Basic engineering mathematics e5

Basic Engineering Mathematics
Fifth edition
John Bird, BSc(Hons), CMath, CEng, CSci, FIMA, FIET, MIEE, FIIE, FCollT
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Contents
Preface ix
Acknowledgements x
Instructor’s Manual xi
1 Basic arithmetic 1
1.1 Introduction 1
1.2 Revision of addition and subtraction 1
1.3 Revision of multiplication and division 3
1.4 Highest common factors and lowest
common multiples 5
1.5 Order of precedence and brackets 6
2 Fractions 9
2.1 Introduction 9
2.2 Adding and subtracting fractions 10
2.3 Multiplication and division of fractions 12
2.4 Order of precedence with fractions 13
Revision Test 1 15
3 Decimals 16
3.1 Introduction 16
3.2 Converting decimals to fractions and
vice-versa 16
3.3 Significant figures and decimal places 17
3.4 Adding and subtracting decimal numbers 18
3.5 Multiplying and dividing decimal numbers 19
4 Using a calculator 22
4.1 Introduction 22
4.2 Adding, subtracting, multiplying and
dividing 22
4.3 Further calculator functions 23
4.4 Evaluation of formulae 28
5 Percentages 33
5.1 Introduction 33
5.2 Percentage calculations 33
5.3 Further percentage calculations 35
5.4 More percentage calculations 36
Revision Test 2 39
6 Ratio and proportion 40
6.1 Introduction 40
6.2 Ratios 40
6.3 Direct proportion 42
6.4 Inverse proportion 45
7 Powers, roots and laws of indices 47
7.1 Introduction 47
7.2 Powers and roots 47
7.3 Laws of indices 48
8 Units, prefixes and engineering notation 53
8.1 Introduction 53
8.2 SI units 53
8.3 Common prefixes 53
8.4 Standard form 56
8.5 Engineering notation 57
Revision Test 3 60
9 Basic algebra 61
9.1 Introduction 61
9.2 Basic operations 61
9.3 Laws of indices 64
10 Further algebra 68
10.1 Introduction 68
10.2 Brackets 68
10.3 Factorization 69
10.4 Laws of precedence 71
11 Solving simple equations 73
11.2 Solving equations 73
11.3 Practical problems involving simple
equations 77
Revision Test 4 82
12 Transposing formulae 83
12.2 Transposing formulae 83
12.3 Further transposing of formulae 85
12.4 More difficult transposing of formulae 87
13 Solving simultaneous equations 90
13.2 Solving simultaneous equations in two
unknowns 90
13.3 Further solving of simultaneous equations 92

vi Contents
13.4 Solving more difﬁcult simultaneous
equations 94
13.5 Practical problems involving simultaneous
equations 96
13.6 Solving simultaneous equations in three
unknowns 99
Revision Test 5 101
14 Solving quadratic equations 102
14.2 Solution of quadratic equations by
factorization 102
‘completing the square’ 105
formula 106
14.5 Practical problems involving quadratic
equations 108
14.6 Solution of linear and quadratic equations
simultaneously 110
15 Logarithms 111
15.1 Introduction to logarithms 111
15.2 Laws of logarithms 113
15.3 Indicial equations 115
15.4 Graphs of logarithmic functions 116
16 Exponential functions 118
16.1 Introduction to exponential functions 118
16.2 The power series for ex 119
16.3 Graphs of exponential functions 120
16.4 Napierian logarithms 122
16.5 Laws of growth and decay 125
Revision Test 6 129
17 Straight line graphs 130
17.1 Introduction to graphs 130
17.2 Axes, scales and co-ordinates 130
17.3 Straight line graphs 132
17.4 Gradients, intercepts and equations
of graphs 134
17.5 Practical problems involving straight line
graphs 141
18 Graphs reducing non-linear laws to linear form 147
18.2 Determination of law 147
18.3 Revision of laws of logarithms 150
18.4 Determination of law involving logarithms 150
19 Graphical solution of equations 155
19.1 Graphical solution of simultaneous
equations 155
19.2 Graphical solution of quadratic equations 156
19.3 Graphical solution of linear and quadratic
equations simultaneously 160
19.4 Graphical solution of cubic equations 161
Revision Test 7 163
20 Angles and triangles 165
20.2 Angular measurement 165
20.3 Triangles 171
20.4 Congruent triangles 175
20.5 Similar triangles 176
20.6 Construction of triangles 179
21 Introduction to trigonometry 181
21.2 The theorem of Pythagoras 181
21.3 Sines, cosines and tangents 183
21.4 Evaluating trigonometric ratios of acute
angles 185
21.5 Solving right-angled triangles 188
21.6 Angles of elevation and depression 191
Revision Test 8 193
22 Trigonometric waveforms 195
22.1 Graphs of trigonometric functions 195
22.2 Angles of any magnitude 196
22.3 The production of sine and cosine waves 198
22.4 Terminology involved with sine and
cosine waves 199
22.5 Sinusoidal form: Asin(ωt ± α) 202
23 Non-right-angled triangles and some practical
applications 205
23.1 The sine and cosine rules 205
23.2 Area of any triangle 205
23.3 Worked problems on the solution of
triangles and their areas 206
23.4 Further worked problems on the solution
of triangles and their areas 207
23.5 Practical situations involving trigonometry 209
23.6 Further practical situations involving
trigonometry 211
24 Cartesian and polar co-ordinates 214
24.2 Changing from Cartesian to polar
co-ordinates 214
24.3 Changing from polar to Cartesian
co-ordinates 216
24.4 Use of Pol/Rec functions on calculators 217

Contents vii
Revision Test 9 218
25 Areas of common shapes 219
25.2 Common shapes 219
25.3 Areas of common shapes 221
25.4 Areas of similar shapes 229
26 The circle 230
26.2 Properties of circles 230
26.3 Radians and degrees 232
26.4 Arc length and area of circles and sectors 233
26.5 The equation of a circle 236
Revision Test 10 238
27 Volumes of common solids 240
27.2 Volumes and surface areas of common
shapes 240
27.3 Summary of volumes and surface areas of
common solids 247
27.4 More complex volumes and surface areas 247
27.5 Volumes and surface areas of frusta of
pyramids and cones 252
27.6 Volumes of similar shapes 256
28 Irregular areas and volumes, and mean values 257
28.1 Areas of irregular figures 257
28.2 Volumes of irregular solids 259
28.3 Mean or average values of waveforms 260
29 Vectors 266
29.2 Scalars and vectors 266
29.3 Drawing a vector 266
29.4 Addition of vectors by drawing 267
29.5 Resolving vectors into horizontal and
vertical components 269
29.6 Addition of vectors by calculation 270
29.7 Vector subtraction 274
29.8 Relative velocity 276
29.9 i, j and k notation 277
30 Methods of adding alternating waveforms 278
30.1 Combining two periodic functions 278
30.2 Plotting periodic functions 278
30.3 Determining resultant phasors by drawing 280
30.4 Determining resultant phasors by the sine
and cosine rules 281
30.5 Determining resultant phasors by
horizontal and vertical components 283
31 Presentation of statistical data 288
31.1 Some statistical terminology 288
31.2 Presentation of ungrouped data 289
31.3 Presentation of grouped data 292
32 Mean, median, mode and standard deviation 299
32.1 Measures of central tendency 299
32.2 Mean, median and mode for discrete data 299
32.3 Mean, median and mode for grouped data 300
32.4 Standard deviation 302
32.5 Quartiles, deciles and percentiles 303
33 Probability 306
33.1 Introduction to probability 306
33.2 Laws of probability 307
34 Introduction to differentiation 313
34.1 Introduction to calculus 313
34.2 Functional notation 313
34.3 The gradient of a curve 314
34.4 Differentiation from first principles 315
34.5 Differentiation of y = axn by the
general rule 315
34.6 Differentiation of sine and cosine functions 318
34.7 Differentiation of eax and lnax 320
34.8 Summary of standard derivatives 321
34.9 Successive differentiation 322
34.10 Rates of change 323
35 Introduction to integration 325
35.1 The process of integration 325
35.2 The general solution of integrals of the
form axn 325
35.3 Standard integrals 326
35.4 Definite integrals 328
35.5 The area under a curve 330
List of formulae 336
Answers to practice exercises 340
Index 356

viii Contents
Website Chapters
(Goto:http://www.booksite.elsevier.com/newnes/bird)
Preface iv
36 Number sequences 1
36.1 Simple sequences 1
36.2 The n’th term of a series 1
36.3 Arithmetic progressions 2
36.4 Geometric progressions 5
37 Binary, octal and hexadecimal 9
37.1 Introduction 9
37.2 Binary numbers 9
37.3 Octal numbers 12
37.4 Hexadecimal numbers 15
38 Inequalities 19
38.1 Introduction to inequalities 19
38.2 Simple inequalities 19
38.3 Inequalities involving a modulus 20
38.4 Inequalities involving quotients 21
38.5 Inequalities involving square functions 22
38.6 Quadratic inequalities 23
39 Graphs with logarithmic scales 25
39.1 Logarithmic scales and logarithmic
graph paper 25
39.2 Graphs of the form y = axn 25
39.3 Graphs of the form y = abx 28
39.4 Graphs of the form y = aekx 29
Revision Test 15 32

Preface
Basic Engineering Mathematics 5th Edition intro-
duces and then consolidates basic mathematical princi-
ples and promotes awareness of mathematical concepts
for students needing a broad base for further vocational
studies.
In thisfifth edition,newmaterial hasbeen added to many
of the chapters, particularly some of the earlier chap-
ters, together with extra practical problems interspersed
throughout the text. The extent of this fifth edition
is such that four chapters from the previous edition
have been removed and placed on the easily available
website http://www.booksite.elsevier.com/newnes/bird.
The chapters removed to the website are ‘Number
sequences’, ‘Binary, octal and hexadecimal’, ‘Inequali-
ties’ and ‘Graphs with logarithmic scales’.
The text is relevant to:
• ‘Mathematics for Engineering Technicians’ for
BTEC First NQF Level 2 – Chapters 1–12, 16–18,
20, 21, 23 and 25–27 are needed for this module.
• The mandatory ‘Mathematics for Technicians’ for
BTEC National Certificate and National Diploma in
Engineering, NQF Level 3 – Chapters 7–10, 14–17,
19, 20–23, 25–27, 31, 32, 34 and 35 are needed
and, in addition, Chapters 1–6, 11 and 12 are helpful
revision for this module.
• Basic mathematics for a wide range of introduc-
tory/access/foundation mathematics courses.
• GCSE revision and for similar mathematics courses
in English-speaking countries worldwide.
Basic Engineering Mathematics 5th Edition provides a
lead into Engineering Mathematics 6th Edition.
Each topic considered in the text is presented in a
way that assumes in the reader little previous know-
ledge of that topic. Each chapter begins with a brief
outline of essential theory, definitions, formulae, laws
and procedures; however, these are kept to a minimum
as problem solving is extensively used to establish and
exemplify the theory. It is intended that readers will gain
real understanding through seeing problems solved and
then solving similar problems themselves.
This textbook contains some 750 worked problems,
followed by over 1550 further problems (all with
answers at the end of the book) contained within some
161 Practice Exercises; each Practice Exercise fol-
lows on directly from the relevant section of work. In
addition, 376 line diagrams enhance understanding of
the theory. Where at all possible, the problems mirror
potential practical situations found in engineering and
science.
Placed at regular intervals throughout the text are
14 Revision Tests (plus another for the website
chapters) to check understanding. For example, Revi-
sion Test 1 covers material contained in Chapters 1
and 2, Revision Test 2 covers the material contained
in Chapters 3–5, and so on. These Revision Tests do
not have answers given since it is envisaged that lec-
turers/instructors could set the tests for students to
attempt as part of their course structure. Lecturers/in-
structors may obtain a complimentary set of solu-
tions of the Revision Tests in an Instructor’s Manual,
available from the publishers via the internet – see
http://www.booksite.elsevier.com/newnes/bird.
At the end of the book a list of relevant formulae con-
tained within the text is included for convenience of
reference.
The principle of learning by example is at the heart of
Basic Engineering Mathematics 5th Edition.
JOHN BIRD
Royal Naval School of Marine Engineering
HMS Sultan, formerly University of Portsmouth
and Highbury College, Portsmouth

Acknowledgements
The publisher wishes to thank CASIO Electronic Co.
Ltd, London for permission to reproduce the image of
the Casio fx-83ES calculator on page 23.
The publishers also wish to thank the AutomobileAsso-
ciation for permission to reproduce a map of Portsmouth
on page 131.

Instructor’s Manual
Full worked solutions and mark scheme for all the
Assignments are contained in this Manual which is
available to lecturers only.
To download the Instructor’s Manual visit http://
www.booksite.elsevier.com/newnes/bird

This page intentionally left blank

Chapter 1
Basic arithmetic
1.1 Introduction
Whole numbers are called integers. +3,+5 and +72
are examples of positive integers; −13,−6 and −51
are examples of negative integers. Between positive
and negative integers is the number 0 which is neither
positive nor negative.
The four basic arithmetic operators are add (+), subtract
(−), multiply (×) and divide (÷).
It is assumed that adding, subtracting, multiplying and
dividing reasonably small numbers can be achieved
without a calculator. However, if revision of this area
is needed then some worked problems are included in
the following sections.
When unlike signs occur together in a calculation, the
overall sign is negative. For example,
3 + (−4) = 3 + −4 = 3 − 4 = −1
and
(+5) × (−2) = −10
Like signs together give an overall positive sign. For
example,
3 − (−4) = 3 − −4 = 3 + 4 = 7
and
(−6) × (−4) = +24
1.2 Revision of addition and
subtraction
You can probably already add two or more numbers
together and subtract one number from another. How-
ever, if you need a revision then the following worked
problems should be helpful.
Problem 1. Determine 735 + 167
H T U
7 3 5
+ 1 6 7
9 0 2
1 1
(i) 5 + 7 = 12. Place 2 in units (U) column. Carry 1
in the tens (T) column.
(ii) 3 + 6 + 1 (carried) = 10. Place the 0 in the tens
column. 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 = 902
Problem 2. Determine 632 − 369
H T U
6 3 2
− 3 6 9
2 6 3
(i) 2 − 9 is not possible; therefore ‘borrow’ 1 from
the tens column (leaving 2 in the tens column). In
the units column, this gives us 12 − 9 = 3.
(ii) Place 3 in the units column.
(iii) 2 − 6 is not possible; therefore ‘borrow’ 1 from
the hundreds column (leaving 5 in the hun-
dreds column). In the tens column, this gives us
12 − 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 = 263
Problem 3. Add 27,−74,81 and −19
This problem is written as 27 − 74 + 81 − 19.
Adding the positive integers: 27
81
Sum of positive integers is 108
Adding the negative integers: 74
19
Sum of negative integers is 93
Taking the sum of the negative integers
from the sum of the positive integers gives 108
−93
15
Thus, 27 − 74 + 81 − 19 = 15
Problem 4. Subtract −74 from 377
This problem is written as 377 − −74. Like signs
together give an overall positive sign, hence
377 − −74 = 377 + 74 3 7 7
+ 7 4
4 5 1
Thus, 377 − −74 = 451
Problem 5. Subtract 243 from 126
The problem is 126 − 243. When the second number is
larger than the ﬁrst, take the smaller number from the
larger and make the result negative. Thus,
126 − 243 = −(243 − 126) 2 4 3
− 1 2 6
1 1 7
Thus, 126 − 243 = −117
Problem 6. Subtract 318 from −269
The problem is −269 − 318. The sum of the negative
integers is
2 6 9
+ 3 1 8
5 8 7
Thus, −269 − 318 = −587
Now try the following Practice Exercise
PracticeExercise 1 Further problems on
addition and subtraction (answers on
page 340)
In Problems 1 to 15, determine the values of the
expressions given, without using a calculator.
1. 67kg − 82 kg + 34kg
2. 73m − 57m
3. 851mm − 372mm
4. 124− 273 + 481 − 398
5. £927 − £114+ £182 − £183 − £247
6. 647 − 872
7. 2417 − 487 + 2424− 1778 − 4712
8. −38419 − 2177 + 2440− 799 + 2834
9. £2715 − £18250+ £11471 − £1509 +
£113274
10. 47 + (−74) − (−23)
11. 813 − (−674)
12. 3151 − (−2763)
13. 4872 g− 4683g
14. −23148 − 47724
15. $53774− $38441
16. Holes are drilled 35.7mm apart in a metal
plate. If a row of 26 holes is drilled, deter-
mine the distance, in centimetres, between
the centres of the ﬁrst and last holes.
17. Calculate the diameter d and dimensions A
and B for the template shown in Figure 1.1.
All dimensions are in millimetres.

Basic arithmetic 3
12
60
50 38
120
110
B
A
d
Figure 1.1
1.3 Revision of multiplication and
division
You can probably already multiply two numbers
together and divide one number by another. However, if
you need a revision then the followingworked problems
should be helpful.
Problem 7. Determine 86 × 7
H T U
8 6
× 7
6 0 2
4
(i) 7 × 6 = 42. Place the 2 in the units (U) column
and ‘carry’ the 4 into the tens (T) column.
(ii) 7 × 8 = 56;56 + 4 (carried) = 60. Place the 0 in
the tens column and the 6 in the hundreds (H)
column.
Hence, 86 × 7 = 602
A good grasp of multiplication tables is needed when
multiplying such numbers; a reminder of the multipli-
cation table up to 12 × 12 is shown below. Conﬁdence
with handling numbers will be greatly improved if this
table is memorized.
Problem 8. Determine 764× 38
7 6 4
× 3 8
6 1 1 2
2 2 9 2 0
2 9 0 3 2
Multiplication table
× 2 3 4 5 6 7 8 9 10 11 12
2 4 6 8 10 12 14 16 18 20 22 24
3 6 9 12 15 18 21 24 27 30 33 36
4 8 12 16 20 24 28 32 36 40 44 48
5 10 15 20 25 30 35 40 45 50 55 60
6 12 18 24 30 36 42 48 54 60 66 72
7 14 21 28 35 42 49 56 63 70 77 84
8 16 24 32 40 48 56 64 72 80 88 96
9 18 27 36 45 54 63 72 81 90 99 108
10 20 30 40 50 60 70 80 90 100 110 120
11 22 33 44 55 66 77 88 99 110 121 132
12 24 36 48 60 72 84 96 108 120 132 144

(i) 8 × 4 = 32. Place the 2 in the units column and
carry 3 into the tens column.
(ii) 8 × 6 = 48;48 + 3 (carried) = 51. Place the 1 in
the tens column and carry the 5 into the hundreds
column.
(iii) 8 × 7 = 56;56 + 5 (carried) = 61. Place 1 in the
hundredscolumn 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 and
carry 1 into the hundreds column.
(vi) 3 × 6 = 18;18 + 1 (carried) = 19. Place the 9 in
the hundreds column and carry the 1 into the
thousands column.
(vii) 3 × 7 = 21;21 + 1 (carried) = 22. Place 2 in the
thousands column and 2 in the ten thousands
column.
(viii) 6112+ 22920 = 29032
Hence, 764 × 38 = 29032
Again, knowing multiplication tables is ratherimportant
when multiplying such numbers.
It is appreciated, of course, that such a multiplication
can,and probably will,beperformed using a calculator.
However, there are times when a calculator may not be
available and it is then useful to be able to calculate the
‘long way’.
Problem 9. Multiply 178 by −46
When the numbers have different signs, the result will
be negative. (With this in mind, the problem can now
be solved by multiplying 178 by 46). Following the
procedure of Problem 8 gives
1 7 8
× 4 6
1 0 6 8
7 1 2 0
8 1 8 8
Thus, 178 × 46 = 8188 and 178 × (−46) = −8188
Problem 10. Determine 1834 ÷ 7
262
7 1834
(i) 7 into 18 goes 2, remainder 4. Place the 2 above
the 8 of 1834 and carry the 4 remainder to the
next digit on the right, making it 43.
(ii) 7 into 43 goes 6, remainder 1. Place the 6 above
the 3 of 1834 and carry the 1 remainder to the
next digit on the right, making it 14.
(iii) 7 into 14 goes 2, remainder 0. Place 2 above the
4 of 1834.
Hence, 1834÷ 7 = 1834/7 =
1834
7
= 262.
The method shown is called short division.
Problem 11. Determine 5796 ÷ 12
483
12 5796
48
99
96
36
36
00
(i) 12 into 5 won’t go. 12 into 57 goes 4; place 4
above 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 =
5796
12
= 483.
The method shown is called long division.

Basic arithmetic 5
multiplication and division (answers on
page 340)
Determine the values of the expressions given in
problems 1 to 9, without using a calculator.
1. (a) 78 × 6 (b) 124 × 7
2. (a) £261 × 7 (b) £462 × 9
3. (a) 783kg × 11 (b) 73kg × 8
4. (a) 27mm × 13 (b) 77mm × 12
5. (a) 448 × 23 (b) 143 × (−31)
6. (a) 288m ÷ 6 (b) 979m ÷ 11
7. (a)
1813
7
(b)
896
16
8. (a)
21424
13
(b) 15900 ÷ 15
9. (a)
88737
11
(b) 46858 ÷ 14
10. A screw has a mass of 15grams. Calculate,
in kilograms, the mass of 1200 such screws
(1kg = 1000g).
1.4 Highest common factors and
lowest common multiples
When two or more numbers are multiplied together, the
individualnumbers are called factors. Thus, a factoris a
number which divides into another number exactly. The
highest common factor (HCF) is the largest number
which divides into two or more numbers exactly.
For example, consider the numbers 12 and 15.
The factors of 12 are 1, 2, 3, 4, 6 and 12 (i.e. all the
numbers that divide into 12).
The factors of 15 are 1, 3, 5 and 15 (i.e. all the numbers
that divide into 15).
1 and 3 are the only common factors; i.e., numbers
which are factors of both 12 and 15.
Hence, the HCF of 12 and 15 is 3 since 3 is the highest
number which divides into both 12 and 15.
A multiple is a number which contains another number
an exact number of times. The smallest number which
is exactly divisible by each of two or more numbers is
called the lowest common multiple (LCM).
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 and
15) and there are no lower common multiples.
Hence, the LCM of 12 and 15 is 60 since 60 is the
lowest 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 numbers
12, 30 and 42
Probably the simplest way of determining an HCF is to
express each number in terms of its lowest factors. This
is achieved by repeatedly dividing by the prime numbers
2, 3, 5, 7, 11, 13, … (where possible) in turn. Thus,
12 = 2 × 2 × 3
30 = 2 × 3 × 5
42 = 2 × 3 × 7
The factors which are common to each of the numbers
are 2 in column 1 and 3 in column 3, shown by the
broken lines. Hence, the HCF is 2 × 3; i.e., 6. That is,
6 is the largest number which will divide into 12, 30
and 42.
Problem 13. Determine the HCF of the numbers
30, 105, 210 and 1155
Using the method shown in Problem 12:
30 = 2 × 3 × 5
105 = 3 × 5 × 7
210 = 2 × 3 × 5 × 7
1155 = 3 × 5 × 7 × 11
The factors which are common to each of the numbers
are 3 in column 2 and 5 in column 3. Hence, the HCF
is 3 × 5 = 15.
Problem 14. Determine the LCM of the numbers
12, 42 and 90

The LCM is obtained by finding the lowest factors of
each of the numbers, as shown in Problems 12 and 13
above, and then selecting the largest group of any of the
factors present. Thus,
12 = 2 × 2 × 3
42 = 2 × 3 × 7
90 = 2 × 3 × 3 × 5
The largest group of any of the factors present is shown
by the broken lines and are 2 × 2 in 12, 3 × 3 in 90, 5 in
90 and 7 in 42.
Hence, the LCM is 2 × 2 × 3 × 3 × 5 × 7 = 1260 and
is the smallest number which 12, 42 and 90 will all
divide into exactly.
Problem 15. Determine the LCM of the numbers
150, 210, 735 and 1365
Using the method shown in Problem 14 above:
150 = 2 × 3 × 5 × 5
210 = 2 × 3 × 5 × 7
735 = 3 × 5 × 7 × 7
1365 = 3 × 5 × 7 × 13
Hence, the LCM is 2 × 3 × 5 × 5 × 7 × 7 × 13 =
95550.
highest common factors and lowest common
multiples (answers on page 340)
Find (a) the HCF and (b) the LCM of the following
groups of numbers.
1. 8, 12 2. 60, 72
3. 50, 70 4. 270, 900
5. 6, 10, 14 6. 12, 30, 45
7. 10, 15, 70, 105 8. 90, 105, 300
9. 196, 210, 462, 910 10. 196, 350, 770
1.5 Order of precedence and brackets
1.5.1 Order of precedence
Sometimes addition, subtraction, multiplication, divi-
sion, powers and brackets may all be involved in a
calculation. For example,
5 − 3 × 4 + 24 ÷ (3 + 5) − 32
This is an extreme example but will demonstrate the
order that is necessary when evaluating.
When we read, we read from left to right. However,
with mathematics there is a definite order of precedence
which we need to adhere to. The order is as follows:
Brackets
Order (or pOwer)
Division
Multiplication
Addition
Subtraction
Notice that the first letters of each word spell BOD-
MAS, a handy aide-mémoire. Order means pOwer. For
example, 42 = 4 × 4 = 16.
5 − 3 × 4 + 24 ÷ (3 + 5) − 32 is evaluated as
follows:
5 − 3 × 4 + 24 ÷ (3 + 5) − 32
= 5 − 3 × 4 + 24 ÷ 8 − 32
(Bracket is removed and
3 + 5 replaced with 8)
= 5 − 3 × 4 + 24 ÷ 8 − 9 (Order means pOwer; in
this 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 performed
before addition. What is important is that the pro-
cessofmultiplicationanddivisionmustbecompleted
before addition and subtraction.
1.5.2 Brackets and operators
The basic laws governing the use of brackets and
operators are shown by the following examples.

Basic arithmetic 7
(a) 2 + 3 = 3 + 2; i.e., the order of numbers when
adding does not matter.
(b) 2 × 3 = 3 × 2; i.e., the order of numbers when
multiplying does not matter.
(c) 2 + (3 + 4) = (2 + 3) + 4; i.e., the use of brackets
when adding does not affect the result.
(d) 2 × (3 × 4) = (2 × 3) × 4; i.e., the use of brackets
when multiplying does not affect the result.
(e) 2 × (3 + 4) = 2(3 + 4) = 2 × 3 + 2 × 4; i.e., a
number placed outside of a bracket indicates
that the whole contents of the bracket must be
multiplied 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 outer
brackets, the inner brackets are removed
ﬁrst.
Here are some further problems in which BODMAS
needs to be used.
Problem 16. Find the value of 6 + 4 ÷ (5 − 3)
The order of precedence of operations is remembered
by the word BODMAS. Thus,
6 + 4 ÷ (5 − 3) = 6 + 4 ÷ 2 (Brackets)
= 6 + 2 (Division)
= 8 (Addition)
Problem 17. Determine the value of
13 − 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. Evaluate
16 ÷(2 + 6) + 18[3 + (4 × 6) − 21]
16 ÷ (2 + 6) + 18[3 + (4 × 6) − 21]
= 16 ÷ (2 + 6) + 18[3 + 24 − 21] (B: inner bracket
is 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 all
that is inside the brackets. In this case,
18[3 + 24 − 21] = 18[6], which means 18 × 6 = 108
Problem 19. Find the value of
23 − 4(2 × 7) +
(144 ÷ 4)
(14 − 8)
23 − 4(2 × 7) +
(144 ÷ 4)
(14 − 8)
= 23 − 4 × 14+
36
6
(B)
= 23 − 4 × 14+ 6 (D)
= 23 − 56 + 6 (M)
= 29 − 56 (A)
= −27 (S)
3 + 52 − 32 + 23
1 + (4 × 6) ÷ (3 × 4)
+
15 ÷ 3 + 2 × 7 − 1
3 ×
√
4+ 8 − 32 + 1
3 + 52 − 32 + 23
1 + (4 × 6) ÷ (3 × 4)
+
15 ÷ 3 + 2 × 7 − 1
3 ×
√
4 + 8 − 32 + 1
=
3 + 4 + 8
1 + 24 ÷ 12
+
15 ÷ 3 + 2 × 7 − 1
3 × 2 + 8 − 9 + 1
=
3 + 4 + 8
1 + 2
+
5 + 2 × 7 − 1
3 × 2 + 8 − 9 + 1
=
15
3
+
5 + 14 − 1
6 + 8 − 9 + 1
= 5 +
18
6
= 5 + 3 = 8

order of precedenceand brackets (answers
on page 340)
Evaluate the following expressions.
1. 14+ 3 × 15
2. 17 − 12 ÷ 4
3. 86 + 24 ÷ (14 − 2)
4. 7(23 − 18) ÷ (12 − 5)
5. 63 − 8(14 ÷ 2) + 26
6.
40
5
− 42 ÷ 6 + (3 × 7)
7.
(50 − 14)
3
+ 7(16 − 7) − 7
8.
(7 − 3)(1 − 6)
4(11 − 6) ÷ (3 − 8)
9.
(3 + 9 × 6) ÷ 3 − 2 ÷ 2
3 × 6 + (4 − 9) − 32 + 5
10.
4 × 32 + 24 ÷ 5 + 9 × 3
2 × 32 − 15 ÷ 3
+
2 + 27 ÷ 3 + 12 ÷ 2 − 32
5 + (13 − 2 × 5) − 4
11.
1 +
√
25 + 3 × 2 − 8 ÷ 2
3 × 4− 32 + 42 + 1
−
(4 × 2 + 7 × 2) ÷ 11
√
9+ 12 ÷ 2 − 23

Chapter 2
Fractions
2.1 Introduction
A mark of 9 out of 14 in an examination may be writ-
ten as
9
14
or 9/14.
9
14
is an example of a fraction. The
number above the line, i.e. 9, is called the numera-
tor. The number below the line, i.e. 14, is called the
denominator.
When the value of the numerator is less than the
value of the denominator, the fraction is called a
proper fraction.
9
14
is an example of a proper
fraction.
When thevalueofthenumeratorisgreaterthan thevalue
of the denominator, the fraction is called an improper
fraction.
5
2
is an example of an improper fraction.
A mixed number is a combination of a whole number
and a fraction. 2
1
2
is an example of a mixed number. In
fact,
5
2
= 2
1
2
.
There are a number of everyday examples in which
fractions are readily referred to. For example, three
people equally sharing a bar of chocolate would have
1
3
each. A supermarket advertises
1
5
off a six-pack of
beer; if the beer normally costs £2 then it will now
cost £1.60.
3
4
of the employees of a company are
women; if the company has 48 employees, then 36 are
women.
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 a
calculator.
Problem 1. Change the following improper
fractions into mixed numbers:
(a)
9
2
(b)
13
4
(c)
28
5
(a)
9
2
means 9 halves and
9
2
= 9 ÷ 2, and 9 ÷ 2 = 4
and 1 half, i.e.
9
2
= 4
1
2
(b)
13
4
means 13 quarters and
13
4
= 13 ÷ 4, and
13 ÷ 4 = 3 and 1 quarter, i.e.
13
4
= 3
1
4
(c)
28
5
means 28 ﬁfths and
28
5
= 28 ÷ 5, and 28 ÷ 5 =
5 and 3 ﬁfths, i.e.
28
5
= 5
3
5
Problem 2. Change the following mixed numbers
into improper fractions:
(a) 5
3
4
(b) 1
7
9
(c) 2
3
7
(a) 5
3
4
means 5 +
3
4
. 5 contains 5 × 4 = 20 quarters.
Thus, 5
3
4
contains 20 + 3 = 23 quarters, i.e.
5
3
4
=
23
4
DOI: 10.1016/B978-1-85617-697-2.00002-8

The quick way to change 5
3
4
into an improper
fraction is
4 × 5 + 3
4
=
23
4
.
(b) 1
7
9
=
9 × 1 + 7
9
=
16
9
.
(c) 2
3
7
=
7 × 2 + 3
7
=
17
7
.
Problem 3. In a school there are 180 students of
which 72 are girls. Express this as a fraction in its
simplest form
The fraction of girls is
72
180
.
Dividing both the numerator and denominator by the
lowest prime number, i.e. 2, gives
72
180
=
36
90
Dividing both the numerator and denominator again by
2 gives
72
180
=
36
90
=
18
45
2 will not divide into both 18 and 45, so dividing both the
numerator and denominator by the next prime number,
i.e. 3, gives
72
180
=
36
90
=
18
45
=
6
15
Dividing both the numerator and denominator again by
3 gives
72
180
=
36
90
=
18
45
=
6
15
=
2
5
So
72
180
=
2
5
in its simplest form.
Thus,
2
5
of the students are girls.
2.2 Adding and subtracting fractions
When the denominators of two (or more) fractions to
be added are the same, the fractions can be added ‘on
sight’.
For example,
2
9
+
5
9
=
7
9
and
3
8
+
1
8
=
4
8
.
In the latter example, dividing both the 4 and the 8 by
4 gives
4
8
=
1
2
, which is the simpliﬁed answer. This is
called cancelling.
Additionand subtraction of fractions is demonstrated
in the following worked examples.
Problem 4. Simplify
1
3
+
1
2
(i) Make the denominators the same for each frac-
tion. The lowest number that both denominators
divideinto is called the lowest commonmultiple
or 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 of
1
3
by 2 gives
1
3
=
2
6
=
(iii) 2 dividesinto 6, 3 times. Multiplyingboth numer-
ator and denominator of
1
2
by 3 gives
1
2
=
3
6
=
(iv) Hence,
1
3
+
1
2
=
2
6
+
3
6
=
5
6
+ =
Problem 5. Simplify
3
4
−
7
16
(i) Make the denominators the same for each frac-
tion. The lowest common multiple (LCM) of 4
and 16 is 16.
(ii) 4 divides into 16, 4 times. Multiplying both
numerator and denominator of
3
4
by 4 gives
3
4
=
12
16
=
(iii)
7
16
already has a denominator of 16.

Fractions 11
(iv) Hence,
3
4
−
7
16
=
12
16
−
7
16
=
5
16
− =
Problem 6. Simplify 4
2
3
− 1
1
6
4
2
3
− 1
1
6
is the same as 4
2
3
− 1
1
6
which is the
same as 4 +
2
3
− 1 +
1
6
which is the same as
4 +
2
3
− 1 −
1
6
which is the same as 3 +
2
3
−
1
6
which
is the same as 3 +
4
6
−
1
6
= 3 +
3
6
= 3 +
1
2
Thus, 4
2
3
− 1
1
6
= 3
1
2
Problem 7. Evaluate 7
1
8
− 5
3
7
7
1
8
− 5
3
7
= 7 +
1
8
− 5 +
3
7
= 7 +
1
8
− 5 −
3
7
= 2 +
1
8
−
3
7
= 2 +
7 × 1 − 8 × 3
56
= 2 +
7 − 24
56
= 2 +
−17
56
= 2 −
17
56
=
112
56
−
17
56
=
112 − 17
56
=
95
56
= 1
39
56
Problem 8. Determine the value of
4
5
8
− 3
1
4
+ 1
2
5
4
5
8
− 3
1
4
+ 1
2
5
= (4 − 3 + 1) +
5
8
−
1
4
+
2
5
= 2 +
5 × 5 − 10× 1 + 8 × 2
40
= 2 +
25 − 10 + 16
40
= 2 +
31
40
= 2
31
40
PracticeExercise 5 Introduction to
fractions (answers on page 340)
1. Change the improper fraction
15
7
into a
mixed number.
2. Change the improper fraction
37
5
into a
mixed number.
3. Change the mixed number 2
4
9
into an
improper fraction.
4. Change the mixed number 8
7
8
into an
improper fraction.
5. A box contains 165 paper clips. 60 clips
are removed from the box. Express this as
a fraction in its simplest form.
6. Order the following fractions from the small-
est to the largest.
4
9
,
5
8
,
3
7
,
1
2
,
3
5
7. A training college has 375 students of which
120 are girls. Express this as a fraction in its
simplest form.
Evaluate, in fraction form, the expressions given in
Problems 8 to 20.
8.
1
3
+
2
5
9.
5
6
−
4
15
10.
1
2
+
2
5
11.
7
16
−
1
4
12.
2
7
+
3
11
13.
2
9
−
1
7
+
2
3
14. 3
2
5
− 2
1
3
15.
7
27
−
2
3
+
5
9
16. 5
3
13
+ 3
3
4
17. 4
5
8
− 3
2
5
18. 10
3
7
− 8
2
3
19. 3
1
4
− 4
4
5
+ 1
5
6
20. 5
3
4
− 1
2
5
− 3
1
2

2.3 Multiplication and division of
fractions
2.3.1 Multiplication
To multiply two or more fractions together, the numer-
ators are ﬁrst multiplied to give a single number
and this becomes the new numerator of the com-
bined fraction. The denominators are then multiplied
together to give the new denominator of the combined
fraction.
For example,
2
3
×
4
7
=
2 × 4
3 × 7
=
8
21
Problem 9. Simplify 7 ×
2
5
7 ×
2
5
=
7
1
×
2
5
=
7 × 2
1 × 5
=
14
5
= 2
4
5
3
7
×
14
15
Dividing numerator and denominator by 3 gives
3
7
×
14
15
=
1
7
×
14
5
=
1 × 14
7 × 5
Dividing numerator and denominator by 7 gives
1 × 14
7 × 5
=
1 × 2
1 × 5
=
2
5
This process of dividingboth the numerator and denom-
inator of a fraction by the same factor(s) is called
cancelling.
Problem 11. Simplify
3
5
×
4
9
3
5
×
4
9
=
1
5
×
4
3
by cancelling
=
4
15
3
5
× 2
1
3
× 3
3
7
Mixed numbers must be expressed as improper frac-
tions before multiplication can be performed. Thus,
1
3
5
× 2
1
3
× 3
3
7
=
5
5
+
3
5
×
6
3
+
1
3
×
21
7
+
3
7
=
8
5
×
7
3
×
24
7
=
8 × 1 × 8
5 × 1 × 1
=
64
5
= 12
4
5
1
5
× 1
2
3
× 2
3
4
The mixed numbers need to be changed to improper
fractions before multiplication can be performed.
3
1
5
× 1
2
3
× 2
3
4
=
16
5
×
5
3
×
11
4
=
4
1
×
1
3
×
11
1
by cancelling
=
4 × 1 × 11
1 × 3 × 1
=
44
3
= 14
2
3
2.3.2 Division
The simple rule for division is change the division
sign into a multiplication sign and invert the second
fraction.
For example,
2
3
÷
3
4
=
2
3
×
4
3
=
8
9
3
7
÷
8
21
3
7
÷
8
21
=
3
7
×
21
8
=
3
1
×
3
8
by cancelling
=
3 × 3
1 × 8
=
9
8
= 1
1
8
Problem 15. Find the value of 5
3
5
÷ 7
1
3
The mixed numbers must be expressed as improper
fractions. Thus,
5
3
5
÷ 7
1
3
=
28
5
÷
22
3
=
28
5
×
3
22
=
14
5
×
3
11
=
42
55
2
3
× 1
3
4
÷ 2
3
4

Fractions 13
Mixed numbersmust beexpressed as improperfractions
before multiplication and division can be performed:
3
2
3
× 1
3
4
÷ 2
3
4
=
11
3
×
7
4
÷
11
4
=
11
3
×
7
4
×
4
11
=
1 × 7 × 1
3 × 1 × 1
by cancelling
=
7
3
= 2
1
3
PracticeExercise 6 Multiplying and
dividing fractions (answers on page 340)
Evaluate the following.
1.
2
5
×
4
7
2. 5 ×
4
9
3.
3
4
×
8
11
4.
3
4
×
5
9
5.
17
35
×
15
68
6.
3
5
×
7
9
× 1
2
7
7.
13
17
× 4
7
11
× 3
4
39
8.
1
4
×
3
11
× 1
5
39
9.
2
9
÷
4
27
10.
3
8
÷
45
64
11.
3
8
÷
5
32
12.
3
4
÷ 1
4
5
13. 2
1
4
× 1
2
3
14. 1
1
3
÷ 2
5
9
15. 2
4
5
÷
7
10
16. 2
3
4
÷ 3
2
3
17.
1
9
×
3
4
× 1
1
3
18. 3
1
4
× 1
3
5
÷
2
5
19. A ship’s crew numbers 105, of which
1
7
are
women. Of the men,
1
6
are officers. How
many male officers are on board?
20. If a storage tank is holding 450 litres when
it is three-quarters full, how much will it
contain when it is two-thirds full?
21. Three people, P, Q and R, contribute to a
fund. 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,
7
10
of the contents are removed, then
3
5
of
the remainder is removed. How much oil is
left in the tank?
2.4 Order of precedence with
fractions
As stated in Chapter 1, sometimes addition, subtraction,
multiplication, division, powers and brackets can all be
involved in a calculation. A definite order of precedence
must be adhered to. The order is:
Brackets
Order (or pOwer)
Division
Multiplication
Addition
Subtraction
This is demonstrated in the followingworked problems.
7
20
−
3
8
×
4
5
7
20
−
3
8
×
4
5
=
7
20
−
3 × 1
2 × 5
by cancelling (M)
=
7
20
−
3
10
(M)
=
7
20
−
6
20
=
1
20
(S)
1
4
− 2
1
5
×
5
8
+
9
10
1
4
− 2
1
5
×
5
8
+
9
10
=
1
4
−
11
5
×
5
8
+
9
10
=
1
4
−
11
1
×
1
8
+
9
10
by cancelling
=
1
4
−
11
8
+
9
10
(M)
=
1 × 10
4 × 10
−
11 × 5
8 × 5
+
9 × 4
10 × 4
(since the LCM of 4, 8 and 10 is 40)

=
10
40
−
55
40
+
36
40
=
10 − 55 + 36
40
(A/S)
= −
9
40
2
1
2
−
2
5
+
3
4
÷
5
8
×
2
3
2
1
2
−
2
5
+
3
4
÷
5
8
×
2
3
=
5
2
−
2 × 4
5 × 4
+
3 × 5
4 × 5
÷
5
8
×
2
3
(B)
=
5
2
−
8
20
+
15
20
÷
5
8
×
2
3
(B)
=
5
2
−
23
20
÷
5
4
×
1
3
by cancelling (B)
=
5
2
−
23
20
÷
5
12
(B)
=
5
2
−
23
20
×
12
5
(D)
=
5
2
−
23
5
×
3
5
by cancelling
=
5
2
−
69
25
(M)
=
5 × 25
2 × 25
−
69 × 2
25 × 2
(S)
=
125
50
−
138
50
(S)
= −
13
50
1
3
of 5
1
2
− 3
3
4
+ 3
1
5
÷
4
5
−
1
2
1
3
of 5
1
2
− 3
3
4
+ 3
1
5
÷
4
5
−
1
2
=
1
3
of 1
3
4
+ 3
1
5
÷
4
5
−
1
2
(B)
=
1
3
×
7
4
+
16
5
÷
4
5
−
1
2
(O)
(Note that the ‘of ’ is replaced with a
multiplication sign.)
=
1
3
×
7
4
+
16
5
×
5
4
−
1
2
(D)
=
1
3
×
7
4
+
4
1
×
1
1
−
1
2
by cancelling
=
7
12
+
4
1
−
1
2
(M)
=
7
12
+
48
12
−
6
12
(A/S)
=
49
12
= 4
1
12
PracticeExercise 7 Order of precedence
with fractions (answers on page 340)
Evaluate the following.
1. 2
1
2
−
3
5
×
20
27
2.
1
3
−
3
4
×
16
27
3.
1
2
+
3
5
÷
9
15
−
1
3
4.
1
5
+ 2
2
3
÷
5
9
−
1
4
5.
4
5
×
1
2
−
1
6
÷
2
5
+
2
3
6.
3
5
−
2
3
−
1
2
÷
5
6
×
3
2
7.
1
2
of 4
2
5
− 3
7
10
+ 3
1
3
÷
2
3
−
2
5
8.
6
2
3
× 1
2
5
−
1
3
6
3
4
÷ 1
1
2
9. 1
1
3
× 2
1
5
÷
2
5
10.
1
4
×
2
5
−
1
5
÷
2
3
+
4
15
11.
2
3
+ 3
1
5
× 2
1
2
+ 1
1
3
8
1
3
÷ 3
1
3
12.
1
13
of 2
9
10
− 1
3
5
+ 2
1
3
÷
2
3
−
3
4

Revision Test 1 : Basic arithmetic and fractions
This assignment covers the material contained in Chapters 1 and 2. The marks available are shown in brackets at the
end of each question.
1. Evaluate
1009cm − 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 such
components. (3)
6. Find (a) the highest common factor and (b) the
lowest 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 ÷ 2
2 × 4 + (5 − 8) − 22 + 3
(5)
12.
(42 × 5 − 8) ÷ 3 + 9 × 8
4 × 32 − 20 ÷ 5
(5)
13. Simplify
(a)
3
4
−
7
15
(b) 1
5
8
− 2
1
3
+ 3
5
6
(8)
14. A training college has 480 students of which 150
are girls. Express this as a fraction in its simplest
form. (2)
15. A tank contains 18000litresof oil. Initially,
7
10
of
the contents are removed, then
2
5
of the remainder
is removed. How much oil is left in the tank? (4)
16. Evaluate
(a) 1
7
9
×
3
8
× 3
3
5
(b) 6
2
3
÷ 1
1
3
(c) 1
1
3
× 2
1
5
÷
2
5
(10)
17. Calculate
(a)
1
4
×
2
5
−
1
5
÷
2
3
+
4
15
(b)
2
3
+ 3
1
5
× 2
1
2
+ 1
1
3
8
1
3
÷ 3
1
3
(8)
18. Simplify
1
13
of 2
9
10
− 1
3
5
+ 2
1
3
÷
2
3
−
3
4
(8)

Chapter 3
Decimals
3.1 Introduction
The decimal system of numbers is based on the digits
0 to 9.
Thereareanumberofeveryday occurrencesin which we
use decimal numbers. For example, a radio is, say, tuned
to 107.5MHz FM; 107.5 is an example of a decimal
number.
In a shop, a pair of trainers cost, say, £57.95; 57.95 is
anotherexampleofadecimal number.57.95 isadecimal
fraction, where a decimal point separates the integer, i.e.
57, from the fractional part, i.e. 0.95
57.95 actually means (5 × 10)+ (7 × 1)
+ 9 ×
1
10
+ 5 ×
1
100
3.2 Converting decimals to fractions
and vice-versa
Converting decimals to fractions and vice-versa is
demonstrated below with worked examples.
Problem 1. Convert 0.375 to a proper fraction in
its simplest form
(i) 0.375 may be written as
0.375 × 1000
1000
i.e.
0.375 =
375
1000
(ii) Dividing both numerator and denominator by 5
gives
375
1000
=
75
200
(iii) Dividing both numerator and denominator by 5
again gives
75
200
=
15
40
(iv) Dividing both numerator and denominator by 5
again gives
15
40
=
3
8
Since both 3 and 8 are only divisible by 1, we cannot
‘cancel’ any further, so
3
8
is the ‘simplest form’ of the
fraction.
Hence, the decimal fraction 0.375 =
3
8
as a proper
fraction.
Problem 2. Convert 3.4375 to a mixed number
(i) 0.4375 may be written as
0.4375 × 10000
10000
i.e.
0.4375 =
4375
10000
(ii) Dividing both numerator and denominator by 25
gives
4375
10000
=
175
400
(iii) Dividing both numerator and denominator by 5
gives
175
400
=
35
80
(iv) Dividing both numerator and denominator by 5
again gives
35
80
=
7
16
Since both 5 and 16 are only divisible by 1, we
cannot ‘cancel’ any further, so
7
16
is the ‘lowest
form’ of the fraction.
(v) Hence, 0.4375 =
7
16
Thus, the decimal fraction 3.4375=3
7
16
as a mixed
number.
DOI: 10.1016/B978-1-85617-697-2.00003-X

Decimals 17
Problem 3. Express
7
8
as a decimal fraction
To convert a proper fraction to a decimal fraction, the
numerator is divided by the denominator.
0.8 7 5
8 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 point
of 7.000
(iii) 8 into 70 goes 8, remainder 6. Place the 8 above
the first zero after the decimal point and carry the
6 remainder to the next digit on the right, making
it 60.
(iv) 8 into 60 goes 7, remainder 4. Place the 7 above
the next zero and carry the 4 remainder to the next
digit on the right, making it 40.
(v) 8 into 40 goes 5, remainder 0. Place 5 above the
next zero.
Hence, the proper fraction
7
8
= 0.875 as a decimal
fraction.
Problem 4. Express 5
13
16
as a decimal fraction
For mixed numbers it is only necessary to convert the
proper fraction part of the mixed number to a decimal
fraction.
0.8 1 2 5
16 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 point
of 13.0000
(iii) 16 into 130 goes 8, remainder 2. Place the 8 above
the first zero after the decimal point and carry the
2 remainder to the next digit on the right, making
it 20.
(iv) 16 into 20 goes 1, remainder 4. Place the 1 above
(v) 16 into 40 goes 2, remainder 8. Place the 2 above
(vi) 16 into 80 goes 5, remainder 0. Place the 5 above
the next zero.
(vii) Hence,
13
16
= 0.8125
Thus, the mixed number 5
13
16
= 5.8125 as a decimal
fraction.
PracticeExercise 8 Converting decimals to
fractions and vice-versa (answers on
page 341)
1. Convert 0.65 to a proper fraction.
5. Convert the following to proper fractions.
(a) 0.65 (b) 0.84 (c) 0.0125
(d) 0.282 (e) 0.024
6. Convert 4.525 to a mixed number.
10. Convert the following to mixed numbers.
(a) 1.82 (b) 4.275 (c) 14.125
(d) 15.35 (e) 16.2125
11. Express
5
8
as a decimal fraction.
12. Express 6
11
16
13. Express
7
32
14. Express 11
3
16
15. Express
9
32
3.3 Significant figures and decimal
places
A number which can be expressed exactly as a
decimal fraction is called a terminating decimal.

For example,
3
3
16
= 3.1825 is a terminating decimal
A number which cannot be expressed exactly as a deci-
mal fraction is called a non-terminating decimal. For
example,
1
5
7
= 1.7142857... is a non-terminating decimal
The answer to a non-terminating decimal may be
expressed in two ways, depending on the accuracy
required:
(a) correct to a number of significant figures, or
(b) correct to a number of decimal places i.e. the
number of figures after the decimal point.
The last digit in the answer is unaltered if the next digit
on the right is in the group of numbers 0, 1, 2, 3 or 4.
For example,
1.714285... = 1.714 correct to 4 significant figures
= 1.714 correct to 3 decimal places
since the next digit on the right in this example is 2.
The last digit in the answer is increased by 1 if the next
digit on the right is in the group of numbers 5, 6, 7, 8 or
9. For example,
1.7142857... = 1.7143 correct to 5 significant figures
since 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 significant
figures.
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 significant
figures.
Note that the zeros to the right of the decimal point do
not count as significant figures.
PracticeExercise 9 Significant figures and
decimal places (answers on page 341)
1. Express 14.1794 correct to 2 decimal places.
2. Express 2.7846 correct to 4 significant figures.
4. Express 43.2746 correct to 4 significant
figures.
figures.
3.4 Adding and subtracting decimal
numbers
When adding or subtracting decimal numbers, care
needs to be taken to ensure that the decimal points
are beneath each other. This is demonstrated in the
following worked examples.
Problem 7. Evaluate 46.8 + 3.06 + 2.4 + 0.09
and give the answer correct to 3 significant figures
The decimal points are placed under each other as
shown. Each column is added, starting from the
right.
46.8
3.06
2.4
+0.09
52.35
11 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 in
the tenths column. Carry the 1 into the units
column.
(iii) 6 + 3 + 2 + 0 + 1 (carried) = 12. Place the 2 in
theunitscolumn.Carry the1 into thetenscolumn.

Decimals 19
(iv) 4 + 1(carried) = 5. Place the 5 in the hundreds
column.
Hence,
46.8 + 3.06 + 2.4 + 0.09 = 52.35
= 52.4,correct to 3
significant figures
Problem 8. Evaluate 64.46 − 28.77 and give the
answer correct to 1 decimal place
As with addition, the decimal points are placed under
each other as shown.
64.46
−28.77
35.69
(i) 6 − 7 is not possible; therefore ‘borrow’ 1 from
the tenths column. This gives 16 − 7 = 9. Place
the 9 in the hundredths column.
(ii) 3 − 7 is not possible; therefore ‘borrow’ 1 from
the unitscolumn. This gives 13 − 7 = 6. Place the
6 in the tenths column.
(iii) 3 − 8 is not possible; therefore ‘borrow’ from the
hundreds column. This gives 13 − 8 = 5. Place
the 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 place
Problem 9. Evaluate 312.64 − 59.826 − 79.66+
38.5 and give the answer correct to 4 significant
figures
The 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 from
the sum of the positive decimal fractions gives
351.140
−139.486
211.654
Hence, 351.140 − 139.486 = 211.654 = 211.7, cor-
rect to 4 significant figures.
PracticeExercise 10 Adding and
subtracting decimal numbers (answers on
page 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 decimal
place.
3. Evaluate 68.92 + 34.84 − 31.223, correct to
4 significant figures.
4. Evaluate 67.841 − 249.55 + 56.883, correct
to 2 decimal places.
5. Evaluate 483.24 − 120.44 − 67.49, correct
to 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 the
length of the shaft shown in Figure 3.1. The
dimensions are in millimetres.
82.92
27.41 8.32 34.67x
Figure 3.1
3.5 Multiplying and dividing decimal
numbers
When multiplying decimal fractions:
(a) the numbers are multiplied as if they were integers,
and
(b) the position of the decimal point in the answer is
such that there are as many digits to the right of
it as the sum of the digits to the right of the dec-
imal points of the two numbers being multiplied
together.
This is demonstrated in the followingworked examples.

Problem 10. Evaluate 37.6 × 5.4
376
×54
1504
18800
20304
(i) 376 × 54 = 20304.
(ii) As there are 1 + 1 = 2 digits to the right of
the decimal points of the two numbers being
multiplied together, 37.6 × 5.4, then
37.6 × 5.4 = 203.04
Problem 11. Evaluate 44.25 ÷ 1.2, correct to
(a) 3 significant figures, (b) 2 decimal places
44.25 ÷ 1.2 =
44.25
1.2
The denominator is multiplied by 10 to change it into an
integer. The numerator is also multiplied by 10 to keep
the fraction the same. Thus,
44.25
1.2
=
44.25 × 10
1.2 × 10
=
442.5
12
The long division is similar to the long division of
integers and the steps are as shown.
36.875
12 442.500
36
82
72
105
96
90
84
60
60
0
(i) 12 into 44 goes 3; place the 3 above the second
4 of 442.500
(ii) 3 × 12 = 36; place the 36 below the 44 of
442.500
(iii) 44 − 36 = 8.
(iv) Bring down the 2 to give 82.
(v) 12 into 82 goes 6; place the 6 above the 2 of
442.500
(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 of
442.500
(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 first
zero 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 above
the second zero of 442.500
(xviii) Hence, 44.25 ÷ 1.2 =
442.5
12
= 36.875
So,
(a) 44.25 ÷ 1.2 = 36.9, correct to 3 significant
figures.
(b) 44.25 ÷ 1.2 = 36.88, correct to 2 decimal
places.
Problem 12. Express 7
2
3
as a decimal fraction,
correct to 4 significant figures
Dividing 2 by 3 gives
2
3
= 0.666666...
and 7
2
3
= 7.666666...
Hence, 7
2
3
= 7.667 correct to 4 significant figures.
Note that 7.6666... is called 7.6 recurring and is
written as 7.
.
6
PracticeExercise 11 Multiplying and
dividing decimal numbers (answers on
page 341)
In Problems 1 to 8, evaluate without using a
calculator.
1. Evaluate 3.57 × 1.4
2. Evaluate 67.92 × 0.7

Decimals 21
3. Evaluate 167.4 × 2.3
4. Evaluate 342.6 × 1.7
5. Evaluate 548.28 ÷ 1.2
6. Evaluate 478.3 ÷ 1.1, correct to 5 significant
figures.
7. Evaluate 563.48 ÷ 0.9, correct to 4 signifi-
cant figures.
8. Evaluate 2387.4 ÷ 1.5
In Problems 9 to 14, express as decimal fractions
to the accuracy stated.
9.
4
9
, correct to 3 significant figures.
10.
17
27
, correct to 5 decimal places.
11. 1
9
16
12. 53
5
11
13. 13
31
37
14. 8
9
13
15. Evaluate 421.8 ÷ 17, (a) correct to 4 signif-
icant figures and (b) correct to 3 decimal
places.
16. Evaluate
0.0147
2.3
, (a) correct to 5 decimal
places and (b) correct to 2 significant figures.
17. Evaluate (a)
12.
.
6
1.5
(b) 5.
.
2 × 12
18. A tank contains 1800 litres of oil. How many
tins containing 0.75 litres can be filled from
this tank?

Chapter 4
Using a calculator
4.1 Introduction
In 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 an
electronic calculator needs to be used.
There are several calculators on the market, many of
which will be satisfactory for our needs. It is essential
to have a scientific notationcalculator which will have
all the necessary functions needed and more.
This chapter assumes you have a CASIO fx-83ES
calculator, or similar, as shown in Figure 4.1.
Besides straightforward addition, subtraction, multipli-
cation and division, which you will already be able
to 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 calculator
which we do not need to concern ourselves with at this
level.
4.2 Adding, subtracting, multiplying
and dividing
Initially, after switching on, press Mode.
Of the three possibilities, use Comp, which is achieved
by pressing 1.
Next, press Shift followed by Setup and, of the eight
possibilities, 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 and
these functions will, no doubt, already have been used
in calculations.
Problem 1. Evaluate 364.7 ÷ 57.5 correct to 3
decimal places
(i) Type in 364.7
(ii) Press ÷.
(iii) Type in 57.5
(iv) Press = and the fraction
3647
575
appears.
(v) Press the S ⇔ D functionand the decimal answer
6.34260869... appears.
Alternatively, after step (iii) press Shift and = and the
decimal will appear.
Hence, 364.7 ÷ 57.5 = 6.343 correct to 3 decimal
places.
Problem 2. Evaluate
12.47 × 31.59
70.45 × 0.052
correct to 4
(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 × 31.59
70.45 × 0.052
= 107.5 correct to 4 significant
figures.
DOI: 10.1016/B978-1-85617-697-2.00004-1

Using a calculator 23
Figure 4.1 A Casio fx-83ES calculator
PracticeExercise 12 Addition, subtraction,
multiplication and division using a calculator
(answers on page 341)
1. Evaluate 378.37 − 298.651 + 45.64 − 94.562
2. Evaluate 25.63 × 465.34 correct to 5 signif-
icant figures.
3. Evaluate 562.6 ÷ 41.3 correct to 2 decimal
places.
4. Evaluate
17.35 × 34.27
41.53 ÷ 3.76
correct to 3 decimal
places.
5. Evaluate 27.48 + 13.72 × 4.15 correct to
6. Evaluate
(4.527 + 3.63)
(452.51 ÷ 34.75)
+ 0.468 correct
7. Evaluate 52.34 −
(912.5 ÷ 41.46)
(24.6 − 13.652)
correct to
3 decimal places.
8. Evaluate
52.14 × 0.347 × 11.23
19.73 ÷ 3.54
correct to
9. Evaluate
451.2
24.57
−
363.8
46.79
correct to 4 signifi-
cant figures.
10. Evaluate
45.6 − 7.35 × 3.61
4.672 − 3.125
correct to 3
decimal places.
4.3 Further calculator functions
4.3.1 Square and cube functions
Locate the x2 and x3 functions on your calculator and
then 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 answer
144
25
appears.
(iv) Press the S ⇔ D function and the fraction
changes to a decimal 5.76
Alternatively, after step (ii) press Shift and = .
Thus, 2.42 = 5.76
Problem 4. Evaluate 0.172 in engineering form
(i) Type in 0.17

(iii) Press Shift and = and the answer 0.0289 appears.
(iv) Press the ENG function and the answer changes
to 28.9 × 10−3, which is engineering form.
Hence, 0.172 = 28.9×10−3 in engineering form. The
ENG function is extremely important in engineering
calculations.
Problem 5. Change 348620 into engineering
form
(i) Type in 348620
(ii) Press = then ENG.
Hence, 348620 = 348.62×103 in engineering form.
Problem 6. Change 0.0000538 into engineering
form
(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
343
125
appears.
changes to a decimal: 2.744
Thus, 1.43 = 2.744.
PracticeExercise 13 Square and cube
functions (answers on page 341)
1. Evaluate 3.52
2. Evaluate 0.192
3. Evaluate 6.852 correct to 3 decimal places.
4. Evaluate (0.036)2 in engineering form.
5. Evaluate 1.5632 correct to 5 significant
figures.
6. Evaluate 1.33
7. Evaluate 3.143 correct to 4 significant
figures.
8. Evaluate (0.38)3 correct to 4 decimal places.
9. Evaluate (6.03)3
correct to 2 decimal places.
4.3.2 Reciprocal and power functions
The reciprocal of 2 is
1
2
, the reciprocal of 9 is
1
9
and the
reciprocal of x is
1
x
, which from indices may be written
as x−1. Locate the reciprocal, i.e. x−1 on the calculator.
Also, locate the power function, i.e. x , on your
calculator and then check the following worked
examples.
Problem 8. Evaluate
1
3.2
(i) Type in 3.2
(ii) Press x−1 and 3.2−1 appears on the screen.
5
16
appears.
Thus,
1
3.2
= 0.3125
Problem 9. Evaluate 1.55 correct to 4 significant
figures
(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 7.59375
appears.
Thus, 1.55 = 7.594 correct to 4 significant figures.
Problem 10. Evaluate 2.46 − 1.94 correct to 3
decimal places
(i) Type in 2.4
(ii) Press x and 2.4 appears on the screen.

(iii) Press 6 and 2.46
appears on the screen.
(iv) The cursor now needs to be moved; this is
achieved by using the cursor key (the large
blue circular function in the top centre of the
calculator). 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 decimal
places.
PracticeExercise 14 Reciprocal and power
1. Evaluate
1
1.75
2. Evaluate
1
0.0250
3. Evaluate
1
7.43
correct to 5 significant figures.
4. Evaluate
1
0.00725
5. Evaluate
1
0.065
−
1
2.341
correct to 4 signifi-
cant figures.
6. Evaluate 2.14
7. Evaluate (0.22)5
correct to 5 significant
figures in engineering form.
8. Evaluate (1.012)7 correct to 4 decimal
places.
10. Evaluate 1.13 + 2.94 − 4.42 correct to 4 sig-
nificant figures.
4.3.3 Root and ×10x functions
Locate the square root function
√
and the
√
function (which is a Shift function located above
the x function) on your calculator. Also, locate the
×10x function and then check the following worked
examples.
√
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.
√
81
(i) Press the
√
function.
(ii) Type in 4 and 4
√
(iii) Press → to move the cursor and then type in 81
and 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 have
to use ×).
(iii) Type in 5
(iv) Press ×
(v) Type in 2
(vi) Press the ×10x function.
(vii) Type in −7
(viii) Press = and the answer
3
25
appears.
(ix) Press the S ⇔ D function and the fraction
Thus, 6 × 105 × 2 × 10−7 = 0.12
PracticeExercise 15 Root and ×10x
1. Evaluate
√
4.76 correct to 3 decimal places.
2. Evaluate
√
123.7 correct to 5 significant
figures.

3. Evaluate
√
34528correct to 2 decimal places.
4. Evaluate
√
figures.
5. Evaluate
√
6. Evaluate 3
√
17 correct to 3 decimal places.
7. Evaluate 4
√
773 correct to 4 significant
figures.
8. Evaluate 5
√
9. Evaluate 3
√
figures.
10. Evaluate
6
√
2451−
4
√
46 correct to 3 decimal
places.
Express the answers to questions 11 to 15 in
engineering form.
11. Evaluate 5 × 10−3 × 7 × 108
12. Evaluate
3 × 10−4
8 × 10−9
13. Evaluate
6 × 103 × 14× 10−4
2 × 106
14. Evaluate
56.43 × 10−3 × 3 × 104
8.349 × 103
correct to
3 decimal places.
15. Evaluate
99 × 105 × 6.7 × 10−3
36.2 × 10−4
correct to 4
significant figures.
4.3.4 Fractions
Locate the and functions on your calculator
(the latter function is a Shift function found above
the function) and then check the following worked
examples.
1
4
+
2
3
(i) Press the function.
(ii) Type in 1
(iii) Press ↓ on the cursor key and type in 4
(iv)
1
4
(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 answer
11
12
appears.
(xi) Press the S ⇔ D function and the fraction
changes to a decimal 0.9166666...
Thus,
1
4
+
2
3
=
11
12
= 0.9167 as a decimal, correct to
4 decimal places.
It is also possible to deal with mixed numbers on the
calculator. Press Shift then the function and
appears.
1
5
− 3
3
4
(i) Press Shift then the function and appears
on 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 5
1
5
(v) Press → on the cursor key.
(vi) Typein – and then pressShift then the function
and 5
1
5
− 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 5
1
5
− 3
3
4
(x) Press = and the answer
29
20
appears.
(xi) Press S ⇔ D function and the fraction changes to
a decimal 1.45
Thus, 5
1
5
− 3
3
4
=
29
20
= 1
9
20
= 1.45 as a decimal.

PracticeExercise 16 Fractions (answers on
page 341)
1. Evaluate
4
5
−
1
3
as a decimal, correct to
4 decimal places.
2. Evaluate
2
3
−
1
6
+
3
7
as a fraction.
3. Evaluate 2
5
6
+ 1
5
8
4. Evaluate 5
6
7
− 3
1
8
5. Evaluate
1
3
−
3
4
×
8
21
as a fraction.
6. Evaluate
3
8
+
5
6
−
1
2
4 decimal places.
7. Evaluate
3
4
×
4
5
−
2
3
÷
4
9
as a fraction.
8. Evaluate 8
8
9
÷ 2
2
3
as a mixed number.
9. Evaluate 3
1
5
× 1
1
3
− 1
7
10
as a decimal, cor-
rect to 3 decimal places.
10. Evaluate
4
1
5
− 1
2
3
3
1
4
× 2
3
5
−
2
9
as a decimal, cor-
rect to 3 significant figures.
4.3.5 Trigonometric functions
Trigonometric ratios will be covered in Chapter 21.
However, very briefly, there are three functions on your
calculator that are involved with trigonometry. They are:
sin which is an abbreviation of sine
cos which is an abbreviation of cosine, and
tan which is an abbreviation of tangent
Exactly what these mean will be explained in
Chapter 21.
There are two main ways that angles are measured, i.e.
in degrees or in radians. Pressing Shift, Setup and 3
shows degrees, and Shift, Setup and 4 shows radians.
Press 3 and your calculator will be in degrees mode,
indicated by a small D appearing at the top of the screen.
Press 4 and your calculator will be in radian mode,
indicated by a small R appearing at the top of the
screen.
Locate the sin, cos and tan functions on your calculator
and then check the following worked examples.
Problem 16. Evaluate sin38◦
(i) Make sure your calculator is in degrees mode.
(ii) Press sin function and sin( appears on the
screen.
(iii) Type in 38 and close the bracket with) and sin (38)
(iv) Press = and the answer 0.615661475... appears.
Thus, sin38◦ = 0.6157, correct to 4 decimal places.
Problem 17. Evaluate 5.3 tan (2.23 rad)
(i) Make sure your calculator is in radian mode by
pressing Shift then Setup then 4 (a small R appears
at the top of the screen).
(ii) Type in 5.3 then press tan function and 5.3 tan(
(iii) Type in 2.23 and close the bracket with) and
5.3 tan (2.23) appears on the screen.
(iv) Press = and the answer −6.84021262... appears.
Thus, 5.3tan(2.23rad) = −6.8402, correct to 4 dec-
imal places.
PracticeExercise 17 Trigonometric
Evaluate the following, each correct to 4 decimal
places.
1. Evaluate sin67◦
2. Evaluate cos43◦
3. Evaluate tan71◦
4. Evaluate sin15.78◦
5. Evaluate cos63.74◦
6. Evaluate tan39.55◦ − sin52.53◦
7. Evaluate sin(0.437 rad)

8. 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 functions
Press Shift and then press the ×10x function key and π
appears on the screen. Either press Shift and = (or =
and S ⇔ D) and the value of π appears in decimal form
as 3.14159265...
Press Shift and then press the ln function key and e
appears on the screen. Enter 1 and then press = and
e1 = e = 2.71828182...
Now check the following worked examples involvingπ
and ex functions.
Problem 18. Evaluate 3.57π
(i) Enter 3.57
(ii) Press Shift and the ×10x key and 3.57π appears
on the screen.
(iii) Either press Shift and = (or = and S ⇔ D)
and the value of 3.57π appears in decimal as
11.2154857...
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 and
e appears on the screen.
(ii) Enter 2.37 and e2.37 appears on the screen.
(iii) Press Shiftand = (or= and S ⇔ D) and the value
of e2.37 appears in decimal as 10.6973922...
Hence, e 2.37 = 10.70 correct to 4 significant figures.
PracticeExercise 18 π and ex functions
Evaluatethefollowing,each correct to 4 significant
figures.
1. 1.59π 2. 2.7(π− 1)
3. π2
√
13 − 1 4. 3eπ
5. 8.5e−2.5 6. 3e2.9 − 1.6
7. 3e(2π−1) 8. 2πe
π
3
9.
5.52π
2e−2 ×
√
26.73
10.
⎡
⎣ e
2−
√
3
π ×
√
8.57
⎤
⎦
4.4 Evaluation of formulae
The statement y = mx + c is called a formula for y in
terms 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 place
of symbols to evaluate engineering quantities.
Just four examples of important formulae are:
1. A straight line graph is of the form y = mx + c (see
Chapter 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 your
calculator that you agree with the working and answers.
Problem 20. In an electrical circuit the voltage V
is given by Ohm’s law, i.e. V = IR. Find, correct to
4 significant figures, the voltage when I = 5.36A
and R = 14.76
V = IR = (5.36)(14.76)
Hence, voltage V = 79.11V, correct to 4 significant
figures.
Problem 21. The surface area A of a hollow cone
is given by A = πrl. Determine, correct to 1
decimal place, the surface area when r = 3.0cm
and l = 8.5cm
A = πrl = π(3.0)(8.5)cm2
Hence, surface area A = 80.1cm2, correct to 1 deci-
mal place.

Problem 22. Velocity v is given by v = u + at. If
u = 9.54m/s, a = 3.67m/s2 and t = 7.82s, find v,
v = u + at = 9.54 + 3.67 × 7.82
= 9.54 + 28.6994 = 38.2394
Hence, velocity v = 38.2m/s, correct to 3 significant
figures.
Problem 23. The area, A, of a circle is given by
A = πr2. Determine the area correct to 2 decimal
places, given radius r = 5.23m
A = πr2
= π(5.23)2
= π(27.3529)
Hence, area, A = 85.93m2, correct to 2 decimal
places.
Problem 24. Density =
mass
volume
. Find the density
when the mass is 6.45kg and the volume is
300 × 10−6 cm3
Density =
mass
volume
=
6.45kg
300 × 10−6 m3
= 21500kg/m3
Problem 25. The power, P watts, dissipated in an
electrical circuit is given by the formula P =
V 2
R
.
Evaluate the power, correct to 4 significant figures,
given that V = 230 V and R = 35.63
P =
V2
R
=
(230)2
35.63
=
52900
35.63
= 1484.70390...
Press ENG and 1.48470390...× 103 appears on the
screen.
Hence, power, P = 1485W or 1.485kW correct to 4
PracticeExercise 19 Evaluation of
formulae (answers on page 341)
1. The area A of a rectangle is given by the
formula A = lb. Evaluate the area when
l = 12.4cm and b = 5.37cm.
2. The circumference C of a circle is given by
theformula C = 2πr. Determine the circum-
ference given r = 8.40mm.
3. A formula used in connection with gases
is R =
PV
T
. Evaluate R when P = 1500,
V = 5 and T = 200.
4. Thevelocity ofabody isgiven byv = u + at.
The initial velocity u is measured when time
t is 15 seconds and found to be 12m/s. If the
acceleration a is 9.81m/s2
calculate the final
velocity v.
5. Calculate the current I in an electrical circuit,
where I = V/R amperes when the voltage V
is measured and found to be 7.2 V and the
resistance R is 17.7 .
6. Find the distance s, given that s =
1
2
gt2
when time t = 0.032seconds and accelera-
tion due to gravity g = 9.81m/s2
. Give the
answer in millimetres.
7. The energy stored in a capacitor is given
by E =
1
2
CV 2 joules. Determine the energy
when capacitance C = 5 × 10−6 farads and
voltage V = 240V.
8. Find the area A of a triangle, given A =
1
2
bh,
when the base length l is 23.42m and the
height h is 53.7m.
9. Resistance R2 is given by R2 = R1(1 + αt).
Find R2, correct to 4 significant figures, when
R1 = 220, α = 0.00027 and t = 75.6
10. Density =
mass
volume
. Find the density when
the mass is 2.462kg and the volume is
173cm3. Give the answer in units of kg/m3
.
11. Velocity = frequency × wavelength. Find
the velocity when the frequency is 1825Hz
and the wavelength is 0.154m.
12. Evaluate resistance RT , given
1
RT
=
1
R1
+
1
R2
+
1
R3
when R1 = 5.5 ,
R2 = 7.42 and R3 = 12.6 .
Here are some further practical examples. Again, check
with your calculator that you agree with the working
and answers.

Problem 26. The volume V cm3
of a right
circular cone is given by V =
1
3
πr2h. Given that
radius r = 2.45cm and height h = 18.7cm, find the
volume, correct to 4 significant figures
V =
1
3
πr2
h =
1
3
π(2.45)2
(18.7)
=
1
3
× 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 the
formula F =
Gm1m2
d2
, where m1 and m2 are
masses, d their distance apart and G is a constant.
Find the value of the force given that
G = 6.67 × 10−11, m1 = 7.36, m2 = 15.5 and
d = 22.6. Express the answer in standard form,
F =
Gm1m2
d2
=
(6.67 × 10−11)(7.36)(15.5)
(22.6)2
=
(6.67)(7.36)(15.5)
(1011)(510.76)
=
1.490
1011
Hence, force F =1.49×10−11 newtons, correct to
Problem 28. The time of swing, t seconds, of a
simple pendulum is given by t = 2π
l
g
Determine the time, correct to 3 decimal places,
given that l = 12.9 and g = 9.81
t = 2π
l
g
= (2π)
12.9
9.81
= 7.20510343...
Hence, time t = 7.205seconds, correct to 3 decimal
places.
Problem 29. Resistance, R , varies with
temperature according to the formula
R = R0(1 + αt). Evaluate R, correct to 3 significant
figures, given R0 = 14.59, α = 0.0043 and t = 80
R = 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 significant
figures.
Problem 30. The current, I amperes, in an a.c.
circuit is given by I =
V
(R2 + X2)
. Evaluate the
current, correct to 2 decimal places, when
V = 250V, R = 25.0 and X = 18.0 .
I=
V
(R2 + X2)
=
250
25.02 + 18.02
= 8.11534341...
Hence, current, I = 8.12A, correct to 2 decimal
places.
PracticeExercise 20 Evaluation of
1. Find the total cost of 37 calculators cost-
ing £12.65 each and 19 drawing sets costing
£6.38 each.
2. Power =
force × distance
time
. Find the power
when a force of 3760N raises an object a
distance of 4.73m in 35s.
3. The potential difference, V volts, available
at battery terminals is given by V = E − Ir.
Evaluate V when E = 5.62, I = 0.70 and
R = 4.30
4. Given force F =
1
2
m(v2 − u2), find F when
m = 18.3,v = 12.7 and u = 8.24
5. The current I amperes flowing in a number
of cells is given by I =
nE
R + nr
. Evaluate the
current when n = 36, E = 2.20, R = 2.80
and r = 0.50
6. The time, t seconds, of oscillation for a
simple pendulum is given by t = 2π
l
g
.
Determine the time when l = 54.32 and
g = 9.81

7. Energy, E joules, is given by the formula
E =
1
2
L I2
. Evaluate the energy when
L = 5.5 and I = 1.2
8. The current I amperes in an a.c. circuit
is given by I =
V
(R2 + X2)
. Evaluate the
current when V =250, R=11.0 and X=16.2
9. Distance s metres is given by the formula
s = ut +
1
2
at2. If u = 9.50, t = 4.60 and
a = −2.50, evaluate the distance.
10. The area, A, of any triangle is given
by A =
√
[s(s − a)(s − b)(s − c)] where
s =
a + b + c
2
. Evaluate the area, given
a = 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 that
v =
ab
c
−
d
e
12. Deduce the following information from the
train timetable shown in Table 4.1.
(a) At what time should a man catch a train
at Fratton to enable him to be in London
Waterloo 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 between
Cosham and Woking is 55miles, calcu-
late the average speed of the train.
(c) A man living at Havant has a meeting
in London at 15.30h. It takes around
25minutes on the underground to reach
his destination from London Waterloo.
What train should he catch from Havant
to comfortably make the meeting?
(d) Nine trains leave Portsmouth harbour
between 12.18h and 13.15h. Which
train should be taken for the shortest
journey time?

Table 4.1 Train timetable from Portsmouth Harbour to London Waterloo
Fratton
Fratton
Hilsea
Hilsea
Cosham
Cosham
Bedhampton
Bedhampton
Havant
Havant
Rowlands Castle
Rowlands Castle
Chichester
Chichester
Barnham
Barnham
Horsham
Horsham
Crawley
Crawley
Three Bridges
Three Bridges
Gatwick Airport
Gatwick Airport
Horley
Horley
Redhill
Redhill
East Croydon
East Croydon
Petersfield
Petersfield
Liss
Liss
Liphook
Liphook
Haslemere
Haslemere
Guildford
Guildford
Portchester
Portchester
Fareham
Fareham
Southampton Central
Southampton Central
Botley
Botley
Hedge End
Hedge End
Eastleigh
Eastleigh
Southampton Airport Parkway
Winchester
Winchester
Micheldever
Micheldever
Basingstoke
Basingstoke
Farnborough
Farnborough
Woking
Woking
Clapham Junction
Vauxhall
Vauxhall
London Waterloo
Clapham Junction
Southampton Airport Parkway
Portsmouth Harbour
S04
dep 12:18SW
12:22GW
12:22GW
12:45SW
12:48
12:50
12:53
12:54
13:03
13:04
13:18
13:31
13:32
13:45
13:47
13:57
13:59
14:23
13:31
13:32
13:45
13:47
13:57
13:59
14:27
13:17
13:18
13:17
13:18
13:30C
13:17
13:03
13:04
13:02
13:04
12:45SW
12:48
12:50
12:53
12:54
12:45SW
12:54SW
13:12SN
13:15SW
13:18
13:20
13:23
13:24
13:33
13:34
13:47
13:48
14:01
14:02
14:15
14:17
14:25
14:26
14:51
13:15
13:16
13:19
13:20
13:29
13:30
13:40
13:41
13:48
13:49
14:16
14:20
14:28
14:29
14:32
14:33
14:37
14:38
14:41
14:41
14:47
14:48
15:00
15:00
15:11C
15:21SW
15:26
15:26
15:31
12:57
12:59
13:02
13:03
13:07
13:07
13:12
13:12
13:17
13:17
13:22
13:23
13:30
13:30
13:34
13:35
13:41
13:42
13:54
14:02
14:02
14:15
14:17
14:30
14:31
14:40
14:41
15:01
15:13
13:53
12:48
12:50
12:53
12:54
12:25
12:27
12:30
12:31
12:25
12:27
12:30
12:31
12:38
12:39
12:38
12:39
12:21
12:24
12:27
12:28
12:32
12:32
12:37
12:37
12:39
12:40
12:46
12:46
12:56
12:57
13:02
13:02
13:09
13:09
12:46
12:47
12:46
12:47
13:08C
13:09SW
13:17
13:18
13:34
13:36
14:12
14:13
14:24
13:00C
13:14C
13:55C
2
14:02SW
13:30SW
13:37
13:47
13:48
14:19
14:21
14:49
13:38
14:11
14:12
14:31
14:32
14:40
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
dep
arr
S03 S08 S02 S03 S04 S04 S01 S02
Saturdays
Portsmouth Harbour - London Waterloo
OUTWARD
Train Alterations
Time Time Time Time Time Time Time Time Time
Portsmouth & Southsea
Portsmouth & Southsea
13:20SW
R 13:36SW
R
14:17SW
R
14:25
14:26
14:51
14:11C

Chapter 5
Percentages
5.1 Introduction
Percentages are used to give a common standard. The
use of percentages is very common in many aspects
of commercial life, as well as in engineering. Interest
rates, sale reductions, pay rises, exams and VAT are all
examples of situations in which percentages are used.
For this chapter you will need to know about decimals
and 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% interest
rate for a year, it will cost you 6.5% of the amount
borrowed to do so,which will need to berepaid along
with the original money you borrowed. If you repay
the loan in 1 year, how much interest will you have
paid?
• A pair of trainers in a shop cost £60. They are adver-
tised in a sale as 20% off. How much will you
pay?
• If you earn £20000 p.a. and you receive a 2.5%
pay rise, how much extra will you have to spend the
following year?
• A book costing £18 can be purchased on the internet
for 30% less. What will be its cost?
When we have completed his chapter on percentages
you will be able to understand how to perform the above
calculations.
Percentages are fractions having 100 as their denom-
inator. For example, the fraction
40
100
is written as 40%
and is read as ‘forty per cent’.
The easiest way to understand percentages is to go
through some worked examples.
5.2 Percentage calculations
5.2.1 To convert a decimal to a percentage
A decimal number is converted to a percentage by
multiplying by 100.
Problem 1. Express 0.015 as a percentage
To express a decimal number as a percentage, merely
multiply by 100, i.e.
0.015 = 0.015 × 100%
= 1.5%
Multiplying a decimal number by 100 means moving
the decimal point 2 places to the right.
Problem 2. Express 0.275 as a percentage
0.275 = 0.275 × 100%
= 27.5%
5.2.2 To convert a percentage to a decimal
A percentage is converted to a decimal number by
dividing by 100.
Problem 3. Express 6.5% as a decimal number
6.5% =
6.5
100
= 0.065
Dividing by 100 means moving the decimal point 2
places to the left.
DOI: 10.1016/B978-1-85617-697-2.00005-3

Problem 4. Express 17.5% as a decimal number
17.5% =
17.5
100
= 0.175
5.2.3 To convert a fraction to a percentage
A fraction is converted to a percentage by multiplying
by 100.
Problem 5. Express
5
8
as a percentage
5
8
=
5
8
× 100% =
500
8
%
= 62.5%
Problem 6. Express
5
19
as a percentage, correct
to 2 decimal places
5
19
=
5
19
× 100%
=
500
19
%
= 26.3157889...by calculator
= 26.32% correct to 2 decimal places
Problem 7. In two successive tests a student
gains marks of 57/79 and 49/67. Is the second
mark better or worse than the ﬁrst?
57/79 =
57
79
=
57
79
× 100% =
5700
79
%
49/67 =
49
67
=
49
67
× 100% =
4900
67
%
Hence, the second test is marginally better than the
ﬁrst test. This question demonstrates how much easier
it is to compare two fractions when they are expressed
as percentages.
5.2.4 To convert a percentage to a fraction
A percentage is converted to a fraction by dividing by
100 and then, by cancelling, reducing it to its simplest
form.
Problem 8. Express 75% as a fraction
75% =
75
100
=
3
4
The fraction
75
100
is reduced to its simplest form by can-
celling, i.e. dividing both numerator and denominator
by 25.
Problem 9. Express 37.5% as a fraction
37.5% =
37.5
100
=
375
1000
by multiplying both numerator and
denominator by 10
=
15
40
by dividing both numerator and
denominator by 25
=
3
8
by dividing both numerator and
denominator by 5
PracticeExercise 21 Percentages (answers
on page 342)
In problems 1 to 5, express the given numbers as
percentages.
1. 0.0032 2. 1.734
3. 0.057 4. 0.374
5. 1.285
6. Express 20% as a decimal number.
7. Express 1.25% as a decimal number.
8. Express
11
16
as a percentage.
9. Express
5
13
as a percentage, correct to 3
decimal places.

Percentages 35
10. Express as percentages, correct to 3 significant
figures,
(a)
7
33
(b)
19
24
(c) 1
11
16
11. Place the following in order of size, the small-
est first, expressing each as a percentage
(a)
12
21
(b)
9
17
(c)
5
9
(d)
6
11
12. Express 65% as a fraction in its simplest form.
13. Express 31.25% as a fraction in its simplest
form.
form.
15. Evaluate A to J in the following table.
Decimal number Fraction Percentage
0.5 A B
C
1
4
D
E F 30
G
3
5
H
I J 85
5.3 Further percentage calculations
5.3.1 Finding a percentage of a quantity
To find apercentageofaquantity,convert thepercentage
to a fraction (by dividing by 100) and remember that ‘of’
means multiply.
Problem 10. Find 27% of £65
27% of £65 =
27
100
× 65
= £17.55 by calculator
Problem 11. In a machine shop, it takes 32
minutes to machine a certain part. Using a new tool,
the time can be reduced by 12.5%. Calculate the
new time taken
12.5% of 32 minutes =
12.5
100
× 32
= 4 minutes
Hence, new time taken = 32 − 4 = 28 minutes.
Alternatively, if the time is reduced by 12.5%, it now
takes 100% − 12.5% = 87.5% of the original time, i.e.
87.5% of 32 minutes =
87.5
100
× 32
= 28 minutes
Problem 12. A 160 GB iPod is advertised as
costing £190 excluding VAT. If VAT is added at
17.5%, what will be the total cost of the iPod?
VAT = 17.5% of £190 =
17.5
100
× 190 = £33.25
Total cost of iPod = £190 + £33.25 = £223.25
A quicker method to determine the total cost is: 1.175 ×
£190 = £223.25
5.3.2 Expressing one quantity as a
percentage of another quantity
To express one quantity as a percentage of another quan-
tity, divide the first quantity by the second then multiply
by 100.
Problem 13. Express 23cm as a percentage of
72cm, correct to the nearest 1%
23cm as a percentage of 72cm =
23
72
× 100%
= 31.94444...%
= 32% correct to
the nearest 1%
Problem 14. Express 47 minutes as a percentage
of 2 hours, correct to 1 decimal place
Note that it is essential that the two quantities are in the
same units.
Working in minute units, 2 hours = 2 × 60
= 120 minutes
47 minutes as a percentage
of 120 min =
47
120
× 100%
= 39.2% correct to
1 decimal place

5.3.3 Percentage change
Percentage change is given by
new value−original value
original value
× 100%.
Problem 15. A box of resistors increases in price
from £45 to £52. Calculate the percentage change
in cost, correct to 3 significant figures
% change =
new value − original value
original value
× 100%
=
52 − 45
45
× 100% =
7
45
× 100
= 15.6% = percentage change in cost
Problem 16. A drilling speed should be set to
400rev/min. The nearest speed available on the
machine is 412rev/min. Calculate the percentage
overspeed
% overspeed =
available speed−correct speed
correct speed
×100%
=
412 − 400
400
×100% =
12
400
×100%
= 3%
PracticeExercise 22 Further percentages
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 are
unsatisfactory. Determinethepercentagethat
is 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 88.2g
of nickel,determine the mass of copper in the
block.
7. An athlete runs 5000m in 15 minutes 20
seconds. With intense training, he is able to
reduce this time by 2.5%. Calculate his new
time.
8. A copper alloy comprises 89% copper, 1.5%
iron and the remainder aluminium. Find the
amount of aluminium, in grams, in a 0.8kg
mass of the alloy.
9. A computer is advertised on the internet at
£520, exclusive of VAT. If VAT is payable at
17.5%, what is the total cost ofthe computer?
10. Express 325mm as a percentage of 867mm,
11. A child sleeps on average 9 hours 25 minutes
per day. Express this as a percentage of the
whole day, correct to 1 decimal place.
12. Express 408g as a percentage of 2.40kg.
13. When signing a new contract, a Premiership
footballer’s pay increases from £15500 to
£21500 per week. Calculate the percentage
pay increase, correct to 3 significant figures.
14. A metal rod 1.80m long is heated and its
length expands by 48.6mm. Calculate the
percentage increase in length.
15. 12.5% of a length of wood is 70cm. What is
the full length?
16. A metal rod, 1.20m long, is heated and
its length expands by 42mm. Calculate the
percentage increase in length.
5.4 More percentage calculations
5.4.1 Percentage error
Percentage error =
error
correct value
× 100%
Problem 17. The length of a component is
measured incorrectly as 64.5mm. The actual length

Percentages 37
is 63mm. What is the percentage error in the
measurement?
% error =
error
correct value
× 100%
=
64.5 − 63
63
× 100%
=
1.5
63
× 100% =
150
63
%
= 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 in
an electrical circuit is calculated as 50V using
Ohm’s law. When measured, the actual voltage is
50.4V. Calculate, correct to 2 decimal places, the
percentage error in the calculation
% error =
error
correct value
× 100%
=
50.4 − 50
50.4
× 100%
=
0.4
50.4
× 100% =
40
50.4
%
= 0.79%
The percentage error in the calculation is 0.79% too
low, which is sometimes written as −0.79% error.
5.4.2 Original value
Original value =
new value
100 ± % change
× 100%
Problem 19. A man pays £149.50 in a sale for a
DVD player which is labelled ‘35% off’. What was
the original price of the DVD player?
In this case, it is a 35% reduction in price, so we
use
new value
100 − % change
× 100, i.e. a minus sign in the
denominator.
Original price =
new value
100 − % change
× 100
=
149.5
100 − 35
× 100
=
149.5
65
× 100 =
14950
65
= £230
Problem 20. A couple buys a flat and make an
18% profit by selling it 3 years later for £153400.
Calculate the original cost of the house
In this case, it is an 18% increase in price, so we
use
new value
100 + % change
× 100, i.e. a plus sign in the
denominator.
Original cost =
new value
100 + % change
× 100
=
153400
100 + 18
× 100
=
153400
118
× 100 =
15340000
118
= £130 000
Problem 21. An electrical store makes 40% profit
on each widescreen television it sells. If the selling
price of a 32 inch HD television is £630, what was
the cost to the dealer?
In this case, it is a 40% mark-up in price, so we
use
new value
100 + % change
× 100, i.e. a plus sign in the
denominator.
Dealer cost =
new value
100 + % change
× 100
=
630
100 + 40
× 100
=
630
140
× 100 =
63000
140
= £450
The dealer buys from the manufacturer for £450 and
sells to his customers for £630.
5.4.3 Percentage increase/decrease and
interest
New value =
100 + % increase
100
× original value
Problem 22. £3600 is placed in an ISA account
which pays 6.25% interest per annum. How much is
the investment worth after 1 year?

Value after 1 year =
100 + 6.25
100
× £3600
=
106.25
100
× £3600
= 1.0625 × £3600
= £3825
Problem 23. The price of a fully installed
combination condensing boileris increased by 6.5%.
It originally cost £2400. What is the new price?
New price =
100 + 6.5
100
× £2,400
=
106.5
100
× £2,400 = 1.065 × £2,400
= £2,556
PracticeExercise 23 Further percentages
1. A machine part has a length of 36mm. The
length 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 from
450μA to 531μA. Determine the percentage
increase in the current.
3. In a shoe shop sale, everything is advertised
as ‘40% off’. If a lady pays £186 for a pair of
Jimmy Choo shoes, what was their original
price?
4. Over a four year period a family home
increases 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 on
all its products. What price does the retailer
pay for a dishwasher which is sold for £351?
6. The cost of a sports car is £23500 inclusive
of VAT at 17.5%. What is the cost of the car
without the VAT added?
7. £8000 is invested in bonds at a building soci-
ety which is offering a rate of 6.75% per
annum. Calculate the value of the investment
after 2 years.
8. An electrical contractor earning £36000 per
annum receives a pay rise of 2.5%. He
pays 22% of his income as tax and 11%
on National Insurance contributions. Calcu-
late the increase he will actually receive per
month.
9. Five mates enjoy a meal out. With drinks, the
total bill comes to £176. They add a 12.5%
tip and divide the amount equally between
them. How much does each pay?
10. In December a shop raises the cost of a 40
inch LCD TV costing £920 by 5%. It does
not sell and in its January sale it reduces
the TV by 5%. What is the sale price of
the 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 the
business?
12. A drilling machine should be set to
250rev/min. The nearest speed available on
the machine is 268rev/min. Calculate the
percentage overspeed.
13. Two kilograms of a compound contain 30%
of element A, 45% of element B and 25% of
element C. Determine the masses of the three
elements present.
14. A concrete mixture contains seven parts by
volume of ballast, four parts by volume of
sand and two parts by volume of cement.
Determine the percentage of each of these
three constituents correct to the nearest 1%
and the mass of cement in a two tonne dry
mix, correct to 1 significant figure.
15. In a sample of iron ore, 18% is iron. How
much ore is needed to produce 3600kg of
iron?
16. A screw’s dimension is 12.5 ± 8%mm. Cal-
culate the maximum and minimum possible
length of the screw.
17. The output power of an engine is 450kW. If
theefficiency oftheengineis75%,determine
the power input.

