Basic Engineering Mathematics

Transcription

1

2 Basic Engineering Mathematics

3 In memory of Elizabeth

4 Basic Engineering Mathematics Third Edition John Bird BSc(Hons), CMath, CEng, FIMA, MIEE, FIIE(Elec), FCollP Newnes OXFORD AMSTERDAM BOSTON LONDON NEW YORK PARIS SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO

5 Newnes An imprint of Elsevier Science Linacre House, Jordan Hill, Oxford OX 8DP 00 Wheeler Rd, Burlington, MA 080 First published 999 Second edition 000 Reprinted 00 Third edition 00 Copyright 999, 000, 00. 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 988 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 988 or under the terms of a licence issued by the Copyright Licensing Agency Ltd, 90 Tottenham Court Rd, London, England WT 4LP. Applications for the copyright holder s written permission to reproduce any part of this publication should be addressed to the publishers British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN For information on all Newnes publications visit our website at Typeset by Laserwords Private Limited, Chennai, India Printed and bound in Great Britain

6 Contents Preface xi. Basic arithmetic. Arithmetic operations. Highest common factors and lowest common multiples. Order of precedence and brackets 4. Fractions, decimals and percentages 6. Fractions 6. Ratio and proportion 8. Decimals 9.4 Percentages Assignment. Indices and standard form 4. Indices 4. Worked problems on indices 4. Further worked problems on indices 6.4 Standard form 7.5 Worked problems on standard form 8.6 Further worked problems on standard form 9 4. Calculations and evaluation of formulae 0 4. Errors and approximations 0 4. Use of calculator 4. Conversion tables and charts Evaluation of formulae 6 Assignment 8 5. Computer numbering systems 9 5. Binary numbers 9 5. Conversion of binary to denary 9 5. Conversion of denary to binary Conversion of denary to binary via octal 5.5 Hexadecimal numbers 6. Algebra 6 6. Basic operations 6 6. Laws of Indices 8

7 vi Contents 6. Brackets and factorization Fundamental laws and precedence Direct and inverse proportionality 44 Assignment Simple equations Expressions, equations and identities Worked problems on simple equations Further worked problems on simple equations Practical problems involving simple equations Further practical problems involving simple equations 5 8. Transposition of formulae 5 8. Introduction to transposition of formulae 5 8. Worked problems on transposition of formulae 5 8. Further worked problems on transposition of formulae Harder worked problems on transposition of formulae 56 Assignment Simultaneous equations Introduction to simultaneous equations Worked problems on simultaneous equations in two unknowns Further worked problems on simultaneous equations More difficult worked problems on simultaneous equations Practical problems involving simultaneous equations Quadratic equations Introduction to quadratic equations Solution of quadratic equations by factorization Solution of quadratic equations by completing the square Solution of quadratic equations by formula Practical problems involving quadratic equations The solution of linear and quadratic equations simultaneously 74 Assignment Straight line graphs 76. Introduction to graphs 76. The straight line graph 76. Practical problems involving straight line graphs 8. Graphical solution of equations 87. Graphical solution of simultaneous equations 87. Graphical solutions of quadratic equations 88. Graphical solution of linear and quadratic equations simultaneously 9.4 Graphical solution of cubic equations 9 Assignment Logarithms 96. Introduction to logarithms 96. Laws of logarithms 96

8 Contents vii. Indicial equations 98.4 Graphs of logarithmic functions Exponential functions The exponential function Evaluating exponential functions The power series for e x Graphs of exponential functions Napierian logarithms Evaluating Napierian logarithms Laws of growth and decay 06 Assignment Reduction of non-linear laws to linear form 0 5. Determination of law 0 5. Determination of law involving logarithms 6. Geometry and triangles 7 6. Angular measurement 7 6. Types and properties of angles 8 6. Properties of triangles Congruent triangles 6.5 Similar triangles 6.6 Construction of triangles 5 Assignment Introduction to trigonometry 8 7. Trigonometry 8 7. The theorem of Pythagoras 8 7. Trigonometric ratios of acute angles Solution of right-angled triangles 7.5 Angles of elevation and depression 7.6 Evaluating trigonometric ratios of any angles 4 8. Trigonometric waveforms 7 8. Graphs of trigonometric functions 7 8. Angles of any magnitude 8 8. The production of a sine and cosine wave Sine and cosine curves Sinusoidal form A sin!t š a 44 Assignment Cartesian and polar co-ordinates Introduction Changing from Cartesian into polar co-ordinates Changing from polar into Cartesian co-ordinates Use of R! P and P! R functions on calculators 5 0. Areas of plane figures 5 0. Mensuration 5 0. Properties of quadrilaterals 5

9 viii Contents 0. Worked problems on areas of plane figures Further worked problems on areas of plane figures Areas of similar shapes 58 Assignment The circle 60. Introduction 60. Properties of circles 60. Arc length and area of a sector 6.4 The equation of a circle 64. Volumes of common solids 66. Volumes and surface areas of regular solids 66. Worked problems on volumes and surface areas of regular solids 66. Further worked problems on volumes and surface areas of regular solids 68.4 Volumes and surface areas of frusta of pyramids and cones 7.5 Volumes of similar shapes 75 Assignment 75. Irregular areas and volumes and mean values of waveforms 77. Areas of irregular figures 77. Volumes of irregular solids 79. The mean or average value of a waveform Triangles and some practical applications Sine and cosine rules Area of any triangle Worked problems on the solution of triangles and their areas Further worked problems on the solution of triangles and their areas Practical situations involving trigonometry Further practical situations involving trigonometry 90 Assignment 9 5. Vectors 9 5. Introduction 9 5. Vector addition 9 5. Resolution of vectors Vector subtraction Relative velocity Number sequences Simple sequences The n 0 th term of a series Arithmetic progressions Worked problems on arithmetic progression Further worked problems on arithmetic progressions Geometric progressions Worked problems on geometric progressions Further worked problems on geometric progressions 06 Assignment 07

10 Contents ix 7. Presentation of statistical data Some statistical terminology Presentation of ungrouped data Presentation of grouped data 8. Measures of central tendency and dispersion 7 8. Measures of central tendency 7 8. Mean, median and mode for discrete data 7 8. Mean, median and mode for grouped data Standard deviation Quartiles, deciles and percentiles 9. Probability 9. Introduction to probability 9. Laws of probability 9. Worked problems on probability Further worked problems on probability 5 Assignment Introduction to differentiation 9 0. Introduction to calculus 9 0. Functional notation 9 0. The gradient of a curve Differentiation from first principles 0.5 Differentiation of y D ax n by the general rule 0.6 Differentiation of sine and cosine functions Differentiation of e ax and ln ax Summary of standard derivatives Successive differentiation Rates of change 7. Introduction to integration 9. The process of integration 9. The general solution of integrals of the form ax n 9. Standard integrals 9.4 Definite integrals 4.5 Area under a curve 4 Assignment 5 47 List of formulae 48 Answers to exercises 5 Index 67

11 Preface Basic Engineering Mathematics, rd edition introduces and then consolidates basic mathematical principles and promotes awareness of mathematical concepts for students needing a broad base for further vocational studies. In this third edition, new material has been added on the introduction to differential and integral calculus. The text covers: (i) the Applied Mathematics content of the GNVQ mandatory unit Applied Science and Mathematics for Engineering at Intermediate level (i.e. GNVQ ) (ii) the mandatory Mathematics for Engineering at Advanced level (i.e. GNVQ ) in Engineering (iii) the optional Applied Mathematics for Engineering at Advanced level (i.e. GNVQ ) in Engineering (iv) the Mathematics content of Applied Science and Mathematics for Technicians for Edexcel/BTEC First Certificate (v) the mandatory Mathematics for Technicians for National Certificate and National Diploma in Engineering (vi) Mathematics for City & Guilds Technician Certificate in Telecommunications and Electronics Engineering (vii) basic mathematics for a wide range of introductory/access/foundation mathematics courses (viii) GCSE revision and for similar mathematics courses in English-speaking countries world-wide. Basic Engineering Mathematics, rd edition provides a lead into Engineering Mathematics. Each topic considered in the text is presented in a way that assumes in the reader little previous knowledge of that topic. Theory is introduced in each chapter by a brief outline of essential theory, definitions, formulae, laws and procedures. However these are 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 solving similar problems themselves. This textbook contains some 575 worked problems, followed by over 000 further problems (all with answers at the end of the book). The further problems are contained within some 8 Exercises; each Exercise follows on directly from the relevant section of work. 60 line diagrams enhance the understanding of the theory. Where at all possible the problems mirror practical situations found in engineering and science. At regular intervals throughout the text are fifteen Assignments to check understanding. For example, Assignment covers material contained in Chapters and, Assignment covers the material contained in Chapters and 4, 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 see below. At the end of the book a list of relevant formulae contained within the text is included for convenience of reference. Learning by Example is at the heart of Early Engineering Mathematics John Bird University of Portsmouth Instructur s Manual 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 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 and enter the book title in the search box, or use the following direct URL:

12 Basic arithmetic. Arithmetic operations Whole numbers are called integers. C, C5, C7 are called positive integers;, 6, 5 are called negative integers. Between positive and negative integers is the number 0 which is neither positive nor negative. The four basic arithmetic operators are: add (C), subtract ( ), multiply (ð) and divide (ł) For addition and subtraction, when unlike signs are together in a calculation, the overall sign is negative. Thus, adding minus 4 to is C 4 and becomes 4 D. Like signs together give an overall positive sign. Thus subtracting minus 4 from is 4 and becomes C 4 D 7. For multiplication and division, when the numbers have unlike signs, the answer is negative, but when the numbers have like signs the answer is positive. Thus ð 4 D, whereas ð 4 DC. Similarly 4 D 4 and 4 DC4 Taking the sum of the negative integers from the sum of the positive integers gives: 08 9 Thus = 5 Problem. Subtract 89 from This is written mathematically as 89 Thus 89 = 4 Problem. Subtract 74 from Problem. Add 7, 74, 8 and 9 This problem is written as Like signs together give an overall positive sign, hence This problem is written as 7 74 C 8 9 Adding the positive integers: 7 8 Sum of positive integers is: 08 Adding the negative integers: 74 9 Sum of negative integers is: D 77 C C74 Thus = 45 Problem 4. Subtract 4 from 6 45 The problem is 6 4. When the second number is larger than the first, take the smaller number from the larger and make the result negative.

13 Basic Engineering Mathematics Thus 6 4 D Thus 6 4 = 7 Problem 5. Subtract 8 from The sum of the negative integers is 69 C8 Thus 69 8 = 587 Problem 6. Multiply 74 by This is written as 74 ð Thus 74 U = 96 Problem ð ð 0 Adding: 96 Multiply by 78 by 46 When the numbers have different signs, the result will be negative. (With this in mind, the problem can now be solved by multiplying 78 by 46) Thus 78 ð 46 D 888 and 78 U ( 46) = 888 Problem 8. Divide l04 by 7 When dividing by the numbers to, it is usual to use a method called short division. 49 ) Step. 7 into 0 goes, remainder. Put above the 0 of 04 and carry the remainder to the next digit on the right, making it 4; Step. 7 into 4 goes 4, remainder 6. Put 4 above the 4 of 04 and carry the 6 remainder to the next digit on the right, making it 6; Step. 7 into 6 goes 9, remainder 0. Put 9 above the of 04. Thus 04 7 = 49 Problem 9. Divide 78 by 4 When dividing by numbers which are larger than, it is usual to use a method called long division. 7 ) 4 78 ð 4! ð 4! 98 Thus 78 4 = 7 00 Problem 0. Divide 5669 by 46 () 4 into 7 goes twice. Put above the 7 of 78. () Subtract. Bring down the 8. 4 into 98 goes 7 times. Put 7 above the 8 of 78. (5) Subtract. This problem may be written as 5669 or 5669 ł 46 or /46 Using the long division method shown in Problem 9 gives: ) As there are no more digits to bring down, =, remainder or 46 Now try the following exercise Exercise Further problems on arithmetic operations (Answers on page 5) In Problems to, determine the values of the expressions given:

14 Basic arithmetic C C C C C C C C (a) 6 ð 7 (b) 46 ð 9 5. (a) 78 ð (b) 7 ð 4 6. (a) 7 ð 8 (b) 77 ð 9 7. (a) 448 ð (b) 4 ð 8. (a) 88 ł 6 (b) 979 ł 9. (a) (a) (a) (b) (b) 5904 ł 56 (b) ł 79. Highest common factors and lowest common multiples When two or more numbers are multiplied together, the individual numbers are called factors. Thus a factor is a number which divides into another number exactly. The highest common factor (HCF) is the largest number which divides into two or more numbers exactly. A multiple is a number which contains another number an exact number of times. The smallest number which is exactly divisible by each of two or more numbers is called the lowest common multiple (LCM). Problem. Determine the HCF of the numbers, 0 and 4 Each number is expressed in terms of its lowest factors. This is achieved by repeatedly dividing by the prime numbers,, 5, 7,,... (where possible) in turn. Thus D ð ð 0 D ð ð 5 4 D ð ð 7 The factors which are common to each of the numbers are in column and in column, shown by the broken lines. Hence the HCF is U, i.e. 6. That is, 6 is the largest number which will divide into, 0 and 4. Problem. Determine the HCF of the numbers 0, 05, 0 and 55 Using the method shown in Problem : 0 D ð ð 5 05 D ð 5 ð 7 0 D ð ð 5 ð 7 55 D ð 5 ð 7ð The factors which are common to each of the numbers are in column and 5 in column. Hence the HCF is U 5 = 5 Problem. Determine the LCM of the numbers, 4 and 90 The LCM is obtained by finding the lowest factors of each of the numbers, as shown in Problems and above, and then selecting the largest group of any of the factors present. Thus D ð ð 4 D ð ð 7 90 D ð ð ð 5 The largest group of any of the factors present are shown by the broken lines and are ð in, ð in90,5in90and 7 in 4. Hence the LCM is U U U U 5 U 7 = 60, and is the smallest number which, 4 and 90 will all divide into exactly.

15 4 Basic Engineering Mathematics Problem 4. Determine the LCM of the numbers 50, 0, 75 and 65 Using the method shown in Problem above: 50 D ð ð 5ð 5 0 D ð ð 5ð 7 75 D ð 5ð 7ð 7 65 D ð 5ð 7 ð The LCM is U U 5 U 5 U 7 U 7 U = (ii) ð D ð, i.e. the order of numbers when multiplying does not matter; (iii) C C 4 D C C 4, i.e. the use of brackets when adding does not affect the result; (iv) ð ð 4 D ð ð 4, i.e. the use of brackets when multiplying does not affect the result; (v) ð C4 D C4 D ðcð4, i.e. a number placed outside of a bracket indicates that the whole contents of the bracket must be multiplied by that number; (vi) C 4 C 5 D 5 9 D 45, i.e. adjacent brackets indicate multiplication; (vii) [ C 4 ð 5 ] D [ C 0] D ð D 46, i.e. when an expression contains inner and outer brackets, the inner brackets are removed first. Now try the following exercise Problem 5. Find the value of 6 C 4 ł 5 Exercise Further problems on highest common factors and lowest common multiples (Answers on page 5) In Problems to 6 find (a) the HCF and (b) the LCM of the numbers given:. 6, 0, 4., 0, 45. 0, 5, 70, , 05, , 0, 90, , 50, 770. Order of precedence and brackets When a particular arithmetic operation is to be performed first, the numbers and the operator(s) are placed in brackets. Thus times the result of 6 minus is written as ð 6 In arithmetic operations, the order in which operations are performed are: (i) to determine the values of operations contained in brackets; (ii) multiplication and division (the word of also means multiply); and (iii) addition and subtraction. This order of precedence can be remembered by the word BODMAS, standing for Brackets, Of, Division, Multiplication, Addition and Subtraction, taken in that order. The basic laws governing the use of brackets and operators are shown by the following examples: (i) C D C, i.e. the order of numbers when adding does not matter; The order of precedence of operations is remembered by the word BODMAS. Thus 6 C 4 ł 5 D 6 C 4 ł Problem 6. D 6 C D 8 Determine the value of ð C 4 ł C 5 (Brackets) (Division) (Addition) ð C 4 ł C 5 D ð C 4 ł 7 (B) Problem 7. Evaluate D ð C D 6 C D 5 6 D 9 6 ł C 6 C 8[ C 4 ð 6 ] 6ł C 6 C 8[ C 4 ð 6 ] D 6 ł C 6 C 8[ C 4 D 6 ł 8 C 8 ð 6 D C 8 ð 6 (D) (M) (A) (S) (B) (B) (D)

16 Basic arithmetic 5 D C 08 D 0 Problem 8. 4 ð 7 C 4 ð 7 C Find the value of 44 ł ł D 4 ð 4 C 6 D 4 ð 4 C 6 D 56 C 6 D 9 56 D 7 (M) (A) (B) (D) (M) (A) (S) Now try the following exercise Exercise Further problems on order of precedence and brackets (Answers on page 5) Simplify the expressions given in Problems to 7:. 4 C ð 5. 7 ł C 4 ł ł ł C ł 7 C ð C

17 Fractions, decimals and percentages. Fractions When is divided by, it may be written as or /. is called a fraction. The number above the line, i.e., is called the numerator and the number below the line, i.e., 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 is a proper fraction. 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 be expressed as a mixed number, that is, an integer and a proper fraction. Thus the improper fraction 7 is equal to the mixed number. 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. Step : the LCM of the two denominators; Step : for the fraction, into goes 7 times, 7 ð the numerator is 7 ð ; Step : for the fraction, 7 into goes times, ð the 7 numerator is ð. Thus C 7 D 7 C 6 D as obtained previously. Problem. Find the value of 6 One method is to split the mixed numbers into integers and their fractional parts. Then ( 6 D C ) ( C ) D C 6 6 Problem. Simplify C 7 The LCM of the two denominators is ð 7, i.e.. Expressing each fraction so that their denominators are, gives: C 7 D ð 7 7 C 7 ð D 7 C 6 D 7 C 6 Alternatively: D Step () Step () # # C 7 ð C ð D 7 " Step () D C D 6 D Another method is to express the mixed numbers as improper fractions. Since D 9,then D 9 C D Similarly, 6 D 6 C 6 D 6 Thus 6 D 6 D 6 6 D 9 6 D as obtained previously. Problem. Evaluate

18 Fractions, decimals and percentages ( 7 D 7 C ) ( 5 C ) D 7 C D C 8 7 D C 7 ð 8 ð 56 D C D C 7 56 D 7 56 D 7 7 D D D 9 56 Problem 4. Determine the value of C C 5 D 4 C C ( C 5 5 ð 5 0 ð C 8 ð D C C 6 D C 40 D C 40 D 40 Problem 5. Find the value of 7 ð 4 5 Dividing numerator and denominator by gives: 7 ð 4 D ð 4 5 D ð 4 7 ð 5 Dividing numerator and denominator by 7 gives: ð 4 7 ð 5 D ð ð 5 D 5 This process of dividing both the numerator and denominator of a fraction by the same factor(s) is called cancelling. Problem 6. Evaluate 5 ð ð 7 Mixed numbers must be expressed as improper fractions before multiplication can be performed. Thus, ) ð 5 ð ð 7 D ( 5 5 C 5 ( 6 C D ð ð 4 8 D 8 ð ð ð ð D 64 5 D 4 5 ) ) ( ð 7 C ) 7 Problem 7. Simplify 7 ł 7 ł D 7 Multiplying both numerator and denominator by the reciprocal of the denominator gives: 7 D 7 ð 4 ð D 4 D 4 This method can be remembered by the rule: invert the second fraction and change the operation from division to multiplication. Thus: 7 ł D 7 ð D as obtained previously. 4 4 Problem 8. Find the value of 5 5 ł 7 The mixed numbers must be expressed as improper fractions. Thus, 5 5 ł 7 D 8 5 ł 8 D4 5 ð D 4 55 Problem 9. Simplify ( C ) ( 5 4 ł ð ) 8 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, ( 5 C 4 ) ł ( 8 ð D 4 ð C 5 ð 0 D 5 0 ð 8 D ð ð 6 D 5 D 7 5 D 4 5 ) ł 4 8 (B) (D) (M) (S)

19 8 Basic Engineering Mathematics Problem of Determine the value of ) C 5 8 ł 6 ( 4 ( 7 6 of ) C ł 6 D 7 6 of 4 C 4 8 ł 6 D 7 6 ð 5 4 C 4 8 ł 6 (B) (O) ð 6 7 C 5 ł of 5 ( 5 ð 5 7 ) C 4 ð ł 5 C 7 ( ð ) ( ł 4 C 4 ( 4 ł 5 ) 6 ) C 5 D 7 6 ð 5 4 C 4 8 ð 6 D 5 4 C 8 5 C 656 D 4 D D 4 D D Now try the following exercise Exercise 4 Further problems on fractions (Answers on page 5) Evaluate the expressions given in Problems to :. (a) C 5. (a) 7 C. (a) 5 C 4 4. (a) (a) 4 ð 5 9 (b) (b) 9 7 C (b) (b) C 5 6 (b) 7 5 ð (a) 5 ð 7 9 ð 7 (b) 7 ð 4 7 ð (a) 4 ð ð 5 9 (b) 4 ł (a) 8 ł (b) ł 5 9 (D) (M) (A) (A) (S). Ratio and proportion 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. Divide 6 in the ratio of 5 to Because the ratio is to be 5 parts to parts, then the total number of parts is 5 C, that is 8. Then, 8 parts correspond to 6 Hence part corresponds to 6 8 D 7, 5 parts correspond to 5 ð 7 D 5 and parts correspond to ð 7 D 9 (Check: the parts must add up to the total 5 C 9 D 6 D the total.) Problem. A piece of timber 7 cm long is cut into three pieces in the ratio of to 7 to. Determine the lengths of the three pieces. The total number of parts is C 7 C, that is,. Hence parts correspond to 7 cm part corresponds to 7 D cm parts correspond to ð D 9 cm 7 parts correspond to 7 ð D 9 cm parts correspond to ð D 4 cm i.e. the lengths of the three pieces are 9 cm, 9 cm and 4 cm. (Check: 9 C 9 C 4 D 7)

20 Fractions, decimals and percentages 9 Problem. A gear wheel having 80 teeth is in mesh with a 5 tooth gear. What is the gear ratio? Gear ratio D 80:5 D 80 5 D 6 5 D. i.e. gear ratio D 6:5 or.: Problem 4. Express 5p as a ratio of 4.5 Working in quantities of the same kind, the required ratio is 5 i.e. 45 That is, 5p is 7 th of 4.5. This may be written either as: 7 5:45: ::7 (stated as 5 is to 45 as is to 7 ) or as 5 45 = 7 Problem 5. An alloy is made up of metals A and B in the ratio.5: by mass. How much of A has to be added to 6 kg of B to make the alloy? Ratio A:B: :.5: i.e. A B D.5 D.5 When B D 6 kg, A D.5 from which, A D 6 ð.5 D 5 kg 6 Problem 6. If people can complete a task in 4 hours, find how long it will 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. people complete the task in 4 hours, person takes three times as long, i.e. 4 ð D hours, 5 people can do it in one fifth of the time that one person takes, that is hours or hours 4 minutes. 5 Now try the following exercise Exercise 5 Further problems on ratio and proportion (Answers on page 5). Divide mm in the ratio of 7 to 7.. Divide 6 cm in the ratio of to 7 to is to be divided between two people in the ratio of 9 to 7. Determine how much each person will receive. 4. When mixing a quantity of paints, dyes of four different colours are used in the ratio of 7::9:5. If the mass of the first dye used is g, determine the total mass of the dyes used. 5. 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: by mass. 6. It takes hours for men to resurface a stretch of road. Find how many men it takes to resurface a similar stretch of road in 50 hours 4 minutes, assuming the work rate remains constant. 7. It takes hours 5 minutes to fly from city A to city B at a constant speed. Find how long the journey takes if (a) the speed is times that of the original speed and (b) if the speed is three-quarters of the original speed.. Decimals The decimal system of numbers is based on the digits 0to 9. A number such as 5.7 is called a decimal fraction, a decimal point separating the integer part, i.e. 5, from the fractional part, i.e. 0.7 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, D.5 is a terminating decimal, but 4 D... is a non-terminating decimal.... can be written as.p, called one point-three 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,,, or 4, but is increased by if the next digit on the right is in the group of numbers 5, 6, 7, 8 or 9. Thus the non-terminating decimal becomes 7.6, correct to 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 becomes 7.68, correct to decimal places, since the next digit on the right is, which is in the group of numbers 0,,, or 4. Problem 7. Evaluate 4.7 C.04 C 8.7 C 0.06

21 0 Basic Engineering Mathematics The numbers are written so that the decimal points are under each other. Each column is added, starting from the right Thus = Problem 8. Take 8.70 from 87. The numbers are written with the decimal points under each other Thus = 5.5 Problem 9. Find the value of C.68 The sum of the positive decimal fractions is.4 C.68 D The sum of the negative decimal fractions is 7.8 C 57.6 D 75.4 Taking the sum of the negative decimal fractions from the sum of the positive decimal fractions gives: i.e D 9.5 Problem 0. Determine the value of 74. ð.8 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 (i) (ii) As there are C D digits to the right of the decimal points of the two numbers being multiplied together, (74. ð.8), then 74. U.8 = 8.4 Problem. Evaluate 7.8 ł.7, correct to (i) 4 significant figures and (ii) 4 decimal places. 7.8 ł.7 D The denominator is changed into an integer by multiplying by 0. The numerator is also multiplied by 0 to keep the fraction the same. Thus 7.8 ð ł.7 D.7 ð 0 D The long division is similar to the long division of integers and the first four steps are as shown:.47.. ) (i) =.4, correct to 4 significant figures, and (ii) =.4, correct to 4 decimal places. Problem. Convert (a) to a proper fraction and (b) 4.85 to a mixed number. without chang- (a) can be written as ing its value, i.e D ð By cancelling D D D 5 80 D 7 6 i.e = 7 6 (b) Similarly, 4.85 D D

22 Fractions, decimals and percentages Problem. (a) 9 6 and (b) Express as decimal fractions: (a) To convert a proper fraction to a decimal fraction, the numerator is divided by the denominator. Division by 6 can be done by the long division method, or, more simply, by dividing by and then 8: 4.50 ) ) Thus, 9 6 = (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 8 gives: ) Thus = i.e. Now try the following exercise Exercise D Further problems on decimals (Answers on page 5) In Problems to 7, determine the values of the expressions given:..6 C C ð ð 4. ð ð ł 7, (a) correct to 4 significant figures and (b) correct to decimal places , (a) correct to 5 decimal places and. (b) correct to significant figures. 8. Convert to proper fractions: (a) 0.65 (b) 0.84 (c) 0.05 (d) 0.8 and (e) Convert to mixed numbers: (a).8 (b) 4.75 (c) 4.5 (d) 5.5 and (e) 6.5 In Problems 0 to 5, express as decimal fractions to the accuracy stated: 0. 4, correct to 5 significant figures. 9. 7, correct to 5 decimal place. 7. 9, correct to 4 significant figures , correct to decimal places. 4., correct to decimal places , correct to significant figures..4 Percentages Percentages areusedtogiveacommonstandardandare fractions having the number 00 as their denominators. For example, 5 per cent means 5 00 i.e. and is written 5%. 4 Problem 4. (b) 0.05 Express as percentages: (a).875 and A decimal fraction is converted to a percentage by multiplying by 00. Thus, (a).875 corresponds to.875 ð 00%, i.e. 87.5% (b) 0.05 corresponds to 0.05 ð 00%, i.e..5% Problem 5. (a) 5 6 and (b) 5 Express as percentages: To convert fractions to percentages, they are (i) converted to decimal fractions and (ii) multiplied by 00 5 (a) By division, 6 D 0.5, hence 5 corresponds to ð 00%, i.e..5% (b) Similarly, 5 fraction. D.4 when expressed as a decimal Hence D.4 ð 00% D 40% 5 Problem 6. It takes 50 minutes to machine a certain part. Using a new type of tool, the time can be reduced by 5%. Calculate the new time taken. 5% of 50 minutes D ð 50 D D 7.5 minutes

23 Basic Engineering Mathematics Hence the new time taken is D 4.5 minutes. Alternatively, if the time is reduced by 5%, then it now takes 85% of the original time, i.e. 85% of 50 D ð50 D 450 D 4.5 minutes, as above. 00 Problem 7. Find.5% of 78.5% of 78 means.5 ð 78, since per cent means per 00 hundred. Hence.5% of 78 D.5 ð 78 D D 47.5 Problem 8. Express 5 minutes as a percentage of hours, correct to the nearest %. Working in minute units, hours D 0 minutes. Hence 5 minutes is 5 ths of hours. 0 5 By cancelling, 0 D 5 4 Expressing 5 4 as a decimal fraction gives 0.08 P Multiplying by 00 to convert the decimal fraction to a percentage gives: 0.08P ð 00 D 0.8P% Thus 5 minutes is % of hours, correct to the nearest %. Problem 9. A German silver alloy consists of 60% copper, 5% zinc and 5% nickel. Determine the masses of the copper, zinc and nickel in a.74 kilogram block of the alloy. Now try the following exercise Exercise 7 Further problems percentages (Answers on page 5). Convert to percentages: (a) (b) 0.74 (c).85. Express as percentages, correct to significant figures: (a) 7 (b) 9 4 (c) 6. Calculate correct to 4 significant figures: (a) 8% of 758 tonnes (b) 47% of 8.4 grams (c) 47% of 4. seconds 4. When 600 bolts are manufactured, 6 are unsatisfactory. Determine the percentage unsatisfactory. 5. Express: (a) 40 kg as a percentage of t (b) 47 s as a percentage of 5 min (c).4 cm as a percentage of.5 m 6. A block of monel alloy consists of 70% nickel and 0% copper. If it contains 88. g of nickel, determine the mass of copper in the block. 7. A drilling machine should be set to 50 rev/min. The nearest speed available on the machine is 68 rev/min. Calculate the percentage overspeed. 8. Two kilograms of a compound contains 0% of element A, 45% of element B and 5% of element C. Determine the masses of the three elements present. 9. 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 % and the mass of cement in a two tonne dry mix, correct to significant figure. By direct proportion: 00% corresponds to.74 kg % corresponds to.74 D kg 00 60% corresponds to 60 ð D.44 kg 5% corresponds to 5 ð D 0.95 kg 5% corresponds to 5 ð D 0.56 kg Thus, the masses of the copper, zinc and nickel are.44 kg, 0.95 kg and 0.56 kg, respectively. (Check:.44 C 0.95 C 0.56 D.74) Assignment This assignment covers the material contained in Chapters and. The marks for each question are shown in brackets at the end of each question.. Evaluate the following: (a) (b) 7 C 5 ł 4

24 Fractions, decimals and percentages (c) 0 5 (d) 5 5 C 5 6 ł 9 C ð 7 (e) C (4). Determine, by long multiplication 7 ð 4 (). Evaluate, by long division 4675 () 4. Find (a) the highest common factor, and (b) the lowest common multiple of the following numbers: (6) 5. Simplify (a) ł (b) ( ( 4 7 ð ) ł C ) C A piece of steel,.69 m long, is cut into three pieces in the ratio to 5 to 6. Determine, in centimeters, the lengths of the three pieces. (4) 7. Evaluate (a) correct to 4 significant figures (b) correct to decimal place (4) 8. Express 7 9 correct to decimal places () Determine, correct to decimal places, 57% of 7.64 g () 0. Express 54.7 mm as a percentage of.5 m, correct to significant figures. () (9)

25 Indices and standard form. Indices The lowest factors of 000 are ð ð ð ð 5 ð 5 ð 5. These factors are written as 4 ð5,whereand5arecalled bases and the numbers 4 and are called indices. When an index is an integer it is called a power. Thus, 4 is called two to the power of four, and has a base of and an index of 4. Similarly, 5 is called five to the power of and has a base of 5 and an index of. Special names may be used when the indices are and, these being called squared and cubed, respectively. Thus 7 is called seven squared and 9 is called nine cubed. When no index is shown, the power is, i.e. means. Reciprocal The reciprocal of a number is when the index is and its value is given by divided by the base. Thus the reciprocal of is and its value is or 0.5. Similarly, the reciprocal of 5 is 5 which means or 0. 5 Square root The square root of a number is when the index is,and the square root of is written as / or p. The value of a square root is the value of the base which when multiplied by itself gives the number. Since ð D 9, then p 9 D. However, ð D 9, so p 9 D. 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 answer to a square root problem. Thus p 9 Dšand 4 / D p 4 Dš, and so on. 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. (i) When multiplying two or more numbers having the same base, the indices are added. Thus ð 4 D C4 D 6 (ii) When a number is divided by a number having the same base, the indices are subtracted. Thus 5 D 5 D (iii) When a number which is raised to a power is raised to a further power, the indices are multiplied. Thus 5 D 5ð D 0 (iv) When a number has an index of 0, its value is. Thus 0 D (v) A number raised to a negative power is the reciprocal of that number raised to a positive power. Thus 4 D 4 Similarly, D (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. Thus 8 / D p 8 D D 4 and 5 / D p 5 D p 5 Dš5 (Note that p p ). Worked problems on indices Problem. Evaluate: (a) 5 ð 5,(b) ð 4 ð and (c) ð ð 5

26 Indices and standard form 5 From law (i): (a) 5 ð 5 D 5 C D 5 5 D 5 ð 5 ð 5 ð 5 ð 5 D 5 (b) ð 4 ð D C4C D 7 D ð ð... to7terms D 87 (c) ð ð 5 D CC5 D 8 D 56 Problem. Find the value of: (a) and (b) From law (ii): (a) (b) D 7 5 D 7 D D D 5 D 5 Problem. Evaluate: (a) 5 ð 5 ł 5 4 and (b) ð 5 ł ð From laws (i) and (ii): (a) 5 ð 5 ł 5 4 D 5 ð 5 D 5 C D D D 5 D 5 (b) ð 5 ł ð D ð 5 ð D C5 C D 6 5 D 6 5 D D Problem 4. Simplify: (a) 4 (b) 5, expressing the answers in index form. From law (iii): (a) 4 D ð4 D (b) 5 D ð5 D 0 Problem 5. Evaluate: From the laws of indices: ð ð 0 D 0 ð 06 D 0 4C 0 D D 0 0 D Problem 6. Find the value of (a) ð 4 7 ð and (b) 5 ð 9 From the laws of indices: (a) (b) ð 4 7 ð D C4 7 D 5 7C5 D 7 D 5 D D 5 ð D ð 6 D 9 C9 D D 4 D D 4 8 Problem 7. Evaluate (a) 4 / (b) 6 /4 (c) 7 / (d) 9 / (a) 4 / D p 4 D ± (b) 6 /4 D 4p 6 D D 8 (Note that it does not matter whether the 4th root of 6 is found first or whether 6 cubed is found first the same answer will result.) (c) 7 / D p 7 D D 9 (d) 9 / D 9 D p D / 9 š D ± Now try the following exercise Exercise 8 Further problems on indices (Answers on page 5) In Problems to, simplify the expressions given, expressing the answers in index form and with positive indices:. (a) ð 4 (b) 4 ð 4 ð 4 4. (a) ð ð (b) 7 ð 7 4 ð 7 ð 7. (a) 4 (b) 7 4. (a) 5 6 ł 5 (b) 7 / (a) 7 (b) 6. (a) 5 5 (b) (a) ð (b) 7 ð (a) ð 5 (b) 5 ð

27 6 Basic Engineering Mathematics 9. (a) 9 ð ð 7 (b) 0. (a) 5 (b) ð ð 4 ð 8. (a) 7 ð 7 (b) ð 4 ð 5 7 ð 7 4 ð ð 6. (a) ð ð 4 ð (b) 5 7 ð ð 5. Further worked problems on indices Problem 8. Evaluate ð ð 4 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: ð ð 4 D 4 ð 57 5 D 4 ð 5 7 D ð 5 4 D 54 D 65 D 08 Problem 9. Find the value of ð 5 ð ð 4 ð ð 5 ð ð 4 ð D 4 ð 5 ð 7 ð 4 D ð ð 7 0 D ð ð D 9 D 4 Alternatively, 4.5 ð 8 / ð D [ ] / ð / D ð /5 ð 5 /5 ð D C D 4 D 6 Problem. Evaluate: ð 5 5 C ð 5 4 ð 5 4 Dividing each term by the HCF (i.e. highest common factor) of the three terms, i.e. ð 5,gives: ð 5 5 C ð 5 4 ð 5 4 D Problem. ð 5 5 ð 5 C ð 5 ð 5 4 ð 5 4 ð 5 D ð 5 5 C ð ð 5 4 D 0 ð 5 C ð 5 0 ð 5 D ð 5 C ð 9 ð 5 Find the value of D 8 45 ð ð 5 4 C ð 5 To simplify the arithmetic, each term is divided by the HCF of all the terms, i.e. ð 5. Thus ð ð 5 4 C ð 5 D D D ð 5 5 ð 5 4 ð 5 4 ð 5 C ð 5 ð 5 ð ð 5 4 C ð 5 0 ð 5 ð 5 C ð 5 0 D 5 45 C D 5 48 Problem 0. Evaluate: 4.5 ð 8 / ð /5 7 ð 4 Problem. Simplify, expressing the ð 7 5 ð 5 answer in index form with positive indices. 4.5 D 4 / D p 4 D D 8, 8 / D p 8 D, D 4 Hence /5 D /5 D 5p D D ð 8 / ð D 8 ð /5 4 ð D 6 D 6 4 Since 7 D 7, D and 7 ð 4 ð 7 5 ð 5 D 4 ð ð 5 7 ð D 5 then D 4C ð 5 D 6 U 5 7 C5 7 8

28 Indices and standard form 7 6 ð 9 Problem 4. Simplify expressing 4 ð ð 8 the answer in index form with positive indices. Expressing the numbers in terms of their lowest prime numbers gives: 6 ð 9 4 ð ð 8 D 4 ð ð ð D 8 ð 4 ð ð 6 D 8 ð 4 ð Dividing each term by the HCF (i.e. )gives: 8 ð 4 ð D 6 ð 4 D 6 4 ( ) Problem 5. ( ) 4 ( ð 5 ( ) 5 Simplify ) giving the answer with positive indices. A fraction raised to a power means that both the numerator and the denominator of the fraction are raised to that power, ( ) 4 i.e. D 4 A fraction raised to a negative power has the same value as the inverse of the fraction raised to a positive power. Thus, Similarly, Thus, ( ) D ( ) 5 D 5 5 ( ) D 5 ( ) 4 ( ) ð 5 ( ) D 5 ( ) 5 D 5 4 ð 5 5 D 4 ð 5 ð 5 D ð 5 D 5 D ð 9 D C ð 5 5 U 5 Now try the following exercise Exercise 9 Further problems on indices (Answers on page 5) In Problems and, simplify the expressions given, expressing the answers in index form and with positive indices:. (a) ð ð 4 (b) 7 ð 5 ð 7 4 ð 7. (a) 4 ð 9 (b) 8 ð 5 ð 4 8 ð 4 5 ð 4 ð 9 ( ). Evaluate (a) (b) ( ) 4 / (c) 6 /4 (d) 9 In problems 4 to 0, evaluate the expressions given ð ð 7 4 C ð 7 5. ð 7 5 ð 7 7. ð 5 ð 7 ( ) ( ) ( ) 9. 5 / ð 8 / ð 4 / ð 9 /.4 Standard form ð 5 ð 8 ð 9 4 ð 4 4 ð 6 ( ) 4 4 ( ) 9 A number written with one digit to the left of the decimal point and multiplied by 0 raised to some power is said to be written in standard form. Thus: 587 is written as 5.87 ð 0 in standard form, and is written as 4.5 ð 0 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 0 has a mantissa of 5.8 and an exponent of 0 (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:. ð 0 4 C.7 ð 0 4 D. C.7 ð 0 4 D 6.0 ð 0 4 and 5.9 ð ð 0 D ð 0 D. ð 0

29 8 Basic Engineering Mathematics When the numbers have different exponents, one way of adding or subtracting the numbers is to express one of the numbers in non-standard form, so that both numbers have the same exponent. Thus:. ð 0 4 C.7 ð 0 D. ð 0 4 C 0.7 ð 0 4 Alternatively, D. C 0.7 ð 0 4 D.67 ð 0 4. ð 0 4 C.7 ð 0 D 000 C 700 D D.67 ð 0 4 (ii) The laws of indices are used when multiplying or dividing numbers given in standard form. For example,.5 ð 0 ð 5 ð 0 D.5 ð 5 ð 0 C D.5 ð 0 5 or.5 ð 0 6 Similarly, 6 ð ð 0 D 6.5 ð 04 D 4 ð 0.5 Worked problems on standard form Problem 6. Express in standard form: (a) 8.7 (b) 746 (c) 0.04 For a number to be in standard form, it is expressed with only one digit to the left of the decimal point. Thus: (a) 8.7 must be divided by 0 to achieve one digit to the left of the decimal point and it must also be multiplied by 0 to maintain the equality, i.e. 8.7 D 8.7 ð 0 D.87 U 0 in standard form 0 (b) 746 D ð 000 D.746 U 0 in standard form. (c) 0.04 D 0.04 ð D.4 00 D.4 U 0 in standard form Problem 7. Express the following numbers, which are in standard form, as decimal numbers: (a).75 ð 0 (b) 5.49 ð 0 4 (c) 9.84 ð 0 0 (a).75 ð 0 D D (b) 5.49 ð 0 4 D 5.49 ð D (c) 9.84 ð 0 0 D 9.84 ð D 9.84 (since 0 0 D ) (a) Problem 8. Express in standard form, correct to significant figures: (a) 8 (b) 9 (c) D 0.75, and expressing it in standard form gives: D.75 U 0 (b) 9 D 9. P6 D.97 U 0 in standard form, correct to significant figures. (c) D D 7.4 U 0 in standard form, correct to significant figures. Problem 9. Express the following numbers, given in standard form, as fractions or mixed numbers: (a).5 ð 0 (b) 6.5 ð 0 (c).54 ð 0 (a).5 ð 0 D.5 0 D 5 00 D 4 (b) 6.5 ð 0 D D D 6 (c).54 ð 0 D 5.4 D D 5 5 Now try the following exercise Exercise 0 Further problems on standard form (Answers on page 5) In Problems to 5, express in standard form:. (a) 7.9 (b) 8.4 (c) (a) 748 (b) 70 (c) 748. (a) 0.40 (b) (c) (a) 70. (b) 0.04 (c) (a) (b) 7 (c) 0 (d) 8 5 In Problems 6 and 7, express the numbers given as integers or decimal fractions: 6. (a).0 ð 0 (b) 9.7 ð 0 (c) 5.4 ð 0 4 (d) 7 ð (a).89 ð 0 (b) 6.74 ð 0 (c) 8 ð 0

30 Indices and standard form 9.6 Further worked problems on standard form Problem 0. Find the value of (a) 7.9 ð ð 0 (b) 8. ð 0 C 5.45 ð 0 and (c) 9.9 ð 0 C. ð 0 expressing the answers in standard form. Numbers having the same exponent can be added or subtracted by adding or subtracting the mantissae and keeping the exponent the same. Thus: (a) 7.9 ð ð 0 D ð 0 D.5 U 0 (b) 8. ð 0 C 5.45 ð 0 D 8. C 5.45 ð 0 D.75 ð 0 D.75 U 0 4 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: 9.9 ð 0 C. ð 0 D 9.9 ð 0 C ð 0 D 9.9 C ð 0 D.9 ð 0 D.9 U 0 in standard form. Alternatively, the numbers can be expressed as decimal fractions, giving: 9.9 ð 0 C. ð 0 D 99. C 00 D 9. D.9 U 0 in standard form as obtained previously. This method is often the safest way of doing this type of problem. Problem. Evaluate (a).75 ð 0 6 ð 0 4 and.5 ð 05 (b) 7 ð 0 expressing answers in standard form. (a).75 ð 0 6 ð 0 4 D.75 ð 6 0 C4 D.50 ð 0 7 (b) D.5 U ð ð 0 D.5 7 ð 05 D 0.5 ð 0 D 5 U 0 Now try the following exercise Exercise Further problems on standard form (Answers on page 5) In Problems to 4, find values of the expressions given, stating the answers in standard form:. (a).7 ð 0 C 9.8 ð 0 (b).4 ð 0 C 7. ð 0 (c) 8.44 ð 0.68 ð 0. (a) 4.8 ð 0 C.4 ð 0 (b).4 ð 0. ð 0 4 (c).8 ð 0 C.47 ð 0 C 5.97 ð 0. (a) 4.5 ð 0 ð 0 (b) ð 5.5 ð (a) 6 ð 0 (b).4 ð 0 ð 0 ð ð Write the following statements in standard form. (a) The density of aluminium is 70 kg m (b) Poisson s ratio for gold is 0.44 (c) The impedance of free space is 76.7 (d) The electron rest energy is 0.5 MeV (e) Proton charge-mass ratio is C kg (f) The normal volume of a perfect gas is 0.04 m mol

31 4 Calculations and evaluation of formulae 4. 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 p D.4 is not strictly correct, but p D.4 correct to 4 significant figures is a true statement. (Actually, p D ) (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 49. ð 8.4 ð. calculation. For example, may 6. ð 8. be approximated to 50 ð 0 ð 0 60 ð 40 and then, by 50 ð 0 ð 0 cancelling, 60 ð 40 D 50. An accurate answer somewhere between 45 and 55 could therefore be expected. Certainly an answer around 500 or 5 would not be expected. Actually, by calculator 49. ð 8.4 ð. D 47., correct to 4 significant 6. ð 8. figures. Problem. The area A of a triangle is given by A D bh. The base b when measured is found to be.6 cm, and the perpendicular height h is 7.5 cm. Determine the area of the triangle. Area of triangle D bh D ð.6 ð 7.5 D.5 cm (by calculator). The approximate value is ð ð 8 D cm,sothereare 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 significant figure more than the least accurate number in the data: this is 7.5 cm, so the result should not have more than significant figures Thus area of triangle =.cm Problem. State which type of error has been made in the following statements: (a) 7 ð.49 D 6.9 (b) 6 ð 0.08 ð 7 D 89.6 (c).74 ð D 0.47, correct to 4 decimal places ð 0.05 (d) D 0., correct to significant.89 figures.

32 Calculations and evaluation of formulae (a) 7 ð.49 D (by calculator), hence a rounding-off error has occurred. The answer should have stated: 7 ð.49 D 6.9, correct to 5 significant figures or 6.9, correct to decimal place. (b) 6 ð 0.08 ð 7 D 4 6 ð 8 ð D ð 7 5 D 8.96 D 4 5 D Hence an order of magnitude error has occurred. (c).74ð is approximately equal to ð9ð0, i.e. about 08 ð 0 or Thus a blunder has been made. (d) 9.74 ð (a) ³ 0 ð 5 ð 0 D 5 0 D 8 D 50 ð 0 or 0.5 hence no order of magnitude error has occurred. However, D 0.8 correct to signifi ð cant figures, which equals 0. correct to significant figures. Hence a rounding-off error has occurred. Problem. Without using a calculator, determine an approximate value of (a).7 ð ð ð ð 5.7 about 4 D 4., correct to signifi- (By calculator, cant figures.) (b).9 ð 0.6 ð 7.9. ð 8.76 is approximately equal to.7 ð ð ð 0.6 ð 7.9 (b) ³. ð 8.76 after cancelling i.e..9 ð 0.6 ð 7.9. ð ð 0 0 ð 5, i.e. ð 00 ð 0 0 ð 0 ³ 80 D ð ð 0.9 ð 0.6 ð 7.9 (By calculator, D 75., correct to. ð 8.76 significant figures.) Now try the following exercise Exercise Further problems on errors (Answers on page 5) In Problems to 5 state which type of error, or errors, have been made:. 5 ð 0.06 ð.4 D ð 6.84 D ð D For a gas pv D c. When pressure p D Pa and V D 0.54 m then c D Pa m. 4.6 ð ð 0.74 D 0.5 In Problems 6 to 8, evaluate the expressions approximately, without using a calculator ð ð ð ð.96 ð ð Use of calculator The most modern aid to calculations is the pocket-sized 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: Problem 4. Evaluate the following, correct to 4 significant figures: (a) C 0.07 (b) (c).9 ð (a) 4.786C0.07 D D 4.80, correct to 4 significant figures. (b) D D 5.88, correct to 4 significant figures. (c).9 ð D D 0.847, correct to 4 significant figures.

33 Basic Engineering Mathematics Evaluate the following, correct to 4 deci- Problem 5. mal places: (a) 46. ð 97.7 ð (c) 6.49 ð 0.07 (b) (a) 46. ð 97.7 ð D D 56.65, correct to 4 decimal places. (b) (c) 4.6 D D 0.945, correct to 4 decimal.76 places ð 0.07 D D 0.574, correct to 4 decimal places. Evaluate the following, correct to deci- Problem 6. mal places: (a) (b) (c) Evaluate the following, correct to deci- Problem 8. mal places: (a) (c) (b) ( ).60 C.9 ( ) D D , correct to decimal places. ( ).60 ( ) 5.40 C D D 8.74, correct.9.45 to decimal places D D 0.4, correct to decimal places. (a) 5.7 (b) (c) 4.9 C.97 Problem 9. Evaluate the following, correct to 4 significant figures: (a) (b) (c) D D 0.09, correct to decimal 5.7 places. D D 6.64, correct to decimal places. 4.9 C D D 0.7, correct to.97 decimal places. (a) p 5.46 (b) p 54.6 (c) p 546. (a) p 5.46 D D.7, correct to 4 significant figures. (b) p 54.6 D D 7.9, correct to 4 significant figures. (c) p 546. D D.7, correct to 4 significant figures. Problem 7. Evaluate the following, expressing the answers in standard form, correct to 4 significant figures. (a) (b) C.95 (c) (a) D.040 ð 0 5 D.04 U 0 5, correct to 4 significant figures. (b) C.95 D D ð 0 D U 0, correct to 4 significant figures. (c) D D.0 U 0, correct to 4 significant figures. Evaluate the following, correct to dec- Problem 0. imal places: (a) p (b) p 5.9 p.76 (c).69 ð 0 4 (a) p D D 0.086, correct to decimal places. (b) p 5.9 p.76 D D.68, correct to decimal places. (c).69 ð 0 4 D p 69 D D 7.66, correct to decimal places.

34 Calculations and evaluation of formulae Problem. Evaluate the following, correct to 4 significant figures: (a) 4.7 (b) (c) (a) 4.7 D D 05., correct to 4 significant figures. (b) D D 0.478, correct to 4 significant figures. (c) D D 70.44, correct to 4 significant figures. Problem. Evaluate the following, correct to significant figures: [ 6.09 ] (a) 5. ð p p (b) (c) 7. C 6.48 C.9 4 [ 6.09 ] (a) 5. ð p D D 0.746, correct to 7 significant figures. (b) p 47.9 D D.6, correct to significant figures. (c) 7. C 6.48 C.9 4 D D 0.8, correct to significant figures. Problem. Evaluate the following, expressing the answers in standard form, correct to 4 decimal places: (a) 5.76 ð 0 (b) (.974 ð 0 ð 8.6 ð 0.46 (c).79 ð 0 4 (a) 5.76ð0 D ð0 5 D.679 U 0 5, correct to 4 decimal places. ) 4 Now try the following exercise Exercise Further problems on use of calculator (Answers on page 5) In Problems to, use a calculator to evaluate the quantities shown correct to 4 significant figures:. (a).49 (b) 7.78 (c).4 (d) (a) p 4.75 (b) p 5.46 (c) p 780. (a) (d) p (b) (c) (d).8 In Problems 4 to, use a calculator to evaluate correct to 4 significant figures: 4. (a) 4.7 ð.9 (b) 54. ð (a) 7.8 ð 0.04 ð 9.8 (b) 5.76 ł (a) (a) 7. (b) (b).8 ð (c) (a).6 (b) (c) (a) p 47. (b) p 76 (c) p 0.07 (d) p ( ) 4.68 ð 0.05 ( 0.68 ð (a) (b) ð.89. (a) (b) ð.6. Evaluate correct to decimal places: 9. (a) (b) p 5.98 p Evaluate correct to 4 significant figures: [ 5.6 ] (a) 9. ð p 0.5 ) 4 (b) (.974 ð 0 ð 8.6 ð 0 ) 4 D D U 0, correct to 4 decimal places (b) 6.9 C Evaluate the following, expressing the answers in standard form, correct to decimal places: (c).79 ð 0 4 D D.87U 0, correct to 4 decimal places. (a) 8.9 ð 0 (b) 7.6 ð 0

35 4 Basic Engineering Mathematics 4. Conversion tables and charts 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. Problem 4. Currency exchange rates for five countries are shown in Table 4. Table 4. France D.50 euros Japan D 75 yen Norway D.5 kronor Switzerland D.0 francs U.S.A. D.5 dollars ($) Calculate: (a) how many French euros 7.80 will buy, (b) the number of Japanese yen which can be bought for, (c) the pounds sterling which can be exchanged for 64.5 Norwegian kronor, (d) the number of American dollars which can be purchased for 90, and (e) the pounds sterling which can be exchanged for 794 Swiss francs. (a) D.50 euros, hence 7.80 D 7.80 ð.50 euros D 4.70 euros (b) D 75 yen, hence D ð 75 yen D 405 yen (c) D.5 kronor, hence 64.5 kronor D D 570 (d) D.5 dollars, hence 90 D 90 ð.5 dollars D $6.80 (e) D.0 Swiss francs, hence 794 franc D D 70 Problem 5. Some approximate imperial to metric conversions are shown in Table 4. Table 4. length inch D.54 cm mile D.6 km weight.lbd kg lbd 6 oz capacity.76 pints D litre 8 pints D gallon 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 00 km, (d) the number of kilograms in 0 pounds weight, (e) the number of pounds and ounces in 4 kilograms (correct to the nearest ounce), (f) the number of litres in 5 gallons, and (g) the number of gallons in 40 litres. (a) 9.5 inches D 9.5 ð.54 cm D 4. cm 4. cm D 4. ð 0 mm D 4.mm (b) 50 m.p.h. D 50 ð.6 km/h D 80.5km=h (c) 00 km D 00 miles D 86. miles.6 (d) 0 lb D 0 kg D.64 kg. (e) 4 kg D 4 ð.lbd 9.4lb 0.4lb D 0.4 ð 6 oz D 6.4oz D 6 oz, correct to the nearest ounce. Thus 4 kg D 9 lb 6 oz, correct to the nearest ounce. (f) 5 gallons D 5 ð 8 pints D 0 pints 0 pints D 0 litres D 68.8 litres.76 (g) 40 litres D 40 ð.76 pints D 70.4 pints 70.4 pints D 70.4 gallons D 8.8 gallons 8 Now try the following exercise Exercise 4 Further problems conversion tables and charts (Answers on page 54). Currency exchange rates listed in a newspaper included the following: Italy D.5 euro Japan D 80 yen Australia D.80 dollars Canada D $.5 Sweden D.90 kronor Calculate (a) how many Italian euros.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 yen, (d) the pounds sterling which can be exchanged for 75.4 Swedish kronor, and (e) the Australian dollars which can be bought for 55

36 Calculations and evaluation of formulae 5. Below is a list of some metric to imperial conversions. Length Weight Capacity.54 cm D inch.6 km D mile kg D.lb lbd 6 ounces litre D.76 pints 8 pints D gallon Use the list to determine (a) the number of millimetres in 5 inches, (b) a speed of 5 mph in km/h, (c) the number of kilometres in 5 miles, (d) the number of pounds and ounces in 4 kg (correct to the nearest ounce), (e) the number of kilograms in 5 lb, (f) the number of litres in gallons and (g) the number of gallons in 5 litres.. Deduce the following information from the BR train timetable shown in Table 4.: (a) At what time should a man catch a train at Mossley Hill to enable him to be in Manchester Piccadilly by 8.5 a.m.? Table 4. Liverpool, Hunt s Cross and Warrington! Manchester Reproduced with permission of British Rail

37 6 Basic Engineering Mathematics (b) A girl leaves Hunts Cross at 8.7 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 0 minutes to walk to his work from Trafford Park station. What time train should he catch from Edge Hill? Problem 9. The area, A, of a circle is given by A D pr. Determine the area correct to decimal places, given radius r D 5. m. A D pr D p 5. D p 7.59 Hence area, A = 85.9 m, correct to decimal places. 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 6. 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 5.6 A and R D V D IR D Hence voltage V figures. = 79. V, correct to 4 significant Problem 7. The surface area A of a hollow cone is given by A D prl. Determine, correct to decimal place, the surface area when r D.0cmandl D 8.5cm. Problem 0. The power P watts dissipated in an electrical circuit may be expressed by the formula P D V R. Evaluate the power, correct to significant figures, given that V D 7.48 V and R D 6.. P D V R D D Hence power, P = 8.46 W, correct to significant figures. Problem. The volume V cm of a right circular cone is given by V D pr h. Given that r D 4. cm and h D 8.5 cm, find the volume, correct to 4 significant figures. V D pr h D p D p Hence volume, V = 58.8cm, correct to 4 significant figures. A D prl D p cm Hence surface area A = 80.cm, correct to decimal place. Problem 8. Velocity v is given by v D u C at. If u D 9.86 m/s, a D 4.5 m/s and t D 6.84 s, find v, correct to significant figures. Problem. Force F newtons is given by the formula F D Gm m,wherem d and m are masses, d their distance apart and G is a constant. Find the value of the force given that G D 6.67 ð 0, m D 7.6,m D 5.5andd D.6. Express the answer in standard form, correct to significant figures. v D u C at D 9.86 C D 9.86 C 9.07 D 8.9 F D Gm m d D 6.67 ð D D Hence velocity v = 8.9m=s, correct to significant figures. Hence force F =.49 U 0 newtons, correct to significant figures.

38 Calculations and evaluation of formulae 7 Problem. The time of swing t seconds, of a simple l pendulum is given by t D p. Determine the time, g correct to decimal places, given that l D.0 and g D 9.8 l.0 t D p g D p 9.8 D p p.459 D p Hence time t = seconds, correct to decimal places. Problem 4. Resistance, R, varies with temperature according to the formula R D R 0 C at. Evaluate R, correct to significant figures, given R 0 D 4.59, a D and t D 80. R D R 0 C at D 4.59[ C ] D 4.59 C 0.44 D Hence resistance, R = 9.6 Z, correct to significant figures. Now try the following exercise Exercise 5 Further problems on evaluation of formulae (Answers on page 54). The area A of a rectangle is given by the formula A D lb. Evaluate the area when l D.4cm and b D 5.7 cm.. The circumference C of a circle is given by the formula C D pr. Determine the circumference given p D.4 and r D 8.40 mm.. A formula used in connection with gases is R D PV /T. EvaluateR when P D 500, V D 5 and T D The velocity of a body is given by v D u C at. The initial velocity u is measured when time t is 5 seconds and found to be m/s. If the acceleration ais9.8m/s calculate the final velocity v. 5. Calculate the current I in an electrical circuit, when I D V/R amperes when the voltage V is measured and found to be 7. V and the resistance R is Find the distance s, given that s D gt. Time t D 0.0 seconds and acceleration due to gravity g D 9.8 m/s. 7. The energy stored in a capacitor is given by E D CV joules. Determine the energy when capacitance C D 5 ð 0 6 farads and voltage V D 40 V. 8. Find the area A of a triangle, given A D bh, when the base length b is.4 m and the height h is 5.7 m. 9. Resistance R is given by R D R C at. Find R, correct to 4 significant figures, when R D 0, a D and t D Density D mass. Find the density when the mass volume is.46 kg and the volume is 7 cm. Give the answer in units of kg/m.. Velocity D frequencyðwavelength. Find the velocity when the frequency is 85 Hz and the wavelength is 0.54 m.. Evaluate resistance R T,given D C C R T R R R when R D 5.5, R D 7.4 and R D.6.. Find the total cost of 7 calculators costing.65 each and 9 drawing sets costing 6.8 each. force ð distance 4. Power D. Find the power when a time force of 760 N raises an object a distance of 4.7 m in 5 s. 5. The potential difference, V volts, available at battery terminals is given by V D E Ir. EvaluateV when E D 5.6, I D 0.70 and R D Given force F D m v u,findfwhen m D 8., v D.7 andu D The current I amperes flowing in a number of cells is given by I D ne. Evaluate the current when R C nr n D 6, E D.0, R D.80 and r D The time, t seconds, of oscillation for a simple l pendulum is given by t D p. Determine the g time when p D.4, l D 54. and g D Energy, E joules, is given by the formula E D LI. Evaluate the energy when L D 5.5 andid. 0. The current I amperes in an a.c. circuit is given V by I D. Evaluate the current when R C X V D 50, R D.0 andxd6.. Distance s metres is given by the formula s D ut C at.ifud9.50, t D 4.60 and a D.50, evaluate the distance.

