Engineering Mathematics

In memory of Elizabeth

Engineering Mathematics

Fourth Edition

JOHN BIRD, BSc(Hons) CMath, FIMA, CEng, MIEE, FCollP, FIIE

Newnes

OXFORD AMSTERDAM BOSTON LONDON NEW YORK PARIS

SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO

Newnes

An imprint of Elsevier Science

Linacre House, Jordan Hill, Oxford OX2 8DP

200 Wheeler Road, Burlington MA 01803

First published 1989

Second edition 1996

Reprinted 1998 (twice), 1999

Third edition 2001

Fourth edition 2003

Copyright 2001, 2003, John Bird. All rights reserved

The right of John Bird to be identiﬁed as the author of this work

has been asserted in accordance with the Copyright, Designs and

Patents Act 1988

No part of this publication may be reproduced in any material

form (including photocopying or storing in any medium by

electronic means and whether or not transiently or incidentally to some

other use of this publication) without the written permission of the

copyright holder except in accordance with the provisions of the Copyright,

Designs and Patents Act 1988 or under the terms of a licence issued by the

Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London,

England W1T 4LP. Applications for the copyright holder’s written

permission to reproduce any part of this publication should be

addressed to the publisher

Permissions may be sought directly from Elsevier’s Science and Technology Rights

Department in Oxford, UK: phone: (+44) (0) 1865 843830; fax: (+44) (0) 1865

853333; e-mail: [email protected] You may also complete your request

on-line via the Elsevier Science homepage (http://www.elsevier.com), by selecting

‘Customer Support’ and then ‘Obtaining Permissions’

British Library Cataloguing in Publication Data

A catalogue record for this book is available from the British Library

ISBN 0 7506 5776 6

For information on all Newnes publications visit our website at www.Newnespress.com

