (b) 0.827 (c) 0.0933 (d) −0.827

For fully worked solutions to each of the problems in Practice Exercises 197 to 199 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 41

The t = tan 2θ substitution θ substitution 2 Sometimes, with an integral containing sin θ and/or cos θ, it is possible, after making a substitution θ t = tan , to obtain an integral which can be determined using partial fractions. This is explained in this 2 chapter where we continue to build the picture of integral calculus, each step building from the previous. A simple substitution can make things so much easier. Why it is important to understand: The t = tan

At the end of this chapter, you should be able to: θ • develop formulae for sin θ, cos θ and dθ in terms of t, where t = tan 2 θ • integrate functions using t = tan substitution 2

41.1

Introduction ∫

1 dθ, where a cos θ + b sin θ + c a, b and c are constants, may be determined by using the θ substitution t = tan . The reason is explained below. 2 If angle A in the right-angled triangle ABC shown in θ Fig. 41.1 is made equal to then, since tangent = 2 θ opposite , if BC = t and AB = 1, then tan = t adjacent 2

Integrals of the form

!1 1 t 2

A

Figure 41.1

u 2 1

√ By Pythagoras’ theorem, AC = 1 + t2 θ θ t 1 Therefore sin = √ and cos = √ Since 2 2 2 1+t 1 + t2 sin 2x = 2 sin x cos x (from double angle formulae, Chapter 15), then θ θ sin θ = 2 sin cos 2 2 ( )( ) t 1 √ =2 √ 1 + t2 1 + t2

C

i.e.

t

Since

B

then

sin θ =

2t (1 + t2 )

(1)

cos 2x = cos2 x − sin2 x cos θ = cos2 ( =

θ θ − sin2 2 2

1 √ 1 + t2

)2

( −

t √ 1 + t2

)2

486 Section H i.e.

cos θ =

1−t2 1+t2

(2)

∫ Thus

θ Also, since t = tan , 2 ( ) 1 dt 1 2θ 2θ = sec = 1 + tan from trigonometric dθ 2 2 2 2 identities, dt 1 = (1 + t2 ) dθ 2

i.e.

dx = cos x =

∫ ∫

1 ( ) 2 dt 1 − t2 1 + t2 1 + t2 2 dt 1 − t2

2 may be resolved into partial fractions (see 1 − t2 Chapter 2). 2 2 = 1 − t2 (1 − t)(1 + t)

Let (3)

=

A B + (1 − t) (1 + t)

Equations (1), (2) and ∫ (3) are used to determine 1 dθ where integrals of the form a cos θ + b sin θ + c a, b or c may be zero.

=

A(1 + t) + B(1 − t) (1 − t)(1 + t)

dθ =

from which,

41.2

2dt 1+t2

Worked problems on the θ t = tan substitution 2 ∫

Problem 1.

Determine

θ 2t 2 dt then sin θ = and dθ = from 2 2 1+t 1 + t2 equations (1) and (3). ∫

Hence

∫

1 dθ sin θ 1 ( ) ∫ 2 dt 2t = 1 + t2 1 + t2 ∫ 1 = dt = ln t + c t ( ) ∫ θ dθ = ln tan +c sinθ 2 dθ = sin θ

∫ Problem 2.

Determine

dx cos x

x 1 − t2 2 dt then cos x = and dx = from 2 2 1+t 1 + t2 equations (2) and (3). If t = tan

When

t = 1, 2 = 2A, from which, A = 1

When

t = −1, 2 = 2B, from which, B = 1 ∫ ∫ 1 1 2 dt = + dt 1 − t2 (1 − t) (1 + t)

Hence

= −ln(1 − t) + ln(1 + t) + c } { (1 + t) +c = ln (1 − t)

dθ sin θ

If t = tan

Thus

2 = A(1 + t) + B(1 − t)

Hence

 x   1 + tan dx 2 +c Thus = ln x  1 − tan  cos x 2 π Note that since tan = 1, the above result may be 4 written as:  π x  ∫  tan + tan  dx 4 2 = ln +c  1 − tan π tan x  cos x 4 2 { ( π x )} + +c = ln tan 4 2 ∫

from compound angles, Chapter 15. ∫ Problem 3.

Determine

dx 1 + cos x

x 1 − t2 2 dt then cos x = and dx = from 2 2 1+t 1 + t2 equations (2) and (3). If t = tan

The t = tan θ2 substitution 487 ∫ Thus

∫ 1 dx = dx 1 + cos x 1 + cos x ( ) ∫ 2 dt 1 = 1 − t2 1 + t2 1+ 1 + t2 ( ) ∫ 2 dt 1 = (1 + t2 ) + (1 − t2 ) 1 + t2 1 + t2 ∫ =

∫

dx 1 − cos x + sin x

2. ∫

dα 3 + 2 cos α

3. ∫

dx 3 sin x − 4 cos x

4.

dt

∫

x dx = t + c = tan + c 1 + cos x 2 ∫ dθ Problem 4. Determine 5 + 4 cos θ

41.3

Further worked problems on the θ t = tan substitution 2

Hence

θ 1−t 2 dt then cos θ = and dθ = 2 1 + t2 1 + t2 from equations (2) and (3). ) ( 2 dt ∫ ∫ dθ 1 + t2 ( ) Thus = 5 + 4 cos θ 1 − t2 5+4 1 + t2 ) ( 2 dt ∫ 1 + t2 = 2 5(1 + t ) + 4(1 − t2 ) (1 + t2 ) ∫ ∫ dt dt =2 = 2 2 2 t +9 t + 32 ( ) 1 −1 t =2 tan + c, 3 3 2

∫ Problem 5.

Determine

If t = tan

If t = tan dx=

dx sin x + cos x

x 2t 1 − t2 then sin x = , cos x = and 2 2 1+t 1 + t2

2 dt from equations (1), (2) and (3). 1 + t2

Thus ∫

dx = sin x + cos x

∫ (

2t 1 + t2

2 dt 2 1 + t( ) +

1 − t2 1 + t2

)

2 dt ∫ 2 dt 1 + t2 = = 1 + 2t − t2 2t + 1 − t2 1 + t2 ∫ ∫ −2 dt −2 dt = = 2 t − 2t − 1 (t − 1)2 − 2 ∫ 2 dt √ = 2 ( 2) − (t − 1)2 [ {√ }] 1 2 + (t − 1) = 2 √ ln √ +c 2 2 2 − (t − 1) ∫

from 12 of Table 39.1, page 468. Hence ( ) ∫ dθ 2 1 θ = tan−1 tan +c 5 + 4 cos θ 3 3 2 Now try the following Practice Exercise

(see Problem 11, Chapter 40, page 483), θ Practice Exercise 201 The t = tan 2 substitution (Answers on page 891) Integrate the following with respect to the variable: ∫ dθ 1. 1 + sin θ

∫ i.e.

dx sin x + cos x √ x  2 − 1 + tan  1 2 +c = √ ln √ 2  2 + 1 − tan x  2

488 Section H ∫Problem 6. Determine dx 7 − 3 sin x + 6 cos x

1 =− 2

dt 3 t2 − t − 1 2 ∫ 1 dt =− )2 ( 2 25 3 − t− 4 16 ∫ 1 dt = )2 ( )2 ( 2 5 3 − t− 4 4 ( )    3  5    + t − 4 1 1 4  +c  ) ( ) ( =  ln   5 3  5 2     − t− 2 4 4 4

From equations (1) and (3), ∫

dx 7 − 3 sin x + 6 cos x 2 dt 1 + t2 ( ( ) ) 2t 1 − t2 7−3 +6 1 + t2 1 + t2

∫ =

2 dt 1 + t2 = 2 7(1 + t ) − 3(2t) + 6(1 − t2 ) 1 + t2 ∫ 2 dt = 7 + 7t2 − 6t + 6 − 6t2 ∫ ∫ 2 dt 2 dt = = t2 − 6t + 13 (t − 3)2 + 22 [ ( )] 1 −1 t − 3 = 2 tan +c 2 2 ∫

from Problem 11, Chapter 40, page 483   1     + t 1 2 = ln +c 5   2−t   ∫ Hence

from 12, Table 39.1, page 468. Hence ∫

dx 7 − 3 sin x + 6 cos x   x tan − 3 2 +c = tan−1  2 ∫

Problem 7.

Determine

dθ 4 cos θ + 3 sin θ

From equations (1) to (3), ∫

∫

or

dθ 4 cos θ + 3 sin θ

  1 θ    + tan 1 2 +c = ln 2 5   2 − tan θ   2   θ    1 + 2 tan 1 2 +c ln 5   4 − 2 tan θ   2

Now try the following Practice Exercise θ 2 substitution (Answers on page 891) Practice Exercise 202

dθ 4 cos θ + 3 sin θ 2 dt 2 1 ( )+ t ( ) = 2 1−t 2t 4 +3 1 + t2 1 + t2 ∫ ∫ dt 2 dt = = 2 4 − 4t + 6t 2 + 3t − 2t2 ∫

The t = tan

In Problems 1 to 4, integrate with respect to the variable. ∫ dθ 1. 5 + 4 sin θ ∫ dx 2. 1 + 2 sin x

The t = tan θ2 substitution 489 ∫ 3. ∫ 4.

dp 3 − 4 sin p + 2 cos p

∫ 6.

0

3 dθ = 3.95, correct to 3 sigcos θ

nificant figures.

dθ 3 − 4 sin θ 7.

5. Show that } {√ ∫ 2 + tan 2t 1 dt +c = √ In √ 1 + 3 cos t 2 2 2 − tan 2t

π/3

Show that

Show that ∫ 0

π/2

dθ π = √ 2 + cos θ 3 3

For fully worked solutions to each of the problems in Practice Exercises 201 and 202 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 12

Maclaurin’s series and further integration

This Revision Test covers the material contained in Chapters 37 to 41. The marks for each question are shown in brackets at the end of each question. 1.

Use Maclaurin’s series to determine a power series for e2x cos 3x as far as the term in x2 . (10)

2.

Show, using Maclaurin’s series, that the first four terms of the power series for cosh 2x is given by: 4 2 cosh 2x = 1 + 2x2 + x4 + x6 3 45

3.

0

4.

5.

6.

Determine the following integrals: ∫ ∫ 3 ln x 7 (a) 5(6t + 5) dt (b) dx x ∫ 2 (c) √ dθ (2θ − 1) Evaluate the following definite integrals: ∫ π ∫ 1 ( 2 2 π) (a) 2 sin 2t + dt (b) 3x e4x −3 dx 3 0 0 Determine the following integrals: ∫ ∫ 2 dx (a) cos3 x sin2 x dx (b) √ (9 − 4x2 ) ∫ 2 (c) √ dx (4x2 − 9)

Evaluate the following definite integrals, correct to 4 significant figures: ∫ π3 ∫ π2 3 sin2 t dt (b) 3 cos 5θ sin 3θ dθ (a) 0

∫ (10)

(c) 0

Expand the function x2 ln(1 + sin x) using Maclaurin’s series and hence evaluate: ∫ 12 x2 ln(1 + sin x) dx correct to 2 significant figures.

7.

8.

(13) 9.

(10)

10. 11.

0 2

5 dx 4 + x2

(14)

Determine: ∫ x − 11 (a) dx x2 − x − 2 ∫ 3−x (b) dx (19) (x2 + 3)(x + 3) ∫ 2 3 Evaluate dx correct to 4 significant 2 (x + 2) x 1 figures. (11) ∫ dx Determine: (7) 2 sin x + cos x ∫ π2 dx Evaluate correct to 3 decimal π 3 − 2 sin x 3 places. (9)

(10)

(12)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 12, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Chapter 42

Integration by parts Why it is important to understand: Integration by parts Integration by parts is a very important technique that is used often in engineering and science. It is frequently used to change the integral of a product of functions into an ideally simpler integral. It is the foundation for the theory of differential equations and is used with Fourier series. We have looked at standard integrals followed by various techniques to change integrals into standard ones; integration by parts is a particularly important technique for integrating a product of two functions.

At the end of this chapter, you should be able to: • appreciate when integration by parts is required • integrate functions using integration by parts • evaluate definite integrals using integration by parts

42.1

Introduction

From the product rule of differentiation: d du dv (uv) = v + u , dx dx dx where u and v are both functions of x. dv d du Rearranging gives: u = (uv) − v dx dx dx Integrating both sides with respect to x gives: ∫

dv u dx = dx

∫

∫ i.e. or

d (uv) dx − dx

∫ v

du dx dx

∫ dv du u dx = uv− v dx dx dx ∫ ∫ u dv = uv − v du

This is known as the integration by parts formula and provides a method of integrating ∫ x such ∫ products of simple functions as xe dx, t sin t dt, ∫ θ ∫ e cos θ dθ and x ln x dx. Given a product of two terms to integrate the initial choice is: ‘which part to make equal to u’ and ‘which part to make equal to v’. The choice must be such that the ‘u part’ becomes a constant after successive differentiation and the ‘dv part’ can be integrated from standard integrals. Invariably, the following rule holds: If a product to be integrated contains an algebraic term (such as x, t2 or 3θ) then this term is chosen as the u part. The one exception to this rule is when a ‘ln x’ term is involved; in this case ln x is chosen as the ‘u part’.

42.2 Worked problems on integration by parts Problem 1.

Determine

∫

x cos x dx

492 Section H ∫

From the integration by parts formula, ∫ ∫ u dv = uv − v du

Problem 3.

du = 1, i.e. du = dx and let dx ∫ dv = cos x dx, from which v = cos x dx = sin x. Expressions for u, du and v are now substituted into the ‘by parts’ formula as shown below. dv

5

x cos x dx

u

v

(x) (sin x)

5

2 2

v

du

(sin x) (dx)

∫ sin θ dθ = −cos θ

v=

Substituting into

∫

u dv = uv −

∫

∫

v du gives:

∫ 2θ sin θ dθ = (2θ)(−cos θ) −

(−cos θ)(2 dθ)

∫ = −2θ cos θ + 2

x cos x dx = x sin x − (−cos x) + c

[This result may be checked by differentiating the righthand side,

cos θ dθ

= −2θ cos θ + 2 sin θ + c

= x sin x + cos x + c

i.e.

2θ sin θ dθ

du Let u = 2θ, from which, = 2, i.e. du = 2 dθ and let dθ dv = sin θ dθ, from which,

∫ i.e.

Evaluate 0

Let u = x, from which

u

π 2

∫

π 2

Hence

2θ sin θ dθ

0

d (x sin x + cos x + c) dx = [(x)(cos x) + (sin x)(1)] − sin x + 0 using the product rule

= [−2θ cos θ + 2 sin θ]02 [ (π ) π π] = −2 cos + 2 sin − [0 + 2 sin 0] 2 2 2

= x cos x, which is the function being integrated.]

= (−0 + 2) − (0 + 0) = 2

Problem 2.

Find

∫

since cos 3te2t dt

du Let u = 3t, from which, = 3, i.e. du = 3 dt and dt ∫ 1 let dv = e2t dt, from which, v = e2t dt = e2t 2 ∫ ∫ Substituting into u dv = uv − v du gives: ( ) ∫ ( ) ∫ 1 2t 1 2t 3te2t dt = (3t) e − e (3 dt) 2 2 ∫ 3 3 = te2t − e2t dt 2 2 ( ) 3 3 e2t = te2t − +c 2 2 2 Hence ∫ 2t

3t e dt =

3 2t 2e

π

( ) t − 12 + c,

which may be checked by differentiating.

∫ Problem 4.

π π = 0 and sin = 1 2 2

1

5xe4x dx, correct to

Evaluate 0

3 significant figures.

du = 5, i.e. du = 5 dx and dx ∫ let dv = e4x dx, from which, v = e4x dx = 14 e4x ∫ ∫ Substituting into u dv = uv − v du gives: Let u = 5x, from which

(

∫ 5xe4x dx = (5x)

e4x 4

)

5 5 = xe4x − 4 4

∫ ( −

∫

e4x 4

e4x dx

( ) 5 5 e4x = xe4x − +c 4 4 4 ( ) 5 1 = e4x x − +c 4 4

) (5 dx)

Integration by parts 493 ∫

1

Now try the following Practice Exercise

5xe4x dx

Hence 0

[ = [ =

Practice Exercise 203 Integration by parts (Answers on page 891)

( )]1 5 4x 1 e x− 4 4 0 ( )] [ ( )] 5 4 1 5 1 e 1− − e0 0 − 4 4 4 4

( =

Determine the integrals in Problems 1 to 5 using integration by parts. ∫ 1. xe2x dx ∫

) ( ) 15 4 5 e − − 16 16

2.

4x dx e3x

∫ = 51.186 + 0.313 = 51.499 = 51.5, correct to 3 significant figures Problem 5.

Determine

∫

x2 sin x dx

du Let u = x2 , from which, = 2x, i.e. du = 2x dx, and let dx dv = sin x dx, from which, ∫ v = sin x dx = −cos x ∫ ∫ Substituting into u dv = uv − v du gives: ∫ ∫ x2 sin x dx = (x2 )(−cos x) − (−cos x)(2x dx) [∫ = −x2 cos x + 2

] x cos x dx

3.

x sin x dx ∫

4.

5θ cos 2θ dθ ∫ 3t2 e2t dt

5.

Evaluate the integrals in Problems 6 to 9, correct to 4 significant figures. ∫ 2 6. 2xex dx 0

∫ 7.

= −x2 cos x + 2{x sin x + cos x} + c = −x2 cos x + 2x sin x + 2 cos x + c

x sin 2x dx

0

∫ 8.

∫ The integral, x cos x dx, is not a ‘standard integral’ and it can only be determined by using the integration by parts formula again. ∫ From Problem 1, x cos x dx = x sin x + cos x ∫ Hence x2 sin x dx

π 4

π 2

t2 cos t dt

0

∫

2

x

3x2 e 2 dx

9. 1

42.3 Further worked problems on integration by parts Problem 6.

Find

∫

x ln x dx

= (2 − x )cos x + 2x sin x + c 2

The logarithmic function is chosen as the ‘u part’. In general, if the algebraic term of a product is of power n, then the integration by parts formula is applied n times.

du 1 dx = , i.e. du = dx x x 2 ∫ x Letting dv = x dx gives v = x dx = 2 Thus when u = ln x, then

494 Section H ∫ ∫ Substituting into u dv = uv − v du gives: ( 2) ∫ ( 2) ∫ x x dx x ln x dx = (ln x) − 2 2 x ∫ 2 x 1 = ln x − x dx 2 2 ( ) x2 1 x2 = ln x − +c 2 2 2 ( ) ∫ x2 1 Hence x ln x dx = ln x − + c or 2 2

∫ 9 √ Hence x ln x dx 1

[ √ ( )]9 2 3 2 = x ln x − 3 3 1 [ √ ( )] [ √ ( )] 2 3 2 2 3 2 = 9 ln 9 − − 1 ln1 − 3 3 3 3 [ ( )] [ ( )] 2 2 2 = 18 ln 9 − − 0− 3 3 3

x2 (2 ln x − 1) + c 4 Problem 7.

Determine

∫

∫

= 27.550 + 0.444 = 27.994 = 28.0,

ln x dx

correct to 3 significant figures.

∫

ln x dx is the same as (1) ln x dx du 1 dx Let u = ln x, from which, = , i.e. du = dx x ∫ x and let dv = 1dx, from which, v = 1 dx = x ∫ ∫ Substituting into u dv = uv − v du gives: ∫ ∫ dx ln x dx = (ln x)(x) − x x ∫ = x ln x − dx = x ln x − x + c ∫ Hence ln x dx = x(ln x − 1) + c

cos bx dx =

x ln x dx, correct to

1

dx Let u = ln x, from which du = x 1 √ and let dv = x dx = x 2 dx, from which, ∫ 1 2 3 v = x 2 dx = x 2 3 ∫ ∫ Substituting into u dv = uv − v du gives: ( ) ∫ ( )( ) ∫ √ 2 3 2 3 dx x ln x dx = (ln x) x2 − x2 3 3 x ∫ √ 1 2 3 2 = x ln x − x 2 dx 3 3 ( ) 2√ 3 2 2 3 = x ln x − x2 + c 3 3 3 [ ] 2√ 3 2 = x ln x − +c 3 3

∫

∫

9√

Evaluate

3 significant figures.

Find

eax cos bx dx

When integrating a product of an exponential and a sine or cosine function it is immaterial which part is made equal to ‘u’ du Let u = eax , from which = aeax , dx i.e. du = aeax dx and let dv = cos bx dx, from which,

v=

∫ Problem 8.

Problem 9.

Substituting into

∫

1 sin bx b

u dv = uv −

∫

v du gives:

∫ eax cos bx dx (

) ∫ ( ) 1 1 = (e ) sin bx − sin bx (aeax dx) b b [∫ ] 1 ax a ax = e sin bx − e sin bx dx (1) b b ax

∫

eax sin bx dx is now determined separately using integration by parts again: Let u = eax then du = aeax dx, and let dv = sin bx dx, from which ∫ v=

1 sin bx dx = − cos bx b

Integration by parts 495 Substituting into the integration by parts formula gives: ( ) ∫ 1 ax ax e sin bx dx = (e ) − cos bx b ) ∫ ( 1 − − cos bx (aeax dx) b ∫ 1 a = − eax cos bx + eax cos bx dx b b Substituting this result into equation (1) gives: [ ∫ 1 a 1 eax cos bx dx = eax sin bx − − eax cos bx b b b ] ∫ a + eax cos bx dx b 1 a = eax sin bx + 2 eax cos bx b b ∫ a2 − 2 eax cos bx dx b The integral on the far right of this equation is the same as the integral on the left hand side and thus they may be combined. ∫ ∫ a2 ax e cos bx dx + 2 eax cos bx dx b 1 a = eax sin bx + 2 eax cos bx b b

Using a similar method to above, that is, integrating by parts twice, the following result may be proved: ∫ eax sin bx dx =

(

b2 + a2 b2

decimal places.

eax cos bx dx =

2

b2

∫ ∫ Comparing et sin 2t dt with eax sin bx dx shows that x = t, a = 1 and b = 2 Hence, substituting into equation (2) gives: ∫

π 4

et sin 2t dt

0

[

et = 2 (1 sin 2t − 2 cos 2t) 1 + 22

b + a2

e

4

0

π (π ) ( π )) e4 ( = sin 2 − 2 cos 2 5 4 4

[ −

]

e0 (sin 0 − 2 cos 0) 5

[

] [ π π ] e4 1 e4 2 = (1 − 0) − (0 − 2) = + 5 5 5 5

)(

ax

e b2

Now try the following Practice Exercise Practice Exercise 204 Integration by parts (Answers on page 891) Determine the integrals in Problems 1 to 5 using integration by parts.

) (b sin bx + a cos bx) + c

∫

ax

a2 + b2

]π

[

2x2 ln x dx

1. =

et sin 2t dt, correct to 4

0

= 0.8387, correct to 4 decimal places.

∫ (

π 4

(2)

eax cos bx dx eax = 2 (b sin bx + a cos bx) b

Hence

(a sin bx − b cos bx) + c

Problem 10. Evaluate

1 a = eax sin bx + 2 eax cos bx b b

)∫

a2 + b2

∫

( )∫ a2 i.e. 1 + 2 eax cos bx dx b

i.e.

eax

∫

(b sin bx + a cos bx) + c 2.

2 ln 3x dx

]

496 Section H ∫

Practice Exercise 205 Multiple-choice questions on integration by parts (Answers on page 892)

x2 sin 3x dx

3. ∫

2e5x cos 2x dx

4.

Each question has only one correct answer ∫ 1. xe 2x dx is:

∫ 2θ sec2 θ dθ

5.

Evaluate the integrals in Problems 6 to 9, correct to 4 significant figures. ∫

2

6.

x ln x dx ∫

1 3x

2e sin 2x dx 0

∫

π 2

8.

et cos 3t dt

0

9.

∫ 4√ x3 ln x dx 1

10. In determining a Fourier series to represent f (x) = x in the range −π to π, Fourier coefficients are given by: ∫ 1 π an = x cos nx dx π −π ∫ 1 π and bn = x sin nx dx π −π where n is a positive integer. Show by using integration by parts that an = 0 and 2 bn = − cos nπ n ∫ 1 11. The equations C = e−0.4θ cos 1.2θ dθ ∫ and

(b) 2e 2x + c (d) 2e 2x (x − 2) + c

1 +c x 1 1 (c) x ln x − 1 + c (d) + 2 + c x x ∫ 3. The indefinite integral x ln x dx is equal to: (a) x(ln x − 1) + c

1

7.

x2 2x e +c 4 e 2x (c) (2x − 1) + c 4 ∫ 2. ln x dx is equal to: (a)

(b)

(a) x(ln 2x − 1) + c ( ) 1 2 (b) x ln 2x − +c 4 ( ) x2 1 (c) ln x − +c 2 2 ( ) x 1 (d) ln 2x − +c 2 4 ∫ π 2 3t sin t dt is equal to: 4. 0

(a) −1.712 (b) 3 (c) 6 ∫ 1 5. 5xe 3x dx is equal to:

(d) −3

0

(a) 1.67 (c) −267.25

(b) 22.32 (d) 22.87

0

S=

1

e−0.4θ sin 1.2θ dθ

0

are involved in the study of damped oscillations. Determine the values of C and S.

For fully worked solutions to each of the problems in Practice Exercises 203 and 204 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 43

Reduction formulae Why it is important to understand: Reduction formulae When an integral contains a power of n, then it may sometimes be rewritten, using integration by parts, in terms of a similar integral containing (n − 1) or (n − 2), and so on. The result is a recurrence relation, i.e. an equation that recursively defines a sequence, once one or more initial terms are given; each further term of the sequence is defined as a function of the preceding terms. It may sound difficult but it’s actually quite straightforward and is just one more technique for integrating certain functions.

At the end of this chapter, you should be able to: • appreciate when reduction formulae might be used ∫ • integrate functions of the form

xn ex dx using reduction formulae ∫ ∫ • integrate functions of the form xn cos x dx and xn sin x dx using reduction formulae ∫ ∫ • integrate functions of the form sinn x dx and cosn x dx using reduction formulae • integrate further functions using reduction formulae

43.1

When using integration ∫ by parts in Chapter 42, an integral such as x2 e∫x dx requires integration by parts twice. Similarly, x3 ex dx requires integration three times.∫ Thus, integrals such as ∫ 5 xby parts ∫ x e dx, x6 cos x dx and x8 sin 2x dx for example, would take a long time to determine using integration by parts. Reduction formulae provide a quicker method for determining such integrals and the method is demonstrated in the following sections.

43.2 Using reduction formulae for ∫ integrals of the form x n e x dx To determine let

du = nxn−1 and du = nxn−1 dx dx

Introduction

∫

xn ex dx using integration by parts, u = xn from which,

and

Thus,

dv = ex dx from which, ∫ v = ex dx = ex ∫ ∫ n x n x x e dx = x e − ex nxn−1 dx

using the integration by parts formula, ∫ n x = x e − n xn−1 ex dx The integral on the far right is seen to be of the same form as the integral on the left-hand side, except that n has been replaced by n − 1 Thus, if we let, ∫ xn ex dx = In

498 Section H ∫ then

xn−1 ex dx = In−1

∫

Now try the following Practice Exercise ∫

xn ex dx = xn ex − n

Hence

xn−1 ex dx

Practice Exercise 206 Using reduction ∫ formulae for integrals of the form xn ex dx (Answers on page 892)

can be written as: In = xn ex − nIn−1

(1)

Equation (1) is an example of a reduction formula since it expresses an integral in n in terms of the same integral in n − 1 Problem 1. Determine reduction formula.

∫

x2 ex dx using a

Using equation (1) with n = 2 gives: ∫ x2 ex dx = I2 = x2 ex − 2I1 and

I1 = x1 ex − 1I0 ∫ ∫ I0 = x0 ex dx = ex dx = ex + c1

Hence

I2 = x2 ex − 2[xex − 1I0 ] = x2 ex − 2[xex − 1(ex + c1 )]

∫

x2 ex dx = x2 ex − 2xex + 2ex + 2c1

i.e.

= ex (x2 − 2x + 2) + c (where c = 2c1 ) As with integration by parts, in the following examples the constant of integration will be added at the last step with indefinite integrals. 2. ∫Problem x3 ex dx

Use a reduction formula to determine

From equation (1), In = xn ex − nIn−1 ∫ Hence x3 ex dx = I3 = x3 ex − 3I2 I2 = x2 ex − 2I1

and

∫

1 x ∫I1 = x e − 1I ∫0 0 x I0 = x e dx = ex dx = ex

x3 ex dx = x3 ex − 3[x2 ex − 2I1 ]

Thus

1. 2. 3.

Use ∫ 4 x a reduction formula to determine x e dx. ∫ Determine t3 e2t dt using a reduction formula. Use result of Problem 2 to evaluate ∫ 1 3the 2t 5t e dt, correct to 3 decimal places. 0

43.3 Using reduction formulae ∫ n for integrals of the form x cos x dx ∫ and xn sin x dx ∫ (a) xn cos x dx ∫ Let In = xn cos x dx then, using integration by parts: du if u = xn then = nxn−1 dx du = nxn−1 dx

from which,

dv = cos x dx then ∫ v = cos x dx = sin x ∫ n Hence In = x sin x − (sin x)nxn−1 dx ∫ = xn sin x − n xn−1 sin x dx and if

Using integration by parts again, this time with u = xn−1 : du = (n − 1)xn−2 , and dv = sin x dx, dx from which, v= Hence

∫ sin x dx = −cos x

[ In = xn sin x − n xn−1 (−cos x)

= x3 ex − 3[x2 ex − 2(xex − I0 )]

∫ i.e.

= x3 ex − 3[x2 ex − 2(xex − ex )] = x3 ex − 3x2 ex + 6(xex − ex ) = x3 ex − 3x2 ex + 6xex − 6ex x3 ex dx = ex (x3 − 3x2 + 6x − 6) + c

]

∫ −

(−cos x)(n − 1)xn−2 dx

= xn sin x + nxn−1 cos x − n(n − 1)

∫ xn−2 cos x dx

Reduction formulae 499 i.e. In = xn sin x + nxn−1 cos x − n(n − 1)In−2 3. ∫Problem x2 cos x dx

(2)

Use a reduction formula to determine

0

Using the reduction formula of equation (2): ∫

From equation (2), x2 cos x dx = I2

In = xn sin x + nxn−1 cos x − n(n − 1)In−2 ∫ π hence xn cos x dx = [xn sin x + nxn−1 cos x]π0

= x2 sin x + 2x1 cos x − 2(1)I0 ∫ I0 = x0 cos x dx ∫ = cos x dx = sin x

and

Hence

Problem ∫ π 5. Determine a reduction formula for xn cos x dx and hence evaluate 0 ∫ π x4 cos x dx, correct to 2 decimal places.

0

− n(n − 1)In−2 = [(π n sin π + nπ n−1 cos π) − (0 + 0)] − n(n − 1)In−2

∫ x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + c ∫

Problem 4.

Let ∫ 3 us firstly t cos t dt.

find

Hence ∫ π x4 cos x dx = I4

2 3

Evaluate

significant figures.

= − nπ n−1 − n(n − 1)In−2

4t cos t dt, correct to 4

0

= −4π 3 − 4(3)I2 since n = 4

1

a

reduction

formula

for

From equation (2), ∫ t3 cos t dt = I3 = t3 sin t + 3t2 cos t − 3(2)I1

When n = 2, ∫ π x2 cos x dx = I2 = −2π 1 − 2(1)I0 0 ∫ π and I0 = x0 cos x dx ∫

0 π

=

cos x dx 0

and

= [sin x]π0 = 0 I1 = t1 sin t + 1t0 cos t − 1(0)In−2 = t sin t + cos t

Hence

0

= −4π 3 + 24π or −48.63,

∫ 3

3

2

t cos t dt = t sin t + 3t cos t − 3(2)[t sin t + cos t] + c = t sin t + 3t cos t − 6t sin t − 6 cos t + c Thus ∫ 2 4t3 cos t dt 3

1

Hence ∫ π x4 cos x dx = −4π 3 − 4(3)[−2π − 2(1)(0)]

2

= [4(t3 sin t + 3t2 cos t − 6t sin t − 6 cos t)]21 = [4(8 sin 2 + 12 cos 2 − 12 sin 2 − 6 cos 2)] − [4(sin 1 + 3 cos 1 − 6 sin 1 − 6 cos 1)] = (−24.53628) − (−23.31305) = −1.223

correct to 2 decimal places. ∫ (b) xn sin x dx ∫ Let In = xn sin x dx Using integration by parts, if u = xn then du = nxn−1 from which, du = nxn−1 dx dx and ∫if dv = sin x dx then v = sin x dx = −cos x. Hence ∫ xn sin x dx ∫ = In = xn (−cos x) − (−cos x)nxn−1 dx ∫ = −xn cos x + n xn−1 cos x dx

500 Section H Using integration by parts again, with u = xn−1 , from du which, = (n − 1)xn−2 dx ∫ and dv = cos x dx, from which, v = cos x dx = sin x. Hence [ n In = −x cos x + n xn−1 (sin x) (sin x)(n − 1)x

= −xn cos x + nxn−1 (sin x)

n−2

0

π 2

θ4 sin θ dθ

0

= 3I4 [ ( ) ] π 3 =3 4 − 4(3)I2 2 ( π )1 I2 = 2 − 2(1)I0 2 ∫ π π 2 and I0 = θ0 sin θ dθ = [−cos x]02

xn−2 sin x dx

0

= [−0 − (−1)] = 1 Hence ∫ π 2 3 θ4 sin θ dθ

Use a reduction formula to determine

0

= 3I4 [ ( ) { ( ) }] π 3 π 1 =3 4 − 4(3) 2 − 2(1)I0 2 2 [ ( ) { ( ) }] π 3 π 1 =3 4 − 4(3) 2 − 2(1)(1) 2 2 [ ( ) ] ( π )1 π 3 =3 4 − 24 + 24 2 2

Using equation (3), ∫ x3 sin x dx = I3 = −x3 cos x + 3x2 sin x − 3(2)I1 I1 = −x1 cos x + 1x0 sin x

and

∫ 3θ4 sin θ dθ = 3

dx

In = −xn cos x + nxn−1 sin x − n(n − 1)In−2 (3)

6. ∫Problem x3 sin x dx

π 2

∫

− n(n − 1) i.e.

∫

]

∫ −

Hence

= −x cos x + sin x Hence ∫ x3 sin x dx = −x3 cos x + 3x2 sin x

= 3(15.503 − 37.699 + 24) = 3(1.804) = 5.41 correct to 2 decimal places

− 6[−x cos x + sin x] = −x3 cos x + 3x2 sin x

Now try the following Practice Exercise

+ 6x cos x − 6 sin x + c ∫ Problem 7.

π 2

Evaluate

2 decimal places.

3θ4 sin θ dθ, correct to

0

Practice Exercise 207 Using reduction ∫ n formulae ∫ n for integrals of the form x cos x dx and x sin x dx (Answers on page 892) 1.

From equation (3), π

In = [−θn cos θ + nθn−1 (sin θ)]02 − n(n − 1)In−2 [( ( ) ) ] ( π )n−1 π n π π = − cos + n sin − (0) 2 2 2 2

2.

places. 3.

− n(n − 1)In−2 =n

( π )n−1 2

− n(n − 1)In−2

Use ∫ 5 a reduction formula to determine x cos x dx ∫ π Evaluate x5 cos x dx, correct to 2 decimal 0

Use a reduction formula to determine ∫ x5 sin x dx ∫

4.

π

x5 sin x dx, correct to 2 decimal

Evaluate places.

0

Reduction formulae 501 8. ∫Problem sin4 x dx

43.4 Using reduction formulae ∫ nfor integrals of the form sin x dx ∫ and cosn x dx ∫

(a) sinn x dx ∫ ∫ Let In = sinn x dx ≡ sinn−1 x sin x dx from laws of indices. Using integration by parts, let u = sinn−1 x, from which, du = (n − 1) sinn−2 x cos x and dx du = (n − 1) sin

n−2

= (sin

∫

− = −sin

Using equation (4), ∫ 1 3 sin4 x dx = I4 = − sin3 x cos x + I2 4 4 1 1 I2 = − sin1 x cos x + I0 2 2 ∫ ∫ I0 = sin0 x dx = 1 dx = x

and

x cos x dx Hence

and∫let dv = sin x dx, from which, v = sin x dx = −cos x. Hence, ∫ In = sinn−1 x sin x dx n−1

∫

1 sin4 x dx = I4 = − sin3 x cos x 4

x)(−cos x) (−cos x)(n − 1) sinn−2 x cos x dx

1 3 = − sin3 x cos x − sin x cos x 4 8

n−1

+

= −sinn−1 x cos x ∫ + (n − 1) (1 − sin2 x) sinn−2 x dx

∫ Problem 9.

In = −sinn−1 x cos x

In + (n − 1)In = −sinn−1 x cos x + (n − 1)In−2

∫

nIn = −sinn−1 x cos x + (n − 1)In−2

from which, ∫ sinn x dx = 1 n−1 In = − sinn−1 xcos x + In−2 n n

1

3 x+c 8

4 sin5 t dt, correct to 3

0

Using equation (4), ∫ 1 4 sin5 t dt = I5 = − sin4 t cos t + I3 5 5 1 2 2 I3 = − sin t cos t + I1 3 3 1 0 and I1 = − sin t cos t + 0 = −cos t 1 Hence

+ (n − 1)In−2 −(n − 1)In

and

Evaluate

significant figures.

= −sinn−1 x cos x {∫ } ∫ n−2 n + (n − 1) sin x dx − sin x dx

i.e.

[ ] 3 1 1 − sin x cos x + (x) 4 2 2

+

x cos x ∫ + (n − 1) cos2 x sinn−2 x dx

i.e.

Use a reduction formula to determine

(4)

1 sin5 t dt = − sin4 t cos t 5 [ ] 4 1 2 + − sin2 t cos t + (−cos t) 5 3 3 1 4 = − sin4 t cos t − sin2 t cos t 5 15 −

8 cos t + c 15

502 Section H ∫ and

t

5

4 sin t dt [ 1 = 4 − sin4 t cos t 5

0

−

4 8 sin2 t cos t − cos t 15 15

]1

[( 1 4 = 4 − sin4 1 cos 1 − sin2 1 cos 1 5 15 ) ( )] 8 8 − cos 1 − −0 − 0 − 15 15 = 4[(−0.054178 − 0.1020196 − 0.2881612) − (−0.533333)] = 4(0.0889745) = 0.356 Problem 10. Determine a reduction formula for ∫ π ∫ π 2 2 n sin x dx and hence evaluate sin6 x dx 0

0

From equation (4), ∫ sinn x dx 1 n−1 = In = − sinn−1 x cos x + In−2 n n hence [ ] π2 ∫ π 2 1 n−1 n−1 n sin x dx = − sin x cos x + In−2 n n 0 0 n−1 = [0 − 0] + In−2 n n−1 i.e. In = In−2 n Hence ∫ π 2 5 sin6 x dx = I6 = I4 6 0 3 1 I4 = I2 , I2 = I0 4 2 ∫ π ∫ π 2 2 π 0 and I0 = sin x dx = 1 dx = 2 0 0 Thus ∫

π 2 0

[ ] 5 5 3 sin6 x dx = I6 = I4 = I2 6 6 4 [ { }] 5 3 1 = I0 6 4 2 [ { }] 5 3 1 [π] 15 = = π 6 4 2 2 96

0

∫ (b) cosn x dx ∫ ∫ Let In = cosn x dx ≡ cosn−1 x cos x dx from laws of indices. Using integration by parts, let u = cosn−1 x from which, du = (n − 1) cosn−2 x(−sin x) dx and

du = (n − 1) cosn−2 x(−sin x) dx

dv = cos x dx ∫ from which, v = cos x dx = sin x and let

Then In = (cosn−1 x)(sin x) ∫ − (sin x)(n − 1) cosn−2 x(−sin x) dx ∫ = (cosn−1 x)(sin x) + (n − 1) sin2 x cosn−2 x dx ∫ n−1 = (cos x)(sin x)+(n − 1) (1 − cos2 x)cosn−2 x dx = (cosn−1 x)(sin x) {∫ } ∫ + (n − 1) cosn−2 x dx − cosn x dx i.e. In = (cosn−1 x)(sin x) + (n − 1)In−2 − (n − 1)In i.e. In + (n − 1)In = (cosn−1 x)(sin x) + (n − 1)In−2 i.e. nIn = (cosn−1 x)(sin x) + (n − 1)In−2 Thus

1 n−1 In = cosn−1 x sin x + In−2 n n

Problem 11. Use a reduction formula to ∫ determine cos4 x dx Using equation (5), ∫ 1 3 cos4 x dx = I4 = cos3 x sin x + I2 4 4 and and

1 1 cos x sin x + I0 2 2 ∫ I0 = cos0 x dx I2 =

∫ =

1 dx = x

(5)

Reduction formulae 503 ∫ cos4 x dx

Hence

=

1 3 cos3 x sin x + 4 4

(

1 1 cos x sin x + x 2 2

)

1 3 3 = cos3 x sin x + cos x sin x + x + c 4 8 8

Thus, from equation (6), ∫ π 2 4 cos5 x dx = I3 , 5 0 ∫

π 2

I1 =

and

2 I3 = I1 3 cos1 x dx

0 π

= [sin x]02 = (1 − 0) = 1

Problem 12. Determine a reduction formula ∫ π ∫ π 2 2 n for cos x dx and hence evaluate cos5 x dx 0

0

∫

π 2

Hence 0

From equation (5), ∫ 1 n−1 cosn x dx = cosn−1 x sin x + In−2 n n and hence [ ] π2 ∫ π 2 1 n n−1 cos x dx = cos x sin x n 0 0 n−1 + In−2 n n−1 In−2 = [0 − 0] + n ∫ π 2 n−1 i.e. cosn x dx = In = In−2 (6) n 0 (Note that this is the same reduction formula as for ∫ π 2 sinn x dx (in Problem 10) and the result is usually

Now try the following Practice Exercise Practice Exercise 208 Using reduction ∫ formulae for integrals of the form sinn x dx ∫ and cosn x dx (Answers on page 892) 1.

2.

∫Use 7 a reduction formula to determine sin x dx ∫ π Evaluate 3 sin3 x dx using a reduction 0

formula.

0

known as Wallis’∗ formula.)

[ ] 4 4 2 cos x dx = I3 = I1 5 5 3 [ ] 4 2 8 = (1) = 5 3 15 5

3.

∫

Evaluate

Determine, using a reduction formula, ∫ cos6 x dx ∫

5.

sin5 x dx using a reduction

0

formula. 4.

π 2

Evaluate

π 2

cos7 x dx

0

43.5

Further reduction formulae

The following worked problems demonstrate further examples where integrals can be determined using reduction formulae. 13. Determine∫a reduction formula for ∫Problem tann x dx and hence find tan7 x dx ∫ ∗

Who was Wallis? John Wallis (23 November 1616–28 October 1703) was an English mathematician partially credited for the development of infinitesimal calculus, and is also credited with introducing the symbol ∞ for infinity. To find out more go to www.routledge.com/cw/bird

tan x dx ≡

Let In = ∫ =

∫ n

tann−2 x tan2 x dx by the laws of indices

tann−2 x(sec2 x − 1) dx since 1 + tan2 x = sec2 x

504 Section H ∫

∫ tann−2 x sec2 x dx −

=

tann−2 x dx

and from equation (6),

∫ tann−2 x sec2 x dx − In−2

=

i.e. In =

tann−1 x − In−2 n−1 and

[ ] 5 5 3 I6 = I4 = I2 6 6 4 [ ( )] 5 3 1 = I0 6 4 2 ∫ π 2 I0 = cos0 t dt

When n = 7, ∫ I7 =

∫

π 2

=

6

tan x tan7 x dx = − I5 6

0

tan4 x tan2 x I5 = − I3 and I3 = − I1 4 2 ∫ I1 = tan x dx = ln(sec x) from Problem 9, Chapter 38, page 464 Thus [ 4 ∫ tan6 x tan x 7 tan x dx = − 6 4

0

Hence

π

1 dt = [x]02 =

5 3 1 π I6 = · · · 6 4 2 2 5π 15π or = 96 32

7 7 5π Similarly, I8 = I6 = · 8 8 32 Thus ∫ π 2 sin2 t cos6 t dt = I6 − I8 0

)] tan2 x − − ln(sec x) 2

(

∫

1 1 1 = tan6 x − tan4 x + tan2 x 6 4 2 − ln(sec x) + c Problem 14. Evaluate, using a reduction ∫ π 2 formula, sin2 t cos6 t dt 0

∫

π 2

∫ 2

π 2

6

sin t cos t dt =

0

(1 − cos2 t) cos6 t dt

0

∫

π 2

=

∫ cos t dt − 6

0

∫

0 π 2

In =

If

cosn t dt

0

then ∫ π 2

0

=

5π 7 5π − · 32 8 32

=

1 5π 5π · = 8 32 256

tan7 x dx

Hence

π 2

cos8 t dt

Problem 15. Use integration by ∫parts to determine a reduction for (ln x)n dx. ∫ formula 3 Hence determine (ln x) dx. ∫ Let In = (ln x)n dx. Using integration by parts, let u = (ln x)n , from which, ( ) du 1 = n(ln x)n−1 dx x ( ) 1 dx and du = n(ln x)n−1 x ∫ and let dv = dx, from which, v = dx = x ∫ Then In = (ln x)n dx ∫ = (ln x) (x) − n

n−1

(x)n(ln x) ∫

= x(ln x)n − n sin2 t cos6 t dt = I6 − I8

π 2

(ln x)n−1 dx

i.e. In = x(ln x)n − nIn−1

( ) 1 dx x

Reduction formulae 505 When n = 3, ∫

Now try the following Practice Exercise (ln x)3 dx = I3 = x(ln x)3 − 3I2 ∫

I2 = x(ln x)2 − 2I1 and I1 = ln x dx = x(ln x − 1) from Problem 7, page 494.

Practice Exercise 209 Reduction formulae (Answers on page 892) ∫ 1.

Hence ∫

π 2

Evaluate

cos2 x sin5 x dx

0

(ln x)3 dx = x(ln x)3 − 3[x(ln x)2 − 2I1 ] + c

∫

2.

= x(ln x)3 − 3[x(ln x)2 − 2[x(ln x − 1)]] + c = x(ln x)3 − 3[x(ln x)2 − 2x ln x + 2x] + c = x(ln x)3 − 3x(ln x)2

tan6 x dx by using reduction for∫ π4 tan6 x dx mulae and hence evaluate Determine

3.

π 2

Evaluate

cos5 x sin4 x dx

0

4.

Use a reduction formula to determine ∫ (ln x)4 dx

5.

Show that

+ 6x ln x − 6x + c

∫

= x[(ln x)3 − 3(ln x)2 + 6 ln x − 6] + c

0

∫

0

π 2

sin3 θ cos4 θ dθ =

2 35

For fully worked solutions to each of the problems in Practice Exercises 206 to 209 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 44

Double and triple integrals Why it is important to understand: Double and triple integrals Double and triple integrals have engineering applications in finding areas, masses and forces of twodimensional regions, and in determining volumes, average values of functions, centres of mass, moments of inertia and surface areas. A multiple integral is a type of definite integral extended to functions of more than one real variable. This chapter explains how to evaluate double and triple integrals and completes the many techniques of integral calculus explained in the preceding chapters.

At the end of this chapter, you should be able to: • evaluate a double integral • evaluate a triple integral

44.1

Double integrals

The procedure ∫ y2 ∫ x2 to determine a double integral of the form: f (x,y) dx dy is: y1

(i)

(ii)

x1

integrate f (x, y) with respect to x between the limits of x = x1 and x = x2 (where y is regarded as being a constant), and integrate the result in (i) with respect to y between the limits of y = y1 and y = y2

It is seen from this procedure that to determine a double integral we start with the innermost integral and then work outwards. Double integrals may be used to determine areas under curves, second moments of area, centroids and moments of inertia. ∫ y2 ∫ x2 (Sometimes f(x, y) dx dy is written as: y1 x1 ∫ y2 ∫ x2 dy f (x, y) dx. All this means is that the y1

x1

right hand side integral is determined first).

Determining double integrals is demonstrated in the following worked problems. ∫ 3∫ Problem 1.

5

Evaluate 1

(2x − 3y) dx dy

2

Following the above procedure: (i)

(2x − 3y) is integrated with respect to x between x = 2 and x = 5, with y regarded as a constant ∫ 5 i.e. (2x − 3y) dx 2

[

]5 [ ]5 2x2 = − (3y)x = x2 − 3xy 2 2 2 [( 2 ) ( 2 )] = 5 − 3(5)y − 2 − 3(2)y = (25 − 15y) − (4 − 6y) = 25 − 15y − 4 + 6y = 21 − 9y

Double and triple integrals 507 ∫ 3∫

5

1

(ii)

2

∫

3

=

) 16 (2x y) dx dy = y dy 3 1 0 1 [ ]3 [( ) ( )] 16y2 16(3)2 16(1)2 = = − 6 1 6 6 ∫ 3∫

(2x − 3y) dx dy

(ii)

(21 − 9y) dy

1

[ ]3 9y2 = 21y − 2 1 [( ) ( )] 9(3)2 9(1)2 = 21(3) − − 21(1) − 2 2

∫ 3(

2

2

= 24 − 2.67 = 21.33 ∫ 3∫

2

1

0

Hence,

(2x2 y) dx dy = 21.33

= (63 − 40.5) − (21 − 4.5) = 63 − 40.5 − 21 + 4.5 = 6 ∫ 3∫ 5 Hence, (2x − 3y) dx dy = 6 1

2

∫ 4∫ Problem 2.

2

Evaluate 0

∫ Problem 4.

1

3x3 = − 2x 3

(ii)

]2

(i) (3x2 − 2) dx dy

1

∫

=

∫

[( ) ( )] = 23 − 2(2) − 13 − 2(1)

1

0

4

4

(5) dy = [5y]0 =

0

[(

) ( )] 5(4) − 5(0)

1

∫ 3∫ Problem 3.

0

∫

[

]2 2x3 y 3 0 [( ) ] 3 ( ) 2(2) y 16 = − 0 = y 3 3

(2x2 y) dx =

∫

3

(ii) 1

∫

2 0

∫ 3(

)

[

(

3

(2x2 y) dx =

dy

dy 1

) 16 y 3

]3 16 16y2 = y dy = 3 6 1 1 [( ) ( )] 16(3)2 16(1)2 = − 6 6 = 24 − 2.67 = 21.33 ∫

3

Hence,

∫ 2 dy (2x2 y) dx = 21.33

1

0

The last two worked problems show that 2

(2x2 y) dx dy

Evaluate 1

2

i.e.

= 20 − 0 = 20 ∫ 4∫ 2 Hence, (3x2 − 2) dx dy = 20

0

Following the above procedure: (i)

(2x2 y) is integrated with respect to x between x = 0 and x = 2,

1

= (8 − 4) − (1 − 2) = 8 − 4 − 1 + 2 = 5 ∫ 4∫ 2 (3x2 − 2) dx dy 0

(2x2 y) dx 0

With this configuration:

(3x2 − 2) is integrated with respect to x between x = 1 and x = 2, ∫ 2 i.e. (3x2 − 2) dx [

2

dy 1

Following the above procedure: (i)

∫

3

Evaluate

(2x2 y) is integrated with respect to x between x = 0 and x = 2, [ 3 ]2 ∫ 2 2x y 2 i.e. (2x y) dx = 3 0 0 [( ) ] ( ) 2(2)3 y 16 = − 0 = y 3 3

∫ 3∫

2

(2x2 y) dx dy gives the same answer as 1

0

∫

∫

3

1

2

(2x2 y) dx

dy 0

∫ 4∫ Problem 5.

π

(2 + sin 2θ) dθ dr

Evaluate 1

0

508 Section H Following the above procedure: (i)

(ii)

Now try the following Practice Exercise

(2 + sin 2θ) is integrated with respect to θ between θ = 0 and θ = π, [ ]π ∫ π 1 i.e. (2 + sin 2θ) dx = 2θ − cos 2θ 2 0 0 [( ) ( )] 1 1 = 2π − cos 2π − 0 − cos 0 2 2 = (2π − 0.5) − (0 − 0.5) = 2π ∫ 4∫ π (2 + sin 2θ) dθ dr 1

0

∫

4(

=

)

2π dr =

1

4 [2πr]1

[( ) ( )] = 2π(4) − 2π(1)

= 8π − 2π = 6π or 18.85 ∫ 4∫

π

(2 + sin 2θ) dθ dr = 18.85

Hence, 1

0

2

∫ 2∫

c

Show that the area is equal to ∫

π 3

∫

Area, A = 2

√ ) c ( 2π + 3 3 6 2

2c cos θ

r dr dθ 0

c

)] )2 ] 2c cos θ [( ( 2c cos θ r2 c2 r dr = = − 2 c 2 2 c ( 2 2 ) ) 4c cos θ c2 c2 ( = − = 4 cos2 θ − 1 2 2 2 ) ∫ π3 ( 2 ) c ( 2 4 cos2 θ − 1 dθ 2 0 ( ) ] π3 [ 1 sin 2θ 2 = c (4) θ+ −θ 2 2 0 [

3

2. 1

(2x − y) dx dy

2

∫ 2∫

3

3. 1

(2x − y) dy dx

2

∫ 5∫ 4.

2

−1

1

∫ 6∫

(x − 5y) dx dy

5

(x2 + 4y) dx dy

5. 1

2

∫ 4∫

0

2c cos θ

Evaluate the double integrals in Problems 1 to 8. ∫ 3∫ 4 1. 2 dx dy

0

Problem 6. The area, A, outside the circle of radius r = c and inside the circle of radius r = 2c cos α is given by the double integral: ∫ π ∫ 2c cos θ A=2 3 r dr dθ

∫

Practice Exercise 210 Double integrals (Answers on page 892)

6. ∫ 7. ∫

2

−3

1

∫

3 −1

π

(3 + sin 2θ) dθ dr 0

3

8.

(3xy2 ) dx dy

∫

4

dx 1

(40 − 2xy) dy

2

9. The volume of a solid, V, bounded by the curve 4 − x − y between the limits x = 0 to x = 1 and y = 0 to y = 2 is given by: ∫ 2∫ 1 V= (4 − x − y) dx dy 0

0

Evaluate V. 10.

The second moment of area, I, of a 5 cm by 3 cm rectangle about an axis through one corner perpendicular to the plane of the figure is given by: ∫ 5∫ 3 I= (x2 + y2 ) dy dx 0

0

from 1 of Table 39.1, page 468 Evaluate I. [( ( ) ) ] 2π ( ) sin π π 3 = c2 2 + − − 0 3 2 3 ) ( ( ( ) √3 √ ) 44.2 Triple integrals 2 2 2π π π 3 2 2 =c + − =c + The procedure to determine a triple integral of the form: 3 2 3 3 2 ∫ z2 ∫ y2 ∫ x2 f(x, y, z) dx dy dz is: √ ) c2 ( z1 y1 x1 Hence, area, A = 2π + 3 3 6

Double and triple integrals 509 (i)

integrate f(x, y, z) with respect to x between the limits of x = x1 and x = x2 (where y and z are regarded as being constants),

(ii)

integrate the result in (i) with respect to y between the limits of y = y1 and y = y2 , and

(iii)

(iii)

(8z − 16) is integrated with respect to z between z = 1 and z = 2 ]2 8z2 (8z − 16) dz = − 16z 2 1 1 [( ) ( )] 8(2)2 8(1)2 = − 16(2) − − 16(1) 2 2 ∫

i.e.

integrate the result in (ii) with respect to z between the limits of z = z1 and z = z2

It is seen from this procedure that to determine a triple integral we start with the innermost integral and then work outwards.

= [(16 − 32) − (4 − 16)] = 16 − 32 − 4 + 16 = −4

Determining triple integrals is demonstrated in the following worked problems.

∫ 2∫

3

1

−1

Hence, Problem 7. Evaluate ∫ 2∫ 3 ∫ 2 (x − 3y + z) dx dy dz 1

−1

0

(i)

(x − 3y + z) is integrated with respect to x between x = 0 and x = 2, with y and z regarded as constants, [ 2 ]2 ∫ 2 x i.e. (x − 3y + z) dx = − (3y)x + (z)x 2 0 0 [( =

0

(i)

(2a2 − b2 + 3c2 ) is integrated with respect to a between a = 0 and a = 1, with b and c regarded as constants, ∫ i.e.

) ] ( ) 22 − (3y)(2) + (z)(2) − 0 2

0

]3 6y = 2y − + (2z)y 2 −1 [( ) 6(3)2 = 2(3) − + (2z)(3) 2 ( )] 6(−1)2 − 2(−1) − + (2z)(−1) 2 = [(6 − 27 + 6z) − (−2 − 3 − 2z)] = 6 − 27 + 6z + 2 + 3 + 2z = 8z − 16

1

(2a2 − b2 + 3c2 ) da [

]1 2a3 2 2 = − (b )a + (3c )a 3 0 [( ) ] ( ) 2 = − (b2 ) + (3c2 ) − 0 3 =

2 − b2 + 3c2 3

(

−1

2

(x − 3y + z) dx dy dz = −4

0

Following the above procedure:

(2 − 6y + 2z) is integrated with respect to y between y = −1 and y = 3, with z regarded as a constant, i.e. ∫ 3 (2 − 6y + 2z) dy [

2

0

= 2 − 6y + 2z (ii)

∫

Problem 8. Evaluate ∫ 3∫ 2∫ 1 (2a2 − b2 + 3c2 ) da db dc 1

Following the above procedure:

[

2

(ii)

) 2 − b2 + 3c2 is integrated with respect to b 3 between b = 0 and b = 2, with c regarded as a constant, i.e. ∫ 2(

) [ ]2 2 2 b3 − b2 + 3c2 db = b − + (3c2 )b 3 3 3 0 0 [( ) ] ( ) 2 (2)3 = (2) − + (3c2 )(2) − 0 3 3 ( ) ( ) 4 8 4 2 = − + 6c − 0 = 6c2 − 3 3 3

510 Section H (iii)

) ( 4 is integrated with respect to c 6c2 − 3 between c = 1 and c = 3 ∫ 3(

) [ 3 ]3 4 6c 4 6c − dc = − c 3 3 3 1 )] [ ( ( ) 4 = 54 − 4 − 2 − 3

i.e. volume of the solid, V = 13.31 cubic units

Now try the following Practice Exercise

2

i.e. 1

= [(50) − (0.67)] = 49.33 ∫ 3∫ 2∫

1

Hence, 1

0

Practice Exercise 211 Triple integrals (Answers on page 892) Evaluate the triple integrals in Problems 1 to 7. ∫ 2∫ 3∫ 1 1. (8xyz) dz dx dy 1

(2a2 − b2 + 3c2 ) da db dc = 49.33

0

√

∫ 2∫

4−x2

∫

∫ 2∫ 3. 0

∫ 3∫

dz dy dx 0

0

4.

0

√ 4−x2

∫ 2∫

∫

∫

0

0

5−xy

√ 4−x2

0

0

( ) ( ) = 5 − xy − 0 = 5 − xy [

] √4−x2 2

( ) xy 5 − xy dy = 5y − 2 0  (√ )2  2 ) x (√ 4 − x   = 5 4 − x2 −  2 [ = 5

(√

4 − x2

)

( )] x 4 − x2 − 2

) 4x x3 4 − x2 − + 2 2 ) ∫ 2 ( (√ ) 3 x 5 4 − x2 − 2x + dx 2 0 [ ( 2 ) ]2 2 x√ x4 −1 x 2 2 = 5 sin + 4−x −x + 2 2 2 8 0 =5

2 −1 1

∫

3(

2

0

0

0

0

−2

(√

from 11 of Table 39.1, page 468 [( ( 2 ) ) ] ( ) 2 2 2√ 24 = 5 sin−1 + 4 − 22 − 22 + − 0 2 2 2 8 ( (π) ) = 10 + 0 − 4 + 2 = 5π − 2 2

(x2 + 5y2 − 2z) dx dy dz

π∫ π∫ π

(xy sin z) dx dy dz ∫ 4∫

6.

−1 ∫ 2

(xy) dx dy dz

∫ 3∫ 2∫ 7. 1

8.

) x + y2 + z3 dx dy dz

1

∫

0

∫

dz dy dx 0

0

5.

5−xy

Volume of the solid, V =

2

−1

2

Calculate the volume, correct to 2 decimal places.

dz = [z] 0

1

(8xyz) dx dy dz 1

5−xy

V=

5−xy

0

2.

Problem 9. It may be shown that the volume of a solid, V, is given by the triple integral:

∫

2

∫ 2∫ 3∫

0

1 1

−1

(xz + y) dx dy dz

A box shape X is described by the triple ∫ 3∫ 2∫ 1 integral: X = (x + y + z) dx dy dz Evaluate X.

0

0

0

Practice Exercise 212 Multiple-choice questions on double and triple integrals (Answers on page 892) Each question has only one correct answer ∫ 3∫ 2 ( 3 ) 1. 4x − 3 dx dy is equal to: 0

1

2.

(a) 36 (b) 72 (c) 108 (d) 84 ∫ 3∫ π ( ) 3 − sin 3t dt dr is equal to:

3.

(a) 8.76 (b) 17.52 (c) −2.00 ∫ 4∫ 2 ( 3 ) 2x y dx dy is equal to:

1

1

0

(d) 18.18

0

(a) 24

(b) 255

(c) 573.25

(d) 60

Double and triple integrals 511 ∫ 1∫ 3∫ 4. 0

(a) 5

1

2(

) 5xyz dx dy dz is equal to:

1

(b) 10

(c) 15

(d) 20

∫ 2∫ 3∫ 5. 0

(a) 6

0

2(

) 2x − y + 2z dx dy dz is equal to:

0

(b) 30

(c) 14

(d) 24

For fully worked solutions to each of the problems in Practice Exercises 210 and 211 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 45

Numerical integration Why it is important to understand: Numerical integration There are two main reasons for why there is a need to do numerical integration − analytical integration may be impossible or unfeasible, or it may be necessary to integrate tabulated data rather than known functions. As has been mentioned before, there are many applications for integration. For example, Maxwell’s equations can be written in integral form; numerical solutions of Maxwell’s equations can be directly used for a huge number of engineering applications. Integration is involved in practically every physical theory in some way − vibration, distortion under weight or one of many types of fluid flow − be it heat flow, air flow (over a wing), or water flow (over a ship’s hull, through a pipe or perhaps even groundwater flow regarding a contaminant), and so on; all these things can be either directly solved by integration (for simple systems) or some type of numerical integration (for complex systems). Numerical integration is also essential for the evaluation of integrals of functions available only at discrete points; such functions often arise in the numerical solution of differential equations or from experimental data taken at discrete intervals. Engineers therefore often require numerical integration and this chapter explains the procedures available.

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

appreciate the need for numerical integration evaluate integrals using the trapezoidal rule evaluate integrals using the mid-ordinate rule evaluate integrals using Simpson’s rule apply numerical integration to practical situations

45.1

Introduction

Even with advanced methods of integration there are many mathematical functions which cannot be integrated by analytical methods and thus approximate methods have then to be used. In many cases, such as in modelling airflow around a car, an exact answer may not even be strictly necessary. Approximate methods of definite integrals may be determined by what is termed numerical integration. It may be shown that determining the value of a definite integral is, in fact, finding the area between a curve,

the horizontal axis and the specified ordinates. Three methods of finding approximate areas under curves are the trapezoidal rule, the mid-ordinate rule and Simpson’s rule, and these rules are used as a basis for numerical integration.

45.2

The trapezoidal rule

∫b Let a required definite integral be denoted by a y dx and be represented by the area under the graph of y = f (x) between the limits x = a and x = b as shown in Fig. 45.1.

Numerical integration 513 y 5 f (x )

y

Problem 1.

(a) Use integration to evaluate, ∫ 3 2 √ dx (b) Use the correct to 3 decimal places, x 1 trapezoidal rule with four intervals to evaluate the integral in part (a), correct to 3 decimal places. ∫

3

(a) 1

y1 y2 y3 y4

O

x5a d

d

∫

3

1

2x− 2 dx

1



(

3 ) −1 2 +1 

[ ]3 1  2x 2 =  = 4x 1 1 − +1 2 1 [√ √ ] [√ ]3 = 4 x 1 = 4 3− 1

yn11

x 5b

2 √ dx = x

x

= 2.928, correct to 3 decimal places.

d

(b) Figure 45.1

Let the range of integration be divided into n equal intervals each of width d, such that nd = b − a, i.e. b−a d= n The ordinates are labelled y1 , y2 , y3 , . . . , yn+1 as shown. An approximation to the area under the curve may be determined by joining the tops of the ordinates by straight lines. Each interval is thus a trapezium, and since the area of a trapezium is given by: 1 area = (sum of parallel sides) (perpendicular 2 distance between them) then ∫ b 1 1 y dx ≈ (y1 + y2 )d + (y2 + y3 )d 2 2 a 1 1 + (y3 + y4 )d + · · · (yn + yn+1 )d 2 2 [ 1 ≈ d y1 + y2 + y3 + y4 + · · · + yn 2 ] 1 + yn+1 2

The range of integration is the difference between the upper and lower limits, i.e. 3 − 1 = 2. Using the trapezoidal rule with four intervals gives an 3−1 interval width d = = 0.5 and ordinates situ4 ated at 1.0, 1.5, 2.0, 2.5 and 3.0. Corresponding 2 values of √ are shown in the table below, each x correct to 4 decimal places (which is one more decimal place than required in the problem). x

2 √ x

1.0

2.0000

1.5

1.6330

2.0

1.4142

2.5

1.2649

3.0

1.1547

From equation (1): { ∫ 3 2 1 √ dx ≈ (0.5) (2.0000 + 1.1547) 2 x 1

}

+ 1.6330 + 1.4142 + 1.2649

i.e. the trapezoidal rule states: ∫

b a

( y dx ≈

){ ( ) 1 first + last width of interval 2 ordinate ( )} sum of remaining + ordinates

= 2.945, correct to 3 decimal places. (1)

This problem demonstrates that even with just 4 intervals a close approximation to the true value of 2.928 (correct to 3 decimal places) is obtained using the trapezoidal rule.

514 Section H Problem 2.

Use the trapezoidal rule with eight ∫ 3 2 √ dx correct to 3 decimal intervals to evaluate x 1 places. 3−1 With eight intervals, the width of each is i.e. 0.25 8 giving ordinates at 1.00, 1.25, 1.50, 1.75, 2.00, 2.25, 2 2.50, 2.75 and 3.00. Corresponding values of √ are x shown in the table below. x

2 √ x

1.00

2.0000

1.25

1.7889

1.50

1.6330

1.75

1.5119

2.00

1.4142

2.25

1.3333

2.50

1.2649

2.75

1.2060

3.00

1.1547

From equation (1): { ∫ 3 2 1 √ dx ≈ (0.25) (2.000 + 1.1547) + 1.7889 2 x 1 + 1.6330 + 1.5119 + 1.4142 } + 1.3333 + 1.2649 + 1.2060 = 2.932, correct to 3 decimal places. This problem demonstrates that the greater the number of intervals chosen (i.e. the smaller the interval width) the more accurate will be the value of the definite integral. The exact value is found when the number of intervals is infinite, which is, of course, what the process of integration is based upon. Problem 3. Use the trapezoidal rule to evaluate ∫ π 2 1 dx using six intervals. Give the 1 + sin x 0 answer correct to 4 significant figures.

π −0 With six intervals, each will have a width of 2 6 π i.e. rad (or 15◦ ) and the ordinates occur at 12 π π π π 5π π 0, , , , , and 12 6 4 3 12 2 1 Corresponding values of are shown in the 1 + sin x table below. x

1 1 + sin x

0

1.0000

π (or 15◦ ) 12

0.79440

π (or 30◦ ) 6

0.66667

π (or 45◦ ) 4

0.58579

π (or 60◦ ) 3

0.53590

5π (or 75◦ ) 12

0.50867

π (or 90◦ ) 2

0.50000

From equation (1): ∫ π ( π ){1 2 1 dx ≈ (1.00000 + 0.50000) 1 + sin x 12 2 0 + 0.79440 + 0.66667 + 0.58579 + 0.53590 } + 0.50867 = 1.006, correct to 4 significant figures.

Now try the following Practice Exercise Practice Exercise 213 Trapezoidal rule (Answers on page 892) In Problems 1 to 4, evaluate the definite integrals using the trapezoidal rule, giving the answers correct to 3 decimal places.

Numerical integration 515 ∫

1

2 dx 1 + x2

(use eight intervals)

2 ln 3x dx

(use eight intervals)

1. ∫

0 3

2. ∫

1 π 3

3. ∫

Problem 4. Use the mid-ordinate rule with (a) four intervals, (b) eight intervals, to evaluate ∫ 3 2 √ dx, correct to 3 decimal places. x 1

√ (sin θ) dθ

(a) With four intervals, each will have a width of 3−1 , i.e. 0.5 and the ordinates will occur at 1.0, 4 1.5, 2.0, 2.5 and 3.0. Hence the mid-ordinates y1 , y2 , y3 and y4 occur at 1.25, 1.75, 2.25 and 2.75. 2 Corresponding values of √ are shown in the x following table.

(use six intervals)

0 1.4

4.

e−x dx 2

(use seven intervals)

0

45.3

The mid-ordinate rule

Let a required definite integral be denoted again ∫b by a y dx and represented by the area under the graph of y = f (x) between the limits x = a and x = b, as shown in Fig. 45.2. y y 5 f (x)

y1

y2

y3

a

O

(b)

yn

b x d

d

d

x

2 √ x

1.25

1.7889

1.75

1.5119

2.25

1.3333

2.75

1.2060

From equation (2): ∫ 3 2 √ dx ≈ (0.5)[1.7889 + 1.5119 x 1 + 1.3333 + 1.2060] = 2.920, correct to 3 decimal places. With eight intervals, each will have a width of 0.25 and the ordinates will occur at 1.00, 1.25, 1.50, 1.75, . . . and thus mid-ordinates at 1.125, 1.375, 1.625, 1.875 . . . 2 Corresponding values of √ are shown in the x following table.

Figure 45.2

With the mid-ordinate rule each interval of width d is assumed to be replaced by a rectangle of height equal to the ordinate at the middle point of each interval, shown as y1 , y2 , y3 , . . . , yn in Fig. 45.2. ∫ b y dx ≈ dy1 + dy2 + dy3 + · · · + dyn Thus a ≈ d( y1 + y2 + y3 + · · · + yn )

x

2 √ x

1.125

1.8856

1.375

1.7056

1.625

1.5689

1.875

1.4606

i.e. the mid-ordinate rule states: ∫ b y dx ≈ (width of interval) (sum

2.125

1.3720

2.375

1.2978

2.625

1.2344

2.875

1.1795

a

of mid-ordinates)

(2)

516 Section H From equation (2): ∫

3 1

Now try the following Practice Exercise Practice Exercise 214 Mid-ordinate rule (Answers on page 892)

2 √ dx ≈ (0.25)[1.8856 + 1.7056 x + 1.5689 + 1.4606 + 1.3720

In Problems 1 to 4, evaluate the definite integrals using the mid-ordinate rule, giving the answers correct to 3 decimal places. ∫ 2 3 dt (use eight intervals) 1. 2 0 1+t

+ 1.2978 + 1.2344 + 1.1795] = 2.926, correct to 3 decimal places.

