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- 1. Engineering Mathematics
- 2. In memory of Elizabeth
- 3. Engineering Mathematics Fourth Edition JOHN BIRD, BSc(Hons) CMath, FIMA, CEng, MIEE, FCollP, FIIE Newnes OXFORD AMSTERDAM BOSTON SAN DIEGO SAN FRANCISCO LONDON NEW YORK SINGAPORE PARIS SYDNEY TOKYO
- 4. 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: permissions@elsevier.co.uk. 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
- 5. Contents Preface xi Part 1 Number and Algebra 1 1 Revision of fractions, decimals and percentages 1 1.1 Fractions 1 1.2 Ratio and proportion 3 1.3 Decimals 4 1.4 Percentages 7 2 Indices and standard form 9 2.1 Indices 9 2.2 Worked problems on indices 9 2.3 Further worked problems on indices 11 2.4 Standard form 13 2.5 Worked problems on standard form 13 2.6 Further worked problems on standard form 14 3 Computer numbering systems 16 3.1 Binary numbers 16 3.2 Conversion of binary to decimal 16 3.3 Conversion of decimal to binary 17 3.4 Conversion of decimal to binary via octal 18 3.5 Hexadecimal numbers 20 4 Calculations and evaluation of formulae 24 4.1 Errors and approximations 24 4.2 Use of calculator 26 4.3 Conversion tables and charts 28 4.4 Evaluation of formulae 30 7 Partial fractions 51 7.1 Introduction to partial fractions 51 7.2 Worked problems on partial fractions with linear factors 51 7.3 Worked problems on partial fractions with repeated linear factors 54 7.4 Worked problems on partial fractions with quadratic factors 55 8 Simple equations 57 8.1 Expressions, equations and identities 57 8.2 Worked problems on simple equations 57 8.3 Further worked problems on simple equations 59 8.4 Practical problems involving simple equations 61 8.5 Further practical problems involving simple equations 62 Assignment 2 64 9 Simultaneous equations 65 9.1 Introduction to simultaneous equations 65 9.2 Worked problems on simultaneous equations in two unknowns 65 9.3 Further worked problems on simultaneous equations 67 9.4 More difﬁcult worked problems on simultaneous equations 69 9.5 Practical problems involving simultaneous equations 70 5 Algebra 34 5.1 Basic operations 34 5.2 Laws of Indices 36 5.3 Brackets and factorisation 38 5.4 Fundamental laws and precedence 40 5.5 Direct and inverse proportionality 42 10 Transposition of formulae 74 10.1 Introduction to transposition of formulae 74 10.2 Worked problems on transposition of formulae 74 10.3 Further worked problems on transposition of formulae 75 10.4 Harder worked problems on transposition of formulae 77 6 Further algebra 44 6.1 Polynomial division 44 6.2 The factor theorem 46 6.3 The remainder theorem 48 11 Quadratic equations 80 11.1 Introduction to quadratic equations 11.2 Solution of quadratic equations by factorisation 80 Assignment 1 33 80
- 6. vi CONTENTS 11.3 Solution of quadratic equations by ‘completing the square’ 82 11.4 Solution of quadratic equations by formula 84 11.5 Practical problems involving quadratic equations 85 11.6 The solution of linear and quadratic equations simultaneously 87 12 Logarithms 89 12.1 Introduction to logarithms 89 12.2 Laws of logarithms 89 12.3 Indicial equations 92 12.4 Graphs of logarithmic functions 93 Assignment 3 94 13 Exponential functions 95 13.1 The exponential function 95 13.2 Evaluating exponential functions 95 13.3 The power series for e x 96 13.4 Graphs of exponential functions 98 13.5 Napierian logarithms 100 13.6 Evaluating Napierian logarithms 100 13.7 Laws of growth and decay 102 14 Number sequences 106 14.1 Arithmetic progressions 106 14.2 Worked problems on arithmetic progression 106 14.3 Further worked problems on arithmetic progressions 107 14.4 Geometric progressions 109 14.5 Worked problems on geometric progressions 110 14.6 Further worked problems on geometric progressions 111 14.7 Combinations and permutations 112 15 The binomial series 114 15.1 Pascal’s triangle 114 15.2 The binomial series 115 15.3 Worked problems on the binomial series 115 15.4 Further worked problems on the binomial series 117 15.5 Practical problems involving the binomial theorem 120 16 Solving equations by iterative 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 Multiple choice questions on chapters 1 to 16 127 Part 2 Mensuration 131 17 Areas of plane ﬁgures 131 17.1 Mensuration 131 17.2 Properties of quadrilaterals 131 17.3 Worked problems on areas of plane ﬁgures 132 17.4 Further worked problems on areas of plane ﬁgures 135 17.5 Worked problems on areas of composite ﬁgures 137 17.6 Areas of similar shapes 138 18 The circle and its properties 139 18.1 Introduction 139 18.2 Properties of circles 139 18.3 Arc length and area of a sector 140 18.4 Worked problems on arc length and sector of a circle 141 18.5 The equation of a circle 143 19 Volumes and surface areas of common solids 145 19.1 Volumes and surface areas of regular solids 145 19.2 Worked problems on volumes and surface areas of regular solids 145 19.3 Further worked problems on volumes and surface areas of regular solids 147 19.4 Volumes and surface areas of frusta of pyramids and cones 151 19.5 The frustum and zone of a sphere 155 19.6 Prismoidal rule 157 19.7 Volumes of similar shapes 159 20 Irregular areas and volumes and mean values of waveforms 161 20.1 Areas of irregular ﬁgures 161 20.2 Volumes of irregular solids 163 20.3 The mean or average value of a waveform 164 Assignment 5 168 Part 3 Trigonometry 171 21 Introduction to trigonometry 171 21.1 Trigonometry 171 21.2 The theorem of Pythagoras 171 21.3 Trigonometric ratios of acute angles 172
- 7. CONTENTS 21.4 Fractional and surd forms of trigonometric ratios 174 21.5 Solution of right-angled triangles 175 21.6 Angles of elevation and depression 176 21.7 Evaluating trigonometric ratios of any angles 178 21.8 Trigonometric approximations for small angles 181 22 Trigonometric waveforms 182 22.1 Graphs of trigonometric functions 182 22.2 Angles of any magnitude 182 22.3 The production of a sine and cosine wave 185 22.4 Sine and cosine curves 185 22.5 Sinusoidal form A sin ωt š ˛ 189 22.6 Waveform harmonics 192 23 Cartesian and polar co-ordinates 194 23.1 Introduction 194 23.2 Changing from Cartesian into polar co-ordinates 194 23.3 Changing from polar into Cartesian co-ordinates 196 23.4 Use of R ! P and P ! R functions on calculators 197 Assignment 6 198 24 Triangles and some practical applications 199 24.1 Sine and cosine rules 199 24.2 Area of any triangle 199 24.3 Worked problems on the solution of triangles and their areas 199 24.4 Further worked problems on the solution of triangles and their areas 201 24.5 Practical situations involving trigonometry 203 24.6 Further practical situations involving trigonometry 205 25 Trigonometric identities and equations 208 25.1 Trigonometric identities 208 25.2 Worked problems on trigonometric identities 208 25.3 Trigonometric equations 209 25.4 Worked problems (i) on trigonometric equations 210 25.5 Worked problems (ii) on trigonometric equations 211 25.6 Worked problems (iii) on trigonometric equations 212 vii 25.7 Worked problems (iv) on trigonometric equations 212 26 Compound angles 214 26.1 Compound angle formulae 214 26.2 Conversion of a sin ωt C b cos ωt into R sin ωt C ˛) 216 26.3 Double angles 220 26.4 Changing products of sines and cosines into sums or differences 221 26.5 Changing sums or differences of sines and cosines into products 222 Assignment 7 224 Multiple choice questions on chapters 17 to 26 225 Part 4 Graphs 231 27 Straight line graphs 231 27.1 Introduction to graphs 231 27.2 The straight line graph 231 27.3 Practical problems involving straight line graphs 237 28 Reduction of non-linear laws to linear form 243 28.1 Determination of law 243 28.2 Determination of law involving logarithms 246 29 Graphs with logarithmic scales 251 29.1 Logarithmic scales 251 29.2 Graphs of the form y D ax n 251 29.3 Graphs of the form y D abx 254 29.4 Graphs of the form y D ae kx 255 30 Graphical solution of equations 258 30.1 Graphical solution of simultaneous equations 258 30.2 Graphical solution of quadratic equations 259 30.3 Graphical solution of linear and quadratic equations simultaneously 263 30.4 Graphical solution of cubic equations 264 31 Functions and their curves 266 31.1 Standard curves 266 31.2 Simple transformations 268 31.3 Periodic functions 273 31.4 Continuous and discontinuous functions 273 31.5 Even and odd functions 273 31.6 Inverse functions 275 Assignment 8 279
- 8. viii CONTENTS 38.3 Worked problems on probability 327 38.4 Further worked problems on probability 329 38.5 Permutations and combinations 331 Part 5 Vectors 281 32 Vectors 281 32.1 Introduction 281 32.2 Vector addition 281 32.3 Resolution of vectors 283 32.4 Vector subtraction 284 33 Combination of waveforms 287 33.1 Combination of two periodic functions 287 33.2 Plotting periodic functions 287 33.3 Determining resultant phasors by calculation 288 Part 6 Complex Numbers 291 34 Complex numbers 291 34.1 Cartesian complex numbers 291 34.2 The Argand diagram 292 34.3 Addition and subtraction of complex numbers 292 34.4 Multiplication and division of complex numbers 293 34.5 Complex equations 295 34.6 The polar form of a complex number 296 34.7 Multiplication and division in polar form 298 34.8 Applications of complex numbers 299 35 De Moivre’s theorem 303 35.1 Introduction 303 35.2 Powers of complex numbers 303 35.3 Roots of complex numbers 304 Assignment 9 306 Part 7 Statistics 307 36 Presentation of statistical data 307 36.1 Some statistical terminology 307 36.2 Presentation of ungrouped data 308 36.3 Presentation of grouped data 312 37 Measures of central tendency and dispersion 319 37.1 Measures of central tendency 319 37.2 Mean, median and mode for discrete data 319 37.3 Mean, median and mode for grouped data 320 37.4 Standard deviation 322 37.5 Quartiles, deciles and percentiles 324 38 Probability 326 38.1 Introduction to probability 38.2 Laws of probability 326 326 39 The binomial and Poisson distribution 333 39.1 The binomial distribution 333 39.2 The Poisson distribution 336 Assignment 10 339 40 The normal distribution 340 40.1 Introduction to the normal distribution 340 40.2 Testing for a normal distribution 344 41 Linear correlation 347 41.1 Introduction to linear correlation 347 41.2 The product-moment formula for determining the linear correlation coefﬁcient 347 41.3 The signiﬁcance of a coefﬁcient of correlation 348 41.4 Worked problems on linear correlation 348 42 Linear regression 351 42.1 Introduction to linear regression 351 42.2 The least-squares regression lines 351 42.3 Worked problems on linear regression 352 43 Sampling and estimation theories 356 43.1 Introduction 356 43.2 Sampling distributions 356 43.3 The sampling distribution of the means 356 43.4 The estimation of population parameters based on a large sample size 359 43.5 Estimating the mean of a population based on a small sample size 364 Assignment 11 368 Multiple choice questions on chapters 27 to 43 369 Part 8 Differential Calculus 375 44 Introduction to differentiation 375 44.1 Introduction to calculus 375 44.2 Functional notation 375 44.3 The gradient of a curve 376 44.4 Differentiation from ﬁrst principles 377
