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## Basic engineering mathematics e5Document Transcript

• Basic Engineering Mathematics
• In memory of Elizabeth
• Basic Engineering MathematicsFifth editionJohn Bird, BSc(Hons), CMath, CEng, CSci, FIMA, FIET, MIEE, FIIE, FCollTAMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORDPARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYONewnes is an imprint of Elsevier
• ContentsPreface ixAcknowledgements xInstructor’s Manual xi1 Basic arithmetic 11.1 Introduction 11.2 Revision of addition and subtraction 11.3 Revision of multiplication and division 31.4 Highest common factors and lowestcommon multiples 51.5 Order of precedence and brackets 62 Fractions 92.1 Introduction 92.2 Adding and subtracting fractions 102.3 Multiplication and division of fractions 122.4 Order of precedence with fractions 13Revision Test 1 153 Decimals 163.1 Introduction 163.2 Converting decimals to fractions andvice-versa 163.3 Signiﬁcant ﬁgures and decimal places 173.4 Adding and subtracting decimal numbers 183.5 Multiplying and dividing decimal numbers 194 Using a calculator 224.1 Introduction 224.2 Adding, subtracting, multiplying anddividing 224.3 Further calculator functions 234.4 Evaluation of formulae 285 Percentages 335.1 Introduction 335.2 Percentage calculations 335.3 Further percentage calculations 355.4 More percentage calculations 36Revision Test 2 396 Ratio and proportion 406.1 Introduction 406.2 Ratios 406.3 Direct proportion 426.4 Inverse proportion 457 Powers, roots and laws of indices 477.1 Introduction 477.2 Powers and roots 477.3 Laws of indices 488 Units, preﬁxes and engineering notation 538.1 Introduction 538.2 SI units 538.3 Common preﬁxes 538.4 Standard form 568.5 Engineering notation 57Revision Test 3 609 Basic algebra 619.1 Introduction 619.2 Basic operations 619.3 Laws of indices 6410 Further algebra 6810.1 Introduction 6810.2 Brackets 6810.3 Factorization 6910.4 Laws of precedence 7111 Solving simple equations 7311.1 Introduction 7311.2 Solving equations 7311.3 Practical problems involving simpleequations 77Revision Test 4 8212 Transposing formulae 8312.1 Introduction 8312.2 Transposing formulae 8312.3 Further transposing of formulae 8512.4 More difﬁcult transposing of formulae 8713 Solving simultaneous equations 9013.1 Introduction 9013.2 Solving simultaneous equations in twounknowns 9013.3 Further solving of simultaneous equations 92
• vi Contents13.4 Solving more difﬁcult simultaneousequations 9413.5 Practical problems involving simultaneousequations 9613.6 Solving simultaneous equations in threeunknowns 99Revision Test 5 10114 Solving quadratic equations 10214.1 Introduction 10214.2 Solution of quadratic equations byfactorization 10214.3 Solution of quadratic equations by‘completing the square’ 10514.4 Solution of quadratic equations byformula 10614.5 Practical problems involving quadraticequations 10814.6 Solution of linear and quadratic equationssimultaneously 11015 Logarithms 11115.1 Introduction to logarithms 11115.2 Laws of logarithms 11315.3 Indicial equations 11515.4 Graphs of logarithmic functions 11616 Exponential functions 11816.1 Introduction to exponential functions 11816.2 The power series for ex 11916.3 Graphs of exponential functions 12016.4 Napierian logarithms 12216.5 Laws of growth and decay 125Revision Test 6 12917 Straight line graphs 13017.1 Introduction to graphs 13017.2 Axes, scales and co-ordinates 13017.3 Straight line graphs 13217.4 Gradients, intercepts and equationsof graphs 13417.5 Practical problems involving straight linegraphs 14118 Graphs reducing non-linear laws to linear form 14718.1 Introduction 14718.2 Determination of law 14718.3 Revision of laws of logarithms 15018.4 Determination of law involving logarithms 15019 Graphical solution of equations 15519.1 Graphical solution of simultaneousequations 15519.2 Graphical solution of quadratic equations 15619.3 Graphical solution of linear and quadraticequations simultaneously 16019.4 Graphical solution of cubic equations 161Revision Test 7 16320 Angles and triangles 16520.1 Introduction 16520.2 Angular measurement 16520.3 Triangles 17120.4 Congruent triangles 17520.5 Similar triangles 17620.6 Construction of triangles 17921 Introduction to trigonometry 18121.1 Introduction 18121.2 The theorem of Pythagoras 18121.3 Sines, cosines and tangents 18321.4 Evaluating trigonometric ratios of acuteangles 18521.5 Solving right-angled triangles 18821.6 Angles of elevation and depression 191Revision Test 8 19322 Trigonometric waveforms 19522.1 Graphs of trigonometric functions 19522.2 Angles of any magnitude 19622.3 The production of sine and cosine waves 19822.4 Terminology involved with sine andcosine waves 19922.5 Sinusoidal form: Asin(ωt ± α) 20223 Non-right-angled triangles and some practicalapplications 20523.1 The sine and cosine rules 20523.2 Area of any triangle 20523.3 Worked problems on the solution oftriangles and their areas 20623.4 Further worked problems on the solutionof triangles and their areas 20723.5 Practical situations involving trigonometry 20923.6 Further practical situations involvingtrigonometry 21124 Cartesian and polar co-ordinates 21424.1 Introduction 21424.2 Changing from Cartesian to polarco-ordinates 21424.3 Changing from polar to Cartesianco-ordinates 21624.4 Use of Pol/Rec functions on calculators 217
• Contents viiRevision Test 9 21825 Areas of common shapes 21925.1 Introduction 21925.2 Common shapes 21925.3 Areas of common shapes 22125.4 Areas of similar shapes 22926 The circle 23026.1 Introduction 23026.2 Properties of circles 23026.3 Radians and degrees 23226.4 Arc length and area of circles and sectors 23326.5 The equation of a circle 236Revision Test 10 23827 Volumes of common solids 24027.1 Introduction 24027.2 Volumes and surface areas of commonshapes 24027.3 Summary of volumes and surface areas ofcommon solids 24727.4 More complex volumes and surface areas 24727.5 Volumes and surface areas of frusta ofpyramids and cones 25227.6 Volumes of similar shapes 25628 Irregular areas and volumes, and mean values 25728.1 Areas of irregular ﬁgures 25728.2 Volumes of irregular solids 25928.3 Mean or average values of waveforms 260Revision Test 11 26429 Vectors 26629.1 Introduction 26629.2 Scalars and vectors 26629.3 Drawing a vector 26629.4 Addition of vectors by drawing 26729.5 Resolving vectors into horizontal andvertical components 26929.6 Addition of vectors by calculation 27029.7 Vector subtraction 27429.8 Relative velocity 27629.9 i, j and k notation 27730 Methods of adding alternating waveforms 27830.1 Combining two periodic functions 27830.2 Plotting periodic functions 27830.3 Determining resultant phasors by drawing 28030.4 Determining resultant phasors by the sineand cosine rules 28130.5 Determining resultant phasors byhorizontal and vertical components 283Revision Test 12 28631 Presentation of statistical data 28831.1 Some statistical terminology 28831.2 Presentation of ungrouped data 28931.3 Presentation of grouped data 29232 Mean, median, mode and standard deviation 29932.1 Measures of central tendency 29932.2 Mean, median and mode for discrete data 29932.3 Mean, median and mode for grouped data 30032.4 Standard deviation 30232.5 Quartiles, deciles and percentiles 30333 Probability 30633.1 Introduction to probability 30633.2 Laws of probability 307Revision Test 13 31234 Introduction to differentiation 31334.1 Introduction to calculus 31334.2 Functional notation 31334.3 The gradient of a curve 31434.4 Differentiation from ﬁrst principles 31534.5 Differentiation of y = axn by thegeneral rule 31534.6 Differentiation of sine and cosine functions 31834.7 Differentiation of eax and lnax 32034.8 Summary of standard derivatives 32134.9 Successive differentiation 32234.10 Rates of change 32335 Introduction to integration 32535.1 The process of integration 32535.2 The general solution of integrals of theform axn 32535.3 Standard integrals 32635.4 Deﬁnite integrals 32835.5 The area under a curve 330Revision Test 14 335List of formulae 336Answers to practice exercises 340Index 356
• viii ContentsWebsite Chapters(Goto:http://www.booksite.elsevier.com/newnes/bird)Preface iv36 Number sequences 136.1 Simple sequences 136.2 The n’th term of a series 136.3 Arithmetic progressions 236.4 Geometric progressions 537 Binary, octal and hexadecimal 937.1 Introduction 937.2 Binary numbers 937.3 Octal numbers 1237.4 Hexadecimal numbers 1538 Inequalities 1938.1 Introduction to inequalities 1938.2 Simple inequalities 1938.3 Inequalities involving a modulus 2038.4 Inequalities involving quotients 2138.5 Inequalities involving square functions 2238.6 Quadratic inequalities 2339 Graphs with logarithmic scales 2539.1 Logarithmic scales and logarithmicgraph paper 2539.2 Graphs of the form y = axn 2539.3 Graphs of the form y = abx 2839.4 Graphs of the form y = aekx 29Revision Test 15 32Answers to practice exercises 33
• AcknowledgementsThe publisher wishes to thank CASIO Electronic Co.Ltd, London for permission to reproduce the image ofthe Casio fx-83ES calculator on page 23.The publishers also wish to thank the AutomobileAsso-ciation for permission to reproduce a map of Portsmouthon page 131.
• Instructor’s ManualFull worked solutions and mark scheme for all theAssignments are contained in this Manual which isavailable to lecturers only.To download the Instructor’s Manual visit http://www.booksite.elsevier.com/newnes/bird
• Chapter 1Basic arithmetic1.1 IntroductionWhole numbers are called integers. +3,+5 and +72are examples of positive integers; −13,−6 and −51are examples of negative integers. Between positiveand negative integers is the number 0 which is neitherpositive nor negative.The four basic arithmetic operators are add (+), subtract(−), multiply (×) and divide (÷).It is assumed that adding, subtracting, multiplying anddividing reasonably small numbers can be achievedwithout a calculator. However, if revision of this areais needed then some worked problems are included inthe following sections.When unlike signs occur together in a calculation, theoverall sign is negative. For example,3 + (−4) = 3 + −4 = 3 − 4 = −1and(+5) × (−2) = −10Like signs together give an overall positive sign. Forexample,3 − (−4) = 3 − −4 = 3 + 4 = 7and(−6) × (−4) = +241.2 Revision of addition andsubtractionYou can probably already add two or more numberstogether and subtract one number from another. How-ever, if you need a revision then the following workedproblems should be helpful.Problem 1. Determine 735 + 167H T U7 3 5+ 1 6 79 0 21 1(i) 5 + 7 = 12. Place 2 in units (U) column. Carry 1in the tens (T) column.(ii) 3 + 6 + 1 (carried) = 10. Place the 0 in the tenscolumn. Carry the 1 in the hundreds (H) column.(iii) 7 + 1 + 1 (carried) = 9. Place the 9 in the hun-dreds column.Hence, 735 + 167 = 902Problem 2. Determine 632 − 369H T U6 3 2− 3 6 92 6 3(i) 2 − 9 is not possible; therefore ‘borrow’ 1 fromthe tens column (leaving 2 in the tens column). Inthe units column, this gives us 12 − 9 = 3.(ii) Place 3 in the units column.(iii) 2 − 6 is not possible; therefore ‘borrow’ 1 fromthe hundreds column (leaving 5 in the hun-dreds column). In the tens column, this gives us12 − 6 = 6.(iv) Place the 6 in the tens column.DOI: 10.1016/B978-1-85617-697-2.00001-6
• 2 Basic Engineering Mathematics(v) 5 − 3 = 2.(vi) Place the 2 in the hundreds column.Hence, 632 − 369 = 263Problem 3. Add 27,−74,81 and −19This problem is written as 27 − 74 + 81 − 19.Adding the positive integers: 2781Sum of positive integers is 108Adding the negative integers: 7419Sum of negative integers is 93Taking the sum of the negative integersfrom the sum of the positive integers gives 108−9315Thus, 27 − 74 + 81 − 19 = 15Problem 4. Subtract −74 from 377This problem is written as 377 − −74. Like signstogether give an overall positive sign, hence377 − −74 = 377 + 74 3 7 7+ 7 44 5 1Thus, 377 − −74 = 451Problem 5. Subtract 243 from 126The problem is 126 − 243. When the second number islarger than the ﬁrst, take the smaller number from thelarger and make the result negative. Thus,126 − 243 = −(243 − 126) 2 4 3− 1 2 61 1 7Thus, 126 − 243 = −117Problem 6. Subtract 318 from −269The problem is −269 − 318. The sum of the negativeintegers is2 6 9+ 3 1 85 8 7Thus, −269 − 318 = −587Now try the following Practice ExercisePracticeExercise 1 Further problems onaddition and subtraction (answers onpage 340)In Problems 1 to 15, determine the values of theexpressions given, without using a calculator.1. 67kg − 82 kg + 34kg2. 73m − 57m3. 851mm − 372mm4. 124− 273 + 481 − 3985. £927 − £114+ £182 − £183 − £2476. 647 − 8727. 2417 − 487 + 2424− 1778 − 47128. −38419 − 2177 + 2440− 799 + 28349. £2715 − £18250+ £11471 − £1509 +£11327410. 47 + (−74) − (−23)11. 813 − (−674)12. 3151 − (−2763)13. 4872 g− 4683g14. −23148 − 4772415. \$53774− \$3844116. Holes are drilled 35.7mm apart in a metalplate. If a row of 26 holes is drilled, deter-mine the distance, in centimetres, betweenthe centres of the ﬁrst and last holes.17. Calculate the diameter d and dimensions Aand B for the template shown in Figure 1.1.All dimensions are in millimetres.
• Basic arithmetic 3126050 38120110BAdFigure 1.11.3 Revision of multiplication anddivisionYou can probably already multiply two numberstogether and divide one number by another. However, ifyou need a revision then the followingworked problemsshould be helpful.Problem 7. Determine 86 × 7H T U8 6× 76 0 24(i) 7 × 6 = 42. Place the 2 in the units (U) columnand ‘carry’ the 4 into the tens (T) column.(ii) 7 × 8 = 56;56 + 4 (carried) = 60. Place the 0 inthe tens column and the 6 in the hundreds (H)column.Hence, 86 × 7 = 602A good grasp of multiplication tables is needed whenmultiplying such numbers; a reminder of the multipli-cation table up to 12 × 12 is shown below. Conﬁdencewith handling numbers will be greatly improved if thistable is memorized.Problem 8. Determine 764× 387 6 4× 3 86 1 1 22 2 9 2 02 9 0 3 2Multiplication table× 2 3 4 5 6 7 8 9 10 11 122 4 6 8 10 12 14 16 18 20 22 243 6 9 12 15 18 21 24 27 30 33 364 8 12 16 20 24 28 32 36 40 44 485 10 15 20 25 30 35 40 45 50 55 606 12 18 24 30 36 42 48 54 60 66 727 14 21 28 35 42 49 56 63 70 77 848 16 24 32 40 48 56 64 72 80 88 969 18 27 36 45 54 63 72 81 90 99 10810 20 30 40 50 60 70 80 90 100 110 12011 22 33 44 55 66 77 88 99 110 121 13212 24 36 48 60 72 84 96 108 120 132 144
• 4 Basic Engineering Mathematics(i) 8 × 4 = 32. Place the 2 in the units column andcarry 3 into the tens column.(ii) 8 × 6 = 48;48 + 3 (carried) = 51. Place the 1 inthe tens column and carry the 5 into the hundredscolumn.(iii) 8 × 7 = 56;56 + 5 (carried) = 61. Place 1 in thehundredscolumn and 6 in thethousandscolumn.(iv) Place 0 in the units column under the 2.(v) 3 × 4 = 12. Place the 2 in the tens column andcarry 1 into the hundreds column.(vi) 3 × 6 = 18;18 + 1 (carried) = 19. Place the 9 inthe hundreds column and carry the 1 into thethousands column.(vii) 3 × 7 = 21;21 + 1 (carried) = 22. Place 2 in thethousands column and 2 in the ten thousandscolumn.(viii) 6112+ 22920 = 29032Hence, 764 × 38 = 29032Again, knowing multiplication tables is ratherimportantwhen multiplying such numbers.It is appreciated, of course, that such a multiplicationcan,and probably will,beperformed using a calculator.However, there are times when a calculator may not beavailable and it is then useful to be able to calculate the‘long way’.Problem 9. Multiply 178 by −46When the numbers have different signs, the result willbe negative. (With this in mind, the problem can nowbe solved by multiplying 178 by 46). Following theprocedure of Problem 8 gives1 7 8× 4 61 0 6 87 1 2 08 1 8 8Thus, 178 × 46 = 8188 and 178 × (−46) = −8188Problem 10. Determine 1834 ÷ 72627 1834(i) 7 into 18 goes 2, remainder 4. Place the 2 abovethe 8 of 1834 and carry the 4 remainder to thenext digit on the right, making it 43.(ii) 7 into 43 goes 6, remainder 1. Place the 6 abovethe 3 of 1834 and carry the 1 remainder to thenext digit on the right, making it 14.(iii) 7 into 14 goes 2, remainder 0. Place 2 above the4 of 1834.Hence, 1834÷ 7 = 1834/7 =18347= 262.The method shown is called short division.Problem 11. Determine 5796 ÷ 1248312 5796489996363600(i) 12 into 5 won’t go. 12 into 57 goes 4; place 4above the 7 of 5796.(ii) 4 × 12 = 48; place the 48 below the 57 of 5796.(iii) 57 − 48 = 9.(iv) Bring down the 9 of 5796 to give 99.(v) 12 into 99 goes 8; place 8 above the 9 of 5796.(vi) 8 × 12 = 96; place 96 below the 99.(vii) 99 − 96 = 3.(viii) Bring down the 6 of 5796 to give 36.(ix) 12 into 36 goes 3 exactly.(x) Place the 3 above the ﬁnal 6.(xi) Place the 36 below the 36.(xii) 36 − 36 = 0.Hence, 5796 ÷ 12 = 5796/12 =579612= 483.The method shown is called long division.
• Basic arithmetic 5Now try the following Practice ExercisePracticeExercise 2 Further problems onmultiplication and division (answers onpage 340)Determine the values of the expressions given inproblems 1 to 9, without using a calculator.1. (a) 78 × 6 (b) 124 × 72. (a) £261 × 7 (b) £462 × 93. (a) 783kg × 11 (b) 73kg × 84. (a) 27mm × 13 (b) 77mm × 125. (a) 448 × 23 (b) 143 × (−31)6. (a) 288m ÷ 6 (b) 979m ÷ 117. (a)18137(b)896168. (a)2142413(b) 15900 ÷ 159. (a)8873711(b) 46858 ÷ 1410. A screw has a mass of 15grams. Calculate,in kilograms, the mass of 1200 such screws(1kg = 1000g).1.4 Highest common factors andlowest common multiplesWhen two or more numbers are multiplied together, theindividualnumbers are called factors. Thus, a factoris anumber which divides into another number exactly. Thehighest common factor (HCF) is the largest numberwhich divides into two or more numbers exactly.For example, consider the numbers 12 and 15.The factors of 12 are 1, 2, 3, 4, 6 and 12 (i.e. all thenumbers that divide into 12).The factors of 15 are 1, 3, 5 and 15 (i.e. all the numbersthat divide into 15).1 and 3 are the only common factors; i.e., numberswhich are factors of both 12 and 15.Hence, the HCF of 12 and 15 is 3 since 3 is the highestnumber which divides into both 12 and 15.A multiple is a number which contains another numberan exact number of times. The smallest number whichis exactly divisible by each of two or more numbers iscalled the lowest common multiple (LCM).For example, the multiples of 12 are 12, 24, 36, 48,60, 72,... and the multiples of 15 are 15, 30, 45,60, 75,...60 is a common multiple (i.e. a multiple of both 12 and15) and there are no lower common multiples.Hence, the LCM of 12 and 15 is 60 since 60 is thelowest number that both 12 and 15 divide into.Here are some further problems involving the determi-nation of HCFs and LCMs.Problem 12. Determine the HCF of the numbers12, 30 and 42Probably the simplest way of determining an HCF is toexpress each number in terms of its lowest factors. Thisis achieved by repeatedly dividing by the prime numbers2, 3, 5, 7, 11, 13, … (where possible) in turn. Thus,12 = 2 × 2 × 330 = 2 × 3 × 542 = 2 × 3 × 7The factors which are common to each of the numbersare 2 in column 1 and 3 in column 3, shown by thebroken lines. Hence, the HCF is 2 × 3; i.e., 6. That is,6 is the largest number which will divide into 12, 30and 42.Problem 13. Determine the HCF of the numbers30, 105, 210 and 1155Using the method shown in Problem 12:30 = 2 × 3 × 5105 = 3 × 5 × 7210 = 2 × 3 × 5 × 71155 = 3 × 5 × 7 × 11The factors which are common to each of the numbersare 3 in column 2 and 5 in column 3. Hence, the HCFis 3 × 5 = 15.Problem 14. Determine the LCM of the numbers12, 42 and 90
• 6 Basic Engineering MathematicsThe LCM is obtained by ﬁnding the lowest factors ofeach of the numbers, as shown in Problems 12 and 13above, and then selecting the largest group of any of thefactors present. Thus,12 = 2 × 2 × 342 = 2 × 3 × 790 = 2 × 3 × 3 × 5The largest group of any of the factors present is shownby the broken lines and are 2 × 2 in 12, 3 × 3 in 90, 5 in90 and 7 in 42.Hence, the LCM is 2 × 2 × 3 × 3 × 5 × 7 = 1260 andis the smallest number which 12, 42 and 90 will alldivide into exactly.Problem 15. Determine the LCM of the numbers150, 210, 735 and 1365Using the method shown in Problem 14 above:150 = 2 × 3 × 5 × 5210 = 2 × 3 × 5 × 7735 = 3 × 5 × 7 × 71365 = 3 × 5 × 7 × 13Hence, the LCM is 2 × 3 × 5 × 5 × 7 × 7 × 13 =95550.Now try the following Practice ExercisePracticeExercise 3 Further problems onhighest common factors and lowest commonmultiples (answers on page 340)Find (a) the HCF and (b) the LCM of the followinggroups of numbers.1. 8, 12 2. 60, 723. 50, 70 4. 270, 9005. 6, 10, 14 6. 12, 30, 457. 10, 15, 70, 105 8. 90, 105, 3009. 196, 210, 462, 910 10. 196, 350, 7701.5 Order of precedence and brackets1.5.1 Order of precedenceSometimes addition, subtraction, multiplication, divi-sion, powers and brackets may all be involved in acalculation. For example,5 − 3 × 4 + 24 ÷ (3 + 5) − 32This is an extreme example but will demonstrate theorder that is necessary when evaluating.When we read, we read from left to right. However,with mathematics there is a deﬁnite order of precedencewhich we need to adhere to. The order is as follows:BracketsOrder (or pOwer)DivisionMultiplicationAdditionSubtractionNotice that the ﬁrst letters of each word spell BOD-MAS, a handy aide-m´emoire. Order means pOwer. Forexample, 42 = 4 × 4 = 16.5 − 3 × 4 + 24 ÷ (3 + 5) − 32 is evaluated asfollows:5 − 3 × 4 + 24 ÷ (3 + 5) − 32= 5 − 3 × 4 + 24 ÷ 8 − 32(Bracket is removed and3 + 5 replaced with 8)= 5 − 3 × 4 + 24 ÷ 8 − 9 (Order means pOwer; inthis case, 32= 3 × 3 = 9)= 5 − 3 × 4 + 3 − 9 (Division: 24 ÷ 8 = 3)= 5 − 12 + 3 − 9 (Multiplication: − 3 × 4 = −12)= 8 − 12 − 9 (Addition: 5 + 3 = 8)= −13 (Subtraction: 8 − 12 − 9 = −13)In practice, it does not matter if multiplicationis per-formed before divisionor if subtraction is performedbefore addition. What is important is that the pro-cessofmultiplicationanddivisionmustbecompletedbefore addition and subtraction.1.5.2 Brackets and operatorsThe basic laws governing the use of brackets andoperators are shown by the following examples.
• Basic arithmetic 7(a) 2 + 3 = 3 + 2; i.e., the order of numbers whenadding does not matter.(b) 2 × 3 = 3 × 2; i.e., the order of numbers whenmultiplying does not matter.(c) 2 + (3 + 4) = (2 + 3) + 4; i.e., the use of bracketswhen adding does not affect the result.(d) 2 × (3 × 4) = (2 × 3) × 4; i.e., the use of bracketswhen multiplying does not affect the result.(e) 2 × (3 + 4) = 2(3 + 4) = 2 × 3 + 2 × 4; i.e., anumber placed outside of a bracket indicatesthat the whole contents of the bracket must bemultiplied by that number.(f) (2 + 3)(4 + 5) = (5)(9) = 5 × 9 = 45; i.e., adja-cent brackets indicate multiplication.(g) 2[3 + (4 × 5)] = 2[3 + 20] = 2 × 23 = 46; i.e.,when an expression contains inner and outerbrackets, the inner brackets are removedﬁrst.Here are some further problems in which BODMASneeds to be used.Problem 16. Find the value of 6 + 4 ÷ (5 − 3)The order of precedence of operations is rememberedby the word BODMAS. Thus,6 + 4 ÷ (5 − 3) = 6 + 4 ÷ 2 (Brackets)= 6 + 2 (Division)= 8 (Addition)Problem 17. Determine the value of13 − 2 × 3 + 14 ÷ (2 + 5)13 − 2 × 3 + 14÷ (2 + 5) = 13 − 2 × 3 + 14 ÷ 7 (B)= 13 − 2 × 3 + 2 (D)= 13 − 6 + 2 (M)= 15 − 6 (A)= 9 (S)Problem 18. Evaluate16 ÷(2 + 6) + 18[3 + (4 × 6) − 21]16 ÷ (2 + 6) + 18[3 + (4 × 6) − 21]= 16 ÷ (2 + 6) + 18[3 + 24 − 21] (B: inner bracketis determined ﬁrst)= 16 ÷ 8 + 18 × 6 (B)= 2 + 18 × 6 (D)= 2 + 108 (M)= 110 (A)Note that a number outside of a bracket multiplies allthat is inside the brackets. In this case,18[3 + 24 − 21] = 18[6], which means 18 × 6 = 108Problem 19. Find the value of23 − 4(2 × 7) +(144 ÷ 4)(14 − 8)23 − 4(2 × 7) +(144 ÷ 4)(14 − 8)= 23 − 4 × 14+366(B)= 23 − 4 × 14+ 6 (D)= 23 − 56 + 6 (M)= 29 − 56 (A)= −27 (S)Problem 20. Evaluate3 + 52 − 32 + 231 + (4 × 6) ÷ (3 × 4)+15 ÷ 3 + 2 × 7 − 13 ×√4+ 8 − 32 + 13 + 52 − 32 + 231 + (4 × 6) ÷ (3 × 4)+15 ÷ 3 + 2 × 7 − 13 ×√4 + 8 − 32 + 1=3 + 4 + 81 + 24 ÷ 12+15 ÷ 3 + 2 × 7 − 13 × 2 + 8 − 9 + 1=3 + 4 + 81 + 2+5 + 2 × 7 − 13 × 2 + 8 − 9 + 1=153+5 + 14 − 16 + 8 − 9 + 1= 5 +186= 5 + 3 = 8
• 8 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 4 Further problems onorder of precedenceand brackets (answerson page 340)Evaluate the following expressions.1. 14+ 3 × 152. 17 − 12 ÷ 43. 86 + 24 ÷ (14 − 2)4. 7(23 − 18) ÷ (12 − 5)5. 63 − 8(14 ÷ 2) + 266.405− 42 ÷ 6 + (3 × 7)7.(50 − 14)3+ 7(16 − 7) − 78.(7 − 3)(1 − 6)4(11 − 6) ÷ (3 − 8)9.(3 + 9 × 6) ÷ 3 − 2 ÷ 23 × 6 + (4 − 9) − 32 + 510.4 × 32 + 24 ÷ 5 + 9 × 32 × 32 − 15 ÷ 3+2 + 27 ÷ 3 + 12 ÷ 2 − 325 + (13 − 2 × 5) − 411.1 +√25 + 3 × 2 − 8 ÷ 23 × 4− 32 + 42 + 1−(4 × 2 + 7 × 2) ÷ 11√9+ 12 ÷ 2 − 23
• Chapter 2Fractions2.1 IntroductionA mark of 9 out of 14 in an examination may be writ-ten as914or 9/14.914is an example of a fraction. Thenumber above the line, i.e. 9, is called the numera-tor. The number below the line, i.e. 14, is called thedenominator.When the value of the numerator is less than thevalue of the denominator, the fraction is called aproper fraction.914is an example of a properfraction.When thevalueofthenumeratorisgreaterthan thevalueof the denominator, the fraction is called an improperfraction.52is an example of an improper fraction.A mixed number is a combination of a whole numberand a fraction. 212is an example of a mixed number. Infact,52= 212.There are a number of everyday examples in whichfractions are readily referred to. For example, threepeople equally sharing a bar of chocolate would have13each. A supermarket advertises15off a six-pack ofbeer; if the beer normally costs £2 then it will nowcost £1.60.34of the employees of a company arewomen; if the company has 48 employees, then 36 arewomen.Calculators are able to handle calculations with frac-tions. However, to understand a little more about frac-tions we will in this chapter show how to add, subtract,multiply and divide with fractions without the use of acalculator.Problem 1. Change the following improperfractions into mixed numbers:(a)92(b)134(c)285(a)92means 9 halves and92= 9 ÷ 2, and 9 ÷ 2 = 4and 1 half, i.e.92= 412(b)134means 13 quarters and134= 13 ÷ 4, and13 ÷ 4 = 3 and 1 quarter, i.e.134= 314(c)285means 28 ﬁfths and285= 28 ÷ 5, and 28 ÷ 5 =5 and 3 ﬁfths, i.e.285= 535Problem 2. Change the following mixed numbersinto improper fractions:(a) 534(b) 179(c) 237(a) 534means 5 +34. 5 contains 5 × 4 = 20 quarters.Thus, 534contains 20 + 3 = 23 quarters, i.e.534=234DOI: 10.1016/B978-1-85617-697-2.00002-8
• 10 Basic Engineering MathematicsThe quick way to change 534into an improperfraction is4 × 5 + 34=234.(b) 179=9 × 1 + 79=169.(c) 237=7 × 2 + 37=177.Problem 3. In a school there are 180 students ofwhich 72 are girls. Express this as a fraction in itssimplest formThe fraction of girls is72180.Dividing both the numerator and denominator by thelowest prime number, i.e. 2, gives72180=3690Dividing both the numerator and denominator again by2 gives72180=3690=18452 will not divide into both 18 and 45, so dividing both thenumerator and denominator by the next prime number,i.e. 3, gives72180=3690=1845=615Dividing both the numerator and denominator again by3 gives72180=3690=1845=615=25So72180=25in its simplest form.Thus,25of the students are girls.2.2 Adding and subtracting fractionsWhen the denominators of two (or more) fractions tobe added are the same, the fractions can be added ‘onsight’.For example,29+59=79and38+18=48.In the latter example, dividing both the 4 and the 8 by4 gives48=12, which is the simpliﬁed answer. This iscalled cancelling.Additionand subtraction of fractions is demonstratedin the following worked examples.Problem 4. Simplify13+12(i) Make the denominators the same for each frac-tion. The lowest number that both denominatorsdivideinto is called the lowest commonmultipleor LCM (see Chapter 1, page 5). In this example,the LCM of 3 and 2 is 6.(ii) 3 divides into 6 twice. Multiplying both numera-tor and denominator of13by 2 gives13=26=(iii) 2 dividesinto 6, 3 times. Multiplyingboth numer-ator and denominator of12by 3 gives12=36=(iv) Hence,13+12=26+36=56+ =Problem 5. Simplify34−716(i) Make the denominators the same for each frac-tion. The lowest common multiple (LCM) of 4and 16 is 16.(ii) 4 divides into 16, 4 times. Multiplying bothnumerator and denominator of34by 4 gives34=1216=(iii)716already has a denominator of 16.
• Fractions 11(iv) Hence,34−716=1216−716=516− =Problem 6. Simplify 423− 116423− 116is the same as 423− 116which is thesame as 4 +23− 1 +16which is the same as4 +23− 1 −16which is the same as 3 +23−16whichis the same as 3 +46−16= 3 +36= 3 +12Thus, 423− 116= 312Problem 7. Evaluate 718− 537718− 537= 7 +18− 5 +37= 7 +18− 5 −37= 2 +18−37= 2 +7 × 1 − 8 × 356= 2 +7 − 2456= 2 +−1756= 2 −1756=11256−1756=112 − 1756=9556= 13956Problem 8. Determine the value of458− 314+ 125458− 314+ 125= (4 − 3 + 1) +58−14+25= 2 +5 × 5 − 10× 1 + 8 × 240= 2 +25 − 10 + 1640= 2 +3140= 23140Now try the following Practice ExercisePracticeExercise 5 Introduction tofractions (answers on page 340)1. Change the improper fraction157into amixed number.2. Change the improper fraction375into amixed number.3. Change the mixed number 249into animproper fraction.4. Change the mixed number 878into animproper fraction.5. A box contains 165 paper clips. 60 clipsare removed from the box. Express this asa fraction in its simplest form.6. Order the following fractions from the small-est to the largest.49,58,37,12,357. A training college has 375 students of which120 are girls. Express this as a fraction in itssimplest form.Evaluate, in fraction form, the expressions given inProblems 8 to 20.8.13+259.56−41510.12+2511.716−1412.27+31113.29−17+2314. 325− 21315.727−23+5916. 5313+ 33417. 458− 32518. 1037− 82319. 314− 445+ 15620. 534− 125− 312
• 12 Basic Engineering Mathematics2.3 Multiplication and division offractions2.3.1 MultiplicationTo multiply two or more fractions together, the numer-ators are ﬁrst multiplied to give a single numberand this becomes the new numerator of the com-bined fraction. The denominators are then multipliedtogether to give the new denominator of the combinedfraction.For example,23×47=2 × 43 × 7=821Problem 9. Simplify 7 ×257 ×25=71×25=7 × 21 × 5=145= 245Problem 10. Find the value of37×1415Dividing numerator and denominator by 3 gives37×1415=17×145=1 × 147 × 5Dividing numerator and denominator by 7 gives1 × 147 × 5=1 × 21 × 5=25This process of dividingboth the numerator and denom-inator of a fraction by the same factor(s) is calledcancelling.Problem 11. Simplify35×4935×49=15×43by cancelling=415Problem 12. Evaluate 135× 213× 337Mixed numbers must be expressed as improper frac-tions before multiplication can be performed. Thus,135× 213× 337=55+35×63+13×217+37=85×73×247=8 × 1 × 85 × 1 × 1=645= 1245Problem 13. Simplify 315× 123× 234The mixed numbers need to be changed to improperfractions before multiplication can be performed.315× 123× 234=165×53×114=41×13×111by cancelling=4 × 1 × 111 × 3 × 1=443= 14232.3.2 DivisionThe simple rule for division is change the divisionsign into a multiplication sign and invert the secondfraction.For example,23÷34=23×43=89Problem 14. Simplify37÷82137÷821=37×218=31×38by cancelling=3 × 31 × 8=98= 118Problem 15. Find the value of 535÷ 713The mixed numbers must be expressed as improperfractions. Thus,535÷ 713=285÷223=285×322=145×311=4255Problem 16. Simplify 323× 134÷ 234
• Fractions 13Mixed numbersmust beexpressed as improperfractionsbefore multiplication and division can be performed:323× 134÷ 234=113×74÷114=113×74×411=1 × 7 × 13 × 1 × 1by cancelling=73= 213Now try the following Practice ExercisePracticeExercise 6 Multiplying anddividing fractions (answers on page 340)Evaluate the following.1.25×472. 5 ×493.34×8114.34×595.1735×15686.35×79× 1277.1317× 4711× 34398.14×311× 15399.29÷42710.38÷456411.38÷53212.34÷ 14513. 214× 12314. 113÷ 25915. 245÷71016. 234÷ 32317.19×34× 11318. 314× 135÷2519. A ship’s crew numbers 105, of which17arewomen. Of the men,16are ofﬁcers. Howmany male ofﬁcers are on board?20. If a storage tank is holding 450 litres whenit is three-quarters full, how much will itcontain when it is two-thirds full?21. Three people, P, Q and R, contribute to afund. P provides 3/5 of the total, Q pro-vides 2/3 of the remainder and R provides£8. Determine (a) the total of the fund and(b) the contributions of P and Q.22. A tank contains 24,000 litres of oil. Initially,710of the contents are removed, then35ofthe remainder is removed. How much oil isleft in the tank?2.4 Order of precedence withfractionsAs stated in Chapter 1, sometimes addition, subtraction,multiplication, division, powers and brackets can all beinvolved in a calculation. A deﬁnite order of precedencemust be adhered to. The order is:BracketsOrder (or pOwer)DivisionMultiplicationAdditionSubtractionThis is demonstrated in the followingworked problems.Problem 17. Simplify720−38×45720−38×45=720−3 × 12 × 5by cancelling (M)=720−310(M)=720−620=120(S)Problem 18. Simplify14− 215×58+91014− 215×58+910=14−115×58+910=14−111×18+910by cancelling=14−118+910(M)=1 × 104 × 10−11 × 58 × 5+9 × 410 × 4(since the LCM of 4, 8 and 10 is 40)
• 14 Basic Engineering Mathematics=1040−5540+3640=10 − 55 + 3640(A/S)= −940Problem 19. Simplify212−25+34÷58×23212−25+34÷58×23=52−2 × 45 × 4+3 × 54 × 5÷58×23(B)=52−820+1520÷58×23(B)=52−2320÷54×13by cancelling (B)=52−2320÷512(B)=52−2320×125(D)=52−235×35by cancelling=52−6925(M)=5 × 252 × 25−69 × 225 × 2(S)=12550−13850(S)= −1350Problem 20. Evaluate13of 512− 334+ 315÷45−1213of 512− 334+ 315÷45−12=13of 134+ 315÷45−12(B)=13×74+165÷45−12(O)(Note that the ‘of ’ is replaced with amultiplication sign.)=13×74+165×54−12(D)=13×74+41×11−12by cancelling=712+41−12(M)=712+4812−612(A/S)=4912= 4112Now try the following Practice ExercisePracticeExercise 7 Order of precedencewith fractions (answers on page 340)Evaluate the following.1. 212−35×20272.13−34×16273.12+35÷915−134.15+ 223÷59−145.45×12−16÷25+236.35−23−12÷56×327.12of 425− 3710+ 313÷23−258.623× 125−13634÷ 1129. 113× 215÷2510.14×25−15÷23+41511.23+ 315× 212+ 113813÷ 31312.113of 2910− 135+ 213÷23−34
• Revision Test 1 : Basic arithmetic and fractionsThis assignment covers the material contained in Chapters 1 and 2. The marks available are shown in brackets at theend of each question.1. Evaluate1009cm − 356cm − 742cm + 94cm. (3)2. Determine £284 × 9. (3)3. Evaluate(a) −11239 − (−4732) + 9639(b) −164 × −12(c) 367 × −19 (6)4. Calculate (a) \$153 ÷ 9 (b) 1397g÷ 11 (4)5. A small component has a mass of 27 grams.Calculate the mass, in kilograms, of 750 suchcomponents. (3)6. Find (a) the highest common factor and (b) thelowest common multiple of the following num-bers: 15 40 75 120. (7)Evaluate the expressions in questions 7 to 12.7. 7 + 20 ÷ (9 − 5) (3)8. 147 − 21(24 ÷ 3) + 31 (3)9. 40 ÷ (1 + 4) + 7[8 + (3 × 8) − 27] (5)10.(7 − 3)(2 − 5)3(9 − 5) ÷ (2 − 6)(3)11.(7 + 4 × 5) ÷ 3 + 6 ÷ 22 × 4 + (5 − 8) − 22 + 3(5)12.(42 × 5 − 8) ÷ 3 + 9 × 84 × 32 − 20 ÷ 5(5)13. Simplify(a)34−715(b) 158− 213+ 356(8)14. A training college has 480 students of which 150are girls. Express this as a fraction in its simplestform. (2)15. A tank contains 18000litresof oil. Initially,710ofthe contents are removed, then25of the remainderis removed. How much oil is left in the tank? (4)16. Evaluate(a) 179×38× 335(b) 623÷ 113(c) 113× 215÷25(10)17. Calculate(a)14×25−15÷23+415(b)23+ 315× 212+ 113813÷ 313(8)18. Simplify113of 2910− 135+ 213÷23−34(8)
• Chapter 3Decimals3.1 IntroductionThe decimal system of numbers is based on the digits0 to 9.Thereareanumberofeveryday occurrencesin which weuse decimal numbers. For example, a radio is, say, tunedto 107.5MHz FM; 107.5 is an example of a decimalnumber.In a shop, a pair of trainers cost, say, £57.95; 57.95 isanotherexampleofadecimal number.57.95 isadecimalfraction, where a decimal point separates the integer, i.e.57, from the fractional part, i.e. 0.9557.95 actually means (5 × 10)+ (7 × 1)+ 9 ×110+ 5 ×11003.2 Converting decimals to fractionsand vice-versaConverting decimals to fractions and vice-versa isdemonstrated below with worked examples.Problem 1. Convert 0.375 to a proper fraction inits simplest form(i) 0.375 may be written as0.375 × 10001000i.e.0.375 =3751000(ii) Dividing both numerator and denominator by 5gives3751000=75200(iii) Dividing both numerator and denominator by 5again gives75200=1540(iv) Dividing both numerator and denominator by 5again gives1540=38Since both 3 and 8 are only divisible by 1, we cannot‘cancel’ any further, so38is the ‘simplest form’ of thefraction.Hence, the decimal fraction 0.375 =38as a properfraction.Problem 2. Convert 3.4375 to a mixed number(i) 0.4375 may be written as0.4375 × 1000010000i.e.0.4375 =437510000(ii) Dividing both numerator and denominator by 25gives437510000=175400(iii) Dividing both numerator and denominator by 5gives175400=3580(iv) Dividing both numerator and denominator by 5again gives3580=716Since both 5 and 16 are only divisible by 1, wecannot ‘cancel’ any further, so716is the ‘lowestform’ of the fraction.(v) Hence, 0.4375 =716Thus, the decimal fraction 3.4375=3716as a mixednumber.DOI: 10.1016/B978-1-85617-697-2.00003-X
• Decimals 17Problem 3. Express78as a decimal fractionTo convert a proper fraction to a decimal fraction, thenumerator is divided by the denominator.0.8 7 58 7.0 0 0(i) 8 into 7 will not go. Place the 0 above the 7.(ii) Place the decimal point above the decimal pointof 7.000(iii) 8 into 70 goes 8, remainder 6. Place the 8 abovethe ﬁrst zero after the decimal point and carry the6 remainder to the next digit on the right, makingit 60.(iv) 8 into 60 goes 7, remainder 4. Place the 7 abovethe next zero and carry the 4 remainder to the nextdigit on the right, making it 40.(v) 8 into 40 goes 5, remainder 0. Place 5 above thenext zero.Hence, the proper fraction78= 0.875 as a decimalfraction.Problem 4. Express 51316as a decimal fractionFor mixed numbers it is only necessary to convert theproper fraction part of the mixed number to a decimalfraction.0.8 1 2 516 13.0 0 0 0(i) 16 into 13 will not go. Place the 0 above the 3.(ii) Place the decimal point above the decimal pointof 13.0000(iii) 16 into 130 goes 8, remainder 2. Place the 8 abovethe ﬁrst zero after the decimal point and carry the2 remainder to the next digit on the right, makingit 20.(iv) 16 into 20 goes 1, remainder 4. Place the 1 abovethe next zero and carry the 4 remainder to the nextdigit on the right, making it 40.(v) 16 into 40 goes 2, remainder 8. Place the 2 abovethe next zero and carry the 8 remainder to the nextdigit on the right, making it 80.(vi) 16 into 80 goes 5, remainder 0. Place the 5 abovethe next zero.(vii) Hence,1316= 0.8125Thus, the mixed number 51316= 5.8125 as a decimalfraction.Now try the following Practice ExercisePracticeExercise 8 Converting decimals tofractions and vice-versa (answers onpage 341)1. Convert 0.65 to a proper fraction.2. Convert 0.036 to a proper fraction.3. Convert 0.175 to a proper fraction.4. Convert 0.048 to a proper fraction.5. Convert the following to proper fractions.(a) 0.65 (b) 0.84 (c) 0.0125(d) 0.282 (e) 0.0246. Convert 4.525 to a mixed number.7. Convert 23.44 to a mixed number.8. Convert 10.015 to a mixed number.9. Convert 6.4375 to a mixed number.10. Convert the following to mixed numbers.(a) 1.82 (b) 4.275 (c) 14.125(d) 15.35 (e) 16.212511. Express58as a decimal fraction.12. Express 61116as a decimal fraction.13. Express732as a decimal fraction.14. Express 11316as a decimal fraction.15. Express932as a decimal fraction.3.3 Signiﬁcant ﬁgures and decimalplacesA number which can be expressed exactly as adecimal fraction is called a terminating decimal.
• 18 Basic Engineering MathematicsFor example,3316= 3.1825 is a terminating decimalA number which cannot be expressed exactly as a deci-mal fraction is called a non-terminating decimal. Forexample,157= 1.7142857... is a non-terminating decimalThe answer to a non-terminating decimal may beexpressed in two ways, depending on the accuracyrequired:(a) correct to a number of signiﬁcant ﬁgures, or(b) correct to a number of decimal places i.e. thenumber of ﬁgures after the decimal point.The last digit in the answer is unaltered if the next digiton the right is in the group of numbers 0, 1, 2, 3 or 4.For example,1.714285... = 1.714 correct to 4 signiﬁcant ﬁgures= 1.714 correct to 3 decimal placessince the next digit on the right in this example is 2.The last digit in the answer is increased by 1 if the nextdigit on the right is in the group of numbers 5, 6, 7, 8 or9. For example,1.7142857... = 1.7143 correct to 5 signiﬁcant ﬁgures= 1.7143 correct to 4 decimal placessince the next digit on the right in this example is 8.Problem 5. Express 15.36815 correct to(a) 2 decimal places, (b) 3 signiﬁcant ﬁgures,(c) 3 decimal places, (d) 6 signiﬁcant ﬁgures(a) 15.36815 = 15.37 correct to 2 decimal places.(b) 15.36815 = 15.4 correct to 3 signiﬁcant ﬁgures.(c) 15.36815 = 15.368 correct to 3 decimal places.(d) 15.36815 = 15.3682 correct to 6 signiﬁcantﬁgures.Problem 6. Express 0.004369 correct to(a) 4 decimal places, (b) 3 signiﬁcant ﬁgures(a) 0.004369 = 0.0044 correct to 4 decimal places.(b) 0.004369 = 0.00437 correct to 3 signiﬁcantﬁgures.Note that the zeros to the right of the decimal point donot count as signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 9 Signiﬁcant ﬁgures anddecimal places (answers on page 341)1. Express 14.1794 correct to 2 decimal places.2. Express 2.7846 correct to 4 signiﬁcant ﬁgures.3. Express 65.3792 correct to 2 decimal places.4. Express 43.2746 correct to 4 signiﬁcantﬁgures.5. Express 1.2973 correct to 3 decimal places.6. Express 0.0005279 correct to 3 signiﬁcantﬁgures.3.4 Adding and subtracting decimalnumbersWhen adding or subtracting decimal numbers, careneeds to be taken to ensure that the decimal pointsare beneath each other. This is demonstrated in thefollowing worked examples.Problem 7. Evaluate 46.8 + 3.06 + 2.4 + 0.09and give the answer correct to 3 signiﬁcant ﬁguresThe decimal points are placed under each other asshown. Each column is added, starting from theright.46.83.062.4+0.0952.3511 1(i) 6 + 9 = 15. Place 5 in the hundredths column.Carry 1 in the tenths column.(ii) 8 + 0 + 4 + 0 + 1 (carried) = 13. Place the 3 inthe tenths column. Carry the 1 into the unitscolumn.(iii) 6 + 3 + 2 + 0 + 1 (carried) = 12. Place the 2 intheunitscolumn.Carry the1 into thetenscolumn.
• Decimals 19(iv) 4 + 1(carried) = 5. Place the 5 in the hundredscolumn.Hence,46.8 + 3.06 + 2.4 + 0.09 = 52.35= 52.4,correct to 3signiﬁcant ﬁguresProblem 8. Evaluate 64.46 − 28.77 and give theanswer correct to 1 decimal placeAs with addition, the decimal points are placed undereach other as shown.64.46−28.7735.69(i) 6 − 7 is not possible; therefore ‘borrow’ 1 fromthe tenths column. This gives 16 − 7 = 9. Placethe 9 in the hundredths column.(ii) 3 − 7 is not possible; therefore ‘borrow’ 1 fromthe unitscolumn. This gives 13 − 7 = 6. Place the6 in the tenths column.(iii) 3 − 8 is not possible; therefore ‘borrow’ from thehundreds column. This gives 13 − 8 = 5. Placethe 5 in the units column.(iv) 5 − 2 = 3. Place the 3 in the hundreds column.Hence,64.46 − 28.77 = 35.69= 35.7 correct to 1 decimal placeProblem 9. Evaluate 312.64 − 59.826 − 79.66+38.5 and give the answer correct to 4 signiﬁcantﬁguresThe sum of the positive decimal fractions= 312.64 + 38.5 = 351.14.The sum of the negative decimal fractions= 59.826 + 79.66 = 139.486.Taking the sum of the negative decimal fractions fromthe sum of the positive decimal fractions gives351.140−139.486211.654Hence, 351.140 − 139.486 = 211.654 = 211.7, cor-rect to 4 signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 10 Adding andsubtracting decimal numbers (answers onpage 341)Determine the following without using a calcula-tor.1. Evaluate 37.69 + 42.6, correct to 3 signiﬁ-cant ﬁgures.2. Evaluate 378.1 − 48.85, correct to 1 decimalplace.3. Evaluate 68.92 + 34.84 − 31.223, correct to4 signiﬁcant ﬁgures.4. Evaluate 67.841 − 249.55 + 56.883, correctto 2 decimal places.5. Evaluate 483.24 − 120.44 − 67.49, correctto 4 signiﬁcant ﬁgures.6. Evaluate 738.22−349.38−427.336+56.779,correct to 1 decimal place.7. Determine the dimension marked x in thelength of the shaft shown in Figure 3.1. Thedimensions are in millimetres.82.9227.41 8.32 34.67xFigure 3.13.5 Multiplying and dividing decimalnumbersWhen multiplying decimal fractions:(a) the numbers are multiplied as if they were integers,and(b) the position of the decimal point in the answer issuch that there are as many digits to the right ofit as the sum of the digits to the right of the dec-imal points of the two numbers being multipliedtogether.This is demonstrated in the followingworked examples.
• 20 Basic Engineering MathematicsProblem 10. Evaluate 37.6 × 5.4376×5415041880020304(i) 376 × 54 = 20304.(ii) As there are 1 + 1 = 2 digits to the right ofthe decimal points of the two numbers beingmultiplied together, 37.6 × 5.4, then37.6 × 5.4 = 203.04Problem 11. Evaluate 44.25 ÷ 1.2, correct to(a) 3 signiﬁcant ﬁgures, (b) 2 decimal places44.25 ÷ 1.2 =44.251.2The denominator is multiplied by 10 to change it into aninteger. The numerator is also multiplied by 10 to keepthe fraction the same. Thus,44.251.2=44.25 × 101.2 × 10=442.512The long division is similar to the long division ofintegers and the steps are as shown.36.87512 442.50036827210596908460600(i) 12 into 44 goes 3; place the 3 above the second4 of 442.500(ii) 3 × 12 = 36; place the 36 below the 44 of442.500(iii) 44 − 36 = 8.(iv) Bring down the 2 to give 82.(v) 12 into 82 goes 6; place the 6 above the 2 of442.500(vi) 6 × 12 = 72; place the 72 below the 82.(vii) 82 − 72 = 10.(viii) Bring down the 5 to give 105.(ix) 12 into 105 goes 8; place the 8 above the 5 of442.500(x) 8 × 12 = 96; place the 96 below the 105.(xi) 105 − 96 = 9.(xii) Bring down the 0 to give 90.(xiii) 12 into 90 goes 7; place the 7 above the ﬁrstzero of 442.500(xiv) 7 × 12 = 84; place the 84 below the 90.(xv) 90 − 84 = 6.(xvi) Bring down the 0 to give 60.(xvii) 12 into 60 gives 5 exactly; place the 5 abovethe second zero of 442.500(xviii) Hence, 44.25 ÷ 1.2 =442.512= 36.875So,(a) 44.25 ÷ 1.2 = 36.9, correct to 3 signiﬁcantﬁgures.(b) 44.25 ÷ 1.2 = 36.88, correct to 2 decimalplaces.Problem 12. Express 723as a decimal fraction,correct to 4 signiﬁcant ﬁguresDividing 2 by 3 gives23= 0.666666...and 723= 7.666666...Hence, 723= 7.667 correct to 4 signiﬁcant ﬁgures.Note that 7.6666... is called 7.6 recurring and iswritten as 7..6Now try the following Practice ExercisePracticeExercise 11 Multiplying anddividing decimal numbers (answers onpage 341)In Problems 1 to 8, evaluate without using acalculator.1. Evaluate 3.57 × 1.42. Evaluate 67.92 × 0.7
• Decimals 213. Evaluate 167.4 × 2.34. Evaluate 342.6 × 1.75. Evaluate 548.28 ÷ 1.26. Evaluate 478.3 ÷ 1.1, correct to 5 signiﬁcantﬁgures.7. Evaluate 563.48 ÷ 0.9, correct to 4 signiﬁ-cant ﬁgures.8. Evaluate 2387.4 ÷ 1.5In Problems 9 to 14, express as decimal fractionsto the accuracy stated.9.49, correct to 3 signiﬁcant ﬁgures.10.1727, correct to 5 decimal places.11. 1916, correct to 4 signiﬁcant ﬁgures.12. 53511, correct to 3 decimal places.13. 133137, correct to 2 decimal places.14. 8913, correct to 3 signiﬁcant ﬁgures.15. Evaluate 421.8 ÷ 17, (a) correct to 4 signif-icant ﬁgures and (b) correct to 3 decimalplaces.16. Evaluate0.01472.3, (a) correct to 5 decimalplaces and (b) correct to 2 signiﬁcant ﬁgures.17. Evaluate (a)12..61.5(b) 5..2 × 1218. A tank contains 1800 litres of oil. How manytins containing 0.75 litres can be ﬁlled fromthis tank?
• Chapter 4Using a calculator4.1 IntroductionIn engineering, calculations often need to be performed.For simple numbers it is useful to be able to use men-tal arithmetic. However, when numbers are larger anelectronic calculator needs to be used.There are several calculators on the market, many ofwhich will be satisfactory for our needs. It is essentialto have a scientiﬁc notationcalculator which will haveall the necessary functions needed and more.This chapter assumes you have a CASIO fx-83EScalculator, or similar, as shown in Figure 4.1.Besides straightforward addition, subtraction, multipli-cation and division, which you will already be ableto do, we will check that you can use squares, cubes,powers, reciprocals, roots, fractions and trigonomet-ric functions (the latter in preparation for Chapter 21).There are several other functions on the calculatorwhich we do not need to concern ourselves with at thislevel.4.2 Adding, subtracting, multiplyingand dividingInitially, after switching on, press Mode.Of the three possibilities, use Comp, which is achievedby pressing 1.Next, press Shift followed by Setup and, of the eightpossibilities, use Mth IO, which is achieved by press-ing 1.By all means experiment with the other menu options –refer to your ‘User’s guide’.All calculators have +, −, × and ÷ functions andthese functions will, no doubt, already have been usedin calculations.Problem 1. Evaluate 364.7 ÷ 57.5 correct to 3decimal places(i) Type in 364.7(ii) Press ÷.(iii) Type in 57.5(iv) Press = and the fraction3647575appears.(v) Press the S ⇔ D functionand the decimal answer6.34260869... appears.Alternatively, after step (iii) press Shift and = and thedecimal will appear.Hence, 364.7 ÷ 57.5 = 6.343 correct to 3 decimalplaces.Problem 2. Evaluate12.47 × 31.5970.45 × 0.052correct to 4signiﬁcant ﬁgures(i) Type in 12.47(ii) Press ×.(iii) Type in 31.59(iv) Press ÷.(v) The denominator must have brackets; i.e. press (.(vi) Type in 70.45 × 0.052 and complete the bracket;i.e. ).(vii) Press = and the answer 107.530518... appears.Hence,12.47 × 31.5970.45 × 0.052= 107.5 correct to 4 signiﬁcantﬁgures.DOI: 10.1016/B978-1-85617-697-2.00004-1
• Using a calculator 23Figure 4.1 A Casio fx-83ES calculatorNow try the following Practice ExercisePracticeExercise 12 Addition, subtraction,multiplication and division using a calculator(answers on page 341)1. Evaluate 378.37 − 298.651 + 45.64 − 94.5622. Evaluate 25.63 × 465.34 correct to 5 signif-icant ﬁgures.3. Evaluate 562.6 ÷ 41.3 correct to 2 decimalplaces.4. Evaluate17.35 × 34.2741.53 ÷ 3.76correct to 3 decimalplaces.5. Evaluate 27.48 + 13.72 × 4.15 correct to4 signiﬁcant ﬁgures.6. Evaluate(4.527 + 3.63)(452.51 ÷ 34.75)+ 0.468 correctto 5 signiﬁcant ﬁgures.7. Evaluate 52.34 −(912.5 ÷ 41.46)(24.6 − 13.652)correct to3 decimal places.8. Evaluate52.14 × 0.347 × 11.2319.73 ÷ 3.54correct to4 signiﬁcant ﬁgures.9. Evaluate451.224.57−363.846.79correct to 4 signiﬁ-cant ﬁgures.10. Evaluate45.6 − 7.35 × 3.614.672 − 3.125correct to 3decimal places.4.3 Further calculator functions4.3.1 Square and cube functionsLocate the x2 and x3 functions on your calculator andthen check the following worked examples.Problem 3. Evaluate 2.42(i) Type in 2.4(ii) Press x2 and 2.42 appears on the screen.(iii) Press = and the answer14425appears.(iv) Press the S ⇔ D function and the fractionchanges to a decimal 5.76Alternatively, after step (ii) press Shift and = .Thus, 2.42 = 5.76Problem 4. Evaluate 0.172 in engineering form(i) Type in 0.17(ii) Press x2 and 0.172 appears on the screen.
• 24 Basic Engineering Mathematics(iii) Press Shift and = and the answer 0.0289 appears.(iv) Press the ENG function and the answer changesto 28.9 × 10−3, which is engineering form.Hence, 0.172 = 28.9×10−3 in engineering form. TheENG function is extremely important in engineeringcalculations.Problem 5. Change 348620 into engineeringform(i) Type in 348620(ii) Press = then ENG.Hence, 348620 = 348.62×103 in engineering form.Problem 6. Change 0.0000538 into engineeringform(i) Type in 0.0000538(ii) Press = then ENG.Hence, 0.0000538 = 53.8×10−6 in engineering form.Problem 7. Evaluate 1.43(i) Type in 1.4(ii) Press x3 and 1.43 appears on the screen.(iii) Press = and the answer343125appears.(iv) Press the S ⇔ D function and the fractionchanges to a decimal: 2.744Thus, 1.43 = 2.744.Now try the following Practice ExercisePracticeExercise 13 Square and cubefunctions (answers on page 341)1. Evaluate 3.522. Evaluate 0.1923. Evaluate 6.852 correct to 3 decimal places.4. Evaluate (0.036)2 in engineering form.5. Evaluate 1.5632 correct to 5 signiﬁcantﬁgures.6. Evaluate 1.337. Evaluate 3.143 correct to 4 signiﬁcantﬁgures.8. Evaluate (0.38)3 correct to 4 decimal places.9. Evaluate (6.03)3correct to 2 decimal places.10. Evaluate (0.018)3 in engineering form.4.3.2 Reciprocal and power functionsThe reciprocal of 2 is12, the reciprocal of 9 is19and thereciprocal of x is1x, which from indices may be writtenas x−1. Locate the reciprocal, i.e. x−1 on the calculator.Also, locate the power function, i.e. x , on yourcalculator and then check the following workedexamples.Problem 8. Evaluate13.2(i) Type in 3.2(ii) Press x−1 and 3.2−1 appears on the screen.(iii) Press = and the answer516appears.(iv) Press the S ⇔ D function and the fractionchanges to a decimal: 0.3125Thus,13.2= 0.3125Problem 9. Evaluate 1.55 correct to 4 signiﬁcantﬁgures(i) Type in 1.5(ii) Press x and 1.5 appears on the screen.(iii) Press 5 and 1.55 appears on the screen.(iv) Press Shift and = and the answer 7.59375appears.Thus, 1.55 = 7.594 correct to 4 signiﬁcant ﬁgures.Problem 10. Evaluate 2.46 − 1.94 correct to 3decimal places(i) Type in 2.4(ii) Press x and 2.4 appears on the screen.
• Using a calculator 25(iii) Press 6 and 2.46appears on the screen.(iv) The cursor now needs to be moved; this isachieved by using the cursor key (the largeblue circular function in the top centre of thecalculator). Press →(v) Press −(vi) Type in 1.9, press x , then press 4.(vii) Press = and the answer 178.07087... appears.Thus, 2.46 − 1.94 = 178.071 correct to 3 decimalplaces.Now try the following Practice ExercisePracticeExercise 14 Reciprocal and powerfunctions (answers on page 341)1. Evaluate11.75correct to 3 decimal places.2. Evaluate10.02503. Evaluate17.43correct to 5 signiﬁcant ﬁgures.4. Evaluate10.00725correct to 1 decimal place.5. Evaluate10.065−12.341correct to 4 signiﬁ-cant ﬁgures.6. Evaluate 2.147. Evaluate (0.22)5correct to 5 signiﬁcantﬁgures in engineering form.8. Evaluate (1.012)7 correct to 4 decimalplaces.9. Evaluate (0.05)6 in engineering form.10. Evaluate 1.13 + 2.94 − 4.42 correct to 4 sig-niﬁcant ﬁgures.4.3.3 Root and ×10x functionsLocate the square root function√and the√function (which is a Shift function located abovethe x function) on your calculator. Also, locate the×10x function and then check the following workedexamples.Problem 11. Evaluate√361(i) Press the√function.(ii) Type in 361 and√361 appears on the screen.(iii) Press = and the answer 19 appears.Thus,√361 = 19.Problem 12. Evaluate 4√81(i) Press the√function.(ii) Type in 4 and 4√appears on the screen.(iii) Press → to move the cursor and then type in 81and 4√81 appears on the screen.(iv) Press = and the answer 3 appears.Thus,4√81 = 3.Problem 13. Evaluate 6 × 105 × 2 × 10−7(i) Type in 6(ii) Press the ×10x function (note, you do not haveto use ×).(iii) Type in 5(iv) Press ×(v) Type in 2(vi) Press the ×10x function.(vii) Type in −7(viii) Press = and the answer325appears.(ix) Press the S ⇔ D function and the fractionchanges to a decimal: 0.12Thus, 6 × 105 × 2 × 10−7 = 0.12Now try the following Practice ExercisePracticeExercise 15 Root and ×10xfunctions (answers on page 341)1. Evaluate√4.76 correct to 3 decimal places.2. Evaluate√123.7 correct to 5 signiﬁcantﬁgures.
• 26 Basic Engineering Mathematics3. Evaluate√34528correct to 2 decimal places.4. Evaluate√0.69 correct to 4 signiﬁcantﬁgures.5. Evaluate√0.025 correct to 4 decimal places.6. Evaluate 3√17 correct to 3 decimal places.7. Evaluate 4√773 correct to 4 signiﬁcantﬁgures.8. Evaluate 5√3.12 correct to 4 decimal places.9. Evaluate 3√0.028 correct to 5 signiﬁcantﬁgures.10. Evaluate6√2451−4√46 correct to 3 decimalplaces.Express the answers to questions 11 to 15 inengineering form.11. Evaluate 5 × 10−3 × 7 × 10812. Evaluate3 × 10−48 × 10−913. Evaluate6 × 103 × 14× 10−42 × 10614. Evaluate56.43 × 10−3 × 3 × 1048.349 × 103correct to3 decimal places.15. Evaluate99 × 105 × 6.7 × 10−336.2 × 10−4correct to 4signiﬁcant ﬁgures.4.3.4 FractionsLocate the and functions on your calculator(the latter function is a Shift function found abovethe function) and then check the following workedexamples.Problem 14. Evaluate14+23(i) Press the function.(ii) Type in 1(iii) Press ↓ on the cursor key and type in 4(iv)14appears on the screen.(v) Press → on the cursor key and type in +(vi) Press the function.(vii) Type in 2(viii) Press ↓ on the cursor key and type in 3(ix) Press → on the cursor key.(x) Press = and the answer1112appears.(xi) Press the S ⇔ D function and the fractionchanges to a decimal 0.9166666...Thus,14+23=1112= 0.9167 as a decimal, correct to4 decimal places.It is also possible to deal with mixed numbers on thecalculator. Press Shift then the function andappears.Problem 15. Evaluate 515− 334(i) Press Shift then the function and appearson the screen.(ii) Type in 5 then → on the cursor key.(iii) Type in 1 and ↓ on the cursor key.(iv) Type in 5 and 515appears on the screen.(v) Press → on the cursor key.(vi) Typein – and then pressShift then the functionand 515− appears on the screen.(vii) Type in 3 then → on the cursor key.(viii) Type in 3 and ↓ on the cursor key.(ix) Type in 4 and 515− 334appears on the screen.(x) Press = and the answer2920appears.(xi) Press S ⇔ D function and the fraction changes toa decimal 1.45Thus, 515− 334=2920= 1920= 1.45 as a decimal.
• 28 Basic Engineering Mathematics8. Evaluate cos(1.42 rad)9. Evaluate tan(5.673 rad)10. Evaluate(sin42.6◦)(tan 83.2◦)cos13.8◦4.3.6 π and e x functionsPress Shift and then press the ×10x function key and πappears on the screen. Either press Shift and = (or =and S ⇔ D) and the value of π appears in decimal formas 3.14159265...Press Shift and then press the ln function key and eappears on the screen. Enter 1 and then press = ande1 = e = 2.71828182...Now check the following worked examples involvingπand ex functions.Problem 18. Evaluate 3.57π(i) Enter 3.57(ii) Press Shift and the ×10x key and 3.57π appearson the screen.(iii) Either press Shift and = (or = and S ⇔ D)and the value of 3.57π appears in decimal as11.2154857...Hence, 3.57 π = 11.22 correct to 4 signiﬁcantﬁgures.Problem 19. Evaluate e2.37(i) Press Shift and then press the ln function key ande appears on the screen.(ii) Enter 2.37 and e2.37 appears on the screen.(iii) Press Shiftand = (or= and S ⇔ D) and the valueof e2.37 appears in decimal as 10.6973922...Hence, e 2.37 = 10.70 correct to 4 signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 18 π and ex functions(answers on page 341)Evaluatethefollowing,each correct to 4 signiﬁcantﬁgures.1. 1.59π 2. 2.7(π− 1)3. π2√13 − 1 4. 3eπ5. 8.5e−2.5 6. 3e2.9 − 1.67. 3e(2π−1) 8. 2πeπ39.5.52π2e−2 ×√26.7310.⎡⎣ e2−√3π ×√8.57⎤⎦4.4 Evaluation of formulaeThe statement y = mx + c is called a formula for y interms of m, x and c.y, m, x and c are called symbols.When given values of m, x and c we can evaluate y.There are a large number of formulae used in engineer-ing and in this section we will insert numbers in placeof symbols to evaluate engineering quantities.Just four examples of important formulae are:1. A straight line graph is of the form y = mx + c (seeChapter 17).2. Ohm’s law states that V = I × R.3. Velocity is expressed as v = u + at.4. Force is expressed as F = m × a.Here are some practical examples. Check with yourcalculator that you agree with the working and answers.Problem 20. In an electrical circuit the voltage Vis given by Ohm’s law, i.e. V = IR. Find, correct to4 signiﬁcant ﬁgures, the voltage when I = 5.36Aand R = 14.76V = IR = (5.36)(14.76)Hence, voltage V = 79.11V, correct to 4 signiﬁcantﬁgures.Problem 21. The surface area A of a hollow coneis given by A = πrl. Determine, correct to 1decimal place, the surface area when r = 3.0cmand l = 8.5cmA = πrl = π(3.0)(8.5)cm2Hence, surface area A = 80.1cm2, correct to 1 deci-mal place.
• Using a calculator 29Problem 22. Velocity v is given by v = u + at. Ifu = 9.54m/s, a = 3.67m/s2 and t = 7.82s, ﬁnd v,correct to 3 signiﬁcant ﬁguresv = u + at = 9.54 + 3.67 × 7.82= 9.54 + 28.6994 = 38.2394Hence, velocity v = 38.2m/s, correct to 3 signiﬁcantﬁgures.Problem 23. The area, A, of a circle is given byA = πr2. Determine the area correct to 2 decimalplaces, given radius r = 5.23mA = πr2= π(5.23)2= π(27.3529)Hence, area, A = 85.93m2, correct to 2 decimalplaces.Problem 24. Density =massvolume. Find the densitywhen the mass is 6.45kg and the volume is300 × 10−6 cm3Density =massvolume=6.45kg300 × 10−6 m3= 21500kg/m3Problem 25. The power, P watts, dissipated in anelectrical circuit is given by the formula P =V 2R.Evaluate the power, correct to 4 signiﬁcant ﬁgures,given that V = 230 V and R = 35.63P =V2R=(230)235.63=5290035.63= 1484.70390...Press ENG and 1.48470390...× 103 appears on thescreen.Hence, power, P = 1485W or 1.485kW correct to 4signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 19 Evaluation offormulae (answers on page 341)1. The area A of a rectangle is given by theformula A = lb. Evaluate the area whenl = 12.4cm and b = 5.37cm.2. The circumference C of a circle is given bytheformula C = 2πr. Determine the circum-ference given r = 8.40mm.3. A formula used in connection with gasesis R =PVT. Evaluate R when P = 1500,V = 5 and T = 200.4. Thevelocity ofabody isgiven byv = u + at.The initial velocity u is measured when timet is 15 seconds and found to be 12m/s. If theacceleration a is 9.81m/s2calculate the ﬁnalvelocity v.5. Calculate the current I in an electrical circuit,where I = V/R amperes when the voltage Vis measured and found to be 7.2 V and theresistance R is 17.7 .6. Find the distance s, given that s =12gt2when time t = 0.032seconds and accelera-tion due to gravity g = 9.81m/s2. Give theanswer in millimetres.7. The energy stored in a capacitor is givenby E =12CV 2 joules. Determine the energywhen capacitance C = 5 × 10−6 farads andvoltage V = 240V.8. Find the area A of a triangle, given A =12bh,when the base length l is 23.42m and theheight h is 53.7m.9. Resistance R2 is given by R2 = R1(1 + αt).Find R2, correct to 4 signiﬁcant ﬁgures, whenR1 = 220, α = 0.00027 and t = 75.610. Density =massvolume. Find the density whenthe mass is 2.462kg and the volume is173cm3. Give the answer in units of kg/m3.11. Velocity = frequency × wavelength. Findthe velocity when the frequency is 1825Hzand the wavelength is 0.154m.12. Evaluate resistance RT , given1RT=1R1+1R2+1R3when R1 = 5.5 ,R2 = 7.42 and R3 = 12.6 .Here are some further practical examples. Again, checkwith your calculator that you agree with the workingand answers.
• 30 Basic Engineering MathematicsProblem 26. The volume V cm3of a rightcircular cone is given by V =13πr2h. Given thatradius r = 2.45cm and height h = 18.7cm, ﬁnd thevolume, correct to 4 signiﬁcant ﬁguresV =13πr2h =13π(2.45)2(18.7)=13× π × 2.452× 18.7= 117.544521...Hence, volume, V =117.5cm3, correct to 4 signiﬁ-cant ﬁgures.Problem 27. Force F newtons is given by theformula F =Gm1m2d2, where m1 and m2 aremasses, d their distance apart and G is a constant.Find the value of the force given thatG = 6.67 × 10−11, m1 = 7.36, m2 = 15.5 andd = 22.6. Express the answer in standard form,correct to 3 signiﬁcant ﬁguresF =Gm1m2d2=(6.67 × 10−11)(7.36)(15.5)(22.6)2=(6.67)(7.36)(15.5)(1011)(510.76)=1.4901011Hence, force F =1.49×10−11 newtons, correct to3 signiﬁcant ﬁgures.Problem 28. The time of swing, t seconds, of asimple pendulum is given by t = 2πlgDetermine the time, correct to 3 decimal places,given that l = 12.9 and g = 9.81t = 2πlg= (2π)12.99.81= 7.20510343...Hence, time t = 7.205seconds, correct to 3 decimalplaces.Problem 29. Resistance, R , varies withtemperature according to the formulaR = R0(1 + αt). Evaluate R, correct to 3 signiﬁcantﬁgures, given R0 = 14.59, α = 0.0043 and t = 80R = R0(1 + αt) = 14.59[1 + (0.0043)(80)]= 14.59(1 + 0.344)= 14.59(1.344)Hence, resistance, R = 19.6 , correct to 3 signiﬁcantﬁgures.Problem 30. The current, I amperes, in an a.c.circuit is given by I =V(R2 + X2). Evaluate thecurrent, correct to 2 decimal places, whenV = 250V, R = 25.0 and X = 18.0 .I=V(R2 + X2)=25025.02 + 18.02= 8.11534341...Hence, current, I = 8.12A, correct to 2 decimalplaces.Now try the following Practice ExercisePracticeExercise 20 Evaluation offormulae (answers on page 341)1. Find the total cost of 37 calculators cost-ing £12.65 each and 19 drawing sets costing£6.38 each.2. Power =force × distancetime. Find the powerwhen a force of 3760N raises an object adistance of 4.73m in 35s.3. The potential difference, V volts, availableat battery terminals is given by V = E − Ir.Evaluate V when E = 5.62, I = 0.70 andR = 4.304. Given force F =12m(v2 − u2), ﬁnd F whenm = 18.3,v = 12.7 and u = 8.245. The current I amperes ﬂowing in a numberof cells is given by I =nER + nr. Evaluate thecurrent when n = 36, E = 2.20, R = 2.80and r = 0.506. The time, t seconds, of oscillation for asimple pendulum is given by t = 2πlg.Determine the time when l = 54.32 andg = 9.81
• Using a calculator 317. Energy, E joules, is given by the formulaE =12L I2. Evaluate the energy whenL = 5.5 and I = 1.28. The current I amperes in an a.c. circuitis given by I =V(R2 + X2). Evaluate thecurrent when V =250, R=11.0 and X=16.29. Distance s metres is given by the formulas = ut +12at2. If u = 9.50, t = 4.60 anda = −2.50, evaluate the distance.10. The area, A, of any triangle is givenby A =√[s(s − a)(s − b)(s − c)] wheres =a + b + c2. Evaluate the area, givena = 3.60cm, b = 4.00cm and c = 5.20cm.11. Given that a = 0.290, b = 14.86, c = 0.042,d = 31.8 and e = 0.650, evaluate v given thatv =abc−de12. Deduce the following information from thetrain timetable shown in Table 4.1.(a) At what time should a man catch a trainat Fratton to enable him to be in LondonWaterloo by 14.23h?(b) A girlleaves Cosham at 12.39h and trav-els to Woking. How long does the jour-ney take? And, if the distance betweenCosham and Woking is 55miles, calcu-late the average speed of the train.(c) A man living at Havant has a meetingin London at 15.30h. It takes around25minutes on the underground to reachhis destination from London Waterloo.What train should he catch from Havantto comfortably make the meeting?(d) Nine trains leave Portsmouth harbourbetween 12.18h and 13.15h. Whichtrain should be taken for the shortestjourney time?
• 32 Basic Engineering MathematicsTable 4.1 Train timetable from Portsmouth Harbour to London WaterlooFrattonFrattonHilseaHilseaCoshamCoshamBedhamptonBedhamptonHavantHavantRowlands CastleRowlands CastleChichesterChichesterBarnhamBarnhamHorshamHorshamCrawleyCrawleyThree BridgesThree BridgesGatwick AirportGatwick AirportHorleyHorleyRedhillRedhillEast CroydonEast CroydonPetersfieldPetersfieldLissLissLiphookLiphookHaslemereHaslemereGuildfordGuildfordPortchesterPortchesterFarehamFarehamSouthampton CentralSouthampton CentralBotleyBotleyHedge EndHedge EndEastleighEastleighSouthampton Airport ParkwayWinchesterWinchesterMicheldeverMicheldeverBasingstokeBasingstokeFarnboroughFarnboroughWokingWokingClapham JunctionVauxhallVauxhallLondon WaterlooClapham JunctionSouthampton Airport ParkwayPortsmouth HarbourS04dep 12:18SW12:22GW12:22GW12:45SW12:4812:5012:5312:5413:0313:0413:1813:3113:3213:4513:4713:5713:5914:2313:3113:3213:4513:4713:5713:5914:2713:1713:1813:1713:1813:30C13:1713:0313:0413:0213:0412:45SW12:4812:5012:5312:5412:45SW12:54SW13:12SN13:15SW13:1813:2013:2313:2413:3313:3413:4713:4814:0114:0214:1514:1714:2514:2614:5113:1513:1613:1913:2013:2913:3013:4013:4113:4813:4914:1614:2014:2814:2914:3214:3314:3714:3814:4114:4114:4714:4815:0015:0015:11C15:21SW15:2615:2615:3112:5712:5913:0213:0313:0713:0713:1213:1213:1713:1713:2213:2313:3013:3013:3413:3513:4113:4213:5414:0214:0214:1514:1714:3014:3114:4014:4115:0115:1313:5312:4812:5012:5312:5412:2512:2712:3012:3112:2512:2712:3012:3112:3812:3912:3812:3912:2112:2412:2712:2812:3212:3212:3712:3712:3912:4012:4612:4612:5612:5713:0213:0213:0913:0912:4612:4712:4612:4713:08C13:09SW13:1713:1813:3413:3614:1214:1314:2413:00C13:14C13:55C214:02SW13:30SW13:3713:4713:4814:1914:2114:4913:3814:1114:1214:3114:3214:40arrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrdeparrS03 S08 S02 S03 S04 S04 S01 S02SaturdaysPortsmouth Harbour - London WaterlooOUTWARDTrain AlterationsTime Time Time Time Time Time Time Time TimePortsmouth & SouthseaPortsmouth & Southsea13:20SWR 13:36SWR14:17SWR14:2514:2614:5114:11C
• Chapter 5Percentages5.1 IntroductionPercentages are used to give a common standard. Theuse of percentages is very common in many aspectsof commercial life, as well as in engineering. Interestrates, sale reductions, pay rises, exams and VAT are allexamples of situations in which percentages are used.For this chapter you will need to know about decimalsand fractions and be able to use a calculator.We are familiar with the symbol for percentage, i.e. %.Here are some examples.• Interest rates indicate the cost at which we can bor-row money. If you borrow £8000 at a 6.5% interestrate for a year, it will cost you 6.5% of the amountborrowed to do so,which will need to berepaid alongwith the original money you borrowed. If you repaythe loan in 1 year, how much interest will you havepaid?• A pair of trainers in a shop cost £60. They are adver-tised in a sale as 20% off. How much will youpay?• If you earn £20000 p.a. and you receive a 2.5%pay rise, how much extra will you have to spend thefollowing year?• A book costing £18 can be purchased on the internetfor 30% less. What will be its cost?When we have completed his chapter on percentagesyou will be able to understand how to perform the abovecalculations.Percentages are fractions having 100 as their denom-inator. For example, the fraction40100is written as 40%and is read as ‘forty per cent’.The easiest way to understand percentages is to gothrough some worked examples.5.2 Percentage calculations5.2.1 To convert a decimal to a percentageA decimal number is converted to a percentage bymultiplying by 100.Problem 1. Express 0.015 as a percentageTo express a decimal number as a percentage, merelymultiply by 100, i.e.0.015 = 0.015 × 100%= 1.5%Multiplying a decimal number by 100 means movingthe decimal point 2 places to the right.Problem 2. Express 0.275 as a percentage0.275 = 0.275 × 100%= 27.5%5.2.2 To convert a percentage to a decimalA percentage is converted to a decimal number bydividing by 100.Problem 3. Express 6.5% as a decimal number6.5% =6.5100= 0.065Dividing by 100 means moving the decimal point 2places to the left.DOI: 10.1016/B978-1-85617-697-2.00005-3
• 34 Basic Engineering MathematicsProblem 4. Express 17.5% as a decimal number17.5% =17.5100= 0.1755.2.3 To convert a fraction to a percentageA fraction is converted to a percentage by multiplyingby 100.Problem 5. Express58as a percentage58=58× 100% =5008%= 62.5%Problem 6. Express519as a percentage, correctto 2 decimal places519=519× 100%=50019%= 26.3157889...by calculator= 26.32% correct to 2 decimal placesProblem 7. In two successive tests a studentgains marks of 57/79 and 49/67. Is the secondmark better or worse than the ﬁrst?57/79 =5779=5779× 100% =570079%= 72.15% correct to 2 decimal places49/67 =4967=4967× 100% =490067%= 73.13% correct to 2 decimal placesHence, the second test is marginally better than theﬁrst test. This question demonstrates how much easierit is to compare two fractions when they are expressedas percentages.5.2.4 To convert a percentage to a fractionA percentage is converted to a fraction by dividing by100 and then, by cancelling, reducing it to its simplestform.Problem 8. Express 75% as a fraction75% =75100=34The fraction75100is reduced to its simplest form by can-celling, i.e. dividing both numerator and denominatorby 25.Problem 9. Express 37.5% as a fraction37.5% =37.5100=3751000by multiplying both numerator anddenominator by 10=1540by dividing both numerator anddenominator by 25=38by dividing both numerator anddenominator by 5Now try the following Practice ExercisePracticeExercise 21 Percentages (answerson page 342)In problems 1 to 5, express the given numbers aspercentages.1. 0.0032 2. 1.7343. 0.057 4. 0.3745. 1.2856. Express 20% as a decimal number.7. Express 1.25% as a decimal number.8. Express1116as a percentage.9. Express513as a percentage, correct to 3decimal places.
• Percentages 3510. Express as percentages, correct to 3 signiﬁcantﬁgures,(a)733(b)1924(c) 1111611. Place the following in order of size, the small-est ﬁrst, expressing each as a percentagecorrect to 1 decimal place.(a)1221(b)917(c)59(d)61112. Express 65% as a fraction in its simplest form.13. Express 31.25% as a fraction in its simplestform.14. Express 56.25% as a fraction in its simplestform.15. Evaluate A to J in the following table.Decimal number Fraction Percentage0.5 A BC14DE F 30G35HI J 855.3 Further percentage calculations5.3.1 Finding a percentage of a quantityTo ﬁnd apercentageofaquantity,convert thepercentageto a fraction (by dividing by 100) and remember that ‘of’means multiply.Problem 10. Find 27% of £6527% of £65 =27100× 65= £17.55 by calculatorProblem 11. In a machine shop, it takes 32minutes to machine a certain part. Using a new tool,the time can be reduced by 12.5%. Calculate thenew time taken12.5% of 32 minutes =12.5100× 32= 4 minutesHence, new time taken = 32 − 4 = 28 minutes.Alternatively, if the time is reduced by 12.5%, it nowtakes 100% − 12.5% = 87.5% of the original time, i.e.87.5% of 32 minutes =87.5100× 32= 28 minutesProblem 12. A 160 GB iPod is advertised ascosting £190 excluding VAT. If VAT is added at17.5%, what will be the total cost of the iPod?VAT = 17.5% of £190 =17.5100× 190 = £33.25Total cost of iPod = £190 + £33.25 = £223.25A quicker method to determine the total cost is: 1.175 ×£190 = £223.255.3.2 Expressing one quantity as apercentage of another quantityTo express one quantity as a percentage of another quan-tity, divide the ﬁrst quantity by the second then multiplyby 100.Problem 13. Express 23cm as a percentage of72cm, correct to the nearest 1%23cm as a percentage of 72cm =2372× 100%= 31.94444...%= 32% correct tothe nearest 1%Problem 14. Express 47 minutes as a percentageof 2 hours, correct to 1 decimal placeNote that it is essential that the two quantities are in thesame units.Working in minute units, 2 hours = 2 × 60= 120 minutes47 minutes as a percentageof 120 min =47120× 100%= 39.2% correct to1 decimal place
• 36 Basic Engineering Mathematics5.3.3 Percentage changePercentage change is given bynew value−original valueoriginal value× 100%.Problem 15. A box of resistors increases in pricefrom £45 to £52. Calculate the percentage changein cost, correct to 3 signiﬁcant ﬁgures% change =new value − original valueoriginal value× 100%=52 − 4545× 100% =745× 100= 15.6% = percentage change in costProblem 16. A drilling speed should be set to400rev/min. The nearest speed available on themachine is 412rev/min. Calculate the percentageoverspeed% overspeed =available speed−correct speedcorrect speed×100%=412 − 400400×100% =12400×100%= 3%Now try the following Practice ExercisePracticeExercise 22 Further percentages(answers on page 342)1. Calculate 43.6% of 50kg.2. Determine 36% of 27m.3. Calculate, correct to 4 signiﬁcant ﬁgures,(a) 18% of 2758 tonnes(b) 47% of 18.42 grams(c) 147% of 14.1 seconds.4. When 1600 bolts are manufactured, 36 areunsatisfactory. Determinethepercentagethatis unsatisfactory.5. Express(a) 140kg as a percentage of 1t.(b) 47s as a percentage of 5min.(c) 13.4cm as a percentage of 2.5m.6. A block of Monel alloy consists of 70%nickel and 30% copper. If it contains 88.2gof nickel,determine the mass of copper in theblock.7. An athlete runs 5000m in 15 minutes 20seconds. With intense training, he is able toreduce this time by 2.5%. Calculate his newtime.8. A copper alloy comprises 89% copper, 1.5%iron and the remainder aluminium. Find theamount of aluminium, in grams, in a 0.8kgmass of the alloy.9. A computer is advertised on the internet at£520, exclusive of VAT. If VAT is payable at17.5%, what is the total cost ofthe computer?10. Express 325mm as a percentage of 867mm,correct to 2 decimal places.11. A child sleeps on average 9 hours 25 minutesper day. Express this as a percentage of thewhole day, correct to 1 decimal place.12. Express 408g as a percentage of 2.40kg.13. When signing a new contract, a Premiershipfootballer’s pay increases from £15500 to£21500 per week. Calculate the percentagepay increase, correct to 3 signiﬁcant ﬁgures.14. A metal rod 1.80m long is heated and itslength expands by 48.6mm. Calculate thepercentage increase in length.15. 12.5% of a length of wood is 70cm. What isthe full length?16. A metal rod, 1.20m long, is heated andits length expands by 42mm. Calculate thepercentage increase in length.5.4 More percentage calculations5.4.1 Percentage errorPercentage error =errorcorrect value× 100%Problem 17. The length of a component ismeasured incorrectly as 64.5mm. The actual length
• Percentages 37is 63mm. What is the percentage error in themeasurement?% error =errorcorrect value× 100%=64.5 − 6363× 100%=1.563× 100% =15063%= 2.38%The percentage measurement error is 2.38% too high,which is sometimes written as + 2.38% error.Problem 18. The voltage across a component inan electrical circuit is calculated as 50V usingOhm’s law. When measured, the actual voltage is50.4V. Calculate, correct to 2 decimal places, thepercentage error in the calculation% error =errorcorrect value× 100%=50.4 − 5050.4× 100%=0.450.4× 100% =4050.4%= 0.79%The percentage error in the calculation is 0.79% toolow, which is sometimes written as −0.79% error.5.4.2 Original valueOriginal value =new value100 ± % change× 100%Problem 19. A man pays £149.50 in a sale for aDVD player which is labelled ‘35% off’. What wasthe original price of the DVD player?In this case, it is a 35% reduction in price, so weusenew value100 − % change× 100, i.e. a minus sign in thedenominator.Original price =new value100 − % change× 100=149.5100 − 35× 100=149.565× 100 =1495065= £230Problem 20. A couple buys a ﬂat and make an18% proﬁt by selling it 3 years later for £153400.Calculate the original cost of the houseIn this case, it is an 18% increase in price, so weusenew value100 + % change× 100, i.e. a plus sign in thedenominator.Original cost =new value100 + % change× 100=153400100 + 18× 100=153400118× 100 =15340000118= £130 000Problem 21. An electrical store makes 40% proﬁton each widescreen television it sells. If the sellingprice of a 32 inch HD television is £630, what wasthe cost to the dealer?In this case, it is a 40% mark-up in price, so weusenew value100 + % change× 100, i.e. a plus sign in thedenominator.Dealer cost =new value100 + % change× 100=630100 + 40× 100=630140× 100 =63000140= £450The dealer buys from the manufacturer for £450 andsells to his customers for £630.5.4.3 Percentage increase/decrease andinterestNew value =100 + % increase100× original valueProblem 22. £3600 is placed in an ISA accountwhich pays 6.25% interest per annum. How much isthe investment worth after 1 year?
• 38 Basic Engineering MathematicsValue after 1 year =100 + 6.25100× £3600=106.25100× £3600= 1.0625 × £3600= £3825Problem 23. The price of a fully installedcombination condensing boileris increased by 6.5%.It originally cost £2400. What is the new price?New price =100 + 6.5100× £2,400=106.5100× £2,400 = 1.065 × £2,400= £2,556Now try the following Practice ExercisePracticeExercise 23 Further percentages(answers on page 342)1. A machine part has a length of 36mm. Thelength is incorrectly measured as 36.9mm.Determine the percentage error in the mea-surement.2. When a resistor is removed from an electri-cal circuit the current ﬂowing increases from450μA to 531μA. Determine the percentageincrease in the current.3. In a shoe shop sale, everything is advertisedas ‘40% off’. If a lady pays £186 for a pair ofJimmy Choo shoes, what was their originalprice?4. Over a four year period a family homeincreases in value by 22.5% to £214375.What was the value of the house 4 years ago?5. An electrical retailer makes a 35% proﬁt onall its products. What price does the retailerpay for a dishwasher which is sold for £351?6. The cost of a sports car is £23500 inclusiveof VAT at 17.5%. What is the cost of the carwithout the VAT added?7. £8000 is invested in bonds at a building soci-ety which is offering a rate of 6.75% perannum. Calculate the value of the investmentafter 2 years.8. An electrical contractor earning £36000 perannum receives a pay rise of 2.5%. Hepays 22% of his income as tax and 11%on National Insurance contributions. Calcu-late the increase he will actually receive permonth.9. Five mates enjoy a meal out. With drinks, thetotal bill comes to £176. They add a 12.5%tip and divide the amount equally betweenthem. How much does each pay?10. In December a shop raises the cost of a 40inch LCD TV costing £920 by 5%. It doesnot sell and in its January sale it reducesthe TV by 5%. What is the sale price ofthe TV?11. A man buys a business and makes a 20%proﬁt when he sells it three years later for£222000. What did he pay originally for thebusiness?12. A drilling machine should be set to250rev/min. The nearest speed available onthe machine is 268rev/min. Calculate thepercentage overspeed.13. Two kilograms of a compound contain 30%of element A, 45% of element B and 25% ofelement C. Determine the masses of the threeelements present.14. A concrete mixture contains seven parts byvolume of ballast, four parts by volume ofsand and two parts by volume of cement.Determine the percentage of each of thesethree constituents correct to the nearest 1%and the mass of cement in a two tonne drymix, correct to 1 signiﬁcant ﬁgure.15. In a sample of iron ore, 18% is iron. Howmuch ore is needed to produce 3600kg ofiron?16. A screw’s dimension is 12.5 ± 8%mm. Cal-culate the maximum and minimum possiblelength of the screw.17. The output power of an engine is 450kW. Iftheefﬁciency oftheengineis75%,determinethe power input.