Revision Test 2 : Decimals, calculators and percentages
This assignment covers the material contained in Chapters 3–5. The marks available are shown in brackets at the
1. Convert 0.048 to a proper fraction. (2)
2. Convert 6.4375 to a mixed number. (3)
3. Express
9
32
as a decimal fraction. (2)
4. Express 0.0784 correct to 2 decimal places. (2)
figures. (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 decimal
place. (3)
9. Evaluate 5.2
·
× 12 (2)
10. Evaluate the following, correct to 4 significant
figures: 3.32 − 2.73 + 1.84 (3)
11. Evaluate
√
6.72 −
3
√
2.54 correct to 3 decimal
places. (3)
12. Evaluate
1
0.0071
−
1
0.065
figures. (2)
13. The potential difference, V volts, available at bat-
tery terminals is given by V = E − Ir. Evaluate
V when E = 7.23, I = 1.37 and r = 3.60 (3)
14. Evaluate
4
9
+
1
5
−
3
8
as a decimal, correct to 3
significant figures. (3)
15. Evaluate
16 × 10−6 × 5 × 109
2 × 107
in engineering
form. (2)
16. Evaluate resistance, R, given
1
R
=
1
R1
+
1
R2
+
1
R3
when R1 = 3.6 k , R2 = 7.2 k and
R3 = 13.6 k . (3)
17. Evaluate 6
2
7
− 4
5
9
as a mixed number and as a
decimal, correct to 3 decimal places. (3)
18. Evaluate, correct to 3 decimal places:
2e1.7 × 3.673
4.61 ×
√
3π
(3)
19. If a = 0.270,b = 15.85,c = 0.038,d = 28.7 and
e = 0.680, evaluate v correct to 3 significant
figures, given that v =
ab
c
−
d
e
(4)
20. Evaluate the following, each correct to 2 decimal
places.
(a)
36.22
× 0.561
27.8 × 12.83
3
(b)
14.692
√
17.42 × 37.98
(4)
21. If 1.6km = 1mile, determine the speed of
45miles/hour in kilometres per hour. (2)
22. The area A of a circle is given by A = πr2. Find
the area of a circle of radius r = 3.73cm, correct
to 2 decimal places. (3)
23. Evaluate B, correct to 3 significant figures, when
W = 7.20,v = 10.0 and g = 9.81, given that
B =
Wv2
2g
(3)
form. (3)
25. 12.5% of a length of wood is 70cm. What is the
full length? (3)
26. A metal rod, 1.20m long, is heated and its length
expands by 42mm. Calculate the percentage inc-
rease in length. (2)
27. A man buys a house and makes a 20% profit when
he sells it three years later for £312000. What did
he pay for it originally? (3)

Chapter 6
Ratio and proportion
6.1 Introduction
Ratio is a way of comparing amounts of something; it
shows how much bigger one thing is than the other.
Some practical examples include mixing paint, sand
and cement, or screen wash. Gears, map scales, food
recipes, scale drawings and metal alloy constituents all
use ratios.
Two quantities are in direct proportion when they
increase or decrease in the same ratio. There are sev-
eral practical engineering laws which rely on direct
proportion. Also, calculating currency exchange rates
and converting imperial to metric units rely on direct
proportion.
Sometimes, as one quantity increases at a particular
rate, another quantity decreases at the same rate; this is
called inverse proportion. For example, the time taken
to do a job is inversely proportional to the number of
people in a team: double the people, half the time.
When we have completed this chapter on ratio and
proportion you will be able to understand, and confi-
dently perform, calculations on the above topics.
For this chapter you will need to know about decimals
and fractions and to be able to use a calculator.
6.2 Ratios
Ratios are generally shown as numbers separated by a
colon (:) so the ratio of 2 and 7 is written as 2:7 and we
read it as a ratio of ‘two to seven.’
Some practical examples which are familiar include:
• Mixing 1 measure of screen wash to 6 measures of
water; 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., the
ratio of red to white paint is 3:1
Ratio is the number of parts to a mix. The paint mix is
4 parts total, with 3 parts red and 1 part white. 3 parts
red paint to 1 part white paint means there is
3
4
red paint to
1
4
white paint
Here are some worked examples to help us understand
more about ratios.
Problem 1. In a class, the ratio of female to male
students is 6:27. Reduce the ratio to its simplest
form
(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 which
means for every 2 females in the class there are 9 male
students.
Problem 2. A gear wheel having 128 teeth is in
mesh with a 48-tooth gear. What is the gear ratio?
Gear ratio = 128:48
A ratio can be simplified by finding common factors.
(i) 128 and 48 can both be divided by 2, i.e. 128:48
is the same as 64:24
(ii) 64 and 24 can both be divided by 8, i.e. 64:24 is
the same as 8:3
(iii) There is no number that divides completely into
both 8 and 3 so 8:3 is the simplest ratio, i.e. the
gear ratio is 8:3
DOI: 10.1016/B978-1-85617-697-2.00006-5

Ratio and proportion 41
Thus, 128:48 is equivalent to 64:24 which is equivalent
to 8:3 and 8:3 is the simplest form.
Problem 3. A wooden pole is 2.08m long. Divide
it in the ratio of 7 to 19
(i) Since the ratio is 7:19, the total number of parts
is 7 + 19 = 26 parts.
(ii) 26 parts corresponds to 2.08m = 208cm, hence,
1 part corresponds to
208
26
= 8.
(iii) Thus, 7 parts corresponds to 7 × 8 = 56 cm and
19 parts corresponds to 19 × 8 = 152 cm.
Hence, 2.08m divides in the ratio of 7:19 as 56 cm
to 152 cm.
(Check: 56 + 152must add up to 208, otherwise an error
would have been made.)
Problem 4. In a competition, prize money of
£828 is to be shared among the first three in the
ratio 5:3:1
(i) Since the ratio is 5:3:1 the total number of parts
is 5 + 3 + 1 = 9 parts.
(ii) 9 parts corresponds to £828.
(iii) 1 part corresponds to
828
9
= £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 to
828, otherwise an error would have been made.)
Problem 5. A map scale is 1:30000. On the map
the distance between two schools is 6cm.
Determine the actual distance between the schools,
giving the answer in kilometres
Actual distance between schools
= 6 × 30000 cm = 180000 cm
=
180,000
100
m = 1800 m
=
1800
1000
m = 1.80 km
(1mile ≈ 1.6km, hence the schools are just over 1mile
apart.)
PracticeExercise 24 Ratios (answers on
page 342)
1. In a box of 333 paper clips, 9 are defective.
Express the number of non-defective paper
clipsasaratio ofthenumberofdefectivepaper
clips, in its simplest form.
2. A gear wheel having 84 teeth is in mesh with
a 24-tooth gear. Determine the gear ratio in its
simplest form.
3. In a box of 2000 nails, 120 are defective.
Express the number of non-defective nails as
a ratio of the number of defective ones, in its
simplest form.
4. A metal pipe 3.36m long is to be cut into two
in the ratio 6 to 15. Calculate the length of each
piece.
5. The instructions for cooking a turkey say that
it needs to be cooked 45 minutes for every
kilogram. How long will it take to cook a 7kg
turkey?
6. In a will, £6440 is to be divided among three
beneficiaries in the ratio 4:2:1. Calculate the
amount each receives.
7. A local map has a scale of 1:22500. The dis-
tance between two motorways is 2.7km. How
far are they apart on the map?
8. Prize money in a lottery totals £3801 and is
shared 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 its
simplest form
(i) Changing both quantities to the same units, i.e. to
pence, gives a ratio of 45:765
(ii) Dividing both quantities by 5 gives
45:765 ≡ 9:153
(iii) Dividing both quantities by 3 gives
9:153 ≡ 3:51
(iv) Dividing both quantities by 3 again gives
3:51 ≡ 1:17

Thus, 45 p as a ratio of £7.65 is 1:17
45:765,9:153,3:51 and 1:17 are equivalent ratios
and 1:17 is the simplest ratio.
Problem 7. A glass contains 30 ml of whisky
which is 40% alcohol. If 45ml of water is added and
the mixture stirred, what is now the alcohol content?
(i) The 30 ml of whisky contains 40%
alcohol =
40
100
× 30 = 12 ml.
(ii) After 45ml of water is added we have 30 + 45
= 75ml of fluid, of which alcohol is 12 ml.
(iii) Fraction of alcohol present =
12
75
(iv) Percentage of alcohol present =
12
75
× 100%
= 16%.
Problem 8. 20 tonnes of a mixture of sand and
gravel is 30% sand. How many tonnes of sand must
be added to produce a mixture which is 40% gravel?
(i) Amount of sand in 20 tonnes = 30% of 20 t
=
30
100
× 20 = 6t.
(ii) If the mixture has 6t of sand then amount of
gravel = 20 − 6 = 14t.
(iii) We want this 14t of gravel to be 40% of the
new mixture. 1% would be
14
40
t and 100% of the
mixture would be
14
40
× 100 t = 35t.
(iv) If there is 14t of gravel then amount of sand
= 35 − 14 = 21t.
(v) We already have 6t of sand, so amount of sand
to be added to produce a mixture with 40%
gravel = 21 − 6 = 15t.
(Note 1tonne = 1000 kg.)
PracticeExercise 25 Further ratios
1. Express 130 g as a ratio of 1.95kg.
2. In a laboratory, acid and water are mixed in the
ratio 2:5. How much acid is needed to make
266ml of the mixture?
3. A glass contains 30 ml of gin which is 40%
alcohol. If 18ml of water is added and the
mixture stirred, determine the new percentage
alcoholic content.
4. A wooden beam 4m long weighs 84kg. Deter-
mine the mass of a similar beam that is 60 cm
long.
5. An alloy is made up of metals P and Q in the
ratio 3.25:1 by mass. How much of P has to
be added to 4.4kg of Q to make the alloy?
6. 15000 kg of a mixture of sand and gravel is
20% sand. Determine the amount of sand that
must be added to produce a mixture with 30%
gravel.
6.3 Direct proportion
Two quantities are in direct proportion when they
increase or decrease in the same ratio. For example,
if 12 cans of lager have a mass of 4kg, then 24 cans of
lager will have a mass of 8kg; i.e., if the quantity of cans
doubles then so does the mass. This is direct proportion.
In the previous section we had an example of mixing
1 shovel of cement to 4 shovels of sand; i.e., the ratio
of cement to sand was 1:4. So, if we have a mix of 10
shovels ofcement and 40 shovels of sand and we wanted
to double the amount of the mix then we would need
to double both the cement and sand, i.e. 20 shovels of
cement and 80 shovels of sand. This is another example
of direct proportion.
Here are three laws in engineering which involve direct
proportion:
(a) Hooke’s law states that, within the elastic limit of
a material, the strain ε produced is directly propor-
tional to the stress σ producing it, i.e. ε ∝ σ (note
than ‘∝’ means ‘is proportional to’).
(b) Charles’s law states that, for a given mass of gas
at constant pressure, the volume V is directly pro-
portional to its thermodynamic temperature T , i.e.
V ∝ T.
(c) Ohm’s law states that the current I flowing
through a fixed resistance is directly proportional
to the applied voltage V , i.e. I ∝ V .

more 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 costs
7.80
3
= £2.60
Hence, 7 light bulbs cost 7 × £2.60 = £18.20
Problem 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 costs
59.92
56
= £1.07
Hence, 32 litres cost 32 × 1.07 = £34.24
Problem 11. Hooke’s law states that stress, σ, is
directly proportional to strain, ε, within the elastic
limit of a material. When, for mild steel, the stress
is 63MPa, the strain is 0.0003. Determine (a) the
value of strain when the stress is 42 MPa, (b) the
value of stress when the strain is 0.00072
(a) Stress is directly proportional to strain.
(i) When the stress is 63MPa, the strain is
0.0003
(ii) Hence, a stress of 1MPa corresponds to a
strain of
0.0003
63
(iii) Thus,the value of strain when the stress is
42 MPa =
0.0003
63
× 42 = 0.0002
(b) Strain is proportional to stress.
(i) When the strain is 0.0003, the stress is
63MPa.
(ii) Hence, a strain of 0.0001 corresponds to
63
3
MPa.
(iii) Thus,the value of stress when the strain is
0.00072 =
63
3
× 7.2 = 151.2 MPa.
Problem 12. Charles’s law states that for a given
mass of gas at constant pressure, the volume is
directly proportional to its thermodynamic
temperature. A gas occupies a volume of 2.4litres
at 600 K. Determine (a) the temperature when the
volume 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 a
temperature of
600
2.4
K.
(iii) Thus,the temperature when the volume is
3.2 litres =
600
2.4
× 3.2 = 800K.
(b) Temperature is proportional to volume.
(i) When the temperature is 600 K, the volume
is 2.4litres.
(ii) Hence, a temperature of 1K corresponds to
a volume of
2.4
600
litres.
(iii) Thus, the volume at a temperature of
540 K =
2.4
600
× 540 = 2.16 litres.
PracticeExercise 26 Direct proportion
1. 3 engine parts cost £208.50. Calculate the cost
of 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 is
57.6kg. Determine the mass of 550 such
bricks.
4. A simple machine has an effort:load ratio of
3:37. Determine the effort, in grams, to lift a
load of 5.55kN.
5. If 16 cans of lager weighs 8.32 kg, what will
28 cans weigh?
6. Hooke’s law states that stress is directly pro-
portional to strain within the elastic limit of
a material. When, for copper, the stress is
60 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

7. Charles’s law states that volume is directly
proportional to thermodynamic temperature
for 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 the
temperature is 396K.
Here are some further worked examples on direct
proportion.
Problem 13. Some guttering on a house has to
decline by 3mm for every 70 cm to allow rainwater
to drain. The gutter spans 8.4m. How much lower
should the low end be?
(i) The guttering has to decline in the ratio 3:700 or
3
700
(ii) If d is the vertical drop in 8.4m or 8400 mm, then
the decline must be in the ratio d :8400 or
d
8400
(iii) Now
d
8400
=
3
700
(iv) Cross-multiplyinggives 700×d =8400×3 from
which, d =
8400 × 3
700
i.e. d = 36mm, which is how much the lower end
should be to allow rainwater to drain.
Problem 14. Ohm’s law state that the current
flowing in a fixed resistance is directly proportional
to the applied voltage. When 90 mV is applied
across a resistor the current flowing is 3A.
Determine (a) the current when the voltage is
60 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 a
current of
3
90
A.
(iii) Thus, when the voltage is 60 mV, the
current = 60 ×
3
90
= 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 a
voltage of
90
3
mV = 30 mV.
(iii) Thus, when the current is 4.2 A, the
voltage = 30 × 4.2 = 126mV.
Problem 15. Some approximate imperial to
metric conversions are shown in Table 6.1. Use the
table to determine
(a) the number of millimetres in 12.5inches
(b) a speed of 50 miles per hour in kilometres
per hour
(c) the number of miles in 300 km
(d) the number of kilograms in 20 pounds weight
(e) the number of pounds and ounces in
56kilograms (correct to the nearest ounce)
(f) the number of litres in 24 gallons
(g) the number of gallons in 60 litres
Table 6.1
length 1inch = 2.54cm
1mile = 1.6km
weight 2.2 lb = 1kg
(1lb = 16oz)
capacity 1.76 pints = 1litre
(8 pints = 1 gallon)
(a) 12.5inches = 12.5 × 2.54cm = 31.75cm
31.73cm = 31.75 × 10 mm = 317.5 mm
(b) 50 m.p.h. = 50 × 1.6km/h = 80 km/h
(c) 300 km =
300
1.6
miles = 186.5 miles
(d) 20 lb =
20
2.2
kg = 9.09 kg
(e) 56kg = 56 × 2.2 lb = 123.2 lb
0.2 lb = 0.2 × 16oz = 3.2 oz = 3oz, correct to
the nearest ounce.
Thus, 56kg = 123 lb 3 oz, correct to the nearest
ounce.
(f) 24 gallons = 24 × 8 pints = 192 pints
192 pints =
192
1.76
litres = 109.1 litres

(g) 60 litres = 60× 1.76 pints = 105.6 pints
105.6 pints =
105.6
8
gallons = 13.2 gallons
Problem 16. Currency exchange rates for five
countries are shown in Table 6.2. Calculate
(a) how many euros £55 will buy
(b) the number of Japanese yen which can be
bought for £23
(c) the number of pounds sterling which can be
exchanged for 6405kronor
(d) the number of American dollars which can be
purchased for £92.50
(e) the number of pounds sterling which can be
exchanged for 2925 Swiss francs
Table 6.2
France £1 = 1.25 euros
Japan £1 = 185 yen
Norway £1 = 10.50kronor
Switzerland £1 = 1.95francs
USA £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 = £
6405
10.50
= £610.
(d) £1 = 1.80 dollars, hence
£92.50 = 92.50 × 1.80 dollars = $166.50
(e) £1 = 1.95 Swiss francs, hence
2925 pesetas = £
2925
1.95
= £1500
PracticeExercise 27 Further direct
proportion (answers on page 342)
1. Ohm’s law states that current is proportional
to p.d. in an electrical circuit. When a p.d. of
60 mV is applied across a circuit a current of
24μAflows. Determine(a)thecurrent flowing
when the p.d. is 5V and (b) the p.d. when the
current is 10 mA.
2. The tourist rate for the Swiss franc is quoted in
a newspaper as £1 = 1.92 fr. How many francs
can 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 the
number of poundsand ounces in 38kg (correct
to the nearest ounce).
5. If 1litre = 1.76 pints and 8 pints = 1 gallon,
determine (a) the number of litres in 35 gallons
and (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 a
material. When for brass the stress is 21MPa,
the strain is 0.00025. Determine the stress
when the strain is 0.00035.
7. If 12 inches = 30.48cm, find the number of
millimetres in 23inches.
8. The tourist rate for the Canadian dollar is
quoted in a newspaper as £1 = 1.84fr. How
many Canadian dollars can be purchased for
£550?
6.4 Inverse proportion
Two variables, x and y, are in inverse proportion to one
another if y is proportional to
1
x
, i.e. y α
1
x
or y =
k
x
or
k = xy where k is a constant, called the coefficient of
proportionality.
Inverse proportion means that, as the value ofone vari-
able increases, the value of another decreases, and that
their 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 journey
is completed in20 minutes, then at 60m.p.h. the journey
would 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 is
inversely proportional to the number of people digging.
So, if 4 men take 3 hours to dig a hole, then 2 men

(working at the same rate) would take 6 hours. Half the
men, 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 four
designers would take a year to develop an
engineering process. How long would three
designers take?
If 4 designers take 1 year, then 1 designer would take
4 yearsto develop theprocess.Hence, 3 designerswould
take
4
3
years, i.e. 1 year 4 months.
Problem 18. A team of five people can deliver
leaflets to every house in a particular area in four
hours. How long will it take a team of three people?
If 5 people take 4 hours to deliver the leaflets, then
1 person would take 5 × 4 = 20 hours. Hence, 3 peo-
ple would take
20
3
hours, i.e. 6
2
3
hours, i.e. 6 hours
40 minutes.
Problem 19. The electrical resistance R of a
piece of wire is inversely proportional to the
cross-sectional area A. When A = 5mm2,
R = 7.02 ohms. Determine (a) the coefficient of
proportionality and (b) the cross-sectional area
when the resistance is 4ohms
(a) Rα
1
A
, i.e. R =
k
A
or k = RA. Hence, when
R = 7.2 and A = 5, the
coefficient of proportionality,k = (7.2)(5) = 36
(b) Since k = RA then A =
k
R
. Hence, when R = 4,
the cross sectional area, A =
36
4
= 9 mm2
Problem 20. Boyle’s law states that, at constant
temperature, the volume V of a fixed mass of gas is
inversely proportional to its absolute pressure p. If
a gas occupies a volume of 0.08m3
at a pressure of
1.5 × 106
pascals, determine (a) the coefficient of
proportionality and (b) the volume if the pressure is
changed to 4 × 106 pascals
(a) V ∝
1
p
i.e. V =
k
p
or k = pV . Hence, the
coefficient of proportionality, k
= (1.5 × 106
)(0.08) = 0.12 × 106
(b) Volume,V =
k
p
=
0.12 × 106
4 × 106
= 0.03m3
PracticeExercise 28 Further inverse
proportion (answers on page 342)
1. A 10 kg bag of potatoes lasts for a week with a
family of 7 people. Assuming all eat the same
amount, how longwill the potatoeslast ifthere
are only two in the family?
2. If 8 men take 5 days to build a wall, how long
would it take 2 men?
3. If y is inversely proportional to x and
y = 15.3 when x = 0.6, determine (a) the
coefficient of proportionality, (b) the value of
y when x is 1.5 and (c) the value of x when y
is 27.2
4. A car travelling at 50 km/h makes a journey in
70 minutes. How long will the journey take at
70 km/h?
5. Boyle’s law states that, for a gas at constant
temperature, the volume of a fixed mass of
gas is inversely proportional to its absolute
pressure. If a gas occupies a volume of 1.5m3
at a pressure of 200 × 103 pascals, determine
(a) the constant of proportionality, (b) the
volume when the pressure is 800 × 103 pas-
cals and (c) the pressure when the volume is
1.25m3.

Chapter 7
Powers, roots and laws of
indices
7.1 Introduction
The manipulationof powers and rootsis a crucial under-
lying skill needed in algebra. In this chapter, powers and
roots of numbers are explained, together with the laws
of indices.
Many worked examplesareincluded to help understand-
ing.
7.2 Powers and roots
7.2.1 Indices
The 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 is
called 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 special
names; i.e. 2 is called ‘squared’ and 3 is called ‘cubed’.
Thus,
42 is called ‘four squared’ rather than ‘4 to the power
of 2’ and
53 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 multiplied
by itself 6 times), and 2 × 2 × 2 × 2 × 2 × 2 = 64
i.e. 26 = 64
(b) 34 means 3 × 3 × 3 × 3 (i.e. 3 multiplied by itself
4 times), and 3 × 3 × 3 × 3 = 81
i.e. 34 = 81
Problem 2. Change the following to index form:
(a) 32 (b) 625
(a) (i) To express 32 in itslowest factors, 32 is initially
divided 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 625
is not divisible by 3 either. The next prime
number 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

× 22
33
× 22
= 3 × 3 × 3 × 2× 2
= 27 × 4
= 108
7.2.2 Square roots
When a number is multiplied by itself the product is
called 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 the
base 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 index
form as 9
1
2 .
9
1
2 ≡
√
9 = ±3
Problem 4. Evaluate
32 × 23 ×
√
36
√
16 × 4
taking only
positive square roots
32 × 23 ×
√
36
√
16 × 4
=
3 × 3 × 2× 2 × 2 × 6
4 × 4
=
9 × 8 × 6
16
=
9 × 1 × 6
2
=
9 × 1 × 3
1
by cancelling
= 27
Problem 5. Evaluate
104 ×
√
100
103
taking the
positive square root only
104 ×
√
100
103
=
10 × 10× 10 × 10 × 10
10 × 10 × 10
=
1 × 1 × 1 × 10× 10
1 × 1 × 1
by cancelling
=
100
1
= 100
PracticeExercise 29 Powers and roots
Evaluate the following without the aid of a calcu-
lator.
1. 33 2. 27
3. 105 4. 24 × 32 × 2 ÷ 3
5. Change 16 to 6. 25
1
2
index form.
7. 64
1
2 8.
105
103
9.
102
× 103
105
10.
25
× 64
1
2 × 32
√
144 × 3
taking positive
square roots only.
7.3 Laws of indices
There are six laws of indices.
(1) From earlier, 22 × 23 = (2 × 2) × (2 × 2 × 2)
= 32
= 25
Hence, 22 × 23 = 25
or 22
× 23
= 22+3
This is the ﬁrst law of indices, which demonstrates
that when multiplying two or more numbers
having the same base, the indices are added.
(2)
25
23
=
2 × 2 × 2 × 2 × 2
2 × 2 × 2
=
1 × 1 × 1 × 2× 2
1 × 1 × 1
=
2 × 2
1
= 4 = 22
Hence,
25
23
= 22
or
25
23
= 25−3
This is the second law of indices, which demon-
strates that when dividing two numbers having
the same base, the index in the denominator is
subtracted from the index in the numerator.
(3) (35)2 = 35×2 = 310 and (22)3 = 22×3 = 26
This is the third law of indices, which demon-
strates that when a number which is raised to
a power is raised to a further power, the indices
are multiplied.

Powers, roots and laws of indices 49
(4) 30
= 1 and 170
= 1
This is the fourth law of indices, which states that
when a number has an index of 0, its value is 1.
(5) 3−4 =
1
34
and
1
2−3
= 23
This isthe ﬁfth law of indices, which demonstrates
that a number raised to a negative power is the
reciprocal of that number raised to a positive
power.
(6) 8
2
3 =
3
√
82 = (2)2 = 4 and
25
1
2 =
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 is
the root of the number and the numerator is the
power.
Here are some worked examples using the laws of
indices.
Problem 6. Evaluate in index form 53 × 5 × 52
53
× 5 × 52
= 53
× 51
× 52
(Note that 5 means 51
)
= 53+1+2
from law (1)
= 56
Problem 7. Evaluate
35
34
35
34
= 35−4
from law (2)
= 31
= 3
Problem 8. Evaluate
24
24
24
24
= 24−4
from law (2)
= 20
But
24
24
=
2 × 2 × 2 × 2
2 × 2 × 2 × 2
=
16
16
= 1
Hence, 20
= 1 from law (4)
Any number raised to the power of zero equals 1. For
example, 60
= 1,1280
= 1,137420
= 1 and so on.
Problem 9. Evaluate
3 × 32
34
3 × 32
34
=
31 × 32
34
=
31+2
34
=
33
34
= 33−4
= 3−1
from laws (1) and (2)
But
33
34
=
3 × 3 × 3
3 × 3 × 3 × 3
=
1 × 1 × 1
1 × 1 × 1 × 3
(by cancelling)
=
1
3
Hence,
3 × 32
34
= 3−1
=
1
3
from law (5)
Similarly, 2−1
=
1
2
,2−5
=
1
25
,
1
54
= 5−4
and so on.
103 × 102
108
103 × 102
108
=
103+2
108
=
105
108
from law (1)
= 105−8
= 10−3
from law (2)
=
1
10+3
=
1
1000
from law (5)
Hence,
103 × 102
108
= 10−3
=
1
1000
= 0.001
Understanding powers of ten is important, especially
when dealing with preﬁxes in Chapter 8. For example,
102
= 100,103
= 1000,104
= 10000,
105
= 100000,106
= 1000000
10−1
=
1
10
= 0.1,10−2
=
1
102
=
1
100
= 0.01
and so on.
Problem 11. Evaluate (a) 52 × 53 ÷ 54
(b) (3 × 35
) ÷ (32
× 33
)
From laws (1) and (2):
(a) 52 × 53 ÷ 54 =
52 × 53
54
=
5(2+3)
54
=
55
54
= 5(5−4) = 51 = 5

(b) (3 × 35) ÷ (32 × 33) =
3 × 35
32 × 33
=
3(1+5)
3(2+3)
=
36
35
= 36−5 = 31 = 3
Problem 12. Simplify (a) (23)4 (b) (32)5,
expressing the answers in index form
From law (3):
(a) (23)4 = 23×4 = 212
(b) (32)5 = 32×5 = 310
Problem 13. Evaluate:
(102)3
104 × 102
From laws (1) to (4):
(102)3
104 × 102
=
10(2×3)
10(4+2)
=
106
106
= 106−6
= 100
= 1
Problem 14. Find the value of (a)
23 × 24
27 × 25
(b)
(32)3
3 × 39
From the laws of indices:
(a)
23 × 24
27 × 25
=
2(3+4)
2(7+5)
=
27
212
= 27−12
= 2−5 =
1
25
=
1
32
(b)
(32)3
3 × 39
=
32×3
31+9
=
36
310
= 36−10
= 3−4 =
1
34
=
1
81
Problem 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 root
of 16 is found ﬁrst or whether 16 cubed is found
ﬁrst – the same answer will result.)
(c) 272/3 =
3
√
272 = (3)2 = 9
(d) 9−1/2 =
1
91/2
=
1
√
9
=
1
±3
= ±
1
3
PracticeExercise 30 Laws of indices
Evaluate the following without the aid of a
calculator.
1. 22 × 2 × 24 2. 35 × 33 × 3
in index form
3.
27
23
4.
33
35
5. 70 6.
23 × 2 × 26
27
7.
10 × 106
105
8. 104 ÷ 10
9.
103 × 104
109
10. 56 × 52 ÷ 57
11. (72)3 in index form 12. (33)2
13.
37 × 34
35
in 14.
(9 × 32)3
(3 × 27)2
in
index form index form
15.
(16 × 4)2
(2 × 8)3
16.
5−2
5−4
17.
32 × 3−4
33
18.
72 × 7−3
7 × 7−4
19.
23 × 2−4 × 25
2 × 2−2 × 26
20.
5−7 × 52
5−8 × 53
Here are some further worked examples using the laws
of indices.
33 × 57
53 × 34
The laws of indices only apply to terms having the
same base. Grouping terms having the same base and
then applying the laws of indices to each of the groups
independently gives
33 × 57
53 × 34
=
33
34
×
57
53
= 3(3−4)
× 5(7−3)
= 3−1
× 54
=
54
31
=
625
3
= 208
1
3

Powers, roots and laws of indices 51
23 × 35 × (72)2
74 × 24 × 33
23 × 35 × (72)2
74 × 24 × 33
= 23−4
× 35−3
× 72×2−4
= 2−1
× 32
× 70
=
1
2
× 32
× 1 =
9
2
= 4
1
2
41.5 × 81/3
22 × 32−2/5
41.5
= 43/2
=
√
43 = 23
= 8, 81/3
=
3
√
8 = 2,
22
= 4, 32−2/5
=
1
322/5
=
1
5
√
322
=
1
22
=
1
4
Hence,
41.5 × 81/3
22 × 32−2/5
=
8 × 2
4 ×
1
4
=
16
1
= 16
Alternatively,
41.5 × 81/3
22 × 32−2/5
=
[(2)2]3/2 × (23)1/3
22 × (25)−2/5
=
23 × 21
22 × 2−2
= 23+1−2−(−2)
= 24
= 16
32 × 55 + 33 × 53
34 × 54
Dividing each termby theHCF(highest common factor)
of the three terms, i.e. 32 × 53, gives
32 × 55 + 33 × 53
34 × 54
=
32 × 55
32 × 53
+
33 × 53
32 × 53
34
× 54
32 × 53
=
3(2−2) × 5(5−3) + 3(3−2) × 50
3(4−2) × 5(4−3)
=
30 × 52 + 31 × 50
32 × 51
=
1 × 25 + 3 × 1
9 × 5
=
28
45
32 × 55
34 × 54 + 33 × 53
To simplify the arithmetic, each term is divided by the
HCF of all the terms, i.e. 32 × 53. Thus,
32 × 55
34 × 54 + 33 × 53
=
32 × 55
32 × 53
34 × 54
32 × 53
+
33 × 53
32 × 53
=
3(2−2) × 5(5−3)
3(4−2) × 5(4−3) + 3(3−2) × 5(3−3)
=
30 × 52
32 × 51 + 31 × 50
=
1 × 52
32 × 5 + 3 × 1
=
25
45 + 3
=
25
48
7−3 × 34
3−2 × 75 × 5−2
expressing the answer in index form with positive
indices
Since 7−3 =
1
73
,
1
3−2
= 32 and
1
5−2
= 52, then
7−3
× 34
3−2 × 75 × 5−2
=
34
× 32
× 52
73 × 75
=
3(4+2) × 52
7(3+5)
=
36 × 52
78
162 × 9−2
4 × 33 − 2−3 × 82
expressing the answer in index form with positive
indices
Expressing the numbers in terms of their lowest prime
numbers gives
162 × 9−2
4 × 33 − 2−3 × 82
=
(24)2 × (32)−2
22 × 33 − 2−3 × (23)2
=
28 × 3−4
22 × 33 − 2−3 × 26
=
28 × 3−4
22 × 33 − 23

Dividing each term by the HCF (i.e. 22
) gives
28 × 3−4
22 × 33 − 23
=
26 × 3−4
33 − 2
=
26
34(33 − 2)
4
3
3
×
3
5
−2
2
5
−3
giving
the answer with positive indices
Raising a fraction to a power means that both the numer-
ator and the denominator of the fraction are raised to that
power, i.e.
4
3
3
=
43
33
A fraction raised to a negative power has the same value
as the inverse of the fraction raised to a positive power.
Thus,
3
5
−2
=
1
3
5
2
=
1
32
52
= 1 ×
52
32
=
52
32
Similarly,
2
5
−3
=
5
2
3
=
53
23
Thus,
4
3
3
×
3
5
−2
2
5
−3
=
43
33
×
52
32
53
23
=
43
33
×
52
32
×
23
53
=
(22)3 × 23
3(3+2) × 5(3−2)
=
29
35 × 5
indices (answers on page 342)
In problems 1 to 4, simplify the expressions given,
expressing the answers in index form and with
positive indices.
1.
33 × 52
54 × 34
2.
7−2 × 3−2
35 × 74 × 7−3
3.
42 × 93
83 × 34
4.
8−2 × 52 × 3−4
252 × 24 × 9−2
In Problems5 to 15,evaluatetheexpressionsgiven.
5.
1
32
−1
6. 810.25
7. 16
−
1
4 8.
4
9
1/2
9.
92 × 74
34 × 74 + 33 × 72
10.
33 × 52
23 × 32 − 82 × 9
11.
33 × 72 − 52 × 73
32 × 5 × 72
12.
(24)2 − 3−2 × 44
23 × 162
13.
1
2
3
−
2
3
−2
3
2
2
14.
4
3
4
2
9
2
15.
(32)3/2 × (81/3)2
(3)2 × (43)1/2 × (9)−1/2

Chapter 8
Units, prefixes and
engineering notation
8.1 Introduction
Of considerable importance in engineering is a knowl-
edge ofunitsof engineering quantities,the preﬁxes used
with units, and engineering notation.
We need to know, for example, that
80kV = 80 × 103 V, which means 80000 volts
and 25mA = 25 × 10−3
A,
which means 0.025 amperes
and 50nF = 50 × 10−9
F,
which means 0.000000050 farads
This is explained in this chapter.
8.2 SI units
The system of units used in engineering and science
is the Système Internationale d’Unités (International
System of Units), usually abbreviated to SI units, and is
based on the metric system. This was introduced in 1960
and has now been adopted by the majority of countries
as the official system of measurement.
The basic seven units used in the SI system are listed in
Table 8.1 with their symbols.
There are, of course, many units other than these seven.
These other units are called derived units and are
defined 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 and
time.
Table 8.1 Basic SI units
Quantity Unit Symbol
Length metre m (1m = 100 cm
= 1000 mm)
Mass kilogramkg (1kg = 1000 g)
Time second s
Electric current ampere A
Thermodynamic
temperature
kelvin K (K = ◦C + 273)
Luminous
intensity
candela cd
Amount of
substance
mole mol
Some derived units are given special names. For exam-
ple, force = mass × acceleration has units of kilogram
metre per second squared, which uses three of the base
units, i.e. kilograms, metres and seconds. The unit of kg
m/s2 is given the special name of a Newton.
Table 8.2 contains a list of some quantities and their
units that are common in engineering.
8.3 Common prefixes
SI units may be made larger or smaller by using prefixes
which denote multiplication or division by a particular
amount.
DOI: 10.1016/B978-1-85617-697-2.00008-9

Table 8.2 Some quantities and their units that are common in engineering
Quantity Unit Symbol
Length metre m
Area square metre m2
Volume cubic metre m3
Mass kilogram kg
Time second s
Electric current ampere A
Speed, velocity metre per second m/s
Acceleration metre per second squared m/s2
Density kilogram per cubic metre kg/m3
Temperature kelvin or Celsius K or ◦C
Angle radian or degree rad or ◦
Angular velocity radian per second rad/s
Frequency hertz Hz
Force newton N
Pressure pascal Pa
Energy, work joule J
Power watt W
Charge, quantity of electricity coulomb C
Electric potential volt V
Capacitance farad F
Electrical resistance ohm
Inductance henry H
Moment of force newton metre Nm
The most common multiples are listed in Table 8.3.
Aknowledgeofindicesisneeded sinceall ofthepreﬁxes
are 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 is
15000000000 hertz,
i.e. 15 gigahertz is written as 15GHz and is equal to 15
thousand million hertz.
(Instead of writing 15000000000 hertz, it is much
neater, takes up less space and prevents errors caused
by having so many zeros, to write the frequency as
15GHz.)
A voltage of 40 MV means 40 × 106 V, which is
40000000 volts,
i.e. 40 megavolts is written as 40MV and is equal to 40
million volts.
An inductance of 12 mH means 12 × 10−3 H or
12
103
H or
12
1000
H, which is 0.012H,
i.e. 12 millihenrys is written as 12mH and is equal to
12 thousandths of a henry.

Units, prefixes and engineering notation 55
Table 8.3 Common SI multiples
Prefix Name Meaning
G giga multiply by 109
i.e. × 1000000000
M mega multiply by 106 i.e. × 1000000
k kilo multiply by 103 i.e. × 1000
m milli multiply by 10−3 i.e. ×
1
103
=
1
1000
= 0.001
μ micro multiply by 10−6 i.e. ×
1
106
=
1
1000000
= 0.000001
n nano multiply by 10−9 i.e. ×
1
109
=
1
1000000000
= 0.000 000001
p pico multiply by 10−12 i.e. ×
1
1012
=
1
1000000000000
=
0.000000000001
A time of 150 ns means 150 × 10−9 s or
150
109
s, which
is 0.000000150s,
i.e. 150 nanoseconds is written as 150ns and is equal to
150 thousand millionths of a second.
A force of 20 kN means 20 × 103 N, which is 20000
newtons,
i.e. 20 kilonewtons is written as 20kN and is equal to
20 thousand newtons.
A charge of 30 μC means 30 × 10−6 C or
30
106
C, which
is 0.000030C,
i.e. 30 microcoulombs is written as 30μC and is equal
to 30 millionths of a coulomb.
A capacitance of 45pF means 45 × 10−12 F or
45
1012
F,
which is 0.000000000045F,
i.e. 45 picofarads is written as 45pF and is equal to 45
million millionths of a farad.
In engineering it is important to understand what such
quantities as 15GHz, 40MV, 12mH, 150ns, 20kN,
30μC and 45pF mean.
PracticeExercise 32 SI units and common
prefixes (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 prefix
milli?

23. What does the prefix p mean?
24. What is thesymbol and meaning ofthe prefix
mega?
8.4 Standard form
A number written with one digit to the left ofthe decimal
point and multiplied by 10 raised to some power is said
to be written in standard form.
For example, 43645 = 4.3645 × 104
in standard form
and 0.0534 = 5.34 × 10−2
in standard form
Problem 1. Express in standard form (a) 38.71
(b) 3746 (c) 0.0124
For a number to be in standard form, it is expressed with
only one digit to the left of the decimal point. Thus,
(a) 38.71 must be divided by 10 to achieve one digit
to the left of the decimal point and it must also be
multiplied by 10 to maintain the equality, i.e.
38.71=
38.71
10
×10=3.871×10 in standard form
(b) 3746 =
3746
1000
× 1000 = 3.746 × 103
in standard
form.
(c) 0.0124 = 0.0124 ×
100
100
=
1.24
100
= 1.24 × 10−2
in 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 =
1.725
100
= 0.01725 (i.e. move the
decimal point 2 places to the left).
(b) 5.491 × 104 = 5.491 × 10000 = 54910 (i.e. move
the 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 3
significant figures, (a)
3
8
(b) 19
2
3
(c) 741
9
16
(a)
3
8
= 0.375, and expressing it in standard form
gives
0.375 = 3.75 × 10−1
(b) 19
2
3
= 19.
.
6 = 1.97 × 10 in standard form, cor-
rect to 3 signiﬁcant figures.
(c) 741
9
16
=741.5625=7.42 × 102
in standard form,
correct to 3 significant figures.
Problem 4. Express the following numbers, given
in 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 =
2.5
10
=
25
100
=
1
4
(b) 6.25 × 10−2 =
6.25
100
=
625
10000
=
1
16
(c) 1.354 × 102 = 135.4 = 135
4
10
= 135
2
5
Problem 5. Evaluate (a) (3.75 × 103)(6 × 104)
(b)
3.5 × 105
7 × 102
, expressing the answers in standard
form
(a) (3.75 × 103)(6 × 104) = (3.75 × 6)(103+4)
= 22.50 × 107
= 2.25 × 108
(b)
3.5 × 105
7 × 102
=
3.5
7
× 105−2 = 0.5 × 103 = 5 × 102
PracticeExercise 33 Standard form
In problems 1 to 5, express in standard form.
1. (a) 73.9 (b) 28.4 (c) 197.62
2. (a) 2748 (b) 33170 (c) 274218
3. (a) 0.2401 (b) 0.0174 (c) 0.00923
4. (a) 1702.3 (b) 10.04 (c) 0.0109
5. (a)
1
2
(b) 11
7
8
(c)
1
32
(d) 130
3
5

In problems 6 and 7, express the numbers given as
integers or decimal fractions.
6. (a) 1.01 × 103 (b) 9.327 × 102
(c) 5.41 × 104
(d) 7 × 100
7. (a) 3.89 × 10−2 (b) 6.741 × 10−1
(c) 8 × 10−3
In 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 × 10−3
3 × 10−5
(b)
(2.4 × 103)(3 × 10−2)
(4.8 × 104)
10. Write the following statements in standard
form.
(a) The density of aluminium is 2710
kgm−3
.
(b) Poisson’s ratio for gold is 0.44
(c) The impedance of free space is
376.73 .
(d) The electron rest energy is 0.511MeV.
(e) Proton charge–mass ratio is 95789700
Ckg−1.
(f) The normal volume of a perfect gas is
0.02241m3 mol−1.
8.5 Engineering notation
In engineering, standard form is not as important as
engineering notation. Engineering notation is similar
to standard form except that the power of 10 is always
a multiple of 3.
For example, 43645 = 43.645 × 103
in engineering notation
and 0.0534 = 53.4 × 10−3
in engineering notation
From the list of engineering prefixes on page 55 it is
apparent that all prefixes involve powers of 10 that are
multiples of 3.
For example, a force of 43645N can rewritten as
43.645 × 103 N and from the list of prefixes can then
be expressed as 43.645kN.
Thus, 43645N ≡ 43.645kN
To help further, on your calculator is an ‘ENG’ button.
Enter the number 43645 into your calculator and then
press =. Now press the ENG button and the answer is
43.645 × 103. We then have to appreciate that 103 is the
prefix ‘kilo’, giving 43645N ≡ 43.645 kN.
In another example, let a current be 0.0745A. Enter
0.0745 into your calculator. Press =. Now press
ENG and the answer is 74.5 × 10−3. We then have
to appreciate that 10−3 is the prefix ‘milli’, giving
0.0745A ≡ 74.5mA.
Problem 6. Express the following in engineering
notation 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 engineering
notation
= 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, 10−6
corresponds to micro.
Hence, 0.000068H = 68 × 10−6 H in engineering
notation
= 68μH in prefix form.
Problem 7. Express the following in engineering
notation and in prefix form:
(a) 42 × 105
(b) 4.7 × 10−10
F
(a) Enter 42 × 105 into the calculator. Press =
Now press ENG and the answer is 4.2 × 106.
sponds to mega.

Hence, 42 × 105
= 4.2 × 106
in engineering
notation
= 4.2 M in prefix form.
(b) Enter 47 ÷ 1010 =
47
10000000000
into the calcu-
lator. Press =
Now press ENG and the answer is 4.7 × 10−9
.
corresponds 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 milli
means ÷1000). Press =
Now press ENG and the answer is 56 × 10−6.
corresponds to micro.
Hence, 0.056mA =
0.056
1000
A = 56 × 10−6 A
= 56μA.
(b) Enter 16700× 1000 into the calculator (since kilo
means ×1000). Press =
Now press ENG and the answer is 16.7 × 106.
sponds to mega.
Hence, 16700kHz = 16700× 1000Hz
= 16.7 × 106
Hz
= 16.7MHz
Problem 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.
sponds to kilo.
Hence, 63 × 104 V = 630 × 103 V = 630kV.
(b) Enter 3100× 10−12
into the calculator. Press =
Now press ENG and the answer is 3.1 × 10−9.
corresponds to nano.
Hence, 3100pF = 31 × 10−12 F = 3.1 × 10−9 F
= 3.1 nF
Problem 10. Rewrite (a) 14700mm in metres
(b) 276cm in metres (c) 3.375kg in grams
(a) 1m = 1000 mm, hence
1mm =
1
1000
=
1
103
= 10−3 m.
Hence, 14700mm = 14700 × 10−3 m = 14.7m.
(b) 1m=100 cm, hence 1cm=
1
100
=
1
102
=10−2 m.
Hence, 276cm = 276 × 10−2
m = 2.76m.
(c) 1kg = 1000 g = 103 g
Hence, 3.375kg = 3.375 × 103 g = 3375g.
PracticeExercise 34 Engineering notation
In problems 1 to 12, express in engineering nota-
tion in preﬁx form.
1. 60000Pa
2. 0.00015W
3. 5 × 107 V
4. 5.5 × 10−8 F
5. 100000W
6. 0.00054A
7. 15 × 105
8. 225 × 10−4 V
9. 35000000000Hz
10. 1.5 × 10−11 F
11. 0.000017A
12. 46200

13. Rewrite 0.003mA in μA
14. Rewrite 2025kHz as MHz
15. Rewrite 5 × 104 N in kN
16. Rewrite 300pF in nF
17. Rewrite 6250cm in metres
18. Rewrite 34.6g in kg
In problems 19 and 20, use a calculator to evaluate
in engineering notation.
19. 4.5 × 10−7
× 3 × 104
20.
1.6 × 10−5 25 × 103
100 × 10−6

Revision Test 3 : Ratio, proportion, powers, roots, indices and units
1. In a box of 1500 nails, 125 are defective. Express
the non-defective nails as a ratio of the defective
ones, in its simplest form. (3)
2. Prize money in a lottery totals £4500 and is shared
among three winners in the ratio 5:3:1. How much
does 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 of
6.15kN. (3)
4. If 15 cans of lager weigh 7.8kg, what will 24 cans
weigh? (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 is
250 × 10−6. Determine the stress when the strain
is 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 when
y = 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) 49
1
2 (4)
9. Evaluate
32 ×
√
36 × 22
3 × 81
1
2
taking positive square
roots only. (3)
10. Evaluate 64 × 6 × 62 in index form. (3)
11. Evaluate
(a)
27
22
(b)
104 × 10× 105
106 × 102
(4)
12. Evaluate
(a)
23 × 2 × 22
24
(b)
23 × 16
2
(8 × 2)3
(c)
1
42
−1
(7)
13. Evaluate
(a) (27)
−
1
3 (b)
3
2
−2
−
2
9
2
3
2
(5)
14. State the SI unit of (a) capacitance (b) electrical
potential (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 in
prefix 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 as
MV (2)

Chapter 9
Basic algebra
9.1 Introduction
We are already familiar with evaluating formulae using
a calculator from Chapter 4.
For example, if the length of a football pitch is L and its
width is b, then the formula for the area A is given by
A = L × b
This is an algebraic equation.
If L = 120m and b = 60m, then the area
A = 120 × 60 = 7200m2.
The total resistance, RT , of resistors R1, R2 and R3
connected in series is given by
RT = R1 + R2 + R3
This is an algebraic equation.
If R1 = 6.3k , R2 = 2.4k and R3 = 8.5k , then
RT = 6.3 + 2.4 + 8.5 = 17.2k
The temperature in Fahrenheit, F, is given by
F =
9
5
C + 32
where C is the temperature in Celsius. This is an
algebraic equation.
If C = 100◦C, then F =
9
5
× 100 + 32
= 180 + 32 = 212◦F.
If you can cope with evaluating formulae then you will
be able to cope with algebra.
9.2 Basic operations
Algebra merely uses letters to represent numbers.
If, say, a,b,c and d represent any four numbers then in
algebra:
(a) a + a + a + a = 4a. For example, if a = 2, then
2 + 2 + 2 + 2 = 4 × 2 = 8.
(b) 5b means 5 × b. For example, if b = 4, then
5b = 5 × 4 = 20.
(c) 2a + 3b + a − 2b = 2a + a + 3b − 2b = 3a + b
Only similar terms can be combined in algebra.
The 2a and the +a can be combined to give 3a
and 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 as
2a + 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 × d
For example, if a = 3,b = −2,c = 1 and d = −5,
then 4abcd = 4 × 3 × −2 × 1 × −5 = 120. (Note
that − × − = +)
(e) (a)(c)(d) means a × c × d
Brackets are often used instead of multiplication
signs. For example, (2)(5)(3) means 2 × 5 × 3 =
30.
(f) ab = ba
If a = 2 and b = 3 then 2 × 3 is exactly the same
as 3 × 2, i.e. 6.
(g) b2
= b × b. For example, if b = 3, then
32 = 3 × 3 = 9.
(h) a3 = a × a × a For example, if a = 2, then
23 = 2 × 2 × 2 = 8.
Here are some worked examples to help get a feel for
basic operations in this introduction to algebra.
DOI: 10.1016/B978-1-85617-697-2.00009-0

9.2.1 Addition and subtraction
Problem 1. Find the sum of 4x,3x,−2x and −x
4x + 3x + −2x + −x = 4x + 3x − 2x − x
(Note that + ×− = −)
= 4x
Problem 2. Find the sum of 5x,3y, z,−3x,−4y
and 6z
5x+3y + z + −3x + −4y + 6z
= 5x + 3y + z − 3x − 4y + 6z
= 5x − 3x + 3y − 4y + z + 6z
= 2x − y + 7z
Note that the order can be changed when terms are sep-
arated by + and − signs. Only similar terms can be
combined.
Problem 3. Simplify 4x2 − x − 2y + 5x + 3y
4x2
− x − 2y + 5x + 3y = 4x2
+ 5x − x + 3y − 2y
= 4x2
+ 4x + y
Problem 4. Simplify 3xy − 7x + 4xy + 2x
3xy − 7x + 4xy + 2x = 3xy + 4xy + 2x − 7x
= 7xy − 5x
PracticeExercise 35 Addition and
subtraction 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 and
7c.
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 − y2
take xy − 2x2
.
9.2.2 Multiplicationand division
Problem 5. Simplify bc × abc
bc × abc = a × b × b × c × c
= a × b2
× c2
= ab2
c2
Problem 6. Simplify −2p × −3p
− × − = + hence,−2p × −3p = 6p2
Problem 7. Simplify ab × b2c × a
ab × b2
c × a = a × a × b × b × b × c
= a2
× b3
× c
= a2
b3
c
Problem 8. Evaluate 3ab + 4bc − abc when
a = 3,b = 2 and c = 5
3ab + 4bc − abc = 3 × a × b + 4 × b × c − a × b × c
= 3 × 3 × 2 + 4 × 2 × 5 − 3 × 2 × 5
= 18 + 40 − 30
= 28
Problem 9. Determine the value of 5pq2r3, given
that p = 2,q =
2
5
and r = 2
1
2
5pq2
r3
= 5 × p × q2
× r3
= 5 × 2 ×
2
5
2
× 2
1
2
3

Basic algebra 63
= 5 × 2 ×
2
5
2
×
5
2
3
since 2
1
2
=
5
2
=
5
1
×
2
1
×
2
5
×
2
5
×
5
2
×
5
2
×
5
2
=
1
1
×
1
1
×
1
1
×
1
1
×
1
1
×
5
1
×
5
1
by cancelling
= 5 × 5
= 25
Problem 10. Multiply 2a + 3b by a + b
Each term in the first expression is multiplied by a, then
each term in the first expression is multiplied by b and
the two results are added. The usual layout is shown
below.
2a + 3b
a + b
Multiplying by a gives 2a2 + 3ab
Multiplying by b gives 2ab + 3b2
Adding gives 2a2 + 5ab + 3b2
Thus, (2a + 3b)(a + b) = 2a2
+ 5ab + 3b2
Problem 11. Multiply 3x − 2y2 + 4xy by
2x − 5y
3x − 2y2 + 4xy
2x − 5y
Multiplying
by 2x → 6x2 − 4xy2 + 8x2 y
Multiplying
by −5y → −20xy2 −15xy + 10y3
Adding gives 6x2 − 24xy2 + 8x2y − 15xy + 10y3
Thus, (3x − 2y2 + 4xy)(2x − 5y)
= 6x2 − 24xy2 + 8x2y − 15xy + 10y3
Problem 12. Simplify 2x ÷ 8xy
2x ÷ 8xy means
2x
8xy
2x
8xy
=
2 × x
8 × x × y
=
1 × 1
4 × 1 × y
by cancelling
=
1
4y
9a2
bc
3ac
9a2bc
3ac
=
9 × a × a × b × c
3 × a × c
= 3 × a × b
= 3ab
Problem 14. Divide 2x2 + x − 3 by x − 1
(i) 2x2 + x − 3 is called the dividend and x − 1 the
divisor. The usual layout is shown below with the
dividend and divisorboth arranged in descending
powers of the symbols.
2x + 3
x − 1 2x2 + x − 3
2x2 − 2x
3x − 3
3x − 3
. .
(ii) Dividing the first term of the dividend by the
first term of the divisor, i.e.
2x2
x
gives 2x, which
is put above the first term of the dividend as
shown.
(iii) The divisor is then multiplied by 2x, i.e.
2x(x − 1) = 2x2 − 2x, which is placed under the
dividend as shown. Subtracting gives 3x − 3.
(iv) The process is then repeated, i.e. the first term
of 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 under
the 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 by
multiplying (2x + 3) by (x − 1), which equals
2x2 + x − 3.)
x3
+ y3
x + y

(i) (iv) (vii)
x2 − xy + y2
x + y x3 + 0 + 0 + y3
x3 + x2 y
−x2
y + y3
−x2 y − xy2
xy2
+ y3
xy2 + y3
. .
(i) x into x3 goes x2. Put x2 above x3.
(ii) x2(x + y) = x3 + x2 y
(iii) Subtract.
(iv) x into −x2
y goes −xy. Put −xy above the
dividend.
(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 + y3
x + y
= x2 − xy + y2.
Thezerosshown in thedividend arenot normally shown,
but are included to clarify the subtraction process and
to keep similar terms in their respective columns.
Problem 16. Divide 4a3 − 6a2b + 5b3 by 2a − b
2a2 − 2ab − b2
2a − b 4a3 − 6a2b + 5b3
4a3−2a2b
−4a2
b + 5b3
−4a2b + 2ab2
−2ab2
+ 5b3
−2ab2 + b3
4b3
Thus,
4a3 − 6a2b + 5b3
2a − b
= 2a2 − 2ab − b2, remain-
der 4b3
.
Alternatively, the answer may be expressed as
4a3 − 6a2b + 5b3
2a − b
= 2a2
− 2ab − b2
+
4b3
2a − b
PracticeExercise 36 Basic operations in
algebra (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 that
x = 2, y = 1
1
2
and z =
2
3
6. If x = 5 and y = 6, evaluate
23(x − y)
y + xy + 2x
7. If a = 4,b = 3,c = 5 and d = 6, evaluate
3a + 2b
3c − 2d
8. Simplify 2x ÷ 14xy.
9. Simplify
25x2
y z3
5xyz
10. 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 indices
The laws of indices with numbers were covered in
Chapter 7; the laws of indices in algebraic terms are
as follows:
(1) am×an =am+n
Forexample,a3 × a4 =a3+4 =a7

Basic algebra 65
(2)
am
an
= am−n For example,
c5
c2
= c5−2 = c3
(3) (am)n = amn For example, d2 3
= d2×3 = d6
(4) a
m
n = n
√
am For example, x
4
3 =
3
√
x4
(5) a−n =
1
an
For example, 3−2 =
1
32
=
1
9
(6) a0 = 1 For example, 170 = 1
Here are some worked examples to demonstrate these
laws of indices.
Problem 17. Simplify a2b3c × ab2c5
a2
b3
c × ab2
c5
= a2
× b3
× c × a × b2
× c5
= a2
× b3
× c1
× a1
× b2
× c5
Grouping together like terms gives
a2 × a1 × b3 × b2 × c1 × c5
Using law (1) of indices gives
a2+1 × b3+2 × c1+5 = a3 × b5 × c6
i.e. a2b3c × ab2c5 = a3 b5 c6
Problem 18. Simplify a
1
3 b
3
2 c−2 × a
1
6 b
1
2 c
Using law (1) of indices,
a
1
3 b
3
2 c−2
× a
1
6 b
1
2 c = a
1
3+
1
6 × b
3
2 +
1
2 × c−2+1
= a
1
2 b2
c−1
x5 y2z
x2 y z3
x5 y2z
x2 yz3
=
x5 × y2 × z
x2 × y × z3
=
x5
x2
×
y2
y1
×
z
z3
= x5−2
× y2−1
× z1−3
by law (2) of indices
= x3
× y1
× z−2
= x3
y z−2
or
x3y
z2
a3
b2
c4
abc−2
and evaluate when
a = 3,b =
1
4
and c = 2
Using law (2) of indices,
a3
a
= a3−1
= a2
,
b2
b
= b2−1
= b and
c4
c−2
= c4−−2
= c6
Thus,
a3
b2
c4
abc−2
= a2bc6
When a =3,b=
1
4
and c = 2,
a2bc6 = (3)2 1
4
(2)6 = (9)
1
4
(64) = 144
Problem 21. Simplify (p3)2(q2)4
(p3
)2
(q2
)4
= p3×2
× q2×4
= p6
q8
(mn2)3
(m1/2n1/4)4
The brackets indicate that each letter in the bracket must
be raised to the power outside. Using law (3) of indices
gives
(mn2)3
(m1/2n1/4)4
=
m1×3n2×3
m(1/2)×4n(1/4)×4
=
m3n6
m2n1
m3n6
m2n1
= m3−2
n6−1
= mn5
Problem 23. Simplify (a3b)(a−4b−2), expressing
the answer with positive indices only
Using law (1) of indices gives a3+−4b1+−2 = a−1b−1
Using law (5) of indices gives a−1b−1 =
1
a+1b+1
=
1
ab
d2e2 f 1/2
(d3/2ef 5/2)2
expressing
the answer with positive indices only

d2e2 f 1/2
(d3/2 e f 5/2)2
=
d2e2 f 1/2
d3e2 f 5
d2−3
e2−2
f
1
2 −5
= d−1
e0
f −
9
2
= d−1
f −
9
2 since e0
= 1 from law
(6) of indices
=
1
df 9/2
from law (5) of indices
In problems 1 to 18, simplify the following,giving
each answer as a power.
1. z2 × z6 2. a × a2 × a5
3. n8 × n−5 4. b4 × b7
5. b2 ÷ b5 6. c5 × c3 ÷ c4
7.
m5
× m6
m4 × m3
8.
(x2
)(x)
x6
9. x3 4
10. y2 −3
11. t × t3 2
12. c−7 −2
13.
a2
a5
3
14.
1
b3
4
15.
b2
b7
−2
16.
1
s3 3
17. p3qr2 × p2q5r × pqr2 18.
x3 y2z
x5 y z3
19. Simplify (x2 y3z)(x3 yz2) and evaluate when
x =
1
2
, y = 2 and z = 3.
20. Simplify
a5bc3
a2b3c2
and evaluate when
a =
3
2
,b =
1
2
and c =
2
3
Here are some further worked examples on the laws of
indices
p1/2q2r2/3
p1/4q1/2r1/6
and evaluate
when p = 16,q = 9 and r = 4, taking positive roots
only
Using law (2) of indices gives p
1
2 −
1
4 q2−
1
2 r
2
3 −
1
6
p
1
2 −
1
4 q2−
1
2 r
2
3 −
1
6 = p
1
4 q
3
2 r
1
2
When p = 16,q = 9 and r = 4,
p
1
4 q
3
2 r
1
2 = 16
1
4 9
3
2 4
1
2
= (
4
√
16)(
√
93)(
√
4) from law (4) of indices
= (2)(33
)(2) = 108
x2 y3 + xy2
xy
Algebraic expressions of the form
a + b
c
can be split
into
a
c
+
b
c
. Thus,
x2 y3 + xy2
xy
=
x2 y3
xy
+
xy2
xy
= x2−1
y3−1
+ x1−1
y2−1
= xy2
+y
(since x0 = 1,fromlaw(6)ofindices).
x2 y
xy2 − xy
The highest common factor (HCF) of each of the three
terms comprising the numerator and denominator is xy.
Dividing each term by xy gives
x2 y
xy2 − xy
=
x2 y
xy
xy2
xy
−
xy
xy
=
x
y − 1
a2b
ab2 − a1/2b3

Basic algebra 67
The HCF of each of the three terms is a1/2
b. Dividing
each term by a1/2b gives
a2b
ab2 − a1/2b3
=
a2b
a1/2b
ab2
a1/2b
−
a1/2b3
a1/2b
=
a3/2
a1/2b−b2
Problem 29. Simplify (a3
√
b
√
c5)(
√
a
3
√
b2c3)
and evaluate when a =
1
4
,b = 6 and c = 1
Using law(4)ofindices,theexpression can bewritten as
(a3
√
b c5)(
√
a
3
b2c3
) = a3
b
1
2 c
5
2 a
1
2 b
2
3 c3
a3
b
1
2 c
5
2 a
1
2 b
2
3 c3
= a3+
1
2 b
1
2 +
2
3 c
5
2 +3
= a
7
2 b
7
6 c
11
2
It is usual to express the answer in the same form as the
question. Hence,
a
7
2 b
7
6 c
11
2 = a7
6
b7
c11
When a =
1
4
,b = 64 and c = 1,
a7 6
b7 c11 =
1
4
7
6
√
647
√
111
=
1
2
7
(2)7
(1) = 1
(x2 y1/2)(
√
x 3
y2)
(x5 y3)1/2
Using laws (3) and (4) of indices gives
x2 y1/2 √
x 3
y2
(x5 y3)1/2
=
x2 y1/2 x1/2 y2/3
x5/2 y3/2
Using laws (1) and (2) of indices gives
x2+
1
2 −
5
2 y
1
2 +
2
3 −
3
2 = x0
y−
1
3 = y−
1
3 or
1
y1/3
or
1
3
√
y
from laws (5) and (6) of indices.
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 + a3b
a2b2
3. (a2)1/2(b2)3 c1/2 3
4.
(abc)2
(a2b−1c−3)3
5.
p3q2
pq2 − p2q
6. (
√
x y3 3
√
z2)(
√
x y3 z3)
7. (e2
f 3
)(e−3
f −5
),expressing theanswerwith
positive indices only.
8.
(a3b1/2c−1/2)(ab)1/3
(
√
a3
√
b c)

Chapter 10
Further algebra
10.1 Introduction
In this chapter, the use of brackets and factorization
with algebra is explained, together with further practice
with the laws of precedence. Understanding of these
topics is often necessary when solving and transposing
equations.
10.2 Brackets
With algebra
(a) 2(a + b) = 2a + 2b
(b) (a + b)(c + d) = a(c + d) + b(c + d)
= ac + ad + bc + bd
Here are some worked examples to help understanding
of brackets with algebra.
Problem 1. Determine 2b(a − 5b)
2b(a − 5b) = 2b × a + 2b × −5b
= 2ba − 10b2
= 2ab − 10b2
(note that 2ba is the
same 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 − 4y2
Problem 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 − 8y
Problem 4. Remove the brackets and simplify the
expression (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 − 6d
Problem 5. Simplify (p + q)(p − q)
(p + q)(p − q) = p(p − q) + q(p − q)
= p2 − pq + qp − q2
= p2 − q2
Problem 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 + 9y2
DOI: 10.1016/B978-1-85617-697-2.00010-7

Further algebra 69
Problem 7. Remove the brackets from the
expression 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 − 4x2
Problem 8. Remove the brackets and simplify the
expression 2a − [3{2(4a − b) − 5(a + 2b)} + 4a]
(i) Removing the innermost brackets gives
2a − [3{8a − 2b − 5a − 10b} + 4a]
(ii) Collecting together similar terms gives
2a − [3{3a − 12b} + 4a]
(iii) Removing the ‘curly’ brackets gives
2a − [9a − 36b + 4a]
(iv) Collecting together similar terms gives
2a − [13a − 36b]
(v) Removing the outer brackets gives
2a − 13a + 36b
(vi) i.e. −11a + 36b or 36b − 11a
PracticeExercise 39 Brackets (answers on
page 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)2
9. (a − c)2 10. (5x + 3)2
11. (2x − 6)2 12. (2x − 3)(2x + 3)
13. (8x + 4)2 14. (rs + t)2
15. 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)2
23. 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 Factorization
The factors of 8 are 1, 2, 4 and 8 because 8 divides by
1, 2, 4 and 8.
The factorsof 24 are 1, 2, 3, 4, 6, 8, 12 and 24 because 24
divides by 1, 2, 3, 4, 6, 8, 12 and 24.
The common factors of 8 and 24 are 1, 2, 4 and 8 since
1, 2, 4 and 8 are factors of both 8 and 24.
The highest common factor (HCF) is the largest
number that divides into two or more terms.
Hence, the HCF of 8 and 24 is 8, as explained in
Chapter 1.
When two or more terms in an algebraic expression con-
tain a common factor, then this factor can be shown
outside of a bracket. For example,
d f + dg = d( f + g)
which is just the reverse of
d( f + g) = d f + dg
This process is called factorization.
of factorizing in algebra.
Problem 9. Factorize ab − 5ac
a is common to both terms ab and −5ac. a is there-
fore taken outside of the bracket. What goes inside the
bracket?

(i) What multiplies a to make ab? Answer: b
(ii) What multiplies a to make −5ac? Answer: −5c
Hence, b − 5c appears in the bracket. Thus,
ab − 5ac = a(b − 5c)
Problem 10. Factorize 2x2 + 14xy3
For the numbers 2 and 14, the highest common factor
(HCF) is 2 (i.e. 2 is the largest number that divides into
both 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 goes
inside the bracket?
(i) What multiplies 2x to make 2x2? Answer: x
(ii) What multiplies 2x to make 14xy3? Answer: 7y3
Hence x + 7y3 appears inside the bracket. Thus,
2x2
+ 14xy3
= 2x(x + 7y3
)
Problem 11. Factorize 3x3
y − 12xy2
+ 15xy
Forthenumbers3,12 and 15,thehighest common factor
is 3 (i.e. 3 is the largest number that divides into 3, 12
and 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 goes
inside 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: 5
Hence, x2 − 4y + 5 appears inside the bracket. Thus,
3x3
y − 12xy2
+ 15xy = 3xy(x2
− 4y + 5)
Problem 12. Factorize 25a2b5 − 5a3b2
For the numbers 25 and 5, the highest common factor
is 5 (i.e. 5 is the largest number that divides into 25
and 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 25a2
b5
and 5a3
b2
is 5a2
b2
.
5a2b2 is therefore taken outside of the bracket. What
goes inside the bracket?
(i) What multiplies 5a2b2 to make 25a2b5? Answer:
5b3
(ii) What multiplies5a2b2 to make −5a3b2? Answer:
−a
Hence, 5b3− a appears in the bracket. Thus,
25a2
b5
− 5a3
b2
= 5a2
b2
(5b3
− a)
Problem 13. Factorize ax − ay + bx − by
The first two terms have a common factor of a and the
last 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 − 3by
a is a common factor of the first two terms and b a
common 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 original
first and third terms and −3y is a common factor of
the 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 before
Problem 15. Factorize x3 + 3x2 − x − 3
x2 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)
PracticeExercise 40 Factorization (answers
on page 344)
Factorize and simplify the following.
1. 2x + 4 2. 2xy − 8xz
3. pb + 2pc 4. 2x + 4xy
5. 4d2 − 12d f 5 6. 4x + 8x2
7. 2q2 + 8qn 8. rs +rp +rt
9. x + 3x2 + 5x3 10. abc + b3c
11. 3x2 y4 − 15xy2 + 18xy
12. 4p3q2 − 10pq3 13. 21a2b2 − 28ab
14. 2xy2 + 6x2 y + 8x3 y
15. 2x2 y − 4xy3 + 8x3 y4
16. 28y + 7y2 + 14xy
17.
3x2 + 6x − 3xy
xy + 2y − y2
18.
abc + 2ab
2c + 4
−
abc
2c
19.
5rs + 15r3t + 20r
6r2t2 + 8t + 2ts
−
r
2t
20. ay + by + a + b 21. px + qx + py + qy
22. ax − ay + bx − by
23. 2ax + 3ay − 4bx − 6by
10.4 Laws of precedence
Sometimes addition, subtraction, multiplication, divi-
sion, powers and brackets can all be involved in an
algebraic expression. With mathematics there is a defi-
nite order of precedence (first met in Chapter 1) which
we need to adhere to.
With the laws of precedence the order is
Brackets
Order (or pOwer)
Division
Multiplication
Addition
Subtraction
The first letter of each word spells BODMAS.
Here are some examples to help understanding of
BODMAS with algebra.
Problem 16. Simplify 2x + 3x × 4x − x
2x + 3x × 4x − x = 2x + 12x2 − x (M)
= 2x − x + 12x2
= x + 12x2 (S)
or x(1 + 12x) by factorizing
Problem 17. Simplify (y + 4y) × 3y − 5y
(y + 4y) × 3y − 5y = 5y × 3y − 5y (B)
= 15y2 − 5y (M)
or 5y(3y − 1) by factorizing
Problem 18. Simplify p + 2p × (4p − 7p)
p + 2p × (4p − 7p) = p + 2p × −3p (B)
= p − 6p2 (M)
or p(1 − 6p) by factorizing
Problem 19. Simplify t ÷ 2t + 3t − 5t
t ÷ 2t + 3t − 5t =
t
2t
+ 3t − 5t (D)
=
1
2
+ 3t − 5t by cancelling
=
1
2
− 2t (S)
Problem 20. Simplify x ÷ (4x + x) − 3x
x ÷ (4x + x) − 3x = x ÷ 5x − 3x (B)
=
x
5x
− 3x (D)
=
1
5
− 3x by cancelling
Problem 21. Simplify 2y ÷ (6y + 3y − 5y)
2y ÷ (6y + 3y − 5y) = 2y ÷ 4y (B)
=
2y
4y
(D)
=
1
2
by cancelling

5a + 3a × 2a + a ÷ 2a − 7a
5a + 3a × 2a + a ÷ 2a − 7a
= 5a + 3a × 2a +
a
2a
− 7a (D)
= 5a + 3a × 2a +
1
2
− 7a by cancelling
= 5a + 6a2
+
1
2
− 7a (M)
= −2a + 6a2 +
1
2
(S)
= 6a2 − 2a +
1
2
(4y + 3y)2y + y ÷ 4y − 6y
(4y + 3y)2y + y ÷ 4y − 6y
= 7y × 2y + y ÷ 4y − 6y (B)
= 7y × 2y +
y
4y
− 6y (D)
= 7y × 2y +
1
4
− 6y by cancelling
= 14y2
+
1
4
− 6y (M)
5b + 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
−3
by cancelling
= 5b + 2b × 3b −
1
3
= 5b + 6b2
−
1
3
(M)
(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
−1
by cancelling
= 6p × −5
= −30p
PracticeExercise 41 Laws of precedence
Simplify the following.
1. 3x + 2x × 4x − x
2. (2y + y) × 4y − 3y
3. 4b + 3b × (b − 6b)
4. 8a ÷ 2a + 6a − 3a
5. 6x ÷ (3x + x) − 4x
6. 4t ÷ (5t − 3t + 2t)
7. 3y + 2y × 5y + 2y ÷ 8y − 6y
8. (x + 2x)3x + 2x ÷ 6x − 4x
9. 5a + 2a × 3a + a ÷ (2a − 9a)
10. (3t + 2t)(5t + t) ÷ (t − 3t)
11. x ÷ 5x − x + (2x − 3x)x
12. 3a + 2a × 5a + 4a ÷ 2a − 6a

Chapter 11
Solving simple equations
11.1 Introduction
3x − 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 expressions
are equal.
Hence, A = πr2 (where A is the area of a circle
of radius r)
F =
9
5
C + 32 (which relates Fahrenheit and
Celsius temperatures)
and y = 3x + 2 (which is the equation of a
straight line graph)
are all examples of equations.
11.2 Solving equations
To ‘solve an equation’ means ‘to ﬁnd the value of the
unknown’. For example, solving 3x − 4 = 2 means that
the 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 to
solve such equations.
Many equations occur in engineering and it is essential
that we can solve them when needed.
Here are some examples to demonstrate how simple
equations are solved.
Problem 1. Solve the equation 4x = 20
Dividing each side of the equation by 4 gives
4x
4
=
20
4
i.e. x = 5 by cancelling, which is the solution to the
equation 4x = 20.
The same operation must be applied to both sides of an
equation so that the equality is maintained.
We can do anything we like to an equation, as long
as we do the same to both sides. This is, in fact, the
only rule to remember when solving simple equations
(and also when transposing formulae, which we do in
Chapter 12).
Problem 2. Solve the equation
2x
5
= 6
Multiplying both sides by 5 gives 5
2x
5
= 5(6)
Cancelling and removing brackets gives 2x = 30
Dividing both sides of the equation by 2 gives
2x
2
=
30
2
Cancelling gives x = 15
which is the solution of the equation
2x
5
= 6.
Problem 3. Solve the equation a − 5 = 8
Adding 5 to both sides of the equation gives
a − 5 + 5 = 8 + 5
i.e. a = 8 + 5
i.e. a = 13
which is the solution of the equation a − 5 = 8.
Note that adding 5 to both sides of the above equation
results in the −5 moving from the LHS to the RHS, but
the sign is changed to +.
Problem 4. Solve the equation x + 3 = 7
Subtracting 3 from both sides gives x + 3 − 3 = 7 − 3
i.e. x = 7 − 3
i.e. x = 4
which is the solution of the equation x + 3 = 7.
DOI: 10.1016/B978-1-85617-697-2.00011-9

Note that subtracting 3 from both sides of the above
equation results in the +3 moving from the LHS to the
RHS, but the sign is changed to –. So, we can move
straight 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 + 9
In such equations the terms containing x are grouped
on one side of the equation and the remaining terms
grouped on the other side of the equation. As in Prob-
lems 3 and 4, changing from one side of an equation to
the other must be accompanied by a change of sign.
Since 6x + 1 = 2x + 9
then 6x − 2x = 9 − 1
i.e. 4x = 8
Dividing both sides by 4 gives
4x
4
=
8
4
Cancelling gives x = 2
which 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,
then
LHS of equation = 6(2) + 1 = 13
RHS of equation = 2(2) + 9 = 13
Since the left hand side (LHS) equals the right hand
side (RHS) then x = 2 must be the correct solution of
the equation.
When solving simple equations, always check your
answers by substituting your solution back into the
original equation.
Problem 6. Solve the equation 4− 3p = 2p − 11
In order to keep the p term positive the terms in p are
moved to the RHS and the constant terms to the LHS.
Similar to Problem 5, if 4 − 3p = 2p − 11
then 4 + 11 = 2p + 3p
i.e. 15 = 5p
15
5
=
5p
5
Cancelling gives 3 = p or p = 3
which is the solution of the equation 4 − 3p = 2p − 11.
By substituting p = 3 into the original equation, the
solution may be checked.
LHS = 4 − 3(3) = 4 − 9 = −5
RHS = 2(3) − 11 = 6 − 11 = −5
Since LHS = RHS, the solution p = 3 must be correct.
If, in this example, the unknown quantities had been
grouped initially on the LHS instead of the RHS, then
−3p − 2p = −11 − 4
i.e. −5p = −15
from which,
−5p
−5
=
−15
−5
and p = 3
as before.
It is often easier, however, to work with positive values
where possible.
Problem 7. Solve the equation 3(x − 2) = 9
Removing the bracket gives 3x − 6 = 9
Rearranging gives 3x = 9 + 6
i.e. 3x = 15
Dividing both sides by 3 gives x = 5
which is the solution of the equation 3(x − 2) = 9.
Theequation may bechecked by substituting x = 5 back
into the original equation.
4(2r − 3) − 2(r − 4) = 3(r − 3) − 1
Removing brackets gives
8r − 12 − 2r + 8 = 3r − 9 − 1
Rearranging gives 8r − 2r − 3r = −9 − 1 + 12 − 8
i.e. 3r = −6
Dividing both sides by 3 gives r =
−6
3
= −2
4(2r − 3) − 2(r − 4) = 3(r − 3) − 1.
The solution may be checked by substituting r = −2
back into the original equation.
LHS = 4(−4 − 3) − 2(−2 − 4) = −28 + 12 = −16
RHS = 3(−2 − 3) − 1 = −15 − 1 = −16
Since LHS = RHS then r = −2 is the correct solution.

Solving simple equations 75
PracticeExercise 42 Solving simple
equations (answers on page 344)
Solve the following equations.
1. 2x + 5 = 7
2. 8 − 3t = 2
3.
2
3
c − 1 = 3
4. 2x − 1 = 5x + 11
5. 7 − 4p = 2p − 5
6. 2.6x − 1.3 = 0.9x + 0.4
7. 2a + 6 − 5a = 0
8. 3x − 2 − 5x = 2x − 4
9. 20d − 3 + 3d = 11d + 5 − 8
10. 2(x − 1) = 4
11. 16 = 4(t + 2)
12. 5( f − 2) − 3(2 f + 5) + 15 = 0
13. 2x = 4(x − 3)
14. 6(2 − 3y) − 42 = −2(y − 1)
15. 2(3g − 5) − 5 = 0
16. 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 solving
simple equations.
4
x
=
2
5
The lowest common multiple (LCM) of the denomina-
tors, i.e. the lowest algebraic expression that both x and
5 will divide into, is 5x.
Multiplying both sides by 5x gives
5x
4
x
= 5x
2
5
Cancelling gives 5(4) = x(2)
i.e. 20 = 2x (1)
20
2
=
2x
2
Cancelling gives 10 = x or x = 10
4
x
=
2
5
When there is just one fraction on each side of the equa-
tion as in this example, there is a quick way to arrive
at equation (1) without needing to ﬁnd the LCM of the
denominators.
We can move from
4
x
=
2
5
to 4 × 5 = 2 × x by what is
called ‘cross-multiplication’.
In general, if
a
b
=
c
d
then ad = bc
We can use cross-multiplication when there is one
fraction only on each side of the equation.
3
t − 2
=
4
3t + 4
Cross-multiplication gives 3(3t + 4) = 4(t − 2)
Removing brackets gives 9t + 12 = 4t − 8
Rearranging gives 9t − 4t = −8 − 12
i.e. 5t = −20
Dividing both sides by 5 gives t =
−20
5
= −4
3
t − 2
=
4
3t + 4
2y
5
+
3
4
+ 5 =
1
20
−
3y
2
The lowest common multiple (LCM) of the denomina-
tors is 20; i.e., the lowest number that 4, 5, 20 and 2 will
divide into.
Multiplying each term by 20 gives
20
2y
5
+ 20
3
4
+ 20(5) = 20
1
20
− 20
3y
2
Cancelling gives 4(2y) + 5(3) + 100 = 1 − 10(3y)
i.e. 8y + 15 + 100 = 1 − 30y
Rearranging gives 8y +30y = 1−15−100

i.e. 38y = −114
38y
38
=
−114
38
Cancelling gives y = −3
2y
5
+
3
4
+ 5 =
1
20
−
3y
2
√
x = 2
Whenever square root signs are involved in an equation,
both sides of the equation must be squared.
Squaring both sides gives
√
x
2
= (2)2
i.e. x = 4
√
x = 2.
Problem 13. Solve the equation 2
√
d = 8
Whenever square roots are involved in an equation, the
square root term needs to be isolated on its own before
squaring both sides.
Cross-multiplying gives √
d =
8
2
Cancelling gives
√
d = 4
√
d
2
= (4)2
i.e. d = 16
which is the solution of the equation 2
√
d = 8.
√
b + 3
√
b
= 2
Cross-multiplying gives
√
b + 3 = 2
√
b
Rearranging gives 3 = 2
√
b −
√
b
i.e. 3 =
√
b
Squaring both sides gives 9 = b
√
b + 3
√
b
= 2.
Problem 15. Solve the equation x2 = 25
Whenever a square term is involved, the square root of
both sides of the equation must be taken.
Taking the square root of both sides gives x2 =
√
25
i.e. x = ±5
which is the solution of the equation x2
= 25.
15
4t2
=
2
3
We need to rearrange the equation to get the t2
term on
its own.
Cross-multiplying gives 15(3) = 2(4t2
)
i.e. 45 = 8t2
Dividing both sides by 8 gives 45
8
=
8t2
8
By cancelling 5.625 = t2
or t2
= 5.625
Taking the square root of both sides gives
t2 =
√
5.625
i.e. t = ±2.372
correct to 4 signiﬁcant ﬁgures, which is the solution of
the equation
15
4t2
=
2
3
PracticeExercise 43 Solving simple
Solve the following equations.
1.
1
5
d + 3 = 4
2. 2 +
3
4
y = 1 +
2
3
y +
5
6
3.
1
4
(2x − 1) + 3 =
1
2
4.
1
5
(2 f − 3) +
1
6
( f − 4) +
2
15
= 0
5.
1
3
(3m − 6) −
1
4
(5m + 4) +
1
5
(2m − 9) = −3
6.
x
3
−
x
5
= 2
7. 1 −
y
3
= 3 +
y
3
−
y
6

8.
2
a
=
3
8
9.
1
3n
+
1
4n
=
7
24
10.
x + 3
4
=
x − 3
5
+ 2
11.
3t
20
=
6 − t
12
+
2t
15
−
3
2
12.
y
5
+
7
20
=
5 − y
4
13.
v − 2
2v − 3
=
1
3
14.
2
a − 3
=
3
2a + 1
15.
x
4
−
x + 6
5
=
x + 3
2
16. 3
√
t = 9
17. 2
√
y = 5
18. 4 =
3
a
+ 3
19.
3
√
x
1 −
√
x
= −6
20. 10 = 5
x
2
− 1
21. 16 =
t2
9
22.
y + 2
y − 2
=
1
2
23.
6
a
=
2a
3
24.
11
2
= 5 +
8
x2
11.3 Practical problems involving
simple equations
There are many practical situations in engineering in
which solving equations is needed. Here are some
worked examples to demonstrate typical practical
situations
Problem 17. Applying the principle of moments
to a beam results in the equation
F × 3 = (7.5 − F) × 2
where F is the force in newtons. Determine the
value of F
Removing brackets gives 3F = 15 − 2F
Rearranging gives 3F + 2F = 15
i.e. 5F = 15
5F
5
=
15
5
from which, force, F = 3N.
Problem 18. A copper wire has a length L of
1.5km, a resistance R of 5 and a resistivity of
17.2 × 10−6 mm. Find the cross-sectional area, a,
of the wire, given that R =
ρL
a
Since R =
ρL
a
then
5 =
(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× 103
and a =
17.2 × 10−6 × 1500× 103
5
=
17.2 × 1500 × 103
106 × 5
=
17.2 × 15
10 × 5
= 5.16
Hence, the cross-sectional area of the wire is 5.16 mm2
.
Problem 19. PV = mRT is the characteristic gas
equation. Find the value of gas constant R when
pressure P = 3 × 106 Pa, volume V = 0.90m3,
mass m = 2.81kg and temperature T = 231K
Dividing both sides of PV = mRT by mT gives
PV
mT
=
mRT
mT
Cancelling gives
PV
mT
= R
Substituting values gives R =
3 × 106 (0.90)
(2.81)(231)

Using a calculator, gas constant, R = 4160J/(kg K),
Problem 20. A rectangular box with square ends
has its length 15cm greater than its breadth and the
total length of its edges is 2.04m. Find the width of
the box and its volume
Let x cm = width = height of box. Then the length of
the box is (x + 15)cm, as shown in Figure 11.1.
(x1
15)
x
x
Figure 11.1
The 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 + 60
204 − 60 = 12x
i.e. 144 = 12x
and x = 12cm
Hence, the width of the box is 12cm.
Volume of box = length × width × height
= (x +15)(x)(x)= (12+15)(12)(12)
= (27)(12)(12)
= 3888cm3
Problem 21. The temperature coefﬁcient of
resistance α may be calculated from the formula
Rt = R0(1 + αt). Find α, given Rt = 0.928,
R0 = 0.80 and t = 40
Since Rt = R0(1 + αt), then
0.928 = 0.80[1 + α(40)]
0.928 = 0.80 + (0.8)(α)(40)
0.928 − 0.80 = 32α
0.128 = 32α
Hence, α =
0.128
32
= 0.004
Problem 22. The distance s metres travelled
in time t seconds is given by the formula
s = ut +
1
2
at2
, where u is the initial velocity in m/s
and a is the acceleration in m/s2. Find the
acceleration of the body if it travels 168m in 6s,
with an initial velocity of 10m/s
s = ut +
1
2
at2
, and s = 168,u = 10 and t = 6
Hence, 168 = (10)(6) +
1
2
a(6)2
168 = 60 + 18a
168 − 60 = 18a
108 = 18a
a =
108
18
= 6
Hence, the acceleration of the body is 6 m/s2.
Problem 23. When three resistors in an electrical
circuit are connected in parallel the total resistance
RT is given by
1
RT
=
1
R1
+
1
R2
+
1
R3
. Find the
total resistance when R1 = 5 , R2 = 10 and
R3 = 30
1
RT
=
1
5
+
1
10
+
1
30
=
6 + 3 + 1
30
=
10
30
=
1
3
Taking the reciprocal of both sides gives RT = 3
Alternatively, if
1
RT
=
1
5
+
1
10
+
1
30
, the LCM of the
denominators is 30RT.
Hence, 30RT
1
RT
= 30RT
1
5
+ 30RT
1
10
+ 30RT
1
30
.
Cancelling gives 30 = 6RT + 3RT + RT
i.e. 30 = 10RT
and RT =
30
10
= 3 , as above.

PracticeExercise 44 Practical problems
involving simple equations (answers on
page 344)
1. A formulaused for calculating resistance of a
cable is R =
ρL
a
. Given R = 1.25, L = 2500
and a = 2 × 10−4, ﬁnd the value of ρ.
2. Force F newtons is given by F = ma, where
m is the mass in kilograms and a is the accel-
eration in metrespersecond squared.Find the
acceleration when a force of 4kN is applied
to a mass of 500kg.
3. PV = mRT is the characteristic gas equa-
tion. Find the value of m when
P = 100 × 103, V = 3.00, R = 288 and
T = 300.
4. When three resistors R1, R2 and R3 are con-
nected in parallel, the total resistance RT is
determined from
1
RT
=
1
R1
+
1
R2
+
1
R3
(a) Find the total resistance when
R1 = 3 , R2 = 6 and R3 = 18 .
(b) Find the value of R3 given that
RT = 3 , R1 = 5 and R2 = 10 .
5. Six digital camera batteries and 3 camcorder
batteries cost £96. If a camcorder battery
costs £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 is
the voltage in volts and R is the resistance
in ohms. A soldering iron takes a current
of 0.30A from a 240V supply. Find the
resistance of the element.
7. The distance, s, travelled in time t seconds is
given by the formula s = ut +
1
2
a t2 where
u is the initial velocity in m/s and a is the
acceleration in m/s2. Calculate the accelera-
tion of the body if it travels 165m in 3s, with
an initial velocity of 10m/s.
Here are some further worked examples on solving
simple equations in practical situations.
Problem 24. The extension x m of an aluminium
tie bar of length l m and cross-sectional area Am2
when carrying a load of F Newtons is given by the
modulus of elasticity E = Fl/Ax. Find the
extension of the tie bar (in mm) if
E = 70 × 109 N/m2, F = 20 × 106 N, A = 0.1m2
and l = 1.4m
E = Fl/Ax, hence
70×109 N
m2
=
(20×106 N)(1.4m)
(0.1m2)(x)
(the unit of x is thus metres)
70 × 109
× 0.1 × x = 20 × 106
× 1.4
x =
20 × 106 × 1.4
70 × 109 × 0.1
Cancelling gives x =
2 × 1.4
7 × 100
m
=
2×1.4
7×100
×1000mm
= 4mm
Hence, the extension of the tie bar, x = 4 mm.
Problem 25. Power in a d.c. circuit is given by
P =
V 2
R
where V is the supply voltage and R is the
circuit resistance. Find the supply voltage if the
circuit resistance is 1.25 and the power measured
is 320W
Since P =
V 2
R
, then 320 =
V 2
1.25
(320)(1.25) = V 2
i.e. V 2
= 400
Supply voltage, V =
√
400 = ±20V
Problem 26. A painter is paid £6.30 per hour for a
basic 36 hour week and overtime is paid at one and
a third times this rate. Determine how many hours
the painter has to work in a week to earn £319.20
Basic rate per hour = £6.30 and overtime rate per hour
= 1
1
3
× £6.30 = £8.40
Let the number of overtime hours worked = x
Then, (36)(6.30) + (x)(8.40) = 319.20
226.80 + 8.40x = 319.20

8.40x = 319.20 − 226.80 = 92.40
x =
92.40
8.40
= 11
Thus, 11 hours overtime would have to be worked to
earn £319.20 per week. Hence, the total number of
hours worked is 36 + 11, i.e. 47 hours.
Problem 27. A formula relating initial and final
states of pressures, P1 and P2, volumes, V1 and V2,
and absolute temperatures, T1 and T2, of an ideal
gas is
P1V1
T1
=
P2V2
T2
. Find the value of P2 given
P1 = 100 × 103, V1 = 1.0, V2 = 0.266, T1 = 423
and T2 = 293
Since
P1V1
T1
=
P2V2
T2
then
(100 × 103)(1.0)
423
=
P2(0.266)
293
(100 × 103
)(1.0)(293) = P2(0.266)(423)
P2 =
(100 × 103)(1.0)(293)
(0.266)(423)
Hence, P2 = 260 × 103
or 2.6 × 105
.
Problem 28. The stress, f , in a material of a
thick cylinder can be obtained from
D
d
=
f + p
f − p
. Calculate the stress, given that
D = 21.5,d = 10.75 and p = 1800
Since
D
d
=
f + p
f − p
then
21.5
10.75
=
f + 1800
f − 1800
i.e.
2 =
f + 1800
f − 1800
Squaring both sides gives 4 =
f + 1800
f − 1800
4( f − 1800) = f + 1800
4 f − 7200 = f + 1800
4 f − f = 1800 + 7200
3 f = 9000
f =
9000
3
= 3000
Hence, stress,f = 3000
Problem 29. 12 workmen employed on a
building site earn between them a total of £4035 per
week. Labourers are paid £275 per week and
craftsmen are paid £380 per week. How many
craftsmen and how many labourers are employed?
Let the number of craftsmen be c. The number of
labourers is therefore (12 − c).
The wage bill equation is
380c + 275(12 − c) = 4035
380c + 3300− 275c = 4035
380c − 275c = 4035 − 3300
105c = 735
c =
735
105
= 7
Hence, there are 7 craftsmen and (12 − 7), i.e. 5
labourers on the site.
involving simple equations (answers on
page 344)
1. A rectangle has a length of 20cm and a width
b cm. When its width is reduced by 4cm its
area becomes 160cm2. Find the original width
and area of the rectangle.
2. Given R2 = R1(1 + αt), find α given
R1 = 5.0, R2 = 6.03 and t = 51.5
3. If v2 = u2 + 2as, find u given v = 24,
a = −40 and s = 4.05
4. The relationship between the temperature on
a Fahrenheit scale and that on a Celsius scale
is given by F =
9
5
C + 32. Express 113◦F in
degrees Celsius.
5. If t = 2π
w
Sg
, find the value of S given
w = 1.219, g = 9.81 and t = 0.3132
6. Two joiners and five mates earn £1824
between them for a particular job. If a joiner
earns £72 more than a mate, calculate the
earnings for a joiner and for a mate.

7. An alloy contains 60% by weight of copper,
the remainder being zinc. How much copper
must be mixed with 50kg of this alloy to give
an alloy containing 75% copper?
8. A rectangular laboratory has a length equal to
one and a half times its width and a perimeter
of 40m. Find its length and width.
9. Applying the principle of moments to a beam
results in the following equation:
F × 3 = (5 − F) × 7
where F is the force in newtons. Determine
the value of F.