39 8 Basic Engineering Mathematics. The area, A, of any triangle is given by A D p [s s a s b s c ]wheres D a C b C c. Evaluate the area given a D.60 cm, b D 4.00 cm and c D 5.0 cm.. Given that a D 0.90, b D 4.86, c D 0.04, d D.8 anded0.650, evaluate v given that ( ab v D c d ) e Assignment This assignment covers the material contained in Chapters and 4. The marks for each question are shown in brackets at the end of each question.. Evaluate the following: ð ð ð 6 ( ) (a) (b) (c) 4 8 ð 4 ( ) (d) 7 / 9 (e) ( ) (4). Express the following in standard form: (a) 6 (b) (c) 45 (6) 5. Determine the value of the following, giving the answer in standard form: (a) 5.9 ð 0 C 7. ð 0 (b).75 ð 0.65 ð 0 (6) 7 ð Is the statement D 4.8 ð 0 correct? If 5.7 not, what should it be? () 5. Evaluate the following, each correct to 4 significant figures: (a) 6. (b) (c) p (6) Evaluate the following, each correct to decimal places: ( 6. ) ð 0.56 (a) 7.8 ð.8 ( 4.69 ) (b) p (7) 7.4 ð If.6km D mile, determine the speed of 45 miles/hour in kilometres per hour. () 8. The area A of a circle is given by A D pr.findthe area of a circle of radius r D.7 cm, correct to decimal places. () 9. Evaluate B, correct to significant figures, when W D 7.0, v D 0.0 andgd9.8, given that B D Wv g ()

40 5 Computer numbering systems 5. Binary numbers 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,,,, 4, 5, 6, 7, 8 and 9) and is said to have a radix or base of 0. The binary system of numbers has a radix of and uses only the digits 0 and. 5. Conversion of binary to denary The denary number 4.5 is equivalent to ð 0 C ð 0 C 4 ð 0 0 C 5 ð 0 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, so 0. is equivalent to: ð C ð C 0 ð C ð 0 C ð Thus the denary number equivalent to the binary number 0. is 8 C 4 C 0 C C, that is.5 i.e. 0. =.5 0, the suffixes and 0 denoting binary and denary systems of numbers respectively. Problem. Convert 0 to a denary number. From above: 0 D ð 4 C ð C 0 ð C ð C ð 0 D 6 C 8 C 0 C C D 7 0 Problem. Convert 0.0 to a decimal fraction. 0.0 D ð C 0 ð C ð C ð 4 Problem. D ð C 0 ð C ð C ð 4 D C 8 C 6 D 0.5 C 0.5 C D Convert 0.00 to a denary number D ð C 0 ð C ð 0 C 0 ð C ð C 0 ð C ð 4 D 4 C 0 C C 0 C 0.5 C 0 C D Now try the following exercise Exercise 6 Further problems on conversion of binary to denary numbers (Answers on page 54) In Problems to 4, convert the binary numbers given to denary numbers.. (a) 0 (b) 0 (c) 0 (d) 00. (a) 00 (b) 00 (c) 00 (d) 00

41 0 Basic Engineering Mathematics. (a) 0.0 (b) 0.00 (c) 0.00 (d) (a) 00. (b) 0.0 (c) 00.0 (d) Conversion of denary to binary An integer denary number can be converted to a corresponding binary number by repeatedly dividing by and noting the remainder at each stage, as shown below for Remainder (most significant bit) 0 0 (least 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 9 0 = 00 The fractional part of a denary number can be converted to a binary number by repeatedly multiplying by, as shown below for the fraction D Remainder Thus 47 0 = 0 Problem 5. 0 Convert to a binary number. From above, repeatedly multiplying by gives: D D D D D. 0 i.e = D Problem 6. Convert to a binary number D. 000 (most significant bit). 0 (least significant bit) For fractions, the most significant bit of the result is the top bit obtained from the integer part of multiplication by. The least significant bit of the result is the bottom bit obtained from the integer part of multiplication by. Thus = 0.0 Problem 4. Convert 47 0 to a binary number. From above, repeatedly dividing by and noting the remainder gives: The integer part is repeatedly divided by, giving: 58 Remainder

42 Computer numbering systems The fractional part is repeatedly multiplied by giving: 0.5 D D D D Thus = Now try the following exercise Exercise 7 Further problems on conversion of denary to binary numbers (Answers on page 54) In Problems to 4, convert the denary numbers given to binary numbers.. (a) 5 (b) 5 (c) 9 (d) 9. (a) (b) 4 (c) 57 (d) 6. (a) 0.5 (b) (c) 0.85 (d) (a) (b) 0.85 (c) (d) Conversion of denary to binary via octal For denary integers containing several digits, repeatedly dividing by can be a lengthy process. In this case, it is usually easier to convert a denary number to a binary number via the octal system of numbers. This system has a radix of 8, using the digits 0,,,, 4, 5, 6 and 7. The denary number equivalent to the octal number 47 8 is 4 ð 8 C ð 8 C ð 8 C 7 ð 8 0 i.e. 4 ð 5 C ð 64 C ð 8 C 7 ð or 55 0 An integer denary 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 Remainder Thus 49 0 D The fractional part of a denary number can be converted to an octal number by repeatedly multiplying by 8, as shown below for the fraction D D For fractions, the most significant bit is the top integer obtained by multiplication of the denary fraction by 8, thus D The natural binary code for digits 0 to 7 is shown in Table 5., and an octal number can be converted to a binary number by writing down the three bits corresponding to the octal digit. Table 5. Octal digit Thus 47 8 D 00 0 Natural binary number and D The 0 on the extreme left does not signify anything, thus D Conversion of denary to binary via octal is demonstrated in the following worked problems. Problem 7. Convert 74 0 to a binary number, via octal. Dividing repeatedly by 8, and noting the remainder gives: 8 74 Remainder

43 Basic Engineering Mathematics From Table 5., 70 8 D i.e = Problem 8. octal. Convert to a binary number, via Multiplying repeatedly by 8, and noting the integer values, gives: D D Thus D From Table 5., D i.e = 0.00 Problem 9. via octal. Convert to a binary number, The integer part is repeatedly divided by 8, noting the remainder, giving: 8 56 Remainder This octal number is converted to a binary number, (see Table 5.) D i.e D The fractional part is repeatedly multiplied by 8, and noting the integer part, giving: D D This octal fraction is converted to a binary number, (see Table 5.) D i.e D 0. 0 Thus, = Problem 0. Convert to a denary number via octal. Grouping the binary number in three s from the binary point gives: Using Table 5. to convert this binary number to an octal number gives: and D ð 8 C 6 ð 8 C ð 8 0 C 4 ð 8 C ð 8 D 9 C 48 C C 0.5 C 0.05 D Now try the following exercise Exercise 8 Further problems on conversion between denary and binary numbers via octal (Answers on page 54) In Problems to, convert the denary numbers given to binary numbers, via octal.. (a) 4 (b) 57 (c) 65. (a) (b) (c) (a) (b) (c) Convert the following binary numbers to denary numbers via octal: (a).0 (b) (c) Hexadecimal numbers The complexity of computers requires higher order numbering systems such as octal (base 8) and hexadecimal (base 6) which are merely extensions of the binary system. A hexadecimal numbering system has a radix of 6 and uses the following 6 distinct digits: 0,,,, 4, 5, 6, 7, 8, 9, A, B, C, D, EandF A corresponds to 0 in the denary system, B to, C to, and so on. To convert from hexadecimal to decimal: For example A 6 D ð 6 C A ð 6 0 i.e. A 6 D 6 0 D ð 6 C 0 ð D 6 C 0 D 6 Similarly, E 6 D ð 6 C E ð 6 0 D ð 6 C 4 ð 6 0 D C 4 D 46 0

44 Computer numbering systems and BF 6 D ð 6 C B ð 6 C F ð 6 0 D ð 6 C ð 6 C 5 ð 6 0 D 56 C 76 C 5 D Table 5. compares decimal, binary, octal and hexadecimal numbers and shows, for example, that 0 D 0 D 7 8 D 7 6 Table 5. Decimal Binary Octal Hexadecimal A 0 B 00 4 C 0 5 D E 5 7 F A 7 0 B C D E 7 F (b) F 6 D ð 6 C F ð 6 0 D ð 6 C 5 ð D 48 C 5 D 6 Thus F 6 = 6 0 Problem. Convert the following hexadecimal numbers into their decimal equivalents: (a) C9 6 (b) BD 6 (a) C9 6 D C ð 6 C 9 ð 6 0 D ð 6 C 9 ð D 9 C 9 D 0 Thus C9 6 = 0 0 (b) BD 6 D B ð 6 C D ð 6 0 D ð 6 C ð D 76 C D 89 Thus BD 6 = 89 0 Problem. Convert A4E 6 into a denary number. A4E 6 D ð 6 C A ð 6 C 4 ð 6 C E ð 6 0 D ð 6 C 0 ð 6 C 4 ð 6 C 4 ð 6 0 D ð 4096 C 0 ð 56 C 4 ð 6 C 4 ð D 4096 C 560 C 64 C 4 D 674 Thus A4E 6 = To convert from decimal to hexadecimal: This is achieved by repeatedly dividing by 6 and noting the remainder at each stage, as shown below for Remainder 6 0 A most significant bit A least significant bit Hence 6 0 = A 6 Similarly, for Problem. Convert the following hexadecimal numbers into their decimal equivalents: (a) 7A 6 (b) F 6 (a) 7A 6 D 7 ð 6 C A ð 6 0 D 7 ð 6 C 0 ð D C 0 D Thus 7A 6 = Remainder F 6 6 B Thus = BF 6 B F

45 4 Basic Engineering Mathematics Problem 4. Convert the following decimal numbers into their hexadecimal equivalents: (a) 7 0 (b) 08 0 (a) 6 7 Remainder 6 5 = = 6 most significant bit 5 least significant bit Hence 7 0 = 5 6 (b) Remainder D C D 6 6 Hence 08 0 = 6C 6 6 C Problem 5. Convert the following decimal numbers into their hexadecimal equivalents: (a) 6 0 (b) 9 0 Problem 6. Convert the following binary numbers into their hexadecimal equivalents: (a) 000 (b) 00 (a) Grouping bits in fours from the right gives: 0 00 and assigning hexadecimal symbols to each group gives: D 6 from Table 5. Thus, 000 = D6 6 (b) Grouping bits in fours from the right gives: 00 0 and assigning hexadecimal symbols to each group gives: 6 7 from Table 5. Thus, 00 = 67 6 (a) Remainder D D A 6 Hence 6 0 = A 6 (b) 6 9 Remainder D F D E 6 Hence 9 0 = EF 6 A E F Problem 7. Convert the following binary numbers into their hexadecimal equivalents: (a) 00 (b) 000 (a) Grouping bits in fours from the right gives: 00 and assigning hexadecimal symbols to each group gives: C F from Table 5. Thus, 00 = CF 6 To convert from binary to hexadecimal: The binary bits are arranged in groups of four, starting from right to left, and a hexadecimal symbol is assigned to each group. For example, the binary number is initially grouped in fours as: and a hexadecimal symbol assigned E 7 A 9 to each group from Table 5. Hence = E7A9 6 To convert from hexadecimal to binary: The above procedure is reversed, thus, for example, 6CF 6 D from Table 5. i.e. 6CF 6 D (b) Grouping bits in fours from the right gives: and assigning hexadecimal symbols to each group gives: 9 E from Table 5. Thus, 000 = 9E 6 Problem 8. Convert the following hexadecimal numbers into their binary equivalents: (a) F 6 (b) A6 6 (a) Spacing out hexadecimal digits gives: F and converting each into binary gives: 00 from Table 5. Thus, F 6 =

46 Computer numbering systems 5 (b) Spacing out hexadecimal digits gives: A 6 and converting each into binary gives: from Table 5. Thus, A6 6 = 0000 Problem 9. Convert the following hexadecimal numbers into their binary equivalents: (a) 7B 6 (b) 7D 6 (a) Spacing out hexadecimal digits gives: 7 B and converting each into binary gives: 0 0 from Table 5. Thus, 7B 6 = 0 (b) Spacing out hexadecimal digits gives: 7 D and converting each into binary gives: from Table 5. Thus, 7D 6 = 00 Now try the following exercise Exercise 9 Further problems on hexadecimal numbers (Answers on page 54) In Problems to 4, convert the given hexadecimal numbers into their decimal equivalents.. E7 6. C F 6 In Problems 5 to 8, convert the given decimal numbers into their hexadecimal equivalents In Problems 9 to, convert the given binary numbers into their hexadecimal equivalents In Problems to 6, convert the given hexadecimal numbers into their binary equivalents ED F 6 6. A 6

47 6 Algebra 6. 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, wherea 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: (i) a C b C c D a C b C c (ii) a bc D ab c (iii) a C b D b C a (iv) ab D ba (v) a b C c D ab C ac a C b (vi) D a c c C b c (vii) a C b c C d D ac C ad C bc C bd Problem. Evaluate ab bc C abc when a D, b D andc D 5 Replacing a, b and c with their numerical values gives: ab bc C abc D ð ð ð ð 5 C ð ð 5 D 9 0 C 5 D 6 Problem. Find the value of 4p qr, given that p D, q D and r D Replacing p, q and r with their numerical values gives: 4p qr D 4 ( )( Problem. ) D 4 ð ð ð ð ð ð D 7 Find the sum of x, x, x and 7x The sum of the positive terms is: x C x D 5x The sum of the negative terms is: x C 7x D 8x Taking the sum of the negative terms from the sum of the positive terms gives: 5x 8x D x Alternatively x C x C x C 7x D x C x x 7x D x Problem 4. and 6c Find the sum of 4a, b, c, a, 5b Each symbol must be dealt with individually. For the a terms: C4a a D a For the b terms: Cb 5b D b For the c terms: Cc C 6c D 7c Thus 4a C b C c C a C 5b C 6c D 4a C b C c a 5b C 6c D a b + 7c Problem 5. Find the sum of 5a b, a C c, 4b 5d and b a C d 4c

48 Algebra 7 The algebraic expressions may be tabulated as shown below, forming columns for the a s, b s, c s and d s. Thus: Adding gives: C5a b Ca C c C 4b 5d a C b 4c C d 6a + b c d p ł 8pq means p. This can be reduced by cancelling as 8pq in arithmetic. Thus: p 8pq D ð p 8 ð p ð q D 4q Problem 0. Divide x C x byx Problem 6. Subtract x C y 4z from x y C 5z x C x iscalledthedividend and x thedivisor. The usual layout is shown below with the dividend and divisor both arranged in descending powers of the symbols. Subtracting gives: x y C 5z x C y 4z x 5y + 9z (Note that C5z 4z DC5z C 4z D 9z) An alternative method of subtracting algebraic expressions is to change the signs of the bottom line and add. Hence: Adding gives: Problem 7. x y C 5z x y C 4z x 5y + 9z Multiply a C b 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, andthetwo results are added. The usual layout is shown below. a C b a C b Multiplying by a! a C ab Multiplying by b! C ab C b Adding gives: a + 5ab + b Problem 8. Multiply x y C 4xy by x 5y x y C 4xy Multiplying x 5y by x! 6x 4xy C 8x y Multiplying by 5y! 0xy 5xy C 0y Adding gives: 6x 4xy + 8x y 5xy + 0y Problem 9. Simplify p ł 8pq x C ) x x C x x x x x 0 0 Dividing the first term of the dividend by the first term of the divisor, i.e. x gives x, which is put above the first term of x the dividend as shown. The divisor is then multiplied by x, i.e. x x D x x, which is placed under the dividend as shown. Subtracting gives x. The process is then repeated, i.e. the first term of the divisor, x, is divided into x, giving, which is placed above the dividend as shown. Then x D x, which is placed under the x. The remainder, on subtraction, is zero, which completes the process. Thus x C x ł x D (x + ) (A check can be made on this answer by multiplying x C by x, which equals x C x ) Problem. Simplify x C y ) x C y 4 7 x xy C y x C y x C 0 C 0 C y x C x y x y C y x y xy xy C y xy C y 0 0 () x into x goes x.putx above x () x x C y D x C x y () Subtract (4) x into x y goes xy. Put xy above dividend (5) xy x C y D x y xy (6) Subtract (7) x into xy goes y.puty above dividend (8) y x C y D xy C y (9) Subtract

49 8 Basic Engineering Mathematics Thus x + y x + y = x xy + y The zeros shown in the dividend are not normally shown, but are included to clarify the subtraction process and to keep similar terms in their respective columns. Problem. Divide 4a 6a b C 5b by a b a ab b ) a b 4a 6a b C 5b 4a a b 4a b C 5b 4a b C ab ab C 5b ab C b 4b Thus 4a 6a b+5b = a ab b, remainder 4b. a b Alternatively, the answer may be expressed as a ab b + 4b a b Now try the following exercise Exercise 0 Further problems on basic operations (Answers on page 54). Find the value of xy C yz xyz, whenx D, y D andz D 4. Evaluate pq r when p D, q D andr D. Find the sum of a, a, 6a, 5a and 4a 4. Simplify 4 c C c 6 c c 5. Find the sum of x, y, 5x, z, y, 4 x 6. Add together acbc4c, 5a bcc, 4a 5b 6c 7. Add together d C 4e, e C f, d f, 4d e C f e 8. From 4x y C z subtract x C y z 9. Subtract a b C c from b 4a c 0. Multiply x C y by x y. Multiply a 5b C c by a C b. Simplify (i) a ł 9ab (ii) 4a b ł a. Divide x C xy y by x C y 4. Divide p pq q by p q 6. Laws of Indices The laws of indices are: (i) a m ð a n D a mcn (ii) a m a D n am n (iii) a m n D a mn (iv) a m/n D np a m (v) a n D a n (vi) a 0 D Problem. Simplify a b c ð ab c 5 Grouping like terms gives: a ð a ð b ð b ð c ð c 5 Using the first law of indices gives: a C ð b C ð c C5 i.e. a 4 ð b 5 ð c 6 D a 4 b 5 c 6 Problem 4. Using the first law of indices, Simplify a / b c ð a /6 b / c a / b c ð a /6 b / c D a / C /6 ð b C / ð c C Problem 5. Simplify a b c 4 a D, b D 8 and c D Using the second law of indices, and Thus a a D a D a, c 4 c D c4 D c 6 a b c 4 abc D a bc 6 When a D, b D and c D, 8 a bc 6 D ( 8 D a = b 5= c abc and evaluate when b b D b D b ) 6 D 9 ( 8 ) 64 D 7

50 Algebra 9 Problem 6. Simplify p/ q r / and evaluate when p /4 q / r/6 p D 6, q D 9andr D 4, taking positive roots only. Using the second law of indices gives: p / /4 q / r / /6 D p =4 q = r = When p D 6, q D 9andr D 4, p /4 q / r / D 6 /4 9 / 4 / D 4p 6 p 9 p 4 D D 08 Problem 7. Simplify x y C xy Algebraic expressions of the form a C b can be split into c a c C b c. Thus x y C xy D x y xy xy C xy xy D x y C x y xy D xy + y (since x 0 D, from the sixth law of indices.) Problem 8. Simplify x y xy 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: x y xy xy D Problem 9. xy xy x y xy Simplify xy xy D x y a b ab a / b The HCF of each of the three terms is a / b. Dividing each term by a / b gives: a b ab a / b D a b a / b ab a / b a/ b a / b D a = a = b b Problem 0. Simplify p / q 4 Using the third law of indices gives: p ð / q ð4 D p (=) q 8 Problem. Simplify mn m / n /4 4 The brackets indicate that each letter in the bracket must be raised to the power outside. Using the third law of indices gives: mn m / n /4 D m ð n ð 4 m / ð4 n D m n 6 /4 ð4 m n Using the second law of indices gives: m n 6 m n D m n 6 D mn 5 Problem. Simplify a p b p c 5 p a p b c and evaluate when a D 4, b D 6andc D Using the fourth law of indices, the expression can be written as: a b / c 5/ a / b / c Using the first law of indices gives: a C / b / C / c 5/ C D a 7/ b 7/6 c / It is usual to express the answer in the same form as the question. Hence a 7/ b 7/6 c / D p a 7 6p b 7p c When a D, b D 64 and c D, 4 p a 7 6p b 7p ( c D ) 7 p p D ( ) 7 7 D Problem. Simplify a b a 4 b, expressing the answer with positive indices only. Using the first law of indices gives: a C 4 b C D a b Using the fifth law of indices gives: a b D a C b D C ab d e f / Problem 4. Simplify expressing the d / ef 5/ answer with positive indices only.

51 40 Basic Engineering Mathematics Using the third law of indices gives: d e f / d / ef 5/ D d e f / d e f 5 Using the second law of indices gives: d e f / 5 D d e 0 f 9/ Problem 5. D d f 9/ since e 0 D from the sixth law of indices D df 9= from the fifth law of indices Simplify x y / p x y x 5 y / Using the third and fourth laws of indices gives: x y / p x y x 5 y / D x y / x / y / x 5/ y / Using the first and second laws of indices gives: x C / 5/ y / C / / D x 0 y / D y = or y = or from the fifth and sixth laws of indices. Now try the following exercise Exercise p y Further problems on laws of indices (Answers on page 55). Simplify x y z x yz and evaluate when x D, y D andzd. Simplify a / bc a / b / c and evaluate when a D, b D 4andcD. Simplify a5 bc a b c and evaluate when a D, b D and c D In Problems 4 to, simplify the given expressions: 0. e f e f 5, expressing the answer with positive indices only. a b / c / ab / p a p bc 6. Brackets and factorization 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 ab C ac D a b C c which is simply the reverse of law (v) of algebra on page 6, and 6px C py 4pz D p x C y z This process is called factorization. Problem 6. Remove the brackets and simplify the expression a C b C b C c 4 c C d Both b and c in the second bracket have to be multiplied by, and c and d in the third bracket by 4 when the brackets are removed. Thus: a C b C b C c 4 c C d D a C b C b C c 4c 4d Collecting similar terms together gives: a + b c 4d Problem 7. Simplify a a ab a b C a When the brackets are removed, both a and ab in the first bracket must be multiplied by and both b and a in the second bracket by a. Thus a a ab a b C a D a a C ab ab a Collecting similar terms together gives: a ab. Since a is a common factor, the answer can be expressed as a( + b) x /5 y / z / x / y / z /6 5. p q pq p q abc a b c a b C a b a b 7. a / b c / 9. p x y z p x y z Problem 8. Simplify a C b a b Each term in the second bracket has to be multiplied by each term in the first bracket. Thus: a C b a b D a a b C b a b D a ab C ab b D a b

52 Algebra 4 Alternatively a C b a b Multiplying by a! a C ab Multiplying by b! ab b Adding gives: a b Problem 9. Remove the brackets from the expression x y x C y x y x C y D x x C y y x C y Problem 0. Simplify x y D x + xy 6xy y x y D x y x y D x x y y x y D 4x 6xy 6xy C 9y D 4x xy + 9y Alternatively, x y x y Multiplying by x! 4x 6xy Multiplying by y! 6xy C 9y Adding gives: 4x xy C 9y Problem. Remove the brackets from the expression [p q C r C q ] Collecting together similar terms gives: a [fa bgc4a] Removing the curly brackets gives: a [9a 6b C 4a] Collecting together similar terms gives: a [a 6b] Removing the outer brackets gives: a a C 6b i.e. a + 6b or 6b a (see law (iii), page 6) Problem. Removing brackets gives: Simplify x x 4y x 4x C y x 4xy 8x xy Collecting together similar terms gives: 6x 6xy Factorizing gives: 6x(x + y) since 6x is common to both terms. Problem 4. Factorize (a) xy xz (b) 4a C 6ab (c) a b 6ab C 5ab For each part of this problem, the HCF of the terms will become one of the factors. Thus: (a) xy xz D x(y z ) (b) 4a C 6ab D 4a(a + 4b ) (c) a b 6ab C 5ab D ab(a b + 5) In this problem there are two brackets and the inner one is removed first. Problem 5. Factorize ax ay C bx by Hence [p q C r C q ] D [p q r C q ] D p 6q 6r + q Problem. Remove the brackets and simplify the expression: a [f 4a b 5 a C b gc4a] Removing the innermost brackets gives: a [f8a b 5a 0bgC4a] The first two terms have a common factor of a and the last two terms a common factor of b. Thus: ax ay C bx by D a x y C b x y The two newly formed terms have a common factor of x y. Thus: a x y C b x y D (x y)(a + b) Problem 6. Factorize ax ay C bx by

53 4 Basic Engineering Mathematics a is a common factor of the first two terms and b a common factor of the last two terms. Thus: ax ay C bx by D a x y C b x y x y is now a common factor thus: a x y C b x y D (x y)(a + b) Alternatively, x is a common factor of the original first and third terms and y is a common factor of the second and fourth terms. Thus: ax ay C bx by D x a C b y a C b a C b is now a common factor thus: as before x a C b y a C b D (a + b)(x y). 5[a a b a b ]. 4p [ 5p q p C q C q] In Problems 4 to 7, factorize: 4. (i) pb C pc (ii) q C 8qn 5. (i) a b 8ab (ii) xy C 6x y C 8x y 6. (i) ay C by C a C b (ii) px C qx C py C qy 7. (i) ax ay C bx by (ii) ax C ay 4bx 6by 6.4 Fundamental laws and precedence The laws of precedence which apply to arithmetic also apply to algebraic expressions. The order is Brackets, Of, Division, Multiplication, Addition and Subtraction (i.e. BODMAS) Problem 7. Factorize x C x x Problem 8. Simplify a C 5a ð a a x is a common factor of the first two terms, thus: x C x x D x x C x is a common factor of the last two terms, thus: x x C x D x x C x C Multiplication is performed before addition and subtraction thus: a C 5a ð a a D a C 5a a D a C 5a D a( + 5a) x C is now a common factor, thus: Problem 9. Simplify a C 5a ð a a x x C x C D (x + )(x ) Now try the following exercise Exercise Further problems on brackets and factorization (Answers on page 55) In Problems to, remove the brackets and simplify where possible:. x C y C x y. 4a C y a y. x y y x 4. x x xy x y x 5. p C q r 4 r q C p C p 6. a C b a C b 7. p C q p q 8. (i) x y (ii) a b 9. a b C c C 4c a b 0. x C [y x C y ]. a C [a a ] The order of precedence is brackets, multiplication, then subtraction. Hence a C 5a ð a a D 6a ð a a D a a D a(4a ) Problem 40. Simplify a C 5a ð a a The order of precedence is brackets, multiplication, then subtraction. Hence a C 5a ð a a D a C 5a ð a D a C 5a Problem 4. D a 5a D a( 5a) Simplify a ł 5a C a a The order of precedence is division, then addition and subtraction. Hence a ł 5a C a a D a C a a 5a D 5 C a a D 5 a

54 Algebra 4 Problem 4. Simplify a ł 5a C a a The order of precedence is brackets, division and subtraction. Hence a ł 5a C a a D a ł 7a a D a 7a a D 7 a Now c c D Multiplying numerator and denominator by gives ð ð i.e. Hence: c C c ð 4c C c c D c C c ð 4c Problem 4. Simplify a ł 5a C a a D c + 8c or The order of precedence is brackets, then division. Hence: a ł 5a C a a D a ł 4a D a 4a D 4 Problem 47. c( + 8c) Simplify c C c 4c C c ł 5c 8c Problem 44. Simplify c C c ð 4c C c ł 5c 8c The order of precedence is division, multiplication, addition and subtraction. Hence: c C c ð 4c C c ł 5c 8c D c C c ð 4c C c 8c 5c D c C 8c C 8c 5 The order of precedence is brackets, division and multiplication. Hence c C c 4c C c ł 5c 8c D 5c ð 5c ł c D 5c ð 5c c D 5c ð 5 D 5 c D 8c 5c + 5 or Problem 48. Simplify a ł 4a C 5 ð 6 a Problem 45. c(8c 5)+ 5 Simplify c C c 4c C c ł 5c 8c The order of precedence is brackets, division, multiplication, addition and subtraction. Hence c C c 4c C c ł 5c 8c D 5c ð 4c C c ł 5c 8c D 5c ð 4c C c 8c 5c D 0c + 5 8c or The bracket around the a shows that both a and have to be divided by 4a, and to remove the bracket the expression is written in fraction form. Hence: a a ł 4a C 5 ð 6 a D C 5 ð 6 a 4a a D C 0 a 4a D a 4a C 0 a 4a D C 0 a 4a D 0 4a a 4c(5c )+ 5 Problem 46. Simplify c C c ð 4c C c ł 5c 8c The order of precedence is brackets, division, multiplication and addition. Hence: c C c ð 4c C c ł 5c 8c D c C c ð 4c C c ł c D c C c ð 4c C c c Problem 49. Simplify of p C 4p p p Applying BODMAS, the expression becomes of p C 4p ð p and changing of to ð gives: ð p C 4p ð p i.e. p + 8p or p( + 8p)

55 44 Basic Engineering Mathematics Now try the following exercise Exercise Further problems on fundamental laws and precedence (Answers on page 55) In Problems to, simplify:. x ł 4x C 6x. x ł 4x C 6x. a a ð 4a C a 4. a a 4a C a 5. a a 4a C a 6. y C 4 ł 6y C ð 4 5y 7. y C 4 ł 6y C ð 4 5y 8. y C 4 ł 6y C 4 5y 9. ł y C ł y C 0. p pq ð p ł 6q C pq. x C x 4 ł x C. of y C y y y Direct and inverse proportionality An expression such as y D x contains two variables. For every value of x there is a corresponding value of y. The variable x is called the independent variable and y is called the dependent variable. When an increase or decrease in an independent variable leads to an increase or decrease of the same proportion in the dependent variable this is termed direct proportion. If y D x then y is directly proportional to x, whichmaybe written as y / x or y D kx, wherek is called the coefficient of proportionality (in this case, k being equal to ). When an increase in an independent variable leads to a decrease of the same proportion in the dependent variable (or vice versa) this is termed inverse proportion. If y is inversely proportional to x then y / or y D k/x. x Alternatively, k D xy, that is, for inverse proportionality the product of the variables is constant. Examples of laws involving direct and inverse proportional in science include: (i) Hooke s law, which states that within the elastic limit of a material, the strain e produced is directly proportional to the stress, s, producing it, i.e. e / s or e D ks. (ii) Charles s law, which states that for a given mass of gas at constant pressure the volume V is directly proportional to its thermodynamic temperature T, i.e. V / T or V D kt. (iii) Ohm s law, which states that the current I flowing through a fixed resistor is directly proportional to the applied voltage V, i.e.i / V or I D kv. (iv) Boyle s law, which states that for a gas at constant temperature, the volume V of a fixed mass of gas is inversely proportional to its absolute pressure p, i.e. p / /V or p D k/v, i.e.pv D k Problem 50. If y is directly proportional to x and y D.48 when x D 0.4, determine (a) the coefficient of proportionality and (b) the value of y when x D 0.65 (a) y / x, i.e. y D kx. If y D.48 when x D 0.4,.48 D k 0.4 Hence the coefficient of proportionality, k D D 6. (b) y D kx, hence, when x D 0.65, y D = 4.0 Problem 5. Hooke s law states that stress s is directly proportional to strain e within the elastic limit of a material. When, for mild steel, the stress is 5 ð 0 6 pascals, the strain is Determine (a) the coefficient of proportionality and (b) the value of strain when the stress is 8 ð 0 6 pascals. (a) s / e, i.e.s D ke, from which k D s/e. Hence the coefficient of proportionality, 5 ð 06 k D D 00 U 09 pascals (The coefficient of proportionality k in this case is called Young s Modulus of Elasticity) (b) Since s D ke, e D s/k Hence when s D 8 ð 0 6,strain e D 8 ð ð 0 9 D Problem 5. The electrical resistance R of a piece of wire is inversely proportional to the cross-sectional area A. WhenA D 5mm, R D 7.0 ohms. Determine (a) the coefficient of proportionality and (b) the cross-sectional area when the resistance is 4 ohms. (a) R /, i.e. R D k/a or k D RA. Hence, when A R D 7. anda D 5, the coefficient of proportionality, k D 7. 5 D 6 (b) Since k D RA then A D k/r When R D 4, the cross sectional area, A D 6 4 D 9mm Problem 5. Boyle s law states that at constant temperature, the volume V of a fixed mass of gas is inversely proportional to its absolute pressure p. Ifa

56 Algebra 45 gas occupies a volume of 0.08 m at a pressure of.5ð0 6 pascals determine (a) the coefficient of proportionality and (b) the volume if the pressure is changed to 4 ð 0 6 pascals.. (a) V /,i.e.vdk/p or k D pv p Hence the coefficient of proportionality, (b) Volume V D k p k D.5 ð D 0. U 0 6 Now try the following exercise Exercise 4 D 0. ð 06 4 ð 0 6 D 0.0 m Further problems on direct and inverse proportionality (Answers on page 55). If p is directly proportional to q and p D 7.5 when q D.5, determine (a) the constant of proportionality and (b) the value of p when q is 5.. Charles s law states that for a given mass of gas at constant pressure the volume is directly proportional to its thermodynamic temperature. A gas occupies a volume of.5 litres at 00 K. Determine (a) the constant of proportionality, (b) the volume at 40 K and (c) the temperature when the volume is.65 litres.. Ohm s law states that the current flowing in a fixed resistor is directly proportional to the applied voltage. When 0 volts is applied across a resistor the current flowing through the resistor is.4 ð 0 amperes. Determine (a) the constant of proportionality, (b) the current when the voltage is 5 volts and (c) the voltage when the current is.6 ð 0 amperes. 4. If y is inversely proportional to x and y D 5. when x D 0.6, determine (a) the coefficient of proportionality, (b) the value of y when x is.5, and (c) the value of x when y is Boyle s law states that for a gas at constant temperature, the volume of a fixed mass of gas is inversely proportional to its absolute pressure. If a gas occupies a volume of.5m at a pressure of 00 ð 0 pascals, determine (a) the constant of proportionality, (b) the volume when the pressure is 800 ð 0 pascals and (c) the pressure when the volume is.5 m. Assignment This assignment covers the material contained in Chapters 5 and 6. The marks for each question are shown in brackets at the end of each question. Convert the following binary numbers to decimal form: (a) 0 (b) (5). Convert the following decimal number to binary form: (a) 7 (b) (6). Convert the following denary numbers to binary, via octal: (a) 479 (b) (7) 4. Convert (a) 5F 6 into its decimal equivalent (b) 0 into its hexadecimal equivalent (c) 000 into its hexadecimal equivalent (6) 5. Evaluate xy z yz when x D 4, y D andz D () 6. Simplify the following: (a) a C b x y (c) xy C x y xy (b) 8a b p c a p b p c (d) x C 4 ł x C 5 ð 4x (0) 7. Remove the brackets in the following expressions and simplify: (a) b a a b b a b (b) x y (c) 4ab [f 4a b C b a g] (7) 8. Factorize x y C 9xy C 6xy () 9. If x is inversely proportional to y and x D when y D 0.4, determine (a) the value of x when y is, and (b) the value of y when x D (4)

57 7 Simple equations 7. Expressions, equations and identities x 5 is an example of an algebraic expression, whereas x 5 D isanexampleofanequation (i.e. it contains an equals sign). An equation is simply a statement that two quantities are equal. For example, m D 000 mm or F D 9 C C or 5 y D mx C c. An identity is a relationship which is true for all values of the unknown, whereas an equation is only true for particular values of the unknown. For example, x 5 D is an equation, since it is only true when x D, whereas x 8x 5x is an identity since it is true for all values of x. (Note means is identical to ). Simple linear equations (or equations of the first degree) are those in which an unknown quantity is raised only to the power. To solve an equation means to find the value of the unknown. Any arithmetic operation may be applied to an equation as long as the equality of the equation is maintained. 7. Worked problems on simple equations Problem. Solve the equation 4x D 0 Dividing each side of the equation by 4 gives: 4x 4 D 0 4 (Note that the same operation has been applied to both the left-hand side (LHS) and the right-hand side (RHS) of the equation so the equality has been maintained) Cancelling gives: x = 5, which is the solution to the equation. Solutions to simple equations should always be checked and this is accomplished by substituting the solution into the original equation. In this case, LHS D 4 5 D 0 D RHS. Problem. Solve x 5 D 6 The LHS is a fraction and this can be removed ( by ) multiplying x both sides of the equation by 5. Hence 5 D Cancelling gives: x D 0 Dividing both sides of the equation by gives: x D 0 i.e. x = 5 Problem. Solve a 5 D 8 Adding 5 to both sides of the equation gives: a 5 C 5 D 8 C 5 i.e. a = The result of the above procedure is to move the 5 from the LHS of the original equation, across the equals sign, to the RHS, but the sign is changed to C Problem 4. Solve x C D 7 Subtracting from both sides of the equation gives: x C D 7 i.e. x = 4 The result of the above procedure is to move the C from the LHS of the original equation, across the equals sign, to

58 Simple equations 47 the RHS, but the sign is changed to. Thus a term can be moved from one side of an equation to the other as long as a change in sign is made. Problem 5. Solve 6x C D x C 9 In such equations the terms containing x are grouped on one side of the equation and the remaining terms grouped on the other side of the equation. As in Problems and 4, changing from one side of an equation to the other must be accompanied by a change of sign. Thus since 6x C D x C 9 then 6x x D 9 4x D 8 4x 4 D 8 4 i.e. x = Check: LHS of original equation D 6 C D RHS of original equation D C 9 D Hence the solution x D is correct. Problem 6. Solve 4 p D p In order to keep the p term positive the terms in p are moved to the RHS and the constant terms to the LHS. Hence 4 C D p C p 5 D 5p 5 5 D 5p 5 Hence = p or p = Check: LHS D 4 D 4 9 D 5 RHS D D 6 D 5 Hence the solution p D is correct. If, in this example, the unknown quantities had been grouped initially on the LHS instead of the RHS then: i.e. and p p D 4 5p D 5 5p 5 D 5 5 p =, as before It is often easier, however, to work with positive values where possible. Problem 7. Solve x D 9 Removing the bracket gives: x 6 D 9 Rearranging gives: x D 9 C 6 x D 5 x D 5 i.e. x = 5 Check: LHS D 5 D D 9 D RHS Hence the solution x D 5 is correct. Problem 8. Solve 4 r r 4 D r Removing brackets gives: 8r r C 8 D r 9 Rearranging gives: 8r r r D 9 C 8 i.e. r D 6 r D 6 D Check: LHS D D 8 C D 6 RHS D D 5 D 6 Hence the solution r D is correct. Now try the following exercise Exercise 5 Further problems on simple equations (Answers on page 55) Solve the following equations:. x C 5 D 7. 8 t D. c D 4. x D 5x C p D p 6..6x. D 0.9x C a C 6 5a D 0 8. x 5x D x d C d D d C x D 4. 6 D 4 t C. 5 f f C 5 C 5 D 0. x D 4 x 4. 6 y 4 D y 5. g 5 5 D x C D 7 x C 4 x C 5

59 48 Basic Engineering Mathematics 7. 0 C r 7 D 6 r C 8. 8 C 4 x 5 x D 5 x 7. Further worked problems on simple equations Problem 9. Solve x D 4 5 The lowest common multiple (LCM) of the denominators, i.e. the lowest algebraic expression that both x and 5 will divide into, is 5x. Multiplying both sides by 5x gives: ( ) ( ) 4 5x D 5x x 5 Cancelling gives: 5 D 4x 5 4 D 4x 4 i.e. x D 5 or 4 4 Check: LHS D D ( ) 4 D D D 4 5 D RHS 4 4 (Note that when there is only one fraction on each side of an equation, cross-multiplication can be applied. In this example, if x D 4 then 5 D 4x, which is a quicker 5 way of arriving at equation () above. Problem 0. Solve y 5 C 4 C 5 D 0 y The LCM of the denominators is 0. Multiplying each term by 0 gives: ( ) ( ) ( ) ( ) y y 0 C 0 C 0 5 D Cancelling gives: 4 y C 5 C 00 D 0 y i.e. 8y C 5 C 00 D 0y Rearranging gives: 8y C 0y D y D 4 y D 4 D 8 Check: LHS D 5 RHS D 0 D 9 0 C 5 D 4 0 C 4 C 5 D 6 5 C 4 C 5 Hence the solution y D is correct. Problem. Solve D 0 C 9 D 4 0 t D 4 t C 4 By cross-multiplication : t C 4 D 4 t Removing brackets gives: 9t C D 4t 8 Rearranging gives: 9t 4t D 8 i.e. 5t D 0 t D 0 D 4 5 Check: LHS D RHS D 4 D 6 D 4 4 C 4 D 4 C 4 D 4 8 D Hence the solution t D 4 is correct. Problem. Solve p x D [ p x D is not a simple equation since the power of x is i.e. p x D x / ; however, it is included here since it occurs often in practise]. Wherever square root signs are involved with the unknown quantity, both sides of the equation must be squared. Hence p x D i.e. x = 4 Problem. Solve p D 8 To avoid possible errors it is usually best to arrange the term containing the square root on its own. Thus i.e. p d D 8 p d D 4 Squaring both sides gives: d = 6, which may be checked in the original equation.

60 Simple equations 49 Problem 4. Solve ( p b C p b ) D ) To remove the fraction each term is multiplied by p b. Hence: ( p p b C b p b D p b Cancelling gives: p p b C D b Rearranging gives: D p b p b D p b Squaring both sides gives: 9 = b p 9 C Check: LHS D p D C D 6 9 D D RHS Problem 5. Solve x D 5 This problem involves a square term and thus is not a simple equation (it is, in fact, a quadratic equation). However the solution of such an equation is often required and is therefore included here for completeness. Whenever a square of the unknown is involved, the square root of both sides of the equation is taken. Hence p x D p 5 i.e. x D 5 However, x D 5 is also a solution of the equation because 5 ð 5 DC5 Therefore, whenever the square root of a number is required there are always two answers, one positive, the other negative. The solution of x D 5 is thus written as x = ±5 Problem 6. Solve 5 4t D Cross-multiplying gives: 5 D 4t i.e. 45 D 8t 45 8 D t i.e. t D 5.65 Hence t D p 5.65 D ±.7, correct to 4 significant figures. Now try the following exercise Exercise 6 Further problems on simple equations (Answers on page 55) Solve the following equations:. 5 d C D 4. C 4 y D C y C x C D 4. 5 f C 6 f 4 C 5 D 0 5. m 6 4 5m C 4 C m 9 D 5 6. x x 5 D 7. y D C y y a D 8 x C 4 D x 5 y 5 C 7 0 D 5 y 4 a D a C 9. C.. 5. n C 4n D 7 4 t 0 D 6 t C t 5 v v D 6. p t D 9 7. p y D 5 ( ) 8. 4 D C 9. a ( x ) 0. 0 D 5 ( ) y C. D. y 4. D 5 C 8 x x 4 x C 6 D x C 5 p x p x D 6. 6 D t 9 6 a D a 7.4 Practical problems involving simple equations Problem 7. A copper wire has a length l of.5 km, a resistance R of 5 and a resistivity of 7. ð 0 6 mm. Find the cross-sectional area, a, of the wire, given that R D rl/a Since R D rl/a then 5 D.7 ð 0 6 mm 500 ð 0 mm a From the units given, a is measured in mm. Thus 5a D 7. ð 0 6 ð 500 ð 0

61 50 Basic Engineering Mathematics and a D 7. ð 0 6 ð 500 ð 0 5 D 7. ð 500 ð ð 5 D 7. ð 5 0 ð 5 D 5.6 Hence the cross-sectional area of the wire is 5.6 mm Problem 8. A rectangular box with square ends has its length 5 cm greater than its breadth and the total length of its edges is.04 m. Find the width of the box and its volume. Let x cm D width D height of box. Then the length of the box is x C 5 cm. The length of the edges of the box is 4x C 4 x C 5 cm. Hence 04 D 4x C 4 x C 5 04 D 8x C 4x C D x i.e. 44 D x and x D cm Hence the width of the box is cm. Volume of box D length ð width ð height D x C 5 x x D 7 D 888 cm Problem 9. The temperature coefficient of resistance a may be calculated from the formula R t D R 0 C at. Find a given R t D 0.98, R 0 D 0.8 andtd40 Since R t D R 0 C at then 0.98 D 0.8[ C a 40 ] 0.98 D 0.8 C 0.8 a D a 0.8 D a Hence a D 0.8 D Problem 0. The distance s metres travelled in time t seconds is given by the formula s D ut C at,where u is the initial velocity in m/s and a is the acceleration in m/s. Find the acceleration of the body if it travels 68 m in 6 s, with an initial velocity of 0 m/s. s D ut C at,ands D 68, u D 0 and t D 6 Hence 68 D 0 6 C a 6 68 D 60 C 8a D 8a 08 D 8a a D 08 8 D 6 Hence the acceleration of the body is 6 m=s Problem. When three resistors in an electrical circuit are connected in parallel the total resistance R T is given by: R T D R C R C R Find the total resistance when R D 5, R D 0 and R D 0. D R T 5 C 0 C 0 D 6 C C 0 D 0 0 D Taking the reciprocal of both sides gives: Alternatively, if denominators is 0 R T Hence ( ) 0R T R T D R T 5 C 0 C 0 ( ) ( ) D 0R T C 0R T 5 0 ( ) C 0R T 0 Cancelling gives: 0 D 6R T C R T C R T 0 D 0R T R T D 0 D Z, as above 0 Now try the following exercise R T = Z the LCM of the Exercise 7 Practical problems involving simple equations (Answers on page 55). A formula used for calculating resistance of a cable is R D rl /a. Given R D.5, l D 500 and a D ð 0 4 find the value of r.. Force F newtons is given by F D ma, wherem is the mass in kilograms and a is the acceleration in metres per second squared. Find the acceleration when a force of 4 kn is applied to a mass of 500 kg.

62 Simple equations 5. PV D mrt is the characteristic gas equation. Find the value of m when P D 00 ð 0, V D.00, R D 88 and T D When three resistors R, R and R are connected in parallel the total resistance R T is determined from R T D R C R C R (a) Find the total resistance when R D, R D 6 and R D 8. (b) Find the value of R given that R T D, R D 5 and R D Five pens and two rulers cost 94p. If a ruler costs 5p more than a pen, find the cost of each. 6. Ohm s law may be represented by I D V/R, wherei is the current in amperes, V is the voltage in volts and R is the resistance in ohms. A soldering iron takes a current of 0.0 A from a 40 V supply. Find the resistance of the element. 7.5 Further practical problems involving simple equations Problem. The extension x m of an aluminium tie bar of length l m and cross-sectional area A m when carrying a load of F newtons is given by the modulus of elasticity E D Fl/Ax. Find the extension of the tie bar (in mm) if E D 70 ð 0 9 N/m, F D 0 ð 0 6 N, A D 0.m and l D.4m E D Fl/Ax, hence 70 ð 0 9 N m D 0 ð 06 N.4m 0.m x (the unit of x is thus metres) 70 ð 0 9 ð 0. ð x D 0 ð 0 6 ð.4 x D 0 ð 06 ð.4 70 ð 0 9 ð 0. Cancelling gives: x D ð.4 7 ð 00 m D ð.4 ð 000 mm 7 ð 00 Hence the extension of the tie bar, x = 4mm Problem. Power in a d.c. circuit is given by P D V where V is the supply voltage and R is R the circuit resistance. Find the supply voltage if the circuit resistance is.5 and the power measured is 0 W Since P D V R then 0 D V D V i.e. V D 400 Supply voltage, V D p 400 D ±0 V Problem 4. A painter is paid 4.0 per hour for a basic 6 hour week, and overtime is paid at one and a third times this rate. Determine how many hours the painter has to work in a week to earn.80 Basic rate per hour = 4.0; overtime rate per hour D ð 4.0 D 5.60 Let the number of overtime hours worked D x Then C x 5.60 D C 5.60x D x D D 6.60 x D D Thus hours overtime would have to be worked to earn.80 per week. Hence the total number of hours worked is 6 C, i.e. 47 hours. Problem 5. A formula relating initial and final states of pressures, P and P, volumes V and V, and absolute temperatures, T and T, of an ideal gas is P V D P V. Find the value of P given T T P D 00 ð 0, V D.0, V D 0.66, T D 4 and T D 9 Since P V D P V then 00 ð 0.0 T T 4 Cross-multiplying gives: 00 ð D P P D 00 ð Hence P = 60 U 0 or.6 U 0 5 D P Problem 6. The stress f in a material of a thick ( cylinder can be obtained from D ) f C p d D. f p Calculate the stress, given that D D.5, d D 0.75 and p D 800

63 5 Basic Engineering Mathematics ( Since D ) ( f C p d D then.5 ) f C 800 f p 0.75 D f 800 ( ) f C 800 i.e. D f 800 Squaring both sides gives: 4 D f C 800 f f 800 D f C 800 4f 700 D f C 800 4f f D 800 C 700 f D 9000 f D 9000 D 000 Hence stress, f = 000 Problem 7. workmen employed on a building site earn between them a total of 45 per week. Labourers are paid 75 per week and craftsmen are paid 0 per week. How many craftsmen and how many labourers are employed?. Let the number of craftsmen be c. The number of labourers is therefore c The wage bill equation is: 0c C 75 c D 45 0c C 00 75c D 45 0c 75c D c D 5 c D 5 45 D 7 Hence there are 7 craftsmen and ( 7), i.e. 5 labourers on the site. Now try the following exercise Exercise 8 Practical problems involving simple equations (Answers on page 55). A rectangle has a length of 0 cm and a width b cm. When its width is reduced by 4 cm its area becomes 60 cm. Find the original width and area of the rectangle.. Given R D R C at, find a given R D 5.0, R D 6.0 and t D 5.5. If v D u C as, findu given v D 4, a D 40 and s D The relationship between the temperature on a Fahrenheit scale and that on a Celsius scale is given by F D 9 C C. Express F in degrees 5 Celsius. 5. If t D p p w/sg, find the value of S given w D.9, g D 9.8 and t D Two joiners and five mates earn 6 between them for a particular job. If a joiner earns 48 more than a mate, calculate the earnings for a joiner and for a mate. 7. An alloy contains 60% by weight of copper, the remainder being zinc. How much copper must be mixed with 50 kg of this alloy to give an alloy containing 75% copper? 8. A rectangular laboratory has a length equal to one and a half times its width and a perimeter of 40 m. Find its length and width.

64 8 Transposition of formulae 8. Introduction to transposition of formulae When a symbol other than the subject is required to be calculated it is usual to rearrange the formula to make a new subject. This rearranging process is called transposing the formula or transposition. The rules used for transposition of formulae are the same as those used for the solution of simple equations (see Chapter 7) basically, that the equality of an equation must be maintained. 8. Worked problems on transposition of formulae Problem. subject. If a C b D w x C y, express x as the Rearranging gives: w x C y D a C b and x D a C b w y Multiplying both sides by gives: x D a C b w y i.e. x D a bcwcy The result of multiplying each side of the equation by is to change all the signs in the equation. It is conventional to express answers with positive quantities first. Hence rather than x D a b C w C y, x = w + y a b, since the order of terms connected by C and signs is immaterial. Problem. subject. Transpose p D q C r C s to make r the Problem. Transpose v D fl to make l the subject. The aim is to obtain r on its own on the left-hand side (LHS) of the equation. Changing the equation around so that r is on the LHS gives: q C r C s D p Subtracting q C s from both sides of the equation gives: q C r C s q C s D p q C s Thus q C r C s q s D p q s i.e. r = p q s It is shown with simple equations, that a quantity can be moved from one side of an equation to the other with an appropriate change of sign. Thus equation () follows immediately from equation () above. Rearranging gives: fl D v Dividing both sides by f gives: fl f D v f, i.e. l = v f Problem 4. When a body falls freely through a height h, the velocity v is given by v D gh. Express this formula with h as the subject. Rearranging gives: gh D v Dividing both sides by g gives: gh g D v v,i.e.h = g g Problem 5. If I D V, rearrange to make V the subject. R

65 54 Basic Engineering Mathematics Rearranging gives: V R D I Multiplying both sides by R gives: ( ) V R D R I R Hence V = IR Now try the following exercise Exercise 9 Further problems on transposition of formulae (Answers on page 56) Make the symbol indicated the subject of each of the formulae shown and express each in its simplest form. Problem 6. Transpose a D F m for m. a C b D c d e (d). x C y D t (y) Rearranging gives: F m D a Multiplying both sides by m gives: ( ) F m D m a m i.e. F D ma Rearranging gives: ma D F Dividing both sides by a gives: ma a D F a i.e. m = F a Problem 7. Rearrange the formula R D rl a (i) a the subject, and (ii) l the subject. (i) Rearranging gives: rl a D R Multiplying both sides by a gives: ( ) rl a D a R i.e. rl D ar a Rearranging gives: ar D rl Dividing both sides by R gives: ar R D rl R i.e. a = rl R rl (ii) a D R Multiplying both sides by a gives: rl D ar Dividing both sides by r gives: rl r D ar r i.e. l = ar r to make. c D pr (r) 4. y D mx C c (x) 5. I D PRT (T) 6. I D E R (R) 7. S D a r (r) 8. F D 9 C C 5 (C) 8. Further worked problems on transposition of formulae Problem 8. Transpose the formula v D u C ft m,to make f the subject. Rearranging gives: u C ft m D v and ft m D v u Multiplying each side by m gives: ( ) ft m D m v u i.e. ft D m v u m Dividing both sides by t gives: ft D m t t v u i.e. f = m (v u) t Problem 9. The final length, l of a piece of wire heated through q C is given by the formula l D l C aq. Make the coefficient of expansion, a, the subject. Rearranging gives: l C aq D l Removing the bracket gives: l C l aq D l Rearranging gives: l aq D l l

66 Transposition of formulae 55 Dividing both sides by l q gives: l aq l q D l l l q i.e. a = l l l q Problem 0. A formula for the distance moved by a body is given by s D v C u t. Rearrange the formula to make u the subject. Rearranging gives: Multiplying both sides by gives: Dividing both sides by t gives: i.e. Hence u = s t v C u t D s v C u t D s v C u t t D s t v C u D s t s vt v or u = t Problem. A formula for kinetic energy is k D mv. Transpose the formula to make v the subject. Rearranging gives: mv D k Whenever the prospective new subject is a squared term, that term is isolated on the LHS, and then the square root of both sides of the equation is taken. Multiplying both sides by gives: mv D k mv Dividing both sides by m gives: m D k m i.e. v D k m Taking the square root of both sides gives: p ( ) k v D m ( ) k i.e. v = m Problem. In a right angled triangle having sides x, y and hypotenuse z, Pythagoras theorem states z D x C y. Transpose the formula to find x. Rearranging gives: x C y D z and x D z y Taking the square root of both sides gives: x = z y Problem. and p. l Given t D p,findg in terms of t, l g Whenever the prospective new subject is within a square root sign, it is best to isolate that term on the LHS and then to square both sides of the equation. l Rearranging gives: p g D t Dividing both sides by p gives: l g D t p Squaring both sides gives: l ( ) t g D D t p 4p Cross-multiplying, i.e. multiplying each term by 4p g,gives: 4p l D gt or Dividing both sides by t gives: i.e. gt D 4p l gt t D 4p l t g = 4p l t Problem 4. The impedance of an a.c. circuit is given by Z D p R C X. Make the reactance, X, the subject. Rearranging gives: R C X D Z Squaring both sides gives: R C X D Z Rearranging gives: X D Z R Taking the square root of both sides gives: X D p Z R Problem 5. The volume V of a hemisphere is given by V D pr.findr in terms of V. Rearranging gives: pr D V Multiplying both sides by gives: pr D V Dividing both sides by p gives: pr p D V i.e. r D V p p Taking the cube root of both sides gives: ( ) p V r D p i.e. ( ) V r = p

67 56 Basic Engineering Mathematics Now try the following exercise Exercise 0 Further problems on transposition of formulae (Answers on page 56) Make the symbol indicated the subject of each of the formulae shown and express each in its simplest form. l x d. y D d (x) F f. A D L (f). y D Ml 8EI (E) 4. R D R 0 C at (t) 5. R D C R R (R ) 6. I D E e R C r (R) 7. y D 4ab c (b) a 8. x C b y D l 9. t D p g (x) (l) 0. v D u C as (u). A D pr q (R) 60 ( ) a C x. N D (a) y Dividing both sides by x C y gives: a x C y x C y D rp x C y i.e. a D rp x C y Taking the square root of both sides gives: ( ) rp a = x + y Problem 7. Make b the subject of the formula a D p x y bd C be Rearranging gives: x y p bd C be D a Multiplying both sides by p bd C be gives: or x y D a p bd C be a p bd C be D x y Dividing both sides by a gives: p bd C be D x y a Squaring both sides gives: ( x y bd C be D a Factorizing the LHS gives: ( x y b d C e D a ( x y Dividing both sides by d C e gives: b D a d C e i.e. b = ) ) ) (x y) a (d + e). Z D R C pfl (L) Problem 8. of a, c and e If cd D d C e ad, express d in terms 8.4 Harder worked problems on transposition of formulae Problem 6. Transpose the formula p D a x C a y r make a the subject. a x C a y Rearranging gives: D p r Multiplying both sides by r gives: a x C a y D rp Factorizing the LHS gives: a x C y D rp to Rearranging to obtain the terms in d on the LHS gives: cd d C ad D e Factorizing the LHS gives: d c C a D e Dividing both sides by c C a gives: e d = c + a Problem 9. If a D b,makeb the subject of the C b formula.

68 Transposition of formulae 57 b Rearranging gives: C b D a Multiplying both sides by C b gives: b D a C b Removing the bracket gives: b D a C ab Rearranging to obtain terms in b on the LHS gives: b ab D a Factorizing the LHS gives: b a D a Dividing both sides by a gives: b D a a Problem 0. Transpose the formula V D Er R C r to make r the subject. Er Rearranging gives: R C r D V Multiplying both sides by R C r gives: Er D V R C r Removing the bracket gives: Er D VR C Vr Rearranging to obtain terms in r on the LHS gives: Er Vr D VR Factorizing gives: r E V D VR Dividing both sides by E V gives: r D VR E V Problem. Transpose the formula y D pq r C q t to make q the subject Rearranging gives: pq r C q t D y pq and r C q D y C t Multiplying both sides by r C q gives: pq D r C q y C t Removing brackets gives: pq D ry C rt C q y C q t Rearranging to obtain terms in q on the LHS gives: pq q y q t D ry C rt Factorizing gives: q p y t D r y C t Dividing both sides by p y t gives: q r y C t D p y t Taking the square root of both sides gives: ( ) r(y + t) q = p y t ( Problem. Given that D ) f C p d D, express p f p in terms of D, d and f. Rearranging gives: Squaring both sides gives: ( ) f C p D D f p d ( ) f C p D D f p d Cross-multiplying, i.e. multiplying each term by d f p, gives: Removing brackets gives: d f C p D D f p d f C d p D D f D p Rearranging, to obtain terms in p on the LHS gives: Factorizing gives: Dividing both sides by (d C D )gives: Now try the following exercise Exercise d p C D p D D f d f p d C D D f D d p = f (D d ) (d D ) Further problems on transposition of formulae (Answers on page 56) Make the symbol indicated the subject of each of the formulae shown in Problems to 7, and express each in its simplest form.. y D a m a n (a) x. M D p R 4 r 4 (R). x C y D r (r) C r

69 58 Basic Engineering Mathematics 4. m D ml L C rcr 5. a D b c b x y D C r p q D (L) (b) (r) r ( ) a C b (b) a b 8. A formula for the focal length, f, of a convex lens is f D u C. Transpose the formula to make v the v subject and evaluate v when f D 5anduD6. 9. The quantity of heat, Q, is given by the formula Q D mc t t.maket the subject of the formula and evaluate t when m D 0, t D 5, c D 4and Q D The velocity, v, of water in a pipe appears in the formula h D 0.0Lv. Express v as the subject of the dg formula and evaluate v when h D 0.7, L D 50, d D 0.0 and g D 9.8. The sag S at the centre of a wire is given by the ( ) d l d formula: S D.Makelthe subject 8 of the formula and evaluate l when d D.75 and S D In an electrical alternating current circuit the imped- { ( ance Z is given by: Z D R C!L ) }.!C Transpose the formula to make C the subject and hence evaluate C when Z D 0, R D 0,! D 4 and L D 0. Assignment 4 This assignment covers the material contained in Chapters 7 and 8. The marks for each question are shown in brackets at the end of each question.. Solve the following equations: (a) t D 5t 4 (b) 4 k k C C 4 D 0 a (c) a 5 D (d) p y D ( ) s C (e) D (0) s. Distance s D ut C at. Find the value of acceleration a when s D 7.5, time t D.5 andud0 (4). A rectangular football pitch has its length equal to twice its width and a perimeter of 60 m. Find its length and width. (4) 4. Transpose the following equations: (a) y D mx C c for m y z (b) x D t for z (c) D C R T R A R B for R A (d) x y D ab for y (e) K D p q C pq for q (8) 5. The passage of sound waves through walls is governed by the equation: v D K C 4 G r Make the shear modulus G the subject of the formula. (4)

70 9 Simultaneous equations 9. Introduction to simultaneous equations Only one equation is necessary when finding the value of a single unknown quantity (as with simple equations in Chapter 7). However, when an equation contains two unknown quantities it has an infinite number of solutions. When two equations are available connecting the same two unknown values then a unique solution is possible. Similarly, for three unknown quantities it is necessary to have three equations in order to solve for a particular value of each of the unknown quantities, and so on. Equations which have to be solved together to find the unique values of the unknown quantities, which are true for each of the equations, are called simultaneous equations. Two methods of solving simultaneous equations analytically are: (a) by substitution, and(b) byelimination. (A graphical solution of simultaneous equations is shown in Chapter.) 9. Worked problems on simultaneous equations in two unknowns Problem. Solve the following equations for x and y, (a) by substitution, and (b) by elimination: x C y D 4x y D 8 (a) By substitution From equation (): x D y Substituting this expression for x into equation () gives: 4 y y D 8 This is now a simple equation in y. Removing the bracket gives: 4 8y y D 8 y D 8 C 4 D y D D Substituting y D into equation () gives: x C D x 4 D x D C4 D Thus x = andy = is the solution to the simultaneous equations. (Check: In equation (), since x D and y D, LHS D 4 D C 6 D 8 D RHS.) (b) By elimination x C y D 4x y D 8 If equation () is multiplied throughout by 4 the coefficient of x will be the same as in equation (), giving: 4x C 8y D 4 Subtracting equation () from equation () gives: 4x y D 8 4x C 8y D 4 0 y D

71 60 Basic Engineering Mathematics Hence y D D (Note, in the above subtraction, 8 4 D 8 C 4 D ) Substituting y D into either equation () or equation () will give x D as in method (a). The solution x =, y = is the only pair of values that satisfies both of the original equations. Problem. Solve, by a substitution method, the simultaneous equations x y D x C y D 7 From equation (), x D 7 y Substituting for x in equation () gives: 7 y y D i.e. 9y y D y D C D Hence y D D Substituting y D in equation () gives: x C D 7 i.e. x 9 D 7 Hence x D 7C9 D Thus x =, y = is the solution of the simultaneous equations. (Such solutions should always be checked by substituting values into each of the original two equations.) Problem. Use an elimination method to solve the simultaneous equations x C 4y D 5 x 5y D If equation () is multiplied throughout by and equation () by, then the coefficient of x will be the same in the newly formed equations. Thus ð equation () gives: 6x C 8y D 0 ð equation () gives: 6x 5y D 6 4 Equation () equation (4) gives: 0 C y D 46 i.e. y D 46 D (Note C8y 5y D 8y C 5y D y and 0 6 D 0 C 6 D 46. Alternatively, change the signs of the bottom line and add.) Substituting y D in equation () gives: x C 4 D 5 from which x D 5 8 D and x D Checking in equation (), left-hand side D 5 D 0 D D right-hand side. Hence x = andy = is the solution of the simultaneous equations. The elimination method is the most common method of solving simultaneous equations. Problem 4. Solve 7x y D 6 6x C 5y D 9 When equation () is multiplied by 5 and equation () by the coefficients of y in each equation are numerically the same, i.e. 0, but are of opposite sign. 5 ð equation () gives: 5x 0y D 0 ð equation () gives: x C 0y D 58 4 Adding equation () and (4) gives: 47x C 0 D 88 Hence x D D 4 [Note that when the signs of common coefficients are different the two equations are added, and when the signs of common coefficients are the same the two equations are subtracted (asinproblemsand)] Substituting x D 4 in equation () gives: 7 4 y D 6 8 y D D y D y Hence y D Checking, by substituting x D 4andy D in equation (), gives: LHS D 6 4 C 5 D 4 C 5 D 9 D RHS Thus the solution is x = 4, y =, since these values maintain the equality when substituted in both equations.