Typeset by Laserwords Private Limited, Chennai, India

Printed and bound in Great Britain

Contents

Preface xi 7 Partial fractions 51

7.1 Introduction to partial fractions 51

Part 1 Number and Algebra 1 7.2 Worked problems on partial fractions

with linear factors 51

1 Revision of fractions, decimals and 7.3 Worked problems on partial fractions

percentages 1 with repeated linear factors 54

1.1 Fractions 1 7.4 Worked problems on partial fractions

1.2 Ratio and proportion 3 with quadratic factors 55

1.3 Decimals 4

1.4 Percentages 7 8 Simple equations 57

8.1 Expressions, equations and

2 Indices and standard form 9 identities 57

2.1 Indices 9 8.2 Worked problems on simple

2.2 Worked problems on indices 9 equations 57

2.3 Further worked problems on 8.3 Further worked problems on simple

indices 11 equations 59

2.4 Standard form 13 8.4 Practical problems involving simple

2.5 Worked problems on standard equations 61

form 13 8.5 Further practical problems involving

2.6 Further worked problems on standard simple equations 62

form 14

Assignment 2 64

3 Computer numbering systems 16

3.1 Binary numbers 16 9 Simultaneous equations 65

3.2 Conversion of binary to decimal 16 9.1 Introduction to simultaneous

3.3 Conversion of decimal to binary 17 equations 65

3.4 Conversion of decimal to binary via 9.2 Worked problems on simultaneous

octal 18 equations in two unknowns 65

3.5 Hexadecimal numbers 20 9.3 Further worked problems on

simultaneous equations 67

4 Calculations and evaluation of 9.4 More difﬁcult worked problems on

formulae 24 simultaneous equations 69

4.1 Errors and approximations 24 9.5 Practical problems involving

4.2 Use of calculator 26 simultaneous equations 70

4.3 Conversion tables and charts 28

4.4 Evaluation of formulae 30 10 Transposition of formulae 74

10.1 Introduction to transposition of

Assignment 1 33 formulae 74

10.2 Worked problems on transposition of

5 Algebra 34 formulae 74

5.1 Basic operations 34 10.3 Further worked problems on

5.2 Laws of Indices 36 transposition of formulae 75

5.3 Brackets and factorisation 38 10.4 Harder worked problems on

5.4 Fundamental laws and precedence 40 transposition of formulae 77

5.5 Direct and inverse proportionality 42

11 Quadratic equations 80

6 Further algebra 44 11.1 Introduction to quadratic equations 80

6.1 Polynomial division 44 11.2 Solution of quadratic equations by

6.2 The factor theorem 46 factorisation 80

6.3 The remainder theorem 48

vi CONTENTS Multiple choice questions on chapters 1 to

16 127

11.3 Solution of quadratic equations by

‘completing the square’ 82 Part 2 Mensuration 131

11.4 Solution of quadratic equations by 17 Areas of plane ﬁgures 131

formula 84 17.1 Mensuration 131

17.2 Properties of quadrilaterals 131

11.5 Practical problems involving quadratic 17.3 Worked problems on areas of plane

equations 85 ﬁgures 132

17.4 Further worked problems on areas of

11.6 The solution of linear and quadratic plane ﬁgures 135

equations simultaneously 87 17.5 Worked problems on areas of

composite ﬁgures 137

12 Logarithms 89 17.6 Areas of similar shapes 138

12.1 Introduction to logarithms 89

12.2 Laws of logarithms 89 18 The circle and its properties 139

12.3 Indicial equations 92 18.1 Introduction 139

12.4 Graphs of logarithmic functions 93 18.2 Properties of circles 139

18.3 Arc length and area of a sector 140

Assignment 3 94 18.4 Worked problems on arc length and

sector of a circle 141

13 Exponential functions 95 18.5 The equation of a circle 143

13.1 The exponential function 95

13.2 Evaluating exponential functions 95 19 Volumes and surface areas of

13.3 The power series for ex 96 common solids 145

13.4 Graphs of exponential functions 98 19.1 Volumes and surface areas of

13.5 Napierian logarithms 100 regular solids 145

13.6 Evaluating Napierian logarithms 100 19.2 Worked problems on volumes and

13.7 Laws of growth and decay 102 surface areas of regular solids 145

19.3 Further worked problems on volumes

14 Number sequences 106 and surface areas of regular

14.1 Arithmetic progressions 106 solids 147

14.2 Worked problems on arithmetic 19.4 Volumes and surface areas of frusta of

progression 106 pyramids and cones 151

14.3 Further worked problems on arithmetic 19.5 The frustum and zone of a sphere 155

progressions 107 19.6 Prismoidal rule 157

14.4 Geometric progressions 109 19.7 Volumes of similar shapes 159

14.5 Worked problems on geometric

progressions 110 20 Irregular areas and volumes and mean

14.6 Further worked problems on geometric values of waveforms 161

progressions 111 20.1 Areas of irregular ﬁgures 161

14.7 Combinations and permutations 112 20.2 Volumes of irregular solids 163

20.3 The mean or average value of a

15 The binomial series 114 waveform 164

15.1 Pascal’s triangle 114

15.2 The binomial series 115 Assignment 5 168

15.3 Worked problems on the binomial

series 115 Part 3 Trigonometry 171

15.4 Further worked problems on the

binomial series 117 21 Introduction to trigonometry 171

15.5 Practical problems involving the 21.1 Trigonometry 171

binomial theorem 120 21.2 The theorem of Pythagoras 171

21.3 Trigonometric ratios of acute

16 Solving equations by iterative angles 172

methods 123

16.1 Introduction to iterative methods 123

16.2 The Newton–Raphson method 123

16.3 Worked problems on the

Newton–Raphson method 123

Assignment 4 126

21.4 Fractional and surd forms of CONTENTS vii

trigonometric ratios 174

25.7 Worked problems (iv) on trigonometric

21.5 Solution of right-angled triangles 175 equations 212

21.6 Angles of elevation and

26 Compound angles 214

depression 176 26.1 Compound angle formulae 214

21.7 Evaluating trigonometric ratios of any 26.2 Conversion of a sin ωt C b cos ωt into

R sin ωt C ˛) 216

angles 178 26.3 Double angles 220

21.8 Trigonometric approximations for small 26.4 Changing products of sines and cosines

into sums or differences 221

angles 181 26.5 Changing sums or differences of sines

and cosines into products 222

22 Trigonometric waveforms 182

22.1 Graphs of trigonometric functions 182 Assignment 7 224

22.2 Angles of any magnitude 182

22.3 The production of a sine and cosine Multiple choice questions on chapters 17

wave 185 to 26 225

22.4 Sine and cosine curves 185

22.5 Sinusoidal form A sin ωt š ˛ 189 Part 4 Graphs 231

22.6 Waveform harmonics 192

27 Straight line graphs 231

23 Cartesian and polar co-ordinates 194 27.1 Introduction to graphs 231

23.1 Introduction 194 27.2 The straight line graph 231

23.2 Changing from Cartesian into polar 27.3 Practical problems involving straight

co-ordinates 194 line graphs 237

23.3 Changing from polar into Cartesian

co-ordinates 196 28 Reduction of non-linear laws to linear

23.4 Use of R ! P and P ! R functions on form 243

calculators 197 28.1 Determination of law 243

28.2 Determination of law involving

Assignment 6 198 logarithms 246

24 Triangles and some practical 29 Graphs with logarithmic scales 251

applications 199 29.1 Logarithmic scales 251

24.1 Sine and cosine rules 199 29.2 Graphs of the form y D axn 251

24.2 Area of any triangle 199 29.3 Graphs of the form y D abx 254

24.3 Worked problems on the solution of 29.4 Graphs of the form y D aekx 255

triangles and their areas 199

24.4 Further worked problems on the 30 Graphical solution of equations 258

solution of triangles and their 30.1 Graphical solution of simultaneous

areas 201 equations 258

24.5 Practical situations involving 30.2 Graphical solution of quadratic

trigonometry 203 equations 259

24.6 Further practical situations involving 30.3 Graphical solution of linear and

trigonometry 205 quadratic equations simultaneously

263

25 Trigonometric identities and 30.4 Graphical solution of cubic equations

equations 208 264

25.1 Trigonometric identities 208

25.2 Worked problems on trigonometric 31 Functions and their curves 266

identities 208 31.1 Standard curves 266

25.3 Trigonometric equations 209 31.2 Simple transformations 268

25.4 Worked problems (i) on trigonometric 31.3 Periodic functions 273

equations 210 31.4 Continuous and discontinuous

25.5 Worked problems (ii) on trigonometric functions 273

equations 211 31.5 Even and odd functions 273

25.6 Worked problems (iii) on trigonometric 31.6 Inverse functions 275

equations 212

Assignment 8 279

viii CONTENTS 38.3 Worked problems on probability 327

38.4 Further worked problems on

Part 5 Vectors 281

probability 329

32 Vectors 281 38.5 Permutations and combinations 331

32.1 Introduction 281

32.2 Vector addition 281 39 The binomial and Poisson distribution 333

32.3 Resolution of vectors 283 39.1 The binomial distribution 333

32.4 Vector subtraction 284 39.2 The Poisson distribution 336

33 Combination of waveforms 287 Assignment 10 339

33.1 Combination of two periodic

functions 287 40 The normal distribution 340

33.2 Plotting periodic functions 287 40.1 Introduction to the normal distribution

33.3 Determining resultant phasors by 340

calculation 288 40.2 Testing for a normal distribution 344

Part 6 Complex Numbers 291 41 Linear correlation 347

41.1 Introduction to linear correlation 347

34 Complex numbers 291 41.2 The product-moment formula for

34.1 Cartesian complex numbers 291 determining the linear correlation

34.2 The Argand diagram 292 coefﬁcient 347

34.3 Addition and subtraction of complex 41.3 The signiﬁcance of a coefﬁcient of

numbers 292 correlation 348

34.4 Multiplication and division of complex 41.4 Worked problems on linear

numbers 293 correlation 348

34.5 Complex equations 295

34.6 The polar form of a complex 42 Linear regression 351

number 296 42.1 Introduction to linear regression 351

34.7 Multiplication and division in polar 42.2 The least-squares regression lines 351

form 298 42.3 Worked problems on linear

34.8 Applications of complex numbers 299 regression 352

35 De Moivre’s theorem 303 43 Sampling and estimation theories 356

35.1 Introduction 303 43.1 Introduction 356

35.2 Powers of complex numbers 303 43.2 Sampling distributions 356

35.3 Roots of complex numbers 304 43.3 The sampling distribution of the

means 356

Assignment 9 306 43.4 The estimation of population

parameters based on a large sample

Part 7 Statistics 307 size 359

43.5 Estimating the mean of a population

36 Presentation of statistical data 307 based on a small sample size 364

36.1 Some statistical terminology 307

36.2 Presentation of ungrouped data 308 Assignment 11 368

36.3 Presentation of grouped data 312

Multiple choice questions on chapters 27

37 Measures of central tendency and to 43 369

dispersion 319

37.1 Measures of central tendency 319 Part 8 Differential Calculus 375

37.2 Mean, median and mode for discrete

data 319 44 Introduction to differentiation 375

37.3 Mean, median and mode for grouped 44.1 Introduction to calculus 375

data 320 44.2 Functional notation 375

37.4 Standard deviation 322 44.3 The gradient of a curve 376

37.5 Quartiles, deciles and percentiles 324 44.4 Differentiation from ﬁrst

principles 377

38 Probability 326

38.1 Introduction to probability 326

38.2 Laws of probability 326

44.5 Differentiation of y D axn by the CONTENTS ix

general rule 379

49.6 Worked problems on integration using

44.6 Differentiation of sine and cosine the tan Â substitution 424

functions 380

Assignment 13 425

44.7 Differentiation of eax and ln ax 382

50 Integration using partial fractions 426

45 Methods of differentiation 384 50.1 Introduction 426

45.1 Differentiation of common functions 50.2 Worked problems on integration using

384 partial fractions with linear

45.2 Differentiation of a product 386 factors 426

45.3 Differentiation of a quotient 387 50.3 Worked problems on integration using

45.4 Function of a function 389 partial fractions with repeated linear

45.5 Successive differentiation 390 factors 427

50.4 Worked problems on integration using

46 Some applications of differentiation 392 partial fractions with quadratic

46.1 Rates of change 392 factors 428

46.2 Velocity and acceleration 393

46.3 Turning points 396 q

46.4 Practical problems involving maximum 51 The t = substitution 430

and minimum values 399

46.5 Tangents and normals 403 2

46.6 Small changes 404 51.1 Introduction 430

Assignment 12 406 Â

51.2 Worked problems on the t D tan

Part 9 Integral Calculus 407

2

47 Standard integration 407 substitution 430

47.1 The process of integration 407 51.3 Further worked problems on the

47.2 The general solution of integrals of the

form axn 407 Â

47.3 Standard integrals 408 t D tan substitution 432

47.4 Deﬁnite integrals 411

2

48 Integration using algebraic substitutions

414 52 Integration by parts 434

48.1 Introduction 414 52.1 Introduction 434

48.2 Algebraic substitutions 414 52.2 Worked problems on integration by

48.3 Worked problems on integration using parts 434

algebraic substitutions 414 52.3 Further worked problems on integration

48.4 Further worked problems on integration by parts 436

using algebraic substitutions 416

48.5 Change of limits 416 53 Numerical integration 439

53.1 Introduction 439

49 Integration using trigonometric 53.2 The trapezoidal rule 439

substitutions 418 53.3 The mid-ordinate rule 441

49.1 Introduction 418 53.4 Simpson’s rule 443

49.2 Worked problems on integration of

sin2 x, cos2 x, tan2 x and cot2 x 418 Assignment 14 447

49.3 Worked problems on powers of sines

and cosines 420 54 Areas under and between curves 448

49.4 Worked problems on integration of 54.1 Area under a curve 448

products of sines and cosines 421 54.2 Worked problems on the area under a

49.5 Worked problems on integration using curve 449

the sin Â substitution 422 54.3 Further worked problems on the area

under a curve 452

54.4 The area between curves 454

55 Mean and root mean square values 457

55.1 Mean or average values 457

55.2 Root mean square values 459

56 Volumes of solids of revolution 461

56.1 Introduction 461

56.2 Worked problems on volumes of solids

of revolution 461

x CONTENTS 59.4 De Morgan’s laws 490

59.5 Karnaugh maps 491

56.3 Further worked problems on volumes 59.6 Logic circuits 495

of solids of revolution 463 59.7 Universal logic circuits 500

57 Centroids of simple shapes 466 60 The theory of matrices and determinants

57.1 Centroids 466 504

57.2 The ﬁrst moment of area 466 60.1 Matrix notation 504

57.3 Centroid of area between a curve and 60.2 Addition, subtraction and multiplication

the x-axis 466 of matrices 504

57.4 Centroid of area between a curve and 60.3 The unit matrix 508

the y-axis 467 60.4 The determinant of a 2 by 2 matrix

57.5 Worked problems on centroids of 508

simple shapes 467 60.5 The inverse or reciprocal of a 2 by 2

57.6 Further worked problems on centroids matrix 509

of simple shapes 468 60.6 The determinant of a 3 by 3 matrix

57.7 Theorem of Pappus 471 510

60.7 The inverse or reciprocal of a 3 by 3

58 Second moments of area 475 matrix 511

58.1 Second moments of area and radius of

gyration 475 61 The solution of simultaneous equations by

58.2 Second moment of area of regular matrices and determinants 514

sections 475 61.1 Solution of simultaneous equations by

58.3 Parallel axis theorem 475 matrices 514

58.4 Perpendicular axis theorem 476 61.2 Solution of simultaneous equations by

58.5 Summary of derived results 476 determinants 516

58.6 Worked problems on second moments 61.3 Solution of simultaneous equations

of area of regular sections 476 using Cramers rule 520

58.7 Worked problems on second moments

of areas of composite areas 480 Assignment 16 521

Assignment 15 482 Multiple choice questions on chapters 44–61

522

Part 10 Further Number and Algebra 483

Answers to multiple choice questions 526

59 Boolean algebra and logic circuits 483

59.1 Boolean algebra and switching circuits Index 527

483

59.2 Simplifying Boolean expressions 488

59.3 Laws and rules of Boolean algebra

488

Preface

This fourth edition of ‘Engineering Mathematics’ 1. Algebraic techniques: 10, 14, 15,

covers a wide range of syllabus requirements. In 28–30, 34, 59–61

particular, the book is most suitable for the latest

National Certiﬁcate and Diploma courses and 2. Trigonometry: 22–24, 26

Vocational Certiﬁcate of Education syllabuses in 3. Calculus: 44–49, 52–58

Engineering. 4. Statistical and probability: 36–43

This text will provide a foundation in mathematical (iii) Applied Mathematics in Engineering, the

principles, which will enable students to solve mathe- compulsory unit for Advanced VCE (for-

matical, scientiﬁc and associated engineering princi- merly Advanced GNVQ), to include all or

ples. In addition, the material will provide engineer- part of the following chapters:

ing applications and mathematical principles neces-

sary for advancement onto a range of Incorporated 1. Number and units: 1, 2, 4

Engineer degree proﬁles. It is widely recognised that 2. Mensuration: 17–20

a students’ ability to use mathematics is a key element 3. Algebra: 5, 8–11

in determining subsequent success. First year under- 4. Functions and graphs: 22, 23, 27

graduates who need some remedial mathematics will 5. Trigonometry: 21, 24

also ﬁnd this book meets their needs.

(iv) Further Mathematics for Engineering, the

In Engineering Mathematics 4th Edition, theory optional unit for Advanced VCE (formerly

is introduced in each chapter by a simple outline of Advanced GNVQ), to include all or part of

essential deﬁnitions, formulae, laws and procedures. the following chapters:

The theory is kept to a minimum, for problem solv-

ing is extensively used to establish and exemplify 1. Algebra and trigonometry: 5, 6,

the theory. It is intended that readers will gain real 12–15, 21, 25

understanding through seeing problems solved and

then through solving similar problems themselves. 2. Graphical and numerical techniques:

20, 22, 26–31

For clarity, the text is divided into ten topic

areas, these being: number and algebra, mensura- 3. Differential and integral calculus:

tion, trigonometry, graphs, vectors, complex num- 44–47, 54

bers, statistics, differential calculus, integral calculus

and further number and algebra. (v) The Mathematics content of Applied Sci-

(vi) ence and Mathematics for Engineering,

This new edition will cover the following syl- (vii) for Intermediate GNVQ

labuses: (viii) Mathematics for Engineering, for Founda-

tion and Intermediate GNVQ

(i) Mathematics for Technicians, the core unit Mathematics 2 and Mathematics 3 for City

for National Certiﬁcate/Diploma courses in & Guilds Technician Diploma in Telecom-

Engineering, to include all or part of the munications and Electronic Engineering

following chapters: Any introductory/access/foundation co-

urse involving Engineering Mathematics at

1. Algebra: 2, 4, 5, 8–13, 17, 19, 27, 30 University, Colleges of Further and Higher

2. Trigonometry: 18, 21, 22, 24 education and in schools.

3. Statistics: 36, 37

4. Calculus: 44, 46, 47, 54 Each topic considered in the text is presented in

a way that assumes in the reader little previous

(ii) Further Mathematics for Technicians, knowledge of that topic.

the optional unit for National Certiﬁ-

cate/Diploma courses in Engineering, to

include all or part of the following chapters:

xii ENGINEERING MATHEMATICS lecturers could set the Assignments for students to

attempt as part of their course structure. Lecturers’

‘Engineering Mathematics 4th Edition’ provides may obtain a complimentary set of solutions of the

a follow-up to ‘Basic Engineering Mathematics’ Assignments in an Instructor’s Manual available

and a lead into ‘Higher Engineering Mathemat- from the publishers via the internet — full worked

ics’. solutions and mark scheme for all the Assignments

are contained in this Manual, which is available to

This textbook contains over 900 worked lecturers only. To obtain a password please e-mail

problems, followed by some 1700 further [email protected] with the following details:

problems (all with answers). The further problems course title, number of students, your job title and

are contained within some 208 Exercises; each work postal address.

Exercise follows on directly from the relevant

section of work, every two or three pages. In To download the Instructor’s Manual visit

addition, the text contains 234 multiple-choice http://www.newnespress.com and enter the book

questions. Where at all possible, the problems title in the search box, or use the following direct

mirror practical situations found in engineering URL: http://www.bh.com/manuals/0750657766/

and science. 500 line diagrams enhance the

understanding of the theory. ‘Learning by Example’ is at the heart of ‘Engi-

neering Mathematics 4th Edition’.