Revision Test 4: Algebra and simple equations
1. Evaluate 3pqr3 − 2p2qr + pqr when p =
1
2
,
q = −2 and r = 1. (3)
In problems 2 to 7, simplify the expressions.
2.
9p2qr3
3pq2r
(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.
a4b2c
ab3c2
(3)
8. Factorize
(a) 2x2 y3 − 10xy2
(b) 21ab2c3 − 7a2bc2 + 28a3bc4 (5)
9. Factorize and simplify
2x2 y + 6xy2
x + 3y
−
x3 y2
x2 y
(5)
10. Remove the brackets and simplify
10a − [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)
4
9
y = 8
(b) 6x − 1 = 4x + 5 (4)
15. Solve the equation
5(t − 2) − 3(4 − t) = 2(t + 3) − 40 (4)
16. Solve the equations:
(a)
3
2x + 1
=
1
4x − 3
(b) 2x2 = 162 (7)
17. Kinetic energy is given by the formula,
Ek =
1
2
mv2 joules, where m is the mass in kilo-
grams and v is the velocity in metres per second.
Evaluate the velocity when Ek = 576 × 10−3J
and the mass is 5kg. (4)
18. An approximate relationship between the num-
ber of teeth T on a milling cutter, the diameter
of the cutter D and the depth of cut d is given
by T =
12.5D
D + 4d
. Evaluate d when T = 10 and
D = 32. (5)
19. The modulus of elasticity E is given by the for-
mula E =
FL
x A
where F is force in newtons, L
is the length in metres, x is the extension in
metres and A the cross-sectional area in square
metres. Evaluate A, in square centimetres, when
E = 80×109 N/m2, x =2mm, F =100×103 N
and L = 2.0m. (5)

Chapter 12
Transposing formulae
12.1 Introduction
In the formula I =
V
R
, I is called the subject of the
formula.
Similarly, in the formula y = mx + c, y is the subject of
the formula.
When a symbol other than the subject is required to
be the subject, the formula needs to be rearranged to
make a new subject. This rearranging process is called
transposing the formula or transposition.
For example, in the above formulae,
if I =
V
R
then V = IR
and if y = mx + c then x =
y − c
m
How did we arrive at these transpositions? This is the
purpose of this chapter — to show how to transpose for-
mulae. A great many equations occur in engineering and
it is essential that we can transpose them when needed.
12.2 Transposing formulae
There are no new rules for transposing formulae.
The same rules as were used for simple equations in
Chapter 11 are used; i.e., the balance of an equation
must be maintained: whatever is done to one side of
an equation must be done to the other.
It is best that you cover simple equations before trying
this chapter.
of transposing formulae.
Problem 1. Transpose p = q +r + s to make r
the subject
The object is to obtain r on its own on the LHS of the
equation. Changing the equation around so that r is on
the LHS gives
q +r + s = p (1)
From Chapter 11 on simple equations, a term can be
moved from one side of an equation to the other side as
long as the sign is changed.
Rearranging gives r = p − q − s.
Mathematically, we have subtracted q + s from both
sides of equation (1).
Problem 2. If a + b = w − x + y, express x as
the subject
As stated in Problem 1, a term can be moved from one
side of an equation to the other side but with a change
of sign.
Hence, rearranging gives x = w + y − a − b
Problem 3. Transpose v = f λ to make λ the
subject
v = f λ relates velocity v, frequency f and wave-
length λ
Rearranging gives f λ = v
Dividing both sides by f gives
f λ
f
=
v
f
Cancelling gives λ =
v
f
Problem 4. When a body falls freely through a
height h, the velocity v is given by v2
= 2gh.
Express this formula with h as the subject
DOI: 10.1016/B978-1-85617-697-2.00012-0

Rearranging gives 2gh = v2
Dividing both sides by 2g gives
2gh
2g
=
v2
2g
Cancelling gives h =
v2
2g
Problem 5. If I =
V
R
, rearrange to make V the
subject
I =
V
R
is Ohm’s law, where I is the current, V is the
voltage and R is the resistance.
Rearranging gives V
R
= I
Multiplying both sides by R gives R
V
R
= R(I)
Cancelling gives V = IR
Problem 6. Transpose a =
F
m
for m
a =
F
m
relates acceleration a, force F and mass m.
Rearranging gives F
m
= a
Multiplying both sides by m gives m
F
m
= m(a)
Cancelling gives F = ma
Rearranging gives ma = F
Dividing both sides by a gives
ma
a
=
F
a
i.e. m =
F
a
Problem 7. Rearrange the formula R =
ρL
A
to
make (a) A the subject and (b) L the subject
R =
ρL
A
relates resistance R of a conductor, resistiv-
ity ρ, conductor length L and conductor cross-sectional
area A.
(a) Rearranging gives
ρL
A
= R
Multiplying both sides by A gives
A
ρL
A
= A(R)
Cancelling gives ρL = AR
Rearranging gives AR = ρl
Dividing both sides by R gives
AR
R
=
ρL
R
Cancelling gives A =
ρL
R
(b) Multiplying both sides of
ρL
A
= R by A gives
ρL = AR
Dividing both sides by ρ gives
ρL
ρ
=
AR
ρ
Cancelling gives L =
AR
ρ
Problem 8. Transpose y = mx + c to make m the
subject
y = mx + c is the equation of a straight line graph,
where y is the vertical axis variable, x is the horizontal
axis variable, m is the gradient of the graph and c is the
y-axis intercept.
Subtracting c from both sides gives y − c = mx
or mx = y − c
Dividing both sides by x gives m =
y − c
x
PracticeExercise 46 Transposing formulae
Make the symbol indicated the subject of each of
theformulaeshown and expresseach in itssimplest
form.
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 85
9. I = PRT (T )
10. XL = 2π fL (L)
11. I =
E
R
(R)
12. y =
x
a
+ 3 (x)
13. F =
9
5
C + 32 (C)
14. XC =
1
2π f C
( f )
12.3 Further transposing of formulae
Here are some more transposition examples to help
us further understand how more difficult formulae are
transposed.
Problem 9. Transpose the formula v = u +
Ft
m
to
make F the subject
v = u +
Ft
m
relates final velocity v, initial velocity u,
force F, mass m and time t.
F
m
is acceleration a.
Rearranging gives u +
Ft
m
= v
and
Ft
m
= v − u
Multiplying each side by m gives
m
Ft
m
= m(v − u)
Cancelling gives Ft = m(v − u)
Dividing both sides by t gives
Ft
t
=
m (v − u)
t
Cancelling gives F =
m(v − u)
t
or F =
m
t
(v − u)
This shows two ways of expressing the answer. There
is often more than one way of expressing a trans-
posed answer. In this case, these equations for F are
equivalent; neither one is more correct than the other.
Problem 10. The final length L2 of a piece of
wire heated through θ◦C is given by the formula
L2 = L1(1 + αθ) where L1 is the original length.
Make the coefficient of expansion α the subject
Rearranging gives L1(1 + αθ) = L2
Removing the bracket gives L1 + L1αθ = L2
Rearranging gives L1αθ = L2 − L1
Dividing both sides by L1θ gives
L1αθ
L1θ
=
L2 − L1
L1θ
Cancelling gives α =
L2 − L1
L1θ
An alternative method of transposing L2 = L1 (1 + αθ)
for α is:
Dividing both sides by L1 gives
L2
L1
= 1 + αθ
Subtracting 1 from both sides gives
L2
L1
− 1 = αθ
or αθ =
L2
L1
− 1
Dividing both sides by θ gives α =
L2
L1
− 1
θ
The two answers α =
L2 − L1
L1θ
and α =
L2
L1
− 1
θ
look
quite different. They are, however, equivalent. The first
answer looks tidier but is no more correct than the
second answer.
Problem 11. A formula for the distance s moved
by a body is given by s =
1
2
(v + u)t. Rearrange the
formula to make u the subject
Rearranging gives
1
2
(v + u)t = s
Multiplying both sides by 2 gives (v + u)t = 2s
Dividing both sides by t gives
(v + u)t
t
=
2s
t
Cancelling gives v + u =
2s
t
Rearranging gives u =
2s
t
− v or u =
2s − vt
t
Problem 12. A formula for kinetic energy is
k =
1
2
mv2. Transpose the formula to make v the
subject
Rearranging gives
1
2
mv2
= k

Whenever the prospective new subject is a squared
term,that termisisolated on theLHSand then thesquare
root of both sides of the equation is taken.
Multiplying both sides by 2 gives mv2
= 2k
Dividing both sides by m gives
mv2
m
=
2k
m
Cancelling gives v2
=
2k
m
√
v2 =
2k
m
i.e. v =
2k
m
Problem 13. In a right-angled triangle having
sides x, y and hypotenuse z, Pythagoras’ theorem
states z2 = x2 + y2. Transpose the formula to ﬁnd x
Rearranging gives x2
+ y2
= z2
and x2
= z2
− y2
x = z2 − y2
Problem 14. Transpose y =
ML2
8EI
to make L the
subject
Multiplying both sides by 8E I gives 8EIy = ML2
Dividing both sides by M gives
8EIy
M
= L2
or L2
=
8EIy
M
√
L2 =
8EIy
M
i.e.
L =
8EIy
M
Problem 15. Given t = 2π
l
g
, ﬁnd g in terms of
t,l and π
Whenever the prospective new subject is withina square
root sign, it is best to isolate that term on the LHS and
then to square both sides of the equation.
Rearranging gives 2π
l
g
= t
Dividing both sides by 2π gives
l
g
=
t
2π
l
g
=
t
2π
2
=
t2
4π2
Cross-multiplying, (i.e. multiplying
each term by 4π2g), gives 4π2
l = gt2
or gt2
= 4π2
l
Dividing both sides by t2 gives
gt2
t2
=
4π2l
t2
Cancelling gives g =
4π2l
t2
Problem 16. The impedance Z of an a.c. circuit
is given by Z =
√
R2 + X2 where R is the
resistance. Make the reactance, X, the subject
Rearranging gives R2 + X2 = Z
Squaring both sides gives R2
+ X2
= Z2
Rearranging gives X2
= Z2
− R2
X = Z2 − R2
Problem 17. The volume V of a hemisphere of
radius r is given by V =
2
3
πr3. (a) Find r in terms
of V. (b) Evaluate the radius when V = 32cm3
(a) Rearranging gives
2
3
πr3
= V
Multiplying both sides by 3 gives 2πr3
= 3V
Dividing both sides by 2π gives
2πr3
2π
=
3V
2π
Cancelling gives r3
=
3V
2π

Taking the cube root of both sides gives
3
√
r3 = 3
3V
2π
i.e. r = 3
3V
2π
(b) When V = 32cm3,
radius r = 3
3V
2π
= 3
3 × 32
2π
= 2.48 cm.
PracticeExercise 47 Further transposing
theformulaeshown in problems1 to 13 and express
each in its simplest form.
1. S =
a
1 −r
(r)
2. y =
λ(x − d)
d
(x)
3. A =
3(F − f )
L
( f )
4. y =
AB2
5CD
(D)
5. R = R0(1 + αt) (t)
6.
1
R
=
1
R1
+
1
R2
(R2)
7. I =
E − e
R +r
(R)
8. y = 4ab2c2 (b)
9.
a2
x2
+
b2
y2
= 1 (x)
10. t = 2π
L
g
(L)
11. v2 = u2 + 2as (u)
12. A =
π R2θ
360
(R)
13. N =
a + x
y
(a)
14. Transpose Z = R2 + (2π f L)2 for L and
evaluate L when Z = 27.82, R = 11.76 and
f = 50.
12.4 More difﬁcult transposing of
formulae
Here are some more transposition examples to help
us further understand how more difﬁcult formulae are
transposed.
Problem 18. (a) Transpose S =
3d (L − d)
8
to
make l the subject. (b) Evaluate L when d = 1.65
and S = 0.82
The formula S =
3d (L − d)
8
represents the sag S at
the centre of a wire.
(a) Squaring both sides gives S2
=
3d(L − d)
8
Multiplying both sides by 8 gives
8S2
= 3d(L − d)
Dividing both sides by 3d gives
8S2
3d
= L − d
Rearranging gives L = d +
8S2
3d
(b) When d = 1.65 and S = 0.82,
L = d +
8 S2
3d
= 1.65 +
8 × 0.822
3 × 1.65
= 2.737
Problem 19. Transpose the formula
p =
a2
x2
+ a2
y
r
to make a the subject
Rearranging gives
a2x2 + a2 y
r
= p
Multiplying both sides by r gives
a2
x + a2
y = rp
Factorizing the LHS gives a2
(x + y) = rp
Dividing both sides by (x + y) gives
a2(x + y)
(x + y)
=
rp
(x + y)

Cancelling gives a2
=
rp
(x + y)
a =
rp
x + y
Whenever the letter required as the new subject
occurs more than once in the original formula, after
rearranging, factorizing will always be needed.
Problem 20. Make b the subject of the formula
a =
x − y
√
bd + be
Rearranging gives
x − y
√
bd + be
= a
Multiplying both sides by
√
bd + be gives
x − y = a
√
bd + be
or a
√
bd + be = x − y
Dividing both sidesbyagives
√
bd + be =
x − y
a
Squaring both sides gives bd + be =
x − y
a
2
Factorizing the LHS gives b(d + e) =
x − y
a
2
Dividing both sides by (d + e) gives
b =
x − y
a
2
(d + e)
or b =
(x − y)2
a2(d + e)
Problem 21. If a =
b
1 + b
, make b the subject of
the formula
Rearranging gives
b
1 + b
= a
Multiplying both sides by (1 + b) gives
b = a(1 + b)
Removing the bracket gives b = a + ab
Rearranging to obtain terms in b on the LHS gives
b − ab = a
Factorizing the LHS gives b(1 − a) = a
Dividing both sides by (1 − a) gives b =
a
1 − a
Problem 22. Transpose the formula V =
Er
R +r
to make r the subject
Rearranging gives
Er
R +r
= V
Multiplying both sides by (R +r) gives
Er = V (R +r)
Removing the bracket gives Er = V R + Vr
Rearranging to obtain terms in r on the LHS gives
Er − Vr = V R
Factorizing gives r(E − V ) = V R
Dividing both sides by (E − V ) gives r =
V R
E − V
Problem 23. Transpose the formula
y =
pq2
r + q2
− t to make q the subject
Rearranging gives
pq2
r + q2
− t = y
and
pq2
r + q2
= y + t
Multiplying both sides by (r + q2) gives
pq2
= (r + q2
)(y + t)
Removing brackets gives pq2
= ry +rt + q2
y + q2
t
Rearranging to obtain terms in q on the LHS gives
pq2
− q2
y − q2
t = ry +rt
Factorizing gives q2
(p − y − t) = r(y + t)
Dividing both sides by (p − y − t) gives
q2
=
r(y + t)
(p − y − t)
q =
r(y + t)
p − y − t
Problem 24. Given that
D
d
=
f + p
f − p
express p in terms of D,d and f

Rearranging gives f + p
f − p
=
D
d
f + p
f − p
=
D2
d2
Cross-multiplying, i.e. multiplying each term
by d2( f − p), gives
d2
( f + p) = D2
( f − p)
Removing brackets gives d2
f + d2
p = D2
f − D2
p
Rearranging, to obtain terms in p on the LHS gives
d2
p + D2
p = D2
f − d2
f
Factorizing gives p(d2
+ D2
) = f (D2
− d2
)
Dividing both sides by (d2 + D2) gives
p =
f (D2 − d2)
(d2 + D2)
PracticeExercise 48 Further transposing
the formulae shown in problems 1 to 7 and express
each in its simplest form.
1. y =
a2m − a2n
x
(a)
2. M = π(R4 −r4) (R)
3. x + y =
r
3 +r
(r)
4. m =
μL
L +rCR
(L)
5. a2 =
b2
− c2
b2
(b)
6.
x
y
=
1 +r2
1 −r2
(r)
7.
p
q
=
a + 2b
a − 2b
(b)
8. A formula for the focal length, f , of a convex
lens is
1
f
=
1
u
+
1
v
. Transpose the formula to
make v the subject and evaluate v when f = 5
and u = 6.
9. Thequantity ofheat, Q,isgiven by theformula
Q = mc(t2 − t1). Make t2 the subject of the
formula and evaluate t2 when m = 10, t1 = 15,
c = 4 and Q = 1600.
10. The velocity, v, of water in a pipe appears
in the formula h =
0.03Lv2
2dg
. Express v
as the subject of the formula and evalu-
ate v when h = 0.712, L = 150,d = 0.30 and
g = 9.81
11. The sag, S, at the centre of a wire is given
by the formula S =
3d(l − d)
8
. Make l
the subject of the formula and evaluate l when
d = 1.75 and S = 0.80.
12. In an electrical alternating current cir-
cuit the impedance Z is given by
Z = R2 + ωL −
1
ωC
2
. Transpose
the formula to make C the subject and hence
evaluate C when Z = 130, R = 120,ω = 314
and L = 0.32
13. An approximate relationship between the
number of teeth, T, on a milling cutter, the
diameter of cutter, D, and the depth of cut, d,
is given by T =
12.5 D
D + 4d
. Determine the value
of D when T = 10 and d = 4mm.
14. Make λ, the wavelength of X-rays, the subject
of the following formula:
μ
ρ
=
CZ4
√
λ5 n
a

Chapter 13
Solving simultaneous
equations
13.1 Introduction
Only oneequation isnecessary when finding thevalueof
a single unknown quantity (as with simple equations
in Chapter 11). However, when an equation contains
two unknown quantities it has an infinite number of
solutions. When two equations are available connecting
the same two unknown values then a unique solution
is possible. Similarly, for three unknown quantities it is
necessary to have three equations in order to solve for a
particular value of each of the unknown quantities, and
so on.
Equations which have to be solved together to find
the unique values of the unknown quantities, which are
true for each of the equations, are called simultaneous
equations.
Two methods of solving simultaneous equations analyt-
ically are:
(a) by substitution, and
(b) by elimination.
(A graphical solution of simultaneous equations is
shown in Chapter 19.)
13.2 Solving simultaneous equations
in two unknowns
The method of solving simultaneous equations is
demonstrated in the following worked problems.
Problem 1. Solve the following equations for x
and y, (a) by substitution and (b) by elimination
x + 2y = −1 (1)
4x − 3y = 18 (2)
(a) By substitution
From equation (1): x = −1 − 2y
Substitutingthis expression for x into equation (2)
gives
4(−1 − 2y) − 3y = 18
This is now a simple equation in y.
Removing the bracket gives
−4 − 8y − 3y = 18
−11y = 18 + 4 = 22
y =
22
−11
= −2
Substituting y = −2 into equation (1) gives
x + 2(−2) = −1
x − 4 = −1
x = −1 + 4 = 3
Thus, x = 3 and y = −2 is the solution to the
simultaneous equations.
Check: in equation (2), since x = 3 and y = −2,
LHS = 4(3) − 3(−2) = 12 + 6 = 18 = RHS
DOI: 10.1016/B978-1-85617-697-2.00013-2

Solving simultaneous equations 91
(b) By elimination
x + 2y = −1 (1)
4x − 3y = 18 (2)
If equation (1) is multiplied throughout by 4, the
coefficient of x will be the same as in equation (2),
giving
4x + 8y = −4 (3)
Subtracting equation (3) from equation (2) gives
4x − 3y = 18 (2)
4x + 8y = −4 (3)
0 − 11y = 22
Hence, 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 solution
x = 3,y = −2 is the only pair of values that satisfies
both of the original equations.
Problem 2. Solve, by a substitution method, the
simultaneous equations
3x − 2y = 12 (1)
x + 3y = −7 (2)
From equation (2), x = −7 − 3y
Substituting for x in equation (1) gives
3(−7 − 3y) − 2y = 12
i.e. −21 − 9y − 2y = 12
−11y = 12 + 21 = 33
Hence, y =
33
−11
= −3
Substituting y = −3 in equation (2) gives
x + 3(−3) = −7
i.e. x − 9 = −7
Hence x = −7 + 9 = 2
Thus, x = 2,y = −3 is the solutionof the simultaneous
equations. (Such solutions should always be checked
by substituting values into each of the original two
equations.)
Problem 3. Use an elimination method to solve
the following simultaneous equations
3x + 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 the
newly formed equations. Thus,
2 × equation (1) gives 6x + 8y = 10 (3)
3 × equation (2) gives 6x − 15y = −36 (4)
Equation (3) – equation (4) gives
0 + 23y = 46
i.e. y =
46
23
= 2
(Note+8y − −15y = 8y + 15y = 23y and 10−−36 =
10 + 36 = 46.)
Substituting y = 2 in equation (1) gives
3x + 4(2) = 5
from which 3x = 5 − 8 = −3
and x = −1
Checking, by substitutingx = −1 and y = 2 in equation
(2), gives
LHS = 2(−1) − 5(2) = −2 − 10 = −12 = RHS
Hence, x = −1 and y = 2 is the solution of the simul-
taneous equations.
The elimination method is the most common method of
solving simultaneous equations.
Problem 4. Solve
7x − 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.

5 × equation (1) gives 35x − 10y = 130 (3)
2 × equation (2) gives 12x + 10y = 58 (4)
Adding equations (3)
and (4) gives 47x + 0 = 188
Hence, x =
188
47
= 4
Note that when the signs of common coefﬁcients are
different the two equations are added and when the
signs of common coefﬁcients are the same the two
equations are subtracted (as in Problems 1 and 3).
Substituting x = 4 in equation (1) gives
7(4) − 2y = 26
28 − 2y = 26
28 − 26 = 2y
2 = 2y
Hence, y = 1
Checking, by substituting x = 4 and y = 1 in equation
(2), gives
LHS = 6(4) + 5(1) = 24 + 5 = 29 = RHS
Thus, the solution is x = 4,y = 1.
PracticeExercise 49 Solving simultaneous
Solve the following simultaneous equations and
verify the results.
1. 2x − y = 6 2. 2x − y = 2
x + y = 6 x − 3y = −9
3. x − 4y = −4 4. 3x − 2y = 10
5x − 2y = 7 5x + y = 21
5. 5p + 4q = 6 6. 7x + 2y = 11
2p − 3q = 7 3x − 5y = −7
7. 2x − 7y = −8 8. a + 2b = 8
3x + 4y = 17 b − 3a = −3
9. a + b = 7 10. 2x + 5y = 7
a − b = 3 x + 3y = 4
11. 3s + 2t = 12 12. 3x − 2y = 13
4s − t = 5 2x + 5y = −4
13. 5m − 3n = 11 14. 8a − 3b = 51
3m + n = 8 3a + 4b = 14
15. 5x = 2y 16. 5c = 1 − 3d
3x + 7y = 41 2d + c + 4 = 0
13.3 Further solving of simultaneous
equations
Here are some further worked problems on solving
simultaneous equations.
Problem 5. Solve
3p = 2q (1)
4p + q + 11 = 0 (2)
Rearranging gives
3p − 2q = 0 (3)
4p + q = −11 (4)
Multiplying equation (4) by 2 gives
8p + 2q = −22 (5)
Adding equations (3) and (5) gives
11p + 0 = −22
p =
−22
11
= −2
Substituting p = −2 into equation (1) gives
3(−2) = 2q
−6 = 2q
q =
−6
2
= −3
Checking, by substituting p = −2 and q = −3 into
equation (2), gives
LHS = 4(−2) + (−3) + 11 = −8 − 3 + 11 = 0 = RHS
Hence, the solution is p = −2,q = −3.

Problem 6. Solve
x
8
+
5
2
= y (1)
13 −
y
3
= 3x (2)
Whenever fractions are involved in simultaneous equa-
tions it is often easier to ﬁrstly remove them. Thus,
multiplying equation (1) by 8 gives
8
x
8
+ 8
5
2
= 8y
i.e. x + 20 = 8y (3)
39 − y = 9x (4)
Rearranging equations (3) and (4) gives
x − 8y = −20 (5)
9x + y = 39 (6)
72x + 8y = 312 (7)
73x + 0 = 292
x =
292
73
= 4
Substituting x = 4 into equation (5) gives
4 − 8y = −20
4 + 20 = 8y
24 = 8y
y =
24
8
= 3
Checking, substituting x = 4 and y = 3 in the original
equations, gives
(1): LHS =
4
8
+
5
2
=
1
2
+ 2
1
2
= 3 = y = RHS
(2): LHS = 13 −
3
3
= 13 − 1 = 12
RHS = 3x = 3(4) = 12
Hence, the solution is x = 4,y = 3.
Problem 7. Solve
2.5x + 0.75 − 3y = 0
1.6x = 1.08 − 1.2y
It is often easier to remove decimal fractions. Thus,
multiplying equations (1) and (2) by 100 gives
250x + 75 − 300y = 0 (1)
160x = 108 − 120y (2)
Rearranging gives
250x − 300y = −75 (3)
160x + 120y = 108 (4)
500x − 600y = −150 (5)
800x + 600y = 540 (6)
1300x + 0 = 390
x =
390
1300
=
39
130
=
3
10
= 0.3
Substituting x = 0.3 into equation (1) gives
250(0.3) + 75 − 300y = 0
75 + 75 = 300y
150 = 300y
y =
150
300
= 0.5
Checking, by substituting x = 0.3 and y = 0.5 in equa-
tion (2), gives
LHS = 160(0.3) = 48
RHS = 108 − 120(0.5) = 108 − 60 = 48
Hence, the solution is x = 0.3,y = 0.5

PracticeExercise 50 Solving simultaneous
Solve the following simultaneous equations and
verify the results.
1. 7p + 11 + 2q = 0 2.
x
2
+
y
3
= 4
−1 = 3q − 5p
x
6
−
y
9
= 0
3.
a
2
− 7 = −2b 4.
3
2
s − 2t = 8
12 = 5a +
2
3
b
s
4
+ 3y = −2
5.
x
5
+
2y
3
=
49
15
6. v − 1 =
u
12
3x
7
−
y
2
+
5
7
= 0 u +
v
4
−
25
2
= 0
7. 1.5x − 2.2y = −18 8. 3b − 2.5a = 0.45
2.4x + 0.6y = 33 1.6a + 0.8b = 0.8
13.4 Solving more difﬁcult
Here are some further worked problems on solving more
difﬁcult simultaneous equations.
Problem 8. Solve
2
x
+
3
y
= 7 (1)
1
x
−
4
y
= −2 (2)
In this type of equation the solution is easier if a
substitution is initially made. Let
1
x
= a and
1
y
= b
Thus equation (1) becomes 2a + 3b = 7 (3)
and equation (2) becomes a − 4b = −2 (4)
2a − 8b = −4 (5)
0 + 11b = 11
i.e. b = 1
Substituting b = 1 in equation (3) gives
2a + 3 = 7
2a = 7 − 3 = 4
i.e. a = 2
Checking, substituting a = 2 and b = 1 in equation (4),
gives
LHS = 2 − 4(1) = 2 − 4 = −2 = RHS
Hence, a = 2 and b = 1.
However, since
1
x
= a, x =
1
a
=
1
2
or 0.5
and since
1
y
= b, y =
1
b
=
1
1
= 1
Hence, the solution is x = 0.5, y = 1.
Problem 9. Solve
1
2a
+
3
5b
= 4 (1)
4
a
+
1
2b
= 10.5 (2)
Let
1
a
= x and
1
b
= y
then x
2
+
3
5
y = 4 (3)
4x +
1
2
y = 10.5 (4)
To remove fractions, equation (3) is multiplied by 10,
giving
10
x
2
+ 10
3
5
y = 10(4)
i.e. 5x + 6y = 40 (5)
8x + y = 21 (6)
48x + 6y = 126 (7)

43x + 0 = 86
x =
86
43
= 2
2
2
+
3
5
y = 4
3
5
y = 4 − 1 = 3
y =
5
3
(3) = 5
Since
1
a
= x, a =
1
x
=
1
2
or 0.5
and since
1
b
= y, b =
1
y
=
1
5
or 0.2
Hence, the solution is a = 0.5, b = 0.2, which may be
checked in the original equations.
Problem 10. Solve
1
x + y
=
4
27
(1)
1
2x − y
=
4
33
(2)
To eliminate fractions, both sides of equation (1) are
multiplied by 27(x + y), giving
27(x + y)
1
x + y
= 27(x + y)
4
27
i.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) gives
60 = 12x and x =
60
12
= 5
Substituting x = 5 in equation (3) gives
27 = 4(5) + 4y
from which 4y = 27 − 20 = 7
and y =
7
4
= 1
3
4
or 1.75
Hence, x = 5, y = 1.75 is the required solution, which
may be checked in the original equations.
Problem 11. Solve
x − 1
3
+
y + 2
5
=
2
15
(1)
1 − x
6
+
5 + y
2
=
5
6
(2)
Before equations (1) and (2) can be simultaneously
solved, the fractions need to be removed and the
equations rearranged.
15
x − 1
3
+ 15
y + 2
5
= 15
2
15
i.e. 5(x − 1) + 3(y + 2) = 2
5x − 5 + 3y + 6 = 2
5x + 3y = 2 + 5 − 6
Hence, 5x + 3y = 1 (3)
6
1 − x
6
+ 6
5 + y
2
= 6
5
6
i.e. (1 − x) + 3(5 + y) = 5
1 − x + 15 + 3y = 5
−x + 3y = 5 − 1 − 15
Hence, −x + 3y = −11 (4)
Thus the initial problem containing fractions can be
expressed as
5x + 3y = 1 (3)
−x + 3y = −11 (4)
6x + 0 = 12
x =
12
6
= 2

5(2) + 3y = 1
10 + 3y = 1
3y = 1 − 10 = −9
y =
−9
3
= −3
Checking, substituting x = 2, y = −3 in equation (4)
gives
LHS = −2 + 3(−3) = −2 − 9 = −11 = RHS
Hence, the solution is x = 2,y = −3.
PracticeExercise 51 Solving more difﬁcult
simultaneous equations (answers on page
345)
In problems 1 to 7, solve the simultaneous equa-
tions and verify the results
1.
3
x
+
2
y
= 14 2.
4
a
−
3
b
= 18
5
x
−
3
y
= −2
2
a
+
5
b
= −4
3.
1
2p
+
3
5q
= 5 4.
5
x
+
3
y
= 1.1
5
p
−
1
2q
=
35
2
3
x
−
7
y
= −1.1
5.
c + 1
4
−
d + 2
3
+ 1 = 0
1 − c
5
+
3 − d
4
+
13
20
= 0
7.
5
x + y
=
20
27
4
2x − y
=
16
336.
3r + 2
5
−
2s − 1
4
=
11
5
3 + 2r
4
+
5 − s
3
=
15
4
8. If 5x −
3
y
= 1 and x +
4
y
=
5
2
, ﬁnd the value
of
xy + 1
y
There are a number of situations in engineering and
science in which the solution of simultaneous equations
is required. Some are demonstrated in the following
worked problems.
Problem 12. The law connecting friction F and
load L for an experiment is of the form F = aL + b
where a and b are constants. When F = 5.6N,
L = 8.0N and when F = 4.4N, L = 2.0N. Find the
values of a and b and the value of F when
L = 6.5N
Substituting F = 5.6 and L = 8.0 into F = aL + b
gives
5.6 = 8.0a + b (1)
Substituting F = 4.4 and L = 2.0 into F = aL + b
gives
4.4 = 2.0a + b (2)
1.2 = 6.0a
a =
1.2
6.0
=
1
5
or 0.2
Substituting a =
1
5
into equation (1) gives
5.6 = 8.0
1
5
+ b
5.6 = 1.6 + b
5.6 − 1.6 = b
i.e. b = 4
Checking, substitutinga =
1
5
and b = 4 in equation (2),
gives
RHS = 2.0
1
5
+ 4 = 0.4 + 4 = 4.4 = LHS
Hence, a =
1
5
and b = 4
When L = 6.5, F = aL + b =
1
5
(6.5) + 4 = 1.3 + 4,
i.e. F = 5.30N.

Problem 13. The equation of a straight line,
of gradient m and intercept on the y-axis c, is
y = mx + c. If a straight line passes through the
point where x = 1 and y = −2, and also through
the point where x = 3.5 and y = 10.5, find the
values of the gradient and the y-axis intercept
Substituting 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 + c
gives
10.5 = 3.5m + c (2)
12.5 = 2.5m, from which, m =
12.5
2.5
= 5
Substituting m = 5 into equation (1) gives
−2 = 5 + c
c = −2 − 5 = −7
Checking, substituting m = 5 and c = −7 in equation
(2), gives
RHS = (3.5)(5) + (−7) = 17.5 − 7 = 10.5 = LHS
Hence, the gradient m = 5 and the y-axis intercept
c = −7.
Problem 14. When Kirchhoff’s laws are applied
to the electrical circuit shown in Figure 13.1, the
currents I1 and I2 are connected by the equations
27 = 1.5I1 + 8(I1 − I2) (1)
−26 = 2I2 − 8(I1 − I2) (2)
27 V 26 V
8 V
1.5 V 2 V
l1 l2
(l12l2)
Figure 13.1
Solve the equations to find the values of currents I1
and I2
Removing the brackets from equation (1) gives
27 = 1.5I1 + 8I1 − 8I2
Rearranging gives
9.5I1 − 8I2 = 27 (3)
Removing the brackets from equation (2) gives
−26 = 2I2 − 8I1 + 8I2
Rearranging gives
−8I1 + 10I2 = −26 (4)
47.5I1 − 40I2 = 135 (5)
−32I1 + 40I2 = −104 (6)
15.5I1 + 0 = 31
I1 =
31
15.5
= 2
Substituting I1 = 2 into equation (3) gives
9.5(2) − 8I1 = 27
19 − 8I2 = 27
19 − 27 = 8I2
−8 = 8I2
and I2 = −1
Hence, the solution is I1 = 2 and I2 = −1 (which may
be checked in the original equations).
Problem 15. The distance s metres from a fixed
point of a vehicle travelling in a straight line with
constant acceleration, a m/s2, is given by
s = ut + 1
2 at2, where u is the initial velocity in m/s
and t the time in seconds. Determine the initial
velocity and the acceleration given that s = 42 m
when t = 2 s, and s = 144m when t = 4s. Also find
the distance travelled after 3s

Substitutings = 42 and t = 2 into s = ut +
1
2
at2 gives
42 = 2u +
1
2
a(2)2
i.e. 42 = 2u + 2a (1)
Substituting s = 144 and t = 4 into s = ut +
1
2
at2
gives
144 = 4u +
1
2
a(4)2
i.e. 144 = 4u + 8a (2)
84 = 4u + 4a (3)
60 = 0 + 4a
and a =
60
4
= 15
Substituting a = 15 into equation (1) gives
42 = 2u + 2(15)
42 − 30 = 2u
u =
12
2
= 6
Substituting a = 15 and u = 6 in equation (2) gives
RHS = 4(6) + 8(15) = 24 + 120 = 144 = LHS
Hence, the initial velocity u = 6 m/s and the acceler-
ation a = 15 m/s2
.
Distance travelled after 3s is given by s = ut +
1
2
at2
where t = 3,u = 6 and a = 15.
Hence, s = (6)(3) +
1
2
(15)(3)2 = 18 + 67.5
i.e. distance travelled after 3 s = 85.5 m.
Problem 16. The resistance R of a length of
wire at t◦C is given by R = R0(1 + αt), where R0
is the resistance at 0◦C and α is the temperature
coefﬁcient of resistance in /◦C. Find the values of α
and R0 if R = 30 at 50◦C and R = 35 at 100◦C
Substituting R = 30 and t = 50 into R = R0(1 + αt)
gives
30 = R0(1 + 50α) (1)
Substituting R = 35 and t = 100 into R = R0(1 + αt)
gives
35 = R0(1 + 100α) (2)
Although these equations may be solved by the conven-
tional substitution method, an easier way is to eliminate
R0 by division. Thus, dividing equation (1) by equation
(2) gives
30
35
=
R0(1 + 50α)
R0(1 + 100α)
=
1 + 50α
1 + 100α
30(1 + 100α) = 35(1 + 50α)
30 + 3000α = 35 + 1750α
3000α − 1750α = 35 − 30
1250α = 5
i.e. α =
5
1250
=
1
250
or 0.004
Substituting α =
1
250
into equation (1) gives
30 = R0 1 + (50)
1
250
30 = R0(1.2)
R0 =
30
1.2
= 25
Checking, substituting α =
1
250
and R0 = 25 in equa-
tion (2), gives
RHS = 25 1 + (100)
1
250
= 25(1.4) = 35 = LHS
Thus, the solution is α = 0.004/◦C and R0 = 25 .
Problem 17. The molar heat capacity of a solid
compound is given by the equation c = a + bT,
where a and b are constants. When c = 52, T = 100
and when c = 172, T = 400. Determine the values
of a and b

When c = 52, T = 100, hence
52 = a + 100b (1)
When c = 172, T = 400, hence
172 = a + 400b (2)
120 = 300b
from which, b =
120
300
= 0.4
Substituting b = 0.4 in equation (1) gives
52 = a + 100(0.4)
a = 52 − 40 = 12
Hence, a = 12 and b = 0.4
involving simultaneous equations (answers
on page 345)
1. In a system of pulleys, the effort P required to
raise a load W is given by P = aW + b, where
a and b are constants. If W = 40 when P = 12
and W = 90 when P = 22, find the values of
a and b.
2. Applying Kirchhoff’s laws to an electrical
circuit produces the following equations:
5 = 0.2I1 + 2(I1 − I2)
12 = 3I2 + 0.4I2 − 2(I1 − I2)
Determine the values of currents I1 and I2
3. 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 the
velocity when t = 3.5
4. Three new cars and 4 new vans supplied to a
dealer together cost £97700 and 5 new cars
and 2 new vans of the same models cost
£103100. Find the respective costs of a car
and a van.
5. y = mx + c is the equation of a straight line
of slope m and y-axis intercept c. If the line
passes through the point where x = 2 and
y = 2, and also through the point where x = 5
and y = 0.5,find theslopeand y-axisintercept
of the straight line.
6. The resistance R ohms of copper wire at t◦C
is given by R = R0(1 + αt), where R0 is the
resistanceat 0◦C and α isthetemperaturecoef-
ficient of resistance. If R = 25.44 at 30◦C
and R = 32.17 at 100◦
C, find α and R0
7. The molar heat capacity of a solid compound
is given by the equation c = a + bT . When
c = 60, T = 100 and when c = 210, T = 400.
Find the values of a and b.
8. In an engineering process, two variables p and
q are related by q = ap + b/p, where a and b
are constants. Evaluate a and b if q = 13 when
p = 2 and q = 22 when p = 5.
9. In a system of forces, the relationship between
two forces F1 and F2 is given by
5F1 + 3F2 + 6 = 0
3F1 + 5F2 + 18 = 0
Solve for F1 and F2
13.6 Solving simultaneous equations
in three unknowns
Equations containing three unknowns may be solved
using exactly the same procedures as those used with
two equations and two unknowns, providing that there
are 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 two
unknowns. For example, multiplying equation (1) by 4
and then subtracting this new equation from equation (2)
will produce an equation with only x and y involved.

4x + 4y + 4z = 16 (4)
−2x − 7y = 17 (5)
Similarly, multiplying equation (3) by 2 and then
adding this new equation to equation (2) will produce
another equation with only x and y involved.
6x − 4y − 4z = 4 (6)
Equation (2)+equation (6) gives
8x − 7y = 37 (7)
Rewriting equation (5) gives
−2x − 7y = 17 (5)
Now we can use the previous method for solving
simultaneous equations in two unknowns.
Equation (7) – equation (5) gives 10x = 20
from which, x = 2
(Note that 8x − −2x = 8x + 2x = 10x)
−4 − 7y = 17
from which, −7y = 17 + 4 = 21
and y = −3
Substituting x = 2 and y = −3 into equation (1) gives
2 − 3 + z = 4
from which, z = 5
Hence, the solution of the simultaneous equations is
x = 2,y = −3 and z = 5.
PracticeExercise 53 Simultaneous
equations in three unknowns (answers on
page 345)
tions in 3 unknowns.
1. x + 2y + 4z = 16 2. 2x + y − z = 0
2x − y + 5z = 18 3x + 2y + z = 4
3x + 2y + 2z = 14 5x + 3y + 2z = 8
3. 3x + 5y + 2z = 6 4. 2x + 4y + 5z = 23
x − y + 3z = 0 3x − y − 2z = 6
2 + 7y + 3z = −3 4x + 2y + 5z = 31
5. 2x + 3y + 4z = 36 6. 4x + y + 3z = 31
3x + 2y + 3z = 29 2x − y + 2z = 10
x + y + z = 11 3x + 3y − 2z = 7
7. 5x + 5y − 4z = 37 8. 6x + 7y + 8z = 13
2x − 2y + 9z = 20 3x + y − z = −11
−4x + y + z = −14 2x − 2y − 2z = −18
9. 3x + 2y + z = 14
7x + 3y + z = 22.5
4x − 4y − z = −8.5
10. Kirchhoff’s laws are used to determine the
current equations in an electrical network and
result in the following:
i1 + 8i2 + 3i3 = −31
3i1 − 2i2 + i3 = −5
2i1 − 3i2 + 2i3 = 6
Determine the values of i1,i2 and i3
11. The forces in three members of a frame-
work are F1, F2 and F3. They are related by
following simultaneous equations.
1.4F1 + 2.8F2 + 2.8F3 = 5.6
4.2F1 − 1.4F2 + 5.6F3 = 35.0
4.2F1 + 2.8F2 − 1.4F3 = −5.6
Find the values of F1, F2 and F3

Revision Test 5 : Transposition and simultaneous equations
This assignment covers the material contained in Chapters 12 and 13. The marks available are shown in brackets at
the end of each question.
1. Transpose p − q +r = a − b for b. (2)
2. Make π the subject of the formula r =
c
2π
(2)
3. Transpose V =
1
3
πr2
h for h. (2)
4. Transpose I =
E − e
R +r
for E. (2)
5. Transpose k =
b
ad − 1
for d. (4)
6. Make g the subject of the formula t = 2π
L
g
(3)
7. Transpose A =
πR2θ
360
for R. (2)
8. Make r the subject of the formula
x + y =
r
3 +r
(5)
9. Make L the subject of the formula
m =
μL
L +rCR
(5)
10. The surface area A of a rectangular prism is
given by the formula A = 2(bh + hl +lb). Eval-
uate b when A = 11750 mm2
,h = 25mm and
l = 75mm. (4)
11. The velocity v of water in a pipe appears in
the formula h =
0.03 Lv2
2dg
. Evaluate v when
h = 0.384,d = 0.20, L = 80 and g = 10. (3)
12. A formula for the focal length f of a convex
lens is
1
f
=
1
u
+
1
v
. Evaluate v when f = 4 and
u = 20. (4)
In problems 13 and 14, solve the simultaneous equa-
tions.
13. (a) 2x + y = 6 (b) 4x − 3y = 11
5x − y = 22 3x + 5y = 30 (9)
14. (a) 3a − 8 +
b
8
= 0
b +
a
2
=
21
4
(b)
2p + 1
5
−
1 − 4q
2
=
5
2
1 − 3p
7
+
2q − 3
5
+
32
35
= 0 (18)
15. In an engineering process two variables x and y
are related by the equation y = ax +
b
x
, where a
and b are constants. Evaluate a and b if y = 15
when x = 1 and y = 13 when x = 3. (5)
16. Kirchhoff’s laws are used to determine the current
equations in an electrical network and result in the
following:
i1 + 8i2 + 3i3 = −31
3i1 − 2i2 + i3 = −5
2i1 − 3i2 + 2i3 = 6
Determine the values of i1, i2 and i3 (10)

Chapter 14
Solving quadratic equations
14.1 Introduction
As stated in Chapter 11, an equation is a statement
that two quantities are equal and to ‘solve an equation’
means ‘to ﬁnd the value of the unknown’. The value of
the unknown is called the root of the equation.
A quadratic equation is one in which the highest
power of the unknown quantity is 2. For example,
x2 − 3x + 1 = 0 is a quadratic equation.
There are four methods of solving quadratic equa-
tions. These are:
(a) by factorization (where possible),
(b) by ‘completing the square’,
(c) by using the ‘quadratic formula’, or
(d) graphically (see Chapter 19).
14.2 Solution of quadratic equations
by factorization
Multiplying out (x + 1)(x − 3) gives x2 − 3x + x − 3
i.e. x2 − 2x − 3. The reverse process of moving from
x2 − 2x − 3 to (x + 1)(x − 3) is called factorizing.
If the quadratic expression can be factorized this
provides the simplest method of solving a quadratic
equation.
For example, if x2 − 2x − 3 = 0, then, by factorizing
(x + 1)(x − 3) = 0
Hence, either (x + 1) = 0, i.e. x = −1
or (x − 3) = 0, i.e. x = 3
Hence, x = −1 and x = 3 are the roots of the quad-
ratic equation x2 − 2x − 3 = 0.
The technique of factorizing is often one of trial and
error.
Problem 1. Solve the equation x2 + x − 6 = 0 by
factorization
The factors of x2 are x and x. These are placed in
brackets: (x )(x )
The factors of −6 are +6 and −1, or −6 and +1, or +3
and −2, or −3 and +2.
The only combination to give a middle term of +x is
+3 and −2,
i.e. x2
+ x − 6 = (x + 3)(x − 2)
The quadratic equation x2 + x − 6 = 0 thus becomes
(x + 3)(x − 2) = 0
Since the only way that this can be true is for either the
ﬁrst or the second or both factors to be zero, then
either (x + 3) = 0, i.e. x = −3
or (x − 2) = 0, i.e. x = 2
Hence, the roots of x2 + x − 6 = 0 are x = −3 and
x = 2.
Problem 2. Solve the equation x2
+ 2x − 8 = 0
by factorization
The factors of x2 are x and x. These are placed in
brackets: (x )(x )
The factors of −8 are +8 and −1, or −8 and +1, or +4
and −2, or −4 and +2.
The only combination to give a middle term of +2x is
+4 and −2,
i.e. x2
+ 2x − 8 = (x + 4)(x − 2)
(Note that the product of the two inner terms (4x) added
to the product of the two outer terms (−2x) must equal
the middle term, +2x in this case.)
DOI: 10.1016/B978-1-85617-697-2.00014-4

Solving quadratic equations 103
The quadratic equation x2
+ 2x − 8 = 0 thus becomes
(x + 4)(x − 2) = 0
Since the only way that this can be true is for either the
ﬁrst or the second or both factors to be zero,
either (x + 4) = 0, i.e. x = −4
or (x − 2) = 0, i.e. x = 2
Hence, the roots of x2 + 2x − 8 = 0 are x = −4 and
x = 2.
Problem 3. Determine the roots of
x2
− 6x + 9 = 0 by factorization
x2 − 6x + 9 = (x − 3)(x − 3), i.e. (x − 3)2 = 0
The LHS is known as a perfect square.
Hence, x = 3 is the only root of the equation
x2 − 6x + 9 = 0.
Problem 4. Solve the equation x2 − 4x = 0
Factorizing gives x(x − 4) = 0
If x(x − 4) = 0,
either x = 0 or x − 4 = 0
i.e. x = 0 or x = 4
These are the two roots of the given equation. Answers
can always be checked by substitution into the original
equation.
Problem 5. Solve the equation x2 + 3x − 4 = 0
Factorizing gives (x − 1)(x + 4) = 0
Hence, either x − 1 = 0 or x + 4 = 0
i.e. x = 1 or x = −4
Problem 6. Determine the roots of 4x2 − 25 = 0
by factorization
The LHS of 4x2 − 25 = 0 is the difference of two
squares, (2x)2 and (5)2.
By factorizing, 4x2 − 25 = (2x + 5)(2x − 5), i.e.
(2x + 5)(2x − 5) = 0
Hence, either (2x + 5) = 0, i.e. x = −
5
2
= −2.5
or (2x − 5) = 0, i.e. x =
5
2
= 2.5
Problem 7. Solve the equation x2
− 5x + 6 = 0
Factorizing gives (x − 3)(x − 2) = 0
Hence, either x − 3 = 0 or x − 2 = 0
i.e. x = 3 or x = 2
Problem 8. Solve the equation x2 = 15 − 2x
+ 2x − 15 = 0
Factorizing gives (x + 5)(x − 3) = 0
Hence, either x + 5 = 0 or x − 3 = 0
i.e. x = −5 or x = 3
Problem 9. Solve the equation 3x2 − 11x − 4 = 0
by factorization
The factors of 3x2 are 3x and x. These are placed in
brackets: (3x )(x )
The factors of −4 are −4 and +1, or +4 and −1, or −2
and 2.
Remembering that the product of the two inner terms
added 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 thus
becomes (3x + 1)(x − 4) = 0
Hence, either (3x + 1) = 0, i.e. x = −
1
3
or (x − 4) = 0, i.e. x = 4
and both solutions may be checked in the original
equation.
Problem 10. Solve the quadratic equation
4x2
+ 8x + 3 = 0 by factorizing
The 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 added
to 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)

Hence, (2x + 3)(2x + 1) = 0, from which either
(2x + 3) = 0 or (2x + 1) = 0.
Thus, 2x = −3, from which x = −
3
2
or −1.5
or 2x = −1, from which x = −
1
2
or −0.5
which may be checked in the original equation.
15x2 + 2x − 8 = 0 by factorizing
The factors of 15x2 are 15x and x or 5x and 3x.
The factors of −8 are −4 are +2, or 4 and −2, or −8
and +1, or 8 and −1.
By trial and error the only combination that works is
15x2
+ 2x − 8 = (5x + 4)(3x − 2)
Hence, (5x + 4)(3x − 2) = 0, from which either 5x +
4 = 0 or 3x − 2 = 0.
Hence, x = −
4
5
or x =
2
3
which may be checked in the original equation.
Problem 12. The roots of a quadratic equation
are
1
3
and −2. Determine the equation in x
If the roots of a quadratic equation are, say, α and β,
then (x − α)(x − β) = 0.
Hence, if α =
1
3
and β = −2,
x −
1
3
(x − (−2)) = 0
x −
1
3
(x + 2) = 0
x2
−
1
3
x + 2x −
2
3
= 0
x2
+
5
3
x −
2
3
= 0
or 3x2
+ 5x − 2 = 0
Problem 13. Find the equation in x whose roots
are 5 and −5
If 5 and −5 are the roots of a quadratic equation then
(x − 5)(x + 5) = 0
i.e. x2
− 5x + 5x − 25 = 0
i.e. x2
− 25 = 0
Problem 14. Find the equation in x whose roots
are 1.2 and −0.4
If 1.2 and −0.4 are the roots of a quadratic equation
then
(x − 1.2)(x + 0.4) = 0
i.e. x2
− 1.2x + 0.4x − 0.48 = 0
i.e. x2
− 0.8x − 0.48 = 0
PracticeExercise 54 Solving quadratic
equations by factorization (answers on
page 346)
In problems 1 to 30, solve the given equations by
factorization.
1. x2 − 16 = 0 2. x2 + 4x − 32 = 0
3. (x + 2)2 = 16 4. 4x2 − 9 = 0
5. 3x2 + 4x = 0 6. 8x2 − 32 = 0
7. x2 − 8x + 16 = 0 8. x2 + 10x + 25 = 0
9. x2 − 2x + 1 = 0 10. x2 + 5x + 6 = 0
11. x2 + 10x + 21 = 0 12. x2 − x − 2 = 0
13. y2 − y − 12 = 0 14. y2 − 9y + 14 = 0
15. x2
+ 8x + 16 = 0 16. x2
− 4x + 4 = 0
17. x2 + 6x + 9 = 0 18. x2 − 9 = 0
19. 3x2 + 8x + 4 = 0 20. 4x2 + 12x + 9 = 0
21. 4z2 −
1
16
= 0 22. x2 + 3x − 28 = 0
23. 2x2 − x − 3 = 0 24. 6x2 − 5x + 1 = 0
25. 10x2 + 3x − 4 = 0 26. 21x2 − 25x = 4
27. 8x2 + 13x − 6 = 0 28. 5x2 + 13x − 6 = 0
29. 6x2 − 5x − 4 = 0 30. 8x2 + 2x − 15 = 0
In problems 31 to 36, determine the quadratic
equations in x whose roots are
31. 3 and 1 32. 2 and −5
33. −1 and −4 34. 2.5 and −0.5
35. 6 and −6 36. 2.4 and −0.7

by ‘completing the square’
An expression such as x2 or (x + 2)2 or (x − 3)2 is
called a perfect square.
If x2 = 3 then x = ±
√
3
If (x + 2)2 = 5 then x + 2 = ±
√
5 and x = −2 ±
√
5
If (x − 3)2 = 8 then x − 3 = ±
√
8 and x = 3 ±
√
8
Hence, if a quadratic equation can be rearranged so that
one side of the equation is a perfect square and the other
side of the equation is a number, then the solution of
the equation is readily obtained by taking the square
roots of each side as in the above examples. The process
of rearranging one side of a quadratic equation into a
perfect square before solving is called ‘completing the
square’.
(x + a)2
= x2
+ 2ax + a2
Thus,inordertomakethequadraticexpressionx2 + 2ax
into a perfect square, it is necessary to add (half the
coefficient of x)2, i.e.
2a
2
2
or a2
For example, x2 + 3x becomes a perfect square by
adding
3
2
2
, i.e.
x2
+ 3x +
3
2
2
= x +
3
2
2
The method of completing the square is demonstrated
in the following worked problems.
Problem 15. Solve 2x2 + 5x = 3 by completing
the square
The procedure is as follows.
(i) Rearrangetheequation so that all termsareon the
same side of the equals sign (and the coefficient
of the x2 term is positive). Hence,
2x2
+ 5x − 3 = 0
(ii) Make the coefficient of the x2 term unity. In this
case thisis achieved by dividingthroughoutby 2.
Hence,
2x2
2
+
5x
2
−
3
2
= 0
i.e. x2
+
5
2
x −
3
2
= 0
(iii) Rearrange the equations so that the x2
and x
terms are on one side of the equals sign and the
constant is on the other side. Hence,
x2
+
5
2
x =
3
2
(iv) Add to both sides of the equation (half the coeffi-
cient of x)2. In this case the coefficient of x is
5
2
Half the coefficient squared is therefore
5
4
2
Thus,
x2
+
5
2
x +
5
4
2
=
3
2
+
5
4
2
The LHS is now a perfect square, i.e.
x +
5
4
2
=
3
2
+
5
4
2
(v) Evaluate the RHS. Thus,
x +
5
4
2
=
3
2
+
25
16
=
24 + 25
16
=
49
16
(vi) Take the square root of both sides of the equation
(remembering that the square root of a number
gives a ± answer). Thus,
x +
5
4
2
=
49
16
i.e. x +
5
4
= ±
7
4
(vii) Solve the simple equation. Thus,
x = −
5
4
±
7
4
i.e. x = −
5
4
+
7
4
=
2
4
=
1
2
or 0.5
and x = −
5
4
−
7
4
= −
12
4
= −3
Hence, x = 0.5 or x = −3; i.e., the roots of the
equation 2x2 + 5x = 3 are 0.5 and −3.
Problem 16. Solve 2x2 + 9x + 8 = 0, correct to 3
significant figures, by completing the square
Making the coefficient of x2 unity gives
x2 +
9
2
x + 4 = 0
Rearranging gives x2 +
9
2
x = −4

Adding to both sides (half the coefficient of x)2
gives
x2
+
9
2
x +
9
4
2
=
9
4
2
− 4
The LHS is now a perfect square. Thus,
x +
9
4
2
=
81
16
− 4 =
81
16
−
64
16
=
17
16
x +
9
4
=
17
16
= ±1.031
Hence, x = −
9
4
± 1.031
i.e. x = −1.22 or −3.28, correct to 3 significant figures.
Problem 17. By completing the square, solve the
quadratic equation 4.6y2 + 3.5y − 1.75 = 0, correct
to 3 decimal places
4.6y2
+ 3.5y − 1.75 = 0
Making the coefficient of y2 unity gives
y2
+
3.5
4.6
y −
1.75
4.6
= 0
and rearranging gives y2
+
3.5
4.6
y =
1.75
4.6
Adding to both sides (half the coefficient of y)2 gives
y2
+
3.5
4.6
y +
3.5
9.2
2
=
1.75
4.6
+
3.5
9.2
2
The LHS is now a perfect square. Thus,
y +
3.5
9.2
2
= 0.5251654
y +
3.5
9.2
=
√
0.5251654 = ±0.7246830
Hence, y = −
3.5
9.2
± 0.7246830
i.e. y = 0.344 or −1.105
equations by completing the square
Solve the following equations correct to 3 decimal
places by completing the square.
1. x2 + 4x + 1 = 0 2. 2x2 + 5x − 4 = 0
3. 3x2 − x − 5 = 0 4. 5x2 − 8x + 2 = 0
5. 4x2 − 11x + 3 = 0 6. 2x2 + 5x = 2
by formula
Let the general form of a quadratic equation be given
by ax2 + bx + c = 0, where a,b and c are constants.
Dividing ax2
+ bx + c = 0 by a gives
x2
+
b
a
x +
c
a
= 0
+
b
a
x = −
c
a
Adding to each side of the equation the square of half
the coefficient of the term in x to make the LHS a perfect
square gives
x2
+
b
a
x +
b
2a
2
=
b
2a
2
−
c
a
Rearranging gives x +
b
a
2
=
b2
4a2
−
c
a
=
b2
− 4ac
4a2
x +
b
2a
=
b2 − 4ac
4a2
=
±
√
b2 − 4ac
2a
Hence, x = −
b
2a
±
√
b2 − 4ac
2a
i.e. the quadratic formula is x =
−b ±
√
b2 − 4ac
2a
(This method of obtainingthe formula is completing the
square − as shown in the previous section.)
In summary,
if ax2
+ bx + c = 0 then x =
−b ±
√
b2 − 4ac
2a
This is known as the quadratic formula.

Problem 18. Solve x2
+ 2x − 8 = 0 by using the
quadratic formula
Comparing x2 + 2x − 8 = 0 with ax2 + bx + c = 0
gives a = 1,b = 2 and c = −8.
Substituting these values into the quadratic formula
x =
−b ±
√
b2 − 4ac
2a
gives
x =
−2 ± 22 − 4(1)(−8)
2(1)
=
−2 ±
√
4 + 32
2
=
−2 ±
√
36
2
=
−2 ± 6
2
=
−2 + 6
2
or
−2 − 6
2
Hence, x =
4
2
or
−8
2
, i.e. x = 2 or x = −4.
Problem 19. Solve 3x2 − 11x − 4 = 0 by using
the quadratic formula
Comparing 3x2 − 11x − 4 = 0 with ax2 + bx + c = 0
gives a = 3,b = −11 and c = −4. Hence,
x =
−(−11) ± (−11)2 − 4(3)(−4)
2(3)
=
+11 ±
√
121 + 48
6
=
11 ±
√
169
6
=
11 ± 13
6
=
11 + 13
6
or
11 − 13
6
Hence, x =
24
6
or
−2
6
, i.e. x = 4 or x = −
1
3
Problem 20. Solve 4x2 + 7x + 2 = 0 giving the
roots correct to 2 decimal places
Comparing 4x2 + 7x + 2 = 0 with ax2 + bx + c gives
a = 4,b = 7 and c = 2. Hence,
x =
−7 ± 72 − 4(4)(2)
2(4)
=
−7 ±
√
17
8
=
−7 ± 4.123
8
=
−7 + 4.123
8
or
−7 − 4.123
8
Hence, x = −0.36 or −1.39, correct to 2 decimal
places.
Problem 21. Use the quadratic formula to solve
x + 2
4
+
3
x − 1
= 7 correct to 4 significant figures
Multiplying throughout by 4(x − 1) gives
4(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 − 28
Hence, x2
− 27x + 38 = 0
Using the quadratic formula,
x =
−(−27) ± (−27)2 − 4(1)(38)
2
=
27 ±
√
577
2
=
27 ± 24.0208
2
Hence, x =
27 + 24.0208
2
= 25.5104
or x =
27 − 24.0208
2
= 1.4896
Hence, x = 25.51 or 1.490, correct to 4 significant
figures.
equations by formula (answers on page 346)
Solve the following equations by using the
quadratic formula, correct to 3 decimal places.
1. 2x2 + 5x − 4 = 0
2. 5.76x2 + 2.86x − 1.35 = 0
3. 2x2
− 7x + 4 = 0
4. 4x + 5 =
3
x
5. (2x + 1) =
5
x − 3
6. 3x2 − 5x + 1 = 0

7. 4x2 + 6x − 8 = 0
8. 5.6x2 − 11.2x − 1 = 0
9. 3x(x + 2) + 2x(x − 4) = 8
10. 4x2 − x(2x + 5) = 14
11.
5
x − 3
+
2
x − 2
= 6
12.
3
x − 7
+ 2x = 7 + 4x
13.
x + 1
x − 1
= x − 3
quadratic equations
There are many practical problems in which a
quadratic equation has first to be obtained, from given
information, before it is solved.
Problem 22. The area of a rectangle is 23.6cm2
and its width is 3.10cm shorter than its length.
Determine the dimensions of the rectangle, correct
to 3 significant figures
Let the length of the rectangle be x cm. Then the width
is (x − 3.10)cm.
Area = length × width = x(x − 3.10) = 23.6
i.e. x2
− 3.10x − 23.6 = 0
x =
−(−3.10) ± (−3.10)2 − 4(1)(−23.6)
2(1)
=
3.10 ±
√
9.61 + 94.4
2
=
3.10 ± 10.20
2
=
13.30
2
or
−7.10
2
Hence, x = 6.65cm or −3.55cm. The latter solution is
neglected 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 to
3 significant figures.)
Problem 23. Calculate the diameter of a solid
cylinder which has a height of 82.0cm and a total
surface area of 2.0m2
Total 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) + 2πr2
i.e. 2πr2
+ 2πr(0.82) − 2.0 = 0
Dividing throughout by 2π gives r2 +0.82r −
1
π
= 0
r =
−0.82 ± (0.82)2 − 4(1) − 1
π
2(1)
=
−0.82 ±
√
1.94564
2
=
−0.82 ± 1.39486
2
= 0.2874 or − 1.1074
Thus, the radius r of the cylinder is 0.2874m (the
negative solution being neglected).
Hence, the diameter of the cylinder
= 2 × 0.2874
= 0.5748m or 57.5cm
Problem 24. The height s metres of a mass
projected vertically upwards at time t seconds is
s = ut −
1
2
gt2. Determine how long the mass will
take after being projected to reach a height of 16m
(a) on the ascent and (b) on the descent, when
u = 30m/s and g = 9.81m/s2
When height s = 16m, 16 = 30t −
1
2
(9.81)t2
i.e. 4.905t2
− 30t + 16 = 0
t =
−(−30) ± (−30)2 − 4(4.905)(16)
2(4.905)
=
30 ±
√
586.1
9.81
=
30 ± 24.21
9.81
= 5.53 or 0.59

Hence, the mass will reach a height of 16m after
0.59s 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 way
around the shed. If the area of the path is 9.50m2,
calculate its width to the nearest centimetre
Figure 14.1 shows a plan view of the shed with its
surrounding path of width t metres.
t
2.0m
4.0m (4.0 12t)
SHED
t
Figure 14.1
Area of path = 2(2.0 × t) + 2t(4.0 + 2t)
i.e. 9.50 = 4.0t + 8.0t + 4t2
or 4t2
+12.0t − 9.50 = 0
Hence,
t =
−(12.0) ± (12.0)2 − 4(4)(−9.50)
2(4)
=
−12.0 ±
√
296.0
8
=
−12.0 ± 17.20465
8
i.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 to
the nearest centimetre.
Problem 26. If the total surface area of a solid
cone is 486.2cm2 and its slant height is 15.3cm,
determine its base diameter.
From Chapter 27, page 245, the total surface area A of
a solid cone is given by A = πrl + πr2, where l is the
slant height and r the base radius.
If A = 482.2 and l = 15.3, then
482.2 = πr(15.3) + πr2
i.e. πr2
+ 15.3πr − 482.2 = 0
or r2
+ 15.3r −
482.2
π
= 0
r =
−15.3 ± (15.3)2 − 4
−482.2
π
2
=
−15.3 ±
√
848.0461
2
=
−15.3 ± 29.12123
2
Hence, radius r = 6.9106cm (or −22.21cm, which is
meaningless and is thus ignored).
Thus, the diameter of the base = 2r = 2(6.9106)
= 13.82cm.
involving quadratic equations (answers on
page 346)
1. The angle a rotating shaft turns through in t
seconds is given by θ = ωt +
1
2
αt2. Deter-
mine the time taken to complete 4 radians if
ω 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 is
the current in amperes. Determine the current
necessary to produce a power of 2.5 watts in
the circuit.
3. The area of a triangle is 47.6cm2 and its
perpendicular height is 4.3cm more than its
base length. Determine the length of the base
4. The sag, l, in metres in a cable stretched
between two supports, distance x m apart, is
given by l =
12
x
+ x. Determine the distance
between the supports when the sag is 20 m.
5. Theacid dissociation constant Ka ofethanoic
acid is 1.8 × 10−5 moldm−3 for a particu-
lar solution. Using the Ostwald dilution law,

Ka =
x2
v(1 − x)
, determine x, the degree of
ionization, given that v = 10dm3.
6. A rectangular building is 15m long by 11m
wide. A concrete path of constant width is
laid all the way around the building. If the
area of the path is 60.0m2, calculate its width
correct to the nearest millimetre.
7. The total surface area of a closed cylindrical
container is 20.0m3. Calculate the radius of
the cylinder if its height is 2.80m.
8. The bending moment M at a point in a beam
is given by M =
3x(20 − x)
2
, where x metres
is the distance from the point of support.
Determine the value of x when the bending
moment is 50Nm.
9. A tennis court measures 24m by 11m. In the
layout ofanumberofcourtsanareaofground
must be allowed for at the ends and at the
sides of each court. If a border of constant
width is allowed around each court and the
total areaofthecourt and itsborderis950m2,
ﬁnd the width of the borders.
10. Two resistors, when connected in series, have
a total resistance of 40 ohms. When con-
nected in parallel their total resistance is 8.4
ohms. If one of the resistors has a resistance
of Rx , ohms,
(a) show that R2
x − 40Rx + 336 = 0 and
(b) calculate the resistance of each.
14.6 Solution of linear and quadratic
equations simultaneously
Sometimes a linear equation and a quadratic equation
need to be solved simultaneously. An algebraic method
of solution is shown in Problem 27; a graphical solution
is shown in Chapter 19, page 160.
Problem 27. Determine the values of x and y
which simultaneously satisfy the equations
y = 5x − 4 − 2x2 and y = 6x − 7
For a simultaneous solution the values of y must be
equal, hence the RHS of each equation is equated.
Thus, 5x − 4 − 2x2 = 6x − 7
Rearranging gives 5x − 4 − 2x2
− 6x + 7 = 0
i.e. −x + 3 − 2x2
= 0
or 2x2
+ x − 3 = 0
Factorizing gives (2x + 3)(x − 1) = 0
i.e. x = −
3
2
or x = 1
In the equation y = 6x − 7,
when x = −
3
2
, y = 6 −
3
2
− 7 = −16
and when x = 1, y = 6 − 7 = −1
(Checking the result in y = 5x − 4 − 2x2:
when x = −
3
2
, y = 5 −
3
2
− 4 − 2 −
3
2
2
= −
15
2
− 4 −
9
2
= −16,as above,
and when x = 1, y = 5 − 4 − 2 = −1,as above.)
Hence, the simultaneous solutions occur when
x = −
3
2
,y = −16 and when x = 1,y = −1.
PracticeExercise 58 Solving linear and
quadratic equations simultaneously
Determine the solutions of the following simulta-
neous equations.
1. y = x2 + x + 1 2. y = 15x2 + 21x − 11
y = 4 − x y = 2x − 1
3. 2x2 + y = 4 + 5x
x + y = 4

Chapter 15
Logarithms
15.1 Introduction to logarithms
With 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 rules
of logarithms.
From Chapter 7, we know that 16 = 24
.
The number 4 is called the power or the exponent or
the index. In the expression 24
, the number 2 is called
the 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 this
expression 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 power
or index in the original expression. It is the base in
the original expression that becomes the base of the
logarithm.
The two statements 16 = 24
and
log216 = 4 are equivalent
If 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 base
of a’, i.e.
if y = ax
then x = loga y
In another example, if we write down that 64 = 82 then
theequivalent statement using logarithmsislog8 64 = 2.
In another example, if we write down that log3 27 = 3
then the equivalent statement using powers is 33 = 27.
So the two sets of statements, one involving powers and
one involving logarithms, are equivalent.
15.1.2 Common logarithms
From the above, if we write down that 1000 = 103, then
3 = log10 1000. This may be checked using the ‘log’
button on your calculator.
Logarithms having a base of 10 are called common
logarithms and log10 is usually abbreviated to lg. The
following values may be checked using a calculator.
lg27.5 = 1.4393...
lg378.1 = 2.5776...
lg0.0204 = −1.6903...
15.1.3 Napierian logarithms
Logarithms having a base of e (where e is a mathemat-
ical constant approximately equal to 2.7183) are called
hyperbolic, Napierian or natural logarithms, and
loge is usually abbreviated to ln. The following values
may be checked using a calculator.
ln3.65 = 1.2947...
ln417.3 = 6.0338...
ln0.182 = −1.7037...
Napierian logarithms are explained further in Chapter
16, following.
DOI: 10.1016/B978-1-85617-697-2.00015-6

Here are some worked problems to help understand-
ing of logarithms.
Problem 1. Evaluate log3 9
Let x = log3 9 then 3x = 9 from the definition of
a logarithm,
i.e. 3x = 32, from which x = 2
Hence, log3 9= 2
Let x = log10 10 then 10x = 10 from the definition
of a logarithm,
i.e. 10x = 101, from which x = 1
Hence, log10 10 = 1 (which may be checked
using a calculator).
Let x = log16 8 then 16x = 8 from the definition
of a logarithm,
i.e. (24)x = 23 i.e. 24x = 23 from the laws
of indices,
from which, 4x = 3 and x =
3
4
Hence, log16 8 =
3
4
Problem 4. Evaluate lg0.001
Let x = lg0.001 = log10 0.001 then 10x = 0.001
i.e. 10x = 10−3
from which, x = −3
Hence, lg 0.001 = −3 (which may be checked
using a calculator)
Problem 5. Evaluate lne
Let x = lne = loge e then ex = e
i.e. ex = e1, from which
x = 1
Hence, lne = 1 (which may be checked
by a calculator)
Problem 6. Evaluate log3
1
81
Let x = log3
1
81
then 3x =
1
81
=
1
34
= 3−4
from which x = −4
Hence, log3
1
81
= −4
Problem 7. Solve the equation lg x = 3
If lg x = 3 then log10 x = 3
and x = 103 i.e. x = 1000
Problem 8. Solve the equation log2 x = 5
If log2 x = 5 then x = 25 = 32
Problem 9. Solve the equation log5 x = −2
If log5 x = −2 then x = 5−2 =
1
52
=
1
25
PracticeExercise 59 Laws of logarithms
In problems1 to 11,evaluatethegiven expressions.
1. log10 10000 2. log2 16 3. log5 125
4. log2
1
8
5. log8 2 6. log7 343
7. lg 100 8. lg 0.01 9. log4 8
10. log27 3 11. ln e2
In problems 12 to 18, solve the equations.
12. log10 x = 4 13. lg x = 5
14. log3 x = 2 15. log4 x = −2
1
2
16. lg x = −2 17. log8 x = −
4
3
18. ln x = 3

Logarithms 113
15.2 Laws of logarithms
There are three laws of logarithms, which apply to any
base:
(1) To multiply two numbers:
log(A × B) = logA + logB
The followingmay be checked by using a calculator.
lg 10 = 1
Also, lg 5 + lg 2 = 0.69897...+ 0.301029... = 1
Hence, lg(5 × 2) = lg 10 = lg 5 + lg 2
(2) To divide two numbers:
log
A
B
= logA −logB
The following may be checked using a calculator.
ln
5
2
= ln2.5 = 0.91629...
Also, ln5 − ln2 = 1.60943...− 0.69314...
= 0.91629...
Hence, ln
5
2
= ln5 − ln2
(3) To raise a number to a power:
logAn
= n logA
The following may be checked using a calculator.
lg52 = lg25 = 1.39794...
Also, 2lg5 = 2 × 0.69897... = 1.39794...
Hence, lg52 = 2lg5
Here are some worked problems to help understanding
of the laws of logarithms.
Problem 10. Write log4 + log7 as the logarithm
of a single number
log4 + log7 = log(7 × 4) by the ﬁrst law of
logarithms
= log28
Problem 11. Write log16 − log2 as the logarithm
of a single number
log16 − log2 = log
16
2
by the second law of
logarithms
= log8
Problem 12. Write 2 log 3 as the logarithm of a
single number
2log 3 = log 32 by the third law of logarithms
= log9
Problem 13. Write
1
2
log 25 as the logarithm of a
single number
1
2
log25 = log25
1
2 by the third law of logarithms
= log
√
25 = log5
Problem 14. Simplify log 64 − log 128 + log 32
64 = 26,128 = 27 and 32 = 25
Hence, log64 − log128 + log32
= log26
− log27
+ log25
= 6log2 − 7log2 + 5log2
by the third law of logarithms
= 4log2
Problem 15. Write
1
2
log16 +
1
3
log27 − 2log5
as the logarithm of a single number
1
2
log16 +
1
3
log27 − 2log5
= log16
1
2 + log27
1
3 − log52
= log
√
16 + log 3
√
27 − log25
by the laws of indices
= log4 + log3 − log25

= log
4 × 3
25
by the first and second
laws of logarithms
= log
12
25
= log0.48
Problem 16. Write (a) log 30 (b) log 450 in terms
of log 2, log 3 and log 5 to any base
(a) log30 = log(2 × 15) = log(2 × 3 × 5)
= log2 + log3 + log5
by the first law
of logarithms
(b) log450 = log(2 × 225) = log(2 × 3 × 75)
= log(2 × 3 × 3 × 25)
= log(2 × 32 × 52)
= log2 + log32 + log52
by the first law
of logarithms
i.e. log450 = log2 + 2log3 + 2log5
Problem 17. Write log
8 ×
4
√
5
81
in terms of
log 2, log 3 and log 5 to any base
log
8 ×
4
√
5
81
= log8 + log
4
√
5 − log81
by the first and second laws
of logarithms
= log23 + log5
1
4 − log34
by the laws of indices
i.e. log
8 ×
4
√
5
81
= 3log2 +
1
4
log5 − 4log3
log25 − log125 +
1
2
log625
3log5
log25 − log125 +
1
2
log625
3log5
=
log52
− log53
+
1
2
log54
3log5
=
2log5 − 3log5 +
4
2
log5
3log5
=
1log5
3log5
=
1
3
log(x − 1) + log(x + 8) = 2log(x + 2)
LHS = log(x − 1) + log(x + 8) = log(x − 1)(x + 8)
from the first
law of logarithms
= log(x2 + 7x − 8)
RHS = 2log(x + 2) = log(x + 2)2
from the first
law of logarithms
= log(x2 + 4x + 4)
Hence, log(x2 + 7x − 8) = log(x2 + 4x + 4)
from which, x2 + 7x − 8 = x2 + 4x + 4
i.e. 7x − 8 = 4x + 4
i.e. 3x = 12
and x = 4
1
2
log4 = logx
1
2
log4 = log4
1
2 from the third law of
logarithms
= log
√
4 from the laws of indices
Hence,
1
2
log4 = logx
becomes log
√
4 = logx
i.e. log2 = logx
from which, 2 = x
i.e. the solution of the equation is x = 2.

Logarithms 115
log x2 − 3 − logx = log2
log x2 − 3 − log x = log
x2 − 3
x
from the second
law of logarithms
Hence, log
x2 − 3
x
= log2
from which,
x2 − 3
x
= 2
Rearranging gives x2 − 3 = 2x
and x2 − 2x − 3 = 0
Factorizing gives (x − 3)(x + 1) = 0
from which, x = 3 or x = −1
x = −1 is not a valid solution since the logarithm of a
negative number has no real root.
Hence, the solution of the equation is x = 3.
PracticeExercise 60 Laws of logarithms
In problems 1 to 11, write as the logarithm of a
single number.
1. log2+ log3 2. log3 + log5
3. log3 + log4 − log6
4. log7 + log21 − log49
5. 2log2 + log3 6. 2log2 + 3log5
7. 2log5 −
1
2
log81 + log36
8.
1
3
log8 −
1
2
log81 + log27
9.
1
2
log4 − 2log3 + log45
10.
1
4
log16 + 2log3 − log18
11. 2log2 + log5 − log10
Simplify the expressions given in problems
12 to 14.
12. log27 − log9 + log81
13. log64+ log32 − log128
14. log8 − log4 + log32
Evaluate the expressions given in problems 15
and 16.
15.
1
2
log16 −
1
3
log8
log4
16.
log9 − log3 +
1
2
log81
2log3
Solve the equations given in problems 17 to 22.
17. logx4 − logx3 = log5x − log2x
18. log2t3 − logt = log16 + logt
19. 2logb2 − 3logb = log8b − log4b
20. log(x + 1) + log(x − 1) = log3
21.
1
3
log27 = log(0.5a)
22. log(x2 − 5) − logx = log4
15.3 Indicial equations
The laws of logarithms may be used to solve
certain equations involving powers, called indicial
equations.
For example, to solve, say, 3x = 27, logarithmsto a base
of 10 are taken of both sides,
i.e. log10 3x
= log10 27
and x log10 3 = log10 27
Rearranging gives x =
log10 27
log10 3
=
1.43136...
0.47712...
= 3 which may be readily
checked.
Note,
log27
log3
is not equal to log
27
3

Problem 22. Solve the equation 2x = 5, correct
to 4 significant figures
Taking logarithms to base 10 of both sides of 2x
= 5
gives
log10 2x
= log10 5
i.e. x log10 2 = log10 5
log10 5
log10 2
=
0.6989700...
0.3010299...
= 2.322,correct to 4
Problem 23. Solve the equation 2x+1 = 32x−5
correct to 2 decimal places
Taking logarithms to base 10 of both sides gives
log10 2x+1
= log10 32x−5
i.e. (x + 1)log10 2 = (2x − 5)log10 3
x log10 2 + log10 2 = 2x log10 3 − 5log10 3
x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771)
i.e. 0.3010x + 0.3010 = 0.9542x − 2.3855
Hence, 2.3855 + 0.3010 = 0.9542x − 0.3010x
2.6865 = 0.6532x
from which x =
2.6865
0.6532
= 4.11,
Problem 24. Solve the equation x2.7 = 34.68,
Taking logarithms to base 10 of both sides gives
log10 x2.7
= log10 34.68
2.7log10 x = log10 34.68
Hence, log10 x =
log10 34.68
2.7
= 0.57040
Thus, x = antilog 0.57040 = 100.57040
= 3.719,
PracticeExercise 61 Indicial equations
In problems 1 to 8, solve the indicial equations for
x, each correct to 4 significant figures.
1. 3x = 6.4 2. 2x = 9
3. 2x−1 = 32x−1 4. x1.5 = 14.91
5. 25.28 = 4.2x 6. 42x−1 = 5x+2
7. x−0.25 = 0.792 8. 0.027x = 3.26
9. The decibel gain n of an amplifier is given
by n = 10log10
P2
P1
, where P1 is the power
input and P2 is the power output. Find the
power gain
P2
P1
when n = 25 decibels.
15.4 Graphs of logarithmic functions
A graph of y = log10 x is shown in Figure 15.1 and a
graph of y = loge x is shown in Figure 15.2. Both can
be seen to be of similar shape; in fact, the same general
shape occurs for a logarithm to any base.
In general, with a logarithm to any base, a, it is noted
that
(a) loga 1 = 0
Let loga = x then ax = 1 from the definition of the
logarithm.
If ax = 1 then x = 0 from the laws of logarithms.
Hence, loga 1 = 0. In the above graphs it is seen
that log101 = 0 and loge 1 = 0.
(b) loga a = 1
Let loga a = x then ax = a from the definition of
a logarithm.
If ax = a then x = 1.

Logarithms 117
y
0.5
0 1 2 3
20.5
21.0
x
x 3
0.48
2
0.30
1
0
0.5
20.30
0.2
20.70
0.1
21.0y5 log10x
Figure 15.1
Hence, loga a = 1. (Check with a calculator that
log10 10 = 1 and loge e = 1.)
(c) loga 0 → −∞
Let loga 0 = x then ax = 0 from the deﬁnition of a
logarithm.
y
2
1
0 1 2 3 4 5 6 x
x 6 5 4 3 2 1 0.5 0.2 0.1
1.79 1.61 1.39 1.10 0.69 0 20.69 21.61 22.30
21
22
y5 logex
Figure 15.2
If ax = 0, and a is a positive real number, then
x 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 16
Exponential functions
16.1 Introduction to exponential
functions
An exponential function is one which contains ex , e
being a constant called the exponent and having an
approximate value of 2.7183. The exponent arises from
the natural laws of growth and decay and is used as a
base for natural or Napierian logarithms.
The most common method of evaluating an exponential
function isby using a scientific notationcalculator. Use
your calculator to check the following values.
e1
= 2.7182818, correct to 8 significant figures,
e−1.618
e0.12
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 4
decimal places, using a calculator:
0.0256(e5.21
− e2.49
)
0.0256(e5.21
− e2.49
)
= 0.0256(183.094058...− 12.0612761...)
Problem 2. Evaluate the following correct to 4
decimal places, using a calculator:
5
e0.25 − e−0.25
e0.25 + e−0.25
5
e0.25 − e−0.25
e0.25 + e−0.25
= 5
1.28402541...− 0.77880078...
1.28402541...+ 0.77880078...
= 5
0.5052246...
2.0628262...
Problem 3. The instantaneous voltage v in a
capacitive circuit is related to time t by the equation
v = V e−t/CR
where V , C and R are constants.
Determine v, correct to 4 significant figures,
when t = 50ms, C = 10μF, R = 47k and
V = 300volts
v = 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.
PracticeExercise 62 Evaluating
exponential functions (answers on page 347)
1. Evaluate the following,correct to 4 significant
figures.
(a) e−1.8 (b) e−0.78 (c) e10
2. Evaluate the following,correct to 5 significant
figures.
(a) e1.629 (b) e−2.7483 (c) 0.62e4.178
DOI: 10.1016/B978-1-85617-697-2.00016-8

Exponential functions 119
In problems 3 and 4, evaluate correct to 5 decimal
places.
3. (a)
1
7
e3.4629 (b) 8.52e−1.2651
(c)
5e2.6921
3e1.1171
4. (a)
5.6823
e−2.1347
(b)
e2.1127 − e−2.1127
2
(c)
4 e−1.7295 − 1
e3.6817
5. The length of a bar, l, at a temperature, θ,
is given by l = l0eαθ , where l0 and α are
constants. Evaluate l, correct to 4 significant
figures, where l0 = 2.587, θ = 321.7 and
α = 1.771 × 10−4.
6. When a chain of length 2L is suspended from
two points, 2D metres apart on the same hor-
izontal level, D = k ln
L +
√
L2 + k2
k
.
Evaluate D when k = 75m and L = 180m.
16.2 The power series for ex
The value of ex can be calculated to any required degree
of accuracy since it is defined in terms of the following
power series:
ex
= 1 + x+
x 2
2!
+
x 3
3!
+
x 4
4!
+ ··· (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 are
added, an actual value for ex (where x is a real number)
is obtained. The more terms that are taken, the closer
will be the value of ex to its actual value. The value of
the exponent e, correct to say 4 decimal places, may be
determined by substituting x = 1 in the power series of
equation (1). Thus,
e1
= 1 + 1 +
(1)2
2!
+
(1)3
3!
+
(1)4
4!
+
(1)5
5!
+
(1)6
6!
+
(1)7
7!
+
(1)8
8!
+ ···
= 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833
+ 0.00139 + 0.00020+ 0.00002 + ···
= 2.71828
i.e. e = 2.7183, correct to 4 decimal places.
The value of e0.05, correct to say 8 significant figures, is
found by substituting x = 0.05 in the power series for
ex . Thus,
e0.05
= 1 + 0.05 +
(0.05)2
2!
+
(0.05)3
3!
+
(0.05)4
4!
+
(0.05)5
5!
+ ···
= 1 + 0.05 + 0.00125 + 0.000020833
+ 0.000000260+ 0.0000000026
i.e. e 0.05
= 1.0512711, correct to 8 significant figures.
In this example, successive terms in the series grow
smaller 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 accurate
result.
If, in the series of equation (1), x is replaced by −x,
then
e−x
= 1 + (−x) +
(−x)2
2!
+
(−x)3
3!
+ ···
i.e. e−x
= 1 − x +
x 2
2!
−
x 3
3!
+ ···
In a similar manner the power series for ex may
be used to evaluate any exponential function of the form
aekx , where a and k are constants.
In the series of equation (1), let x be replaced by kx.
Then
aekx
= a 1 + (kx) +
(kx)2
2!
+
(kx)3
3!
+ ···
Thus, 5e2x
= 5 1 + (2x) +
(2x)2
2!
+
(2x)3
3!
+ ···
= 5 1 + 2x +
4x2
2
+
8x3
6
+ ···
i.e. 5e2x
= 5 1 + 2x + 2x2
+
4
3
x3
+ ···
Problem 4. Determine the value of 5e0.5, correct
to 5 significant figures, by using the power series
for ex

From equation (1),
ex
= 1 + x +
x2
2!
+
x3
3!
+
x4
4!
+ ···
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
+ 0.0000217
i.e. e0.5
= 1.64872, correct to 6 significant
figures
Hence, 5e0.5
= 5(1.64872) = 8.2436, correct to 5
Problem 5. Determine the value of 3e−1, correct
to 4 decimal places, using the power series for ex
Substituting x = −1 in the power series
ex
= 1 + x +
x2
2!
+
x3
3!
+
x4
4!
+ ···
gives e−1
= 1 + (−1) +
(−1)2
2!
+
(−1)3
3!
+
(−1)4
4!
+ ···
= 1 − 1 + 0.5 − 0.166667 + 0.041667
− 0.008333 + 0.001389
− 0.000198 + ···
Hence, 3e−1 = (3)(0.367858) = 1.1036, correct to 4
decimal places.
Problem 6. Expand ex (x2 − 1) as far as the term
in x5
The power series for ex is
ex
= 1 + x +
x2
2!
+
x3
3!
+
x4
4!
+
x5
5!
+ ···
Hence,
ex
(x2
− 1)
= 1 + x +
x2
2!
+
x3
3!
+
x4
4!
+
x5
5!
+ ··· (x2
− 1)
= x2
+ x3
+
x4
2!
+
x5
3!
+ ···
− 1 + x +
x2
2!
+
x3
3!
+
x4
4!
+
x5
5!
+ ···
Grouping like terms gives
ex
(x2
− 1)
= −1 − x + x2
−
x2
2!
+ x3
−
x3
3!
+
x4
2!
−
x4
4!
+
x5
3!
−
x5
5!
+ ···
= −1 − x +
1
2
x2
+
5
6
x3
+
11
24
x4
+
19
120
x5
when expanded as far as the term in x5.
PracticeExercise 63 Power series for ex
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−0.3
and 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 functions
Values of ex and e−x obtained from a calculator, correct
to 2 decimal places, over a range x = −3 to x = 3, are
shown in Table 16.1.

Table 16.1
x −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0 0.5 1.0 1.5 2.0 2.5 3.0
ex
0.05 0.08 0.14 0.22 0.37 0.61 1.00 1.65 2.72 4.48 7.39 12.18 20.09
e−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 0.05
Figure 16.1 shows graphs of y = ex and y = e−x .
y
20
16
y ϭ e xy ϭ eϪx
12
8
4
Ϫ1 0 1 2 3 xϪ2Ϫ3
Figure 16.1
Problem 7. Plot a graph of y = 2e0.3x over a
range of x = −2 to x = 3. Then determine the value
of y when x = 2.2 and the value of x when y = 1.6
A table of values is drawn up as shown below.
x −3 −2 −1 0 1 2 3
2e0.3 x 0.81 1.10 1.48 2.00 2.70 3.64 4.92
A graph of y = 2e0.3x is shown plotted in Figure 16.2.
From the graph, when x = 2.2, y = 3.87 and when
y = 1.6, x = −0.74
Problem 8. Plot a graph of y =
1
3
e−2x over the
range x = −1.5 to x = 1.5. Determine from the
graph the value of y when x = −1.2 and the value
of x when y = 1.4
x −1.5 −1.0 −0.5 0 0.5 1.0 1.5
1
3
e−2x 6.70 2.46 0.91 0.33 0.12 0.05 0.02
y
5
y ϭ 2e 0.3x
4
3
2
1.6
3.87
1
Ϫ0.74 2.2
Ϫ1 0 1 2 3 xϪ2Ϫ3
Figure 16.2
A graph of
1
3
e−2x is shown in Figure 16.3.
From the graph, when x = −1.2, y = 3.67 and when
y = 1.4, x = −0.72
y
6
5
4
2
3
1
7
0.5 x1.0 1.5Ϫ0.5
Ϫ1.2 Ϫ0.72
Ϫ1.0Ϫ1.5
3.67
1.4
y ϭ 1
3
e Ϫ2x
Figure 16.3
Problem 9. The decay of voltage, v volts, across
a capacitor at time t seconds is given by
v = 250e−t/3. Draw a graph showing the natural
decay curve over the ﬁrst 6seconds. From the
graph, ﬁnd (a) the voltage after 3.4s and (b) the
time when the voltage is 150V

t 0 1 2 3
e−t/3
1.00 0.7165 0.5134 0.3679
v = 250e−t/3 250.0 179.1 128.4 91.97
t 4 5 6
e−t/3 0.2636 0.1889 0.1353
v = 250e−t/3 65.90 47.22 33.83
The natural decay curve of v = 250e−t/3 is shown in
Figure 16.4.
250
200
150
Voltagev(volts)
100
80
50
Time t (seconds)
v 5 250e 2t /3
0 1 1.5 2 3 3.4 4 5 6
Figure 16.4
From the graph,
(a) when time t = 3.4s, voltage v = 80V, and
(b) when voltage v = 150V, time t = 1.5s.
PracticeExercise 64 Exponential graphs
1. Plot a graph of y = 3e0.2x over the range
x = −3 to x = 3. Hence determine the value
of y when x = 1.4 and the value of x when
y = 4.5
2. Plot a graph of y =
1
2
e−1.5x over a range
x = −1.5 to x = 1.5 and then determine the
value of y when x = −0.8 and the value of x
when y = 3.5
3. In a chemical reaction the amount of starting
material C cm3 left after t minutes is given
by C = 40e−0.006t. Plot a graph of C against
t and determine
(a) the concentration C after 1hour.
(b) the time taken for the concentration to
decrease by half.
4. The rate at which a body cools is given by
θ = 250e−0.05t where the excess of tempera-
ture of a body above its surroundings at time t
minutes 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 logarithms
Logarithms having a base of e are called hyperbolic,
Napierian or natural logarithms and the Napierian
logarithm of x is written as loge x or, more commonly,
as ln x. Logarithms were invented by John Napier, a
Scotsman (1550–1617).
The most common method of evaluating a Napierian
logarithmisby ascientificnotationcalculator.Useyour
calculator to check the following values:
ln 4.328 = 1.46510554... = 1.4651, correct to 4
decimal places
ln 1.812 = 0.59443, correct to 5 significant figures
ln 1 = 0
ln 527 = 6.2672, correct to 5 significant figures
ln 0.17 = −1.772, correct to 4 significant figures
ln 0.00042 = −7.77526, correct to 6 significant
figures
ln e3
= 3
ln e1 = 1
From the last two examples we can conclude that
loge ex
= x
This is useful when solving equations involving expo-
nential functions. For example, to solve e3x = 7, take
Napierian logarithms of both sides, which gives

ln e3x
= ln7
i.e. 3x = ln7
from which x =
1
3
ln7 = 0.6486,
Problem 10. Evaluate the following, each correct
to 5 significant figures: (a)
1
2
ln4.7291
(b)
ln7.8693
7.8693
(c)
3.17ln24.07
e−0.1762
(a)
1
2
ln4.7291 =
1
2
(1.5537349...) = 0.77687,
(b)
ln7.8693
7.8693
=
2.06296911...
7.8693
= 0.26215, correct
(c)
3.17ln24.07
e−0.1762
=
3.17(3.18096625...)
0.83845027...
= 12.027,
Problem 11. Evaluate the following: (a)
lne2.5
lg100.5
(b)
5e2.23 lg2.23
ln2.23
(correct to 3 decimal places)
(a)
lne2.5
lg100.5
=
2.5
0.5
= 5
(b)
5e2.23 lg2.23
ln2.23
=
5(9.29986607...)(0.34830486...)
(0.80200158...)
Problem 12. Solve the equation 9 = 4e−3x to find
x, correct to 4 significant figures
Rearranging 9 = 4e−3x gives
9
4
= e−3x
Taking the reciprocal of both sides gives
4
9
=
1
e−3x
= e3x
Taking Napierian logarithms of both sides gives
ln
4
9
= ln(e3x
)
Since loge eα = α, then ln
4
9
= 3x
Hence, x =
1
3
ln
4
9
=
1
3
(−0.81093) = −0.2703,
Problem 13. Given 32 = 70 1 − e−
t
2 ,
determine the value of t, correct to 3 significant
figures
Rearranging 32 = 70 1 − e−
t
2 gives
32
70
= 1 − e−
t
2
and
e−
t
2 = 1 −
32
70
=
38
70
e
t
2 =
70
38
lne
t
2 = ln
70
38
i.e.
t
2
= ln
70
38
from which, t = 2ln
70
38
= 1.22, correct to 3 signifi-
cant figures.
2.68 = ln
4.87
x
to find x
From the definition of a logarithm, since
2.68 = ln
4.87
x
then e2.68 =
4.87
x
4.87
e2.68
= 4.87e−2.68
i.e. x = 0.3339,
Problem 15. Solve
7
4
= e3x correct to 4

Taking natural logs of both sides gives
ln
7
4
= lne3x
ln
7
4
= 3x lne
Since lne = 1, ln
7
4
= 3x
i.e. 0.55962 = 3x
i.e. x = 0.1865,
Problem 16. Solve ex−1 = 2e3x−4 correct to 4
Taking natural logarithms of both sides gives
ln ex−1
= ln 2e3x−4
and by the first law of logarithms,
ln ex−1
= ln2 + ln e3x−4
i.e. x − 1 = ln2 + 3x − 4
Rearranging gives
4 − 1 − ln2 = 3x − x
i.e. 3 − ln2 = 2x
from which, x =
3 − ln2
2
= 1.153
Problem 17. Solve, correct to 4 significant
figures, ln(x − 2)2 = ln(x − 2) − ln(x + 3) + 1.6
Rearranging gives
ln(x − 2)2
− ln(x − 2) + ln(x + 3) = 1.6
and by the laws of logarithms,
ln
(x − 2)2(x + 3)
(x − 2)
= 1.6
Cancelling gives
ln{(x − 2)(x + 3)} = 1.6
and (x − 2)(x + 3) = e1.6
i.e. x2
+ x − 6 = e1.6
or x2
+ x − 6 − e1.6
= 0
i.e. x2
+ x − 10.953 = 0
x =
−1 ± 12 − 4(1)(−10.953)
2
=
−1 ±
√
44.812
2
=
−1 ± 6.6942
2
i.e. x = 2.847 or − 3.8471
x = −3.8471 is not valid since the logarithm of a
negative number has no real root.
Hence, the solution of the equation is x = 2.847
PracticeExercise 65 Evaluating Napierian
logarithms (answers on page 347)
In problems 1 and 2, evaluate correct to 5 signifi-
cant figures.
1. (a)
1
3
ln5.2932 (b)
ln82.473
4.829
(c)
5.62ln321.62
e1.2942
2. (a)
1.786lne1.76
lg101.41
(b)
5e−0.1629
2ln0.00165
(c)
ln4.8629 − ln2.4711
5.173
In problems 3 to 16, solve the given equations, each
3. 1.5 = 4e2t
4. 7.83 = 2.91e−1.7x
5. 16 = 24 1 − e−
t
2
6. 5.17 = ln
x
4.64
7. 3.72ln
1.59
x
= 2.43
8. lnx = 2.40
9. 24 + e2x = 45
10. 5 = ex+1 − 7

11. 5 = 8 1 − e
−x
2
12. 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 the
subject.
18. If
P
Q
= 10log10
R1
R2
, find the value of R1
when P = 160, Q = 8 and R2 = 5.
19. If U2 = U1e
W
PV , make W the subject of the
formula.
16.5 Laws of growth and decay
Laws of exponential growth and decay are of the form
y = Ae−kx and y = A(1 − e−kx ), where A and k are
constants. When plotted, the form of these equations is
as shown in Figure 16.5.
y
A
0
y 5 Ae 2kx
y 5 A(12e 2kx)
x
y
A
0 x
Figure 16.5
The laws occur frequently in engineering and science
and examples of quantities related by a natural law
include:
(a) Linear expansion l = l0eαθ
(b) Change in electrical resistance with temperature
Rθ = 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 circuit
i = Ie−Rt/L
(j) Growth of current in a capacitive circuit
i = I(1 − e−t/CR)
Here are some worked problems to demonstrate the laws
of growth and decay.
Problem 18. The resistance R of an electrical
conductor at temperature θ◦C is given by
R = R0eαθ , where α is a constant and R0 = 5k .
Determine the value of α correct to 4 significant
figures when R = 6k and θ = 1500◦C. Also, find
the temperature, correct to the nearest degree, when
the resistance R is 5.4k
Transposing R = R0eαθ gives
R
R0
= eαθ
ln
R
R0
= lneαθ
= αθ
Hence, α =
1
θ
ln
R
R0
=
1
1500
ln
6 × 103
5 × 103
=
1
1500
(0.1823215...)
= 1.215477...× 10−4
Hence, α = 1.215 × 10−4
figures.
From above, ln
R
R0
= αθ hence θ =
1
α
ln
R
R0

When R = 5.4 × 103
, α = 1.215477...× 10−4
and
R0 = 5 × 103
θ =
1
1.215477...× 10−4
ln
5.4 × 103
5 × 103
=
104
1.215477...
(7.696104... × 10−2
)
= 633◦
C correct to the nearest degree.
Problem 19. In an experiment involving
Newton’s law of cooling, the temperature θ(◦
C) is
given by θ = θ0e−kt . Find the value of constant k
when θ0 = 56.6◦C, θ = 16.5◦C and
t = 79.0seconds
Transposing θ = θ0e−kt gives
θ
θ0
= e−kt , from which
θ0
θ
=
1
e−kt
= ekt
ln
θ0
θ
= kt
from which,
k =
1
t
ln
θ0
θ
=
1
79.0
ln
56.6
16.5
=
1
79.0
(1.2326486...)
Hence, k = 0.01560 or 15.60×10−3.
Problem 20. The current i amperes ﬂowing in a
capacitor at time t seconds is given by
i = 8.0(1 − e−
t
CR ), where the circuit resistance R is
25k 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 reach
6.0A. Sketch the graph of current against time
(a) Current i = 8.0 1−e−
t
CR
= 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−
t
CR gives
i
8.0
= 1 − e−
t
CR
from which, e−
t
CR = 1 −
i
8.0
=
8.0 − i
8.0
e
t
CR =
8.0
8.0 − i
t
CR
= ln
8.0
8.0 − i
Hence,
t = CRln
8.0
8.0 − i
When i = 6.0A,
t = (16 × 10−6
)(25 × 103
)ln
8.0
8.0 − 6.0
i.e. t =
400
103
ln
8.0
2.0
= 0.4ln4.0
= 0.4(1.3862943...)
= 0.5545s
= 555ms correct to the nearest ms.
A graph of current against time is shown in
Figure 16.6.
8
6
5.71
0.555
4
2
i 58.0(12e 2t/CR)
t(s)
i (A)
0 0.5 1.0 1.5
Figure 16.6
Problem 21. The temperature θ2 of a winding
which is being heated electrically at time t is given
by θ2 = θ1(1 − e−
t
τ ), where θ1 is the temperature
(in degrees Celsius) at time t = 0 and τ is a
constant. Calculate

(a) θ1, correct to the nearest degree, when θ2 is
50◦C, t is 30s and τ is 60s and
(b) the time t, correct to 1 decimal place, for θ2 to
be half the value of θ1
(a) Transposing the formula to make θ1 the subject
gives
θ1 =
θ2
1 − e−t/τ
=
50
1 − e−30/60
=
50
1 − e−0.5
=
50
0.393469...
i.e. θ1 = 127◦C correct to the nearest degree.
(b) Transposing to make t the subject of the formula
gives
θ2
θ1
= 1 − e−
t
τ
from which, e−
t
τ = 1 −
θ2
θ1
Hence, −
t
τ
= ln 1 −
θ2
θ1
i.e. t = −τ ln 1 −
θ2
θ1
Since θ2 =
1
2
θ1
t = −60ln 1 −
1
2
= −60ln0.5
= 41.59s
Hence, the time for the temperature θ2 to be one half
of the value of θ1 is 41.6 s, correct to 1 decimal place.
PracticeExercise 66 Laws of growth and
decay (answers on page 347)
1. The temperature, T ◦C, of a cooling object
varies with time, t minutes, according
to 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 metres
above ground level is given by p = p0e−h/C,
where p0 is the pressure at ground level
and C is a constant. Find pressure p when
p0 = 1.012 × 105 Pa, height h = 1420m and
C = 71500.
3. The voltage drop, v volts, across an inductor
L henrys at time t seconds is given by
v = 200e−
Rt
L , where R = 150 and
L = 12.5 × 10−3 H. Determine (a) the
voltage when t = 160 × 10−6 s and (b) the
time 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 l0
and α are constants. Determine (a) the value
of l when l0 = 1.894, α = 2.038 × 10−4 and
t = 250◦C and (b) the value of l0 when
l = 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 by
θ2 = 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 for
θ2 to fall to half the value of θ1 if T remains
at 80s.
6. Abelt isin contact with apulley forasectorof
θ = 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 the
belt, T newtons, when tension on the slack
side is given by T0 = 22.7newtons, given
that these quantities are related by the law
T = T0eμθ .
7. The instantaneous current i at time t is
given by i = 10e−t/CR
when a capacitor
is being charged. The capacitance C is
7×10−6 farads and the resistance R is
0.3×106 ohms. Determine (a) the instanta-
neous current when t is 2.5seconds and (b)
thetimefortheinstantaneouscurrent to fall to
5amperes. Sketch a curve of current against
time from t = 0 to t = 6seconds.
8. The amount of product x (in mol/cm3)
found in a chemical reaction starting
with 2.5mol/cm3
of reactant is given by
x = 2.5(1 − e−4t ) where t is the time, in
minutes, to form product x. Plot a graph
at 30second intervals up to 2.5minutes and
determine x after 1minute.

9. The current i ﬂowing in a capacitor at
time t is given by i = 12.5(1 − e−t/CR),
where resistance R is 30k and the
capacitance C is 20μF. Determine (a)
the current ﬂowing after 0.5seconds and
(b) the time for the current to reach
10amperes.
10. The amount A after n years of a sum invested
P is given by the compound interest law
A = Pern/100, when the per unit interest rate
r is added continuously. Determine, correct
to the nearest pound, the amount after 8 years
for a sum of £1500 invested if the interest
rate is 6% per annum.

