Mat1500 Sb Colour

MAT1500

Engineering Mathematics IFaculty of Sciences

S tudy book

Edited by:D. Mander, H. Butler & A. FrederiksDepartment of Mathematics & ComputingUniversity of Southern Queensland

Based on materials developed within the Department of Mathematics & Computing andLTSU, with contributions by: H. Butler, P. Dunn, L. Galligan, D. Mander, C. McDonald,J. Mohr, J. Taylor, and others.

c© The University of Southern Queensland, August 27, 2010.

Distributed by

The University of Southern QueenslandToowoomba, Queensland 4350Australiahttp://www.usq.edu.au

Copyrighted materials reproduced herein are used under the provisions of theCopyright Act 1968 as amended, or as a result of application to the copyright owner.

No part of this publication may be reproduced, stored in a retrieval system ortransmitted in any form or by any means electronic, mechanical, photocopying,recording or otherwise without prior permission.

Produced with LATEX by the author(s) using the Department of Mathematics &Computing (http://www.sci.usq.edu.au) StudyBook class. Adapted from the LATEX“report“ class.

http://www.usq.edu.au

http://www.sci.usq.edu.au

Table of Contents

1 Arithmetic and Algebra 1

1.1 A few preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Powers and indices . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Scientific notation and units . . . . . . . . . . . . . . . . . . . . . . 13

1.1.3 Number calculations . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.1.4 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.2 Algebraic notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.3 Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.4 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.5 Factorising Cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1.6 Algebraic fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.7 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.7.1 Quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.7.2 Simultaneous equations . . . . . . . . . . . . . . . . . . . . . . . . 63

1.8 Inequalities and absolute values . . . . . . . . . . . . . . . . . . . . . . . 79

1.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

1.10 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

1.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

1.11.1 Answers to selected Exercises . . . . . . . . . . . . . . . . . . . . . 92

1.11.2 Post-test solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

2 Functions and Graphing 103

2.1 The concept of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

2.1.1 What is a function? . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

2.1.2 Functional notation and representation . . . . . . . . . . . . . . . 105

2.1.3 Inverse function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

c© USQ, August 27, 2010 i

MAT1500 Table of Contents

2.2 Linear functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

2.2.1 Slope-point equation . . . . . . . . . . . . . . . . . . . . . . . . . . 124

2.2.2 Point-point equation . . . . . . . . . . . . . . . . . . . . . . . . . . 126

2.2.3 Some special lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

2.2.4 Distance between two points . . . . . . . . . . . . . . . . . . . . . 129

2.3 Quadratic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

2.3.1 Completing the square . . . . . . . . . . . . . . . . . . . . . . . . . 137

2.3.2 Quadratic Equation Formula . . . . . . . . . . . . . . . . . . . . . 138

2.3.3 Sketching parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

2.4 Rational functions — the rectangular hyperbola . . . . . . . . . . . . . 144

2.4.1 Graphing rational functions . . . . . . . . . . . . . . . . . . . . . . 144

2.5 Circle — not a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

2.6 When two functions meet . . . . . . . . . . . . . . . . . . . . . . . . . . 158

2.7 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

2.8 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

2.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172



3 Exponential, logarithmic and trigonometric functions 185

3.1 Exponential functions (ex, 10x and ax) . . . . . . . . . . . . . . . . . . . 187

3.2 Logarithm as an inverse function . . . . . . . . . . . . . . . . . . . . . . 192

3.2.1 Properties of log functions . . . . . . . . . . . . . . . . . . . . . . . 199

3.2.2 Working with logarithmic equations . . . . . . . . . . . . . . . . . 203

3.2.3 Working with exponential equations . . . . . . . . . . . . . . . . . 205

3.3 Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

3.4 Trigonometric identities . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

3.5 Triangle solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

3.6 Compound angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

3.7 Radian measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

3.8 Graphs involving trigonometric functions . . . . . . . . . . . . . . . . . 249

3.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

3.10 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

3.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

ii c© USQ, August 27, 2010

Table of Contents MAT1500



4 Vectors 279

4.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280

4.2 Displacement Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

4.3 Vectors in General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

4.4 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

4.5 Angle between vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

4.6 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

4.7 Post Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294

4.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295



5 Matrices 303

5.1 What is a matrix? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

5.2 Matrix addition and subtraction . . . . . . . . . . . . . . . . . . . . . . . 309

5.3 Multiplication of a matrix by a scalar . . . . . . . . . . . . . . . . . . . . 313

5.4 Transpose of a matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

5.5 Multiplication of a matrix by a matrix . . . . . . . . . . . . . . . . . . . 316

5.6 Some special matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

5.6.1 Identity matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

5.6.2 The inverse matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 334

5.7 Linear equations in matrix notation . . . . . . . . . . . . . . . . . . . . . 340

5.8 Solution of linear equations using the inverse matrix . . . . . . . . . . . 345

5.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354

5.10 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355

5.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357



6 Calculus – Differentiation 363

6.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364

6.2 Rate of change - the problem of the curve . . . . . . . . . . . . . . . . . 366

c© USQ, August 27, 2010 iii

MAT1500 Table of Contents

6.3 Instantaneous rate of change - the derivative . . . . . . . . . . . . . . . 369

6.4 Finding the derivative algebraically . . . . . . . . . . . . . . . . . . . . 378

6.5 Derivatives of compound functions . . . . . . . . . . . . . . . . . . . . . 386

6.6 Application: maxima and minima . . . . . . . . . . . . . . . . . . . . . 392

6.7 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

6.8 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

6.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405



A Technical communication 415

A.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

A.2 Writing Mathematics and numerical data . . . . . . . . . . . . . . . . . 416

A.2.1 Whole numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416

A.2.2 Decimals and fractions . . . . . . . . . . . . . . . . . . . . . . . . . 418