At regular intervals throughout the text are some

16 Assignments to check understanding. For exam- John Bird

ple, Assignment 1 covers material contained in

Chapters 1 to 4, Assignment 2 covers the material University of Portsmouth

in Chapters 5 to 8, and so on. These Assignments

do not have answers given since it is envisaged that

Part 1 Number and Algebra

1

Revision of fractions, decimals

and percentages

1.1 Fractions Alternatively:

Step (2) Step (3)

When 2 is divided by 3, it may be written as 2 or ##

3

2 1 2 7ð1 C 3ð2

2/3. 3 is called a fraction. The number above the CD

37 21

line, i.e. 2, is called the numerator and the number

"

below the line, i.e. 3, is called the denominator.

Step (1)

When the value of the numerator is less than

the value of the denominator, the fraction is called Step 1: the LCM of the two denominators;

Step 2:

a proper fraction; thus 2 is a proper fraction. 1

3 Step 3: 3

When the value of the numerator is greater than for the fraction , 3 into 21 goes 7 times,

the denominator, the fraction is called an improper 7 ð the numerator is 7 ð 1;

fraction. Thus 7 is an improper fraction and can also for the fraction 2 , 7 into 21 goes 3 times,

3 7

be expressed as a mixed number, that is, an integer

3 ð the numerator is 3 ð 2.

7

and a proper fraction. Thus the improper fraction 3

is equal to the mixed number 2 1 . 1 2 7 C 6 13

3 Thus C D D as obtained previously.

When a fraction is simpliﬁed by dividing the 3 7 21 21

numerator and denominator by the same number,

the process is called cancelling. Cancelling by 0 is 21

Problem 2. Find the value of 3 2

not permissible.

36

12 One method is to split the mixed numbers into

Problem 1. Simplify C integers and their fractional parts. Then

37 21 2 1

3 2 D 3C 2C

The lowest common multiple (i.e. LCM) of the two 36 3

denominators is 3 ð 7, i.e. 21 6

Expressing each fraction so that their denomina- 2 1

tors are 21, gives: D3C 2

12 1723 7 6 3 6

CDðCðD C 4 131

D1C

3 7 3 7 7 3 21 21 6 D1 D1

7 C 6 13 662

DD Another method is to express the mixed numbers as

21 21 improper fractions.

2 ENGINEERING MATHEMATICS 8 1 7 24 8 8 ð 1 ð 8

D ðð D

9 2 9 2 11

Since 3 D , then 3 D C D 5 13 71 5ð1ð1

64 4

3 3 33 3 D D 12

1 12 1 13 55

Similarly, 2 D C D

6 66 6 3 12

2 1 11 13 22 13 9 1 Problem 6. Simplify ł

Thus 3 2 D D D D 1

3 63 6 6 66 2 7 21

as obtained previously.

Problem 3. Determine the value of 3

512

3 ł 12 D 7

4 3 C1 7 21 12

845

21

512 512 Multiplying both numerator and denominator by the

4 3 C1 D 4 3C1 C C reciprocal of the denominator gives:

845 845

5 ð 5 10 ð 1 C 8 ð 2 3 1 3 21 3 3

D2C ð

D 17 12 4 D 4 D 3

40 7 1 12 21 1 1 4

25 10 C 16 12

D2C

21 ð

40 1 21 12 1

31 31

D2C D2 This method can be remembered by the rule: invert

40 40 the second fraction and change the operation from

division to multiplication. Thus:

3 14 3 12 1 3 21 3 3

Problem 4. Find the value of ð ł D ð D as obtained previously.

7 15 7 21 1 7 12 4 4

Dividing numerator and denominator by 3 gives: 31

Problem 7. Find the value of 5 ł 7

1 3 14 1 14 1 ð 14

ð Dð D 53

7 15 5 7 5 7 ð 5 The mixed numbers must be expressed as improper

Dividing numerator and denominator by 7 gives: fractions. Thus,

1 ð 14 2 1 ð 2 2 3 1 28 22 14 28 3 42

DD 5 ł7 D ł D ð D

5 3 5 3 5 22 11 55

17ð5 1ð5 5

This process of dividing both the numerator and

denominator of a fraction by the same factor(s) is

called cancelling.

Problem 8. Simplify

313 1 21 31

Problem 5. Evaluate 1 ð 2 ð 3 Cłð

3 54 83

537

Mixed numbers must be expressed as improper The order of precedence of operations for problems

containing fractions is the same as that for inte-

fractions before multiplication can be performed. gers, i.e. remembered by BODMAS (Brackets, Of,

Division, Multiplication, Addition and Subtraction).

Thus, Thus,

313

1 ð2 ð3

537

53 61 21 3 1 21 31

DCðCð C Cłð

55 33 77 3 54 83

REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 3

1 4ð2C5ð1 31 23 2 12

Dł (B) 2. (a) C (b) C

(D) 7 11 9

3 20 24 8 (M) 73

1 13 8 2 (S) 43 47

Dð (b)

3 5 20 1 (a)

1 26 77 63

D

35 3 2 145

5 ð 1 3 ð 26 3. (a) 10 8 (b) 3 4 C 1

D

7 3 456

15 16 17

73 13

D D −4 (a) 1 (b)

15 15 21 60

35 17 15

4. (a) ð (b) ð

49 35 119

Problem 9. Determine the value of 53

(a) (b)

12 49

7 11 131 37 2 13 7 4

of 3 2 C 5 ł

6 24 8 16 2 5. (a) ð ð 1 (b) ð 4 ð 3

59 7 17 11 39

7 11 13 1 3

of 3 2 C 5 ł 2 (a) (b) 11

5

6 24 8 16 3 45 15

7 1 41 3 1 6. (a) ł (b) 1 ł 2

D of 1 C ł 8 64 39

(B)

6 4 8 16 2 (O) 8 12

7 5 41 3 1 (D) (a) (b)

DðC ł (M)

6 4 8 16 2 (A) 15 23

(A)

7 5 41 16 2 1 (S) 13 8 1 7

DðC ð 2 7. C ł 1

6 4 18 3 2 5 15 3 24

35 82 1

DC 7 5 3 15 4

24 3 2 8. of 15 ð C ł 5

35 C 656 1

D 15 7 4 16 5

24 2 12132 13

691 1 9. ð łC 126

D 43357

24 2

691 12 21 21 3 28

D 10. ð 1 ł C C 1 2

34 34 5 55

24

679 7 1.2 Ratio and proportion

D D 28

24 24 The ratio of one quantity to another is a fraction, and

is the number of times one quantity is contained in

Now try the following exercise another quantity of the same kind. If one quantity is

directly proportional to another, then as one quan-

Exercise 1 Further problems on fractions tity doubles, the other quantity also doubles. When a

Evaluate the following: quantity is inversely proportional to another, then

as one quantity doubles, the other quantity is halved.

12 7 1

1. (a) C (b) 4 Problem 10. A piece of timber 273 cm

25 16 long is cut into three pieces in the ratio of 3

9 to 7 to 11. Determine the lengths of the three

(a) 3 pieces

(b)

10

16

4 ENGINEERING MATHEMATICS 1 person takes three times as long, i.e.

4 ð 3 D 12 hours,

The total number of parts is 3 C 7 C 11, that is, 21.

Hence 21 parts correspond to 273 cm 5 people can do it in one ﬁfth of the time that

12

273

1 part corresponds to D 13 cm one person takes, that is hours or 2 hours

5

21

3 parts correspond to 3 ð 13 D 39 cm 24 minutes.

7 parts correspond to 7 ð 13 D 91 cm

11 parts correspond to 11 ð 13 D 143 cm Now try the following exercise

i.e. the lengths of the three pieces are 39 cm, Exercise 5 Further problems on ratio and

91 cm and 143 cm. proportion

(Check: 39 C 91 C 143 D 273) 1. Divide 621 cm in the ratio of 3 to 7 to 13.

[81 cm to 189 cm to 351 cm]

Problem 11. A gear wheel having 80 teeth

is in mesh with a 25 tooth gear. What is the 2. When mixing a quantity of paints, dyes of

gear ratio?

four different colours are used in the ratio

80 16

Gear ratio D 80:25 D D D 3.2 of 7:3:19:5. If the mass of the ﬁrst dye

25 5 used is 3 1 g, determine the total mass of

i.e. gear ratio D 16 : 5 or 3.2 : 1 2

the dyes used. [17 g]

Problem 12. An alloy is made up of

metals A and B in the ratio 2.5 : 1 by mass. 3. Determine how much copper and how

How much of A has to be added to 6 kg of much zinc is needed to make a 99 kg

B to make the alloy? brass ingot if they have to be in the

proportions copper : zinc: :8 : 3 by mass.

Ratio A : B: :2.5 : 1 (i.e. A is to B as 2.5 is to 1)

A 2.5 [72 kg : 27 kg]

or D D 2.5 4. It takes 21 hours for 12 men to resurface

B1

A a stretch of road. Find how many men

When B D 6 kg, D 2.5 from which, it takes to resurface a similar stretch of

6

road in 50 hours 24 minutes, assuming

A D 6 ð 2.5 D 15 kg

the work rate remains constant. [5]

Problem 13. If 3 people can complete a

task in 4 hours, how long will it take 5 5. It takes 3 hours 15 minutes to ﬂy from

people to complete the same task, assuming city A to city B at a constant speed. Find

the rate of work remains constant how long the journey takes if

(a) the speed is 1 1 times that of the

2

original speed and

(b) if the speed is three-quarters of the

original speed.

[(a) 2 h 10 min (b) 4 h 20 min]

1.3 Decimals

The more the number of people, the more quickly The decimal system of numbers is based on the

the task is done, hence inverse proportion exists. digits 0 to 9. A number such as 53.17 is called

a decimal fraction, a decimal point separating the

3 people complete the task in 4 hours, integer part, i.e. 53, from the fractional part, i.e. 0.17

A number which can be expressed exactly as REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 5

a decimal fraction is called a terminating deci- 87.23

81.70

mal and those which cannot be expressed exactly 5.53

as a decimal fraction are called non-terminating Thus 87.23 − 81.70 = 5.53

decimals. Thus, 3 D 1.5 is a terminating decimal, Problem 16. Find the value of

2 23.4 17.83 57.6 C 32.68

4

but 3 D 1.33333. . . is a non-terminating decimal. The sum of the positive decimal fractions is

23.4 C 32.68 D 56.08

1.33333. . . can be written as 1.3P, called ‘one point-

The sum of the negative decimal fractions is

three recurring’. 17.83 C 57.6 D 75.43

The answer to a non-terminating decimal may be Taking the sum of the negative decimal fractions

from the sum of the positive decimal fractions gives:

expressed in two ways, depending on the accuracy

56.08 75.43

required: i.e. 75.43 56.08 D −19.35

(i) correct to a number of signiﬁcant ﬁgures, that Problem 17. Determine the value of

is, ﬁgures which signify something, and 74.3 ð 3.8

(ii) correct to a number of decimal places, that is,

the number of ﬁgures after the decimal point.

