1. Engineering Mathematics

2. In memory of Elizabeth 3. Engineering MathematicsFourth EditionJOHN BIRD, BSc(Hons) CMath, FIMA, CEng, MIEE, FCollP, FIIENewnesOXFORD AMSTERDAM BOSTON LONDON NEW YORK PARISSAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO 4. NewnesAn imprint of Elsevier ScienceLinacre House, Jordan Hill, Oxford OX2 8DP200 Wheeler Road, Burlington MA 01803First published 1989Second edition 1996Reprinted 1998 (twice), 1999Third edition 2001Fourth edition 2003Copyright 2001, 2003, John Bird. All rights reservedThe right of John Bird to be identified as the author of this workhas been asserted in accordance with the Copyright, Designs andPatents Act 1988No part of this publication may be reproduced in any materialform (including photocopying or storing in any medium byelectronic means and whether or not transiently or incidentally to someother use of this publication) without the written permission of thecopyright holder except in accordance with the provisions of the Copyright,Designs and Patents Act 1988 or under the terms of a licence issued by theCopyright Licensing Agency Ltd, 90 Tottenham Court Road, London,England W1T 4LP. Applications for the copyright holders writtenpermission to reproduce any part of this publication should beaddressed to the publisherPermissions may be sought directly from Elseviers Science and Technology RightsDepartment in Oxford, UK: phone: (+44) (0) 1865 843830; fax: (+44) (0) 1865853333; e-mail: permissions@elsevier.co.uk. You may also complete your requeston-line via the Elsevier Science homepage (http://www.elsevier.com), by selectingCustomer Support and then Obtaining PermissionsBritish Library Cataloguing in Publication DataA catalogue record for this book is available from the British LibraryISBN 0 7506 5776 6For information on all Newnes publications visit our website at www.Newnespress.comTypeset by Laserwords Private Limited, Chennai, IndiaPrinted and bound in Great Britain 5. ContentsPreface xiPart 1 Number and Algebra 11 Revision of fractions, decimals andpercentages 11.1 Fractions 11.2 Ratio and proportion 31.3 Decimals 41.4 Percentages 72 Indices and standard form 92.1 Indices 92.2 Worked problems on indices 92.3 Further worked problems onindices 112.4 Standard form 132.5 Worked problems on standardform 132.6 Further worked problems on standardform 143 Computer numbering systems 163.1 Binary numbers 163.2 Conversion of binary to decimal 163.3 Conversion of decimal to binary 173.4 Conversion of decimal to binary viaoctal 183.5 Hexadecimal numbers 204 Calculations and evaluation offormulae 244.1 Errors and approximations 244.2 Use of calculator 264.3 Conversion tables and charts 284.4 Evaluation of formulae 30Assignment 1 335 Algebra 345.1 Basic operations 345.2 Laws of Indices 365.3 Brackets and factorisation 385.4 Fundamental laws and precedence 405.5 Direct and inverse proportionality 426 Further algebra 446.1 Polynomial division 446.2 The factor theorem 466.3 The remainder theorem 487 Partial fractions 517.1 Introduction to partial fractions 517.2 Worked problems on partial fractionswith linear factors 517.3 Worked problems on partial fractionswith repeated linear factors 547.4 Worked problems on partial fractionswith quadratic factors 558 Simple equations 578.1 Expressions, equations andidentities 578.2 Worked problems on simpleequations 578.3 Further worked problems on simpleequations 598.4 Practical problems involving simpleequations 618.5 Further practical problems involvingsimple equations 62Assignment 2 649 Simultaneous equations 659.1 Introduction to simultaneousequations 659.2 Worked problems on simultaneousequations in two unknowns 659.3 Further worked problems onsimultaneous equations 679.4 More difficult worked problems onsimultaneous equations 699.5 Practical problems involvingsimultaneous equations 7010 Transposition of formulae 7410.1 Introduction to transposition offormulae 7410.2 Worked problems on transposition offormulae 7410.3 Further worked problems ontransposition of formulae 7510.4 Harder worked problems ontransposition of formulae 7711 Quadratic equations 8011.1 Introduction to quadratic equations 8011.2 Solution of quadratic equations byfactorisation 80 6. vi CONTENTS11.3 Solution of quadratic equations bycompleting the square 8211.4 Solution of quadratic equations byformula 8411.5 Practical problems involving quadraticequations 8511.6 The solution of linear and quadraticequations simultaneously 8712 Logarithms 8912.1 Introduction to logarithms 8912.2 Laws of logarithms 8912.3 Indicial equations 9212.4 Graphs of logarithmic functions 93Assignment 3 9413 Exponential functions 9513.1 The exponential function 9513.2 Evaluating exponential functions 9513.3 The power series for ex 9613.4 Graphs of exponential functions 9813.5 Napierian logarithms 10013.6 Evaluating Napierian logarithms 10013.7 Laws of growth and decay 10214 Number sequences 10614.1 Arithmetic progressions 10614.2 Worked problems on arithmeticprogression 10614.3 Further worked problems on arithmeticprogressions 10714.4 Geometric progressions 10914.5 Worked problems on geometricprogressions 11014.6 Further worked problems on geometricprogressions 11114.7 Combinations and permutations 11215 The binomial series 11415.1 Pascals triangle 11415.2 The binomial series 11515.3 Worked problems on the binomialseries 11515.4 Further worked problems on thebinomial series 11715.5 Practical problems involving thebinomial theorem 12016 Solving equations by iterativemethods 12316.1 Introduction to iterative methods 12316.2 The NewtonRaphson method 12316.3 Worked problems on theNewtonRaphson method 123Assignment 4 126Multiple choice questions on chapters 1 to16 127Part 2 Mensuration 13117 Areas of plane figures 13117.1 Mensuration 13117.2 Properties of quadrilaterals 13117.3 Worked problems on areas of planefigures 13217.4 Further worked problems on areas ofplane figures 13517.5 Worked problems on areas ofcomposite figures 13717.6 Areas of similar shapes 13818 The circle and its properties 13918.1 Introduction 13918.2 Properties of circles 13918.3 Arc length and area of a sector 14018.4 Worked problems on arc length andsector of a circle 14118.5 The equation of a circle 14319 Volumes and surface areas ofcommon solids 14519.1 Volumes and surface areas ofregular solids 14519.2 Worked problems on volumes andsurface areas of regular solids 14519.3 Further worked problems on volumesand surface areas of regularsolids 14719.4 Volumes and surface areas of frusta ofpyramids and cones 15119.5 The frustum and zone of a sphere 15519.6 Prismoidal rule 15719.7 Volumes of similar shapes 15920 Irregular areas and volumes and meanvalues of waveforms 16120.1 Areas of irregular figures 16120.2 Volumes of irregular solids 16320.3 The mean or average value of awaveform 164Assignment 5 168Part 3 Trigonometry 17121 Introduction to trigonometry 17121.1 Trigonometry 17121.2 The theorem of Pythagoras 17121.3 Trigonometric ratios of acuteangles 172 7. CONTENTS vii21.4 Fractional and surd forms oftrigonometric ratios 17421.5 Solution of right-angled triangles 17521.6 Angles of elevation anddepression 17621.7 Evaluating trigonometric ratios of anyangles 17821.8 Trigonometric approximations for smallangles 18122 Trigonometric waveforms 18222.1 Graphs of trigonometric functions 18222.2 Angles of any magnitude 18222.3 The production of a sine and cosinewave 18522.4 Sine and cosine curves 18522.5 Sinusoidal form A sint 18922.6 Waveform harmonics 19223 Cartesian and polar co-ordinates 19423.1 Introduction 19423.2 Changing from Cartesian into polarco-ordinates 19423.3 Changing from polar into Cartesianco-ordinates 19623.4 Use of R ! P and P ! R functions oncalculators 197Assignment 6 19824 Triangles and some practicalapplications 19924.1 Sine and cosine rules 19924.2 Area of any triangle 19924.3 Worked problems on the solution oftriangles and their areas 19924.4 Further worked problems on thesolution of triangles and theirareas 20124.5 Practical situations involvingtrigonometry 20324.6 Further practical situations involvingtrigonometry 20525 Trigonometric identities andequations 20825.1 Trigonometric identities 20825.2 Worked problems on trigonometricidentities 20825.3 Trigonometric equations 20925.4 Worked problems (i) on trigonometricequations 21025.5 Worked problems (ii) on trigonometricequations 21125.6 Worked problems (iii) on trigonometricequations 21225.7 Worked problems (iv) on trigonometricequations 21226 Compound angles 21426.1 Compound angle formulae 21426.2 Conversion of a sin t C b cos t intoR sint C ) 21626.3 Double angles 22026.4 Changing products of sines and cosinesinto sums or differences 22126.5 Changing sums or differences of sinesand cosines into products 222Assignment 7 224Multiple choice questions on chapters 17to 26 225Part 4 Graphs 23127 Straight line graphs 23127.1 Introduction to graphs 23127.2 The straight line graph 23127.3 Practical problems involving straightline graphs 23728 Reduction of non-linear laws to linearform 24328.1 Determination of law 24328.2 Determination of law involvinglogarithms 24629 Graphs with logarithmic scales 25129.1 Logarithmic scales 25129.2 Graphs of the form y D axn 25129.3 Graphs of the form y D abx 25429.4 Graphs of the form y D aekx 25530 Graphical solution of equations 25830.1 Graphical solution of simultaneousequations 25830.2 Graphical solution of quadraticequations 25930.3 Graphical solution of linear andquadratic equations simultaneously26330.4 Graphical solution of cubic equations26431 Functions and their curves 26631.1 Standard curves 26631.2 Simple transformations 26831.3 Periodic functions 27331.4 Continuous and discontinuousfunctions 27331.5 Even and odd functions 27331.6 Inverse functions 275Assignment 8 279 8. viii CONTENTSPart 5 Vectors 28132 Vectors 28132.1 Introduction 28132.2 Vector addition 28132.3 Resolution of vectors 28332.4 Vector subtraction 28433 Combination of waveforms 28733.1 Combination of two periodicfunctions 28733.2 Plotting periodic functions 28733.3 Determining resultant phasors bycalculation 288Part 6 Complex Numbers 29134 Complex numbers 29134.1 Cartesian complex numbers 29134.2 The Argand diagram 29234.3 Addition and subtraction of complexnumbers 29234.4 Multiplication and division of complexnumbers 29334.5 Complex equations 29534.6 The polar form of a complexnumber 29634.7 Multiplication and division in polarform 29834.8 Applications of complex numbers 29935 De Moivres theorem 30335.1 Introduction 30335.2 Powers of complex numbers 30335.3 Roots of complex numbers 304Assignment 9 306Part 7 Statistics 30736 Presentation of statistical data 30736.1 Some statistical terminology 30736.2 Presentation of ungrouped data 30836.3 Presentation of grouped data 31237 Measures of central tendency anddispersion 31937.1 Measures of central tendency 31937.2 Mean, median and mode for discretedata 31937.3 Mean, median and mode for groupeddata 32037.4 Standard deviation 32237.5 Quartiles, deciles and percentiles 32438 Probability 32638.1 Introduction to probability 