A.2.3 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

A.2.4 Mathematical equations . . . . . . . . . . . . . . . . . . . . . . . . 421

A.3 Tables and illustrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

A.3.1 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

A.3.2 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428

A.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431

References 433

iv c© USQ, August 27, 2010

Chapter 1

Arithmetic and Algebra

Chapter contents1.1 A few preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Powers and indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Scientific notation and units . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.1.3 Number calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.1.4 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.2 Algebraic notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.3 Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.4 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.5 Factorising Cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1.6 Algebraic fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.7 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.7.1 Quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.7.2 Simultaneous equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

1.8 Inequalities and absolute values . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

1.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

1.10 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

1.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

1.11.1 Answers to selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 92

1.11.2 Post-test solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Objectives

When you have successfully completed this chapter, you should be able to:

• apply the index laws to simplify problems involving either algebraic symbols, dec-imal numbers or numbers in scientific notation;

c© USQ, August 27, 2010 1

MAT1500

• change a number expressed in an inappropriate metric unit to the correct unit andprefix;

• factorise and expand algebraic expressions;

• convert algebraic fractions to a single term;

• manipulate algebraic equations and inequations in order to change the subject ofthe equation;

• express inequalities in interval notation;

• translate word problems into a mathematical formulation;

• recognise the general form of a quadratic equation;

• determine whether a quadratic equation has no solution, one solution or two dif-ferent solutions using the discriminant;

• find the solutions (if any) of any quadratic equation; and

• solve two simultaneous equations using either the substitution or the eliminationmethod.

Introduction

Arithmetic and algebra are the building blocks upon which most of mathematics is con-structed. As you embark on the task of developing your own understanding of mathe-matics, you will find that you continually rely on the skills and concepts you have de-veloped in this chapter. We cover a number of very useful tools for your further studiesincluding the concept of raising quantities to powers, applying this technique to scien-tific notation and metric units, the use of brackets and their expansion in algebraic ex-pressions, the reverse process of factorisation, the extension of fractions from numericalto algebraic, and the solution of equations and inequations.

Hint How to read maths:

• skim the materials first – get an idea of the major concepts and topics;

• circle or note in a summary book words you do not understand;

• re-read the material, concentrating fully;

• stop at examples and go through step by step (if steps are skipped, writethem in now);

• do activities when you come to them; and

• remember maths is learnt by doing, not just reading.

2 c© USQ, August 27, 2010

1.1. A few preliminaries MAT1500

1.1 A few preliminaries

1.1.1 Powers and indices

Powers are everywhere. In arithmetic, they help to represent large and small numbers.Look at this example from Australian Organizational Behaviour by Sherman (1985).

Ethnic background Number of AustraliansBritish 4.7 millionIrish 3.5 millionGerman/Austrian 1.2 millionItalian 1 millionGreek 6.5× 105

Maltese 4× 105

Yugoslav 2.5× 105

Dutch 2× 105

Aboriginal 1.6× 105

Spanish speaking 1.5× 105

Arabic speaking 1.5× 105

Polish 1× 105

In business calculations you might see formulas like,

S = p(1 + t)n − 1

i,

PV = p1− (1 + i)−n

t, or

G = pi

(1 + i)1k − 1

,

which are aspects of annuity calculations.

In statistics you will see,

S =

√∑n

i=1(xi − x)2

n− 1,

which is the formula for standard deviation of a sample.

In chemistry you might come across

y =( m

2πkT

) 12

c2e−mc22kT ,

which is the Maxwell-Boltzmann distribution function for molecular speeds. In all theseexamples the numbers written as superscripts are called indices.

Do not be put off by these equations. They may look a bit daunting but they are just ex-amples at this stage. Do not learn them. But what do these positive, negative and fractional


MAT1500 1.1. A few preliminaries

indices mean? Let us look at a more familiar example.

The area of a square is the length of the side multiplied by itself. Consequently we say anumber is ‘squared’ whenever it is multiplied by itself. We write this as A = L× L = L2 .Multiplying this again by L gives the volume of a cube with length L. That is, the volumeequals the length ‘cubed’.

This is written V = L× L× L = L3 .

We call the superscripts 2 and 3 above powers of the quantity L. This pattern continueswith

L× L× L× L = L4 ,

described as ‘L to the fourth’ or ‘L to the power four’ and so on for higher powers. In general‘L to the n’ is

Ln = L× L× L× · · · × L︸︷︷︸n L’s

.

Notice that

102 = 10× 10 = 100 ,

103 = 10× 10× 10 = 1 000 ,

104 = 10× 10× 10× 10 = 10 000 ,

105 = 10× 10× 10× 10× 10 = 100 000 .