The last digit in the answer is unaltered if the next

digit on the right is in the group of numbers 0, 1,

2, 3 or 4, but is increased by 1 if the next digit

on the right is in the group of numbers 5, 6, 7, 8

or 9. Thus the non-terminating decimal 7.6183. . .

becomes 7.62, correct to 3 signiﬁcant ﬁgures, since

the next digit on the right is 8, which is in the group

of numbers 5, 6, 7, 8 or 9. Also 7.6183. . . becomes

7.618, correct to 3 decimal places, since the next

digit on the right is 3, which is in the group of

numbers 0, 1, 2, 3 or 4.

Problem 14. Evaluate When multiplying decimal fractions: (i) the numbers

42.7 C 3.04 C 8.7 C 0.06 are multiplied as if they are integers, and (ii) the

position of the decimal point in the answer is such

The numbers are written so that the decimal points that there are as many digits to the right of it as the

are under each other. Each column is added, starting sum of the digits to the right of the decimal points

from the right. of the two numbers being multiplied together. Thus

42.7 (i) 743

3.04 38

8.7

0.06 5 944

22 290

54.50

28 234

Thus 42.7 Y 3.04 Y 8.7 Y 0.06 = 54.50

(ii) As there are 1 C 1 D 2 digits to the right of

Problem 15. Take 81.70 from 87.23 the decimal points of the two numbers being

multiplied together, (74.3 ð 3.8), then

The numbers are written with the decimal points

under each other. 74.3 × 3.8 = 282.34

Problem 18. Evaluate 37.81 ł 1.7, correct

to (i) 4 signiﬁcant ﬁgures and (ii) 4 decimal

places

6 ENGINEERING MATHEMATICS

37.81 Problem 20. Express as decimal fractions:

37.81 ł 1.7 D 97

1.7 (a) and (b) 5

16 8

The denominator is changed into an integer by

multiplying by 10. The numerator is also multiplied

by 10 to keep the fraction the same. Thus

37.81 ł 1.7 D 37.81 ð 10 378.1 (a) To convert a proper fraction to a decimal frac-

D tion, the numerator is divided by the denomi-

1.7 ð 10 17 nator. Division by 16 can be done by the long

division method, or, more simply, by dividing

The long division is similar to the long division of by 2 and then 8:

integers and the ﬁrst four steps are as shown:

22.24117.. 4.50 0.5625

2 9.00 8 4.5000

17 378.100000

34 9

Thus, = 0.5625

38

34 16

41 (b) For mixed numbers, it is only necessary to

34

convert the proper fraction part of the mixed

70

68 number to a decimal fraction. Thus, dealing

20 with the 7 gives:

8

0.875 7

8 7.000 i.e. D 0.875

8

(i) 37.81 ÷ 1.7 = 22.24, correct to 4 signiﬁcant 7

ﬁgures, and Thus 5 = 5.875

8

(ii) 37.81 ÷ 1.7 = 22.2412, correct to 4 decimal Now try the following exercise

places.

Exercise 3 Further problems on decimals

Problem 19. Convert (a) 0.4375 to a proper In Problems 1 to 6, determine the values of

fraction and (b) 4.285 to a mixed number the expressions given:

0.4375 ð 10 000 1. 23.6 C 14.71 18.9 7.421 [11.989]

(a) 0.4375 can be written as

2. 73.84 113.247 C 8.21 0.068

10 000 [ 31.265]

without changing its value,

3. 3.8 ð 4.1 ð 0.7 [10.906]

4375

i.e. 0.4375 D 4. 374.1 ð 0.006 [2.2446]

10 000 5. 421.8 ł 17, (a) correct to 4 signiﬁcant

By cancelling ﬁgures and (b) correct to 3 decimal

places.

4375 875 175 35 7

D D DD [(a) 24.81 (b) 24.812]

10 000 2000 400 80 16 0.0147

7 6. , (a) correct to 5 decimal places

i.e. 0.4375 = 2.3

16 and (b) correct to 2 signiﬁcant ﬁgures.

285 57

[(a) 0.00639 (b) 0.0064]

(b) Similarly, 4.285 D 4 D 4

1000 200

7. Convert to proper fractions: REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 7

(a) 0.65 (b) 0.84 (c) 0.0125 (d) 0.282 (a) 1.875 corresponds to 1.875 ð 100%, i.e.

and (e) 0.024 187.5%

13 21 1 141 3 (b) 0.0125 corresponds to 0.0125 ð 100%, i.e.

(a) (b) (c) (d) (e) 1.25%

20 25 80 500 125 Problem 22. Express as percentages:

52

8. Convert to mixed numbers:

(a) and (b) 1

(a) 1.82 (b) 4.275 (c) 14.125 (d) 15.35 16 5

and (e) 16.2125

41 11 1 To convert fractions to percentages, they are (i) con-

(a) 1 (b) 4 (c) 14 8 verted to decimal fractions and (ii) multiplied by 100

50

7 40 55

(d) 15 17 (a) By division, D 0.3125, hence corre-

20 (e) 16

80 16 16

sponds to 0.3125 ð 100%, i.e. 31.25%

In Problems 9 to 12, express as decimal frac-

tions to the accuracy stated: 2

(b) Similarly, 1 D 1.4 when expressed as a

4

9. , correct to 5 signiﬁcant ﬁgures. 5

decimal fraction.

9

[0.44444] 2

Hence 1 D 1.4 ð 100% D 140%

17

10. , correct to 5 decimal place. 5

27 Problem 23. It takes 50 minutes to machine

[0.62963] a certain part. Using a new type of tool, the

time can be reduced by 15%. Calculate the

9 new time taken

11. 1 , correct to 4 signiﬁcant ﬁgures.

16

[1.563]

31 15% of 50 minutes 15 750

12. 13 , correct to 2 decimal places. D ð 50 D

37 100 100

[13.84] D 7.5 minutes.

hence the new time taken is

50 7.5 D 42.5 minutes.

1.4 Percentages Alternatively, if the time is reduced by 15%, then

Percentages are used to give a common standard it now takes 85% of the original time, i.e. 85% of

and are fractions having the number 100 as their

85 4250 D 42.5 minutes, as above.

25 50 D ð 50 D

denominators. For example, 25 per cent means 100 100

100 Problem 24. Find 12.5% of £378

1

i.e. and is written 25%. 12.5% of £378 means 12.5 ð 378, since per cent

4

means ‘per hundred’. 100

Problem 21. Express as percentages:

(a) 1.875 and (b) 0.0125 12.51 1

Hence 12.5% of £378 D ð 378 D ð 378 D

A decimal fraction is converted to a percentage by 100 8 8

multiplying by 100. Thus,

378 D £47.25

8

8 ENGINEERING MATHEMATICS

Problem 25. Express 25 minutes as a 2. Express as percentages, correct to 3

percentage of 2 hours, correct to the signiﬁcant ﬁgures:

nearest 1%

7 19 11

(a) (b) (c) 1

33 24 16

Working in minute units, 2 hours D 120 minutes.

25 [(a) 21.2% (b) 79.2% (c) 169%]

Hence 25 minutes is ths of 2 hours. By can- 3. Calculate correct to 4 signiﬁcant ﬁgures:

120 (a) 18% of 2758 tonnes (b) 47% of

25 5 18.42 grams (c) 147% of 14.1 seconds

celling, D

[(a) 496.4 t (b) 8.657 g (c) 20.73 s]

120 24

4. When 1600 bolts are manufactured, 36

Expressing 5 as a decimal fraction gives 0.2083P

24 are unsatisfactory. Determine the percent-

Multiplying by 100 to convert the decimal fraction age unsatisfactory. [2.25%]

to a percentage gives:

5. Express: (a) 140 kg as a percentage of

0.2083P ð 100 D 20.83P% 1 t (b) 47 s as a percentage of 5 min

(c) 13.4 cm as a percentage of 2.5 m

Thus 25 minutes is 21% of 2 hours, correct to the

nearest 1%. [(a) 14% (b) 15.67% (c) 5.36%]

Problem 26. A German silver alloy consists 6. A block of monel alloy consists of 70%

of 60% copper, 25% zinc and 15% nickel.

Determine the masses of the copper, zinc and

nickel in a 3.74 kilogram block of the alloy

nickel and 30% copper. If it contains

By direct proportion: 88.2 g of nickel, determine the mass of

copper in the block. [37.8 g]

100% corresponds to 3.74 kg 7. A drilling machine should be set to

1% corresponds to 3.74 D 0.0374 kg 250 rev/min. The nearest speed available

100 on the machine is 268 rev/min. Calculate

60% corresponds to 60 ð 0.0374 D 2.244 kg the percentage over speed. [7.2%]

25% corresponds to 25 ð 0.0374 D 0.935 kg 8. Two kilograms of a compound contains

30% of element A, 45% of element B and

15% corresponds to 15 ð 0.0374 D 0.561 kg 25% of element C. Determine the masses

of the three elements present.

Thus, the masses of the copper, zinc and nickel are

2.244 kg, 0.935 kg and 0.561 kg, respectively. [A 0.6 kg, B 0.9 kg, C 0.5 kg]

(Check: 2.244 C 0.935 C 0.561 D 3.74) 9. A concrete mixture contains seven parts

by volume of ballast, four parts by vol-

Now try the following exercise ume of sand and two parts by volume of

cement. Determine the percentage of each

Exercise 4 Further problems percentages of these three constituents correct to the

1. Convert to percentages: nearest 1% and the mass of cement in a

two tonne dry mix, correct to 1 signiﬁcant

(a) 0.057 (b) 0.374 (c) 1.285 ﬁgure.

[(a) 5.7% (b) 37.4% (c) 128.5%]

[54%, 31%, 15%, 0.3 t]

2

Indices and standard form

2.1 Indices (i) When multiplying two or more numbers hav-

ing the same base, the indices are added. Thus

The lowest factors of 2000 are 2ð2ð2ð2ð5ð5ð5.

These factors are written as 24 ð 53, where 2 and 5 32 ð 34 D 32C4 D 36

are called bases and the numbers 4 and 3 are called

indices. (ii) When a number is divided by a number having

the same base, the indices are subtracted. Thus

When an index is an integer it is called a power.

Thus, 24 is called ‘two to the power of four’, and 35 D 35 2 D 33

has a base of 2 and an index of 4. Similarly, 53 is 32

called ‘ﬁve to the power of 3’ and has a base of 5

and an index of 3. (iii) When a number which is raised to a power

is raised to a further power, the indices are

Special names may be used when the indices are multiplied. Thus

2 and 3, these being called ‘squared’ and ‘cubed’,

respectively. Thus 72 is called ‘seven squared’ and 35 2 D 35ð2 D 310

93 is called ‘nine cubed’. When no index is shown,

the power is 1, i.e. 2 means 21.

Reciprocal (iv) When a number has an index of 0, its value

is 1. Thus 30 D 1

The reciprocal of a number is when the index is (v) A number raised to a negative power is the

1 and its value is given by 1, divided by the base. reciprocal of that number raised to a positive

Thus the reciprocal of 2 is 2 1 and its value is 1 power. Thus 3 4D 1 Similarly, 1 D 23

2 34 23

or 0.5. Similarly, the reciprocal of 5 is 5 1 which

means 1 or 0.2 (vi) When a number is raised to a fractional power

5 the denominator of the fraction is the root of

the number and the numerator is the power.