Revision Test 6 : Quadratics, logarithms and exponentials
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 decimal
places.
(a) 5x2 +7x −3=0 (b) 3a2 +4a −5=0
(8)
3. Solvetheequation 3x2
− x − 4 = 0 by completing
the square. (6)
4. Determine the quadratic equation in x whose roots
are 1 and −3. (3)
5. The bending moment M at a point in a beam
is given by M =
3x(20 − x)
2
where x metres is
the distance from the point of support. Deter-
mine the value of x when the bending moment is
50Nm. (5)
6. The current i flowing through an electronic device
is given by i = (0.005v2 + 0.014v) amperes
where v is the voltage. Calculate the values of v
when i = 3 × 10−3. (6)
7. Evaluate the following, correct to 4 significant
figures.
(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 log16
1
8
(4)
10. Write the following as the logarithm of a single
number.
(a) 3log2 + 2log5 −
1
2
log16
(b) 3log3 +
1
4
log16 −
1
3
log27 (8)
11. Solve the equation
log(x2 + 8) − log(2x) = log3. (5)
12. Evaluate the following, each correct to 3 decimal
places.
(a) ln462.9
(b) ln0.0753
(c)
ln3.68 − ln2.91
4.63
(3)
13. Expand xe3x to six terms. (5)
14. Evaluate v given that v = E 1 − e−
t
CR volts
when E = 100V,C = 15μF, R = 50k and
t = 1.5s. Also, determine the time when the
voltage is 60V. (8)
15. Plot a graph of y =
1
2
e−1.2x over the range
x = −2 to x = +1 and hence determine, correct to
1 decimal place,
(a) the value of y when x = −0.75, and
(b) the value of x when y = 4.0 (8)

Chapter 17
Straight line graphs
17.1 Introduction to graphs
A graph is a visual representation of information,
showing how one quantity varies with another related
quantity.
We often see graphs in newspapers or in business
reports, 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, Fizzy
Pops, is shown in Figure 17.1.
Generally, we see that the share price increases to a high
of400p in June,but dipsdown to around 280p in August
before recovering slightly in September.
A graph should convey information more quickly to the
reader than if the same information was explained in
words.
When this chapter is completed you should be able to
draw up a table of values, plot co-ordinates, determine
the gradient and state the equation of a straight line
graph. Some typical practical examples are included in
which straight lines are used.
400
Fizzy Pops
350
300
250
Apr 07 May 07 Jun 07 Jul 07 Aug 07 Sep 07
Figure 17.1
17.2 Axes, scales and co-ordinates
We are probably all familiar with reading a map to locate
a town, or a local map to locate a particular street. For
example, a street map of central Portsmouth is shown in
Figure 17.2. Notice the squares drawn horizontally and
vertically on the map; this is called a grid and enables
us to locate a place of interest or a particular road. Most
maps contain such a grid.
We locate places ofinterest on the map by stating a letter
and a number – this is called the grid reference.
For example, on the map, the Portsmouth & Southsea
station is in square D2, King’s Theatre is in square E5,
HMS Warrior is in square A2, Gunwharf Quays is in
square B3 and High Street is in square B4.
Portsmouth & Southsea station is located by moving
horizontally along the bottom of the map until the
square labelled D is reached and then moving vertically
upwards 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 of
DOI: 10.1016/B978-1-85617-697-2.00017-X

Straight line graphs 131
1
2
3
4
5
A B C D E F
Figure 17.2 Reprinted with permission from AA Media Ltd.
a point on a map. If you are familiar with using a
map in this way then you should have no difﬁculties
with graphs, because similar co-ordinates are used with
graphs.
As stated earlier, a graph is a visual representation
of information, showing how one quantity varies with
another related quantity. The most common method of
showing a relationship between two sets of data is to
use a pair of reference axes – these are two lines drawn
at right angles to each other (often called Cartesian or
rectangular 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 0
and y is 0 is called the origin.
x values have scales that are positive to the right of
the origin and negative to the left. y values have scales
that are positive up from the origin and negative down
from the origin.
Co-ordinates are written with brackets and a comma
in between two numbers. For example, point A is shown
with co-ordinates (3, 2) and is located by starting at the
A(3, 2)
4
24 23 22 21 0
3
2
1
21
22
23
24
Origin
C(23, 22)
B (24, 3)
y
x1 3 42
Figure 17.3
origin and moving 3 units in the positive x direction
(i.e. to the right) and then 2 units in the positive y
direction (i.e. up).
When co-ordinates are stated the ﬁrst number is
always the x value and the second number is always

the 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 graphs
The distances travelled by a car in certain peri-
ods of time are shown in the following table of
values.
Time(s) 10 20 30 40 50 60
Distance travelled (m) 50 100 150 200 250 300
We will plot time on the horizontal (or x) axis with a
scale of 1cm = 10s.
We will plot distance on the vertical (or y) axis with a
scale of 1cm = 50m.
(When choosing scales it is better to choose ones such as
1cm = 1unit,1cm = 2unitsor 1cm = 10unitsbecause
doing so makes reading values between these values
easier.)
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 17.4
using crosses. (Alternatively, a dot or a dot and circle
may be used, as shown in Figure 17.3.)
A straight line is drawn through the plotted co-
ordinates as shown in Figure 17.4.
300
DistanceTravelled(m)
Time (s)
250
200
150
100
50
10 20 30 40 50 60
Distance/time graph
Figure 17.4
Student task
The following table gives the force F newtons which,
when applied to a lifting machine, overcomes a
corresponding load of L newtons.
F (Newtons) 19 35 50 93 125 147
L (Newtons) 40 120 230 410 540 680
1. Plot L horizontally and F vertically.
2. Scales are normally chosen such that the graph
occupies as much space as possible on the
graph paper. So in this case, the following
scales are chosen.
Horizontal axis (i.e. L): 1cm = 50N
Vertical axis (i.e. F): 1cm = 10N
3. Draw the axes and label them L (newtons) for
the horizontal axis and F (newtons) for the
vertical 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, and
so 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 linethrough
the points. You will notice that not all of the
pointslieexactly on a straight line.This is quite
normal with experimental values. In a practi-
cal situation it would be surprising if all of the
points lay exactly on a straight line.
9. Extend the straight line at each end.
10. From the graph, determine the force applied
when the load is 325N. It should be close
to 75N. This process of ﬁnding an equivalent
value within the given data is called interpola-
tion. Similarly, determine the load that a force
of 45N will overcome. It should be close to
170N.
11. From the graph, determine the force needed to
overcome a 750N load. It should be close to
161N. This process of ﬁnding an equivalent

value outside the given data is called extrapo-
lation.To extrapolateweneed to haveextended
thestraight linedrawn.Similarly,determinethe
force applied when theload iszero. It shouldbe
close to 11N. The point where the straight line
crosses the vertical axis is called the vertical-
axis intercept. So, in this case, the vertical-axis
intercept = 11N at co-ordinates (0, 11).
The graph you have drawn should look something
like Figure 17.5 shown below.
10
1000 200 300 400 500 600 700 800
L (Newtons)
F(Newtons)
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
Graph of F against L
Figure 17.5
In another example, let the relationship between two
variables x and y be y = 3x + 2.
When x = 0, y = 0 + 2 = 2
When x = 1, y = 3 + 2 = 5
When x = 2, y = 6 + 2 = 8, and so on.
The co-ordinates (0, 2), (1, 5) and (2, 8) have been
produced and are plotted, with others, as shown in
Figure 17.6.
When the points are joined together a straight line
graph results, i.e. y = 3x + 2 is a straight line graph.
17.3.1 Summary of general rules to be
applied when drawing graphs
(a) Give the graph a title clearly explaining what is
being illustrated.
(b) Choose scales such that the graph occupies as
much space as possible on the graph paper
being used.
1 2
yϭ3xϩ 2
y
x0
2
4
6
8
Ϫ1
Figure 17.6 Graph of y/x
(c) Choose scales so that interpolation is made as
easy as possible. Usually scales such as 1cm =
1unit,1cm = 2 units or 1cm = 10 units are used.
Awkward scalessuch as1cm = 3 unitsor1cm = 7
units should not be used.
(d) The scales need not start at zero, particularly when
starting at zero produces an accumulation ofpoints
within a small area of the graph paper.
(e) The co-ordinates, or points, should be clearly
marked. This is achieved by a cross, or a dot and
circle, or just by a dot (see Figure 17.3).
(f) A statement should be made next to each axis
explaining the numbers represented with their
appropriate units.
(g) Sufﬁcient numbers should be written next to each
axis without cramping.
Problem 1. Plot the graph y = 4x + 3 in the
range x = −3 to x = +4. From the graph, ﬁnd
(a) the value of y when x = 2.2 and (b) the value
of x when y = −3
Whenever an equation is given and a graph is required,
a table giving corresponding values of the variable is
necessary. The table is achieved as follows:
When x = −3, y = 4x + 3 = 4(−3) + 3
= −12 + 3 = −9
When 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 4
y −9 −5 −1 3 7 11 15 19
The co-ordinates (−3,−9), (−2,−5), (−1,−1), and
so on, are plotted and joined together to produce the

20
15
10
11.8
5
Ϫ5
Ϫ3
10
0Ϫ2 Ϫ1
Ϫ1.5
Ϫ3 1
2.2
2 43 x
y
Figure 17.7
straight line shown in Figure 17.7. (Note that the scales
used 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 = −1.5
PracticeExercise 67 Straight line graphs
1. Assuming graph paper measuring 20cm by
20cm is available, suggest suitable scales for
the following ranges of values.
(a) Horizontal axis: 3V to 55V; vertical
axis: 10 to 180 .
(b) Horizontal axis: 7m to 86m; vertical
axis: 0.3V to 1.69V.
(c) Horizontal axis: 5N to 150N; vertical
axis: 0.6mm to 3.4mm.
2. Corresponding values obtained experimen-
tally for two quantities are
x −5 −3 −1 0 2 4
y −13 −9 −5 −3 1 5
Plot 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 graph
paper with the origin somewhere in the centre
of 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 = 7
3. Corresponding values obtained experimen-
tally for two quantities are
x −2.0 −0.5 0 1.0 2.5 3.0 5.0
y −13.0 −5.5 −3.0 2.0 9.5 12.0 22.0
Use a horizontal scale for x of 1cm = 1
2 unit
and a vertical scale for y of 1cm = 2 units and
draw a graph of x against y. Label the graph
and each of its axes. By interpolation, find
from the graph the value of y when x is 3.5.
4. Draw a graph of y − 3x + 5 = 0 over a range
of x = −3 to x = 4. Hence determine
(a) the value of y when x = 1.3
(b) the value of x when y = −9.2
5. The speed n rev/min of a motor changes when
the voltage V across the armature is varied.
The results are shown in the following table.
n (rev/min) 560 720 900 1010 1240 1410
V (volts) 80 100 120 140 160 180
It is suspected that one of the readings taken of
the speed is inaccurate. Plot a graph of speed
(horizontally) against voltage (vertically) and
find 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 and
equations of graphs
17.4.1 Gradients
The gradient or slope of a straight line is the ratio of
the change in the value of y to the change in the value of
x between any two points on the line. If, as x increases,
(→), y also increases, (↑), then the gradient is positive.

(a)
8
7
6
5
4
3
2
1
021 1
A B
C
y
2 3 4
(b)
11
10
8
6
4
2
24 23 22 21
F
E D
0 xx
yy523x 12
y5 2x 1 1
y
2
1
0 2 3
3
(c)
1 x
y53
Figure 17.8
In Figure 17.8(a), a straight line graph y = 2x + 1 is
shown. To ﬁnd the gradient of this straight line, choose
two points on the straight line graph, such as A and C.
Then construct a right-angled triangle, such as ABC,
where BC is vertical and AB is horizontal.
Then, gradient of AC =
change in y
change in x
=
CB
BA
=
7 − 3
3 − 1
=
4
2
= 2
In Figure 17.8(b), a straight line graph y = −3x + 2 is
shown. To ﬁnd the gradient of this straight line, choose
two points on the straight line graph, such as D and F.
Then construct a right-angled triangle, such as DEF,
where EF is vertical and DE is horizontal.
Then, gradient of DF =
change in y
change in x
=
FE
ED
=
11 − 2
−3 − 0
=
9
−3
= −3
Figure 17.8(c) shows a straight line graph y = 3.
Since the straight line is horizontal the gradient
is zero.
17.4.2 The y-axis intercept
The value of y when x = 0 is called the y-axis inter-
cept. In Figure 17.8(a) the y-axis intercept is 1 and in
Figure 17.8(b) the y-axis intercept is 2.
17.4.3 The equation of a straight line graph
The general equation of a straight line graph is
y = mx + c
where m is the gradient and c is the y-axis intercept.
Thus, as we have found in Figure 17.8(a),
y = 2x + 1 represents a straight line of gradient 2 and
y-axis intercept 1. So, given the equation y = 2x + 1,
we are able to state, on sight, that the gradient = 2
and the y-axis intercept = 1, without the need for any
analysis.
Similarly, in Figure 17.8(b), y = −3x + 2 represents a
straight line of gradient −3 and y-axis intercept 2.
In Figure 17.8(c), y = 3 may be rewritten as y = 0x + 3
and therefore represents a straight line of gradient 0 and
y-axis intercept 3.
Here are some worked problems to help understanding
of gradients, intercepts and equations of graphs.
Problem 2. Plot the following graphs on the same
axes in the range x = −4 to x = +4 and determine
the gradient of each.
(a) y = x (b) y = x + 2
(c) y = x + 5 (d) y = x − 3
A table of co-ordinates is produced for each graph.
(a) y = x
x −4 −3 −2 −1 0 1 2 3 4
y −4 −3 −2 −1 0 1 2 3 4
(b) y = x + 2
x −4 −3 −2 −1 0 1 2 3 4
y −2 −1 0 1 2 3 4 5 6
(c) y = x + 5
x −4 −3 −2 −1 0 1 2 3 4
y 1 2 3 4 5 6 7 8 9
(d) y = x − 3
x −4 −3 −2 −1 0 1 2 3 4
y −7 −6 −5 −4 −3 −2 −1 0 1
The co-ordinates are plotted and joined for each graph.
Theresultsareshown in Figure17.9.Each ofthestraight
lines produced is parallel to the others; i.e., the slope or
gradient is the same for each.

Ϫ4 Ϫ3 Ϫ2 Ϫ1 1 2 3 4 x
A
BC
F
E
D
yϭ
xϩ
5
yϭ
xϩ
2
yϭ
xϪ
3
yϭ
x
9
y
8
7
6
5
4
3
2
1
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Figure 17.9
To ﬁnd the gradient of any straight line, say, y = x − 3,
a horizontal and vertical component needs to be con-
structed. In Figure 17.9, AB is constructed vertically at
x = 4 and BC is constructed horizontally at y = −3.
The gradient of AC =
AB
BC
=
1 − (−3)
4 − 0
=
4
4
= 1
i.e. the gradient of the straight line y = x − 3 is 1, which
could have been deduced ‘on sight’ since y = 1x − 3
represents a straight line graph with gradient 1 and
y-axis intercept of −3.
The actual positioning of AB and BC is unimportant
because the gradient is also given by
DE
EF
=
−1 − (−2)
2 − 1
=
1
1
= 1
The slope or gradient of each of the straight lines in
Figure 17.9 is thus 1 since they are parallel to each
other.
Problem 3. Plot the following graphs on the same
axes between the values x = −3 to x = +3 and
determine the gradient and y-axis intercept of each.
(a) y = 3x (b) y = 3x + 7
(c) y = −4x + 4 (d) y = −4x − 5
A table of co-ordinatesis drawn up for each equation.
(a) y = 3x
x −3 −2 −1 0 1 2 3
y −9 −6 −3 0 3 6 9
(b) y = 3x + 7
x −3 −2 −1 0 1 2 3
y −2 1 4 7 10 13 16
(c) y = −4x + 4
x −3 −2 −1 0 1 2 3
y 16 12 8 4 0 −4 −8
(d) y = −4x − 5
x −3 −2 −1 0 1 2 3
y 7 3 −1 −5 −9 −13 −17
Each of the graphs is plotted as shown in Figure 17.10
and each is a straight line. y = 3x and y = 3x + 7 are
parallel to each other and thus have the same gradient.
The gradient of AC is given by
CB
BA
=
16 − 7
3 − 0
=
9
3
= 3
y5
3x1
7
y5
2
4x2
5
y5
2
4x1
4
y5
3x
16
12
8
4
23 22 0
28
212
216
21 1 2 3 x
24
B
C
A
F
E D
y
Figure 17.10
Hence, the gradients of both y = 3x and y = 3x + 7
are 3, which could have been deduced ‘on sight’.
y = −4x + 4 and y = −4x − 5 are parallel to each other
and thus have the same gradient. The gradient of DF is

given by
FE
ED
=
−5 − (−17)
0 − 3
=
12
−3
= −4
Hence, the gradient of both y = −4x + 4 and
y = −4x − 5 is −4, which, again, could have been
deduced ‘on sight’.
The y-axis intercept means the value of y where the
straight line cuts the y-axis. From Figure 17.10,
y = 3x cuts the y-axis at y = 0
y = 3x + 7 cuts the y-axis at y = +7
y = −4x + 4 cuts the y-axis at y = +4
y = −4x − 5 cuts the y-axis at y = −5
Some general conclusionscan be drawn from the graphs
shown in Figures 17.9 and 17.10. When an equation is
of 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 be
deduced ‘on sight’ that its gradient is +3 and its y-axis
intercept is +7, as shown in Figure 17.10. Similarly, if
y = −4x − 5, the gradient is−4 and the y-axis intercept
is −5, as shown in Figure 17.10.
When plottinga graph of the form y = mx + c, only two
co-ordinatesneed bedetermined.When theco-ordinates
are plotted a straight line is drawn between the two
points. Normally, three co-ordinates are determined, the
third one acting as a check.
Problem 4. Plot the graph 3x + y + 1 = 0 and
2y − 5 = x on the same axes and ﬁnd their point of
intersection
Rearranging 3x + y + 1 = 0 gives y = −3x − 1
Rearranging 2y − 5 = x gives 2y = x + 5 and
y =
1
2
x + 2
1
2
Since both equations are of the form y = mx + c, both
are straight lines. Knowing an equation is a straight
line means that only two co-ordinates need to be plot-
ted and a straight line drawn through them. A third
co-ordinate is usually determined to act as a check.
A table ofvalues is produced for each equation as shown
below.
x 1 0 −1
−3x − 1 −4 −1 2
x 2 0 −3
1
2 x + 21
2 31
2 21
2 1
The graphs are plotted as shown in Figure 17.11.
The two straight lines are seen to intersect at (−1,2).
4
24 23 22 21 0
3
2
1
21
22
23
24
y523x 21
y
x1 3 42
y5 x11
2
5
2
Figure 17.11
Problem 5. If graphs of y against x were to be
plotted for each of the following, state (i) the
gradient 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 = x
If y = mx + c then m = gradient and c = y-axis
intercept.
(a) If y = 9x + 2, then (i) gradient = 9
(ii) y-axis intercept = 2
(b) If y = −4x + 7, then (i) gradient = −4
(c) If y = 3x i.e. y = 3x + 0,
then (i) gradient = 3
i.e. the straight line passes through the origin.

(d) If y = −5x − 3, then (i) gradient = −5
(ii) y-axis intercept = −3
(e) If y = 6 i.e. y = 0x + 6,
i.e. y = 6 is a straight horizontal line.
(f) If y = x i.e. y = 1x + 0,
Since y = x, as x increases, y increases by the same
amount; i.e., y is directly proportional to x.
Problem 6. Without drawing graphs, determine
the gradient and y-axis intercept for each of the
following equations.
(a) y + 4 = 3x (b) 2y + 8x = 6 (c) 3x = 4y + 7
If y = mx + c then m = gradient and c = y-axis
intercept.
(a) Transposing y + 4 = 3x gives y = 3x − 4
Hence, gradient = 3 and y-axis intercept = −4.
(b) Transposing 2y + 8x = 6 gives 2y = −8x + 6
Dividing both sides by 2 gives y = −4x + 3
Hence, gradient = −4 and y-axis intercept = 3.
(c) Transposing 3x = 4y + 7 gives 3x − 7 = 4y
or 4y = 3x − 7
Dividing both sides by 4 gives y=
3
4
x −
7
4
or y=0.75x − 1.75
Hence, gradient = 0.75 and y-axis intercept
= −1.75
Problem 7. Without plotting graphs, determine
the gradient and y-axis intercept values of the
(a) y = 7x − 3 (b) 3y = −6x + 2
(c) y − 2 = 4x + 9 (d)
y
3
=
x
3
−
1
5
(e) 2x + 9y + 1 = 0
(a) y = 7x − 3 is of the form y = mx + c
Hence, gradient, m = 7 and y-axis intercept,
c = −3.
(b) Rearranging 3y = −6x + 2 gives y = −
6x
3
+
2
3
,
i.e. y = −2x +
2
3which is of the form y = mx + c
Hence, gradient m = −2 and y-axis intercept,
c =
2
3
(c) Rearranging y − 2 = 4x + 9 gives y = 4x + 11.
Hence, gradient = 4 and y-axis intercept = 11.
(d) Rearranging
y
3
=
x
2
−
1
5
gives y = 3
x
2
−
1
5
=
3
2
x −
3
5
Hence, gradient =
3
2
and
y-axis intercept = −
3
5
(e) Rearranging 2x +9y +1=0 gives 9y = − 2x −1,
i.e. y = −
2
9
x −
1
9
Hence, gradient = −
2
9
and
y-axis intercept = −
1
9
Problem 8. Determine for the straight line shown
in Figure 17.12 (a) the gradient and (b) the equation
of the graph
23
20
24 23 22 0
15
10
8
5
25
210
215
220
y
x1 3 4221
Figure 17.12
(a) A right-angled triangle ABC is constructed on the
graph as shown in Figure 17.13.
Gradient =
AC
CB
=
23 − 8
4 − 1
=
15
3
= 5

23
20
24 23 22 0
15
10
8
5
25
210
215
220
y
x
C
B
A
1 3 4221
Figure 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 where
m = 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 the
straight line shown in Figure 17.14.
4
24 23 22 0
3
2
1
21
22
23
24
y
x
EF
D
1 3 4221
Figure 17.14
The triangleDEF is shown constructed in Figure17.14.
Gradient of DE =
DF
FE
=
3 − (−3)
−1 − 2
=
6
−3
= −2
and the y-axis intercept = 1.
Hence, the equation of the straight line is y = mx + c
i.e. y = −2x + 1.
Problem 10. The velocity of a body was
measured at various times and the results obtained
were as follows:
Velocity v (m/s) 8 10.5 13 15.5 18 20.5 23
Time t (s) 1 2 3 4 5 6 7
Plot a graph of velocity (vertically) against time
(horizontally) and determine the equation of the
graph
Suitable scales are chosen and the co-ordinates (1, 8),
(2, 10.5), (3, 13), and so on, are plotted as shown in
Figure 17.15.
0
4
5.5
6
8 Q R
P
10
12
14
16
Velocity(y),inmetrespersecond
18
20
22
1 2 3
Time (t), in seconds
4 5 6 7
Figure 17.15
The right-angled triangle PRQ is constructed on the
graph as shown in Figure 17.15.
Gradient of PQ =
PR
RQ
=
18 − 8
5 − 1
=
10
4
= 2.5
The vertical axis intercept is at v = 5.5 m/s.
The equation of a straight line graph is y = mx + c. In
this case, t corresponds to x and v corresponds to y.
Hence, the equation of the graph shown in Figure 17.15

is v = mt + c. But, from above, gradient, m = 2.5 and
v-axis intercept, c = 5.5.
Hence, the equation of the graph is v = 2.5t + 5.5
Problem 11. Determine the gradient of the
straight line graph passing through the co-ordinates
(a) (−2,5) and (3,4), and (b) (−2,−3) and (−1,3)
From Figure 17.16, a straight line graph passing
through co-ordinates (x1, y1) and (x2, y2) has a gradient
given by
m =
y2 − y1
x2 − x1
(x1, y1)
(x2, y2)
(x2Ϫx1)
(y2Ϫ y1)
0
y
y2
y1
x1 x2 x
Figure 17.16
(a) A straight line passes through (−2,5) and (3,4),
hence x1 = −2, y1 = 5, x2 = 3 and y2 = 4, hence,
gradient, m =
y2 − y1
x2 − x1
=
4 − 5
3 − (−2)
= −
1
5
(b) A straight line passes through (−2,−3) and
(−1,3), hence x1 = −2, y1 = −3, x2 = −1 and
y2 = 3, hence, gradient,
m =
y2 − y1
x2 − x1
=
3 − (−3)
−1 − (−2)
=
3 + 3
−1 + 2
=
6
1
= 6
PracticeExercise 68 Gradients, intercepts
and equations of graphs (answers on page
347)
1. The equation ofa lineis 4y = 2x + 5. A table
of corresponding values is produced and is
shown below. Complete the table and plot a
graph of y against x. Find the gradient of the
graph.
x −4 −3 −2 −1 0 1 2 3 4
y −0.25 1.25 3.25
2. Determine the gradient and intercept on the
y-axis for each of the following equations.
(a) y = 4x − 2 (b) y = −x
(c) y = −3x − 4 (d) y = 4
3. Find the gradient and intercept on the y-axis
for each of the following equations.
(a) 2y − 1 = 4x (b) 6x − 2y = 5
(c) 3(2y − 1) =
x
4
Determine the gradient and y-axis intercept
for each of the equations in problems 4 and
5 and sketch the graphs.
4. (a) y = 6x − 3 (b) y = −2x + 4
(c) y = 3x (d) y = 7
5. (a) 2y + 1 = 4x (b) 2x + 3y + 5 = 0
(c) 3(2y − 4) =
x
3
(d) 5x −
y
2
−
7
3
= 0
6. Determine the gradient of the straight line
graphs passing through the co-ordinates:
(a) (2, 7) and (−3,4)
(b) (−4,−1) and (−5,3)
(c)
1
4
,−
3
4
and −
1
2
,
5
8
7. State which of the following equations will
produce graphs which are parallel to one
another.
(a) y − 4 = 2x (b) 4x = −(y + 1)
(c) x =
1
2
(y + 5) (d) 1 +
1
2
y =
3
2
x
(e) 2x =
1
2
(7 − y)
8. Draw on the same axes the graphs of
y = 3x − 5 and 3y + 2x = 7. Find the co-
ordinates of the point of intersection. Check
the result obtained by solving the two simul-
taneous equations algebraically.
9. Plot the graphs y = 2x + 3 and 2y = 15 − 2x
on the same axes and determine their point of
intersection.
10. Draw on the same axes the graphs of
y = 3x − 1 and y + 2x = 4. Find the co-
ordinates of the point of intersection.

11. A piece of elastic is tied to a support so that it
hangs vertically and a pan, on which weights
can be placed, is attached to the free end. The
length of the elastic is measured as various
weights are added to the pan and the results
obtained are as follows:
Load, W (N) 5 10 15 20 25
Length, l (cm) 60 72 84 96 108
Plot a graph of load (horizontally) against
length (vertically) and determine
(a) the value of length when the load is 17N.
(b) the value of load when the length is
74cm.
(c) its gradient.
(d) the equation of the graph.
12. The following table gives the effort P to lift
a load W with a small lifting machine.
W (N) 10 20 30 40 50 60
P (N) 5.1 6.4 8.1 9.6 10.9 12.4
Plot W horizontally against P vertically and
show that the values lie approximately on a
straight line. Determine the probable rela-
tionship connecting P and W in the form
P = aW + b.
13. In an experiment the speeds N rpm of a fly-
wheel slowly coming to rest were recorded
against the time t in minutes. Plot the results
and show that N and t are connected by
an equation of the form N = at + b. Find
probable values of a and b.
t (min) 2 4 6 8 10 12 14
N (rev/min) 372 333 292 252 210 177 132
straight line graphs
When a set of co-ordinate values are given or are
obtained experimentally and it is believed that they
follow a law of the form y = mx + c, if a straight line
can be drawn reasonably close to most oftheco-ordinate
values when plotted, this verifies that a law of the
form y = mx + c exists. From the graph, constants
m (i.e. gradient) and c (i.e. y-axis intercept) can be
determined.
Here are some worked problems in which practical
situations are featured.
Problem 12. The temperature in degrees Celsius
and the corresponding values in degrees Fahrenheit
are shown in the table below. Construct rectangular
axes, choose suitable scales and plot a graph of
degrees Celsius (on the horizontal axis) against
degrees Fahrenheit (on the vertical scale).
◦C 10 20 40 60 80 100
◦F 50 68 104 140 176 212
From the graph find (a) the temperature in degrees
Fahrenheit at 55◦C, (b) the temperature in degrees
Celsius at 167◦F, (c) the Fahrenheit temperature at
0◦C and (d) the Celsius temperature at 230◦F
The co-ordinates (10, 50), (20, 68), (40, 104), and so
on are plotted as shown in Figure 17.17. When the
co-ordinates are joined, a straight line is produced.
Since a straight line results, there is a linear relationship
between degrees Celsius and degrees Fahrenheit.
0
40
32
80
120
DegreesFahrenheit(8F)
131
160
167
200
240
230
20 40 55
Degrees Celsius (8C)
7560 80 120110100
y
D
E
F
G
x
A
B
Figure 17.17
(a) To find the Fahrenheit temperature at 55◦C, a verti-
cal line AB is constructed from the horizontal axis
to meet the straight line at B. The point where the

horizontalline BD meets the vertical axis indicates
the equivalent Fahrenheit temperature.
Hence, 55◦C is equivalent to 131◦F.
This process of finding an equivalent value in
between the given information in the above table
is called interpolation.
(b) To find the Celsius temperature at 167◦F, a
horizontal line EF is constructed as shown in
Figure 17.17. The point where the vertical line FG
cuts the horizontal axis indicates the equivalent
Celsius temperature.
Hence, 167◦F is equivalent to 75◦C.
(c) If the graph is assumed to be linear even outside of
the given data, the graph may be extended at both
ends (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 outside
of 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 measured
for various temperatures T◦C. The results are
shown below.
V m3
25.0 25.8 26.6 27.4 28.2 29.0
T ◦C 60 65 70 75 80 85
Plot a graph of volume (vertical) against
temperature (horizontal) and from it find (a) the
temperature when the volume is 28.6m3 and (b) the
volume when the temperature is 67◦C
If a graph is plotted with both the scales starting at zero
then the result is as shown in Figure 17.18. All of the
points lie in the top right-hand corner of the graph,
making interpolation difficult. A more accurate graph
is obtained if the temperature axis starts at 55◦C and
the volume axis starts at 24.5m3. The axes correspond-
ing to these values are shown by the broken lines in
Figure 17.18 and are called false axes, since the origin
is not now at zero. A magnified version of this relevant
part of the graph is shown in Figure 17.19. From the
graph,
(a) When the volume is 28.6m3, the equivalent tem-
perature is 82.5◦
C.
(b) When the temperature is 67◦C, the equivalent
volume is 26.1m3.
30
25
20
15
Volume(m3)
10
5
0 20 40 60 80 100
Temperature (8C)
y
x
Figure 17.18
29
28.6
28
27
Volume(m3
)
26
25
55 60 65 67 70 75 80 82.5 85
26.1
Temperature (ЊC)
y
x
Figure 17.19
Problem 14. In an experiment demonstrating
Hooke’s law, the strain in an aluminium wire was
measured for various stresses. The results were:
Stress (N/mm2) 4.9 8.7 15.0
Strain 0.00007 0.00013 0.00021
Stress (N/mm2) 18.4 24.2 27.3
Strain 0.00027 0.00034 0.00039

Plot a graph of stress (vertically) against strain
(horizontally). Find (a) Young’s modulus of
elasticity for aluminium, which is given by the
gradient of the graph, (b) the value of the strain at a
stress of 20N/mm2 and (c) the value of the stress
when the strain is 0.00020
The co-ordinates (0.00007, 4.9), (0.00013, 8.7), and
so on, are plotted as shown in Figure 17.20. The
graph produced is the best straight line which can be
drawn corresponding to these points. (With experimen-
tal results it is unlikely that all the points will lie exactly
on a straight line.) The graph, and each of its axes, are
labelled. Since the straight line passes through the ori-
gin, stress is directly proportional to strain for the given
range of values.
Stress(N/mm2)
0.00005
0
4
8
12
14
16
20
24
28
0.00015 0.00025
Strain
0.000285
0.00035
y
x
A
B
C
Figure 17.20
(a) The gradient of the straight line AC is given by
AB
BC
=
28 − 7
0.00040 − 0.00010
=
21
0.00030
=
21
3 × 10−4
=
7
10−4
= 7 × 104
= 70000N/mm2
Thus, 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 ×109
N/m2 (or pascals).
From Figure 17.20,
(b) The value of the strain at a stress of 20N/mm2 is
0.000285
(c) The value of the stress when the strain is 0.00020
is 14 N/mm2.
Problem 15. The following values of resistance
R ohms and corresponding voltage V volts are
obtained from a test on a ﬁlament lamp.
R ohms 30 48.5 73 107 128
V volts 16 29 52 76 94
Choose suitable scales and plot a graph with R
representing the vertical axis and V the horizontal
axis. Determine (a) the gradient of the graph, (b) the
R axis intercept value, (c) the equation of the graph,
(d) the value of resistance when the voltage is 60V
and (e) the value of the voltage when the resistance
is 40 ohms. (f) If the graph were to continue in the
same manner, what value of resistance would be
obtained at 110V?
The co-ordinates (16, 30), (29, 48.5), and so on are
shown plotted in Figure 17.21, where the best straight
line is drawn through the points.
147
140
120
85
80
60
100
ResistanceR(ohms)
40
10
20
240 20 40 60 80 100 110 120
Voltage V (volts)
y
C
B
A
x
Figure 17.21
(a) The slope or gradient of the straight line AC is
given by
AB
BC
=
135 − 10
100 − 0
=
125
100
= 1.25

(Note that the vertical line AB and the horizon-
tal line BC may be constructed anywhere along
the length of the straight line. However, calcula-
tions are made easier if the horizontal line BC is
carefully 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, when
y is plotted on the vertical axis and x on the hori-
zontal axis. m represents the gradient and c the
y-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, the
resistance is 147 .
Problem 16. Experimental tests to determine the
breaking stress σ of rolled copper at various
temperatures t gave the following results.
Stress σ (N/cm2) 8.46 8.04 7.78
Temperature t(◦
C) 70 200 280
Stress σ (N/cm2) 7.37 7.08 6.63
Temperature t(◦C) 410 500 640
Show that the values obey the law σ = at + b,
where a and b are constants, and determine
approximate values for a and b. Use the law to
determine the stress at 250◦
C and the temperature
when the stress is 7.54N/cm2.
The co-ordinates (70, 8.46), (200, 8.04), and so on, are
plotted as shown in Figure 17.22. Since the graph is a
straight line then the values obey the law σ = at + b,
and the gradient of the straight line is
a =
AB
BC
=
8.36 − 6.76
100 − 600
=
1.60
−500
= −0.0032
Vertical axis intercept, b = 8.68
Hence, the law of the graph is σ = 0.0032t + 8.68
When the temperature is 250◦C, stress σ is given by
σ = −0.0032(250) + 8.68 = 7.88N/cm2
Temperature t (8C)
Stress␴(N/cm2)
8.68
8.36
8.50
8.00
7.50
7.00
6.50
0 100 200 300 400 500 600 700
6.76
y
x
CB
A
Figure 17.22
Rearranging σ = −0.0032t + 8.68 gives
0.0032t = 8.68 − σ, i.e. t =
8.68 − σ
0.0032
Hence, when the stress, σ = 7.54N/cm2,
temperature, t =
8.68 − 7.54
0.0032
= 356.3◦
C
involving straight line graphs (answers on
page 347)
1. The resistance R ohms of a copper winding
is measured at various temperatures t ◦C and
the results are as follows:
R (ohms) 112 120 126 131 134
t◦C 20 36 48 58 64
Plot a graph of R (vertically) against t (hori-
zontally) and ﬁnd from it (a) the temperature
when the resistance is 122 and (b) the
resistance when the temperature is 52◦C.
2. The speed of a motor varies with armature
voltage as shown by the following experi-
mental results.

n (rev/min) 285 517 615
V (volts) 60 95 110
n (rev/min) 750 917 1050
V (volts) 130 155 175
Plot a graph of speed (horizontally) against
voltage(vertically) and draw the best straight
line through the points. Find from the graph
(a) the speed at a voltage of 145V and (b) the
voltage at a speed of 400rev/min.
3. The following table gives the force F
newtons which, when applied to a lifting
machine, overcomes a corresponding load of
L newtons.
Force F (newtons) 25 47 64
Load L (newtons) 50 140 210
Force F (newtons) 120 149 187
Load L (newtons) 430 550 700
Choose suitable scales and plot a graph of
F (vertically) against L (horizontally).Draw
the best straight line through the points.
Determine from the graph
(a) the gradient,
(b) the F-axis intercept,
(c) the equation of the graph,
(d) the force applied when the load is
310N, and
(e) the load that a force of 160N will
overcome.
(f) If the graph were to continue in the
same manner, what value of force will
be needed to overcome a 800N load?
4. The following table gives the results of tests
carried out to determine the breaking stress
σ of rolled copper at varioustemperatures, t.
Stress σ (N/cm2) 8.51 8.07 7.80
Stress σ (N/cm2) 7.47 7.23 6.78
Plot a graph of stress (vertically) against
temperature (horizontally). Draw the best
straight linethrough the plotted co-ordinates.
Determine the slope of the graph and the
vertical axis intercept.
5. The velocity v of a body after varying time
intervals t was measured as follows:
t (seconds) 2 5 8
v (m/s) 16.9 19.0 21.1
t (seconds) 11 15 18
v (m/s) 23.2 26.0 28.1
Plot v vertically and t horizontally and draw
a graph of velocity against time. Determine
from the graph (a) the velocity after 10s,
(b) the time at 20m/s and (c) the equation
of the graph.
6. The mass m of a steel joist varies with length
L as follows:
mass, m (kg) 80 100 120 140 160
length, L (m) 3.00 3.74 4.48 5.23 5.97
Plot a graph of mass (vertically) against
length (horizontally). Determine the equa-
tion of the graph.
7. The crushing strength of mortar varies with
the percentage of water used in its prepara-
tion, as shown below.
Crushing strength, 1.67 1.40 1.13
F (tonnes)
% of water used, w% 6 9 12
Crushing strength, 0.86 0.59 0.32
F (tonnes)
% of water used, w% 15 18 21

Plot a graph of F (vertically) against w
(horizontally).
(a) Interpolate and determine the crushing
strength when 10% water is used.
(b) Assuming the graph continues in the
same manner, extrapolate and deter-
mine the percentage of water used when
the 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 for
various stresses. The results were
Stress 10.6 × 106 18.2 × 106 24.0 × 106
(pascals)
Strain 0.00011 0.00019 0.00025
Stress 30.7 × 106 39.4 × 106
(pascals)
Strain 0.00032 0.00041
Plot a graph of stress (vertically) against
strain (horizontally). Determine
(a) Young’s modulus of elasticity for cop-
per, which is given by the gradient of
the graph,
(b) the value of strain at a stress of
21 × 106 Pa,
(c) the value of stress when the strain is
0.00030,
9. An experiment with a set of pulley blocks
gave the following results.
Effort, E (newtons) 9.0 11.0 13.6
Load, L (newtons) 15 25 38
Effort, E (newtons) 17.4 20.8 23.6
Load, L (newtons) 57 74 88
Plot 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 with
temperature T is believed to follow a law of
the form p = aT + b, where a and b are con-
stants. Verify this law for the results given
below and determine the approximate values
of a and b. Hence, determine the pressures
at temperatures of 285K and 310K and the
temperature at a pressure of 250kPa.
Pressure, p (kPa) 244 247 252
Temperature, T (K) 273 277 282
Pressure, p (kPa) 258 262 267
Temperature, T (K) 289 294 300

Chapter 18
Graphs reducing non-linear
laws to linear form
18.1 Introduction
In Chapter 17 we discovered that the equation of a
straight line graph is of the form y = mx + c, where
m is the gradient and c is the y-axis intercept. This
chapterexplainshowthelawofagraph can still bedeter-
mined even when it is not of thelinear form y = mx + c.
The method used is called determination of law and is
explained in the following sections.
18.2 Determination of law
Frequently, the relationship between two variables, say
x and y, is not a linear one; i.e., when x is plottedagainst
y a curve results. In such cases the non-linear equation
may be modiﬁed to the linear form, y = mx + c, so that
the constants, and thus the law relating the variables, can
be determined. This technique is called ‘determination
of law’.
Some examples of the reduction of equations to linear
form include
(i) y = ax2 + b compares with Y = mX + c, where
m = a, c = b and X = x2.
Hence, y is plotted vertically against x2 horizon-
tally to produce a straight line graph of gradient
a and y-axis intercept b.
(ii) y =
a
x
+ b, i.e. y = a
1
x
+ b
y is plotted vertically against
1
x
horizontally to
produce a straight line graph of gradient a and
y-axis intercept b.
(iii) y = ax2 + bx
Dividing both sides by x gives
y
x
= ax + b.
Comparing with Y = mX + c shows that
y
x
is
plotted vertically against x horizontally to pro-
duce a straight line graph of gradient a and
y
x
axis 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 law
y = ax2 + b. By plotting a suitable graph, verify
this law and determine approximate values of
a and b
x 1 2 3 4 5
y 9.8 15.2 24.2 36.5 53.0
If y is plotted against x a curve results and it is not
possible to determine the values of constants a and b
from the curve.
Comparing y = ax2 + b with Y = mX + c shows that
y is to be plotted vertically against x2
horizontally. A
table of values is drawn up as shown below.
x 1 2 3 4 5
x2 1 4 9 16 25
y 9.8 15.2 24.2 36.5 53.0
DOI: 10.1016/B978-1-85617-697-2.00018-1

A graph of y against x2
is shown in Figure 18.1, with
the best straight line drawn through the points. Since a
straight line graph results, the law is veriﬁed.
53
y
A
BC
50
40
30
20
10
0 5 10 15 20 25
x2
17
8
Figure 18.1
From the graph, gradient, a =
AB
BC
=
53 − 17
25 − 5
=
36
20
= 1.8
and the y-axis intercept, b = 8.0
Hence, the law of the graph is y = 1.8x2 + 8.0
Problem 2. Values of load L newtons and
distance d metres obtained experimentally are
shown in the following table.
Load, L(N) 32.3 29.6 27.0 23.2
Distance, d(m) 0.75 0.37 0.24 0.17
Load, L(N) 18.3 12.8 10.0 6.4
Distance, d(m) 0.12 0.09 0.08 0.07
(a) Verify that load and distance are related by a
law of the form L =
a
d
+ b and determine
approximate values of a and b.
(b) Hence, calculate the load when the distance is
0.20m and the distance when the load is 20N.
(a) Comparing L =
a
d
+ b i.e. L = a
1
d
+ b with
Y = mX + c shows that L is to be plotted ver-
tically against
1
d
horizontally. Another table of
values is drawn up as shown below.
L 32.3 29.6 27.0 23.2 18.3 12.8 10.0 6.4
d 0.75 0.37 0.24 0.17 0.12 0.09 0.08 0.07
1
d
1.33 2.70 4.17 5.88 8.33 11.11 12.50 14.29
A graph of L against
1
d
is shown in Figure 18.2. A
straight line can be drawn through the points,
which veriﬁes that load and distance are related
by a law of the form L =
a
d
+ b.
A
B C
L
35
31
25
20
30
15
10
11
5
0 2 4 6 8 10 12 14
d
1
Figure 18.2
Gradient of straight line, a =
AB
BC
=
31 − 11
2 − 12
=
20
−10
= −2.
L-axis intercept, b = 35.
Hence, the law of the graph is L = −
2
d
+ 35.
(b) When the distance d = 0.20m,
load,L =
−2
0.20
+ 35 = 25.0N.
Rearranging L = −
2
d
+ 35 gives
2
d
= 35 − L and d =
2
35 − L

Graphs reducing non-linear laws to linear form 149
Hence, when the load L = 20N, distance,
d =
2
35 − 20
=
2
15
= 0.13m.
Problem 3. The solubility s of potassium chlorate
is shown by the following table.
t ◦C 10 20 30 40 50 60 80 100
s 4.9 7.6 11.1 15.4 20.4 26.4 40.6 58.0
The relationship between s and t is thought to be of
the form s = 3 + at + bt2. Plot a graph to test the
supposition and use the graph to find approximate
values of a and b. Hence, calculate the solubility of
potassium chlorate at 70◦C
Rearranging s = 3 + at + bt2 gives
s − 3 = at + bt2
and s − 3
t
= a + bt
or s − 3
t
= bt + a
which is of the form Y = mX + c
This shows that
s − 3
t
is to be plotted vertically and
t 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 80 100
s 4.9 7.6 11.1 15.4 20.4 26.4 40.6 58.0
s − 3
t
0.19 0.23 0.27 0.31 0.35 0.39 0.47 0.55
A graph of
s − 3
t
against t is shown plotted in Figure
18.3. A straight line fits the points, which shows that
s and t are related by s = 3 + at + bt2
.
Gradient of straight line, b =
AB
BC
=
0.39 − 0.19
60 − 10
=
0.20
50
= 0.004
Vertical axis intercept, a = 0.15
Hence, the law of the graph is s = 3 + 0.15t + 0.004t2.
The solubility of potassium chlorate at 70◦C is
given by
s = 3 + 0.15(70) + 0.004(70)2
= 3 + 10.5 + 19.6 = 33.1
C
B
A
0.6
0.5
0.4
0.39
0.19
0.15
0.3
0.2
0.1
0 20 40 60
t 8C
80 100
s23
t
Figure 18.3
PracticeExercise 70 Graphs reducing
non-linear laws to linear form (answers on
page 348)
In problems 1 to 5, x and y are two related vari-
ables and all other letters denote constants. For the
stated laws to be verified it is necessary to plot
graphs of the variables in a modified form. State
for each, (a) what should be plotted on the vertical
axis, (b) what should be plotted on the horizon-
tal axis, (c) the gradient and (d) the vertical axis
intercept.
1. y = d + cx2 2. y − a = b
√
x
3. y − e =
f
x
4. y − cx = bx2
5. y =
a
x
+ bx
6. In an experiment the resistance of wire is mea-
sured for wires of different diameters with the
following results.
R (ohms) 1.64 1.14 0.89 0.76 0.63
d (mm) 1.10 1.42 1.75 2.04 2.56
It is thought that R is related to d by the law
R =
a
d2
+ b, where a and b are constants. Ver-
ify this and find the approximate values for

a and b. Determine the cross-sectional area
needed for a resistance reading of 0.50ohms.
7. Corresponding experimental values of two
quantities x and y are given below.
x 1.5 3.0 4.5 6.0 7.5 9.0
y 11.5 25.0 47.5 79.0 119.5 169.0
By plotting a suitable graph, verify that y
and x are connected by a law of the form
y = kx2 + c, where k and c are constants.
Determine the law of the graph and hence ﬁnd
the value of x when y is 60.0
8. Experimental results of the safe load L kN,
applied to girders of varying spans, d m, are
shown below.
Span, d (m) 2.0 2.8 3.6 4.2 4.8
Load, L (kN) 475 339 264 226 198
It isbelieved that therelationship between load
and 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 are
believed to be related by a law of the form
y = ax2 + bx, where a and b are constants.
x 33.86 55.54 72.80 84.10 111.4 168.1
y 3.4 5.2 6.5 7.3 9.1 12.4
Verify the law and determine approximate val-
ues of a and b. Hence, determine (a) the value
of y when x is 8.0 and (b) the value of x when
y is 146.5
18.3 Revision of laws of logarithms
The laws of logarithms were stated in Chapter 15 as
follows:
log(A × B) = logA + logB (1)
log
A
B
= logA − logB (2)
logAn
= n × logA (3)
Also, lne = 1 and if, say, lg x = 1.5,
then x = 101.5
= 31.62
Further, if 3x = 7 then lg 3x = lg 7 and x lg 3 = lg 7,
from which x =
lg 7
lg 3
= 1.771
Theselawsand techniquesareused whenevernon-linear
laws of the form y = axn, y = abx and y = aebx are
reduced to linear form with the values ofaand b needing
to be calculated. This is demonstrated in the following
section.
18.4 Determination of laws involving
logarithms
Examples of the reduction of equations to linear form
involving logarithms include
(a) y = axn
Taking logarithms to a base of 10 of both sides
gives
lg y = lg(axn
)
= lga + lgxn
by law (1)
i.e. lg y = n lgx + lga by law (3)
which compares with Y = mX + c
and shows that lgy is plotted vertically against
lgx horizontally to produce a straight line graph
of gradient n and lgy-axis intercept lg a.
See worked Problems 4 and 5 to demonstrate how
this law is determined.
(b) y = abx
Taking logarithms to a base of 10 of both sides
gives
lg y = lg(abx
)
i.e. lg y = lga + lgbx
by law (1)
lg y = lga + x lgb by law (3)
i.e. lg y = x lgb + lga
or lg y = (lgb)x + lga
which compares with
Y = mX + c

and shows that lgy is plotted vertically against
x horizontally to produce a straight line graph
of gradient lgb and lgy-axis intercept lg a.
See worked Problem 6 to demonstrate how this law
is determined.
(c) y = aebx
Taking logarithms to a base of e of both sides
gives
ln y = ln(aebx
)
i.e. ln y = lna + lnebx
by law (1)
i.e. ln y = lna + bx lne by law (3)
i.e. ln y = bx + lna since lne = 1
which compares with
Y = mX + c
and shows that lny is plotted vertically against
x horizontally to produce a straight line graph
of gradient b and ln y-axis intercept ln a.
See worked Problem 7 to demonstrate how this law is
determined.
Problem 4. The current flowing in, and the
power dissipated by, a resistor are measured
experimentally for various values and the results are
as shown below.
Current, I (amperes) 2.2 3.6 4.1 5.6 6.8
Power, P (watts) 116 311 403 753 1110
Show that the law relating current and power is of
the form P = RIn , where R and n are constants,
and determine the law
Taking logarithms to a base of 10 of both sides of
P = RIn gives
lg P = lg(RIn
) = lg R + lg In
= lg R + n lg I
by the laws of logarithms
i.e. lg P = n lg I + lg R
which is of the form Y = mX + c,
showing that lg P is to be plotted vertically against lg I
horizontally.
A table of values for lg I and lg P is drawn up as shown
below.
I 2.2 3.6 4.1 5.6 6.8
lg I 0.342 0.556 0.613 0.748 0.833
P 116 311 403 753 1110
lg P 2.064 2.493 2.605 2.877 3.045
A graph of lg P against lg I is shown in Figure 18.4
and, since a straight line results, the law P = RIn is
verified.
C
B
A
D
2.0
0.30 0.40 0.50 0.60 0.70 0.80 0.90
2.18
2.5
lgP
lg L
2.78
2.98
3.0
Figure 18.4
Gradient of straight line, n =
AB
BC
=
2.98 − 2.18
0.8 − 0.4
=
0.80
0.4
= 2
It is not possible to determine the vertical axis intercept
on sight since the horizontal axis scale does not start
at zero. Selecting any point from the graph, say point
D, where lg I = 0.70 and lg P = 2.78 and substituting
values into
lg P = n lg I + lg R
gives 2.78 = (2)(0.70) + lg R
from which, lg R = 2.78 − 1.40 = 1.38
Hence, R = antilog 1.38 = 101.38
= 24.0
Hence, the law of the graph is P = 24.0I2
Problem 5. The periodic time, T, of oscillation of
a pendulum is believed to be related to its length, L,
by a law of the form T = kLn, where k and n are
constants. Values of T were measured for various
lengths of the pendulum and the results are as
shown below.
Periodic time, T(s) 1.0 1.3 1.5 1.8 2.0 2.3
Length, L(m) 0.25 0.42 0.56 0.81 1.0 1.32
Show that the law is true and determine the
approximate values of k and n. Hence find the
periodic time when the length of the pendulum is
0.75m

From para (a), page 150, if T = kLn
then lgT = n lgL + lgk
and comparing with Y = mX + c
shows that lgT is plotted vertically against lgL hor-
izontally, with gradient n and vertical-axis intercept
lgk.
A table of values for lgT and lg L is drawn up as shown
below.
T 1.0 1.3 1.5 1.8 2.0 2.3
lgT 0 0.114 0.176 0.255 0.301 0.362
L 0.25 0.42 0.56 0.81 1.0 1.32
lgL −0.602 −0.377 −0.252 −0.092 0 0.121
A graph of lgT against lg L is shown in Figure 18.5 and
the law T = kLn is true since a straight line results.
A
BC
020.4020.5020.60 20.3020.20 0.20
0.10
0.20
0.30
0.25
0.05
lg L
lg T
0.40
20.10 0.10
Figure 18.5
From the graph, gradient of straight line,
n =
AB
BC
=
0.25 − 0.05
−0.10 − (−0.50)
=
0.20
0.40
=
1
2
Vertical axis intercept, lgk = 0.30. Hence,
k = antilog 0.30 = 100.30 = 2.0
Hence, the law of the graph is T = 2.0L1/2 or
T = 2.0
√
L.
When length L = 0.75m, T = 2.0
√
0.75 = 1.73s
Problem 6. Quantities x and y are believed to be
related by a law of the form y = abx, where a and b
are constants. The values of x and corresponding
values of y are
x 0 0.6 1.2 1.8 2.4 3.0
y 5.0 9.67 18.7 36.1 69.8 135.0
Verify the law and determine the approximate
values of a and b. Hence determine (a) the value of
y when x is 2.1 and (b) the value of x when y is 100
From para (b), page 150, if y = abx
then lg y = (lgb)x + lga
and comparing with Y = mX + c
shows 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 3.0
y 5.0 9.67 18.7 36.1 69.8 135.0
lg y 0.70 0.99 1.27 1.56 1.84 2.13
A graph of lg y against x is shown in Figure 18.6
and, since a straight line results, the law y = abx
is
veriﬁed.
C
B
A
lgy
x
2.50
2.13
2.00
1.50
1.00
1.17
0.50
0 1.0 2.0 3.0
0.70
Figure 18.6

Gradient of straight line,
lgb =
AB
BC
=
2.13 − 1.17
3.0 − 1.0
=
0.96
2.0
= 0.48
Hence, b = antilog 0.48 = 100.48 = 3.0, correct to 2
Vertical axis intercept, lga = 0.70,from which
a = antilog 0.70
= 100.70
= 5.0, correct to
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 which
100/5.0 = (3.0)x
i.e. 20 = (3.0)x
Taking logarithms of both sides gives
lg20 = lg(3.0)x
= x lg3.0
Hence, x =
lg20
lg3.0
=
1.3010
0.4771
= 2.73
Problem 7. The current i mA ﬂowing in a
capacitor which is being discharged varies with
time t ms, as shown below.
i (mA) 203 61.14 22.49 6.13 2.49 0.615
t (ms) 100 160 210 275 320 390
Show that these results are related by a law of the
form i = Iet/T , where I and T are constants.
Determine the approximate values of I and T
Taking Napierian logarithms of both sides of
i = Iet/T
gives lni = ln(Iet/T
) = ln I + lnet/T
= ln I +
t
T
lne
i.e. lni = ln I +
t
T
since lne = 1
or lni =
1
T
t + ln I
which compares with y = mx + c
showing that lni is plotted vertically against t hor-
izontally, with gradient 1
T and vertical-axis intercept
ln I.
Another table of values is drawn up as shown below.
t 100 160 210 275 320 390
i 203 61.14 22.49 6.13 2.49 0.615
lni 5.31 4.11 3.11 1.81 0.91 −0.49
A graph of lni against t is shown in Figure 18.7
and, since a straight line results, the law i = Iet/T is
veriﬁed.
A
B
C
lni
5.0
4.0
3.0
3.31
2.0
1.0
100 200 300 400 t (ms)
D (200, 3.31)
21.0
1.30
0
Figure 18.7
Gradient of straight line,
1
T
=
AB
BC
=
5.30 − 1.30
100 − 300
=
4.0
−200
= −0.02
Hence, T =
1
−0.02
= −50
Selecting any point on the graph, say point D, where
t = 200 and lni = 3.31, and substituting
into lni =
1
T
t + ln I
gives 3.31 = −
1
50
(200) + ln I
from which, ln I = 3.31 + 4.0 = 7.31
and I = antilog 7.31 = e7.31 = 1495 or 1500
Hence, the law of the graph is i = 1500e−t/50

Practice Exercise 71 Determination of law
involving 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 the
stated laws to be verified it is necessary to plot
graphs of the variables in a modified form. State
for each, (a) what should be plotted on the vertical
axis, (b) what should be plotted on the horizon-
tal axis, (c) the gradient and (d) the vertical axis
intercept.
1. y = bax 2. y = kxL 3.
y
m
= enx
4. The luminosity I of a lamp varies with
the applied voltage V and the relationship
between I and V is thought to be I = kV n.
Experimental results obtained are
I(candelas) 1.92 4.32 9.72
V (volts) 40 60 90
I(candelas) 15.87 23.52 30.72
V (volts) 115 140 160
Verify that the law is true and determine
the law of the graph. Also determine the
luminosity when 75V is applied across the
lamp.
5. The head of pressure h and the flow veloc-
ity v are measured and are believed to be
connected by the law v = ahb, where a and
b are constants. The results are as shown
below.
h 10.6 13.4 17.2 24.6 29.3
v 9.77 11.0 12.44 14.88 16.24
Verify that the law is true and determine values
of a and b.
6. Experimental values of x and y are measured
as follows.
x 0.4 0.9 1.2 2.3 3.8
y 8.35 13.47 17.94 51.32 215.20
The law relating x and y is believed to be of
the 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 and
the value of x when y is 100.
7. The activity of a mixture of radioactive iso-
topes is believed to vary according to the law
R = R0t−c, where R0 and c are constants.
Experimental results are shown below.
R 9.72 2.65 1.15 0.47 0.32 0.23
t 2 5 9 17 22 28
Verify that the law is true and determine
approximate values of R0 and c.
8. Determine the law of the form y = aekx
which
relates the following values.
y 0.0306 0.285 0.841 5.21 173.2 1181
x –4.0 5.3 9.8 17.4 32.0 40.0
9. The tension T in a belt passing round a pul-
ley wheel and in contact with the pulley over
an angle of θ radians is given by T = T0eμθ ,
where T0 and μ are constants. Experimental
results obtained are
T (newtons) 47.9 52.8 60.3 70.1 80.9
θ (radians) 1.12 1.48 1.97 2.53 3.06
Determine approximate values of T0 and μ.
Hence, find the tension when θ is 2.25 radians
and the value of θ when the tension is 50.0
newtons.

Chapter 19
Graphical solution of
equations
19.1 Graphical solution of
Linear simultaneous equations in two unknowns may
be 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 the
required solution.
Here are some worked problems to demonstrate the
graphical solution of simultaneous equations.
Problem 1. Solve graphically the simultaneous
equations
2x − y = 4
x + y = 5
Rearranging each equation into y = mx + c form gives
y = 2x − 4
y = −x + 5
Only three co-ordinates need be calculated for each
graph since both are straight lines.
x 0 1 2
y = 2x − 4 −4 −2 0
x 0 1 2
y = −x + 5 5 4 3
Each of the graphs is plotted as shown in Figure 19.1.
The point of intersection is at (3, 2) and since this is the
only point which lies simultaneously on both lines
then x = 3,y = 2 is the solution of the simultaneous
equations.
24 23 22 21 1
1
2
3
4
5
y
y 52x1 5
y 52x2 4
2 3 4 x
21
22
23
24
0
Figure 19.1
Problem 2. Solve graphically the equations
1.20x + y = 1.80
x − 5.0y = 8.50
DOI: 10.1016/B978-1-85617-697-2.00019-3

Rearranging each equation into y = mx + c form gives
y = −1.20x + 1.80 (1)
y =
x
5.0
−
8.5
5.0
i.e. y = 0.20x − 1.70 (2)
Three co-ordinates are calculated for each equation as
shown below.
x 0 1 2
y = −1.20x + 1.80 1.80 0.60 −0.60
x 0 1 2
y = 0.20x − 1.70 −1.70 −1.50 −1.30
The two sets of co-ordinates are plotted as shown in
Figure 19.2. The point of intersection is (2.50,−1.20).
Hence, the solution of the simultaneous equations is
x = 2.50,y = −1.20
(It is sometimes useful to initially sketch the two straight
lines to determine the region where the point of inter-
section is. Then, for greater accuracy, a graph having a
smaller range of values can be drawn to ‘magnify’ the
point of intersection.)
21
21
22
23
21.20
2223
3
1
0 1 2 3
2.50
4
y
x
y 5 0.20x 21.70
y 521.20x 11.80
Figure 19.2
PracticeExercise 72 Graphical solution of
simultaneous equations (Answers on
page 347)
tions graphically.
1. y = 3x − 2 2. x + y = 2
y = −x + 6 3y − 2x = 1
3. y = 5 − x 4. 3x + 4y = 5
x − y = 2 2x − 5y + 12 = 0
5. 1.4x − 7.06 = 3.2y 6. 3x − 2y = 0
2.1x − 6.7y = 12.87 4x + y + 11 = 0
7. The friction force F newtons and load L
newtons are connected by a law of the form
F = aL + b, where a and b are constants.
When F = 4N, L = 6N and when F = 2.4N,
L = 2N. Determine graphically the values of a
and b.
19.2 Graphical solution of quadratic
equations
A general quadratic equation is of the form
y = ax2 + bx + c, where a, b and c are constants and
a is not equal to zero.
A graph of a quadratic equation always produces a shape
called a parabola.
The gradients of the curves between 0 and A and
between B and C in Figure 19.3 are positive, whilst
the gradient between A and B is negative. Points such
as A and B are called turning points. At A the gradi-
ent is zero and, as x increases, the gradient of the curve
changesfrompositivejust before Ato negativejust after.
Such a point is called a maximum value. At B the gra-
dient is also zero and, as x increases, the gradient of the
curve changes from negative just before B to positive
just after. Such a point is called a minimum value.
y
x
A
C
B
0
Figure 19.3
Following are three examples of solutions using
quadratic graphs.
(a) y = ax2
Graphs of y = x2, y = 3x2 and y =
1
2
x2 are
shown in Figure 19.4. All have minimum values at
the origin (0, 0).

Graphical solution of equations 157
(a)
y
y5 x2
x
2
1
21 0 1
(b)
y53x2
y
x
2
1
21 0 1
y
x
2
1
21 0 1
(c)
y 5 x2
2
1
Figure 19.4
Graphs of y = −x2, y = −3x2 and y = −
1
2
x2 are
shown in Figure 19.5. All have maximum values
at the origin (0, 0).
(a)
y
y 52x2
x21
21
22
0
1
(b)
y523x2
21
22
y
x21
0
1
21
22
y
x21
0
1
(c)
y52 x2
2
1
Figure 19.5
When y = ax2
,
(i) curves are symmetrical about the y-axis,
(ii) the magnitude of a affects the gradient of the
curve, and
(iii) the sign of a determines whether it has a
maximum or minimum value.
(b) y = ax2 + c
Graphs of y = x2 + 3, y = x2 − 2, y = −x2 + 2
and 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 the
curve, and
(iii) the constant c is the y-axis intercept.
(c) y = ax2 + bx + c
Whenever b has a value other than zero the curve
is displaced to the right or left of the y-axis.
When b/a is positive, the curve is displaced b/2a
to the left of the y-axis, as shown in Figure19.7(a).
When b/a is negative, the curve is displaced
b/2a to the right of the y-axis, as shown in
Figure 19.7(b).
Quadratic equations of the form ax2 + bx + c = 0
may 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).
y
y 5x213
y 52x21 2
y 522x221
y 5 x222
x21 0
(a)
3
1
y
x
22
21 1
(b)
2
0
y
x21
(c)
0
2
1
y
x21
21
24
(d)
0
1
Figure 19.6
(a)
25 24 23 22 21 1
y
x
12
10
6
4
2
0
y5 x216x111
(b)
y 5x2 25x14
21
y
x0
2
4
6
22
1 2 3 4
8
Figure 19.7
The x values of the points of intersection give the
required solutions since at these points both y = 0 and
ax2 + bx + c = 0.
The number of solutions, or roots, of a quadratic equa-
tion depends on how many times the curve cuts the
x-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 in
Figure 19.7(b).
graphical solution of quadratic equations.
4x2 + 4x − 15 = 0 graphically, given that the
solutions lie in the range x = −3 to x = 2.
Determine also the co-ordinates and nature of the
turning point of the curve

Let y = 4x2
+ 4x − 15. A table of values is drawn up as
shown below.
x −3 −2 −1 0 1 2
y = 4x2 + 4x − 15 9 −7 −15 −15 −7 9
y 54x2
14x 21512
y
8
4
24
28
212
216
20.5
1.5
0 1 2 x
A B
22.5
23 22 21
Figure 19.8
A graph of y = 4x2 + 4x − 15 is shown in Figure 19.8.
The only pointswhere y = 4x2 + 4x − 15 and y = 0 are
the points marked A and B. This occurs at x = −2.5
and x = 1.5 and these are the solutions of the quadratic
equation 4x2 + 4x − 15 = 0.
By substituting x = −2.5 and x = 1.5 into the original
equation the solutions may be checked.
The curve has a turning point at (−0.5,−16) and the
nature of the point is a minimum.
An alternative graphical method of solving
4x2 + 4x − 15 = 0 is to rearrange the equation as
4x2
= −4x + 15 and then plot two separate graphs
− in this case, y = 4x2 and y = −4x + 15. Their
points of intersection give the roots of the equation
4x2 = −4x + 15, i.e. 4x2 + 4x − 15 = 0. This is shown
in Figure 19.9, where the roots are x = −2.5 and
x = 1.5, as before.
Problem 4. Solve graphically the quadratic
equation −5x2 + 9x + 7.2 = 0 given that the
solutions lie between x = −1 and x = 3. Determine
also the co-ordinates of the turning point and state
its nature
y 524x 115
30
25
20
15
10
5
y
10 2 3 x2122
22.5 1.5
23
y 54x 2
Figure 19.9
Let y = −5x2 + 9x + 7.2. A table of values is drawn up
as shown below.
x −1 −0.5 0 1
y = −5x2 + 9x + 7.2 −6.8 1.45 7.2 11.2
x 2 2.5 3
y = −5x2 + 9x + 7.2 5.2 −1.55 −10.8
A graph of y = −5x2 + 9x + 7.2 is shown plotted in
Figure 19.10. The graph crosses the x-axis (i.e. where
y = 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 hence
solve the equations
(a) 2x2 − 8 = 0 (b) 2x2 − x − 3 = 0
A graph of y = 2x2
is shown in Figure 19.11.
(a) Rearranging 2x2 − 8 = 0 gives 2x2 = 8 and the
solution ofthisequation isobtained fromthepoints
of 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 solutions
of 2x2
− 8 = 0 are x = −2 and x = +2.
(b) Rearranging 2x2 − x − 3 = 0 gives 2x2 = x + 3
and the solution of this equation is obtained
from the points of intersection of y = 2x2 and
y = x + 3; i.e., at C and D in Figure 19.11. Hence,
the solutions of 2x2 − x − 3 = 0 are x = −1 and
x = 1.5

Ϫ10
Ϫ8
Ϫ6
Ϫ4
Ϫ2
Ϫ0.6
yϭϪ5x2ϩ 9x ϩ7.2
0.9 2.4
0Ϫ1 1 2 3
2
4
6
8
10
11.25
12
y
x
Figure 19.10
yϭ2x 2
yϭ8
yϭxϩ3
Ϫ2 Ϫ1 0
2
4
6
8
10
A
C
D
B
y
1 1.5 2 x
Figure 19.11
Problem 6. Plot the graph of y = −2x2 + 3x + 6
for values of x from x = −2 to x = 4. Use the
graph 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 = 0
A table of values for y = −2x2 + 3x + 6 is drawn up as
shown below.
x −2 −1 0 1 2 3 4
y −8 1 6 7 4 −3 −14
A graph of y = −2x2
+ 3x + 6 is shown in
Figure 19.12.
28
26
24
22
21.1321.35
22
21.5
G
BA
DC
H
F y 523
y 54
y 52x 11
y 522x21 3x 16
x
E
1 2
1.85 2.63
30
2
4
6
8
y
20.521
Figure 19.12
(a) The parabola y = −2x2 + 3x + 6 and the straight
line y = 0 intersect at A and B, where x = −1.13
and x = 2.63, and these are the roots of the
equation −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 of
equation (2), the RHS of both equations will
be the same. Hence, 4 = −2x2 + 3x + 6. The
solution of this equation is found from the
points of intersection of the line y = 4 and
the parabola y = −2x2 + 3x + 6; i.e., points C
and 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 of
this equation is obtained from the points
of intersection of the line y = −3 and the
parabola y = −2x2 + 3x + 6; i.e., at points E
and F in Figure 19.12. Hence, the roots of
−2x2 + 3x + 9 = 0 are x = −1.5 and x = 3.

(d) Comparing y = −2x2
+ 3x + 6 (3)
with 0 = −2x2
+ x + 5 (4)
shows that, if 2x + 1 is added to both sides
of equation (4), the RHS of both equations
will be the same. Hence, equation (4) may be
written as 2x + 1 = −2x2 + 3x + 6. The solu-
tion of this equation is found from the points
of intersection of the line y = 2x + 1 and the
parabola y = −2x2 + 3x + 6; i.e., points G
and H in Figure 19.12. Hence, the roots of
−2x2 + x + 5 = 0 are x = −1.35 and x = 1.85
equations graphically (answers on page 348)
1. Sketch the following graphs and state the
nature and co-ordinates of their respective
turning points.
(a) y = 4x2 (b) y = 2x2 − 1
(c) y = −x2 + 3 (d) y = −
1
2
x2 − 1
Solve graphically the quadratic equations in prob-
lems 2 to 5 by plotting the curves between the given
limits. Give answers correct to 1 decimal place.
2. 4x2 − x − 1 = 0; x = −1 to x = 1
3. x2 − 3x = 27; x = −5 to x = 8
4. 2x2 − 6x − 9 = 0; x = −2 to x = 5
5. 2x(5x − 2) = 39.6; x = −2 to x = 3
6. Solve the quadratic equation
2x2 + 7x + 6 = 0 graphically, given that
the solutions lie in the range x = −3 to
x = 1. Determine also the nature and
co-ordinates of its turning point.
7. Solve graphically the quadratic equation
10x2 − 9x − 11.2 = 0,given that therootslie
between x = −1 and x = 2.
8. Plot a graph of y = 3x2 and hence solve the
(a) 3x2 − 8 = 0 (b) 3x2 − 2x − 1 = 0
9. Plot the graphs y = 2x2 and y = 3 − 4x on
the same axes and ﬁnd the co-ordinates of the
points of intersection. Hence, determine the
roots of the equation 2x2 + 4x − 3 = 0.
10. Plot a graph of y = 10x2 − 13x − 30 for
values of x between x = −2 and x = 3.
Solve the equation 10x2 − 13x − 30 = 0 and
from 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 equation
10x2 − 15x − 18 = 0.
19.3 Graphical solution of linear
and quadratic equations
simultaneously
The solutionof linear and quadratic equations simul-
taneously may be achieved graphically by
(a) plotting the straight line and parabola on the same
axes, and
(b) noting the points of intersection.
The co-ordinates of the points of intersection give the
required solutions.
Here is a worked problem to demonstrate the simulta-
neous solution of a linear and quadratic equation.
Problem 7. Determine graphically the values of
x and y which simultaneously satisfy the equations
y = 2x2 − 3x − 4 and y = 2 − 4x
y = 2x2 − 3x − 4 is a parabola and a table of values is
drawn up as shown below.
x −2 −1 0 1 2 3
y 10 1 −4 −5 −2 5
y = 2 − 4x is a straight line and only three co-ordinates
need be calculated:
x 0 1 2
y 2 −2 −6
The two graphs are plotted in Figure 19.13 and the
points 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 and
when x = 1.5, y = −4.
These solutions may be checked by substituting into
each of the original equations.

Ϫ4
0Ϫ1Ϫ2 1
B
A
yϭ2Ϫ 4x
yϭ2x2 Ϫ3x Ϫ4
2 3 x
2
4
6
8
10
y
Ϫ2
Figure 19.13
PracticeExercise 74 Solving linear and
quadratic equations simultaneously
1. Determine graphically the values of x and
y which simultaneously satisfy the equations
y = 2(x2 − 2x − 4) and y + 4 = 3x.
2. Plot the graph of y = 4x2 − 8x − 21 for values
of x from −2 to +4. Use the graph to ﬁnd the
roots of the following equations.
(a) 4x2 − 8x − 21 = 0
(b) 4x2 − 8x − 16 = 0
(c) 4x2
− 6x − 18 = 0
19.4 Graphical solution of cubic
equations
A cubic equation of the form ax3 + bx2 + cx + d = 0
may 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 the
required solution since at these points both y = 0 and
ax3 + bx2 + cx + d = 0.
The number of solutions, or roots, of a cubic equation
depends on how many times the curve cuts the x-axis
and there can be one, two or three possible roots, as
shown in Figure 19.14.
(a)
y
x
(b)
y
x
(c)
y
x
Figure 19.14
graphical solution of cubic equations.
Problem 8. Solve graphically the cubic equation
4x3 − 8x2 − 15x + 9 = 0, given that the roots lie
between x = −2 and x = 3. Determine also the
co-ordinates of the turning points and distinguish
between them
Let y = 4x3
− 8x2
− 15x + 9.Atableofvaluesisdrawn
up as shown below.
x −2 −1 0 1 2 3
y −25 12 9 −10 −21 0
A graph of y = 4x3 − 8x2 − 15x + 9 is shown in
Figure 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 the
cubic equation 4x3 − 8x2 − 15x + 9 = 0.
The turning points occur at (−0.6,14.2), which is a
maximum, and (2,−21), which is a minimum.
Problem 9. Plot the graph of
y = 2x3 − 7x2 + 4x + 4 for values of x between
x = −1 and x = 3. Hence, determine the roots of
the equation 2x3 − 7x2 + 4x + 4 = 0
x −1 0 1 2 3
y −9 4 3 0 7

22 21 20.6
24
4
8
12
16
y
14.2
28
212
216
220
221
224
0 1 2 3 x
y 54x32 8x2215x19
Figure 19.15
A graph of y = 2x3 − 7x2 + 4x + 4 is shown in
Figure 19.16. The graph crosses the x-axis at x = −0.5
and touches the x-axis at x = 2.
28
26
24
22
21 1 2 30
2
4
6
8
y
x
y 52x327x214x 14
Figure 19.16
Hence the solutions of the equation
2x3 − 7x2 + 4x + 4 = 0 are x = −0.5 and x = 2.
PracticeExercise 75 Solving cubic
1. Plot the graph y = 4x3 + 4x2 − 11x − 6
between x = −3 and x = 2 and use
the graph to solve the cubic equation
4x3 + 4x2 − 11x − 6 = 0.
2. By plotting a graph of y = x3 − 2x2 − 5x + 6
between x = −3 and x = 4, solve the equa-
tion x3 − 2x2 − 5x + 6 = 0. Determine also
the co-ordinates of the turning points and
distinguish between them.
In problems 3 to 6, solve graphically the cubic
equations given, each correct to 2 signiﬁcant
ﬁgures.
3. x3 − 1 = 0
4. x3 − x2 − 5x + 2 = 0
5. x3 − 2x2 = 2x − 2
6. 2x3 − x2 − 9.08x + 8.28 = 0
7. Show that the cubic equation
8x3 + 36x2 + 54x + 27 = 0 has only one real
root and determine its value.

Revision Test 7 : Graphs
1. Determine the value of P in the following table of
values.
x 0 1 4
y = 3x − 5 −5 −2 P (2)
2. Assuming graph paper measuring 20cm by 20cm
is available, suggest suitable scales for the follow-
ing ranges of values.
Horizontal axis: 5N to 70N; vertical axis: 20mm
to 190mm. (2)
3. Corresponding valuesobtained experimentally for
two quantities are:
x −5 −3 −1 0 2 4
y −17 −11 −5 −2 4 10
Plot a graph of y (vertically) against x (hori-
zontally) to scales of 1cm = 1 for the horizontal
x-axis and 1cm = 2 for the vertical y-axis. From
the 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 each
of the following, state (i) the gradient, and (ii) the
y-axis intercept.
(a) y = −5x + 3 (b) y = 7x
(c) 2y + 4 = 5x (d) 5x + 2y = 6
(e) 2x −
y
3
=
7
6
(10)
5. Theresistance R ohmsofacopperwinding ismea-
sured at various temperatures t◦C and the results
are as follows.
R ( ) 38 47 55 62 72
t (◦C) 16 34 50 64 84
Plot 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 other
letters denote constants. For the stated laws to
be verified it is necessary to plot graphs of the
variables in a modified form. State for each
(a) what should be plotted on the vertical axis,
(b) what should be plotted on the horizontal axis,
(c) the gradient,
(d) the vertical axis intercept.
(i) y = p +rx2
(ii) y =
a
x
+ bx (4)
7. The following results give corresponding values
of two quantities x and y which are believed to be
related by a law of the form y = ax2 + bx where
a and b are constants.
y 33.9 55.5 72.8 84.1 111.4 168.1
x 3.4 5.2 6.5 7.3 9.1 12.4
Verify the law and determine approximate values
of a and b.
Hence determine (i) the value of y when x is 8.0
and (ii) the value of x when y is 146.5 (18)
8. By taking logarithms of both sides of y = k xn ,
show that lg y needs to be plotted vertically and
lgx needs to be plotted horizontally to produce
a straight line graph. Also, state the gradient and
vertical-axis intercept. (6)
9. By taking logarithms of both sides of y = aek x
show that ln y needs to be plotted vertically and
x needs to be plotted horizontally to produce a
straight line graph. Also, state the gradient and
vertical-axis intercept. (6)
10. Show from the following results of voltage V and
admittance Y of an electrical circuit that the law
connecting the quantities is of the form V =kYn
and determine the values of k and n.
Voltage
V (volts) 2.88 2.05 1.60 1.22 0.96
Admittance,
Y(siemens) 0.52 0.73 0.94 1.23 1.57
(12)

11. The current i ﬂowing in a discharging capacitor
varies with time t as shown.
i (mA) 50.0 17.0 5.8 1.7 0.58 0.24
t (ms) 200 255 310 375 425 475
Show that these results are connected by the
law of the form i = Ie
t
T where I is the initial
current ﬂowing and T is a constant. Determine
approximate values of constants I and T. (15)
12. Solve, correct to 1 decimal place, the quadratic
equation 2x2 − 6x − 9 = 0 by plotting values of x
from x = −2 to x = 5. (8)
13. Plot the graph of y = x3 + 4x2 + x − 6 for
values of x between x = −4 and x = 2.
Hence determine the roots of the equation
x3
+ 4x2
+ x − 6 = 0. (9)
14. Plot a graph of y = 2x2 from x = −3 to x = +3
and hence solve the following equations.
(a) 2x2 − 8 = 0
(b) 2x2 − 4x − 6 = 0 (10)

Chapter 20
Angles and triangles
20.1 Introduction
Trigonometry is a subject that involves the measurement
of sides and angles of triangles and their relationship to
each other. This chapter involves the measurement of
angles and introduces types of triangle.
20.2 Angular measurement
An angle is the amount of rotation between two straight
lines. Angles may be measured either in degrees or in
radians.
If a circle is divided into 360 equal parts, then each part
is called 1 degree and is written as 1◦
i.e. 1 revolution = 360◦
or 1 degree is
1
360
th of a revolution
Some angles are given special names.
• Any angle between 0◦
and 90◦
is called an acute
angle.
• An angle equal to 90◦ is called a right angle.
• Any angle between 90◦ and 180◦ is called an obtuse
angle.
• Any angle greater than 180◦ and less than 360◦ is
called 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 thesame
plane and never meet. Such lines are denoted by
arrows, as in Figure 20.1.
• A straight line which crosses two parallel lines is
called a transversal (see MN in Figure 20.1).
P
R
M
N
g
h e
f
ad
bc
Q
S
Figure 20.1
With reference to Figure 20.1,
(a) a = c, b = d, e = g and f = h. Such pairs of
angles are called vertically opposite angles.
(b) a = e, b = f , c = g and d = h. Such pairs of
angles are called corresponding angles.
(c) c = e and b = h. Such pairs of angles are called
alternate angles.
(d) b + e = 180◦ and c + h = 180◦. Such pairs of
angles are called interior angles.
20.2.1 Minutes and seconds
One degree may be sub-divided into 60 parts, called
minutes.
i.e. 1 degree = 60 minutes
which is written as 1◦
= 60 .
DOI: 10.1016/B978-1-85617-697-2.00020-X

41 degrees and 29 minutes is written as 41◦
29 . 41◦
29
is equivalent to 41
29◦
60
= 41.483◦ as a decimal, correct
to 3 decimal places by calculator.
1 minute further subdivides into 60 seconds,
i.e. 1 minute = 60 seconds
which is written as 1 = 60 .
(Notice that for minutes, 1 dash is used and for seconds,
2 dashes are used.)
For example, 56 degrees, 36 minutes and 13 seconds is
written as 56◦36 13 .
20.2.2 Radians and degrees
Oneradian isdeﬁned astheanglesubtended at thecentre
of a circle by an arc equal in length to the radius. (For
more on circles, see Chapter 26.)
With reference to Figure 20.2, for arc length s,
θ radians =
s
r
r
r
S
O
␪
Figure 20.2
When s is the whole circumference, i.e. when
s = 2πr,
θ =
s
r
=
2πr
r
= 2π
In one revolution, θ = 360◦. Hence, the relationship
between degrees and radians is
360◦
= 2π radians or 180◦
= π rad
i.e. 1 rad =
180◦
π
≈ 57.30◦
Here are some worked examples on angular measure-
ment.
Problem 1. Evaluate 43◦29 + 27◦43
43◦ 29
+ 27◦ 43
71◦ 12
1◦
(i) 29 + 43 = 72
(ii) Since 60 = 1◦,72 = 1◦12
(iii) The 12 is placed in the minutes column and 1◦ is
carried in the degrees column.
(iv) 43◦ + 27◦ + 1◦ (carried) = 71◦. Place 71◦ in the
degrees column.
This answer can be obtained using the calculator as
follows.
1. Enter 43 2. Press ◦
’ ’ ’ 3. Enter 29
4. Press ◦ ’ ’ ’ 5. Press + 6. Enter 27
7. Press ◦ ’ ’ ’ 8. Enter 43 9. Press ◦ ’ ’ ’
10. Press = Answer = 71◦
12
Thus, 43◦
29 + 27◦
43 = 71◦
12 .
Problem 2. Evaluate 84◦13 − 56◦39
84◦ 13
− 56◦ 39
27◦ 34
(i) 13 − 39 cannot be done.
(ii) 1◦
or 60 is ‘borrowed’ from the degrees column,
which leaves 83◦ in that column.
(iii) (60 + 13 ) − 39 = 34 , which is placed in the
minutes column.
(iv) 83◦ − 56◦ = 27◦, which is placed in the degrees
column.
follows.
’ ’ ’ 3. Enter 13
4. Press ◦ ’ ’ ’ 5. Press − 6. Enter 56
34
Thus, 84◦
13 − 56◦
39 = 27◦
34 .
Problem 3. Evaluate 19◦
51 47 + 63◦
27 34
19◦ 51 47
+ 63◦ 27 34
83◦ 19 21
1◦ 1

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 1
is 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 the
degrees column.
follows.
’ ’ ’ 3. Enter 51
7. Press + 8. Enter 63 9. Press ◦ ’ ’ ’
10. Enter 27 11. Press ◦ ’ ’ ’
12. Enter 34 13. Press ◦ ’ ’ ’
19 21
Thus, 19◦
51 47 + 63◦
27 34 = 83◦
19 21 .
Problem 4. Convert 39◦ 27 to degrees in decimal
form
39◦
27 = 39
27◦
60
27◦
60
= 0.45◦
by calculator
Hence, 39◦
27 = 39
27◦
60
= 39.45◦
follows.
1. Enter 39 2. Press ◦ ’ ’ ’ 3. Enter 27
4. Press ◦ ’ ’ ’ 5. Press = 6. Press ◦ ’ ’ ’
Answer = 39.45◦
Problem 5. Convert 63◦ 26 51 to degrees in
decimal form, correct to 3 decimal places
63◦
26 51 = 63◦
26
51
60
= 63◦
26.85
63◦
26.85 = 63
26.85◦
60
= 63.4475◦
Hence, 63◦
26 51 = 63.448◦
correct to 3 decimal
places.
follows.
1. Enter 63 2. Press ◦ ’ ’ ’ 3. Enter 26
7. Press = 8. Press ◦ ’ ’ ’ Answer = 63.4475◦
Problem 6. Convert 53.753◦ to degrees, minutes
and seconds
0.753◦
= 0.753 × 60 = 45.18
0.18 = 0.18 × 60 = 11
to the nearest second
Hence, 53.753◦
= 53◦
45 11
follows.
1. Enter 53.753 2. Press =
3. Press ◦ ’ ’ ’ Answer = 53◦
45 10.8
PracticeExercise 76 Angular measurement
1. Evaluate 52◦39 + 29◦48
2. Evaluate 76◦31 − 48◦37
3. Evaluate 77◦22 + 41◦36 − 67◦47
4. Evaluate 41◦37 16 + 58◦29 36
5. Evaluate 54◦37 42 − 38◦53 25
6. Evaluate79◦26 19 −45◦58 56 +53◦21 38
7. Convert 72◦33 to degrees in decimal form.
8. Convert 27◦45 15 to degrees correct to 3
decimal places.
9. Convert 37.952◦ to degrees and minutes.
10. Convert 58.381◦ to degrees, minutes and
seconds.
Here are some further worked examples on angular
measurement.
Problem 7. State the general name given to the
following angles: (a) 157◦ (b) 49◦ (c) 90◦ (d) 245◦
(a) Any angle between 90◦ and 180◦ is called an
obtuse angle.
Thus, 157◦ is an obtuse angle.
(b) Any angle between 0◦ and 90◦ is called an acute
angle.

Thus, 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◦ is
called a reflex angle.
Thus, 245◦
is a reflex angle.
Problem 8. Find the angle complementary to
48◦ 39
If two angles add up to 90◦ they are called comple-
mentary angles. Hence, the angle complementary to
48◦
39 is
90◦
− 48◦
39 = 41◦
21
Problem 9. Find the angle supplementary to
74◦25
If two angles add up to 180◦ they are called supple-
mentary angles. Hence, the angle supplementary to
74◦
25 is
180◦
− 74◦
25 = 105◦
35
Problem 10. Evaluate angle θ in each of the
diagrams shown in Figure 20.3
(a) (b)
418
538 448␪
␪
Figure 20.3
(a) The symbol shown in Figure 20.4 is called a right
angle and it equals 90◦.
Figure 20.4
Hence, 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 diagram
shown in Figure 20.5
39Њ
58Њ
64Њ
108Њ
␪
Figure 20.5
There are 360◦ in a complete revolution of a circle.
Thus, 360◦ = 58◦ + 108◦ + 64◦ + 39◦ + θ
fromwhich, θ = 360◦ −58◦ −108◦ −64◦ −39◦ =91◦
Problem 12. Two straight lines AB and CD
intersect at 0. If ∠AOC is 43◦, find ∠AOD,
∠DOB and ∠BOC
From Figure 20.6, ∠AOD is supplementary to ∠AOC.
Hence, ∠AOD = 180◦ − 43◦ = 137◦
.
When two straight linesintersect, the vertically opposite
angles are equal.
Hence, ∠DOB = 43◦
and ∠BOC 137◦
A
D
B
C
438
0
Figure 20.6

Problem 13. Determine angle β in Figure 20.7
1338
␣
␤
Figure 20.7
α = 180◦ − 133◦ = 47◦ (i.e. supplementary angles).
α = β = 47◦
(corresponding angles between parallel
lines).
Problem 14. Determine the value of angle θ in
Figure 20.8
23Њ37Ј
35Њ49Ј
A
F
C
B
G
E
D
␪
Figure 20.8
Let a straight line FG be drawn through E such that FG
is parallel to AB and CD.
∠BAE = ∠AEF (alternate angles between parallel lines
AB and FG), hence ∠AEF = 23◦37 .
∠ECD = ∠FEC (alternateanglesbetween parallel lines
FG and CD), hence ∠FEC = 35◦49 .
Angle θ = ∠AEF + ∠FEC = 23◦
37 + 35◦
49
= 59◦
26
Problem 15. Determine angles c and d in
Figure 20.9
d
b a
c
468
Figure 20.9
a = b = 46◦
(corresponding angles between parallel
lines).
Also, b + c + 90◦ = 180◦ (angles on a straight line).
Hence, 46◦ + c + 90◦ = 180◦, from which, c = 44◦
.
b and d are supplementary, hence d = 180◦ − 46◦
= 134◦
.
Alternatively, 90◦ + c = d (vertically opposite angles).
Problem 16. Convert the following angles to
radians, correct to 3 decimal places.
(a) 73◦ (b) 25◦37
Although wemay bemorefamiliarwith degrees,radians
is the SI unit of angular measurement in engineering
(1 radian ≈ 57.3◦).
(a) Since 180◦
= π rad then 1◦
=
π
180
rad.
Hence, 73◦
= 73 ×
π
180
rad = 1.274 rad.
(b) 25◦37 = 25
37◦
60
= 25.616666...
Hence, 25◦
37 = 25.616666...◦
= 25.616666...×
π
180
rad
= 0.447 rad.
Problem 17. Convert 0.743 rad to degrees and
minutes
Since 180◦ = π rad then 1 rad =
180◦
π
Hence, 0.743 rad = 0.743 ×
180◦
π
= 42.57076...◦
= 42◦
34
Since π rad = 180◦, then
π
2
rad = 90◦,
π
4
rad = 45◦,
π
3
rad = 60◦ and
π
6
rad = 30◦
PracticeExercise 77 Further angular
measurement (answers on page 348)
1. State the general name given to an angle of
197◦.
136◦.
49◦.

90◦
.
5. Determine the angles complementary to the
following.
(a) 69◦ (b) 27◦37 (c) 41◦3 43
6. Determine the angles supplementary to
(a) 78◦
(b) 15◦
(c) 169◦
41 11
7. Find the values of angle θ in diagrams
(a) to (i) of Figure 20.10.
1208
(a)
␪
558 508
(c)
␪
608 808
(f)
␪
␪
(e)
708
1508
␪
378
(d)
␪
(g)
50.78 ␪
(h)
1118
688
458
578
␪
(i)
368
␪
708
(b)
␪
Figure 20.10
8. With reference to Figure 20.11, what is the
name given to the line XY? Give examples of
each of the following.
x
y
1
4
5
8
3
2
7
6
Figure 20.11
(a) vertically opposite angles
(b) supplementary angles
(c) corresponding angles
(d) alternate angles
9. In Figure 20.12, ﬁnd angle α.
137Њ29Ј
16Њ49Ј
␣
Figure 20.12
10. In Figure 20.13, ﬁnd angles a,b and c.
a
c
b
298
698
Figure 20.13
11. Find angle β in Figure 20.14.
133Њ
98Њ
␤
Figure 20.14
12. Convert 76◦ to radians, correct to 3 decimal
places.
13. Convert 34◦40 to radians, correct to 3 deci-
mal places.
14. Convert 0.714 rad to degrees and minutes.

20.3 Triangles
A triangle is a ﬁgure enclosed by three straight lines.
The sum of the three angles of a triangle is equal
to 180◦.
20.3.1 Types of triangle
An acute-angled triangle is one in which all the angles
are acute; i.e., all the angles are less than 90◦. An
example is shown in triangle ABC in Figure 20.15(a).
A right-angled triangle is one which contains a right
angle; i.e., one in which one of the angles is 90◦. An
example is shown in triangle DEF in Figure 20.15(b).
678 598
508
408
548
A
B C
(a) (b)
D
E F
Figure 20.15
An obtuse-angled triangle is one which contains an
obtuse angle; i.e., one angle which lies between 90◦
and 180◦. An example is shown in triangle PQR in
Figure 20.16(a).
An equilateral triangleis one in which all the sides and
all the angles are equal; i.e., each is 60◦. An example is
shown in triangle ABC in Figure 20.16(b).
P
Q R
1318
278
228
(a)
608
608 608
A
B C
(b)
Figure 20.16
An isosceles triangle is one in which two angles and
two sides are equal. An example is shown in triangle
EFG in Figure 20.17(a).
A scalene triangle is one with unequal angles and
therefore unequal sides. An example of an acute
angled scalene triangle is shown in triangle ABC in
Figure 20.17(b).
(a) (b)
A
B
C
438
828
558
F
E G
308
758758
Figure 20.17
B
A
C
c b
a
␪
Figure 20.18
With reference to Figure 20.18,
(a) Angles A, B and C are called interior angles of
the triangle.
(b) Angle θ is called an exterior angle of the triangle
and is equal to the sum of the two opposite interior
angles; i.e., θ = A + C.
(c) a + b + c is called the perimeter of the triangle.
A
C B
b
a
c
Figure 20.19

A 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 are
labelled as A, B and C, respectively. The right angle
is angle C. The side opposite the right angle is given
the special name of the hypotenuse. The hypotenuse,
length AB in Figure 20.19, is always the longest side of
a right-angled triangle. With reference to angle B, AC
is the opposite side and BC is called the adjacent side.
With reference to angle A, BC is the opposite side and
AC is the adjacent side.
Often sides of a triangle are labelled with lower case
letters, a being the side opposite angle A,b being the
side opposite angle B and c being the side opposite
angle C. So, in the triangle ABC, length AB = c, length
BC = a and length AC = b. Thus, c is the hypotenuse
in the triangle ABC.
∠ is the symbol used for ‘angle’. For example, in the
triangle shown, ∠C = 90◦. Another way of indicating
an angle is to use all three letters. For example, ∠ABC
actually means ∠B; i.e., we take the middle letter as the
angle. Similarly, ∠BAC means ∠A and ∠ACB means
∠C.
more about triangles.
Problem 18. Name the types of triangle shown in
Figure 20.20
2
2.6
2.8
2.5
2.5
2.1
2.1
398
1078
518
2
(b)
(d) (e)
(c)(a)
2
Figure 20.20
(a) Equilateral triangle (since all three sides are
equal).
(b) Acute-angled scalene triangle (since all the
angles 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 the
angles lies between 90◦ and 180◦).
(e) Isosceles triangle (since two sides are equal).
Problem 19. In the triangle ABC shown in
Figure 20.21, with reference to angle θ, which side
is the adjacent?
B
A
␪
C
Figure 20.21
The triangle is right-angled; thus, side AC is the
hypotenuse. With reference to angle θ, the opposite side
is BC. The remaining side, AB, is the adjacent side.
Problem 20. In the triangle shown in
Figure 20.22, determine angle θ
568
␪
Figure 20.22
The sum of the three angles of a triangle is equal to
180◦
.
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 in
Figure 20.23

62Њ
15Њ
A
B
D
E
C ␪
␣
Figure 20.23
In triangle ABC,∠A + ∠B + ∠C = 180◦ (the angles in
a 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 is
equal 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 in
which the unequal angle BAC is 56◦. AB is extended
to D as shown in Figure 20.24. Find, for the
triangle, ∠ABC and ∠ACB. Also, calculate ∠DBC
568
A
B C
D
Figure 20.24
Since triangle ABC is isosceles, two sides – i.e. AB and
AC – 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 angle
equals sum of two interior opposite angles),
i.e. ∠DBC = 56◦ + 62◦ = 118◦
.
Problem 23. Find angles a,b,c,d and e in
Figure 20.25
558
628
b
e
cd
a
Figure 20.25
a = 62◦
and c = 55◦
(alternate angles between parallel
lines).
55◦ + b + 62◦ = 180◦ (angles in a triangle add up to
180◦); hence, b = 180◦ − 55◦ − 62◦ = 63◦
.
b = d = 63◦ (alternate angles between parallel lines).
e + 55◦ + 63◦ = 180◦ (angles in a triangle add up to
180◦); hence, e = 180◦ − 55◦ − 63◦ = 62◦
.
Check: e = a = 62◦ (corresponding angles between
parallel lines).
PracticeExercise 78 Triangles (answers on
page 348)
1. Namethetypesoftriangleshown in diagrams
(a) to (f) in Figure 20.26.
(a) (b)
(e) (f)
(c)
81Њ
15Њ
48Њ
66Њ
45Њ
45Њ
(d)
97Њ
39Њ
60Њ
55
53Њ
37Њ
Figure 20.26

2. Find the angles a to f in Figure 20.27.
1058
1058
d
e
f
(c)
328
114°
b c
(b)
578
838 a
(a)
Figure 20.27
3. In the triangle DEF of Figure 20.28, which
side is the hypotenuse? With reference to
angle D, which side is the adjacent?
388
D
E F
Figure 20.28
4. In triangle DEF of Figure 20.28, determine
angle D.
5. MNO is an isosceles triangle in which
the unequal angle is 65◦ as shown in
Figure 20.29. Calculate angle θ.
658
O
N
M
P
␪
Figure 20.29
6. Determine ∠φ and ∠x in Figure 20.30.
A
B C
E
D
x
588
198
␾
Figure 20.30
7. In Figure 20.31(a) and (b), ﬁnd angles w,x, y
and z. What is the name given to the types of
triangle shown in (a) and (b)?
(a) (b)
1108 1108x y
708
z
2cm
2cm
␻
Figure 20.31
8. Find the values of angles a to g in
Figure 20.32(a) and (b).

(a) (b)
568299
148419
a
b
688
1318
c
d e
g
f
Figure 20.32
9. Find the unknown angles a to k in
Figure 20.33.
998
1258
228 db
a
c
k
e g h
f
ij
Figure 20.33
10. Triangle ABC has a right angle at B and
∠BAC is 34◦. BC is produced to D. If the
bisectors of ∠ABC and ∠ACD meet at E,
determine ∠BEC.
11. Ifin Figure20.34 triangleBCD isequilateral,
ﬁnd the interior angles of triangle ABE.
978
A
B
C
D
E
Figure 20.34
20.4 Congruent triangles
Two triangles are said to be congruent if they are equal
in all respects; i.e., three angles and three sides in one
triangle are equal to three angles and three sides in the
other triangle. Two triangles are congruent if
(a) the three sides of one are equal to the three sides
of the other (SSS),
(b) two sides of one are equal to two sides of the other
and the angles included by these sides are equal
(SAS),
(c) two angles of the one are equal to two angles of
the other and any side of the ﬁrst is equal to the
corresponding side of the other (ASA), or
(d) their hypotenuses are equal and one other side of
one is equal to the corresponding side of the other
(RHS).
Problem 24. State which of the pairs of triangles
shown in Figure 20.35 are congruent and name their
sequence
(c)
M
N
P
O
Q
R
(d)
S
T
UV
X
W
(e)
F
D E
B
A
C
(a) (b)
A
B
C
E D
F K
L
JH
G
I
Figure 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 that
any side coincides.
(e) Congruent ABC, FED (side, side, side; i.e.,
SSS).