- 9. CONTENTS 44.5 Differentiation of y D ax n by the general rule 379 44.6 Differentiation of sine and cosine functions 380 44.7 Differentiation of e ax and ln ax 382 45 Methods of differentiation 384 45.1 Differentiation of common functions 384 45.2 Differentiation of a product 386 45.3 Differentiation of a quotient 387 45.4 Function of a function 389 45.5 Successive differentiation 390 46 Some 46.1 46.2 46.3 46.4 applications of differentiation 392 Rates of change 392 Velocity and acceleration 393 Turning points 396 Practical problems involving maximum and minimum values 399 46.5 Tangents and normals 403 46.6 Small changes 404 Assignment 12 406 Part 9 Integral Calculus 407 47 Standard integration 407 47.1 The process of integration 407 47.2 The general solution of integrals of the form axn 407 47.3 Standard integrals 408 47.4 Deﬁnite integrals 411 48 Integration using algebraic substitutions 414 48.1 Introduction 414 48.2 Algebraic substitutions 414 48.3 Worked problems on integration using algebraic substitutions 414 48.4 Further worked problems on integration using algebraic substitutions 416 48.5 Change of limits 416 49 Integration using trigonometric substitutions 418 49.1 Introduction 418 49.2 Worked problems on integration of sin2 x, cos2 x, tan2 x and cot2 x 418 49.3 Worked problems on powers of sines and cosines 420 49.4 Worked problems on integration of products of sines and cosines 421 49.5 Worked problems on integration using the sin Â substitution 422 ix 49.6 Worked problems on integration using the tan Â substitution 424 Assignment 13 425 50 Integration using partial fractions 426 50.1 Introduction 426 50.2 Worked problems on integration using partial fractions with linear factors 426 50.3 Worked problems on integration using partial fractions with repeated linear factors 427 50.4 Worked problems on integration using partial fractions with quadratic factors 428 q 51 The t = substitution 430 2 51.1 Introduction 430 Â 51.2 Worked problems on the t D tan 2 substitution 430 51.3 Further worked problems on the Â t D tan substitution 432 2 52 Integration by parts 434 52.1 Introduction 434 52.2 Worked problems on integration by parts 434 52.3 Further worked problems on integration by parts 436 53 Numerical integration 439 53.1 Introduction 439 53.2 The trapezoidal rule 439 53.3 The mid-ordinate rule 441 53.4 Simpson’s rule 443 Assignment 14 447 54 Areas under and between curves 448 54.1 Area under a curve 448 54.2 Worked problems on the area under a curve 449 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
- 10. x CONTENTS 56.3 Further worked problems on volumes of solids of revolution 463 57 Centroids of simple shapes 466 57.1 Centroids 466 57.2 The ﬁrst moment of area 466 57.3 Centroid of area between a curve and the x-axis 466 57.4 Centroid of area between a curve and the y-axis 467 57.5 Worked problems on centroids of simple shapes 467 57.6 Further worked problems on centroids of simple shapes 468 57.7 Theorem of Pappus 471 58 Second moments of area 475 58.1 Second moments of area and radius of gyration 475 58.2 Second moment of area of regular sections 475 58.3 Parallel axis theorem 475 58.4 Perpendicular axis theorem 476 58.5 Summary of derived results 476 58.6 Worked problems on second moments of area of regular sections 476 58.7 Worked problems on second moments of areas of composite areas 480 Assignment 15 482 Part 10 Further Number and Algebra 483 59 Boolean algebra and logic circuits 483 59.1 Boolean algebra and switching circuits 483 59.2 Simplifying Boolean expressions 488 59.3 Laws and rules of Boolean algebra 488 59.4 59.5 59.6 59.7 De Morgan’s laws 490 Karnaugh maps 491 Logic circuits 495 Universal logic circuits 500 60 The theory of matrices and determinants 504 60.1 Matrix notation 504 60.2 Addition, subtraction and multiplication of matrices 504 60.3 The unit matrix 508 60.4 The determinant of a 2 by 2 matrix 508 60.5 The inverse or reciprocal of a 2 by 2 matrix 509 60.6 The determinant of a 3 by 3 matrix 510 60.7 The inverse or reciprocal of a 3 by 3 matrix 511 61 The solution of simultaneous equations by matrices and determinants 514 61.1 Solution of simultaneous equations by matrices 514 61.2 Solution of simultaneous equations by determinants 516 61.3 Solution of simultaneous equations using Cramers rule 520 Assignment 16 521 Multiple choice questions on chapters 44–61 522 Answers to multiple choice questions 526 Index 527
- 11. Preface This fourth edition of ‘Engineering Mathematics’ covers a wide range of syllabus requirements. In particular, the book is most suitable for the latest National Certiﬁcate and Diploma courses and Vocational Certiﬁcate of Education syllabuses in Engineering. This text will provide a foundation in mathematical principles, which will enable students to solve mathematical, scientiﬁc and associated engineering principles. In addition, the material will provide engineering applications and mathematical principles necessary for advancement onto a range of Incorporated Engineer degree proﬁles. It is widely recognised that a students’ ability to use mathematics is a key element in determining subsequent success. First year undergraduates who need some remedial mathematics will also ﬁnd this book meets their needs. In Engineering Mathematics 4th Edition, theory is introduced in each chapter by a simple outline of essential deﬁnitions, formulae, laws and procedures. The theory is kept to a minimum, for problem solving is extensively used to establish and exemplify the theory. It is intended that readers will gain real understanding through seeing problems solved and then through solving similar problems themselves. For clarity, the text is divided into ten topic areas, these being: number and algebra, mensuration, trigonometry, graphs, vectors, complex numbers, statistics, differential calculus, integral calculus and further number and algebra. This new edition will cover the following syllabuses: (i) Mathematics for Technicians, the core unit for National Certiﬁcate/Diploma courses in Engineering, to include all or part of the following chapters: 1. 2. 3. 4. (ii) Algebra: 2, 4, 5, 8–13, 17, 19, 27, 30 Trigonometry: 18, 21, 22, 24 Statistics: 36, 37 Calculus: 44, 46, 47, 54 Further Mathematics for Technicians, the optional unit for National Certiﬁcate/Diploma courses in Engineering, to include all or part of the following chapters: 1. 2. 3. 4. (iii) Applied Mathematics in Engineering, the compulsory unit for Advanced VCE (formerly Advanced GNVQ), to include all or part of the following chapters: 1. 2. 3. 4. 5. (iv) Algebraic techniques: 10, 14, 15, 28–30, 34, 59–61 Trigonometry: 22–24, 26 Calculus: 44–49, 52–58 Statistical and probability: 36–43 Number and units: 1, 2, 4 Mensuration: 17–20 Algebra: 5, 8–11 Functions and graphs: 22, 23, 27 Trigonometry: 21, 24 Further Mathematics for Engineering, the optional unit for Advanced VCE (formerly Advanced GNVQ), to include all or part of the following chapters: 1. Algebra and trigonometry: 5, 6, 12–15, 21, 25 2. Graphical and numerical techniques: 20, 22, 26–31 3. Differential and integral calculus: 44–47, 54 (v) The Mathematics content of Applied Science and Mathematics for Engineering, for Intermediate GNVQ (vi) Mathematics for Engineering, for Foundation and Intermediate GNVQ (vii) Mathematics 2 and Mathematics 3 for City & Guilds Technician Diploma in Telecommunications and Electronic Engineering (viii) Any introductory/access/foundation course involving Engineering Mathematics at University, Colleges of Further and Higher education and in schools. Each topic considered in the text is presented in a way that assumes in the reader little previous knowledge of that topic.
- 12. xii ENGINEERING MATHEMATICS ‘Engineering Mathematics 4th Edition’ provides a follow-up to ‘Basic Engineering Mathematics’ and a lead into ‘Higher Engineering Mathematics’. This textbook contains over 900 worked problems, followed by some 1700 further problems (all with answers). The further problems are contained within some 208 Exercises; each Exercise follows on directly from the relevant section of work, every two or three pages. In addition, the text contains 234 multiple-choice questions. Where at all possible, the problems mirror practical situations found in engineering and science. 500 line diagrams enhance the understanding of the theory. At regular intervals throughout the text are some 16 Assignments to check understanding. For example, Assignment 1 covers material contained in Chapters 1 to 4, Assignment 2 covers the material in Chapters 5 to 8, and so on. These Assignments do not have answers given since it is envisaged that lecturers could set the Assignments for students to attempt as part of their course structure. Lecturers’ may obtain a complimentary set of solutions of the Assignments in an Instructor’s Manual available from the publishers via the internet — full worked solutions and mark scheme for all the Assignments are contained in this Manual, which is available to lecturers only. To obtain a password please e-mail j.blackford@elsevier.com with the following details: course title, number of students, your job title and work postal address. To download the Instructor’s Manual visit http://www.newnespress.com and enter the book title in the search box, or use the following direct URL: http://www.bh.com/manuals/0750657766/ ‘Learning by Example’ is at the heart of ‘Engineering Mathematics 4th Edition’. John Bird University of Portsmouth
- 13. Part 1 Number and Algebra 1 Revision of fractions, decimals and percentages Alternatively: 1.1 Fractions Step (2) Step (3) # # 7ð1 C 3ð2 1 2 C D 21 3 7 " Step (1) 2 3 When 2 is divided by 3, it may be written as or 2/3. 2 is called a fraction. The number above the 3 line, i.e. 2, is called the numerator and the number below the line, i.e. 3, is called the denominator. When the value of the numerator is less than the value of the denominator, the fraction is called a proper fraction; thus 2 is a proper fraction. 3 When the value of the numerator is greater than the denominator, the fraction is called an improper fraction. Thus 7 is an improper fraction and can also 3 be expressed as a mixed number, that is, an integer and a proper fraction. Thus the improper fraction 7 3 is equal to the mixed number 2 1 . 3 When a fraction is simpliﬁed by dividing the numerator and denominator by the same number, the process is called cancelling. Cancelling by 0 is not permissible. Problem 1. Simplify 1 2 C 3 7 The lowest common multiple (i.e. LCM) of the two denominators is 3 ð 7, i.e. 21 Expressing each fraction so that their denominators are 21, gives: 1 2 1 7 2 3 7 6 C D ð C ð D C 3 7 3 7 7 3 21 21 13 7C6 D D 21 21 Step 1: the LCM of the two denominators; Step 2: for the fraction 1 , 3 into 21 goes 7 times, 3 7 ð the numerator is 7 ð 1; Step 3: for the fraction 2 , 7 into 21 goes 3 times, 7 3 ð the numerator is 3 ð 2. Thus 7C6 13 1 2 C D D as obtained previously. 3 7 21 21 Problem 2. Find the value of 3 2 3 2 1 6 One method is to split the mixed numbers into integers and their fractional parts. Then 2 2 1 1 3 2C 2 D 3C 3 6 3 6 1 2 2 D3C 3 6 4 1 3 1 D1C D1 D1 6 6 6 2 Another method is to express the mixed numbers as improper fractions.
- 14. 2 ENGINEERING MATHEMATICS 2 9 2 11 9 , then 3 D C D 3 3 3 3 3 1 12 1 13 Similarly, 2 D C D 6 6 6 6 2 1 11 13 22 13 9 1 Thus 3 2 D D D D1 3 6 3 6 6 6 6 2 8 1 7 24 8 8 ð 1 ð 8 ð ð D 5 13 71 5 ð 1 ð 1 4 64 D 12 D 5 5 Since 3 D as obtained previously. D Problem 6. 5 8 5 1 2 C 8 4 5 5 ð 5 10 ð 1 C 8 ð 2 D2C 40 25 10 C 16 D2C 40 31 31 D2 D2C 40 40 1 2 3 C1 D 4 4 5 Problem 4. 3C1 C Find the value of 3 14 ð 7 15 Dividing numerator and denominator by 3 gives: 13 14 1 ð 14 1 14 ð D D ð 7 15 5 7 5 7ð5 Dividing numerator and denominator by 7 gives: 2 1 ð 14 2 1 ð 2 D D 7ð 5 1ð5 5 1 This process of dividing both the numerator and denominator of a fraction by the same factor(s) is called cancelling. Multiplying both numerator and denominator by the reciprocal of the denominator gives: 13 21 3 3 3 ð 7 D 1 7 12 4 D 4 D 3 12 1 12 21 1 1 4 ð 21 12 1 1 21 This method can be remembered by the rule: invert the second fraction and change the operation from division to multiplication. Thus: 3 12 1 3 21 3 3 D as obtained previously. ł D ð 7 21 1 7 12 4 4 Problem 7. 3 1 3 Evaluate 1 ð 2 ð 3 5 3 7 Mixed numbers must be expressed as improper fractions before multiplication can be performed. Thus, 3 1 3 1 ð2 ð3 5 3 7 5 3 6 1 21 3 ð ð C C C D 5 5 3 3 7 7 3 1 Find the value of 5 ł 7 5 3 The mixed numbers must be expressed as improper fractions. Thus, 1 28 22 14 28 3 42 3 ł D ð D 5 ł7 D 5 3 5 3 5 55 22 11 Problem 8. Problem 5. 3 12 ł 7 21 3 3 12 ł D 7 12 7 21 21 Problem 3. Determine the value of 5 1 2 4 3 C1 8 4 5 4 Simplify 1 3 2 1 C 5 4 Simplify ł 3 1 ð 8 3 The order of precedence of operations for problems containing fractions is the same as that for integers, i.e. remembered by BODMAS (Brackets, Of, Division, Multiplication, Addition and Subtraction). Thus, 1 2 1 3 1 ł C ð 3 5 4 8 3