Square root

p

The square root of a number is when the indpex is 1 , Thus 82/3 D 3 82 D 2 2 D 4

2 and

pp

and the square root of 2 is written as 21/2 or 2. The 251/2 D 2 251 D 251 D š5

value of a square root is the value of the base which pp

(Note that Á 2 )

when multipliedpby itself gives the number. Since

3ðp3 D 9, then 9 D 3. However, 3 ð 3 D 9,

so 9 D 3. There are always two answers when

ﬁnding the square root of a number and this is shown 2.2 Worked problems on indices

by putting both a C and a sign in frpont of the Problem 1. Evaluate: (a) 52 ð 53,

answer to apsquare root problem. Thus 9 D š3 (b) 32 ð 34 ð 3 and (c) 2 ð 22 ð 25

and 41/2 D 4 D š2, and so on.

From law (i):

Laws of indices (a) 52ð53 D 5 2C3 D 55 D 5ð5ð5ð5ð5 D 3125

When simplifying calculations involving indices,

certain basic rules or laws can be applied, called

the laws of indices. These are given below.

10 ENGINEERING MATHEMATICS Problem 6. Find the value of

(b) 32 ð 34 ð 3 D 3 2C4C1 D 37 23 ð 24 32 3

D 3 ð 3 ð Ð Ð Ð to 7 terms (a) 27 ð 25 and (b) 3 ð 39

D 2187

(c) 2 ð 22 ð 25 D 2 1C2C5 D 28 D 256

Problem 2. Find the value of: From the laws of indices:

75 57

(a) 23 ð 24 2 3C4 D 27 D 27 12 D 2 5

(a) 73 and (b) 54 27 ð 25 D 2 7C5 212

From law (ii): 11

DD

25 32

(b) 32 3 32ð3 36 D 36 10 D 3 4

3 ð 39 D D

(a) 75 D75 3 D 72 D 49

73 31C9 310

57 11

54 D 34 D 81

(b) D57 4 D 53 D 125

Now try the following exercise

Problem 3. Evaluate: (a) 52 ð 53 ł 54 and Exercise 5 Further problems on indices

(b) 3 ð 35 ł 32 ð 33

In Problems 1 to 10, simplify the expressions

From laws (i) and (ii): given, expressing the answers in index form

and with positive indices:

52 ð 53 5 2C3

(a) 52 ð 53 ł 54 D 54 D 54 1. (a) 33 ð 34 (b) 42 ð 43 ð 44

[(a) 37 (b) 49]

D 55 D55 4 D 51 D 5

54 2. (a) 23 ð 2 ð 22 (b) 72 ð 74 ð 7 ð 73

[(a) 26 (b) 710]

(b) 3 ð 35 ł 32 ð 33 3 ð 35 3 1C5

DD

24 37

32 ð 33 3 2C3 3. (a) 23 (b) 32

D 36 D 36 5 D 31 D 3 [(a) 2 (b) 35]

35

Problem 4. Simplify: (a) 23 4 (b) 32 5, 4. (a) 56 ł 53 (b) 713/710

expressing the answers in index form. [(a) 53 (b) 73]

5. (a) 72 3 (b) 33 2 [(a) 76 (b) 36]

From law (iii): 22 ð 23 37 ð 34

(a) 23 4 D 23ð4 D 212 (b) 32 5 D 32ð5 D 310 6. (a) 24 (b) 35

[(a) 2 (b) 36]

102 3 57 135

Problem 5. Evaluate: 104 ð 102 7. (a) 52 ð 53 (b) 13 ð 132

[(a) 52 (b) 132]

From the laws of indices: 9 ð 32 3 16 ð 4 2

102 3 10 2ð3 106 8. (a) 3 ð 27 2 (b) 2 ð 8 3

104 ð 102 D 10 4C2 D 106

[(a) 34 (b) 1]

D 106 6 D 100 D 1

INDICES AND STANDARD FORM 11

5 2 32 ð 3 4 (Note that it does not matter whether the 4th root

9. (a) 5 4 (b) 33 of 16 is found ﬁrst or whether 16 cubed is found

ﬁrst — the same answer will result).

(a) 52 1

(b) 35 p

(c) 272/3 D 3 272 D 3 2 D 9

72 ð 7 3 23 ð 2 4 ð 25 (d) 9 1/2 D 1 D 1 D 1 1

10. (a) 7 ð 7 4 (b) 2 ð 2 2 ð 26 91/2 p š3 D±

9 3

(a) 72 1 41.5 ð 81/3

(b) Problem 10. Evaluate: 22 ð 32 2/5

2

2.3 Further worked problems on p

indices 41.5 D 43/2 D 43 D 23 D 8,

p

81/3 D 3 8 D 2, 22 D 4

and 32 2/5 D 1 1 1 D 1

322/5 Dp D 22 4

33 ð 57

Problem 7. Evaluate: 53 ð 34 5 322

Hence 41.5 ð 81/3 8 ð 2 16 D 16

Alternatively, D D

22 ð 32 1 1

2/5 4ð

The laws of indices only apply to terms having the 4

same base. Grouping terms having the same base,

and then applying the laws of indices to each of the 41.5 ð 81/3 [ 2 2]3/2 ð 23 1/3 23 ð 21

groups independently gives: 22 ð 32 2/5 D 22 ð 25 2/5 D 22 ð 2 2

33 ð 57 D 33 57 D33 4 ð57 3 D 23C1 2 2 D 24 D 16

53 ð 34 ð

34 53

54 625 1 Problem 11. Evaluate: 32 ð 55 C 33 ð 53

31 3 D 208

D3 1 ð 54 D D 34 ð 54

3

Problem 8. Find the value of Dividing each term by the HCF (i.e. highest com-

23 ð 35 ð 72 2 mon factor) of the three terms, i.e. 32 ð 53, gives:

74 ð 24 ð 33 32 ð 55 33 ð 53

32 ð 53 C 32 ð 53

32 ð 55 C 33 ð 53 D 34 ð 54

34 ð 54

23 ð 35 ð 72 2 D 23 4 ð 35 3 ð 72ð2 4 32 ð 53

74 ð 24 ð 33

3 2 2 ð 5 5 3 C 3 3 2 ð 50

D 2 1 ð 32 ð 70 D

34 2 ð54 3

D 1 ð 32 ð 1 D 9 D 1 30 ð 52 C 31 ð 50

4 D 32 ð 51

2 22

1 ð 25 C 3 ð 1 28

DD

Problem 9. Evaluate: 9ð5 45

(a) 41/2 (b) 163/4 (c) 272/3 (d) 9 1/2

Problem 12. Find the value of

p 32 ð 55

(a) 41/2 D 4 D ±2

34 ð 54 C 33 ð 53

p

(b) 163/4 D 4 163 D š2 3 D ±8

12 ENGINEERING MATHEMATICS

To simplify the arithmetic, each term is divided by Now try the following exercise

the HCF of all the terms, i.e. 32 ð 53. Thus

32 ð 55 Exercise 6 Further problems on indices

34 ð 54 C 33 ð 53 In Problems 1 and 2, simplify the expressions

given, expressing the answers in index form

32 ð 55 and with positive indices:

D 32 ð 53

34 ð 54 33 ð 53

C

32 ð 53 32 ð 53 33 ð 52 7 2ð3 2

1. (a) 54 ð 34 (b) 35 ð 74 ð 7 3

32 2 ð55 3

D 34 2 ð54 3 C33 2 ð53 3 11

(a) 3 ð 52 (b) 73 ð 37

30 ð 52 25 25

D DD

32 ð 51 C 31 ð 50 45 C 3 48

42 ð 93 8 2 ð 52 ð 3 4

2. (a) 83 ð 34 (b) 252 ð 24 ð 9 2

43 2

ð 3

Problem 13. Simplify: 32 1

3 5 (a) 25 (b) 210 ð 52

2

3

5 3. Evaluate a 11 b 810.25

giving the answer with positive indices 32 4 1/2

A fraction raised to a power means that both the c 16 1/4 d

9

numerator and the denominator of the fraction are

4 3 43

raised to that power, i.e. D 33 12

3 (a) 9 (b) š3 (c) š (d) š

23

A fraction raised to a negative power has the In Problems 4 to 8, evaluate the expressions

same value as the inverse of the fraction raised to a given.

positive power.

32 1 1 52 52 92 ð 74 147

Thus, 5 D 3 2 D 32 D 1 ð 32 D 32 4. 34 ð 74 C 33 ð 72 148

5 52 24 2 3 2 ð 44 1

5. 23 ð 162 9

23 5 3 53

Similarly, D D 23 13 22

5 2

65

43 3 2 43 52 6. 23 5

ð 32

ð 72

35 53

Thus, 23 D 33 32 5

5 23 44

43 52 23

D 33 ð 32 ð 53 7. 3 [64]

22

22 3 ð 23 9

D

32 3/2 ð 81/3 2 1

3 3C2 ð 5 3 2 8. 3 2 ð 43 1/2 ð 9 1/2 4

29 2

D 35 × 5

INDICES AND STANDARD FORM 13

2.4 Standard form Similarly,

A number written with one digit to the left of the 6 ð 104 6 104 2 D 4 ð 102

decimal point and multiplied by 10 raised to some 1.5 ð 102 D 1.5 ð

power is said to be written in standard form. Thus:

5837 is written as 5.837 ð 103 in standard form, 2.5 Worked problems on standard

and 0.0415 is written as 4.15 ð 10 2 in standard form

form.

Problem 14. Express in standard form:

When a number is written in standard form, the (a) 38.71 (b) 3746 (c) 0.0124

ﬁrst factor is called the mantissa and the second

factor is called the exponent. Thus the number For a number to be in standard form, it is expressed

5.8 ð 103 has a mantissa of 5.8 and an exponent with only one digit to the left of the decimal point.

of 103. Thus:

(i) Numbers having the same exponent can be (a) 38.71 must be divided by 10 to achieve one

added or subtracted in standard form by adding digit to the left of the decimal point and it

or subtracting the mantissae and keeping the must also be multiplied by 10 to maintain the

exponent the same. Thus: equality, i.e.