Problem 25. In Figure 20.36, triangle PQR is
isosceles with Z, the mid-point of PQ. Prove that
triangles PXZ and QYZ are congruent and that
triangles RXZ and RYZ are congruent. Determine
the values of angles RPZ and RXZ
P Q
R
Y678
288288
X
Z
Figure 20.36
Since triangle PQR is isosceles, PR = RQ and thus
∠QPR = ∠RQP.
∠RXZ = ∠QPR + 28◦ and ∠RYZ = ∠RQP + 28◦ (ext-
erioranglesofatriangleequal thesumofthetwo interior
opposite 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 since
PR = 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 opposite
angles).
PracticeExercise 79 Congruent triangles
1. State which of the pairs of triangles in
Figure 20.37 are congruent and name their
sequences.
A
C
B D
M
(a) (b) (c)
(e)(d)
R
Q S
T
V
W
U
X
Y
Z
K
L
O
P
N
E
I
G
H
J
F
Figure 20.37
2. In a triangle ABC, AB = BC and D and E
are points on AB and BC, respectively, such
that AD = CE. Show that triangles AEB and
CDB are congruent.
20.5 Similar triangles
Two triangles are said to be similar if the angles of one
triangle are equal to the angles ofthe other triangle. With
reference to Figure 20.38, triangles ABC and PQR are
similar and the corresponding sides are in proportion to
each other,
i.e.
p
a
=
q
b
=
r
c
658 588
578
c
a CB
A
b
658 588
578r
p RQ
P
q
Figure 20.38
Problem 26. In Figure 20.39, ﬁnd the length of
side a

508
708
c 512.0cm
B
a
C
A
508
608
f 55.0cm
d5 4.42cm
E F
D
Figure 20.39
In 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 their
angles are the same. Since corresponding sides are in
proportion to each other,
a
d
=
c
f
i.e.
a
4.42
=
12.0
5.0
Hence, side, a =
12.0
5.0
(4.42) = 10.61cm.
Problem 27. In Figure 20.40, ﬁnd the dimensions
marked r and p
558
358
x 57.44cm
q56.82cm
z5
12.97cm
y510.63cm
Y
QP
R
Z
X
p
r
Figure 20.40
In triangle PQR,∠Q = 180◦ − 90◦ − 35◦ = 55◦.
In triangle XYZ,∠X = 180◦
− 90◦
− 55◦
= 35◦
.
Hence, triangles PQR and ZYX are similar since their
angles are the same. The triangles may he redrawn as
358
558
y510.63 cmq 56.82cm
x57.44cm
z5
12.97cm
Z X
Y
358
558
P R
pr
Q
Figure 20.41
By proportion:
p
z
=
r
x
=
q
y
i.e.
p
12.97
=
r
7.44
=
6.82
10.63
from which, r = 7.44
6.82
10.63
= 4.77cm
By proportion:
p
z
=
q
y
i.e.
p
12.97
=
6.82
10.63
Hence, p = 12.97
6.82
10.63
= 8.32cm
Problem 28. In Figure 20.42, show that triangles
CBD and CAE are similar and hence ﬁnd the length
of CD and BD
B
10
6
9
12
C
D
E
A
Figure 20.42
Since BD is parallel to AE then ∠CBD = ∠CAE and
∠CDB = ∠CEA (corresponding angles between paral-
lel lines). Also, ∠C is common to triangles CBD and
CAE.
Since the angles in triangle CBD are the same as in
triangle CAE, the triangles are similar. Hence,
by proportion:
CB
CA
=
CD
CE
=
BD
AE
i.e.
9
6 + 9
=
CD
12
, from which
CD = 12
9
15
= 7.2cm
Also,
9
15
=
BD
10
, from which
BD = 10
9
15
= 6cm

Problem 29. A rectangular shed 2m wide and
3m high stands against a perpendicular building of
height 5.5m. A ladder is used to gain access to the
roof of the building. Determine the minimum
distance between the bottom of the ladder and the
shed
A side view is shown in Figure 20.43, where AF
is the minimum length of the ladder. Since BD
and CF are parallel, ∠ADB = ∠DFE (correspond-
ing angles between parallel lines). Hence, triangles
BAD and EDF are similar since their angles are the
same.
AB = AC − BC = AC − DE = 5.5 − 3 = 2.5m
By proportion:
AB
DE
=
BD
EF
i.e.
2.5
3
=
2
EF
Hence, EF = 2
3
2.5
= 2.4m = minimum distance
from bottom of ladder to the shed.
3m
2m 5.5m
Shed
D
E C
B
A
F
Figure 20.43
PracticeExercise 80 Similar triangles
1. In Figure 20.44, find the lengths x and y.
111Њ
32Њ
32Њ 37Њ
25.69mm
4.74mm
7.36mm
14.58mm
x
y
Figure 20.44
2. 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 with
QR. If PS is 9cm, find the length of ST. X
is a point on ST between S and T such that
the line PX is the bisector of ∠SPT. Find the
length 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 E
C
BA
Figure 20.45
4. In Figure 20.46, AF = 8m,AB = 5m and
BC = 3m. Find the length of BD.
D
E
C
B
A F
Figure 20.46

20.6 Construction of triangles
To 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 triangle
construction.
Problem 30. Construct a triangle whose sides are
6cm,5cm and 3cm
D
F
E
C G
BA 6cm
Figure 20.47
With reference to Figure 20.47:
(i) Draw a straight line of any length and, with a pair
of compasses, mark out 6cm length and label it
AB.
(ii) Set compass to 5cm and with centre at A describe
arc DE.
(iii) Set compass to 3cm and with centre at B describe
arc FG.
(iv) The intersection of the two curves at C is the ver-
tex of the required triangle. Join AC and BC by
straight lines.
It may be proved by measurement that the ratio of the
angles of a triangle is not equal to the ratio of the sides
(i.e., in this problem, the angle opposite the 3cm side is
not equal to half the angle opposite the 6cm side).
Problem 31. Construct a triangle ABC such that
a = 6cm,b = 3cm and ∠C = 60◦
C
A
B a56cm
608
b53cm
Figure 20.48
(i) Draw a line BC,6cm long.
(ii) Using a protractor centred at C, make an angle of
60◦ 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 that
QR = 5cm,∠Q = 70◦ and ∠R = 44◦
Q R
P
Q9
R9
708 448
5cm
Figure 20.49
(i) Draw a straight line 5cm long and label it QR.
(ii) Use a protractor centred at Q and make an angle
of 70◦. Draw QQ .
(iii) Use a protractor centred at R and make an angle
of 44◦. Draw RR .
(iv) The intersection of QQ and RR forms the vertex
P of the triangle.
Problem 33. Construct a triangle XYZ given that
XY = 5cm, the hypotenuse YZ = 6.5cm and
∠X = 90◦

A9B A X
Q
C
P S
Z
U
V
R
Y
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 XA
and XA is arbitrary.) With compass centred at A
draw the arc PQ. With the same compass setting
and centred at A , draw the arc RS. Join the inter-
section of the arcs, C to X, and a right angle to
XY is produced at X. (Alternatively, a protractor
can be used to construct a 90◦ angle.)
(iii) Thehypotenuseisalwaysoppositetheright angle.
Thus, YZ is opposite ∠X. Using a compass
centred at Y and set to 6.5cm, describe the
arc UV.
(iv) The intersection of the arc UV withXC produced,
forms the vertex Z of the required triangle. Join
YZ with a straight line.
PracticeExercise 81 Construction of
triangles (answers on page 349)
In the following, construct the triangles ABC for
the 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 21
Introduction to trigonometry
21.1 Introduction
Trigonometry is a subject that involves the measurement
of sides and angles of triangles and their relationship to
each other.
The theorem of Pythagoras and trigonometric ratios
are used with right-angled triangles only. However,
there are many practical examples in engineering
where knowledge of right-angled triangles is very
important.
In this chapter, three trigonometric ratios – i.e. sine,
cosine and tangent – are deﬁned and then evaluated
using a calculator. Finally, solving right-angled trian-
gle problems using Pythagoras and trigonometric ratios
is demonstrated, together with some practical examples
involving angles of elevation and depression.
21.2 The theorem of Pythagoras
The theorem of Pythagoras states:
In any right-angled triangle, the square of the
hypotenuse is equal to the sum of the squares of the
other two sides.
In the right-angled triangle ABC shown in Figure 21.1,
this means
b2
= a2
+ c2
(1)
B
A
Ca
bc
Figure 21.1
If the lengths of any two sides of a right-angled triangle
are known, the length of the third side may be calculated
by Pythagoras’ theorem.
From equation (1): b = a2 + c2
Transposing equation (1) for a gives a2 = b2 − c2, from
which a =
√
b2 − c2
Transposing equation (1) for c gives c2 = b2 − a2, from
which c =
√
b2 − a2
theorem of Pythagoras.
Problem 1. In Figure 21.2, ﬁnd the length of BC
BA
C
ab54cm
c53 cm
Figure 21.2
From Pythagoras, a2
= b2
+ c2
i.e. a2
= 42
+ 32
= 16 + 9 = 25
Hence, a =
√
25 = 5cm.
√
25 = ±5 but in a practical example likethisan answer
of a = −5cm has no meaning, so we take only the
positive answer.
Thus a = BC = 5cm.
DOI: 10.1016/B978-1-85617-697-2.00021-1

PQR is a 3, 4, 5 triangle. There are not many right-
angled triangles which have integer values (i.e. whole
numbers) for all three sides.
Problem 2. In Figure 21.3, find the length of EF
E d F
D
f55cm
e513cm
Figure 21.3
By Pythagoras’ theorem, e2
= d2
+ f 2
Hence, 132
= d2
+ 52
169 = d2
+ 25
d2
= 169 − 25 = 144
Thus, d =
√
144 = 12cm
i.e. d = EF = 12cm
DEF is a 5, 12, 13 triangle, another right-angled
triangle which has integer values for all three sides.
Problem 3. Two aircraft leave an airfield at the
same time. One travels due north at an average
speed of 300km/h and the other due west at an
average speed of 220km/h. Calculate their distance
apart after 4hours
After 4 hours, the first aircraft has travelled
4 × 300 = 1200km due north
and the second aircraft has travelled
4 × 220 = 880km due west,
as shown in Figure 21.4. The distance apart after
4hours = BC.
A
BN
EW
S
C
1200km
880km
Figure 21.4
From Pythagoras’ theorem,
BC2
= 12002
+ 8802
= 1440000+ 774400 = 2214400
and BC =
√
2214400 = 1488km.
Hence, distance apart after 4 hours = 1488km.
PracticeExercise 82 Theorem of
Pythagoras (answers on page 350)
1. Find the length of side x in Figure 21.5.
41cm
40cm
x
Figure 21.5
2. Find the length of side x in Figure 21.6(a).
3. Find the length of side x in Figure 21.6(b),
25 m
(a)
7 m
4.7 mm
8.3 mm
(b)
x
x
Figure 21.6
4. In a triangle ABC, AB = 17cm, BC = 12 cm
and ∠ABC = 90◦. Determine the length of
AC, correct to 2 decimal places.
5. A tent peg is 4.0m away from a 6.0m high
tent. What length of rope, correct to the

Introduction to trigonometry 183
nearest centimetre, runs from the top of the
tent to the peg?
6. In a triangle ABC, ∠B is a right angle,
AB = 6.92 cm and BC = 8.78cm. Find the
length 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 and
17cm it is right-angled.
9. Triangle PQR is isosceles, Q being a right
angle. If the hypotenuse is 38.46cm ﬁnd (a)
the lengths of sides PQ and QR and (b) the
value of ∠QPR.
10. Aman cycles24kmduesouth and then 20km
due east. Another man, starting at the same
time as the ﬁrst man, cycles 32km due east
and then 7km due south. Find the distance
between the two men.
11. A ladder 3.5m long is placed against a per-
pendicular wall with its foot 1.0m from the
wall. How far up the wall (to the nearest cen-
timetre) does the ladder reach? If the foot of
the ladder is now moved 30cm further away
from the wall, how far does the top of the
ladder fall?
12. Two ships leave a port at the same time. One
travels due west at 18.4knots and the other
due south at 27.6knots. If 1knot = 1 nautical
mile per hour, calculate how far apart the two
ships are after 4 hours.
13. Figure 21.7 shows a bolt rounded off at one
end. Determine the dimension h.
R5 45mm
h
r516mm
Figure 21.7
14. 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, calculate
the dimension x.
72mm
␾74mm
x
Figure 21.8
21.3 Sines, cosines and tangents
With reference to angle θ in the right-angled triangle
ABC shown in Figure 21.9,
sineθ =
opposite side
hypotenuse
‘Sine’ is abbreviated to ‘sin’, thus sinθ =
BC
AC
Hypotenuse
AdjacentA
␪
B
C
Opposite
Figure 21.9
Also, cosineθ =
adjacent side
hypotenuse
‘Cosine’ is abbreviated to ‘cos’, thus cosθ =
AB
AC
Finally, tangentθ =
opposite side
adjacent side
‘Tangent’ is abbreviated to ‘tan’, thus tanθ =
BC
AB
These three trigonometric ratios only apply to right-
angled triangles. Remembering these three equations
is very important and the mnemonic ‘SOH CAH TOA’
is one way of remembering them.

SOH indicates sin = opposite ÷ hypotenuse
CAH indicates cos = adjacent ÷ hypotenuse
TOA indicates tan = opposite ÷ adjacent
Here are some worked problems to help familiarize
ourselves with trigonometric ratios.
Problem 4. In triangle PQR shown in
Figure 21.10, determine sin θ, cos θ and tan θ
␪
P
Q R
5
12
13
Figure 21.10
sinθ =
opposite side
hypotenuse
=
PQ
PR
=
5
13
= 0.3846
cosθ =
adjacent side
hypotenuse
=
QR
PR
=
12
13
= 0.9231
tanθ =
opposite side
adjacent side
=
PQ
QR
=
5
12
= 0.4167
Problem 5. In triangle ABC of Figure 21.11,
determine length AC,sin C,cosC,tanC,sin A,
cos A and tan A
A
B C
3.47 cm
4.62 cm
Figure 21.11
By Pythagoras, AC2
= AB2
+ BC2
i.e. AC2
= 3.472
+ 4.622
from which AC = 3.472 + 4.622 = 5.778cm
sinC =
opposite side
hypotenuse
=
AB
AC
=
3.47
5.778
= 0.6006
cosC =
adjacent side
hypotenuse
=
BC
AC
=
4.62
5.778
= 0.7996
tanC =
opposite side
adjacent side
=
AB
BC
=
3.47
4.62
= 0.7511
sinA =
opposite side
hypotenuse
=
BC
AC
=
4.62
5.778
= 0.7996
cosA =
adjacent side
hypotenuse
=
AB
AC
=
3.47
5.778
= 0.6006
tanA =
opposite side
adjacent side
=
BC
AB
=
4.62
3.47
= 1.3314
Problem 6. If tan B =
8
15
, determine the value of
sin B,cos B,sin A and tan A
A right-angled triangle ABC is shown in Figure 21.12.
If tan B =
8
15
, then AC = 8 and BC = 15.
A
CB
8
15
Figure 21.12
By Pythagoras, AB2
= AC2
+ BC2
i.e. AB2
= 82
+ 152
from which AB = 82 + 152 = 17
sinB =
AC
AB
=
8
17
or 0.4706
cosB =
BC
AB
=
15
17
or 0.8824
sinA =
BC
AB
=
15
17
or 0.8824
tanA =
BC
AC
=
15
8
or 1.8750
Problem 7. Point A lies at co-ordinate (2, 3) and
point B at (8, 7). Determine (a) the distance AB and
(b) the gradient of the straight line AB

B
A
20 4
(a) (b)
6 8
8
f(x)
7
6
4
3
2
B
CA
20 4 6 8
8
f(x)
7
6
4
3
2
␪
Figure 21.13
(a) Points A and B are shown in Figure 21.13(a).
In Figure 21.13(b), the horizontal and vertical
lines AC and BC are constructed. Since ABC is
a right-angled triangle, and AC = (8 − 2) = 6 and
BC = (7 − 3) = 4, by Pythagoras’ theorem,
AB2
= AC2
+ BC2
= 62
+ 42
and AB = 62 + 42 =
√
52
= 7.211 correct to 3
decimal places.
(b) The gradient of AB is given by tanθ, i.e.
gradient = tanθ =
BC
AC
=
4
6
=
2
3
PracticeExercise 83 Trigonometric ratios
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, find
sin A,cos A,tan A,sin B,cos B and tan B.
B
C
35
A
Figure 21.14
3. If cos A =
15
17
, find sin A and tan A, in fraction
form.
4. If tan X =
15
112
, find sin X and cos X, in frac-
tion form.
5. For the right-angled triangle shown in
Figure 21.15, find (a) sinα (b) cosθ (c) tanθ.
␪
␣ 178
15
Figure 21.15
6. If tanθ =
7
24
, find sinθ and cosθ in fraction
form.
7. Point P lies at co-ordinate (−3,1) and point
Q at (5,−4). Determine
(a) the distance PQ.
(b) the gradient of the straight line PQ.
21.4 Evaluating trigonometric ratios
of acute angles
The easiest way to evaluate trigonometric ratios of any
angle is to use a calculator. Use a calculator to check the
following (each correct to 4 decimal places).
sin29◦ = 0.4848 sin53.62◦ = 0.8051
cos67◦ = 0.3907 cos83.57◦ = 0.1120
tan34◦ = 0.6745 tan67.83◦ = 2.4541
sin67◦
43 = sin67
43
60
◦
= sin67.7166666...◦
= 0.9253
cos13◦
28 = cos13
28
60
◦
= cos13.466666...◦
= 0.9725
tan56◦
54 = tan56
54
60
◦
= tan56.90◦
= 1.5340
If we know the value ofa trigonometricratio and need to
find the angle we use the inverse function on our calcu-
lators. For example, using shift and sin on our calculator
gives sin−1
(
If, for example, we know the sine of an angle is 0.5 then
the value of the angle is given by
sin−1
0.5 = 30◦
(Check that sin30◦
= 0.5)
Similarly, if
cosθ = 0.4371 then θ = cos−1
0.4371 = 64.08◦

and if
tan A = 3.5984 then A = tan−1
3.5984 = 74.47◦
each correct to 2 decimal places.
Use your calculator to check the following worked
problems.
Problem 8. Determine, correct to 4 decimal
places, sin43◦
39
sin43◦
39 = sin43
39
60
◦
= sin43.65◦
= 0.6903
follows:
1. Press sin 2. Enter 43 3. Press◦ ”’
4. Enter 39 5. Press◦ ”’ 6. Press )
7. Press= Answer=0.6902512...
Problem 9. Determine, correct to 3 decimal
places, 6 cos62◦12
6 cos 62◦
12 = 6 cos 62
12
60
◦
= 6cos62.20◦
= 2.798
follows:
1. Enter 6 2. Press cos 3. Enter 62
4. Press◦ ”’ 5. Enter 12 6. Press◦ ”’
7. Press ) 8. Press= Answer=2.798319...
Problem 10. Evaluate sin 1.481, correct to 4
sin 1.481 means the sine of 1.481 radians. (If there is no
degreessign,i.e.◦ ,then radiansareassumed).Therefore
a calculator needs to be on the radian function.
Hence, sin1.481 = 0.9960
Problem 11. Evaluate cos(3π/5), correct to 4
As in Problem 10, 3π/5 is in radians.
Hence, cos(3π/5) = cos1.884955... = −0.3090
Since, from page 166, πradians = 180◦,
3π/5 rad =
3
5
× 180◦ = 108◦.
i.e. 3π/5 rad = 108◦. Check with your calculator that
cos108◦ = −0.3090
Problem 12. Evaluate tan 2.93, correct to 4
Again, since there is no degrees sign, 2.93 means 2.93
radians.
Hence, tan2.93 = −0.2148
It is important to know when to have your calculator on
either degrees mode or radian mode. A lot of mistakes
can arise from this if we are not careful.
Problem 13. Find the acute angle sin−1
0.4128 in
degrees, correct to 2 decimal places
sin−1
0.4128 means ‘the angle whose sine is 0.4128’.
Using a calculator,
1. Press shift 2. Press sin 3. Enter 0.4128
4. Press ) 5. Press =
The answer 24.380848... is displayed.
Hence, sin−1
0.4128 = 24.38◦ correct to 2 decimal
places.
Problem 14. Find the acute angle cos−1
0.2437 in
degrees and minutes
cos−1 0.2437 means ‘the angle whose cosine is 0.2437’.
Using a calculator,
1. Press shift 2. Press cos 3. Enter 0.2437
6. Press ◦ ”’ and 75◦ 53 41.93 is displayed.
Hence, cos−1 0.2437 = 75.89◦ = 77◦54 correct to the
nearest minute.
Problem 15. Find the acute angle tan−1 7.4523 in
degrees and minutes
tan−1 7.4523 means‘theanglewhosetangent is7.4523’.
Using a calculator,
1. Press shift 2. Press tan 3. Enter 7.4523
6. Press ◦ ”’ and 82◦ 21 26.35 is displayed.
Hence, tan−1 7.4523 = 82.36◦ = 82◦21 correct to the
nearest minute.

Problem 16. In triangle EFG in Figure 21.16,
calculate angle G
E
F G
2.30cm
8.71cm
Figure 21.16
With reference to ∠G, the two sides of the trianglegiven
are the oppositeside EF and the hypotenuse EG; hence,
sine is used,
i.e. sin G =
2.30
8.71
= 0.26406429...
from which, G = sin−1
0.26406429...
i.e. G = 15.311360◦
Hence, ∠G = 15.31◦
or 15◦
19 .
Problem 17. Evaluate the following expression,
4.2tan49◦26 − 3.7sin66◦1
7.1cos29◦34
By calculator:
tan49◦
26 = tan 49
26
60
◦
= 1.1681,
sin66◦
1 = 0.9137 and cos29◦
34 = 0.8698
Hence, 4.2tan49◦26 − 3.7sin66◦1
7.1cos29◦34
=
(4.2 × 1.1681) − (3.7 × 0.9137)
(7.1 × 0.8698)
=
4.9060 − 3.3807
6.1756
=
1.5253
6.1756
= 0.2470 = 0.247,
PracticeExercise 84 Evaluating
trigonometric 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) sine
2π
3
(b) cos1.681 (c) tan3.672
5. Find the acute angle sin−1 0.6734 in degrees,
6. Find the acute angle cos−1 0.9648 in degrees,
7. Find the acute angle tan−1 3.4385 in degrees,
8. Find the acute angle sin−1
0.1381 in degrees
and minutes.
9. Find the acute angle cos−1 0.8539 in degrees
and minutes.
10. Find the acute angle tan−1 0.8971 in degrees
and minutes.
11. In the triangle shown in Figure 21.17, deter-
mine angle θ, correct to 2 decimal places.
␪
5
9
Figure 21.17
12. In the triangle shown in Figure 21.18, deter-
mine angle θ in degrees and minutes.
␪
8
23
Figure 21.18

13. 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 the
unknown 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 least
three facts are known, the other three can usually be
calculated.
solution of right-angled triangles.
Problem 18. In the triangle ABC shown in
Figure 21.19, find the lengths AC and AB
A
B C
428
6.2mm
Figure 21.19
There is usually more than one way to solve such a
triangle.
In triangle ABC,
tan42◦
=
AC
BC
=
AC
6.2
(Remember SOH CAH TOA)
Transposing gives
AC = 6.2tan42◦
= 5.583 mm
cos42◦
=
BC
AB
=
6.2
AB
, from which
AB =
6.2
cos42◦
= 8.343 mm
Alternatively, by Pythagoras, AB2 = AC2 + BC2
from which AB =
√
AC2 + BC2 =
√
5.5832 + 6.22
=
√
69.609889 = 8.343 mm.
Problem 19. Sketch a right-angled triangle ABC
such that B = 90◦, AB = 5cm and BC = 12 cm.
Determine the length of AC and hence evaluate
sin A,cosC and tan A
Triangle ABC is shown in Figure 21.20.
A
B C
5cm
12cm
Figure 21.20
By Pythagoras’ theorem, AC = 52 + 122 = 13
By definition: sin A =
opposite side
hypotenuse
=
12
13
or 0.9231
(Remember SOH CAH TOA)
cosC =
adjacent side
hypotenuse
=
12
13
or 0.9231
and tan A =
opposite side
adjacent side
=
12
5
or 2.400
Problem 20. In triangle PQR shown in
Figure 21.21, find the lengths of PQ and PR
Q R
P
7.5cm
388
Figure 21.21
tan38◦
=
PQ
QR
=
PQ
7.5
, hence
PQ = 7.5tan38◦
= 7.5(0.7813) = 5.860 cm

cos38◦
=
QR
PR
=
7.5
PR
, hence
PR =
7.5
cos38◦
=
7.5
0.7880
= 9.518 cm
Check: using Pythagoras’ theorem,
(7.5)2 + (5.860)2 = 90.59 = (9.518)2.
Problem 21. Solve the triangle ABC shown in
Figure 21.22
A
C
B
37mm
35 mm
Figure 21.22
To ‘solve the triangle ABC’ means ‘to ﬁnd the length
AC and angles B and C’.
sinC =
35
37
= 0.94595, hence
C = sin−1
0.94595 = 71.08◦
B = 180◦ − 90◦ − 71.08◦ = 18.92◦ (since the angles in
a triangle add up to 180◦)
sin B =
AC
37
, hence
AC = 37sin18.92◦
= 37(0.3242) = 12.0 mm
or, using Pythagoras’ theorem, 372 = 352 + AC2 , from
which AC = (372 − 352) = 12.0 mm.
Problem 22. Solve triangle XYZ given
∠X = 90◦, ∠Y = 23◦17 and YZ = 20.0 mm
It is always advisable to make a reasonably accurate
sketch so as to visualize the expected magnitudes of
unknown sides and angles. Such a sketch is shown in
Figure 21.23.
∠Z = 180◦
− 90◦
− 23◦
17 = 66◦
43
X Y
Z
20.0mm
238179
Figure 21.23
sin23◦
17 =
XZ
20.0
, hence XZ = 20.0sin23◦
17
= 20.0(0.3953) = 7.906mm
cos23◦
17 =
XY
20.0
, hence XY = 20.0cos23◦
17
= 20.0(0.9186) = 18.37mm
Check: using Pythagoras’ theorem,
(18.37)2 + (7.906)2 = 400.0 = (20.0)2,
PracticeExercise 85 Solving right-angled
triangles (answers on page 349)
1. Calculate the dimensions shown as x in
Figures 21.24(a) to (f), each correct to 4
298
x
(c)
17.0
708
228
x
13.0
(a)
(b)
x
15.0
Figure 21.24

598
(d)
x
4.30
438
x
(e)
6.0
(f)
538
x
7.0
Figure 21.24
2. Find theunknown sidesand anglesin theright-
angled triangles shown in Figure 21.25. The
dimensions shown are in centimetres.
5.0
3.0
A
B
C
(a)
Figure 21.25
8.0
4.0
E
D
F
(b)
(c)
H
G
J
288
12.0
(d)
278
15.0
K
L
M
(e)
N
P
O
4.0
648
(f)
R S
Q
5.0
418
Figure 21.25

3. A ladder rests against the top of the perpendi-
cular wall of a building and makes an angle of
73◦ with the ground. If the foot of the ladder is
2m from the wall, calculate the height of the
building.
4. Determine the length x in Figure 21.26.
x
568
10mm
Figure 21.26
21.6 Angles of elevation and
depression
If, in Figure21.27, BC represents horizontal ground and
AB a vertical flagpole, the angle of elevation of the top
of the flagpole, A, from the point C is the angle that the
imaginary straight line AC must be raised (or elevated)
from the horizontal CB; i.e., angle θ.
A
BC
␪
Figure 21.27
P
Q R
␾
Figure 21.28
If, in Figure 21.28, PQ represents a vertical cliff and
R a ship at sea, the angle of depression of the ship
from point P is the angle through which the imaginary
straight line PR must be lowered (or depressed) from
the horizontal to the ship; i.e., angle φ. (Note, ∠PRQ is
also φ − alternate angles between parallel lines.)
Problem 23. An electricity pylon stands on
horizontal ground. At a point 80m from the base of
the pylon, the angle of elevation of the top of the
pylon is 23◦. Calculate the height of the pylon to the
nearest metre
Figure 21.29 shows the pylon AB and the angle of
elevation of A from point C is 23◦.
80 m
238
A
BC
Figure 21.29
tan23◦
=
AB
BC
=
AB
80
Hence, height of pylon AB = 80tan23◦
= 80(0.4245) = 33.96m
= 34m to the nearest metre.
Problem 24. A surveyor measures the angle of
elevation of the top of a perpendicular building as
19◦. He moves 120m nearer to the building and
finds the angle of elevation is now 47◦. Determine
the height of the building
The building PQ and the angles of elevation are shown
in Figure 21.30.
P
Q
h
x
R
S
120
478 198
Figure 21.30
In triangle PQS, tan19◦ =
h
x + 120
Hence, h = tan19◦
(x + 120)
i.e. h = 0.3443(x + 120) (1)
In triangle PQR, tan47◦
=
h
x
Hence, h = tan47◦
(x) i.e. h = 1.0724x (2)
Equating equations (1) and (2) gives
0.3443(x + 120) = 1.0724x
0.3443x + (0.3443)(120) = 1.0724x
(0.3443)(120) = (1.0724 − 0.3443)x
41.316 = 0.7281x
x =
41.316
0.7281
= 56.74m

From equation (2), height of building,
h = 1.0724x = 1.0724(56.74) = 60.85m.
Problem 25. The angle of depression of a ship
viewed at a particular instant from the top of a 75m
vertical cliff is 30◦. Find the distance of the ship
from the base of the cliff at this instant. The ship is
sailing away from the cliff at constant speed and 1
minute later its angle of depression from the top of
the cliff is 20◦. Determine the speed of the ship in
km/h
Figure 21.31 shows the cliff AB, the initial position of
the ship at C and the final position at D. Since the angle
of depression is initially 30◦, ∠ACB = 30◦ (alternate
angles between parallel lines).
x
75 m
308
208
308
208
A
B DC
Figure 21.31
tan30◦
=
AB
BC
=
75
BC
hence,
BC =
75
tan30◦
= 129.9m = initial position
of ship from base of cliff
In triangle ABD,
tan20◦ =
AB
BD
=
75
BC + CD
=
75
129.9 + x
Hence, 129.9 + x =
75
tan20◦
= 206.06m
from which x = 206.06 − 129.9 = 76.16m
Thus, the ship sails 76.16m in 1 minute; i.e., 60s,
Hence, speed of ship =
distance
time
=
76.16
60
m/s
=
76.16 × 60× 60
60 × 1000
km/h = 4.57km/h.
PracticeExercise 86 Angles of elevation
and depression (answers on page 349)
1. A vertical tower stands on level ground. At
a point 105m from the foot of the tower the
angle of elevation of the top is 19◦. Find the
height of the tower.
2. If the angle of elevation of the top of a vertical
30m high aerial is 32◦, how far is it to the
aerial?
3. From the top of a vertical cliff 90.0m high
the angle of depression of a boat is 19◦
50 .
Determine the distance of the boat from the
cliff.
4. From the top of a vertical cliff 80.0m high the
angles of depression of two buoys lying due
west of the cliff are 23◦
and 15◦
, respectively.
How far apart are the buoys?
5. From a point on horizontal ground a surveyor
measures the angle of elevation of the top of
a flagpole as 18◦40 . He moves 50m nearer
to the flagpole and measures the angle of ele-
vation as 26◦22 . Determine the height of the
flagpole.
6. A flagpole stands on the edge of the top of a
building. At a point 200m from the building
the 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 of
the top and bottom of a vertical lighthouse
standing on the edge of a vertical cliff are
31◦
and 26◦
, respectively. If the lighthouse is
25.0m high, calculate the height of the cliff.
8. From a window 4.2m above horizontal ground
the angle of depression of the foot of a building
acrosstheroad is24◦ and theangleofelevation
of the top of the building is 34◦
. Determine,
correct to the nearest centimetre, the width of
the road and the height of the building.
9. The elevation of a tower from two points, one
due west of the tower and the other due east
of it are 20◦ and 24◦, respectively, and the two
points of observation are 300m apart. Find the
height of the tower to the nearest metre.

Revision Test 8 : Angles, triangles and trigonometry
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 decimal
places. (3)
5. Determine angle θ in Figure RT8.1. (3)
538
478
1248 298
␪
Figure RT8.1
6. Determine angle θ in the triangle in Figure RT8.2.
(2)
␪
␪
508
Figure RT8.2
7. Determine angle θ in the triangle in Figure RT8.3.
(2)
438
688
␪
Figure RT8.3
8. In Figure RT8.4, if triangle ABC is equilateral,
determine ∠CDE. (3)
A
B
C
D
E
928
Figure RT8.4
9. Find angle J in Figure RT8.5. (2)
H
G
J
288
Figure RT8.5
10. Determine angle θ in Figure RT8.6. (3)
␪
328
588
Figure RT8.6
11. State the angle (a) supplementary to 49◦ (b) com-
plementary to 49◦. (2)

12. In Figure RT8.7, determine angles x, y and z.
(3)
598
y z
x
378
Figure RT8.7
13. In Figure RT8.8, determine angles a to e. (5)
c
e
a b d
608
1258
Figure RT8.8
14. In Figure RT8.9, determine the length of AC.
(4)
D A
C
E
B
3m
10m
8m
Figure RT8.9
15. 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 decimal
places. (4)
K
LJ
11.60
5.91
Figure RT8.10
16. Two ships leave a port at the same time. Ship X
travels due west at 30km/h and ship Y travels due
north.After4 hoursthetwo shipsare130kmapart.
Calculate the velocity of ship Y. (4)
17. If sin A =
12
37
, find tan A in fraction form. (3)
18. Evaluate 5tan62◦11 correct to 3 significant
figures. (2)
19. Determinetheacuteanglecos−1 0.3649 in degrees
and minutes. (2)
20. In triangle PQR in Figure RT8.11, find angle P
in decimal form, correct to 2 decimal places. (2)
P Q
R
15.229.6
Figure RT8.11
21. Evaluate, correct to 3 significant figures,
3tan81.27◦ − 5cos7.32◦ − 6sin54.81◦. (2)
22. In triangle ABC in Figure RT8.12, find lengths
AB and AC, correct to 2 decimal places. (4)
A
B C
6
408 359
Figure RT8.12
23. From a point P, the angle of elevation of a 40m
high electricity pylon is 20◦. How far is point P
from the base of the pylon, correct to the nearest
metre? (3)

Chapter 22
Trigonometric waveforms
22.1 Graphs of trigonometric
functions
By drawing up tables of values from 0◦ to 360◦, graphs
of 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 plotting
graphs), using 30◦ intervals, are shown below, with the
respective graphs shown in Figure 22.1.
(a) y = sinA
A 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦
sin A 0 0.500 0.866 1.000 0.866 0.500 0
A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦
sin A −0.500 −0.866 −1.000 −0.866 −0.500 0
(b) y = cosA
A 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦
cos A 1.000 0.866 0.500 0 −0.500 −0.866 −1.000
A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦
cos A −0.866 −0.500 0 0.500 0.866 1.000
(c) y = tanA
A 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦
tan A 0 0.577 1.732 ∞ −1.732 −0.577 0
A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦
tan A 0.577 1.732 ∞ −1.732 −0.577 0
21.0
24
20.5
21.0
22
0.5
1.0
20.5
0 30 60 90 120 150
180
210 240 270 300 330 360
30 60 90 120 180 210 240 270 300 360
30 60 90 120 150 180 210 240 270 300 330 360
(a)
1.0
0.5
0
(b)
(c)
4
2
0
150 330
y
y
y
y 5tan A
y 5cos A
A8
A8
A8
y 5sin A
Figure 22.1
From Figure 22.1 it is seen that
(a) Sine and cosine graphs oscillate between peak
values of ±1.
(b) The cosine curve is the same shape as the sine
curve but displaced by 90◦.
(c) Thesineand cosinecurvesare continuousand they
repeat at intervals of 360◦ and the tangent curve
appearsto bediscontinuousand repeatsat intervals
of 180◦.
DOI: 10.1016/B978-1-85617-697-2.00022-3

22.2 Angles of any magnitude
Figure 22.2 shows rectangular axes XX and YY
intersecting at origin 0. As with graphical work, mea-
surements made to the right and above 0 are positive,
while those to the left and downwards are negative.
Let 0A be free to rotate about 0. By convention,
when 0A moves anticlockwise angular measurement is
considered positive, and vice versa.
3608
2708
1808
908
X9 X
Y9
Y
08
A
Quadrant 2
Quadrant 3 Quadrant 4
Quadrant 1
0
2 1
1
2
1
2
Figure 22.2
Let 0A be rotated anticlockwise so that θ1 is any
angle in the first quadrant and let perpendicular AB
be constructed to form the right-angled triangle 0AB
in Figure 22.3. Since all three sides of the trian-
gle are positive, the trigonometric ratios sine, cosine
and tangent will all be positive in the first quadrant.
(Note: 0A is always positive since it is the radius of a
circle.)
Let 0A be further rotated so that θ2 is any angle in the
second quadrant and let AC be constructed to form the
1808
908
2708
3608
08␪2
␪3 ␪4
␪1
Quadrant 2
Quadrant 4Quadrant 3
Quadrant 1
0
A
B
AA
C
D
E
A
2
2 2
1
1
1
1
1
1 1
Figure 22.3
right-angled triangle 0AC. Then,
sinθ2 =
+
+
= + cosθ2 =
−
+
= −
tanθ2 =
+
−
= −
third quadrant and let AD be constructed to form the
right-angled triangle 0AD. Then,
sinθ3 =
−
+
= − cosθ3 =
−
+
= −
tanθ3 =
−
−
= +
fourth quadrant and let AE be constructed to form the
right-angled triangle 0AE. Then,
sinθ4 =
−
+
= − cosθ4 =
+
+
= +
tanθ4 =
−
+
= −
The above results are summarized in Figure 22.4, in
which all three trigonometric ratios are positive in the
first quadrant, only sine is positive in the second quad-
rant, only tangent is positive in the third quadrant and
only cosine is positive in the fourth quadrant.
The underlined letters in Figure 22.4 spell the word
CAST when starting in the fourth quadrant and moving
in an anticlockwise direction.
908
1808
2708
3608
08
Sine
Tangent Cosine
All positive
Figure 22.4
It is seen that, in the first quadrant of Figure 22.1,
all of the curves have positive values; in the second only
sine is positive; in the third only tangent is positive;
and in the fourth only cosine is positive – exactly as
summarized in Figure 22.4.

Trigonometric waveforms 197
A knowledge of angles of any magnitude is needed
when ﬁnding, for example, all the angles between 0◦
and 360◦ whose sine is, say, 0.3261. If 0.3261 is entered
into a calculator and then the inverse sine key pressed (or
sin−1 key) the answer 19.03◦ appears. However, there
is a second angle between 0◦ and 360◦ which the cal-
culator does not give. Sine is also positive in the second
quadrant (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◦ whose
sine is 0.3261 (check that sin160.97◦ = 0.3261 on your
calculator).
19.038 19.038
1808
2708
3608
08
␪
908
S A
T C
Figure 22.5
Be careful! Your calculator only gives you one of
these answers. The second answer needs to be deduced
from a knowledge of angles of any magnitude, as shown
in the following worked problems.
Problem 1. Determine all of the angles between
0◦ and 360◦ whose sine is −0.4638
The angles whose sine is −0.4638 occur in the third
and fourth quadrants since sine is negative in these
quadrants – see Figure 22.6.
1.0
21.0
20.4638
0 908 1808 2708 3608
332.378207.638
x
y 5sinxy
Figure 22.6
From 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 gives
one answer: −27.632588◦.)
T
S A
C
908
1808
2708
3608
08
␪␪
Figure 22.7
Problem 2. Determine all of the angles between
0◦ and 360◦ whose tangent is 1.7629
A tangent is positive in the ﬁrst and third quadrants –
see Figure 22.8.
1.7629
60.448 240.448
0 360827081808908
y 5tan x
y
x
Figure 22.8
From 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◦
1808
2708
3608
908
08
CT
S A
␪
␪
Figure 22.9
cos−1
(−0.2348) = α for angles of α between 0◦
and 360◦

S
1808
2708
08
3608
908
T
A
C
␪
␪
Figure 22.10
Cosine is positive in the ﬁrst and fourth quadrants and
thus negative in the second and third quadrants – see
Figure 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 cosine
is −0.2348 are α = 180◦ − 76.42◦, i.e. 103.58◦
and
α = 180◦ + 76.42◦, i.e. 256.42◦
PracticeExercise 87 Angles of any
magnitude (answers on page 349)
1. Determine all of the angles between 0◦ and
360◦ whose sine is
(a) 0.6792 (b) −0.1483
2. Solve the following equations for values of x
between 0◦ and 360◦.
(a) x = cos−1 0.8739
(b) x = cos−1(−0.5572)
3. Find the angles between 0◦ to 360◦ whose
tangent is
(a) 0.9728 (b) −2.3420
In problems 4 to 6, solve the given equations in the
range 0◦ to 360◦, giving the answers in degrees and
minutes.
4. cos−1(−0.5316) = t
5. sin−1(−0.6250) = α
6. tan−1 0.8314 = θ
22.3 The production of sine and
cosine waves
In Figure22.11, let OR be a vector 1 unitlong and free to
rotate anticlockwiseabout 0. In one revolutiona circle is
produced and isshown with 15◦ sectors.Each radiusarm
has a vertical and a horizontal component. For example,
at 30◦, the vertical component is TS and the horizontal
component is OS.
From triangle OST,
sin30◦ =
TS
TO
=
TS
1
i.e. TS = sin30◦
and cos30◦ =
OS
TO
=
OS
1
i.e. OS = cos30◦
1208
908
608
3608
3308 20.5
21.0
1.0
0.5T
y
R
S
S9
T9
y 5sin x
Angle x8
308 608 1208 2108 2708 3308
3008
2708
2408
2108
1808
1508
O
Figure 22.11

20.5
0.5
21.0
1.0
y
S
R
T S9
O9
y 5cosx
Angle x8
308 608 1208 1808 2408 3008 3608
458
08158
3308
3158
2858
2558
08
2258
2108
1808
1508
1208
908
608
O
Figure 22.12
22.3.1 Sine waves
The vertical component TS may be projected across to
T S , which is the corresponding value of 30◦ on the
graph of y against angle x◦. If all such vertical compo-
nents as TS are projected on to the graph, a sine wave
is produced as shown in Figure 22.11.
22.3.2 Cosine waves
If all horizontal components such as OS are projected
on to a graph of y against angle x◦, a cosine wave is
produced. It is easier to visualize these projections by
redrawing the circle with the radius arm OR initially in
a vertical position as shown in Figure 22.12.
It is seen from Figures 22.11 and 22.12 that a cosine
curve is of the same form as the sine curve but is dis-
placed by 90◦ (or π/2 radians). Both sine and cosine
waves repeat every 360◦.
22.4 Terminology involved with sine
and cosine waves
Sine waves are extremely important in engineering, with
examples occurring with alternating currents and volt-
ages – the mains supply is a sine wave – and with simple
harmonic motion.
22.4.1 Cycle
When a sine wave has passed through a complete
series of values, both positive and negative, it is said
to have completed one cycle. One cycle of a sine
wave is shown in Figure 22.1(a) on page 195 and in
Figure 22.11.
22.4.2 Amplitude
The amplitude is the maximum value reached in a half
cycle by a sine wave. Another name for amplitude is
peak value or maximum value.
A sine wave y = 5sinx has an amplitude of 5, a
sine wave v = 200sin314t has an amplitude of 200 and
the sine wave y = sin x shown in Figure 22.11 has an
amplitude of 1.
22.4.3 Period
The 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◦ (from
Figure 22.1(c)).
A graph of y = 3sin2A, as shown in Figure 22.13, has
an amplitude of 3 and period 180◦.
A graph of y = sin3A, as shown in Figure 22.14, has an
amplitude of 1 and period of 120◦.
A graph of y = 4cos2x, as shown in Figure 22.15, has
an amplitude of 4 and a period of 180◦.
In general, if y = Asinpx or y = Acospx,
amplitude = A and period =
360◦
p
y
3
23
0 A8
y 53 sin 2A
2708 36081808908
Figure 22.13

y
1.0
Ϫ1.0
0 90Њ 270Њ AЊ180Њ 360Њ
yϭsin 3A
Figure 22.14
y
90Њ 180Њ 270Њ 360Њ xЊ0
Ϫ4
4 y ϭ4 cos 2x
Figure 22.15
Problem 4. Sketch y = 2sin
3
5
A over one cycle
Amplitude = 2; period =
360◦
3
5
=
360◦ × 5
3
= 600◦
A sketch of y = 2sin
3
5
A is shown in Figure 22.16.
180Њ 360Њ 540Њ 600Њ
y
AЊ0
Ϫ2
2
yϭ2 sin A
3
5
Figure 22.16
22.4.4 Periodic time
In practice, the horizontal axis of a sine wave will be
time. The time taken for a sine wave to complete one
cycle 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 10V
and the periodic time is 20ms; i.e., T = 20 ms.
v
10 20 t (ms)0
210
10
Figure 22.17
22.4.5 Frequency
The number of cycles completed in one second is called
the frequency f and is measured in hertz, Hz.
f =
1
T
or T =
1
f
Problem 5. Determine the frequency of the sine
wave shown in Figure 22.17
In the sine wave shown in Figure 22.17, T = 20 ms,
hence
frequency, f =
1
T
=
1
20 × 10−3
= 50Hz
Problem 6. If a waveform has a frequency of
200kHz, determine the periodic time
If a waveform has a frequency of 200kHz, the periodic
time T is given by
periodic time, T =
1
f
=
1
200 × 103
= 5 × 10−6
s = 5μs
22.4.6 Lagging and leading angles
A sine or cosine curve may not always start at 0◦.
To show this, a periodic function is represented by
y = Asin(x ± α) where α is a phase displacement
compared with y = Asin x. For example, y = sin A is
shown by the broken line in Figure 22.18 and, on
the same axes, y = sin(A − 60◦) is shown. The graph
y = 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◦) is
said to lead y = cosA by 45◦.

908 2708
y
A80
21.0
1.0
y 5sin (A 2608)
y 5sin A
608
608
1808 3608
Figure 22.18
1808
458
3608 A80
y
21.0
y5cos (A 1458)
y 5 cos A
458
2708908
Figure 22.19
Problem 7. Sketch y = 5sin(A + 30◦) from
A = 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 in
Figure 22.20.
Problem 8. Sketch y = 7sin(2A − π/3) in the
range 0 ≤ A ≤ 360◦
Amplitude = 7 and period = 2π/2 = π radians
In 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 in
Figure 22.21.
908 2708 A80
5
25
y 55 sin (A 1308)
y55 sin A
308
308
1808 3608
y
Figure 22.20
0
7
y
A83608
y57sin 2A
y57sin (2A2␲/3)
␲/6
␲/6
2␲␲
7
27081808908
3␲/2␲/2
Figure 22.21
Problem 9. Sketch y = 2cos(ωt − 3π/10) over
one cycle
Amplitude = 2 and period = 2π/ω rad
2cos(ωt − 3π/10) lags 2cosωt by 3π/10ω seconds.
A sketch of y = 2cos(ωt − 3π/10) is shown in
Figure 22.22.
0
y
t␲/2␻ ␲/␻ 3␲/2␻ 2␲/␻
y52cos ␻t
y52cos(␻t23␲/10)
22
2
3␲/10␻ rads
Figure 22.22

PracticeExercise 88 Trigonometric
waveforms (answers on page 350)
1. A sine wave is given by y = 5sin3x. State its
peak value.
2. A sine wave is given by y = 4sin2x. State its
period in degrees.
3. Aperiodicfunction isgiven by y = 30cos5x.
State its maximum value.
4. Aperiodicfunction isgiven by y = 25cos3x.
State its period in degrees.
In problems 5 to 11, state the amplitude and period
of the waveform and sketch the curve between 0◦
and 360◦.
5. y = cos3A 6. y = 2sin
5x
2
7. y = 3sin4t 8. y = 5cos
θ
2
9. y =
7
2
sin
3x
8
10. y = 6sin(t − 45◦)
11. y = 4cos(2θ + 30◦)
12. The frequency of a sine wave is 200Hz.
Calculate the periodic time.
13. Calculate the frequency of a sine wave that
has a periodic time of 25ms.
14. Calculate the periodic time for a sine wave
having a frequency of 10kHz.
15. An alternating current completes 15 cycles in
24ms. Determine its frequency.
16. Graphs of y1 = 2sinx and
y2 = 3sin(x + 50◦) are drawn on the same
axes. Is y2 lagging or leading y1?
17. Graphs of y1 = 6sinx and
y2 = 5sin(x − 70◦) are drawn on the same
axes. Is y1 lagging or leading y2?
22.5 Sinusoidal form: Asin(ωt ± α)
If a sine wave is expressed in the form
y = Asin(ωt ± α) then
(a) A = amplitude.
(b) ω = angular velocity = 2π f rad/s.
(c) frequency, f =
ω
2π
hertz.
(d) periodic time, T =
2π
ω
seconds i.e. T =
1
f
.
(e) α = angle of lead or lag (compared with
y = Asinωt).
Here are some worked problems involving the sinu-
soidal form Asin(ωt ± α).
Problem 10. An alternating current is given by
i = 30sin(100πt + 0.35) amperes. Find the
(a) amplitude, (b) frequency, (c) periodic time and
(d) phase angle (in degrees and minutes)
(a) i = 30sin(100πt + 0.35)A; hence,
amplitude = 30A.
(b) Angular velocity, ω = 100π,rad/s, hence
frequency,f =
ω
2π
=
100π
2π
= 50Hz
(c) Periodic time, T =
1
f
=
1
50
= 0.02 s or 20 ms.
(d) 0.35 is the angle in radians. The relationship
between radians and degrees is
360◦
= 2π radians or 180◦
= πradians
from which,
1◦
=
π
180
rad and 1rad =
180◦
π
(≈ 57.30◦
)
Hence, phase angle, α = 0.35 rad
= 0.35 ×
180
π
◦
= 20.05◦ or 20◦3 leading
i = 30sin(100πt).
Problem 11. An oscillating mechanism has a
maximum displacement of 2.5m and a frequency of
60Hz. At time t = 0 the displacement is 90cm.
Express the displacement in the general form
Asin(ωt ± α)
Amplitude = maximum displacement = 2.5m.
Angular velocity, ω = 2π f = 2π(60) = 120π rad/s.
Hence, displacement = 2.5sin(120πt + α)m.
When t = 0, displacement = 90 cm = 0.90 m.

hence, 0.90 = 2.5sin(0 + α)
i.e. sinα =
0.90
2.5
= 0.36
Hence, α = sin−1
0.36 = 21.10◦
= 21◦
6 = 0.368rad.
Thus,displacement = 2.5sin(120πt + 0.368) m.
Problem 12. The instantaneous value of voltage
in an a.c. circuit at any time t seconds is given by
v = 340sin(50πt − 0.541) volts. Determine the
(a) amplitude, frequency, periodic time and phase
angle (in degrees), (b) value of the voltage when
t = 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 V
Angular velocity, ω = 50π
Frequency, f =
ω
2π
=
50π
2π
= 25Hz
Periodic time, T =
1
f
=
1
25
= 0.04s or 40 ms
Phase angle = 0.541rad = 0.541 ×
180
π
◦
= 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)
200
340
= sin(50πt − 0.541)
Hence, (50πt −0.541) = sin−1 200
340
= 36.03◦ or 0.628875rad
50πt = 0.628875 + 0.541
= 1.169875
Hence, when v = 200 V,
time, t =
1.169875
50π
= 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−1
1 = 90◦
or 1.5708 rad
50πt = 1.5708 + 0.541 = 2.1118
Hence, time, t =
2.1118
50π
= 13.44ms
(f ) A sketch of v = 340sin(50πt − 0.541) volts is
v5340sin(50 ␲t 20.541)
v5340sin 50␲t
0 t (ms)10 30 40
7.448 13.44
2340
2175.1
200
291.4
340
Voltage v
20
Figure 22.23
PracticeExercise 89 Sinusoidal form
Asin(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) of
the alternating quantities given.
1. i = 40sin(50πt + 0.29)mA
2. y = 75sin(40t − 0.54)cm

3. v = 300sin(200πt − 0.412)V
4. A sinusoidal voltage has a maximum value of
120V and a frequency of 50Hz. At time t = 0,
the voltage is (a) zero and (b) 50V. Express
the instantaneous voltage v in the form
v = Asin(ωt ± α).
5. An alternating current has a periodic time of
25ms and a maximum value of 20A. When
time = 0, current i = −10 amperes. Express
the 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 form
Asin(ωt ± α)
7. The current in an a.c. circuit at any time t
seconds is given by
i = 5sin(100πt − 0.432) amperes
Determine the
(a) amplitude, frequency, periodic time and
phase 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 23
Non-right-angled triangles
and some practical
applications
23.1 The sine and cosine rules
To ‘solve a triangle’ means ‘to ﬁnd the values of
unknown sides and angles’. If a triangle is right-angled,
trigonometric ratios and the theorem of Pythagoras
may be used for its solution, as shown in Chapter 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 cosine
rule, are used.
23.1.1 The sine rule
With reference to triangle ABC of Figure 23.1, the sine
rule states
a
sinA
=
b
sinB
=
c
sinC
a
bc
B C
A
Figure 23.1
The 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) are
initially given.
23.1.2 The cosine rule
With reference to triangle ABC of Figure 23.1, the
cosine rule states
a2
= b2
+ c2
− 2bc cos A
or b2
= a2
+ c2
− 2ac cos B
or c2
= a2
+ b2
− 2ab cosC
The rule may be used only when
(a) 2 sides and the included angle are initially
given, or
(b) 3 sides are initially given.
23.2 Area of any triangle
The area of any triangle such as ABC of Figure 23.1
is given by
(a)
1
2
× base × perpendicular height
DOI: 10.1016/B978-1-85617-697-2.00023-5

or (b)
1
2
ab sinC or
1
2
ac sinB or
1
2
bc sinA
or (c) [s(s − a)(s − b)(s − c)] where s =
a + b + c
2
23.3 Worked problems on the solution
of triangles and their areas
Problem 1. In a triangle XYZ, ∠X = 51◦,
∠Y = 67◦ and YZ = 15.2 cm. Solve the triangle and
find its area
The triangle XYZ is shown in Figure 23.2. Solving the
triangle means finding ∠Z and sides XZ and XY.
X
Y Zx 515.2cm
yz
518
678
Figure 23.2
Since the angles in a triangle add up to 180◦,
Z = 180◦ − 51◦ − 67◦ = 62◦
Applying the sine rule,
15.2
sin51◦
=
y
sin67◦
=
z
sin62◦
Using
15.2
sin51◦
=
y
sin67◦
and transposing gives y =
15.2sin67◦
sin51◦
= 18.00cm = XZ
Using
15.2
sin51◦
=
z
sin62◦
and transposing gives z =
15.2sin62◦
sin51◦
= 17.27cm = XY
Area of triangle XYZ =
1
2
xy sin Z
=
1
2
(15.2)(18.00)sin 62◦
= 120.8cm2
(or area =
1
2
xz sinY =
1
2
(15.2)(17.27) sin67◦
= 120.8cm2
)
It is always worth checking with triangle problems that
the longest side is opposite the largest angle and vice-
versa. In this problem, Y is the largest angle and X Z is
the longest of the three sides.
Problem 2. Solve the triangle ABC given
B = 78◦51 ,AC = 22.31mm and AB = 17.92 mm.
Also find its area
Triangle ABC is shown in Figure 23.3. Solving the
triangle means finding angles A and C and side BC.
A
B a C
b522.31 mm
c517.92mm
788519
Figure 23.3
22.31
sin78◦51
=
17.92
sinC
from which sinC =
17.92sin78◦51
22.31
= 0.7881
Hence, C = sin−1
0.7881 = 52◦
0 or 128◦
0
Since B = 78◦51 , C cannot be 128◦0 , since 128◦0 +
78◦51 is greater than 180◦. Thus, only C = 52◦0 is
valid.
Angle A = 180◦ − 78◦51 − 52◦0 = 49◦9 .
Applying the sine rule, a
sin49◦9
=
22.31
sin78◦51
from which a =
22.31sin49◦9
sin78◦51
= 17.20mm
Hence, A = 49◦
9 ,C = 52◦
0 and BC = 17.20mm.
Area of triangle ABC =
1
2
ac sin B
=
1
2
(17.20)(17.92)sin78◦
51
= 151.2mm2
Problem 3. Solve the triangle PQR and find its
area given that QR = 36.5mm,PR = 29.6mm and
∠Q = 36◦

Non-right-angled triangles and some practical applications 207
Triangle PQR is shown in Figure 23.4.
q 529.6 mm
p5 36.5 mm
368
r
Q R
P
Figure 23.4
29.6
sin36◦
=
36.5
sin P
from which sin P =
36.5sin36◦
29.6
= 0.7248
Hence, P = sin−1
0.7248 = 46.45◦
or 133.55◦
When P = 46.45◦ and Q = 36◦ then
R = 180◦ − 46.45◦ − 36◦ = 97.55◦
When P = 133.55◦ and Q = 36◦ then
R = 180◦ − 133.55◦ − 36◦ = 10.45◦
Thus, in this problem, there are two separate sets of
results and both are feasible solutions. Such a situation
is called the ambiguous case.
Case 1. P = 46.45◦, Q = 36◦, R = 97.55◦,
p = 36.5mm and q = 29.6mm
From the sine rule,
r
sin97.55◦
=
29.6
sin36◦
from which r =
29.6sin97.55◦
sin36◦
= 49.92mm = PQ
Area of PQR =
1
2
pq sin R =
1
2
(36.5)(29.6)sin 97.55◦
= 535.5mm2
Case 2. P = 133.55◦, Q = 36◦, R = 10.45◦,
p = 36.5mm and q = 29.6mm
From the sine rule,
r
sin10.45◦
=
29.6
sin36◦
from which r =
29.6sin10.45◦
sin36◦
= 9.134mm = PQ
Area of PQR =
1
2
pq sin R =
1
2
(36.5)(29.6)sin 10.45◦
= 97.98mm2
The triangle PQR for case 2 is shown in Figure 23.5.
368 10.458
133.558
9.134 mm 29.6 mm
36.5 mmQ
P
R
Figure 23.5
PracticeExercise 90 Solution of triangles
and their areas (answers on page 350)
In problems 1 and 2, use the sine rule to solve the
triangles ABC and find their areas.
1. A = 29◦, B = 68◦,b = 27mm
2. B = 71◦26 ,C = 56◦32 ,b = 8.60 cm
triangles DEF and find their areas.
3. d = 17cm, f = 22 cm, F = 26◦
4. d = 32.6mm,e = 25.4mm, D = 104◦
22
triangles 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 the
solution of triangles and their
areas
Problem 4. Solve triangle DEF and find its area
given 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. Since
two sides and the angle in between the two sides are
given, the cosine needs to be used.
648
D
FE
e
d 535.0mm
f525.0mm
Figure 23.6
Applying the cosine rule, e2
= d2
+ f 2
− 2d f cos E
i.e. e2
= (35.0)2
+ (25.0)2
− [2(35.0)(25.0)cos64◦
]
= 1225 + 625 − 767.15
= 1082.85

from which e =
√
1082.85
= 32.91mm = DF
32.91
sin64◦
=
25.0
sin F
from which sin F =
25.0sin64◦
32.91
= 0.6828
Thus, ∠F = sin−1
0.6828 = 43◦
4 or 136◦
56
F = 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=
1
2
d f sin E
=
1
2
(35.0)(25.0)sin 64◦ = 393.2mm2
Problem 5. A triangle ABC has sides
a = 9.0 cm,b = 7.5cm and c = 6.5cm. Determine
its three angles and its area
Triangle ABC is shown in Figure 23.7. It is usual ﬁrst
to calculate the largest angle to determine whether the
triangle is acute or obtuse. In this case the largest angle
is A (i.e. opposite the longest side).
A
B Ca59.0cm
c56.5cm b5 7.5cm
Figure 23.7
Applying the cosine rule, a2
= b2
+ c2
− 2bccos A
from which 2bccos A = b2
+ c2
− a2
and cos A =
b2 + c2 − a2
2bc
=
7.52 + 6.52 − 9.02
2(7.5)(6.5)
= 0.1795
Hence, A = cos−1
0.1795 = 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◦.)
9.0
sin79.67◦
=
7.5
sin B
from which sin B =
7.5sin79.67◦
9.0
= 0.8198
Hence, B = sin−1
0.8198 = 55.07◦
and C = 180◦
− 79.67◦
− 55.07◦
= 45.26◦
Area =
√
[s(s − a)(s − b)(s − c)], where
s =
a + b + c
2
=
9.0 + 7.5 + 6.5
2
= 11.5cm
Hence,
area = [11.5(11.5 − 9.0)(11.5 − 7.5)(11.5 − 6.5)]
= [11.5(2.5)(4.0)(5.0)] = 23.98cm2
Alternatively,area =
1
2
ac sin B
=
1
2
(9.0)(6.5)sin 55.07◦
= 23.98cm2
Problem 6. Solve triangle XYZ, shown in
Figure 23.8, and ﬁnd its area given that
Y = 128◦,XY = 7.2 cm and YZ = 4.5cm
1288
x 54.5cm
z57.2cm
y
X
Y Z
Figure 23.8
Applying 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 = 112.0
y =
√
112.0 = 10.58cm = XZ
10.58
sin128◦
=
7.2
sin Z
from which sin Z =
7.2sin128◦
10.58
= 0.5363
Hence, Z = sin−1
0.5363 = 32.43◦
(or 147.57◦ which is not possible)
Thus, X = 180◦
− 128◦
− 32.43◦
= 19.57◦
Area of XYZ =
1
2
xz sinY =
1
2
(4.5)(7.2)sin 128◦
= 12.77cm2

PracticeExercise 91 Solution of triangles
and their areas (answers on page 350)
In problems 1 and 2, use the cosine and sine rules
to 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 rules
to solve the triangles XYZ and find their areas.
3. x = 10.0 cm, y = 8.0 cm, z = 7.0 cm
4. x = 21mm, y = 34mm, z = 42 mm
23.5 Practical situations involving
trigonometry
There are a number of practical situations in which the
use of trigonometryis needed to find unknown sides and
anglesoftriangles.Thisisdemonstrated in thefollowing
worked problems.
Problem 7. A room 8.0 m wide has a span roof
which slopes at 33◦ on one side and 40◦ on the
other. Find the length of the roof slopes, correct to
the nearest centimetre
A section of the roof is shown in Figure 23.9.
B
A C8.0 m
338 408
Figure 23.9
Angle at ridge, B = 180◦ − 33◦ − 40◦ = 107◦
From the sine rule,
8.0
sin107◦
=
a
sin33◦
from which a =
8.0sin33◦
sin107◦
= 4.556m = BC
Also from the sine rule,
8.0
sin107◦
=
c
sin40◦
from which c =
8.0sin40◦
sin107◦
= 5.377m = AB
Hence, the roof slopes are 4.56 m and 5.38 m, correct
to the nearest centimetre.
Problem 8. A man leaves a point walking at
6.5km/h in a direction E 20◦ N (i.e. a bearing of
70◦). A cyclist leaves the same point at the same
time in a direction E 40◦ S (i.e. a bearing of 130◦)
travelling at a constant speed. Find the average
speed of the cyclist if the walker and cyclist are
80 km apart after 5 hours
After5 hoursthewalkerhastravelled 5 × 6.5 = 32.5km
(shown as AB in Figure 23.10). If AC is the distance the
cyclist travels in 5 hours then BC = 80 km.
b
A
B
C
80 km
W
S
408
E
N
208
32.5km
Figure 23.10
80
sin60◦
=
32.5
sinC
from which sinC =
32.5sin60◦
80
= 0.3518
Hence, C = sin−1
0.3518 = 20.60◦
(or 159.40◦, which is not possible)
and B = 180◦
− 60◦
− 20.60◦
= 99.40◦
Applying the sine rule again,
80
sin60◦
=
b
sin99.40◦
from which b =
80sin99.40◦
sin60◦
= 91.14km
Since the cyclist travels 91.14km in 5 hours,
average speed =
distance
time
=
91.14
5
= 18.23km/h
Problem 9. Two voltage phasors are shown in
Figure 23.11. If V1 = 40 V and V2 = 100 V,
determine the value of their resultant (i.e. length
OA) and the angle the resultant makes with V1

458
V25100V
V1540 V
A
BO
Figure 23.11
Angle OBA = 180◦
− 45◦
= 135◦
Applying the cosine rule,
OA2
= V 2
1 + V 2
2 − 2V1V2 cosOBA
= 402
+ 1002
− {2(40)(100)cos135◦
}
= 1600 + 10000− {−5657}
= 1600 + 10000+ 5657 = 17257
Thus, resultant,OA =
√
17257 = 131.4 V
Applying the sine rule
131.4
sin135◦
=
100
sinAOB
from which sinAOB =
100sin135◦
131.4
= 0.5381
Hence, angle AOB = sin−1 0.5381 = 32.55◦ (or
147.45◦, which is not possible)
Hence, the resultant voltage is 131.4 volts at 32.55◦
to V1
Problem 10. In Figure 23.12, PR represents the
inclined jib of a crane and is 10.0 m long. PQ is
4.0 m long. Determine the inclination of the jib to
the vertical and the length of tie QR.
P
Q
R
1208
4.0 m 10.0m
Figure 23.12
PR
sin120◦
=
PQ
sin R
from which sin R =
PQsin120◦
PR
=
(4.0)sin 120◦
10.0
= 0.3464
Hence, ∠R = sin−1
0.3464 = 20.27◦
(or 159.73◦
,
which is not possible)
∠P = 180◦ − 120◦ − 20.27◦ = 39.73◦, which is the
inclination of the jib to the vertical
10.0
sin120◦
=
QR
sin39.73◦
from which length of tie,QR =
10.0sin39.73◦
sin120◦
= 7.38m
PracticeExercise 92 Practical situations
involving trigonometry (answers on
page 350)
1. A ship P sails at a steady speed of 45km/h
in a direction of W 32◦ N (i.e. a bearing of
302◦
) from a port. At the same time another
ship Q leaves the port at a steady speed of
35km/h in a direction N 15◦ E (i.e. a bearing
of 015◦). Determine their distance apart after
4 hours.
2. Two sides of a triangular plot of land are
52.0 m and 34.0 m, respectively. If the area
of the plot is 620 m2, ﬁnd (a) the length
of 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 the
tie rod PR is 8.0 m long and PQ is 4.5m long,
determine (a) the length of jib RQ and (b) the
angle between the jib and the tie rod.
R
1308 P
Q
Figure 23.13
4. A building site is in the form of a quadri-
lateral, as shown in Figure 23.14, and its
area is 1510 m2. Determine the length of the
perimeter of the site.

28.5m
34.6m
728
52.4m 758
Figure 23.14
5. Determine the length of members BF and EB
in the roof truss shown in Figure 23.15.
4 m
2.5 m
508
4m
2.5m
508
5 m 5 m
E
A B C
DF
Figure 23.15
6. A laboratory 9.0 m wide has a span roof which
slopes 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 the
alternating 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 PQ
to form triangle PQR) is 14.0 A and is at an
angle of 35◦ to the horizontal. Determine the
resultant phasor PR and the angle it makes
with phasor PQ.
23.6 Further practical situations
involving trigonometry
Problem 11. A vertical aerial stands on
horizontal ground. A surveyor positioned due east
of the aerial measures the elevation of the top as
48◦. He moves due south 30.0 m and measures the
elevation as 44◦. Determine the height of the aerial
In Figure 23.16, DC represents the aerial, A is the initial
position of the surveyor and B his ﬁnal position.
From triangle ACD, tan48◦ =
DC
AC
from which
AC =
DC
tan48◦
Similarly, from triangle BCD, BC =
DC
tan44◦
B
C
D
A
30.0 m448
488
Figure 23.16
For triangle ABC, using Pythagoras’ theorem,
BC2
= AB2
+ AC2
DC
tan44◦
2
= (30.0)2
+
DC
tan48◦
2
DC2 1
tan2 44◦
−
1
tan2 48◦
= 30.02
DC2
(1.072323 − 0.810727) = 30.02
DC2
=
30.02
0.261596
= 3440.4
Hence, height of aerial, DC =
√
3340.4 = 58.65 m.
Problem 12. A crank mechanism of a petrol
engine is shown in Figure 23.17. Arm OA is
10.0 cm long and rotates clockwise about O. The
connecting rod AB is 30.0 cm long and end B is
constrained to move horizontally.
(a) For the position shown in Figure 23.17,
determine the angle between the connecting
rod AB and the horizontal, and the length
of OB.
(b) How far does B move when angle AOB
changes from 50◦ to 120◦?
508
O
30.0cm
10.0 cm
A
B
Figure 23.17

(a) Applying the sine rule,
AB
sin50◦
=
AO
sin B
from which sin B =
AOsin50◦
AB
=
10.0sin50◦
30.0
= 0.2553
Hence, B = sin−1 0.2553 = 14.79◦ (or 165.21◦,
which is not possible)
Hence, the connecting rod AB makes an angle
of 14.79◦ with the horizontal.
Angle OAB = 180◦ − 50◦ − 14.79◦ = 115.21◦
Applying the sine rule:
30.0
sin50◦
=
OB
sin115.21◦
from which OB =
30.0sin115.21◦
sin50◦
= 35.43cm
(b) Figure 23.18 shows the initial and final positions
of the crank mechanism.
508
O
30.0 cm
1208
B B9
A9A
10.0cm
Figure 23.18
In triangle OA B , applying the sine rule,
30.0
sin120◦
=
10.0
sin AB O
from which sin A BO =
10.0sin120◦
30.0
= 0.28868
Hence, AB O = sin−1 0.28868 = 16.78◦ (or
163.22◦, which is not possible)
Angle OA B = 180◦ − 120◦ − 16.78◦ = 43.22◦
30.0
sin120◦
=
OB
sin43.22◦
from whichOB =
30.0sin43.22◦
sin120◦
= 23.72 cm
Since OB = 35.43cm and OB = 23.72 cm,
BB = 35.43 − 23.72 = 11.71cm
Hence, B moves 11.71 cm when angle AOB
changes from 50◦ to 120◦.
Problem 13. The area of a field is in the form of a
quadrilateral ABCD as shown in Figure 23.19.
Determine its area
62.3m
21.4 m
39.8 m
1148
568
B
D
C
A
42.5 m
Figure 23.19
A diagonal drawn from B to D divides the quadrilateral
into two triangles.
Area of quadrilateral ABCD
= area of triangle ABD
+ area of triangle BCD
=
1
2
(39.8)(21.4)sin 114◦
+
1
2
(42.5)(62.3)sin 56◦
= 389.04 + 1097.5
= 1487m2
PracticeExercise 93 More practical
situations involving trigonometry (answers
on page 350)
1. Three forces acting on a fixed point are repre-
sented by the sides of a triangle of dimensions
7.2 cm,9.6cm and 11.0 cm. Determine the
anglesbetween thelinesofaction and thethree
forces.
2. A vertical aerial AB, 9.60 m high, stands on
ground which is inclined 12◦ to the horizontal.
Astay connectsthetop oftheaerial Ato apoint
C on the ground 10.0 m downhill from B, the
foot of the aerial. Determine (a) the length of
the stay and (b) the angle the stay makes with
the ground.
3. A reciprocating engine mechanism is shown
in Figure 23.20. The crank AB is 12.0cm long
and the connecting rod BC is 32.0cm long.
For the position shown determine the length
of AC and the angle between the crank and the
connecting rod.

B
C
A
408
Figure 23.20
4. From Figure 23.20, determine how far C
moves, correct to the nearest millimetre, when
angle CAB changes from 40◦ to 160◦, B
moving in an anticlockwise direction.
5. A surveyor standing W 25◦
S of a tower mea-
sures the angle of elevation of the top of
the tower as 46◦30 . From a position E 23◦S
from the tower the elevation of the top is
37◦15 .Determinetheheight ofthetowerifthe
distance between the two observationsis75m.
6. Calculate, correct to 3 significant figures, the
co-ordinates x and y to locate the hole centre
at P shown in Figure 23.21.
100 mm
1168 1408
P
x
y
Figure 23.21
7. An idler gear, 30 mm in diameter, has to be
fitted between a 70 mm diameter driving gear
and a 90 mm diameter driven gear, as shown
in Figure 23.22. Determine the value of angle
θ between the centre lines.
90mm
dia
30mm
dia
70mm
dia
␪99.78mm
Figure 23.22
8. 16 holes are equally spaced on a pitch circle
of 70 mm diameter. Determine the length of
the chord joining the centres of two adjacent
holes.

Chapter 24
Cartesian and polar
co-ordinates
24.1 Introduction
There are two ways in which the position of a point in
a 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 radius
from a ﬁxed point and θ is an angle from a ﬁxed
point.
24.2 Changing from Cartesian to
polar co-ordinates
In Figure 24.1, if lengths x and y are known then
the length of r can be obtained from Pythagoras’ the-
orem (see Chapter 21) since OPQ is a right-angled
triangle.
y
P
Q xO
x
r y
␪
Figure 24.1
Hence, r2 = (x2 + y2), from which r = x2 + y2
From trigonometric ratios (see Chapter 21), tanθ =
y
x
from which θ = tan−1 y
x
r = x2 + y2 and θ = tan−1 y
x
are the two formulae
we 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 is
suggested that when changing from Cartesian to polar
co-ordinates a diagram should always be sketched.
Problem 1. Change the Cartesian co-ordinates
(3, 4) into polar co-ordinates
A diagram representing the point (3, 4) is shown in
Figure 24.2.
P
4
3
y
xO
r
␪
Figure 24.2
From Pythagoras’ theorem, r =
√
32 + 42 = 5 (note
that −5 has no meaning in this context).
By trigonometric ratios, θ = tan−1 4
3
= 53.13◦ or
0.927rad.
Note that 53.13◦ = 53.13 ×
π
180
rad = 0.927rad.
DOI: 10.1016/B978-1-85617-697-2.00024-7

Cartesian and polar co-ordinates 215
Hence, (3, 4) in Cartesian co-ordinates corre-
sponds to (5, 53.13◦) or (5, 0.927 rad) in polar
co-ordinates.
Problem 2. Express in polar co-ordinates the
position (−4, 3)
A diagram representing the point using the Cartesian
co-ordinates (−4, 3) is shown in Figure 24.3.
y
P
3
4
xO
r
␪␣
Figure 24.3
From Pythagoras’ theorem, r =
√
42 + 32 = 5
By trigonometric ratios, α = tan−1 3
4
= 36.87◦ or
0.644 rad
Hence, θ = 180◦ − 36.87◦ = 143.13◦
or θ = π − 0.644 = 2.498 rad
Hence, the position of point P in polar co-ordinate
form is (5, 143.13◦) or (5, 2.498 rad).
Problem 3. Express (−5, −12) in polar
co-ordinates
A sketch showing the position (−5, −12) is shown in
Figure 24.4.
y
P
12
5
xO
r
␪
␣
Figure 24.4
r =
√
52 + 122 = 13 and α = tan−1 12
5
= 67.38◦
or 1.176 rad
Hence, θ = 180◦
+ 67.38◦
= 247.38◦
or θ = π + 1.176 = 4.318 rad.
Thus, (−5, −12) in Cartesian co-ordinates corre-
sponds to (13, 247.38◦) or (13, 4.318 rad) in polar
co-ordinates.
Problem 4. Express (2, −5) in polar co-ordinates
A sketch showing the position (2, −5) is shown in
Figure 24.5.
y
xO
5
2
r
P
␪
␣
Figure 24.5
r =
√
22 + 52 =
√
29 = 5.385, correct to 3 decimal
places
α = tan−1 5
2
= 68.20◦ or 1.190 rad
Hence, θ = 360◦ − 68.20◦ = 291.80◦
or θ = 2π − 1.190 = 5.093 rad.
Thus, (2, −5) in Cartesian co-ordinates corresponds
to (5.385, 291.80◦
) or (5.385, 5.093 rad) in polar co-
ordinates.
PracticeExercise 94 Changing from
Cartesian to polar co-ordinates (answers
on page 350)
In problems 1 to 8, express the given Cartesian co-
ordinatesaspolarco-ordinates,correct to 2 decimal
places, in both degrees and radians.
1. (3, 5) 2. (6.18, 2.35)
3. (−2, 4) 4. (−5.4, 3.7)
5. (−7, −3) 6. (−2.4, −3.6)
7. (5, −3) 8. (9.6, −12.4)

24.3 Changing from polar to Cartesian
co-ordinates
y
y
Q x
x
O
P
r
␪
Figure 24.6
From the right-angled triangle OPQ in Figure 24.6,
cosθ =
x
r
and sinθ =
y
r
from trigonometric ratios
Hence, x = r cosθ and y = r sinθ.
If lengths r and angle θ are known then x = r cosθ and
y = r sinθ are the two formulae we need to change from
polar to Cartesian co-ordinates.
Problem 5. Change (4, 32◦) into Cartesian
co-ordinates
A sketch showing the position (4, 32◦) is shown in
Figure 24.7.
y
y
O x
x
r ϭ 4
␪ ϭ32Њ
Figure 24.7
Now x = r cosθ = 4cos32◦ = 3.39
and y = r sinθ = 4sin32◦ = 2.12
Hence, (4, 32◦) in polar co-ordinates corresponds to
(3.39, 2.12) in Cartesian co-ordinates.
Problem 6. Express (6, 137◦) in Cartesian
co-ordinates
A sketch showing the position (6, 137◦) is shown in
Figure 24.8.
B
OA
y
x
r ϭ 6
␪ϭ137Њ
Figure 24.8
x = r cosθ = 6cos137◦
= −4.388
which corresponds to length OA in Figure 24.8.
y = r sinθ = 6sin137◦
= 4.092
which 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. Use
of x = r cos θ and y = r sin θ automatically produces
the correct values and signs.)
Problem 7. Express (4.5, 5.16 rad) in Cartesian
co-ordinates
A sketch showing the position (4.5, 5.16 rad) is shown
in Figure 24.9.
y
A
B
O x
␪55.16 rad
r 5 4.5
Figure 24.9
x = r cosθ = 4.5cos5.16 = 1.948
which corresponds to length OA in Figure 24.9.
y = r sinθ = 4.5sin5.16 = −4.057
which 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.

Cartesian and polar co-ordinates 217
PracticeExercise 95 Changing polar to
Cartesian co-ordinates (answers on
page 350)
In problems 1 to 8, express the given polar co-
ordinates as Cartesian co-ordinates, correct to 3
decimal places.
1. (5, 75◦) 2. (4.4, 1.12rad)
3. (7, 140◦
) 4. (3.6, 2.5rad)
5. (10.8, 210◦) 6. (4, 4rad)
7. (1.5, 300◦) 8. (6, 5.5rad)
9. Figure 24.10 shows 5 equally spaced holes
on an 80mm pitch circle diameter. Calculate
their co-ordinates relative to axes Ox and Oy
in (a) polar form, (b) Cartesian form.
10. In Figure 24.10, calculate the shortest dis-
tance between the centres of two adjacent
holes.
y
xO
Figure 24.10
24.4 Use of Pol/Rec functions on
calculators
Another name for Cartesian co-ordinates is rectangular
co-ordinates. Many scientiﬁc notation calculators have
Pol and Rec functions. ‘Rec’ is an abbreviation of ‘rect-
angular’ (i.e. Cartesian) and ‘Pol’ is an abbreviation of
‘polar’. Check the operation manual for your particular
calculator to determine how to use these two func-
tions. They make changing from Cartesian to polar co-
ordinates, and vice-versa, so much quicker and easier.
For example, with the Casio fx-83ES calculator, or sim-
ilar, to change the Cartesian number (3, 4) into polar
form, the following procedure is adopted.
1. Press ‘shift’ 2. Press ‘Pol’ 3. Enter 3
4. Enter ‘comma’ (obtained by ‘shift’ then ) )
5. Enter 4 6. Press ) 7. Press =
The answer is r = 5,θ = 53.13◦
Hence, (3, 4) in Cartesian form is the same as
(5, 53.13◦) in polar form.
If the angle is required in radians, then before repeating
the above procedure press ‘shift’, ‘mode’ and then 4 to
change your calculator to radian mode.
Similarly, to change the polar form number (7, 126◦)
into Cartesian or rectangular form, adopt the following
procedure.
1. Press ‘shift’ 2. Press ‘Rec’
3. Enter 7 4. Enter ‘comma’
5. Enter 126 (assuming your calculator is in degrees
mode)
The answer is X = −4.11 and, scrolling across,
Y = 5.66, correct to 2 decimal places.
Hence, (7, 126◦) in polar form is the same as
(−4.11, 5.66) in rectangular or Cartesian form.
Now return to Practice Exercises 94 and 95 in
this chapter and use your calculator to determine the
answers, and see how much more quickly they may be
obtained.

Revision Test 9 : Trigonometric waveforms and practical trigonometry
1. A sine wave is given by y = 8sin4x. State itspeak
value 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. Calculate
the periodic time in milliseconds. (2)
4. Calculate the frequency of a sine wave that has a
periodic time of 40μs. (2)
5. Calculate the periodic time for a sine wave having
a frequency of 20kHz. (2)
6. An alternating current completes 12 cycles in
16ms. What is its frequency? (3)
7. A sinusoidal voltage is given by
e = 150sin(500πt − 0.25) volts. Determine the
(a) amplitude,
(b) frequency,
(c) periodic time,
(d) phase angle (stating whether it is leading or
lagging 150sin500πt). (4)
8. Determinetheacuteanglesin degrees,degreesand
minutes, and radians.
(a) sin−1 0.4721 (b) cos−1 0.8457
(c) tan−1 1.3472 (9)
9. Sketch the following curves, labelling relevant
points.
(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 by
i = 120sin(100πt + 0.274) amperes. Determine
(a) the amplitude, frequency, periodic time and
phase 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 in
Figure RT9.1. Solve the triangle and determine
its area. (10)
A
B
C
15m
15.4m
718
Figure RT9.1
12. A car is travelling 20m above sea level. It then
travels 500m up a steady slope of 17◦. Determine,
correct to the nearest metre, how high the car is
now above sea level. (3)
13. Figure RT9.2 shows a roof truss PQR with rafter
PQ = 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 roof
truss. (11)
P
Q RP9
3m
408 328
Figure RT9.2
14. Solve triangle ABC given b = 10cm,c = 15cm
and ∠A = 60◦. (10)
15. Change the following Cartesian co-ordinates into
polar co-ordinates, correct to 2 decimal places, in
both degrees and in radians.
(a) (−2.3,5.4) (b) (7.6,−9.2) (10)
16. Change the following polar co-ordinates into
Cartesian co-ordinates, correct to 3 decimal
places.
(a) (6.5,132◦) (b) (3,3rad) (6)

Chapter 25
Areas of common shapes
25.1 Introduction
Area is a measure of the size or extent of a plane surface.
Area ismeasured in square units such as mm2, cm2 and
m2. This chapter deals with finding the areas of common
shapes.
In engineering it is often important to be able to cal-
culate simple areas of various shapes. In everyday life
its important to be able to measure area to, say, lay a car-
pet, order sufficient paint for a decorating job or order
sufficient 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, trapeziums
and circles.
25.2 Common shapes
25.2.1 Polygons
A polygon is a closed plane figure bounded by straight
lines. A polygon which has
3 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 Quadrilaterals
There are five types of quadrilateral, these being rect-
angle, square, parallelogram, rhombus and trapezium.
If the opposite corners of any quadrilateral are joined
by a straight line, two triangles are produced. Since the
sum of the angles of a triangle is 180◦, the sum of the
angles of a quadrilateral is 360◦.
Rectangle
In 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 bisect
one another.
Square
In 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 bisect
one another at right angles.
Parallelogram
In 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.
Rhombus
In 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 right
angles.
DOI: 10.1016/B978-1-85617-697-2.00025-9

(a) (c)(b)
(d) (e) (f)
Figure 25.1
BA
CD
Figure 25.2
QP
RS
QP
RS
Figure 25.3
W
Z Y
X
Figure 25.4
A
␣ ␣
␤
␤
D C
B
Figure 25.5
E
H G
F
Figure 25.6
Trapezium
In the trapezium EFGH shown in Figure 25.6,
(a) only one pair of sides is parallel.
Problem 1. State the types of quadrilateral shown
in Figure 25.7 and determine the angles marked a
to l
(a) (b) (c)
(d) (e)
BA
x
x
a
CD
E
H G
F
d
c
b
408
308
KJ
x
x
LM
e
f
O
Q Pj
i
hg
N
658
528
R
U
T
S
k
358
758
1158
l
Figure 25.7
(a) ABCD is a square
The diagonals of a square bisect each of the right
angles, hence
a =
90◦
2
= 45◦
(b) EFGH is a rectangle

Areas of common shapes 221
In triangle FGH, 40◦
+ 90◦
+ b = 180◦
, since the
angles in a triangle add up to 180◦, from which
b = 50◦. Also, c = 40◦ (alternate angles between
parallel lines EF and HG). (Alternatively, b and c
are complementary; i.e., add up to 90◦.)
d = 90◦ + c (external angle of a triangle equals
the sum of the interior opposite angles), hence
d = 90◦ + 40◦ = 130◦ (or ∠EFH = 50◦ and
d = 180◦ − 50◦ = 130◦).
(c) JKLM is a rhombus
The diagonals of a rhombus bisect the interior
angles 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 parallelogram
g = 52◦ since the opposite interior angles of a
parallelogram are equal.
In triangle NOQ, g + h + 65◦ = 180◦ (the angles
in a triangle add up to 180◦), from which
h = 180◦ − 65◦ − 52◦ = 63◦.
i = 65◦ (alternate angles between parallel lines
NQ and OP).
j = 52◦ + i = 52◦ + 65◦ = 117◦ (the external
angle of a triangle equals the sum of the interior
opposite angles). (Alternatively, ∠PQO = h =
63◦; hence, j = 180◦ − 63◦ = 117◦.)
(e) RSTU is a trapezium
35◦ + k = 75◦ (external angle of a triangle equals
the sum of the interior opposite angles), hence,
k = 40◦.
∠STR = 35◦
(alternate angles between parallel
lines RU and ST). l + 35◦ = 115◦ (external angle
of a triangle equals the sum of the interioropposite
angles), hence, l = 115◦ − 35◦ = 80◦.
PracticeExercise 96 Common shapes
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).
758
388
1258
628
958
578
478
408
p
q s
r
t
(a) (b) (c)
Figure 25.8
25.3 Areas of common shapes
The formulae forthe areas of common shapes are shown
in Table 25.1.
Here are some worked problems to demonstrate how
the formulae are used to determine the area of common
shapes.
Problem 2. Calculate the area and length of the
perimeter of the square shown in Figure 25.9
4.0 cm
4.0 cm
Figure 25.9
Area of square = x2
= (4.0)2
= 4.0cm × 4.0cm
= 16.0cm2
(Note the unit of area is cm×cm = cm2; i.e., square
centimetres or centimetres squared.)
Perimeter of square = 4.0cm + 4.0cm + 4.0cm
+ 4.0cm = 16.0cm
Problem 3. Calculate the area and length of the
perimeter of the rectangle shown in Figure 25.10
7.0 cm
4.5 cm
A
D C
B
Figure 25.10

Table 25.1 Formulae for the areas of common shapes
Area of plane ﬁgures
Square x
x
Area = x2
Rectangle
I
b
Area = l × b
Parallelogram
h
b
Area = b × h
Triangle
h
b
Area =
1
2
× b × h
Trapezium
h
b
a
Area =
1
2
(a + b)h
Circle
r
Area = πr2 or
πd2
4
Circumference = 2πr
Radian measure: 2π radians = 360 degrees
Sector of circle
s
r
␪
Area =
θ◦
360
(πr2)

Area of rectangle = l × b = 7.0 × 4.5
= 31.5cm2
Perimeter of rectangle = 7.0cm + 4.5cm
+ 7.0cm + 4.5cm
= 23.0cm
Problem 4. Calculate the area of the
parallelogram shown in Figure 25.11
21mm
16mm
GH
E F
9mm
Figure 25.11
Area of a parallelogram = base × perpendicular height
The perpendicular height h is not shown in Figure 25.11
but may be found using Pythagoras’ theorem (see
Chapter 21).
From Figure 25.12, 92 = 52 + h2, from which
h2 = 92 − 52 = 81 − 25 = 56.
Hence, perpendicular height,
h =
√
56 = 7.48mm.
9mm
E
G
h
F
H
16 mm 5mm
Figure 25.12
Hence, area of parallelogram EFGH
= 16mm × 7.48mm
= 120mm2
.
Problem 5. Calculate the area of the triangle
I
K
J
5.68cm
1.92cm
Figure 25.13
Area of triangle IJK =
1
2
×base×perpendicular height
=
1
2
× IJ × JK
To ﬁnd JK, Pythagoras’ theorem is used; i.e.,
5.682
= 1.922
+ JK 2
, from which
JK = 5.682 − 1.922 = 5.346cm
Hence, area of triangle IJK =
1
2
× 1.92 × 5.346
= 5.132cm2
.
Problem 6. Calculate the area of the trapezium
27.4mm
8.6 mm
5.5 mm
Figure 25.14
Area of a trapezium =
1
2
× (sum of parallel sides)
×(perpendicular distance
between the parallel sides)
Hence, area of trapezium LMNO
=
1
2
× (27.4 + 8.6) × 5.5
=
1
2
× 36 × 5.5 = 99mm2

Problem 7. A rectangular tray is 820mm long
and 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
, or
1mm2
=
1
100
cm2
= 0.01cm2
Hence, 328000 mm2
= 328000× 0.01cm2
= 3280cm2
.
(c) Since 1m = 100cm,1m2
= 1m × 1m
= 100cm × 100cm = 10000cm2
, or
1cm2
=
1
10000
m2
= 0.0001m2
Hence, 3280cm2
= 3280 × 0.0001m2
= 0.3280m2
.
Problem 8. The outside measurements of a
picture frame are 100cm by 50cm. If the frame is
4cm wide, ﬁnd the area of the wood used to make
the frame
A sketch of the frame is shown shaded in Figure 25.15.
100 cm
50 cm 42 cm
92 cm
Figure 25.15
Area of wood = area of large rectangle − area of
small rectangle
= (100 × 50) − (92 × 42)
= 5000 − 3864
= 1136cm2
Problem 9. Find the cross-sectional area of the
girder shown in Figure 25.16
5mm
50 mm
B
C
A
8mm
70 mm
75mm
6mm
Figure 25.16
Thegirdermay bedivided into threeseparaterectangles,
as shown.
Area of rectangle A = 50 × 5 = 250mm2
Area of rectangle B = (75 − 8 − 5) × 6
= 62 × 6 = 372mm2
Area of rectangle C = 70 × 8 = 560mm2
Total area of girder = 250 + 372 + 560
= 1182mm2
or 11.82cm2
Problem 10. Figure 25.17 shows the gable end of
a building. Determine the area of brickwork in the
gable end
6m
5m 5m
A
DC
B
8m
Figure 25.17
The shape is that of a rectangle and a triangle.
Area of rectangle = 6 × 8 = 48m2
Area of triangle =
1
2
× base × height
CD = 4m and AD = 5m, hence AC = 3m (since it is a
3, 4, 5 triangle – or by Pythagoras).
Hence, area of triangle ABD =
1
2
× 8 × 3 = 12m2
.