72 Simultaneous equations 6 Now try the following exercise Exercise Further problems on simultaneous equations (Answers on page 56) Solve the following simultaneous equations and verify the results.. a C b D 7. x C 5y D 7 a b D x C y D 4. s C t D 4. x y D 4s t D 5 x C 5y D m n D 6. 8a b D 5 m C n D 8 a C 4b D x D y 8. 5c D d x C 7y D 4 d C c C 4 D 0 Problem 6. Solve x 8 C 5 D y y D x Whenever fractions are involved in simultaneous equations it is usual to firstly remove them. Thus, multiplying equation () by 8 gives: ( x ) ( ) 5 8 C 8 D 8y 8 i.e. x C 0 D 8y Multiplying equation () by gives: 9 y D 9x 4 9. Further worked problems on simultaneous equations Problem 5. Solve p D q 4p C q C D 0 Rearranging equations () and (4) gives: x 8y D 0 9x C y D 9 Multiplying equation (6) by 8 gives: 7x C 8y D Adding equations (5) and (7) gives: Rearranging gives: p q D 0 4p C q D Multiplying equation (4) by gives: 8p C q D Adding equations () and (5) gives: p C 0 D p D D Substituting p D into equation () gives: D q 6 D q q D 6 D Checking, by substituting p D andqd into equation () gives: LHS D 4 C C D 8 C D 0 D RHS Hence the solution is p =, q = 4 5 7x C 0 D 9 x D 9 7 D 4 Substituting x D 4 into equation (5) gives: 4 8y D 0 4 C 0 D 8y 4 D 8y y D 4 8 D Checking: substituting x D 4, y D in the original equations, gives: Equation (): LHS D 4 8 C 5 D C D D y D RHS Equation (): LHS D D D RHS D x D 4 D Hence the solution is x = 4, y =

73 6 Basic Engineering Mathematics Problem 7. Solve.5x C 0.75 y D 0.6x D.08.y It is often easier to remove decimal fractions. Thus multiplying equations () and () by 00 gives: 50x C 75 00y D 0 60x D 08 0y Rearranging gives: 50x 00y D 75 60x C 0y D 08 Multiplying equation () by gives: 500x 600y D 50 Multiplying equation (4) by 5 gives: 800x C 600y D 540 Adding equations (5) and (6) gives: 00x C 0 D 90 x D D 9 0 D 0 D 0. Substituting x D 0. into equation () gives: C 75 00y D 0 75 C 75 D 00y 50 D 00y y D D 0.5 Checking x D 0., y D 0.5 in equation () gives: LHS D D 48 RHS D D D 48 Hence the solution is x = 0., y = 0.5 Now try the following exercise Exercise Further problems on simultaneous equations (Answers on page 56) Solve the following simultaneous equations and verify the results.. 7p C C q D 0. x C y D 4 D q 5p x 6 y 9 D a 7 D b 4. s t D 8 D 5a C b s C y D 4 x 5 C y D v D u x 7 y C 5 7 D 0 u C v 5 4 D x.y D 8 8. b.5a D x C 0.6y D.6a C 0.8b D More difficult worked problems on simultaneous equations Problem 8. Solve x C y D 7 x 4 y D In this type of equation the solution is easier if a substitution is initially made. Let x D a and y D b Thus equation () becomes: a C b D 7 and equation () becomes: a 4b D 4 Multiplying equation (4) by gives: a 8b D 4 5 Subtracting equation (5) from equation () gives: 0 C b D i.e. b D Substituting b D in equation () gives: a C D 7 a D 7 D 4 i.e. a D Checking, substituting a D andb D in equation (4) gives: LHS D 4 D 4 D DRHS Hence a D andb D However, since x D a then x D a D and since y D b then y D b D D

74 Simultaneous equations 6 Hence the solution is x =, y =, which may be checked in the original equations. Problem 9. Solve a C 5b D 4 4 a C b D 0.5 Let a D x and b D y then x C 5 y D 4 4x C y D 0.5 To remove fractions, equation () is multiplied by 0 giving: ( x ) ( ) 0 C 0 5 y D 0 4 i.e. 5x C 6y D 40 5 Multiplying equation (4) by gives: 8x C y D Multiplying equation (6) by 6 gives: 4 6 Problem 0. Solve x C y D 4 7 x y D 4 To eliminate fractions, both sides of equation () are multiplied by 7 x C y giving: i.e. ( ) 7 x C y x C y Similarly, in equation (): ( ) 4 D 7 x C y 7 7 D 4 x C y 7 D 4x C 4y D 4 x y i.e. D 8x 4y 4 Equation () + equation (4) gives: 60 D x, i.e. x D 60 D 5 Substituting x D 5 in equation () gives: 7 D 4 5 C 4y from which 4y D 7 0 D 7 and y D 7 4 D 4 48x C 6y D 6 Subtracting equation (5) from equation (7) gives: 7 Hence x = 5, y = is the required solution, which may 4 be checked in the original equations. 4x C 0 D 86 x D 86 4 D Substituting x D into equation () gives: Since C 5 y D 4 5 y D 4 D y D 5 D 5 a D x then a D x D and since b D y then b D y D 5 Hence the solution is a =, b =, which may be checked 5 in the original equations. Problem. x x 6 Solve C y C 5 C 5 C y D 5 D 5 6 Before equations () and () can be simultaneously solved, the fractions need to be removed and the equations rearranged. Multiplying equation () by 5 gives: ( ) ( ) x y C 5 C 5 5 i.e. 5 x C y C D 5x 5 C y C 6 D ( ) D 5 5 5x C y D C 5 6 Hence 5x C y D

75 64 Basic Engineering Mathematics Multiplying equation () by 6 gives: ( ) ( ) ( ) x 5 C y 5 6 C 6 D i.e. x C 5 C y D 5 x C 5 C y D 5 x C y D 5 5 Hence x C y D 4 Thus the initial problem containing fractions can be expressed as: 5x C y D x C y D 4 Subtracting equation (4) from equation () gives: 6x C 0 D x D 6 D Substituting x D into equation () gives: 5 C y D 0 C y D y D 0 D 9 y D 9 D Checking, substituting x D, y D in equation (4) gives: LHS D C D 9 D D RHS Hence the solution is x =, y =, which may be checked in the original equations. Now try the following exercise Exercise 4 Further more difficult problems on simultaneous equations (Answers on page 56) In problems to 7, solve the simultaneous equations and verify the results.. x C y D 4. 4 a b D 8 5 x y D a C 5 b D 4 p C 5q D x C y D. 5 p p D 5 x 7 y D c C 4 c 5 r C 5 C r 4 d C C d 4 s 4 C 5 s 5 x C y D x y D 6 C D 0 C 0 D 0 D 5 D If 5x y D andx C 4 y D Practical problems involving simultaneous equations find the value of xy C y There are a number of situations in engineering and science where the solution of simultaneous equations is required. Some are demonstrated in the following worked problems. Problem. The law connecting friction F and load L for an experiment is of the form F D al Cb, wherea and b are constants. When F D 5.6, L D 8.0 andwhen F D 4.4, L D.0. Find the values of a and b and the value of F when L D 6.5 Substituting F D 5.6, L D 8.0 intof D al C b gives: 5.6 D 8.0a C b Substituting F D 4.4, L D.0 intof D al C b gives: 4.4 D.0a C b Subtracting equation () from equation () gives:. D 6.0a a D. 6.0 D 5 Substituting a D into equation () gives: D 8.0 ( ) 5 C b 5.6 D.6 C b D b i.e. b D 4

76 Simultaneous equations 65 Checking, substituting a D 5 gives: and b D 4 in equation (), l l RHS D.0 ( 5 ) C 4 D 0.4 C 4 D 4.4 D LHS 7V (l l ) 6V Hence a = 5 and b = 4 When L = 6.5, F D al C b D 6.5 C 4 D. C 4, i.e. 5 F = 5.0.5Ω 8Ω Ω Problem. The equation of a straight line, of gradient m and intercept on the y-axis c, isy D mx C c. Ifa straight line passes through the point where x D and y D, and also through the point where x D and y D 0, find the values of the gradient and the y-axis intercept. Substituting x D andy D intoy D mx C c gives: D m C c Substituting x D and y D 0 into y D mx C c gives: 0 D m C c Subtracting equation () from equation () gives: D m from which, m D Substituting m D 5 into equation () gives: D 5 C c c D 5 D 7 D 5 Checking, substituting m D 5andc D 7 in equation (), gives: RHS D ( ) 5 C 7 D 7 7 D 0 D LHS Hence the gradient, m = 5 and the y-axis intercept, c = 7 Problem 4. When Kirchhoff s laws are applied to the electrical circuit shown in Figure 9. the currents I and I are connected by the equations: 7 D.5I C 8 I I 6 D I 8 I I Solve the equations to find the values of currents I and I Fig. 9. Removing the brackets from equation () gives: 7 D.5I C 8I 8I Rearranging gives: 9.5I 8I D 7 Removing the brackets from equation () gives: 6 D I 8I C 8I Rearranging gives: 8I C 0I D 6 4 Multiplying equation () by 5 gives: 47.5I 40I D 5 5 Multiplying equation (4) by 4 gives: I C 40I D 04 6 Adding equations (5) and (6) gives: 5.5I C 0 D I D 5.5 D Substituting I D into equation () gives: 9.5 8I D 7 9 8I D D 8I 8 D 8I I D Hence the solution is I = andi = (which may be checked in the original equations.) Problem 5. The distance s metres from a fixed point of a vehicle travelling in a straight line with constant acceleration, a m/s,isgivenbysdut C at,whereu

77 66 Basic Engineering Mathematics is the initial velocity in m/s and t the time in seconds. Determine the initial velocity and the acceleration given that s D 4 m when t D sands D 44 m when t D 4s. Find also the distance travelled after s. Substituting s D 4, t D intos D ut C at gives: 4 D u C a i.e. 4 D u C a Substituting s D 44, t D 4intosDut C at gives: 44 D 4u C a 4 i.e. 44 D 4u C 8a Multiplying equation () by gives: 84 D 4u C 4a Subtracting equation () from equation () gives: 60 D 0 C 4a a D 60 4 D 5 Substituting a D 5 into equation () gives: 4 D u C D u u D D 6 Substituting a D 5, u D 6 in equation () gives: RHS D 4 6 C 8 5 D 4 C 0 D 44 D LHS Hence the initial velocity, u = 6 m/s and the acceleration, a = 5 m/s Distance travelled after s is given by s D ut C at where t D, u D 6andaD5 Hence s D 6 C 5 D 8 C 67 i.e. distance travelled after s = 85 m Problem 6. A craftsman and 4 labourers together earn 865 per week, whilst 4 craftsmen and 9 labourers earn 40 basic per week. Determine the basic weekly wage of a craftsman and a labourer. Let C represent the wage of a craftsman and L that of a labourer. Thus C C 4L D 865 4C C 9L D 40 Multiplying equation () by 4 gives: 4C C 6L D 460 Subtracting equation () from equation () gives: 7L D 0 L D 0 D 60 7 Substituting L D 8 into equation () gives: C C 4 60 D 865 C C 640 D 865 C D D 5 Checking, substituting C D 5 and L D 60 into equation (), gives: LHS D 4 5 C 9 60 D 900 C 440 D 40 D RHS Thus the basic weekly wage of a craftsman is 5 and that of a labourer is 60 Problem 7. The resistance R of a length of wire at t C isgivenbyr D R 0 C at, wherer 0 is the resistance at 0 C anda is the temperature coefficient of resistance in / C. Find the values of a and R 0 if R D 0 at 50 C andr D 5 at 00 C. Substituting R D 0, t D 50 into R D R 0 C at gives: 0 D R 0 C 50a Substituting R D 5, t D 00 into R D R 0 C at gives: 5 D R 0 C 00a Although these equations may be solved by the conventional substitution method, an easier way is to eliminate R 0 by division. Thus, dividing equation () by equation () gives: 0 5 D R 0 C 50a R 0 C 00a D C 50a C 00a Cross-multiplying gives: 0 C 00a D 5 C 50a 0 C 000a D 5 C 750a 000a 750a D a D 5 i.e. a D 5 50 D or

78 Simultaneous equations 67 Substituting a D into equation () gives: 50 ( )} 0 D R 0 { C D R 0. R 0 D 0. D 5 Checking, substituting a D 50 and R 0 D 5 in equation () gives: { ( )} RHS D 5 C D 5.4 D 5 D LHS Thus the solution is a = 0.004= CandR 0 = 5 Z Problem 8. The molar heat capacity of a solid compound is given by the equation c D a C bt, wherea and b are constants. When c D 5, T D 00 and when c D 7, T D 400. Determine the values of a and b. When c D 5, T D 00, hence 5 D a C 00b When c D 7, T D 400, hence 7 D a C 400b Equation () equation () gives: 0 D 00b from which, b D 0 00 D 0.4 Substituting b D 0.4 in equation () gives: 5 D a C a D 5 40 D Hence a= and b = 0.4 Now try the following exercise Exercise 5 Further practical problems involving simultaneous equations (Answers on page 56). In a system of pulleys, the effort P required to raise a load W is given by P D aw C b, wherea and b are constants. If W D 40 when P D and W D 90 when P D, find the values of a and b.. Applying Kirchhoff s laws to an electrical circuit produces the following equations: 5 D 0.I C I I D I C 0.4I I I Determine the values of currents I and I. Velocity v is given by the formula v D ucat.ifv D 0 when t D andvd40 when t D 7 find the values of u and a. Hence find the velocity when t D.5 4. Three new cars and 4 new vans supplied to a dealer together cost and 5 new cars and new vans of the same models cost Find the cost of a car and a van. 5. y D mx C c is the equation of a straight line of slope m and y-axis intercept c. If the line passes through the point where x D andy D, and also through the point where x D 5andy D, find the slope and y-axis intercept of the straight line. 6. The resistance R ohms of copper wire at t C isgiven by R D R 0 C at, wherer 0 is the resistance at 0 C and a is the temperature coefficient of resistance. If R D 5.44 at 0 C andr D.7 at 00 C, find a and R 0 7. The molar heat capacity of a solid compound is given by the equation c D a C bt. Whenc D 5, T D 00 and when c D 7, T D 400. Find the values of a and b 8. In an engineering process two variables p and q are related by: q D apcb/p, wherea and b are constants. Evaluate a and b if q D when p D andqd when p D 5

79 0 Quadratic equations 0. Introduction to quadratic equations As stated in Chapter 7, an equation is a statement that two quantities are equal and to solve an equation means to find the value of the unknown. The value of the unknown is called the root of the equation. A quadratic equation is one in which the highest power of the unknown quantity is. For example, x x C D 0 is a quadratic equation. There are four methods of solving quadratic equations. These are: (i) by factorization (where possible) (ii) by completing the square (iii) by using the quadratic formula or (iv) graphically (see Chapter ). 0. Solution of quadratic equations by factorization Multiplying out x C x gives x 6x C x, i.e. x 5x. The reverse process of moving from x 5x to x C x is called factorizing. If the quadratic expression can be factorized this provides the simplest method of solving a quadratic equation. For example, if x 5x D 0, then, by factorizing: x C x D 0 Hence either x C D 0 i.e. x D or x D 0 i.e. x D The technique of factorizing is often one of trial and error. Problem. Solve the equations (a) x C x 8 D 0 (b) x x 4 D 0 by factorization. (a) x C x 8 D 0. The factors of x are x and x. These are placed in brackets thus: (x )(x ) The factors of 8 arec8 and, or 8 andc, or C4 and, or 4 andc. The only combination to give a middle term of Cx is C4 and, i.e. x C x 8 D x C 4 x (Note that the product of the two inner terms added to the product of the two outer terms must equal the middle term, Cx in this case.) The quadratic equation x C x 8 D 0 thus becomes x C 4 x D 0. Since the only way that this can be true is for either the first or the second, or both factors to be zero, then either x C 4 D 0 i.e. x D 4 or x D 0 i.e. x D Hence the roots of x + x 8 = 0 are x = 4 and (b) x x 4 D 0 The factors of x are x and x. These are placed in brackets thus: (x )(x ) The factors of 4 are 4 andc, or C4 and, or and. Remembering that the product of the two inner terms added to the product of the two outer terms must equal x, the only combination to give this is C and 4, i.e. x x 4 D x C x 4

80 Quadratic equations 69 The quadratic equation x x 4 D 0 thus becomes x C x 4 D 0 Hence, either x C D 0 i.e. x = or x 4 D 0 i.e. x = 4 and both solutions may be checked in the original equation. Problem. Determine the roots of (a) x 6xC9 D 0, and (b) 4x 5 D 0, by factorization. (a) x 6xC9 D 0. Hence x x D 0, i.e. x D 0 (the left-hand side is known as a perfect square). Hence x = is the only root of the equation x 6x C 9 D 0. (b) 4x 5 D 0 (the left-hand side is the difference of two squares, x and 5 ). Thus x C 5 x 5 D 0 Hence either x C 5 D 0 i.e. x = 5 or x 5 D 0 i.e. x= 5 Problem. Solve the following quadratic equations by factorizing: (a) 4x C 8x C D 0 (b) 5x C x 8 D 0. (a) 4x C 8x C D 0. The factors of 4x are 4x and x or x and x. The factors of are and, or and. Remembering that the product of the inner terms added to the product of the two outer terms must equal C8x, the only combination that is true (by trial and error) is 4x C 8x C D x C x C Hence x C x C D 0 from which, either x C D 0or x C D 0 Thus x D, from which x = or x D, from which x = which may be checked in the original equation. (b) 5x C x 8 D 0. The factors of 5x are 5x and x or 5x and x. The factors of 8 are 4 andc, or 4 and, or 8 andc, or 8 and. By trial and error the only combination that works is 5x C x 8 D 5x C 4 x Hence 5x C 4 x D 0 from which either 5x C 4 D 0 or x D 0 Hence x = 4 5 or x = which may be checked in the original equation. Problem 4. The roots of a quadratic equation are and. Determine the equation. If the roots of a quadratic equation are a and b then x a x b D 0 Hence if a D and b D, then ( x ) x D 0 ( x ) x C D 0 x x C x D 0 x C 5 x D 0 Hence x + 5x = 0 Problem 5. Find the equations in x whose roots are (a) 5 and 5 (b). and 0.4 (a) If 5 and 5 are the roots of a quadratic equation then x 5 x C 5 D 0 i.e. x 5x C 5x 5 D 0 i.e. x 5 = 0 (b) If. and 0.4 are the roots of a quadratic equation then x. x C 0.4 D 0 i.e. x.x C 0.4x 0.48 D 0 i.e. x 0.8x 0.48 = 0 Now try the following exercise Exercise 6 Further problems on solving quadratic equations by factorization (Answers on page 56) In Problems to, solve the given equations by factorization.. x C 4x D 0. x 6 D 0. x C D 6 4. x x D x 5x C D x C x 4 D 0 7. x 4x C 4 D 0 8. x 5x D x C x 6 D x C x 6 D 0

81 70 Basic Engineering Mathematics. 6x 5x 4 D 0. 8x C x 5 D 0 In Problems to 8, determine the quadratic equations in x whose roots are. and 4. and 5 5. and 4 6. and 7. 6 and and Solution of quadratic equations by completing the square An expression such as x or x C or x is called a perfect square. If x D thenxdš p If x C D 5thenxCDš p 5andxD š p 5 If x D 8thenx Dš p 8andxDš p 8 Hence if a quadratic equation can be rearranged so that one side of the equation is a perfect square and the other side of the equation is a number, then the solution of the equation is readily obtained by taking the square roots of each side as in the above examples. The process of rearranging one side of a quadratic equation into a perfect square before solving is called completing the square. x C a D x C ax C a Thus in order to make the quadratic expression x C ax into a perfect square it is necessary to add (half the coefficient of ( ) a x i.e. or a For example, x C x becomes a perfect square by adding ( ),i.e. x C x C ( ) ( D x C ) The method is demonstrated in the following worked problems.. Make the coefficient of the x term unity. In this case this is achieved by dividing throughout by. Hence x C 5x D 0 i.e. x C 5 x D 0. Rearrange the equations so that the x and x terms are on one side of the equals sign and the constant is on the other side. Hence x C 5 x D 4. Add to both sides of the equation (half the coefficient of x. In this case the coefficient of x is 5.Halfthe ( ) 5 coefficient squared is therefore. Thus 4 x C 5 x C ( 5 4 ) D C ( ) 5 4 The LHS is now a perfect square, i.e. ( x C 5 ) D 4 C ( ) Evaluate the RHS. Thus ( x C 5 ) D 4 C 5 4 C 5 D D Taking the square root of both sides of the equation (remembering that the square root of a number gives a š answer). Thus ( x C 5 ) D 4 ( ) 49 6 i.e. x C 5 4 Dš Solve the simple equation. Thus x D 5 4 š 7 4 Problem 6. square. Solve x C 5x D by completing the i.e. x D 5 4 C 7 4 D 4 D and x D D 4 D Theprocedureisasfollows:. Rearrange the equation so that all terms are on the same side of the equals sign (and the coefficient of the x term is positive). Hence x C 5x D 0 Hence x = x C 5x D or are the roots of the equation Problem 7. Solve x C 9x C 8 D 0, correct to significant figures, by completing the square.

82 Quadratic equations 7 Making the coefficient of x unity gives: x C 9 x C 4 D 0 and rearranging gives: x C 9 x D 4 Adding to both sides (half the coefficient of x gives: x C 9 ( ) 9 ( ) 9 x C D The LHS is now a perfect square, thus ( x C 9 ) D D Taking the square root of both sides gives: ( x C 9 ) 7 4 D Dš.0 6 Hence x D 9 4 š.0 Now try the following exercise Exercise 7 Further problems on solving quadratic equations by completing the square (Answers on page 57) In Problems to 6, solve the given equations by completing the square, each correct to decimal places.. x C 4x C D 0. x C 5x 4 D 0. x x 5 D x 8x C D x x C D 0 6. x C 5x D 0.4 Solution of quadratic equations by formula Let the general form of a quadratic equation be given by: i.e. x =. or.8, correct to significant figures. ax C bx C c D 0 where a, b and c are constants. Problem 8. By completing the square, solve the quadratic equation 4.6y C.5y.75 D 0, correct to decimal places. 4.6y C.5y.75 D 0 Making the coefficient of y unity gives: y C.5.75 y D 0 and rearranging gives: y C y D Adding to both sides (half the coefficient of y gives: y C.5 ( ) y C D.75 ( ) C 9. The LHS is now a perfect square, thus ( y C.5 ) D Taking the square root of both sides gives: y C.5 9. D p Dš Hence y D.5 9. š i.e. y = 0.44 or.05 Dividing ax C bx C c D 0bya gives: x C b a x C c a D 0 Rearranging gives: x C b a x D c a Adding to each side of the equation the square of half the coefficient of the term in x to make the LHS a perfect square gives: x C b a x C ( b a ) D ( ) b c a a Rearranging gives: ( x C b ) D b a 4a c a D b 4ac 4a Taking the square root of both sides gives: ( x C b b ) a D 4ac D šp b 4ac 4a a Hence x D b a š p b 4ac a i.e. the quadratic formula is x D b š p b 4ac a (This method of solution is completing the square as shown in Section 0..)

83 7 Basic Engineering Mathematics Summarizing: Problem. Use the quadratic formula to solve if ax C bx C c D 0 then x = b ± p b 4ac a This is known as the quadratic formula. Problem 9. Solve (a) x C x 8 D 0and (b) x x 4 D 0 by using the quadratic formula. (a) Comparing x C x 8 D 0 with ax C bx C c D 0gives a D, b D andc D 8 Substituting these values into the quadratic formula x D b š p b 4ac a gives: x D š 4 8 D š p 6 D C 6 or D š 6 6 D š p 4 C Hence x = 4 8 = or = 4 (as in Problem (a)). (b) Comparing x x 4 D 0 with ax C bx C c D 0 gives a D, b D and c D 4. Hence x D š 4 4 D C š p C 48 6 D Hence x = 4 6 š 6 D C 6 D š p 69 6 or 6 = 4 or 6 = (as in Problem (b)). Problem 0. Solve 4x C 7x C D 0 giving the roots correct to decimal places. Comparing 4x C 7x C D 0 with ax C bx C c gives a D 4, b D 7andc D. Hence x D 7 š [ ] 4 D 7 š 4. 8 D 7 C 4. 8 D 7 š p 7 8 or Hence x = 0.6 or.9, correct to decimal places. x C 4 C D 7 correct to 4 significant figures. x Multiplying throughout by 4 x gives: i.e. x C 4 x 4 C 4 x D 4 x 7 x x x C C 4 D 8 x x C x C D 8x 8 Hence x 7x C 8 D 0 Using the quadratic formula: Hence x D or x D x D 7 š D 7 š p 577 D 7 C š D D.4896 Hence x = 5.5 or.490, correct to 4 significant figures. Now try the following exercise Exercise 8 Further problems on solving quadratic equations by formula (Answers on page 57) In Problems to 6 solve the given equations by using the quadratic formula, correct to decimal places.. x C 5x 4 D x C.86x.5 D 0. x 7x C 4 D x C 5 D x 5. x C D 5 x C 6. x x D x 0.5 Practical problems involving quadratic equations There are many practical problems where a quadratic equation has first to be obtained, from given information, before it is solved.

84 Quadratic equations 7 Problem. The area of a rectangle is.6 cm and its width is.0 cm shorter than its length. Determine the dimensions of the rectangle, correct to significant figures. Let the length of the rectangle be x cm. Then the width is x.0 cm. Area D length ð width D x x.0 D.6 i.e. x.0x.6 D 0 Using the quadratic formula, x D.0 š D.0 š p 9.6 C 94.4 D.0 or 7.0 D.0 š 0.0 Hence x D 6.65 cm or.55 cm. The latter solution is neglected since length cannot be negative. Thus length x D 6.65 cm and width D x.0 D D.55 cm. Hence the dimensions of the rectangle are 6.65 cm by.55 cm. (Check: Area D 6.65 ð.55 D.6cm, correct to significant figures.) Problem. Calculate the diameter of a solid cylinder which has a height of 8.0 cm and a total surface area of.0 m Total surface area of a cylinder D curved surface area C circular ends D prh C pr (where r D radius and h D height) Since the total surface area D.0m and the height h D 8 cm or 0.8 m, then.0 D pr 0.8 C pr i.e. pr C pr D 0 Dividing throughout by p gives: r C 0.8r p D 0 Thus the radius r of the cylinder is m (the negative solution being neglected). Hence the diameter of the cylinder D ð = m or 57.5 cm correct to significant figures Problem 4. The height s metres of a mass projected vertically upwards at time t seconds is s D ut gt. Determine how long the mass will take after being projected to reach a height of 6 m (a) on the ascent and (b) on the descent, when u D 0 m/s and g D 9.8 m/s. When height s D 6 m, 6 D 0 t 9.8 t i.e t 0t C 6 D 0 Using the quadratic formula: t D 0 š D 0 š p D 0 š D 5.5 or 0.59 Hence the mass will reach a height of 6 m after 0.59 s on the ascent and after 5.5 s on the descent. Problem 5. A shed is 4.0 m long and.0 m wide. A concrete path of constant width is laid all the way around the shed. If the area of the path is 9.50 m calculate its width to the nearest centimetre. Figure 0. shows a plan view of the shed with its surrounding path of width t metres. t t.0 m Using the quadratic formula: 0.8 š r D D 0.8 š p.9456 D D or.074 ( ) p 0.8 š.948 Fig m SHED (4.0+t)

85 74 Basic Engineering Mathematics Area of path D.0 ð t C t 4.0 C t i.e D 4.0t C 8.0t C 4t or 4t C.0t 9.50 D 0 Hence t D.0 š D.0 š p D.0 š Hence t D m or m Neglecting the negative result which is meaningless, the width of the path, t = 0.65 m or 65 cm, correct to the nearest centimetre. Problem 6. If the total surface area of a solid cone is 486. cm and its slant height is 5. cm, determine its base diameter. From Chapter, page 66, the total surface area A of a solid cone is given by: A D prl C pr where l is the slant height and r thebaseradius. If A D 48. andld5., then 48. D pr 5. C pr i.e. pr C 5.pr 48. D 0 or r C 5.r 48. D 0 p Using the quadratic formula, [ ( )] š 5. 4 p r D D 5. š p D 5. š 9. Hence radius r D cm (or. cm, which is meaningless, and is thus ignored). Thus the diameter of the base Now try the following exercise Exercise 9 D r D D.8 cm Further practical problems involving quadratic equations (Answers on page 57). The angle a rotating shaft turns through in t seconds is given by q D!t C at. Determine the time taken to complete 4 radians if! is.0 rad/s and a is 0.60 rad/s.. The power P developed in an electrical circuit is given by P D 0I 8I,whereI is the current in amperes. Determine the current necessary to produce a power of.5 watts in the circuit.. The area of a triangle is 47.6cm and its perpendicular height is 4. cm more than its base length. Determine the length of the base correct to significant figures. 4. The sag l metres in a cable stretched between two supports,distance x m apart is given by: l D x C x Determine the distance between supports when the sag is 0 m. 5. The acid dissociation constant K a of ethanoic acid is.8 ð 0 5 mol dm for a particular solution. Using x the Ostwald dilution law K a D v l x determine x, the degree of ionization, given that v D 0 dm. 6. A rectangular building is 5 m long by m wide. A concrete path of constant width is laid all the way around the building. If the area of the path is 60.0 m, calculate its width correct to the nearest millimetre. 7. The total surface area of a closed cylindrical container is 0.0 m. Calculate the radius of the cylinder if its height is.80 m. 8. The bending moment M at a point in a beam is given x 0 x by M D where x metres is the distance from the point of support. Determine the value of x when the bending moment is 50 Nm. 9. A tennis court measures 4 m by m. In the layout of a number of courts an area of ground must be allowed for at the ends and at the sides of each court. If a border of constant width is allowed around each court and the total area of the court and its border is 950 m, find the width of the borders. 0. Two resistors, when connected in series, have a total resistance of 40 ohms. When connected in parallel their total resistance is 8.4 ohms. If one of the resistors has a resistance R x ohms: (a) show that Rx 40R x C 6 D 0and (b) calculate the resistance of each 0.6 The solution of linear and quadratic equations simultaneously Sometimes a linear equation and a quadratic equation need to be solved simultaneously. An algebraic method of solution

86 Quadratic equations 75 is shown in Problem 7; a graphical solution is shown in Chapter, page 9. Problem 7. Determine the values of x and y which simultaneously satisfy the equations: y D 5x 4 x and y D 6x 7. y D 5x C x y D x. x C y D 4 C 5x x C y D 4 For a simultaneous solution the values of y must be equal, hence the RHS of each equation is equated. Thus 5x 4 x D 6x 7 Rearranging gives: 5x 4 x 6x C 7 D 0 i.e. x C x D 0 or x C x D 0 Factorizing gives: x C x D 0 i.e. x D or x D In the equation y D 6x 7, when x D ( ),y D 6 7 D 6 and when x D,y D 6 7 D [Checking the result in y D 5x 4 x : when x D (, y D 5 ) ( 4 ) D D 6 as above; and when x D, y D 5 4 D as above] Hence the simultaneous solutions occur when x =, y = 6 and when x =, y = Now try the following exercise Exercise 40 Further problems on solving linear and quadratic equations simultaneously (Answers on page 57) In Problems to determine the solutions of the simultaneous equations.. y D x C x C y D 4 x Assignment 5 This assignment covers the material contained in Chapters 9 and 0. The marks for each question are shown in brackets at the end of each question.. Solve the following pairs of simultaneous equations: (a) 7x y D x C 4y D 8 (b) a 8 C b 8 D 0 b C a D 4 p C 4q (c) D 5 5 p q C C D 0 (0). In an engineering process two variables x and y are related by the equation y D ax C b where a and b are x constants. Evaluate a and b if y D 5 when x D and y D when x D (5). Solve the following equations by factorization: (a) x 9 D 0 (b) x 5x D 0 (6) 4. Determine the quadratic equation in x whose roots are and (4) 5. Solve the equation x x 4 D 0bycompletingthe square. (5) 6. Solve the equation 4x 9x C D 0 correct to decimal places. (5) 7. The current i flowing through an electronic device is given by: i D v C 0.04 v where v is the voltage. Calculate the values of v when i D ð 0 (5)

87 Straight line graphs. Introduction to graphs A graph is a pictorial representation of information showing how one quantity varies with another related quantity. The most common method of showing a relationship between two sets of data is to use Cartesian or rectangular axes as shown in Fig... B ( 4, ) 4 A (, ) C (, ) Origin y 4 Abscissa A Ordinate x. The straight line graph Let a relationship between two variables x and y be y D x C. When x D 0, y D 0 C D. When x D, y D C D 5. When x D, y D C D 8, and so on. Thus co-ordinates (0, ), (, 5) and (, 8) have been produced from the equation by selecting arbitrary values of x, and are shown plotted in Fig... When the points are joined together a straight-line graph results. y y = x + x Fig.. The points on a graph are called co-ordinates. Point A in Fig.. has the co-ordinates (, ), i.e. units in the x direction and units in the y direction. Similarly, point B has co-ordinates ( 4, ) and C has co-ordinates (, ). The origin has co-ordinates (0, 0). The horizontal distance of a point from the vertical axis is called the abscissa and the vertical distance from the horizontal axis is called the ordinate. Fig.. The gradient or slope of a straight line is the ratio of the change in the value of y to the change in the value of x between any two points on the line. If, as x increases, (!), y also increases ("), then the gradient is positive. In Fig..(a), the gradient of AC D change in y change in x D CB BA D 7 D 4 D

88 Straight line graphs 77 y A 0 Fig.. (a) C B 4 x y y = x + 0 (c) y = y = x + F E 4 x (b) y D If as x increases (!), y decreases (#), then the gradient is negative. In Fig..(b), the gradient of DF change in y D change in x D FE D ED 0 D 9 D Figure.(c) shows a straight line graph y D. Since the straight line is horizontal the gradient is zero. The value of y when x D 0 is called the y-axis intercept. In Fig..(a) the y-axis intercept is and in Fig..(b) is. If the equation of a graph is of the form y = mx + c, where m and c are constants, the graph will always be a straight line, m representing the gradient and c the y-axis intercept. Thus y D 5x C represents a straight line of gradient 5 and y-axis intercept. Similarly, y D x 4represents a straight line of gradient andy-axis intercept 4 Summary of general rules to be applied when drawing graphs (i) Give the graph a title clearly explaining what is being illustrated. (ii) Choose scales such that the graph occupies as much space as possible on the graph paper being used. (iii) Choose scales so that interpolation is made as easy as possible. Usually scales such as cm D unit, or cm D units, or cm D 0 units are used. Awkward scales such as cm D units or cm D 7 units should not be used. x (iv) The scales need not start at zero, particularly when starting at zero produces an accumulation of points within a small area of the graph paper. (v) The co-ordinates, or points, should be clearly marked. This may be done either by a cross, or a dot and circle, or just by a dot (see Fig..). (vi) A statement should be made next to each axis explaining the numbers represented with their appropriate units. (vii) Sufficient numbers should be written next to each axis without cramping. Problem. Plot the graph y D 4x C in the range x D toxdc4. From the graph, find (a) the value of y when x D., and (b) the value of x when y D Whenever an equation is given and a graph is required, a table giving corresponding values of the variable is necessary. The table is achieved as follows: When x D, y D 4x C D 4 C D C D 9. When x D, y D 4 C D 8C D 5, and so on. Such a table is shown below: x 0 4 y The co-ordinates (, 9), (, 5), (, ), and so on, are plotted and joined together to produce the straight line shown in Fig..4. (Note that the scales used on the x and y axes do not have to be the same) From the graph: (a) when x D., y =.8, and (b) when y D, x =.5 y Fig x.

89 78 Basic Engineering Mathematics Problem. Plot the following graphs on the same axes between the range x D 4tox DC4, and determine the gradient of each. (a) y D x (b) y D x C (c) y D x C 5 (d) y D x A table of co-ordinates is produced for each graph. (a) y D x x y (b) y D x C x y (c) y D x C 5 Fig..5 x y y x D C F E y = x + 5 y = x + y = x y = x A B (d) y D x x y The co-ordinates are plotted and joined for each graph. The results are shown in Fig..5. Each of the straight lines produced are parallel to each other, i.e. the slope or gradient is the same for each. To find the gradient of any straight line, say, y D x a horizontal and vertical component needs to be constructed. In Fig..5, AB is constructed vertically at x D 4andBC constructed horizontally at y D. The gradient of AC D AB D D 4 BC D i.e. the gradient of the straight line y D x is. The actual positioning of AB and BC is unimportant for the gradient is also given by DE D D EF D The slope or gradient of each of the straight lines in Fig..5 is thus since they are all parallel to each other. Problem. Plot the following graphs on the same axes between the values x D tox DC and determine the gradient and y-axis intercept of each. (a) y D x (b) y D x C 7 (c) y D 4x C 4 (d) y D 4x 5 A table of co-ordinates is drawn up for each equation. (a) y D x x 0 y (b) y D x C 7 x 0 y (c) y D 4x C 4 x 0 y (d) y D 4x 5 x 0 y

90 Straight line graphs 79 Each of the graphs is plotted as shown in Fig..6, and each is a straight line. y D x and y D x C 7 are parallel to each other and thus have the same gradient. The gradient of AC is given by Fig..6 CB 6 7 D BA 0 D 9 D y= 4x+4 y= 4x 5 y x F 8 6 E A 4 y=x+7 y =x Hence the gradient of both y = x and y = x + 7is. y D 4x C 4 and y D 4x 5 are parallel to each other and thus have the same gradient. The gradient of DF is given by FE 5 7 D D ED 0 D 4 Hence the gradient of both y = 4x + 4 and y = 4x 5is 4 The y-axis intercept means the value of y where the straight line cuts the y-axis. From Fig..6, y D x cuts the y-axis at y D 0 y D x C 7 cuts the y-axis at y DC7 y D 4xC4 cuts the y-axis at y DC4 and y D 4x 5 cuts the y-axis at y D 5 Some general conclusions can be drawn from the graphs shown in Figs..4,.5 and.6. When an equation is of the form y D mx C c, wherem and c are constants, then C B D (i) a graph of y against x produces a straight line, (ii) m represents the slope or gradient of the line, and (iii) c represents the y-axis intercept. Thus, given an equation such as y D x C 7, it may be deduced on sight that its gradient is C and its y-axis intercept is C7, as shown in Fig..6. Similarly, if y D 4x 5, then the gradient is 4 andthey-axis intercept is 5, as shown in Fig..6. When plotting a graph of the form y D mx C c, only two co-ordinates need be determined. When the co-ordinates are plotted a straight line is drawn between the two points. Normally, three co-ordinates are determined, the third one acting as a check. Problem 4. The following equations represent straight lines. Determine, without plotting graphs, the gradient and y-axis intercept for each. (a) y D (b) y D x (c) y D 5x (d) x C y D (a) y D (which is of the form y D 0x C ) represents a horizontal straight line intercepting the y-axis at. Since the line is horizontal its gradient is zero. (b) y D x is of the form y D mx C c, wherec is zero. Hence gradient = and y-axis intercept = 0 (i.e. the origin). (c) y D 5x is of the form y D mx C c. Hence gradient = 5 and y-axis intercept = (d) x C y D is not in the form y D mx C c as it stands. Transposing to make y the subject gives y D x, i.e. x y D D i.e. y D x C x which is of the form y D mx C c. Hence gradient = and y-axis intercept =+ Problem 5. Without plotting graphs, determine the gradient and y-axis intercept values of the following equations: (a) y D 7x (b) y D 6xC (c) y D 4x C 9 (d) y D x 5 (e) x C 9y C D 0 (a) y D 7x is of the form y D mx C c, hence gradient, m = 7 and y-axis intercept, c =.

91 80 Basic Engineering Mathematics (b) Rearranging y D 6xCgives y D 6x C i.e. y D xc which is of the form y D mx C c. Hence gradient m = and y-axis intercept, c = (c) Rearranging y D 4x C 9givesy D 4x C, hence gradient = 4 and y-axis intercept = (d) Rearranging y D x 5 gives ( x y D ) D 5 x 5 Hence gradient = and y-axis intercept = 5 (e) Rearranging x C 9y C D 0gives 9y D x, i.e. y D 9 x 9 Hence gradient = 9 and y-axis intercept = 9 Problem 6. Determine the gradient of the straight line graph passing through the co-ordinates (a) (, 5) and (, 4), and (b) (, ) and (, ) A straight line graph passing through co-ordinates (x, y ) and (x, y ) has a gradient given by: Fig..7 m D y y x x (see Fig..7) y y (x, y ) y (x x ) (x, y ) 0 x x x (y y ) (a) A straight line passes through (, 5) and (, 4), hence x D, y D 5, x D andy D 4, hence gradient m D y y x x D 4 5 = 5 (b) A straight line passes through (, ) and (, ), hence x D, y D, x D andy D, hence gradient, m D y y D x x D C C D 6 D 6 Problem 7. Plot the graph x C y C D 0 and y 5 D x on the same axes and find their point of intersection. Rearranging x C y C D 0givesyD x Rearranging y 5 D x gives y D x C 5andyD x C Since both equations are of the form y D mx C c both are straight lines. Knowing an equation is a straight line means that only two co-ordinates need to be plotted and a straight line drawn through them. A third co-ordinate is usually determined to act as a check. A table of values is produced for each equation as shown below. x 0 x 4 x 0 x C The graphs are plotted as shown in Fig..8 Fig..8 y = x y x 4 y = x + The two straight lines are seen to intersect at (, ) 5

92 Straight line graphs 8 Now try the following exercise Exercise 4 Further problems on straight line graphs (Answers on page 57). Corresponding values obtained experimentally for two quantities are: x y Use a horizontal scale for x of cm D unit and a vertical scale for y of cm D units and draw a graph of x against y. Label the graph and each of its axes. By interpolation, find from the graph the value of y when x is.5. The equation of a line is 4y D x C 5. A table of corresponding values is produced and is shown below. Complete the table and plot a graph of y against x. Find the gradient of the graph. x y Determine the gradient and intercept on the y-axis for each of the following equations: (a) y D 4x (b) y D x (c) y D x 4 (d) y D 4 4. Find the gradient and intercept on the y-axis for each of the following equations: (a) y D 4x (b) 6x y D 5 (c) y D x 4 Determine the gradient and y-axis intercept for each of the equations in Problems 5 and 6 and sketch the graphs. 5. (a) y D 6x (b) y D x C 4 (c) y D x (d) y D 7 6. (a) y C D 4x (b) x C y C 5 D 0 (c) y 4 D x (d) 5x y 7 D 0 7. Determine the gradient of the straight line graphs passing through the co-ordinates: (a) (, 7) and (, 4) (b) ( 4, ) and ( 5, ) ( (c) 4, ) ( and 4, 5 ) 8 8. State which of the following equations will produce graphs which are parallel to one another: (a) y 4 D x (b) 4x D y C (c) x D y C 5 (d) C y D x (e) x D 7 y 9. Draw a graph of y x C 5 D 0 over a range of x D tox D 4. Hence determine (a) the value of y when x D. and (b) the value of x when y D Draw on the same axes the graphs of y D x 5 and y C x D 7. Find the co-ordinates of the point of intersection. Check the result obtained by solving the two simultaneous equations algebraically.. Plot the graphs y D x C andy D 5 x on the same axes and determine their point of intersection.. Practical problems involving straight line graphs When a set of co-ordinate values are given or are obtained experimentally and it is believed that they follow a law of the form y D mx C c, then if a straight line can be drawn reasonably close to most of the co-ordinate values when plotted, this verifies that a law of the form y D mx C c exists. From the graph, constants m (i.e. gradient) and c (i.e. y-axis intercept) can be determined. This technique is called determination of law (see Chapter 5). Problem 8. The temperature in degrees Celsius and the corresponding values in degrees Fahrenheit are shown in the table below. Construct rectangular axes, choose a suitable scale and plot a graph of degrees Celsius (on the horizontal axis) against degrees Fahrenheit (on the vertical scale). C F From the graph find (a) the temperature in degrees Fahrenheit at 55 C, (b) the temperature in degrees Celsius at 67 F, (c) the Fahrenheit temperature at 0 C, and (d) the Celsius temperature at 0 F. The co-ordinates (0, 50), (0, 68), (40, 04), and so on are plotted as shown in Fig..9. When the co-ordinates are joined, a straight line is produced. Since a straight line results there is a linear relationship between degrees Celsius and degrees Fahrenheit.

93 8 Basic Engineering Mathematics Degrees Fahrenheit ( F) y Fig..9 E D B A x Degrees Celsius ( C) (a) To find the Fahrenheit temperature at 55 C a vertical line AB is constructed from the horizontal axis to meet the straight line at B. The point where the horizontal line BD meets the vertical axis indicates the equivalent Fahrenheit temperature. Hence 55 C is equivalent to F. This process of finding an equivalent value in between the given information in the above table is called interpolation. (b) To find the Celsius temperature at 67 F, a horizontal line EF is constructed as shown in Fig..9. The point where the vertical line FG cuts the horizontal axis indicates the equivalent Celsius temperature. Hence 67 F is equivalent to 75 C. (c) If the graph is assumed to be linear even outside of the given data, then the graph may be extended at both ends (shown by broken line in Fig..9). From Fig..9, 0 C corresponds to F. (d) 0 F is seen to correspond to 0 C. The process of finding equivalent values outside of the given range is called extrapolation. Problem 9. In an experiment on Charles s law, the value of the volume of gas, V m, was measured for various temperatures T C. Results are shown below. V m T C F G Plot a graph of volume (vertical) against temperature (horizontal) and from it find (a) the temperature when the volume is 8.6 m, and (b) the volume when the temperature is 67 C. If a graph is plotted with both the scales starting at zero then the result is as shown in Fig..0. All of the points lie in the top right-hand corner of the graph, making interpolation difficult. A more accurate graph is obtained if the temperature axis starts at 55 C and the volume axis starts at 4.5 m.the axes corresponding to these values is shown by the broken lines in Fig..0 and are called false axes, since the origin is not now at zero. A magnified version of this relevant part of the graph is shown in Fig... From the graph: Volume (m ) Fig..0 y x Temperature ( C) (a) when the volume is 8.6 m, the equivalent temperature is 8.5 C,and (b) when the temperature is 67 C, the equivalent volume is 6. m Problem 0. In an experiment demonstrating Hooke s law, the strain in an aluminium wire was measured for various stresses. The results were: Stress N/mm Strain Stress N/mm Strain Plot a graph of stress (vertically) against strain (horizontally). Find:

94 Straight line graphs 8 (a) Young s Modulus of Elasticity for aluminium which is given by the gradient of the graph, (b) the value of the strain at a stress of 0 N/mm,and (c) the value of the stress when the strain is y 8 4 A y Stress (N/mm ) C B x Strain Volume (m ) Fig.. Thus Young s Modulus of Elasticity for aluminium is N/mm Since m D 0 6 mm, N/mm is equivalent to ð 0 6 N/m,i.e.70 U 0 9 N=m (or Pascals) From Fig..: (b) the value of the strain at a stress of 0 N/mm is , and (c) the value of the stress when the strain is is 4 N/mm 5 Fig x Temperature ( C) The co-ordinates ( , 4.9), (0.000, 8.7), and so on, are plotted as shown in Fig... The graph produced is the best straight line which can be drawn corresponding to these points. (With experimental results it is unlikely that all the points will lie exactly on a straight line.) The graph, and each of its axes, are labelled. Since the straight line passes through the origin, then stress is directly proportional to strain for the given range of values. (a) The gradient of the straight line AC is given by AB BC D D D ð 0 D D 7 ð 0 4 D N/mm Problem. The following values of resistance R ohms and corresponding voltage V volts are obtained from a test on a filament lamp. R ohms V volts Choose suitable scales and plot a graph with R representing the vertical axis and V the horizontal axis. Determine (a) the gradient of the graph, (b) the R axis intercept value, (c) the equation of the graph, (d) the value of resistance when the voltage is 60 V, and (e) the value of the voltage when the resistance is 40 ohms. (f) If the graph were to continue in the same manner, what value of resistance would be obtained at 0 V? The co-ordinates (6, 0), (9, 48.5), and so on, are shown plotted in Fig.. where the best straight line is drawn through the points. (a) The slope or gradient of the straight line AC is given by AB 5 0 D BC 00 0 D 5 00 D.5 (Note that the vertical line AB and the horizontal line BC may be constructed anywhere along the length of the

95 84 Basic Engineering Mathematics y A Show that the values obey the law s D at C b, wherea and b are constants and determine approximate values for a and b. Use the law to determine the stress at 50 C and the temperature when the stress is 7.54 N/cm. Resistance R ohms C B The co-ordinates (70, 8.46), (00, 8.04), and so on, are plotted as shown in Fig..4. Since the graph is a straight line then the values obey the law s D at C b, and the gradient of the straight line is a D AB D BC D D 0.00 y A Fig x Voltage V volts straight line. However, calculations are made easier if the horizontal line BC is carefully chosen, in this case, 00). (b) The R-axis intercept is at R = 0 ohms (by extrapolation). (c) The equation of a straight line is y D mx C c, when y is plotted on the vertical axis and x on the horizontal axis. m represents the gradient and c the y-axis intercept. In this case, R corresponds to y, V corresponds to x, m D.5 and c D 0. Hence the equation of the graph is R =(.5 V + 0)Z. From Fig.., Stress σ N/cm Fig B C x Temperature t C (d) when the voltage is 60 V, the resistance is 85 Z (e) when the resistance is 40 ohms, the voltage is 4 V, and (f) by extrapolation, when the voltage is 0 V, the resistance is 47 Z. Problem. Experimental tests to determine the breaking stress s of rolled copper at various temperatures t gave the following results. Stress s N/cm Temperature t C Stress s N/cm Temperature t C Vertical axis intercept, b = 8.68 Hence the law of the graph is s = 0.00t When the temperature is 50 C, stress s is given by s D C 8.68 D 7.88 N=cm Rearranging s D 0.00t C 8.68 gives i.e. 0.00t D 8.68 s, t D 8.68 s 0.00 Hence when the stress s D 7.54 N/cm, temperature t D D 56. C

96 Straight line graphs 85 Now try the following exercise Exercise 4 Further practical problems involving straight line graphs (Answers on page 57). The resistance R ohms of a copper winding is measured at various temperatures t C and the results are as follows: R ohms t C Plot a graph of R (vertically) against t (horizontally) and find from it (a) the temperature when the resistance is Z and (b) the resistance when the temperature is 5 C.. The speed of a motor varies with armature voltage as shown by the following experimental results: n (rev/min) V volts Plot a graph of speed (horizontally) against voltage (vertically) and draw the best straight line through the points. Find from the graph (a) the speed at a voltage of 45 V, and (b) the voltage at a speed of 400 rev/min.. The following table gives the force F newtons which, when applied to a lifting machine, overcomes a corresponding load of L newtons. Force F newtons Load L newtons Choose suitable scales and plot a graph of F (vertically) against L (horizontally). Draw the best straight line through the points. Determine from the graph (a) the gradient, (b) the F-axis intercept, (c) the equation of the graph, (d) the force applied when the load is 0 N, and (e) the load that a force of 60 N will overcome. (f) If the graph were to continue in the same manner, what value of force will be needed to overcome a 800 N load? 4. The following table gives the results of tests carried out to determine the breaking stress s of rolled copper at various temperatures, t: Stress s (N/cm ) Temperature t( C) Stress s (N/cm ) Temperature t( C) Plot a graph of stress (vertically) against temperature (horizontally). Draw the best straight line through the plotted co-ordinates. Determine the slope of the graph and the vertical axis intercept. 5. The velocity v of a body after varying time intervals t was measured as follows: t (seconds) v (m/s) Plot v vertically and t horizontally and draw a graph of velocity against time. Determine from the graph (a) the velocity after 0 s, (b) the time at 0 m/s and (c) the equation of the graph. 6. The mass m of a steel joint varies with length L as follows: mass, m (kg) length, L (m) Plot a graph of mass (vertically) against length (horizontally). Determine the equation of the graph. 7. The crushing strength of mortar varies with the percentage of water used in its preparation, as shown below. Crushing strength, F (tonnes) % of water used, w% Plot a graph of F (vertically) against w (horizontally). (a) Interpolate and determine the crushing strength when 0% of water is used. (b) Assuming the graph continues in the same manner extrapolate and determine the percentage of water used when the crushing strength is 0.5 tonnes. (c) What is the equation of the graph? 8. In an experiment demonstrating Hooke s law, the strain in a copper wire was measured for various stresses. The results were: Stress (pascals) 0.6 ð ð ð 0 6 Strain

97 86 Basic Engineering Mathematics Stress (pascals) 0.7 ð ð 0 6 Strain Plot a graph of stress (vertically) against strain (horizontally). Determine (a) Young s Modulus of Elasticity for copper, which is given by the gradient of the graph, (b) the value of strain at a stress of ð0 6 Pa, (c) the value of stress when the strain is An experiment with a set of pulley blocks gave the following results: Plot a graph of effort (vertically) against load (horizontally) and determine (a) the gradient, (b) the vertical axis intercept, (c) the law of the graph, (d) the effort when the load is 0 N and (e) the load when the effort is 9 N 0. The variation of pressure p in a vessel with temperature T is believed to follow a law of the form p D at C b, wherea and b are constants. Verify this law for the results given below and determine the approximate values of a and b. Hence determine the pressures at temperatures of 85 K and 0 K and the temperature at a pressure of 50 kpa. Effort, E (newtons) Load, L (newtons) Pressure, p kpa Temperature, T K

98 Graphical solution of equations. Graphical solution of simultaneous equations Linear simultaneous equations in two unknowns may be solved graphically by: (i) plotting the two straight lines on the same axes, and (ii) noting their point of intersection. The co-ordinates of the point of intersection give the required solution. Problem. equations x y D 4 x C y D 5 Solve graphically the simultaneous Rearranging each equation into y D mx C c form gives: y D x 4 y D xc5 Only three co-ordinates need be calculated for each graph since both are straight lines. x 0 y D x x 0 y D xc5 5 4 Each of the graphs is plotted as shown in Fig... The point of intersection is at (,) and since this is the only point which lies simultaneously on both lines then x =, y = is the solution of the simultaneous equations. Fig Problem..0x C y D.80 x 5.0y D 8.50 y y = x +5 y = x 4 4 x Solve graphically the equations Rearranging each equation into y D mx C c form gives: y D.0x C.80 y D x i.e. y D 0.0x.70 Three co-ordinates are calculated for each equation as shown below.

99 88 Basic Engineering Mathematics x 0 y D.0x C x 0 y D 0.0x The two lines are plotted as shown in Fig... The point of intersection is (.50,.0). Hence the solution of the simultaneous equation is x =.50, y =.0. (It is sometimes useful initially to sketch the two straight lines to determine the region where the point of intersection is. Then, for greater accuracy, a graph having a smaller range of values can be drawn to magnify the point of intersection). y x.0 y = 0.0x +.70 y =.0x +.80 a and b are constants. When F D 4 newtons, L D 6 newtons and when F D.4 newtons, L D newtons. Determine graphically the values of a and b.. Graphical solutions of quadratic equations A general quadratic equation is of the form y D ax C bx C c, where a, b and c are constants and a is not equal to zero. A graph of a quadratic equation always produces a shape called a parabola. The gradient of the curve between 0 and A and between B and C in Fig.. is positive, whilst the gradient between A and B is negative. Points such as A and B are called turning points. AtA the gradient is zero and, as x increases, the gradient of the curve changes from positive just before A to negative just after. Such a point is called a maximum value. At B the gradient is also zero, and, as x increases, the gradient of the curve changes from negative just before B to positive just after. Such a point is called a minimum value. y A C Fig.. Now try the following exercise Exercise 4 Further problems on the graphical solution of simultaneous equations (Answers on page 57) In Problems to 5, solve the simultaneous equations graphically.. x C y D. y D 5 x y x D x y D. x C 4y D x 7.06 D.y x 5y C D 0.x 6.7y D x y D 0 4x C y C D 0 6. The friction force F newtons and load L newtons are connected by a law of the form F D al C b, where Fig.. Quadratic graphs (i) y = ax B 0 x Graphs of y D x, y D x and y D x are shown in Fig..4. y 0 Fig..4 y =x x y y = x 0 x (a) (b) (c) y y = x 0 x

100 Graphical solution of equations 89 All have minimum values at the origin (0,0). Graphs of y D x, y D x and y D x are shown in Fig x 0 x x 0 y = x y = x y = x y (a) Fig..5 y (b) y (c) When y D ax C c: (a) curves are symmetrical about the y-axis, (b) the magnitude of a affects the gradient of the curve, and (c) the constant c isthey-axis intercept. (iii) y = ax + bx + c Whenever b has a value other than zero the curve is displaced to the right or left of the y-axis. When b/a is positive, the curve is displaced b/a to the left of the y-axis, as shown in Fig..7(a). When b/a is negative the curve is displaced b/a to the right of the y-axis, as shown in Fig..7(b). All have maximum values at the origin (0,0). When y D ax : (a) curves are symmetrical about the y-axis, (b) the magnitude of a affects the gradient of the curve, and (c) the sign of a determines whether it has a maximum or minimum value. (ii) y = ax + c Graphs of y D x C, y D x, y D x C and y D x are shown in Fig..6. y 8 y y =x +6x y =x 5x x 0 4 x (a) (b) Fig..7 y y =x + 0 x (a) y 0 x y= x + (c) Fig..6 y y =x 0 x (b) y 0 x 4 y= x (d) Quadratic equations of the form ax C bx C c D 0maybe solved graphically by: (i) plotting the graph y D ax C bx C c, and (ii) noting the points of intersection on the x-axis (i.e. where y D 0). The x values of the points of intersection give the required solutions since at these points both y D 0and ax C bx C c D 0 The number of solutions, or roots of a quadratic equation, depends on how many times the curve cuts the x-axis and there can be no real roots (as in Fig..7(a)) or one root (as in Figs..4 and.5) or two roots (as in Fig..7(b)). Problem. Solve the quadratic equation 4x C 4x 5 D 0 graphically given that the solutions lie in the range x D tox D. Determine also the co-ordinates and nature of the turning point of the curve. Let y D 4x C 4x 5. A table of values is drawn up as shown below.