• Revision Test 2 : Decimals, calculators and percentagesThis assignment covers the material contained in Chapters 3–5. The marks available are shown in brackets at theend of each question.1. Convert 0.048 to a proper fraction. (2)2. Convert 6.4375 to a mixed number. (3)3. Express932as a decimal fraction. (2)4. Express 0.0784 correct to 2 decimal places. (2)5. Express 0.0572953 correct to 4 signiﬁcantﬁgures. (2)6. Evaluate(a) 46.7 + 2.085 + 6.4 + 0.07(b) 68.51 − 136.34 (4)7. Determine 2.37 × 1.2 (3)8. Evaluate 250.46 ÷ 1.1 correct to 1 decimalplace. (3)9. Evaluate 5.2·× 12 (2)10. Evaluate the following, correct to 4 signiﬁcantﬁgures: 3.32 − 2.73 + 1.84 (3)11. Evaluate√6.72 −3√2.54 correct to 3 decimalplaces. (3)12. Evaluate10.0071−10.065correct to 4 signiﬁcantﬁgures. (2)13. The potential difference, V volts, available at bat-tery terminals is given by V = E − Ir. EvaluateV when E = 7.23, I = 1.37 and r = 3.60 (3)14. Evaluate49+15−38as a decimal, correct to 3signiﬁcant ﬁgures. (3)15. Evaluate16 × 10−6 × 5 × 1092 × 107in engineeringform. (2)16. Evaluate resistance, R, given1R=1R1+1R2+1R3when R1 = 3.6 k , R2 = 7.2 k andR3 = 13.6 k . (3)17. Evaluate 627− 459as a mixed number and as adecimal, correct to 3 decimal places. (3)18. Evaluate, correct to 3 decimal places:2e1.7 × 3.6734.61 ×√3π(3)19. If a = 0.270,b = 15.85,c = 0.038,d = 28.7 ande = 0.680, evaluate v correct to 3 signiﬁcantﬁgures, given that v =abc−de(4)20. Evaluate the following, each correct to 2 decimalplaces.(a)36.22× 0.56127.8 × 12.833(b)14.692√17.42 × 37.98(4)21. If 1.6km = 1mile, determine the speed of45miles/hour in kilometres per hour. (2)22. The area A of a circle is given by A = πr2. Findthe area of a circle of radius r = 3.73cm, correctto 2 decimal places. (3)23. Evaluate B, correct to 3 signiﬁcant ﬁgures, whenW = 7.20,v = 10.0 and g = 9.81, given thatB =Wv22g(3)24. Express 56.25% as a fraction in its simplestform. (3)25. 12.5% of a length of wood is 70cm. What is thefull length? (3)26. A metal rod, 1.20m long, is heated and its lengthexpands by 42mm. Calculate the percentage inc-rease in length. (2)27. A man buys a house and makes a 20% proﬁt whenhe sells it three years later for £312000. What didhe pay for it originally? (3)
• Chapter 6Ratio and proportion6.1 IntroductionRatio is a way of comparing amounts of something; itshows how much bigger one thing is than the other.Some practical examples include mixing paint, sandand cement, or screen wash. Gears, map scales, foodrecipes, scale drawings and metal alloy constituents alluse ratios.Two quantities are in direct proportion when theyincrease or decrease in the same ratio. There are sev-eral practical engineering laws which rely on directproportion. Also, calculating currency exchange ratesand converting imperial to metric units rely on directproportion.Sometimes, as one quantity increases at a particularrate, another quantity decreases at the same rate; this iscalled inverse proportion. For example, the time takento do a job is inversely proportional to the number ofpeople in a team: double the people, half the time.When we have completed this chapter on ratio andproportion you will be able to understand, and conﬁ-dently perform, calculations on the above topics.For this chapter you will need to know about decimalsand fractions and to be able to use a calculator.6.2 RatiosRatios are generally shown as numbers separated by acolon (:) so the ratio of 2 and 7 is written as 2:7 and weread it as a ratio of ‘two to seven.’Some practical examples which are familiar include:• Mixing 1 measure of screen wash to 6 measures ofwater; i.e., the ratio of screen wash to water is 1:6• Mixing 1 shovel of cement to 4 shovels of sand; i.e.,the ratio of cement to sand is 1:4• Mixing 3 parts of red paint to 1 part white, i.e., theratio of red to white paint is 3:1Ratio is the number of parts to a mix. The paint mix is4 parts total, with 3 parts red and 1 part white. 3 partsred paint to 1 part white paint means there is34red paint to14white paintHere are some worked examples to help us understandmore about ratios.Problem 1. In a class, the ratio of female to malestudents is 6:27. Reduce the ratio to its simplestform(i) Both 6 and 27 can be divided by 3.(ii) Thus, 6:27 is the same as 2:9.6:27 and 2:9 are called equivalent ratios.It is normal to express ratios in their lowest, or simplest,form. In this example, the simplest form is 2:9 whichmeans for every 2 females in the class there are 9 malestudents.Problem 2. A gear wheel having 128 teeth is inmesh with a 48-tooth gear. What is the gear ratio?Gear ratio = 128:48A ratio can be simpliﬁed by ﬁnding common factors.(i) 128 and 48 can both be divided by 2, i.e. 128:48is the same as 64:24(ii) 64 and 24 can both be divided by 8, i.e. 64:24 isthe same as 8:3(iii) There is no number that divides completely intoboth 8 and 3 so 8:3 is the simplest ratio, i.e. thegear ratio is 8:3DOI: 10.1016/B978-1-85617-697-2.00006-5
• Ratio and proportion 41Thus, 128:48 is equivalent to 64:24 which is equivalentto 8:3 and 8:3 is the simplest form.Problem 3. A wooden pole is 2.08m long. Divideit in the ratio of 7 to 19(i) Since the ratio is 7:19, the total number of partsis 7 + 19 = 26 parts.(ii) 26 parts corresponds to 2.08m = 208cm, hence,1 part corresponds to20826= 8.(iii) Thus, 7 parts corresponds to 7 × 8 = 56 cm and19 parts corresponds to 19 × 8 = 152 cm.Hence, 2.08m divides in the ratio of 7:19 as 56 cmto 152 cm.(Check: 56 + 152must add up to 208, otherwise an errorwould have been made.)Problem 4. In a competition, prize money of£828 is to be shared among the ﬁrst three in theratio 5:3:1(i) Since the ratio is 5:3:1 the total number of partsis 5 + 3 + 1 = 9 parts.(ii) 9 parts corresponds to £828.(iii) 1 part corresponds to8289= £92, 3 parts cor-responds to 3 × £92 = £276 and 5 parts corre-sponds to 5 × £92 = £460.Hence, £828 divides in the ratio of 5:3:1 as £460 to£276 to £92. (Check: 460 + 276 + 92 must add up to828, otherwise an error would have been made.)Problem 5. A map scale is 1:30000. On the mapthe distance between two schools is 6cm.Determine the actual distance between the schools,giving the answer in kilometresActual distance between schools= 6 × 30000 cm = 180000 cm=180,000100m = 1800 m=18001000m = 1.80 km(1mile ≈ 1.6km, hence the schools are just over 1mileapart.)Now try the following Practice ExercisePracticeExercise 24 Ratios (answers onpage 342)1. In a box of 333 paper clips, 9 are defective.Express the number of non-defective paperclipsasaratio ofthenumberofdefectivepaperclips, in its simplest form.2. A gear wheel having 84 teeth is in mesh witha 24-tooth gear. Determine the gear ratio in itssimplest form.3. In a box of 2000 nails, 120 are defective.Express the number of non-defective nails asa ratio of the number of defective ones, in itssimplest form.4. A metal pipe 3.36m long is to be cut into twoin the ratio 6 to 15. Calculate the length of eachpiece.5. The instructions for cooking a turkey say thatit needs to be cooked 45 minutes for everykilogram. How long will it take to cook a 7kgturkey?6. In a will, £6440 is to be divided among threebeneﬁciaries in the ratio 4:2:1. Calculate theamount each receives.7. A local map has a scale of 1:22500. The dis-tance between two motorways is 2.7km. Howfar are they apart on the map?8. Prize money in a lottery totals £3801 and isshared among three winners in the ratio 4:2:1.How much does the ﬁrst prize winner receive?Here are some further worked examples on ratios.Problem 6. Express 45p as a ratio of £7.65 in itssimplest form(i) Changing both quantities to the same units, i.e. topence, gives a ratio of 45:765(ii) Dividing both quantities by 5 gives45:765 ≡ 9:153(iii) Dividing both quantities by 3 gives9:153 ≡ 3:51(iv) Dividing both quantities by 3 again gives3:51 ≡ 1:17
• Ratio and proportion 43Here are some worked examples to help us understandmore about direct proportion.Problem 9. 3 energy saving light bulbs cost£7.80. Determine the cost of 7 such light bulbs(i) 3 light bulbs cost £7.80(ii) Therefore, 1 light bulb costs7.803= £2.60Hence, 7 light bulbs cost 7 × £2.60 = £18.20Problem 10. If 56litres of petrol costs £59.92,calculate the cost of 32 litres(i) 56litres of petrol costs £59.92(ii) Therefore, 1 litre of petrol costs59.9256= £1.07Hence, 32 litres cost 32 × 1.07 = £34.24Problem 11. Hooke’s law states that stress, σ, isdirectly proportional to strain, ε, within the elasticlimit of a material. When, for mild steel, the stressis 63MPa, the strain is 0.0003. Determine (a) thevalue of strain when the stress is 42 MPa, (b) thevalue of stress when the strain is 0.00072(a) Stress is directly proportional to strain.(i) When the stress is 63MPa, the strain is0.0003(ii) Hence, a stress of 1MPa corresponds to astrain of0.000363(iii) Thus,the value of strain when the stress is42 MPa =0.000363× 42 = 0.0002(b) Strain is proportional to stress.(i) When the strain is 0.0003, the stress is63MPa.(ii) Hence, a strain of 0.0001 corresponds to633MPa.(iii) Thus,the value of stress when the strain is0.00072 =633× 7.2 = 151.2 MPa.Problem 12. Charles’s law states that for a givenmass of gas at constant pressure, the volume isdirectly proportional to its thermodynamictemperature. A gas occupies a volume of 2.4litresat 600 K. Determine (a) the temperature when thevolume is 3.2 litres, (b) the volume at 540 K(a) Volume is directly proportionalto temperature.(i) When the volume is 2.4litres, the tempera-ture is 600 K.(ii) Hence, a volume of 1 litre corresponds to atemperature of6002.4K.(iii) Thus,the temperature when the volume is3.2 litres =6002.4× 3.2 = 800K.(b) Temperature is proportional to volume.(i) When the temperature is 600 K, the volumeis 2.4litres.(ii) Hence, a temperature of 1K corresponds toa volume of2.4600litres.(iii) Thus, the volume at a temperature of540 K =2.4600× 540 = 2.16 litres.Now try the following Practice ExercisePracticeExercise 26 Direct proportion(answers on page 342)1. 3 engine parts cost £208.50. Calculate the costof 8 such parts.2. If 9litres of gloss white paint costs £24.75,calculate the cost of 24litres of the same paint.3. The total mass of 120 household bricks is57.6kg. Determine the mass of 550 suchbricks.4. A simple machine has an effort:load ratio of3:37. Determine the effort, in grams, to lift aload of 5.55kN.5. If 16 cans of lager weighs 8.32 kg, what will28 cans weigh?6. Hooke’s law states that stress is directly pro-portional to strain within the elastic limit ofa material. When, for copper, the stress is60 MPa, the strain is 0.000625. Determine(a) the strain when the stress is 24MPa and(b) the stress when the strain is 0.0005
• 44 Basic Engineering Mathematics7. Charles’s law states that volume is directlyproportional to thermodynamic temperaturefor a given mass of gas at constant pressure.A gas occupies a volume of 4.8litres at 330 K.Determine (a) the temperature when the vol-ume is 6.4litres and (b) the volume when thetemperature is 396K.Here are some further worked examples on directproportion.Problem 13. Some guttering on a house has todecline by 3mm for every 70 cm to allow rainwaterto drain. The gutter spans 8.4m. How much lowershould the low end be?(i) The guttering has to decline in the ratio 3:700 or3700(ii) If d is the vertical drop in 8.4m or 8400 mm, thenthe decline must be in the ratio d :8400 ord8400(iii) Nowd8400=3700(iv) Cross-multiplyinggives 700×d =8400×3 fromwhich, d =8400 × 3700i.e. d = 36mm, which is how much the lower endshould be to allow rainwater to drain.Problem 14. Ohm’s law state that the currentﬂowing in a ﬁxed resistance is directly proportionalto the applied voltage. When 90 mV is appliedacross a resistor the current ﬂowing is 3A.Determine (a) the current when the voltage is60 mV and (b) the voltage when the current is 4.2 A(a) Current is directly proportional to the voltage.(i) When voltage is 90 mV, the current is 3A.(ii) Hence, a voltage of 1mV corresponds to acurrent of390A.(iii) Thus, when the voltage is 60 mV, thecurrent = 60 ×390= 2A.(b) Voltage is directly proportional to the current.(i) When current is 3A, the voltage is 90 mV.(ii) Hence, a current of 1A corresponds to avoltage of903mV = 30 mV.(iii) Thus, when the current is 4.2 A, thevoltage = 30 × 4.2 = 126mV.Problem 15. Some approximate imperial tometric conversions are shown in Table 6.1. Use thetable to determine(a) the number of millimetres in 12.5inches(b) a speed of 50 miles per hour in kilometresper hour(c) the number of miles in 300 km(d) the number of kilograms in 20 pounds weight(e) the number of pounds and ounces in56kilograms (correct to the nearest ounce)(f) the number of litres in 24 gallons(g) the number of gallons in 60 litresTable 6.1length 1inch = 2.54cm1mile = 1.6kmweight 2.2 lb = 1kg(1lb = 16oz)capacity 1.76 pints = 1litre(8 pints = 1 gallon)(a) 12.5inches = 12.5 × 2.54cm = 31.75cm31.73cm = 31.75 × 10 mm = 317.5 mm(b) 50 m.p.h. = 50 × 1.6km/h = 80 km/h(c) 300 km =3001.6miles = 186.5 miles(d) 20 lb =202.2kg = 9.09 kg(e) 56kg = 56 × 2.2 lb = 123.2 lb0.2 lb = 0.2 × 16oz = 3.2 oz = 3oz, correct tothe nearest ounce.Thus, 56kg = 123 lb 3 oz, correct to the nearestounce.(f) 24 gallons = 24 × 8 pints = 192 pints192 pints =1921.76litres = 109.1 litres
• Ratio and proportion 45(g) 60 litres = 60× 1.76 pints = 105.6 pints105.6 pints =105.68gallons = 13.2 gallonsProblem 16. Currency exchange rates for ﬁvecountries are shown in Table 6.2. Calculate(a) how many euros £55 will buy(b) the number of Japanese yen which can bebought for £23(c) the number of pounds sterling which can beexchanged for 6405kronor(d) the number of American dollars which can bepurchased for £92.50(e) the number of pounds sterling which can beexchanged for 2925 Swiss francsTable 6.2France £1 = 1.25 eurosJapan £1 = 185 yenNorway £1 = 10.50kronorSwitzerland £1 = 1.95francsUSA £1 = 1.80 dollars(a) £1 = 1.25 euros, hence £55 = 55 × 1.25 euros= 68.75 euros.(b) £1 = 185 yen, hence £23 = 23 × 185 yen= 4255 yen.(c) £1 = 10.50kronor, hence 6405 lira = £640510.50= £610.(d) £1 = 1.80 dollars, hence£92.50 = 92.50 × 1.80 dollars = \$166.50(e) £1 = 1.95 Swiss francs, hence2925 pesetas = £29251.95= £1500Now try the following Practice ExercisePracticeExercise 27 Further directproportion (answers on page 342)1. Ohm’s law states that current is proportionalto p.d. in an electrical circuit. When a p.d. of60 mV is applied across a circuit a current of24μAﬂows. Determine(a)thecurrent ﬂowingwhen the p.d. is 5V and (b) the p.d. when thecurrent is 10 mA.2. The tourist rate for the Swiss franc is quoted ina newspaper as £1 = 1.92 fr. How many francscan be purchased for £326.40?3. If 1inch = 2.54cm, ﬁnd the number of mil-limetres in 27inches.4. If 2.2 lb = 1kg and 1lb = 16oz, determine thenumber of poundsand ounces in 38kg (correctto the nearest ounce).5. If 1litre = 1.76 pints and 8 pints = 1 gallon,determine (a) the number of litres in 35 gallonsand (b) the number of gallons in 75litres.6. Hooke’s law states that stress is directly pro-portional to strain within the elastic limit of amaterial. When for brass the stress is 21MPa,the strain is 0.00025. Determine the stresswhen the strain is 0.00035.7. If 12 inches = 30.48cm, ﬁnd the number ofmillimetres in 23inches.8. The tourist rate for the Canadian dollar isquoted in a newspaper as £1 = 1.84fr. Howmany Canadian dollars can be purchased for£550?6.4 Inverse proportionTwo variables, x and y, are in inverse proportion to oneanother if y is proportional to1x, i.e. y α1xor y =kxork = xy where k is a constant, called the coefﬁcient ofproportionality.Inverse proportion means that, as the value ofone vari-able increases, the value of another decreases, and thattheir product is always the same.For example, the time for a journey is inversely propor-tional to the speed of travel. So, if at 30m.p.h. a journeyis completed in20 minutes, then at 60m.p.h. the journeywould be completed in 10 minutes. Double the speed,half the journey time. (Note that 30 × 20 = 60× 10.)In another example, the time needed to dig a hole isinversely proportional to the number of people digging.So, if 4 men take 3 hours to dig a hole, then 2 men
• 46 Basic Engineering Mathematics(working at the same rate) would take 6 hours. Half themen, twice the time. (Note that 4 × 3 = 2 × 6.)Here are some worked examples on inverse proportion.Problem 17. It is estimated that a team of fourdesigners would take a year to develop anengineering process. How long would threedesigners take?If 4 designers take 1 year, then 1 designer would take4 yearsto develop theprocess.Hence, 3 designerswouldtake43years, i.e. 1 year 4 months.Problem 18. A team of ﬁve people can deliverleaﬂets to every house in a particular area in fourhours. How long will it take a team of three people?If 5 people take 4 hours to deliver the leaﬂets, then1 person would take 5 × 4 = 20 hours. Hence, 3 peo-ple would take203hours, i.e. 623hours, i.e. 6 hours40 minutes.Problem 19. The electrical resistance R of apiece of wire is inversely proportional to thecross-sectional area A. When A = 5mm2,R = 7.02 ohms. Determine (a) the coefﬁcient ofproportionality and (b) the cross-sectional areawhen the resistance is 4ohms(a) Rα1A, i.e. R =kAor k = RA. Hence, whenR = 7.2 and A = 5, thecoefﬁcient of proportionality,k = (7.2)(5) = 36(b) Since k = RA then A =kR. Hence, when R = 4,the cross sectional area, A =364= 9 mm2Problem 20. Boyle’s law states that, at constanttemperature, the volume V of a ﬁxed mass of gas isinversely proportional to its absolute pressure p. Ifa gas occupies a volume of 0.08m3at a pressure of1.5 × 106pascals, determine (a) the coefﬁcient ofproportionality and (b) the volume if the pressure ischanged to 4 × 106 pascals(a) V ∝1pi.e. V =kpor k = pV . Hence, thecoefﬁcient of proportionality, k= (1.5 × 106)(0.08) = 0.12 × 106(b) Volume,V =kp=0.12 × 1064 × 106= 0.03m3Now try the following Practice ExercisePracticeExercise 28 Further inverseproportion (answers on page 342)1. A 10 kg bag of potatoes lasts for a week with afamily of 7 people. Assuming all eat the sameamount, how longwill the potatoeslast ifthereare only two in the family?2. If 8 men take 5 days to build a wall, how longwould it take 2 men?3. If y is inversely proportional to x andy = 15.3 when x = 0.6, determine (a) thecoefﬁcient of proportionality, (b) the value ofy when x is 1.5 and (c) the value of x when yis 27.24. A car travelling at 50 km/h makes a journey in70 minutes. How long will the journey take at70 km/h?5. Boyle’s law states that, for a gas at constanttemperature, the volume of a ﬁxed mass ofgas is inversely proportional to its absolutepressure. If a gas occupies a volume of 1.5m3at a pressure of 200 × 103 pascals, determine(a) the constant of proportionality, (b) thevolume when the pressure is 800 × 103 pas-cals and (c) the pressure when the volume is1.25m3.
• Chapter 7Powers, roots and laws ofindices7.1 IntroductionThe manipulationof powers and rootsis a crucial under-lying skill needed in algebra. In this chapter, powers androots of numbers are explained, together with the lawsof indices.Many worked examplesareincluded to help understand-ing.7.2 Powers and roots7.2.1 IndicesThe number 16 is the same as 2× 2 × 2 × 2, and 2 × 2 ×2 × 2 can be abbreviated to 24. When written as 24, 2 iscalled the base and the 4 is called the index or power.24 is read as ‘two to the power of four’.Similarly, 35 is read as ‘three to the power of 5’.When the indices are 2 and 3 they are given specialnames; i.e. 2 is called ‘squared’ and 3 is called ‘cubed’.Thus,42 is called ‘four squared’ rather than ‘4 to the powerof 2’ and53 is called ‘ﬁve cubed’ rather than ‘5 to the power of 3’When no index is shown, the power is 1. For example,2 means 21.Problem 1. Evaluate (a) 26 (b) 34(a) 26 means 2 × 2 × 2 × 2× 2 × 2 (i.e. 2 multipliedby itself 6 times), and 2 × 2 × 2 × 2 × 2 × 2 = 64i.e. 26 = 64(b) 34 means 3 × 3 × 3 × 3 (i.e. 3 multiplied by itself4 times), and 3 × 3 × 3 × 3 = 81i.e. 34 = 81Problem 2. Change the following to index form:(a) 32 (b) 625(a) (i) To express 32 in itslowest factors, 32 is initiallydivided by the lowest prime number, i.e. 2.(ii) 32 ÷ 2 = 16, hence 32 = 2 × 16.(iii) 16 is also divisible by 2, i.e. 16 = 2 × 8. Thus,32 = 2 × 2× 8.(iv) 8 is also divisible by 2, i.e. 8 = 2 × 4. Thus,32 = 2 × 2× 2 × 4.(v) 4 is also divisible by 2, i.e. 4 = 2 × 2. Thus,32 = 2 × 2 × 2 × 2 × 2.(vi) Thus, 32 = 25.(b) (i) 625 is not divisible by the lowest prime num-ber, i.e. 2. The next prime number is 3 and 625is not divisible by 3 either. The next primenumber is 5.(ii) 625 ÷ 5 = 125, i.e. 625 = 5 × 125.(iii) 125 is also divisible by 5, i.e. 125 = 5 × 25.Thus, 625 = 5 × 5 × 25.(iv) 25 is also divisible by 5, i.e. 25 = 5 × 5. Thus,625 = 5 × 5 × 5 × 5.(v) Thus, 625 = 54.DOI: 10.1016/B978-1-85617-697-2.00007-7
• 48 Basic Engineering MathematicsProblem 3. Evaluate 33× 2233× 22= 3 × 3 × 3 × 2× 2= 27 × 4= 1087.2.2 Square rootsWhen a number is multiplied by itself the product iscalled a square.For example, the square of 3 is 3 × 3 = 32= 9.A square root is the reverse process; i.e., the value of thebase which when multiplied by itself gives the number;i.e., the square root of 9 is 3.The symbol√is used to denote a square root. Thus,√9 = 3. Similarly,√4 = 2 and√25 = 5.Because −3 × −3 = 9,√9 also equals −3. Thus,√9 =+3 or −3 which is usually written as√9 = ±3. Simi-larly,√16 = ±4 and√36 = ±6.The square root of, say, 9 may also be written in indexform as 912 .912 ≡√9 = ±3Problem 4. Evaluate32 × 23 ×√36√16 × 4taking onlypositive square roots32 × 23 ×√36√16 × 4=3 × 3 × 2× 2 × 2 × 64 × 4=9 × 8 × 616=9 × 1 × 62=9 × 1 × 31by cancelling= 27Problem 5. Evaluate104 ×√100103taking thepositive square root only104 ×√100103=10 × 10× 10 × 10 × 1010 × 10 × 10=1 × 1 × 1 × 10× 101 × 1 × 1by cancelling=1001= 100Now try the following Practice ExercisePracticeExercise 29 Powers and roots(answers on page 342)Evaluate the following without the aid of a calcu-lator.1. 33 2. 273. 105 4. 24 × 32 × 2 ÷ 35. Change 16 to 6. 2512index form.7. 6412 8.1051039.102× 10310510.25× 6412 × 32√144 × 3taking positivesquare roots only.7.3 Laws of indicesThere are six laws of indices.(1) From earlier, 22 × 23 = (2 × 2) × (2 × 2 × 2)= 32= 25Hence, 22 × 23 = 25or 22× 23= 22+3This is the ﬁrst law of indices, which demonstratesthat when multiplying two or more numbershaving the same base, the indices are added.(2)2523=2 × 2 × 2 × 2 × 22 × 2 × 2=1 × 1 × 1 × 2× 21 × 1 × 1=2 × 21= 4 = 22Hence,2523= 22or2523= 25−3This is the second law of indices, which demon-strates that when dividing two numbers havingthe same base, the index in the denominator issubtracted from the index in the numerator.(3) (35)2 = 35×2 = 310 and (22)3 = 22×3 = 26This is the third law of indices, which demon-strates that when a number which is raised toa power is raised to a further power, the indicesare multiplied.
• Powers, roots and laws of indices 49(4) 30= 1 and 170= 1This is the fourth law of indices, which states thatwhen a number has an index of 0, its value is 1.(5) 3−4 =134and12−3= 23This isthe ﬁfth law of indices, which demonstratesthat a number raised to a negative power is thereciprocal of that number raised to a positivepower.(6) 823 =3√82 = (2)2 = 4 and2512 =2√251 =√251 = ±5(Note that√≡ 2√)This is the sixth law of indices, which demon-strates that when a number is raised to a frac-tional power the denominator of the fraction isthe root of the number and the numerator is thepower.Here are some worked examples using the laws ofindices.Problem 6. Evaluate in index form 53 × 5 × 5253× 5 × 52= 53× 51× 52(Note that 5 means 51)= 53+1+2from law (1)= 56Problem 7. Evaluate35343534= 35−4from law (2)= 31= 3Problem 8. Evaluate24242424= 24−4from law (2)= 20But2424=2 × 2 × 2 × 22 × 2 × 2 × 2=1616= 1Hence, 20= 1 from law (4)Any number raised to the power of zero equals 1. Forexample, 60= 1,1280= 1,137420= 1 and so on.Problem 9. Evaluate3 × 32343 × 3234=31 × 3234=31+234=3334= 33−4= 3−1from laws (1) and (2)But3334=3 × 3 × 33 × 3 × 3 × 3=1 × 1 × 11 × 1 × 1 × 3(by cancelling)=13Hence,3 × 3234= 3−1=13from law (5)Similarly, 2−1=12,2−5=125,154= 5−4and so on.Problem 10. Evaluate103 × 102108103 × 102108=103+2108=105108from law (1)= 105−8= 10−3from law (2)=110+3=11000from law (5)Hence,103 × 102108= 10−3=11000= 0.001Understanding powers of ten is important, especiallywhen dealing with preﬁxes in Chapter 8. For example,102= 100,103= 1000,104= 10000,105= 100000,106= 100000010−1=110= 0.1,10−2=1102=1100= 0.01and so on.Problem 11. Evaluate (a) 52 × 53 ÷ 54(b) (3 × 35) ÷ (32× 33)From laws (1) and (2):(a) 52 × 53 ÷ 54 =52 × 5354=5(2+3)54=5554= 5(5−4) = 51 = 5
• 50 Basic Engineering Mathematics(b) (3 × 35) ÷ (32 × 33) =3 × 3532 × 33=3(1+5)3(2+3)=3635= 36−5 = 31 = 3Problem 12. Simplify (a) (23)4 (b) (32)5,expressing the answers in index formFrom law (3):(a) (23)4 = 23×4 = 212(b) (32)5 = 32×5 = 310Problem 13. Evaluate:(102)3104 × 102From laws (1) to (4):(102)3104 × 102=10(2×3)10(4+2)=106106= 106−6= 100= 1Problem 14. Find the value of (a)23 × 2427 × 25(b)(32)33 × 39From the laws of indices:(a)23 × 2427 × 25=2(3+4)2(7+5)=27212= 27−12= 2−5 =125=132(b)(32)33 × 39=32×331+9=36310= 36−10= 3−4 =134=181Problem 15. Evaluate (a) 41/2 (b) 163/4 (c) 272/3(d) 9−1/2(a) 41/2 =√4 = ±2(b) 163/4 =4√163 = (2)3 = 8(Note that it does not matter whether the 4th rootof 16 is found ﬁrst or whether 16 cubed is foundﬁrst – the same answer will result.)(c) 272/3 =3√272 = (3)2 = 9(d) 9−1/2 =191/2=1√9=1±3= ±13Now try the following Practice ExercisePracticeExercise 30 Laws of indices(answers on page 342)Evaluate the following without the aid of acalculator.1. 22 × 2 × 24 2. 35 × 33 × 3in index form3.27234.33355. 70 6.23 × 2 × 26277.10 × 1061058. 104 ÷ 109.103 × 10410910. 56 × 52 ÷ 5711. (72)3 in index form 12. (33)213.37 × 3435in 14.(9 × 32)3(3 × 27)2inindex form index form15.(16 × 4)2(2 × 8)316.5−25−417.32 × 3−43318.72 × 7−37 × 7−419.23 × 2−4 × 252 × 2−2 × 2620.5−7 × 525−8 × 53Here are some further worked examples using the lawsof indices.Problem 16. Evaluate33 × 5753 × 34The laws of indices only apply to terms having thesame base. Grouping terms having the same base andthen applying the laws of indices to each of the groupsindependently gives33 × 5753 × 34=3334×5753= 3(3−4)× 5(7−3)= 3−1× 54=5431=6253= 20813
• Powers, roots and laws of indices 51Problem 17. Find the value of23 × 35 × (72)274 × 24 × 3323 × 35 × (72)274 × 24 × 33= 23−4× 35−3× 72×2−4= 2−1× 32× 70=12× 32× 1 =92= 412Problem 18. Evaluate41.5 × 81/322 × 32−2/541.5= 43/2=√43 = 23= 8, 81/3=3√8 = 2,22= 4, 32−2/5=1322/5=15√322=122=14Hence,41.5 × 81/322 × 32−2/5=8 × 24 ×14=161= 16Alternatively,41.5 × 81/322 × 32−2/5=[(2)2]3/2 × (23)1/322 × (25)−2/5=23 × 2122 × 2−2= 23+1−2−(−2)= 24= 16Problem 19. Evaluate32 × 55 + 33 × 5334 × 54Dividing each termby theHCF(highest common factor)of the three terms, i.e. 32 × 53, gives32 × 55 + 33 × 5334 × 54=32 × 5532 × 53+33 × 5332 × 5334× 5432 × 53=3(2−2) × 5(5−3) + 3(3−2) × 503(4−2) × 5(4−3)=30 × 52 + 31 × 5032 × 51=1 × 25 + 3 × 19 × 5=2845Problem 20. Find the value of32 × 5534 × 54 + 33 × 53To simplify the arithmetic, each term is divided by theHCF of all the terms, i.e. 32 × 53. Thus,32 × 5534 × 54 + 33 × 53=32 × 5532 × 5334 × 5432 × 53+33 × 5332 × 53=3(2−2) × 5(5−3)3(4−2) × 5(4−3) + 3(3−2) × 5(3−3)=30 × 5232 × 51 + 31 × 50=1 × 5232 × 5 + 3 × 1=2545 + 3=2548Problem 21. Simplify7−3 × 343−2 × 75 × 5−2expressing the answer in index form with positiveindicesSince 7−3 =173,13−2= 32 and15−2= 52, then7−3× 343−2 × 75 × 5−2=34× 32× 5273 × 75=3(4+2) × 527(3+5)=36 × 5278Problem 22. Simplify162 × 9−24 × 33 − 2−3 × 82expressing the answer in index form with positiveindicesExpressing the numbers in terms of their lowest primenumbers gives162 × 9−24 × 33 − 2−3 × 82=(24)2 × (32)−222 × 33 − 2−3 × (23)2=28 × 3−422 × 33 − 2−3 × 26=28 × 3−422 × 33 − 23
• 52 Basic Engineering MathematicsDividing each term by the HCF (i.e. 22) gives28 × 3−422 × 33 − 23=26 × 3−433 − 2=2634(33 − 2)Problem 23. Simplify433×35−225−3givingthe answer with positive indicesRaising a fraction to a power means that both the numer-ator and the denominator of the fraction are raised to thatpower, i.e.433=4333A fraction raised to a negative power has the same valueas the inverse of the fraction raised to a positive power.Thus,35−2=1352=13252= 1 ×5232=5232Similarly,25−3=523=5323Thus,433×35−225−3=4333×52325323=4333×5232×2353=(22)3 × 233(3+2) × 5(3−2)=2935 × 5Now try the following Practice ExercisePracticeExercise 31 Further problems onindices (answers on page 342)In problems 1 to 4, simplify the expressions given,expressing the answers in index form and withpositive indices.1.33 × 5254 × 342.7−2 × 3−235 × 74 × 7−33.42 × 9383 × 344.8−2 × 52 × 3−4252 × 24 × 9−2In Problems5 to 15,evaluatetheexpressionsgiven.5.132−16. 810.257. 16−14 8.491/29.92 × 7434 × 74 + 33 × 7210.33 × 5223 × 32 − 82 × 911.33 × 72 − 52 × 7332 × 5 × 7212.(24)2 − 3−2 × 4423 × 16213.123−23−232214.43429215.(32)3/2 × (81/3)2(3)2 × (43)1/2 × (9)−1/2
• Chapter 8Units, prefixes andengineering notation8.1 IntroductionOf considerable importance in engineering is a knowl-edge ofunitsof engineering quantities,the preﬁxes usedwith units, and engineering notation.We need to know, for example, that80kV = 80 × 103 V, which means 80000 voltsand 25mA = 25 × 10−3A,which means 0.025 amperesand 50nF = 50 × 10−9F,which means 0.000000050 faradsThis is explained in this chapter.8.2 SI unitsThe system of units used in engineering and scienceis the Système Internationale d’Unités (InternationalSystem of Units), usually abbreviated to SI units, and isbased on the metric system. This was introduced in 1960and has now been adopted by the majority of countriesas the official system of measurement.The basic seven units used in the SI system are listed inTable 8.1 with their symbols.There are, of course, many units other than these seven.These other units are called derived units and aredefined in terms of the standard units listed in the table.For example, speed is measured in metres per second,therefore using two of the standard units, i.e. length andtime.Table 8.1 Basic SI unitsQuantity Unit SymbolLength metre m (1m = 100 cm= 1000 mm)Mass kilogramkg (1kg = 1000 g)Time second sElectric current ampere AThermodynamictemperaturekelvin K (K = ◦C + 273)Luminousintensitycandela cdAmount ofsubstancemole molSome derived units are given special names. For exam-ple, force = mass × acceleration has units of kilogrammetre per second squared, which uses three of the baseunits, i.e. kilograms, metres and seconds. The unit of kgm/s2 is given the special name of a Newton.Table 8.2 contains a list of some quantities and theirunits that are common in engineering.8.3 Common prefixesSI units may be made larger or smaller by using prefixeswhich denote multiplication or division by a particularamount.DOI: 10.1016/B978-1-85617-697-2.00008-9
• 54 Basic Engineering MathematicsTable 8.2 Some quantities and their units that are common in engineeringQuantity Unit SymbolLength metre mArea square metre m2Volume cubic metre m3Mass kilogram kgTime second sElectric current ampere ASpeed, velocity metre per second m/sAcceleration metre per second squared m/s2Density kilogram per cubic metre kg/m3Temperature kelvin or Celsius K or ◦CAngle radian or degree rad or ◦Angular velocity radian per second rad/sFrequency hertz HzForce newton NPressure pascal PaEnergy, work joule JPower watt WCharge, quantity of electricity coulomb CElectric potential volt VCapacitance farad FElectrical resistance ohmInductance henry HMoment of force newton metre NmThe most common multiples are listed in Table 8.3.Aknowledgeofindicesisneeded sinceall ofthepreﬁxesare powers of 10 with indices that are a multiple of 3.Here are some examples of preﬁxes used with engineer-ing units.A frequency of 15 GHz means 15 × 109 Hz, which is15000000000 hertz,i.e. 15 gigahertz is written as 15GHz and is equal to 15thousand million hertz.(Instead of writing 15000000000 hertz, it is muchneater, takes up less space and prevents errors causedby having so many zeros, to write the frequency as15GHz.)A voltage of 40 MV means 40 × 106 V, which is40000000 volts,i.e. 40 megavolts is written as 40MV and is equal to 40million volts.An inductance of 12 mH means 12 × 10−3 H or12103H or121000H, which is 0.012H,i.e. 12 millihenrys is written as 12mH and is equal to12 thousandths of a henry.
• Units, prefixes and engineering notation 55Table 8.3 Common SI multiplesPrefix Name MeaningG giga multiply by 109i.e. × 1000000000M mega multiply by 106 i.e. × 1000000k kilo multiply by 103 i.e. × 1000m milli multiply by 10−3 i.e. ×1103=11000= 0.001μ micro multiply by 10−6 i.e. ×1106=11000000= 0.000001n nano multiply by 10−9 i.e. ×1109=11000000000= 0.000 000001p pico multiply by 10−12 i.e. ×11012=11000000000000=0.000000000001A time of 150 ns means 150 × 10−9 s or150109s, whichis 0.000000150s,i.e. 150 nanoseconds is written as 150ns and is equal to150 thousand millionths of a second.A force of 20 kN means 20 × 103 N, which is 20000newtons,i.e. 20 kilonewtons is written as 20kN and is equal to20 thousand newtons.A charge of 30 μC means 30 × 10−6 C or30106C, whichis 0.000030C,i.e. 30 microcoulombs is written as 30μC and is equalto 30 millionths of a coulomb.A capacitance of 45pF means 45 × 10−12 F or451012F,which is 0.000000000045F,i.e. 45 picofarads is written as 45pF and is equal to 45million millionths of a farad.In engineering it is important to understand what suchquantities as 15GHz, 40MV, 12mH, 150ns, 20kN,30μC and 45pF mean.Now try the following Practice ExercisePracticeExercise 32 SI units and commonprefixes (answers on page 343)1. State the SI unit of volume.2. State the SI unit of capacitance.3. State the SI unit of area.4. State the SI unit of velocity.5. State the SI unit of density.6. State the SI unit of energy.7. State the SI unit of charge.8. State the SI unit of power.9. State the SI unit of angle.10. State the SI unit of electric potential.11. State which quantity has the unit kg.12. State which quantity has the unit symbol .13. State which quantity has the unit Hz.14. State which quantity has the unit m/s2.15. State which quantity has the unit symbol A.16. State which quantity has the unit symbol H.17. State which quantity has the unit symbol m.18. State which quantity has the unit symbol K.19. State which quantity has the unit Pa.20. State which quantity has the unit rad/s.21. What does the prefix G mean?22. What isthe symbol and meaning of the prefixmilli?
• 56 Basic Engineering Mathematics23. What does the prefix p mean?24. What is thesymbol and meaning ofthe prefixmega?8.4 Standard formA number written with one digit to the left ofthe decimalpoint and multiplied by 10 raised to some power is saidto be written in standard form.For example, 43645 = 4.3645 × 104in standard formand 0.0534 = 5.34 × 10−2in standard formProblem 1. Express in standard form (a) 38.71(b) 3746 (c) 0.0124For a number to be in standard form, it is expressed withonly one digit to the left of the decimal point. Thus,(a) 38.71 must be divided by 10 to achieve one digitto the left of the decimal point and it must also bemultiplied by 10 to maintain the equality, i.e.38.71=38.7110×10=3.871×10 in standard form(b) 3746 =37461000× 1000 = 3.746 × 103in standardform.(c) 0.0124 = 0.0124 ×100100=1.24100= 1.24 × 10−2in standard form.Problem 2. Express the following numbers,which are in standard form, as decimal numbers:(a) 1.725 × 10−2(b) 5.491 × 104(c) 9.84 × 100(a) 1.725 × 10−2 =1.725100= 0.01725 (i.e. move thedecimal point 2 places to the left).(b) 5.491 × 104 = 5.491 × 10000 = 54910 (i.e. movethe decimal point 4 places to the right).(c) 9.84 × 100 = 9.84 × 1 = 9.84 (since 100 = 1).Problem 3. Express in standard form, correct to 3significant figures, (a)38(b) 1923(c) 741916(a)38= 0.375, and expressing it in standard formgives0.375 = 3.75 × 10−1(b) 1923= 19..6 = 1.97 × 10 in standard form, cor-rect to 3 signiﬁcant figures.(c) 741916=741.5625=7.42 × 102in standard form,correct to 3 significant figures.Problem 4. Express the following numbers, givenin standard form, as fractions or mixed numbers,(a) 2.5 × 10−1(b) 6.25 × 10−2(c) 1.354 × 102(a) 2.5 × 10−1 =2.510=25100=14(b) 6.25 × 10−2 =6.25100=62510000=116(c) 1.354 × 102 = 135.4 = 135410= 13525Problem 5. Evaluate (a) (3.75 × 103)(6 × 104)(b)3.5 × 1057 × 102, expressing the answers in standardform(a) (3.75 × 103)(6 × 104) = (3.75 × 6)(103+4)= 22.50 × 107= 2.25 × 108(b)3.5 × 1057 × 102=3.57× 105−2 = 0.5 × 103 = 5 × 102Now try the following Practice ExercisePracticeExercise 33 Standard form(answers on page 343)In problems 1 to 5, express in standard form.1. (a) 73.9 (b) 28.4 (c) 197.622. (a) 2748 (b) 33170 (c) 2742183. (a) 0.2401 (b) 0.0174 (c) 0.009234. (a) 1702.3 (b) 10.04 (c) 0.01095. (a)12(b) 1178(c)132(d) 13035
• Units, prefixes and engineering notation 57In problems 6 and 7, express the numbers given asintegers or decimal fractions.6. (a) 1.01 × 103 (b) 9.327 × 102(c) 5.41 × 104(d) 7 × 1007. (a) 3.89 × 10−2 (b) 6.741 × 10−1(c) 8 × 10−3In problems 8 and 9, evaluate the given expres-sions, stating the answers in standard form.8. (a) (4.5 × 10−2)(3 × 103)(b) 2 × (5.5 × 104)9. (a)6 × 10−33 × 10−5(b)(2.4 × 103)(3 × 10−2)(4.8 × 104)10. Write the following statements in standardform.(a) The density of aluminium is 2710kgm−3.(b) Poisson’s ratio for gold is 0.44(c) The impedance of free space is376.73 .(d) The electron rest energy is 0.511MeV.(e) Proton charge–mass ratio is 95789700Ckg−1.(f) The normal volume of a perfect gas is0.02241m3 mol−1.8.5 Engineering notationIn engineering, standard form is not as important asengineering notation. Engineering notation is similarto standard form except that the power of 10 is alwaysa multiple of 3.For example, 43645 = 43.645 × 103in engineering notationand 0.0534 = 53.4 × 10−3in engineering notationFrom the list of engineering prefixes on page 55 it isapparent that all prefixes involve powers of 10 that aremultiples of 3.For example, a force of 43645N can rewritten as43.645 × 103 N and from the list of preﬁxes can thenbe expressed as 43.645kN.Thus, 43645N ≡ 43.645kNTo help further, on your calculator is an ‘ENG’ button.Enter the number 43645 into your calculator and thenpress =. Now press the ENG button and the answer is43.645 × 103. We then have to appreciate that 103 is theprefix ‘kilo’, giving 43645N ≡ 43.645 kN.In another example, let a current be 0.0745A. Enter0.0745 into your calculator. Press =. Now pressENG and the answer is 74.5 × 10−3. We then haveto appreciate that 10−3 is the preﬁx ‘milli’, giving0.0745A ≡ 74.5mA.Problem 6. Express the following in engineeringnotation and in prefix form:(a) 300000W (b) 0.000068H(a) Enter 300000 into the calculator. Press =Now press ENG and the answer is 300 × 103.From the table of preﬁxes on page 55, 103 corre-sponds to kilo.Hence, 300000W = 300 × 103 W in engineeringnotation= 300 kW in prefix form.(b) Enter 0.000068 into the calculator. Press =Now press ENG and the answer is 68 × 10−6.From the table of preﬁxes on page 55, 10−6corresponds to micro.Hence, 0.000068H = 68 × 10−6 H in engineeringnotation= 68μH in prefix form.Problem 7. Express the following in engineeringnotation and in prefix form:(a) 42 × 105(b) 4.7 × 10−10F(a) Enter 42 × 105 into the calculator. Press =Now press ENG and the answer is 4.2 × 106.From the table of preﬁxes on page 55, 106 corre-sponds to mega.
• 58 Basic Engineering MathematicsHence, 42 × 105= 4.2 × 106in engineeringnotation= 4.2 M in prefix form.(b) Enter 47 ÷ 1010 =4710000000000into the calcu-lator. Press =Now press ENG and the answer is 4.7 × 10−9.From the table of preﬁxes on page 55, 10−9corresponds to nano.Hence, 47 ÷1010 F = 4.7 × 10−9 F in engineer-ing notation= 4.7 nF in prefix form.Problem 8. Rewrite (a) 0.056mA in μA(b) 16700kHz as MHz(a) Enter 0.056 ÷ 1000 into the calculator (since millimeans ÷1000). Press =Now press ENG and the answer is 56 × 10−6.From the table of preﬁxes on page 55, 10−6corresponds to micro.Hence, 0.056mA =0.0561000A = 56 × 10−6 A= 56μA.(b) Enter 16700× 1000 into the calculator (since kilomeans ×1000). Press =Now press ENG and the answer is 16.7 × 106.From the table of preﬁxes on page 55, 106 corre-sponds to mega.Hence, 16700kHz = 16700× 1000Hz= 16.7 × 106Hz= 16.7MHzProblem 9. Rewrite (a) 63 × 104 V in kV(b) 3100pF in nF(a) Enter 63 × 104 into the calculator. Press =Now press ENG and the answer is 630 × 103.From the table of preﬁxes on page 55, 103 corre-sponds to kilo.Hence, 63 × 104 V = 630 × 103 V = 630kV.(b) Enter 3100× 10−12into the calculator. Press =Now press ENG and the answer is 3.1 × 10−9.From the table of preﬁxes on page 55, 10−9corresponds to nano.Hence, 3100pF = 31 × 10−12 F = 3.1 × 10−9 F= 3.1 nFProblem 10. Rewrite (a) 14700mm in metres(b) 276cm in metres (c) 3.375kg in grams(a) 1m = 1000 mm, hence1mm =11000=1103= 10−3 m.Hence, 14700mm = 14700 × 10−3 m = 14.7m.(b) 1m=100 cm, hence 1cm=1100=1102=10−2 m.Hence, 276cm = 276 × 10−2m = 2.76m.(c) 1kg = 1000 g = 103 gHence, 3.375kg = 3.375 × 103 g = 3375g.Now try the following Practice ExercisePracticeExercise 34 Engineering notation(answers on page 343)In problems 1 to 12, express in engineering nota-tion in preﬁx form.1. 60000Pa2. 0.00015W3. 5 × 107 V4. 5.5 × 10−8 F5. 100000W6. 0.00054A7. 15 × 1058. 225 × 10−4 V9. 35000000000Hz10. 1.5 × 10−11 F11. 0.000017A12. 46200
• Units, prefixes and engineering notation 5913. Rewrite 0.003mA in μA14. Rewrite 2025kHz as MHz15. Rewrite 5 × 104 N in kN16. Rewrite 300pF in nF17. Rewrite 6250cm in metres18. Rewrite 34.6g in kgIn problems 19 and 20, use a calculator to evaluatein engineering notation.19. 4.5 × 10−7× 3 × 10420.1.6 × 10−5 25 × 103100 × 10−6
• Revision Test 3 : Ratio, proportion, powers, roots, indices and unitsThis assignment covers the material contained in Chapters 6–8. The marks available are shown in brackets at theend of each question.1. In a box of 1500 nails, 125 are defective. Expressthe non-defective nails as a ratio of the defectiveones, in its simplest form. (3)2. Prize money in a lottery totals £4500 and is sharedamong three winners in the ratio 5:3:1. How muchdoes the ﬁrst prize winner receive? (3)3. A simple machine has an effort:load ratio of 3:41.Determine the effort, in newtons, to lift a load of6.15kN. (3)4. If 15 cans of lager weigh 7.8kg, what will 24 cansweigh? (3)5. Hooke’s law states that stress is directly propor-tional to strain within theelasticlimitofamaterial.When for brass the stress is 21MPa, the strain is250 × 10−6. Determine the stress when the strainis 350 × 10−6. (3)6. If 12 inches = 30.48cm, ﬁnd the number of mil-limetres in 17inches. (3)7. If x is inversely proportional to yand x = 12 wheny = 0.4, determine(a) the value of x when y is 3.(b) the value of y when x = 2. (3)8. Evaluate(a) 3 × 23 × 22(b) 4912 (4)9. Evaluate32 ×√36 × 223 × 8112taking positive squareroots only. (3)10. Evaluate 64 × 6 × 62 in index form. (3)11. Evaluate(a)2722(b)104 × 10× 105106 × 102(4)12. Evaluate(a)23 × 2 × 2224(b)23 × 162(8 × 2)3(c)142−1(7)13. Evaluate(a) (27)−13 (b)32−2−29232(5)14. State the SI unit of (a) capacitance (b) electricalpotential (c) work (3)15. State the quantity that has an SI unit of(a) kilograms (b) henrys (c) hertz (d) m3 (4)16. Express the following in engineering notation inpreﬁx form.(a) 250000 J (b) 0.05H(c) 2 × 108 W (d) 750 × 10−8 F (4)17. Rewrite (a) 0.0067mA in μA (b) 40 × 104 kV asMV (2)
• Chapter 9Basic algebra9.1 IntroductionWe are already familiar with evaluating formulae usinga calculator from Chapter 4.For example, if the length of a football pitch is L and itswidth is b, then the formula for the area A is given byA = L × bThis is an algebraic equation.If L = 120m and b = 60m, then the areaA = 120 × 60 = 7200m2.The total resistance, RT , of resistors R1, R2 and R3connected in series is given byRT = R1 + R2 + R3This is an algebraic equation.If R1 = 6.3k , R2 = 2.4k and R3 = 8.5k , thenRT = 6.3 + 2.4 + 8.5 = 17.2kThe temperature in Fahrenheit, F, is given byF =95C + 32where C is the temperature in Celsius. This is analgebraic equation.If C = 100◦C, then F =95× 100 + 32= 180 + 32 = 212◦F.If you can cope with evaluating formulae then you willbe able to cope with algebra.9.2 Basic operationsAlgebra merely uses letters to represent numbers.If, say, a,b,c and d represent any four numbers then inalgebra:(a) a + a + a + a = 4a. For example, if a = 2, then2 + 2 + 2 + 2 = 4 × 2 = 8.(b) 5b means 5 × b. For example, if b = 4, then5b = 5 × 4 = 20.(c) 2a + 3b + a − 2b = 2a + a + 3b − 2b = 3a + bOnly similar terms can be combined in algebra.The 2a and the +a can be combined to give 3aand the 3b and −2b can be combined to give 1b,which is written as b.In addition,with termsseparated by +and −signs,the order in which they are written does not matter.In this example, 2a + 3b + a − 2b is the same as2a + a + 3b − 2b, which is the same as 3b + a +2a − 2b, and so on. (Note that the ﬁrst term, i.e.2a, means +2a.)(d) 4abcd = 4 × a × b × c × dFor example, if a = 3,b = −2,c = 1 and d = −5,then 4abcd = 4 × 3 × −2 × 1 × −5 = 120. (Notethat − × − = +)(e) (a)(c)(d) means a × c × dBrackets are often used instead of multiplicationsigns. For example, (2)(5)(3) means 2 × 5 × 3 =30.(f) ab = baIf a = 2 and b = 3 then 2 × 3 is exactly the sameas 3 × 2, i.e. 6.(g) b2= b × b. For example, if b = 3, then32 = 3 × 3 = 9.(h) a3 = a × a × a For example, if a = 2, then23 = 2 × 2 × 2 = 8.Here are some worked examples to help get a feel forbasic operations in this introduction to algebra.DOI: 10.1016/B978-1-85617-697-2.00009-0
• 62 Basic Engineering Mathematics9.2.1 Addition and subtractionProblem 1. Find the sum of 4x,3x,−2x and −x4x + 3x + −2x + −x = 4x + 3x − 2x − x(Note that + ×− = −)= 4xProblem 2. Find the sum of 5x,3y, z,−3x,−4yand 6z5x+3y + z + −3x + −4y + 6z= 5x + 3y + z − 3x − 4y + 6z= 5x − 3x + 3y − 4y + z + 6z= 2x − y + 7zNote that the order can be changed when terms are sep-arated by + and − signs. Only similar terms can becombined.Problem 3. Simplify 4x2 − x − 2y + 5x + 3y4x2− x − 2y + 5x + 3y = 4x2+ 5x − x + 3y − 2y= 4x2+ 4x + yProblem 4. Simplify 3xy − 7x + 4xy + 2x3xy − 7x + 4xy + 2x = 3xy + 4xy + 2x − 7x= 7xy − 5xNow try the following Practice ExercisePracticeExercise 35 Addition andsubtraction in algebra (answers on page 343)1. Find the sum of 4a,−2a,3a and −8a.2. Find the sum of 2a,5b,−3c,−a,−3b and7c.3. Simplify 2x − 3x2 − 7y + x + 4y − 2y2.4. Simplify 5ab − 4a + ab + a.5. Simplify 2x − 3y + 5z − x − 2y + 3z + 5x.6. Simplify 3 + x + 5x − 2 − 4x.7. Add x − 2y + 3 to 3x + 4y − 1.8. Subtract a − 2b from 4a + 3b.9. From a + b − 2c take 3a + 2b − 4c.10. From x2+ xy − y2take xy − 2x2.9.2.2 Multiplicationand divisionProblem 5. Simplify bc × abcbc × abc = a × b × b × c × c= a × b2× c2= ab2c2Problem 6. Simplify −2p × −3p− × − = + hence,−2p × −3p = 6p2Problem 7. Simplify ab × b2c × aab × b2c × a = a × a × b × b × b × c= a2× b3× c= a2b3cProblem 8. Evaluate 3ab + 4bc − abc whena = 3,b = 2 and c = 53ab + 4bc − abc = 3 × a × b + 4 × b × c − a × b × c= 3 × 3 × 2 + 4 × 2 × 5 − 3 × 2 × 5= 18 + 40 − 30= 28Problem 9. Determine the value of 5pq2r3, giventhat p = 2,q =25and r = 2125pq2r3= 5 × p × q2× r3= 5 × 2 ×252× 2123
• Basic algebra 63= 5 × 2 ×252×523since 212=52=51×21×25×25×52×52×52=11×11×11×11×11×51×51by cancelling= 5 × 5= 25Problem 10. Multiply 2a + 3b by a + bEach term in the ﬁrst expression is multiplied by a, theneach term in the ﬁrst expression is multiplied by b andthe two results are added. The usual layout is shownbelow.2a + 3ba + bMultiplying by a gives 2a2 + 3abMultiplying by b gives 2ab + 3b2Adding gives 2a2 + 5ab + 3b2Thus, (2a + 3b)(a + b) = 2a2+ 5ab + 3b2Problem 11. Multiply 3x − 2y2 + 4xy by2x − 5y3x − 2y2 + 4xy2x − 5yMultiplyingby 2x → 6x2 − 4xy2 + 8x2 yMultiplyingby −5y → −20xy2 −15xy + 10y3Adding gives 6x2 − 24xy2 + 8x2y − 15xy + 10y3Thus, (3x − 2y2 + 4xy)(2x − 5y)= 6x2 − 24xy2 + 8x2y − 15xy + 10y3Problem 12. Simplify 2x ÷ 8xy2x ÷ 8xy means2x8xy2x8xy=2 × x8 × x × y=1 × 14 × 1 × yby cancelling=14yProblem 13. Simplify9a2bc3ac9a2bc3ac=9 × a × a × b × c3 × a × c= 3 × a × b= 3abProblem 14. Divide 2x2 + x − 3 by x − 1(i) 2x2 + x − 3 is called the dividend and x − 1 thedivisor. The usual layout is shown below with thedividend and divisorboth arranged in descendingpowers of the symbols.2x + 3x − 1 2x2 + x − 32x2 − 2x3x − 33x − 3. .(ii) Dividing the ﬁrst term of the dividend by theﬁrst term of the divisor, i.e.2x2xgives 2x, whichis put above the ﬁrst term of the dividend asshown.(iii) The divisor is then multiplied by 2x, i.e.2x(x − 1) = 2x2 − 2x, which is placed under thedividend as shown. Subtracting gives 3x − 3.(iv) The process is then repeated, i.e. the ﬁrst termof the divisor, x, is divided into 3x, giving +3,which is placed above the dividend as shown.(v) Then 3(x − 1) = 3x − 3, which is placed underthe 3x − 3. The remainder, on subtraction, is zero,which completes the process.Thus, (2x2 + x − 3) ÷ (x − 1)=(2x + 3).(A check can be made on this answer bymultiplying (2x + 3) by (x − 1), which equals2x2 + x − 3.)Problem 15. Simplifyx3+ y3x + y
• 64 Basic Engineering Mathematics(i) (iv) (vii)x2 − xy + y2x + y x3 + 0 + 0 + y3x3 + x2 y−x2y + y3−x2 y − xy2xy2+ y3xy2 + y3. .(i) x into x3 goes x2. Put x2 above x3.(ii) x2(x + y) = x3 + x2 y(iii) Subtract.(iv) x into −x2y goes −xy. Put −xy above thedividend.(v) −xy(x + y) = −x2 y − xy2(vi) Subtract.(vii) x into xy2 goes y2. Put y2 above the dividend.(viii) y2(x + y) = xy2 + y3(ix) Subtract.Thus,x3 + y3x + y= x2 − xy + y2.Thezerosshown in thedividend arenot normally shown,but are included to clarify the subtraction process andto keep similar terms in their respective columns.Problem 16. Divide 4a3 − 6a2b + 5b3 by 2a − b2a2 − 2ab − b22a − b 4a3 − 6a2b + 5b34a3−2a2b−4a2b + 5b3−4a2b + 2ab2−2ab2+ 5b3−2ab2 + b34b3Thus,4a3 − 6a2b + 5b32a − b= 2a2 − 2ab − b2, remain-der 4b3.Alternatively, the answer may be expressed as4a3 − 6a2b + 5b32a − b= 2a2− 2ab − b2+4b32a − bNow try the following Practice ExercisePracticeExercise 36 Basic operations inalgebra (answers on page 343)1. Simplify pq × pq2r.2. Simplify −4a × −2a.3. Simplify 3 × −2q × −q.4. Evaluate 3pq − 5qr − pqr when p = 3,q = −2 and r = 4.5. Determine the value of 3x2 yz3, given thatx = 2, y = 112and z =236. If x = 5 and y = 6, evaluate23(x − y)y + xy + 2x7. If a = 4,b = 3,c = 5 and d = 6, evaluate3a + 2b3c − 2d8. Simplify 2x ÷ 14xy.9. Simplify25x2y z35xyz10. Multiply 3a − b by a + b.11. Multiply 2a − 5b + c by 3a + b.12. Simplify 3a ÷ 9ab.13. Simplify 4a2b ÷ 2a.14. Divide 6x2 y by 2xy.15. Divide 2x2 + xy − y2 by x + y.16. Divide 3p2 − pq − 2q2 by p − q.17. Simplify (a + b)2 + (a − b)2.9.3 Laws of indicesThe laws of indices with numbers were covered inChapter 7; the laws of indices in algebraic terms areas follows:(1) am×an =am+nForexample,a3 × a4 =a3+4 =a7
• Basic algebra 65(2)aman= am−n For example,c5c2= c5−2 = c3(3) (am)n = amn For example, d2 3= d2×3 = d6(4) amn = n√am For example, x43 =3√x4(5) a−n =1anFor example, 3−2 =132=19(6) a0 = 1 For example, 170 = 1Here are some worked examples to demonstrate theselaws of indices.Problem 17. Simplify a2b3c × ab2c5a2b3c × ab2c5= a2× b3× c × a × b2× c5= a2× b3× c1× a1× b2× c5Grouping together like terms givesa2 × a1 × b3 × b2 × c1 × c5Using law (1) of indices givesa2+1 × b3+2 × c1+5 = a3 × b5 × c6i.e. a2b3c × ab2c5 = a3 b5 c6Problem 18. Simplify a13 b32 c−2 × a16 b12 cUsing law (1) of indices,a13 b32 c−2× a16 b12 c = a13+16 × b32 +12 × c−2+1= a12 b2c−1Problem 19. Simplifyx5 y2zx2 y z3x5 y2zx2 yz3=x5 × y2 × zx2 × y × z3=x5x2×y2y1×zz3= x5−2× y2−1× z1−3by law (2) of indices= x3× y1× z−2= x3y z−2orx3yz2Problem 20. Simplifya3b2c4abc−2and evaluate whena = 3,b =14and c = 2Using law (2) of indices,a3a= a3−1= a2,b2b= b2−1= b andc4c−2= c4−−2= c6Thus,a3b2c4abc−2= a2bc6When a =3,b=14and c = 2,a2bc6 = (3)2 14(2)6 = (9)14(64) = 144Problem 21. Simplify (p3)2(q2)4Using law (3) of indices gives(p3)2(q2)4= p3×2× q2×4= p6q8Problem 22. Simplify(mn2)3(m1/2n1/4)4The brackets indicate that each letter in the bracket mustbe raised to the power outside. Using law (3) of indicesgives(mn2)3(m1/2n1/4)4=m1×3n2×3m(1/2)×4n(1/4)×4=m3n6m2n1Using law (2) of indices givesm3n6m2n1= m3−2n6−1= mn5Problem 23. Simplify (a3b)(a−4b−2), expressingthe answer with positive indices onlyUsing law (1) of indices gives a3+−4b1+−2 = a−1b−1Using law (5) of indices gives a−1b−1 =1a+1b+1=1abProblem 24. Simplifyd2e2 f 1/2(d3/2ef 5/2)2expressingthe answer with positive indices only
• 66 Basic Engineering MathematicsUsing law (3) of indices givesd2e2 f 1/2(d3/2 e f 5/2)2=d2e2 f 1/2d3e2 f 5Using law (2) of indices givesd2−3e2−2f12 −5= d−1e0f −92= d−1f −92 since e0= 1 from law(6) of indices=1df 9/2from law (5) of indicesNow try the following Practice ExercisePracticeExercise 37 Laws of indices(answers on page 343)In problems 1 to 18, simplify the following,givingeach answer as a power.1. z2 × z6 2. a × a2 × a53. n8 × n−5 4. b4 × b75. b2 ÷ b5 6. c5 × c3 ÷ c47.m5× m6m4 × m38.(x2)(x)x69. x3 410. y2 −311. t × t3 212. c−7 −213.a2a5314.1b3415.b2b7−216.1s3 317. p3qr2 × p2q5r × pqr2 18.x3 y2zx5 y z319. Simplify (x2 y3z)(x3 yz2) and evaluate whenx =12, y = 2 and z = 3.20. Simplifya5bc3a2b3c2and evaluate whena =32,b =12and c =23Here are some further worked examples on the laws ofindicesProblem 25. Simplifyp1/2q2r2/3p1/4q1/2r1/6and evaluatewhen p = 16,q = 9 and r = 4, taking positive rootsonlyUsing law (2) of indices gives p12 −14 q2−12 r23 −16p12 −14 q2−12 r23 −16 = p14 q32 r12When p = 16,q = 9 and r = 4,p14 q32 r12 = 1614 932 412= (4√16)(√93)(√4) from law (4) of indices= (2)(33)(2) = 108Problem 26. Simplifyx2 y3 + xy2xyAlgebraic expressions of the forma + bccan be splitintoac+bc. Thus,x2 y3 + xy2xy=x2 y3xy+xy2xy= x2−1y3−1+ x1−1y2−1= xy2+y(since x0 = 1,fromlaw(6)ofindices).Problem 27. Simplifyx2 yxy2 − xyThe highest common factor (HCF) of each of the threeterms comprising the numerator and denominator is xy.Dividing each term by xy givesx2 yxy2 − xy=x2 yxyxy2xy−xyxy=xy − 1Problem 28. Simplifya2bab2 − a1/2b3
• Basic algebra 67The HCF of each of the three terms is a1/2b. Dividingeach term by a1/2b givesa2bab2 − a1/2b3=a2ba1/2bab2a1/2b−a1/2b3a1/2b=a3/2a1/2b−b2Problem 29. Simplify (a3√b√c5)(√a3√b2c3)and evaluate when a =14,b = 6 and c = 1Using law(4)ofindices,theexpression can bewritten as(a3√b c5)(√a3b2c3) = a3b12 c52 a12 b23 c3Using law (1) of indices givesa3b12 c52 a12 b23 c3= a3+12 b12 +23 c52 +3= a72 b76 c112It is usual to express the answer in the same form as thequestion. Hence,a72 b76 c112 = a76b7c11When a =14,b = 64 and c = 1,a7 6b7 c11 =1476√647√111=127(2)7(1) = 1Problem 30. Simplify(x2 y1/2)(√x 3y2)(x5 y3)1/2Using laws (3) and (4) of indices givesx2 y1/2 √x 3y2(x5 y3)1/2=x2 y1/2 x1/2 y2/3x5/2 y3/2Using laws (1) and (2) of indices givesx2+12 −52 y12 +23 −32 = x0y−13 = y−13 or1y1/3or13√yfrom laws (5) and (6) of indices.Now try the following Practice ExercisePracticeExercise 38 Laws of indices(answers on page 343)1. Simplify a3/2bc−3 a1/2b−1/2c and eval-uate when a = 3,b = 4 and c = 2.In problems 2 to 5, simplify the given expressions.2.a2b + a3ba2b23. (a2)1/2(b2)3 c1/2 34.(abc)2(a2b−1c−3)35.p3q2pq2 − p2q6. (√x y3 3√z2)(√x y3 z3)7. (e2f 3)(e−3f −5),expressing theanswerwithpositive indices only.8.(a3b1/2c−1/2)(ab)1/3(√a3√b c)
• Chapter 10Further algebra10.1 IntroductionIn this chapter, the use of brackets and factorizationwith algebra is explained, together with further practicewith the laws of precedence. Understanding of thesetopics is often necessary when solving and transposingequations.10.2 BracketsWith algebra(a) 2(a + b) = 2a + 2b(b) (a + b)(c + d) = a(c + d) + b(c + d)= ac + ad + bc + bdHere are some worked examples to help understandingof brackets with algebra.Problem 1. Determine 2b(a − 5b)2b(a − 5b) = 2b × a + 2b × −5b= 2ba − 10b2= 2ab − 10b2(note that 2ba is thesame as 2ab)Problem 2. Determine (3x + 4y)(x − y)(3x + 4y)(x − y) = 3x(x − y) + 4y(x − y)= 3x2 − 3xy + 4yx − 4y2= 3x2 − 3xy + 4xy − 4y2(note that 4yx is the same as 4xy)= 3x2 + xy − 4y2Problem 3. Simplify 3(2x − 3y) − (3x − y)3(2x − 3y) − (3x − y) = 3 × 2x − 3 × 3y − 3x − −y(Note that − (3x − y) = −1(3x − y) and the−1 multiplies both terms in the bracket)= 6x − 9y − 3x + y(Note: − ×− = +)= 6x − 3x + y − 9y= 3x − 8yProblem 4. Remove the brackets and simplify theexpression (a − 2b) + 5(b − c) − 3(c + 2d)(a − 2b) + 5(b − c) − 3(c + 2d)= a − 2b + 5 × b + 5 × −c − 3 × c − 3 × 2d= a − 2b + 5b − 5c − 3c − 6d= a + 3b − 8c − 6dProblem 5. Simplify (p + q)(p − q)(p + q)(p − q) = p(p − q) + q(p − q)= p2 − pq + qp − q2= p2 − q2Problem 6. Simplify (2x − 3y)2(2x − 3y)2 = (2x − 3y)(2x − 3y)= 2x(2x − 3y) − 3y(2x − 3y)= 2x × 2x + 2x × −3y − 3y × 2x−3y × −3y= 4x2 − 6xy − 6xy + 9y2(Note: + ×− = − and − ×− = +)= 4x2 − 12xy + 9y2DOI: 10.1016/B978-1-85617-697-2.00010-7
• Further algebra 69Problem 7. Remove the brackets from theexpression and simplify 2[x2 − 3x(y + x) + 4xy]2[x2 − 3x(y + x) + 4xy] = 2[x2 − 3xy − 3x2 + 4xy](Whenever more than one type of brackets is involved,always start with the inner brackets)= 2[−2x2 + xy]= −4x2 + 2xy= 2xy − 4x2Problem 8. Remove the brackets and simplify theexpression 2a − [3{2(4a − b) − 5(a + 2b)} + 4a](i) Removing the innermost brackets gives2a − [3{8a − 2b − 5a − 10b} + 4a](ii) Collecting together similar terms gives2a − [3{3a − 12b} + 4a](iii) Removing the ‘curly’ brackets gives2a − [9a − 36b + 4a](iv) Collecting together similar terms gives2a − [13a − 36b](v) Removing the outer brackets gives2a − 13a + 36b(vi) i.e. −11a + 36b or 36b − 11aNow try the following Practice ExercisePracticeExercise 39 Brackets (answers onpage 344)Expand the brackets in problems 1 to 28.1. (x + 2)(x + 3) 2. (x + 4)(2x + 1)3. (2x + 3)2 4. (2 j − 4)( j + 3)5. (2x + 6)(2x + 5) 6. (pq +r)(r + pq)7. (a + b)(a + b) 8. (x + 6)29. (a − c)2 10. (5x + 3)211. (2x − 6)2 12. (2x − 3)(2x + 3)13. (8x + 4)2 14. (rs + t)215. 3a(b − 2a) 16. 2x(x − y)17. (2a − 5b)(a + b)18. 3(3p − 2q) − (q − 4p)19. (3x − 4y) + 3(y − z) − (z − 4x)20. (2a + 5b)(2a − 5b)21. (x − 2y)2 22. (3a − b)223. 2x + [y − (2x + y)]24. 3a + 2[a − (3a − 2)]25. 4[a2− 3a(2b + a) + 7ab]26. 3[x2 − 2x(y + 3x) + 3xy(1 + x)]27. 2 − 5[a(a − 2b) − (a − b)2]28. 24p − [2{3(5p − q) − 2(p + 2q)} + 3q]10.3 FactorizationThe factors of 8 are 1, 2, 4 and 8 because 8 divides by1, 2, 4 and 8.The factorsof 24 are 1, 2, 3, 4, 6, 8, 12 and 24 because 24divides by 1, 2, 3, 4, 6, 8, 12 and 24.The common factors of 8 and 24 are 1, 2, 4 and 8 since1, 2, 4 and 8 are factors of both 8 and 24.The highest common factor (HCF) is the largestnumber that divides into two or more terms.Hence, the HCF of 8 and 24 is 8, as explained inChapter 1.When two or more terms in an algebraic expression con-tain a common factor, then this factor can be shownoutside of a bracket. For example,d f + dg = d( f + g)which is just the reverse ofd( f + g) = d f + dgThis process is called factorization.Here are some worked examples to help understandingof factorizing in algebra.Problem 9. Factorize ab − 5aca is common to both terms ab and −5ac. a is there-fore taken outside of the bracket. What goes inside thebracket?
• 70 Basic Engineering Mathematics(i) What multiplies a to make ab? Answer: b(ii) What multiplies a to make −5ac? Answer: −5cHence, b − 5c appears in the bracket. Thus,ab − 5ac = a(b − 5c)Problem 10. Factorize 2x2 + 14xy3For the numbers 2 and 14, the highest common factor(HCF) is 2 (i.e. 2 is the largest number that divides intoboth 2 and 14).For the x terms, x2 and x, the HCF is x.Thus, the HCF of 2x2 and 14xy3 is 2x.2x is therefore taken outside of the bracket. What goesinside the bracket?(i) What multiplies 2x to make 2x2? Answer: x(ii) What multiplies 2x to make 14xy3? Answer: 7y3Hence x + 7y3 appears inside the bracket. Thus,2x2+ 14xy3= 2x(x + 7y3)Problem 11. Factorize 3x3y − 12xy2+ 15xyForthenumbers3,12 and 15,thehighest common factoris 3 (i.e. 3 is the largest number that divides into 3, 12and 15).For the x terms, x3, x and x, the HCF is x.For the y terms, y, y2 and y, the HCF is y.Thus, the HCF of 3x3 y and 12xy2 and 15xy is 3xy.3xy is therefore taken outside of the bracket. What goesinside the bracket?(i) What multiplies 3xy to make 3x3 y? Answer: x2(ii) What multiplies 3xy to make −12xy2? Answer:−4y(iii) What multiplies 3xy to make 15xy? Answer: 5Hence, x2 − 4y + 5 appears inside the bracket. Thus,3x3y − 12xy2+ 15xy = 3xy(x2− 4y + 5)Problem 12. Factorize 25a2b5 − 5a3b2For the numbers 25 and 5, the highest common factoris 5 (i.e. 5 is the largest number that divides into 25and 5).For the a terms, a2 and a3, the HCF is a2.For the b terms, b5 and b2, the HCF is b2.Thus, the HCF of 25a2b5and 5a3b2is 5a2b2.5a2b2 is therefore taken outside of the bracket. Whatgoes inside the bracket?(i) What multiplies 5a2b2 to make 25a2b5? Answer:5b3(ii) What multiplies5a2b2 to make −5a3b2? Answer:−aHence, 5b3− a appears in the bracket. Thus,25a2b5− 5a3b2= 5a2b2(5b3− a)Problem 13. Factorize ax − ay + bx − byThe ﬁrst two terms have a common factor of a and thelast two terms a common factor of b. Thus,ax − ay + bx − by = a(x − y) + b(x − y)The two newly formed terms have a common factor of(x − y). Thus,a(x − y) + b(x − y) = (x − y)(a + b)Problem 14. Factorize 2ax − 3ay + 2bx − 3bya is a common factor of the ﬁrst two terms and b acommon factor of the last two terms. Thus,2ax − 3ay + 2bx − 3by = a(2x − 3y) + b(2x − 3y)(2x − 3y) is now a common factor. Thus,a(2x − 3y) + b(2x − 3y) = (2x − 3y)(a + b)Alternatively, 2x is a common factor of the originalﬁrst and third terms and −3y is a common factor ofthe second and fourth terms. Thus,2ax − 3ay + 2bx − 3by = 2x(a + b) − 3y(a + b)(a + b) is now a common factor. Thus,2x(a + b) − 3y(a + b) = (a + b)(2x − 3y)as beforeProblem 15. Factorize x3 + 3x2 − x − 3x2 is a common factor of the ﬁrst two terms. Thus,x3+ 3x2− x − 3 = x2(x + 3) − x − 3−1 is a common factor of the last two terms. Thus,x2(x + 3) − x − 3 = x2(x + 3) − 1(x + 3)
• Further algebra 71(x + 3) is now a common factor. Thus,x2(x + 3) − 1(x − 3) = (x + 3)(x2− 1)Now try the following Practice ExercisePracticeExercise 40 Factorization (answerson page 344)Factorize and simplify the following.1. 2x + 4 2. 2xy − 8xz3. pb + 2pc 4. 2x + 4xy5. 4d2 − 12d f 5 6. 4x + 8x27. 2q2 + 8qn 8. rs +rp +rt9. x + 3x2 + 5x3 10. abc + b3c11. 3x2 y4 − 15xy2 + 18xy12. 4p3q2 − 10pq3 13. 21a2b2 − 28ab14. 2xy2 + 6x2 y + 8x3 y15. 2x2 y − 4xy3 + 8x3 y416. 28y + 7y2 + 14xy17.3x2 + 6x − 3xyxy + 2y − y218.abc + 2ab2c + 4−abc2c19.5rs + 15r3t + 20r6r2t2 + 8t + 2ts−r2t20. ay + by + a + b 21. px + qx + py + qy22. ax − ay + bx − by23. 2ax + 3ay − 4bx − 6by10.4 Laws of precedenceSometimes addition, subtraction, multiplication, divi-sion, powers and brackets can all be involved in analgebraic expression. With mathematics there is a deﬁ-nite order of precedence (ﬁrst met in Chapter 1) whichwe need to adhere to.With the laws of precedence the order isBracketsOrder (or pOwer)DivisionMultiplicationAdditionSubtractionThe ﬁrst letter of each word spells BODMAS.Here are some examples to help understanding ofBODMAS with algebra.Problem 16. Simplify 2x + 3x × 4x − x2x + 3x × 4x − x = 2x + 12x2 − x (M)= 2x − x + 12x2= x + 12x2 (S)or x(1 + 12x) by factorizingProblem 17. Simplify (y + 4y) × 3y − 5y(y + 4y) × 3y − 5y = 5y × 3y − 5y (B)= 15y2 − 5y (M)or 5y(3y − 1) by factorizingProblem 18. Simplify p + 2p × (4p − 7p)p + 2p × (4p − 7p) = p + 2p × −3p (B)= p − 6p2 (M)or p(1 − 6p) by factorizingProblem 19. Simplify t ÷ 2t + 3t − 5tt ÷ 2t + 3t − 5t =t2t+ 3t − 5t (D)=12+ 3t − 5t by cancelling=12− 2t (S)Problem 20. Simplify x ÷ (4x + x) − 3xx ÷ (4x + x) − 3x = x ÷ 5x − 3x (B)=x5x− 3x (D)=15− 3x by cancellingProblem 21. Simplify 2y ÷ (6y + 3y − 5y)2y ÷ (6y + 3y − 5y) = 2y ÷ 4y (B)=2y4y(D)=12by cancelling
• 72 Basic Engineering MathematicsProblem 22. Simplify5a + 3a × 2a + a ÷ 2a − 7a5a + 3a × 2a + a ÷ 2a − 7a= 5a + 3a × 2a +a2a− 7a (D)= 5a + 3a × 2a +12− 7a by cancelling= 5a + 6a2+12− 7a (M)= −2a + 6a2 +12(S)= 6a2 − 2a +12Problem 23. Simplify(4y + 3y)2y + y ÷ 4y − 6y(4y + 3y)2y + y ÷ 4y − 6y= 7y × 2y + y ÷ 4y − 6y (B)= 7y × 2y +y4y− 6y (D)= 7y × 2y +14− 6y by cancelling= 14y2+14− 6y (M)Problem 24. Simplify5b + 2b × 3b + b ÷ (4b − 7b)5b + 2b × 3b + b ÷ (4b − 7b)= 5b + 2b × 3b + b ÷ −3b (B)= 5b + 2b × 3b +b−3b(D)= 5b + 2b × 3b +1−3by cancelling= 5b + 2b × 3b −13= 5b + 6b2−13(M)Problem 25. Simplify(5p + p)(2p + 3p) ÷ (4p − 5p)(5p + p)(2p + 3p) ÷ (4p − 5p)= (6p)(5p) ÷ (−p) (B)= 6p × 5p ÷ −p= 6p ×5p−p(D)= 6p ×5−1by cancelling= 6p × −5= −30pNow try the following Practice ExercisePracticeExercise 41 Laws of precedence(answers on page 344)Simplify the following.1. 3x + 2x × 4x − x2. (2y + y) × 4y − 3y3. 4b + 3b × (b − 6b)4. 8a ÷ 2a + 6a − 3a5. 6x ÷ (3x + x) − 4x6. 4t ÷ (5t − 3t + 2t)7. 3y + 2y × 5y + 2y ÷ 8y − 6y8. (x + 2x)3x + 2x ÷ 6x − 4x9. 5a + 2a × 3a + a ÷ (2a − 9a)10. (3t + 2t)(5t + t) ÷ (t − 3t)11. x ÷ 5x − x + (2x − 3x)x12. 3a + 2a × 5a + 4a ÷ 2a − 6a
• Chapter 11Solving simple equations11.1 Introduction3x − 4 is an example of an algebraic expression.3x − 4 = 2 is an example of an algebraic equation (i.e.it contains an ‘=’ sign).An equation is simply a statement that two expressionsare equal.Hence, A = πr2 (where A is the area of a circleof radius r)F =95C + 32 (which relates Fahrenheit andCelsius temperatures)and y = 3x + 2 (which is the equation of astraight line graph)are all examples of equations.11.2 Solving equationsTo ‘solve an equation’ means ‘to ﬁnd the value of theunknown’. For example, solving 3x − 4 = 2 means thatthe value of x is required.In this example, x = 2. How did we arrive at x = 2?This is the purpose of this chapter – to show how tosolve such equations.Many equations occur in engineering and it is essentialthat we can solve them when needed.Here are some examples to demonstrate how simpleequations are solved.Problem 1. Solve the equation 4x = 20Dividing each side of the equation by 4 gives4x4=204i.e. x = 5 by cancelling, which is the solution to theequation 4x = 20.The same operation must be applied to both sides of anequation so that the equality is maintained.We can do anything we like to an equation, as longas we do the same to both sides. This is, in fact, theonly rule to remember when solving simple equations(and also when transposing formulae, which we do inChapter 12).Problem 2. Solve the equation2x5= 6Multiplying both sides by 5 gives 52x5= 5(6)Cancelling and removing brackets gives 2x = 30Dividing both sides of the equation by 2 gives2x2=302Cancelling gives x = 15which is the solution of the equation2x5= 6.Problem 3. Solve the equation a − 5 = 8Adding 5 to both sides of the equation givesa − 5 + 5 = 8 + 5i.e. a = 8 + 5i.e. a = 13which is the solution of the equation a − 5 = 8.Note that adding 5 to both sides of the above equationresults in the −5 moving from the LHS to the RHS, butthe sign is changed to +.Problem 4. Solve the equation x + 3 = 7Subtracting 3 from both sides gives x + 3 − 3 = 7 − 3i.e. x = 7 − 3i.e. x = 4which is the solution of the equation x + 3 = 7.DOI: 10.1016/B978-1-85617-697-2.00011-9
• 74 Basic Engineering MathematicsNote that subtracting 3 from both sides of the aboveequation results in the +3 moving from the LHS to theRHS, but the sign is changed to –. So, we can movestraight from x + 3 = 7 to x = 7 − 3.Thus, a term can be moved from one side of an equa-tion to the other as long as a change in sign is made.Problem 5. Solve the equation 6x + 1 = 2x + 9In such equations the terms containing x are groupedon one side of the equation and the remaining termsgrouped on the other side of the equation. As in Prob-lems 3 and 4, changing from one side of an equation tothe other must be accompanied by a change of sign.Since 6x + 1 = 2x + 9then 6x − 2x = 9 − 1i.e. 4x = 8Dividing both sides by 4 gives4x4=84Cancelling gives x = 2which is the solution of the equation 6x + 1 = 2x + 9.In the above examples, the solutions can be checked.Thus, in Problem 5, where 6x + 1 = 2x + 9, if x = 2,thenLHS of equation = 6(2) + 1 = 13RHS of equation = 2(2) + 9 = 13Since the left hand side (LHS) equals the right handside (RHS) then x = 2 must be the correct solution ofthe equation.When solving simple equations, always check youranswers by substituting your solution back into theoriginal equation.Problem 6. Solve the equation 4− 3p = 2p − 11In order to keep the p term positive the terms in p aremoved to the RHS and the constant terms to the LHS.Similar to Problem 5, if 4 − 3p = 2p − 11then 4 + 11 = 2p + 3pi.e. 15 = 5pDividing both sides by 5 gives155=5p5Cancelling gives 3 = p or p = 3which is the solution of the equation 4 − 3p = 2p − 11.By substituting p = 3 into the original equation, thesolution may be checked.LHS = 4 − 3(3) = 4 − 9 = −5RHS = 2(3) − 11 = 6 − 11 = −5Since LHS = RHS, the solution p = 3 must be correct.If, in this example, the unknown quantities had beengrouped initially on the LHS instead of the RHS, then−3p − 2p = −11 − 4i.e. −5p = −15from which,−5p−5=−15−5and p = 3as before.It is often easier, however, to work with positive valueswhere possible.Problem 7. Solve the equation 3(x − 2) = 9Removing the bracket gives 3x − 6 = 9Rearranging gives 3x = 9 + 6i.e. 3x = 15Dividing both sides by 3 gives x = 5which is the solution of the equation 3(x − 2) = 9.Theequation may bechecked by substituting x = 5 backinto the original equation.Problem 8. Solve the equation4(2r − 3) − 2(r − 4) = 3(r − 3) − 1Removing brackets gives8r − 12 − 2r + 8 = 3r − 9 − 1Rearranging gives 8r − 2r − 3r = −9 − 1 + 12 − 8i.e. 3r = −6Dividing both sides by 3 gives r =−63= −2which is the solution of the equation4(2r − 3) − 2(r − 4) = 3(r − 3) − 1.The solution may be checked by substituting r = −2back into the original equation.LHS = 4(−4 − 3) − 2(−2 − 4) = −28 + 12 = −16RHS = 3(−2 − 3) − 1 = −15 − 1 = −16Since LHS = RHS then r = −2 is the correct solution.
• Solving simple equations 75Now try the following Practice ExercisePracticeExercise 42 Solving simpleequations (answers on page 344)Solve the following equations.1. 2x + 5 = 72. 8 − 3t = 23.23c − 1 = 34. 2x − 1 = 5x + 115. 7 − 4p = 2p − 56. 2.6x − 1.3 = 0.9x + 0.47. 2a + 6 − 5a = 08. 3x − 2 − 5x = 2x − 49. 20d − 3 + 3d = 11d + 5 − 810. 2(x − 1) = 411. 16 = 4(t + 2)12. 5( f − 2) − 3(2 f + 5) + 15 = 013. 2x = 4(x − 3)14. 6(2 − 3y) − 42 = −2(y − 1)15. 2(3g − 5) − 5 = 016. 4(3x + 1) = 7(x + 4) − 2(x + 5)17. 11 + 3(r − 7) = 16 − (r + 2)18. 8 + 4(x − 1) − 5(x − 3) = 2(5 − 2x)Here are some further worked examples on solvingsimple equations.Problem 9. Solve the equation4x=25The lowest common multiple (LCM) of the denomina-tors, i.e. the lowest algebraic expression that both x and5 will divide into, is 5x.Multiplying both sides by 5x gives5x4x= 5x25Cancelling gives 5(4) = x(2)i.e. 20 = 2x (1)Dividing both sides by 2 gives202=2x2Cancelling gives 10 = x or x = 10which is the solution of the equation4x=25When there is just one fraction on each side of the equa-tion as in this example, there is a quick way to arriveat equation (1) without needing to ﬁnd the LCM of thedenominators.We can move from4x=25to 4 × 5 = 2 × x by what iscalled ‘cross-multiplication’.In general, ifab=cdthen ad = bcWe can use cross-multiplication when there is onefraction only on each side of the equation.Problem 10. Solve the equation3t − 2=43t + 4Cross-multiplication gives 3(3t + 4) = 4(t − 2)Removing brackets gives 9t + 12 = 4t − 8Rearranging gives 9t − 4t = −8 − 12i.e. 5t = −20Dividing both sides by 5 gives t =−205= −4which is the solution of the equation3t − 2=43t + 4Problem 11. Solve the equation2y5+34+ 5 =120−3y2The lowest common multiple (LCM) of the denomina-tors is 20; i.e., the lowest number that 4, 5, 20 and 2 willdivide into.Multiplying each term by 20 gives202y5+ 2034+ 20(5) = 20120− 203y2Cancelling gives 4(2y) + 5(3) + 100 = 1 − 10(3y)i.e. 8y + 15 + 100 = 1 − 30yRearranging gives 8y +30y = 1−15−100
• 76 Basic Engineering Mathematicsi.e. 38y = −114Dividing both sides by 38 gives38y38=−11438Cancelling gives y = −3which is the solution of the equation2y5+34+ 5 =120−3y2Problem 12. Solve the equation√x = 2Whenever square root signs are involved in an equation,both sides of the equation must be squared.Squaring both sides gives√x2= (2)2i.e. x = 4which is the solution of the equation√x = 2.Problem 13. Solve the equation 2√d = 8Whenever square roots are involved in an equation, thesquare root term needs to be isolated on its own beforesquaring both sides.Cross-multiplying gives √d =82Cancelling gives√d = 4Squaring both sides gives√d2= (4)2i.e. d = 16which is the solution of the equation 2√d = 8.Problem 14. Solve the equation√b + 3√b= 2Cross-multiplying gives√b + 3 = 2√bRearranging gives 3 = 2√b −√bi.e. 3 =√bSquaring both sides gives 9 = bwhich is the solution of the equation√b + 3√b= 2.Problem 15. Solve the equation x2 = 25Whenever a square term is involved, the square root ofboth sides of the equation must be taken.Taking the square root of both sides gives x2 =√25i.e. x = ±5which is the solution of the equation x2= 25.Problem 16. Solve the equation154t2=23We need to rearrange the equation to get the t2term onits own.Cross-multiplying gives 15(3) = 2(4t2)i.e. 45 = 8t2Dividing both sides by 8 gives 458=8t28By cancelling 5.625 = t2or t2= 5.625Taking the square root of both sides givest2 =√5.625i.e. t = ±2.372correct to 4 signiﬁcant ﬁgures, which is the solution ofthe equation154t2=23Now try the following Practice ExercisePracticeExercise 43 Solving simpleequations (answers on page 344)Solve the following equations.1.15d + 3 = 42. 2 +34y = 1 +23y +563.14(2x − 1) + 3 =124.15(2 f − 3) +16( f − 4) +215= 05.13(3m − 6) −14(5m + 4) +15(2m − 9) = −36.x3−x5= 27. 1 −y3= 3 +y3−y6
• Solving simple equations 778.2a=389.13n+14n=72410.x + 34=x − 35+ 211.3t20=6 − t12+2t15−3212.y5+720=5 − y413.v − 22v − 3=1314.2a − 3=32a + 115.x4−x + 65=x + 3216. 3√t = 917. 2√y = 518. 4 =3a+ 319.3√x1 −√x= −620. 10 = 5x2− 121. 16 =t2922.y + 2y − 2=1223.6a=2a324.112= 5 +8x211.3 Practical problems involvingsimple equationsThere are many practical situations in engineering inwhich solving equations is needed. Here are someworked examples to demonstrate typical practicalsituationsProblem 17. Applying the principle of momentsto a beam results in the equationF × 3 = (7.5 − F) × 2where F is the force in newtons. Determine thevalue of FRemoving brackets gives 3F = 15 − 2FRearranging gives 3F + 2F = 15i.e. 5F = 15Dividing both sides by 5 gives5F5=155from which, force, F = 3N.Problem 18. A copper wire has a length L of1.5km, a resistance R of 5 and a resistivity of17.2 × 10−6 mm. Find the cross-sectional area, a,of the wire, given that R =ρLaSince R =ρLathen5 =(17.2 × 10−6 mm)(1500 × 103 mm)a.From the units given, a is measured in mm2.Thus, 5a = 17.2 × 10−6× 1500× 103and a =17.2 × 10−6 × 1500× 1035=17.2 × 1500 × 103106 × 5=17.2 × 1510 × 5= 5.16Hence, the cross-sectional area of the wire is 5.16 mm2.Problem 19. PV = mRT is the characteristic gasequation. Find the value of gas constant R whenpressure P = 3 × 106 Pa, volume V = 0.90m3,mass m = 2.81kg and temperature T = 231KDividing both sides of PV = mRT by mT givesPVmT=mRTmTCancelling givesPVmT= RSubstituting values gives R =3 × 106 (0.90)(2.81)(231)
• 78 Basic Engineering MathematicsUsing a calculator, gas constant, R = 4160J/(kg K),correct to 4 signiﬁcant ﬁgures.Problem 20. A rectangular box with square endshas its length 15cm greater than its breadth and thetotal length of its edges is 2.04m. Find the width ofthe box and its volumeLet x cm = width = height of box. Then the length ofthe box is (x + 15)cm, as shown in Figure 11.1.(x115)xxFigure 11.1The length of the edges of the box is 2(4x) + 4(x + 15)cm, which equals 2.04m or 204cm.Hence, 204 = 2(4x) + 4(x + 15)204 = 8x + 4x + 60204 − 60 = 12xi.e. 144 = 12xand x = 12cmHence, the width of the box is 12cm.Volume of box = length × width × height= (x +15)(x)(x)= (12+15)(12)(12)= (27)(12)(12)= 3888cm3Problem 21. The temperature coefﬁcient ofresistance α may be calculated from the formulaRt = R0(1 + αt). Find α, given Rt = 0.928,R0 = 0.80 and t = 40Since Rt = R0(1 + αt), then0.928 = 0.80[1 + α(40)]0.928 = 0.80 + (0.8)(α)(40)0.928 − 0.80 = 32α0.128 = 32αHence, α =0.12832= 0.004Problem 22. The distance s metres travelledin time t seconds is given by the formulas = ut +12at2, where u is the initial velocity in m/sand a is the acceleration in m/s2. Find theacceleration of the body if it travels 168m in 6s,with an initial velocity of 10m/ss = ut +12at2, and s = 168,u = 10 and t = 6Hence, 168 = (10)(6) +12a(6)2168 = 60 + 18a168 − 60 = 18a108 = 18aa =10818= 6Hence, the acceleration of the body is 6 m/s2.Problem 23. When three resistors in an electricalcircuit are connected in parallel the total resistanceRT is given by1RT=1R1+1R2+1R3. Find thetotal resistance when R1 = 5 , R2 = 10 andR3 = 301RT=15+110+130=6 + 3 + 130=1030=13Taking the reciprocal of both sides gives RT = 3Alternatively, if1RT=15+110+130, the LCM of thedenominators is 30RT.Hence, 30RT1RT= 30RT15+ 30RT110+ 30RT130.Cancelling gives 30 = 6RT + 3RT + RTi.e. 30 = 10RTand RT =3010= 3 , as above.