That is, the power of 10 is the number of 0’s in the number.



Hint You must be able to do calculations with powers on your calculator. Ifyou have trouble using your calculator, contact The Learning Centre or view theFoundation Mathematics website for access to calculator booklets which haveinstructions on what to do.

Example 1.1: Use your calculator to evaluate the following.

(a) 54

(b) 2.733

(c) 06

(d) (−4)2

(e) −42

Solution:

(a) 625.

(b) 20.346 417.

(c) 0. Note that zero raised to any power is zero with the exception of 00 whichcannot be determined. (Try finding 00 on your calculator. You will see that itcomes up with an ERROR message in the display.)

(d) You should use brackets on your calculator to assist with this calculation oth-erwise you will end up calculating the answer to part (e). Recall that

(−4)2 = −4×−4 = 16 .

(e) −42 = −1× 42 = −1× 4× 4 = −16 . This type of calculation can be donedirectly on most modern scientific calculators.

Multiplication and division by powers

What happens if we multiply two numbers together where both are powers of the samenumber? Say,

L3 × L5 = L× L× L︸︷︷︸L3

× L× L× L× L× L︸︷︷︸L5

= L8

Clearly the number of L’s multiplied together is just the sum of the powers of the L3 andL5. In general, we could say that for any number a, and any integers n and m,

an × am = an+m .



What happens when we divide one number by another, where both are powers of thesame number? Using a similar example to above we could say

L5

L3 =L× L× L× L× L

L× L× L

=L× L×L×L×LL×L×L

= L× L

= L2 .

Clearly, the pairs of L’s cancel, so that the number of L’s remaining is the difference be-tween the two powers. In general, we could say that for any number a, and any integersn and m,

an

am = an−m .

Powers of powers

Suppose we have a number that is the power of another number, L3 say, and that thisnumber is then multiplied by itself 5 times. That is, the number L3 is raised to the power5. We write this (L3)5. When we expand this expression we have

(L3)5

= L3 × L3 × L3 × L3 × L3 .

Since L× L× L could replace each L3, we have

(L3)5

= L× L× L × L× L× L × L× L× L × L× L× L

×L× L× L

= L15 .

In general we could write(an)m = an×m = anm .

Negative powers

Observe the pattern in the following table as we divide each preceding row by L startingwith the number L4.

Using power notation Using multiplication Power

L4 L× L× L× L 4L3 L× L× L 3L2 L× L 2L1 L 1L0 1 0



Notice that L to the power zero is 1, irrespective of the value of L. In general a0 = 1 forany value of a (except 00, that is). Let us continue the table further:

Using power notation Using multiplication Power

L4 L× L× L× L 4L3 L× L× L 3L2 L× L 2L1 L 1L0 1 0

L−1 1L

−1

L−2 1L× L

=1L2 −2

L−3 1L× L× L

=1L3 −3

In general the negative power indicates the reciprocal of that power. That is,

L−n =1Ln .

Similarly,1

L−n = 1÷ L−n = 1÷ 1Ln = 1× Ln = Ln .

Power of a product

Look at what happens when we take a number, which is the product of two numbers,and raise it to a power. For example,

(LG)3 = (L× G)3

= L× G × L× G × L× G= L× L× L × G× G× G Since the order we multiply numbers does

not matter.= L3 × G3

= L3G3 .

Of course there was nothing special about the number 3 in this example so in general

(ab)n = an × bn = anbn ,

and ( ab

)n=

an

bn .

Be careful though, (a + b)n is not an + bn (as we shall see in section 1.2). For example ifa = 1 and b = 2 , then (1 + 2)2 = 32 = 9 , but 12 + 22 = 1 + 4 = 5 , so (1 + 2)2 6= 12 + 22 .



(Note that 6= means ‘not equal to’.)

Fractional index

If area A = L2, we say that A is the square of L, and conversely, L is the square root of A.We could also say, given the ‘power of a power’ property, that

A =(

A12

)2.

Thus A12 is the same as L, that is, the square root of A. By similar argument A

13 is the

cube root of A, and in general a1n is the nth root of a.

Exploring amn , we see that this could be written

(a

1n

)m=(

n√

a)m or (am)

1n = n√

am .

Now might be a good time to make a summary of the index laws for future reference.

Index Laws

an × am = an+m

an

am = an−m

(an)m = an×m = anm

(ab)n = an × bn = anbn

a−n =1an

a1n = n√

a

a0 = 1

Hint Now is the time to start preparing for the exam by:

• starting a summary of important rules or formulas; and

• developing a glossary of terms you are not familiar with.

Remember both of these items may be taken into the open book exam.



Example 1.2: Use index laws to simplify and evaluate, if possible, without a calculator.

(a)8a−2

a3

(b) 3−2.5 × 33.5

(c)(

x2y−4) 1

2

(d) 823 × 9

32

(e)8(a3b6)− 1

3

b−5

(f)

(9x−2y4) 1

2

3 (x−2y−4)32

Solution:

(a)8a−2

a3 = 8× a−2−3 = 8a−5 .

(b) 3−2.5 × 33.5 = 3−2.5+3.5 = 31 = 3 .