101 90 Basic Engineering Mathematics x 0 4x x y 0 5 y =4x y D 4x C 4x A graph of y D 4x C 4x 5 is shown in Fig..8. The only points where y D 4x C 4x 5 and y D 0are the points marked A and B. This occurs at x =.5 and x =.5 and these are the solutions of the quadratic equation 4x C 4x 5 D 0. (By substituting x D.5 andx D.5 into the original equation the solutions may be checked.) The curve has a turning point at ( 0.5, 6) and the nature of the point is a minimum y = 4x +5 x y Fig..9 A 8 4 y = 4x +4x B 0 x y y = 5x +9x x Fig..8 An alternative graphical method of solving 4x C 4x 5 D is to rearrange the equation as 4x D 4x C 5 and then plot two separate graphs in this case y D 4x and y D 4xC5. Their points of intersection give the roots of equation 4x D 4x C 5, i.e. 4x C 4x 5 D 0. This is shown in Fig..9, where the roots are x D.5 andx D.5 as before. Problem 4. Solve graphically the quadratic equation 5x C9xC7. D 0 given that the solutions lie between x D andxd. Determine also the co-ordinates of the turning point and state its nature. Fig..0 Let y D 5x C 9x C 7.. A table of values is drawn up as shown on page 9. A graph of y D 5x C9x C7. is shown plotted in Fig..0. The graph crosses the x-axis (i.e. where y D 0) at x = 0.6 andx =.4 and these are the solutions of the quadratic equation 5x C 9x C 7. D 0 The turning point is a maximum having co-ordinates (0.9,.5)

102 Graphical solution of equations 9 x x C9x C y D 5x C 9x C x.5 5x C9x C y D 5x C 9x C Problem 5. Plot a graph of y D x and hence solve the equations: (a) x 8 D 0and(b)x x D 0 Problem 6. Plot the graph of y D x C x C 6for values of x from x D tox D 4. Use the graph to find the roots of the following equations: (a) x C x C 6 D 0 (b) x C x C D 0 (c) x C x C 9 D 0 (d) x C x C 5 D 0 A table of values is drawn up as shown below. x 0 4 x Cx C y A graph of x C x C 6 is shown in Fig... A graph of y D x is shown in Fig... y y 8 y = x +x +6 y = x A 0 8 B y = 8 C 6 4 H D y = x + y = 4 C x D y =x +.5. A B G x Fig.. E 4 F y = (a) Rearranging x 8 D 0 gives x D 8 and the solution of this equation is obtained from the points of intersection of y D x and y D 8, i.e. at co-ordinates (,8) and (,8), shown as A and B, respectively, in Fig... Hence the solutions of x 8 D 0 are x = andx =+ (b) Rearranging x x D 0givesx D x C andthe solution of this equation is obtained from the points of intersection of y D x and y D x C, i.e. at C and D in Fig... Hence the solutions of x x D 0are x = andx =.5 Fig (a) The parabola y D x C x C 6 and the straight line y D 0 intersect at A and B, where x =. and x =.6 and these are the roots of the equation x C x C 6 D 0

103 9 Basic Engineering Mathematics (b) Comparing y D x C x C 6 with 0 D x C x C shows that if 4 is added to both sides of equation (), the right-hand side of both equations will be the same. Hence 4 D x C x C 6. The solution of this equation is found from the points of intersection of the line y D 4 and the parabola y D x C x C 6, i.e. points C and D in Fig... Hence the roots of x C x C D 0 are x = 0.5 andx = (c) x C x C 9 D 0 may be rearranged as x C x C 6 D and the solution of this equation is obtained from the points of intersection of the line y D and the parabola y D x Cx C6x, i.e. at points E and F in Fig... Hence the roots of x C x C 9 D 0arex =.5 and x = (d) Comparing y D x C x C 6 with 0 D x C x C 5 4 shows that if xc is added to both sides of equation (4) the right-hand side of both equations will be the same. Hence equation (4) may be written as x C D x C x C 6 The solution of this equation is found from the points of intersection of the line y D x C and the parabola y D x C x C 6, i.e. points G and H in Fig... Hence the roots of x C x C 5 D 0arex =.5 and x =.85 Now try the following exercise Exercise 44 Further problems on solving quadratic equations graphically (Answers on page 57). Sketch the following graphs and state the nature and co-ordinates of their turning points (a) y D 4x (b) y D x (c) y D x C (d) y D x Solve graphically the quadratic equations in Problems to 5 by plotting the curves between the given limits. Give answers correct to decimal place.. 4x x D 0; x D toxd. x x D 7; x D 5toxD8 4. x 6x 9 D 0; x D toxd5 5. x 5x D 9.6; x D toxd 6. Solve the quadratic equation x C 7x C 6 D 0 graphically, given that the solutions lie in the range x D tox D. Determine also the nature and co-ordinates of its turning point. 7. Solve graphically the quadratic equation 0x 9x. D 0 given that the roots lie between x D andxd 8. Plot a graph of y D x and hence solve the equations. (a) x 8 D 0and(b)x x D 0 9. Plot the graphs y D x and y D 4x on the same axes and find the co-ordinates of the points of intersection. Hence determine the roots of the equation x C 4x D 0 0. Plot a graph of y D 0x x 0 for values of x between x D andxd. Solve the equation 0x x 0 D 0 and from the graph determine (a) the value of y when x is., (b) the value of x when y is 0 and (c) the roots of the equation 0x 5x 8 D 0. Graphical solution of linear and quadratic equations simultaneously The solution of linear and quadratic equations simultaneously may be achieved graphically by: (i) plotting the straight line and parabola on the same axes, and (ii) noting the points of intersection. The co-ordinates of the points of intersection give the required solutions. Problem 7. Determine graphically the values of x and y which simultaneously satisfy the equations y D x x 4andy D 4x y D x x 4 is a parabola and a table of values is drawn up as shown below: x 0 x x y y D 4x is a straight line and only three co-ordinates need be calculated: x 0 y 6 The two graphs are plotted in Fig.. and the points of intersection, shown as A and B, are at co-ordinates (,0)

104 Graphical solution of equations 9 y A y = x x 4 the required solution since at these points both y D 0and ax C bx C cx C d D 0. The number of solutions, or roots of a cubic equation depends on how many times the curve cuts the x-axis and there can be one, two or three possible roots, as shown in Fig y y y x x x 0 x 4 B Fig..4 (a) (b) (c) Fig.. y = 4x and ( 4). Hence the simultaneous solutions occur when x =, y = 0 and when x =, y = 4 (These solutions may be checked by substituting into each of the original equations) Now try the following exercise Exercise 45 Further problems on solving linear and quadratic equations simultaneously (Answers on page 58). Determine graphically the values of x and y which simultaneously satisfy the equations y D x x 4 and y C 4 D x. Plot the graph of y D 4x 8x for values of x from toc4. Use the graph to find the roots of the following equations: (a) 4x 8x D 0 (b) 4x 8x 6 D 0 (c) 4x 6x 8 D 0.4 Graphical solution of cubic equations A cubic equation of the form ax C bx C cx C d D 0may be solved graphically by: (i) plotting the graph y D ax C bx C cx C d and (ii) noting the points of intersection on the x-axis (i.e. where y D 0). The x-values of the points of intersection give Problem 8. Solve graphically the cubic equation 4x 8x 5x C 9 D 0 given that the roots lie between x D andx D. Determine also the co-ordinates of the turning points and distinguish between them. Let y D 4x 8x 5x C 9. A table of values is drawn up as shown below. x 0 4x x x C y A graph of y D 4x 8x 5x C 9 is shown in Fig..5. The graph crosses the x-axis (where y D 0) at x =, x = and x = and these are the solutions to the cubic equation 4x 8x 5x C 9 D 0. The turning points occur at ( 0.6,4.), whichisamaximum, and(, ), whichis a minimum. Problem 9. Plot the graph of y D x 7x C4xC4for values of x between x D andx D. Hence determine the roots of the equation x 7x C 4x C 4 D 0 A table of values is drawn up as shown below. x 0 x x C4x C y

105 94 Basic Engineering Mathematics y y = 4x 8x 5x x By plotting a graph of y D x x 5x C 6 between x D and x D 4 solve the equation x x 5x C6 D 0. Determine also the co-ordinates of the turning points and distinguish between them. In Problems to 6, solve graphically the cubic equations given, each correct to significant figures.. x D 0 4. x x 5x C D 0 5. x x D x 6. x x 9.08x C 8.8 D 0 7. Show that the cubic equation 8x C 6x C 54x C 7 D 0 has only one real root and determine its value. Fig..5 Fig..6 y y =x 7x +4x+4 0 x A graph of y D x 7x C 4x C 4 is shown in Fig..6. The graph crosses the x-axis at x D 0.5 and touches the x-axis at x D. Hence the solutions of the equation x 7x C4x C4 D 0 are x = 0.5 andx = Now try the following exercise Exercise 46 Further problems on solving cubic equations (Answers on page 58). Plot the graph y D 4x C 4x x 6 between x D andx D and use the graph to solve the cubic equation 4x C 4x x 6 D 0 Assignment 6 This assignment covers the material contained in Chapters and. The marks for each question are shown in brackets at the end of each question.. Determine the gradient and intercept on the y-axis for the following equations: (a) y D 5xC (b) x C y C D 0 (5). The equation of a line is y D 4x C 7. A table of corresponding values is produced and is as shown below. Complete the table and plot a graph of y against x. Determine the gradient of the graph. (6) x 0 y Plot the graphs y D x C and y Cx D 6onthesame axes and determine the co-ordinates of their point of intersection. (7) 4. The velocity v of a body over varying time intervals t was measured as follows: t seconds v m/s Plot a graph with velocity vertical and time horizontal. Determine from the graph (a) the gradient, (b) the

106 Graphical solution of equations 95 vertical axis intercept, (c) the equation of the graph, (d) the velocity after.5 s, and (e) the time when the velocity is 8 m/s. (9) 5. Solve, correct to decimal place, the quadratic equation x 6x 9 D 0 by plotting values of x from x D toxd5 (7) 6. Plot the graph of y D x C 4x C x 6 for values of x between x D 4andx D. Hence determine the roots of the equation x C 4x C x 6 D 0 (7) 7. Plot a graph of y D x from x D toxdcand hence solve the equations: (a) x 8 D 0 (b) x 4x 6 D 0 (9)

107 Logarithms. Introduction to logarithms With the use of calculators firmly established, logarithmic tables are now rarely used for calculation. However, the theory of logarithms is important, for there are several scientific and engineering laws that involve the rules of logarithms. If a number y can be written in the form a x, then the index x is called the logarithm of y to the base of a, i.e. if y = a x then x = log a y Thus, since 000 D 0,thenDlog Check this using the log button on your calculator. (a) Logarithms having a base of 0 are called common logarithms and log 0 is usually abbreviated to lg. The following values may be checked by using a calculator: lg 7.9 D.58...,lg 46.7 D and lg 0.07 D (b) Logarithms having a base of e (where e is a mathematical constant approximately equal to.78) are called hyperbolic, Napierian or natural logarithms, andlog e is usually abbreviated to ln. The following values may be checked by using a calculator: ln.5 D ,ln 6.7 D and ln 0.56 D For more on Napierian logarithms see Chapter 4.. Laws of logarithms There are three laws of logarithms, which apply to any base: (i) To multiply two numbers: log (A U B) =log A + log B The following may be checked by using a calculator: lg 0 D, also lg 5 C lg D C D Hence lg 5 ð D lg 0 D lg 5 C lg (ii) To divide two numbers: ( ) A log = log A log B B The following may be checked using a calculator: ( ) 5 ln D ln.5 D Also ln 5 ln D D ( ) 5 Hence ln D ln 5 ln (iii) To raise a number to a power: lg A n = n log A The following may be checked using a calculator: lg 5 D lg 5 D Also lg5d ð D Hence lg 5 D lg5 Problem. Evaluate (a) log 9 (b) log 0 0 (c) log 6 8

108 Logarithms 97 (a) Let x D log 9then x D 9 from the definition of a logarithm, i.e. x D, from which x D. Hence log 9 = (b) Let x D log 0 0 then 0 x D 0 from the definition of a logarithm, i.e. 0 x D 0, from which x D. Hence log 0 0 = (which may be checked by a calculator). (c) Let x =log 6 8then6 x D 8, from the definition of a logarithm, i.e. ( 4 x D,i.e. 4x D from the laws of indices, from which, 4x D andxd 4 Hence log 6 8 = 4 Problem. Evaluate (a) lg 0.00 (b) ln e (c) log 8 (a) Let x D lg 0.00 D log then 0 x D 0.00, i.e. 0 x D 0, from which x D Hence lg 0.00 = (which may be checked by a calculator). (b) Let x D ln e D log e ethene x D e, i.e. e x D e from which x D Hence ln e = (which may be checked by a calculator). (c) Let x D log 8 then x D 8 D D 4 4, from which x D 4 Hence log 8 = 4 Problem. Solve the following equations: (a) lg x D (b) log x D (c) log 5 x D (a) If lg x D thenlog 0 x D andxd0,i.e.x = 000 (b) If log x D thenx = = 8 (c) If log 5 x D thenx D 5 D 5 D 5 (a) Problem 4. Write (a) log 0 (b) log 450 in terms of log, log and log 5 to any base. log 0 D log ð 5 D log ð ð 5 D log + log + log 5 by the first law of logarithms (b) log 450 D log ð 5 D log ð ð 75 D log ð ð ð 5 D log ð ð 5 D log C log C log 5 by the first law of logarithms i.e log 450 D log + log+ log5 by the third law of logarithms ( p ) 8 ð 4 5 Problem 5. Write log in terms of log, 8 logandlog5toanybase. ( p ) 8 ð 4 5 log D log 8 C log 4p 5 log 8, by the first 8 and second laws of logarithms D log C log 5 /4 log 4 by the laws of indices ( p ) 8 ð 4 5 i.e. log D log+ log 5 4log 8 4 by the third law of logarithms Problem 6. Simplify log 64 log 8 C log 64 D 6, 8 D 7 and D 5 Hence log 64 log 8 C log D log 6 log 7 C log 5 D 6log 7logC 5log by the third law of logarithms D 4 log Problem 7. Evaluate log 5 log 5 C log 65 log5 log 5 log 5 C log 65 log5 Problem 8. Solve the equation: D log 5 log 5 C log 54 log5 D log5 log5c 4 log 5 log5 D log5 log5 D log x C log x C D log x C log x C log x C D log x x C from the first law of logarithms D log x log x C D log x C D log x C 4x C 4

109 98 Basic Engineering Mathematics Hence if log x D log x C 4x C 4 then x D x C 4x C 4 i.e. D 4x C 4 i.e. 5 D 4x i.e. x = 5 4 or 4 Solve the equations given in Problems 8 to 0: 8. log x 4 log x D log 5x log x 9. log t log t D log 6 C log t 0. log b logb D log 8b log 4b Now try the following exercise Exercise 47 Further problems on the laws of logarithms (Answers on page 58) In Problems to, evaluate the given expression:. log log 6. log log 8 5. log 8 6. log lg lg log log 7. ln e In Problems to 8 solve the equations:. log 0 x D 4. log x D 5 4. log x D 5. log 4 x D 6. lg x D 7. log 8 x D 4 8. ln x D In Problems 9 to write the given expressions in terms of log, log and log 5 to any base: 9. log log 00 ( p ) ( 6 ð 4 5 p ) 5 ð 4 6. log. log 7 4p 8 Simplify the expressions given in Problems to 5:. log 7 log 9 C log 8 4. log 64 C log log 8 5. log 8 log 4 C log Evaluate the expressions given in Problems 6 and 7: log 6 log 8 log 4 log 9 log C log 8 log. Indicial equations The laws of logarithms may be used to solve certain equations involving powers called indicial equations. For example, to solve, say, x D 7, logarithms to a base of 0 are taken of both sides, i.e. log 0 x D log 0 7 and x log 0 D log 0 7 by the third law of logarithms. Rearranging gives x D log D D which log may be readily checked. (Note, (log 8/ log ) is not equal to lg(8/)) Problem 9. Solve the equation x D, correct to 4 significant figures. Taking logarithms to base 0 of both sides of x D gives: log 0 x D log 0 i.e. x log 0 D log 0 Rearranging gives: x D log D log D.585 correct to 4 significant figures. Problem 0. Solve the equation xc D x 5 correct to decimal places. Taking logarithms to base 0 of both sides gives: log 0 xc D log 0 x 5 i.e. x C log 0 D x 5 log 0 x log 0 C log 0 D x log 0 5log 0 x 0.00 C 0.00 D x i.e. 0.00x C 0.00 D 0.954x.855 Hence.855 C 0.00 D 0.954x 0.00x.6865 D 0.65x

110 Logarithms 99 from which x D D 4. correct to decimal places. y 0.5 Problem. Solve the equation x. D 4.5, correct to 4 significant figures. Taking logarithms to base 0 of both sides gives: x x y = log 0 x log 0 x. D log log 0 x D log Hence log 0 x D log D Thus x D antilog D D.95 correct to 4 significant figures. Now try the following exercise Exercise 48 Indicial equations (Answers on page 58) Solve the following indicial equations for x, each correct to 4 significant figures:. x D 6.4. x D 9. x D x 4. x.5 D D 4. x 6. 4 x D 5 xc 7. x 0.5 D x D.6 Fig.. y x Fig.. x y = log e x Graphs of logarithmic functions A graph of y D log 0 x is shown in Fig.. and a graph of y D log e x is shown in Fig... Both are seen to be of similar shape; in fact, the same general shape occurs for a logarithm to any base. In general, with a logarithm to any base a, it is noted that: (i) log a = 0 Let log a D x, thena x D from the definition of the logarithm. If a x D thenxd0 from the laws of logarithms. Hence log a D 0. In the above graphs it is seen that log 0 D 0andlog e D 0 (ii) log a a = Let log a a D x then a x D a from the definition of a logarithm. If a x D a then x D Hence log a a D. (Check with a calculator that log 0 0 D andlog e e D ) (iii) log a 0 X Let log a 0 D x then a x D 0 from the definition of a logarithm. If a x D 0, and a is a positive real number, then x must approach minus infinity. (For example, check with a calculator, D 0.5, 0 D 9.54 ð 0 7, 00 D 6. ð 0 6, and so on.) Hence log a 0!

111 4 Exponential functions 4. The exponential function An exponential function is one which contains e x,ebeing a constant called the exponent and having an approximate value of.78. The exponent arises from the natural laws of growth and decay and is used as a base for natural or Napierian logarithms. 4. Evaluating exponential functions The value of e x may be determined by using: (a) a calculator, or (b) the power series for e x (see Section 4.), or (c) tables of exponential functions. The most common method of evaluating an exponential function is by using a scientific notation calculator, this now having replaced the use of tables. Most scientific notation calculators contain an e x function which enables all practical values of e x and e x to be determined, correct to 8 or 9 significant figures. For example e D.7888 e.4 D.076 e.68 D correct to 8 significant figures. In practical situations the degree of accuracy given by a calculator is often far greater than is appropriate. The accepted convention is that the final result is stated to one significant figure greater than the least significant measured value. Use your calculator to check the following values: e 0. D.75, correct to 5 significant figures e.47 D 0.99, correct to 5 decimal places e 0.4 D , correct to 4 decimal places e 9. D 59, correct to 5 significant figures e.785 D , correct to 7 decimal places Problem. Using a calculator, evaluate, correct to 5 significant figures: (a) e.7 (b) e.6 (c) 5 e5.5 (a) e.7 D D 5.48, correct to 5 significant figures. (b) e.6 D D 0.044, correct to 5 significant figures. (c) 5 e5.5 D D 8.56, correct to 5 significant figures. Problem. Use a calculator to determine the following, each correct to 4 significant figures: (a).7e 0.8 (b) 5.e.4 (c) 5 e7 (a).7e 0.8 D D 4.454, correct to 4 significant figures. (b) 5.e.4 D D., correct to 4 significant figures. 5 (c) e7 D D 44.94, correct to 4 significant figures.

112 Exponential functions 0 Problem. Evaluate the following correct to 4 decimal places, using a calculator: ( e (a) e 5. e e 0.5 ) (b) 5 e 0.5 C e 0.5 (a) e 5. e.49 D D 4.784, correct to 4 decimal places ( e 0.5 e 0.5 ) (b) 5 e 0.5 C e 0.5 ( ) D C ( ) D D.46, correct to 4 decimal places Problem 4. The instantaneous voltage v in a capacitive circuit is related to time t by the equation v D Ve t/cr where V, C and R are constants. Determine v, correct to 4 significant figures, when t D 0 ð 0 seconds, C D 0 ð 0 6 farads, R D 47 ð 0 ohms and V D 00 volts. v D Ve t/cr D 00e 0ð0 / 0ð0 6 ð47ð0 Using a calculator, v D 00e Now try the following exercise Exercise 49 D D 87.6 volts Further problems on evaluating exponential functions (Answers on page 58) In Problems and use a calculator to evaluate the given functions correct to 4 significant figures:. (a) e 4.4 (b) e 0.5 (c) e 0.9. (a) e.8 (b) e 0.78 (c) e 0. Evaluate, correct to 5 significant figures: (a).5e.8 (b) 6 5 e.5 (c).6e Use a calculator to evaluate the following, correct to 5 significant figures: (a) e.69 (b) e.748 (c) 0.6e 4.78 In Problems 5 and 6, evaluate correct to 5 decimal places. 5. (a) 6. (a) (c) 7 e.469 (b) 8.5e.65 (c) 5.68 e.47 4 e.795 e.687 (b) 5e.69 e.7 e.7 e.7 7. The length of a bar, l, at a temperature q is given by l D l 0 e aq,wherel 0 and a are constants. Evaluate l, correct to 4 significant figures, when l 0 D.587, q D.7 anda D.77 ð The power series for e x The value of e x can be calculated to any required degree of accuracy since it is defined in terms of the following power series: e x D C x C x! C x! C x4 CÐÐÐ 4! (where! D ð ð and is called factorial ) The series is valid for all values of x. The series is said to converge, i.e.ifallthetermsare added, an actual value for e x (where x is a real number) is obtained. The more terms that are taken, the closer will be the value of e x to its actual value. The value of the exponent e, correct to say 4 decimal places, may be determined by substituting x D in the power series of equation (). Thus i.e. e D C C! C 7 7! C 8 8! C! CÐÐÐ C 4 4! C 5 5! C 6 6! D C C 0.5 C C C C C C CÐÐÐ D.788 e D.78 correct to 4 decimal places The value of e 0.05, correct to say 8 significant figures, is found by substituting x D 0.05 in the power series for e x. Thus e 0.05 D C 0.05 C 0.05! C ! C ! C 0.05! CÐÐÐ

113 0 Basic Engineering Mathematics and by adding, D C 0.05 C C C C e 0.05 D.057, correct to 8 significant figures In this example, successive terms in the series grow smaller very rapidly and it is relatively easy to determine the value of e 0.05 to a high degree of accuracy. However, when x is nearer to unity or larger than unity, a very large number of terms are required for an accurate result. If in the series of equation (), x is replaced by x, then e x D C x C x! C x! CÐÐÐ e x D x C x x!! CÐÐÐ In a similar manner the power series for e x may be used to evaluate any exponential function of the form ae kx,wherea and k are constants. In the series of equation (), let x be replaced by kx. Then ae kx D a { C kx C kx! Thus 5e x D 5 { C x C x! D 5 { C x C 4x C kx! C x! C 8x 6 CÐÐÐ { i.e. 5e x D 5 C x C x C 4 x CÐÐÐ } CÐÐÐ } CÐÐÐ Problem 5. Determine the value of 5e 0.5, correct to 5 significant figures by using the power series for e x. e x D C x C x! C x! C x4 4! CÐÐÐ Hence e 0.5 D C 0.5 C 0.5 C 0.5 i.e. C } } C C D C 0.5 C 0.5 C C C C e 0.5 D.6487 correct to 6 significant figures Hence 5e 0.5 D D 8.46, correct to 5 significant figures. Problem 6. Determine the value of e, correct to 4 decimal places, using the power series for e x. Substituting x D in the power series e x D C x C x! C x! C x4 4! CÐÐÐ gives e D C C! C! C 4 4! CÐÐÐ D C C C CÐÐÐ D correct to 6 decimal places Hence e D D.06 correct to 4 decimal places. Problem 7. Expand e x x as far as the term in x 5 The power series for e x is e x D C x C x! C x! C x4 4! C x5 5! CÐÐÐ Hence ) e x x D ( C x C x! C x! C x4 4! C x5 5! CÐÐÐ x ) D (x C x C x4! C x5! CÐÐÐ ) ( C x C x! C x! C x4 4! C x5 5! CÐÐÐ Grouping like terms gives: ) ) e x x D xc (x x C (x x!! ( x 4 ) ( x4 x 5 ) x5 C C CÐÐÐ! 4!! 5! D x + x x + 4 x x 5 when expanded as far as the term in x 5 Now try the following exercise Exercise 50 Further problems on the power series for e x (Answers on page 58). Evaluate 5.6e, correct to 4 decimal places, using the power series for e x.

114 Exponential functions 0. Use the power series for e x to determine, correct to 4 significant figures, (a) e (b) e 0. and check your result by using a calculator.. Expand x e x as far as the term in x Expand e x x / ) to six terms. A table of values is drawn up as shown below. x 0 0.x e 0.x e 0.x A graph of y D e 0.x is shown plotted in Fig Graphs of exponential functions y Values of e x and e x obtained from a calculator, correct to decimal places, over a range x D tox D, are shown in the following table y = e 0.x x e x e x x e x e x Figure 4. shows graphs of y D e x and y D e x x y Fig. 4. y = e x 0 6 y = e x From the graph, when x =., y =.87 and when y =.6, x = 0.74 Problem 9. Plot a graph of y D e x over the range x D.5 tox D.5. Determine from the graph the value of y when x D.and the value of x when y D A table of values is drawn up as shown below. x x 0 e x e x Fig x Problem 8. Plot a graph of y D e 0.x over a range of x D tox D. Hence determine the value of y when x D. and the value of x when y D.6 A graph of e x is shown in Fig. 4.. From the graph, when x =., y =.67 and when y =.4, x = 0.7 Problem 0. The decay of voltage, v volts, across a capacitor at time t seconds is given by v D 50e t/. Draw a graph showing the natural decay curve over the first 6 seconds. From the graph, find (a) the voltage after.4 s, and (b) the time when the voltage is 50 V.

115 04 Basic Engineering Mathematics y = e x y From the graph: (a) when time t =.4 s, voltage v = 80 volts and (b) when voltage v = 50 volts, time t =.5 seconds. Now try the following exercise 4.67 Exercise 5 Further problems on exponential graphs (Answers on page 58).4. Plot a graph of y D e 0.x over the range x D to x D. Hence determine the value of y when x D.4 and the value of x when y D 4.5 Fig A table of values is drawn up as shown below x t 0 e t/ v D 50e t/ t e t/ v D 50e t/ Plot a graph of y D e.5x over a range x D.5 to x D.5 and hence determine the value of y when x D 0.8 and the value of x when y D.5. In a chemical reaction the amount of starting material C cm left after t minutes is given by C D 40e 0.006t. Plot a graph of C against t and determine (a) the concentration C after hour, and (b) the time taken for the concentration to decrease by half. 4. The rate at which a body cools is given by q D 50e 0.05t where the excess of temperature of a body above its surroundings at time t minutes is q C. Plot a graph showing the natural decay curve for the first hour of cooling. Hence determine (a) the temperature after 5 minutes, and (b) the time when the temperature is 95 C The natural decay curve of v D 50e t/ Fig is shown in 4.5 Napierian logarithms 50 v = 50e t/ Logarithms having a base of e are called hyperbolic, Napierian or natural logarithms and the Napierian logarithm of x is written as log e x, or more commonly, ln x. Voltage v (volts) Fig Time t (seconds) 4.6 Evaluating Napierian logarithms The value of a Napierian logarithm may be determined by using: (a) a calculator, or (b) a relationship between common and Napierian logarithms, or (c) Napierian logarithm tables. The most common method of evaluating a Napierian logarithm is by a scientific notation calculator, this now having replaced the use of four-figure tables, and also the

116 Exponential functions 05 relationship between common and Napierian logarithms, log e y D.06 log 0 y Most scientific notation calculators contain a ln x function which displays the value of the Napierian logarithm of a number when the appropriate key is pressed. Using a calculator, ln 4.69 D D.5459, correct to 4 decimal places and ln 5.78 D D.5774, correct to 4 decimal places Use your calculator to check the following values: ln.7 D , correct to 5 significant figures ln D 0 ln 0.5 D 0.659, correct to 4 decimal places ln 59 D 6.85, correct to 5 significant figures ln 750 D , correct to 4 decimal places ln 0.7 D.77, correct to 4 significant figures ln D , correct to 6 significant figures ln e D ln e D From the last two examples we can conclude that log e e x = x This is useful when solving equations involving exponential functions. For example, to solve e x D 8, take Napierian logarithms of both sides, which gives ln e x D ln 8 i.e. x D ln 8 from which x D ln 8 D 0.69, correct to 4 decimal places Problem. Using a calculator evaluate correct to 5 significant figures: (a) ln 47.9 (b) ln 0.06 (c). ln 76.9 (a) ln 47.9 D D.856, correct to 5 significant figures. (b) ln 0.06 D D.7785, correct to 5 significant figures. (c).ln76.9 D D.9, correct to 5 significant figures. Problem. Use a calculator to evaluate the following, each correct to 5 significant figures: (a) (a) (c) 4 ln 4.79 (b) ln ln 4.07 e ln 4.79 D D 0.884, correct to 5 4 significant figures. ln (b) D D 0.65, correct to significant figures. 5.9 ln (c) D D 0.070, correct to 5 significant e figures. (a) (b) Problem. (a) ln e.5 lg Evaluate the following: (b) ln e.5.5 D lg D 5 4e. lg. ln. 4e. lg. ln. (correct to decimal places) D D 6.56, correct to decimal places Problem 4. Solve the equation 7 D 4e x to find x, correct to 4 significant figures. Rearranging 7 D 4e x gives: 7 4 D e x Taking the reciprocal of both sides gives: 4 7 D D ex e x Taking Napierian logarithms of both sides gives: ( ) 4 ln D ln e x 7 ( ) 4 Since log e e a D a, thenln D x 7 Hence x D ( ) 4 ln D D 0.865, correct 7 to 4 significant figures. Problem 5. Given 0 D 60 e t/ determine the value of t, correct to significant figures.

117 06 Basic Engineering Mathematics Rearranging 0 D 60 e t/ gives: 0 D e / 60 and e t/ D 0 60 D Taking the reciprocal of both sides gives: e t/ D Taking Napierian logarithms of both sides gives: 5. (a) (c).946 ln e.76 lg 0.4 ln ln (b) 5e 0.69 ln In Problems 6 to 0 solve the given equations, each correct to 4 significant figures D 4e t D.9e.7x ( x ) 8. 6 D 4 e t/ D ln ( ) ln D.4 x ln e t/ D ln i.e. t D ln from which, t D ln D 0.88, correct to significant figures. Problem 6. find x. ( ) 5.4 Solve the equation.7 D ln to x ( ) 5.4 From the definition of a logarithm, since.7 D x then e.7 D 5.4 x Rearranging gives: x D 5.4 D 5.4e.7 e.7 i.e. x = 0.46, correct to 4 significant figures Now try the following exercise Exercise 5 Further problems on evaluating Napierian logarithms (Answers on page 58) In Problems to use a calculator to evaluate the given functions, correct to 4 decimal places.. (a) ln.7 (b) ln 5.4 (c) ln 9.4. (a) ln 7. (b) ln 54. (c) ln 94. (a) ln 0.7 (b) ln (c) ln In Problems 4 and 5, evaluate correct to 5 significant figures. 4. (a) 6 ln 5.9 (b) ln (c) 5.6 ln.6 e Laws of growth and decay The laws of exponential growth and decay are of the form y D Ae kx and y D A e kx,wherea and k are constants. When plotted, the form of each of these equations is as shown in Fig The laws occur frequently in engineering and science and examples of quantities related by a natural law include: (i) Linear expansion l D l 0 e aq (ii) Change in electrical resistance with temperature R q D R 0 e aq (iii) Tension in belts (iv) Newton s law of cooling (v) Biological growth (vi) Discharge of a capacitor (vii) Atmospheric pressure (viii) Radioactive decay (ix) Decay of current in an inductive circuit T D T 0 e mq q D q 0 e kt y D y 0 e kt q D Qe t/cr p D p 0 e h/c N D N 0 e lt i D Ie Rt/L (x) Growth of current in a capacitive circuit i D I e t/cr Problem 7. The resistance R of an electrical conductor at temperature q C isgivenbyr D R 0 e aq,where a is a constant and R 0 D 5 ð 0 ohms. Determine the value of a, correct to 4 significant figures, when R D 6 ð 0 ohms and q D 500 C. Also, find the temperature, correct to the nearest degree, when the resistance R is 5.4 ð 0 ohms. Transposing R D R 0 e aq gives R R 0 D e aq Taking Napierian logarithms of both sides gives: ln R R 0 D ln e aq D aq

118 Exponential functions 07 y A Transposing q D q 0 e kt gives q q 0 D e kt from which y =Ae kx 0 x y A y = A( e kx ) 0 x q 0 q D D e kt e kt Taking Napierian logarithms of both sides gives: ln q 0 q D kt from which, k D t ln q 0 q D 8.0 ln D Hence k =.485 U 0 ( ) Problem 9. Thecurrent i amperes flowing in a capacitor at time t seconds is given by i D 8.0 e t/cr, where the circuit resistance R is 5 ð 0 ohms and capacitance C is 6 ð 0 6 farads. Determine (a) the current i after 0.5 seconds and (b) the time, to the nearest millisecond, for the current to reach 6.0 A. Sketch the graph of current against time. Fig. 4.5 (a) Current i D 8.0 e t/cr Hence a D q ln R D ( 6 ð 0 ) R ln 5 ð 0 D D ð Hence a =.5 U 0 4, correct to 4 significant figures. From above, ln R R 0 D aq hence q D a ln R R 0 When R D 5.4 ð 0, a D ð 0 4 and R 0 D 5 ð 0 ( 5.4 ð 0 ) q D ln ð ð D ð 0 D 6 C correct to the nearest degree. Problem 8. In an experiment involving Newton s law of cooling, the temperature q C is given by q D q 0 e kt. Find the value of constant k when q 0 D 56.6 C, q D 6.5 C andt D 8.0 seconds. D 8.0[ e 0.5/ 6ð0 6 5ð0 ] D 8.0 e.5 D D D 5.7 amperes (b) Transposing i D 8.0 e t/cr i gives D e t/cr 8.0 from which, e t/cr D i 8.0 i D Taking the reciprocal of both sides gives: e t/cr D i Taking Napierian logarithms of both sides gives: ( ) t 8.0 CR D ln 8.0 i ( ) 8.0 Hence t D CR ln 8.0 i ( ) 8.0 D 6 ð ð 0 ln when i D 6.0 amperes,

119 08 Basic Engineering Mathematics i.e. t D 400 ( ) ln D 0.4ln4.0.0 D D s D 555 ms, to the nearest millisecond A graph of current against time is shown in Fig Since q D q ( t D 60 ln ) D 60 ln 0.5 D 4.59 s Hence the time for the temperature q to be one half of the value of q is 4.6 s, correct to decimal place i(a) Fig i = 8.0( e t/cr ) t(s) Problem 0. The temperature q of a winding which is being heated electrically at time t is given by: q D q e t/t where q is the temperature (in degrees Celsius) at time t D 0andt is a constant. Calculate. (a) q, correct to the nearest degree, when q is 50 C, t is 0 s and t is 60 s (b) the time t, correct to decimal place, for q to be half the value of q (a) Transposing the formula to make q the subject gives: q D D q e t/t D 50 e 0/60 50 e D i.e. q = 7 C, correct to the nearest degree. (b) Transposing to make t the subject of the formula gives: q q D e t/t from which, e t/t D q q Hence t ( t D ln q ) i.e. q ( t D tln q ) q Now try the following exercise Exercise 5 Further problems on the laws of growth and decay (Answers on page 58). The pressure p pascals at height h metres above ground level is given by p D p 0 e h/c, where p 0 is the pressure at ground level and C is a constant. Find pressure p when p 0 D.0 ð 0 5 Pa, height h D 40 m and C D The voltage drop, v volts, across an inductor L henrys at time t seconds is given by v D 00e Rt/L,where R D 50 and L D.5 ð 0 H. Determine (a) the voltage when t D 60 ð 0 6 s, and (b) the time for the voltage to reach 85 V.. The length l metres of a metal bar at temperature t C isgivenbyl D l 0 e at,wherel 0 and a are constants. Determine (a) the value of l when l 0 D.894, a D.08 ð 0 4 and t D 50 C, and (b) the value of l 0 when l D.46, t D 0 C anda D.68 ð The temperature q C of an electrical conductor at time t seconds is given by q D q e t/t,whereq is the initial temperature and T seconds is a constant. Determine (a) q when q D 59.9 C, t D 0 s and T D 80 s, and (b) the time t for q to fall to half the value of q if T remains at 80 s. 5. A belt is in contact with a pulley for a sector of q D. radians and the coefficient of friction between these two surfaces is m D 0.6. Determine the tension on the taut side of the belt, T newtons, when tension on the slack side is given by T 0 D.7 newtons, given that these quantities are related by the law T D T 0 e mq 6. The instantaneous current i at time t is given by: i D 0e t/cr when a capacitor is being charged. The capacitance C is 7 ð 0 6 farads and the resistance R is 0. ð 0 6 ohms. Determine: (a) the instantaneous current when t is.5 seconds, and (b) the time for the instantaneous current to fall to 5 amperes.

120 Exponential functions 09 Sketch a curve of current against time from t D 0to t D 6 seconds. 7. The amount of product x (in mol/cm ) found in a chemical reaction starting with.5 mol/cm of reactant is given by x D.5 e 4t where t is the time, in minutes, to form product x. Plotagraphat 0 second intervals up to.5 minutes and determine x after minute. 8. The current i flowing in a capacitor at time t is given by: i D.5 e t/cr where resistance R is 0 kilohms and the capacitance C is 0 microfarads. Determine (a) the current flowing after 0.5 seconds, and (b) the time for the current to reach 0 amperes. 9. The amount A after n years of a sum invested P is given by the compound interest law: A D Pe rn/00 when the per unit interest rate r is added continuously. Determine, correct to the nearest pound, the amount after 8 years for a sum of 500 invested if the interest rate is 6% per annum. Assignment 7 This assignment covers the material contained in Chapters and 4. The marks for each question are shown in brackets at the end of each question.. Evaluate log 6 8 (). Solve (a) log x D (b) log x C log x D log log x (6). Solve the following equations, each correct to significant figures: (a) x D 5.5 (b) t D 7 tc (c) e x D 4. () 4. Evaluate the following, each correct to 4 significant figures: (a) e 0.68 (b).4e.6 5 e.7 (c) 6 e Use the power series for e x to evaluate e.5, correct to 4 significant figures. (4) 6. Expand xe x to six terms. (5) 7. Plot a graph of y D e.x over the range x D to x DC and hence determine, correct to decimal place, (a) the value of y when x D 0.75, and (b) the value of x when y D 4.0 (6) 8. Evaluate the following, each correct to decimal places: (a) ln 46.9 (b) ln0.075 (c) ln.68 ln (6) 9. Two quantities x and y are related by the equation y D ae kx,wherea and k are constants. Determine, correct to decimal place, the value of y when a D.4, k D.0 and x D.49 ()

121 5 Reduction of non-linear laws to linear form 5. Determination of law Frequently, the relationship between two variables, say x and y, is not a linear one, i.e. when x is plotted against y a curve results. In such cases the non-linear equation may be modified to the linear form, y D mx C c, so that the constants, and thus the law relating the variables can be determined. This technique is called determination of law. Some examples of the reduction of equations to linear form include: (i) y D ax C b compares with Y D mx C c, wherem D a, c D b and X D x. Hence y is plotted vertically against x horizontally to produce a straight line graph of gradient a andy-axis intercept b (ii) y D a x C b y is plotted vertically against horizontally to produce a straight line graph of gradient a andy-axis x intercept b (iii) y D ax C bx Dividing both sides by x gives y D ax C b. x Comparing with Y D mx C c shows that y is plotted x vertically against x horizontally to produce a straight line graph of gradient a and y axis intercept b x Problem. Experimental values of x and y, shown below, are believed to be related by the law y D ax C b. By plotting a suitable graph verify this law and determine approximate values of a and b. x 4 5 y If y is plotted against x a curve results and it is not possible to determine the values of constants a and b from the curve. Comparing y D ax C b with Y D mx C c shows that y is to be plotted vertically against x horizontally. A table of values is drawn up as shown below. x 4 5 x y A graph of y against x is shown in Fig. 5., with the best straight line drawn through the points. Since a straight line graph results, the law is verified. y Fig. 5. C x A B

122 Reduction of non-linear laws to linear form From the graph, gradient a D AB 5 7 D BC 5 5 D 6 0 D.8 and the y-axis intercept, b = 8.0 Hence the law of the graph is y =.8x A Problem. Values of load L newtons and distance d metres obtained experimentally are shown in the following table. L 5 0 Load, L N distance, d m Load, L N distance, d m B C Verify that load and distance are related by a law of the form L D a C b and determine approximate values of d a and b. Hence calculate the load when the distance is 0.0 m and the distance when the load is 0 N. Comparing L D a ( ) d C b i.e. L D a C b with Y D mx C c d shows that L is to be plotted vertically against d horizontally. Another table of values is drawn up as shown below. L d d A graph of L against is shown in Fig. 5.. A straight line d can be drawn through the points, which verifies that load and distance are related by a law of the form L D a d C b. Gradient of straight line, a D AB D BC D 0 0 D L-axis intercept, b = 5 Hence the law of the graph is L = d + 5 When the distance d D 0.0 m, load L D C 5 D 5.0N 0.0 Rearranging L D C 5 gives d d D 5 L and d D 5 L Fig d Hence when the load L D 0 N, distance d D 5 0 D 5 D 0. m Problem. The solubility s of potassium chlorate is shown by the following table: t C s The relationship between s and t is thought to be of the form s D CatCbt. Plot a graph to test the supposition and use the graph to find approximate values of a and b. Hence calculate the solubility of potassium chlorate at 70 C. Rearranging s D C at C bt gives s D at C bt and s t D a C bt or s t D bt C a which is of the form Y D mx C c, showing that s is t to be plotted vertically and t horizontally. Another table of values is drawn up as shown below. t s s t A graph of s t against t is shown plotted in Fig. 5..

123 Basic Engineering Mathematics S t Fig C A B t C A straight line fits the points which shows that s and t are related by s D C at C bt Gradient of straight line, b D AB D D 0.0 BC D Vertical axis intercept, a = 0.5 Hence the law of the graph is s = + 0.5t t The solubility of potassium chlorate at 70 C isgivenby s D C C D C 0.5 C 9.6 D. Now try the following exercise Exercise 54 Further problems on reducing non-linear laws to linear form (Answers on page 59) In Problems to 5, x and y are two related variables and all other letters denote constants. For the stated laws to be verified it is necessary to plot graphs of the variables in a modified form. State for each (a) what should be plotted on the vertical axis, (b) what should be plotted on the horizontal axis, (c) the gradient and (d) the vertical axis intercept.. y D d C cx. y a D b p x 5. y D a x C bx 6. In an experiment the resistance of wire is measured for wires of different diameters with the following results. R ohms d mm It is thought that R is related to d by the law R D a/d Cb,whereaand b are constants. Verify this and find the approximate values for a and b. Determine the cross-sectional area needed for a resistance reading of 0.50 ohms. 7. Corresponding experimental values of two quantities x and y are given below. x y By plotting a suitable graph verify that y and x are connected by a law of the form y D kx C c, where k and c are constants. Determine the law of the graph and hence find the value of x when y is Experimental results of the safe load L kn, applied to girders of varying spans, d m, are shown below. Span, d m Load, L kn It is believed that the relationship between load and span is L D c/d, wherec is a constant. Determine (a) the value of constant c and (b) the safe load for a span of.0 m. 9. The following results give corresponding values of two quantities x and y which are believed to be related by a law of the form y D ax C bx where a and b are constants. y x Verify the law and determine approximate values of a and b. Hence determine (i) the value of y when x is 8.0 and (ii) the value of x when y is Determination of law involving logarithms. y e D f x 4. y cx D bx Examples of reduction of equations to linear form involving logarithms include:

124 Reduction of non-linear laws to linear form (i) y D ax n Taking logarithms to a base of 0 of both sides gives: lg y D lg ax n D lg a C lg x n i.e. lg y D n lg x C lg a by the laws of logarithms which compares with Y D mx C c and shows that lg y is plotted vertically against lg x horizontally to produce a straight line graph of gradient n and lg y-axis intercept lg a (ii) y D ab x Taking logarithms to a base of 0 of the both sides gives: lg y D lg ab x i.e. lg y D lg a C lg b x i.e. lg y D x lg b C lg a by the laws of logarithms or lg y D lg b x C lg a which compares with Y D mx C c and shows that lg y is plotted vertically against x horizontally to produce a straight line graph of gradient lg b and lg y-axis intercept lg a (iii) y D ae bx Taking logarithms to a base of e of both sides gives: ln y D ln ae bx i.e. ln y D ln a C ln e bx i.e. ln y D ln a C bx ln e i.e. ln y D bx C ln a since ln e D which compares with Y D mx C c and shows that ln y is plotted vertically against x horizontally to produce a straight line graph of gradient b and ln y-axis intercept ln a Problem 4. The current flowing in, and the power dissipated by, a resistor are measured experimentally for various values and the results are as shown below. Current, I amperes Power, P watts Show that the law relating current and power is of the form P D RI n,wherer and n are constants, and determine the law. Taking logarithms to a base of 0 of both sides of P D RI n gives: lg P D lg RI n D lg R C lg I n D lg R C n lg I by the laws of logarithms. i.e. lg P D n lg I C lg R, which is of the form Y D mx C c, showing that lg P is to be plotted vertically against lg I horizontally. A table of values for lg I and lg P is drawn up as shown below. I lg I P lg P A graph of lg P against lg I is shown in Fig. 5.4 and since a straight line results the law P D RI n is verified. lg P Fig. 5.4 C Gradient of straight line, lg I D A B n D AB.98.8 D BC D D It is not possible to determine the vertical axis intercept on sight since the horizontal axis scale does not start at zero. Selecting any point from the graph, say point D, where lg I D 0.70 and lg P D.78, and substituting values into lg P D n lg I C lg R gives.78 D 0.70 C lg R from which lg R D D.8 Hence R D antilog.8 D 0.8 D 4.0 Hence the law of the graph is P = 4.0I Problem 5. The periodic time, T, of oscillation of a pendulum is believed to be related to its length, l, by a law of the form T D kl n, where k and n

125 4 Basic Engineering Mathematics are constants. Values of T were measured for various lengths of the pendulum and the results are as shown below. Periodic time, T s Length, l m Show that the law is true and determine the approximate values of k and n. Hence find the periodic time when the length of the pendulum is 0.75 m. From para (i), if T D kl n then lg T D n lg l C lg k and comparing with Y D mx C c shows that lg T is plotted vertically against lg l horizontally. A table of values for lg T and lg l is drawn up as shown below. T lg T l lg l A graph of lg T against lg l is shown in Fig. 5.5 and the law T D kl n is true since a straight line results. A 0.40 lg T From the graph, gradient of straight line, n D AB BC D D D Vertical axis intercept, lg k D 0.0 Hence k D antilog 0.0 D D.0 Hence the law of the graph is T =.0 l = or T =.0 p l When length l D 0.75 m then T D.0 p 0.75 D.7 s Problem 6. Quantities x and y are believed to be related by a law of the form y D ab x,wherea and b are constants. Values of x and corresponding values of y are: x y Verify the law and determine the approximate values of a and b. Hence determine (a) the value of y when x is. and (b) the value of x when y is 00 From para (ii), if y D ab x then lg y D lg b x C lg a and comparing with Y D mx C c shows that lg y is plotted vertically and x horizontally. Another table is drawn up as shown below. x y lg y A graph of lg y against x is shown in Fig. 5.6 and since a straight line results, the law y D ab x is verified. Gradient of straight line, lg b D AB..7 D BC.0.0 D D 0.48 Hence b D antilog 0.48 D D.0, correct to significant figures. Vertical axis intercept, lg a D 0.70, C Fig. 5.5 B lg I from which a D antilog 0.70 D D 5.0, correct to significant figures Hence the law of the graph is y = 5.0(.0) x (a) When x D., y D D 50.

126 Reduction of non-linear laws to linear form A which compares with y D mx C c, showing that ln i is plotted vertically against t horizontally. (For methods of evaluating Napierian logarithms see Chapter 4.) Another table of values is drawn up as shown below. t i ln i lg y.50 A graph of ln i against t is shown in Fig. 5.7 and since a straight line results the law i D Ie t/t is verified C B 5.0 A x D(00,.) ln i Fig (b) When y D 00, 00 D x,.0.0 B C from which 00/5.0 D.0 x, i.e. 0 D.0 x t (ms) Taking logarithms of both sides gives Hence x = lg 0 D lg.0 x D x lg.0 lg 0 lg.0 = =.7 Problem 7. The current i ma flowing in a capacitor which is being discharged varies with time t ms as shown below. i ma t ms Show that these results are related by a law of the form i D Ie t/t,wherei and T are constants. Determine the approximate values of I and T. Taking Napierian logarithms of both sides of i D Ie t/t gives ln i D ln Ie t/t D ln I C ln e t/t i.e. ln i D ln I C t T or ln i D ( ) t C ln I T (since ln e D.0 Fig. 5.7 Gradient of straight line, T D AB D BC D D 0.0 Hence T D 0.0 D 50 Selecting any point on the graph, say point D, wheret ( ) D 00 and ln i D., and substituting into ln i D t C ln I T gives. D 00 C ln I 50 from which ln I D. C 4.0 D 7. and I D antilog 7. D e 7. D 495 or 500 correct to significant figures. Hence the law of the graph is i = 500e t=50

127 6 Basic Engineering Mathematics Now try the following exercise Exercise 55 Further problems on reducing non-linear laws to linear form (Answers on page 59) In Problems to, x and y are two related variables and all other letters denote constants. For the stated laws to be verified it is necessary to plot graphs of the variables in a modified form. State for each (a) what should be plotted on the vertical axis, (b) what should be plotted on the horizontal axis, (c) the gradient and (d) the vertical axis intercept.. y D ba x. y D kx l y. m D enx 4. The luminosity I of a lamp varies with the applied voltage V and the relationship between I and V is thought to be I D kv n. Experimental results obtained are: I candelas V volts Verify that the law is true and determine the law of the graph. Determine also the luminosity when 75 V is applied across the lamp. 5. The head of pressure h and the flow velocity v are measured and are believed to be connected by the law v D ah b,wherea and b are constants. The results are as shown below. h v Verify that the law is true and determine values of a and b 6. Experimental values of x and y are measured as follows. x y The law relating x and y is believed to be of the form y D ab x,wherea and b are constants. Determine the approximate values of a and b. Hence find the value of y when x is.0 and the value of x when y is The activity of a mixture of radioactive isotope is believed to vary according to the law R D R 0 t c, where R 0 and c are constants. Experimental results are shown below. R t Verify that the law is true and determine approximate values of R 0 and c. 8. Determine the law of the form y D ae kx which relates the following values. y x The tension T in a belt passing round a pulley wheel and in contact with the pulley over an angle of q radiansisgivenbyt D T 0 e mq,wheret 0 and m are constants. Experimental results obtained are: T newtons q radians Determine approximate values of T 0 and m. Hence find the tension when q is.5 radians and the value of q when the tension is 50.0 newtons.

128 6 Geometry and triangles 6. Angular measurement Geometry is a part of mathematics in which the properties of points, lines, surfaces and solids are investigated. An angle is the amount of rotation between two straight lines. Angles may be measured in either degrees or radians (see Section.). revolution D 60 degrees, thus degree D th of one revolution. Also minute D 60 th of a degree and 60 second D 60 th of a minute. minute is written as 0 and second is written as 00 Thus = 60 and = 60 Problem. Add and C 9 0 D 7 0.Since60 0 D, 7 0 D 0. Thus the 0 is placed in the minutes column and is carried in the degrees column. Then 4 C 7 C (carried) D 5 Thus = 5 Problem. Subtract from cannot be done. Hence or 60 0 is borrowed from the degrees column, which leaves 7 in that column. Now 60 0 C D 6 0, which is placed in the minutes column. 7 5 D, which is placed in the degrees column. Thus = 6 Problem. Determine (a) C (b) (a) Adding: (b) Subtracting: Problem 4. Convert (a) (b) to degrees and decimals of a degree. (a) Since minute D th of a degree, ( ) D D Hence 4 4 = 4.70

129 8 Basic Engineering Mathematics (b) Since second D th of a minute, 60 ( ) D D Hence D P 0 ( ) 5.4 P D D , correct to 4 decimal places. Hence = 78.6, correct to 4 significant places. Problem 5. seconds. Convert 45.7 into degrees, minutes and Since D 60 0,0.7 D 0.7 ð 60 0 D.6 0 Since: 0 D 60 00,0.6 0 D 0.6 ð D D 6 00 to the nearest second. Hence 45.7 = 45 6 Now try the following exercise Exercise 56 Further problems on angular measurement (Answers on page 59). Add together the following angles: (a) 9 0 and (b) 9 4 0, and 6 54 (c) and (d) , and Determine (a) (b) (c) (d) Convert the following angles to degrees and decimals of a degree, correct to decimal places: (a) 5 0 (b) (c) (d) Convert the following angles into degrees, minutes and seconds: (a) 5.4 (b) 6.48 (c) (d) Types and properties of angles (a) (i) Any angle between 0 and 90 is called an acute angle. (ii) An angle equal to 90 is called a right angle. (iii) Any angle between 90 and 80 is called an obtuse angle. (iv) Any angle greater than 80 and less than 60 is called a reflex angle. (b) (i) An angle of 80 lies on a straight line. (ii) If two angles add up to 90 they are called complementary angles. (iii) If two angles add up to 80 they are called supplementary angles. (iv) Parallel lines are straight lines which are in the same plane and never meet. (Such lines are denoted by arrows, as in Fig. 6.). (v) A straight line which crosses two parallel lines is called a transversal (see MN in Fig. 6.). Fig. 6. P R M h e g f d a c b (c) With reference to Fig. 6.: (i) a D c, b D d, e D g and f D h. Such pairs of angles are called vertically opposite angles. (ii) a D e, b D f, c D g and d D h. Such pairs of angles are called corresponding angles. (iii) c D e and b D h. Such pairs of angles are called alternate angles. (iv) b C e D 80 and c C h D 80. Such pairs of angles are called interior angles. Problem 6. State the general name given to the following angles: (a) 59 (b) 6 (c) 90 (d) 7 (a) 59 lies between 90 and 80 and is therefore called an obtuse angle. (b) 6 lies between 0 and 90 and is therefore called an acute angle. (c) 90 is called a right angle. N Q S

130 Geometry and triangles 9 (d) 7 is greater than 80 and less than 60 and is therefore called a reflex angle. Problem 7. Find the angles complementary to (a) 4 (b) (a) The complement of 4 is 90 4, i.e.49 (b) The complement of is ,i.e. Problem 8. Find the angles supplementary to (a) 7 (b) 0 (a) The supplement of 7 is 80 7, i.e.5 (b) The supplement of 0 is 80 0,i.e Problem 9. Two straight lines AB and CD intersect at 0. If 6 AOC is 4, find 6 AOD, 6 DOB and 6 BOC From Fig. 6., 6 AOD is supplementary to 6 AOC. Hence AOD D 80 4 D 7. When two straight lines intersect the vertically opposite angles are equal. Hence DOB = 4 and BOC = 7 Fig. 6. A C 4 Problem 0. Determine angle b in Fig D B Problem. Determine the value of angle q in Fig A 7 B F θ E G Fig. 6.4 C 5 49 Let a straight line FG be drawn through E such that FG is parallel to AB and CD. 6 BAE D 6 AEF (alternate angles between parallel lines AB and FG), hence 6 AEF D ECD D 6 FEC (alternate angles between parallel lines FG and CD), hence 6 FEC D Angle q D 6 AEF C 6 FEC D 7 0 C D 59 6 Problem. Determine angles c and d in Fig Fig. 6.5 d b c 46 b D 46 (corresponding angles between parallel lines). Also b C c C 90 D 80 (angles on a straight line). Hence 46 C c C 90 D 80, from which c = 44. b and d are supplementary, hence d D D 4. Alternatively, 90 C c D d (vertically opposite angles). a D Now try the following exercise α β Fig. 6. a D 80 D 47 (i.e. supplementary angles). a D b = 47 (corresponding angles between parallel lines). Exercise 57 Further problems on types and properties of angles (Answers on page 59). State the general name given to the (a) 6 (b) 47 (c) 50. Determine the angles complementary to the following: (a) 69 (b) (c) Determine the angles supplementary to (a) 78 (b) 5 (c)

131 0 Basic Engineering Mathematics 4. With reference to Fig. 6.6, what is the name given to the line XY. Give examples of each of the following: (a) vertically opposite angles. (b) supplementary angles. (c) corresponding angles. (d) alternate angles. x 4 Fig In Fig. 6.7, find angle a y 6. Properties of triangles A triangle is a figure enclosed by three straight lines. The sum of the three angles of a triangle is equal to 80. Types of triangles: (i) An acute-angled triangle is one in which all the angles are acute, i.e. all the angles are less than 90. (ii) A right-angled triangle is one which contains a right angle. (iii) An obtuse-angled triangle is one which contains an obtuse angle, i.e. one angle which lies between 90 and 80. (iv) An equilateral triangle is one in which all the sides and all the angles are equal (i.e. each 60 ). (v) An isosceles triangle is one in which two angles and two sides are equal. (vi) A scalene triangle is one with unequal angles and therefore unequal sides. With reference to Fig. 6.0: 7 9 A α 6 49 c b Fig In Fig. 6.8, find angles a, b and c. Fig. 6.0 θ B a C c 9 69 b a (i) Angles A, B and C are called interior angles of the triangle. (ii) Angle q is called an exterior angle of the triangle and is equal to the sum of the two opposite interior angles, i.e. q D A C C (iii) a C b C c is called the perimeter of the triangle. Problem. Fig. 6.. Name the types of triangles shown in Fig Find angle b in Fig (a) (b).6.8. (c) 9 5 β (d) (e).5 Fig. 6.9 Fig. 6.