• Solving simple equations 79Now try the following Practice ExercisePracticeExercise 44 Practical problemsinvolving simple equations (answers onpage 344)1. A formulaused for calculating resistance of acable is R =ρLa. Given R = 1.25, L = 2500and a = 2 × 10−4, ﬁnd the value of ρ.2. Force F newtons is given by F = ma, wherem is the mass in kilograms and a is the accel-eration in metrespersecond squared.Find theacceleration when a force of 4kN is appliedto a mass of 500kg.3. PV = mRT is the characteristic gas equa-tion. Find the value of m whenP = 100 × 103, V = 3.00, R = 288 andT = 300.4. When three resistors R1, R2 and R3 are con-nected in parallel, the total resistance RT isdetermined from1RT=1R1+1R2+1R3(a) Find the total resistance whenR1 = 3 , R2 = 6 and R3 = 18 .(b) Find the value of R3 given thatRT = 3 , R1 = 5 and R2 = 10 .5. Six digital camera batteries and 3 camcorderbatteries cost £96. If a camcorder batterycosts £5 more than a digital camera battery,ﬁnd the cost of each.6. Ohm’s law may be represented by I = V/R,where I is the current in amperes, V isthe voltage in volts and R is the resistancein ohms. A soldering iron takes a currentof 0.30A from a 240V supply. Find theresistance of the element.7. The distance, s, travelled in time t seconds isgiven by the formula s = ut +12a t2 whereu is the initial velocity in m/s and a is theacceleration in m/s2. Calculate the accelera-tion of the body if it travels 165m in 3s, withan initial velocity of 10m/s.Here are some further worked examples on solvingsimple equations in practical situations.Problem 24. The extension x m of an aluminiumtie bar of length l m and cross-sectional area Am2when carrying a load of F Newtons is given by themodulus of elasticity E = Fl/Ax. Find theextension of the tie bar (in mm) ifE = 70 × 109 N/m2, F = 20 × 106 N, A = 0.1m2and l = 1.4mE = Fl/Ax, hence70×109 Nm2=(20×106 N)(1.4m)(0.1m2)(x)(the unit of x is thus metres)70 × 109× 0.1 × x = 20 × 106× 1.4x =20 × 106 × 1.470 × 109 × 0.1Cancelling gives x =2 × 1.47 × 100m=2×1.47×100×1000mm= 4mmHence, the extension of the tie bar, x = 4 mm.Problem 25. Power in a d.c. circuit is given byP =V 2Rwhere V is the supply voltage and R is thecircuit resistance. Find the supply voltage if thecircuit resistance is 1.25 and the power measuredis 320WSince P =V 2R, then 320 =V 21.25(320)(1.25) = V 2i.e. V 2= 400Supply voltage, V =√400 = ±20VProblem 26. A painter is paid £6.30 per hour for abasic 36 hour week and overtime is paid at one anda third times this rate. Determine how many hoursthe painter has to work in a week to earn £319.20Basic rate per hour = £6.30 and overtime rate per hour= 113× £6.30 = £8.40Let the number of overtime hours worked = xThen, (36)(6.30) + (x)(8.40) = 319.20226.80 + 8.40x = 319.20
• 80 Basic Engineering Mathematics8.40x = 319.20 − 226.80 = 92.40x =92.408.40= 11Thus, 11 hours overtime would have to be worked toearn £319.20 per week. Hence, the total number ofhours worked is 36 + 11, i.e. 47 hours.Problem 27. A formula relating initial and ﬁnalstates of pressures, P1 and P2, volumes, V1 and V2,and absolute temperatures, T1 and T2, of an idealgas isP1V1T1=P2V2T2. Find the value of P2 givenP1 = 100 × 103, V1 = 1.0, V2 = 0.266, T1 = 423and T2 = 293SinceP1V1T1=P2V2T2then(100 × 103)(1.0)423=P2(0.266)293Cross-multiplying gives(100 × 103)(1.0)(293) = P2(0.266)(423)P2 =(100 × 103)(1.0)(293)(0.266)(423)Hence, P2 = 260 × 103or 2.6 × 105.Problem 28. The stress, f , in a material of athick cylinder can be obtained fromDd=f + pf − p. Calculate the stress, given thatD = 21.5,d = 10.75 and p = 1800SinceDd=f + pf − pthen21.510.75=f + 1800f − 1800i.e.2 =f + 1800f − 1800Squaring both sides gives 4 =f + 1800f − 1800Cross-multiplying gives4( f − 1800) = f + 18004 f − 7200 = f + 18004 f − f = 1800 + 72003 f = 9000f =90003= 3000Hence, stress,f = 3000Problem 29. 12 workmen employed on abuilding site earn between them a total of £4035 perweek. Labourers are paid £275 per week andcraftsmen are paid £380 per week. How manycraftsmen and how many labourers are employed?Let the number of craftsmen be c. The number oflabourers is therefore (12 − c).The wage bill equation is380c + 275(12 − c) = 4035380c + 3300− 275c = 4035380c − 275c = 4035 − 3300105c = 735c =735105= 7Hence, there are 7 craftsmen and (12 − 7), i.e. 5labourers on the site.Now try the following Practice ExercisePracticeExercise 45 Practical problemsinvolving simple equations (answers onpage 344)1. A rectangle has a length of 20cm and a widthb cm. When its width is reduced by 4cm itsarea becomes 160cm2. Find the original widthand area of the rectangle.2. Given R2 = R1(1 + αt), ﬁnd α givenR1 = 5.0, R2 = 6.03 and t = 51.53. If v2 = u2 + 2as, ﬁnd u given v = 24,a = −40 and s = 4.054. The relationship between the temperature ona Fahrenheit scale and that on a Celsius scaleis given by F =95C + 32. Express 113◦F indegrees Celsius.5. If t = 2πwSg, ﬁnd the value of S givenw = 1.219, g = 9.81 and t = 0.31326. Two joiners and ﬁve mates earn £1824between them for a particular job. If a joinerearns £72 more than a mate, calculate theearnings for a joiner and for a mate.
• Solving simple equations 817. An alloy contains 60% by weight of copper,the remainder being zinc. How much coppermust be mixed with 50kg of this alloy to givean alloy containing 75% copper?8. A rectangular laboratory has a length equal toone and a half times its width and a perimeterof 40m. Find its length and width.9. Applying the principle of moments to a beamresults in the following equation:F × 3 = (5 − F) × 7where F is the force in newtons. Determinethe value of F.
• Revision Test 4: Algebra and simple equationsThis assignment covers the material contained in Chapters 9–11. The marks available are shown in brackets at theend of each question.1. Evaluate 3pqr3 − 2p2qr + pqr when p =12,q = −2 and r = 1. (3)In problems 2 to 7, simplify the expressions.2.9p2qr33pq2r(3)3. 2(3x − 2y) − (4y − 3x) (3)4. (x − 2y)(2x + y) (3)5. p2q−3r4 × pq2r−3 (3)6. (3a − 2b)2 (3)7.a4b2cab3c2(3)8. Factorize(a) 2x2 y3 − 10xy2(b) 21ab2c3 − 7a2bc2 + 28a3bc4 (5)9. Factorize and simplify2x2 y + 6xy2x + 3y−x3 y2x2 y(5)10. Remove the brackets and simplify10a − [3(2a − b) − 4(b − a) + 5b] (4)11. Simplify x ÷ 5x − x + (2x − 3x)x (4)12. Simplify 3a + 2a × 5a + 4a ÷ 2a − 6a (4)13. Solve the equations(a) 3a = 39(b) 2x − 4 = 9 (3)14. Solve the equations(a)49y = 8(b) 6x − 1 = 4x + 5 (4)15. Solve the equation5(t − 2) − 3(4 − t) = 2(t + 3) − 40 (4)16. Solve the equations:(a)32x + 1=14x − 3(b) 2x2 = 162 (7)17. Kinetic energy is given by the formula,Ek =12mv2 joules, where m is the mass in kilo-grams and v is the velocity in metres per second.Evaluate the velocity when Ek = 576 × 10−3Jand the mass is 5kg. (4)18. An approximate relationship between the num-ber of teeth T on a milling cutter, the diameterof the cutter D and the depth of cut d is givenby T =12.5DD + 4d. Evaluate d when T = 10 andD = 32. (5)19. The modulus of elasticity E is given by the for-mula E =FLx Awhere F is force in newtons, Lis the length in metres, x is the extension inmetres and A the cross-sectional area in squaremetres. Evaluate A, in square centimetres, whenE = 80×109 N/m2, x =2mm, F =100×103 Nand L = 2.0m. (5)
• Chapter 12Transposing formulae12.1 IntroductionIn the formula I =VR, I is called the subject of theformula.Similarly, in the formula y = mx + c, y is the subject ofthe formula.When a symbol other than the subject is required tobe the subject, the formula needs to be rearranged tomake a new subject. This rearranging process is calledtransposing the formula or transposition.For example, in the above formulae,if I =VRthen V = IRand if y = mx + c then x =y − cmHow did we arrive at these transpositions? This is thepurpose of this chapter — to show how to transpose for-mulae. A great many equations occur in engineering andit is essential that we can transpose them when needed.12.2 Transposing formulaeThere are no new rules for transposing formulae.The same rules as were used for simple equations inChapter 11 are used; i.e., the balance of an equationmust be maintained: whatever is done to one side ofan equation must be done to the other.It is best that you cover simple equations before tryingthis chapter.Here are some worked examples to help understandingof transposing formulae.Problem 1. Transpose p = q +r + s to make rthe subjectThe object is to obtain r on its own on the LHS of theequation. Changing the equation around so that r is onthe LHS givesq +r + s = p (1)From Chapter 11 on simple equations, a term can bemoved from one side of an equation to the other side aslong as the sign is changed.Rearranging gives r = p − q − s.Mathematically, we have subtracted q + s from bothsides of equation (1).Problem 2. If a + b = w − x + y, express x asthe subjectAs stated in Problem 1, a term can be moved from oneside of an equation to the other side but with a changeof sign.Hence, rearranging gives x = w + y − a − bProblem 3. Transpose v = f λ to make λ thesubjectv = f λ relates velocity v, frequency f and wave-length λRearranging gives f λ = vDividing both sides by f givesf λf=vfCancelling gives λ =vfProblem 4. When a body falls freely through aheight h, the velocity v is given by v2= 2gh.Express this formula with h as the subjectDOI: 10.1016/B978-1-85617-697-2.00012-0
• 84 Basic Engineering MathematicsRearranging gives 2gh = v2Dividing both sides by 2g gives2gh2g=v22gCancelling gives h =v22gProblem 5. If I =VR, rearrange to make V thesubjectI =VRis Ohm’s law, where I is the current, V is thevoltage and R is the resistance.Rearranging gives VR= IMultiplying both sides by R gives RVR= R(I)Cancelling gives V = IRProblem 6. Transpose a =Fmfor ma =Fmrelates acceleration a, force F and mass m.Rearranging gives Fm= aMultiplying both sides by m gives mFm= m(a)Cancelling gives F = maRearranging gives ma = FDividing both sides by a givesmaa=Fai.e. m =FaProblem 7. Rearrange the formula R =ρLAtomake (a) A the subject and (b) L the subjectR =ρLArelates resistance R of a conductor, resistiv-ity ρ, conductor length L and conductor cross-sectionalarea A.(a) Rearranging givesρLA= RMultiplying both sides by A givesAρLA= A(R)Cancelling gives ρL = ARRearranging gives AR = ρlDividing both sides by R givesARR=ρLRCancelling gives A =ρLR(b) Multiplying both sides ofρLA= R by A givesρL = ARDividing both sides by ρ givesρLρ=ARρCancelling gives L =ARρProblem 8. Transpose y = mx + c to make m thesubjecty = mx + c is the equation of a straight line graph,where y is the vertical axis variable, x is the horizontalaxis variable, m is the gradient of the graph and c is they-axis intercept.Subtracting c from both sides gives y − c = mxor mx = y − cDividing both sides by x gives m =y − cxNow try the following Practice ExercisePracticeExercise 46 Transposing formulae(answers on page 344)Make the symbol indicated the subject of each oftheformulaeshown and expresseach in itssimplestform.1. a + b = c − d − e (d)2. y = 7x (x)3. pv = c (v)4. v = u + at (a)5. V = IR (R)6. x + 3y = t (y)7. c = 2πr (r)8. y = mx + c (x)
• Transposing formulae 859. I = PRT (T )10. XL = 2π fL (L)11. I =ER(R)12. y =xa+ 3 (x)13. F =95C + 32 (C)14. XC =12π f C( f )12.3 Further transposing of formulaeHere are some more transposition examples to helpus further understand how more difﬁcult formulae aretransposed.Problem 9. Transpose the formula v = u +Ftmtomake F the subjectv = u +Ftmrelates ﬁnal velocity v, initial velocity u,force F, mass m and time t.Fmis acceleration a.Rearranging gives u +Ftm= vandFtm= v − uMultiplying each side by m givesmFtm= m(v − u)Cancelling gives Ft = m(v − u)Dividing both sides by t givesFtt=m (v − u)tCancelling gives F =m(v − u)tor F =mt(v − u)This shows two ways of expressing the answer. Thereis often more than one way of expressing a trans-posed answer. In this case, these equations for F areequivalent; neither one is more correct than the other.Problem 10. The ﬁnal length L2 of a piece ofwire heated through θ◦C is given by the formulaL2 = L1(1 + αθ) where L1 is the original length.Make the coefﬁcient of expansion α the subjectRearranging gives L1(1 + αθ) = L2Removing the bracket gives L1 + L1αθ = L2Rearranging gives L1αθ = L2 − L1Dividing both sides by L1θ givesL1αθL1θ=L2 − L1L1θCancelling gives α =L2 − L1L1θAn alternative method of transposing L2 = L1 (1 + αθ)for α is:Dividing both sides by L1 givesL2L1= 1 + αθSubtracting 1 from both sides givesL2L1− 1 = αθor αθ =L2L1− 1Dividing both sides by θ gives α =L2L1− 1θThe two answers α =L2 − L1L1θand α =L2L1− 1θlookquite different. They are, however, equivalent. The ﬁrstanswer looks tidier but is no more correct than thesecond answer.Problem 11. A formula for the distance s movedby a body is given by s =12(v + u)t. Rearrange theformula to make u the subjectRearranging gives12(v + u)t = sMultiplying both sides by 2 gives (v + u)t = 2sDividing both sides by t gives(v + u)tt=2stCancelling gives v + u =2stRearranging gives u =2st− v or u =2s − vttProblem 12. A formula for kinetic energy isk =12mv2. Transpose the formula to make v thesubjectRearranging gives12mv2= k
• 86 Basic Engineering MathematicsWhenever the prospective new subject is a squaredterm,that termisisolated on theLHSand then thesquareroot of both sides of the equation is taken.Multiplying both sides by 2 gives mv2= 2kDividing both sides by m givesmv2m=2kmCancelling gives v2=2kmTaking the square root of both sides gives√v2 =2kmi.e. v =2kmProblem 13. In a right-angled triangle havingsides x, y and hypotenuse z, Pythagoras’ theoremstates z2 = x2 + y2. Transpose the formula to ﬁnd xRearranging gives x2+ y2= z2and x2= z2− y2Taking the square root of both sides givesx = z2 − y2Problem 14. Transpose y =ML28EIto make L thesubjectMultiplying both sides by 8E I gives 8EIy = ML2Dividing both sides by M gives8EIyM= L2or L2=8EIyMTaking the square root of both sides gives√L2 =8EIyMi.e.L =8EIyMProblem 15. Given t = 2πlg, ﬁnd g in terms oft,l and πWhenever the prospective new subject is withina squareroot sign, it is best to isolate that term on the LHS andthen to square both sides of the equation.Rearranging gives 2πlg= tDividing both sides by 2π giveslg=t2πSquaring both sides giveslg=t2π2=t24π2Cross-multiplying, (i.e. multiplyingeach term by 4π2g), gives 4π2l = gt2or gt2= 4π2lDividing both sides by t2 givesgt2t2=4π2lt2Cancelling gives g =4π2lt2Problem 16. The impedance Z of an a.c. circuitis given by Z =√R2 + X2 where R is theresistance. Make the reactance, X, the subjectRearranging gives R2 + X2 = ZSquaring both sides gives R2+ X2= Z2Rearranging gives X2= Z2− R2Taking the square root of both sides givesX = Z2 − R2Problem 17. The volume V of a hemisphere ofradius r is given by V =23πr3. (a) Find r in termsof V. (b) Evaluate the radius when V = 32cm3(a) Rearranging gives23πr3= VMultiplying both sides by 3 gives 2πr3= 3VDividing both sides by 2π gives2πr32π=3V2πCancelling gives r3=3V2π
• Transposing formulae 87Taking the cube root of both sides gives3√r3 = 33V2πi.e. r = 33V2π(b) When V = 32cm3,radius r = 33V2π= 33 × 322π= 2.48 cm.Now try the following Practice ExercisePracticeExercise 47 Further transposingformulae (answers on page 345)Make the symbol indicated the subject of each oftheformulaeshown in problems1 to 13 and expresseach in its simplest form.1. S =a1 −r(r)2. y =λ(x − d)d(x)3. A =3(F − f )L( f )4. y =AB25CD(D)5. R = R0(1 + αt) (t)6.1R=1R1+1R2(R2)7. I =E − eR +r(R)8. y = 4ab2c2 (b)9.a2x2+b2y2= 1 (x)10. t = 2πLg(L)11. v2 = u2 + 2as (u)12. A =π R2θ360(R)13. N =a + xy(a)14. Transpose Z = R2 + (2π f L)2 for L andevaluate L when Z = 27.82, R = 11.76 andf = 50.12.4 More difﬁcult transposing offormulaeHere are some more transposition examples to helpus further understand how more difﬁcult formulae aretransposed.Problem 18. (a) Transpose S =3d (L − d)8tomake l the subject. (b) Evaluate L when d = 1.65and S = 0.82The formula S =3d (L − d)8represents the sag S atthe centre of a wire.(a) Squaring both sides gives S2=3d(L − d)8Multiplying both sides by 8 gives8S2= 3d(L − d)Dividing both sides by 3d gives8S23d= L − dRearranging gives L = d +8S23d(b) When d = 1.65 and S = 0.82,L = d +8 S23d= 1.65 +8 × 0.8223 × 1.65= 2.737Problem 19. Transpose the formulap =a2x2+ a2yrto make a the subjectRearranging givesa2x2 + a2 yr= pMultiplying both sides by r givesa2x + a2y = rpFactorizing the LHS gives a2(x + y) = rpDividing both sides by (x + y) givesa2(x + y)(x + y)=rp(x + y)
• 88 Basic Engineering MathematicsCancelling gives a2=rp(x + y)Taking the square root of both sides givesa =rpx + yWhenever the letter required as the new subjectoccurs more than once in the original formula, afterrearranging, factorizing will always be needed.Problem 20. Make b the subject of the formulaa =x − y√bd + beRearranging givesx − y√bd + be= aMultiplying both sides by√bd + be givesx − y = a√bd + beor a√bd + be = x − yDividing both sidesbyagives√bd + be =x − yaSquaring both sides gives bd + be =x − ya2Factorizing the LHS gives b(d + e) =x − ya2Dividing both sides by (d + e) givesb =x − ya2(d + e)or b =(x − y)2a2(d + e)Problem 21. If a =b1 + b, make b the subject ofthe formulaRearranging givesb1 + b= aMultiplying both sides by (1 + b) givesb = a(1 + b)Removing the bracket gives b = a + abRearranging to obtain terms in b on the LHS givesb − ab = aFactorizing the LHS gives b(1 − a) = aDividing both sides by (1 − a) gives b =a1 − aProblem 22. Transpose the formula V =ErR +rto make r the subjectRearranging givesErR +r= VMultiplying both sides by (R +r) givesEr = V (R +r)Removing the bracket gives Er = V R + VrRearranging to obtain terms in r on the LHS givesEr − Vr = V RFactorizing gives r(E − V ) = V RDividing both sides by (E − V ) gives r =V RE − VProblem 23. Transpose the formulay =pq2r + q2− t to make q the subjectRearranging givespq2r + q2− t = yandpq2r + q2= y + tMultiplying both sides by (r + q2) givespq2= (r + q2)(y + t)Removing brackets gives pq2= ry +rt + q2y + q2tRearranging to obtain terms in q on the LHS givespq2− q2y − q2t = ry +rtFactorizing gives q2(p − y − t) = r(y + t)Dividing both sides by (p − y − t) givesq2=r(y + t)(p − y − t)Taking the square root of both sides givesq =r(y + t)p − y − tProblem 24. Given thatDd=f + pf − pexpress p in terms of D,d and f
• Transposing formulae 89Rearranging gives f + pf − p=DdSquaring both sides givesf + pf − p=D2d2Cross-multiplying, i.e. multiplying each termby d2( f − p), givesd2( f + p) = D2( f − p)Removing brackets gives d2f + d2p = D2f − D2pRearranging, to obtain terms in p on the LHS givesd2p + D2p = D2f − d2fFactorizing gives p(d2+ D2) = f (D2− d2)Dividing both sides by (d2 + D2) givesp =f (D2 − d2)(d2 + D2)Now try the following Practice ExercisePracticeExercise 48 Further transposingformulae (answers on page 345)Make the symbol indicated the subject of each ofthe formulae shown in problems 1 to 7 and expresseach in its simplest form.1. y =a2m − a2nx(a)2. M = π(R4 −r4) (R)3. x + y =r3 +r(r)4. m =μLL +rCR(L)5. a2 =b2− c2b2(b)6.xy=1 +r21 −r2(r)7.pq=a + 2ba − 2b(b)8. A formula for the focal length, f , of a convexlens is1f=1u+1v. Transpose the formula tomake v the subject and evaluate v when f = 5and u = 6.9. Thequantity ofheat, Q,isgiven by theformulaQ = mc(t2 − t1). Make t2 the subject of theformula and evaluate t2 when m = 10, t1 = 15,c = 4 and Q = 1600.10. The velocity, v, of water in a pipe appearsin the formula h =0.03Lv22dg. Express vas the subject of the formula and evalu-ate v when h = 0.712, L = 150,d = 0.30 andg = 9.8111. The sag, S, at the centre of a wire is givenby the formula S =3d(l − d)8. Make lthe subject of the formula and evaluate l whend = 1.75 and S = 0.80.12. In an electrical alternating current cir-cuit the impedance Z is given byZ = R2 + ωL −1ωC2. Transposethe formula to make C the subject and henceevaluate C when Z = 130, R = 120,ω = 314and L = 0.3213. An approximate relationship between thenumber of teeth, T, on a milling cutter, thediameter of cutter, D, and the depth of cut, d,is given by T =12.5 DD + 4d. Determine the valueof D when T = 10 and d = 4mm.14. Make λ, the wavelength of X-rays, the subjectof the following formula:μρ=CZ4√λ5 na
• Chapter 13Solving simultaneousequations13.1 IntroductionOnly oneequation isnecessary when ﬁnding thevalueofa single unknown quantity (as with simple equationsin Chapter 11). However, when an equation containstwo unknown quantities it has an inﬁnite number ofsolutions. When two equations are available connectingthe same two unknown values then a unique solutionis possible. Similarly, for three unknown quantities it isnecessary to have three equations in order to solve for aparticular value of each of the unknown quantities, andso on.Equations which have to be solved together to ﬁndthe unique values of the unknown quantities, which aretrue for each of the equations, are called simultaneousequations.Two methods of solving simultaneous equations analyt-ically are:(a) by substitution, and(b) by elimination.(A graphical solution of simultaneous equations isshown in Chapter 19.)13.2 Solving simultaneous equationsin two unknownsThe method of solving simultaneous equations isdemonstrated in the following worked problems.Problem 1. Solve the following equations for xand y, (a) by substitution and (b) by eliminationx + 2y = −1 (1)4x − 3y = 18 (2)(a) By substitutionFrom equation (1): x = −1 − 2ySubstitutingthis expression for x into equation (2)gives4(−1 − 2y) − 3y = 18This is now a simple equation in y.Removing the bracket gives−4 − 8y − 3y = 18−11y = 18 + 4 = 22y =22−11= −2Substituting y = −2 into equation (1) givesx + 2(−2) = −1x − 4 = −1x = −1 + 4 = 3Thus, x = 3 and y = −2 is the solution to thesimultaneous equations.Check: in equation (2), since x = 3 and y = −2,LHS = 4(3) − 3(−2) = 12 + 6 = 18 = RHSDOI: 10.1016/B978-1-85617-697-2.00013-2
• Solving simultaneous equations 91(b) By eliminationx + 2y = −1 (1)4x − 3y = 18 (2)If equation (1) is multiplied throughout by 4, thecoefﬁcient of x will be the same as in equation (2),giving4x + 8y = −4 (3)Subtracting equation (3) from equation (2) gives4x − 3y = 18 (2)4x + 8y = −4 (3)0 − 11y = 22Hence, y =22−11= −2(Note: in the above subtraction,18 − −4 = 18 + 4 = 22.)Substituting y = −2 into either equation (1) or equa-tion (2) will give x = 3 as in method (a). The solutionx = 3,y = −2 is the only pair of values that satisﬁesboth of the original equations.Problem 2. Solve, by a substitution method, thesimultaneous equations3x − 2y = 12 (1)x + 3y = −7 (2)From equation (2), x = −7 − 3ySubstituting for x in equation (1) gives3(−7 − 3y) − 2y = 12i.e. −21 − 9y − 2y = 12−11y = 12 + 21 = 33Hence, y =33−11= −3Substituting y = −3 in equation (2) givesx + 3(−3) = −7i.e. x − 9 = −7Hence x = −7 + 9 = 2Thus, x = 2,y = −3 is the solutionof the simultaneousequations. (Such solutions should always be checkedby substituting values into each of the original twoequations.)Problem 3. Use an elimination method to solvethe following simultaneous equations3x + 4y = 5 (1)2x − 5y = −12 (2)If equation (1) is multiplied throughout by 2 and equa-tion (2) by 3, the coefﬁcient of x will be the same in thenewly formed equations. Thus,2 × equation (1) gives 6x + 8y = 10 (3)3 × equation (2) gives 6x − 15y = −36 (4)Equation (3) – equation (4) gives0 + 23y = 46i.e. y =4623= 2(Note+8y − −15y = 8y + 15y = 23y and 10−−36 =10 + 36 = 46.)Substituting y = 2 in equation (1) gives3x + 4(2) = 5from which 3x = 5 − 8 = −3and x = −1Checking, by substitutingx = −1 and y = 2 in equation(2), givesLHS = 2(−1) − 5(2) = −2 − 10 = −12 = RHSHence, x = −1 and y = 2 is the solution of the simul-taneous equations.The elimination method is the most common method ofsolving simultaneous equations.Problem 4. Solve7x − 2y = 26 (1)6x + 5y = 29 (2)When equation (1) is multiplied by 5 and equation (2)by 2, the coefﬁcients of y in each equation are numeri-cally the same, i.e. 10, but are of opposite sign.
• 92 Basic Engineering Mathematics5 × equation (1) gives 35x − 10y = 130 (3)2 × equation (2) gives 12x + 10y = 58 (4)Adding equations (3)and (4) gives 47x + 0 = 188Hence, x =18847= 4Note that when the signs of common coefﬁcients aredifferent the two equations are added and when thesigns of common coefﬁcients are the same the twoequations are subtracted (as in Problems 1 and 3).Substituting x = 4 in equation (1) gives7(4) − 2y = 2628 − 2y = 2628 − 26 = 2y2 = 2yHence, y = 1Checking, by substituting x = 4 and y = 1 in equation(2), givesLHS = 6(4) + 5(1) = 24 + 5 = 29 = RHSThus, the solution is x = 4,y = 1.Now try the following Practice ExercisePracticeExercise 49 Solving simultaneousequations (answers on page 345)Solve the following simultaneous equations andverify the results.1. 2x − y = 6 2. 2x − y = 2x + y = 6 x − 3y = −93. x − 4y = −4 4. 3x − 2y = 105x − 2y = 7 5x + y = 215. 5p + 4q = 6 6. 7x + 2y = 112p − 3q = 7 3x − 5y = −77. 2x − 7y = −8 8. a + 2b = 83x + 4y = 17 b − 3a = −39. a + b = 7 10. 2x + 5y = 7a − b = 3 x + 3y = 411. 3s + 2t = 12 12. 3x − 2y = 134s − t = 5 2x + 5y = −413. 5m − 3n = 11 14. 8a − 3b = 513m + n = 8 3a + 4b = 1415. 5x = 2y 16. 5c = 1 − 3d3x + 7y = 41 2d + c + 4 = 013.3 Further solving of simultaneousequationsHere are some further worked problems on solvingsimultaneous equations.Problem 5. Solve3p = 2q (1)4p + q + 11 = 0 (2)Rearranging gives3p − 2q = 0 (3)4p + q = −11 (4)Multiplying equation (4) by 2 gives8p + 2q = −22 (5)Adding equations (3) and (5) gives11p + 0 = −22p =−2211= −2Substituting p = −2 into equation (1) gives3(−2) = 2q−6 = 2qq =−62= −3Checking, by substituting p = −2 and q = −3 intoequation (2), givesLHS = 4(−2) + (−3) + 11 = −8 − 3 + 11 = 0 = RHSHence, the solution is p = −2,q = −3.
• Solving simultaneous equations 93Problem 6. Solvex8+52= y (1)13 −y3= 3x (2)Whenever fractions are involved in simultaneous equa-tions it is often easier to ﬁrstly remove them. Thus,multiplying equation (1) by 8 gives8x8+ 852= 8yi.e. x + 20 = 8y (3)Multiplying equation (2) by 3 gives39 − y = 9x (4)Rearranging equations (3) and (4) givesx − 8y = −20 (5)9x + y = 39 (6)Multiplying equation (6) by 8 gives72x + 8y = 312 (7)Adding equations (5) and (7) gives73x + 0 = 292x =29273= 4Substituting x = 4 into equation (5) gives4 − 8y = −204 + 20 = 8y24 = 8yy =248= 3Checking, substituting x = 4 and y = 3 in the originalequations, gives(1): LHS =48+52=12+ 212= 3 = y = RHS(2): LHS = 13 −33= 13 − 1 = 12RHS = 3x = 3(4) = 12Hence, the solution is x = 4,y = 3.Problem 7. Solve2.5x + 0.75 − 3y = 01.6x = 1.08 − 1.2yIt is often easier to remove decimal fractions. Thus,multiplying equations (1) and (2) by 100 gives250x + 75 − 300y = 0 (1)160x = 108 − 120y (2)Rearranging gives250x − 300y = −75 (3)160x + 120y = 108 (4)Multiplying equation (3) by 2 gives500x − 600y = −150 (5)Multiplying equation (4) by 5 gives800x + 600y = 540 (6)Adding equations (5) and (6) gives1300x + 0 = 390x =3901300=39130=310= 0.3Substituting x = 0.3 into equation (1) gives250(0.3) + 75 − 300y = 075 + 75 = 300y150 = 300yy =150300= 0.5Checking, by substituting x = 0.3 and y = 0.5 in equa-tion (2), givesLHS = 160(0.3) = 48RHS = 108 − 120(0.5) = 108 − 60 = 48Hence, the solution is x = 0.3,y = 0.5
• 94 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 50 Solving simultaneousequations (answers on page 345)Solve the following simultaneous equations andverify the results.1. 7p + 11 + 2q = 0 2.x2+y3= 4−1 = 3q − 5px6−y9= 03.a2− 7 = −2b 4.32s − 2t = 812 = 5a +23bs4+ 3y = −25.x5+2y3=49156. v − 1 =u123x7−y2+57= 0 u +v4−252= 07. 1.5x − 2.2y = −18 8. 3b − 2.5a = 0.452.4x + 0.6y = 33 1.6a + 0.8b = 0.813.4 Solving more difﬁcultsimultaneous equationsHere are some further worked problems on solving moredifﬁcult simultaneous equations.Problem 8. Solve2x+3y= 7 (1)1x−4y= −2 (2)In this type of equation the solution is easier if asubstitution is initially made. Let1x= a and1y= bThus equation (1) becomes 2a + 3b = 7 (3)and equation (2) becomes a − 4b = −2 (4)Multiplying equation (4) by 2 gives2a − 8b = −4 (5)Subtracting equation (5) from equation (3) gives0 + 11b = 11i.e. b = 1Substituting b = 1 in equation (3) gives2a + 3 = 72a = 7 − 3 = 4i.e. a = 2Checking, substituting a = 2 and b = 1 in equation (4),givesLHS = 2 − 4(1) = 2 − 4 = −2 = RHSHence, a = 2 and b = 1.However, since1x= a, x =1a=12or 0.5and since1y= b, y =1b=11= 1Hence, the solution is x = 0.5, y = 1.Problem 9. Solve12a+35b= 4 (1)4a+12b= 10.5 (2)Let1a= x and1b= ythen x2+35y = 4 (3)4x +12y = 10.5 (4)To remove fractions, equation (3) is multiplied by 10,giving10x2+ 1035y = 10(4)i.e. 5x + 6y = 40 (5)Multiplying equation (4) by 2 gives8x + y = 21 (6)Multiplying equation (6) by 6 gives48x + 6y = 126 (7)
• Solving simultaneous equations 95Subtracting equation (5) from equation (7) gives43x + 0 = 86x =8643= 2Substituting x = 2 into equation (3) gives22+35y = 435y = 4 − 1 = 3y =53(3) = 5Since1a= x, a =1x=12or 0.5and since1b= y, b =1y=15or 0.2Hence, the solution is a = 0.5, b = 0.2, which may bechecked in the original equations.Problem 10. Solve1x + y=427(1)12x − y=433(2)To eliminate fractions, both sides of equation (1) aremultiplied by 27(x + y), giving27(x + y)1x + y= 27(x + y)427i.e. 27(1) = 4(x + y)27 = 4x + 4y (3)Similarly, in equation (2) 33 = 4(2x − y)i.e. 33 = 8x − 4y (4)Equation (3)+equation (4) gives60 = 12x and x =6012= 5Substituting x = 5 in equation (3) gives27 = 4(5) + 4yfrom which 4y = 27 − 20 = 7and y =74= 134or 1.75Hence, x = 5, y = 1.75 is the required solution, whichmay be checked in the original equations.Problem 11. Solvex − 13+y + 25=215(1)1 − x6+5 + y2=56(2)Before equations (1) and (2) can be simultaneouslysolved, the fractions need to be removed and theequations rearranged.Multiplying equation (1) by 15 gives15x − 13+ 15y + 25= 15215i.e. 5(x − 1) + 3(y + 2) = 25x − 5 + 3y + 6 = 25x + 3y = 2 + 5 − 6Hence, 5x + 3y = 1 (3)Multiplying equation (2) by 6 gives61 − x6+ 65 + y2= 656i.e. (1 − x) + 3(5 + y) = 51 − x + 15 + 3y = 5−x + 3y = 5 − 1 − 15Hence, −x + 3y = −11 (4)Thus the initial problem containing fractions can beexpressed as5x + 3y = 1 (3)−x + 3y = −11 (4)Subtracting equation (4) from equation (3) gives6x + 0 = 12x =126= 2
• 96 Basic Engineering MathematicsSubstituting x = 2 into equation (3) gives5(2) + 3y = 110 + 3y = 13y = 1 − 10 = −9y =−93= −3Checking, substituting x = 2, y = −3 in equation (4)givesLHS = −2 + 3(−3) = −2 − 9 = −11 = RHSHence, the solution is x = 2,y = −3.Now try the following Practice ExercisePracticeExercise 51 Solving more difﬁcultsimultaneous equations (answers on page345)In problems 1 to 7, solve the simultaneous equa-tions and verify the results1.3x+2y= 14 2.4a−3b= 185x−3y= −22a+5b= −43.12p+35q= 5 4.5x+3y= 1.15p−12q=3523x−7y= −1.15.c + 14−d + 23+ 1 = 01 − c5+3 − d4+1320= 07.5x + y=202742x − y=16336.3r + 25−2s − 14=1153 + 2r4+5 − s3=1548. If 5x −3y= 1 and x +4y=52, ﬁnd the valueofxy + 1y13.5 Practical problems involvingsimultaneous equationsThere are a number of situations in engineering andscience in which the solution of simultaneous equationsis required. Some are demonstrated in the followingworked problems.Problem 12. The law connecting friction F andload L for an experiment is of the form F = aL + bwhere a and b are constants. When F = 5.6N,L = 8.0N and when F = 4.4N, L = 2.0N. Find thevalues of a and b and the value of F whenL = 6.5NSubstituting F = 5.6 and L = 8.0 into F = aL + bgives5.6 = 8.0a + b (1)Substituting F = 4.4 and L = 2.0 into F = aL + bgives4.4 = 2.0a + b (2)Subtracting equation (2) from equation (1) gives1.2 = 6.0aa =1.26.0=15or 0.2Substituting a =15into equation (1) gives5.6 = 8.015+ b5.6 = 1.6 + b5.6 − 1.6 = bi.e. b = 4Checking, substitutinga =15and b = 4 in equation (2),givesRHS = 2.015+ 4 = 0.4 + 4 = 4.4 = LHSHence, a =15and b = 4When L = 6.5, F = aL + b =15(6.5) + 4 = 1.3 + 4,i.e. F = 5.30N.
• Solving simultaneous equations 97Problem 13. The equation of a straight line,of gradient m and intercept on the y-axis c, isy = mx + c. If a straight line passes through thepoint where x = 1 and y = −2, and also throughthe point where x = 3.5 and y = 10.5, ﬁnd thevalues of the gradient and the y-axis interceptSubstituting x = 1 and y = −2 into y = mx + c gives−2 = m + c (1)Substituting x = 3.5 and y = 10.5 into y = mx + cgives10.5 = 3.5m + c (2)Subtracting equation (1) from equation (2) gives12.5 = 2.5m, from which, m =12.52.5= 5Substituting m = 5 into equation (1) gives−2 = 5 + cc = −2 − 5 = −7Checking, substituting m = 5 and c = −7 in equation(2), givesRHS = (3.5)(5) + (−7) = 17.5 − 7 = 10.5 = LHSHence, the gradient m = 5 and the y-axis interceptc = −7.Problem 14. When Kirchhoff’s laws are appliedto the electrical circuit shown in Figure 13.1, thecurrents I1 and I2 are connected by the equations27 = 1.5I1 + 8(I1 − I2) (1)−26 = 2I2 − 8(I1 − I2) (2)27 V 26 V8 V1.5 V 2 Vl1 l2(l12l2)Figure 13.1Solve the equations to ﬁnd the values of currents I1and I2Removing the brackets from equation (1) gives27 = 1.5I1 + 8I1 − 8I2Rearranging gives9.5I1 − 8I2 = 27 (3)Removing the brackets from equation (2) gives−26 = 2I2 − 8I1 + 8I2Rearranging gives−8I1 + 10I2 = −26 (4)Multiplying equation (3) by 5 gives47.5I1 − 40I2 = 135 (5)Multiplying equation (4) by 4 gives−32I1 + 40I2 = −104 (6)Adding equations (5) and (6) gives15.5I1 + 0 = 31I1 =3115.5= 2Substituting I1 = 2 into equation (3) gives9.5(2) − 8I1 = 2719 − 8I2 = 2719 − 27 = 8I2−8 = 8I2and I2 = −1Hence, the solution is I1 = 2 and I2 = −1 (which maybe checked in the original equations).Problem 15. The distance s metres from a ﬁxedpoint of a vehicle travelling in a straight line withconstant acceleration, a m/s2, is given bys = ut + 12 at2, where u is the initial velocity in m/sand t the time in seconds. Determine the initialvelocity and the acceleration given that s = 42 mwhen t = 2 s, and s = 144m when t = 4s. Also ﬁndthe distance travelled after 3s
• 98 Basic Engineering MathematicsSubstitutings = 42 and t = 2 into s = ut +12at2 gives42 = 2u +12a(2)2i.e. 42 = 2u + 2a (1)Substituting s = 144 and t = 4 into s = ut +12at2gives144 = 4u +12a(4)2i.e. 144 = 4u + 8a (2)Multiplying equation (1) by 2 gives84 = 4u + 4a (3)Subtracting equation (3) from equation (2) gives60 = 0 + 4aand a =604= 15Substituting a = 15 into equation (1) gives42 = 2u + 2(15)42 − 30 = 2uu =122= 6Substituting a = 15 and u = 6 in equation (2) givesRHS = 4(6) + 8(15) = 24 + 120 = 144 = LHSHence, the initial velocity u = 6 m/s and the acceler-ation a = 15 m/s2.Distance travelled after 3s is given by s = ut +12at2where t = 3,u = 6 and a = 15.Hence, s = (6)(3) +12(15)(3)2 = 18 + 67.5i.e. distance travelled after 3 s = 85.5 m.Problem 16. The resistance R of a length ofwire at t◦C is given by R = R0(1 + αt), where R0is the resistance at 0◦C and α is the temperaturecoefﬁcient of resistance in /◦C. Find the values of αand R0 if R = 30 at 50◦C and R = 35 at 100◦CSubstituting R = 30 and t = 50 into R = R0(1 + αt)gives30 = R0(1 + 50α) (1)Substituting R = 35 and t = 100 into R = R0(1 + αt)gives35 = R0(1 + 100α) (2)Although these equations may be solved by the conven-tional substitution method, an easier way is to eliminateR0 by division. Thus, dividing equation (1) by equation(2) gives3035=R0(1 + 50α)R0(1 + 100α)=1 + 50α1 + 100αCross-multiplying gives30(1 + 100α) = 35(1 + 50α)30 + 3000α = 35 + 1750α3000α − 1750α = 35 − 301250α = 5i.e. α =51250=1250or 0.004Substituting α =1250into equation (1) gives30 = R0 1 + (50)125030 = R0(1.2)R0 =301.2= 25Checking, substituting α =1250and R0 = 25 in equa-tion (2), givesRHS = 25 1 + (100)1250= 25(1.4) = 35 = LHSThus, the solution is α = 0.004/◦C and R0 = 25 .Problem 17. The molar heat capacity of a solidcompound is given by the equation c = a + bT,where a and b are constants. When c = 52, T = 100and when c = 172, T = 400. Determine the valuesof a and b
• Solving simultaneous equations 99When c = 52, T = 100, hence52 = a + 100b (1)When c = 172, T = 400, hence172 = a + 400b (2)Equation (2) – equation (1) gives120 = 300bfrom which, b =120300= 0.4Substituting b = 0.4 in equation (1) gives52 = a + 100(0.4)a = 52 − 40 = 12Hence, a = 12 and b = 0.4Now try the following Practice ExercisePracticeExercise 52 Practical problemsinvolving simultaneous equations (answerson page 345)1. In a system of pulleys, the effort P required toraise a load W is given by P = aW + b, wherea and b are constants. If W = 40 when P = 12and W = 90 when P = 22, ﬁnd the values ofa and b.2. Applying Kirchhoff’s laws to an electricalcircuit produces the following equations:5 = 0.2I1 + 2(I1 − I2)12 = 3I2 + 0.4I2 − 2(I1 − I2)Determine the values of currents I1 and I23. Velocity v is given by the formula v = u + at.If v = 20 when t = 2 and v = 40 when t = 7,ﬁnd the values of u and a. Then, ﬁnd thevelocity when t = 3.54. Three new cars and 4 new vans supplied to adealer together cost £97700 and 5 new carsand 2 new vans of the same models cost£103100. Find the respective costs of a carand a van.5. y = mx + c is the equation of a straight lineof slope m and y-axis intercept c. If the linepasses through the point where x = 2 andy = 2, and also through the point where x = 5and y = 0.5,ﬁnd theslopeand y-axisinterceptof the straight line.6. The resistance R ohms of copper wire at t◦Cis given by R = R0(1 + αt), where R0 is theresistanceat 0◦C and α isthetemperaturecoef-ﬁcient of resistance. If R = 25.44 at 30◦Cand R = 32.17 at 100◦C, ﬁnd α and R07. The molar heat capacity of a solid compoundis given by the equation c = a + bT . Whenc = 60, T = 100 and when c = 210, T = 400.Find the values of a and b.8. In an engineering process, two variables p andq are related by q = ap + b/p, where a and bare constants. Evaluate a and b if q = 13 whenp = 2 and q = 22 when p = 5.9. In a system of forces, the relationship betweentwo forces F1 and F2 is given by5F1 + 3F2 + 6 = 03F1 + 5F2 + 18 = 0Solve for F1 and F213.6 Solving simultaneous equationsin three unknownsEquations containing three unknowns may be solvedusing exactly the same procedures as those used withtwo equations and two unknowns, providing that thereare three equations to work with. The method is demon-strated in the following worked problem.Problem 18. Solve the simultaneous equations.x + y + z = 4 (1)2x − 3y + 4z = 33 (2)3x − 2y − 2z = 2 (3)There are a number of ways of solving these equations.One method is shown below.The initial object is to produce two equations with twounknowns. For example, multiplying equation (1) by 4and then subtracting this new equation from equation (2)will produce an equation with only x and y involved.
• 100 Basic Engineering MathematicsMultiplying equation (1) by 4 gives4x + 4y + 4z = 16 (4)Equation (2) – equation (4) gives−2x − 7y = 17 (5)Similarly, multiplying equation (3) by 2 and thenadding this new equation to equation (2) will produceanother equation with only x and y involved.Multiplying equation (3) by 2 gives6x − 4y − 4z = 4 (6)Equation (2)+equation (6) gives8x − 7y = 37 (7)Rewriting equation (5) gives−2x − 7y = 17 (5)Now we can use the previous method for solvingsimultaneous equations in two unknowns.Equation (7) – equation (5) gives 10x = 20from which, x = 2(Note that 8x − −2x = 8x + 2x = 10x)Substituting x = 2 into equation (5) gives−4 − 7y = 17from which, −7y = 17 + 4 = 21and y = −3Substituting x = 2 and y = −3 into equation (1) gives2 − 3 + z = 4from which, z = 5Hence, the solution of the simultaneous equations isx = 2,y = −3 and z = 5.Now try the following Practice ExercisePracticeExercise 53 Simultaneousequations in three unknowns (answers onpage 345)In problems 1 to 9, solve the simultaneous equa-tions in 3 unknowns.1. x + 2y + 4z = 16 2. 2x + y − z = 02x − y + 5z = 18 3x + 2y + z = 43x + 2y + 2z = 14 5x + 3y + 2z = 83. 3x + 5y + 2z = 6 4. 2x + 4y + 5z = 23x − y + 3z = 0 3x − y − 2z = 62 + 7y + 3z = −3 4x + 2y + 5z = 315. 2x + 3y + 4z = 36 6. 4x + y + 3z = 313x + 2y + 3z = 29 2x − y + 2z = 10x + y + z = 11 3x + 3y − 2z = 77. 5x + 5y − 4z = 37 8. 6x + 7y + 8z = 132x − 2y + 9z = 20 3x + y − z = −11−4x + y + z = −14 2x − 2y − 2z = −189. 3x + 2y + z = 147x + 3y + z = 22.54x − 4y − z = −8.510. Kirchhoff’s laws are used to determine thecurrent equations in an electrical network andresult in the following:i1 + 8i2 + 3i3 = −313i1 − 2i2 + i3 = −52i1 − 3i2 + 2i3 = 6Determine the values of i1,i2 and i311. The forces in three members of a frame-work are F1, F2 and F3. They are related byfollowing simultaneous equations.1.4F1 + 2.8F2 + 2.8F3 = 5.64.2F1 − 1.4F2 + 5.6F3 = 35.04.2F1 + 2.8F2 − 1.4F3 = −5.6Find the values of F1, F2 and F3
• Revision Test 5 : Transposition and simultaneous equationsThis assignment covers the material contained in Chapters 12 and 13. The marks available are shown in brackets atthe end of each question.1. Transpose p − q +r = a − b for b. (2)2. Make π the subject of the formula r =c2π(2)3. Transpose V =13πr2h for h. (2)4. Transpose I =E − eR +rfor E. (2)5. Transpose k =bad − 1for d. (4)6. Make g the subject of the formula t = 2πLg(3)7. Transpose A =πR2θ360for R. (2)8. Make r the subject of the formulax + y =r3 +r(5)9. Make L the subject of the formulam =μLL +rCR(5)10. The surface area A of a rectangular prism isgiven by the formula A = 2(bh + hl +lb). Eval-uate b when A = 11750 mm2,h = 25mm andl = 75mm. (4)11. The velocity v of water in a pipe appears inthe formula h =0.03 Lv22dg. Evaluate v whenh = 0.384,d = 0.20, L = 80 and g = 10. (3)12. A formula for the focal length f of a convexlens is1f=1u+1v. Evaluate v when f = 4 andu = 20. (4)In problems 13 and 14, solve the simultaneous equa-tions.13. (a) 2x + y = 6 (b) 4x − 3y = 115x − y = 22 3x + 5y = 30 (9)14. (a) 3a − 8 +b8= 0b +a2=214(b)2p + 15−1 − 4q2=521 − 3p7+2q − 35+3235= 0 (18)15. In an engineering process two variables x and yare related by the equation y = ax +bx, where aand b are constants. Evaluate a and b if y = 15when x = 1 and y = 13 when x = 3. (5)16. Kirchhoff’s laws are used to determine the currentequations in an electrical network and result in thefollowing:i1 + 8i2 + 3i3 = −313i1 − 2i2 + i3 = −52i1 − 3i2 + 2i3 = 6Determine the values of i1, i2 and i3 (10)
• Solving quadratic equations 103The quadratic equation x2+ 2x − 8 = 0 thus becomes(x + 4)(x − 2) = 0Since the only way that this can be true is for either theﬁrst or the second or both factors to be zero,either (x + 4) = 0, i.e. x = −4or (x − 2) = 0, i.e. x = 2Hence, the roots of x2 + 2x − 8 = 0 are x = −4 andx = 2.Problem 3. Determine the roots ofx2− 6x + 9 = 0 by factorizationx2 − 6x + 9 = (x − 3)(x − 3), i.e. (x − 3)2 = 0The LHS is known as a perfect square.Hence, x = 3 is the only root of the equationx2 − 6x + 9 = 0.Problem 4. Solve the equation x2 − 4x = 0Factorizing gives x(x − 4) = 0If x(x − 4) = 0,either x = 0 or x − 4 = 0i.e. x = 0 or x = 4These are the two roots of the given equation. Answerscan always be checked by substitution into the originalequation.Problem 5. Solve the equation x2 + 3x − 4 = 0Factorizing gives (x − 1)(x + 4) = 0Hence, either x − 1 = 0 or x + 4 = 0i.e. x = 1 or x = −4Problem 6. Determine the roots of 4x2 − 25 = 0by factorizationThe LHS of 4x2 − 25 = 0 is the difference of twosquares, (2x)2 and (5)2.By factorizing, 4x2 − 25 = (2x + 5)(2x − 5), i.e.(2x + 5)(2x − 5) = 0Hence, either (2x + 5) = 0, i.e. x = −52= −2.5or (2x − 5) = 0, i.e. x =52= 2.5Problem 7. Solve the equation x2− 5x + 6 = 0Factorizing gives (x − 3)(x − 2) = 0Hence, either x − 3 = 0 or x − 2 = 0i.e. x = 3 or x = 2Problem 8. Solve the equation x2 = 15 − 2xRearranging gives x2+ 2x − 15 = 0Factorizing gives (x + 5)(x − 3) = 0Hence, either x + 5 = 0 or x − 3 = 0i.e. x = −5 or x = 3Problem 9. Solve the equation 3x2 − 11x − 4 = 0by factorizationThe factors of 3x2 are 3x and x. These are placed inbrackets: (3x )(x )The factors of −4 are −4 and +1, or +4 and −1, or −2and 2.Remembering that the product of the two inner termsadded to the product of the two outer terms must equal−11x, the only combination to give this is +1 and −4,i.e. 3x2−11x − 4 = (3x + 1)(x − 4)The quadratic equation 3x2 − 11x − 4 = 0 thusbecomes (3x + 1)(x − 4) = 0Hence, either (3x + 1) = 0, i.e. x = −13or (x − 4) = 0, i.e. x = 4and both solutions may be checked in the originalequation.Problem 10. Solve the quadratic equation4x2+ 8x + 3 = 0 by factorizingThe factors of 4x2 are 4x and x or 2x and 2x.The factors of 3 are 3 and 1, or −3 and −1.Remembering that the product of the inner terms addedto the product of the two outer terms must equal +8x,the only combination that is true (by trial and error) is(4x2+ 8x + 3) = (2x + 3)(2x + 1)
• 104 Basic Engineering MathematicsHence, (2x + 3)(2x + 1) = 0, from which either(2x + 3) = 0 or (2x + 1) = 0.Thus, 2x = −3, from which x = −32or −1.5or 2x = −1, from which x = −12or −0.5which may be checked in the original equation.Problem 11. Solve the quadratic equation15x2 + 2x − 8 = 0 by factorizingThe factors of 15x2 are 15x and x or 5x and 3x.The factors of −8 are −4 are +2, or 4 and −2, or −8and +1, or 8 and −1.By trial and error the only combination that works is15x2+ 2x − 8 = (5x + 4)(3x − 2)Hence, (5x + 4)(3x − 2) = 0, from which either 5x +4 = 0 or 3x − 2 = 0.Hence, x = −45or x =23which may be checked in the original equation.Problem 12. The roots of a quadratic equationare13and −2. Determine the equation in xIf the roots of a quadratic equation are, say, α and β,then (x − α)(x − β) = 0.Hence, if α =13and β = −2,x −13(x − (−2)) = 0x −13(x + 2) = 0x2−13x + 2x −23= 0x2+53x −23= 0or 3x2+ 5x − 2 = 0Problem 13. Find the equation in x whose rootsare 5 and −5If 5 and −5 are the roots of a quadratic equation then(x − 5)(x + 5) = 0i.e. x2− 5x + 5x − 25 = 0i.e. x2− 25 = 0Problem 14. Find the equation in x whose rootsare 1.2 and −0.4If 1.2 and −0.4 are the roots of a quadratic equationthen(x − 1.2)(x + 0.4) = 0i.e. x2− 1.2x + 0.4x − 0.48 = 0i.e. x2− 0.8x − 0.48 = 0Now try the following Practice ExercisePracticeExercise 54 Solving quadraticequations by factorization (answers onpage 346)In problems 1 to 30, solve the given equations byfactorization.1. x2 − 16 = 0 2. x2 + 4x − 32 = 03. (x + 2)2 = 16 4. 4x2 − 9 = 05. 3x2 + 4x = 0 6. 8x2 − 32 = 07. x2 − 8x + 16 = 0 8. x2 + 10x + 25 = 09. x2 − 2x + 1 = 0 10. x2 + 5x + 6 = 011. x2 + 10x + 21 = 0 12. x2 − x − 2 = 013. y2 − y − 12 = 0 14. y2 − 9y + 14 = 015. x2+ 8x + 16 = 0 16. x2− 4x + 4 = 017. x2 + 6x + 9 = 0 18. x2 − 9 = 019. 3x2 + 8x + 4 = 0 20. 4x2 + 12x + 9 = 021. 4z2 −116= 0 22. x2 + 3x − 28 = 023. 2x2 − x − 3 = 0 24. 6x2 − 5x + 1 = 025. 10x2 + 3x − 4 = 0 26. 21x2 − 25x = 427. 8x2 + 13x − 6 = 0 28. 5x2 + 13x − 6 = 029. 6x2 − 5x − 4 = 0 30. 8x2 + 2x − 15 = 0In problems 31 to 36, determine the quadraticequations in x whose roots are31. 3 and 1 32. 2 and −533. −1 and −4 34. 2.5 and −0.535. 6 and −6 36. 2.4 and −0.7
• Solving quadratic equations 10514.3 Solution of quadratic equationsby ‘completing the square’An expression such as x2 or (x + 2)2 or (x − 3)2 iscalled a perfect square.If x2 = 3 then x = ±√3If (x + 2)2 = 5 then x + 2 = ±√5 and x = −2 ±√5If (x − 3)2 = 8 then x − 3 = ±√8 and x = 3 ±√8Hence, if a quadratic equation can be rearranged so thatone side of the equation is a perfect square and the otherside of the equation is a number, then the solution ofthe equation is readily obtained by taking the squareroots of each side as in the above examples. The processof rearranging one side of a quadratic equation into aperfect square before solving is called ‘completing thesquare’.(x + a)2= x2+ 2ax + a2Thus,inordertomakethequadraticexpressionx2 + 2axinto a perfect square, it is necessary to add (half thecoefﬁcient of x)2, i.e.2a22or a2For example, x2 + 3x becomes a perfect square byadding322, i.e.x2+ 3x +322= x +322The method of completing the square is demonstratedin the following worked problems.Problem 15. Solve 2x2 + 5x = 3 by completingthe squareThe procedure is as follows.(i) Rearrangetheequation so that all termsareon thesame side of the equals sign (and the coefﬁcientof the x2 term is positive). Hence,2x2+ 5x − 3 = 0(ii) Make the coefﬁcient of the x2 term unity. In thiscase thisis achieved by dividingthroughoutby 2.Hence,2x22+5x2−32= 0i.e. x2+52x −32= 0(iii) Rearrange the equations so that the x2and xterms are on one side of the equals sign and theconstant is on the other side. Hence,x2+52x =32(iv) Add to both sides of the equation (half the coefﬁ-cient of x)2. In this case the coefﬁcient of x is52Half the coefﬁcient squared is therefore542Thus,x2+52x +542=32+542The LHS is now a perfect square, i.e.x +542=32+542(v) Evaluate the RHS. Thus,x +542=32+2516=24 + 2516=4916(vi) Take the square root of both sides of the equation(remembering that the square root of a numbergives a ± answer). Thus,x +542=4916i.e. x +54= ±74(vii) Solve the simple equation. Thus,x = −54±74i.e. x = −54+74=24=12or 0.5and x = −54−74= −124= −3Hence, x = 0.5 or x = −3; i.e., the roots of theequation 2x2 + 5x = 3 are 0.5 and −3.Problem 16. Solve 2x2 + 9x + 8 = 0, correct to 3signiﬁcant ﬁgures, by completing the squareMaking the coefﬁcient of x2 unity givesx2 +92x + 4 = 0Rearranging gives x2 +92x = −4
• 106 Basic Engineering MathematicsAdding to both sides (half the coefﬁcient of x)2givesx2+92x +942=942− 4The LHS is now a perfect square. Thus,x +942=8116− 4 =8116−6416=1716Taking the square root of both sides givesx +94=1716= ±1.031Hence, x = −94± 1.031i.e. x = −1.22 or −3.28, correct to 3 signiﬁcant ﬁgures.Problem 17. By completing the square, solve thequadratic equation 4.6y2 + 3.5y − 1.75 = 0, correctto 3 decimal places4.6y2+ 3.5y − 1.75 = 0Making the coefﬁcient of y2 unity givesy2+3.54.6y −1.754.6= 0and rearranging gives y2+3.54.6y =1.754.6Adding to both sides (half the coefﬁcient of y)2 givesy2+3.54.6y +3.59.22=1.754.6+3.59.22The LHS is now a perfect square. Thus,y +3.59.22= 0.5251654Taking the square root of both sides givesy +3.59.2=√0.5251654 = ±0.7246830Hence, y = −3.59.2± 0.7246830i.e. y = 0.344 or −1.105Now try the following Practice ExercisePracticeExercise 55 Solving quadraticequations by completing the square(answers on page 346)Solve the following equations correct to 3 decimalplaces by completing the square.1. x2 + 4x + 1 = 0 2. 2x2 + 5x − 4 = 03. 3x2 − x − 5 = 0 4. 5x2 − 8x + 2 = 05. 4x2 − 11x + 3 = 0 6. 2x2 + 5x = 214.4 Solution of quadratic equationsby formulaLet the general form of a quadratic equation be givenby ax2 + bx + c = 0, where a,b and c are constants.Dividing ax2+ bx + c = 0 by a givesx2+bax +ca= 0Rearranging gives x2+bax = −caAdding to each side of the equation the square of halfthe coefﬁcient of the term in x to make the LHS a perfectsquare givesx2+bax +b2a2=b2a2−caRearranging gives x +ba2=b24a2−ca=b2− 4ac4a2Taking the square root of both sides givesx +b2a=b2 − 4ac4a2=±√b2 − 4ac2aHence, x = −b2a±√b2 − 4ac2ai.e. the quadratic formula is x =−b ±√b2 − 4ac2a(This method of obtainingthe formula is completing thesquare − as shown in the previous section.)In summary,if ax2+ bx + c = 0 then x =−b ±√b2 − 4ac2aThis is known as the quadratic formula.
• Solving quadratic equations 107Problem 18. Solve x2+ 2x − 8 = 0 by using thequadratic formulaComparing x2 + 2x − 8 = 0 with ax2 + bx + c = 0gives a = 1,b = 2 and c = −8.Substituting these values into the quadratic formulax =−b ±√b2 − 4ac2agivesx =−2 ± 22 − 4(1)(−8)2(1)=−2 ±√4 + 322=−2 ±√362=−2 ± 62=−2 + 62or−2 − 62Hence, x =42or−82, i.e. x = 2 or x = −4.Problem 19. Solve 3x2 − 11x − 4 = 0 by usingthe quadratic formulaComparing 3x2 − 11x − 4 = 0 with ax2 + bx + c = 0gives a = 3,b = −11 and c = −4. Hence,x =−(−11) ± (−11)2 − 4(3)(−4)2(3)=+11 ±√121 + 486=11 ±√1696=11 ± 136=11 + 136or11 − 136Hence, x =246or−26, i.e. x = 4 or x = −13Problem 20. Solve 4x2 + 7x + 2 = 0 giving theroots correct to 2 decimal placesComparing 4x2 + 7x + 2 = 0 with ax2 + bx + c givesa = 4,b = 7 and c = 2. Hence,x =−7 ± 72 − 4(4)(2)2(4)=−7 ±√178=−7 ± 4.1238=−7 + 4.1238or−7 − 4.1238Hence, x = −0.36 or −1.39, correct to 2 decimalplaces.Problem 21. Use the quadratic formula to solvex + 24+3x − 1= 7 correct to 4 signiﬁcant ﬁguresMultiplying throughout by 4(x − 1) gives4(x − 1)(x + 2)4+ 4(x − 1)3(x − 1)= 4(x − 1)(7)Cancelling gives (x − 1)(x + 2) + (4)(3) = 28(x − 1)x2+ x − 2 + 12 = 28x − 28Hence, x2− 27x + 38 = 0Using the quadratic formula,x =−(−27) ± (−27)2 − 4(1)(38)2=27 ±√5772=27 ± 24.02082Hence, x =27 + 24.02082= 25.5104or x =27 − 24.02082= 1.4896Hence, x = 25.51 or 1.490, correct to 4 signiﬁcantﬁgures.Now try the following Practice ExercisePracticeExercise 56 Solving quadraticequations by formula (answers on page 346)Solve the following equations by using thequadratic formula, correct to 3 decimal places.1. 2x2 + 5x − 4 = 02. 5.76x2 + 2.86x − 1.35 = 03. 2x2− 7x + 4 = 04. 4x + 5 =3x5. (2x + 1) =5x − 36. 3x2 − 5x + 1 = 0
• 108 Basic Engineering Mathematics7. 4x2 + 6x − 8 = 08. 5.6x2 − 11.2x − 1 = 09. 3x(x + 2) + 2x(x − 4) = 810. 4x2 − x(2x + 5) = 1411.5x − 3+2x − 2= 612.3x − 7+ 2x = 7 + 4x13.x + 1x − 1= x − 314.5 Practical problems involvingquadratic equationsThere are many practical problems in which aquadratic equation has ﬁrst to be obtained, from giveninformation, before it is solved.Problem 22. The area of a rectangle is 23.6cm2and its width is 3.10cm shorter than its length.Determine the dimensions of the rectangle, correctto 3 signiﬁcant ﬁguresLet the length of the rectangle be x cm. Then the widthis (x − 3.10)cm.Area = length × width = x(x − 3.10) = 23.6i.e. x2− 3.10x − 23.6 = 0Using the quadratic formula,x =−(−3.10) ± (−3.10)2 − 4(1)(−23.6)2(1)=3.10 ±√9.61 + 94.42=3.10 ± 10.202=13.302or−7.102Hence, x = 6.65cm or −3.55cm. The latter solution isneglected since length cannot be negative.Thus, length x = 6.65cm and width = x − 3.10 =6.65 − 3.10 = 3.55cm, i.e. the dimensions of the rect-angle are 6.65cm by 3.55cm.(Check: Area = 6.65 × 3.55 = 23.6cm2, correct to3 signiﬁcant ﬁgures.)Problem 23. Calculate the diameter of a solidcylinder which has a height of 82.0cm and a totalsurface area of 2.0m2Total surface area of a cylinder= curved surface area + 2 circular ends= 2πrh + 2πr2(where r = radius and h = height)Since the total surface area = 2.0m2 and the height h =82cm or 0.82m,2.0 = 2πr(0.82) + 2πr2i.e. 2πr2+ 2πr(0.82) − 2.0 = 0Dividing throughout by 2π gives r2 +0.82r −1π= 0Using the quadratic formula,r =−0.82 ± (0.82)2 − 4(1) − 1π2(1)=−0.82 ±√1.945642=−0.82 ± 1.394862= 0.2874 or − 1.1074Thus, the radius r of the cylinder is 0.2874m (thenegative solution being neglected).Hence, the diameter of the cylinder= 2 × 0.2874= 0.5748m or 57.5cmcorrect to 3 signiﬁcant ﬁgures.Problem 24. The height s metres of a massprojected vertically upwards at time t seconds iss = ut −12gt2. Determine how long the mass willtake after being projected to reach a height of 16m(a) on the ascent and (b) on the descent, whenu = 30m/s and g = 9.81m/s2When height s = 16m, 16 = 30t −12(9.81)t2i.e. 4.905t2− 30t + 16 = 0Using the quadratic formula,t =−(−30) ± (−30)2 − 4(4.905)(16)2(4.905)=30 ±√586.19.81=30 ± 24.219.81= 5.53 or 0.59
• Solving quadratic equations 109Hence, the mass will reach a height of 16m after0.59s on the ascent and after 5.53s on the descent.Problem 25. A shed is 4.0m long and 2.0m wide.A concrete path of constant width is laid all the wayaround the shed. If the area of the path is 9.50m2,calculate its width to the nearest centimetreFigure 14.1 shows a plan view of the shed with itssurrounding path of width t metres.t2.0m4.0m (4.0 12t)SHEDtFigure 14.1Area of path = 2(2.0 × t) + 2t(4.0 + 2t)i.e. 9.50 = 4.0t + 8.0t + 4t2or 4t2+12.0t − 9.50 = 0Hence,t =−(12.0) ± (12.0)2 − 4(4)(−9.50)2(4)=−12.0 ±√296.08=−12.0 ± 17.204658i.e. t = 0.6506m or − 3.65058m.Neglecting the negative result, which is meaningless,the width of the path, t = 0.651m or 65cm correct tothe nearest centimetre.Problem 26. If the total surface area of a solidcone is 486.2cm2 and its slant height is 15.3cm,determine its base diameter.From Chapter 27, page 245, the total surface area A ofa solid cone is given by A = πrl + πr2, where l is theslant height and r the base radius.If A = 482.2 and l = 15.3, then482.2 = πr(15.3) + πr2i.e. πr2+ 15.3πr − 482.2 = 0or r2+ 15.3r −482.2π= 0Using the quadratic formula,r =−15.3 ± (15.3)2 − 4−482.2π2=−15.3 ±√848.04612=−15.3 ± 29.121232Hence, radius r = 6.9106cm (or −22.21cm, which ismeaningless and is thus ignored).Thus, the diameter of the base = 2r = 2(6.9106)= 13.82cm.Now try the following Practice ExercisePracticeExercise 57 Practical problemsinvolving quadratic equations (answers onpage 346)1. The angle a rotating shaft turns through in tseconds is given by θ = ωt +12αt2. Deter-mine the time taken to complete 4 radians ifω is 3.0 rad/s and α is 0.60 rad/s2.2. The power P developed in an electrical cir-cuit is given by P = 10I − 8I2, where I isthe current in amperes. Determine the currentnecessary to produce a power of 2.5 watts inthe circuit.3. The area of a triangle is 47.6cm2 and itsperpendicular height is 4.3cm more than itsbase length. Determine the length of the basecorrect to 3 signiﬁcant ﬁgures.4. The sag, l, in metres in a cable stretchedbetween two supports, distance x m apart, isgiven by l =12x+ x. Determine the distancebetween the supports when the sag is 20 m.5. Theacid dissociation constant Ka ofethanoicacid is 1.8 × 10−5 moldm−3 for a particu-lar solution. Using the Ostwald dilution law,
• 110 Basic Engineering MathematicsKa =x2v(1 − x), determine x, the degree ofionization, given that v = 10dm3.6. A rectangular building is 15m long by 11mwide. A concrete path of constant width islaid all the way around the building. If thearea of the path is 60.0m2, calculate its widthcorrect to the nearest millimetre.7. The total surface area of a closed cylindricalcontainer is 20.0m3. Calculate the radius ofthe cylinder if its height is 2.80m.8. The bending moment M at a point in a beamis given by M =3x(20 − x)2, where x metresis the distance from the point of support.Determine the value of x when the bendingmoment is 50Nm.9. A tennis court measures 24m by 11m. In thelayout ofanumberofcourtsanareaofgroundmust be allowed for at the ends and at thesides of each court. If a border of constantwidth is allowed around each court and thetotal areaofthecourt and itsborderis950m2,ﬁnd the width of the borders.10. Two resistors, when connected in series, havea total resistance of 40 ohms. When con-nected in parallel their total resistance is 8.4ohms. If one of the resistors has a resistanceof Rx , ohms,(a) show that R2x − 40Rx + 336 = 0 and(b) calculate the resistance of each.14.6 Solution of linear and quadraticequations simultaneouslySometimes a linear equation and a quadratic equationneed to be solved simultaneously. An algebraic methodof solution is shown in Problem 27; a graphical solutionis shown in Chapter 19, page 160.Problem 27. Determine the values of x and ywhich simultaneously satisfy the equationsy = 5x − 4 − 2x2 and y = 6x − 7For a simultaneous solution the values of y must beequal, hence the RHS of each equation is equated.Thus, 5x − 4 − 2x2 = 6x − 7Rearranging gives 5x − 4 − 2x2− 6x + 7 = 0i.e. −x + 3 − 2x2= 0or 2x2+ x − 3 = 0Factorizing gives (2x + 3)(x − 1) = 0i.e. x = −32or x = 1In the equation y = 6x − 7,when x = −32, y = 6 −32− 7 = −16and when x = 1, y = 6 − 7 = −1(Checking the result in y = 5x − 4 − 2x2:when x = −32, y = 5 −32− 4 − 2 −322= −152− 4 −92= −16,as above,and when x = 1, y = 5 − 4 − 2 = −1,as above.)Hence, the simultaneous solutions occur whenx = −32,y = −16 and when x = 1,y = −1.Now try the following Practice ExercisePracticeExercise 58 Solving linear andquadratic equations simultaneously(answers on page 346)Determine the solutions of the following simulta-neous equations.1. y = x2 + x + 1 2. y = 15x2 + 21x − 11y = 4 − x y = 2x − 13. 2x2 + y = 4 + 5xx + y = 4
• Chapter 15Logarithms15.1 Introduction to logarithmsWith the use of calculators ﬁrmly established, logarith-mic tables are now rarely used forcalculation. However,the theory of logarithms is important, for there are sev-eral scientiﬁc and engineering laws that involve the rulesof logarithms.From Chapter 7, we know that 16 = 24.The number 4 is called the power or the exponent orthe index. In the expression 24, the number 2 is calledthe base.In another example, we know that 64 = 82.In this example, 2 is the power, or exponent, or index.The number 8 is the base.15.1.1 What is a logarithm?Consider the expression 16 = 24.An alternative, yet equivalent, way of writing thisexpression is log2 16 = 4.This is stated as ‘log to the base 2 of 16 equals 4’.We see that the logarithm is the same as the poweror index in the original expression. It is the base inthe original expression that becomes the base of thelogarithm.The two statements 16 = 24andlog216 = 4 are equivalentIf we write either of them, we are automatically imply-ing the other.In general, if a number y can be written in the form ax,then the index x is called the ‘logarithm of y to the baseof a’, i.e.if y = axthen x = loga yIn another example, if we write down that 64 = 82 thentheequivalent statement using logarithmsislog8 64 = 2.In another example, if we write down that log3 27 = 3then the equivalent statement using powers is 33 = 27.So the two sets of statements, one involving powers andone involving logarithms, are equivalent.15.1.2 Common logarithmsFrom the above, if we write down that 1000 = 103, then3 = log10 1000. This may be checked using the ‘log’button on your calculator.Logarithms having a base of 10 are called commonlogarithms and log10 is usually abbreviated to lg. Thefollowing values may be checked using a calculator.lg27.5 = 1.4393...lg378.1 = 2.5776...lg0.0204 = −1.6903...15.1.3 Napierian logarithmsLogarithms having a base of e (where e is a mathemat-ical constant approximately equal to 2.7183) are calledhyperbolic, Napierian or natural logarithms, andloge is usually abbreviated to ln. The following valuesmay be checked using a calculator.ln3.65 = 1.2947...ln417.3 = 6.0338...ln0.182 = −1.7037...Napierian logarithms are explained further in Chapter16, following.DOI: 10.1016/B978-1-85617-697-2.00015-6
• 112 Basic Engineering MathematicsHere are some worked problems to help understand-ing of logarithms.Problem 1. Evaluate log3 9Let x = log3 9 then 3x = 9 from the deﬁnition ofa logarithm,i.e. 3x = 32, from which x = 2Hence, log3 9= 2Problem 2. Evaluate log10 10Let x = log10 10 then 10x = 10 from the deﬁnitionof a logarithm,i.e. 10x = 101, from which x = 1Hence, log10 10 = 1 (which may be checkedusing a calculator).Problem 3. Evaluate log16 8Let x = log16 8 then 16x = 8 from the deﬁnitionof a logarithm,i.e. (24)x = 23 i.e. 24x = 23 from the lawsof indices,from which, 4x = 3 and x =34Hence, log16 8 =34Problem 4. Evaluate lg0.001Let x = lg0.001 = log10 0.001 then 10x = 0.001i.e. 10x = 10−3from which, x = −3Hence, lg 0.001 = −3 (which may be checkedusing a calculator)Problem 5. Evaluate lneLet x = lne = loge e then ex = ei.e. ex = e1, from whichx = 1Hence, lne = 1 (which may be checkedby a calculator)Problem 6. Evaluate log3181Let x = log3181then 3x =181=134= 3−4from which x = −4Hence, log3181= −4Problem 7. Solve the equation lg x = 3If lg x = 3 then log10 x = 3and x = 103 i.e. x = 1000Problem 8. Solve the equation log2 x = 5If log2 x = 5 then x = 25 = 32Problem 9. Solve the equation log5 x = −2If log5 x = −2 then x = 5−2 =152=125Now try the following Practice ExercisePracticeExercise 59 Laws of logarithms(answers on page 346)In problems1 to 11,evaluatethegiven expressions.1. log10 10000 2. log2 16 3. log5 1254. log2185. log8 2 6. log7 3437. lg 100 8. lg 0.01 9. log4 810. log27 3 11. ln e2In problems 12 to 18, solve the equations.12. log10 x = 4 13. lg x = 514. log3 x = 2 15. log4 x = −21216. lg x = −2 17. log8 x = −4318. ln x = 3
• Logarithms 11315.2 Laws of logarithmsThere are three laws of logarithms, which apply to anybase:(1) To multiply two numbers:log(A × B) = logA + logBThe followingmay be checked by using a calculator.lg 10 = 1Also, lg 5 + lg 2 = 0.69897...+ 0.301029... = 1Hence, lg(5 × 2) = lg 10 = lg 5 + lg 2(2) To divide two numbers:logAB= logA −logBThe following may be checked using a calculator.ln52= ln2.5 = 0.91629...Also, ln5 − ln2 = 1.60943...− 0.69314...= 0.91629...Hence, ln52= ln5 − ln2(3) To raise a number to a power:logAn= n logAThe following may be checked using a calculator.lg52 = lg25 = 1.39794...Also, 2lg5 = 2 × 0.69897... = 1.39794...Hence, lg52 = 2lg5Here are some worked problems to help understandingof the laws of logarithms.Problem 10. Write log4 + log7 as the logarithmof a single numberlog4 + log7 = log(7 × 4) by the ﬁrst law oflogarithms= log28Problem 11. Write log16 − log2 as the logarithmof a single numberlog16 − log2 = log162by the second law oflogarithms= log8Problem 12. Write 2 log 3 as the logarithm of asingle number2log 3 = log 32 by the third law of logarithms= log9Problem 13. Write12log 25 as the logarithm of asingle number12log25 = log2512 by the third law of logarithms= log√25 = log5Problem 14. Simplify log 64 − log 128 + log 3264 = 26,128 = 27 and 32 = 25Hence, log64 − log128 + log32= log26− log27+ log25= 6log2 − 7log2 + 5log2by the third law of logarithms= 4log2Problem 15. Write12log16 +13log27 − 2log5as the logarithm of a single number12log16 +13log27 − 2log5= log1612 + log2713 − log52by the third law of logarithms= log√16 + log 3√27 − log25by the laws of indices= log4 + log3 − log25
• 114 Basic Engineering Mathematics= log4 × 325by the ﬁrst and secondlaws of logarithms= log1225= log0.48Problem 16. Write (a) log 30 (b) log 450 in termsof log 2, log 3 and log 5 to any base(a) log30 = log(2 × 15) = log(2 × 3 × 5)= log2 + log3 + log5by the ﬁrst lawof logarithms(b) log450 = log(2 × 225) = log(2 × 3 × 75)= log(2 × 3 × 3 × 25)= log(2 × 32 × 52)= log2 + log32 + log52by the ﬁrst lawof logarithmsi.e. log450 = log2 + 2log3 + 2log5by the third law of logarithmsProblem 17. Write log8 ×4√581in terms oflog 2, log 3 and log 5 to any baselog8 ×4√581= log8 + log4√5 − log81by the ﬁrst and second lawsof logarithms= log23 + log514 − log34by the laws of indicesi.e. log8 ×4√581= 3log2 +14log5 − 4log3by the third law of logarithmsProblem 18. Evaluatelog25 − log125 +12log6253log5log25 − log125 +12log6253log5=log52− log53+12log543log5=2log5 − 3log5 +42log53log5=1log53log5=13Problem 19. Solve the equationlog(x − 1) + log(x + 8) = 2log(x + 2)LHS = log(x − 1) + log(x + 8) = log(x − 1)(x + 8)from the ﬁrstlaw of logarithms= log(x2 + 7x − 8)RHS = 2log(x + 2) = log(x + 2)2from the ﬁrstlaw of logarithms= log(x2 + 4x + 4)Hence, log(x2 + 7x − 8) = log(x2 + 4x + 4)from which, x2 + 7x − 8 = x2 + 4x + 4i.e. 7x − 8 = 4x + 4i.e. 3x = 12and x = 4Problem 20. Solve the equation12log4 = logx12log4 = log412 from the third law oflogarithms= log√4 from the laws of indicesHence,12log4 = logxbecomes log√4 = logxi.e. log2 = logxfrom which, 2 = xi.e. the solution of the equation is x = 2.
• Logarithms 115Problem 21. Solve the equationlog x2 − 3 − logx = log2log x2 − 3 − log x = logx2 − 3xfrom the secondlaw of logarithmsHence, logx2 − 3x= log2from which,x2 − 3x= 2Rearranging gives x2 − 3 = 2xand x2 − 2x − 3 = 0Factorizing gives (x − 3)(x + 1) = 0from which, x = 3 or x = −1x = −1 is not a valid solution since the logarithm of anegative number has no real root.Hence, the solution of the equation is x = 3.Now try the following Practice ExercisePracticeExercise 60 Laws of logarithms(answers on page 346)In problems 1 to 11, write as the logarithm of asingle number.1. log2+ log3 2. log3 + log53. log3 + log4 − log64. log7 + log21 − log495. 2log2 + log3 6. 2log2 + 3log57. 2log5 −12log81 + log368.13log8 −12log81 + log279.12log4 − 2log3 + log4510.14log16 + 2log3 − log1811. 2log2 + log5 − log10Simplify the expressions given in problems12 to 14.12. log27 − log9 + log8113. log64+ log32 − log12814. log8 − log4 + log32Evaluate the expressions given in problems 15and 16.15.12log16 −13log8log416.log9 − log3 +12log812log3Solve the equations given in problems 17 to 22.17. logx4 − logx3 = log5x − log2x18. log2t3 − logt = log16 + logt19. 2logb2 − 3logb = log8b − log4b20. log(x + 1) + log(x − 1) = log321.13log27 = log(0.5a)22. log(x2 − 5) − logx = log415.3 Indicial equationsThe laws of logarithms may be used to solvecertain equations involving powers, called indicialequations.For example, to solve, say, 3x = 27, logarithmsto a baseof 10 are taken of both sides,i.e. log10 3x= log10 27and x log10 3 = log10 27by the third law of logarithmsRearranging gives x =log10 27log10 3=1.43136...0.47712...= 3 which may be readilychecked.Note,log27log3is not equal to log273
• 116 Basic Engineering MathematicsProblem 22. Solve the equation 2x = 5, correctto 4 signiﬁcant ﬁguresTaking logarithms to base 10 of both sides of 2x= 5giveslog10 2x= log10 5i.e. x log10 2 = log10 5by the third law of logarithmsRearranging gives x =log10 5log10 2=0.6989700...0.3010299...= 2.322,correct to 4signiﬁcant ﬁgures.Problem 23. Solve the equation 2x+1 = 32x−5correct to 2 decimal placesTaking logarithms to base 10 of both sides giveslog10 2x+1= log10 32x−5i.e. (x + 1)log10 2 = (2x − 5)log10 3x log10 2 + log10 2 = 2x log10 3 − 5log10 3x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771)i.e. 0.3010x + 0.3010 = 0.9542x − 2.3855Hence, 2.3855 + 0.3010 = 0.9542x − 0.3010x2.6865 = 0.6532xfrom which x =2.68650.6532= 4.11,correct to 2 decimal places.Problem 24. Solve the equation x2.7 = 34.68,correct to 4 signiﬁcant ﬁguresTaking logarithms to base 10 of both sides giveslog10 x2.7= log10 34.682.7log10 x = log10 34.68Hence, log10 x =log10 34.682.7= 0.57040Thus, x = antilog 0.57040 = 100.57040= 3.719,correct to 4 signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 61 Indicial equations(answers on page 346)In problems 1 to 8, solve the indicial equations forx, each correct to 4 signiﬁcant ﬁgures.1. 3x = 6.4 2. 2x = 93. 2x−1 = 32x−1 4. x1.5 = 14.915. 25.28 = 4.2x 6. 42x−1 = 5x+27. x−0.25 = 0.792 8. 0.027x = 3.269. The decibel gain n of an ampliﬁer is givenby n = 10log10P2P1, where P1 is the powerinput and P2 is the power output. Find thepower gainP2P1when n = 25 decibels.15.4 Graphs of logarithmic functionsA graph of y = log10 x is shown in Figure 15.1 and agraph of y = loge x is shown in Figure 15.2. Both canbe seen to be of similar shape; in fact, the same generalshape occurs for a logarithm to any base.In general, with a logarithm to any base, a, it is notedthat(a) loga 1 = 0Let loga = x then ax = 1 from the deﬁnition of thelogarithm.If ax = 1 then x = 0 from the laws of logarithms.Hence, loga 1 = 0. In the above graphs it is seenthat log101 = 0 and loge 1 = 0.(b) loga a = 1Let loga a = x then ax = a from the deﬁnition ofa logarithm.If ax = a then x = 1.
• Logarithms 117y0.50 1 2 320.521.0xx 30.4820.30100.520.300.220.700.121.0y5 log10xFigure 15.1Hence, loga a = 1. (Check with a calculator thatlog10 10 = 1 and loge e = 1.)(c) loga 0 → −∞Let loga 0 = x then ax = 0 from the deﬁnition of alogarithm.y210 1 2 3 4 5 6 xx 6 5 4 3 2 1 0.5 0.2 0.11.79 1.61 1.39 1.10 0.69 0 20.69 21.61 22.302122y5 logexFigure 15.2If ax = 0, and a is a positive real number, thenx must approach minus inﬁnity. (For example,check with a calculator, 2−2 = 0.25,2−20 =9.54 × 10−7,2−200 = 6.22 × 10−61, and so on.)Hence, loga 0 → −∞
• Chapter 16Exponential functions16.1 Introduction to exponentialfunctionsAn exponential function is one which contains ex , ebeing a constant called the exponent and having anapproximate value of 2.7183. The exponent arises fromthe natural laws of growth and decay and is used as abase for natural or Napierian logarithms.The most common method of evaluating an exponentialfunction isby using a scientiﬁc notationcalculator. Useyour calculator to check the following values.e1= 2.7182818, correct to 8 signiﬁcant ﬁgures,e−1.618= 0.1982949, correct to 7 signiﬁcant ﬁgures,e0.12= 1.1275, correct to 5 signiﬁcant ﬁgures,e−1.47= 0.22993, correct to 5 decimal places,e−0.431= 0.6499, correct to 4 decimal places,e9.32= 11159, correct to 5 signiﬁcant ﬁgures,e−2.785= 0.0617291, correct to 7 decimal places.Problem 1. Evaluate the following correct to 4decimal places, using a calculator:0.0256(e5.21− e2.49)0.0256(e5.21− e2.49)= 0.0256(183.094058...− 12.0612761...)= 4.3784, correct to 4 decimal places.Problem 2. Evaluate the following correct to 4decimal places, using a calculator:5e0.25 − e−0.25e0.25 + e−0.255e0.25 − e−0.25e0.25 + e−0.25= 51.28402541...− 0.77880078...1.28402541...+ 0.77880078...= 50.5052246...2.0628262...= 1.2246, correct to 4 decimal places.Problem 3. The instantaneous voltage v in acapacitive circuit is related to time t by the equationv = V e−t/CRwhere V , C and R are constants.Determine v, correct to 4 signiﬁcant ﬁgures,when t = 50ms, C = 10μF, R = 47k andV = 300voltsv = V e−t/CR= 300e(−50×10−3)/(10×10−6×47×103)Using a calculator, v = 300e−0.1063829...= 300(0.89908025...)= 269.7volts.Now try the following Practice ExercisePracticeExercise 62 Evaluatingexponential functions (answers on page 347)1. Evaluate the following,correct to 4 signiﬁcantﬁgures.(a) e−1.8 (b) e−0.78 (c) e102. Evaluate the following,correct to 5 signiﬁcantﬁgures.(a) e1.629 (b) e−2.7483 (c) 0.62e4.178DOI: 10.1016/B978-1-85617-697-2.00016-8
• Exponential functions 119In problems 3 and 4, evaluate correct to 5 decimalplaces.3. (a)17e3.4629 (b) 8.52e−1.2651(c)5e2.69213e1.11714. (a)5.6823e−2.1347(b)e2.1127 − e−2.11272(c)4 e−1.7295 − 1e3.68175. The length of a bar, l, at a temperature, θ,is given by l = l0eαθ , where l0 and α areconstants. Evaluate l, correct to 4 signiﬁcantﬁgures, where l0 = 2.587, θ = 321.7 andα = 1.771 × 10−4.6. When a chain of length 2L is suspended fromtwo points, 2D metres apart on the same hor-izontal level, D = k lnL +√L2 + k2k.Evaluate D when k = 75m and L = 180m.16.2 The power series for exThe value of ex can be calculated to any required degreeof accuracy since it is deﬁned in terms of the followingpower series:ex= 1 + x+x 22!+x 33!+x 44!+ ··· (1)(where 3!= 3 × 2 × 1 and is called ‘factorial 3’).The series is valid for all values of x.The series is said to converge; i.e., if all the terms areadded, an actual value for ex (where x is a real number)is obtained. The more terms that are taken, the closerwill be the value of ex to its actual value. The value ofthe exponent e, correct to say 4 decimal places, may bedetermined by substituting x = 1 in the power series ofequation (1). Thus,e1= 1 + 1 +(1)22!+(1)33!+(1)44!+(1)55!+(1)66!+(1)77!+(1)88!+ ···= 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833+ 0.00139 + 0.00020+ 0.00002 + ···= 2.71828i.e. e = 2.7183, correct to 4 decimal places.The value of e0.05, correct to say 8 signiﬁcant ﬁgures, isfound by substituting x = 0.05 in the power series forex . Thus,e0.05= 1 + 0.05 +(0.05)22!+(0.05)33!+(0.05)44!+(0.05)55!+ ···= 1 + 0.05 + 0.00125 + 0.000020833+ 0.000000260+ 0.0000000026i.e. e 0.05= 1.0512711, correct to 8 signiﬁcant ﬁgures.In this example, successive terms in the series growsmaller very rapidly and it is relatively easy to deter-mine the value of e0.05 to a high degree of accuracy.However, when x is nearer to unity or larger than unity,a very large number of terms are required for an accurateresult.If, in the series of equation (1), x is replaced by −x,thene−x= 1 + (−x) +(−x)22!+(−x)33!+ ···i.e. e−x= 1 − x +x 22!−x 33!+ ···In a similar manner the power series for ex maybe used to evaluate any exponential function of the formaekx , where a and k are constants.In the series of equation (1), let x be replaced by kx.Thenaekx= a 1 + (kx) +(kx)22!+(kx)33!+ ···Thus, 5e2x= 5 1 + (2x) +(2x)22!+(2x)33!+ ···= 5 1 + 2x +4x22+8x36+ ···i.e. 5e2x= 5 1 + 2x + 2x2+43x3+ ···Problem 4. Determine the value of 5e0.5, correctto 5 signiﬁcant ﬁgures, by using the power seriesfor ex
• 120 Basic Engineering MathematicsFrom equation (1),ex= 1 + x +x22!+x33!+x44!+ ···Hence, e0.5= 1 + 0.5 +(0.5)2(2)(1)+(0.5)3(3)(2)(1)+(0.5)4(4)(3)(2)(1)+(0.5)5(5)(4)(3)(2)(1)+(0.5)6(6)(5)(4)(3)(2)(1)= 1 + 0.5 + 0.125 + 0.020833+ 0.0026042+ 0.0002604+ 0.0000217i.e. e0.5= 1.64872, correct to 6 signiﬁcantﬁguresHence, 5e0.5= 5(1.64872) = 8.2436, correct to 5signiﬁcant ﬁgures.Problem 5. Determine the value of 3e−1, correctto 4 decimal places, using the power series for exSubstituting x = −1 in the power seriesex= 1 + x +x22!+x33!+x44!+ ···gives e−1= 1 + (−1) +(−1)22!+(−1)33!+(−1)44!+ ···= 1 − 1 + 0.5 − 0.166667 + 0.041667− 0.008333 + 0.001389− 0.000198 + ···= 0.367858 correct to 6 decimal placesHence, 3e−1 = (3)(0.367858) = 1.1036, correct to 4decimal places.Problem 6. Expand ex (x2 − 1) as far as the termin x5The power series for ex isex= 1 + x +x22!+x33!+x44!+x55!+ ···Hence,ex(x2− 1)= 1 + x +x22!+x33!+x44!+x55!+ ··· (x2− 1)= x2+ x3+x42!+x53!+ ···− 1 + x +x22!+x33!+x44!+x55!+ ···Grouping like terms givesex(x2− 1)= −1 − x + x2−x22!+ x3−x33!+x42!−x44!+x53!−x55!+ ···= −1 − x +12x2+56x3+1124x4+19120x5when expanded as far as the term in x5.Now try the following Practice ExercisePracticeExercise 63 Power series for ex(answers on page 347)1. Evaluate 5.6e−1, correct to 4 decimal places,using the power series for ex .2. Use the power series for ex to determine, cor-rect to 4 signiﬁcant ﬁgures, (a) e2 (b) e−0.3and check your results using a calculator.3. Expand (1 − 2x)e2x as far as the term in x4.4. Expand (2ex2)(x1/2) to six terms.16.3 Graphs of exponential functionsValues of ex and e−x obtained from a calculator, correctto 2 decimal places, over a range x = −3 to x = 3, areshown in Table 16.1.