∫

As previously, the greater the number of intervals the nearer the result is to the true value (of 2.928, correct to 3 decimal places).

2. 0

∫ ∫ Problem 5.

2.4

Evaluate

e

−x2 3

3. 1

dx, correct to 4

∫

0

significant figures, using the mid-ordinate rule with six intervals. 2.4 − 0 , i.e. 6 0.40 and the ordinates will occur at 0, 0.40, 0.80, 1.20, 1.60, 2.00 and 2.40 and thus mid-ordinates at 0.20, 0.60, −x2 1.00, 1.40, 1.80 and 2.20. Corresponding values of e 3 are shown in the following table.

π 2

4.

3

1 dθ (use six intervals) 1 + sin θ

ln x dx x

π 3

(use ten intervals)

√ (cos3 x)dx (use six intervals)

0

With six intervals each will have a width of

−x 2 3

x

e

0.20

0.98676

0.60

0.88692

1.00

0.71653

1.40

0.52031

1.80

0.33960

2.20

0.19922

45.4

The approximation made with the trapezoidal rule is to join the top of two successive ordinates by a straight line, i.e. by using a linear approximation of the form a + bx. With Simpson’s∗ rule, the approximation made is to join the tops of three successive ordinates by a parabola, i.e. by using a quadratic approximation of the form a + bx + cx2 . Fig. 45.3 shows a parabola y = a + bx + cx2 with ordinates y1 , y2 and y3 at x = −d, x = 0 and x = d respectively. Thus the width of each of the two intervals is d. The area enclosed by the parabola, the x-axis and ordinates x = −d and x = d is given by:

2.4

e

−x2 3

dx ≈ (0.40)[0.98676 + 0.88692

= 1.460, correct to 4 significant figures.

d

2

0

+ 0.71653 + 0.52031 + 0.33960 + 0.19922]

[

]d bx2 cx3 (a + bx + cx )dx = ax + + 2 3 −d −d ( ) bd2 cd3 = ad + + 2 3 ( ) bd2 cd3 − −ad + − 2 3

∫

From equation (2): ∫

Simpson’s rule

∗

Who was Simpson? Thomas Simpson FRS (20 August 1710–14 May 1761) was the British mathematician who invented Simpson’s rule to approximate definite integrals. For an image of Simpson, see page 214. To find out more go to www.routledge.com/cw/bird

Numerical integration 517 y

y y5

a 1 bx 1cx 2 y 5f(x)

y1

y2

y3 y2

y1 2d

d

O

y4

y2n11

x

a

O

Figure 45.3

b d

d

x

d

Figure 45.4

2 = 2ad + cd3 or 3 1 d(6a + 2cd2 ) 3

y3

∫ (3)

b

Then

y dx a

1 1 d(y1 + 4y2 + y3 ) + d(y3 + 4y4 + y5 ) 3 3 1 + d(y2n−1 + 4y2n + y2n+1 ) 3 1 ≈ d[(y1 + y2n+1 ) + 4(y2 + y4 + · · · + y2n ) 3 + 2(y3 + y5 + · · · + y2n−1 )] ≈

Since

y = a + bx + cx2 ,

at

x = −d, y1 = a − bd + cd2

at

x = 0, y2 = a

and at

x = d, y3 = a + bd + cd2

Hence y1 + y3 = 2a + 2cd2 and

y1 + 4y2 + y3 = 6a + 2cd2

(4)

Thus the area under the parabola between x = −d and x=d in Fig. 45.3 may be expressed as 13 d(y1 + 4y2 + y3 ), from equations (3) and (4), and the result is seen to be independent of the position of the origin. ∫b Let a definite integral be denoted by a y dx and represented by the area under the graph of y = f (x) between the limits x = a and x = b, as shown in Fig. 45.4. The range of integration, b − a, is divided into an even number of intervals, say 2n, each of width d. Since an even number of intervals is specified, an odd number of ordinates, 2n + 1, exists. Let an approximation to the curve over the first two intervals be a parabola of the form y = a + bx + cx2 which passes through the tops of the three ordinates y1 , y2 and y3 . Similarly, let an approximation to the curve over the next two intervals be the parabola which passes through the tops of the ordinates y3 , y4 and y5 , and so on.

i.e. Simpson’s rule states: ( ) {( ) ∫ b 1 width of first + last y dx ≈ ordinate 3 interval a ( ) sum of even +4 ordinates ( )} sum of remaining +2 odd ordinates

(5)

Note that Simpson’s rule can only be applied when an even number of intervals is chosen, i.e. an odd number of ordinates. Problem 6.

Use Simpson’s rule with (a) four ∫ 3 2 √ dx, intervals, (b) eight intervals, to evaluate x 1 correct to 3 decimal places. (a) With

four intervals, each will have a 3−1 , i.e. 0.5 and the ordinates width of 4 will occur at 1.0, 1.5, 2.0, 2.5 and 3.0. The values

518 Section H of the ordinates are as shown in the table of Problem 1(b), page 513. Thus, from equation (5): ∫

3 1

2 1 √ dx ≈ (0.5) [(2.0000 + 1.1547) 3 x + 4(1.6330 + 1.2649) + 2(1.4142)]

π rad (or 10◦ ), and the ordinates will occur at 18 π π π 2π 5π π 0, , , , , and 18 9 6 9 18 3 √( ) 1 Corresponding values of 1 − sin2 θ are shown 3 in the table below.

1 = (0.5)[3.1547 + 11.5916 3

θ + 2.8284]

= 2.929, correct to 3 decimal places. (b)

With eight intervals, each will have a width of 3−1 , i.e. 0.25 and the ordinates occur at 1.00, 8 1.25, 1.50, 1.75, . . . , 3.0. The values of the ordinates are as shown in the table in Problem 2, page 514. Thus, from equation (5): ∫

3 1

2 1 √ dx ≈ (0.25) [(2.0000 + 1.1547) 3 x + 4(1.7889 + 1.5119 + 1.3333 + 1.2060) + 2(1.6330 + 1.4142 + 1.2649)] 1 = (0.25)[3.1547 + 23.3604 3 + 8.6242] = 2.928, correct to 3 decimal places.

It is noted that the latter answer is exactly the same as that obtained by integration. In general, Simpson’s rule is regarded as the most accurate of the three approximate methods used in numerical integration. Problem 7.

Evaluate ∫ 0

π 3

√(

) 1 1 − sin2 θ dθ 3

correct to 3 decimal places, using Simpson’s rule with six intervals. π −0 With six intervals, each will have a width of 3 i.e. 6

π 18

0

π 9

π 6

(or 10◦ ) (or 20◦ ) (or 30◦ )

√( ) 1 2 1 − sin θ 1.0000 0.9950 0.9803 0.9574 3 2π 9

5π 18

π 3

(or 40◦ )

(or 50◦ )

(or 60◦ )

0.9286

0.8969

0.8660

θ √(

1 1 − sin2 θ 3

)

From equation (5) ) ∫ π √( 3 1 1 − sin2 θ dθ 3 0 ≈

1( π ) [(1.0000 + 0.8660) + 4(0.9950 3 18 + 0.9574 + 0.8969)

+ 2(0.9803 + 0.9286)] ( ) 1 π = [1.8660 + 11.3972 + 3.8178] 3 18 = 0.994, correct to 3 decimal places. Problem 8. An alternating current i has the following values at equal intervals of 2.0 milliseconds: Time (ms)

Current i (A)

0

0

2.0

3.5

4.0

8.2

6.0

10.0

8.0

7.3

10.0

2.0

12.0

0

Numerical integration 519

Charge q, in millicoulombs, is given by ∫ 12.0 q = 0 i dt. Use Simpson’s rule to determine the approximate charge in the 12 millisecond period.

From equation (5): ∫

12.0

Charge, q = 0

1 i dt ≈ (2.0) [(0 + 0) + 4(3.5 3 +10.0 + 2.0) + 2(8.2 + 7.3)]

∫ 2

∫

0.1

Practice Exercise 215 Simpson’s rule (Answers on page 892)

1.6

(use six intervals)

0 1.0

3. 0.2

∫

π 2

4.

1 dθ 1 + θ4

(use eight intervals)

sin θ dθ θ

(use eight intervals)

x cos x dx

(use six intervals)

∫

π 3

velocity v (ms−1 ) 0

1.0

0.4

2.0

1.0

3.0

1.7

4.0

2.9

5.0

4.1

6.0

6.2

7.0

8.0

8.0

9.4

The distance travelled in 8.0 s is given by ∫ 8.0 v dt 0

0

5.

A vehicle starts from rest and its velocity is measured every second for 8 s, with values as follows:

0

0

2. ∫

√ (sin x) dx

1 √ dy (1 − y2 )

time t (s)

In Problems 1 to 5, evaluate the definite integrals using Simpson’s rule, giving the answers correct to 3 decimal places.

∫

0.7

9.

Now try the following Practice Exercise

∫

(Use 8 intervals)

0

10.

1.

1 √ dx (2x − 1)

In Problems 8 and 9 evaluate the definite integrals using (a) the trapezoidal rule, (b) the mid-ordinate rule, (c) Simpson’s rule. Use 6 intervals in each case and give answers correct to 3 decimal places. ∫ 3√ 8. (1 + x4 ) dx

= 62 mC

π 2

6

7.

2

ex sin 2x dx (use ten intervals)

Estimate this distance using Simpson’s rule.

0

11. In Problems 6 and 7 evaluate the definite integrals using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate rule, (d) Simpson’s rule. Give answers correct to 3 decimal places. ∫

4

6. 1

4 dx x3

(Use 6 intervals)

A pin moves along a straight guide so that its velocity v (m/s) when it is a distance x(m) from the beginning of the guide at time t(s) is given in the table below. Use Simpson’s rule with eight intervals to determine the approximate total distance travelled by the pin in the 4.0 s period, giving the answer correct to 3 significant figures.

520 Section H

t (s)

v (m/s)

0

0

0.5

0.052

1.0

0.082

1.5

0.125

2.0

0.162

2.5

0.175

3.0

0.186

3.5

0.160

4.0

0

45.5 Accuracy of numerical integration For a function with an increasing gradient, the trapezoidal rule will tend to over-estimate and the midordinate rule will tend to under-estimate (but by half as much). The appropriate combination of the two in Simpson’s rule eliminates this error term, giving a rule which will perfectly model anything up to a cubic, and have a proportionately lower error for any function of greater complexity. In general, for a given number of strips, Simpson’s rule is considered the most accurate of the three numerical methods.

Practice Exercise 216 Multiple-choice questions on numerical integration (Answers on page 892) Each question has only one correct answer 1.

An alternating current i has the following values at equal intervals of 2 ms: Time t (ms) 0 2.0 4.0 6.0 8.0 10.0 12.0 Current I(A) 0 4.6 7.4 10.8 8.5 3.7 0 Charge q (in millicoulombs) is given by ∫ 12.0 i dt. Using the trapezoidal rule, the q= 0

approximate charge in the 12 ms period is: (a) 70 mC (b) 72.1 mC (c) 35 mC (d) 216.4 mC 2.

Using the mid-ordinate rule with 8 intervals, ) ∫ 4( 5 the value of the integral dx, cor1 + x2 0 rect to 3 decimal places is: (a) 13.258 (c) 6.630

3.

(b) 9.000 (d) 4.276

Using Simpson’s rule with 8 intervals, the ∫ 8 ( ) value of ln x2 + 3 dx, correct to 4 signif0

icant figures is: (a) 25.88 (b) 67.03 (c) 22.34 (d) 33.52

For fully worked solutions to each of the problems in Practice Exercises 213 to 215 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 13

Further integration

This Revision Test covers the material contained in Chapters 42 to 45. The marks for each question are shown in brackets at the end of each question. 1.

∫ 3∫

Determine the following integrals: ∫ ∫ (a) 5x e2x dx (b) t2 sin 2t dt

6.

∫

(14)

Evaluate correct to 3 decimal places: ∫ 4 √ x ln x dx

(11)

1

3.

Use reduction formulae to determine: ∫ ∫ 3 3x (a) x e dx (b) t4 sin t dt ∫

4.

Evaluate formula.

5.

π 2

(14)

using

a

reduction (6)

π

(1 + sin 3θ)dθdr correct to 4 sig-

Evaluate 1

0

nificant figures.

∫

2

(2x + y2 + 4z3 ) dx dy dz

(7)

(9)

1

3

5 dx using (a) integration (b) the 2 1 x trapezoidal rule (c) the mid-ordinate rule (d) Simpson’s rule. In each of the approximate methods use eight intervals and give the answers correct to 3 decimal places. (19) Evaluate

An alternating current i has the following values at equal intervals of 5 ms: Time t(ms)

0 5

Current i(A) 0 4.8 cos6 x dx

0

∫ 3∫

8.

1 −1

0

7. 2.

Evaluate

10

15

20

9.1 12.7

25

8.8

3.5

30 0

Charge q, in coulombs, is given by ∫ 30×10−3 q= i dt 0

Use Simpson’s rule to determine the approximate charge in the 30 ms period. (5)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 13, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Section I

Differential equations

Chapter 46

Introduction to differential equations Why it is important to understand: Introduction to differential equations Differential equations play an important role in modelling virtually every physical, technical, or biological process, from celestial motion, to bridge design, to interactions between neurons. Further applications are found in fluid dynamics with the design of containers and funnels, in heat conduction analysis with the design of heat spreaders in microelectronics, in rigid-body dynamic analysis, with falling objects and in exponential growth of current in an R–L circuit, to name but a few. This chapter introduces first-order differential equations – the subject is clearly of great importance in many different areas of engineering.

At the end of this chapter, you should be able to: • sketch a family of curves given a simple derivative • define a differential equation – first-order, second-order, general solution, particular solution, boundary conditions dy • solve a differential equation of the form = f (x) dx dy • solve a differential equation of the form = f (y) dx dy • solve a differential equation of the form = f (x) · f (y) dx

46.1

Family of curves

dy Integrating both sides of the derivative = 3 with dx ∫ respect to x gives y = 3 dx, i.e., y = 3x + c, where c is an arbitrary constant.

y = 3x + c represents a family of curves, each of the curves in the family depending on the value of c. Examples include y = 3x + 8, y = 3x + 3, y = 3x and y = 3x − 10 and these are shown in Fig. 46.1. Each are straight lines of gradient 3. A particular curve of a family may be determined when a point on the

526 Section I y

y

y 5 3x

30

8 y 5 3x 2 10

24 23 22 21 0 24

1

2

3

4

20

y5

4

26

12

2x 2

y 5 3x 1 3

y5

16

y5 y5 2 2x 2 2x 2 x 2 1 11 8 5

y 5 3x 1 8

x 10

28 212 24

216

23

22

21

0

1

2

3

4

x

210

Figure 46.1 Figure 46.2

curve is specified. Thus, if y = 3x + c passes through the point (1, 2) then 2 = 3(1) + c, from which, c = −1. The equation of the curve passing through (1, 2) is therefore y = 3x − 1

Problem 1.

Sketch the family of curves given by dy the equation = 4x and determine the equation of dx one of these curves which passes through the point (2, 3) dy Integrating both sides of = 4x with respect to x dx gives: ∫

dy dx = dx

∫ 4x dx,

Now try the following Practice Exercise Practice Exercise 217 Families of curves (Answers on page 893) 1.

2.

Sketch a family of curves represented by each of the following differential equations: dy dy dy (a) = 6 (b) = 3x (c) = x+2 dx dx dx Sketch the family of curves given by the equady tion = 2x + 3 and determine the equation of dx one of these curves which passes through the point (1, 3)

i.e., y = 2x2 + c

46.2 Some members of the family of curves having an equation y = 2x2 + c include y = 2x2 + 15, y = 2x2 + 8, y = 2x2 and y = 2x2 − 6, and these are shown in Fig. 46.2. To determine the equation of the curve passing through the point (2, 3), x = 2 and y = 3 are substituted into the equation y = 2x2 + c Thus 3 = 2(2)2 + c, from which c = 3 − 8 = −5 Hence the equation of the curve passing through the point (2, 3) is y = 2x2 − 5

Differential equations

A differential equation is one that contains differential coefficients. Examples include

(i)

dy = 7x dx

and (ii)

d2 y dy +5 + 2y = 0 dx2 dx

Differential equations are classified according to the highest derivative which occurs in them. Thus example (i) above is a first-order differential equation, and example (ii) is a second-order differential equation.

Introduction to differential equations 527 The degree of a differential equation is that of the highest power of the highest differential which the equation contains after simplification. ( 2 )3 ( )5 dx dx = 7 is a second-order differenThus +2 dt2 dt tial equation of degree three. Starting with a differential equation it is possible, by integration and by being given sufficient data to determine unknown constants, to obtain the original function. This process is called ‘solving the differential equation’. A solution to a differential equation which contains one or more arbitrary constants of integration is called the general solution of the differential equation. When additional information is given so that constants may be calculated the particular solution of the differential equation is obtained. The additional information is called boundary conditions. It was shown in Section 46.1 that y = 3x + c is the general solution of dy =3 the differential equation dx Given the boundary conditions x = 1 and y = 2, produces the particular solution of y = 3x − 1 Equations which can be written in the form dy dy dy = f(x), = f (y) and = f (x) · f( y) dx dx dx can all be solved by integration. In each case it is possible to separate the y’s to one side of the equation and the x’s to the other. Solving such equations is therefore known as solution by separation of variables.

46.3

The solution of equations of the dy form = f(x) dx

dy A differential equation of the form = f (x) is solved dx by integrating both sides with respect to x, ∫ i.e. y = f(x) dx

Problem 2. dy x = 2 − 4x3 dx Rearranging x

Determine the general solution of

Integrating both sides with respect to x gives: ) ∫ ( 2 − 4x2 dx y= x 4 3 i.e. y = 2 ln x − x + c, 3 which is the general solution. Problem 3.

Find the particular solution of the dy differential equation 5 + 2x = 3, given the dx 2 boundary conditions y = 1 when x = 2 5 dy 3 − 2x 3 2x dy + 2x = 3 then = = − dx dx 5 5 5 ) ∫ ( 3 2x Hence y = − dx 5 5 3x x2 i.e. y = − + c, 5 5 which is the general solution. Substituting the boundary conditions y = 1 25 and x = 2 to evaluate c gives: 1 52 = 65 − 54 + c, from which, c = 1 3x x2 Hence the particular solution is y = − + 1 5 5 Since 5

) 4. Solve the equation (Problem dθ = 5, given θ = 2 when t = 1 2t t − dt Rearranging gives: dθ 5 dθ 5 = and =t− dt 2t dt 2t Integrating both sides with respect to t gives: ) ∫ ( 5 θ= t− dt 2t t2 5 i.e. θ = − ln t + c, 2 2 which is the general solution. t−

When θ = 2, t = 1, thus 2 = 21 − 52 ln 1 + c from which, c = 32 Hence the particular solution is:

dy = 2 − 4x3 gives: dx

dy 2 − 4x3 2 4x3 2 = = − = − 4x2 dx x x x x

t2 5 3 − ln t + 2 2 2 1 2 θ = (t − 5 ln t + 3) 2 θ=

i.e.

528 Section I Problem 5. The bending moment M of a beam dM is given by = −w(l − x), where w and x are dx constants. Determine M in terms of x given: M = 12 wl 2 when x = 0

5.

1 dy + 2 = x − 3 , given y = 1 when x = 0 ex dx

6.

The gradient of a curve is given by: dy x2 + = 3x dx 2

dM = −w(l − x) = −wl + wx dx

Find the equation( of )the curve if it passes through the point 1, 13

Integrating both sides with respect to x gives: M = −wlx +

wx2 +c 2

7.

The acceleration a of a body is equal to its rate dv of change of velocity, . Find an equation for dt v in terms of t, given that when t = 0, velocity v=u

8.

An object is thrown vertically upwards with an initial velocity u of 20 m/s. The motion of the object follows the differential equation ds = u − gt, where s is the height of the object dt in metres at time t seconds and g = 9.8 m/s2 . Determine the height of the object after three seconds if s = 0 when t = 0

which is the general solution. When M = 12 wl 2 , x = 0 1 w(0)2 Thus wl 2 = −wl(0) + +c 2 2 1 2 from which, c = wl 2 Hence the particular solution is: M = −wlx +

w(x)2 1 2 + wl 2 2

1 i.e. M = w(l2 − 2lx + x2 ) 2 1 or M = w(l − x)2 2

Now try the following Practice Exercise

Practice Exercise 218 Solving equations of dy the form = f(x) (Answers on page 893) dx In Problems 1 to 5, solve the differential equations. 1.

dy = cos 4x − 2x dx

46.4

The solution of equations of the dy form = f( y) dx

dy A differential equation of the form = f(y) is initially dx dy rearranged to give dx = and then the solution is f(y) obtained by direct integration, ∫

Problem 6.

dy 2. 2x = 3 − x3 dx 3.

dy + x = 3, given y = 2 when x = 1 dx

4. 3

dy 2 π + sin θ = 0, given y = when θ = dθ 3 3

∫ dx =

i.e.

Rearranging

Find the general solution of dy = 3 + 2y dx

dy = 3 + 2y gives: dx dx =

dy f( y )

dy 3 + 2y

Introduction to differential equations 529 Integrating both sides gives: ∫ ∫ dy dx = 3 + 2y Thus, by using the substitution u = (3 + 2y) – see Chapter 38, x = 12 ln (3 + 2y) + c

(1)

It is possible to give the general solution of a differential equation in a different form. For example, if c = ln k, where k is a constant, then:

(a)

1 2

x = ln(3 + 2y) + ln k, i.e. or

Problem 8. (a) The variation of resistance, R ohms, of an aluminium conductor with dR temperature θ◦ C is given by = αR, where α dθ is the temperature coefficient of resistance of aluminium. If R = R0 when θ = 0◦ C, solve the equation for R. (b) If α = 38 × 10−4 /◦ C, determine the resistance of an aluminium conductor at 50◦ C, correct to 3 significant figures, when its resistance at 0◦ C is 24.0 Ω.

1

x = ln(3 + 2y) 2 + ln k √ x = ln [k (3 + 2y)]

by the laws of logarithms, from which, √ ex = k (3 + 2y)

(2)

(3)

dR dy = αR is of the form = f(y) dθ dx dR Rearranging gives: dθ = αR Integrating both sides gives: ∫ ∫ dR dθ = αR 1 ln R + c, α which is the general solution. Substituting the boundary conditions R = R0 when θ = 0 gives: i.e.

Equations (1), (2) and (3) are all acceptable general solutions of the differential equation dy = 3 + 2y dx

θ=

1 ln R0 + c α 1 from which c = − ln R0 α 0=

Problem 7. Determine the particular solution dy 1 of (y2 − 1) = 3y given that y = 1 when x = 2 dx 6

Hence the particular solution is

Rearranging gives: ( 2 ) ( ) y −1 y 1 dx = dy = − dy 3y 3 3y

1 1 1 ln R − ln R0 = ( ln R − ln R0 ) α α α ( ) ( ) 1 R R i.e. θ = ln or αθ = ln α R0 R0

Integrating gives: ) ∫ ∫ ( y 1 dx = − dy 3 3y y2 1 x = − ln y + c, i.e. 6 3 which is the general solution. 1 When y = 1, x = 2 6 , thus 2 16 = 16 − 13 ln 1 + c, from which, c = 2

Hence eαθ =

Hence the particular solution is: x=

y2 1 − ln y + 2 6 3

θ=

(b)

R from which, R = R0 eαθ R0

Substituting α = 38 × 10−4 , R0 = 24.0 and θ = 50 into R = R0 eαθ gives the resistance at 50◦ C, i.e. −4 R50 = 24.0 e(38×10 ×50) = 29.0 ohms

Now try the following Practice Exercise Practice Exercise 219 Solving equations of dy the form = f (y) (Answers on page 893) dx In Problems 1 to 3, solve the differential equations.

530 Section I

1.

dy = 2 + 3y dx

2.

dy = 2 cos2 y dx

3. (y2 + 2)

8.

dy 1 = 5y, given y = 1 when x = dx 2

4. The current in an electric circuit is given by the equation Ri + L

di =0 dt

where L and R are constants. Show that i = Ie

−Rt L ,

given that i = I when t = 0

46.5 The solution of equations of the dy form = f(x) · f( y) dx dy = f (x) · f (y), dx where f(x) is a function of x only and f (y) is a funcdy tion of y only, may be rearranged as = f (x) dx, and f (y) then the solution is obtained by direct integration, i.e. A differential equation of the form

∫

5. The velocity of a chemical reaction is given dx by = k(a − x), where x is the amount transdt ferred in time t, k is a constant and a is the concentration at time t = 0 when x = 0. Solve the equation and determine x in terms of t. 6.

(a) Charge Q coulombs at time t seconds is given by the differential equation dQ Q R + = 0, where C is the capacidt C tance in farads and R the resistance in ohms. Solve the equation for Q given that Q = Q0 when t = 0 (b)

The rate of cooling of a body is given by dθ = kθ, where k is a constant. If θ = 60◦ C dt when t = 2 minutes and θ = 50◦ C when t = 5 minutes, determine the time taken for θ to fall to 40◦ C, correct to the nearest second.

A circuit possesses a resistance of 250 × 103 Ω and a capacitance of 8.5 × 10−6 F, and after 0.32 seconds the charge falls to 8.0 C. Determine the initial charge and the charge after one second, each correct to 3 significant figures.

7. A differential equation relating the difference in tension T, pulley contact angle θ and coefdT ficient of friction µ is = µT. When θ = 0, dθ T = 150 N, and µ = 0.30 as slipping starts. Determine the tension at the point of slipping when θ = 2 radians. Determine also the value of θ when T is 300 N.

Problem 9.

∫ dy = f (x) dx f( y)

Solve the equation 4xy

dy = y2 − 1 dx

Separating the variables gives: ( ) 4y 1 dy = dx y2 − 1 x Integrating both sides gives: ) ∫ ( ∫ ( ) 4y 1 dy = dx 2 y −1 x Using the substitution u = y2 − 1, the general solution is: 2 ln ( y 2 − 1) = ln x + c

(1)

ln( y2 − 1)2 − ln x = c { 2 } (y − 1)2 =c from which, ln x or

and

( y2 − 1)2 = ec x

(2)

If in equation (1), c = ln A, where A is a different constant, then ln(y2 − 1)2 = ln x + ln A i.e. ln(y2 − 1)2 = ln Ax

Introduction to differential equations 531 i.e. ( y 2 − 1)2 = Ax

(3)

Equations (1) to (3) are thus three valid solutions of the differential equation

When x = 0, y = 1 thus c=0

1 2

ln 1 = ln 1 + c, from which,

Hence the particular solution is 12 ln(1 + x2 ) = ln y 1

dy 4xy = y2 − 1 dx Problem 10. Determine the particular solution dθ of = 2e3t−2θ , given that t = 0 when θ = 0 dt dθ = 2e3t−2θ = 2(e3t )(e−2θ ), dt by the laws of indices. Separating the variables gives: dθ = 2e3t dt, e−2θ i.e. e2θ dθ = 2e3t dt Integrating both sides gives: ∫ ∫ e2θ dθ = 2e3t dt Thus the general solution is:

1

i.e. ln(1 + x2 ) 2 = ln y, from which, (1 + x2 ) 2 = y √ Hence the equation of the curve is y = (1 + x2 ) Problem 12. The current i in an electric circuit containing resistance R and inductance L in series with a constant voltage source ( ) E is given by di = Ri. Solve the the differential equation E − L dt equation and find i in terms of time t given that when t = 0, i = 0 In the R−L series circuit shown in Fig. 46.3, the supply p.d., E, is given by E = VR + VL VR = iR and VL = L Hence

E = iR + L

from which

E−L

1 2θ 2 3t e = e +c 2 3

di dt

di dt

di = Ri dt L

R

When t = 0, θ = 0, thus: 1 0 2 0 e = e +c 2 3 1 2 1 from which, c = − = − 2 3 6 Hence the particular solution is: 1 2θ 2 3t 1 e = e − 2 3 6 or

3e2θ = 4e3t − 1

Problem 11. Find the curve which satisfies dy the equation xy = (1 + x2 ) and passes through the dx point (0, 1) Separating the variables gives: x dy dx = (1 + x2 ) y Integrating both sides gives: 1 2

2

ln (1 + x ) = ln y + c

VR

VL

i E

Figure 46.3

Most electrical circuits can be reduced to a differential equation. di di E − Ri Rearranging E − L = Ri gives = dt dt L and separating the variables gives: di dt = E − Ri L Integrating both sides gives: ∫ ∫ di dt = E − Ri L Hence the general solution is: 1 t − ln (E − Ri) = + c R L

532 Section I (by making Chapter 38).

a

substitution

u = E − Ri,

see where Cp and Cv are constants. Given n =

1 When t = 0, i = 0, thus − ln E = c R Thus the particular solution is:

that pVn = constant.

Cp show Cv

Separating the variables gives:

1 t 1 − ln (E − Ri) = − ln E R L R

Cv

Transposing gives: 1 1 t − ln (E − Ri) + ln E = R R L

dp dV = −Cp p V

Integrating both sides gives: ∫

t 1 [ln E − ln (E − Ri)] = R L ( ) E Rt ln = E − Ri L

Cv i.e.

dp = −Cp p

∫

dV V

Cv ln p = −Cp ln V + k

E Rt =e L E − Ri E − Ri −Rt =e L Hence E −Rt −Rt and E − Ri = Ee L and Ri = E − Ee L

Dividing throughout by constant Cv gives:

Hence current, ( ) −Rt E L i= 1−e R

Since

which represents the law of growth of current in an inductive circuit as shown in Fig. 46.4.

i.e. ln p + ln Vn = K or ln pVn = K, by the laws of logarithms.

from which

ln p = −

Cp k ln V + Cv Cv

Cp = n, then ln p + n ln V = K, Cv

where K =

k Cv

Hence pVn = eK , i.e. pVn = constant. i

Now try the following Practice Exercise

E R

Practice Exercise 220 Solving equations of dy the form = f(x) · f(y) (Answers on page dx 893)

i 5 RE (12e2Rt/L )

0

Time t

In Problems 1 to 4, solve the differential equations. 1.

dy = 2y cos x dx

2.

(2y − 1)

Figure 46.4

Problem 13. gas Cv

For an adiabatic expansion of a

dp dV + Cp =0 p V

3. 4.

dy = (3x2 + 1), given x = 1 when y = 2 dx

dy = e2x−y , given x = 0 when y = 0 dx dy 2y(1 − x) + x(1 + y) = 0, given dx when y = 1

x=1

Introduction to differential equations 533

5. Show that the solution of the equation y2 + 1 y dy = is of the form x2 + 1 x dx √( ) y2 + 1 = constant. x2 + 1 dy 6. Solve xy = (1 − x2 ) for y, given x = 0 dx when y = 1 7. Determine the equation of the curve which dy satisfies the equation xy = x2 − 1, and dx which passes through the point (1, 2) 8. The p.d., V, between the plates of a capacitor C charged by a steady voltage E through a resisdV tor R is given by the equation CR + V = E dt (a)

Solve the equation for V given that at t = 0, V = 0

(b)

Calculate V, correct to 3 significant figures, when E = 25V, C = 20 ×10−6 F, R = 200 ×103 Ω and t = 3.0 s

3.

4.

5.

6.

7.

9. Determine the value of p, given that dy x3 = p − x, and that y = 0 when x = 2 and dx when x = 6 8. Practice Exercise 221 Multiple-choice questions on the introduction to differential equations (Answers on page 893) Each question has only one correct answer 1. The order and degree of the differential ( )3 d4 y 3 dy equation 3x 4 + 4x − 5xy = 0 is: dx dx (a) third order, first degree (b) fourth order, first degree (c) first order, fourth degree (d) first order, third degree. dy = 2x 2. Solving the differential equation dx gives: (a) y = 2 ln x + c (b) y = 2x2 + c (c) y = 2 (d) y = x2 + c

Which of the following equations is a variable-separable differential equation? (a) (x + x2 y)dy = (3x + xy2 )dx (b) (x + y)dx − 5ydy = 0 (c) 3ydx = (x2 + 1)dy (d) y2 dx + (4x − 3y)dy = 0 Given that dy = x2 dx, the equation of y in terms of x if the curve passes through (1, 1) is: (a) x2 − 3y + 3 = 0 (b) x3 − 3y + 2 = 0 (c) x3 + 3y2 + 2 = 0 (d) 2y + x3 − 2 = 0 The solution of the differential equation dy − xdx = 0, if the curve passes through (1, 0), is: (a) 3x2 + 2y − 3 = 0 (b) 2y2 + x2 − 1 = 0 (c) x2 − 2y − 1 = 0 (d) 2x2 + 2y − 2 = 0 dy = 2y Solving the differential equation dx gives: (a) ln y = 2x + c (b) ln x = y2 + c (c) y = e 2x+c (d) y = ce 2x The velocity of a particle moving in a straight line is described by the differential equation: ds = 4t. At time t = 2, the position s = 12. dt The particular solution is given by: (a) s = 2t2 − 12 (b) s = 2t2 − 280 (c) s = 2t2 + 4 (d) s = 4 Which of the following solutions to the differ( ) dy ential equation = y 2x − 3 is incorrect? dx 2 (a) ln y = x2 − 3x + c (b) y = e x − 3x + c 2 2 (c) y = e x −3x+c (d) y = ce x −3x

9.

The particular solution of the differential dy equation + 3x = 4, given that when x = 1, dx y = 5, is: (a) 2y + 3x2 = 9 (b) 2y = 8x + 3x2 − 5 2 (c) 2y − 8x = 3x − 5 (d) 2y + 3x2 = 8x + 5

10.

Solving the differential equation dy = 3e 2x−3y , given that x = 0 when y = 0, dx gives: (a) 2e 3y = 9e 2x − 7 (b) y = − e(2x−3y ) (c) 2e 3y − 9e 2x = 7 (d) 2y = 3 e 2x−3y − 1

For fully worked solutions to each of the problems in Practice Exercises 217 to 220 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 47

Homogeneous first-order differential equations Why it is important to understand: Homogeneous first-order differential equations As was previously stated, differential equations play a prominent role in engineering, physics, economics and other disciplines. Applications are many and varied; for example, they are involved in combined heat conduction and convection with the design of heating and cooling chambers, in fluid mechanics analysis, in heat transfer analysis, in kinematic analysis of rigid body dynamics, with exponential decay in radioactive material, Newton’s law of cooling and in mechanical oscillations. This chapter explains how to solve a particular type of differential equation – the homogeneous first-order type.

At the end of this chapter, you should be able to: • Recognise a homogeneous differential equation dy • solve a differential equations of the form P = Q dx

47.1

Introduction

Certain first-order differential equations are not of the ‘variable-separable’ type, but can be made separable by changing the variable. dy An equation of the form P = Q, where P and Q are dx functions of both x and y of the same degree throughout, is said to be homogeneous in y and x. For example, f (x, y) = x2 + 3xy + y2 is a homogeneous function since each of the three terms are of degree 2. However, x2 − y f(x, y) = 2 is not homogeneous since the term in y 2x + y2 in the numerator is of degree 1 and the other three terms are of degree 2.

47.2 Procedure to solve differential dy equations of the form P = Q dx (i) (ii)

(iii)

dy dy Q = Q into the form = dx dx P Make the substitution y = vx (where v is a funcdy dv tion of x), from which, = v(1) + x , by the dx dx product rule. Rearrange P

dy Substitute for both y and in the equation dx dy Q = . Simplify, by cancelling, and an equation dx P results in which the variables are separable.

Homogeneous first-order differential equations 535 (iv) Separate the variables and solve using the method shown in Chapter 46. (v)

y Substitute v = to solve in terms of the original x variables.

47.3 Worked problems on homogeneous first-order differential equations

Problem 2. Find the particular solution of the dy x2 + y2 , given the boundary equation: x = dx y conditions that y = 4 when x = 1 Using the procedure of section 47.2: (i)

(i)

Rearranging y − x = x

dy gives: dx

Let y = vx then

(iii)

Substituting for y and dy x2 + y2 = gives: dx xy

which is homogeneous in x and y (ii)

dy dv Let y = vx, then =v+x dx dx

(iii)

Substituting for y and

dy gives: dx

v+x

dv vx − x x(v − 1) = = =v − 1 dx x x (iv) Separating the variables gives:

(v)

∫

1 − dx x

Hence, v = −ln x + c y y Replacing v by gives: = −ln x + c, which is x x the general solution. 2 When x = 1, y = 2, thus: = − ln 1 + c from 1 which, c = 2 y Thus, the particular solution is: = − ln x + 2 x or y = −x(ln x − 2) or y = x(2 − ln x)

dv 1 + v2 1 + v2 − v2 1 = − v= = dx v v v

1 Hence, v dv = dx x Integrating both sides gives:

Integrating both sides gives: ∫

dv x2 + v2 x2 x2 + v2 x2 1 + v2 = = = dx x(vx) vx2 v

x

dv 1 = v − 1 − v = −1, i.e. dv = − dx dx x

dv =

dy in the equation dx

(iv) Separating the variables gives:

v+x

∫

dy dv =v+x dx dx

(ii)

dy y − x = dx x

x

dy x2 + y2 = gives: dx y

dy x2 + y2 = which is homogeneous in x and y dx xy since each of the three terms on the right-hand side are of the same degree (i.e. degree 2).

Problem 1. Solve the differential equation: dy y − x = x , given x = 1 when y = 2 dx Using the above procedure:

Rearranging x

∫ v dv =

(v)

1 v2 dx i.e. = ln x + c x 2

y2 y gives: 2 = ln x + c, which is x 2x the general solution.

Replacing v by

When x = 1, y = 4, thus: which, c = 8

16 = ln 1 + c from 2

Hence, the particular solution is: or y2 = 2x2 (8 + ln x)

y2 = ln x + 8 2x2

536 Section I Now try the following Practice Exercise

(iv) Separating the variables gives: x

Practice Exercise 222 Homogeneous first-order differential equations (Answers on page 893) 1. Find the general solution of: x2 = y2

dv 2 + 12v − 10v2 = −v dx 7 − 7v

dy dx

=

(2 + 12v − 10v2 ) − v(7 − 7v) 7 − 7v

=

2 + 5v − 3v2 7 − 7v

2. Find the general solution of: dy x−y+x =0 dx

Hence,

3. Find the particular solution of the differential equation: (x2 + y2 )dy = xy dx, given that x = 1 when y = 1

Integrating both sides gives: ) ∫ ( ∫ 7 − 7v 1 dv = dx 2 + 5v − 3v2 x

4. Solve the differential equation:

7 − 7v into partial fractions 2 + 5v − 3v2 4 1 )−( ) (see Chapter 2) gives: ( 1 + 3v 2−v ) ∫ ( ∫ 4 1 1 ( )−( ) dv = Hence, dx x 1 + 3v 2−v

x + y dy = y − x dx

Resolving

5. Find the particular solution of the differential ( ) 2y − x dy equation: = 1 given that y = 3 y + 2x dx when x = 2

47.4 Further worked problems on homogeneous first-order differential equations Problem 3. Solve the equation: 7x(x − y)dy = 2(x2 + 6xy − 5y2 )dx given that x = 1 when y = 0

dx 7 − 7v dv = 2 + 5v − 3v2 x

4 ln(1 + 3v) + ln(2 − v) = ln x + c 3 y Replacing v by gives: x ( ) ( 4 3y y) ln 1 + + ln 2 − = ln x+ c 3 x x ( ) ( ) 4 x + 3y 2x − y or ln + ln = ln x+ c 3 x x which is the general solution i.e.

(v)

Using the procedure of Section 47.2: dy 2x2 + 12xy − 10y2 (i) Rearranging gives: = dx 7x2 − 7xy which is homogeneous in x and y since each of the terms on the right-hand side is of degree 2. dy dv =v+x dx dx dy (iii) Substituting for y and gives: dx (ii) Let y = vx then

( )2 dv 2x2 + 12x(vx) − 10 vx v+x = dx 7x2 − 7x(vx) =

2 + 12v − 10v2 7 − 7v

When x = 1, y = 0, thus:

4 ln 1 + ln 2 = ln 1 + c 3 from which, c = ln 2

Hence, the particular solution is: ( ) ( ) 4 x + 3y 2x − y ln + ln = ln x + ln 2 3 x x (

) 2x − y = ln(2x) x from the laws of logarithms ( )43 ( ) x + 3y 2x − y i.e. = 2x x x i.e. ln

x + 3y x

)34 (

Homogeneous first-order differential equations 537 Hence, the particular solution is: ) ( 2 y ln 2 − 1 = ln x + ln 8 = ln 8x x by the laws of logarithms ) ( 2 y2 y − 1 = 8x i.e. = 8x + 1 and Hence, x2 x2 ( ) y2 = x2 8x + 1 √ i.e. y = x (8x + 1)

Problem 4.

Show that the solution of the dy differential equation: x2 − 3y2 + 2xy = 0 is: dx √( ) y = x 8x + 1 , given that y = 3 when x = 1 Using the procedure of Section 47.2: (i)

Rearranging gives: 2xy

(ii) (iii)

dy = 3y2 − x2 dx

and

dy 3y2 − x2 = dx 2xy

dy dv =v+x dx dx dy gives: Substituting for y and dx ( )2 dv 3 vx − x2 3v2 − 1 v+x = = dx 2x(vx) 2v

(iv) Separating the variables gives: x

dv 3v2 − 1 3v2 − 1 − 2v2 v2 − 1 = − v= = dx 2v 2v 2v

2v 1 dv = dx v2 − 1 x Integrating both sides gives: ∫ ∫ 2v 1 dv = dx 2 v −1 x Hence,

(v)

Now try the following Practice Exercise

Let y = vx then

i.e. ln(v2 − 1) = ln x + c y Replacing v by gives: x ) ( 2 y ln 2 − 1 = ln x + c, x which is the general solution. ( ) 9 When y = 3, x = 1, thus: ln − 1 = ln 1 + c 1 from which, c = ln 8

Practice Exercise 223 Homogeneous first-order differential equations (Answers on page 893) 1. 2. 3.

4.

Solve the differential equation: xy3 dy = (x4 + y4 )dx dy Solve: (9xy − 11xy) = 11y2 − 16xy + 3x2 dx Solve the differential equation: dy 2x = x + 3y, given that when x = 1, y = 1 dx Show that the solution of the differential equady tion: 2xy = x2 + y2 can be expressed as: dx x = K(x2 − y2 ), where K is a constant.

5.

Determine the particular solution of dy x3 + y3 = , given that x = 1 when y = 4 dx xy2

6.

Show that the solution of the differential dy y3 − xy2 − x2 y − 5x3 equation = is of the dx xy2 − x2 y − 2x3 form: ( ) y2 4y y − 5x + + 18 ln = ln x + 42, 2x2 x x when x = 1 and y = 6

For fully worked solutions to each of the problems in Practice Exercises 222 and 223 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 48

Linear first-order differential equations Why it is important to understand: Linear first-order differential equations As has been stated in previous chapters, differential equations have many applications in engineering and science. For example, first-order differential equations model phenomena of cooling, population growth, radioactive decay, mixture of salt solutions, series circuits, survivability with AIDS, draining a tank, economics and finance, drug distribution, pursuit problem and harvesting of renewable natural resources. This chapter explains how to solve another specific type of equation, the linear first-order differential equation.

At the end of this chapter, you should be able to: • recognise a linear differential equation • solve a differential equation of the form

48.1

dy + Py = Q where P and Q are functions of x only dx

(iii)

Introduction

dy An equation of the form + Py = Q, where P and dx Q are functions of x only is called a linear differential equation since y and its derivatives are of the first degree. dy (i) The solution of + Py = Q is obtained by muldx tiplying throughout by what is termed an integrating factor. dy (ii) Multiplying + Py = Q by say R, a function of dx x only, gives: R

dy + RPy = RQ dx

(1)

The differential coefficient of a product Ry is obtained using the product rule, i.e.

d dy dR (Ry) = R + y dx dx dx

which is the same as the left-hand side of equation (1), when R is chosen such that RP =

dR dx

dR = RP, then separating the variables dx dR gives = P dx R

(iv) If

Linear first-order differential equations 539 Integrating both sides gives: ∫ ∫ ∫ dR = P dx i.e. ln R = P dx + c R from which, ∫

R=e

P dx+c

i.e. R = Ae (v)

(vi)

∫

∫

=e

P dx

P dx c

e from the laws of indices

, where A = ec = a constant.

∫

which may be checked by differentiating ye with respect to x, using the product rule. (vii)

1 dy Problem 1. Solve + 4y = 2 given the x dx boundary conditions x = 0 when y = 4 Using the above procedure:

∫

Substituting R = Ae P dx in equation (1) gives: ( ) ∫ ∫ ∫ P dx dy Ae + Ae P dx Py = Ae P dx Q dx ( ) ∫ ∫ ∫ dy i.e. e P dx + e P dx Py = e P dx Q (2) dx The left-hand side of equation (2) is d ( ∫ P dx ) ye dx

48.3 Worked problems on linear first-order differential equations

(ii)

dy + 4xy = 2x, which is of the Rearranging gives dx dy form + Py = Q where P = 4x and Q = 2x dx ∫ ∫ Pdx = 4xdx = 2x2

(iii)

Integrating factor e

(i)

e

P dx

(v)

Hence the general solution is: 2

by using the substitution u = 2x2 When x = 0, y = 4, thus 4e0 = 21 e0 + c, from which, c = 72 Hence the particular solution is

(ii) (iii)

Determine the integrating factor e

∫

(v)

P dx

2

( ) 2 2 or y = 12 + 72 e−2x or y = 12 1 + 7e−2x

Rearrange the differential equation into the form dy + Py = Q, where P and Q are functions of x dx ∫ Determine P dx ∫

2

ye2x = 12 e2x + c,

is the integrating factor.

(iv) Substitute e

2

ye2x = 12 e2x + 72

48.2 Procedure to solve differential equations of the form dy + Py = Q dx (i)

= e2x

P dx

2

(viii)

P dx

(iv) Substituting into equation (3) gives: ∫ 2 2x2 ye = e2x (2x) dx

From equation (2), ∫ d ( ∫ P dx ) ye = e P dx Q dx Integrating both sides with respect to x gives: ∫ ∫ ∫ P dx ye = e P dx Q dx (3) ∫

∫

P dx

into equation (3)

Integrate the right-hand side of equation (3) to give the general solution of the differential equation. Given boundary conditions, the particular solution may be determined.

Problem 2. Show that the solution of the dy y 3 − x2 equation + 1 = − is given by y = , given dx x 2x x = 1 when y = 1 Using the procedure of Section 48.2: ( ) 1 dy + (i) Rearranging gives: y = −1, which is dx x 1 dy + Py = Q, where P = and of the form dx x Q = −1. (Note that Q can be considered to be −1x0 , i.e. a function of x) ∫ ∫ 1 (ii) P dx = dx = ln x x (iii)

∫

Integrating factor e P dx = eln x = x (from the definition of a logarithm).

540 Section I (iv) Substituting into equation (3) gives: ∫ yx = x(−1) dx (v)

Now try the following Practice Exercise Practice Exercise 224 Linear first-order differential equations (Answers on page 893)

Hence the general solution is:

Solve the following differential equations.

−x2 yx = +c 2

1.

−1 When x = 1, y = 1, thus 1 = + c, from 2 3 which, c = 2 Hence the particular solution is: yx = i.e.

−x2 3 + 2 2

2yx = 3 − x2 and y =

3 − x2 2x

5. 6.

Using the procedure of Section 48.2:

(ii)

dy Rearranging gives + y = x, which is of the dx dy + P, = Q, where P = 1 and Q = x. form dx (In this case P can be considered to be 1x0 , i.e. a function of x). ∫ ∫ P dx = 1dx = x

(iii)

Integrating factor e

∫

P dx

= ex

(iv) Substituting in equation (3) gives: ∫ yex = ex (x) dx (v)

3. 4.

Problem 3. Determine the particular solution of dy − x + y = 0, given that x = 0 when y = 2 dx

(i)

2.

(4)

∫

ex (x) dx is determined using integration by parts (see Chapter 42). ∫ xex dx = xex − ex + c

dy =3−y dx dy = x(1 − 2y) dx dy t −5t = −y dt ) ( dy + 1 = x3 − 2y, given x = 1 when x dx y=3 1 dy +y=1 x dx dy + x = 2y dx

x

48.4 Further worked problems on linear first-order differential equations Problem 4. Solve the differential equation dy = sec θ + y tan θ given the boundary conditions dθ y = 1 when θ = 0 Using the procedure of Section 48.2: dy (i) Rearranging gives − (tan θ)y = sec θ, which is dθ dy of the form + Py = Q where P = −tan θ and dθ Q = sec θ ∫ ∫ (ii) P dθ = − tan θdθ = − ln(sec θ) = ln(sec θ)−1 = ln(cos θ) ∫

Integrating factor e P dθ = eln(cos θ) = cos θ (from the definition of a logarithm).

Hence from equation (4): ye = xe − e + c, which is the general solution.

(iii)

When x = 0, y = 2 thus 2e0 = 0 − e0 + c, from which, c = 3 Hence the particular solution is:

(iv) Substituting in equation (3) gives:

x

x

x

∫ y cos θ =

−x

ye = xe − e + 3 or y = x − 1 + 3e x

x

x

cos θ(sec θ) dθ ∫

i.e.

y cos θ =

dθ

Linear first-order differential equations 541 (v)

= 2 ln (x + 1) + ln(x − 2)

Integrating gives: y cos θ = θ + c, which is the general solution. When θ = 0, y = 1, thus 1 cos 0 = 0 + c, from which, c = 1 Hence the particular solution is:

= ln [(x + 1)2 (x − 2)] (iii)

∫

y cos θ = θ + 1 or y = (θ + 1) sec θ

e

(b)

dy 3(x − 1) + y=1 dx (x + 1)

(v)

(b)

dy 3(x − 1) 1 + y= dx (x + 1)(x − 2) (x − 2)

(ii)

which is of the form dy 3(x − 1) + Py = Q, where P = dx (x + 1)(x − 2) 1 and Q = (x − 2) ∫ ∫ 3(x − 1) P dx = dx, which is inte(x + 1)(x − 2) grated using partial fractions. 3x − 3 Let (x + 1)(x − 2) A B ≡ + (x + 1) (x − 2) ≡

A(x − 2) + B(x + 1) (x + 1)(x − 2)

from which, 3x − 3 = A(x − 2) + B(x + 1) When x = −1, −6 = −3A, from which, A = 2 When x = 2, 3 = 3B, from which, B = 1 ∫ 3x − 3 dx Hence (x + 1)(x − 2) ] ∫ [ 2 1 = + dx x+1 x−2

Hence the general solution is: y(x + 1)2 (x − 2) = 13 (x + 1)3 + c

(a) Using the procedure of Section 48.2: Rearranging gives:

2

= eln[(x+1) (x−2)] = (x + 1)2 (x − 2) y(x + 1)2 (x − 2) ∫ 1 = (x + 1)2 (x − 2) dx x − 2 ∫ = (x + 1)2 dx

Given the boundary conditions that y = 5 when x = −1, find the particular solution of the equation given in (a).

(i)

P dx

(iv) Substituting in equation (3) gives:

Problem 5. (a) Find the general solution of the equation (x − 2)

Integrating factor

When x = −1, y = 5 thus 5(0)(−3) = 0 + c, from which, c = 0 Hence y(x + 1)2 (x − 2) = 13 (x + 1)3 i.e. y =

(x + 1)3 3(x + 1)2 (x − 2)

and hence the particular solution is y=

(x + 1) 3(x − 2)

Now try the following Practice Exercise Practice Exercise 225 Linear first-order differential equations (Answers on page 893) In problems 1 and 2, solve the differential equations. dy π 1. cot x = 1 − 2y, given y = 1 when x = dx 4 dθ 2. t + sec t(t sin t + cos t)θ = sec t, given dt t = π when θ = 1 2 dy − y show 3. Given the equation x = dx x + 2 2 that the particular solution is y = ln (x + 2), x given the boundary conditions that x = −1 when y = 0 4.

Show that the solution of the differential equation 4 dy − 2(x + 1)3 = y dx (x + 1)

542 Section I di i is given by: Ri + L = E0 sin ωt. Given i = 0 dt when t = 0, show that the solution of the equation is given by: ) ( E0 (R sin ωt − ωL cos ωt) i= R 2 + ω 2 L2 ( ) E0 ωL + e−Rt/L R2 + ω 2 L2

is y = (x + 1)4 ln(x + 1)2 , given that x = 0 when y=0 5. Show that the solution of the differential equation dy + ky = a sin bx dx is given by: ( ) a y= (k sin bx − b cos bx) k 2 + b2 ( 2 ) k + b2 + ab −kx + e , k 2 + b2

8.

given y = 1 when x = 0 dv 6. The equation = −(av + bt), where a and dt b are constants, represents an equation of motion when a particle moves in a resisting medium. Solve the equation for v given that v = u when t = 0 7. In an alternating current circuit containing resistance R and inductance L the current

The concentration C of impurities of an oil purifier varies with time t and is described dC = b + dm − Cm, where a, by the equation a dt b, d and m are constants. Given C = c0 when t = 0, solve the equation and show that: ( C=

9.

)( ) b + d 1 − e−mt/α + c0 e−mt/α m

The equation of motion of a train is given dv by: m = mk(1 − e−t ) − mcv, where v is the dt speed, t is the time and m, k and c are constants. Determine the speed v given v = 0 at t = 0.

For fully worked solutions to each of the problems in Practice Exercises 224 and 225 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 49

Numerical methods for first-order differential equations Why it is important to understand: Numerical methods for first-order differential equations Most physical systems can be described in mathematical terms through differential equations. Specific types of differential equation have been solved in the preceding chapters, i.e. the separable-variable type, the homogeneous type and the linear type. However, differential equations such as those used to solve real-life problems may not necessarily be directly solvable, i.e. do not have closed form solutions. Instead, solutions can be approximated using numerical methods and in science and engineering, a numeric approximation to the solution is often good enough to solve a problem. Various numerical methods are explained in this chapter.

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

state the reason for solving differential equations using numerical methods obtain a numerical solution to a first-order differential equation using Euler’s method obtain a numerical solution to a first-order differential equation using the Euler–Cauchy method obtain a numerical solution to a first-order differential equation using the Runge–Kutta method

49.1

Introduction

Not all first-order differential equations may be solved by separating the variables (as in Chapter 46) or by the integrating factor method (as in Chapter 48). A number of other analytical methods of solving

differential equations exist. However the differential equations that can be solved by such analytical methods is fairly restricted. Where a differential equation and known boundary conditions are given, an approximate solution may be obtained by applying a numerical method. There are a number of such numerical methods

544 Section I available and the simplest of these is called Euler’s∗ method.

49.2

y

P

Euler’s method

f (h) f (0)

From Chapter 37, Maclaurin’s series may be stated as: f (x) = f (0) + x f ′ (0) +

x2 ′′ f (0) + · · · 2!

0

x h

Figure 49.1

Hence at some point f (h) in Fig. 49.1:

y

h2 f(h) = f (0) + h f (0) + f ′′ (0) + · · · 2!

P

If the y-axis and origin are moved a units to the left, as shown in Fig. 49.2, the equation of the same curve relative to the new axis becomes y = f (a + x) and the function value at P is f (a). At point Q in Fig. 49.2: h2 2!

f ′′ (a) + · · ·

y 5 f (a 1 h)

Q

′

f(a + h) = f(a) + h f ′ (a) +

y 5 f (x )

Q

(1)

f (a)

f (a 1x)

0 a

h

x

Figure 49.2

which is a statement called Taylor’s∗ series.

∗ ∗

Who was Euler? Leonhard Euler (15 April 1707–18 September 1783) was a pioneering Swiss mathematician and physicist who made important discoveries in infinitesimal calculus and graph theory. He also introduced much of the modern mathematical terminology and notation. To find out more go to www.routledge.com/cw/bird

Who was Taylor? Brook Taylor (18 August 1685–29 December 1731) solved the problem of the centre of oscillation of a body. In 1712 Taylor was elected to the Royal Society. Taylor added a new branch to mathematics now called the ‘calculus of finite differences’, invented integration by parts, and discovered the celebrated series known as Taylor’s expansion. To find out more go to www.routledge.com/cw/bird

Numerical methods for first-order differential equations 545 If h is the interval between two new ordinates y0 and y1 , as shown in Fig. 49.3, and if f (a) = y0 and y1 = f (a + h), then Euler’s method states:

With x0 = 1 and y0 = 4, ( y′ )0 = 3(1 + 1) − 4 = 2 By Euler’s method:

f (a + h) = f(a) + hf ′ (a) y1 = y0 +h ( y′ )0

i.e.

dy = y′ = 3(1 + x) − y dx

y1 = y0 + h(y′ )0 , from equation (2)

(2) Hence

y1 = 4 + (0.2)(2) = 4.4, since h = 0.2

At point Q in Fig. 49.4, x1 = 1.2, y1 = 4.4 y

and (y′ )1 = 3(1 + x1 ) − y1 y 5 f (x)

Q

i.e. ( y′ )1 = 3(1 + 1.2) − 4.4 = 2.2

P y y0

y1 Q

4.4 0

(a 1 h) a

x

P

4

h

Figure 49.3

y0

0

The approximation used with Euler’s method is to take only the first two terms of Taylor’s series shown in equation (1). Hence if y0 , h and (y′ )0 are known, y1 , which is an approximate value for the function at Q in Fig. 49.3, can be calculated. Euler’s method is demonstrated in the worked problems following.

49.3 Worked problems on Euler’s method

y1

x0 5 1

x1 5 1.2

x

h

Figure 49.4

If the values of x, y and y′ found for point Q are regarded as new starting values of x0 , y0 and (y′ )0 , the above process can be repeated and values found for the point R shown in Fig. 49.5.

y

R Q P

Problem 1. Obtain a numerical solution of the differential equation dy = 3(1 + x) − y dx given the initial conditions that x = 1 when y = 4, for the range x = 1.0 to x = 2.0 with intervals of 0.2 Draw the graph of the solution.

y0

0

1.0

y1

x0 5 1.2

x1 5 1.4 h

Figure 49.5

x

546 Section I Thus at point R, y1 = y0 + h(y′ )0

from equation (2)

(As the range is 1.0 to 2.0 there is no need to calculate (y′ )0 in line 6). The particular solution is given by the value of y against x. dy = 3(1 + x) − y with initial dx conditions x = 1 and y = 4 is shown in Fig. 49.6. In practice it is probably best to plot the graph as each calculation is made, which checks that there is a smooth progression and that no calculation errors have occurred.

A graph of the solution of

= 4.4 + (0.2)(2.2) = 4.84 When x1 = 1.4 and y1 = 4.84, ( y′ )0 = 3(1 + 1.4) − 4.84 = 2.36 This step-by-step Euler’s method can be continued and it is easiest to list the results in a table, as shown in Table 49.1. The results for lines 1 to 3 have been produced above.

y

Table 49.1 x0

(y′ )0

y0

1.

1

4

2

2.

1.2

4.4

2.2

3.

1.4

4.84

2.36

4.

1.6

5.312

2.488

5.

1.8

5.8096

2.5904

6.

2.0

6.32768

6.0

5.0

4.0 1.0

1.2

1.4

1.6

1.8

2.0

x

For line 4, where x0 = 1.6: y1 = y0 + h(y′ )0

Figure 49.6

= 4.84 + (0.2)(2.36) = 5.312 and

(y′ )0 = 3(1 + 1.6) − 5.312 = 2.488

For line 5, where x0 = 1.8: y1 = y0 + h(y′ )0

Problem 2. Use Euler’s method to obtain a numerical solution of the differential equation dy + y = 2x, given the initial conditions that at dx x = 0, y = 1, for the range x = 0(0.2)1.0. Draw the graph of the solution in this range.

= 5.312 + (0.2)(2.488) = 5.8096 and

( y′ )0 = 3(1 + 1.8) − 5.8096 = 2.5904

x = 0(0.2)1.0 means that x ranges from 0 to 1.0 in equal intervals of 0.2 (i.e. h = 0.2 in Euler’s method). dy + y = 2x, dx

For line 6, where x0 = 2.0: y1 = y0 + h(y′ )0 = 5.8096 + (0.2)(2.5904) = 6.32768

hence

dy = 2x − y, dx

i.e. y′ = 2x − y

Numerical methods for first-order differential equations 547 If initially x0 = 0 and y0 = 1, then

For line 6, where x0 = 1.0: y1 = y0 + h(y′ )0

(y′ )0 = 2(0) − 1 = −1 Hence line 1 in Table 49.2 can be completed with x = 0, y = 1 and y′ (0) = −1

Table 49.2 x0

= 0.8288 + (0.2)(0.7712) = 0.98304 As the range is 0 to 1.0, (y′ )0 in line 6 is not needed. dy A graph of the solution of + y = 2x, with initial dx conditions x = 0 and y = 1 is shown in Fig. 49.7.

(y′ )0

y0

1.

0

1

−1

2.

0.2

0.8

−0.4

3.

0.4

0.72

0.08

4.

0.6

0.736

0.464

5.

0.8

0.8288

0.7712

6.

1.0

0.98304

y 1.0

0.5

For line 2, where x0 = 0.2 and h = 0.2: y1 = y0 + h(y′ )0

(y′ )0 = 2x0 − y0 = 2(0.2) − 0.8 = −0.4

For line 3, where x0 = 0.4: y1 = y0 + h(y′ )0 = 0.8 + (0.2)(−0.4) = 0.72 and

0.4

0.6

0.8

1.0

x

(y′ )0 = 2x0 − y0 = 2(0.4) − 0.72 = 0.08

Figure 49.7

Problem 3. (a) Obtain a numerical solution, using Euler’s method, of the differential equation dy = y − x, with the initial conditions that at dx x = 0, y = 2, for the range x = 0(0.1)0.5. Draw the graph of the solution. (b)

For line 4, where x0 = 0.6:

= 0.72 + (0.2)(0.08) = 0.736 (y′ )0 = 2x0 − y0 = 2(0.6) − 0.736 = 0.464

For line 5, where x0 = 0.8: y1 = y0 + h(y′ )0 = 0.736 + (0.2)(0.464) = 0.8288 and (y′ )0 = 2x0 − y0 = 2(0.8) − 0.8288 = 0.7712

By an analytical method (using the integrating factor method of Chapter 48), the solution of the above differential equation is given by y = x + 1 + ex Determine the percentage error at x = 0.3

y1 = y0 + h(y′ )0

and

0.2

from equation (2)

= 1 + (0.2)(−1) = 0.8 and

0

(a)

dy = y′ = y − x dx If initially x0 = 0 and y0 = 2 then (y′ )0 = y0 − x0 = 2 − 0 = 2 Hence line 1 of Table 49.3 is completed.

For line 2, where x0 = 0.1: y1 = y0 + h(y′ )0 ,

from equation (2),

= 2 + (0.1)(2) = 2.2

548 Section I Table 49.3

y

x0

y0

′

(y )0

1.

0

2

2

2.

0.1

2.2

2.1

3.

0.2

2.41

2.21

4.

0.3

2.631

2.331

5.

0.4

2.8641

2.4641

6.

0.5

3.11051

3.0

2.5

2.0 0

and (y′ )0 = y0 − x0 = 2.2 − 0.1 = 2.1

0.1

0.2

0.3

0.4

0.5

x

Figure 49.8

For line 3, where x0 = 0.2: y1 = y0 + h(y′ )0 = 2.2 + (0.1)(2.1) = 2.41 and (y′ )0 = y0 − x0 = 2.41 − 0.2 = 2.21

By Euler’s method, when x = 0.3 (i.e. line 4 in Table 49.3), y = 2.631 Percentage error ( =

For line 4, where x0 = 0.3: y1 = y0 + h(y′ )0

( =

= 2.41 + (0.1)(2.21) = 2.631 and (y′ )0 = y0 − x0

actual − estimated actual 2.649859 − 2.631 2.649859

) × 100%

) × 100%

= 0.712%

= 2.631 − 0.3 = 2.331 For line 5, where x0 = 0.4: y1 = y0 + h(y′ )0 = 2.631 + (0.1)(2.331) = 2.8641

Euler’s method of numerical solution of differential equations is simple, but approximate. The method is most useful when the interval h is small.

′

and (y )0 = y0 − x0 = 2.8641 − 0.4 = 2.4641

Now try the following Practice Exercise

For line 6, where x0 = 0.5: y1 = y0 + h(y′ )0 = 2.8641 + (0.1)(2.4641) = 3.11051 dy = y − x with x = 0, y = 2 A graph of the solution of dx is shown in Fig. 49.8. (b)

If the solution of the differential equation dy = y − x is given by y = x + 1 + ex , then when dx x = 0.3, y = 0.3 + 1 + e0.3 = 2.649859

Practice Exercise 226 Euler’s method (Answers on page 894) 1.

Use Euler’s method to obtain a numerical solution of the differential equation dy y = 3 − , with the initial conditions that dx x x = 1 when y = 2, for the range x = 1.0 to x = 1.5 with intervals of 0.1. Draw the graph of the solution in this range.

Numerical methods for first-order differential equations 549

2. Obtain a numerical solution of the differen1 dy tial equation + 2y = 1, given the initial x dx conditions that x = 0 when y = 1, in the range x = 0(0.2)1.0 dy y 3. (a) The differential equation +1 = − dx x has the initial conditions that y = 1 at x = 2. Produce a numerical solution of the differential equation in the range x = 2.0(0.1)2.5 (b)

If the solution of the differential equation by an analytical method is given 4 x by y = − , determine the percentage x 2 error at x = 2.2

to obtain an approximate value of y1 at point Q. QR in Fig. 49.9 is the resulting error in the result. In an improved Euler method, called the Euler– Cauchy∗ method, the gradient at P(x0 , y0 ) across half the interval is used and then continues with a line whose gradient approximates to the gradient of the curve at x1 , shown in Fig. 49.10. Let yP1 be the predicted value at point R using Euler’s method, i.e. length RZ, where yP1 = y0 + h( y′ )0

(3)

The error shown as QT in Fig. 49.10 is now less than the error QR used in the basic Euler method and the calculated results will be of greater accuracy. The corrected value, yC1 in the improved Euler method is given by: yC1 = y0 + 12 h[( y′ )0 + f (x1 , yP1 )]

4. (a) Use Euler’s method to obtain a numer-

(4)

ical solution of the differential equation dy 2y = x − , given the initial conditions dx x that y = 1 when x = 2, in the range x = 2.0(0.2)3.0. (b)

49.4

If the solution of the differential equation x2 is given by y = , determine the per4 centage error by using Euler’s method when x = 2.8

The Euler–Cauchy method

In Euler’s method of Section 49.2, the gradient (y′ )0 at P(x0 , y0 ) in Fig. 49.9 across the whole interval h is used

y Q R P ∗

y0 0

x0

x1 h

Figure 49.9

x

Who was Cauchy? Baron Augustin-Louis Cauchy (21 August 1789–23 May 1857) was a French mathematician who became an early pioneer of analysis. A prolific writer; he wrote approximately 800 research articles. To find out more go to www.routledge.com/cw/bird ∗ Who was Euler? For image and resumé of Leonhard Euler, see page 544. To find out more go to www.rout ledge.com/cw/bird

550 Section I y

Table 49.4 x

Q T R

P

S

Z 0

x0

x0 1 1 h 2

x1

x

(y′ )0

y

1.

0

2

2

2.

0.1

2.205

2.105

3.

0.2

2.421025

2.221025

4.

0.3

2.649232625

2.349232625

5.

0.4

2.890902051

2.490902051

6.

0.5

3.147446766

h

Figure 49.10

For line 3, x1 = 0.2

The following worked problems demonstrate how equations (3) and (4) are used in the Euler–Cauchy method.

yP1 = y0 + h(y′ )0 = 2.205 + (0.1)(2.105) = 2.4155 yC1 = y0 + 12 h[(y′ )0 + f(x1 , yP1 )]

Problem 4. Apply the Euler–Cauchy method to solve the differential equation

= 2.205 + 12 (0.1)[2.105 + (2.4155 − 0.2)] = 2.421025

dy = y−x dx

(y′ )0 = yC1 − x1 = 2.421025 − 0.2 = 2.221025

in the range 0(0.1)0.5, given the initial conditions that at x = 0, y = 2

For line 4, x1 = 0.3 yP1 = y0 + h(y′ )0

dy = y′ = y − x dx Since the initial conditions are x0 = 0 and y0 = 2 then (y′ )0 = 2 − 0 = 2. Interval h = 0.1, hence x1 = x0 + h = 0 + 0.1 = 0.1 From equation (3), yP1 = y0 + h(y′ )0 = 2 + (0.1)(2) = 2.2

= 2.421025 + (0.1)(2.221025) = 2.6431275 yC1 = y0 + 12 h[(y′ )0 + f (x1 , yP1 )] = 2.421025 + 12 (0.1)[2.221025 + (2.6431275 − 0.3)]

From equation (4), yC1 = y0 + 21 h[(y′ )0 + f(x1 , yP1 )]

= 2.649232625

= y0 + 12 h[(y′ )0 + (yP1 − x1 )], in this case = 2 + 12 (0.1)[2 + (2.2 − 0.1)] = 2.205 (y′ )0 = yC1 − x1 = 2.205 − 0.1 = 2.105 If we produce a table of values, as in Euler’s method, we have so far determined lines 1 and 2 of Table 49.4. The results in line 2 are now taken as x0 , y0 and (y′ )0 for the next interval and the process is repeated.

(y′ )0 = yC1 − x1 = 2.649232625 − 0.3 = 2.349232625 For line 5, x1 = 0.4 yP1 = y0 + h(y′ )0 = 2.649232625 + (0.1)(2.349232625) = 2.884155888

Numerical methods for first-order differential equations 551 yC1 = y0 + 12 h[(y′ )0 + f (x1 , yP1 )]

Table 49.6 x

= 2.649232625 + 21 (0.1)[2.349232625 + (2.884155888 − 0.4)]

Error in Euler method

Error in Euler–Cauchy method

0

0

0

0.1

0.234%

0.00775%

0.2

0.471%

0.0156%

0.3

0.712%

0.0236%

For line 6, x1 = 0.5

0.4

0.959%

0.0319%

yP1 = y0 + h(y′ )0

0.5

1.214%

0.0405%

= 2.890902051 (y′ )0 = yC1 − x1 = 2.890902051 − 0.4 = 2.490902051

= 2.890902051 + (0.1)(2.490902051) % error with Euler method

= 3.139992256 yC1 = y0 +

′ 1 2 h[(y )0

( =

+ f (x1 , yP1 )]

(

= 2.890902051 + 21 (0.1)[2.490902051 =

+ (3.139992256 − 0.5)] = 3.147446766

actual − estimated actual

) × 100%

2.649858808 − 2.631 2.649858808

) × 100%

= 0.712% % error with Euler–Cauchy method

Problem 4 is the same example as Problem 3 and Table 49.5 shows a comparison of the results, i.e. it compares the results of Tables 49.3 and 49.4. dy = y − x may be solved analytically by the intedx grating factor method of Chapter 48 with the solution y = x + 1 + ex . Substituting values of x of 0, 0.1, 0.2, . . . give the exact values shown in Table 49.5. The percentage error for each method for each value of x is shown in Table 49.6. For example when x = 0.3,

( =

2.649858808 − 2.649232625 2.649858808

1.