32638.2 Laws of probability 32638.3 Worked problems on probability 32738.4 Further worked problems onprobability 32938.5 Permutations and combinations 33139 The binomial and Poisson distribution 33339.1 The binomial distribution 33339.2 The Poisson distribution 336Assignment 10 33940 The normal distribution 34040.1 Introduction to the normal distribution34040.2 Testing for a normal distribution 34441 Linear correlation 34741.1 Introduction to linear correlation 34741.2 The product-moment formula fordetermining the linear correlationcoefficient 34741.3 The significance of a coefficient ofcorrelation 34841.4 Worked problems on linearcorrelation 34842 Linear regression 35142.1 Introduction to linear regression 35142.2 The least-squares regression lines 35142.3 Worked problems on linearregression 35243 Sampling and estimation theories 35643.1 Introduction 35643.2 Sampling distributions 35643.3 The sampling distribution of themeans 35643.4 The estimation of populationparameters based on a large samplesize 35943.5 Estimating the mean of a populationbased on a small sample size 364Assignment 11 368Multiple choice questions on chapters 27to 43 369Part 8 Differential Calculus 37544 Introduction to differentiation 37544.1 Introduction to calculus 37544.2 Functional notation 37544.3 The gradient of a curve 37644.4 Differentiation from firstprinciples 377 9. CONTENTS ix44.5 Differentiation of y D axn by thegeneral rule 37944.6 Differentiation of sine and cosinefunctions 38044.7 Differentiation of eax and ln ax 38245 Methods of differentiation 38445.1 Differentiation of common functions38445.2 Differentiation of a product 38645.3 Differentiation of a quotient 38745.4 Function of a function 38945.5 Successive differentiation 39046 Some applications of differentiation 39246.1 Rates of change 39246.2 Velocity and acceleration 39346.3 Turning points 39646.4 Practical problems involving maximumand minimum values 39946.5 Tangents and normals 40346.6 Small changes 404Assignment 12 406Part 9 Integral Calculus 40747 Standard integration 40747.1 The process of integration 40747.2 The general solution of integrals of theform axn 40747.3 Standard integrals 40847.4 Definite integrals 41148 Integration using algebraic substitutions41448.1 Introduction 41448.2 Algebraic substitutions 41448.3 Worked problems on integration usingalgebraic substitutions 41448.4 Further worked problems on integrationusing algebraic substitutions 41648.5 Change of limits 41649 Integration using trigonometricsubstitutions 41849.1 Introduction 41849.2 Worked problems on integration ofsin2 x, cos2 x, tan2 x and cot2 x 41849.3 Worked problems on powers of sinesand cosines 42049.4 Worked problems on integration ofproducts of sines and cosines 42149.5 Worked problems on integration usingthe sinsubstitution 42249.6 Worked problems on integration usingthe tansubstitution 424Assignment 13 42550 Integration using partial fractions 42650.1 Introduction 42650.2 Worked problems on integration usingpartial fractions with linearfactors 42650.3 Worked problems on integration usingpartial fractions with repeated linearfactors 42750.4 Worked problems on integration usingpartial fractions with quadraticfactors 42851 The t =q2substitution 43051.1 Introduction 43051.2 Worked problems on the t D tan2substitution 43051.3 Further worked problems on thet D tan2substitution 43252 Integration by parts 43452.1 Introduction 43452.2 Worked problems on integration byparts 43452.3 Further worked problems on integrationby parts 43653 Numerical integration 43953.1 Introduction 43953.2 The trapezoidal rule 43953.3 The mid-ordinate rule 44153.4 Simpsons rule 443Assignment 14 44754 Areas under and between curves 44854.1 Area under a curve 44854.2 Worked problems on the area under acurve 44954.3 Further worked problems on the areaunder a curve 45254.4 The area between curves 45455 Mean and root mean square values 45755.1 Mean or average values 45755.2 Root mean square values 45956 Volumes of solids of revolution 46156.1 Introduction 46156.2 Worked problems on volumes of solidsof revolution 461 10. x CONTENTS56.3 Further worked problems on volumesof solids of revolution 46357 Centroids of simple shapes 46657.1 Centroids 46657.2 The first moment of area 46657.3 Centroid of area between a curve andthe x-axis 46657.4 Centroid of area between a curve andthe y-axis 46757.5 Worked problems on centroids ofsimple shapes 46757.6 Further worked problems on centroidsof simple shapes 46857.7 Theorem of Pappus 47158 Second moments of area 47558.1 Second moments of area and radius ofgyration 47558.2 Second moment of area of regularsections 47558.3 Parallel axis theorem 47558.4 Perpendicular axis theorem 47658.5 Summary of derived results 47658.6 Worked problems on second momentsof area of regular sections 47658.7 Worked problems on second momentsof areas of composite areas 480Assignment 15 482Part 10 Further Number and Algebra 48359 Boolean algebra and logic circuits 48359.1 Boolean algebra and switching circuits48359.2 Simplifying Boolean expressions 48859.3 Laws and rules of Boolean algebra48859.4 De Morgans laws 49059.5 Karnaugh maps 49159.6 Logic circuits 49559.7 Universal logic circuits 50060 The theory of matrices and determinants50460.1 Matrix notation 50460.2 Addition, subtraction and multiplicationof matrices 50460.3 The unit matrix 50860.4 The determinant of a 2 by 2 matrix50860.5 The inverse or reciprocal of a 2 by 2matrix 50960.6 The determinant of a 3 by 3 matrix51060.7 The inverse or reciprocal of a 3 by 3matrix 51161 The solution of simultaneous equations bymatrices and determinants 51461.1 Solution of simultaneous equations bymatrices 51461.2 Solution of simultaneous equations bydeterminants 51661.3 Solution of simultaneous equationsusing Cramers rule 520Assignment 16 521Multiple choice questions on chapters 4461522Answers to multiple choice questions 526Index 527 11. PrefaceThis fourth edition of Engineering Mathematicscovers a wide range of syllabus requirements. Inparticular, the book is most suitable for the latestNational Certificate and Diploma courses andVocational Certificate of Education syllabuses inEngineering.This text will provide a foundation inmathematicalprinciples, which will enable students to solve mathe-matical,scientific and associated engineering princi-ples.In addition, the material will provide engineer-ingapplications and mathematical principles neces-saryfor advancement onto a range of IncorporatedEngineer degree profiles. It is widely recognised thata students ability to usemathematics is a key elementin determining subsequent success. First year under-graduateswho need some remedial mathematics willalso find this book meets their needs.In Engineering Mathematics 4th Edition, theoryis introduced in each chapter by a simple outline ofessential definitions, formulae, laws and procedures.The theory is kept to a minimum, for problem solv-ingis extensively used to establish and exemplifythe theory. It is intended that readers will gain realunderstanding through seeing problems solved andthen through solving similar problems themselves.For clarity, the text is divided into ten topicareas, these being: number and algebra, mensura-tion,trigonometry, graphs, vectors, complex num-bers,statistics, differential calculus, integral calculusand further number and algebra.This new edition will cover the following syl-labuses:(i) Mathematics for Technicians, the core unitfor National Certificate/Diploma courses inEngineering, to include all or part of thefollowing chapters:1. Algebra: 2, 4, 5, 813, 17, 19, 27, 302. Trigonometry: 18, 21, 22, 243. Statistics: 36, 374. Calculus: 44, 46, 47, 54(ii) Further Mathematics for Technicians,the optional unit for National Certifi-cate/Diploma courses in Engineering, toinclude all or part of the following chapters:1. Algebraic techniques: 10, 14, 15,2830, 34, 59612. Trigonometry: 2224, 263. Calculus: 4449, 52584. Statistical and probability: 3643(iii) Applied Mathematics in Engineering, thecompulsory unit for Advanced VCE (for-merlyAdvanced GNVQ), to include all orpart of the following chapters:1. Number and units: 1, 2, 42. Mensuration: 17203. Algebra: 5, 8114. Functions and graphs: 22, 23, 275. Trigonometry: 21, 24(iv) Further Mathematics for Engineering, theoptional unit for Advanced VCE (formerlyAdvanced GNVQ), to include all or part ofthe following chapters:1. Algebra and trigonometry: 5, 6,1215, 21, 252. Graphical and numerical techniques:20, 22, 26313. Differential and integral calculus:4447, 54(v) The Mathematics content of Applied Sci-enceand Mathematics for Engineering,for Intermediate GNVQ(vi) Mathematics for Engineering, for Founda-tionand Intermediate GNVQ(vii) Mathematics 2 and Mathematics 3 for City Guilds Technician Diploma in Telecom-municationsand Electronic Engineering(viii) Any introductory/access/foundation co-urseinvolving Engineering Mathematics atUniversity, Colleges of Further and Highereducation and in schools.Each topic considered in the text is presented ina way that assumes in the reader little previousknowledge of that topic. 12. xii ENGINEERING MATHEMATICSEngineering Mathematics 4th Edition providesa follow-up to Basic Engineering Mathematicsand a lead into Higher Engineering Mathemat-ics.This textbook contains over 900 workedproblems, followed by some 1700 furtherproblems (all with answers). The further problemsare contained within some 208 Exercises; eachExercise follows on directly from the relevantsection of work, every two or three pages. Inaddition, the text contains 234 multiple-choicequestions. Where at all possible, the problemsmirror practical situations found in engineeringand science. 500 line diagrams enhance theunderstanding of the theory.At regular intervals throughout the text are some16 Assignments to check understanding. For exam-ple,Assignment 1 covers material contained inChapters 1 to 4, Assignment 2 covers the materialin Chapters 5 to 8, and so on. These Assignmentsdo not have answers given since it is envisaged thatlecturers could set the Assignments for students toattempt as part of their course structure. Lecturersmay obtain a complimentary set of solutions of theAssignments in an Instructors Manual availablefrom the publishers via the internetfull workedsolutions and mark scheme for all the Assignmentsare contained in this Manual, which is available tolecturers only. To obtain a password please e-mailj.blackford@elsevier.com with the following details:course title, number of students, your job title andwork postal address.To download the Instructors Manual visithttp://www.newnespress.com and enter the booktitle in the search box, or use the following directURL: http://www.bh.com/manuals/0750657766/Learning by Example is at the heart of Engi-neeringMathematics 4th Edition.John BirdUniversity of Portsmouth 13. Part 1 Number and Algebra1Revision of fractions, decimalsand percentages1.1 FractionsWhen 2 is divided by 3, it may be written as 23or2/3. 