• Exponential functions 121Table 16.1x −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0 0.5 1.0 1.5 2.0 2.5 3.0ex0.05 0.08 0.14 0.22 0.37 0.61 1.00 1.65 2.72 4.48 7.39 12.18 20.09e−x 20.09 12.18 7.39 4.48 2.72 1.65 1.00 0.61 0.37 0.22 0.14 0.08 0.05Figure 16.1 shows graphs of y = ex and y = e−x .y2016y ϭ e xy ϭ eϪx1284Ϫ1 0 1 2 3 xϪ2Ϫ3Figure 16.1Problem 7. Plot a graph of y = 2e0.3x over arange of x = −2 to x = 3. Then determine the valueof y when x = 2.2 and the value of x when y = 1.6A table of values is drawn up as shown below.x −3 −2 −1 0 1 2 32e0.3 x 0.81 1.10 1.48 2.00 2.70 3.64 4.92A graph of y = 2e0.3x is shown plotted in Figure 16.2.From the graph, when x = 2.2, y = 3.87 and wheny = 1.6, x = −0.74Problem 8. Plot a graph of y =13e−2x over therange x = −1.5 to x = 1.5. Determine from thegraph the value of y when x = −1.2 and the valueof x when y = 1.4A table of values is drawn up as shown below.x −1.5 −1.0 −0.5 0 0.5 1.0 1.513e−2x 6.70 2.46 0.91 0.33 0.12 0.05 0.02y5y ϭ 2e 0.3x4321.63.871Ϫ0.74 2.2Ϫ1 0 1 2 3 xϪ2Ϫ3Figure 16.2A graph of13e−2x is shown in Figure 16.3.From the graph, when x = −1.2, y = 3.67 and wheny = 1.4, x = −0.72y65423170.5 x1.0 1.5Ϫ0.5Ϫ1.2 Ϫ0.72Ϫ1.0Ϫ1.53.671.4y ϭ 13e Ϫ2xFigure 16.3Problem 9. The decay of voltage, v volts, acrossa capacitor at time t seconds is given byv = 250e−t/3. Draw a graph showing the naturaldecay curve over the ﬁrst 6seconds. From thegraph, ﬁnd (a) the voltage after 3.4s and (b) thetime when the voltage is 150V
• 122 Basic Engineering MathematicsA table of values is drawn up as shown below.t 0 1 2 3e−t/31.00 0.7165 0.5134 0.3679v = 250e−t/3 250.0 179.1 128.4 91.97t 4 5 6e−t/3 0.2636 0.1889 0.1353v = 250e−t/3 65.90 47.22 33.83The natural decay curve of v = 250e−t/3 is shown inFigure 16.4.250200150Voltagev(volts)1008050Time t (seconds)v 5 250e 2t /30 1 1.5 2 3 3.4 4 5 6Figure 16.4From the graph,(a) when time t = 3.4s, voltage v = 80V, and(b) when voltage v = 150V, time t = 1.5s.Now try the following Practice ExercisePracticeExercise 64 Exponential graphs(answers on page 347)1. Plot a graph of y = 3e0.2x over the rangex = −3 to x = 3. Hence determine the valueof y when x = 1.4 and the value of x wheny = 4.52. Plot a graph of y =12e−1.5x over a rangex = −1.5 to x = 1.5 and then determine thevalue of y when x = −0.8 and the value of xwhen y = 3.53. In a chemical reaction the amount of startingmaterial C cm3 left after t minutes is givenby C = 40e−0.006t. Plot a graph of C againstt and determine(a) the concentration C after 1hour.(b) the time taken for the concentration todecrease by half.4. The rate at which a body cools is given byθ = 250e−0.05t where the excess of tempera-ture of a body above its surroundings at time tminutes is θ ◦C. Plot a graph showing the nat-ural decay curve for the ﬁrst hour of cooling.Then determine(a) the temperature after 25 minutes.(b) the time when the temperature is 195◦C.16.4 Napierian logarithmsLogarithms having a base of e are called hyperbolic,Napierian or natural logarithms and the Napierianlogarithm of x is written as loge x or, more commonly,as ln x. Logarithms were invented by John Napier, aScotsman (1550–1617).The most common method of evaluating a Napierianlogarithmisby ascientiﬁcnotationcalculator.Useyourcalculator to check the following values:ln 4.328 = 1.46510554... = 1.4651, correct to 4decimal placesln 1.812 = 0.59443, correct to 5 signiﬁcant ﬁguresln 1 = 0ln 527 = 6.2672, correct to 5 signiﬁcant ﬁguresln 0.17 = −1.772, correct to 4 signiﬁcant ﬁguresln 0.00042 = −7.77526, correct to 6 signiﬁcantﬁguresln e3= 3ln e1 = 1From the last two examples we can conclude thatloge ex= xThis is useful when solving equations involving expo-nential functions. For example, to solve e3x = 7, takeNapierian logarithms of both sides, which gives
• Exponential functions 123ln e3x= ln7i.e. 3x = ln7from which x =13ln7 = 0.6486,correct to 4 decimal places.Problem 10. Evaluate the following, each correctto 5 signiﬁcant ﬁgures: (a)12ln4.7291(b)ln7.86937.8693(c)3.17ln24.07e−0.1762(a)12ln4.7291 =12(1.5537349...) = 0.77687,correct to 5 signiﬁcant ﬁgures.(b)ln7.86937.8693=2.06296911...7.8693= 0.26215, correctto 5 signiﬁcant ﬁgures.(c)3.17ln24.07e−0.1762=3.17(3.18096625...)0.83845027...= 12.027,correct to 5 signiﬁcant ﬁgures.Problem 11. Evaluate the following: (a)lne2.5lg100.5(b)5e2.23 lg2.23ln2.23(correct to 3 decimal places)(a)lne2.5lg100.5=2.50.5= 5(b)5e2.23 lg2.23ln2.23=5(9.29986607...)(0.34830486...)(0.80200158...)= 20.194, correct to 3 decimal places.Problem 12. Solve the equation 9 = 4e−3x to ﬁndx, correct to 4 signiﬁcant ﬁguresRearranging 9 = 4e−3x gives94= e−3xTaking the reciprocal of both sides gives49=1e−3x= e3xTaking Napierian logarithms of both sides givesln49= ln(e3x)Since loge eα = α, then ln49= 3xHence, x =13ln49=13(−0.81093) = −0.2703,correct to 4 signiﬁcant ﬁgures.Problem 13. Given 32 = 70 1 − e−t2 ,determine the value of t, correct to 3 signiﬁcantﬁguresRearranging 32 = 70 1 − e−t2 gives3270= 1 − e−t2ande−t2 = 1 −3270=3870Taking the reciprocal of both sides giveset2 =7038Taking Napierian logarithms of both sides giveslnet2 = ln7038i.e.t2= ln7038from which, t = 2ln7038= 1.22, correct to 3 signiﬁ-cant ﬁgures.Problem 14. Solve the equation2.68 = ln4.87xto ﬁnd xFrom the deﬁnition of a logarithm, since2.68 = ln4.87xthen e2.68 =4.87xRearranging gives x =4.87e2.68= 4.87e−2.68i.e. x = 0.3339,correct to 4 signiﬁcant ﬁgures.Problem 15. Solve74= e3x correct to 4signiﬁcant ﬁgures
• 124 Basic Engineering MathematicsTaking natural logs of both sides givesln74= lne3xln74= 3x lneSince lne = 1, ln74= 3xi.e. 0.55962 = 3xi.e. x = 0.1865,correct to 4 signiﬁcant ﬁgures.Problem 16. Solve ex−1 = 2e3x−4 correct to 4signiﬁcant ﬁguresTaking natural logarithms of both sides givesln ex−1= ln 2e3x−4and by the ﬁrst law of logarithms,ln ex−1= ln2 + ln e3x−4i.e. x − 1 = ln2 + 3x − 4Rearranging gives4 − 1 − ln2 = 3x − xi.e. 3 − ln2 = 2xfrom which, x =3 − ln22= 1.153Problem 17. Solve, correct to 4 signiﬁcantﬁgures, ln(x − 2)2 = ln(x − 2) − ln(x + 3) + 1.6Rearranging givesln(x − 2)2− ln(x − 2) + ln(x + 3) = 1.6and by the laws of logarithms,ln(x − 2)2(x + 3)(x − 2)= 1.6Cancelling givesln{(x − 2)(x + 3)} = 1.6and (x − 2)(x + 3) = e1.6i.e. x2+ x − 6 = e1.6or x2+ x − 6 − e1.6= 0i.e. x2+ x − 10.953 = 0Using the quadratic formula,x =−1 ± 12 − 4(1)(−10.953)2=−1 ±√44.8122=−1 ± 6.69422i.e. x = 2.847 or − 3.8471x = −3.8471 is not valid since the logarithm of anegative number has no real root.Hence, the solution of the equation is x = 2.847Now try the following Practice ExercisePracticeExercise 65 Evaluating Napierianlogarithms (answers on page 347)In problems 1 and 2, evaluate correct to 5 signiﬁ-cant ﬁgures.1. (a)13ln5.2932 (b)ln82.4734.829(c)5.62ln321.62e1.29422. (a)1.786lne1.76lg101.41(b)5e−0.16292ln0.00165(c)ln4.8629 − ln2.47115.173In problems 3 to 16, solve the given equations, eachcorrect to 4 signiﬁcant ﬁgures.3. 1.5 = 4e2t4. 7.83 = 2.91e−1.7x5. 16 = 24 1 − e−t26. 5.17 = lnx4.647. 3.72ln1.59x= 2.438. lnx = 2.409. 24 + e2x = 4510. 5 = ex+1 − 7
• Exponential functions 12511. 5 = 8 1 − e−x212. ln(x + 3) − ln x = ln(x − 1)13. ln(x − 1)2− ln3 = ln(x − 1)14. ln(x + 3) + 2 = 12 − ln(x − 2)15. e(x+1) = 3e(2x−5)16. ln(x + 1)2 = 1.5 − ln(x − 2) + ln(x + 1)17. Transpose b= lnt − a ln D to make t thesubject.18. IfPQ= 10log10R1R2, ﬁnd the value of R1when P = 160, Q = 8 and R2 = 5.19. If U2 = U1eWPV , make W the subject of theformula.16.5 Laws of growth and decayLaws of exponential growth and decay are of the formy = Ae−kx and y = A(1 − e−kx ), where A and k areconstants. When plotted, the form of these equations isas shown in Figure 16.5.yA0y 5 Ae 2kxy 5 A(12e 2kx)xyA0 xFigure 16.5The laws occur frequently in engineering and scienceand examples of quantities related by a natural lawinclude:(a) Linear expansion l = l0eαθ(b) Change in electrical resistance with temperatureRθ = R0eαθ(c) Tension in belts T1 = T0eμθ(d) Newton’s law of cooling θ = θ0e−kt(e) Biological growth y = y0ekt(f) Discharge of a capacitor q = Qe−t/CR(g) Atmospheric pressure p = p0e−h/c(h) Radioactive decay N = N0e−λt(i) Decay of current in an inductive circuiti = Ie−Rt/L(j) Growth of current in a capacitive circuiti = I(1 − e−t/CR)Here are some worked problems to demonstrate the lawsof growth and decay.Problem 18. The resistance R of an electricalconductor at temperature θ◦C is given byR = R0eαθ , where α is a constant and R0 = 5k .Determine the value of α correct to 4 signiﬁcantﬁgures when R = 6k and θ = 1500◦C. Also, ﬁndthe temperature, correct to the nearest degree, whenthe resistance R is 5.4kTransposing R = R0eαθ givesRR0= eαθTaking Napierian logarithms of both sides giveslnRR0= lneαθ= αθHence, α =1θlnRR0=11500ln6 × 1035 × 103=11500(0.1823215...)= 1.215477...× 10−4Hence, α = 1.215 × 10−4correct to 4 signiﬁcantﬁgures.From above, lnRR0= αθ hence θ =1αlnRR0
• 126 Basic Engineering MathematicsWhen R = 5.4 × 103, α = 1.215477...× 10−4andR0 = 5 × 103θ =11.215477...× 10−4ln5.4 × 1035 × 103=1041.215477...(7.696104... × 10−2)= 633◦C correct to the nearest degree.Problem 19. In an experiment involvingNewton’s law of cooling, the temperature θ(◦C) isgiven by θ = θ0e−kt . Find the value of constant kwhen θ0 = 56.6◦C, θ = 16.5◦C andt = 79.0secondsTransposing θ = θ0e−kt givesθθ0= e−kt , from whichθ0θ=1e−kt= ektTaking Napierian logarithms of both sides giveslnθ0θ= ktfrom which,k =1tlnθ0θ=179.0ln56.616.5=179.0(1.2326486...)Hence, k = 0.01560 or 15.60×10−3.Problem 20. The current i amperes ﬂowing in acapacitor at time t seconds is given byi = 8.0(1 − e−tCR ), where the circuit resistance R is25k and capacitance C is 16μF. Determine(a) the current i after 0.5seconds and (b) the time,to the nearest millisecond, for the current to reach6.0A. Sketch the graph of current against time(a) Current i = 8.0 1−e−tCR= 8.0[1 − e−0.5/(16×10−6)(25×103)]= 8.0(1 − e−1.25)= 8.0(1 − 0.2865047...)= 8.0(0.7134952...)= 5.71amperes(b) Transposing i = 8.0 1 − e−tCR givesi8.0= 1 − e−tCRfrom which, e−tCR = 1 −i8.0=8.0 − i8.0Taking the reciprocal of both sides givesetCR =8.08.0 − iTaking Napierian logarithms of both sides givestCR= ln8.08.0 − iHence,t = CRln8.08.0 − iWhen i = 6.0A,t = (16 × 10−6)(25 × 103)ln8.08.0 − 6.0i.e. t =400103ln8.02.0= 0.4ln4.0= 0.4(1.3862943...)= 0.5545s= 555ms correct to the nearest ms.A graph of current against time is shown inFigure 16.6.865.710.55542i 58.0(12e 2t/CR)t(s)i (A)0 0.5 1.0 1.5Figure 16.6Problem 21. The temperature θ2 of a windingwhich is being heated electrically at time t is givenby θ2 = θ1(1 − e−tτ ), where θ1 is the temperature(in degrees Celsius) at time t = 0 and τ is aconstant. Calculate
• Exponential functions 127(a) θ1, correct to the nearest degree, when θ2 is50◦C, t is 30s and τ is 60s and(b) the time t, correct to 1 decimal place, for θ2 tobe half the value of θ1(a) Transposing the formula to make θ1 the subjectgivesθ1 =θ21 − e−t/τ=501 − e−30/60=501 − e−0.5=500.393469...i.e. θ1 = 127◦C correct to the nearest degree.(b) Transposing to make t the subject of the formulagivesθ2θ1= 1 − e−tτfrom which, e−tτ = 1 −θ2θ1Hence, −tτ= ln 1 −θ2θ1i.e. t = −τ ln 1 −θ2θ1Since θ2 =12θ1t = −60ln 1 −12= −60ln0.5= 41.59sHence, the time for the temperature θ2 to be one halfof the value of θ1 is 41.6 s, correct to 1 decimal place.Now try the following Practice ExercisePracticeExercise 66 Laws of growth anddecay (answers on page 347)1. The temperature, T ◦C, of a cooling objectvaries with time, t minutes, accordingto the equation T = 150e−0.04t. Deter-mine the temperature when (a) t = 0,(b) t = 10 minutes.2. The pressure p pascals at height h metresabove ground level is given by p = p0e−h/C,where p0 is the pressure at ground leveland C is a constant. Find pressure p whenp0 = 1.012 × 105 Pa, height h = 1420m andC = 71500.3. The voltage drop, v volts, across an inductorL henrys at time t seconds is given byv = 200e−RtL , where R = 150 andL = 12.5 × 10−3 H. Determine (a) thevoltage when t = 160 × 10−6 s and (b) thetime for the voltage to reach 85V.4. The length l metres of a metal bar at tem-perature t ◦C is given by l = l0eαt , where l0and α are constants. Determine (a) the valueof l when l0 = 1.894, α = 2.038 × 10−4 andt = 250◦C and (b) the value of l0 whenl = 2.416,t =310◦C and α =1.682×10−4.5. The temperature θ2◦C of an electrical con-ductor at time t seconds is given byθ2 = θ1(1 − e−t/T ), where θ1 is the ini-tial temperature and T seconds is a con-stant. Determine (a) θ2 when θ1 = 159.9◦C,t = 30s and T = 80s and (b) the time t forθ2 to fall to half the value of θ1 if T remainsat 80s.6. Abelt isin contact with apulley forasectorofθ = 1.12 radians and the coefﬁcient of fric-tion between these two surfaces is μ = 0.26.Determine the tension on the taut side of thebelt, T newtons, when tension on the slackside is given by T0 = 22.7newtons, giventhat these quantities are related by the lawT = T0eμθ .7. The instantaneous current i at time t isgiven by i = 10e−t/CRwhen a capacitoris being charged. The capacitance C is7×10−6 farads and the resistance R is0.3×106 ohms. Determine (a) the instanta-neous current when t is 2.5seconds and (b)thetimefortheinstantaneouscurrent to fall to5amperes. Sketch a curve of current againsttime from t = 0 to t = 6seconds.8. The amount of product x (in mol/cm3)found in a chemical reaction startingwith 2.5mol/cm3of reactant is given byx = 2.5(1 − e−4t ) where t is the time, inminutes, to form product x. Plot a graphat 30second intervals up to 2.5minutes anddetermine x after 1minute.
• 128 Basic Engineering Mathematics9. The current i ﬂowing in a capacitor attime t is given by i = 12.5(1 − e−t/CR),where resistance R is 30k and thecapacitance C is 20μF. Determine (a)the current ﬂowing after 0.5seconds and(b) the time for the current to reach10amperes.10. The amount A after n years of a sum investedP is given by the compound interest lawA = Pern/100, when the per unit interest rater is added continuously. Determine, correctto the nearest pound, the amount after 8 yearsfor a sum of £1500 invested if the interestrate is 6% per annum.
• Revision Test 6 : Quadratics, logarithms and exponentialsThis assignment covers the material contained in Chapters 14–16. The marks available are shown in brackets at theend of each question.1. Solve the following equations by factorization.(a) x2 − 9 = 0 (b) x2 + 12x + 36 = 0(c) x2 + 3x − 4 = 0 (d) 3z2 − z − 4 = 0(9)2. Solvethefollowing equations,correct to 3 decimalplaces.(a) 5x2 +7x −3=0 (b) 3a2 +4a −5=0(8)3. Solvetheequation 3x2− x − 4 = 0 by completingthe square. (6)4. Determine the quadratic equation in x whose rootsare 1 and −3. (3)5. The bending moment M at a point in a beamis given by M =3x(20 − x)2where x metres isthe distance from the point of support. Deter-mine the value of x when the bending moment is50Nm. (5)6. The current i ﬂowing through an electronic deviceis given by i = (0.005v2 + 0.014v) ampereswhere v is the voltage. Calculate the values of vwhen i = 3 × 10−3. (6)7. Evaluate the following, correct to 4 signiﬁcantﬁgures.(a) 3.2ln4.92−5lg17.9 (b)5(1−e−2.65)e1.73(4)8. Solve the following equations.(a) lgx = 4 (b) lnx = 2(c) log2 x = 5 (d) 5x = 2(e) 32t−1 = 7t+2 (f) 3e2x = 4.2 (18)9. Evaluate log1618(4)10. Write the following as the logarithm of a singlenumber.(a) 3log2 + 2log5 −12log16(b) 3log3 +14log16 −13log27 (8)11. Solve the equationlog(x2 + 8) − log(2x) = log3. (5)12. Evaluate the following, each correct to 3 decimalplaces.(a) ln462.9(b) ln0.0753(c)ln3.68 − ln2.914.63(3)13. Expand xe3x to six terms. (5)14. Evaluate v given that v = E 1 − e−tCR voltswhen E = 100V,C = 15μF, R = 50k andt = 1.5s. Also, determine the time when thevoltage is 60V. (8)15. Plot a graph of y =12e−1.2x over the rangex = −2 to x = +1 and hence determine, correct to1 decimal place,(a) the value of y when x = −0.75, and(b) the value of x when y = 4.0 (8)
• Chapter 17Straight line graphs17.1 Introduction to graphsA graph is a visual representation of information,showing how one quantity varies with another relatedquantity.We often see graphs in newspapers or in businessreports, in travel brochures and government publica-tions. For example, a graph of the share price (in pence)over a six month period for a drinks company, FizzyPops, is shown in Figure 17.1.Generally, we see that the share price increases to a highof400p in June,but dipsdown to around 280p in Augustbefore recovering slightly in September.A graph should convey information more quickly to thereader than if the same information was explained inwords.When this chapter is completed you should be able todraw up a table of values, plot co-ordinates, determinethe gradient and state the equation of a straight linegraph. Some typical practical examples are included inwhich straight lines are used.400Fizzy Pops350300250Apr 07 May 07 Jun 07 Jul 07 Aug 07 Sep 07Figure 17.117.2 Axes, scales and co-ordinatesWe are probably all familiar with reading a map to locatea town, or a local map to locate a particular street. Forexample, a street map of central Portsmouth is shown inFigure 17.2. Notice the squares drawn horizontally andvertically on the map; this is called a grid and enablesus to locate a place of interest or a particular road. Mostmaps contain such a grid.We locate places ofinterest on the map by stating a letterand a number – this is called the grid reference.For example, on the map, the Portsmouth & Southseastation is in square D2, King’s Theatre is in square E5,HMS Warrior is in square A2, Gunwharf Quays is insquare B3 and High Street is in square B4.Portsmouth & Southsea station is located by movinghorizontally along the bottom of the map until thesquare labelled D is reached and then moving verticallyupwards until square 2 is met.The letter/number, D2, is referred to as co-ordinates;i.e., co-ordinates are used to locate the position ofDOI: 10.1016/B978-1-85617-697-2.00017-X
• Straight line graphs 13112345A B C D E FFigure 17.2 Reprinted with permission from AA Media Ltd.a point on a map. If you are familiar with using amap in this way then you should have no difﬁcultieswith graphs, because similar co-ordinates are used withgraphs.As stated earlier, a graph is a visual representationof information, showing how one quantity varies withanother related quantity. The most common method ofshowing a relationship between two sets of data is touse a pair of reference axes – these are two lines drawnat right angles to each other (often called Cartesian orrectangular axes), as shown in Figure 17.3.The horizontal axis is labelled the x-axis and the ver-tical axis is labelled the y-axis. The point where x is 0and y is 0 is called the origin.x values have scales that are positive to the right ofthe origin and negative to the left. y values have scalesthat are positive up from the origin and negative downfrom the origin.Co-ordinates are written with brackets and a commain between two numbers. For example, point A is shownwith co-ordinates (3, 2) and is located by starting at theA(3, 2)424 23 22 21 032121222324OriginC(23, 22)B (24, 3)yx1 3 42Figure 17.3origin and moving 3 units in the positive x direction(i.e. to the right) and then 2 units in the positive ydirection (i.e. up).When co-ordinates are stated the ﬁrst number isalways the x value and the second number is always
• 132 Basic Engineering Mathematicsthe y value. In Figure 17.3, point B has co-ordinates(−4,3) and point C has co-ordinates (−3,−2).17.3 Straight line graphsThe distances travelled by a car in certain peri-ods of time are shown in the following table ofvalues.Time(s) 10 20 30 40 50 60Distance travelled (m) 50 100 150 200 250 300We will plot time on the horizontal (or x) axis with ascale of 1cm = 10s.We will plot distance on the vertical (or y) axis with ascale of 1cm = 50m.(When choosing scales it is better to choose ones such as1cm = 1unit,1cm = 2unitsor 1cm = 10unitsbecausedoing so makes reading values between these valueseasier.)With the above data, the (x, y) co-ordinates become(time, distance) co-ordinates; i.e., the co-ordinates are(10, 50), (20, 100), (30, 150), and so on.The co-ordinates are shown plotted in Figure 17.4using crosses. (Alternatively, a dot or a dot and circlemay be used, as shown in Figure 17.3.)A straight line is drawn through the plotted co-ordinates as shown in Figure 17.4.300DistanceTravelled(m)Time (s)2502001501005010 20 30 40 50 60Distance/time graphFigure 17.4Student taskThe following table gives the force F newtons which,when applied to a lifting machine, overcomes acorresponding load of L newtons.F (Newtons) 19 35 50 93 125 147L (Newtons) 40 120 230 410 540 6801. Plot L horizontally and F vertically.2. Scales are normally chosen such that the graphoccupies as much space as possible on thegraph paper. So in this case, the followingscales are chosen.Horizontal axis (i.e. L): 1cm = 50NVertical axis (i.e. F): 1cm = 10N3. Draw the axes and label them L (newtons) forthe horizontal axis and F (newtons) for thevertical axis.4. Label the origin as 0.5. Writeonthehorizontal scalingat 100,200,300,and so on, every 2cm.6. Write on the vertical scaling at 10, 20, 30, andso on, every 1cm.7. Plot on the graph the co-ordinates (40, 19),(120, 35), (230, 50), (410, 93), (540, 125) and(680, 147), marking each with a cross or a dot.8. Using aruler,drawthebest straight linethroughthe points. You will notice that not all of thepointslieexactly on a straight line.This is quitenormal with experimental values. In a practi-cal situation it would be surprising if all of thepoints lay exactly on a straight line.9. Extend the straight line at each end.10. From the graph, determine the force appliedwhen the load is 325N. It should be closeto 75N. This process of ﬁnding an equivalentvalue within the given data is called interpola-tion. Similarly, determine the load that a forceof 45N will overcome. It should be close to170N.11. From the graph, determine the force needed toovercome a 750N load. It should be close to161N. This process of ﬁnding an equivalent
• Straight line graphs 133value outside the given data is called extrapo-lation.To extrapolateweneed to haveextendedthestraight linedrawn.Similarly,determinetheforce applied when theload iszero. It shouldbeclose to 11N. The point where the straight linecrosses the vertical axis is called the vertical-axis intercept. So, in this case, the vertical-axisintercept = 11N at co-ordinates (0, 11).The graph you have drawn should look somethinglike Figure 17.5 shown below.101000 200 300 400 500 600 700 800L (Newtons)F(Newtons)2030405060708090100110120130140150160Graph of F against LFigure 17.5In another example, let the relationship between twovariables x and y be y = 3x + 2.When x = 0, y = 0 + 2 = 2When x = 1, y = 3 + 2 = 5When x = 2, y = 6 + 2 = 8, and so on.The co-ordinates (0, 2), (1, 5) and (2, 8) have beenproduced and are plotted, with others, as shown inFigure 17.6.When the points are joined together a straight linegraph results, i.e. y = 3x + 2 is a straight line graph.17.3.1 Summary of general rules to beapplied when drawing graphs(a) Give the graph a title clearly explaining what isbeing illustrated.(b) Choose scales such that the graph occupies asmuch space as possible on the graph paperbeing used.1 2yϭ3xϩ 2yx02468Ϫ1Figure 17.6 Graph of y/x(c) Choose scales so that interpolation is made aseasy as possible. Usually scales such as 1cm =1unit,1cm = 2 units or 1cm = 10 units are used.Awkward scalessuch as1cm = 3 unitsor1cm = 7units should not be used.(d) The scales need not start at zero, particularly whenstarting at zero produces an accumulation ofpointswithin a small area of the graph paper.(e) The co-ordinates, or points, should be clearlymarked. This is achieved by a cross, or a dot andcircle, or just by a dot (see Figure 17.3).(f) A statement should be made next to each axisexplaining the numbers represented with theirappropriate units.(g) Sufﬁcient numbers should be written next to eachaxis without cramping.Problem 1. Plot the graph y = 4x + 3 in therange x = −3 to x = +4. From the graph, ﬁnd(a) the value of y when x = 2.2 and (b) the valueof x when y = −3Whenever an equation is given and a graph is required,a table giving corresponding values of the variable isnecessary. The table is achieved as follows:When x = −3, y = 4x + 3 = 4(−3) + 3= −12 + 3 = −9When x = −2, y = 4(−2) + 3= −8 + 3 = −5,and so on.Such a table is shown below.x −3 −2 −1 0 1 2 3 4y −9 −5 −1 3 7 11 15 19The co-ordinates (−3,−9), (−2,−5), (−1,−1), andso on, are plotted and joined together to produce the
• 134 Basic Engineering Mathematics20151011.85Ϫ5Ϫ3100Ϫ2 Ϫ1Ϫ1.5Ϫ3 12.22 43 xyFigure 17.7straight line shown in Figure 17.7. (Note that the scalesused on the x and y axes do not have to be the same.)From the graph:(a) when x = 2.2, y = 11.8, and(b) when y = −3, x = −1.5Now try the following Practice ExercisePracticeExercise 67 Straight line graphs(answers on page 347)1. Assuming graph paper measuring 20cm by20cm is available, suggest suitable scales forthe following ranges of values.(a) Horizontal axis: 3V to 55V; verticalaxis: 10 to 180 .(b) Horizontal axis: 7m to 86m; verticalaxis: 0.3V to 1.69V.(c) Horizontal axis: 5N to 150N; verticalaxis: 0.6mm to 3.4mm.2. Corresponding values obtained experimen-tally for two quantities arex −5 −3 −1 0 2 4y −13 −9 −5 −3 1 5Plot a graph of y (vertically) against x (hori-zontally)to scales of 2cm = 1 for the horizon-tal x-axis and 1cm = 1 for the vertical y-axis.(This graph will need the whole of the graphpaper with the origin somewhere in the centreof the paper).From the graph, ﬁnd(a) the value of y when x = 1(b) the value of y when x = −2.5(c) the value of x when y = −6(d) the value of x when y = 73. Corresponding values obtained experimen-tally for two quantities arex −2.0 −0.5 0 1.0 2.5 3.0 5.0y −13.0 −5.5 −3.0 2.0 9.5 12.0 22.0Use a horizontal scale for x of 1cm = 12 unitand a vertical scale for y of 1cm = 2 units anddraw a graph of x against y. Label the graphand each of its axes. By interpolation, ﬁndfrom the graph the value of y when x is 3.5.4. Draw a graph of y − 3x + 5 = 0 over a rangeof x = −3 to x = 4. Hence determine(a) the value of y when x = 1.3(b) the value of x when y = −9.25. The speed n rev/min of a motor changes whenthe voltage V across the armature is varied.The results are shown in the following table.n (rev/min) 560 720 900 1010 1240 1410V (volts) 80 100 120 140 160 180It is suspected that one of the readings taken ofthe speed is inaccurate. Plot a graph of speed(horizontally) against voltage (vertically) andﬁnd this value. Find also(a) the speed at a voltage of 132V.(b) the voltage at a speed of 1300rev/min.17.4 Gradients, intercepts andequations of graphs17.4.1 GradientsThe gradient or slope of a straight line is the ratio ofthe change in the value of y to the change in the value ofx between any two points on the line. If, as x increases,(→), y also increases, (↑), then the gradient is positive.
• Straight line graphs 137given byFEED=−5 − (−17)0 − 3=12−3= −4Hence, the gradient of both y = −4x + 4 andy = −4x − 5 is −4, which, again, could have beendeduced ‘on sight’.The y-axis intercept means the value of y where thestraight line cuts the y-axis. From Figure 17.10,y = 3x cuts the y-axis at y = 0y = 3x + 7 cuts the y-axis at y = +7y = −4x + 4 cuts the y-axis at y = +4y = −4x − 5 cuts the y-axis at y = −5Some general conclusionscan be drawn from the graphsshown in Figures 17.9 and 17.10. When an equation isof the form y = mx + c, where m and c are constants,then(a) a graph of y against x produces a straight line,(b) m represents the slope or gradient of the line, and(c) c represents the y-axis intercept.Thus, given an equation such as y = 3x + 7, it may bededuced ‘on sight’ that its gradient is +3 and its y-axisintercept is +7, as shown in Figure 17.10. Similarly, ify = −4x − 5, the gradient is−4 and the y-axis interceptis −5, as shown in Figure 17.10.When plottinga graph of the form y = mx + c, only twoco-ordinatesneed bedetermined.When theco-ordinatesare plotted a straight line is drawn between the twopoints. Normally, three co-ordinates are determined, thethird one acting as a check.Problem 4. Plot the graph 3x + y + 1 = 0 and2y − 5 = x on the same axes and ﬁnd their point ofintersectionRearranging 3x + y + 1 = 0 gives y = −3x − 1Rearranging 2y − 5 = x gives 2y = x + 5 andy =12x + 212Since both equations are of the form y = mx + c, bothare straight lines. Knowing an equation is a straightline means that only two co-ordinates need to be plot-ted and a straight line drawn through them. A thirdco-ordinate is usually determined to act as a check.A table ofvalues is produced for each equation as shownbelow.x 1 0 −1−3x − 1 −4 −1 2x 2 0 −312 x + 212 312 212 1The graphs are plotted as shown in Figure 17.11.The two straight lines are seen to intersect at (−1,2).424 23 22 21 032121222324y523x 21yx1 3 42y5 x11252Figure 17.11Problem 5. If graphs of y against x were to beplotted for each of the following, state (i) thegradient and (ii) the y-axis intercept.(a) y = 9x + 2 (b) y = −4x + 7(c) y = 3x (d) y = −5x − 3(e) y = 6 (f) y = xIf y = mx + c then m = gradient and c = y-axisintercept.(a) If y = 9x + 2, then (i) gradient = 9(ii) y-axis intercept = 2(b) If y = −4x + 7, then (i) gradient = −4(ii) y-axis intercept = 7(c) If y = 3x i.e. y = 3x + 0,then (i) gradient = 3(ii) y-axis intercept = 0i.e. the straight line passes through the origin.
• Straight line graphs 139232024 23 22 015108525210215220yxCBA1 3 4221Figure 17.13(b) The y-axis intercept at x = 0 is seen to be at y = 3.y = mx + c is a straight line graph wherem = gradient and c = y-axis intercept.From above, m = 5 and c = 3.Hence, the equation of the graph is y = 5x + 3.Problem 9. Determine the equation of thestraight line shown in Figure 17.14.424 23 22 032121222324yxEFD1 3 4221Figure 17.14The triangleDEF is shown constructed in Figure17.14.Gradient of DE =DFFE=3 − (−3)−1 − 2=6−3= −2and the y-axis intercept = 1.Hence, the equation of the straight line is y = mx + ci.e. y = −2x + 1.Problem 10. The velocity of a body wasmeasured at various times and the results obtainedwere as follows:Velocity v (m/s) 8 10.5 13 15.5 18 20.5 23Time t (s) 1 2 3 4 5 6 7Plot a graph of velocity (vertically) against time(horizontally) and determine the equation of thegraphSuitable scales are chosen and the co-ordinates (1, 8),(2, 10.5), (3, 13), and so on, are plotted as shown inFigure 17.15.045.568 Q RP10121416Velocity(y),inmetrespersecond1820221 2 3Time (t), in seconds4 5 6 7Figure 17.15The right-angled triangle PRQ is constructed on thegraph as shown in Figure 17.15.Gradient of PQ =PRRQ=18 − 85 − 1=104= 2.5The vertical axis intercept is at v = 5.5 m/s.The equation of a straight line graph is y = mx + c. Inthis case, t corresponds to x and v corresponds to y.Hence, the equation of the graph shown in Figure 17.15
• Straight line graphs 14111. A piece of elastic is tied to a support so that ithangs vertically and a pan, on which weightscan be placed, is attached to the free end. Thelength of the elastic is measured as variousweights are added to the pan and the resultsobtained are as follows:Load, W (N) 5 10 15 20 25Length, l (cm) 60 72 84 96 108Plot a graph of load (horizontally) againstlength (vertically) and determine(a) the value of length when the load is 17N.(b) the value of load when the length is74cm.(c) its gradient.(d) the equation of the graph.12. The following table gives the effort P to lifta load W with a small lifting machine.W (N) 10 20 30 40 50 60P (N) 5.1 6.4 8.1 9.6 10.9 12.4Plot W horizontally against P vertically andshow that the values lie approximately on astraight line. Determine the probable rela-tionship connecting P and W in the formP = aW + b.13. In an experiment the speeds N rpm of a ﬂy-wheel slowly coming to rest were recordedagainst the time t in minutes. Plot the resultsand show that N and t are connected byan equation of the form N = at + b. Findprobable values of a and b.t (min) 2 4 6 8 10 12 14N (rev/min) 372 333 292 252 210 177 13217.5 Practical problems involvingstraight line graphsWhen a set of co-ordinate values are given or areobtained experimentally and it is believed that theyfollow a law of the form y = mx + c, if a straight linecan be drawn reasonably close to most oftheco-ordinatevalues when plotted, this veriﬁes that a law of theform y = mx + c exists. From the graph, constantsm (i.e. gradient) and c (i.e. y-axis intercept) can bedetermined.Here are some worked problems in which practicalsituations are featured.Problem 12. The temperature in degrees Celsiusand the corresponding values in degrees Fahrenheitare shown in the table below. Construct rectangularaxes, choose suitable scales and plot a graph ofdegrees Celsius (on the horizontal axis) againstdegrees Fahrenheit (on the vertical scale).◦C 10 20 40 60 80 100◦F 50 68 104 140 176 212From the graph ﬁnd (a) the temperature in degreesFahrenheit at 55◦C, (b) the temperature in degreesCelsius at 167◦F, (c) the Fahrenheit temperature at0◦C and (d) the Celsius temperature at 230◦FThe co-ordinates (10, 50), (20, 68), (40, 104), and soon are plotted as shown in Figure 17.17. When theco-ordinates are joined, a straight line is produced.Since a straight line results, there is a linear relationshipbetween degrees Celsius and degrees Fahrenheit.0403280120DegreesFahrenheit(8F)13116016720024023020 40 55Degrees Celsius (8C)7560 80 120110100yDEFGxABFigure 17.17(a) To ﬁnd the Fahrenheit temperature at 55◦C, a verti-cal line AB is constructed from the horizontal axisto meet the straight line at B. The point where the
• 142 Basic Engineering Mathematicshorizontalline BD meets the vertical axis indicatesthe equivalent Fahrenheit temperature.Hence, 55◦C is equivalent to 131◦F.This process of ﬁnding an equivalent value inbetween the given information in the above tableis called interpolation.(b) To ﬁnd the Celsius temperature at 167◦F, ahorizontal line EF is constructed as shown inFigure 17.17. The point where the vertical line FGcuts the horizontal axis indicates the equivalentCelsius temperature.Hence, 167◦F is equivalent to 75◦C.(c) If the graph is assumed to be linear even outside ofthe given data, the graph may be extended at bothends (shown by broken lines in Figure 17.17).From Figure 17.17, 0◦C corresponds to 32◦F.(d) 230◦F is seen to correspond to 110◦C.The process of ﬁnding equivalent values outsideof the given range is called extrapolation.Problem 13. In an experiment on Charles’s law,the value of the volume of gas, V m3, was measuredfor various temperatures T◦C. The results areshown below.V m325.0 25.8 26.6 27.4 28.2 29.0T ◦C 60 65 70 75 80 85Plot a graph of volume (vertical) againsttemperature (horizontal) and from it ﬁnd (a) thetemperature when the volume is 28.6m3 and (b) thevolume when the temperature is 67◦CIf a graph is plotted with both the scales starting at zerothen the result is as shown in Figure 17.18. All of thepoints lie in the top right-hand corner of the graph,making interpolation difﬁcult. A more accurate graphis obtained if the temperature axis starts at 55◦C andthe volume axis starts at 24.5m3. The axes correspond-ing to these values are shown by the broken lines inFigure 17.18 and are called false axes, since the originis not now at zero. A magniﬁed version of this relevantpart of the graph is shown in Figure 17.19. From thegraph,(a) When the volume is 28.6m3, the equivalent tem-perature is 82.5◦C.(b) When the temperature is 67◦C, the equivalentvolume is 26.1m3.30252015Volume(m3)1050 20 40 60 80 100Temperature (8C)yxFigure 17.182928.62827Volume(m3)262555 60 65 67 70 75 80 82.5 8526.1Temperature (ЊC)yxFigure 17.19Problem 14. In an experiment demonstratingHooke’s law, the strain in an aluminium wire wasmeasured for various stresses. The results were:Stress (N/mm2) 4.9 8.7 15.0Strain 0.00007 0.00013 0.00021Stress (N/mm2) 18.4 24.2 27.3Strain 0.00027 0.00034 0.00039
• Straight line graphs 143Plot a graph of stress (vertically) against strain(horizontally). Find (a) Young’s modulus ofelasticity for aluminium, which is given by thegradient of the graph, (b) the value of the strain at astress of 20N/mm2 and (c) the value of the stresswhen the strain is 0.00020The co-ordinates (0.00007, 4.9), (0.00013, 8.7), andso on, are plotted as shown in Figure 17.20. Thegraph produced is the best straight line which can bedrawn corresponding to these points. (With experimen-tal results it is unlikely that all the points will lie exactlyon a straight line.) The graph, and each of its axes, arelabelled. Since the straight line passes through the ori-gin, stress is directly proportional to strain for the givenrange of values.Stress(N/mm2)0.000050481214162024280.00015 0.00025Strain0.0002850.00035yxABCFigure 17.20(a) The gradient of the straight line AC is given byABBC=28 − 70.00040 − 0.00010=210.00030=213 × 10−4=710−4= 7 × 104= 70000N/mm2Thus, Young’s modulus of elasticity for alu-minium is70 000 N/mm2.Since1m2 = 106 mm2,70000N/mm2 is equivalent to 70000 ×106 N/m2,i.e. 70 ×109N/m2 (or pascals).From Figure 17.20,(b) The value of the strain at a stress of 20N/mm2 is0.000285(c) The value of the stress when the strain is 0.00020is 14 N/mm2.Problem 15. The following values of resistanceR ohms and corresponding voltage V volts areobtained from a test on a ﬁlament lamp.R ohms 30 48.5 73 107 128V volts 16 29 52 76 94Choose suitable scales and plot a graph with Rrepresenting the vertical axis and V the horizontalaxis. Determine (a) the gradient of the graph, (b) theR axis intercept value, (c) the equation of the graph,(d) the value of resistance when the voltage is 60Vand (e) the value of the voltage when the resistanceis 40 ohms. (f) If the graph were to continue in thesame manner, what value of resistance would beobtained at 110V?The co-ordinates (16, 30), (29, 48.5), and so on areshown plotted in Figure 17.21, where the best straightline is drawn through the points.147140120858060100ResistanceR(ohms)401020240 20 40 60 80 100 110 120Voltage V (volts)yCBAxFigure 17.21(a) The slope or gradient of the straight line AC isgiven byABBC=135 − 10100 − 0=125100= 1.25
• 144 Basic Engineering Mathematics(Note that the vertical line AB and the horizon-tal line BC may be constructed anywhere alongthe length of the straight line. However, calcula-tions are made easier if the horizontal line BC iscarefully chosen; in this case, 100.)(b) The R-axis intercept is at R = 10ohms (by extra-polation).(c) The equation of a straight line is y = mx + c, wheny is plotted on the vertical axis and x on the hori-zontal axis. m represents the gradient and c they-axis intercept. In this case, R corresponds to y,V corresponds to x, m = 1.25 and c = 10. Hence,the equation of the graph is R = (1.25V+ 10) .From Figure 17.21,(d) When the voltage is 60V, the resistance is 85 .(e) When theresistanceis40ohms,thevoltageis24V.(f) By extrapolation, when the voltage is 110V, theresistance is 147 .Problem 16. Experimental tests to determine thebreaking stress σ of rolled copper at varioustemperatures t gave the following results.Stress σ (N/cm2) 8.46 8.04 7.78Temperature t(◦C) 70 200 280Stress σ (N/cm2) 7.37 7.08 6.63Temperature t(◦C) 410 500 640Show that the values obey the law σ = at + b,where a and b are constants, and determineapproximate values for a and b. Use the law todetermine the stress at 250◦C and the temperaturewhen the stress is 7.54N/cm2.The co-ordinates (70, 8.46), (200, 8.04), and so on, areplotted as shown in Figure 17.22. Since the graph is astraight line then the values obey the law σ = at + b,and the gradient of the straight line isa =ABBC=8.36 − 6.76100 − 600=1.60−500= −0.0032Vertical axis intercept, b = 8.68Hence, the law of the graph is σ = 0.0032t + 8.68When the temperature is 250◦C, stress σ is given byσ = −0.0032(250) + 8.68 = 7.88N/cm2Temperature t (8C)Stress␴(N/cm2)8.688.368.508.007.507.006.500 100 200 300 400 500 600 7006.76yxCBAFigure 17.22Rearranging σ = −0.0032t + 8.68 gives0.0032t = 8.68 − σ, i.e. t =8.68 − σ0.0032Hence, when the stress, σ = 7.54N/cm2,temperature, t =8.68 − 7.540.0032= 356.3◦CNow try the following Practice ExercisePracticeExercise 69 Practical problemsinvolving straight line graphs (answers onpage 347)1. The resistance R ohms of a copper windingis measured at various temperatures t ◦C andthe results are as follows:R (ohms) 112 120 126 131 134t◦C 20 36 48 58 64Plot a graph of R (vertically) against t (hori-zontally) and ﬁnd from it (a) the temperaturewhen the resistance is 122 and (b) theresistance when the temperature is 52◦C.2. The speed of a motor varies with armaturevoltage as shown by the following experi-mental results.
• Straight line graphs 145n (rev/min) 285 517 615V (volts) 60 95 110n (rev/min) 750 917 1050V (volts) 130 155 175Plot a graph of speed (horizontally) againstvoltage(vertically) and draw the best straightline through the points. Find from the graph(a) the speed at a voltage of 145V and (b) thevoltage at a speed of 400rev/min.3. The following table gives the force Fnewtons which, when applied to a liftingmachine, overcomes a corresponding load ofL newtons.Force F (newtons) 25 47 64Load L (newtons) 50 140 210Force F (newtons) 120 149 187Load L (newtons) 430 550 700Choose suitable scales and plot a graph ofF (vertically) against L (horizontally).Drawthe 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 is310N, and(e) the load that a force of 160N willovercome.(f) If the graph were to continue in thesame manner, what value of force willbe needed to overcome a 800N load?4. The following table gives the results of testscarried out to determine the breaking stressσ of rolled copper at varioustemperatures, t.Stress σ (N/cm2) 8.51 8.07 7.80Temperature t(◦C) 75 220 310Stress σ (N/cm2) 7.47 7.23 6.78Temperature t(◦C) 420 500 650Plot a graph of stress (vertically) againsttemperature (horizontally). Draw the beststraight linethrough the plotted co-ordinates.Determine the slope of the graph and thevertical axis intercept.5. The velocity v of a body after varying timeintervals t was measured as follows:t (seconds) 2 5 8v (m/s) 16.9 19.0 21.1t (seconds) 11 15 18v (m/s) 23.2 26.0 28.1Plot v vertically and t horizontally and drawa graph of velocity against time. Determinefrom the graph (a) the velocity after 10s,(b) the time at 20m/s and (c) the equationof the graph.6. The mass m of a steel joist varies with lengthL as follows:mass, m (kg) 80 100 120 140 160length, L (m) 3.00 3.74 4.48 5.23 5.97Plot a graph of mass (vertically) againstlength (horizontally). Determine the equa-tion of the graph.7. The crushing strength of mortar varies withthe percentage of water used in its prepara-tion, as shown below.Crushing strength, 1.67 1.40 1.13F (tonnes)% of water used, w% 6 9 12Crushing strength, 0.86 0.59 0.32F (tonnes)% of water used, w% 15 18 21
• 146 Basic Engineering MathematicsPlot a graph of F (vertically) against w(horizontally).(a) Interpolate and determine the crushingstrength when 10% water is used.(b) Assuming the graph continues in thesame manner, extrapolate and deter-mine the percentage of water used whenthe crushing strength is 0.15 tonnes.(c) What is the equation of the graph?8. In an experiment demonstrating Hooke’slaw,the strain in a copper wire was measured forvarious stresses. The results wereStress 10.6 × 106 18.2 × 106 24.0 × 106(pascals)Strain 0.00011 0.00019 0.00025Stress 30.7 × 106 39.4 × 106(pascals)Strain 0.00032 0.00041Plot a graph of stress (vertically) againststrain (horizontally). Determine(a) Young’s modulus of elasticity for cop-per, which is given by the gradient ofthe graph,(b) the value of strain at a stress of21 × 106 Pa,(c) the value of stress when the strain is0.00030,9. An experiment with a set of pulley blocksgave the following results.Effort, E (newtons) 9.0 11.0 13.6Load, L (newtons) 15 25 38Effort, E (newtons) 17.4 20.8 23.6Load, L (newtons) 57 74 88Plot a graph of effort (vertically)against load(horizontally). Determine(a) the gradient,(b) the vertical axis intercept,(c) the law of the graph,(d) the effort when the load is 30N,(e) the load when the effort is 19N.10. The variation of pressure p in a vessel withtemperature T is believed to follow a law ofthe form p = aT + b, where a and b are con-stants. Verify this law for the results givenbelow and determine the approximate valuesof a and b. Hence, determine the pressuresat temperatures of 285K and 310K and thetemperature at a pressure of 250kPa.Pressure, p (kPa) 244 247 252Temperature, T (K) 273 277 282Pressure, p (kPa) 258 262 267Temperature, T (K) 289 294 300
• Chapter 18Graphs reducing non-linearlaws to linear form18.1 IntroductionIn Chapter 17 we discovered that the equation of astraight line graph is of the form y = mx + c, wherem is the gradient and c is the y-axis intercept. Thischapterexplainshowthelawofagraph can still bedeter-mined even when it is not of thelinear form y = mx + c.The method used is called determination of law and isexplained in the following sections.18.2 Determination of lawFrequently, the relationship between two variables, sayx and y, is not a linear one; i.e., when x is plottedagainsty a curve results. In such cases the non-linear equationmay be modiﬁed to the linear form, y = mx + c, so thatthe constants, and thus the law relating the variables, canbe determined. This technique is called ‘determinationof law’.Some examples of the reduction of equations to linearform include(i) y = ax2 + b compares with Y = mX + c, wherem = a, c = b and X = x2.Hence, y is plotted vertically against x2 horizon-tally to produce a straight line graph of gradienta and y-axis intercept b.(ii) y =ax+ b, i.e. y = a1x+ by is plotted vertically against1xhorizontally toproduce a straight line graph of gradient a andy-axis intercept b.(iii) y = ax2 + bxDividing both sides by x givesyx= ax + b.Comparing with Y = mX + c shows thatyxisplotted vertically against x horizontally to pro-duce a straight line graph of gradient a andyxaxis intercept b.Here are some worked problems to demonstrate deter-mination of law.Problem 1. Experimental values of x and y,shown below, are believed to be related by the lawy = ax2 + b. By plotting a suitable graph, verifythis law and determine approximate values ofa and bx 1 2 3 4 5y 9.8 15.2 24.2 36.5 53.0If y is plotted against x a curve results and it is notpossible to determine the values of constants a and bfrom the curve.Comparing y = ax2 + b with Y = mX + c shows thaty is to be plotted vertically against x2horizontally. Atable of values is drawn up as shown below.x 1 2 3 4 5x2 1 4 9 16 25y 9.8 15.2 24.2 36.5 53.0DOI: 10.1016/B978-1-85617-697-2.00018-1
• Graphs reducing non-linear laws to linear form 149Hence, when the load L = 20N, distance,d =235 − 20=215= 0.13m.Problem 3. The solubility s of potassium chlorateis shown by the following table.t ◦C 10 20 30 40 50 60 80 100s 4.9 7.6 11.1 15.4 20.4 26.4 40.6 58.0The relationship between s and t is thought to be ofthe form s = 3 + at + bt2. Plot a graph to test thesupposition and use the graph to ﬁnd approximatevalues of a and b. Hence, calculate the solubility ofpotassium chlorate at 70◦CRearranging s = 3 + at + bt2 givess − 3 = at + bt2and s − 3t= a + btor s − 3t= bt + awhich is of the form Y = mX + cThis shows thats − 3tis to be plotted vertically andt horizontally, with gradient b and vertical axis inter-cept a.Another table of values is drawn up as shown below.t 10 20 30 40 50 60 80 100s 4.9 7.6 11.1 15.4 20.4 26.4 40.6 58.0s − 3t0.19 0.23 0.27 0.31 0.35 0.39 0.47 0.55A graph ofs − 3tagainst t is shown plotted in Figure18.3. A straight line ﬁts the points, which shows thats and t are related by s = 3 + at + bt2.Gradient of straight line, b =ABBC=0.39 − 0.1960 − 10=0.2050= 0.004Vertical axis intercept, a = 0.15Hence, the law of the graph is s = 3 + 0.15t + 0.004t2.The solubility of potassium chlorate at 70◦C isgiven bys = 3 + 0.15(70) + 0.004(70)2= 3 + 10.5 + 19.6 = 33.1CBA0.60.50.40.390.190.150.30.20.10 20 40 60t 8C80 100s23tFigure 18.3Now try the following Practice ExercisePracticeExercise 70 Graphs reducingnon-linear laws to linear form (answers onpage 348)In problems 1 to 5, x and y are two related vari-ables and all other letters denote constants. For thestated laws to be veriﬁed it is necessary to plotgraphs of the variables in a modiﬁed form. Statefor each, (a) what should be plotted on the verticalaxis, (b) what should be plotted on the horizon-tal axis, (c) the gradient and (d) the vertical axisintercept.1. y = d + cx2 2. y − a = b√x3. y − e =fx4. y − cx = bx25. y =ax+ bx6. In an experiment the resistance of wire is mea-sured for wires of different diameters with thefollowing results.R (ohms) 1.64 1.14 0.89 0.76 0.63d (mm) 1.10 1.42 1.75 2.04 2.56It is thought that R is related to d by the lawR =ad2+ b, where a and b are constants. Ver-ify this and ﬁnd the approximate values for
• 150 Basic Engineering Mathematicsa and b. Determine the cross-sectional areaneeded for a resistance reading of 0.50ohms.7. Corresponding experimental values of twoquantities x and y are given below.x 1.5 3.0 4.5 6.0 7.5 9.0y 11.5 25.0 47.5 79.0 119.5 169.0By plotting a suitable graph, verify that yand x are connected by a law of the formy = kx2 + c, where k and c are constants.Determine the law of the graph and hence ﬁndthe value of x when y is 60.08. Experimental results of the safe load L kN,applied to girders of varying spans, d m, areshown below.Span, d (m) 2.0 2.8 3.6 4.2 4.8Load, L (kN) 475 339 264 226 198It isbelieved that therelationship between loadand span is L = c/d, where c is a constant.Determine (a) the value of constant c and (b)the safe load for a span of 3.0m.9. The following results give corresponding val-ues of two quantities x and y which arebelieved to be related by a law of the formy = ax2 + bx, where a and b are constants.x 33.86 55.54 72.80 84.10 111.4 168.1y 3.4 5.2 6.5 7.3 9.1 12.4Verify the law and determine approximate val-ues of a and b. Hence, determine (a) the valueof y when x is 8.0 and (b) the value of x wheny is 146.518.3 Revision of laws of logarithmsThe laws of logarithms were stated in Chapter 15 asfollows:log(A × B) = logA + logB (1)logAB= logA − logB (2)logAn= n × logA (3)Also, lne = 1 and if, say, lg x = 1.5,then x = 101.5= 31.62Further, if 3x = 7 then lg 3x = lg 7 and x lg 3 = lg 7,from which x =lg 7lg 3= 1.771Theselawsand techniquesareused whenevernon-linearlaws of the form y = axn, y = abx and y = aebx arereduced to linear form with the values ofaand b needingto be calculated. This is demonstrated in the followingsection.18.4 Determination of laws involvinglogarithmsExamples of the reduction of equations to linear forminvolving logarithms include(a) y = axnTaking logarithms to a base of 10 of both sidesgiveslg y = lg(axn)= lga + lgxnby law (1)i.e. lg y = n lgx + lga by law (3)which compares with Y = mX + cand shows that lgy is plotted vertically againstlgx horizontally to produce a straight line graphof gradient n and lgy-axis intercept lg a.See worked Problems 4 and 5 to demonstrate howthis law is determined.(b) y = abxTaking logarithms to a base of 10 of both sidesgiveslg y = lg(abx)i.e. lg y = lga + lgbxby law (1)lg y = lga + x lgb by law (3)i.e. lg y = x lgb + lgaor lg y = (lgb)x + lgawhich compares withY = mX + c
• Graphs reducing non-linear laws to linear form 151and shows that lgy is plotted vertically againstx horizontally to produce a straight line graphof gradient lgb and lgy-axis intercept lg a.See worked Problem 6 to demonstrate how this lawis determined.(c) y = aebxTaking logarithms to a base of e of both sidesgivesln y = ln(aebx)i.e. ln y = lna + lnebxby law (1)i.e. ln y = lna + bx lne by law (3)i.e. ln y = bx + lna since lne = 1which compares withY = mX + cand shows that lny is plotted vertically againstx horizontally to produce a straight line graphof gradient b and ln y-axis intercept ln a.See worked Problem 7 to demonstrate how this law isdetermined.Problem 4. The current ﬂowing in, and thepower dissipated by, a resistor are measuredexperimentally for various values and the results areas shown below.Current, I (amperes) 2.2 3.6 4.1 5.6 6.8Power, P (watts) 116 311 403 753 1110Show that the law relating current and power is ofthe form P = RIn , where R and n are constants,and determine the lawTaking logarithms to a base of 10 of both sides ofP = RIn giveslg P = lg(RIn) = lg R + lg In= lg R + n lg Iby the laws of logarithmsi.e. lg P = n lg I + lg Rwhich is of the form Y = mX + c,showing that lg P is to be plotted vertically against lg Ihorizontally.A table of values for lg I and lg P is drawn up as shownbelow.I 2.2 3.6 4.1 5.6 6.8lg I 0.342 0.556 0.613 0.748 0.833P 116 311 403 753 1110lg P 2.064 2.493 2.605 2.877 3.045A graph of lg P against lg I is shown in Figure 18.4and, since a straight line results, the law P = RIn isveriﬁed.CBAD2.00.30 0.40 0.50 0.60 0.70 0.80 0.902.182.5lgPlg L2.782.983.0Figure 18.4Gradient of straight line, n =ABBC=2.98 − 2.180.8 − 0.4=0.800.4= 2It is not possible to determine the vertical axis intercepton sight since the horizontal axis scale does not startat zero. Selecting any point from the graph, say pointD, where lg I = 0.70 and lg P = 2.78 and substitutingvalues intolg P = n lg I + lg Rgives 2.78 = (2)(0.70) + lg Rfrom which, lg R = 2.78 − 1.40 = 1.38Hence, R = antilog 1.38 = 101.38= 24.0Hence, the law of the graph is P = 24.0I2Problem 5. The periodic time, T, of oscillation ofa pendulum is believed to be related to its length, L,by a law of the form T = kLn, where k and n areconstants. Values of T were measured for variouslengths of the pendulum and the results are asshown below.Periodic time, T(s) 1.0 1.3 1.5 1.8 2.0 2.3Length, L(m) 0.25 0.42 0.56 0.81 1.0 1.32Show that the law is true and determine theapproximate values of k and n. Hence ﬁnd theperiodic time when the length of the pendulum is0.75m
• 152 Basic Engineering MathematicsFrom para (a), page 150, if T = kLnthen lgT = n lgL + lgkand comparing with Y = mX + cshows that lgT is plotted vertically against lgL hor-izontally, with gradient n and vertical-axis interceptlgk.A table of values for lgT and lg L is drawn up as shownbelow.T 1.0 1.3 1.5 1.8 2.0 2.3lgT 0 0.114 0.176 0.255 0.301 0.362L 0.25 0.42 0.56 0.81 1.0 1.32lgL −0.602 −0.377 −0.252 −0.092 0 0.121A graph of lgT against lg L is shown in Figure 18.5 andthe law T = kLn is true since a straight line results.ABC020.4020.5020.60 20.3020.20 0.200.100.200.300.250.05lg Llg T0.4020.10 0.10Figure 18.5From the graph, gradient of straight line,n =ABBC=0.25 − 0.05−0.10 − (−0.50)=0.200.40=12Vertical axis intercept, lgk = 0.30. Hence,k = antilog 0.30 = 100.30 = 2.0Hence, the law of the graph is T = 2.0L1/2 orT = 2.0√L.When length L = 0.75m, T = 2.0√0.75 = 1.73sProblem 6. Quantities x and y are believed to berelated by a law of the form y = abx, where a and bare constants. The values of x and correspondingvalues of y arex 0 0.6 1.2 1.8 2.4 3.0y 5.0 9.67 18.7 36.1 69.8 135.0Verify the law and determine the approximatevalues of a and b. Hence determine (a) the value ofy when x is 2.1 and (b) the value of x when y is 100From para (b), page 150, if y = abxthen lg y = (lgb)x + lgaand comparing with Y = mX + cshows that lg y is plotted vertically and x horizontally,with gradient lgb and vertical-axis intercept lga.Another table is drawn up as shown below.x 0 0.6 1.2 1.8 2.4 3.0y 5.0 9.67 18.7 36.1 69.8 135.0lg y 0.70 0.99 1.27 1.56 1.84 2.13A graph of lg y against x is shown in Figure 18.6and, since a straight line results, the law y = abxisveriﬁed.CBAlgyx2.502.132.001.501.001.170.500 1.0 2.0 3.00.70Figure 18.6
• Graphs reducing non-linear laws to linear form 153Gradient of straight line,lgb =ABBC=2.13 − 1.173.0 − 1.0=0.962.0= 0.48Hence, b = antilog 0.48 = 100.48 = 3.0, correct to 2signiﬁcant ﬁgures.Vertical axis intercept, lga = 0.70,from whicha = antilog 0.70= 100.70= 5.0, correct to2 signiﬁcant ﬁgures.Hence, the law of the graph is y = 5.0(3.0)x(a) When x = 2.1,y = 5.0(3.0)2.1 = 50.2(b) When y = 100,100 = 5.0(3.0)x , from which100/5.0 = (3.0)xi.e. 20 = (3.0)xTaking logarithms of both sides giveslg20 = lg(3.0)x= x lg3.0Hence, x =lg20lg3.0=1.30100.4771= 2.73Problem 7. The current i mA ﬂowing in acapacitor which is being discharged varies withtime t ms, as shown below.i (mA) 203 61.14 22.49 6.13 2.49 0.615t (ms) 100 160 210 275 320 390Show that these results are related by a law of theform i = Iet/T , where I and T are constants.Determine the approximate values of I and TTaking Napierian logarithms of both sides ofi = Iet/Tgives lni = ln(Iet/T) = ln I + lnet/T= ln I +tTlnei.e. lni = ln I +tTsince lne = 1or lni =1Tt + ln Iwhich compares with y = mx + cshowing that lni is plotted vertically against t hor-izontally, with gradient 1T and vertical-axis interceptln I.Another table of values is drawn up as shown below.t 100 160 210 275 320 390i 203 61.14 22.49 6.13 2.49 0.615lni 5.31 4.11 3.11 1.81 0.91 −0.49A graph of lni against t is shown in Figure 18.7and, since a straight line results, the law i = Iet/T isveriﬁed.ABClni5.04.03.03.312.01.0100 200 300 400 t (ms)D (200, 3.31)21.01.300Figure 18.7Gradient of straight line,1T=ABBC=5.30 − 1.30100 − 300=4.0−200= −0.02Hence, T =1−0.02= −50Selecting any point on the graph, say point D, wheret = 200 and lni = 3.31, and substitutinginto lni =1Tt + ln Igives 3.31 = −150(200) + ln Ifrom which, ln I = 3.31 + 4.0 = 7.31and I = antilog 7.31 = e7.31 = 1495 or 1500correct to 3 signiﬁcant ﬁgures.Hence, the law of the graph is i = 1500e−t/50
• 154 Basic Engineering MathematicsNow try the following Practice ExercisePractice Exercise 71 Determination of lawinvolving logarithms (answers on page 348)In problems 1 to 3, x and y are two related vari-ables and all other letters denote constants. For thestated laws to be veriﬁed it is necessary to plotgraphs of the variables in a modiﬁed form. Statefor each, (a) what should be plotted on the verticalaxis, (b) what should be plotted on the horizon-tal axis, (c) the gradient and (d) the vertical axisintercept.1. y = bax 2. y = kxL 3.ym= enx4. The luminosity I of a lamp varies withthe applied voltage V and the relationshipbetween I and V is thought to be I = kV n.Experimental results obtained areI(candelas) 1.92 4.32 9.72V (volts) 40 60 90I(candelas) 15.87 23.52 30.72V (volts) 115 140 160Verify that the law is true and determinethe law of the graph. Also determine theluminosity when 75V is applied across thelamp.5. The head of pressure h and the ﬂow veloc-ity v are measured and are believed to beconnected by the law v = ahb, where a andb are constants. The results are as shownbelow.h 10.6 13.4 17.2 24.6 29.3v 9.77 11.0 12.44 14.88 16.24Verify that the law is true and determine valuesof a and b.6. Experimental values of x and y are measuredas follows.x 0.4 0.9 1.2 2.3 3.8y 8.35 13.47 17.94 51.32 215.20The law relating x and y is believed to be ofthe form y = abx, where a and b are constants.Determine the approximate values of a and b.Hence, ﬁnd the value of y when x is 2.0 andthe value of x when y is 100.7. The activity of a mixture of radioactive iso-topes is believed to vary according to the lawR = R0t−c, where R0 and c are constants.Experimental results are shown below.R 9.72 2.65 1.15 0.47 0.32 0.23t 2 5 9 17 22 28Verify that the law is true and determineapproximate values of R0 and c.8. Determine the law of the form y = aekxwhichrelates the following values.y 0.0306 0.285 0.841 5.21 173.2 1181x –4.0 5.3 9.8 17.4 32.0 40.09. The tension T in a belt passing round a pul-ley wheel and in contact with the pulley overan angle of θ radians is given by T = T0eμθ ,where T0 and μ are constants. Experimentalresults obtained areT (newtons) 47.9 52.8 60.3 70.1 80.9θ (radians) 1.12 1.48 1.97 2.53 3.06Determine approximate values of T0 and μ.Hence, ﬁnd the tension when θ is 2.25 radiansand the value of θ when the tension is 50.0newtons.
• Chapter 19Graphical solution ofequations19.1 Graphical solution ofsimultaneous equationsLinear simultaneous equations in two unknowns maybe solved graphically by(a) plotting the two straight lines on the same axes,and(b) noting their point of intersection.The co-ordinates of the point of intersection give therequired solution.Here are some worked problems to demonstrate thegraphical solution of simultaneous equations.Problem 1. Solve graphically the simultaneousequations2x − y = 4x + y = 5Rearranging each equation into y = mx + c form givesy = 2x − 4y = −x + 5Only three co-ordinates need be calculated for eachgraph since both are straight lines.x 0 1 2y = 2x − 4 −4 −2 0x 0 1 2y = −x + 5 5 4 3Each of the graphs is plotted as shown in Figure 19.1.The point of intersection is at (3, 2) and since this is theonly point which lies simultaneously on both linesthen x = 3,y = 2 is the solution of the simultaneousequations.24 23 22 21 112345yy 52x1 5y 52x2 42 3 4 x212223240Figure 19.1Problem 2. Solve graphically the equations1.20x + y = 1.80x − 5.0y = 8.50DOI: 10.1016/B978-1-85617-697-2.00019-3
• Graphical solution of equations 157(a)yy5 x2x2121 0 1(b)y53x2yx2121 0 1yx2121 0 1(c)y 5 x221Figure 19.4Graphs of y = −x2, y = −3x2 and y = −12x2 areshown in Figure 19.5. All have maximum valuesat the origin (0, 0).(a)yy 52x2x21212201(b)y523x22122yx21012122yx2101(c)y52 x221Figure 19.5When y = ax2,(i) curves are symmetrical about the y-axis,(ii) the magnitude of a affects the gradient of thecurve, and(iii) the sign of a determines whether it has amaximum or minimum value.(b) y = ax2 + cGraphs of y = x2 + 3, y = x2 − 2, y = −x2 + 2and y = −2x2 − 1 are shown in Figure 19.6.When y = ax2 + c,(i) curves are symmetrical about the y-axis,(ii) the magnitude of a affects the gradient of thecurve, and(iii) the constant c is the y-axis intercept.(c) y = ax2 + bx + cWhenever b has a value other than zero the curveis displaced to the right or left of the y-axis.When b/a is positive, the curve is displaced b/2ato the left of the y-axis, as shown in Figure19.7(a).When b/a is negative, the curve is displacedb/2a to the right of the y-axis, as shown inFigure 19.7(b).Quadratic equations of the form ax2 + bx + c = 0may be solved graphically by(a) plotting the graph y = ax2 + bx + c, and(b) noting the points of intersection on the x-axis (i.e.where y = 0).yy 5x213y 52x21 2y 522x221y 5 x222x21 0(a)31yx2221 1(b)20yx21(c)021yx212124(d)01Figure 19.6(a)25 24 23 22 21 1yx12106420y5 x216x111(b)y 5x2 25x1421yx0246221 2 3 48Figure 19.7The x values of the points of intersection give therequired solutions since at these points both y = 0 andax2 + bx + c = 0.The number of solutions, or roots, of a quadratic equa-tion depends on how many times the curve cuts thex-axis. There can be no real roots, as in Figure 19.7(a),one root, as in Figures 19.4 and 19.5, or two roots, as inFigure 19.7(b).Here are some worked problems to demonstrate thegraphical solution of quadratic equations.Problem 3. Solve the quadratic equation4x2 + 4x − 15 = 0 graphically, given that thesolutions lie in the range x = −3 to x = 2.Determine also the co-ordinates and nature of theturning point of the curve
• 158 Basic Engineering MathematicsLet y = 4x2+ 4x − 15. A table of values is drawn up asshown below.x −3 −2 −1 0 1 2y = 4x2 + 4x − 15 9 −7 −15 −15 −7 9y 54x214x 21512y84242821221620.51.50 1 2 xA B22.523 22 21Figure 19.8A graph of y = 4x2 + 4x − 15 is shown in Figure 19.8.The only pointswhere y = 4x2 + 4x − 15 and y = 0 arethe points marked A and B. This occurs at x = −2.5and x = 1.5 and these are the solutions of the quadraticequation 4x2 + 4x − 15 = 0.By substituting x = −2.5 and x = 1.5 into the originalequation the solutions may be checked.The curve has a turning point at (−0.5,−16) and thenature of the point is a minimum.An alternative graphical method of solving4x2 + 4x − 15 = 0 is to rearrange the equation as4x2= −4x + 15 and then plot two separate graphs− in this case, y = 4x2 and y = −4x + 15. Theirpoints of intersection give the roots of the equation4x2 = −4x + 15, i.e. 4x2 + 4x − 15 = 0. This is shownin Figure 19.9, where the roots are x = −2.5 andx = 1.5, as before.Problem 4. Solve graphically the quadraticequation −5x2 + 9x + 7.2 = 0 given that thesolutions lie between x = −1 and x = 3. Determinealso the co-ordinates of the turning point and stateits naturey 524x 11530252015105y10 2 3 x212222.5 1.523y 54x 2Figure 19.9Let y = −5x2 + 9x + 7.2. A table of values is drawn upas shown below.x −1 −0.5 0 1y = −5x2 + 9x + 7.2 −6.8 1.45 7.2 11.2x 2 2.5 3y = −5x2 + 9x + 7.2 5.2 −1.55 −10.8A graph of y = −5x2 + 9x + 7.2 is shown plotted inFigure 19.10. The graph crosses the x-axis (i.e. wherey = 0) at x = −0.6 and x = 2.4 and these are the solu-tions of the quadratic equation −5x2 + 9x + 7.2 = 0.The turning point is a maximum, having co-ordinates(0.9,11.25).Problem 5. Plot a graph of y = 2x2 and hencesolve the equations(a) 2x2 − 8 = 0 (b) 2x2 − x − 3 = 0A graph of y = 2x2is shown in Figure 19.11.(a) Rearranging 2x2 − 8 = 0 gives 2x2 = 8 and thesolution ofthisequation isobtained fromthepointsof intersection of y = 2x2 and y = 8; i.e., at co-ordinates (−2,8) and (2,8), shown as A and B,respectively, in Figure 19.11. Hence, the solutionsof 2x2− 8 = 0 are x = −2 and x = +2.(b) Rearranging 2x2 − x − 3 = 0 gives 2x2 = x + 3and the solution of this equation is obtainedfrom the points of intersection of y = 2x2 andy = x + 3; i.e., at C and D in Figure 19.11. Hence,the solutions of 2x2 − x − 3 = 0 are x = −1 andx = 1.5
• Graphical solution of equations 159Ϫ10Ϫ8Ϫ6Ϫ4Ϫ2Ϫ0.6yϭϪ5x2ϩ 9x ϩ7.20.9 2.40Ϫ1 1 2 324681011.2512yxFigure 19.10yϭ2x 2yϭ8yϭxϩ3Ϫ2 Ϫ1 0246810ACDBy1 1.5 2 xFigure 19.11Problem 6. Plot the graph of y = −2x2 + 3x + 6for values of x from x = −2 to x = 4. Use thegraph to ﬁnd the roots of the following equations.(a) −2x2 + 3x + 6 = 0 (b) −2x2 + 3x + 2 = 0(c) −2x2 + 3x + 9 = 0 (d) −2x2 + x + 5 = 0A table of values for y = −2x2 + 3x + 6 is drawn up asshown below.x −2 −1 0 1 2 3 4y −8 1 6 7 4 −3 −14A graph of y = −2x2+ 3x + 6 is shown inFigure 19.12.2826242221.1321.352221.5GBADCHF y 523y 54y 52x 11y 522x21 3x 16xE1 21.85 2.63302468y20.521Figure 19.12(a) The parabola y = −2x2 + 3x + 6 and the straightline y = 0 intersect at A and B, where x = −1.13and x = 2.63, and these are the roots of theequation −2x2 + 3x + 6 = 0.(b) Comparing y = −2x2+ 3x + 6 (1)with 0 = −2x2+ 3x + 2 (2)shows that, if 4 is added to both sides ofequation (2), the RHS of both equations willbe the same. Hence, 4 = −2x2 + 3x + 6. Thesolution of this equation is found from thepoints of intersection of the line y = 4 andthe parabola y = −2x2 + 3x + 6; i.e., points Cand D in Figure 19.12. Hence, the roots of−2x2 + 3x + 2 = 0 are x = −0.5 and x = 2.(c) −2x2 + 3x + 9 = 0 may be rearranged as−2x2 + 3x + 6 = −3 and the solution ofthis equation is obtained from the pointsof intersection of the line y = −3 and theparabola y = −2x2 + 3x + 6; i.e., at points Eand F in Figure 19.12. Hence, the roots of−2x2 + 3x + 9 = 0 are x = −1.5 and x = 3.
• 160 Basic Engineering Mathematics(d) Comparing y = −2x2+ 3x + 6 (3)with 0 = −2x2+ x + 5 (4)shows that, if 2x + 1 is added to both sidesof equation (4), the RHS of both equationswill be the same. Hence, equation (4) may bewritten as 2x + 1 = −2x2 + 3x + 6. The solu-tion of this equation is found from the pointsof intersection of the line y = 2x + 1 and theparabola y = −2x2 + 3x + 6; i.e., points Gand H in Figure 19.12. Hence, the roots of−2x2 + x + 5 = 0 are x = −1.35 and x = 1.85Now try the following Practice ExercisePracticeExercise 73 Solving quadraticequations graphically (answers on page 348)1. Sketch the following graphs and state thenature and co-ordinates of their respectiveturning points.(a) y = 4x2 (b) y = 2x2 − 1(c) y = −x2 + 3 (d) y = −12x2 − 1Solve graphically the quadratic equations in prob-lems 2 to 5 by plotting the curves between the givenlimits. Give answers correct to 1 decimal place.2. 4x2 − x − 1 = 0; x = −1 to x = 13. x2 − 3x = 27; x = −5 to x = 84. 2x2 − 6x − 9 = 0; x = −2 to x = 55. 2x(5x − 2) = 39.6; x = −2 to x = 36. Solve the quadratic equation2x2 + 7x + 6 = 0 graphically, given thatthe solutions lie in the range x = −3 tox = 1. Determine also the nature andco-ordinates of its turning point.7. Solve graphically the quadratic equation10x2 − 9x − 11.2 = 0,given that therootsliebetween x = −1 and x = 2.8. Plot a graph of y = 3x2 and hence solve thefollowing equations.(a) 3x2 − 8 = 0 (b) 3x2 − 2x − 1 = 09. Plot the graphs y = 2x2 and y = 3 − 4x onthe same axes and ﬁnd the co-ordinates of thepoints of intersection. Hence, determine theroots of the equation 2x2 + 4x − 3 = 0.10. Plot a graph of y = 10x2 − 13x − 30 forvalues of x between x = −2 and x = 3.Solve the equation 10x2 − 13x − 30 = 0 andfrom the graph determine(a) the value of y when x is 1.3,(b) the value of x when y is 10,(c) the roots of the equation10x2 − 15x − 18 = 0.19.3 Graphical solution of linearand quadratic equationssimultaneouslyThe solutionof linear and quadratic equations simul-taneously may be achieved graphically by(a) plotting the straight line and parabola on the sameaxes, and(b) noting the points of intersection.The co-ordinates of the points of intersection give therequired solutions.Here is a worked problem to demonstrate the simulta-neous solution of a linear and quadratic equation.Problem 7. Determine graphically the values ofx and y which simultaneously satisfy the equationsy = 2x2 − 3x − 4 and y = 2 − 4xy = 2x2 − 3x − 4 is a parabola and a table of values isdrawn up as shown below.x −2 −1 0 1 2 3y 10 1 −4 −5 −2 5y = 2 − 4x is a straight line and only three co-ordinatesneed be calculated:x 0 1 2y 2 −2 −6The two graphs are plotted in Figure 19.13 and thepoints of intersection, shown as A and B, are at co-ordinates (−2,10) and (1.5,−4). Hence, the simul-taneous solutions occur when x = −2, y = 10 andwhen x = 1.5, y = −4.These solutions may be checked by substituting intoeach of the original equations.
• Graphical solution of equations 161Ϫ40Ϫ1Ϫ2 1BAyϭ2Ϫ 4xyϭ2x2 Ϫ3x Ϫ42 3 x246810yϪ2Figure 19.13Now try the following Practice ExercisePracticeExercise 74 Solving linear andquadratic equations simultaneously(answers on page 348)1. Determine graphically the values of x andy which simultaneously satisfy the equationsy = 2(x2 − 2x − 4) and y + 4 = 3x.2. Plot the graph of y = 4x2 − 8x − 21 for valuesof x from −2 to +4. Use the graph to ﬁnd theroots of the following equations.(a) 4x2 − 8x − 21 = 0(b) 4x2 − 8x − 16 = 0(c) 4x2− 6x − 18 = 019.4 Graphical solution of cubicequationsA cubic equation of the form ax3 + bx2 + cx + d = 0may be solved graphically by(a) plotting the graph y = ax3 + bx2 + cx + d, and(b) noting the points of intersection on the x-axis (i.e.where y = 0).The x-values of the points of intersection give therequired solution since at these points both y = 0 andax3 + bx2 + cx + d = 0.The number of solutions, or roots, of a cubic equationdepends on how many times the curve cuts the x-axisand there can be one, two or three possible roots, asshown in Figure 19.14.(a)yx(b)yx(c)yxFigure 19.14Here are some worked problems to demonstrate thegraphical solution of cubic equations.Problem 8. Solve graphically the cubic equation4x3 − 8x2 − 15x + 9 = 0, given that the roots liebetween x = −2 and x = 3. Determine also theco-ordinates of the turning points and distinguishbetween themLet y = 4x3− 8x2− 15x + 9.Atableofvaluesisdrawnup as shown below.x −2 −1 0 1 2 3y −25 12 9 −10 −21 0A graph of y = 4x3 − 8x2 − 15x + 9 is shown inFigure 19.15.Thegraph crossesthe x-axis(where y = 0)at x = −1.5,x = 0.5 and x = 3 and these are the solutions to thecubic equation 4x3 − 8x2 − 15x + 9 = 0.The turning points occur at (−0.6,14.2), which is amaximum, and (2,−21), which is a minimum.Problem 9. Plot the graph ofy = 2x3 − 7x2 + 4x + 4 for values of x betweenx = −1 and x = 3. Hence, determine the roots ofthe equation 2x3 − 7x2 + 4x + 4 = 0A table of values is drawn up as shown below.x −1 0 1 2 3y −9 4 3 0 7
• 162 Basic Engineering Mathematics22 21 20.624481216y14.2282122162202212240 1 2 3 xy 54x32 8x2215x19Figure 19.15A graph of y = 2x3 − 7x2 + 4x + 4 is shown inFigure 19.16. The graph crosses the x-axis at x = −0.5and touches the x-axis at x = 2.2826242221 1 2 302468yxy 52x327x214x 14Figure 19.16Hence the solutions of the equation2x3 − 7x2 + 4x + 4 = 0 are x = −0.5 and x = 2.Now try the following Practice ExercisePracticeExercise 75 Solving cubicequations (answers on page 348)1. Plot the graph y = 4x3 + 4x2 − 11x − 6between x = −3 and x = 2 and usethe graph to solve the cubic equation4x3 + 4x2 − 11x − 6 = 0.2. By plotting a graph of y = x3 − 2x2 − 5x + 6between x = −3 and x = 4, solve the equa-tion x3 − 2x2 − 5x + 6 = 0. Determine alsothe co-ordinates of the turning points anddistinguish between them.In problems 3 to 6, solve graphically the cubicequations given, each correct to 2 signiﬁcantﬁgures.3. x3 − 1 = 04. x3 − x2 − 5x + 2 = 05. x3 − 2x2 = 2x − 26. 2x3 − x2 − 9.08x + 8.28 = 07. Show that the cubic equation8x3 + 36x2 + 54x + 27 = 0 has only one realroot and determine its value.
• Revision Test 7 : GraphsThis assignment covers the material contained in Chapters 17–19. The marks available are shown in brackets at theend of each question.1. Determine the value of P in the following table ofvalues.x 0 1 4y = 3x − 5 −5 −2 P (2)2. Assuming graph paper measuring 20cm by 20cmis available, suggest suitable scales for the follow-ing ranges of values.Horizontal axis: 5N to 70N; vertical axis: 20mmto 190mm. (2)3. Corresponding valuesobtained experimentally fortwo quantities are:x −5 −3 −1 0 2 4y −17 −11 −5 −2 4 10Plot a graph of y (vertically) against x (hori-zontally) to scales of 1cm = 1 for the horizontalx-axis and 1cm = 2 for the vertical y-axis. Fromthe graph, ﬁnd(a) the value of y when x = 3,(b) the value of y when x = −4,(c) the value of x when y = 1,(d) the value of x when y = −20. (8)4. If graphs of y against x were to be plotted for eachof the following, state (i) the gradient, and (ii) they-axis intercept.(a) y = −5x + 3 (b) y = 7x(c) 2y + 4 = 5x (d) 5x + 2y = 6(e) 2x −y3=76(10)5. Theresistance R ohmsofacopperwinding ismea-sured at various temperatures t◦C and the resultsare as follows.R ( ) 38 47 55 62 72t (◦C) 16 34 50 64 84Plot a graph of R (vertically) against t (horizon-tally) and ﬁnd from it(a) the temperature when the resistance is 50 ,(b) the resistance when the temperature is 72◦C,(c) the gradient,(d) the equation of the graph. (10)6. x and y are two related variables and all otherletters denote constants. For the stated laws tobe veriﬁed it is necessary to plot graphs of thevariables 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,(d) the vertical axis intercept.(i) y = p +rx2(ii) y =ax+ bx (4)7. The following results give corresponding valuesof two quantities x and y which are believed to berelated by a law of the form y = ax2 + bx wherea and b are constants.y 33.9 55.5 72.8 84.1 111.4 168.1x 3.4 5.2 6.5 7.3 9.1 12.4Verify the law and determine approximate valuesof a and b.Hence determine (i) the value of y when x is 8.0and (ii) the value of x when y is 146.5 (18)8. By taking logarithms of both sides of y = k xn ,show that lg y needs to be plotted vertically andlgx needs to be plotted horizontally to producea straight line graph. Also, state the gradient andvertical-axis intercept. (6)9. By taking logarithms of both sides of y = aek xshow that ln y needs to be plotted vertically andx needs to be plotted horizontally to produce astraight line graph. Also, state the gradient andvertical-axis intercept. (6)10. Show from the following results of voltage V andadmittance Y of an electrical circuit that the lawconnecting the quantities is of the form V =kYnand determine the values of k and n.VoltageV (volts) 2.88 2.05 1.60 1.22 0.96Admittance,Y(siemens) 0.52 0.73 0.94 1.23 1.57(12)
• 164 Basic Engineering Mathematics11. The current i ﬂowing in a discharging capacitorvaries with time t as shown.i (mA) 50.0 17.0 5.8 1.7 0.58 0.24t (ms) 200 255 310 375 425 475Show that these results are connected by thelaw of the form i = IetT where I is the initialcurrent ﬂowing and T is a constant. Determineapproximate values of constants I and T. (15)12. Solve, correct to 1 decimal place, the quadraticequation 2x2 − 6x − 9 = 0 by plotting values of xfrom x = −2 to x = 5. (8)13. Plot the graph of y = x3 + 4x2 + x − 6 forvalues of x between x = −4 and x = 2.Hence determine the roots of the equationx3+ 4x2+ x − 6 = 0. (9)14. Plot a graph of y = 2x2 from x = −3 to x = +3and hence solve the following equations.(a) 2x2 − 8 = 0(b) 2x2 − 4x − 6 = 0 (10)
• Chapter 20Angles and triangles20.1 IntroductionTrigonometry is a subject that involves the measurementof sides and angles of triangles and their relationship toeach other. This chapter involves the measurement ofangles and introduces types of triangle.20.2 Angular measurementAn angle is the amount of rotation between two straightlines. Angles may be measured either in degrees or inradians.If a circle is divided into 360 equal parts, then each partis called 1 degree and is written as 1◦i.e. 1 revolution = 360◦or 1 degree is1360th of a revolutionSome angles are given special names.• Any angle between 0◦and 90◦is called an acuteangle.• An angle equal to 90◦ is called a right angle.• Any angle between 90◦ and 180◦ is called an obtuseangle.• Any angle greater than 180◦ and less than 360◦ iscalled a reﬂex angle.• An angle of 180◦ lies on a straight line.• If two angles add up to 90◦ they are called comple-mentary angles.• If two angles add up to 180◦ they are called supple-mentary angles.• Parallel lines are straight lineswhich are in thesameplane and never meet. Such lines are denoted byarrows, as in Figure 20.1.• A straight line which crosses two parallel lines iscalled a transversal (see MN in Figure 20.1).PRMNgh efadbcQSFigure 20.1With reference to Figure 20.1,(a) a = c, b = d, e = g and f = h. Such pairs ofangles are called vertically opposite angles.(b) a = e, b = f , c = g and d = h. Such pairs ofangles are called corresponding angles.(c) c = e and b = h. Such pairs of angles are calledalternate angles.(d) b + e = 180◦ and c + h = 180◦. Such pairs ofangles are called interior angles.20.2.1 Minutes and secondsOne degree may be sub-divided into 60 parts, calledminutes.i.e. 1 degree = 60 minuteswhich is written as 1◦= 60 .DOI: 10.1016/B978-1-85617-697-2.00020-X
• Angles and triangles 167(i) 47 + 34 = 81(ii) Since 60 = 1 ,81 = 1 21(iii) The 21 is placed in the seconds column and 1is carried in the minutes column.(iv) 51 + 27 + 1 = 79(v) Since 60 = 1◦,79 = 1◦19(vi) The 19 is placed in the minutes column and 1◦is carried in the degrees column.(vii) 19◦ + 63◦ + 1◦ (carried) = 83◦. Place 83◦ in thedegrees column.This answer can be obtained using the calculator asfollows.1. Enter 19 2. Press ◦’ ’ ’ 3. Enter 514. Press ◦ ’ ’ ’ 5. Enter 47 6. Press ◦ ’ ’ ’7. Press + 8. Enter 63 9. Press ◦ ’ ’ ’10. Enter 27 11. Press ◦ ’ ’ ’12. Enter 34 13. Press ◦ ’ ’ ’14. Press = Answer = 83◦19 21Thus, 19◦51 47 + 63◦27 34 = 83◦19 21 .Problem 4. Convert 39◦ 27 to degrees in decimalform39◦27 = 3927◦6027◦60= 0.45◦by calculatorHence, 39◦27 = 3927◦60= 39.45◦This answer can be obtained using the calculator asfollows.1. Enter 39 2. Press ◦ ’ ’ ’ 3. Enter 274. Press ◦ ’ ’ ’ 5. Press = 6. Press ◦ ’ ’ ’Answer = 39.45◦Problem 5. Convert 63◦ 26 51 to degrees indecimal form, correct to 3 decimal places63◦26 51 = 63◦265160= 63◦26.8563◦26.85 = 6326.85◦60= 63.4475◦Hence, 63◦26 51 = 63.448◦correct to 3 decimalplaces.This answer can be obtained using the calculator asfollows.1. Enter 63 2. Press ◦ ’ ’ ’ 3. Enter 264. Press ◦ ’ ’ ’ 5. Enter 51 6. Press ◦ ’ ’ ’7. Press = 8. Press ◦ ’ ’ ’ Answer = 63.4475◦Problem 6. Convert 53.753◦ to degrees, minutesand seconds0.753◦= 0.753 × 60 = 45.180.18 = 0.18 × 60 = 11to the nearest secondHence, 53.753◦= 53◦45 11This answer can be obtained using the calculator asfollows.1. Enter 53.753 2. Press =3. Press ◦ ’ ’ ’ Answer = 53◦45 10.8Now try the following Practice ExercisePracticeExercise 76 Angular measurement(answers on page 348)1. Evaluate 52◦39 + 29◦482. Evaluate 76◦31 − 48◦373. Evaluate 77◦22 + 41◦36 − 67◦474. Evaluate 41◦37 16 + 58◦29 365. Evaluate 54◦37 42 − 38◦53 256. Evaluate79◦26 19 −45◦58 56 +53◦21 387. Convert 72◦33 to degrees in decimal form.8. Convert 27◦45 15 to degrees correct to 3decimal places.9. Convert 37.952◦ to degrees and minutes.10. Convert 58.381◦ to degrees, minutes andseconds.Here are some further worked examples on angularmeasurement.Problem 7. State the general name given to thefollowing angles: (a) 157◦ (b) 49◦ (c) 90◦ (d) 245◦(a) Any angle between 90◦ and 180◦ is called anobtuse angle.Thus, 157◦ is an obtuse angle.(b) Any angle between 0◦ and 90◦ is called an acuteangle.
• 168 Basic Engineering MathematicsThus, 49◦is an acute angle.(c) An angle equal to 90◦ is called a right angle.(d) Any angle greater than 180◦ and less than 360◦ iscalled a reﬂex angle.Thus, 245◦is a reﬂex angle.Problem 8. Find the angle complementary to48◦ 39If two angles add up to 90◦ they are called comple-mentary angles. Hence, the angle complementary to48◦39 is90◦− 48◦39 = 41◦21Problem 9. Find the angle supplementary to74◦25If two angles add up to 180◦ they are called supple-mentary angles. Hence, the angle supplementary to74◦25 is180◦− 74◦25 = 105◦35Problem 10. Evaluate angle θ in each of thediagrams shown in Figure 20.3(a) (b)418538 448␪␪Figure 20.3(a) The symbol shown in Figure 20.4 is called a rightangle and it equals 90◦.Figure 20.4Hence, from Figure 20.3(a),θ + 41◦ = 90◦from which, θ = 90◦ −41◦ = 49◦(b) An angle of 180◦lies on a straight line.Hence, from Figure 20.3(b),180◦ = 53◦ + θ + 44◦from which, θ = 180◦ − 53◦ − 44◦ = 83◦Problem 11. Evaluate angle θ in the diagramshown in Figure 20.539Њ58Њ64Њ108Њ␪Figure 20.5There are 360◦ in a complete revolution of a circle.Thus, 360◦ = 58◦ + 108◦ + 64◦ + 39◦ + θfromwhich, θ = 360◦ −58◦ −108◦ −64◦ −39◦ =91◦Problem 12. Two straight lines AB and CDintersect at 0. If ∠AOC is 43◦, ﬁnd ∠AOD,∠DOB and ∠BOCFrom Figure 20.6, ∠AOD is supplementary to ∠AOC.Hence, ∠AOD = 180◦ − 43◦ = 137◦.When two straight linesintersect, the vertically oppositeangles are equal.Hence, ∠DOB = 43◦and ∠BOC 137◦ADBC4380Figure 20.6
• 170 Basic Engineering Mathematics4. State the general name given to an angle of90◦.5. Determine the angles complementary to thefollowing.(a) 69◦ (b) 27◦37 (c) 41◦3 436. Determine the angles supplementary to(a) 78◦(b) 15◦(c) 169◦41 117. Find the values of angle θ in diagrams(a) to (i) of Figure 20.10.1208(a)␪558 508(c)␪608 808(f)␪␪(e)7081508␪378(d)␪(g)50.78 ␪(h)1118688458578␪(i)368␪708(b)␪Figure 20.108. With reference to Figure 20.11, what is thename given to the line XY? Give examples ofeach of the following.xy14583276Figure 20.11(a) vertically opposite angles(b) supplementary angles(c) corresponding angles(d) alternate angles9. In Figure 20.12, ﬁnd angle α.137Њ29Ј16Њ49Ј␣Figure 20.1210. In Figure 20.13, ﬁnd angles a,b and c.acb298698Figure 20.1311. Find angle β in Figure 20.14.133Њ98Њ␤Figure 20.1412. Convert 76◦ to radians, correct to 3 decimalplaces.13. Convert 34◦40 to radians, correct to 3 deci-mal places.14. Convert 0.714 rad to degrees and minutes.