0

2.

Euler method y

× 100%

= 0.0236% This calculation and the others listed in Table 49.6 show the Euler–Cauchy method to be more accurate than the Euler method.

Table 49.5 x

)

Euler–Cauchy method y

Exact value y = x + 1 + ex

2

2

2

0.1

2.2

2.205

2.205170918

3.

0.2

2.41

2.421025

2.421402758

4.

0.3

2.631

2.649232625

2.649858808

5.

0.4

2.8641

2.890902051

2.891824698

6.

0.5

3.11051

3.147446766

3.148721271

552 Section I Problem 5. Obtain a numerical solution of the differential equation dy = 3(1 + x) − y dx

For line 3, x1 = 1.4 yP1 = y0 + h(y′ )0 = 4.42 + 0.2(2.18) = 4.856 yC1 = y0 + 21 h[(y′ )0 + (3 + 3x1 − yP1 )]

in the range 1.0(0.2)2.0, using the Euler–Cauchy method, given the initial conditions that x = 1 when y=4

= 4.42 + 12 (0.2)[2.18 + (3 + 3(1.4) − 4.856)] = 4.8724

This is the same as Problem 1 on page 545, and a comparison of values may be made.

(y′ )1 = 3 + 3x1 − yP1 = 3 + 3(1.4) − 4.8724 = 2.3276

dy = y′ = 3(1 + x) − y i.e. y′ = 3 + 3x − y dx

For line 4, x1 = 1.6 yP1 = y0 + h(y′ )0 = 4.8724 + 0.2(2.3276)

x0 = 1.0, y0 = 4 and h = 0.2 (y′ )0 = 3 + 3x0 − y0 = 3 + 3(1.0) − 4 = 2

= 5.33792 yC1 = y0 + 21 h[(y′ )0 + (3 + 3x1 − yP1 )]

x1 = 1.2 and from equation (3), yP1 = y0 + h(y′ )0 = 4 + 0.2(2) = 4.4

= 4.8724 + 12 (0.2)[2.3276

yC1 = y0 + 12 h[(y′ )0 + f(x1 , yP1 )] from equation (4)

+ (3 + 3(1.6) − 5.33792)] = 5.351368

= y0 + 12 h[(y′ )0 + (3 + 3x1 − yP1 )] = 4 + 12 (0.2)[2 + (3 + 3(1.2) − 4.4)]

(y′ )1 = 3 + 3x1 − yP1 = 3 + 3(1.6) − 5.351368

= 4.42 (y′ )0 = 3 + 3x1 − yP1 = 3 + 3(1.2) − 4.42 = 2.18 Thus the first two lines of Table 49.7 have been completed.

= 2.448632 For line 5, x1 = 1.8 yP1 = y0 + h(y′ )0 = 5.351368 + 0.2(2.448632) = 5.8410944 yC1 = y0 + 12 h[(y′ )0 + (3 + 3x1 − yP1 )]

Table 49.7 x0

(y′ )0

y0

1.

1.0

4

2

2.

1.2

4.42

2.18

3.

1.4

4.8724

2.3276

4.

1.6

5.351368

2.448632

5.

1.8

5.85212176

2.54787824

6.

2.0

6.370739843

= 5.351368 + 21 (0.2)[2.448632 + (3 + 3(1.8) − 5.8410944)] = 5.85212176 (y′ )1 = 3 + 3x1 − yP1 = 3 + 3(1.8) − 5.85212176 = 2.54787824

Numerical methods for first-order differential equations 553 For line 6, x1 = 2.0

The Euler–Cauchy method is seen to be more accurate than the Euler method when x = 1.6.

yP1 = y0 + h(y′ )0 Now try the following Practice Exercise

= 5.85212176 + 0.2(2.54787824) = 6.361697408

Practice Exercise 227 Euler–Cauchy method (Answers on page 894)

yC1 = y0 + 12 h[(y′ )0 + (3 + 3x1 − yP1 )] = 5.85212176 +

1.

1 2 (0.2)[2.54787824

Apply the Euler–Cauchy method to solve the differential equation dy y = 3− dx x

+ (3 + 3(2.0) − 6.361697408)] = 6.370739843

for the range 1.0(0.1)1.5, given the initial conditions that x = 1 when y = 2

Problem 6. Using the integrating factor method the solution of the differential equation dy = 3(1 + x) − y of Problem 5 is y = 3x + e1 − x . dx When x = 1.6, compare the accuracy, correct to 3 decimal places, of the Euler and the Euler–Cauchy methods. When x = 1.6, y = 3x + e1−x = 3(1.6) + e1−1.6 = −0.6 4.8 + e = 5.348811636 From Table 49.1, page 546, by Euler’s method, when x = 1.6, y = 5.312

2.

Solving the differential equation in Problem 1 by the integrating factor method gives 3 1 y = x + . Determine the percentage error, 2 2x correct to 3 significant figures, when x = 1.3 using (a) Euler’s method and (b) the Euler– Cauchy method.

3. (a) Apply the Euler-Cauchy method to solve the differential equation dy −x = y dx for the range x = 0 to x = 0.5 in increments of 0.1, given the initial conditions that when x = 0, y = 1

% error in the Euler method (

5.348811636 − 5.312 = 5.348811636

) × 100%

(b)

= 0.688% From Table 49.7 of Problem 5, by the Euler–Cauchy method, when x = 1.6, y = 5.351368 % error in the Euler–Cauchy method ( =

5.348811636 − 5.351368 5.348811636

= −0.048%

) × 100%

4.

The solution of the differential equation in part (a) is given by y = 2ex − x − 1. Determine the percentage error in the Euler-Cauchy method, correct to 3 decimal places, when x = 0.4

Obtain a numerical solution of the differential equation 1 dy + 2y = 1 x dx using the Euler–Cauchy method in the range x = 0(0.2)1.0, given the initial conditions that x = 0 when y = 1

554 Section I 49.5

The Runge–Kutta method

The Runge–Kutta∗ method for solving first-order differential equations is widely used and provides a high degree of accuracy. Again, as with the two previous methods, the Runge–Kutta method is a step-by-step process where results are tabulated for a range of values of x. Although several intermediate calculations are needed at each stage, the method is fairly straightforward. The seven-step procedure for the Runge–Kutta method, without proof, is as follows: dy = f (x, y) given the To solve the differential equation dx initial condition y = y0 at x = x0 for a range of values of x = x0 (h)xn :

1.

Identify x0 , y0 and h, and values of x1 , x2 , x3 , . . . .

2.

Evaluate k1 = f (xn , yn ) starting with n = 0

3.

( ) h h Evaluate k2 = f xn + , yn + k1 2 2

4.

) ( h h Evaluate k3 = f xn + , yn + k2 2 2

5. 6.

( ) Evaluate k4 = f xn + h, yn + hk3 Use the values determined from steps 2 to 5 to evaluate: h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 } 6

7.

∗

Who was Runge? Carl David Tolmé Runge (1856–1927) was a German mathematician, physicist, and spectroscopist. He was co-developer of the Runge–Kutta method in the field of numerical analysis. The Runge crater on the Moon is named after him. To find out more go to www.routledge.com/cw/bird

∗

Repeat steps 2 to 6 for n = 1, 2, 3, . . .

Who was Kutta? Martin Wilhelm Kutta (3 November 1867–25 December 1944) wrote a thesis that contains the now famous Runge–Kutta method for solving ordinary differential equations. To find out more go to www.routledge.com/cw/bird

Numerical methods for first-order differential equations 555 Thus, step 1 is given, and steps 2 to 5 are intermediate steps leading to step 6. It is usually most convenient to construct a table of values. The Runge–Kutta method is demonstrated in the following worked problems.

6.

h y1 = y0 + {k1 + 2k2 + 2k3 + k4 } 6 = 2+

Problem 7. Use the Runge–Kutta method to solve the differential equation:

Using the above procedure: 1.

x0 = 0, y0 = 2 and since h = 0.1, and the range is from x = 0 to x = 0.5, then x1 = 0.1, x2 = 0.2, x3 = 0.3, x4 = 0.4, and x5 = 0.5

= 2+

k1 = f (x0 , y0 ) = f(0, 2);

3.

dy since = y − x, f(0, 2) = 2 − 0 = 2 dx ) ( h h k2 = f x0 + , y0 + k1 2 2 ( ) 0.1 0.1 = f 0+ , 2+ (2) 2 2 = f(0.05, 2.1) = 2.1 − 0.05 = 2.05

4.

2.

3.

4.

5.

k4 = f(x1 + h, y1 + hk3 )

= f(0.2, 2.421490) = 2.421490 − 0.2 = 2.221490

= f(0 + 0.1, 2 + 0.1(2.0525))

= 2.20525 − 0.1 = 2.10525

= 2.31042955 − 0.15 = 2.160430 ( ) h h k3 = f x1 + , y1 + k2 2 2 ( ) 0.1 0.1 = f 0.1 + , 2.205171 + (2.160430) 2 2 ( ) = f 0.15, 2.3131925 = 2.3131925 − 0.15

= f(0.1 + 0.1, 2.205171 + 0.1(2.163193))

k4 = f(x0 + h, y0 + hk3 )

= f(0.1, 2.20525)

dy = y − x, f (0.1, 2.205171) dx = 2.205171 − 0.1 = 2.105171 ) ( h h k2 = f x1 + , y1 + k1 2 2 ( ) 0.1 0.1 = f 0.1 + , 2.205171 + (2.105171) 2 2

= 2.163193

= 2.1025 − 0.05 = 2.0525

5.

k1 = f (x1 , y1 ) = f (0.1, 2.205171); since

= f(0.15, 2.31042955)

( ) h h k3 = f x0 + , y0 + k2 2 2 ( ) 0.1 0.1 = f 0+ , 2+ (2.05) 2 2 = f (0.05, 2.1025)

0.1 {12.31025} = 2.205171 6

A table of values may be constructed as shown in Table 49.8. The working has been shown for the first two rows. Let n = 1 to determine y2 :

Let n = 0 to determine y1 : 2.

0.1 {2 + 2(2.05) + 2(2.0525) 6 + 2.10525}

dy =y − x dx in the range 0(0.1)0.5, given the initial conditions that at x = 0, y = 2

h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 } and when n = 6 0:

6.

h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 } 6 and when n = 1:

556 Section I Table 49.8 n

xn

k1

k2

k3

k4

yn

0

0

1

0.1

2.0

2.05

2.0525

2.10525

2.205171

2

0.2

2.105171

2.160430

2.163193

2.221490

2.421403

3

0.3

2.221403

2.282473

2.285527

2.349956

2.649859

4

0.4

2.349859

2.417352

2.420727

2.491932

2.891825

5

0.5

2.491825

2.566416

2.570146

2.648840

3.148721

2

h y2 = y1 + {k1 + 2k2 + 2k3 + k4 } 6 0.1 = 2.205171+ {2.105171+2(2.160430) 6 + 2(2.163193) + 2.221490} = 2.205171 +

0.1 {12.973907} = 2.421403 6

This completes the third row of Table 49.8. In a similar manner y3 , y4 and y5 can be calculated and the results are as shown in Table 49.8. Such a table is best produced by using a spreadsheet, such as Microsoft Excel. This problem is the same as Problem 3, page 547 which used Euler’s method, and Problem 4, page 550 which used the improved Euler’s method, and a comparison of results can be made. dy The differential equation = y − x may be solved dx analytically using the integrating factor method of Chapter 48, with the solution: y=x+1+e

x

Using the above procedure: 1.

Let n = 0 to determine y1 : 2.

3.

k1 = f (x0 , y0 ) = f (1.0, 4.0); since dy = 3(1 + x) − y, dx f(1.0, 4.0) = 3(1 + 1.0) − 4.0 = 2.0 ( ) h h k2 = f x0 + , y0 + k1 2 2 ) ( 0.2 0.2 , 4.0 + (2) = f 1.0 + 2 2 = f (1.1, 4.2) = 3(1 + 1.1) − 4.2 = 2.1

4.

Substituting values of x of 0, 0.1, 0.2, . . ., 0.5 will give the exact values. A comparison of the results obtained by Euler’s method, the Euler–Cauchy method and the Runga–Kutta method, together with the exact values is shown in Table 49.9. It is seen from Table 49.9 that the Runge–Kutta method is exact, correct to 5 decimal places. Problem 8. Obtain a numerical solution of the dy differential equation: = 3(1 + x) − y in the range dx 1.0(0.2)2.0, using the Runge–Kutta method, given the initial conditions that x = 1.0 when y = 4.0

x0 = 1.0, y0 = 4.0 and since h = 0.2, and the range is from x = 1.0 to x = 2.0, then x1 = 1.2, x2 = 1.4, x3 = 1.6, x4 = 1.8, and x5 = 2.0

( ) h h k3 = f x0 + , y0 + k2 2 2 ( ) 0.2 0.2 = f 1.0 + , 4.0 + (2.1) 2 2 ( ) = f 1.1, 4.21 = 3(1 + 1.1) − 4.21 = 2.09

5.

k4 = f(x0 + h, y0 + hk3 ) = f(1.0 + 0.2, 4.1 + 0.2(2.09)) = f(1.2, 4.418) = 3(1 + 1.2) − 4.418 = 2.182

Numerical methods for first-order differential equations 557 Table 49.9

6.

x

Euler’s method y

Euler–Cauchy method y

Runge–Kutta method y

Exact value y = x + 1 + ex

0

2

2

2

2

0.1

2.2

2.205

2.205171

2.205170918

0.2

2.41

2.421025

2.421403

2.421402758

0.3

2.631

2.649232625

2.649859

2.649858808

0.4

2.8641

2.890902051

2.891825

2.891824698

0.5

3.11051

3.147446766

3.148721

3.148721271

h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 } and when 6 n = 0:

( ) h h k2 = f x1 + , y1 + k1 2 2 ( ) 0.2 0.2 , 4.418733 + (2.181267) = f 1.2 + 2 2 ( ) = f 1.3, 4.636860

3.

h y1 = y0 + {k1 + 2k2 + 2k3 + k4 } 6 = 4.0 +

0.2 {2.0 + 2(2.1) + 2(2.09) + 2.182} 6

= 4.0 +

0.2 {12.562} = 4.418733 6

= 3(1 + 1.3) − 4.636860 = 2.263140 ) ( h h k3 = f x1 + , y1 + k2 2 2 ( ) 0.2 0.2 = f 1.2 + , 4.418733 + (2.263140) 2 2 ( ) = f 1.3, 4.645047 = 3(1 + 1.3) − 4.645047

4.

A table of values is compiled in Table 49.10. The working has been shown for the first two rows. Let n = 1 to determine y2 : 2.

k1 = f(x1 , y1 ) = f(1.2, 4.418733); since

= 2.254953

dy = 3(1 + x) − y, f (1.2, 4.418733) dx

k4 = f(x1 + h, y1 + hk3 )

5.

= 3(1 + 1.2) − 4.418733 = 2.181267

= f(1.2 + 0.2, 4.418733 + 0.2(2.254953))

Table 49.10 n

xn

k1

k2

k3

k4

yn

0

1.0

1

1.2

2.0

2.1

2.09

2.182

4.418733

2

1.4

2.181267

2.263140

2.254953

2.330276

4.870324

3

1.6

2.329676

2.396708

2.390005

2.451675

5.348817

4

1.8

2.451183

2.506065

2.500577

2.551068

5.849335

5

2.0

2.550665

2.595599

2.591105

2.632444

6.367886

4.0

558 Section I = f (1.4, 4.869724) = 3(1 + 1.4) − 4.869724

6.

= 2.330276 h yn+1 = yn + {k1 + 2k2 + 2k3 + k4 } and when 6 n = 1: h y2 = y1 + {k1 + 2k2 + 2k3 + k4 } 6 = 4.418733 +

0.2 {2.181267 + 2(2.263140) 6 + 2(2.254953) + 2.330276}

= 4.418733 +

0.2 {13.547729} = 4.870324 6

This completes the third row of Table 49.10. In a similar manner y3 , y4 and y5 can be calculated and the results are as shown in Table 49.10. As in the previous problem such a table is best produced by using a spreadsheet. This problem is the same as Problem 1, page 545 which used Euler’s method, and Problem 5, page 552 which used the Euler–Cauchy method, and a comparison of results can be made. dy The differential equation = 3(1 + x) − y may be dx solved analytically using the integrating factor method of Chapter 48, with the solution: y = 3x + e1−x

Runga–Kutta method, together with the exact values is shown in Table 49.11. It is seen from Table 49.11 that the Runge–Kutta method is exact, correct to 4 decimal places. The percentage error in the Runge–Kutta method when, say, x = 1.6 is: (

) 5.348811636 − 5.348817 ×100% = −0.0001% 5.348811636

From Problem 6, page 553, when x = 1.6, the percentage error for the Euler method was 0.688%, and for the Euler–Cauchy method −0.048%. Clearly, the Runge–Kutta method is the most accurate of the three methods.

Now try the following Practice Exercise Practice Exercise 228 Runge–Kutta method (Answers on page 894) 1.

Apply the Runge–Kutta method to solve the dy y differential equation: = 3 − for the range dx x 1.0(0.1)1.5, given the initial conditions x = 1 when y = 2

2.

Obtain a numerical solution of the differen1 dy tial equation: + 2y = 1 using the Runge– x dx Kutta method in the range x = 0(0.2)1.0, given the initial conditions that x = 0 when y = 1

Substituting values of x of 1.0, 1.2, 1.4, . . ., 2.0 will give the exact values. A comparison of the results obtained by Euler’s method, the Euler–Cauchy method and the

Table 49.11

x

Euler’s method y

Euler–Cauchy method y

Runge–Kutta method y

Exact value y = 3x + e1−x

1.0

4

4

4

4

1.2

4.4

4.42

4.418733

4.418730753

1.4

4.84

4.8724

4.870324

4.870320046

1.6

5.312

5.351368

5.348817

5.348811636

1.8

5.8096

5.85212176

5.849335

5.849328964

2.0

6.32768

6.370739843

6.367886

6.367879441

Numerical methods for first-order differential equations 559

y dy +1 = − 3. (a) The differential equation: dx x has the initial conditions that y = 1 at x = 2. Produce a numerical solution of the differential equation, correct to 6 decimal places, using the Runge–Kutta method in the range x = 2.0(0.1)2.5

(b)

If the solution of the differential equation by an analytical method is given by: 4 x y = − determine the percentage error at x 2 x = 2.2

For fully worked solutions to each of the problems in Practice Exercises 226 to 228 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 14

First-order differential equations

This Revision Test covers the material contained in Chapters 46 to 49. The marks for each question are shown in brackets at the end of each question. 1. 2.

3.

dy Solve the differential equation: x + x2 = 5 given dx that y = 2.5 when x = 1 (4)

6.

dy y = + x2 − 2 dx x

Determine the equation of the curve which satisfies dy the differential equation 2xy = x2 + 1 and which dx passes through the point (1, 2) (5)

given the initial conditions that x = 1 when y = 3, for the range x = 1.0 (0.1) 1.5 (b) Apply the Euler–Cauchy method to the differential equation given in part (a) over the same range.

A capacitor C is charged by applying a steady voltage E through a resistance R. The p.d. between the plates V is given by the differential equation: CR

dV +V = E dt

(c) Apply the integrating factor method to solve the differential equation in part (a) analytically.

(a) Solve the equation for V given that when time t = 0, V = 0 (b)

(d)

Evaluate voltage V when E=50 V, C=10 µF, R = 200 kΩ and t = 1.2 s (13)

4.

Show that the solution to the differential equation: dy x2 + y2 4x = is of the form dx y ( ) √ √ 3y2 = x x3 − 1 given that y = 0 when x=1 (12)

5.

Show that the solution to the differential equation

(a) Use Euler’s method to obtain a numerical solution of the differential equation:

7.

Determine the percentage error, correct to 3 significant figures, in each of the two numerical methods when x = 1.2 (29)

Use the Runge–Kutta method to solve the difdy y = + x2 − 2 in the range ferential equation: dx x 1.0(0.1)1.5, given the initial conditions that at x = 1, y = 3. Work to an accuracy of 6 decimal places. (27)

dy + (x sin x + cos x)y = 1 dx is given by: xy = sin x + k cos x where k is a constant. (10) x cos x

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 14, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

Chapter 50

Second-order differential equations of the form 2y d dy adx2 + bdx + cy = 0 d2 y dy + b + cy = 0 dx2 dx Second-order differential equations have many engineering applications. These include free vibration analysis with simple and damped mass-spring systems, resonant and non-resonant vibration analysis, with modal analysis, time-varying mechanical forces or pressure, fluid-induced vibration such as intermittent wind, forced electrical and mechanical oscillations, tidal waves, acoustics, ultrasonic and random movements of support. This chapter explains the procedure to solve second-order differential equations d2 y dy of the form a 2 + b + cy = 0 dx dx Why it is important to understand: Second-order differential equations of the form a

At the end of this chapter, you should be able to: • identify and solve the auxiliary equation of a second-order differential equation d2 y dy • solve a second-order differential equation of the form a 2 + b + cy = 0 dx dx

50.1

Introduction

d2 y dy An equation of the form a 2 + b + cy = 0, where dx dx a, b and c are constants, is called a linear second-order differential equation with constant coefficients.

When the right-hand side of the differential equation is zero, it is referred to as a homogeneous differential equation. When the right-hand side is not equal to zero (as in Chapter 51) it is referred to as a nonhomogeneous differential equation. There are numerous engineering examples of secondorder differential equations. Three examples are:

562 Section I (i)

(ii)

(iii)

d2 q dq 1 L 2 + R + q = 0, representing an equation dt dt C for charge q in an electrical circuit containing resistance R, inductance L and capacitance C in series. ds d2 s m 2 + a + ks = 0, defining a mechanical sysdt dt tem, where s is the distance from a fixed point after t seconds, m is a mass, a the damping factor and k the spring stiffness.

a as (b)

d2 y dy Substituting these values into a 2 + b + cy = 0 dx dx gives: a(Am 2 emx ) + b(Amemx ) + c(Aemx ) = 0 i.e.

Aemx (am 2 + bm + c) = 0

Thus y = Aemx is a solution of the given equation provided that (am 2 + bm + c) = 0. am 2 + bm + c = 0 is called the auxiliary equation, and since the equation is a quadratic, m may be obtained either by factorising or by using the quadratic formula. Since, in the auxiliary equation, a, b and c are real values, then the equation may have either

(aD2 + bD + c)y = 0

Substitute m for D and solve the auxiliary equation am 2 + bm + c = 0 for m

(c) If the roots of the auxiliary equation are: (i)

d2 y P + y = 0, representing an equation for the dx2 EI deflected profile y of a pin-ended uniform strut of length l subjected to a load P. E is Young’s modulus and I is the second moment of area.

d d2 If D represents and D2 represents 2 then the above dx dx equation may be stated as (aD2 + bD + c)y = 0. This equation is said to be in ‘D-operator’ form. dy d2 y If y = Aemx then = Amemx and 2 = Am 2 emx dx dx

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

real and different, say m = α and m = β, then the general solution is y = Aeαx + Beβx

(ii)

real and equal, say m = α twice, then the general solution is y = (Ax + B)eαx

(iii)

complex, say m = α ± jβ, then the general solution is y = eαx {A cos βx + B sin βx}

(d)

Given boundary conditions, constants A and B, may be determined and the particular solution of the differential equation obtained.

The particular solutions obtained in the worked problems of Section 50.3 may each be verified by substidy d2 y tuting expressions for y, and 2 into the original dx dx equation.

50.3

Worked problems on differential equations of d2 y dy the form a 2 + b + cy = 0 dx dx

(i) two different real roots (when b2 > 4ac) or (ii) two equal real roots (when b2 = 4ac) or (iii) two complex roots (when b2 < 4ac)

50.2

Procedure to solve differential equations of the form d2 y dy a 2 + b + cy = 0 dx dx

(a) Rewrite the differential equation

Problem 1. By applying Kirchhoff’s voltage law to a circuit the following differential equation is d2 y dy obtained: 2 2 + 5 − 3y = 0. Determine the dx dx general solution. Find also the particular solution dy given that when x = 0, y = 4 and =9 dx Using the above procedure: d2 y dy + 5 − 3y = 0 in D-operator form is 2 dx dx d (2D2 + 5D − 3)y = 0, where D ≡ dx

(a) 2

Second-order differential equations 563 (b)

Substituting m for D gives the auxiliary equation

dy =3 dt 4 thus 3= (0 + B) e0 + Ae0 3 4 i.e. 3 = B + A from which, A = −1, since 3 B=3 Hence the particular solution is When t = 0,

2m + 5m − 3 = 0 2

Factorising gives: (2m − 1)(m + 3) = 0, from which, m = 21 or m = −3 (c) Since the roots are real and different the general 1 solution is y = Ae 2 x + Be−3x (d)

When x = 0, y = 4,

4

hence

4= A+B

Since

y = Ae 2 x + Be−3x 1

dy 1 1 x = Ae 2 − 3Be−3x then dx 2 dy =9 When x = 0, dx 1 thus 9 = A − 3B (2) 2 Solving the simultaneous equations (1) and (2) gives A = 6 and B = −2

Problem 3. Solve the differential equation d2 y dy + 6 + 13y = 0, given that when x = 0, y = 3 dx2 dx dy =7 and dx Using the procedure of Section 50.2: (a)

Hence the particular solution is y = 6e 2 x − 2e−3x 1

(b)

Problem 2. Find the general solution of d2 y dy 9 2 − 24 + 16y = 0 and also the particular dt dt solution given the boundary conditions that when dy t = 0, y = = 3 dt

d2 y dy (a) 9 2 − 24 + 16y = 0 in D-operator form is dt dt d (9D2 − 24D + 16)y = 0 where D ≡ dt (b) Substituting m for D gives the auxiliary equation 9m 2 − 24m + 16 = 0 Factorising gives: (3m − 4)(3m − 4) = 0, i.e. m = 43 twice. (c) Since the roots are real and equal, the general 4 solution is y = (At + B)e 3 t 0

When t = 0, y = 3 hence 3 = (0 + B)e , i.e. B = 3. 4

Since y = (At + B)e 3 t ( ) dy 4 4t 4 then = (At + B) e 3 + Ae 3 t , by the product dt 3 rule.

d2 y dy + 6 + 13y = 0 in D-operator form is dx2 dx d 2 (D + 6D + 13)y = 0, where D ≡ dx Substituting m for D gives the auxiliary equation m 2 + 6m + 13 = 0 Using the quadratic formula: √ −6 ± [(6)2 − 4(1)(13)] m= 2(1) √ −6 ± (−16) = 2

Using the procedure of Section 50.2:

(d)

4

y = (−t + 3)e 3 t or y = (3 − t)e 3 t

(1)

−6 ± j4 = −3 ± j2 2 (c) Since the roots are complex, the general solution is i.e. m=

y = e−3x (A cos 2x + B sin 2x) (d)

When x = 0, y = 3, hence 3 = e0 (A cos 0 + B sin 0), i.e. A = 3 Since y = e−3x (A cos 2x + B sin 2x) then

dy = e−3x (−2A sin 2x + 2B cos 2x) dx − 3e−3x (A cos 2x + B sin 2x), by the product rule, =e

−3x

[(2B − 3A) cos 2x − (2A + 3B) sin 2x]

564 Section I When x = 0,

dy = 7, dx

hence 7 = e0 [(2B − 3A) cos 0 − (2A + 3B) sin 0] i.e. 7 = 2B − 3A, from which, B = 8, since A = 3 Hence the particular solution is y = e−3x (3 cos 2x + 8 sin 2x) Since, from Chapter 15 , page 175, a cos√ ωt + b sin ωt = R sin(ωt + α), where a R = (a2 + b2 ) and α = tan−1 then b 3 cos 2x + 8 sin 2x √ = (32 + 82 ) sin(2x + tan−1 38 ) √ = 73 sin(2x + 20.56◦ ) √ = 73 sin(2x + 0.359)

5.

6.

d (9D2 + 30D + 25)y = 0, where D ≡ ; when dx dy x = 0, y = 0 and =2 dx d2 x dx − 6 + 9x = 0; when t = 0, x = 2 and 2 dt dt dx =0 dt

7.

d2 y dy + 6 + 13y = 0; when x = 0, y = 4 and dx2 dx dy =0 dx

8.

9.

Thus the particular solution may also be expressed as y=

d2 y dy 4 2 − 5 + y = 0; when t = 0, y = 1 and dt dt dy = −2 dt

d (4D2 + 20D + 125)θ = 0, where D ≡ ; when dt dθ t = 0, θ = 3 and = 2.5 dt

√ 73 e−3x sin(2x + 0.359)

50.4 Now try the following Practice Exercise Practice Exercise 229 Second-order differential equations of the form d2 y dy a 2 + b + cy = 0 (Answers on page 895) dx dx

Problem 4. The equation of motion of a body oscillating on the end of a spring is d2 x + 100x = 0 dt2

In Problems 1 to 3, determine the general solution of the given differential equations. 1. 6

d2 y dy − − 2y = 0 dt2 dt

2. 4

d2 θ dθ +4 +θ=0 dt2 dt

d2 y dy + 2 + 5y = 0 2 dx dx In Problems 4 to 9, find the particular solution of the given differential equations for the stated boundary conditions. 3.

2

dy dy 4. 6 2 + 5 − 6y = 0; when x = 0, y = 5 and dx dx dy = −1 dx

Further worked problems on practical differential equations d2 y dy of the form a 2 + b + cy = 0 dx dx

where x is the displacement in metres of the body from its equilibrium position after time t seconds. Determine x in terms of t given that at time t = 0, dx x = 2m and =0 dt d2 x An equation of the form 2 + m 2 x = 0 is a differential dt equation representing simple harmonic motion (SHM). Using the procedure of Section 50.2: (a)

(b)

d2 x + 100x = 0 in D-operator form is dt2 2 (D + 100)x = 0 The auxiliary equation is m 2 + 100 = 0, i.e. √ 2 m = −100 and m = (−100), i.e. m = ± j10

Second-order differential equations 565 (c) Since the roots are complex, the general solution is x = e0 (A cos 10t + B sin 10t),

Hence the particular solution is V = 5eωt + 2e−ωt

i.e. x = (A cos 10t + B sin 10t) metres (d)

When t = 0, x = 2, thus 2 = A dx = −10A sin 10t + 10B cos 10t dt dx When t = 0, =0 dt thus 0 = −10A sin 0 + 10B cos 0, i.e. B = 0

Since

sinh ωt = 21 (eωt − e−ωt )

and

cosh ωt = 12 (eωt + e−ωt )

then

sinh ωt + cosh ωt = eωt

and

cosh ωt − sinh ωt = e−ωt from Chapter 12.

Hence the particular solution may also be written as

Hence the particular solution is

V = 5(sinh ωt + cosh ωt)

x = 2 cos 10t metres

+ 2(cosh ωt − sinh ωt) i.e. V = (5 + 2) cosh ωt + (5 − 2) sinh ωt

Problem 5. Given the differential equation d2 V = ω 2 V, where ω is a constant, show that its dt2 solution may be expressed as:

i.e. V = 7 cosh ωt + 3 sinh ωt Problem 6.

V = 7 cosh ωt + 3 sinh ωt

d2 i R di 1 + + i=0 dt2 L dt LC

given the boundary conditions that when t = 0, V = 7 and

dV = 3ω dt

Using the procedure of Section 50.2: (a)

(b)

d2 V d2 V = ω 2 V, i.e. − ω 2 V = 0 in D-operator 2 dt dt2 d form is (D2 − ω 2 )V = 0, where D ≡ dt The auxiliary equation is m 2 − ω 2 = 0, from which, m 2 = ω 2 and m = ±ω

represents a current i flowing in an electrical circuit containing resistance R, inductance L and capacitance C connected in series. If R = 200 ohms, L = 0.20 henry and C = 20 × 10−6 farads, solve the equation for i given the boundary conditions that di when t = 0, i = 0 and = 100 dt Using the procedure of Section 50.2: (a)

d2 i R di 1 + + i = 0 in D-operator form is dt2 L dt LC ( ) R 1 d 2 D + D+ i = 0 where D ≡ L LC dt

(c) Since the roots are real and different, the general solution is V = Aeωt + Be−ωt (d)

When t = 0, V = 7 hence 7 = A + B

(1)

(b)

R 1 The auxiliary equation is m 2 + m + =0 L LC v[ u ( )2 ( )] R u 1 R − ±t − 4(1) L L LC

dV = Aωeωt − Bωe−ωt dt When thus i.e.

t = 0,

dV = 3ω, dt

Hence m =

From equations (1) and (2), A = 5 and B = 2

2

When R = 200, L = 0.20 and C = 20 × 10−6 , then

3ω = Aω − Bω, 3 = A−B

The equation

v[ ] u ( )2 200 u 200 4 t − ± − 0.20 0.20 (0.20)(20 × 10−6 )

(2) m=

2

566 Section I =

−1000 ± 2

√

0

thus

= −500

A = 12 and B = −8 Hence the particular solution is x = 12e−2t − 8e−4t

When t = 0, i = 0, hence B = 0 di = (At + B)(−500e−500t ) + (e−500t )(A), dt by the product rule di When t = 0, = 100, thus 100 = −500B + A dt i.e. A = 100, since B = 0

i.e. displacement, x = 4(3e−2t − 2e−4t ) cm Now try the following Practice Exercise Practice Exercise 230 Second-order differential equations of the form d2 y dy a 2 + b + cy = 0 (Answers on page 895) dx dx

Hence the particular solution is i = 100te−500t A Problem 7. The oscillations of a heavily damped pendulum satisfy the differential equation d2 x dx + 6 + 8x = 0, where x cm is the 2 dt dt displacement of the bob at time t seconds. The initial displacement )is equal to +4 cm and the ( dx is 8 cm/s. Solve the initial velocity i.e. dt equation for x.

1.

The charge q on a capacitor in a certain electrical circuit satisfies the differential equad2 q dq tion 2 + 4 + 5q = 0. Initially (i.e. when dt dt dq t = 0), q = Q and = 0. Show that the dt charge √ in the circuit can be expressed as: q = 5 Qe−2t sin(t + 0.464)

2.

A body moves in a straight line so that its distance s metres from the origin after time d2 s t seconds is given by 2 + a2 s = 0, where a is dt a constant. Solve the equation for s given that ds 2π s = c and = 0 when t = dt a

3.

The motion of the pointer of a galvanometer about its position of equilibrium is represented by the equation

Using the procedure of Section 50.2: (a)

(b)

d2 x dx + 6 + 8x = 0 in D-operator form is dt2 dt d 2 (D + 6D + 8)x = 0, where D ≡ dt The auxiliary equation is m 2 + 6m + 8 = 0 Factorising gives: (m + 2)(m + 4) = 0, from which, m = −2 or m = −4

d2 θ dθ + K + Fθ = 0 dt2 dt If I, the moment of inertia of the pointer about its pivot, is 5 × 10−3 , K, the resistance due to friction at unit angular velocity, is 2 × 10−2 and F, the force on the spring necessary to produce unit displacement, is 0.20, solve the equation for θ in terms of t given that when dθ t = 0, θ = 0.3 and =0 dt I

(c) Since the roots are real and different, the general solution is x = Ae−2t + Be−4t (d)

Initial displacement means that time t = 0. At this instant, x = 4 Thus 4 = A + B

(1)

Velocity, dx = −2Ae−2t − 4Be−4t dt dx = 8 cm/s when t = 0, dt

(2)

From equations (1) and (2),

(c) Since the two roots are real and equal (i.e. −500 twice, since for a second-order differential equation there must be two solutions), the general solution is i = (At + B)e−500t (d)

8 = −2A − 4B

4.

Determine an expression for x for a differd2 x dx ential equation 2 + 2n + n 2 x = 0 which dt dt

Second-order differential equations 567

represents a critically damped oscillator, dx given that at time t = 0, x = s and =u dt di 1 d2 i 5. L 2 + R + i = 0 is an equation repredt dt C senting current i in an electric circuit. If inductance L is 0.25 henry, capacitance C is 29.76 × 10−6 farads and R is 250 ohms, solve the equation for i given the boundary condidi tions that when t = 0, i = 0 and = 34 dt

2.

A differential equation is given by d2 y dy − 4 + 4y = 0. The particular solution, dt2 dt dy given that when t = 0, y = 5 and = 13, is: ( ) 2t ( dt ) − 2t (a) y = (5t + 3) e (b) y = (2t + 5) e (c) y = 3t + 5 e 2t (d) y = 3t + 5 e − 2t

3.

A differential equation is given by d2 x dx + 10 + 29x = 0. The particular dt2 dt solution, given that when t = 0, x = 2 and dx = 2, is: dt ( ) (a) x = 2e − 5t (cos 2t − 2 sin 2t ) (b) x = 2e − 2t ( cos 5t + 3 sin 2t) (c) x = 2e − 5t( cos 2t + 3 sin 2t ) (d) x = 2e 2t cos 5t − 2 sin 5t

6. The displacement s of a body in a damped mechanical system, with no external forces, satisfies the following differential equation: 2

ds d2 s + 6 + 4.5s = 0 2 dt dt

where t represents time. If initially, when

4.

ds t = 0, s = 0 and = 4, solve the differential dt equation for s in terms of t.

Practice Exercise 231 Multiple-choice questions on second order differential d2 y dy equations of the form a 2 + b + cy = 0 dx dx (Answers on page 895) Each question has only one correct answer 1. A differential equation is given by d2 y dy − − 6y = 0. The particular soludx2 dx tion, given that when x = 0, y = 3 and dy = −1, is: dx (a) y = −7e 3x + 10e − 2x (b) y = e 3x + 2e − 2x (c) y = e − 3x + 2e 2x (d) y = 2e 3x + e − 2x

5.

d2 x + 64x = 0 is an equation representing dt2 SHM. Given that at time t = 0 s, x = 4 m and dx = 0, the solution of the equation for x in dt terms of t is: (a) x = 4(cos 8t + sin 8t) (b) x = 4 sin 8t (c) x = 4(cos 8t − sin 8t) (d) x = 4 cos 8t A differential equation is given by d2 y dy − − 2y = 0. The particular soludx2 dx tion, given that when x = 0, y = 1 and dy = −7, is: dx (a) y = 3e − x − 2e 2x (b) y = 3e 2x − 2e − x −x 2x (c) y = 3e + e (d) y = 2e − x − 3e 2x

For fully worked solutions to each of the problems in Practice Exercises 229 and 230 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 51

Second-order differential equations of the form 2y d dy adx2 + bdx + cy = f(x) d2 y dy + b + cy = f (x) dx2 dx Second-order differential equations have many engineering applications. Differential equations govern the fundamental operation of important areas such as automobile dynamics, tyre dynamics, aerodynamics, acoustics, active control systems, including speed control, engine performance and emissions control, climate control, ABS control systems, airbag deployment systems, structural dynamics of buildings, bridges and dams, for example, earthquake and wind engineering, industrial process control, control and operation of automation (robotic) systems, the operation of the electric power grid, electric power generation, orbital dynamics of satellite systems, heat transfer from electrical equipment (including computer chips), economic systems, biological systems, chemical systems and so on. This chapter explains the d2 y dy procedure to solve second-order differential equations of the form a 2 + b + cy = f (x) for different dx dx functions f (x). Equations of this form are referred to as non-homogeneous linear second order differential equations because the right-hand side is not equal to zero. Why it is important to understand: Second-order differential equations of the form a

At the end of this chapter, you should be able to: • identify the complementary function and particular integral of a second-order differential equation d2 y dy • solve a second order differential equation of the form a 2 + b + cy = f(x) where f(x) is a constant or a dx dx polynomial function, an exponential function, a sine or cosine function, or a sum or product

Second-order differential equations 569 If we let

51.1 Complementary function and particular integral

a

If in the differential equation

a

d2 u du + b + cu = 0 2 dx dx

(3)

a (1)

the substitution y = u + v is made then: d2 (u + v) d(u + v) +b + c(u + v) = f (x) dx2 dx

Rearranging gives: (

(2)

then

dy d2 y + b + cy = f (x) dx2 dx

a

d2 v dv + b + cv = f (x) dx2 dx

) ( 2 ) d2 u du dv dv a 2 + b + cu + a 2 + b +cv = f (x) dx dx dx dx

The general solution, u, of equation (3) will contain two unknown constants, as required for the general solution of equation (1). The method of solution of equation (3) is shown in Chapter 50. The function u is called the complementary function (C.F.). If the particular solution, v, of equation (2) can be determined without containing any unknown constants then y = u + v will give the general solution of equation (1). The function v is called the particular integral (P.I.). Hence the general solution of equation (1) is given by: y = C.F. + P.I.

Table 51.1 Form of particular integral for different functions Type

Straightforward cases Try as particular integral:

‘Snag’ cases Try as particular integral:

(a) f (x) = a constant

v=k

v = kx (used when C.F. contains a constant)

(b) f (x) = polynomial (i.e.

v = a + bx + cx2 + · · ·

See problem 1, 2 3

f(x) = L + Mx + Nx + · · · 2

where any of the coefficients may be zero) (c) f (x) = an exponential function

v = keax

(i) v = kxeax (used when eax

ax

(i.e. f (x) = Ae )

4, 5

appears in the C.F.) (ii) v = kx2 eax (used when eax and xeax both appear in the C.F.)

(d) f (x) = a sine or cosine function

v = A sin px + B cos px

v = x(A sin px + B cos px)

(i.e. f (x) = a sin px + b cos px,

(used when sin px and/or

where a or b may be zero)

cos px appears in the C.F.)

(e) f (x) = a sum e.g.

6

7, 8

9

(i) f (x) = 4x − 3 sin 2x 2

2

(i) v = ax + bx + c + d sin 2x + e cos 2x

(ii) f (x) = 2 − x + e

3x

(f ) f(x) = a product e.g. x

f (x) = 2e cos 2x

(ii) v = ax + b + ce3x v = ex (A sin 2x + B cos 2x)

10

570 Section I 51.2

(ii)

Procedure to solve differential equations of the form dy d2 y a 2 + b + cy = f(x) dx dx

(iii)

(i)

Rewrite the given differential equation as (aD2 + bD + c)y = f (x).

(ii)

Substitute m for D, and solve the auxiliary equation am 2 + bm + c = 0 for m.

(iii)

(vi)

Since the roots are real and different, the C.F., u = Aex + Be−2x (iv) Since the term on the right-hand side of the given equation is a constant, i.e. f (x) = 4, let the P.I. also be a constant, say v = k (see Table 51.1(a)). (v)

Substituting v = k into (D2 + D − 2)v = 4 gives (D2 + D − 2)k = 4. Since D(k) = 0 and D2 (k) = 0 then −2k = 4, from which, k = −2. Hence the P.I., v = −2

(vi)

The general solution is given by y = u + v, i.e. y = Aex + Be−2x − 2

Obtain the complementary function, u, which is achieved using the same procedure as in Section 50.2(c), page 562.

(iv) To determine the particular integral, v, firstly assume a particular integral which is suggested by f(x), but which contains undetermined coefficients. Table 51.1 gives some suggested substitutions for different functions f(x). (v)

Substitute the suggested P.I. into the differential equation (aD2 + bD + c)v = f (x) and equate relevant coefficients to find the constants introduced.

Problem 2.

Determine the particular solution of d2 y dy the equation 2 − 3 = 9, given the boundary dx dx dy conditions that when x = 0, y = 0 and =0 dx Using the procedure of Section 51.2: (i)

The general solution is given by y = C.F. + P.I., i.e. y = u + v

(vii)

Given boundary conditions, arbitrary constants in the C.F. may be determined and the particular solution of the differential equation obtained.

51.3

Using the procedure of Section 51.2: d2 y dy + − 2y = 4 dx2 dx 2 (D + D − 2)y = 4

in

D-operator

form

d2 y dy − 3 =9 dx2 dx is (D2 − 3D)y = 9

in

D-operator

form

(ii)

Substituting m for D gives the auxiliary equation m 2 − 3m = 0. Factorising gives: m(m − 3) = 0, from which, m = 0 or m = 3

(iii)

Since the roots are real and different, the C.F., u = Ae0 + Be3x , i.e. u = A + Be3x

(iv) Since the C.F. contains a constant (i.e. A) then let the P.I., v = kx (see Table 51.1(a)).

Differential equations of the d2 y dy form a 2 + b + cy = f(x) dx dx where f(x) is a constant or polynomial

Problem 1. Solve the differential equation d2 y dy + − 2y = 4 dx2 dx

(i)

Substituting m for D gives the auxiliary equation m 2 + m − 2 = 0. Factorising gives: (m − 1)(m + 2) = 0, from which m = 1 or m = −2

is

(v)

Substituting v = kx into (D2 − 3D)v = 9 gives (D2 − 3D)kx = 9 D(kx) = k and D2 (kx) = 0 Hence (D2 − 3D)kx = 0 − 3k = 9, from which, k = −3 Hence the P.I., v = −3x

(vi)

The general solution is given by y = u + v, i.e. y = A + Be3x −3x

(vii)

When x = 0, y = 0, thus 0 = A + Be0 − 0, i.e. 0=A+B (1) dy dy = 3Be3x − 3; = 0 when x = 0, thus dx dx 0 = 3Be0 − 3 from which, B = 1. From equation (1), A = −1

Second-order differential equations 571 Hence the particular solution is y = −1 + 1e3x − 3x, i.e. y = e3x − 3x − 1 Problem 3. Solve the differential equation dy d2 y 2 2 − 11 + 12y = 3x − 2 dx dx Using the procedure of Section 51.2: (i)

2

d2 y dy − 11 + 12y = 3x − 2 in D-operator form is dx dx2

Now try the following Practice Exercise Practice Exercise 232 Second-order differential equations of the form dy d2 y + cy = f(x) where f (x) is a a 2 +b dx dx constant or polynomial (Answers on page 895) In Problems 1 and 2, find the general solutions of the given differential equations. 1.

2

d2 y dy + 5 − 3y = 6 dx2 dx

2.

6

d2 y dy + 4 − 2y = 3x − 2 dx2 dx

(2D2 − 11D + 12)y = 3x − 2 (ii)

(iii)

Substituting m for D gives the auxiliary equation 2m 2 − 11m + 12 = 0. Factorising gives: (2m − 3)(m − 4) = 0, from which, m = 32 or m=4 Since the roots are real and different, the C.F.,

In Problems 3 and 4 find the particular solutions of the given differential equations. 3.

3

u = Ae 2 x + Be4x (iv) Since f (x) = 3x − 2 is a polynomial, let the P.I., v = ax + b (see Table 51.1(b)). (v)

4.

Substituting v = ax + b into (2D2 − 11D + 12)v = 3x − 2 gives: (2D2 − 11D + 12)(ax + b) = 3x − 2,

In a galvanometer the deflection θ satisfies dθ d2 θ the differential equation 2 + 4 + 4 θ = 8. dt dt Solve the equation for θ given that when t = 0, dθ θ= =2 dt

Equating the coefficients of x gives: 12a = 3, from which, a = 41

11 3 1 = i.e. b = 4 4 16 1 1 Hence the P.I., v = ax + b = x + 4 16 The general solution is given by y = u + v, i.e. 12b = −2 +

(vi)

3 1 1 y = Ae 2 x + Be4x + x + 4 16

x = 0,

6.

0 − 11a + 12ax + 12b = 3x − 2

−11a + 12b = −2 ( ) i.e. −11 14 + 12b = −2 from which,

when

The charge q in an electric circuit at time t dq 1 d2 q satisfies the equation L 2 + R + q = E, dt dt C where L, R, C and E are constants. Solve the equation given L = 2H, C = 200 × 10−6 F and E = 250 V, when (a) R = 200 Ω and (b) R is negligible. Assume that when t = 0, q = 0 and dq =0 dt

+ 12(ax + b) = 3x − 2

Equating the constant terms gives:

d2 y dy − 12 + 4y = 3x − 1; dx2 dx 4 dy =− y = 0 and dx 3 9

5.

i.e. 2D2 (ax + b) − 11D(ax + b) i.e.

d2 y dy 3 2 + − 4y = 8; when x = 0, y = 0 and dx dx dy =0 dx

51.4

Differential equations of the d2 y dy form a 2 + b + cy = f(x) dx dx where f(x) is an exponential function

572 Section I Using the procedure of Section 51.2:

Problem 4. Solve the equation dy d2 y − 2 + y = 3e4x given the boundary 2 dx dx dy conditions that when x = 0, y = − 23 and = 4 13 dx

2

(ii)

Substituting m for D gives the auxiliary equation 2m 2 − m − 3 = 0. Factorising gives: (2m − 3)(m + 1) = 0, from which, m = 32 or m = −1. Since the roots are real and different

Using the procedure of Section 51.2: (i) (ii)

(iii)

d2 y dy − 2 + y = 3e4x in D-operator form is dx2 dx (D2 − 2D + 1)y = 3e4x

3

then the C.F., u = Ae 2 x + Be−x 3

Since e 2 x appears in the C.F. and in the righthand side of the differential equation, let the

Substituting m for D gives the auxiliary equation m 2 − 2m + 1 = 0. Factorising gives: (m − 1)(m − 1) = 0, from which, m = 1 twice.

(iii)

Since the roots are real and equal the C.F., u = (Ax + B)ex

(iv) Substituting v = kxe 2 x into (2D2 − D − 3)v = 5e 2 x

(iv) Let the particular Table 51.1(c)). (v)

3 d2 y dy − − 3y = 5e 2 x in D-operator form is 2 dx dx 3 2 (2D − D − 3)y = 5e 2 x

(i)

integral,

v = ke4x

3

P.I., v = kxe 2 x (see Table 51.1(c), snag case (i)). 3

3

(see

by the product rule, 3 ( ) = ke 2 x 32 x + 1 ) [ ] ( 3 ( 3 ) D2 kxe 2 x = D ke 2 x 32 x + 1

(D2 − 2D + 1)v = 3e4x gives: (D2 − 2D + 1)ke4x = 3e4x i.e. D2 (ke4x ) − 2D(ke4x ) + 1(ke4x ) = 3e4x

( ) 3 (3) x = ke 2 2

16ke4x − 8ke4x + ke4x = 3e4x 4x

4x

Hence 9ke = 3e , from which, k =

) + 2x + 1 (3

1 3

Hence the P.I., v = ke4x = 13 e4x (vi) (vii)

3

gives: (2D2 − D − 3)kxe 2 x = 5e 2 x ) ( ) ( ) ( 3 3 3 x 3 2x 2 = (kx) 2 e + e 2 x (k), D kxe

Substituting v = ke4x into

i.e.

3

3

= ke 2 x

The general solution is given by y = u + v, i.e. y = (Ax + B)ex + 31 e4x

(9

)

)

4x + 3

) ( 3 x Hence (2D − D − 3) kxe 2 [ ] [ ] 3 ( 3 ( ) ) x x 9 3 = 2 ke 2 4 x + 3 − ke 2 2 x + 1 [ ] 3 3 − 3 kxe 2 x = 5e 2 x 3

3

3

3

i.e. 92 xke 2 x + 6ke 2 x − 32 kxe 2 x − ke 2 x 3

3

− 3kxe 2 x = 5e 2 x

3

y = (4x − 1)ex + 13 e4x

Equating coefficients of e 2 x gives: 5k = 5, from which, k = 1 3

Solve the differential equation 3 d2 y dy 2 2− − 3y = 5e 2 x dx dx

3 3 2x 2 ke

2

When x = 0, y = − 32 thus − 23 = (0 + B)e0 + 31 e0 , from which, B = −1 dy = (Ax + B)ex + ex (A) + 34 e4x dx dy 1 13 4 When x = 0, = 4 , thus =B+A+ dx 3 3 3 from which, A = 4, since B = −1 Hence the particular solution is:

Problem 5.

(

3

Hence the P.I., v = kxe 2 x = xe 2 x (v)

The

general

3 y = Ae 2 x

solution

3 + Be−x + xe 2 x

is

y = u + v,

i.e.

Second-order differential equations 573

Problem 6.

Solve

d2 y dy − 4 + 4y = 3e2x 2 dx dx

Using the procedure of Section 51.2: (i)

d2 y dy − 4 + 4y = 3e2x in D-operator form is 2 dx dx (D2 − 4D + 4)y = 3e2x

(ii)

Substituting m for D gives the auxiliary equation m 2 − 4m + 4 = 0. Factorising gives: (m − 2)(m − 2) = 0, from which, m = 2 twice.

(iii)

Since the roots are real and equal, the C.F., u = (Ax + B)e2x

1.

d2 y dy − − 6y = 2ex dx2 dx

2.

d2 y dy − 3 − 4y = 3e−x dx2 dx

3.

d2 y + 9y = 26e2x dx2

t d2 y dy − 6 + y = 12e 3 2 dt dt In problems 5 and 6 find the particular solutions of the given differential equations.

4.

5.

(iv) Since e2x and xe2x both appear in the C.F. let the P.I., v = kx2 e2x (see Table 51.1(c), snag case (ii)). (v)

Substituting v = kx2 e2x into (D2 − 4D + 4)v = 3e2x gives: (D2 − 4D + 4)(kx2 e2x ) = 3e2x

9

dy 1 d2 y + 9 − 2y = 3ex ; when x = 0, y = dx2 dx 4 dy and =0 dx 5

d2 y dy − 6 + 9y = 4e3t ; when t = 0, y = 2 and 2 dt dt dy =0 dt

6.

D(kx2 e2x ) = (kx2 )(2e2x ) + (e2x )(2kx) = 2ke2x (x2 + x) D2 (kx2 e2x ) = D[2ke2x (x2 + x)] 2x

51.5 2

2x

= (2ke )(2x + 1) + (x + x)(4ke ) = 2ke2x (4x + 1 + 2x2 ) Hence (D2 − 4D + 4)(kx2 e2x ) = [2ke2x (4x + 1 + 2x2 )]

Differential equations of the d2 y dy form a 2 + b + cy = f(x) dx dx where f(x) is a sine or cosine function

− 4[2ke2x (x2 + x)] + 4[kx2 e2x ] = 3e2x from which, 2ke2x = 3e2x and k = 23 Hence the P.I., v = kx2 e2x = 23 x2 e2x (vi)

The general solution, y = u + v, i.e. y = (Ax + B)e2x + 23 x2 e2x

Now try the following Practice Exercise

Problem 7. Solve the differential equation d2 y dy 2 2 + 3 − 5y = 6 sin 2x which represents the dx dx motion of a vibrating mass. Using the procedure of Section 51.2: d2 y dy + 3 − 5y = 6 sin 2x in D-operator form dx2 dx is (2D2 + 3D − 5)y = 6 sin 2x

(i)

2

(ii)

The auxiliary equation is 2m 2 + 3m − 5 = 0, from which, (m − 1)(2m + 5) = 0,

Practice Exercise 233 Second-order differential equations of the form d2 y dy a 2 + b +cy = f(x) where f(x) is an dx dx exponential function (Answers on page 895)

(iii)

In Problems 1 to 4, find the general solutions of the given differential equations.

(iv) Let the P.I., Table 51.1(d)).

i.e. m = 1 or m = −52 Since the roots are real and different the C.F., 5 u = Aex + Be− 2 x v = A sin 2x + B cos 2x

(see

574 Section I (v)

Substituting v = A sin 2x + B cos 2x into (2D2 + 3D − 5)v = 6 sin 2x gives: (2D2 + 3D−5)(A sin 2x + B cos 2x) = 6 sin 2x

Using the procedure of Section 51.2: (i)

(D2 + 16)y = 10 cos 4x

D(A sin 2x + B cos 2x) = 2A cos 2x − 2B sin 2x

(ii)

D2 (A sin 2x + B cos 2x)

(iii)

= D(2A cos 2x − 2B sin 2x) = −4A sin 2x − 4B cos 2x Hence

= −8A sin 2x − 8B cos 2x + 6A cos 2x − 6B sin 2x − 5A sin 2x − 5B cos 2x = 6 sin 2x Equating coefficients, of sin 2x gives: −13A − 6B = 6

6 × (1)gives : −78A − 36B = 36 78A − 169B = 0

(1)

(vi)

(2)

(D2 + 16)[x(C sin 4x + D cos 4x)] = 10 cos 4x

= x(4C cos 4x − 4D sin 4x) + (C sin 4x + D cos 4x)(1),

(4)

by the product rule

−36 Substituting B = into equation (1) or (2) 205 −78 gives A = 205 −78 36 Hence the P.I., v = sin 2x − cos 2x 205 205 The general solution, y = u + v, i.e. x

D2 [x(C sin 4x + D cos 4x)] = x(−16C sin 4x − 16D cos 4x) + (4C cos 4x − 4D sin 4x)(1) + (4C cos 4x − 4D sin 4x) Hence (D2 + 16)[x(C sin 4x + D cos 4x)] = −16Cx sin 4x−16Dx cos 4x + 4C cos 4x − 4D sin 4x + 4C cos 4x − 4D sin 4x

− 52 x

y = Ae + Be 2 − (39 sin 2x + 18 cos 2x) 205

d2 y Problem 8. Solve 2 + 16y = 10 cos 4x given dx dy y = 3 and = 4 when x = 0 dx

into

(3)

−36 B= 205

from which,

Substituting v = x(C sin 4x + D cos 4x) (D2 + 16)v = 10 cos 4x gives:

D[x(C sin 4x + D cos 4x)]

− 205B = 36

(3) + (4)gives :

Since the roots are complex the C.F., u = e0 (A cos 4x + B sin 4x)

(iv) Since sin 4x occurs in the C.F. and in the right-hand side of the given differential equation, let the P.I., v = x(C sin 4x + D cos 4x) (see Table 51.1(d), snag case – constants C and D are used since A and B have already been used in the C.F.). (v)

Equating coefficients of cos 2x gives: 6A − 13B = 0

2 The auxiliary √ equation is m + 16 = 0, from which m = −16 = ± j4

i.e. u = Acos 4x + B sin 4x

(2D2 + 3D − 5)(A sin 2x + B cos 2x)

13 × (2)gives :

d2 y + 16y = 10 cos 4x in D-operator form is dx2

+ 16Cx sin 4x + 16Dx cos 4x = 10 cos 4x, i.e. −8D sin 4x + 8C cos 4x = 10 cos 4x Equating coefficients of cos 4x gives: 10 5 8C = 10, from which, C = = 8 4 Equating coefficients of sin 4x gives: −8D = 0, from which, D = 0

Second-order differential equations 575 Hence the P.I., v = x (vi)

(5 4

sin 4x

) equation (given n ̸= 0 and p 2 ̸= n 2 ) given that dy when t = 0, y = = 0 dt

The general solution, y = u + v, i.e. y = A cos 4x + B sin 4x + 54 x sin 4x

(vii)

When x = 0, y = 3, thus 3 = A cos 0 + B sin 0 + 0, i.e. A = 3

6.

dy = −4A sin 4x + 4B cos 4x dx

The motion of a vibrating mass is given by d2 y dy + 8 + 20y = 300 sin 4t. Show that the dt2 dt general solution of the differential equation is given by: y = e−4t (A cos 2t + B sin 2t) 15 (sin 4t − 8 cos 4t) + 13

+ 45 x(4 cos 4x) + 45 sin 4x When x = 0,

dy = 4, thus dx

4 = −4A sin 0 + 4B cos 0 + 0 + 45 sin 0

7.

i.e. 4 = 4B, from which, B = 1 Hence the particular solution is y = 3 cos 4x + sin 4x + 54 x sin 4x Now try the following Practice Exercise Practice Exercise 234 Second order differential equations of the form d2 y dy a 2 + b + cy = f (x ) where f(x ) is a sine dx dx or cosine function (Answers on page 895)

51.6

In Problems 1 to 3, find the general solutions of the given differential equations. 1. 2

d2 y dy − − 3y = 25 sin 2x dx2 dx

dy dy − 4 + 4y = 5 cos x dx2 dx

3.

d2 y + y = 4 cos x dx2

4. Find the particular solution of the differend2 y dy tial equation 2 − 3 − 4y = 3 sin x; when dx dx dy x = 0, y = 0 and =0 dx 5. A differential equation representing the d2 y + n 2 y = k sin pt, motion of a body is dt2 where k, n and p are constants. Solve the

Differential equations of the d2 y dy form a 2 + b + cy = f(x) dx dx where f(x) is a sum or a product

Problem 9.

2

2.

d2 q dq 1 L 2 + R + q = V0 sin ωt represents the dt dt C variation of capacitor charge in an electric circuit. Determine an expression for q at time t seconds given that R = 40 Ω, L = 0.02 H, C = 50 × 10−6 F, V0 = 540.8 V and ω = 200 rad/s and given the boundary condidq tions that when t = 0, q = 0 and = 4.8 dt

Solve d2 y dy + − 6y = 12x − 50 sin x dx2 dx

Using the procedure of Section 51.2: (i)

d2 y dy + − 6y = 12x − 50 sin x dx2 dx form is

in

D-operator

(D2 + D − 6)y = 12x − 50 sin x (ii)

The auxiliary equation is (m 2 + m − 6) = 0, from which, (m − 2)(m + 3) = 0, i.e. m = 2 or m = −3

576 Section I (iii)

Since the roots are real and different, the C.F., u = Ae2x + Be−3x

(iv) Since the right-hand side of the given differential equation is the sum of a polynomial and a sine function let the P.I. v = ax + b + c sin x + d cos x (see Table 51.1(e)). (v)

Substituting v into (D2 + D − 6)v = 12x − 50 sin x gives: (D2 + D − 6)(ax + b + c sin x + d cos x)

(vi)

i.e.

− 13 + 7 sin x + cos x Problem 10. Solve the differential equation d2 y dy − 2 + 2y = 3ex cos 2x, given that when x = 0, 2 dx dx dy y = 2 and =3 dx

(i)

D(ax + b + c sin x + d cos x) = a + c cos x − d sin x

= −c sin x − d cos x

(iii)

Since the roots are complex, the C.F., u = ex (A cos x + B sin x)

− d sin x) − 6(ax + b + c sin x + d cos x) = 12x − 50 sin x Equating constant terms gives: (1)

Equating coefficients of x gives: −6a = 12, from which, a = −2 Hence, from (1), b = − 13

−d + c − 6d = 0 (2)

(3)

Solving equations (2) and (3) gives: c = 7 and d=1 Hence the P.I., υ = −2x − + 7 sin x + cos x

(D2 − 2D + 2)[ex (C sin 2x + D cos 2x)] = 3ex cos 2x + ex (C sin 2x + D cos 2x)

−c − d − 6c = −50

1 3

Substituting v into (D2 − 2D + 2)v = 3ex cos 2x gives:

D(v) = ex (2C cos 2x − 2D sin 2x)

Equating the coefficients of sin x gives: − 7c − d = −50

(iv) Since the right-hand side of the given differential equation is a product of an exponential and a cosine function, let the P.I., v = ex (C sin 2x + D cos 2x) (see Table 51.1(f) – again, constants C and D are used since A and B have already been used for the C.F.). (v)

Equating the coefficients of cos x gives:

i.e.

D-operator

The auxiliary equation is m 2 − 2m + 2 = 0 Using the quadratic formula, √ 2 ± [4 − 4(1)(2)] m= 2 √ 2 ± −4 2 ± j2 = = i.e. m = 1 ± j1 2 2

(D2 + D − 6)(v)

c − 7d = 0

in

(ii)

= (−c sin x − d cos x) + (a + c cos x

i.e.

dy d2 y − 2 + 2y = 3ex cos 2x dx2 dx form is

(D2 − 2D + 2)y = 3ex cos 2x

D2 (ax + b + c sin x + d cos x)

a − 6b = 0

y = Ae2x + Be−3x − 2x

Using the procedure of Section 51.2:

= 12x − 50 sin x

Hence

The general solution, y = u + v,

≡ ex {(2C + D) cos 2x + (C − 2D) sin 2x} D2 (v) = ex (−4C sin 2x − 4D cos 2x) + ex (2C cos 2x − 2D sin 2x) + ex (2C cos 2x − 2D sin 2x) + ex (C sin 2x + D cos 2x)

Second-order differential equations 577 ≡ ex {(−3C − 4D) sin 2x + (4C − 3D) cos 2x}

i.e.

Hence the particular solution is

= ex {(−3C − 4D) sin 2x

y = ex (3 cos x + sin x) − ex cos 2x

+ (4C − 3D) cos 2x} + (C − 2D) sin 2x} + 2ex (C sin 2x + D cos 2x) = 3ex cos 2x Equating coefficients of ex sin 2x gives: −3C − 4D − 2C + 4D + 2C = 0 i.e. −3C = 0, from which, C = 0 Equating coefficients of ex cos 2x gives: 4C − 3D − 4C − 2D + 2D = 3 i.e. −3D = 3, from which, D = −1

Now try the following Practice Exercise Practice Exercise 235 Second-order differential equations of the form dy d2 y + cy = f(x) where f(x) is a sum a 2 +b dx dx or product (Answers on page 896) In Problems 1 to 4, find the general solutions of the given differential equations.

The general solution, y = u + v, i.e. y = ex (A cos x + B sin x) − ex cos 2x

(vii)

8

2.

dy d2 y − 3 + 2y = 2 sin 2 θ − 4 cos 2 θ 2 dθ dθ

3.

d2 y dy + − 2y = x2 + e2x dx2 dx

4.

d2 y dy − 2 + 2y = et sin t dt2 dt

When x = 0, y = 2 thus 2 = e0 (A cos 0 + B sin 0) − e0 cos 0 i.e.

2 = A − 1, from which, A = 3 dy = ex (−A sin x + B cos x) dx + ex (A cos x + B sin x)

In Problems 5 to 6 find the particular solutions of the given differential equations. 5.

− [ex (−2 sin 2x) + ex cos 2x] When thus

dy =3 dx 3 = e0 (−A sin 0 + B cos 0) x = 0,

+ e0 (A cos 0 + B sin 0) − e0 (−2 sin 0) − e0 cos 0

d2 y dy − 6 + y = 2x + 40 sin x dx2 dx

1.

Hence the P.I., υ = ex (−cos 2x) (vi)

from which,

B = 1, since A = 3

Hence (D2 − 2D + 2)v

− 2ex {(2C + D) cos 2x

3 = B + A − 1,

6.

d2 y dy − 7 + 10y = e2x + 20; when x = 0, y = 0 2 dx dx dy 1 and =− dx 3 d2 y dy − − 6y = 6ex cos x; dx2 dx 21 dy 20 y = − and = −6 29 dx 29 2

For fully worked solutions to each of the problems in Practice Exercises 232 to 235 in this chapter, go to the website: www.routledge.com/cw/bird

when

x = 0,

Chapter 52

Power series methods of solving ordinary differential equations Why it is important to understand: Power series methods of solving ordinary differential equations The differential equations studied so far have all had closed form solutions, that is, their solutions could be expressed in terms of elementary functions, such as exponential, trigonometric, polynomial and logarithmic functions, and most such elementary functions have expansions in terms of power series. However, there are a whole class of functions which are not elementary functions and which occur frequently in mathematical physics and engineering. These equations can sometimes be solved by discovering a power series that satisfies the differential equation, but the solution series may not be summable to an elementary function. In this chapter the methods of solution to such equations are explained.

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

appreciate the reason for using power series methods to solve differential equations determine higher order differential coefficients as a series use Leibniz’s theorem to obtain the nth derivative of a given function obtain a power series solution of a differential equation by the Leibniz−Maclaurin method obtain a power series solution of a differential equation by the Frobenius method determine the general power series solution of Bessel’s equation express Bessel’s equation in terms of gamma functions determine the general power series solution of Legendre’s equation determine Legendre polynomials determine Legendre polynomials using Rodrigues’ formula

52.1

Introduction

Second-order ordinary differential equations that cannot be solved by analytical methods (as shown in Chapters 50 and 51), i.e. those involving variable

coefficients, can often be solved in the form of an infinite series of powers of the variable. This chapter looks at some of the methods that make this possible – by the Leibniz–Maclaurin and Frobenius methods, involving Bessel’s and Legendre’s equations, Bessel and gamma

Power series methods of solving ordinary differential equations 579 ( nπ ) y(n) = an sin ax + 2 For example, if

functions and Legendre’s polynomials. Before introducing Leibniz’s theorem, some trends with higher differential coefficients are considered. To better understand this chapter it is necessary to be able to:

In general,

d5 y = y(5) dx5 ( ) ( 5π π) 5 = 3 sin 3x + = 35 sin 3x + 2 2

y = sin 3x, then

(i) differentiate standard functions (as explained in Chapters 25 and 30), (ii) appreciate the binomial theorem (as explained in Chapter 5), and (iii) use Maclaurin’s Chapter 37).

theorem

(as

explained

(iii)

The following is an extension of successive differentiation (see page 337), but looking for trends, or series, as the differential coefficient of common functions rises. (i)

If y = eax , then

dy d2 y = aeax , 2 = a2 eax , and so on. dx dx

dy d2 y If we abbreviate as y′ , as y′′ , … and 2 dx dx dn y as y(n) , then y′ = aeax , y′′ = a2 eax , and the dxn emerging pattern gives:

y(n) = an eax

For example, if y = 3e2x , then d7 y = y(7) = 3(27 ) e2x = 384e2x dx7 (ii)

= 243 cos 3x

in

52.2 Higher order differential coefficients as series

If y = sin ax,

( π) y′ = a cos ax = a sin ax + 2

y′′ = −a2 sin ax = a2 sin(ax + π) ( ) 2π = a2 sin ax + 2 y′′′ = −a3 cos ax ( ) 3π = a3 sin ax + and so on. 2

(2)

If y = cosax,

( π) y′ = −a sin ax = a cos ax + 2 ) ( 2π y′′ = −a2 cos ax = a2 cos ax + 2

( ) 3π y′′′ = a3 sin ax = a3 cos ax + and so on. 2 ( nπ ) (3) In general, y(n) = an cos ax + 2 For example, if y = 4 cos 2x, then

( ) d6 y 6π (6) 6 = y = 4(2 ) cos 2x + dx6 2 = 4(26 ) cos (2x + 3π) = 4(26 ) cos (2x + π)

(1)

= −256 cos 2x (iv) If y = x a , y′ = axa−1 , y′′ = a(a − 1)xa−2 , y′′′ = a(a − 1)(a − 2)xa−3 , and y(n) = a(a − 1)(a − 2) . . . (a − n + 1) xa−n or y(n) =

a! xa−n (a − n)!

where a is a positive integer. For example, if y = 2x6 , then = (2)

(4) d4 y = y(4) dx4

6! x6−4 (6 − 4)!