(c)(x2y−4) 1

2 =(

x2) 12 ×

(y−4) 1

2 = x1 × y−2 = x1 × 1y2 =

xy2 .

(d) 823 × 9

32 =

(23) 2

3 ×(32) 3

2 = 22 × 33 = 4× 27 = 108 .

(e)

8(a3b6)− 1

3

b−5 =8(a3)− 1

3(b6)− 1

3

b−5

=8a−1b−2

b−5

= 8a−1b−2−−5

= 8× 1a× b−2+5

=8a× b3

=8b3

a.



(f)

(9x−2y4) 1

2

3 (x−2y−4)32

=9

12 × x−2× 1

2 × y4× 12

3× x−2× 32 × y−4× 3

2

=3x−1y2

3x−3y−6

= x−1−−3y2−−6

= x2y8 .

Example 1.3: Simplify, using the index laws.

(a)5−4 × 5

72

12513

(b)

(m3 × n−2)2

mn−3

(c) (−8)13 +

4√

16

(d) 2723 −

(13

)−3

(e) a4.2 − a3.1 + a0.9

(f)√

x2 + x4

Solution:

(a)5−4 × 5

72

12513

=5−4 × 5

72

(53)13

=5−4 × 5

72

51 = 5−4+ 72−1 = 5−

32 =

1

532

,

or1

51512=

15√

5.

(b) (m3 × n−2)2

mn−3 =m3×2n−2×2

m1 × n−3 =m6n−4

m1 × n−3 = m6−1n−4−−3

= m5n−1 orm5

n.

(c)

(−8)13 +

4√

16 =((−2)3

) 13+(

24) 1

4= (−2)1 + 21 = 0 .

Unlike the square root, you can find the cube root of a negative number.

(d)

2723 −

(13

)−3

=(33) 2

3 −(

3−1)−3

= 32 − 33 = 9− 27 = −18 .

The expression cannot be simplified further using the index laws.

10 c© USQ, August 27, 2010


(e)a4.2 − a3.1 + a0.9

As the terms are added rather than multiplied, no simplification is possibleusing the index laws.

(f) √x2 + x4 =

√x2 (1 + x2) =

√x2√

1 + x2 = x√

1 + x2

or(

x2 + x4) 1

2=(

x2 (1 + x2)) 12 = x2× 1

2(1 + x2) 1

2 = x√

1 + x2 .

Note, as terms are added inside the square root, we cannot take the square rootof each term.

Exercise 1.4: Reduce the following expressions to simplified forms without using yourcalculator.

(a)16

12 × 3

32

23 × 2−1

(b)9

32 × 4

52

32 × 23

(c)

(42)3 × 92

642 × 2743

(d)73 × 54

352

(e)63 × 22

32 × 23

(f)121

52 × 49

32

113 × 72

(g)8a2b3

(ab)4

(h)

√16a2b

a3b32

(i)3√

27a4b2√

81a3b4

(j)

(a3b−3)3

ab−2

(k) (−27)13 +

4√

81

(l) 12132 − 64

43

c© USQ, August 27, 2010 11


Application: science A dimension is the qualitative nature of a physical quan-tity (for example length, mass, and time). Dimensional analysis aids in deriving orchecking formulas by treating dimensions as algebraic quantities. Quantities canbe added or subtracted only if they have the same dimensions, and quantities ontwo sides of an equation must have the same dimensions. Dimensional analysiscannot give numerical factors.

Quantity DimensionArea [A] = L2

Volume [V] = L3

Velocity [v] = L/TAcceleration [a] = L/T2

Mass [m] = M

For example: The distance (x) travelled by a car in a given time (t), starting fromrest and moving with constant acceleration (a), is given by x = at2 . We cancheck this equation with dimensional analysis:

LHS → [x] = L ,

RHS →[

12

at2]= [a][t2] =

LT2 × T2 = L .

Since the dimension of the left hand side (LHS) of the equations is the same as thaton the right hand side (RHS), the equation is said to be dimensionally consistent.

Points to remember

• an = a× a× a . . .× a︸︷︷︸n times

.

• an × am = an+m.

• an

am = an−m.

• (an)m = an×m = anm.

• (ab)n = an × bn = anbn.

• a−n =1an .

• a1n = n√

a.

• a0 = 1.

12 c© USQ, August 27, 2010


1.1.2 Scientific notation and units

Scientific notation

Consider the following measurements:

Breaking stress of steel = 430 000 000 pascals.Wavelength of red light = 0.000 000 644 metres.

Mass of a virus = 0.000 000 000 000 000 000 3 kilograms.

These very large and very small numbers involving long strings of zeros are too difficultto handle as they stand, particularly using calculators. To overcome this, notice that eachtime we multiply a number by 10, we effectively move the decimal point one place to theright. For example, 12.35× 10 = 123.5 . Similarly each time we divide a number by 10(which is the same as multiplying by 10−1), we move the decimal point one place to theleft. For example 123.5× 10−1 = 12.35 .

Thus,

430 000 000 = 4.3× 10× 10× 10× 10× 10× 10× 10× 10

= 4.3× 108 .