23is called a fraction. The number above theline, i.e. 2, is called the numerator and the numberbelow the line, i.e. 3, is called the denominator.When the value of the numerator is less thanthe value of the denominator, the fraction is calleda proper fraction; thus 23is a proper fraction.73When the value of the numerator is greater thanthe denominator, the fraction is called an improperfraction. Thus is an improper fraction and can also1373be expressed as a mixed number, that is, an integerand a proper fraction. Thus the improper fraction is equal to the mixed number 2.When a fraction is simplified by dividing thenumerator and denominator by the same number,the process is called cancelling. Cancelling by 0 isnot permissible.Problem 1. Simplify13C27The lowest common multiple (i.e. LCM) of the twodenominators is 3 7, i.e. 21Expressing each fraction so that their denomina-torsare 21, gives:121CD37377C2733D721C621D7 C 621D1321Alternatively:13C27DStep (2)#7 1 CStep (3)#3 221Step (1)Step 1: the LCM of the two denominators;Step 2: for the fraction 13, 3 into 21 goes 7 times,7 the numerator is 7 1;Step 3: for the fraction 27, 7 into 21 goes 3 times,3 the numerator is 3 2.Thus13C27D7 C 621D1321as obtained previously.Problem 2. Find the value of 323 216One method is to split the mixed numbers intointegers and their fractional parts. Then323 216D3 C232 C16D 3 C23 2 16D 1 C4616D 136D 112Another method is to express the mixed numbers asimproper fractions. 14. 2 ENGINEERING MATHEMATICSSince 3 D93, then 323D93C23D113Similarly, 216D126C16D136Thus 323 216D113136D226136D96D 112as obtained previously.Problem 3. Determine the value of458 314C 125458 314C 125D 43 C 1 C5814C25D 2 C5 510 1 C 8 240D 2 C2510 C 1640D 2 C3140D 23140Problem 4. Find the value of371415Dividing numerator and denominator by 3 gives:1 371415 5D17145D1 147 5Dividing numerator and denominator by 7 gives:1 14 21 7 5D1 21 5D25This process of dividing both the numerator anddenominator of a fraction by the same factor(s) iscalled cancelling.Problem 5. Evaluate 135 213 337Mixed numbers must be expressed as improperfractions before multiplication can be performed.Thus,135 213 337D55C3563C13217C37D851 71 324 87 1D8 1 85 1 1D645D 1245Problem 6. Simplify371221371221D371221Multiplying both numerator and denominator by thereciprocal of the denominator gives:371221D1 31 721 312 41 121 21D21 112 1341D34This method can be remembered by the rule: invertthe second fraction and change the operation fromdivision to multiplication. Thus:3712211 3D1 721 312 4D34as obtained previously.Problem 7. Find the value of 535 713The mixed numbers must be expressed as improperfractions. Thus,535 713D28522314 28D5322 11D4255Problem 8. Simplify1213C354813The order of precedence of operations for problemscontaining fractions is the same as that for inte-gers,i.e. remembered by BODMAS (Brackets, Of,Division, Multiplication, Addition and Subtraction).Thus,1325C143813 15. REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 3D134 2 C 5 1203 124 8(B)D13135 208 21(D)D13265(M)D5 13 2615(S)D7315D 41315Problem 9. Determine the value of76of312 214C 5183161276of312 214C 51831612D76of 114C41831612(B)D7654C41831612(O)D7654C411 816 2312(D)D3524C82312(M)D35 C 6562412(A)D6912412(A)D6911224(S)D67924D 28724Now try the following exerciseExercise 1 Further problems on fractionsEvaluate the following:1. (a)12C25(b)71614(a)910(b)3162. (a)27C311(b)2917C23(a)4377(b)47633. (a) 1037 823(b) 314 445C 156(a) 11621(b)17604. (a)3459(b)173515119(a)512(b)3495. (a)3579 127(b)131774 4 31139 (a)35(b) 116. (a)384564(b) 113 25 9(a)815(b)12237.12C358151317248.715of15 57C341516 5459.14231335C271312610.23 11423C14C 135228551.2 Ratio and proportionThe ratio of one quantity to another is a fraction, andis the number of times one quantity is contained inanother quantity of the same kind. If one quantity isdirectly proportional to another, then as one quan-titydoubles, the other quantity also doubles. When aquantity is inversely proportional to another, thenas one quantity doubles, the other quantity is halved.Problem 10. A piece of timber 273 cmlong is cut into three pieces in the ratio of 3to 7 to 11. Determine the lengths of the threepieces 16. 4 ENGINEERING MATHEMATICSThe total number of parts is 3 C 7 C 11, that is, 21.Hence 21 parts correspond to 273 cm1 part corresponds to27321D 13 cm3 parts correspond to 3 13 D 39 cm7 parts correspond to 7 13 D 91 cm11 parts correspond to 11 13 D 143 cmi.e. the lengths of the three pieces are 39 cm,91 cm and 143 cm.(Check: 39 C 91 C 143 D 273)Problem 11. A gear wheel having 80 teethis in mesh with a 25 tooth gear. What is thegear ratio?Gear ratio D 80:25 D8025D165D 3.2i.e. gear ratio D 16 : 5 or 3.2 : 1Problem 12. An alloy is made up ofmetals A and B in the ratio 2.5 : 1 by mass.How much of A has to be added to 6 kg ofB to make the alloy?Ratio A : B: :2.5 : 1 (i.e. A is to B as 2.5 is to 1)orABD2.51D 2.5When B D 6 kg,A6D 2.5 from which,A D 6 2.5 D 15 kgProblem 13. If 3 people can complete atask in 4 hours, how long will it take 5people to complete the same task, assumingthe rate of work remains constantThe more the number of people, the more quicklythe task is done, hence inverse proportion exists.3 people complete the task in 4 hours,1 person takes three times as long, i.e.4 3 D 12 hours,5 people can do it in one fifth of the time thatone person takes, that is125hours or 2 hours24 minutes.Now try the following exerciseExercise 5 Further problems on ratio andproportion1. 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 of12four different colours are used in the ratioof 7:3:19:5. If the mass of the first dyeused is 3g, determine the total mass ofthe dyes used. [17 g]3. Determine how much copper and howmuch zinc is needed to make a 99 kgbrass ingot if they have to be in theproportions copper : zinc: :8 : 3 by mass.[72 kg : 27 kg]4. It takes 21 hours for 12 men to resurfacea stretch of road. Find how many menit takes to resurface a similar stretch ofroad in 50 hours 24 minutes, assumingthe work rate remains constant. [5]5. It takes 3 hours 15 minutes to fly fromcity A to city B at a constant speed. Findhow long the journey takes if(a) the speed is 112times that of theoriginal speed and(b) if the speed is three-quarters of theoriginal speed.[(a) 2 h 10 min (b) 4 h 20 min]1.3 DecimalsThe decimal system of numbers is based on thedigits 0 to 9. A number such as 53.17 is calleda decimal fraction, a decimal point separating theinteger part, i.e. 53, from the fractional part, i.e. 0.17 17. REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 5A number which can be expressed exactly asa decimal fraction is called a terminating deci-maland those which cannot be expressed exactly32as a decimal fraction are called non-terminatingdecimals. Thus, D 1.5 is a terminating decimal,but 43D 1.33333. . . is a non-terminating decimal.1.33333. . . can be written as 1.P3, called one point-threerecurring.The answer to a non-terminating decimal may beexpressed in two ways, depending on the accuracyrequired:(i) correct to a number of significant figures, thatis, 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 nextdigit on the right is in the group of numbers 0, 1,2, 3 or 4, but is increased by 1 if the next digiton the right is in the group of numbers 5, 6, 7, 8or 9. Thus the non-terminating decimal 7.6183. . .becomes 7.62, correct to 3 significant figures, sincethe next digit on the right is 8, which is in the groupof numbers 5, 6, 7, 8 or 9. Also 7.6183. . . becomes7.618, correct to 3 decimal places, since the nextdigit on the right is 3, which is in the group ofnumbers 0, 1, 2, 3 or 4.Problem 14. Evaluate42.7 C 3.04 C 8.7 C 0.06The numbers are written so that the decimal pointsare under each other. Each column is added, startingfrom the right.42.73.048.70.0654.50Thus 42.7 Y 3.04 Y 8.7 Y 0.06 = 54.50Problem 15. Take 81.70 from 87.23The numbers are written with the decimal pointsunder each other.87.2381.705.53Thus 87.23 81.70 = 5.53Problem 16. Find the value of23.417.8357.6 C 32.68The sum of the positive decimal fractions is23.4 C 32.68 D 56.08The sum of the negative decimal fractions is17.83 C 57.6 D 75.43Taking the sum of the negative decimal fractionsfrom the sum of the positive decimal fractions gives:56.0875.43i.e. 75.4356.08 D 19.35Problem 17. Determine the value of74.3 3.8When multiplying decimal fractions: (i) the numbersare multiplied as if they are integers, and (ii) theposition of the decimal point in the answer is suchthat there are as many digits to the right of it as thesum of the digits to the right of the decimal pointsof the two numbers being multiplied together. Thus(i) 743385 94422 29028 234(ii) As there are 1 C 1 D 2 digits to the right ofthe decimal points of the two numbers beingmultiplied together, (74.3 3.8), then74.3 3.8 = 282.34Problem 18. Evaluate 37.81 1.7, correctto (i) 4 significant figures and (ii) 4 decimalplaces 18. 6 ENGINEERING MATHEMATICS37.81 1.7 D37.811.7The denominator is changed into an integer bymultiplying by 10. The numerator is also multipliedby 10 to keep the fraction the same. Thus37.81 1.7 D37.81 101.7 10D378.117The long division is similar to the long division ofintegers and the first four steps are as shown:22.24117..17378.1000003438344134706820(i) 37.81 1.7 = 22.24, correct to 4 significantfigures, and(ii) 37.81 1.7 = 22.2412, correct to 4 decimalplaces.Problem 19. Convert (a) 0.4375 to a properfraction and (b) 4.285 to a mixed number(a) 0.4375 can be written as0.4375 10 00010 000without changing its value,i.e. 0.4375 D437510 000By cancelling437510 000D8752000D175400D3580D716i.e. 0.4375 =716(b) Similarly, 4.285 D 42851000D 457200Problem 20. Express as decimal fractions:(a)916and (b) 578(a) To convert a proper fraction to a decimal frac-tion,the numerator is divided by the denomi-nator.Division by 16 can be done by the longdivision method, or, more simply, by dividingby 2 and then 8:4.5029.000.562584.5000Thus,916= 0.5625(b) For mixed numbers, it is only necessary toconvert the proper fraction part of the mixednumber to a decimal fraction. Thus, dealingwith the 78gives:0.87587.000i.e.78D 0.875Thus 578= 5.875Now try the following exerciseExercise 3 Further problems on decimalsIn Problems 1 to 6, determine the values ofthe expressions given:1. 23.6 C 14.7118.97.421 [11.989]2. 73.84113.247 C 8.210.068[31.265]3. 3.8 4.1 0.7 [10.906]4. 374.1 0.006 [2.2446]5. 421.8 17, (a) correct to 4 significantfigures and (b) correct to 3 decimalplaces.[(a) 24.81 (b) 24.812]6.0.01472.3, (a) correct to 5 decimal placesand (b) correct to 2 significant figures.[(a) 0.00639 (b) 0.0064] 19. REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 77. Convert to proper fractions:(a) 0.65 (b) 0.84 (c) 0.0125 (d) 0.282and (e) 0.024(a)1320(b)2125(c)180(d)141500(e)31258. Convert to mixed numbers:(a) 1.82 (b) 4.275 (c) 14.125 (d) 15.35and (e) 16.2125(a) 14150(b) 41140(c) 1418(d) 15720(e) 161780In Problems 9 to 12, express as decimal frac-tionsto the accuracy stated:9.49, correct to 5 significant figures.[0.44444]10.1727, correct to 5 decimal place.[0.62963]11. 