2.3 ð 104 C 3.7 ð 104 38.71 D 38.71 ð 10 D 3.871 × 10 in standard

D 2.3 C 3.7 ð 104 D 6.0 ð 104

form 10

and 5.9 ð 10 2 4.6 ð 10 2

D 5.9 4.6 ð 10 2 D 1.3 ð 10 2 (b) 3746 D 3746 ð 1000 D 3.746 × 103 in stan-

1000

When the numbers have different exponents,

one way of adding or subtracting the numbers dard form

is to express one of the numbers in non-

standard form, so that both numbers have the 100 1.24

same exponent. Thus: (c) 0.0124 D 0.0124 ð D

2.3 ð 104 C 3.7 ð 103 100 100

D 2.3 ð 104 C 0.37 ð 104 D 1.24 × 10−2 in standard form

D 2.3 C 0.37 ð 104 D 2.67 ð 104

Problem 15. Express the following

Alternatively, numbers, which are in standard form, as

decimal numbers: (a) 1.725 ð 10 2

2.3 ð 104 C 3.7 ð 103 (b) 5.491 ð 104 (c) 9.84 ð 100

D 23 000 C 3700 D 26 700 (a) 1.725 ð 10 2 D 1.725 D 0.01725

D 2.67 ð 104 100

(ii) The laws of indices are used when multiplying (b) 5.491 ð 104 D 5.491 ð 10 000 D 54 910

or dividing numbers given in standard form. (c) 9.84 ð 100 D 9.84 ð 1 D 9.84 (since 100 D 1)

For example,

Problem 16. Express in standard form,

2.5 ð 103 ð 5 ð 102 correct to 3 signiﬁcant ﬁgures:

D 2.5 ð 5 ð 103C2

D 12.5 ð 105 or 1.25 ð 106 32 9

(a) (b) 19 (c) 741

83 16

14 ENGINEERING MATHEMATICS

3 6. (a) 3.89 ð 10 2 (b) 6.741 ð 10 1

(a) D 0.375, and expressing it in standard form (c) 8 ð 10 3

[(a) 0.0389 (b) 0.6741 (c) 0.008]

8

gives: 0.375 D 3.75 × 10−1 2.6 Further worked problems on

standard form

(b) 2 D 19.6P D 1.97 × 10 in standard form,

19

3

correct to 3 signiﬁcant ﬁgures

(c) 9 D 741.5625 D 7.42 × 102 in standard

741

16

form, correct to 3 signiﬁcant ﬁgures

Problem 17. Express the following Problem 18. Find the value of:

numbers, given in standard form, as fractions (a) 7.9 ð 10 2 5.4 ð 10 2

or mixed numbers: (a) 2.5 ð 10 1 (b) 8.3 ð 103 C 5.415 ð 103 and

(b) 6.25 ð 10 2 (c) 1.354 ð 102 (c) 9.293 ð 102 C 1.3 ð 103 expressing the

(a) 2.5 ð 10 1 D 2.5 D 25 1 answers in standard form.

D

10 100 4 Numbers having the same exponent can be added

or subtracted by adding or subtracting the mantissae

(b) 6.25 ð 10 2 D 6.25 D 625 1 and keeping the exponent the same. Thus:

D

100 10 000 16 (a) 7.9 ð 10 2 5.4 ð 10 2

D 7.9 5.4 ð 10 2 D 2.5 × 10−2

(c) 1.354 ð 102 D 135.4 D 135 4 2

D 135 (b) 8.3 ð 103 C 5.415 ð 103

10 5 D 8.3 C 5.415 ð 103 D 13.715 ð 103

D 1.3715 × 104 in standard form

Now try the following exercise

(c) Since only numbers having the same exponents

Exercise 7 Further problems on standard can be added by straight addition of the man-

form tissae, the numbers are converted to this form

before adding. Thus:

In Problems 1 to 4, express in standard form: 9.293 ð 102 C 1.3 ð 103

D 9.293 ð 102 C 13 ð 102

1. (a) 73.9 (b) 28.4 (c) 197.72 D 9.293 C 13 ð 102

D 22.293 ð 102 D 2.2293 × 103

(a) 7.39 ð 10 (b) 2.84 ð 10

(c) 1.9762 ð 102 in standard form.

Alternatively, the numbers can be expressed as

2. (a) 2748 (b) 33170 (c) 274218 decimal fractions, giving:

(a) 2.748 ð 103 (b) 3.317 ð 104

(c) 2.74218 ð 105 9.293 ð 102 C 1.3 ð 103

3. (a) 0.2401 (b) 0.0174 (c) 0.00923 D 929.3 C 1300 D 2229.3

(a) 2.401 ð 10 1 (b) 1.74 ð 10 2 D 2.2293 × 103

(c) 9.23 ð 10 3 in standard form as obtained previously. This

method is often the ‘safest’ way of doing this

17 31 type of problem.

4. (a) (b) 11 (c) 130 (d)

28 5 32

(a) 5 ð 10 1 (b) 1.1875 ð 10

(c) 1.306 ð 102 (d) 3.125 ð 10 2

In Problems 5 and 6, express the numbers

given as integers or decimal fractions:

5. (a) 1.01 ð 103 (b) 9.327 ð 102

(c) 5.41 ð 104 (d) 7 ð 100

[(a) 1010 (b) 932.7 (c) 54 100 (d) 7]

INDICES AND STANDARD FORM 15

Problem 19. Evaluate 3.5 ð 105 3. (a) 4.5 ð 10 2 3 ð 103

and (b) 7 ð 102 (b) 2 ð 5.5 ð 104

(a) 3.75 ð 103 6 ð 104

[(a) 1.35 ð 102 (b) 1.1 ð 105]

expressing answers in standard form 6 ð 10 3 2.4 ð 103 3 ð 10 2

4. (a) 3 ð 10 5 (b) 4.8 ð 104

(a) 3.75 ð 103 6 ð 104 D 3.75 ð 6 103C4 [(a) 2 ð 102 (b) 1.5 ð 10 3]

D 22.50 ð 107 5. Write the following statements in stan-

dard form:

D 2.25 × 108

(b) 3.5 ð 105 D 3.5 ð 105 2 (a) The density of aluminium is

7 ð 102 7 2710 kg m 3

D 0.5 ð 103 D 5 × 102 [2.71 ð 103 kg m 3]

Now try the following exercise (b) Poisson’s ratio for gold is 0.44

[4.4 ð 10 1]

Exercise 8 Further problems on standard (c) The impedance of free space is

form

376.73 [3.7673 ð 102 ]

In Problems 1 to 4, ﬁnd values of the expres-

sions given, stating the answers in standard (d) The electron rest energy is

form: 0.511 MeV [5.11 ð 10 1 MeV]

1. (a) 3.7 ð 102 C 9.81 ð 102 (e) Proton charge-mass ratio is

(b) 1.431 ð 10 1 C 7.3 ð 10 1 9 5 789 700 C kg 1

[(a) 1.351 ð 103 (b) 8.731 ð 10 1]

[9.57897 ð 107 C kg 1]

2. (a) 4.831 ð 102 C 1.24 ð 103

(b) 3.24 ð 10 3 1.11 ð 10 4 (f) The normal volume of a perfect gas

[(a) 1.7231 ð 103 (b) 3.129 ð 10 3] is 0.02241 m3 mol 1

[2.241 ð 10 2 m3 mol 1]

3

Computer numbering systems

3.1 Binary numbers Problem 2. Convert 0.10112 to a decimal

fraction

The system of numbers in everyday use is the

denary or decimal system of numbers, using 0.10112 D 1 ð 2 1 C 0 ð 2 2 C 1 ð 2 3

the digits 0 to 9. It has ten different digits C1ð2 4

(0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) and is said to have a 11 1

radix or base of 10.

D 1 ð 2 C 0 ð 22 C 1 ð 23

The binary system of numbers has a radix of 2 1

and uses only the digits 0 and 1.

C 1 ð 24

3.2 Conversion of binary to decimal 11 1

DCC

The decimal number 234.5 is equivalent to 2 8 16

D 0.5 C 0.125 C 0.0625

2 ð 102 C 3 ð 101 C 4 ð 100 C 5 ð 10 1 D 0.687510

i.e. is the sum of terms comprising: (a digit) multi- Problem 3. Convert 101.01012 to a decimal

plied by (the base raised to some power). number

In the binary system of numbers, the base is 2, so 101.01012 D 1 ð 22 C 0 ð 21 C 1 ð 20

1101.1 is equivalent to: C0ð2 1C1ð2 2

C0ð2 3C1ð2 4

1 ð 23 C 1 ð 22 C 0 ð 21 C 1 ð 20 C 1 ð 2 1

D 4 C 0 C 1 C 0 C 0.25

Thus the decimal number equivalent to the binary C 0 C 0.0625

number 1101.1 is

D 5.312510

1

8 C 4 C 0 C 1 C , that is 13.5 Now try the following exercise

2 Exercise 9 Further problems on conver-

sion of binary to decimal num-

i.e. 1101.12 = 13.510, the sufﬁxes 2 and 10 denot- bers

ing binary and decimal systems of numbers respec-

tively.

Problem 1. Convert 110112 to a decimal

number

From above: 110112 D 1 ð 24 C 1 ð 23 C 0 ð 22 In Problems 1 to 4, convert the binary num-

C 1 ð 21 C 1 ð 20 bers given to decimal numbers.

D 16 C 8 C 0 C 2 C 1 1. (a) 110 (b) 1011 (c) 1110 (d) 1001

D 2710 [(a) 610 (b) 1110 (c) 1410 (d) 910]

2. (a) 10101 (b) 11001 (c) 101101 COMPUTER NUMBERING SYSTEMS 17

(d) 110011

[(a) 2110 (b) 2510 (c) 4510 (d) 5110] For fractions, the most signiﬁcant bit of the result

is the top bit obtained from the integer part of

3. (a) 0.1101 (b) 0.11001 (c) 0.00111 multiplication by 2. The least signiﬁcant bit of the

(d) 0.01011 result is the bottom bit obtained from the integer

part of multiplication by 2.

(a) 0.812510 (b) 0.7812510

(c) 0.2187510 (d) 0.3437510 Thus 0.62510 = 0.1012

4. (a) 11010.11 (b) 10111.011

(c) 110101.0111 (d) 11010101.10111 Problem 4. Convert 4710 to a binary

(a) 26.7510 (b) 23.37510 number

(c) 53.437510 (d) 213.7187510

From above, repeatedly dividing by 2 and noting the

remainder gives:

2 47 Remainder

2 23 1

2 11 1

3.3 Conversion of decimal to binary 25 1

An integer decimal number can be converted to a 22 1

corresponding binary number by repeatedly dividing

by 2 and noting the remainder at each stage, as 21 0

shown below for 3910

01

2 39 Remainder 1 0 1 11 1

2 19 1

29 1 Thus 4710 = 1011112

24 1

22 0 Problem 5. Convert 0.4062510 to a binary

21 0 number

1

0

(most → 1 0 0 1 1 1 ← (least From above, repeatedly multiplying by 2 gives:

significant bit) significant bit) 0.40625 × 2 = 0. 8125

The result is obtained by writing the top digit of 0.8125 × 2 = 1. 625

the remainder as the least signiﬁcant bit, (a bit is a

binary digit and the least signiﬁcant bit is the one 0.625 × 2 = 1. 25

on the right). The bottom bit of the remainder is the

most signiﬁcant bit, i.e. the bit on the left.