Similarly,

0.000 000 644 = 6.44× 10−1 × 10−1 × 10−1 × 10−1 × 10−1 × 10−1 × 10−1

= 6.44× 10−7 .

So the breaking stress of steel becomes 4.3× 108 Pa. Similarly, the wavelength of red lightis 6.44× 10−7 m and the mass of a virus is 3× 10−19 kg.

This way of expressing numbers is called scientific notation. Notice that its form is anumber between 1 and 10 (called the mantissa), multiplied by a power of 10.

Beware of negative indices, for example 10−7 is a larger number than 10−19 but both arestill positive numbers.

When calculating with scientific numbers, you must use the index laws to find the powersof 10.

Application: engineering Engineering notation is a term that refers to theapplication of scientific notation in which the powers of ten are limited to multiplesof three. In scientific notation, a quantity is expressed as a product of a positivenumber between 1 and 10 and a power of 10. For example, the quantity 250 000would be expressed in scientific notation as 2.5× 105. But in engineering notation,it would be 250× 103. Although this notation is not used widely in Australia, somecalculators and some electronic instruments, such as the multimeter, are set up inranges to accommodate engineering notation.

c© USQ, August 27, 2010 13


Example 1.5: Convert the following to scientific notation.

(a) 1 072 000

(b) 0.003 71

(c) 2 421.372

Solution:

(a) 1.072× 106 .

(b) 3.71× 10−3 .

(c) 2.421 372× 103 .

Example 1.6: Change the decimal point to make the number fall in the range 1 to 10.

(a) 741.3× 107

(b) 0.034× 10−2

(c) 1 820× 10−5

Solution:

(a) 7.413× 109 .

(b) 3.4× 10−4 .

(c) 1.820× 10−2 .

Example 1.7: Simplify the following, using scientific notation throughout.

(a) 4.37× 10105 + 2.41× 10104

(b) (2.5× 10−3)× (4.4× 1014)

(c)7.8× 104

2× 10−3

(d)√

1.6× 10−5

Solution:

(a) Recall that numbers with different indices cannot be added or subtracted. Onemust be converted to the same index as the other. Therefore,

4.37× 10105 + 2.41× 10104 = 4.37× 10105 + 0.241× 10105

= (4.37 + 0.241)× 10105

= 4.611× 10105 .

(b)

(2.5× 10−3)× (4.4× 1014) = (2.5× 4.4)× (10−3 × 1014)

= 11× 10−3+14

= 11× 1011

= 1.1× 1012 .

14 c© USQ, August 27, 2010


(c) Note that the index of −3 refers only to 10 and not to 2. Thus when 10−3 isbrought to the top to become 103, the number 2 remains underneath.

7.8× 104

2× 10−3 =7.82× 104 × 103

= 3.9× 104+3

= 3.9× 107 .

(d) √1.6× 10−5 =

√1.6×

√10−5 = 1.264 9× 10−2.5 .

Although this is correct it is not a scientific number since the index shouldindicate the number of places the decimal point was moved. The index mustbe a whole number. Thus we must change the problem first, for example,

√1.6× 10−5 =

√16× 10−6 =

√16×

√10−6 = 4× 10−3 ,

or √1.6× 10−5 =

√0.16× 10−4 =

√0.16×

√10−4 = 0.4× 10−2 .

Exercise 1.8: Convert the following numbers to scientific notation.

(a) 221.6× 106

(b) 0.042 1× 10−5

(c) 3 124.8× 10−4

(d) 16.185× 104

(e) 11.214× 10−2

(f) 1 121.4

(g) 3 124.8× 104

(h) 0.001 618 5

(i) 0.000 45× 106

(j) 0.000 45× 10−6

Exercise 1.9: Simplify the following using scientific notation throughout.

(a) 5.684× 103 + 0.218× 102

(b) 1.68× 103 + 0.64× 102

(c) 25.782 4× 1060 + 327.987× 1059 − 3 204.68× 1058

(d)√

1.44× 10−4

(e) 3√

2.7× 107

(f)√

1.68× 105 + 0.42× 106 + 41.2× 104

c© USQ, August 27, 2010 15


Application: computer science In 1999, the fastest computer in the worldcould perform a single calculation in about 0.000 000 000 000 26 second (twenty-six hundred trillionths of a second). How long would it take this computer to do7 billion calculations? The computer would take

0.000 000 000 000 26× 7 000 000 000 = 2.6× 10−13 × 7× 109

= 18.2× 10−4

= 1.82× 10−3 seconds.

Metric units

Another way of reducing the size of a large number, or increasing the size of a verysmall number is to change the size of the units we are using. You may sometimes hearpoliticians or engineers bragging that their new dam contains so many times the capacityof Sydney Harbour. Because the quantity is so large it is not appropriate or convenientto express it in litres. Similarly you may hear of small electronic components being somefraction of a human hair. The metric system is a more rigorous system of units thanSydney Harbours and human hairs. The metric system uses a number of units for length,mass, time, and so on, to which are attached various prefixes to indicate the size of theunit. The most common units include:

Table 1.1: Most common units.

Measurement Unit SymbolLength metre mMass gram gVolume litre LTime second sForce newton NPower watt WPressure pascal Pa

The system of prefixes is based on powers of 103. Any prefix may be attached to any unit.