• Angles and triangles 17120.3 TrianglesA triangle is a ﬁgure enclosed by three straight lines.The sum of the three angles of a triangle is equalto 180◦.20.3.1 Types of triangleAn acute-angled triangle is one in which all the anglesare acute; i.e., all the angles are less than 90◦. Anexample is shown in triangle ABC in Figure 20.15(a).A right-angled triangle is one which contains a rightangle; i.e., one in which one of the angles is 90◦. Anexample is shown in triangle DEF in Figure 20.15(b).678 598508408548AB C(a) (b)DE FFigure 20.15An obtuse-angled triangle is one which contains anobtuse angle; i.e., one angle which lies between 90◦and 180◦. An example is shown in triangle PQR inFigure 20.16(a).An equilateral triangleis one in which all the sides andall the angles are equal; i.e., each is 60◦. An example isshown in triangle ABC in Figure 20.16(b).PQ R1318278228(a)608608 608AB C(b)Figure 20.16An isosceles triangle is one in which two angles andtwo sides are equal. An example is shown in triangleEFG in Figure 20.17(a).A scalene triangle is one with unequal angles andtherefore unequal sides. An example of an acuteangled scalene triangle is shown in triangle ABC inFigure 20.17(b).(a) (b)ABC438828558FE G308758758Figure 20.17BACc ba␪Figure 20.18With reference to Figure 20.18,(a) Angles A, B and C are called interior angles ofthe triangle.(b) Angle θ is called an exterior angle of the triangleand is equal to the sum of the two opposite interiorangles; i.e., θ = A + C.(c) a + b + c is called the perimeter of the triangle.AC BbacFigure 20.19
• 172 Basic Engineering MathematicsA right-angled triangle ABC is shown in Figure 20.19.The point of intersection of two lines is called a vertex(plural vertices); the three vertices of the triangle arelabelled as A, B and C, respectively. The right angleis angle C. The side opposite the right angle is giventhe special name of the hypotenuse. The hypotenuse,length AB in Figure 20.19, is always the longest side ofa right-angled triangle. With reference to angle B, ACis the opposite side and BC is called the adjacent side.With reference to angle A, BC is the opposite side andAC is the adjacent side.Often sides of a triangle are labelled with lower caseletters, a being the side opposite angle A,b being theside opposite angle B and c being the side oppositeangle C. So, in the triangle ABC, length AB = c, lengthBC = a and length AC = b. Thus, c is the hypotenusein the triangle ABC.∠ is the symbol used for ‘angle’. For example, in thetriangle shown, ∠C = 90◦. Another way of indicatingan angle is to use all three letters. For example, ∠ABCactually means ∠B; i.e., we take the middle letter as theangle. Similarly, ∠BAC means ∠A and ∠ACB means∠C.Here are some worked examples to help us understandmore about triangles.Problem 18. Name the types of triangle shown inFigure 20.2022.62.82.52.52.12.139810785182(b)(d) (e)(c)(a)2Figure 20.20(a) Equilateral triangle (since all three sides areequal).(b) Acute-angled scalene triangle (since all theangles are less than 90◦).(c) Right-angled triangle (39◦ + 51◦ = 90◦; hence,the third angle must be 90◦, since there are 180◦in a triangle).(d) Obtuse-angled scalene triangle (since one of theangles lies between 90◦ and 180◦).(e) Isosceles triangle (since two sides are equal).Problem 19. In the triangle ABC shown inFigure 20.21, with reference to angle θ, which sideis the adjacent?BA␪CFigure 20.21The triangle is right-angled; thus, side AC is thehypotenuse. With reference to angle θ, the opposite sideis BC. The remaining side, AB, is the adjacent side.Problem 20. In the triangle shown inFigure 20.22, determine angle θ568␪Figure 20.22The sum of the three angles of a triangle is equal to180◦.The triangle is right-angled. Hence,90◦ + 56◦ + ∠θ = 180◦from which, ∠θ = 180◦− 90◦− 56◦= 34◦.Problem 21. Determine the value of θ and α inFigure 20.23
• Angles and triangles 17362Њ15ЊABDEC ␪␣Figure 20.23In triangle ABC,∠A + ∠B + ∠C = 180◦ (the angles ina triangle add up to 180◦).Hence, ∠C = 180◦ − 90◦ − 62◦ = 28◦. Thus,∠DCE = 28◦ (vertically opposite angles).θ = ∠DCE + ∠DEC (the exterior angle of a triangle isequal to the sum of the two opposite interior angles).Hence, ∠θ = 28◦ + 15◦ = 43◦.∠α and ∠DEC are supplementary; thus,α = 180◦ − 15◦ = 165◦.Problem 22. ABC is an isosceles triangle inwhich the unequal angle BAC is 56◦. AB is extendedto D as shown in Figure 20.24. Find, for thetriangle, ∠ABC and ∠ACB. Also, calculate ∠DBC568AB CDFigure 20.24Since triangle ABC is isosceles, two sides – i.e. AB andAC – are equal and two angles – i.e. ∠ABC and ∠ACB –are equal.The sum of the three angles of a triangle is equal to 180◦.Hence, ∠ABC + ∠ACB = 180◦ − 56◦ = 124◦.Since ∠ABC = ∠ACB then∠ABC = ∠ACB =124◦2= 62◦.An angle of 180◦ lies on a straight line; hence,∠ABC + ∠DBC = 180◦ from which,∠DBC = 180◦ − ∠ABC = 180◦ − 62◦ = 118◦.Alternatively, ∠DBC = ∠A + ∠C (exterior angleequals sum of two interior opposite angles),i.e. ∠DBC = 56◦ + 62◦ = 118◦.Problem 23. Find angles a,b,c,d and e inFigure 20.25558628becdaFigure 20.25a = 62◦and c = 55◦(alternate angles between parallellines).55◦ + b + 62◦ = 180◦ (angles in a triangle add up to180◦); hence, b = 180◦ − 55◦ − 62◦ = 63◦.b = d = 63◦ (alternate angles between parallel lines).e + 55◦ + 63◦ = 180◦ (angles in a triangle add up to180◦); hence, e = 180◦ − 55◦ − 63◦ = 62◦.Check: e = a = 62◦ (corresponding angles betweenparallel lines).Now try the following Practice ExercisePracticeExercise 78 Triangles (answers onpage 348)1. Namethetypesoftriangleshown in diagrams(a) to (f) in Figure 20.26.(a) (b)(e) (f)(c)81Њ15Њ48Њ66Њ45Њ45Њ(d)97Њ39Њ60Њ5553Њ37ЊFigure 20.26
• 174 Basic Engineering Mathematics2. Find the angles a to f in Figure 20.27.10581058def(c)328114°b c(b)578838 a(a)Figure 20.273. In the triangle DEF of Figure 20.28, whichside is the hypotenuse? With reference toangle D, which side is the adjacent?388DE FFigure 20.284. In triangle DEF of Figure 20.28, determineangle D.5. MNO is an isosceles triangle in whichthe unequal angle is 65◦ as shown inFigure 20.29. Calculate angle θ.658ONMP␪Figure 20.296. Determine ∠φ and ∠x in Figure 20.30.AB CEDx588198␾Figure 20.307. In Figure 20.31(a) and (b), ﬁnd angles w,x, yand z. What is the name given to the types oftriangle shown in (a) and (b)?(a) (b)1108 1108x y708z2cm2cm␻Figure 20.318. Find the values of angles a to g inFigure 20.32(a) and (b).
• Angles and triangles 175(a) (b)568299148419ab6881318cd egfFigure 20.329. Find the unknown angles a to k inFigure 20.33.9981258228 dbacke g hfijFigure 20.3310. Triangle ABC has a right angle at B and∠BAC is 34◦. BC is produced to D. If thebisectors of ∠ABC and ∠ACD meet at E,determine ∠BEC.11. Ifin Figure20.34 triangleBCD isequilateral,ﬁnd the interior angles of triangle ABE.978ABCDEFigure 20.3420.4 Congruent trianglesTwo triangles are said to be congruent if they are equalin all respects; i.e., three angles and three sides in onetriangle are equal to three angles and three sides in theother triangle. Two triangles are congruent if(a) the three sides of one are equal to the three sidesof the other (SSS),(b) two sides of one are equal to two sides of the otherand the angles included by these sides are equal(SAS),(c) two angles of the one are equal to two angles ofthe other and any side of the ﬁrst is equal to thecorresponding side of the other (ASA), or(d) their hypotenuses are equal and one other side ofone is equal to the corresponding side of the other(RHS).Problem 24. State which of the pairs of trianglesshown in Figure 20.35 are congruent and name theirsequence(c)MNPOQR(d)STUVXW(e)FD EBAC(a) (b)ABCE DF KLJHGIFigure 20.35(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 thatany side coincides.(e) Congruent ABC, FED (side, side, side; i.e.,SSS).
• 176 Basic Engineering MathematicsProblem 25. In Figure 20.36, triangle PQR isisosceles with Z, the mid-point of PQ. Prove thattriangles PXZ and QYZ are congruent and thattriangles RXZ and RYZ are congruent. Determinethe values of angles RPZ and RXZP QRY678288288XZFigure 20.36Since triangle PQR is isosceles, PR = RQ and thus∠QPR = ∠RQP.∠RXZ = ∠QPR + 28◦ and ∠RYZ = ∠RQP + 28◦ (ext-erioranglesofatriangleequal thesumofthetwo interioropposite angles). Hence, ∠RXZ = ∠RYZ.∠PXZ = 180◦ − ∠RXZ and ∠QYZ = 180◦ − ∠RYZ.Thus, ∠PXZ = ∠QYZ.Triangles PXZ and QYZ are congruent since∠XPZ = ∠YQZ, PZ = ZQ and ∠XZP = ∠YZQ (ASA).Hence, XZ = YZ.Triangles PRZ and QRZ are congruent sincePR = RQ, ∠RPZ = ∠RQZ and PZ = ZQ (SAS).Hence, ∠RZX = ∠RZY.Triangles RXZ and RYZ are congruent since∠RXZ = ∠RYZ, XZ = YZ and ∠RZX = ∠RZY (ASA).∠QRZ = 67◦ and thus ∠PRQ = 67◦ + 67◦ = 134◦.Hence, ∠RPZ = ∠RQZ =180◦ − 134◦2= 23◦.∠RXZ = 23◦ + 28◦ = 51◦(external angle of a tri-angle equals the sum of the two interior oppositeangles).Now try the following Practice ExercisePracticeExercise 79 Congruent triangles(answers on page 349)1. State which of the pairs of triangles inFigure 20.37 are congruent and name theirsequences.ACB DM(a) (b) (c)(e)(d)RQ STVWUXYZKLOPNEIGHJFFigure 20.372. In a triangle ABC, AB = BC and D and Eare points on AB and BC, respectively, suchthat AD = CE. Show that triangles AEB andCDB are congruent.20.5 Similar trianglesTwo triangles are said to be similar if the angles of onetriangle are equal to the angles ofthe other triangle. Withreference to Figure 20.38, triangles ABC and PQR aresimilar and the corresponding sides are in proportion toeach other,i.e.pa=qb=rc658 588578ca CBAb658 588578rp RQPqFigure 20.38Problem 26. In Figure 20.39, ﬁnd the length ofside a
• Angles and triangles 177508708c 512.0cmBaCA508608f 55.0cmd5 4.42cmE FDFigure 20.39In triangle ABC,50◦ + 70◦ + ∠C = 180◦, from which∠C = 60◦.In triangle DEF,∠E = 180◦ − 50◦ − 60◦ = 70◦.Hence, triangles ABC and DEF are similar, since theirangles are the same. Since corresponding sides are inproportion to each other,ad=cfi.e.a4.42=12.05.0Hence, side, a =12.05.0(4.42) = 10.61cm.Problem 27. In Figure 20.40, ﬁnd the dimensionsmarked r and p558358x 57.44cmq56.82cmz512.97cmy510.63cmYQPRZXprFigure 20.40In triangle PQR,∠Q = 180◦ − 90◦ − 35◦ = 55◦.In triangle XYZ,∠X = 180◦− 90◦− 55◦= 35◦.Hence, triangles PQR and ZYX are similar since theirangles are the same. The triangles may he redrawn asshown in Figure 20.41.358558y510.63 cmq 56.82cmx57.44cmz512.97cmZ XY358558P RprQFigure 20.41By proportion:pz=rx=qyi.e.p12.97=r7.44=6.8210.63from which, r = 7.446.8210.63= 4.77cmBy proportion:pz=qyi.e.p12.97=6.8210.63Hence, p = 12.976.8210.63= 8.32cmProblem 28. In Figure 20.42, show that trianglesCBD and CAE are similar and hence ﬁnd the lengthof CD and BDB106912CDEAFigure 20.42Since BD is parallel to AE then ∠CBD = ∠CAE and∠CDB = ∠CEA (corresponding angles between paral-lel lines). Also, ∠C is common to triangles CBD andCAE.Since the angles in triangle CBD are the same as intriangle CAE, the triangles are similar. Hence,by proportion:CBCA=CDCE=BDAEi.e.96 + 9=CD12, from whichCD = 12915= 7.2cmAlso,915=BD10, from whichBD = 10915= 6cm
• 178 Basic Engineering MathematicsProblem 29. A rectangular shed 2m wide and3m high stands against a perpendicular building ofheight 5.5m. A ladder is used to gain access to theroof of the building. Determine the minimumdistance between the bottom of the ladder and theshedA side view is shown in Figure 20.43, where AFis the minimum length of the ladder. Since BDand CF are parallel, ∠ADB = ∠DFE (correspond-ing angles between parallel lines). Hence, trianglesBAD and EDF are similar since their angles are thesame.AB = AC − BC = AC − DE = 5.5 − 3 = 2.5mBy proportion:ABDE=BDEFi.e.2.53=2EFHence, EF = 232.5= 2.4m = minimum distancefrom bottom of ladder to the shed.3m2m 5.5mShedDE CBAFFigure 20.43Now try the following Practice ExercisePracticeExercise 80 Similar triangles(answers on page 349)1. In Figure 20.44, ﬁnd the lengths x and y.111Њ32Њ32Њ 37Њ25.69mm4.74mm7.36mm14.58mmxyFigure 20.442. PQR is an equilateral triangle of side 4cm.When PQ and PR are produced to S and T,respectively, ST is found to be parallel withQR. If PS is 9cm, ﬁnd the length of ST. Xis a point on ST between S and T such thatthe line PX is the bisector of ∠SPT. Find thelength of PX.3. In Figure 20.45, ﬁnd(a) the length of BC when AB = 6cm,DE = 8cm and DC = 3cm,(b) the length of DE when EC = 2cm,AC = 5cm and AB = 10cm.D ECBAFigure 20.454. In Figure 20.46, AF = 8m,AB = 5m andBC = 3m. Find the length of BD.DECBA FFigure 20.46
• Angles and triangles 17920.6 Construction of trianglesTo construct any triangle, the following drawing instru-ments are needed:(a) ruler and/or straight edge(b) compass(c) protractor(d) pencil.Here are some worked problems to demonstrate triangleconstruction.Problem 30. Construct a triangle whose sides are6cm,5cm and 3cmDFEC GBA 6cmFigure 20.47With reference to Figure 20.47:(i) Draw a straight line of any length and, with a pairof compasses, mark out 6cm length and label itAB.(ii) Set compass to 5cm and with centre at A describearc DE.(iii) Set compass to 3cm and with centre at B describearc FG.(iv) The intersection of the two curves at C is the ver-tex of the required triangle. Join AC and BC bystraight lines.It may be proved by measurement that the ratio of theangles of a triangle is not equal to the ratio of the sides(i.e., in this problem, the angle opposite the 3cm side isnot equal to half the angle opposite the 6cm side).Problem 31. Construct a triangle ABC such thata = 6cm,b = 3cm and ∠C = 60◦With reference to Figure 20.48:CAB a56cm608b53cmFigure 20.48(i) Draw a line BC,6cm long.(ii) Using a protractor centred at C, make an angle of60◦ to BC.(iii) From C measure a length of 3cm and label A.(iv) Join B to A by a straight line.Problem 32. Construct a triangle PQR given thatQR = 5cm,∠Q = 70◦ and ∠R = 44◦Q RPQ9R9708 4485cmFigure 20.49With reference to Figure 20.49:(i) Draw a straight line 5cm long and label it QR.(ii) Use a protractor centred at Q and make an angleof 70◦. Draw QQ .(iii) Use a protractor centred at R and make an angleof 44◦. Draw RR .(iv) The intersection of QQ and RR forms the vertexP of the triangle.Problem 33. Construct a triangle XYZ given thatXY = 5cm, the hypotenuse YZ = 6.5cm and∠X = 90◦
• 180 Basic Engineering MathematicsA9B A XQCP SZUVRYFigure 20.50With reference to Figure 20.50:(i) Draw a straight line 5cm long and label it XY.(ii) Produce XY any distance to B. With compass cen-tred at X make an arc at A and A . (The length XAand XA is arbitrary.) With compass centred at Adraw the arc PQ. With the same compass settingand centred at A , draw the arc RS. Join the inter-section of the arcs, C to X, and a right angle toXY is produced at X. (Alternatively, a protractorcan be used to construct a 90◦ angle.)(iii) Thehypotenuseisalwaysoppositetheright angle.Thus, YZ is opposite ∠X. Using a compasscentred at Y and set to 6.5cm, describe thearc UV.(iv) The intersection of the arc UV withXC produced,forms the vertex Z of the required triangle. JoinYZ with a straight line.Now try the following Practice ExercisePracticeExercise 81 Construction oftriangles (answers on page 349)In the following, construct the triangles ABC forthe given sides/angles.1. a = 8cm,b = 6cm and c = 5cm.2. a = 40mm,b = 60mm and C = 60◦.3. a = 6cm,C = 45◦ and B = 75◦.4. c = 4cm, A = 130◦ and C = 15◦.5. a = 90mm, B = 90◦,hypotenuse = 105mm.
• Chapter 21Introduction to trigonometry21.1 IntroductionTrigonometry is a subject that involves the measurementof sides and angles of triangles and their relationship toeach other.The theorem of Pythagoras and trigonometric ratiosare used with right-angled triangles only. However,there are many practical examples in engineeringwhere knowledge of right-angled triangles is veryimportant.In this chapter, three trigonometric ratios – i.e. sine,cosine and tangent – are deﬁned and then evaluatedusing a calculator. Finally, solving right-angled trian-gle problems using Pythagoras and trigonometric ratiosis demonstrated, together with some practical examplesinvolving angles of elevation and depression.21.2 The theorem of PythagorasThe theorem of Pythagoras states:In any right-angled triangle, the square of thehypotenuse is equal to the sum of the squares of theother two sides.In the right-angled triangle ABC shown in Figure 21.1,this meansb2= a2+ c2(1)BACabcFigure 21.1If the lengths of any two sides of a right-angled triangleare known, the length of the third side may be calculatedby Pythagoras’ theorem.From equation (1): b = a2 + c2Transposing equation (1) for a gives a2 = b2 − c2, fromwhich a =√b2 − c2Transposing equation (1) for c gives c2 = b2 − a2, fromwhich c =√b2 − a2Here are some worked problems to demonstrate thetheorem of Pythagoras.Problem 1. In Figure 21.2, ﬁnd the length of BCBACab54cmc53 cmFigure 21.2From Pythagoras, a2= b2+ c2i.e. a2= 42+ 32= 16 + 9 = 25Hence, a =√25 = 5cm.√25 = ±5 but in a practical example likethisan answerof a = −5cm has no meaning, so we take only thepositive answer.Thus a = BC = 5cm.DOI: 10.1016/B978-1-85617-697-2.00021-1
• 182 Basic Engineering MathematicsPQR is a 3, 4, 5 triangle. There are not many right-angled triangles which have integer values (i.e. wholenumbers) for all three sides.Problem 2. In Figure 21.3, ﬁnd the length of EFE d FDf55cme513cmFigure 21.3By Pythagoras’ theorem, e2= d2+ f 2Hence, 132= d2+ 52169 = d2+ 25d2= 169 − 25 = 144Thus, d =√144 = 12cmi.e. d = EF = 12cmDEF is a 5, 12, 13 triangle, another right-angledtriangle which has integer values for all three sides.Problem 3. Two aircraft leave an airﬁeld at thesame time. One travels due north at an averagespeed of 300km/h and the other due west at anaverage speed of 220km/h. Calculate their distanceapart after 4hoursAfter 4 hours, the ﬁrst aircraft has travelled4 × 300 = 1200km due northand the second aircraft has travelled4 × 220 = 880km due west,as shown in Figure 21.4. The distance apart after4hours = BC.ABNEWSC1200km880kmFigure 21.4From Pythagoras’ theorem,BC2= 12002+ 8802= 1440000+ 774400 = 2214400and BC =√2214400 = 1488km.Hence, distance apart after 4 hours = 1488km.Now try the following Practice ExercisePracticeExercise 82 Theorem ofPythagoras (answers on page 350)1. Find the length of side x in Figure 21.5.41cm40cmxFigure 21.52. Find the length of side x in Figure 21.6(a).3. Find the length of side x in Figure 21.6(b),correct to 3 signiﬁcant ﬁgures.25 m(a)7 m4.7 mm8.3 mm(b)xxFigure 21.64. In a triangle ABC, AB = 17cm, BC = 12 cmand ∠ABC = 90◦. Determine the length ofAC, correct to 2 decimal places.5. A tent peg is 4.0m away from a 6.0m hightent. What length of rope, correct to the
• Introduction to trigonometry 183nearest centimetre, runs from the top of thetent to the peg?6. In a triangle ABC, ∠B is a right angle,AB = 6.92 cm and BC = 8.78cm. Find thelength of the hypotenuse.7. In a triangle CDE, D = 90◦,CD = 14.83mm and CE = 28.31mm.Determine the length of DE.8. Show that if a triangle has sides of 8, 15 and17cm it is right-angled.9. Triangle PQR is isosceles, Q being a rightangle. If the hypotenuse is 38.46cm ﬁnd (a)the lengths of sides PQ and QR and (b) thevalue of ∠QPR.10. Aman cycles24kmduesouth and then 20kmdue east. Another man, starting at the sametime as the ﬁrst man, cycles 32km due eastand then 7km due south. Find the distancebetween the two men.11. A ladder 3.5m long is placed against a per-pendicular wall with its foot 1.0m from thewall. How far up the wall (to the nearest cen-timetre) does the ladder reach? If the foot ofthe ladder is now moved 30cm further awayfrom the wall, how far does the top of theladder fall?12. Two ships leave a port at the same time. Onetravels due west at 18.4knots and the otherdue south at 27.6knots. If 1knot = 1 nauticalmile per hour, calculate how far apart the twoships are after 4 hours.13. Figure 21.7 shows a bolt rounded off at oneend. Determine the dimension h.R5 45mmhr516mmFigure 21.714. Figure 21.8 shows a cross-section of a com-ponent that is to be made from a round bar.If the diameter of the bar is 74mm, calculatethe dimension x.72mm␾74mmxFigure 21.821.3 Sines, cosines and tangentsWith reference to angle θ in the right-angled triangleABC shown in Figure 21.9,sineθ =opposite sidehypotenuse‘Sine’ is abbreviated to ‘sin’, thus sinθ =BCACHypotenuseAdjacentA␪BCOppositeFigure 21.9Also, cosineθ =adjacent sidehypotenuse‘Cosine’ is abbreviated to ‘cos’, thus cosθ =ABACFinally, tangentθ =opposite sideadjacent side‘Tangent’ is abbreviated to ‘tan’, thus tanθ =BCABThese three trigonometric ratios only apply to right-angled triangles. Remembering these three equationsis very important and the mnemonic ‘SOH CAH TOA’is one way of remembering them.
• 184 Basic Engineering MathematicsSOH indicates sin = opposite ÷ hypotenuseCAH indicates cos = adjacent ÷ hypotenuseTOA indicates tan = opposite ÷ adjacentHere are some worked problems to help familiarizeourselves with trigonometric ratios.Problem 4. In triangle PQR shown inFigure 21.10, determine sin θ, cos θ and tan θ␪PQ R51213Figure 21.10sinθ =opposite sidehypotenuse=PQPR=513= 0.3846cosθ =adjacent sidehypotenuse=QRPR=1213= 0.9231tanθ =opposite sideadjacent side=PQQR=512= 0.4167Problem 5. In triangle ABC of Figure 21.11,determine length AC,sin C,cosC,tanC,sin A,cos A and tan AAB C3.47 cm4.62 cmFigure 21.11By Pythagoras, AC2= AB2+ BC2i.e. AC2= 3.472+ 4.622from which AC = 3.472 + 4.622 = 5.778cmsinC =opposite sidehypotenuse=ABAC=3.475.778= 0.6006cosC =adjacent sidehypotenuse=BCAC=4.625.778= 0.7996tanC =opposite sideadjacent side=ABBC=3.474.62= 0.7511sinA =opposite sidehypotenuse=BCAC=4.625.778= 0.7996cosA =adjacent sidehypotenuse=ABAC=3.475.778= 0.6006tanA =opposite sideadjacent side=BCAB=4.623.47= 1.3314Problem 6. If tan B =815, determine the value ofsin B,cos B,sin A and tan AA right-angled triangle ABC is shown in Figure 21.12.If tan B =815, then AC = 8 and BC = 15.ACB815Figure 21.12By Pythagoras, AB2= AC2+ BC2i.e. AB2= 82+ 152from which AB = 82 + 152 = 17sinB =ACAB=817or 0.4706cosB =BCAB=1517or 0.8824sinA =BCAB=1517or 0.8824tanA =BCAC=158or 1.8750Problem 7. Point A lies at co-ordinate (2, 3) andpoint B at (8, 7). Determine (a) the distance AB and(b) the gradient of the straight line AB
• Introduction to trigonometry 185BA20 4(a) (b)6 88f(x)76432BCA20 4 6 88f(x)76432␪Figure 21.13(a) Points A and B are shown in Figure 21.13(a).In Figure 21.13(b), the horizontal and verticallines AC and BC are constructed. Since ABC isa right-angled triangle, and AC = (8 − 2) = 6 andBC = (7 − 3) = 4, by Pythagoras’ theorem,AB2= AC2+ BC2= 62+ 42and AB = 62 + 42 =√52= 7.211 correct to 3decimal places.(b) The gradient of AB is given by tanθ, i.e.gradient = tanθ =BCAC=46=23Now try the following Practice ExercisePracticeExercise 83 Trigonometric ratios(answers on page 349)1. Sketch a triangle XYZ such that∠Y = 90◦, XY = 9cm and YZ = 40 cm.Determine sin Z,cos Z,tan X and cos X.2. In triangle ABC shown in Figure 21.14, ﬁndsin A,cos A,tan A,sin B,cos B and tan B.BC35AFigure 21.143. If cos A =1517, ﬁnd sin A and tan A, in fractionform.4. If tan X =15112, ﬁnd sin X and cos X, in frac-tion form.5. For the right-angled triangle shown inFigure 21.15, ﬁnd (a) sinα (b) cosθ (c) tanθ.␪␣ 17815Figure 21.156. If tanθ =724, ﬁnd sinθ and cosθ in fractionform.7. Point P lies at co-ordinate (−3,1) and pointQ at (5,−4). Determine(a) the distance PQ.(b) the gradient of the straight line PQ.21.4 Evaluating trigonometric ratiosof acute anglesThe easiest way to evaluate trigonometric ratios of anyangle is to use a calculator. Use a calculator to check thefollowing (each correct to 4 decimal places).sin29◦ = 0.4848 sin53.62◦ = 0.8051cos67◦ = 0.3907 cos83.57◦ = 0.1120tan34◦ = 0.6745 tan67.83◦ = 2.4541sin67◦43 = sin674360◦= sin67.7166666...◦= 0.9253cos13◦28 = cos132860◦= cos13.466666...◦= 0.9725tan56◦54 = tan565460◦= tan56.90◦= 1.5340If we know the value ofa trigonometricratio and need toﬁnd the angle we use the inverse function on our calcu-lators. For example, using shift and sin on our calculatorgives sin−1(If, for example, we know the sine of an angle is 0.5 thenthe value of the angle is given bysin−10.5 = 30◦(Check that sin30◦= 0.5)Similarly, ifcosθ = 0.4371 then θ = cos−10.4371 = 64.08◦
• Introduction to trigonometry 187Problem 16. In triangle EFG in Figure 21.16,calculate angle GEF G2.30cm8.71cmFigure 21.16With reference to ∠G, the two sides of the trianglegivenare the oppositeside EF and the hypotenuse EG; hence,sine is used,i.e. sin G =2.308.71= 0.26406429...from which, G = sin−10.26406429...i.e. G = 15.311360◦Hence, ∠G = 15.31◦or 15◦19 .Problem 17. Evaluate the following expression,correct to 3 signiﬁcant ﬁgures4.2tan49◦26 − 3.7sin66◦17.1cos29◦34By calculator:tan49◦26 = tan 492660◦= 1.1681,sin66◦1 = 0.9137 and cos29◦34 = 0.8698Hence, 4.2tan49◦26 − 3.7sin66◦17.1cos29◦34=(4.2 × 1.1681) − (3.7 × 0.9137)(7.1 × 0.8698)=4.9060 − 3.38076.1756=1.52536.1756= 0.2470 = 0.247,correct to 3 signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 84 Evaluatingtrigonometric ratios (answers on page 349)1. Determine, correct to 4 decimal places,3sin66◦41 .2. Determine, correct to 3 decimal places,5cos14◦15 .3. Determine, correct to 4 signiﬁcant ﬁgures,7tan79◦9 .4. Determine(a) sine2π3(b) cos1.681 (c) tan3.6725. Find the acute angle sin−1 0.6734 in degrees,correct to 2 decimal places.6. Find the acute angle cos−1 0.9648 in degrees,correct to 2 decimal places.7. Find the acute angle tan−1 3.4385 in degrees,correct to 2 decimal places.8. Find the acute angle sin−10.1381 in degreesand minutes.9. Find the acute angle cos−1 0.8539 in degreesand minutes.10. Find the acute angle tan−1 0.8971 in degreesand minutes.11. In the triangle shown in Figure 21.17, deter-mine angle θ, correct to 2 decimal places.␪59Figure 21.1712. In the triangle shown in Figure 21.18, deter-mine angle θ in degrees and minutes.␪823Figure 21.18
• 188 Basic Engineering Mathematics13. Evaluate, correct to 4 decimal places,4.5cos67◦34 − sin90◦2tan45◦14. Evaluate, correct to 4 signiﬁcant ﬁgures,(3sin37.83◦)(2.5tan 57.48◦)4.1cos12.56◦21.5 Solving right-angled triangles‘Solving a right-angled triangle’ means ‘ﬁnding theunknown sides and angles’. This is achieved using(a) the theorem of Pythagoras and/or(b) trigonometric ratios.Six piecesofinformation describeatrianglecompletely;i.e., three sides and three angles. As long as at leastthree facts are known, the other three can usually becalculated.Here are some worked problems to demonstrate thesolution of right-angled triangles.Problem 18. In the triangle ABC shown inFigure 21.19, ﬁnd the lengths AC and ABAB C4286.2mmFigure 21.19There is usually more than one way to solve such atriangle.In triangle ABC,tan42◦=ACBC=AC6.2(Remember SOH CAH TOA)Transposing givesAC = 6.2tan42◦= 5.583 mmcos42◦=BCAB=6.2AB, from whichAB =6.2cos42◦= 8.343 mmAlternatively, by Pythagoras, AB2 = AC2 + BC2from which AB =√AC2 + BC2 =√5.5832 + 6.22=√69.609889 = 8.343 mm.Problem 19. Sketch a right-angled triangle ABCsuch that B = 90◦, AB = 5cm and BC = 12 cm.Determine the length of AC and hence evaluatesin A,cosC and tan ATriangle ABC is shown in Figure 21.20.AB C5cm12cmFigure 21.20By Pythagoras’ theorem, AC = 52 + 122 = 13By deﬁnition: sin A =opposite sidehypotenuse=1213or 0.9231(Remember SOH CAH TOA)cosC =adjacent sidehypotenuse=1213or 0.9231and tan A =opposite sideadjacent side=125or 2.400Problem 20. In triangle PQR shown inFigure 21.21, ﬁnd the lengths of PQ and PRQ RP7.5cm388Figure 21.21tan38◦=PQQR=PQ7.5, hencePQ = 7.5tan38◦= 7.5(0.7813) = 5.860 cm
• Introduction to trigonometry 189cos38◦=QRPR=7.5PR, hencePR =7.5cos38◦=7.50.7880= 9.518 cmCheck: using Pythagoras’ theorem,(7.5)2 + (5.860)2 = 90.59 = (9.518)2.Problem 21. Solve the triangle ABC shown inFigure 21.22ACB37mm35 mmFigure 21.22To ‘solve the triangle ABC’ means ‘to ﬁnd the lengthAC and angles B and C’.sinC =3537= 0.94595, henceC = sin−10.94595 = 71.08◦B = 180◦ − 90◦ − 71.08◦ = 18.92◦ (since the angles ina triangle add up to 180◦)sin B =AC37, henceAC = 37sin18.92◦= 37(0.3242) = 12.0 mmor, using Pythagoras’ theorem, 372 = 352 + AC2 , fromwhich AC = (372 − 352) = 12.0 mm.Problem 22. Solve triangle XYZ given∠X = 90◦, ∠Y = 23◦17 and YZ = 20.0 mmIt is always advisable to make a reasonably accuratesketch so as to visualize the expected magnitudes ofunknown sides and angles. Such a sketch is shown inFigure 21.23.∠Z = 180◦− 90◦− 23◦17 = 66◦43X YZ20.0mm238179Figure 21.23sin23◦17 =XZ20.0, hence XZ = 20.0sin23◦17= 20.0(0.3953) = 7.906mmcos23◦17 =XY20.0, hence XY = 20.0cos23◦17= 20.0(0.9186) = 18.37mmCheck: using Pythagoras’ theorem,(18.37)2 + (7.906)2 = 400.0 = (20.0)2,Now try the following Practice ExercisePracticeExercise 85 Solving right-angledtriangles (answers on page 349)1. Calculate the dimensions shown as x inFigures 21.24(a) to (f), each correct to 4signiﬁcant ﬁgures.298x(c)17.0708228x13.0(a)(b)x15.0Figure 21.24
• 190 Basic Engineering Mathematics598(d)x4.30438x(e)6.0(f)538x7.0Figure 21.242. Find theunknown sidesand anglesin theright-angled triangles shown in Figure 21.25. Thedimensions shown are in centimetres.5.03.0ABC(a)Figure 21.258.04.0EDF(b)(c)HGJ28812.0(d)27815.0KLM(e)NPO4.0648(f)R SQ5.0418Figure 21.25
• Introduction to trigonometry 1913. A ladder rests against the top of the perpendi-cular wall of a building and makes an angle of73◦ with the ground. If the foot of the ladder is2m from the wall, calculate the height of thebuilding.4. Determine the length x in Figure 21.26.x56810mmFigure 21.2621.6 Angles of elevation anddepressionIf, in Figure21.27, BC represents horizontal ground andAB a vertical ﬂagpole, the angle of elevation of the topof the ﬂagpole, A, from the point C is the angle that theimaginary straight line AC must be raised (or elevated)from the horizontal CB; i.e., angle θ.ABC␪Figure 21.27PQ R␾Figure 21.28If, in Figure 21.28, PQ represents a vertical cliff andR a ship at sea, the angle of depression of the shipfrom point P is the angle through which the imaginarystraight line PR must be lowered (or depressed) fromthe horizontal to the ship; i.e., angle φ. (Note, ∠PRQ isalso φ − alternate angles between parallel lines.)Problem 23. An electricity pylon stands onhorizontal ground. At a point 80m from the base ofthe pylon, the angle of elevation of the top of thepylon is 23◦. Calculate the height of the pylon to thenearest metreFigure 21.29 shows the pylon AB and the angle ofelevation of A from point C is 23◦.80 m238ABCFigure 21.29tan23◦=ABBC=AB80Hence, height of pylon AB = 80tan23◦= 80(0.4245) = 33.96m= 34m to the nearest metre.Problem 24. A surveyor measures the angle ofelevation of the top of a perpendicular building as19◦. He moves 120m nearer to the building andﬁnds the angle of elevation is now 47◦. Determinethe height of the buildingThe building PQ and the angles of elevation are shownin Figure 21.30.PQhxRS120478 198Figure 21.30In triangle PQS, tan19◦ =hx + 120Hence, h = tan19◦(x + 120)i.e. h = 0.3443(x + 120) (1)In triangle PQR, tan47◦=hxHence, h = tan47◦(x) i.e. h = 1.0724x (2)Equating equations (1) and (2) gives0.3443(x + 120) = 1.0724x0.3443x + (0.3443)(120) = 1.0724x(0.3443)(120) = (1.0724 − 0.3443)x41.316 = 0.7281xx =41.3160.7281= 56.74m
• 192 Basic Engineering MathematicsFrom equation (2), height of building,h = 1.0724x = 1.0724(56.74) = 60.85m.Problem 25. The angle of depression of a shipviewed at a particular instant from the top of a 75mvertical cliff is 30◦. Find the distance of the shipfrom the base of the cliff at this instant. The ship issailing away from the cliff at constant speed and 1minute later its angle of depression from the top ofthe cliff is 20◦. Determine the speed of the ship inkm/hFigure 21.31 shows the cliff AB, the initial position ofthe ship at C and the ﬁnal position at D. Since the angleof depression is initially 30◦, ∠ACB = 30◦ (alternateangles between parallel lines).x75 m308208308208AB DCFigure 21.31tan30◦=ABBC=75BChence,BC =75tan30◦= 129.9m = initial positionof ship from base of cliffIn triangle ABD,tan20◦ =ABBD=75BC + CD=75129.9 + xHence, 129.9 + x =75tan20◦= 206.06mfrom which x = 206.06 − 129.9 = 76.16mThus, the ship sails 76.16m in 1 minute; i.e., 60s,Hence, speed of ship =distancetime=76.1660m/s=76.16 × 60× 6060 × 1000km/h = 4.57km/h.Now try the following Practice ExercisePracticeExercise 86 Angles of elevationand depression (answers on page 349)1. A vertical tower stands on level ground. Ata point 105m from the foot of the tower theangle of elevation of the top is 19◦. Find theheight of the tower.2. If the angle of elevation of the top of a vertical30m high aerial is 32◦, how far is it to theaerial?3. From the top of a vertical cliff 90.0m highthe angle of depression of a boat is 19◦50 .Determine the distance of the boat from thecliff.4. From the top of a vertical cliff 80.0m high theangles of depression of two buoys lying duewest of the cliff are 23◦and 15◦, respectively.How far apart are the buoys?5. From a point on horizontal ground a surveyormeasures the angle of elevation of the top ofa ﬂagpole as 18◦40 . He moves 50m nearerto the ﬂagpole and measures the angle of ele-vation as 26◦22 . Determine the height of theﬂagpole.6. A ﬂagpole stands on the edge of the top of abuilding. At a point 200m from the buildingthe angles of elevation of the top and bot-tom of the pole are 32◦ and 30◦ respectively.Calculate the height of the ﬂagpole.7. From a ship at sea, the angles of elevation ofthe top and bottom of a vertical lighthousestanding on the edge of a vertical cliff are31◦and 26◦, respectively. If the lighthouse is25.0m high, calculate the height of the cliff.8. From a window 4.2m above horizontal groundthe angle of depression of the foot of a buildingacrosstheroad is24◦ and theangleofelevationof the top of the building is 34◦. Determine,correct to the nearest centimetre, the width ofthe road and the height of the building.9. The elevation of a tower from two points, onedue west of the tower and the other due eastof it are 20◦ and 24◦, respectively, and the twopoints of observation are 300m apart. Find theheight of the tower to the nearest metre.
• Revision Test 8 : Angles, triangles and trigonometryThis assignment covers the material contained in Chapters 20 and 21. The marks available are shown in brackets atthe end of each question.1. Determine 38◦48 + 17◦23 . (2)2. Determine 47◦43 12 − 58◦35 53 + 26◦17 29 .(3)3. Change 42.683◦ to degrees and minutes. (2)4. Convert 77◦42 34 to degrees correct to 3 decimalplaces. (3)5. Determine angle θ in Figure RT8.1. (3)5384781248 298␪Figure RT8.16. Determine angle θ in the triangle in Figure RT8.2.(2)␪␪508Figure RT8.27. Determine angle θ in the triangle in Figure RT8.3.(2)438688␪Figure RT8.38. In Figure RT8.4, if triangle ABC is equilateral,determine ∠CDE. (3)ABCDE928Figure RT8.49. Find angle J in Figure RT8.5. (2)HGJ288Figure RT8.510. Determine angle θ in Figure RT8.6. (3)␪328588Figure RT8.611. State the angle (a) supplementary to 49◦ (b) com-plementary to 49◦. (2)
• 194 Basic Engineering Mathematics12. In Figure RT8.7, determine angles x, y and z.(3)598y zx378Figure RT8.713. In Figure RT8.8, determine angles a to e. (5)cea b d6081258Figure RT8.814. In Figure RT8.9, determine the length of AC.(4)D ACEB3m10m8mFigure RT8.915. In triangle J K L in Figure RT8.10, ﬁnd(a) the length K J correct to 3 signiﬁcant ﬁgures.(b) sin L and tan K, each correct to 3 decimalplaces. (4)KLJ11.605.91Figure RT8.1016. Two ships leave a port at the same time. Ship Xtravels due west at 30km/h and ship Y travels duenorth.After4 hoursthetwo shipsare130kmapart.Calculate the velocity of ship Y. (4)17. If sin A =1237, ﬁnd tan A in fraction form. (3)18. Evaluate 5tan62◦11 correct to 3 signiﬁcantﬁgures. (2)19. Determinetheacuteanglecos−1 0.3649 in degreesand minutes. (2)20. In triangle PQR in Figure RT8.11, ﬁnd angle Pin decimal form, correct to 2 decimal places. (2)P QR15.229.6Figure RT8.1121. Evaluate, correct to 3 signiﬁcant ﬁgures,3tan81.27◦ − 5cos7.32◦ − 6sin54.81◦. (2)22. In triangle ABC in Figure RT8.12, ﬁnd lengthsAB and AC, correct to 2 decimal places. (4)AB C6408 359Figure RT8.1223. From a point P, the angle of elevation of a 40mhigh electricity pylon is 20◦. How far is point Pfrom the base of the pylon, correct to the nearestmetre? (3)
• Chapter 22Trigonometric waveforms22.1 Graphs of trigonometricfunctionsBy drawing up tables of values from 0◦ to 360◦, graphsof y = sin A, y = cos A and y = tan A may be plotted.Values obtained with a calculator (correct to 3 deci-mal places – which is more than sufﬁcient for plottinggraphs), using 30◦ intervals, are shown below, with therespective graphs shown in Figure 22.1.(a) y = sinAA 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦sin A 0 0.500 0.866 1.000 0.866 0.500 0A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦sin A −0.500 −0.866 −1.000 −0.866 −0.500 0(b) y = cosAA 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦cos A 1.000 0.866 0.500 0 −0.500 −0.866 −1.000A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦cos A −0.866 −0.500 0 0.500 0.866 1.000(c) y = tanAA 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦tan A 0 0.577 1.732 ∞ −1.732 −0.577 0A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦tan A 0.577 1.732 ∞ −1.732 −0.577 021.02420.521.0220.51.020.50 30 60 90 120 150180210 240 270 300 330 36030 60 90 120 180 210 240 270 300 36030 60 90 120 150 180 210 240 270 300 330 360(a)1.00.50(b)(c)420150 330yyyy 5tan Ay 5cos AA8A8A8y 5sin AFigure 22.1From Figure 22.1 it is seen that(a) Sine and cosine graphs oscillate between peakvalues of ±1.(b) The cosine curve is the same shape as the sinecurve but displaced by 90◦.(c) Thesineand cosinecurvesare continuousand theyrepeat at intervals of 360◦ and the tangent curveappearsto bediscontinuousand repeatsat intervalsof 180◦.DOI: 10.1016/B978-1-85617-697-2.00022-3
• Trigonometric waveforms 197A knowledge of angles of any magnitude is neededwhen ﬁnding, for example, all the angles between 0◦and 360◦ whose sine is, say, 0.3261. If 0.3261 is enteredinto a calculator and then the inverse sine key pressed (orsin−1 key) the answer 19.03◦ appears. However, thereis a second angle between 0◦ and 360◦ which the cal-culator does not give. Sine is also positive in the secondquadrant (either from CAST or from Figure 22.1(a)).The other angle is shown in Figure 22.5 as angle θ,where θ = 180◦ − 19.03◦ = 160.97◦. Thus, 19.03◦and 160.97◦ are the angles between 0◦ and 360◦ whosesine is 0.3261 (check that sin160.97◦ = 0.3261 on yourcalculator).19.038 19.03818082708360808␪908S AT CFigure 22.5Be careful! Your calculator only gives you one ofthese answers. The second answer needs to be deducedfrom a knowledge of angles of any magnitude, as shownin the following worked problems.Problem 1. Determine all of the angles between0◦ and 360◦ whose sine is −0.4638The angles whose sine is −0.4638 occur in the thirdand fourth quadrants since sine is negative in thesequadrants – see Figure 22.6.1.021.020.46380 908 1808 2708 3608332.378207.638xy 5sinxyFigure 22.6From Figure 22.7, θ = sin−1 0.4638 = 27.63◦. Mea-sured from 0◦, the two angles between 0◦ and 360◦whose sine is −0.4638 are 180◦+ 27.63◦i.e. 207.63◦and 360◦ – 27.63◦, i.e. 332.37◦. (Note that if a calcu-lator is used to determine sin−1(−0.4638) it only givesone answer: −27.632588◦.)TS AC90818082708360808␪␪Figure 22.7Problem 2. Determine all of the angles between0◦ and 360◦ whose tangent is 1.7629A tangent is positive in the ﬁrst and third quadrants –see Figure 22.8.1.762960.448 240.4480 360827081808908y 5tan xyxFigure 22.8From Figure 22.9, θ = tan−1 1.7629 = 60.44◦. Mea-sured from 0◦, the two angles between 0◦ and 360◦whose tangent is 1.7629 are 60.44◦and 180◦+ 60.44◦,i.e. 240.44◦18082708360890808CTS A␪␪Figure 22.9Problem 3. Solve the equationcos−1(−0.2348) = α for angles of α between 0◦and 360◦
• 198 Basic Engineering MathematicsS18082708083608908TAC␪␪Figure 22.10Cosine is positive in the ﬁrst and fourth quadrants andthus negative in the second and third quadrants – seeFigure 22.10 or from Figure 22.1(b).In Figure 22.10, angle θ = cos−1(0.2348) = 76.42◦.Measured from 0◦, the two angles whose cosineis −0.2348 are α = 180◦ − 76.42◦, i.e. 103.58◦andα = 180◦ + 76.42◦, i.e. 256.42◦Now try the following Practice ExercisePracticeExercise 87 Angles of anymagnitude (answers on page 349)1. Determine all of the angles between 0◦ and360◦ whose sine is(a) 0.6792 (b) −0.14832. Solve the following equations for values of xbetween 0◦ and 360◦.(a) x = cos−1 0.8739(b) x = cos−1(−0.5572)3. Find the angles between 0◦ to 360◦ whosetangent is(a) 0.9728 (b) −2.3420In problems 4 to 6, solve the given equations in therange 0◦ to 360◦, giving the answers in degrees andminutes.4. cos−1(−0.5316) = t5. sin−1(−0.6250) = α6. tan−1 0.8314 = θ22.3 The production of sine andcosine wavesIn Figure22.11, let OR be a vector 1 unitlong and free torotate anticlockwiseabout 0. In one revolutiona circle isproduced and isshown with 15◦ sectors.Each radiusarmhas a vertical and a horizontal component. For example,at 30◦, the vertical component is TS and the horizontalcomponent is OS.From triangle OST,sin30◦ =TSTO=TS1i.e. TS = sin30◦and cos30◦ =OSTO=OS1i.e. OS = cos30◦120890860836083308 20.521.01.00.5TyRSS9T9y 5sin xAngle x8308 608 1208 2108 2708 3308300827082408210818081508OFigure 22.11
• Trigonometric waveforms 19920.50.521.01.0ySRT S9O9y 5cosxAngle x8308 608 1208 1808 2408 3008 36084580815833083158285825580822582108180815081208908608OFigure 22.1222.3.1 Sine wavesThe vertical component TS may be projected across toT S , which is the corresponding value of 30◦ on thegraph of y against angle x◦. If all such vertical compo-nents as TS are projected on to the graph, a sine waveis produced as shown in Figure 22.11.22.3.2 Cosine wavesIf all horizontal components such as OS are projectedon to a graph of y against angle x◦, a cosine wave isproduced. It is easier to visualize these projections byredrawing the circle with the radius arm OR initially ina vertical position as shown in Figure 22.12.It is seen from Figures 22.11 and 22.12 that a cosinecurve is of the same form as the sine curve but is dis-placed by 90◦ (or π/2 radians). Both sine and cosinewaves repeat every 360◦.22.4 Terminology involved with sineand cosine wavesSine waves are extremely important in engineering, withexamples occurring with alternating currents and volt-ages – the mains supply is a sine wave – and with simpleharmonic motion.22.4.1 CycleWhen a sine wave has passed through a completeseries of values, both positive and negative, it is saidto have completed one cycle. One cycle of a sinewave is shown in Figure 22.1(a) on page 195 and inFigure 22.11.22.4.2 AmplitudeThe amplitude is the maximum value reached in a halfcycle by a sine wave. Another name for amplitude ispeak value or maximum value.A sine wave y = 5sinx has an amplitude of 5, asine wave v = 200sin314t has an amplitude of 200 andthe sine wave y = sin x shown in Figure 22.11 has anamplitude of 1.22.4.3 PeriodThe waveforms y = sin x and y = cosx repeat them-selves every 360◦. Thus, for each, the period is 360◦.A waveform of y = tan x has a period of 180◦ (fromFigure 22.1(c)).A graph of y = 3sin2A, as shown in Figure 22.13, hasan amplitude of 3 and period 180◦.A graph of y = sin3A, as shown in Figure 22.14, has anamplitude of 1 and period of 120◦.A graph of y = 4cos2x, as shown in Figure 22.15, hasan amplitude of 4 and a period of 180◦.In general, if y = Asinpx or y = Acospx,amplitude = A and period =360◦py3230 A8y 53 sin 2A2708 36081808908Figure 22.13
• 200 Basic Engineering Mathematicsy1.0Ϫ1.00 90Њ 270Њ AЊ180Њ 360Њyϭsin 3AFigure 22.14y90Њ 180Њ 270Њ 360Њ xЊ0Ϫ44 y ϭ4 cos 2xFigure 22.15Problem 4. Sketch y = 2sin35A over one cycleAmplitude = 2; period =360◦35=360◦ × 53= 600◦A sketch of y = 2sin35A is shown in Figure 22.16.180Њ 360Њ 540Њ 600ЊyAЊ0Ϫ22yϭ2 sin A35Figure 22.1622.4.4 Periodic timeIn practice, the horizontal axis of a sine wave will betime. The time taken for a sine wave to complete onecycle is called the periodic time, T.In the sine wave of voltage v (volts) against time t (mil-liseconds) shown in Figure 22.17, the amplitude is 10Vand the periodic time is 20ms; i.e., T = 20 ms.v10 20 t (ms)021010Figure 22.1722.4.5 FrequencyThe number of cycles completed in one second is calledthe frequency f and is measured in hertz, Hz.f =1Tor T =1fProblem 5. Determine the frequency of the sinewave shown in Figure 22.17In the sine wave shown in Figure 22.17, T = 20 ms,hencefrequency, f =1T=120 × 10−3= 50HzProblem 6. If a waveform has a frequency of200kHz, determine the periodic timeIf a waveform has a frequency of 200kHz, the periodictime T is given byperiodic time, T =1f=1200 × 103= 5 × 10−6s = 5μs22.4.6 Lagging and leading anglesA sine or cosine curve may not always start at 0◦.To show this, a periodic function is represented byy = Asin(x ± α) where α is a phase displacementcompared with y = Asin x. For example, y = sin A isshown by the broken line in Figure 22.18 and, onthe same axes, y = sin(A − 60◦) is shown. The graphy = sin(A − 60◦) is said to lag y = sinA by 60◦.In another example, y = cos A is shown by the bro-ken line in Figure 22.19 and, on the same axes, y =cos(A+45◦) is shown. The graph y = cos(A+45◦) issaid to lead y = cosA by 45◦.
• Trigonometric waveforms 201908 2708yA8021.01.0y 5sin (A 2608)y 5sin A6086081808 3608Figure 22.1818084583608 A80y21.0y5cos (A 1458)y 5 cos A4582708908Figure 22.19Problem 7. Sketch y = 5sin(A + 30◦) fromA = 0◦ to A = 360◦Amplitude = 5 and period = 360◦/1 = 360◦5sin(A + 30◦) leads 5sin A by 30◦ (i.e. starts 30◦earlier).A sketch of y = 5sin(A + 30◦) is shown inFigure 22.20.Problem 8. Sketch y = 7sin(2A − π/3) in therange 0 ≤ A ≤ 360◦Amplitude = 7 and period = 2π/2 = π radiansIn general, y = sin(pt − α) lags y = sinpt by α/p,hence 7sin(2A − π/3) lags 7sin2A by (π/3)/2, i.e.π/6 rad or 30◦A sketch of y = 7sin(2A − π/3) is shown inFigure 22.21.908 2708 A80525y 55 sin (A 1308)y55 sin A3083081808 3608yFigure 22.2007yA83608y57sin 2Ay57sin (2A2␲/3)␲/6␲/62␲␲7270818089083␲/2␲/2Figure 22.21Problem 9. Sketch y = 2cos(ωt − 3π/10) overone cycleAmplitude = 2 and period = 2π/ω rad2cos(ωt − 3π/10) lags 2cosωt by 3π/10ω seconds.A sketch of y = 2cos(ωt − 3π/10) is shown inFigure 22.22.0yt␲/2␻ ␲/␻ 3␲/2␻ 2␲/␻y52cos ␻ty52cos(␻t23␲/10)2223␲/10␻ radsFigure 22.22
• Trigonometric waveforms 203hence, 0.90 = 2.5sin(0 + α)i.e. sinα =0.902.5= 0.36Hence, α = sin−10.36 = 21.10◦= 21◦6 = 0.368rad.Thus,displacement = 2.5sin(120πt + 0.368) m.Problem 12. The instantaneous value of voltagein an a.c. circuit at any time t seconds is given byv = 340sin(50πt − 0.541) volts. Determine the(a) amplitude, frequency, periodic time and phaseangle (in degrees), (b) value of the voltage whent = 0, (c) value of the voltage when t = 10 ms,(d) time when the voltage ﬁrst reaches 200V and(e) time when the voltage is a maximum. Also,(f ) sketch one cycle of the waveform(a) Amplitude = 340 VAngular velocity, ω = 50πFrequency, f =ω2π=50π2π= 25HzPeriodic time, T =1f=125= 0.04s or 40 msPhase angle = 0.541rad = 0.541 ×180π◦= 31◦lagging v = 340sin(50πt)(b) When t = 0,v = 340sin(0 − 0.541)= 340sin(−31◦) = −175.1V(c) When t = 10ms,v = 340sin(50π × 10 × 10−3 − 0.541)= 340sin(1.0298)= 340sin59◦ = 291.4 volts(d) When v = 200 volts,200 = 340sin(50πt − 0.541)200340= sin(50πt − 0.541)Hence, (50πt −0.541) = sin−1 200340= 36.03◦ or 0.628875rad50πt = 0.628875 + 0.541= 1.169875Hence, when v = 200 V,time, t =1.16987550π= 7.448ms(e) When the voltage is a maximum, v = 340 V.Hence 340 = 340sin(50πt − 0.541)1 = sin(50πt − 0.541)50πt − 0.541 = sin−11 = 90◦or 1.5708 rad50πt = 1.5708 + 0.541 = 2.1118Hence, time, t =2.111850π= 13.44ms(f ) A sketch of v = 340sin(50πt − 0.541) volts isshown in Figure 22.23.v5340sin(50 ␲t 20.541)v5340sin 50␲t0 t (ms)10 30 407.448 13.4423402175.1200291.4340Voltage v20Figure 22.23Now try the following Practice ExercisePracticeExercise 89 Sinusoidal formAsin(ωt ± α) (answers on page 350)In problems 1 to 3 ﬁnd the (a) amplitude,(b)frequency, (c) periodictime and (d)phase angle(stating whether it is leading or lagging sinωt) ofthe alternating quantities given.1. i = 40sin(50πt + 0.29)mA2. y = 75sin(40t − 0.54)cm
• 204 Basic Engineering Mathematics3. v = 300sin(200πt − 0.412)V4. A sinusoidal voltage has a maximum value of120V and a frequency of 50Hz. At time t = 0,the voltage is (a) zero and (b) 50V. Expressthe instantaneous voltage v in the formv = Asin(ωt ± α).5. An alternating current has a periodic time of25ms and a maximum value of 20A. Whentime = 0, current i = −10 amperes. Expressthe current i in the form i = Asin(ωt ± α).6. An oscillating mechanismhasamaximumdis-placement of 3.2m and a frequency of 50Hz.At time t = 0 the displacement is 150cm.Express the displacement in the general formAsin(ωt ± α)7. The current in an a.c. circuit at any time tseconds is given byi = 5sin(100πt − 0.432) amperesDetermine the(a) amplitude, frequency, periodic time andphase angle (in degrees),(b) value of current at t = 0,(c) value of current at t = 8ms,(d) time when the current is ﬁrst a maximum,(e) time when the current ﬁrst reaches 3A.Also,(f ) sketch one cycle of the waveform show-ing relevant points.
• Chapter 23Non-right-angled trianglesand some practicalapplications23.1 The sine and cosine rulesTo ‘solve a triangle’ means ‘to ﬁnd the values ofunknown sides and angles’. If a triangle is right-angled,trigonometric ratios and the theorem of Pythagorasmay be used for its solution, as shown in Chapter 21.However, for a non-right-angled triangle, trigono-metric ratios and Pythagoras’ theorem cannot be used.Instead, two rules, called the sine rule and the cosinerule, are used.23.1.1 The sine ruleWith reference to triangle ABC of Figure 23.1, the sinerule statesasinA=bsinB=csinCabcB CAFigure 23.1The rule may be used only when(a) 1 side and any 2 angles are initially given, or(b) 2 sides and an angle (not the included angle) areinitially given.23.1.2 The cosine ruleWith reference to triangle ABC of Figure 23.1, thecosine rule statesa2= b2+ c2− 2bc cos Aor b2= a2+ c2− 2ac cos Bor c2= a2+ b2− 2ab cosCThe rule may be used only when(a) 2 sides and the included angle are initiallygiven, or(b) 3 sides are initially given.23.2 Area of any triangleThe area of any triangle such as ABC of Figure 23.1is given by(a)12× base × perpendicular heightDOI: 10.1016/B978-1-85617-697-2.00023-5
• 206 Basic Engineering Mathematicsor (b)12ab sinC or12ac sinB or12bc sinAor (c) [s(s − a)(s − b)(s − c)] where s =a + b + c223.3 Worked problems on the solutionof triangles and their areasProblem 1. In a triangle XYZ, ∠X = 51◦,∠Y = 67◦ and YZ = 15.2 cm. Solve the triangle andﬁnd its areaThe triangle XYZ is shown in Figure 23.2. Solving thetriangle means ﬁnding ∠Z and sides XZ and XY.XY Zx 515.2cmyz518678Figure 23.2Since the angles in a triangle add up to 180◦,Z = 180◦ − 51◦ − 67◦ = 62◦Applying the sine rule,15.2sin51◦=ysin67◦=zsin62◦Using15.2sin51◦=ysin67◦and transposing gives y =15.2sin67◦sin51◦= 18.00cm = XZUsing15.2sin51◦=zsin62◦and transposing gives z =15.2sin62◦sin51◦= 17.27cm = XYArea of triangle XYZ =12xy sin Z=12(15.2)(18.00)sin 62◦= 120.8cm2(or area =12xz sinY =12(15.2)(17.27) sin67◦= 120.8cm2)It is always worth checking with triangle problems thatthe longest side is opposite the largest angle and vice-versa. In this problem, Y is the largest angle and X Z isthe longest of the three sides.Problem 2. Solve the triangle ABC givenB = 78◦51 ,AC = 22.31mm and AB = 17.92 mm.Also ﬁnd its areaTriangle ABC is shown in Figure 23.3. Solving thetriangle means ﬁnding angles A and C and side BC.AB a Cb522.31 mmc517.92mm788519Figure 23.3Applying the sine rule,22.31sin78◦51=17.92sinCfrom which sinC =17.92sin78◦5122.31= 0.7881Hence, C = sin−10.7881 = 52◦0 or 128◦0Since B = 78◦51 , C cannot be 128◦0 , since 128◦0 +78◦51 is greater than 180◦. Thus, only C = 52◦0 isvalid.Angle A = 180◦ − 78◦51 − 52◦0 = 49◦9 .Applying the sine rule, asin49◦9=22.31sin78◦51from which a =22.31sin49◦9sin78◦51= 17.20mmHence, A = 49◦9 ,C = 52◦0 and BC = 17.20mm.Area of triangle ABC =12ac sin B=12(17.20)(17.92)sin78◦51= 151.2mm2Problem 3. Solve the triangle PQR and ﬁnd itsarea given that QR = 36.5mm,PR = 29.6mm and∠Q = 36◦
• Non-right-angled triangles and some practical applications 207Triangle PQR is shown in Figure 23.4.q 529.6 mmp5 36.5 mm368rQ RPFigure 23.4Applying the sine rule,29.6sin36◦=36.5sin Pfrom which sin P =36.5sin36◦29.6= 0.7248Hence, P = sin−10.7248 = 46.45◦or 133.55◦When P = 46.45◦ and Q = 36◦ thenR = 180◦ − 46.45◦ − 36◦ = 97.55◦When P = 133.55◦ and Q = 36◦ thenR = 180◦ − 133.55◦ − 36◦ = 10.45◦Thus, in this problem, there are two separate sets ofresults and both are feasible solutions. Such a situationis called the ambiguous case.Case 1. P = 46.45◦, Q = 36◦, R = 97.55◦,p = 36.5mm and q = 29.6mmFrom the sine rule,rsin97.55◦=29.6sin36◦from which r =29.6sin97.55◦sin36◦= 49.92mm = PQArea of PQR =12pq sin R =12(36.5)(29.6)sin 97.55◦= 535.5mm2Case 2. P = 133.55◦, Q = 36◦, R = 10.45◦,p = 36.5mm and q = 29.6mmFrom the sine rule,rsin10.45◦=29.6sin36◦from which r =29.6sin10.45◦sin36◦= 9.134mm = PQArea of PQR =12pq sin R =12(36.5)(29.6)sin 10.45◦= 97.98mm2The triangle PQR for case 2 is shown in Figure 23.5.368 10.458133.5589.134 mm 29.6 mm36.5 mmQPRFigure 23.5Now try the following Practice ExercisePracticeExercise 90 Solution of trianglesand their areas (answers on page 350)In problems 1 and 2, use the sine rule to solve thetriangles ABC and ﬁnd their areas.1. A = 29◦, B = 68◦,b = 27mm2. B = 71◦26 ,C = 56◦32 ,b = 8.60 cmIn problems 3 and 4, use the sine rule to solve thetriangles DEF and ﬁnd their areas.3. d = 17cm, f = 22 cm, F = 26◦4. d = 32.6mm,e = 25.4mm, D = 104◦22In problems 5 and 6, use the sine rule to solve thetriangles JKL and ﬁnd their areas.5. j = 3.85cm,k = 3.23cm, K = 36◦6. k = 46mm,l = 36mm, L = 35◦23.4 Further worked problems on thesolution of triangles and theirareasProblem 4. Solve triangle DEF and ﬁnd its areagiven that EF = 35.0 mm,DE = 25.0 mm and∠E = 64◦Triangle DEF is shown in Figure 23.6. Solving the trian-gle means ﬁnding angles D and F and side DF. Sincetwo sides and the angle in between the two sides aregiven, the cosine needs to be used.648DFEed 535.0mmf525.0mmFigure 23.6Applying the cosine rule, e2= d2+ f 2− 2d f cos Ei.e. e2= (35.0)2+ (25.0)2− [2(35.0)(25.0)cos64◦]= 1225 + 625 − 767.15= 1082.85
• 208 Basic Engineering Mathematicsfrom which e =√1082.85= 32.91mm = DFApplying the sine rule,32.91sin64◦=25.0sin Ffrom which sin F =25.0sin64◦32.91= 0.6828Thus, ∠F = sin−10.6828 = 43◦4 or 136◦56F = 136◦56 is not possible in this case since 136◦56 +64◦ is greater than 180◦. Thus, only F = 43◦4 is valid.Then ∠D = 180◦ − 64◦ − 43◦4 = 72◦56 .Area of triangle DEF=12d f sin E=12(35.0)(25.0)sin 64◦ = 393.2mm2Problem 5. A triangle ABC has sidesa = 9.0 cm,b = 7.5cm and c = 6.5cm. Determineits three angles and its areaTriangle ABC is shown in Figure 23.7. It is usual ﬁrstto calculate the largest angle to determine whether thetriangle is acute or obtuse. In this case the largest angleis A (i.e. opposite the longest side).AB Ca59.0cmc56.5cm b5 7.5cmFigure 23.7Applying the cosine rule, a2= b2+ c2− 2bccos Afrom which 2bccos A = b2+ c2− a2and cos A =b2 + c2 − a22bc=7.52 + 6.52 − 9.022(7.5)(6.5)= 0.1795Hence, A = cos−10.1795 = 79.67◦(or 280.33◦, which is clearly 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., would lie between 90◦ and 180◦.)Applying the sine rule,9.0sin79.67◦=7.5sin Bfrom which sin B =7.5sin79.67◦9.0= 0.8198Hence, B = sin−10.8198 = 55.07◦and C = 180◦− 79.67◦− 55.07◦= 45.26◦Area =√[s(s − a)(s − b)(s − c)], wheres =a + b + c2=9.0 + 7.5 + 6.52= 11.5cmHence,area = [11.5(11.5 − 9.0)(11.5 − 7.5)(11.5 − 6.5)]= [11.5(2.5)(4.0)(5.0)] = 23.98cm2Alternatively,area =12ac sin B=12(9.0)(6.5)sin 55.07◦= 23.98cm2Problem 6. Solve triangle XYZ, shown inFigure 23.8, and ﬁnd its area given thatY = 128◦,XY = 7.2 cm and YZ = 4.5cm1288x 54.5cmz57.2cmyXY ZFigure 23.8Applying the cosine rule,y2= x2+ z2− 2 xz cosY= 4.52+ 7.22− [2(4.5)(7.2)cos 128◦]= 20.25 + 51.84− [−39.89]= 20.25 + 51.84+ 39.89 = 112.0y =√112.0 = 10.58cm = XZApplying the sine rule,10.58sin128◦=7.2sin Zfrom which sin Z =7.2sin128◦10.58= 0.5363Hence, Z = sin−10.5363 = 32.43◦(or 147.57◦ which is not possible)Thus, X = 180◦− 128◦− 32.43◦= 19.57◦Area of XYZ =12xz sinY =12(4.5)(7.2)sin 128◦= 12.77cm2
• Non-right-angled triangles and some practical applications 209Now try the following Practice ExercisePracticeExercise 91 Solution of trianglesand their areas (answers on page 350)In problems 1 and 2, use the cosine and sine rulesto solve the triangles PQR and ﬁnd their areas.1. q = 12 cm,r = 16cm, P = 54◦2. q = 3.25m,r = 4.42 m, P = 105◦In problems 3 and 4, use the cosine and sine rulesto solve the triangles XYZ and ﬁnd their areas.3. x = 10.0 cm, y = 8.0 cm, z = 7.0 cm4. x = 21mm, y = 34mm, z = 42 mm23.5 Practical situations involvingtrigonometryThere are a number of practical situations in which theuse of trigonometryis needed to ﬁnd unknown sides andanglesoftriangles.Thisisdemonstrated in thefollowingworked problems.Problem 7. A room 8.0 m wide has a span roofwhich slopes at 33◦ on one side and 40◦ on theother. Find the length of the roof slopes, correct tothe nearest centimetreA section of the roof is shown in Figure 23.9.BA C8.0 m338 408Figure 23.9Angle at ridge, B = 180◦ − 33◦ − 40◦ = 107◦From the sine rule,8.0sin107◦=asin33◦from which a =8.0sin33◦sin107◦= 4.556m = BCAlso from the sine rule,8.0sin107◦=csin40◦from which c =8.0sin40◦sin107◦= 5.377m = ABHence, the roof slopes are 4.56 m and 5.38 m, correctto the nearest centimetre.Problem 8. A man leaves a point walking at6.5km/h in a direction E 20◦ N (i.e. a bearing of70◦). A cyclist leaves the same point at the sametime in a direction E 40◦ S (i.e. a bearing of 130◦)travelling at a constant speed. Find the averagespeed of the cyclist if the walker and cyclist are80 km apart after 5 hoursAfter5 hoursthewalkerhastravelled 5 × 6.5 = 32.5km(shown as AB in Figure 23.10). If AC is the distance thecyclist travels in 5 hours then BC = 80 km.bABC80 kmWS408EN20832.5kmFigure 23.10Applying the sine rule,80sin60◦=32.5sinCfrom which sinC =32.5sin60◦80= 0.3518Hence, C = sin−10.3518 = 20.60◦(or 159.40◦, which is not possible)and B = 180◦− 60◦− 20.60◦= 99.40◦Applying the sine rule again,80sin60◦=bsin99.40◦from which b =80sin99.40◦sin60◦= 91.14kmSince the cyclist travels 91.14km in 5 hours,average speed =distancetime=91.145= 18.23km/hProblem 9. Two voltage phasors are shown inFigure 23.11. If V1 = 40 V and V2 = 100 V,determine the value of their resultant (i.e. lengthOA) and the angle the resultant makes with V1
• 210 Basic Engineering Mathematics458V25100VV1540 VABOFigure 23.11Angle OBA = 180◦− 45◦= 135◦Applying the cosine rule,OA2= V 21 + V 22 − 2V1V2 cosOBA= 402+ 1002− {2(40)(100)cos135◦}= 1600 + 10000− {−5657}= 1600 + 10000+ 5657 = 17257Thus, resultant,OA =√17257 = 131.4 VApplying the sine rule131.4sin135◦=100sinAOBfrom which sinAOB =100sin135◦131.4= 0.5381Hence, angle AOB = sin−1 0.5381 = 32.55◦ (or147.45◦, which is not possible)Hence, the resultant voltage is 131.4 volts at 32.55◦to V1Problem 10. In Figure 23.12, PR represents theinclined jib of a crane and is 10.0 m long. PQ is4.0 m long. Determine the inclination of the jib tothe vertical and the length of tie QR.PQR12084.0 m 10.0mFigure 23.12Applying the sine rule,PRsin120◦=PQsin Rfrom which sin R =PQsin120◦PR=(4.0)sin 120◦10.0= 0.3464Hence, ∠R = sin−10.3464 = 20.27◦(or 159.73◦,which is not possible)∠P = 180◦ − 120◦ − 20.27◦ = 39.73◦, which is theinclination of the jib to the verticalApplying the sine rule,10.0sin120◦=QRsin39.73◦from which length of tie,QR =10.0sin39.73◦sin120◦= 7.38mNow try the following Practice ExercisePracticeExercise 92 Practical situationsinvolving trigonometry (answers onpage 350)1. A ship P sails at a steady speed of 45km/hin a direction of W 32◦ N (i.e. a bearing of302◦) from a port. At the same time anothership Q leaves the port at a steady speed of35km/h in a direction N 15◦ E (i.e. a bearingof 015◦). Determine their distance apart after4 hours.2. Two sides of a triangular plot of land are52.0 m and 34.0 m, respectively. If the areaof the plot is 620 m2, ﬁnd (a) the lengthof fencing required to enclose the plot and(b) the angles of the triangular plot.3. A jib crane is shown in Figure 23.13. If thetie rod PR is 8.0 m long and PQ is 4.5m long,determine (a) the length of jib RQ and (b) theangle between the jib and the tie rod.R1308 PQFigure 23.134. A building site is in the form of a quadri-lateral, as shown in Figure 23.14, and itsarea is 1510 m2. Determine the length of theperimeter of the site.
• Non-right-angled triangles and some practical applications 21128.5m34.6m72852.4m 758Figure 23.145. Determine the length of members BF and EBin the roof truss shown in Figure 23.15.4 m2.5 m5084m2.5m5085 m 5 mEA B CDFFigure 23.156. A laboratory 9.0 m wide has a span roof whichslopes at 36◦on one side and 44◦on the other.Determine the lengths of the roof slopes.7. PQ and QR are the phasors representing thealternating currents in two branches of a cir-cuit. Phasor PQ is 20.0 A and is horizontal.Phasor QR (which is joined to the end of PQto form triangle PQR) is 14.0 A and is at anangle of 35◦ to the horizontal. Determine theresultant phasor PR and the angle it makeswith phasor PQ.23.6 Further practical situationsinvolving trigonometryProblem 11. A vertical aerial stands onhorizontal ground. A surveyor positioned due eastof the aerial measures the elevation of the top as48◦. He moves due south 30.0 m and measures theelevation as 44◦. Determine the height of the aerialIn Figure 23.16, DC represents the aerial, A is the initialposition of the surveyor and B his ﬁnal position.From triangle ACD, tan48◦ =DCACfrom whichAC =DCtan48◦Similarly, from triangle BCD, BC =DCtan44◦BCDA30.0 m448488Figure 23.16For triangle ABC, using Pythagoras’ theorem,BC2= AB2+ AC2DCtan44◦2= (30.0)2+DCtan48◦2DC2 1tan2 44◦−1tan2 48◦= 30.02DC2(1.072323 − 0.810727) = 30.02DC2=30.020.261596= 3440.4Hence, height of aerial, DC =√3340.4 = 58.65 m.Problem 12. A crank mechanism of a petrolengine is shown in Figure 23.17. Arm OA is10.0 cm long and rotates clockwise about O. Theconnecting rod AB is 30.0 cm long and end B isconstrained to move horizontally.(a) For the position shown in Figure 23.17,determine the angle between the connectingrod AB and the horizontal, and the lengthof OB.(b) How far does B move when angle AOBchanges from 50◦ to 120◦?508O30.0cm10.0 cmABFigure 23.17
• 212 Basic Engineering Mathematics(a) Applying the sine rule,ABsin50◦=AOsin Bfrom which sin B =AOsin50◦AB=10.0sin50◦30.0= 0.2553Hence, B = sin−1 0.2553 = 14.79◦ (or 165.21◦,which is not possible)Hence, the connecting rod AB makes an angleof 14.79◦ with the horizontal.Angle OAB = 180◦ − 50◦ − 14.79◦ = 115.21◦Applying the sine rule:30.0sin50◦=OBsin115.21◦from which OB =30.0sin115.21◦sin50◦= 35.43cm(b) Figure 23.18 shows the initial and ﬁnal positionsof the crank mechanism.508O30.0 cm1208B B9A9A10.0cmFigure 23.18In triangle OA B , applying the sine rule,30.0sin120◦=10.0sin AB Ofrom which sin A BO =10.0sin120◦30.0= 0.28868Hence, AB O = sin−1 0.28868 = 16.78◦ (or163.22◦, which is not possible)Angle OA B = 180◦ − 120◦ − 16.78◦ = 43.22◦Applying the sine rule,30.0sin120◦=OBsin43.22◦from whichOB =30.0sin43.22◦sin120◦= 23.72 cmSince OB = 35.43cm and OB = 23.72 cm,BB = 35.43 − 23.72 = 11.71cmHence, B moves 11.71 cm when angle AOBchanges from 50◦ to 120◦.Problem 13. The area of a ﬁeld is in the form of aquadrilateral ABCD as shown in Figure 23.19.Determine its area62.3m21.4 m39.8 m1148568BDCA42.5 mFigure 23.19A diagonal drawn from B to D divides the quadrilateralinto two triangles.Area of quadrilateral ABCD= area of triangle ABD+ area of triangle BCD=12(39.8)(21.4)sin 114◦+12(42.5)(62.3)sin 56◦= 389.04 + 1097.5= 1487m2Now try the following Practice ExercisePracticeExercise 93 More practicalsituations involving trigonometry (answerson page 350)1. Three forces acting on a ﬁxed point are repre-sented by the sides of a triangle of dimensions7.2 cm,9.6cm and 11.0 cm. Determine theanglesbetween thelinesofaction and thethreeforces.2. A vertical aerial AB, 9.60 m high, stands onground which is inclined 12◦ to the horizontal.Astay connectsthetop oftheaerial Ato apointC on the ground 10.0 m downhill from B, thefoot of the aerial. Determine (a) the length ofthe stay and (b) the angle the stay makes withthe ground.3. A reciprocating engine mechanism is shownin Figure 23.20. The crank AB is 12.0cm longand the connecting rod BC is 32.0cm long.For the position shown determine the lengthof AC and the angle between the crank and theconnecting rod.
• Non-right-angled triangles and some practical applications 213BCA408Figure 23.204. From Figure 23.20, determine how far Cmoves, correct to the nearest millimetre, whenangle CAB changes from 40◦ to 160◦, Bmoving in an anticlockwise direction.5. A surveyor standing W 25◦S of a tower mea-sures the angle of elevation of the top ofthe tower as 46◦30 . From a position E 23◦Sfrom the tower the elevation of the top is37◦15 .Determinetheheight ofthetowerifthedistance between the two observationsis75m.6. Calculate, correct to 3 signiﬁcant ﬁgures, theco-ordinates x and y to locate the hole centreat P shown in Figure 23.21.100 mm1168 1408PxyFigure 23.217. An idler gear, 30 mm in diameter, has to beﬁtted between a 70 mm diameter driving gearand a 90 mm diameter driven gear, as shownin Figure 23.22. Determine the value of angleθ between the centre lines.90mmdia30mmdia70mmdia␪99.78mmFigure 23.228. 16 holes are equally spaced on a pitch circleof 70 mm diameter. Determine the length ofthe chord joining the centres of two adjacentholes.
• Chapter 24Cartesian and polarco-ordinates24.1 IntroductionThere are two ways in which the position of a point ina plane can be represented. These are(a) Cartesian co-ordinates, i.e. (x, y).(b) Polar co-ordinates, i.e. (r, θ), where r is a radiusfrom a ﬁxed point and θ is an angle from a ﬁxedpoint.24.2 Changing from Cartesian topolar co-ordinatesIn Figure 24.1, if lengths x and y are known thenthe length of r can be obtained from Pythagoras’ the-orem (see Chapter 21) since OPQ is a right-angledtriangle.yPQ xOxr y␪Figure 24.1Hence, r2 = (x2 + y2), from which r = x2 + y2From trigonometric ratios (see Chapter 21), tanθ =yxfrom which θ = tan−1 yxr = x2 + y2 and θ = tan−1 yxare the two formulaewe need to change from Cartesian to polar co-ordinates.The angle θ, which may be expressed in degrees or radi-ans, must always be measured from the positive x-axis;i.e., measured from the line OQ in Figure 24.1. It issuggested that when changing from Cartesian to polarco-ordinates a diagram should always be sketched.Problem 1. Change the Cartesian co-ordinates(3, 4) into polar co-ordinatesA diagram representing the point (3, 4) is shown inFigure 24.2.P43yxOr␪Figure 24.2From Pythagoras’ theorem, r =√32 + 42 = 5 (notethat −5 has no meaning in this context).By trigonometric ratios, θ = tan−1 43= 53.13◦ or0.927rad.Note that 53.13◦ = 53.13 ×π180rad = 0.927rad.DOI: 10.1016/B978-1-85617-697-2.00024-7
• 216 Basic Engineering Mathematics24.3 Changing from polar to Cartesianco-ordinatesyyQ xxOPr␪Figure 24.6From the right-angled triangle OPQ in Figure 24.6,cosθ =xrand sinθ =yrfrom trigonometric ratiosHence, x = r cosθ and y = r sinθ.If lengths r and angle θ are known then x = r cosθ andy = r sinθ are the two formulae we need to change frompolar to Cartesian co-ordinates.Problem 5. Change (4, 32◦) into Cartesianco-ordinatesA sketch showing the position (4, 32◦) is shown inFigure 24.7.yyO xxr ϭ 4␪ ϭ32ЊFigure 24.7Now x = r cosθ = 4cos32◦ = 3.39and y = r sinθ = 4sin32◦ = 2.12Hence, (4, 32◦) in polar co-ordinates corresponds to(3.39, 2.12) in Cartesian co-ordinates.Problem 6. Express (6, 137◦) in Cartesianco-ordinatesA sketch showing the position (6, 137◦) is shown inFigure 24.8.BOAyxr ϭ 6␪ϭ137ЊFigure 24.8x = r cosθ = 6cos137◦= −4.388which corresponds to length OA in Figure 24.8.y = r sinθ = 6sin137◦= 4.092which corresponds to length AB in Figure 24.8.Thus, (6, 137◦) in polar co-ordinates corresponds to(−4.388, 4.092) in Cartesian co-ordinates.(Note that when changing from polar to Cartesian co-ordinates it is not quiteso essential to draw a sketch. Useof x = r cos θ and y = r sin θ automatically producesthe correct values and signs.)Problem 7. Express (4.5, 5.16 rad) in Cartesianco-ordinatesA sketch showing the position (4.5, 5.16 rad) is shownin Figure 24.9.yABO x␪55.16 radr 5 4.5Figure 24.9x = r cosθ = 4.5cos5.16 = 1.948which corresponds to length OA in Figure 24.9.y = r sinθ = 4.5sin5.16 = −4.057which corresponds to length AB in Figure 24.9.Thus, (1.948, −4.057) in Cartesian co-ordinates cor-responds to (4.5, 5.16 rad) in polar co-ordinates.
• Revision Test 9 : Trigonometric waveforms and practical trigonometryThis assignment covers the material contained in Chapters 22–24. The marks available are shown in brackets at theend of each question.1. A sine wave is given by y = 8sin4x. State itspeakvalue and its period, in degrees. (2)2. A periodic function is given by y = 15tan2x.State its period in degrees. (2)3. The frequency of a sine wave is 800Hz. Calculatethe periodic time in milliseconds. (2)4. Calculate the frequency of a sine wave that has aperiodic time of 40μs. (2)5. Calculate the periodic time for a sine wave havinga frequency of 20kHz. (2)6. An alternating current completes 12 cycles in16ms. What is its frequency? (3)7. A sinusoidal voltage is given bye = 150sin(500πt − 0.25) volts. Determine the(a) amplitude,(b) frequency,(c) periodic time,(d) phase angle (stating whether it is leading orlagging 150sin500πt). (4)8. Determinetheacuteanglesin degrees,degreesandminutes, and radians.(a) sin−1 0.4721 (b) cos−1 0.8457(c) tan−1 1.3472 (9)9. Sketch the following curves, labelling relevantpoints.(a) y = 4cos(θ + 45◦) (b) y = 5sin(2t − 60◦)(8)10. The current in an alternating current cir-cuit at any time t seconds is given byi = 120sin(100πt + 0.274) amperes. Determine(a) the amplitude, frequency, periodic time andphase angle (with reference to 120sin100πt),(b) the value of current when t = 0,(c) the value of current when t = 6ms.Sketch one cycle of the oscillation. (16)11. A triangular plot of land ABC is shown inFigure RT9.1. Solve the triangle and determineits area. (10)ABC15m15.4m718Figure RT9.112. A car is travelling 20m above sea level. It thentravels 500m up a steady slope of 17◦. Determine,correct to the nearest metre, how high the car isnow above sea level. (3)13. Figure RT9.2 shows a roof truss PQR with rafterPQ = 3m. Calculate the length of(a) the roof rise PP ,(b) rafter PR,(c) the roof span QR.Find also (d) the cross-sectional area of the rooftruss. (11)PQ RP93m408 328Figure RT9.214. Solve triangle ABC given b = 10cm,c = 15cmand ∠A = 60◦. (10)15. Change the following Cartesian co-ordinates intopolar co-ordinates, correct to 2 decimal places, inboth degrees and in radians.(a) (−2.3,5.4) (b) (7.6,−9.2) (10)16. Change the following polar co-ordinates intoCartesian co-ordinates, correct to 3 decimalplaces.(a) (6.5,132◦) (b) (3,3rad) (6)
• Chapter 25Areas of common shapes25.1 IntroductionArea is a measure of the size or extent of a plane surface.Area ismeasured in square units such as mm2, cm2 andm2. This chapter deals with ﬁnding the areas of commonshapes.In engineering it is often important to be able to cal-culate simple areas of various shapes. In everyday lifeits important to be able to measure area to, say, lay a car-pet, order sufﬁcient paint for a decorating job or ordersufﬁcient bricks for a new wall.On completing this chapter you will be able to recog-nize common shapes and be able to ﬁnd the areas of rect-angles, squares, parallelograms, triangles, trapeziumsand circles.25.2 Common shapes25.2.1 PolygonsA polygon is a closed plane ﬁgure bounded by straightlines. A polygon which has3 sides is called a triangle – see Figure 25.1(a)4 sides is called a quadrilateral – see Figure 25.1(b)5 sides is called a pentagon – see Figure 25.1(c)6 sides is called a hexagon – see Figure 25.1(d)7 sides is called a heptagon – see Figure 25.1(e)8 sides is called an octagon – see Figure 25.1(f)25.2.2 QuadrilateralsThere are ﬁve types of quadrilateral, these being rect-angle, square, parallelogram, rhombus and trapezium.If the opposite corners of any quadrilateral are joinedby a straight line, two triangles are produced. Since thesum of the angles of a triangle is 180◦, the sum of theangles of a quadrilateral is 360◦.RectangleIn the rectangle ABCD shown in Figure 25.2,(a) all four angles are right angles,(b) the opposite sides are parallel and equal in length,and(c) diagonals AC and BD are equal in length and bisectone another.SquareIn the square PQRS shown in Figure 25.3,(a) all four angles are right angles,(b) the opposite sides are parallel,(c) all four sides are equal in length, and(d) diagonals PR and QS are equal in length and bisectone another at right angles.ParallelogramIn the parallelogram WXYZ shown in Figure 25.4,(a) opposite angles are equal,(b) opposite sides are parallel and equal in length, and(c) diagonals WY and XZ bisect one another.RhombusIn the rhombus ABCD shown in Figure 25.5,(a) opposite angles are equal,(b) opposite angles are bisected by a diagonal,(c) opposite sides are parallel,(d) all four sides are equal in length, and(e) diagonals AC and BD bisect one another at rightangles.DOI: 10.1016/B978-1-85617-697-2.00025-9
• 220 Basic Engineering Mathematics(a) (c)(b)(d) (e) (f)Figure 25.1BACDFigure 25.2QPRSQPRSFigure 25.3WZ YXFigure 25.4A␣ ␣␤␤D CBFigure 25.5EH GFFigure 25.6TrapeziumIn the trapezium EFGH shown in Figure 25.6,(a) only one pair of sides is parallel.Problem 1. State the types of quadrilateral shownin Figure 25.7 and determine the angles marked ato l(a) (b) (c)(d) (e)BAxxaCDEH GFdcb408308KJxxLMefOQ PjihgN658528RUTSk3587581158lFigure 25.7(a) ABCD is a squareThe diagonals of a square bisect each of the rightangles, hencea =90◦2= 45◦(b) EFGH is a rectangle
• Areas of common shapes 221In triangle FGH, 40◦+ 90◦+ b = 180◦, since theangles in a triangle add up to 180◦, from whichb = 50◦. Also, c = 40◦ (alternate angles betweenparallel lines EF and HG). (Alternatively, b and care complementary; i.e., add up to 90◦.)d = 90◦ + c (external angle of a triangle equalsthe sum of the interior opposite angles), henced = 90◦ + 40◦ = 130◦ (or ∠EFH = 50◦ andd = 180◦ − 50◦ = 130◦).(c) JKLM is a rhombusThe diagonals of a rhombus bisect the interiorangles and the opposite internal angles are equal.Thus, ∠JKM=∠MKL = ∠JMK = ∠LMK = 30◦,hence, e = 30◦.In triangle KLM, 30◦ + ∠KLM + 30◦ = 180◦(the angles in a triangle add up to 180◦), hence,∠KLM = 120◦. The diagonal JL bisects ∠KLM,hence, f =120◦2= 60◦.(d) NOPQ is a parallelogramg = 52◦ since the opposite interior angles of aparallelogram are equal.In triangle NOQ, g + h + 65◦ = 180◦ (the anglesin a triangle add up to 180◦), from whichh = 180◦ − 65◦ − 52◦ = 63◦.i = 65◦ (alternate angles between parallel linesNQ and OP).j = 52◦ + i = 52◦ + 65◦ = 117◦ (the externalangle of a triangle equals the sum of the interioropposite angles). (Alternatively, ∠PQO = h =63◦; hence, j = 180◦ − 63◦ = 117◦.)(e) RSTU is a trapezium35◦ + k = 75◦ (external angle of a triangle equalsthe sum of the interior opposite angles), hence,k = 40◦.∠STR = 35◦(alternate angles between parallellines RU and ST). l + 35◦ = 115◦ (external angleof a triangle equals the sum of the interioroppositeangles), hence, l = 115◦ − 35◦ = 80◦.Now try the following Practice ExercisePracticeExercise 96 Common shapes(answers on page 351)1. Find the angles p and q in Figure 25.8(a).2. Find the angles r and s in Figure 25.8(b).3. Find the angle t in Figure 25.8(c).7583881258628958578478408pq srt(a) (b) (c)Figure 25.825.3 Areas of common shapesThe formulae forthe areas of common shapes are shownin Table 25.1.Here are some worked problems to demonstrate howthe formulae are used to determine the area of commonshapes.Problem 2. Calculate the area and length of theperimeter of the square shown in Figure 25.94.0 cm4.0 cmFigure 25.9Area of square = x2= (4.0)2= 4.0cm × 4.0cm= 16.0cm2(Note the unit of area is cm×cm = cm2; i.e., squarecentimetres or centimetres squared.)Perimeter of square = 4.0cm + 4.0cm + 4.0cm+ 4.0cm = 16.0cmProblem 3. Calculate the area and length of theperimeter of the rectangle shown in Figure 25.107.0 cm4.5 cmAD CBFigure 25.10
• 222 Basic Engineering MathematicsTable 25.1 Formulae for the areas of common shapesArea of plane ﬁguresSquare xxArea = x2RectangleIbArea = l × bParallelogramhbArea = b × hTrianglehbArea =12× b × hTrapeziumhbaArea =12(a + b)hCirclerArea = πr2 orπd24Circumference = 2πrRadian measure: 2π radians = 360 degreesSector of circlesr␪Area =θ◦360(πr2)
• Areas of common shapes 223Area of rectangle = l × b = 7.0 × 4.5= 31.5cm2Perimeter of rectangle = 7.0cm + 4.5cm+ 7.0cm + 4.5cm= 23.0cmProblem 4. Calculate the area of theparallelogram shown in Figure 25.1121mm16mmGHE F9mmFigure 25.11Area of a parallelogram = base × perpendicular heightThe perpendicular height h is not shown in Figure 25.11but may be found using Pythagoras’ theorem (seeChapter 21).From Figure 25.12, 92 = 52 + h2, from whichh2 = 92 − 52 = 81 − 25 = 56.Hence, perpendicular height,h =√56 = 7.48mm.9mmEGhFH16 mm 5mmFigure 25.12Hence, area of parallelogram EFGH= 16mm × 7.48mm= 120mm2.Problem 5. Calculate the area of the triangleshown in Figure 25.13IKJ5.68cm1.92cmFigure 25.13Area of triangle IJK =12×base×perpendicular height=12× IJ × JKTo ﬁnd JK, Pythagoras’ theorem is used; i.e.,5.682= 1.922+ JK 2, from whichJK = 5.682 − 1.922 = 5.346cmHence, area of triangle IJK =12× 1.92 × 5.346= 5.132cm2.Problem 6. Calculate the area of the trapeziumshown in Figure 25.1427.4mm8.6 mm5.5 mmFigure 25.14Area of a trapezium =12× (sum of parallel sides)×(perpendicular distancebetween the parallel sides)Hence, area of trapezium LMNO=12× (27.4 + 8.6) × 5.5=12× 36 × 5.5 = 99mm2
• 224 Basic Engineering MathematicsProblem 7. A rectangular tray is 820mm longand 400mm wide. Find its area in (a) mm2 (b) cm2(c) m2(a) Area of tray = length × width = 820 × 400= 328000mm2(b) Since 1cm = 10mm,1cm2= 1cm × 1cm= 10mm × 10mm = 100mm2, or1mm2=1100cm2= 0.01cm2Hence, 328000 mm2= 328000× 0.01cm2= 3280cm2.(c) Since 1m = 100cm,1m2= 1m × 1m= 100cm × 100cm = 10000cm2, or1cm2=110000m2= 0.0001m2Hence, 3280cm2= 3280 × 0.0001m2= 0.3280m2.Problem 8. The outside measurements of apicture frame are 100cm by 50cm. If the frame is4cm wide, ﬁnd the area of the wood used to makethe frameA sketch of the frame is shown shaded in Figure 25.15.100 cm50 cm 42 cm92 cmFigure 25.15Area of wood = area of large rectangle − area ofsmall rectangle= (100 × 50) − (92 × 42)= 5000 − 3864= 1136cm2Problem 9. Find the cross-sectional area of thegirder shown in Figure 25.165mm50 mmBCA8mm70 mm75mm6mmFigure 25.16Thegirdermay bedivided into threeseparaterectangles,as shown.Area of rectangle A = 50 × 5 = 250mm2Area of rectangle B = (75 − 8 − 5) × 6= 62 × 6 = 372mm2Area of rectangle C = 70 × 8 = 560mm2Total area of girder = 250 + 372 + 560= 1182mm2or 11.82cm2Problem 10. Figure 25.17 shows the gable end ofa building. Determine the area of brickwork in thegable end6m5m 5mADCB8mFigure 25.17The shape is that of a rectangle and a triangle.Area of rectangle = 6 × 8 = 48m2Area of triangle =12× base × heightCD = 4m and AD = 5m, hence AC = 3m (since it is a3, 4, 5 triangle – or by Pythagoras).Hence, area of triangle ABD =12× 8 × 3 = 12m2.