132 Geometry and triangles (a) Equilateral triangle. (b) Acute-angled scalene triangle. (c) Right-angled triangle. (d) Obtuse-angled scalene triangle. (e) Isosceles triangle. Problem 4. Fig. 6.. A 6 B Fig. 6. Determine the value of q and a in In triangle ABC, 6 A C 6 B C 6 C D 80 (angles in a triangle addupto80 ), hence 6 C D D 8. Thus 6 DCE D 8 (vertically opposite angles). q D 6 DCE C 6 DEC (exterior angle of a triangle is equal to the sum of the two opposite interior angles). Hence q D 8 C 5 D 4 6 a and 6 DEC are supplementary, thus a D 80 5 D 65 C D θ 5 α E Problem 6. Find angles a, b, c, d and e in Fig Fig. 6.4 e 55 d c a b 6 a = 6 and c = 55 (alternate angles between parallel lines) 55 C b C 6 D 80 (angles in a triangle add up to 80 ), hence b D D 6 b = d = 6 (alternate angles between parallel lines). e C 55 C 6 D 80 (angles in a triangle add up to 80 ), hence e D D 6 [Check: e D a D 6 (corresponding angles between parallel lines)]. Now try the following exercise Exercise 58 Further problems on properties of triangles (Answers on page 59). In Fig. 6.5, (i) and (ii), find angles w, x, y and z. What is the name given to the types of triangle shown in (i) and (ii)? Problem 5. ABC is an isosceles triangle in which the unequal angle BAC is 56. AB is extended to D as shown in Fig. 6.. Determine the angle DBC. ω cm cm 70 A 0 x y 0 z 56 (i) (ii) Fig. 6.5 D B C. Find the values of angles a to g in Fig. 6.6 (i) and (ii). Fig. 6. Since the three interior angles of a triangle add up to 80 then 56 C 6 BC 6 C D 80,i.e. 6 BC 6 C D D 4. Triangle ABC is isosceles hence 6 B D 6 C D 4 D 6. 6 DBC D 6 A C 6 C (exterior angle equals sum of two interior opposite angles), i.e. DBC D 56 C 6 D 8 [Alternatively, 6 DBC C 6 ABC D 80 (i.e. supplementary angles)]. Fig b (i) 4 4 a 68 c d f e g (ii)

133 Basic Engineering Mathematics. Find the unknown angles a to k in Fig Fig e g h b a c d j i k Triangle ABC has a right angle at B and 6 BAC is 4. BC is produced to D. If the bisectors of 6 ABC and 6 ACD meet at E, determine 6 BEC 5. If in Fig. 6.8, triangle BCD is equilateral, find the interior angles of triangle ABE. Fig. 6.8 A E Congruent triangles Two triangles are said to be congruent if they are equal in all respects, i.e. three angles and three sides in one triangle are equal to three angles and three sides in the other triangle. Two triangles are congruent if: (i) the three sides of one are equal to the three sides of the other (SSS), (ii) they have two sides of the one equal to two sides of the other, and if the angles included by these sides are equal (SAS), (iii) two angles of the one are equal to two angles of the other and any side of the first is equal to the corresponding side of the other (ASA), or (iv) their hypotenuses are equal and if one other side of one is equal to the corresponding side of the other (RHS). Problem 7. State which of the pairs of triangles shown in Fig. 6.9 are congruent and name their sequence. B f C D B A M N P (c) Fig. 6.9 C E (a) O Q R D F H S (d) G W (a) Congruent ABC, FDE (Angle, side, angle, i.e. ASA). (b) Congruent GIH, JLK (Side, angle, side, i.e. SAS). (c) Congruent MNO, RQP (Right-angle, hypotenuse, side, i.e. RHS). (d) Not necessarily congruent. It is not indicated that any side coincides. (e) Congruent ABC, FED (Side, side, side, i.e. SSS). Problem 8. In Fig. 6.0, triangle PQR is isosceles with Z the mid-point of PQ. Prove that triangle PXZ and QYZ are congruent, and that triangles RXZ and RYZ are congruent. Determine the values of angles RPZ and RXZ. Fig. 6.0 P V J I X (b) T R X 67 Y 8 8 Z Since triangle PQR is isosceles PR D RQ and thus 6 QPR D 6 RQP 6 RXZ D 6 QPR C 8 and 6 RYZ D 6 RQP C 8 (exterior angles of a triangle equal the sum of the two interior opposite angles). Hence 6 RXZ D 6 RYZ. 6 PXZ D 80 6 RXZ and 6 QYZ D 80 6 RYZ. Thus 6 PXZ D 6 QYZ. Triangles PXZ and QYZ are congruent since 6 XPZ D 6 YQZ, PZ D ZQ and 6 XZP D 6 YZQ (ASA) Hence XZ D YZ Triangles PRZ and QRZ are congruent since PR D RQ, 6 RPZ D 6 RQZ and PZ D ZQ (SAS). Hence 6 RZX D 6 RZY U K D B Q L A F E C (e)

134 Geometry and triangles Triangles RXZ and RYZ are congruent since 6 RXZ D 6 RYZ, XZ D YZ and 6 RZX D 6 RZY (ASA). 6 QRZ D 67 and thus 6 PRQ D 67 C 67 D 4. Hence RPZ D 6 RQZ D 80 4 D RXZ D C 8 D 5 (external angle of a triangle equals the sum of the two interior opposite angles). Now try the following exercise i.e. p a D q b D r c c A 57 b B a C Q p R P r 57 q Exercise 59 Further problems on congruent triangles (Answers on page 59). State which of the pairs of triangles in Fig. 6. are congruent and name their sequence. B Q (d) A C E G I J (a) (b) (c) V R T D S H U X F N Z Y K W (e) O L P M Fig. 6. Problem 9. In Fig. 6., find the length of side a. A c =.0 cm B Fig a 50 C D f = 5.0 cm E F d = 4.4 cm In triangle ABC, 50 C 70 C 6 C D 80, from which 6 C D 60 In triangle DEF, 6 E D D 70. Hence triangles ABC and DEF are similar, since their angles are the same. Since corresponding sides are in proportion to each other then: a d D c f i.e. a 4.4 D Hence a D D 0.6 cm 5.0 Fig. 6.. In a triangle ABC, AB D BC and D and E are points on AB and BC, respectively, such that AD D CE. Show that triangles AEB and CDB are congruent. 6.5 Similar triangles Two triangles are said to be similar if the angles of one triangle are equal to the angles of the other triangle. With reference to Fig. 6.: Triangles ABC and PQR are similar and the corresponding sides are in proportion to each other, Problem 0. r and p P q =6.8 cm 5 R Fig. 6.4 In Fig. 6.4, find the dimensions marked Y 55 r Q p X z =.97 cm y =0.6 cm x =7.44 cm Z

135 4 Basic Engineering Mathematics In triangle PQR, 6 Q D D 55 In triangle XYZ, 6 X D D 5 Hence triangles PQR and ZYX are similar since their angles are the same. The triangles may be redrawn as shown in Fig r P Q 55 Fig. 6.5 p 5 q =6.8 cm By proportion: R x =7.44 cm Y Z p z D r x D q y 55 z =.97 cm y =0.6 cm p Hence z D r 7.44 D ( ) 6.8 from which, r D 7.44 D 4.77 cm 0.6 By proportion: p z D q p i.e. y.97 D Hence ( ) 6.8 p D.97 D 8. cm Problem. In Fig. 6.6, show that triangles CBD and CAE are similar and hence find the length of CD and BD. X lines). Also 6 C is common to triangles CBD and CAE. Since the angles in triangle CBD are the same as in triangle CAE the triangles are similar. Hence, by proportion: i.e. Also, CB CA D CD CE ( D BD ) AE 9 6 C 9 D CD, from which CD D ( ) 9 5 D 7.cm 9 5 D BD ( ) 9, from which BD D D 6cm Problem. A rectangular shed m wide and m high stands against a perpendicular building of height 5.5 m. A ladder is used to gain access to the roof of the building. Determine the minimum distance between the bottom of the ladder and the shed. A side view is shown in Fig. 6.7, where AF is the minimum length of ladder. Since BD and CF are parallel, 6 ADB D 6 DFE (corresponding angles between parallel lines). Hence triangles BAD and EDF are similar since their angles are the same. AB D AC BC D AC DE D 5.5 D.5m. AB By proportion: DE D BD EF i.e..5 D EF ( ) Hence EF D D.4 m= minimum.5 distance from bottom of ladder to the shed A 6 A B 0 E D 9 C D m m Shed B 5.5m Fig. 6.6 F E C Since BD is parallel to AE then 6 CBD D 6 CAE and 6 CDB D 6 CEA (corresponding angles between parallel Fig. 6.7

136 Geometry and triangles 5 Now try the following exercise C Exercise 60 Further problems on similar triangles (Answers on page 59) B D. In Fig. 6.8, find the lengths x and y. A E F 4.58 mm Fig mm x Fig mm y mm 6.6 Construction of triangles To construct any triangle the following drawing instruments are needed: (i) ruler and/or straight edge, (ii) compass, (iii) protractor, (iv) pencil. For actual constructions, see Problems to 6 which follow. Problem. Construct a triangle whose sides are 6cm, 5cm and cm.. PQR is an equilateral triangle of side 4 cm. When PQ and PR are produced to S and T, respectively, ST is found to be parallel with QR. IfPS is 9 cm, find the length of ST. X is a point on ST between S and T such that the line PX is the bisector of 6 SPT. Find the length of PX.. In Fig. 6.9, find (a) the length of BC when AB D 6cm, DE D 8cm and DC D cm, (b) the length of DE when EC D cm, AC D 5cm and AB D 0 cm. Fig. 6.9 A D C 4. In Fig. 6.0, AF D 8m, AB D 5m and BC D m. Find the length of BD. B E With reference to Fig. 6.: Fig. 6. A F 6 cm (i) Draw a straight line of any length, and with a pair of compasses, mark out 6 cm length and label it AB. (ii) Set compass to 5 cm and with centre at A describe arc DE. (iii) Set compass to cm and with centre at B describe arc FG. (iv) The intersection of the two curves at C is the vertex of the required triangle. Join AC and BC by straight lines. It may be proved by measurement that the ratio of the angles of a triangle is not equal to the ratio of the sides (i.e. in this problem, the angle opposite the cm side is not equal to half the angle opposite the 6 cm side). Problem 4. Construct a triangle ABC such that a D 6cm, b D cm and 6 C D 60 D C E G B

137 6 Basic Engineering Mathematics A b = cm P Z U C V S Fig. 6. B With reference to Fig. 6.: a = 6 cm 60 (i) Draw a line BC, 6 cm long. (ii) Using a protractor centred at C make an angle of 60 to BC. (iii) From C measure a length of cm and label A. (iv) Join B to A by a straight line. Problem 5. Construct a triangle PQR given that QR D 5cm, 6 Q D 70 and 6 R D 44 With reference to Fig. 6.: C Fig. 6.4 B A R Q X A (ii) Produce XY any distance to B. With compass centred at X make an arc at A and A 0. (The length XA and XA 0 is arbitrary.) With compass centred at A draw the arc PQ. With the same compass setting and centred at A 0,draw the arc RS. Join the intersection of the arcs, C, tox, and a right angle to XY is produced at X. (Alternatively, a protractor can be used to construct a 90 angle). (iii) The hypotenuse is always opposite the right angle. Thus YZ is opposite 6 X. Using a compass centred at Y and set to 6.5 cm, describe the arc UV. (iv) The intersection of the arc UV with XC produced, forms the vertex Z of the required triangle. Join YZ by a straight line. Y R P Q Now try the following exercise Exercise 6 Further problems on the construction of triangles (Answers on page 59) In problems to 5, construct the triangles ABC for the given sides/angles. Fig. 6. Q cm (i) Draw a straight line 5 cm long and label it QR. (ii) Use a protractor centred at Q and make an angle of 70. Draw QQ 0 (iii) Use a protractor centred at R and make an angle of 44. Draw RR 0. (iv) The intersection of QQ 0 and RR 0 forms the vertex P of the triangle. Problem 6. Construct a triangle XYZ given that XY D 5 cm, the hypotenuse YZ D 6.5cm and 6 X D 90 With reference to Fig. 6.4: (i) Draw a straight line 5 cm long and label it XY. R. a D 8cm, b D 6cm and c D 5cm. a D 40 mm, b D 60 mm and C D 60. a D 6cm, C D 45 and B D c D 4cm, A D 0 and C D 5 5. a D 90 mm, B D 90, hypotenuse D 05 mm Assignment 8 This assignment covers the material contained in chapters 5 and 6. The marks for each question are shown in brackets at the end of each question.. In the following equations, x and y are two related variables and k and t are constants. For the stated equations to be verified it is necessary to plot graphs of the variables in modified form. State for each (a) what shouldbeplottedonthehorizontalaxis,(b)whatshould

138 Geometry and triangles 7 be plotted on the vertical axis, (c) the gradient, and (d) the vertical axis intercept. (i) y k x D t (ii) y k D xt (8). The following experimental values of x and y are believed to be related by the law y D ax C b, where a and b are constants. By plotting a suitable graph verify this law and find the approximate values of a and b. x y In Fig. A8., determine angles a to e (5) c e a b d 5 60 (9). Determine the law of the form y D ae kx which relates the following values: y x (9) 4. Evaluate: C () 5. Convert 47.9 to degrees, minutes and seconds () Fig. A8. 9. In Fig. A8., determine the length of AC (4) B 6. State the angle (a) supplementary to 49 (b) complementary to In Fig. A8., determine angles x, y and z () 0 m D m A C 59 7 y z 8 m x Fig. A8. Fig. A8. 0. Construct a triangle PQR given PQ D 5cm, 6 QPR D 0 and 6 PRQ D 5 (5)

139 7 Introduction to trigonometry 7. Trigonometry Trigonometry is the branch of mathematics which deals with the measurement of sides and angles of triangles, and their relationship with each other. There are many applications in engineering where a knowledge of trigonometry is needed. By Pythagoras theorem, a D b C c i.e. a D 4 C D 6 C 9 D 5 Hence a D p 5 Dš5 5 hasno meaning in this context and is thus ignored) Thus bc = 5cm 7. The theorem of Pythagoras With reference to Fig. 7., the side opposite the right angle (i.e. side b) is called the hypotenuse. The theorem of Pythagoras states: In any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Hence b = a + c A Problem. Fig. 7. In Fig. 7., find the length of EF D f = 5 cm e = cm E d F Fig. 7. Problem. c b B a C In Fig. 7., find the length of BC C By Pythagoras theorem: e D d C f Hence D d C 5 69 D d C 5 d D 69 5 D 44 Thus d D p 44 D cm i.e. EF = cm b = 4 cm A a c = cm B Problem. Two aircraft leave an airfield at the same time. One travels due north at an average speed of 00 km/h and the other due west at an average speed of 0 km/h. Calculate their distance apart after 4 hours. Fig. 7. After 4 hours, the first aircraft has travelled 4 ð 00 D 00 km, due north, and the second aircraft has travelled

140 Introduction to trigonometry 9 4 ð 0 D 880 km due west, as shown in Fig Distance apart after 4 hours D BC From Pythagoras theorem: and BC D 00 C 880 D C BC D W N E B 7. Trigonometric ratios of acute angles (a) With reference to the right-angled triangle shown in Fig. 7.5: (i) (ii) sine q D cosine q D (iii) tangent q D opposite side hypotenuse, adjacent side hypotenuse, opposite side adjacent side, i.e. sin q = b c i.e. cos q = a c i.e. tan q = b a S 00 km C 880 km A c b Fig. 7.4 θ a Hence distance apart after 4 hours = 488 km. Fig. 7.5 Now try the following exercise Exercise 6 Further problems on the theorem of Pythagoras (Answers on page 60). In a triangle ABC, 6 B is a right angle, AB D 6.9 cm and BC D 8.78 cm. Find the length of the hypotenuse.. In a triangle CDE, D D 90, CD D 4.8 mm and CE D 8. mm. Determine the length of DE.. Show that if a triangle has sides of 8, 5 and 7 cm it is right angled. 4. Triangle PQR is isosceles, Q being a right angle. If the hypotenuse is 8.47 cm find (a) the lengths of sides PQ and QR, and (b) the value of 6 QPR. 5. A man cycles 4 km due south and then 0 km due east. Another man, starting at the same time as the first man, cycles km due east and then 7 km due south. Find the distance between the two men. 6. A ladder.5 m long is placed against a perpendicular wall with its foot.0 m from the wall. How far up the wall (to the nearest centimetre) does the ladder reach? If the foot of the ladder is now moved 0 cm further away from the wall, how far does the top of the ladder fall? 7. Two ships leave a port at the same time. One travels due west at 8.4 km/h and the other due south at 7.6 km/h. Calculate how far apart the two ships are after 4 hours. Problem 4. tan F Fig. 7.6 From Fig. 7.6, find sin D, cosd and D 7 By Pythagoras theorem, 7 D 8 C EF from which, EF D 7 8 D 5 sin D D EF DF D 5 or cos D D DE DF D 8 7 tan F D DE EF D 8 5 F E or or 0.5 Problem 5. Determine the values of sin q, cosq and tan q for the right-angled triangle ABC shown in Fig. 7.7.

141 0 Basic Engineering Mathematics Fig. 7.7 A θ B 5 C f(x) A B f(x) A θ B C By definition: sin q D cos q D tan q D opposite side hypotenuse D 5 D adjacent side hypotenuse D D 0.9 opposite side adjacent side D 5 D Problem 6. IfcosX D 9 determine the value of sin X 4 and tan X. Figure 7.8 shows a right-angled triangle XYZ. Z (a) Fig (b) Since ABC is a right-angled triangle, and AC D 8 D 6 and BC D 7 D 4, then by Pythagoras theorem AB D AC C BC D 6 C 4 and AB D p 6 C 4 D p 5 D 7., correct to decimal places. (b) The gradient of AB is given by tan q, i.e. gradient D tan q D BC AC D 4 6 D (c) The angle AB makes with the horizontal is given by tan D.69 Fig. 7.8 X 4 Since cos X D 9,thenXY D 9 units and XZ D 4 units. 4 Using Pythagoras theorem: 4 D 9 C YZ from which YZ D p 4 9 D 40 units. Thus sin X = Y and tan X = 40 9 = Problem 7. Point A lies at co-ordinate (,) and point B at (8,7). Determine (a) the distance AB, (b) the gradient of the straight line AB, and (c) the angle AB makes with the horizontal. (a) Points A and B are shown in Fig. 7.9(a). In Fig. 7.9(b), the horizontal and vertical lines AC and BC are constructed. Now try the following exercise Exercise 6 Further problems on trigonometric ratios of acute angles (Answers on page 60). Sketch a triangle XYZ such that 6 Y D 90, XY D 9cm and YZ D 40 cm. Determine sin Z, cosz, tanx and cos X.. In triangle ABC shown in Fig. 7.0, find sin A, cosa, tan A, sinb, cosb and tan B. Fig. 7.0 A 5. If cos A D 5 find sin A and tan A, in fraction form If tan X D 5,findsinX and cos X, in fraction form. B C

142 Introduction to trigonometry 5. For the right-angled triangle shown in Fig. 7., find: (a) sina (b) cosq (c) tan q Problem 9. In triangle PQR shown in Fig. 7., find the lengths of PQ and PR. 8 α 5 7 θ P Fig If tan q D 7,findsinq and cos q in fraction form Point P lies at co-ordinate (,) and point Q at (5, 4). Determine (a) the distance PQ, (b) the gradient of the straight line PQ and, (c) the angle PQ makes with the horizontal. 7.4 Solution of right-angled triangles To solve a right-angled triangle means to find the unknown sides and angles. This is achieved by using (i) the theorem of Pythagoras, and/or (ii) trigonometric ratios. This is demonstrated in the following problems. Problem 8. Sketch a right-angled triangle ABC such that B D 90, AB D 5cm and BC D cm. Determine the length of AC and hence evaluate sin A, cosc and tan A. Triangle ABC is shown in Fig. 7.. By Pythagoras theorem, AC D p 5 C D Fig. 7. Q cm tan 8 D PQ QR D PQ 7.5 hence PQ D 7.5tan8 D D cm cos 8 D QR PR D 7.5 PR hence PR D 7.5 cos D 7.5 D 9.58 cm [Check: Using Pythagoras theorem 7.5 C D D 9.58 ] Problem 0. Solve the triangle ABC shown in Fig Fig. 7.4 A C 5 mm 7 mm R B By definition: sin A D and Fig. 7. cos C D tan A D A 5 cm B opposite side hypotenuse D adjacent side hypotenuse D opposite side adjacent side D 5 cm C or 0.9 or 0.9 or.400 To solve triangle ABC means to find the length AC and angles B and C 0 sin C D 5 7 D hence C D sin D 7.08 or 7 5 B D D 8 55 (since angles in a triangle addupto80 ) sin B D AC 7 hence AC D 7 sin D D.0mm or, using Pythagoras theorem, 7 D 5 C AC, from which, AC D 7 5 D.0mm.

143 Basic Engineering Mathematics Problem. Solve triangle XYZ given 6 X D 90, 6 Y D 7 0 and YZ D 0.0 mm. Determine also its area.. Solve triangle GHI in Fig. 7.6(iii). 4. Solve the triangle JKL in Fig. 7.7(i) and find its area. It is always advisable to make a reasonably accurate sketch so as to visualize the expected magnitudes of unknown sides and angles. Such a sketch is shown in Fig Z D D 66 4 sin 7 0 D XZ 0.0 cos 7 0 D XY 0.0 Fig. 7.5 Z X hence XZ D 0.0 sin 7 0 D D mm hence XY D 0.0 cos mm 7 (Check: Using Pythagoras theorem D D 8.7 mm 8.7 C D D 0.0 Area of triangle XYZ D (base)(perpendicular height) D XY XZ D D 7.6 mm Y J 6.7 mm K Fig. 7.7 M.0 mm m Q P 5 R L O (i) (ii) (iii) 8.75 m 5. Solve the triangle MNO in Fig. 7.7(ii) and find its area. 6. Solve the triangle PQR in Fig. 7.7(iii) and find its area. 7. A ladder rests against the top of the perpendicular wall of a building and makes an angle of 7 with the ground. If the foot of the ladder is m from the wall, calculate the height of the building. 7.5 Angles of elevation and depression (a) If, in Fig. 7.8, BC represents horizontal ground and AB a vertical flagpole, then the angle of elevation of the top of the flagpole, A, from the point C is the angle that the imaginary straight line AC must be raised (or elevated) from the horizontal CB, i.e. angle q. N A Now try the following exercise C θ B Exercise 64 Further problems on the solution of right-angled triangles (Answers on page 60). Solve triangle ABC in Fig. 7.6(i). Fig. 7.8 P φ. Solve triangle DEF in Fig. 7.6(ii). Q R A cm (i) Fig. 7.6 B C 4 cm F D (ii) cm E G H mm I (iii) Fig. 7.9 (b) If, in Fig. 7.9, PQ represents a vertical cliff and R a ship at sea, then the angle of depression of the ship from point P is the angle through which the imaginary straight line PR must be lowered (or depressed) from the horizontal to the ship, i.e. angle f.

144 Introduction to trigonometry (Note, 6 PRQ is also f alternate angles between parallel lines.) Problem. An electricity pylon stands on horizontal ground. At a point 80 m from the base of the pylon, the angle of elevation of the top of the pylon is. Calculate the height of the pylon to the nearest metre. Figure 7.0 shows the pylon AB and the angle of elevation of A from point C is tan D AB BC D AB 80 Hence height of pylon AB D 80 tan D D.96 m D 4 m to the nearest metre A Equating equations () and () gives: 0.44 x C 0 D.074x 0.44x C D.074x D x 4.6 D 0.78x x D D m From equation (), height of building, h D.074x D D m Problem 4. The angle of depression of a ship viewed at a particular instant from the top of a 75 m vertical cliff is 0. Find the distance of the ship from the base of the cliff at this instant. The ship is sailing away from the cliff at constant speed and minute later its angle of depression from the top of the cliff is 0. Determine the speed of the ship in km/h. Fig. 7.0 C 80 m Problem. A surveyor measures the angle of elevation of the top of a perpendicular building as 9. He moves 0 m nearer the building and finds the angle of elevation is now 47. Determine the height of the building. The building PQ and the angles of elevation are shown in Fig. 7.. h In triangle PQS, tan9 D x C 0 hence h D tan 9 x C 0, i.e. h D 0.44 x C 0 B Figure 7. shows the cliff AB, the initial position of the ship at C and the final position at D. Since the angle of depression is initially 0 then 6 ACB D 0 (alternate angles between parallel lines). tan 0 D AB BC D 75 BC Fig. 7. hence BC D 75 tan 0 D D 9.9mD initial position of ship from base of cliff A 0 75 m B C x D Fig. 7. h Q P 47 R 9 x 0 In triangle PQR, tan47 D h x hence h D tan 47 x, i.e. h D.074x S In triangle ABD, tan0 D AB BD D 75 BC C CD D C x Hence 9.9 C x D 75 tan D D 06.0m from which x D D 76.m Thus the ship sails 76. m in minute, i.e. 60 s, hence speed of ship D distance D 76. time 60 m/s 76. ð 60 ð 60 D km/h D 4.57 km/h 60 ð 000

145 4 Basic Engineering Mathematics Now try the following exercise Exercise 65 Further problems on angles of elevation and depression (Answers on page 60). A vertical tower stands on level ground. At a point 05 m from the foot of the tower the angle of elevation of the top is 9. Find the height of the tower.. If the angle of elevation of the top of a vertical 0 m high aerial is, how far is it to the aerial?. From the top of a vertical cliff 90.0 m high the angle of depression of a boat is Determine the distance of the boat from the cliff. 4. From the top of a vertical cliff 80.0 m high the angles of depression of two buoys lying due west of the cliff are and 5, respectively. How far are the buoys apart? 5. From a point on horizontal ground a surveyor measures the angle of elevation of the top of a flagpole as He moves 50 m nearer to the flagpole and measures the angle of elevation as 6 0. Determine the height of the flagpole. 6. A flagpole stands on the edge of the top of a building. At a point 00 m from the building the angles of elevation of the top and bottom of the pole are and 0 respectively. Calculate the height of the flagpole. 7. From a ship at sea, the angles of elevation of the top and bottom of a vertical lighthouse standing on the edge of a vertical cliff are and 6, respectively. If the lighthouse is 5.0 m high, calculate the height of the cliff. 8. From a window 4. m above horizontal ground the angle of depression of the foot of a building across the road is 4 and the angle of elevation of the top of the building is 4. Determine, correct to the nearest centimetre, the width of the road and the height of the building. 9. The elevation of a tower from two points, one due east of the tower and the other due west of it are 0 and 4, respectively, and the two points of observation are 00 m apart. Find the height of the tower to the nearest metre. 7.6 Evaluating trigonometric ratios of any angles Four-figure tables are available which gives sines, cosines, and tangents, for angles between 0 and 90. However, the easiest method of evaluating trigonometric functions of any angle is by using a calculator. The following values, correct to 4 decimal places, may be checked: sine 8 D cosine 56 D tangent 9 D sine 7 D 0.9 cosine 5 D 0.46 tangent 78 D sine 4.6 D cosine.78 D 0.88 tangent 96.4 D.07 To evaluate, say, sine 4 0 using a calculator means finding sine 4 since there are 60 minutes in degree D 0.8 P, thus 4 0 D 4.8P Thus sine 4 0 D sine 4.8 D 0.674, correct to 4 decimal places. Similarly, cosine D cosine 7 8 D 0.985, correct 60 to 4 decimal places. Problem 5. Evaluate correct to 4 decimal places: (a) sine (b) sine.68 (c) sine (a) sine D (b) sine.68 D (c) sine D sine D 0.98 Problem 6. Evaluate, correct to 4 decimal places: (a) cosine (b) cosine 59. (c) cosine 4 0 (a) cosine D (b) cosine 59. D (c) cosine 4 0 D cosine 4 60 D Problem 7. Evaluate, correct to 4 significant figures: (a) tangent 76 (b) tangent.9 (c) tangent (a) tangent 76 D 9.54 (b) tangent.9 D.9 (c) tangent D tan D 4.0

146 Introduction to trigonometry 5 Problem 8. Evaluate, correct to 4 significant figures: (a) sin.48 (b) cos p/5 (c) tan.9 (a) sin.48 means the sine of.48 radians. Hence a calculator needs to be on the radian function. Hence sin.48 D (b) cos p/5 D cos d (c) tan.9 D 0.48 Problem. (a) sin (c) tangent 7.9 Evaluate correct to 4 decimal places: (b) cosine (a) Positive angles are considered by convention to be anticlockwise and negative angles as clockwise. From Fig. 7., is actually the same as C48 (i.e. 60 ). Problem 9. Determine the acute angles: 90 (a) sin 0.7 (b) cos (c) tan.4695 (a) sin q is an abbreviation for the angle whose sine is equal to q. 0.7 is entered into a calculator and then the inverse sine (or sin ) key is pressed. Hence sin 0.7 D Subtracting 47 leaves and multiplying this by 60 gives 4 0 to the nearest minute. Hence sin 0.7 D or 47 4 Alternatively, in radians, sin 0.7 D 0.8 radians. (b) cos D 65. or 65 0 or.40 radians. (c) tan.4695 D or or 0.97 radians. Problem 0. Evaluate the following expression, correct to 4 significant figures: 4.tan sin cos By calculator: ( tan D tan 49 6 ) D.68, 60 sin 66 0 D 0.97 and cos D Hence D 4.tan sin cos D 4. ð.68.7 ð ð D D D 0.47, D correct to significant figures. Fig Hence, by calculator, sin D sin 48 D 0.97 ( (b) cosine D cosine 9 6 ) D (c) tangent 7.9 D (which is the same as tan , i.e.tan4.7 ) Now try the following exercise Exercise 66 Further problems on evaluating trigonometric ratios of any angle (Answers on page 60) In Problems to 4, evaluate correct to 4 decimal places:. (a) sine 7 (b) sine 7.4 (c) sine (a) cosine 4 (b) cosine.46 (c) cosine (a) tangent 45 (b) tangent 0.59 (c) tangent (a) sine p (b) cos.68 (c) tan.67 In Problems 5 to 7, determine the acute angle in degrees (correct to decimal places), degrees and minutes, and in radians (correct to decimal places). 5. sin cos tan 0.806

147 6 Basic Engineering Mathematics In problems 8 to 0, evaluate correct to 4 significant figures cos sin tan49 0 sin 90 cos 45 5sin86 0 tan4 9 0 cos 9 0. Determine the acute angle, in degrees and minutes, correct to the nearest minute, given by ( 4. sin 4 6 sin 0 ) Evaluate sin cos 69 0 tan5 9 0 correct to 4 significant figures. Evaluate correct to 4 decimal places: (a) sine 5 (b) tan 4 (c) cos

148 8 Trigonometric waveforms 8. Graphs of trigonometric functions By drawing up tables of values from 0 to 60, graphs of y D sin A, y D cos A and y D tan A may be plotted. Values obtained with a calculator (correct to decimal places which is more than sufficient for plotting graphs), using 0 intervals, are shown below, with the respective graphs shown in Fig. 8.. (a) y = sin A A sin A A sin A (b) y = cos A A cos A A cos A (c) y = tan A A tan A A tan A From Fig. 8. it is seen that: (i) Sine and cosine graphs oscillate between peak values of š Fig. 8. (ii) The cosine curve is the same shape as the sine curve but displaced by 90. (iii) The sine and cosine curves are continuous and they repeat at intervals of 60 ; the tangent curve appears to be discontinuous and repeats at intervals of 80.

149 8 Basic Engineering Mathematics 8. Angles of any magnitude Figure 8. shows rectangular axes XX 0 and YY 0 intersecting at origin 0. As with graphical work, measurements made to the right and above 0 are positive, while those to the left and downwards are negative. Let 0A be free to rotate about 0. By convention, when 0A moves anticlockwise angular measurement is considered positive, and vice versa. right-angled triangle 0AC. Then sin q D C C DC tan q D C D cos q D C D Let 0A be further rotated so that q is any angle in the third quadrant and let AD be constructed to form the right-angled triangle 0AD. Then sin q D C D cos q D C D tan q D DC Let 0A be further rotated so that q 4 is any angle in the fourth quadrant and let AE be constructed to form the right-angled triangle 0AE. Then sin q 4 D C D cos q 4 D C C DC tan q 4 D C D Fig. 8. The above results are summarized in Fig The letters underlined spell the word CAST when starting in the fourth quadrant and moving in an anticlockwise direction. Let 0A be rotated anticlockwise so that q is any angle in the first quadrant and let perpendicular AB be constructed to form the right-angled triangle 0AB in Fig. 8.. Since all three sides of the triangle are positive, the trigonometric ratios sine, cosine and tangent will all be positive in the first quadrant. (Note: 0A is always positive since it is the radius of a circle). Let 0A be further rotated so that q is any angle in the second quadrant and let AC be constructed to form the Fig. 8.4 Fig. 8. In the first quadrant of Fig. 8. all of the curves have positive values; in the second only sine is positive; in the third only tangent is positive; in the fourth only cosine is positive exactly as summarized in Fig A knowledge of angles of any magnitude is needed when finding, for example, all the angles between 0 and 60 whose sine is, say, 0.6. If 0.6 is entered into a calculator and then the inverse sine key pressed (or sin-key) the answer

150 Trigonometric waveforms appears. However, there is a second angle between 0 and 60 which the calculator does not give. Sine is also positive in the second quadrant [either from CAST or from Fig. 8.(a)]. The other angle is shown in Fig. 8.5 as angle q where q D D Thus 9.0 and are the angles between 0 and 60 whose sine is 0.6 (check that sin D 0.6 on your calculator). Fig. 8.7 (Note that a calculator only gives one answer, i.e ) Problem. Determine all the angles between 0 and 60 whose tangent is.769 A tangent is positive in the first and third quadrants see Fig From Fig. 8.9, q D tan.769 D Fig. 8.5 Be careful! Your calculator only gives you one of these answers. The second answer needs to be deduced from a knowledge of angles of any magnitude, as shown in the following worked problems. Problem. Determine all the angles between 0 and 60 whosesineis The angles whose sine is occurs in the third and fourth quadrants since sine is negative in these quadrants see Fig Fig. 8.8 Fig. 8.9 Fig. 8.6 From Fig. 8.7, q D sin D 7.6. Measured from 0, the two angles between 0 and 60 whose sine is are 80 C7.6, i.e.07.6 and , i.e..7 Measured from 0, the two angles between 0 and 60 whose tangent is.769 are and 80 C 60.44, i.e Problem. Solve the equation cos 0.48 D a for angles of a between 0 and 60.

151 40 Basic Engineering Mathematics Cosine is positive in the first and fourth quadrants and thus negative in the second and third quadrants from Fig. 8.5 or from Fig. 8.(b). In Fig. 8.0, angle q D cos 0.48 D Find the angles between 0 to 60 whose tangent is: (a) (b) S A 8. The production of a sine and cosine wave 80 T θ θ 70 C 0 60 In Fig. 8., let OR be a vector unit long and free to rotate anticlockwise about O. In one revolution a circle is produced and is shown with 5 sectors. Each radius arm has a vertical and a horizontal component. For example, at 0, the vertical component is TS and the horizontal component is OS. From trigonometric ratios, Fig. 8.0 Measured from 0, the two angles whose cosine is 0.48 are a D i.e and a D 80 C 76.4, i.e Now try the following exercise Exercise 67 Further problems on angles of any magnitude (Answers on page 60). Determine all of the angles between 0 and 60 whose sine is: (a) (b) Solve the following equations for values of x between 0 and 60 : (a) x D cos (b) x D cos sin 0 D TS TO D TS, i.e. TS D sin 0 and cos 0 D OS TO D OS, i.e. OS D cos 0 The vertical component TS may be projected across to T 0 S 0, which is the corresponding value of 0 on the graph of y against angle x. If all such vertical components as TS are projected on to the graph, then a sine wave is produced as shown in Fig. 8.. If all horizontal components such as OS are projected on to a graph of y against angle x,thenacosine wave is produced. It is easier to visualize these projections by redrawing the circle with the radius arm OR initially in a vertical position as shown in Fig. 8.. From Figs. 8. and 8. it is seen that a cosine curve is of the same form as the sine curve but is displaced by 90 (or p/ radians). Fig. 8.

152 Trigonometric waveforms 4 Fig Sine and cosine curves Graphs of sine and cosine waveforms (i) A graph of y D sin A is shown by the broken line in Fig. 8. and is obtained by drawing up a table of values as in Section 8.. A similar table may be produced for y D sin A. A A sin A (ii) A graph of y D sin A is shown in Fig. 8.4 using the following table of values. A A sin A A A sin A A A sin A A A sin A A graph of y D sin A is shown in Fig. 8.. Fig. 8.4 (iii) A graph of y D cos A is shown by the broken line in Fig. 8.5 and is obtained by drawing up a table of values. A similar table may be produced for y D cos A with the result as shown. (iv) A graph of y D cos A is shown in Fig. 8.6 which may be produced by drawing up a table of values, similar to above. Periodic time and period Fig. 8. (i) Each of the graphs shown in Figs. 8. to 8.6 will repeat themselves as angle A increases and are thus called periodic functions.

153 4 Basic Engineering Mathematics Amplitude D and period D 60 / D 0. AsketchofyDsin A is shown in Fig Fig. 8.5 Fig. 8.7 Problem 5. radians Sketchy D sina from A D 0toA D p Amplitude D and period D p/ D p rads (or 80 ) AsketchofyDsinA is shown in Fig Fig. 8.6 (ii) y D sina and y D cos A repeat themselves every 60 (or p radians); thus 60 is called the period of these waveforms. y D sin A and y D cos A repeat themselves every 80 (or p radians); thus 80 is the period of these waveforms. (iii) In general, if y D sin pa or y D cos pa (where p is a constant) then the period of the waveform is 60 /p (or p/p rad). Hence if y D sin A then the period is 60/, i.e. 0, andify D cos 4A then the period is 60/4, i.e. 90 Amplitude Amplitude is the name given to the maximum or peak value of a sine wave. Each of the graphs shown in Figs. 8. to 8.6 has an amplitude of C (i.e. they oscillate between C and ). However, if y D 4sinA, each of the values in the table is multiplied by 4 and the maximum value, and thus amplitude, is 4. Similarly, if y D 5cosA, the amplitude is 5 and the period is 60 /, i.e. 80 Problem 4. A D 60 Sketch y D sin A between A D 0 and Fig. 8.8 Problem 6. Sketch y D 4cosx from x D 0 to x D 60 Amplitude D 4 and period D 60 / D 80. AsketchofyD4cosx is shown in Fig Problem 7. Amplitude D ; period D 60 Sketch y D sin A over one cycle. 5 5 D 60 ð 5 D 600. AsketchofyDsin A is shown in Fig

154 Trigonometric waveforms 4 (iii) By drawing up a table of values, a graph of y D cos A C 45 may be plotted as shown in Fig. 8.. If y D cos A is assumed to start at 0 then y D cos A C 45 starts 45 earlier (i.e. has a maximum value 45 earlier).thus y D cos A C 45 is said to lead y D cos A by 45 Fig. 8.9 Fig. 8. Fig. 8.0 Lagging and leading angles (i) A sine or cosine curve may not always start at 0. To show this a periodic function is represented by y D sin A š a or y D cos A š a where a is a phase displacement compared with y D sin A or y D cos A. (ii) By drawing up a table of values, a graph of y D sin A 60 may be plotted as shown in Fig. 8.. If y D sin A isassumedtostartat0 then y D sin A 60 starts 60 later (i.e. has a zero value 60 later). Thus y D sin A 60 is said to lag y D sin A by 60 (iv) Generally, a graph of y D sin A a lags y D sin A by angle a, and a graph of y D sin A C a leads y D sin A by angle a (v) A cosine curve is the same shape as a sine curve but starts 90 earlier, i.e. leads by 90. Hence cos A D sin A C 90 Problem 8. A D 60 Sketch y D 5sin A C 0 from A D 0 to Amplitude D 5 and period D 60 / D 60. 5sin A C 0 leads 5 sina by 0 (i.e. starts 0 earlier). AsketchofyD5sin A C 0 is shown in Fig. 8.. Fig. 8. Fig. 8.

155 44 Basic Engineering Mathematics Problem 9. cycle. Sketch y D 7sin A p/ over one. y D cos A. y D sin 5x Amplitude D 7 and period D p/ D p radians. In general, y = sin(pt a) lags y = sin pt by a=p, hence 7sin A p/ lags 7 sin A by p/ /, i.e. p/6 rad or 0. AsketchofyD7sin A p/ is shown in Fig y D sin4t 4. y D cos q 5. y D 7 x sin 8 6. y D 6sin t y D 4cos q C Sinusoidal form A sin(!t ± a) Fig. 8.4 Problem 0. Sketch y D cos!t p/0 over one cycle. Amplitude D and period D p/! rad. cos!t p/0 lags cos!t by p/0! seconds. AsketchofyDcos!t p/0 is shown in Fig In Fig. 8.6, let OR represent a vector that is free to rotate anti-clockwise about O at a velocity of! rad/s. A rotating vector is called a phasor. After a time t seconds OR will have turned through an angle!t radians (shown as angle TOR in Fig. 8.6). If ST is constructed perpendicular to OR, thensin!t D ST/OT, i.e.st D OT sin!t. If all such vertical components are projected on to a graph of y against!t, a sine wave results of amplitude OR (as shown in Section 8.). If phasor OR makes one revolution (i.e. p radians) in T seconds, then the angular velocity,! D p/t rad/s, from which, T = p=! seconds T is known as the periodic time. The number of complete cycles occurring per second is called the frequency, f Frequency D number of cycles second D T D! p Hz i.e. f =! p Hz Hence angular velocity,! = pf rad/s Fig. 8.5 Now try the following exercise Exercise 68 Further problems on sine and cosine curves (Answers on page 60) In Problems to 7 state the amplitude and period of the waveform and sketch the curve between 0 and 60. Amplitude is the name given to the maximum or peak value of a sine wave, as explained in Section 8.4. The amplitude of the sine wave shown in Fig. 8.6 has an amplitude of. A sine or cosine wave may not always start at 0. To show this a periodic function is represented by y D sin!t š a or y D cos!t š a, wherea is a phase displacement compared with y D sin A or y D cos A. Agraphofy D sin!t a lags y D sin!t by angle a, and a graph of y D sin!t C a leads y D sin!t by angle a. The angle!t is measured in radians ( i.e.! rad s ) t s D!t radians hence angle a should also be in radians. The relationship between degrees and radians is: 60 D p radians or 80 = p radians

156 Trigonometric waveforms 45 Fig. 8.6 Hence rad D 80 p D 57.0 and, for example, 7 D 7 ð p D.9 rad 80 Given a general sinusoidal function y = A sin(!t ± a), then (i) A D amplitude (ii)! D angular velocity D pf rad/s p (iii) D periodic time T seconds!! (iv) D frequency, f hertz p (v) a D angle of lead or lag (compared with y D A sin!t) Problem. An alternating current is given by i D 0 sin 00pt C 0.7 amperes. Find the amplitude, periodic time, frequency and phase angle (in degrees and minutes). i D 0 sin 00pt C 0.7 A, hence amplitude = 0 A Angular velocity! D 00p, hence periodic time, T D p! D p 00p D D 0.0 s or 0 ms 50 Frequency, f D T D D 50 Hz 0.0 Phase angle, ( a D 0.7 rad D 0.7 ð 80 p D 5.47 or 5 8 leading ) i= 0 sin(00pt) Problem. An oscillating mechanism has a maximum displacement of.5 m and a frequency of 60 Hz. At time t D 0 the displacement is 90 cm. Express the displacement in the general form A sin!t š a. Amplitude D maximum displacement D.5m Angular velocity,! D pf D p 60 D 0p rad/s Hence displacement D.5sin 0pt C a m When t D 0, displacement D 90 cm D 0.90 m Hence 0.90 D.5sin 0 C a i.e. sin a D D 0.6 Hence a D sin 0.6 D.0 D 6 0 D 0.68 rad Thus displacement =.5sin(0pt ) m Problem. The instantaneous value of voltage in an a.c. circuit at any time t seconds is given by v D 40 sin 50pt 0.54 volts. Determine: (a) the amplitude, periodic time, frequency and phase angle (in degrees) (b) the value of the voltage when t D 0 (c) the value of the voltage when t D 0 ms (d) the time when the voltage first reaches 00 V, and (e) the time when the voltage is a maximum Sketch one cycle of the waveform. (a) Amplitude = 40 V Angular velocity,! D 50p Hence periodic time, T D p! D p 50p D 5 = 0.04 s or 40 ms Frequency f D T D D 5 Hz 0.04 ( Phase angle D 0.54 rad D 0.54 ð 80 ) p D lagging v D 40 sin 50pt

157 46 Basic Engineering Mathematics (b) When t = 0, v D 40 sin D 40 sin D 75.V ( (c) When t = 0 ms then v D 40 sin 50p 0 ) D 40 sin.098 D 40 sin 59 D 9.4 volts (d) When v D 00 volts then 00 D 40 sin 50pt D sin 50pt Hence 50pt 0.54 D sin 40 D 6.0 or Hence when v D 00 V, time, t D p rad 50pt D C 0.54 D.698 D ms (e) When the voltage is a maximum, v D 40 V Hence 40 D 40 sin 50pt 0.54 D sin 50pt pt 0.54 D sin D 90 or.5708 rad 50pt D.5708 C 0.54 D.8 Hence time, t D.8 50p D.44 m AsketchofvD40 sin 50pt 0.54 volts is shown in Fig. 8.7 Now try the following exercise Exercise 69 Further problems on the sinusoidal form A sin(!t ± a) (Answers on page 60) In Problems to find the amplitude, periodic time, frequency and phase angle (stating whether it is leading or lagging sin!t) of the alternating quantities given.. i D 40 sin 50pt C 0.9 ma. y D 75 sin 40t 0.54 cm. v D 00 sin 00pt 0.4 V 4. A sinusoidal voltage has a maximum value of 0 V and a frequency of 50 Hz. At time t D 0, the voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form v D A sin!t š a 5. An alternating current has a periodic time of 5 ms and a maximum value of 0 A. When time t D 0, current i D 0 amperes. Express the current i in the form i D A sin!t š a 6. An oscillating mechanism has a maximum displacement of. m and a frequency of 50 Hz. At time t D 0 the displacement is 50 cm. Express the displacement in the general form A sin!t š a 7. The current in an a.c. circuit at any time t seconds is given by: i D 5sin 00pt 0.4 amperes Determine (a) the amplitude, periodic time, frequency and phase angle (in degrees) (b) the value of current at t D 0 (c) the value of current at t D 8ms (d) the time when the current is first a maximum (e) the time when the current first reaches A. Sketch one cycle of the waveform showing relevant points. Assignment 9 This assignment covers the material in Chapters 7 and 8. The marks for each question are shown in brackets at the end of each question. Fig Figure A9. shows a plan view of a kite design. Calculate the lengths of the dimensions shown as a and b (4). In Fig. A9., evaluate (a) angle q (b) angle a (6). Determine the area of the plan view of a kite shown in Fig. A9.. (4)

158 Trigonometric waveforms Evaluate, each correct to 4 significant figures: 0.0 cm a 4.0 cm α (a) sin.78 (b) cos (c) tan p () 8 6. (a) Determine the acute angle cos 0.47 (i) in degrees and minutes and (ii) in radians (correct to decimal places). (b) Calculate the other angle (in degrees) between 0 and 60 which satisfies cos 0.47 (4) 7. Sketch the following curves labelling relevant points: b 60.0 cm (a) y D 4cos q C 45 (b) y D 5sin t 60 (8) 8. The current in an alternating current circuit at any time t seconds is given by: θ Fig. A9. 4. If the angle of elevation of the top of a 5 m perpendicular building from point A is measured as 7, determine the distance to the building. Calculate also the angle of elevation at a point B, 0 m closer to the building than point A. (5) i D 0 sin 00pt C 0.74 amperes Determine (a) the amplitude, periodic time, frequency and phase angle (with reference to 0 sin 00pt) (b) the value of current when t D 0 (c) the value of current when t D 6 ms Sketch one cycle of the oscillation. (6)

159 9 Cartesian and polar co-ordinates 9. Introduction There are two ways in which the position of a point in a plane can be represented. These are (a) by Cartesian co-ordinates, i.e.(x, y), and (b) by polar co-ordinates, i.e.(r, q), where r is a radius from a fixed point and q is an angle from a fixed point. 9. Changing from Cartesian into polar co-ordinates In Fig. 9., if lengths x and y are known, then the length of r can be obtained from Pythagoras theorem (see Chapter 7) since OPQ is a right-angled triangle. Hence r D x C y From trigonometric ratios (see Chapter 7), from which tan q D y x q = tan y x r D x C y and q D tan y x arethetwoformulaeweneed to change from Cartesian to polar co-ordinates. The angle q, which may be expressed in degrees or radians, must always be measured from the positive x-axis, i.e. measured from the line OQ in Fig. 9.. It is suggested that when changing from Cartesian to polar co-ordinates a diagram should always be sketched. Problem. Change the Cartesian co-ordinates (, 4) into polar co-ordinates. A diagram representing the point (, 4) is shown in Fig. 9.. from which, r = x + y Fig. 9. Fig. 9. From Pythagoras theorem, r D p C 4 D 5 (note that 5 has no meaning in this context). By trigonometric ratios, q D tan 4 D 5. or 0.97 rad.

160 Cartesian and polar co-ordinates 49 [note that 5. D 5. ð p/80 rad D 0.97 rad.] Hence (, 4) in Cartesian co-ordinates corresponds to (5, 5. ) or (5, 0.97 rad) in polar co-ordinates. Express in polar co-ordinates the posi- Problem. tion ( 4, ) A diagram representing the point using the Cartesian coordinates ( 4, ) is shown in Fig. 9.. Fig. 9.4 Fig. 9. From Pythagoras theorem, r D p 4 C D 5 By trigonometric ratios, a D tan 4 D 6.87 or rad. Hence q D D 4. or q D p D.498 rad Hence the position of point P in polar co-ordinate form is (5, 4. ) or (5,.498 rad) Problem. Express ( 5, ) in polar co-ordinates. A sketch showing the position ( 5, ) is shown in Fig r D 5 C D and a D tan 5 D 67.8 or.76 rad Hence q D 80 C 67.8 D 47.8 or q D p C.76 D 4.8 rad Thus ( 5, ) in Cartesian co-ordinates corresponds to (, 47.8 ) or (, 4.8 rad) in polar co-ordinates. Problem 4. Express (, 5) in polar co-ordinates. A sketch showing the position (, 5) is shown in Fig r D C 5 D p 9 D 5.85 correct to decimal places a D tan 5 D 68.0 or.90 rad Fig. 9.5 Hence q D D 9.80 or q D p.90 D 5.09 rad Thus (, 5) in Cartesian co-ordinates corresponds to (5.85, 9.80 ) or (5.85, 5.09 rad) in polar co-ordinates. Now try the following exercise Exercise 70 Further problems on changing from Cartesian into polar co-ordinates (Answers on page 60) In Problems to 8, express the given Cartesian coordinates as polar co-ordinates, correct to decimal places, in both degrees and in radians.., ,.5., , , 6..4, , ,.4

161 50 Basic Engineering Mathematics 9. Changing from polar into Cartesian co-ordinates From the right-angled triangle OPQ in Fig A sketch showing the position (6, 7 ) is shown in Fig x D r cos q D 6 cos 7 D 4.88 cos q D x r and sin q D y, from trigonometric ratios r Hence x = r cos q and y = r sin q Fig. 9.8 which corresponds to length OA in Fig y D r sin q D 6 sin 7 D 4.09 Fig. 9.6 If lengths r and angle q are known then x D r cos q and y D r sin q are the two formulae we need to change from polar to Cartesian co-ordinates. which corresponds to length AB in Fig Thus (6, 7 ) in polar co-ordinates corresponds to ( 4.88, 4.09) in Cartesian co-ordinates. (Note that when changing from polar to Cartesian coordinates it is not quite so essential to draw a sketch. Use of x D r cos q and y D r sin q automatically produces the correct signs.) Problem 5. Change (4, ) into Cartesian co-ordinates. Problem 7. co-ordinates Express (4.5, 5.6 rad) in Cartesian A sketch showing the position (4, ) is shown in Fig Now x D r cos q D 4 cos D.9 and y D r sin q D 4sin D. A sketch showing the position (4.5, 5.6 rad) is shown in Fig x D r cos q D 4.5cos5.6 D.948 Fig. 9.7 Hence (4, ) in polar co-ordinates corresponds (.9,.) in Cartesian co-ordinates. Problem 6. Express (6, 7 ) in Cartesian co-ordinates. to Fig. 9.9 which corresponds to length OA in Fig y D r sin q D 4.5sin5.6 D which corresponds to length AB in Fig. 9.9.

162 Cartesian and polar co-ordinates 5 Thus (.948, 4.057) in Cartesian co-ordinates corresponds to (4.5, 5.6 rad) in polar co-ordinates. 9.4 Use of R P and P R functions on calculators Another name for Cartesian co-ordinates is rectangular coordinates. Many scientific notation calculators possess R! P and P! R functions. The R is the first letter of the word rectangular and the P is the first letter of the word polar. Check the operation manual for your particular calculator to determine how to use these two functions. They make changing from Cartesian to polar co-ordinates, and vice-versa, so much quicker and easier. Now try the following exercise Exercise 7 Further problems on changing polar into Cartesian co-ordinates (Answers on page 6) In Problems to 8, express the given polar co-ordinates as Cartesian co-ordinates, correct to decimal places.. 5, ,. rad. 7, ,.5 rad , , 4rad 7..5, , 5.5 rad

163 0 Areas of plane figures 0. Mensuration Mensuration is a branch of mathematics concerned with the determination of lengths, areas and volumes. A D B C 0. Properties of quadrilaterals Polygon A polygon is a closed plane figure bounded by straight lines. A polygon which has: (i) sides is called a triangle (ii) 4 sides is called a quadrilateral (iii) 5 sides is called a pentagon (iv) 6 sides is called a hexagon (v) 7 sides is called a heptagon (vi) 8 sides is called an octagon There are five types of quadrilateral, these being: (i) rectangle (ii) square (iii) parallelogram (iv) rhombus (v) trapezium (The properties of these are given below). If the opposite corners of any quadrilateral are joined by a straight line, two triangles are produced. Since the sum of the angles of a triangle is 80, the sum of the angles of a quadrilateral is 60. In a rectangle, shown in Fig. 0.: (i) all four angles are right angles, (ii) opposite sides are parallel and equal in length, and (iii) diagonals AC and BD are equal in length and bisect one another. Fig. 0. Fig. 0. In a square, shown in Fig. 0.: P S (i) all four angles are right angles, (ii) opposite sides are parallel, (iii) all four sides are equal in length, and (iv) diagonals PR and QS are equal in length and bisect one another at right angles. In a parallelogram, shown in Fig. 0.: (i) opposite angles are equal, (ii) opposite sides are parallel and equal in length, and (iii) diagonals WY and XZ bisect one another. In a rhombus, shown in Fig. 0.4: (i) opposite angles are equal, (ii) opposite angles are bisected by a diagonal, (iii) opposite sides are parallel, (iv) all four sides are equal in length, and (v) diagonals AC and BD bisect one another at right angles. Q R

164 Areas of plane figures 5 W X Table 0. (Continued) Z Y Fig. 0. A α α B β β D C Fig. 0.4 Problem. State the types of quadrilateral shown in Fig. 0.6 and determine the angles marked a to l. Fig. 0.5 H In a trapezium, shown in Fig. 0.5: (i) only one pair of sides is parallel E 0. Worked problems on areas of plane figures F G A N D E x B x c a b C H (i) (ii) d F 40 J e G M 5 S R l O g h 5 i 65 5 U Q k j P 75 T x 0 f L (iii) x K Table 0. (iv) (v) Fig. 0.6 (i) ABCD is a square The diagonals of a square bisect each of the right angles, hence a D 90 D 45 (ii) EFGH is a rectangle In triangle FGH, 40 C 90 C b D 80 (angles in a triangle add up to 80 ) from which, b = 50.Also c = 40 (alternate angles between parallel lines EF and HG). (Alternatively, b and c are complementary, i.e. add up to 90 ) d D 90 C c (external angle of a triangle equals the sum of the interior opposite angles), hence d D 90 C 40 D 0

165 54 Basic Engineering Mathematics (iii) JKLM is a rhombus The diagonals of a rhombus bisect the interior angles and opposite internal angles are equal. Thus 6 JKM D 6 MKL D 6 JMK D 6 LMK D 0, hence e = 0 In triangle KLM, 0 C 6 KLM C 0 D 80 (angles in a triangle add up to 80 ), hence 6 KLM D 0. The diagonal JL bisects 6 KLM, hence f D 0 D 60 (iv) NOPQ is a parallelogram g = 5 (since opposite interior angles of a parallelogram are equal). In triangle NOQ, gchc65 D 80 (angles in a triangle addupto80 ), from which, h D D 6 i = 65 (alternate angles between parallel lines NQ and OP). j D 5 C i D 5 C 65 D 7 (external angle of a triangle equals the sum of the interior opposite angles). (v) RSTU is a trapezium 5 C k D 75 (external angle of a triangle equals the sum of the interior opposite angles), hence k = 40 6 STR D 5 (alternate angles between parallel lines RU and ST). l C 5 D 5 (external angle of a triangle equals the sum of the interior opposite angles), hence 8 mm 5 mm Fig mm A B C 70 mm (a) 6 mm 75 mm m 5 m (a) The girder may be divided into three separate rectangles as shown. (b) Area of rectangle A D 50 ð 5 D 50 mm Area of rectangle B D ð 6 D 6 ð 6 D 7 mm Area of rectangle C D 70 ð 8 D 560 mm Total area of girder D 50 C 7 C 560 D 8 mm or.8 cm (b) Area of path D area of large rectangle area of small rectangle 0 m D 5 ð 0 ð 6 D D 64 m Problem 4. Find the area of the parallelogram shown in Fig. 0.8 (dimensions are in mm). l D 5 5 D 80 A B Problem. A rectangular tray is 80 mm long and 400 mm wide. Find its area in (a) mm,(b)cm,(c)m. (a) Area D length ð width D 80 ð 400 D mm (b) cm D 00 mm. Hence mm D (c) m D cm. Hence cm D 80 cm 80 cm D m D 0.80 m Problem. Find (a) the cross-sectional area of the girder shown in Fig. 0.7(a) and (b) the area of the path shown in Fig. 0.7(b). Fig D 5 C E 4 Area of parallelogram D base ð perpendicular height. The perpendicular height h is found using Pythagoras theorem. BC D CE C h i.e. 5 D 4 5 C h h D 5 9 D 5 8 D 44 Hence, h D p 44 D mm ( can be neglected). Hence, area of ABCD D 5 ð D 00 mm h

166 Areas of plane figures 55 Problem 5. Figure 0.9 shows the gable end of a building. Determine the area of brickwork in the gable end. Fig. 0.9 B 6 m A 5 m 5 m C 8 m The shape is that of a rectangle and a triangle. Area of rectangle D 6 ð 8 D 48 m Area of triangle D ð base ð height. CD D 4m, AD D 5 m, hence AC D m (since it is a, 4, 5 triangle). Hence, area of triangle ABD D ð 8 ð D m Total area of brickwork D 48 C D 60 m Problem 6. Fig Determine the area of the shape shown in D (c) Circumference, c D pr, hence r D c p D 70 p D 5 p mm ( ) 5 Area of circle D pr D p D 5 p p D 89.9mm or.899 cm Problem 8. Calculate the areas of the following sectors of circles: (a) having radius 6 cm with angle subtended at centre 50 (b) having diameter 80 mm with angle subtended at centre (c) having radius 8 cm with angle subtended at centre.5 radians. Area of sector of a circle D q 60 pr or r q (q in radians). (a) Area of sector D ð p ð 6 60 p6 D D 5p D 5.7 cm 60 (b) If diameter D 80 mm, then radius, r D 40 mm, and area of sector D p40 D p40 D p40 D 504 mm or 5.04 cm Fig mm 7.4 mm 8.6 mm (c) Area of sector D r q D ð 8 ð.5 D 6.8cm Problem 9. A hollow shaft has an outside diameter of 5.45 cm and an inside diameter of.5 cm. Calculate the cross-sectional area of the shaft. The shape shown is a trapezium. Area of trapezium D (sum of parallel sides)(perpendicular distance between them) D 7.4 C The cross-sectional area of the shaft is shown by the shaded part in Fig. 0. (often called an annulus). D ð 6 ð 5.5 D 99 mm Problem 7. Find the areas of the circles having (a) a radius of 5 cm, (b) a diameter of 5 mm, (c) a circumference of 70 mm. Area of a circle D pr or pd 4 (a) Area D pr D p 5 D 5p D cm (b) Area D pd 4 D p 5 D 5p D 76.7mm 4 4 Fig. 0. d =.5 cm d = 5.45 cm Area of shaded part D area of large circle area of small circle D pd 4 pd 4 D p 4 D d D p D 9.5 cm

167 56 Basic Engineering Mathematics Now try the following exercise Exercise 7 Further problems on areas of plane figures (Answers on page 6). A rectangular plate is 85 mm long and 4 mm wide. Find its area in square centimetres.. A rectangular field has an area of. hectares and a length of 50 m. Find (a) its width and (b) the length of a diagonal ( hectare D m ).. Determine the area of each of the angle iron sections shown in Fig cm cm cm (a) Fig. 0. cm cm 5 mm 0 mm 0 mm 50 mm (b) 8 mm 6 mm 4. A rectangular garden measures 40 m by 5 m. A m flower border is made round the two shorter sides and one long side. A circular swimming pool of diameter 8 m is constructed in the middle of the garden. Find, correct to the nearest square metre, the area remaining. 5. The area of a trapezium is.5 cm and the perpendicular distance between its parallel sides is cm. If the length of one of the parallel sides is 5.6 cm, find the length of the other parallel side. 6. Find the angles p, q, r, s and t in Fig. 0.(a) to (c). r 5 4 cm 5.94 cm (i).5 cm Fig mm 6 mm 0 mm (ii) 8 mm 0 mm 6 cm cm 6 cm 0 cm 8. Determine the area of circles having (a) a radius of 4 cm (b) a diameter of 0 mm (c) a circumference of 00 mm. 9. An annulus has an outside diameter of 60 mm and an inside diameter of 0 mm. Determine its area. 0. If the area of a circle is 0 mm, find (a) its diameter, and (b) its circumference.. Calculate the areas of the following sectors of circles: (a) radius 9 cm, angle subtended at centre 75, (b) diameter 5 mm, angle subtended at centre , (c) diameter 5 cm, angle subtended at centre.9 radians.. Determine the area of the template shown in Fig Fig mm 90 mm (iii) (iv) 80 mm radius. An archway consists of a rectangular opening topped by a semi-circular arch as shown in Fig Determine the area of the opening if the width is m and the greatest height is m. 40 p q s 8 6 t (a) (b) (c) m Fig Name the types of quadrilateral shown in Fig. 0.4(i) to (iv), and determine (a) the area, and (b) the perimeter of each. Fig. 0.6 m

168 Areas of plane figures Calculate the area of the steel plate shown in Fig Dimensions in mm A hexagon is a 6-sided polygon which may be divided into 6 equal triangles as shown in Fig The angle subtended at the centre of each triangle is 60 /6 D 60. The other two angles in the triangle add up to 0 and are equal to each other. Hence each of the triangles is equilateral with each angle 60 and each side 8 cm. h 4 cm 8 cm 5 60 Fig Further worked problems on areas of plane figures Problem 0. Calculate the area of a regular octagon, if each side is 5 cm and the width across the flats is cm. An octagon is an 8-sided polygon. If radii are drawn from the centre of the polygon to the vertices then 8 equal triangles are produced (see Fig. 0.8). Fig cm Area of one triangle D ð base ð height D ð 8 ð h. h is calculated using Pythagoras theorem: 8 D h C 4 from which h D 8 4 D 6.98 cm Hence area of one triangle D ð 8 ð 6.98 D 7.7 cm Area of hexagon D 6 ð 7.7 D 66.cm Problem. Figure 0.0 shows a plan of a floor of a building which is to be carpeted. Calculate the area of the floor in square metres. Calculate the cost, correct to the nearest pound, of carpeting the floor with carpet costing 6.80 per m, assuming 0% extra carpet is required due to wastage in fitting. cm L.5 m M 5 cm m 0.6 m K J 4 m Fig. 0.8 Area of one triangle D ð base ð height 0.6 m H m 0.8 m G l 0.8 m F E m m A 0 B 60 C D m m B D ð 5 ð D 5 cm Fig. 0.0 Area of octagon D 8 ð 5 D 0 cm Problem. Determine the area of a regular hexagon which has sides 8 cm long. Area of floor plan D area of triangle ABC C area of semicircle C area of rectangle CGLM C area of rectangle CDEF area of trapezium HIJK

169 58 Basic Engineering Mathematics Triangle ABC is equilateral since AB D BC D m and hence angle B 0 CB D 60 sin B 0 CB D BB 0 /, i.e. BB 0 D sin60 D.598 m Area of triangle ABC D AC BB0 D.598 D.897 m Area of semicircle D pr D p.5 D 9.87 m Area of CGLM D 5 ð 7 D 5 m Area of CDEF D 0.8 ð D.4m Area of HIJK D KH C IJ 0.8 Since MC D 7m then LG D 7 m, hence JI D 7 5. D.8m 0.5 Areas of similar shapes The areas of similar shapes are proportional to the squares of corresponding linear dimensions. For example, Fig. 0. shows two squares, one of which has sides three times as long as the other. Area of Figure 0.(a) D x x D x Area of Figure 0.(b) D x x D 9x Hence Figure 0.(b) has an area (), i.e. 9 times the area of Figure 0.(a). Hence area of HIJK D C D.9 m Total floor area D.897 C 9.87 C 5 C.4.9 D m To allow for 0% wastage, amount of carpet required D. ð D 6.95 m Cost of carpet at 6.80 per m D 6.95 ð 6.80 D 074, correct to the nearest pound. Fig. 0. x (a) x x (b) x Now try the following exercise Exercise 7 Further problems on areas of plane figures (Answers on page 6). Calculate the area of a regular octagon if each side is 0 mm and the width across the flats is 48. mm.. Determine the area of a regular hexagon which has sides 5 mm.. A plot of land is in the shape shown in Fig. 0.. Determine (a) its area in hectares ( ha D 0 4 m ), and (b) the length of fencing required, to the nearest metre, to completely enclose the plot of land. 0 m 0 m 0 m Problem. A rectangular garage is shown on a building plan having dimensions 0 mm by 0 mm. If the plan is drawn to a scale of to 50, determine the true area of the garage in square metres. Area of garage on the plan D 0 mm ð 0 mm D 00 mm. Since the areas of similar shapes are proportional to the squares of corresponding dimensions then: True area of garage D 00 ð 50 D.5 ð 0 6 mm D.5 ð m D.5m 0 m 0 m 0 m Now try the following exercise 0 m 0 m Exercise 74 Further problems on areas of similar shapes (Answers on page 6) Fig m 40 m 5 m 0 m. The area of a park on a map is 500 mm. If the scale of the map is to determine the true area of the park in hectares ( hectare D 0 4 m ).. A model of a boiler is made having an overall height of 75 mm corresponding to an overall height of the actual boiler of 6 m. If the area of metal required for the model is 500 mm determine, in square metres, the area of metal required for the actual boiler.

170 Areas of plane figures 59 Assignment 0 0 mm This assignment covers the material contained in Chapters 9 and 0. The marks for each question are shown in brackets at the end of each question. 0 mm 45 mm. Change the following Cartesian co-ordinates into polar co-ordinates, correct to decimal places, in both degrees and in radians: (a) (., 5.4) (b) (7.6, 9.) (0). Change the following polar co-ordinates into Cartesian co-ordinates, correct to decimal places: (a) (6.5, ) (b) (, rad) (6). A rectangular park measures 50 m by 70 m. A m flower border is constructed round the two longer sides and one short side. A circular fish pond of diameter 5 m is in the centre of the park and the remainder of the park is grass. Calculate, correct to the nearest square metre, the area of (a) the fish pond, (b) the flower borders, (c) the grass. (0) 4. A swimming pool is 55 m long and 0 m wide. The perpendicular depth at the deep end is 5 m and at the shallow end is.5 m, the slope from one end to the other being uniform. The inside of the pool needs two coats of a protective paint before it is filled with water. Determine how many litres of paint will be needed if litre covers 0 m. (7) 5. Find the area of an equilateral triangle of side 0.0 cm. (4) 6. A steel template is of the shape shown in Fig. A0., the circular area being removed. Determine the area of the template, in square centimetres, correct to decimal place. (7) 70 mm Fig. A0. 50 mm dia. 70 mm 50 mm 60 mm 0 mm 7. The area of a plot of land on a map is 400 mm.if the scale of the map is to , determine the true area of the land in hectares ( hectare D 0 4 m ). () 8. Determine the shaded area in Fig. A0., correct to the nearest square centimetre. () Fig. A0. 0 cm cm

171 The circle. Introduction A circle is a plain figure enclosed by a curved line, every point on which is equidistant from a point within, called the centre.. Properties of circles (i) The distance from the centre to the curve is called the radius, r, of the circle (see OP in Fig..). Fig.. P R C O Q B A (v) A semicircle is one half of the whole circle. (vi) A quadrant is one quarter of a whole circle. (vii) A tangent to a circle is a straight line which meets the circle in one point only and does not cut the circle when produced. AC in Fig.. is a tangent to the circle since it touches the curve at point B only. If radius OB is drawn, then angle ABO is a right angle. (viii) A sector of a circle is the part of a circle between radii (for example, the portion OXY of Fig.. is a sector). If a sector is less than a semicircle it is called a minor sector, if greater than a semicircle it is called a major sector. Fig.. S O X R Y T (ii) The boundary of a circle is called the circumference, c. (iii) Any straight line passing through the centre and touching the circumference at each end is called the diameter, d (see QR in Fig..). Thus d = r (iv) The ratio circumference D a constant for any circle. diameter This constant is denoted by the Greek letter p (pronounced pie ), where p D.459, correct to 5 decimal places. Hence c/d D p or c = pd or c = pr (ix) A chord of a circle is any straight line which divides the circle into two parts and is terminated at each end by the circumference. ST, in Fig.. is a chord. (x) A segment is the name given to the parts into which a circle is divided by a chord. If the segment is less than a semicircle it is called a minor segment (see shaded area in Fig..). If the segment is greater than a semicircle it is called a major segment (see the unshaded area in Fig..). (xi) An arc is a portion of the circumference of a circle. The distance SRT in Fig.. is called a minor arc and the distance SXYT is called a major arc.