Table 1.2: Prefixes used with units.

Prefix Factor Symbolgiga- 109 Gmega- 106 Mkilo- 103 k(base unit) 100 = 1milli- 10−3 mmicro- 10−6 µnano- 10−9 npico- 10−12 pcenti-† 10−2 c

16 c© USQ, August 27, 2010


†‘Centi’ is not properly part of the system as it is not a power of 103. However, it iscommonly used and is therefore included here.

Numbers should always be expressed in the range from 1 to 999. This is achieved byselecting the appropriate prefix, for example, 2 487 mL (millilitres) becomes 2.487 L (litres)and 0.047 Mg (megagrams, often called tonne or metric ton) becomes 47 kg (kilograms).When changing prefixes, you should remember that if you make the number smaller, theprefix becomes larger and vice versa. Since the prefixes are based on powers of 103, thedecimal point moves in steps of three places.

Thus in appropriate metric units,

breaking stress of steel = 430 000 000 pascals

= 430 MPa (mega-pascals)

wavelength of red light = 0.000 000 644 metres

= 644 nm (nanometres)

mass of a virus = 0.000 000 000 000 000 000 3 kilograms

= 3× 10−4 pg (pico-grams)

Example 1.10: Complete the following unit conversions.

(a) Convert 0.012 4 kilograms to grams.

(b) Convert 472× 104 millimetres to kilometres.

(c) 2 472 kilo-pascals + 1.34 mega-pascals (express your answer as mega-pascals).

(d) Convert 12 000× 120 microseconds to seconds.

(e) Convert 1 784 centimetres to metres.

(f) Convert 2 000 000 mm2 to m2

Solution:

(a)

0.012 4 kg = 0.012 4× 103 g

= 12.4 g .

(b)

472× 104 mm = 4 720 000 mm

= 4 720 000× 10−3 m

= 4 720 m

= 4 720× 10−3 km

= 4.72 km.

c© USQ, August 27, 2010 17


(c)

2 472 kPa + 1.34 MPa = 2.472 MPa + 1.34 MPa

= 3.812 MPa.

(d)

12 000× 120 µs = 1 440 000 µs

= 1 440 ms

= 1.44 s.

(e) 1 784 cm = 17.84 m. (Note that for ‘centi’ the decimal moves only 2 places.)

(f) If 1 m is equivalent to 1 000 mm , then 1 m2 must be equivalent to 1 000 ×1 000 mm2. So 2 000 000 mm2 must be equivalent to

2 000 0001 000× 1 000

= 2 m2.

Exercise 1.11: Complete the following conversions.

(a) 240 800 g to kg

(b) 16 800 m to km

(c) 1 910 L to kL

(d) 0.678 kg to g

(e) 0.068 m to mm

(f) 0.62 L to mL

(g) 16 425 mg to g

(h) 15× 103 mm to m

(i) 26× 106 mL to kL

(j) 0.000 625 µg to pg

(k) 0.006 nm to pm

(l) 0.024 9 kL to L

(m) 300 000 mm2 to m2

Application: physics In many of the physical sciences all calculations must bedone in one of the fundamental SI units (International Systems of Units). Thereare seven fundamental units defined in this system by international agreement in1971. The fundamental SI units are as below.

18 c© USQ, August 27, 2010


Table 1.3: A list of the fundamental SI units.

Quantity Name SymbolLength metre mMass kilogram kgTime second sTemperature kelvin KElectric current ampere ANumber of particles mole molLuminous intensity candela cd

For example, a car travels 2.5 km in 3.0 minutes. Its average speed, v, would becalculated as follows:

v =2.5 km3.0min ×

1min60 s ×

103 m1 km

=2.5× 103

3× 60ms−1

= 13.9ms−1 .

Uncertainty in measurement

In any scientific measurement or calculation there is a limit to the accuracy of the result.The degree of accuracy is dependent on the reliability of the measuring instrument, theappropriateness of the method used and the skill of the observer. If we perform a calcu-lation involving a number of measurements we have to consider the level of accuracy ofeach measurement and understand how this could affect the accuracy of the final answer.

For example, suppose we were asked to find the volume of a container about the size andshape of a shoe box. The measuring instrument is a metre stick, graduated in millimetresand the measurements we obtain are:

8.2 cm× 15.9 cm× 30.1 cm

These measurements are not exact but could all be plus or minus half a millimetre.

Now, if we find the volume by multiplying the three dimensions together, we get 3924.438cubic centimetres. How accurate is this? Let us consider the two worst cases.

If each measurement we made was 0.5 mm too small, the actual volume of the containerwould be

8.15 cm× 15.85 cm× 30.05 cm ≈ 3 881.78 cubic centimetres

If each measurement we made was 0.5 mm too large, the actual volume of the containerwould be

8.25 cm× 15.95 cm× 30.15 cm ≈ 3 967.36 cubic centimetres

c© USQ, August 27, 2010 19


It is clear that because of the errors associated with the level of accuracy of the measuringinstrument, only the first two figures in our calculation have any significance. This exam-ple illustrates a general rule as a guide to determining significant figures in a calculation.

When multiplying or dividing several quantities, the number of significant figuresin the final result is the same as the number of significant figures in the leastaccurate of the quantities being multiplied.