• Areas of common shapes 225Total area of brickwork = 48 + 12= 60m2.Now try the following Practice ExercisePracticeExercise 97 Areas of commonshapes (answers on page 351)1. Name the types of quadrilateral shown inFigure25.18(i)to (iv)and determine for each(a) the area and (b) the perimeter.120 mm30 mm(iii)26 cm12 cm16 cm10 cm(iv)4 cm(i)5.94 cm3.5cm6 mm(ii)30mm38mmFigure 25.182. A rectangular plate is 85mm long and42mm wide. Find its area in square centi-metres.3. Arectangular ﬁeld hasan areaof1.2hectaresand a length of 150m. If 1hectare =10000m2,ﬁnd (a)theﬁeld’swidth and (b)thelength of a diagonal.4. Find the area of a triangle whose base is8.5cm and perpendicular height is 6.4cm.5. A square has an area of 162cm2. Determinethe length of a diagonal.6. A rectangular picture has an area of 0.96m2.If one of the sides has a length of 800mm,calculate, in millimetres, the length of theother side.7. Determine the area of each of the angle ironsections shown in Figure 25.19.7cm1cm1cm(a) (b)2cm2cm30mm25mm8mm10mm50mm6mmFigure 25.198. Figure 25.20 shows a 4m wide path aroundthe outside of a 41m by 37m garden. Calcu-late the area of the path.414 37Figure 25.209. The area of a trapezium is 13.5cm2 and theperpendicular distance between its parallelsides is 3cm. If the length of one of the par-allel sides is 5.6cm, ﬁnd the length of theother parallel side.10. Calculate the area of the steel plate shown inFigure 25.21.2560140Dimensionsin mm1002525Figure 25.21
• 226 Basic Engineering Mathematics11. Determine the area of an equilateral triangleof side 10.0cm.12. If paving slabs are produced in 250mm by250mm squares, determine the number ofslabs required to cover an area of 2m2.Here are some further worked problems on ﬁnding theareas of common shapes.Problem 11. Find the area of a circle having aradius of 5cmAreaof circle = πr2= π(5)2= 25π = 78.54cm2Problem 12. Find the area of a circle having adiameter of 15mmAreaof circle =πd24=π(15)24=225π4= 176.7mm2Problem 13. Find the area of a circle having acircumference of 70mmCircumference, c = 2πr,henceradius,r =c2π=702π=35πmmArea of circle = πr2= π35π2=352π= 389.9mm2or 3.899cm2Problem 14. Calculate the area of the sector of acircle having radius 6cm with angle subtended atcentre 50◦Area of sector =θ2360(πr2) =50360(π62)=50 × π × 36360= 15.71cm2Problem 15. Calculate the area of the sector of acircle having diameter 80mm with angle subtendedat centre 107◦42If diameter = 80mm then radius, r = 40mm, andarea of sector =107◦42360(π402) =1074260360(π402)=107.7360(π402)= 1504mm2or 15.04cm2Problem 16. A hollow shaft has an outsidediameter of 5.45cm and an inside diameter of2.25cm. Calculate the cross-sectional area of theshaftThe cross-sectional area of the shaft is shown by theshaded part in Figure 25.22 (often called an annulus).d 52.25 cmD 55.45 cmFigure 25.22Area of shaded part = area of large circle – area ofsmall circle=π D24−πd24=π4(D2− d2)=π4(5.452− 2.252)= 19.35cm2Now try the following Practice ExercisePracticeExercise 98 Areas of commonshapes (answers on page 351)1. A rectangular garden measures 40m by 15m.A 1m ﬂower border is made round the twoshorter sides and one long side. A circularswimming pool of diameter 8m is constructed
• Areas of common shapes 227in the middleof the garden. Find, correct to thenearest square metre, the area remaining.2. Determine the area of circles having (a) aradius of 4cm (b) a diameter of 30mm (c) acircumference of 200mm.3. An annulus has an outside diameter of 60mmand an inside diameter of 20mm. Determineits area.4. If the area of a circle is 320mm2, ﬁnd (a) itsdiameter and (b) its circumference.5. Calculate the areas of the following sectors ofcircles.(a) radius 9cm, angle subtended at centre75◦.(b) diameter 35mm, angle subtended atcentre 48◦37 .6. Determine the shaded area of the templateshown in Figure 25.23.120mm90mm80mmradiusFigure 25.237. An archway consists of a rectangular openingtopped by a semi-circular arch, as shown inFigure 25.24. Determine the area of the open-ing if the width is 1m and the greatest heightis 2m.1 m2 mFigure 25.24Here are some further worked problems of commonshapes.Problem 17. Calculate the area of a regularoctagon if each side is 5cm and the width across theﬂats is 12cmAn octagon is an 8-sided polygon. If radii are drawnfrom the centre of the polygon to the vertices then 8equal triangles are produced, as shown in Figure 25.25.12cm5 mFigure 25.25Area of one triangle =12× base × height=12× 5 ×122= 15cm2Area of octagon = 8 × 15 = 120cm2Problem 18. Determine the area of a regularhexagon which has sides 8cm longA hexagon is a 6-sided polygon which may be dividedinto 6 equal triangles as shown in Figure 25.26. Theangle subtended at the centre of each triangle is 360◦ ÷6 = 60◦. The other two angles in the triangle add up to120◦ and are equal to each other. Hence, each of thetriangles is equilateral with each angle 60◦ and eachside 8cm.4cm8cm8cm608hFigure 25.26Area of one triangle =12× base × height=12× 8 × h
• 228 Basic Engineering Mathematicsh is calculated using Pythagoras’ theorem:82= h2+ 42from which h = 82 − 42 = 6.928cmHence,Area of one triangle =12× 8 × 6.928 = 27.71cm2Area of hexagon = 6 × 27.71= 166.3cm2Problem 19. Figure 25.27 shows a plan of a ﬂoorof a building which is to be carpeted. Calculate thearea of the ﬂoor in square metres. Calculate the cost,correct to the nearest pound, of carpeting the ﬂoorwith carpet costing £16.80perm2, assuming 30%extra carpet is required due to wastage in ﬁtting2.5mMLKJ4m2m0.6m0.6m0.8 m2 m0.8 m30Њ60Њ3m3m2 m 3 mIHGFABЈ BCDEFigure 25.27Area of ﬂoor plan= area of triangle ABC + area of semicircle+ area of rectangle CGLM+ area of rectangle CDEF− area of trapezium HIJKTriangle ABC is equilateral since AB = BC = 3m and,hence, angle B CB = 60◦.sinB CB = BB /3i.e. BB = 3sin60◦= 2.598m.Area of triangle ABC =12(AC)(BB )=12(3)(2.598) = 3.897m2Area of semicircle =12πr2=12π(2.5)2= 9.817m2Area of CGLM = 5 × 7 = 35m2Area of CDEF = 0.8 × 3 = 2.4m2Area of HIJK =12(KH + IJ)(0.8)Since MC = 7m then LG = 7m, henceJI = 7 − 5.2 = 1.8m. Hence,Area of HIJK =12(3 + 1.8)(0.8) = 1.92m2Total ﬂoor area = 3.897 + 9.817 + 35 + 2.4 − 1.92= 49.194m2To allow for 30% wastage, amount of carpet required= 1.3 × 49.194 = 63.95m2Cost of carpet at £16.80 perm2= 63.95 × 16.80 = £1074, correct to the nearest pound.Now try the following Practice ExercisePracticeExercise 99 Areas of commonshapes (answers on page 351)1. Calculate the area of a regular octagon if eachside is 20mm and the width across the ﬂats is48.3mm.2. Determine the area of a regular hexagon whichhas sides 25mm.3. A plot of land is in the shape shown inFigure 25.28. Determine20m20 m20 m30 m10 m20 m30m20 m15m40 m15 m20 mFigure 25.28
• Areas of common shapes 229(a) its area in hectares (1 ha = 104 m2).(b) the length of fencing required, to the near-est metre, to completely enclose the plotof land.25.4 Areas of similar shapesFigure 25.29 shows two squares, one of which has sidesthree times as long as the other.3x3xxx(a) (b)Figure 25.29Area of Figure 25.29(a) = (x)(x) = x2Area of Figure 25.29(b) = (3x)(3x) = 9x2Hence, Figure 25.29(b) has an area (3)2; i.e., 9 times thearea of Figure 25.29(a).In summary, the areas of similar shapes are pro-portional to the squares of corresponding lineardimensions.Problem 20. A rectangular garage is shown on abuilding plan having dimensions 10mm by 20mm.If the plan is drawn to a scale of 1 to 250, determinethe true area of the garage in square metresArea of garage on the plan = 10mm × 20mm= 200mm2Since the areas of similar shapes are proportional to thesquares of corresponding dimensions,True area of garage = 200 × (250)2= 12.5 × 106mm2=12.5 × 106106m2since 1m2 = 106 mm2= 12.5m2Now try the following Practice ExercisePracticeExercise 100 Areas of similarshapes (answers on page 351)1. The area of a park on a map is 500mm2. Ifthe scale of the map is 1 to 40000, deter-mine the true area of the park in hectares(1hectare = 104 m2).2. A model of a boiler is made having an overallheight of 75mm corresponding to an overallheight of the actual boiler of 6m. If the area ofmetal required for the model is 12500mm2,determine, in square metres, the area of metalrequired for the actual boiler.3. The scale of an Ordnance Survey map is1:2500. A circular sports ﬁeld has a diam-eter of 8cm on the map. Calculate its areain hectares, giving your answer correct to3 signiﬁcant ﬁgures. (1 hectare = 104 m2.)
• Chapter 26The circle26.1 IntroductionA circle is a plain ﬁgure enclosed by a curved line,everypoint on which is equidistant from a point within, calledthe centre.In Chapter25,worked problemson theareasofcirclesand sectors were demonstrated. In this chapter, proper-ties of circles are listed and arc lengths are calculated,together with more practical examples on the areas ofsectors of circles. Finally, the equation of a circle isexplained.26.2 Properties of circles(a) The distance from the centre to the curve is calledtheradius,r,ofthecircle(see OP in Figure26.1).CBQOPRAFigure 26.1(b) The boundary of a circle is called the circumfer-ence, c.(c) Any straight line passing through the centre andtouching the circumference at each end is calledthe diameter, d (see QR in Figure 26.1). Thus,d = 2r.(d) The ratiocircumferencediameteris a constant for any cir-cle. This constant is denoted by the Greek letter π(pronounced ‘pie’), where π = 3.14159, correctto 5 decimal places (check with your calculator).Hence,cd= π or c = πd or c = 2πr.(e) A semicircle is one half of a whole circle.(f) A quadrant is one quarter of a whole circle.(g) A tangent to a circle is a straight linewhich meetsthecircleat onepoint only and doesnot cut thecir-cle when produced. AC in Figure 26.1 isa tangentto the circle since it touches the curve at point Bonly. If radius OB is drawn, angleABO is a rightangle.(h) The sector of a circle is the part of a circlebetween radii (for example, the portion OXY ofFigure 26.2 is a sector). If a sector is less than asemicircle it is called a minor sector; if greaterthan a semicircle it is called a major sector.XYTSROFigure 26.2(i) The chord of a circle is any straight line whichdivides the circle into two parts and is termi-nated at each end by the circumference. ST, inFigure 26.2, is a chord.(j) Segment is the name given to the parts into whicha circle is dividedby a chord. If thesegment is lessthan a semicircle it is called a minor segment(see shaded area in Figure 26.2). If the segmentDOI: 10.1016/B978-1-85617-697-2.00026-0
• 236 Basic Engineering Mathematics19. Determine the length of steel strip requiredto make the clip shown in Figure 26.13.125mmrad130Њ100mm100mmFigure 26.1320. A 50◦ tapered hole is checked with a 40mmdiameter ball as shown in Figure 26.14.Determine the length shown as x.70 mmx50Њ40mmFigure 26.1426.5 The equation of a circleThe simplest equation of a circle, centre at the originand radius r, is given byx2+ y2= r2For example, Figure 26.15 shows a circle x2 + y2 =9.332x2ϩy2ϭ9xy2110Ϫ1Ϫ1Ϫ2Ϫ2Ϫ3Ϫ3Figure 26.15More generally, the equation of a circle, centre (a,b)and radius r, is given by(x − a)2+ (y − b)2= r2(1)Figure 26.16 shows a circle (x − 2)2 + (y − 3)2 = 4.r52y5420 2 4 xb5 3a5 2Figure 26.16The general equation of a circle isx2+ y2+ 2ex + 2 f y + c = 0 (2)Multiplying out the bracketed terms in equation (1)givesx2− 2ax + a2+ y2− 2by + b2= r2Comparing this with equation (2) gives2e = −2a, i.e. a = −2e2and 2 f = −2b, i.e. b = −2f2and c = a2+ b2−r2, i.e. r = a2 + b2 − cThus, for example, the equationx2+ y2− 4x − 6y + 9 = 0represents a circle with centre,a = −−42,b = −−62i.e., at (2, 3) andradius, r = 22 + 32 − 9 = 2Hence, x2 + y2 − 4x − 6y + 9 = 0 isthecircleshown inFigure 26.16 (which may be checked by multiplyingoutthe brackets in the equation (x − 2)2 + (y − 3)2 = 4).Problem 18. Determine (a) the radius and (b) theco-ordinates of the centre of the circle given by theequation x2 + y2 + 8x − 2y + 8 = 0
• The circle 237x2+ y2+ 8x − 2y + 8 = 0 is of the form shown inequation (2),where a = −82= −4,b = −−22= 1and r = (−4)2+ 12 − 8 =√9 = 3.Hence, x2 + y2 + 8x − 2y + 8 = 0 represents a circlecentre (−4,1) and radius 3, as shown in Figure 26.17.a= Ϫ4b=1Ϫ224yϪ4Ϫ6Ϫ8 0r=3xFigure 26.17Alternatively, x2 + y2 + 8x − 2y + 8 = 0 may be rear-ranged as(x + 4)2+ (y − 1)2− 9 = 0i.e. (x + 4)2+ (y − 1)2= 32which represents a circle, centre (−4,1) and radius 3,as stated above.Problem 19. Sketch the circle given by theequation x2 + y2 − 4x + 6y − 3 = 0The equation of a circle, centre (a,b), radius r isgiven by(x − a)2+ (y − b)2= r2The general equation of a circle isx2 + y2 + 2ex + 2 f y + c = 0From above a = −2e2,b= −2 f2and r =√a2 + b2 − cHence, if x2 + y2 − 4x + 6y − 3 = 0then a = −−42= 2,b = −62= −3 andr = 22 + (−3)2− (−3) =√16 = 4Thus, the circle has centre (2,−3) and radius 4, asshown in Figure 26.18.Ϫ4Ϫ22341yϪ4Ϫ5Ϫ7Ϫ8Ϫ3Ϫ2 2 4 6 x0r = 4Figure 26.18Alternatively, x2 + y2 − 4x + 6y − 3 = 0 may be rear-ranged as(x − 2)2+ (y + 3)2− 3 − 13 = 0i.e. (x − 2)2+ (y + 3)2= 42which represents a circle, centre (2,−3) and radius 4,as stated above.Now try the following Practice ExercisePracticeExercise 104 The equation of acircle (answers on page 351)1. Determine (a) the radius and (b) theco-ordinates of the centre of the circle givenby the equation x2 + y2 − 6x + 8y + 21 = 0.2. Sketch the circle given by the equationx2+ y2− 6x + 4y − 3 = 0.3. Sketch the curve x2+ (y − 1)2− 25 = 0.4. Sketch the curve x = 6 1 −y62
• Revision Test 10 : Areas and circlesThis assignment covers the material contained in Chapters 25 and 26. The marks available are shown in brackets atthe end of each question.1. A rectangular metal platehas an area of 9600cm2.If the length of the plate is 1.2m, calculate thewidth, in centimetres. (3)2. Calculate the cross-sectional area of the angle ironsection shown in Figure RT10.1, the dimensionsbeing in millimetres. (4)282751934Figure RT10.13. Find the area of the trapezium M N O P shown inFigure RT10.2 when a = 6.3cm,b = 11.7cm andh = 5.5cm. (3)M NOPhabFigure RT10.24. Find the area of the triangle DE F in FigureRT10.3, correct to 2 decimal places. (4)5. Arectangularpark measures150m by 70m.A2mﬂower border is constructed round the two longersides and one short side. A circular ﬁsh pond ofdiameter 15m is in the centre of the park and theremainder of the park is grass. Calculate, correctto the nearest square metre, the area of (a) theﬁsh pond, (b) the ﬂower borders and (c) the grass.(6)6. A swimming pool is 55mlong and 10mwide. Theperpendicular depth at the deep end is 5m and at8.75cm12.44cmFEDFigure RT10.3the shallow end is 1.5m, the slope from one endto the other being uniform. The inside of the poolneeds two coats of a protective paint before it isﬁlled with water. Determine how many litres ofpaint will be needed if 1 litre covers 10m2. (7)7. Find the area of an equilateral triangle of side20.0cm. (4)8. A steel template is of the shape shown inFigure RT10.4, the circular area being removed.Determine the area of the template, in squarecentimetres, correct to 1 decimal place. (8)30mm45mm130mm70mm70mm 150mm60mm30 mm50mmdia.Figure RT10.49. The area of a plot of land on a map is400mm2. If the scale of the map is 1 to 50000,
• Revision Test 10 : Areas and circles 239determine the true area of the land in hectares(1 hectare = 104 m2). (4)10. Determine the shaded area in Figure RT10.5,correct to the nearest square centimetre. (3)20cm2 cmFigure RT10.511. Determine the diameter of a circle, correct tothe nearest millimetre, whose circumference is178.4cm. (2)12. Calculate the area of a circle of radius 6.84cm,correct to 1 decimal place. (2)13. The circumference of a circle is 250mm. Find itsarea, correct to the nearest square millimetre. (4)14. Find the area of the sector of a circle having aradius of 50.0mm, with angle subtended at centreof 120◦. (3)15. Determine the total area of the shape shown inFigure RT10.6, correct to 1 decimal place. (7)7.0 m10.0m6.0 mFigure RT10.616. The radius of a circular cricket ground is75m. Theboundary is painted with white paint and 1 tin ofpaint will paint a line 22.5m long. How many tinsof paint are needed? (3)17. Find the area of a 1.5m wide path surrounding acircular plot of land 100m in diameter. (3)18. A cyclometer shows 2530 revolutions in a dis-tance of 3.7km. Find the diameter of the wheelin centimetres, correct to 2 decimal places. (4)19. The minute hand of a wall clock is 10.5cm long.How far does the tip travel in the course of24 hours? (4)20. Convert(a) 125◦47 to radians.(b) 1.724 radians to degrees and minutes. (4)21. Calculate the length of metal strip needed tomake the clip shown in Figure RT10.7. (7)30mm rad15mm rad15mm rad70mm70mm75mmFigure RT10.722. A lorry has wheels of radius 50cm. Calculate thenumber of complete revolutions a wheel makes(correct to the nearest revolution) when travelling3 miles (assume 1mile = 1.6km). (4)23. The equation of a circle isx2 + y2 + 12x − 4y + 4 = 0. Determine(a) the diameter of the circle.(b) the co-ordinates of the centre of the circle.(7)
• Chapter 27Volumes of common solids27.1 IntroductionThe volume of any solid is a measure of the space occu-pied by the solid. Volume is measured in cubic unitssuch as mm3, cm3 and m3.This chapter deals with ﬁnding volumes of commonsolids; in engineering it is often important to be able tocalculatevolumeorcapacity to estimate,say,theamountof liquid, such as water, oil or petrol, in different shapedcontainers.A prism is a solid with a constant cross-section andwith two ends parallel. The shape of the end is used todescribe the prism. For example, there are rectangularprisms (called cuboids), triangular prisms and circularprisms (called cylinders).On completing this chapter you will be able to cal-culate the volumes and surface areas of rectangular andother prisms, cylinders, pyramids, cones and spheres,together with frusta of pyramids and cones. Volumes ofsimilar shapes are also considered.27.2 Volumes and surface areas ofcommon shapes27.2.1 Cuboids or rectangular prismsA cuboid is a solid ﬁgure bounded by six rectangularfaces; all angles are right angles and opposite faces areequal. A typical cuboid is shown in Figure 27.1 withlength l, breadth b and height h.Volume of cuboid = l × b × handsurface area = 2bh + 2hl + 2lb = 2(bh + hl + lb)hblFigure 27.1A cube is a square prism. If all the sides of a cube are xthenVolume = x3and surface area = 6x2Problem 1. A cuboid has dimensions of 12cm by4cm by 3cm. Determine (a) its volume and (b) itstotal surface areaThe cuboid is similar to that in Figure 27.1, withl = 12cm,b = 4cm and h = 3cm.(a) Volume of cuboid= l × b × h = 12 × 4 × 3= 144 cm3(b) Surface area = 2(bh + hl +lb)= 2(4 × 3 + 3 × 12+ 12 × 4)= 2(12 + 36 + 48)= 2 × 96 = 192 cm2Problem 2. An oil tank is the shape of a cube,each edge being of length 1.5m. Determine (a) themaximum capacity of the tank in m3 and litres and(b) its total surface area ignoring input and outputoriﬁcesDOI: 10.1016/B978-1-85617-697-2.00027-2
• Volumes of common solids 241(a) Volume of oil tank=volume of cube= 1.5m × 1.5m× 1.5m= 1.53m3= 3.375 m31m3 = 100cm × 100cm× 100cm = 106 cm3.Hence,volume of tank = 3.375 × 106cm31litre = 1000cm3, hence oil tank capacity=3.375 × 1061000litres = 3375 litres(b) Surface area of one side = 1.5m × 1.5m= 2.25m2.A cube has six identical sides, hencetotal surface area of oil tank = 6 × 2.25= 13.5 m2Problem 3. A water tank is the shape of arectangular prism having length 2m, breadth 75cmand height 500mm. Determine the capacity of thetank in (a) m3 (b) cm3 (c) litresCapacity means volume; when dealing with liquids, theword capacity is usually used.The water tank is similar in shape to that in Figure 27.1,with l = 2m,b = 75cm and h = 500mm.(a) Capacity of water tank = l × b × h. To use this for-mula, all dimensions must be in the same units.Thus, l = 2m,b = 0.75m and h = 0.5m (since1m = 100cm = 1000mm). Hence,capacity of tank = 2 × 0.75 × 0.5 = 0.75 m3(b) 1m3 = 1m× 1m× 1m= 100cm × 100cm× 100cmi.e., 1 m3= 1 000 000 =106cm3. Hence,capacity = 0.75m3= 0.75 × 106cm3= 750 000 cm3(c) 1litre = 1000cm3. Hence,750 000 cm3=750,0001000= 750 litres27.2.2 CylindersA cylinder is a circular prism. A cylinder of radiusr andheight h is shown in Figure 27.2.hrFigure 27.2Volume = πr2hCurved surface area = 2πrhTotal surface area = 2πrh + 2πr2Total surface area means the curved surface area plusthe area of the two circular ends.Problem 4. A solid cylinder has a base diameterof 12cm and a perpendicular height of 20cm.Calculate (a) the volume and (b) the total surfacearea(a) Volume = πr2h = π ×1222× 20= 720π = 2262 cm3(b) Total surface area= 2πrh + 2πr2= (2 × π × 6 × 20) + (2 × π × 62)= 240π + 72π = 312π = 980 cm2Problem 5. A copper pipe has the dimensionsshown in Figure 27.3. Calculate the volume ofcopper in the pipe, in cubic metres.2.5m12cm25cmFigure 27.3
• 242 Basic Engineering MathematicsOuter diameter, D = 25cm = 0.25m and inner diame-ter, d = 12cm = 0.12m.Area of cross-section of copper=π D24−πd24=π(0.25)24−π(0.12)24= 0.0491 − 0.0113 = 0.0378m2Hence, volume of copper= (cross-sectional area) × length of pipe= 0.0378 × 2.5 = 0.0945 m327.2.3 More prismsA right-angled triangular prism is shown in Figure 27.4with dimensions b,h and l.IbhFigure 27.4Volume =12bhlandsurface area = area of each end+ area of three sidesNotice that the volume is given by the area of the end(i.e. area of triangle = 12 bh) multiplied by the length l.In fact, the volume of any shaped prism is given by thearea of an end multiplied by the length.Problem 6. Determine the volume (in cm3) of theshape shown in Figure 27.512 mm16 mm40 mmFigure 27.5The solid shown in Figure 27.5 is a triangular prism.The volume V of any prism is given by V = Ah, whereA is the cross-sectional area and h is the perpendicularheight. Hence,volume =12× 16 × 12× 40 = 3840mm3= 3.840 cm3(since 1cm3 = 1000mm3)Problem 7. Calculate the volume of theright-angled triangular prism shown in Figure 27.6.Also, determine its total surface area6 cm40cmAB C8 cmFigure 27.6Volume of right-angled triangular prism=12bhl =12× 8 × 6 × 40i.e. volume = 960 cm3
• Volumes of common solids 243Total surface area = area of each end + area of threesides.In triangle ABC, AC2= AB2+ BC2from which, AC = AB2 + BC2 = 62 + 82= 10cmHence, total surface area= 212bh + (AC × 40) + (BC × 40) + (AB × 40)= (8 × 6) + (10 × 40) + (8 × 40) + (6 × 40)= 48 + 400 + 320 + 240i.e. total surface area = 1008 cm2Problem 8. Calculate the volume and totalsurface area of the solid prism shown in Figure 27.715cm5cm5cm5cm4cm11cmFigure 27.7The solid shown in Figure 27.7 is a trapezoidal prism.Volume of prism = cross-sectional area × height=12(11 + 5)4 × 15 = 32 × 15= 480 cm3Surface area of prism= sum of two trapeziums + 4 rectangles= (2 × 32) + (5 × 15) + (11 × 15) + 2(5 × 15)= 64 + 75 + 165 + 150 = 454 cm2Now try the following Practice ExercisePracticeExercise 105 Volumes and surfaceareas of common shapes (answers onpage 351)1. Change a volume of 1200000cm3 to cubicmetres.2. Change a volume of 5000mm3 to cubiccentimetres.3. A metal cube has a surface area of 24cm2.Determine its volume.4. A rectangular block of wood has dimensionsof 40mm by 12mm by 8mm. Determine(a) its volume, in cubic millimetres(b) its total surface area in square millime-tres.5. Determine the capacity, in litres, ofa ﬁsh tankmeasuring 90cm by 60cm by 1.8m, given1litre = 1000cm3.6. A rectangular block of metal has dimensionsof 40mm by 25mm by 15mm. Determine itsvolume in cm3. Find also its mass if the metalhas a density of 9g/cm3.7. Determine the maximum capacity, in litres,of a ﬁsh tank measuring 50cm by 40cm by2.5m(1litre = 1000cm3).8. Determine how many cubic metres of con-crete are required for a 120m long path,150mm wide and 80mm deep.9. A cylinder has a diameter 30mm and height50mm. Calculate(a) its volume in cubic centimetres, correctto 1 decimal place(b) the total surface area in square centime-tres, correct to 1 decimal place.10. Find (a) the volume and (b) the total sur-face area of a right-angled triangular prismof length 80cm and whose triangular endhas a base of 12cm and perpendicularheight 5cm.11. A steel ingot whose volume is 2m2 is rolledout into a plate which is 30mm thick and1.80m wide. Calculate the length of the platein metres.
• 244 Basic Engineering Mathematics12. The volume of a cylinder is 75cm3. If itsheight is 9.0cm, ﬁnd its radius.13. Calculate the volume of a metal tube whoseoutside diameter is 8cm and whose insidediameter is 6cm, if the length of the tube is4m.14. The volume of a cylinder is 400cm3. Ifits radius is 5.20cm, ﬁnd its height. Alsodetermine its curved surface area.15. A cylinder is cast from a rectangular piece ofalloy 5cm by 7cm by 12cm. If the length ofthe cylinder is to be 60cm, ﬁnd its diameter.16. Find the volume and the total surface areaof a regular hexagonal bar of metal of length3m if each side of the hexagon is 6cm.17. A block of lead 1.5m by 90cm by 750mmis hammered out to make a square sheet15mm thick. Determine the dimensions ofthe square sheet, correct to the nearest cen-timetre.18. How long will it take a tap dripping at a rateof 800mm3/s to ﬁll a 3-litre can?19. A cylinder is cast from a rectangular pieceof alloy 5.20cm by 6.50cm by 19.33cm. Ifthe height of the cylinder is to be 52.0cm,determine its diameter, correct to the nearestcentimetre.20. How much concrete is required for the con-struction of the path shown in Figure 27.8, ifthe path is 12cm thick?2 m1.2m8.5 m2.4m3.1 mFigure 27.827.2.4 PyramidsVolume of any pyramid=13×area of base × perpendicular heightA square-based pyramid is shown in Figure 27.9 withbase dimension x by x and the perpendicular height ofthe pyramid h. For the square-base pyramid shown,volume =13x2hhxxFigure 27.9Problem 9. A square pyramid has aperpendicular height of 16cm. If a side of the baseis 6cm, determine the volume of a pyramidVolume of pyramid=13× area of base × perpendicular height=13× (6 × 6) × 16= 192 cm3Problem 10. Determine the volume and the totalsurface area of the square pyramid shown inFigure 27.10 if its perpendicular height is 12cm.Volume of pyramid=13(area of base) × perpendicular height=13(5 × 5) × 12= 100 cm3
• Volumes of common solids 2455cm5cmDCEABFigure 27.10The total surface area consists of a square base and 4equal triangles.Area of triangle ADE=12× base × perpendicular height=12× 5 × ACThe length AC may be calculated using Pythagorastheorem on triangle ABC, where AB = 12cm andBC = 12 × 5 = 2.5cm.AC = AB2 + BC2 = 122 + 2.52 = 12.26cmHence,area of triangle ADE =12× 5 × 12.26 = 30.65cm2Total surface area of pyramid= (5 × 5) + 4(30.65)= 147.6 cm2Problem 11. A rectangular prism of metal havingdimensions of 5cm by 6cm by 18cm is melteddown and recast into a pyramid having arectangular base measuring 6cm by 10cm.Calculate the perpendicular height of the pyramid,assuming no waste of metalVolume of rectangular prism= 5 × 6 × 18 = 540cm3Volume of pyramid=13× area of base × perpendicular heightHence, 540 =13× (6 × 10) × hfrom which, h =3 × 5406 × 10= 27cmi.e. perpendicular height of pyramid = 27 cm27.2.5 ConesA cone is a circular-based pyramid. A cone of baseradius r and perpendicular height h is shown inFigure 27.11.Volume =13× area of base × perpendicular heighthrlFigure 27.11i.e. Volume =13πr2hCurved surface area = πrlTotal surface area = πrl + πr2Problem 12. Calculate the volume, in cubiccentimetres, of a cone of radius 30mm andperpendicular height 80mmVolume of cone=13πr2h =13× π × 302× 80= 75398.2236... mm31cm = 10mm and1cm3 = 10mm × 10mm× 10mm = 103 mm3, or1 mm3= 10−3cm3Hence, 75398.2236... mm3= 75398.2236...× 10−3 cm3i.e.,volume = 75.40 cm3
• 246 Basic Engineering MathematicsAlternatively, fromthe question,r = 30mm = 3cm andh = 80mm = 8cm. Hence,volume =13πr2h =13× π × 32× 8 = 75.40 cm3Problem 13. Determine the volume and totalsurface area of a cone of radius 5cm andperpendicular height 12cmThe cone is shown in Figure 27.12.hϭ12cmrϭ5 cmlFigure 27.12Volume of cone =13πr2h =13× π × 52× 12= 314.2 cm3Total surfacearea=curved surfacearea+areaofbase= πrl + πr2From Figure 27.12, slant height l may be calculatedusing Pythagoras’ theorem:l = 122 + 52 = 13cmHence, total surface area= (π × 5 × 13) + (π × 52)= 282.7 cm2.27.2.6 SpheresFor the sphere shown in Figure 27.13:Volume =43πr3and surface area = 4πr2rFigure 27.13Problem 14. Find the volume and surface area ofa sphere of diameter 10cmSince diameter= 10cm, radius, r = 5cm.Volume of sphere =43πr3=43× π × 53= 523.6 cm3Surface area of sphere = 4πr2= 4 × π × 52= 314.2 cm2Problem 15. The surface area of a sphere is201.1cm2. Find the diameter of the sphere andhence its volumeSurface area of sphere= 4πr2.Hence, 201.1cm2 = 4 × π × r2,from which r2=201.14 × π= 16.0and radius, r =√16.0 = 4.0cmfrom which, diameter = 2 × r = 2 × 4.0 = 8.0 cmVolume of sphere =43πr3=43× π × (4.0)3= 268.1 cm3Now try the following Practice ExercisePracticeExercise 106 Volumes and surfaceareas of common shapes (answers onpage 351)1. If a cone has a diameter of 80mm and aperpendicular height of 120mm, calculateits volume in cm3 and its curved surfacearea.2. A square pyramid has a perpendicular heightof 4cm. If a side of the base is 2.4cm long,ﬁnd the volume and total surface area of thepyramid.
• Volumes of common solids 2473. A sphere has a diameter of 6cm. Determineits volume and surface area.4. If the volume of a sphere is 566cm3, ﬁnd itsradius.5. A pyramid having a square base has a per-pendicular height of 25cm and a volume of75cm3. Determine, in centimetres, thelengthof each side of the base.6. A cone has a base diameter of 16mm and aperpendicular height of 40mm. Find its vol-ume correct to the nearest cubic millimetre.7. Determine (a) the volume and (b) the surfacearea of a sphere of radius 40mm.8. The volume of a sphere is 325cm3. Deter-mine its diameter.9. Given the radius of the earth is 6380km,calculate, in engineering notation(a) its surface area in km2.(b) its volume in km3.10. An ingot whose volume is 1.5m3 is to bemade into ball bearings whose radii are8.0cm. How many bearings will be producedfrom the ingot, assuming 5% wastage?27.3 Summary of volumes and surfaceareas of common solidsA summary of volumes and surface areas of regularsolids is shown in Table 27.1.Table 27.1 Volumes and surface areas of regularsolidsRectangular prism(or cuboid)hbl Volume = l × b × hSurface area = 2(bh + hl + lb)CylinderhrVolume = πr2hTotal surface area = 2πrh + 2πr2Triangular prismIbhVolume =12bhlSurface area = area of each end +area of three sidesPyramidhAVolume =13× A × hTotal surface area =sum of areas of trianglesforming sides + area of baseConehrlVolume =13πr2hCurved surface area = πrlTotal surface area = πrl + πr2SphererVolume =43πr3Surface area = 4πr227.4 More complex volumes andsurface areasHere are some worked problems involving more com-plex and composite solids.Problem 16. A wooden section is shown inFigure 27.14. Find (a) its volume in m3 and(b) its total surface area
• 248 Basic Engineering Mathematics3m rr5 8cm12cmFigure 27.14(a) The section of wood is a prism whose end com-prises a rectangle and a semicircle. Since theradius of the semicircle is 8cm, the diameter is16cm. Hence, the rectangle has dimensions12cmby 16cm.Area of end = (12 × 16) +12π82= 292.5cm2Volume of wooden section= area of end × perpendicular height= 292.5 × 300 = 87 750 cm3=87750106m3,since 1m3= 106cm3= 0.08775 m3(b) The total surface area comprises the two ends(each of area 292.5cm2), three rectangles and acurved surface (which is half a cylinder). Hence,total surface area= (2 × 292.5) + 2(12 × 300)+ (16 × 300) +12(2π × 8 × 300)= 585 + 7200+ 4800 + 2400π= 20 125 cm2or 2.0125 m2Problem 17. A pyramid has a rectangular base3.60cm by 5.40cm. Determine the volume and totalsurface area of the pyramid if each of its slopingedges is 15.0cmThe pyramid is shown in Figure 27.15. To calculate thevolume of the pyramid, the perpendicular height EF isrequired. Diagonal BD is calculated using Pythagoras’theorem,i.e. BD = 3.602 + 5.402 = 6.490cmHence, EB =12BD =6.4902= 3.245cmCFDGAEHB15.0cm15.0cm15.0cm15.0cm5.40cm3.60cmFigure 27.15Using Pythagoras’ theorem on triangle BEF givesBF2= E B2+ E F2from which EF = BF2 − EB2= 15.02 − 3.2452 = 14.64cmVolume of pyramid=13(area of base)(perpendicular height)=13(3.60 × 5.40)(14.64) = 94.87 cm3Area of triangle ADF (which equals triangle BCF)= 12(AD)(FG), where G is the midpoint of AD.Using Pythagoras’ theorem on triangle FGA givesFG = 15.02 − 1.802 = 14.89cmHence, area of triangleADF =12(3.60)(14.89)= 26.80cm2Similarly, if H is the mid-point of AB,FH = 15.02 − 2.702 = 14.75cmHence, area of triangle ABF (which equals triangleCDF) =12(5.40)(14.75) = 39.83cm2
• Volumes of common solids 249Total surface area of pyramid= 2(26.80) + 2(39.83) + (3.60)(5.40)= 53.60 + 79.66 + 19.44= 152.7 cm2Problem 18. Calculate the volume and totalsurface area of a hemisphere of diameter 5.0cmVolume of hemisphere =12(volume of sphere)=23πr3=23π5.023= 32.7 cm3Total surface area= curved surface area + area of circle=12(surface area of sphere) + πr2=12(4πr2) + πr2= 2πr2+ πr2= 3πr2= 3π5.022= 58.9cm2Problem 19. A rectangular piece of metal havingdimensions 4cm by 3cm by 12cm is melted downand recast into a pyramid having a rectangular basemeasuring 2.5cm by 5cm. Calculate theperpendicular height of the pyramidVolume of rectangular prism of metal = 4 × 3 × 12= 144cm3Volume of pyramid=13(area of base)(perpendicular height)Assuming no waste of metal,144 =13(2.5 × 5)(height)i.e. perpendicular height of pyramid=144 × 32.5 × 5= 34.56 cmProblem 20. A rivet consists of a cylindricalhead, of diameter 1cm and depth 2mm, and a shaftof diameter 2mm and length 1.5cm. Determine thevolume of metal in 2000 such rivetsRadius of cylindrical head=12cm = 0.5cm andheight of cylindrical head= 2mm = 0.2cm.Hence, volume of cylindrical head= πr2h = π(0.5)2(0.2) = 0.1571cm3Volume of cylindrical shaft= πr2h = π0.222(1.5) = 0.0471cm3Total volume of 1 rivet= 0.1571 + 0.0471= 0.2042cm3Volume of metal in 2000 such rivets= 2000 × 0.2042 = 408.4 cm3Problem 21. A solid metal cylinder of radius6cm and height 15cm is melted down and recastinto a shape comprising a hemisphere surmountedby a cone. Assuming that 8% of the metal is wastedin the process, determine the height of the conicalportion if its diameter is to be 12cmVolume of cylinder = πr2h = π × 62× 15= 540π cm3If 8% of metal is lost then 92% of 540π gives thevolume of the new shape, shown in Figure 27.16.hr12cmFigure 27.16
• 250 Basic Engineering MathematicsHence, the volume of (hemisphere + cone)= 0.92 × 540π cm3i.e. 1243πr3+13πr2h = 0.92 × 540πDividing throughout by π gives23r3+13r2h = 0.92 × 540Sincethediameterofthenewshapeisto be12cm,radiusr = 6cm,hence 23(6)3+13(6)2h = 0.92 × 540144 + 12h = 496.8i.e. height of conical portion,h =496.8 − 14412= 29.4 cmProblem 22. A block of copper having a mass of50kg is drawn out to make 500m of wire ofuniform cross-section. Given that the density ofcopper is 8.91g/cm3, calculate (a) the volume ofcopper, (b) the cross-sectional area of the wire and(c) the diameter of the cross-section of the wire(a) A density of 8.91g/cm3 means that 8.91g of cop-per has a volume of 1cm3, or 1g of copper has avolume of (1 ÷ 8.91)cm3.Density =massvolumefrom which volume =massdensityHence, 50kg, i.e. 50000g, has avolume =massdensity=500008.91cm3= 5612 cm3(b) Volume of wire = area of circular cross-section× length of wire.Hence, 5612cm3= area × (500 × 100cm)from which, area =5612500 × 100cm2= 0.1122 cm2(c) Area of circle = πr2 orπd24hence, 0.1122=πd24from which, d =4 × 0.1122π=0.3780cmi.e. diameter of cross-section is 3.780 mm.Problem 23. A boiler consists of a cylindricalsection of length 8m and diameter 6m, on one endof which is surmounted a hemispherical section ofdiameter 6m and on the other end a conical sectionof height 4m and base diameter 6m. Calculate thevolume of the boiler and the total surface areaThe boiler is shown in Figure 27.17.8m4mCIBRAQP3m6mFigure 27.17Volume of hemisphere, P =23πr3=23× π × 33= 18π m3Volume of cylinder, Q = πr2h = π × 32× 8= 72π m3Volume of cone, R =13πr2h =13× π × 32× 4= 12π m3Total volume of boiler = 18π + 72π + 12π= 102π = 320.4 m3Surface area of hemisphere, P =12(4πr2)= 2 × π × 32= 18πm2
• Volumes of common solids 251Curved surface area of cylinder, Q = 2πrh= 2 × π × 3 × 8= 48π m2The slant height of the cone, l, is obtained by Pythago-ras’ theorem on triangle ABC, i.e.l = 42 + 32 = 5Curved surface area of cone,R = πrl = π × 3 × 5 = 15π m2Total surface area of boiler= 18π + 48π + 15π= 81π = 254.5 m2Now try the following Practice ExercisePracticeExercise 107 More complexvolumes and surface areas (answers onpage 351)1. Find the total surface area of a hemisphere ofdiameter 50mm.2. Find (a) the volume and (b) the total surfacearea of a hemisphere of diameter 6cm.3. Determine the mass of a hemispherical cop-per container whose external and internalradii are 12cm and 10cm, assuming that1cm3 of copper weighs 8.9g.4. A metal plumb bob comprises a hemispheresurmounted by a cone. If the diameter of thehemisphere and cone are each 4cm and thetotal length is 5cm, ﬁnd its total volume.5. A marquee is in the form of a cylinder sur-mounted by acone.Thetotal height is6mandthe cylindrical portion has a height of 3.5mwith a diameter of 15m. Calculate the surfacearea of material needed to make the marqueeassuming 12% of the material is wasted inthe process.6. Determine (a) the volume and (b) the totalsurface area of the following solids.(i) a cone of radius 8.0cm and perpendi-cular height 10cm.(ii) a sphere of diameter 7.0cm.(iii) a hemisphere of radius 3.0cm.(iv) a 2.5cm by 2.5cm square pyramid ofperpendicular height 5.0cm.(v) a 4.0cm by 6.0cm rectangular pyra-mid of perpendicular height 12.0cm.(vi) a 4.2cm by 4.2cm square pyramidwhose sloping edges are each 15.0cm(vii) a pyramid having an octagonal base ofside 5.0cm and perpendicular height20cm.7. A metal sphere weighing 24kg is melteddown and recast into a solid cone of baseradius 8.0cm. If the density of the metal is8000kg/m3 determine(a) the diameter of the metal sphere.(b) the perpendicular height of the cone,assuming that 15% of the metal is lostin the process.8. Find the volume of a regular hexagonal pyra-midiftheperpendicularheight is16.0cmandthe side of the base is 3.0cm.9. A buoy consists of a hemisphere surmountedby a cone. The diameter of the cone andhemisphere is 2.5m and the slant height ofthe cone is 4.0m. Determine the volume andsurface area of the buoy.10. A petrol container is in the form of a centralcylindrical portion 5.0m long with a hemi-spherical section surmounted on each end. Ifthe diameters of the hemisphere and cylinderare both 1.2m, determine the capacity of thetank in litres (1litre = 1000cm3).11. Figure 27.18 shows a metal rod section.Determine its volume and total surface area.1.00cmradius 1.00m2.50cmFigure 27.18
• 252 Basic Engineering Mathematics12. Find the volume (in cm3) of the die-castingshown in Figure 27.19. The dimensions arein millimetres.6030 rad2550100Figure 27.1913. The cross-section of part of a circular ven-tilation shaft is shown in Figure 27.20, endsAB and CD being open. Calculate(a) the volume of the air, correct tothe nearest litre, contained in thepart of the system shown, neglect-ing the sheet metal thickness (given1litre = 1000cm3).(b) the cross-sectional area of the sheetmetal used to make the system, insquare metres.(c) thecost ofthesheet metal ifthematerialcosts £11.50 per square metre, assum-ing that 25% extra metal is required dueto wastage.500mmAB2m1.5m1.5m800mmDCFigure 27.2027.5 Volumes and surface areas offrusta of pyramids and conesThe frustum of a pyramid or cone is the portionremain-ing when a part containing the vertex is cut off by aplane parallel to the base.The volume of a frustum of a pyramid or coneis givenby the volume of the whole pyramid or cone minus thevolume of the small pyramid or cone cut off.The surface area of the sides of a frustum of a pyra-mid or cone is given by the surface area of the wholepyramid or cone minus the surface area of the smallpyramid or cone cut off. This gives the lateral surfacearea of the frustum. If the total surface area of the frus-tum is required then the surface area of the two parallelends are added to the lateral surface area.There is an alternative method for ﬁnding the volumeand surface area of a frustum of a cone. With referenceto Figure 27.21,hRIrFigure 27.21Volume =13πh(R2+ Rr + r2)Curved surface area = πl(R + r)Total surface area = πl(R + r) + πr2+ πR2Problem 24. Determine the volume of a frustumof a cone if the diameter of the ends are 6.0cm and4.0cm and its perpendicular height is 3.6cm(i) Method 1A section through the vertex of a complete cone isshown in Figure 27.22.Using similar triangles,APDP=DRBRHence, AP2.0=3.61.0from which AP =(2.0)(3.6)1.0= 7.2cm
• Volumes of common solids 2534.0cmPEADRBQC2.0cm3.6cm1.0cm3.0cm6.0cmFigure 27.22The height of the large cone= 3.6 + 7.2= 10.8cmVolume of frustum of cone= volume of large cone− volume of small cone cut off=13π(3.0)2(10.8) −13π(2.0)2(7.2)= 101.79 − 30.16 = 71.6 cm3(ii) Method 2From above, volume of the frustum of a cone=13πh(R2+ Rr +r2)where R = 3.0cm,r = 2.0cm and h = 3.6cmHence, volume of frustum=13π(3.6) (3.0)2+ (3.0)(2.0) + (2.0)2=13π(3.6)(19.0) = 71.6 cm3Problem 25. Find the total surface area of thefrustum of the cone in Problem 24.(i) Method 1Curved surface area of frustum = curved surfacearea of large cone − curved surface area of smallcone cut off.From Figure 27.22, using Pythagoras’ theorem,AB2= AQ2+ BQ2from which AB = 10.82 + 3.02 = 11.21cmand AD2= AP2+ DP2from which AD = 7.22 + 2.02 = 7.47cmCurved surface area of large cone = πrl= π(BQ)(AB) = π(3.0)(11.21)= 105.65cm2and curved surface area of small cone= π(DP)(AD) = π(2.0)(7.47) = 46.94cm2Hence, curved surface area of frustum= 105.65 − 46.94= 58.71cm2Total surface area of frustum= curved surface area+ area of two circular ends= 58.71 + π(2.0)2+ π(3.0)2= 58.71 + 12.57 + 28.27 = 99.6 cm2(ii) Method 2From page 252, total surface area of frustum= πl(R +r) + πr2+ π R2where l = BD = 11.21 − 7.47 = 3.74cm,R = 3.0cm and r = 2.0cm. Hence,total surface area of frustum= π(3.74)(3.0 + 2.0) + π(2.0)2+ π(3.0)2= 99.6 cm2
• 254 Basic Engineering MathematicsProblem 26. A storage hopper is in the shape of afrustum of a pyramid. Determine its volume if theends of the frustum are squares of sides 8.0m and4.6m, respectively, and the perpendicular heightbetween its ends is 3.6mThe frustum is shown shaded in Figure 27.23(a) as partof a complete pyramid. A section perpendicular to thebase through the vertex is shown in Figure 27.23(b).8.0m4.6cm8.0m(a)B2.3m1.7m 2.3m(b)4.0m2.3m 3.6mCAEDGFH4.6cmFigure 27.23By similar trianglesCGBG=BHAHFrom which, heightCG = BGBHAH=(2.3)(3.6)1.7= 4.87mHeight of complete pyramid = 3.6 + 4.87 = 8.47mVolume of large pyramid =13(8.0)2(8.47)= 180.69m3Volume of small pyramid cut off =13(4.6)2(4.87)= 34.35m3Hence, volume of storage hopper= 180.69 − 34.35= 146.3 m3Problem 27. Determine the lateral surface area ofthe storage hopper in Problem 26The lateral surface area of the storage hopper consistsof four equal trapeziums. From Figure 27.24,Area of trapezium PRSU =12(PR + SU)(QT )4.6m8.0mPUTSO8.0mQR4.6mFigure 27.24OT = 1.7m (same as AH in Figure 27.23(b) andOQ = 3.6m. By Pythagoras’ theorem,QT = OQ2 + OT 2 = 3.62 + 1.72 = 3.98mArea of trapezium PRSU=12(4.6 + 8.0)(3.98) = 25.07m2Lateral surface area of hopper = 4(25.07)= 100.3 m2Problem 28. A lampshade is in the shape of afrustum of a cone. The vertical height of the shadeis 25.0cm and the diameters of the ends are 20.0cmand 10.0cm, respectively. Determine the area of thematerial needed to form the lampshade, correct to 3signiﬁcant ﬁguresThe curved surface area of a frustum of a cone= πl(R +r) from page 252. Since the diameters ofthe ends of the frustum are 20.0cm and 10.0cm, fromFigure 27.25,r5 5.0cmIh525.0cmR510.0cm5.0cmFigure 27.25
• Volumes of common solids 255r = 5.0cm, R = 10.0cmand l = 25.02 + 5.02 = 25.50cmfrom Pythagoras’ theorem.Hence, curved surface area= π(25.50)(10.0 + 5.0) = 1201.7cm2i.e., the area of material needed to form the lampshadeis 1200 cm2, correct to 3 signiﬁcant ﬁgures.Problem 29. A cooling tower is in the form of acylinder surmounted by a frustum of a cone, asshown in Figure 27.26. Determine the volume of airspace in the tower if 40% of the space is used forpipes and other structures12.0m25.0m12.0m30.0mFigure 27.26Volume of cylindrical portion = πr2h= π25.022(12.0)= 5890m3Volume of frustum of cone =13πh(R2+ Rr +r2)where h = 30.0 − 12.0 = 18.0m,R = 25.0 ÷ 2 = 12.5mand r = 12.0 ÷ 2 = 6.0m.Hence, volume of frustum of cone=13π(18.0) (12.5)2+ (12.5)(6.0) + (6.0)2= 5038m3Total volume of cooling tower = 5890 + 5038= 10 928m3If 40% of space is occupied thenvolume of air space= 0.6 × 10928 = 6557 m3Now try the following Practice ExercisePracticeExercise 108 Volumes and surfaceareas of frusta of pyramidsand cones(answers on page 352)1. The radii of the faces of a frustum of a coneare 2.0cm and 4.0cm and the thickness of thefrustum is 5.0cm. Determine its volume andtotal surface area.2. A frustum of a pyramid has square ends,the squares having sides 9.0cm and 5.0cm,respectively. Calculate the volume and totalsurface area of the frustum if the perpendiculardistance between its ends is 8.0cm.3. A cooling tower is in the form of a frustum ofa cone. The base has a diameter of 32.0m, thetop has a diameter of 14.0m and the verticalheight is 24.0m. Calculate the volume of thetower and the curved surface area.4. A loudspeaker diaphragm is in the form of afrustum of a cone. If the end diameters are28.0cm and 6.00cm and the vertical distancebetween the ends is 30.0cm, ﬁnd the area ofmaterial needed to cover the curved surface ofthe speaker.5. A rectangular prism of metal having dimen-sions 4.3cm by 7.2cm by 12.4cm is melteddown and recast into a frustum of a squarepyramid, 10% of the metal being lost in theprocess. If the ends of the frustum are squaresof side 3cm and 8cm respectively, ﬁnd thethickness of the frustum.6. Determine the volume and total surface areaof a bucket consisting of an inverted frustumof a cone, of slant height 36.0cm and enddiameters 55.0cm and 35.0cm.7. A cylindrical tank of diameter 2.0m and per-pendicular height 3.0m is to be replaced bya tank of the same capacity but in the formof a frustum of a cone. If the diameters ofthe ends of the frustum are 1.0m and 2.0m,respectively, determine the vertical heightrequired.
• 256 Basic Engineering Mathematics27.6 Volumes of similar shapesFigure 27.27 shows two cubes, one of which has sidesthree times as long as those of the other.3xxxx3x3x(b)(a)Figure 27.27Volume of Figure 27.27(a) = (x)(x)(x) = x3Volume of Figure 27.27(b) = (3x)(3x)(3x) = 27x3Hence, Figure 27.27(b) has a volume (3)3, i.e. 27, timesthe volume of Figure 27.27(a).Summarizing, the volumes of similar bodies areproportional to the cubes of corresponding lineardimensions.Problem 30. A car has a mass of 1000kg.A model of the car is made to a scale of 1 to 50.Determine the mass of the model if the car and itsmodel are made of the same materialVolume of modelVolume of car=1503since the volume of similar bodies are proportional tothe cube of corresponding dimensions.Mass=density×volume and, since both car and modelare made of the same material,Mass of modelMass of car=1503Hence, mass of model= (mass of car)1503=1000503= 0.008 kg or 8 gNow try the following Practice ExercisePracticeExercise 109 Volumes of similarshapes (answers on page 352)1. The diameter of two spherical bearings are inthe ratio 2:5. What is the ratio of their vol-umes?2. An engineering component has a mass of400g. If each of its dimensions are reducedby 30%, determine its new mass.
• Chapter 28Irregular areas and volumes,and mean values28.1 Areas of irregular figuresAreas of irregular plane surfaces may be approximatelydetermined by using(a) a planimeter,(b) the trapezoidal rule,(c) the mid-ordinate rule, or(d) Simpson’s rule.Such methods may be used by, for example, engineersestimating areas ofindicatordiagrams ofsteam engines,surveyors estimating areas of plots of land or navalarchitects estimating areas of water planes or transversesections of ships.(a) A planimeter is an instrument for directly mea-suring small areas bounded by an irregular curve.There are many different kinds of planimeters butall operate in a similar way. A pointer on theplanimeter is used to trace around the boundaryof the shape. This induces a movement in anotherpart of the instrument and a reading of this is usedto establish the area of the shape.(b) Trapezoidal ruleTo determine the area PQRS in Figure 28.1,(i) Divide base PS into any number of equalintervals, each of width d (the greater thenumber of intervals, the greater the accu-racy).(ii) Accurately measure ordinates y1, y2, y3, etc.y1y2y3y4y5y6y7RQPd d d d d dSFigure 28.1(iii) Area PQRS= dy1 + y72+ y2 + y3 + y4 + y5 + y6 .In general, the trapezoidal rule statesArea =width ofinterval12ﬁrst + lastordinate+⎛⎝sum ofremainingordinates⎞⎠⎤⎦(c) Mid-ordinate ruley1y2y3y4y5y6CBAd d d d d dDFigure 28.2To determine the area ABCD of Figure 28.2,DOI: 10.1016/B978-1-85617-697-2.00028-4
• 258 Basic Engineering Mathematics(i) Divide base AD into any number of equalintervals, each of width d (the greater thenumberofintervals,thegreatertheaccuracy).(ii) Erect ordinates in the middle of each interval(shown by broken lines in Figure 28.2).(iii) Accurately measure ordinates y1, y2, y3, etc.(iv) Area ABCD= d(y1 + y2 + y3 + y4 + y5 + y6).In general, the mid-ordinate rule statesArea =(width of interval)(sum ofmid-ordinates)(d) Simpson’s ruleTo determine the area PQRS of Figure 28.1,(i) Divide base PS into an even number of inter-vals, each of width d (the greater the numberof intervals, the greater the accuracy).(ii) Accurately measure ordinates y1, y2, y3, etc.(iii) Area PQRS =d3[(y1 + y7) + 4(y2 + y4 + y6)+2(y3 + y5)]In general, Simpson’s rule statesArea =13width ofintervalﬁrst + lastordinate+ 4sum of evenordinates+ 2sum of remainingodd ordinatesProblem 1. A car starts from rest and its speed ismeasured every second for 6s.Time t (s) 0 1 2 3 4 5 6Speed v (m/s) 0 2.5 5.5 8.75 12.5 17.5 24.0Determine the distance travelled in 6 seconds (i.e.the area under the v/t graph), using (a) thetrapezoidal rule (b) the mid-ordinate rule(c) Simpson’s ruleA graph of speed/time is shown in Figure 28.3.(a) Trapezoidal rule (see (b) above)The time base is divided into 6 strips, each ofwidth 1s, and the length of the ordinates measured.Thus,area = (1)0 + 24.02+ 2.5 + 5.5+ 8.75 + 12.5 + 17.5 = 58.75 mGraph of speed/timeSpeed(m/s)302520151052.54.05.57.08.7510.7512.515.017.520.2524.00 1 2 3Time (seconds)4 5 61.25Figure 28.3(b) Mid-ordinate rule (see (c) above)The time base is divided into 6 strips each ofwidth 1s. Mid-ordinates are erected as shown inFigure28.3 by the broken lines. The length of eachmid-ordinate is measured. Thus,area = (1)[1.25 + 4.0 + 7.0 + 10.75 + 15.0+ 20.25] = 58.25m(c) Simpson’s rule (see (d) above)The time base is divided into 6 strips each of width1s and the length of the ordinatesmeasured. Thus,area =13(1)[(0 + 24.0) + 4(2.5 + 8.75+ 17.5) + 2(5.5 + 12.5)] = 58.33mProblem 2. A river is 15m wide. Soundings ofthe depth are made at equal intervals of 3m acrossthe river and are as shown below.Depth (m) 0 2.2 3.3 4.5 4.2 2.4 0Calculate the cross-sectional area of the ﬂow ofwater at this point using Simpson’s ruleFrom (d) above,Area =13(3)[(0+0) + 4(2.2+4.5+2.4)+2(3.3+4.2)]= (1)[0 + 36.4 + 15] = 51.4m2
• Irregular areas and volumes, and mean values 259Now try the following Practice ExercisePractice Exercise 110 Areas of irregularﬁgures (answers on page 352)1. Plot a graph of y = 3x − x2 by completinga table of values of y from x = 0 to x = 3.Determine the area enclosed by the curve,the x-axis and ordinates x = 0 and x = 3 by(a) the trapezoidal rule (b) the mid-ordinaterule (c) Simpson’s rule.2. Plot the graph of y = 2x2 + 3 between x = 0and x = 4. Estimate the area enclosed by thecurve, the ordinates x = 0 and x = 4 and thex-axis by an approximate method.3. The velocity of a car at one second intervals isgiven in the following table.Time t (s) 0 1 2 3 4 5 6v (m/s) 0 2.0 4.5 8.0 14.0 21.0 29.0VelocityDetermine the distance travelled in 6 sec-onds (i.e. the area under the v/t graph) usingSimpson’s rule.4. The shape of a piece of land is shown inFigure 28.4. To estimate the area of the land,a surveyor takes measurements at intervalsof 50m, perpendicular to the straight portionwith the results shown (the dimensions beingin metres). Estimate the area of the land inhectares (1ha = 104 m2).50 50200 190 180 13016014050 50 50 50Figure 28.45. The deck of a ship is 35m long. At equal inter-vals of 5m the width is given by the followingtable.Width (m) 0 2.8 5.2 6.5 5.8 4.1 3.0 2.3Estimate the area of the deck.28.2 Volumes of irregular solidsIf the cross-sectional areas A1, A2, A3,... of an irreg-ular solid bounded by two parallel planes are known atequal intervals of width d (as shown in Figure 28.5), bySimpson’s ruleVolume, V =d3[(A1 + A7) + 4(A2 + A4 + A6)+2(A3 + A5)]ddddddA6 A7A2A1 A3 A4 A5Figure 28.5Problem 3. A tree trunk is 12m in length and hasa varying cross-section. The cross-sectional areas atintervals of 2m measured from one end are0.52, 0.55, 0.59, 0.63, 0.72, 0.84 and 0.97m2.Estimate the volume of the tree trunkA sketch of the tree trunk is similar to that shownin Figure 28.5 above, where d = 2m, A1 = 0.52m2,A2 = 0.55m2, and so on.Using Simpson’s rule for volumes givesVolume =23[(0.52 + 0.97) + 4(0.55 + 0.63 + 0.84)+ 2(0.59 + 0.72)]=23[1.49 + 8.08 + 2.62] = 8.13 m3Problem 4. The areas of seven horizontalcross-sections of a water reservoir at intervals of10m are 210, 250, 320, 350, 290, 230 and 170m2.Calculate the capacity of the reservoir in litresUsing Simpson’s rule for volumes givesVolume =103[(210 + 170) + 4(250 + 350 + 230)+ 2(320 + 290)]=103[380+ 3320 + 1220] = 16 400m3
• 260 Basic Engineering Mathematics16400m3= 16400× 106cm3.Since 1 litre = 1000cm3,capacity of reservoir =16400× 1061000litres= 16400000 = 16.4 × 106litresNow try the following Practice ExercisePracticeExercise 111 Volumes of irregularsolids (answers on page 352)1. The areas of equidistantly spaced sections ofthe underwater form of a small boat are as fol-lows:1.76, 2.78, 3.10, 3.12, 2.61, 1.24 and 0.85m2.Determine the underwater volume if thesections are 3m apart.2. To estimate the amount of earth to beremoved when constructing a cutting, thecross-sectional area at intervals of 8m wereestimated as follows:0, 2.8, 3.7, 4.5, 4.1, 2.6 and 0m3.Estimate the volume of earth to be excavated.3. Thecircumferenceofa 12mlong log oftimberof varying circular cross-section is measuredat intervals of 2m along its length and theresults are as follows. Estimate the volume ofthe timber in cubic metres.Distance fromone end(m) 0 2 4 6Circumference(m) 2.80 3.25 3.94 4.32Distance fromone end(m) 8 10 12Circumference(m) 5.16 5.82 6.3628.3 Mean or average values ofwaveformsThe mean or average value, y, of the waveform shownin Figure 28.6 is given byy =area under curvelength of base, bIf the mid-ordinaterule is used to ﬁnd the area under thecurve, theny =sum of mid-ordinatesnumber of mid-ordinates=y1 + y2 + y3 + y4 + y5 + y6 + y77for Figure 28.6y1y2y3y4 y5 y6 y7d d d dbd d dyFigure 28.6For a sine wave, the mean or average value(a) over one complete cycle is zero (seeFigure 28.7(a)),(b) over half a cycle is 0.637 × maximum value or2π× maximum value,(c) of a full-wave rectiﬁed waveform (seeFigure 28.7(b)) is 0.637 × maximum value,(d) of a half-wave rectiﬁed waveform (seeFigure 28.7(c)) is 0.318 × maximum valueor1π× maximum value.V0 tVmV0(a) (b)tVm(c)V0 tVmFigure 28.7
• Irregular areas and volumes, and mean values 261Problem 5. Determine the average values overhalf a cycle of the periodic waveforms shown inFigure 28.8(c)Voltage(V)0102 4 6 8210t (ms)(a)0Voltage(V)201 2 3 4220t (ms)(b)0Current(A)3211 2 3 4232221 t (s)5 6Figure 28.8(a) Area undertriangularwaveform (a) for a half cycleis given byArea =12(base)(perpendicular height)=12(2 × 10−3)(20) = 20 × 10−3VsAverage value of waveform=area under curvelength of base=20 × 10−3Vs2 × 10−3s= 10V(b) Area under waveform (b) for a half cycle= (1 × 1) + (3 × 2) = 7AsAverage value of waveform =area under curvelength of base=7As3s= 2.33A(c) A half cycle of the voltage waveform (c) iscompleted in 4ms.Area under curve =12{(3 − 1)10−3}(10)= 10 × 10−3VsAverage value of waveform =area under curvelength of base=10 × 10−3Vs4 × 10−3s= 2.5VProblem 6. Determine the mean value of currentover one complete cycle of the periodic waveformsshown in Figure 28.90Current(mA)54 8 12 16 20 24 28t (ms)0Current(A)22 4 6 8 10 12 t (ms)Figure 28.9(a) One cycle of the trapezoidal waveform (a) is com-pleted in 10ms (i.e. the periodic time is 10ms).Area under curve = area of trapezium=12(sum of parallel sides)(perpendiculardistance between parallel sides)=12{(4 + 8) ×10−3}(5 × 10−3)= 30 × 10−6AsMean value over one cycle =area under curvelength of base=30 × 10−6 As10 × 10−3 s= 3mA(b) One cycle of the saw-tooth waveform (b) is com-pleted in 5ms.Area under curve =12(3 × 10−3)(2)= 3 × 10−3AsMean value over one cycle =area under curvelength of base=3 × 10−3 As5 × 10−3 s= 0.6A
• 262 Basic Engineering MathematicsProblem 7. The power used in a manufacturingprocess during a 6 hour period is recorded atintervals of 1 hour as shown below.Time (h) 0 1 2 3 4 5 6Power (kW) 0 14 29 51 45 23 0Plot a graph of power against time and, by using themid-ordinate rule, determine (a) the area under thecurve and (b) the average value of the powerThe graph of power/time is shown in Figure 28.10.Graph of power/timePower(kW)50403020100 1 2 3Time (hours)4 5 621.57.0 42.0 49.5 37.0 10.0Figure 28.10(a) Thetimebaseisdivided into 6 equal intervals,eachof width 1 hour. Mid-ordinates are erected (shownby broken lines in Figure 28.10) and measured.The values are shown in Figure 28.10.Area under curve= (width of interval)(sum of mid-ordinates)= (1)[7.0 + 21.5 + 42.0 + 49.5 + 37.0 + 10.0]= 167kWh (i.e. a measure of electrical energy)(b) Average value of waveform =area under curvelength of base=167kWh6h= 27.83 kWAlternatively, average value=sum of mid-ordinatesnumber of mid-ordinatesProblem 8. Figure 28.11 shows a sinusoidaloutput voltage of a full-wave rectiﬁer. Determine,using the mid-ordinate rule with 6 intervals, themean output voltage0 308608908 1808 2708 36082␲10Voltage(V)3␲2␲2␲␪Figure 28.11Onecycleoftheoutput voltageiscompleted inπ radiansor 180◦. The base is divided into 6 intervals, each ofwidth 30◦. The mid-ordinate of each interval will lie at15◦,45◦,75◦, etc.At 15◦ the height of the mid-ordinate is10sin 15◦ = 2.588V,At 45◦ the height of the mid-ordinate is10sin 45◦= 7.071V, and so on.The results are tabulated below.Mid-ordinate Height of mid-ordinate15◦10sin15◦= 2.588V45◦ 10sin45◦ = 7.071V75◦ 10sin75◦ = 9.659V105◦ 10sin105◦ = 9.659V135◦ 10sin135◦ = 7.071V165◦ 10sin165◦ = 2.588VSum of mid-ordinates = 38.636VMean or average value of output voltage=sum of mid-ordinatesnumber of mid-ordinates=38.6366= 6.439V(With a larger number of intervals a more accurateanswer may be obtained.)For a sine wave the actual mean value is0.637 × maximum value, which in this problemgives 6.37V.Problem 9. An indicator diagram for a steamengine is shown in Figure 28.12. The base line has
• Irregular areas and volumes, and mean values 263been divided into 6 equally spaced intervals and thelengths of the 7 ordinates measured with the resultsshown in centimetres. Determine(a) the area of the indicator diagram usingSimpson’s rule(b) the mean pressure in the cylinder given that1cm represents 100kPa.12.0cm3.6 4.0 3.5 2.9 2.2 1.7 1.6Figure 28.12(a) The width of each interval is12.06cm. UsingSimpson’s rule,area =13(2.0)[(3.6 + 1.6) + 4(4.0 + 2.9 + 1.7)+ 2(3.5 + 2.2)]=23[5.2 + 34.4 + 11.4] = 34cm2(b) Mean height of ordinates =area of diagramlength of base=3412= 2.83cmSince 1cm represents 100kPa,mean pressure in the cylinder= 2.83cm × 100kPa/cm = 283kPaNow try the following Practice ExercisePracticeExercise 112 Mean or averagevalues of waveforms (answers on page 352)1. Determine the mean value of the periodicwaveforms shown in Figure 28.13 over a halfcycle.2. Find the average value of the periodic wave-forms shown in Figure 28.14 over one com-plete cycle.(a)0Current(A)210 2022t (ms)(b)0Voltage(V)1005 102100t (ms)(c)Current(A)0515 3025t (ms)Figure 28.13Voltage(mV)0102 4 6 8 10 t(ms)Current(A)052 4 6 8 10 t(ms)Figure 28.143. An alternating current hasthefollowing valuesat equal intervals of 5ms:Time (ms) 0 5 10 15 20 25 30Current (A) 0 0.9 2.6 4.9 5.8 3.5 0Plot a graph of current against time and esti-mate the area under the curve over the 30msperiod, using the mid-ordinate rule, and deter-mine its mean value.4. Determine, using an approximate method, theaverage value of a sine wave of maximumvalue 50V for (a) a half cycle (b) a completecycle.5. An indicator diagram of a steam engine is12cm long. Seven evenly spaced ordinates,including the end ordinates, are measured asfollows:5.90, 5.52, 4.22, 3.63, 3.32, 3.24 and 3.16cm.Determine the area of the diagram and themean pressure in the cylinder if 1cm repre-sents 90kPa.
• Revision Test 11 : Volumes, irregular areas and volumes, and mean valuesThis assignment covers the material contained in Chapters 27 and 28. The marks available are shown in brackets atthe end of each question.1. A rectangular block of alloy has dimensions of60mm by 30mm by 12mm. Calculate the volumeof the alloy in cubic centimetres. (3)2. Determine how many cubic metres of concrete arerequired for a 120m long path, 400mm wide and10cm deep. (3)3. Find the volume of a cylinder of radius 5.6cmand height 15.5cm. Give the answer correct to thenearest cubic centimetre. (3)4. A garden roller is 0.35m wide and has a diame-ter of 0.20m. What area will it roll in making 40revolutions? (4)5. Find the volume of a cone of height 12.5cm andbase diameter 6.0cm, correct to 1 decimal place.(3)6. Find (a) the volume and (b) the total surfacearea of the right-angled triangular prism shown inFigure RT11.1. (9)9.70cm4.80cm11.6cmFigure RT11.17. A pyramid having a square base has a volumeof 86.4cm3. If the perpendicular height is 20cm,determine the length of each side of the base. (4)8. A copper pipe is 80m long. It has a bore of 80mmand an outside diameter of 100mm. Calculate, incubic metres, the volume of copper in the pipe. (4)9. Find (a) the volume and (b) the surface area of asphere of diameter 25mm. (4)10. A piece of alloy with dimensions 25mm by60mm by 1.60m is melted down and recast intoa cylinder whose diameter is 150mm. Assum-ing no wastage, calculate the height of thecylinder in centimetres, correct to 1 decimalplace. (4)11. Determine the volume (in cubic metres) and thetotal surface area (in square metres) of a solidmetal cone of base radius 0.5m and perpendicularheight 1.20m. Give answers correct to 2 decimalplaces. (6)12. A rectangular storage container has dimensions3.2m by 90cm by 60cm. Determine its volume in(a) m3 (b) cm3. (4)13. Calculate (a) the volume and (b) the total surfacearea of a 10cm by 15cm rectangular pyramid ofheight 20cm. (8)14. A water container is of the form of a central cylin-drical part 3.0m long and diameter 1.0m, with ahemispherical section surmounted at each end asshown in Figure RT11.2. Determine the maximumcapacity of the container, correct to the nearestlitre. (1 litre = 1000cm3.)3.0m1.0mFigure RT11.2(5)15. Find the total surface area of a bucket consist-ing of an inverted frustum of a cone of slantheight 35.0cm and end diameters 60.0cm and40.0cm. (4)16. A boat has a mass of 20000kg. A model of theboat is made to a scale of 1 to 80. If the model ismade of the same material as the boat, determinethe mass of the model (in grams). (3)
• Revision Test 11 : Volumes, irregular areas and volumes, and mean values 26517. Plot a graph of y = 3x2+ 5 from x = 1 to x = 4.Estimate,correct to 2 decimal places,using 6 inter-vals, the area enclosed by the curve, the ordinatesx = 1 and x = 4, and the x-axis by(a) the trapezoidal rule(b) the mid-ordinate rule(c) Simpson’s rule. (16)18. A circular cooling tower is 20m high. The insidediameter of the tower at different heights is givenin the following table.Height (m) 0 5.0 10.0 15.0 20.0Diameter (m) 16.0 13.3 10.7 8.6 8.0Determine the area corresponding to each diame-ter and hence estimate the capacity of the tower incubic metres. (7)19. A vehicle starts from rest and its velocity ismeasured every second for 6 seconds, with thefollowing results.Time t (s) 0 1 2 3 4 5 6Velocity v (m/s) 0 1.2 2.4 3.7 5.2 6.0 9.2Using Simpson’s rule, calculate(a) the distance travelled in 6s (i.e. the area underthe v/t graph),(b) the average speed over this period. (6)
• Chapter 29Vectors29.1 IntroductionThis chapter initially explains the difference betweenscalar and vector quantities and shows how a vector isdrawn and represented.Any object that is acted upon by an external forcewill respond to that force by moving in the line ofthe force. However, if two or more forces act simul-taneously, the result is more difﬁcult to predict; theability to add two or more vectors then becomesimportant.This chapter thus shows how vectors are added andsubtracted, both by drawing and by calculation, and howﬁnding the resultant of two or more vectors has manyuses in engineering. (Resultant means the single vectorwhich would have the same effect as the individual vec-tors.) Relative velocities and vector i,j,k notation arealso brieﬂy explained.29.2 Scalars and vectorsThe time taken to ﬁll a water tank may be measured as,say, 50s. Similarly, the temperature in a room may bemeasured as, say, 16◦C or the mass of a bearing may bemeasured as, say, 3kg. Quantities such as time, temper-ature and mass are entirely deﬁned by a numerical valueand are called scalars or scalar quantities.Not all quantities are like this. Some are deﬁnedby more than just size; some also have direction. Forexample, the velocity of a car may be 90km/h duewest, a force of 20N may act vertically downwards,or an acceleration of 10m/s2 may act at 50◦ to thehorizontal.Quantities such as velocity, force and acceleration,which have both a magnitude and a direction, arecalled vectors.Now try the following Practice ExercisePracticeExercise 113 Scalar and vectorquantities (answers on page 352)1. State the difference between scalar and vectorquantities.In problems 2 to 9, state whether the quantitiesgiven are scalar or vector.2. A temperature of 70◦C3. 5m3 volume4. A downward force of 20N5. 500J of work6. 30cm2 area7. A south-westerly wind of 10knots8. 50m distance9. An acceleration of 15m/s2 at 60◦ to thehorizontal29.3 Drawing a vectorA vector quantity can be represented graphically by aline, drawn so that(a) the length of the line denotes the magnitude of thequantity, and(b) the direction of the line denotes the direction inwhich the vector quantity acts.An arrow is used to denote the sense, or direction, of thevector.The arrow end of a vector is called the ‘nose’ and theother end the ‘tail’. For example, a force of 9N actingDOI: 10.1016/B978-1-85617-697-2.00029-6
• Vectors 267at 45◦to the horizontal is shown in Figure 29.1. Notethat an angle of +45◦ is drawn from the horizontal andmoves anticlockwise.9N0a45ЊFigure 29.1A velocity of 20m/s at −60◦ is shown in Figure 29.2.Note that an angle of −60◦ is drawn from the horizontaland moves clockwise.60Њ20m/s0bFigure 29.229.3.1 Representing a vectorThere are a number of ways of representing vectorquantities. These include(a) Using bold print.(b)−→AB where an arrow above two capital lettersdenotes the sense of direction, where A is thestarting point and B the end point of the vector.(c) AB or a; i.e., a line over the top of letter.(d) a; i.e., underlined letter.The force of 9N at 45◦ shown in Figure 29.1 may berepresented as0a or−→0a or 0aThe magnitude of the force is 0a.Similarly, the velocity of 20m/s at −60◦ shown inFigure 29.2 may be represented as0b or−→0b or 0bThe magnitude of the velocity is 0b.In this chapter a vector quantity is denoted by boldprint.29.4 Addition of vectors by drawingAdding two or more vectors by drawing assumes thata ruler, pencil and protractor are available. Resultsobtained by drawing are naturally not as accurate asthose obtained by calculation.(a) Nose-to-tail methodTwo force vectors, F1 and F2, are shown inFigure 29.3. When an object is subjected to morethan one force, the resultant of the forces is foundby the addition of vectors.␪F2F1Figure 29.3To add forces F1 and F2,(i) Force F1 is drawn to scale horizontally,shown as 0a in Figure 29.4.(ii) From the nose of F1, force F2 is drawn atangle θ to the horizontal, shown as ab.(iii) The resultant force is given by length 0b,which may be measured.This procedure is called the ‘nose-to-tail’ or‘triangle’ method.␪F2F1ab0Figure 29.4(b) Parallelogram methodTo add the two force vectors, F1 and F2 ofFigure 29.3,(i) A line cb is constructed which is parallel toand equal in length to 0a (see Figure 29.5).(ii) A line ab is constructed which is parallel toand equal in length to 0c.(iii) The resultant force is given by the diagonalof the parallelogram; i.e., length 0b.This procedure is called the ‘parallelogram’method.
• 268 Basic Engineering Mathematics0cF1F2abFigure 29.5Problem 1. A force of 5N is inclined at an angleof 45◦ to a second force of 8N, both forces acting ata point. Find the magnitude of the resultant of thesetwo forces and the direction of the resultant withrespect to the 8N force by (a) the nose-to-tailmethod and (b) the parallelogram methodThe two forces are shown in Figure 29.6. (Although the8N force is shown horizontal, it could have been drawnin any direction.)45Њ5N8NFigure 29.6(a) Nose-to-tail method(i) The 8N force is drawn horizontally 8 unitslong, shown as 0a in Figure 29.7.(ii) From the nose of the 8N force, the 5N forceis drawn 5 units long at an angle of 45◦ to thehorizontal, shown as ab.(iii) The resultant force is given by length 0b andis measured as 12 N and angle θ is measuredas 17◦.␪ 4585N8N a0bFigure 29.7(b) Parallelogram method(i) In Figure 29.8, a line is constructed whichis parallel to and equal in length to the 8Nforce.(ii) A line is constructed which is parallel to andequal in length to the 5N force.␪45Њ5N8Nb0Figure 29.8(iii) The resultant force is given by the diagonalof the parallelogram, i.e. length 0b, and ismeasured as 12 N and angle θ is measuredas 17◦.Thus, the resultant of the two force vectors inFigure 29.6 is 12 N at 17◦ to the 8 N force.Problem 2. Forces of 15 and 10N are at an angleof 90◦ to each other as shown in Figure 29.9. Find,by drawing, the magnitude of the resultant of thesetwo forces and the direction of the resultant withrespect to the 15N force15N10NFigure 29.9Using the nose-to-tail method,(i) The 15N force is drawn horizontally 15 unitslong, as shown in Figure 29.10.10NR15N␪Figure 29.10(ii) From the nose of the 15N force, the 10N forceis drawn 10 units long at an angle of 90◦ to thehorizontal as shown.
• Vectors 269(iii) The resultant force is shown as R and is measuredas 18N and angle θ is measured as 34◦.Thus, the resultant of the two force vectors is 18 N at34◦ to the 15 N force.Problem 3. Velocities of 10m/s, 20m/s and15m/s act as shown in Figure 29.11. Determine, bydrawing, the magnitude of the resultant velocity andits direction relative to the horizontal15Њ␯3␯2␯130Њ10m/s20m/s15m/sFigure 29.11When more than 2 vectors are being added the nose-to-tail method is used. The order in which the vectors areadded does not matter. In this case the order taken is ν1,then ν2, then ν3. However, if a different order is takenthe same result will occur.(i) ν1 is drawn 10 units long at an angle of 30◦ to thehorizontal, shown as 0a in Figure 29.12.b195Њ105Њ030Њ␯1␯3␯2raFigure 29.12(ii) From the nose of ν1, ν2 is drawn 20 units long atan angle of 90◦ to the horizontal, shown as ab.(iii) From the nose of ν2, ν3 is drawn 15 units long atan angle of 195◦ to the horizontal, shown as br.(iv) The resultant velocity is given by length 0r andis measured as 22m/s and the angle measured tothe horizontal is 105◦.Thus, the resultant of the three velocities is 22 m/s at105◦ to the horizontal.Worked Examples 1 to 3 have demonstrated how vec-tors are added to determine their resultant and theirdirection. However, drawing to scale is time-consumingand not highly accurate. The following sections demon-strate how to determine resultant vectors by calculationusing horizontal and vertical components and, wherepossible, by Pythagoras’ theorem.29.5 Resolving vectors into horizontaland vertical componentsA force vector F is shown in Figure 29.13 at angle θto the horizontal. Such a vector can be resolved intotwo components such that the vector addition of thecomponents is equal to the original vector.␪FFigure 29.13The two components usually taken are a horizontalcomponent and a vertical component.Ifaright-angledtriangle is constructed as shown in Figure 29.14, 0a iscalled the horizontal component of F and ab is calledthe vertical component of F.0aFb␪Figure 29.14From trigonometry(see Chapter 21 and remember SOHCAH TOA),cosθ =0a0b, from which 0a = 0bcosθ = F cosθ
• 270 Basic Engineering Mathematicsi.e. the horizontal component of F = F cosθ, andsinθ =ab0bfrom which,ab = 0bsinθ = F sinθi.e. the vertical component of F = F sinθ.Problem 4. Resolve the force vector of 50N at anangle of 35◦ to the horizontal into its horizontal andvertical componentsThe horizontal component of the 50N force,0a = 50cos35◦ = 40.96N.The vertical component of the 50N force,ab = 50sin35◦ = 28.68N.The horizontal and vertical components are shown inFigure 29.15.358040.96N28.68N50NabFigure 29.15(To check: by Pythagoras,0b = 40.962 + 28.682 = 50Nand θ = tan−1 28.6840.96= 35◦Thus, the vector addition of components 40.96N and28.68N is 50N at 35◦.)Problem 5. Resolve the velocity vector of 20m/sat an angle of −30◦ to the horizontal into horizontaland vertical componentsThe horizontal component of the 20m/s velocity,0a = 20cos(−30◦) = 17.32m/s.The vertical component of the 20m/s velocity,ab = 20sin(−30◦) = −10m/s.The horizontal and vertical components are shown inFigure 29.16.Problem 6. Resolve the displacement vector of40m at an angle of 120◦into horizontal and verticalcomponents30820m/s210m/s17.32m/sba0Figure 29.16The horizontal component of the 40m displacement,0a = 40cos120◦ = −20.0m.The vertical component of the 40m displacement,ab = 40sin120◦ = 34.64m.The horizontal and vertical components are shown inFigure 29.17.220.0N40 N1208034.64NabFigure 29.1729.6 Addition of vectors bycalculationTwo forcevectors, F1 and F2,areshown in Figure29.18,F1 being at an angle of θ1 and F2 at an angle of θ2.F1 F2F1sin␪1F2sin␪2HV␪1␪2F2 cos ␪2F1 cos ␪1Figure 29.18A method of adding two vectors together is to usehorizontal and vertical components.The horizontal component of force F1 is F1 cosθ1 andthe horizontal component of force F2 is F2 cosθ2. The
• Vectors 271total horizontal component of the two forces,H = F1 cosθ1 + F2 cosθ2The vertical component of force F1 is F1 sinθ1 and thevertical component of force F2 is F2 sinθ2. The totalvertical component of the two forces,V = F1 sinθ1 + F2 sinθ2Since we have H and V , the resultant of F1 and F2is obtained by using the theorem of Pythagoras. FromFigure 29.19,0b2= H2+ V 2i.e. resultant = H2 + V 2 at an anglegiven by θ = tan−1 VHVbResultantaH0␪Figure 29.19Problem 7. A force of 5N is inclined at an angleof 45◦ to a second force of 8N, both forces acting ata point. Calculate the magnitude of the resultant ofthese two forces and the direction of the resultantwith respect to the 8N forceThe two forces are shown in Figure 29.20.4588N5NFigure 29.20The horizontal component of the 8N force is 8cos0◦and the horizontal component of the 5N force is5cos45◦. The total horizontal component of the twoforces,H = 8cos0◦+ 5cos45◦= 8 + 3.5355 = 11.5355The vertical component of the 8N force is 8sin0◦andthe vertical component of the 5N force is 5sin45◦. Thetotal vertical component of the two forces,V = 8sin0◦+ 5sin45◦= 0 + 3.5355 = 3.5355␪ResultantH ϭ11.5355 NV ϭ3.5355NFigure 29.21From Figure 29.21, magnitude of resultant vector= H2 + V 2= 11.53552 + 3.53552 = 12.07NThe direction of the resultant vector,θ = tan−1 VH= tan−1 3.535511.5355= tan−10.30648866... = 17.04◦Thus, the resultant of the two forces is a single vectorof 12.07 N at 17.04◦ to the 8 N vector.Problem 8. Forces of 15N and 10N are at anangle of 90◦ to each other as shown in Figure 29.22.Calculate the magnitude of the resultant of thesetwo forces and its direction with respect to the15N force10N15 NFigure 29.22The horizontal component of the 15N force is 15cos0◦and the horizontal component of the 10N force is10cos90◦. The total horizontal component of the twovelocities,H = 15cos0◦+ 10cos90◦= 15 + 0 = 15
• 272 Basic Engineering MathematicsThe vertical component of the 15N force is 15sin0◦andthe vertical component of the 10N force is 10sin90◦.The total vertical component of the two velocities,V = 15sin0◦+ 10sin90◦= 0 + 10 = 10Magnitude of resultant vector=√H2 + V 2 =√152 + 102 = 18.03NThe direction of the resultant vector,θ = tan−1 VH= tan−1 1015= 33.69◦Thus, the resultant of the two forces is a single vectorof 18.03 N at 33.69◦ to the 15 N vector.There is an alternative method of calculating the resul-tant vector in this case. If we used the triangle method,the diagram would be as shown in Figure 29.23.15 N10NR␪Figure 29.23Since a right-angled triangle results, we could usePythagoras’ theorem without needing to go through theprocedure for horizontal and vertical components. Infact, the horizontal and vertical components are 15Nand 10N respectively.This is, of course, a special case. Pythagoras can onlybeusedwhenthere isanangleof90◦ betweenvectors.This is demonstrated in worked Problem 9.Problem 9. Calculate the magnitude anddirection of the resultant of the two accelerationvectors shown in Figure 29.24.15m/s228m/s2Figure 29.24The 15m/s2acceleration is drawn horizontally, shownas 0a in Figure 29.25.015a28bR␣ ␪Figure 29.25From the nose of the 15m/s2 acceleration, the 28m/s2acceleration isdrawn at an angleof 90◦ to thehorizontal,shown as ab.The resultant acceleration, R, is given by length 0b.Since a right-angled triangle results, the theorem ofPythagoras may be used.0b = 152 + 282 = 31.76m/s2and α = tan−1 2815= 61.82◦Measuring from the horizontal,θ = 180◦ − 61.82◦ = 118.18◦Thus, the resultant of the two accelerations is a singlevector of 31.76 m/s2 at 118.18◦ to the horizontal.Problem 10. Velocities of 10m/s, 20m/s and15m/s act as shown in Figure 29.26. Calculate themagnitude of the resultant velocity and its directionrelative to the horizontal20 m/s10 m/s15m/s158308␯1␯2␯3Figure 29.26The horizontal component of the 10m/s velocity= 10cos30◦ = 8.660 m/s,
• Vectors 273the horizontal component of the 20m/s velocity is20cos90◦ = 0 m/sand the horizontal component of the 15m/s velocity is15cos195◦ = −14.489m/s.The total horizontal component of the three velocities,H = 8.660 + 0 − 14.489 = −5.829m/sThe vertical component of the 10m/s velocity= 10sin30◦ = 5m/s,the vertical component of the 20m/s velocity is20sin90◦ = 20 m/sand the vertical component of the 15m/s velocity is15sin195◦= −3.882 m/s.The total vertical component of the three forces,V = 5 + 20 − 3.882 = 21.118m/s5.82921.118R␣ ␪Figure 29.27From Figure 29.27, magnitude of resultant vector,R = H2 + V 2 = 5.8292 + 21.1182 = 21.91m/sThe direction of the resultant vector,α = tan−1 VH= tan−1 21.1185.829= 74.57◦Measuring from the horizontal,θ = 180◦ − 74.57◦ = 105.43◦.Thus, the resultant of the three velocities is a singlevector of 21.91 m/s at 105.43◦ to the horizontal.Now try the following Practice ExercisePracticeExercise 114 Addition of vectorsby calculation (answers on page 352)1. A force of 7N is inclined at an angle of 50◦ toa second force of 12N, both forces acting ata point. Calculate the magnitude of the resul-tant of the two forces and the direction of theresultant with respect to the 12N force.2. Velocities of 5m/s and 12m/s act at a pointat 90◦ to each other. Calculate the resultantvelocity and itsdirectionrelativeto the12m/svelocity.3. Calculate the magnitude and direction of theresultant of the two force vectors shown inFigure 29.28.10N13NFigure 29.284. Calculate the magnitude and direction of theresultant of the two force vectors shown inFigure 29.29.22N18NFigure 29.295. A displacement vector s1 is 30m at 0◦. Asecond displacement vector s2 is 12m at 90◦.Calculate the magnitude and direction of theresultant vector s1 + s26. Three forces of 5N, 8N and 13N act asshown in Figure 29.30. Calculate the mag-nitude and direction of the resultant force.5 N13N8 N708608Figure 29.30
• 274 Basic Engineering Mathematics7. If velocity v1 = 25m/s at 60◦ andv2 = 15m/sat −30◦,calculatethemagnitudeand direction of v1 + v2.8. Calculate the magnitude and direction of theresultant vector of the force system shown inFigure 29.31.3081586086 N8 N4 NFigure 29.319. Calculate the magnitude and direction ofthe resultant vector of the system shown inFigure 29.32.1584582m/s4m/s3.5m/s308Figure 29.3210. An object is acted upon by two forces of mag-nitude10N and 8N at an angle of 60◦ to eachother. Determine the resultant force on theobject.11. A ship heads in a direction of E 20◦S at aspeed of 20knots while the current is 4knotsin a directionof N 30◦E. Determine the speedand actual direction of the ship.29.7 Vector subtractionIn Figure 29.33, a force vector F is represented byoa. The vector (−oa) can be obtained by drawing avector from o in the opposite sense to oa but havingthe same magnitude, shown as ob in Figure 29.33; i.e.,ob = (−oa).b2FFaoFigure 29.33For two vectors acting at a point, as shown inFigure 29.34(a), the resultant of vector addition isos = oa + obFigure 29.33(b) shows vectors ob + (−oa) that is,ob − oa and the vector equation is ob − oa = od. Com-paring od in Figure 29.34(b) with the broken line abin Figure 29.34(a) shows that the second diagonal ofthe parallelogram method of vector addition gives themagnitude and direction of vector subtraction of oafrom ob.(b)(a)a2ad bb sao oFigure 29.34Problem 11. Accelerations of a1 = 1.5m/s2 at90◦and a2 = 2.6m/s2at 145◦act at a point. Finda1 + a2 and a1 − a2 by (a) drawing a scale vectordiagram and (b) calculation(a) The scale vector diagram is shown in Figure 29.35.By measurement,a1 + a2 = 3.7 m/s2at 126◦a1 − a2 = 2.1 m/s2at 0◦(b) Resolving horizontally and vertically givesHorizontal component of a1 + a2,H = 1.5cos90◦ + 2.6cos145◦ = −2.13
• Vectors 275a12a2a11a22.6m/s21268a12a2a2a11.5m/s21458Figure 29.35Vertical component of a1 + a2,V = 1.5sin90◦ + 2.6sin145◦ = 2.99From Figure 29.36, the magnitude of a1 + a2,R = (−2.13)2 + 2.992 = 3.67m/s2In Figure 29.36, α = tan−1 2.992.13= 54.53◦ andθ = 180◦ − 54.53◦ = 125.47◦Thus, a1 + a2 = 3.67m/s2at 125.47◦R02.132.99␣␪Figure 29.36Horizontal component of a1 − a2= 1.5cos90◦− 2.6cos145◦= 2.13Vertical component of a1 − a2= 1.5sin90◦ − 2.6sin145◦ = 0Magnitude of a1 − a2 =√2.132 + 02= 2.13m/s2Direction of a1 − a2 = tan−1 02.13= 0◦Thus, a1 − a2 = 2.13m/s2at 0◦Problem 12. Calculate the resultant of(a) v1 − v2 + v3 and (b) v2 − v1 − v3 whenv1 = 22 units at 140◦, v2 = 40 units at 190◦ andv3 = 15 units at 290◦(a) The vectors are shown in Figure 29.37.1540221408190829082H 1H1V2VFigure 29.37The horizontal component of v1 − v2 + v3= (22cos140◦) − (40cos190◦) + (15cos290◦)= (−16.85) − (−39.39) + (5.13) = 27.67 unitsThe vertical component of v1 −v2 +v3= (22sin140◦) − (40sin190◦) + (15sin290◦)= (14.14) − (−6.95) + (−14.10) = 6.99 unitsThe magnitude of the resultant,R =√27.672 + 6.992 = 28.54 unitsThe direction of the resultantR = tan−1 6.9927.67= 14.18◦Thus, v1 − v2 + v3 = 28.54 units at 14.18◦(b) The horizontal component of v2 − v1 − v3= (40cos190◦) − (22cos140◦) − (15cos290◦)= (−39.39) − (−16.85) − (5.13) =−27.67 unitsThe vertical component of v2 − v1 − v3= (40sin190◦) − (22sin140◦) − (15sin290◦)= (−6.95) − (14.14) − (−14.10) = −6.99 unitsFrom Figure 29.38, the magnitude of the resul-tant, R = (−27.67)2 + (−6.99)2 = 28.54 unitsand α = tan−1 6.9927.67= 14.18◦, from which,θ = 180◦ + 14.18◦ = 194.18◦
• 276 Basic Engineering MathematicsR027.676.99␪␣Figure 29.38Thus, v2 − v1 − v3 = 28.54 units at 194.18◦This result is as expected, since v2 − v1 − v3= −(v1 − v2 + v3) and the vector 28.54 units at194.18◦ is minus times (i.e. is 180◦ out of phasewith) the vector 28.54 units at 14.18◦Now try the following Practice ExercisePracticeExercise 115 Vector subtraction(answers on page 352)1. Forces of F1 = 40 N at 45◦ and F2 = 30 N at125◦act at a point. Determine by drawing andby calculation (a) F1 + F2 (b) F1 − F22. Calculate the resultant of (a) v1 + v2 − v3(b) v3 − v2 + v1 when v1 = 15m/s at 85◦,v2 = 25m/s at 175◦ and v3 = 12 m/s at 235◦.29.8 Relative velocityFor relative velocity problems, some ﬁxed datum pointneeds to be selected. This is often a ﬁxed point on theearth’s surface. In any vector equation, only the startand ﬁnish points affect the resultant vector of a system.Two different systems are shown in Figure 29.39, but,in each of the systems, the resultant vector is ad.a db(a)adbc(b)Figure 29.39The vector equation of the system shown inFigure 29.39(a) isad = ab + bdand that for the system shown in Figure 29.39(b) isad = ab + bc + cdThus, in vector equations of this form, only the ﬁrst andlast letters, a and d, respectively, ﬁx the magnitude anddirection of the resultant vector. This principle is usedin relative velocity problems.Problem 13. Two cars, P and Q, are travellingtowards the junction of two roads which are at rightangles to one another. Car P has a velocity of45km/h due east and car Q a velocity of 55km/hdue south. Calculate (a) the velocity of car Prelative to car Q and (b) the velocity of car Qrelative to car P(a) The directions of the cars are shown inFigure 29.40(a), which is called a space diagram.The velocity diagram is shown in Figure 29.40(b),in which pe is taken as the velocity of car P rela-tive to point e on the earth’s surface. The velocityof P relative to Q is vector pq and the vector equa-tion is pq = pe + eq. Hence, the vector directionsare as shown, eq being in the opposite directionto qe.(a) (b) (c)QPENWS55 km/h45km/hpeqpeqFigure 29.40From the geometry of the vector triangle, the mag-nitude of pq =√452 + 552 = 71.06km/h and thedirection of pq = tan−1 5545= 50.71◦That is, the velocity of car P relative to car Q is71.06 km/h at 50.71◦(b) The velocity of car Q relative to car P is given bythe vector equation qp = qe + ep and the vectordiagram is as shown in Figure 29.40(c), having epopposite in direction to pe.