172 The circle 6 (xii) The angle at the centre of a circle, subtended by an arc, is double the angle at the circumference subtended by the same arc. With reference to Fig.., Angle AOC = U angle ABC. Fig.. Q A P C (xiii) The angle in a semicircle is a right angle (see angle BQP in Fig..). Problem. Find the circumference of a circle of radius.0 cm. Circumference, O c D ð p ð radius D pr D p.0 D cm Problem. If the diameter of a circle is 75 mm, find its circumference. Circumference, c D p ð diameter D pd D p 75 D 5.6mm B Pythagoras theorem: AO D AB C OB from which, AO D AB C OB D 50 C 40 D 55.mm Now try the following exercise Exercise 75 Further problems on properties of a circle (Answers on page 6). Calculate the length of the circumference of a circle of radius 7. cm.. If the diameter of a circle is 8.6 mm, calculate the circumference of the circle.. Determine the radius of a circle whose circumference is 6.5 cm. 4. Find the diameter of a circle whose perimeter is 49.8 cm.. Arc length and area of a sector One radian is defined as the angle subtended at the centre of a circle by an arc equal in length to the radius. With reference to Fig..5, for arc length s, q radians D s/r or arc length, s = rq where q is in radians. Problem. Determine the radius of a circle if its perimeter is cm. r o θ r s Perimeter D circumference, c D pr Hence radius r D c p D D 7.8 cm p Problem 4. In Fig..4, AB is a tangent to the circle at B. If the circle radius is 40 mm and AB D 50 mm, calculate the length AO. Fig..4 A A tangent to a circle is at right angles to a radius drawn from the point of contact, i.e. ABO D 90. Hence, using B r O Fig..5 When s D whole circumference (D pr) then q D s/r D pr/r D p i.e. p radians D 60 or p radians = 80 Thus rad D 80 /p D 57.0, correct to decimal places. Since p rad D 80, then p/ D 90, p/ D 60, p/4 D 45, and so on. Area of a sector D q 60 (pr ) when q is in degrees D q p pr D r q when q is in radians

173 6 Basic Engineering Mathematics Problem 5. Convert to radians: (a) 5 (b) (a) Since 80 D p rad then D p/80 rad, therefore ( p ) c 5 D 5 D.8 radians 80 (Note that c means circular measure and indicates radian measure.) (b) D D ( p ) c D D.8 radians 80 Problem 6. Convert to degrees and minutes: (a) radians (b) p/4radians (a) Since p rad D 80 then rad D 80 /p, therefore ( ) D D 4.95 p 0.95 D 0.95 ð 60 0 D 55 0, correct to the nearest minute, hence radians = 4 55 ( ) 80 (b) Since rad D then p D 4 80 D 5 p 4 rad D p 4 ( 80 Problem 7. Express in radians, in terms of p (a) 50 (b) 70 (c) 7.5 Since 80 D p radthen D 80/p, hence ( p ) (a) 50 D 50 rad D 5p 80 6 rad ( p ) (b) 70 D 70 rad D p 80 rad ( p ) (c) 7.5 D 7.5 rad D 75p 5p rad D rad Problem 8. Find the length of arc of a circle of radius 5.5 cm when the angle subtended at the centre is.0 radians. From equation (), length of arc, s D rq, where q is in radians, hence s D D 6.60 cm p ) Problem 9. Determine the diameter and circumference of a circle if an arc of length 4.75 cm subtends an angle of 0.9 radians. Since s D rq then r D s q D 4.75 D 5. cm. 0.9 Diameter D ð radius D ð 5. D 0.44 cm. Circumference, c D pd D p 0.44 D.80 cm. Problem 0. If an angle of 5 is subtended by an arc of a circle of radius 8.4 cm, find the length of (a) the minor arc, and (b) the major arc, correct to significant figures. ( p ) Since 80 D p rad then D rad 80 ( p ) and 5 D 5 rad 80 Length of minor arc, ( p ) s D rq D D 8.cm 80 correct to significant figures. Length of major arc D (circumference minor arc) D p D 4.5cm, correct to significant figures. (Alternatively, major arc D rq D p/80 D 4.5cm.) Problem. Determine the angle, in degrees and minutes, subtended at the centre of a circle of diameter 4 mm by an arc of length 6 mm. Calculate also the area of the minor sector formed. Since length of arc, s D rq then q D s/r Radius, r D diameter D 4 D mm hence q D s r D 6 D.74 radians.74 rad D.74 ð 80/p D 98. D 98 D angle subtended at centre of circle. From equation (), area of sector D r q D.74 D 78 mm Problem. A football stadium floodlight can spread its illumination over an angle of 45 to a distance of 55 m. Determine the maximum area that is floodlit. Floodlit area D area of sector D r q D 55 ( 45 ð = 88 m p ) 80 from equation ()

174 The circle 6 Problem. An automatic garden spray produces a spray to a distance of.8 m and revolves through an angle a which may be varied. If the desired spray catchment area is to be.5m, to what should angle a be set, correct to the nearest degree. Length BC D 0 mm (i.e. the radius of the circle), and AB D 0 0. D 7.88 mm from Fig..6. Hence sin q D and q ( ) 0 D sin D and angle q = 68 Area of sector D r q, hence.5 D.8 a from which, a D.5 ð D.54 radians.8 (.54 rad D.54 ð 80 ) D 88.4 p Hence angle a = 88, correct to the nearest degree. Problem 4. The angle of a tapered groove is checked using a 0 mm diameter roller as shown in Fig..6. If the roller lies. mm below the top of the groove, determine the value of angle q.. mm Fig..6 0 mm θ 0 mm In Fig..7, triangle ABC is right-angled at C (see Section.(vii), page 60).. mm Fig..7 B A θ 0 mm C 0 mm Now try the following exercise Exercise 76 Further problems on arc length and area of a sector (Answers on page 6). Convert to radians in terms of p: (a) 0 (b) 75 (c) 5. Convert to radians: (a) 48 (b) (c) 5 0. Convert to degrees: (a) 5p 6 rad (b) 4p 7p rad (c) 9 rad 4. Convert to degrees and minutes: (a) 0.05 rad (b).69 rad (c) 7.4 rad 5. Find the length of an arc of a circle of radius 8. cm when the angle subtended at the centre is.4 radians. Calculate also the area of the minor sector formed. 6. If the angle subtended at the centre of a circle of diameter 8 mm is.46 rad, find the lengths of the (a) minor arc (b) major arc. 7. A pendulum of length.5 m swings through an angle of 0 in a single swing. Find, in centimetres, the length of the arc traced by the pendulum bob. 8. Determine the length of the radius and circumference of a circle if an arc length of.6 cm subtends an angle of.76 radians. 9. Determine the angle of lap, in degrees and minutes, if 80 mm of a belt drive are in contact with a pulley of diameter 50 mm. 0. Determine the number of complete revolutions a motorcycle wheel will make in travelling km, if the wheel s diameter is 85. cm.. The floodlights at a sports ground spread its illumination over an angle of 40 to a distance of 48 m. Determine (a) the angle in radians, and (b) the maximum area that is floodlit.. Determine (a) the shaded area in Fig..8 (b) the percentage of the whole sector that the shaded area represents.

175 64 Basic Engineering Mathematics 0.75 rad mm 50 mm y x + y = 9 0 x Fig..8. Determine the length of steel strip required to make the clip shown in Fig mm Fig.. More generally, the equation of a circle, centre (a, b), radius r, is given by: x a C y b D r Figure. shows a circle x C y D mm rad y 5 4 r = 00 mm b = Fig A 50 tapered hole is checked with a 40 mm diameter ball as shown in Fig..0. Determine the length shown as x. x Fig mm 40 mm 50.4 The equation of a circle The simplest equation of a circle, centre at the origin, radius r, is given by: x C y D r For example, Fig.. shows a circle x C y D 9. Fig x a = The general equation of a circle is: x C y C ex C fy C c D 0 Multiplying out the bracketed terms in equation () gives: x ax C a C y by C b D r Comparing this with equation () gives: e D a, i.e. a = e and f D b, i.e. b = f and c D a C b r, i.e. a + b c Thus, for example, the equation x C y 4x 6y C 9 D 0 represents a circle with centre ( ) ( ) 4 6 a D, b D i.e., at (, ) and radius r D p C 9 D

176 The circle 65 Hence x C y 4x 6y C 9 D 0 is the circle shown in Fig.. (which may be checked by multiplying out the brackets in the equation x C y D 4 Problem 5. Determine (a) the radius, and (b) the coordinates of the centre of the circle given by the equation: x C y C 8x y C 8 D 0 x Cy C8x yc8 D 0 is of the form shown in equation (), ( ) ( ) 8 where a D D 4, bd D Hence if x C y 4x C 6y D 0. ( ) ( ) 4 6 then a D D, b D D and r D C D p 6 D 4 Thus thecirclehascentre(, ) and radius 4, as shown in Fig..4. y 4 and r D 4 C 8 D p 9 D x Hence x C y C 8x y C 8 0 represents a circle centre ( 4, ) and radius, as shown in Fig... y 4 4 r = 4 r = 8 b = x Fig..4 Now try the following exercise Fig.. Problem 6. a = 4 Sketch the circle given by the equation: x C y 4x C 6y D 0 The equation of a circle, centre (a, b), radius r is given by: x a C y b D r The general equation of a circle is x C y C ex C fy C c D 0 From above a D e,bd f and r D a C b c Exercise 77 Further problems on the equation of a circle (Answers on page 6). Determine (a) the radius, and (b) the co-ordinates of the centre of the circle given by the equation x C y C 8x y C 8 D 0. Sketch the circle given by the equation x C y 6x C 4y D 0. Sketch the curve x C y 5 D 0 4. Sketch the curve [ ( y ) ] x D 6 6

177 Volumes of common solids. Volumes and surface areas of regular solids A summary of volumes and surface areas of regular solids is shown in Table.. Table. (i) Rectangular prism (or cuboid). Worked problems on volumes and surface areas of regular solids Problem. A water tank is the shape of a rectangular prism having length m, breadth 75 cm and height 50 cm. Determine the capacity of the tank in (a) m (b) cm (c) litres. h (ii) Cylinder (iii) Pyramid b r h Volume = πr h Volume = l b h Surface area = (bh + hl + lb) Total surface area = πrh + πr h l Volume = A h where A = area of base and h = perpendicular height Total surface area = (sum of areas of triangles forming sides) + (area of base) Volume of rectangular prism D l ð b ð h (see Table.) (a) Volume of tank D ð 0.75 ð 0.5 D 0.75 m (b) m D 0 6 cm, hence 0.75 m D 0.75 ð 0 6 cm (c) litre D 000 cm, hence cm D D cm litres D 750 litres Problem. Find the volume and total surface area of a cylinder of length 5 cm and diameter 8 cm. (iv) Cone h l Volume = πr h Volume of cylinder D pr h (see Table.) Since diameter D 8 cm, then radius r D 4cm (v) Sphere r r Curved surface area = πrl Total surface area = πrl + πr 4 Volume = πr Surface area = 4πr Hence volume D p ð 4 ð 5 D 754 cm Total surface area (i.e. including the two ends) D prh C pr D ð p ð 4 ð 5 C ð p ð 4 D 477.5cm

178 Volumes of common solids 67 Problem. Determine the volume (in cm ) of the shape shown in Fig... Problem 5. Determine the volume and the total surface area of the square pyramid shown in Fig.. if its perpendicular height is cm. 6 mm A mm 40 mm Fig.. The solid shown in Fig.. is a triangular prism. The volume V of any prism is given by: V D Ah, wherea is the cross-sectional area and h is the perpendicular height. Hence volume D ð 6 ð ð 40 D 840 mm D.840 cm (since cm D 000 mm Problem 4. Calculate the volume and total surface area of the solid prism shown in Fig... 4 cm cm 5 cm Fig.. 5 cm D B E C 5 cm Volume of pyramid D (area of base) (perpendicular height) D 5 ð 5 ð D 00 cm The total surface area consists of a square base and 4 equal triangles. Area of triangle ADE D ð base ð perpendicular height D ð 5 ð AC The length AC may be calculated using Pythagoras theorem on triangle ABC, whereab D cm, BC D ð 5 D.5cm AC D AB C BC D C.5 D.6 cm Hence area of triangle ADE D ð 5 ð.6 D 0.65 cm Total surface area of pyramid D 5 ð 5 C D 47.6cm 5 cm 5 cm 5 cm Problem 6. Determine the volume and total surface area of a cone of radius 5 cm and perpendicular height cm. Fig.. The solid shown in Fig.. is a trapezoidal prism. Volume D cross-sectional area ð height D C 5 4 ð 5 D ð 5 D 480 cm Surface area D sum of two trapeziums C 4 rectangles D ð C 5 ð 5 C ð 5 C 5 ð 5 D 64 C 75 C 65 C 50 D 454 cm The cone is shown in Fig..4. Volume of cone D pr h D ð p ð 5 ð D 4.cm Total surface area D curved surface area C area of base D prl C pr From Fig..4, slant height l may be calculated using Pythagoras theorem l D C 5 D cm Hence total surface area D p ð 5 ð C p ð 5 D 8.7cm

179 r = 8 mm 68 Basic Engineering Mathematics Fig..4 h = cm l r = 5 cm Problem 7. Find the volume and surface area of a sphere of diameter 8 cm. Since diameter D 8 cm, then radius, r D 4cm. Volume of sphere D 4 pr D 4 ð p ð 4 D 68.cm 9. A square pyramid has a perpendicular height of 4 cm. If a side of the base is.4 cm long find the volume and total surface area of the pyramid. 0. A sphere has a diameter of 6 cm. Determine its volume and surface area.. Find the total surface area of a hemisphere of diameter 50 mm.. Further worked problems on volumes and surface areas of regular solids Problem 8. A wooden section is shown in Fig..5. Find (a) its volume (in m ), and (b) its total surface area. Surface area of sphere D 4pr D 4 ð p ð 4 D 0.cm Now try the following exercise Exercise 78 Further problems on volumes and surface areas of regular solids (Answers on page 6). A rectangular block of metal has dimensions of 40 mm by 5 mm by 5 mm. Determine its volume. Find also its mass if the metal has a density of 9g/cm. Determine the maximum capacity, in litres, of a fish tank measuring 50 cm by 40 cm by.5 m ( litre D 000 cm.. Determine how many cubic metres of concrete are required for a 0 m long path, 50 mm wide and 80 mm deep. 4. Calculate the volume of a metal tube whose outside diameter is 8 cm and whose inside diameter is 6 cm, if the length of the tube is 4 m. 5. The volume of a cylinder is 400 cm. If its radius is 5.0 cm, find its height. Determine also its curved surface area. 6. If a cone has a diameter of 80 mm and a perpendicular height of 0 mm calculate its volume in cm and its curved surface area. 7. A cylinder is cast from a rectangular piece of alloy 5 cm by 7 cm by cm. If the length of the cylinder is to be 60 cm, find its diameter. 8. Find the volume and the total surface area of a regular hexagonal bar of metal of length m if each side of the hexagon is 6 cm. Fig..5 m r cm (a) The section of wood is a prism whose end comprises a rectangle and a semicircle. Since the radius of the semicircle is 8 cm, the diameter is 6 cm. Hence the rectangle has dimensions cm by 6 cm. Area of end D ð 6 C p8 D 9.5cm Volume of wooden section D area of end ð perpendicular height D 9.5 ð 00 D cm D D m m 0 6 (b) The total surface area comprises the two ends (each of area 9.5 cm ), three rectangles and a curved surface (which is half a cylinder), hence total surface area D ð 9.5 C ð 00 C 6 ð 00 C p ð 8 ð 00 D 585 C 700 C 4800 C 400p D 05 cm or.05 m

180 Volumes of common solids 69 Problem 9. A pyramid has a rectangular base.60 cm by 5.40 cm. Determine the volume and total surface area of the pyramid if each of its sloping edges is 5.0 cm. The pyramid is shown in Fig..6. To calculate the volume of the pyramid the perpendicular height EF is required. Diagonal BD is calculated using Pythagoras theorem, i.e. BD D [.60 C 5.40 ] D cm F Similarly, if H is the mid-point of AB, then FH D D 4.75 cm, hence area of triangle ABF (which equals triangle CDF) D D 9.8 cm Total surface area of pyramid D 6.80 C 9.8 C D 5.60 C C 9.44 D 5.7cm Problem 0. Calculate the volume and total surface area of a hemisphere of diameter 5.0 cm. Volume of hemisphere D (volume of sphere) 5.0 cm 5.0 cm 5.0 cm 5.0 cm Total surface area D pr D ( ) 5.0 p D.7cm C D curved surface area C area of circle D D (surface area of sphere) C pr.60 cm G A H E 5.40 cm B D 4pr C pr ( ) 5.0 D pr C pr D pr D p D 58.9cm Fig..6 Hence EB D BD D D.45 cm. Using Pythagoras theorem on triangle BEF gives from which, BF D EB C EF EF D BF EB D D 4.64 cm. Volume of pyramid D (area of base)(perpendicular height) D.60 ð D cm Area of triangle ADF (which equals triangle BCF D AD FG, where G is the midpoint of AD. Using Pythagoras theorem on triangle FGA gives FG D [ ] D 4.89 cm Hence area of triangle ADF D D 6.80 cm Problem. A rectangular piece of metal having dimensions 4 cm by cm by cm is melted down and recast into a pyramid having a rectangular base measuring.5 cm by 5 cm. Calculate the perpendicular height of the pyramid. Volume of rectangular prism of metal D 4ðð D 44 cm Volume of pyramid D (area of base)(perpendicular height) Assuming no waste of metal, 44 D.5 ð 5 (height) i.e. perpendicular height D 44 ð D 4.56 cm.5 ð 5 Problem. A rivet consists of a cylindrical head, of diameter cm and depth mm, and a shaft of diameter mm and length.5 cm. Determine the volume of metal in 000 such rivets.

181 70 Basic Engineering Mathematics Radius of cylindrical head D cm D 0.5cm and height of cylindrical head D mmd 0.cm Hence, volume of cylindrical head D pr h D p D 0.57 cm Volume of cylindrical shaft ( ) 0. D pr h D p.5 D cm Total volume of rivet D 0.57 C D 0.04 cm Volume of metal in 000 such rivets D 000 ð 0.04 D 408.4cm Problem. A solid metal cylinder of radius 6 cm and height 5 cm is melted down and recast into a shape comprising a hemisphere surmounted by a cone. Assuming that 8% of the metal is wasted in the process, determine the height of the conical portion, if its diameter is to be cm. Volume of cylinder D pr h D p ð 6 ð 5 D 540p cm If 8% of metal is lost then 9% of 540p gives the volume of the new shape (shown in Fig..7). Fig..7 h r cm Hence the volume of (hemisphereccone) D 0.9ð540p cm, ( ) 4 i.e. pr C pr h D 0.9 ð 540p Dividing throughout by p gives: r C r h D 0.9 ð 540 Since the diameter of the new shape is to be cm, then radius r D 6cm, hence 6 C 6 h D 0.9 ð C h D i.e. height of conical portion, h D D 9.4cm Problem 4. A block of copper having a mass of 50 kg is drawn out to make 500 m of wire of uniform cross-section. Given that the density of copper is 8.9 g/cm, calculate (a) the volume of copper, (b) the cross-sectional area of the wire, and (c) the diameter of the cross-section of the wire. (a) A density of 8.9 g/cm means that 8.9 g of copper has a volume of cm, or g of copper has a volume of (/8.9) cm Hence 50 kg, i.e g, has a volume (b) Volume of wire cm D 56 cm D area of circular cross-section ð length of wire. Hence 56 cm D area ð 500 ð 00 cm, from which, area D ð 00 cm D 0. cm (c) Area of circle D pr or pd pd, hence 0. D 4 4 from which ( ) 4 ð 0. d D D cm p i.e. diameter of cross-section is.780 mm. Problem 5. A boiler consists of a cylindrical section of length 8 m and diameter 6 m, on one end of which is surmounted a hemispherical section of diameter 6 m, and on the other end a conical section of height 4 m and base diameter 6 m. Calculate the volume of the boiler and the total surface area. The boiler is shown in Fig..8. Volume of hemisphere, P D pr D ð p ð D 8p m Volume of cylinder, Q D pr h D p ð ð 8 D 7p m Volume of cone, R D pr h D ð p ð ð 4 D p m

182 Volumes of common solids 7 Fig..8 8 m P Q A m B 4 m R I C 6 m Total volume of boiler D 8p C 7p C p D 0p D 0.4m Surface area of hemisphere, P D 4pr D ð p ð D 8p m Curved surface area of cylinder, Q D prh D ð p ð ð 8 D 48p m The slant height of the cone, l, is obtained by Pythagoras theorem on triangle ABC, i.e.l D 4 C D 5 Curved surface area of cone, R D prl D p ð ð 5 D 5p m Total surface area of boiler D 8p C 48p C 5p Now try the following exercise Exercise 79 D 8p D 54.5m Further problems on volumes and surface areas of regular solids (Answers on page 6). Determine the mass of a hemispherical copper container whose external and internal radii are cm and 0 cm. Assuming that cm of copper weighs 8.9 g.. If the volume of a sphere is 566 cm, find its radius.. A metal plumb bob comprises a hemisphere surmounted by a cone. If the diameter of the hemisphere and cone are each 4 cm and the total length is 5 cm, find its total volume. 4. A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 m and the cylindrical portion has a height of.5 m, with a diameter of 5 m. Calculate the surface area of material needed to make the marquee assuming % of the material is wasted in the process. 5. Determine (a) the volume and (b) the total surface area of the following solids: (i) a cone of radius 8.0 cm and perpendicular height 0 cm (ii) a sphere of diameter 7.0 cm (iii) a hemisphere of radius.0 cm (iv) a.5 cm by.5 cm square pyramid of perpendicular height 5.0 cm (v) a 4.0 cm by 6.0 cm rectangular pyramid of perpendicular height.0 cm (vi) a 4. cm by 4. cm square pyramid whose sloping edges are each 5.0 cm (vii) a pyramid having an octagonal base of side 5.0 cm and perpendicular height 0 cm. 6. The volume of a sphere is 5 cm. Determine its diameter. 7. A metal sphere weighing 4 kg is melted down and recast into a solid cone of base radius 8.0 cm. If the density of the metal is 8000 kg/m determine (a) the diameter of the metal sphere and (b) the perpendicular height of the cone, assuming that 5% of the metal is lost in the process. 8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 6.0 cm and the side of base is.0 cm. 9. A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere is.5 m and the slant height of the cone is 4.0 m. Determine the volume and surface area of the buoy. 0. A petrol container is in the form of a central cylindrical portion 5.0 m long with a hemispherical section surmounted on each end. If the diameters of the hemisphere and cylinder are both. m determine the capacity of the tank in litres litre D 000 cm.. Figure.9 shows a metal rod section. Determine its volume and total surface area. Fig cm radius.50 cm.00 m

183 7 Basic Engineering Mathematics.4 Volumes and surface areas of frusta of pyramids and cones The frustum of a pyramid or cone is the portion remaining when a part containing the vertex is cut off by a plane parallel to the base. The volume of a frustum of a pyramid or cone is given by the volume of the whole pyramid or cone minus the volume of the small pyramid or cone cut off. The surface area of the sides of a frustum of a pyramid or cone is given by the surface area of the whole pyramid or cone minus the surface area of the small pyramid or cone cut off. This gives the lateral surface area of the frustum. If the total surface area of the frustum is required then the surface area of the two parallel ends are added to the lateral surface area. There is an alternative method for finding the volume and surface area of a frustum of a cone. With reference to Fig..0: D 4.0 cm.0 cm A P E.6 cm Volume D ph(r + Rr + r ) Curved surface area D pl(r + r) Total surface area D pl(r + r)+pr + pr B R.0 cm.0 cm Q C I h r 6.0 cm Fig.. Volume of frustum of cone R D volume of large cone volume of small cone cut off Fig..0 D p p.0 7. D D 7.6cm Problem 6. Determine the volume of a frustum of a cone if the diameter of the ends are 6.0 cm and 4.0 cm and its perpendicular height is.6 cm. Method A section through the vertex of a complete cone is shown in Fig... Using similar triangles AP DP D DR BR AP Hence.0 D.6.0 from which AP D.0.6 D 7.cm.0 The height of the large cone D.6 C 7. D 0.8cm. Method From above, volume of the frustum of a cone D ph R C Rr C r, where R D.0cm, r D.0cm and h D.6cm Hence volume of frustum D p.6 [.0 C.0.0 C.0 ] D p D 7.6cm Problem 7. Find the total surface area of the frustum of the cone in Problem 6. Method Curved surface area of frustum D curved surface area of large cone curved surface area of small cone cut off.

184 Volumes of common solids 7 From Fig.., using Pythagoras theorem: AB D AQ C BQ, from which AB D [0.8 C.0 ] D. cm and AD D AP C DP, from which AD D [7. C.0 ] D 7.47 cm Curved surface area of large cone D prl D p BQ AB D p.0. D cm and curved surface area of small cone D p DP AD D p D cm Hence, curved surface area of frustum D D 58.7 cm Total surface area of frustum D curved surface area C area of two circular ends D 58.7 C p.0 C p.0 D 58.7 C.57 C 8.7 D 99.6cm Method From page 7, total surface area of frustum D pl R C r C pr C pr, 8.0 m Fig cm 8.0 m (a) 4.6 cm.7 m Volume of small pyramid cut off A B D D 4.5 m Hence volume of storage hopper D D 46.m. m. m.6 m H. m C (b) G 4.0 m Problem 9. Determine the lateral surface area of the storage hopper in Problem 8. The lateral surface area of the storage hopper consists of four equal trapeziums. F D E where l D BD D D.74 cm, R D.0cm and r D.0cm. Hence total surface area of frustum D p.74.0 C.0 C p.0 C p.0 D 99.6cm 4.6 m P 4.6 m R Q S Problem 8. A storage hopper is in the shape of a frustum of a pyramid. Determine its volume if the ends of the frustum are squares of sides 8.0 m and 4.6 m, respectively, and the perpendicular height between its ends is.6 m. 8.0 m U 0 T 8.0 m The frustum is shown shaded in Fig..(a) as part of a complete pyramid. A section perpendicular to the base through the vertex is shown in Fig..(b). By similar triangles: CG BG D BH AH Height CG D BG ( BH AH ) D..6.7 D 4.87 m Height of complete pyramid D.6 C 4.87 D 8.47 m Volume of large pyramid D D m Fig.. From Fig.., area of trapezium PRSU D PR C SU QT OT D.7m (same as AH in Fig..(b) and OQ D.6m. By Pythagoras theorem, QT D OQ C OT D [.6 C.7 ] D.98 m Area of trapezium PRSU D 4.6 C D 5.07 m Lateral surface area of hopper D D 00.m

185 74 Basic Engineering Mathematics Problem 0. A lampshade is in the shape of a frustum of a cone. The vertical height of the shade is 5.0 cm and the diameters of the ends are 0.0 cm and 0.0 cm, respectively. Determine the area of the material needed to form the lampshade, correct to significant figures. The curved surface area of a frustum of a cone D pl R C r from page 7. Since the diameters of the ends of the frustum are 0.0 cm and 0.0 cm, then from Fig..4, r D 5.0cm,RD0.0cm and l D [5.0 C 5.0 ] D 5.50 cm, from Pythagoras theorem. r = 5.0 cm Volume of cylindrical portion ( ) 5.0 D pr h D p.0 D 5890 m Volume of frustum of cone D ph R C Rr C r where h D D 8.0m, R D 5.0/ D.5m and r D.0/ D 6.0m. Hence volume of frustum of cone D p 8.0 [.5 C C 6.0 ] D 508 m Total volume of cooling tower D 5890 C 508 D 0 98 m. If 40% of space is occupied then volume of air space D 0.6 ð 0 98 D 6557 m Fig..4 h = 5.0 cm R = 0.0 cm I 5.0 cm Hence curved surface area D p C 5.0 D 0.7cm, i.e. the area of material needed to form the lampshade is 00 cm, correct to significant figures. Problem. A cooling tower is in the form of a cylinder surmounted by a frustum of a cone as shown in Fig..5. Determine the volume of air space in the tower if 40% of the space is used for pipes and other structures. Fig..5.0 m 5.0 m.0 m 0.0 m Now try the following exercise Exercise 80 Further problems on volumes and surface areas of frustra of pyramids and cones (Answers on page 6). The radii of the faces of a frustum of a cone are.0 cm and 4.0 cm and the thickness of the frustum is 5.0 cm. Determine its volume and total surface area.. A frustum of a pyramid has square ends, the squares having sides 9.0 cm and 5.0 cm, respectively. Calculate the volume and total surface area of the frustum if the perpendicular distance between its ends is 8.0 cm.. A cooling tower is in the form of a frustum of a cone. The base has a diameter of.0 m, the top has a diameter of 4.0 m and the vertical height is 4.0 m. Calculate the volume of the tower and the curved surface area. 4. A loudspeaker diaphragm is in the form of a frustum of a cone. If the end diameters are 8.0 cm and 6.00 cm and the vertical distance between the ends is 0.0 cm, find the area of material needed to cover the curved surface of the speaker. 5. A rectangular prism of metal having dimensions 4. cm by 7. cm by.4 cm is melted down and recast into a frustum of a square pyramid, 0% of the metal being lost in the process. If the ends of the frustum are squares of side cm and 8 cm respectively, find the thickness of the frustum. 6. Determine the volume and total surface area of a bucket consisting of an inverted frustum of a cone, of slant height 6.0 cm and end diameters 55.0 cm and 5.0 cm.

186 Volumes of common solids A cylindrical tank of diameter.0 m and perpendicular height.0 m is to be replaced by a tank of the same capacity but in the form of a frustum of a cone. If the diameters of the ends of the frustum are.0 m and.0 m, respectively, determine the vertical height required..5 Volumes of similar shapes The volumes of similar bodies are proportional to the cubes of corresponding linear dimensions. For example, Fig..6 shows two cubes, one of which has sides three times as long as those of the other. ( ) Hence mass of model D (mass of car) 50 D 000 D kg or 8g 50 Now try the following exercise Exercise 8 Further problems on volumes of similar shapes (Answers on page 6). The diameter of two spherical bearings are in the ratio :5. What is the ratio of their volumes?. An engineering component has a mass of 400 g. If each of its dimensions are reduced by 0% determine its new mass. Assignment x Fig..6 x (a) x x x (b) Volume of Fig..6(a) D x x x D x x Volume of Fig..6(b) D x x x D 7x Hence Fig..6(b) has a volume (), i.e. 7 times the volume of Fig..6(a). Problem. A car has a mass of 000 kg. A model of the car is made to a scale of to 50. Determine the mass of the model if the car and its model are made of the same material. Volume of model Volume of car D ( ) 50 since the volume of similar bodies are proportional to the cube of corresponding dimensions. Mass D density ð volume, and since both car and model are made of the same material then: Mass of model Mass of car D ( 50 ) This assignment covers the material contained in Chapters and. The marks for each question are shown in brackets at the end of each question.. Determine the diameter of a circle whose circumference is 78.4 cm. (). Convert (a) to radians (b).74 radians to degrees and minutes (4). Calculate the length of metal strip needed to make the clip shown in Fig. A.. (6) 75 mm 5 mm rad 0 mm rad 5 mm rad 70 mm 70 mm Fig. A. 4. A lorry has wheels of radius 50 cm. Calculate the number of complete revolutions a wheel makes (correct to the nearest revolution) when travelling miles (assume mile D.6km). (5) 5. The equation of a circle is: x C y C x 4y C 4 D 0 Determine (a) the diameter of the circle, and (b) the co-ordinates of the centre of the circle. (5)

187 76 Basic Engineering Mathematics 6. Determine the volume (in cubic metres) and the total surface area (in square metres) of a solid metal cone of base radius 0.5 m and perpendicular height.0 m. Give answers correct to decimal places. (5) 7. A rectangular storage container has dimensions. m by 90 cm by 60 cm. Determine its volume in (a) m (b) cm (4) 8. Calculate (a) the volume, and (b) the total surface area of a 0 cm by 5 cm rectangular pyramid of height 0 cm. (7) 9. A water container is of the form of a central cylindrical part.0 m long and diameter.0 m, with a hemispherical section surmounted at each end as shown in Fig. A.. Determine the maximum capacity of the container, correct to the nearest litre. ( litre D 000 cm ) (5) Fig. A..0 m.0 m 0. Find the total surface area of a bucket consisting of an inverted frustum of a cone, of slant height 5.0 cm and end diameters 60.0 cm and 40.0 cm. (4). A boat has a mass of kg. A model of the boat is made to a scale of to 80. If the model is made of the same material as the boat, determine the mass of the model (in grams). ()

188 Irregular areas and volumes and mean values of waveforms. Areas of irregular figures Areas of irregular plane surfaces may be approximately determined by using (a) a planimeter, (b) the trapezoidal rule, (c) the mid-ordinate rule, and (d) Simpson s rule. Such methods may be used, for example, by engineers estimating areas of indicator diagrams of steam engines, surveyors estimating areas of plots of land or naval architects estimating areas of water planes or transverse sections of ships. In general, the trapezoidal rule states: Area = ( width of interval ) ( ) first + last ordinate + (c) Mid-ordinate rule To determine the area ABCD of Fig..: sum of remaining ordinates (a) Aplanimeteris an instrument for directly measuring small areas bounded by an irregular curve. (b) Trapezoidal rule To determine the areas PQRS in Fig..: B y y y y 4 y 5 y 6 C Q R y y y y 4 y 5 y 6 y 7 Fig.. A d d d d d d D Fig.. P d d d d d d (i) Divide base PS into any number of equal intervals, each of width d (the greater the number of intervals, the greater the accuracy). (ii) Accurately measure ordinates y, y, y,etc. (iii) Area PQRS [ ] y C y 7 D d C y C y C y 4 C y 5 C y 6 S (i) Divide base AD into any number of equal intervals, each of width d (the greater the number of intervals, the greater the accuracy). (ii) Erect ordinates in the middle of each interval (shown by broken lines in Fig..). (iii) Accurately measure ordinates y, y, y,etc. (iv) Area ABCD D d y C y C y C y 4 C y 5 C y 6. In general, the mid-ordinate rule states: Area D (width of interval)(sum of mid-ordinates)

189 78 Basic Engineering Mathematics (d) Simpson s rule To determine the area PQRS of Fig..: (i) Divide base PS into an even number of intervals, each of width d (the greater the number of intervals, the greater the accuracy). (ii) Accurately measure ordinates y, y, y,etc. (iii) Area PQRS D d [ y C y 7 C 4 y C y 4 C y 6 C y C y 5 ] In general, Simpson s rule states: Area = (width of interval) [(first + last ordinate) +4 (sum of even ordinates) + (sum of remaining odd ordinates)] Problem. A car starts from rest and its speed is measured every second for 6 s: Time t (s) Speed v (m/s) Determine the distance travelled in 6 seconds (i.e. the area under the v/t graph), by (a) the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson s rule. A graph of speed/time is shown in Fig... (a) Trapezoidal rule (see para. (b) above). The time base is divided into 6 strips each of width s, and the length of the ordinates measured. Thus [( ) 0 C 4.0 area D C.5 C 5.5 C 8.75 ] C.5 C 7.5 D m (b) Mid-ordinate rule (see para. (c) above). The time base is divided into 6 strips each of width second. Mid-ordinates are erected as shown in Fig.. by the broken lines. The length of each mid-ordinate is measured. Thus area D [.5 C 4.0 C 7.0 C 0.75 C 5.0 C 0.5] D 58.5 m (c) Simpson s rule (see para. (d) above). The time base is divided into 6 strips each of width s, and the length of the ordinates measured. Thus area D [ 0 C 4.0 C 4.5 C 8.75 C 7.5 C 5.5 C.5 ] D 58. m Problem. A river is 5 m wide. Soundings of the depth are made at equal intervals of m across the river and are as shown below. Depth (m) Calculate the cross-sectional area of the flow of water at this point using Simpson s rule. From para. (d) above, Area D [ 0 C 0 C 4. C 4.5 C.4 C. C 4. ] 0 5 Graph of speed/time D [0 C 6.4 C 5] D 5.4m Now try the following exercise Speed (m/s) Fig Time (seconds) Exercise 8 Further problems on areas of irregular figures (Answers on page 6). Plot a graph of y D x x by completing a table of values of y from x D 0tox D. Determine the area enclosed by the curve, the x-axis and ordinate x D 0andx D by (a) the trapezoidal rule, (b) the mid-ordinate rule and (c) by Simpson s rule.. Plot the graph of y D x C between x D 0and x D 4. Estimate the area enclosed by the curve, the ordinates x D 0andx D 4, and the x-axis by an approximate method.

190 Irregular areas and volumes and mean values of waveforms 79. The velocity of a car at one second intervals is given in the following table: time t (s) velocity v (m/s) Determine the distance travelled in 6 seconds (i.e. the area under the v/t graph) using an approximate method. 4. The shape of a piece of land is shown in Fig..4. To estimate the area of the land, a surveyor takes measurements at intervals of 50 m, perpendicular to the straight portion with the results shown (the dimensions being in metres). Estimate the area of the land in hectares ( ha D 0 4 m ). Problem. A tree trunk is m in length and has a varying cross-section. The cross-sectional areas at intervals of m measured from one end are: 0.5, 0.55, 0.59, 0.6, 0.7, 0.84, 0.97 m Estimate the volume of the tree trunk. A sketch of the tree trunk is similar to that shown in Fig..5, where d D m, A D 0.5 m, A D 0.55 m, and so on. Using Simpson s rule for volumes gives: Volume D [ 0.5 C 0.97 C C 0.6 C 0.84 C 0.59 C 0.7 ] D [.49 C 8.08 C.6] D 8. m Problem 4. The areas of seven horizontal crosssections of a water reservoir at intervals of 0 m are: 0, 50, 0, 50, 90, 0, 70 m Calculate the capacity of the reservoir in litres. Fig The deck of a ship is 5 m long. At equal intervals of 5 m the width is given by the following table: Width (m) Estimate the area of the deck.. Volumes of irregular solids If the cross-sectional areas A, A, A,..of an irregular solid bounded by two parallel planes are known at equal intervals of width d (as shown in Fig..5), then by Simpson s rule: Volume, V= d [(A + A 7 )+4(A + A 4 + A 6 ) +(A + A 5 )] A A A A 4 A 5 A 6 A 7 Using Simpson s rule for volumes gives: Volume D 0 [ 0 C 70 C 4 50 C 50 C 0 C 0 C 90 ] D 0 [80 C 0 C 0] D 6400 m 6400 m D 6400 ð 0 6 cm Since litre D 000 cm, capacity of reservoir Now try the following exercise Exercise ð 06 D litres 000 D D.64 U 0 7 litres Further problems on volumes of irregular solids (Answers on page 6). The areas of equidistantly spaced sections of the underwater form of a small boat are as follows:.76,.78,.0,.,.6,.4, 0.85 m Fig..5 d d d d d d Determine the underwater volume if the sections are m apart.. To estimate the amount of earth to be removed when constructing a cutting the cross-sectional area

191 80 Basic Engineering Mathematics at intervals of 8 m were estimated as follows: 0,.8,.7, 4.5, 4.,.6, 0 m Estimate the volume of earth to be excavated.. The circumference of a m long log of timber of varying circular cross-section is measured at intervals of m along its length and the results are: V V m 0 t (a) V V m V V m 0 t (b) Distance from one end (m) Circumference (m) Estimate the volume of the timber in cubic metres. 0 t Fig..7 (c). The mean or average value of a waveform Problem 5. Determine the average values over half a cycle of the periodic waveforms shown in Fig..8. The mean or average value, y, of the waveform shown in Fig..6 is given by: area under curve y = length of base, b Voltage (V) t (ms) 0 (a) Current (A) t (s) (b) y y y y 4 y 5 y 6 y 7 y Voltage (V) t (ms) d d d d d d d b 0 (c) Fig..6 If the mid-ordinate rule is used to find the area under the curve, then: sum of mid-ordinates y= number of mid-ordinates ( D y ) C y C y C y 4 C y 5 C y 6 C y 7 for Fig..6 7 For a sine wave, the mean or average value: (i) over one complete cycle is zero (see Fig..7(a)), (ii) over half a cycle is 0.67 U maximum value, or =pumaximum value, (iii) of a full-wave rectified waveform (see Fig..7(b)) is 0.67 U maximum value, (iv) of a half-wave rectified waveform (see Fig..7(c)) is 0.8 U maximum value, or=pu maximum value. Fig..8 (a) Area under triangular waveform (a) for a half cycle is given by: Area D (base)(perpendicular height) D ð 0 0 D 0 ð 0 Vs area under curve Average value of waveform D length of base D 0 ð 0 Vs D 0 V ð 0 s (b) Area under waveform (b) for a half cycle D ð C ð D 7As area under curve Average value of waveform D length of base D 7As s D. A

192 Irregular areas and volumes and mean values of waveforms 8 (c) A half cycle of the voltage waveform (c) is completed in 4 ms. Area under curve D f 0 g 0 D 0 ð 0 Vs area under curve Average value of waveform D length of base D 0 ð 0 Vs D.5V 4 ð 0 s Problem 6. Determine the mean value of current over one complete cycle of the periodic waveforms shown in Fig..9. hour as shown below. Time (h) Power (kw) Plot a graph of power against time and, by using the mid-ordinate rule, determine (a) the area under the curve and (b) the average value of the power. The graph of power/time is shown in Fig Graph of power/time Fig..9 Current (ma) Current (A) t (ms) t (ms) Power (kw) Time ( hours) (a) One cycle of the trapezoidal waveform (a) is completed in 0 ms (i.e. the periodic time is 0 ms). Area under curve D area of trapezium D (sum of parallel sides)(perpendicular distance between parallel sides) D f 4 C 8 ð 0 g 5 ð 0 D 0 ð 0 6 As area under curve Mean value over one cycle D length of base D 0 ð 0 6 As 0 ð 0 s D ma (b) One cycle of the sawtooth waveform (b) is completed in 5ms. Area under curve D ð 0 D ð 0 As area under curve Mean value over one cycle D length of base D ð 0 As 5 ð 0 s D 0.6A Problem 7. The power used in a manufacturing process during a 6 hour period is recorded at intervals of Fig..0 (a) The time base is divided into 6 equal intervals, each of width hour. Mid-ordinates are erected (shown by broken lines in Fig..0) and measured. The values are shown in Fig..0. Area under curve D (width of interval) (sum of mid-ordinates) D [7.0 C.5 C 4.0 C 49.5 C 7.0 C 0.0] D 67 kwh (i.e. a measure of electrical energy) (b) Average value of waveform area under curve D D length of base Alternatively, average value D Sum of mid-ordinates number of mid-ordinate 67 kwh 6h D 7.8 kw Problem 8. Figure. shows a sinusoidal output voltage of a full-wave rectifier. Determine, using the mid-ordinate rule with 6 intervals, the mean output voltage.

193 8 Basic Engineering Mathematics Voltage (V) 0 Fig π π π π θ (a) The width of each interval is.0 cm. Using Simpson s 6 rule, area D.0 [.6 C.6 C C.9 C.7 C.5 C. ] D [5. C 4.4 C.4] D 4 cm One cycle of the output voltage is completed in p radians or 80. The base is divided into 6 intervals, each of width 0. The mid-ordinate of each interval will lie at 5,45,75,etc. At 5 the height of the mid-ordinate is 0 sin 5 D.588 V. At 45 the height of the mid-ordinate is 0 sin 45 D 7.07 V, andsoon. The results are tabulated below: Mid-ordinate Height of mid-ordinate 5 0 sin 5 D.588 V 45 0 sin 45 D 7.07 V 75 0 sin 75 D V 05 0 sin 05 D V 5 0 sin 5 D 7.07 V 65 0 sin 65 D.588 V Sum of mid-ordinates D 8.66 V area of diagram (b) Mean height of ordinates D length of base D 4 D.8 cm Since cm represents 00 kpa, the mean pressure in the cylinder D.8 cm ð 00 kpa/cm D 8 kpa Now try the following exercise Exercise 84 Further problems on mean or average values of waveforms (Answers on page 6). Determine the mean value of the periodic waveforms shown in Fig.. over a half cycle. Mean or average value of output voltage Sum of mid-ordinates D number of mid-ordinate D 8.66 D 6.49 V 6 (With a larger number of intervals a more accurate answer may be obtained.) For a sine wave the actual mean value is 0.67 ð maximum value, which in this problem gives 6.7 V. Problem 9. An indicator diagram for a steam engine is shown in Fig... The base line has been divided into 6 equally spaced intervals and the lengths of the 7 ordinates measured with the results shown in centimetres. Determine (a) the area of the indicator diagram using Simpson s rule, and (b) the mean pressure in the cylinder given that cm represents 00 kpa. Current (A) Current (A) 0 0 0t (ms) (a) t (ms) 5 (c) Fig.. Voltage (V) t (ms) 00 (b) cm Fig Find the average value of the periodic waveforms shown in Fig..4 over one complete cycle.. An alternating current has the following values at equal intervals of 5 ms. Time (ms) Current (A)

194 Irregular areas and volumes and mean values of waveforms 8 Fig..4 Voltage (mv) Current (A) t (ms) t (ms) Plot a graph of current against time and estimate the area under the curve over the 0 ms period using the mid-ordinate rule and determine its mean value. 4. Determine, using an approximate method, the average value of a sine wave of maximum value 50 V for (a) a half cycle and (b) a complete cycle. 5. An indicator diagram of a steam engine is cm long. Seven evenly spaced ordinates, including the end ordinates, are measured as follows: 5.90, 5.5, 4.,.6,.,.4,.6 cm Determine the area of the diagram and the mean pressure in the cylinder if cm represents 90 kpa.

195 4 Triangles and some practical applications 4. Sine and cosine rules To solve a triangle means to find the values of unknown sides and angles. If a triangle is right angled, trigonometric ratios and the theorem of Pythagoras may be used for its solution, as shown in Chapter 7. However, for a nonright-angled triangle, trigonometric ratios and Pythagoras theorem cannot be used. Instead, two rules, called the sine rule and the cosine rule, areused. Sine rule With reference to triangle ABC of Fig. 4., the sine rule states: Fig. 4. a sin A D b sin B D B c sin C The rule may be used only when: c A a (i) side and any angles are initially given, or (ii) sides and an angle (not the included angle) are initially given. Cosine rule With reference to triangle ABC of Fig. 4., the cosine rule states: b C or or a = b + c bc cos A b = a + c ac cos B c = a + b ab cos C The rule may be used only when: (i) sides and the included angle are initially given, or (ii) sides are initially given. 4. Area of any triangle The area of any triangle such as ABC of Fig. 4. is given by: (i) U baseu perpendicular height, or (ii) ab sin C or ac sin B or bc sin A, or (iii) [s(s a)(s b)(s c)], where s D a C b C c 4. Worked problems on the solution of triangles and their areas Problem. In a triangle XYZ, 6 X D 5, 6 Y D 67 and YZ D 5. cm. Solve the triangle and find its area.

196 Triangles and some practical applications 85 Fig. 4. X 5 z 67 y Y x = 5. cm Z The triangle XYZ is shown in Fig. 4.. Since the angles in a triangle add up to 80, thenz D D 6. Applying the sine rule: 5. sin D y 5 sin D z 67 sin 6 5. Using sin D y and transposing gives: 5 sin 67 y D 5.sin67 D 8.00 cm = XZ sin 5 5. Using sin D z and transposing gives: 5 sin 6 z D 5.sin6 D 7.7 cm = XY sin 5 Area of triangle XYZ D xy sin Z D sin 6 D 0.8cm Applying the sine rule:. D 7.9 sin sin C from which, sin C D 7.9 sin D Hence C D sin D or (see Chapters 7 and 8). Since B D , C cannot be 8 0 0,since C is greater than 80. Thus only C D is valid. Angle A D D Applying the sine rule: a D. sin sin from which, a D. sin D 7.0 mm sin Hence A = 49 9, C = 5 0 and BC = 7.0 mm. Area of triangle ABC D ac sin B D sin D 5.mm Problem. Solve the triangle PQR and find its area given that QR D 6.5 mm, PR D 9.6mm and 6 Q D 6 (or aread xz sin YD sin 67 D 0.8cm ) It is always worth checking with triangle problems that the longest side is opposite the largest angle, and vice-versa. In this problem, Y is the largest angle and XZ is the longest of the three sides. Problem. Solve the triangle ABC given B D , AC D. mm and AB D 7.9 mm. Find also its area. Triangle ABC is shown in Fig. 4.. Fig. 4. c = 7.9 mm B A 78 5 a b =. mm C Triangle PQR is shown in Fig Fig. 4.4 r 6 Q p = 6.5 mm R P q = 9.6 mm Applying the sine rule: 9.6 sin D sin P from which, sin P D 6.5sin6 D Hence P D sin D or 0 When P D and Q D 6 then R D D 97 0 When P D 0 and Q D 6 then R D D 0 7 0

197 86 Basic Engineering Mathematics Thus, in this problem, there are two separate sets of results and both are feasible solutions. Such a situation is called the ambiguous case. Case. P D , Q D 6, R D 97 0, p D 6.5mm and q D 9.6 mm. From the sine rule: r D 9.6 sin 97 0 sin 6 from which, r D 9.6sin97 0 D 49.9 mm sin 6 Area D pq sin RD sin 97 0 D55.5mm. Case. P D 0, Q D 6, R D 0 7 0, p D 6.5mm and q D 9.6 mm. From the sine rule: r D 9.6 sin sin 6 from which, r D 9.6sin0 7 0 D 9.4 mm sin 6 Area D pq sin R D sin D mm. Triangle PQR for case is shown in Fig Fig. 4.5 P 9.4 mm 9.6 mm Q R 6.5 mm Now try the following exercise Exercise 85 Further problems on the solution of triangles and their areas (Answers on page 6) In Problems and, use the sine rule to solve the triangles ABC and find their areas.. A D 9, B D 68, b D 7 mm.. B D 7 6 0, C D 56 0, b D 8.60 cm. In Problems and 4, use the sine rule to solve the triangles DEF and find their areas.. d D 7 cm, f D cm, F D 6 4. d D.6 mm, e D 5.4 mm, D D 04 0 In Problems 5 and 6, use the sine rule to solve the triangles JKL and find their areas. 5. j D.85 cm, k D. cm, K D 6 6. k D 46 mm, l D 6 mm, L D Further worked problems on the solution of triangles and their areas Problem 4. Solve triangle DEF and find its area given that EF D 5.0 mm, DE D 5.0mm and 6 E D 64 Triangle DEF is shown in Fig Fig. 4.6 Applying the cosine rule: D f = 5.0 e mm 64 E d = 5.0 mm F e D d C f df cos E i.e. e D 5.0 C 5.0 [ cos 64 ] D 5 C D 08 from which, e D p 08 D.9 mm Applying the sine rule:.9 sin D sin F from which, sin F D 5.0sin64 D Thus 6 F D sin D or F D is not possible in this case since C 64 is greater than 80. Thus only F = 4 4 is valid. D D D 7 56 Area of triangle DEF D df sin E D sin 64 D 9.mm Problem 5. A triangle ABC has sides a D 9.0cm, b D 7.5cm and c D 6.5 cm. Determine its three angles and its area. Triangle ABC is shown in Fig It is usual first to calculate the largest angle to determine whether the triangle is acute or obtuse. In this case the largest angle is A (i.e. opposite the longest side).

198 Triangles and some practical applications 87 Fig. 4.7 c = 6.5 cm B Applying the cosine rule: A a = 9.0 cm b = 7.5 cm a D b C c bc cos A from which, bc cos A D b C c a and cos A D b C c a bc D 7.5 C D Hence A D cos D (or ,whichis obviously impossible). The triangle is thus acute angled since cos A is positive. (If cos A had been negative, angle A would be obtuse, i.e. lie between 90 and 80 ). Applying the sine rule: 9.0 D 7.5 sin sin B from which, sin B D 7.5sin D Hence B D sin D 55 4 and C D D 45 6 Area D p [s s a s b s c ], where s D a C b C c D C 9.0 C 7.5 C 6.5 D.5cm Hence area D [ ] D [ ] D.98 cm Alternatively, area D ab sin C D sin D.98 cm Problem 6. Solve triangle XYZ, shown in Fig. 4.8, and find its area given that Y D 8, XY D 7.cmand YZ D 4.5cm. X y Applying the cosine rule: y D x C z xz cos Y D 4.5 C 7. [ cos 8 ] D 0.5 C 5.84 [ 9.89] D 0.5 C 5.84 C 9.89 D.0 y D p.0 D 0.58 cm Applying the sine rule: 0.58 sin D 7. 8 sin Z from which, sin Z D 7. sin 8 D Hence Z D sin 0.56 D 6 (or which, here, is impossible). X D D 9 4 Area D xz sin Y D sin 8 D.77 cm Now try the following exercise Exercise 86 Further problems on the solution of triangles and their areas (Answers on page 6) In Problems and, use the cosine and sine rules to solve the triangles PQR and find their areas.. q D cm, r D 6 cm, P D 54. q D.5 m, r D 4.4 m, P D 05 In Problems and 4, use the cosine and sine rules to solve the triangles XYZ and find their areas.. x D 0.0cm, y D 8.0cm, z D 7.0cm. 4. x D mm, y D 4 mm, z D 4 mm. 4.5 Practical situations involving trigonometry There are a number of practical situations where the use of trigonometry is needed to find unknown sides and angles of triangles. This is demonstrated in the following worked problems. Fig. 4.8 z = 7. cm 8 Y Z x = 4.5 cm Problem 7. A room 8.0 m wide has a span roof which slopes at on one side and 40 on the other. Find the length of the roof slopes, correct to the nearest centimetre.

199 88 Basic Engineering Mathematics Fig. 4.9 A B m A section of the roof is shown in Fig Angle at ridge, B D D 07 From the sine rule: 8.0 sin D a 07 sin from which, a D 8.0sin sin 07 Also from the sine rule: 8.0 sin D c 07 sin 40 C D m from which, c D 8.0sin40 D 5.77 m sin 07 Hence the roof slopes are 4.56 m and 5.8 m, correct to the nearest centimetre. Problem 8. A man leaves a point walking at 6.5 km/h in a direction E 0 N (i.e. a bearing of 70 ). A cyclist leaves the same point at the same time in a direction E40 S (i.e. a bearing of 0 ) travelling at a constant speed. Find the average speed of the cyclist if the walker and cyclist are 80 km apart after 5 hours. After 5 hours the walker has travelled 5 ð 6.5 D.5km (shown as AB in Fig. 4.0). If AC is the distance the cyclist travels in 5 hours then BC D 80 km. Fig. 4.0 W 0 N A S.5 km E 40 b B 80 km C Hence C D sin 0.58 D (or ,whichis impossible in this case). B D D Applying the sine rule again: 80 sin D b 60 sin from which, b D 80 sin D 9.4 km sin 60 Since the cyclist travels 9.4 km in 5 hours then average speed D distance time D D 8. km=h Problem 9. Two voltage phasors are shown in Fig. 4.. If V D 40 V and V D 00 V determine the value of their resultant (i.e. length OA) and the angle the resultant makes with V Fig C B V = 40 V Angle OBA D D 5 A V = 00 V Applying the cosine rule: OA D V C V V V cos OBA D 40 C 00 f cos 5 g D 600 C f 5657g D 600 C C 5657 D 7 57 The resultant OA D p 7 57 D.4V Applying the sine rule:.4 sin D 00 5 sin AOB from which, sin AOB D 00 sin 5 D Hence angle AOB D sin 0.58 D 0 (or , which is impossible in this case). Applying the sine rule: 80 sin D.5 60 sin C from which, sin C D.5sin60 80 D 0.58 Hence the resultant voltage is.4 volts at to V Problem 0. In Fig. 4., PR represents the inclined jib of a crane and is 0.0 m long. PQ is 4.0 m long. Determine the inclination of the jib to the vertical and the length of tie QR.

200 Triangles and some practical applications 89 R R Q m 0.0 m 0 P Fig. 4. P Q Applying the sine rule: PR sin D PQ 0 sin R from which, sin R D PQ sin 0 PR D 4.0 sin 0 D Hence 6 R D sin D (or ,whichis impossible in this case). P D D 9 44, which is the inclination of the jib to the vertical. Applying the sine rule: 0.0 sin D QR 0 sin from which, length of tie, QR D 0.0sin sin 0 D 7.8 m Now try the following exercise Exercise 87 Further problems on practical situations involving trigonometry (Answers on page 6). A ship P sails at a steady speed of 45 km/h in a direction of W N (i.e. a bearing of 0 ) from a port. At the same time another ship Q leaves the port at a steady speed of 5 km/h in a direction N 5 E (i.e. a bearing of 05 ). Determine their distance apart after 4 hours.. Two sides of a triangular plot of land are 5.0 m and 4.0 m, respectively. If the area of the plot is 60 m find (a) the length of fencing required to enclose the plot and (b) the angles of the triangular plot.. A jib crane is shown in Fig. 4.. If the tie rod PR is 8.0 long and PQ is 4.5 m long determine (a) the length of jib RQ and (b) the angle between the jib and the tie rod. Fig A building site is in the form of a quadrilateral as shown in Fig. 4.4, and its area is 50 m. Determine the length of the perimeter of the site. Fig m 8.5 m m Determine the length of members BF and EB in the roof truss shown in Fig Fig. 4.5 E 4 m 4 m F D.5 m.5 m A 5 m B 5 m C 6. A laboratory 9.0 m wide has a span roof which slopes at 6 on one side and 44 on the other. Determine the lengths of the roof slopes. 7. PQ and QR are the phasors representing the alternating currents in two branches of a circuit. Phasor PQ is 0.0 A and is horizontal. Phasor QR (which is joined to the end of PQ to form triangle PQR) is 4.0 A and is at an angle of 5 to the horizontal. Determine the resultant phasor PR and the angle it makes with phasor PQ.