Total area of brickwork = 48 + 12
= 60m2
.
PracticeExercise 97 Areas of common
shapes (answers on page 351)
1. Name the types of quadrilateral shown in
Figure25.18(i)to (iv)and determine for each
(a) the area and (b) the perimeter.
120 mm
30 mm
(iii)
26 cm
12 cm
16 cm
10 cm
(iv)
4 cm
(i)
5.94 cm
3.5cm
6 mm
(ii)
30mm
38mm
Figure 25.18
2. A rectangular plate is 85mm long and
42mm wide. Find its area in square centi-
metres.
3. Arectangular field hasan areaof1.2hectares
and a length of 150m. If 1hectare =
10000m2,find (a)thefield’swidth and (b)the
length of a diagonal.
4. Find the area of a triangle whose base is
8.5cm and perpendicular height is 6.4cm.
5. A square has an area of 162cm2. Determine
the 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 the
other side.
7. Determine the area of each of the angle iron
sections shown in Figure 25.19.
7cm
1cm
1cm
(a) (b)
2cm2cm
30mm
25mm
8mm
10mm
50mm
6mm
Figure 25.19
8. Figure 25.20 shows a 4m wide path around
the outside of a 41m by 37m garden. Calcu-
late the area of the path.
41
4 37
Figure 25.20
9. The area of a trapezium is 13.5cm2 and the
perpendicular distance between its parallel
sides is 3cm. If the length of one of the par-
allel sides is 5.6cm, find the length of the
other parallel side.
10. Calculate the area of the steel plate shown in
Figure 25.21.
25
60
140
Dimensions
in mm
100
25
25
Figure 25.21

11. Determine the area of an equilateral triangle
of side 10.0cm.
12. If paving slabs are produced in 250mm by
250mm squares, determine the number of
slabs required to cover an area of 2m2.
Here are some further worked problems on ﬁnding the
areas of common shapes.
Problem 11. Find the area of a circle having a
radius of 5cm
Areaof circle = πr2
= π(5)2
= 25π = 78.54cm2
diameter of 15mm
Areaof circle =
πd2
4
=
π(15)2
4
=
225π
4
= 176.7mm2
circumference of 70mm
Circumference, c = 2πr,hence
radius,r =
c
2π
=
70
2π
=
35
π
mm
Area of circle = πr2
= π
35
π
2
=
352
π
= 389.9mm2
or 3.899cm2
Problem 14. Calculate the area of the sector of a
circle having radius 6cm with angle subtended at
centre 50◦
Area of sector =
θ2
360
(πr2
) =
50
360
(π62
)
=
50 × π × 36
360
= 15.71cm2
Problem 15. Calculate the area of the sector of a
circle having diameter 80mm with angle subtended
at centre 107◦42
If diameter = 80mm then radius, r = 40mm, and
area of sector =
107◦42
360
(π402
) =
107
42
60
360
(π402
)
=
107.7
360
(π402
)
= 1504mm2
or 15.04cm2
Problem 16. A hollow shaft has an outside
diameter of 5.45cm and an inside diameter of
2.25cm. Calculate the cross-sectional area of the
shaft
The cross-sectional area of the shaft is shown by the
shaded part in Figure 25.22 (often called an annulus).
d 5
2.25 cm
D 55.45 cm
Figure 25.22
Area of shaded part = area of large circle – area of
small circle
=
π D2
4
−
πd2
4
=
π
4
(D2
− d2
)
=
π
4
(5.452
− 2.252
)
= 19.35cm2
1. A rectangular garden measures 40m by 15m.
A 1m ﬂower border is made round the two
shorter sides and one long side. A circular
swimming pool of diameter 8m is constructed

in the middleof the garden. Find, correct to the
nearest square metre, the area remaining.
2. Determine the area of circles having (a) a
radius of 4cm (b) a diameter of 30mm (c) a
circumference of 200mm.
3. An annulus has an outside diameter of 60mm
and an inside diameter of 20mm. Determine
its area.
4. If the area of a circle is 320mm2, ﬁnd (a) its
diameter and (b) its circumference.
5. Calculate the areas of the following sectors of
circles.
(a) radius 9cm, angle subtended at centre
75◦.
(b) diameter 35mm, angle subtended at
centre 48◦37 .
6. Determine the shaded area of the template
120mm
90mm
80mm
radius
Figure 25.23
7. An archway consists of a rectangular opening
topped by a semi-circular arch, as shown in
Figure 25.24. Determine the area of the open-
ing if the width is 1m and the greatest height
is 2m.
1 m
2 m
Figure 25.24
Here are some further worked problems of common
shapes.
Problem 17. Calculate the area of a regular
octagon if each side is 5cm and the width across the
ﬂats is 12cm
An octagon is an 8-sided polygon. If radii are drawn
from the centre of the polygon to the vertices then 8
equal triangles are produced, as shown in Figure 25.25.
12cm
5 m
Figure 25.25
Area of one triangle =
1
2
× base × height
=
1
2
× 5 ×
12
2
= 15cm2
Area of octagon = 8 × 15 = 120cm2
Problem 18. Determine the area of a regular
hexagon which has sides 8cm long
A hexagon is a 6-sided polygon which may be divided
into 6 equal triangles as shown in Figure 25.26. The
angle subtended at the centre of each triangle is 360◦ ÷
6 = 60◦. The other two angles in the triangle add up to
120◦ and are equal to each other. Hence, each of the
triangles is equilateral with each angle 60◦ and each
side 8cm.
4cm
8cm
8cm
608
h
Figure 25.26
1
2
× base × height
=
1
2
× 8 × h

h is calculated using Pythagoras’ theorem:
82
= h2
+ 42
from which h = 82 − 42 = 6.928cm
Hence,
1
2
× 8 × 6.928 = 27.71cm2
Area of hexagon = 6 × 27.71
= 166.3cm2
Problem 19. Figure 25.27 shows a plan of a floor
of a building which is to be carpeted. Calculate the
area of the floor in square metres. Calculate the cost,
correct to the nearest pound, of carpeting the floor
with carpet costing £16.80perm2, assuming 30%
extra carpet is required due to wastage in fitting
2.5m
ML
K
J
4m
2m
0.6m
0.6m
0.8 m
2 m
0.8 m
30Њ
60Њ
3m
3m
2 m 3 m
I
H
G
F
A
BЈ B
C
DE
Figure 25.27
Area of floor plan
= area of triangle ABC + area of semicircle
+ area of rectangle CGLM
+ area of rectangle CDEF
− area of trapezium HIJK
Triangle ABC is equilateral since AB = BC = 3m and,
hence, angle B CB = 60◦.
sinB CB = BB /3
i.e. BB = 3sin60◦
= 2.598m.
Area of triangle ABC =
1
2
(AC)(BB )
=
1
2
(3)(2.598) = 3.897m2
Area of semicircle =
1
2
πr2
=
1
2
π(2.5)2
= 9.817m2
Area of CGLM = 5 × 7 = 35m2
Area of CDEF = 0.8 × 3 = 2.4m2
Area of HIJK =
1
2
(KH + IJ)(0.8)
Since MC = 7m then LG = 7m, hence
JI = 7 − 5.2 = 1.8m. Hence,
Area of HIJK =
1
2
(3 + 1.8)(0.8) = 1.92m2
Total floor area = 3.897 + 9.817 + 35 + 2.4 − 1.92
= 49.194m2
To allow for 30% wastage, amount of carpet required
= 1.3 × 49.194 = 63.95m2
Cost of carpet at £16.80 perm2
= 63.95 × 16.80 = £1074, correct to the nearest pound.
1. Calculate the area of a regular octagon if each
side is 20mm and the width across the flats is
48.3mm.
2. Determine the area of a regular hexagon which
has sides 25mm.
3. A plot of land is in the shape shown in
Figure 25.28. Determine
20m
20 m
20 m
30 m
10 m
20 m
30m
20 m
15m
40 m
15 m
20 m
Figure 25.28

(a) its area in hectares (1 ha = 104 m2).
(b) the length of fencing required, to the near-
est metre, to completely enclose the plot
of land.
25.4 Areas of similar shapes
Figure 25.29 shows two squares, one of which has sides
three times as long as the other.
3x
3x
x
x
(a) (b)
Figure 25.29
Area of Figure 25.29(a) = (x)(x) = x2
Area of Figure 25.29(b) = (3x)(3x) = 9x2
Hence, Figure 25.29(b) has an area (3)2; i.e., 9 times the
area of Figure 25.29(a).
In summary, the areas of similar shapes are pro-
portional to the squares of corresponding linear
dimensions.
Problem 20. A rectangular garage is shown on a
building plan having dimensions 10mm by 20mm.
If the plan is drawn to a scale of 1 to 250, determine
the true area of the garage in square metres
Area of garage on the plan = 10mm × 20mm
= 200mm2
Since the areas of similar shapes are proportional to the
squares of corresponding dimensions,
True area of garage = 200 × (250)2
= 12.5 × 106
mm2
=
12.5 × 106
106
m2
since 1m2 = 106 mm2
= 12.5m2
PracticeExercise 100 Areas of similar
1. The area of a park on a map is 500mm2. If
the 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 overall
height of 75mm corresponding to an overall
height of the actual boiler of 6m. If the area of
metal required for the model is 12500mm2,
determine, in square metres, the area of metal
required for the actual boiler.
3. The scale of an Ordnance Survey map is
1:2500. A circular sports field has a diam-
eter of 8cm on the map. Calculate its area
in hectares, giving your answer correct to
3 significant figures. (1 hectare = 104 m2.)

Chapter 26
The circle
26.1 Introduction
A circle is a plain ﬁgure enclosed by a curved line,every
point on which is equidistant from a point within, called
the centre.
In Chapter25,worked problemson theareasofcircles
and 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 of
sectors of circles. Finally, the equation of a circle is
explained.
26.2 Properties of circles
(a) The distance from the centre to the curve is called
theradius,r,ofthecircle(see OP in Figure26.1).
C
B
Q
O
P
R
A
Figure 26.1
(b) The boundary of a circle is called the circumfer-
ence, c.
(c) Any straight line passing through the centre and
touching the circumference at each end is called
the diameter, d (see QR in Figure 26.1). Thus,
d = 2r.
(d) The ratio
circumference
diameter
is a constant for any cir-
cle. This constant is denoted by the Greek letter π
(pronounced ‘pie’), where π = 3.14159, correct
to 5 decimal places (check with your calculator).
Hence,
c
d
= 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 meets
thecircleat onepoint only and doesnot cut thecir-
cle when produced. AC in Figure 26.1 isa tangent
to the circle since it touches the curve at point B
only. If radius OB is drawn, angleABO is a right
angle.
(h) The sector of a circle is the part of a circle
between radii (for example, the portion OXY of
Figure 26.2 is a sector). If a sector is less than a
semicircle it is called a minor sector; if greater
than a semicircle it is called a major sector.
X
Y
TS
R
O
Figure 26.2
(i) The chord of a circle is any straight line which
divides the circle into two parts and is termi-
nated at each end by the circumference. ST, in
Figure 26.2, is a chord.
(j) Segment is the name given to the parts into which
a circle is dividedby a chord. If thesegment is less
than a semicircle it is called a minor segment
(see shaded area in Figure 26.2). If the segment
DOI: 10.1016/B978-1-85617-697-2.00026-0

The circle 231
is greater than a semicircle it is called a major
segment (see the un-shaded area in Figure 26.2).
(k) An arc is a portion of the circumference of a cir-
cle. The distance SRT in Figure 26.2 is called
a minor arc and the distance SXYT is called a
major arc.
(l) The angle at the centre of a circle, subtended
by an arc, is double the angle at the circumfer-
ence subtended by the same arc. With reference
to Figure 26.3,
Angle AOC = 2×angle ABC
Q
A
P
C
O
B
Figure 26.3
(m) Theanglein asemicircleisaright angle(seeangle
BQP in Figure 26.3).
Problem 1. Find the circumference of a circle of
radius 12.0cm
Circumference,c = 2 × π × radius = 2πr = 2π(12.0)
= 75.40cm
Problem 2. If the diameter of a circle is 75mm,
ﬁnd its circumference
Circumference,c = π × diameter = πd = π(75)
= 235.6mm
Problem 3. Determine the radius of a circular
pond if its perimeter is 112m
Perimeter = circumference, c = 2πr
Hence, radius of pond, r =
c
2π
=
112
2π
= 17.83cm
Problem 4. In Figure 26.4, AB is a tangent to the
circle at B. If the circle radius is 40mm and
AB = 150 mm, calculate the length AO
A
B
r
O
Figure 26.4
A tangent to a circle is at right angles to a radius drawn
fromthepoint ofcontact; i.e.,ABO = 90◦.Hence,using
Pythagoras’ theorem,
AO2
= AB2
+ OB2
from which, AO = AB2 + OB2
= 1502 + 402 = 155.2mm
PracticeExercise 101 Properties of a circle
1. Calculate the length of the circumference of
a circle of radius 7.2cm.
2. If the diameter of a circle is 82.6mm, calcu-
late the circumference of the circle.
3. Determine the radius of a circle whose cir-
cumference is 16.52cm.
4. Find the diameter of a circle whose perimeter
is 149.8cm.
5. A crank mechanism is shown in Figure 26.5,
where XY is a tangent to the circle at point
X. If the circle radius OX is 10cm and length
OY is 40cm, determine the length of the
connecting rod XY.
X
Y
O 40 cm
Figure 26.5
6. If the circumference of the earth is 40000km
at the equator, calculate its diameter.

7. Calculate the length of wire in the paper clip
shown in Figure 26.6. The dimensions are in
millimetres.
2.5rad
2.5rad
3 rad
12
6
32
Figure 26.6
26.3 Radians and degrees
One radian is deﬁned as the angle subtended at the
centre of a circle by an arc equal in length to the radius.
With reference to Figure 26.7, for arc length s,
θ radians =
s
r
s
r
O r
␪
Figure 26.7
When s = whole circumference (= 2πr) then
θ =
s
r
=
2πr
r
= 2π
i.e. 2π radians = 360◦
or π radians = 180◦
Thus, 1rad =
180◦
π
= 57.30◦
, correct to 2 decimal
places.
Since π rad = 180◦, then
π
2
= 90◦,
π
3
= 60◦,
π
4
= 45◦,
and so on.
Problem 5. Convert to radians: (a) 125◦
(b) 69◦
47
(a) Since 180◦ = π rad, 1◦ =
180
π
rad, therefore
125◦
= 125
π
180
rad = 2.182 radians.
(b) 69◦47 = 69
47◦
60
= 69.783◦ (or, with your calcu-
lator, enter 69◦47 using ◦ ’ ’ ’ function, press =
and press ◦ ’ ’ ’ again).
and 69.783◦ = 69.783
π
180
rad
= 1.218 radians.
Problem 6. Convert to degrees and minutes:
(a) 0.749 radians (b) 3π/4 radians
(a) Since π rad = 180◦
,1rad =
180◦
π
therefore 0.749rad= 0.749
180
π
◦
= 42.915◦
0.915◦ = (0.915 × 60) = 55 , correct to the nea-
rest minute,
Hence, 0.749 radians = 42◦55
(b) Since 1rad =
180
π
o
then
3π
4
rad =
3π
4
180
π
o
=
3
4
(180)◦ = 135◦
Problem 7. Express in radians, in terms of π,
(a) 150◦ (b) 270◦ (c) 37.5◦
Since 180◦ = πrad,1◦ =
π
180
rad
(a) 150◦ = 150
π
180
rad =
5π
6
rad
(b) 270◦ = 270
π
180
rad =
3π
2
rad
(c) 37.5◦ = 37.5
π
180
rad =
75π
360
rad =
5π
24
rad
PracticeExercise 102 Radians and degrees
1. Convert to radians in terms of π:
(a) 30◦
(b) 75◦
(c) 225◦
2. Convert to radians, correct to 3 decimal
places:
(a) 48◦ (b) 84◦51 (c) 232◦15

The circle 233
3. Convert to degrees:
(a)
7π
6
rad (b)
4π
9
rad (c)
7π
12
rad
4. Convert to degrees and minutes:
(a) 0.0125rad (b) 2.69rad
(c) 7.241rad
26.4 Arc length and area of circles and
sectors
26.4.1 Arc length
From the definition of the radian in the previous section
and Figure 26.7,
arc length,s = rθ where θ is in radians
26.4.2 Area of a circle
From Chapter 25, for any circle, area = π × (radius)2
i.e. area = πr2
Since r =
d
2
, area = πr2
or
πd2
4
26.4.3 Area of a sector
Area of a sector =
θ
360
(πr2
) when θ is in degrees
=
θ
2π
πr2
=
1
2
r2
θ when θ is in radians
Problem 8. A hockey pitch has a semicircle of
radius 14.63m around each goal net. Find the area
enclosed by the semicircle, correct to the nearest
square metre
Area of a semicircle =
1
2
πr2
When r = 14.63m, area =
1
2
π (14.63)2
i.e. area of semicircle = 336m2
Problem 9. Find the area of a circular metal plate
having a diameter of 35.0mm, correct to the nearest
square millimetre
Area of a circle = πr2
=
πd2
4
When d = 35.0mm, area =
π (35.0)2
4
i.e. area of circular plate = 962mm2
circumference of 60.0mm
Circumference,c = 2πr
from which radius,r =
c
2π
=
60.0
2π
=
30.0
π
Area of a circle = πr2
i.e. area = π
30.0
π
2
=286.5mm2
Problem 11. Find the length of the arc of a circle
of radius 5.5cm when the angle subtended at the
centre is 1.20 radians
Length of arc, s = rθ, where θ is in radians.
Hence, arc length, s = (5.5)(1.20) = 6.60cm.
Problem 12. Determine the diameter and
circumference of a circle if an arc of length 4.75cm
subtends an angle of 0.91 radians
Since arc length, s = rθ then
radius, r =
s
θ
=
4.75
0.91
= 5.22 cm
Diameter = 2×radius = 2 × 5.22 = 10.44 cm
Circumference, c = πd = π(10.44) = 32.80 cm
Problem 13. If an angle of 125◦ is subtended by
an arc of a circle of radius 8.4cm, find the length of
(a) the minor arc and (b) the major arc, correct to
3 significant figures
Since 180◦ = π rad then 1◦ =
π
180
rad and
125◦ = 125
π
180
rad
(a) Length of minor arc,
s = rθ = (8.4)(125)
π
180
= 18.3cm,

(b) Length of major arc = (circumference − minor
arc) = 2π(8.4)−18.3 = 34.5cm, correct to 3
(Alternatively, major arc = rθ
= 8.4(360 − 125)
π
180
= 34.5cm.)
Problem 14. Determine the angle, in degrees and
minutes, subtended at the centre of a circle of
diameter 42mm by an arc of length 36mm.
Calculate also the area of the minor sector formed
Since length of arc, s = rθ then θ =
s
r
Radius, r =
diameter
2
=
42
2
= 21mm,
hence θ =
s
r
=
36
21
= 1.7143 radians.
1.7143rad = 1.7143 ×
180
π
◦
= 98.22◦ = 98◦13 =
angle subtended at centre of circle.
From equation (2),
area of sector =
1
2
r2
θ =
1
2
(21)2
(1.7143)
= 378mm2
.
Problem 15. A football stadium ﬂoodlight can
spread its illumination over an angle of 45◦ to a
distance of 55m. Determine the maximum area that
is ﬂoodlit.
Floodlit area = area of sector =
1
2
r2
θ
=
1
2
(55)2
45 ×
π
180
= 1188m2
Problem 16. An automatic garden sprayer
produces spray to a distance of 1.8m and revolves
through an angle α which may be varied. If the
desired spray catchment area is to be 2.5m2, to what
should angle α be set, correct to the nearest degree?
Area of sector =
1
2
r2θ, hence 2.5 =
1
2
(1.8)2
α
from which, α =
2.5 × 2
1.82
= 1.5432 radians
1.5432rad = 1.5432 ×
180
π
◦
= 88.42◦
Hence, angle α = 88◦, correct to the nearest degree.
Problem 17. The angle of a tapered groove is
checked using a 20mm diameter roller as shown in
Figure 26.8. If the roller lies 2.12mm below the top
of the groove, determine the value of angle θ
2.12mm
20mm
30mm
␪
Figure 26.8
In Figure 26.9, triangle ABC is right-angled at C (see
property (g) in Section 26.2).
2.12mm
2
10mm
B
A
C
30mm
␪
Figure 26.9
Length BC = 10 mm (i.e. the radius of the circle), and
AB = 30 − 10 − 2.12 = 17.88mm, from Figure 26.9.
Hence, sin
θ
2
=
10
17.88
and
θ
2
= sin−1 10
17.88
= 34◦
and angle θ = 68◦
.
PracticeExercise 103 Arc length and area
of circles and sectors (answers on page 351)
1. Calculate the area of a circle of radius 6.0cm,
correct to the nearest square centimetre.
2. The diameter of a circle is 55.0mm. Deter-
mine its area, correct to the nearest square
millimetre.

The circle 235
3. The perimeter of a circle is 150mm. Find its
area, correct to the nearest square millimetre.
4. Find the area of the sector, correct to the
nearest square millimetre, of a circle having
a radius of 35mm with angle subtended at
centre of 75◦.
5. An annulus has an outside diameter of
49.0mm and an inside diameter of 15.0mm.
Find its area correct to 4 significant figures.
6. Find the area, correct to the nearest square
metre, of a 2m wide path surrounding a
circular plot of land 200m in diameter.
7. A rectangular park measures 50m by 40m.
A 3m flower bed is made round the two
longersidesand oneshort side.Acircularfish
pond of diameter 8.0m is constructed in the
centre of the park. It is planned to grass the
remaining area. Find, correct to the nearest
square metre, the area of grass.
8. With reference to Figure26.10, determine (a)
the perimeter and (b) the area.
17cm
28cm
Figure 26.10
9. Find the area of the shaded portion of
Figure 26.11.
10m
Figure 26.11
10. Find the length of an arc of a circle of radius
8.32cm when the angle subtended at the cen-
tre is 2.14 radians. Calculate also the area of
the minor sector formed.
11. If the angle subtended at the centre of a cir-
cle of diameter 82mm is 1.46rad, find the
lengthsof the (a) minorarc and (b) major arc.
12. A pendulum of length 1.5m swings through
an angle of 10◦ in a single swing. Find, in
centimetres, the length of the arc traced by
the pendulum bob.
13. Determine the length of the radius and cir-
cumference of a circle if an arc length of
32.6cm subtends an angle of 3.76 radians.
14. Determine the angle of lap, in degrees and
minutes, if 180mm of a belt drive are in
contact with a pulley of diameter 250mm.
15. Determine the number of complete revo-
lutions a motorcycle wheel will make in
travelling 2km if the wheel’s diameter is
85.1cm.
16. A floodlightat a sports ground spread its illu-
mination over an angle of 40◦ to a distance
of 48m. Determine (a) the angle in radians
and (b) the maximum area that is floodlit.
17. Find the area swept out in 50 minutes by the
minute hand of a large floral clock if the hand
is 2m long.
18. Determine(a)theshaded areain Figure26.12
and (b)thepercentageofthewholesectorthat
the shaded area represents.
50mm
0.75 rad
12mm
Figure 26.12

19. Determine the length of steel strip required
to make the clip shown in Figure 26.13.
125mm
rad
130Њ
100mm
100mm
Figure 26.13
20. A 50◦ tapered hole is checked with a 40mm
diameter ball as shown in Figure 26.14.
Determine the length shown as x.
70 mm
x
50Њ
40mm
Figure 26.14
26.5 The equation of a circle
The simplest equation of a circle, centre at the origin
and radius r, is given by
x2
+ y2
= r2
For example, Figure 26.15 shows a circle x2 + y2 =9.
3
3
2
x2
ϩy2
ϭ9
x
y
2
1
10
Ϫ1
Ϫ1
Ϫ2
Ϫ2
Ϫ3
Ϫ3
Figure 26.15
More 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.
r5
2
y
5
4
2
0 2 4 x
b5 3
a5 2
Figure 26.16
The general equation of a circle is
x2
+ y2
+ 2ex + 2 f y + c = 0 (2)
Multiplying out the bracketed terms in equation (1)
gives
x2
− 2ax + a2
+ y2
− 2by + b2
= r2
Comparing this with equation (2) gives
2e = −2a, i.e. a = −
2e
2
and 2 f = −2b, i.e. b = −
2f
2
and c = a2
+ b2
−r2
, i.e. r = a2 + b2 − c
Thus, for example, the equation
x2
+ y2
− 4x − 6y + 9 = 0
represents a circle with centre,
a = −
−4
2
,b = −
−6
2
i.e., at (2, 3) and
radius, r = 22 + 32 − 9 = 2
Hence, x2 + y2 − 4x − 6y + 9 = 0 isthecircleshown in
Figure 26.16 (which may be checked by multiplyingout
the brackets in the equation (x − 2)2 + (y − 3)2 = 4).
Problem 18. Determine (a) the radius and (b) the
co-ordinates of the centre of the circle given by the
equation x2 + y2 + 8x − 2y + 8 = 0

The circle 237
x2
+ y2
+ 8x − 2y + 8 = 0 is of the form shown in
equation (2),
where a = −
8
2
= −4,b = −
−2
2
= 1
and r = (−4)2
+ 12 − 8 =
√
9 = 3.
Hence, x2 + y2 + 8x − 2y + 8 = 0 represents a circle
centre (−4,1) and radius 3, as shown in Figure 26.17.
a= Ϫ4
b=1
Ϫ2
2
4
y
Ϫ4Ϫ6Ϫ8 0
r=
3
x
Figure 26.17
Alternatively, x2 + y2 + 8x − 2y + 8 = 0 may be rear-
ranged as
(x + 4)2
+ (y − 1)2
− 9 = 0
i.e. (x + 4)2
+ (y − 1)2
= 32
which represents a circle, centre (−4,1) and radius 3,
as stated above.
Problem 19. Sketch the circle given by the
equation x2 + y2 − 4x + 6y − 3 = 0
The equation of a circle, centre (a,b), radius r is
given by
(x − a)2
+ (y − b)2
= r2
The general equation of a circle is
x2 + y2 + 2ex + 2 f y + c = 0
From above a = −
2e
2
,b= −
2 f
2
and r =
√
a2 + b2 − c
Hence, if x2 + y2 − 4x + 6y − 3 = 0
then a = −
−4
2
= 2,b = −
6
2
= −3 and
r = 22 + (−3)2
− (−3) =
√
16 = 4
Thus, the circle has centre (2,−3) and radius 4, as
Ϫ4
Ϫ2
2
3
4
1
y
Ϫ4
Ϫ5
Ϫ7
Ϫ8
Ϫ3
Ϫ2 2 4 6 x0
r = 4
Figure 26.18
Alternatively, x2 + y2 − 4x + 6y − 3 = 0 may be rear-
ranged as
(x − 2)2
+ (y + 3)2
− 3 − 13 = 0
i.e. (x − 2)2
+ (y + 3)2
= 42
which represents a circle, centre (2,−3) and radius 4,
as stated above.
PracticeExercise 104 The equation of a
circle (answers on page 351)
1. Determine (a) the radius and (b) the
co-ordinates of the centre of the circle given
by the equation x2 + y2 − 6x + 8y + 21 = 0.
2. Sketch the circle given by the equation
x2
+ y2
− 6x + 4y − 3 = 0.
3. Sketch the curve x2
+ (y − 1)2
− 25 = 0.
4. Sketch the curve x = 6 1 −
y
6
2

Revision Test 10 : Areas and circles
1. A rectangular metal platehas an area of 9600cm2.
If the length of the plate is 1.2m, calculate the
width, in centimetres. (3)
2. Calculate the cross-sectional area of the angle iron
section shown in Figure RT10.1, the dimensions
being in millimetres. (4)
28
27
5
19
3
4
Figure RT10.1
3. Find the area of the trapezium M N O P shown in
Figure RT10.2 when a = 6.3cm,b = 11.7cm and
h = 5.5cm. (3)
M N
O
P
h
a
b
Figure RT10.2
4. Find the area of the triangle DE F in Figure
RT10.3, correct to 2 decimal places. (4)
5. Arectangularpark measures150m by 70m.A2m
flower border is constructed round the two longer
sides and one short side. A circular fish pond of
diameter 15m is in the centre of the park and the
remainder of the park is grass. Calculate, correct
to the nearest square metre, the area of (a) the
fish pond, (b) the flower borders and (c) the grass.
(6)
6. A swimming pool is 55mlong and 10mwide. The
perpendicular depth at the deep end is 5m and at
8.75cm
12.44cm
F
E
D
Figure RT10.3
the shallow end is 1.5m, the slope from one end
to the other being uniform. The inside of the pool
needs two coats of a protective paint before it is
filled with water. Determine how many litres of
paint will be needed if 1 litre covers 10m2. (7)
7. Find the area of an equilateral triangle of side
20.0cm. (4)
8. A steel template is of the shape shown in
Figure RT10.4, the circular area being removed.
Determine the area of the template, in square
centimetres, correct to 1 decimal place. (8)
30mm
45mm
130mm
70mm
70mm 150mm
60mm
30 mm
50mm
dia.
Figure RT10.4
9. The area of a plot of land on a map is
400mm2. If the scale of the map is 1 to 50000,

Revision Test 10 : Areas and circles 239
determine 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)
20cm
2 cm
Figure RT10.5
11. Determine the diameter of a circle, correct to
the nearest millimetre, whose circumference is
178.4cm. (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 its
area, correct to the nearest square millimetre. (4)
14. Find the area of the sector of a circle having a
radius of 50.0mm, with angle subtended at centre
of 120◦. (3)
15. Determine the total area of the shape shown in
Figure RT10.6, correct to 1 decimal place. (7)
7.0 m
10.0m
6.0 m
Figure RT10.6
16. The radius of a circular cricket ground is75m. The
boundary is painted with white paint and 1 tin of
paint will paint a line 22.5m long. How many tins
of paint are needed? (3)
17. Find the area of a 1.5m wide path surrounding a
circular 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 wheel
in 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 of
24 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 to
make the clip shown in Figure RT10.7. (7)
30mm rad
15mm rad
15mm rad
70mm70mm
75mm
Figure RT10.7
22. A lorry has wheels of radius 50cm. Calculate the
number of complete revolutions a wheel makes
(correct to the nearest revolution) when travelling
3 miles (assume 1mile = 1.6km). (4)
23. The equation of a circle is
x2 + y2 + 12x − 4y + 4 = 0. Determine
(a) the diameter of the circle.
(b) the co-ordinates of the centre of the circle.
(7)

Chapter 27
Volumes of common solids
27.1 Introduction
The volume of any solid is a measure of the space occu-
pied by the solid. Volume is measured in cubic units
such as mm3, cm3 and m3.
This chapter deals with finding volumes of common
solids; in engineering it is often important to be able to
calculatevolumeorcapacity to estimate,say,theamount
of liquid, such as water, oil or petrol, in different shaped
containers.
A prism is a solid with a constant cross-section and
with two ends parallel. The shape of the end is used to
describe the prism. For example, there are rectangular
prisms (called cuboids), triangular prisms and circular
prisms (called cylinders).
On completing this chapter you will be able to cal-
culate the volumes and surface areas of rectangular and
other prisms, cylinders, pyramids, cones and spheres,
together with frusta of pyramids and cones. Volumes of
similar shapes are also considered.
27.2 Volumes and surface areas of
common shapes
27.2.1 Cuboids or rectangular prisms
A cuboid is a solid figure bounded by six rectangular
faces; all angles are right angles and opposite faces are
equal. A typical cuboid is shown in Figure 27.1 with
length l, breadth b and height h.
Volume of cuboid = l × b × h
and
surface area = 2bh + 2hl + 2lb = 2(bh + hl + lb)
h
b
l
Figure 27.1
A cube is a square prism. If all the sides of a cube are x
then
Volume = x3
and surface area = 6x2
Problem 1. A cuboid has dimensions of 12cm by
4cm by 3cm. Determine (a) its volume and (b) its
total surface area
The cuboid is similar to that in Figure 27.1, with
l = 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 cm2
Problem 2. An oil tank is the shape of a cube,
each edge being of length 1.5m. Determine (a) the
maximum capacity of the tank in m3 and litres and
(b) its total surface area ignoring input and output
orifices
DOI: 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
= 1.53
m3
= 3.375 m3
1m3 = 100cm × 100cm× 100cm = 106 cm3.
Hence,
volume of tank = 3.375 × 106
cm3
1litre = 1000cm3, hence oil tank capacity
=
3.375 × 106
1000
litres = 3375 litres
(b) Surface area of one side = 1.5m × 1.5m
= 2.25m2.
A cube has six identical sides, hence
total surface area of oil tank = 6 × 2.25
= 13.5 m2
Problem 3. A water tank is the shape of a
rectangular prism having length 2m, breadth 75cm
and height 500mm. Determine the capacity of the
tank in (a) m3 (b) cm3 (c) litres
Capacity means volume; when dealing with liquids, the
word 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 (since
1m = 100cm = 1000mm). Hence,
capacity of tank = 2 × 0.75 × 0.5 = 0.75 m3
(b) 1m3 = 1m× 1m× 1m
= 100cm × 100cm× 100cm
i.e., 1 m3
= 1 000 000 =106
cm3. Hence,
capacity = 0.75m3
= 0.75 × 106
cm3
= 750 000 cm3
(c) 1litre = 1000cm3. Hence,
750 000 cm3
=
750,000
1000
= 750 litres
27.2.2 Cylinders
A cylinder is a circular prism. A cylinder of radiusr and
height h is shown in Figure 27.2.
h
r
Figure 27.2
Volume = πr2
h
Curved surface area = 2πrh
Total surface area = 2πrh + 2πr2
Total surface area means the curved surface area plus
the area of the two circular ends.
Problem 4. A solid cylinder has a base diameter
of 12cm and a perpendicular height of 20cm.
Calculate (a) the volume and (b) the total surface
area
(a) Volume = πr2h = π ×
12
2
2
× 20
= 720π = 2262 cm3
(b) Total surface area
= 2πrh + 2πr2
= (2 × π × 6 × 20) + (2 × π × 62
)
= 240π + 72π = 312π = 980 cm2
Problem 5. A copper pipe has the dimensions
shown in Figure 27.3. Calculate the volume of
copper in the pipe, in cubic metres.
2.5m
12cm
25cm
Figure 27.3

Outer diameter, D = 25cm = 0.25m and inner diame-
ter, d = 12cm = 0.12m.
Area of cross-section of copper
=
π D2
4
−
πd2
4
=
π(0.25)2
4
−
π(0.12)2
4
= 0.0491 − 0.0113 = 0.0378m2
Hence, volume of copper
= (cross-sectional area) × length of pipe
= 0.0378 × 2.5 = 0.0945 m3
27.2.3 More prisms
A right-angled triangular prism is shown in Figure 27.4
with dimensions b,h and l.
I
b
h
Figure 27.4
Volume =
1
2
bhl
and
surface area = area of each end
+ area of three sides
Notice that the volume is given by the area of the end
(i.e. area of triangle = 1
2 bh) multiplied by the length l.
In fact, the volume of any shaped prism is given by the
area of an end multiplied by the length.
Problem 6. Determine the volume (in cm3
) of the
shape shown in Figure 27.5
12 mm
16 mm
40 mm
Figure 27.5
The solid shown in Figure 27.5 is a triangular prism.
The volume V of any prism is given by V = Ah, where
A is the cross-sectional area and h is the perpendicular
height. Hence,
volume =
1
2
× 16 × 12× 40 = 3840mm3
= 3.840 cm3
(since 1cm3 = 1000mm3)
Problem 7. Calculate the volume of the
right-angled triangular prism shown in Figure 27.6.
Also, determine its total surface area
6 cm
40cm
A
B C
8 cm
Figure 27.6
Volume of right-angled triangular prism
=
1
2
bhl =
1
2
× 8 × 6 × 40
i.e. volume = 960 cm3

Total surface area = area of each end + area of three
sides.
In triangle ABC, AC2
= AB2
+ BC2
from which, AC = AB2 + BC2 = 62 + 82
= 10cm
Hence, total surface area
= 2
1
2
bh + (AC × 40) + (BC × 40) + (AB × 40)
= (8 × 6) + (10 × 40) + (8 × 40) + (6 × 40)
= 48 + 400 + 320 + 240
i.e. total surface area = 1008 cm2
Problem 8. Calculate the volume and total
surface area of the solid prism shown in Figure 27.7
15cm
5cm
5cm
5cm
4cm
11cm
Figure 27.7
The solid shown in Figure 27.7 is a trapezoidal prism.
Volume of prism = cross-sectional area × height
=
1
2
(11 + 5)4 × 15 = 32 × 15
= 480 cm3
Surface area of prism
= sum of two trapeziums + 4 rectangles
= (2 × 32) + (5 × 15) + (11 × 15) + 2(5 × 15)
= 64 + 75 + 165 + 150 = 454 cm2
PracticeExercise 105 Volumes and surface
areas of common shapes (answers on
page 351)
1. Change a volume of 1200000cm3 to cubic
metres.
2. Change a volume of 5000mm3 to cubic
centimetres.
3. A metal cube has a surface area of 24cm2.
Determine its volume.
4. A rectangular block of wood has dimensions
of 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 tank
measuring 90cm by 60cm by 1.8m, given
1litre = 1000cm3.
6. A rectangular block of metal has dimensions
of 40mm by 25mm by 15mm. Determine its
volume in cm3. Find also its mass if the metal
has a density of 9g/cm3.
7. Determine the maximum capacity, in litres,
of a ﬁsh tank measuring 50cm by 40cm by
2.5m(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 height
50mm. Calculate
(a) its volume in cubic centimetres, correct
to 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 prism
of length 80cm and whose triangular end
has a base of 12cm and perpendicular
height 5cm.
11. A steel ingot whose volume is 2m2 is rolled
out into a plate which is 30mm thick and
1.80m wide. Calculate the length of the plate
in metres.

12. The volume of a cylinder is 75cm3. If its
height is 9.0cm, find its radius.
13. Calculate the volume of a metal tube whose
outside diameter is 8cm and whose inside
diameter is 6cm, if the length of the tube is
4m.
14. The volume of a cylinder is 400cm3. If
its radius is 5.20cm, find its height. Also
determine its curved surface area.
15. A cylinder is cast from a rectangular piece of
alloy 5cm by 7cm by 12cm. If the length of
the cylinder is to be 60cm, find its diameter.
16. Find the volume and the total surface area
of a regular hexagonal bar of metal of length
3m if each side of the hexagon is 6cm.
17. A block of lead 1.5m by 90cm by 750mm
is hammered out to make a square sheet
15mm thick. Determine the dimensions of
the square sheet, correct to the nearest cen-
timetre.
18. How long will it take a tap dripping at a rate
of 800mm3/s to fill a 3-litre can?
19. A cylinder is cast from a rectangular piece
of alloy 5.20cm by 6.50cm by 19.33cm. If
the height of the cylinder is to be 52.0cm,
determine its diameter, correct to the nearest
centimetre.
20. How much concrete is required for the con-
struction of the path shown in Figure 27.8, if
the path is 12cm thick?
2 m
1.2m
8.5 m
2.4m
3.1 m
Figure 27.8
27.2.4 Pyramids
Volume of any pyramid
=
1
3
×area of base × perpendicular height
A square-based pyramid is shown in Figure 27.9 with
base dimension x by x and the perpendicular height of
the pyramid h. For the square-base pyramid shown,
volume =
1
3
x2
h
h
x
x
Figure 27.9
Problem 9. A square pyramid has a
perpendicular height of 16cm. If a side of the base
is 6cm, determine the volume of a pyramid
Volume of pyramid
=
1
3
× area of base × perpendicular height
=
1
3
× (6 × 6) × 16
= 192 cm3
Problem 10. Determine the volume and the total
surface area of the square pyramid shown in
Figure 27.10 if its perpendicular height is 12cm.
Volume of pyramid
=
1
3
(area of base) × perpendicular height
=
1
3
(5 × 5) × 12
= 100 cm3

5cm
5cm
D
C
E
A
B
Figure 27.10
The total surface area consists of a square base and 4
equal triangles.
Area of triangle ADE
=
1
2
=
1
2
× 5 × AC
The length AC may be calculated using Pythagoras
theorem on triangle ABC, where AB = 12cm and
BC = 1
2 × 5 = 2.5cm.
AC = AB2 + BC2 = 122 + 2.52 = 12.26cm
Hence,
area of triangle ADE =
1
2
× 5 × 12.26 = 30.65cm2
Total surface area of pyramid= (5 × 5) + 4(30.65)
= 147.6 cm2
Problem 11. A rectangular prism of metal having
dimensions of 5cm by 6cm by 18cm is melted
down and recast into a pyramid having a
rectangular base measuring 6cm by 10cm.
Calculate the perpendicular height of the pyramid,
assuming no waste of metal
Volume of rectangular prism= 5 × 6 × 18 = 540cm3
Volume of pyramid
=
1
3
Hence, 540 =
1
3
× (6 × 10) × h
from which, h =
3 × 540
6 × 10
= 27cm
i.e. perpendicular height of pyramid = 27 cm
27.2.5 Cones
A cone is a circular-based pyramid. A cone of base
radius r and perpendicular height h is shown in
Figure 27.11.
Volume =
1
3
h
r
l
Figure 27.11
i.e. Volume =
1
3
πr2
h
Curved surface area = πrl
Total surface area = πrl + πr2
Problem 12. Calculate the volume, in cubic
centimetres, of a cone of radius 30mm and
perpendicular height 80mm
Volume of cone=
1
3
πr2
h =
1
3
× π × 302
× 80
= 75398.2236... mm3
1cm = 10mm and
1cm3 = 10mm × 10mm× 10mm = 103 mm3, or
1 mm3
= 10−3
cm3
Hence, 75398.2236... mm3
= 75398.2236...× 10−3 cm3
i.e.,
volume = 75.40 cm3

Alternatively, fromthe question,r = 30mm = 3cm and
h = 80mm = 8cm. Hence,
volume =
1
3
πr2
h =
1
3
× π × 32
× 8 = 75.40 cm3
Problem 13. Determine the volume and total
surface area of a cone of radius 5cm and
perpendicular height 12cm
The cone is shown in Figure 27.12.
hϭ
12cm
rϭ5 cm
l
Figure 27.12
Volume of cone =
1
3
πr2
h =
1
3
× π × 52
× 12
= 314.2 cm3
Total surfacearea=curved surfacearea+areaofbase
= πrl + πr2
From Figure 27.12, slant height l may be calculated
using Pythagoras’ theorem:
l = 122 + 52 = 13cm
Hence, total surface area= (π × 5 × 13) + (π × 52)
= 282.7 cm2
.
27.2.6 Spheres
For the sphere shown in Figure 27.13:
Volume =
4
3
πr3
and surface area = 4πr2
r
Figure 27.13
Problem 14. Find the volume and surface area of
a sphere of diameter 10cm
Since diameter= 10cm, radius, r = 5cm.
Volume of sphere =
4
3
πr3
=
4
3
× π × 53
= 523.6 cm3
Surface area of sphere = 4πr2
= 4 × π × 52
= 314.2 cm2
Problem 15. The surface area of a sphere is
201.1cm2. Find the diameter of the sphere and
hence its volume
Surface area of sphere= 4πr2.
Hence, 201.1cm2 = 4 × π × r2,
from which r2
=
201.1
4 × π
= 16.0
and radius, r =
√
16.0 = 4.0cm
from which, diameter = 2 × r = 2 × 4.0 = 8.0 cm
Volume of sphere =
4
3
πr3
=
4
3
× π × (4.0)3
= 268.1 cm3
areas of common shapes (answers on
page 351)
1. If a cone has a diameter of 80mm and a
perpendicular height of 120mm, calculate
its volume in cm3 and its curved surface
area.
2. A square pyramid has a perpendicular height
of 4cm. If a side of the base is 2.4cm long,
ﬁnd the volume and total surface area of the
pyramid.

3. A sphere has a diameter of 6cm. Determine
its volume and surface area.
4. If the volume of a sphere is 566cm3, ﬁnd its
radius.
5. A pyramid having a square base has a per-
pendicular height of 25cm and a volume of
75cm3. Determine, in centimetres, thelength
of each side of the base.
6. A cone has a base diameter of 16mm and a
perpendicular height of 40mm. Find its vol-
ume correct to the nearest cubic millimetre.
7. Determine (a) the volume and (b) the surface
area 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 be
made into ball bearings whose radii are
8.0cm. How many bearings will be produced
from the ingot, assuming 5% wastage?
27.3 Summary of volumes and surface
areas of common solids
A summary of volumes and surface areas of regular
solids is shown in Table 27.1.
Table 27.1 Volumes and surface areas of regular
solids
Rectangular prism
(or cuboid)
h
b
l Volume = l × b × h
Surface area = 2(bh + hl + lb)
Cylinder
h
r
Volume = πr2h
Triangular prism
I
b
h
Volume =
1
2
bhl
Surface area = area of each end +
area of three sides
Pyramid
h
A
Volume =
1
3
× A × h
Total surface area =
sum of areas of triangles
forming sides + area of base
Cone
h
r
l
Volume =
1
3
πr2h
Curved surface area = πrl
Total surface area = πrl + πr2
Sphere
r
Volume =
4
3
πr3
Surface area = 4πr2
27.4 More complex volumes and
surface areas
Here are some worked problems involving more com-
plex and composite solids.
Problem 16. A wooden section is shown in
Figure 27.14. Find (a) its volume in m3 and
(b) its total surface area

3m r
r5 8cm
12cm
Figure 27.14
(a) The section of wood is a prism whose end com-
prises a rectangle and a semicircle. Since the
radius of the semicircle is 8cm, the diameter is
16cm. Hence, the rectangle has dimensions12cm
by 16cm.
Area of end = (12 × 16) +
1
2
π82
= 292.5cm2
Volume of wooden section
= area of end × perpendicular height
= 292.5 × 300 = 87 750 cm3
=
87750
106
m3
,since 1m3
= 106
cm3
= 0.08775 m3
(b) The total surface area comprises the two ends
(each of area 292.5cm2), three rectangles and a
curved surface (which is half a cylinder). Hence,
total surface area
= (2 × 292.5) + 2(12 × 300)
+ (16 × 300) +
1
2
(2π × 8 × 300)
= 585 + 7200+ 4800 + 2400π
= 20 125 cm2
or 2.0125 m2
Problem 17. A pyramid has a rectangular base
3.60cm by 5.40cm. Determine the volume and total
surface area of the pyramid if each of its sloping
edges is 15.0cm
The pyramid is shown in Figure 27.15. To calculate the
volume of the pyramid, the perpendicular height EF is
required. Diagonal BD is calculated using Pythagoras’
theorem,
i.e. BD = 3.602 + 5.402 = 6.490cm
Hence, EB =
1
2
BD =
6.490
2
= 3.245cm
C
F
D
G
A
E
H
B
15.0cm
15.0cm
15.0cm
15.0cm
5.40cm
3.60cm
Figure 27.15
Using Pythagoras’ theorem on triangle BEF gives
BF2
= E B2
+ E F2
from which EF = BF2 − EB2
= 15.02 − 3.2452 = 14.64cm
Volume of pyramid
=
1
3
(area of base)(perpendicular height)
=
1
3
(3.60 × 5.40)(14.64) = 94.87 cm3
Area of triangle ADF (which equals triangle BCF)
= 1
2(AD)(FG), where G is the midpoint of AD.
Using Pythagoras’ theorem on triangle FGA gives
FG = 15.02 − 1.802 = 14.89cm
Hence, area of triangleADF =
1
2
(3.60)(14.89)
= 26.80cm2
Similarly, if H is the mid-point of AB,
FH = 15.02 − 2.702 = 14.75cm
Hence, area of triangle ABF (which equals triangle
CDF) =
1
2
(5.40)(14.75) = 39.83cm2

Total surface area of pyramid
= 2(26.80) + 2(39.83) + (3.60)(5.40)
= 53.60 + 79.66 + 19.44
= 152.7 cm2
Problem 18. Calculate the volume and total
surface area of a hemisphere of diameter 5.0cm
Volume of hemisphere =
1
2
(volume of sphere)
=
2
3
πr3
=
2
3
π
5.0
2
3
= 32.7 cm3
Total surface area
= curved surface area + area of circle
=
1
2
(surface area of sphere) + πr2
=
1
2
(4πr2
) + πr2
= 2πr2
+ πr2
= 3πr2
= 3π
5.0
2
2
= 58.9cm2
Problem 19. A rectangular piece of metal having
dimensions 4cm by 3cm by 12cm is melted down
and recast into a pyramid having a rectangular base
measuring 2.5cm by 5cm. Calculate the
perpendicular height of the pyramid
Volume of rectangular prism of metal = 4 × 3 × 12
= 144cm3
Volume of pyramid
=
1
3
(area of base)(perpendicular height)
Assuming no waste of metal,
144 =
1
3
(2.5 × 5)(height)
i.e. perpendicular height of pyramid=
144 × 3
2.5 × 5
= 34.56 cm
Problem 20. A rivet consists of a cylindrical
head, of diameter 1cm and depth 2mm, and a shaft
of diameter 2mm and length 1.5cm. Determine the
volume of metal in 2000 such rivets
Radius of cylindrical head=
1
2
cm = 0.5cm and
height of cylindrical head= 2mm = 0.2cm.
Hence, volume of cylindrical head
= πr2
h = π(0.5)2
(0.2) = 0.1571cm3
Volume of cylindrical shaft
= πr2
h = π
0.2
2
2
(1.5) = 0.0471cm3
Total volume of 1 rivet= 0.1571 + 0.0471
= 0.2042cm3
Volume of metal in 2000 such rivets
= 2000 × 0.2042 = 408.4 cm3
Problem 21. A solid metal cylinder of radius
6cm and height 15cm is melted down and recast
into a shape comprising a hemisphere surmounted
by a cone. Assuming that 8% of the metal is wasted
in the process, determine the height of the conical
portion if its diameter is to be 12cm
Volume of cylinder = πr2
h = π × 62
× 15
= 540π cm3
If 8% of metal is lost then 92% of 540π gives the
volume of the new shape, shown in Figure 27.16.
h
r
12cm
Figure 27.16

Hence, the volume of (hemisphere + cone)
= 0.92 × 540π cm3
i.e. 1
2
4
3
πr3
+
1
3
πr2
h = 0.92 × 540π
Dividing throughout by π gives
2
3
r3
+
1
3
r2
h = 0.92 × 540
Sincethediameterofthenewshapeisto be12cm,radius
r = 6cm,
hence 2
3
(6)3
+
1
3
(6)2
h = 0.92 × 540
144 + 12h = 496.8
i.e. height of conical portion,
h =
496.8 − 144
12
= 29.4 cm
Problem 22. A block of copper having a mass of
50kg is drawn out to make 500m of wire of
uniform cross-section. Given that the density of
copper is 8.91g/cm3, calculate (a) the volume of
copper, (b) the cross-sectional area of the wire and
(c) the diameter of the cross-section of the wire
(a) A density of 8.91g/cm3 means that 8.91g of cop-
per has a volume of 1cm3, or 1g of copper has a
volume of (1 ÷ 8.91)cm3.
Density =
mass
volume
from which volume =
mass
density
Hence, 50kg, i.e. 50000g, has a
volume =
mass
density
=
50000
8.91
cm3
= 5612 cm3
(b) Volume of wire = area of circular cross-section
× length of wire.
Hence, 5612cm3
= area × (500 × 100cm)
from which, area =
5612
500 × 100
cm2
= 0.1122 cm2
(c) Area of circle = πr2 or
πd2
4
hence, 0.1122=
πd2
4
from which, d =
4 × 0.1122
π
=0.3780cm
i.e. diameter of cross-section is 3.780 mm.
Problem 23. A boiler consists of a cylindrical
section of length 8m and diameter 6m, on one end
of which is surmounted a hemispherical section of
diameter 6m and on the other end a conical section
of height 4m and base diameter 6m. Calculate the
volume of the boiler and the total surface area
The boiler is shown in Figure 27.17.
8m
4m
C
I
B
R
A
Q
P
3m
6m
Figure 27.17
Volume of hemisphere, P =
2
3
πr3
=
2
3
× π × 33
= 18π m3
Volume of cylinder, Q = πr2
h = π × 32
× 8
= 72π m3
Volume of cone, R =
1
3
πr2
h =
1
3
× π × 32
× 4
= 12π m3
Total volume of boiler = 18π + 72π + 12π
= 102π = 320.4 m3
Surface area of hemisphere, P =
1
2
(4πr2
)
= 2 × π × 32
= 18πm2

Curved surface area of cylinder, Q = 2πrh
= 2 × π × 3 × 8
= 48π m2
The slant height of the cone, l, is obtained by Pythago-
ras’ theorem on triangle ABC, i.e.
l = 42 + 32 = 5
Curved surface area of cone,
R = πrl = π × 3 × 5 = 15π m2
Total surface area of boiler= 18π + 48π + 15π
= 81π = 254.5 m2
PracticeExercise 107 More complex
volumes and surface areas (answers on
page 351)
1. Find the total surface area of a hemisphere of
diameter 50mm.
2. Find (a) the volume and (b) the total surface
area of a hemisphere of diameter 6cm.
3. Determine the mass of a hemispherical cop-
per container whose external and internal
radii are 12cm and 10cm, assuming that
1cm3 of copper weighs 8.9g.
4. A metal plumb bob comprises a hemisphere
surmounted by a cone. If the diameter of the
hemisphere and cone are each 4cm and the
total length is 5cm, ﬁnd its total volume.
5. A marquee is in the form of a cylinder sur-
mounted by acone.Thetotal height is6mand
the cylindrical portion has a height of 3.5m
with a diameter of 15m. Calculate the surface
area of material needed to make the marquee
assuming 12% of the material is wasted in
the process.
6. Determine (a) the volume and (b) the total
surface 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 of
perpendicular 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 pyramid
whose sloping edges are each 15.0cm
(vii) a pyramid having an octagonal base of
side 5.0cm and perpendicular height
20cm.
7. A metal sphere weighing 24kg is melted
down and recast into a solid cone of base
radius 8.0cm. If the density of the metal is
8000kg/m3 determine
(a) the diameter of the metal sphere.
(b) the perpendicular height of the cone,
assuming that 15% of the metal is lost
in the process.
8. Find the volume of a regular hexagonal pyra-
midiftheperpendicularheight is16.0cmand
the side of the base is 3.0cm.
9. A buoy consists of a hemisphere surmounted
by a cone. The diameter of the cone and
hemisphere is 2.5m and the slant height of
the cone is 4.0m. Determine the volume and
surface area of the buoy.
10. A petrol container is in the form of a central
cylindrical portion 5.0m long with a hemi-
spherical section surmounted on each end. If
the diameters of the hemisphere and cylinder
are both 1.2m, determine the capacity of the
tank in litres (1litre = 1000cm3).
11. Figure 27.18 shows a metal rod section.
Determine its volume and total surface area.
1.00cm
radius 1.00m
2.50cm
Figure 27.18

12. Find the volume (in cm3) of the die-casting
shown in Figure 27.19. The dimensions are
in millimetres.
60
30 rad
25
50
100
Figure 27.19
13. The cross-section of part of a circular ven-
tilation shaft is shown in Figure 27.20, ends
AB and CD being open. Calculate
(a) the volume of the air, correct to
the nearest litre, contained in the
part of the system shown, neglect-
ing the sheet metal thickness (given
1litre = 1000cm3).
(b) the cross-sectional area of the sheet
metal used to make the system, in
square metres.
(c) thecost ofthesheet metal ifthematerial
costs £11.50 per square metre, assum-
ing that 25% extra metal is required due
to wastage.
500mm
A
B
2m
1.5m
1.5m
800mm
DC
Figure 27.20
27.5 Volumes and surface areas of
frusta of pyramids and cones
The frustum of a pyramid or cone is the portionremain-
ing when a part containing the vertex is cut off by a
plane parallel to the base.
The volume of a frustum of a pyramid or coneis given
by the volume of the whole pyramid or cone minus the
volume of the small pyramid or cone cut off.
The surface area of the sides of a frustum of a pyra-
mid or cone is given by the surface area of the whole
pyramid or cone minus the surface area of the small
pyramid or cone cut off. This gives the lateral surface
area of the frustum. If the total surface area of the frus-
tum is required then the surface area of the two parallel
ends are added to the lateral surface area.
There is an alternative method for ﬁnding the volume
and surface area of a frustum of a cone. With reference
to Figure 27.21,
h
R
I
r
Figure 27.21
Volume =
1
3
πh(R2
+ Rr + r2
)
Curved surface area = πl(R + r)
Total surface area = πl(R + r) + πr2
+ πR2
Problem 24. Determine the volume of a frustum
of a cone if the diameter of the ends are 6.0cm and
4.0cm and its perpendicular height is 3.6cm
(i) Method 1
A section through the vertex of a complete cone is
Using similar triangles,
AP
DP
=
DR
BR
Hence, AP
2.0
=
3.6
1.0
from which AP =
(2.0)(3.6)
1.0
= 7.2cm

4.0cm
P
E
A
D
R
B
Q
C
2.0cm
3.6cm
1.0cm
3.0cm
6.0cm
Figure 27.22
The height of the large cone= 3.6 + 7.2
= 10.8cm
Volume of frustum of cone
= volume of large cone
− volume of small cone cut off
=
1
3
π(3.0)2
(10.8) −
1
3
π(2.0)2
(7.2)
= 101.79 − 30.16 = 71.6 cm3
(ii) Method 2
From above, volume of the frustum of a cone
=
1
3
πh(R2
+ Rr +r2
)
where R = 3.0cm,
r = 2.0cm and h = 3.6cm
Hence, volume of frustum
=
1
3
π(3.6) (3.0)2
+ (3.0)(2.0) + (2.0)2
=
1
3
π(3.6)(19.0) = 71.6 cm3
Problem 25. Find the total surface area of the
frustum of the cone in Problem 24.
(i) Method 1
Curved surface area of frustum = curved surface
area of large cone − curved surface area of small
cone cut off.
From Figure 27.22, using Pythagoras’ theorem,
AB2
= AQ2
+ BQ2
from which AB = 10.82 + 3.02 = 11.21cm
and AD2
= AP2
+ DP2
from which AD = 7.22 + 2.02 = 7.47cm
Curved surface area of large cone = πrl
= π(BQ)(AB) = π(3.0)(11.21)
= 105.65cm2
and curved surface area of small cone
= π(DP)(AD) = π(2.0)(7.47) = 46.94cm2
Hence, curved surface area of frustum
= 105.65 − 46.94
= 58.71cm2
Total 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 2
From page 252, total surface area of frustum
= πl(R +r) + πr2
+ π R2
where 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

Problem 26. A storage hopper is in the shape of a
frustum of a pyramid. Determine its volume if the
ends of the frustum are squares of sides 8.0m and
4.6m, respectively, and the perpendicular height
between its ends is 3.6m
The frustum is shown shaded in Figure 27.23(a) as part
of a complete pyramid. A section perpendicular to the
base through the vertex is shown in Figure 27.23(b).
8.0m
4.6cm
8.0m
(a)
B
2.3m
1.7m 2.3m
(b)
4.0m
2.3m 3.6m
C
A
E
DG
FH
4.6cm
Figure 27.23
By similar triangles
CG
BG
=
BH
AH
From which, height
CG = BG
BH
AH
=
(2.3)(3.6)
1.7
= 4.87m
Height of complete pyramid = 3.6 + 4.87 = 8.47m
Volume of large pyramid =
1
3
(8.0)2
(8.47)
= 180.69m3
Volume of small pyramid cut off =
1
3
(4.6)2
(4.87)
= 34.35m3
Hence, volume of storage hopper= 180.69 − 34.35
= 146.3 m3
Problem 27. Determine the lateral surface area of
the storage hopper in Problem 26
The lateral surface area of the storage hopper consists
of four equal trapeziums. From Figure 27.24,
Area of trapezium PRSU =
1
2
(PR + SU)(QT )
4.6m
8.0m
P
U
T
S
O
8.0m
Q
R
4.6m
Figure 27.24
OT = 1.7m (same as AH in Figure 27.23(b) and
OQ = 3.6m. By Pythagoras’ theorem,
QT = OQ2 + OT 2 = 3.62 + 1.72 = 3.98m
Area of trapezium PRSU
=
1
2
(4.6 + 8.0)(3.98) = 25.07m2
Lateral surface area of hopper = 4(25.07)
= 100.3 m2
Problem 28. A lampshade is in the shape of a
frustum of a cone. The vertical height of the shade
is 25.0cm and the diameters of the ends are 20.0cm
and 10.0cm, respectively. Determine the area of the
material needed to form the lampshade, correct to 3
The curved surface area of a frustum of a cone
= πl(R +r) from page 252. Since the diameters of
the ends of the frustum are 20.0cm and 10.0cm, from
Figure 27.25,
r5 5.0cm
I
h525.0cm
R510.0cm
5.0cm
Figure 27.25

r = 5.0cm, R = 10.0cm
and l = 25.02 + 5.02 = 25.50cm
from Pythagoras’ theorem.
Hence, curved surface area
= π(25.50)(10.0 + 5.0) = 1201.7cm2
i.e., the area of material needed to form the lampshade
is 1200 cm2
Problem 29. A cooling tower is in the form of a
cylinder surmounted by a frustum of a cone, as
shown in Figure 27.26. Determine the volume of air
space in the tower if 40% of the space is used for
pipes and other structures
12.0m
25.0m
12.0m
30.0m
Figure 27.26
Volume of cylindrical portion = πr2
h
= π
25.0
2
2
(12.0)
= 5890m3
Volume of frustum of cone =
1
3
πh(R2
+ Rr +r2
)
where h = 30.0 − 12.0 = 18.0m,
R = 25.0 ÷ 2 = 12.5m
and r = 12.0 ÷ 2 = 6.0m.
Hence, volume of frustum of cone
=
1
3
π(18.0) (12.5)2
+ (12.5)(6.0) + (6.0)2
= 5038m3
Total volume of cooling tower = 5890 + 5038
= 10 928m3
If 40% of space is occupied then
volume of air space= 0.6 × 10928 = 6557 m3
areas of frusta of pyramidsand cones
1. The radii of the faces of a frustum of a cone
are 2.0cm and 4.0cm and the thickness of the
frustum is 5.0cm. Determine its volume and
total 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 total
surface area of the frustum if the perpendicular
distance between its ends is 8.0cm.
3. A cooling tower is in the form of a frustum of
a cone. The base has a diameter of 32.0m, the
top has a diameter of 14.0m and the vertical
height is 24.0m. Calculate the volume of the
tower and the curved surface area.
4. A loudspeaker diaphragm is in the form of a
frustum of a cone. If the end diameters are
28.0cm and 6.00cm and the vertical distance
between the ends is 30.0cm, ﬁnd the area of
material needed to cover the curved surface of
the speaker.
5. A rectangular prism of metal having dimen-
sions 4.3cm by 7.2cm by 12.4cm is melted
down and recast into a frustum of a square
pyramid, 10% of the metal being lost in the
process. If the ends of the frustum are squares
of side 3cm and 8cm respectively, ﬁnd the
thickness of the frustum.
6. Determine the volume and total surface area
of a bucket consisting of an inverted frustum
of a cone, of slant height 36.0cm and end
diameters 55.0cm and 35.0cm.
7. A cylindrical tank of diameter 2.0m and per-
pendicular height 3.0m is to be replaced by
a tank of the same capacity but in the form
of a frustum of a cone. If the diameters of
the ends of the frustum are 1.0m and 2.0m,
respectively, determine the vertical height
required.

27.6 Volumes of similar shapes
Figure 27.27 shows two cubes, one of which has sides
three times as long as those of the other.
3x
x
x
x
3x
3x
(b)(a)
Figure 27.27
Volume of Figure 27.27(a) = (x)(x)(x) = x3
Volume of Figure 27.27(b) = (3x)(3x)(3x) = 27x3
Hence, Figure 27.27(b) has a volume (3)3, i.e. 27, times
the volume of Figure 27.27(a).
Summarizing, the volumes of similar bodies are
proportional to the cubes of corresponding linear
dimensions.
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 its
model are made of the same material
Volume of model
Volume of car
=
1
50
3
since the volume of similar bodies are proportional to
the cube of corresponding dimensions.
Mass=density×volume and, since both car and model
are made of the same material,
Mass of model
Mass of car
=
1
50
3
Hence, mass of model
= (mass of car)
1
50
3
=
1000
503
= 0.008 kg or 8 g
PracticeExercise 109 Volumes of similar
1. The diameter of two spherical bearings are in
the ratio 2:5. What is the ratio of their vol-
umes?
2. An engineering component has a mass of
400g. If each of its dimensions are reduced
by 30%, determine its new mass.

Chapter 28
Irregular areas and volumes,
and mean values
28.1 Areas of irregular figures
Areas of irregular plane surfaces may be approximately
determined by using
(a) a planimeter,
(b) the trapezoidal rule,
(c) the mid-ordinate rule, or
(d) Simpson’s rule.
Such methods may be used by, for example, engineers
estimating areas ofindicatordiagrams ofsteam engines,
surveyors estimating areas of plots of land or naval
architects estimating areas of water planes or transverse
sections 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 but
all operate in a similar way. A pointer on the
planimeter is used to trace around the boundary
of the shape. This induces a movement in another
part of the instrument and a reading of this is used
to establish the area of the shape.
(b) Trapezoidal rule
To determine the area PQRS in Figure 28.1,
(i) Divide base PS into any number of equal
intervals, each of width d (the greater the
number of intervals, the greater the accu-
racy).
(ii) Accurately measure ordinates y1, y2, y3, etc.
y1
y2
y3
y4
y5
y6
y7
RQ
P
d d d d d d
S
Figure 28.1
(iii) Area PQRS
= d
y1 + y7
2
+ y2 + y3 + y4 + y5 + y6 .
In general, the trapezoidal rule states
Area =
width of
interval
1
2
ﬁrst + last
ordinate
+
⎛
⎝
sum of
remaining
ordinates
⎞
⎠
⎤
⎦
(c) Mid-ordinate rule
y1
y2
y3
y4
y5
y6
C
B
A
d d d d d d
D
Figure 28.2
To determine the area ABCD of Figure 28.2,
DOI: 10.1016/B978-1-85617-697-2.00028-4

(i) Divide base AD into any number of equal
intervals, each of width d (the greater the
numberofintervals,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 states
Area =(width of interval)(sum of
mid-ordinates)
(d) Simpson’s rule
To 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 number
of intervals, the greater the accuracy).
(ii) Accurately measure ordinates y1, y2, y3, etc.
(iii) Area PQRS =
d
3
[(y1 + y7) + 4(y2 + y4 + y6)
+2(y3 + y5)]
In general, Simpson’s rule states
Area =
1
3
width of
interval
ﬁrst + last
ordinate
+ 4
sum of even
ordinates
+ 2
sum of remaining
odd ordinates
Problem 1. A car starts from rest and its speed is
measured every second for 6s.
Time t (s) 0 1 2 3 4 5 6
Speed v (m/s) 0 2.5 5.5 8.75 12.5 17.5 24.0
Determine the distance travelled in 6 seconds (i.e.
the area under the v/t graph), using (a) the
trapezoidal rule (b) the mid-ordinate rule
(c) Simpson’s rule
A 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 of
width 1s, and the length of the ordinates measured.
Thus,
area = (1)
0 + 24.0
2
+ 2.5 + 5.5
+ 8.75 + 12.5 + 17.5 = 58.75 m
Graph of speed/time
Speed(m/s)
30
25
20
15
10
5
2.5
4.0
5.5
7.0
8.75
10.75
12.5
15.0
17.5
20.25
24.0
0 1 2 3
Time (seconds)
4 5 6
1.25
Figure 28.3
(b) Mid-ordinate rule (see (c) above)
The time base is divided into 6 strips each of
width 1s. Mid-ordinates are erected as shown in
Figure28.3 by the broken lines. The length of each
mid-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 width
1s and the length of the ordinatesmeasured. Thus,
area =
1
3
(1)[(0 + 24.0) + 4(2.5 + 8.75
+ 17.5) + 2(5.5 + 12.5)] = 58.33m
Problem 2. A river is 15m wide. Soundings of
the depth are made at equal intervals of 3m across
the river and are as shown below.
Depth (m) 0 2.2 3.3 4.5 4.2 2.4 0
Calculate the cross-sectional area of the ﬂow of
water at this point using Simpson’s rule
From (d) above,
Area =
1
3
(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 259
Practice Exercise 110 Areas of irregular
ﬁgures (answers on page 352)
1. Plot a graph of y = 3x − x2 by completing
a 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-ordinate
rule (c) Simpson’s rule.
2. Plot the graph of y = 2x2 + 3 between x = 0
and x = 4. Estimate the area enclosed by the
curve, the ordinates x = 0 and x = 4 and the
x-axis by an approximate method.
3. The velocity of a car at one second intervals is
given in the following table.
Time t (s) 0 1 2 3 4 5 6
v (m/s) 0 2.0 4.5 8.0 14.0 21.0 29.0
Velocity
Determine the distance travelled in 6 sec-
onds (i.e. the area under the v/t graph) using
Simpson’s rule.
4. The shape of a piece of land is shown in
Figure 28.4. To estimate the area of the land,
a surveyor takes measurements at intervals
of 50m, perpendicular to the straight portion
with the results shown (the dimensions being
in metres). Estimate the area of the land in
hectares (1ha = 104 m2).
50 50
200 190 180 130160140
50 50 50 50
Figure 28.4
5. The deck of a ship is 35m long. At equal inter-
vals of 5m the width is given by the following
table.
Width (m) 0 2.8 5.2 6.5 5.8 4.1 3.0 2.3
Estimate the area of the deck.
28.2 Volumes of irregular solids
If the cross-sectional areas A1, A2, A3,... of an irreg-
ular solid bounded by two parallel planes are known at
equal intervals of width d (as shown in Figure 28.5), by
Simpson’s rule
Volume, V =
d
3
[(A1 + A7) + 4(A2 + A4 + A6)
+2(A3 + A5)]
dddddd
A6 A7A2A1 A3 A4 A5
Figure 28.5
Problem 3. A tree trunk is 12m in length and has
a varying cross-section. The cross-sectional areas at
intervals of 2m measured from one end are
0.52, 0.55, 0.59, 0.63, 0.72, 0.84 and 0.97m2
.
Estimate the volume of the tree trunk
A sketch of the tree trunk is similar to that shown
in Figure 28.5 above, where d = 2m, A1 = 0.52m2,
A2 = 0.55m2, and so on.
Using Simpson’s rule for volumes gives
Volume =
2
3
[(0.52 + 0.97) + 4(0.55 + 0.63 + 0.84)
+ 2(0.59 + 0.72)]
=
2
3
[1.49 + 8.08 + 2.62] = 8.13 m3
Problem 4. The areas of seven horizontal
cross-sections of a water reservoir at intervals of
10m are 210, 250, 320, 350, 290, 230 and 170m2.
Calculate the capacity of the reservoir in litres
Using Simpson’s rule for volumes gives
Volume =
10
3
[(210 + 170) + 4(250 + 350 + 230)
+ 2(320 + 290)]
=
10
3
[380+ 3320 + 1220] = 16 400m3

16400m3
= 16400× 106
cm3
.
Since 1 litre = 1000cm3,
capacity of reservoir =
16400× 106
1000
litres
= 16400000 = 16.4 × 106
litres
PracticeExercise 111 Volumes of irregular
solids (answers on page 352)
1. The areas of equidistantly spaced sections of
the 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 the
sections are 3m apart.
2. To estimate the amount of earth to be
removed when constructing a cutting, the
cross-sectional area at intervals of 8m were
estimated 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 oftimber
of varying circular cross-section is measured
at intervals of 2m along its length and the
results are as follows. Estimate the volume of
the timber in cubic metres.
Distance from
one end(m) 0 2 4 6
Circumference(m) 2.80 3.25 3.94 4.32
Distance from
one end(m) 8 10 12
Circumference(m) 5.16 5.82 6.36
28.3 Mean or average values of
waveforms
The mean or average value, y, of the waveform shown
in Figure 28.6 is given by
y =
area under curve
length of base, b
If the mid-ordinaterule is used to find the area under the
curve, then
y =
sum of mid-ordinates
number of mid-ordinates
=
y1 + y2 + y3 + y4 + y5 + y6 + y7
7
for Figure 28.6
y1
y2
y3
y4 y5 y6 y7
d d d d
b
d d d
y
Figure 28.6
For a sine wave, the mean or average value
(a) over one complete cycle is zero (see
Figure 28.7(a)),
(b) over half a cycle is 0.637 × maximum value or
2
π
× maximum value,
(c) of a full-wave rectified waveform (see
Figure 28.7(b)) is 0.637 × maximum value,
(d) of a half-wave rectified waveform (see
Figure 28.7(c)) is 0.318 × maximum value
or
1
π
× maximum value.
V
0 t
Vm
V
0
(a) (b)
t
Vm
(c)
V
0 t
Vm
Figure 28.7

Problem 5. Determine the average values over
half a cycle of the periodic waveforms shown in
Figure 28.8
(c)
Voltage(V)
0
10
2 4 6 8
210
t (ms)
(a)
0
Voltage(V)
20
1 2 3 4
220
t (ms)
(b)
0
Current(A)
3
2
1
1 2 3 4
23
22
21 t (s)5 6
Figure 28.8
(a) Area undertriangularwaveform (a) for a half cycle
is given by
Area =
1
2
(base)(perpendicular height)
=
1
2
(2 × 10−3
)(20) = 20 × 10−3
Vs
Average value of waveform
=
area under curve
length of base
=
20 × 10−3Vs
2 × 10−3s
= 10V
(b) Area under waveform (b) for a half cycle
= (1 × 1) + (3 × 2) = 7As
Average value of waveform =
area under curve
length of base
=
7As
3s
= 2.33A
(c) A half cycle of the voltage waveform (c) is
completed in 4ms.
Area under curve =
1
2
{(3 − 1)10−3
}(10)
= 10 × 10−3
Vs
Average value of waveform =
area under curve
length of base
=
10 × 10−3
Vs
4 × 10−3s
= 2.5V
Problem 6. Determine the mean value of current
over one complete cycle of the periodic waveforms
0
Current(mA)
5
4 8 12 16 20 24 28t (ms)
0
Current(A)
2
2 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
=
1
2
(sum of parallel sides)(perpendicular
distance between parallel sides)
=
1
2
{(4 + 8) ×10−3
}(5 × 10−3
)
= 30 × 10−6
As
Mean value over one cycle =
area under curve
length of base
=
30 × 10−6 As
10 × 10−3 s
= 3mA
(b) One cycle of the saw-tooth waveform (b) is com-
pleted in 5ms.
Area under curve =
1
2
(3 × 10−3
)(2)
= 3 × 10−3
As
Mean value over one cycle =
area under curve
length of base
=
3 × 10−3 As
5 × 10−3 s
= 0.6A

Problem 7. The power used in a manufacturing
process during a 6 hour period is recorded at
intervals of 1 hour as shown below.
Time (h) 0 1 2 3 4 5 6
Power (kW) 0 14 29 51 45 23 0
Plot a graph of power against time and, by using the
mid-ordinate rule, determine (a) the area under the
curve and (b) the average value of the power
The graph of power/time is shown in Figure 28.10.
Graph of power/time
Power(kW)
50
40
30
20
10
0 1 2 3
Time (hours)
4 5 6
21.57.0 42.0 49.5 37.0 10.0
Figure 28.10
(a) Thetimebaseisdivided into 6 equal intervals,each
of width 1 hour. Mid-ordinates are erected (shown
by 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 curve
length of base
=
167kWh
6h
= 27.83 kW
Alternatively, average value
=
Problem 8. Figure 28.11 shows a sinusoidal
output voltage of a full-wave rectiﬁer. Determine,
using the mid-ordinate rule with 6 intervals, the
mean output voltage
0 308608908 1808 2708 3608
2␲
10
Voltage(V)
3␲
2
␲
2
␲
␪
Figure 28.11
Onecycleoftheoutput voltageiscompleted inπ radians
or 180◦. The base is divided into 6 intervals, each of
width 30◦. The mid-ordinate of each interval will lie at
15◦,45◦,75◦
, etc.
At 15◦ the height of the mid-ordinate is
10sin 15◦ = 2.588V,
At 45◦ the height of the mid-ordinate is
10sin 45◦
= 7.071V, and so on.
The results are tabulated below.
Mid-ordinate Height of mid-ordinate
15◦
10sin15◦
= 2.588V
45◦ 10sin45◦ = 7.071V
75◦ 10sin75◦ = 9.659V
105◦ 10sin105◦ = 9.659V
135◦ 10sin135◦ = 7.071V
165◦ 10sin165◦ = 2.588V
Sum of mid-ordinates = 38.636V
Mean or average value of output voltage
=
=
38.636
6
= 6.439V
(With a larger number of intervals a more accurate
answer may be obtained.)
For a sine wave the actual mean value is
0.637 × maximum value, which in this problem
gives 6.37V.
Problem 9. An indicator diagram for a steam
engine is shown in Figure 28.12. The base line has

been divided into 6 equally spaced intervals and the
lengths of the 7 ordinates measured with the results
shown in centimetres. Determine
(a) the area of the indicator diagram using
Simpson’s rule
(b) the mean pressure in the cylinder given that
1cm represents 100kPa.
12.0cm
3.6 4.0 3.5 2.9 2.2 1.7 1.6
Figure 28.12
(a) The width of each interval is
12.0
6
cm. Using
Simpson’s rule,
area =
1
3
(2.0)[(3.6 + 1.6) + 4(4.0 + 2.9 + 1.7)
+ 2(3.5 + 2.2)]
=
2
3
[5.2 + 34.4 + 11.4] = 34cm2
(b) Mean height of ordinates =
area of diagram
length of base
=
34
12
= 2.83cm
Since 1cm represents 100kPa,
mean pressure in the cylinder
= 2.83cm × 100kPa/cm = 283kPa
PracticeExercise 112 Mean or average
values of waveforms (answers on page 352)
1. Determine the mean value of the periodic
waveforms shown in Figure 28.13 over a half
cycle.
2. Find the average value of the periodic wave-
forms shown in Figure 28.14 over one com-
plete cycle.
(a)
0
Current(A)
2
10 20
22
t (ms)
(b)
0
Voltage(V)
100
5 10
2100
t (ms)
(c)
Current(A)
0
5
15 30
25
t (ms)
Figure 28.13
Voltage(mV)
0
10
2 4 6 8 10 t(ms)
Current(A)
0
5
2 4 6 8 10 t(ms)
Figure 28.14
3. An alternating current hasthefollowing values
at equal intervals of 5ms:
Time (ms) 0 5 10 15 20 25 30
Current (A) 0 0.9 2.6 4.9 5.8 3.5 0
Plot a graph of current against time and esti-
mate the area under the curve over the 30ms
period, using the mid-ordinate rule, and deter-
mine its mean value.
4. Determine, using an approximate method, the
average value of a sine wave of maximum
value 50V for (a) a half cycle (b) a complete
cycle.
5. An indicator diagram of a steam engine is
12cm long. Seven evenly spaced ordinates,
including the end ordinates, are measured as
follows:
5.90, 5.52, 4.22, 3.63, 3.32, 3.24 and 3.16cm.
Determine the area of the diagram and the
mean pressure in the cylinder if 1cm repre-
sents 90kPa.

Revision Test 11 : Volumes, irregular areas and volumes, and mean values
1. A rectangular block of alloy has dimensions of
60mm by 30mm by 12mm. Calculate the volume
of the alloy in cubic centimetres. (3)
2. Determine how many cubic metres of concrete are
required for a 120m long path, 400mm wide and
10cm deep. (3)
3. Find the volume of a cylinder of radius 5.6cm
and height 15.5cm. Give the answer correct to the
nearest 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 40
revolutions? (4)
5. Find the volume of a cone of height 12.5cm and
base diameter 6.0cm, correct to 1 decimal place.
(3)
6. Find (a) the volume and (b) the total surface
area of the right-angled triangular prism shown in
Figure RT11.1. (9)
9.70cm
4.80cm
11.6cm
Figure RT11.1
7. A pyramid having a square base has a volume
of 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 80mm
and an outside diameter of 100mm. Calculate, in
cubic metres, the volume of copper in the pipe. (4)
9. Find (a) the volume and (b) the surface area of a
sphere of diameter 25mm. (4)
10. A piece of alloy with dimensions 25mm by
60mm by 1.60m is melted down and recast into
a cylinder whose diameter is 150mm. Assum-
ing no wastage, calculate the height of the
cylinder in centimetres, correct to 1 decimal
place. (4)
11. Determine the volume (in cubic metres) and the
total surface area (in square metres) of a solid
metal cone of base radius 0.5m and perpendicular
height 1.20m. Give answers correct to 2 decimal
places. (6)
12. A rectangular storage container has dimensions
3.2m by 90cm by 60cm. Determine its volume in
(a) m3 (b) cm3. (4)
13. Calculate (a) the volume and (b) the total surface
area of a 10cm by 15cm rectangular pyramid of
height 20cm. (8)
14. A water container is of the form of a central cylin-
drical part 3.0m long and diameter 1.0m, with a
hemispherical section surmounted at each end as
shown in Figure RT11.2. Determine the maximum
capacity of the container, correct to the nearest
litre. (1 litre = 1000cm3.)
3.0m
1.0m
Figure RT11.2
(5)
15. Find the total surface area of a bucket consist-
ing of an inverted frustum of a cone of slant
height 35.0cm and end diameters 60.0cm and
40.0cm. (4)
16. A boat has a mass of 20000kg. A model of the
boat is made to a scale of 1 to 80. If the model is
made of the same material as the boat, determine
the mass of the model (in grams). (3)

Revision Test 11 : Volumes, irregular areas and volumes, and mean values 265
17. 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 ordinates
x = 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 inside
diameter of the tower at different heights is given
in the following table.
Height (m) 0 5.0 10.0 15.0 20.0
Diameter (m) 16.0 13.3 10.7 8.6 8.0
Determine the area corresponding to each diame-
ter and hence estimate the capacity of the tower in
cubic metres. (7)
19. A vehicle starts from rest and its velocity is
measured every second for 6 seconds, with the
following results.
Time t (s) 0 1 2 3 4 5 6
Velocity v (m/s) 0 1.2 2.4 3.7 5.2 6.0 9.2
Using Simpson’s rule, calculate
(a) the distance travelled in 6s (i.e. the area under
the v/t graph),
(b) the average speed over this period. (6)

Chapter 29
Vectors
29.1 Introduction
This chapter initially explains the difference between
scalar and vector quantities and shows how a vector is
drawn and represented.
Any object that is acted upon by an external force
will respond to that force by moving in the line of
the force. However, if two or more forces act simul-
taneously, the result is more difficult to predict; the
ability to add two or more vectors then becomes
important.
This chapter thus shows how vectors are added and
subtracted, both by drawing and by calculation, and how
finding the resultant of two or more vectors has many
uses in engineering. (Resultant means the single vector
which would have the same effect as the individual vec-
tors.) Relative velocities and vector i,j,k notation are
also briefly explained.
29.2 Scalars and vectors
The time taken to fill a water tank may be measured as,
say, 50s. Similarly, the temperature in a room may be
measured as, say, 16◦C or the mass of a bearing may be
measured as, say, 3kg. Quantities such as time, temper-
ature and mass are entirely defined by a numerical value
and are called scalars or scalar quantities.
Not all quantities are like this. Some are defined
by more than just size; some also have direction. For
example, the velocity of a car may be 90km/h due
west, a force of 20N may act vertically downwards,
or an acceleration of 10m/s2 may act at 50◦ to the
horizontal.
Quantities such as velocity, force and acceleration,
which have both a magnitude and a direction, are
called vectors.
PracticeExercise 113 Scalar and vector
quantities (answers on page 352)
1. State the difference between scalar and vector
quantities.
In problems 2 to 9, state whether the quantities
given are scalar or vector.
2. A temperature of 70◦C
3. 5m3 volume
4. A downward force of 20N
5. 500J of work
6. 30cm2 area
7. A south-westerly wind of 10knots
8. 50m distance
9. An acceleration of 15m/s2 at 60◦ to the
horizontal
29.3 Drawing a vector
A vector quantity can be represented graphically by a
line, drawn so that
(a) the length of the line denotes the magnitude of the
quantity, and
(b) the direction of the line denotes the direction in
which the vector quantity acts.
An arrow is used to denote the sense, or direction, of the
vector.
The arrow end of a vector is called the ‘nose’ and the
other end the ‘tail’. For example, a force of 9N acting
DOI: 10.1016/B978-1-85617-697-2.00029-6

Vectors 267
at 45◦
to the horizontal is shown in Figure 29.1. Note
that an angle of +45◦ is drawn from the horizontal and
moves anticlockwise.
9N
0
a
45Њ
Figure 29.1
A velocity of 20m/s at −60◦ is shown in Figure 29.2.
Note that an angle of −60◦ is drawn from the horizontal
and moves clockwise.
60Њ
20m/s
0
b
Figure 29.2
29.3.1 Representing a vector
There are a number of ways of representing vector
quantities. These include
(a) Using bold print.
(b)
−→
AB where an arrow above two capital letters
denotes the sense of direction, where A is the
starting 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 be
represented as
0a or
−→
0a or 0a
The magnitude of the force is 0a.
Similarly, the velocity of 20m/s at −60◦ shown in
Figure 29.2 may be represented as
0b or
−→
0b or 0b
The magnitude of the velocity is 0b.
In this chapter a vector quantity is denoted by bold
print.
29.4 Addition of vectors by drawing
Adding two or more vectors by drawing assumes that
a ruler, pencil and protractor are available. Results
obtained by drawing are naturally not as accurate as
those obtained by calculation.
(a) Nose-to-tail method
Two force vectors, F1 and F2, are shown in
Figure 29.3. When an object is subjected to more
than one force, the resultant of the forces is found
by the addition of vectors.
␪
F2
F1
Figure 29.3
To 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 at
angle 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.
␪
F2
F1
a
b
0
Figure 29.4
(b) Parallelogram method
To add the two force vectors, F1 and F2 of
Figure 29.3,
(i) A line cb is constructed which is parallel to
and equal in length to 0a (see Figure 29.5).
(ii) A line ab is constructed which is parallel to
and equal in length to 0c.
(iii) The resultant force is given by the diagonal
of the parallelogram; i.e., length 0b.
This procedure is called the ‘parallelogram’
method.

0
c
F1
F2
a
b
Figure 29.5
Problem 1. A force of 5N is inclined at an angle
of 45◦ to a second force of 8N, both forces acting at
a point. Find the magnitude of the resultant of these
two forces and the direction of the resultant with
respect to the 8N force by (a) the nose-to-tail
method and (b) the parallelogram method
The two forces are shown in Figure 29.6. (Although the
8N force is shown horizontal, it could have been drawn
in any direction.)
45Њ
5N
8N
Figure 29.6
(a) Nose-to-tail method
(i) The 8N force is drawn horizontally 8 units
long, shown as 0a in Figure 29.7.
(ii) From the nose of the 8N force, the 5N force
is drawn 5 units long at an angle of 45◦ to the
horizontal, shown as ab.
(iii) The resultant force is given by length 0b and
is measured as 12 N and angle θ is measured
as 17◦.
␪ 458
5N
8N a
0
b
Figure 29.7
(b) Parallelogram method
(i) In Figure 29.8, a line is constructed which
is parallel to and equal in length to the 8N
force.
(ii) A line is constructed which is parallel to and
equal in length to the 5N force.
␪
45Њ
5N
8N
b
0
Figure 29.8
(iii) The resultant force is given by the diagonal
of the parallelogram, i.e. length 0b, and is
measured as 12 N and angle θ is measured
as 17◦.
Thus, the resultant of the two force vectors in
Figure 29.6 is 12 N at 17◦ to the 8 N force.
Problem 2. Forces of 15 and 10N are at an angle
of 90◦ to each other as shown in Figure 29.9. Find,
by drawing, the magnitude of the resultant of these
two forces and the direction of the resultant with
respect to the 15N force
15N
10N
Figure 29.9
Using the nose-to-tail method,
(i) The 15N force is drawn horizontally 15 units
long, as shown in Figure 29.10.
10N
R
15N
␪
Figure 29.10
(ii) From the nose of the 15N force, the 10N force
is drawn 10 units long at an angle of 90◦ to the
horizontal as shown.

Vectors 269
(iii) The resultant force is shown as R and is measured
as 18N and angle θ is measured as 34◦.
Thus, the resultant of the two force vectors is 18 N at
34◦ to the 15 N force.
Problem 3. Velocities of 10m/s, 20m/s and
15m/s act as shown in Figure 29.11. Determine, by
drawing, the magnitude of the resultant velocity and
its direction relative to the horizontal
15Њ
␯3
␯2
␯1
30Њ
10m/s
20m/s
15m/s
Figure 29.11
When more than 2 vectors are being added the nose-to-
tail method is used. The order in which the vectors are
added does not matter. In this case the order taken is ν1,
then ν2, then ν3. However, if a different order is taken
the same result will occur.
(i) ν1 is drawn 10 units long at an angle of 30◦ to the
horizontal, shown as 0a in Figure 29.12.
b
195Њ
105Њ
0
30Њ
␯1
␯3
␯2
r
a
Figure 29.12
(ii) From the nose of ν1, ν2 is drawn 20 units long at
an angle of 90◦ to the horizontal, shown as ab.
(iii) From the nose of ν2, ν3 is drawn 15 units long at
an angle of 195◦ to the horizontal, shown as br.
(iv) The resultant velocity is given by length 0r and
is measured as 22m/s and the angle measured to
the horizontal is 105◦.
Thus, the resultant of the three velocities is 22 m/s at
105◦ to the horizontal.
Worked Examples 1 to 3 have demonstrated how vec-
tors are added to determine their resultant and their
direction. However, drawing to scale is time-consuming
and not highly accurate. The following sections demon-
strate how to determine resultant vectors by calculation
using horizontal and vertical components and, where
possible, by Pythagoras’ theorem.
29.5 Resolving vectors into horizontal
and vertical components
A force vector F is shown in Figure 29.13 at angle θ
to the horizontal. Such a vector can be resolved into
two components such that the vector addition of the
components is equal to the original vector.
␪
F
Figure 29.13
The two components usually taken are a horizontal
component and a vertical component.Ifaright-angled
triangle is constructed as shown in Figure 29.14, 0a is
called the horizontal component of F and ab is called
the vertical component of F.
0
a
F
b
␪
Figure 29.14
From trigonometry(see Chapter 21 and remember SOH
CAH TOA),
cosθ =
0a
0b
, from which 0a = 0bcosθ = F cosθ

i.e. the horizontal component of F = F cosθ, and
sinθ =
ab
0b
from which,ab = 0bsinθ = F sinθ
i.e. the vertical component of F = F sinθ.
Problem 4. Resolve the force vector of 50N at an
angle of 35◦ to the horizontal into its horizontal and
vertical components
The 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 in
Figure 29.15.
358
0
40.96N
28.68N
50N
a
b
Figure 29.15
(To check: by Pythagoras,
0b = 40.962 + 28.682 = 50N
and θ = tan−1 28.68
40.96
= 35◦
Thus, the vector addition of components 40.96N and
28.68N is 50N at 35◦.)
Problem 5. Resolve the velocity vector of 20m/s
at an angle of −30◦ to the horizontal into horizontal
and vertical components
The 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.
Figure 29.16.
Problem 6. Resolve the displacement vector of
40m at an angle of 120◦
into horizontal and vertical
components
308
20m/s
210m/s
17.32m/s
b
a
0
Figure 29.16
The horizontal component of the 40m displacement,
0a = 40cos120◦ = −20.0m.
The vertical component of the 40m displacement,
ab = 40sin120◦ = 34.64m.
Figure 29.17.
220.0N
40 N
1208
0
34.64N
a
b
Figure 29.17
29.6 Addition of vectors by
calculation
Two forcevectors, F1 and F2,areshown in Figure29.18,
F1 being at an angle of θ1 and F2 at an angle of θ2.
F1 F2
F1sin␪1
F2sin␪2
H
V
␪1
␪2
F2 cos ␪2
F1 cos ␪1
Figure 29.18
A method of adding two vectors together is to use
horizontal and vertical components.
The horizontal component of force F1 is F1 cosθ1 and
the horizontal component of force F2 is F2 cosθ2. The

Vectors 271
total horizontal component of the two forces,
H = F1 cosθ1 + F2 cosθ2
The vertical component of force F1 is F1 sinθ1 and the
vertical component of force F2 is F2 sinθ2. The total
vertical component of the two forces,
V = F1 sinθ1 + F2 sinθ2
Since we have H and V , the resultant of F1 and F2
is obtained by using the theorem of Pythagoras. From
Figure 29.19,
0b2
= H2
+ V 2
i.e. resultant = H2 + V 2 at an angle
given by θ = tan−1 V
H
V
b
Resultant
a
H
0
␪
Figure 29.19
Problem 7. A force of 5N is inclined at an angle
of 45◦ to a second force of 8N, both forces acting at
a point. Calculate the magnitude of the resultant of
these two forces and the direction of the resultant
with respect to the 8N force
The two forces are shown in Figure 29.20.
458
8N
5N
Figure 29.20
The horizontal component of the 8N force is 8cos0◦
and the horizontal component of the 5N force is
5cos45◦. The total horizontal component of the two
forces,
H = 8cos0◦
+ 5cos45◦
= 8 + 3.5355 = 11.5355
The vertical component of the 8N force is 8sin0◦
and
the vertical component of the 5N force is 5sin45◦. The
total vertical component of the two forces,
V = 8sin0◦
+ 5sin45◦
= 0 + 3.5355 = 3.5355
␪
Resultant
H ϭ11.5355 N
V ϭ3.5355N
Figure 29.21
From Figure 29.21, magnitude of resultant vector
= H2 + V 2
= 11.53552 + 3.53552 = 12.07N
The direction of the resultant vector,
θ = tan−1 V
H
= tan−1 3.5355
11.5355
= tan−1
0.30648866... = 17.04◦
Thus, the resultant of the two forces is a single vector
of 12.07 N at 17.04◦ to the 8 N vector.
Problem 8. Forces of 15N and 10N are at an
angle of 90◦ to each other as shown in Figure 29.22.
Calculate the magnitude of the resultant of these
two forces and its direction with respect to the
15N force
10N
15 N
Figure 29.22
The horizontal component of the 15N force is 15cos0◦
and the horizontal component of the 10N force is
10cos90◦. The total horizontal component of the two
velocities,
H = 15cos0◦
+ 10cos90◦
= 15 + 0 = 15

The vertical component of the 15N force is 15sin0◦
and
the vertical component of the 10N force is 10sin90◦.
The total vertical component of the two velocities,
V = 15sin0◦
+ 10sin90◦
= 0 + 10 = 10
Magnitude of resultant vector
=
√
H2 + V 2 =
√
152 + 102 = 18.03N
θ = tan−1 V
H
= tan−1 10
15
= 33.69◦
Thus, the resultant of the two forces is a single vector
of 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 N
10N
R
␪
Figure 29.23
Since a right-angled triangle results, we could use
Pythagoras’ theorem without needing to go through the
procedure for horizontal and vertical components. In
fact, the horizontal and vertical components are 15N
and 10N respectively.
This is, of course, a special case. Pythagoras can only
beusedwhenthere isanangleof90◦ betweenvectors.
This is demonstrated in worked Problem 9.
Problem 9. Calculate the magnitude and
direction of the resultant of the two acceleration
vectors shown in Figure 29.24.
15m/s2
28m/s2
Figure 29.24
The 15m/s2
acceleration is drawn horizontally, shown
as 0a in Figure 29.25.
015a
28
b
R
␣ ␪
Figure 29.25
From the nose of the 15m/s2 acceleration, the 28m/s2
acceleration 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 of
Pythagoras may be used.
0b = 152 + 282 = 31.76m/s2
and α = tan−1 28
15
= 61.82◦
Measuring from the horizontal,
θ = 180◦ − 61.82◦ = 118.18◦
Thus, the resultant of the two accelerations is a single
vector of 31.76 m/s2 at 118.18◦ to the horizontal.
Problem 10. Velocities of 10m/s, 20m/s and
15m/s act as shown in Figure 29.26. Calculate the
magnitude of the resultant velocity and its direction
relative to the horizontal
20 m/s
10 m/s
15m/s
158
308
␯1
␯2
␯3
Figure 29.26
The horizontal component of the 10m/s velocity
= 10cos30◦ = 8.660 m/s,

Vectors 273
the horizontal component of the 20m/s velocity is
20cos90◦ = 0 m/s
and the horizontal component of the 15m/s velocity is
15cos195◦ = −14.489m/s.
The total horizontal component of the three velocities,
H = 8.660 + 0 − 14.489 = −5.829m/s
The vertical component of the 10m/s velocity
= 10sin30◦ = 5m/s,
the vertical component of the 20m/s velocity is
20sin90◦ = 20 m/s
and the vertical component of the 15m/s velocity is
15sin195◦
= −3.882 m/s.
The total vertical component of the three forces,
V = 5 + 20 − 3.882 = 21.118m/s
5.829
21.118
R
␣ ␪
Figure 29.27
From Figure 29.27, magnitude of resultant vector,
R = H2 + V 2 = 5.8292 + 21.1182 = 21.91m/s
α = tan−1 V
H
= tan−1 21.118
5.829
= 74.57◦
Measuring from the horizontal,
θ = 180◦ − 74.57◦ = 105.43◦.
Thus, the resultant of the three velocities is a single
vector of 21.91 m/s at 105.43◦ to the horizontal.
PracticeExercise 114 Addition of vectors
by calculation (answers on page 352)
1. A force of 7N is inclined at an angle of 50◦ to
a second force of 12N, both forces acting at
a point. Calculate the magnitude of the resul-
tant of the two forces and the direction of the
resultant with respect to the 12N force.
2. Velocities of 5m/s and 12m/s act at a point
at 90◦ to each other. Calculate the resultant
velocity and itsdirectionrelativeto the12m/s
velocity.
3. Calculate the magnitude and direction of the
resultant of the two force vectors shown in
Figure 29.28.
10N
13N
Figure 29.28
resultant of the two force vectors shown in
Figure 29.29.
22N
18N
Figure 29.29
5. A displacement vector s1 is 30m at 0◦. A
second displacement vector s2 is 12m at 90◦
.
Calculate the magnitude and direction of the
resultant vector s1 + s2
6. Three forces of 5N, 8N and 13N act as
shown in Figure 29.30. Calculate the mag-
nitude and direction of the resultant force.
5 N
13N
8 N
708
608
Figure 29.30

7. If velocity v1 = 25m/s at 60◦ and
v2 = 15m/sat −30◦,calculatethemagnitude
and direction of v1 + v2.
resultant vector of the force system shown in
Figure 29.31.
308
158
608
6 N
8 N
4 N
Figure 29.31
9. Calculate the magnitude and direction of
the resultant vector of the system shown in
Figure 29.32.
158
458
2m/s
4m/s
3.5m/s
308
Figure 29.32
10. An object is acted upon by two forces of mag-
nitude10N and 8N at an angle of 60◦ to each
other. Determine the resultant force on the
object.
11. A ship heads in a direction of E 20◦S at a
speed of 20knots while the current is 4knots
in a directionof N 30◦E. Determine the speed
and actual direction of the ship.
29.7 Vector subtraction
In Figure 29.33, a force vector F is represented by
oa. The vector (−oa) can be obtained by drawing a
vector from o in the opposite sense to oa but having
the same magnitude, shown as ob in Figure 29.33; i.e.,
ob = (−oa).
b
2F
F
a
o
Figure 29.33
For two vectors acting at a point, as shown in
Figure 29.34(a), the resultant of vector addition is
os = oa + ob
Figure 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 ab
in Figure 29.34(a) shows that the second diagonal of
the parallelogram method of vector addition gives the
magnitude and direction of vector subtraction of oa
from ob.
(b)(a)
a2a
d bb s
ao o
Figure 29.34
Problem 11. Accelerations of a1 = 1.5m/s2 at
90◦
and a2 = 2.6m/s2
at 145◦
act at a point. Find
a1 + a2 and a1 − a2 by (a) drawing a scale vector
diagram and (b) calculation
(a) The scale vector diagram is shown in Figure 29.35.
By measurement,
a1 + a2 = 3.7 m/s2
at 126◦
a1 − a2 = 2.1 m/s2
at 0◦
(b) Resolving horizontally and vertically gives
Horizontal component of a1 + a2,
H = 1.5cos90◦ + 2.6cos145◦ = −2.13

Vectors 275
a12a2
a11a2
2.6m/s2
1268
a1
2a2
a2
a1
1.5m/s2
1458
Figure 29.35
Vertical component of a1 + a2,
V = 1.5sin90◦ + 2.6sin145◦ = 2.99
From Figure 29.36, the magnitude of a1 + a2,
R = (−2.13)2 + 2.992 = 3.67m/s2
In Figure 29.36, α = tan−1 2.99
2.13
= 54.53◦ and
θ = 180◦ − 54.53◦ = 125.47◦
Thus, a1 + a2 = 3.67m/s2
at 125.47◦
R
02.13
2.99
␣
␪
Figure 29.36
Horizontal component of a1 − a2
= 1.5cos90◦
− 2.6cos145◦
= 2.13
Vertical component of a1 − a2
= 1.5sin90◦ − 2.6sin145◦ = 0
Magnitude of a1 − a2 =
√
2.132 + 02
= 2.13m/s2
Direction of a1 − a2 = tan−1 0
2.13
= 0◦
Thus, a1 − a2 = 2.13m/s2
at 0◦
Problem 12. Calculate the resultant of
(a) v1 − v2 + v3 and (b) v2 − v1 − v3 when
v1 = 22 units at 140◦, v2 = 40 units at 190◦ and
v3 = 15 units at 290◦
(a) The vectors are shown in Figure 29.37.
15
40
22
1408
1908
2908
2H 1H
1V
2V
Figure 29.37
The horizontal component of v1 − v2 + v3
= (22cos140◦) − (40cos190◦) + (15cos290◦)
= (−16.85) − (−39.39) + (5.13) = 27.67 units
The vertical component of v1 −v2 +v3
= (22sin140◦) − (40sin190◦) + (15sin290◦)
= (14.14) − (−6.95) + (−14.10) = 6.99 units
The magnitude of the resultant,
R =
√
27.672 + 6.992 = 28.54 units
The direction of the resultant
R = tan−1 6.99
27.67
= 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 units
The vertical component of v2 − v1 − v3
= (40sin190◦) − (22sin140◦) − (15sin290◦)
= (−6.95) − (14.14) − (−14.10) = −6.99 units
From Figure 29.38, the magnitude of the resul-
tant, R = (−27.67)2 + (−6.99)2 = 28.54 units
and α = tan−1 6.99
27.67
= 14.18◦, from which,
θ = 180◦ + 14.18◦ = 194.18◦

R
0
27.67
6.99
␪
␣
Figure 29.38
Thus, 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 at
194.18◦ is minus times (i.e. is 180◦ out of phase
with) the vector 28.54 units at 14.18◦
PracticeExercise 115 Vector subtraction
1. Forces of F1 = 40 N at 45◦ and F2 = 30 N at
125◦
act at a point. Determine by drawing and
by calculation (a) F1 + F2 (b) F1 − F2
2. 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 velocity
For relative velocity problems, some fixed datum point
needs to be selected. This is often a fixed point on the
earth’s surface. In any vector equation, only the start
and finish points affect the resultant vector of a system.
Two different systems are shown in Figure 29.39, but,
in each of the systems, the resultant vector is ad.
a d
b
(a)
a
d
b
c
(b)
Figure 29.39
The vector equation of the system shown in
Figure 29.39(a) is
ad = ab + bd
and that for the system shown in Figure 29.39(b) is
ad = ab + bc + cd
Thus, in vector equations of this form, only the first and
last letters, a and d, respectively, fix the magnitude and
direction of the resultant vector. This principle is used
in relative velocity problems.
Problem 13. Two cars, P and Q, are travelling
towards the junction of two roads which are at right
angles to one another. Car P has a velocity of
45km/h due east and car Q a velocity of 55km/h
due south. Calculate (a) the velocity of car P
relative to car Q and (b) the velocity of car Q
relative to car P
(a) The directions of the cars are shown in
Figure 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 velocity
of P relative to Q is vector pq and the vector equa-
tion is pq = pe + eq. Hence, the vector directions
are as shown, eq being in the opposite direction
to qe.
(a) (b) (c)
Q
P
E
N
W
S
55 km/h
45km/h
p
e
q
p
e
q
Figure 29.40
From the geometry of the vector triangle, the mag-
nitude of pq =
√
452 + 552 = 71.06km/h and the
direction of pq = tan−1 55
45
= 50.71◦
That is, the velocity of car P relative to car Q is
71.06 km/h at 50.71◦
(b) The velocity of car Q relative to car P is given by
the vector equation qp = qe + ep and the vector
diagram is as shown in Figure 29.40(c), having ep
opposite in direction to pe.