1916, correct to 4 significant figures.[1.563]12. 133137, correct to 2 decimal places.[13.84]1.4 PercentagesPercentages are used to give a common standardand are fractions having the number 100 as theirdenominators. For example, 25 per cent means25100i.e.14and is written 25%.Problem 21. Express as percentages:(a) 1.875 and (b) 0.0125A decimal fraction is converted to a percentage bymultiplying by 100. Thus,(a) 1.875 corresponds to 1.875 100%, i.e.187.5%(b) 0.0125 corresponds to 0.0125 100%, i.e.1.25%Problem 22. Express as percentages:(a)516and (b) 125To convert fractions to percentages, they are (i) con-vertedto decimal fractions and (ii) multiplied by 100(a) By division,516D 0.3125, hence516corre-spondsto 0.3125 100%, i.e. 31.25%(b) Similarly, 125D 1.4 when expressed as adecimal fraction.Hence 125D 1.4 100% D 140%Problem 23. It takes 50 minutes to machinea certain part. Using a new type of tool, thetime can be reduced by 15%. Calculate thenew time taken15% of 50 minutes D15100 50 D750100D 7.5 minutes.hence the new time taken is507.5 D 42.5 minutes.Alternatively, if the time is reduced by 15%, thenit now takes 85% of the original time, i.e. 85% of50 D85100 50 D4250100D 42.5 minutes, as above.Problem 24. Find 12.5% of 37812.5% of 378 means12.5100 378, since per centmeans per hundred.Hence 12.5% of 378 D12.51100 8 378 D18 378 D3788D 47.25 20. 8 ENGINEERING MATHEMATICSProblem 25. Express 25 minutes as apercentage of 2 hours, correct to thenearest 1%Working in minute units, 2 hours D 120 minutes.Hence 25 minutes is25120ths of 2 hours. By can-celling,25120D524Expressing524as a decimal fraction gives 0.208P3Multiplying by 100 to convert the decimal fractionto a percentage gives:0.208P3 100 D 20.8P3%Thus 25 minutes is 21% of 2 hours, correct to thenearest 1%.Problem 26. A German silver alloy consistsof 60% copper, 25% zinc and 15% nickel.Determine the masses of the copper, zinc andnickel in a 3.74 kilogram block of the alloyBy direct proportion:100% corresponds to 3.74 kg1% corresponds to3.74100D 0.0374 kg60% corresponds to 60 0.0374 D 2.244 kg25% corresponds to 25 0.0374 D 0.935 kg15% corresponds to 15 0.0374 D 0.561 kgThus, the masses of the copper, zinc and nickel are2.244 kg, 0.935 kg and 0.561 kg, respectively.(Check: 2.244 C 0.935 C 0.561 D 3.74)Now try the following exerciseExercise 4 Further problems percentages1. Convert to percentages:(a) 0.057 (b) 0.374 (c) 1.285[(a) 5.7% (b) 37.4% (c) 128.5%]2. Express as percentages, correct to 3significant figures:(a)733(b)1924(c) 11116[(a) 21.2% (b) 79.2% (c) 169%]3. Calculate correct to 4 significant figures:(a) 18% of 2758 tonnes (b) 47% of18.42 grams (c) 147% of 14.1 seconds[(a) 496.4 t (b) 8.657 g (c) 20.73 s]4. When 1600 bolts are manufactured, 36are unsatisfactory. Determine the percent-ageunsatisfactory. [2.25%]5. Express: (a) 140 kg as a percentage of1 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 contains88.2 g of nickel, determine the mass ofcopper in the block. [37.8 g]7. A drilling machine should be set to250 rev/min. The nearest speed availableon the machine is 268 rev/min. Calculatethe percentage over speed. [7.2%]8. Two kilograms of a compound contains30% of element A, 45% of element B and25% of element C. Determine the massesof the three elements present.[A 0.6 kg, B 0.9 kg, C 0.5 kg]9. A concrete mixture contains seven partsby volume of ballast, four parts by vol-umeof sand and two parts by volume ofcement. Determine the percentage of eachof these three constituents correct to thenearest 1% and the mass of cement in atwo tonne dry mix, correct to 1 significantfigure.[54%, 31%, 15%, 0.3 t] 21. 2Indices and standard form2.1 IndicesThe lowest factors of 2000 are 2222555.These factors are written as 24 53, where 2 and 5are called bases and the numbers 4 and 3 are calledindices.When an index is an integer it is called a power.Thus, 24 is called two to the power of four, andhas a base of 2 and an index of 4. Similarly, 53 iscalled five to the power of 3 and has a base of 5and an index of 3.Special names may be used when the indices are2 and 3, these being called squared and cubed,respectively. Thus 72 is called seven squared and93 is called nine cubed. When no index is shown,the power is 1, i.e. 2 means 21.15ReciprocalThe 12reciprocal of a number is when the index is1 and its value is given by 1, divided by the base.Thus the reciprocal of 2 is 21 and its value is or 0.5. Similarly, the reciprocal of 5 is 51 whichmeans or 0.2Square rootThe square root of a number is when the index is 12,and the square root of 2 is written as 21/2 orp2. Thevalue of a square root is the value of the base whichwhen multiplied by itself gives the number. Since33 D 9, thenp9 D 3. However, 33 D 9,sop9 D 3. There are always two answers whenfinding the square root of a number and this is shownby putting both a C and asign in front of theanswer to a square root problem. Thusp9 D 3and 41/2 Dp4 D 2, and so on.Laws of indicesWhen simplifying calculations involving indices,certain basic rules or laws can be applied, calledthe laws of indices. These are given below.(i) When multiplying two or more numbers hav-ingthe same base, the indices are added. Thus32 34 D 32C4 D 36(ii) When a number is divided by a number havingthe same base, the indices are subtracted. Thus3532D 352 D 33(iii) When a number which is raised to a poweris raised to a further power, the indices aremultiplied. Thus352 D 352 D 310(iv) When a number has an index of 0, its valueis 1. Thus 30 D 1(v) A number raised to a negative power is thereciprocal of that number raised to a positivepower. Thus 34 D134 Similarly,123 D 23(vi) When a number is raised to a fractional powerthe denominator of the fraction is the root ofthe number and the numerator is the power.Thus 82/3 D3 p82 D 22 D 4and 251/2 D2 p251 Dp251 D 5(Note thatp2 p)2.2 Worked problems on indicesProblem 1. Evaluate: (a) 52 53,(b) 32 34 3 and (c) 2 22 25From law (i):(a) 5253 D 52C3 D 55 D 55555 D 3125 22. 10 ENGINEERING MATHEMATICS(b) 32 34 3 D 32C4C1 D 37D 3 3 to 7 termsD 2187(c) 2 22 25 D 21C2C5 D 28 D 256Problem 2. Find the value of:(a)7573 and (b)5754From law (ii):(a)7573D 753 D 72 D 49(b)5754D 574 D 53 D 125Problem 3. Evaluate: (a) 52 53 54 and(b) 3 35 32 33From laws (i) and (ii):(a) 52 53 54 D52 5354 D52C354D5554D 554 D 51 D 5(b) 3 35 32 33 D3 3532 33D31C532C3D3635D 365 D 31 D 3Problem 4. Simplify: (a) 234 (b) 325,expressing the answers in index form.From law (iii):(a) 234 D 234 D 212 (b) 325 D 325 D 310Problem 5. Evaluate:1023104 102From the laws of indices:1023104 102 D1023104C2D106106D 1066 D 100 D 1Problem 6. Find the value of(a)23 2427 25 and (b)3233 39From the laws of indices:(a)23 2427 25D23C427C5D27212D 2712 D 25D125D132(b)3233 39D32331C9D36310D 3610 D 34D134D181Now try the following exerciseExercise 5 Further problems on indicesIn Problems 1 to 10, simplify the expressionsgiven, expressing the answers in index formand with positive indices:1. (a) 33 34 (b) 42 43 44[(a) 37 (b) 49]2. (a) 23 2 22 (b) 72 74 7 73[(a) 26 (b) 710]3. (a)2423 (b)3732 [(a) 2 (b) 35]4. (a) 56 53 (b) 713/710[(a) 53 (b) 73]5. (a) 723 (b) 332 [(a) 76 (b) 36]6. (a)22 2324 (b)37 3435[(a) 2 (b) 36]7. (a)5752 53 (b)13513 132[(a) 52 (b) 132]8. (a)9 3233 272 (b)16 422 83[(a) 34 (b) 1] 23. INDICES AND STANDARD FORM 119. (a)5254 (b)32 3433 (a) 52 (b)13510. (a)72 737 74 (b)23 24 252 22 26(a) 72 (b)122.3 Further worked problems onindicesProblem 7. Evaluate:33 5753 34The laws of indices only apply to terms having thesame base. Grouping terms having the same base,and then applying the laws of indices to each of thegroups independently gives:33 5753 34D33345753D 334 573D 31 54 D5431 D6253D 20813Problem 8. Find the value of23 35 72274 24 3323 35 72274 24 33 D 234 353 7224D 21 32 70D12 32 1 D92D 412Problem 9. Evaluate:(a) 41/2 (b) 163/4 (c) 272/3 (d) 91/2(a) 41/2 Dp4 D 2(b) 163/4 D4 p 163 D 23 D 8(Note that it does not matter whether the 4th rootof 16 is found first or whether 16 cubed is foundfirstthe same answer will result).(c) 272/3 D3 p272 D 32 D 9(d) 91/2 D191/2D1p9D13D 13Problem 10. Evaluate:41.5 81/322 322/541.5 D 43/2 Dp43 D 23 D 8,81/3 D3 p8 D 2, 22 D 4and 322/5 D1322/5 D15 p322D122 D14Hence41.5 81/322 322/5D8 24 14D161D 16Alternatively,41.5 81/322 322/5D[22]3/2 231/322 252/5D23 2122 22D 23C122 D 24 D 16Problem 11. Evaluate:32 55 C 33 5334 54Dividing each term by the HCF (i.e. highest com-monfactor) of the three terms, i.e. 32 53, gives:32 55 C 33 5334 54D32 5532 53C33 5332 5334 5432 53D322 553 C 332 50342 543D30 52 C 31 5032 51D1 25 C 3 19 5D2845Problem 12. Find the value of32 5534 54 C 33 53 24. 12 ENGINEERING MATHEMATICSTo simplify the arithmetic, each term is divided bythe HCF of all the terms, i.e. 32 53. Thus32 5534 54 C 33 53D32 5532 5334 5432 53C33 5332 53D322 553342 543 C 332 533D30 5232 51 C 31 50D2545 C 3D2548Problem 13. Simplify:433352253giving the answer with positive indicesA fraction raised to a power means that both thenumerator and the denominator of the fraction areraised to that power, i.e.433D4333A fraction raised to a negative power has thesame value as the inverse of the fraction raised to apositive power.Thus,352D1352 D13252D 1 5232 D5232Similarly,253D523D5323Thus,433352253 D433352325323D4333 5232 2353D223 2333C2 532D2935 5Now try the following exerciseExercise 6 Further problems on indicesIn Problems 1 and 2, simplify the expressionsgiven, expressing the answers in index formand with positive indices:1. (a)33 5254 34 (b)72 3235 74 73(a)13 52 (b)173 372. (a)42 9383 34 (b)82 52 34252 24 92(a)3225 (b)1210 523. Evaluate a1321b 810.25c 161/4 d491/2(a) 9 (b) 3 (c)12(d) 23In Problems 4 to 8, evaluate the expressionsgiven.4.92 7434 74 C 33 721471485.24232 4423 162196.123232352565727.434292 [64]8.323/2 81/3232 431/2 91/2412 25. INDICES AND STANDARD FORM 132.4 Standard formA number written with one digit to the left of thedecimal point and multiplied by 10 raised to somepower is said to be written in standard form. Thus:5837 is written as 5.837 103 in standard form,and 0.0415 is written as 4.15 102 in standardform.When a number is written in standard form, thefirst factor is called the mantissa and the secondfactor is called the exponent. Thus the number5.8 103 has a mantissa of 5.8 and an exponentof 103.(i) Numbers having the same exponent can beadded or subtracted in standard form by addingor subtracting the mantissae and keeping theexponent the same. Thus:2.3 104 C 3.7 104D 2.3 C 3.7 104 D 6.0 104and 5.9 1024.6 102D 5.94.6 102 D 1.3 102When the numbers have different exponents,one way of adding or subtracting the numbersis to express one of the numbers in non-standardform, so that both numbers have thesame exponent. Thus:2.3 104 C 3.7 103D 2.3 104 C 0.37 104D 2.3 C 0.37 104 D 2.67 104Alternatively,2.3 104 C 3.7 103D 23 000 C 3700 D 26 700D 2.67 104(ii) The laws of indices are used when multiplyingor dividing numbers given in standard form.For example,2.5 103 5 102D 2.5 5 103C2D 12.5 105 or 1.25 106Similarly,6 1041.5 102D61.5 1042 D 4 1022.5 Worked problems on standardformProblem 14. Express in standard form:(a) 38.71 (b) 3746 (c) 0.0124For a number to be in standard form, it is expressedwith only one digit to the left of the decimal point.Thus:(a) 38.71 must be divided by 10 to achieve onedigit to the left of the decimal point and itmust also be multiplied by 10 to maintain theequality, i.e.38.71 D38.711010 D 3.871 10 in standardform(b) 3746 D37461000 1000 D 3.746 103 in stan-dardform(c) 0.0124 D 0.0124 100100D1.24100D 1.24 102 in standard formProblem 15. Express the followingnumbers, which are in standard form, asdecimal numbers: (a) 1.725 102(b) 5.491 104 (c) 9.84 100(a) 1.725 102 D1.725100D 0.01725(b) 5.491 104 D 5.491 10 000 D 54 910(c) 9.84 100 D 9.84 1 D 9.84 (since 100 D 1)Problem 16. Express in standard form,correct to 3 significant figures:(a)38(b) 1923(c) 741916 26. 