6×5×4×3×2×1 2 x 2×1 = 720x2 = (2)

580 Section I (v)

If y = sinh ax, y′ = a cosh ax y′′ = a2 sinh ax y′′′ = a3 cosh ax, and so on Since sinh ax is not periodic (see graph on page 154), it is more difficult to find a general statement for y(n) . However, this is achieved with the following general series: y(n) =

an {[1 + (−1)n ] sinh ax 2 + [1 − (−1)n ] cosh ax}

1 1 2 (vii) If y = ln ax, y′ = , y′′ = − 2 , y′′′ = 3 , and so x x x on. (n − 1)! (7) In general, y(n) = (−1)n−1 xn For example, if y = ln( 5x, then ) d6 y 5! 120 (6) 6−1 = y = (−1) =− 6 6 6 dx x x 1 Note that if y = ln x, y′ = ; if in equation (7), x ′ 0 (0)! n = 1 then y = (−1) 1 x 1 (−1)0 = 1 and if y′ = then (0)! = 1 (check that x (−1)0 = 1 and (0)! = 1 on a calculator).

(5)

For example, if y = sinh 2x, then =

Now try the following Practice Exercise

d5 y = y(5) dx5

Exercise 236 Higher-order differential coefficients as series (Answers on page 896)

25 {[1 + (−1)5 ] sinh 2x 2 + [1 − (−1)5 ] cosh 2x}

Determine the following derivatives:

5

2 {[0] sinh 2x + [2] cosh 2x} 2 = 32 cosh 2x =

(vi)

(a) y(4) when y = e2x (b) y(5) when y = 8 e 2

2.

(a) y(4) when y = sin 3t 1 (b) y(7) when y = sin 5θ 50 (a) y(8) when y = cos 2x 2 (b) y(9) when y = 3 cos t 3

3.

If y = cosh ax, y′ = a sinh ax y′′ = a2 cosh ax

4.

y′′′ = a3 sinh ax, and so on

5. 6.

Since cosh ax is not periodic (see graph on page 154), again it is more difficult to find a general statement for y(n) . However, this is achieved with the following general series: y(n) =

an {[1 − (−1)n ] sinh ax 2 + [1 + (−1)n ] cosh ax}

1 For example, if y = cosh 3x, 9 ( ) 7 7 d y 1 3 then 7 = y(7) = (2 sinh 3x) dx 9 2 = 243 sinh 3x

(6)

t

1.

7.

t7 (a) y(7) when y = 2x9 (b) y(6) when y = 8 1 (7) (a) y when y = sinh 2x 4 (b) y(6) when y = 2 sinh 3x (a) y(7) when y = cosh 2x 1 (b) y(8) when y = cosh 3x 9 (a) y(4) when y = 2ln 3θ 1 (b) y(7) when y = ln 2t 3

52.3 If

Leibniz’s theorem y = uv

(8)

where u and v are each functions of x, then by using the product rule, y′ = uv′ + vu′

(9)

Power series methods of solving ordinary differential equations 581 y′′ = uv′′ + v′ u′ + vu′′ + u′ v′ = u′′ v + 2u′ v′ + uv′′

(10)

y′′′ = u′′ v′ + vu′′′ + 2u′ v′′ + 2v′ u′′ + uv′′′ + v′′ u′ = u′′′ v + 3u′′ v′ + 3u′ v′′ + uv′′′

(11)

(ii) v is the one whose derivative reduces to zero after a few stages of differentiation. Thus, when y = x2 e3x , v = x2 , since its third derivative is zero, and u = e3x since the nth derivative is known from equation (1), i.e. 3n eax Using Leinbiz’s theorem (equation (13)),

y(4) = u(4) v + 4u(3) v(1) + 6u(2) v(2) (1) (3)

+ 4u v

(4)

+ uv

y(n) = u(n) v + nu(n−1) v(1) +

(12)

From equations (8) to (12) it is seen that

+

(a) the nth derivative of u decreases by 1 moving from left to right, (b)

the nth derivative of v increases by 1 moving from left to right,

(c) the coefficients 1, 4, 6, 4, 1 are the normal binomial coefficients (see page 49). (n)

n(n − 1)(n − 2) (n−3) (3) u v + ··· 3!

where in this case v = x2 , v(1) = 2x, v(2) = 2 and v(3) = 0 Hence,

(n)

In fact, (uv) may be obtained by expanding (u + v) using the binomial theorem (see page 50), where the ‘powers’ are interpreted as derivatives. Thus, expanding (u + v)(n) gives:

y(n) = (3n e3x )(x2 ) + n(3n−1 e3x )(2x) +

n(n − 1) n−2 3x (3 e )(2) 2!

+

n(n − 1)(n − 2) n−3 3x (3 e )(0) 3!

= 3n−2 e3x (32 x2 + n(3)(2x) + n(n − 1) + 0)

y(n) = (uv)(n) = u(n) v + nu(n−1) v(1) +

n(n − 1) (n−2) (2) u v 2!

+

n(n − 1)(n − 2) (n−3) (3) u v +··· 3!

i.e.

(13)

Equation (13) is a statement of Leibniz’s theorem, ∗ which can be used to differentiate a product n times. The theorem is demonstrated in the following worked problems. Problem 1.

Differentiating each term of x2 y′′ + 2xy′ + y = 0 n times, using Leibniz’s theorem of equation (13), gives: { } n(n − 1) (n) y(n+2) x2 + n y(n+1) (2x) + y (2) + 0 2!

Determine y(n) when y = x2 e3x

u is the one whose nth derivative can readily be determined (from equations (1) to (7)),

y(n) = e3x 3n−2 (9x2 + 6nx + n(n − 1))

Problem 2. If x2 y′′ + 2xy′ + y = 0 show that: xy(n+2) + 2(n + 1)xy(n+1) + (n 2 + n + 1)y(n) = 0

+ {y(n+1) (2x) + n y(n) (2) + 0} + {y(n) } = 0

For a product y = uv, the function taken as (i)

n(n − 1) (n−2) (2) u v 2!

i.e.

x2 y(n+2) + 2n xy(n+1) + n(n − 1)y(n) + 2xy(n+1) + 2n y(n) + y(n) = 0

i.e. ∗

Who was Leibniz? For image and resume, go to www.routledge.com/cw/bird

x2 y(n+2) + 2(n + 1)xy(n+1) + (n 2 − n + 2n + 1)y(n) = 0

582 Section I x2 y(n+2) + 2(n + 1) x y(n+1)

or

+

+ (n2 + n + 1)y(n) = 0

( ) (5)(4) 3π (12x2 ) sin x + 2 2

+ Problem 3. Differentiate the following differential equation n times: (1 + x2 )y′′ + 2xy′ − 3y = 0 By Leibniz’s equation, equation (13), { } n(n −1) (n) y(n+2)(1 + x2 ) + ny(n+1)(2x)+ y (2)+ 0 2! + 2{y(n+1) (x) + n y(n) (1) + 0} − 3{y(n) } = 0

+

(1 + x2 )y(n+2) + 2n xy(n+1) + n(n − 1)y(n)

(1 + x )y

or

(n+2)

+ 2(n + 1)xy

(n+1)

+ (n 2 − n + 2n − 3)y(n) = 0 i.e.

Problem 4.

Find the fifth derivative of y = x4 sin x

If y = x4 sin x, then using Leibniz’s equation with u = sin x and v = x4 gives: y

+ 240x(−sin x) + 120 cos x i.e. y(5) = (x4 − 120x2 + 120)cos x + (20x3 − 240x) sin x

(1 + x2 )y(n+2) + 2(n + 1)xy(n+1) + (n2 + n − 3)y(n) = 0

(n)

sin (x + π) ≡ −sin x,

then y(5) = x4 cos x + 20x3 sin x + 120x2 (−cos x)

+ 2xy(n+1) + 2 ny(n) − 3y(n) = 0 2

( (5)(4)(3)(2) π) (24) sin x + (4)(3)(2) 2

( ) ( 5π π) Since sin x + ≡ sin x + ≡ cos x, 2 2 ( ) 3π sin(x + 2π) ≡ sin x, sin x + ≡ −cos x, 2 and

i.e.

(5)(4)(3) (24x) sin (x + π) (3)(2)

[ ( nπ ) 4 ] = sin x + x 2 [ ( ) ] (n − 1)π + n sin x + 4x3 2 [ ( ) ] n(n − 1) (n − 2)π + sin x + 12x2 2! 2 [ ( ) ] n(n − 1)(n − 2) (n − 3)π + sin x + 24x 3! 2 [ ( n(n − 1)(n − 2)(n − 3) + sin x 4! ) ] (n − 4)π + 24 2

( ) 5π and y(5) = x4 sin x + + 20x3 sin(x + 2π) 2

Now try the following Practice Exercise Practice Exercise 237 Leibniz’s theorem (Answers on page 896) Use the theorem of Leibniz in the following problems: 1.

Obtain the nth derivative of: x2 y

2.

If y = x3 e2x find y(n) and hence y(3)

3.

Determine the fourth derivative of: y = 2x3 e−x

4.

If y = x3 cos x determine the fifth derivative

5.

Find an expression for y(4) if y = e−t sin t

6.

If y = x5 ln 2x find y(3)

7.

Given 2x 2 y ′′ + xy ′ + 3y = 0 show that 2x 2 y(n+2) + (4n + 1)xy(n+1) + (2n 2 − n + 3)y(n) = 0

8.

If y = (x3 + 2x2 )e2x determine an expansion for y(5)

Power series methods of solving ordinary differential equations 583 (ii) At x = 0, equation (14) becomes:

52.4 Power series solution by the Leibniz–Maclaurin method

y(n+2) + (n + 2) y(n) = 0

For second-order differential equations that cannot be solved by algebraic methods, the Leibniz–Maclaurin∗ method produces a solution in the form of infinite series of powers of the unknown variable. The following simple five-step procedure may be used in the Leibniz–Maclaurin method: (i) Differentiate the given equation n times, using the Leibniz theorem of equation (13), (ii) rearrange the result to obtain the recurrence relation at x = 0, (iii) determine the values of the derivatives at x = 0, i.e. find ( y)0 and (y′ )0 ,

from which, y(n+2) = −(n + 2) y(n) This equation is called a recurrence relation or recurrence formula, because each recurring term depends on a previous term. (iii) Substituting n = 0, 1, 2, 3, … will produce a set of relationships between the various coefficients. For n = 0, ( y′′ )0 = −2(y)0 n = 1, (y′′′ )0 = −3(y′ )0 n = 2, ( y(4) )0 = −4(y′′ )0 = −4{−2( y)0 } = 2 × 4( y)0 n = 3, ( y )0 = −5(y′′′ )0 = −5{−3( y′ )0 } (5)

(iv) substitute in the Maclaurin expansion for y = f (x) (see page 452, equation (5)), (v) simplify the result where possible and apply boundary conditions, (if given).

= 3 × 5( y′ )0 n = 4,(y(6) )0 = −6(y(4) )0 = −6{2 × 4(y)0 } = −2 × 4 × 6(y)0 n = 5,(y )0 = −7(y(5) )0 = −7{3×5(y′ )0 } (7)

The Leibniz–Maclaurin method is demonstrated, using the above procedure, in the following worked problems.

= −3 × 5 × 7(y′ )0 n = 6,(y(8) )0 = −8(y(6) )0 =

Problem 5. Determine the power series solution of the differential equation: d2 y dy + x + 2y = 0 using the Leibniz–Maclaurin 2 dx dx method, given the boundary conditions that at x = 0, dy y = 1 and =2 dx

−8{−2 × 4 × 6(y)0} = 2×4×6×8(y)0 (iv) Maclaurin’s theorem from page 452 may be written as: y = (y)0 + x( y′ )0 +

x2 ′′ x3 (y )0 + (y′′′ )0 2! 3!

x4 (4) (y )0 + · · · 4! Substituting the above values into Maclaurin’s theorem gives: +

Following the above procedure: (i) The differential equation is rewritten as: y′′ + xy′ + 2y = 0 and from the Leibniz theorem of equation (13), each term is differentiated n times, which gives: y(n+2) +{y(n+1) (x)+n y(n) (1)+0}+2 y(n) = 0 i.e.

y(n+2) + xy(n+1) + (n + 2) y(n) = 0 (14)

∗

Who was Maclaurin? For image and resume of Maclaurin, see page 451. To find out more go to www.routledge.com/cw/bird

x2 {−2( y)0 } 2! x3 x4 + {−3( y′ )0 } + {2 × 4( y)0 } 3! 4! 5 x x6 + {3 × 5( y′ )0 } + {−2 × 4 × 6(y)0 } 5! 6! 7 x + {−3 × 5 × 7(y′ )0 } 7! x8 + {2 × 4 × 6 × 8( y)0 } + · · · . 8!

y = (y)0 + x(y′ )0 +

584 Section I (v)

y(n+2) + y(n+1) + y(n) (x) + n y(n−1) (1) + 0 = 0

Collecting similar terms together gives: { 2x2 2 × 4x4 y = ( y)0 1 − + 2! 4! 2 × 4 × 6x 2 × 4 × 6 × 8x + 6! 8! } { 3 3x 3 × 5x5 − · · · + ( y′)0 x − + 3! 5! 6

−

(15)

8

3 × 5 × 7x7 − + ··· 7!

(ii) At x = 0, equation (15) becomes: y(n+2) + y(n+1) + n y(n−1) = 0

′

+ (y )0 ×

{

from which, y(n+2) = −{y(n+1) + n y(n−1) }

}

This is the recurrence relation and applies for n≥1

{ x2 x4 x6 i.e. y = ( y)0 1 − + − 1 1×3 3×5 +

y(n+2) + y(n+1) + xy(n) + n y(n−1) = 0

i.e.

x8 − ··· 3×5×7

x x3 x5 − + 1 1×2 2×4 x7 − + ··· 2×4×6

}

For n = 1,

}

The boundary conditions are that at x = 0, y = 1 dy and = 2, i.e. (y)0 = 1 and (y′ )0 = 2 dx Hence, the power series solution of the differend2 y dy tial equation: 2 + x + 2y = 0 is: dx dx { 2 4 x x x6 y = 1− + − 1 1×3 3×5 } { x8 x x3 + −··· +2 − 3×5×7 1 1×2 x5 x7 + − +··· 2×4 2×4×6

(iii) Substituting n = 1, 2, 3, . . . will produce a set of relationships between the various coefficients. ( y′′′ )0 = −{(y′′ )0 + (y)0 }

n = 2,

( y(4) )0 = −{(y′′′ )0 + 2(y′ )0 }

n = 3,

( y(5) )0 = −{(y(4) )0 + 3(y′′ )0 }

n = 4,

( y(6) )0 = −{(y(5) )0 + 4(y′′′ )0 }

n = 5,

( y(7) )0 = −{(y(6) )0 + 5(y(4) )0 }

n = 6,

( y(8) )0 = −{(y(7) )0 + 6(y(5) )0 }

From the given boundary conditions, at x = 0, dy y = 0, thus (y)0 = 0, and at x = 0, = 1, thus dx ′ ( y )0 = 1 From the given differential equation, y′′ + y′ + xy = 0, and, at x = 0, ( y′′ )0 + (y′ )0 + (0)y = 0 from which, ( y′′ )0 = −(y′ )0 = −1 }

Thus, (y)0 = 0, (y′ )0 = 1, (y′′ )0 = −1, (y′′′ )0 = −{( y′′ )0 + (y)0 } = −(−1 + 0) = 1 (y(4) )0 = −{( y′′′ )0 + 2( y′ )0 }

Problem 6. Determine the power series solution of the differential equation: d2 y dy + + xy = 0 given the boundary conditions dx2 dx dy that at x = 0, y = 0 and = 1, using the dx Leibniz–Maclaurin method. Following the above procedure: (i)

The differential equation is rewritten as: y′′ + y′ + xy = 0 and from the Leibniz theorem of equation (13), each term is differentiated n times, which gives:

= −[1 + 2(1)] = −3 (y(5) )0 = −{( y(4) )0 + 3(y′′ )0 } = −[−3 + 3(−1)] = 6 (y(6) )0 = −{( y(5) )0 + 4(y′′′ )0 } = −[6 + 4(1)] = −10 (y )0 = −{( y(6) )0 + 5(y(4) )0 } (7)

= −[−10 + 5(−3)] = 25 (y )0 = −{( y(7) )0 + 6(y(5) )0 } (8)

= −[25 + 6(6)] = −61

Power series methods of solving ordinary differential equations 585 (iv) Maclaurin’s theorem states: y = ( y)0 + x( y′ )0 +

and substituting the above Maclaurin’s theorem gives: y = 0 + x(1) +

the boundary conditions that at x = 0, y = 1 and dy =1 dx

x2 ′′ x3 (y )0 + (y′′′ )0 2! 3! x4 (4) + (y )0 + · · · 4! values

4. into

x3 x4 x2 {−1} + {1} + {−3} 2! 3! 4! x5 x6 x7 + {6} + {−10} + {25} 5! 6! 7! x8 + {−61} + · · · 8!

(v) Simplifying, the power series solution of the difd2 y dy ferential equation: 2 + + xy = 0 is given by: dx dx y = x−

x2 x3 3x4 6x5 10x6 + − + − 2! 3! 4! 5! 6! 25x7 61x8 + − +··· 7! 8!

Use the Leibniz–Maclaurin method to determine the power series solution for the differd2 y dy + xy = 0 given that ential equation: x 2 + dx dx dy at x = 0, y = 1 and =2 dx

52.5 Power series solution by the Frobenius method A differential equation of the form y ′′ + Py ′ + Qy = 0, where P and Q are both functions of x, such that the equation can be represented by a power series, may be solved by the Frobenius method.∗

Now try the following Practice Exercise Practice Exercise 238 Power series solutions by the Leibniz-Maclaurin method (Answers on page 896) 1. Determine the power series solution of the difd2 y dy ferential equation: 2 + 2x + y = 0 using dx dx the Leibniz–Maclaurin method, given that at dy x = 0, y = 1 and =2 dx 2. Show that the power series solution of the ( ) d2 y differential equation: x + 1 + (x − 1) dx2 dy − 2y = 0, using the Leibniz–Maclaurin dx method, is given by: y = 1 + x2 + e−x given the boundary conditions that at x = 0, y = 2 dy = −1 and dx 3. Find the particular solution of the differd2 y dy ential equation: (x2 + 1) 2 + x − 4y = 0 dx dx using the Leibniz–Maclaurin method, given

∗

Who was Frobenius? Ferdinand Georg Frobenius (26 October 1849–3 August 1917) was a German mathematician best known for his contributions to the theory of elliptic functions, differential equations and to group theory. He is also known for determinantal identities, known as Frobenius–Stickelberger formulae. To find out more go to www.routledge.com/cw/bird

586 Section I The following four-step procedure may be used in the Frobenius method: (i)

Assume a trial solution of the form y = { } xc a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · ·

(ii)

differentiate the trial series,

(iii)

substitute the results in the given differential equation,

(iii) Substituting y, y′ and y′′ into each term of the given equation 3xy′′ + y′ − y = 0 gives: 3xy′′ = 3a0 c(c − 1)xc−1 + 3a1 c(c + 1)xc + 3a2 (c + 1)(c + 2)xc+1 + · · · + 3ar (c + r − 1)(c+r)xc+r−1 +· · · (a) y′ = a0 cxc−1 +a1 (c + 1)xc +a2 (c + 2)xc+1

(iv) equate coefficients of corresponding powers of the variable on each side of the equation; this enables index c and coefficients a1 , a2 , a3 , … from the trial solution, to be determined.

+ · · · + ar (c + r)xc+r−1 + · · · −y = −a0 xc − a1 xc+1 − a2 xc+2 − a3 xc+3

This introductory treatment of the Frobenius method covering the simplest cases is demonstrated, using the above procedure, in the following worked problems.

The differential equation may be rewritten as: 3xy′′ + y′ − y = 0

i.e. (16)

where a0 ̸= 0, i.e. y = a0 x + a1 x

+ a2 x

+ · · · + ar x

c+2

c+r

(ii)

c+3

+ a3 x

+ ···

(17)

Differentiating equation (17) gives: y′ = a0 cxc−1 + a1 (c + 1)xc

and

a1 (3c2 + 3c + c + 1) − a0 = a1 (3c2 + 4c + 1) − a0 = 0

or

a1 (3c + 1)(c + 1) − a0 = 0

(19)

In each of series (a), (b) and (c) an xc term is involved, after which, a general relationship can be obtained for xc+r , where r ≥ 0 In series (a) and (b), terms in xc+r−1 are present; replacing r by (r + 1) will give the corresponding terms in xc+r , which occurs in all three equations, i.e. in series (a),

3ar+1 (c + r)(c + r + 1)xc+r

+ a2 (c + 2)xc+1 + · · ·

in series (b),

ar+1 (c + r + 1)xc+r

+ ar (c + r)xc+r−1 + · · ·

in series (c), −ar xc+r

y′′ = a0 c(c − 1)xc−2 + a1 c(c + 1)xc−1 + a2 (c + 1)(c + 2)xc + · · · + ar (c + r − 1)(c + r)x

(18)

The coefficient of xc is equated to zero giving: 3a1 c(c + 1) + a1 (c + 1) − a0 = 0

{ y = xc a0 + a1 x + a2 x2 + a3 x3 + · · · } + ar xr + · · ·

c+1

(c)

or a0 c[3c − 3 + 1] = a0 c(3c − 2) = 0

Let a trial solution be of the form

c

− · · · − ar xc+r − · · ·

(iv) The sum of these three terms forms the left-hand side of the equation. Since the right-hand side is zero, the coefficients of each power of x can be equated to zero. For example, the coefficient of xc−1 is equated to zero giving: 3a0 c(c − 1) + a0 c = 0

Problem 7. Determine, using the Frobenius method, the general power series solution of the d2 y dy differential equation: 3x 2 + − y = 0 dx dx

(i)

(b)

c+r−2

Equating the total coefficients of xc+r to zero gives: 3ar+1 (c + r)(c + r + 1) + ar+1 (c + r + 1)

+ ···

− ar = 0

Power series methods of solving ordinary differential equations 587 which simplifies to:

(b) When c =

ar+1 {(c + r + 1)(3c + 3r + 1)} − ar = 0

(20)

Equation (18), which was formed from the coefficients of the lowest power of x, i.e. xc−1 , is called the indicial equation, from which the value of c is obtained. From equation (18), since 2 a0 ̸= 0, then c = 0 or c = 3

(a) When c = 0 From equation (19), if c = 0, a1 (1 × 1) − a0 = 0, i.e. a1 = a0

2 3

( ) 2 5 From equation (19), if c = , a1 (3) − a0 = 0, i.e. 3 3 a0 a1 = 5 2 From equation (20), if c = 3 ( ) 2 ar+1 + r + 1 (2 + 3r + 1) − ar = 0, 3 ( ) 5 i.e. ar+1 r + (3r + 3) − ar 3 = ar+1 (3r 2 + 8r + 5) − ar = 0,

From equation (20), if c = 0, ar+1 (r + 1)(3r + 1) − ar = 0, ar i.e. ar+1 = r≥0 (r + 1)(3r + 1) a1 a0 Thus, when r = 1, a2 = = (2 × 4) (2 × 4) since a1 = a0 a2 a0 when r = 2, a3 = = (3 × 7) (2 × 4)(3 × 7) a0 or (2 × 3)(4 × 7) a3 when r = 3, a4 = (4 × 10) a0 = (2 × 3 × 4)(4 × 7 × 10)

i.e. ar+1 =

ar (r + 1)(3r + 5)

r≥0

a1 a0 = (2 × 8) (2 × 5 × 8) a0 since a1 = 5 a2 when r = 2, a3 = (3 × 11)

Thus, when r = 1, a2 =

a0 (2 × 3)(5 × 8 × 11)

=

a3 (4 × 14)

when r = 3, a4 =

a0 (2×3×4)(5×8×11×14)

=

and so on.

and so on.

From equation (16), the trial solution was:

From equation (16), the trial solution was:

y = x {a0 + a1 x + a2 x + a3 x + · · · + ar x + · · · } c

2

3

r

Substituting c = 0 and the above values of a1 , a2 , a3 , … into the trial solution gives: { ( y = x0 a0 + a0 x +

) a0 x2 (2 × 4) ( ) a0 + x3 (2 × 3)(4 × 7) ( ) } a0 4 + x + ··· (2 × 3 × 4)(4 × 7 × 10) { x2 x3 )+( )( ) i.e. y = a0 1 + x + ( 2×4 2×3 4×7 } x4 )( ) + ··· +( (21) 2 × 3 × 4 4 × 7 × 10

y = xc {a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · · } 2 Substituting c = and the above values of a1 , a2 , 3 a3 , … into the trial solution gives: { 2

y = x3

a0 + (

+ (

(a ) 0

5

( x+

) a0 x2 2×5×8

) a0 x3 (2 × 3)(5 × 8 × 11)

) } a0 4 + x + ··· (2 × 3 × 4)(5 × 8 × 11 × 14)

588 Section I { x x2 i.e. y = a0 x 1 + + 5 (2 × 5 × 8)

and y′′ = a0 c(c − 1)xc−2 + a1 c(c + 1)xc−1

2 3

+ a2 (c + 1)(c + 2)xc + · · · + ar (c + r − 1)(c + r)xc+r−2 + · · ·

+

x3 (2 × 3)(5 × 8 × 11)

+

x4 + ··· (2 × 3 × 4)(5 × 8 × 11 × 14)

} (22)

Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different. Let a0 = A in equation (21), and a0 = B in equation (22). Also, if the first solution is denoted by u(x) and the second by v(x), then the general solution of the given differential equation is y = u(x) + v(x). Hence, { x2 x3 y = A 1+x+ + (2 × 4) (2 × 3)(4 × 7) } x4 + +··· (2 × 3 × 4)(4 × 7 × 10) { 2 x x2 +Bx3 1+ + 5 (2 × 5 × 8)

(iii) Substituting y, y′ and y′′ into each term of the given equation 2x2 y′′ − xy′ + (1 − x)y = 0 gives: 2x2 y′′ = 2a0 c(c − 1)xc + 2a1 c(c + 1)xc+1 + 2a2 (c + 1)(c + 2)xc+2 + · · · + 2ar (c + r − 1)(c + r)xc+r + · · · (a) −xy′ = −a0 cxc − a1 (c + 1)xc+1 − a2 (c + 2)xc+2 − · · · − ar (c + r)xc+r − · · · (1 − x)y = (1 − x)(a0 xc + a1 xc+1 + a2 xc+2 + a3 xc+3 + · · · + ar xc+r + · · · ) = a0 xc + a1 xc+1 + a2 xc+2 + a3 xc+3 + · · · + ar xc+r + · · ·

3

+ +

x (2 × 3)(5 × 8 × 11) x4 +··· (2 × 3 × 4)(5 × 8 × 11 × 14)

− a0 xc+1 − a1 xc+2 − a2 xc+3

}

− a3 xc+4 − · · · − ar xc+r+1 − · · · (c)

Problem 8. Use the Frobenius method to determine the general power series solution of the differential equation: d2 y dy 2x2 2 − x + (1 − x)y = 0 dx dx The differential equation may be rewritten as: 2x2 y′′ − xy′ + (1 − x)y = 0 (i)

Let a trial solution be of the form y = xc {a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · · }

(23)

where a0 ̸= 0, i.e.

y = a0 xc + a1 xc+1 + a2 xc+2 + a3 xc+3 + · · · + ar xc+r + · · ·

(ii)

(b)

(24)

(iv) The indicial equation, which is obtained by equating the coefficient of the lowest power of x to zero, gives the value(s) of c. Equating the total coefficients of xc (from equations (a) to (c)) to zero gives: i.e. i.e.

2a0 c(c − 1) − a0 c + a0 = 0 a0 [2c(c − 1) − c + 1] = 0 a0 [2c2 − 2c − c + 1] = 0

i.e. i.e.

a0 [2c2 − 3c + 1] = 0 a0 [(2c − 1)(c − 1)] = 0

1 from which, c = 1 or c = 2 Equating the coefficient of the general term, i.e. xc+r , (from equations (a) to (c)) to zero gives: 2ar (c + r − 1)(c + r) − ar (c + r) + ar − ar−1 = 0

Differentiating equation (24) gives: y′ = a0 cxc−1 + a1 (c + 1)xc + a2 (c + 2)xc+1 + · · · + ar (c + r)xc+r−1 + · · ·

from which, ar [2(c + r − 1)(c + r) − (c + r) + 1] = ar−1

Power series methods of solving ordinary differential equations 589 ar−1 and ar = 2(c+r−1)(c+r)−(c+r)+1

(25)

{ i.e. y = a0 x 1+ 1

ar−1 2(r)(1 + r) − (1 + r) +1 ar−1 = 2 2r + 2r − 1 − r + 1 ar−1 ar−1 = 2 = 2r + r r (2r + 1)

(a) When c = 1, ar =

+ +

x x2 + (1×3) (1 × 2) × (3 × 5)

x3 (1 × 2 × 3) × (3 × 5 × 7) x4 (1×2×3×4)×(3×5×7×9) } + ··· (26)

Thus, when r = 1, a1 =

a0 a0 = 1(2 + 1) 1 × 3

(b) When c =

when r = 2,

1 2

ar−1 2(c + r − 1)(c + r) − (c + r) + 1 from equation (25) a )( r−1 ) ( ) i.e. ar = ( 1 1 1 2 +r−1 +r − + r +1 2 2 2 ar =

a1 a1 = 2(4 + 1) (2 × 5) a0 a0 = or (1 × 3)(2 × 5) (1 × 2) × (3 × 5)

a2 =

when r = 3,

a ) ( r−1) = ( 1 1 1 2 r− r+ − −r+1 2 2 2

a2 a2 = 3(6 + 1) (3 × 7) a0 = (1 × 2 × 3) × (3 × 5 × 7)

a3 =

ar−1 ) = ( 1 1 2 2 r − − −r+1 4 2

when r = 4, a3 a3 = 4(8 + 1) 4 × 9 a0 = (1 × 2 × 3 × 4) × (3 × 5 × 7 × 9) and so on.

a4 =

From equation (23), the trial solution was: { y = xc a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · ·

ar−1 ar−1 = 2 1 1 2r − r 2r 2 − − − r + 1 2 2 ar−1 = r(2r − 1) =

}

Substituting c = 1 and the above values of a1 , a2 , a3 , … into the trial solution gives: { a0 a0 y = x1 a0 + x+ x2 (1×3) (1×2)×(3×5) a0 x3 (1 × 2 × 3) × (3 × 5 × 7) a0 + x4 (1×2×3×4)×(3×5×7×9) +

+ ···

a0 a0 = 1(2 − 1) 1 × 1 a1 a1 when r = 2, a2 = = 2(4 − 1) (2 × 3) a0 = (2 × 3) a2 a2 when r = 3, a3 = = 3(6 − 1) 3 × 5 a0 = (2 × 3) × (3 × 5) a3 a3 when r = 4, a4 = = 4(8 − 1) 4 × 7 a0 = (2×3×4)×(3×5×7)

Thus, when r = 1, a1 =

}

and so on.

590 Section I From equation (23), the trial solution was: { y = xc a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · ·

Problem 9. Use the Frobenius method to determine the general power series solution of the d2 y differential equation: 2 − 2y = 0 dx

}

1 Substituting c = and the above values of a1 , a2 , 2 a3 , … into the trial solution gives: { a0 2 a0 y=x a0 +a0 x+ x+ x3 (2×3) (2×3)×(3×5) } a0 + x4 + · · · (2 × 3 × 4) × (3 × 5 × 7) 1 2

{ 1

i.e. y = a0 x 2

1+x+ +

x x2 + (1 × 3) (1 × 2) × (3 × 5)

where a0 ̸= 0,

(29)

y′ = a0 cxc−1 + a1 (c + 1)xc + a2 (c + 2)xc+1 + · · · + ar (c + r)xc+r−1 + · · · and y′′ = a0 c(c − 1)xc−2 + a1 c(c + 1)xc−1 + a2 (c + 1)(c + 2)xc + · · · + ar (c + r − 1)(c + r)xc+r−2 + · · · (iii) Replacing r by (r + 2) in ar (c + r − 1)(c + r) xc+r−2 gives: ar+2 (c + r + 1)(c + r + 2)xc+r Substituting y and y′′ into each term of the given equation y′′ − 2y = 0 gives:

+ [a2 (c+1)(c + 2)−2a0 ]xc +· · · + [ar+2 (c + r + 1)(c + r + 2)

x4 + (1 × 2 × 3×4)×(3×5×7×9) } { 1 x2 2 +··· +Bx 1+x+ (2 × 3)

x4 +··· (2 × 3 × 4) ×(3 × 5 × 7)

(28)

y′′ − 2y = a0 c(c − 1)xc−2 + a1 c(c + 1)xc−1

x3 + (1 × 2 × 3) × (3 × 5 × 7)

+

}

(ii) Differentiating equation (29) gives:

Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different. Let a0 = A in equation (26), and a0 = B in equation (27). Also, if the first solution is denoted by u(x) and the second by v(x), then the general solution of the given differential equation is y = u(x) + v(x),

x3 (2 × 3) × (3 × 5)

+ ar xr + · · ·

+ · · · + ar xc+r + · · ·

x3 (2 × 3) × (3 × 5)

+

(i) Let a trial solution be of the form { y = xc a0 + a1 x + a2 x2 + a3 x3 + · · ·

i.e. y = a0 xc + a1 xc+1 + a2 xc+2 + a3 xc+3

x2 (2 × 3)

x4 + (2 × 3 × 4) × (3 × 5 × 7) } + ··· (27)

{ i.e. y = A x 1 +

The differential equation may be rewritten as: y′′ − 2y = 0

− 2ar ] xc+r + · · · = 0

(30)

(iv) The indicial equation is obtained by equating the coefficient of the lowest power of x to zero. Hence, a0 c(c − 1) = 0 from which, c = 0 or c = 1 since a0 ̸= 0 For the term in xc−1 , i.e. a1 c(c + 1) = 0

}

With c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined with the zero value of c would make the product zero.

Power series methods of solving ordinary differential equations 591 For the term in xc ,

Since

a2 (c + 1)(c + 2) − 2a0 = 0 from which, 2a0 a2 = (31) (c + 1)(c + 2)

a3 =

ar+2 (c + r + 1)(c + r + 2) − 2ar = 0

a4 = (32)

a5 =

In general, ar + 2 =

from equation (28) { } 2x2 4x4 = a0 1 + + + ··· 2! 4! { } 3 2x 4x5 + a1 x + + + ··· 3! 5! Since a0 and a1 are arbitrary constants depending on boundary conditions, let a0 = P and a1 = Q, then: { } 2x2 4x4 y=P 1 + + +··· 2! 4! { } 2x3 4x5 +Q x+ + +··· (33) 3! 5! (b) When c = 1: a1 = 0, and from equation (31), a2 =

2a0 2a0 = (2 × 3) 3!

2a3 =0 (5 × 6)

Hence, when c = 1, { } 2a0 2 4a0 4 1 y = x a0 + x + x + ··· 3! 5! from}equation (28) { 2x3 4x5 i.e. y = a0 x + + + ... 3! 5!

2a0 2a0 = (1 × 2) 2!

2ar and (r + 1)(r + 2) 2a1 2a1 2a1 when r = 1, a3 = = = (2 × 3) (1 × 2 × 3) 3! 2a2 4a0 when r = 2, a4 = = (3 × 4) 4! { 2a0 2 2a1 3 Hence, y = x0 a0 + a1 x + x + x 2! 3! } 4a0 4 + x + ··· 4!

2a2 2 2a0 4a0 = × = (4 × 5) (4 × 5) 3! 5!

when r = 3,

(a) When c = 0: a1 is indeterminate, and from equation (31) a2 =

2a1 = 0 since a1 = 0 (3 × 4)

when r = 2,

from which, 2ar (c + r + 1)(c + r + 2)

2ar (c + r + 1)(c + r + 2) 2ar = (r + 2)(r + 3)

ar+2 =

from equation (32) and when r = 1,

For the term in xc+r ,

ar+2 =

c = 1,

Again, a0 is an arbitrary constant; let a0 = K, { } 2x3 4x5 then y=K x+ + +··· 3! 5! However, this latter solution is not a separate solution, for it is the same form as the second series in equation (33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solution. This is always the case when the two values of c differ by an integer (i.e. whole number). From the above three worked problems, the following can be deduced, and in future assumed: (i)

if two solutions of the indicial equation differ by a quantity not an integer, then two independent solutions y = u(x) + v(x) result, the general solution of which is y = Au + Bv (note: Problem 7 had 2 1 c = 0 and and Problem 8 had c = 1 and ; in 3 2 neither case did c differ by an integer)

(ii)

if two solutions of the indicial equation do differ by an integer, as in Problem 9 where c = 0 and 1, and if one coefficient is indeterminate, as with when c = 0, then the complete solution is always given by using this value of c. Using the second value of c, i.e. c = 1 in Problem 9, always gives a series which is one of the series in the first solution.

592 Section I Now try the following Practice Exercise

Using the Frobenius method from page 586: (i) Let a trial solution be of the form

Practice Exercise 239 Power series solutions by the Frobenius method (Answers on page 896)

y = xc {a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · · }

1. Produce, using the Frobenius method, a power series solution for the differential equation: d2 y dy −y = 0 2x 2 + dx dx 2. Use the Frobenius method to determine the general power series solution of the differend2 y tial equation: 2 + y = 0 dx

(34)

where a0 ̸= 0, i.e.

y = a0 xc + a1 xc+1 + a2 xc+2 + a3 xc+3 + · · · + ar xc+r + · · ·

(35)

(ii) Differentiating equation (35) gives:

3. Determine the power series solution of the d2 y dy differential equation: 3x 2 + 4 − y = 0 dx dx using the Frobenius method.

y′ = a0 cxc−1 + a1 (c + 1)xc + a2 (c + 2)xc+1 + · · · + ar (c + r)xc+r−1 + · · ·

4. Show, using the Frobenius method, that the power series solution of the differential d2 y equation: − y = 0 may be expressed as dx2 y = P cosh x + Q sinh x, where P and Q are constants. [Hint: check the series expansions for cosh x and sinh x on page 159].

52.6 Bessel’s equation and Bessel’s functions One of the most important differential equations in applied mathematics is Bessel’s∗equation and is of the form: x2

d2 y dy + x + (x2 − v2 )y = 0 2 dx dx

where v is a real constant. The equation, which has applications in electric fields, vibrations and heat conduction, may be solved using Frobenius’ method of the previous section. Problem 10. Determine the general power series solution of Bessel’s equation. d2 y dy + x + (x2 − v2 )y = 0 may dx2 dx be rewritten as: x2 y′′ + xy′ + (x2 − v2 )y = 0 Bessel’s equation x2

∗

Who was Bessel? Friedrich Wilhelm Bessel (22 July 1784– 17 March 1846) was a German mathematician, astronomer, and the systematiser of the Bessel functions. Bessel produced a refinement on the orbital calculations for Halley’s Comet and produced precise positions for 3222 stars. To find out more go to www.routledge.com/cw/bird

Power series methods of solving ordinary differential equations 593 and y′′ = a0 c(c − 1)xc−2 + a1 c(c + 1)xc−1 + a2 (c + 1)(c + 2)xc + · · · + ar (c + r − 1)(c + r)xc+r−2 + · · · (iii) Substituting y, y′ and y′′ into each term of the given equation: x2 y′′ + xy′ + (x2 − v2 )y = 0 gives: a0 c(c − 1)xc + a1 c(c + 1)xc+1

i.e.

a1 [(c + 1)2 − v2 ] = 0

but if c = v

a1 [(v + 1)2 − v2 ] = 0 a1 [2v + 1] = 0

i.e.

Similarly, if c = −va1 [1 − 2v] = 0 The terms (2v + 1) and (1 − 2v) cannot both be zero since v is a real constant, hence a1 = 0 Since a1 = 0, then a3 = a5 = a7 = . . . = 0

+ a2 (c + 1)(c + 2)xc+2 + · · ·

when r = 2, a2 =

a0 v2 − (c + 2)2

when r = 4, a4

a0 2 − (c + 2) ][v2

+ a1 (c + 1)xc+1 + a2 (c + 2)xc+2 + · · · + ar (c + r)xc+r + · · · + a0 xc+2 + a1 xc+3 + a2 xc+4 + · · · + ar xc+r+2 + · · · − a0 v2 xc

(iv) The indicial equation is obtained by equating the coefficient of the lowest power of x to zero. a0 c(c − 1) + a0 c − a0 v2 = 0

a0 [c2 − v2 ] = 0

When c = +v, a0 a0 a2 = 2 = 2 v − (v + 2)2 v − v2 − 4v − 4

a4 =

from which, c = +v or c = −v since a0 ̸= 0 For the term in xc+r , ar (c + r − 1)(c + r) + ar (c + r) + ar−2 − ar v2 = 0

i.e.

ar [(c + r)(c + r − 1 + 1) − v ] = −ar−2 ar [(c + r)2 − v2 ] = −ar−2

i.e.

ar =

ar−2 v2 − (c + r)2

for

r≥2

=

a0 25 (v + 1)(v + 2)

=

a0 24 × 2(v + 1)(v + 2)

= (37) =

For the term in xc+1 , a1 [c(c + 1) + (c + 1) − v2 ] = 0

a0 [v2 − (v + 2)2 ] [v2 − (v + 4)2 ] a0 [−22 (v + 1)][−23 (v + 2)]

a6 =

i.e. the recurrence relation is:

−a0 −a0 = 4 + 4v 22 (v + 1)

=

ar [(c + r − 1)(c + r) + (c + r) − v2 ] = −ar−2 2

− (c + 4)2 ]

and so on.

=

a0 [c2 − c + c − v2 ] = 0

[v2

a0 [v2 − (c + 2)2 ][v2 −(c + 4)2 ][v2 − (c + 6)2 ]

(36)

i.e.

(37)

when r = 6, a6 =

− a1 v2 xc+1 − · · · − ar v2 xc+r − · · · = 0

from which,

equation

and

+ ar (c + r − 1)(c + r)xc+r + · · · + a0 cxc

Hence,

from

=

a0 [v2 −(v+2)2 ][v2 −(v+4)2 ][v2 −(v+6)2 ] a0 [24 × 2(v + 1)(v + 2)][−12(v + 3)] 24

−a0 × 2(v + 1)(v + 2) × 22 × 3(v + 3)

−a0 and so on. 26 × 3!(v + 1)(v + 2)(v + 3)

594 Section I The resulting solution for c = +v is given by: y=u= { v A x 1−

+Bx x2 x4 + 22 (v +1) 24 × 2!(v +1)(v +2)

x6 − 6 +··· 2 × 3!(v +1)(v + 2)(v + 3)

}

(38) which is valid provided v is not a negative integer and where A is an arbitrary constant. When c = −v, a2 =

a0 a0 = 2 v2 − (−v + 2)2 v − (v2 − 4v + 4)

26

x4 24 × 2!(v − 1)(v − 2)

x6 + 6 +· · · 2 × 3!(v−1)(v−2)(v−3)

} (39)

The gamma function The solution of the Bessel equation of Problem 10 may be expressed in terms of gamma functions. Γ is the upper case Greek letter gamma, and the gamma function Γ(x) is defined by the integral ∫ ∞ Γ(x) = tx−1 e−t dt (40)

From equation (40),

∫

∞

Γ(x + 1) =

tx e−t dt

0

and by using integration by parts (see page 491): [ ( )] ( ) e−t ∞ Γ(x + 1) = tx −1 0 ∫ ∞ ( −t ) e − x tx−1 dt −1 0 ∫ ∞ = (0 − 0) + x e−t tx−1 dt

a0 × 3!(v−1)(v−2)(v−3)

0

y=w= { −v Bx 1+

x2 x4 + 22 (v−1) 24 ×2!(v−1)(v−2) } x6 + 6 + ··· 2 × 3!(v − 1)(v − 2)(v − 3)

which is valid provided v is not a positive integer and where B is an arbitrary constant. The complete solution of Bessel’s equation: ) d2 y dy ( x2 2 + x + x2 − v2 y = 0 is: dx dx y = u+w =

−

x2 22 (v − 1)

and is convergent for x > 0

Hence,

{ v Ax 1−

{ 1+

0

a0 a0 ) = = 2( 4v − 4 2 v − 1 a0 a4 = 2 [2 (v − 1)][v2 − (−v + 4)2 ] a0 = 2 [2 (v − 1)][23 (v − 2)] a0 = 4 2 × 2(v − 1)(v − 2) Similarly, a6 =

+

−v

2

x

22 (v + 1)

4

+

x

= xΓ(x)

from equation (40)

This is an important recurrence relation for gamma functions. Thus, since

Γ(x + 1) = xΓ(x)

then similarly,

Γ(x + 2) = (x + 1)Γ(x + 1) = (x + 1)xΓ(x)

and

(41)

Γ(x + 3) = (x + 2)Γ(x + 2) = (x + 2)(x + 1)xΓ(x), and so on.

These relationships involving gamma functions are used with Bessel functions.

24 × 2!(v + 1)(v + 2)

x6 +··· 26 × 3!(v + 1)(v + 2)(v + 3)

}

Bessel functions The power series solution of the Bessel equation may be written in terms of gamma functions as shown in Problem 11 below.

Power series methods of solving ordinary differential equations 595 Problem 11. Show that the power series solution of the Bessel equation of Problem 10 may be written in terms of the Bessel functions Jv (x) and J−v (x) as: AJv (x) + BJ−v (x) ( x )v { 1 x2 = − 2 2 Γ(v + 1) 2 (1!)Γ(v + 2) x4 + 4 − ··· 2 (2!)Γ(v + 4) +

( x )−v { 2

a2k =

(−1)k 2v+2k (k!)Γ(v + k + 1)

Hence, it is possible to write the new form for equation (38) as: }

1 x2 − 2 Γ(1 − v) 2 (1!)Γ(2 − v) } x4 + 4 − ··· 2 (2!)Γ(3 − v)

{

1 x2 − 2v Γ(v + 1) 2v+2 (1!)Γ(v + 2)

v

y = Ax

x4 + v+4 − ··· 2 (2!)Γ(v + 3)

Jv (x) =

i.e.

( x )v { 2

1 2v Γ(v + 1)

−1 −1 = 22 (v + 1) 2v Γ(v + 1) 2v+2 (v + 1)Γ(v + 1) −1 = v+2 from equation (41) 2 Γ(v + 2)

a2 =

a2 from equation (37) v2 − (c + 4)2 a2 a2 = = (v − c − 4)(v + c + 4) −4(2v + 4) since c = v −a2 −1 −1 = 3 = 2 (v + 2) 23 (v + 2) 2v+2 Γ(v + 2)

For the second solution, when c = −v, replacing v by −v in equation (42) above gives: a2k =

(−1)k ( ) 22k−v k! Γ(k − v + 1) (

from =

which, 1

2−v Γ(1 − v)

a6 =

−1 and so on. 2v+6 (3!)Γ(v + 4)

since 0! = 1 (see page 580) (−1)1 ( ) 22−v 1! Γ(1 − v + 1) −1 = 2−v 2 (1!)Γ(2 − v)

when k = 1, a2 =

when k = 2, a4 = =

The recurrence relation is: (−1)r/2 (r ) ( ) ar = r 2v+r ! Γ v + + 1 2 2

)0 −1 k = 0, a0 = −v 2 (0!)Γ(1 − v)

when

since (v + 2)Γ(v + 2) = Γ(v + 3) and

1 x2 − 2 Γ(v + 1) 2 (1!)Γ(v + 2) } x4 + 4 −··· 2 (2!)Γ(v + 3)

provided v is not a negative integer.

Similarly, a4 =

1 2v+4 (2!)Γ(v + 3)

}

This is called the Bessel function of the first-order kind, of order v, and is denoted by Jv (x),

then

=

(42) for k = 1, 2, 3, . . .

From Problem 10 above, when c = +v, −a0 a2 = 2 2 (v + 1) If we let a0 =

and if we let r = 2k, then

when k = 3, a6 = =

(−1)2 24−v (2!)Γ(2 − v + 1) 1 24−v (2!)Γ(3 − v) 26−v

(−1)3 ( ) 3! Γ(3 − v + 1) −1

26−v (3!)Γ(4 − v)

and so on.

596 Section I −v

{

Hence, y = Bx

1 x2 − 2−v Γ(1 − v) 22−v (1!)Γ(2 − v) x4 + 4−v − ··· 2 (2!)Γ(3 − v)

i.e. J−v (x) =

( x )−v { 2

}

x2 1 − 2 Γ(1 − v) 2 (1!)Γ(2 − v) x4 −· · · + 4 2 (2!)Γ(3 − v)

Jv (x) and J−v (x) are two independent solutions of the Bessel equation; the complete solution is: y = AJv (x) + BJ−v (x) where A and B are constants i.e. y = AJv (x) + BJ−v (x) =A

2

1 x2 − 2 Γ(v + 1) 2 (1!)Γ(v + 2) x4 + 4 −··· 2 (2!)Γ(v + 3)

+B

( x )−v { 2

y = J1(x)

0

2

4

6

8

10

12

14 x

–0.5

Figure 52.1

From this series two commonly used functions are derived, 1 ( x )2 1 ( x )4 1 − + i.e. J0 (x) = (0!) (1!)2 2 (2!)2 2 1 ( x )6 − + ··· (3!)2 2 = 1−

}

x2 22 (1!)2

x and J1 (x) = 2

{

+

x4 24 (2!)2

−

x6 26 (3!)2

+···

1 1 ( x )2 − (1!) (1!)(2!) 2 1 ( x )4 + − ··· (2!)(3!) 2

1 x − 2 Γ(1 − v) 2 (1!)Γ(2 − v)

In general terms: Jv (x) =

=

}

( x )v ∑ ∞

(−1)k x2k 2 k=0 22k (k!)Γ(v+k+1)

( x )−v ∑ ∞ 2

0.5

2

x4 + 4 −··· 2 (2!)Γ(3 − v)

and J−v (x) =

y = J0(x)

}

provided v is not a positive integer.

( x )v {

y 1

(−1)k x2k 2k k=0 2 (k!)Γ(k − v + 1)

}

x x3 x5 − 3 + 5 2 2 (1!)(2!) 2 (2!)(3!) −

x7 27 (3!)(4!)

+···

Tables of Bessel functions are available for a range of values of n and x, and in these, J0 (x) and J1 (x) are most commonly used. Graphs of J0 (x), which looks similar to a cosine, and J1 (x), which looks similar to a sine, are shown in Figure 52.1.

Another Bessel function It may be shown that another series for Jn (x) is given by: ( x )n { 1 ( x )2 1 Jn (x) = − 2 n! (n + 1)! 2 } ( x )4 1 + − ··· (2!)(n + 2)! 2

Now try the following Practice Exercise Practice Exercise 240 Bessel’s equation and Bessel’s functions (Answers on page 896) 1.

Determine the power series solution of Besd2 y dy sel’s equation: x2 2 + x + (x2 −v2 )y = 0 dx dx when v = 2, up to and including the term in x6 .

Power series methods of solving ordinary differential equations 597

2. Find the power series (solution) of the Bessel function: x2 y′′ + xy′ + x2 − v2 y = 0 in terms of the Bessel function J3 (x) when v = 3. Give the answer up to and including the term in x7 .

To solve Legendre’s equation (1 − x2 )y′′ − 2xy′ + k(k + 1)y = 0 using the Frobenius method: (i)

3. Evaluate the Bessel functions J0 (x) and J1 (x) when x = 1, correct to 3 decimal places.

Let a trial solution be of the form { y = xc a0 + a1 x + a2 x2 + a3 x3 } + · · · + ar xr + · · · (43) where a0 = ̸ 0, i.e.

y = a0 xc + a1 xc+1 + a2 xc+2 + a3 xc+3 + · · · + ar xc+r + · · ·

52.7 Legendre’s equation and Legendre polynomials

(ii)

(44)

Differentiating equation (44) gives: y′ = a0 cxc−1 + a1 (c + 1)xc

Another important differential equation in physics and engineering applications is Legendre’s∗ equady d2 y tion of the form: (1 − x2 ) 2 − 2x + k(k + 1)y = 0 or dx dx (1 − x2 )y′′ − 2xy′ + k(k + 1)y = 0 where k is a real constant.

+ a2 (c + 2)xc+1 + · · · + ar (c + r)xc+r−1 + · · · and y′′ = a0 c(c − 1)xc−2 + a1 c(c + 1)xc−1 + a2 (c + 1)(c + 2)xc + · · ·

Problem 12. Determine the general power series solution of Legendre’s equation.

+ ar (c + r − 1)(c + r)xc+r−2 + · · · (iii)

Substituting y, y′ and y′′ into each term of the given equation: ( ) 1 − x2 y′′ − 2xy′ + k(k + 1)y = 0 gives: a0 c(c − 1)xc−2 + a1 c(c + 1)xc−1 + a2 (c + 1)(c + 2)xc + · · · + ar (c + r − 1)(c + r)xc+r−2 + · · · − a0 c(c − 1)xc − a1 c(c + 1)xc+1 − a2 (c + 1)(c + 2)xc+2 − · · · − ar (c + r − 1)(c + r)xc+r − · · · − 2a0 cxc − 2a1 (c + 1)xc+1 − 2a2 (c + 2)xc+2 − · · · − 2ar (c + r)xc+r − · · · + k 2 a0 xc + k 2 a1 xc+1 + k 2 a2 xc+2 + · · · + k 2 ar xc+r + · · · + ka0 xc + ka1 xc+1 + · · ·

∗

Who was Legendre? Adrien-Marie Legendre (18 September 1752–10 January 1833) was a French mathematician. Legendre developed the least squares method, which is used in linear regression, signal processing, statistics, and curve fitting. To find out more go to www.routledge.com/cw/bird

+ kar xc+r + · · · = 0

(45)

(iv) The indicial equation is obtained by equating the coefficient of the lowest power of x (i.e. xc−2 )

598 Section I to zero. Hence, a0 c(c − 1) = 0 from which, c = 0 or c = 1 since a0 ̸= 0 c−1

For the term in x , i.e. a1 c(c + 1) = 0, with c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined with the zero value of c would make the product zero. For the term in xc+r , ar+2 (c + r + 1)(c + r + 2) −ar (c + r − 1) (c + r) − 2ar (c + r) + k 2 ar + kar = 0 from which, [

] ar (c+r−1)(c+r)+2(c+r)−k −k ar+2 = (c+r+1)(c+r+2) =

2

ar [(c + r)(c + r + 1) − k(k + 1)] (c + r + 1)(c + r + 2)

When c = 0, ar+2 =

ar [r(r + 1) − k(k + 1)] (r + 1)(r + 2)

(46)

For r = 3, a5 =

a3 [(3)(4) − k(k + 1)] −a3 [k 2 + k − 12] = (4)(5) (4)(5)

=

−a3 (k + 4)(k − 3) (4)(5)

=

−(k + 4)(k − 3) −a1 (k − 1)(k + 2) . (4)(5) (2)(3)

=

a1 (k − 1)(k − 3)(k + 2)(k + 4) and so on. 5!

Substituting values into equation (43) gives: { a0 k(k + 1) 2 0 y = x a0 + a1 x − x 2! a1 (k − 1)(k + 2) 3 x 3! a0 k(k + 1)(k − 2)(k + 3) 4 + x 4! −

+

a1 (k − 1)(k − 3)(k + 2)(k + 4) 5 x 5!

For r = 0, a2 =

a0 [−k(k + 1)] (1)(2)

For r = 1, a1 [(1)(2) − k(k + 1)] a3 = (2)(3) −a1 [k 2 + k − 2] −a1 (k − 1)(k + 2) = = 3! 3! For r = 2, [ ] a2 [(2)(3) − k(k + 1)] −a2 k 2 + k − 6 a4 = = (3)(4) (3)(4) =

−a2 (k + 3)(k − 2) (3)(4)

=

−(k + 3)(k − 2) a0 [−k(k + 1)] . (3)(4) (1)(2)

=

a0 k(k + 1)(k + 3)(k −2) 4!

} + ··· { k(k + 1) 2 i.e. y = a0 1 − x 2! +

k(k + 1)(k − 2)(k + 3) 4 x −··· 4!

}

{ (k − 1)(k + 2) 3 + a1 x − x 3! +

} (k − 1)(k − 3)(k + 2)(k + 4) 5 x −··· 5! (47)

From page 591, it was stated that if two solutions of the indicial equation differ by an integer, as in this case, where c = 0 and 1, and if one coefficient is indeterminate, as with when c = 0, then the complete solution is always given by using this value of c. Using the second value of c, i.e. c = 1 in this problem, will give a series which is one of the series in the first solution. (This may be checked for c = 1 and where a1 = 0; the result will be the second part of equation (47) above.)

Power series methods of solving ordinary differential equations 599 Legendre’s polynomials

Rodrigues’ formula

(A polynomial is an expression of the form: f(x) = a + bx + cx2 + dx3 + · · · .) When k in equation (47) above is an integer, say, n, one of the solution series terminates after a finite number of terms. For example, if k = 2, then the first series terminates after the term in x2 . The resulting polynomial in x, denoted by Pn (x), is called a Legendre polynomial. Constants a0 and a1 are chosen so that y = 1 when x = 1. This is demonstrated in the following worked problems.

An alternative method of determining Legendre polynomials is by using Rodrigues’∗ formula, which states: ( )n 1 dn x2 − 1 Pn (x) = n (48) 2 n! dxn

Problem 13. Determine the Legendre polynomial P2 (x) Since in P2 (x), n = k = 2, then from the first part of equation (47), i.e. the even powers of x: } { 2(3) 2 x + 0 = a0 {1 − 3x2 } y = a0 1 − 2!

This is demonstrated in the following worked problems. Problem 15. Determine the Legendre polynomial P2 (x) using Rodrigues’ formula. ( )n 1 dn x2 − 1 In Rodrigues’ formula, Pn (x) = n 2 n! dxn and when n = 2, P2 (x) =

1 d2 (x2 − 1)2 22 2! dx2

a0 is chosen to make y = 1 when x = 1 i.e. 1 = a0 {1 − 3(1)2 } = −2a0 , from which, a0 = − Hence, P2 (x) = −

) 1 1( 1 − 3x2 = (3x2 − 1) 2 2

1 2

Problem 14. Determine the Legendre polynomial P3 (x) Since in P3 (x), n = k = 3, then from the second part of equation (47), i.e. the odd powers of x: { (k − 1)(k + 2) 3 y = a1 x − x 3! } (k − 1)(k − 3)(k + 2)(k + 4) 5 + x − ··· 5! { } (2)(5) 3 (2)(0)(5)(7) 5 x + x i.e. y = a1 x − 3! 5! { } 5 = a1 x − x3 + 0 3 a1 is chosen to make y = 1 when x = 1. { } ( ) 5 2 3 i.e. 1 = a1 1 − = a1 − from which, a1 = − 3 3 2 ( ) 3 5 3 1 Hence, P3 (x) =− x− x or P3 (x) = (5x3 − 3x) 2 3 2

∗

Who was Rodrigues? Benjamin Olinde Rodrigues (1795–1851), was a French banker, mathematician, and social reformer. Rodrigues is remembered for three results: Rodrigues’ rotation formula for vectors; the Rodrigues formula about series of orthogonal polynomials; and the Euler–Rodrigues parameters. To find out more go to www.routledge.com/cw/bird

600 Section I ( ) d x6 −3x4 +3x2 −1 = 6x5 − 12x3 + 6x dx

1 d2 (x4 − 2x2 + 1) = 3 2 dx2 d 4 (x − 2x2 + 1) = 4x3 − 4x dx ( ) d2 x4 − 2x2 + 1 d(4x3 − 4x) and = = 12x2 − 4 2 dx dx ( ) ) 1 d2 x4 −2x2 +1 1( Hence, P2 (x) = 3 = 12x2 − 4 2 2 dx 8 i.e. P2 (x) =

) 1( 2 3x − 1 , the same as in Problem 13. 2

Problem 16. Determine the Legendre polynomial P3 (x) using Rodrigues’ formula. ( )n 1 dn x2 − 1 and In Rodrigues’ formula, Pn (x) = n 2 n! dxn when n = 3, ( )3 1 d3 x2 − 1 P3 (x) = 3 2 3! dx3

( ) d 6x5 −12x3 +6x = 30x4 − 36x2 + 6 dx ( ) d 30x4 − 36x2 + 6 = 120x3 − 72x and dx ( ) 1 d3 x6 − 3x4 + 3x2 − 1 Hence, P3 (x) = (8)(6) dx3 ) 1( ) 1 ( = 120x3 − 72x = 20x3 − 12x (8)(6) 8 ) 1( 3 i.e. P3 (x) = 5x − 3x , the same as in Problem 14. 2 Now try the following Practice Exercise Practice Exercise 241 Legendre’s equation and Legendre polynomials (Answers on page 897) 1.

Determine the power series solution of the Legendre equation: ( ) 1 − x2 y′′ − 2xy′ + k(k + 1)y = 0 when (a) k = 0 (b) k = 2, up to and including the term in x5 .

2.

Find the following Legendre polynomials: (a) P1 (x) (b) P4 (x) (c) P5 (x)

( )( ) 1 d3 x2 − 1 x4 − 2x2 + 1 = 3 2 (6) dx3 ( ) 1 d3 x6 − 3x4 + 3x2 − 1 = (8)(6) dx3

For fully worked solutions to each of the problems in Practice Exercises 236 to 241 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 53

An introduction to partial differential equations Why it is important to understand: An introduction to partial differential equations In engineering, physics and economics, quantities are frequently encountered − for example energy − that depend on many variables, such as position, velocity and temperature. Usually this dependency is expressed through a partial differential equation, and solving these equations is important for understanding these complex relationships. Solving ordinary differential equations involves finding a function (or a set of functions) of one independent variable, but partial differential equations are for functions of two or more variables. Examples of physical models using partial differential equations are the heat equation for the evolution of the temperature distribution in a body, the wave equation for the motion of a wave front, the flow equation for the flow of fluids and Laplace’s equation for an electrostatic potential or elastic strain field. In such cases, not only are the initial conditions needed, but also boundary conditions for the region in which the model applies; thus boundary value problems have to be solved. This chapter provides an introduction to the often complex subject of partial differential equations.

At the end of this chapter, you should be able to: • recognise some important engineering partial differential equations • solve a partial differential equation by direct partial integration • solve differential equations by separating the variables • solve the wave equation

∂2u 1 ∂2u = ∂x2 c2 ∂t2

• solve the heat conduction equation • solve Laplace’s equation

∂2u 1 ∂u = 2 2 ∂x c ∂t

∂2u ∂2u + =0 ∂x2 ∂y2

602 Section I ∫

53.1

Introduction

and u =

∂u ∂u +b =c ∂x ∂y

a

(ii)

∂2u 1 ∂u = 2 ∂x2 c ∂t

= −5 cos x cos t + f(x) ∂2u = 6x2 cos 2y is integrated partially ∂x∂y with respect to y, ∫ ∫ ( 2) ∂u 2 cos 2y dy then = 6x cos 2y dy = 6x ∂x ( ) ( ) 1 = 6x2 sin 2y + f (x) 2

Similarly, if

(known as the heat conduction equation) (iii)

∂2u ∂2u + =0 ∂x2 ∂y2

= 3x2 sin 2y + f(x)

(known as Laplace’s equation) Equation (i) is a first-order partial differential equation, and equations (ii) and (iii) are second-order partial differential equations since the highest power of the differential is 2. Partial differential equations occur in many areas of engineering and technology; electrostatics, heat conduction, magnetism, wave motion, hydrodynamics and aerodynamics all use models that involve partial differential equations. Such equations are difficult to solve, but techniques have been developed for the simpler types. In fact, for all but the simplest cases, there are a number of numerical methods of solutions of partial differential equations available. To be able to solve simple partial differential equations knowledge of the following is required: (a) partial integration, (b)

first-and second-order partial differentiation – as explained in Chapter 32, and

(c) the solution of ordinary differential equations – as explained in Chapters 46 to 51. It should be appreciated that whole books have been written on partial differential equations and their solutions. This chapter does no more than introduce the topic.

53.2

sin t dt

= (5 cos x)(−cos t) + c

A partial differential equation is an equation that contains one or more partial derivatives. Examples include: (i)

∫ 5 cos x sin t dt = (5 cos x)

Partial integration

Integration is the reverse process of differentiation. ∂u Thus, if, for example, = 5 cos x sin t is integrated ∂t partially with respect to t, then the 5 cos x term is considered as a constant,

and integrating ∫ u=

∂u partially with respect to x gives: ∂x

[3x2 sin 2y + f(x)] dx

= x3 sin 2y + (x) f(x) + g(y) f(x) and g(y) are functions that may be determined if extra information, called boundary conditions or initial conditions, are known.

53.3

Solution of partial differential equations by direct partial integration

The simplest form of partial differential equations occurs when a solution can be determined by direct partial integration. This is demonstrated in the following worked problems. Problem 1. Solve the differential equation ( ) ∂2u = 6x2 2y − 1 given the boundary conditions 2 ∂x ∂u = sin 2y and u = cos y that at x = 0, ∂x ( ) ∂2u Since 2 = 6x2 2y − 1 then integrating partially with ∂x respect to x gives: ∫ ∫ ∂u = 6x2 (2y − 1)dx = (2y − 1) 6x2 dx ∂x = (2y − 1)

6x3 + f (y) 3

= 2x3 (2y − 1) + f (y) where f (y) is an arbitrary function.

An introduction to partial differential equations 603 From the boundary conditions, when x = 0, ∂u = sin 2y ∂x Hence,

sin 2y = 2(0)3 (2y − 1) + f (y)

from which,

f (y) = sin 2y

Integrating partially with respect to x gives: ∫ u = [sin(x + y) + 2 − sin x]dx = −cos(x + y) + 2x + cos x + f (y) From the boundary conditions, u = y2 when x = 0, hence

∂u = 2x3 (2y − 1) + sin 2y ∂x

Now

y2 = −cos y + 0 + cos 0 + f(y) = 1 − cos y + f (y)

Integrating partially with respect to x gives: ∫ [2x (2y − 1) + sin 2y]dx 3

u=

2x4 = (2y − 1) + x(sin 2y) + F(y) 4

from which, f(y) = y2 − 1 + cos y ∂2u = cos(x + y) for the given ∂x∂y boundary conditions is: Hence, the solution of

From the boundary conditions, when x = 0, u = cos y, hence cos y =

(0)4 (2y − 1) + (0)sin 2y + F(y) 2

from which, F(y) = cos y ∂2u Hence, the solution of 2 = 6x2 (2y − 1) for the given ∂x boundary conditions is: 4

u=

x (2y − 1) + x sin 2 y + cos y 2

Problem 2. Solve the differential equation: ∂2u ∂u = cos(x + y) given that = 2 when y = 0, ∂x∂y ∂x and u = y2 when x = 0

u = −cos(x + y) + 2x + cos x + y2 − 1 + cos y Problem 3.

Verify that 1 ϕ(x, y, z) = √ satisfies the partial x2 + y2 + z2 ∂2ϕ ∂2ϕ ∂2ϕ differential equation: 2 + 2 + 2 = 0 ∂x ∂y ∂z The partial differential equation ∂2ϕ ∂2ϕ ∂2ϕ + + = 0 is called Laplace’s equation. ∂x2 ∂y2 ∂z2 1 1 If ϕ(x, y, z) = √ = (x2 + y2 + z2 )− 2 x2 + y2 + z2 then differentiating partially with respect to x gives: 3 ∂ϕ 1 = − (x2 + y2 + z2 )− 2 (2x) ∂x 2

∂2u = cos(x + y) then integrating partially with ∂x∂y respect to y gives: Since

∂u = ∂x

∫

3

and cos(x + y)dy = sin(x + y) + f (x)

∂u From the boundary conditions, = 2 when y = 0, ∂x hence 2 = sin x + f (x) from which,

f(x) = 2 − sin x

i.e.

∂u = sin(x + y) + 2 − sin x ∂x

= −x(x2 + y2 + z2 )− 2 [ ] ∂2ϕ 3 2 2 2 − 52 = (−x) − (x + y + z ) (2x) ∂x2 2 + (x2 + y2 + z2 )− 2 (−1) 3

by the product rule =

=

3x2 (x2

+ y2

5 + z2 ) 2

−

1 (x2

(3x2 ) − (x2 + y2 + z2 ) 5

(x2 + y2 + z2 ) 2

3

+ y2 + z2 ) 2

604 Section I Similarly, it may be shown that ∂2u = 8ey sin 2x given that at y = 0, ∂x∂y ∂u π = sin x, and at x = , u = 2y2 ∂x 2

5. Solve

∂ ϕ (3y ) − (x + y + z ) = 5 ∂y2 (x2 + y2 + z2 ) 2 2

and

2

2

2

2

∂ 2 ϕ (3z2 ) − (x2 + y2 + z2 ) = 5 ∂z2 (x2 + y2 + z2 ) 2

∂2u = y(4x2 − 1) given that at x = 0, ∂x2 ∂u u = sin y and = cos 2y ∂x

6. Solve

Thus, ∂ 2 ϕ ∂ 2 ϕ ∂ 2 ϕ (3x2 ) − (x2 + y2 + z2 ) + 2 + 2 = 5 ∂x2 ∂y ∂z (x2 + y2 + z2 ) 2 (3y ) − (x + y + z ) 2

+

+ 

2

2

2

5

(x2 + y2 + z2 ) 2 (3z2 ) − (x2 + y2 + z2 ) (x2 + y2 + z2 )

2x

   + 3y2 − (x2 + y2 + z2 )   =

+ 3z − (x + y + z ) 2

2

8. Show that u(x, y) = xy +

5 2



3x2 − (x2 + y2 + z2 )

2

∂2u ∂u = sin(x + t) given that =1 ∂x∂t ∂x when t = 0, and when u = 2t when x = 0

7. Solve

2

5

(x2 + y2 + z2 ) 2

=0

1 Thus, √ satisfies the Laplace equation 2 x + y 2 + z2 ∂2ϕ ∂2ϕ ∂2ϕ + + =0 ∂x2 ∂y2 ∂z2

9. Find the particular solution of the differen∂2u tial equation = cos x cos y given the ini∂x∂y ∂u tial conditions that when y = π, = x, and ∂x when x = π, u = 2 cos y 10.