- 15. REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES D 4ð2C5ð1 31 ł 20 24 8 1 3 (B) 13 82 ð 1 5 20 1 26 D 3 5 5ð1 3 ð 26 D 15 73 13 D D −4 15 15 D 1 3 D D D D D D D D 1 2 2 1 4 7 16 3. (a) 10 (a) 3 7 1 2 C 7 3 43 (a) 77 2 3 (b) 3 (b) 8 3 16 1 2 1 2 1 2 3 7 2 ð ð1 5 9 7 6. (a) 3 45 ł 8 64 (B) 1 4 4 5 4 C1 5 6 16 17 (a) 1 (b) 21 60 (b) 7. 8 1 3 C ł 2 5 15 8. 7 of 15 (A) 9. 1 2 ð 4 3 (A) 10. (M) 15 ð 2 1 ð1 3 4 3 49 (b) 1 5 (b) 1 ł 2 3 9 (O) (D) 5 12 13 7 4 ð4 ð3 17 11 39 3 (b) 11 (a) 5 (a) 8 15 1 3 5 7 1 C 3 15 ł 4 16 1 3 2 ł C 3 5 7 ł 12 23 (b) 2 1 C 3 4 7 24 5 4 5 13 126 C1 3 5 2 28 55 (S) 1.2 Ratio and proportion 1 4 (a) (a) 1 2 47 63 (b) 17 15 ð 35 119 3 5 ð 4 9 5. Evaluate the following: (b) 2 9 (a) Exercise 1 Further problems on fractions 1 2 C 2 5 (b) (S) Now try the following exercise 1. (a) 3 2 C 7 11 4. (M) 1 C5 ł 8 1 41 3 7 of 1 C ł 6 4 8 16 7 5 41 3 ð C ł 6 4 8 16 7 5 41 16 2 ð C ð 6 4 18 3 35 82 1 C 24 3 2 35 C 656 1 24 2 691 1 24 2 691 12 24 7 679 D 28 24 24 3 (a) (D) Problem 9. Determine the value of 1 1 7 1 3 1 C5 ł of 3 2 6 2 4 8 16 2 7 of 6 2. 3 9 10 (b) 3 16 The ratio of one quantity to another is a fraction, and is the number of times one quantity is contained in another quantity of the same kind. If one quantity is directly proportional to another, then as one quantity doubles, the other quantity also doubles. When a quantity is inversely proportional to another, then as one quantity doubles, the other quantity is halved. Problem 10. A piece of timber 273 cm long is cut into three pieces in the ratio of 3 to 7 to 11. Determine the lengths of the three pieces
- 16. 4 ENGINEERING MATHEMATICS The total number of parts is 3 C 7 C 11, that is, 21. Hence 21 parts correspond to 273 cm 273 D 13 cm 21 3 parts correspond to 3 ð 13 D 39 cm 1 part corresponds to 7 parts correspond to 7 ð 13 D 91 cm 11 parts correspond to 11 ð 13 D 143 cm i.e. the lengths of the three pieces are 39 cm, 91 cm and 143 cm. 1 person takes three times as long, i.e. 4 ð 3 D 12 hours, 5 people can do it in one ﬁfth of the time that 12 one person takes, that is hours or 2 hours 5 24 minutes. Now try the following exercise Exercise 5 Further problems on ratio and proportion (Check: 39 C 91 C 143 D 273) 1. Problem 11. A gear wheel having 80 teeth is in mesh with a 25 tooth gear. What is the gear ratio? Gear ratio D 80:25 D 80 16 D D 3.2 25 5 Divide 621 cm in the ratio of 3 to 7 to 13. [81 cm to 189 cm to 351 cm] 2. When mixing a quantity of paints, dyes of four different colours are used in the ratio of 7:3:19:5. If the mass of the ﬁrst dye used is 3 1 g, determine the total mass of 2 the dyes used. [17 g] 3. Determine how much copper and how much zinc is needed to make a 99 kg brass ingot if they have to be in the proportions copper : zinc: :8 : 3 by mass. [72 kg : 27 kg] 4. It takes 21 hours for 12 men to resurface a stretch of road. Find how many men it takes to resurface a similar stretch of road in 50 hours 24 minutes, assuming the work rate remains constant. [5] 5. It takes 3 hours 15 minutes to ﬂy from city A to city B at a constant speed. Find how long the journey takes if i.e. gear ratio D 16 : 5 or 3.2 : 1 Problem 12. An alloy is made up of metals A and B in the ratio 2.5 : 1 by mass. How much of A has to be added to 6 kg of B to make the alloy? Ratio A : B: :2.5 : 1 (i.e. A is to B as 2.5 is to 1) A 2.5 or D D 2.5 B 1 A D 2.5 from which, 6 A D 6 ð 2.5 D 15 kg When B D 6 kg, Problem 13. If 3 people can complete a task in 4 hours, how long will it take 5 people to complete the same task, assuming the rate of work remains constant The more the number of people, the more quickly the task is done, hence inverse proportion exists. 3 people complete the task in 4 hours, (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 decimal system of numbers is based on the digits 0 to 9. A number such as 53.17 is called a decimal fraction, a decimal point separating the integer part, i.e. 53, from the fractional part, i.e. 0.17
- 17. REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES A number which can be expressed exactly as a decimal fraction is called a terminating decimal and those which cannot be expressed exactly as a decimal fraction are called non-terminating decimals. Thus, 3 D 1.5 is a terminating decimal, 2 but 4 D 1.33333. . . is a non-terminating decimal. 3 P 1.33333. . . can be written as 1.3, called ‘one pointthree recurring’. The answer to a non-terminating decimal may be expressed in two ways, depending on the accuracy required: (i) correct to a number of signiﬁcant ﬁgures, that is, ﬁgures which signify something, and (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 42.7 C 3.04 C 8.7 C 0.06 The numbers are written so that the decimal points are under each other. Each column is added, starting from the right. 87.23 81.70 5.53 Thus 87.23 − 81.70 = 5.53 Problem 16. Find the value of 23.4 17.83 57.6 C 32.68 The sum of the positive decimal fractions is 23.4 C 32.68 D 56.08 The sum of the negative decimal fractions is 17.83 C 57.6 D 75.43 Taking the sum of the negative decimal fractions from the sum of the positive decimal fractions gives: 56.08 i.e. 75.43 Problem 17. 74.3 ð 3.8 75.43 56.08 D −19.35 Determine the value of When multiplying decimal fractions: (i) the numbers are multiplied as if they are integers, and (ii) the position of the decimal point in the answer is such that there are as many digits to the right of it as the sum of the digits to the right of the decimal points of the two numbers being multiplied together. Thus 743 38 (i) 42.7 3.04 8.7 0.06 5 944 22 290 28 234 54.50 (ii) Thus 42.7 Y 3.04 Y 8.7 Y 0.06 = 54.50 As there are 1 C 1 D 2 digits to the right of the decimal points of the two numbers being multiplied together, (74.3 ð 3.8), then 74.3 × 3.8 = 282.34 Problem 15. 5 Take 81.70 from 87.23 The numbers are written with the decimal points under each other. Problem 18. Evaluate 37.81 ł 1.7, correct to (i) 4 signiﬁcant ﬁgures and (ii) 4 decimal places
- 18. 6 ENGINEERING MATHEMATICS 37.81 ł 1.7 D 37.81 1.7 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 ð 10 378.1 37.81 ł 1.7 D D 1.7 ð 10 17 The long division is similar to the long division of integers and the ﬁrst four steps are as shown: Problem 20. (a) Express as decimal fractions: 7 9 and (b) 5 16 8 (a) To convert a proper fraction to a decimal fraction, the numerator is divided by the denominator. Division by 16 can be done by the long division method, or, more simply, by dividing by 2 and then 8: 0.5625 4.50 22.24117.. 17 2 378.100000 34 Thus, 38 34 0.875 8 20 (i) 37.81 ÷ 1.7 = 22.24, correct to 4 signiﬁcant ﬁgures, and (ii) 37.81 ÷ 1.7 = 22.2412, correct to 4 decimal places. Thus 5 Problem 19. Convert (a) 0.4375 to a proper fraction and (b) 4.285 to a mixed number 7.000 7 D 0.875 8 Now try the following exercise Further problems on decimals In Problems 1 to 6, determine the values of the expressions given: 2. 73.84 3.8 ð 4.1 ð 0.7 [10.906] 374.1 ð 0.006 [2.2446] 5. By cancelling 23.6 C 14.71 4. 4375 10 000 1. 3. 0.4375 ð 10 000 (a) 0.4375 can be written as 10 000 without changing its value, 421.8 ł 17, (a) correct to 4 signiﬁcant ﬁgures and (b) correct to 3 decimal places. [(a) 24.81 (b) 24.812] 4375 875 175 35 7 D D D D 10 000 2000 400 80 16 0.4375 = i.e. 7 = 5.875 8 Exercise 3 (b) 4.5000 9 = 0.5625 16 70 68 i.e. 8 (b) For mixed numbers, it is only necessary to convert the proper fraction part of the mixed number to a decimal fraction. Thus, dealing with the 7 gives: 8 41 34 i.e. 0.4375 D 9.00 7 16 Similarly, 4.285 D 4 6. 285 57 D4 1000 200 18.9 7.421 113.247 C 8.21 [11.989] 0.068 [ 31.265] 0.0147 , (a) correct to 5 decimal places 2.3 and (b) correct to 2 signiﬁcant ﬁgures. [(a) 0.00639 (b) 0.0064]
- 19. REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 7. Convert to proper fractions: (a) 0.65 (b) 0.84 (c) 0.0125 (d) 0.282 and (e) 0.024 (a) (a) 1.875 corresponds to 1.875 ð 100%, i.e. 187.5% (b) 0.0125 corresponds to 0.0125 ð 100%, i.e. 1.25% 21 1 141 3 13 (b) (c) (d) (e) 20 25 80 500 125 Problem 22. 8. Convert to mixed numbers: (a) 1.82 (b) 4.275 (c) 14.125 (d) 15.35 and (e) 16.2125 41 11 1 (b) 4 (c) 14 (a) 1 50 40 8 7 17 (d) 15 (e) 16 20 80 (a) 9. 10. 17 , correct to 5 decimal place. 27 [0.62963] 11. 1 9 , correct to 4 signiﬁcant ﬁgures. 16 [1.563] 12. 13 31 , correct to 2 decimal places. 37 [13.84] Express as percentages: 2 5 and (b) 1 16 5 To convert fractions to percentages, they are (i) converted to decimal fractions and (ii) multiplied by 100 5 5 D 0.3125, hence corre16 16 sponds to 0.3125 ð 100%, i.e. 31.25% (a) By division, In Problems 9 to 12, express as decimal fractions to the accuracy stated: 4 , correct to 5 signiﬁcant ﬁgures. 9 [0.44444] 7 (b) 2 Similarly, 1 D 1.4 when expressed as a 5 decimal fraction. 2 Hence 1 D 1.4 ð 100% D 140% 5 Problem 23. It takes 50 minutes to machine a certain part. Using a new type of tool, the time can be reduced by 15%. Calculate the new time taken 15% of 50 minutes 15 750 ð 50 D 100 100 D 7.5 minutes. D hence the new time taken is 50 1.4 Percentages Percentages are used to give a common standard and are fractions having the number 100 as their 25 denominators. For example, 25 per cent means 100 1 i.e. and is written 25%. 4 Problem 21. Express as percentages: (a) 1.875 and (b) 0.0125 7.5 D 42.5 minutes. Alternatively, if the time is reduced by 15%, then it now takes 85% of the original time, i.e. 85% of 85 4250 50 D ð 50 D D 42.5 minutes, as above. 100 100 Problem 24. Find 12.5% of £378 12.5 12.5% of £378 means ð 378, since per cent 100 means ‘per hundred’. Hence 12.5% of £378 D A decimal fraction is converted to a percentage by multiplying by 100. Thus, 378 D £47.25 8 1 12.51 ð 378 D ð 378 D 100 8 8
- 20. 8 ENGINEERING MATHEMATICS Problem 25. Express 25 minutes as a percentage of 2 hours, correct to the nearest 1% 2. Express as percentages, correct to 3 signiﬁcant ﬁgures: (a) Working in minute units, 2 hours D 120 minutes. 25 Hence 25 minutes is ths of 2 hours. By can120 5 25 D celling, 120 24 5 P Expressing as a decimal fraction gives 0.2083 24 Multiplying by 100 to convert the decimal fraction to a percentage gives: Problem 26. A German silver alloy consists 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 3. Exercise 4 Further problems percentages 1. Convert to percentages: (a) 0.057 (b) 0.374 (c) 1.285 [(a) 5.7% (b) 37.4% (c) 128.5%] (b) 79.2% (c) 169%] Calculate correct to 4 signiﬁcant ﬁgures: When 1600 bolts are manufactured, 36 are unsatisfactory. Determine the percentage unsatisfactory. [2.25%] 5. Express: (a) 140 kg as a percentage of 1 t (b) 47 s as a percentage of 5 min (c) 13.4 cm as a percentage of 2.5 m [(a) 14% (b) 15.67% (c) 5.36%] 6. A block of monel alloy consists of 70% nickel and 30% copper. If it contains 88.2 g of nickel, determine the mass of copper in the block. [37.8 g] 7. A drilling machine should be set to 250 rev/min. The nearest speed available on the machine is 268 rev/min. Calculate the percentage over speed. [7.2%] 8. Two kilograms of a compound contains 30% of element A, 45% of element B and 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. Now try the following exercise 11 16 4. 15% corresponds to 15 ð 0.0374 D 0.561 kg (Check: 2.244 C 0.935 C 0.561 D 3.74) (c) 1 [(a) 496.4 t (b) 8.657 g (c) 20.73 s] 3.74 1% corresponds to D 0.0374 kg 100 60% corresponds to 60 ð 0.0374 D 2.244 kg 25% corresponds to 25 ð 0.0374 D 0.935 kg 19 24 (a) 18% of 2758 tonnes (b) 47% of 18.42 grams (c) 147% of 14.1 seconds By direct proportion: 100% corresponds to 3.74 kg (b) [(a) 21.2% P P 0.2083 ð 100 D 20.83% Thus 25 minutes is 21% of 2 hours, correct to the nearest 1%. 7 33 [A 0.6 kg, 9. B 0.9 kg, C 0.5 kg] A concrete mixture contains seven parts by volume of ballast, four parts by volume of sand and two parts by volume of cement. Determine the percentage of each of these three constituents correct to the nearest 1% and the mass of cement in a two tonne dry mix, correct to 1 signiﬁcant ﬁgure. [54%, 31%, 15%, 0.3 t]