Thus 3910 = 1001112 0.25 × 2 = 0. 5

The fractional part of a decimal number can be con- 0.5 × 2 = 1. 0

verted to a binary number by repeatedly multiplying

by 2, as shown below for the fraction 0.625

0.625 × 2 = 1. 250 .0 1 1 0 1

0.250 × 2 = 0. 500 i.e. 0.4062510 = 0.011012

0.500 × 2 = 1. 000 Problem 6. Convert 58.312510 to a binary

number

(most significant bit) .1 0 1 (least significant bit)

18 ENGINEERING MATHEMATICS 3.4 Conversion of decimal to binary

via octal

The integer part is repeatedly divided by 2, giving:

For decimal integers containing several digits, repe-

2 58 Remainder atedly dividing by 2 can be a lengthy process. In

2 29 0 this case, it is usually easier to convert a decimal

2 14 1 number to a binary number via the octal system of

27 0 numbers. This system has a radix of 8, using the

23 1 digits 0, 1, 2, 3, 4, 5, 6 and 7. The denary number

21 1 equivalent to the octal number 43178 is

01 4 ð 83 C 3 ð 82 C 1 ð 81 C 7 ð 80

1 1 1 01 0 i.e. 4 ð 512 C 3 ð 64 C 1 ð 8 C 7 ð 1 or 225510

The fractional part is repeatedly multiplied by 2 An integer decimal number can be converted to a

giving: corresponding octal number by repeatedly dividing

by 8 and noting the remainder at each stage, as

0.3125 × 2 = 0.625 shown below for 49310

0.625 × 2 = 1.25

0.25 × 2 = 8 493 Remainder

0.5 × 2 = 0.5

8 61 5

.0 1 0 1 1.0

87 5

Thus 58.312510 = 111010.01012

07

Now try the following exercise

Exercise 10 Further problems on conver- 755

sion of decimal to binary

numbers Thus 49310 = 7558

The fractional part of a decimal number can be con-

In Problems 1 to 4, convert the decimal verted to an octal number by repeatedly multiplying

numbers given to binary numbers. by 8, as shown below for the fraction 0.437510

1. (a) 5 (b) 15 (c) 19 (d) 29 0.4375 × 8 = 3 . 5

(a) 1012 (b) 11112 0.5 × 8 = 4 . 0

(c) 100112 (d) 111012

.3 4

2. (a) 31 (b) 42 (c) 57 (d) 63

For fractions, the most signiﬁcant bit is the top

(a) 111112 (b) 1010102 integer obtained by multiplication of the decimal

(c) 1110012 (d) 1111112 fraction by 8, thus

3. (a) 0.25 (b) 0.21875 (c) 0.28125 0.437510 D 0.348

(d) 0.59375

The natural binary code for digits 0 to 7 is shown

(a) 0.012 (b) 0.001112 in Table 3.1, and an octal number can be converted

to a binary number by writing down the three bits

(c) 0.010012 (d) 0.100112 corresponding to the octal digit.

Thus 4378 D 100 011 1112

4. (a) 47.40625 (b) 30.8125 and 26.358 D 010 110.011 1012

(c) 53.90625 (d) 61.65625

(a) 101111.011012 (b) 11110.11012

(c) 110101.111012 (d) 111101.101012

COMPUTER NUMBERING SYSTEMS 19

Table 3.1 Natural Problem 9. Convert 5613.9062510 to a

binary number binary number, via octal

Octal digit

000 The integer part is repeatedly divided by 8, noting

0 001 the remainder, giving:

1 010

2 011 8 5613 Remainder

3 100

4 101 8 701 5

5 110

6 111 8 87 5

7

8 10 7

81 2

The ‘0’ on the extreme left does not signify any- 01

thing, thus 26.358 D 10 110.011 1012

12755

Conversion of decimal to binary via octal is demon-

strated in the following worked problems. This octal number is converted to a binary number,

(see Table 3.1)

Problem 7. Convert 371410 to a binary

number, via octal 127558 D 001 010 111 101 1012

i.e. 561310 D 1 010 111 101 1012

Dividing repeatedly by 8, and noting the remainder

gives: The fractional part is repeatedly multiplied by 8, and

noting the integer part, giving:

8 3714 Remainder 0.90625 × 8 = 7.25

0.25 × 8 = 2.00

8 464 2

8 58 0 .7 2

87 2

07 This octal fraction is converted to a binary number,

(see Table 3.1)

7202

0.728 D 0.111 0102

From Table 3.1, 72028 D 111 010 000 0102 i.e. 0.9062510 D 0.111 012

i.e. 371410 = 111 010 000 0102

Thus, 5613.9062510 = 1 010 111 101 101.111 012

Problem 8. Convert 0.5937510 to a binary Problem 10. Convert 11 110 011.100 012

number, via octal to a decimal number via octal

Multiplying repeatedly by 8, and noting the integer Grouping the binary number in three’s from the

values, gives: binary point gives: 011 110 011.100 0102

0.59375 × 8 = 4.75 Using Table 3.1 to convert this binary number to

0.75 × 8 = 6.00 an octal number gives: 363.428 and

.4 6 363.428 D 3 ð 82 C 6 ð 81 C 3 ð 80

C4ð8 1C2ð8 2

Thus 0.5937510 D 0.468

From Table 3.1, 0.468 D 0.100 1102 D 192 C 48 C 3 C 0.5 C 0.03125

i.e. 0.5937510 = 0.100 112

D 243.5312510

20 ENGINEERING MATHEMATICS

Now try the following exercise To convert from hexadecimal to decimal:

Exercise 11 Further problems on con- For example

version between decimal and 1A16 D 1 ð 161 C A ð 160

binary numbers via octal D 1 ð 161 C 10 ð 1 D 16 C 10 D 26

In Problems 1 to 3, convert the decimal i.e. 1A16 D 2610

numbers given to binary numbers, via octal. Similarly,

1. (a) 343 (b) 572 (c) 1265 2E16 D 2 ð 161 C E ð 160

D 2 ð 161 C 14 ð 160 D 32 C 14 D 4610

(a) 1010101112 (b) 10001111002

(c) 100111100012 and 1BF16 D 1 ð 162 C B ð 161 C F ð 160

D 1 ð 162 C 11 ð 161 C 15 ð 160

2. (a) 0.46875 (b) 0.6875 (c) 0.71875 D 256 C 176 C 15 D 44710

(a) 0.011112 (b) 0.10112 Table 3.2 compares decimal, binary, octal and hex-

(c) 0.101112 adecimal numbers and shows, for example, that

3. (a) 247.09375 (b) 514.4375 2310 D 101112 D 278 D 1716

(c) 1716.78125

(a) 11110111.000112 Problem 11. Convert the following

(b) 1000000010.01112 hexadecimal numbers into their decimal

(c) 11010110100.110012 equivalents: (a) 7A16 (b) 3F16

4. Convert the following binary numbers to (a) 7A16 D 7 ð 161 C A ð 160 D 7 ð 16 C 10 ð 1

decimal numbers via octal: D 112 C 10 D 122

(a) 111.011 1 (b) 101 001.01 Thus 7A16 = 12210

(c) 1 110 011 011 010.001 1 (b) 3F16 D 3 ð 161 C F ð 160 D 3 ð 16 C 15 ð 1

(a) 7.437510 (b) 41.2510 D 48 C 15 D 63

Thus, 3F16 = 6310

(c) 7386.187510

Problem 12. Convert the following

3.5 Hexadecimal numbers hexadecimal numbers into their decimal

equivalents: (a) C916 (b) BD16

The complexity of computers requires higher order

numbering systems such as octal (base 8) and hex- (a) C916 D C ð 161 C 9 ð 160 D 12 ð 16 C 9 ð 1

adecimal (base 16), which are merely extensions D 192 C 9 D 201

of the binary system. A hexadecimal numbering

system has a radix of 16 and uses the following 16 Thus C916 = 20110

distinct digits: (b) BD16 D B ð 161 C D ð 160 D 11 ð 16 C 13 ð 1

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F D 176 C 13 D 189

Thus BD16 = 18910

‘A’ corresponds to 10 in the denary system, B to

11, C to 12, and so on. Problem 13. Convert 1A4E16 into a denary

number

COMPUTER NUMBERING SYSTEMS 21

Table 3.2 Hence 2610 = 1A16

Decimal Binary Octal Hexadecimal Similarly, for 44710

0 0000 0 0 16 447 Remainder

1 0001 1 1

2 0010 2 2 16 27 15 ≡ F16

3 0011 3 3 16 1

4 0100 4 4 11 ≡ B16

5 0101 5 5 0 1 ≡ 116

6 0110 6 6

7 0111 7 7 1 BF

8 1000 10 8

9 1001 11 9 Thus 44710 = 1BF16

10 1010 12 A

11 1011 13 B Problem 14. Convert the following decimal

12 1100 14 C numbers into their hexadecimal equivalents:

13 1101 15 D (a) 3710 (b) 10810

14 1110 16 E

15 1111 17 F (a) 16 37 Remainder

16 10000 20 10 16 2 5 = 516

17 10001 21 11 0 2 = 216

18 10010 22 12

19 10011 23 13 most significant bit → 2 5 ← least significant bit

20 10100 24 14

21 10101 25 15 Hence 3710 = 2516

22 10110 26 16

23 10111 27 17 (b) 16 108 Remainder

24 11000 30 18

25 11001 31 19 16 6 12 = C16

26 11010 32 1A 0 6 = 616

27 11011 33 1B

28 11100 34 1C 6C

29 11101 35 1D

30 11110 36 1E Hence 10810 = 6C16

31 11111 37 1F

32 100000 40 20 Problem 15. Convert the following decimal

numbers into their hexadecimal equivalents:

(a) 16210 (b) 23910

1A4E16 (a) 16 162 Remainder

D 1 ð 163 C A ð 162 C 4 ð 161 C E ð 160 16 10 2 = 216

D 1 ð 163 C 10 ð 162 C 4 ð 161 C 14 ð 160 0 10 = A16

A2

D 1 ð 4096 C 10 ð 256 C 4 ð 16 C 14 ð 1

Hence 16210 = A216

D 4096 C 2560 C 64 C 14 D 6734 (b) 16 239 Remainder

Thus, 1A4E16 = 673410 16 14 15 = F16

0 14 = E16

To convert from decimal to hexadecimal: EF

This is achieved by repeatedly dividing by 16 and Hence 23910 = EF16

noting the remainder at each stage, as shown below

for 2610 To convert from binary to hexadecimal:

16 26 Remainder The binary bits are arranged in groups of four,

16 1 starting from right to left, and a hexadecimal symbol

10 ≡ A16

0 1 ≡ 116

most significant bit → 1 A ← least significant bit

22 ENGINEERING MATHEMATICS

is assigned to each group. For example, the binary (b) Grouping bits in fours from

number 1110011110101001 is initially grouped in

the right gives: 0001 1001 1110

fours as: 1110 0111 1010 1001 and assigning hexadecimal

and a hexadecimal symbol symbols to each group gives: 1 9 E

from Table 3.2

assigned to each group as E 7 A 9

from Table 3.2 Thus, 1100111102 = 19E16

Hence 11100111101010012 = E7A916.