14 ENGINEERING MATHEMATICS(a)38D 0.375, and expressing it in standard formgives: 0.375 D 3.75 101(b) 1923D 19.P6 D 1.97 10 in standard form,correct to 3 significant figures(c) 741916D 741.5625 D 7.42 102 in standardform, correct to 3 significant figuresProblem 17. Express the followingnumbers, given in standard form, as fractionsor mixed numbers: (a) 2.5 101(b) 6.25 102 (c) 1.354 102(a) 2.5 101 D2.510D25100D14(b) 6.25 102 D6.25100D62510 000D116(c) 1.354 102 D 135.4 D 135410D 13525Now try the following exerciseExercise 7 Further problems on standardformIn Problems 1 to 4, express in standard form:1. (a) 73.9 (b) 28.4 (c) 197.72(a) 7.39 10 (b) 2.84 10(c) 1.9762 1022. (a) 2748 (b) 33170 (c) 274218(a) 2.748 103 (b) 3.317 104(c) 2.74218 1053. (a) 0.2401 (b) 0.0174 (c) 0.00923(a) 2.401 101 (b) 1.74 102(c) 9.23 1034. (a)12(b) 1178(c) 13035(d)1 32(a) 5 101 (b) 1.1875 10(c) 1.306 102 (d) 3.125 102In Problems 5 and 6, express the numbersgiven as integers or decimal fractions:5. (a) 1.01 103 (b) 9.327 102(c) 5.41 104 (d) 7 100[(a) 1010 (b) 932.7 (c) 54 100 (d) 7]6. (a) 3.89 102 (b) 6.741 101(c) 8 103[(a) 0.0389 (b) 0.6741 (c) 0.008]2.6 Further worked problems onstandard formProblem 18. Find the value of:(a) 7.9 1025.4 102(b) 8.3 103 C 5.415 103 and(c) 9.293 102 C 1.3 103 expressing theanswers in standard form.Numbers having the same exponent can be addedor subtracted by adding or subtracting the mantissaeand keeping the exponent the same. Thus:(a) 7.9 1025.4 102D 7.95.4 102 D 2.5 102(b) 8.3 103 C 5.415 103D 8.3 C 5.415 103 D 13.715 103D 1.3715 104 in standard form(c) Since only numbers having the same exponentscan be added by straight addition of the man-tissae,the numbers are converted to this formbefore adding. Thus:9.293 102 C 1.3 103D 9.293 102 C 13 102D 9.293 C 13 102D 22.293 102 D 2.2293 103in standard form.Alternatively, the numbers can be expressed asdecimal fractions, giving:9.293 102 C 1.3 103D 929.3 C 1300 D 2229.3D 2.2293 103in standard form as obtained previously. Thismethod is often the safest way of doing thistype of problem. 27. INDICES AND STANDARD FORM 15Problem 19. Evaluate(a) 3.75 1036 104 and (b)3.5 1057 102expressing answers in standard form(a) 3.75 1036 104 D 3.75 6103C4D 22.50 107D 2.25 108(b)3.5 1057 102D3.57 1052D 0.5 103 D 5 102Now try the following exerciseExercise 8 Further problems on standardformIn Problems 1 to 4, find values of the expres-sionsgiven, stating the answers in standardform:1. (a) 3.7 102 C 9.81 102(b) 1.431 101 C 7.3 101[(a) 1.351 103 (b) 8.731 101]2. (a) 4.831 102 C 1.24 103(b) 3.24 1031.11 104[(a) 1.7231 103 (b) 3.129 103]3. (a) 4.5 1023 103(b) 2 5.5 104[(a) 1.35 102 (b) 1.1 105]4. (a)6 1033 105 (b)2.4 1033 1024.8 104[(a) 2 102 (b) 1.5 103]5. Write the following statements in stan-dardform:(a) The density of aluminium is2710 kg m3[2.71 103 kg m3](b) Poissons ratio for gold is 0.44[4.4 101](c) The impedance of free space is376.73[3.7673 102 ](d) The electron rest energy is0.511 MeV [5.11 101 MeV](e) Proton charge-mass ratio is9 5 789 700 C kg1[9.57897 107 C kg1](f) The normal volume of a perfect gasis 0.02241 m3 mol1[2.241 102 m3 mol1] 28. 3Computer numbering systems3.1 Binary numbersThe system of numbers in everyday use is thedenary or decimal system of numbers, usingthe 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 aradix or base of 10.The binary system of numbers has a radix of 2and uses only the digits 0 and 1.3.2 Conversion of binary to decimalThe decimal number 234.5 is equivalent to2 102 C 3 101 C 4 100 C 5 101i.e. is the sum of terms comprising: (a digit) multi-pliedby (the base raised to some power).In the binary system of numbers, the base is 2, so1101.1 is equivalent to:1 23 C 1 22 C 0 21 C 1 20 C 1 21Thus the decimal number equivalent to the binarynumber 1101.1 is8 C 4 C 0 C 1 C12, that is 13.5i.e. 1101.12 = 13.510, the suffixes 2 and 10 denot-ingbinary and decimal systems of numbers respec-tively.Problem 1. Convert 110112 to a decimalnumberFrom above: 110112 D 1 24 C 1 23 C 0 22C 1 21 C 1 20D 16 C 8 C 0 C 2 C 1D 2710Problem 2. Convert 0.10112 to a decimalfraction0.10112 D 1 21 C 0 22 C 1 23C 1 24D 1 12C 0 122C 1 123C 1 124D12C18C116D 0.5 C 0.125 C 0.0625D 0.687510Problem 3. Convert 101.01012 to a decimalnumber101.01012 D 1 22 C 0 21 C 1 20C 0 21 C 1 22C 0 23 C 1 24D 4 C 0 C 1 C 0 C 0.25C 0 C 0.0625D 5.312510Now try the following exerciseExercise 9 Further problems on conver-sionof binary to decimal num-bersIn Problems 1 to 4, convert the binary num-bersgiven to decimal numbers.1. (a) 110 (b) 1011 (c) 1110 (d) 1001[(a) 610 (b) 1110 (c) 1410 (d) 910] 29. COMPUTER NUMBERING SYSTEMS 172. (a) 10101 (b) 11001 (c) 101101(d) 110011[(a) 2110 (b) 2510 (c) 4510 (d) 5110]3. (a) 0.1101 (b) 0.11001 (c) 0.00111(d) 0.01011(a) 0.812510 (b) 0.7812510(c) 0.2187510 (d) 0.34375104. (a) 11010.11 (b) 10111.011(c) 110101.0111 (d) 11010101.10111(a) 26.7510 (b) 23.37510(c) 53.437510 (d) 213.71875103.3 Conversion of decimal to binaryAn integer decimal number can be converted to acorresponding binary number by repeatedly dividingby 2 and noting the remainder at each stage, asshown below for 39102 39 Remainder2 19 12 9 12 4 12 2 02 1 00 1(most significant bit)1 0 0 1 1 1 (leastsignificant bit)The result is obtained by writing the top digit ofthe remainder as the least significant bit, (a bit is abinary digit and the least significant bit is the oneon the right). The bottom bit of the remainder is themost significant bit, i.e. the bit on the left.Thus 3910 = 1001112The fractional part of a decimal number can be con-vertedto a binary number by repeatedly multiplyingby 2, as shown below for the fraction 0.6250.625 2 = 1. 2500.250 2 = 0. 5000.500 2 = 1. 000(most significant bit) .1 0 1 (least significant bit)For fractions, the most significant bit of the resultis the top bit obtained from the integer part ofmultiplication by 2. The least significant bit of theresult is the bottom bit obtained from the integerpart of multiplication by 2.Thus 0.62510 = 0.1012Problem 4. Convert 4710 to a binarynumberFrom above, repeatedly dividing by 2 and noting theremainder gives:2 47 Remainder2 23 12 11 12 5 12 2 12 1 00 11 0 1 1 1 1Thus 4710 = 1011112Problem 5. Convert 0.4062510 to a binarynumberFrom above, repeatedly multiplying by 2 gives:0.40625 2 = 0. 81250.8125 2 = 1. 6250.625 2 = 1. 250.25 2 = 0. 50.5 2 = 1. 0. 0 1 1 0 1i.e. 0.4062510 = 0.011012Problem 6. Convert 58.312510 to a binarynumber 30. 18 ENGINEERING MATHEMATICSThe integer part is repeatedly divided by 2, giving:2 58 Remainder2 29 02 14 12 7 02 3 12 1 10 11 1 1 0 1 0The fractional part is repeatedly multiplied by 2giving:0.3125 2 = 0.6250.625 2 = 1.250.25 2 = 0.50.5 2 = 1.0. 0 1 0 1Thus 58.312510 = 111010.01012Now try the following exerciseExercise 10 Further problems on conver-sionof decimal to binarynumbersIn Problems 1 to 4, convert the decimalnumbers given to binary numbers.1. (a) 5 (b) 15 (c) 19 (d) 29(a) 1012 (b) 11112(c) 100112 (d) 1110122. (a) 31 (b) 42 (c) 57 (d) 63(a) 111112 (b) 1010102(c) 1110012 (d) 11111123. (a) 0.25 (b) 0.21875 (c) 0.28125(d) 0.59375(a) 0.012 (b) 0.001112(c) 0.010012 (d) 0.1001124. (a) 47.40625 (b) 30.8125(c) 53.90625 (d) 61.65625(a) 101111.011012 (b) 11110.11012(c) 110101.111012 (d) 111101.1010123.4 Conversion of decimal to binaryvia octalFor decimal integers containing several digits, repe-atedlydividing by 2 can be a lengthy process. Inthis case, it is usually easier to convert a decimalnumber to a binary number via the octal system ofnumbers. This system has a radix of 8, using thedigits 0, 1, 2, 3, 4, 5, 6 and 7. The denary numberequivalent to the octal number 43178 is4 83 C 3 82 C 1 81 C 7 80i.e. 4 512 C 3 64 C 1 8 C 7 1 or 225510An integer decimal number can be converted to acorresponding octal number by repeatedly dividingby 8 and noting the remainder at each stage, asshown below for 493108 493 Remainder8 61 58 7 50 77 5 5Thus 49310 = 7558The fractional part of a decimal number can be con-vertedto an octal number by repeatedly multiplyingby 8, as shown below for the fraction 0.4375100.4375 8 = 3 . 50.5 8 =4 . 0. 3 4For fractions, the most significant bit is the topinteger obtained by multiplication of the decimalfraction by 8, thus0.437510 D 0.348The natural binary code for digits 0 to 7 is shownin Table 3.1, and an octal number can be convertedto a binary number by writing down the three bitscorresponding to the octal digit.Thus 4378 D 100 011 1112and 26.358 D 010 110.011 1012 31. COMPUTER NUMBERING SYSTEMS 19Table 3.1Octal digit Naturalbinary number0 0001 0012 0103 0114 1005 1016 1107 111The 0 on the extreme left does not signify any-thing,thus 26.358 D 10 110.011 1012Conversion of decimal to binary via octal is demon-stratedin the following worked problems.Problem 7. Convert 371410 to a binarynumber, via octalDividing repeatedly by 8, and noting the remaindergives:8 3714 Remainder8 464 28 58 00 77 2 0 28 7 2From Table 3.1, 72028 D 111 010 000 0102i.e. 371410 = 111 010 000 0102Problem 8. Convert 0.5937510 to a binarynumber, via octalMultiplying repeatedly by 8, and noting the integervalues, gives:0.59375 8 = 4.750.75 8 = 6.00. 4 6Thus 0.5937510 D 0.468From Table 3.1, 0.468 D 0.100 1102i.e. 0.5937510 = 0.100 112Problem 9. Convert 5613.9062510 to abinary number, via octalThe integer part is repeatedly divided by 8, notingthe remainder, giving:8 5613 Remainder8 701 5887 58108 107211 2 7 5 5This octal number is converted to a binary number,(see Table 3.1)127558 D 001 010 111 101 1012i.e. 561310 D 1 010 111 101 1012The fractional part is repeatedly multiplied by 8, andnoting the integer part, giving:0.90625 8 = 7.250.25 8 = 2.00. 7 2This octal fraction is converted to a binary number,(see Table 3.1)0.728 D 0.111 0102i.e. 0.9062510 D 0.111 012Thus, 5613.9062510 = 1 010 111 101 101.111 012Problem 10. Convert 11 110 011.100 012to a decimal number via octalGrouping the binary number in threes from thebinary point gives: 011 110 011.100 0102Using Table 3.1 to convert this binary number toan octal number gives: 363.428 and363.428 D 3 82 C 6 81 C 3 80C 4 81 C 2 82D 192 C 48 C 3 C 0.5 C 0.03125D 243.5312510 32. 