- 21. 2 Indices and standard form 2.1 Indices The lowest factors of 2000 are 2ð2ð2ð2ð5ð5ð5. These factors are written as 24 ð 53 , where 2 and 5 are called bases and the numbers 4 and 3 are called indices. When an index is an integer it is called a power. Thus, 24 is called ‘two to the power of four’, and has a base of 2 and an index of 4. Similarly, 53 is called ‘ﬁve to the power of 3’ and has a base of 5 and an index of 3. Special names may be used when the indices are 2 and 3, these being called ‘squared’ and ‘cubed’, respectively. Thus 72 is called ‘seven squared’ and 93 is called ‘nine cubed’. When no index is shown, the power is 1, i.e. 2 means 21 . (i) When multiplying two or more numbers having the same base, the indices are added. Thus 32 ð 34 D 32C4 D 36 (ii) When a number is divided by a number having the same base, the indices are subtracted. Thus 35 D 35 32 (iii) 2 D 33 When a number which is raised to a power is raised to a further power, the indices are multiplied. Thus 35 2 D 35ð2 D 310 (iv) When a number has an index of 0, its value is 1. Thus 30 D 1 (v) A number raised to a negative power is the reciprocal of that number raised to a positive 1 1 power. Thus 3 4 D 4 Similarly, D 23 3 2 3 (vi) When a number is raised to a fractional power the denominator of the fraction is the root of the number and the numerator is the power. p 3 Thus 82/3 D 82 D 2 2 D 4 p p 2 and 251/2 D 251 D 251 D š5 p p (Note that Á 2 ) Reciprocal The reciprocal of a number is when the index is 1 and its value is given by 1, divided by the base. Thus the reciprocal of 2 is 2 1 and its value is 1 2 or 0.5. Similarly, the reciprocal of 5 is 5 1 which means 1 or 0.2 5 Square root The square root of a number is when the index is 1 , 2 p and the square root of 2 is written as 21/2 or 2. The value of a square root is the value of the base which when multiplied by itself gives the number. Since p 3ð3 D 9, then 9 D 3. However, 3 ð 3 D 9, p so 9 D 3. There are always two answers when ﬁnding the square root of a number and this is shown by putting both a C and a sign in front of the p answer to ap square root problem. Thus 9 D š3 and 41/2 D 4 D š2, and so on. 2.2 Worked problems on indices Problem 1. Evaluate: (a) 52 ð 53 , (b) 32 ð 34 ð 3 and (c) 2 ð 22 ð 25 Laws of indices When simplifying calculations involving indices, certain basic rules or laws can be applied, called the laws of indices. These are given below. From law (i): (a) 52 ð53 D 5 2C3 D 55 D 5ð5ð5ð5ð5 D 3125
- 22. 10 (b) ENGINEERING MATHEMATICS 32 ð 34 ð 3 D 3 2C4C1 D 37 D 3 ð 3 ð Ð Ð Ð to 7 terms D 2187 Problem 6. 3 (a) (c) 2 ð 22 ð 25 D 2 1C2C5 D 28 D 256 Problem 2. 75 57 (a) 3 and (b) 4 7 5 (a) From law (ii): (b) 75 D75 73 3 D 72 D 49 (b) 57 D57 54 4 2 ð2 32 3 and (b) 27 ð 25 3 ð 39 From the laws of indices: Find the value of: (a) Find the value of 4 23 ð 24 2 3C4 27 D 7C5 D 12 D 27 12 D 2 5 27 ð 25 2 2 1 1 D 5 D 2 32 32ð3 36 32 3 D 1C9 D 10 D 36 10 D 3 4 3 ð 39 3 3 1 1 D 4 D 3 81 D 53 D 125 Now try the following exercise Problem 3. Evaluate: (a) 52 ð 53 ł 54 and (b) 3 ð 35 ł 32 ð 33 Exercise 5 Further problems on indices In Problems 1 to 10, simplify the expressions given, expressing the answers in index form and with positive indices: From laws (i) and (ii): 52 ð 53 5 2C3 D (a) 52 ð 53 ł 54 D 54 54 5 5 D 4 D 5 5 4 D 51 D 5 5 1. (a) 33 ð 34 (b) 42 ð 43 ð 44 [(a) 37 3 ð 35 3 1C5 D 2C3 32 ð 33 3 6 3 D 5 D 36 5 D 31 D 3 3 3. (a) (a) 56 ł 53 (b) 72 ð 74 ð 7 ð 73 [(a) 26 (b) 710 ] 3 ð 35 ł 32 ð 33 D 3 4 2 5 Problem 4. Simplify: (a) 2 (b) 3 expressing the answers in index form. 23 4 Evaluate: (b) 32 5 102 3 104 ð 102 (b) 37 32 [(a) 2 (b) 35 ] (b) 713 /710 [(a) 53 5. (a) 72 6. D 23ð4 D 212 Problem 5. 24 23 , From law (iii): (a) (a) 23 ð 2 ð 22 4. (b) 2. (a) D 32ð5 D 310 3 (b) 33 22 ð 23 24 (b) 2 (b) 73 ] [(a) 76 (b) 36 ] 37 ð 34 35 [(a) 2 (b) 36 ] 7. (a) 57 52 ð 53 (b) 135 13 ð 132 [(a) 52 From the laws of indices: 10 2ð3 106 102 3 D D 6 104 ð 102 10 4C2 10 6 6 D 10 D 100 D 1 (b) 49 ] 8. (a) 9 ð 32 3 ð 27 3 2 (b) 16 ð 4 2 2ð8 3 [(a) 34 (b) 132 ] (b) 1]
- 23. INDICES AND STANDARD FORM 2 5 5 9. (a) (b) 4 32 ð 3 33 4 (a) 52 72 ð 7 7ð7 10. (a) 3 4 (b) 1 35 23 ð 2 4 ð 25 2 ð 2 2 ð 26 (b) (a) 72 (Note that it does not matter whether the 4th root of 16 is found ﬁrst or whether 16 cubed is found ﬁrst — the same answer will result). p 3 (c) 272/3 D 272 D 3 2 D 9 1/2 (d) 9 (b) 1 2 D 33 ð 57 53 ð 34 D3 Problem 8. 1 4 ð57 Problem 11. 1/2 (a) 4 3 ð 72ð2 4 3 (b) 163/4 D ±8 32 ð 55 C 33 ð 53 34 ð 54 32 2 ð55 3 C33 34 2 ð54 3 30 ð 52 C 31 ð 50 32 ð 51 28 1 ð 25 C 3 ð 1 D D 9ð5 45 1/2 2 3 D 24 D 16 D p 4 D ±2 p 4 D 163 D š2 Evaluate: D 0 Problem 12. (a) 41/2 D 2 3 Evaluate: (b) 163/4 (c) 272/3 (d) 9 2 32 ð 55 33 ð 53 C 2 2 3 3 3 ð5 C3 ð5 D 3 ð 5 4 34 ð 5 4 ð 54 3 3 ð5 32 ð 53 5 D2 ð3 ð7 1 1 9 D ð 32 ð 1 D D 4 2 2 2 Problem 9. 1 1 1 1 Dp D 2 D 5 2/5 32 2 4 322 Dividing each term by the HCF (i.e. highest common factor) of the three terms, i.e. 32 ð 53 , gives: 2 2 D D 23C1 23 ð 35 ð 72 2 74 ð 24 ð 33 1 2/5 41.5 ð 81/3 16 8ð2 D D D 16 2 ð 32 2/5 1 2 1 4ð 4 3 Find the value of ð 35 41.5 ð 81/3 22 ð 32 2/5 [ 2 2 ]3/2 ð 23 1/3 23 ð 21 41.5 ð 81/3 D D 2 2 ð 32 2/5 2 ð 25 2/5 2 2 2 ð2 2 1 54 625 D 208 ð5 D 1 D 3 3 3 4 Evaluate: Alternatively, 4 23 ð 35 ð 72 2 D 23 74 ð 24 ð 33 32 Hence The laws of indices only apply to terms having the same base. Grouping terms having the same base, and then applying the laws of indices to each of the groups independently gives: 33 ð 57 33 57 D 4 ð 3 D33 53 ð 34 3 5 1 1 1 Dp D D± š3 3 9 p 41.5 D 43/2 D 43 D 23 D 8, p 3 81/3 D 8 D 2, 22 D 4 and Evaluate: 1 91/2 Problem 10. 2.3 Further worked problems on indices Problem 7. 11 34 5 Find the value of 3 ð5 ð 54 C 33 ð 53 2 ð 50
- 24. 12 ENGINEERING MATHEMATICS To simplify the arithmetic, each term is divided by the HCF of all the terms, i.e. 32 ð 53 . Thus 34 32 ð 55 ð 54 C 33 ð 53 Exercise 6 34 32 ð54 2 0 2 3 ð55 C33 1. 3 2 ð53 33 ð 52 54 ð 34 (a) 3 (a) 2. 3 4 3 Simplify: 3 5 ð 42 ð 93 83 ð 34 (a) 2 2 3 5 giving the answer with positive indices A fraction raised to a negative power has the same value as the inverse of the fraction raised to a positive power. Thus, 4 3 1 D D 3 ð 2 5 2 3 5 3 2 5 Similarly, Thus, 2 3 5 3 1 52 52 D 2 D1ð 2 D 2 3 3 3 52 3 5 2 D 2 3. 2 4 5 ð 2 3 D 3 33 5 23 3 4 52 23 D 3ð 2ð 3 3 3 5 D 22 3 3C2 9 D ð 23 ð53 2 32 25 (b) 73 (b) 1 ð 37 4 2 210 1 ð 52 1 b 810.25 1/4 (a) 9 (b) š3 4 9 d (c) š 1 2 1/2 (d) š 2 3 In Problems 4 to 8, evaluate the expressions given. 4. 5. 6. 7. 3 2 35 × 5 1 32 Evaluate a 3 8 2 ð 52 ð 3 252 ð 24 ð 9 (b) c 16 53 23 3 1 3 ð 52 (a) A fraction raised to a power means that both the numerator and the denominator of the fraction are 4 3 43 raised to that power, i.e. D 3 3 3 3 5 7 2ð3 2 35 ð 74 ð 7 (b) 2 3 ð5 25 25 D 2 D D 1 C 31 ð 50 3 ð5 45 C 3 48 Problem 13. Further problems on indices In Problems 1 and 2, simplify the expressions given, expressing the answers in index form and with positive indices: 32 ð 55 2 3 D 4 34 ð 5 3 3 ð5 3 ð 53 C 2 32 ð 53 3 ð 53 D Now try the following exercise 8. 34 92 ð 74 ð 74 C 33 ð 72 24 1 2 2 3 2 ð 44 23 ð 162 3 2 3 3 5 4 3 2 9 147 148 1 9 2 5 2 65 72 4 [64] 2 32 3/2 ð 81/3 2 3 2 ð 43 1/2 ð 9 1/2 4 1 2
- 25. INDICES AND STANDARD FORM Similarly, 2.4 Standard form A number written with one digit to the left of the decimal point and multiplied by 10 raised to some power is said to be written in standard form. Thus: 5837 is written as 5.837 ð 103 in standard form, and 0.0415 is written as 4.15 ð 10 2 in standard form. When a number is written in standard form, the ﬁrst factor is called the mantissa and the second factor is called the exponent. Thus the number 5.8 ð 103 has a mantissa of 5.8 and an exponent of 103 . (i) Numbers having the same exponent can be added or subtracted in standard form by adding or subtracting the mantissae and keeping the exponent the same. Thus: 4 4 2.3 ð 10 C 3.7 ð 10 D 2.3 C 3.7 ð 104 D 6.0 ð 104 and 5.9 ð 10 D 5.9 2 13 4.6 ð 10 2 D 1.3 ð 10 D 4 ð 102 Problem 14. Express in standard form: (a) 38.71 (b) 3746 (c) 0.0124 For a number to be in standard form, it is expressed with only one digit to the left of the decimal point. Thus: (a) 38.71 must be divided by 10 to achieve one digit to the left of the decimal point and it must also be multiplied by 10 to maintain the equality, i.e. 38.71 D form 2 When the numbers have different exponents, one way of adding or subtracting the numbers is to express one of the numbers in nonstandard form, so that both numbers have the same exponent. Thus: 2 2.5 Worked problems on standard form 2 4.6 ð 10 6 ð 104 6 D ð 104 1.5 ð 102 1.5 (b) 38.71 ð 10 D 3.871 × 10 in standard 10 3746 ð 1000 D 3.746 × 103 in stan3746 D 1000 dard form (c) 0.0124 D 0.0124 ð 100 1.24 D 100 100 D 1.24 × 10−2 in standard form 2.3 ð 104 C 3.7 ð 103 D 2.3 ð 104 C 0.37 ð 104 Problem 15. Express the following numbers, which are in standard form, as decimal numbers: (a) 1.725 ð 10 2 (b) 5.491 ð 104 (c) 9.84 ð 100 D 2.3 C 0.37 ð 104 D 2.67 ð 104 Alternatively, 2.3 ð 104 C 3.7 ð 103 1.725 D 0.01725 100 5.491 ð 104 D 5.491 ð 10 000 D 54 910 D 23 000 C 3700 D 26 700 D 2.67 ð 104 (ii) (a) 1.725 ð 10 (b) The laws of indices are used when multiplying or dividing numbers given in standard form. For example, 2.5 ð 103 ð 5 ð 102 D 2.5 ð 5 ð 103C2 D 12.5 ð 105 or 1.25 ð 106 2 D (c) 9.84 ð 100 D 9.84 ð 1 D 9.84 (since 100 D 1) Problem 16. Express in standard form, correct to 3 signiﬁcant ﬁgures: (a) 2 9 3 (b) 19 (c) 741 8 3 16
- 26. 14 ENGINEERING MATHEMATICS 3 D 0.375, and expressing it in standard form 8 gives: 0.375 D 3.75 × 10−1 2 P (b) 19 D 19.6 D 1.97 × 10 in standard form, 3 correct to 3 signiﬁcant ﬁgures 9 (c) 741 D 741.5625 D 7.42 × 102 in standard 16 form, correct to 3 signiﬁcant ﬁgures (a) 6. (a) 3.89 ð 10 (c) 8 ð 10 2 (b) 6.741 ð 10 1 3 [(a) 0.0389 (b) 0.6741 (c) 0.008] 2.6 Further worked problems on standard form Problem 17. Express the following numbers, given in standard form, as fractions or mixed numbers: (a) 2.5 ð 10 1 (b) 6.25 ð 10 2 (c) 1.354 ð 102 Problem 18. Find the value of: (a) 7.9 ð 10 2 2 5.4 ð 10 (b) 8.3 ð 103 C 5.415 ð 103 and (c) 9.293 ð 102 C 1.3 ð 103 expressing the answers in standard form. 2.5 25 1 D D 10 100 4 625 1 6.25 D D (b) 6.25 ð 10 2 D 100 10 000 16 4 2 (c) 1.354 ð 102 D 135.4 D 135 D 135 10 5 Numbers having the same exponent can be added or subtracted by adding or subtracting the mantissae and keeping the exponent the same. Thus: Now try the following exercise (a) (a) 2.5 ð 10 1 D 7.9 ð 10 D 7.9 Exercise 7 Further problems on standard form In Problems 1 to 4, express in standard form: 1. (a) 73.9 (b) 28.4 (c) 197.72 (a) 7.39 ð 10 (b) 2.84 ð 10 (c) 1.9762 ð 102 2. (a) 2748 (b) 33170 (c) 274218 (a) 2.748 ð 103 (b) 3.317 ð 104 (c) 2.74218 ð 105 3. (a) 0.2401 (b) 0.0174 (c) 0.00923 (a) 2.401 ð 10 1 (b) 1.74 ð 10 2 (c) 9.23 ð 10 3 1 7 3 1 4. (a) (b) 11 (c) 130 (d) 2 8 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. 3 2 (a) 1.01 ð 10 (b) 9.327 ð 10 (c) 5.41 ð 104 (d) 7 ð 100 [(a) 1010 (b) 932.7 (c) 54 100 (d) 7] (b) 2 2 5.4 ð 10 5.4 ð 10 2 D 2.5 × 10−2 8.3 ð 103 C 5.415 ð 103 D 8.3 C 5.415 ð 103 D 13.715 ð 103 D 1.3715 × 104 in standard form (c) Since only numbers having the same exponents can be added by straight addition of the mantissae, the numbers are converted to this form before adding. Thus: 9.293 ð 102 C 1.3 ð 103 D 9.293 ð 102 C 13 ð 102 D 9.293 C 13 ð 102 D 22.293 ð 102 D 2.2293 × 103 in standard form. Alternatively, the numbers can be expressed as decimal fractions, giving: 9.293 ð 102 C 1.3 ð 103 D 929.3 C 1300 D 2229.3 D 2.2293 × 103 in standard form as obtained previously. This method is often the ‘safest’ way of doing this type of problem.