Verify that ϕ(x, y) = x cos y + e x sin y satisfies the differential equation ∂2ϕ ∂2ϕ + + x cos y = 0 ∂x2 ∂y2

Now try the following Practice Exercise Practice Exercise 242 The solution of partial differential equations by direct partial integration (Answers on page 897) 1. Determine the general solution of

∂2u ∂2u + y 2 = 2x ∂x∂y ∂y

x is a solution of y

∂u = 4ty ∂y

∂u 2. Solve = 2t cos θ given that u = 2t when ∂t θ=0 3. Verify that u(θ, t) = θ2 + θt is a solution of ∂u ∂u −2 =t ∂θ ∂t 4. Verify that u = e−y cos x is a solution of ∂2u ∂2u + =0 ∂x2 ∂y2

53.4 Some important engineering partial differential equations There are many types of partial differential equations. Some typically found in engineering and science include: (a) The wave equation, where the equation of motion is given by: ∂2u 1 ∂2u = 2 2 2 ∂x c ∂t T , with T being the tension in a string ρ and ρ being the mass/unit length of the string. where c2 =

An introduction to partial differential equations 605 (b)

The heat conduction equation is of the form: ∂ 2 u 1 ∂u = ∂x2 c2 ∂t h , with h being the thermal conducσρ tivity of the material, σ the specific heat of the material, and ρ the mass/unit length of material. where c2 =

(c)

Laplace’s∗ equation, used extensively with electrostatic fields is of the form: ∂2u ∂2u ∂2u + + =0 ∂x2 ∂y2 ∂z2

(d)

The transmission equation, where the potential u in a transmission cable is of the form: ∂2u ∂2u ∂u = A + B + Cu where A, B and C are ∂x2 ∂t2 ∂t constants.

Some of these equations are used in the next sections.

53.5

Separating the variables

Let u(x, t) = X(x)T(t), where X(x) is a function of x only and T(t) is a function of t only, be a trial solution to the ∂2u 1 ∂2u wave equation 2 = 2 2 . If the trial solution is sim∂x c ∂t ∂u ∂2u plified to u = XT, then = X′T and = X′′T. Also ∂x ∂x2 ∂u ∂2u = XT′ and 2 = XT′′ ∂t ∂t Substituting into the partial differential equation ∂2u 1 ∂2u = 2 2 gives: 2 ∂x c ∂t 1 X′′T = 2 XT′′ c Separating the variables gives: X′′ X

=

1 T′′ c2 T

X′′ 1 T′′ = where µ is a constant. X c2 T X′′ Thus, since µ = (a function of x only), it must be X 1 T′′ (a function of t independent of t; and, since µ = 2 c T only), it must be independent of x. If µ is independent of x and t, it can only be a conX′′ stant. If µ = then X′′ = µX or X′′ − µX = 0 and if X ′′ 1 T µ= 2 then T′′ = c2 µT or T′′ − c2 µT = 0 c T Such ordinary differential equations are of the form found in Chapter 50, and their solutions will depend on whether µ > 0, µ = 0 or µ < 0 Let µ =

Problem 4 will be a reminder of solving ordinary differential equations of this type. Problem 4. Find the general solution of the following differential equations: (a) X′′ − 4X = 0

(b) T′′ + 4T = 0

(a) If X′′ − 4X = 0 then the auxiliary equation (see Chapter 50) is: ∗

Who was Laplace? Pierre-Simon, marquis de Laplace (23 March 1749–5 March 1827) was a French mathematician and astronomer who formulated Laplace’s equation, and pioneered the Laplace transform which appears in many branches of mathematical physics. To find out more go to www.routledge.com/cw/bird

m 2 − 4 = 0 i.e. m 2 = 4 from which, m = +2 or m = −2 Thus, the general solution is: X = Ae2x + Be−2x

606 Section I (b)

If T′′ + 4T = 0 then the auxiliary equation is:

(i)

m 2 + 4 = 0 i.e. m 2 = −4 from which, √ m = −4 = ±j2

The string is fixed at both ends, i.e. x = 0 and x = L for all values of time t. Hence, u(x, t) becomes: } u(0, t) = 0 for all values of t ≥ 0 u(L, t) = 0

Thus, the general solution is: T = e0 {A cos 2t + B sin 2t} = A cos 2t + B sin 2t

Now try the following Practice Exercise Practice Exercise 243 Revising the solution of ordinary differential equations (Answers on page 897) 1. Solve T′′ = c2 µT given c = 3 and µ = 1 2. Solve T′′ − c2 µT = 0 given c = 3 and µ = −1

(ii)

If the initial deflection of P at t = 0 is denoted by f(x) then u(x, 0) = f (x)

(iii)

Let the initial velocity of P be g(x), then [ ] ∂u = g(x) ∂t t=0

Initially a trial solution of the form u(x, t) = X(x)T(t) is assumed, where X(x) is a function of x only and T(t) is a function of t only. The trial solution may be simplified to u = XT and the variables separated as explained in the previous section to give:

3. Solve X′′ = µX given µ= 1

X′′ 1 T′′ = X c2 T

′′

4. Solve X − µX = 0 given µ = −1

53.6

When both sides are equated to a constant µ this results in two ordinary differential equations:

The wave equation

T′′ − c2 µT = 0

An elastic string is a string with elastic properties, i.e. the string satisfies Hooke’s law. Fig. 53.1 shows a flexible elastic string stretched between two points at x = 0 and x = L with uniform tension T. The string will vibrate if the string is displaced slightly from its initial position of rest and released, the end points remaining fixed. The position of any point P on the string depends on its distance from one end, and on the instant in time. Its displacement u at any time t can be expressed as u = f(x, t), where x is its distance from 0 The equation of motion is as stated in Section 53.4 (a), ∂2u 1 ∂2u i.e. 2 = 2 2 ∂x c ∂t The boundary and initial conditions are:

u 5 f (x, t )

y

u(x, t ) L x

Figure 53.1

X′′ − µX = 0

Three cases are possible, depending on the value of µ.

Case 1: µ > 0 For convenience, let µ = p 2 , where p is a real constant. Then the equations X′′ − p 2 X = 0

and T′′ − c2 p 2 T = 0

have solutions: X = Aepx + Be−px and T = Cecpt + De−cpt where A, B, C and D are constants. But X = 0 at x = 0, hence 0 = A + B i.e. B = −A and X = 0 at x = L, hence 0 = AepL + Be−pL = A(epL − e−pL ) Assuming (epL – e−pL ) is not zero, then A = 0 and since B = −A, then B = 0 also. This corresponds to the string being stationary; since it is non-oscillatory, this solution will be disregarded.

Case 2: µ = 0

P

0

and

x

In this case, since µ = p 2 = 0, T′′ = 0 and X′′ = 0. We will assume that T(t) = ̸ 0. Since X′′ = 0, X′ = a and X = ax + b where a and b are constants. But X = 0 at x = 0, hence b = 0 and X = ax and X = 0 at x = L, hence a = 0. Thus, again, the solution is non-oscillatory and is also disregarded.

An introduction to partial differential equations 607 Case 3: µ < 0

To find An and Bn we put in the initial conditions not yet taken into account.

For convenience, let µ = −p 2 then X′′ + p 2 X = 0 from which, X = A cos px + B sin px and T′′ + c2 p 2 T = 0

(i) (1)

At t = 0, u(x, 0) = f (x) for 0 ≤ x ≤ L Hence, from equation (5), u(x, 0) = f (x) =

from which,

T = C cos cpt + D sin cpt

[

(2) (ii)

(see Problem 4 above).

Also at t = 0,

u = {A cos px + B sin px}{C cos cpt + D sin cpt}

∞ ∂u ∑ { nπx ( ( cnπ cnπt ) = sin An − sin ∂t L L L n=1 ( cnπ cnπt ))} + Bn cos L L

(3) Applying the boundary conditions: (i) u = 0 when x = 0 for all values of t,

and when t = 0,

thus 0 = {A cos 0 + B sin 0}{C cos cpt + D sin cpt}

g(x) =

0 = A{C cos cpt + D sin cpt}

nπx cnπ } Bn L L

cπ ∑ { nπx } Bn n sin L L

(7)

n=1

u = {B sin px}{C cos cpt (4)

(ii) u = 0 when x = L for all values of t Hence,

sin

∞

i.e. g(x) =

+ D sin cpt} ̸= 0) + D sin cpt}

∞ { ∑ n=1

from which, A = 0 (since {C cos cpt Hence,

= g(x) for 0 ≤ x ≤ L t=0

Differentiating equation (5) with respect to t gives:

Thus, the suggested solution u = XT now becomes:

i.e.

(6)

n=1

]

∂u ∂t

∞ { ∑ nπx } An sin L

0 = {B sin pL}{C cos cpt + D sin cpt}

Now B ̸= 0 or u(x, t) would be identically zero. nπ Thus sin pL = 0 i.e. pL = nπ or p = for integer L values of n. Substituting in equation (4) gives: { nπx}{ cnπt cnπt} u = B sin C cos + D sin L L L nπx { cnπt cnπt} i.e. u = sin An cos + Bn sin L L L (where constants An = BC and Bn = BD). There will be many solutions, depending on the value of n. Thus, more generally, ( ∞{ ∑ nπx cnπt un (x, t) = sin An cos L L n=1 )} cnπt + Bn sin L (5)

From Fourier series (see page 691) it may be shown that: nπx An is twice the mean value of f(x) sin between x = 0 L and x = L ∫ 2 L nπx i.e. An = f(x)sin dx for n = 1, 2, 3, . . . (8) L 0 L ( cnπ ) and Bn is twice the mean value of L nπx g(x)sin between x = 0 and x = L L ( )∫ L L 2 nπx i.e. Bn = g(x)sin dx cnπ L L 0 or

Bn =

2

∫

cnπ

L

g(x)sin 0

nπx L

dx

(9)

Summary of solution of the wave equation The above may seem complicated; however a practical problem may be solved using the following eight-point procedure: 1.

Identify clearly conditions.

the

initial

and

boundary

608 Section I Assume a solution of the form u = XT and express the equations in terms of X and T and their derivatives.

2.

3.

u(x, 0) = f (x) =

Let µ = −p 2 to give an oscillatory solution.

5.

The two solutions are of the form:

6.

u(x, t) = {A cos px + B sin px}{C cos cpt + D sin cpt}

Apply the boundary conditions to determine constants A and B.

7.

Determine the general solution as an infinite sum.

8.

Apply the remaining initial and boundary conditions and determine the coefficients An and Bn from equations (8) and (9), using Fourier series techniques.

Problem 5. Fig. 53.2 shows a stretched string of length 50 cm which is set oscillating by displacing its mid-point a distance of 2 cm from its rest position and releasing it with zero velocity. ∂2u 1 ∂2u Solve the wave equation: 2 = 2 2 where ∂x c ∂t c2 = 1, to determine the resulting motion u(x, t)

2.

3.

u(x, 0 )

Assuming a solution u = XT, where X is a function of x only, and T is a function of t only, ∂u ∂2u ∂u then = X′T and = X′′T and = XT′ and ∂x ∂x2 ∂t ∂2u = XT′′. Substituting into the partial differential ∂t2 ∂2u 1 ∂2u equation, 2 = 2 2 gives: ∂x c ∂t 1 X′′T = 2 XT′′ i.e. X′′T = XT′′ since c2 = 1 c X′′ T′′ Separating the variables gives: = X T Let a constant, X′′ T′′ X′′ T′′ = then µ = and µ = X T X T from which, X′′ − µX = 0 and

4

4.

u 5 f (x )

2

0

25

T′′ − µT = 0

Letting µ = −p 2 to give an oscillatory solution gives: X′′ + p 2 X = 0

50 x (cm)

Figure 53.2

and T′′ + p 2 T = 0

The auxiliary equation for each is: m 2 + p 2 = 0 √ from which, m = −p 2 = ±jp

Following the above procedure, The boundary and initial conditions given are: u(0, t) = 0

100 − 2x 2 x+4 = 25 25

µ=

y

1.

0 ≤ x ≤ 25

(Note: y = mx + c is a straight line graph, so the gradient, m, between 0 and 25 is 2/25 and the y-axis 2 intercept is zero, thus y = f (x) = x + 0; between 25 25 and 50, the gradient = −2/25 and the y-axis 2 intercept is at 4, thus f(x) = − x + 4) 25 [ ] ∂u = 0 i.e. zero initial velocity. ∂t t=0

T = C cos cpt + D sin cpt.

Then

x

25 ≤ x ≤ 50

X = A cos px + B sin px and

25

=−

Separate the variables by transposing the equation and equate each side to a constant, say, µ; two separate equations are obtained, one in x and the other in t.

4.

2

}

u(50, t) = 0

i.e. fixed end points

5.

Solving each equation gives: X = A cos px + B sin px, and T = C cos pt + D sin pt Thus, u(x, t) = {A cos px+B sin px}{C cos pt+D sin pt}

An introduction to partial differential equations 609 6.

Applying the boundary conditions to determine constants A and B gives:

Substituting into equation (b) gives:

u(0, t) = 0, hence 0 = A{C cos pt + D sin pt} from which we conclude that A = 0 Therefore,

(i)

u(x, t) = B sin px{C cos pt + D sin pt}

(a)

nπx { nπt nπt} C cos + D sin 50 50 50

un (x, t) =

∞ ∑

sin

n=1

nπt nπx { An cos 50 50 nπt } + Bn sin 50

(b)

where An = BC and Bn = BD 8.

Hence,

2 L

∞ 16 ∑ 1

π2

n=1

n2

sin

nπx 50

sin

nπ 2

cos

nπt 50

For stretched string problems as in Problem 5 above, the main parts of the procedure are:

2.

Determine An from equation (8). ∫ 2 L nπx Note that dx is always equal to f (x) sin L 0 L 8d nπ sin (see Fig. 53.3) 2 2 n π 2 Determine Bn from equation (9)

3.

Substitute in equation (5) to determine u(x, t)

From equation (8),

y y 5 f (x )

∫

d

L

nπx dx L 0 [∫ ( ) 25 2 nπx 2 x sin dx = 50 0 25 50 ] ) ∫ 50 ( 100 − 2x nπx + sin dx 25 50 25

An =

nπt } 50 { ∞ ∑ 16 nπ nπt nπx sin cos = sin 50 n 2 π 2 2 50 n=1 nπt } + (0) sin 50

u(x, t) =

1.

or, more generally,

nπt nπx { An cos 50 50

+ Bn sin

u(50, t) = 0, hence 0 = B sin 50p{C cos pt + D sin pt}. B= ̸ 0, hence sin 50p = 0 from which, 50p = nπ and nπ p= 50 Substituting in equation (a) gives: u(x, t) = B sin

sin

n=1

(ii)

7.

∞ ∑

un (x, t) =

f(x) sin

0

L 2

L

x

Figure 53.3

Now try the following Practice Exercise Each integral is determined using integration by parts (see Chapter 42, page 491) with the result: An =

16 nπ sin n2 π2 2

1.

From equation (9),

Bn = [

∂u ∂t

]

2 cnπ

∫

L

g(x) sin 0

nπx dx L

= 0 = g(x) thus, Bn = 0 t=0

Practice Exercise 244 The wave equation (Answers on page 897) An elastic string is stretched between two points 40 cm apart. Its centre point is displaced 1.5 cm from its position of rest at right angles to the original direction of the string and then released with zero velocity. Determine the subsequent motion u(x, t) by ∂2u 1 ∂2u applying the wave equation = 2 2 2 ∂x c ∂t with c2 = 9

610 Section I

2. The centre point of an elastic string between two points P and Q, 80 cm apart, is deflected a distance of 1 cm from its position of rest perpendicular to PQ and released initially with zero velocity. Apply the wave ∂2u 1 ∂2u equation 2 = 2 2 where c = 8, to deter∂x c ∂t mine the motion of a point distance x from P at time t.

As with the wave equation, a solution of the form u(x, t) = X(x)T(t) is assumed, where X is a function of x only and T is a function of t only. If the trial solution is simplified to u = XT, then ∂2u ∂u ∂u = X′T = X′′T and = XT′ 2 ∂x ∂x ∂t Substituting into the partial differential equation, ∂2u 1 ∂u = 2 gives: ∂x2 c ∂t X′′T =

53.7

The heat conduction equation

∂ 2 u 1 ∂u is solved in The heat conduction equation 2 = 2 ∂x c ∂t a similar manner to that for the wave equation; the equation differs only in that the right-hand side contains a first partial derivative instead of the second. The conduction of heat in a uniform bar depends on the initial distribution of temperature and on the physical properties of the bar, i.e. the thermal conductivity, h, the specific heat of the material, σ, and the mass per unit length, ρ, of the bar. In the above equation, h c2 = σρ With a uniform bar insulated, except at its ends, any heat flow is along the bar and, at any instant, the temperature u at a point P is a function of its distance x from one end, and of the time t. Consider such a bar, shown in Fig. 53.4, where the bar extends from x = 0 to x = L, the temperature of the ends of the bar is maintained at zero, and the initial temperature distribution along the bar is defined by f (x). Thus, the boundary conditions can be expressed as: u(0, t) = 0

}

u(L, t) = 0

for all values of t ≥ 0

u(x, 0) = f (x) for 0 ≤ x ≤ L

and

u 5 f (x, t )

y

P u (x, t )

0

L

1 ′ XT c2

Separating the variables gives: X′′ X

=

1 T′ c2 T

Let −p 2 =

X′′ If −p 2 = then X′′ = −p 2 X or X′′ + p 2 X = 0, giving X X = A cos px + B sin px 1 T′ T′ and if −p 2 = 2 then = −p 2 c2 and integrating c T T with respect to t gives: ∫

T′ dt = T

∫ −p 2 c2 dt

from which, ln T = −p 2 c2 t + c1 The left-hand integral is obtained by an algebraic substitution (see Chapter 38). If ln T = −p 2 c2 t + c1 then 2 2 2 2 2 2 T = e−p c t+c1 = e−p c t ec1 i.e. T = k e−p c t (where constant k = ec1 ). 2 2 Hence, u(x, t) = XT = {A cos px + B sin px}k e−p c t i.e. 2 2 u(x, t) = {P cos px + Q sin px}e−p c t where P = Ak and Q = Bk Applying the boundary conditions u(0, t) = 0 gives: 2 2 2 2 0={P cos 0+Q sin 0}e−p c t =P e−p c t from which, 2 2 P = 0 and u(x, t) = {Q sin px} e−p c t 2 2 Also, u(L, t) = 0 thus, 0 = Q sin pL e−p c t and since nπ Q= ̸ 0 then sin pL = 0 from which, pL = nπ or p = L where n = 1, 2, 3, . . . There are therefore many values of u(x, t). Thus, in general,

x

x

Figure 53.4

1 T′ X′′ = 2 where −p 2 is a constant. X c T

u(x, t) =

∞ { ∑ 2 2 nπx } Qn e−p c t sin L n=1

An introduction to partial differential equations 611 Applying the remaining boundary condition, that when t = 0, u(x, t) = f(x) for 0 ≤ x ≤ L, gives: f (x) =

∞ { ∑ nπx } Qn sin L

X = A cos px + B sin px and

n=1

From Fourier series, Qn = 2 × mean value of nπx f(x) sin from x to L L ∫ 2 L nπx Hence, Qn = f(x) sin dx L 0 L

∞

2∑ L

{(∫

L 0

n=1

T = k e−p

2 2

c t

Thus, the general solution is given by: u(x, t) = {P cos px + Q sin px}e−p u(0, t) = 0 thus 0 = P e−p

2 2

c t

2 2

c t

from which, P = 0 and u(x, t) = {Q sin px}e−p

u(x, t) =

Thus,

Assuming a solution of the form u = XT, then, from above,

Also, u(1, t) = 0 thus 0 = {Q sin p}e−p

) } nπx nπx −p 2 c2 t dx e sin f (x) sin L L

This method of solution is demonstrated in the following worked problem. Problem 6. A metal bar, insulated along its sides, is 1 m long. It is initially at room temperature of 15◦ C and at time t = 0, the ends are placed into ice at 0◦ C. Find an expression for the temperature at a point P at a distance x m from one end at any time t seconds after t = 0 The temperature u along the length of bar is shown in Fig. 53.5. ∂ 2 u 1 ∂u The heat conduction equation is = and the ∂x2 c2 ∂t given boundary conditions are: u(0, t) = 0, u(1, t) = 0

2 2

c t

2 2

c t

Since Q ̸= 0, sin p = 0 from which, p = nπ where n = 1, 2, 3, . . . ∞ { } ∑ 2 2 Hence, u(x, t) = Qn e−p c t sin nπx n=1

The final initial condition given was that at t = 0, u = 15, i.e. u(x, 0) = f(x) = 15 ∞ ∑ { } Qn sin nπx where, from Fourier Hence, 15 = n=1

coefficients, Qn = 2 × mean value of 15 sin nπx from x = 0 to x = 1, ∫ [ cos nπx ]1 2 1 i.e. Qn = 15 sin nπx dx = 30 − 1 0 nπ 0 30 [cos nπ − cos 0] nπ ) 30 ( = 1 − cos nπ nπ =−

and u(x, 0) = 15

= 0 (when n is even) and

60 (when n is odd) nπ

Hence, the required solution is: ∞ { } ∑ 2 2 u(x, t) = Qn e−p c t sin nπx

u (x, 0 )

15

n=1

= 0

1

x (m )

60 π

∞ ∑ n(odd)=1

1 n

(sin nπx) e−n

2

π 2 c2 t

u (x, t )

Now try the following Practice Exercise Practice Exercise 245 The heat conduction equation (Answers on page 897)

P u (x, t ) 0

1 x

Figure 53.5

x (m )

1.

A metal bar, insulated along its sides, is 4 m long. It is initially at a temperature of 10◦ C

612 Section I y

z

and at time t = 0, the ends are placed into ice at 0◦ C. Find an expression for the temperature at a point P at a distance x m from one end at any time t seconds after t = 0

3. The ends of an insulated rod PQ, 20 units long, are maintained at 0◦ C. At time t = 0, the temperature within the rod rises uniformly from each end, reaching 4◦ C at the mid-point of PQ. Find an expression for the temperature u(x, t) at any point in the rod, distance x from P at any time t after t = 0. Assume the heat ∂ 2 u 1 ∂u = and conduction equation to be ∂x2 c2 ∂t 2 take c = 1

53.8

Laplace’s equation

The distribution of electrical potential, or temperature, over a plane area subject to certain boundary conditions, can be described by Laplace’s∗ equation. The potential at a point P in a plane (see Fig. 53.6) can be indicated by an ordinate axis and is a function of its position, i.e. z = u(x, y), where u(x, y) is the solution of the Laplace ∂2u ∂2u two-dimensional equation 2 + 2 = 0 ∂x ∂y The method of solution of Laplace’s equation is similar to the previous examples, as shown below. Fig. 53.7 shows a rectangle 0PQR bounded by the lines x = 0, y = 0, x = a, and y = b, for which we are required to ∂2u ∂2u find a solution of the equation 2 + 2 = 0. The solu∂x ∂y tion z = (x, y) will give, say, the potential at any point within the rectangle 0PQR. The boundary conditions are:

∗

Who was Laplace? See page 605 for image and resume of Pierre-Simon, marquis de Laplace. To find out more go to www.routledge.com/cw/bird

0

x

Figure 53.6 z y R y5b

Q

u (x, y )

2. An insulated uniform metal bar, 8 m long, has the temperature of its ends maintained at 0◦ C, and at time t = 0 the temperature distribution f (x) along the bar is defined by f (x) = x(8 − x). If c2 = 1, solve the heat conduction equation ∂ 2 u 1 ∂u = to determine the temperature u ∂x2 c2 ∂t at any point in the bar at time t.

P

P x5a

0

x

Figure 53.7

u = 0 when x = 0 i.e. u(0, y) = 0

for

0≤y≤b

u = 0 when x = a i.e. u(a, y) = 0

for

0≤y≤b

u = 0 when y = b i.e. u(x, b) = 0

for

0≤x≤a

u = f(x) when y = 0 i.e. u(x, 0) = f (x) for

0≤x≤a

As with previous partial differential equations, a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and Y is a function of y only. Simplifying to u = XY, determining partial derivatives, and ∂2u ∂2u substituting into 2 + 2 = 0 gives: X′′ Y + XY′′ = 0 ∂x ∂y Y′′ X′′ =− Separating the variables gives: X Y Letting each side equal a constant, −p 2 , gives the two equations: X′′ + p 2 X = 0

and Y′′ − p 2 Y = 0

from which, X = A cos px + B sin px and Y = C epy + D e−py or Y = C cosh py + D sinh py (see Problem 5, page 565 for this conversion). This latter form can also be expressed as: Y = E sinh p( y + ϕ) by using compound angles. Hence u(x, y) = XY = {A cos px + B sin px}{E sinh p(y + ϕ)}

An introduction to partial differential equations 613 or u(x, y) = {P cos px + Q sin px}{sinh p( y + ϕ)} where P = AE and Q = BE The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + ϕ) from which, P = 0. Hence, u(x, y) = Q sin px sinh p(y + ϕ) The second boundary condition is: u(a, y) = 0, hence 0 = Q sin pa sinh p(y + ϕ) from which, nπ sin pa = 0, hence, pa = nπ or p = for n = 1, 2, 3, . . . a The third boundary condition is: u(x, b) = 0, hence, 0 = Q sin px sinh p(b + ϕ) from which, sinh p(b + ϕ) = 0 and ϕ = −b Hence, u(x, y) = Q sin px sinh p(y − b) = Q1 sin px sinh p(b − y) where Q1 = −Q Since there are many solutions for integer values of n, u(x, y) =

∞ ∑

=

Qn sin px sinh p(b − y)

Qn sin

n=1

nπx nπ sinh (b − y) a a

The fourth boundary condition is: u(x, 0) = f (x), hence, f (x) =

∞ ∑ n=1

i.e.

u = 0 when x = 0

0≤y≤1

u = 0 when x = 1

0 ≤ y ≤1

u = 0 when y = 0

0≤x≤1

u = 4 when y = 1

0≤x≤1

Initially a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and Y is a function of y only. Simplifying to u = XY, determining ∂2u ∂2u partial derivatives, and substituting into 2 + 2 = 0 ∂x ∂y gives: X′′ Y + XY′′ = 0 X′′ Y′′ Separating the variables gives: =− X Y Letting each side equal a constant, −p 2 , gives the two equations: X′′ + p 2 X = 0

n=1 ∞ ∑

distribution u(x, y) over the plate, subject to the following boundary conditions:

nπx nπb Qn sin sinh a a

) ∞ ( ∑ nπb nπx f (x) = Qn sinh sin a a n=1

From Fourier series coefficients, ( ) nπb Qn sinh = 2 × the mean value of a nπx f(x) sin from x = 0 to x = a a ∫ a nπx = f(x) sin dx from which, a 0 Qn may be determined. This is demonstrated in the following worked problem. Problem 7. A square plate is bounded by the lines x = 0, y = 0, x = 1 and y = 1. Apply the Laplace ∂2u ∂2u equation 2 + 2 = 0 to determine the potential ∂x ∂y

and

Y′′ − p 2 Y = 0

from which, X = A cos px + B sin px and or or

Y = Ce py + De−py Y = C cosh py + D sinh py Y = E sinh p(y + ϕ)

Hence

u(x, y) = XY

= {A cos px + B sin px}{E sinh p(y + ϕ)} or

u(x, y) = {P cos px + Q sin px}{sinh p(y + ϕ)}

where P = AE and Q = BE The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + ϕ) from which, P = 0 Hence, u(x, y) = Q sin px sinh p(y + ϕ) The second boundary condition is: u(1, y) = 0, hence 0 = Q sin p(1) sinh p(y + ϕ) from which, sin p = 0, hence, p = nπ for n = 1, 2, 3, . . . The third boundary condition is: u(x, 0) = 0, hence, 0 = Q sin px sinh p(ϕ) from which, sinh p(ϕ) = 0 and ϕ = 0 Hence, u(x, y) = Q sin px sinh py Since there are many solutions for integer values of n, ∞ ∑ u(x, y) = Qn sin px sinh py n=1

=

∞ ∑ n=1

Qn sin nπx sinh nπy

(a)

614 Section I The fourth boundary condition is: u(x, 1) = 4 = f (x), ∞ ∑ hence, f (x) = Qn sin nπx sinh nπ(1). n=1

From Fourier series coefficients, Qn sinh nπ = 2 × the mean value of

Now try the following Practice Exercise Practice Exercise 246 The Laplace equation (Answers on page 897) 1.

f(x) sin nπx from x = 0 to x = 1 ∫

i.e.

2 1 4 sin nπx dx 1 0 [ cos nπx ]1 =8 − nπ 0 ( ) 8 =− cos nπ − cos 0 nπ ) 8 ( = 1 − cos nπ nπ

=

= 0 (for even values of n), 16 (for odd values of n) nπ 16 16 Hence, Qn = = cosech nπ nπ(sinh nπ) nπ =

Hence, from equation (a), ∞ ∑ u(x, y) = Qn sin nπx sinh nπy n=1

=

∞ 16 ∑ 1 (cosech nπ sin nπx sinh nπy) π n

A rectangular plate is bounded by the lines x = 0, y = 0, x = 1 and y = 3. Apply the Laplace ∂2u ∂2u + = 0 to determine the equation ∂x2 ∂y2 potential distribution u(x, y) over the plate, subject to the following boundary conditions: u = 0 when x = 0 0 ≤ y ≤ 2 u = 0 when x = 1 0 ≤ y ≤ 2 u = 0 when y = 2 0 ≤ x ≤ 1 u = 5 when y = 3 0 ≤ x ≤ 1

2.

A rectangular plate is bounded by the lines x = 0, y = 0, x = 3, y = 2. Determine the potential distribution u(x, y) over the rectangle using the Laplace equation ∂2u ∂2u + = 0, subject to the following ∂x2 ∂y2 boundary conditions: u(0, y) = 0 u(3, y) = 0 u(x, 2) = 0 u(x, 0) = x(3 − x)

0≤y≤2 0≤y≤2 0≤x≤3 0≤x≤3

n(odd)=1

For fully worked solutions to each of the problems in Practice Exercises 242 to 246 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 15

Second-order differential equations, power series methods and partial differential equations

This Revision Test covers the material contained in Chapters 50 to 53. The marks for each question are shown in brackets at the end of each question. 1.

Find the particular solution of the following differential equations: d2 y (a) 12 2 − 3y = 0 given that when t = 0, y = 3 and dt dy 1 = dt 2 dy d2 y + 2 + 2y = 10ex given that when x = 0, (b) dx2 dx dy y = 0 and =1 (20) dx

2.

In a galvanometer the deflection θ satisfies the differential equation:

6.

x2

3.

Determine y(n) when y = 2x3 e4x

4.

Determine the power series solution of the difd2 y dy ferential equation: + 2x + y = 0 using the dx2 dx Leibniz–Maclaurin method, given the boundary dy =1 (20) conditions that at x = 0, y = 2 and dx

5.

(10)

Use the Frobenius method to determine the general power series solution of the differential d2 y (21) equation: 2 + 4y = 0 dx

dy d2 y + x + (x2 − v2 )y = 0 dx2 dx

and hence state the series up to and including the term in x6 when v = +3 (26) 7.

8.

d2 θ dθ +2 +θ = 4 dt2 dt Solve the equation for θ given that when t = 0, dθ θ = 0 and =0 (12) dt

Determine the general power series solution of Bessel’s equation:

9.

Determine the general solution of

∂u = 5xy ∂x

(2)

∂2u Solve the differential equation = x2 (y − 3) ∂x2 given the boundary conditions that at x = 0, ∂u = sin y and u = cos y (6) ∂x Figure RT15.1 shows a stretched string of length 40 cm which is set oscillating by displacing its mid-point a distance of 1 cm from its rest position and releasing it with zero velocity. Solve ∂2u 1 ∂2u the wave equation: 2 = 2 2 where c2 = 1, to ∂x c ∂t determine the resulting motion u(x, t)

(23)

u (x,0) 1 0

20

Figure RT15.1

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 15, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

40 x (cm)

Section J

Laplace transforms

Chapter 54

Introduction to Laplace transforms Why it is important to understand: Introduction to Laplace transforms The Laplace transform is a very powerful mathematical tool applied in various areas of engineering and science. With the increasing complexity of engineering problems, Laplace transforms help in solving complex problems with a very simple approach; the transform is an integral transform method which is particularly useful in solving linear ordinary differential equations. It has very wide applications in various areas of physics, electrical engineering, control engineering, optics, mathematics and signal processing. This chapter just gets us started in understanding some standard Laplace transforms.

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

define a Laplace transform recognise common notations used for the Laplace transform derive Laplace transforms of elementary functions use a standard list of Laplace transforms to determine the transform of common functions

54.1

Introduction

The solution of most electrical circuit problems can be reduced ultimately to the solution of differential equations. The use of Laplace∗ transforms provides an alternative method to those discussed in Chapters 46 to 51 for solving linear differential equations.

54.2

Definition of a Laplace transform

The Laplace transform of the function f (t) is defined ∫∞ by the integral 0 e−st f (t) dt, where s is a parameter assumed to be a real number.

Common notations used for the Laplace transform There are various commonly used notations for the Laplace transform of f (t) and these include: (i)

∗

Who was Laplace? See page 605 for image and resume of Pierre-Simon, marquis de Laplace. To find out more go to www.routledge.com/cw/bird

L{ f (t)} or L{ f (t)}

(ii) L( f) or L f (iii) f (s) or f(s)

620 Section J Also, the letter p is sometimes used instead of s as the parameter. The notation adopted in this book will be f(t) for the original function and L{ f (t)} for its Laplace transform. Hence, from above: ∫

∞

L{ f(t)} =

e

−st

f(t) d t

(1)

0

(b) f (t) = k. From equation (2), L{k} = kL{1} ( ) k 1 Hence L{k} = k = , from (a) above. s s (c) f (t) = eat (where a is a real constant ̸= 0). From equation (1), ∫ ∞ ∫ ∞ at −st at L{e } = e (e ) dt = e−(s−a)t dt,

54.3 Linearity property of the Laplace transform

0

[

e−(s−a)t = −(s − a)

From equation (1), ∫ L{kf (t)} =

∞

e−st k f (t) dt

=

0

∫

∞

=k

e−st f(t) dt

0

i.e.

L{k f (t)} = kL{f (t)}

(2)

where k is any constant. Similarly,

0

]∞

from the laws of indices, =

0

1 (0 − 1) −(s − a)

1 s−a (provided (s − a) > 0, i.e. s > a)

(d) f (t) = cos at (where a is a real constant). From equation (1), ∫ ∞ L{cos at} = e−st cos at dt 0

∫

∞

L{a f (t) + bg(t)} =

[

e−st = 2 (a sin at − s cos at) s + a2

e−st (a f (t) + bg(t)) dt

0

∫

=a

∞

e−st f(t) dt

0

∫

∞

+b

e−st g(t) dt

0

i.e. L{af (t) + bg(t)} = aL{ f(t)} + bL{g(t)},

(3)

where a and b are any real constants. The Laplace transform is termed a linear operator because of the properties shown in equations (2) and (3).

54.4 Laplace transforms of elementary functions Using the definition of the Laplace transform in equation (1) a number of elementary functions may be transformed. For example: (a) f(t) = 1. From equation (1), [ −st ]∞ ∫ ∞ e −st L{1} = e (1) dt = −s 0 0 1 1 = − [e−s(∞) − e0 ] = − [0 − 1] s s 1 = (provided s > 0) s

]∞ 0

by integration by parts twice (see page 494), [ −s(∞) e = 2 (a sin a(∞) − s cos a(∞)) s + a2 ] e0 − 2 (a sin 0 − s cos 0) s + a2 s = 2 ( provided s > 0) s + a2 (e) f (t) = t. From equation (1), [ −st ∫ −st ]∞ ∫ ∞ te e −st L{t} = e t dt = − dt −s −s 0 0 [

te−st e−st = − 2 −s s

]∞ 0

by integration by parts ] [ ] ∞e e e0 = − − 0 − −s s2 s2 ( ) 1 = (0 − 0) − 0 − 2 s [

−s(∞)

−s(∞)

since (∞ × 0) = 0 =

1 (provided s > 0) s2

Introduction to Laplace transforms 621 (f)

f(t) = tn (where n = 0, 1, 2, 3, …).

Table 54.1 Elementary standard Laplace transforms Function f (t)

By a similar method to (e) it may be shown 2 (3)(2) 3! that L{t2 } = 3 and L{t3 } = = 4 . These s s4 s results can be extended to n being any positive integer. n! Thus L{tn } = n+1 provided s > 0) s (g)

(i)

Laplace transforms ∫∞ L{ f (t)} = 0 e−st f (t) dt 1 s

1

k s

(ii) k

f(t) = sinh at.

1 From Chapter 12, sinh at = (eat − e−at ). 2 Hence, { } 1 at 1 −at L{sinh at} = L e − e 2 2 1 1 = L{eat } − L{e−at } 2 2 from equations (2) and (3) [ ] [ ] 1 1 1 1 = − 2 s−a 2 s+a [

]

from (c) above

1 s−a a 2 s + a2 s 2 s + a2

(iii) eat (iv) sin at (v) cos at

1 s2

(vi) t

(viii) tn (n = 1, 2, 3, . . .)

1 1 1 − = 2 s−a s+a a = 2 (provided s > a) s − a2

(x) sinh at

1 2 1 = + 2− s s 3

54.5 Worked problems on standard Laplace transforms

{

(a)

1 L 1 + 2t − t4 3

}

1 = L{1} + 2L{t} − L{t4 } 3 from equations (2) and (3) ( ) ( ) 1 1 1 4! = +2 2 − s s 3 s4+1 from (i), (vi) and (viii) of Table 54.1

sn+1 s s2 − a2 a s2 − a2

(ix) cosh at

A list of elementary standard Laplace transforms are summarised in Table 54.1.

Problem 1. Using a standard list of Laplace transforms, determine { } the following: 14 (a) L 1 + 2t − t (b) L{5e2t − 3e−t } 3

2! s3 n!

(vii) t2

= (b)

(

4.3.2.1 s5

)

1 2 8 + 2− 5 s s s

L{5e2t − 3e−t } = 5L(e2t ) − 3L{e−t } from equations (2) and (3) ( ) ( ) 1 1 −3 =5 s−2 s − (−1) from (iii) of Table 54.1 =

5 3 − s−2 s+1

=

5(s + 1) − 3(s − 2) (s − 2)(s + 1)

=

2s + 11 s2 − s − 2

622 Section J Problem 2. Find the Laplace transforms of: (a) 6 sin 3t − 4 cos 5t (b) 2 cosh 2θ − sinh 3θ (a)

(b) From equation (1), ∫

L{6 sin 3t − 4 cos 5t}

[ =

= 6L{sin 3t} − 4L{cos 5t} ) ( ) ( s 3 − 4 =6 2 s + 32 s2 + 52 from (iv) and (v) of Table 54.1 =

(b)

s2

18 4s − + 9 s2 + 25

∞

L{t } = 2

e−st t2 dt

0

t2 e−st 2te−st 2e−st − 2 − 3 −s s s

by integration by parts twice )] ( 2 = (0 − 0 − 0) − 0 − 0 − 3 s 2 (provided s > 0) s3

(c) From equation (1), { L{cosh at} = L

= 2L{cosh 2θ} − L{sinh 3θ} ) ( ) ( 3 s − =2 2 s − 22 s2 − 3 2 from (ix) and (x) of Table 54.1

Problem 3.

L{sin at} =

e [

=

−st

1 2

(

1 s−a

)

equations (2) and (3) ( ) 1 1 + 2 s − (−a)

from (iii) of Table 54.1 ] 1 1 1 = + 2 s−a s+a [ ] 1 (s + a) + (s − a) = 2 (s − a)(s + a) [

(a) From equation (1), ∞

}

1 1 = L{eat } + L{e−at } from 2 2

= Prove that 2 a (b) L{t2 } = 3 (a) L{sin at} = 2 s + a2 s s (c) L{cosh at} = 2 s − a2

1 at (e + e−at ) 2

from Chapter 12

2s 3 − s2 − 4 s2 − 9

∫

0

[

=

L{2 cosh 2θ − sinh 3θ}

=

]∞

= sin at dt

0

e−st (−s sin at − a cos at) 2 s + a2

s s2 − a2

(provided s > a)

]∞ 0

Problem 4. Determine the Laplace transforms of: (a) sin2 t (b) cosh2 3x

by integration by parts twice (a) Since from Chapter 19, cos 2t = 1 − 2sin2 t then 1 sin2 t = (1 − cos2t). 2 Hence,

1 [e−s(∞) (−s sin a(∞) = 2 s + a2 − a cos a(∞)) − e0 (−s sin 0 −a cos 0)] 1 = 2 [(0) − 1(0 − a)] s + a2 =

s2

a (provided s > 0) + a2

{ L{sin2 t} = L

1 (1 − cos 2t) 2

}

1 1 = L{1} − L{cos 2t} 2 2

Introduction to Laplace transforms 623 =

1 2

( ) ( ) 1 1 s − s 2 s2 + 22 from (i) and (v) of Table 54.1

= =

(s2 + 4) − s2 4 = 2s(s2 + 4) 2s(s2 + 4) 2 s(s + 4) 2

(b) Since cosh 2x = 2 cosh2 x − 1 then 1 cosh2 x = (1 + cosh 2x) from Chapter 12. 2 1 Hence cosh2 3x = (1 + cosh 6x) 2 { } 1 Thus L{cosh2 3x} = L (1 + cosh 6x) 2 1 1 = L{1} + L{cosh 6x} 2 2 ( ) ( ) 1 1 1 s = + 2 s 2 s2 − 62 from (i) and (ix) of Table 54.1 2s2 − 36 s2 − 18 = = 2s(s2 − 36) s(s2 − 36)

Now try the following Practice Exercise Practice Exercise 247 Introduction to Laplace transforms (Answers on page 897) Determine the Laplace transforms in Problems 1 to 9. 1. (a) 2t − 3 (b) 5t2 + 4t − 3 2. (a)

t3 t5 t2 − 3t + 2 (b) − 2t4 + 24 15 2

3. (a) 5e3t (b) 2e−2t 4. (a) 4 sin 3t (b) 3 cos 2t 5. (a) 7 cosh 2x (b)

1 sinh 3t 3

6. (a) 2 cos2 t (b) 3 sin2 2x 7. (a) cosh2 t (b) 2 sinh2 2θ 8. 4 sin(at + b), where a and b are constants. 9. 3 cos(ωt − α), where ω and α are constants. 10.

Show that L(cos2 3t − sin2 3t) =

s s2 + 36

Problem 5. Find the Laplace transform of 3 sin(ωt + α), where ω and α are constants. Using the compound angle formula for sin(A + B), from Chapter 15, sin(ωt + α) may be expanded to (sin ωt cos α + cos ωt sin α). Hence,

Practice Exercise 248 Multiple-choice questions on the introduction to Laplace transforms (Answers on page 897) Each question has only one correct answer

L{3sin (ωt + α)} = L{3(sin ωt cos α + cos ωt sin α)} = 3 cos αL{sin ωt} + 3 sin αL{cos ωt}, since α is a constant ( ) ( ) ω s = 3 cos α 2 + 3 sin α s + ω2 s2 + ω 2

1.

L{2e −3t } is equal to: (a) 2e − 3t

2.

2 s−3

(c) 2e 3t

L{5 sin 2t} is equal to: (a)

from (iv) and (v) of Table 54.1 3 = 2 (ω cos α + s sin α) (s + ω 2 )

(b)

(c)

10 +4

(b)

2 s2 + 25

(d)

s2

s2

5 +4

5 s2 − 4

(d)

2 s+3

624 Section J {

3. L{3 cosh 2t} is equal to: 3 6 (b) 2 (a) 2 s +4 s −4 3s 3s (c) 2 (d) 2 s −4 s +9 4. L{4t4 } is equal to: 96 24 24 (a) 3 (b) 5 (c) 5 s s s

6.

7. 16 (d) 5 s

} 16 L t is equal to: 5 48 144 24 (a) 7 (b) 5 (c) 7 s s s { } 8 L−1 is equal to: 2s + 1 1

(a) 4e − 2 t

1

(b) 8e 2 t

(d)

1

(c) 4e 2 t

144 s7

(d) 4e 2t

5. L{3 cos 4t} is equal to: 3s 3 (b) 2 (a) 2 s + 16 s + 16 12 3s (c) 2 (d) 2 s +4 s − 16

For fully worked solutions to each of the problems in Practice Exercise 247 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 55

Properties of Laplace transforms Why it is important to understand: Properties of Laplace transforms As stated in the preceding chapter, the Laplace transform is a widely used integral transform with many applications in engineering, where it is used for analysis of linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices and mechanical systems. The Laplace transform is also a valuable tool in solving differential equations, such as in electronic circuits, and in feedback control systems, such as in stability and control of aircraft systems. This chapter considers further transforms together with the Laplace transform of derivatives that are needed when solving differential equations.

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

derive the Laplace transform of eat f (t) use a standard list of Laplace transforms to determine transforms of the form eat f (t) derive the Laplace transforms of derivatives state and use the initial and final value theorems

55.1

The Laplace transform of eat f(t)

From Chapter 54, the definition of the Laplace transform of f (t) is: ∫

∞

L{f(t)} =

e−st f (t) dt

(1)

0

∫

∞

Thus L{eat f(t)} =

e−st (eat f(t)) dt

∞

=

e(s−a)t f(t) dt

55.2 Laplace transforms of the form eat f(t) From equation (2), Laplace transforms of the form eat f (t) may be deduced. For example:

0

∫

Hence the substitution of (s − a) for s in the transform shown in equation (1) corresponds to the multiplication of the original function f (t) by eat . This is known as a shift theorem.

(2)

0

(where a is a real constant)

(i)

L{eat tn } n! Since L{tn } = n+1 from (viii) of Table 54.1, s page 621.

626 Section J n! from equation (2) (s − a)n+1 above (provided s > a) then L{eat tn } =

(ii)

(b)

L{4e3t cos 5t} = 4L{e3t cos 5t} ) ( s−3 =4 (s − 3)2 + 52

L{eat sin ωt} Since L{sin ωt} = 54.1, page 621.

From (iii) of Table 55.1,

ω from (iv) of Table s2 + ω 2

ω from equa(s − a)2 + ω 2 tion (2) (provided s > a)

=

4(s − 3) s2 − 6s + 9 + 25

=

4(s − 3) s2 − 6s + 34

then L{eat sin ωt} =

(iii)

L{eat cosh ωt} s Since L{cosh ωt} = 2 from (ix) of Table s − ω2 54.1, page 621. s−a then L{e cosh ωt} = from equa(s − a)2 − ω 2 tion (2) (provided s > a)

Problem 2. Determine (a) L{e−2t sin 3t} (b) L{3eθ cosh 4θ} (a) From (ii) of Table 55.1,

at

L{e−2t sin 3t} =

A summary of Laplace transforms of the form eat f (t) is shown in Table 55.1. Problem 1. Determine (a) L{2t4 e3t } (b) L{4e3t cos 5t}

=

(b)

3 3 = 2 + 4s + 4 + 9 s + 4s + 13

From (v) of Table 55.1,

(

4! L{2t e } = 2L{t e } = 2 (s − 3)4+1

)

=

4 3t

=

s2

L{3eθ cosh 4θ} = 3L{eθ cosh 4θ} =

(a) From (i) of Table 55.1, 4 3t

3 3 = 2 2 (s − (−2)) + 3 (s + 2)2 + 9

2(4)(3)(2) 48 = (s − 3)5 (s − 3)5

s2

3(s − 1) (s − 1)2 − 42

3(s − 1) 3(s − 1) = 2 − 2s + 1 − 16 s − 2s − 15

Problem 3. Determine the Laplace transforms of (a) 5e−3t sinh 2t (b) 2e3t (4 cos 2t − 5 sin 2t) (a) From (iv) of Table 55.1,

Table 55.1 Laplace transforms of the form eat f(t) Function eat f (t) (a is a real constant) (i) eat tn (ii) eat sin ωt (iii) eat cos ωt (iv) eat sinh ωt (v)

eat cosh ωt

L{5e−3t sinh 2t} = 5L{e−3t sinh 2t} ( ) 2 =5 (s − (−3))2 − 22

Laplace transform L{eat f(t)} n! (s − a)n+1 ω (s − a)2 + ω 2 s−a (s − a)2 + ω 2 ω (s − a)2 − ω 2 s−a (s − a)2 − ω 2

(b)

=

10 10 = 2 (s + 3)2 − 22 s + 6s+9 − 4

=

10 s2 + 6s + 5

L{2e3t (4 cos 2t − 5 sin 2t)} = 8L{e3t cos 2t} − 10L{e3t sin 2t}

Properties of Laplace transforms 627 =

8(s − 3) 10(2) − 2 2 (s − 3) + 2 (s − 3)2 + 22 from (iii) and (ii) of Table 55.1

=

8s − 44 8(s − 3) − 10(2) = 2 2 2 (s − 3) + 2 s − 6s + 13

Problem 4. Show that { } 1 48 −2x 2 L 3e sin x = (2s + 1)(4s2 + 4s + 17) 1 Since cos 2x = 1 − 2 sin x, sin x = (1 − cos 2x) 2 Hence, { } 1 L 3e− 2 x sin2 x { } 1 1 = L 3e− 2 x (1 − cos 2x) 2 { } { } 1 1 3 3 −2x −2x = L e − L e cos 2x 2 2  ( ( ))    1 s− −   3 1 3 2  −  ( ) =    ( ( )) 2 1  2 2  1 2 s− − s− − +2 2 2 2

Now try the following Practice Exercise Practice Exercise 249 Laplace transforms of the form eat f(t) (Answers on page 897) Determine the Laplace transforms of the following functions: 1.

(a) 2te2t

2.

(a) 4t3 e−2t

3.

(a) et cos t (b) 3e2t sin 2t

4.

(a) 5e−2t cos 3t (b) 4e−5t sin t

5.

(a) 2et sin2 t

6.

(a) et sinh t (b) 3e2t cosh 4t

7.

(a) 2e−t sinh 3t

8.

(a) 2et (cos 3t − 3 sin 3t)

2

from (iii) of Table 54.1 (page 621) and (iii) of Table 55.1 above, ( ) 1 3 s+ 3 2 ) − [( ] = ( )2 1 1 2 s+ 2 2 s+ +2 2 2 3 6s + 3 ) = − ( 1 2s + 1 2 4 s +s+ +4 4 3 6s + 3 = − 2s + 1 4s2 + 4s + 17 =

3(4s2 + 4s + 17) − (6s + 3)(2s + 1) (2s + 1)(4s2 + 4s + 17)

=

12s2 + 12s + 51 − 12s2 − 6s − 6s − 3 (2s + 1)(4s2 + 4s + 17)

=

48 (2s + 1)(4s2 + 4s + 17)

(b) t2 et 1 (b) t4 e−3t 2

1 (b) e3t cos2 t 2 1 (b) e−3t cosh 2t 4

(b) 3e−2t (sinh 2t − 2 cosh 2t)

55.3 The Laplace transforms of derivatives (a) First derivative Let the first derivative of f (t) be f ′ (t) then, from equation (1), L{f ′ (t)} =

∫

∞

e−st f ′ (t) dt

0

From Chapter 42, when integrating by parts ∫ ∫ dv du u dt = uv − v dt dt dt When evaluating

∫∞ 0

let u = e−st and

e−st f ′ (t) dt,

dv = f ′ (t) dt

from which, du = −se−st and v = dt

∫

f ′ (t) dt = f (t)

628 Section J ∫

∞

Hence 0

e−st f ′ (t) dt

[ ]∞ = e−st f(t) 0 −

∫

∫ = [0 − f (0)] + s

∞

0 ∞

(c) L{e−at } = f (t)(−se−st ) dt

From equation (3), L{ f ′ (t)} = sL{f (t)} − f(0) −st

e

f(t) dt

(a) Let f (t) = k, then f ′ (t) = 0 and f(0) = k

0

= −f(0) + sL{f (t)}

Substituting into equation (3) gives:

−st

assuming e f (t) → 0 as t → ∞, and f (0) is the value of f(t) at t = 0. Hence,  L{ f ′ (t)} = sL{ f (t)} − f(0)  { } (3) dy or L = sL{ y} − y(0)  dx

L{0}= sL{k} − k k = sL{k} k Hence L{k} = s ′ Let f (t) = 2t then f (t) = 2 and f(0) = 0 i.e.

(b)

Substituting into equation (3) gives:

where y(0) is the value of y at x = 0

(b) Second derivative Let the second derivative of f(t) be f ′′ (t), then from equation (1), ∫ ∞ ′′ L{ f (t)} = e−st f ′′ (t) dt 0

Integrating by parts gives: ∫ ∞ ∫ [ ]∞ e−st f ′′ (t) dt = e−st f ′ (t) 0 + s 0

e−st f ′ (t) dt

= [0 − f ′ (0)] + sL{ f ′ (t)} assuming e−st f ′ (t) → 0 as t → ∞, and f ′ (0) is the value of f ′ (t) at t = 0. Hence {f ′′ (t)} = −f ′ (0) + s[s(f (t)) − f (0)], from equation (3), i.e.  L{ f ′′ (t)}     = s2 L{ f (t)} − sf (0) − f ′ (0)    { 2 } (4) d y  or L  2  dx     2 ′ = s L{ y} − sy(0) − y (0) dy at x = 0 dx

Equations (3) and (4) are important and are used in the solution of differential equations (see Chapter 58) and simultaneous differential equations (Chapter 59). Problem 5. Use the Laplace transform of the first derivative to derive: (a) L{k} =

k s

(b) L{2t} =

2 s2

i.e. Hence

L{2} = sL{2t} − 0 2 = sL{2t} s 2 L{2t} = 2 s

(c) Let f (t) = e−at then f ′ (t) = −ae−at and f (0) = 1 Substituting into equation (3) gives:

∞

0

where y′ (0) is the value of

1 s+a

L{−ae−at } = sL{e−at } − 1 −aL{e−at } = sL{e−at } − 1 1 = sL{e−at } + aL{e−at } 1 = (s + a)L{e−at } Hence L{e−at } =

1 s+a

Problem 6. Use the Laplace transform of the second derivative to derive s L{cos at} = 2 s + a2 From equation (4), L{ f ′′ (t)} = s2 L{ f(t)} − sf(0) − f ′ (0) Let f(t) = cos at, then f ′ (t) = −a sin at and f ′′ (t) = −a2 cos at, f (0) = 1 and f ′ (0) = 0 Substituting into equation (4) gives:

Properties of Laplace transforms 629 L{−a2 cos at} = s2 {cos at} − s(1) − 0

For example, if f (t) = 3e4t then

−a2 L{cos at} = s2 L{cos at} − s

i.e.

L{3e4t } =

s = (s2 + a2 )L{cos at}

Hence

from which, L{cos at} =

s s2 + a2

from (iii) of Table 54.1, page 621. By the initial value theorem, [ ( )] 3 limit[3e4t ] = limit s s→∞ t→0 s−4 ( ) 3 i.e. 3e0 = ∞ ∞−4

Now try the following Practice Exercise Practice Exercise 250 Laplace transforms of derivatives (Answers on page 898) 1. Derive the Laplace transform of the first derivative from the definition of a Laplace transform. Hence derive the transform L{1} =

1 s

2. Use the Laplace transform of the first derivative to derive the transforms: 1 6 (a) L{eat } = (b) L{3t2 } = 3 s−a s 3. Derive the Laplace transform of the second derivative from the definition of a Laplace transform. Hence derive the transform a L{sin at} = 2 s + a2 4. Use the Laplace transform of the second derivative to derive the transforms: a (a) L{sinh at} = 2 s − a2 s (b) L{cosh at} = 2 s − a2

i.e.

55.4 The initial and final value theorems There are several Laplace transform theorems used to simplify and interpret the solution of certain problems. Two such theorems are the initial value theorem and the final value theorem.

(a) The initial value theorem states: limit [ f (t)] = limit [sL{ f(t)}] s→∞

3 = 3, which illustrates the theorem.

Problem 7. Verify the initial value theorem for the voltage function (5 + 2 cos 3t) volts, and state its initial value. Let

f (t) = 5 + 2 cos 3t

L{f(t)} = L{5 + 2 cos 3t} =

2s 5 + s s2 + 9

from (ii) and (v) of Table 54.1, page 621. By the initial value theorem, limit[ f (t)] = limit[sL{ f (t)}] t→0

s→∞

[ ( )] 5 2s i.e. limit[5 + 2 cos 3t] = limit s + s→∞ t→0 s s2 + 9 [ ] 2s2 = limit 5 + 2 s→∞ s +9 2∞2 = 5+2 ∞2 + 9 i.e. 7 = 7, which verifies the theorem in this case.

i.e.

t→0

3 s−4

5 + 2(1) = 5 +

The initial value of the voltage is thus 7 V Problem 8. Verify the initial value theorem for the function (2t − 3)2 and state its initial value. Let Let

f (t) = (2t − 3)2 = 4t2 − 12t + 9 L{ f (t)} = L(4t2 − 12t + 9) ( ) 2 12 9 =4 3 − 2 + s s s

from (vii), (vi) and (ii) of Table 54.1, page 621.

630 Section J By the initial value theorem,

[ ( )] 8 12 9 2 limit[(2t − 3) ] = limit s 3 − 2 + s→∞ t→0 s s s ] [ 8 12 +9 = limit 2 − s→∞ s s 8 12 i.e. (0 − 3)2 = 2 − +9 ∞ ∞

i.e. 9 = 9, which verifies the theorem in this case. The initial value of the given function is thus 9

(b) The final value theorem states: limit [ f (t)] = limit [sL{ f(t)}]

t→∞

s→0

For example, if f (t) = 3e−4t then: [ ( )] 3 limit[3e−4t ] = limit s t→∞ s→0 s)+ 4 ( 3 i.e. 3e−∞ = (0) 0+4

i.e. limit[2 + 3e−2t sin 4t] t→∞ )] [ ( 12 2 + = limit s s→0 s (s + 2)2 + 16 [ ] 12s = limit 2 + s→0 (s + 2)2 + 16 i.e. 2 + 0 = 2 + 0 i.e. 2 = 2, which verifies the theorem in this case. The final value of the displacement is thus 2 cm. The initial and final value theorems are used in pulse circuit applications where the response of the circuit for small periods of time, or the behaviour immediately after the switch is closed, are of interest. The final value theorem is particularly useful in investigating the stability of systems (such as in automatic aircraft-landing systems) and is concerned with the steady state response for large values of time t, i.e. after all transient effects have died away. Now try the following Practice Exercise

i.e. 0 = 0, which illustrates the theorem. Problem 9. Verify the final value theorem for the function (2 + 3e−2t sin 4t) cm, which represents the displacement of a particle. State its final steady value. f (t) = 2 + 3e−2t sin 4t

Let

L{ f (t)} = L{2 + 3e−2t sin 4t} ( ) 2 4 = +3 s (s − (−2))2 + 42 2 12 = + s (s + 2)2 + 16 from (ii) of Table 54.1, page 621 and (ii) of Table 55.1 on page 626. By the final value theorem, limit[ f (t)] = limit[sL{ f (t)}] t→∞

s→0

Practice Exercise 251 Initial and final value theorems (Answers on page 898) 1.

State the initial value theorem. Verify the theorem for the functions (a) 3 − 4 sin t (b) (t − 4)2 and state their initial values.

2.

Verify the initial value theorem for the voltage functions: (a) 4 + 2 cos t (b) t − cos 3t and state their initial values.

3.

(a) State the final value theorem and state a practical application where it is of use. (b) Verify the theorem for the function 4 + e−2t (sin t + cos t) representing a displacement and state its final value.

4.

Verify the final value theorem for the function 3t2 e−4t and determine its steady state value.

Properties of Laplace transforms 631 Practice Exercise 252 Multiple-choice questions on the properties of Laplace transforms (Answers on page 898)

3.

L{3e 2t cosh 4t} is equal to: 4 3(s + 2) (a) (b) 2 (s − 3) + 4 (s + 2)2 + 16 4 3(s − 2) (c) (d) (s − 2)2 − 16 (s − 2)2 − 16

4.

L{4e − 3t cos 2t} is equal to: 4(s − 3) 4(s + 3) (a) (b) (s − 3)2 + 4 (s + 3)2 − 4 4(s − 3) 4(s + 3) (d) (c) 2 (s + 3) + 4 (s − 3)2 − 4

5.

L{5e − 4t cosh 3t} is equal to: 5(s − 4) 5(s + 4) (a) (b) 2 (s − 4) + 9 (s + 4)2 + 9 5(s + 4) 5(s − 4) (d) (c) (s − 4)2 − 9 (s + 4)2 − 9

Each question has only one correct answer 1. L{2e −3t t4 } is equal to: 24 48 (b) (a) (s + 3)5 (s − 3)5 4 48 (c) (d) 5 (s + 3) (s − 3)5 2. L{5e −2t sin 3t} is equal to: 15 2 (b) (a) 2 (s + 3) + 4 (s + 2)2 + 9 15 10 (d) (c) (s − 2)2 + 9 (s − 3)2 + 4

For fully worked solutions to each of the problems in Practice Exercises 249 to 251 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 56

Inverse Laplace transforms Why it is important to understand: Inverse Laplace transforms Laplace transforms and their inverses are a mathematical technique which allows us to solve differential equations, by primarily using algebraic methods. This simplification in the solving of equations, coupled with the ability to directly implement electrical components in their transformed form, makes the use of Laplace transforms widespread in both electrical engineering and control systems engineering. Laplace transforms have many further applications in mathematics, physics, optics, signal processing and probability. This chapter specifically explains how the inverse Laplace transform is determined, which can also involve the use of partial fractions. In addition, poles and zeros of transfer functions are briefly explained; these are of importance in stability and control systems.

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

define the inverse Laplace transform use a standard list to determine the inverse Laplace transforms of simple functions determine inverse Laplace transforms using partial fractions define a pole and a zero determine poles and zeros for transfer functions, showing them on a pole–zero diagram

56.1 Definition of the inverse Laplace transform If the Laplace transform of a function f (t) is F(s), i.e. L{ f(t)} = F(s), then f(t) is called the inverse Laplace transform of F(s) and is written as f(t) = L−1 {F(s)} { } 1 −1 1 For example, since L{1} = then L =1 s s a Similarly, since L{sin at} = 2 then s + a2 { } a −1 L = sin at, and so on. s2 + a 2

56.2 Inverse Laplace transforms of simple functions Tables of Laplace transforms, such as the tables in Chapters 54 and 55 (see pages 621 and 626) may be used to find inverse Laplace transforms. However, for convenience, a summary of inverse Laplace transforms is shown in Table 56.1. Problem 1. Find the following inverse Laplace transforms: { } { } 1 5 (a) L−1 2 (b) L−1 s +9 3s − 1

Inverse Laplace transforms 633 Table 56.1 Inverse Laplace transforms L−1 {F(s)} = f (t)

F(s) = L{ f (t)} (i)

1 s

1

(ii)

k s

k

(iii)

1 s−a

eat

(iv) (v)

s2 + a2 s

cos at

s2 + a2

t

(vii)

2! s3

t2

n! sn+1

tn (n = 1, 2, 3, . . .)

(ix)

a s2 − a2

sinh at

(x)

s s2 − a2

cosh at

(xi)

n! (s − a)n+1

eat tn

(xiii) (xiv) (xv)

   

   

5 ( )  1    3 s−  3         5 1 1 5 −1 ( ) = e3t = L  1  3 3    s−  3 = L−1

sin at

1 s2

(xii)

5 3s − 1

}

from (iii) of Table 56.1

a

(vi)

(viii)

(b) L−1

{

Problem 2. Find the following inverse Laplace transforms: { } { } 3 6 (a) L−1 3 (b) L−1 4 s s {

} 2 = t2 s3 { } { } 6 2 Hence L−1 3 = 3L−1 3 = 3t2 s s

(a) From (vii) of Table 56.1, L−1

ω

(s − a)2 + ω 2 s−a (s − a)2 + ω 2 ω

(s − a)2 − ω 2 s−a (s − a)2 − ω 2

eat sin ω t

From (viii) of Table 56.1, if s is to have a power of 4 then n = 3 { } { } 6 3! Thus L−1 4 = t3 i.e. L−1 4 = t3 s s

eat cos ω t

Hence L−1

{

3 s4

}

1 = L−1 2

{

6 s4

}

1 = t3 2

eat sinh ω t eat cosh ω t

(a) From (iv) of Table 56.1, { } a = sin at, L−1 2 s + a2 { } { } 1 1 −1 −1 Hence L =L s2 + 9 s2 + 32 { } 1 3 = L−1 2 3 s + 32 =

(b)

1 sin 3t 3

Problem 3. Determine { } { } 7s 4s (a) L−1 2 (b) L−1 2 s +4 s − 16 { −1

(a) L

} { } 7s s −1 = 7L = 7 cos 2t s2 + 4 s2 + 22 from (v) of Table 56.1

{ −1

(b) L

4s s2 − 16

}

{ −1

= 4L

s s2 − 42

}

= 4 cosh 4t from (x) of Table 56.1

634 Section J Problem 4. Find { } { } 3 2 −1 −1 (a) L (b) L s2 − 7 (s − 3)5

(b) L−1

−1

{

L

3 s2 − 7

}

{

From (xi) of Table 56.1, { } n! L−1 = eat tn (s − a)n+1 { } 1 1 −1 Thus L = eat tn (s − a)n+1 n! { } 2 −1 and comparing with L shows that (s − 3)5 n = 4 and a = 3

−1

{

(a) L

5 s2 + 2s − 3

L−1

2 (s − 3)5

{

{ (b)

−1

Problem 5. Determine { } 3 −1 (a) L s2 − 4s + 13 { −1

(b) L

(a) L−1

{

2(s + 1) 2 s + 2s + 10

3 s2 − 4s + 13

}

}

}

L

4s − 3 s2 − 4s − 5 {

=L

}

1 (s − 3)5 ( ) 1 3t 4 1 =2 e t = e3t t4 4! 12 = 2L−1

2(s + 1) (s + 1)2 + 32

}

} 5 =L (s + 1)2 − 22   5     (2) −1 2 =L 2 2   (s + 1) − 2   −1

{

from (xiv) of Table 56.1

−1

}

{

5 = e−t sinh 2t 2

Hence {

= L−1

Problem 6. Determine { } 5 −1 (a) L s2 + 2s − 3 { } 4s − 3 (b) L−1 2 s − 4s − 5

√ 3 = √ sinh 7t 7 (b)

}

from (xiii) of Table 56.1

} 1 √ = 3L s2 − ( 7)2 { } √ 7 3 −1 √ =√ L 7 s2 − ( 7)2 −1

2(s + 1) 2 s + 2s + 10

= 2e−t cos 3t

(a) From (ix) of Table 56.1, { } a −1 L = sinh at s2 − a2 Thus

{

−1

=L

{

}

{ −1

=L

4(s − 2) + 5 (s − 2)2 − 32 4(s − 2) (s − 2)2 − 32

4s − 3 (s − 2)2 − 32

}

} } {

5 +L (s − 2)2 − 32   5     (3) 2t −1 3 = 4e cosh 3t + L 2 2   (s − 2) − 3   −1

}

from (xv) of Table 56.1

= L−1

{

3 (s − 2)2 + 32

}

= e2t sin 3t from (xii) of Table 56.1

= 4e2t cosh 3t +

5 2t e sinh 3t 3 from (xiv) of Table 56.1

Inverse Laplace transforms 635 Now try the following Practice Exercise Practice Exercise 253 Inverse Laplace transforms of simple functions (Answers on page 898) Determine the inverse Laplace transforms of the following: 2 s−5

1. (a)

7 s

2. (a)

3 2s + 1

3. (a)

1 s2 + 25

4. (a)

5s 2s2 + 18

5. (a)

5 s3

6. (a)

7. (a)

(b)

(b)

(b)

2s s2 + 4

(b)

15

(b)

8. (a)

1 (s + 2)4

9. (a)

s+1 s2 + 2s + 10

(b)

2(s − 3)

10.

(a)

11.

2s + 5 (a) 2 s + 4s − 5

−1

Determine L

6 s2

s2 − 6s + 13

4s − 5 s2 − s − 2

}

4s − 5 4s − 5 A B ≡ ≡ + s2 − s − 2 (s − 2)(s + 1) (s − 2) (s + 1) ≡

(b)

3s2 − 27

2s − 3 s(s − 3)

cannot be inverted on sight from Table 56.1. How2s − 3 1 1 ever, by using partial fractions, ≡ + s(s − 3) s s−3 which may be inverted as 1 + e3t from (i) and (iii) of Table 54.1. Partial fractions are discussed in Chapter 2, and a summary of the forms of partial fractions is given in Table 2.1 on page 18. Problem 7.

8 s4

3s 1 2 s −8 2

F(s) =

{

4 s2 + 9

(b)

fractions which may be inverted on sight. For example, the function,

7 s2 − 16 4 (s − 1)3 3 (s − 3)5 (b)

(b)

3 s2 + 6s + 13

Hence 4s − 5 ≡ A(s + 1) + B(s − 2) When s = 2, 3 = 3A, from which, A = 1 When s = −1, −9 = −3B, from which, B = 3 { } 4s − 5 −1 Hence L s2 − s − 2 { } 1 3 −1 ≡L + s−2 s+1 { } { } 1 3 = L−1 + L−1 s−2 s+1 = e2t + 3e−t , from (iii) of Table 56.1

7 s2 − 8s + 12

3s + 2 (b) 2 s − 8s + 25

Problem 8. The unit step input for a closed loop transfer function for a control system is given by: T(s) =

56.3 Inverse Laplace transforms using partial fractions Sometimes the function whose inverse is required is not recognisable as a standard type, such as those listed in Table 56.1. In such cases it may be possible, by using partial fractions, to resolve the function into simpler

A(s+1) + B(s−2) (s − 2)(s + 1)

2 s2 (s − 4)

Find the time domain response { } if this response is 2 −1 given by: L s2 (s − 4) A B C + 2+ s s (s − 4) As(s − 4) + B(s − 4) + Cs2 = s2 (s − 4) from which, 2 = As(s − 4) + B(s − 4) + Cs2 Let

2

s2 (s − 4)

≡

636 Section J 1 2 1 Let s = 4, then 2 = 0 + 0 + 16C from which, C = 8 Equating s2 coefficients gives: 0 = A + C from which, 1 A=− 8 Hence,   1 1  1  { }   − − 2 8 + 2 + 8 L−1 2 = L−1  s (s − 4) s2 (s − 4)   s  Let s = 0, then

3 = 2e3t + e−t − 4e−t t + e−t t2 2

2 = 0 − 4B + 0 from which, B = −

from (iii) and (xi) of Table 56.1 { Problem 10. Determine L

{ Problem 9.