When adding or subtracting several quantities, the number of significant figuresin the final result should equal the smallest number of decimal places of anyterm involved.

In doing any scientific calculation you must keep in mind the rules about significantfigures. It is no use giving an answer to five decimal places when it is beyond the accuracyof the measuring instrument.

Points to remember

• In scientific notation a quantity is expressed as a number between1 and 10 (called the mantissa), multiplied by a power of 10. (e.g.2.35× 103).

• Become familiar with the Units and Prefixes given in Ta-bles 1.1 and 1.2.

1.1.3 Number calculations

As an Scientist etc., you will continually need to perform calculations as a matter ofcourse. There are some small traps for the unwary in using a calculator to perform eventhe simplest of calculations. Firstly you need to be aware of the conventions that applyin writing arithmetic and algebraic expressions.

Order of operations

When performing calculations there are conventions regulating the priorities of the op-erations. These conventions can be remembered by the mnemonic

BOMDAS − BO MD AS.

These letters should be considered in pairs, as within each pair the order of operations isnot significant. Let’s take them one pair at a time.

BO: Operations inside brackets (B) should be performed first. The ‘O’ stands for ‘over’which is considered equivalent to a bracket.

20 c© USQ, August 27, 2010


Example 1.12:

6× (2 + 3) is 6× (5) = 30 Operations inside brackets are completed first.That is we do not multiply 6 by 2 and then add3.

12− 76

is56≈ 1.166 6 Over is considered equivalent to a bracket and

therefore completed first. We do not divide 12by 6 and then subtract 7.

MD: stands for Multiply and Divide. These operations should be actioned next. It doesnot matter which of them is first.

Example 1.13:

3 + 6× 2 is 3 + 12 = 15 The multiplication of 6 by 2 is completed beforeadding 3. We do not add 3 to 6 and then multi-ply by 2.

7− 126

is 7− 2 = 5 12 is divided by 6 before it is subtracted from 7.We do not subtract 12 from 7 and then divideby 6.

AS: Addition and Subtraction are the last operations completed. Again it does notmatter which order they are performed.

Example 1.14:

2 + 3− 1 = 5− 1 = 4 The addition 2 + 3 is performed first and thenthe subtraction, −1.

2 + 3− 1 = 2 + 2 = 4 The subtraction 3 − 1 is performed first andthen the result, 2, added to 2.

Now consider a more complicated example:

3(1 + 22) + 45

− 6 .

The fraction line, i.e. OVER, means that the calculation above the line should be per-formed before dividing by 5 or subtracting 6.

The calculation above the line includes a bracket, so the contents of this should be com-pleted before multiplying by 3 or adding 4.

Within the bracket, the power of 2 is in fact a multiplication (2 multiplied by itself) andshould therefore be done first. Let’s see how that all works out.

c© USQ, August 27, 2010 21


3(1 + 22) + 45

− 6

=3(1 + 4) + 4

5− 6 evaluate 22;

=3× 5 + 4

5− 6 add 1 and 4 (note that everything in the brackets is now complete);

=15 + 4

5− 6 multiply 3 and 5;

=195− 6 add 15 and 4 (note that everything over the line is now complete);

= 3.8− 6 divide 19 by 5;

= −2.2 subtract 6 from 3.8.

Calculators

Make sure you are familiar with the conventions that your calculator uses. Unfortunatelythere is some variation between makes as to how it interprets your input. You shouldtherefore try some simple calculations to see if you get the answers you expect.

Rounding

Rounding is the process of reducing the number of significant digits in a number. Forexample, 25.3 could be rounded to the nearest whole number, which would be 25. If thenumber 25.6 were rounded to the nearest whole number the result would have been 26.The rule used is that if the digit immediately to the right of the place that we are roundingto is 0 to 4 inclusive, the figure in that place remains the same; if the digit immediately tothe right of the place that we are rounding to is 5 to 9 inclusive, the figure in that place isincreased by one.

22 c© USQ, August 27, 2010


Original number Round to: Rounded number Reason

465.49 Whole number 465 The digit immediately tothe right of the wholenumber place is the 4 inthe tenths place. Since thisis < 5, the whole numberdigit remains unchanged.

465.65 First decimal place 465.7 The digit immediately tothe right of the first dec-imal place is the 5 in thehundredths place. Sincethis is 5 or above, the fig-ure in the first decimalplace, 6, is increased by 1to 7.

465.49 Nearest hundred 500 The hundreds place isoriginally 4. Immediatelyto the right is the 6 inthe tens place. Becausethis is 5 or above, the 4 isincreased by 1.

465.497 3 Second decimal place 465.50 The second decimal placeis held by the 9. Immedi-ately to the right is the 7in the thousandths place.Since this is 5 or above,the 9 is increased by 1.But since increasing 9 by 1is 10, we need to changethe .49 to .50. It is veryimportant to retain the 0(i.e. not 465.5) as this indi-cates the number is signifi-cant to the second decimalplace.