To convert from hexadecimal to binary: Problem 18. Convert the following

hexadecimal numbers into their binary

The above procedure is reversed, thus, for example, equivalents: (a) 3F16 (b) A616

6CF316 D 0110 1100 1111 0011

from Table 3.2 (a) Spacing out hexadecimal

i.e. 6CF316= 1101100111100112 digits gives: 3F

and converting each into

Problem 16. Convert the following binary binary gives: 0011 1111

numbers into their hexadecimal equivalents: from Table 3.2

(a) 110101102 (b) 11001112 Thus, 3F16 = 1111112

(b) Spacing out hexadecimal digits

(a) Grouping bits in fours from the

gives: A6

right gives: 1101 0110

and converting each into binary

and assigning hexadecimal symbols

gives: 1010 0110

to each group gives: D6 from Table 3.2

from Table 3.2

Thus, A616 = 101001102

Thus, 110101102 = D616

(b) Grouping bits in fours from the Problem 19. Convert the following

hexadecimal numbers into their binary

right gives: 0110 0111 equivalents: (a) 7B16 (b) 17D16

and assigning hexadecimal symbols

to each group gives: 67 (a) Spacing out hexadecimal

Thus, 11001112 = 6716 from Table 3.2

digits gives: 7B

Problem 17. Convert the following binary and converting each into

numbers into their hexadecimal equivalents:

binary gives: 0111 1011

(a) 110011112 (b) 1100111102 from Table 3.2

Thus, 7B16 = 11110112

(a) Grouping bits in fours from the (b) Spacing out hexadecimal

right gives: 1100 1111 digits gives: 17D

and assigning hexadecimal and converting each into

symbols to each group gives: C F binary gives: 0001 0111 1101

from Table 3.2 from Table 3.2

Thus, 110011112 = CF16 Thus, 17D16 = 1011111012

COMPUTER NUMBERING SYSTEMS 23

Now try the following exercise 9. 110101112 [D716]

10. 111010102 [EA16]

Exercise 12 Further problems on hexa- 11. 100010112 [8B16]

decimal numbers 12. 101001012 [A516]

In Problems 1 to 4, convert the given hexadec- In Problems 13 to 16, convert the given hex-

imal numbers into their decimal equivalents. adecimal numbers into their binary equiva-

lents.

1. E716 [23110] 2. 2C16 [4410]

13. 3716 [1101112]

3. 9816 [15210] 4. 2F116 [75310]

In Problems 5 to 8, convert the given decimal 14. ED16 [111011012]

numbers into their hexadecimal equivalents.

15. 9F16 [100111112]

5. 5410 [3616] 6. 20010 [C816]

16. A2116 [1010001000012]

7. 9110 [5B16] 8. 23810 [EE16]

In Problems 9 to 12, convert the given binary

numbers into their hexadecimal equivalents.

4

Calculations and evaluation of

formulae

4.1 Errors and approximations 55 could therefore be expected. Certainly

an answer around 500 or 5 would not be

(i) In all problems in which the measurement of expected. Actually, by calculator

distance, time, mass or other quantities occurs, 49.1 ð 18.4 ð 122.1

an exact answer cannot be given; only an

answer which is correct to a stated degree of D 47.31, correct to

accuracy can be given. To take account of this 61.2 ð 38.1

an error due to measurement is said to exist. 4 signiﬁcant ﬁgures.

(ii) To take account of measurement errors it Problem 1. The area A of a triangle is

is usual to limit answers so that the result 1

given is not more than one signiﬁcant ﬁgure

greater than the least accurate number given by A D bh. The base b when

given in the data. 2

(iii) Rounding-off errors can exist with decimal measured is found to be 3.26 cm, and the

fractions. For example, to state that D perpendicular height h is 7.5 cm. Determine

3.142 is not strictly correct, but ‘ D 3.142 the area of the triangle.

correct to 4 signiﬁcant ﬁgures’ is a true state-

ment. (Actually, D 3.14159265 . . .) 11

Area of triangle D bh D ð 3.26 ð 7.5 D

(iv) It is possible, through an incorrect procedure,

to obtain the wrong answer to a calculation. 22

This type of error is known as a blunder. 12.225 cm2 (by calculator).

(v) An order of magnitude error is said to exist The approximate value is 1 ð 3 ð 8 D 12 cm2, so

if incorrect positioning of the decimal point 2

occurs after a calculation has been completed.

there are no obvious blunder or magnitude errors.

(vi) Blunders and order of magnitude errors can However, it is not usual in a measurement type

be reduced by determining approximate val- problem to state the answer to an accuracy greater

ues of calculations. Answers which do not than 1 signiﬁcant ﬁgure more than the least accurate

seem feasible must be checked and the cal- number in the data: this is 7.5 cm, so the result

culation must be repeated as necessary. should not have more than 3 signiﬁcant ﬁgures

An engineer will often need to make a Thus, area of triangle = 12.2 cm2

quick mental approximation for a calcula- Problem 2. State which type of error has

been made in the following statements:

tion. For example, 49.1 ð 18.4 ð 122.1 may

(a) 72 ð 31.429 D 2262.9

61.2 ð 38.1

(b) 16 ð 0.08 ð 7 D 89.6

be approximated to 50 ð 20 ð 120 and then,

(c) 11.714 ð 0.0088 D 0.3247 correct to

60 ð 40 4 decimal places.

by cancelling, 50 ð1 20 ð 120 2 1 29.74 ð 0.0512

D 50. An (d) D 0.12, correct to

1 60 ð 40 2 1

11.89

2 signiﬁcant ﬁgures.

accurate answer somewhere between 45 and

CALCULATIONS AND EVALUATION OF FORMULAE 25

(a) 72 ð 31.429 D 2262.888 (by calculator), 2.19 ð 203.6 ð 17.91

hence a rounding-off error has occurred. The i.e. ³ 80

answer should have stated:

12.1 ð 8.76

(By calculator, 2.19 ð 203.6 ð 17.91 D 75.3,

72 ð 31.429 D 2262.9, correct to 5 signiﬁcant 12.1 ð 8.76

ﬁgures or 2262.9, correct to 1 decimal place.

correct to 3 signiﬁcant ﬁgures.)

8 32 ð 7

(b) 16 ð 0.08 ð 7 D 16 ð ð 7 D Now try the following exercise

100 25 Exercise 13 Further problems on errors

224 24

D D 8 D 8.96

25 25

Hence an order of magnitude error has In Problems 1 to 5 state which type of error,

occurred. or errors, have been made:

(c) 11.714 ð 0.0088 is approximately equal to 1. 25 ð 0.06 ð 1.4 D 0.21

12 ð 9 ð 10 3, i.e. about 108 ð 10 3 or 0.108. [order of magnitude error]

Thus a blunder has been made.

(d) 29.74 ð 0.0512 30 ð 5 ð 10 2 2. 137 ð 6.842 D 937.4

³

11.89 12 Rounding-off error–should add ‘correct

to 4 signiﬁcant ﬁgures’ or ‘correct to

150 15 1 1 decimal place’

D 12 ð 102 D 120 D 8 or 0.125

24 ð 0.008 [Blunder]

3. D 10.42

hence no order of magnitude error has

12.6

occurred. However, 29.74 ð 0.0512 D 0.128 4. For a gas pV D c. When pressure

p D 1 03 400 Pa and V D 0.54 m3 then

11.89 c D 55 836 Pa m3.

correct to 3 signiﬁcant ﬁgures, which equals

Measured values, hence

0.13 correct to 2 signiﬁcant ﬁgures. c D 55 800 Pa m3

Hence a rounding-off error has occurred.

Problem 3. Without using a calculator, 4.6 ð 0.07

determine an approximate value of: 5. D 0.225

52.3 ð 0.274

Order of magnitude error and rounding-

11.7 ð 19.1 2.19 ð 203.6 ð 17.91 off error–should be 0.0225, correct to

3 signiﬁcant ﬁgures or 0.0225,

(a) (b) 12.1 ð 8.76

9.3 ð 5.7

correct to 4 decimal places

11.7 ð 19.1 In Problems 6 to 8, evaluate the expressions

(a) is approximately equal to approximately, without using a calculator.

9.3 ð 5.7 6. 4.7 ð 6.3 [³30 (29.61, by calculator)]

10 ð 20

2.87 ð 4.07

, i.e. about 4 7.

10 ð 5

6.12 ð 0.96

(By calculator, 11.7 ð 19.1 D 4.22, correct to

³2 (1.988, correct to 4 s.f., by

9.3 ð 5.7 calculator)

3 signiﬁcant ﬁgures.)

2.19 ð 203.6 ð 17.91 2 ð 20 200 ð 20 2 72.1 ð 1.96 ð 48.6

(b) ³ 8.

12.1 ð 8.76 1 10 ð 10 1

139.3 ð 5.2

D 2 ð 20 ð 2 after cancelling,

³10 (9.481, correct to 4 s.f., by

calculator)

26 ENGINEERING MATHEMATICS 1

(a) D 0.01896453 . . . D 0.019, correct to 3

4.2 Use of calculator

52.73

The most modern aid to calculations is the pocket- decimal places

sized electronic calculator. With one of these, cal-

culations can be quickly and accurately performed, 1

correct to about 9 signiﬁcant ﬁgures. The scientiﬁc (b) D 36.3636363 . . . D 36.364, correct to

type of calculator has made the use of tables and

logarithms largely redundant. 0.0275

3 decimal places

To help you to become competent at using your

calculator check that you agree with the answers to 11

the following problems: (c) C D 0.71086624 . . . D 0.711, cor-

4.92 1.97

rect to 3 decimal places

Problem 4. Evaluate the following, correct Problem 7. Evaluate the following,

to 4 signiﬁcant ﬁgures: expressing the answers in standard form,

correct to 4 signiﬁcant ﬁgures.

(a) 4.7826 C 0.02713 (b) 17.6941 11.8762

(c) 21.93 ð 0.012981 (a) 0.00451 2 (b) 631.7 6.21 C 2.95 2

(a) 4.7826 C 0.02713 D 4.80973 D 4.810, correct (c) 46.272 31.792

to 4 signiﬁcant ﬁgures

(a) 0.00451 2 D 2.03401ð10 5 D 2.034 × 10−5,

(b) 17.6941 11.8762 D 5.8179 D 5.818, correct correct to 4 signiﬁcant ﬁgures

to 4 signiﬁcant ﬁgures

(b) 631.7 6.21 C 2.95 2 D 547.7944 D

(c) 21.93 ð 0.012981 D 0.2846733 . . . D 0.2847, 5.477944 ð 102 D 5.478 × 102, correct to 4

correct to 4 signiﬁcant ﬁgures signiﬁcant ﬁgures

Problem 5. Evaluate the following, correct (c) 46.272 31.792 D 1130.3088 D 1.130 × 103,

to 4 decimal places: correct to 4 signiﬁcant ﬁgures

(a) 46.32 ð 97.17 ð 0.01258 4.621 Problem 8. Evaluate the following, correct

1 (b) to 3 decimal places:

(c) 62.49 ð 0.0172 23.76

2

2.37 2 3.60 2 5.40 2

(a) (b) C

0.0526 1.92 2.45

(a) 46.32 ð 97.17 ð 0.01258 D 56.6215031 . . . D

56.6215, correct to 4 decimal places 15

4.621 (c) 7.62 4.82

(b) D 0.19448653 . . . D 0.1945, correct to 2.37 2

23.76 (a) D 106.785171 . . . D 106.785, correct

4 decimal places

1 0.0526

to 3 decimal places

(c) 62.49 ð 0.0172 D 0.537414 D 0.5374,

2 3.60 2 5.40 2

correct to 4 decimal places

(b) C D 8.37360084 . . . D

1.92 2.45

Problem 6. Evaluate the following, correct 8.374, correct to 3 decimal places

to 3 decimal places:

1 1 11 15

(a) (b) (c) C (c) 7.62 4.82 D 0.43202764 . . . D 0.432, cor-

52.73 0.0275 4.92 1.97

rect to 3 decimal places