- 27. INDICES AND STANDARD FORM Problem 19. 3. Evaluate 3.5 ð 105 7 ð 102 expressing answers in standard form (a) 3.75 ð 103 6 ð 104 and (b) [(a) 1.35 ð 102 4. (a) 3 3.75 ð 10 4 6 ð 10 (a) 3C4 D 2.25 × 10 8 5 3.5 ð 10 3.5 ð 105 D 7 ð 102 7 2 6 ð 10 3 ð 10 3 5 (b) 5. (b) 1.1 ð 105 ] 2.4 ð 103 3 ð 10 4.8 ð 104 [(a) 2 ð 102 D 3.75 ð 6 10 D 22.50 ð 107 (b) (a) 4.5 ð 10 2 3 ð 103 (b) 2 ð 5.5 ð 104 2 (b) 1.5 ð 10 3 ] Write the following statements in standard form: (a) The density 2710 kg m 3 of aluminium is [2.71 ð 103 kg m 3 ] D 0.5 ð 103 D 5 × 102 (b) Poisson’s ratio for gold is 0.44 Now try the following exercise [4.4 ð 10 1 ] Exercise 8 Further problems on standard form (c) The impedance of free space is 376.73 [3.7673 ð 102 ] In Problems 1 to 4, ﬁnd values of the expressions given, stating the answers in standard form: (d) 1. (a) 3.7 ð 102 C 9.81 ð 102 (b) 1.431 ð 10 1 C 7.3 ð 10 1 [(a) 1.351 ð 103 (b) 8.731 ð 10 1 ] 2. (a) 4.831 ð 102 C 1.24 ð 103 (b) 3.24 ð 10 3 1.11 ð 10 4 [(a) 1.7231 ð 103 (b) 3.129 ð 10 3 ] The electron rest energy is 0.511 MeV [5.11 ð 10 1 MeV] (e) Proton charge-mass 9 5 789 700 C kg 1 ratio is [9.57897 ð 107 C kg 1 ] (f) The normal volume of a perfect gas is 0.02241 m3 mol 1 [2.241 ð 10 2 m3 mol 1 ] 15
- 28. 3 Computer numbering systems 3.1 Binary numbers Problem 2. fraction The system of numbers in everyday use is the denary or decimal system of numbers, using the digits 0 to 9. It has ten different digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) and is said to have a radix or base of 10. The binary system of numbers has a radix of 2 and uses only the digits 0 and 1. 0.10112 D 1 ð 2 The decimal number 234.5 is equivalent to 1 Problem 3. number i.e. 1101.12 = 13.510 , the sufﬁxes 2 and 10 denoting binary and decimal systems of numbers respectively. 110112 D 1 ð 24 C 1 ð 23 C 0 ð 22 D 2710 1 C1ð2 2 3 C1ð2 4 D 4 C 0 C 1 C 0 C 0.25 C 0 C 0.0625 D 5.312510 Now try the following exercise Convert 110112 to a decimal D 16 C 8 C 0 C 2 C 1 3 Convert 101.01012 to a decimal C0ð2 Exercise 9 C 1 ð 21 C 1 ð 20 C1ð2 C0ð2 1 8 C 4 C 0 C 1 C , that is 13.5 2 From above: 2 101.01012 D 1 ð 22 C 0 ð 21 C 1 ð 20 1 Thus the decimal number equivalent to the binary number 1101.1 is Problem 1. number C0ð2 D 0.687510 i.e. is the sum of terms comprising: (a digit) multiplied by (the base raised to some power). In the binary system of numbers, the base is 2, so 1101.1 is equivalent to: 1 ð 23 C 1 ð 22 C 0 ð 21 C 1 ð 20 C 1 ð 2 1 C1ð2 4 1 1 1 D1ð C0ð 2 C1ð 3 2 2 2 1 C1ð 4 2 1 1 1 D C C 2 8 16 D 0.5 C 0.125 C 0.0625 3.2 Conversion of binary to decimal 2 ð 102 C 3 ð 101 C 4 ð 100 C 5 ð 10 Convert 0.10112 to a decimal Further problems on conversion of binary to decimal numbers In Problems 1 to 4, convert the binary numbers given to decimal numbers. 1. (a) 110 (b) 1011 [(a) 610 (c) 1110 (d) 1001 (b) 1110 (c) 1410 (d) 910 ]
- 29. COMPUTER NUMBERING SYSTEMS 2. (a) 10101 (b) 11001 (c) 101101 (d) 110011 [(a) 2110 (b) 2510 (c) 4510 3. (a) 0.1101 (b) 0.11001 (d) 5110 ] (c) 0.00111 For fractions, the most signiﬁcant bit of the result is the top bit obtained from the integer part of multiplication by 2. The least signiﬁcant bit of the result is the bottom bit obtained from the integer part of multiplication by 2. Thus 0.62510 = 0.1012 (d) 0.01011 (a) 0.812510 (c) 0.2187510 17 (b) 0.7812510 (d) 0.3437510 Problem 4. number Convert 4710 to a binary 4. (a) 11010.11 (b) 10111.011 (c) 110101.0111 (d) 11010101.10111 (a) 26.7510 (b) 23.37510 (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 2 5 1 An integer decimal number can be converted to a corresponding binary number by repeatedly dividing by 2 and noting the remainder at each stage, as shown below for 3910 2 2 1 2 1 0 0 1 2 2 2 2 2 2 1 39 Remainder 19 1 9 1 4 1 2 0 1 0 0 1 0 1 1 1 1 Thus 4710 = 1011112 Problem 5. number (most → 1 0 0 1 1 1 ← (least significant bit) significant bit) The result is obtained by writing the top digit of the remainder as the least signiﬁcant bit, (a bit is a binary digit and the least signiﬁcant bit is the one on the right). The bottom bit of the remainder is the most signiﬁcant bit, i.e. the bit on the left. Thus 3910 = 1001112 The fractional part of a decimal number can be converted to a binary number by repeatedly multiplying by 2, as shown below for the fraction 0.625 0.625 × 2 = 0. 500 0.500 × 2 = 1. 000 From above, repeatedly multiplying by 2 gives: 0.40625 × 2 = (most significant bit) . 1 0 1 (least significant bit) 0. 8125 0.8125 × 2= 1. 625 0.625 × 2= 1. 25 0.25 × 2= 0. 5 0.5 × 2 = 1. 0 1. 250 0.250 × 2 = Convert 0.4062510 to a binary . 0 1 1 0 1 i.e. 0.4062510 = 0.011012 Problem 6. number Convert 58.312510 to a binary
- 30. 18 ENGINEERING MATHEMATICS The integer part is repeatedly divided by 2, giving: 2 2 2 2 2 2 58 Remainder 0 29 1 14 0 7 1 3 1 1 0 1 1 1 1 For decimal integers containing several digits, repeatedly dividing by 2 can be a lengthy process. In this case, it is usually easier to convert a decimal number to a binary number via the octal system of numbers. This system has a radix of 8, using the digits 0, 1, 2, 3, 4, 5, 6 and 7. The denary number equivalent to the octal number 43178 is 0 1 0 The fractional part is repeatedly multiplied by 2 giving: 0.3125 × 2 = 0.625 × 2 = 0.25 × 2 = 0.5 ×2= 3.4 Conversion of decimal to binary via octal 0.625 1.25 0.5 1.0 4 ð 83 C 3 ð 82 C 1 ð 81 C 7 ð 80 i.e. 4 ð 512 C 3 ð 64 C 1 ð 8 C 7 ð 1 or 225510 An integer decimal number can be converted to a corresponding octal number by repeatedly dividing by 8 and noting the remainder at each stage, as shown below for 49310 8 493 Remainder .0 1 0 1 8 61 5 8 7 5 0 Thus 58.312510 = 111010.01012 7 Now try the following exercise Exercise 10 Further problems on conversion of decimal to binary numbers In Problems 1 to 4, convert the decimal numbers given to binary numbers. 7 5 5 Thus 49310 = 7558 The fractional part of a decimal number can be converted to an octal number by repeatedly multiplying by 8, as shown below for the fraction 0.437510 1. (a) 5 (b) 15 (c) 19 (d) 29 (a) 1012 (c) 100112 (b) 11112 (d) 111012 2. (a) 31 (b) 42 (c) 57 (d) 63 (a) 111112 (c) 1110012 3. (a) 0.25 (b) 0.21875 (d) 0.59375 (c) 0.28125 (a) 0.012 (c) 0.010012 4. (a) 47.40625 (c) 53.90625 (b) 1010102 (d) 1111112 (b) 0.001112 (d) 0.100112 (b) 30.8125 (d) 61.65625 (a) 101111.011012 (b) 11110.11012 (c) 110101.111012 (d) 111101.101012 0.4375 × 8 = 3. 5 ×8= 4. 0 0.5 .3 4 For fractions, the most signiﬁcant bit is the top integer obtained by multiplication of the decimal fraction by 8, thus 0.437510 D 0.348 The natural binary code for digits 0 to 7 is shown in Table 3.1, and an octal number can be converted to a binary number by writing down the three bits corresponding to the octal digit. Thus 4378 D 100 011 1112 and 26.358 D 010 110.011 1012
- 31. COMPUTER NUMBERING SYSTEMS 19 Table 3.1 Octal digit Natural binary number 0 1 2 3 4 5 6 7 000 001 010 011 100 101 110 111 Problem 9. Convert 5613.9062510 to a binary number, via octal The integer part is repeatedly divided by 8, noting the remainder, giving: 8 5613 8 701 87 8 10 8 1 8 0 The ‘0’ on the extreme left does not signify anything, thus 26.358 D 10 110.011 1012 Conversion of decimal to binary via octal is demonstrated in the following worked problems. Problem 7. Convert 371410 to a binary number, via octal Dividing repeatedly by 8, and noting the remainder gives: 8 3714 Remainder 8 464 2 8 58 0 8 7 2 0 7 1 2 7 5 5 This octal number is converted to a binary number, (see Table 3.1) 127558 D 001 010 111 101 1012 i.e. i.e. 72028 D 111 010 000 0102 371410 = 111 010 000 0102 Problem 8. Convert 0.5937510 to a binary number, via octal Multiplying repeatedly by 8, and noting the integer values, gives: 0.59375 × 8 = 0.75 × 8= 4.75 6.00 .4 6 Thus 0.5937510 D 0.468 561310 D 1 010 111 101 1012 The fractional part is repeatedly multiplied by 8, and noting the integer part, giving: 0.90625 × 8 = 0.25 × 8= 7.25 2.00 .7 2 This octal fraction is converted to a binary number, (see Table 3.1) 7 2 0 2 From Table 3.1, Remainder 5 5 7 2 1 0.728 D 0.111 0102 i.e. 0.9062510 D 0.111 012 Thus, 5613.9062510 = 1 010 111 101 101.111 012 Problem 10. Convert 11 110 011.100 012 to a decimal number via octal Grouping the binary number in three’s from the binary point gives: 011 110 011.100 0102 Using Table 3.1 to convert this binary number to an octal number gives: 363.428 and 363.428 D 3 ð 82 C 6 ð 81 C 3 ð 80 C4ð8 1 C2ð8 2 From Table 3.1, 0.468 D 0.100 1102 D 192 C 48 C 3 C 0.5 C 0.03125 i.e. D 243.5312510 0.5937510 = 0.100 112
- 32. 20 ENGINEERING MATHEMATICS To convert from hexadecimal to decimal: Now try the following exercise For example Exercise 11 Further problems on conversion between decimal and binary numbers via octal In Problems 1 to 3, convert the decimal numbers given to binary numbers, via octal. 1. (a) 343 1A16 D 1 ð 161 C A ð 160 D 1 ð 161 C 10 ð 1 D 16 C 10 D 26 i.e. Similarly, 2E16 D 2 ð 161 C E ð 160 (b) 572 (c) 1265 (a) 1010101112 D 2 ð 161 C 14 ð 160 D 32 C 14 D 4610 (b) 10001111002 and 1BF16 D 1 ð 162 C B ð 161 C F ð 160 (c) 100111100012 D 1 ð 162 C 11 ð 161 C 15 ð 160 2. (a) 0.46875 (b) 0.6875 (c) 0.71875 (a) 0.011112 (b) 0.10112 (c) 0.101112 3. (a) 247.09375 1A16 D 2610 D 256 C 176 C 15 D 44710 Table 3.2 compares decimal, binary, octal and hexadecimal numbers and shows, for example, that 2310 D 101112 D 278 D 1716 (b) 514.4375 (c) 1716.78125 (a) 11110111.000112 (b) 1000000010.01112 (c) 11010110100.110012 Problem 11. Convert the following hexadecimal numbers into their decimal equivalents: (a) 7A16 (b) 3F16 (a) 7A16 D 7 ð 161 C A ð 160 D 7 ð 16 C 10 ð 1 4. Convert the following binary numbers to decimal numbers via octal: (a) 111.011 1 (b) 101 001.01 (c) 1 110 011 011 010.001 1 (a) 7.437510 D 112 C 10 D 122 Thus 7A16 = 12210 (b) 3F16 D 3 ð 161 C F ð 160 D 3 ð 16 C 15 ð 1 D 48 C 15 D 63 (b) 41.2510 Thus, 3F16 = 6310 (c) 7386.187510 Problem 12. Convert the following hexadecimal numbers into their decimal equivalents: (a) C916 (b) BD16 3.5 Hexadecimal numbers (a) C916 D C ð 161 C 9 ð 160 D 12 ð 16 C 9 ð 1 The complexity of computers requires higher order numbering systems such as octal (base 8) and hexadecimal (base 16), which are merely extensions of the binary system. A hexadecimal numbering system has a radix of 16 and uses the following 16 distinct digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F ‘A’ corresponds to 10 in the denary system, B to 11, C to 12, and so on. D 192 C 9 D 201 Thus C916 = 20110 (b) BD16 D B ð 161 C D ð 160 D 11 ð 16 C 13 ð 1 D 176 C 13 D 189 Thus BD16 = 18910 Problem 13. number Convert 1A4E16 into a denary