201 90 Basic Engineering Mathematics 4.6 Further practical situations involving trigonometry Problem. A vertical aerial stands on horizontal ground. A surveyor positioned due east of the aerial measures the elevation of the top as 48. He moves due south 0.0 m and measures the elevation as 44. Determine the height of the aerial. In Fig. 4.6, DC represents the aerial, A is the initial position of the surveyor and B his final position. From triangle ACD, tan48 D DC, from which AC D DC AC tan 48 Fig. 4.6 C D B A 0.0 m Similarly, from triangle BCD, BC D DC tan 44 For triangle ABC, using Pythagoras theorem: BC D AB C AC ( ) DC ( ) DC D 0.0 C tan 44 tan 48 ( ) DC tan D tan 48 DC D 0.0 DC D D Hence, height of aerial, DC = p 40.4 = m. Problem. A crank mechanism of a petrol engine is shown in Fig Arm OA is 0.0 cm long and rotates clockwise about 0. The connecting rod AB is 0.0 cm long and end B is constrained to move horizontally. (a) Fig. 4.7 B 0.0 cm A 0.0 cm 50 O (a) For the position shown in Fig. 4.7 determine the angle between the connecting rod AB and the horizontal and the length of OB. (b) How far does B move when angle AOB changes from 50 to 0? Applying the sine rule: AB sin D AO 50 sin B from which, sin B D AO sin 50 AB D 0.55 D 0.0sin Hence B D sin 0.55 D (or 65 0,whichis impossible in this case). Hence the connecting rod AB makes an angle of 4 47 with the horizontal. Angle OAB D D 5 0 Applying the sine rule: 0.0 sin D OB 50 sin 5 0 from which, OB D 0.0 sin 5 0 D 5.4 cm sin 50 (b) Figure 4.8 shows the initial and final positions of the crank mechanism. In triangle OA 0 B 0, applying the sine rule: 0.0 sin D sin A 0 B 0 O from which, sin A 0 B 0 O D 0.0 sin 0 D Fig. 4.8 B 0.0 cm B A A cm O Hence A 0 B 0 O D sin D (or 6 0 which is impossible in this case). Angle OA 0 B 0 D D 4 0

202 Triangles and some practical applications 9 Applying the sine rule: 0.0 sin D OB 0 0 sin 4 0 from which, OB 0 D 0.0sin4 0 D.7 cm sin 0 Since OB D 5.4 cm and OB 0 D.7 cm then BB 0 D D.7 cm Hence B moves.7 cm when angle AOB changes from 50 to 0 Problem. The area of a field is in the form of a quadrilateral ABCD as shown in Fig Determine its area. 9.8 m B A 4.4 m D 4.5 m m C Fig. 4.0 B A 40 connecting rod BC is.0 cm long. For the position shown determine the length of AC and the angle between the crank and the connecting rod. 4. From Fig..0, determine how far C moves, correct to the nearest millimetre when angle CAB changes from 40 to 60, B moving in an anticlockwise direction. 5. A surveyor, standing W 5 S of a tower measures the angle of elevation of the top of the tower as From a position E S from the tower the elevation of the top is Determine the height of the tower if the distance between the two observations is 75 m. 6. Calculate, correct to significant figures, the coordinates x and y to locate the hole centre at P shown in Fig. 4.. P C Fig. 4.9 A diagonal drawn from B to D divides the quadrilateral into two triangles. Area of quadrilateral ABCD D area of triangle ABD C area of triangle BCD D sin 4 C sin 56 y D C D 487 m Fig. 4. x mm Now try the following exercise Exercise 88 Further problems on practical situations involving trigonometry (Answers on page 6). Three forces acting on a fixed point are represented by the sides of a triangle of dimensions 7. cm, 9.6 cm and.0 cm. Determine the angles between the lines of action and the three forces.. A vertical aerial AB, 9.60m high, stands on ground which is inclined to the horizontal. A stay connects the top of the aerial A to a point C on the ground 0.0 m downhill from B, the foot of the aerial. Determine (a) the length of the stay, and (b) the angle the stay makes with the ground.. A reciprocating engine mechanism is shown in Fig The crank AB is.0 cm long and the 7. An idler gear, 0 mm in diameter, has to be fitted between a 70 mm diameter driving gear and a 90 mm diameter driven gear as shown in Fig. 4.. Determine the value of angle q between the centre lines. Fig mm dia 0 mm dia mm θ 70 mm dia

203 9 Basic Engineering Mathematics Assignment This assignment covers the material in Chapters and 4. The marks for each question are shown in brackets at the end of each question.. Plot a graph of y D x C 5 from x D tox D 4. Estimate, correct to decimal places, using 6 intervals, the area enclosed by the curve, the ordinates x D and x D 4, and the x-axis by (a) the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson s rule. (). A circular cooling tower is 0 m high. The inside diameter of the tower at different heights is given in the following table: Height (m) Diameter (m) Determine the area corresponding to each diameter and hence estimate the capacity of the tower in cubic metres. (7). A vehicle starts from rest and its velocity is measured every second for 6 seconds, with the following results: 4. A triangular plot of land ABC is shown in Fig. A.. Solve the triangle and determine its area. (0) Fig. A. B 5 m m 5. A car is travelling 0 m above sea level. It then travels 500m up a steady slope of 7. Determine, correct to the nearest metre, how high the car is now above see level. () 6. Figure A. shows a roof truss PQR with rafter PQ D m. Calculate the length of (a) the roof rise PP 0, (b) rafter PR, and (c) the roof span QR. Find also (d) the cross-sectional area of the roof truss. () P A C Time t (s) Velocity v (m/s) Using Simpson s rule, calculate (a) the distance travelled in 6 s (i.e. the area under the v/t graph) and (b) the average speed over this period. (6) Q Fig. A. m 40 P R

204 5 Vectors 5. Introduction Some physical quantities are entirely defined by a numerical value and are called scalar quantities or scalars. Examples of scalars include time, mass, temperature, energy and volume. Other physical quantities are defined by both a numerical value and a direction in space and these are called vector quantities or vectors. Examples of vectors include force, velocity, moment and displacement. 5. Vector addition (i) bold print, (ii) two capital letters with an arrow above them to denote the sense of direction, e.g.! AB, wherea is the starting point and B the end point of the vector, (iii) a line over the top of letters, e.g. AB or a, (iv) letters with an arrow above, e.g. Ea, EA, (v) underlined letters, e.g. a, (vi) xi C jy, wherei and j are axes at right-angles to each other; for example, i C 4j means units in the i direction and 4 units in the j direction, as shown in Fig. 5.. A vector may be represented by a straight line, the length of line being directly proportional to the magnitude of the quantity and the direction of the line being in the same direction as the line of action of the quantity. An arrow is used to denote the sense of the vector, that is, for a horizontal vector, say, whether it acts from left to right or vice-versa. The arrow is positioned at the end of the vector and this position is called the nose of the vector. Figure 5. shows a velocity of 0 m/s at an angle of 45 to the horizontal and may be depicted by oa D 0 m/s at 45 to the horizontal. j 4 A(,4) a O i Fig. 5. o 0 m/s 45 Fig. 5. ( ) a (vii) a column matrix ; for example, the vector OA b ( ) shown in Fig. 5. could be represented by 4 To distinguish between vector and scalar quantities, various ways are used. These include: Thus, in Fig. 5., OA! ( ) OA OA i C 4j 4

205 94 Basic Engineering Mathematics The one adopted in this text is to denote vector quantities in bold print. Thus, oa represents a vector quantity, but oa is the magnitude of the vector oa. Also, positive angles are measured in an anticlockwise direction from a horizontal, right facing line and negative angles in a clockwise direction from this line as with graphical work. Thus 90 is a line vertically upwards and 90 is a line vertically downwards. The resultant of adding two vectors together, say V at an angle q and V at angle q, as shown in Fig. 5.(a), can be obtained by drawing oa to represent V andthendrawing ar to represent V. The resultant of V + V is given by or. This is shown in Fig. 5.(b), the vector equation being oa + ar = or. This is called the nose-to-tail method of vector addition. o 7 N Scale in Newtons N O 45 7 N (a) r 4 N 45 a (b) O 4 N 45 7 N (c) R Fig. 5.4 V θ θ V Fig. 5. o a V r θ O θ V (a) (b) (c) Alternatively, by drawing lines parallel to V and V from the noses of V and V, respectively, and letting the point of intersection of these parallel lines be R, givesor as the magnitude and direction of the resultant of adding V and V, as shown in Fig. 5.(c). This is called the parallelogram method of vector addition. Problem. A force of 4 N is inclined at an angle of 45 to a second force of 7 N, both forces acting at a point. Find the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 7 N force by both the triangle and the parallelogram methods. The forces are shown in Fig. 5.4(a). Although the 7 N force is shown as a horizontal line, it could have been drawn in any direction. Using the nose-to-tail method, a line 7 units long is drawn horizontally to give vector oa in Fig. 5.4(b). To the nose of this vector ar is drawn 4 units long at an angle of 45 to oa. The resultant of vector addition is or and by measurement is 0. units long and at an angle of 6 to the 7 N force. Figure 5.4(c) uses the parallelogram method in which lines are drawn parallel to the 7 N and 4 N forces from the noses of the 4 N and 7 N forces, respectively. These intersect R at R. Vector OR gives the magnitude and direction of the resultant of vector addition and as obtained by the nose-totail method is 0. units long at an angle of 6 to the 7 N force. Problem. Use a graphical method to determine the magnitude and direction of the resultant of the three velocities shown in Fig v Fig v 7 m/s 5 m/s 0 m/s 0 Often it is easier to use the nose-to-tail method when more than two vectors are being added. The order in which the vectors are added is immaterial. In this case the order taken is v,thenv,thenv but just the same result would have been obtained if the order had been, say, v, v and finally v. v is drawn 0 units long at an angle of 0 to the horizontal, shown by oa in Fig v is added to v by drawing a line 5 units long vertically upwards from a, shown as ab. Finally, v is added to v C v by drawing a line 7 units long at an angle at 90 from b, shown as br. The resultant of vector addition is or and by measurement is 7.5 units long at an angle of 8 to the horizontal. Thus v + v + v = 7.5m=s at8 to the horizontal. v

206 Vectors 95 r 0 b Having obtained H and V, the magnitude of the resultant vector R is given by p H + V and its angle to the horizontal is given by tan (V =H ) Fig Scale in m/s o 8 5. Resolution of vectors 0 A vector can be resolved into two component parts such that the vector addition of the component parts is equal to the original vector. The two components usually taken are a horizontal component and a vertical component. For the vector shown as F in Fig. 5.7, the horizontal component is F cos q and the vertical component is F sin q. a Problem. Resolve the acceleration vector of 7 m/s at an angle of 0 to the horizontal into a horizontal and a vertical component. For a vector A at angle q to the horizontal, the horizontal component is given by A cos q and the vertical component by A sin q. Any convention of signs may be adopted, in this case horizontally from left to right is taken as positive and vertically upwards is taken as positive. Horizontal component H D 7 cos 0 D 8.50 m=s, acting from left to right. Vertical component V D 7 sin 0 D 4.7 m=s, acting vertically upwards. These component vectors are shown in Fig V F sin θ F 7 m/s θ F cos θ 4.7 m/s Fig. 5.7 For the vectors F and F shown in Fig. 5.8, the horizontal component of vector addition is: H 8.50 m/s 0 +H H D F cos q C F cos q and the vertical component of vector addition is: V D F sin q C F sin q V V Fig. 5.9 Problem 4. Calculate the resultant force of the two forces given in Problem. F sin θ F sin θ F θ θ F With reference to Fig. 5.4(a): Horizontal component of force, H D 7cos0 C 4 cos 45 F cos θ H D 7 C.88 D 9.88 N F cos θ Vertical component of force, V D 7sin0 C 4sin45 Fig. 5.8 D 0 C.88 D.88 N

207 96 Basic Engineering Mathematics The magnitude of the resultant of vector addition D H C V D 9.88 C.88 D p D 0. N The direction of the resultant of vector addition ( ) ( ) V.88 D tan D tan D 6.05 H 9.88 Thus, the resultant of the two forces is a single vector of 0. N at 6.05 to the 7 N vector.. Calculate the magnitude and direction of velocities of m/s at 8 and 7 m/s at 5 when acting simultaneously on a point. 4. Three forces of N, N and 4 N act as shown in Fig Calculate the magnitude of the resultant force and its direction relative to the N force. 4 N Problem 5. Calculate the resultant velocity of the three velocities given in Problem. With reference to Fig. 5.5: Horizontal component of the velocity, N H D 0 cos 0 C 5 cos 90 C 7 cos 90 D 9.97 C 0 C D.50 m=s Vertical component of the velocity, V D 0 sin 0 C 5 sin 90 C 7 sin 90 D.40 C 5 C.6 D 7.04 m=s Magnitude of the resultant of vector addition D H C V D.50 C 7.04 D p 0.4 D 7.9 m=s Direction of the resultant of vector addition ( ) ( ) V 7.04 D tan D tan H.50 D tan D 8.7 Thus, the resultant of the three velocities is a single vector of 7.9 m/s at 8.7 to the horizontal Now try the following exercise Exercise 89 Further problems on vectors (Answers on page 6). Forces of N and 4 N act at a point and are inclined at 90 to each other. Find, by drawing, the resultant force and its direction relative to the 4 N force.. Forces A, B and C are coplanar and act at a point. Force A is kn at 90, B is 5 kn at 80 and C is kn at 9. Determine graphically the resultant force. Fig. 5.0 N 5. A load of 5.89 N is lifted by two strings, making angles of 0 and 5 with the vertical. Calculate the tensions in the strings. [For a system such as this, the vectors representing the forces form a closed triangle when the system is in equilibrium]. 6. The acceleration of a body is due to four component, coplanar accelerations. These are m/s due north, m/s due east, 4 m/s to the south-west and 5 m/s to the south-east. Calculate the resultant acceleration and its direction. 7. A current phasor i is 5 A and horizontal. A second phasor i is 8 A and is at 50 to the horizontal. Determine the resultant of the two phasors, i C i, and the angle the resultant makes with current i. 5.4 Vector subtraction In Fig. 5., a force vector F is represented by oa. The vector ( oa) can be obtained by drawing a vector from o in the opposite sense to oa but having the same magnitude, shown as ob in Fig. 5., i.e. ob =( oa). For two vectors acting at a point, as shown in Fig. 5.(a), the resultant of vector addition is os = oa + ob. Fig. 5.(b) shows vectors ob +( oa), thatis,ob oa and the vector equation is ob oa = od. Comparing od in Fig. 5.(b) with the broken line ab in Fig. 5.(a) shows that the second diagonal of the parallelogram method of vector addition gives the magnitude and direction of vector subtraction of oa from ob.

208 Vectors 97 b Fig. 5. o b Fig. 5. (a) F a s a Problem 6. Accelerations of a D.5m/s at 90 and a D.6m/s at 45 act at a point. Find a + a and a a by (i) drawing a scale vector diagram and (ii) by calculation. o (i) The scale vector diagram is shown in Fig. 5.. By measurement, a a + a d a + a =.7m=s at 6 a a =.m=s at 0 o (b) F a Scale in m/s b 0 a a Vertical component of a + a, V D.5sin90 C.6 sin 45 D.99 Magnitude of a + a D. C.99 D.67 m=s ( ).99 Direction of a + a D tan and must lie. in the second quadrant since H is negative and V is positive. ( ).99 tan D 54.5, and for this to be in the. second quadrant, the true angle is 80 displaced, i.e or 5.47 Thus a + a =.67 m=s at Horizontal component of a a,thatis,a +( a ) D.5 cos 90 C.6cos D.6cos 5 D. Vertical component of a a,thatis,a +( a ) D.5sin90 C.6sin 5 D 0 Magnitude of a a D p. C 0 D. m/s ( ) 0 Direction of a a D tan D 0. Thus a a =. m/s at 0 Problem 7. Calculate the resultant of (i) v v + v and (ii) v v v when v D units at 40, v D 40 units at 90 and v D 5 units at 90 (i) The vectors are shown in Fig V Fig m/s.5 m/s 45 6 a a a H 90 +H 40 5 (ii) Resolving horizontally and vertically gives: Horizontal component of a + a, V H D.5 cos 90 C.6 cos 45 D. Fig. 5.4

209 98 Basic Engineering Mathematics The horizontal component of v v + v D cos cos 90 C 5 cos 90 D C 5. D 7.67 units The vertical component of v v + v D sin sin 90 C 5 sin 90 D C 4.0 D 6.99 units The magnitude of the resultant, R, which can be represented by the mathematical symbol for the modulus of as jv v C v j is given by: jrj D 7.67 C 6.99 D 8.54 units The direction of the resultant, R, which can be represented by the mathematical symbol for the argument of as arg v v C v is given by: ( ) 6.99 arg R D tan D Thus v v + v = 8.54 units at 4.8 (ii) The horizontal component of v v v D 40 cos 90 cos 40 5 cos 90 D D 7.67 units The vertical component of v v v D 40 sin 90 sin 40 5 sin 90 D D 6.99 units Let R D v v v then jrj D 7.67 ( ) C D 8.54 units and arg R D tan and must 7.67 lie in the third quadrant since both H and V are negative quantities. ( ) 6.99 Tan D 4.8, hence the required angle is C 4.8 D 94.8 Thus v v v = 8.54 units at 94.8 This result is as expected, since v v v = (v v + v ) and the vector 8.54 units at 94.8 is minus times the vector 8.54 units at 4.8 Now try the following exercise Exercise 90 Further problems on vectors (Answers on page 6). Forces of F D 40 N at 45 and F D 0 N at 5 act at a point. Determine by drawing and by calculation (a) F + F (b) F F. Calculate the resultant of (a) v + v v (b) v v + v when v D 5 m/s at 85, v D 5 m/s at 75 and v D m/s at Relative velocity For relative velocity problems, some fixed datum point needs to be selected. This is often a fixed point on the earth s surface. In any vector equation, only the start and finish points affect the resultant vector of a system. Two different systems are shown in Fig. 5.5, but in each of the systems, the resultant vector is ad. The vector equation of the system shown in Fig. 5.5(a) is: ad = ab + bd and that for the system shown in Fig. 5.5(b) is: a Fig. 5.5 ad = ab + bc + cd b (a) d Thus in vector equations of this form, only the first and last letters, a and d, respectively, fix the magnitude and direction of the resultant vector. This principle is used in relative velocity problems. Problem 8. Two cars, P and Q, are travelling towards the junction of two roads which are at right angles to one another. Car P has a velocity of 45 km/h due east and car Q a velocity of 55 km/h due south. Calculate (i) the velocity of car P relative to car Q, and (ii) the velocity of car Q relative to car P. (i) The directions of the cars are shown in Fig. 5.6 (a), called a space diagram. The velocity diagram is shown in Fig. 5.6 (b), in which pe is taken as the velocity of car P relative to point e on the earth s surface. The velocity of P relative to Q is vector pq and the vector equation is pq = pe + eq. Hence the vector directions are as shown, eq being in the opposite direction to qe. From the geometry of the vector triangle, jpq D 45 C 55 D 7.06 km/h and ( ) 55 arg pq D tan D i.e., the velocity of car P relative to car Q is 7.06 km/h at 50.7 (ii) The velocity of car Q relative to car P is given by the vector equation qp = qe + ep and the vector diagram a b (b) c d

210 Vectors 99 W N S E q q but must lie in the third quadrant, i.e., the required angle is 80 C 50.7 D 0.7 Thus the velocity of car Q relative to car P is 7.06 km/h at 0.7 P Q 45 km/h 55 km/h p e p e Now try the following exercise (a) (b) (c) Fig. 5.6 is as shown in Fig. 5.6(c), having ep opposite in direction to pe. From the geometry of this vector triangle: jqp D 45 C 55 D 7.06 km/h and ( ) 55 arg qp D tan D Exercise 9 Further problems on relative velocity (Answers on page 6). A car is moving along a straight horizontal road at 79. km/h and rain is falling vertically downwards at 6.4 km/h. Find the velocity of the rain relative to the driver of the car.. Calculate the time needed to swim across a river 4 m wide when the swimmer can swim at km/h in still water and the river is flowing at km/h.

211 6 Number sequences 6. Simple sequences A set of numbers which are connected by a definite law is called a series or a sequence of numbers. Each of the numbers in the series is called a term of the series. Forexample,,,5,7,... is a series obtained by adding to the previous term, and, 8,, 8,... is a sequence obtained by multiplying the previous term by 4. Problem. Determine the next two terms in the series:, 6, 9,,... We notice that the sequence, 6, 9,,... progressively increases by, thus the next two terms will be 5 and 8 Now try the following exercise Exercise 9 Further problems on simple sequences (Answers on page 6) Determine the next two terms in each of the following series:. 5, 9,, 7,...., 6,, 4,...., 56, 8,... 4., 7,,... 5., 5, 0, 7, 6, 7,... 6., 0., 0.0, , 9, 9, 4,... Problem. Find the next three terms in the series: 9, 5,,... We notice that each term in the series 9, 5,,... progressively decreases by 4, thus the next two terms will be 4, i.e. and 4, i.e. 7 Problem. Determine the next two terms in the series:, 6, 8, 54,... We notice that the second term, 6, is three times the first term, the third term, 8, is three times the second term, and that the fourth term, 54, is three times the third term. Hence the fifth term will be ð 54 D 6 and the sixth term will be ð 6 D The n th term of a series If a series is represented by a general expression, say, n C, where n is an integer (i.e. a whole number), then by substituting n D,,,... the terms of the series can be determined; in this example, the first three terms will be: C, C, C,..., i.e., 5, 7,... What is the n 0 th term of the sequence,, 5, 7,...? Firstly, we notice that the gap between each term is, hence the law relating the numbers is: n C something The second term, D n C something, hence when n D (i.e. the second term of the series), then D 4 C something and the something must be. Thus the n th term of,, 5, 7, ::: is n. Hence the fifth

212 Number sequences 0 term is given by 5 D 9, and the twentieth term is 0 D 9, and so on. Problem 4. The n 0 th term of a sequence is given by n C. Write down the first four terms. The first four terms of the series n C will be: C, C, C and 4 C i.e. 4, 7, 0 and Problem 5. Then 0 th term of a series is given by 4n. Write down the first four terms. The first four terms of the series 4n will be: 4, 4, 4 and 4 4 i.e., 7, and 5 Problem 6. Find the n 0 th term of the series:, 4, 7,... We notice that the gap between each of the given three terms is, hence the law relating the numbers is: n C something The second term, 4 D n C something, so when n D, then 4 D 6 C something, so the something must be (from simple equations). Thus the n th term of the series, 4, 7, ::: is: n Problem 7. Find the n 0 th term of the sequence:, 9, 5,,... Hence determine the 5 th term of the series. We notice that the gap between each of the given four terms is 6, hence the law relating the numbers is: 6n C something The second term, 9 D 6n C something, so when n D, then 9 D C something, so the something must be Thus the n th term of the series, 9, 5,, ::: is: 6n The 5 th term of the series is given by 6n whennd5. Hence the 5 th term of the series, 9, 5,, ::: is: 6 5 D 87 Problem 8. Find the n 0 th term of the series:, 4, 9, 6, 5,... This is a special series and does not follow the pattern of the previous examples. Each of the terms in the given series are square numbers, i.e., 4, 9, 6, 5,...,,, 4, 5,... Hence the n th term is: n Now try the following exercise Exercise 9 Further problems on the n th term of a series (Answers on page 6). The n 0 th term of a sequence is given by n. Write down the first four terms.. The n 0 th term of a sequence is given by n C 4. Write down the first five terms.. Write down the first four terms of the sequence given by 5n C Find the n 0 th term in the following series: 4. 5, 0, 5, 0, , 0, 6,,... 6., 5, 7, 9,... 7., 6, 0, 4, ,, 5, 8,... 9., 8, 7, 64, 5, Arithmetic progressions When a sequence has a constant difference between successive terms it is called an arithmetic progression (often abbreviated to AP). Examples include: (i), 4, 7, 0,,... where the common difference is, and (ii) a, a C d, a C d, a C d,... where the common difference is d. If the first term of an AP is a and the common difference is d then the n 0 th term is : a +(n )d In example (i) above, the 7th term is given by C 7 D 9, which may be readily checked. The sum S of an AP can be obtained by multiplying the average of all the terms by the number of terms. The average of all the terms D a C,where a isthefirst term and l is the last term, i.e. l D a C n d, forn terms.

213 0 Basic Engineering Mathematics Hence the sum of n terms, ( ) a C S n D n i.e. S n = n [a +(n )d] D n fa C [a C n d]g For example, the sum of the first 7 terms of the series, 4, 7, 0,,... is given by S 7 D 7 [ C 7 ], since a D anddd D 7 [ C 8] D 7 [0] D Worked problems on arithmetic progression Problem 9. Determine (a) the ninth, and (b) the sixteenth term of the series, 7,, 7,..., 7,, 7,... is an arithmetic progression with a common difference, d, of5 (a) The n 0 th term of an AP is given by a C n d Since the first term a D,dD 5andnD9 then the 9th term is: C 9 5 D C 8 5 D C 40 D 4 (b) The 6th term is: C 6 5 D C 5 5 D C 75 D 77 Problem 0. The 6th term of an AP is 7 and the th term is 8. Determine the 9th term. The n 0 th term of an AP is a C n d The 6th term is: a C 5d D 7 The th term is: a C d D 8 Equation () equation () gives: 7d D, from which, d D 7 D Substituting in equation () gives: a C 5 D 7, from which, a D Hence the 9th term is: a C n d D C 9 D C 8 D C 54 D 56 Problem. Determine the number of the term whose valueisintheseries, 4, 5, 7,..., 4, 5, 7,... is an AP where a D and d D Hence if the n 0 th term is then: a C n d D i.e. ( C n ) D ( n ) D D 9 n D 9 D and n D C D 4 i.e. the 4th term of the AP is Problem. Find the sum of the first terms of the series 5, 9,, 7,... 5, 9,, 7,... is an AP where a D 5anddD4 The sum of n terms of an AP, S n D n [a C n d] Hence the sum of the first terms, S D [ 5 C 4] D 6[0 C 44] D 6 54 D 4 Problem. Find the sum of the first terms of the series.5, 4., 4.7, 5.,....5, 4., 4.7, 5.,... is an AP where a D.5 anddd0.6 The sum of the first terms, S D [a C n d] D [.5 C 0.6] D [7 C ] D 9 D 99 D 99.5 Now try the following exercise Exercise 94 Further problems on arithmetic progressions (Answers on page 6). Find the th term of the series 8, 4, 0, 6,.... Find the 7th term of the series, 0.7, 0.4, 0.,.... The seventh term of a series is 9 and the eleventh term is 54. Determine the sixteenth term.

214 Number sequences 0 4. Find the 5th term of an arithmetic progression of which the first term is and the tenth term is Determine the number of the term which is 9 in the series 7, 9.,.4,.6, Find the sum of the first terms of the series 4, 7, 0,, Determine the sum of the series 6.5, 8.0, 9.5,.0,..., 6.5 Further worked problems on arithmetic progressions Problem 4. The sum of 7 terms of an AP is 5 and the common difference is.. Determine the first term of the series. n D 7,dD. ands 7 D 5 Since the sum of n terms of an AP is given by S n D n [a C n d, then 5 D 7 [a C 7.] D 7 [a C 7.] Hence 5 ð D a C D a C 7. Thus a D 0 7. D.8, from which a D.8 D.4 i.e. the first term, a =.4 Problem 5. Three numbers are in arithmetic progression. Their sum is 5 and their product is 80. Determine the three numbers. Let the three numbers be (a d), a and (a C d) Then a d C a C a C d D 5, i.e. a D 5, from which, a D 5 Also, a a d a C d D 80, i.e. a a d D 80 Since a D 5, 5 5 d D d D D 5d 45 D 5d from which, d D 45 5 D 9. Hence d D p 9 Dš The three numbers are thus 5, 5and 5 C, i.e., 5 and 8 Problem 6. Find the sum of all the numbers between 0 and 07 which are exactly divisible by. The series, 6, 9,, is an AP whose first term a D and common difference d D The last term is a C n d D 07 i.e. C n D 07, from which 07 n D D 68 Hence n D 68 C D 69 The sum of all 69 terms is given by S 69 D n [a C n d] D 69 [ C 69 ] D 69 [6 C 04] D 69 0 D 745 Problem 7. The first, twelfth and last term of an arithmetic progression are 4,, and 76 respectively. Determine (a) the number of terms in the series, (b) the sum of all the terms and (c) the 80 th term. (a) Let the AP be a, a C d, a C d,...,ac n d, where a D 4 The th term is: a C d D i.e. 4 C d D, from which, d D 4 D 7 Hence d D 7 D The last term is a C n d i.e. 4 C n ( ) D 76 n D 76 4 D 7 Hence the number of terms in the series, n = 49 + = 50 (b) Sum of all the terms, S 50 D n [a C n d] D 50 D 75 [ ( 4 C 50 )] [ 8 C 49 D D D 49 ( )] D 85[8 C 7.5

215 04 Basic Engineering Mathematics (c) The 80th term is: a C n d D 4 C 80 ( ) D 4 C 79 ( ) Now try the following exercise Exercise 95 D 4 C 97.5 D 0 Further problems on arithmetic progressions (Answers on page 6). The sum of 5 terms of an arithmetic progression is 0.5 and the common difference is. Find the first term of the series.. Three numbers are in arithmetic progression. Their sum is 9 and their product is 0. Determine the three 4 numbers.. Find the sum of all the numbers between 5 and 50 which are exactly divisible by 4 4. Find the number of terms of the series 5, 8,,... of which the sum is Insert four terms between 5 and to form an arithmetic progression. 6. The first, tenth and last terms of an arithmetic progression are 9, 40.5, and 45.5 respectively. Find (a) the number of terms, (b) the sum of all the terms and (c) the 70th term. 7. On commencing employment a man is paid a salary of 700 per annum and receives annual increments of 50. Determine his salary in the 9th year and calculate the total he will have received in the first years. 8. An oil company bores a hole 80 m deep. Estimate the cost of boring if the cost is 0 for drilling the first metre with an increase in cost of per metre for each succeeding metre. 6.6 Geometric progressions When a sequence has a constant ratio between successive terms it is called a geometric progression (often abbreviated to GP). The constant is called the. common ratio, r Examples include (i),,4,8,... where the common ratio is, and (ii) a, ar, ar,ar,... where the common ratio is r If the first term of a GP is a and the common ratio is r, then the n 0 th term is : ar n which can be readily checked from the above examples. For example, the 8th term of the GP,, 4, 8,... is 7 D 8, sincea D andrd Let a GP be a, ar, ar,ar,...ar n then the sum of n terms, S n D a C ar C ar C ar C...C ar n... Multiplying throughout by r gives: rs n D ar C ar C ar C ar 4 C...ar n C ar n... Subtracting equation () from equation () gives: i.e. S n rs n D a ar n S n r D a r n Thus the sum of n terms, S n = a( r n ) which is valid ( r) when r< Subtracting equation () from equation () gives S n = a(r n ) (r ) which is valid when r> For example, the sum of the first 8 terms of the GP,, 4, 8, 6,... is given by S 8 D 8, since a D andr D 56 i.e. S 8 D D 55 When the common ratio r of a GP is less than unity, the sum of n terms, S n D a rn, which may be written as r a S n D r arn r Since r<,r n becomes less as n increases, i.e. r n! 0 as n! ar n Hence! 0 as n! r Thus S n! a as n! r a The quantity r is called the sum to infinity, S,andis the limiting value of the sum of an infinite number of terms, i.e. S X = a ( r) which is valid when <r<

216 Number sequences 05 For example, the sum to infinity of the GP,,,... is 4 S D,sinceaDandrD,i.e.S D 6.7 Worked problems on geometric progressions Problem 8. Determine the tenth term of the series, 6,, 4,..., 6,, 4,... is a geometric progression with a common ratio r of The n 0 th term of a GP is ar n,wherea is the first term. Hence the 0th term is: 0 D 9 D 5 D 56 Problem 9. Find the sum of the first 7 terms of the series,,, 4,,...,, 4,,... is a GP with a common ratio r D The sum of n terms, S n D a rn r Hence S 7 D 7 D 87 D 546 Problem 0. The first term of a geometric progression is and the fifth term is 55. Determine the 8 th term and the th term. The 5th term is given by ar 4 D 55, where the first term a D ( Hence r 4 D 55 a D 55 ) 55 and r D 4 D The 8th term is ar 7 D D 7. The th term is ar 0 D D 59.7 Problem. Which term of the series 87, 79, 4,... is 9? 87, 79, 4,... is a GP with a common ratio r D and first term a D 87 The n 0 th term of a GP is given by: ar n Hence 9 D 87 ( ) n from which ( ) n D 9 87 D D 7 D 9 Thus n D 9, from which, n D 9 C D 0 i.e. is the 0th term of the GP 9 ( ) 9 Problem. Find the sum of the first 9 terms of the series 7.0, 57.6, 46.08,... The common ratio, r D ar a D 57.6 (also 7.0 D 0.8 ar ar D ) 57.6 D 0.8 The sum of 9 terms, S 9 D a rn r D D D.7 Problem. Find the sum to infinity of the series,,,...,,,... is a GP of common ratio, r D The sum to infinity, S D a r D D Now try the following exercise Exercise 96 D 9 D 4 Further problems on geometric progressions (Answers on page 6). Find the 0th term of the series 5, 0, 0, 40,.... Determine the sum of the first 7 terms of the series,,, 6, The first term of a geometric progression is 4 and the 6th term is 8. Determine the 8th and th terms. 4. Which term of the series, 9, 7,... is 59049? 5. Find the sum of the first 7 terms of the series, 5,,... (correct to 4 significant figures). 6. Determine the sum to infinity of the series 4,,, Find the sum to infinity of the series, 4, 5 8,...

217 06 Basic Engineering Mathematics 6.8 Further worked problems on geometric progressions Problem 4. In a geometric progression the sixth term is 8 times the third term and the sum of the seventh and eighth terms is 9. Determine (a) the common ratio, (b) the first term, and (c) the sum of the fifth to eleventh terms, inclusive. (a) Let the GP be a, ar, ar,ar,...,ar n The rd term D ar and the sixth term D ar 5 The 6th term is 8 times the rd Hence ar 5 D 8ar from which, r D 8,r D p 8 i.e. the common ratio r = (b) The sum of the 7th and 8th terms is 9. Hence ar 6 C ar 7 D 9 Since r D, then 64a C 8a D 9 9a D 9 from which, a, the first term = (c) The sum of the 5th to th terms (inclusive) is given by: S S 4 D a r r D a r4 r 4 D 4 D 4 D D 0 Problem 5. A hire tool firm finds that their net return from hiring tools is decreasing by 0% per annum. If their net gain on a certain tool this year is 400, find the possible total of all future profits from this tool (assuming the tool lasts for ever). The net gain forms a series: 400 C 400 ð 0.9 C 400 ð 0.9 C..., which is a GP with a D 400 and r D 0.9 The sum to infinity, a S D r D D 4000 = total future profits Problem 6. If 00 is invested at compound interest of 8% per annum, determine (a) the value after 0 years, (b) the time, correct to the nearest year, it takes to reach more than 00 (a) Let the GP be a, ar, ar,...ar n The first term a D 00 The common ratio r D.08 Hence the second term is ar D D 08, which is the value after year, the third term is ar D D 6.64, which is the value after years, and so on. Thus the value after 0 years D ar 0 D D 5.89 (b) When 00 has been reached, 00 D ar n i.e. and 00 D n D.08 n Taking logarithms to base 0 of both sides gives: lg D lg.08 n D n lg.08, by the laws of logarithms from which, n D lg lg.08 D 4. Hence it will take 5 years to reach more than 00 Problem 7. A drilling machine is to have 6 speeds ranging from 50 rev/min to 750 rev/min. If the speeds form a geometric progression determine their values, each correct to the nearest whole number. Let the GP of n terms be given by a, ar, ar,...ar n The first term a D 50 rev/min. The 6th term is given by ar 6, which is 750 rev/min, i.e., ar 5 D 750 from which r 5 D 750 a D D 5 Thus the common ratio, r D 5p 5 D.788 The first term is a D 50 rev/min the second term is ar D D 85.94, the third term is ar D D 47.7, the fourth term is ar D D 5.89, the fifth term is ar 4 D D 46.9, the sixth term is ar 5 D D Hence, correct to the nearest whole number, the 6 speeds of the drilling machine are 50, 86, 48, 54, 46 and 750 rev/min.

218 Number sequences 07 Now try the following exercise F = 0 N Exercise 97 Further problems on geometric progressions (Answers on page 6). In a geometric progression the 5th term is 9 times the rd term and the sum of the 6th and 7th terms is 944. Determine (a) the common ratio, (b) the first term and (c) the sum of the 4th to 0th terms inclusive.. The value of a lathe originally valued at 000 depreciates 5% per annum. Calculate its value after 4 years. The machine is sold when its value is less than 550. After how many years is the lathe sold?. If the population of Great Britain is 55 million and is decreasing at.4% per annum, what will be the population in 5 years time? g of a radioactive substance disintegrates at a rate of % per annum. How much of the substance is left after years? 5. If 50 is invested at compound interest of 6% per annum determine (a) the value after 5 years, (b) the time, correct to the nearest year, it takes to reach A drilling machine is to have 8 speeds ranging from 00 rev/min to 000 rev/min. If the speeds form a geometric progression determine their values, each correct to the nearest whole number. Assignment This assignment covers the material in chapters 5 and 6. The marks for each question are shown in brackets at the end of each question.. Forces of 0 N, 6 N and 0 N act as shown in Figure A.. Determine the magnitude of the resultant force and its direction relative to the 6 N force (a) by scaled drawing, and (b) by calculation. (9). For the three forces shown in Fig. A., calculate the resultant of F F F and its direction relative to force F (6) Fig. A F = 6 N F = 0 N. Two cars, A and B, are travelling towards crossroads. A has a velocity of 60 km/h due south and B a velocity of 75 km/h due west. Calculate the velocity of A relative to B. (6) 4. Determine the 0 th term of the series 5.6, 5, 4.4,.8,... () 5. The sum of terms of an arithmetic progression is 86 and the common difference is. Determine the first term of the series. (4) 6. An engineer earns 000 per annum and receives annual increments of 600. Determine the salary in the 9 th year and calculate the total earnings in the first years. (5) 7. Determine the th term of the series.5,, 6,,... () 8. Find the sum of the first eight terms of the series,, 6,..., correct to decimal place. (4) 4 9. Determine the sum to infinity of the series 5,,,... 5 () 0. A machine is to have seven speeds ranging from 5 rev/min to 500 rev/min. If the speeds form a geometric progression, determine their value, each correct to the nearest whole number. (8)

219 7 Presentation of statistical data 7. Some statistical terminology Data are obtained largely by two methods: (a) by counting for example, the number of stamps sold by a post office in equal periods of time, and (b) by measurement for example, the heights of a group of people. When data are obtained by counting and only whole numbers are possible, the data are called discrete. Measured data can have any value within certain limits and are called continuous (see Problem ). A set is a group of data and an individual value within the set is called a member of the set. Thus, if the masses of five people are measured correct to the nearest 0. kilogram and are found to be 5. kg, 59.4 kg, 6. kg, 77.8 kg and 64.4 kg, then the set of masses in kilograms for these five people is: f5., 59.4, 6., 77.8, 64.4g and one of the members of the set is 59.4 A set containing all the members is called a population. Some members selected at random from a population are called a sample. Thus all car registration numbers form a population, but the registration numbers of, say, 0 cars taken at random throughout the country are a sample drawn from that population. The number of times that the value of a member occurs in a set is called the frequency of that member. Thus in the set: f,, 4, 5, 4,, 4, 7, 9g, member 4 has a frequency of three, member has a frequency of and the other members have a frequency of one. The relative frequency with which any member of a set occurs is given by the ratio: frequency of member total frequency of all members For the set: f,, 5, 4, 7, 5, 6,, 8g, the relative frequency of member 5 is 9 Often, relative frequency is expressed as a percentage and the percentage relative frequency is: relative frequency ð 00 % Problem. Data are obtained on the topics given below. State whether they are discrete or continuous data. (a) The number of days on which rain falls in a month for each month of the year. (b) The mileage travelled by each of a number of salesmen. (c) The time that each of a batch of similar batteries lasts. (d) The amount of money spent by each of several families on food. (a) The number of days on which rain falls in a given month must be an integer value and is obtained by counting the number of days. Hence, these data are discrete. (b) A salesman can travel any number of miles (and parts of a mile) between certain limits and these data are measured. Hence the data are continuous. (c) The time that a battery lasts is measured and can have any value between certain limits. Hence these data are continuous. (d) The amount of money spent on food can only be expressed correct to the nearest pence, the amount being counted. Hence, these data are discrete. Now try the following exercise Exercise 98 Further problems on discrete and continuous data (Answers on page 6) In Problems and, state whether data relating to the topics given are discrete or continuous.

220 Presentation of statistical data 09. (a) The amount of petrol produced daily, for each of days, by a refinery. (b) The amount of coal produced daily by each of 5 miners. (c) The number of bottles of milk delivered daily by each of 0 milkmen. (d) The size of 0 samples of rivets produced by a machine.. (a) The number of people visiting an exhibition on each of 5 days. (b) The time taken by each of athletes to run 00 metres. (c) The value of stamps sold in a day by each of 0 post offices. (d) The number of defective items produced in each of 0 one-hour periods by a machine. 7. Presentation of ungrouped data Ungrouped data can be presented diagrammatically in several ways and these include: (a) pictograms, in which pictorial symbols are used to represent quantities (see Problem ), (b) horizontal bar charts, having data represented by equally spaced horizontal rectangles (see Problem ), and (c) vertical bar charts, in which data are represented by equally spaced vertical rectangles (see Problem 4). Trends in ungrouped data over equal periods of time can be presented diagrammatically by a percentage component bar chart. In such a chart, equally spaced rectangles of any width, but whose height corresponds to 00%, are constructed. The rectangles are then subdivided into values corresponding to the percentage relative frequencies of the members (see Problem 5). A pie diagram is used to show diagrammatically the parts making up the whole. In a pie diagram, the area of a circle represents the whole, and the areas of the sectors of the circle are made proportional to the parts which make up the whole (see Problem 6). Problem. The number of television sets repaired in a workshop by a technician in six, one-month periods is as shown below. Present these data as a pictogram. Month January February March April May June Number repaired Each symbol shown in Fig. 7. represents two television sets repaired. Thus, in January, 5 symbols are used to Fig. 7. represent the sets repaired, in February, symbols are used to represent the 6 sets repaired, and so on. Problem. The distance in miles travelled by four salesmen in a week are as shown below. Salesmen P Q R S Distance traveled (miles) Use a horizontal bar chart to represent these data diagrammatically. Equally spaced horizontal rectangles of any width, but whose length is proportional to the distance travelled, are used. Thus, the length of the rectangle for salesman P is proportional to 4 miles, and so on. The horizontal bar chart depicting these data is shown in Fig. 7.. Salesmen S R Q P Fig Distance travelled, miles Problem 4. The number of issues of tools or materials from a store in a factory is observed for seven, onehour periods in a day, and the results of the survey are as follows: Period Number of issues Present these data on a vertical bar chart. In a vertical bar chart, equally spaced vertical rectangles of any width, but whose height is proportional to the quantity

221 0 Basic Engineering Mathematics being represented, are used. Thus the height of the rectangle for period is proportional to 4 units, and so on. The vertical bar chart depicting these data is shown in Fig. 7.. Fig. 7. Number of issues Periods Problem 5. The numbers of various types of dwellings sold by a company annually over a three-year period are as shown below. Draw percentage component bar charts to present these data. Year Year Year 4-roomed bungalows roomed bungalows roomed houses roomed houses roomed houses A table of percentage relative frequency values, correct to the nearest %, is the first requirement. Since, percentage relative frequency D frequency of member ð 00 total frequency then for 4-roomed bungalows in year : 4 ð 00 percentage relative frequency D 4 C 8 C 44 C 64 C 0 D % The percentage relative frequencies of the other types of dwellings for each of the three years are similarly calculated and the results are as shown in the table below. Year Year Year 4-roomed bungalows % 7% % 5-roomed bungalows 9% 8% 4% 4-roomed houses % 0% 5% 5-roomed houses % % 4% 6-roomed houses 5% % 7% The percentage component bar chart is produced by constructing three equally spaced rectangles of any width, corresponding to the three years. The heights of the rectangles correspond to 00% relative frequency, and are subdivided into the values in the table of percentages shown above. A key is used (different types of shading or different colour schemes) to indicate corresponding percentage values in the rows of the table of percentages. The percentage component bar chart is shown in Fig Percentage relative frequency Fig. 7.4 Year Key 6-roomed houses 5-roomed houses 4-roomed houses 5-roomed bungalows 4-roomed bungalows Problem 6. The retail price of a product costing is made up as follows: materials 0p, labour 0p, research and development 40p, overheads 70p, profit 60p. Present these data on a pie diagram. A circle of any radius is drawn, and the area of the circle represents the whole, which in this case is. The circle is subdivided into sectors so that the areas of the sectors are proportional to the parts, i.e. the parts which make up the total retail price. For the area of a sector to be proportional to a part, the angle at the centre of the circle must be proportional to that part. The whole, or 00p, corresponds to 60. Therefore, 0p corresponds to 60 ð 0 degrees, i.e p corresponds to 60 ð 0 degrees, i.e and so on, giving the angles at the centre of the circle for the parts of the retail price as: 8, 6, 7, 6 and 08, respectively. The pie diagram is shown in Fig. 7.5 Problem 7. (a) Using the data given in Fig. 7. only, calculate the amount of money paid to each salesman

222 Presentation of statistical data for travelling expenses, if they are paid an allowance of 7p per mile. (b) Using the data presented in Fig. 7.4, comment on the housing trends over the three-year period. (c) Determine the profit made by selling 700 units of the product shown in Fig. 7.5 Fig. 7.5 Research and development Labour Materials Overheads 08 lp.8 Profit (a) By measuring the length of rectangle P the mileage covered by salesman P is equivalent to 4 miles. Hence salesman P receives a travelling allowance of 4 ð 7 i.e Similarly, for salesman Q, the miles travelled are 64 and his allowance is 64 ð 7 i.e Salesman R travels 597 miles and he receives 597 ð 7 i.e Finally, salesman S receives 4 ð 7 i.e (b) An analysis of Fig. 7.4 shows that 5-roomed bungalows and 5-roomed houses are becoming more popular, the greatest change in the three years being a 5% increase in the sales of 5-roomed bungalows. (c) Since.8 corresponds to p and the profit occupies 08 of the pie diagram, then the profit per unit is 08 ð i.e. 60p.8 The profit when selling 700 units of the product is 700 ð 60 i.e Now try the following exercise Exercise 99 Further problems on presentation of ungrouped data (Answers on page 6). The number of vehicles passing a stationary observer on a road in six ten-minute intervals is as shown. Draw a pictogram to represent these data. Period of Time Number of Vehicles The number of components produced by a factory in a week is as shown below: Day Mon Tues Wed Thur Fri Number of Components Show these data on a pictogram.. For the data given in Problem above, draw a horizontal bar chart. 4. Present the data given in Problem above on a horizontal bar chart. 5. For the data given in Problem above, construct a vertical bar chart. 6. Depict the data given in Problem above on a vertical bar chart. 7. A factory produces three different types of components. The percentages of each of these components produced for three, one-month periods are as shown below. Show this information on percentage component bar charts and comment on the changing trend in the percentages of the types of component produced. Month Component P Component Q Component R A company has five distribution centres and the mass of goods in tonnes sent to each centre during four, one-week periods, is as shown. Week 4 Centre A Centre B Centre C Centre D Centre E Use a percentage component bar chart to present these data and comment on any trends.

223 Basic Engineering Mathematics 9. The employees in a company can be split into the following categories: managerial, supervisory 9, craftsmen, semi-skilled 67, others 44. Show these data on a pie diagram. 0. The way in which an apprentice spent his time over a one-month period is as follows: drawing office 44 hours, production 64 hours, training hours, at college 8 hours. Use a pie diagram to depict this information.. (a) With reference to Fig. 7.5, determine the amount spent on labour and materials to produce 650 units of the product. (b) If in year of Fig. 7.4, % corresponds to.5 dwellings, how many bungalows are sold in that year.. (a) If the company sell 500 units per annum of the product depicted in Fig. 7.5, determine the cost of their overheads per annum. (b) If % of the dwellings represented in year of Fig. 7.4 corresponds to dwellings, find the total number of houses sold in that year. 7. Presentation of grouped data When the number of members in a set is small, say ten or less, the data can be represented diagrammatically without further analysis, by means of pictograms, bar charts, percentage components bar charts or pie diagrams (as shown in Section 7.). For sets having more than ten members, those members having similar values are grouped together in classes to form a frequency distribution. To assist in accurately counting members in the various classes, a tally diagram is used (see Problems 8 and ). A frequency distribution is merely a table showing classes and their corresponding frequencies (see Problems 8 and ). The new set of values obtained by forming a frequency distribution is called grouped data. The terms used in connection with grouped data are shown in Fig. 7.6(a). The size or range of a class is given by the upper class boundary value minus the lower class boundary value, and in Fig. 7.6 is , i.e The class interval for the class shown in Fig. 7.6(b) is 7.4 to 7.6 and the class mid-point value is given by (upper class boundary value) C (lower class boundary value) 7.65 C 7.5 andinfig.7.6is, i.e. 7.5 One of the principal ways of presenting grouped data diagrammatically is by using a histogram, inwhichtheareas Fig. 7.6 (a) (b) Lower class boundary to 7. Class interval Class mid-point 7.4 to 7.6 Upper class boundary to of vertical, adjacent rectangles are made proportional to frequencies of the classes (see Problem 9). When class intervals are equal, the heights of the rectangles of a histogram are equal to the frequencies of the classes. For histograms having unequal class intervals, the area must be proportional to the frequency. Hence, if the class interval of class A is twice the class interval of class B, then for equal frequencies, the height of the rectangle representing A is half that of B (see Problem ). Another method of presenting grouped data diagrammatically is by using a frequency polygon, which is the graph produced by plotting frequency against class mid-point values and joining the co-ordinates with straight lines (see Problem ). A cumulative frequency distribution is a table showing the cumulative frequency for each value of upper class boundary. The cumulative frequency for a particular value of upper class boundary is obtained by adding the frequency of the class to the sum of the previous frequencies. A cumulative frequency distribution is formed in Problem. The curve obtained by joining the co-ordinates of cumulative frequency (vertically) against upper class boundary (horizontally) is called an ogive or a cumulative frequency distribution curve (see Problem ). Problem 8. The data given below refer to the gain of each of a batch of 40 transistors, expressed correct to the nearest whole number. Form a frequency distribution for these data having seven classes The range of the data is the value obtained by taking the value of the smallest member from that of the largest member. Inspection of the set of data shows that, range D 89 7 D 8. The size of each class is given approximately by range divided by the number of classes. Since 7 classes

224 Presentation of statistical data are required, the size of each class is 8/7, that is, approximately. To achieve seven equal classes spanning a range of values from 7 to 89, the class intervals are selected as: 70 7, 7 75, and so on. To assist with accurately determining the number in each class, a tally diagram is produced, as shown in Table 7.(a). This is obtained by listing the classes in the left-hand column, and then inspecting each of the 40 members of the set in turn and allocating them to the appropriate classes by putting s in the appropriate rows. Every fifth allocated to a particular row is shown as an oblique line crossing the four previous s, to help with final counting. A frequency distribution for the data is shown in Table 7.(b) and lists classes and their corresponding frequencies, obtained from the tally diagram. (Class mid-point values are also shown in the table, since they are used for constructing the histogram for these data (see Problem 9)). Table 7.(a) Class Table 7.(b) Tally Class Class mid-point Frequency Problem 9. Construct a histogram for the data given in Table 7.(b). The histogram is shown in Fig The width of the rectangles correspond to the upper class boundary values minus the lower class boundary values and the heights of the rectangles correspond to the class frequencies. The easiest way to draw a histogram is to mark the class mid-point values on the horizontal scale and draw the rectangles symmetrically about the appropriate class mid-point values and touching one another. Frequency Fig Class mid-point values Problem 0. The amount of money earned weekly by 40 people working part-time in a factory, correct to the nearest 0, is shown below. Form a frequency distribution having 6 classes for these data Inspection of the set given shows that the majority of the members of the set lie between 80 and 0 and that there are a much smaller number of extreme values ranging from 0 to 70. If equal class intervals are selected, the frequency distribution obtained does not give as much information as one with unequal class intervals. Since the majority of members are between 80 and 00, the class intervals in this range are selected to be smaller than those outside of this range. There is no unique solution and one possible solution is shown in Table 7.. Problem. Table 7.. Table 7. Class Frequency Draw a histogram for the data given in When dealing with unequal class intervals, the histogram must be drawn so that the areas, (and not the heights), of the rectangles are proportional to the frequencies of the

225 4 Basic Engineering Mathematics classes. The data given are shown in columns and of Table 7.. Columns and 4 give the upper and lower class boundaries, respectively. In column 5, the class ranges (i.e. upper class boundary minus lower class boundary values) are listed. The heights of the rectangles are proportional to the ratio frequency, as shown in column 6. The histogram is class range shown in Fig. 7.8 Table Class Frequency Upper Lower Class Height class class range of boundary boundary rectangle Frequency per unit class range /5 0/5 8/5 6/5 4/5 /5 Fig D D 5 0 D D D 5 0 D Class mid-point values Problem. The masses of 50 ingots in kilograms are measured correct to the nearest 0. kg and the results are as shown below. Produce a frequency distribution having about 7 classes for these data and then present the grouped data as (a) a frequency polygon and (b) a histogram The range of the data is the member having the largest value minus the member having the smallest value. Inspection of the set of data shows that: range D D.0 The size of each class is given approximately by range number of classes Since about seven classes are required, the size of each class is.0/7, that is approximately 0., and thus the class limits are selected as 7. to 7., 7.4 to 7.6, 7.7 to 7.9, and so on. The class mid-point for the 7. to 7. class is 7.5 C 7.05, 7.65 C 7.5, i.e. 7.5, and i.e. 7., for the 7.4 to 7.6 class is so on. To assist with accurately determining the number in each class, a tally diagram is produced as shown in Table 7.4. This is obtained by listing the classes in the left-hand column and then inspecting each of the 50 members of the set of data in turn and allocating it to the appropriate class by putting a in the appropriate row. Each fifth allocated to a particular row is marked as an oblique line to help with final counting. Table 7.4 Class Tally 7. to to to to8. 8.to to to9. A frequency distribution for the data is shown in Table 7.5 and lists classes and their corresponding frequencies. Class mid-points are also shown in this table, since they are used when constructing the frequency polygon and histogram. Table 7.5 Class Class mid-point Frequency 7.to to to to to to to9. 9.0

226 Presentation of statistical data 5 A frequency polygon is shown in Fig. 7.9, the coordinates corresponding to the class mid-point/frequency values, given in Table 7.5. The co-ordinates are joined by straight lines and the polygon is anchored-down at each end by joining to the next class mid-point value and zero frequency. Frequency Fig. 7.9 Frequency polygon Class mid-point values A histogram is shown in Fig. 7.0, the width of a rectangle corresponding to (upper class boundary value lower class boundary value) and height corresponding to the class frequency. The easiest way to draw a histogram is to mark class mid-point values on the horizontal scale and to draw the rectangles symmetrically about the appropriate class midpoint values and touching one another. A histogram for the data given in Table 7.5 is shown in Fig Frequency Fig. 7.0 Class mid-point values Histogram Problem. The frequency distribution for the masses in kilograms of 50 ingots is: 7.to7. 7.4to to to to to to9. Form a cumulative frequency distribution for these data and draw the corresponding ogive. A cumulative frequency distribution is a table giving values of cumulative frequency for the values of upper class boundaries, and is shown in Table 7.6. Columns and show the classes and their frequencies. Column lists the upper class boundary values for the classes given in column. Column 4 gives the cumulative frequency values for all frequencies less than the upper class boundary values given in column. Thus, for example, for the 7.7 to 7.9 class shown in row, the cumulative frequency value is the sum of all frequencies having values of less than 7.95, i.e. C 5 C 9 D 7, and so on. The ogive for the cumulative frequency distribution given in Table 7.6 is shown in Fig. 7.. The co-ordinates corresponding to each upper class boundary/cumulative frequency value are plotted and the co-ordinates are joined by straight lines ( not the best curve drawn through the co-ordinates as in experimental work.) The ogive is anchored at its start by adding the co-ordinate (7.05, 0). Table Class Frequency Upper class Cumulative boundary frequency Less than Cumulative frequency Fig Upper class boundary values in kilograms

227 6 Basic Engineering Mathematics Now try the following exercise Exercise 00 Further problems on presentation of grouped data (Answers on page 64). The mass in kilograms, correct to the nearest one-tenth of a kilogram, of 60 bars of metal are as shown. Form a frequency distribution of about 8 classes for these data Draw a histogram for the frequency distribution given in the solution of Problem.. The information given below refers to the value of resistance in ohms of a batch of 48 resistors of similar value. Form a frequency distribution for the data, having about 6 classes and draw a frequency polygon and histogram to represent these data diagramatically The time taken in hours to the failure of 50 specimens of a metal subjected to fatigue failure tests are as shown. Form a frequency distribution, having about 8 classes and unequal class intervals, for these data Form a cumulative frequency distribution and hence draw the ogive for the frequency distribution given in the solution to Problem. 6. Draw a histogram for the frequency distribution given in the solution to Problem The frequency distribution for a batch of 48 resistors of similar value, measured in ohms, is: Form a cumulative frequency distribution for these data. 8. Draw an ogive for the data given in the solution of Problem The diameter in millimetres of a reel of wire is measured in 48 places and the results are as shown (a) Form a frequency distribution of diameters having about 6 classes. (b) Draw a histogram depicting the data. (c) Form a cumulative frequency distribution. (d) Draw an ogive for the data.