• Vectors 277From the geometry of thisvector triangle, the mag-nitude of qp =√452 + 552 = 71.06m/s and thedirection of qp = tan−1 5545= 50.71◦ but mustlie in the third quadrant; i.e., the required angle is180◦ + 50.71◦ = 230.71◦That is, the velocity of car Q relative to car P is71.06 m/s at 230.71◦Now try the following Practice ExercisePracticeExercise 116 Relative velocity(answers on page 352)1. A car is moving along a straight horizontalroad at 79.2km/h and rain is falling verticallydownwards at 26.4km/h. Find the velocity ofthe rain relative to the driver of the car.2. Calculate the time needed to swim across ariver 142m wide when the swimmer can swimat 2km/h in still water and the river is ﬂowingat 1km/h. At what angle to the bank shouldthe swimmer swim?3. A ship is heading in a direction N 60◦E at aspeed which in still water would be 20km/h.It is carried off course by a current of 8km/hin a direction of E 50◦S. Calculate the ship’sactual speed and direction.29.9 i, j and k notationA method of completely specifying the direction of avector in space relative to some reference point is to usethree unit vectors, i,j and k, mutually at right anglesto each other, as shown in Figure 29.41.yxzkojiFigure 29.41Calculations involving vectors given in i,j,k nota-tion are carried out in exactly the same way as standardalgebraic calculations, as shown in the worked examplesbelow.Problem 14. Determine(3i + 2j + 2k) − (4i − 3j + 2k)(3i + 2j + 2k) − (4i − 3j + 2k)= 3i + 2j + 2k − 4i + 3j − 2k= −i + 5jProblem 15. Given p = 3i + 2k,q = 4i − 2j + 3k and r = −3i + 5j − 4k,determine(a) −r (b) 3p (c) 2p + 3q (d) −p + 2r(e) 0.2p + 0.6q − 3.2r(a) −r = −(−3i + 5j − 4k) = +3i − 5j + 4k(b) 3p = 3(3i + 2k) = 9i + 6k(c) 2p + 3q = 2(3i + 2k) + 3(4i − 2j + 3k)= 6i + 4k + 12i − 6j + 9k= 18i − 6j + 13k(d) −p + 2r = −(3i + 2k) + 2(−3i + 5j − 4k)= −3i − 2k + (−6i + 10j − 8k)= −3i − 2k − 6i + 10j − 8k= −9i + 10j − 10k(e) 0.2p +0.6q − 3.2r= 0.2(3i + 2k) + 0.6(4i − 2j + 3k)−3.2(−3i + 5j − 4k)= 0.6i + 0.4k + 2.4i − 1.2j + 1.8k + 9.6i−16j + 12.8k= 12.6i − 17.2j + 15kNow try the following Practice ExercisePracticeExercise 117 i,j,k notation(answers on page 352)Given that p = 2i + 0.5j − 3k,q = −i + j + 4kand r = 6j − 5k, evaluate and simplify the follow-ing vectors in i, j,k form.1. −q 2. 2p3. q + r 4. −q + 2p5. 3q + 4r 6. q − 2p7. p + q + r 8. p + 2q + 3r9. 2p + 0.4q + 0.5r 10. 7r − 2q
• Chapter 30Methods of addingalternating waveforms30.1 Combining two periodicfunctionsThere are a number of instances in engineering and sci-ence where waveforms have to be combined and whereit is required to determine the single phasor (calledthe resultant) that could replace two or more separatephasors. Uses are found in electrical alternating cur-rent theory, in mechanical vibrations, in the addition offorces and with sound waves.There are a number of methods of determining theresultant waveform. These include(a) Drawing the waveforms and adding graphically.(b) Drawing the phasors and measuring the resultant.(c) Using the cosine and sine rules.(d) Using horizontal and vertical components.30.2 Plotting periodic functionsThis may be achieved by sketching the separate func-tions on the same axes and then adding (or subtracting)ordinates at regular intervals. This is demonstrated inthe following worked problems.Problem 1. Plot the graph of y1 = 3sin A fromA = 0◦ to A = 360◦. On the same axes ploty2 = 2cos A. By adding ordinates, plotyR = 3sin A + 2cos A and obtain a sinusoidalexpression for this resultant waveformy1 = 3sin A and y2 = 2cos A are shown plotted inFigure 30.1. Ordinates may be added at, say, 15◦intervals. For example,at 0◦, y1 + y2 = 0 + 2 = 2at 15◦, y1 + y2 = 0.78 + 1.93 = 2.71at 120◦, y1 + y2 = 2.60 + −1 = 1.6at 210◦, y1 + y2 = −1.50 − 1.73 = −3.23, andso on.yy15 3 sin Ay25 2 cos AyR 53.6 sin(A 1348)A023222133.621348908 1808 2708 3608Figure 30.1The resultant waveform, shown by the broken line,has the same period, i.e. 360◦, and thus the same fre-quency as the single phasors. The maximum value,or amplitude, of the resultant is 3.6. The resultantDOI: 10.1016/B978-1-85617-697-2.00030-2
• 280 Basic Engineering Mathematics2␲ angle ␻t19Њ19Њi1 ϭ 20 sin␻tiR ϭ 20 sin ␻t ϩ10 sin(␻tϩ )90Њ 180Њ 270Њ 360ЊϪ30Ϫ20Ϫ10102026.5303␲2␲2␲3␲i2 ϭ 10 sin(␻tϩ )3␲Figure 30.43. Express 12sinωt + 5cosωt in the formAsin(ωt ± α) by drawing and measurement.30.3 Determining resultant phasorsby drawingThe resultant of two periodic functions may be foundfrom their relative positions when the time is zero.For example, if y1 = 4sinωt and y2 = 3sin(ωt − π/3)then each may be represented as phasors as shown inFigure 30.5, y1 being 4 units long and drawn horizon-tally and y2 being 3 unitslong,lagging y1 by π/3 radiansor60◦.To determinetheresultant of y1 + y2, y1 isdrawnhorizontally as shown in Figure 30.6 and y2 is joined tothe end of y1 at 60◦ to the horizontal. The resultantis given by yR. This is the same as the diagonal of aparallelogram that is shown completed in Figure 30.7.608 or ␲/3 radsy15 4y25 3Figure 30.5Resultant yR, in Figures 30.6 and 30.7, may be deter-mined by drawing the phasors and their directions toscale and measuring using a ruler and protractor. In thisy15 4y253␾ 608yRFigure 30.6y1ϭ 4y2ϭ 3␾yRFigure 30.7example, yR is measured as 6 units long and angle φ ismeasured as 25◦.25◦= 25 ×π180radians = 0.44 radHence, summarizing, by drawing,yR = y1 + y2 = 4sinωt + 3sin(ωt − π/3)= 6sin(ωt − 0.44).If the resultant phasor, yR = y1 − y2 is required theny2 is still 3 units long but is drawn in the oppositedirection, as shown in Figure 30.8.Problem 5. Two alternating currents are given byi1 = 20sinωt amperes and i2 = 10sin ωt +π3amperes. Determine i1 + i2 by drawing phasors
• Methods of adding alternating waveforms 28160Њ60Њ␾y1ϭ 4Ϫy2 ϭ 3yRy2Figure 30.8The relative positions of i1 and i2 at time t = 0 areshown as phasors in Figure 30.9, whereπ3rad = 60◦.The phasor diagram in Figure 30.10 is drawn to scalewith a ruler and protractor.i15 20Ai2 5 10A608Figure 30.9i25 10Ai15 20AiR608␾Figure 30.10The resultant iR is shown and is measured as 26A andangle φ as 19◦ or 0.33rad leading i1. Hence, by drawingand measuring,iR = i1 + i2 = 26sin(ωt + 0.33)AProblem 6. For the currents in Problem 5,determine i1 − i2 by drawing phasorsAt time t = 0, current i1 is drawn 20 units long hori-zontally as shown by 0a in Figure 30.11. Current i2 isshown, drawn 10 units long in a broken line and leadingby 60◦. The current −i2 is drawn in the opposite direc-tion to the broken line ofi2, shown as ab in Figure 30.11.The resultant iR is given by 0b lagging by angle φ.i15 20Ai2 5 10 AiRab02i2608␾Figure 30.11By measurement, iR = 17A and φ = 30◦ or 0.52rad.Hence, by drawing phasors,iR = i1 − i2 = 17sin(ωt − 0.52)ANow try the following Practice ExercisePracticeExercise 119 Determiningresultant phasors by drawing (answers onpage 352)1. Determine a sinusoidal expression for2sinθ +4cosθ by drawing phasors.2. If v1 = 10sinωt volts andv2 = 14sin(ωt + π/3) volts, determine bydrawing phasors sinusoidal expressions for(a) v1 + v2 (b) v1 − v23. Express 12sinωt + 5cosωt in the formAsin(ωt ± α) by drawing phasors.30.4 Determining resultant phasorsby the sine and cosine rulesAs stated earlier, the resultant of two periodic func-tions may be found from their relative positions whenthe time is zero. For example, if y1 = 5sinωt andy2 = 4sin(ωt − π/6) then each may be represented byphasors as shown in Figure 30.12, y1 being 5 unitslong and drawn horizontally and y2 being 4 units long,lagging y1 by π/6 radians or 30◦. To determine theresultant of y1 + y2, y1 is drawn horizontally as shownin Figure 30.13 and y2 is joined to the end of y1 at π/6radians; i.e., 30◦ to the horizontal. The resultant is givenby yR.Using the cosine rule on triangle 0ab of Figure 30.13givesy2R = 52+ 42− [2(5)(4)cos150◦]= 25 + 16 − (−34.641) = 75.641
• 282 Basic Engineering Mathematicsy15 5y2 54308Figure 30.12y1ϭ 5y2 ϭ4␾0yRab30Њ150ЊFigure 30.13from which yR =√75.641 = 8.697Using the sine rule,8.697sin150◦=4sinφfrom which sinφ =4sin150◦8.697= 0.22996and φ = sin−10.22996= 13.29◦or 0.232radHence, yR = y1 + y2 = 5sinωt + 4sin(ωt − π/6)= 8.697sin(ωt − 0.232)Problem 7. Given y1 = 2sinωt andy2 = 3sin(ωt + π/4), obtain an expression, bycalculation, for the resultant, yR = y1 + y2When time t = 0, the position of phasors y1 and y2 areas shown in Figure 30.14(a). To obtain the resultant, y1is drawn horizontally, 2 units long, and y2 is drawn 3units long at an angle of π/4rad or 45◦ and joined tothe end of y1 as shown in Figure 30.14(b).From Figure 30.14(b), and using the cosine rule,y2R = 22+ 32− [2(2)(3)cos135◦]= 4 + 9 − [−8.485] = 21.485Hence, yR =√21.485 = 4.6352Using the sine rule3sinφ=4.6352sin135◦y25 3y15 2␲/4 or 458(a)y15 2y25 31358458␾yR(b)Figure 30.14from which sinφ =3sin135◦4.6352= 0.45765Hence, φ = sin−10.45765= 27.24◦or 0.475radThus, by calculation, yR = 4.635sin(ωt + 0.475)Problem 8. Determine20sinωt + 10sin ωt +π3using the cosine and sine rulesFrom the phasor diagram of Figure 30.15 and using thecosine rule,i2R = 202+ 102− [2(20)(10)cos 120◦] = 700i2 5 10 Ai1 5 20AiR6081208␾Figure 30.15Hence, iR =√700 = 26.46AUsing the sine rule gives10sinφ=26.46sin120◦
• Methods of adding alternating waveforms 283from which sinφ =10sin120◦26.46= 0.327296and φ = sin−10.327296 = 19.10◦= 19.10 ×π180= 0.333radHence, by cosine and sine rules,iR = i1 + i2 = 26.46sin(ωt + 0.333)ANow try the following Practice ExercisePracticeExercise 120 Resultant phasors bythe sine and cosine rules (answers onpage 352)1. Determine, using the cosine and sine rules, asinusoidal expression fory = 2sin A + 4cos A.2. Given v1 =10sinωt volts andv2 =14sin(ωt+ π/3) volts, use the cosineand sine rules to determine sinusoidal expres-sions for (a) v1 + v2 (b) v1 − v2In problems 3 to 5, express the given expressionsin the form Asin(ωt ± α) by using the cosine andsine rules.3. 12sinωt + 5cosωt4. 7sinωt + 5sin ωt +π45. 6sinωt + 3sin ωt −π630.5 Determining resultant phasorsby horizontal and verticalcomponentsIf a right-angled triangle is constructed as shown inFigure 30.16, 0a is called the horizontal component ofF and ab is called the vertical component of F.0HaVbF␪Figure 30.16From trigonometry(see Chapter 21 and remember SOHCAH TOA),cosθ =0a0b, from which 0a = 0bcosθ = F cosθi.e. the horizontal component of F,H = F cosθ,andsinθ =ab0b, from which ab = 0bsinθ = F sinθi.e. the vertical component of F,V = F sinθ.Determining resultant phasorsby horizontal and verticalcomponents is demonstrated in the following workedproblems.Problem 9. Two alternating voltages are given byv1 = 15sinωt volts and v2 = 25sin(ωt − π/6)volts. Determine a sinusoidal expression for theresultant vR = v1 + v2 by ﬁnding horizontal andvertical componentsThe relative positions of v1 and v2 at time t = 0 areshown in Figure 30.17(a) and the phasor diagram isshown in Figure 30.17(b).v15 15 V(a)v25 25V␲/6 or 308(b)␾0vRv2v13081508Figure 30.17The horizontal component of vR,H = 15cos0◦ + 25cos(−30◦) = 36.65VThe vertical component of vR,V = 15sin0◦ + 25sin(−30◦) = −12.50V
• 284 Basic Engineering MathematicsHence, vR = 36.652 + (−12.50)2by Pythagoras’ theorem= 38.72 voltstanφ =VH=−12.5036.65= −0.3411from which φ = tan−1(−0.3411)= −18.83◦or −0.329 radians.Hence, vR = v1 + v2 = 38.72sin(ωt − 0.329)VProblem 10. For the voltages in Problem 9,determine the resultant vR = v1 − v2 usinghorizontal and vertical componentsThe horizontal component of vR,H = 15cos0◦ − 25cos(−30◦) = −6.65VThe vertical component of vR,V = 15sin0◦ − 25sin(−30◦) = 12.50VHence, vR = (−6.65)2 + (12.50)2by Pythagoras’ theorem= 14.16 voltstanφ =VH=12.50−6.65= −1.8797from which φ = tan−1(−1.8797)= 118.01◦or 2.06 radians.Hence, vR = v1 − v2 = 14.16sin(ωt + 2.06)VThe phasor diagram is shown in Figure 30.18.v15 15V2v25 25Vv25 25V␾vR308308Figure 30.18Problem 11. Determine20sinωt + 10sin ωt +π3using horizontal and vertical componentsi15 20Ai2 5 10A608Figure 30.19From the phasors shown in Figure 30.19,Total horizontal component,H = 20cos0◦+ 10cos60◦= 25.0Total vertical component,V =20sin0◦ +10sin60◦ = 8.66By Pythagoras, the resultant,iR = 25.02+8.662 = 26.46APhase angle, φ = tan−1 8.6625.0= 19.11◦ or 0.333radHence, by using horizontal and vertical components,20sinωt + 10sin ωt +π3= 26.46sin(ωt + 0.333)Now try the following Practice ExercisePracticeExercise 121 Resultant phasors byhorizontal and vertical components (answerson page 353)In problems 1 to 5, express the combination ofperiodic functions in the form Asin(ωt ± α) byhorizontal and vertical components.1. 7sinωt + 5sin ωt +π42. 6sinωt + 3sin ωt −π63. i = 25sinωt − 15sin ωt +π34. v = 8sinωt − 5sin ωt −π45. x = 9sin ωt +π3− 7sin ωt −3π86. The voltage drops across two componentswhen connected in series across an a.c.supply are v1 = 200sin314.2t andv2 = 120sin(314.2t − π/5) voltsrespectively. Determine(a) the voltage of the supply (given byv1 + v2) in the form Asin(ωt ± α).
• Methods of adding alternating waveforms 285(b) the frequency of the supply.7. If the supply to a circuit is v = 20sin628.3tvolts and the voltage drop across one ofthe components is v1 = 15sin(628.3t − 0.52)volts, calculate(a) the voltage drop across the remainder ofthe circuit, given by v − v1, in the formAsin(ωt ± α)(b) the supply frequency(c) the periodic time of the supply.8. The voltages across three components in aseries circuit when connected across an a.c.supply are v1 = 25sin 300πt +π6volts,v2 = 40sin 300πt −π4volts andv3 = 50sin 300πt +π3volts. Calculate(a) the supply voltage, in sinusoidal form, inthe form Asin(ωt ± α)(b) the frequency of the supply(c) the periodic time.
• Revision Test 12 : Vectors and adding waveformsThis assignment covers the material contained in Chapters 29 and 30. The marks available are shown in brackets atthe end of each question.1. State the difference between scalar and vectorquantities. (2)2. State whether the following are scalar or vectorquantities.(a) A temperature of 50◦C.(b) 2m3 volume.(c) A downward force of 10N.(d) 400J of work.(e) 20cm2 area.(f) A south-easterly wind of 20 knots.(g) 40m distance.(h) An acceleration of 25m/s2 at 30◦ to thehorizontal. (8)3. A velocity vector of 16m/s acts at an angle of−40◦ to the horizontal. Calculate its horizontaland vertical components, correct to 3 signiﬁcantﬁgures. (4)4. Calculate the resultant and direction of the dis-placement vectors shown in Figure RT12.1, cor-rect to 2 decimal places. (6)41 m26 mFigure RT12.15. Calculate the resultant and direction of the forcevectors shown in Figure RT12.2, correct to 2decimal places. (6)6. If acceleration a1 = 11m/s2at 70◦anda2 = 19m/s2 at −50◦, calculate the magnitudeand direction of a1 + a2, correct to 2 decimalplaces. (8)7. If velocity v1 = 36m/s at 52◦ and v2 = 17m/s at−15◦, calculate the magnitude and direction ofv1 − v2, correct to 2 decimal places. (8)5N7NFigure RT12.28. Forces of 10N, 16N and 20N act as shown inFigure RT12.3. Determine the magnitude of theresultant force and its direction relative to the 16Nforce(a) by scaled drawing.(b) by calculation. (13)F1510NF2 516NF3 520N458608Figure RT12.39. For the three forces shown in Figure RT12.3,calculate the resultant of F1 − F2 − F3 and itsdirection relative to force F2. (9)10. Two cars, A and B, are travelling towards cross-roads. A has a velocity of 60km/h due south andB a velocity of 75km/h due west. Calculate thevelocity of A relative to B. (6)
• Revision Test 12 : Vectors and adding waveforms 28711. Given a = −3i + 3j + 5k, b = 2i − 5j + 7k andc = 3i + 6j − 4k, determine the following:(i)−4b(ii) a + b − c (iii) 5b − 3a (6)12. Calculate the magnitude and direction of the resul-tant vector of the displacement system shown inFigure RT12.4. (9)7 m12 m 10 m308608208Figure RT12.413. The instantaneous values of two alternating volt-ages are given byv1 = 150sin ωt +π3voltsandv2 = 90sin ωt −π6voltsPlot the two voltages on the same axes to scalesof 1cm = 50 volts and 1cm = π/6. Obtain a sinu-soidal expression for the resultant of v1 and v2 inthe form Rsin (ωt + α) (a) by adding ordinates atintervals and (b) by calculation. (15)
• Chapter 31Presentation ofstatistical data31.1 Some statistical terminology31.1.1 Discrete and continuous dataData are obtained largely by two methods:(a) By counting – for example, the number of stampssold by a post ofﬁce in equal periods of time.(b) By measurement – for example, the heights of agroup of people.When data are obtained by counting and only wholenumbers are possible, the data are called discrete. Mea-sured data can have any value within certain limits andare called continuous.Problem 1. Data are obtained on the topics givenbelow. State whether they are discrete or continuousdata.(a) The number of days on which rain falls in amonth for each month of the year.(b) The mileage travelled by each of a number ofsalesmen.(c) The time that each of a batch of similarbatteries lasts.(d) The amount of money spent by each of severalfamilies on food.(a) The number of days on which rain falls in a givenmonth must be an integer value and is obtained bycounting the number of days. Hence, these dataare discrete.(b) A salesman can travel any number of miles(and parts of a mile) between certain limits andthese data are measured. Hence, the data arecontinuous.(c) The time that a battery lasts is measured andcan have any value between certain limits. Hence,these data are continuous.(d) The amount of money spent on food can only beexpressed correct to the nearest pence, the amountbeing counted. Hence, these data are discrete.Now try the following Practice ExercisePracticeExercise 122 Discrete andcontinuous data (answers on page 353)In the followingproblems, state whether data relat-ing to the topics given are discrete or continuous.1. (a) The amount of petrol produced daily, foreach of 31 days, by a reﬁnery.(b) The amount of coal produced daily byeach of 15 miners.(c) The number of bottles of milk delivereddaily by each of 20 milkmen.(d) Thesizeof10 samplesofrivetsproducedby a machine.2. (a) The number of people visiting an exhi-bition on each of 5 days.(b) The time taken by each of 12 athletes torun 100 metres.DOI: 10.1016/B978-1-85617-697-2.00031-4
• Presentation of statistical data 289(c) The value of stamps sold in a day by eachof 20 post ofﬁces.(d) The number of defective items producedin each of 10 one-hour periods by amachine.31.1.2 Further statisticalterminologyA set is a group of data and an individual value withinthe set is called a member of the set. Thus, if themasses of ﬁve people are measured correct to the near-est 0.1kilogram and are found to be 53.1kg, 59.4kg,62.1kg, 77.8kg and 64.4kg then the set of masses inkilograms for these ﬁve people is{53.1,59.4,62.1,77.8,64.4}and one of the members of the set is 59.4A set containing all the members is called a pop-ulation. Some members selected at random from apopulation are called a sample. Thus, all car registrationnumbers form a population but the registration numbersof, say, 20 cars taken at random throughout the countryare a sample drawn from that population.The number of times that the value of a memberoccurs in a set is called the frequency of that mem-ber. Thus, in the set {2,3,4,5,4,2,4,7,9}, member 4has a frequency of three, member 2 has a frequency of2 and the other members have a frequency of one.The relative frequency with which any member of aset occurs is given by the ratiofrequency of membertotal frequency of all membersFor the set {2,3,5,4,7,5,6,2,8}, the relative fre-quency of member 5 is29. Often, relative frequency isexpressed as a percentage and the percentage relativefrequency is(relative frequency × 100)%31.2 Presentation of ungrouped dataUngrouped data can be presented diagrammatically inseveral ways and these include(a) pictograms, in which pictorial symbols are usedto represent quantities (see Problem 2),(b) horizontal bar charts, having data representedby equally spaced horizontal rectangles (see Prob-lem 3), and(c) vertical bar charts, in which data are repre-sented by equally spaced vertical rectangles (seeProblem 4).Trends in ungrouped data over equal periods of timecan be presented diagrammatically by a percent-age component bar chart. In such a chart, equallyspaced rectangles of any width, but whose height cor-responds to 100%, are constructed. The rectanglesare then subdivided into values corresponding to thepercentage relative frequencies of the members (seeProblem 5).A pie diagram is used to show diagrammatically theparts making up the whole. In a pie diagram, the area ofa circle represents the whole and the areas of the sectorsof the circle are made proportional to the parts whichmake up the whole (see Problem 6).Problem 2. The number of television setsrepaired in a workshop by a technician in sixone-month periods is as shown below. Present thesedata as a pictogramMonth January February MarchNumber repaired 11 6 15Month April May JuneNumber repaired 9 13 8Each symbol shown in Figure 31.1 represents two tele-vision sets repaired. Thus, in January, 512symbols areused to represent the 11 sets repaired; in February, 3symbols are used to represent the 6 sets repaired, andso on.JanuaryFebruaryMarchMonth Number of TV sets repaired ;2 setsAprilMayJuneFigure 31.1
• 290 Basic Engineering MathematicsProblem 3. The distance in miles travelled byfour salesmen in a week are as shown below.Salesman P Q R SDistance travelled (miles) 413 264 597 143Use a horizontal bar chart to represent these datadiagrammaticallyEqually spaced horizontal rectangles of any width, butwhose length is proportional to the distance travelled,are used. Thus, the length of the rectangle for sales-man P is proportional to 413 miles, and so on. Thehorizontal bar chart depicting these data is shown inFigure 31.2.0PQSalesmenRS100 200 300Distance travelled, miles400 500 600Figure 31.2Problem 4. The number of issues of tools ormaterials from a store in a factory is observed forseven one-hour periods in a day and the results ofthe survey are as follows.Period 1 2 3 4 5 6 7Number of issues 34 17 9 5 27 13 6Present these data on a vertical bar chart11020Numberofissues30402 3 4 5 6Periods7Figure 31.3In a vertical bar chart, equally spaced vertical rect-angles of any width, but whose height is proportionalto the quantity being represented, are used. Thus, theheight of the rectangle for period 1 is proportional to 34units, and so on. The vertical bar chart depicting thesedata is shown in Figure 31.3.Problem 5. The numbers of various types ofdwellings sold by a company annually over athree-year period are as shown below. Drawpercentage component bar charts to presentthese dataYear 1 Year 2 Year 34-roomed bungalows 24 17 75-roomed bungalows 38 71 1184-roomed houses 44 50 535-roomed houses 64 82 1476-roomed houses 30 30 25A table of percentage relative frequency values, correctto the nearest 1%, is the ﬁrst requirement. Sincepercentage relative frequency=frequency of member × 100total frequencythen for 4-roomed bungalows in year 1percentage relative frequency=24 × 10024 + 38 + 44 + 64 + 30= 12%The percentage relative frequencies of the other typesof dwellings for each of the three years are similarlycalculated and theresultsareasshown in thetablebelow.Year 1 Year 2 Year 34-roomed bungalows 12% 7% 2%5-roomed bungalows 19% 28% 34%4-roomed houses 22% 20% 15%5-roomed houses 32% 33% 42%6-roomed houses 15% 12% 7%Thepercentagecomponent barchart isproduced by con-structing three equally spaced rectangles of any width,corresponding to the three years. The heightsof the rect-angles correspond to 100% relative frequency and are
• Presentation of statistical data 291subdivided into the values in the table of percentagesshown above. A key is used (different types of shadingor different colour schemes) to indicate correspondingpercentage values in the rows of the table of percent-ages. The percentage component bar chart is shown inFigure 31.4.11020Percentagerelativefrequency304050607080901002 3YearKey6-roomed houses5-roomed houses4-roomed houses5-roomed bungalows4-roomed bungalowsFigure 31.4Problem 6. The retail price of a product costing£2 is made up as follows: materials 10p, labour20p, research and development 40p, overheads70p, proﬁt 60p. Present these data on a pie diagramA circle of any radius is drawn. The area of the circlerepresents the whole, which in this case is £2. The circleis subdivided into sectors so that the areas of the sectorsare proportional to the parts; i.e., the parts which makeup the total retail price. For the area of a sector to beproportionalto a part, the angle at the centre of the circlemust be proportional to that part. The whole, £2 or 200p,corresponds to 360◦. Therefore,10p corresponds to 360×10200degrees, i.e. 18◦20p corresponds to 360×20200degrees, i.e. 36◦and so on, giving the angles at the centre of the circlefor the parts of the retail price as 18◦,36◦,72◦,126◦ and108◦, respectively.The pie diagram is shown in Figure 31.5.Problem 7.(a) Using the data given in Figure 31.2 only,calculate the amount of money paid to eachsalesman for travelling expenses if they arepaid an allowance of 37p per mile.1088Ip ; 1.881883687281268OverheadsProfitLabourResearch anddevelopmentMaterialsFigure 31.5(b) Using the data presented in Figure 31.4,comment on the housing trends over thethree-year period.(c) Determine the proﬁt made by selling 700 unitsof the product shown in Figure 31.5(a) By measuring the length of rectangle P, themileage covered by salesman P is equivalent to413 miles. Hence salesman P receives a travellingallowance of£413 × 37100i.e. £152.81Similarly, for salesman Q, the miles travelled are264 and his allowance is£264 × 37100i.e. £97.68Salesman R travels 597 miles and he receives£597 × 37100i.e. £220.89Finally, salesman S receives£143 × 37100i.e. £52.91(b) An analysis of Figure 31.4 shows that 5-roomedbungalows and 5-roomed houses are becomingmorepopular,thegreatest changein thethreeyearsbeing a 15% increase in the sales of 5-roomedbungalows.(c) Since 1.8◦ corresponds to 1p and the proﬁt occu-pies 108◦ of the pie diagram, the proﬁt per unit is108 × 11.8i.e. 60pThe proﬁt when selling 700 units of the prod-uct is£700 × 60100i.e. £420
• 292 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 123 Presentation ofungrouped data (answers on page 352)1. The number of vehicles passing a stationaryobserver on a road in six ten-minute intervalsis as shown. Draw a pictogram to representthese data.Period of time 1 2 3 4 5 6Number ofvehicles 35 44 62 68 49 412. The number of components produced by afactory in a week is as shown below.Day Mon Tues Wed Thur FriNumber ofcomponents 1580 2190 1840 2385 1280Show these data on a pictogram.3. For the data given in Problem 1 above, drawa horizontal bar chart.4. Present the data given in Problem 2 above ona horizontal bar chart.5. For the data given in Problem 1 above,construct a vertical bar chart.6. Depict the data given in Problem 2 above ona vertical bar chart.7. A factory produces three different types ofcomponents. The percentages of each ofthese components produced for three one-month periods are as shown below. Show thisinformation on percentage component barcharts and comment on the changing trendin the percentages of the types of componentproduced.Month 1 2 3Component P 20 35 40Component Q 45 40 35Component R 35 25 258. A company has ﬁve distribution centres andthe mass of goods in tonnes sent to eachcentre during four one-week periods is asshown.Week 1 2 3 4Centre A 147 160 174 158Centre B 54 63 77 69Centre C 283 251 237 211Centre D 97 104 117 144Centre E 224 218 203 194Use a percentage component bar chart to pre-sent these data and comment on any trends.9. The employees in a company can be splitinto the following categories: managerial 3,supervisory 9, craftsmen 21, semi-skilled 67,others 44. Show these data on a pie diagram.10. The way in which an apprentice spent histime over a one-month period is as follows:drawing ofﬁce44 hours,production 64 hours,training 12 hours, at college 28 hours. Use apie diagram to depict this information.11. (a) With reference to Figure 31.5, determinethe amount spent on labourand materialsto produce 1650 units of the product.(b) If in year 2 of Figure 31.4 1% cor-responds to 2.5 dwellings, how manybungalows are sold in that year?12. (a) If the company sells 23500 units perannum of the product depicted inFigure 31.5, determine the cost of theiroverheads per annum.(b) If 1% of the dwellings represented inyear 1 of Figure 31.4 corresponds to2 dwellings, ﬁnd the total number ofhouses sold in that year.31.3 Presentation of grouped dataWhen the number of members in a set is small, say ten orless, the data can be represented diagrammatically with-out further analysis, by means of pictograms, bar charts,percentage components bar charts or pie diagrams (asshown in Section 31.2).For sets having more than ten members, those mem-bers having similar values are grouped together in
• Presentation of statistical data 293classes to form a frequency distribution. To assist inaccurately counting members in the various classes, atally diagram is used (see Problems 8 and 12).A frequency distribution is merely a table show-ing classes and their corresponding frequencies (seeProblems 8 and 12). The new set of values obtainedby forming a frequency distribution is called groupeddata. The terms used in connection with grouped dataare shown in Figure 31.6(a). The size or range of a classis given by the upper class boundary value minus thelower class boundary value and in Figure 31.6(b) is7.65−7.35; i.e., 0.30. The class interval for the classshown in Figure 31.6(b) is 7.4 to 7.6 and the classmid-point value is given by(upper class boundary value) + (lower class boundary value)2and in Figure 31.6(b) is7.65 + 7.352i.e. 7.5Class interval(a)(b)LowerclassboundaryUpperclassboundaryClassmid-point7.35to 7.3 7.4 to 7.6 7.7 to7.657.5Figure 31.6One of the principal ways of presenting grouped datadiagrammatically is to use a histogram, in which theareas of vertical, adjacent rectangles are made propor-tional to frequencies of the classes (see Problem 9).When class intervals are equal, the heights of the rect-angles of a histogram are equal to the frequencies of theclasses. 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 inter-val of class B, then for equal frequencies the heightof the rectangle representing A is half that of B (seeProblem 11).Anothermethod of presenting grouped data diagram-matically is to use a frequency polygon, which is thegraph produced by plottingfrequency against class mid-point values and joining the co-ordinates with straightlines (see Problem 12).A cumulative frequency distribution is a tableshowing the cumulative frequency for each value ofupper class boundary. The cumulative frequency for aparticular value of upper class boundary is obtained byadding the frequency of the class to the sum of the pre-vious frequencies. A cumulative frequency distributionis formed in Problem 13.The curve obtained by joining the co-ordinates ofcumulative frequency (vertically) against upper classboundary (horizontally) is called an ogive or a cumu-lative frequency distribution curve (see Problem 13).Problem 8. The data given below refer to the gainof each of a batch of 40 transistors, expressedcorrect to the nearest whole number. Form afrequency distribution for these data having sevenclasses81 83 87 74 76 89 82 8486 76 77 71 86 85 87 8884 81 80 81 73 89 82 7981 79 78 80 85 77 84 7883 79 80 83 82 79 80 77The range of the data is the value obtained by tak-ing the value of the smallest member from that of thelargest member. Inspection of the set of data shows thatrange = 89 − 71 = 18. The size of each class is givenapproximately by the range divided by the number ofclasses. Since 7 classes are required, the size of eachclass is 18÷ 7; that is, approximately 3. To achieveseven equal classes spanning a range of values from 71to 89, the class intervals are selected as 70–72, 73–75,and so on.To assist with accurately determining the number ineach class, a tally diagram is produced, as shown inTable 31.1(a). This is obtained by listing the classesin the left-hand column and then inspecting each of the40 members of the set in turn and allocating them tothe appropriate classes by putting ‘1’s in the appropri-ate rows. Every ﬁfth ‘1’ allocated to a particular row isshown as an obliqueline crossing the four previous ‘1’s,to help with ﬁnal counting.A frequency distribution for the data is shown inTable 31.1(b) and lists classes and their correspond-ing frequencies, obtained from the tally diagram. (Classmid-point values are also shown in the table, since theyare used for constructing the histogram for these data(see Problem 9).)
• 294 Basic Engineering MathematicsTable 31.1(a)Class Tally70–72 173–75 1176–78 1111 1179–81 1111 1111 1182–84 1111 111185–87 1111 188–90 111Table 31.1(b)Class Class mid-point Frequency70–72 71 173–75 74 276–78 77 779–81 80 1282–84 83 985–87 86 688–90 89 3Problem 9. Construct a histogram for the datagiven in Table 31.1(b)The histogram is shown in Figure 31.7. The width ofthe rectangles corresponds to the upper class boundaryvalues minus the lower class boundary values and theheightsoftherectanglescorrespond to theclassfrequen-cies. The easiest way to draw a histogram is to mark the71426Frequency10812141674 77 80 83Class mid-point values8986Figure 31.7class mid-point values on the horizontal scale and drawthe rectangles symmetrically about the appropriateclassmid-point values and touching one another.Problem 10. The amount of money earnedweekly by 40 people working part-time in a factory,correct to the nearest £10, is shown below. Form afrequency distribution having 6 classes for thesedata80 90 70 110 90 160 110 80140 30 90 50 100 110 60 10080 90 110 80 100 90 120 70130 170 80 120 100 110 40 11050 100 110 90 100 70 110 80Inspection of the set given shows that the majority ofthe members of the set lie between £80 and £110 andthat there is a much smaller number of extreme val-ues ranging from £30 to £170. If equal class intervalsare selected, the frequency distribution obtained doesnot give as much information as one with unequal classintervals. Since the majority of the members lie between£80 and £100, the class intervals in this range areselected to be smaller than those outside of this range.There is no unique solution and one possible solution isshown in Table 31.2.Table 31.2Class Frequency20–40 250–70 680–90 12100–110 14120–140 4150–170 2Problem 11. Draw a histogram for the data givenin Table 31.2When dealing with unequal class intervals, the his-togram must be drawn so that the areas (and notthe heights) of the rectangles are proportional to the
• Presentation of statistical data 295Table 31.31 2 3 4 5 6Class Frequency Upper class Lower class Class range Height ofboundary boundary rectangle20–40 2 45 15 30230=11550–70 6 75 45 30630=31580–90 12 95 75 201220=915100–110 14 115 95 201420=101215120–140 4 145 115 30430=215150–170 2 175 145 30230=115frequencies of the classes. The data given are shownin columns 1 and 2 of Table 31.3. Columns 3 and 4give the upper and lower class boundaries, respectively.In column 5, the class ranges (i.e. upper class bound-ary minus lower class boundary values) are listed. Theheights of the rectangles are proportional to the ratiofrequencyclass range, as shown in column 6. The histogram isshown in Figure 31.8.304/152/156/15Frequencyperunitclassrange10/158/1512/1560 85Class mid-point values130 160105Figure 31.8Problem 12. The masses of 50 ingots inkilograms are measured correct to the nearest 0.1kgand the results are as shown below. Produce afrequency distribution having about 7 classes forthese data and then present the grouped data as afrequency polygon and a histogram8.0 8.6 8.2 7.5 8.0 9.1 8.5 7.6 8.2 7.88.3 7.1 8.1 8.3 8.7 7.8 8.7 8.5 8.4 8.57.7 8.4 7.9 8.8 7.2 8.1 7.8 8.2 7.7 7.58.1 7.4 8.8 8.0 8.4 8.5 8.1 7.3 9.0 8.67.4 8.2 8.4 7.7 8.3 8.2 7.9 8.5 7.9 8.0The range of the data is the member having the largestvalue minus the member having the smallest value.Inspection of the set of data shows that range = 9.1 −7.1 = 2.0.The size of each class is given approximately byrangenumber of classesSince about seven classes are required, the size of eachclassis2.0 ÷ 7,i.e.approximately 0.3,and thusthe classlimits are selected as 7.1 to 7.3, 7.4 to 7.6, 7.7 to 7.9,and so on. The class mid-pointfor the 7.1 to 7.3 class is7.35 + 7.052i.e. 7.2the class midpoint for the 7.4 to 7.6 class is7.65 + 7.352i.e. 7.5and so on.
• 296 Basic Engineering MathematicsTo assist with accurately determining the numberin each class, a tally diagram is produced as shownin Table 31.4. This is obtained by listing the classes inthe left-hand column and then inspecting each of the 50members of the set of data in turn and allocating it to theappropriate class by puttinga ‘1’ in the appropriate row.Each ﬁfth ‘1’ allocated to a particular row is marked asan oblique line to help with ﬁnal counting.Table 31.4Class Tally7.1 to 7.3 1117.4 to 7.6 11117.7 to 7.9 1111 11118.0 to 8.2 1111 1111 11118.3 to 8.5 1111 1111 18.6 to 8.8 1111 18.9 to 9.1 11A frequency distribution for the data is shown inTable 31.5 and lists classes and their corresponding fre-quencies. Class mid-points are also shown in this tablesince they are used when constructing the frequencypolygon and histogram.Table 31.5Class Class mid-point Frequency7.1 to 7.3 7.2 37.4 to 7.6 7.5 57.7 to 7.9 7.8 98.0 to 8.2 8.1 148.3 to 8.5 8.4 118.6 to 8.8 8.7 68.9 to 9.1 9.0 2A frequency polygon is shown in Figure 31.9,the co-ordinates corresponding to the class mid-point/frequency values given in Table 31.5. The co-ordinates are joined by straight lines and the polygonis ‘anchored-down’ at each end by joining to the nextclass mid-point value and zero frequency.A histogram is shown in Figure 31.10, the widthof a rectangle corresponding to (upper class boundary7.24206Frequency12108147.87.5 8.1Class mid-point values8.7 9.08.4Frequency polygonFigure 31.9value – lower class boundary value) and height corre-sponding to the class frequency. The easiest way to drawahistogramisto mark classmid-point valueson thehori-zontal scale and to draw the rectangles symmetricallyabout the appropriate class mid-point values and touch-ing one another. A histogram for the data given inTable 31.5 is shown in Figure 31.10.7.24206Frequency10812147.57.357.657.958.258.558.859.157.8 8.1 8.4Class mid-point values9.08.7HistogramFigure 31.10Problem 13. The frequency distribution for themasses in kilograms of 50 ingots is7.1 to 7.3 37.4 to 7.6 57.7 to 7.9 98.0 to 8.2 148.3 to 8.5 118.6 to 8.8 68.9 to 9.1 2Form a cumulative frequency distribution for thesedata and draw the corresponding ogive
• Presentation of statistical data 297A cumulative frequency distribution is a table givingvalues of cumulative frequency for the values of upperclass boundaries and is shown in Table 31.6. Columns1 and 2 show the classes and their frequencies. Column3 lists the upper class boundary values for the classesgiven in column 1. Column 4 gives the cumulative fre-quency values for all frequencies less than the upperclass boundary values given in column 3. Thus, forexample, for the 7.7 to 7.9 class shown in row 3, thecumulative frequency value is the sum of all frequen-cies having values of less than 7.95, i.e. 3 + 5 + 9 = 17,and so on.Table 31.61 2 3 4Class Frequency Upper class Cumulativeboundary less than frequency7.1–7.3 3 7.35 37.4–7.6 5 7.65 87.7–7.9 9 7.95 178.0–8.2 14 8.25 318.3–8.5 11 8.55 428.6–8.8 6 8.85 488.9–9.1 2 9.15 50The ogive for the cumulative frequency distribu-tion given in Table 31.6 is shown in Figure 31.11.The co-ordinates corresponding to each upper classboundary/cumulative frequency value are plotted and7.0510Cumulativefrequency403020507.957.35 7.65 8.25Upper class boundary values in kilograms8.85 9.158.55Figure 31.11the co-ordinates are joined by straight lines (not the bestcurve drawn throughtheco-ordinates as in experimentalwork). The ogive is ‘anchored’ at its start by adding theco-ordinate (7.05, 0).Now try the following Practice ExercisePracticeExercise 124 Presentation ofgrouped data (answers on page 353)1. The mass in kilograms, correct to the nearestone-tenth of a kilogram, of 60 bars of metalare as shown. Form a frequency distributionof about 8 classes for these data.39.8 40.1 40.3 40.0 40.6 39.7 40.0 40.4 39.6 39.339.6 40.7 40.2 39.9 40.3 40.2 40.4 39.9 39.8 40.040.2 40.1 40.3 39.7 39.9 40.5 39.9 40.5 40.0 39.940.1 40.8 40.0 40.0 40.1 40.2 40.1 40.0 40.2 39.939.7 39.8 40.4 39.7 39.9 39.5 40.1 40.1 39.9 40.239.5 40.6 40.0 40.1 39.8 39.7 39.5 40.2 39.9 40.32. Draw a histogram for the frequency distribu-tion given in the solution of Problem 1.3. The information given below refers to thevalue of resistance in ohms of a batch of48 resistors of similar value. Form a fre-quency distribution for the data, having about6 classes, and draw a frequency polygon andhistogram to represent these data diagram-matically.21.0 22.4 22.8 21.5 22.6 21.1 21.6 22.322.9 20.5 21.8 22.2 21.0 21.7 22.5 20.723.2 22.9 21.7 21.4 22.1 22.2 22.3 21.322.1 21.8 22.0 22.7 21.7 21.9 21.1 22.621.4 22.4 22.3 20.9 22.8 21.2 22.7 21.622.2 21.6 21.3 22.1 21.5 22.0 23.4 21.24. The time taken in hours to the failure of 50specimens of a metal subjected to fatigue fail-ure tests are as shown. Form a frequency dis-tribution, having about 8 classes and unequalclass intervals, for these data.
• 298 Basic Engineering Mathematics28 22 23 20 12 24 37 28 21 2521 14 30 23 27 13 23 7 26 1924 22 26 3 21 24 28 40 27 2420 25 23 26 47 21 29 26 22 3327 9 13 35 20 16 20 25 18 225. Form a cumulative frequency distribution andhence draw the ogive for the frequency distri-bution given in the solution to Problem 3.6. Draw a histogram for the frequency distribu-tion given in the solution to Problem 4.7. The frequency distribution for a batch of50 capacitors of similar value, measured inmicrofarads, is10.5–10.9 211.0–11.4 711.5–11.9 1012.0–12.4 1212.5–12.9 1113.0–13.4 8Form a cumulative frequency distribution forthese data.8. Drawan ogiveforthedatagiven in thesolutionof Problem 7.9. The diameter in millimetres of a reel of wireis measured in 48 places and the results are asshown.2.10 2.29 2.32 2.21 2.14 2.222.28 2.18 2.17 2.20 2.23 2.132.26 2.10 2.21 2.17 2.28 2.152.16 2.25 2.23 2.11 2.27 2.342.24 2.05 2.29 2.18 2.24 2.162.15 2.22 2.14 2.27 2.09 2.212.11 2.17 2.22 2.19 2.12 2.202.23 2.07 2.13 2.26 2.16 2.12(a) Form a frequency distribution of diame-ters having about 6 classes.(b) Draw a histogram depicting the data.(c) Form a cumulative frequency distribu-tion.(d) Draw an ogive for the data.
• Chapter 32Mean, median, mode andstandard deviation32.1 Measures of central tendencyA single value, which is representative of a set of values,may be used to give an indication of the general size ofthe members in a set, the word ‘average’ often beingused to indicate the single value. The statistical termused for ‘average’ is the ‘arithmetic mean’ or just the‘mean’.Other measures of central tendency may be used andthese include the median and the modal values.32.2 Mean, median and mode fordiscrete data32.2.1 MeanThe arithmeticmean value isfound by adding togetherthe values of the members of a set and dividing by thenumber of members in the set. Thus, the mean of the setof numbers {4, 5, 6, 9} is4 + 5 + 6 + 94i.e. 6In general, the mean of the set {x1, x2, x3,...xn} isx =x1 + x2 + x3 + ···+ xnnwritten asxnwhere is the Greek letter ‘sigma’ and means ‘the sumof ’ and x (called x-bar) is used to signify a mean value.32.2.2 MedianThe median value often gives a better indication of thegeneral size of a set containing extreme values. The set{7, 5, 74, 10} has a mean value of 24, which is not reallyrepresentative ofany of the values of the members of theset. 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 setscontaining an odd number of members or ﬁndingthe value of the mean of the two middle membersfor sets containing an even number of members.For example, the set {7, 5, 74, 10} is ranked as{5, 7, 10, 74} and, since it contains an even numberof members (four in this case), the mean of 7 and 10 istaken, giving a median value of 8.5. Similarly, the set{3, 81, 15, 7, 14} is ranked as {3, 7, 14, 15, 81} andthe median value is the value of the middle member,i.e. 14.32.2.3 ModeThe modal value, or mode, is the most commonlyoccurring value in a set. If two values occur withthe same frequency, the set is ‘bi-modal’. The set{5, 6, 8, 2, 5, 4, 6, 5, 3} has a modal value of 5, since themember having a value of 5 occurs the most, i.e. threetimes.Problem 1. Determine the mean, median andmode for the set {2, 3, 7, 5, 5, 13, 1, 7, 4, 8, 3, 4, 3}DOI: 10.1016/B978-1-85617-697-2.00032-6
• 300 Basic Engineering MathematicsThe mean value is obtained by adding together thevalues of the members of the set and dividing by thenumber of members in the set. Thus,mean value,x=2 + 3 + 7 + 5 + 5 + 13 + 1 + 7 + 4 + 8 + 3 + 4 + 313=6513= 5To obtain the median value the set is ranked, that is,placed in ascending order of magnitude, and since theset contains an odd number of members the value of themiddle member is the median value. Ranking the setgives {1, 2, 3, 3, 3, 4, 4, 5, 5, 7, 7, 8, 13}. The middleterm is the seventh member; i.e., 4. Thus, the medianvalue is 4.The modal value is the value of the most commonlyoccurring member and is 3, which occurs three times,all other members only occurring once or twice.Problem 2. The following set of data refers to theamount of money in £s taken by a news vendor for6 days. Determine the mean, median and modalvalues of the set{27.90, 34.70, 54.40, 18.92, 47.60, 39.68}Mean value=27.90 + 34.70 + 54.40 + 18.92 + 47.60 + 39.686= £37.20The ranked set is{18.92, 27.90, 34.70, 39.68, 47.60, 54.40}.Since the set has an even number of members, the meanof the middle two members is taken to give the medianvalue; i.e.,median value =34.70 + 39.682= £37.19Since no two members have the same value, this set hasno mode.Now try the following Practice ExercisePracticeExercise 125 Mean, median andmode for discrete data (answers onpage 353)In problems 1 to 4, determine the mean, medianand modal values for the sets given.1. {3, 8, 10, 7, 5, 14, 2, 9, 8}2. {26, 31, 21, 29, 32, 26, 25, 28}3. {4.72, 4.71, 4.74, 4.73, 4.72, 4.71, 4.73, 4.72}4. {73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9}32.3 Mean, median and mode forgrouped dataThe mean value for a set of grouped data is found bydetermining thesumofthe(frequency× classmid-pointvalues) and dividing by the sum of the frequencies; i.e.,mean value,x =f1x1 + f2x2 + ···fnxnf1 + f2 + ···+ fn=( fx)fwhere f is the frequency of the class having a mid-pointvalue of x, and so on.Problem 3. The frequency distribution for thevalue of resistance in ohms of 48 resistors is asshown. Determine the mean value of resistance20.5–20.9 321.0–21.4 1021.5–21.9 1122.0–22.4 1322.5–22.9 923.0–23.4 2The class mid-point/frequency values are 20.7 3, 21.210, 21.7 11, 22.2 13, 22.7 9 and 23.2 2.For grouped data, the mean value is given byx =( f x)fwhere f is the class frequency and x is the class mid-point value. Hence mean value,(3 × 20.7) + (10 × 21.2) + (11 × 21.7)x =+(13 × 22.2) + (9 × 22.7) + (2 × 23.2)48=1052.148= 21.919...i.e. the mean value is 21.9 ohms, correct to 3 signiﬁcantﬁgures.32.3.1 HistogramsThe mean, median and modal values for grouped datamay be determined from a histogram. In a histogram,
• Mean, median, mode and standard deviation 301frequency values are represented vertically and variablevalueshorizontally.Themean valueisgiven by thevalueof the variable corresponding to a vertical line drawnthroughthe centroid of the histogram.The median valueis obtained by selecting a variable value such that thearea of the histogram to the left of a vertical line drawnthroughthe selected variable value is equal to the area ofthe histogram on the right of the line. The modal valueis the variable value obtained by dividing the width ofthe highest rectangle in the histogram in proportion tothe heights of the adjacent rectangles. The method ofdetermining the mean, median and modal values froma histogram is shown in Problem 4.Problem 4. The time taken in minutes toassemble a device is measured 50 times and theresults are as shown. Draw a histogram depictingthe data and hence determine the mean, median andmodal values of the distribution14.5–15.5 516.5–17.5 818.5–19.5 1620.5–21.5 1222.5–23.5 624.5–25.5 3The histogram is shown in Figure 32.1. The mean valuelies at the centroid of the histogram. With reference toany arbitrary axis, say YY shown at a time of 14 min-utes, the position of the horizontal value of the centroidcan be obtained from the relationship AM = (am),where A is the area of the histogram, M is the hor-izontal distance of the centroid from the axis YY, ais the area of a rectangle of the histogram and m isthe distance of the centroid of the rectangle from YY.The areas of the individual rectangles are shown circled14 15 16 17 18 19 20 21 22 23 24 25626 27426Frequency108121614Time in minutes1224321610EDAYY5.6ModeMedian MeanFCBFigure 32.1on the histogram giving a total area of 100 squareunits.The positions,m, of the centroids of the individualrectangles are 1,3,5,... units from YY. Thus100M = (10 × 1) + (16 × 3) + (32 × 5) + (24 × 7)+ (12 × 9) + (6 × 11)i.e. M =560100= 5.6 units from YYThus, the position of the mean with reference to thetime scale is 14 + 5.6, i.e. 19.6 minutes.The median is the value of time corresponding to avertical line dividing the total area of the histogram intotwo equal parts. The total area is 100 square units, hencethe vertical line must be drawn to give 50 units of area oneach side. To achieve this with reference to Figure 32.1,rectangle ABFE must be split so that 50 − (10 + 16)units of area lie on one side and 50− (24 + 12 + 6) unitsof area lie on the other. This shows that the area of ABFEis split so that 24 units of area lie to the left ofthe line and8 units of area lie to the right; i.e., the vertical line mustpass through 19.5 minutes. Thus, the median value ofthe distribution is 19.5 minutes.The mode is obtained by dividing the line AB, whichis the height of the highest rectangle, proportionally tothe heights of the adjacent rectangles. With referenceto Figure 32.1, this is achieved by joining AC and BDand drawing a vertical line through the point of inter-section of these two lines. This gives the mode of thedistribution, which is 19.3 minutes.Now try the following Practice ExercisePracticeExercise 126 Mean, median andmode for grouped data (answers onpage 354)1. 21 bricks have a mean mass of 24.2kg and29 similar bricks have a mass of 23.6kg.Determine the mean mass of the 50 bricks.2. The frequency distribution given below refersto the heights in centimetres of 100 people.Determine the mean value of the distribution,correct to the nearest millimetre.150–156 5157–163 18164–170 20171–177 27178–184 22185–191 8
• 302 Basic Engineering Mathematics3. The gain of 90 similar transistors is measuredand the results are as shown. By drawing ahistogramofthisfrequency distribution,deter-mine the mean, median and modal values ofthe distribution.83.5–85.5 686.5–88.5 3989.5–91.5 2792.5–94.5 1595.5–97.5 34. The diameters, in centimetres, of 60 holesbored in engine castings are measured andthe results are as shown. Draw a histogramdepicting these results and hence determinethe mean, median and modal values of thedistribution.2.011–2.014 72.016–2.019 162.021–2.024 232.026–2.029 92.031–2.034 532.4 Standard deviation32.4.1 Discrete dataThestandard deviation ofaset ofdatagivesan indicationof the amount of dispersion, or the scatter, of membersof the set from the measure of central tendency. Its valueis the root-mean-square value of the members of the setand for discrete data is obtained as follows.(i) Determine the measure of central tendency, usu-ally the mean value, (occasionally the median ormodal values are speciﬁed).(ii) Calculate the deviation of each member of the setfrom the mean, giving(x1 − x),(x2 − x),(x3 − x),...(iii) Determine the squares of these deviations; i.e.,(x1 − x)2,(x2 − x)2,(x3 − x)2,...(iv) Find the sum of the squares of the deviations, i.e.(x1 − x)2+ (x2 − x)2+ (x3 − x)2,...(v) Divide by the number of members in the set, n,giving(x1 − x)2+ (x2 − x)2+ x3 − x2+ ···n(vi) Determine the square root of (v).The standard deviation is indicated by σ (the Greekletter small ‘sigma’) and is written mathematically asstandard deviation,σ =(x − x)2nwhere x is a member of the set, x is the mean value ofthe set and n is the number of members in the set. Thevalue of standard deviation gives an indication of thedistance of the members of a set from the mean value.The set {1, 4, 7, 10, 13} has a mean value of 7 and astandard deviation of about 4.2. The set {5, 6, 7, 8, 9}also has a mean value of 7 but the standard deviation isabout 1.4. This shows that the members of the secondset are mainly much closer to the mean value than themembers of the ﬁrst set. The method of determining thestandard deviation for a set of discrete data is shown inProblem 5.Problem 5. Determine the standard deviationfrom the mean of the set of numbers{5, 6, 8, 4, 10, 3}, correct to 4 signiﬁcant ﬁguresThe arithmetic mean, x =xn=5 + 6 + 8 + 4 + 10 + 36= 6Standard deviation, σ =(x − x)2nThe (x − x)2 values are (5 − 6)2,(6 − 6)2,(8 − 6)2,(4 − 6)2,(10 − 6)2and (3 − 6)2.The sum of the (x − x)2 values,i.e. (x − x)2, is 1 + 0 + 4 + 4 + 16 + 9 = 34and(x − x)2n=346= 5.·6since there are 6 members in the set.Hence, standard deviation,σ =(x − x)2n= 5.·6=2.380,correct to 4 signiﬁcant ﬁgures.
• Mean, median, mode and standard deviation 30332.4.2 Grouped dataFor grouped data,standard deviation,σ ={ f (x − x)2}fwhere f is the class frequency value, x is the class mid-point value and x is the mean value of the grouped data.The method of determining the standard deviation for aset of grouped data is shown in Problem 6.Problem 6. The frequency distribution for thevalues of resistance in ohms of 48 resistors is asshown. Calculate the standard deviation from themean of the resistors, correct to 3 signiﬁcant ﬁgures20.5–20.9 321.0–21.4 1021.5–21.9 1122.0–22.4 1322.5–22.9 923.0–23.4 2The standard deviation for grouped data is given byσ ={ f (x − x)2}fFrom Problem 3, the distribution mean value isx = 21.92, correct to 2 signiﬁcant ﬁgures.The ‘x-values’ are the class mid-point values, i.e.20.7,21.2,21.7,...Thus, the (x − x)2 values are (20.7 − 21.92)2,(21.2 − 21.92)2, (21.7 − 21.92)2,...and the f (x − x)2values are 3(20.7 − 21.92)2,10(21.2 − 21.92)2,11(21.7 − 21.92)2,...The f (x − x)2values are4.4652 + 5.1840+ 0.5324 + 1.0192+5.4756 + 3.2768 = 19.9532f (x − x)2f=19.953248= 0.41569and standard deviation,σ ={ f (x − x)2}f=√0.41569= 0.645,correct to 3 signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 127 Standard deviation(answers on page 354)1. Determine the standard deviation from themean of the set of numbers{35, 22, 25, 23, 28, 33, 30},correct to 3 signiﬁcant ﬁgures.2. The values of capacitances, in microfarads, often capacitors selected at random from a largebatch of similar capacitors are34.3, 25.0, 30.4, 34.6, 29.6, 28.7,33.4, 32.7, 29.0 and 31.3.Determine the standard deviation from themean for these capacitors, correct to 3 signif-icant ﬁgures.3. The tensile strength in megapascals for 15samplesoftin weredetermined and found to be34.61, 34.57, 34.40, 34.63, 34.63, 34.51,34.49, 34.61, 34.52, 34.55, 34.58, 34.53,34.44, 34.48 and 34.40.Calculate the mean and standard deviationfrom the mean for these 15 values, correct to4 signiﬁcant ﬁgures.4. Calculate the standard deviation from themean for the mass of the 50 bricks given inproblem 1 of Practice Exercise 126, page 301,correct to 3 signiﬁcant ﬁgures.5. Determine the standard deviation from themean, correct to 4 signiﬁcant ﬁgures, for theheights of the 100 people given in problem 2of Practice Exercise 126, page 301.6. Calculate the standard deviation from themean for the data given in problem 4 ofPractice Exercise 126, page 302, correct to 3decimal places.32.5 Quartiles, deciles and percentilesOther measures of dispersion which are sometimesused are the quartile, decile and percentile values. Thequartile values of a set of discrete data are obtained byselecting the values of members which divide the set
• 304 Basic Engineering Mathematicsinto four equal parts. Thus, for the set {2, 3, 4, 5, 5, 7,9, 11, 13, 14, 17} there are 11 members and the valuesof the members dividing the set into four equal partsare 4, 7 and 13. These values are signiﬁed by Q1, Q2and Q3 and called the ﬁrst, second and third quartilevalues, respectively. It can be seen that the second quar-tile value, Q2, is the value of the middle member andhence is the median value of the set.For grouped data the ogive may be used to determinethequartilevalues.In thiscase,pointsareselected on thevertical cumulative frequency values of the ogive, suchthat they divide the total value of cumulative frequencyinto four equal parts. Horizontal lines are drawn fromthese values to cut the ogive. The values of the variablecorresponding to these cutting points on the ogive givethe quartile values (see Problem 7).When a set contains a large number of members, theset can be split into ten parts, each containing an equalnumber of members. These ten parts are then calleddeciles.Forsetscontaining avery largenumberofmem-bers, the set may be split into one hundred parts, eachcontaining an equal number of members. One of theseparts is called a percentile.Problem 7. The frequency distribution givenbelow refers to the overtime worked by a group ofcraftsmen during each of 48 working weeks in ayear. Draw an ogive for these data and hencedetermine the quartile values.25–29 530–34 435–39 740–44 1145–49 1250–54 855–59 1The cumulative frequency distribution (i.e. upper classboundary/cumulative frequency values) is29.5 5, 34.5 9, 39.5 16, 44.5 27,49.5 39, 54.5 47, 59.5 48.The ogive is formed by plotting these values on a graph,as shown in Figure 32.2. The total frequency is dividedinto four equal parts, each having a range of 48 ÷ 4,i.e. 12. This gives cumulative frequency values of 0 to12 corresponding to the ﬁrst quartile, 12 to 24 corre-sponding to the second quartile, 24 to 36 correspondingto the third quartile and 36 to 48 corresponding to thefourth quartile of the distribution; i.e., the distribution2510Cumulativefrequency403020504030 35Q1 Q2 Q345Upper class boundary values, hours55 6050Figure 32.2is divided into four equal parts. The quartile valuesare those of the variable corresponding to cumulativefrequency values of 12, 24 and 36, marked Q1, Q2 andQ3 in Figure 32.2. These values, correct to the nearesthour, are 37 hours, 43 hours and 48 hours, respec-tively. The Q2 value is also equal to the median valueof the distribution. One measure of the dispersion of adistribution is called the semi-interquartile range andis given by (Q3 − Q1) ÷ 2 and is (48 − 37) ÷ 2 in thiscase; i.e., 512hours.Problem 8. Determine the numbers contained inthe (a) 41st to 50th percentile group and (b) 8thdecile group of the following set of numbers.14 22 17 21 30 28 37 7 23 3224 17 20 22 27 19 26 21 15 29The set is ranked, giving7 14 15 17 17 19 20 21 21 2222 23 24 26 27 28 29 30 32 37(a) Thereare20 numbersin theset,hence theﬁrst 10%will be the two numbers 7 and 14, the second 10%will be 15 and 17, and so on. Thus, the 41st to 50thpercentile group will be the numbers 21 and 22.(b) The ﬁrst decile group is obtained by splitting theranked set into 10 equal groups and selecting theﬁrst group; i.e., the numbers 7 and 14. The seconddecile group is the numbers 15 and 17, and so on.Thus, the 8th decile group contains the numbers27 and 28.
• Mean, median, mode and standard deviation 305Now try the following Practice ExercisePracticeExercise 128 Quartiles, decilesand percentiles (answers on page 354)1. The number of working days lost due to acci-dents for each of 12 one-monthly periods areas shown. Determine the median and ﬁrst andthird quartile values for this data.27 37 40 28 23 30 35 24 30 32 31 282. The number of faults occurring on a produc-tion line in a nine-week period are as shownbelow. Determine the median and quartilevalues for the data.30 27 25 24 27 37 31 27 353. Determine the quartile values and semi-interquartile range for the frequency distribu-tion given in problem 2 of Practice Exercise126, page 301.4. Determine the numbers contained in the 5thdecile group and in the 61st to 70th percentilegroups for the following set of numbers.40 46 28 32 37 42 50 31 48 4532 38 27 33 40 35 25 42 38 415. Determine the numbers in the 6th decile groupand in the 81st to 90th percentile group for thefollowing set of numbers.43 47 30 25 15 51 17 21 37 33 44 56 40 49 2236 44 33 17 35 58 51 35 44 40 31 41 55 50 16
• Chapter 33Probability33.1 Introduction to probability33.1.1 ProbabilityThe probability of something happening is the likeli-hood or chance of it happening. Values of probabilityliebetween 0 and 1, where 0 represents an absolute impos-sibilityand 1 represents an absolute certainty. The prob-ability of an event happening usually lies somewherebetween these two extreme values and is expressedas either a proper or decimal fraction. Examples ofprobability arethat a length of copper wirehas zero resistance at 100◦C 0that a fair, six-sided dicewill stop with a 3 upwards16or 0.1667that a fair coin will landwith a head upwards12or 0.5that a length of copper wirehas some resistance at 100◦C 1If p is the probability of an event happening and q is theprobability of the same event not happening, then thetotal probability is p + q and is equal to unity, since itis an absolute certainty that the event either will or willnot occur; i.e., p + q = 1.Problem 1. Determine the probabilities ofselecting at random (a) a man and (b) a womanfrom a crowd containing 20 men and 33 women(a) The probability of selecting at random a man, p,is given by the rationumber of mennumber in crowdi.e. p =2020 + 33=2053or 0.3774(b) The probability of selecting at random a woman,q, is given by the rationumber of womennumber in crowdi.e. q =3320 + 33=3353or 0.6226(Check: the total probability should be equal to 1:p =2053and q =3353, thus the total probability,p + q =2053+3353= 1hence no obvious error has been made.)33.1.2 ExpectationThe expectation, E, of an event happening is deﬁnedin general terms as the product of the probability p ofan event happening and the number of attempts made,n; i.e., E = pn.Thus, since the probability of obtaining a 3 upwardswhen rolling a fair dice is 1/6, the expectation of gettinga 3 upwards on four throws of the dice is16× 4,i.e.23Thus expectation is the average occurrence of anevent.DOI: 10.1016/B978-1-85617-697-2.00033-8
• Probability 307Problem 2. Find the expectation of obtaining a 4upwards with 3 throws of a fair diceExpectation is the average occurrence of an event and isdeﬁned as the probability times the number of attempts.The probability, p, of obtaining a 4 upwards for onethrow of the dice is 1/6.If 3 attempts are made, n = 3 and the expectation, E, ispn, i.e.E =16× 3 =12or 0.5033.1.3 Dependent eventsA dependent event is one in which the probability ofan event happening affects the probability of anotherever happening. Let 5 transistors be taken at randomfrom a batch of 100 transistors for test purposes and theprobability of there being a defective transistor, p1, bedetermined. At some later time, let another 5 transistorsbe taken at random from the 95 remaining transistorsin the batch and the probability of there being a defec-tive transistor, p2, be determined. The value of p2 isdifferent from p1 since the batch size has effectivelyaltered from 100 to 95; i.e., probability p2 is depen-dent on probability p1. Since transistors are drawn andthen another 5 transistors are drawn without replacingthe ﬁrst 5, the second random selection is said to bewithout replacement.33.1.4 Independent eventsAn independent event is one in which the probabilityof an event happening does not affect the probabilityof another event happening. If 5 transistors are taken atrandom from a batch oftransistorsand the probabilityofa defective transistor, p1, is determined and the processis repeated after the original 5 have been replaced inthe batch to give p2, then p1 is equal to p2. Since the5 transistors are replaced between draws, the secondselection is said to be with replacement.33.2 Laws of probability33.2.1 The addition law of probabilityThe addition law of probability is recognized by theword ‘or’ joining the probabilities.If pA is the probability of event A happening and pBis the probability of event B happening, the probabilityof event A or event B happening is given by pA + pB.Similarly, the probability of events A or B or C or...N happening is given bypA + pB + pC + ··· + pN33.2.2 The multiplicationlaw of probabilityThe multiplication law of probability is recognized bythe word ‘and’ joining the probabilities.If pA is the probability of event A happening and pB isthe probability of event B happening, the probability ofevent A and event B happening is given by pA × pB.Similarly, the probabilityof events A and B and C and...N happening is given bypA × pB × pC × ...× pNHere are some worked problems to demonstrate proba-bility.Problem 3. Calculate the probabilities ofselecting at random(a) the winning horse in a race in which 10 horsesare running and(b) the winning horses in both the ﬁrst and secondraces if there are 10 horses in each race(a) Since only one of the ten horses can win, theprobability of selecting at random the winninghorse isnumber of winnersnumber of horses,i.e.110or 0.10(b) The probability of selecting the winning horse inthe ﬁrst race is110The probability of selecting the winning horse inthe second race is110The probability of selecting the winning horsesin the ﬁrst and second race is given by themultiplication law of probability; i.e.,probability =110×110=1100or 0.01Problem 4. The probability of a componentfailing in one year due to excessive temperature is1/20, that due to excessive vibration is 1/25 and thatdue to excessive humidity is 1/50. Determine theprobabilities that during a one-year period acomponent(a) fails due to excessive temperature andexcessive vibration,
• 308 Basic Engineering Mathematics(b) fails due to excessive vibration or excessivehumidity,(c) will not fail because of both excessivetemperature and excessive humidityLet pA be the probability of failure due to excessivetemperature, thenpA =120and pA =1920(where pA is the probability of not failing)Let pB be the probability of failure due to excessivevibration, thenpB =125and pB =2425Let pC be the probability of failure due to excessivehumidity, thenpC =150and pC =4950(a) The probability of a component failing due toexcessive temperature and excessive vibration isgiven bypA × pB =120×125=1500or 0.002(b) The probability of a component failing due toexcessive vibration or excessive humidity ispB + pC =125+150=350or 0.06(c) The probability that a component will not fail dueexcessive temperature and will not fail due toexcess humidity ispA × pC =1920×4950=9311000or 0.931Problem 5. A batch of 100 capacitors contains 73which are within the required tolerance values and17 which are below the required tolerance values,the remainder being above the required tolerancevalues. Determine the probabilities that, whenrandomly selecting a capacitor and then a secondcapacitor,(a) both are within the required tolerance valueswhen selecting with replacement,(b) the ﬁrst one drawn is below and the secondone drawn is above the required tolerancevalue, when selection is without replacement(a) The probability of selecting a capacitor withinthe required tolerance values is 73/100. The ﬁrstcapacitor drawn is now replaced and a second oneis drawn from the batch of 100. The probabilityof this capacitor being within the required toler-ance values is also 73/100.Thus, the probabilityofselecting a capacitor within the required tolerancevalues for both the ﬁrst and the second draw is73100×73100=532910000or 0.5329(b) The probability of obtaining a capacitor belowthe required tolerance values on the ﬁrst drawis 17/100. There are now only 99 capacitorsleft in the batch, since the ﬁrst capacitor is notreplaced. The probability of drawing a capacitorabove the required tolerance values on the seconddraw is 10/99, since there are (100 − 73 − 17),i.e. 10, capacitors above the required tolerancevalue. Thus, the probability of randomly select-ing a capacitor below the required tolerance valuesand subsequently randomly selecting a capacitorabove the tolerance values is17100×1099=1709900=17990or 0.0172Now try the following Practice ExercisePracticeExercise 129 Laws of probability(answers on page 354)1. In a batch of 45 lamps 10 are faulty. If onelamp is drawn at random, ﬁnd the probabilityof it being (a) faulty (b) satisfactory.2. A box of fuses are all of the same shape andsize and comprises 23 2A fuses, 47 5A fusesand 69 13A fuses. Determine the probabilityof selecting at random (a) a 2A fuse (b) a 5Afuse (c) a 13A fuse.3. (a) Find the probability of having a 2upwards when throwing a fair 6-sideddice.(b) Find the probability of having a 5upwards when throwing a fair 6-sideddice.(c) Determine the probability of having a 2and then a 5 on two successive throws ofa fair 6-sided dice.
• Probability 3094. Determine the probability that the total scoreis 8 when two like dice are thrown.5. The probability of event A happening is35and the probability of event B happening is23. Calculate the probabilities of(a) both A and B happening.(b) only event A happening, i.e. event Ahappening and event B not happening.(c) only event B happening.(d) either A, or B, or A and B happening.6. When testing 1000 soldered joints, 4 failedduring a vibration test and 5 failed due tohaving a high resistance. Determine the prob-ability of a joint failing due to(a) vibration.(b) high resistance.(c) vibration or high resistance.(d) vibration and high resistance.Here are some further worked problems on probability.Problem 6. A batch of 40 components contains 5which are defective. A component is drawn atrandom from the batch and tested and then a secondcomponent is drawn. Determine the probability thatneither of the components is defective when drawn(a) with replacement and (b) without replacement(a) With replacementThe probabilitythat the component selected on theﬁrst draw is satisfactory is 35/40 i.e. 7/8. The com-ponent is nowreplaced and a second draw is made.The probability that this component is also satis-factory is 7/8. Hence, the probability that both theﬁrst component drawn and the second componentdrawn are satisfactory is78×78=4964or 0.7656(b) Without replacementThe probability that the ﬁrst component drawnis satisfactory is 7/8. There are now only 34satisfactory components left in the batch and thebatch number is 39. Hence, the probability ofdrawing a satisfactory component on the sec-ond draw is 34/39. Thus, the probability that theﬁrst component drawn and the second componentdrawn are satisfactory i.e., neither is defective is78×3439=238312or 0.7628Problem 7. A batch of 40 components contains 5which are defective. If a component is drawn atrandom from the batch and tested and then a secondcomponent is drawn at random, calculate theprobability of having one defective component, both(a) with replacement and (b) without replacementThe probability of having one defective component canbe achieved in two ways. If p is the probability of draw-ing a defective component and q is the probability ofdrawing a satisfactory component, then the probabilityof having one defective component is given by drawinga satisfactory component and then a defective compo-nent or by drawing a defective component and then asatisfactory one; i.e., by q × p + p × q.(a) With replacementp =540=18and q =3540=78Hence, the probability of having one defectivecomponent is18×78+78×18i.e.764+764=732or 0.2188(b) Without replacementp1 =18and q1 =78on the ﬁrst of the two drawsThe batch number is now 39 for the second draw,thus,p2 =539and q2 =3539p1q2 + q1 p2 =18×3539+78×539=35 + 35312=70312or 0.2244
• 310 Basic Engineering MathematicsProblem 8. A box contains 74 brass washers,86 steel washers and 40 aluminium washers. Threewashers are drawn at random from the box withoutreplacement. Determine the probability that allthree are steel washersAssume, for clarity of explanation, that a washer isdrawn at random, then a second, then a third (althoughthis assumption does not affect the results obtained).The total number of washers is74 + 86 + 40, i.e. 200The probability of randomly selecting a steel washer onthe ﬁrst draw is 86/200. There are now 85 steel washersin a batch of 199. The probability of randomly selectinga steel washer on the second draw is 85/199. There arenow 84 steel washers in a batch of 198. The probabilityof randomly selecting a steel washer on the third drawis 84/198. Hence, the probability of selecting a steelwasher on the ﬁrst draw and the second draw and thethird draw is86200×85199×84198=6140407880400= 0.0779Problem 9. For the box of washers given inProblem 8 above, determine the probability thatthere are no aluminium washers drawn when threewashers are drawn at random from the box withoutreplacementThe probability of not drawing an aluminium washeron the ﬁrst draw is 1 −40200i.e., 160/200. There arenow 199 washers in the batch of which 159 are not madeof aluminium. Hence, the probability of not drawingan aluminium washer on the second draw is 159/199.Similarly, the probability of not drawing an aluminiumwasher on the third draw is 158/198. Hence the proba-bility of not drawing an aluminium washer on the ﬁrstand second and third draws is160200×159199×158198=40195207880400= 0.5101Problem 10. For the box of washers in Problem 8above, ﬁnd the probability that there are two brasswashers and either a steel or an aluminium washerwhen three are drawn at random, withoutreplacementTwo brass washers (A) and one steel washer (B) can beobtained in any of the following ways.1st draw 2nd draw 3rd drawA A BA B AB A ATwo brass washers and one aluminium washer (C) canalso be obtained in any of the following ways.1st draw 2nd draw 3rd drawA A CA C AC A AThus, there are six possible ways of achieving the com-binations speciﬁed. If A represents a brass washer,B a steel washer and C an aluminium washer, thecombinations and their probabilities are as shown.Draw ProbabilityFirst Second ThirdA A B74200×73199×86198= 0.0590A B A74200×86199×73198= 0.0590B A A86200×74199×73198= 0.0590A A C74200×73199×40198= 0.0274A C A74200×40199×73198= 0.0274C A A40200×74199×73198= 0.0274The probability of having the ﬁrst combination or thesecond or the third, and so on, is given by the sum ofthe probabilities; i.e., by 3 × 0.0590 + 3 × 0.0274, i.e.0.2592
• Probability 311Now try the following Practice ExercisePracticeExercise 130 Laws of probability(answers on page 354)1. The probabilitythat component A will operatesatisfactorily for 5 years is 0.8 and that B willoperate satisfactorily over that same period oftime is 0.75. Find the probabilities that in a 5year period(a) both components will operatesatisfactorily.(b) only component A will operatesatisfactorily.(c) only component B will operatesatisfactorily.2. In a particular street, 80% of the houses havelandline telephones. If two houses selected atrandom are visited, calculate the probabilitiesthat(a) they both have a telephone.(b) one has a telephone but the other doesnot.3. Veroboard pins are packed in packets of 20by a machine. In a thousand packets, 40 haveless than 20 pins. Find the probability that if 2packets are chosen at random, one will containless than 20 pins and the other will contain 20pins or more.4. A batch of 1kW ﬁre elements contains 16which are within a power tolerance and 4which are not.If 3 elements are selected at ran-dom from the batch, calculate the probabilitiesthat(a) all three are within the power tolerance.(b) two are within but one is not within thepower tolerance.5. An ampliﬁer is made up of three transis-tors, A, B and C. The probabilities of A, Bor C being defective are 1/20, 1/25 and1/50, respectively. Calculate the percentage ofampliﬁers produced(a) which work satisfactorily.(b) which have just one defective transistor.6. Abox contains14 40Wlamps,28 60Wlampsand 58 25W lamps, all the lamps being of thesame shape and size. Three lamps are drawn atrandom from the box, ﬁrst one, then a second,then a third. Determine the probabilities of(a) getting one25W,one40Wand one60Wlamp with replacement.(b) getting one25W,one40Wand one60Wlamp without replacement.(c) getting either one 25W and two 40Wor one 60W and two 40 Wlamps withreplacement.
• Revision Test 13 : Statistics and probabilityThis assignment covers the material contained in Chapters 31–33. The marks available are shown in brackets at theend of each question.1. A company produces ﬁve products in the follow-ing proportions:Product A 24Product B 6Product C 15Product D 9Product E 18Draw (a) a horizontal bar chart and (b) a piediagram to represent these data visually. (9)2. State whether the data obtained on the followingtopics 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. (3)3. Draw a histogram, frequency polygon and ogivefor the data given below which refers to thediameter of 50 components produced by amachine.Class intervals Frequency1.30–1.32mm 41.33–1.35mm 71.36–1.38mm 101.39–1.41mm 121.42–1.44mm 81.45–1.47mm 51.48–1.50mm 4(16)4. Determine the mean, median and modal values forthe following lengths given in metres:28,20,44,30,32,30,28,34,26,28 (6)5. The length in millimetres of 100 bolts is as shownbelow.50–56 657–63 1664–70 2271–77 3078–84 1985–91 7Determine for the sample(a) the mean value.(b) the standard deviation, correct to 4 signiﬁcantﬁgures. (10)6. The number of faulty components in a factory ina 12 week period is14 12 16 15 10 13 15 11 16 19 17 19Determine the median and the ﬁrst and thirdquartile values. (7)7. Determine the probability of winning a prize ina lottery by buying 10 tickets when there are 10prizes and a total of 5000 tickets sold. (4)8. A sample of 50 resistors contains 44 which arewithin the required tolerance value, 4 which arebelow and the remainder which are above. Deter-mine the probability of selecting from the samplea resistor which is(a) below the required tolerance.(b) above the required tolerance.Now two resistors are selected at random fromthe sample. Determine the probability, correct to3 decimal places, that neither resistor is defectivewhen drawn(c) with replacement.(d) without replacement.(e) If a resistor is drawn at random from thebatch and tested and then a second resistor isdrawn from those left, calculate the probabil-ity of having one defective component whenselection is without replacement. (15)
• Chapter 34Introduction todifferentiation34.1 Introduction to calculusCalculus is a branch of mathematics involving or lead-ing to calculations dealing with continuously varyingfunctions such as velocity and acceleration, rates ofchange and maximum and minimum values of curves.Calculus has widespread applications in science andengineering and is used to solve complicated problemsfor which algebra alone is insufﬁcient.Calculus is a subject that falls into two parts:(a) differential calculus (or differentiation),(b) integral calculus (or integration).This chapter provides an introduction to differentiationand appliesdifferentiation to ratesofchange.Chapter35introduces integration and applies it to determine areasunder curves.Further applications of differentiation and integrationare explored in Engineering Mathematics (Bird, 2010).34.2 Functional notationIn an equation such as y = 3x2 + 2x − 5, y is said to bea function of x and may be written as y = f (x).An equation written in the form f (x) = 3x2 + 2x − 5is termed functional notation. The value of f (x)when x = 0 is denoted by f (0), and the value of f (x)when x = 2 is denoted by f (2), and so on. Thus, whenf (x) = 3x2 + 2x − 5,f (0) = 3(0)2+ 2(0) − 5 = −5and f (2) = 3(2)2+ 2(2) − 5 = 11,and so on.Problem 1. If f (x) = 4x2 − 3x + 2, ﬁndf (0), f (3), f (−1) and f (3) − f (−1)f (x) = 4x2− 3x + 2f (0) = 4(0)2− 3(0) + 2 = 2f (3) = 4(3)2− 3(3) + 2 = 36 − 9 + 2 = 29f (−1) = 4(−1)2− 3(−1) + 2 = 4 + 3 + 2 = 9f (3) − f (−1) = 29 − 9 = 20Problem 2. Given that f (x) = 5x2 + x − 7,determine (a) f (−2) (b) f (2) ÷ f (1)(a) f (−2) = 5(−2)2 + (−2) − 7 = 20 − 2 − 7 = 11(b) f (2) = 5(2)2 + 2 − 7 = 15f (1) = 5(1)2 + 1 − 7 = −1f (2) ÷ f (1) =15−1= −15Now try the following Practice ExercisePracticeExercise 131 Functional notation(answers on page 354)1. If f (x) = 6x2 − 2x + 1, ﬁnd f (0), f (1),f (2), f (−1) and f (−3).2. If f (x) = 2x2 + 5x − 7, ﬁnd f (1), f (2),f (−1), f (2) − f (−1).DOI: 10.1016/B978-1-85617-697-2.00034-X
• Introduction to differentiation 315chord JK. By moving J nearer and nearer toK, determine the gradient of the tangent of thecurve at K.34.4 Differentiation from ﬁrstprinciplesIn Figure 34.4, A and B are two points very closetogether on a curve, δx (delta x) and δy (delta y) rep-resenting small increments in the x and y directions,respectively.0yf(x)f(x1␦x)␦y␦xxA(x, y)B (x 1␦x, y1␦y)Figure 34.4Gradient of chord AB =δyδxhowever, δy = f (x + δx) − f (x)Hence,δyδx=f (x + δx) − f (x)δxAs δx approaches zero,δyδxapproaches a limiting valueand the gradient of the chord approaches the gradient ofthe tangent at A.When determining the gradient of a tangent to a curvethere are two notations used. The gradient of the curveat A in Figure 34.4 can either be written aslimitδx→0δyδxor limitδx→0f (x + δx) − f (x)δxIn Leibniz notation,dydx= limitδx→0δyδxIn functional notation,f (x) = limitδx→0f (x + δx) − f (x)δxdydxis the same as f (x) and is called the differentialcoefﬁcient or the derivative. The process of ﬁnding thedifferential coefﬁcient is called differentiation.Summarizing, the differential coefﬁcient,dydx= f (x) = limitδx→0δyδx= limitδx→0f (x + δx) − f (x)δxProblem 3. Differentiate from ﬁrst principlesf (x) = x2To ‘differentiate from ﬁrst principles’ means ‘to ﬁndf (x)’ using the expressionf (x) = limitδx→0f (x + δx) − f (x)δxf (x) = x2 and substituting (x + δx) for x givesf (x + δx) = (x + δx)2 = x2 + 2xδx + δx2, hence,f (x) = limitδx→0(x2 + 2xδx + δx2) − (x2)δx= limitδx→02xδx + δx2δx= limitδx→0{2x + δx}As δx → 0,{2x + δx} → {2x + 0}.Thus, f (x) = 2x i.e. the differential coefﬁcient of x2is 2x.This means that the general equation for the gradient ofthe curve f (x) = x2 is 2x. If the gradient is required at,say, x = 3, then gradient = 2(3) = 6.Differentiation from ﬁrst principles can be a lengthyprocess and we do not want to have to go through thisprocedureevery timewewant to differentiateafunction.In reality we do not have to because from the aboveprocedure has evolved a set general rule, which weconsider in the following section.34.5 Differentiation of y = axn by thegeneral ruleFrom differentiation by ﬁrst principles, a general rulefor differentiating axn emerges where a and n are anyconstants. This rule isif y = axnthendydx= anxn−1or if f (x) = axnthen f (x) = anxn−1
• 316 Basic Engineering MathematicsWhen differentiating, results can be expressed in anumber of ways. For example,(a) if y = 3x2 thendydx= 6x(b) if f (x) = 3x2 then f (x) = 6x(c) the differential coefﬁcient of 3x2 is 6x(d) the derivative of 3x2is 6x(e)ddx(3x2) = 6x34.5.1 Revision of some laws of indices1xa= x−aFor example,1x2= x−2and x−5=1x5√x = x12 For example,√5 = 512 and1612 =√16 = ±4 and1√x=1x12= x−12a√xb = xba For example,3√x5 = x53 and x43 =3√x4and13√x7=1x73= x−73x0 = 1 For example, 70 = 1 and 43.50 = 1Here are some worked problems to demonstrate thegeneral rule for differentiating y = axn.Problem 4. Differentiate the following withrespect to x: y = 4x7Comparing y = 4x7 with y = axn shows that a = 4 andn = 7. Using the general rule,dydx= anxn−1= (4)(7)x7−1= 28x6Problem 5. Differentiate the following withrespect to x: y =3x2y =3x2= 3x−2, hence a = 3 and n = −2 in the generalrule.dydx= anxn−1= (3)(−2)x−2−1= −6x−3= −6x3Problem 6. Differentiate the following withrespect to x: y = 5√xy = 5√x = 5x12 , hence a = 5 and n =12in thegeneral rule.dydx= anxn−1= (5)12x12 −1=52x−12 =52x12=52√xProblem 7. Differentiate y = 4y = 4 may be written as y = 4x0; i.e., in the general rulea = 4 and n = 0. Hence,dydx= (4)(0)x0−1= 0The equation y = 4 represents a straight horizontalline and the gradient of a horizontal line is zero, hencethe result could have been determined on inspection.In general, the differential coefﬁcient of a constant isalways zero.Problem 8. Differentiate y = 7xSince y = 7x, i.e. y = 7x1, in the general rule a = 7 andn = 1. Hence,dydx= (7)(1)x1−1= 7x0= 7 since x0= 1The gradient of the line y = 7x is 7 (fromy = mx + c), hence the result could have been obtainedby inspection.In general, the differential coefﬁcient ofkx, where k is a constant, is always k.Problem 9. Find the differential coefﬁcient ofy =23x4 −4x3+ 9y =23x4−4x3+ 9i.e. y =23x4− 4x−3+ 9dydx=23(4)x4−1− (4)(−3)x−3−1+ 0=83x3+ 12x−4
• Introduction to differentiation 317i.e.dydx=83x3+12x4Problem 10. If f (t) = 4t +1√t3ﬁnd f (t)f (t) = 4t +1√t3= 4t +1t32= 4t1+ t− 32Hence, f (t) = (4)(1)t1−1+ −32t− 32 −1= 4t0−32t− 52i.e. f (t) = 4 −32t52= 4 −32√t5Problem 11. Determinedydxgiven y =3x2 − 5x2xy =3x2 − 5x2x=3x22x−5x2x=32x −52Hence,dydx=32or 1.5Problem 12. Find the differential coefﬁcient ofy =25x3−4x3+ 4√x5 + 7y =25x3−4x3+ 4 x5 + 7i.e. y =25x3− 4x−3+ 4x52 + 7dydx=25(3)x3−1− (4)(−3)x−3−1+ (4)52x52 −1+ 0=65x2+ 12x−4+ 10x32i.e.dydx=65x2+12x4+ 10 x3Problem 13. Differentiate y =(x + 2)2xwithrespect to xy =(x + 2)2x=x2 + 4x + 4x=x2x+4xx+4xi.e. y = x1+ 4 + 4x−1Hence,dydx= 1x1−1+ 0 + (4)(−1)x−1−1= x0− 4x−2= 1 −4x2(since x0= 1)Problem 14. Find the gradient of the curvey = 2x2−3xat x = 2y = 2x2−3x= 2x3− 3x−1Gradient =dydx= (2)(2)x2−1− (3)(−1)x−1−1= 4x + 3x−2= 4x +3x2When x = 2, gradient = 4x +3x2= 4(2) +3(2)2= 8 +34= 8.75Problem 15. Find the gradient of the curvey = 3x4 − 2x2 + 5x − 2 at the points (0,−2)and (1,4)The gradient of a curve at a given point is given by thecorresponding value of the derivative.Thus, since y = 3x4 − 2x2 + 5x − 2,the gradient =dydx= 12x3 − 4x + 5.At the point (0,−2), x = 0, thusthe gradient = 12(0)3 − 4(0) + 5 = 5At the point (1,4), x = 1, thusthe gradient = 12(1)3 − 4(1) + 5 = 13Now try the following Practice ExercisePracticeExercise 133 Differentiation ofy = axnby the general rule (answers onpage 354)In problems 1 to 20, determine the differentialcoefﬁcients with respect to the variable.1. y = 7x4 2. y = 2x + 13. y = x2 − x 4. y = 2x3 − 5x + 6
• Introduction to differentiation 319y0(a)(b)0y 5cos xx radiansx radians122␲21dydx2␲␲ 3␲2␲22␲␲ 3␲2(cos x)52sin xddxFigure 34.6If each point on the curve y = sin x (as shown inFigure 34.5(a)) were to be made negative (i.e. +π2made −π2,−3π2made +3π2, and so on) then the graphshown in Figure 34.6(b) would result. This latter graphtherefore represents the curve of −sin x.Thus,if y = cosx,dydx= −sinxIt may also be shown thatif y = cosax,dydx= −asinax (3)(wherea isaconstant)and if y = cos(ax + α),dydx= −asin(ax + α) (4)(where a and α are constants).Problem 16. Find the differential coefﬁcient ofy = 7sin2x − 3cos4xdydx= (7)(2cos2x) − (3)(−4sin 4x)from equations (1) and (3)= 14cos2x + 12sin4xProblem 17. Differentiate the following withrespect to the variable (a) y = 2sin5θ(b) f (t) = 3cos2t(a) y = 2sin5θdydθ= (2)(5cos5θ) = 10cos5θfrom equation (1)(b) f (t) = 3cos2tf (t) = (3)(−2sin 2t) = −6sin2tfrom equation (3)Problem 18. Differentiate the following withrespect to the variable(a) f (θ) = 5sin(100πθ − 0.40)(b) f (t) = 2cos(5t + 0.20)(a) If f (θ) = 5sin(100πθ − 0.40)f (θ) = 5[100π cos(100πθ − 0.40)]from equation (2), where a = 100π= 500π cos(100πθ − 0.40)(b) If f (t) = 2cos(5t + 0.20)f (t) = 2[−5sin(5t + 0.20)]from equation (4), where a = 5= −10sin(5t + 0.20)Problem 19. An alternating voltage is given byv = 100sin200t volts, where t is the time inseconds. Calculate the rate of change of voltagewhen (a) t = 0.005s and (b) t = 0.01sv = 100sin200t volts. The rate of change of v is givenbydvdtdvdt= (100)(200cos200t) = 20000cos200t(a) When t = 0.005s,dvdt= 20000cos(200)(0.005) = 20000cos1.cos1 means ‘the cosine of 1 radian’ (make sureyourcalculator is on radians, not degrees). Hence,dvdt= 10 806 volts per second(b) When t = 0.01s,dvdt= 20000cos(200)(0.01) = 20000cos2.Hence,dvdt= −8323 volts per second
• 320 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 134 Differentiation ofsine and cosine functions (answers onpage 354)1. Differentiatewith respect to x: (a) y = 4sin3x(b) y = 2cos6x.2. Given f (θ) = 2sin3θ − 5cos2θ, ﬁnd f (θ).3. Find the gradient of the curve y = 2cos12x atx =π24. Determine the gradient of the curvey = 3sin2x at x =π35. An alternating current is given byi = 5sin100t amperes, where t is the timein seconds. Determine the rate of change ofcurrent i.e.didtwhen t = 0.01 seconds.6. v = 50sin40t volts represents an alternatingvoltage, v, where t is the time in seconds. Ata time of 20 × 10−3 seconds, ﬁnd the rate ofchange of voltage i.e.dvdt.7. If f (t) = 3sin(4t + 0.12) − 2cos(3t − 0.72),determine f (t).34.7 Differentiation of eax and lnaxA graph of y = ex is shown in Figure 34.7(a). The gra-dient of the curve at any point is given bydydxand iscontinually changing. By drawing tangents to the curveat many points on the curve and measuring the gradientofthetangents,valuesofdydxforcorresponding valuesofx may be obtained. These values are shown graphicallyin Figure 34.7(b).The graph ofdydxagainst x is identical to the originalgraph of y = ex. It follows thatif y = ex, thendydx= exIt may also be shown thatif y = eax, thendydx= aeax3y5exx215101520y212223 0(a)3dydx5exdydxx215101520y212223 0(b)Figure 34.7Therefore,if y = 2e6x,thendydx= (2)(6e6x) = 12e6xA graph of y = lnx is shown in Figure 34.8(a). The gra-dient of the curve at any point is given bydydxand iscontinually changing. By drawing tangents to the curveat many points on the curve and measuring the gradientof the tangents, values ofdydxfor corresponding valuesof x may be obtained. These values are shown graphi-cally in Figure 34.8(b).The graph ofdydxagainst x is the graph ofdydx=1xIt follows thatif y = lnx, thendydx=1xIt may also be shown thatif y = lnax, thendydx=1x(Note that, in the latter expression, the constant a doesnot appear in thedydxterm.) Thus,if y = ln4x, thendydx=1x
• Introduction to differentiation 3211.00.50 1 2 3(b)(a)y 5 In x4 5 6 x1.52y02122121 2 3 4 5 6 xdy5dx1xdydxFigure 34.8Problem 20. Differentiate the following withrespect to the variable (a) y = 3e2x (b) f (t) =43e5t(a) If y = 3e2x thendydx= (3)(2e2x ) = 6e2x(b) If f (t) =43e5t=43e−5t , thenf (t) =43(−5e−5t) = −203e−5t= −203e5tProblem 21. Differentiate y = 5ln3xIf y = 5ln3x, thendydx= (5)1x=5xNow try the following Practice ExercisePracticeExercise 135 Differentiation of eaxand lnax (answers on page 354)1. Differentiate with respect to x: (a) y = 5e3x(b) y =27e2x2. Given f (θ) = 5ln2θ − 4ln3θ, determinef (θ).3. If f (t) = 4lnt + 2, evaluate f (t) whent = 0.254. Find the gradient of the curvey = 2ex −14ln2x at x =12correct to 2decimal places.5. Evaluatedydxwhen x = 1, giveny = 3e4x −52e3x+ 8ln5x. Give the answercorrect to 3 signiﬁcant ﬁgures.34.8 Summary of standardderivativesThe standard derivatives used in this chapter are sum-marized in Table 34.1 and are true for all real valuesof x.Table 34.1y or f (x)dydxor f (x)axnanxn−1sinax a cosaxcosax −a sinaxeax aeaxlnax1xProblem 22. Find the gradient of the curvey = 3x2− 7x + 2 at the point (1,−2)If y = 3x2 − 7x + 2, then gradient =dydx= 6x − 7At the point (1,−2), x = 1,hence gradient = 6(1) − 7 = −1Problem 23. If y =3x2− 2sin4x +2ex+ ln5x,determinedydxy =3x2− 2sin4x +2ex+ ln5x= 3x−2− 2sin4x + 2e−x+ ln5x
• 322 Basic Engineering Mathematicsdydx= 3(−2x−3) − 2(4cos4x) + 2(−e−x) +1x= −6x3− 8cos4x −2ex+1xNow try the following Practice ExercisePracticeExercise 136 Standard derivatives(answers on page 354)1. Find the gradient of the curvey = 2x4 + 3x3 − x + 4 at the points(a) (0,4) (b) (1,8)2. Differentiate with respect to x:y =2x2+ 2ln2x−2(cos5x + 3sin2x)−2e3x34.9 Successive differentiationWhen a function y = f (x) is differentiated with respectto x,thedifferential coefﬁcient iswritten asdydxor f (x).If the expression is differentiated again, the second dif-ferential coefﬁcient is obtained and is written asd2 ydx2(pronounced dee two y by dee x squared) or f (x)(pronounced f double-dash x). By successive differen-tiation further higher derivatives such asd3 ydx3andd4 ydx4may be obtained. Thus,if y = 5x4,dydx= 20x3,d2 ydx2= 60x2,d3 ydx3= 120x,d4 ydx4= 120 andd5 ydx5= 0Problem 24. If f (x) = 4x5 − 2x3 + x − 3, ﬁndf (x)f (x) = 4x5− 2x3+ x − 3f (x) = 20x4− 6x2+ 1f (x) = 80x3− 12x or 4x(20x2− 3)Problem 25. Given y =23x3 −4x2+12x−√x,determined2 ydx2y =23x3−4x2+12x−√x=23x3− 4x−2+12x−1− x12dydx=233x2− 4 − 2x−3+12− 1x−2−12x−12i.e. dydx= 2x2+ 8x−3−12x−2−12x−12d2ydx2= 4x + (8)(−3x−4) −12− 2x−3−12−12x−32= 4x − 24x−4+ 1x−3+14x−32i.e.d2ydx2= 4x −24x4+1x3+14√x3Now try the following Practice ExercisePracticeExercise 137 Successivedifferentiation (answers on page 354)1. If y = 3x4 + 2x3 − 3x + 2, ﬁnd (a)d2 ydx2(b)d3 ydx32. If y = 4x2 +1xﬁndd2 ydx23. (a) Given f (t) =25t2 −1t3+3t−√t + 1,determine f (t).(b) Evaluate f (t) in part (a) when t = 1.4. If y = 3sin2t + cost, ﬁndd2 ydx25. If f (θ) = 2ln4θ, show that f (θ) = −2θ2
• Introduction to differentiation 32334.10 Rates of changeIf a quantity y depends on and varies with a quantityx then the rate of change of y with respect to x isdydx.Thus, for example, the rate of change of pressure p withheight h isdpdhA rate of change with respect to time is usually justcalled ‘the rate of change’, the ‘with respect to time’being assumed. Thus, for example, a rate of change ofcurrent, i, isdidtand a rate of change of temperature, θ,isdθdt, and so on.Here are some worked problems to demonstrate practi-cal examples of rates of change.Problem 26. The length L metres of a certainmetal rod at temperature t◦C is given byL = 1 + 0.00003t + 0.0000004t2. Determine therate of change of length, in mm/◦C, when thetemperature is (a) 100◦C (b) 250◦CThe rate of change of length meansdLdtSince length L = 1 + 0.00003t + 0.0000004t2, thendLdt= 0.00003 + 0.0000008t.(a) When t = 100◦C,dLdt= 0.00003 + (0.0000008)(100)= 0.00011m/◦C = 0.11mm/◦C.(b) When t = 250◦C,dLdt= 0.00003 + (0.0000008)(250)= 0.00023m/◦C = 0.23mm/◦C.Problem 27. The luminous intensity I candelasof a lamp at varying voltage V is given byI = 5 × 10−4V2. Determine the voltage at whichthe light is increasing at a rate of 0.4 candelasper voltThe rate of change of light with respect to voltage isgiven bydIdVSince I = 5 × 10−4 V2,dIdV= (5 × 10−4)(2V ) = 10 × 10−4 V = 10−3 V.When the lightis increasing at 0.4 candelas per volt then+0.4 = 10−3 V, from whichvoltage,V =0.410−3= 0.4 × 10+3= 400 voltsProblem 28. Newton’s law of cooling is given byθ = θ0e−kt , where the excess of temperature at zerotime is θ0◦C and at time t seconds is θ◦C.Determine the rate of change of temperature after50s, given that θ0 = 15◦C and k = −0.02The rate of change of temperature isdθdtSince θ = θ0e−kt, thendθdt= (θ0)(−ke−kt)= −kθ0e−ktWhen θ0 = 15,k = −0.02 and t = 50, thendθdt= −(−0.02)(15)e−(−0.02)(50)= 0.30 e1= 0.815◦C/sProblem 29. The pressure p of the atmosphere atheight h above ground level is given byp = p0 e−h/c, where p0 is the pressure at groundlevel and c is a constant. Determine the rateof change of pressure with height whenp0 = 105 pascals and c = 6.2 × 104 at 1550 metresThe rate of change of pressure with height isdpdhSince p = p0e−h/c, thendpdh= (p0) −1ce−h/c= −p0ce−h/cWhen p0 = 105,c = 6.2 × 104 and h = 1550, thenrate of change of pressure,dpdh= −1056.2 × 104e− 1550/6.2×104= −106.2e−0.025= −1.573Pa/m
• 324 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 138 Rates of change(answers on page 355)1. An alternating current, i amperes, is given byi = 10sin2πft, where f is the frequency inhertz and t is the time in seconds. Determinethe rate of change of current when t = 12 ms,given that f = 50 Hz.2. The luminous intensity, I candelas, of a lampis given by I = 8 × 10−4 V2, where V is thevoltage. Find(a) the rate of change of luminous intensitywith voltage when V = 100 volts.(b) the voltage at which the light is increas-ing at a rate of 0.5 candelas per volt.3. The voltage across the plates of a capacitor atany time t seconds is given by v = V e−t/CR,where V,C and R are constants. GivenV = 200V,C = 0.10μF and R = 2M , ﬁnd(a) the initial rate of change of voltage.(b) the rate of change of voltage after 0.2s4. The pressure p of the atmosphere at height habove ground level is given by p = p0e−h/c,where p0 is the pressure at ground level and cis a constant. Determine the rate of change ofpressure with height when p0 = 1.013 × 105pascals and c = 6.05 × 104 at 1450 metres.