20 ENGINEERING MATHEMATICSNow try the following exerciseExercise 11 Further problems on con-versionbetween decimal andbinary numbers via octalIn Problems 1 to 3, convert the decimalnumbers given to binary numbers, via octal.1. (a) 343 (b) 572 (c) 1265(a) 1010101112 (b) 10001111002(c) 1001111000122. (a) 0.46875 (b) 0.6875 (c) 0.71875(a) 0.011112 (b) 0.10112(c) 0.1011123. (a) 247.09375 (b) 514.4375(c) 1716.78125(a) 11110111.000112(b) 1000000010.01112(c) 11010110100.1100124. Convert the following binary numbers todecimal numbers via octal:(a) 111.011 1 (b) 101 001.01(c) 1 110 011 011 010.001 1(a) 7.437510 (b) 41.2510(c) 7386.1875103.5 Hexadecimal numbersThe complexity of computers requires higher ordernumbering systems such as octal (base 8) and hex-adecimal(base 16), which are merely extensionsof the binary system. A hexadecimal numberingsystem has a radix of 16 and uses the following 16distinct digits:0, 1, 2, 3, 4, 5, 6, 7, 8, 9,A, B,C,D,E and FA corresponds to 10 in the denary system, B to11, C to 12, and so on.To convert from hexadecimal to decimal:For example1A16 D 1 161 C A 160D 1 161 C 10 1 D 16 C 10 D 26i.e. 1A16 D 2610Similarly,2E16 D 2 161 C E 160D 2 161 C 14 160 D 32 C 14 D 4610and 1BF16 D 1 162 C B 161 C F 160D 1 162 C 11 161 C 15 160D 256 C 176 C 15 D 44710Table 3.2 compares decimal, binary, octal and hex-adecimalnumbers and shows, for example, that2310 D 101112 D 278 D 1716Problem 11. Convert the followinghexadecimal numbers into their decimalequivalents: (a) 7A16 (b) 3F16(a) 7A16 D 7 161 C A 160 D 7 16 C 10 1D 112 C 10 D 122Thus 7A16 = 12210(b) 3F16 D 3 161 C F 160 D 3 16 C 15 1D 48 C 15 D 63Thus, 3F16 = 6310Problem 12. Convert the followinghexadecimal numbers into their decimalequivalents: (a) C916 (b) BD16(a) C916 D C 161 C 9 160 D 12 16 C 9 1D 192 C 9 D 201Thus C916 = 20110(b) BD16 D B161 CD160 D 1116C131D 176 C 13 D 189Thus BD16 = 18910Problem 13. Convert 1A4E16 into a denarynumber 33. COMPUTER NUMBERING SYSTEMS 21Table 3.2Decimal Binary Octal Hexadecimal0 0000 0 01 0001 1 12 0010 2 23 0011 3 34 0100 4 45 0101 5 56 0110 6 67 0111 7 78 1000 10 89 1001 11 910 1010 12 A11 1011 13 B12 1100 14 C13 1101 15 D14 1110 16 E15 1111 17 F16 10000 20 1017 10001 21 1118 10010 22 1219 10011 23 1320 10100 24 1421 10101 25 1522 10110 26 1623 10111 27 1724 11000 30 1825 11001 31 1926 11010 32 1A27 11011 33 1B28 11100 34 1C29 11101 35 1D30 11110 36 1E31 11111 37 1F32 100000 40 201A4E16D 1 163 C A 162 C 4 161 C E 160D 1 163 C 10 162 C 4 161 C 14 160D 1 4096 C 10 256 C 4 16 C 14 1D 4096 C 2560 C 64 C 14 D 6734Thus, 1A4E16 = 673410To convert from decimal to hexadecimal:This is achieved by repeatedly dividing by 16 andnoting the remainder at each stage, as shown belowfor 261016 26 Remainder16 1 10 A160 1 116most significant bit 1 A least significant bitHence 2610 = 1A16Similarly, for 44710447 Remainder2711616 11 B160 1 1161 BF1615 F16Thus 44710 = 1BF16Problem 14. Convert the following decimalnumbers into their hexadecimal equivalents:(a) 3710 (b) 10810(a) 16Remainder537216 = 5160 2= 216most significant bit 2 5 least significant bitHence 3710 = 2516(b) 1610816 = C160 = 616CRemainder12666Hence 10810 = 6C16Problem 15. Convert the following decimalnumbers into their hexadecimal equivalents:(a) 16210 (b) 23910(a) 16Remainder21016216 = 2160 = A16A102Hence 16210 = A216(b) 1623916 = F160 = E16ERemainder151414FHence 23910 = EF16To 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 MATHEMATICSis assigned to each group. For example, the binarynumber 1110011110101001 is initially grouped infours as: 1110 0111 1010 1001and a hexadecimal symbolassigned to each group as E 7 A 9from Table 3.2Hence 11100111101010012 = E7A916.To convert from hexadecimal to binary:The above procedure is reversed, thus, for example,6CF316 D 0110 1100 1111 0011from Table 3.2i.e. 6CF316= 1101100111100112Problem 16. Convert the following binarynumbers into their hexadecimal equivalents:(a) 110101102 (b) 11001112(a) Grouping bits in fours from theright gives: 1101 0110and assigning hexadecimal symbolsto each group gives: D 6from Table 3.2Thus, 110101102 = D616(b) Grouping bits in fours from theright gives: 0110 0111and assigning hexadecimal symbolsto each group gives: 6 7from Table 3.2Thus, 11001112 = 6716Problem 17. Convert the following binarynumbers into their hexadecimal equivalents:(a) 110011112 (b) 1100111102(a) Grouping bits in fours from theright gives: 1100 1111and assigning hexadecimalsymbols to each group gives: C Ffrom Table 3.2Thus, 110011112 = CF16(b) Grouping bits in fours fromthe right gives: 0001 1001 1110and assigning hexadecimalsymbols to each group gives: 1 9 Efrom Table 3.2Thus, 1100111102 = 19E16Problem 18. Convert the followinghexadecimal numbers into their binaryequivalents: (a) 3F16 (b) A616(a) Spacing out hexadecimaldigits gives: 3 Fand converting each intobinary gives: 0011 1111from Table 3.2Thus, 3F16 = 1111112(b) Spacing out hexadecimal digitsgives: A 6and converting each into binarygives: 1010 0110from Table 3.2Thus, A616 = 101001102Problem 19. Convert the followinghexadecimal numbers into their binaryequivalents: (a) 7B16 (b) 17D16(a) Spacing out hexadecimaldigits gives: 7 Band converting each intobinary gives: 0111 1011from Table 3.2Thus, 7B16 = 11110112(b) Spacing out hexadecimaldigits gives: 1 7 Dand converting each intobinary gives: 0001 0111 1101from Table 3.2Thus, 17D16 = 1011111012 35. COMPUTER NUMBERING SYSTEMS 23Now try the following exerciseExercise 12 Further problems on hexa-decimalnumbersIn Problems 1 to 4, convert the given hexadec-imalnumbers into their decimal equivalents.1. E716 [23110] 2. 2C16 [4410]3. 9816 [15210] 4. 2F116 [75310]In Problems 5 to 8, convert the given decimalnumbers into their hexadecimal equivalents.5. 5410 [3616] 6. 20010 [C816]7. 9110 [5B16] 8. 23810 [EE16]In Problems 9 to 12, convert the given binarynumbers into their hexadecimal equivalents.9. 110101112 [D716]10. 111010102 [EA16]11. 100010112 [8B16]12. 101001012 [A516]In Problems 13 to 16, convert the given hex-adecimalnumbers into their binary equiva-lents.13. 3716 [1101112]14. ED16 [111011012]15. 9F16 [100111112]16. A2116 [1010001000012] 36. 4Calculations and evaluation offormulae4.1 Errors and approximations(i) In all problems in which the measurement ofdistance, time, mass or other quantities occurs,an exact answer cannot be given; only ananswer which is correct to a stated degree ofaccuracy can be given. To take account of thisan error due to measurement is said to exist.(ii) To take account of measurement errors itis usual to limit answers so that the resultgiven is not more than one significant figuregreater than the least accurate numbergiven in the data.(iii) Rounding-off errors can exist with decimalfractions. For example, to state thatD3.142 is not strictly correct, but D 3.142correct to 4 significant figures is a true state-ment.(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 existif incorrect positioning of the decimal pointoccurs after a calculation has been completed.(vi) Blunders and order of magnitude errors canbe reduced by determining approximate val-uesof calculations. Answers which do notseem feasible must be checked and the cal-culationmust be repeated as necessary.An engineer will often need to make aquick mental approximation for a calcula-tion.For example,49.1 18.4 122.161.2 38.1maybe approximated to50 20 12060 40and then,by cancelling,50 1 20 120 2 11 60 40 2 1D 50. Anaccurate answer somewhere between 45 and55 could therefore be expected. Certainlyan answer around 500 or 5 would not beexpected. Actually, by calculator49.1 18.4 122.161.2 38.1D 47.31, correct to4 significant figures.Problem 1. The area A of a triangle isgiven by A D12bh. The base b whenmeasured is found to be 3.26 cm, and theperpendicular height h is 7.5 cm. Determinethe area of the triangle.Area of triangle D12bh D12 3.26 7.5 D12.225 cm2 (by calculator).The approximate value is1238 D 12 cm2, sothere are no obvious blunder or magnitude errors.However, it is not usual in a measurement typeproblem to state the answer to an accuracy greaterthan 1 significant figure more than the least accuratenumber in the data: this is 7.5 cm, so the resultshould not have more than 3 significant figuresThus, area of triangle = 12.2 cm2Problem 2. State which type of error hasbeen made in the following statements:(a) 72 31.429 D 2262.9(b) 16 0.08 7 D 89.6(c) 11.714 0.0088 D 0.3247 correct to4 decimal places.(d)29.74 0.051211.89D 0.12, correct to2 significant figures. 37. CALCULATIONS AND EVALUATION OF FORMULAE 25(a) 72 31.429 D 2262.888 (by calculator),hence a rounding-off error has occurred. Theanswer should have stated:72 31.429 D 2262.9, correct to 5 significantfigures or 2262.9, correct to 1 decimal place.(b) 16 0.08 7 D 16 8100 7 D32 725D22425D 82425D 8.96Hence an order of magnitude error hasoccurred.(c) 11.714 0.0088 is approximately equal to129103, i.e. about 108103 or 0.108.Thus a blunder has been made.(d) 29.74 0.051211.8930 5 10212D15012 102D15120D18or 0.125hence no order of magnitude error has29.74 0.0512occurred. However,11.89D 0.128correct to 3 significant figures, which equals0.13 correct to 2 significant figures.Hence a rounding-off error has occurred.Problem 3. Without using a calculator,determine an approximate value of:(a)11.7 19.19.3 5.7(b)2.19 203.6 17.9112.1 8.76(a)11.7 19.19.3 5.7is approximately equal to10 2010 5, i.e. about 4(By calculator,11.7 19.19.3 5.7D 4.22, correct to3 significant figures.)(b)2.19 203.6 17.9112.1 8.762 20 200 20 21 10 10 1D 2 20 2 after cancelling,i.e.2.19 203.6 17.9112.1 8.76 80(By calculator,2.19 203.6 17.9112.1 8.76D 75.3,correct to 3 significant figures.)Now try the following exerciseExercise 13 Further problems on errorsIn Problems 1 to 5 state which type of error,or errors, have been made:1. 25 0.06 1.4 D 0.21[order of magnitude error]2. 137 6.842 D 937.4 Rounding-off errorshould add correctto 4 significant figures or correct to1 decimal place3.24 0.00812.6D 10.42 [Blunder]4. For a gas pV D c. When pressurep D 1 03 400 Pa and V D 0.54 m3 thenc D 55 836 Pa m3.Measured values, hencec D 55 800 Pa m35.4.6 0.0752.3 0.274D 0.225Order of magnitude error and rounding-offerrorshould be 0.0225, correct to3 significant figures or 0.0225,correct to 4 decimal placesIn Problems 6 to 8, evaluate the expressionsapproximately, without using a calculator.6. 4.7 6.3 [30 (29.61, by calculator)]7.2.87 4.076.12 0.962 (1.988, correct to 4 s.f., bycalculator)8.72.1 1.96 48.6139.3 5.210 (9.481, correct to 4 s.f., bycalculator) 38. 