Vectors 277
From the geometry of thisvector triangle, the mag-
nitude of qp =
√
452 + 552 = 71.06m/s and the
direction of qp = tan−1 55
45
= 50.71◦ but must
lie in the third quadrant; i.e., the required angle is
180◦ + 50.71◦ = 230.71◦
That is, the velocity of car Q relative to car P is
71.06 m/s at 230.71◦
PracticeExercise 116 Relative velocity
1. A car is moving along a straight horizontal
road at 79.2km/h and rain is falling vertically
downwards at 26.4km/h. Find the velocity of
the rain relative to the driver of the car.
2. Calculate the time needed to swim across a
river 142m wide when the swimmer can swim
at 2km/h in still water and the river is ﬂowing
at 1km/h. At what angle to the bank should
the swimmer swim?
3. A ship is heading in a direction N 60◦
E at a
speed which in still water would be 20km/h.
It is carried off course by a current of 8km/h
in a direction of E 50◦S. Calculate the ship’s
actual speed and direction.
29.9 i, j and k notation
A method of completely specifying the direction of a
vector in space relative to some reference point is to use
three unit vectors, i,j and k, mutually at right angles
to each other, as shown in Figure 29.41.
y
x
z
k
o
j
i
Figure 29.41
Calculations involving vectors given in i,j,k nota-
tion are carried out in exactly the same way as standard
algebraic calculations, as shown in the worked examples
below.
Problem 14. Determine
(3i + 2j + 2k) − (4i − 3j + 2k)
(3i + 2j + 2k) − (4i − 3j + 2k)
= 3i + 2j + 2k − 4i + 3j − 2k
= −i + 5j
Problem 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 + 15k
PracticeExercise 117 i,j,k notation
Given that p = 2i + 0.5j − 3k,q = −i + j + 4k
and r = 6j − 5k, evaluate and simplify the follow-
ing vectors in i, j,k form.
1. −q 2. 2p
3. q + r 4. −q + 2p
5. 3q + 4r 6. q − 2p
7. p + q + r 8. p + 2q + 3r
9. 2p + 0.4q + 0.5r 10. 7r − 2q

Chapter 30
Methods of adding
alternating waveforms
30.1 Combining two periodic
functions
There are a number of instances in engineering and sci-
ence where waveforms have to be combined and where
it is required to determine the single phasor (called
the resultant) that could replace two or more separate
phasors. Uses are found in electrical alternating cur-
rent theory, in mechanical vibrations, in the addition of
forces and with sound waves.
There are a number of methods of determining the
resultant 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 functions
This 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 in
the following worked problems.
Problem 1. Plot the graph of y1 = 3sin A from
A = 0◦ to A = 360◦. On the same axes plot
y2 = 2cos A. By adding ordinates, plot
yR = 3sin A + 2cos A and obtain a sinusoidal
expression for this resultant waveform
y1 = 3sin A and y2 = 2cos A are shown plotted in
Figure 30.1. Ordinates may be added at, say, 15◦
intervals. For example,
at 0◦, y1 + y2 = 0 + 2 = 2
at 15◦, y1 + y2 = 0.78 + 1.93 = 2.71
at 120◦, y1 + y2 = 2.60 + −1 = 1.6
at 210◦, y1 + y2 = −1.50 − 1.73 = −3.23, and
so on.
y
y15 3 sin A
y25 2 cos A
yR 53.6 sin(A 1348)
A0
23
22
21
3
3.6
2
1
348
908 1808 2708 3608
Figure 30.1
The 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 resultant
DOI: 10.1016/B978-1-85617-697-2.00030-2

Methods of adding alternating waveforms 279
waveform leads y1 = 3sin A by 34◦ or 34 ×
π
180
rad =
0.593rad.
The sinusoidal expression for the resultant waveform is
yR = 3.6sin(A + 34◦
) or yR = 3.6sin(A + 0.593)
Problem 2. Plot the graphs of y1 = 4sinωt and
y2 = 3sin(ωt − π/3) on the same axes, over one
cycle. By adding ordinates at intervals plot
yR = y1 + y2 and obtain a sinusoidal expression for
the resultant waveform
y1 = 4sinωt and y2 = 3sin(ωt − π/3) are shown plot-
ted in Figure 30.2.
908
y
y154 sin ␻t
y253 sin(␻t2␲/3)
0
26
24
22
6
6.1
4
2
258
258
yR 5y1 1y2
␻t1808 2708 3608
␲/2 ␲ 3␲/2 2␲
Figure 30.2
Ordinates are added at 15◦ intervals and the resultant is
shown by the broken line. The amplitude of the resultant
is 6.1 and it lags y1 by 25◦ or 0.436rad.
Hence, the sinusoidal expression for the resultant wave-
form is
yR = 6.1sin(ωt − 0.436)
Problem 3. Determine a sinusoidal expression
for y1 − y2 when y1 = 4sinωt and
y2 = 3sin(ωt − π/3)
y1 and y2 are shown plotted in Figure 30.3. At 15◦
intervals y2 is subtracted from y1. For example,
at 0◦, y1 − y2 = 0 − (−2.6) = +2.6
at 30◦, y1 − y2 = 2 − (−1.5) = +3.5
at 150◦, y1 − y2 = 2 − 3 = −1, and so on.
The amplitude, or peak, value of the resultant (shown by
the broken line) is 3.6 and it leads y1 by 45◦ or 0.79 rad.
908
y
y1
0
24
22
4
2
3.6
458
y2
y12y2
␻t1808 2708 3608
␲/2 ␲ 3␲/2 2␲
Figure 30.3
Hence,
y1 − y2 = 3.6sin(ωt + 0.79)
Problem 4. Two alternating currents are given by
i1 = 20sinωt amperes and i2 = 10sin ωt +
π
3
amperes. By drawing the waveforms on the same
axes and adding, determine the sinusoidal
expression for the resultant i1 + i2
i1 and i2 are shown plotted in Figure 30.4. The resultant
waveform for i1 + i2 is shown by the broken line. It has
the same period, and hence frequency, as i1 and i2.
The amplitude or peak value is 26.5A. The resultant
waveform leads the waveform of i1 = 20sinωt by 19◦
or 0.33rad.
Hence, the sinusoidal expression for the resultant
i1 + i2 is given by
iR = i1 + i2 = 26.5sin(ωt + 0.33)A
PracticeExercise 118 Plotting periodic
1. Plot the graph of y = 2sin A from A =
0◦ to A = 360◦. On the same axes plot
y = 4cos A. By adding ordinates at inter-
vals plot y = 2sin A + 4cos A and obtain a
sinusoidal expression for the waveform.
2. Two alternating voltages are given by
v1 = 10sinωt volts and v2 = 14sin(ωt +
π/3) volts. By plotting v1 and v2 on the
same axes over one cycle obtain a sinusoidal
expression for (a) v1 + v2 (b) v1 − v2

2␲ angle ␻t
19Њ
19Њ
i1 ϭ 20 sin␻t
iR ϭ 20 sin ␻t ϩ10 sin(␻tϩ )
90Њ 180Њ 270Њ 360Њ
Ϫ30
Ϫ20
Ϫ10
10
20
26.5
30
3␲
2
␲
2
␲
3
␲
i2 ϭ 10 sin(␻tϩ )
3
␲
Figure 30.4
3. Express 12sinωt + 5cosωt in the form
Asin(ωt ± α) by drawing and measurement.
30.3 Determining resultant phasors
by drawing
The resultant of two periodic functions may be found
from 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 in
Figure 30.5, y1 being 4 units long and drawn horizon-
tally and y2 being 3 unitslong,lagging y1 by π/3 radians
or60◦.To determinetheresultant of y1 + y2, y1 isdrawn
horizontally as shown in Figure 30.6 and y2 is joined to
the end of y1 at 60◦ to the horizontal. The resultant
is given by yR. This is the same as the diagonal of a
parallelogram that is shown completed in Figure 30.7.
608 or ␲/3 rads
y15 4
y25 3
Figure 30.5
Resultant yR, in Figures 30.6 and 30.7, may be deter-
mined by drawing the phasors and their directions to
scale and measuring using a ruler and protractor. In this
y15 4
y2
53
␾ 608
yR
Figure 30.6
y1ϭ 4
y2ϭ 3
␾
yR
Figure 30.7
example, yR is measured as 6 units long and angle φ is
measured as 25◦.
25◦
= 25 ×
π
180
radians = 0.44 rad
Hence, summarizing, by drawing,
yR = y1 + y2 = 4sinωt + 3sin(ωt − π/3)
= 6sin(ωt − 0.44).
If the resultant phasor, yR = y1 − y2 is required then
y2 is still 3 units long but is drawn in the opposite
direction, as shown in Figure 30.8.
Problem 5. Two alternating currents are given by
i1 = 20sinωt amperes and i2 = 10sin ωt +
π
3
amperes. Determine i1 + i2 by drawing phasors

60Њ
60Њ
␾
y1ϭ 4
Ϫy2 ϭ 3
yR
y2
Figure 30.8
The relative positions of i1 and i2 at time t = 0 are
shown as phasors in Figure 30.9, where
π
3
rad = 60◦.
The phasor diagram in Figure 30.10 is drawn to scale
with a ruler and protractor.
i15 20A
i2 5 10A
608
Figure 30.9
i25 10A
i15 20A
iR
608␾
Figure 30.10
The resultant iR is shown and is measured as 26A and
angle φ as 19◦ or 0.33rad leading i1. Hence, by drawing
and measuring,
iR = i1 + i2 = 26sin(ωt + 0.33)A
Problem 6. For the currents in Problem 5,
determine i1 − i2 by drawing phasors
At time t = 0, current i1 is drawn 20 units long hori-
zontally as shown by 0a in Figure 30.11. Current i2 is
shown, drawn 10 units long in a broken line and leading
by 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 20A
i2 5 10 A
iR
a
b
0
2i2
608
␾
Figure 30.11
By measurement, iR = 17A and φ = 30◦ or 0.52rad.
Hence, by drawing phasors,
iR = i1 − i2 = 17sin(ωt − 0.52)A
PracticeExercise 119 Determining
resultant phasors by drawing (answers on
page 352)
1. Determine a sinusoidal expression for
2sinθ +4cosθ by drawing phasors.
2. If v1 = 10sinωt volts and
v2 = 14sin(ωt + π/3) volts, determine by
drawing phasors sinusoidal expressions for
(a) v1 + v2 (b) v1 − v2
3. Express 12sinωt + 5cosωt in the form
Asin(ωt ± α) by drawing phasors.
by the sine and cosine rules
As stated earlier, the resultant of two periodic func-
tions may be found from their relative positions when
the time is zero. For example, if y1 = 5sinωt and
y2 = 4sin(ωt − π/6) then each may be represented by
phasors as shown in Figure 30.12, y1 being 5 units
long and drawn horizontally and y2 being 4 units long,
lagging y1 by π/6 radians or 30◦. To determine the
resultant of y1 + y2, y1 is drawn horizontally as shown
in Figure 30.13 and y2 is joined to the end of y1 at π/6
radians; i.e., 30◦ to the horizontal. The resultant is given
by yR.
Using the cosine rule on triangle 0ab of Figure 30.13
gives
y2
R = 52
+ 42
− [2(5)(4)cos150◦
]
= 25 + 16 − (−34.641) = 75.641

y15 5
y2 54
308
Figure 30.12
y1ϭ 5
y2 ϭ4
␾0
yR
a
b
30Њ
150Њ
Figure 30.13
from which yR =
√
75.641 = 8.697
Using the sine rule,
8.697
sin150◦
=
4
sinφ
from which sinφ =
4sin150◦
8.697
= 0.22996
and φ = sin−1
0.22996
= 13.29◦
or 0.232rad
Hence, yR = y1 + y2 = 5sinωt + 4sin(ωt − π/6)
= 8.697sin(ωt − 0.232)
Problem 7. Given y1 = 2sinωt and
y2 = 3sin(ωt + π/4), obtain an expression, by
calculation, for the resultant, yR = y1 + y2
When time t = 0, the position of phasors y1 and y2 are
as shown in Figure 30.14(a). To obtain the resultant, y1
is drawn horizontally, 2 units long, and y2 is drawn 3
units long at an angle of π/4rad or 45◦ and joined to
the end of y1 as shown in Figure 30.14(b).
From Figure 30.14(b), and using the cosine rule,
y2
R = 22
+ 32
− [2(2)(3)cos135◦
]
= 4 + 9 − [−8.485] = 21.485
Hence, yR =
√
21.485 = 4.6352
Using the sine rule
3
sinφ
=
4.6352
sin135◦
y25 3
y15 2
␲/4 or 458
(a)
y15 2
y25 3
1358
458␾
yR
(b)
Figure 30.14
from which sinφ =
3sin135◦
4.6352
= 0.45765
Hence, φ = sin−1
0.45765
= 27.24◦
or 0.475rad
Thus, by calculation, yR = 4.635sin(ωt + 0.475)
20sinωt + 10sin ωt +
π
3
using the cosine and sine rules
From the phasor diagram of Figure 30.15 and using the
cosine rule,
i2
R = 202
+ 102
− [2(20)(10)cos 120◦
] = 700
i2 5 10 A
i1 5 20A
iR
6081208␾
Figure 30.15
Hence, iR =
√
700 = 26.46A
Using the sine rule gives
10
sinφ
=
26.46
sin120◦

from which sinφ =
10sin120◦
26.46
= 0.327296
and φ = sin−1
0.327296 = 19.10◦
= 19.10 ×
π
180
= 0.333rad
Hence, by cosine and sine rules,
iR = i1 + i2 = 26.46sin(ωt + 0.333)A
PracticeExercise 120 Resultant phasors by
the sine and cosine rules (answers on
page 352)
1. Determine, using the cosine and sine rules, a
sinusoidal expression for
y = 2sin A + 4cos A.
2. Given v1 =10sinωt volts and
v2 =14sin(ωt+ π/3) volts, use the cosine
and sine rules to determine sinusoidal expres-
sions for (a) v1 + v2 (b) v1 − v2
In problems 3 to 5, express the given expressions
in the form Asin(ωt ± α) by using the cosine and
sine rules.
3. 12sinωt + 5cosωt
4. 7sinωt + 5sin ωt +
π
4
5. 6sinωt + 3sin ωt −
π
6
by horizontal and vertical
components
If a right-angled triangle is constructed as shown in
Figure 30.16, 0a is called the horizontal component of
F and ab is called the vertical component of F.
0
H
a
V
b
F
␪
Figure 30.16
From trigonometry(see Chapter 21 and remember SOH
CAH TOA),
cosθ =
0a
0b
, from which 0a = 0bcosθ = F cosθ
i.e. the horizontal component of F,H = F cosθ,
and
sinθ =
ab
0b
, from which ab = 0bsinθ = F sinθ
i.e. the vertical component of F,V = F sinθ.
Determining resultant phasorsby horizontal and vertical
components is demonstrated in the following worked
problems.
Problem 9. Two alternating voltages are given by
v1 = 15sinωt volts and v2 = 25sin(ωt − π/6)
volts. Determine a sinusoidal expression for the
resultant vR = v1 + v2 by ﬁnding horizontal and
vertical components
The relative positions of v1 and v2 at time t = 0 are
shown in Figure 30.17(a) and the phasor diagram is
shown in Figure 30.17(b).
v15 15 V
(a)
v25 25V
␲/6 or 308
(b)
␾0
vR
v2
v1
308
1508
Figure 30.17
The horizontal component of vR,
H = 15cos0◦ + 25cos(−30◦) = 36.65V
The vertical component of vR,
V = 15sin0◦ + 25sin(−30◦) = −12.50V

Hence, vR = 36.652 + (−12.50)2
by Pythagoras’ theorem
= 38.72 volts
tanφ =
V
H
=
−12.50
36.65
= −0.3411
from which φ = tan−1
(−0.3411)
= −18.83◦
or −0.329 radians.
Hence, vR = v1 + v2 = 38.72sin(ωt − 0.329)V
Problem 10. For the voltages in Problem 9,
determine the resultant vR = v1 − v2 using
horizontal and vertical components
The horizontal component of vR,
H = 15cos0◦ − 25cos(−30◦) = −6.65V
The vertical component of vR,
V = 15sin0◦ − 25sin(−30◦) = 12.50V
Hence, vR = (−6.65)2 + (12.50)2
by Pythagoras’ theorem
= 14.16 volts
tanφ =
V
H
=
12.50
−6.65
= −1.8797
from which φ = tan−1
(−1.8797)
= 118.01◦
or 2.06 radians.
Hence, vR = v1 − v2 = 14.16sin(ωt + 2.06)V
The phasor diagram is shown in Figure 30.18.
v15 15V
2v25 25V
v25 25V
␾
vR
308
308
Figure 30.18
π
3
using horizontal and vertical components
i15 20A
i2 5 10A
608
Figure 30.19
From the phasors shown in Figure 30.19,
Total horizontal component,
H = 20cos0◦
+ 10cos60◦
= 25.0
Total vertical component,
V =20sin0◦ +10sin60◦ = 8.66
By Pythagoras, the resultant,
iR = 25.02+8.662 = 26.46A
Phase angle, φ = tan−1 8.66
25.0
= 19.11◦ or 0.333rad
Hence, by using horizontal and vertical components,
π
3
= 26.46sin(ωt + 0.333)
PracticeExercise 121 Resultant phasors by
horizontal and vertical components (answers
on page 353)
In problems 1 to 5, express the combination of
periodic functions in the form Asin(ωt ± α) by
horizontal and vertical components.
1. 7sinωt + 5sin ωt +
π
4
2. 6sinωt + 3sin ωt −
π
6
3. i = 25sinωt − 15sin ωt +
π
3
4. v = 8sinωt − 5sin ωt −
π
4
5. x = 9sin ωt +
π
3
− 7sin ωt −
3π
8
6. The voltage drops across two components
when connected in series across an a.c.
supply are v1 = 200sin314.2t and
v2 = 120sin(314.2t − π/5) volts
respectively. Determine
(a) the voltage of the supply (given by
v1 + v2) in the form Asin(ωt ± α).

(b) the frequency of the supply.
7. If the supply to a circuit is v = 20sin628.3t
volts and the voltage drop across one of
the components is v1 = 15sin(628.3t − 0.52)
volts, calculate
(a) the voltage drop across the remainder of
the circuit, given by v − v1, in the form
Asin(ωt ± α)
(b) the supply frequency
(c) the periodic time of the supply.
8. The voltages across three components in a
series circuit when connected across an a.c.
supply are v1 = 25sin 300πt +
π
6
volts,
v2 = 40sin 300πt −
π
4
volts and
v3 = 50sin 300πt +
π
3
volts. Calculate
(a) the supply voltage, in sinusoidal form, in
the form Asin(ωt ± α)
(b) the frequency of the supply
(c) the periodic time.

Revision Test 12 : Vectors and adding waveforms
1. State the difference between scalar and vector
quantities. (2)
2. State whether the following are scalar or vector
quantities.
(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 the
horizontal. (8)
3. A velocity vector of 16m/s acts at an angle of
−40◦ to the horizontal. Calculate its horizontal
and vertical components, correct to 3 signiﬁcant
ﬁgures. (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 m
26 m
Figure RT12.1
5. Calculate the resultant and direction of the force
vectors shown in Figure RT12.2, correct to 2
decimal places. (6)
6. If acceleration a1 = 11m/s2
at 70◦
and
a2 = 19m/s2 at −50◦, calculate the magnitude
and direction of a1 + a2, correct to 2 decimal
places. (8)
7. If velocity v1 = 36m/s at 52◦ and v2 = 17m/s at
−15◦, calculate the magnitude and direction of
v1 − v2, correct to 2 decimal places. (8)
5N
7N
Figure RT12.2
8. Forces of 10N, 16N and 20N act as shown in
Figure RT12.3. Determine the magnitude of the
resultant force and its direction relative to the 16N
force
(a) by scaled drawing.
(b) by calculation. (13)
F1510N
F2 516N
F3 520N
458
608
Figure RT12.3
9. For the three forces shown in Figure RT12.3,
calculate the resultant of F1 − F2 − F3 and its
direction 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 and
B a velocity of 75km/h due west. Calculate the
velocity of A relative to B. (6)

Revision Test 12 : Vectors and adding waveforms 287
11. Given a = −3i + 3j + 5k, b = 2i − 5j + 7k and
c = 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 in
Figure RT12.4. (9)
7 m
12 m 10 m
308
608
208
Figure RT12.4
13. The instantaneous values of two alternating volt-
ages are given by
v1 = 150sin ωt +
π
3
volts
and
v2 = 90sin ωt −
π
6
volts
Plot the two voltages on the same axes to scales
of 1cm = 50 volts and 1cm = π/6. Obtain a sinu-
soidal expression for the resultant of v1 and v2 in
the form Rsin (ωt + α) (a) by adding ordinates at
intervals and (b) by calculation. (15)

Chapter 31
Presentation of
statistical data
31.1 Some statistical terminology
31.1.1 Discrete and continuous data
Data are obtained largely by two methods:
(a) By counting – for example, the number of stamps
sold by a post ofﬁce in equal periods of time.
(b) By measurement – for example, the heights of a
group of people.
When data are obtained by counting and only whole
numbers are possible, the data are called discrete. Mea-
sured data can have any value within certain limits and
are called continuous.
Problem 1. Data are obtained on the topics given
below. State whether they are discrete or continuous
data.
(a) The number of days on which rain falls in a
month for each month of the year.
(b) The mileage travelled by each of a number of
salesmen.
(c) The time that each of a batch of similar
batteries lasts.
(d) The amount of money spent by each of several
families on food.
(a) The number of days on which rain falls in a given
month must be an integer value and is obtained by
counting the number of days. Hence, these data
are discrete.
(b) A salesman can travel any number of miles
(and parts of a mile) between certain limits and
these data are measured. Hence, the data are
continuous.
(c) The time that a battery lasts is measured and
can have any value between certain limits. Hence,
these data are continuous.
(d) The amount of money spent on food can only be
expressed correct to the nearest pence, the amount
being counted. Hence, these data are discrete.
PracticeExercise 122 Discrete and
continuous 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, for
each of 31 days, by a reﬁnery.
(b) The amount of coal produced daily by
each of 15 miners.
(c) The number of bottles of milk delivered
daily by each of 20 milkmen.
(d) Thesizeof10 samplesofrivetsproduced
by 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 to
run 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 each
of 20 post offices.
(d) The number of defective items produced
in each of 10 one-hour periods by a
machine.
31.1.2 Further statisticalterminology
A set is a group of data and an individual value within
the set is called a member of the set. Thus, if the
masses of five people are measured correct to the 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 in
kilograms for these five people is
{53.1,59.4,62.1,77.8,64.4}
and one of the members of the set is 59.4
A set containing all the members is called a pop-
ulation. Some members selected at random from a
population are called a sample. Thus, all car registration
numbers form a population but the registration numbers
of, say, 20 cars taken at random throughout the country
are a sample drawn from that population.
The number of times that the value of a member
occurs in a set is called the frequency of that mem-
ber. Thus, in the set {2,3,4,5,4,2,4,7,9}, member 4
has a frequency of three, member 2 has a frequency of
2 and the other members have a frequency of one.
The relative frequency with which any member of a
set occurs is given by the ratio
frequency of member
total frequency of all members
For the set {2,3,5,4,7,5,6,2,8}, the relative fre-
quency of member 5 is
2
9
. Often, relative frequency is
expressed as a percentage and the percentage relative
frequency is
(relative frequency × 100)%
31.2 Presentation of ungrouped data
Ungrouped data can be presented diagrammatically in
several ways and these include
(a) pictograms, in which pictorial symbols are used
to represent quantities (see Problem 2),
(b) horizontal bar charts, having data represented
by equally spaced horizontal rectangles (see Prob-
lem 3), and
(c) vertical bar charts, in which data are repre-
sented by equally spaced vertical rectangles (see
Problem 4).
Trends in ungrouped data over equal periods of time
can be presented diagrammatically by a percent-
age component bar chart. In such a chart, equally
spaced rectangles of any width, but whose height cor-
responds to 100%, are constructed. The rectangles
are then subdivided into values corresponding to the
percentage relative frequencies of the members (see
Problem 5).
A pie diagram is used to show diagrammatically the
parts making up the whole. In a pie diagram, the area of
a circle represents the whole and the areas of the sectors
of the circle are made proportional to the parts which
make up the whole (see Problem 6).
Problem 2. The number of television sets
repaired in a workshop by a technician in six
one-month periods is as shown below. Present these
data as a pictogram
Month January February March
Number repaired 11 6 15
Month April May June
Number repaired 9 13 8
Each symbol shown in Figure 31.1 represents two tele-
vision sets repaired. Thus, in January, 5
1
2
symbols are
used to represent the 11 sets repaired; in February, 3
symbols are used to represent the 6 sets repaired, and
so on.
January
February
March
Month Number of TV sets repaired ;2 sets
April
May
June
Figure 31.1

Problem 3. The distance in miles travelled by
four salesmen in a week are as shown below.
Salesman P Q R S
Distance travelled (miles) 413 264 597 143
Use a horizontal bar chart to represent these data
diagrammatically
Equally spaced horizontal rectangles of any width, but
whose length is proportional to the distance travelled,
are used. Thus, the length of the rectangle for sales-
man P is proportional to 413 miles, and so on. The
horizontal bar chart depicting these data is shown in
Figure 31.2.
0
P
Q
Salesmen
R
S
100 200 300
Distance travelled, miles
400 500 600
Figure 31.2
Problem 4. The number of issues of tools or
materials from a store in a factory is observed for
seven one-hour periods in a day and the results of
the survey are as follows.
Period 1 2 3 4 5 6 7
Number of issues 34 17 9 5 27 13 6
Present these data on a vertical bar chart
1
10
20
Numberofissues
30
40
2 3 4 5 6
Periods
7
Figure 31.3
In a vertical bar chart, equally spaced vertical rect-
angles of any width, but whose height is proportional
to the quantity being represented, are used. Thus, the
height of the rectangle for period 1 is proportional to 34
units, and so on. The vertical bar chart depicting these
data is shown in Figure 31.3.
Problem 5. The numbers of various types of
dwellings sold by a company annually over a
three-year period are as shown below. Draw
percentage component bar charts to present
these data
Year 1 Year 2 Year 3
4-roomed bungalows 24 17 7
5-roomed bungalows 38 71 118
4-roomed houses 44 50 53
A table of percentage relative frequency values, correct
to the nearest 1%, is the ﬁrst requirement. Since
percentage relative frequency
=
frequency of member × 100
total frequency
then for 4-roomed bungalows in year 1
percentage relative frequency
=
24 × 100
24 + 38 + 44 + 64 + 30
= 12%
The percentage relative frequencies of the other types
of dwellings for each of the three years are similarly
calculated and theresultsareasshown in thetablebelow.
Year 1 Year 2 Year 3
4-roomed bungalows 12% 7% 2%
5-roomed bungalows 19% 28% 34%
4-roomed houses 22% 20% 15%
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

subdivided into the values in the table of percentages
shown above. A key is used (different types of shading
or different colour schemes) to indicate corresponding
percentage values in the rows of the table of percent-
ages. The percentage component bar chart is shown in
Figure 31.4.
1
10
20
Percentagerelativefrequency
30
40
50
60
70
80
90
100
2 3
Year
Key
6-roomed houses
5-roomed houses
4-roomed houses
5-roomed bungalows
4-roomed bungalows
Figure 31.4
Problem 6. The retail price of a product costing
£2 is made up as follows: materials 10p, labour
20p, research and development 40p, overheads
70p, profit 60p. Present these data on a pie diagram
A circle of any radius is drawn. The area of the circle
represents the whole, which in this case is £2. The circle
is subdivided into sectors so that the areas of the sectors
are proportional to the parts; i.e., the parts which make
up the total retail price. For the area of a sector to be
proportionalto a part, the angle at the centre of the circle
must be proportional to that part. The whole, £2 or 200p,
corresponds to 360◦. Therefore,
10p corresponds to 360×
10
200
degrees, i.e. 18◦
20p corresponds to 360×
20
200
degrees, i.e. 36◦
and so on, giving the angles at the centre of the circle
for the parts of the retail price as 18◦,36◦,72◦,126◦ and
108◦
, 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 each
salesman for travelling expenses if they are
paid an allowance of 37p per mile.
1088
Ip ; 1.88
188
368728
1268
Overheads
Profit
Labour
Research and
development
Materials
Figure 31.5
(b) Using the data presented in Figure 31.4,
comment on the housing trends over the
three-year period.
(c) Determine the profit made by selling 700 units
of the product shown in Figure 31.5
(a) By measuring the length of rectangle P, the
mileage covered by salesman P is equivalent to
413 miles. Hence salesman P receives a travelling
allowance of
£413 × 37
100
i.e. £152.81
Similarly, for salesman Q, the miles travelled are
264 and his allowance is
£264 × 37
100
i.e. £97.68
Salesman R travels 597 miles and he receives
£597 × 37
100
i.e. £220.89
Finally, salesman S receives
£143 × 37
100
i.e. £52.91
(b) An analysis of Figure 31.4 shows that 5-roomed
bungalows and 5-roomed houses are becoming
morepopular,thegreatest changein thethreeyears
being a 15% increase in the sales of 5-roomed
bungalows.
(c) Since 1.8◦ corresponds to 1p and the profit occu-
pies 108◦ of the pie diagram, the profit per unit is
108 × 1
1.8
i.e. 60p
The profit when selling 700 units of the prod-
uct is
£
700 × 60
100
i.e. £420

PracticeExercise 123 Presentation of
ungrouped data (answers on page 352)
1. The number of vehicles passing a stationary
observer on a road in six ten-minute intervals
is as shown. Draw a pictogram to represent
these data.
Period of time 1 2 3 4 5 6
Number of
vehicles 35 44 62 68 49 41
2. The number of components produced by a
factory in a week is as shown below.
Day Mon Tues Wed Thur Fri
Number of
components 1580 2190 1840 2385 1280
Show these data on a pictogram.
3. For the data given in Problem 1 above, draw
a horizontal bar chart.
4. Present the data given in Problem 2 above on
a 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 on
a vertical bar chart.
7. A factory produces three different types of
components. The percentages of each of
these components produced for three one-
month periods are as shown below. Show this
information on percentage component bar
charts and comment on the changing trend
in the percentages of the types of component
produced.
Month 1 2 3
Component P 20 35 40
Component Q 45 40 35
Component R 35 25 25
8. A company has five distribution centres and
the mass of goods in tonnes sent to each
centre during four one-week periods is as
shown.
Week 1 2 3 4
Centre A 147 160 174 158
Centre B 54 63 77 69
Centre C 283 251 237 211
Centre D 97 104 117 144
Centre E 224 218 203 194
Use a percentage component bar chart to pre-
sent these data and comment on any trends.
9. The employees in a company can be split
into 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 his
time over a one-month period is as follows:
drawing office44 hours,production 64 hours,
training 12 hours, at college 28 hours. Use a
pie diagram to depict this information.
11. (a) With reference to Figure 31.5, determine
the amount spent on labourand materials
to produce 1650 units of the product.
(b) If in year 2 of Figure 31.4 1% cor-
responds to 2.5 dwellings, how many
bungalows are sold in that year?
12. (a) If the company sells 23500 units per
annum of the product depicted in
Figure 31.5, determine the cost of their
overheads per annum.
(b) If 1% of the dwellings represented in
year 1 of Figure 31.4 corresponds to
2 dwellings, find the total number of
houses sold in that year.
31.3 Presentation of grouped data
When the number of members in a set is small, say ten or
less, the data can be represented diagrammatically with-
out further analysis, by means of pictograms, bar charts,
percentage components bar charts or pie diagrams (as
shown in Section 31.2).
For sets having more than ten members, those mem-
bers having similar values are grouped together in

classes to form a frequency distribution. To assist in
accurately counting members in the various classes, a
tally diagram is used (see Problems 8 and 12).
A frequency distribution is merely a table show-
ing classes and their corresponding frequencies (see
Problems 8 and 12). The new set of values obtained
by forming a frequency distribution is called grouped
data. The terms used in connection with grouped data
are shown in Figure 31.6(a). The size or range of a class
is given by the upper class boundary value minus the
lower class boundary value and in Figure 31.6(b) is
7.65−7.35; i.e., 0.30. The class interval for the class
shown in Figure 31.6(b) is 7.4 to 7.6 and the class
mid-point value is given by
(upper class boundary value) + (lower class boundary value)
2
and in Figure 31.6(b) is
7.65 + 7.35
2
i.e. 7.5
Class interval(a)
(b)
Lower
class
boundary
Upper
class
boundary
Class
mid-point
7.35
to 7.3 7.4 to 7.6 7.7 to
7.657.5
Figure 31.6
One of the principal ways of presenting grouped data
diagrammatically is to use a histogram, in which the
areas 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 the
classes. For histograms having unequal class intervals,
the area must be proportional to the frequency. Hence,
if the class interval of class A is twice the class inter-
val of class B, then for equal frequencies the height
of the rectangle representing A is half that of B (see
Problem 11).
Anothermethod of presenting grouped data diagram-
matically is to use a frequency polygon, which is the
graph produced by plottingfrequency against class mid-
point values and joining the co-ordinates with straight
lines (see Problem 12).
A cumulative frequency distribution is a table
showing the cumulative frequency for each value of
upper class boundary. The cumulative frequency for a
particular value of upper class boundary is obtained by
adding the frequency of the class to the sum of the pre-
vious frequencies. A cumulative frequency distribution
is formed in Problem 13.
The curve obtained by joining the co-ordinates of
cumulative frequency (vertically) against upper class
boundary (horizontally) is called an ogive or a cumu-
lative frequency distribution curve (see Problem 13).
Problem 8. The data given below refer to the gain
of each of a batch of 40 transistors, expressed
correct to the nearest whole number. Form a
frequency distribution for these data having seven
classes
81 83 87 74 76 89 82 84
86 76 77 71 86 85 87 88
84 81 80 81 73 89 82 79
81 79 78 80 85 77 84 78
83 79 80 83 82 79 80 77
The range of the data is the value obtained by tak-
ing the value of the smallest member from that of the
largest member. Inspection of the set of data shows that
range = 89 − 71 = 18. The size of each class is given
approximately by the range divided by the number of
classes. Since 7 classes are required, the size of each
class is 18÷ 7; that is, approximately 3. To achieve
seven equal classes spanning a range of values from 71
to 89, the class intervals are selected as 70–72, 73–75,
and so on.
To assist with accurately determining the number in
each class, a tally diagram is produced, as shown in
Table 31.1(a). This is obtained by listing the classes
in the left-hand column and then inspecting each of the
40 members of the set in turn and allocating them to
the appropriate classes by putting ‘1’s in the appropri-
ate rows. Every ﬁfth ‘1’ allocated to a particular row is
shown as an obliqueline crossing the four previous ‘1’s,
to help with ﬁnal counting.
A frequency distribution for the data is shown in
Table 31.1(b) and lists classes and their correspond-
ing frequencies, obtained from the tally diagram. (Class
mid-point values are also shown in the table, since they
are used for constructing the histogram for these data
(see Problem 9).)

Table 31.1(a)
Class Tally
70–72 1
73–75 11
76–78 1111 11
79–81 1111 1111 11
82–84 1111 1111
85–87 1111 1
88–90 111
Table 31.1(b)
Class Class mid-point Frequency
70–72 71 1
73–75 74 2
76–78 77 7
79–81 80 12
82–84 83 9
85–87 86 6
88–90 89 3
Problem 9. Construct a histogram for the data
given in Table 31.1(b)
The histogram is shown in Figure 31.7. The width of
the rectangles corresponds to the upper class boundary
values minus the lower class boundary values and the
heightsoftherectanglescorrespond to theclassfrequen-
cies. The easiest way to draw a histogram is to mark the
71
4
2
6
Frequency
10
8
12
14
16
74 77 80 83
Class mid-point values
8986
Figure 31.7
class mid-point values on the horizontal scale and draw
the rectangles symmetrically about the appropriateclass
mid-point values and touching one another.
Problem 10. The amount of money earned
weekly by 40 people working part-time in a factory,
correct to the nearest £10, is shown below. Form a
frequency distribution having 6 classes for these
data
80 90 70 110 90 160 110 80
140 30 90 50 100 110 60 100
80 90 110 80 100 90 120 70
130 170 80 120 100 110 40 110
50 100 110 90 100 70 110 80
Inspection of the set given shows that the majority of
the members of the set lie between £80 and £110 and
that there is a much smaller number of extreme val-
ues ranging from £30 to £170. If equal class intervals
are selected, the frequency distribution obtained does
not give as much information as one with unequal class
intervals. Since the majority of the members lie between
£80 and £100, the class intervals in this range are
selected to be smaller than those outside of this range.
There is no unique solution and one possible solution is
shown in Table 31.2.
Table 31.2
Class Frequency
20–40 2
50–70 6
80–90 12
100–110 14
120–140 4
150–170 2
Problem 11. Draw a histogram for the data given
in Table 31.2
When dealing with unequal class intervals, the his-
togram must be drawn so that the areas (and not
the heights) of the rectangles are proportional to the

Table 31.3
1 2 3 4 5 6
Class Frequency Upper class Lower class Class range Height of
boundary boundary rectangle
20–40 2 45 15 30
2
30
=
1
15
50–70 6 75 45 30
6
30
=
3
15
80–90 12 95 75 20
12
20
=
9
15
100–110 14 115 95 20
14
20
=
101
2
15
120–140 4 145 115 30
4
30
=
2
15
150–170 2 175 145 30
2
30
=
1
15
frequencies of the classes. The data given are shown
in columns 1 and 2 of Table 31.3. Columns 3 and 4
give 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. The
heights of the rectangles are proportional to the ratio
frequency
class range
, as shown in column 6. The histogram is
30
4/15
2/15
6/15
Frequencyperunit
classrange
10/15
8/15
12/15
60 85
130 160105
Figure 31.8
Problem 12. The masses of 50 ingots in
kilograms are measured correct to the nearest 0.1kg
and the results are as shown below. Produce a
frequency distribution having about 7 classes for
these data and then present the grouped data as a
frequency polygon and a histogram
8.0 8.6 8.2 7.5 8.0 9.1 8.5 7.6 8.2 7.8
8.3 7.1 8.1 8.3 8.7 7.8 8.7 8.5 8.4 8.5
7.7 8.4 7.9 8.8 7.2 8.1 7.8 8.2 7.7 7.5
8.1 7.4 8.8 8.0 8.4 8.5 8.1 7.3 9.0 8.6
7.4 8.2 8.4 7.7 8.3 8.2 7.9 8.5 7.9 8.0
The range of the data is the member having the largest
value minus the member having the smallest value.
Inspection of the set of data shows that range = 9.1 −
7.1 = 2.0.
The size of each class is given approximately by
range
number of classes
Since about seven classes are required, the size of each
classis2.0 ÷ 7,i.e.approximately 0.3,and thusthe class
limits 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 is
7.35 + 7.05
2
i.e. 7.2
the class midpoint for the 7.4 to 7.6 class is
7.65 + 7.35
2
i.e. 7.5
and so on.

To assist with accurately determining the number
in each class, a tally diagram is produced as shown
in Table 31.4. This is obtained by listing the classes in
the left-hand column and then inspecting each of the 50
members of the set of data in turn and allocating it to the
appropriate class by puttinga ‘1’ in the appropriate row.
Each ﬁfth ‘1’ allocated to a particular row is marked as
an oblique line to help with ﬁnal counting.
Table 31.4
Class Tally
7.1 to 7.3 111
7.4 to 7.6 1111
7.7 to 7.9 1111 1111
8.0 to 8.2 1111 1111 1111
8.3 to 8.5 1111 1111 1
8.6 to 8.8 1111 1
8.9 to 9.1 11
A frequency distribution for the data is shown in
Table 31.5 and lists classes and their corresponding fre-
quencies. Class mid-points are also shown in this table
since they are used when constructing the frequency
polygon and histogram.
Table 31.5
Class Class mid-point Frequency
7.1 to 7.3 7.2 3
7.4 to 7.6 7.5 5
7.7 to 7.9 7.8 9
8.0 to 8.2 8.1 14
8.3 to 8.5 8.4 11
8.6 to 8.8 8.7 6
8.9 to 9.1 9.0 2
A 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 polygon
is ‘anchored-down’ at each end by joining to the next
class mid-point value and zero frequency.
A histogram is shown in Figure 31.10, the width
of a rectangle corresponding to (upper class boundary
7.2
4
2
0
6
Frequency
12
10
8
14
7.87.5 8.1
8.7 9.08.4
Frequency polygon
Figure 31.9
value – lower class boundary value) and height corre-
sponding to the class frequency. The easiest way to draw
ahistogramisto mark classmid-point valueson thehori-
zontal scale and to draw the rectangles symmetrically
about the appropriate class mid-point values and touch-
ing one another. A histogram for the data given in
Table 31.5 is shown in Figure 31.10.
7.2
4
2
0
6
Frequency
10
8
12
14
7.5
7.35
7.65
7.95
8.25
8.55
8.85
9.15
7.8 8.1 8.4
9.08.7
Histogram
Figure 31.10
Problem 13. The frequency distribution for the
masses in kilograms of 50 ingots is
7.1 to 7.3 3
7.4 to 7.6 5
7.7 to 7.9 9
8.0 to 8.2 14
8.3 to 8.5 11
8.6 to 8.8 6
8.9 to 9.1 2
Form a cumulative frequency distribution for these
data and draw the corresponding ogive

A cumulative frequency distribution is a table giving
values of cumulative frequency for the values of upper
class boundaries and is shown in Table 31.6. Columns
1 and 2 show the classes and their frequencies. Column
3 lists the upper class boundary values for the classes
given in column 1. Column 4 gives the cumulative fre-
quency values for all frequencies less than the upper
class boundary values given in column 3. Thus, for
example, for the 7.7 to 7.9 class shown in row 3, the
cumulative 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 31.6
1 2 3 4
Class Frequency Upper class Cumulative
boundary less than frequency
7.1–7.3 3 7.35 3
7.4–7.6 5 7.65 8
7.7–7.9 9 7.95 17
8.0–8.2 14 8.25 31
8.3–8.5 11 8.55 42
8.6–8.8 6 8.85 48
8.9–9.1 2 9.15 50
The 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 class
boundary/cumulative frequency value are plotted and
7.05
10
Cumulativefrequency
40
30
20
50
7.957.35 7.65 8.25
Upper class boundary values in kilograms
8.85 9.158.55
Figure 31.11
the co-ordinates are joined by straight lines (not the best
curve drawn throughtheco-ordinates as in experimental
work). The ogive is ‘anchored’ at its start by adding the
co-ordinate (7.05, 0).
PracticeExercise 124 Presentation of
grouped data (answers on page 353)
1. The mass in kilograms, correct to the nearest
one-tenth of a kilogram, of 60 bars of metal
are as shown. Form a frequency distribution
of about 8 classes for these data.
39.8 40.1 40.3 40.0 40.6 39.7 40.0 40.4 39.6 39.3
39.6 40.7 40.2 39.9 40.3 40.2 40.4 39.9 39.8 40.0
40.2 40.1 40.3 39.7 39.9 40.5 39.9 40.5 40.0 39.9
40.1 40.8 40.0 40.0 40.1 40.2 40.1 40.0 40.2 39.9
39.7 39.8 40.4 39.7 39.9 39.5 40.1 40.1 39.9 40.2
39.5 40.6 40.0 40.1 39.8 39.7 39.5 40.2 39.9 40.3
2. Draw a histogram for the frequency distribu-
tion given in the solution of Problem 1.
3. The information given below refers to the
value of resistance in ohms of a batch of
48 resistors of similar value. Form a fre-
quency distribution for the data, having about
6 classes, and draw a frequency polygon and
histogram to represent these data diagram-
matically.
21.0 22.4 22.8 21.5 22.6 21.1 21.6 22.3
22.9 20.5 21.8 22.2 21.0 21.7 22.5 20.7
23.2 22.9 21.7 21.4 22.1 22.2 22.3 21.3
22.1 21.8 22.0 22.7 21.7 21.9 21.1 22.6
21.4 22.4 22.3 20.9 22.8 21.2 22.7 21.6
22.2 21.6 21.3 22.1 21.5 22.0 23.4 21.2
4. The time taken in hours to the failure of 50
specimens of a metal subjected to fatigue fail-
ure tests are as shown. Form a frequency dis-
tribution, having about 8 classes and unequal
class intervals, for these data.

28 22 23 20 12 24 37 28 21 25
21 14 30 23 27 13 23 7 26 19
24 22 26 3 21 24 28 40 27 24
20 25 23 26 47 21 29 26 22 33
27 9 13 35 20 16 20 25 18 22
5. Form a cumulative frequency distribution and
hence 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 of
50 capacitors of similar value, measured in
microfarads, is
10.5–10.9 2
11.0–11.4 7
11.5–11.9 10
12.0–12.4 12
12.5–12.9 11
13.0–13.4 8
Form a cumulative frequency distribution for
these data.
8. Drawan ogiveforthedatagiven in thesolution
of Problem 7.
9. The diameter in millimetres of a reel of wire
is measured in 48 places and the results are as
shown.
2.10 2.29 2.32 2.21 2.14 2.22
2.28 2.18 2.17 2.20 2.23 2.13
2.26 2.10 2.21 2.17 2.28 2.15
2.16 2.25 2.23 2.11 2.27 2.34
2.24 2.05 2.29 2.18 2.24 2.16
2.15 2.22 2.14 2.27 2.09 2.21
2.11 2.17 2.22 2.19 2.12 2.20
2.23 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 32
Mean, median, mode and
standard deviation
32.1 Measures of central tendency
A single value, which is representative of a set of values,
may be used to give an indication of the general size of
the members in a set, the word ‘average’ often being
used to indicate the single value. The statistical term
used for ‘average’ is the ‘arithmetic mean’ or just the
‘mean’.
Other measures of central tendency may be used and
these include the median and the modal values.
32.2 Mean, median and mode for
discrete data
32.2.1 Mean
The arithmeticmean value isfound by adding together
the values of the members of a set and dividing by the
number of members in the set. Thus, the mean of the set
of numbers {4, 5, 6, 9} is
4 + 5 + 6 + 9
4
i.e. 6
In general, the mean of the set {x1, x2, x3,...xn} is
x =
x1 + x2 + x3 + ···+ xn
n
written as
x
n
where is the Greek letter ‘sigma’ and means ‘the sum
of ’ and x (called x-bar) is used to signify a mean value.
32.2.2 Median
The median value often gives a better indication of the
general size of a set containing extreme values. The set
{7, 5, 74, 10} has a mean value of 24, which is not really
representative ofany of the values of the members of the
set. The median value is obtained by
(a) ranking the set in ascending order of magnitude,
and
(b) selecting the value of the middle member for sets
containing an odd number of members or ﬁnding
the value of the mean of the two middle members
for sets containing an even number of members.
For example, the set {7, 5, 74, 10} is ranked as
{5, 7, 10, 74} and, since it contains an even number
of members (four in this case), the mean of 7 and 10 is
taken, giving a median value of 8.5. Similarly, the set
{3, 81, 15, 7, 14} is ranked as {3, 7, 14, 15, 81} and
the median value is the value of the middle member,
i.e. 14.
32.2.3 Mode
The modal value, or mode, is the most commonly
occurring value in a set. If two values occur with
the 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 the
member having a value of 5 occurs the most, i.e. three
times.
Problem 1. Determine the mean, median and
mode 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

The mean value is obtained by adding together the
values of the members of the set and dividing by the
number of members in the set. Thus,
mean value,x
=
2 + 3 + 7 + 5 + 5 + 13 + 1 + 7 + 4 + 8 + 3 + 4 + 3
13
=
65
13
= 5
To obtain the median value the set is ranked, that is,
placed in ascending order of magnitude, and since the
set contains an odd number of members the value of the
middle member is the median value. Ranking the set
gives {1, 2, 3, 3, 3, 4, 4, 5, 5, 7, 7, 8, 13}. The middle
term is the seventh member; i.e., 4. Thus, the median
value is 4.
The modal value is the value of the most commonly
occurring 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 the
amount of money in £s taken by a news vendor for
6 days. Determine the mean, median and modal
values 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 + 39.68
6
= £37.20
The 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 mean
of the middle two members is taken to give the median
value; i.e.,
median value =
34.70 + 39.68
2
= £37.19
Since no two members have the same value, this set has
no mode.
PracticeExercise 125 Mean, median and
mode for discrete data (answers on
page 353)
In problems 1 to 4, determine the mean, median
and 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 for
grouped data
The mean value for a set of grouped data is found by
determining thesumofthe(frequency× classmid-point
values) and dividing by the sum of the frequencies; i.e.,
mean value,x =
f1x1 + f2x2 + ···fnxn
f1 + f2 + ···+ fn
=
( fx)
f
where f is the frequency of the class having a mid-point
value of x, and so on.
value of resistance in ohms of 48 resistors is as
shown. Determine the mean value of resistance
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
The class mid-point/frequency values are 20.7 3, 21.2
10, 21.7 11, 22.2 13, 22.7 9 and 23.2 2.
For grouped data, the mean value is given by
x =
( f x)
f
where 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
=
1052.1
48
= 21.919...
i.e. the mean value is 21.9 ohms, correct to 3 signiﬁcant
ﬁgures.
32.3.1 Histograms
The mean, median and modal values for grouped data
may be determined from a histogram. In a histogram,

Mean, median, mode and standard deviation 301
frequency values are represented vertically and variable
valueshorizontally.Themean valueisgiven by thevalue
of the variable corresponding to a vertical line drawn
throughthe centroid of the histogram.The median value
is obtained by selecting a variable value such that the
area of the histogram to the left of a vertical line drawn
throughthe selected variable value is equal to the area of
the histogram on the right of the line. The modal value
is the variable value obtained by dividing the width of
the highest rectangle in the histogram in proportion to
the heights of the adjacent rectangles. The method of
determining the mean, median and modal values from
a histogram is shown in Problem 4.
Problem 4. The time taken in minutes to
assemble a device is measured 50 times and the
results are as shown. Draw a histogram depicting
the data and hence determine the mean, median and
modal values of the distribution
14.5–15.5 5
16.5–17.5 8
18.5–19.5 16
20.5–21.5 12
22.5–23.5 6
24.5–25.5 3
The histogram is shown in Figure 32.1. The mean value
lies at the centroid of the histogram. With reference to
any arbitrary axis, say YY shown at a time of 14 min-
utes, the position of the horizontal value of the centroid
can 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, a
is the area of a rectangle of the histogram and m is
the distance of the centroid of the rectangle from YY.
The areas of the individual rectangles are shown circled
14 15 16 17 18 19 20 21 22 23 24 25
6
26 27
4
2
6
Frequency
10
8
12
16
14
Time in minutes
12
24
32
16
10
E
D
AY
Y
5.6
Mode
Median Mean
F
C
B
Figure 32.1
on the histogram giving a total area of 100 square
units.The positions,m, of the centroids of the individual
rectangles are 1,3,5,... units from YY. Thus
100M = (10 × 1) + (16 × 3) + (32 × 5) + (24 × 7)
+ (12 × 9) + (6 × 11)
i.e. M =
560
100
= 5.6 units from YY
Thus, the position of the mean with reference to the
time scale is 14 + 5.6, i.e. 19.6 minutes.
The median is the value of time corresponding to a
vertical line dividing the total area of the histogram into
two equal parts. The total area is 100 square units, hence
the vertical line must be drawn to give 50 units of area on
each 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) units
of area lie on the other. This shows that the area of ABFE
is split so that 24 units of area lie to the left ofthe line and
8 units of area lie to the right; i.e., the vertical line must
pass through 19.5 minutes. Thus, the median value of
the distribution is 19.5 minutes.
The mode is obtained by dividing the line AB, which
is the height of the highest rectangle, proportionally to
the heights of the adjacent rectangles. With reference
to Figure 32.1, this is achieved by joining AC and BD
and drawing a vertical line through the point of inter-
section of these two lines. This gives the mode of the
distribution, which is 19.3 minutes.
PracticeExercise 126 Mean, median and
mode for grouped data (answers on
page 354)
1. 21 bricks have a mean mass of 24.2kg and
29 similar bricks have a mass of 23.6kg.
Determine the mean mass of the 50 bricks.
2. The frequency distribution given below refers
to the heights in centimetres of 100 people.
Determine the mean value of the distribution,
correct to the nearest millimetre.
150–156 5
157–163 18
164–170 20
171–177 27
178–184 22
185–191 8

3. The gain of 90 similar transistors is measured
and the results are as shown. By drawing a
histogramofthisfrequency distribution,deter-
mine the mean, median and modal values of
the distribution.
83.5–85.5 6
86.5–88.5 39
89.5–91.5 27
92.5–94.5 15
95.5–97.5 3
4. The diameters, in centimetres, of 60 holes
bored in engine castings are measured and
the results are as shown. Draw a histogram
depicting these results and hence determine
the mean, median and modal values of the
distribution.
2.011–2.014 7
2.016–2.019 16
2.021–2.024 23
2.026–2.029 9
2.031–2.034 5
32.4 Standard deviation
32.4.1 Discrete data
Thestandard deviation ofaset ofdatagivesan indication
of the amount of dispersion, or the scatter, of members
of the set from the measure of central tendency. Its value
is the root-mean-square value of the members of the set
and for discrete data is obtained as follows.
(i) Determine the measure of central tendency, usu-
ally the mean value, (occasionally the median or
modal values are specified).
(ii) Calculate the deviation of each member of the set
from 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 − x
2
+ ···
n
(vi) Determine the square root of (v).
The standard deviation is indicated by σ (the Greek
letter small ‘sigma’) and is written mathematically as
standard deviation,σ =
(x − x)2
n
where x is a member of the set, x is the mean value of
the set and n is the number of members in the set. The
value of standard deviation gives an indication of the
distance of the members of a set from the mean value.
The set {1, 4, 7, 10, 13} has a mean value of 7 and a
standard deviation of about 4.2. The set {5, 6, 7, 8, 9}
also has a mean value of 7 but the standard deviation is
about 1.4. This shows that the members of the second
set are mainly much closer to the mean value than the
members of the first set. The method of determining the
standard deviation for a set of discrete data is shown in
Problem 5.
Problem 5. Determine the standard deviation
from the mean of the set of numbers
{5, 6, 8, 4, 10, 3}, correct to 4 significant figures
The arithmetic mean, x =
x
n
=
5 + 6 + 8 + 4 + 10 + 3
6
= 6
Standard deviation, σ =
(x − x)2
n
The (x − x)2 values are (5 − 6)2,(6 − 6)2,(8 − 6)2,
(4 − 6)2
,(10 − 6)2
and (3 − 6)2
.
The sum of the (x − x)2 values,
i.e. (x − x)2
, is 1 + 0 + 4 + 4 + 16 + 9 = 34
and
(x − x)2
n
=
34
6
= 5.
·
6
since there are 6 members in the set.
Hence, standard deviation,
σ =
(x − x)2
n
= 5.
·
6=2.380,

32.4.2 Grouped data
For grouped data,
standard deviation,σ =
{ f (x − x)2
}
f
where 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 a
set of grouped data is shown in Problem 6.
values of resistance in ohms of 48 resistors is as
shown. Calculate the standard deviation from the
mean of the resistors, correct to 3 significant figures
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
The standard deviation for grouped data is given by
σ =
{ f (x − x)2
}
f
From Problem 3, the distribution mean value is
x = 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)2
values are 3(20.7 − 21.92)2
,
10(21.2 − 21.92)2,11(21.7 − 21.92)2,...
The f (x − x)2
values are
4.4652 + 5.1840+ 0.5324 + 1.0192
+5.4756 + 3.2768 = 19.9532
f (x − x)2
f
=
19.9532
48
= 0.41569
and standard deviation,
σ =
{ f (x − x)2
}
f
=
√
0.41569
= 0.645,correct to 3 significant figures.
PracticeExercise 127 Standard deviation
1. Determine the standard deviation from the
mean of the set of numbers
{35, 22, 25, 23, 28, 33, 30},
2. The values of capacitances, in microfarads, of
ten capacitors selected at random from a large
batch of similar capacitors are
34.3, 25.0, 30.4, 34.6, 29.6, 28.7,
33.4, 32.7, 29.0 and 31.3.
Determine the standard deviation from the
mean for these capacitors, correct to 3 signif-
icant figures.
3. The tensile strength in megapascals for 15
samplesoftin weredetermined and found to be
34.61, 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 deviation
from the mean for these 15 values, correct to
4. Calculate the standard deviation from the
mean for the mass of the 50 bricks given in
problem 1 of Practice Exercise 126, page 301,
5. Determine the standard deviation from the
mean, correct to 4 significant figures, for the
heights of the 100 people given in problem 2
of Practice Exercise 126, page 301.
6. Calculate the standard deviation from the
mean for the data given in problem 4 of
Practice Exercise 126, page 302, correct to 3
decimal places.
32.5 Quartiles, deciles and percentiles
Other measures of dispersion which are sometimes
used are the quartile, decile and percentile values. The
quartile values of a set of discrete data are obtained by
selecting the values of members which divide the set

into four equal parts. Thus, for the set {2, 3, 4, 5, 5, 7,
9, 11, 13, 14, 17} there are 11 members and the values
of the members dividing the set into four equal parts
are 4, 7 and 13. These values are signified by Q1, Q2
and Q3 and called the first, second and third quartile
values, respectively. It can be seen that the second quar-
tile value, Q2, is the value of the middle member and
hence is the median value of the set.
For grouped data the ogive may be used to determine
thequartilevalues.In thiscase,pointsareselected on the
vertical cumulative frequency values of the ogive, such
that they divide the total value of cumulative frequency
into four equal parts. Horizontal lines are drawn from
these values to cut the ogive. The values of the variable
corresponding to these cutting points on the ogive give
the quartile values (see Problem 7).
When a set contains a large number of members, the
set can be split into ten parts, each containing an equal
number of members. These ten parts are then called
deciles.Forsetscontaining avery largenumberofmem-
bers, the set may be split into one hundred parts, each
containing an equal number of members. One of these
parts is called a percentile.
Problem 7. The frequency distribution given
below refers to the overtime worked by a group of
craftsmen during each of 48 working weeks in a
year. Draw an ogive for these data and hence
determine the quartile values.
25–29 5
30–34 4
35–39 7
40–44 11
45–49 12
50–54 8
55–59 1
The cumulative frequency distribution (i.e. upper class
boundary/cumulative frequency values) is
29.5 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 divided
into four equal parts, each having a range of 48 ÷ 4,
i.e. 12. This gives cumulative frequency values of 0 to
12 corresponding to the first quartile, 12 to 24 corre-
sponding to the second quartile, 24 to 36 corresponding
to the third quartile and 36 to 48 corresponding to the
fourth quartile of the distribution; i.e., the distribution
25
10
Cumulativefrequency
40
30
20
50
4030 35Q1 Q2 Q345
Upper class boundary values, hours
55 6050
Figure 32.2
is divided into four equal parts. The quartile values
are those of the variable corresponding to cumulative
frequency values of 12, 24 and 36, marked Q1, Q2 and
Q3 in Figure 32.2. These values, correct to the nearest
hour, are 37 hours, 43 hours and 48 hours, respec-
tively. The Q2 value is also equal to the median value
of the distribution. One measure of the dispersion of a
distribution is called the semi-interquartile range and
is given by (Q3 − Q1) ÷ 2 and is (48 − 37) ÷ 2 in this
case; i.e., 5
1
2
hours.
Problem 8. Determine the numbers contained in
the (a) 41st to 50th percentile group and (b) 8th
decile group of the following set of numbers.
14 22 17 21 30 28 37 7 23 32
24 17 20 22 27 19 26 21 15 29
The set is ranked, giving
7 14 15 17 17 19 20 21 21 22
22 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 50th
percentile group will be the numbers 21 and 22.
(b) The first decile group is obtained by splitting the
ranked set into 10 equal groups and selecting the
first group; i.e., the numbers 7 and 14. The second
decile group is the numbers 15 and 17, and so on.
Thus, the 8th decile group contains the numbers
27 and 28.

PracticeExercise 128 Quartiles, deciles
and percentiles (answers on page 354)
1. The number of working days lost due to acci-
dents for each of 12 one-monthly periods are
as shown. Determine the median and ﬁrst and
third quartile values for this data.
27 37 40 28 23 30 35 24 30 32 31 28
2. The number of faults occurring on a produc-
tion line in a nine-week period are as shown
below. Determine the median and quartile
values for the data.
30 27 25 24 27 37 31 27 35
3. Determine the quartile values and semi-
interquartile range for the frequency distribu-
tion given in problem 2 of Practice Exercise
126, page 301.
4. Determine the numbers contained in the 5th
decile group and in the 61st to 70th percentile
groups for the following set of numbers.
40 46 28 32 37 42 50 31 48 45
32 38 27 33 40 35 25 42 38 41
5. Determine the numbers in the 6th decile group
and in the 81st to 90th percentile group for the
following set of numbers.
43 47 30 25 15 51 17 21 37 33 44 56 40 49 22
36 44 33 17 35 58 51 35 44 40 31 41 55 50 16

Chapter 33
Probability
33.1 Introduction to probability
33.1.1 Probability
The probability of something happening is the likeli-
hood or chance of it happening. Values of probabilitylie
between 0 and 1, where 0 represents an absolute impos-
sibilityand 1 represents an absolute certainty. The prob-
ability of an event happening usually lies somewhere
between these two extreme values and is expressed
as either a proper or decimal fraction. Examples of
probability are
that a length of copper wire
has zero resistance at 100◦
C 0
that a fair, six-sided dice
will stop with a 3 upwards
1
6
or 0.1667
that a fair coin will land
with a head upwards
1
2
or 0.5
that a length of copper wire
has some resistance at 100◦C 1
If p is the probability of an event happening and q is the
probability of the same event not happening, then the
total probability is p + q and is equal to unity, since it
is an absolute certainty that the event either will or will
not occur; i.e., p + q = 1.
Problem 1. Determine the probabilities of
selecting at random (a) a man and (b) a woman
from a crowd containing 20 men and 33 women
(a) The probability of selecting at random a man, p,
is given by the ratio
number of men
number in crowd
i.e. p =
20
20 + 33
=
20
53
or 0.3774
(b) The probability of selecting at random a woman,
q, is given by the ratio
number of women
number in crowd
i.e. q =
33
20 + 33
=
33
53
or 0.6226
(Check: the total probability should be equal to 1:
p =
20
53
and q =
33
53
, thus the total probability,
p + q =
20
53
+
33
53
= 1
hence no obvious error has been made.)
33.1.2 Expectation
The expectation, E, of an event happening is deﬁned
in general terms as the product of the probability p of
an event happening and the number of attempts made,
n; i.e., E = pn.
Thus, since the probability of obtaining a 3 upwards
when rolling a fair dice is 1/6, the expectation of getting
a 3 upwards on four throws of the dice is
1
6
× 4,i.e.
2
3
Thus expectation is the average occurrence of an
event.
DOI: 10.1016/B978-1-85617-697-2.00033-8

Probability 307
Problem 2. Find the expectation of obtaining a 4
upwards with 3 throws of a fair dice
Expectation is the average occurrence of an event and is
defined as the probability times the number of attempts.
The probability, p, of obtaining a 4 upwards for one
throw of the dice is 1/6.
If 3 attempts are made, n = 3 and the expectation, E, is
pn, i.e.
E =
1
6
× 3 =
1
2
or 0.50
33.1.3 Dependent events
A dependent event is one in which the probability of
an event happening affects the probability of another
ever happening. Let 5 transistors be taken at random
from a batch of 100 transistors for test purposes and the
probability of there being a defective transistor, p1, be
determined. At some later time, let another 5 transistors
be taken at random from the 95 remaining transistors
in the batch and the probability of there being a defec-
tive transistor, p2, be determined. The value of p2 is
different from p1 since the batch size has effectively
altered from 100 to 95; i.e., probability p2 is depen-
dent on probability p1. Since transistors are drawn and
then another 5 transistors are drawn without replacing
the first 5, the second random selection is said to be
without replacement.
33.1.4 Independent events
An independent event is one in which the probability
of an event happening does not affect the probability
of another event happening. If 5 transistors are taken at
random from a batch oftransistorsand the probabilityof
a defective transistor, p1, is determined and the process
is repeated after the original 5 have been replaced in
the batch to give p2, then p1 is equal to p2. Since the
5 transistors are replaced between draws, the second
selection is said to be with replacement.
33.2 Laws of probability
33.2.1 The addition law of probability
The addition law of probability is recognized by the
word ‘or’ joining the probabilities.
If pA is the probability of event A happening and pB
is the probability of event B happening, the probability
of 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 by
pA + pB + pC + ··· + pN
33.2.2 The multiplicationlaw of probability
The multiplication law of probability is recognized by
the word ‘and’ joining the probabilities.
If pA is the probability of event A happening and pB is
the probability of event B happening, the probability of
event A and event B happening is given by pA × pB.
Similarly, the probabilityof events A and B and C and
...N happening is given by
pA × pB × pC × ...× pN
Here are some worked problems to demonstrate proba-
bility.
Problem 3. Calculate the probabilities of
selecting at random
(a) the winning horse in a race in which 10 horses
are running and
(b) the winning horses in both the first and second
races if there are 10 horses in each race
(a) Since only one of the ten horses can win, the
probability of selecting at random the winning
horse is
number of winners
number of horses
,i.e.
1
10
or 0.10
(b) The probability of selecting the winning horse in
the first race is
1
10
The probability of selecting the winning horse in
the second race is
1
10
The probability of selecting the winning horses
in the first and second race is given by the
multiplication law of probability; i.e.,
probability =
1
10
×
1
10
=
1
100
or 0.01
Problem 4. The probability of a component
failing in one year due to excessive temperature is
1/20, that due to excessive vibration is 1/25 and that
due to excessive humidity is 1/50. Determine the
probabilities that during a one-year period a
component
(a) fails due to excessive temperature and
excessive vibration,

(b) fails due to excessive vibration or excessive
humidity,
(c) will not fail because of both excessive
temperature and excessive humidity
Let pA be the probability of failure due to excessive
temperature, then
pA =
1
20
and pA =
19
20
(where pA is the probability of not failing)
Let pB be the probability of failure due to excessive
vibration, then
pB =
1
25
and pB =
24
25
Let pC be the probability of failure due to excessive
humidity, then
pC =
1
50
and pC =
49
50
(a) The probability of a component failing due to
excessive temperature and excessive vibration is
given by
pA × pB =
1
20
×
1
25
=
1
500
or 0.002
(b) The probability of a component failing due to
excessive vibration or excessive humidity is
pB + pC =
1
25
+
1
50
=
3
50
or 0.06
(c) The probability that a component will not fail due
excessive temperature and will not fail due to
excess humidity is
pA × pC =
19
20
×
49
50
=
931
1000
or 0.931
Problem 5. A batch of 100 capacitors contains 73
which are within the required tolerance values and
17 which are below the required tolerance values,
the remainder being above the required tolerance
values. Determine the probabilities that, when
randomly selecting a capacitor and then a second
capacitor,
(a) both are within the required tolerance values
when selecting with replacement,
(b) the first one drawn is below and the second
one drawn is above the required tolerance
value, when selection is without replacement
(a) The probability of selecting a capacitor within
the required tolerance values is 73/100. The first
capacitor drawn is now replaced and a second one
is drawn from the batch of 100. The probability
of this capacitor being within the required toler-
ance values is also 73/100.Thus, the probabilityof
selecting a capacitor within the required tolerance
values for both the first and the second draw is
73
100
×
73
100
=
5329
10000
or 0.5329
(b) The probability of obtaining a capacitor below
the required tolerance values on the first draw
is 17/100. There are now only 99 capacitors
left in the batch, since the first capacitor is not
replaced. The probability of drawing a capacitor
above the required tolerance values on the second
draw is 10/99, since there are (100 − 73 − 17),
i.e. 10, capacitors above the required tolerance
value. Thus, the probability of randomly select-
ing a capacitor below the required tolerance values
and subsequently randomly selecting a capacitor
above the tolerance values is
17
100
×
10
99
=
170
9900
=
17
990
or 0.0172
PracticeExercise 129 Laws of probability
1. In a batch of 45 lamps 10 are faulty. If one
lamp is drawn at random, find the probability
of it being (a) faulty (b) satisfactory.
2. A box of fuses are all of the same shape and
size and comprises 23 2A fuses, 47 5A fuses
and 69 13A fuses. Determine the probability
of selecting at random (a) a 2A fuse (b) a 5A
fuse (c) a 13A fuse.
3. (a) Find the probability of having a 2
upwards when throwing a fair 6-sided
dice.
(b) Find the probability of having a 5
upwards when throwing a fair 6-sided
dice.
(c) Determine the probability of having a 2
and then a 5 on two successive throws of
a fair 6-sided dice.

Probability 309
4. Determine the probability that the total score
is 8 when two like dice are thrown.
5. The probability of event A happening is
3
5
and the probability of event B happening is
2
3
. Calculate the probabilities of
(a) both A and B happening.
(b) only event A happening, i.e. event A
happening and event B not happening.
(c) only event B happening.
(d) either A, or B, or A and B happening.
6. When testing 1000 soldered joints, 4 failed
during a vibration test and 5 failed due to
having 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 5
which are defective. A component is drawn at
random from the batch and tested and then a second
component is drawn. Determine the probability that
neither of the components is defective when drawn
(a) with replacement and (b) without replacement
(a) With replacement
The probabilitythat the component selected on the
first 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 the
first component drawn and the second component
drawn are satisfactory is
7
8
×
7
8
=
49
64
or 0.7656
(b) Without replacement
The probability that the first component drawn
is satisfactory is 7/8. There are now only 34
satisfactory components left in the batch and the
batch number is 39. Hence, the probability of
drawing a satisfactory component on the sec-
ond draw is 34/39. Thus, the probability that the
first component drawn and the second component
drawn are satisfactory i.e., neither is defective is
7
8
×
34
39
=
238
312
or 0.7628
Problem 7. A batch of 40 components contains 5
which are defective. If a component is drawn at
random from the batch and tested and then a second
component is drawn at random, calculate the
probability of having one defective component, both
(a) with replacement and (b) without replacement
The probability of having one defective component can
be achieved in two ways. If p is the probability of draw-
ing a defective component and q is the probability of
drawing a satisfactory component, then the probability
of having one defective component is given by drawing
a satisfactory component and then a defective compo-
nent or by drawing a defective component and then a
satisfactory one; i.e., by q × p + p × q.
(a) With replacement
p =
5
40
=
1
8
and q =
35
40
=
7
8
Hence, the probability of having one defective
component is
1
8
×
7
8
+
7
8
×
1
8
i.e.
7
64
+
7
64
=
7
32
or 0.2188
(b) Without replacement
p1 =
1
8
and q1 =
7
8
on the first of the two draws
The batch number is now 39 for the second draw,
thus,
p2 =
5
39
and q2 =
35
39
p1q2 + q1 p2 =
1
8
×
35
39
+
7
8
×
5
39
=
35 + 35
312
=
70
312
or 0.2244

Problem 8. A box contains 74 brass washers,
86 steel washers and 40 aluminium washers. Three
washers are drawn at random from the box without
replacement. Determine the probability that all
three are steel washers
Assume, for clarity of explanation, that a washer is
drawn at random, then a second, then a third (although
this assumption does not affect the results obtained).
The total number of washers is
74 + 86 + 40, i.e. 200
The probability of randomly selecting a steel washer on
the first draw is 86/200. There are now 85 steel washers
in a batch of 199. The probability of randomly selecting
a steel washer on the second draw is 85/199. There are
now 84 steel washers in a batch of 198. The probability
of randomly selecting a steel washer on the third draw
is 84/198. Hence, the probability of selecting a steel
washer on the first draw and the second draw and the
third draw is
86
200
×
85
199
×
84
198
=
614040
7880400
= 0.0779
Problem 9. For the box of washers given in
Problem 8 above, determine the probability that
there are no aluminium washers drawn when three
washers are drawn at random from the box without
replacement
The probability of not drawing an aluminium washer
on the first draw is 1 −
40
200
i.e., 160/200. There are
now 199 washers in the batch of which 159 are not made
of aluminium. Hence, the probability of not drawing
an aluminium washer on the second draw is 159/199.
Similarly, the probability of not drawing an aluminium
washer on the third draw is 158/198. Hence the proba-
bility of not drawing an aluminium washer on the first
and second and third draws is
160
200
×
159
199
×
158
198
=
4019520
7880400
= 0.5101
Problem 10. For the box of washers in Problem 8
above, find the probability that there are two brass
washers and either a steel or an aluminium washer
when three are drawn at random, without
replacement
Two brass washers (A) and one steel washer (B) can be
obtained in any of the following ways.
1st draw 2nd draw 3rd draw
A A B
A B A
B A A
Two brass washers and one aluminium washer (C) can
also be obtained in any of the following ways.
1st draw 2nd draw 3rd draw
A A C
A C A
C A A
Thus, 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, the
combinations and their probabilities are as shown.
Draw Probability
First Second Third
A A B
74
200
×
73
199
×
86
198
= 0.0590
A B A
74
200
×
86
199
×
73
198
= 0.0590
B A A
86
200
×
74
199
×
73
198
= 0.0590
A A C
74
200
×
73
199
×
40
198
= 0.0274
A C A
74
200
×
40
199
×
73
198
= 0.0274
C A A
40
200
×
74
199
×
73
198
= 0.0274
The probability of having the first combination or the
second or the third, and so on, is given by the sum of
the probabilities; i.e., by 3 × 0.0590 + 3 × 0.0274, i.e.
0.2592

Probability 311
PracticeExercise 130 Laws of probability
1. The probabilitythat component A will operate
satisfactorily for 5 years is 0.8 and that B will
operate satisfactorily over that same period of
time is 0.75. Find the probabilities that in a 5
year period
(a) both components will operate
satisfactorily.
(b) only component A will operate
satisfactorily.
(c) only component B will operate
satisfactorily.
2. In a particular street, 80% of the houses have
landline telephones. If two houses selected at
random are visited, calculate the probabilities
that
(a) they both have a telephone.
(b) one has a telephone but the other does
not.
3. Veroboard pins are packed in packets of 20
by a machine. In a thousand packets, 40 have
less than 20 pins. Find the probability that if 2
packets are chosen at random, one will contain
less than 20 pins and the other will contain 20
pins or more.
4. A batch of 1kW fire elements contains 16
which are within a power tolerance and 4
which are not.If 3 elements are selected at ran-
dom from the batch, calculate the probabilities
that
(a) all three are within the power tolerance.
(b) two are within but one is not within the
power tolerance.
5. An amplifier is made up of three transis-
tors, A, B and C. The probabilities of A, B
or C being defective are 1/20, 1/25 and
1/50, respectively. Calculate the percentage of
amplifiers produced
(a) which work satisfactorily.
(b) which have just one defective transistor.
6. Abox contains14 40Wlamps,28 60Wlamps
and 58 25W lamps, all the lamps being of the
same shape and size. Three lamps are drawn at
random from the box, first one, then a second,
then a third. Determine the probabilities of
(a) getting one25W,one40Wand one60W
lamp with replacement.
(b) getting one25W,one40Wand one60W
lamp without replacement.
(c) getting either one 25W and two 40W
or one 60W and two 40 Wlamps with
replacement.

Revision Test 13 : Statistics and probability
1. A company produces five products in the follow-
ing proportions:
Product A 24
Product B 6
Product C 15
Product D 9
Product E 18
Draw (a) a horizontal bar chart and (b) a pie
diagram to represent these data visually. (9)
2. State whether the data obtained on the following
topics are likely to be discrete or continuous.
(a) the number of books in a library.
(b) the speed of a car.
(c) the time to failure of a light bulb. (3)
3. Draw a histogram, frequency polygon and ogive
for the data given below which refers to the
diameter of 50 components produced by a
machine.
Class intervals Frequency
1.30–1.32mm 4
1.33–1.35mm 7
1.36–1.38mm 10
1.39–1.41mm 12
1.42–1.44mm 8
1.45–1.47mm 5
1.48–1.50mm 4
(16)
4. Determine the mean, median and modal values for
the 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 shown
below.
50–56 6
57–63 16
64–70 22
71–77 30
78–84 19
85–91 7
Determine for the sample
(a) the mean value.
(b) the standard deviation, correct to 4 significant
figures. (10)
6. The number of faulty components in a factory in
a 12 week period is
14 12 16 15 10 13 15 11 16 19 17 19
Determine the median and the first and third
quartile values. (7)
7. Determine the probability of winning a prize in
a lottery by buying 10 tickets when there are 10
prizes and a total of 5000 tickets sold. (4)
8. A sample of 50 resistors contains 44 which are
within the required tolerance value, 4 which are
below and the remainder which are above. Deter-
mine the probability of selecting from the sample
a resistor which is
(a) below the required tolerance.
(b) above the required tolerance.
Now two resistors are selected at random from
the sample. Determine the probability, correct to
3 decimal places, that neither resistor is defective
when drawn
(c) with replacement.
(d) without replacement.
(e) If a resistor is drawn at random from the
batch and tested and then a second resistor is
drawn from those left, calculate the probabil-
ity of having one defective component when
selection is without replacement. (15)

Chapter 34
Introduction to
differentiation
34.1 Introduction to calculus
Calculus is a branch of mathematics involving or lead-
ing to calculations dealing with continuously varying
functions such as velocity and acceleration, rates of
change and maximum and minimum values of curves.
Calculus has widespread applications in science and
engineering and is used to solve complicated problems
for 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 differentiation
and appliesdifferentiation to ratesofchange.Chapter35
introduces integration and applies it to determine areas
under curves.
Further applications of differentiation and integration
are explored in Engineering Mathematics (Bird, 2010).
34.2 Functional notation
In an equation such as y = 3x2 + 2x − 5, y is said to be
a function of x and may be written as y = f (x).
An equation written in the form f (x) = 3x2 + 2x − 5
is 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, when
f (x) = 3x2 + 2x − 5,
f (0) = 3(0)2
+ 2(0) − 5 = −5
and f (2) = 3(2)2
+ 2(2) − 5 = 11,and so on.
Problem 1. If f (x) = 4x2 − 3x + 2, find
f (0), f (3), f (−1) and f (3) − f (−1)
f (x) = 4x2
− 3x + 2
f (0) = 4(0)2
− 3(0) + 2 = 2
f (3) = 4(3)2
− 3(3) + 2 = 36 − 9 + 2 = 29
f (−1) = 4(−1)2
− 3(−1) + 2 = 4 + 3 + 2 = 9
f (3) − f (−1) = 29 − 9 = 20
Problem 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 = 15
f (1) = 5(1)2 + 1 − 7 = −1
f (2) ÷ f (1) =
15
−1
= −15
PracticeExercise 131 Functional notation
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

3. Given f (x) = 3x3 + 2x2 − 3x + 2, prove that
f (1) =
1
7
f (2).
34.3 The gradient of a curve
If a tangent is drawn at a point P on a curve, the gradient
of this tangent is said to be the gradient of the curve
at P. In Figure 34.1, the gradient of the curve at P is
equal to the gradient of the tangent PQ.
0 x
Q
P
f(x)
Figure 34.1
For the curve shown in Figure 34.2, let the points
A and B have co-ordinates (x1, y1) and (x2, y2),
respectively. In functional notation, y1 = f (x1) and
y2 = f (x2), as shown.
0
B
A
E D
C
f(x2)
f(x1)
f(x)
xx1 x2
Figure 34.2
The gradient of the chord AB
=
BC
AC
=
BD − CD
ED
=
f (x2) − f (x1)
(x2 − x1)
For the curve f (x) = x2 shown in Figure 34.3,
(a) the gradient of chord AB
=
f (3) − f (1)
3 − 1
=
9 − 1
2
= 4
0 1 1.5 2 3
2
4
6
8
10
f(x)
x
A
D
C
B f(x)5x2
Figure 34.3
(b) the gradient of chord AC
=
f (2) − f (1)
2 − 1
=
4 − 1
1
= 3
(c) the gradient of chord AD
=
f (1.5) − f (1)
1.5 − 1
=
2.25 − 1
0.5
= 2.5
(d) if E is the point on the curve (1.1, f (1.1)) then the
gradient of chord AE
=
f (1.1) − f (1)
1.1 − 1
=
1.21 − 1
0.1
= 2.1
(e) if F is the point on the curve (1.01, f (1.01)) then
the gradient of chord AF
=
f (1.01) − f (1)
1.01 − 1
=
1.0201 − 1
0.01
= 2.01
Thus, as point B moves closer and closer to point A,
the gradient of the chord approaches nearer and nearer
to the value 2. This is called the limiting value of the
gradient of the chord AB and when B coincides with A
the chord becomes the tangent to the curve.
PracticeExercise 132 The gradient of a
curve (answers on page 354)
1. Plot the curve f (x) = 4x2 − 1 for values
of x from x = −1 to x = +4. Label the
co-ordinates (3, f (3)) and (1, f (1)) as J
and K, respectively. Join points J and K to
form the chord JK. Determine the gradient of

Introduction to differentiation 315
chord JK. By moving J nearer and nearer to
K, determine the gradient of the tangent of the
curve at K.
34.4 Differentiation from first
principles
In Figure 34.4, A and B are two points very close
together on a curve, δx (delta x) and δy (delta y) rep-
resenting small increments in the x and y directions,
respectively.
0
y
f(x)
f(x1␦x)
␦y
␦x
x
A(x, y)
B (x 1␦x, y1␦y)
Figure 34.4
Gradient of chord AB =
δy
δx
however, δy = f (x + δx) − f (x)
Hence,
δy
δx
=
f (x + δx) − f (x)
δx
As δx approaches zero,
δy
δx
approaches a limiting value
and the gradient of the chord approaches the gradient of
the tangent at A.
When determining the gradient of a tangent to a curve
there are two notations used. The gradient of the curve
at A in Figure 34.4 can either be written as
limit
δx→0
δy
δx
or limit
δx→0
f (x + δx) − f (x)
δx
In Leibniz notation,
dy
dx
= limit
δx→0
δy
δx
In functional notation,
f (x) = limit
δx→0
f (x + δx) − f (x)
δx
dy
dx
is the same as f (x) and is called the differential
coefficient or the derivative. The process of finding the
differential coefficient is called differentiation.
Summarizing, the differential coefficient,
dy
dx
= f (x) = limit
δx→0
δy
δx
= limit
δx→0
f (x + δx) − f (x)
δx
Problem 3. Differentiate from first principles
f (x) = x2
To ‘differentiate from first principles’ means ‘to find
f (x)’ using the expression
f (x) = limit
δx→0
f (x + δx) − f (x)
δx
f (x) = x2 and substituting (x + δx) for x gives
f (x + δx) = (x + δx)2 = x2 + 2xδx + δx2, hence,
f (x) = limit
δx→0
(x2 + 2xδx + δx2) − (x2)
δx
= limit
δx→0
2xδx + δx2
δx
= limit
δx→0
{2x + δx}
As δx → 0,{2x + δx} → {2x + 0}.
Thus, f (x) = 2x i.e. the differential coefficient of x2
is 2x.
This means that the general equation for the gradient of
the 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 lengthy
process and we do not want to have to go through this
procedureevery timewewant to differentiateafunction.
In reality we do not have to because from the above
procedure has evolved a set general rule, which we
consider in the following section.
34.5 Differentiation of y = axn by the
general rule
From differentiation by first principles, a general rule
for differentiating axn emerges where a and n are any
constants. This rule is
if y = axn
then
dy
dx
= anxn−1
or if f (x) = axn
then f (x) = anxn−1

When differentiating, results can be expressed in a
number of ways. For example,
(a) if y = 3x2 then
dy
dx
= 6x
(b) if f (x) = 3x2 then f (x) = 6x
(c) the differential coefficient of 3x2 is 6x
(d) the derivative of 3x2
is 6x
(e)
d
dx
(3x2) = 6x
34.5.1 Revision of some laws of indices
1
xa
= x−a
For example,
1
x2
= x−2
and x−5
=
1
x5
√
x = x
1
2 For example,
√
5 = 5
1
2 and
16
1
2 =
√
16 = ±4 and
1
√
x
=
1
x
1
2
= x−
1
2
a
√
xb = x
b
a For example,
3
√
x5 = x
5
3 and x
4
3 =
3
√
x4
and
1
3
√
x7
=
1
x
7
3
= x−
7
3
x0 = 1 For example, 70 = 1 and 43.50 = 1
general rule for differentiating y = axn.
Problem 4. Differentiate the following with
respect to x: y = 4x7
Comparing y = 4x7 with y = axn shows that a = 4 and
n = 7. Using the general rule,
dy
dx
= anxn−1
= (4)(7)x7−1
= 28x6
respect to x: y =
3
x2
y =
3
x2
= 3x−2, hence a = 3 and n = −2 in the general
rule.
dy
dx
= anxn−1
= (3)(−2)x−2−1
= −6x−3
= −
6
x3
respect to x: y = 5
√
x
y = 5
√
x = 5x
1
2 , hence a = 5 and n =
1
2
in the
general rule.
dy
dx
= anxn−1
= (5)
1
2
x
1
2 −1
=
5
2
x−
1
2 =
5
2x
1
2
=
5
2
√
x
Problem 7. Differentiate y = 4
y = 4 may be written as y = 4x0
; i.e., in the general rule
a = 4 and n = 0. Hence,
dy
dx
= (4)(0)x0−1
= 0
The equation y = 4 represents a straight horizontal
line and the gradient of a horizontal line is zero, hence
the result could have been determined on inspection.
In general, the differential coefficient of a constant is
always zero.
Problem 8. Differentiate y = 7x
Since y = 7x, i.e. y = 7x1
, in the general rule a = 7 and
n = 1. Hence,
dy
dx
= (7)(1)x1−1
= 7x0
= 7 since x0
= 1
The gradient of the line y = 7x is 7 (from
y = mx + c), hence the result could have been obtained
by inspection.In general, the differential coefficient of
kx, where k is a constant, is always k.
Problem 9. Find the differential coefficient of
y =
2
3
x4 −
4
x3
+ 9
y =
2
3
x4
−
4
x3
+ 9
i.e. y =
2
3
x4
− 4x−3
+ 9
dy
dx
=
2
3
(4)x4−1
− (4)(−3)x−3−1
+ 0
=
8
3
x3
+ 12x−4

i.e.
dy
dx
=
8
3
x3
+
12
x4
Problem 10. If f (t) = 4t +
1
√
t3
ﬁnd f (t)
f (t) = 4t +
1
√
t3
= 4t +
1
t
3
2
= 4t1
+ t− 3
2
Hence, f (t) = (4)(1)t1−1
+ −
3
2
t− 3
2 −1
= 4t0
−
3
2
t− 5
2
i.e. f (t) = 4 −
3
2t
5
2
= 4 −
3
2
√
t5
dy
dx
given y =
3x2 − 5x
2x
y =
3x2 − 5x
2x
=
3x2
2x
−
5x
2x
=
3
2
x −
5
2
Hence,
dy
dx
=
3
2
or 1.5
y =
2
5
x3
−
4
x3
+ 4
√
x5 + 7
y =
2
5
x3
−
4
x3
+ 4 x5 + 7
i.e. y =
2
5
x3
− 4x−3
+ 4x
5
2 + 7
dy
dx
=
2
5
(3)x3−1
− (4)(−3)x−3−1
+ (4)
5
2
x
5
2 −1
+ 0
=
6
5
x2
+ 12x−4
+ 10x
3
2
i.e.
dy
dx
=
6
5
x2
+
12
x4
+ 10 x3
Problem 13. Differentiate y =
(x + 2)2
x
with
respect to x
y =
(x + 2)2
x
=
x2 + 4x + 4
x
=
x2
x
+
4x
x
+
4
x
i.e. y = x1
+ 4 + 4x−1
Hence,
dy
dx
= 1x1−1
+ 0 + (4)(−1)x−1−1
= x0
− 4x−2
= 1 −
4
x2
(since x0
= 1)
Problem 14. Find the gradient of the curve
y = 2x2
−
3
x
at x = 2
y = 2x2
−
3
x
= 2x3
− 3x−1
Gradient =
dy
dx
= (2)(2)x2−1
− (3)(−1)x−1−1
= 4x + 3x−2
= 4x +
3
x2
When x = 2, gradient = 4x +
3
x2
= 4(2) +
3
(2)2
= 8 +
3
4
= 8.75
y = 3x4 − 2x2 + 5x − 2 at the points (0,−2)
and (1,4)
The gradient of a curve at a given point is given by the
corresponding value of the derivative.
Thus, since y = 3x4 − 2x2 + 5x − 2,
the gradient =
dy
dx
= 12x3 − 4x + 5.
At the point (0,−2), x = 0, thus
the gradient = 12(0)3 − 4(0) + 5 = 5
At the point (1,4), x = 1, thus
the gradient = 12(1)3 − 4(1) + 5 = 13
PracticeExercise 133 Differentiation of
y = axn
by the general rule (answers on
page 354)
In problems 1 to 20, determine the differential
coefﬁcients with respect to the variable.
1. y = 7x4 2. y = 2x + 1
3. y = x2 − x 4. y = 2x3 − 5x + 6

5. y =
1
x
6. y = 12
7. y = x −
1
x2
8. y = 3x5 − 2x4 + 5x3 + x2 − 1
9. y =
2
x3
10. y = 4x(1 − x)
11. y =
√
x 12. y =
√
t3
13. y = 6 +
1
x3
14. y = 3x −
1
√
x
+
1
x
15. y = (x + 1)2 16. y = x + 3
√
x
17. y = (1 − x)2 18. y =
5
x2
−
1
√
x7
+ 2
19. y = 3(t − 2)2 20. y =
(x + 2)2
x
21. Find the gradient of the following curves at
the given points.
(a) y = 3x2
at x = 1
(b) y =
√
x at x = 9
(c) y = x3 + 3x − 7 at x = 0
(d) y =
1
√
x
at x = 4
(e) y =
1
x
at x = 2
(f) y = (2x + 3)(x − 1) at x = −2
22. Differentiate f (x) = 6x2 − 3x + 5 and ﬁnd
the gradient of the curve at
(a) x = −1 (b) x = 2
23. Find the differential coefﬁcient of
y = 2x3 + 3x2 − 4x − 1 and determine the
gradient of the curve at x = 2.
24. Determine the derivative of
y = −2x3 + 4x + 7 and determine the
gradient of the curve at x = −1.5
34.6 Differentiation of sine and
cosine functions
Figure 34.5(a) shows a graph of y = sin x. The gradient
is continually changing as the curve moves from 0 to
y
0
(a)
(b)
0
y 5sin x
x radians
x radians
1
2
2
A
A9
09
C9
B9
D9
B D
C
␲
2
d
dx
1
dy
dx
2␲
2␲
3␲
2
␲
␲ 3␲
2
(sinx) 5cosx
␲
2
Figure 34.5
A to B to C to D. The gradient, given by
dy
dx
, may be
plotted in a corresponding position below y = sin x, as
shown in Figure 34.5(b).
At 0, the gradient is positiveand is at itssteepest. Hence,
0 is a maximum positive value. Between 0 and A the
gradient is positive but is decreasing in value until at A
the gradient is zero, shown as A . Between A and B the
gradient is negative but is increasing in value until at B
the gradient is at its steepest. Hence B is a maximum
negative value.
If the gradient of y = sin x is further investigated
between B and C and C and D then the resulting graph
of
dy
dx
is seen to be a cosine wave.
Hence the rate of change of sinx is cosx, i.e.
if y = sinx then
dy
dx
= cosx
It may also be shown that
if y = sinax,
dy
dx
= acosax (1)
(wherea isaconstant)
and if y = sin(ax + α),
dy
dx
= acos(ax + α) (2)
(where a and α are constants).
If a similar exercise is followed for y = cosx then the
graphs of Figure 34.6 result, showing
dy
dx
to be a graph
of sin x but displaced by π radians.

y
0
(a)
(b)
0
y 5cos x
x radians
x radians
1
2
2
␲
2
1
dy
dx
2␲␲ 3␲
2
␲
2
2␲␲ 3␲
2
(cos x)52sin x
d
dx
Figure 34.6
If each point on the curve y = sin x (as shown in
Figure 34.5(a)) were to be made negative (i.e. +
π
2
made −
π
2
,−
3π
2
made +
3π
2
, and so on) then the graph
shown in Figure 34.6(b) would result. This latter graph
therefore represents the curve of −sin x.
Thus,
if y = cosx,
dy
dx
= −sinx
if y = cosax,
dy
dx
= −asinax (3)
(wherea isaconstant)
and if y = cos(ax + α),
dy
dx
= −asin(ax + α) (4)
(where a and α are constants).
y = 7sin2x − 3cos4x
dy
dx
= (7)(2cos2x) − (3)(−4sin 4x)
from equations (1) and (3)
= 14cos2x + 12sin4x
respect to the variable (a) y = 2sin5θ
(b) f (t) = 3cos2t
(a) y = 2sin5θ
dy
dθ
= (2)(5cos5θ) = 10cos5θ
from equation (1)
(b) f (t) = 3cos2t
f (t) = (3)(−2sin 2t) = −6sin2t
from equation (3)
respect 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 by
v = 100sin200t volts, where t is the time in
seconds. Calculate the rate of change of voltage
when (a) t = 0.005s and (b) t = 0.01s
v = 100sin200t volts. The rate of change of v is given
by
dv
dt
dv
dt
= (100)(200cos200t) = 20000cos200t
(a) When t = 0.005s,
dv
dt
= 20000cos(200)(0.005) = 20000cos1.
cos1 means ‘the cosine of 1 radian’ (make sure
yourcalculator is on radians, not degrees). Hence,
dv
dt
= 10 806 volts per second
(b) When t = 0.01s,
dv
dt
= 20000cos(200)(0.01) = 20000cos2.
Hence,
dv
dt
= −8323 volts per second

PracticeExercise 134 Differentiation of
sine and cosine functions (answers on
page 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 = 2cos
1
2
x at
x =
π
2
4. Determine the gradient of the curve
y = 3sin2x at x =
π
3
5. An alternating current is given by
i = 5sin100t amperes, where t is the time
in seconds. Determine the rate of change of
current i.e.
di
dt
when t = 0.01 seconds.
6. v = 50sin40t volts represents an alternating
voltage, v, where t is the time in seconds. At
a time of 20 × 10−3 seconds, ﬁnd the rate of
change of voltage i.e.
dv
dt
.
7. If f (t) = 3sin(4t + 0.12) − 2cos(3t − 0.72),
determine f (t).
34.7 Differentiation of eax and lnax
A graph of y = ex is shown in Figure 34.7(a). The gra-
dient of the curve at any point is given by
dy
dx
and is
continually changing. By drawing tangents to the curve
at many points on the curve and measuring the gradient
ofthetangents,valuesof
dy
dx
forcorresponding valuesof
x may be obtained. These values are shown graphically
in Figure 34.7(b).
The graph of
dy
dx
against x is identical to the original
graph of y = ex
. It follows that
if y = ex
, then
dy
dx
= ex
if y = eax
, then
dy
dx
= aeax
3
y5ex
x21
5
10
15
20
y
212223 0
(a)
3
dy
dx
5ex
dy
dx
x21
5
10
15
20
y
212223 0
(b)
Figure 34.7
Therefore,
if y = 2e6x
,then
dy
dx
= (2)(6e6x
) = 12e6x
A graph of y = lnx is shown in Figure 34.8(a). The gra-
dient of the curve at any point is given by
dy
dx
and is
continually changing. By drawing tangents to the curve
at many points on the curve and measuring the gradient
of the tangents, values of
dy
dx
for corresponding values
of x may be obtained. These values are shown graphi-
cally in Figure 34.8(b).
The graph of
dy
dx
against x is the graph of
dy
dx
=
1
x
It follows that
if y = lnx, then
dy
dx
=
1
x
if y = lnax, then
dy
dx
=
1
x
(Note that, in the latter expression, the constant a does
not appear in the
dy
dx
term.) Thus,
if y = ln4x, then
dy
dx
=
1
x

1.0
0.5
0 1 2 3
(b)
(a)
y 5 In x
4 5 6 x
1.5
2
y
0
21
22
1
2
1 2 3 4 5 6 x
dy
5
dx
1
x
dy
dx
Figure 34.8
respect to the variable (a) y = 3e2x (b) f (t) =
4
3e5t
(a) If y = 3e2x then
dy
dx
= (3)(2e2x ) = 6e2x
(b) If f (t) =
4
3e5t
=
4
3
e−5t , then
f (t) =
4
3
(−5e−5t
) = −
20
3
e−5t
= −
20
3e5t
Problem 21. Differentiate y = 5ln3x
If y = 5ln3x, then
dy
dx
= (5)
1
x
=
5
x
PracticeExercise 135 Differentiation of eax
and lnax (answers on page 354)
1. Differentiate with respect to x: (a) y = 5e3x
(b) y =
2
7e2x
2. Given f (θ) = 5ln2θ − 4ln3θ, determine
f (θ).
3. If f (t) = 4lnt + 2, evaluate f (t) when
t = 0.25
4. Find the gradient of the curve
y = 2ex −
1
4
ln2x at x =
1
2
correct to 2
decimal places.
5. Evaluate
dy
dx
when x = 1, given
y = 3e4x −
5
2e3x
+ 8ln5x. Give the answer
34.8 Summary of standard
derivatives
The standard derivatives used in this chapter are sum-
marized in Table 34.1 and are true for all real values
of x.
Table 34.1
y or f (x)
dy
dx
or f (x)
axn
anxn−1
sinax a cosax
cosax −a sinax
eax aeax
lnax
1
x
y = 3x2
− 7x + 2 at the point (1,−2)
If y = 3x2 − 7x + 2, then gradient =
dy
dx
= 6x − 7
At the point (1,−2), x = 1,
hence gradient = 6(1) − 7 = −1
Problem 23. If y =
3
x2
− 2sin4x +
2
ex
+ ln5x,
determine
dy
dx
y =
3
x2
− 2sin4x +
2
ex
+ ln5x
= 3x−2
− 2sin4x + 2e−x
+ ln5x

dy
dx
= 3(−2x−3
) − 2(4cos4x) + 2(−e−x
) +
1
x
= −
6
x3
− 8cos4x −
2
ex
+
1
x
PracticeExercise 136 Standard derivatives
y = 2x4 + 3x3 − x + 4 at the points
(a) (0,4) (b) (1,8)
2. Differentiate with respect to x:
y =
2
x2
+ 2ln2x−2(cos5x + 3sin2x)−
2
e3x
34.9 Successive differentiation
When a function y = f (x) is differentiated with respect
to x,thedifferential coefficient iswritten as
dy
dx
or f (x).
If the expression is differentiated again, the second dif-
ferential coefficient is obtained and is written as
d2 y
dx2
(pronounced dee two y by dee x squared) or f (x)
(pronounced f double-dash x). By successive differen-
tiation further higher derivatives such as
d3 y
dx3
and
d4 y
dx4
may be obtained. Thus,
if y = 5x4
,
dy
dx
= 20x3
,
d2 y
dx2
= 60x2
,
d3 y
dx3
= 120x,
d4 y
dx4
= 120 and
d5 y
dx5
= 0
Problem 24. If f (x) = 4x5 − 2x3 + x − 3, find
f (x)
f (x) = 4x5
− 2x3
+ x − 3
f (x) = 20x4
− 6x2
+ 1
f (x) = 80x3
− 12x or 4x(20x2
− 3)
Problem 25. Given y =
2
3
x3 −
4
x2
+
1
2x
−
√
x,
determine
d2 y
dx2
y =
2
3
x3
−
4
x2
+
1
2x
−
√
x
=
2
3
x3
− 4x−2
+
1
2
x−1
− x
1
2
dy
dx
=
2
3
3x2
− 4 − 2x−3
+
1
2
− 1x−2
−
1
2
x−
1
2
i.e. dy
dx
= 2x2
+ 8x−3
−
1
2
x−2
−
1
2
x−
1
2
d2y
dx2
= 4x + (8)(−3x−4
) −
1
2
− 2x−3
−
1
2
−
1
2
x−
3
2
= 4x − 24x−4
+ 1x−3
+
1
4
x−
3
2
i.e.
d2y
dx2
= 4x −
24
x4
+
1
x3
+
1
4
√
x3
PracticeExercise 137 Successive
differentiation (answers on page 354)
1. If y = 3x4 + 2x3 − 3x + 2, find (a)
d2 y
dx2
(b)
d3 y
dx3
2. If y = 4x2 +
1
x
find
d2 y
dx2
3. (a) Given f (t) =
2
5
t2 −
1
t3
+
3
t
−
√
t + 1,
determine f (t).
(b) Evaluate f (t) in part (a) when t = 1.
4. If y = 3sin2t + cost, find
d2 y
dx2
5. If f (θ) = 2ln4θ, show that f (θ) = −
2
θ2

34.10 Rates of change
If a quantity y depends on and varies with a quantity
x then the rate of change of y with respect to x is
dy
dx
.
Thus, for example, the rate of change of pressure p with
height h is
dp
dh
A rate of change with respect to time is usually just
called ‘the rate of change’, the ‘with respect to time’
being assumed. Thus, for example, a rate of change of
current, i, is
di
dt
and a rate of change of temperature, θ,
is
dθ
dt
, 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 certain
metal rod at temperature t◦C is given by
L = 1 + 0.00003t + 0.0000004t2. Determine the
rate of change of length, in mm/◦C, when the
temperature is (a) 100◦C (b) 250◦C
The rate of change of length means
dL
dt
Since length L = 1 + 0.00003t + 0.0000004t2
, then
dL
dt
= 0.00003 + 0.0000008t.
(a) When t = 100◦C,
dL
dt
= 0.00003 + (0.0000008)(100)
= 0.00011m/◦C = 0.11mm/◦C.
(b) When t = 250◦C,
dL
dt
= 0.00003 + (0.0000008)(250)
= 0.00023m/◦C = 0.23mm/◦C.
Problem 27. The luminous intensity I candelas
of a lamp at varying voltage V is given by
I = 5 × 10−4
V2
. Determine the voltage at which
the light is increasing at a rate of 0.4 candelas
per volt
The rate of change of light with respect to voltage is
given by
dI
dV
Since I = 5 × 10−4 V2,
dI
dV
= (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 which
voltage,V =
0.4
10−3
= 0.4 × 10+3
= 400 volts
Problem 28. Newton’s law of cooling is given by
θ = θ0e−kt , where the excess of temperature at zero
time is θ0
◦C and at time t seconds is θ◦C.
Determine the rate of change of temperature after
50s, given that θ0 = 15◦C and k = −0.02
The rate of change of temperature is
dθ
dt
Since θ = θ0e−kt
, then
dθ
dt
= (θ0)(−ke−kt
)
= −kθ0e−kt
When θ0 = 15,k = −0.02 and t = 50, then
dθ
dt
= −(−0.02)(15)e−(−0.02)(50)
= 0.30 e1
= 0.815◦
C/s
Problem 29. The pressure p of the atmosphere at
height h above ground level is given by
p = p0 e−h/c, where p0 is the pressure at ground
level and c is a constant. Determine the rate
of change of pressure with height when
p0 = 105 pascals and c = 6.2 × 104 at 1550 metres
The rate of change of pressure with height is
dp
dh
Since p = p0e−h/c
, then
dp
dh
= (p0) −
1
c
e−h/c
= −
p0
c
e−h/c
When p0 = 105,c = 6.2 × 104 and h = 1550, then
rate of change of pressure,
dp
dh
= −
105
6.2 × 104
e− 1550/6.2×104
= −
10
6.2
e−0.025
= −1.573Pa/m

PracticeExercise 138 Rates of change
1. An alternating current, i amperes, is given by
i = 10sin2πft, where f is the frequency in
hertz and t is the time in seconds. Determine
the rate of change of current when t = 12 ms,
given that f = 50 Hz.
2. The luminous intensity, I candelas, of a lamp
is given by I = 8 × 10−4 V2, where V is the
voltage. Find
(a) the rate of change of luminous intensity
with 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 at
any time t seconds is given by v = V e−t/CR,
where V,C and R are constants. Given
V = 200V,C = 0.10μF and R = 2M , ﬁnd
(a) the initial rate of change of voltage.
(b) the rate of change of voltage after 0.2s
4. The pressure p of the atmosphere at height h
above ground level is given by p = p0e−h/c,
where p0 is the pressure at ground level and c
is a constant. Determine the rate of change of
pressure with height when p0 = 1.013 × 105
pascals and c = 6.05 × 104 at 1450 metres.