Find L

−1

3s3 + s2 + 12s + 2 (s − 3)(s + 1)3

3s3 + s2 + 12s + 2 (s − 3)(s + 1)3 A B C D + + + s − 3 s + 1 (s + 1)2 (s + 1)3 ( ) A(s + 1)3 + B(s − 3)(s + 1)2 + C(s − 3)(s + 1) + D(s − 3) ≡ (s − 3)(s + 1)3

≡

Hence 5s2 + 8s − 1 ≡ A(s2 + 1) + (Bs + C)(s + 3) When s = −3, 20 = 10A, from which, A = 2 Equating s2 terms gives: 5 = A + B, from which, B = 3, since A = 2 Equating s terms gives: 8 = 3B + C, from which, C = −1, since B = 3 { 2 } 5s + 8s − 1 Hence L−1 (s + 3)(s2 + 1) } 2 3s − 1 ≡L + s + 3 s2 + 1 { } { } 2 3s ≡ L−1 + L−1 2 s+3 s +1 −1

Hence

{

3s3 + s2 + 12s + 2 ≡ A(s + 1)3 + B(s − 3)(s + 1)2

−L

+ C(s − 3)(s + 1) + D(s − 3) When s = 3,

{ −1

Problem 11. Find L

2 = A − 3B − 3C − 3D i.e.

2 = 2 − 3 − 3C − 9

from which, 3C = −12 and C = − 4 Hence { 3 2 } −1 3s + s + 12s + 2 L (s − 3)(s + 1)3 { } 1 4 2 3 −1 ≡L + − + s − 3 s + 1 (s + 1)2 (s + 1)3

{

1 s2 + 1

}

from (iii), (v) and (iv) of Table 56.1

When s = −1, −12 = −4D, from which, D = 3

Equating constant terms gives:

−1

= 2e−3t + 3 cos t − sin t

128 = 64A, from which, A = 2

Equating s3 terms gives: 3 = A + B from which, B = 1

}

A(s2 + 1) + (Bs + C)(s + 3) (s + 3)(s2 + 1)

≡

}

5s2 + 8s − 1 (s + 3)(s2 + 1)

A Bs + C 5s2 + 8s − 1 ≡ + 2 2 (s + 3)(s + 1) s + 3 (s + 1)

1 1 1 = − − t + e 4t 8 2 8 1 1 1 i.e. the time domain response = e 4t − − t 8 8 2

−1

s(s2

7s + 13 s(s2 + 4s + 13)

}

A Bs + C 7s + 13 ≡ + 2 + 4s + 13) s s + 4s + 13 ≡

A(s2 + 4s + 13) + (Bs + C)(s) s(s2 + 4s + 13)

Hence 7s + 13 ≡ A(s2 + 4s + 13) + (Bs + C)(s). When s = 0, 13 = 13A, from which, A = 1 Equating s2 terms gives: 0 = A + B, from which, B = −1 Equating s terms gives: 7 = 4A + C, from which, C = 3

Inverse Laplace transforms 637 } 7s + 13 s(s2 + 4s + 13) { } −s + 3 −1 1 ≡L + s s2 + 4s + 13 { } { } −s + 3 1 + L−1 ≡ L−1 s (s + 2)2 + 32 { } { } −1 1 −1 −(s + 2) + 5 ≡L +L s (s + 2)2 + 32 { } { } s+2 −1 1 −1 ≡L −L s (s + 2)2 + 32 } { 5 + L−1 (s + 2)2 + 32 {

Hence L−1

5 ≡ 1 − e−2t cos 3t + e−2t sin 3t 3 from (i), (xiii) and (xii) of Table 56.1

Now try the following Practice Exercise Practice Exercise 254 Inverse Laplace transforms using partial fractions (Answers on page 898) Use partial fractions to find the inverse Laplace transforms of the following functions: 1.

11 − 3s s2 + 2s − 3

2.

2s2 − 9s − 35 (s + 1)(s − 2)(s + 3)

3.

5s2 − 2s − 19 (s + 3)(s − 1)2

4.

3s2 + 16s + 15 (s + 3)3

5.

7s2 + 5s + 13 (s2 + 2)(s + 1)

6.

3 + 6s + 4s2 − 2s3 s2 (s2 + 3)

7.

26 − s2 s(s2 + 4s + 13)

56.4

Poles and zeros

It was seen in the previous section that Laplace transϕ(s) . This is forms, in general, have the form f(s) = θ(s) the same form as most transfer functions for engineering systems, a transfer function being one that relates the response at a given pair of terminals to a source or stimulus at another pair of terminals. Let a function in the s domain be given by: ϕ(s) where ϕ(s) is of less degree f (s) = (s − a)(s − b)(s − c) than the denominator. Poles: The values a, b, c, … that makes the denominator zero, and hence f (s) infinite, are called the system poles of f (s). If there are no repeated factors, the poles are simple poles. If there are repeated factors, the poles are multiple poles. Zeros: Values of s that make the numerator ϕ(s) zero, and hence f(s) zero, are called the system zeros of f(s). s−4 has simple poles at s = −1 (s + 1)(s − 2) and s = +2, and a zero at s = 4. s+3 5 has a simple pole at s = − and double (s + 1)2 (2s + 5) 2 poles at s = −1, and a zero at s = −3. s+2 has simple poles at s = 0, +1, −4, s(s − 1)(s + 4)(2s + 1) 1 and − and a zero at s = −2 2 For example:

Pole–zero diagram The poles and zeros of a function are values of complex frequency s and can therefore be plotted on the complex frequency or s-plane. The resulting plot is the pole–zero diagram or pole–zero map. On the rectangular axes, the real part is labelled the σ-axis and the imaginary part the jω-axis. The location of a pole in the s-plane is denoted by a cross (×) and the location of a zero by a small circle (o). This is demonstrated in the following examples. From the pole–zero diagram it may be determined that the magnitude of the transfer function will be larger when it is closer to the poles and smaller when it is close to the zeros. This is important in understanding

638 Section J what the system does at various frequencies and is crucial in the study of stability and control theory in general. Problem 12. function: R(s) =

For the numerator to be zero, (s + 3) = 0 and (s − 2) = 0, hence zeros occur at s = −3 and at s = +2 Poles occur when the denominator is zero, i.e. when (s + 4) = 0, i.e. s = −4, and when s2 + 2s + 2 = 0

Determine for the transfer i.e. s =

400 (s + 10) s (s + 25)(s2 + 10s + 125)

(a) the zero and (b) the poles. Show the poles and zero on a pole–zero diagram.

=

√ √ 22 − 4(1)(2) − 2 ± −4 = 2 2

− 2 ± j2 = (−1 + j) or (−1 − j) 2

The poles and zeros are shown on the pole–zero map of F(s) in Figure 56.2.

(a) For the numerator to be zero, (s + 10) = 0 Hence, s = −10 is a zero of R(s) (b)

−2 ±

jv

For the denominator to be zero, s = 0 or s = −25 or s2 + 10s + 125 = 0

j

Using the quadratic formula, √ √ −10 ± 102 −4(1)(125) −10 ± −400 s= = 2 2 =

−10 ± j20 2

24

23

22

21

0

1

2

3

Figure 56.2

= (−5 ± j10) Hence, poles occur at s = 0, s = −25, (−5 + j10) and (−5 − j10) The pole–zero diagram is shown in Figure 56.1.

It is seen from these problems that poles and zeros are always real or complex conjugate. Now try the following Practice Exercise

jv

Practice Exercise 255 Poles and zeros (Answers on page 898)

j10

1.

225

220

215

210

25

0

2j10

Problem 13.

3.

Determine the poles and zeros

for the function: F(s) =

(s + 3)(s − 2) (s + 4)(s2 + 2s + 2)

and plot them on a pole–zero map.

Determine

for the transfer 50 (s + 4) R(s) = s (s + 2)(s2 − 8s + 25)

function:

(a) the zero and (b) the poles. Show the poles and zeros on a pole–zero diagram.

s

2.

Figure 56.1

s

2j

4.

Determine the poles and zeros for the func(s − 1)(s + 2) tion: F(s) = and plot them (s + 3)(s2 − 2s + 5) on a pole–zero map. s−1 (s + 2)(s2 + 2s + 5) determine the poles and zeros and show them on a pole–zero diagram. For the function G(s) =

Find the poles and zeros for the transfer funcs2 − 5s − 6 tion: H(s) = and plot the results in s(s2 + 4) the s-plane.

Inverse Laplace transforms 639 Practice Exercise 256 Multiple-choice questions on inverse Laplace transforms (Answers on page 898) Each question has only one correct answer { } 12 is equal to: 1. L−1 s4 1 (a) 2t3 (b) 6t4 (c) t3 (d) t3 2 { } 3 2. L−1 ( )4 is equal to: s−2 1 1 (b) e 2t t3 (a) e −3t t2 2 2 (c) 2e 2t t3 (d) 2e −2t t3

{ 3.

−1

L

6 2 s −9

(a) 2 sinh 3t

} is equal to: (b)

1 cosh 3t 2

1 sinh 9t (d) 2 cosh 9t 2 } { 6 is equal to: L−1 2s − 1 (c)

4.

1

5.

(a) 6e −t (b) 3e − 2 t (c) 6e 2t { } 8 is equal to: L−1 2 s +4 (a) 8 sin 2t (b) 8 cosh 4t (c) 4 sin 2t (d) 4 sinh 2t

1

(d) 3e 2 t

For fully worked solutions to each of the problems in Practice Exercises 253 to 255 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 57

The Laplace transform of the Heaviside function Why it is important to understand: The Laplace transform of the Heaviside function The Heaviside unit step function is used in the mathematics of control theory and signal processing to represent a signal that switches on at a specified time and stays switched on indefinitely. It is also used in structural mechanics to describe different types of structural loads. The Heaviside function has applications in engineering where periodic functions are represented. In many physical situations things change suddenly; brakes are applied, a switch is thrown, collisions occur. The Heaviside unit function is very useful for representing sudden change.

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

define the Heaviside unit step function use a standard list to determine the Laplace transform of H(t − c) use a standard list to determine the Laplace transform of H(t − c) · f (t − c) determine the inverse transforms of Heaviside functions

57.1

Heaviside unit step function

In engineering applications, functions are frequently encountered whose values change abruptly at specified values of time t. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time t. The switching process can be described mathematically by the function called the Unit Step Function – otherwise known as the Heaviside unit step function. Fig. 57.1 shows a function that maintains a zero value for all values of t up to t = c and a value of 1 for all values of t ≥ c. This is the Heaviside unit step function and is denoted by: f (t) = H(t − c) or u(t − c)

f(t) 1

f(t) = H (t – c)

0

c

t

Figure 57.1

where the c indicates the value of t at which the function changes from a value of zero to a value of unity (i.e. 1).

The Laplace transform of the Heaviside function 641 It follows that f (t) = H(t − 5) is as shown in Fig. 57.2 and f (t) = 3H(t − 4) is as shown in Fig. 57.3.

f(t )

12

f(t ) = t 2

f(t)

8

1

f(t) = H (t – 5) 4

–4 (a) 0

5

–3

–2

–1

t

0

1

2

3

4 t

f(t)

12

Figure 57.2

8 f(t)

f(t) = H(t –2).t 2

4 3

f(t) = 3 H (t – 4) –4 (b)

–3

–2

–1

0

1

2

3

4 t

Figure 57.5 0

4

t

and removed later, say at t = b. Such a situation is written using step functions as: V(t) = H(t − a) − H(t − b)

Figure 57.3

If the unit step occurs at the origin, then c = 0 and f(t) = H(t − 0), i.e. H(t) as shown in Fig. 57.4.

For example, Fig. 57.6 f (t) = H(t − 2) − H(t − 5)

shows

the

function

f(t)

f(t) 1 f(t) = H (t )

f(t) = H(t − 2) − H(t − 5)

1

0

t 0

Figure 57.4

Fig. 57.5(a) shows a graph of f (t) = t2 ; the graph shown in Fig. 57.5(b) is f (t) = H(t − 2) · t2 where for t < 2, H(t − 2)t2 = 0 and when t ≥ 2, H(t − 2) · t2 = t2 . The function H(t − 2) suppresses the function t2 for all values of t up to t = 2 and then ‘switches on’ the function t2 at t = 2. A common situation in an electrical circuit is for a voltage V to be applied at a particular time, say, t = a,

2

5

t

Figure 57.6

Representing the Heaviside unit step function is further explored in the following worked problems. Problem 1. A 12 V source is switched on at time t = 3 s. Sketch the waveform and write the function in terms of the Heaviside step function.

642 Section J f(t)

The function is shown sketched in Fig. 57.7. v(t) (volts)

5

12 0

0

3

t (s)

t

2

Figure 57.9

A function 5H(t − 2) has a maximum value of 5 and starts when t = 2, as shown in Fig. 57.9. Figure 57.7

Problem 4.

The Heaviside step function is: V(t) = 12H(t − 3) Problem 2. Write the function { 1 for 0 < t < a V(t) = 0 for t > a in terms of the Heaviside step function and sketch the waveform.

Sketch the graph of f (t) = H(t − π/3). sin t

Fig. 57.10(a) shows a graph of f (t) = sin t; the graph shown in Fig. 57.10(b) is f (t) = H(t − π/3) · sin t where the graph of sin t does not ‘switch on’ until t = π/3 f(t) 1 f(t) = sin t

The voltage has a value of 1 up until time t = a; then it is turned off. The function is shown sketched in Fig. 57.8.

0

p 2

p

3p 2

2p

t

–1 (a)

v(t)

f(t) 1

1

f(t) = H(t – p ).sin t 3

0 0

a

p p 3 2

p

3p 2

2p

t

t –1 (b)

Figure 57.8

Figure 57.10

The Heaviside step function is: V(t) = H(t) − H(t − a) Problem 3. Sketch the graph of f(t) = 5H(t − 2) A function H(t − 2) has a maximum value of 1 and starts when t = 2.

Problem 5. Sketch the graph of f (t) = 2H(t − 2π/3) · sin(t − π/6) Fig. 57.11(a) shows a graph of f (t) = 2 sin(t − π/6); the graph shown in Fig. 57.11(b) is f (t) = 2H(t − 2π/3) · sin(t −π /6) where the graph of 2 sin(t − π/6) does not ‘switch on’ until t = 2π/3

The Laplace transform of the Heaviside function 643 f(t)

f(t) f(t) = 2 sin t

2

f(t) = 2 sin (t –

0

p 6

p 2

3p 2

p

1

p ) 6 2p

t

0.5 –2 (a)

f(t ) = H(t – 1).e–t

f(t) f(t) = 2 H(t –

2

2p p ). sin (t – ) 3 6

0

1

2

3

4

t

(b) p 2p 2 3

0

(b)

3p 2

p

2p

t

f(t ) 1

–2

Figure 57.11 0.5

Problem 6. Sketch the graphs of (a) f (t) = H(t − 1).e−t (b) f (t) = [H(t − 1) − H(t − 3)].e−t

f(t ) = [H (t – 1) – H(t – 3)].e–t

Fig. 57.12(a) shows a graph of f (t) = e−t

0

(a) The graph shown in Fig. 57.12(b) is f (t) = H(t − 1) · e−t where the graph of e−t does not ‘switch on’ until t = 1. (b)

Fig. 57.12(c) shows the graph of f (t) = [H(t − 1) − H(t − 3)] · e−t where the graph of e−t does not ‘switch on’ until t = 1, but then ‘switches off’ at t = 3. f(t)

0.75

f(t ) = e–t

0.25

0 (a)

Figure 57.12

2

3

4

t

Figure 57.12 (Continued)

Now try the following Practice Exercise Practice Exercise 257 Heaviside unit step function (Answers on page 898) 1. A 6 V source is switched on at time t = 4 s. Write the function in terms of the Heaviside step function and sketch the waveform. { 2 for 0 < t < 5 2. Write the function V(t) = 0 for t > 5 in terms of the Heaviside step function and sketch the waveform.

1

0.5

1

(c)

In problems 3 to 12, sketch graphs of the given functions. 1

2

3

4

t

3. f (t) = H(t − 2) 4. f (t) = 2H(t) 5. f (t) = 4H(t − 1)

644 Section J This is demonstrated in the following worked problems. 6. f (t) = 7H(t − 5) ( π) · cos t 7. f (t) = H t − 4 ) ( ( π) π · cos t − 8. f (t) = 3H t − 2 6 ( ) 2 9. f (t) = H t − 1 · t 10. f (t) = H(t − 2) · e− 2 t

11. f (t) = [H(t − 2) − H(t − 5)] · e− 4 ( ( π) π) 12. f (t) = 5H t − · sin t + 3 4 t

57.2

Laplace transform of H(t – c)

From the definition of a Laplace transform, ∫ ∞ L{H(t − c)} = e−st H(t − c) dt

Problem 7.

From above, L{H(t − c) · f (t − c)} = e−c s F(s) where in this case, F(s) = L{4} and c = ( 5 ) −5 s 4 from (ii) of Hence, L{4H(t − 5)} = e s Table 54.1, page 621 −5 s 4e = s Problem 8.

{ However, e

∫

∞

Hence, L{H(t − c)} = ∫ =

= 0 e−st

H(t − c) =

−st

e 0

∞

e−st dt =

c

[

e−st −s

for 0 < t < c for t ≥ c

H(t − c) dt

]∞

[

] e−s(∞) e−s c − −s −s c [ ] −s c e e−sc = 0− = −s s =

When c = 0 (i.e. a unit step at the origin), e−s(0) 1 L{H(t)} = = s s 1 e−c s Summarising, L{H(t)} = and L{H(t − c)} = s s From the definition of H(t): L{1} = {1 · H(t)} L{t} = {t · H(t)} L{f(t)} = {f (t) · H(t)}

and

Problem 9.

It may be shown that: L{H(t − c) · f (t − c)} = e−cs L{f (t)} = e−cs F(s)

2 e−3s s3

Determine L{H(t − 2). sin(t − 2)}

From above, L{H(t − c) · f (t − c)} = e−c s F(s) where in this case, F(s) = L{sin t} and c = 2 ( ) 1 Hence, L{H(t − 2) · sin(t − 2)} = e−2 s 2 from s + 12 (iv) of Table 54.1, page 621 =

e−2s s2 + 1

Problem 10. Determine L{H(t − 1) · sin 4(t − 1)} From above, L{H(t − c) · f (t − c)} = e−c s F(s) where in this case, F(s) = L{sin 4t} and c = 1 ( ) 4 Hence, L{H(t − 1) · sin 4(t − 1)} = e−s 2 s + 42 from (iv) of Table 54.1, page 621

57.3 Laplace transform of H(t – c) · f(t – c)

where F(s) = L{f (t)}

Determine L{H(t − 3) · (t − 3)2 }

From above, L{H(t − c) · f (t − c)} = e−c s F(s) where in this case, F(s) = L{t2 } and c = 3 Note that F(s) is the transform of t2 and not of (t − 3)2 ( ) 2 −3 s 2! from (vii) Hence, L{H(t − 3) · (t − 3) } = e s3 of Table 54.1, page 621

0

−st

Determine L{4H(t − 5)}

= Problem 11.

4 e−s s2 + 16

Determine L{H(t − 3) · et−3 }

From above, L{H(t − c) · f (t − c)} = e−c s F(s) where in this case, F(s) = L{et } and c = 3

The Laplace transform of the Heaviside function 645 ) 1 from (iii) of s−1 Table 54.1, page 621

( Hence, L{H(t − 3) · et−3 } = e−3 s −3s

=

e s−1

Problem 12. Determine { ( ( π) π )} L H t− · cos 3 t − 2 2 From above, L{H(t − c) · f (t − c)} = e−c s F(s) where in π this case, F(s) = L{cos 3t} and c = 2 ( { ( π )} π) · cos 3 t − Hence, L H t − 2 ( )2 s π = e− 2 s 2 from (v) of Table 54.1, page 621 s + 32 π s e− 2 s = 2 s +9 Now try the following Practice Exercise Practice Exercise 258 Laplace transform of H (t – c) · f (t – c) (Answers on page 900) 1. Determine L{H(t − 1)} 2. Determine L{7H(t − 3)} 3. Determine L{H(t − 2) · (t − 2)2 } 4. Determine L{H(t − 3) · sin(t − 3)} 5. Determine L{H(t − 4) · et−4 }

Written in reverse, this becomes: if F(s) = L{f(t)}, then e−c s F(s) = L{H(t − c) · f (t − c)} This is known as the second shift theorem and is used when finding inverse Laplace transforms, as demonstrated in the following worked problems. { −1

Problem 13. Determine L

Part of the numerator corresponds to e−c s where c = 2 This indicates H(t − 2) 3 Then = F(s) = L{3} from (ii) of Table 54.1, s page 621 { } 3 e−2s −1 Hence, L = 3H(t − 2) s

Problem 14. Determine the inverse of

8. Determine L{H(t − 6) · cos 3(t − 6)} 9. Determine L{5H(t − 5) · sinh 2(t − 5)} ( ( π) π) 10. Determine L{H t − · cos 2 t − } 3 3 11. Determine L{2H(t − 3) · et−3 } 12. Determine L{3H(t − 2) · cosh(t − 2)}

57.4 Inverse Laplace transforms of Heaviside functions In the previous section it was stated that: L{H(t−c).f(t−c)} = e−cs F(s) where F(s) = L{f (t)}

e−3 s s2

The numerator corresponds to e−c s where c = 3. This indicates H(t − 3) 1 = F(s) = L{t} from (vi) of Table 54.1, s2 page 621 { −3 s } −1 e Hence L = H(t − 3) · (t − 3) s2

6. Determine L{H(t − 5) · sin 3(t − 5)} 7. Determine L{H(t − 1) · (t − 1)3 }

}

3 e− 2 s s

Problem 15. Determine L−1

{

8 e−4s s2 + 4

}

Part of the numerator corresponds to e−c s where c = 4. This indicates H(t − 4) ( ) 8 2 may be written as: 4 s2 + 4 s2 + 22 ( ) 2 Then 4 2 = F(s) = L{4 sin 2t} from (iv) of s + 22 Table 54.1, page 621 } { −4s 8 e = H(t − 4) · 4 sin 2(t − 4) Hence, L−1 2 s +4 = 4H(t − 4) · sin 2(t − 4)

646 Section J

Problem 16.

Determine L−1

{

5 s e− 2 s s2 + 9

}

Part of the numerator corresponds to e−c s where c = 2. This indicates H(t − 2) ( ) s 5s may be written as: 5 s2 + 9 s2 + 32 ) ( s = F(s) = L{5 cos 3t} from (v) of Then 5 2 s + 32 Table 54.1, page 621 { } 5 se−2s −1 Hence, L = H(t − 2) · 5 cos 3(t − 2) s2 + 9 = 5H(t − 2) · cos 3(t − 2)

Problem 17.

Determine L−1

{

} 7 e− 3 s s2 − 1

Part of the numerator corresponds to e−c s where c = 3. This indicates H(t − 3) ( ) 7 1 may be written as: 7 s2 − 1 s2 − 12 ( ) 1 Then 7 2 = F(s) = L{7 sinh t} from (ix) of s − 12 Table 54.1, page 621 { −3s } −1 7 e Hence, L = H(t − 3) · 7sinh(t − 3) s2 − 1 = 7H(t − 3) · sinh(t − 3)

Now try the following Practice Exercise Practice Exercise 259 Inverse Laplace transformer of Heaviside functions (Answers on page 901) { −9s } e 1. Determine L−1 s { −3s } 4e 2. Determine L−1 s { −2s } 2e 3. Determine L−1 s2

{ 4. Determine L−1 { −1

5. Determine L

{ 6. Determine L−1 { 7. Determine L−1 { 8. Determine L−1 { −1

9. Determine L

{ 10.

−1

Determine L

5 e−2s s2 + 1

}

3 s e−4s s2 + 16 6 e−2s s2 − 1 3 e−6s s3

}

} }

2 s e−4s s2 − 16

}

2 s e− 2 s s2 + 5 1

4 e−7s s−1

}

}

Practice Exercise 260 Multiple-choice questions on the Laplace transform of the Heaviside function (Answers on page 901) Each question has only one correct answer 1.

L{H(t − 3). sin(t − 3)} is equal to: e 3s e −3s (a) 2 (b) 2 s +3 s +9 e −3s e −3s (c) 2 (d) 2 s −3 s +1

2.

L{H(t − 2). sin 3(t − 2)} is equal to: 3e −2s e 2s (a) 2 (b) 2 s +9 s +9 3e −2s e 2s (c) 2 (d) 2 s −4 s +4 { } 7s e −3s is equal to: L−1 2 s + 16 (a) 7H(t − 3). cos(t − 3) (b) 7H(t − 3). cos 4(t − 3)

3.

(c) 7H(t + 3). cos 4(t + 3) (d) H(t − 3). cos 2(t − 3)

The Laplace transform of the Heaviside function 647 4. L{H(t − 4).e t−4 } is equal to: e 4s e − 4s (a) (b) s−1 s−1 − 4s e 4s e (d) (c) s+1 s+1

{

5.

} 6 e −4s L is equal to: s2 + 9 (a) 2H(t + 4). sin 3(t + 4) −1

(b) 2H(t + 3). sin 4(t + 3) (c) 2H(t − 4). sin 3(t − 4) (d) H(t − 4). sin 2(t − 4)

For fully worked solutions to each of the problems in Practice Exercises 257 to 259 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 58

The solution of differential equations using Laplace transforms Why it is important to understand: The solution of differential equations using Laplace transforms Laplace transforms and their inverses are a mathematical technique which allows us to solve differential equations, by primarily using algebraic methods. This simplification in the solving of equations, coupled with the ability to directly implement electrical components in their transformed form, makes the use of Laplace transforms widespread in both electrical engineering and control systems engineering. The procedures explained in previous chapters are used in this chapter which demonstrates how differential equations are solved using Laplace transforms.

At the end of this chapter, you should be able to: • understand the procedure to solve differential equations using Laplace transforms • solve differential equations using Laplace transforms

58.1

for the Laplace transforms of derivatives (i.e. equations (3) and (4) of Chapter 55) and, where necessary, using a list of standard Laplace transforms, such as Tables 54.1 and 55.1 on pages 621 and 626.

Introduction

An alternative method of solving differential equations to that used in Chapters 46 to 51 is possible by using Laplace transforms.

58.2

(i)

Procedure to solve differential equations by using Laplace transforms

Take the Laplace transform of both sides of the differential equation by applying the formulae

(ii)

Put in the given initial conditions, i.e. y(0) and y′ (0).

(iii)

Rearrange the equation to make L{y} the subject.

(iv) Determine y by using, where necessary, partial fractions, and taking the inverse of each term by using Table 56.1 on page 633.

The solution of differential equations using Laplace transforms 649 58.3

Worked problems on solving differential equations using Laplace transforms

≡

B A + 2s − 1 s + 3

≡

A(s + 3) + B(2s − 1) (2s − 1)(s + 3)

Hence 8s + 38 = A(s + 3) + B(2s − 1) Problem 1. Use Laplace transforms to solve the differential equation dy d2 y 2 2 + 5 − 3y = 0, given that when x = 0, dx dx dy y = 4 and =9 dx

1 1 When s = , 42 = 3 A, from which, A = 12 2 2 When s = −3, 14 = −7B, from which, B = −2 −1

Hence y = L

This is the same problem as Problem 1 of Chapter 50, page 562 and a comparison of methods can be made. Using the above procedure: { (i)

2

d y dx2

2L

}

{ + 5L

{

−1

{

=L

{

}

−1

=L

dy − 3L{y} = L{0} dx

8s + 38 2s2 + 5s − 3

}

12 2 − 2s − 1 s + 3 12 ( ) 2 s − 12

}

} − L−1

{

2 s+3

}

1

Hence y = 6e 2 x − 2e−3x , from (iii) of

′

2[s L{y} − sy(0) − y (0)] + 5[sL{y} 2

Table 56.1.

− y(0)] − 3L{y} = 0, from equations (3) and (4) of Chapter 55. (ii)

y(0) = 4 and y′ (0) = 9 Thus 2[s2 L{y} − 4s − 9] + 5[sL{y} − 4] − 3L{y} = 0 i.e.

2s2 L{y} − 8s − 18 + 5sL{y} − 20 − 3L{y} = 0

(iii)

Problem 2. Use Laplace transforms to solve the differential equation: d2 y dy + 6 + 13y = 0, given that when x = 0, y = 3 dx2 dx dy and =7 dx This is the same as Problem 3 of Chapter 50, page 563. Using the above procedure: {

Rearranging gives: (i)

L

(2s2 + 5s − 3)L{y} = 8s + 38

i.e.

{ (iv) y = L

Hence

8s + 38 L{y} = 2 2s + 5s − 3 −1

8s + 38 2 2s + 5s − 3

}

{ + 6L

dy dx

} + 13L{y} = L{0}

[s2 L{y} − sy(0) − y′ (0)] + 6[sL{y} − y(0)] + 13L{y} = 0,

}

8s + 38 8s + 38 ≡ 2s2 + 5s − 3 (2s − 1)(s + 3)

d2 y dx2

from equations (3) and (4) of Chapter 55. (ii)

y(0) = 3 and y′ (0) = 7 Thus

s2 L{y} − 3s − 7 + 6sL{y} − 18 + 13L{y} = 0

650 Section J { (iii)

(iv) y = L−1

Rearranging gives: (s2 + 6s + 13)L{y} = 3s + 25 L{y} =

i.e.

}

A B C 9 ≡ + 2+ s2 (s − 3) s s s−3

3s + 25 s2 + 6s + 13

≡

{

} 3s + 25 (iv) y = L s2 + 6s + 13 { } 3s + 25 = L−1 (s + 3)2 + 22 −1

{

9 s2 (s − 3)

A(s)(s − 3) + B(s − 3) + Cs2 s2 (s − 3)

Hence 9 ≡ A(s)(s − 3) + B(s − 3) + Cs2 When s = 0, 9 = −3B, from which, B = −3 When s = 3, 9 = 9C, from which, C = 1

}

3(s + 3) + 16 (s + 3)2 + 22 { } 3(s + 3) −1 =L (s + 3)2 + 22 { } 8(2) −1 +L (s + 3)2 + 22

Equating s2 terms gives: 0 = A + C, from which, A = −1, since C = 1. Hence, } { } { 1 3 1 9 = L−1 − − 2 + L−1 2 s (s − 3) s s s−3

= 3e−3x cos 2x + 8e−3x sin 2x, from (xiii)

i.e.

= L−1

= −1 − 3x + e3x , from (i), (vi) and (iii) of Table 56.1. y = e3x − 3x − 1

and (xii) of Table 56.1 Hence y = e−3x (3 cos 2x + 8 sin 2x) Problem 3. Use Laplace transforms to solve the differential equation: d2 y dy − 3 = 9, given that when x = 0, y = 0 and dx2 dx dy =0 dx This is the same problem as Problem 2 of Chapter 51, page 570. Using the procedure: { } { } d2 y dy (i) L − 3L = L{9} 2 dx dx Hence

[s2 L{y} − sy(0) − y′ (0)] − 3[sL{y} − y(0)] =

(ii)

d2 y dy − 7 + 10y = e2x + 20, given that when x = 0, dx2 dx dy 1 y = 0 and =− dx 3 Using the procedure: { } { } d2 y dy (i) L − 7L + 10L{y} = L{ e2x + 20} dx2 dx Hence [s2 L{y} − sy(0) − y′ (0)] − 7[sL{y} − y(0)] + 10L{y} = (ii)

9 s

9 s

Rearranging gives:

i.e.

9 (s2 − 3s)L{y} = s 9 9 L{y} = 2 = 2 s(s − 3s) s (s − 3)

1 3 ( ) 1 s2 L{y} − 0 − − − 7sL{y} + 0 3 + 10L{y} =

(iii)

1 20 + s−2 s

y(0) = 0 and y′ (0) = − Hence

y(0) = 0 and y′ (0) = 0 Hence s2 L{y} − 3sL{y} =

(iii)

Problem 4. Use Laplace transforms to solve the differential equation:

(s2 − 7s + 10)L{y} =

21s − 40 s(s − 2)

21s − 40 1 − s(s − 2) 3

=

3(21s − 40) − s(s − 2) 3s(s − 2)

The solution of differential equations using Laplace transforms 651 =

−s2 + 65s − 120 3s(s − 2)

−s2 + 65s − 120 3s(s − 2)(s2 − 7s + 10) [ ] 1 −s2 + 65s − 120 = 3 s(s − 2)(s − 2)(s − 5)

Hence L{y} =

[ ] 1 −s2 + 65s − 120 = 3 s(s − 5)(s − 2)2 { 2 } 1 −1 −s + 65s − 120 (iv) y = L 3 s(s − 5)(s − 2)2

constants. Use Laplace transforms to solve the equation for current i given that when t = 0, i=0 Using the procedure: { } di (i) L{Ri} + L L = L{E} dt RL{i} + L[sL{i} − i(0)] =

i.e. (ii)

i(0) = 0, hence RL{i} + LsL{i} =

(iii)

Rearranging gives:

−s + 65s − 120 s(s − 5)(s − 2)2 2

≡

≡

A B C D + + + s s − 5 s − 2 (s − 2)2 ( ) A(s − 5)(s − 2)2 + B(s)(s − 2)2 + C(s)(s − 5)(s − 2) + D(s)(s − 5)

(R + Ls)L{i} = L{i} =

i.e.

{ −1

(iv) i = L

Hence

E s(R + Ls)

≡

≡A(s − 5)(s − 2)2 + B(s)(s − 2)2

When s = 5, 180 = 45B, from which, B = 4 When s = 2, 6 = −6D, from which, D = −1 Equating s3 terms gives: 0 = A + B + C, from which, C = −10 { 2 } 1 −s + 65s − 120 Hence L−1 3 s(s − 5)(s − 2)2 { } 1 −1 6 4 10 1 = L + − − 3 s s − 5 s − 2 (s − 2)2 1 = [6 + 4 e5x − 10 e2x − x e2x ] 3 4 10 x Thus y = 2 + e5x − e2x − e2x 3 3 3 Problem 5. The current flowing in an electrical circuit is given by the differential equation Ri + L( di/dt) = E, where E, L and R are

E s

E s(R + Ls) }

−s2 + 65s − 120

When s = 0, −120 = − 20A, from which, A = 6

E s

E A B ≡ + s(R + Ls) s R + Ls

s(s − 5)(s − 2)2

+ C(s)(s − 5)(s − 2) + D(s)(s − 5)

E s

A(R + Ls) + Bs s(R + Ls)

Hence

E = A(R + Ls) + Bs

When

s = 0, E = AR,

from which,

A=

E R

When

( ) R R s=− , E = B − L L

from which,

B=−

{

EL R }

E s(R + Ls) { } E/R −EL/R = L−1 + s R + Ls { } E EL −1 =L − Rs R(R + Ls)      E (1) E   1  = L−1 −     R R R s +s  L −1

Hence L

652 Section J    

   

1 E −1 1 ) L −(  R  R s     s+ L ( ) Rt E −L Hence current i = 1−e R =

7.

8.

Now try the following Practice Exercise 9. Practice Exercise 261 Solving differential equations using Laplace transforms (Answers on page 901)

10.

1. A first-order differential equation involving current i in a series R − L circuit is given by: E di + 5i = and i = 0 at time t = 0 dt 2 Use Laplace transforms to solve for i when (a) E = 20 (b) E = 40 e−3t and (c) E = 50 sin 5t In Problems 2 to 9, use Laplace transforms to solve the given differential equations.

4.

d2 i di + 1000 + 250 000i = 0, given dt2 dt i(0) = 0 and i′ (0) = 100

5.

d2 x dx + 6 + 8x = 0, given x(0) = 4 and 2 dt dt x′ (0) = 8

6.

d2 y dy 2 − 2 + y = 3 e4x , given y(0) = − and dx2 dx 3 1 ′ y (0) = 4 3

d2 y dy − 2 + 2y = 3 e x cos 2x, given 2 dx dx y(0) = 2 and y′ (0) = 5 The free oscillations of a lightly damped elastic system are given by the equation:

where y is the displacement from the equilibrium position. If when time t = 0, y = 2 dy and = 0, determine an expression for the dt displacement. 11.

The potential difference, VC , between the plates of a capacitor C charged by a steady voltage V through a resistance R is given by dVC the differential equation: VC + CR =V dt Use Laplace transforms to solve for VC , given that at time t = 0, VC = 0

12.

Solve, using Laplace transforms, Problems 4 to 9 of Exercise 229, page 564 and Problems 1 to 6 of Exercise 230, page 566.

13.

Solve, using Laplace transforms, Problems 3 to 6 of Exercise 232, page 571, Problems 5 and 6 of Exercise 233, page 573, Problems 4 and 7 of Exercise 234, page 575 and Problems 5 and 6 of Exercise 235, page 577.

d2 y dy − 24 + 16y = 0, given y(0) = 3 and 2 dt dt y′ (0) = 3 d2 x + 100x = 0, given x(0) = 2 and dt2 x′ (0) = 0

d2 y dy + − 2y = 3 cos 3x − 11 sin 3x, given dx2 dx y(0) = 0 and y′ (0) = 6

dy d2 y +2 + 5y = 0 2 dt dt

2. 9

3.

d2 y dy − 3 − 4y = 3 sin x, given y(0) = 0 dx2 ′ dx and y (0) = 0

For fully worked solutions to each of the problems in Practice Exercise 261 in this chapter, go to the website: www.routledge.com/cw/bird

Chapter 59

The solution of simultaneous differential equations using Laplace transforms Why it is important to understand: The solution of simultaneous differential equations using Laplace transforms As stated in previous chapters, Laplace transforms have many applications in mathematics, physics, optics, electrical engineering, control engineering, signal processing and probability, and Laplace transforms and their inverses are a mathematical technique which allows us to solve differential equations, by primarily using algebraic methods. Specifically, this chapter explains the procedure for solving simultaneous differential equations; this requires all of the knowledge gained in the preceding chapters.

At the end of this chapter, you should be able to: • understand the procedure to solve simultaneous differential equations using Laplace transforms • solve simultaneous differential equations using Laplace transforms

59.1

Introduction

It is sometimes necessary to solve simultaneous differential equations. An example occurs when two electrical circuits are coupled magnetically where the equations relating the two currents i1 and i2 are typically: L1

di2 di1 +M + R1 i1 = E1 dt dt

L2

di2 di1 +M + R2 i2 = 0 dt dt

where L represents inductance, R resistance, M mutual inductance and E1 the p.d. applied to one of the circuits.

59.2 Procedure to solve simultaneous differential equations using Laplace transforms (i)

Take the Laplace transform of both sides of each simultaneous equation by applying the formulae for the Laplace transforms of derivatives (i.e. equations (3) and (4) of Chapter 55, page 628) and

654 Section J using a list of standard Laplace transforms, as in Table 54.1, page 621 and Table 55.1, page 626. ′

(ii) Put in the initial conditions, i.e. x(0), y(0), x (0), y′ (0) (iii) Solve the simultaneous equations for L{y} and L{x} by the normal algebraic method.

and equation (2′ ) becomes: 4 s−1 4 or −L{y} + sL{x} = − s−1 sL{x} − L{y} = −

(iii) 1 × equation (1′′ ) and s × equation (2′′ ) gives:

(iv) Determine y and x by using, where necessary, partial fractions, and taking the inverse of each term.

59.3 Worked problems on solving simultaneous differential equations by using Laplace transforms

sL{y} + L{x} =

(3) 4s s−1

(4)

Adding equations (3) and (4) gives: 1 4s − s s−1 (s − 1) − s(4s) = s(s − 1)

(s2 + 1)L{x} =

−4s2 + s − 1 s(s − 1)

= from which,

dy +x = 1 dt

−4s2 + s − 1 s(s − 1)(s2 + 1)

L{x} =

(5)

Using partial fractions

dx − y + 4et = 0 dt

−4s2 + s − 1 A B Cs + D ≡ + + s(s − 1)(s2 + 1) s (s − 1) (s2 + 1) ( ) A(s − 1)(s2 + 1) + Bs(s2 + 1)

given that at t = 0, x = 0 and y = 0 Using the above procedure: { } dy (i) L + L{x} = L{1} dt { } dx L − L{y} + 4L{et } = 0 dt

=

[sL{y} − y(0)] + L{x} =

s(s − 1)(s2 + 1)

Hence (2)

−4s2 + s − 1 = A(s − 1)(s2 + 1) + Bs(s2 + 1) + (Cs + D)s(s − 1)

1 s

(1′ )

from equation (3), page 628 and Table 54.1, page 621. Equation (2) becomes: 4 [sL{x} − x(0)] − L{y} = − s−1

When s = 0,

−1 = −A

hence A = 1

When s = 1,

−4 = 2B

hence B = −2

Equating s3 coefficients: 0 = A+B+C

(2′ )

hence C = 1

(since A = 1 and B = −2) Equating s2 coefficients:

x(0) = 0 and y(0) = 0 hence

−4 = −A + D − C

Equation (1′ ) becomes: 1 s

+ (Cs + D)s(s − 1)

(1)

Equation (1) becomes:

sL{y} + L{x} =

1 s

−sL{y} + s2 L{x} = −

Problem 1. Solve the following pair of simultaneous differential equations

(ii)

(2′′ )

(1′′ )

hence

D = −2

(since A = 1 and C = 1)

The solution of simultaneous differential equations using Laplace transforms 655 Equation (1) becomes: L{x} =

Thus

= (iv) Hence −1

{

x=L

−1

=L i.e.

{

−4s2 + s − 1 s(s − 1)(s2 + 1)

3[sL{x} − x(0)] − 5[sL{y} − y(0)] + 2L{x} =

1 2 s−2 − + 2 s (s − 1) (s + 1)

1 2 s−2 − + 2 s (s − 1) (s + 1)

from equation (3), page 628, and Table 54.1, page 621.

}

i.e. 3sL{x} − 3x(0) − 5sL{y}

1 2 s 2 − + 2 − 2 s (s − 1) (s + 1) (s + 1)

}

i.e. (3s + 2)L{x} − 3x(0) − 5sL{y} + 5y(0) =

From the second equation given in the question,

2[sL{y} − y(0)] − [sL{x} − x(0)] 1 s

from equation (3), page 628, and Table 54.1, page 621,

from which, dx + 4 et dt

i.e. 2sL{y} − 2y(0) − sL{x} + x(0) − L{y} = −

d (1 − 2 et + cos t − 2 sin t) + 4 et dt

= −2 e − sin t − 2 cos t + 4 e

1 s

i.e. (2s − 1)L{y} − 2y(0) − sL{x}

t

+ x(0) = −

i.e. y = 2et − sin t − 2 cos t [Alternatively, to determine equations (1′′ ) and (2′′ )]

(1′ )

− L{y} = −

dx − y + 4 et = 0 dt

t

6 s

Equation (2) becomes:

from Table 56.1, page 633

=

6 s

+ 5y(0) + 2L{x} =

x = 1 − 2et + cos t − 2 sin t,

y=

6 s

1 s

(2′ )

(ii) x(0) = 8 and y(0) = 3, hence equation (1′ ) becomes y,

return

to

(3s + 2)L{x} − 3(8) − 5sL{y} + 5(3) =

Problem 2. Solve the following pair of simultaneous differential equations

(1′′ )

1 s

(2′′ )

and equation (2′ ) becomes

dx dy − 5 + 2x = 6 dt dt dy dx 2 − − y = −1 dt dt

(2s − 1)L{y} − 2(3) − sL{x}

3

+8=−

given that at t = 0, x = 8 and y = 3 Using the above procedure: { } { } (i) 3L dx − 5L dy + 2L{x} = L{6} dt dt { } { } dy dx 2L −L − L{y} = L{−1} dt dt

6 s

(1) (2)

Rearranging equations (1′′ ) and (2′′ ) gives:  (3s + 2)L{x} − 5sL{y}       6  = +9 (1′′′ )   s  − sL{x} + (2s − 1)L{y}       1 ′′′  = − −2 (2 )  s

(A)

656 Section J (iii) s × equation (1′′′ ) and (3s + 2) × equation (2′′′ ) gives: ( ) 6 s(3s + 2)L{x} − 5s2 L{y} = s +9 (3) s

Returning to equations (A) to determine L{x} and hence x: (2s − 1) × equation (1′′′ ) and 5s × (2′′′ ) gives: (2s − 1)(3s + 2)L{x} − 5s(2s − 1)L{y} ( ) 6 = (2s − 1) +9 s

−s(3s + 2)L{x} + (3s + 2)(2s − 1)L{y} ( ) 1 = (3s + 2) − − 2 (4) s s(3s + 2)L{x} − 5s2 L{y} = 6 + 9s

i.e.

(3′ )

−s(5s)L{x} + 5s(2s − 1)L{y} ( ) 1 = 5s − − 2 s

and

−s(3s + 2)L{x} + (6s2 + s − 2)L{y} 2 = −6s − − 7 s Adding equations (3′ ) and (4′ ) gives: (s2 + s − 2)L{y} = −1 + 3s −

from which,

= 12 + 18s −

=

−s + 3s2 − 2 s

L{y} =

3s2 − s − 2 s(s2 + s − 2)

− 5s2 L{x} + 5s(2s − 1)L{y}

and

=

A B C + + s (s + 2) (s − 1)

A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2) = s(s + 2)(s − 1)

=

+ Bs(s − 1) + Cs(s + 2) When s = 0, −2 = −2A, hence A = 1

1 2 + s s+2

≡

A B C + + s (s + 2) (s − 1)

=

A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2) s(s + 2)(s − 1)

B=2

i.e. = 1 + 2e−2t

8s2 − 2s − 6 s(s + 2)(s − 1)

8s2 − 2s − 6 s(s + 2)(s − 1)

3s2 − s − 2 1 2 L{y} = 2 = + s(s + s − 2) s (s + 2) y = L−1

8s2 − 2s − 6 s(s2 + s − 2)

Using partial fractions

When s = 1, 0 = 3C, hence C = 0

}

6 s

−2s + 8s2 − 6 s

from which, L{x} =

i.e. 3s2 − s − 2 = A(s + 2)(s − 1)

(iv) Hence

(6′ )

(s2 + s − 2)L{x} = −2 + 8s −

{

(5′ )

Adding equations (5′ ) and (6′ ) gives:

3s − s − 2 s(s2 + s − 2)

Thus

6 −9 s

= −5 − 10s

2

When s = −2, 12 = 6B, hence

(6)

(6s2 + s − 2)L{x} − 5s(2s − 1)L{y}

i.e.

2 s

Using partial fractions

≡

(4′ )

(5)

8s2 − 2s − 6 = A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2)

The solution of simultaneous differential equations using Laplace transforms 657 When s = 0, −6 = −2A, When s = 1, 0 = 3C,

hence A = 3

Equation (3) × (s2 + 1) and equation (4) × 1 gives:

hence C = 0

When s = −2, 30 = 6B,

(s2 + 1)(s2 − 1)L{x} − (s2 + 1)L{y}

hence B = 5

= (s2 + 1)2s

8s − 2s − 6 3 5 Thus L{x} = = + s(s + 2)(s − 1) s (s + 2) 2

Hence

x = L−1

{

3 5 + s s+2

}

L{x} + (s2 + 1)L{y} = −s

and

x = 3 + 5e−2t

[(s2 + 1)(s2 − 1) + 1]L{x} = (s2 + 1)2s − s i.e.

s4 L{x} = 2s3 + s = s(2s2 + 1)

from which, L{x} =

(These solutions may be checked by substituting the expressions for x and y into the original equations.) Problem 3. Solve the following pair of simultaneous differential equations

d y + y = −x dt2

Returning to equations (3) and (4) to determine y:

dx given that at t = 0, x = 2, y = −1, =0 dt dy and =0 dt

1 × equation (3) and (s2 − 1) × equation (4) gives: (s2 − 1)L{x} − L{y} = 2s

Using the procedure: [s2 L{x} − sx(0) − x′ (0)] − L{x} = L{y}

(1)

[s2 L{y} − sy(0) − y′ (0)] + L{y} = −L{x}

(2)

x(0) = 2, y(0) = −1, x′ (0) = 0 and y′ (0) = 0 hence s2 L{x} − 2s − L{x} = L{y} s2 L{y} + s + L{y} = −L{x}

(iii)

(7)

(s2 − 1)L{x} + (s2 − 1)(s2 + 1)L{y}

from equation (4), page 628 (ii)

2s2 1 2 1 + 3= + 3 3 s s s s { } 2 1 x = L−1 + 3 s s 1 x = 2 + t2 2

i.e.

2

s(2s2 + 1) 2s2 + 1 = s4 s3

=

(iv) Hence

d2 x −x = y dt2

(i)

(6)

Adding equations (5) and (6) gives:

= 3 + 5e−2t

Therefore the solutions of the given simultaneous differential equations are y = 1 + 2e−2t

(5)

(1′ ) (2′ )

= −s(s2 − 1) Equation (7) − equation (8) gives:

[−1 − (s2 − 1)(s2 + 1)]L{y} = 2s + s(s2 − 1) i.e.

−s4 L{y} = s3 + s

and

L{y} =

from which,

Rearranging gives: (s2 − 1)L{x} − L{y} = 2s

(3)

L{x} + (s2 + 1)L{y} = −s

(4)

(8)

i.e.

s3 + s 1 1 =− − 3 −s4 s s { } 1 1 y = L−1 − − 3 s s 1 y = −1 − t2 2

658 Section J Now try the following Practice Exercise Practice Exercise 262 Solving simultaneous differential equations using Laplace transforms (Answers on page 901) Solve the following pairs of simultaneous differential equations: 1.

dx dy + = 5 et dt dt dy dx −3 = 5 dt dt

2

3.

d2 x + 2x = y dt2 d2 y + 2y = x dt2 given that at t = 0, x = 4, y = 2, and

dx =0 dt

dy =0 dt

given that when t = 0, x = 0 and y = 0 2.

2

dx dy −y+x+ − 5 sin t = 0 dt dt dy dx 3 + x − y + 2 − et = 0 dt dt

given that at t = 0, x = 0 and y = 0

For fully worked solutions to each of the problems in Practice Exercise 262 in this chapter, go to the website: www.routledge.com/cw/bird

Revision Test 16

Laplace transforms

This Revision Test covers the material contained in Chapters 54 to 59. The marks for each question are shown in brackets at the end of each question. 1.

Find the Laplace transforms of the following functions: (a) 2t − 4t + 5 3

(c) 3 cosh 2t (e) 5e2t cos 3t 2.

(c) (e) (g)

4.

− 4 sin 2t

6.

4 −3t

(d) 2t e (f) 2e3t sinh 4t

5 2s + 1 4s 2 s +9 3 (s + 2)4 8 s2 − 4s + 3

Sketch the graphs of (a) f (t) = 4H(t − 3) (b) f (t) = 3[H(t − 2) − H(t − 5)]

(5)

Determine (a) L{H(t − 2).e t−2 } (b) L{5H(t − 1). sin(t − 1)}

(6)

(16)

Find the inverse Laplace transforms of the following functions: (a)

3.

(b) 3e

−2t

5.

12 (b) 5 s 5 (d) 2 s −9 s−4 (f) 2 s − 8s − 20

{ 7.

Determine (a) L−1

3 e−2s s2

}

{ (b) L−1

3se−5s s2 + 4

} (9)

8.

In a galvanometer the deflection θ satisfies the differential equation: d2 θ dθ +2 +θ = 4 2 dt dt

(17)

Use partial fractions to determine the following: { } 5s − 1 (a) L−1 2 s −s−2 { 2 } 2s + 11s − 9 −1 (b) L s(s − 1)(s + 3) { } 13 − s2 (c) L−1 (24) s(s2 + 4s + 13)

9.

Use Laplace transforms to solve the equation for θ dθ given that when t = 0, θ = 0 and = 0 (13) dt Solve the following pair of simultaneous differential equations: 3

dx = 3x + 2y dt

2

dy + 3x = 6y dt

given that when t = 0, x = 1 and y = 3

Determine the poles and zeros for the transfer func(s + 2)(s − 3) tion: F(s) = and plot them on a (s + 3)(s2 + 2s + 5) pole–zero diagram. (10)

For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 16, together with a full marking scheme, are available at the website: www.routledge.com/cw/bird

(20)

Section K

Fourier series

Chapter 60

Fourier series for periodic functions of period 2π Why it is important to understand: Fourier series for periodic functions of periodic 2π A Fourier series changes a periodic function into an infinite expansion of a function in terms of sines and cosines. In engineering and physics, expanding functions in terms of sines and cosines is useful because it makes it possible to more easily manipulate functions that are just too difficult to represent analytically. The fields of electronics, quantum mechanics and electrodynamics all make great use of Fourier series. The Fourier series has become one of the most widely used and useful mathematical tools available to any scientist. This chapter introduces and explains Fourier series.

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

describe a Fourier series understand periodic functions state the formula for a Fourier series and Fourier coefficients obtain Fourier series for given functions

664 Section K f (x)

60.1

Introduction

Fourier∗ series provides a method of analysing periodic functions into their constituent components. Alternating currents and voltages, displacement, velocity and acceleration of slider-crank mechanisms and acoustic waves are typical practical examples in engineering and science where periodic functions are involved and often require analysis.

60.2

Periodic functions

As stated in Chapter 16, a function f (x) is said to be periodic if f (x + T) = f (x) for all values of x, where T is some positive number. T is the interval between two successive repetitions and is called the period of the function f (x). For example, y = sin x is periodic in x with period 2π since sin x = sin(x + 2π) = sin(x + 4π), and so on. In general, if y = sin ωt then the period of the waveform is 2π/ω. The function shown in Fig. 60.1 is also periodic of period 2π and is

1

22p

2p

2p

p

0

x

21

Figure 60.1

defined by:

{

f (x) =

−1, when 1, when

−π < x < 0 0 1 (z − 1)4 z for | z | > | a | z−a az ( )2 for | z | > | a | z−a az(z + a) ( )3 for | z | > | a | z−a z z − e −a z sin a z2 − 2z cos a + 1 z(z − cos a) z2 − 2z cos a + 1

{ } { } 6. ak = 1, a, a2 , a3 , ... { } { } 7. kak = 0, a, 2a2 , 3a3 , ... { } { } 8. k 2 ak = 0, a, 4a2 , 9a3 , ... { } { } 9. e −ak = e −a , e −2a , e −3a , ... { } 10. sin ak = sin a, sin 2a, ... { } 11. cos ak = cos a, cos 2a, ... { } 12. e −ak sin bk = e −a sin b, e −2a sin 2b, ...

ze −a sin b z2 − 2ze −a cos b + e −2a

{ } 13. e −ak cos bk = e −a cos b, e −2a cos 2b, ...

z2 − ze −a cos b z2 − 2ze −a cos b + e −2a

These formulae are available for downloading at the website: www.routledge.com/cw/bird

Answers

Answers to Practice Exercises 6. x2 + 3x − 10 = 0 7. (a) 0.637, −3.137 8. x = 2.602 m

Chapter 1 Exercise 1 (page 4) 1. −16 3. 3x − 5y + 5z 1 5. x5 y4 z3 , 13 2 1+a 7. b

2. −8 4. 6a2 − 13ab + 3ac − 5b2 + bc 1 6. ±4 √ 2 √ 6 11 1 3 a11 3 b − 8. a 6 b 3 c 2 or √ c3

Exercise 5 (page 11) 1. 2x − y

5. 2xy(y + 3x + 4x2 ) 7.

5 −1 y

Exercise 6 (page 13) 1. (x − 1)(x + 3) 2. (x + 1)(x + 2)(x − 2) ( )( ) 3. x + 1 2x2 + 3x − 7 4. (x − 1)(x + 3)(2x − 5)

8. ab

5. x3 + 4x2 + x − 6 = (x − 1)(x + 2)(x + 3) x = 1, x = −2 and x = −3 6. x = 1, x = 2 and x = −1

Exercise 3 (page 7) 1 2 1 3. − 8

3F − yL yL 5. f = or f = F − 3 3 7. L =

mrCR µ−m

2.

−3

4.

4

6.

ℓ=

8.

9. 500 MN/m2

r=

gt2 4π2 √(

Exercise 7 (page 14) x−y x+y

10. 0.0559 kg/m2

5. (a) 4, −8 (b)

2. a = 2, b = −3 4. r = 0.258, ω = 32.3 5 3 ,− 4 2

)

1. (a) 6 (b) 9 2. (a) −39 (b) −29 3. (x − 1)(x − 2)(x − 3) 4. x = −1, x = −2 and x = −4 5. a = −3 6. x = 1, x = −2 and x = 1.5

Exercise 4 (page 8) 1. x = 6, y = −1 3. x = 3, y = 4

3. 5x − 2

8 5. x2 + 2xy + y2 6. 5x + 4 + x−1 2 7. 3x2 − 4x + 3 − x+2 481 3 2 8. 5x + 18x + 54x + 160 + x−3

2. x2 − xy − 2y2 4. 7ab(3ab − 4) 2 6. + 12 − 3y 3y

1.

2. 3x − 1

4. 7x + 1

Exercise 2 (page 5) 1. −5p + 10q − 6r 3. 11q − 2p

(b) 2.443, 0.307 9. x = 1229 m or 238.9 m

864 Higher Engineering Mathematics Exercise 8 (page 15) 1. (b) 8. (d) 15. (a) 22. (c) 29. (b)

2. (a) 9. (d) 16. (d) 23. (b) 30. (d)

3. (a) 10. (c) 17. (a) 24. (c)

Exercise 13 (page 28)

4. (c) 11. (a) 18. (c) 25. (b)

5. (c) 12. (d) 19. (a) 26. (d)

6. (d) 13. (a) 20. (b) 27. (b)

7. (b) 14. (c) 21. (a) 28. (c)

1. log 6 4. log 3 7. log 100 10. log 1 = 0

2. log 15 5. log 12 8. log 6 11. log 2

3. log 2 6. log 500 9. log 10

12. log 243 or log 35 or 5 log 3 13. log 16 or log 24 or 4 log 2

Chapter 2

14. log 64 or log 26 or 6 log 2

Exercise 9 (page 20) 2 2 1. − (x − 3) (x + 3)

5 1 2. − (x + 1) (x − 3)

15. 0.5 18. t = 8

16. 1.5 19. b = 2

17. x = 2.5 20. x = 2

21. a = 6 22. x = 5 3 2 4 7 3 2 3. + − 4. − − x (x − 2) (x − 1) (x + 4) (x + 1) (2x − 1) 2 6 3 2 Exercise 14 (page 30) 5. 1 + + 6. 1 + − (x + 3) (x − 2) (x + 1) (x − 3) 1. 1.690 2. 3.170 3. 0.2696 1 5 7. 3x − 2 + − 5. 2.251 6. 3.959 7. 2.542 (x − 2) (x + 2) 9. 316.2 10. 0.057 m3

4. 6.058 8. −0.3272

Exercise 10 (page 22) 4 7 2 1 1 − 2. + − (x + 1) (x + 1)2 x x2 (x + 3) 5 10 4 3. − + (x − 2) (x − 2)2 (x − 2)3 2 3 4 4. − + (x − 5) (x + 2) (x + 2)2 1.

1. (c) 2. (c) 3. (d) 4. (b) 5. (a) 6. (b) 7. (c) 8. (b) 9. (d) 10. (c) 11. (d) 12. (d) 13. (a) 14. (a) 15. (b)

Chapter 4

Exercise 11 (page 23) 1.

Exercise 15 (page 31)

2x + 3 1 1 2−x − 2. + (x2 + 7) (x − 2) (x − 4) (x2 + 3)

Exercise 16 (page 33)

1 3 2 − 5x 3 2 1 − 2x 3. + + 4. + + x x2 (x2 + 5) (x − 1) (x − 1)2 (x2 + 8) 5. Proof

Chapter 3

1. (a) 0.1653 (b) 0.4584 (c) 22030 2. (a) 5.0988 (b) 0.064037 (c) 40.446 3. (a) 4.55848 (b) 2.40444 (c) 8.05124 4. (a) 48.04106 5. 2.739

(b) 4.07482

(c) −0.08286

6. 120.7 m

Exercise 12 (page 26) 1. 5.

4 1 3

1 9. 1 2 13. 100,000 1 17. 16

2.

4

3.

3

4.

−3

6.

3

7.

2

8.

−2

1 10. 3 14. 9 18. e3

11. 2 1 15. 32

12. 10,000 16. 0.01

Exercise 17 (page 35) 1. 2.0601

2. (a) 7.389 (b) 0.7408 8 3. 1 − 2x2 − x3 − 2x4 3 1 1 1 4. 2x1/2 + 2x5/2 + x9/2 + x13/2 + x17/2 + x21/2 3 12 60

Answers to Practice Exercises 865 Exercise 18 (page 36)

Exercise 22 (page 46)

1. 3.95, 2.05 2. 1.65, −1.30

1. (b) 7. (b)

2. (b) 8. (d)

3. (a) 9. (d)

4. (c) 10. (c)

5. (c)

6. (a)

3. (a) 28 cm3 (b) 116 min 4. (a) 70◦ C (b) 5 minutes

Chapter 5 Exercise 23 (page 50)

Exercise 19 (page 39) 1. (a) 0.55547 (b) 0.91374 (c) 8.8941 2. (a) 2.2293 (b) −0.33154 (c) 0.13087 3. 8.166 6. −0.4904 9. 816.2 12. 3 15. 4.901

4. 1.522 7. −0.5822

5. 1.485 8. 2.197

10. 0.8274 13. 4 16. 3.095

1. x7 − 7x6 y + 21x5 y2 − 35x4 y3 + 35x3 y4 −21x2 y5 + 7xy6 − y7 2. 32a5 + 240a4 b + 720a3 b2 + 1080a2 b3 + 810ab4 +243b5

11. 1.962 14. 147.9 a

17. t = eb+a ln D = eb ea ln D = eb eln D i.e. t = eb Da 18. 500 ( ) U2 19. W = PV ln U1 20. p2 = 348.5 Pa 21. 992 m/s

Exercise 24 (page 51) 1. a4 + 8a3 x + 24a2 x2 + 32ax3 + 16x4 2. 64 − 192x + 240x2 − 160x3 + 60x4 − 12x5 + x6 3. 16x4 − 96x3 y + 216x2 y2 − 216xy3 + 81y4 320 160 32 + 3 + 5 x x x 5. p11 + 22p10 q + 210p9 q2 + 1320p8 q3 + 5280p7 q4 4. 32x5 + 160x3 + 320x +

6. 34 749p8 q5

Exercise 20 (page 42) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

7. 700 000a4 b4

(a) 150◦ C (b) 100.5◦ C 99.21 kPa (a) 29.32 volts (b) 71.31 × 10−6 s (a) 2.038 × 10−4 (b) 2.293 m (a) 50◦ C (b) 55.45 s 30.4 N, 0.807 rad (a) 3.04 A (b) 1.46 s 2.45 mol/cm3 (a) 7.07 A (b) 0.966 s (a) 100% (b) 67.03% (c) 1.83% 2.45 mA 142 ms 99.752% 20 min 38 s

Exercise 25 (page 53) 1. 1 + x + x2 + x3 + . . ., |x| < 1 2. 1 − 2x + 3x2 − 4x3 + . . ., |x| < 1 [ ] 1 3 3 5 3. 1 − x + x2 − x3 + . . . , |x| < 2 8 2 2 4 ( ) √ x x2 x3 4. 2 1 + − + − · · · , |x| < 2 4 32 128 or −2 < x < 2 3 27 135 3 1 5. 1 − x + x2 − x , |x| < 2 8 16 3 [ ] 1 189 2 2 6. 1 − 9x + x + . . . , |x| < 64 4 3 7. Proofs

Exercise 21 (page 45) −5

1. a = 76, k = −7 × 10−5 , p = 76e−7×10 2. θ0 = 152, k = −0.05

h

, 37.74 kPa

8. 4 −

31 x 15

9. (a) 1 − x +

x2 , |x| < 1 2

7 1 (b) 1 − x − x2 , |x|< 2 3

866 Higher Engineering Mathematics ( ) Z = C. A.B + A

Exercise 26 (page 55)

2.

1. 3. 4. 6. 8. 10. 12. 14.

A

B

C

A

B

A·B

A·B+A

( ) Z = C· A·B+A

0

0

0

1

1

0

1

0

0

0

1

1

1

0

1

1

0

1

0

1

0

0

1

0

0

1

1

1

0

0

1

1

1

0

0

0

1

1

1

0

1

0

1

0

1

1

1

1

1

1

0

0

0

0

0

0

1

1

1

0

0

0

0

0

0.6% decrease 2. 3.5% decrease (a) 4.5% increase (b) 3.0% increase 2.2% increase 5. 4.5% increase Proof 7. 7.5% decrease 2.5% increase 9. 0.9% too small +7% 11. Proof 5.5% 13. +1.5% 5% increase

Exercise 27 (page 56) 1. (d)

2. (b)

3. (b)

4. (c)

5. (a)

3.

Chapter 6

A B C A B C B·C B·C A·B

Exercise 28 (page 61) 1. 1.19

2. 1.146

3. 1.20

4. 3.146

5. 1.849

Exercise 29 (page 64) 1. −3.36, 1.69 3. −1.53, 1.68

2. −2.686 4. −12.01, 1.000

Exercise 30 (page 64) 1. (b)

2. (d)

3. (c)

4. (c)

5. (a)

0

0

0

0

0 0 1 1 1 0

0

1

0

1

0

0 1 0 1 0 1

1

0

1

1

0

0 1 1 1 0 0

0

0

1

1

0

1 0 0 0 1 1

0

0

0

0

0

1 0 1 0 1 0

0

1

0

1

0

1 1 0 0 0 1

1

0

0

1

1

1 1 1 0 0 0

0

0

0

0

0

C

A·B

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

( ) A B C A C B·C·A B+C A· B+C

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

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

0

4.

Exercise 31 (page 69)

( B·C +

0 0 0 1 1 1

Chapter 7

A

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

A A·B A·B + A·B

(

Z=

C· A·B + A·B 0

0

0

0

1

0

0

0

0

0

1

0

1

0

0

0

0

1

0

0

1

1

1

0

0

1

1

0

1

1

1

1

1

0

0

0

0

0

0

0

1

0

1

0

0

0

0

0

1

1

0

1

0

0

1

0

1

1

1

1

0

0

1

1

)

B · C · A+ Z ( ) A· B+C

0

0 0 1 1

0

1

0

0

0

0

0 1 1 0

0

0

0

0

0

0

1 0 1 1

0

1

0

0

0

0

1 1 1 0

1

1

0

1

1

1

0 0 0 1

0

1

1

1

0

1

0 1 0 0

0

0

0

0

0

1

1 0 0 1

0

1

1

1

0

1

1 1 0 0

0

1

1

1

1

Answers to Practice Exercises 867 C

A

5.

Exercise 32 (page 71)

Input

Output

A

C

B

A

B

1. 3. 5. 7. 9. 11.

P G P.Q F.G.H + G.H G P.R + P.R

2. 4. 6. 8. 10. 12.

P + P.Q F ( ) H. F + F.G Q.R + P.Q.R F.H + G.H P + R.Q

A

6. Input

A

B

Output

C

B

1. A.B 4. 1

C

A

7. Input

Output

A A B C

3. A.B + A.B.C

1. Y

2. X + Y

3. P.Q ( ) 5. R. P + Q ( ) 7. A.C. B + D ( ) 9. D. A + B.C

4. B + A.C + A.C 6. P.(Q + R) + P.Q.R ( ) 8. B.C. A + D

10. A.D + A.B.C.D ( ) 11. A.C + A.C.D + B.D. A + C

8. A · B · C + A · B · C A

2. A + B + C ( ) 5. P. Q + R

Exercise 34 (page 77)

C

B

Exercise 33 (page 73)

B

C

Input

Output A

B

C

Exercise 35 (page 80)

9. A.B.C + A.B.C + A.B.C

1. A Input

A

B

B

C

C

B

B

1 &

Z5 A 1B · C

C

Output

2. A

A

A B

C

& 1 &

Z5 A · B1 B · C

C

10. A.B.C + A.B.C + A.B.C + A.B.C A

B

3.

C

A

&

B

1

Z 5A · B · C 1A · B · C

&

Z5(A1 B) · (C1D)

& A

B

C

C

Input

Output

A

B

C

4.

A

1

B A

B

C

C D

1

868 Higher Engineering Mathematics 5. Z1 = A.B + C A

&

Exercise 36 (page 83) 1

B

1.

Z15A · B1 C

&

A

&

Z

B·C

A

B

C

&

C

6. Z2 = A.B + B.C 2.

A &

B

1

Z2 5A · B1 B · C

&

C

7. Z3 = A.C + B 3. A B &

C

1

Z3 5A · C1 B

8. Z = P + Q

4.

P

&

B

&

C

&

A

&

B

&

C

&

A

&

Z

&

Z5A · B · C1A · B · C

A·B

B·C

&

& &

1

1

1

Z (A B) · (C D)

B C

Z 5 P1 Q

1

A

1

1

D

Q

( ) 9. R. P + Q

5.

P 1

Q

&

Z 5 R · (P1 Q)

A

1

B

1

C

1

1 1

1

1

Z A·B B·C C·D

1

D

R

6.

( ) 10. Q. P + R P

P

1

Q

1

R

1

1 &

Q

Z 5 Q · (P1 R )

R

7. (

11. D. A.C + B A

)

1 1

1

1

Z P · Q P · (Q R)

1

P

&

Z

P · (Q R )

Q 1

R

&

1

B

8.

C &

D

Z 5D · (A · C1 B )

Z

A · (B

C

D)

1

C D

P

R

&

B

( ) 12. P. Q + R

Q

A

& 1

Z5 P · (Q1R )

9.

A

1

B

1

C

1

1 1 1

1

1 Z A · (B C )

B·C

Answers to Practice Exercises 869 10.

12. 51.92◦ , 51◦ 55′ , 0.906 rad

A B

&

C

&

&

&

&

Z A · (B

C)

13. 23.69◦ , 23◦ 41′ , 0.413 rad 14. 27.01◦ , 27◦ 1′ , 0.471 rad 15. 29.05◦ 16. 20◦ 21′

Exercise 37 (page 84) 1. (b) 2. (d) 3. (c)

17. 1.097

4. (b) 5. (c) 6. (d) 7. (a)

8. (c) 9. (d) 10. (a)

18. 5.805 19. −5.325 20. 0.7199 21. 21◦ 42′

Chapter 8

22. 1.8258, 1.1952, 0.6546

Exercise 38 (page 91)

23. (a) −0.8192

1. 24.11 mm 2. (a) 27.20 cm each

24. (a) −1.6616 (b) −0.32492 25. β = −68.37◦

(b) 45◦

3. 20.81 km

4. 3.35 m, 10 cm

5. 132.7 nautical miles

6. 2.94 mm

(b) −1.8040

(c) 0.6528 (c) 2.5985

Exercise 41 (page 98)

7. 24 mm

1. BC = 3.50 cm, AB = 6.10 cm, ∠B = 55◦

Exercise 39 (page 93)

2. FE = 5 cm, ∠E = 53.13◦ , ∠F = 36.87◦

3 4 1. sin A = , cos A = , tan A = 5 5 3 4 cos B = , tan B = 5 3 8 8 2. sin A = , tan A = 17 15 15 15 8 3. (a) (b) (c) 17 17 15 4. (a) 9.434 (b) −0.625 (c)

3 4 , sin B = , 4 5

3. GH = 9.841 mm, GI = 11.32 mm, ∠H = 49◦ 4. KL = 5.43 cm, JL = 8.62 cm, ∠J = 39◦ , area = 18.19 cm2 5. MN = 28.86 mm, NO = 13.82 mm, ∠O = 64◦ 25′ , area = 199.4 mm2

◦

32

6. PR = 7.934 m, ∠Q = 65.06◦ , ∠R = 24.94◦ , area = 14.64 m2 7. 6.54 m

Exercise 40 (page 96) (b) 0.1321

(c) −0.8399

1.