26 ENGINEERING MATHEMATICS4.2 Use of calculatorThe most modern aid to calculations is the pocket-sizedelectronic calculator. With one of these, cal-culationscan be quickly and accurately performed,correct to about 9 significant figures. The scientifictype of calculator has made the use of tables andlogarithms largely redundant.To help you to become competent at using yourcalculator check that you agree with the answers tothe following problems:Problem 4. Evaluate the following, correctto 4 significant figures:(a) 4.7826 C 0.02713 (b) 17.694111.8762(c) 21.93 0.012981(a) 4.7826 C 0.02713 D 4.80973 D 4.810, correctto 4 significant figures(b) 17.694111.8762 D 5.8179 D 5.818, correctto 4 significant figures(c) 21.93 0.012981 D 0.2846733 . . . D 0.2847,correct to 4 significant figuresProblem 5. Evaluate the following, correctto 4 decimal places:(a) 46.32 97.17 0.01258 (b)4.62123.76(c)1262.49 0.0172 (a) 46.32 97.17 0.01258 D 56.6215031 . . . D56.6215, correct to 4 decimal places(b)4.62123.76D 0.19448653 . . . D 0.1945, correct to4 decimal places(c)1262.49 0.0172D 0.537414 D 0.5374,correct to 4 decimal placesProblem 6. Evaluate the following, correctto 3 decimal places:(a)152.73(b)10.0275(c)14.92C11.97(a)152.73D 0.01896453 . . . D 0.019, correct to 3decimal places(b)10.0275D 36.3636363 . . . D 36.364, correct to3 decimal places(c)14.92C11.97D 0.71086624 . . . D 0.711, cor-rectto 3 decimal placesProblem 7. Evaluate the following,expressing the answers in standard form,correct to 4 significant figures.(a) 0.00451 2 (b) 631.76.21 C 2.95 2(c) 46.27231.792(a) 0.00451 2 D 2.03401105 D 2.034 105,correct to 4 significant figures(b) 631.76.21 C 2.95 2 D 547.7944 D5.477944 102 D 5.478 102, correct to 4significant figures(c) 46.27231.792 D 1130.3088 D 1.130 103,correct to 4 significant figuresProblem 8. Evaluate the following, correctto 3 decimal places:(a)2.37 20.0526(b)3.601.92 39. 2C5.402.45 40. 2(c)157.624.82(a)2.37 20.0526D 106.785171 . . . D 106.785, correctto 3 decimal places(b)3.601.92 41. 2C5.402.45 42. 2D 8.37360084 . . . D8.374, correct to 3 decimal places(c)157.624.82 D 0.43202764 . . . D 0.432, cor-rectto 3 decimal places 43. CALCULATIONS AND EVALUATION OF FORMULAE 27Problem 9. Evaluate the following, correctto 4 significant figures:ppp(a)5.462 (b)54.62 (c)546.2(a)p5.462 D 2.3370922 . . . D 2.337, correct to 4significant figures(b)p54.62 D 7.39053448 . . . D 7.391, correct to4 significant figures(c)p546.2 D 23.370922 . . . D 23.37, correct to 4significant figuresProblem 10. Evaluate the following, correctto 3 decimal places:(a)p0.007328 (b)p52.91 p31.76(c)p1.6291 104(a)p0.007328 D 0.08560373 D 0.086, correct to3 decimal places(b)p52.91 p31.76 D 1.63832491 . . . D 1.638,correct to 3 decimal places(c)p1.6291 104 Dp16291 D 127.636201. . . D 127.636, correct to 3 decimal placesProblem 11. Evaluate the following, correctto 4 significant figures:(a) 4.723 (b) 0.8316 4p(c)76.21229.102(a) 4.723 D 105.15404 . . . D 105.2, correct to 4significant figures(b) 0.8316 4 D 0.47825324 . . . D 0.4783, correctto 4 significant figures(c)p76.21229.102 D 70.4354605 . . . D 70.44,correct to 4 significant figuresProblem 12. Evaluate the following, correctto 3 significant figures:(a) 6.09225.2 p7(b) 3 p47.291(c)p7.2132 C 6.4183 C 3.2914(a) 6.09225.2 p7D 0.74583457 . . . D 0.746, cor-rectto 3 significant figures(b) 3 p47.291 D 3.61625876 . . . D 3.62, correct to3 significant figures(c)p7.2132 C 6.4183 C 3.2914 D 20.8252991. . . D 20.8, correct to 3 significant figuresProblem 13. Evaluate the following,expressing the answers in standard form,correct to 4 decimal places:(a) 5.176 103 2 1.974 101 8.61 102(b)3.462 44. 4(c)p1.792 104(a) 5.176 103 2 D 2.679097 . . . 105 D2.6791 105, correct to 4 decimal places(b)1.974 101 8.61 1023.462 45. 4D 0.05808887 . . .D 5.8089 102,correct to 4 decimal places(c)p1.792 104 D 0.0133865 . . .D 1.3387 102, correct to 4 decimal placesNow try the following exerciseExercise 14 Further problems on use ofcalculatorIn Problems 1 to 9, use a calculator to evaluatethe quantities shown correct to 4 significantfigures:1. (a) 3.2492 (b) 73.782 (c) 311.42(d) 0.06392(a) 10.56 (b) 5443 (c) 96 970(d) 0.0040832. (a)p4.735 (b)p35.46 (c)p73 280(d)p0.0256(a) 2.176 (b) 5.955 (c) 270.7(d) 0.1600 46. 28 ENGINEERING MATHEMATICS3. (a)17.768(b)148.46(c)10.0816(d)11.118(a) 0.1287 (b) 0.02064(c) 12.25 (d) 0.89454. (a) 127.8 0.0431 19.8(b) 15.76 4.329[(a) 109.1 (b) 3.641]5. (a)137.6552.9(b)11.82 1.7360.041[(a) 0.2489 (b) 500.5]6. (a) 13.63 (b) 3.4764 (c) 0.1245[(a) 2515 (b) 146.0 (c) 0.00002932]7. (a)24.68 0.05327.412 47. 3(b)0.2681 41.2232.6 11.89 48. 4[(a) 0.005559 (b) 1.900]8. (a)14.32321.682 (b)4.821317.33215.86 11.6[(a) 6.248 (b) 0.9630]9. (a) 15.62 229.21 p10.52(b)p6.9212 C 4.81632.1614[(a) 1.605 (b) 11.74]10. Evaluate the following, expressing theanswers in standard form, correct to3 decimal places: (a) 8.291 102 2(b)p7.623 103[(a) 6.874 103 (b) 8.731 102]4.3 Conversion tables and chartsIt is often necessary to make calculations from vari-ousconversion tables and charts. Examples includecurrency exchange rates, imperial to metric unitconversions, train or bus timetables, productionschedules and so on.Problem 14. Currency exchange rates forfive countries are shown in Table 4.1Table 4.1France 1 D 1.46 eurosJapan 1 D 190 yenNorway 1 D 10.90 kronorSwitzerland 1 D 2.15 francsU.S.A. 1 D 1.52 dollars ($)Calculate:(a) how many French euros 27.80 will buy(b) the number of Japanese yen which canbe bought for 23(c) the pounds sterling which can beexchanged for 6409.20Norwegian kronor(d) the number of American dollars whichcan be purchased for 90, and(e) the pounds sterling which can beexchanged for 2795 Swiss francs(a) 1 D 1.46 euros, hence27.80 D 27.80 1.46 euros D 40.59 euros(b) 1 D 190 yen, hence23 D 23 190 yen D 4370 yen(c) 1 D 10.90 kronor, hence6409.20 kronor D 6409.2010.90D 588(d) 1 D 1.52 dollars, hence90 D 90 1.52 dollars D $136.80(e) 1 D 2.15 Swiss francs, hence2795 franc D 27952.15D 1300Problem 15. Some approximate imperial tometric conversions are shown in Table 4.2Table 4.2length 1 inch D 2.54 cm1 mile D 1.61 kmweight 2.2 lb D 1kg1 lb D 16 oz capacity 1.76 pints D 1 litre8 pints D 1 gallon 49. CALCULATIONS AND EVALUATION OF FORMULAE 29Use the table to determine:(a) the number of millimetres in 9.5 inches,(b) a speed of 50 miles per hour inkilometres per hour,(c) the number of miles in 300 km,(d) the number of kilograms in 30 poundsweight,(e) the number of pounds and ounces in42 kilograms (correct to the nearestounce),(f) the number of litres in 15 gallons, and(g) the number of gallons in 40 litres.(a) 9.5 inches D 9.5 2.54 cm D 24.13 cm24.13 cm D 24.13 10 mm D 241.3 mm(b) 50 m.p.h. D 50 1.61 km/h D 80.5 km=h(c) 300 km D3001.61miles D 186.3 miles(d) 30 lb D302.2kg D 13.64 kg(e) 42 kg D 42 2.2 lb D 92.4 lb0.4 lb D 0.4 16 oz D 6.4 oz D 6 oz, correctto the nearest ounceThus 42 kg D 92 lb 6 oz, correct to the near-estounce.(f) 15 gallons D 15 8 pints D 120 pints120 pints D1201.76litres D 68.18 litres(g) 40 litres D 40 1.76 pints D 70.4 pints70.4 pints D70.48gallons D 8.8 gallonsNow try the following exerciseExercise 15 Further problems conversiontables and charts1. Currency exchange rates listed in a news-paperincluded the following:Italy 1 D 1.48 euroJapan 1 D 185 yenAustralia 1 D 2.70 dollarsCanada 1 D $2.40Sweden 1 D 13.25 kronorCalculate (a) how many Italian euros32.50 will buy, (b) the number ofCanadian dollars that can be purchasedfor 74.80, (c) the pounds sterling whichcan be exchanged for 14 040 yen, (d) thepounds sterling which can be exchangedfor 1754.30 Swedish kronor, and (e) theAustralian dollars which can be boughtfor 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 imperialconversions.Length 2.54 cm D 1 inch1.61 km D 1 mileWeight 1 kg D 2.2 lb 1 lb D 16 ounces Capacity 1 litre D 1.76 pints8 pints D 1 gallon Use the list to determine (a) the numberof millimetres in 15 inches, (b) a speed of35 mph in km/h, (c) the number of kilo-metresin 235 miles, (d) the number ofpounds and ounces in 24 kg (correct tothe nearest ounce), (e) the number of kilo-gramsin 15 lb, (f) the number of litres in12 gallons and (g) the number of gallonsin 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 gallons3. Deduce the following information fromthe BR train timetable shown in Table 4.3:(a) At what time should a man catch atrain at Mossley Hill to enable himto be in Manchester Piccadilly by8.15 a.m.?(b) A girl leaves Hunts Cross at8.17 a.m. and travels to ManchesterOxford Road. How long does thejourney take. What is the averagespeed of the journey?(c) A man living at Edge Hill has to beat work at Trafford Park by 8.45 a.m.It takes him 10 minutes to walk to 50. 30 ENGINEERING MATHEMATICSTable 4.3 Liverpool, Hunts Cross and Warrington! ManchesterReproduced with permission of British Railhis work from Trafford Park sta-tion.What time train should he catchfrom Edge Hill? (a) 7.09 a.m.(b) 51 minutes, 32 m.p.h.(c) 7.04 a.m.]4.4 Evaluation of formulaeThe statement v D u C at is said to be a formulafor 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 theequation, v, is called the subject of the formulae.Provided values are given for all the symbols ina formula except one, the remaining symbol canbe made the subject of the formula and may beevaluated by using a calculator.Problem 16. In an electrical circuit thevoltage V is given by Ohms law, i.e.V D IR. Find, correct to 4 significant figures,the voltage when I D 5.36 A andR D 14.76 .V D IR D 5.36 14.76 51. CALCULATIONS AND EVALUATION OF FORMULAE 31Hence, voltage V = 79.11 V, correct to 4 signifi-cantfigures.Problem 17. The surface area A of a hollowcone is given by A D rl. Determine, correctto 1 decimal place, the surface area whenr D 3.0 cm and l D 8.5 cm.A D rl D 3.0 8.5cm2Hence, surface area A = 80.1 cm2, correct to 1decimal place.Problem 18. Velocity v is given byv D u C at. If u D 9.86 m/s, a D 4.25 m/s2and t D 6.84 s, find v, correct to 3 significantfigures.v D u C at D 9.86 C 4.25 6.84 D 9.86 C 29.07 D 38.93Hence, velocity v = 38.9 m=s, correct to 3 signi-ficantfigures.Problem 19. The power, P watts, dissipatedin an electrical circuit may be expressed bythe formula P DV2R. Evaluate the power,correct to 3 significant figures, given thatV D 17.48 V and R D 36.12 .P DV2RD17.48 236.12D305.550436.12Hence power, P = 8.46 W, correct to 3 signifi-cantfigures.Problem 20. The volume V cm3 of a rightcircular cone is given by V D13r2h. Giventhat r D 4.321 cm and h D 18.35 cm, findthe volume, correct to 4 significant figures.V D13r2h D134.321 218.35 D1318.671041 18.35 Hence volume, V = 358.8 cm3, correct to 4 sig-nificantfigures.Problem 21. Force F newtons is given bythe formula F DGm1m2d2 , where m1 and m2are masses, d their distance apart and G is aconstant. Find the value of the force giventhat G D 6.671011, m1 D 7.36,m2 D 15.5and d D 22.6. Express the answer in standardform, correct to 3 significant figures.F DGm1m2d2D6.67 1011 7.36 15.5 22.6 2D6.67 7.36 15.5 1011 510.76 D1.4901011Hence force F = 1.49 1011 newtons, correctto 3 significant figures.Problem 22. The time of swing t seconds,of a simplependulum is given byt D 2lg. Determine the time, correct to 3decimal places, given that l D 12.0 andg D 9.81t D 2 lgD 2 12.09.81D 2 p1.22324159D 2 1.106002527 Hence time t = 6.950 seconds, correct to3 decimal places.Problem 23. Resistance, R, varies withtemperature according to the formulaR D R01 C t . Evaluate R, correct to 3significant figures, given R0 D 14.59, D 0.0043 and t D 80.R D R01 C tD 14.59[1 C 0.0043 80 ]D 14.591 C 0.344 D 14.591.344 Hence, resistance, R = 19.6 Z, correct to 3 sig-nificantfigures. 52. 