Chapter 35
Introduction to integration
35.1 The process of integration
The process of integration reverses the process of
differentiation. In differentiation, if f (x) = 2x2 then
f (x) = 4x. Thus, the integral of 4x is 2x2; i.e., inte-
gration is the process of moving from f (x) to f (x). By
similar reasoning, the integral of 2t is t2
.
Integration is a process of summation or adding parts
together and an elongated S, shown as , is used to
replace the words ‘the integral of’. Hence, from above,
4x = 2x2 and 2t is t2.
In differentiation, the differential coefficient
dy
dx
indi-
cates that a function of x is being differentiated with
respect to x, the dx indicating that it is ‘with respect
to x’.
In integration the variable of integration is shown
by adding d (the variable) after the function to be
integrated. Thus,
4x dx means ‘the integral of 4x with respect to x’,
and 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 also
be equal to 2x2 + 7. To allow for the possible presence
of a constant, whenever the process of integration is
performed a constant c is added to the result. Thus,
4x dx = 2x2
+ c and 2t dt = t2
+ c
c is called the arbitrary constant of integration.
35.2 The general solution of integrals
of the form axn
The general solution of integrals of the form axn dx,
where a and n are constants and n = −1 is given by
axn
dx =
axn+1
n + 1
+ c
Using this rule gives
(i) 3x4
dx =
3x4+1
4 + 1
+ c =
3
5
x5
+ c
(ii)
4
9
t3
dt dx =
4
9
t3+1
3 + 1
+ c =
4
9
t4
4
+ c
=
1
9
t4 + c
(iii)
2
x2
dx = 2x−2
dx =
2x−2+1
−2 + 1
+ c
=
2x−1
−1
+ c = −
2
x
+ c
(iv)
√
x dx = x
1
2 dx =
x
1
2 +1
1
2 + 1
+ c =
x
3
2
3
2
+ c
=
2
3
√
x3 + c
Each of these results may be checked by differentia-
tion.
(a) The integral of a constant k is kx + c. For
example,
8dx = 8x + c and 5dt = 5t + c
DOI: 10.1016/B978-1-85617-697-2.00035-1

(b) When a sum of several terms is integrated the
result is the sum of the integrals of the separate
terms. For example,
(3x + 2x2
− 5)dx
= 3x dx + 2x2
dx − 5dx
=
3x2
2
+
2x3
3
− 5x + c
35.3 Standard integrals
From Chapter 34,
d
dx
(sin ax) = a cos ax. Since inte-
gration is the reverse process of differentiation, it
follows that
a cos ax dx = sin ax + c
or cos ax dx =
1
a
sin ax + c
By similar reasoning
sin ax dx = −
1
a
cos ax + c
eax
dx =
1
a
eax
+ c
and
1
x
dx = lnx + c
From above, axn dx =
axn+1
n + 1
+ c except when
n = −1
When n = −1, x−1 dx =
1
x
dx = ln x + c
A list of standard integrals is summarized in
Table 35.1.
Table 35.1 Standard integrals
y y dx
1. axn axn+1
n + 1
+ c (except when n = −1)
2. cos ax dx
1
a
sin ax + c
3. sin ax dx −
1
a
cos ax + c
4. eax dx
1
a
eax + c
5.
1
x
dx ln x + c
Problem 1. Determine 7x2
dx
The standard integral, axn
dx =
axn+1
n + 1
+ c
When a = 7 and n = 2,
7x2
dx =
7x2+1
2 + 1
+ c =
7x3
3
+ c or
7
3
x3
+ c
Problem 2. Determine 2t3 dt
2t3
dt =
2t3+1
3 + 1
+ c =
2t4
4
+ c =
1
2
t4
+ c
Note that each of the results in worked examples 1 and
2 may be checked by differentiating them.
Problem 3. Determine 8dx
8dx is the same as 8x0 dx and, using the general
rule when a = 8 and n = 0, gives
8x0
dx =
8x0+1
0 + 1
+ c = 8x + c
In general, if k is a constant then kdx = kx + c.
Problem 4. Determine 2x dx
2x dx = 2x1
dx =
2x1+1
1 + 1
+ c =
2x2
2
+ c
= x2
+ c
Problem 5. Determine 3 +
2
5
x − 6x2
dx
3 +
2
5
x − 6x2 dx may be written as
3dx+
2
5
x dx − 6x2 dx
i.e., each term is integrated separately. (This splitting
up of terms only applies, however, for addition and

Introduction to integration 327
subtraction.) Hence,
3 +
2
5
x − 6x2
dx
= 3x +
2
5
x1+1
1 + 1
− (6)
x2+1
2 + 1
+ c
= 3x +
2
5
x2
2
− (6)
x3
3
+ c = 3x +
1
5
x2
− 2x3
+ c
Note that when an integral contains more than one term
there is no need to have an arbitrary constant for each;
just a single constant c at the end is sufﬁcient.
2x3 − 3x
4x
dx
Rearranging into standard integral form gives
2x3 − 3x
4x
dx =
2x3
4x
−
3x
4x
dx
=
1
2
x2
−
3
4
dx =
1
2
x2+1
2 + 1
−
3
4
x + c
=
1
2
x3
3
−
3
4
x + c =
1
6
x 3
−
3
4
x + c
Problem 7. Determine 1 − t
2
dt
Rearranging (1 − t)2
dt gives
(1 − 2t + t2
)dt = t −
2t1+1
1 + 1
+
t2+1
2 + 1
+ c
= t −
2t2
2
+
t3
3
+ c
= t − t2
+
1
3
t3
+ c
This example shows that functionsoften have to be rear-
ranged into the standard form of axndx before it is
possible to integrate them.
5
x2
dx
5
x2
dx = 5x−2
dx
Using the standard integral, axn
dx, when a = 5 and
n = −2, gives
5x−2
dx =
5x−2+1
−2 + 1
+ c =
5x−1
−1
+ c
= −5x−1
+ c = −
5
x
+ c
Problem 9. Determine 3
√
xdx
For fractional powers it is necessary to appreciate
n
√
am = a
m
n
3
√
x dx = 3x
1
2 dx =
3x
1
2 +1
1
2
+ 1
+ c =
3x
3
2
3
2
+ c
= 2x
3
2 + c = 2 x3 + c
−5
9
4
√
t3
dt
−5
9
4
√
t3
dt =
−5
9t
3
4
dt = −
5
9
t−
3
4 dt
= −
5
9
t−
3
4+1
−
3
4
+ 1
+ c
= −
5
9
t
1
4
1
4
+ c = −
5
9
4
1
t
1
4 + c
= −
20
9
4
√
t + c
Problem 11. Determine 4cos 3x dx
From 2 of Table 35.1,
4cos 3x dx = (4)
1
3
sin3x + c
=
4
3
sin 3x + c
Problem 12. Determine 5sin2θdθ

5sin2θdθ = (5) −
1
2
cos2θ + c
= −
5
2
cos 2θ+c
Problem 13. Determine 5e3xdx
5e3x
dx = (5)
1
3
e3x
+ c
=
5
3
e3x
+ c
2
3e4t
dt
2
3e4t
dt =
2
3
e−4t
dt
=
2
3
−
1
4
e−4t
+ c
= −
1
6
e−4t
+ c = −
1
6e4t
+ c
3
5x
dx
3
5x
dx =
3
5
1
x
dx
=
3
5
lnx + c
2x2 + 1
x
dx
2x2 + 1
x
dx =
2x2
x
+
1
x
dx
= 2x +
1
x
dx =
2x2
2
+ ln x + c
= x2
+ lnx + c
PracticeExercise 139 Standard integrals
Determine the following integrals.
1. (a) 4dx (b) 7x dx
2. (a) 5x3 dx (b) 3t7 dt
3. (a)
2
5
x2
dx (b)
5
6
x3
dx
4. (a) (2x4 − 3x)dx (b) (2 − 3t3)dt
5. (a)
3x2 − 5x
x
dx (b) (2 + θ)2dθ]
6. (a) (2 + θ)(3θ − 1)dθ
(b) (3x − 2)(x2 + 1)dx
7. (a)
4
3x2
dx (b)
3
4x4
dx
8. (a) 2
√
x3dx (b)
1
4
4
√
x5 dx
9. (a)
−5
√
t3
dt (b)
3
7
5
√
x4
dx
10. (a) 3cos2x dx (b) 7sin3θ dθ
11. (a) 3sin
1
2
x dx (b) 6cos
1
3
x dx
12. (a)
3
4
e2x dx (b)
2
3
dx
e5x
13. (a)
2
3x
dx (b)
u2 − 1
u
du
14. (a)
(2 + 3x)2
√
x
dx (b)
1
t
+ 2t
2
dt
35.4 Definite integrals
Integrals containing an arbitrary constant c in their
results are called indefinite integrals since theirprecise
value cannot be determined without furtherinformation.
Definite integrals are those in which limits are applied.
Ifan expression iswritten as[x]b
a, b iscalled the upper
limit and a the lower limit. The operation of applying
the limits is defined as [x]b
a = (b) − (a).

For example, the increase in the value of the integral
x2 as x increases from 1 to 3 is written as
3
1 x2dx.
Applying the limits gives
3
1
x2
dx =
x3
3
+ c
3
1
=
33
3
+ c −
13
3
+ c
= (9 + c) −
1
3
+ c = 8
2
3
Note that the c term always cancels out when limits are
applied and it need not be shown with deﬁnite integrals.
2
1
3xdx
2
1
3xdx =
3x2
2
2
1
=
3
2
(2)2
−
3
2
(1)2
= 6 − 1
1
2
= 4
1
2
3
−2
(4 − x2
) dx
3
−2
(4 − x2
)dx = 4x −
x3
3
3
−2
= 4(3) −
(3)3
3
− 4(−2) −
(−2)3
3
= {12 − 9} − −8 −
−8
3
= {3} − −5
1
3
= 8
1
3
2
0
x(3 + 2x)dx
2
0
x(3+2x)dx =
2
0
(3x +2x2
)dx =
3x2
2
+
2x3
3
2
0
=
3(2)2
2
+
2(2)3
3
− {0 + 0}
= 6 +
16
3
= 11
1
3
or 11.33
1
−1
x4 − 5x2 + x
x
dx
1
−1
x4 − 5x2 + x
x
dx
=
1
−1
x4
x
−
5x2
x
+
x
x
dx
=
1
−1
x3
− 5x + 1 dx =
x4
4
−
5x2
2
+ x
1
−1
=
1
4
−
5
2
+ 1 −
(−1)4
4
−
5(−1)2
2
+ (−1)
=
1
4
−
5
2
+ 1 −
1
4
−
5
2
− 1 = 2
2
1
1
x2
+
2
x
dx correct
to 3 decimal places
2
1
1
x2
+
2
x
dx
=
2
1
x−2
+ 2
1
x
dx =
x−2+1
−2 + 1
+ 2ln x
2
1
=
x−1
−1
+ 2lnx
2
1
= −
1
x
+ 2lnx
2
1
= −
1
2
+ 2ln2 − −
1
1
+ 2 ln1 = 1.886
π/2
0
3sin2x dx
π/2
0
3sin2x dx
= (3) −
1
2
cos2x
π/2
0
= −
3
2
cos2x
π/2
0
= −
3
2
cos2
π
2
− −
3
2
cos2(0)
= −
3
2
cosπ − −
3
2
cos0

= −
3
2
(−1) − −
3
2
(1)
=
3
2
+
3
2
= 3
2
1
4cos3t dt
2
1
4cos3t dt = (4)
1
3
sin3t
2
1
=
4
3
sin3t
2
1
=
4
3
sin6 −
4
3
sin3
Note that limits of trigonometric functions are always
expressed in radians – thus, for example, sin 6 means
the sine of 6 radians = −0.279415... Hence,
2
1
4cos3t dt=
4
3
(−0.279415 ...) −
4
3
(0.141120 ...)
= (−0.37255) − (0.18816)
= −0.5607
2
1
4e2x
dx correct to
4 significant figures
2
1
4e2x
dx =
4
2
e2x
2
1
= 2 e2x
2
1
= 2[e4
− e2
]
= 2[54.5982− 7.3891]
= 94.42
4
1
3
4u
du correct to 4
4
1
3
4u
du =
3
4
lnu
4
1
=
3
4
[ln4 − ln 1]
=
3
4
[1.3863 − 0] = 1.040
PracticeExercise 140 Definite integrals
In problems 1 to 10, evaluate the definite integrals
(where necessary, correct to 4 significant figures).
1. (a)
2
1
x dx (b)
2
1
(x − 1)dx
2. (a)
4
1
5x2
dx (b)
1
−1
−
3
4
t2
dt
3. (a)
2
−1
(3 − x2
)dx (b)
3
1
(x2
− 4x + 3)dx
4. (a)
2
1
(x3
− 3x)dx (b)
2
1
(x2
− 3x + 3)dx
5. (a)
4
0
2
√
x dx (b)
3
2
1
x2
dx
6. (a)
π
0
3
2
cosθ dθ (b)
π/2
0
4cosθ dθ
7. (a)
π/3
π/6
2sin2θ dθ (b)
2
0
3sint dt
8. (a)
1
0
5cos3x dx
(b)
π/2
π/4
(3sin2x − 2cos3x)dx
9. (a)
1
0
3e3t
dt (b)
2
−1
2
3e2x
dx
10. (a)
3
2
2
3x
dx (b)
3
1
2x2 + 1
x
dx
35.5 The area under a curve
The 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 (see
Chapter 28) or, more precisely, by using integration.
0 x 5 a x 5 b
y 5 f(x)
y
x
Figure 35.1

The shaded area in Figure 35.1 is given by
shaded area =
b
a
y dx =
b
a
f(x) dx
Thus, determining the area under a curve by integration
merely involves evaluating a deﬁnite integral, as shown
in Section 35.4.
There are several instances in engineering and science
where the area beneath a curve needs to be accurately
determined. For example, the areas between the limits
of a
velocity/time graph gives distance travelled,
force/distance graph gives work done,
voltage/current graph gives power, and so on.
Should a curve drop below the x-axis then y(= f (x))
becomes negative and f (x)dx is negative. When
determining such areas by integration, a negative sign
is placed before the integral. For the curve shown in
Figure 35.2, the total shaded area is given by (area E+
area F +area G).
E
0 F
G
y
a b c d x
y ϭ f(x)
Figure 35.2
By integration,
total shaded area =
b
a
f(x) dx −
c
b
f(x) dx
+
d
c
f(x) dx
(Note that this is not the same as
d
a
f (x) dx)
It is usually necessary to sketch a curve in order to check
whether it crosses the x-axis.
Problem 26. Determine the area enclosed by
y = 2x + 3, the x-axis and ordinates x = 1 and
x = 4
y = 2x + 3 is a straight line graph as shown in
Figure 35.3, in which the required area is shown shaded.
12
10
8
6
4
2
10 2 3 4 5 x
y5 2x 13
y
Figure 35.3
By integration,
shaded area =
4
1
y dx =
4
1
(2x + 3)dx =
2x2
2
+ 3x
4
1
= [(16 + 12)−(1 + 3)] = 24 square units
(This answer may be checked since the shaded area
is a trapezium: area of trapezium =
1
2
(sum of paral-
lel sides)(perpendiculardistance between parallel sides)
=
1
2
(5 + 11)(3) = 24 square units.)
Problem 27. The velocity v of a body t seconds
after a certain instant is given by v = 2t2 + 5 m/s.
Find by integration how far it moves in the interval
from t = 0 to t = 4s
Since 2t2 + 5 is a quadratic expression, the curve
v = 2t2 + 5 is a parabola cutting the v-axis at v = 5,
as shown in Figure 35.4.
The distance travelled is given by the area under the v/t
curve (shown shaded in Figure 35.4). By integration,
shaded area =
4
0
vdt =
4
0
(2t2
+ 5)dt =
2t3
3
+ 5t
4
0
=
2(43)
3
+ 5(4) − (0)
i.e. distance travelled = 62.67 m

40
30
20
10
5
1 2 3 40
v (m/s)
t(s)
v ϭ 2t2 ϩ 5
Figure 35.4
Problem 28. Sketch the graph
y = x3 + 2x2 − 5x − 6 between x = −3 and
x = 2 and determine the area enclosed by
the curve and the x-axis
A table of values is produced and the graph sketched as
shown in Figure 35.5, in which the area enclosed by the
curve and the x-axis is shown shaded.
x −3 −2 −1 0 1 2
y 0 4 0 −6 −8 0
y
x
6
0 1212223
y 5 x3 1 2x2 2 5x 2 6
2
Figure 35.5
Shaded area =
−1
−3
y dx −
2
−1
y dx, the minus sign
before the second integral being necessary since the
enclosed area is below the x-axis. Hence,
shaded area =
−1
−3
(x3
+ 2x2
− 5x − 6)dx
−
2
−1
(x3
+ 2x2
− 5x − 6)dx
=
x4
4
+
2x3
3
−
5x2
2
− 6x
−1
−3
−
x4
4
+
2x3
3
−
5x2
2
− 6x
2
−1
=
1
4
−
2
3
−
5
2
+ 6 −
81
4
− 18 −
45
2
+ 18
− 4 +
16
3
− 10 − 12 −
1
4
−
2
3
−
5
2
+ 6
= 3
1
12
− −2
1
4
− −12
2
3
− 3
1
12
= 5
1
3
− −15
3
4
= 21
1
12
or 21.08 square units
Problem 29. Determine the area enclosed by the
curve y = 3x2 + 4, the x-axis and ordinates x = 1
and x = 4 by (a) the trapezoidal rule, (b) the
mid-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 rule
area =
width of
interval
1
2
ﬁrst + last
ordinate
+
sum of
remaining
ordinates

x 0 1.0 1.5 2.0 2.5 3.0 3.5 4.0
y 4 7 10.75 16 22.75 31 40.75 52
0 1 2 3 4 x
4
50
40
30
20
10
y
y 5 3x2 1 4
Figure 35.6
Selecting 6 intervals each of width 0.5 gives
area = (0.5)
1
2
(7 + 52) + 10.75 + 16
+ 22.75 + 31 + 40.75
= 75.375 square units
(b) By the mid-ordinate rule
area = (width of interval)(sum of mid-ordinates)
Selecting 6 intervals, each of width 0.5, gives the
mid-ordinates as shown by the broken lines in
Figure 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 rule
area =
1
3
width of
interval
ﬁrst + last
ordinates
+ 4
sum of even
ordinates
+2
sum of remaining
odd ordinates
Selecting 6 intervals, each of width 0.5, gives
area =
1
3
(0.5)[(7 + 52) + 4(10.75 + 22.75
+ 40.75) + 2(16 + 31)]
= 75 square units
(d) By integration
shaded area =
4
1
ydx
=
4
1
(3x2
+ 4)dx = x3
+ 4x
4
1
= (64 + 16) − (1 + 4)
= 75 square units
Integration gives the precise value for the area under
a curve. In this case, Simpson’s rule is seen to be the
most accurate of the three approximate methods.
Problem 30. Find the area enclosed by the curve
y = sin2x, the x-axis and the ordinates x = 0 and
x =
π
3
A sketch of y = sin2x is shown in Figure 35.7. (Note
that y = sin2x has a period of
2π
2
i.e., π radians.)
1
0 ␲/2 ␲␲/3
y 5 sin 2x
x
y
Figure 35.7

Shaded area =
π/3
0
y dx
=
π/3
0
sin2x dx = −
1
2
cos2x
π/3
0
= −
1
2
cos
2π
3
− −
1
2
cos0
= −
1
2
−
1
2
− −
1
2
(1)
=
1
4
+
1
2
=
3
4
or 0.75 square units
PracticeExercise 141 Area under curves
Unless otherwise stated all answers are in square
units.
1. Show by integration that the area of a rect-
angle formed by the line y = 4, the ordinates
x = 1 and x = 6 and the x-axis is 20 square
units.
2. Show by integration that the area of the trian-
gle formed by the line y = 2x, the ordinates
x = 0 and x = 4 and the x-axis is 16 square
units.
3. Sketch the curve y = 3x2 + 1 between
x = −2 and x = 4. Determine by integration
the area enclosed by the curve, the x-axis and
ordinates x = −1 and x = 3. Use an approx-
imate method to find the area and compare
your result with that obtained by integration.
4. The force F newtons acting on a body at a
distance x metres from a fixed point is given
by F = 3x + 2x2. If work done =
x2
x1
F dx,
determine the work done when the body
moves from the position where x1 = 1 m to
that when x2 = 3m.
In problems 5 to 9, sketch graphs of thegiven equa-
tions and then find the area enclosed between the
curves, the horizontal axis and the given ordinates.
5. y = 5x; x = 1, x = 4
6. y = 2x2 − x + 1; x = −1, x = 2
7. y = 2sin2x; x = 0, x =
π
4
8. y = 5cos3t; t = 0,t =
π
6
9. y = (x − 1)(x − 3); x = 0, x = 3
10. The velocity v of a vehicle t seconds after a
certain instant isgiven by v = 3t2
+ 4 m/s.
Determine how far it moves in the interval
from t = 1s to t = 5s.

Revision Test 14 : Differentiation and integration
1. Differentiate the following functions with respect
to x.
(a) y = 5x2 − 4x + 9 (b) y = x4 − 3x2 − 2
(4)
2. Given y = 2(x − 1)2, find
dy
dx
(3)
3. If y =
3
x
determine
dy
dx
(2)
4. Given f (t) =
√
t5, find f (t). (2)
5. Determine the derivative of y = 5 − 3x +
4
x2
(3)
6. Calculate the gradient of the curve y = 3cos
x
3
at
x =
π
4
, correct to 3 decimal places. (4)
f (x) = 7x2 − 4x + 2 at the point (1, 5) (3)
8. If y = 5sin3x − 2cos4x find
dy
dx
(2)
9. Determine the value of the differential coefficient
of y = 5ln2x −
3
e2x
when x = 0.8, correct to 3
significant figures. (4)
10. If y = 5x4 − 3x3 + 2x2 − 6x + 5, find (a)
dy
dx
(b)
d2 y
dx2
(4)
11. Newton’s law of cooling is given by θ = θ0e−kt ,
where the excess of temperature at zero time is
θ ◦
0 C and at time t seconds is θ◦C. Determine
the rate of change of temperature after 40s, cor-
rect to 3 decimal places, given that θ0 = 16◦C and
k = −0.01 (4)
In problems 12 to 15, determine the indefinite integrals.
12. (a) (x2
+ 4)dx (b)
1
x3
dx (4)
13. (a)
2
√
x
+ 3
√
x dx (b) 3 t5dt (4)
14. (a)
2
3
√
x2
dx (b) e0.5x
+
1
3x
− 2 dx
(6)
15. (a) (2 + θ)2
dθ
(b) cos
1
2
x +
3
x
− e2x
dx (6)
Evaluate the integrals in problems 16 to 19, each, where
necessary, correct to 4 significant figures.
16. (a)
3
1
(t2
− 2t)dt (b)
2
−1
2x3
− 3x2
+ 2 dx
(6)
17. (a)
π/3
0
3sin2tdt (b)
3π/4
π/4
cos
1
3
xdx (7)
18. (a)
2
1
2
x2
+
1
x
+
3
4
dx
(b)
2
1
3
x
−
1
x3
dx (8)
19. (a)
1
0
√
x + 2ex
dx (b)
2
1
r3
−
1
r
dr
(6)
In Problems 20 to 22, find the area bounded by the curve,
the x-axis and the given ordinates. Assume answers are
in square units.Give answers correct to 2 decimal places
where 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 curve
y =
√
x +
1
√
x
, the horizontal axis and the ordi-
nates x = 1 and x = 4. Give the answer correct to
2 decimal places. (5)
24. The force F newtons acting on a body at a dis-
tance x metres from a fixed point is given by
F = 2x + 3x2. If work done =
x2
x1
Fdx, deter-
mine the work done when the body moves
from the position when x = 1m to that when
x = 4m. (3)

List of formulae
Laws of indices:
am × an = am+n am
an
= am−n (am)n
= amn
am/n = n
√
am a−n =
1
an
a0 = 1
Quadratic formula:
If ax2 + bx + c = 0 then x =
−b ±
√
b2 − 4ac
2a
Equation of a straight line:
y = mx + c
Deﬁnition of a logarithm:
If y = ax
then x = loga y
Laws of logarithms:
log(A × B) = log A + log B
log
A
B
= log A − log B
log An
= n × log A
Exponential series:
ex = 1 + x +
x2
2!
+
x3
3!
+ ··· (valid for all values of x)
Theorem of Pythagoras:
b2
= a2
+ c2
A
B Ca
c
b
Areas of plane ﬁgures:
(i) Rectangle Area = l × b
b
l
(ii) Parallelogram Area = b × h
b
h
(iii) Trapezium Area =
1
2
(a + b)h
a
h
b
(iv) Triangle Area =
1
2
× b × h
h
b

(v) Circle Area = πr2
Circumference = 2πr
r
r
s
␪
Radian measure: 2π radians = 360degrees
For a sector of circle:
arc length, s =
θ◦
360
(2πr) = rθ (θ in rad)
shaded area =
θ◦
360
(πr2) =
1
2
r2θ (θ in rad)
Equation of a circle, centre at origin, radius r:
x2
+ y2
= r2
Equation of a circle, centre at (a, b), radius r:
(x − a)2
+ (y − b)2
= r2
Volumes and surface areas of regular
solids:
(i) Rectangular prism (or cuboid)
Volume = l × b × h
Surface area = 2(bh + hl +lb)
h
b
l
(ii) Cylinder
Volume = πr2
h
r
h
(iii) Pyramid
If area of base = A and
perpendicular height = h then:
Volume =
1
3
× A × h
h
Total surface area = sum of areas of triangles
forming sides + area of base
(iv) Cone
Volume =
1
3
πr2
h
Curved Surface area = πrl
Total Surface area = πrl + πr2
l
h
r
(v) Sphere
Volume =
4
3
πr3
Surface area = 4πr2
r

Areas of irregular figures by approximate
methods:
Trapezoidal rule
Area ≈
width of
interval
1
2
first + last
ordinate
+ sum of remaining ordinates
Mid-ordinate rule
Area ≈ (width of interval)(sum of mid-ordinates)
Simpson’s rule
Area ≈
1
3
width of
interval
first + last
ordinate
+ 4
sum of even
ordinates
+ 2
sum of remaining
odd ordinates
Mean or average value of a waveform:
mean value, y =
area under curve
length of base
=
Triangle formulae:
Sine rule:
a
sin A
=
b
sin B
=
c
sinC
Cosine rule: a2
= b2
+ c2
− 2bccos A
A
CB a
c b
Area of any triangle
=
1
2
=
1
2
absinC or
1
2
acsin B or
1
2
bcsin A
= [s (s − a)(s − b)(s − c)] where s =
a + b + c
2
For a general sinusoidal function y= Asin(ωt ± α),
then
A = amplitude
ω = angular velocity = 2π f rad/s
ω
2π
= frequency, f hertz
2π
ω
= periodic time Tseconds
α = angle of lead or lag (compared with
y = Asinωt)
Cartesian and polar co-ordinates:
If co-ordinate (x, y) = (r,θ) then
r = x2 + y2 and θ = tan−1 y
x
If co-ordinate (r,θ) = (x, y) then
x = r cosθ and y = r sinθ
Arithmetic progression:
If a = first term and d = common difference, then the
arithmetic progression is: a,a + d,a + 2d,...
The n’th term is: a + (n − 1)d
Sum of n terms, Sn =
n
2
[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−1
Sum of n terms, Sn =
a (1 −rn )
(1 −r)
or
a (rn − 1)
(r − 1)
If − 1 < r < 1, S∞ =
a
(1 −r)
Statistics:
Discrete data:
mean, ¯x =
x
n
standard deviation, σ =
(x − ¯x)2
n

Grouped data:
mean, ¯x =
f x
f
standard deviation, σ =
f (x − ¯x)2
f
Standard derivatives
y or f(x)
dy
dx
= or f (x)
axn anxn−1
sinax a cosax
cosax −a sin ax
eax aeax
ln ax
1
x
Standard integrals
y y dx
axn a
xn+1
n + 1
+ c(except when n = −1)
cosax
1
a
sinax + c
sinax −
1
a
cosax + c
eax 1
a
eax + c
1
x
lnx + c

Answers
Answers to practice exercises
Chapter 1
Exercise 1 (page 2)
1. 19kg 2. 16m 3. 479mm
4. −66 5. £565 6. −225
7. −2136 8. −36121 9. £107701
10. −4 11. 1487 12. 5914
13. 189g 14. −70872 15. $15333
16. 89.25cm
17. d = 64mm, A = 136mm, B = 10 mm
Exercise 2 (page 5)
1. (a) 468 (b) 868 2. (a) £1827 (b) £4158
3. (a) 8613kg (b) 584kg
4. (a) 351mm (b) 924mm
5. (a) 10304 (b) −4433 6. (a) 48m (b) 89m
7. (a) 259 (b) 56 8. (a) 1648 (b) 1060
9. (a) 8067 (b) 3347 10. 18kg
Exercise 3 (page 6)
1. (a) 4 (b) 24 2. (a) 12 (b) 360
3. (a) 10 (b) 350 4. (a) 90 (b) 2700
5. (a) 2 (b) 210 6. (a) 3 (b) 180
7. (a) 5 (b) 210 8. (a) 15 (b) 6300
9. (a) 14 (b) 420420 10. (a) 14 (b) 53900
Exercise 4 (page 8)
1. 59 2. 14 3. 88 4. 5 5. −33
6. 22 7. 68 8. 5 9. 2 10. 5
11. −1
Chapter 2
Exercise 5 (page 11)
1. 2
1
7
2. 7
2
5
3.
22
9
4.
71
8
5.
4
11
6.
3
7
,
4
9
,
1
2
,
3
5
,
5
8
7.
8
25
8.
11
15
9.
17
30
10.
9
10
11.
3
16
12.
3
16
13.
43
77
14. 1
1
15
15.
4
27
16. 8
51
52
17. 1
9
40
18. 1
16
21
19.
17
60
20.
17
20
1.
8
35
2. 2
2
9
3.
6
11
4.
5
12
5.
3
28
6.
3
5
7. 11 8.
1
13
9. 1
1
2
10.
8
15
11. 2
2
5
12.
5
12
13. 3
3
4
14.
12
23
15. 4
16.
3
4
17.
1
9
18. 13 19. 15 20. 400 litres
21. (a) £60, P£36, Q£16 22. 2880 litres
1. 2
1
18
2. −
1
9
3. 1
1
6
4. 4
3
4
5.
13
20
6.
7
15
7. 4
19
20
8. 2 9. 7
1
3
10.
1
15
11. 4 12. 2
17
20
DOI: 10.1016/B978-1-85617-697-2.00040-5

Chapter 3
1.
13
20
2.
9
250
3.
7
40
4.
6
125
5. (a)
13
20
(b)
21
25
(c)
1
80
(d)
141
500
(e)
3
125
6. 4
21
40
7. 23
11
25
8. 10
3
200
9. 6
7
16
10. (a) 1
41
50
(b) 4
11
40
(c) 14
1
8
(d) 15
7
20
(e) 16
17
80
11. 0.625 12. 6.6875 13. 0.21875 14. 11.1875
15. 0.28125
1. 14.18 2. 2.785 3. 65.38 4. 43.27
5. 1.297 6. 0.000528
1. 80.3 2. 329.3 3. 72.54 4. −124.83
5. 295.3 6. 18.3 7. 12.52 mm
1. 4.998 2. 47.544 3. 385.02 4. 582.42
5. 456.9 6. 434.82 7. 626.1 8. 1591.6
9. 0.444 10. 0.62963 11. 1.563 12. 53.455
13. 13.84 14. 8.69 15. (a) 24.81 (b) 24.812
16. (a) 0.00639 (b) 0.0064 17. (a) 8.˙4 (b) 62.˙6
18. 2400
Chapter 4
1. 30.797 2. 11927 3. 13.62 4. 53.832
5. 84.42 6. 1.0944 7. 50.330 8. 36.45
9. 10.59 10. 12.325
1. 12.25 2. 0.0361 3. 46.923 4. 1.296 × 10−3
5. 2.4430 6. 2.197 7. 30.96 8. 0.0549
9. 219.26 10. 5.832 × 10−6
1. 0.571 2. 40 3. 0.13459 4. 137.9
5. 14.96 6. 19.4481 7. 515.36 × 10−6
8. 1.0871 9. 15.625 × 10−9 10. 52.70
1. 2.182 2. 11.122 3. 185.82
4. 0.8307 5. 0.1581 6. 2.571
7. 5.273 8. 1.256 9. 0.30366
10. 1.068 11. 3.5×106 12. 37.5×103
13. 4.2×10−6 14. 202.767×10−3 15. 18.32×106
1. 0.4667 2.
13
14
3. 4.458 4. 2.732
5.
1
21
6. 0.7083 7. −
9
10
8. 3
1
3
9. 2.567 10. 0.0776
1. 0.9205 2. 0.7314 3. 2.9042 4. 0.2719
5. 0.4424 6. 0.0321 7. 0.4232 8. 0.1502
9. −0.6992 10. 5.8452
1. 4.995 2. 5.782 3. 25.72 4. 69.42
5. 0.6977 6. 52.92 7. 591.0 8. 17.90
9. 3.520 10. 0.3770
1. A = 66.59cm2 2. C = 52.78mm 3. R = 37.5
4. 159m/s 5. 0.407A 6. 5.02 mm
7. 0.144J 8. 628.8m2 9. 224.5
10. 14230 kg/m3 11. 281.1m/s 12. 2.526
1. £589.27 2. 508.1W 3. V = 2.61V
4. F = 854.5 5. I = 3.81A 6. T = 14.79s
7. E = 3.96J 8. I = 12.77A 9. s = 17.25m
10. A = 7.184cm2 11. V = 7.327
12. (a) 12.53h (b) 1h 40min, 33 m.p.h.
(c) 13.02h (d) 13.15h

Chapter 5
1. 0.32% 2. 173.4% 3. 5.7% 4. 37.4%
5. 128.5% 6. 0.20 7. 0.0125 8. 0.6875
9. 38.462% 10. (a) 21.2% (b) 79.2% (c) 169%
11. (b), (d), (c), (a) 12.
13
20
13.
5
16
14.
9
16
15. A =
1
2
, B = 50%, C = 0.25, D = 25%, E = 0.30,
F =
3
10
, G = 0.60, H = 60%, I = 0.85, J =
17
20
1. 21.8kg 2. 9.72 m
3. (a) 496.4t (b) 8.657g (c) 20.73s 4. 2.25%
5. (a) 14% (b) 15.67% (c) 5.36% 6. 37.8g
7. 14 minutes 57 seconds 8. 76g 9. £611
10. 37.49% 11. 39.2% 12. 17% 13. 38.7%
14. 2.7% 15. 5.60 m 16. 3.5%
1. 2.5% 2. 18% 3. £310 4. £175000
5. £260 6. £20000 7. £9116.45 8. £50.25
9. £39.60 10. £917.70 11. £185000 12. 7.2%
13. A 0.6kg, B 0.9kg,C 0.5kg
14. 54%,31%,15%,0.3t
15. 20000 kg (or 20 tonnes)
16. 13.5mm,11.5mm 17. 600 kW
Chapter 6
1. 36 : 1 2. 3.5 : 1 or 7 : 2 3. 47 : 3
4. 96cm,240 cm 5. 5
1
4
hours or 5 hours 15 minutes
6. £3680, £1840, £920 7. 12 cm 8. £2172
1. 1 : 15 2. 76ml 3. 25% 4. 12.6kg
5. 14.3kg 6. 25000 kg
1. £556 2. £66 3. 264kg 4. 450 g 5. 14.56kg
6. (a) 0.00025 (b) 48MPa 7. (a) 440 K (b) 5.76litre
1. (a) 2 mA (b) 25V 2. 170 fr 3. 685.8mm
4. 83lb10 oz 5. (a) 159.1litres (b) 16.5gallons
6. 29.4MPa 7. 584.2 mm 8. $1012
1. 3.5 weeks 2. 20 days
3. (a) 9.18 (b) 6.12 (c) 0.3375 4. 50 minutes
5. (a) 300 × 103 (b) 0.375m2 (c) 24 × 103 Pa
Chapter 7
1. 27 2. 128 3. 100000 4. 96 5. 24
6. ±5 7. ±8 8. 100 9. 1 10. 64
1. 128 2. 39
3. 16 4.
1
9
5. 1 6. 8
7. 100 8. 1000 9.
1
100
or 0.01 10. 5 11. 76
12. 36 13. 36 14. 34 15. 1 16. 25
17.
1
35
or
1
243
18. 49 19.
1
2
or 0.5 20. 1
1.
1
3 × 52
2.
1
73 × 37
3.
32
25
4.
1
210 × 52
5. 9 6. ±3 7. ±
1
2
8. ±
2
3
9.
147
148
10. −1
19
56
11. −3
13
45
12.
1
9
13. −
17
18
14. 64 15. 4
1
2

Chapter 8
1. cubic metres, m3 2. farad
3. square metres, m2 4. metres per second, m/s
5. kilogram per cubic metre, kg/m3
6. joule 7. coulomb 8. watt
9. radian or degree 10. volt 11. mass
12. electrical resistance 13. frequency
14. acceleration 15. electric current
16. inductance 17. length
18. temperature 19. pressure
20. angular velocity 21. ×109 22. m,×10−3
23. ×10−12
24. M,×106
1. (a) 7.39 × 10 (b) 2.84 × 10 (c) 1.9762 × 102
2. (a) 2.748 × 103 (b) 3.317 × 104 (c) 2.74218 × 105
3. (a) 2.401 × 10−1
(b) 1.74 × 10−2
(c) 9.23 × 10−3
4. (a) 1.7023 × 103 (b) 1.004 × 10 (c) 1.09 × 10−2
5. (a) 5 × 10−1 (b) 1.1875 × 10
(c) 1.306 × 102 (d) 3.125 × 10−2
6. (a) 1010 (b) 932.7 (c) 54100 (d) 7
7. (a) 0.0389 (b) 0.6741 (c) 0.008
8. (a) 1.35 × 102 (b) 1.1 × 105
9. (a) 2 × 102 (b) 1.5 × 10−3
10. (a) 2.71 × 103 kgm−3 (b) 4.4 × 10−1
(c) 3.7673 × 102
(d) 5.11 × 10−1
MeV
(e) 9.57897×107 Ckg−1
(f) 2.241×10−2 m3 mol−1
1. 60 kPa 2. 150μW or 0.15mW
3. 50 MV 4. 55nF
5. 100 kW 6. 0.54mA or 540μA
7. 1.5M 8. 22.5mV
9. 35GHz 10. 15pF
11. 17μA 12. 46.2 k
13. 3μA 14. 2.025MHz
15. 50 kN 16. 0.3nF
17. 62.50 m 18. 0.0346kg
19. 13.5 × 10−3 20. 4 × 103
Chapter 9
1. −3a 2. a + 2b + 4c
3. 3x − 3x2 − 3y − 2y2 4. 6ab − 3a
5. 6x − 5y + 8z 6. 1 + 2x
7. 4x + 2y + 2 8. 3z + 5b
9. −2a − b + 2c 10. −x2
− y2
1. p2q3r 2. 8a2 3. 6q2
4. 46 5. 5
1
3
6. −
1
2
7. 6 8.
1
7y
9. 5xz2
10. 3a2
+ 2ab − b2
11. 6a2 − 13ab + 3ac − 5b2 + bc
12.
1
3b
13. 3x 14. 2ab
15. 2x − y 16. 3p + 2q 17. 2a2 + 2b2
1. z8 2. a8 3. n3
4. b11 5. b−3 6. c4
7. m4 8. x−3 or
1
x3
9. x12
10. y−6
or
1
x6
11. t8
12. c14
13. a−9 or
1
x9
14. b−12 or
1
x12
15. b10 16. s−9
17. p6q7r5 18. x−2y z−2 or
y
x2 z2
19. x5 y4z3,13
1
2
20. a2b−2c or
a2c
b2
,9
1. a2 b1/2c−2, 4
1
2
2.
1 + a
b
3. a b6 c3/2
4. a−4 b5 c11 5.
p2q
q − p
6. x y3 6
√
z13
7.
1
ef 2
8. a11/6 b1/3 c−3/2 or
6
√
a11 3
√
b
√
c3

Chapter 10
1. x2 + 5x + 6 2. 2x2 + 9x + 4
3. 4x2 + 12x + 9 4. 2 j2 + 2 j − 12
5. 4x2 + 22x + 30 6. 2pqr + p2q2 +r2
7. a2 + 2ab + b2 8. x2 + 12x + 36
9. a2 − 2ac + c2 10. 25x2 + 30x + 9
11. 4x2 − 24x + 36 12. 4x2 − 9
13. 64x2 + 64x + 16 14. r2s2 + 2rst + t2
15. 3ab − 6a2 16. 2x2 − 2xy
17. 2a2 − 3ab − 5b2 18. 13p − 7q
19. 7x − y − 4z 20. 4a2 − 25b2
21. x2 − 4xy + 4y2 22. 9a2 − 6ab + b2
23. 0 24. 4 − a
25. 4ab − 8a2 26. 3xy + 9x2 y − 15x2
27. 2 + 5b2 28. 11q − 2p
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 − 5q
13. 7ab(3ab − 4) 14. 2xy(y + 3x + 4x2)
15. 2xy x − 2y2 + 4x2 y3 16. 7y(4 + y + 2x)
17.
3x
y
18. 0 19.
2r
t
20. (a + b)(y + 1) 21. (p + q)(x + y)
22. (x − y)(a + b) 23. (a − 2b)(2x + 3y)
1. 2x + 8x2 2. 12y2 − 3y
3. 4b − 15b2 4. 4 + 3a
5.
3
2
− 4x 6. 1
7. 10y2
− 3y +
1
4
8. 9x2
+
1
3
− 4x
9. 6a2 + 5a −
1
7
10. −15t
11.
1
5
− x − x2 12. 10a2 − 3a + 2
Chapter 11
1. 1 2. 2 3. 6 4. −4 5. 2
6. 1 7. 2 8.
1
2
9. 0 10. 3
11. 2 12. −10 13. 6 14. −2 15. 2.5
16. 2 17. 6 18. −3
1. 5 2. −2 3. −4
1
2
4. 2 5. 12
6. 15 7. −4 8. 5
1
3
9. 2 10. 13
11. −10 12. 2 13. 3 14. −11 15. −6
16. 9 17. 6
1
4
18. 3 19. 4 20. 10
21. ±12 22. −3
1
3
23. ±3 24. ±4
1. 10−7 2. 8m/s2 3. 3.472
4. (a) 1.8 (b) 30
5. digital camera battery £9, camcorder battery £14
6. 800 7. 30 m/s2
1. 12 cm,240 cm2 2. 0.004 3. 30
4. 45◦
C 5. 50 6. £312,£240
7. 30 kg 8. 12 m,8m 9. 3.5N
Chapter 12
1. d = c − e − a − b 2. x =
y
7
3. v =
c
p
4. a =
v − u
t
5. R =
V
I
6. y =
1
3
(t − x)
7. r =
c
2π
8. x =
y − c
m

9. T =
I
PR
10. L =
XL
2π f
11. R =
E
I
12. x = a(y − 3)
13. C =
5
9
(F − 32) 14. f =
1
2π CXC
1. r =
S − a
S
or 1 −
a
S
2. x =
d
λ
(y + λ) or d +
yd
λ
3. f =
3F − AL
3
or f = F −
AL
3
4. D =
AB2
5Ey
5. t =
R − R0
R0α
6. R2 =
RR1
R1 − R
7. R =
E − e − Ir
I
or R =
E − e
I
−r
8. b =
y
4ac2
9. x =
ay
(y2 − b2)
10. L =
t2g
4π2
11. u =
√
v2 − 2as
12. R =
360A
πθ
13. a = N2 y − x
14. L =
√
Z2 − R2
2π f
,0.080
1. a =
xy
m − n
2. R = 4 M
π
+r4
3. r =
3(x + y)
(1 − x − y)
4. L =
mrCR
μ − m
5. b =
c
√
1 − a2
6. r =
x − y
x + y
7. b =
a(p2 − q2)
2(p2 + q2)
8. v =
u f
u − f
,30
9. t2 = t1 +
Q
mc
,55 10. v =
2dgh
0.03L
,0.965
11. L =
8S2
3d
+ d,2.725
12. C =
1
ω ωL −
√
Z2 − R2
,63.1 × 10−6
13. 64mm 14. λ =
5 aμ
ρCZ4n
2
Chapter 13
1. x = 4, y = 2 2. x = 3, y = 4
3. x = 2, y = 1.5 4. x = 4, y = 1
5. p = 2,q = −1 6. x = 1, y = 2
7. x = 3, y = 2 8. a = 2,b = 3
9. a = 5,b = 2 10. x = 1, y = 1
11. s = 2,t = 3 12. x = 3, y = −2
13. m = 2.5,n = 0.5 14. a = 6,b = −1
15. x = 2, y = 5 16. c = 2,d = −3
1. p = −1, q = −2 2. x = 4, y = 6
3. a = 2, b = 3 4. s = 4,t = −1
5. x = 3, y = 4 6. u = 12,v = 2
7. x = 10, y = 15 8. a = 0.30,b = 0.40
1. x =
1
2
, y =
1
4
2. a =
1
3
,b = −
1
2
3. p =
1
4
,q =
1
5
4. x = 10, y = 5
5. c = 3,d = 4 6. r = 3,s =
1
2
7. x = 5, y = 1
3
4
8. 1
1. a = 0.2,b = 4 2. I1 = 6.47, I2 = 4.62
3. u = 12,a = 4,v = 26 4. £15500,£12800
5. m = −0.5,c = 3
6. α = 0.00426, R0 = 22.56 7. a = 12,b = 0.40
8. a = 4,b = 10 9. F1 = 1.5, F2 = −4.5
1. x = 2, y = 1, z = 3 2. x = 2, y = −2, z = 2
3. x = 5, y = −1, z = −2 4. x = 4, y = 0, z = 3

5. x = 2, y = 4, z = 5 6. x = 1, y = 6, z = 7
7. x = 5, y = 4, z = 2 8. x = −4, y = 3, z = 2
9. x = 1.5, y = 2.5, z = 4.5
10. i1 = −5,i2 = −4,i3 = 2
11. F1 = 2, F2 = −3 F3 = 4
Chapter 14
1. 4 or −4 2. 4 or −8 3. 2 or −6
4. −1.5 or 1.5 5. 0 or −
4
3
6. 2 or −2
7. 4 8. −5 9. 1
10. −2 or −3 11. −3 or −7 12. 2 or −1
13. 4 or −3 14. 2 or 7 15. −4
16. 2 17. −3 18. 3 or −3
19. −2 or −
2
3
20. −1.5 21.
1
8
or −
1
8
22. 4 or −7 23. −1 or 1.5 24.
1
2
or
1
3
25.
1
2
or −
4
5
26. 1
1
3
or −
1
7
27.
3
8
or −2
28.
2
5
or −3 29.
4
3
or −
1
2
30.
5
4
or −
3
2
31. x2
− 4x + 3 = 0 32. x2
+ 3x − 10 = 0
33. x2 + 5x + 4 = 0 34. 4x2 − 8x − 5 = 0
35. x2 − 36 = 0 36. x2 − 1.7x − 1.68 = 0
1. −3.732 or −0.268 2. −3.137 or 0.637
3. 1.468 or −1.135 4. 1.290 or 0.310
5. 2.443 or 0.307 6. −2.851 or 0.351
1. 0.637 or −3.137 2. 0.296 or −0.792
3. 2.781 or 0.719 4. 0.443 or −1.693
5. 3.608 or −1.108 6. 1.434 or 0.232
7. 0.851 or −2.351 8. 2.086 or −0.086
9. 1.481 or −1.081 10. 4.176 or −1.676
11. 4 or 2.167 12. 7.141 or −3.641
13. 4.562 or 0.438
1. 1.191s 2. 0.345A or 0.905A 3. 7.84cm
4. 0.619m or 19.38m 5. 0.0133
6. 1.066m 7. 86.78cm
8. 1.835m or 18.165m 9. 7m
10. 12 ohms,28ohms
1. x = 1, y = 3 and x = −3, y = 7
2. x =
2
5
, y = −
1
5
and −1
2
3
, y = −4
1
3
3. x = 0, y = 4 and x = 3, y = 1
Chapter 15
1. 4 2. 4 3. 3 4. −3 5.
1
3
6. 3 7. 2 8. −2 9. 1
1
2
10.
1
3
11. 2 12. 10000 13. 100000 14. 9 15.
1
32
16. 0.01 17.
1
16
18. e3
1. log6 2. log15 3. log2 4. log3
5. log12 6. log500 7. log100 8. log6
9. log10 10. log1 = 0 11. log2
12. log243 or log35 or 5log3
15. 0.5 16. 1.5 17. x = 2.5 18. t = 8
19. b = 2 20. x = 2 21. a = 6 22. x = 5
1. 1.690 2. 3.170 3. 0.2696 4. 6.058 5. 2.251
6. 3.959 7. 2.542 8. −0.3272 9. 316.2

Chapter 16
1. (a) 0.1653 (b) 0.4584 (c) 22030
2. (a) 5.0988 (b) 0.064037 (c) 40.446
3. (a) 4.55848 (b) 2.40444 (c) 8.05124
4. (a) 48.04106 (b) 4.07482 (c) −0.08286
5. 2.739 6. 120.7m
1. 2.0601 2. (a) 7.389 (b) 0.7408
3. 1 − 2x2 −
8
3
x3 − 2x4
4. 2x1/2 + 2x5/2 + x9/2 +
1
3
x13/2
+
1
12
x17/2
+
1
60
x21/2
1. 3.95,2.05 2. 1.65,−1.30
3. (a) 28cm3 (b) 116min 4. (a) 70◦C (b) 5 minutes
1. (a) 0.55547 (b) 0.91374 (c) 8.8941
2. (a) 2.2293 (b) −0.33154 (c) 0.13087
3. −0.4904 4. −0.5822 5. 2.197 6. 816.2
7. 0.8274 8. 11.02 9. 1.522 10. 1.485
11. 1.962 12. 3 13. 4
14. 147.9 15. 4.901 16. 3.095
17. t = eb+a ln D
= eb
ea ln D
= eb
eln Da
i.e. t = eb
Da
18. 500 19. W = PV ln
U2
U1
1. (a) 150◦C (b) 100.5◦C 2. 99.21kPa
3. (a) 29.32 volts (b) 71.31 × 10−6 s
4. (a) 1.993m (b) 2.293m 5. (a) 50◦C (b) 55.45s
6. 30.37N 7. (a) 3.04A (b) 1.46s
8. 2.45mol/cm3 9. (a) 7.07A (b) 0.966s
10. £2424
Chapter 17
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 mm
2. (a) −1 (b) −8 (c) −1.5 (d) 5 3. 14.5
4. (a) −1.1 (b) −1.4
5. The 1010 rev/min reading should be 1070 rev/min;
(a) 1000 rev/min (b) 167V
1. Missing values: −0.75,0.25,0.75,2.25,2.75;
Gradient =
1
2
2. (a) 4,−2 (b) −1,0 (c) −3,−4 (d) 0,4
3. (a) 2,
1
2
(b) 3,−2
1
2
(c)
1
24
,
1
2
4. (a) 6,−3 (b) −2,4 (c) 3,0 (d) 0,7
5. (a) 2,−
1
2
(b) −
2
3
,−1
2
3
(c)
1
18
,2 (d) 10,−4
2
3
6. (a)
3
5
(b) −4 (c) −1
5
6
7. (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 + 48
12. P = 0.15 W + 3.5 13. a = −20,b = 412
1. (a) 40◦C (b) 128
2. (a) 850 rev/min (b) 77.5V
3. (a) 0.25 (b) 12 (c) F = 0.25L + 12
(d) 89.5N (e) 592 N (f) 212 N
4. −0.003,8.73
5. (a) 22.5m/s (b) 6.43s (c) v = 0.7t +15.5
6. m = 26.9L − 0.63
7. (a) 1.31t (b) 22.89% (c) F = −0.09W + 2.21
8. (a) 96 × 109
Pa (b) 0.00022 (c) 28.8 × 106
Pa

9. (a)
1
5
(b) 6 (c) E =
1
5
L + 6 (d) 12 N (e) 65N
10. a = 0.85,b = 12,254.3kPa,275.5kPa,280 K
Chapter 18
1. (a) y (b) x2
(c) c (d) d 2. (a) y (b)
√
x (c) b (d) a
3. (a) y (b)
1
x
(c) f (d) e 4. (a)
y
x
(b) x (c) b (d) c
5. (a)
y
x
(b)
1
x2
(c) a (d) b
6. a = 1.5,b = 0.4,11.78mm2 7. y = 2x2 + 7,5.15
8. (a) 950 (b) 317kN
9. a = 0.4,b = 8.6 (i) 94.4 (ii) 11.2
1. (a) lg y (b) x (c) lga (d) lgb
2. (a) lg y (b) lgx (c) L (d) lgk
3. (a) ln y (b) x (c) n (d) lnm
4. I = 0.0012 V2
,6.75 candelas
5. a = 3.0,b = 0.5
6. a = 5.7,b = 2.6,38.53,3.0
7. R0 = 26.0,c = 1.42 8. y = 0.08e0.24x
9. T0 = 35.4N,μ = 0.27,65.0 N,1.28 radians
Chapter 19
1. x = 2, y = 4 2. x = 1, y = 1
3. x = 3.5, y = 1.5 4. x = −1, y = 2
5. x = 2.3, y = −1.2 6. x = −2, y = −3
7. a = 0.4,b = 1.6
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 6.9
4. −1.1 or 4.1 5. −1.8 or 2.2
6. 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 −0.3
9. (−2.6,13.2),(0.6,0.8);x = −2.6 or 0.6
10. x = −1.2 or 2.5 (a) −30 (b) 2.75 and −1.50
(c) 2.3 or −0.8
1. x = 4, y = 8 and x = −0.5, y = −5.5
2. (a) x = −1.5 or 3.5 (b) x = −1.24 or 3.24
(c) x = −1.5 or 3.0
1. x = −2.0,−0.5 or 1.5
2. 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 2.6
5. x = 0.7 or 2.5
6. x = −2.3,1.0 or 1.8 7. x = −1.5
Chapter 20
1. 82◦
27 2. 27◦
54 3. 51◦
11 4. 100◦
6 52
5. 15◦44 17 6. 86◦49 1 7. 72.55◦ 8. 27.754◦
9. 37◦57 10. 58◦22 52
1. reﬂex 2. obtuse 3. acute 4. right angle
5. (a) 21◦ (b) 62◦23 (c) 48◦56 17
6. (a) 102◦ (b) 165◦ (c) 10◦18 49
7. (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 & 8
9. 59◦20 10. a = 69◦,b = 21◦,c = 82◦ 11. 51◦
12. 1.326rad 13. 0.605rad 14. 40◦55
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

2. 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◦, isosceles
8. 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◦
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. proof
1. x = 16.54mm, y = 4.18mm 2. 9cm,7.79cm
3. (a) 2.25cm (b) 4cm 4. 3m
1–5. Constructions – see similar constructions in
worked problems 30 to 33 on pages 179–180.
Chapter 21
1. 9cm 2. 24m 3. 9.54mm
4. 20.81cm 5. 7.21m 6. 11.18cm
7. 24.11mm 8. 82 + 152 = 172
9. (a) 27.20 cm each (b) 45◦ 10. 20.81km
11. 3.35m,10 cm 12. 132.7 nautical miles
13. 2.94mm 14. 24mm
1. sin Z =
9
41
,cos Z =
40
41
,tan X =
40
9
,cos X =
9
41
2. sin A =
3
5
,cos A =
4
5
,tan A =
3
4
,sin B =
4
5
,
cos B =
3
5
,tan B =
4
3
3. sin A =
8
17
,tan A =
8
15
4. sin X =
15
113
,cos X =
112
113
5. (a)
15
17
(b)
15
17
(c)
8
15
6. (a) sinθ =
7
25
(b) cosθ =
24
25
7. (a) 9.434 (b) −0.625
1. 2.7550 2. 4.846 3. 36.52
4. (a) 0.8660 (b) −0.1010 (c) 0.5865
5. 42.33◦ 6. 15.25◦ 7. 73.78◦ 8. 7◦56
9. 31◦22 10. 41◦54 11. 29.05◦ 12. 20◦21
13. 0.3586 14. 1.803
1. (a) 12.22 (b) 5.619 (c) 14.87 (d) 8.349
(e) 5.595 (f ) 5.275
2. (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 = 6.625cm
3. 6.54m 4. 9.40 mm
1. 36.15m 2. 48m 3. 249.5m 4. 110.1m
5. 53.0 m 6. 9.50 m 7. 107.8m
8. 9.43m,10.56m 9. 60 m
Chapter 22
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◦

4. t = 122◦
7 and 237◦
53
5. α = 218◦41 and 321◦19
6. θ = 39◦44 and 219◦44
1. 5 2. 180◦ 3. 30 4. 120◦
5. 1,120◦ 6. 2,144◦ 7. 3,90◦ 8. 5,720◦
9.
7
2
,960◦ 10. 6,360◦ 11. 4,180◦ 12. 5ms
13. 40 Hz 14. 100μs or 0.1ms
15. 625Hz 16. leading 17. leading
1. (a) 40 (b) 25Hz (c) 0.04s or 40 ms (c) 25Hz
(d) 0.29rad (or 16.62◦) leading 40sin50πt
2. (a) 75cm (b) 6.37Hz (c) 0.157s
(d) 0.54rad (or 30.94◦) lagging 75sin40t
3. (a) 300 V (b) 100 Hz (c) 0.01s or 10 ms
(d) 0.412 rad (or 23.61◦) lagging 300sin200πt
4. (a) v = 120sin100πt volts
(b) v = 120sin(100πt + 0.43) volts
5. i = 20sin 80πt −
π
6
A or
i = 20sin(80πt − 0.524)A
6. 3.2sin(100πt + 0.488)m
7. (a) 5A,50 Hz,20 ms,24.75◦
lagging
(b) −2.093A (c) 4.363A (d) 6.375ms (e) 3.423ms
Chapter 23
1. C = 83◦,a = 14.1mm,c = 28.9mm,
area = 189mm2
2. A = 52◦2 ,c = 7.568cm,a = 7.152 cm,
area = 25.65cm2
3. D = 19◦48 , E = 134◦12 ,e = 36.0 cm,
area = 134cm2
4. E = 49◦0 , F = 26◦38 , f = 15.08mm,
area = 185.6mm2
5. J = 44◦29 , L = 99◦31 ,l = 5.420 cm,
area = 6.132 cm2, or, J = 135◦31 , L = 8◦29 ,
l = 0.811cm, area = 0.917cm2
6. K = 47◦8 , J = 97◦52 , j = 62.2 mm,
area = 820.2 mm2 or K = 132◦52 , J = 12◦8 ,
j = 13.19mm, area = 174.0 mm2
1. p = 13.2 cm, Q = 47.35◦, R = 78.65◦,
area = 77.7cm2
2. p = 6.127m, Q = 30.83◦
, R = 44.17◦
,
area = 6.938m2
3. X = 83.33◦,Y = 52.62◦, Z = 44.05◦,
area = 27.8cm2
4. Z = 29.77◦,Y = 53.50◦, Z = 96.73◦,
area = 355mm2
1. 193km 2. (a) 122.6m (b) 94.80◦,40.66◦,44.54◦
3. (a) 11.4m (b) 17.55◦ 4. 163.4m
5. BF = 3.9m, EB = 4.0 m 6. 6.35m,5.37m
7. 32.48A,14.31◦
1. 80.42◦,59.38◦,40.20◦ 2. (a) 15.23m (b) 38.07◦
3. 40.25cm,126.05◦ 4. 19.8cm 5. 36.2 m
6. x = 69.3mm, y = 142 mm 7. 130◦ 8. 13.66mm
Chapter 24
1. (5.83,59.04◦) or (5.83,1.03rad)
2. (6.61,20.82◦) or (6.61,0.36rad)
3. (4.47,116.57◦) or (4.47,2.03rad)
4. (6.55,145.58◦) or (6.55,2.54rad)
5. (7.62,203.20◦
) or (7.62,3.55rad)
6. (4.33,236.31◦) or (4.33,4.12 rad)
7. (5.83,329.04◦) or (5.83,5.74rad)
8. (15.68,307.75◦) or (15.68,5.37rad)
1. (1.294,4.830) 2. (1.917,3.960)
3. (−5.362,4.500) 4. (−2.884,2.154)
5. (−9.353,−5.400) 6. (−2.615,−3.207)
7. (0.750,−1.299) 8. (4.252,−4.233)
9. (a) 40∠18◦,40∠90◦,40∠162◦,40∠234◦,40∠306◦
(b) (38.04,12.36),(0,40),(−38.04,12.36),
(−23.51,−32.36),(23.51,−32.36)
10. 47.0 mm

Chapter 25
1. p = 105◦,q = 35◦ 2. r = 142◦,s = 95◦
3. t = 146◦
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) 62.91cm
2. 35.7cm2 3. (a) 80 m (b) 170 m 4. 27.2 cm2
5. 18cm 6. 1200 mm
7. (a) 29cm2 (b) 650 mm2 8. 560 m2
9. 3.4cm 10. 6750 mm2 11. 43.30 cm2
12. 32
1. 482 m2
2. (a) 50.27cm2 (b) 706.9mm2 (c) 3183mm2
3. 2513mm2 4. (a) 20.19mm (b) 63.41mm
5. (a) 53.01cm2 (b) 129.9mm2 6. 5773mm2
7. 1.89m2
1. 1932 mm2 2. 1624mm2 3. (a) 0.918ha (b) 456m
1. 80 ha 2. 80 m2 3. 3.14ha
Chapter 26
1. 45.24cm 2. 259.5mm 3. 2.629cm 4. 47.68cm
5. 38.73cm 6. 12730 km 7. 97.13mm
1. (a)
π
6
(b)
5π
12
(c)
5π
4
2. (a) 0.838 (b) 1.481 (c) 4.054
3. (a) 210◦ (b) 80◦ (c) 105◦
4. (a) 0◦43 (b) 154◦8 (c) 414◦53
1. 113cm2 2. 2376mm2 3. 1790 mm2
4. 802 mm2 5. 1709mm2 6. 1269m2
7. 1548m2 8. (a) 106.0 cm (b) 783.9cm2
9. 21.46m2 10. 17.80 cm,74.07cm2
11. (a) 59.86mm (b) 197.8mm 12. 26.2 cm
13. 8.67cm,54.48cm 14. 82.5◦
15. 748
16. (a) 0.698rad (b) 804.2 m2 17. 10.47m2
18. (a) 396mm2 (b) 42.24% 19. 701.8mm
20. 7.74mm
1. (a) 2 (b) (3,−4) 2. Centre at (3,−2), radius 4
3. Circle, centre (0, 1), radius 5
4. Circle, centre (0, 0), radius 6
Chapter 27
1. 1.2 m3 2. 5cm3 3. 8cm3
4. (a) 3840 mm3 (b) 1792 mm2
5. 972 litres 6. 15cm3,135g 7. 500 litres
8. 1.44m3 9. (a) 35.3cm3 (b) 61.3cm2
10. (a) 2400 cm3 (b) 2460 cm2 11. 37.04m
12. 1.63cm 13. 8796cm3
14. 4.709cm,153.9cm2
15. 2.99cm 16. 28060 cm3,1.099m2
17. 8.22 m by 8.22 m 18. 62.5min
19. 4cm 20. 4.08m3
1. 201.1cm3,159.0 cm2 2. 7.68cm3,25.81cm2
3. 113.1cm3,113.1cm2 4. 5.131cm 5. 3cm
6. 2681mm3 7. (a) 268083mm3 or 268.083cm3
(b) 20106mm2 or 201.06cm2
8. 8.53cm
9. (a) 512 × 106 km2 (b) 1.09 × 1012 km3 10. 664
1. 5890 mm2 or 58.90 cm2
2. (a) 56.55cm3 (b) 84.82 cm2 3. 13.57kg
4. 29.32 cm3 5. 393.4m2
6. (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

(b) 146cm2 (vi) (a) 86.5cm3 (b) 142 cm2
(vii) (a) 805cm3 (b) 539cm2
7. (a) 17.9cm (b) 38.0 cm 8. 125cm3
9. 10.3m3,25.5m2 10. 6560 litres
11. 657.1cm3,1027cm2 12. 220.7cm3
13. (a) 1458litres (b) 9.77m2 (c) £140.45
1. 147cm3,164cm2 2. 403cm3,337cm2
3. 10480 m3,1852 m2 4. 1707cm2
5. 10.69cm 6. 55910 cm3,6051cm2
7. 5.14m
1. 8 : 125 2. 137.2 g
Chapter 28
1. 4.5 square units 2. 54.7 square units 3. 63.33m
4. 4.70 ha 5. 143m2
1. 42.59m3 2. 147m3 3. 20.42 m3
1. (a) 2 A (b) 50 V (c) 2.5A 2. (a) 2.5V (b) 3A
3. 0.093As, 3.1A 4. (a) 31.83V (b) 0
5. 49.13cm2
,368.5kPa
Chapter 29
1. A scalar quantity has magnitude only; a vector
quantity has both magnitude and direction.
2. scalar 3. scalar 4. vector 5. scalar
6. scalar 7. vector 8. scalar 9. vector
1. 17.35N at 18.00◦ to the 12 N force
2. 13m/s at 22.62◦ to the 12 m/s velocity
3. 16.40 N at 37.57◦ to the 13N force
4. 28.43N at 129.30◦ to the 18N force
5. 32.31m at 21.80◦ to the 30 m displacement
6. 14.72 N at −14.72◦
to the 5N force
7. 29.15m/s at 29.04◦ to the horizontal
8. 9.28N at 16.70◦ 9. 6.89m/s at 159.56◦
10. 15.62 N at 26.33◦ to the 10 N force
11. 21.07knots, E 9.22◦S
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◦
1. 83.5km/h at 71.6◦
to the vertical
2. 4minutes 55seconds, 60◦
3. 22.79km/h, E 9.78◦ N
1. i − j − 4k 2. 4i + j − 6k
3. −i + 7j − k 4. 5i − 10k
5. −3i + 27j − 8k 6. −5i + 10k
7. i + 7.5j − 4k 8. 20.5j − 10k
9. 3.6i + 4.4j − 6.9k 10. 2i + 40j − 43k
Chapter 30
1. 4.5sin(A + 63.5◦)
2. (a) 20.9sin(ωt + 0.62) volts
(b) 12.5sin(ωt − 1.33) volts
3. 13sin(ωt + 0.395)
1. 4.5sin(A + 63.5◦)
2. (a) 20.9sin(ωt + 0.62) volts
(b) 12.5sin(ωt − 1.33) volts
3. 13sin(ωt + 0.395)
1. 4.5sin(A + 63.5◦)
2. (a) 20.9sin(ωt + 0.62) volts
(b) 12.5sin(ωt − 1.33) volts
3. 13sin(ωt + 0.395) 4. 11.11sin(ωt + 0.324)
5. 8.73sin(ωt − 0.173)

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 Hz
7. (a) 10.21sin(628.3t + 0.818)V (b) 100 Hz
(c) 10 ms
8. (a) 79.83sin(300πt + 0.352)V (b) 150 Hz
(c) 6.667ms
Chapter 31
1. (a) continuous (b) continuous (c) discrete
(d) continuous
2. (a) discrete (b) continuous (c) discrete (d) discrete
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, working
correct to the nearest 100 components gives: Mon 8,
Tues 11, Wed 9, Thurs 12 and Fri 6.5.
3. 6 equally spaced horizontal rectangles, whose
lengths are proportional to 35, 44, 62, 68, 49 and
41, respectively.
4. 5 equally spaced horizontal rectangles, whose
lengths are proportional to 1580, 2190, 1840, 2385
and 1280 units, respectively.
5. 6 equally spaced vertical rectangles, whose heights
are proportional to 35, 44, 62, 68, 49 and 41 units,
respectively.
6. 5 equally spaced vertical rectangles, whose heights
are proportional to 1580, 2190, 1840, 2385 and 1280
units, respectively.
7. Three rectangles of equal height, subdivided in the
percentages 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 of
about 8% in C, an increase of about 7% in D and a
reduction 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) 138
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,... and
heights of 1,5,9,17,...
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.
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 48
6. Rectangles, touching one another, having mid-
points of 5.5, 15, 21, 24, 27, 33.5 and 43.5. The
heights of the rectangles (frequency per unit class
range) 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), (12.95
42), (13.45, 50)
8. A graph of cumulative frequency against upper class
boundary having co-ordinates given in the answer
to 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 the
solution to part (a) gives 2.0953; 2.14513;
2.19524; 2.24537; 2.29546; 2.34548.
(d) A graph of cumulative frequency against upper
class boundary having the co-ordinates given
in part (c).
Chapter 32
1. Mean 7.33, median 8, mode 8
2. Mean 27.25, median 27, mode 26

3. Mean 4.7225, median 4.72, mode 4.72
4. Mean 115.2, median 126.4, no mode
1. 23.85kg 2. 171.7cm
3. Mean 89.5, median 89, mode 89.2
4. Mean 2.02158cm, median 2.02152 cm, mode
2.02167cm
1. 4.60 2. 2.83μF
3. Mean 34.53MPa, standard deviation 0.07474MPa
4. 0.296kg 5. 9.394cm 6. 0.00544cm
1. 30, 25.5, 33.5 days 2. 27, 26, 33 faults
3. Q1 = 164.5cm, Q2 = 172.5cm, Q3 = 179cm,
7.25cm
4. 37 and 38; 40 and 41 5. 40, 40, 41; 50, 51, 51
Chapter 33
1. (a)
2
9
or 0.2222 (b)
7
9
or 0.7778
2. (a)
23
139
or 0.1655 (b)
47
139
or 0.3381
(c)
69
139
or 0.4964
3. (a)
1
6
(b)
1
6
(c)
1
36
4.
5
36
5. (a)
2
5
(b)
1
5
(c)
4
15
(d)
13
15
6. (a)
1
250
(b)
1
200
(c)
9
1000
(d)
1
50000
1. (a) 0.6 (b) 0.2 (c) 0.15 2. (a) 0.64 (b) 0.32
3. 0.0768 4. (a) 0.4912 (b) 0.4211
5. (a) 89.38% (b) 10.25%
6. (a) 0.0227 (b) 0.0234 (c) 0.0169
Chapter 34
1. 1, 5, 21, 9, 61 2. 0, 11, −10, 21 3. proof
1. 16, 8
1. 28x3 2. 2 3. 2x − 1
4. 6x2 − 5 5. −
1
x2
6. 0
7. 1 +
2
x3
8. 15x4 − 8x3 + 15x2 + 2x
9. −
6
x4
10. 4 − 8x 11.
1
2
√
x
12.
3
2
√
t 13. −
3
x4
16. 1 +
3
2
√
x
17. 2x − 2
18. −
10
x3
+
7
2
√
x9
19. 6t − 12 20. 1 −
4
x2
21. (a) 6 (b)
1
6
(c) 3 (d) −
1
16
(e) −
1
4
(f ) −7
22. 12x − 3 (a) −15 (b) 21
23. 6x2 + 6x − 4,32 24. −6x2 + 4,−9.5
1. (a) 12cos3x (b) −12sin6x
2. 6cos3θ + 10sin2θ 3. −0.707 4. −3
5. 270.2A/s 6. 1393.4V/s
7. 12cos(4t + 0.12) + 6sin(3t − 0.72)
1. (a) 15e3x (b) −
4
7e2x
2.
5
θ
−
4
θ
=
1
θ
3. 16
4. 2.80 5. 664
1. (a) −1 (b) 16
2. −
4
x3
+
2
x
+ 10sin5x − 12cos2x +
6
e3x
1. (a) 36x2 + 12x (b) 72x + 12 2. 8 +
2
x3
3. (a)
4
5
−
12
t5
+
6
t3
+
1
4
√
t3
(b) −4.95
4. −12sin2t − cost 5. −
2
θ2

1. −2542A/s 2. (a) 0.16cd/V (b) 312.5V
3. (a) −1000 V/s (b) −367.9V/s
4. −1.635Pa/m
Chapter 35
1. (a) 4x + c (b)
7x2
2
+ c
2. (a)
5
4
x4
+ c (b)
3
8
t8
+ c
3. (a)
2
15
x3
+ c (b)
5
24
x4
+ c
4. (a)
2
5
x5 −
3
2
x2 + c (b) 2t −
3
4
t4 + c
5. (a)
3x2
2
− 5x + c (b) 4θ + 2θ2 +
θ3
3
+ c
6. (a)
5
2
θ2 − 2θ + θ3 + c
(b)
3
4
x4 −
2
3
x3 +
3
2
x2 − 2x + c
7. (a) −
4
3x
+ c (b) −
1
4x3
+ c
8. (a)
4
5
√
x5 + c (b)
1
9
4
√
x9 + c
9. (a)
10
√
t
+ c (b)
15
7
5
√
x + c
10. (a)
3
2
sin2x + c (b) −
7
3
cos3θ + c
11. (a) −6cos
1
2
x + c (b) 18sin
1
3
x + c
12. (a)
3
8
e2x + c (b)
−2
15e5x
+ c
13. (a)
2
3
ln x + c (b)
u2
2
− lnu + c
14. (a) 8
√
x + 8
√
x3 +
18
5
√
x5 + c
(b) −
1
t
+ 4t +
4t3
3
+ c
1. (a) 1.5 (b) 0.5 2. (a) 105 (b) −0.5
3. (a) 6 (b) −1.333 4. (a) −0.75 (b) 0.833
5. (a) 10.67 (b) 0.1667 6. (a) 0 (b) 4
7. (a) 1 (b) 4.248 8. (a) 0.2352 (b) 2.638
9. (a) 19.09 (b) 2.457 10. (a) 0.2703 (b) 9.099
1. proof 2. proof 3. 32 4. 29.33Nm
5. 37.5 6. 7.5 7. 1
8. 1.67 9. 2.67 10. 140 m

Index
Acute angle, 165
Acute angled triangle, 171
Adding waveforms, 278
Addition in algebra, 62
Addition law of probability, 307
Addition of fractions, 10
numbers, 1, 18
two periodic functions, 278
vectors, 267
by calculation, 270
Algebra, 61, 68
Algebraic equation, 61, 73
expression, 73
Alternate angles, 165, 191
Ambiguous case, 207
Amplitude, 199
Angle, 165
Angle, lagging and leading, 200
types and properties of, 165
Angles of any magnitude, 196
depression, 191
elevation, 191
Angular measurement, 165
velocity, 202
Annulus, 226
Arbitrary constant of integration, 325
Arc, 231
Arc length, 233
Area, 219
Area of common shapes, 219, 221
under a curve, 330
Area of circle, 222, 233
common shapes, 219
irregular figures, 257
sector, 222, 233
similar shapes, 229
triangles, 205
Arithmetic, basic, 1
Average, 299
value of waveform, 260
Axes, 130
Bar charts, 289
Base, 47
Basic algebraic operations, 61
BODMAS with algebra, 71
fractions, 13
numbers, 6
Boyle’s law, 46
Brackets, 6, 68
Calculation of resultant phasors, 281,
283
Calculations, 22, 28
Calculator, 22
addition, subtraction, multiplication
and division, 22
fractions, 26
π and ex functions, 28, 118
reciprocal and power functions, 24
roots and ×10x functions, 25
square and cube functions, 23
trigonometric functions, 27
Calculus, 313
Cancelling, 10
Cartesian axes, 131
co-ordinates, 214
Charles’s law, 42, 142
Chord, 230
Circle, 222, 230, 233
equation of, 236
properties of, 230
Circumference, 230
Classes, 293
Class interval, 293
limits, 295
mid-point, 293, 295
Coefficient of proportionality, 45
Combination of two periodic functions,
278
Common factors, 69
logarithms, 111
prefixes, 53
shapes, 219
Complementary angles, 165
Completing the square, 105
Cone, 245
frustum of, 252
Congruent triangles, 175
Construction of triangles, 179
Continuous data, 288
Co-ordinates, 130, 131
Corresponding angles, 165
Cosine, 27, 183
graph of, 195
Cosine rule, 205, 281
wave, 195
Cross-multiplication, 75
Cube root, 23
Cubic equation, 161
graphs, 161
units, 240
Cuboid, 240
Cumulative frequency distribution,
293, 297
curve, 293
Cycle, 199
Cylinder, 241
Deciles, 304
Decimal fraction, 216
places, 13, 18
Decimals, 16
addition and subtraction, 19
multiplication and division, 19
Definite integrals, 328
Degrees, 27, 165, 166, 232
Denominator, 9
Dependent event, 307
Depression, angle of, 191
Derivatives, 315
standard list, 321
Derived units, 53
Determination of law, 147
involving logarithms, 150
Diameter, 230
Difference of two squares, 103
Differential calculus, 313
coefficient, 315
Differentiation, 313, 315
from first principles, 315
of axn, 315
of eax and ln ax, 320
of sine and cosine functions, 318
successive, 322
Direct proportion, 40, 42
Discrete data, 288
standard deviation, 302
Dividend, 63
Division in algebra, 62
Division of fractions, 12
numbers, 3, 4, 19
Divisor, 63
Drawing vectors, 266

Index 357
Elevation, angle of, 191
Engineering notation, 57
Equation of a graph, 135
Equations, 73
circles, 236
cubic, 161
indicial, 115
linear and quadratic, simultaneously,
110
quadratic, 102
simple, 73
simultaneous, 90
Equilateral triangle, 171
Evaluation of formulae, 28
trigonometric ratios, 185
Expectation, 306
Exponential functions, 118
graphs of, 120
Expression, 73
Exterior angle of triangle, 171
Extrapolation, 133
Factorial, 119
Factorization, 69
to solve quadratic equations, 102
Factors, 5, 69
False axes, 142
Formula, 28
quadratic, 106
Formulae, evaluation of, 28
list of, 336
transposition of, 83
Fractions, 9
addition and subtraction, 10
multiplication and division, 12
on calculator, 26
Frequency, 200, 289
relative, 289
Frequency distribution, 293, 296
polygon, 293, 296
Frustum, 252
Full wave rectified waveform, 260
Functional notation, 313, 315
Gradient, 134
of a curve, 314
Graph drawing rules, 133
Graphical solution of equations, 155
cubic, 161
linear and quadratic, simultaneously,
110
quadratic, 156
simultaneous, 155
Graphs, 130
exponential functions, 120
logarithmic functions, 116
reducing non-linear to linear form,
147
sine and cosine, 199
straight lines, 130, 132
trigonometric functions, 195
Grid, 130
reference, 130
Grouped data, 292
mean, median and mode, 300
standard deviation, 302
Growth and decay, laws of, 125
Half-wave rectified waveform, 260
Hemisphere, 249
Heptagon, 219
Hexagon, 219
Highest common factor (HCF), 5, 51,
66, 69
Histogram, 293–296, 300
Hooke’s law, 42, 142
Horizontal bar chart, 289
component, 269, 283
Hyperbolic logarithms, 111, 122
Hypotenuse, 172
i, j,k notation, 277
Improper fraction, 9
Indefinite integrals, 328
Independent event, 307
Index, 47
Indices, 47, 64
laws of, 48, 64
Indicial equations, 115
Integers, 1
Integral calculus, 313
Integrals, 325
definite, 328
standard, 326
Integration, 313, 325
of axn, 325
Intercept, y-axis, 135
Interest, 37
Interior angles, 165, 171
Interpolation, 132
Inverse proportion, 40, 45
trigonometric function, 185
Irregular areas, 257
volumes, 259
Isosceles triangle, 171
Lagging angle, 200
Laws of algebra, 61
growth and decay, 125
indices, 48, 64, 316
logarithms, 113, 150
precedence, 6, 71
probability, 307
Leading angle, 200
Leibniz notation, 315
Limiting value, 314
Linear and quadratic equations
simultaneously, 110
graphical solution, 160
Logarithms, 111
graphs involving, 116
laws of, 113, 150
Long division, 4
Lower class boundary, 293
Lowest common multiple (LCM), 5,
10, 75
Major arc, 231
sector, 230
segment, 231
Maximum value, 156, 199
Mean, 299, 300
value of waveform, 260
Measures of central tendency, 299
Median, 299
Member of set, 289
Mid-ordinate rule, 257
Minimum value, 156
Minor arc, 231
sector, 230
segment, 230
Mixed number, 9
Mode, 299
Multiple, 5
Multiplication in algebra, 62
law of probability, 307
of fractions, 12
of number, 3, 19
Table, 3
Napierian logarithms, 111, 122
Natural logarithms, 111, 122
Newton, 53
Non right-angled triangles, 205
Non-terminating decimals, 18
Nose-to-tail method, 267
Numerator, 9
Obtuse angle, 165
Obtuse-angled triangle, 171
Octagon, 219
Ogive, 293, 297
Ohm’s law, 42
Order of precedence, 6, 13, 71
with fractions, 13
with numbers, 6
Origin, 131

358 Index
Parabola, 156
Parallel lines, 165
Parallelogram, 219
method, 267
Peak value, 199
Pentagon, 219
Percentage component bar chart, 289
error, 36
relative frequency, 289
Percentages, 33
Percentile, 304
Perfect square, 105
Perimeter, 171
Period, 199
Periodic function, 200
plotting, 238
Periodic time, 200
Phasor, 280
Pictograms, 289
Pie diagram, 289
Planimeter, 257
Plotting periodic functions, 238
Polar co-ordinates, 214
Pol/Rec function on calculator, 217
Polygon, 210
frequency, 293, 296
Population, 289
Power, 47
series for ex, 119
Powers and roots, 47
Practical problems
quadratic equations, 108
simple equations, 77
simultaneous equations, 96
straight line graphs, 141
trigonometry, 209
Precedence, 6, 71
Prefixes, 53
Presentation of grouped data, 292
statistical data, 288
Prism, 240, 242
Probability, 306
laws of, 307
Production of sine and cosine waves,
198
Proper fraction, 9
Properties of circles, 230
triangles, 171
Proportion, 40
Pyramid, 244
volumes and surface area of frustum
of, 252
Pythagoras’ theorem, 181
Quadrant, 230
Quadratic equations, 102
by completing the square, 105
factorization, 102
formula, 106
graphically, 156
practical problems, 108
Quadratic formula, 106
graphs, 156
Quadrilaterals, 219
properties of, 219
Quartiles, 303
Radians, 27, 165, 166, 232
Radius, 230
Range, 295
Ranking, 299
Rates of change, 323
Ratio and proportion, 40
Ratios, 40
Reciprocal, 24
Rectangle, 219
Rectangular axes, 131
co-ordinates, 131
prism, 240
Reduction of non-linear laws to linear
form, 147
Reflex angle, 165
Relative frequency, 289
velocity, 276
Resolution of vectors, 269
Resultant phasors, by drawing, 280
horizontal and vertical components,
283
plotting, 278
sine and cosine rules, 281
Rhombus, 219
Right angle, 165
Right angled triangle, 171
solution of, 188
Sample, 289
Scalar quantities, 266
Scalene triangle, 171
Scales, 131
Sector, 222, 230
area of, 233
Segment, 230
Semicircle, 230
Semi-interquartile range, 304
Set, 289
Short division, 4
Significant figures, 17, 18
Similar shapes, 229, 256
triangles, 176
Simple equations, 73
Simpson’s rule, 258
Simultaneous equations, 90
graphical solution, 155
in three unknowns, 99
in two unknowns, 90
Sine, 27, 183
graph of, 195
Sine rule, 205, 281
wave, 198, 260
mean value, 260
Sinusoidal form Asin(ωt ± α), 202
SI units, 53
Slope, 134
Solution of linear and quadratic
equations simultaneously, 110
Solving right-angled triangles, 188
simple equations, 73
Space diagram, 276
Sphere, 246
Square, 23, 219
numbers, 23
root, 25, 48
units, 219
Standard deviation, 302
discrete data, 302
grouped data, 303
Standard differentials, 321
form, 56
integrals, 326
Statistical data, presentation of, 288
terminology, 288
Straight line, 165
equation of, 135
Straight line graphs, 132
Subject of formulae, 83
Subtraction in algebra, 62
Subtraction of fractions, 10
numbers, 1, 18
vectors, 274
Successive differentiation, 322
Supplementary angles, 165
Surface areas of frusta of pyramids and
cones, 252
of solids, 247
Symbols, 28
Tally diagram, 293, 296
Tangent, 27, 183, 230
graph of, 195
Terminating decimal, 17
Theorem of Pythagoras, 181
Transposition of formulae, 83
Transversal, 165
Trapezium, 220
Trapezoidal rule, 257

Index 359
Triangle, 171, 219
Triangles, area of, 205
congruent, 175
construction of, 179
properties of, 171
similar, 176
Trigonometric functions, 27
Trigonometric ratios, 183
evaluation of, 185
graphs of, 195
waveforms, 195
Trigonometry, 181
practical situations, 209
Turning points, 156
Ungrouped data, 289
Units, 53
Upper class boundary, 293
Use of calculator, 22
Vector addition, 267
subtraction, 274
Vectors, 266
addition of, 267
by calculation, 267
by horizontal and vertical
components, 269
drawing, 266
subtraction of, 274
Velocity, relative, 276
Vertical axis intercept, 133
bar chart, 289
component, 269, 283
Vertically opposite angles, 165
Vertices of triangle, 172
Volumes of common solids, 240
frusta of pyramids and cones, 252
irregular solids, 259
pyramids, 244
similar shapes, 256
Waveform addition, 278
y-axis intercept, 135
Young’s modulus of elasticity, 143

Basic engineering mathematics e5

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Basic engineering mathematics e5