• Chapter 35Introduction to integration35.1 The process of integrationThe process of integration reverses the process ofdifferentiation. In differentiation, if f (x) = 2x2 thenf (x) = 4x. Thus, the integral of 4x is 2x2; i.e., inte-gration is the process of moving from f (x) to f (x). Bysimilar reasoning, the integral of 2t is t2.Integration is a process of summation or adding partstogether and an elongated S, shown as , is used toreplace the words ‘the integral of’. Hence, from above,4x = 2x2 and 2t is t2.In differentiation, the differential coefﬁcientdydxindi-cates that a function of x is being differentiated withrespect to x, the dx indicating that it is ‘with respectto x’.In integration the variable of integration is shownby adding d (the variable) after the function to beintegrated. Thus,4x dx means ‘the integral of 4x with respect to x’,and 2t dt means ‘the integral of 2t with respect to t’As stated above, the differential coefﬁcient of 2x2 is 4x,hence; 4x dx = 2x2. However, the differential coefﬁ-cient of 2x2 + 7 is also 4x. Hence, 4x dx could alsobe equal to 2x2 + 7. To allow for the possible presenceof a constant, whenever the process of integration isperformed a constant c is added to the result. Thus,4x dx = 2x2+ c and 2t dt = t2+ cc is called the arbitrary constant of integration.35.2 The general solution of integralsof the form axnThe general solution of integrals of the form axn dx,where a and n are constants and n = −1 is given byaxndx =axn+1n + 1+ cUsing this rule gives(i) 3x4dx =3x4+14 + 1+ c =35x5+ c(ii)49t3dt dx =49t3+13 + 1+ c =49t44+ c=19t4 + c(iii)2x2dx = 2x−2dx =2x−2+1−2 + 1+ c=2x−1−1+ c = −2x+ c(iv)√x dx = x12 dx =x12 +112 + 1+ c =x3232+ c=23√x3 + cEach of these results may be checked by differentia-tion.(a) The integral of a constant k is kx + c. Forexample,8dx = 8x + c and 5dt = 5t + cDOI: 10.1016/B978-1-85617-697-2.00035-1
• 326 Basic Engineering Mathematics(b) When a sum of several terms is integrated theresult is the sum of the integrals of the separateterms. For example,(3x + 2x2− 5)dx= 3x dx + 2x2dx − 5dx=3x22+2x33− 5x + c35.3 Standard integralsFrom Chapter 34,ddx(sin ax) = a cos ax. Since inte-gration is the reverse process of differentiation, itfollows thata cos ax dx = sin ax + cor cos ax dx =1asin ax + cBy similar reasoningsin ax dx = −1acos ax + ceaxdx =1aeax+ cand1xdx = lnx + cFrom above, axn dx =axn+1n + 1+ c except whenn = −1When n = −1, x−1 dx =1xdx = ln x + cA list of standard integrals is summarized inTable 35.1.Table 35.1 Standard integralsy y dx1. axn axn+1n + 1+ c (except when n = −1)2. cos ax dx1asin ax + c3. sin ax dx −1acos ax + c4. eax dx1aeax + c5.1xdx ln x + cProblem 1. Determine 7x2dxThe standard integral, axndx =axn+1n + 1+ cWhen a = 7 and n = 2,7x2dx =7x2+12 + 1+ c =7x33+ c or73x3+ cProblem 2. Determine 2t3 dtWhen a = 2 and n = 3,2t3dt =2t3+13 + 1+ c =2t44+ c =12t4+ cNote that each of the results in worked examples 1 and2 may be checked by differentiating them.Problem 3. Determine 8dx8dx is the same as 8x0 dx and, using the generalrule when a = 8 and n = 0, gives8x0dx =8x0+10 + 1+ c = 8x + cIn general, if k is a constant then kdx = kx + c.Problem 4. Determine 2x dxWhen a = 2 and n = 1,2x dx = 2x1dx =2x1+11 + 1+ c =2x22+ c= x2+ cProblem 5. Determine 3 +25x − 6x2dx3 +25x − 6x2 dx may be written as3dx+25x dx − 6x2 dxi.e., each term is integrated separately. (This splittingup of terms only applies, however, for addition and
• Introduction to integration 327subtraction.) Hence,3 +25x − 6x2dx= 3x +25x1+11 + 1− (6)x2+12 + 1+ c= 3x +25x22− (6)x33+ c = 3x +15x2− 2x3+ cNote that when an integral contains more than one termthere is no need to have an arbitrary constant for each;just a single constant c at the end is sufﬁcient.Problem 6. Determine2x3 − 3x4xdxRearranging into standard integral form gives2x3 − 3x4xdx =2x34x−3x4xdx=12x2−34dx =12x2+12 + 1−34x + c=12x33−34x + c =16x 3−34x + cProblem 7. Determine 1 − t2dtRearranging (1 − t)2dt gives(1 − 2t + t2)dt = t −2t1+11 + 1+t2+12 + 1+ c= t −2t22+t33+ c= t − t2+13t3+ cThis example shows that functionsoften have to be rear-ranged into the standard form of axndx before it ispossible to integrate them.Problem 8. Determine5x2dx5x2dx = 5x−2dxUsing the standard integral, axndx, when a = 5 andn = −2, gives5x−2dx =5x−2+1−2 + 1+ c =5x−1−1+ c= −5x−1+ c = −5x+ cProblem 9. Determine 3√xdxFor fractional powers it is necessary to appreciaten√am = amn3√x dx = 3x12 dx =3x12 +112+ 1+ c =3x3232+ c= 2x32 + c = 2 x3 + cProblem 10. Determine−594√t3dt−594√t3dt =−59t34dt = −59t−34 dt= −59t−34+1−34+ 1+ c= −59t1414+ c = −5941t14 + c= −2094√t + cProblem 11. Determine 4cos 3x dxFrom 2 of Table 35.1,4cos 3x dx = (4)13sin3x + c=43sin 3x + cProblem 12. Determine 5sin2θdθ
• 328 Basic Engineering MathematicsFrom 3 of Table 35.1,5sin2θdθ = (5) −12cos2θ + c= −52cos 2θ+cProblem 13. Determine 5e3xdxFrom 4 of Table 35.1,5e3xdx = (5)13e3x+ c=53e3x+ cProblem 14. Determine23e4tdt23e4tdt =23e−4tdt=23−14e−4t+ c= −16e−4t+ c = −16e4t+ cProblem 15. Determine35xdxFrom 5 of Table 35.1,35xdx =351xdx=35lnx + cProblem 16. Determine2x2 + 1xdx2x2 + 1xdx =2x2x+1xdx= 2x +1xdx =2x22+ ln x + c= x2+ lnx + cNow try the following Practice ExercisePracticeExercise 139 Standard integrals(answers on page 355)Determine the following integrals.1. (a) 4dx (b) 7x dx2. (a) 5x3 dx (b) 3t7 dt3. (a)25x2dx (b)56x3dx4. (a) (2x4 − 3x)dx (b) (2 − 3t3)dt5. (a)3x2 − 5xxdx (b) (2 + θ)2dθ]6. (a) (2 + θ)(3θ − 1)dθ(b) (3x − 2)(x2 + 1)dx7. (a)43x2dx (b)34x4dx8. (a) 2√x3dx (b)144√x5 dx9. (a)−5√t3dt (b)375√x4dx10. (a) 3cos2x dx (b) 7sin3θ dθ11. (a) 3sin12x dx (b) 6cos13x dx12. (a)34e2x dx (b)23dxe5x13. (a)23xdx (b)u2 − 1udu14. (a)(2 + 3x)2√xdx (b)1t+ 2t2dt35.4 Deﬁnite integralsIntegrals containing an arbitrary constant c in theirresults are called indeﬁnite integrals since theirprecisevalue cannot be determined without furtherinformation.Deﬁnite integrals are those in which limits are applied.Ifan expression iswritten as[x]ba, b iscalled the upperlimit and a the lower limit. The operation of applyingthe limits is deﬁned as [x]ba = (b) − (a).
• Introduction to integration 329For example, the increase in the value of the integralx2 as x increases from 1 to 3 is written as31 x2dx.Applying the limits gives31x2dx =x33+ c31=333+ c −133+ c= (9 + c) −13+ c = 823Note that the c term always cancels out when limits areapplied and it need not be shown with deﬁnite integrals.Problem 17. Evaluate213xdx213xdx =3x2221=32(2)2−32(1)2= 6 − 112= 412Problem 18. Evaluate3−2(4 − x2) dx3−2(4 − x2)dx = 4x −x333−2= 4(3) −(3)33− 4(−2) −(−2)33= {12 − 9} − −8 −−83= {3} − −513= 813Problem 19. Evaluate20x(3 + 2x)dx20x(3+2x)dx =20(3x +2x2)dx =3x22+2x3320=3(2)22+2(2)33− {0 + 0}= 6 +163= 1113or 11.33Problem 20. Evaluate1−1x4 − 5x2 + xxdx1−1x4 − 5x2 + xxdx=1−1x4x−5x2x+xxdx=1−1x3− 5x + 1 dx =x44−5x22+ x1−1=14−52+ 1 −(−1)44−5(−1)22+ (−1)=14−52+ 1 −14−52− 1 = 2Problem 21. Evaluate211x2+2xdx correctto 3 decimal places211x2+2xdx=21x−2+ 21xdx =x−2+1−2 + 1+ 2ln x21=x−1−1+ 2lnx21= −1x+ 2lnx21= −12+ 2ln2 − −11+ 2 ln1 = 1.886Problem 22. Evaluateπ/203sin2x dxπ/203sin2x dx= (3) −12cos2xπ/20= −32cos2xπ/20= −32cos2π2− −32cos2(0)= −32cosπ − −32cos0
• 330 Basic Engineering Mathematics= −32(−1) − −32(1)=32+32= 3Problem 23. Evaluate214cos3t dt214cos3t dt = (4)13sin3t21=43sin3t21=43sin6 −43sin3Note that limits of trigonometric functions are alwaysexpressed in radians – thus, for example, sin 6 meansthe sine of 6 radians = −0.279415... Hence,214cos3t dt=43(−0.279415 ...) −43(0.141120 ...)= (−0.37255) − (0.18816)= −0.5607Problem 24. Evaluate214e2xdx correct to4 signiﬁcant ﬁgures214e2xdx =42e2x21= 2 e2x21= 2[e4− e2]= 2[54.5982− 7.3891]= 94.42Problem 25. Evaluate4134udu correct to 4signiﬁcant ﬁgures4134udu =34lnu41=34[ln4 − ln 1]=34[1.3863 − 0] = 1.040Now try the following Practice ExercisePracticeExercise 140 Deﬁnite integrals(answers on page 355)In problems 1 to 10, evaluate the deﬁnite integrals(where necessary, correct to 4 signiﬁcant ﬁgures).1. (a)21x dx (b)21(x − 1)dx2. (a)415x2dx (b)1−1−34t2dt3. (a)2−1(3 − x2)dx (b)31(x2− 4x + 3)dx4. (a)21(x3− 3x)dx (b)21(x2− 3x + 3)dx5. (a)402√x dx (b)321x2dx6. (a)π032cosθ dθ (b)π/204cosθ dθ7. (a)π/3π/62sin2θ dθ (b)203sint dt8. (a)105cos3x dx(b)π/2π/4(3sin2x − 2cos3x)dx9. (a)103e3tdt (b)2−123e2xdx10. (a)3223xdx (b)312x2 + 1xdx35.5 The area under a curveThe area shown shaded in Figure 35.1 may be deter-mined using approximate methods such as the trape-zoidal rule, the mid-ordinate rule or Simpson’s rule (seeChapter 28) or, more precisely, by using integration.0 x 5 a x 5 by 5 f(x)yxFigure 35.1
• 332 Basic Engineering Mathematics4030201051 2 3 40v (m/s)t(s)v ϭ 2t2 ϩ 5Figure 35.4Problem 28. Sketch the graphy = x3 + 2x2 − 5x − 6 between x = −3 andx = 2 and determine the area enclosed bythe curve and the x-axisA table of values is produced and the graph sketched asshown in Figure 35.5, in which the area enclosed by thecurve and the x-axis is shown shaded.x −3 −2 −1 0 1 2y 0 4 0 −6 −8 0yx60 1212223y 5 x3 1 2x2 2 5x 2 62Figure 35.5Shaded area =−1−3y dx −2−1y dx, the minus signbefore the second integral being necessary since theenclosed area is below the x-axis. Hence,shaded area =−1−3(x3+ 2x2− 5x − 6)dx−2−1(x3+ 2x2− 5x − 6)dx=x44+2x33−5x22− 6x−1−3−x44+2x33−5x22− 6x2−1=14−23−52+ 6 −814− 18 −452+ 18− 4 +163− 10 − 12 −14−23−52+ 6= 3112− −214− −1223− 3112= 513− −1534= 21112or 21.08 square unitsProblem 29. Determine the area enclosed by thecurve y = 3x2 + 4, the x-axis and ordinates x = 1and x = 4 by (a) the trapezoidal rule, (b) themid-ordinate rule, (c) Simpson’s rule and(d) integration.The curve y = 3x2 + 4 is shown plotted in Figure 35.6.The trapezoidal rule, the mid-ordinate rule and Simp-son’s rule are discussed in Chapter 28, page 257.(a) By the trapezoidal rulearea =width ofinterval12ﬁrst + lastordinate+sum ofremainingordinates
• Introduction to integration 333x 0 1.0 1.5 2.0 2.5 3.0 3.5 4.0y 4 7 10.75 16 22.75 31 40.75 520 1 2 3 4 x45040302010yy 5 3x2 1 4Figure 35.6Selecting 6 intervals each of width 0.5 givesarea = (0.5)12(7 + 52) + 10.75 + 16+ 22.75 + 31 + 40.75= 75.375 square units(b) By the mid-ordinate rulearea = (width of interval)(sum of mid-ordinates)Selecting 6 intervals, each of width 0.5, gives themid-ordinates as shown by the broken lines inFigure 35.6. Thus,area = (0.5)(8.7 + 13.2 + 19.2 + 26.7+ 35.7 + 46.2)= 74.85 square units(c) By Simpson’s rulearea =13width ofintervalﬁrst + lastordinates+ 4sum of evenordinates+2sum of remainingodd ordinatesSelecting 6 intervals, each of width 0.5, givesarea =13(0.5)[(7 + 52) + 4(10.75 + 22.75+ 40.75) + 2(16 + 31)]= 75 square units(d) By integrationshaded area =41ydx=41(3x2+ 4)dx = x3+ 4x41= (64 + 16) − (1 + 4)= 75 square unitsIntegration gives the precise value for the area undera curve. In this case, Simpson’s rule is seen to be themost accurate of the three approximate methods.Problem 30. Find the area enclosed by the curvey = sin2x, the x-axis and the ordinates x = 0 andx =π3A sketch of y = sin2x is shown in Figure 35.7. (Notethat y = sin2x has a period of2π2i.e., π radians.)10 ␲/2 ␲␲/3y 5 sin 2xxyFigure 35.7
• 334 Basic Engineering MathematicsShaded area =π/30y dx=π/30sin2x dx = −12cos2xπ/30= −12cos2π3− −12cos0= −12−12− −12(1)=14+12=34or 0.75 square unitsNow try the following Practice ExercisePracticeExercise 141 Area under curves(answers on page 355)Unless otherwise stated all answers are in squareunits.1. Show by integration that the area of a rect-angle formed by the line y = 4, the ordinatesx = 1 and x = 6 and the x-axis is 20 squareunits.2. Show by integration that the area of the trian-gle formed by the line y = 2x, the ordinatesx = 0 and x = 4 and the x-axis is 16 squareunits.3. Sketch the curve y = 3x2 + 1 betweenx = −2 and x = 4. Determine by integrationthe area enclosed by the curve, the x-axis andordinates x = −1 and x = 3. Use an approx-imate method to ﬁnd the area and compareyour result with that obtained by integration.4. The force F newtons acting on a body at adistance x metres from a ﬁxed point is givenby F = 3x + 2x2. If work done =x2x1F dx,determine the work done when the bodymoves from the position where x1 = 1 m tothat when x2 = 3m.In problems 5 to 9, sketch graphs of thegiven equa-tions and then ﬁnd the area enclosed between thecurves, the horizontal axis and the given ordinates.5. y = 5x; x = 1, x = 46. y = 2x2 − x + 1; x = −1, x = 27. y = 2sin2x; x = 0, x =π48. y = 5cos3t; t = 0,t =π69. y = (x − 1)(x − 3); x = 0, x = 310. The velocity v of a vehicle t seconds after acertain instant isgiven by v = 3t2+ 4 m/s.Determine how far it moves in the intervalfrom t = 1s to t = 5s.
• Revision Test 14 : Differentiation and integrationThis assignment covers the material contained in Chapters 34 and 35. The marks available are shown in brackets atthe end of each question.1. Differentiate the following functions with respectto x.(a) y = 5x2 − 4x + 9 (b) y = x4 − 3x2 − 2(4)2. Given y = 2(x − 1)2, ﬁnddydx(3)3. If y =3xdeterminedydx(2)4. Given f (t) =√t5, ﬁnd f (t). (2)5. Determine the derivative of y = 5 − 3x +4x2(3)6. Calculate the gradient of the curve y = 3cosx3atx =π4, correct to 3 decimal places. (4)7. Find the gradient of the curvef (x) = 7x2 − 4x + 2 at the point (1, 5) (3)8. If y = 5sin3x − 2cos4x ﬁnddydx(2)9. Determine the value of the differential coefﬁcientof y = 5ln2x −3e2xwhen x = 0.8, correct to 3signiﬁcant ﬁgures. (4)10. If y = 5x4 − 3x3 + 2x2 − 6x + 5, ﬁnd (a)dydx(b)d2 ydx2(4)11. Newton’s law of cooling is given by θ = θ0e−kt ,where the excess of temperature at zero time isθ ◦0 C and at time t seconds is θ◦C. Determinethe rate of change of temperature after 40s, cor-rect to 3 decimal places, given that θ0 = 16◦C andk = −0.01 (4)In problems 12 to 15, determine the indeﬁnite integrals.12. (a) (x2+ 4)dx (b)1x3dx (4)13. (a)2√x+ 3√x dx (b) 3 t5dt (4)14. (a)23√x2dx (b) e0.5x+13x− 2 dx(6)15. (a) (2 + θ)2dθ(b) cos12x +3x− e2xdx (6)Evaluate the integrals in problems 16 to 19, each, wherenecessary, correct to 4 signiﬁcant ﬁgures.16. (a)31(t2− 2t)dt (b)2−12x3− 3x2+ 2 dx(6)17. (a)π/303sin2tdt (b)3π/4π/4cos13xdx (7)18. (a)212x2+1x+34dx(b)213x−1x3dx (8)19. (a)10√x + 2exdx (b)21r3−1rdr(6)In Problems 20 to 22, ﬁnd the area bounded by the curve,the x-axis and the given ordinates. Assume answers arein square units.Give answers correct to 2 decimal placeswhere necessary.20. y = x2; x = 0, x = 2 (3)21. y = 3x − x2; x = 0, x = 3 (3)22. y = (x − 2)2; x = 1, x = 2 (4)23. Find the area enclosed between the curvey =√x +1√x, the horizontal axis and the ordi-nates x = 1 and x = 4. Give the answer correct to2 decimal places. (5)24. The force F newtons acting on a body at a dis-tance x metres from a ﬁxed point is given byF = 2x + 3x2. If work done =x2x1Fdx, deter-mine the work done when the body movesfrom the position when x = 1m to that whenx = 4m. (3)
• List of formulaeLaws of indices:am × an = am+n aman= am−n (am)n= amnam/n = n√am a−n =1ana0 = 1Quadratic formula:If ax2 + bx + c = 0 then x =−b ±√b2 − 4ac2aEquation of a straight line:y = mx + cDeﬁnition of a logarithm:If y = axthen x = loga yLaws of logarithms:log(A × B) = log A + log BlogAB= log A − log Blog An= n × log AExponential series:ex = 1 + x +x22!+x33!+ ··· (valid for all values of x)Theorem of Pythagoras:b2= a2+ c2AB CacbAreas of plane ﬁgures:(i) Rectangle Area = l × bbl(ii) Parallelogram Area = b × hbh(iii) Trapezium Area =12(a + b)hahb(iv) Triangle Area =12× b × hhb
• List of formulae 337(v) Circle Area = πr2Circumference = 2πrrrs␪Radian measure: 2π radians = 360degreesFor a sector of circle:arc length, s =θ◦360(2πr) = rθ (θ in rad)shaded area =θ◦360(πr2) =12r2θ (θ in rad)Equation of a circle, centre at origin, radius r:x2+ y2= r2Equation of a circle, centre at (a, b), radius r:(x − a)2+ (y − b)2= r2Volumes and surface areas of regularsolids:(i) Rectangular prism (or cuboid)Volume = l × b × hSurface area = 2(bh + hl +lb)hbl(ii) CylinderVolume = πr2hTotal surface area = 2πrh + 2πr2rh(iii) PyramidIf area of base = A andperpendicular height = h then:Volume =13× A × hhTotal surface area = sum of areas of trianglesforming sides + area of base(iv) ConeVolume =13πr2hCurved Surface area = πrlTotal Surface area = πrl + πr2lhr(v) SphereVolume =43πr3Surface area = 4πr2r
• 338 Basic Engineering MathematicsAreas of irregular ﬁgures by approximatemethods:Trapezoidal ruleArea ≈width ofinterval12ﬁrst + lastordinate+ sum of remaining ordinatesMid-ordinate ruleArea ≈ (width of interval)(sum of mid-ordinates)Simpson’s ruleArea ≈13width ofintervalﬁrst + lastordinate+ 4sum of evenordinates+ 2sum of remainingodd ordinatesMean or average value of a waveform:mean value, y =area under curvelength of base=sum of mid-ordinatesnumber of mid-ordinatesTriangle formulae:Sine rule:asin A=bsin B=csinCCosine rule: a2= b2+ c2− 2bccos AACB ac bArea of any triangle=12× base × perpendicular height=12absinC or12acsin B or12bcsin A= [s (s − a)(s − b)(s − c)] where s =a + b + c2For a general sinusoidal function y= Asin(ωt ± α),thenA = amplitudeω = angular velocity = 2π f rad/sω2π= frequency, f hertz2πω= periodic time Tsecondsα = angle of lead or lag (compared withy = Asinωt)Cartesian and polar co-ordinates:If co-ordinate (x, y) = (r,θ) thenr = x2 + y2 and θ = tan−1 yxIf co-ordinate (r,θ) = (x, y) thenx = r cosθ and y = r sinθArithmetic progression:If a = ﬁrst term and d = common difference, then thearithmetic progression is: a,a + d,a + 2d,...The n’th term is: a + (n − 1)dSum of n terms, Sn =n2[2a + (n − 1)d]Geometric progression:If a = ﬁrst term and r = common ratio, then the geom-etric progression is: a,ar,ar2,...The n’th term is: arn−1Sum of n terms, Sn =a (1 −rn )(1 −r)ora (rn − 1)(r − 1)If − 1 < r < 1, S∞ =a(1 −r)Statistics:Discrete data:mean, ¯x =xnstandard deviation, σ =(x − ¯x)2n
• List of formulae 339Grouped data:mean, ¯x =f xfstandard deviation, σ =f (x − ¯x)2fStandard derivativesy or f(x)dydx= or f (x)axn anxn−1sinax a cosaxcosax −a sin axeax aeaxln ax1xStandard integralsy y dxaxn axn+1n + 1+ c(except when n = −1)cosax1asinax + csinax −1acosax + ceax 1aeax + c1xlnx + c
• AnswersAnswers to practice exercisesChapter 1Exercise 1 (page 2)1. 19kg 2. 16m 3. 479mm4. −66 5. £565 6. −2257. −2136 8. −36121 9. £10770110. −4 11. 1487 12. 591413. 189g 14. −70872 15. \$1533316. 89.25cm17. d = 64mm, A = 136mm, B = 10 mmExercise 2 (page 5)1. (a) 468 (b) 868 2. (a) £1827 (b) £41583. (a) 8613kg (b) 584kg4. (a) 351mm (b) 924mm5. (a) 10304 (b) −4433 6. (a) 48m (b) 89m7. (a) 259 (b) 56 8. (a) 1648 (b) 10609. (a) 8067 (b) 3347 10. 18kgExercise 3 (page 6)1. (a) 4 (b) 24 2. (a) 12 (b) 3603. (a) 10 (b) 350 4. (a) 90 (b) 27005. (a) 2 (b) 210 6. (a) 3 (b) 1807. (a) 5 (b) 210 8. (a) 15 (b) 63009. (a) 14 (b) 420420 10. (a) 14 (b) 53900Exercise 4 (page 8)1. 59 2. 14 3. 88 4. 5 5. −336. 22 7. 68 8. 5 9. 2 10. 511. −1Chapter 2Exercise 5 (page 11)1. 2172. 7253.2294.7185.4116.37,49,12,35,587.8258.11159.173010.91011.31612.31613.437714. 111515.42716. 8515217. 194018. 1162119.176020.1720Exercise 6 (page 13)1.8352. 2293.6114.5125.3286.357. 11 8.1139. 11210.81511. 22512.51213. 33414.122315. 416.3417.1918. 13 19. 15 20. 400 litres21. (a) £60, P£36, Q£16 22. 2880 litresExercise 7 (page 14)1. 21182. −193. 1164. 4345.13206.7157. 419208. 2 9. 71310.11511. 4 12. 21720DOI: 10.1016/B978-1-85617-697-2.00040-5
• Answers to practice exercises 341Chapter 3Exercise 8 (page 17)1.13202.92503.7404.61255. (a)1320(b)2125(c)180(d)141500(e)31256. 421407. 2311258. 1032009. 671610. (a) 14150(b) 41140(c) 1418(d) 15720(e) 16178011. 0.625 12. 6.6875 13. 0.21875 14. 11.187515. 0.28125Exercise 9 (page 18)1. 14.18 2. 2.785 3. 65.38 4. 43.275. 1.297 6. 0.000528Exercise 10 (page 19)1. 80.3 2. 329.3 3. 72.54 4. −124.835. 295.3 6. 18.3 7. 12.52 mmExercise 11 (page 20)1. 4.998 2. 47.544 3. 385.02 4. 582.425. 456.9 6. 434.82 7. 626.1 8. 1591.69. 0.444 10. 0.62963 11. 1.563 12. 53.45513. 13.84 14. 8.69 15. (a) 24.81 (b) 24.81216. (a) 0.00639 (b) 0.0064 17. (a) 8.˙4 (b) 62.˙618. 2400Chapter 4Exercise 12 (page 23)1. 30.797 2. 11927 3. 13.62 4. 53.8325. 84.42 6. 1.0944 7. 50.330 8. 36.459. 10.59 10. 12.325Exercise 13 (page 24)1. 12.25 2. 0.0361 3. 46.923 4. 1.296 × 10−35. 2.4430 6. 2.197 7. 30.96 8. 0.05499. 219.26 10. 5.832 × 10−6Exercise 14 (page 25)1. 0.571 2. 40 3. 0.13459 4. 137.95. 14.96 6. 19.4481 7. 515.36 × 10−68. 1.0871 9. 15.625 × 10−9 10. 52.70Exercise 15 (page 25)1. 2.182 2. 11.122 3. 185.824. 0.8307 5. 0.1581 6. 2.5717. 5.273 8. 1.256 9. 0.3036610. 1.068 11. 3.5×106 12. 37.5×10313. 4.2×10−6 14. 202.767×10−3 15. 18.32×106Exercise 16 (page 27)1. 0.4667 2.13143. 4.458 4. 2.7325.1216. 0.7083 7. −9108. 3139. 2.567 10. 0.0776Exercise 17 (page 27)1. 0.9205 2. 0.7314 3. 2.9042 4. 0.27195. 0.4424 6. 0.0321 7. 0.4232 8. 0.15029. −0.6992 10. 5.8452Exercise 18 (page 28)1. 4.995 2. 5.782 3. 25.72 4. 69.425. 0.6977 6. 52.92 7. 591.0 8. 17.909. 3.520 10. 0.3770Exercise 19 (page 29)1. A = 66.59cm2 2. C = 52.78mm 3. R = 37.54. 159m/s 5. 0.407A 6. 5.02 mm7. 0.144J 8. 628.8m2 9. 224.510. 14230 kg/m3 11. 281.1m/s 12. 2.526Exercise 20 (page 30)1. £589.27 2. 508.1W 3. V = 2.61V4. F = 854.5 5. I = 3.81A 6. T = 14.79s7. E = 3.96J 8. I = 12.77A 9. s = 17.25m10. A = 7.184cm2 11. V = 7.32712. (a) 12.53h (b) 1h 40min, 33 m.p.h.(c) 13.02h (d) 13.15h
• 342 Basic Engineering MathematicsChapter 5Exercise 21 (page 34)1. 0.32% 2. 173.4% 3. 5.7% 4. 37.4%5. 128.5% 6. 0.20 7. 0.0125 8. 0.68759. 38.462% 10. (a) 21.2% (b) 79.2% (c) 169%11. (b), (d), (c), (a) 12.132013.51614.91615. A =12, B = 50%, C = 0.25, D = 25%, E = 0.30,F =310, G = 0.60, H = 60%, I = 0.85, J =1720Exercise 22 (page 36)1. 21.8kg 2. 9.72 m3. (a) 496.4t (b) 8.657g (c) 20.73s 4. 2.25%5. (a) 14% (b) 15.67% (c) 5.36% 6. 37.8g7. 14 minutes 57 seconds 8. 76g 9. £61110. 37.49% 11. 39.2% 12. 17% 13. 38.7%14. 2.7% 15. 5.60 m 16. 3.5%Exercise 23 (page 38)1. 2.5% 2. 18% 3. £310 4. £1750005. £260 6. £20000 7. £9116.45 8. £50.259. £39.60 10. £917.70 11. £185000 12. 7.2%13. A 0.6kg, B 0.9kg,C 0.5kg14. 54%,31%,15%,0.3t15. 20000 kg (or 20 tonnes)16. 13.5mm,11.5mm 17. 600 kWChapter 6Exercise 24 (page 41)1. 36 : 1 2. 3.5 : 1 or 7 : 2 3. 47 : 34. 96cm,240 cm 5. 514hours or 5 hours 15 minutes6. £3680, £1840, £920 7. 12 cm 8. £2172Exercise 25 (page 42)1. 1 : 15 2. 76ml 3. 25% 4. 12.6kg5. 14.3kg 6. 25000 kgExercise 26 (page 43)1. £556 2. £66 3. 264kg 4. 450 g 5. 14.56kg6. (a) 0.00025 (b) 48MPa 7. (a) 440 K (b) 5.76litreExercise 27 (page 45)1. (a) 2 mA (b) 25V 2. 170 fr 3. 685.8mm4. 83lb10 oz 5. (a) 159.1litres (b) 16.5gallons6. 29.4MPa 7. 584.2 mm 8. \$1012Exercise 28 (page 46)1. 3.5 weeks 2. 20 days3. (a) 9.18 (b) 6.12 (c) 0.3375 4. 50 minutes5. (a) 300 × 103 (b) 0.375m2 (c) 24 × 103 PaChapter 7Exercise 29 (page 48)1. 27 2. 128 3. 100000 4. 96 5. 246. ±5 7. ±8 8. 100 9. 1 10. 64Exercise 30 (page 50)1. 128 2. 393. 16 4.195. 1 6. 87. 100 8. 1000 9.1100or 0.01 10. 5 11. 7612. 36 13. 36 14. 34 15. 1 16. 2517.135or124318. 49 19.12or 0.5 20. 1Exercise 31 (page 52)1.13 × 522.173 × 373.32254.1210 × 525. 9 6. ±3 7. ±128. ±239.14714810. −1195611. −3134512.1913. −171814. 64 15. 412
• Answers to practice exercises 343Chapter 8Exercise 32 (page 55)1. cubic metres, m3 2. farad3. square metres, m2 4. metres per second, m/s5. kilogram per cubic metre, kg/m36. joule 7. coulomb 8. watt9. radian or degree 10. volt 11. mass12. electrical resistance 13. frequency14. acceleration 15. electric current16. inductance 17. length18. temperature 19. pressure20. angular velocity 21. ×109 22. m,×10−323. ×10−1224. M,×106Exercise 33 (page 56)1. (a) 7.39 × 10 (b) 2.84 × 10 (c) 1.9762 × 1022. (a) 2.748 × 103 (b) 3.317 × 104 (c) 2.74218 × 1053. (a) 2.401 × 10−1(b) 1.74 × 10−2(c) 9.23 × 10−34. (a) 1.7023 × 103 (b) 1.004 × 10 (c) 1.09 × 10−25. (a) 5 × 10−1 (b) 1.1875 × 10(c) 1.306 × 102 (d) 3.125 × 10−26. (a) 1010 (b) 932.7 (c) 54100 (d) 77. (a) 0.0389 (b) 0.6741 (c) 0.0088. (a) 1.35 × 102 (b) 1.1 × 1059. (a) 2 × 102 (b) 1.5 × 10−310. (a) 2.71 × 103 kgm−3 (b) 4.4 × 10−1(c) 3.7673 × 102(d) 5.11 × 10−1MeV(e) 9.57897×107 Ckg−1(f) 2.241×10−2 m3 mol−1Exercise 34 (page 58)1. 60 kPa 2. 150μW or 0.15mW3. 50 MV 4. 55nF5. 100 kW 6. 0.54mA or 540μA7. 1.5M 8. 22.5mV9. 35GHz 10. 15pF11. 17μA 12. 46.2 k13. 3μA 14. 2.025MHz15. 50 kN 16. 0.3nF17. 62.50 m 18. 0.0346kg19. 13.5 × 10−3 20. 4 × 103Chapter 9Exercise 35 (page 62)1. −3a 2. a + 2b + 4c3. 3x − 3x2 − 3y − 2y2 4. 6ab − 3a5. 6x − 5y + 8z 6. 1 + 2x7. 4x + 2y + 2 8. 3z + 5b9. −2a − b + 2c 10. −x2− y2Exercise 36 (page 64)1. p2q3r 2. 8a2 3. 6q24. 46 5. 5136. −127. 6 8.17y9. 5xz210. 3a2+ 2ab − b211. 6a2 − 13ab + 3ac − 5b2 + bc12.13b13. 3x 14. 2ab15. 2x − y 16. 3p + 2q 17. 2a2 + 2b2Exercise 37 (page 66)1. z8 2. a8 3. n34. b11 5. b−3 6. c47. m4 8. x−3 or1x39. x1210. y−6or1x611. t812. c1413. a−9 or1x914. b−12 or1x1215. b10 16. s−917. p6q7r5 18. x−2y z−2 oryx2 z219. x5 y4z3,131220. a2b−2c ora2cb2,9Exercise 38 (page 67)1. a2 b1/2c−2, 4122.1 + ab3. a b6 c3/24. a−4 b5 c11 5.p2qq − p6. x y3 6√z137.1ef 28. a11/6 b1/3 c−3/2 or6√a11 3√b√c3
• 344 Basic Engineering MathematicsChapter 10Exercise 39 (page 69)1. x2 + 5x + 6 2. 2x2 + 9x + 43. 4x2 + 12x + 9 4. 2 j2 + 2 j − 125. 4x2 + 22x + 30 6. 2pqr + p2q2 +r27. a2 + 2ab + b2 8. x2 + 12x + 369. a2 − 2ac + c2 10. 25x2 + 30x + 911. 4x2 − 24x + 36 12. 4x2 − 913. 64x2 + 64x + 16 14. r2s2 + 2rst + t215. 3ab − 6a2 16. 2x2 − 2xy17. 2a2 − 3ab − 5b2 18. 13p − 7q19. 7x − y − 4z 20. 4a2 − 25b221. x2 − 4xy + 4y2 22. 9a2 − 6ab + b223. 0 24. 4 − a25. 4ab − 8a2 26. 3xy + 9x2 y − 15x227. 2 + 5b2 28. 11q − 2pExercise 40 (page 71)1. 2(x + 2) 2. 2x(y − 4z)3. p(b + 2c) 4. 2x(1 + 2y)5. 4d(d − 3 f 5) 6. 4x(1 + 2x)7. 2q(q + 4n) 8. x(1 + 3x + 5x2)9. bc(a + b2) 10. r(s + p + t)11. 3xy(xy3 − 5y + 6) 12. 2 p q2 2p2 − 5q13. 7ab(3ab − 4) 14. 2xy(y + 3x + 4x2)15. 2xy x − 2y2 + 4x2 y3 16. 7y(4 + y + 2x)17.3xy18. 0 19.2rt20. (a + b)(y + 1) 21. (p + q)(x + y)22. (x − y)(a + b) 23. (a − 2b)(2x + 3y)Exercise 41 (page 72)1. 2x + 8x2 2. 12y2 − 3y3. 4b − 15b2 4. 4 + 3a5.32− 4x 6. 17. 10y2− 3y +148. 9x2+13− 4x9. 6a2 + 5a −1710. −15t11.15− x − x2 12. 10a2 − 3a + 2Chapter 11Exercise 42 (page 75)1. 1 2. 2 3. 6 4. −4 5. 26. 1 7. 2 8.129. 0 10. 311. 2 12. −10 13. 6 14. −2 15. 2.516. 2 17. 6 18. −3Exercise 43 (page 76)1. 5 2. −2 3. −4124. 2 5. 126. 15 7. −4 8. 5139. 2 10. 1311. −10 12. 2 13. 3 14. −11 15. −616. 9 17. 61418. 3 19. 4 20. 1021. ±12 22. −31323. ±3 24. ±4Exercise 44 (page 79)1. 10−7 2. 8m/s2 3. 3.4724. (a) 1.8 (b) 305. digital camera battery £9, camcorder battery £146. 800 7. 30 m/s2Exercise 45 (page 80)1. 12 cm,240 cm2 2. 0.004 3. 304. 45◦C 5. 50 6. £312,£2407. 30 kg 8. 12 m,8m 9. 3.5NChapter 12Exercise 46 (page 84)1. d = c − e − a − b 2. x =y73. v =cp4. a =v − ut5. R =VI6. y =13(t − x)7. r =c2π8. x =y − cm
• Answers to practice exercises 3459. T =IPR10. L =XL2π f11. R =EI12. x = a(y − 3)13. C =59(F − 32) 14. f =12π CXCExercise 47 (page 87)1. r =S − aSor 1 −aS2. x =dλ(y + λ) or d +ydλ3. f =3F − AL3or f = F −AL34. D =AB25Ey5. t =R − R0R0α6. R2 =RR1R1 − R7. R =E − e − IrIor R =E − eI−r8. b =y4ac29. x =ay(y2 − b2)10. L =t2g4π211. u =√v2 − 2as12. R =360Aπθ13. a = N2 y − x14. L =√Z2 − R22π f,0.080Exercise 48 (page 89)1. a =xym − n2. R = 4 Mπ+r43. r =3(x + y)(1 − x − y)4. L =mrCRμ − m5. b =c√1 − a26. r =x − yx + y7. b =a(p2 − q2)2(p2 + q2)8. v =u fu − f,309. t2 = t1 +Qmc,55 10. v =2dgh0.03L,0.96511. L =8S23d+ d,2.72512. C =1ω ωL −√Z2 − R2,63.1 × 10−613. 64mm 14. λ =5 aμρCZ4n2Chapter 13Exercise 49 (page 92)1. x = 4, y = 2 2. x = 3, y = 43. x = 2, y = 1.5 4. x = 4, y = 15. p = 2,q = −1 6. x = 1, y = 27. x = 3, y = 2 8. a = 2,b = 39. a = 5,b = 2 10. x = 1, y = 111. s = 2,t = 3 12. x = 3, y = −213. m = 2.5,n = 0.5 14. a = 6,b = −115. x = 2, y = 5 16. c = 2,d = −3Exercise 50 (page 94)1. p = −1, q = −2 2. x = 4, y = 63. a = 2, b = 3 4. s = 4,t = −15. x = 3, y = 4 6. u = 12,v = 27. x = 10, y = 15 8. a = 0.30,b = 0.40Exercise 51 (page 96)1. x =12, y =142. a =13,b = −123. p =14,q =154. x = 10, y = 55. c = 3,d = 4 6. r = 3,s =127. x = 5, y = 1348. 1Exercise 52 (page 99)1. a = 0.2,b = 4 2. I1 = 6.47, I2 = 4.623. u = 12,a = 4,v = 26 4. £15500,£128005. m = −0.5,c = 36. α = 0.00426, R0 = 22.56 7. a = 12,b = 0.408. a = 4,b = 10 9. F1 = 1.5, F2 = −4.5Exercise 53 (page 100)1. x = 2, y = 1, z = 3 2. x = 2, y = −2, z = 23. x = 5, y = −1, z = −2 4. x = 4, y = 0, z = 3
• 346 Basic Engineering Mathematics5. x = 2, y = 4, z = 5 6. x = 1, y = 6, z = 77. x = 5, y = 4, z = 2 8. x = −4, y = 3, z = 29. x = 1.5, y = 2.5, z = 4.510. i1 = −5,i2 = −4,i3 = 211. F1 = 2, F2 = −3 F3 = 4Chapter 14Exercise 54 (page 104)1. 4 or −4 2. 4 or −8 3. 2 or −64. −1.5 or 1.5 5. 0 or −436. 2 or −27. 4 8. −5 9. 110. −2 or −3 11. −3 or −7 12. 2 or −113. 4 or −3 14. 2 or 7 15. −416. 2 17. −3 18. 3 or −319. −2 or −2320. −1.5 21.18or −1822. 4 or −7 23. −1 or 1.5 24.12or1325.12or −4526. 113or −1727.38or −228.25or −3 29.43or −1230.54or −3231. x2− 4x + 3 = 0 32. x2+ 3x − 10 = 033. x2 + 5x + 4 = 0 34. 4x2 − 8x − 5 = 035. x2 − 36 = 0 36. x2 − 1.7x − 1.68 = 0Exercise 55 (page 106)1. −3.732 or −0.268 2. −3.137 or 0.6373. 1.468 or −1.135 4. 1.290 or 0.3105. 2.443 or 0.307 6. −2.851 or 0.351Exercise 56 (page 107)1. 0.637 or −3.137 2. 0.296 or −0.7923. 2.781 or 0.719 4. 0.443 or −1.6935. 3.608 or −1.108 6. 1.434 or 0.2327. 0.851 or −2.351 8. 2.086 or −0.0869. 1.481 or −1.081 10. 4.176 or −1.67611. 4 or 2.167 12. 7.141 or −3.64113. 4.562 or 0.438Exercise 57 (page 109)1. 1.191s 2. 0.345A or 0.905A 3. 7.84cm4. 0.619m or 19.38m 5. 0.01336. 1.066m 7. 86.78cm8. 1.835m or 18.165m 9. 7m10. 12 ohms,28ohmsExercise 58 (page 110)1. x = 1, y = 3 and x = −3, y = 72. x =25, y = −15and −123, y = −4133. x = 0, y = 4 and x = 3, y = 1Chapter 15Exercise 59 (page 112)1. 4 2. 4 3. 3 4. −3 5.136. 3 7. 2 8. −2 9. 11210.1311. 2 12. 10000 13. 100000 14. 9 15.13216. 0.01 17.11618. e3Exercise 60 (page 115)1. log6 2. log15 3. log2 4. log35. log12 6. log500 7. log100 8. log69. log10 10. log1 = 0 11. log212. log243 or log35 or 5log313. log16 or log24 or 4log214. log64 or log26 or 6log215. 0.5 16. 1.5 17. x = 2.5 18. t = 819. b = 2 20. x = 2 21. a = 6 22. x = 5Exercise 61 (page 116)1. 1.690 2. 3.170 3. 0.2696 4. 6.058 5. 2.2516. 3.959 7. 2.542 8. −0.3272 9. 316.2
• Answers to practice exercises 347Chapter 16Exercise 62 (page 118)1. (a) 0.1653 (b) 0.4584 (c) 220302. (a) 5.0988 (b) 0.064037 (c) 40.4463. (a) 4.55848 (b) 2.40444 (c) 8.051244. (a) 48.04106 (b) 4.07482 (c) −0.082865. 2.739 6. 120.7mExercise 63 (page 120)1. 2.0601 2. (a) 7.389 (b) 0.74083. 1 − 2x2 −83x3 − 2x44. 2x1/2 + 2x5/2 + x9/2 +13x13/2+112x17/2+160x21/2Exercise 64 (page 122)1. 3.95,2.05 2. 1.65,−1.303. (a) 28cm3 (b) 116min 4. (a) 70◦C (b) 5 minutesExercise 65 (page 124)1. (a) 0.55547 (b) 0.91374 (c) 8.89412. (a) 2.2293 (b) −0.33154 (c) 0.130873. −0.4904 4. −0.5822 5. 2.197 6. 816.27. 0.8274 8. 11.02 9. 1.522 10. 1.48511. 1.962 12. 3 13. 414. 147.9 15. 4.901 16. 3.09517. t = eb+a ln D= ebea ln D= ebeln Dai.e. t = ebDa18. 500 19. W = PV lnU2U1Exercise 66 (page 127)1. (a) 150◦C (b) 100.5◦C 2. 99.21kPa3. (a) 29.32 volts (b) 71.31 × 10−6 s4. (a) 1.993m (b) 2.293m 5. (a) 50◦C (b) 55.45s6. 30.37N 7. (a) 3.04A (b) 1.46s8. 2.45mol/cm3 9. (a) 7.07A (b) 0.966s10. £2424Chapter 17Exercise 67 (page 134)1. (a) Horizontal axis: 1cm = 4V (or 1cm = 5V),vertical axis: 1cm = 10(b) Horizontal axis: 1cm = 5m, vertical axis:1cm = 0.1V(c) Horizontal axis: 1cm = 10 N, vertical axis:1cm = 0.2 mm2. (a) −1 (b) −8 (c) −1.5 (d) 5 3. 14.54. (a) −1.1 (b) −1.45. The 1010 rev/min reading should be 1070 rev/min;(a) 1000 rev/min (b) 167VExercise 68 (page 140)1. Missing values: −0.75,0.25,0.75,2.25,2.75;Gradient =122. (a) 4,−2 (b) −1,0 (c) −3,−4 (d) 0,43. (a) 2,12(b) 3,−212(c)124,124. (a) 6,−3 (b) −2,4 (c) 3,0 (d) 0,75. (a) 2,−12(b) −23,−123(c)118,2 (d) 10,−4236. (a)35(b) −4 (c) −1567. (a) and (c), (b) and (e)8. (2, 1) 9. (1.5, 6) 10. (1, 2)11. (a) 89cm (b) 11N (c) 2.4 (d) l = 2.4 W + 4812. P = 0.15 W + 3.5 13. a = −20,b = 412Exercise 69 (page 144)1. (a) 40◦C (b) 1282. (a) 850 rev/min (b) 77.5V3. (a) 0.25 (b) 12 (c) F = 0.25L + 12(d) 89.5N (e) 592 N (f) 212 N4. −0.003,8.735. (a) 22.5m/s (b) 6.43s (c) v = 0.7t +15.56. m = 26.9L − 0.637. (a) 1.31t (b) 22.89% (c) F = −0.09W + 2.218. (a) 96 × 109Pa (b) 0.00022 (c) 28.8 × 106Pa
• 348 Basic Engineering Mathematics9. (a)15(b) 6 (c) E =15L + 6 (d) 12 N (e) 65N10. a = 0.85,b = 12,254.3kPa,275.5kPa,280 KChapter 18Exercise 70 (page 149)1. (a) y (b) x2(c) c (d) d 2. (a) y (b)√x (c) b (d) a3. (a) y (b)1x(c) f (d) e 4. (a)yx(b) x (c) b (d) c5. (a)yx(b)1x2(c) a (d) b6. a = 1.5,b = 0.4,11.78mm2 7. y = 2x2 + 7,5.158. (a) 950 (b) 317kN9. a = 0.4,b = 8.6 (i) 94.4 (ii) 11.2Exercise 71 (page 154)1. (a) lg y (b) x (c) lga (d) lgb2. (a) lg y (b) lgx (c) L (d) lgk3. (a) ln y (b) x (c) n (d) lnm4. I = 0.0012 V2,6.75 candelas5. a = 3.0,b = 0.56. a = 5.7,b = 2.6,38.53,3.07. R0 = 26.0,c = 1.42 8. y = 0.08e0.24x9. T0 = 35.4N,μ = 0.27,65.0 N,1.28 radiansChapter 19Exercise 72 (page 156)1. x = 2, y = 4 2. x = 1, y = 13. x = 3.5, y = 1.5 4. x = −1, y = 25. x = 2.3, y = −1.2 6. x = −2, y = −37. a = 0.4,b = 1.6Exercise 73 (page 160)1. (a) Minimum (0, 0) (b) Minimum (0,−1)(c) Maximum (0, 3) (d) Maximum (0,−1)2. −0.4 or 0.6 3. −3.9 or 6.94. −1.1 or 4.1 5. −1.8 or 2.26. x = −1.5 or −2, Minimum at (−1.75,−0.1)7. x = −0.7 or 1.6 8. (a) ±1.63 (b) 1 or −0.39. (−2.6,13.2),(0.6,0.8);x = −2.6 or 0.610. x = −1.2 or 2.5 (a) −30 (b) 2.75 and −1.50(c) 2.3 or −0.8Exercise 74 (page 161)1. x = 4, y = 8 and x = −0.5, y = −5.52. (a) x = −1.5 or 3.5 (b) x = −1.24 or 3.24(c) x = −1.5 or 3.0Exercise 75 (page 162)1. x = −2.0,−0.5 or 1.52. x = −2,1 or 3, Minimum at (2.1,−4.1),Maximum at (−0.8,8.2)3. x = 1 4. x = −2.0,0.4 or 2.65. x = 0.7 or 2.56. x = −2.3,1.0 or 1.8 7. x = −1.5Chapter 20Exercise 76 (page 167)1. 82◦27 2. 27◦54 3. 51◦11 4. 100◦6 525. 15◦44 17 6. 86◦49 1 7. 72.55◦ 8. 27.754◦9. 37◦57 10. 58◦22 52Exercise 77 (page 169)1. reﬂex 2. obtuse 3. acute 4. right angle5. (a) 21◦ (b) 62◦23 (c) 48◦56 176. (a) 102◦ (b) 165◦ (c) 10◦18 497. (a) 60◦ (b) 110◦ (c) 75◦ (d) 143◦ (e) 140◦(f ) 20◦(g) 129.3◦(h) 79◦(i) 54◦8. Transversal (a) 1 & 3, 2 & 4, 5 & 7, 6 & 8(b) 1 & 2, 2 & 3, 3 & 4, 4 & 1, 5 & 6, 6 & 7,7 & 8, 8 & 5, 3 & 8, 1 & 6, 4 & 7 or 2 & 5(c) 1 & 5, 2 & 6, 4 & 8, 3 & 7 (d) 3 & 5 or 2 & 89. 59◦20 10. a = 69◦,b = 21◦,c = 82◦ 11. 51◦12. 1.326rad 13. 0.605rad 14. 40◦55Exercise 78 (page 173)1. (a) acute-angled scalene triangle(b) isosceles triangle (c) right-angled triangle(d) obtuse-angled scalene triangle(e) equilateral triangle (f) right-angled triangle
• Answers to practice exercises 3492. a = 40◦,b = 82◦,c = 66◦,d = 75◦, e = 30◦, f = 75◦3. DF, DE 4. 52◦ 5. 122.5◦6. φ = 51◦, x = 161◦7. 40◦,70◦,70◦,125◦, isosceles8. a = 18◦50 ,b = 71◦10 ,c = 68◦,d = 90◦,e = 22◦, f = 49◦, g = 41◦9. a = 103◦,b = 55◦,c = 77◦,d = 125◦,e = 55◦, f = 22◦, g = 103◦,h = 77◦,i = 103◦, j = 77◦,k = 81◦10. 17◦11. A = 37◦, B = 60◦, E = 83◦Exercise 79 (page 176)1. (a) congruent BAC, DAC (SAS)(b) congruent FGE, JHI (SSS)(c) not necessarily congruent(d) congruent QRT, SRT (RHS)(e) congruent UVW, XZY (ASA)2. proofExercise 80 (page 178)1. x = 16.54mm, y = 4.18mm 2. 9cm,7.79cm3. (a) 2.25cm (b) 4cm 4. 3mExercise 81 (page 180)1–5. Constructions – see similar constructions inworked problems 30 to 33 on pages 179–180.Chapter 21Exercise 82 (page 182)1. 9cm 2. 24m 3. 9.54mm4. 20.81cm 5. 7.21m 6. 11.18cm7. 24.11mm 8. 82 + 152 = 1729. (a) 27.20 cm each (b) 45◦ 10. 20.81km11. 3.35m,10 cm 12. 132.7 nautical miles13. 2.94mm 14. 24mmExercise 83 (page 185)1. sin Z =941,cos Z =4041,tan X =409,cos X =9412. sin A =35,cos A =45,tan A =34,sin B =45,cos B =35,tan B =433. sin A =817,tan A =8154. sin X =15113,cos X =1121135. (a)1517(b)1517(c)8156. (a) sinθ =725(b) cosθ =24257. (a) 9.434 (b) −0.625Exercise 84 (page 187)1. 2.7550 2. 4.846 3. 36.524. (a) 0.8660 (b) −0.1010 (c) 0.58655. 42.33◦ 6. 15.25◦ 7. 73.78◦ 8. 7◦569. 31◦22 10. 41◦54 11. 29.05◦ 12. 20◦2113. 0.3586 14. 1.803Exercise 85 (page 189)1. (a) 12.22 (b) 5.619 (c) 14.87 (d) 8.349(e) 5.595 (f ) 5.2752. (a) AC = 5.831cm,∠A = 59.04◦,∠C = 30.96◦(b) DE = 6.928cm,∠D = 30◦,∠F = 60◦(c) ∠J = 62◦,HJ = 5.634cm,GH = 10.59cm(d) ∠L = 63◦,LM = 6.810 cm,KM = 13.37cm(e) ∠N = 26◦,ON = 9.125cm,NP = 8.201cm(f ) ∠S = 49◦,RS = 4.346cm,QS = 6.625cm3. 6.54m 4. 9.40 mmExercise 86 (page 192)1. 36.15m 2. 48m 3. 249.5m 4. 110.1m5. 53.0 m 6. 9.50 m 7. 107.8m8. 9.43m,10.56m 9. 60 mChapter 22Exercise 87 (page 198)1. (a) 42.78◦ and 137.22◦ (b) 188.53◦ and 351.47◦2. (a) 29.08◦ and 330.92◦ (b) 123.86◦ and 236.14◦3. (a) 44.21◦ and 224.21◦ (b) 113.12◦ and 293.12◦
• Answers to practice exercises 351Chapter 25Exercise 96 (page 221)1. p = 105◦,q = 35◦ 2. r = 142◦,s = 95◦3. t = 146◦Exercise 97 (page 225)1. (i) rhombus (a) 14cm2(b) 16cm (ii) parallelogram(a) 180 mm2 (b) 80 mm (iii) rectangle (a) 3600 mm2(b) 300 mm (iv) trapezium (a) 190 cm2 (b) 62.91cm2. 35.7cm2 3. (a) 80 m (b) 170 m 4. 27.2 cm25. 18cm 6. 1200 mm7. (a) 29cm2 (b) 650 mm2 8. 560 m29. 3.4cm 10. 6750 mm2 11. 43.30 cm212. 32Exercise 98 (page 226)1. 482 m22. (a) 50.27cm2 (b) 706.9mm2 (c) 3183mm23. 2513mm2 4. (a) 20.19mm (b) 63.41mm5. (a) 53.01cm2 (b) 129.9mm2 6. 5773mm27. 1.89m2Exercise 99 (page 228)1. 1932 mm2 2. 1624mm2 3. (a) 0.918ha (b) 456mExercise 100 (page 229)1. 80 ha 2. 80 m2 3. 3.14haChapter 26Exercise 101 (page 231)1. 45.24cm 2. 259.5mm 3. 2.629cm 4. 47.68cm5. 38.73cm 6. 12730 km 7. 97.13mmExercise 102 (page 232)1. (a)π6(b)5π12(c)5π42. (a) 0.838 (b) 1.481 (c) 4.0543. (a) 210◦ (b) 80◦ (c) 105◦4. (a) 0◦43 (b) 154◦8 (c) 414◦53Exercise 103 (page 234)1. 113cm2 2. 2376mm2 3. 1790 mm24. 802 mm2 5. 1709mm2 6. 1269m27. 1548m2 8. (a) 106.0 cm (b) 783.9cm29. 21.46m2 10. 17.80 cm,74.07cm211. (a) 59.86mm (b) 197.8mm 12. 26.2 cm13. 8.67cm,54.48cm 14. 82.5◦15. 74816. (a) 0.698rad (b) 804.2 m2 17. 10.47m218. (a) 396mm2 (b) 42.24% 19. 701.8mm20. 7.74mmExercise 104 (page 237)1. (a) 2 (b) (3,−4) 2. Centre at (3,−2), radius 43. Circle, centre (0, 1), radius 54. Circle, centre (0, 0), radius 6Chapter 27Exercise 105 (page 243)1. 1.2 m3 2. 5cm3 3. 8cm34. (a) 3840 mm3 (b) 1792 mm25. 972 litres 6. 15cm3,135g 7. 500 litres8. 1.44m3 9. (a) 35.3cm3 (b) 61.3cm210. (a) 2400 cm3 (b) 2460 cm2 11. 37.04m12. 1.63cm 13. 8796cm314. 4.709cm,153.9cm215. 2.99cm 16. 28060 cm3,1.099m217. 8.22 m by 8.22 m 18. 62.5min19. 4cm 20. 4.08m3Exercise 106 (page 246)1. 201.1cm3,159.0 cm2 2. 7.68cm3,25.81cm23. 113.1cm3,113.1cm2 4. 5.131cm 5. 3cm6. 2681mm3 7. (a) 268083mm3 or 268.083cm3(b) 20106mm2 or 201.06cm28. 8.53cm9. (a) 512 × 106 km2 (b) 1.09 × 1012 km3 10. 664Exercise 107 (page 251)1. 5890 mm2 or 58.90 cm22. (a) 56.55cm3 (b) 84.82 cm2 3. 13.57kg4. 29.32 cm3 5. 393.4m26. (i) (a) 670 cm3 (b) 523cm2 (ii) (a) 180 cm3(b) 154cm2 (iii) (a) 56.5cm3 (b) 84.8cm2(iv) (a) 10.4cm3(b) 32.0 cm2(v) (a) 96.0 cm3
• 352 Basic Engineering Mathematics(b) 146cm2 (vi) (a) 86.5cm3 (b) 142 cm2(vii) (a) 805cm3 (b) 539cm27. (a) 17.9cm (b) 38.0 cm 8. 125cm39. 10.3m3,25.5m2 10. 6560 litres11. 657.1cm3,1027cm2 12. 220.7cm313. (a) 1458litres (b) 9.77m2 (c) £140.45Exercise 108 (page 255)1. 147cm3,164cm2 2. 403cm3,337cm23. 10480 m3,1852 m2 4. 1707cm25. 10.69cm 6. 55910 cm3,6051cm27. 5.14mExercise 109 (page 256)1. 8 : 125 2. 137.2 gChapter 28Exercise 110 (page 259)1. 4.5 square units 2. 54.7 square units 3. 63.33m4. 4.70 ha 5. 143m2Exercise 111 (page 260)1. 42.59m3 2. 147m3 3. 20.42 m3Exercise 112 (page 263)1. (a) 2 A (b) 50 V (c) 2.5A 2. (a) 2.5V (b) 3A3. 0.093As, 3.1A 4. (a) 31.83V (b) 05. 49.13cm2,368.5kPaChapter 29Exercise 113 (page 266)1. A scalar quantity has magnitude only; a vectorquantity has both magnitude and direction.2. scalar 3. scalar 4. vector 5. scalar6. scalar 7. vector 8. scalar 9. vectorExercise 114 (page 273)1. 17.35N at 18.00◦ to the 12 N force2. 13m/s at 22.62◦ to the 12 m/s velocity3. 16.40 N at 37.57◦ to the 13N force4. 28.43N at 129.30◦ to the 18N force5. 32.31m at 21.80◦ to the 30 m displacement6. 14.72 N at −14.72◦to the 5N force7. 29.15m/s at 29.04◦ to the horizontal8. 9.28N at 16.70◦ 9. 6.89m/s at 159.56◦10. 15.62 N at 26.33◦ to the 10 N force11. 21.07knots, E 9.22◦SExercise 115 (page 276)1. (a) 54.0 N at 78.16◦(b) 45.64N at 4.66◦2. (a) 31.71m/s at 121.81◦ (b) 19.55m/s at 8.63◦Exercise 116 (page 277)1. 83.5km/h at 71.6◦to the vertical2. 4minutes 55seconds, 60◦3. 22.79km/h, E 9.78◦ NExercise 117 (page 277)1. i − j − 4k 2. 4i + j − 6k3. −i + 7j − k 4. 5i − 10k5. −3i + 27j − 8k 6. −5i + 10k7. i + 7.5j − 4k 8. 20.5j − 10k9. 3.6i + 4.4j − 6.9k 10. 2i + 40j − 43kChapter 30Exercise 118 (page 279)1. 4.5sin(A + 63.5◦)2. (a) 20.9sin(ωt + 0.62) volts(b) 12.5sin(ωt − 1.33) volts3. 13sin(ωt + 0.395)Exercise 119 (page 281)1. 4.5sin(A + 63.5◦)2. (a) 20.9sin(ωt + 0.62) volts(b) 12.5sin(ωt − 1.33) volts3. 13sin(ωt + 0.395)Exercise 120 (page 283)1. 4.5sin(A + 63.5◦)2. (a) 20.9sin(ωt + 0.62) volts(b) 12.5sin(ωt − 1.33) volts3. 13sin(ωt + 0.395) 4. 11.11sin(ωt + 0.324)5. 8.73sin(ωt − 0.173)
• Answers to practice exercises 353Exercise 121 (page 284)1. 11.11sin(ωt + 0.324) 2. 8.73sin(ωt − 0.173)3. i = 21.79sin(ωt − 0.639)4. v = 5.695sin(ωt + 0.670)5. x = 14.38sin(ωt + 1.444)6. (a) 305.3sin(314.2t − 0.233)V (b) 50 Hz7. (a) 10.21sin(628.3t + 0.818)V (b) 100 Hz(c) 10 ms8. (a) 79.83sin(300πt + 0.352)V (b) 150 Hz(c) 6.667msChapter 31Exercise 122 (page 288)1. (a) continuous (b) continuous (c) discrete(d) continuous2. (a) discrete (b) continuous (c) discrete (d) discreteExercise 123 (page 292)1. If one symbol is used to represent 10 vehicles, work-ing correct to the nearest 5 vehicles, gives 3.5, 4.5,6, 7, 5 and 4 symbols respectively.2. If one symbol represents 200 components, workingcorrect to the nearest 100 components gives: Mon 8,Tues 11, Wed 9, Thurs 12 and Fri 6.5.3. 6 equally spaced horizontal rectangles, whoselengths are proportional to 35, 44, 62, 68, 49 and41, respectively.4. 5 equally spaced horizontal rectangles, whoselengths are proportional to 1580, 2190, 1840, 2385and 1280 units, respectively.5. 6 equally spaced vertical rectangles, whose heightsare proportional to 35, 44, 62, 68, 49 and 41 units,respectively.6. 5 equally spaced vertical rectangles, whose heightsare proportional to 1580, 2190, 1840, 2385 and 1280units, respectively.7. Three rectangles of equal height, subdivided in thepercentages shown in the columns of the question.P increases by 20% at the expense of Q and R.8. Four rectangles of equal height, subdivided as fol-lows: week 1: 18%, 7%, 35%, 12%, 28%; week 2:20%,8%,32%,13%,27%; week 3: 22%,10%,29%,14%, 25%; week 4: 20%, 9%, 27%, 19%, 25%.Little change in centres A and B, a reduction ofabout 8% in C, an increase of about 7% in D and areduction of about 3% in E.9. A circle of any radius, subdivided into sectors hav-ing angles of 7.5◦,22.5◦,52.5◦,167.5◦ and 110◦,respectively.10. A circle of any radius, subdivided into sectors hav-ing angles of 107◦,156◦,29◦ and 68◦, respectively.11. (a) £495 (b) 88 12. (a) £16 450 (b) 138Exercise 124 (page 297)1. There is no unique solution, but one solution is:39.3–39.4 1; 39.5–39.6 5; 39.7–39.8 9;39.9–40.0 17; 40.1–40.2 15; 40.3–40.4 7;40.5–40.6 4; 40.7–40.8 2.2. Rectangles, touching one another, having mid-points of 39.35,39.55,39.75,39.95,... andheights of 1,5,9,17,...3. There is no unique solution, but one solution is:20.5–20.9 3; 21.0–21.4 10; 21.5–21.9 11;22.0–22.4 13; 22.5–22.9 9; 23.0–23.4 2.4. There is no unique solution, but one solution is:1–10 3; 11–19 7; 20–22 12; 23–25 11;26–28 10; 29–38 5; 39–48 2.5. 20.95 3; 21.45 13; 21.95 24; 22.45 37; 22.95 46;23.45 486. Rectangles, touching one another, having mid-points of 5.5, 15, 21, 24, 27, 33.5 and 43.5. Theheights of the rectangles (frequency per unit classrange) are 0.3, 0.78, 4, 4.67, 2.33, 0.5 and 0.2.7. (10.95 2), (11.45 9), (11.95 19), (12.45 31), (12.9542), (13.45, 50)8. A graph of cumulative frequency against upper classboundary having co-ordinates given in the answerto problem 7.9. (a) There is no uniquesolution,but one solutionis:2.05–2.09 3; 2.10–2.14 10; 2.15–2.19 11;2.20–2.24 13; 2.25–2.29 9; 2.30–2.34 2.(b) Rectangles, touching one another, having mid-pointsof 2.07,2.12,... and heights of 3,10,...(c) Using the frequency distribution given in thesolution to part (a) gives 2.0953; 2.14513;2.19524; 2.24537; 2.29546; 2.34548.(d) A graph of cumulative frequency against upperclass boundary having the co-ordinates givenin part (c).Chapter 32Exercise 125 (page 300)1. Mean 7.33, median 8, mode 82. Mean 27.25, median 27, mode 26
• 354 Basic Engineering Mathematics3. Mean 4.7225, median 4.72, mode 4.724. Mean 115.2, median 126.4, no modeExercise 126 (page 301)1. 23.85kg 2. 171.7cm3. Mean 89.5, median 89, mode 89.24. Mean 2.02158cm, median 2.02152 cm, mode2.02167cmExercise 127 (page 303)1. 4.60 2. 2.83μF3. Mean 34.53MPa, standard deviation 0.07474MPa4. 0.296kg 5. 9.394cm 6. 0.00544cmExercise 128 (page 304)1. 30, 25.5, 33.5 days 2. 27, 26, 33 faults3. Q1 = 164.5cm, Q2 = 172.5cm, Q3 = 179cm,7.25cm4. 37 and 38; 40 and 41 5. 40, 40, 41; 50, 51, 51Chapter 33Exercise 129 (page 308)1. (a)29or 0.2222 (b)79or 0.77782. (a)23139or 0.1655 (b)47139or 0.3381(c)69139or 0.49643. (a)16(b)16(c)1364.5365. (a)25(b)15(c)415(d)13156. (a)1250(b)1200(c)91000(d)150000Exercise 130 (page 311)1. (a) 0.6 (b) 0.2 (c) 0.15 2. (a) 0.64 (b) 0.323. 0.0768 4. (a) 0.4912 (b) 0.42115. (a) 89.38% (b) 10.25%6. (a) 0.0227 (b) 0.0234 (c) 0.0169Chapter 34Exercise 131 (page 313)1. 1, 5, 21, 9, 61 2. 0, 11, −10, 21 3. proofExercise 132 (page 314)1. 16, 8Exercise 133 (page 317)1. 28x3 2. 2 3. 2x − 14. 6x2 − 5 5. −1x26. 07. 1 +2x38. 15x4 − 8x3 + 15x2 + 2x9. −6x410. 4 − 8x 11.12√x12.32√t 13. −3x416. 1 +32√x17. 2x − 218. −10x3+72√x919. 6t − 12 20. 1 −4x221. (a) 6 (b)16(c) 3 (d) −116(e) −14(f ) −722. 12x − 3 (a) −15 (b) 2123. 6x2 + 6x − 4,32 24. −6x2 + 4,−9.5Exercise 134 (page 320)1. (a) 12cos3x (b) −12sin6x2. 6cos3θ + 10sin2θ 3. −0.707 4. −35. 270.2A/s 6. 1393.4V/s7. 12cos(4t + 0.12) + 6sin(3t − 0.72)Exercise 135 (page 321)1. (a) 15e3x (b) −47e2x2.5θ−4θ=1θ3. 164. 2.80 5. 664Exercise 136 (page 322)1. (a) −1 (b) 162. −4x3+2x+ 10sin5x − 12cos2x +6e3xExercise 137 (page 322)1. (a) 36x2 + 12x (b) 72x + 12 2. 8 +2x33. (a)45−12t5+6t3+14√t3(b) −4.954. −12sin2t − cost 5. −2θ2
• Answers to practice exercises 355Exercise 138 (page 324)1. −2542A/s 2. (a) 0.16cd/V (b) 312.5V3. (a) −1000 V/s (b) −367.9V/s4. −1.635Pa/mChapter 35Exercise 139 (page 328)1. (a) 4x + c (b)7x22+ c2. (a)54x4+ c (b)38t8+ c3. (a)215x3+ c (b)524x4+ c4. (a)25x5 −32x2 + c (b) 2t −34t4 + c5. (a)3x22− 5x + c (b) 4θ + 2θ2 +θ33+ c6. (a)52θ2 − 2θ + θ3 + c(b)34x4 −23x3 +32x2 − 2x + c7. (a) −43x+ c (b) −14x3+ c8. (a)45√x5 + c (b)194√x9 + c9. (a)10√t+ c (b)1575√x + c10. (a)32sin2x + c (b) −73cos3θ + c11. (a) −6cos12x + c (b) 18sin13x + c12. (a)38e2x + c (b)−215e5x+ c13. (a)23ln x + c (b)u22− lnu + c14. (a) 8√x + 8√x3 +185√x5 + c(b) −1t+ 4t +4t33+ cExercise 140 (page 330)1. (a) 1.5 (b) 0.5 2. (a) 105 (b) −0.53. (a) 6 (b) −1.333 4. (a) −0.75 (b) 0.8335. (a) 10.67 (b) 0.1667 6. (a) 0 (b) 47. (a) 1 (b) 4.248 8. (a) 0.2352 (b) 2.6389. (a) 19.09 (b) 2.457 10. (a) 0.2703 (b) 9.099Exercise 141 (page 334)1. proof 2. proof 3. 32 4. 29.33Nm5. 37.5 6. 7.5 7. 18. 1.67 9. 2.67 10. 140 m
• IndexAcute angle, 165Acute angled triangle, 171Adding waveforms, 278Addition in algebra, 62Addition law of probability, 307Addition of fractions, 10numbers, 1, 18two periodic functions, 278vectors, 267by calculation, 270Algebra, 61, 68Algebraic equation, 61, 73expression, 73Alternate angles, 165, 191Ambiguous case, 207Amplitude, 199Angle, 165Angle, lagging and leading, 200types and properties of, 165Angles of any magnitude, 196depression, 191elevation, 191Angular measurement, 165velocity, 202Annulus, 226Arbitrary constant of integration, 325Arc, 231Arc length, 233Area, 219Area of common shapes, 219, 221under a curve, 330Area of circle, 222, 233common shapes, 219irregular ﬁgures, 257sector, 222, 233similar shapes, 229triangles, 205Arithmetic, basic, 1Average, 299value of waveform, 260Axes, 130Bar charts, 289Base, 47Basic algebraic operations, 61BODMAS with algebra, 71fractions, 13numbers, 6Boyle’s law, 46Brackets, 6, 68Calculation of resultant phasors, 281,283Calculations, 22, 28Calculator, 22addition, subtraction, multiplicationand division, 22fractions, 26π and ex functions, 28, 118reciprocal and power functions, 24roots and ×10x functions, 25square and cube functions, 23trigonometric functions, 27Calculus, 313Cancelling, 10Cartesian axes, 131co-ordinates, 214Charles’s law, 42, 142Chord, 230Circle, 222, 230, 233equation of, 236properties of, 230Circumference, 230Classes, 293Class interval, 293limits, 295mid-point, 293, 295Coefﬁcient of proportionality, 45Combination of two periodic functions,278Common factors, 69logarithms, 111preﬁxes, 53shapes, 219Complementary angles, 165Completing the square, 105Cone, 245frustum of, 252Congruent triangles, 175Construction of triangles, 179Continuous data, 288Co-ordinates, 130, 131Corresponding angles, 165Cosine, 27, 183graph of, 195Cosine rule, 205, 281wave, 195Cross-multiplication, 75Cube root, 23Cubic equation, 161graphs, 161units, 240Cuboid, 240Cumulative frequency distribution,293, 297curve, 293Cycle, 199Cylinder, 241Deciles, 304Decimal fraction, 216places, 13, 18Decimals, 16addition and subtraction, 19multiplication and division, 19Deﬁnite integrals, 328Degrees, 27, 165, 166, 232Denominator, 9Dependent event, 307Depression, angle of, 191Derivatives, 315standard list, 321Derived units, 53Determination of law, 147involving logarithms, 150Diameter, 230Difference of two squares, 103Differential calculus, 313coefﬁcient, 315Differentiation, 313, 315from ﬁrst principles, 315of axn, 315of eax and ln ax, 320of sine and cosine functions, 318successive, 322Direct proportion, 40, 42Discrete data, 288standard deviation, 302Dividend, 63Division in algebra, 62Division of fractions, 12numbers, 3, 4, 19Divisor, 63Drawing vectors, 266
• Index 357Elevation, angle of, 191Engineering notation, 57Equation of a graph, 135Equations, 73circles, 236cubic, 161indicial, 115linear and quadratic, simultaneously,110quadratic, 102simple, 73simultaneous, 90Equilateral triangle, 171Evaluation of formulae, 28trigonometric ratios, 185Expectation, 306Exponential functions, 118graphs of, 120Expression, 73Exterior angle of triangle, 171Extrapolation, 133Factorial, 119Factorization, 69to solve quadratic equations, 102Factors, 5, 69False axes, 142Formula, 28quadratic, 106Formulae, evaluation of, 28list of, 336transposition of, 83Fractions, 9addition and subtraction, 10multiplication and division, 12on calculator, 26Frequency, 200, 289relative, 289Frequency distribution, 293, 296polygon, 293, 296Frustum, 252Full wave rectiﬁed waveform, 260Functional notation, 313, 315Gradient, 134of a curve, 314Graph drawing rules, 133Graphical solution of equations, 155cubic, 161linear and quadratic, simultaneously,110quadratic, 156simultaneous, 155Graphs, 130exponential functions, 120logarithmic functions, 116reducing non-linear to linear form,147sine and cosine, 199straight lines, 130, 132trigonometric functions, 195Grid, 130reference, 130Grouped data, 292mean, median and mode, 300standard deviation, 302Growth and decay, laws of, 125Half-wave rectiﬁed waveform, 260Hemisphere, 249Heptagon, 219Hexagon, 219Highest common factor (HCF), 5, 51,66, 69Histogram, 293–296, 300Hooke’s law, 42, 142Horizontal bar chart, 289component, 269, 283Hyperbolic logarithms, 111, 122Hypotenuse, 172i, j,k notation, 277Improper fraction, 9Indeﬁnite integrals, 328Independent event, 307Index, 47Indices, 47, 64laws of, 48, 64Indicial equations, 115Integers, 1Integral calculus, 313Integrals, 325deﬁnite, 328standard, 326Integration, 313, 325of axn, 325Intercept, y-axis, 135Interest, 37Interior angles, 165, 171Interpolation, 132Inverse proportion, 40, 45trigonometric function, 185Irregular areas, 257volumes, 259Isosceles triangle, 171Lagging angle, 200Laws of algebra, 61growth and decay, 125indices, 48, 64, 316logarithms, 113, 150precedence, 6, 71probability, 307Leading angle, 200Leibniz notation, 315Limiting value, 314Linear and quadratic equationssimultaneously, 110graphical solution, 160Logarithms, 111graphs involving, 116laws of, 113, 150Long division, 4Lower class boundary, 293Lowest common multiple (LCM), 5,10, 75Major arc, 231sector, 230segment, 231Maximum value, 156, 199Mean, 299, 300value of waveform, 260Measures of central tendency, 299Median, 299Member of set, 289Mid-ordinate rule, 257Minimum value, 156Minor arc, 231sector, 230segment, 230Mixed number, 9Mode, 299Multiple, 5Multiplication in algebra, 62law of probability, 307of fractions, 12of number, 3, 19Table, 3Napierian logarithms, 111, 122Natural logarithms, 111, 122Newton, 53Non right-angled triangles, 205Non-terminating decimals, 18Nose-to-tail method, 267Numerator, 9Obtuse angle, 165Obtuse-angled triangle, 171Octagon, 219Ogive, 293, 297Ohm’s law, 42Order of precedence, 6, 13, 71with fractions, 13with numbers, 6Origin, 131
• 358 IndexParabola, 156Parallel lines, 165Parallelogram, 219method, 267Peak value, 199Pentagon, 219Percentage component bar chart, 289error, 36relative frequency, 289Percentages, 33Percentile, 304Perfect square, 105Perimeter, 171Period, 199Periodic function, 200plotting, 238Periodic time, 200Phasor, 280Pictograms, 289Pie diagram, 289Planimeter, 257Plotting periodic functions, 238Polar co-ordinates, 214Pol/Rec function on calculator, 217Polygon, 210frequency, 293, 296Population, 289Power, 47series for ex, 119Powers and roots, 47Practical problemsquadratic equations, 108simple equations, 77simultaneous equations, 96straight line graphs, 141trigonometry, 209Precedence, 6, 71Preﬁxes, 53Presentation of grouped data, 292statistical data, 288Prism, 240, 242Probability, 306laws of, 307Production of sine and cosine waves,198Proper fraction, 9Properties of circles, 230triangles, 171Proportion, 40Pyramid, 244volumes and surface area of frustumof, 252Pythagoras’ theorem, 181Quadrant, 230Quadratic equations, 102by completing the square, 105factorization, 102formula, 106graphically, 156practical problems, 108Quadratic formula, 106graphs, 156Quadrilaterals, 219properties of, 219Quartiles, 303Radians, 27, 165, 166, 232Radius, 230Range, 295Ranking, 299Rates of change, 323Ratio and proportion, 40Ratios, 40Reciprocal, 24Rectangle, 219Rectangular axes, 131co-ordinates, 131prism, 240Reduction of non-linear laws to linearform, 147Reﬂex angle, 165Relative frequency, 289velocity, 276Resolution of vectors, 269Resultant phasors, by drawing, 280horizontal and vertical components,283plotting, 278sine and cosine rules, 281Rhombus, 219Right angle, 165Right angled triangle, 171solution of, 188Sample, 289Scalar quantities, 266Scalene triangle, 171Scales, 131Sector, 222, 230area of, 233Segment, 230Semicircle, 230Semi-interquartile range, 304Set, 289Short division, 4Signiﬁcant ﬁgures, 17, 18Similar shapes, 229, 256triangles, 176Simple equations, 73practical problems, 77Simpson’s rule, 258Simultaneous equations, 90graphical solution, 155in three unknowns, 99in two unknowns, 90practical problems, 96Sine, 27, 183graph of, 195Sine rule, 205, 281wave, 198, 260mean value, 260Sinusoidal form Asin(ωt ± α), 202SI units, 53Slope, 134Solution of linear and quadraticequations simultaneously, 110Solving right-angled triangles, 188simple equations, 73Space diagram, 276Sphere, 246Square, 23, 219numbers, 23root, 25, 48units, 219Standard deviation, 302discrete data, 302grouped data, 303Standard differentials, 321form, 56integrals, 326Statistical data, presentation of, 288terminology, 288Straight line, 165equation of, 135Straight line graphs, 132practical problems, 141Subject of formulae, 83Subtraction in algebra, 62Subtraction of fractions, 10numbers, 1, 18vectors, 274Successive differentiation, 322Supplementary angles, 165Surface areas of frusta of pyramids andcones, 252of solids, 247Symbols, 28Tally diagram, 293, 296Tangent, 27, 183, 230graph of, 195Terminating decimal, 17Theorem of Pythagoras, 181Transposition of formulae, 83Transversal, 165Trapezium, 220Trapezoidal rule, 257
• Index 359Triangle, 171, 219Triangles, area of, 205congruent, 175construction of, 179properties of, 171similar, 176Trigonometric functions, 27Trigonometric ratios, 183evaluation of, 185graphs of, 195waveforms, 195Trigonometry, 181practical situations, 209Turning points, 156Ungrouped data, 289Units, 53Upper class boundary, 293Use of calculator, 22Vector addition, 267subtraction, 274Vectors, 266addition of, 267by calculation, 267by horizontal and verticalcomponents, 269drawing, 266subtraction of, 274Velocity, relative, 276Vertical axis intercept, 133bar chart, 289component, 269, 283Vertically opposite angles, 165Vertices of triangle, 172Volumes of common solids, 240frusta of pyramids and cones, 252irregular solids, 259pyramids, 244similar shapes, 256Waveform addition, 278y-axis intercept, 135Young’s modulus of elasticity, 143