32 ENGINEERING MATHEMATICSNow try the following exerciseExercise 16 Further problems on evalua-tionof formulae1. A formula used in connection with gasesis R D PV /T. Evaluate R whenP D 1500, V D 5 and T D 200.[R D 37.5]2. The velocity of a body is given byv D u C at. The initial velocity u is mea-suredwhen time t is 15 seconds andfound to be 12 m/s. If the accelera-tiona is 9.81 m/s2 calculate the finalvelocity v. [159 m/s]3. Find the distance s, given that s D 12gt2,time t D 0.032 seconds and accelerationdue to gravity g D 9.81 m/s2.[0.00502 m or 5.02 mm]4. The energy stored in a capacitor is givenby E D 12CV2 joules. Determine theenergy when capacitance C D 5 106 farads and voltage V D 240V.[0.144 J]5. Resistance R2 is given byR2 D R11 C t . Find R2, correct to4 significant figures, when R1 D 220, D 0.00027 and t D 75.6 [224.5]6. Density Dmassvolume. Find the densitywhen the mass is 2.462 kg and the vol-umeis 173 cm3. Give the answer in unitsof kg/m3. [14 230 kg/m3]7. Velocity D frequency wavelength.Find the velocity when the frequency is1825 Hz and the wavelength is 0.154 m.[281.1 m/s]8. Evaluate resistance RT, given1RTD1R1C1R2C1R3when R1 D 5.5 ,R2 D 7.42and R3 D 12.6 .[2.526 ]9. Power Dforce distancetime. Find thepower when a force of 3760 N raises anobject a distance of 4.73 m in 35 s.[508.1 W]10. The potential difference, V volts, avail-ableat battery terminals is given byV D EIr. 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 12mv2u2 , find Fwhen m D 18.3, v D 12.7 and u D 8.24[F D 854.5]12. The current I amperes flowing in a num-berof cells is given by I DnER 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]13. The time, t seconds, of oscillation for asimple pendulumis given byt D 2lg. Determine the time when 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 formulaE D 12LI2. Evaluate the energy whenL D 5.5 and I D 1.2 [E D 3.96 J]15. The current I amperes in an a.c. circuitis given by I DVpR2 C X2. Evaluate thecurrent when V D 250, R D 11.0 andX D 16.2 [I D 12.77 A]16. Distance s metres is given by the for-mulas D ut C 12at2. If u D 9.50,t D 4.60 and a D 2.50, evaluate thedistance.[s D 17.25 m]17. The area, A, of any triangle is givenby A Dpssa sb scwheres Da C b C c2. Evaluate the area givena D 3.60 cm, b D 4.00 cm andc 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,abevaluate v, given that v Dcde[v D 7.327] 53. CALCULATIONS AND EVALUATION OF FORMULAE 33Assignment 1This assignment covers the material con-tainedin Chapters 1 to 4. The marks foreach question are shown in brackets atthe end of each question.1. Simplify (a) 223 313(b) 147 214 54. 13C15 55. C 2724(9)2. A piece of steel, 1.69 m long, is cutinto three pieces in the ratio 2 to 5 to6. Determine, in centimetres, the lengthsof the three pieces. (4)3. Evaluate576.2919.3(a) correct to 4 significant figures(b) correct to 1 decimal place (2)4. Determine, correct to 1 decimal places,57% of 17.64 g (2)5. Express 54.7 mm as a percentage of1.15 m, correct to 3 significant figures.(3)6. Evaluate the following:(a)23 2 2224 (b)23 16 28 2 3(c)142 56. 1(d) (27)13(e)32 57. 22923 58. 2 (14)7. Express the following in standard form:(a) 1623 (b) 0.076 (c) 14525(3)8. Determine the value of the following,giving the answer in standard form:(a) 5.9 102 C 7.31 102(b) 2.75 1022.65 103 (4)9. Convert the following binary numbers todecimal form:(a) 1101 (b) 101101.0101 (5)10. Convert the following decimal numberto binary form:(a) 27 (b) 44.1875 (6)11. Convert the following decimal numbersto binary, via octal:(a) 479 (b) 185.2890625 (6)12. Convert (a) 5F16 into its decimal equiv-alent(b) 13210 into its hexadecimalequivalent (c) 1101010112 into its hex-adecimalequivalent (6)13. Evaluate the following, each correct to 4significant figures:(a) 61.222 (b)10.0419(c)p0.0527(3)14. Evaluate the following, each correct to 2decimal places:(a)36.22 0.56127.8 12.83 59. 3(b) 14.692p17.42 37.98(7)15. If 1.6 km D 1 mile, determine the speedof 45 miles/hour in kilometres per hour.(3)16. Evaluate B, correct to 3 significantfigures, when W D 7.20, v D 10.0 andg D 9.81, given that B DWv22g. (3) 60. 5Algebra5.1 Basic operationsAlgebra is that part of mathematics in which therelations and properties of numbers are investigatedby means of general symbols. For example, the areaof a rectangle is found by multiplying the lengthby the breadth; this is expressed algebraically asA D lb, where A represents the area, l the lengthand b the breadth.The basic laws introduced in arithmetic are gen-eralizedin algebra.Let a, b, c and d represent any four numbers.Then:(i) a C b C c D a C b C c(ii) abc D abc(iii) a C b D b C a(iv) ab D ba(v) ab C c D ab C ac(vi)a C bcDacCbc(vii) a C bc C d D ac C ad C bc C bdProblem 1. Evaluate: 3ab2bc C abcwhen a D 1, b D 3 and c D 5Replacing a, b and c with their numerical valuesgives:3ab2bc C abc D 3 1 32 3 5C 1 3 5D 930 C 15 D 6Problem 2. Find the value of 4p2qr3, giventhat p D 2, q D12and r D 112Replacing p, q and r with their numerical valuesgives:4p2qr3 D 42212323D 4 2 2 12323232D 27Problem 3. Find the sum of: 3x, 2x, xand 7xThe sum of the positive terms is: 3x C 2x D 5xThe sum of the negative terms is: x C 7x D 8xTaking the sum of the negative terms from the sumof the positive terms gives:5x8x D 3xAlternatively3x C 2x C x C 7x D 3x C 2xx7xD 3xProblem 4. Find the sum of 4a, 3b, c, 2a,5b and 6cEach symbol must be dealt with individually.For the a terms: C4a2a D 2aFor the b terms: C3b5b D 2bFor the c terms: Cc C 6c D 7cThus4a C 3b C c C 2a C 5b C 6cD 4a C 3b C c2a5b C 6cD 2a 2b Y 7cProblem 5. Find the sum of: 5a2b,2a C c, 4b5d and ba C 3d4c 61. ALGEBRA 35The algebraic expressions may be tabulated asshown below, forming columns for the as, bs, csand ds. Thus:C5a2bC2a C cC 4b5d a C b4c C 3dAdding gives: 6a Y 3b 3c 2dProblem 6. Subtract 2x C 3y4z fromx2y C 5zx2y C 5z2x C 3y4zSubtracting gives: x 5y Y 9z(Note that C5z4z D C5z C 4z D 9z)An alternative method of subtracting algebraicexpressions is to change the signs of the bottomline and add. Hence:x2y C 5z2x3y C 4zAdding gives: x 5y Y 9zProblem 7. Multiply 2a C 3b by a C bEach term in the first expression is multiplied by a,then each term in the first expression is multipliedby b, and the two results are added. The usual layoutis shown below.2a C 3ba C bMultiplying by a ! 2a2 C 3abMultiplying by b ! C 2ab C 3b2Adding gives: 2a2 Y 5ab Y 3b2Problem 8. Multiply 3x2y2 C 4xy by2x5y3x2y2 C 4xy2x5yMultiplyingby 2x ! 6x2 4xy2 C 8x2yMultiplyingby5y! 20xy215xy C 10y3Adding gives: 6x2 24xy2 Y 8x2y 15xy Y 10y3Problem 9. Simplify: 2p 8pq2p 8pq means2p8pq. This can be reduced bycancelling as in arithmetic.Thus:2p8pqD2 p8 p qD14qNow try the following exerciseExercise 17 Further problems on basicoperations1. Find the value of 2xy C3yzxyz, whenx D 2, y D 2 and z D4 [16]2. Evaluate 3pq2r3 when p D23, q D 2and r D 1 [8]3. Find the sum of 3a, 2a, 6a, 5a and4a [4a]4. Add together 2aC3bC4c, 5a2bCc,4a5b6c [a4bc]5. Add together 3dC4e, 2eCf, 2d3f,4de C 2f3e [9d2e]6. From 4x3y C 2z subtract x C 2y3z[3x5y C 5z]7. Subtract32a b3C c fromb2 4a3c512a C56b4c8. Multiply 3x C 2y by xy[3x2xy2y2]9. Multiply 2a5b C c by 3a C b[6a213ab C 3ac5b2 C bc]10. Simplify (i) 3a 9ab (ii) 4a2b 2a i13bii 2ab 62. 36 ENGINEERING MATHEMATICS5.2 Laws of IndicesThe laws of indices are:(i) am an D amCn (ii)amanD amn(iii) (amn D amn (iv) am/n D n pam(v) an D1an (vi) a0 D 1Problem 10. Simplify: a3b2c ab3c5Grouping like terms gives:a3 a b2 b3 c c5Using the first law of indices gives:a3C1 b2C3 c1C5i.e. a4 b5 c6 D a4b5c6Problem 11. Simplify:a1/2b2c2 a1/6b1/2cUsing the first law of indices,a1/2b2c2 a1/6b1/2cD a1/2C1/6 b2C1/2 c2C1D a2=3b5=2c1Problem 12. Simplify:a3b2c4abc2 and evaluatewhen a D 3, b D18and c D 2Using the second law of indices,a3aD a31 D a2,b2bD b21 D bandc4c2 D c42 D c6Thusa3b2c4abc2D a2bc6When a D 3, b D18and c D 2,a2bc6 D 321826 D 91864 D 72Problem 13. Simplify:p1/2q2r2/3p1/4q1/2r1/6 andevaluate when p D 16, q D 9 and r D 4,taking positive roots onlyUsing the second law of indices gives:p1/21/4q21/2r2/31/6 D p1=4q3=2r1=2When p D 16, q D 9 and r D 4,p1/4q3/2r1/2 D 161/493/241/2D4 p16p93p4D 2332 D 108Problem 14. Simplify:x2y3 C xy2xyAlgebraic expressions of the forma C bccan be splitintoacCbc. Thusx2y3 C xy2xyDx2y3xyCxy2xyD x21y31 C x11y21D xy2 Y y(since x0 D 1, from the sixth law of indices)Problem 15. Simplify:x2yxy2xyThe highest common factor (HCF) of each of thethree terms comprising the numerator and denomi-natoris xy. Dividing each term by xy gives:x2yxy2xyDx2yxyxy2xyxyxyDxy 1Problem 16. Simplify: p31/2q24Using the third law of indices gives:p31/2q24 D p.3=2/q8 63. ALGEBRA 37Problem 17. Simplify:mn23m1/2n1/44The brackets indicate that each letter in the bracketmust be raised to the power outside. Using the thirdlaw of indices gives:mn23m1/2n1/44 Dm13n23m1/24n1/44 Dm3n6m2n1Using the second law of indices gives:m3n6m2n1D m32n61 D mn5Problem 18. Simplify:ppa3pbc5a 3 pb2 c3and evaluatewhen a D14, b D 6 and c D 1Using the fourth law of indices, the expression canbe written as:a3b1/2c5/2a1/2b2/3c3Using the first law of indices gives:a3C1/2b1/2C2/3c5/2C3 D a7/2b7/6c11/2It is usual to express the answer in the same formas the question. Hencea7/2b7/6c11/2 Dpa7 6 pb7pc11When a D14, b D 64 and c D 1,pa7 6 pb7pc11 D147 6 p647p111D127271 D 1Problem 19. Simplify:d2e2f1/2d3/2ef5/22expressing the answer with positive indicesonlyUsing the third law of indices gives:d2e2f1/2d3/2ef5/22Dd2e2f1/2d3e2f5Using the second law of indices gives:d23e22f1/25 D d1e0f9/2D d1f9/2 since e0 D 1from the sixth law of indices=1df 9=2from the fifth law of indicesProblem 20. Simplify:x2y1/2px 3y2x5y31/2Using the third and fourth laws of indices gives:x2y1/2px 3y2x5y31/2Dx2y1/2x1/2y2/3x5/2y3/2Using the first and second laws of indices gives:x2C1/25/2y1/2C2/33/2 D x0y1/3D y1=3or1y1=3 or13 pyfrom the fifth and sixth laws of indices.Now try the following exerciseExercise 18 Further problems on laws ofindices1. Simplify x2y3zx3yz2 and evaluatewhen x D12, y D 2 and z D 3x5y4z3, 13122. Simplify (a3/2bc3a1/2b1/2c andevaluate when a D 3, b D 4 and c D 2a2b1/2c2, 412 64. 38 ENGINEERING MATHEMATICS3. Simplify:a5bc3a2b3c2 and evaluate when a D32, b D12and c D23a3b2c,916In Problems 4 to 10, simplify the givenexpressions:4.x1/5y1/2z1/3x1/2y1/3z1/6 [x7/10y1/6z1/2]5.a2b C a3ba2b21 C ab6.p3q2pq2p2qp2qqp7. a21/2b23c1/23 [ab6c3/2]8.abc2a2b1c33 [a4b5c11]9. (px y3 3 pz2px y3pz3)[xy3 6 pz13]10.a3b1/2c1/2ab1/3pa3pb ca11/6b1/3c3/2 or6 pa11 3 pbpc3 65. 5.3 Brackets and factorisationWhen two or more terms in an algebraic expressioncontain a common factor, then this factor can beshown outside of a bracket. For exampleab C ac D ab C cwhich is simply the reverse of law (v) of algebra onpage 34, and6px C 2py4pz D 2p3x C y2zThis process is called factorisation.Problem 21. Remove the brackets andsimplify the expression:3a C b C 2b C c4c C dBoth b and c in the second bracket have to bemultiplied by 2, and c and d in the third bracketby 4 when the brackets are removed. Thus:3a C b C 2b C c4c C dD 3a C b C 2b C 2c4c4dCollecting similar terms together gives:3a Y 3b 2c 4dProblem 22. Simplify:a22aaba3b C