- 33. COMPUTER NUMBERING SYSTEMS Table 3.2 Hence 2610 = 1A16 Decimal Binary Octal Hexadecimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111 100000 0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 40 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20 1A4E16 D 1 ð 163 C A ð 162 C 4 ð 161 C E ð 160 D 1 ð 163 C 10 ð 162 C 4 ð 161 C 14 ð 160 D 1 ð 4096 C 10 ð 256 C 4 ð 16 C 14 ð 1 D 4096 C 2560 C 64 C 14 D 6734 Thus, 1A4E16 = 673410 To convert from decimal to hexadecimal: This is achieved by repeatedly dividing by 16 and noting the remainder at each stage, as shown below for 2610 16 26 16 1 0 21 Remainder 10 ≡ A16 1 ≡ 116 most significant bit → 1 A ← least significant bit Similarly, for 44710 16 447 Remainder 16 27 15 ≡ F16 16 1 11 ≡ B16 0 1 ≡ 116 1 BF Thus 44710 = 1BF16 Problem 14. Convert the following decimal numbers into their hexadecimal equivalents: (a) 3710 (b) 10810 (a) 16 37 Remainder 16 2 5 = 516 0 2 = 216 most significant bit → 2 5 ← least significant bit Hence 3710 = 2516 (b) 16 108 Remainder 16 6 12 = C16 0 6 = 616 6C Hence 10810 = 6C16 Problem 15. Convert the following decimal numbers into their hexadecimal equivalents: (a) 16210 (b) 23910 (a) 16 162 Remainder 16 10 2 = 216 0 10 = A16 A2 Hence 16210 = A216 (b) 16 239 Remainder 16 14 15 = F16 0 14 = E16 EF Hence 23910 = EF16 To convert from binary to hexadecimal: The binary bits are arranged in groups of four, starting from right to left, and a hexadecimal symbol
- 34. 22 ENGINEERING MATHEMATICS is assigned to each group. For example, the binary number 1110011110101001 is initially grouped in fours as: (b) the right gives: 1110 0111 1010 1001 0001 1001 1110 and assigning hexadecimal and a hexadecimal symbol assigned to each group as E Grouping bits in fours from symbols to each group gives: 7 A 9 from Table 3.2 1 9 E from Table 3.2 Thus, 1100111102 = 19E16 Hence 11100111101010012 = E7A916 . Problem 18. Convert the following hexadecimal numbers into their binary equivalents: (a) 3F16 (b) A616 To convert from hexadecimal to binary: The above procedure is reversed, thus, for example, 6CF316 D 0110 1100 1111 0011 from Table 3.2 i.e. (a) Spacing out hexadecimal digits gives: 6CF316 = 1101100111100112 binary gives: right gives: gives: D 6 from Table 3.2 6 1010 0110 from Table 3.2 Thus, A616 = 101001102 Thus, 110101102 = D616 Grouping bits in fours from the Problem 19. Convert the following hexadecimal numbers into their binary equivalents: (a) 7B16 (b) 17D16 0110 0111 and assigning hexadecimal symbols 6 7 from Table 3.2 A and converting each into binary and assigning hexadecimal symbols to each group gives: Spacing out hexadecimal digits gives: 1101 0110 right gives: 0011 1111 from Table 3.2 Thus, 3F16 = 1111112 (b) (a) Grouping bits in fours from the (b) F and converting each into Problem 16. Convert the following binary numbers into their hexadecimal equivalents: (a) 110101102 (b) 11001112 to each group gives: 3 (a) Spacing out hexadecimal Thus, 11001112 = 6716 digits gives: 7 B and converting each into Problem 17. Convert the following binary numbers into their hexadecimal equivalents: (a) 110011112 (b) 1100111102 binary gives: 0111 1011 from Table 3.2 Thus, 7B16 = 11110112 (b) (a) Grouping bits in fours from the right gives: 1100 1111 Thus, 110011112 = CF16 digits gives: 1 7 D and converting each into and assigning hexadecimal symbols to each group gives: Spacing out hexadecimal C F from Table 3.2 binary gives: Thus, 17D16 = 1011111012 0001 0111 1101 from Table 3.2
- 35. COMPUTER NUMBERING SYSTEMS 9. Now try the following exercise Exercise 12 Further problems on hexadecimal numbers In Problems 1 to 4, convert the given hexadecimal numbers into their decimal equivalents. 1. E716 [23110 ] 2. 2C16 [4410 ] 3. 9816 [15210 ] 4. 2F116 [75310 ] In Problems 5 to 8, convert the given decimal numbers into their hexadecimal equivalents. 110101112 [D716 ] 10. 111010102 [EA16 ] 11. 100010112 [8B16 ] 12. 101001012 [A516 ] In Problems 13 to 16, convert the given hexadecimal numbers into their binary equivalents. 13. 3716 14. ED16 [111011012 ] [100111112 ] 5. 5410 [3616 ] 6. 20010 [C816 ] 7. 9110 [5B16 ] 8. 23810 [EE16 ] 15. 9F16 In Problems 9 to 12, convert the given binary numbers into their hexadecimal equivalents. 16. A2116 [1101112 ] [1010001000012 ] 23
- 36. 4 Calculations and evaluation of formulae 55 could therefore be expected. Certainly an answer around 500 or 5 would not be expected. Actually, by calculator 4.1 Errors and approximations (i) In all problems in which the measurement of distance, time, mass or other quantities occurs, an exact answer cannot be given; only an answer which is correct to a stated degree of accuracy can be given. To take account of this an error due to measurement is said to exist. (ii) To take account of measurement errors it is usual to limit answers so that the result given is not more than one signiﬁcant ﬁgure greater than the least accurate number given in the data. (iii) Rounding-off errors can exist with decimal fractions. For example, to state that D 3.142 is not strictly correct, but ‘ D 3.142 correct to 4 signiﬁcant ﬁgures’ is a true statement. (Actually, D 3.14159265 . . .) (iv) It is possible, through an incorrect procedure, to obtain the wrong answer to a calculation. This type of error is known as a blunder. (v) An order of magnitude error is said to exist if incorrect positioning of the decimal point occurs after a calculation has been completed. (vi) Blunders and order of magnitude errors can be reduced by determining approximate values of calculations. Answers which do not seem feasible must be checked and the calculation must be repeated as necessary. An engineer will often need to make a quick mental approximation for a calcula49.1 ð 18.4 ð 122.1 may tion. For example, 61.2 ð 38.1 50 ð 20 ð 120 and then, be approximated to 60 ð 40 49.1 ð 18.4 ð 122.1 D 47.31, correct to 61.2 ð 38.1 4 signiﬁcant ﬁgures. Problem 1. The area A of a triangle is 1 given by A D bh. The base b when 2 measured is found to be 3.26 cm, and the perpendicular height h is 7.5 cm. Determine the area of the triangle. 1 1 bh D ð 3.26 ð 7.5 D 2 2 2 12.225 cm (by calculator). 1 The approximate value is ð 3 ð 8 D 12 cm2 , so 2 there are no obvious blunder or magnitude errors. However, it is not usual in a measurement type problem to state the answer to an accuracy greater than 1 signiﬁcant ﬁgure more than the least accurate number in the data: this is 7.5 cm, so the result should not have more than 3 signiﬁcant ﬁgures Area of triangle D Thus, area of triangle = 12.2 cm2 Problem 2. State which type of error has been made in the following statements: (a) 72 ð 31.429 D 2262.9 (b) (c) 11.714 ð 0.0088 D 0.3247 correct to 4 decimal places. 1 50 ð1 20 ð 120 2 D 50. An 1 60 ð 40 2 1 accurate answer somewhere between 45 and by cancelling, 16 ð 0.08 ð 7 D 89.6 (d) 29.74 ð 0.0512 D 0.12, correct to 11.89 2 signiﬁcant ﬁgures.
- 37. CALCULATIONS AND EVALUATION OF FORMULAE (a) 72 ð 31.429 D 2262.888 (by calculator), hence a rounding-off error has occurred. The answer should have stated: 2.19 ð 203.6 ð 17.91 ³ 80 12.1 ð 8.76 2.19 ð 203.6 ð 17.91 D 75.3, (By calculator, 12.1 ð 8.76 correct to 3 signiﬁcant ﬁgures.) i.e. 72 ð 31.429 D 2262.9, correct to 5 signiﬁcant ﬁgures or 2262.9, correct to 1 decimal place. (b) 8 32 ð 7 ð7D 100 25 24 224 D8 D 8.96 D 25 25 16 ð 0.08 ð 7 D 16 ð Now try the following exercise Exercise 13 Hence an order of magnitude error has occurred. (c) 11.714 ð 0.0088 is approximately equal to 12 ð 9 ð 10 3 , i.e. about 108 ð 10 3 or 0.108. Thus a blunder has been made. (d) In Problems 1 to 5 state which type of error, or errors, have been made: 1. 25 ð 0.06 ð 1.4 D 0.21 [order of magnitude error] 137 ð 6.842 D 937.4 Rounding-off error–should add ‘correct to 4 signiﬁcant ﬁgures’ or ‘correct to 1 decimal place’ 3. hence no order of magnitude error has 29.74 ð 0.0512 D 0.128 occurred. However, 11.89 correct to 3 signiﬁcant ﬁgures, which equals 0.13 correct to 2 signiﬁcant ﬁgures. 4. Hence a rounding-off error has occurred. 5. Problem 3. Without using a calculator, determine an approximate value of: (a) (a) equal to 11.7 ð 19.1 D 4.22, correct to 9.3 ð 5.7 3 signiﬁcant ﬁgures.) (b) D 2 ð 20 ð 2 after cancelling, [Blunder] For a gas pV D c. When pressure p D 1 03 400 Pa and V D 0.54 m3 then c D 55 836 Pa m3 . Measured values, hence c D 55 800 Pa m3 4.6 ð 0.07 D 0.225 52.3 ð 0.274 Order of magnitude error and rounding off error–should be 0.0225, correct to 3 signiﬁcant ﬁgures or 0.0225, correct to 4 decimal places In Problems 6 to 8, evaluate the expressions approximately, without using a calculator. 6. 7. (By calculator, 2 ð 20 200 ð 20 2 2.19 ð 203.6 ð 17.91 ³ 12.1 ð 8.76 1 10 ð 10 1 24 ð 0.008 D 10.42 12.6 11.7 ð 19.1 2.19 ð 203.6 ð 17.91 (b) 9.3 ð 5.7 12.1 ð 8.76 11.7 ð 19.1 is approximately 9.3 ð 5.7 10 ð 20 , i.e. about 4 10 ð 5 Further problems on errors 2. 29.74 ð 0.0512 30 ð 5 ð 10 2 ³ 11.89 12 150 15 1 D D D or 0.125 2 12 ð 10 120 8 25 8. 4.7 ð 6.3 [³30 (29.61, by calculator)] 2.87 ð 4.07 6.12 ð 0.96 ³2 (1.988, correct to 4 s.f., by calculator) 72.1 ð 1.96 ð 48.6 139.3 ð 5.2 ³10 (9.481, correct to 4 s.f., by calculator)
- 38. 26 ENGINEERING MATHEMATICS 4.2 Use of calculator (a) The most modern aid to calculations is the pocketsized electronic calculator. With one of these, calculations can be quickly and accurately performed, correct to about 9 signiﬁcant ﬁgures. The scientiﬁc type of calculator has made the use of tables and logarithms largely redundant. To help you to become competent at using your calculator check that you agree with the answers to the following problems: (b) (c) 1 D 36.3636363 . . . D 36.364, correct to 0.0275 3 decimal places 1 1 C D 0.71086624 . . . D 0.711, cor4.92 1.97 rect to 3 decimal places Problem 4. Evaluate the following, correct to 4 signiﬁcant ﬁgures: (a) 4.7826 C 0.02713 (b) 17.6941 (c) 21.93 ð 0.012981 1 D 0.01896453 . . . D 0.019, correct to 3 52.73 decimal places Problem 7. Evaluate the following, expressing the answers in standard form, correct to 4 signiﬁcant ﬁgures. 11.8762 (a) 0.00451 (c) 46.272 (a) 4.7826 C 0.02713 D 4.80973 D 4.810, correct to 4 signiﬁcant ﬁgures 2 31.792 (b) 631.7 6.21 C 2.95 2 17.6941 11.8762 D 5.8179 D 5.818, correct to 4 signiﬁcant ﬁgures (a) 0.00451 2 D 2.03401ð10 5 D 2.034 × 10−5 , correct to 4 signiﬁcant ﬁgures (c) 21.93 ð 0.012981 D 0.2846733 . . . D 0.2847, correct to 4 signiﬁcant ﬁgures (b) 631.7 6.21 C 2.95 2 D 547.7944 D 5.477944 ð 102 D 5.478 × 102 , correct to 4 signiﬁcant ﬁgures (b) Problem 5. Evaluate the following, correct to 4 decimal places: (c) 46.272 31.792 D 1130.3088 D 1.130 × 103 , correct to 4 signiﬁcant ﬁgures 4.621 (a) 46.32 ð 97.17 ð 0.01258 (b) 23.76 1 (c) 62.49 ð 0.0172 2 Problem 8. Evaluate the following, correct to 3 decimal places: (a) (a) 46.32 ð 97.17 ð 0.01258 D 56.6215031 . . . D 56.6215, correct to 4 decimal places (b) (c) 4.621 D 0.19448653 . . . D 0.1945, correct to 23.76 4 decimal places (c) (a) 1 62.49 ð 0.0172 D 0.537414 D 0.5374, 2 correct to 4 decimal places (b) Problem 6. Evaluate the following, correct to 3 decimal places: (a) 1 1 1 1 (b) (c) C 52.73 0.0275 4.92 1.97 (c) 2.37 2 (b) 0.0526 3.60 1.92 2 C 5.40 2.45 2 15 7.62 4.82 2.37 2 D 106.785171 . . . D 106.785, correct 0.0526 to 3 decimal places 3.60 2 5.40 2 C D 8.37360084 . . . D 1.92 2.45 8.374, correct to 3 decimal places 15 D 0.43202764 . . . D 0.432, cor4.82 rect to 3 decimal places 7.62