(a) 0.4540

2.

(a) −0.5592 (b) 0.9307

3.

(a) −0.7002 (b) −1.1671

4.

(a) 3.4203

5.

(a) −1.8361 (b) 3.7139

(c) −1.3421

6.

(a) 0.3443

(b) −1.8510

(c) −1.2519

7.

(a) 0.8660

(b) −0.1100

(c) 0.5865

8.

(a) 1.0824

(b) 5.5675

9.

13.54◦ , 13◦ 32′ , 0.236 rad

(b) 3.5313

10. 34.20◦ , 34◦ 12′ , 0.597 rad 11. 39.03◦ , 39◦ 2′ , 0.681 rad

(c) 0.2447 (c) 1.1612

(c) −1.0974

(c) −1.7083

Exercise 42 (page 100) 1. 48 m

2. 110.1 m

3. 53.0 m

4. 9.50 m

5. 107.8 m

6. 9.43 m, 10.56 m

7. 60 m

Exercise 43 (page 102) 1. C = 83◦ , a = 14.1 mm, c = 28.9 mm, area = 189 mm2 2. A = 52◦ 2′ , c = 7.568 cm, a = 7.152 cm, area = 25.65 cm2

870 Higher Engineering Mathematics 3. D = 19.80◦ , E = 134.20◦ , e = 36.0 cm, 2

area = 134 cm

4. E = 49◦ 0′ , F = 26◦ 38′ , f = 15.09 mm, area = 185.6 mm2 5. J = 44◦ 29′ , L = 99◦ 31′ , l = 5.420 cm,

Exercise 47 (page 108) 1. (d) 8. (d) 15. (b) 22. (a)

2. (a) 9. (d) 16. (c) 23. (c)

3. (b) 10. (d) 17. (d) 24. (a)

4. (c) 5. (c) 6. (b) 7. (c) 11. (a) 12. (b) 13. (d) 14. (b) 18. (c) 19. (d) 20. (a) 21. (a) 25. (b)

area = 6.133 cm2 OR J = 135◦ 31′ , L = 8◦ 29′ , l = 0.811 cm, area = 0.917 cm2 6. K = 47◦ 8′ , J = 97◦ 52′ , j = 62.2 mm, area = 820.2 mm

Chapter 9 Exercise 48 (page 114) 1. (5.83, 59.04◦ ) or (5.83, 1.03 rad)

2

OR K = 132◦ 52′ , J = 12◦ 8′ , j = 13.19 mm,

2. (6.61, 20.82◦ ) or (6.61, 0.36 rad)

area = 174.0 mm2

3. (4.47, 116.57◦ ) or (4.47, 2.03 rad) 4. (6.55, 145.58◦ ) or (6.55, 2.54 rad) 5. (7.62, 203.20◦ ) or (7.62, 3.55 rad)

Exercise 44 (page 103)

6. (4.33, 236.31◦ ) or (4.33, 4.12 rad)

1. p = 13.2 cm, Q = 47.34◦ , R = 78.66◦ ,

7. (5.83, 329.04◦ ) or (5.83, 5.74 rad)

area = 77.7 cm2

8. (15.68, 307.75◦ ) or (15.68, 5.37 rad) ◦

◦

2. p = 6.127 m, Q = 30.83 , R = 44.17 , area = 6.938 m2 3. X = 83.33◦ , Y = 52.62◦ , Z = 44.05◦ , 2

area = 27.8 cm ◦

◦

◦

4. X = 29.77 , Y = 53.50 , Z = 96.73 , area = 355 mm2

Exercise 45 (page 105) 1. 2. 3. 5.

193 km (a) 122.6 m (b) 94.80◦ , 40.66◦ , 44.54◦ (a) 11.4 m (b) 17.55◦ 4. 163.4 m BF = 3.9 m, EB = 4.0 m 6. 6.35 m, 5.37 m

Exercise 49 (page 115) 1. 3. 5. 7. 9.

(1.294, 4.830) 2. (1.917, 3.960) (−5.362, 4.500) 4. (−2.884, 2.154) (−9.353, −5.400) 6. (−2.615, −3.027) (0.750, −1.299) 8. (4.252, −4.233) (a) 40∠18◦ , 40∠90◦ , 40∠162◦ , 40∠234◦ , 40∠306◦ (b) (38.04, 12.36), (0, 40), (−38.04, 12.36), (−23.51, −32.36), (23.51, −32.36) (c) 47.02 mm

Exercise 50 (page 116) 1. (a)

2. (d)

3. (b)

4. (a)

5. (c)

Exercise 46 (page 107) 1. 32.48 A, 14.31◦

2. 80.42◦ , 59.38◦ , 40.20◦

3. x = 69.3 mm, y = 142 mm 4. 130◦

Chapter 10

5. 40.25 cm, 126.05◦

6. 19.8 cm

Exercise 51 (page 118)

7. 36.2 m

8. 13.95◦ , 829.9 km/h

1. 259.5 mm

2. 47.68 cm

4. 12 730 km

5. 97.13 mm

9. 13.66 mm

3. 38.73 cm

Answers to Practice Exercises 871 Exercise 52 (page 120)

Exercise 57 (page 128)

π 5π 5π (b) (c) 6 12 4 2. (a) 0.838 (b) 1.481 (c) 4.054

1. (d) 7. (c)

1. (a)

3. (a) 210◦

(b) 80◦

4. (a) 0◦ 43′

(b) 154◦ 8′

2. (c) 8. (b)

3. (b) 9. (d)

4. (b) 10. (a)

5. (a)

6. (c)

(c) 105◦ (c) 414◦ 53′

Chapter 11

5. 104.72 rad/s

Exercise 58 (page 135)

Exercise 53 (page 122)

1. 227.06◦ and 312.94◦

2. 23.27◦ and 156.73◦

3. 122.26◦ and 302.26◦

4. t = 122.11◦ and 237.89◦

1. 113 cm2

2. 2376 mm2

3. 1790 mm2

4. 802 mm2

5. 1709 mm2

6. 1269 m2

7. 1548 m2

Exercise 59 (page 141)

8. 17.80 cm, 74.07 cm2 9. (a) 59.86 mm

(b) 197.8 mm

10. 26.2 cm 11. 8.67 cm, 54.48 cm 13. 15. 17. 19.

19.63 m2 2880 mm2 8.48 m 60 mm

21. (a) 0.698 rad 22. (a) 396 mm2

12. 82◦ 30′ 14. 16. 18. 20.

1. 1, 120◦

2. 2, 144◦

3. 3, 90◦

4. 3, 720◦

5. 3.5, 960◦

6. 6, 360◦

7. 4, 180

2107 mm2 4.49 m2 9.55 m 748

◦

8. 2, 90

◦

9. 5, 120◦

Exercise 60 (page 144) 1. (a) 40 mA

(b) 804.2 m2

(b) 25 Hz

(c) 0.04 s or 40 ms

◦

(d) 0.29 rad (or 16.62 ) leading 40 sin 50πt

(b) 42.24%

23. 701.8 mm

5. x = 64.42◦ and 295.58◦ 6. θ = 39.74◦ and 219.74◦

24. 7.74 mm

2. (a) 75 cm

(b) 6.37 Hz

(c) 0.157 s

◦

(d) 0.54 rad (or 30.94 ) lagging 75 sin 40t

Exercise 54 (page 125) 1. (a) 6

3. (a) 300 V

(b) 100 Hz

(c) 0.01 s or 10 ms

◦

(d) 0.412 rad (or 23.61 ) lagging 300 sin 200πt

(b) (−3, 1)

2. Centre at (3, −2), radius 4

4. (a) v = 120 sin 100πt volts

3. Circle, centre (0, 1), radius 5 4. Circle, centre (0, 0), radius 6

(b) v = 120 sin(100πt + 0.43) volts ( π) 5. i = 20 sin 80πt − A or 6 i = 20 sin(80πt − 0.524) A

Exercise 55 (page 126)

6. 3.2 sin(100πt + 0.488) m

1. ω = 90 rad/s, v = 13.5 m/s

7. (a) 5 A, 50 Hz, 20 ms, 24.75◦ lagging (b) −2.093 A (c) 4.363 A

2. v = 10 m/s, ω = 40 rad/s

(d) 6.375 ms

(e) 3.423 ms

3. (a) 75 rad/s, 716.2 rev/min (b) 1074 revs

Exercise 61 (page 149) Exercise 56 (page 128) 1. 2 N

2. 988 N, 5.14 km/h

1. (a) i = (70.71 sin 628.3t + 16.97 sin 1885t)A 3. 1.49 m/s2

2. (a) v = 300 sin 314.2t + 90 sin(628.3t − π/2) +30 sin(1256.6t + π/3) V

872 Higher Engineering Mathematics 3. Sketch 4. i = (16 sin 2π103 t + 3.2 sin 6π103 t + 1.6 sin π104 t)A 5. (a) 60 Hz, 180 Hz, 300 Hz

(b) 40%

(c) 10%

Chapter 13 Exercise 67 (page 163) 1 – 6. Proofs

Exercise 62 (page 149) 1. (d)

2. (a)

3. (a)

Exercise 68 (page 165)

4. (c)

5. (d)

1. 3. 5. 7. 9. 11.

Chapter 12 Exercise 63 (page 153) 1. 3. 5. 7. 9.

(a) 0.6846 (a) 0.5717 (a) 0.9285 56.38 5.042

(b) 4.376 (b) 0.9478 (b) 0.1859

2. 4. 6. 8.

(a) 1.271 (a) 1.754 (a) 2.398 30.71

(b) 5.910 (b) 0.08849 (b) 1.051

Exercise 64 (page 157) 1–4. Proofs

5. P = 2, Q = −4

6. A = 9, B = 1

θ = 34.85◦ or 145.15◦ 2. t = 66.75◦ or 246.75◦ 4. 59◦ , 239◦ 6. ◦ ±131.81 8. −30◦ , −150◦ 10. 101.31◦ , 281.31◦

A = 213.06◦ or 326.94◦ 60◦ , 300◦ 41.81◦ , 138.19◦ 39.81◦ , −140.19◦ 33.69◦ , 213.69◦

Exercise 69 (page 166) 1. 2. 3. 4. 5.

y = 50.77◦ , 129.23◦ , 230.77◦ or 309.23◦ θ = 60◦ , 120◦ , 240◦ or 300◦ θ = 60◦ , 120◦ , 240◦ or 300◦ D = 90◦ or 270◦ θ = 32.31◦ , 147.69◦ , 212.31◦ or 327.69◦

Exercise 70 (page 166) Exercise 65 (page 159) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

(a) 0.8814 (b) −1.6209 (a) ±1.2384 (b) ±0.9624 (a) −0.9962 (b) 1.0986 (a) ±2.1272 (b) ±0.4947 (a) 0.6442 (b) 0.9832 (a) 0.4162 (b) −0.6176 −0.8959 0.6389 or −2.2484 0.2554 (a) 67.30 (b) ±26.42

Exercise 66 (page 160) 1. (a) 2.3524 (b) 1.3374 2. (a) 0.5211 (b) 3.6269 9 81 2 3. (a) 3x + x3 + x5 (b) 1 + 2x2 + x4 2 40 3 4 – 5. Proofs

1. 2. 3. 4.

A = 19.47◦ , 160.53◦ , 203.58◦ or 336.42◦ θ = 51.34◦ , 123.69◦ , 231.34◦ or 303.69◦ t = 14.48◦ , 165.52◦ , 221.81◦ or 318.19◦ θ = 60◦ or 300◦

Exercise 71 (page 167) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

θ = 90◦ , 210◦ , 330◦ t = 190.10◦ , 349.90◦ θ = 38.67◦ , 321.33◦ θ = 0◦ , 60◦ , 300◦ , 360◦ θ = 48.19◦ , 138.59◦ , 221.41◦ or 311.81◦ x = 52.94◦ or 307.06◦ A = 90◦ t = 107.83◦ or 252.17◦ a = 27.83◦ or 152.17◦ β = 60.17◦ , 161.02◦ , 240.17◦ or 341.02◦ θ = 51.83◦ , 308.17◦ θ = 30◦ , 150◦

Answers to Practice Exercises 873 ( π) 12.x = 7.07 sin 2t + cm 4

Exercise 72 (page 168) 1. (c) 6. (d)

2. (a) 7. (b)

3. (c) 8. (d)

4. (a)

5. (b)

Exercise 77 (page 180)

Chapter 14

) V2 ( 1 + cos 2t 2R 2. Proofs 1.

Exercise 73 (page 170) 1 – 5. Proofs

3. cos 3θ = 4 cos3 θ − 3 cos θ 4. −90◦ , 30◦ , 150◦

Exercise 74 (page 172)

5. −160.53◦ , −90◦ , −19.47◦ , 90◦

1. 1 − tanh2 θ = sech2 θ

6. −150◦ , −90◦ , −30◦ , 90◦

2. cosh(θ + ϕ) = cosh θ cosh ϕ + sinh θ sinh ϕ

7. −90◦

3. sinh(θ − ϕ) = sinh θ cosh ϕ − cosh θ sinh ϕ 4. 5. 6. 7.

8. 45◦ , −135◦

2 tanh θ tanh 2θ = 1 + tanh2 θ 1 cosh θ sinh ϕ = [sinh(θ + ϕ) − sinh(θ − ϕ)] 2 1 3 3 sinh θ = sinh 3θ − sinh θ ( 4 ) 4 coth2 θ 1 − sec h2 θ = 1

Exercise 78 (page 182) 1 [sin 9t + sin 5t] 2 3. cos 4t − cos 10t 3[ π π] 5. sin + sin 2 2 6 1.

Chapter 15

1 [sin 10x − sin 6x] 2 4. 2[cos 4θ + cos 2θ] 2.

6. 30◦ , 90◦ and 150◦

Exercise 75 (page 175) 1. (a) sin 58◦

2. (a) cos 104◦

(b) sin 4t

3. Proof

(b) cos

4. Proof (c) −2.4678

5. (a) 0.3136 (b) 0.9495 6. 64.72◦ or 244.72◦

7. 67.52◦ or 247.52◦

Exercise 76 (page 179) 1. 9.434 sin(ωt + 1.012)

2. 5 sin(ωt − 0.644)

3. 8.062 sin(ωt + 2.622)

4. 6.708 sin(ωt − 2.034)

5. (a) 74.44◦ or 338.70◦

(b) 64.69◦ or 189.05◦

6. (a) 72.74◦ or 354.64◦

(b) 11.15◦ or 311.98◦

7. (a) 90◦ or 343.74◦

(b) 0◦ or 53.14◦

8. (a) 82.92◦ or 296◦ (b) 32.36◦ , 97◦ , 152.36◦ , 217◦ , 272.36◦ or 337◦ 9. 8.13 sin(3θ + 2.584) 10. x = 4.0 sin(ωt + 0.927) m 11. 9.434 sin(ωt + 2.583) V

π 12

Exercise 79 (page 183) 2. cos 8θ sin θ 1 4. − sin 3t sin 2t 4 6. Proofs

1. 2 sin 2x cos x

3. 2 cos 4t cos t 7π π 5. cos cos 24 24 7. 22.5◦ , 45◦ , 67.5◦ , 112.5◦ , 135◦ , 157.5◦ 8. 0◦ , 45◦ , 135◦ , 180◦

9. 21.47◦ or 158.53◦

10. 0◦ , 60◦ , 90◦ , 120◦ , 180◦ , 240◦ , 270◦ , 300◦ , 360◦

Exercise 80 (page 186) 1. (c)

2. (b)

3. (c)

4. (d)

5. (a)

874 Higher Engineering Mathematics Chapter 16

5. y

Exercise 81 (page 198) 1. y

15

10 10 y 5(x24) 212

5 y 5 3x 25 0

1

2

3

5 x 0

2

4

6

8

x

25

2. y 6. 4

y 0.50

2 y 5x 2x 2

0.25 0

1

2

22

3

x 0

y 5 23x 14

1

x

3. y

8

7.

6

y 5 x 213

y

4 10

2 22

21

1

0

2

x

22

21

0

4. y

25

210

8 y 5(x 23)2 4

0

2

4

6

x

y 5 x 3 12

5

1

2

x

Answers to Practice Exercises 875 8.

Exercise 83 (page 203)

y

1. f −1 (x) = x − 1

y 5 11 2 cos 3x

3 2

3. f −1 (x) =

1

5. −

√ 3 x−1

π or −1.5708 rad 2

π or 0.7854 rad 4 9. 0.4115 rad π 11. or 0.7854 rad 4 13. 1.533 rad 7.

0

3p 2

p

p 2

2p

x

21

1 2. f −1 (x) = (x + 1) 5 1 −1 4. f (x) = x−2 π 6. or 1.0472 rad 3 8. 0.4636 rad 10. 0.8411 rad 12. 0.257 rad

9. y 6 y 5 3 2 2 sin(x 1

p ) 4

4 2

Exercise 84 (page 207) 1. y = 1, x = −1 2. x = 3, y = 1 and y = −1 3. x = −1, x = −2 and y = 1 4. x = 0, y = x and y = −x

0

3p 2

p

p 2

2p

10.

x

5. y = 4, y = −4 and x = 0 6. x = −1, y = x − 2 (see Fig. A on page 876) 7. x = 0, y = 0, y = x (see Fig. B on page 876)

y

Exercise 85 (page 211)

3

2

y 5 2 ln x

1

0

1

2

3

4 x

21

22

Exercise 82 (page 201) 1. (a) even (b) odd (c) neither (d) even 2. (a) odd (b) even (c) odd (d) neither 3. (a) even (b) odd

1. (a) Parabola with minimum value at (−1.5, −5) and passing through (0, 1.75) (b) Parabola with maximum value at (2, 70) and passing through (0, 50) 2. Circle, centre (0, 0), radius 4 units 3. Parabola, symmetrical about x-axis, vertex at (0, 0) 4. Hyperbola, symmetrical about x- and y- axes, distance between vertices 8 units along x-axis 5. Ellipse, centre (0, 0),√major axis 10 units along x-axis, minor axis 2 10 units along y-axis 6. Hyperbola, symmetrical about x- and y- axes, distance between vertices 6 units along x-axis 7. Rectangular hyperbola, lying in first and third quadrants only 8. Ellipse, centre (0, 0),√major axis 4 units along x-axis, minor axis 2 2 units along y-axis 9. Circle, centre (2, −5), radius 2 units √ 10. Ellipse, centre (0, 0), major axis 2 3 units along y-axis, minor axis 2 units along x-axis 11. Hyperbola, symmetrical about x- and y- axes, vertices 2 units apart along x-axis 12. Circle, centre (0, 0), radius 3 units

876 Higher Engineering Mathematics 13. Rectangular hyperbola, lying in first and third quadrants, symmetrical about x- and y- axes 14. Parabola, vertex at (0, 0), symmetrical about the x-axis √ 15. Ellipse, centre (0, 0), major axis 2 8 units along y-axis, minor axis 4 units along x-axis

Chapter 17 Exercise 87 (page 215) 1. 4.5 square units

2. 54.7 square units

3. 63.33 m 5. 143 m2

4. 4.70 ha

Exercise 86 (page 211) 1. (c)

2. (c)

3. (a)

4. (b)

5. (d)

Exercise 88 (page 217) 1. 42.59 m3

y

2. 147 m3

3. 20.42 m3

6 x 521

2 2

Exercise 89 (page 220) y5 x

4

24

y5

x 2 2x2 4 x 11

22

0

y5

22

6 x

4

2

(b) 50 V

(c) 2.5 A

2. (a) 2.5 mV (b) 3 A 3. 0.093 As, 3.1 A 4. (a) 31.83 V (b) 0

2

26

1. (a) 2 A

5. 49.13 cm2 , 368.5 kPa

x 2 2x 24 x 11

Exercise 90 (page 221) 1. (c)

24

26

2. (d)

3. (b)

4. (c)

5. (a)

Chapter 18 Exercise 91 (page 229) ±j5 2. x = 1 ± j 3. x = 2 ± j x = 3±j 5. x = 0.5 ± j0.5 6. x = 2 ± j2 x = 0.2 ± j0.2 √ 3 23 8. x = − ± j or x = −0.750 ± j1.199 4 √ 4 5 87 9. x = ± j or x = 0.625 ± j1.166 8 8 10. (a) 1 (b) −j (c) −j2 1. 4. 7.

Figure A y

xy 2 2 x 2y 1 2x 2y 5 5 6

y5 x

4

2

26 xy 2

2

x 2y

24

22

0

4

6

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

1 2x 2y 5 5 22

24

26

Figure B

2

x

Exercise 92 (page 232) 1. (a) 8 + j (b) −5 + j8 2. (a) 3 − j4 (b) 2 + j 3. (a) 5 (b) 1 − j2 (c) j5 (e) 5 (f) 3 + j4 4. (a) 7 − j4 (b) −2 − j6 5. (a) 10 + j5 (b) 13 − j 13

(d) 16 + j3

Answers to Practice Exercises 877 6. (a) −13 − j2 (b) −35 + j20 2 11 19 43 7. (a) − + j (b) − + j 25 25 85 85 3 41 45 9 8. (a) +j (b) −j 26 26 26 26 1 1 9. (a) −j (b) − j 2 2 10. Proof

mh 2π 9. (a) 922 km/h at 77.47◦ 8. ±

10. (a) 3.770∠8.17◦ 12. 353.6∠ − 45◦

13. 275∠ − 36.87◦ mA

3 1 2. x = , y = − 2 2 4. x = 3, y = 1

3. a = −5, b = −12 5. 10 + j13.75

Exercise 94 (page 235) 1. (a) 4.472, 63.43

◦

(b) 5.385, −158.20

(c) 2.236, 63.43 √ 2. (a) 13∠56.31◦

4. (a) (c) 5. (a) (c) 6. (a) 7. (a)

Exercise 96 (page 239) 1. (b) 2. (d) 3. (c) 4. (d) 5. (a) 6. (d) 7. (a) 8. (c) 9. (b) 10. (a) 11. (b) 12. (c) 13. (b) 14. (d) 15. (a) 16. (a) 17. (b) 18. (c) 19. (c) 20. (d)

Chapter 19 ◦

◦

3. (a)

(b) 1.488∠100.37◦

11. Proof

Exercise 93 (page 233) 1. a = 8, b = −1

(b) 922 km/h at −102.53◦

Exercise 97 (page 243)

1. (a) 7.594∠75◦ (b) 125∠20.61◦ √ ◦ (b) 4∠180 (c) 37∠170.54 2. (a) 81∠164◦ , −77.86 + j22.33 √ √ 3∠ − 90◦ (b) 125∠100.30◦ (c) 2∠ − 135◦ (b) 55.90∠ − 47.18◦ , 38 − j41 √ 4.330 + j2.500 (b) 1.500 + j2.598 3. 10∠ − 18.43◦ , 3162∠ − 129◦ 4.950 + j4.950 4. 476.4∠119.42◦ , −234 + j415 −3.441 + j4.915 (b) −4.000 + j0 5. 45 530∠12.78◦ , 44 400 + j10 070 −1.750 − j3.031 6. 2809∠63.78◦ , 1241 + j2520 45∠65◦ (b) 10.56∠44◦ ( ) 7. 38.27 × 106 ∠176.15◦ , 106 (−38.18 + j2.570) ◦ ◦ 3.2∠42 (b) 2∠150

8. (a) 6.986∠26.79◦

◦

(b) 7.190∠85.77◦

Exercise 98 (page 244) 1. (a) ±(1.099 + j0.455) (b) ±(0.707 + j0.707)

Exercise 95 (page 238)

2. (a) ±(2 − j)

1. (a) R = 3Ω, L = 25.5 mH

(b) ±(0.786 − j1.272)

3. (a) ±(2.291 + j1.323) (b) ±(−2.449 + j2.449)

(b) R = 2Ω, C = 1061 µF

4. Modulus 1.710, arguments 17.71◦ , 137.71◦

(c) R = 0, L = 44.56 mH (d) R = 4Ω, C = 459.4 µF 2. 15.76 A, 23.20◦ lagging

and 257.71◦ 5. Modulus 1.223, arguments 38.36◦ , 128.36◦ , 218.36◦ and 308.36◦

3. 27.25 A, 3.37◦ lagging 4. 14.42 A, 43.85◦ lagging, 0.721 5. 14.6 A, 2.51◦ leading 6. 8.394 N, 208.68◦ from force A ◦

7. (10 + j20)Ω, 22.36∠63.43 Ω

6. Modulus 2.795, arguments 109.90◦ and 289.90◦ 7. Modulus 0.3420, arguments 24.58◦ , 144.58◦ and 264.58◦ 8. Z0 = 390.2∠ − 10.43◦ Ω, γ = 0.1029∠61.92◦

878 Higher Engineering Mathematics Exercise 99 (page 246) 1. 5.83e j0.54 2. 4.89e j2.11

10. (a) x2 + 2x + 2y + y2 = 0 or (x + 1)2 + (y + 1)2 = 2 √ (b) a circle, centre (−1, −1) and radius 2

5. −4.52 − j3.38

11. (a) y = 2x + 1.5 (b) a straight line 12. (a) y = 2x − 3 (b) a straight line 1 13. (a) x = (b) a straight line 2

6. (a) ln 7 + j2.1 (b) 2.86∠47.18◦ or 2.86∠0.82 rad

Exercise 101 (page 249)

3. −1.50 + j3.27 4. 34.79 + j20.09

7. 3.51∠ − 0.61 or 3.51∠ − 34.72◦ 8. (a) 2.06∠35.25◦ or 2.06∠0.615 rad (b) 4.11∠66.96◦ or 4.11∠1.17 rad (√ ) ht − 4mf − h2 9. Ae 2m cos t 2m − a

1. (a) 2. (b) 3. (c) 4. (d) 5. (b) 6. (c) 7. (d) 8. (a)

Chapter 20 Exercise 102 (page 257) ( 1. (

Exercise 100 (page 249)

3. (

1. (a) x2 + y2 = 4 (b) a circle, centre (0, 0) and radius 2 2. (a) x2 + y2 = 25 (b) a circle, centre (0, 0) and radius 5 √ ( ) 3. (a) y = 3 x − 2 (b) a straight line ) 1 ( 4. (a) y = √ x + 1 (b) a straight line 3 5. (a) x2 − 4x − 12 + y2 = 0 or (x − 2)2 + y2 = 42 (b) a circle, centre (2, 0) and radius 4 6. (a) x2 + 6x − 16 + y2 = 0 or (x + 3)2 + y2 = 52 (b) a circle, centre (−3, 0) and radius 5 7. (a) 2x2 − 5x + 2 + 2y2 = 0 ( )2 ( )2 5 3 or x − + y2 = 4 4 ( ) 5 3 (b) a circle, centre , 0 and radius 4 4 8. (a) x2 + 2x − 1 + y2 = 0 or (x + 1)2 + y2 = 2 √ (b) a circle, centre (−1, 0) and radius 2 9. (a) x2 − x − y + y2 = 0 ( )2 ( )2 1 1 1 or x − + y− = 2 2 2 ( ) 1 1 1 (b) a circle, centre , and radius √ 2 2 2

5. 

8 1 −5 13 −2 −3 −3 1 45 7 −26 71



) 2. ) 4. )

 7 −1 8  3 1 7  4 7 −2 ( ) 9.3 −6.4 −7.5 16.9

 4.6 −5.6 −7.6 6.  17.4 −16.2 28.6  −14.2 0.4 17.2 ( ) ( ) −11 16 0 7. 8. 43 −27 34   ( ) 135 −6.4 26.1 9. 10.  −52  22.7 −56.9 −85   5 6 11.  12 −3  1 0   55.4 3.4 10.1 12.  −12.6 10.4 −20.4  −16.9 25.0 37.9 ( ) −6.4 26.1 13. A × C = 22.7 −56.9 ( ) −33.5 53.1 C×A = 23.1 −29.8 Hence, A × C ̸= C × A

Exercise 103 (page 258) 1. 4. 5. 6.

17 2. −3 3. −13.43 −5 + j3 (−19.75 + j19.79) or 27.96∠134.94◦ x = 6 or x = −1

Answers to Practice Exercises 879 Exercise 104 (page 259)  7 1   17 17  1.   4 3 17 17 ( ) 0.290 0.551 3. 0.186 0.097



5  77 2.   2 −4 7



4 7   3  −6 7 8

Exercise 107 (page 263) 1. (c) 2. (b) 3. (b) 4. (c) 5. (d) 6. (c) 7. (a) 8. (d) 9. (a) 10. (d) 11. (a) 12. (c) 13. (d) 14. (b) 15. (a) 16. (b) 17. (d) 18. (b) 19. (c) 20. (a)

Chapter 21 Exercise 108 (page 267)

Exercise 105 (page 261) 

−16 1.  −14 −24  −16 2.  14 −24 3. −212 6. −2 − j

 8 −34 −46 63  12 2  −8 −34 −46 −63  −12 2 4. −328 5. −242.83

7. 26.94∠ − 139.52◦ or (−20.49 − j17.49) 8. (a) λ = 3 or 4 (b) λ = 1 or 2 or 3

Exercise 106 (page 262) 

1.

2.

3.

4.

5.

6.

 4 −2 5  −7 4 7  6 0 −4   3 5 −1    6 −2 0    3  1 3  7 5  2  −16 14 −24  −8 −46 −12  −34 −63 2   2 3 1 − −3 42  5 5 3     3 1   −10 2  −18  10 2      2 − −6 −32 3  −16 14 −24 1  −8 −46 −12  − 212 −34 −63 2   2 3 1 − −3 42  5 5 3   15  3 1   −  −10 2 −18  923  10 2    2 − −6 −32 3

x = 4, y = −3 p = 1.2, q = −3.4 x = 1, y = −1, z = 2 a = 2.5, b = 3.5, c = 6.5 p = 4.1, q = −1.9, r = −2.7 I1 = 2, I2 = −3 s = 2, v = −3, a = 4 8. ¨x = 0.5, x˙ = 0.77, x = 1.4 1. 2. 3. 4. 5. 6. 7.

Exercise 109 (page 271) 1.

x = −1.2, y = 2.8

2.

m = −6.4, n = −4.9

3.

x = 1, y = 2, z = −1

4.

p = 1.5, q = 4.5, r = 0.5

5.

x=

6.

7 17 5 ,y= ,z=− 20 40 24 F1 = 1.5, F2 = −4.5

7.

I1 = 10.77∠19.23◦ A, I2 = 10.45∠ − 56.73◦ A

8.

i1 = −5, i2 = −4, i3 = 2

9.

F1 = 2, F2 = −3, F3 = 4

10. I1 = 3.317∠22.57◦ A, I2 = 1.963∠40.97◦ A, I3 = 1.010∠ − 148.32◦ A

Exercise 110 (page 272) Answers to Exercises 108 and 109 are as above

Exercise 111 (page 274) 1. ¨x = −0.30, x˙ = 0.60, x = 1.20 2. T1 = 0.8, T2 = 0.4, T3 = 0.2 3. Answers to Exercise 108 are as above 4. Answers to Exercise 109 are as above

880 Higher Engineering Mathematics Exercise 112 (page 275) 

Exercise 116 (page 293)



180.7 2.41 0 1. (Q0 ) =  2.41 8.032 0  in GPa 0 0 5

1. 17.35 N at 18.00◦ to the 12 N vector 2. 13 m/s at 22.62◦ to the 12 m/s velocity 3. 16.40 N at 37.57◦ to the 13 N force

Exercise 113 (page 280)

(

) ( ) −4 1 1. (a) λ1 = 3, λ2 = −2 (b) , 1 1 ( ) ( ) −3 2 2. (a) λ1 = 1, λ2 = 6 (b) , 1 1 ( ) ( ) 1 1 3. (a) λ1 = 1, λ2 = 2 (b) , −2 −1

7. 29.15 m/s at 29.04◦ to the horizontal 8. 9.28 N at 16.70◦ to the horizontal 9. 6.89 m/s at 159.56◦ to the horizontal

Exercise 117 (page 297) 1. (a) 54.0 N at 78.16◦

(b) 45.64 N at 4.66◦

2. (a) 31.71 m/s at 121.81◦

1. 83.5 km/h at 71.6◦ to the vertical 2. 4 minutes 55 seconds, 60◦ 3. 22.79 km/h. E 9.78◦ N

Exercise 119 (page 298)

Exercise 114 (page 280) 4. (b)

5. (c)

1. i − j − 4k 3. −i + 7j − k

2. 4i + j − 6k 4. 5i − 10k

Exercise 115 (page 286)

5. −3i + 27j − 8k 7. i + 7.5j − 4k 9. 3.6i + 4.4j − 6.9k

1. A scalar quantity has magnitude only; a vector quantity has both magnitude and direction. 2. Scalar 3. Scalar 4. Vector 5. Scalar

Exercise 120 (page 299)

6. Scalar 9. Vector

1. (c) 7. (d)

Chapter 22

7. Vector

(b) 19.55 m/s at 8.63◦

Exercise 118 (page 298)

7. (a) λ1 = 1, λ2 = 2, λ3 = 3       0 1 2 (b)  2 ,  1 ,  2  −1 0 1

3. (a)

6. 14.72 N at – 14.72◦ to the 5 N force

11. 21.07 knots, E 9.22◦ S

6. (a) λ1 = 1, λ2 = 2, λ3 = 4       −2 −2 0 (b)  1 ,  1 ,  1  0 1 1

2. (a)

5. 32.31 N at 21.80◦ to the 30 N displacement

10. 15.62 N at 26.33◦ to the 10 N force

4. (a) λ1 = 2, λ2 = 6, λ3 = −2       1 0 1 (b)  −2 ,  1  ,  1  1 1 0 5. (a) λ1 = 0, λ2 = 1, λ3 = 3       1 1 1 (b)  1 ,  0  ,  −2  1 −1 1

1. (d)

4. 28.43 N at 129.29◦ to the horizontal

8. Scalar

2. (c) 8. (a)

3. (a) 9. (b)

6. −5i + 10k 8. 20.5j − 10k 10. 2i + 40j − 43k

4. (a) 10. (c)

5. (d)

6. (b)

Answers to Practice Exercises 881 Exercise 125 (page 310)

Chapter 23

1.

12.07 sin(ωt + 0.297) V

2.

14.51 sin(ωt − 0.315) A

1. 4.5 sin(A + 63.5 ) 2. (a) 20.9 sin(ωt + 0.63) volts

3.

9.173 sin(ωt + 0.396) V

(b) 12.5 sin(ωt − 1.36) volts 3. 13 sin(ωt + 0.393)

4.

16.168 sin(ωt + 1.451) m

5.

(a) 371.95 sin(314.2t − 0.239) V

6.

(a) 11.44 sin(200πt + 0.715) V

Exercise 121 (page 303) ◦

(b) 50 Hz (b) 100 Hz

(c) 10 ms

Exercise 122 (page 305)

7.

(a) 79.73 sin(300πt − 0.536) V (c) 6.667 ms

1. 4.5 sin(θ + 63.5◦ ) 2. (a) 20.9 sin(ωt + 0.62) volts (b) 12.5 sin(ωt − 1.33) volts 3. 13 sin(ωt + 0.40)

(b) 150 Hz

(d) 56.37 V

8.

IN = 354.6∠32.41◦ A

9.

s = 85 sin(ωt + 0.49) mm

10. 15 V at 0.927 rad or 53.13◦

Exercise 126 (page 311)

Exercise 123 (page 306)

1. (b) 2. (d) 3. (c) 4. (b) 5. (a)

1. 4.472 sin(A + 63.44◦ ) 2. (a) 20.88 sin(ωt + 0.62) volts

Chapter 24

(b) 12.50 sin(ωt − 1.33) volts 3. 13 sin(ωt + 0.395)

Exercise 127 (page 317)

4. 11.11 sin(ωt + 0.324) 5. 8.73 sin(ωt − 0.173)

1. (a) 7 (b) 0

6. 1.01 sin(ωt − 0.698)A

3. (a) 11 (b) 11 5. (a) −16

(b) 38

7. (a) 143.82◦

Exercise 124 (page 308)

2. (a) −12 (b) −4 √ √ 4. (a) 13 (b) 14 √ 6. (a) 19 (b) 7.347

(b) 44.52◦

8. (a) 0.555, −0.832, 0 (b) 0, 0.970, −0.243 (c) 0.267, 0.535, −0.802 9. 11.54◦ 10. 66.40◦

1. 11.11 sin(ωt + 0.324) A 2. 8.73 sin(ωt − 0.173) V 3. i = 21.79 sin(ωt − 0.639) A

11. 53 N m

4. v = 5.695 sin(ωt + 0.695) V

Exercise 128 (page 320)

5. x = 14.38 sin(ωt + 1.444) m 6. (a) 305.3 sin(314.2t − 0.233) V

(b) 50 Hz

1. (a) 4i − 7j − 6k (b) −4i + 7j + 6k

7. (a) 10.21 sin(628.3t + 0.818) V

(b) 100 Hz

2. (a) 11.92

(b) 150 Hz

3. (a) −36i − 30j + 54k (b) 11i + 4j − k 4. (a) −22i − j + 33k (b) 18i + 162j + 102k

(c) 10 ms 8. (a) 79.83 sin(300πt + 0.352) V (c) 6.667 ms 9. 150.6 sin(ωt − 0.247) V

(b) 13.96

5. (i) −15 (ii) −4i + 4j + 10k (iv) 4i − 4j − 10k (v) 142.55

(iii) 11.49 ◦

6. (i) −62.5 (ii) −1.5i − 4j + 11k (iii) 11.80 (iv) 1.5i + 4j − 11k (v) 169.31◦

882 Higher Engineering Mathematics 7. 10 N m

Exercise 132 (page 333)

8. M = (5i + 8j − 2k) N m, |M| = 9.64 N m

1. x cos x + sin x 2. 2xe2x (x + 1) ( ) 3. x(1 + 2 ln x) 4. 6x2 cos 3x − x sin 3x ( ) ( ) √ 3 5. x 1 + ln 3x 6. e3t 4 cos 4t + 3 sin 4t 2 ( ) 1 4θ 7. e + 4 ln 3θ θ {( ) } 1 8. et + ln t cos t − ln t sin t t 9. 8.732 10. 32.31

9. v = −14i + 7j + 12k, [v] = 19.72 m/s 10. 6i − 10j − 14k, 18.22 m/s

Exercise 129 (page 322) 1. r = (5 + 2λ)i + (7λ − 2)j + (3 − 4λ)k, r = 9i + 12j − 5k x−5 y+2 3−z 2. = = =λ 2 7 4 ( ) ( ) ( ) 1 1 3. r = 1 + 4λ i + 2λ − 1 j + 4 − 3λ k 3 5 ) ) ) 1( 1( 1( 4. r = λ − 1 i + 1 − 5λ j + 1 + 4λ k 2 4 3

Exercise 130 (page 322) 1. (b)

2. (c)

3. (a)

4. (d)

5. (a)

6. (b)

Chapter 25 Exercise 131 (page 332) 1. (a) 25x4

(b) 8.4x2.5

(c) −

1 x2

8 2. (a) 3 (b) 0 (c) 2 x √ 1 2 3 3. (a) √ (b) 5 x2 (c) − √ x x3 1 4. (a) √ (b) 2(x − 1) (c) 6 cos 3x 3 4 x 15 5. (a) 8 sin 2x (b) 12e6x (c) − 5x e 4 ex + e−x 1 1 6. (a) (b) (c) − 2 + √ x 2 x 2 x3 7. −1, 16 ( ) 1 3 8. , 2 4 4 2 6 9. (a) − 3 + + 10 sin 5θ − 12 cos 2θ + 3θ θ θ e (b) 22.30 10. 3.29 mg k 12. 27.0 volts 11. x =

Exercise 133 (page 335) ) 6 ( x sin 3x + cos 3x x4 cos x √ √ + x sin x 2 2(1 − x ) 2 x 3. ( 4. )2 cos2 x x2 + 1 √ { } 3 θ 3 sin 2θ − 4θ cos 2θ 5. 4 sin2 2θ ( ) 1 1 6. √ 1 − ln 2t 2 t3 4x {( ) } 2e 7. 1 + 4x sin x − x cos x 2 sin x 8. −18 9. 3.82 1.

x cos x − sin x x2

2. −

Exercise 134 (page 336) ( )5 1. 12 2x − 1 ( ) 3. 6 cos 3θ − 2 ( ) 5 2 − 3x2 5. ( )6 x3 − 2x + 1( ) 7. −20t cosec2 5t2 + 3 9. 2 sec2 θetan θ 11. (a) 24.21 mm/s

( )4 ( ) 2. 5 2x3 − 5x 6x2 − 5 4. −10 cos4 α sin α 6. 10e2t+1 ( ) 8. 18 sec2 3y + 1 10. 1.86

(b) −70.46 mm/s

Exercise 135 (page 338) 1. (a) 36x2 + 12x (b) 72x + 12 4 12 6 1 2. (a) − 5 + 3 + √ (b) −4.95 5 t t 4 t3 t V V − t 3. (a) − e− CR (b) e CR R CR2 2 4. (a) −(12 sin 2t + cos t) (b) − 2 θ ( ) ( )2 5. (a) 4 sin2 x − cos2 x (b) 48 2x − 3 6. 18

Answers to Practice Exercises 883 7. 8. 9.

Proof Proof Proof

10. M = −(PL +

Exercise 140 (page 351) WL2 Wx2 ) + x(P − WL) − 2 2

1. 2. 3. 4. 5.

Exercise 136 (page 338) 1. (b) 2. (d) 3. (b) 4. (c) 5. (b) 6. (c) 7. (d) 8. (c) 9. (a) 10. (b) 11. (a) 12. (a) 13. (d) 14. (d) 15. (a) 16. (d) 17. (a) 18. (c) 19. (b) 20. (c)

6. 7. 8. 9. 10. 11. 12.

Chapter 26 Exercise 137 (page 341) 1. 3000π A/s

2. (a) 0.24 cd/V

(b) 250 V

3. (a) −625 V/s (b) −220.5 V/s 4. −1.635 Pa/m 5. −390 m3 /min

13.

(3, −9) Minimum (1, 9) Maximum (2, −1) Minimum (0, 3) Minimum, (2, 7) Maximum ( ) 2 2 Minimum at , 3 3 (3, 9) Maximum (2, −88) Minimum, (−2.5, 94.25) Maximum (0.4000, 3.8326) Minimum (0.6931, −0.6137) Maximum ( ) 2 22 (1, 2.5) Minimum, − , 4 Maximum 3 27 (0.5, 6) Minimum Maximum of 13 at 337.38◦ , Minimum of – 13 at 157.38◦ Proof

Exercise 141 (page 354) 1. 54 km/h 4. 11.42 m2

Exercise 138 (page 344) 100 m/s (b) 4 s (c) 200 m (d) −100 m/s 90 km/h (b) 62.5 m 4 s (b) 3 rads 3 m/s, −1 m/s2 (b) 6 m/s, −4 m/s2 (c) 0.75 s ω = 1.40 rad/s (b) α = −0.37 rad/s2 t = 6.28 s 6. 6 m/s, −23 m/s2 (b) 117 m/s, 97 m/s2 1 (c) 0.75 s or 0.4 s (d) 1.5 s (e) 75 m 6 7. 3 s 1 8. (a) 12 rad/s (b) 48 rad/s2 (c) 0 or s 3 9. 162 kJ 10. 2.378 m/s 11. (a) 959.2 m (b) 132.3 m/s (c) 3.97 m/s2 1. 2. 3. 4. 5.

(a) (a) (a) (a) (a) (c) (a)

2. 90 000 m2

3.

48 m

5. Radius = 4.607 cm, height = 9.212 cm 6. 6.67 cm 7. Proof 8. Height = 5.42 cm, radius = 2.71 cm 9. 44.72 10. 42.72 volts 11. 50.0 miles/gallon, 52.6 miles/hour 12. 45◦

13. 0.607

14. 1.028

Exercise 142 (page 357) (

) ( ) 1 1 1. , −1 2. − , 4 3. (0, 0) 2 4 4. (3, −100) 5. (2, 0.541) ( ) ( ) 6. Max ( at ) 0, 10 , Min at 2, −2 , point of inflexion at 1, 4

Exercise 143 (page 358) Exercise 139 (page 347) 1. 3. 5. 7. 9.

−2.742, 4.742 −1.721, 2.648 1.147 2.05 4.19

2. 4. 6. 8. 10.

2.313 −1.386, 1.491 −1.693, −0.846, 0.744 0.0399 2.9143

1. (a) y = 4x − 2 (b) 4y + x = 9 2. (a) y = 10x − 12 (b) 10y + x = 82 3 3. (a) y = x + 1 (b) 6y + 4x + 7 = 0 2 4. (a) y = 5x + 5 (b) 5y + x + 27 = 0 2 5. (a) 9θ + t = 6 (b) θ = 9t − 26 or 3θ = 27t − 80 3

884 Higher Engineering Mathematics Exercise 144 (page 359)

2. (a)

1. (a) −0.03 (b) −0.008 3. (a) 60 cm3

2. −0.032, −1.6%

3. (a) 6 cos 2θ

(b) 12 cm2

4. (a) −6.03 cm2

5 dt 2t dx

(b) −18.10 cm3

4. (a) −

5. 12.5%

(b) dθ dy

3 2y+1 dy e 2 dx

(c) 6 sec2 3y

√ dx (b) 6 x dy

6 dx 2 (3x + 1) du

(c) −

dy dx

2 dt et dy

(b) 6 sec 2θ tan 2θ

dθ du

1 dy (c) − √ y3 du

Exercise 145 (page 360) 1. (c) 7. (d) 12. (a)

2. (d) 8. (d) 13. (c)

3. (c) 9. (b) 14. (c)

4. (b) 10. (a) 15. (b)

5. (a) 11. (a)

Exercise 146 (page 366) ) 1( 2t − 1 3 1 3. (a) − cot θ 4 4. 4

2. 2

1 7. y = − x + 5 4

Exercise 147 (page 367) 1. (a) 3.122 (b) −14.43 3. y = −x + π 5. (a) 13.14 (b) 5.196

(

) ( ) dy 2 dy 1. 3xy 3x + 2y 2. x −y dx 5x2 dx ( ) ( ) 3 dv cos 3x dy √ 3. v − u 4. 3 − 9 y sin 3x √ 4v2 du 2 y dx ( ) dx 2 x 5. 2x + 3 ln y y dy

Exercise 151 (page 373)

1 (b) − cos ec3 θ 16 5. −6.25

6. y = −1.155x + 4

Exercise 150 (page 371) 2

Chapter 27

1.

6. (d)

2x + 4 1. 3 − 2y −(x + sin 4y) 4. 4x cos 4y 1 − 2 ln y 7. 2x 3+ − 4y3 y 10. ±1.5

3. (c)

4. (b)

5. (a)

Exercise 149 (page 370) (b) −8 sin 4θ

2. (a)

9. ± 0.5774

11. −6

3. (c)

4. (b)

5. (c)

Exercise 153 (page 376) 2 2x − 5 10(x + 1) 4. 5x2 + 10x − 7 3 7. x

Chapter 28

dy dx

5

Chapter 29

1.

1. (a) 15y4

8.

Exercise 152 (page 374)

Exercise 148 (page 368) 2. (d)

5.

√ 5 3. − 2 x(4y + 9x) 6. cos y − 2x2

2. y = −2x + 3 4. 0.02975 1. (d)

1. (a)

2.

3 1 − 6y2 4x − y 3y + x

dθ dx

(c)

1 dk √ 2 k dx

2. −3 tan 3x 5.

1 x

8. 2 cot x

9x2 + 1 3x3 + x 2x 6. 2 x −1 12x2 − 12x + 3 9. 4x3 − 6x2 + 3x 3.

Answers to Practice Exercises 885 Exercise 154 (page 378) (x − 2)(x + 1) 1. (x − 1)(x + 3)

2.

3.

4. 5. 6. 7. 8.

Chapter 30

{

1 1 + (x − 2) (x + 1) } 1 1 − − (x − 1) (x + 3) { (x + 1)(2x + 1)3 1 6 + (x − 3)2 (x + 2)4 (x + 1) (2x + 1) } 2 4 − − (x − 3) (x + 2) √( ) (2x − 1) x + 2 { 2 1 √( )3 (2x + 1) + 2(x + 2) (x − 3) x + 1 } 1 3 − − (x − 3) 2(x + 1) { } e2x cos 3x 1 √ 2 − 3 tan 3x − 2(x −}4) (x − 4) { 1 3θ sin θ cos θ + cos θ − tan θ θ { } 2x4 tan x 4 1 1 + −2− e2x ln 2x x sin x cos x x ln 2x 13 16 −6.71

Exercise 155 (page 379)

( ) 1. 2x2x 1 + ln x { } ( )x 2x 2. 2x − 1 + ln(2x − 1) 2x − 1 { } √ 1 ln(x + 3) 3. x (x + 3) − x(x + 3) x2 ( ) 1 4. 3x4x+1 4 + + 4 ln x x 5. Proof 1 6. 3 7. Proof

2. (a)

3. (a)

4. (d)

1. (a) 6 ch 2x

10 5 t t sec h5x tanh 5x (b) − cosech coth 3 16 2 2 −14cosech2 7θ 3 θ θ 2 coth x (b) sec h cosech 8 2 2 ( ) ( ) 2 sh2 2x + ch2 2x (b) 6e2x sec h2 2x + th2x ( ) 2 cos 2tsh2t + ch2t sin 2t 12xch4x − 9sh4x (b) 2x4 cos2 2t

2. (a) − (c) 3. (a) 4. (a) 5. (a)

Exercise 158 (page 384) 1. (b)

2. (a)

3. (d)

4. (b)

5. (c)

Chapter 31 Exercise 159 (page 390) 1. (a) √(

4

)

(b) √(

1

) 1 − 16x2 4 − x2 −3 −2 2. (a) √( ) (b) √( ) 1 − 9x2 3 9 − x2 6 1 3. (a) (b) √ 2 1 + 4x 4 x(1 + x)

6. (a)

−6 1 + 4t2

(b)

4 √( ) x 9x2 − 16 −2 (b) √( ) x x4 − 1 (b)

−1 √( ) θ θ2 − 1

1 + x2 ) 1 − x2 + x4 6x −1 8. (a) √( ) + 2 sin 3x 2 1 − 9x t −1 (b) √( ) + 2t sec 2t 4t2 − 1 ( ) 2θ2 9. (a) 2θ cos−1 θ2 − 1 − √( ) 2 − θ2 ( ) 1 − x2 (b) − 2xtan−1 x 1 + x2 7. (

5. (c)

(c) 36 sec h2 9t

(b) 10 sh 5θ

2 4. (a) √( ) t 4t2 − 1 −5 5. (a) √( ) θ θ2 − 4

Exercise 156 (page 380) 1. (b)

Exercise 157 (page 384)

886 Higher Engineering Mathematics √) −2 t 1 + √ cot−1 t 1 + t2 t √ 1 cosec−1 x − √ 2 (x − 1)    1  3x −1 √ − 2 sin 3x  x3  (1 − 9x2 ) ( −1 ) x −1 + √ cos x 1 − x2 1 − x2

( 10. (a) (b) 11. (a)

(b)

[ ] 1 1 −1 √ √ + 4sech t t3 1−t 1 + 2x tanh−1 x (b) ( )2 1 − x2 12. 2x 11. (a) −

x 3 2x + c (b) sinh−1 +c 3 2 5 x t 14. (a) cosh−1 + c (b) cosh−1 √ + c 4 5 1 −1 θ 3 x 15. (a) tan + c (b) √ tanh−1 √ + c 6 6 2 8 8 13. (a) sinh−1

Exercise 160 (page 392) 1. (a) 0.4812

(b) 2.0947

(c) 0.8089

2. (a) 0.6931

(b) 1.7627

(c) 2.1380

3. (a) 0.2554

(b) 0.7332

(c) 0.8673

Exercise 161 (page 395) 1 4 1. (a) √( ) (b) √( ) x2 + 9 16x2 + 1 2 1 2. (a) √( ) (b) √( ) t2 − 9 4θ2 − 1 10 9 3. (a) (b) 25 − 4x2 1 − 9x2 −4 1 √( 4. (a) √( ) (b) ) x 16 − 9x2 2x 1 − 4x2 −4 −1 √( 5. (a) √( ) (b) ) 2 x x + 16 2x 16x2 + 1 14 3 ) 6. (a) (b) ( 2 49 − 4x 4 1 − 9t2 2 1 7. (a) √( ) (b) √( ) x2 − 1 2 x2 + 1 −1 √ 8. (a) (b) 1 (x − 1) [x(2 − x)] −1 )√ 9. (a) ( (b) −cosec x t − 1 2t − 1 θ −1 10. (a) √( ) + sinh θ 2 θ +1 √ x cosh−1 x √ (b) √ + 2 x x2 − 1

Exercise 162 (page 396) 1. (b)

2. (d)

3. (c)

4. (a)

5. (c)

6. (b)

Chapter 32 Exercise 163 (page 399) 1. 2. 3. 4. 5. 6. 7. 8. 9.

∂z ∂z = 2y, = 2x ∂x ∂y ∂z ∂z = 3x2 − 2y, = −2x + 2y ∂x ∂y ∂z 1 ∂z x = , =− 2 ∂x y ∂y y ∂z ∂z = 4 cos(4x + 3y), = 3 cos(4x + 3y) ∂x ∂y ∂z 2y ∂z 1 1 = 3x2 y2 + 3 , = 2x3 y − 2 − 2 ∂x x ∂y x y ∂z ∂z = −3 sin 3x sin 4y, = 4 cos 3x cos 4y ∂x ∂y ∂V 1 2 ∂V 2 = πr , = πrh ∂h 3 ∂r 3 Proof ( ) ( ) { ( ) ∂z nπb nπ nπb = sin x c cos t ∂x L L L ( )} nπb −k sin t L { ( ) ( nπ ) ∂z ( nπ ) nπb = cos x k cos t ∂y L L L ( )} nπb +c sin t L

Answers to Practice Exercises 887 ∂k A∆H T∆S−∆H = e RT ∂T RT2 ∂A k∆H ∆H−T∆S (b) =− e RT ∂T RT2 ∂(∆S) ∆H (c) =− 2 ∂T T ( ) ∂(∆H) k (d) = ∆S − R ln ∂T A

10. (a)

Exercise 167 (page 407) 1. +226.2 cm3 /s 3. 515.5 cm/s 5. 17.4 cm2 /s

2. 2520 units/s 4. 1.35 cm3 /s

Exercise 168 (page 409) Exercise 164 (page 402) (b) 18 (c) −12 (d) −12 2 2 2. (a) − 2 (b) − 2 (c) 0 (d) 0 x y 4y 4x 3. (a) − (b) (x + 3)3 (x + 3)3 2(x − y) 2(x − y) (c) (d) (x + y)3 (x + y)3 1. (a) 8

4. (a) sinh x cosh 2y

(b) 4 sinh x cosh 2y

(c) 2 cosh x sinh 2y (d) 2 cosh x sinh 2y ( ) 5. (a) 2 − x2 sin(x − 2y) + 4x cos(x − 2y) (b) −4x2 sin(x − 2y) ( ) −x −x 1 1 6. √( )3 , √y2 − x2 y2 + (y2 − x2 ) , 2 2 y −x ∂2z ∂2z y = = √( )3 ∂x∂y ∂y∂x y2 − x2 1 7. − √ or −0.7071 2 8. Proof

Exercise 165 (page 403)

1. +21 watts 4. +1.35 cm4 7. +2.2%

2. +2% 5. −0.179 cm

3. −1% 6. +6%

Exercise 169 (page 410) 1. (b)

2. (c)

3. (a)

4. (b)

5. (d)

Chapter 34 Exercise 170 (page 415) 1. Minimum at (0, 0) 2. (a) Minimum at (1, −2) (b) Saddle point at (1, 2) (c) Maximum at (0, 1) 3. Maximum point at (0, 0), saddle point at (4, 0) 4. Minimum at (0, 0) ( ) 1 1 5. Saddle point at (0, 0), minimum at , 3 3

1. (c) 2. (b) 3. (c) 4. (d) 5. (a)

Chapter 33 Exercise 166 (page 405) 1. 3x2 dx + 2ydy 2. (2y + sin x)dx + 2xdy ( ) ( ) 2y 2x x 3. dx − dy 4. ln y dx + dy (x + y)2 (x + y)2 y ( ) ( √ ) 1 x 5. y + √ dx + x − 2 dy y 2y x 6. b(2 + c)da + (2a − 6bc + ac)db + b(a − 3b)dc 7. du = cot(xy)[y dx + x dy]

Exercise 171 (page 418) 1. Minimum at (−4, 4) 2. 4 m by 4 m by 2 m, surface area = 48 m2 3. Minimum at (1, 0), minimum at (−1, 0), saddle point at (0, 0) 4. Maximum at (0, 0), saddle point at (4, 0) 5. Minimum at (1, 2), maximum at (−1, −2), saddle points at (1, −2) and (−1, 2) 6. 150 m2

888 Higher Engineering Mathematics Chapter 35

Chapter 36

Exercise 172 (page 427)

Exercise 175 (page 433)

7x 2 +c 2 2 3 5 4 2. (a) x + c (b) x +c 15 24 3x2 θ3 3. (a) − 5x + c (b) 4θ + 2θ2 + + c 2 3 4 1 4. (a) − + c (b) − 3 + c 3x 4x 4√ 5 1√ 4 9 5. (a) x + c (b) x +c 5 9 10 15 √ 5 6. (a) √ + c (b) x+c 7 t 3 7 7. (a) sin 2x + c (b) − cos 3θ + c 2 3 1 1 8. (a) tan 3x + c (b) − cot 4θ + c 4 2 5 1 9. (a) − cosec 2t + c (b) sec 4t + c 2 3 3 2x −2 10. (a) e + c (b) +c 8 15e5x 2 u2 11. (a) ln x + c (b) − ln u + c 3 2 √ √ 18 √ 5 12. (a) 8 x + 8 x3 + x +c 5 3 1 4t (b) − + 4t + +c t 3

1 1. 1 square units 3 1 3. 2 square units 2

1. (a) 4x + c

(b)

5 2. 20 square units 6

Exercise 176 (page 434) 2 1. 2 km 3 3. 15.92 mA, 17.68 mA

2. 2.198 √ E21 + E23 4. 2

5. 0

Exercise 177 (page 436) 1. 1.5π cubic units 3. (a) 329.4π (b) 81π 4. (b) 0.352

2 2. 170 π cubic units 3

(c) 0.419 square units

(d) 0.222 K

Exercise 173 (page 428) (b) −0.5

1. (a) 105 3. (a) 0

(b) 4

5. (a) 0.2352 7. (a) 19.09

2. (a) 6 4. (a) 1

(b) 2.598 (b) 2.457

9. 55.65 J/K

1 3 (b) 4.248 (b) −1

6. (a) 0.2527

(b) 2.638

8. (a) 0.2703

(b) 9.099

Exercise 178 (page 437) 1. (2.50, 4.75) 4. (1, −0.4)

2. (3.036, 24.36) 5. (2.4, 0)

3. (2, 1.6)

10. Proof 12. 77.7 m3

11. 7.26

Exercise 174 (page 429) 1. (d)

2. (c)

3. (a)

4. (a)

5. (c)

6. (d)

7. (b)

8. (b)

9. (b)

10. (d)

Exercise 179 (page 440) 1. 189.6 cm3 2. On the centre line, distance 2.40 cm from the centre, i.e. at co-ordinates (1.70, 1.70) 3. (a) 45 square units (b)(i) 1215π cubic units (ii) 202.5π cubic units (c) (2.25, 13.5) 4. 64.90 cm3 , 16.86%, 506.2 g

Answers to Practice Exercises 889 Exercise 180 (page 446) 1. (a) 72 cm4 , 1.73 cm

Exercise 183 (page 457)

(b) 128 cm4 , 2.31 cm

1. 1.784

2. 0.88

3. 0.53

4. 0.061

(c) 512 cm4 , 4.62 cm 2. (a) 729 cm4 , 3.67 cm

(b) 2187 cm4 , 6.36 cm

(c) 243 cm4 , 2.12 cm 3. (a) 201 cm4 , 2.0 cm

Exercise 184 (page 459) 1 9 1 6. 2 1.

(b) 1005 cm4 , 4.47 cm

4. 3927 mm4 , 5.0 mm 5. (a) 335 cm4 , 4.73 cm

2. 1

3. 1

4. −1

1 3

8. 1

9.

7.

5.

1 2

1 3

(b) 22 030 cm4 , 14.3 cm

(c) 628 cm4 , 7.07 cm

Exercise 185 (page 459)

6. 0.866 m

1. (c) 2. (d) 3. (c) 4. (d) 5. (b) 6. (b) 7. (a)

7. 0.245 m4 , 0.559 m 8. 14 280 cm4 , 5.96 cm 9. (a) 12 190 mm4 , 10.9 mm (b) 549.5 cm4 , 4.18 cm 10. (a) IAA = 4224 cm4 , (b) IBB = 6718 cm4 , ICC = 37 300 cm

4

11. 1350 cm4 , 5.67 cm

Exercise 181 (page 447) 1. (d) 2. (a) 3. (c) 7. (b) 8. (a) 9. (d) 13. (a) 14. (b) 15. (b)

4. (d) 5. (b) 6. (d) 10. (c) 11. (c) 12. (c)

Chapter 38 Exercise 186 (page 463) 1 1. − cos(4x + 9) + c 2 4 3. tan(3t + 1) + c 3 3 5. − ln(2x − 1) + c 2 7. 227.5 9. 0.9428

3 sin(2θ − 5) + c 2 1 4. (5x − 3)7 + c 70 2.

6. e3θ+5 + c 8. 4.333 10. 0.7369

11. 1.6 years

Chapter 37

Exercise 187 (page 465)

Exercise 182 (page 455) 1. 2. 3. 4. 5. 6. 7. 8. 9.

4 4 8 7 sin 2x = 2x − x3 + x5 − x 3 15 315 9 27 81 cosh 3x = 1 + x2 + x4 + x6 2 8 80 x x2 ln 2 + + 2 8 32 256 6 1 − 8t2 + t4 − t 3 45 3 9 9 1 + x2 + x4 + x6 2 8 16 10 1 + 2x2 + x4 3 5 2 1 + 2θ − θ 2 1 4 2 2 x − x + x6 3 45 81 + 216t + 216t2 + 96t3 + 16t4

1. 3. 4. 5. 7.

)6 1 ( 2 5 2x − 3 + c 2. − cos6 t + c 12 6 1 1 2 2 sec 3x + c or tan 3x + c 2 2 √( )3 2 3t2 − 1 + c 9 1 ( )2 3 ln θ + c 6. ln(sec 2t) + c 2√ 2 ( ) 4 e4 + 4 + c 8. 1.763

9. 0.6000 {√( ) } 11. 2πσ 92 + r2 − r 13. Proof

10. 0.09259 8π2 IkT h2 14. 11 min 50 s 12.

Exercise 188 (page 466) 1. (d)

2. (d)

3. (a)

4. (b)

5. (c)

890 Higher Engineering Mathematics Exercise 193 (page 474)

Chapter 39

1.

Exercise 189 (page 470)

( ) ( ) 1 sin 4x 3 sin 2t x− +c 2. t+ +c 2 4 2 2 ( ) 1 3. 5 tan 3θ − θ + c 4. −(cot 2t + 2t) + c 3 π π 5. or 1.571 6. or 0.3927 2 8 7. −4.185 8. 0.6311

3 −1 t tan +c 2 2

3. 2.356

2.

5 3θ tan−1 +c 12 4

4. 2.457

1.

Exercise 190 (page 471)

Exercise 194 (page 476)

√ x 3 5 −1 √ 1. 2 sinh +c 2. sinh x+c 4 3 5 9 x x√ 2 3. sinh−1 + [x + 9] + c 2 3 2 25 2t t √ 2 4. sinh−1 + [4t + 25] + c 4 5 2 −1

cos3 θ sin3 2x +c 2. sin 2x − +c 3 3 7 5 2 2 cos x cos x 5. 3.525 6. 4.348 3. − cos3 t + cos5 t + c 4. − + +c 3 5 5 7 3θ 1 1 5. − sin 4θ + sin 8θ + c 4 4 32 t 1 Exercise 195 (page 477) 6. − sin 4t + c 8 32 t 3 2x 1. cosh−1 + c 2. cosh−1 +c 4 2 3 Exercise 191 (page 472) ) 9 θ √( 2 θ ( ) 3. θ − 9 − cosh−1 + c 1 cos 7t cos 3t sin 2x sin 4x 2 2 3 √( 1. − + + c 2. − +c ) 2 7 3 2 4 25 25 2θ 2 [ ] 4. θ θ − − cosh−1 +c 4 4 5 3 sin 7x sin 5x 3. + +c 2 7 5 5. 2.634 6. 1.429 ( ) 1 cos 2θ cos 6θ 4. − +c 4 2 6 3 5. or 0.4286 6. 0.5973 7 Exercise 196 (page 478) 1. − cos θ +

7. 0.2474

8. −0.1999

1. (a)

2. (a)

3. (b)

4. (c)

5. (d)

Exercise 192 (page 473) t x +c 2. 3 sin−1 + c 2 3 ) x x √( 3. 2 sin−1 + 4 − x2 + c 2 2 ) 8 3t t √( 4. sin−1 + 16 − 9t2 + c 3 4 2 π 5. or 1.571 6. 2.760 2 1. 5 sin−1

Chapter 40 Exercise 197 (page 480)

(

1. 2 ln(x − 3) − 2 ln(x + 3) + c or ln { 2. 5 ln(x + 1) − ln(x − 3) + c or ln

x−3 x+3

)2

(x + 1)5 (x − 3)

3. 7 ln(x + 4) − 3 ln(x + 1) − ln(2x − 1) + c ( ) ( )7 x+4 or ln ( )3 ( ) +c x + 1 2x − 1

} +c

Answers to Practice Exercises 891 4.

x + 2 ln(x + 3) + 6 ln(x − 2) + c { ( )6 } or x + ln (x + 3)2 x − 2 +c

Exercise 202 (page 488) 

1.

2

6.

3x − 2x + ln(x − 2) − 5 ln(x + 2) + c 2 0.6275

7.

0.8122

8.

1 3

9.

Proof

5.

2 −1  tan  3 

2.

3.

10. 19.05 ms 4.

Exercise 198 (page 482) 7 +c (x + 1) 10 2 2. 5 ln(x − 2) + − +c (x − 2) (x − 2)2 3. 1.663 4. 1.089 5. Proof 1. 4 ln(x + 1) +

5. 7.

5 tan

 θ +4  2 +c 3

√  x + 2 − 3 1 2 √ ln  √ +c x 3 tan + 2 + 3 2  √  p  tan − 4 − 11  1 2 √ ln +c √ 11  tan p − 4 + 11  2  √  θ 3 tan − 4 − 7 1   2 √ ln  +c √ θ 7 3 tan − 4 + 7 √ 2  t   2 + tan  1 2 +c √ ln √ 6. Proof t 2 2   2 − tan   2 Proof tan

Chapter 42 Exercise 199 (page 483)

Exercise 203 (page 493)

( ) 3 x 1. ln x2 + 7 + √ tan−1 √ − ln(x − 2) + c 7 7

1.

2. 0.5880

3. 0.2939

4. 0.1865

5. Proof

Exercise 200 (page 483) 1. (a)

2. (c)

3. (d)

4. (b)

5. (c)

Chapter 41 Exercise 201 (page 487) 1. −

2

+c θ 1 + tan 2( ) 2 1 α −1 √ tan 3. √ tan +c 2 5 5   x   2 tan − 1 1 2 4. ln +c 5  tan x + 2  2

( ) ( ) e2x 1 4 1 x− +c 2. − e−3x x + +c 2 2 3 3 3. −x(cos x + sin x + c ) 5 1 4. θ sin 2θ + cos 2θ + c 2 ( 2) 3 2t 2 1 5. e t − t + +c 6. 16.78 2 2 7. 0.2500

8. 0.4674

9. 15.78

Exercise 204 (page 495)  x   tan  2 2. ln +c x  1 + tan  2

( ) 2 3 1 x ln x − +c 2. 2x(ln 3x − 1) + c 3 3 ) 2 cos 3x ( 3. 2 − 9x2 + x sin 3x + c 27 9 2 5x 4. e (2 sin 2x + 5 cos 2x) + c 29 1.

5. 2 [θ tan θ − ln(sec θ)] + c

6. 0.6363

7. 11.31

8. −1.543

9. 12.78 11. C = 0.66, S = 0.41

10. Proof

892 Higher Engineering Mathematics Exercise 205 (page 496) 1. (c)

2. (a)

3. (c)

4. (b)

5. (d)

Chapter 44 Exercise 210 (page 508)

Chapter 43 Exercise 206 (page 498) 1. ex (x4 − 4x3 + 12x2 − 24x + 24) + c ( ) 13 32 3 3 2. e2t t − t + t− +c 3. 6.493 2 4 4 8

1. 4. 7. 9.

12 2. 3.5 −174 5. 405 15π or 47.12 5 10. 170 cm4

Exercise 211 (page 510) 1. 15 4

5.

Exercise 207 (page 500) 1. x5 sin x + 5x4 cos x − 20x3 sin x − 60x2 cos x + 120x sin x + 120 cos x + c 2. −134.87 3. −x5 cos x + 5x4 sin x + 20x3 cos x − 60x2 sin x − 120x cos x + 120 sin x + c 4. 62.89

3. 0.5 6. −157.5 8. 112

π or 48.70 2

2. 15

3. 60

4. −8

6. −9

7. 8

8. 18

Exercise 212 (page 510) 1. (a)

2. (b)

3. (d)

4. (c)

5. (b)

Chapter 45 Exercise 213 (page 514)

Exercise 208 (page 503) 1 6 8 1. − sin6 x cos x − sin4 x cos x − sin2 x cos x 7 35 35 16 − cos x + c 35 2. 4 8 15 1 5 5 4. cos5 x sin x+ cos3 x sin x+ cos x sin x 6 24 16 5 + x+c 16 16 5. 35 3.

Exercise 209 (page 505)

1. 1.569

2. 6.979

3. 0.672

4. 0.843

Exercise 214 (page 516) 1. 3.323

2. 0.997

3. 0.605

4. 0.799

Exercise 215 (page 519) 1.

1.187

4. 6. 7.

0.571 5. 1.260 (a) 1.875 (b) 2.107 (c) 1.765 (a) 1.585 (b) 1.588 (c) 1.583

2.