228 8 Measures of central tendency and dispersion 8. Measures of central tendency A single value, which is representative of a set of values, may be used to give an indication of the general size of the members in a set, the word average often being used to indicate the single value. The statistical term used for average is the arithmetic mean or just the mean. Other measures of central tendency may be used and these include the median and the modal values. 8. Mean, median and mode for discrete data Mean The arithmetic mean value is found by adding together the values of the members of a set and dividing by the number of members in the set. Thus, the mean of the set of numbers: f4, 5, 6, 9g is: 4 C 5 C 6 C 9, i.e. 6 4 In general, the mean of the set: fx,x,x,...x n g is x x D x C x C x CÐÐÐCx n, written as n n where is the Greek letter sigma and means the sum of, and x (called x-bar) is used to signify a mean value. Median The median value often gives a better indication of the general size of a set containing extreme values. The set: f7, 5, 74, 0g has a mean value of 4, which is not really representative of any of the values of the members of the set. The median value is obtained by: (a) ranking the set in ascending order of magnitude, and (b) selecting the value of the middle member for sets containing an odd number of members, or finding the value of the mean of the two middle members for sets containing an even number of members. For example, the set: f7, 5, 74, 0g is ranked as f5, 7, 0, 74g, and since it contains an even number of members (four in this case), the mean of 7 and 0 is taken, giving a median value of 8.5. Similarly, the set: f, 8, 5, 7, 4g is ranked as f, 7, 4, 5, 8g and the median value is the value of the middle member, i.e. 4 Mode The modal value, ormode, is the most commonly occurring value in a set. If two values occur with the same frequency, the set is bi-modal. The set: f5, 6, 8,, 5, 4, 6, 5, g has a modal value of 5, since the member having a value of 5 occurs three times. Problem. the set: Determine the mean, median and mode for f,, 7, 5, 5,,, 7, 4, 8,, 4, g The mean value is obtained by adding together the values of the members of the set and dividing by the number of members in the set. Thus, mean value, x D C C 7 C 5 C 5 C C C 7 C 4 C 8 C C 4 C D 65 D 5

229 8 Basic Engineering Mathematics To obtain the median value the set is ranked, that is, placed in ascending order of magnitude, and since the set contains an odd number of members the value of the middle member is the median value. Ranking the set gives: f,,,,, 4, 4, 5, 5, 7, 7, 8, g The middle term is the seventh member, i.e. 4, thus the median value is 4. Themodal value is the value of the most commonly occurring member and is, which occurs three times, all other members only occurring once or twice. Problem. The following set of data refers to the amount of money in s taken by a news vendor for 6 days. Determine the mean, median and modal values of the set: f7.90, 4.70, 54.40, 8.9, 47.60, 9.68g Mean value 7.90 C 4.70 C C 8.9 C C 9.68 D 6 D 7.0 The ranked set is: f8.9, 7.90, 4.70, 9.68, 47.60, 54.40g Since the set has an even number of members, the mean of the middle two members is taken to give the median value, i.e C 9.68 median value D D 7.9 Since no two members have the same value, this set has no mode. values) and dividing by the sum of the frequencies, i.e. mean value x D f x C f x CÐÐÐf n x n f C f CÐÐÐCf n D fx f where f is the frequency of the class having a mid-point value of x, and so on. Problem. The frequency distribution for the value of resistance in ohms of 48 resistors is as shown. Determine the mean value of resistance ,.0.4 0,.5.9,.0.4,.5.9 9,.0.4 The class mid-point/frequency values are: 0.7,. 0,.7,.,.7 9 and. For grouped data, the mean value is given by: fx x D f where f is the class frequency and x is the class mid-point value. Hence mean value, ð 0.7 C 0 ð. C ð.7 C ð. C 9 ð.7 C ð. x D 48 D 05. D i.e. the mean value is.9 ohms, correct to significant figures. Now try the following exercise Exercise 0 Further problems on mean, median and mode for discrete data (Answers on page 64) In Problems to 4, determine the mean, median and modal values for the sets given.. f, 8, 0, 7, 5, 4,, 9, 8g. f6,,, 9,, 6, 5, 8g. f4.7, 4.7, 4.74, 4.7, 4.7, 4.7, 4.7, 4.7g 4. f7.8, 6.4, 40.7, 4.7, 8.5, 7.4, 57.9g 8. Mean, median and mode for grouped data The mean value for a set of grouped data is found by determining the sum of the (frequency ð class mid-point Histogram The mean, median and modal values for grouped data may be determined from a histogram. In a histogram, frequency values are represented vertically and variable values horizontally. The mean value is given by the value of the variable corresponding to a vertical line drawn through the centroid of the histogram. The median value is obtained by selecting a variable value such that the area of the histogram to the left of a vertical line drawn through the selected variable value is equal to the area of the histogram on the right of the line. The modal value is the variable value obtained by dividing the width of the highest rectangle in the histogram in proportion to the heights of the adjacent rectangles. The method of determining the mean, median and modal values from a histogram is shown in Problem 4. Problem 4. The time taken in minutes to assemble a device is measured 50 times and the results are as shown. Draw a histogram depicting this data and hence

230 Measures of central tendency and dispersion 9 determine the mean, median and modal values of the distribution , , , 0.5.5,.5.5 6, The histogram is shown in Fig. 8.. The mean value lies at the centroid of the histogram. With reference to any arbitrary axis, say YY shown at a time of 4 minutes, the position of the horizontal value of the centroid can be obtained from the relationship AM D am, wherea is the area of the histogram, M is the horizontal distance of the centroid from the axis YY, a is the area of a rectangle of the histogram and m is the distance of the centroid of the rectangle from YY. The areas of the individual rectangles are shown circled on the histogram giving a total area of 00 square units. The positions, m, of the centroids of the individual rectangles are,, 5,... units from YY. Thus 00 M D 0 ð C 6 ð C ð 5 C 4 ð 7 C ð 9 C 6 ð i.e. M D 560 D 5.6 units from YY 00 right, i.e. the vertical line must pass through 9.5 minutes. Thus the median value of the distribution is 9.5 minutes. The mode is obtained by dividing the line AB, which is the height of the highest rectangle, proportionally to the heights of the adjacent rectangles. With reference to Fig. 8., this is done by joining AC and BD and drawing a vertical line through the point of intersection of these two lines. This gives the mode of the distribution and is 9. minutes. Now try the following exercise Exercise 0 Further problems on mean, median and mode for grouped data (Answers on page 64). The frequency distribution given below refers to the heights in centimetres of 00 people. Determine the mean value of the distribution, correct to the nearest millimetre , , , 78 84, The gain of 90 similar transistors is measured and the results are as shown. Frequency Fig Y Y Mean Median Mode A B 5.6 C 4 D E F Time in minutes , , , , By drawing a histogram of this frequency distribution, determine the mean, median and modal values of the distribution.. The diameters, in centimetres, of 60 holes bored in engine castings are measured and the results are as shown. Draw a histogram depicting these results and hence determine the mean, median and modal values of the distribution , ,.0.04, , Thus the position of the mean with reference to the time scale is 4 C 5.6, i.e. 9.6 minutes. The median is the value of time corresponding to a vertical line dividing the total area of the histogram into two equal parts. The total area is 00 square units, hence the vertical line must be drawn to give 50 units of area on each side. To achieve this with reference to Fig. 8., rectangle ABFE must be split so that 50 0 C 6 units of area lie on one side and 50 4 C C 6 units of area lie on the other. This shows that the area of ABFE is split so that 4 units of area lie to the left of the line and 8 units of area lie to the 8.4 Standard deviation (a) Discrete data The standard deviation of a set of data gives an indication of the amount of dispersion, or the scatter, of members of the set from the measure of central tendency. Its value is the root-mean-square value of the members of the set and for discrete data is obtained as follows:

231 0 Basic Engineering Mathematics (a) determine the measure of central tendency, usually the mean value, (occasionally the median or modal values are specified), (b) calculate the deviation of each member of the set from the mean, giving x x, x x, x x,..., (c) determine the squares of these deviations, i.e. x x, x x, x x,..., x x and D 4 n 6 D 5. P6 since there are 6 members in the set. Hence, standard deviation, { x x } s D D 5.P6 D.80 n correct to 4 significant figures. (d) find the sum of the squares of the deviations, that is x x C x x C x x,..., (e) divide by the number of members in the set, n, giving x x C x x C x x CÐÐÐ n (f) determine the square root of (e). The standard deviation is indicated by s (the Greek letter small sigma ) and is written mathematically as: { (x x) } standard deviation, s = n where x is a member of the set, x is the mean value of the set and n is the number of members in the set. The value of standard deviation gives an indication of the distance of the members of a set from the mean value. The set: f, 4, 7, 0, g has a mean value of 7 and a standard deviation of about 4.. The set f5, 6, 7, 8, 9g also has a mean value of 7, but the standard deviation is about.4. This shows that the members of the second set are mainly much closer to the mean value than the members of the first set. The method of determining the standard deviation for a set of discrete data is shown in Problem 5. Problem 5. Determine the standard deviation from the mean of the set of numbers: f5, 6, 8, 4, 0, g, correct to 4 significant figures. x The arithmetic mean, x D n D 5 C 6 C 8 C 4 C 0 C D 6 6 { x x } Standard deviation, s D n The x x values are: 5 6, 6 6, 8 6, 4 6, 0 6 and 6. The sum of the x x values, i.e. x x D C 0 C 4 C 4 C 6 C 9 D 4 (b) Grouped data For grouped data, standard deviation { {f (x x) } } s = f where f is the class frequency value, x is the class midpoint value and x is the mean value of the grouped data. The method of determining the standard deviation for a set of grouped data is shown in Problem 6. Problem 6. The frequency distribution for the values of resistance in ohms of 48 resistors is as shown. Calculate the standard deviation from the mean of the resistors, correct to significant figures ,.0.4 0,.5.9,.0.4,.5.9 9,.0.4 The standard deviation for grouped data is given by: { ff x x } g s D f From Problem, the distribution mean value, x D.9, correct to 4 significant figures. The x-values are the class mid-point values, i.e. 0.7,.,.7,... Thus the x x values are 0.7.9,..9,.7.9,..., and the f x x values are 0.7.9, 0..9,.7.9,... The f x x values are C C 0.54 C.09 C C.768 D 9.95 ff x x g D 9.95 D f 48

232 Measures of central tendency and dispersion and standard deviation, { ff x x } g s D D p f D 0.645, correct to significant figures Now try the following exercise Exercise 0 Further problems on standard deviation (Answers on page 64). Determine the standard deviation from the mean of the set of numbers: f5,, 5,, 8,, 0g correct to significant figures.. The values of capacitances, in microfarads, of ten capacitors selected at random from a large batch of similar capacitors are: 4., 5.0, 0.4, 4.6, 9.6, 8.7,.4,.7, 9.0 and. Determine the standard deviation from the mean for these capacitors, correct to significant figures.. The tensile strength in megapascals for 5 samples of tin were determined and found to be: 4.6, 4.57, 4.40, 4.6, 4.6, 4.5, 4.49, 4.6, 4.5, 4.55, 4.58, 4.5, 4.44, 4.48 and 4.40 Calculate the mean and standard deviation from the mean for these 5 values, correct to 4 significant figures. 4. Determine the standard deviation from the mean, correct to 4 significant figures, for the heights of the 00 people given in Problem of Exercise 0, page Calculate the standard deviation from the mean for the data given in Problem of Exercise 0, page 9, correct to decimal places. 8.5 Quartiles, deciles and percentiles Other measures of dispersion which are sometimes used are the quartile, decile and percentile values. The quartile values of a set of discrete data are obtained by selecting the values of members which divide the set into four equal parts. Thus for the set: f,, 4, 5, 5, 7, 9,,, 4, 7g there are members and the values of the members dividing the set into four equal parts are 4, 7, and. These values are signified by Q, Q and Q and called the first, second and third quartile values, respectively. It can be seen that the second quartile value, Q, is the value of the middle member and hence is the median value of the set. For grouped data the ogive may be used to determine the quartile values. In this case, points are selected on the vertical cumulative frequency values of the ogive, such that they divide the total value of cumulative frequency into four equal parts. Horizontal lines are drawn from these values to cut the ogive. The values of the variable corresponding to these cutting points on the ogive give the quartile values (see Problem 7). When a set contains a large number of members, the set can be split into ten parts, each containing an equal number of members. These ten parts are then called deciles. For sets containing a very large number of members, the set may be split into one hundred parts, each containing an equal number of members. One of these parts is called a percentile. Problem 7. The frequency distribution given below refers to the overtime worked by a group of craftsmen during each of 48 working weeks in a year , 0 4 4, 5 9 7, 40 44, 45 49, , Draw an ogive for this data and hence determine the quartile values. The cumulative frequency distribution (i.e. upper class boundary/cumulative frequency values) is: 9.5 5, 4.5 9, 9.5 6, , , , The ogive is formed by plotting these values on a graph, as shown in Fig. 6.. The total frequency is divided into four equal parts, each having a range of 48/4, i.e.. This gives cumulative frequency values of 0 to corresponding to the first quartile, to 4 corresponding to the second quartile, 4 to 6 corresponding to the third quartile and 6 to 48 corresponding to the fourth quartile of the distribution, i.e. the distribution is divided into four equal parts. The quartile values are those of the variable corresponding to cumulative frequency values of, 4 and 6, marked Q, Q and Q in Fig. 8.. These values, correct to the nearest hour, are 7 hours, 4 hours and 48 hours, respectively. The Q value is also equal to the median value of the distribution. One measure of the dispersion of a distribution is called the semi-interquartile range and is given by Q Q /, and is 48 7 / in this case, i.e. 5 hours. Problem 8. Determine the numbers contained in the (a) 4st to 50th percentile group, and (b) 8th decile group of the set of numbers shown below:

233 Basic Engineering Mathematics 50 Now try the following exercise Cumulative frequency Fig Q 40 Q 45 Q Upper class boundary values, hours The set is ranked, giving: (a) There are 0 numbers in the set, hence the first 0% will be the two numbers 7 and 4, the second 0% will be 5 and 7, and so on. Thus the 4st to 50th percentile group will be the numbers and. (b) The first decile group is obtained by splitting the ranked set into 0 equal groups and selecting the first group, i.e. the numbers 7 and 4. The second decile group are the numbers 5 and 7, and so on. Thus the 8th decile group contains the numbers 7 and 8. Exercise 04 Further problems on quartiles, deciles and percentiles (Answers on page 64). The number of working days lost due to accidents for each of one-monthly periods are as shown. Determine the median and first and third quartile values for this data The number of faults occurring on a production line in a nine-week period are as shown below. Determine the median and quartile values for the data Determine the quartile values and semi-interquartile range for the frequency distribution given in Problem of Exercise 0, page Determine the numbers contained in the 5th decile group and in the 6st to 70th percentile groups for the set of numbers: Determine the numbers in the 6th decile group and in the 8st to 90th percentile group for the set of numbers:

234 9 Probability 9. Introduction to probability Probability The probability of something happening is the likelihood or chance of it happening. Values of probability lie between 0 and, where 0 represents an absolute impossibility and represents an absolute certainty. The probability of an event happening usually lies somewhere between these two extreme values and is expressed either as a proper or decimal fraction. Examples of probability are: that a length of copper wire has zero resistance at 00 C that a fair, six-sided dice will stop with a upwards that a fair coin will land with a head upwards that a length of copper wire has some resistance at 00 C 0 or or 0.5 If p is the probability of an event happening and q is the probability of the same event not happening, then the total probability is pcq and is equal to unity, since it is an absolute certainty that the event either does or does not occur, i.e. p + q = Dependent event A dependent event is one in which the probability of an event happening affects the probability of another ever happening. Let 5 transistors be taken at random from a batch of 00 transistors for test purposes, and the probability of there being a defective transistor, p, be determined. At some later time, let another 5 transistors be taken at random from the 95 remaining transistors in the batch and the probability of there being a defective transistor, p, be determined. The value of p is different from p since batch size has effectively altered from 00 to 95, i.e. probability p is dependent on probability p. Since transistors are drawn, and then another 5 transistors drawn without replacing the first 5, the second random selection is said to be without replacement. Independent event An independent event is one in which the probability of an event happening does not affect the probability of another event happening. If 5 transistors are taken at random from a batch of transistors and the probability of a defective transistor p is determined and the process is repeated after the original 5 have been replaced in the batch to give p,then p is equal to p. Since the 5 transistors are replaced between draws, the second selection is said to be with replacement. Expectation The expectation, E, of an event happening is defined in general terms as the product of the probability p of an event happening and the number of attempts made, n, i.e.e = pn. Thus, since the probability of obtaining a upwards when rolling a fair dice is, the expectation of getting a upwards 6 on four throws of the dice is ð 4, i.e. 6 Thus expectation is the average occurrence of an event. 9. Laws of probability The addition law of probability The addition law of probability is recognised by the word or joining the probabilities. If p A is the probability of event A happening and p B is the probability of event B happening, the probability of event A or event B happening is given by

235 4 Basic Engineering Mathematics p A C p B. Similarly, the probability of events A or B or C or ::: N happening is given by p A + p B C p C + + p N The multiplication law of probability The multiplication law of probability is recognised by the word and joining the probabilities. If p A is the probability of event A happening and p B is the probability of event B happening, the probability of event A and event B happening is given by p A ð p B. Similarly, the probability of events A and B and C and ::: N happening is given by p A U p B ð p C U U p N 9. Worked problems on probability Problem. Determine the probabilities of selecting at random (a) a man, and (b) a woman from a crowd containing 0 men and women. (a) The probability of selecting at random a man, p, isgiven by the ratio number of men number in crowd i.e. p D 0 0 C D 0 or (b) The probability of selecting at random a women, q, is given by the ratio number of women number in crowd i.e. q D 0 C D or (Check: the total probability should be equal to ; p D 0 and q D 5 5 thus the total probability, p C q D 0 5 C 5 D hence no obvious error has been made). Problem. Find the expectation of obtaining a 4 upwards with throws of a fair dice. Expectation is the average occurrence of an event and is defined as the probability times the number of attempts. The probability, p, of obtaining a 4 upwards for one throw of the dice is 6 Also, attempts are made, hence n D and the expectation, E, ispn, i.e.e D ð D or Problem. random: Calculate the probabilities of selecting at (a) the winning horse in a race in which 0 horses are running, (b) the winning horses in both the first and second races if there are 0 horses in each race. (a) Since only one of the ten horses can win, the probability of selecting at random the winning horse is number of winners number of horses,i.e. or (b) The probability of selecting the winning horse in the first race is. The probability of selecting the winning horse 0 in the second race is. The probability of selecting the 0 winning horses in the first and second race is given by the multiplication law of probability, i.e. probability = 0 ð 0 = or Problem 4. The probability of a component failing in one year due to excessive temperature is, due to 0 excessive vibration is and due to excessive humidity 5 is. Determine the probabilities that during a oneyear period a component: (a) fails due to excessive 50 temperature and excessive vibration, (b) fails due to excessive vibration or excessive humidity, and (c) will not fail because of both excessive temperature and excessive humidity. Let p A be the probability of failure due to excessive temperature, then p A D 0 and p A D 9 0 (where p A is the probability of not failing.) Let p B be the probability of failure due to excessive vibration, then p B D and p B D Let p C be the probability of failure due to excessive humidity, then p C D 50 and p C D 49 50

236 Probability 5 (a) The probability of a component failing due to excessive temperature and excessive vibration is given by: p A ð p B D 0 ð 5 D or (b) The probability of a component failing due to excessive vibration or excessive humidity is: p B C p C D 5 C 50 D or (c) The probability that a component will not fail due excessive temperature and will not fail due to excess humidity is: p A ð p C D 9 0 ð D or 0.9 Problem 5. A batch of 00 capacitors contains 7 which are within the required tolerance values, 7 which are below the required tolerance values, and the remainder are above the required tolerance values. Determine the probabilities that when randomly selecting a capacitor and then a second capacitor: (a) both are within the required tolerance values when selecting with replacement, and (b) the first one drawn is below and the second one drawn is above the required tolerance value, when selection is without replacement. (a) The probability of selecting a capacitor within the 7 required tolerance values is. The first capacitor 00 drawn is now replaced and a second one is drawn from the batch of 00. The probability of this capacitor being within the required tolerance values is also Thus, the probability of selecting a capacitor within the required tolerance values for both the first and the second draw is 7 00 ð 7 00 D 59 or (b) The probability of obtaining a capacitor below the required tolerance values on the first draw is 7 00.There are now only 99 capacitors left in the batch, since the first capacitor is not replaced. The probability of drawing a capacitor above the required tolerance values on the second draw is 0, since there are , i.e capacitors above the required tolerance value. Thus, the probability of randomly selecting a capacitor below the required tolerance values and followed by randomly selecting a capacitor above the tolerance values is 7 00 ð 0 99 D D or 0.07 Now try the following exercise Exercise 05 Further problems on probability (Answers on page 64). In a batch of 45 lamps there are 0 faulty lamps. If one lamp is drawn at random, find the probability of it being (a) faulty and (b) satisfactory.. A box of fuses are all of the same shape and size and comprises A fuses, 47 5 A fuses and 69 A fuses. Determine the probability of selecting at random (a) a A fuse, (b) a 5 A fuse and (c) a A fuse.. (a) Find the probability of having a upwards when throwing a fair 6-sided dice. (b) Find the probability of having a 5 upwards when throwing a fair 6-sided dice. (c) Determine the probability of having a and then a 5 on two successive throws of a fair 6-sided dice. 4. The probability of event A happening is and the 5 probability of event B happening is. Calculate the probabilities of (a) both A and B happening, (b) only event A happening, i.e. event A happening and event B not happening, (c) only event B happening, and (d) either A, orb, ora and B happening. 5. When testing 000 soldered joints, 4 failed during a vibration test and 5 failed due to having a high resistance. Determine the probability of a joint failing due to (a) vibration, (b) high resistance, (c) vibration or high resistance and (d) vibration and high resistance. 9.4 Further worked problems on probability Problem 6. A batch of 40 components contains 5 which are defective. A component is drawn at random from the batch and tested and then a second component is drawn. Determine the probability that neither of the components is defective when drawn (a) with replacement, and (b) without replacement. (a) With replacement The probability that the component selected on the first draw is satisfactory is 5 40,i.e. 7. The component is now replaced 8 and a second draw is made. The probability that this component is also satisfactory is 7. Hence, the probability that 8 both the first component drawn and the second component drawn are satisfactory is: 7 8 ð 7 8 D 49 or

237 6 Basic Engineering Mathematics (b) Without replacement The probability that the first component drawn is satisfactory is 7. There are now only 4 satisfactory components left in 8 the batch and the batch number is 9. Hence, the probability of drawing a satisfactory component on the second draw is 4. Thus the probability that the first component drawn and 9 the second component drawn are satisfactory, i.e. neither is defective, is: 7 8 ð 4 9 D 8 or Problem 7. A batch of 40 components contains 5 which are defective. If a component is drawn at random from the batch and tested and then a second component is drawn at random, calculate the probability of having one defective component, both with and without replacement. The probability of having one defective component can be achieved in two ways. If p is the probability of drawing a defective component and q is the probability of drawing a satisfactory component, then the probability of having one defective component is given by drawing a satisfactory component and then a defective component or by drawing a defective component and then a satisfactory one, i.e. by q ð p C p ð q With replacement: p D 5 40 D and q D D 7 8 Hence, probability of having one defective component is: 8 ð 7 8 C 7 8 ð 8 7 i.e. 64 C 7 64 D 7 or 0.88 Without replacement: p D and q 8 D 7 on the first of the two draws. The batch 8 number is now 9 for the second draw, thus, p D 5 9 and q D 5 9 p q C q p D 8 ð 5 9 C 7 8 ð 5 5 C 5 D 9 D 70 or 0.44 Problem 8. A box contains 74 brass washers, 86 steel washers and 40 aluminium washers. Three washers are drawn at random from the box without replacement. Determine the probability that all three are steel washers. Assume, for clarity of explanation, that a washer is drawn at random, then a second, then a third (although this assumption does not affect the results obtained). The total number of washers is 74 C 86 C 40, i.e. 00. The probability of randomly selecting a steel washer on the first draw is 86. There are now 85 steel washers in a batch 00 of 99. The probability of randomly selecting a steel washer on the second draw is 85. There are now 84 steel washers 99 in a batch of 98. The probability of randomly selecting a steel washer on the third draw is 84. Hence the probability 98 of selecting a steel washer on the first draw and the second draw and the third draw is: ð ð D D Problem 9. For the box of washers given in Problem 8 above, determine the probability that there are no aluminium washers drawn, when three washers are drawn at random from the box without replacement. The probability( of not) drawing an aluminium washer on the 40 first draw is,i.e. 60. There are now 99 washers in the batch of which 59 are not aluminium washers. Hence, the probability of not drawing an aluminium washer on the second draw is 59. Similarly, the probability of not drawing 99 an aluminium washer on the third draw is 58. Hence the 98 probability of not drawing an aluminium washer on the first and second and third draws is ð ð D D 0.50 Problem 0. For the box of washers in Problem 8 above, find the probability that there are two brass washers and either a steel or an aluminium washer when three are drawn at random, without replacement. Two brass washers (A) and one steel washer (B) can be obtained in any of the following ways: st draw nd draw rd draw A A B A B A B A A

238 Probability 7 Two brass washers and one aluminium washer (C) can also be obtained in any of the following ways: st draw nd draw rd draw A A C A C A C A A Thus there are six possible ways of achieving the combinations specified. If A represents a brass washer, B a steel washer and C an aluminium washer, then the combinations and their probabilities are as shown: DRAW First Second Third A A B A B A B A A A A C A C A C A A PROBABILITY ð 7 99 ð D ð ð 7 98 D ð ð 7 98 D ð 7 99 ð D ð ð 7 98 D ð ð 7 98 D The probability of having the first combination or the second, or the third, and so on, is given by the sum of the probabilities, i.e. by ð C ð 0.074, that is, 0.59 Now try the following exercise Exercise 06 Further problems on probability (Answers on page 64). The probability that component A will operate satisfactorily for 5 years is 0.8 and that B will operate satisfactorily over that same period of time is Find the probabilities that in a 5 year period: (a) both components operate satisfactorily, (b) only component A will operate satisfactorily, and (c) only component B will operate satisfactorily.. In a particular street, 80% of the houses have telephones. If two houses selected at random are visited, calculate the probabilities that (a) they both have a telephone and (b) one has a telephone but the other does not have telephone.. Veroboard pins are packed in packets of 0 by a machine. In a thousand packets, 40 have less than 0 pins. Find the probability that if packets are chosen at random, one will contain less than 0 pins and the other will contain 0 pins or more. 4. A batch of kw fire elements contains 6 which are within a power tolerance and 4 which are not. If elements are selected at random from the batch, calculate the probabilities that (a) all three are within the power tolerance and (b) two are within but one is not within the power tolerance. 5. An amplifier is made up of three transistors, A, B and C. The probabilities of A, B or C being defective are 0, 5 and, respectively. Calculate the percentage 50 of amplifiers produced (a) which work satisfactorily and (b) which have just one defective transistor. 6. A box contains 4 40 W lamps, 8 60 W lamps and 58 5 W lamps, all the lamps being of the same shape and size. Three lamps are drawn at random from the box, first one, then a second, then a third. Determine the probabilities of: (a) getting one 5 W, one 40 W and one 60 W lamp, with replacement, (b) getting one 5 W, one 40 W and one 60 W lamp without replacement, and (c) getting either one 5 W and two 40 W or one 60 W and two 40 W lamps with replacement. Assignment 4 This assignment covers the material in Chapters 7 to 9. The marks for each question are shown in brackets at the end of each question.. A company produces five products in the following proportions: Product A 4 Product B 6 Product C 5 Product D 9 Product E 8 Draw (a) a horizontal bar chart, and (b) a pie diagram to represent these data visually. (7). State whether the data obtained on the following topics are likely to be discrete or continuous: (a) the number of books in a library (b) the speed of a car (c) the time to failure of a light bulb (). Draw a histogram, frequency polygon and ogive for the data given below which refers to the diameter of 50 components produced by a machine.

239 8 Basic Engineering Mathematics Class intervals Frequency.0. mm 4..5 mm mm mm.4.44 mm mm mm 4 (0) 4. Determine the mean, median and modal values for the lengths given in metres: 8, 0, 44, 0,, 0, 8, 4, 6, 8 () 5. The length in millimetres of 00 bolts is as shown below: Determine for the sample (a) the mean value, and (b) the standard deviation, correct to 4 significant figures. (9) 6. The number of faulty components in a factory in a week period is: Determine the median and the first and third quartile values. () 7. Determine the probability of winning a prize in a lottery by buying 0 tickets when there are 0 prizes and a total of 5000 tickets sold. () 8. A sample of 50 resistors contains 44 which are within the required tolerance value, 4 which are below and the remainder which are above. Determine the probability of selecting from the sample a resistor which is (a) below the required tolerance, and (b) above the required tolerance. Now two resistors are selected at random from the sample. Determine the probability, correct to decimal places, that neither resistor is defective when drawn (c) with replacement, and (d) without replacement. (e) If a resistor is drawn at random from the batch and tested, and then a second resistor is drawn from those left, calculate the probability of having one defective component when selection is without replacement. ()

240 0 Introduction to differentiation 0. Introduction to calculus Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions. Calculus is a subject that falls into two parts: (i) differential calculus (or differentiation) and (ii) integral calculus (or integration). Differentiation is used in calculations involving rates of change (see section 0.0), velocity and acceleration, and maximum and minimum values of curves (see Engineering Mathematics ). 0. Functional notation In an equation such as y D x C x 5, y is said to be a function of x and may be written as y D f x. An equation written in the form f x D x C x 5is termed functional notation. The value of f x when x D 0 is denoted by f 0, and the value of f x when x D is denoted by f andsoon. Thus when f x D x C x 5, then f 0 D 0 C 0 5 D 5 and f D C 5 D and so on. Problem. If f x D 4x x C find: f 0, f, f and f f f x D 4x x C f 0 D C D f D 4 C D 6 9 C D 9 f D 4 C D 4 C C D 9 f f D 9 9 D 0 Problem. (i) f ł f (iii) f C a f Given that f x D 5x C x 7 determine: (ii) f C a (iv) f C a f a f x D 5x C x 7 (i) f D 5 C 7 D 5 f D 5 C 7 D f ł f D 5 D 5 (ii) f C a D 5 C a C C a 7 D 5 9 C 6a C a C C a 7 D 45C0aC5a CCa 7 D 4 + a + 5a (iii) f D 5 C 7 D 4 f Ca f D 4CaC5a 4 D a + 5a (iv) f C a f a D a C 5a a Now try the following exercise Exercise 07 D + 5a Further problems on functional notation (Answers on page 65). If f x D 6x x C findf 0, f, f, f and f. If f x D x C 5x 7findf, f, f, f f

241 0 Basic Engineering Mathematics. Given f x D x C x x C prove that f D 7 f f (x) 0 B f (x) = x 4. If f x D x C x C 6findf, f C a, f C a f f C a f and a The gradient of a curve (a) If a tangent is drawn at a point P on a curve, then the gradient of this tangent is said to be the gradient of the curve at P. In Figure 0., the gradient of the curve at P is equal to the gradient of the tangent PQ. f (x) P Q 0 x Fig. 0. (b) For the curve shown in Figure 0., let the points A and B have co-ordinates (x, y and (x, y, respectively. In functional notation, y D f x and y D f x as shown. f (x) f (x ) A E B C D f (x ) 0 x x x Fig. 0. The gradient of the chord AB D BC BD CD D AC ED D f x f x x x (c) For the curve f x D x shown in Figure 0.: (i) the gradient of chord AB D f f D 9 D 4 4 A 0.5 Fig. 0. D (ii) the gradient of chord AC D f f C D 4 D (iii) the gradient of chord AD D f.5 f D.5 D (iv) if E is the point on the curve (., f. ) then the gradient of chord AE f. f D.. D D. 0. (v) if F is the point on the curve (.0, f.0 ) then the gradient of chord AF D f.0 f.0.00 D D Thus as point B moves closer and closer to point A the gradient of the chord approaches nearer and nearer to the value. This is called the limiting value of the gradient of the chord AB and when B coincides with A the chord becomes the tangent to the curve. Now try the following exercise Exercise 08 A further problem on the gradient of a curve (Answers on page 65). Plot the curve f x D 4x for values of x from x D tox DC4. Label the co-ordinates (, f ) and (, f ) asj and K, respectively. Join points J and K x

242 Introduction to differentiation to form the chord JK. Determine the gradient of chord JK. By moving J nearer and nearer to K determine the gradient of the tangent of the curve at K. 0.4 Differentiation from first principles (i) In Figure 0.4, A and B are two points very close together on a curve, dx (delta x) and dy (delta y) representing small increments in the x and y directions, respectively. y Summarising, the differential coefficient, { } dy dx D dy f x C dx f x f0 x D limit dx!0 dx D limit dx!0 dx Problem. Differentiate from first principles f x D x and determine the value of the gradient of the curve at x D To differentiate from first principles means to find f 0 x by using the expression { } f x C dx f x f 0 x D limit dx!0 dx f x D x Fig. 0.4 B(x+dx, y+dy) dy A(x,y) f (x+dx) f (x) dx 0 x Substituting x C dx for x gives f x C dx D x C dx D x C xdx C dx, hence { x f 0 C xdx C dx x } x D limit dx!0 dx { xdx C dx } D limit D limit fx C dxg dx!0 dx dx!0 As dx! 0, fx C dxg! fx C 0g. Thus f (x) =x, i.e. the differential coefficient of x is x. At x D, the gradient of the curve, f 0 x D D 4 Gradient of chord AB D dy dx however, dy D f x C dx f x Hence dy f x C dx f x D dx dx As dx approaches zero, dy approaches a limiting value dx and the gradient of the chord approaches the gradient of the tangent at A. (ii) When determining the gradient of a tangent to a curve there are two notations used. The gradient of the curve at A in Figure 0.4 can either be written as: (iii) dy limit dx!0 dx or limit dx!0 { } f x C dx f x dx dy In Leibniz notation, dx = limit dy dx 0 dx In functional notation, f (x) =limit dx 0 { } f (x + dx) f (x) dy dx is the same as f0 x and is called the differential coefficient or the derivative. The process of finding the differential coefficient is called differentiation. dx Problem 4. Find the differential coefficient of y D 5x By definition, dy { } f x C dx f x dx D f0 x D limit dx!0 dx The function being differentiated is y D f x D 5x. Substituting (x C dx) forx gives: f x C dx D 5 x C dx D 5x C 5dx. Hence dy dx D f0 x D limit dx!0 D limit dx!0 { } 5x C 5dx 5x dx { } 5dx D limit dx f5g dx!0 Since the term dx does not appear in f5g the limiting value as dx! 0off5gis 5. Thus dy = 5, i.e.the differential coefficient of 5x is 5. dx The equation y D 5x represents a straight line of gradient 5 (see Chapter ). The differential coefficient (i.e. dy dx or f0 x ) means the gradient of the curve, and since the gradient of the line y D 5x is 5 this result can be obtained by inspection. Hence, in general, if y = kx (where k is a constant), then the gradient of the line is k and dy dx or f (x) =k.

243 Basic Engineering Mathematics Problem 5. Find the derivative of y D 8 y D f x D 8. Since there are no x-values in the original equation, substituting (x C dx) forx still gives f x C dx D 8. Hence { } dy f x C dx f x dx D f0 x D limit dx!0 dx { } 8 8 D limit D 0 dx!0 dx Thus, when y D 8, dy dx = 0 The equation y D 8 represents a straight horizontal line and the gradient of a horizontal line is zero, hence the result could have been determined by inspection. Finding the derivative means finding the gradient, hence, in general, for any horizontal line if y = k (where k is a constant) then dy dx =0. Problem 6. Differentiate from first principles f x D x Substituting (x C dx) forx gives f x C dx D x C dx D x C dx x C xdx C dx D x C x dx C xdx C dx D x C 6x dx C 6xdx C dx { } dy f x C dx f x dx D f0 x D limit dx!0 dx { x C 6x dx C 6xdx C dx x } D limit dx!0 dx { 6x dx C 6xdx C dx } D limit dx!0 dx { D limit 6x C 6xdx C dx } dx!0 Hence f (x) D 6x,i.e.the differential coefficient of x is 6x. Problem 7. Find the differential coefficient of y D 4x C 5x and determine the gradient of the curve at x D y D f x D 4x C 5x f x C dx D 4 x C dx C 5 x C dx D 4 x C xdx C dx C 5x C 5dx D 4x C 8xdx C 4dx C 5x C 5dx { } dy f x C dx f x dx D f0 x D limit dx!0 dx 4x C 8xdx C 4dx C 5x C 5dx 4x D limit C 5x dx!0 dx D limit dx!0 { 8xdx C 4dx } C 5dx dx D limit f8x C 4dx C 5g dx!0 i.e. dy dx = f (x) =8x + 5 At x D, the gradient of the curve D dy dx D f0 x D 8 C 5 D 9 Now try the following exercise Exercise 09 Further problems on differentiation from first principles (Answers on page 65) In Problems to, differentiate from first principles.. y D x. y D 7x. y D 4x 4. y D 5x 5. y D x C x 6. y D 7. f x D 9x 8. f x D x 9. f x D 9x 0. f x D 7x. f x D x C 5x 4. f x D 4. Determine d dx 4x from first principles 4. Find d dx x C 5 from first principles 0.5 Differentiation of y = ax n by the general rule From differentiation by first principles, a general rule for differentiating ax n emerges where a and n are any constants. This rule is: if y = ax n then dy dx = anx n or, if f(x) = ax n then f (x) =anx n (Each of the results obtained in worked problems and 7 may be deduced by using this general rule). When differentiating, results can be expressed in a number of ways.

244 Introduction to differentiation For example: (i) if y D x then dy dx D 6x, (ii) if f x D x then f 0 x D 6x, (iii) the differential coefficient of x is 6x, (iv) the derivative of x is 6x, and d (v) dx x D 6x Problem 8. Using the general rule, differentiate the following with respect to x: (a) y D 5x 7 (b) y D p x (c) y D 4 x (a) Comparing y D 5x 7 with y D ax n shows that a D 5and n D 7. Using the general rule, dy dx D anxn D 5 7 x 7 D 5x 6 (b) y D p x D x. Hence a D andnd dy dx D anxn D x D x D x D p x (c) y D 4 x D 4x. Hence a D 4andn D dy dx D anxn D 4 x D 8x D 8 x Problem 9. Find the differential coefficient of y D 5 x 4 x C 4p x 5 C 7 Problem 0. If f t D 5t C p t f t D 5t C p t D 5t C t find f0 t D 5t C t ( Hence f 0 t D 5 t C ) t D 5t 0 5 t i.e. f (t) D 5 i.e. Problem. to x Hence y D t 5 D 5 p t 5 since t 0 D Differentiate y D x C x y D x C 4 C 4x D x C 4x C 4 x x C x with respect D x x C 4x x C 4 x dy dx D x C 0 C 4 x D x 0 4x D 4 x since x 0 D Now try the following exercise Exercise 0 Further problems on differentiation of y = ax n by the general rule (Answers on page 65) In Problems to 8, determine the differential coefficients with respect to the variable.. y D 7x 4. y D p x i.e. i.e. y D 5 x 4 x C 4p x 5 C 7 y D 5 x 4x C 4x 5/ C 7 from the laws of dy dx D indices (see Chapter ) ( ) x 4 x 5 D 6 5 x C x 4 C 0x / dy dx = 6 5 x + x 4 + 0p x C 4 ( ) 5 x 5/ C 0. y D p t 4. y D 6 C x 5. y D x p C 6. y D 5 x x x p C x 7 7. y D t 8. y D x C 9. Using the general rule for ax n check the results of Problems to of Exercise 09, page. 0. Differentiate f x D 6x x C 5andfindthe gradient of the curve at (a) x D, and (b) x D.. Find the differential coefficient of y D x C x 4x and determine the gradient. of the curve at x D.. Determine the derivative of y D x C 4x C 7and determine the gradient of the curve at x D.5.

245 4 Basic Engineering Mathematics 0.6 Differentiation of sine and cosine functions Figure 0.5(a) shows a graph of y D sin q. The gradient is continually changing as the curve moves from 0 to A to B to C to D. The gradient, given by dy, may be plotted dq in a corresponding position below y D sin q, as shown in Figure 0.5(b). If a similar exercise is followed for y D cos q then the graphs of Figure 0.6 result, showing dy to be a graph of dq sin q, but displaced by p radians. y y = cos q + + y A y = sin q (a) 0 p p p p q radians (a) (b) + dy dx 0 0 p B p d dq (sin q) = cos q A p p p C C p D p D p q radians q radians (b) Fig dy dq 0 p p p d dq (cos q) = sin q p q radians Fig. 0.5 B (i) At 0, the gradient is positive and is at its steepest. Hence 0 0 is a maximum positive value. (ii) Between 0 and A the gradient is positive but is decreasing in value until at A the gradient is zero, shown as A 0. (iii) Between A and B the gradient is negative but is increasing in value until at B the gradient is at its steepest. Hence B 0 is a maximum negative value. (iv) If the gradient of y D sin q is further investigated between B and C and C and D then the resulting graph of dy is seen to be a cosine wave. dq Hence the rate of change of sin q is cos q, i.e. if y= sin q then It may also be shown that: if and if y= sin aq, dy dq = cos q dy = a cos aq dq (where a is a constant) dy y= sin(aq + a), = a cos(aq + a) dq (where a and a are constants) If each point on the curve y D sin q (as shown in Figure 0.5(a)) were to be made negative, (i.e. C p is made p, p is made C p, and so on) then the graph shown in Figure 0.6(b) would result. This latter graph therefore represents the curve of sin q. Thus, if y = cos q, dy dq = sin q It may also be shown that: if y = cos aq, dy = a sin aq dq (where a is a constant) and dy if y = cos(aq + a), = a sin(aq + a) dq 4 (where a and a are constants) Problem. Differentiate the following with respect to the variable: (a) y D sin 5q (b) f t D cos t (a) y D sin 5q dy D 5cos 5q D 0 cos 5q dq (b) f t D cos t f 0 t D sin t D 6sin t from equation from equation

246 Introduction to differentiation 5 Problem. Find the differential coefficient of y D 7sin x cos 4x y D 7sin x cos 4x dy D 7 cos x 4sin 4x from dx equations () and () = 4 cos x + sin 4x Problem 4. Differentiate the following with respect to the variable: (a) f q D 5sin 00pq 0.40 (b) f t D cos 5t C 0.0 (a) If f q D 5sin 00pq 0.40 f (q) D 5[00p cos 00pq 0.40 ] from equation (), where a D 00p = 500p cos(00pq 0.40) (b) If f t D cos 5t C 0.0 f (t) D [ 5sin 5t C 0.0 ] from equation (4), where a D 5 = 0 sin(5t + 0.0) Problem 5. An alternating voltage is given by: v D 00 sin 00t volts, where t is the time in seconds. Calculate the rate of change of voltage when (a) t D s and (b) t D 0.0 s v D 00 sin 00t volts. The rate of change of v is given by dv dt. dv D cos 00t D 0000 cos 00t dt (a) When t D s, dv D cos D cos dt cos means the cosine of radian (make sure your calculator is on radians not degrees). Hence dv = volts per second dt (b) When t D 0.0 s, dv D cos D 0000 cos dt Hence dv dt = 8 volts per second Now try the following exercise Exercise Further problems on the differentiation of sine and cosine functions (Answers on page 65). Differentiate with respect to x: (a) y D 4sin x (b) y D cos6x. Given f q D sinq 5cos q, findf 0 q. An alternating current is given by i D 5 sin 00t amperes, where t is the time in seconds. Determine the rate of change of current (i.e. di when t D 0.0 seconds. dt 4. v D 50 sin 40t volts represents an alternating voltage, v, where t is the time in seconds. At a time of 0 ð 0 seconds, find the rate of change of voltage (i.e. dv dt. 5. If f t D sin 4tC0. cos t 0.7 determine f 0 t. 0.7 Differentiation of e ax and ln ax A graph of y D e x is shown in Figure 0.7(a). The gradient of the curve at any point is given by dy and is continually dx changing. By drawing tangents to the curve at many points on the curve and measuring the gradient of the tangents, values of dy for corresponding values of x may be obtained. These dx values are shown graphically in Figure 0.7(b). The graph of dy dx against x is identical to the original graph of y D ex.it follows that: if y = e x dy, then dx = ex It may also be shown that if y = e ax dy, then dx = aeax Therefore if y D e 6x,then dy dx D 6e6x D e 6x A graph of y D ln x is shown in Figure 0.8(a). The gradient of the curve at any point is given by dy and is continually dx changing. By drawing tangents to the curve at many points on the curve and measuring the gradient of the tangents, values of dy for corresponding values of x may be obtained. These dx values are shown graphically in Figure 0.8(b). The graph of dy dy against x is the graph of dx dx D. It follows that: x dy if y = ln x, then dx = x

247 6 Basic Engineering Mathematics y y = e x Thus if y D ln 4x, then dy dx D x Problem 6. Differentiate the following with respect to the variable: (a) y D e x (b) f t D 4 e 5t Fig x (a) dy dx dy dx = e x 0 x (b) dy dx y x dy = dx x (a) y = ln x (a) If y D e x then dy dx D ex D 6e x (b) If f t D 4 e 5t D 4 e 5t,then f (t) D 4 5e 5t D 0 e 5t D 0 e 5t Problem 7. Differentiate y D 5ln x If y D 5ln x, then dy ( ) dx D 5 D 5 x x Now try the following exercise Exercise Further problems on the differentiation of e ax and ln ax (Answers on page 65). Differentiate with respect to x: (a) y D 5e x (b) y D 7e x. Given f q D 5ln q 4ln q, determine f 0 q. If f t D 4ln t C, evaluate f 0 t when t D Evaluate dy dx when x D, given y D e4x 5 e C x 8ln5x. Give the answer correct to significant figures. 0.8 Summary of standard derivatives The five main differential coefficients used in this chapter are summarised in Table 0.. Fig. 0.8 It may also be shown that x dy if y = ln ax, then dx = x (Note that in the latter expression, the constant a does not appear in the dy dx term). (b) Table 0. y or f(x) ax n sin ax cos ax e ax ln ax dy dx or f (x) anx n a cos ax a sin ax ae ax x

248 Introduction to differentiation 7 Problem 8. Find the gradient of the curve y D x 7x C at the point (, ) If y D x 7x C, then gradient D dy D 6x 7 dx At the point (, ), x D, hence gradient D 6 7 D Problem 9. If y D x sin 4x C ex C ln 5x, determine dy dx f x D 4x 5 x C x f 0 x D 0x 4 6x C f (x) =80x x or 4x(0x ) Problem. Given y D x 4 x C x p x, determine d y dx y D x 4 x C x p x y D x sin 4x C C ln 5x ex D x sin 4x C e x C ln 5x dy dx D x 4cos 4x C ( e x) C x D 6 x 8cos 4x e x + x Now try the following exercise Exercise Further problems on standard derivatives (Answers on page 65). Find the gradient of the curve y D x 4 C x x C 4 at the points (a) (0, 4) and (b) (, 8) i.e. i.e. D x 4x C x x ( ) dy (x dx D ) 4 x C dy dx D x C 8x x x d ( ) y ( x dx D 4x C 8 x 4 ) D 4x 4x 4 C x C 4 x d y 4 = 4x dx x + 4 x + 4 p x ( ) ( x ) x ( )( ) x. Differentiate: y D C ln x cos 5x C sin x x e x 0.9 Successive differentiation When a function y D f x is differentiated with respect to x the differential coefficient is written as dy dx or f0 x. If the expression is differentiated again, the second differential coefficient is obtained and is written as d y (pronounced dee dx two y by dee x squared) or f 00 x (pronounced f doubledash x). By successive differentiation further higher derivatives such as d y dx and d4 y dx 4 Thus if y D 5x 4, may be obtained. dy dx D 0x, d 4 y dx 4 D 0 and d5 y dx 5 D 0 Problem 0. d y dx D 60x, d y dx D 0x, If f x D 4x 5 x C x, find f 00 x Now try the following exercise Exercise 4 Further problems on successive differentiation (Answers on page 65). If y D x 4 C x x C, find (a) d y dx (b) d y dx. (a) Given f t D 5 t t C t p t C, determine f 00 t (b) Evaluate f 00 t in part (a) when t D. If y D sin t C cos t, find d y dt 4. If f q D ln 4q, show that f 00 q D q 0.0 Rates of change If a quantity y depends on and varies with a quantity x then the rate of change of y with respect to x is dy. Thus, for dx example, the rate of change of pressure p with height h is dp dh.

249 8 Basic Engineering Mathematics A rate of change with respect to time is usually just called the rate of change, the with respect to time being assumed. Thus, for example, a rate of change of current, i, is di dt and a rate of change of temperature, q, is dq, and so on. dt Problem. The length L metres of a certain metal rod at temperature t C is given by: L D C t C t. Determine the rate of change of length, in mm/ C, when the temperature is (a) 00 C and (b) 50 C The rate of change of length means dl dt Since length L D C t C t,then dl D C t dt (a) When t D 00 C, dl dt D C D m/ C D 0. mm= C (b) When t D 50 C, dl D C dt D m/ C D 0. mm= C Problem. The luminous intensity I candelas of a lamp at varying voltage V is given by: I D 5ð0 4 V. Determine the voltage at which the light is increasing at a rate of 0.4 candelas per volt. The rate of change of light with respect to voltage is given by di dv. Since I D 5 ð 0 4 V, di dv D 5 ð 0 4 V D 0 ð 0 4 V D 0 V When the light is increasing at 0.4 candelas per volt then C0.4 D 0 V, from which, voltage V D D 0.4 ð 0C D 400 volts Problem 4. Newton s law of cooling is given by: q D q 0 e kt, where the excess of temperature at zero time is q 0 C and at time t seconds is q C. Determine the rate of change of temperature after 50 s, given that q 0 D 5 C andk D 0.0 The rate of change of temperature is dq dt. Since q D q 0 e kt then dq dt D q 0 ke kt D kq 0 e kt When q 0 D 5, k D 0.0 and t D 50, then dq dt D e D 0.0e D 0.85 C=s Problem 5. The pressure p of the atmosphere at height h above ground level is given by p D p 0 e h/c, where p 0 is the pressure at ground level and c is a constant. Determine the rate of change of pressure with height when p 0 D 0 5 Pascals and c D 6. ð 0 4 at 550 metres. The rate of change of pressure with height is dp dh. Since p D p 0 e h/c then dp dh D p 0 ( c ) e h/c D p 0 c e h/c When p 0 D 0 5, c D 6. ð 0 4 and h D 550, then rate of change of pressure, dp 05 D dh 6. ð 0 4 e 550/6.ð04 D 0 6. e 0.05 D.57 Pa=m Now try the following exercise Exercise 5 Further problems on rates of change (Answers on page 65). An alternating current, i amperes, is given by i D 0 sin pft, wheref is the frequency in hertz and t the time in seconds. Determine the rate of change of current when t D ms, given that f D 50 Hz.. The luminous intensity, I candelas, of a lamp is given by I D 8 ð 0 4 V,whereV is the voltage. Find (a) the rate of change of luminous intensity with voltage when V D 00 volts, and (b) the voltage at which the light is increasing at a rate of 0.5 candelas per volt.. The voltage across the plates of a capacitor at any time t seconds is given by v D V e t/cr,wherev, C and R are constants. Given V D 00 volts, C D 0.0 ð 0 6 farads and R D ð 0 6 ohms, find (a) the initial rate of change of voltage, and (b) the rate of change of voltage after 0. s. 4. The pressure p of the atmosphere at height h above ground level is given by p D p 0 e h/c, where p 0 is the pressure at ground level and c is a constant. Determine the rate of change of pressure with height when p 0 D.0 ð 0 5 Pascals and c D 6.05 ð 0 4 at 450 metres.

250 Introduction to integration. The process of integration The process of integration reverses the process of differentiation. In differentiation, if f x D x then f 0 x D 4x. Thus, the integral of 4x is x, i.e. integration is the process of moving from f 0 x to f x. By similar reasoning, the integral of t is t. Integration is a process of summation or adding parts together and an elongated S, shown as, is used to replace the words the integral of. Hence, from above, 4x D x and t is t. In differentiation, the differential coefficient dy dx indicates that a function of x is being differentiated with respect to x, the dx indicating that it is with respect to x. In integration the variable of integration is shown by adding d (the variable) after the function to be integrated. Thus 4x dx means the integral of 4x with respect to x, and t dt means the integral of t with respect to t As stated above, the differential coefficient of x is 4x, hence 4x dx D x. However, the differential coefficient of x C7 is also 4x. Hence 4x dx could also be equal to x C 7. To allow for the possible presence of a constant, whenever the process of integration is performed, a constant c is added to the result. Thus 4x dx D x C c and t dt D t C c c is called the arbitrary constant of integration.. The general solution of integrals of the form ax n ax n dx = ax n+ n + +c Using this rule gives: (i) x 4 dx D x4c 4 C C c D 5 x 5 + c 4 (ii) 9 t dt dx D 4 ( t C ) C c 9 C D 4 ( t 4 ) C c D t 4 + c Both of these results may be checked by differentiation.. Standard integrals d From Chapter 0, sin ax D a cos ax dx Since integration is the reverse process of differentiation it follows that: a cos ax dx D sin ax C c or cos ax dx = sin ax + c a By similar reasoning sin ax dx = cos ax + c a e ax dx = a eax + c The general solution of integrals of the form ax n dx, where a and n are constants and n 6D, is given by: and x dx = ln x + c

251 40 Basic Engineering Mathematics From above, ax n dx D axnc C c except when n = n C When n D, then x dx D dx D ln x C c x A list of standard integrals is summarised in Table.. Table. Standard integrals (i) ax n dx D axnc n C C c (ii) (iii) (iv) (v) cos ax dx D sin ax C c a sin ax dx D cos ax C c a x Problem. The general rule is e ax dx D a eax C c dx D ln x C c Determine: (a) x dx ax n dx D axnc n C C c (except when n D ) (b) t dt (a) When a D andndthen x dx D xc C C c D x + c (b) When a D andndthen t dt D tc C C c D t4 4 C c D t 4 + c Each of these results may be checked by differentiating them. Problem. ( C 5 7 x 6x ) Determine: ( C 57 ) x 6x dx dx may be written as: 5 dxc x dx 6x dx 7 i.e. each term is integrated separately. (This splitting up of terms only applies for addition and subtraction). Hence ( C 57 ) x 6x dx ( ) 5 x C D x C 7 C ( ) 5 x D x C 7 6 xc C C c 6 x C c = x x x + c Note that when an integral contains more than one term there is no need to have an arbitrary constant for each; just a single constant at the end is sufficient. Problem 4. Determine: dx x dx D x dx. Using the standard integral, ax n dx when a D andn D gives: x x dx D x C C C c D x C c D x C c D x + c Problem. Determine (a) 8dx (b) x dx Problem 5. Determine: p x dx (a) 8dx is the same as 8x 0 dx and, using the general rule when a D 8andnD0gives: 8x 0 dx D 8x0C 0 C C c D 8x + c In general, if k is a constant then k dx = kx + c (b) When a D andnd, then x dx D x dx D xc C C c D x C c = x + c For fractional powers it is necessary to appreciate that np am D a m n, from which, p x D x. Hence, Problem 6. p x dx D D dx D x C x C c D C c C ( ) x D x C c D p x + c x Determine: 5 p x dx

252 Introduction to integration 4 5 p dx D x D 5 5x Problem 7. Determine: ( x ) x (a) dx x dx D 5x C C c D 5x C ( ) x C c D 0 p x + c (b) x dx (a) Rearranging into standard integral form gives: ( x ) ( x x ) x dx D dx x x x ( x D ) dx ( ) x C D C x C c ( ) x D x C c = 9 x x + c (b) Rearranging x dx gives: x C x dx D x xc C C xc C C c D x x C x C c D x x + x + c C c This problem shows that functions often have to be rearranged into the standard form of ax n dx before it is possible to integrate them. (b) From Table.(iii), sinx dx D Problem 9. sin x dx D ( ) cos x C c D cos x + c Determine: (a) 5e x dx 6 (b) (a) From Table.(iv), ( ) 5e x dx D 5 e x dx D 5 ex C c (b) 6 dx D ex D 5 ex + c D 6 6e x dx D 6 e x dx ( ) e x C c D e x C c D e x + c Problem 0. Determine: ( x ) (a) 5x dx (b) dx x (a) 5x dx D ( 5 )( ) x dx D 5 ( ) x D ln x + c (from Table.(v)) 5 ( x ) ( x (b) dx D x x ) dx x ( D x )dx D x x C ln x C c dx e x dx Problem 8. Determine: (a) 5cos x dx (b) sin x dx Now try the following exercise D x + ln x + c (a) From Table.(ii), 5cos x dx D 5 cos x dx ( ) D 5 sin x C c = 5 sin x + c Exercise 6 Further problems on standard integrals (Answers on page 65) Determine the following integrals:. (a) 4dx (b) 7x dx. (a) 5 x dx (b) 5 6 x dx

253 4 Basic Engineering Mathematics. (a) C x 4x dx (b) x C 5x dx ( x ) 5x 4. (a) dx (b) C x dx x 4 5. (a) dx (b) x 4x dx 4 px 6. (a) dx (b) p dx x 7. (a) cos x dx (b) 7sin x dx dx 8. (a) 4 ex dx (b) e 5x ( x ) 9. (a) x dx (b) dx x.4 Definite integrals Integrals containing an arbitrary constant c in their results are called indefinite integrals since their precise value cannot be determined without further information. Definite integrals are those in which limits are applied. If an expression is written as [x] b a b is called the upper limit and a the lower limit. The operation of applying the limits is defined as: [x] b a D b a The increase in the value of the integral x as x increases from to is written as x dx Applying the limits gives: [ x x ] ( ) ( ) dx D C c D C c C c ( ) D 9 C c C c D 8 Note that the c term always cancels out when limits are applied and it need not be shown with definite integrals. (a) Problem. (a) x dx Evaluate: (b) [ x x dx D ] D 4 x dx { } { } (b) (a) (b) ] 4 x dx D [4x x } } D {4 {4 { D f 9g 8 8 } { D fg 5 } D 8 or 8. Problem. (a) 0 0 x C x dx Evaluate: x C x dx D 0 (b) { D ( x 4 5x ) C x dx x [ x x C x dx D D 6 C 6 = ( x 4 5x ) C x dx D x D Problem. p/ 0 x 5x C dx C x } C f0c0g ( x 4 x or. 5x x [ x 4 ] 5x D 4 C x { D 4 5 } { 4 C 5 4 { D 4 5 } { C 4 5 } D Evaluate: p/ 0 sin x dx C x x ) ] dx 0 } C [ sinx dx D ( )] p/ cos x 0 D [ ] p/ cos x D [cos x]p/ 0 0 D [{ ( p )} ] cos fcos 0 g D [ cos p cos 0 ] D 6 D 4 or 4.5 D [ ] D D

254 Introduction to integration 4 Problem 4. Evaluate: 4cos t dt D 4cos t dt [ ( )] 4 sin t D 4 [sin t] D 4 [sin 6 sin ] 6. (a) 7. (a) 8. (a) 0 0 5cos x dx e t dt x dx (b) p/ p/4 (b) (b) sin x cos x dx e x dx ( x ) C dx x Note that limits of trigonometric functions are always expressed in radians thus, for example, sin 6 means the sine of 6 radians D Hence 4cos t dt D 4 [ ] D 4 [ ] D Area under a curve The area shown shaded in Figure. may be determined using approximate methods such as the trapezoidal rule, the mid-ordinate rule or Simpson s rule (see Chapter ) or, more precisely, by using integration. y Problem 5. Evaluate: (a) 4e x dx each correct to 4 significant figures. 4 (b) 4x du, y = f(x) (a) (b) 4 [ ] 4 4e x dx D ex D [e x ] D [e4 e ] D [ ] D 94.4 [ ] 4 4x du D 4 ln x D [ln 4 ln ] 4 D [.86 0] D Now try the following exercise Exercise 7 Further problems on definite integrals (Answers on page 65) Evaluate the following definite integrals (where necessary, correct to 4 significant figures):. (a). (a). (a) 4. (a) 5. (a) 4 p 0 p/ 5x dx x dx x C 4x dx p/6 (b) (b) (b) cos q dq (b) sin x dx (b) p/ 4 t dt x 4x C dx ( x C x ) x dx x 0 0 4cos x dx sin t dt Fig.. 0 x =a x =b The shaded area in Figure. is given by: shaded area = b a y dx = b a x f (x) dx Thus, determining the area under a curve by integration merely involves evaluating a definite integral, as shown in Section.4. There are several instances in engineering and science where the area beneath a curve needs to be accurately determined. For example, the areas between limits of a: velocity/time graph gives distance travelled, force/distance graph gives work done, voltage/current graph gives power, and so on. Should a curve drop below the x-axis, then y D f x becomes negative and f x dx is negative. When determining such areas by integration, a negative sign is placed before the integral. For the curve shown in Figure., the total shaded area is given by (area E C area F C area G). By integration, total shaded area = b a f (x) dx c (Note that this is not the same as d a f x dx). b + f (x) dx d c f (x) dx

255 44 Basic Engineering Mathematics y [This answer may be checked since the shaded area is a trapezium. y = f(x) Area of trapezium D (sum of parallel sides) 0 a E b F c G d x (perpendicular distance between parallel sides) D 5 C D 4 square units] Fig.. It it usually necessary to sketch a curve in order to check whether it crosses the x-axis. Problem 6. Determine the area enclosed by y D x C, the x-axis and ordinates x D andxd4. Problem 7. The velocity v of a body t seconds after a certain instant is: (t C 5)m/s. Find by integration how far it moves in the interval from t D 0totD4s. Since t C 5 is a quadratic expression, the curve v D t C 5 is a parabola cutting the v-axis at v D 5, as shown in Figure.4. y D x C is a straight line graph as shown in Figure., where the required area is shown shaded. y 0 8 y = x + v (m/s) v = t x 0 4 t (s) Fig.. By integration, shaded area 4 D y dx 4 D x C dx [ x ] 4 D C x D [ 6 C C ] D 4 square units Fig..4 The distance travelled is given by the area under the v/t curve (shown shaded in Figure.4). By integration, shaded area i.e. D 4 distance travelled = 6.67 m 0 v dt D 4 0 t C 5 dt [ t ] 4 D C 5t 0 ( 4 ) D C 5 4 0

256 Introduction to integration 45 Problem 8. Sketch the graph y D x C x 5x 6 between x D andx D and determine the area enclosed by the curve and the x-axis. A table of values is produced and the graph sketched as shown in Figure.5 where the area enclosed by the curve and the x-axis is shown shaded. x 0 x x x y y [{ D 4 5 } C 6 [{ 4 C 6 D D [{ } [ 5 ] [ 5 ] 4 D or { 8 4 } { }] [{ 4.08 square units }] C 8 { 4 5 }] C 6 } }] { Problem 9. Determine the area enclosed by the curve y D x C 4, the x-axis and ordinates x D andxd4 by (a) the trapezoidal rule, (c) Simpson s rule, and (b) the mid-ordinate rule, (d) integration. The curve y D x C 4 is shown plotted in Figure.6. y = x + x x 5x 6 x y y y = x + 4 x Fig..5 Shaded area D y dx y dx, the minus sign before the second integral being necessary since the enclosed area is below the x-axis. Hence, shaded area D [ x 4 D 4 C x x C x 5x 6 dx 5x x C x 5x 6 dx ] [ x 4 6x 4 C x 5x 6x ] Fig x The trapezoidal rule, the mid-ordinate rule and Simpson s rule are discussed in Chapter, page 77. (a) By the trapezoidal rule, area D (width of interval) [ ( ) first C last C sum of ordinate remaining ordinates]

257 46 Basic Engineering Mathematics Selecting 6 intervals each of width 0.5 gives: area D 0.5 [ 7 C 5 C 0.75 C 6 C.75 C C 40.75] D square units y y =sin x (b) By the mid-ordinate rule, area D (width of interval)(sum of mid-ordinates). Selecting 6 intervals, each of width 0.5 gives the midordinates as shown by the broken lines in Figure.6. Thus, area D C. C 9. C 6.7 C5.7 C 46. D square units (c) By Simpson s rule, area D ( )[( ) widthof first C last interval ordinates ( ) ( )] sum of even sum of remaining C4 C ordinates odd ordinates Selecting 6 intervals, each of width 0.5 gives: area D 0.5 [ 7 C 5 C C.75 C C 6 C ] D 75 square units (d) By integration, shaded area D 4 y dx D 4 x C 4 dx D [x C 4x] 4 D 64 C 6 C 4 D 75 square units Integration gives the precise value for the area under a curve. In this case Simpson s rule is seen to be the most accurate of the three approximate methods. Problem 0. Find the area enclosed by the curve y D sin x, thex-axis and the ordinates x D 0and x D p Asketchofy D sin x is shown in Figure.7. (Note that y D sin x has a period of p,i.e.p radians). p/ p/ Shaded area D y dx D sin x dx D [ ] p/ cos x D 0 { cos p 0 } { cos 0 } 0 Fig..7 0 p/ p/ p x D { ( )} { } D 4 C D or 0.75 square units 4 Now try the following exercise Exercise 8 Further problems on area under curves (Answers on page 65). Show by integration that the area of a rectangle formed by the line y D 4, the ordinates x D and x D 6andthex-axis is 0 square units. Show by integration that the area of the triangle formed by the line y D x, the ordinates x D 0 and x D 4andthex-axis is 6 square units.. Sketch the curve y D x C between x D and x D 4. Determine by integration the area enclosed by the curve, the x-axis and ordinates x D and x D. Use an approximate method to find the area and compare your result with that obtained by integration. 4. The force F newtons acting on a body at a distance x metres from a fixed point is given by: F D x C x. If work done D x x F dx, determine the work done when the body moves from the position where x D m to that when x D m. In Problems 5 to 9, sketch graphs of the given equations, and then find the area enclosed between the curves, the horizontal axis and the given ordinates. 5. y D 5x; x D, x D 4 6. y D x x C ; x D, x D 7. y D sin x; x D 0, x D p 4

258 Introduction to integration y D 5cos t; t D 0, t D p 6 9. y D x x ; x D 0, x D 0. The velocity v of a vehicle t seconds after a certain instant is given by: v D t C4 m/s. Determine how far it moves in the interval from t D stotd5s. 6. Given y D x x C p x find d y dx (6) 7. Newtons law of cooling is given by: q D q 0 e kt, where the excess of temperature at zero time is q 0 C and at time t seconds is q C. Determine the rate of change of temperature after 40 s, correct to decimal places, given that q 0 D 6 C andk D 0.0 (4) Assignment 5 This assignment covers the material contained in Chapters 0 and. The marks for each question are shown in brackets at the end of each question 8. Determine the following: (a) x C 5x dx (b) 4cos x p x dx 4. Differentiate from first principles f x D x C x (5). If y D 4x 5 p x determine dy (4) dx. Given f x D x x find f0 x (4) 4. Differentiate the following with respect to the variable: (a) y D 4sin x 5cos x (b) y D e t ln 7t (4) 5. Find the gradient of the curve y D 5x C x at the point (, 4) (4) 9. Evaluate the following definite integrals, correct to 4 significant figures: ( (a) x ) dx x (b) (c) p/ 0 0 sin x dx (e x px ) dx 0 0. Sketch the curve y D x C 5 between x D and x D. Determine, by integration, the area enclosed by the curve, the x-axis and ordinates x D and x D. (5)

259 List of formulae Laws of indices: a m ð a n D a mcn a m a n D am n a m n D a mn a m/n D np a m a n D a n a 0 D Exponential series: e x D C x C x! C x CÐÐÐ (valid for all values of x)! Theorem of Pythagoras: Quadratic formula: b D a C c If ax C bx C c D 0 then x D b š p b 4ac a A c b Equation of a straight line: y D mx C c Fig. F B a C Definition of a logarithm: Areas of plane figures: If y D a x then x D log a y (i) Rectangle Area D l ð b Laws of logarithms: b log A ð B D log A C log B ( ) A log D log A log B B l log A n D n ð log A Fig. F

260 List of formulae 49 (ii) Parallelogram Area D b ð h h For a sector of circle: arc length, s D q pr D rq q in rad 60 shaded area D q 60 pr D r q q in rad Equation of a circle, centre at origin, radius r: x C y D r b Equation of a circle, centre at (a, b), radius r: Fig. F x a C y b D r (iii) Trapezium Area D a C b h a Volumes and surface areas of regular solids: (i) Rectangular prism (or cuboid) h Volume D l ð b ð h Surface area D bh C hl C lb b Fig. F4 (iv) Triangle Area D ð b ð h h l h Fig. F7 b Fig. F5 b (ii) Cylinder Volume D pr h Total surface area D prh C pr (v) Circle Area D pr Circumference D pr r h r θ s r Fig. F8 Fig. F6 Radian measure: p radians D 60 degrees (iii) Pyramid If area of base D A and perpendicular height D h then : Volume D ð A ð h

261 50 List of formulae Fig. F9 h Total surface area D sum of areas of triangles forming sides C area of base (iv) Cone Volume D pr h Curved surface area D prl Total surface area D prl C pr l h Mid-ordinate rule Area ³ (width of interval)(sum of mid-ordinates) Simpson s rule Area ³ ( )[( ) width of first C last interval ordinate ( ) ( )] sum of even sum of remaining C 4 C ordinates odd ordinates Mean or average value of a waveform mean value, y D D area under curve length of base sum of mid-ordinates number of mid-ordinates Fig. F0 (v) Sphere Volume D 4 pr Surface area D 4pr r Triangle formulae: Sine rule: Cosine rule: a sin A D b sin B D c sin C a D b C c bc cos A c A b B a C Fig. F r Area of any triangle D ð base ð perpendicular height Fig. F Areas of irregular figures by approximate methods: Trapezoidal rule Area ³ ( width of interval )[ ( ) first C last ordinate ] C sum of remaining ordinates D ab sin C or ac sin B or bc sin A D [s s a s b s c ] where s D a C b C c For a general sinusoidal function y = Asin(! t ± a), then A D amplitude! D angular velocity D pf rad/s p D periodic time T seconds!! D frequency,f hertz p a D angle of lead or lag (compared with y D A sin!t