- 39. CALCULATIONS AND EVALUATION OF FORMULAE Problem 9. Evaluate the following, correct to 4 signiﬁcant ﬁgures: p p p (a) 5.462 (b) 54.62 (c) 546.2 (a) p 5.462 D 2.3370922 . . . D 2.337, correct to 4 signiﬁcant ﬁgures p 54.62 D 7.39053448 . . . D 7.391, correct to (b) 4 signiﬁcant ﬁgures p 546.2 D 23.370922 . . . D 23.37, correct to 4 (c) signiﬁcant ﬁgures 6.092 p D 0.74583457 . . . D 0.746, cor25.2 ð 7 rect to 3 signiﬁcant ﬁgures p (b) 3 47.291 D 3.61625876 . . . D 3.62, correct to 3 signiﬁcant ﬁgures p 7.2132 C 6.4183 C 3.2914 D 20.8252991 (c) . . . D 20.8, correct to 3 signiﬁcant ﬁgures (a) Problem 13. Evaluate the following, expressing the answers in standard form, correct to 4 decimal places: (a) 5.176 ð 10 3 2 1.974 ð 101 ð 8.61 ð 10 (b) 3.462 p (c) 1.792 ð 10 4 Problem 10. Evaluate the following, correct to 3 decimal places: p p p (a) p0.007328 (b) 52.91 31.76 4 (c) 1.6291 ð 10 (a) (a) p 0.007328 D 0.08560373 D 0.086, correct to 3 decimal places p p (b) 52.91 31.76 D 1.63832491 . . . D 1.638, correct to 3 decimal places p p (c) 1.6291 ð 104 D 16291 D 127.636201 . . . D 127.636, correct to 3 decimal places Problem 11. Evaluate the following, correct to 4 signiﬁcant ﬁgures: (a) 4.723 p (c) 76.212 (b) 0.8316 (a) 4.723 D 105.15404 . . . D 105.2, correct to 4 signiﬁcant ﬁgures 0.8316 4 D 0.47825324 . . . D 0.4783, correct to 4 signiﬁcant ﬁgures p (c) 76.212 29.102 D 70.4354605 . . . D 70.44, correct to 4 signiﬁcant ﬁgures (b) 5.176 ð 10 3 2 D 2.679097 . . . ð 10 5 D 2.6791 × 10−5 , correct to 4 decimal places 1.974 ð 101 ð 8.61 ð 10 3.462 (b) 2 4 D 0.05808887 . . . D 5.8089 × 10−2 , correct to 4 decimal places (c) p 1.792 ð 10 4 D 0.0133865 . . . D 1.3387 ×10−2 , correct to 4 decimal places Now try the following exercise Exercise 14 Further problems on use of calculator In Problems 1 to 9, use a calculator to evaluate the quantities shown correct to 4 signiﬁcant ﬁgures: 1. (a) 3.2492 (b) 73.782 (d) 0.06392 (a) 10.56 Problem 12. Evaluate the following, correct to 3 signiﬁcant ﬁgures: p 6.092 p (b) 3 47.291 (a) p 25.2 ð 7 (c) 7.2132 C 6.4183 C 3.2914 4 2 4 29.102 27 2. (c) 311.42 (b) 5443 (d) 0.004083 p p (a) p4.735 (b) 35.46 (d) 0.0256 (a) 2.176 (b) 5.955 (d) 0.1600 (c) 96 970 (c) p 73 280 (c) 270.7
- 40. 28 ENGINEERING MATHEMATICS 1 7.768 1 (d) 1.118 3. (a) (b) 1 48.46 (c) 1 0.0816 Problem 14. Currency exchange rates for ﬁve countries are shown in Table 4.1 Table 4.1 (a) 0.1287 (b) 0.02064 (c) 12.25 (d) 0.8945 France Japan Norway Switzerland U.S.A. 4. (a) 127.8 ð 0.0431 ð 19.8 (b) 15.76 ł 4.329 £1 D 1.46 euros £1 D 190 yen £1 D 10.90 kronor £1 D 2.15 francs £1 D 1.52 dollars ($) [(a) 109.1 (b) 3.641] 137.6 5. (a) 552.9 [(a) 0.2489 6. (a) 13.63 [(a) 2515 7. (a) (b) (b) 3.4764 (a) how many French euros £27.80 will buy (b) the number of Japanese yen which can be bought for £23 (c) the pounds sterling which can be exchanged for 6409.20 Norwegian kronor (d) the number of American dollars which can be purchased for £90, and (e) the pounds sterling which can be exchanged for 2795 Swiss francs (b) 500.5] (c) 0.1245 (b) 146.0 (c) 0.00002932] 24.68 ð 0.0532 7.412 0.2681 ð 41.22 32.6 ð 11.89 3 4 [(a) 0.005559 (b) 1.900] (a) £1 D 1.46 euros, hence £27.80 D 27.80 ð 1.46 euros D 40.59 euros (b) 4.8213 17.332 15.86 ð 11.6 (b) [(a) 6.248 (b) 0.9630] 8. (a) 14.323 21.682 Calculate: 11.82 ð 1.736 (b) 0.041 (c) £1 D 10.90 kronor, hence 6409.20 D £588 6409.20 kronor D £ 10.90 15.62 2 p 29.21 ð 10.52 p (b) 6.9212 C 4.8163 2.1614 9. (a) [(a) 1.605 (b) 11.74] 10. Evaluate the following, expressing the answers in standard form, correct to 3 decimal places: (a) 8.291 ð 10 2 2 p (b) 7.623 ð 10 3 [(a) 6.874 ð 10 3 (b) 8.731 ð 10 2 ] (d) £1 D 190 yen, hence £23 D 23 ð 190 yen D 4370 yen £1 D 1.52 dollars, hence £90 D 90 ð 1.52 dollars D $136.80 (e) £1 D 2.15 Swiss francs, hence 2795 2795 franc D £ D £1300 2.15 Problem 15. Some approximate imperial to metric conversions are shown in Table 4.2 Table 4.2 4.3 Conversion tables and charts length 1 inch D 2.54 cm 1 mile D 1.61 km It is often necessary to make calculations from various conversion tables and charts. Examples include currency exchange rates, imperial to metric unit conversions, train or bus timetables, production schedules and so on. weight 2.2 lb D 1 kg 1 lb D 16 oz capacity 1.76 pints D 1 litre 8 pints D 1 gallon
- 41. CALCULATIONS AND EVALUATION OF FORMULAE Use the table to determine: (a) the number of millimetres in 9.5 inches, (b) a speed of 50 miles per hour in kilometres per hour, (c) the number of miles in 300 km, (d) the number of kilograms in 30 pounds weight, (e) the number of pounds and ounces in 42 kilograms (correct to the nearest ounce), (f) the number of litres in 15 gallons, and (g) the number of gallons in 40 litres. Calculate (a) how many Italian euros £32.50 will buy, (b) the number of Canadian dollars that can be purchased for £74.80, (c) the pounds sterling which can be exchanged for 14 040 yen, (d) the pounds sterling which can be exchanged for 1754.30 Swedish kronor, and (e) the Australian dollars which can be bought for £55 [(a) 48.10 euros (b) $179.52 (c) £75.89 (d) £132.40 (e) 148.50 dollars] 2. Below is a list of some metric to imperial conversions. Length 2.54 cm D 1 inch 1.61 km D 1 mile Weight (a) 9.5 inches D 9.5 ð 2.54 cm D 24.13 cm 1 kg D 2.2 lb 1 lb D 16 ounces 24.13 cm D 24.13 ð 10 mm D 241.3 mm (b) (c) (d) (e) (f) (g) 50 m.p.h. D 50 ð 1.61 km/h D 80.5 km=h 300 300 km D miles D 186.3 miles 1.61 30 kg D 13.64 kg 30 lb D 2.2 42 kg D 42 ð 2.2 lb D 92.4 lb 0.4 lb D 0.4 ð 16 oz D 6.4 oz D 6 oz, correct to the nearest ounce Thus 42 kg D 92 lb 6 oz, correct to the nearest ounce. 15 gallons D 15 ð 8 pints D 120 pints 120 120 pints D litres D 68.18 litres 1.76 40 litres D 40 ð 1.76 pints D 70.4 pints 70.4 gallons D 8.8 gallons 70.4 pints D 8 Now try the following exercise Exercise 15 Further problems conversion tables and charts 1. Currency exchange rates listed in a newspaper included the following: Italy Japan Australia Canada Sweden £1 D 1.48 euro £1 D 185 yen £1 D 2.70 dollars £1 D $2.40 £1 D 13.25 kronor Capacity 1 litre D 1.76 pints 8 pints D 1 gallon Use the list to determine (a) the number of millimetres in 15 inches, (b) a speed of 35 mph in km/h, (c) the number of kilometres in 235 miles, (d) the number of pounds and ounces in 24 kg (correct to the nearest ounce), (e) the number of kilograms in 15 lb, (f) the number of litres in 12 gallons and (g) the number of gallons in 25 litres. (a) 381 mm (b) 56.35 km/h (c) 378.35 km (d) 52 lb 13 oz (e) 6.82 kg (f) 54.55 l (g) 5.5 gallons 3. Deduce the following information from the BR train timetable shown in Table 4.3: (a) At what time should a man catch a train at Mossley Hill to enable him to be in Manchester Piccadilly by 8.15 a.m.? (b) A girl leaves Hunts Cross at 8.17 a.m. and travels to Manchester Oxford Road. How long does the journey take. What is the average speed of the journey? (c) A man living at Edge Hill has to be at work at Trafford Park by 8.45 a.m. It takes him 10 minutes to walk to 29
- 42. 30 ENGINEERING MATHEMATICS Table 4.3 Liverpool, Hunt’s Cross and Warrington ! Manchester Reproduced with permission of British Rail his work from Trafford Park station. What time train should he catch from Edge Hill? (a) 7.09 a.m. (b) 51 minutes, 32 m.p.h. (c) 7.04 a.m.] 4.4 Evaluation of formulae The statement v D u C at is said to be a formula for v in terms of u, a and t. v, u, a and t are called symbols. The single term on the left-hand side of the equation, v, is called the subject of the formulae. Provided values are given for all the symbols in a formula except one, the remaining symbol can be made the subject of the formula and may be evaluated by using a calculator. Problem 16. In an electrical circuit the voltage V is given by Ohm’s law, i.e. V D IR. Find, correct to 4 signiﬁcant ﬁgures, the voltage when I D 5.36 A and R D 14.76 . V D IR D 5.36 14.76
- 43. CALCULATIONS AND EVALUATION OF FORMULAE Hence, voltage V = 79.11 V, correct to 4 signiﬁcant ﬁgures. Problem 17. The surface area A of a hollow cone is given by A D rl. Determine, correct to 1 decimal place, the surface area when r D 3.0 cm and l D 8.5 cm. A D rl D 3.0 8.5 cm2 Hence, surface area A = 80.1 cm2 , correct to 1 decimal place. Hence volume, V = 358.8 cm3 , correct to 4 signiﬁcant ﬁgures. Problem 21. Force F newtons is given by Gm1 m2 the formula F D , where m1 and m2 d2 are masses, d their distance apart and G is a constant. Find the value of the force given that G D 6.67 ð 10 11 , m1 D 7.36, m2 D 15.5 and d D 22.6. Express the answer in standard form, correct to 3 signiﬁcant ﬁgures. FD Problem 18. Velocity v is given by v D u C at. If u D 9.86 m/s, a D 4.25 m/s2 and t D 6.84 s, ﬁnd v, correct to 3 signiﬁcant ﬁgures. 31 Gm1 m2 6.67 ð 10 11 7.36 15.5 D d2 22.6 2 6.67 7.36 15.5 1.490 D 1011 510.76 1011 D Hence force F = 1.49 × 10−11 newtons, correct to 3 signiﬁcant ﬁgures. v D u C at D 9.86 C 4.25 6.84 D 9.86 C 29.07 D 38.93 Hence, velocity v = 38.9 m=s, correct to 3 signiﬁcant ﬁgures. Problem 19. The power, P watts, dissipated in an electrical circuit may be expressed by V2 . Evaluate the power, the formula P D R correct to 3 signiﬁcant ﬁgures, given that V D 17.48 V and R D 36.12 . PD 17.48 2 305.5504 V2 D D R 36.12 36.12 Hence power, P = 8.46 W, correct to 3 signiﬁcant ﬁgures. The volume V cm3 of a right 1 2 circular cone is given by V D r h. Given 3 that r D 4.321 cm and h D 18.35 cm, ﬁnd the volume, correct to 4 signiﬁcant ﬁgures. Problem 20. Problem 22. The time of swing t seconds, of a simple pendulum is given by l tD2 . Determine the time, correct to 3 g decimal places, given that l D 12.0 and g D 9.81 tD2 12.0 9.81 l D 2 g p D 2 1.22324159 D 2 1.106002527 Hence time t = 6.950 seconds, correct to 3 decimal places. Problem 23. Resistance, R , varies with temperature according to the formula R D R0 1 C ˛t . Evaluate R, correct to 3 signiﬁcant ﬁgures, given R0 D 14.59, ˛ D 0.0043 and t D 80. R D R0 1 C ˛t D 14.59[1 C 0.0043 80 ] D 14.59 1 C 0.344 1 2 1 VD r hD 3 3 1 D 3 4.321 2 18.35 18.671041 18.35 D 14.59 1.344 Hence, resistance, R = 19.6 Z, correct to 3 signiﬁcant ﬁgures.