1.034

8.

(a) 10.194

(b) 10.007

3. 0.747

(c) 10.070

9. (a) 0.677 (b) 0.674 (c) 0.675 10. 28.8 m 11. 0.485 m

8 13 π 8 2. − or 0.08127 3. 105 15 4 315 4. x(ln x)4 − 4x(ln x)3 + 12x(ln x)2 − 24x ln x + 24x + c

Exercise 216 (page 520)

5. Proof

1. (a)

1.

2. (c)

3. (c)

(d) 1.916 (d) 1.585

Answers to Practice Exercises 893 Chapter 46

Chapter 47 Exercise 222 (page 536)

Exercise 217 (page 526) 1. Sketches

2. y = x2 + 3x − 1

Exercise 218 (page 528) 1 3 x3 sin 4x − x2 + c 2. y = ln x − + c 4 2 6 x2 1 1 1 3. y = 3x − − 4. y = cos θ + 2 3 2 ( 2 ) 1 2 2 3 2 x3 5. y = x − 4x + x + 4 6. y = x − − 1 6 e 2 6 7. v = u + at 8. 15.9 m 1. y =

Exercise 219 (page 529) 1. x =

1 ln(2 + 3y) + c 3

y2 3. + 2 ln y = 5x − 2 2 ( ) 5. x = a 1 − e−kt t

6. (a) Q = Q0 e− CR

2. tan y = 2x + c

( 3 ) 1 x − y3 1. − ln = ln x + c 3 x3 2. y = x(c − ln x) ( ) 1 3. x2 = 2y2 ln y + 2 ) ( 1 2y y2 4. − ln 1 + − 2 = ln x + c or x2 + 2xy − y2 = k 2 x x 5. x2 + xy − y2 = 1

Exercise 223 (page 537) 1. y4 = 4x4 (ln x + c) [ ( ) ( )] 1 3 13y − 3x y−x 2. ln − ln = ln x + c 5 13 x x 3. (x + y)2 = 4x3 ( ) 5. y3 = x3 3 ln x + 64

Chapter 48

(b) 9.30 C, 5.81 C

Exercise 224 (page 540)

8. 8 m 40 s

1. y = 3 +

Exercise 220 (page 532)

3. y =

2. y2 − y = x3 + x

1. ln y = 2 sin x + c 1 1 3. ey = e2x + 2 2

) 4. ln x2 y = 2x − y − 1

5. Proof

6. y = √(

(

7. y2 = x2 − 2 ln x + 3

1

)

1 − x2 ( t ) 8. (a) V = E 1 − e− CR (b) 13.2 V

9. 3

Exercise 221 (page 533) 2. (d) 7. (c)

6. Proof

4. Proof

7. 273.3 N, 2.31 rad

1. (b) 6. (a)

4. Proof

3. (c) 8. (b)

4. (b) 9. (d)

5. (c) 10. (a)

c x

2. y =

5t c + 2 t

1 2 + ce−x 2

x3 x 47 − + 5 3 15x2 1 1 6. y = x + + ce2x 2 4 4. y =

5. y = 1 + ce−x /2 2

Exercise 225 (page 541) ) 1 1( + cos2 x 2. θ = sin t − π cos t 2 t 3. Proof 4. Proof ( ) b bt b 5. Proof 6. v = 2 − + u − 2 e−at a a a 7. Proof   mt mt ( ) − − b 8. C = + d 1 − e a  + c0 e a m { } 1 e−t e−ct 9. v = k − + c c − 1 c(c − 1) 1. y =

894 Higher Engineering Mathematics Exercise 227 (page 553)

Chapter 49

1.

Exercise 226 (page 548) 1.

x

2.

3. (a)

y

x

(y′ )0

y

1.0

2

1

1.1

2.10454546

1.08677686

1.2

2.216666672

1.152777773

1.3

2.33461539

1.204142008 1.244897956

1.0

2

1.1

2.1

1.2

2.209091

1.3

2.325000

1.4

2.457142862

1.4

2.446154

1.5

2.5883333338

1.5

2.571429

x

y

( y ′ )0

0

1

0

0.2

1

0.4

2. (a) 0.412% (b) −0.000000214% 3. (a) x y (y′ )0 0

1

1

−0.2

0.1

1.11

1.21

0.96

−0.368

0.2

1.24205

1.44205

0.6

0.8864

−0.46368

0.3

1.398465

1.698465

0.8

0.793664

−0.469824

0.4

1.581804

1.981804

1.0

0.699692

0.5

1.794893

(b) 0.117% x

y

2.0

1

2.1

0.85

2.2

4.

x

(y′ )0

y

0

1

0.709524

0.2

0.98

−0.192

2.3

0.577273

0.4

0.925472

−0.3403776

2.4

0.452174

0.6

0.84854666

−0.41825599

2.5

0.333333

0.8

0.76433779

−0.42294046

1.0

0.68609380

(b) 1.206% 4. (a) x

y

2.0

1

2.2

1.2

2.4

0

Exercise 228 (page 558) 1.

n

xn

1.421818

0

1.0

2.0

2.6

1.664849

1

1.1

2.104545

2.8

1.928718

3.0

2.213187

2

1.2

2.216666

3

1.3

2.334615

4

1.4

2.457143

5

1.5

2.585153

(b) 1.596%

yn

Answers to Practice Exercises 895 2.

Exercise 231 (page 567)

n

xn

0

0

1.0

1

0.2

0.980395

2

0.4

0.926072

3

0.6

0.848838

4

0.8

0.763649

5

1.0

0.683952

3. (a)

yn

1. (b)

n

xn

0

2.0

1.0

1

2.1

0.854762

2

2.2

0.718182

3

2.3

0.589131

4

2.4

0.464923

5

2.5

0.350000

Exercise 232 (page 571) 1

1. y = Ae 2 x + Be−3x − 2 1 3 2. y = Ae 3 x + Be−x − 2 − x ( ) 2 4 2 −3x 3. y = 3e + 4ex − 2 7( ) 2 3 3 4. y = − 2 + x e 3 x + 2 + x 4 4 ( ) 1 5 1 5. (a) q = − t+ e−50t 20 2 20 ) 1 ( (b) q = 1 − cos 50t 20 ( −2t ) 6. θ = 2 te + 1

yn

Exercise 233 (page 573) 1 1. y = Ae3x + Be−2x − ex 3 3 −x 4x −x 2. y = Ae + Be − xe 5 3. y = A cos 3x + B sin 3x + 2e2x

Chapter 50 Exercise 229 (page 564) 2 1 ( ) 1 1. y = Ae 3 t + Be− 2 t 2. θ = At + B e− 2 t { } 3. y = e−x A cos 2x + B sin 2x 3

4. y = 3e 3 x + 2e− 2 x

1

5. y = 4e 4 t − 3et 5 ( ) 6. y = 2xe− 3{x 7. x =} 2 1 − 3t e3t 8. y = 2e−3x 2 cos 2x + 3 sin 2x { } 9. θ = e−2.5t 3 cos 5t + 2 sin 5t

Exercise 230 (page 566) 2. s = c cos at { } 3. θ = e−2t 0.3 cos 6t + 0.1 sin 6t { ( )} 4. x = s + u + ns t e−nt 1. Proof

5. i = 6. s =

) 1 ( −160t e − e−840t A 20

3 4te− 2 t

5. (a)

Chapter 51

(b) No error

2

2. (c) 3. (c) 4. (d)

( ) t 2 t 4. y = At + B e 3 + t2 e 3 3 ( ) 1 5 1 5. y = e−2x − e 5 x + ex 44 4 ( ) 3t 2 6. y = 2e 1 − 3t + t

Exercise 234 (page 575) 3 ) 1( 1. y = Ae 2 x + Be−x − 11 sin 2x − 2 cos 2x 5 4 3 2. y = (Ax + B)e2x − sin x + cos x 5 5 3. y = A cos x + B sin x + 2x sin x ) 1 ( ) 1 ( 4x 4. y = 6e − 51e−x − 15 sin x − 9 cos x 170 34 ( ( ) k p 2 ) p sin pt − sin nt 5. y = ( 4 n n − p4 ) +n 2 (cos pt − cos nt)

6. Proof ( ) 7. q = 10t + 0.01 e−1000t + 0.024 sin 200t −0.010 cos 200t

896 Higher Engineering Mathematics Exercise 235 (page 577) ) 8 ( 6 cos x − 7 sin x 17 ) 1( y = Ae2θ + Beθ + sin 2θ + cos 2θ 2 3 1 1 1 x −2x y = Ae + Be − − x − x2 + e2x 4 2 2 4 ( ) t t t y = e A cos t + B sin t − e cos t 2 4 5x 10 2x 1 2x y = e − e − xe + 2 3 3 3 x ( 3 ) 3e −2x y = 2e − 2e2x + 3 sin x − 7 cos x 29 1

1

1. y = Ae 4 x + Be 2 x + 2x + 12 + 2. 3. 4. 5. 6.

x3 x5 x7 3. y = 1 + x + 2x2 + − + + ······ 3 8 16 { } 1 2 1 1 4 6 4. y = 1 − 2 x + 2 x − 2 x + ··· 2 2 × 42 2 × 42 × 62 { } x3 x5 x7 +2 x − 2 + 2 − + · · · 3 × 52 32 × 52 × 72 3

Exercise 239 (page 591) {

x2 x3 )+( )( ) 1. y = A 1 + x + ( 2×3 2×3 3×5

} x4 )( ) + ... +( 2×3×4 3×5×7 { 1 x x2 +Bx 2 1 + + (1 × 3) (1 × 2)(3 × 5)

Chapter 52 Exercise 236 (page 580) 1 1t e2 4 2. (a) 81 sin 3t (b) −1562.5 cos 5θ 1. (a) 16e2x

(b)

29 2 3. (a) 256 cos 2x (b) − 8 sin t 3 3 ( ) 4. (a) 9! x2 (b) 630 t 5. (a) 32 cosh 2x (b) 1458 sinh 3x 6. (a) 128 sinh 2x 7. (a) −

12 θ4

(b)

(b) 729 cosh 3x 240 t7

.+

x3 (1 × 2 × 3)(3 × 5 × 7)

} x4 + ··· (1 × 2 × 3 × 4)(3 × 5 × 7 × 9) { } { } x2 x4 x3 x5 2. y = A 1 − + − · · · + B x − + − . . . 2! 4! 3! 5! +

= A cos x + B sin x { x x2 )+( )( ) 3. y = A 1+ ( 1×4 1×2 4×7

} x3 )( ) + ··· +( 1 × 2 × 3 4 × 7 × 10 { 1 x x2 −3 +Bx 1+ + (1 × 2) (1 × 2)(2 × 5) } x3 + + ··· (1 × 2 × 3)(2 × 5 × 8)

Exercise 237 (page 582) 1. x2 y(n) + 2n xy(n−1) + n(n − 1)y(n−2) 2. y(n) = e2x 2n−3 {8x3 + 12nx2 + n(n − 1)(6x) +n(n − 1)(n − 2)}

4. Proof

y(3) = e2x (8x3 + 36x2 + 36x + 6) 3. y(4) = 2e−x (x3 − 12x2 + 36x − 24)

Exercise 240 (page 596)

4. y(5) = (60x − x3 ) sin x + (15x2 − 60) cos x 5. y(4) = −4 e−t sin t 6. y(3) = x2 (47 + 60 ln 2x) 7. Proof

8. y(5) = e2x 24 (2x3 + 19x2 + 50x + 35)

Exercise 238 (page 585)

{ } x2 5x4 5 × 9x6 5 × 9 × 13x8 1. y = 1− + − + − ··· 4! 6! 8! {2! } 3x3 3 × 7x5 3 × 7 × 11x7 +2 x − + − + ··· 3! 5! 7! 2. Proof

{

} x2 x4 1. y =Ax 1 − + − ··· 12 384 { } x4 x6 or A x2 − + − ··· 12 384 } ( x )3 { 1 x2 x4 2. J3 (x) = − 2 + 5 − ··· 2 Γ4 2 Γ5 2 Γ6 x3 x5 x7 or − 5 + 8 − ··· 8Γ4 2 Γ5 2 Γ6 3. J0 (x) = 0.765, J1 (x) = 0.440 2

Answers to Practice Exercises 897 Exercise 241 (page 600)

( ) x3 x5 1. (a) y = a0 + a1 x + + + · · · 3 5 ( ) ( ) 2 1 (b) y = a0 1 − 3x2 + a1 x − x3 − x5 − · · · 3 5 ) 1( 4 35x − 30x2 + 3 8

(c)

Exercise 247 (page 623)

( ) 2. u = t2 cos θ − 1 + 2t 4. Proof

1. 3.

u = 2ty2 + f(t) Proof

5.

7.

u = −4ey cos 2x − cos x + 4 cos 2x + 2y2 − 4ey + 4 ( 4 ) x x2 u=y − + x cos 2y + sin y 3( 2 ) u = − sin x + t + x + sin x + 2t + sin t

8.

Proof

9.

u = sin x sin y +

x2 π2 + 2 cos y − 2 2

2. 3.

5.

10. Proof

6.

Exercise 243 (page 606) −3t

3. X = Aex + Be−x

2 3 10 4 3 − (b) 3 + 2 − 2 s s s s s 1 3 2 8 48 1 (a) − + (b) 6 − 5 + 3 4s4 s2 s s s s 5 2 (a) (b) s−3 s+2 12 3s (a) 2 (b) 2 s +9 s +4 7s 1 (a) 2 (b) s (− 4 ) s3 − 9 2 s2 + 2 24 ) (b) ( ) (a) ( 2 s s +4 s s2 + 16 s2 − 2 16 ) (b) ( ) (a) ( 2 2 s s −4 s s − 16 ) 4 ( a cos b + s sin b 2 2 s +a ( ) 3 s cos α + ω sin α s 2 + ω2 Proof

1. (a)

4.

1. T = Ae + Be

∞ 216 ∑ 1 2nπ nπx cosech sin π 3 n(odd))=1 n 3 3 3 nπ sinh (2 − y) 3

Chapter 54

Exercise 242 (page 604)

3t

∞ 20 ∑ 1 cosech nπ sin nπx ( ) π n odd) =1 n

sinh nπ(y − 2)

) 1( 5 63x − 70x3 + 15x 8

Chapter 53

6.

1. u(x, y) =

2. u(x, y) =

2. (a) x (b)

Exercise 246 (page 614)

7.

2. T = A cos 3t + B sin 3t 4. X = A cos x + B sin x

8. 9.

Exercise 244 (page 609)

10.

∞ 1 12 ∑ nπ nπx 3nπt 1. u(x, t) = 2 sin sin cos π n=1 n 2 2 40 40 ∞ 8 ∑ 1 nπ nπx nπt 2. u(x, t) = 2 sin sin cos π n=1 n 2 2 80 10

Exercise 248 (page 623) 1. (d) 7. (a)

2. (a)

3. (c)

4. (b)

Exercise 245 (page 611)

Chapter 55

∞ 40 ∑ 1 − n 2 π2 c2 t nπx 16 e sin π n(odd)=1 n 4 ( )3 ∞ ∑ 1 − n 2 π2 t 8 nπx 2. u(x, t) = e 64 sin π n(odd)=1 n 3 8

1. u(x, t) =

(

∞ 32 ∑ 1 nπ nπx − 3. u(x, t) = 2 sin sin e 2 π n(odd)=1 n 2 20

Exercise 249 (page 627) 1. (a) ( n 2 π2 t 400

)

2

)2 s−2 24 2. (a) ( )4 s+2

(b) (

2

)3 s−1 12 (b) ( )5 s+3

5. (b)

6. (d)

898 Higher Engineering Mathematics 3. (a) 4. (a) 5. (a) (b) 6. (a) 7. (a) 8. (a)

s−1 6 (b) 2 − 2s + 2 s − 4s + 8 ( ) 5 s+2 4 (b) 2 s2 + 4s + 13 s + 10s + 26 1 s−1 − s − 1 s2 − 2s + 5 ( ) 1 1 s−3 + 2 4 s − 3 s − 6s + 13 ( ) 3 s−2 1 (b) 2 s(s − 2) s − 4s − 12 6 s+3 ) (b) ( 2 2 s + 2s − 8 4 s + 6s + 5 ( ) ( ) 2 s − 10 −6 s + 1 ( ) (b) 2 s − 2s + 10 s s+4

1 11. (a) 2e−2t cosh 3t + e−2t sinh 3t 3 14 (b) 3e4t cos 3t + e4t sin 3t 3

s2

Exercise 254 (page 637) 1. 2et − 5e−3t 2. 4e−t − 3e2t + e−3t 3. 2e−3t + 3et − 4et t 4. e−3t (3 − 2t − 3t2 ) √ √ 3 5. 2 cos 2t + √ sin 2t + 5e−t 2 √ √ √ 6. 2 + t + 3 sin 3t − 4 cos 3t 2 7. 2 − 3e−2t cos 3t − e−2t sin 3t 3

Exercise 250 (page 629) 1. Proof

2. Proof

Exercise 255 (page 638)

3. Proof

4. Proof

Exercise 251 (page 630) 1. (a) 3 (b) 16 3. (a) See page 630

(b) 4

2. (a) 6 4. 0

(b) –1

3. Poles at s = −2, s = −1 + j2, s = −1 − j2, zero at s = +1

Exercise 252 (page 631) 1. (a)

2. (b)

3. (d)

4. (c)

1. (a) s = −4 (b) s = 0, s = −2, s = 4 + j3, s = 4 − j3 2. Poles at s = −3, s = 1 + j2, s = 1 − j2, zeros at s = +1, s = −2

5. (d)

4. Poles at s = 0, s = +j2, s = −j2, zeros at s = −1, s = +6

Exercise 256 (page 639)

Chapter 56

1. (a)

2. (b)

3. (a)

4. (d)

5. (c)

Exercise 253 (page 635) (b) 2e5t 1 3 − t e 2 (b) 2 cos 2t 2 1 4 sin 5t (b) sin 3t 5 3 5 cos 3t (b) 6t 2 52 4 t (b) t3 2 3 7 6 cosh 4t (b) sinh 4t 4 5 sinh 3t (b) 2et t2 3 1 −2t 3 1 e t (b) e3t t4 6 8 3 −t e cos 3t (b) e−3t sin 2t 2 7 2e3t cos 2t (b) e4t sinh 2t 2

1.

(a) 7

2.

(a)

3.

(a)

4.

(a)

5.

(a)

6.

(a)

7. (a) 8. (a) 9. (a) 10. (a)

Chapter 57 Exercise 257 (page 643) 1. V(t) = 6H(t − 4)

v(t)

6

0 (a)

4

t

Answers to Practice Exercises 899 2. 2H(t) − H(t − 5)

5.

f(t)

v(t)

4

2

0

0

5

1

t

t

(e)

(b)

6.

3.

f(t) f(t)

7

1

0

0 2

5

t

t (f)

(c)

7.

4.

f(t)

1

f(t)

1 0

0 (d)

–1 t (g)

p 4

p 2

p

3p 2

2p t

900 Higher Engineering Mathematics 8.

11. f(t) f(t) 1 p 6

3

0.5 0

p 2

3p 2

p

2p t

–3

0

(h)

2

1

3

4

5

6

t

(k)

12.

9. f(t)

f(t)

5 8

4 0 –2

–1

0

2

1

3

p 3

3p 2

p

p 2

p 4 –5

t

(i) (l)

10. f(t)

Exercise 258 (page 645)

1

e−s s e−3s 4. 2 s +1 6e−s 7. s4 1.

0.5

7e−3s s e−4s 5. s−1 se−6s 8. 2 s +9 2.

π

(j)

0

1

2

3

4

5

t

se− 3 s 10. 2 s +4

11.

2e−3s s−1

2e−2s s3 3e−5s 6. 2 s +9 10e−5s 9. 2 s −4 3.

12.

3se−2s s2 − 1

2p t

Answers to Practice Exercises 901 Exercise 259 (page 646)

Chapter 59

1.

H(t − 9)

2.

4H(t − 3)

Exercise 262 (page 658)

3.

2H(t − 2).(t − 2)

1. x = et − t − 1 and y = 3et + 2t − 3

4.

5H(t − 2). sin(t − 2)

5.

3H(t − 4). cos 4(t − 4)

6.

6H(t − 2). sinh(t − 2) ( )2 1.5H(t − 6). t − 6

7.

2. x = 5 cos t + 5 sin t − e2t − et − 3 and y = e2t + 2et − 3 − 5 sin t (√ ) (√ ) 3. x = 3 cos t + cos 3t and y = 3 cos t − cos 3t

2H(t − 4). cosh 4(t − 4) ( ( ) ) √ 1 1 9. 2H t − . cos 5 t − 2 2 t−7 10. 4H(t − 7).e 8.

Chapter 60 Exercise 263 (page 668) 1. 2.

Exercise 260 (page 646) 1. (d)

2. (a)

3. (b)

4. (b)

5. (c)

Chapter 58

4.

Exercise 261 (page 652) ( ) 1. (a) 2 1 − e−5t

3.

(

−3t

(b) 10 e

−e

−5t

)

) 5 ( −5t e + sin 5t − cos 5t 2 ( ) 4 2. y = 3 − t e 3 t (c) i =

3. x = 2 cos 10t

( ) 1 1 8 sin x + sin 3x + sin 5x + . . . f(x) = π 3 5 π 1 1 1 = 1 − + − + .... 4 3 5 7 ( ) 1 2 1 1 (a) f(x) = + cos x − cos 3x + cos 5x + . . . 2 π 3 5 1 (b) 2 Graph sketching

2 sin 3x 3π ( 2 1 1 6. f (x) = cos t − cos 3t + cos 5t − . . . + sin 2t π 3 5 ) 1 1 + sin 6t + sin 10t + . . . 3 5 ( ) 2 1 cos 2θ cos 4θ cos 6θ 7. f(θ) = − − − − ... π 2 (3) (3)(5) (5)(7) 5.

4. i = 100te−500t ( ) 5. y = 4 3e−2t − 2e−4t ( ) 1 6. y = 4x − 1 ex + e4x 3 3 4x 3 −x 9 15 7. y = e − e + cos x − sin x 85 10 34 34 8. y = ex − e−2x + sin 3x ( ) 9. y = 3ex cos x + sin x − ex cos 2x ( ) 10. y = e−t 2 cos 2t + sin 2t ( t ) 11. VC = V 1 − e− CR 12. See answers to Exercises 229 and 230 of Chapter 50 13. See answers to Exercises 232 to 235 of Chapter 51

Exercise 264 (page 669) 1. (b)

2. (d)

3. (b)

4. (c)

5. (a)

Chapter 61 Exercise 265 (page 674) 1 1 sin 2x + sin 3x 2 3 1 1 1 + sin 4x + sin 5x + sin 6x + . . .) 4 5 6

1. f(x) = π − 2(sin x +

902 Higher Engineering Mathematics 2. f(t) =

( ) π 4 cos 3t cos 5t +1− cos t + 2 + 2 + . . . 2 π 3 5

1 1 sin 2x + sin 3x 2 3 1 1 1 − sin 4x + sin 5x − sin 6x + . . .) 4 5 6 ( ) π 4 cos 3x cos 5x f(x) = − cos x + + + ... 2 π( 32 52 ) 2 π 1 1 f(θ) = − 4 cos θ − 2 cos 2θ + 2 cos 3θ − · · · 3 2 3 ∞ 1 ∑ π2 = 2 6 n=1 n ( 8 1 1 f(x) = 2 cos x + 2 cos 3x + 2 cos 5x π 3 5 ) 1 + 2 cos 7x + . . . 7

3. f(x) = π + 2(sin x −

4. 5. 6. 7.

8.

π2 1 1 1 1 = 1 + 2 + 2 + 2 + 2 + ... 8 3 5 7 9

Exercise 266 (page 675) 1. (c)

2. (c)

3. (a)

4. (b)

5. (d)

π2 1 1 1 = 1 + 2 + 2 + 2 + ... 8 3 5 7 ( 8 1 1 5. y = 2 sin θ − 2 sin 3θ + 2 sin 5θ π 3 5 4.

−

1 sin 7θ + . . . 72

)

Exercise 268 (page 682) 2 1. f(x) = π

( sin x +

π 1 sin 2x− sin 3x 4 9 π − sin 4x + . . . 8

)

( ) 1 2 1 1 2. (a) f(t) = − cos t − cos 3t + cos 5t − . . . 2 π 3 5 ( 2 1 1 (b) f(t) = sin t − sin 2t + sin 3t + sin 5t π 3 5 ) 1 − sin 6t + . . . 3 ( 8 sin x sin 3x sin 5x 2 3. f(x) = sin x = − − π (1)(3) (1)(3)(5) (3)(5)(7) ) sin 7x − − ... (5)(7)(9) ( ) π 2 cos 6x cos 10x 4. f(x) = − cos 2x + + + . . . 4 π 32 52

Exercise 269 (page 682)

Chapter 62 Exercise 267 (page 679)

( 4 1 1 1. f(x) = cos x − cos 3x + cos 5x π 3 5 ) 1 − cos 7x + . . . 7 ( 1 1 2. f(t) = −2 sin t + sin 2t + sin 3t 2 3 ) 1 + sin 4t + . . . 4 ( π 4 1 3. f(x) = + 1 − cos x + 2 cos 3x 2 π 3 ) 1 + 2 cos 5x + . . . 5

1. (b)

2. (c)

3. (a)

4. (b)

5. (a)

6. (d)

Chapter 63 Exercise 270 (page 687)

[ ( ) ( ) 5 10 πt 1 3πt + sin + sin 2 π 10 3 10 ( ) ] 1 5πt + sin + ... 5 10 [ ( ) ( ) 5 5 2πx 1 4πx 2. f(x) = − sin + sin 2 π 5 2 5 ( ) ] 1 6πx + sin + ... 3 5 1. v(t) =

Answers to Practice Exercises 903 { ( ) ( ) 12 πx 1 3πx sin + sin π 2 3 2 ( ) } 1 5πx + sin + ... 5 2 ( V V 2V cos 2t cos 4t 4. f(t) = + sin t− + π 2 π (1)(3) (3)(5) ) cos 6t + + ... (5)(7) 3. f(x) =

Exercise 273 (page 695) 1. (a) Only odd cosine terms present (b) Only even sine terms present 2. y = 9.4 + 17.2 cos θ − 24.1 sin θ + 0.92 cos 2θ− 0.14 sin 2θ + 0.83 cos 3θ + 0.67 sin 3θ 3. i = 4.00 − 4.67 cos 2θ + 1.00 cos 4θ− 0.66 cos 6θ + . . . 4. y = 1.83 − 25.67 cos θ + 83.89 sin θ +1.0 cos 2θ − 0.29 sin 2θ + 15.83 cos 3θ +10.5 sin 3θ

Exercise 271 (page 689)

{ ( ) ( ) 3 12 πx 1 3πx − 2 cos + 2 cos 2 π 3 3 3 ( ) } 1 5πx + 2 cos + ... 5 3 { ( ) ( ) ( ) 6 πx 1 2πx 1 3πx 2. f(x) = sin − sin + sin π 3 2 3 3 3 ( ) } 1 4πx − sin + ... 4 3 { ( ) ( ) 8 πt 1 3πt 3. f(t) = 2 sin − 2 sin π 2 3 2 ( ) } 1 5πt + 2 sin − ... 5 2 1. f(x) =

{ ( ) ( ) 16 64 πθ 1 2πθ 4. f(θ) = − 2 cos − 2 cos 3 π 4 2 4 ( ) } 1 3πθ + 2 cos − ... 3 4

Chapter 65 Exercise 274 (page 703) ) j ( cos nπ − 1 e jnt nπ n=−∞ ( ) 2 1 1 = 1−j e jt + e j3t + e j5t + . . . π 3 5 ( ) 2 −jt 1 −j3t 1 −j5t +j e + e + e + ... π 3 5 2. Proof 1. f(t) =

∞ ∑

3. Proof

( (2−jπn) ) ∞ 1 ∑ e − e−(2−jπn) jπnt 4. f(t) = e 2 n=−∞ 2 − jπn

Exercise 275 (page 706)

Chapter 64

1. f(x) =

{

∞ ∑ n=−∞

Exercise 272 (page 694) 1. y = 23.92 + 7.81 cos θ + 14.61 sin θ+ 0.17 cos 2θ + 2.31 sin 2θ − 0.33 cos 3θ+ 0.50 sin 3θ 2. y = 5.00 − 10.78 cos θ + 6.83 sin θ+ 0.13 cos 2θ + 0.79 sin 2θ + 0.58 cos 3θ− 1.08 sin 3θ 3. i = 0.64 + 1.58 cos θ − 2.73 sin θ− 0.23 cos 2θ − 0.42 sin 2θ + 0.27 cos 3θ+ 0.05 sin 3θ

8 2. f(x) = π

3. f(t) =

(

cos x −

∞ ∑ n=−∞

(

( nπ )} 4 sin e jnx πn 2

(

1 1 cos 3x+ cos 5x 3 5

) j2 cos nπ e jnt n

1 − cos 7x + . . . 7

) 1 1 1 4. f(t) = 4 sin t − sin 2t + sin 3t − sin 4t + ... 2 3 4

)

904 Higher Engineering Mathematics Exercise 276 (page 710) 1. (a) 2e j4t + 2e−j4t , 2∠0◦ anticlockwise, ◦ 2∠0 clockwise, each with ω = 4 rad/s (b) 2e j4t e jπ/2 + 2e−j4t e−jπ/2 , 2∠π/2 anticlockwise, 2∠ − π/2 clockwise, each with ω = 4 rad/s ( ) ( ) 2. 5 + j6 e j2t + 5 − j6 e−j2t , 7.81 ∠0.88 rotating anticlockwise, 7.81∠ − 0.88 rotating clockwise each with ω = 2 rad/s ( ) ( ) 3. 2 − j3 e jt + 2 + j3 e−jt , 3.61∠ − 0.98 rotating anticlockwise, 3.61∠0.98 rotating clockwise, each with ω = 1 rad/s

3.

3z(sin 2 − z + cos 2) z2 − 2z cos 2 + 1

4.

6z(z − 1) z2 − 9

( ) z3 − z2 + z 1 − e −1 z ) or ( 5. ( )2 + ( )2 ( ) z − e −1 z−1 z − 1 z − e −1 z

Exercise 279 (page 718) 1.

8z z−2

4.

z(7z − 12) z2 − 5z + 6

2.

Exercise 280 (page 719)

Chapter 66

1.

2z (z − 1)2

2.

3z(z + 1) (z − 1)3

1 (z − 1) 1 4. 2 z (z − a)

3.

4z(z2 + 4z + 1) (z − 1)4

4.

z sin 3 z2 − 2z cos 3 + 1

Exercise 281 (page 720)

5.

z z−2

6.

2z (z − 2)2

1.

8.

3z z − e −2

4. 20/9 or 2

5z(z − cos 2) − 2z cos 2 + 1

z2

1 2 9. 1 2 z − 2z cos + 1 2 z 11. z+4 3z 13. z+5 3z 15. z − e5 z sin

1 z2 (z − 1) 1 5. 3 z (z − 3)

1.

Exercise 277 (page 716)

7.

(

z2 − ze 3 cos 2 10. 4 2 z − 2ze 3 cos 2 + e 6 6z(z + 2) 12. (z − 2)3 −3z 14. (z + 3)2 2ze −4 sin 2 16. 2 z − 2ze −4 cos 2 + e −8

2.

2z (z − 2)2

2. 2 9

4z (z − 4)2

1.

5z (z − 1)2

( ) z 7z + 1 2. ( )3 z−1

}

{

−

4z z − e −3

}

( ) −4z3 + 13z2 − z 5e −3 + 4 ( )2 ( ) z − 1 z − e −3

6.

2z (z − 2)2

Exercise 282 (page 722) 1.

1

2.

(2)k

3.

(−1)k ( )k 1 3 ( )k 1 5

4.

(−4)k ( )k 1 − 4

e 3k

10. 3k 1 12. (2)k 4 } 3{ 14. 1 − (−3)k 4 { ( )k } 1 1 k 16. (3) − − 7 2 18. 10(3)k − 7(2)k 3 { k ( )k } 20. (2) − −5 7

5. 7.

} 1{ k (2) − (−1)k 3 } 1{ 15. 1 − (−2)k 3 17. (2)k − 1 {( ) } k ( )k 2 1 19. − −1 3 2 13.

or

1 z(z − a)

3. 0

5. 1

11. 5k 2

Exercise 278 (page 717)

3.

)

9.

{

6z2 3. ( )2 z−1

4z z−4

6. 8.

e −5k

Answers to Practice Exercises 905 Exercise 283 (page 725) ( )k 1. 4 3 3.

} 1{ 10 − (4)k 3

5. (−1)k − (−2)k

2. 3 − 2(3)k { ( )k } 1 1 k 4. (2) + 5 3 2 ( )k 7 k 5 6. − 6 2 + (3) 2 2

9.

10.

Exercise 284 (page 725) 1. (c)

2. (b)

3. (d)

4. (b)

5. (c)

6. (b)

7. (a)

8. (d)

9. (c)

10. (a)

11. 12.

Week 2: 20%, 8%, 32%, 13%, 27% Week 3: 22%, 10%, 29%, 14%, 25% Week 4: 20%, 9%, 27%, 19%, 25%. Little change in centres A and B, there is a reduction of around 8% in centre C, an increase of around 7% in centre D and a reduction of about 3% in centre E. A circle of any radius, subdivided into sectors, having angles of 7.5◦ , 22.5 ◦ , 52.5 ◦ , 167.5 ◦ and 110 ◦ , respectively. A circle of any radius, subdivided into sectors, having angles of 107◦ , 156◦ , 29◦ and 68◦ , respectively. (a) £495 (b) 88 (a) £16 450 (b) 138

Exercise 287 (page 741)

Chapter 67 Exercise 285 (page 732) 1. (a) (d) 2. (a) (d)

continuous (b) continuous (c) discrete continuous discrete (b) continuous (c) discrete discrete

Exercise 286 (page 735) 1. If one symbol is used to represent ten vehicles, working correct to the nearest five vehicles, gives 3.5, 4.5, 6, 7, 5 and 4 symbols respectively. 2. If one symbol represents 200 components, working correct to the nearest 100 components gives: Mon 8, Tues 11, Wed 9, Thurs 12 and Fri 6.5 3. Six equally spaced horizontal rectangles, whose lengths are proportional to 35, 44, 62, 68, 49 and 41 units respectively 4. Five equally spaced horizontal rectangles, whose lengths are proportional to 1580, 2190, 1840, 2385 and 1280 units, respectively 5. Six equally spaced vertical rectangles, whose heights are proportional to 35, 44, 62, 68, 49 and 41 units, respectively 6. Six equally spaced vertical rectangles, whose heights are proportional to 1580, 2190, 1840, 2385 and 1280 units, respectively 7. Three rectangles of equal height, subdivided in the percentages shown; P increases by 20% at the expense of Q and R 8. Four rectangles of equal heights, subdivided as follows: Week 1: 18%, 7%, 35%, 12%, 28%

1. There is no unique solution, but one solution is: 39.3 – 39.4 1; 39.5 – 39.6 5; 39.7 – 39.8 9; 39.9 – 40.0 17; 40.1 – 40.2 15; 40.3 – 40.4 7; 40.5 – 40.6 4; 40.7 – 40.8 2; 2. Rectangles, touching one another, having midpoints of 39.35, 39.55, 39.75, 39.95.. . . and heights of 1, 5, 9, 17,. . . 3. There is no unique solution, but one solution is: 20.5 – 20.9 3; 21.0 – 21.4 10; 21.5 – 21.9 11; 22.0 – 22.4 13; 22.5 – 22.9 9; 23.0 – 23.4 2 4. There is no unique solution, but one solution is: 1 – 10 3; 11 – 19 7; 20 – 22 12; 23 – 25 11; 26 – 28 10; 29 – 38 5; 39 – 48 2 5. 20.95 3; 21.45 13; 21.95 24; 22.45 37; 22.95 46; 23.45 48 6. Rectangles, touching one another, having midpoints of 5.5, 15, 21, 24, 33.5 and 43.5. The heights of the rectangles (frequency per unit class range) are 0.3, 0.78, 4, 3.67, 3.33, 0.5 and 0.2 7. (10.95 2), (11.45 9), (11.95 19), (12.45 31), (12.95 42), (13.45 50) 8. Ogive 9. (a) There is no unique solution, but one solution is: 2.05 – 2.09 3; 2.10 – 2.14 10; 2.15 – 2.19 11; 2.20 – 2.24 13; 2.25 – 2.29 9; 2.30 – 2.34 2. (b) Rectangles, touching one another, having mid-points of 2.07, 2.12 .. . . and heights of 3, 10 . . . (c) Using the frequency distribution given in the solution to part (a) gives: 2.095 3; 2.145 13; 2.195 24; 2.245 37; 2.295 46; 2.345 48. (d) A graph of cumulative frequency against upper class boundary having the co-ordinates given in part (c).

906 Higher Engineering Mathematics Exercise 288 (page 742) 1. (a)

2. (d)

3. (c)

4. (c)

5. (b)

Chapter 69 Exercise 294 (page 755) 2 7 or 0.2222 (b) or 0.7778 9 9 23 47 2. (a) or 0.1655 (b) or 0.3381 139 139 69 (c) or 0.4964 139 1 1 1 3. (a) (b) (c) 6 6 36 4. 0.7 or 70% 1. (a)

Chapter 68 Exercise 289 (page 745) 1 1. Mean 7 , median 8, mode 8 3 2. Mean 27.25, median 27, mode 26 3. Mean 4.7225, median 4.72, mode 4.72 4. Mean 115.2, median 126.4, no mode

5.

5 36

2 1 4 13 (b) (c) (d) 5 5 15 15 1 1 9 1 7. (a) (b) (c) (d) 250 200 1000 50 000 6. (a)

Exercise 290 (page 746) 1. 23.85 kg 2. 171.7 cm 3. Mean 89.5, median 89, mode 88.2 4. Mean 2.02158 cm, median 2.02152 cm, Mode 2.02167 cm

Exercise 291 (page 748) 1. 3. 4. 6.

4.60 2. 2.83 µF Mean 34.53 MPa, standard deviation 0.07474 MPa 0.296 kg 5. 9.394 cm 0.00544 cm

Exercise 295 (page 757) 1. 2. 3. 4. 5. 6.

(a) 0.6 (b) 0.2 (c) 0.15 (a) 0.64 (b) 0.32 0.0768 (a) 0.4912 (b) 0.4211 (a) 89.38% (b) 10.25% (a) 0.0227 (b) 0.0234 (c) 0.0169

Exercise 296 (page 759) 1. (a) 210 3. 210

Exercise 292 (page 749) 1. 30, 27.5, 33.5 days 2. 27, 26, 33 faults 3. Q1 = 164.5 cm, Q2 = 172.5 cm, Q3 = 179 cm, 7.25 cm 4. 37 and 38; 40 and 41 5. 40, 40, 41; 50, 51, 51

2. (a) 7. (b)

3. (b) 8. (c)

5.

(b) 15

4. 3003

10 49 C

2. (a) 792

= 6

10 1 = or 715 × 10−9 13 98 3816 13 98382

Exercise 297 (page 760) 1. 53.26%

2. 70.32%

3. 7.48%

Exercise 298 (page 761)

Exercise 293 (page 750) 1. (d) 6. (b)

(b) 3024

4. (c) 9. (a)

5. (d) 10. (d)

1. (c) 6. (d)

2. (d) 7. (a)

3. (c) 8. (a)

4. (b) 9. (b)

5. (b) 10. (c)

Answers to Practice Exercises 907 Exercise 303 (page 778)

Chapter 70

1. Graphically, x = 27.1, σ = 0.3; by calculation, x = 27.079, σ = 0.3001;

Exercise 299 (page 767)

2. (a) x = 23.5 kN, σ = 2.9 kN (b) x = 23.364 kN, σ = 2.917 kN

1. (a) 0.0186 (b) 0.9976 2. (a) 0.2316 (b) 0.1408 3. (a) 0.7514 (b) 0.0019 4. (a) 0.9655 (b) 0.0028 5. Vertical adjacent rectangles, whose heights are proportional to 0.0313, 0.1563, 0.3125, 0.3125, 0.1563 and 0.0313 6. Vertical adjacent rectangles, whose heights are proportional to 0.0280, 0.1306, 0.2613, 0.2903, 0.1935, 0.0774, 0.0172 and 0.0016 7. 0.0574

Exercise 304 (page 779) 1. (b)

2. (c)

3. (a)

4. (d)

5. (a)

Chapter 72 Exercise 305 (page 783)

Exercise 300 (page 770) 1. 0.0613

1. 0.999

2. – 0.916

3. 0.422

5. –0.962

6. 0.632

7. 0.937

2. 0.5768

3. (a) 0.1771 (b) 0.5153 4. 0.9856 5. The probabilities of the demand for 0, 1, 2,. . .. . tools are 0.0067, 0.0337, 0.0842, 0.1404, 0.1755, 0.1755, 0.1462, 0.1044, 0.0653, . . . This shows that the probability of wanting a tool eight times a day is 0.0653, i.e. less than 10%. Hence seven should be kept in the store 6. Vertical adjacent rectangles having heights proportional to 0.4966, 0.3476, 0.1217, 0.0284, 0.0050 and 0.0007

Chapter 73 Exercise 306 (page 789) 1. Y = −256 + 80.6X

2. Y = 0.0477 + 0.216X

3. X = 3.20 + 0.0124Y

4. X = −0.047 + 4.556Y

5. Y = 1.142 + 2.268X

6. X = −0.483 + 0.440Y

7. (a) 7.95 V

(b) 17.1 mA 8. Y = 0.881 − 0.0290X

9. X = 30.194 − 34.039Y

10. (a) 0.417 s (b) 21.7 N

Chapter 74 Exercise 301 (page 770)

Exercise 307 (page 796)

1. (b) 2. (c) 3. (a) 4. (c) 5. (d)

1. µx = µ = 22.4 cm σx = 0.0080 cm 2. σx = 0.0079 cm 3. (a) µx = 1.70 cm, σx = 2.91 × 10−3 cm

Chapter 71

(b) µx = 1.70 cm, σx = 2.89 × 10−3 cm

Exercise 302 (page 775) 1. 6 3. (a) 479 5. (a) 131 7. (a) 65

4. 0.999

2. 22 (b) 63 (c) 21 4. 4 (b) 553 6. (a) 15 (b) 209

(c) 89 8. (a) 1

4. 0.023 5. 0.497 (b) 4 (b) 85

6. 0.0038 (c) 13 7. (a) 0.0179

(b) 0.740

(c) 0.242

908 Higher Engineering Mathematics Exercise 308 (page 800)

Exercise 312 (page 820)

1. 66.89 and 68.01 mm, 66.72 and 68.18 mm 2. (a) 2.355Mg to 2.445Mg; 2.341Mg to 2.459Mg

1. Take x as 24 + 15, i.e. 39 hours, z = −1.28, z0.05 , one-tailed test = −1.645, hence hypothesis is accepted 2. z = 2.357, z0.05 , two-tailed test = ±1.96, hence the null hypothesis is rejected 3. x1 = 23.7, s1 = 1.73, σ1 = 1.93, x2 = 25.7, s2 = 2.50, σ2 = 2.80, |t| = 1.32, t0.99 , ν8 = 2.90 hence hypothesis is accepted 4. z (sample) = 1.99, (a) z0.05 , two-tailed test = ±1.96, no significance, (b) z0.01 , two-tailed test = ±2.58, significant difference 5. Assuming null hypothesis of no difference, σ = 0.397, |t| = 1.85, (a) t0.99 , ν8 = −2.51, alternative hypothesis rejected, (b) t0.95 , ν22 = −1.72, alternative hypothesis accepted 6. σ = 0.571, |t| = 3.32, t0.95 , ν8 = 1.86, hence null hypothesis is rejected

(b) 86% 3. 12.73 × 10−4 m◦ C−1 to 12.89 × 10−4 m◦ C−1 4. (a) at least 68 (b) at least 271 5. 10.91t to 11.27t 6. 45.6 seconds

Exercise 309 (page 804) 1. 5.133 MPa to 5.207 MPa 2. 5.125 MPa to 5.215 MPa 3. 1.10 Ω m−1 to 1.15 Ω m−1 4. 95%

Chapter 75

Chapter 76

Exercise 310 (page 812) 1. (a) 28.1% 2. (a) 55.2% 3. (a) 35.3%

(b) 4.09% (b) 4.65% (b) 18.5%

4. (a) 32.3% (b) 20.1%

Exercise 313 (page 823) (c) 0.19% (c) 0.07% (c) 8.4%

1. 10.2

2. 3.16

(c) 11.9%

Exercise 314 (page 829) Exercise 311 (page 816) 1. z (sample) = 3.54, zα = 2.33, hence the null hypothesis is rejected, where zα is the z-value corresponding to a level of significance of α 2. t0.95 , ν8 = 1.86, |t| = 1.89, hence null hypothesis rejected 3. z (sample) = 2.85, zα = ±2.58, hence the null hypothesis is rejected 4. x = 10.38, s = 0.33, t0.95 , ν19 = 1.73, |t| = 1.72, hence the null hypothesis is accepted 5. |t| = 3.00, (a) t0.975 , ν9 = 2.26, hence the hypothesis rejected, (b) t0.995 , ν9 = 3.25, hence the null hypothesis is accepted 6. |t| = 3.08, (a) t0.950 , ν5 = 1.48, hence claim supported, (b) t0.98 , ν5 = 2.83, hence claim supported

1. Expected frequencies: 7, 33, 65, 73, 48, 19, 4, 0; χ2 -value = 3.62, χ20.95 , ν7 = 14.1, hence null hypothesis accepted. χ20.10 , ν7 = 2.83, hence data is not ‘too good’ 2. λ = 2.404; expected frequencies: 5, 11, 14, 11, 7, 3, 1 χ2 -value = 1.86; χ20.95 , ν6 = 12.6, hence the data does fit a Poisson distribution at a level of significance of 0.05 3. x = 1.32MΩ, s = 0.0180MΩ; expected frequencies, 6, 17, 36, 55, 65, 55, 36, 17, 6; χ2 -value = 5.98; χ20.95 ν6 = 12.6, hence the null hypothesis is accepted, i.e. the data does correspond to a normal distribution 4. λ = 0.896; expected frequencies are 102, 91, 41, 12, 3, 0, 0; χ2 -value = 5.10; χ20.95 , ν6 = 12.6, hence this data fits a Poisson distribution at a level of significance of 0.05

Answers to Practice Exercises 909 5. x = 10.09 MN; s = 0.733 MN; expected frequencies, 2, 5, 12, 16, 14, 8, 3, 1; χ2 -value = 0.563; χ20.95 , ν5 = 11.1. Hence null hypothesis accepted. χ20.05 , ν5 = 1.15, hence the results are ‘too good to be true’

3. H0 : N = R, H1 : N ̸= R, T = 5. From Table 76.4, with n = 10 (since two differences are zero), T ≤ 8, Hence there is a significant difference in the drying times

Exercise 315 (page 833)

Exercise 317 (page 843)

1. H0 : t = 15h, H1 : t ̸= 15h S = 7. From Table 76.3, S ≤ 2, hence accept H0 2. S = 6. From Table 76.3, S ≤ 4, hence null hypothesis accepted 3. H0 : mean A = mean B, H1 : mean A ̸= mean B, S = 4. From Table 76.3, S ≤ 4, hence H1 is accepted

Exercise 316 (page 837) 1. H0 : t = 220h, H1 : t ̸= 220h, T = 74. From Table 76.4, T ≤ 29, hence H0 is accepted 2. H0 : s = 150, H1 : s ̸= 150, T = 59.5. From Table 76.4, T ≤ 40, hence null hypothesis H0 is accepted

1. H0 : TA = TB , H1 : TA = ̸ TB , U = 30. From Table 76.5, U ≤ 17, hence accept H0 , i.e. there is no difference between the brands 2. H0 : B.S.A = B.S.B , H1 : B.S.A ̸= B.S.B , α2 = 10%, U = 28. From Table 76.5, U ≤ 15, hence accept H0 , i.e. there is no difference between the processes 3. H0 : A = B, H1 : A = ̸ B, α2 = 5%, U = 4. From Table 76.5, U ≤ 8, hence null hypothesis is rejected, i.e. the two methods are not equally effective 4. H0 : meanA = meanB , H1 : meanA ̸= meanB , α2 = 5%, U = 90. From Table 76.5, U ≤ 99, hence H0 is rejected and H1 accepted

Index Accuracy of numerical integration, 520 Acute angles, trigonometric, 91 Adding alternating waveforms, 301 Adjoint of matrix, 261 Algebra, 3 Algebraic method of successive approximations, 61 substitution, integration, 461 Ambiguous case, 102 Amplitude, 138, 142 And-function, 66 -gate, 78 Angle between two vectors, 314 of any magnitude, 133 of banking, 95 of depression, 99 of elevation, 99 Angular velocity, 125, 142 Applications of complex numbers, 236–239 differentiation, 340 Newton-Raphson method, 345 rates of change, 340 small changes, 358 tangents and normals, 357 turning points, 348–351 velocity and acceleration, 342 integration, 431 areas, 432 centroids, 436 mean value, 433 r.m.s. value, 433 second moment of area, 440–447 volumes, 435 matrices and determinants, 265, 366 Arbitrary constant of integration, 423 Arc, 118 length, 120 Area of circle, 120 irregular figures, 213 sector, 120 triangle, 101 under curve, 432 Argand diagram, 229 Argand, Jean-Robert, 229 Argument, 233 Arithmetic mean, 744 Array of matrix, 254

Astroid, 364 Asymptotes, 203–208 Auxiliary equation, 562 Average, 743 value of waveform, 217 Base, 24 Bayes’ theorem, 759 Belt-driven pulley, 293 Benedict-Webb-Rubine equation, 402 Bernoulli’s equation, 6 Bessel, Friedrich Wilhelm, 592 Bessel functions, 592, 594, 596 Bessel’s correction, 814 equation, 592 Binomial distribution, 764–767, 810 expression, 48 series/theorem, 48, 50–53 practical problems, 54 Bisection method, 58 BODMAS, 5 Boolean algebra, 65, 66 laws and rules of, 70 Boole, George, 66 Boundary conditions, 527, 602 Brackets, 4 Calculator, 32, 37, 93, 235, 260 Calculus, 327 Cardioid, 364 Cartesian complex numbers, 228 co-ordinates, 111–116 Catenary, 153 Cauchy, Baron Augustin-Louis, 549 Centre of area, 436 gravity, 436 mass, 436 Centripetal acceleration, 127 force, 126, 127 Centroids, 436 Chain rule, 335 Change of limits, 464 Characteristic determinant, 275 equation, 275 impedance of transmission line, 244 value, 275 vectors, 275 Chi-square distribution, 825 values, 822

Chord, 118 Circle, 117, 193 area, 120 equation of, 123, 210 properties of, 118 Circumference, 117 Class interval, 736 Coefficient of correlation, 780 significance of, 781 Cofactor of matrix, 259 Combinational logic networks, 79 Combination of periodic functions, 301 Combinations, 758 Common logarithms, 25 Comparing two sample means, 817 Complementary function, 569 Complex numbers, 227 addition and subtraction, 229 applications of, 236 Cartesian form, 228 coefficients, 699 conjugate, 231 equations, 232 exponential form, 245 form of Fourier series, 698 multiplication and division, 230, 235 phasors, 308–310 polar form, 228, 233 powers of, 242 roots of, 243 Complex wave, 144, 690–697 considerations, 694 Compound angle formulae, 173–175 angles, 173 Conditional probability, 752 Confidence coefficients, 797 intervals, 797 levels, 797 limits of the mean of a population, 798 Continuous data, 732 function, 199, 664 Contour map, 414 Convergence, 33 Conversion of a sin ωt + b cos ωt into R sin(ωt + α), 175–179 Correlation, linear, 780–784 Cosecant, 92

Index Cosh, 151 series, 159 Cosh θ substitution, 476 Coshec, 152 Cosine, 91 curves, 137 rule, 101, 305 wave production, 136 Cotangent, 92 Coth, 152 Couple, 73 Cramer, Gabriel, 271 Cramer’s rule, 271 Crank mechanism, 106, 108, 118 Critical regions, 808 values, 808 Cross product, 317–320 Cubic equations, 182 Cumulative frequency curve, 737 distribution, 737, 740 Curve sketching, 208–211 Cycles of log graph paper, 43 Cycloid, 364 Deciles, 748 Definite integrals, 427 Definition of z-transform, 714 Degree of differential equation, 527 Degrees of freedom, 801 De Moivre, Abraham, 242 De Moivre’s theorem, 242, 243 De Morgan, Augustus, 72 De Morgan’s laws, 72 Dependent event, 752 Depression, angle of, 99 Derivative, 329 Laplace transforms of, 627 z-transforms, 719 Descartes, René 112, 228 Determinant, 257, 259 to solve simultaneous equations, 265, 268 Determination of law, 43 Diameter, 118 Difference equations, 722 Differential calculus, 327 coefficient, 329 Differential equations, 525, 526 dy d2 x a 2 + b + cy = 0 type, 561–567 dx dy dy d2 x + b + cy = f(x) type, dx dy2 568–577 dy = f(x) type, 527 dx

dy = f(x).f(y) type, 530 dx dy = f(y) type, 528 dx dy + Py = Q type, 538–542 dx degree of, 527 first-order, separation of variables, 527 homogeneous first-order, 534–537 numerical methods, 543–559 partial, 601, 602 dy P = Q, 533–537 dx power series methods, 578–600 simultaneous, using Laplace transforms, 653–658 using Laplace transforms, 648–652 Differentiation, 327, 329 applications, 341 from first principles, 328 function of a function, 335, 370 implicit, 369–374 inverse hyperbolic function, 386, 393–396 trigonometric function, 386, 388–391 logarithmic, 375–380 methods of, 327 of common functions, 329 of hyperbolic functions, 382–385 of parametric equations, 363–368 partial, 397–403 first-order, 397 second-order, 400 product, 332 quotient, 334 successive, 337 Direction cosines, 316 Discontinuous function, 199, 664 Discrete data, 732 Distribution curve, 771 Distribution-free tests, 830 Mann-Whitney test, 837–843 sign test, 830–833 Wilcoxon signed-rank test, 833–837 Dividend, 10 Divisor, 10 D-operator form, 562 Dot product, 313 Double angles, 179–181 integral, 506–508

a

Eigenvalues, 275–280 Eigenvectors, 275–280 Elastic string, 606

911

Element of matrix, 254 Elevation, angle of, 99 Ellipse, 193, 211, 364 Equations, Bessel’s, 592 circle, 123 complex, 232 exponential functions, 40 heat conduction, 605, 610–612 hyperbolic, 157 indicial, 28 iterative methods, 57 Laplace, 603, 605, 612–614 Legendre’s, 597 Newton-Raphson, 345 normals, 726 of circle, 123 quadratic, 8 simple, 5 simultaneous, 7 solving by iterative methods, 57 tangents, 357 transmission, 605 trigonometric, 163–168 wave, 604, 606–610 Estimation of population parameters based on a large sample size, 796 the means of a population based on a small sample size, 801 Euler-Cauchy method, 549–553 Euler’s formula, 245, 708 Euler, Leonhard, 544, 708 Euler’s method, 544–549 Evaluation of trigonometric ratios, 93 Even function, 152, 153, 200, 676 Expectation, 752 Explicit function, 369 Exponent, 24 Exponential form of complex number, 245 Fourier series, 698–710 Exponential functions, 32, 194 graphs of, 35, 194 power series, 33 Extension of solid bar, 121 Extrapolation, 786 Factorisation, 4 Factor theorem, 11 Family of curves, 525 Final value theorem, 629, 719 First moment of area, 440 order differential equations, 526 order partial differential equations, 602 shift theorem, 717

912 Index Fitting data to theoretical distributions, 824 Force, centripetal, 126 Formulae, essential mathematical, 846–863 Fourier coefficients, 665 Fourier, Jean Baptiste Joseph, 664 Fourier series, 144, 663, 664 cosine, 676 exponential form, 698–710 half range, 680, 688 non-periodic over range 2π, 670–675 over any range, 684–689 periodic of period 2π, 663–669 sine, 676 Free vector, 321 Frequency, 142, 732 curve, 771 distribution, 736, 737, 740 domain, 706 polygon, 737, 740 relative, 732 spectrum, 706 Frobenius, Ferdinand Georg, 585 Frobenius method, 585–592 Functional notation, 329 Function of a function, 335, 370 Functions and their curves, 191 of two variables, 411 Fundamental, 665 Gamma function, 594 Gaussian elimination, 272 Gauss, Johann Carl Friedrich, 272 General solution of a differential equation, 527 Gradient of a curve, 327, 331 Graphs of exponential functions, 35 hyperbolic functions, 153 logarithmic functions, 30 trigonometric functions, 132, 137 Grouped data, 736 Growth and decay laws, 40 Half range Fourier series, 680, 688 Half-wave rectifier, 147, 217 Harmonic analysis, 144–149, 690 synthesis, 144 Heat conduction equation, 605, 610–612 Heaviside unit step function, 640 Heron’s formula, 101 Hero’s formula, 101 Higher order differentials, 579 Histogram, 737, 738, 740, 745 of probabilities, 766, 769 Homogeneous, 534

first-order differential equations, 534–537 Horizontal bar chart, 733 component, 289, 306 Hyperbola, 193, 211, 364 rectangular, 193, 211, 364 Hyperbolic functions, 151, 382 graphs of, 153 properties of, 152 differentiation of, 382–385 inverse, 386 solving equations, 157 Hyperbolic identities, 155–157, 170–172 logarithms, 25, 36 Hypotenuse, 90 Hypotheses, 805 Identities, hyperbolic, 155–157 trigonometric, 161–163 Idler gear, 107 i, j, k notation, 298 Imaginary part, 228 Implicit differentiation, 369–374 function, 369 Indefinite integrals, 427 Independent event, 752 Index, 24 Indicator diagram, 220 Indices, laws of, 3 Indicial equations, 28, 588 Industrial inspection, 766 Inflexion, points of, 348, 355 Initial conditions, 602 value theorem, 629, 719 Integral calculus, 327 Integrals, double, 506–508 triple, 508–511 Integrating factor, 538 Integration, 327, 423 algebraic substitution, 461–466 applications of, 431–448 areas, 432 centroids, 436 mean value, 433 r.m.s. value, 433 second moment of area, 440–447 volumes, 435 by partial fractions, 479–484 by parts, 491–496 by t = tan θ/2 substitution, 485–489 change of limits, 464 cosh θ substitution, 476 definite, 427 hyperbolic substitutions, 467 numerical, 455, 512

reduction formulae, 497–505 sine θ substitution, 472 sinh θ substitution, 474 standard, 423 tan θ substitution, 474 trigonometric substitutions, 467 Interpolation, 786 Inverse functions, 94, 201, 386 hyperbolic functions, 386 differentiation of, 393–396 trigonometric functions, 202, 386 differentiation of, 386, 388–391 Inverse Laplace transforms, 632–635 of Heaviside functions, 645 using partial fractions, 635 Inverse matrix, 258, 261 z-transform, 720 Invert-gate, 78 Irregular areas, 213 volumes, 216 Iterative methods, 57 Karnaugh maps, 73–77 Karnaugh, Maurice, 73 Kinetic energy, 55 Kutta, Martin Wilhelm, 554 Lagging angle, 139, 142 Lamina, 440 Laplace, Pierre-Simon, 605, 619 Laplace’s equation, 603, 605, 612–614 Laplace transforms, 619 common notations, 619 definition, 619 derivatives, 627 for differential equations, 648–652 for simultaneous differential equations, 663–658 Heaviside function, 640–644 Inverse of, 632–635 using partial fractions, 635–637 linearity property, 620 of elementary functions, 620–627 of H(t – c), 644 of H(t – c).(t – c), 644 properties of, 625 Latent roots, 275 Laws of Boolean algebra, 70 De Morgan, 72 growth and decay, 40 indices, 3 logarithms, 26, 375 probability, 752 Leading angle, 139, 142 Least-squares regression lines, 786 Legendre, Adrien-Marie, 597

Index Leibniz, Gottfried Wilhelm, 581 Leibniz-Maclaurin method, 583–585 Leibniz notation, 329 theorem, 581 Legendre polynomials, 597, 599 Legendre’s equation, 597 Level of significance, 808 L’Hopital, Gillaume Francois Antoine, 458 L’Hopital’s rule, 458 Limiting values, 328, 457 Linear correlation, 780–784 extrapolation, 786 factors, partial fractions, 18, 479 first-order differential equation, 538–542 interpolation, 786 regression, 785–790 second-order differential equation, 561 velocity, 125 Locus problems, 246–249 Logarithmic differentiation, 375–380 forms of inverse hyperbolic functions, 391 graphs, 30 scale, 43 Logarithms, 24 common, 25 graphs of, 30, 193 hyperbolic, 25 laws of, 26, 375 Napierian, 25 natural, 25 Logic circuits, 77 Log-linear graph paper, 44 Log-log graph paper, 43 Lower class boundary value, 736 Maclaurin, Colin, 451, 583 Maclaurin’s series/theorem, 450–457 numerical integration, 455 Mann-Whitney test, 837–843 Matrices, 253 addition and subtraction, 254 multiplication, 255 to solve simultaneous equations, 265, 266 Matrix, 253 adjoint, 261 determinant of, 257, 259 inverse, 258, 261 notation, 253 reciprocal, 258, 261 transpose, 261 unit, 257

Maxima, minima and saddle points, 411–419 Maximum point, 348 practical problems, 351–355 Mean value, 433, 743–745 of waveform, 217–221 Measures of central tendency, 743 Median, 743, 744 Member of set, 732 Mid-ordinate rule, 214, 515 Minimum point, 348 practical problems, 351–355 Minor of matrix, 259 Mode, 743, 744 Modulus, 233 Moment of a force, 319 Nand-gate, 78 Napierian logarithms, 25, 36 Napier, John, 37 Natural logarithms, 25, 36 Newton-Raphson method, 345 Newton, Sir Isaac, 345 Newtons law of cooling, 41, 341 Non-homogeneous differential equation, 561 Non-parametric tests, 822, 830 Non-right-angled triangles, 101 Nor-gate, 78 Norm, 315 Normal, 357 approximation to a binomial distribution, 806, 810 distribution, 771–776 equations, 786 probability curve, 772 paper, 776 standard variate, 772 Nose-to-tail method, 287 Not-function, 66 -gate, 77 Numerical integration, 455, 512 accuracy of, 520 method of harmonic analysis, 144, 690–697 methods for first-order differential equations, 543–559 Odd function, 152, 153, 200, 676 Ogive, 737, 741 Order of precedence, 4 Or-function, 66 -gate, 77 Osborne’s rule, 155, 171

913

Pappus of Alexandria, 438 Pappus theorem, 438 Parabola, 210, 364 Parallel axis theorem, 441 Parallelogram method, 287 Parameter, 363 Parametric equations, 363, 364 differentiation of, 364–368 Partial differential equations, 601 Partial derivatives, 397 differentiation, 397–403 Partial differential equations, 601–614 Partial fractions, 17 inverse Laplace transforms, 635–637 integration, using, 479 linear factors, 18, 479 quadratic factors, 22, 482 repeated linear factors, 20, 481 Partial integration, 602 Particular integral, 569 solution of differential equation, 527, 562 Pascal, Blaise, 49 Pascal’s triangle, 48 Pearson product-moment correlation coefficient, 780 Percentage component bar chart, 733, 734 relative frequency, 732 Percentile, 748 Percentile values for Chi-square distribution, 825 Student’s t distribution, 801, 803 Period, 138, 664 Periodic function, 138, 199, 302, 664 combination of, 301–311 Periodic time, 142 Permutations, 758 Perpendicular axis theorem, 441 Phasor, 141, 303, 707, 708 Pictogram, 773 Pie diagram, 733, 735 Planimeter, 213 Plotting periodic functions, 302 Point and interval estimates, 786 Point of contraflexure, 9 inflexion, 348, 355 Poiseuille’s equation, 359 Poisson distribution, 767–770, 810 Poisson, Simeon Denis, 768 Polar co-ordinates, 111–116 curves, 194 form, 228–233 Poles, 637 Pole-zero diagram, 637, 638 Pol/Rec function, 115

914 Index Polynomial division, 9 Polynomial, Legendre’s, 597, 599 Polytropic gas law, 29 Population, 732 Power, 24 Power series for ex , 33 cosh x and sinh x, 159 Power series methods of solving differential equations, 578–600 by Frobenius’s method, 585–592 by Leibniz-Maclaurin method, 583–585 Powers of complex numbers, 242 Power waveforms, 183–186 Practical binomial theorem, 54–56 maximum and minimum, 351–355 trigonometry, 104–108 Precedence, 4 Presentation of grouped data, 736 statistical data, 731 ungrouped data, 733 Principal value, 234 Probability, 751–758 laws of, 752 paper, 776 Production of sine and cosine waves, 136 Product-moment formula, 780 Product rule of differentiation, 332 Projectile path, 9 Properties of circles, 117 z-transforms, 717 Pythagoras of Samos, 90 Pythagoras, theorem of, 90 Quadrant, 118 Quadratic equations, 8 factors, partial fractions, 22, 482 graphs, 192 Quartiles, 748 Quotient rule of differentiation, 334 Radian, 119, 142 Radius, 117 of curvature, 367 of gyration, 440 Ranking, 744 Raphson, Joseph, 345 Rates of change, 340, 405–408 Reaction moment of cantilever, 6 Real part of complex number, 228 Reciprocal matrix, 258, 261 motion, 9 ratios, 92 Rectangular hyperbola, 193, 211, 364

Recurrence formula, 583 relation, 583 Reduction formulae, 497–505 of exponential laws to linear form, 43 Regression, 785 coefficients, 786 linear, 785–790 Relation between trigonometric and hyperbolic functions, 169–172 Relative frequency, 732 velocity, 297 Reliability, 796 Remainder theorem, 13 Repeated linear factors, partial fractions, 20, 481 Resolution of vectors, 289 Resultant phasor by complex numbers, 308–310 horizontal and vertical components, 306 phasor diagrams, 303 plotting, 302, 303 sine and cosine rules, 305 Right-angled triangles, 97 Rocket velocity, 40 Rodrigues, Benjamin Olinde, 599 Rodrigue’s formula, 599 R.m.s. values, 433 Roots of complex numbers, 243 Runge, Carl David Tolme, 554 Runge-Kutta method, 554–559 Saddle point, 412 Sample, 732, 792 Sampling and estimation theories, 792 distribution of the means, 793 distributions, 792, 793 statistics, 797 Scalar product, 312, 313 application of, 316 Scalar multiplication, 244 quantity, 285 Scatter diagram, 780 Schering bridge, 239 Secant, 92 Sech, 152 Second moment of area, 440–447 order differential equations, 526, 568–577 shift theorem, 718 Sector, 118 area of, 120 Segment, 118 Semicircle, 118 Semi-interquartile range, 749

Separation of variables, 527, 605 Sequences, 714–716 Series, binomial, 48, 50 exponential, 33 Maclaurin’s, 451 sinh and cosh, 159 Set, 732 Shift theorem, 625 Sign test, 830–833 table of critical values, 832 Significance testing, 805 tests for population means, 812 Simple equations, 5 Simpson’s rule, 214, 516 Simpson, Thomas, 214, 516 Simultaneous differential equations by Laplace transforms, 653–658 Simultaneous equations, 7 by Cramers rule, 271 by determinants, 268 by Gaussian elimination, 272 by matrices, 266 Sine, 91 curves, 137 rule, 101, 305 wave, 136, 217 wave production, 136 Sine θ substitution, 472 Sinh, 151 series, 159 Sinh θ substitution, 474 Sinusoidal form, A sin(ωt ± α), 141 Small changes, 358, 408 Solution of any triangle, 100 right-angled triangles, 97 Space diagram, 297 Spectrum of waveform, 706 Square roots of complex number, 243 wave, 145 Standard curves, 191 derivatives, 330–332 deviation, 746, 747 error of the means, 793 integration, 423–430 Stationary points, 348 Statistical table of normal curve, 774 terminology, 732 Steel bar cooling, 39 Stiffness matrix, 274 Straight line, 192 Strain, 7 Stress, 8 Student’s t distribution, 801, 803 Successive approximations, 61 differentiation, 337 Surface tension of liquid, 367

Index Switching circuits, 60 Symmetry relationships, 703–706 Table of Laplace transforms, 621, 626 normal curve, 774 z-transforms, 716 Tally diagram, 736, 737, 739 Tangent, 91, 118, 357 Tangential velocity, 320 Tanh, 151 Tan θ substitution, 474 Tapered groove, 121 Taylor, Brook, 544 Taylor’s series, 544 Testing for a normal distribution, 776–778 Theodolite, 99 Theorems: binomial, 48, 50–53 Maclaurin’s, 450–457 Pappus, 438 Parallel axis, 441 Perpendicular axis, 441 Pythagoras, 90 Total differential, 404 Transfer function, 239, 637 Transformations, 194–199 Translation of z-transform, 719 Transmission equation, 605 line current, 239 matrix, 257 Transpose of matrix, 261 Transposition of formulae, 6 Trapezoidal rule, 213, 512, 691 Trial solution, 606

Triangle, area of, 101 Trigonometric ratios, 91 evaluation of, 93 and hyperbolic substitutions, integration, 467 equations, 163–168 functions, 132, 192 identities, 161–163 inverse function, 202 waveforms, 132 Trigonometry, 89, 90 practical situations, 104–108 Triple integrals, 508–511 Truth table, 66 t = tan θ/2 substitution, 485–489 Turning points, 348–351 Type 1 and type 2 errors, 806 Ungrouped data, 733–736 Unit matrix, 257 step function, 640 triad, 312 Universal logic gates, 81–83 Upper class boundary value, 736 Variance, 747 Vector addition, 287–295 nose-to-tail method, 287 parallelogram method, 287 Vector drawing, 286 equation of a line, 321 products, 312, 317–320 applications of, 319 quantities, 285 Vectors, 285 characteristic, 275

Vector subtraction, 295–297 Velocity and acceleration, 342 Velocity, angular, 125 linear, 125 relative, 297 Vertical bar chart, 733 component, 289, 306 Viscosity, 56 Volumes of irregular solids, 216 solids of revolution, 435 Young’s modulus, 7 Wallis, John, 503 Wallis’s formula, 503 Wave equation, 604, 606–610 Waveform analyser, 144 Wilcoxon signed-rank test, 833–837 Work done, 316 in isothermal expansion, 39 Zeros (and poles), 637 Z-transforms, 713 definition of, 714 derivative of, 719 final value theorem, 719 first shift theorem, 717 initial value theorem, 719 inverse, 720 properties of, 717 second shift theorem, 718 sequences, 714–716 table of, 716 to solve difference equations, 722 translation, 719

915