- 44. 32 ENGINEERING MATHEMATICS 10. The potential difference, V volts, available at battery terminals is given by V D E Ir. Evaluate V when E D 5.62, I D 0.70 and R D 4.30 [V D 2.61 V] 11. Given force F D 1 m v2 u2 , ﬁnd F 2 when m D 18.3, v D 12.7 and u D 8.24 [F D 854.5] 2. The velocity of a body is given by v D u C at. The initial velocity u is measured when time t is 15 seconds and found to be 12 m/s. If the acceleration a is 9.81 m/s2 calculate the ﬁnal velocity v. [159 m/s] 12. The current I amperes ﬂowing in a numnE . ber of cells is given by I D R C nr Evaluate the current when n D 36. E D 2.20, R D 2.80 and r D 0.50 [I D 3.81 A] 3. Find the distance s, given that s D 1 gt2 , 2 time t D 0.032 seconds and acceleration due to gravity g D 9.81 m/s2 . [0.00502 m or 5.02 mm] 13. The time, t seconds, of oscillation for a simple pendulum is given by l t D 2 . Determine the time when g D 3.142, l D 54.32 and g D 9.81 [t D 14.79 s] 14. Energy, E joules, is given by the formula E D 1 LI2 . Evaluate the energy when 2 L D 5.5 and I D 1.2 [E D 3.96 J] 15. The current I amperes in an a.c. circuit V . Evaluate the is given by I D p 2 C X2 R current when V D 250, R D 11.0 and X D 16.2 [I D 12.77 A] 16. Distance s metres is given by the formula s D ut C 1 at2 . If u D 9.50, 2 t D 4.60 and a D 2.50, evaluate the distance. Now try the following exercise Exercise 16 Further problems on evaluation of formulae 1. A formula used in connection with gases is R D PV /T. Evaluate R when P D 1500, V D 5 and T D 200. [R D 37.5] 4. The energy stored in a capacitor is given by E D 1 CV2 joules. Determine the 2 energy when capacitance C D 5 ð 10 6 farads and voltage V D 240V. [0.144 J] 5. Resistance R2 is given by R2 D R1 1 C ˛t . Find R2 , correct to 4 signiﬁcant ﬁgures, when R1 D 220, ˛ D 0.00027 and t D 75.6 [224.5] mass 6. Density D . Find the density volume when the mass is 2.462 kg and the volume is 173 cm3 . Give the answer in units [14 230 kg/m3 ] of kg/m3 . 7. Velocity D frequency ð wavelength. Find the velocity when the frequency is 1825 Hz and the wavelength is 0.154 m. [281.1 m/s] 17. 8. Evaluate resistance RT , given 1 1 1 1 D C C when R1 D 5.5 RT R1 R2 R3 R2 D 7.42 and R3 D 12.6 . [2.526 The area, A, of any triangle is given p by A D s s a s b s c where aCbCc sD . Evaluate the area given 2 a D 3.60 cm, b D 4.00 cm and c D 5.20 cm. [A D 7.184 cm2 ] 18. Given that a D 0.290, b D 14.86, c D 0.042, d D 31.8 and e D 0.650, ab d evaluate v, given that v D c e [v D 7.327] [s D 17.25 m] , ] force ð distance 9. Power D . Find the time power when a force of 3760 N raises an object a distance of 4.73 m in 35 s. [508.1 W]
- 45. CALCULATIONS AND EVALUATION OF FORMULAE 7. Assignment 1 This assignment covers the material contained in Chapters 1 to 4. The marks for each question are shown in brackets at the end of each question. 8. Express the following in standard form: 2 (a) 1623 (b) 0.076 (c) 145 (3) 5 Determine the value of the following, giving the answer in standard form: (a) 5.9 ð 102 C 7.31 ð 102 (b) 2.75 ð 10 1. 2. 3. 1 ł 1 4 ð2 7 4 1 1 C 3 5 C2 Convert the following decimal number to binary form: (9) (a) 27 (b) correct to 1 decimal place (2) 4. 5. Express 54.7 mm as a percentage of 1.15 m, correct to 3 signiﬁcant ﬁgures. (3) Convert the following decimal numbers to binary, via octal: (a) 479 (b) 185.2890625 (6) Convert (a) 5F16 into its decimal equivalent (b) 13210 into its hexadecimal equivalent (c) 1101010112 into its hexadecimal equivalent (6) 13. Evaluate the following, each correct to 4 signiﬁcant ﬁgures: p 1 (a) 61.222 (b) (c) 0.0527 0.0419 (3) 14. Evaluate the following, each correct to 2 decimal places: (a) Evaluate the following: (e) 2 3 2 2 3 2 9 2 (6) 11. Determine, correct to 1 decimal places, 57% of 17.64 g (2) (c) (b) 44.1875 12. 576.29 Evaluate 19.3 23 ð 2 ð 22 24 1 1 42 (4) 10. 7 24 A piece of steel, 1.69 m long, is cut into three pieces in the ratio 2 to 5 to 6. Determine, in centimetres, the lengths of the three pieces. (4) (a) 3 Convert the following binary numbers to decimal form: (a) 1101 (b) 101101.0101 (5) (a) correct to 4 signiﬁcant ﬁgures 6. 2.65 ð 10 9. 2 1 Simplify (a) 2 ł 3 3 3 (b) 2 (b) (d) 23 ð 16 2 8ð2 3 (27) (b) 1 3 15. (14) 36.22 ð 0.561 27.8 ð 12.83 p 3 14.692 17.42 ð 37.98 (7) If 1.6 km D 1 mile, determine the speed of 45 miles/hour in kilometres per hour. (3) 16. Evaluate B, correct to 3 signiﬁcant ﬁgures, when W D 7.20, v D 10.0 and Wv2 . (3) g D 9.81, given that B D 2g 33
- 46. 5 Algebra Replacing p, q and r with their numerical values gives: 5.1 Basic operations Algebra is that part of mathematics in which the relations and properties of numbers are investigated by means of general symbols. For example, the area of a rectangle is found by multiplying the length by the breadth; this is expressed algebraically as A D l ð b, where A represents the area, l the length and b the breadth. The basic laws introduced in arithmetic are generalized in algebra. Let a, b, c and d represent any four numbers. Then: 4p2 qr 3 D 4 2 1 2 2 D4ð2ð2ð Problem 3. and 7x 3 3 2 1 3 3 3 ð ð ð D 27 2 2 2 2 Find the sum of: 3x, 2x, x The sum of the positive terms is: 3x C 2x D 5x x C 7x D 8x (i) aC bCc D aCb Cc The sum of the negative terms is: (ii) a bc D ab c (iii) aCbDbCa Taking the sum of the negative terms from the sum of the positive terms gives: (iv) ab D ba 5x 8x D −3x Alternatively (v) a b C c D ab C ac 3x C 2x C (vi) a b aCb D C c c c (vii) a C b c C d D ac C ad C bc C bd x C 7x D 3x C 2x D −3x Problem 1. Evaluate: 3ab 2bc C abc when a D 1, b D 3 and c D 5 Problem 4. Find the sum of 4a, 3b, c, 5b and 6c Replacing a, b and c with their numerical values gives: 2bc C abc D 3 ð 1 ð 3 2ð3ð5 D9 C4a 2a D 2a For the ‘b’ terms: C3b 5b D For the ‘c’ terms: 30 C 15 D −6 2b Cc C 6c D 7c Thus 4a C 3b C c C C1ð3ð5 2a C D 4a C 3b C c 2a 5b C 6c 5b C 6c D 2a − 2b Y 7c Problem 2. 2 3 Find the value of 4p qr , given 1 1 that p D 2, q D and r D 1 2 2 2a, Each symbol must be dealt with individually. For the ‘a’ terms: 3ab x Problem 5. 2a C c, 4b Find the sum of: 5a 2b, 5d and b a C 3d 4c 7x
- 47. ALGEBRA The algebraic expressions may be tabulated as shown below, forming columns for the a’s, b’s, c’s and d’s. Thus: C5a C2a 2b C c C 4b aC b 5d 4c C 3d 6a Y 3b − 3c − 2d Adding gives: Problem 6. Subtract 2x C 3y x 2y C 5z 4z from x 2y C 5z 2x C 3y 4z −x − 5y Y 9z Subtracting gives: Multiplying by 2x ! Multiplying by 5y ! Adding gives: Problem 9. 3x 2x 2y 2 C 4xy 5y 6x 2 4xy 2 C 8x 2 y 20xy 2 15xy C 10y 3 6x 2 − 24xy 2 Y 8x 2 y − 15xy Y 10y 3 Simplify: 2p ł 8pq 2p . This can be reduced by 8pq cancelling as in arithmetic. 2p 2ðp 1 Thus: D D 8pq 8ðpðq 4q 2p ł 8pq means Now try the following exercise Exercise 17 (Note that C5z 4z D C5z C 4z D 9z) An alternative method of subtracting algebraic expressions is to ‘change the signs of the bottom line and add’. Hence: 1. Further problems on basic operations Find the value of 2xy C 3yz x D 2, y D 2 and z D 4 xyz, when [ 16] 2 Evaluate 3pq2 r 3 when p D , q D 2 3 and r D 1 [ 8] 2y C 5z 3y C 4z 2. −x − 5y Y 9z 3. Find the sum of 3a, 4a 4. Add together 2aC3bC4c, 5a 2bCc, 4a 5b 6c [a 4b c] 5. Add together 3dC4e, 4d e C 2f 3e 6. From 4x x 2x Adding gives: Problem 7. Multiply 2a C 3b by a C b Each term in the ﬁrst expression is multiplied by a, then each term in the ﬁrst expression is multiplied by b, and the two results are added. The usual layout is shown below. 8. Adding gives: 2a 2 Y 5ab Y 3b 2 9. 10. 2y C 4xy by 2eCf, 2d 3f, [9d 2e] Multiply 3x C 2y by x 2a2 C 3ab C 2ab C 3b2 Multiply 3x 6a, 5a and [4a] 7. Multiplying by a ! Multiplying by b ! Problem 8. 2x 5y 2a, 3y C 2z subtract x C 2y 3z [3x 5y C 5z] 3 b b Subtract a C c from 4a 3c 2 3 2 5 1 5 a C b 4c 2 6 2a C 3b a C b 2 35 y [3x 2 xy 2y 2 ] Multiply 2a 5b C c by 3a C b [6a2 13ab C 3ac 5b2 C bc] Simplify (i) 3a ł 9ab (ii) 4a2 b ł 2a 1 i ii 2ab 3b
- 48. 36 ENGINEERING MATHEMATICS 5.2 Laws of Indices The laws of indices are: (i) am ð an D amCn (iii) (am (v) a n n D amn 1 D n a p1/2 q2 r 2/3 and p1/4 q1/2 r 1/6 evaluate when p D 16, q D 9 and r D 4, taking positive roots only Problem 13. (ii) (iv) (vi) am D am n an p am/n D n am Simplify: Using the second law of indices gives: 0 a D1 p 1/2 1/4 q2 1/2 r 2/3 1/6 D p 1=4 q 3=2 r 1=2 When p D 16, q D 9 and r D 4, Simplify: a3 b2 c ð ab3 c5 Problem 10. p1/4 q3/2 r 1/2 D 16 1/4 9 3/2 4 1/2 p p p 4 D 16 93 4 Grouping like terms gives: D 2 33 2 D 108 a3 ð a ð b2 ð b3 ð c ð c 5 Using the ﬁrst law of indices gives: Problem 14. Simplify: a3C1 ð b2C3 ð c1C5 i.e. x 2 y 3 C xy 2 xy a4 ð b5 ð c 6 D a 4 b 5 c 6 Algebraic expressions of the form Problem 11. Simplify: a1/2 b2 c 2 ð a1/6 b1/2 c into x2y 3 xy 2 x 2 y 3 C xy 2 D C xy xy xy Using the ﬁrst law of indices, a1/2 b2 c 2 ð a 1/6 b 1/2 c D x2 1y 3 D a 1/2 C 1/6 ð b2C 1/2 ð c 2C1 D a 2=3 b 5=2 c −1 a3 b2 c 4 Simplify: and evaluate abc 2 1 when a D 3, b D and c D 2 8 Using the second law of indices, a3 D a3 a 1 D a2 , b2 D b2 b 1 2 C x1 1y 2 1 D xy 2 Y y x2y xy 2 x2y xy 2 x y Thus Simplify: xy The highest common factor (HCF) of each of the three terms comprising the numerator and denominator is xy. Dividing each term by xy gives: xy 2 D c6 a3 b2 c 4 D a 2 bc 6 abc 2 1 When a D 3, b D and c D 2, 8 1 2 6 2 1 26D 9 a bc D 3 8 8 Problem 15. Db 4 c D c4 c 2 1 (since x 0 D 1, from the sixth law of indices) Problem 12. and a b C . Thus c c aCb can be split c xy Problem 16. D 2 xy xy xy xy D Simplify: p3 x y −1 1/2 q2 Using the third law of indices gives: 64 D 72 p3ð 1/2 q2ð4 D p .3=2/ q 8 4
- 49. ALGEBRA mn2 3 Simplify: m1/2 n1/4 Problem 17. Using the third law of indices gives: d2 e2 f1/2 d3/2 ef5/2 4 The brackets indicate that each letter in the bracket must be raised to the power outside. Using the third law of indices gives: mn2 3 1/2 n1/4 m 4 D D mn 5 d2 e2 f1/2 d3 e2 f5 Using the second law of indices gives: d2 3 e2 2 f 1/2 5 D d 1 e0 f D d 1f 9/2 9/2 since e0 D 1 from the sixth law of indices 1 = 9=2 df Using the second law of indices gives: 1 D 2 m1ð3 n2ð3 m 3 n6 D 2 1 1/2 ð4 n 1/4 ð4 m m n m 3 n6 D m 3 2 n6 m 2 n1 from the ﬁfth law of indices p 3 2 x 2 y 1/2 x y Simplify: 5 y 3 1/2 x Problem 20. Problemp Simplify: p 18. p p 3 a3 b c 5 a b2 c3 and evaluate 1 when a D , b D 6 and c D 1 4 Using the third and fourth laws of indices gives: p 3 2 x 2 y 1/2 x y x 2 y 1/2 x 1/2 y 2/3 D x 5 y 3 1/2 x 5/2 y 3/2 Using the fourth law of indices, the expression can be written as: Using the ﬁrst and second laws of indices gives: a3 b1/2 c5/2 a1/2 b2/3 c3 x 2C 1/2 Using the ﬁrst law of indices gives: 5/2 y 1/2 C 2/3 3/2 D x0y or It is usual to express the answer in the same form as the question. Hence p p p 6 a7/2 b7/6 c11/2 D a 7 b 7 c 11 1 When a D , b D 64 and c D 1, 4 D 1 4 1 2 7 647 p 111 7 Simplify: Further problems on laws of indices x 2 y 3 z x 3 yz2 and evaluate 1 when x D , y D 2 and z D 3 2 1 x 5 y 4 z3 , 13 2 1. Simplify 2. Simplify (a3/2 bc 3 a1/2 b 1/2 c and evaluate when a D 3, b D 4 and c D 2 1 D1 d2 e2 f1/2 d3/2 ef5/2 2 expressing the answer with positive indices only Problem 19. Now try the following exercise 7 2 1 1 or p 3 y y 1=3 from the ﬁfth and sixth laws of indices. Exercise 18 p 6 1/3 D y −1=3 a3C 1/2 b 1/2 C 2/3 c 5/2 C3 D a7/2 b7/6 c11/2 p p p 6 a7 b7 c11 D 37 a2 b1/2 c 2 , 4 1 2
- 50. 38 ENGINEERING MATHEMATICS a5 bc3 and evaluate when a D a2 b3 c 2 1 2 3 9 , b D and c D a3 b 2 c, 2 2 3 16 3. Simplify: Both b and c in the second bracket have to be multiplied by 2, and c and d in the third bracket by 4 when the brackets are removed. Thus: 3a C b C 2 b C c 4 cCd D 3a C b C 2b C 2c In Problems 4 to 10, simplify the given expressions: 4c 4d Collecting similar terms together gives: 4. x 1/5 y 1/2 z1/3 x 1/2 y 1/3 z 1/6 5. a2 b C a3 b a2 b2 1Ca b 6. p3 q 2 pq2 p2 q p2 q q p 7. a2 1/2 b2 3 c1/2 [x 7/10 y 1/6 z1/2 ] 3 abc 2 2b 1c 3 3 a p p p 9. ( x y 3 3 z2 x [ab6 c3/2 ] [a 4 b5 c11 ] 8. y3 p z3 ) [xy 10. a3 b1/2 c 1/2 ab p p a3 b c p 3 6 z13 ] 1/3 p 6 a11/6 b1/3 c 3/2 or a11 p p 3 3a Y 3b − 2c − 4d Problem 22. Simplify: 2a ab a 3b C a a2 When the brackets are removed, both 2a and ab in the ﬁrst bracket must be multiplied by 1 and both 3b and a in the second bracket by a. Thus a2 2a D a2 Problem 23. a2 3ab Simplify: a C b a b b c3 Each term in the second bracket has to be multiplied by each term in the ﬁrst bracket. Thus: b Da a D a2 5.3 Brackets and factorisation When two or more terms in an algebraic expression contain a common factor, then this factor can be shown outside of a bracket. For example b Cb a b b2 ab C ab D a2 − b2 a C b a b Alternatively Multiplying by a ! Multiplying by b! ab C ac D a b C c which is simply the reverse of law (v) of algebra on page 34, and 4pz D 2p 3x C y 2a C ab Collecting similar terms together gives: 2a 2ab Since 2a is a common factor, the answer can be expressed as: −2a .1 Y b / aCb a 6px C 2py a 3b C a ab 2z a2 C ab ab b2 Adding gives: a2 b2 Problem 24. Simplify: 3x 3y 2 This process is called factorisation. 2x Problem 21. Remove the brackets and simplify the expression: 3a C b C 2 b C c 4 cCd 3y 2 D 2x D 2x 2x D 4x 2 3y 2x 3y 3y 3y 2x 6xy 6xy C 9y 2 D 4x 2 − 12xy Y 9y 2 3y

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