Warning: The process of rounding is always replacing the original number with an ap-proximation and is therefore introducing a ‘small’ error. For this reason when perform-ing a calculation, you should avoid rounding until you have reached the final answer.By rounding early in the calculation, the error that you introduce can be substantiallymagnified by subsequent operations leading to an inaccurate result.

c© USQ, August 27, 2010 23


1.1.4 Geometry

The following geometric results are given without proof. You are expected to be alreadyfamiliar with these. If not you should follow up with your lecturer to find some refer-ences. You will not be assessed directly on this material, however any material in thisbook or questions in assessments may assume knowledge of these.

Table 1.4: Angles and straight lines.

Angles on a straightline sum to 180.

Vertically opposite an-gles are equal.

Table 1.5: Properties of parallel lines.

Alternative angles are equal.

Corresponding angles areequal.

Co-interior angles sum to180.

24 c© USQ, August 27, 2010


Table 1.6: Properties of triangles.

Type Example Properties

All Triangles

Three sides, three angles.Angles sum to 180.Can be acute angled (all angles < 90) or ob-tuse angled (one angle > 90) or right angled(one angle equal 90).

Area = =12× base × perpendicular height.

Exterior angle = sum of opposite angles, i.e.θ = α + β.

Right angled triangle

One angle equals 90.Other two angles sum to 90.Pythagoras theorem applies: a2 + b2 = c2,where a, b and c are the lengths of the sides,and c is the side opposite the right angle calledthe hypotenuse.

Equilateral triangle

All sides are equal.All angles are equal (60).Line bisecting each angle bisects opposite sideat right angles.

Isosceles triangle

Two sides equal.Two angles opposite the equal sides are alsoequal.Line bisecting angle between equal sides bi-sects opposite side at a right angle.

c© USQ, August 27, 2010 25

MAT1500 1.2. Algebraic notation

Table 1.7: Properties of Quadrilaterals.

Type Example Properties

All quadrilaterals

Four sides, four angles.Angles sum to 360.Diagonals are the lines joining oppositeangles.

Trapezium (calledTrapezoid in USA)

One pair of sides parallel.Can be isosceles where non parallel sidesare equal in which case angles at each endof the parallel sides are equal.

Kite

Two pairs of adjacent sides are equal.Two angles between unequal sides areequal.Diagonals intersect at right angles.One diagonal bisects the other.

Parallelogram

Opposite sides parallel.Opposite sides equal.Opposite angles equal.Adjacent angles add to180.Diagonals bisect each other.Area = base× perpendicular height .

Rhombus

Opposite sides parallel.All sides equal.Opposite angles equal.Diagonals bisect each other at right angle.

Rectangle

Opposite sides parallel.Opposite sides equal.All angles = 90.Area = a× b where a and b are length andwidth.Diagonals bisect each other.

Square

Opposite sides parallel.All sides equal.All angles = 90.Area = a2 where a is the length of eachside.Diagonals bisect each other at right an-gles.

1.2 Algebraic notation

In algebra actual numbers or quantities are replaced by letters or pronumerals. Thesegeneralized quantities may assume any number of a set of values. This enables us to exam-ine the general properties of numbers without referring to specific examples.

Other good things about algebra:

• Algebraic solutions are usually more compact than a set of numerical solutions.

26 c© USQ, August 27, 2010

1.2. Algebraic notation MAT1500

• Algebraic solutions give more direct information about the relationship betweenthe variables than a host of figures.

Just like any other language or tool, algebra comes with a set of operating instructions orconventions. We can use these to simplify expressions and later, to solve equations andinequations. Before we go any further, however, we need to recall two things.

1. Names for parts of an expression

Consider the expression:−2x + 3p2 − 4 .

(Note that this could be written as −2x + 3p2 +−4.)

• This expression has three terms, −2x, 3p2, −4. Terms are separated by an ad-dition sign.

• The expression has two variables, x and p. (Note p2 is not the variable as this isreally p× p.)

• −2 and 3 are called coefficients, as they are associated with the variable by mul-tiplication. −2 is the coefficient of x and 3 is the coefficient of p2.

• 4 is called ‘the constant’, as it is a constant term that is not associated with avariable.

2. Arithmetic properties that also apply to algebra

Arithmetic properties Algebraic properties Name2 + 3 = 3 + 2 a + b = b + a Commutative law

for addition

(2 + 3) + 1 = 2 + (3 + 1) (a + b) + c = a + (b + c) Associative lawfor addition

2× 3 = 3× 2 a× b = b× a Commutative lawor ab = ba for multiplication

(2× 3)× 1 = 2× (3× 1) (a× b)× c = a× (b× c) Associative lawfor multiplication

2 + 0 = 0 + 2 = 2 a + 0 = 0 + a = a Identity lawfor addition

2× 1 = 1× 2 = 2 a× 1 = 1× a = a Identity lawfor multiplication

2× (3 + 1) = 2× 3 + 2× 1 a× (b + c) = a× b + a× c Distributive lawor 2(3 + 1) = 2× 3 + 2× 1 or a(b + c) = ab + ac (see below)

2× 0 = 0× 2 = 0 a× 0 = 0× a = 0 Multiplying byzero

c© USQ, August 27, 2010 27

MAT1500 1.3. Expansion

Hint Your mistakes help you to learn.

Much of our learning comes from our mistakes. The more mistakes we make, themore we learn. Use this to your advantage and when you make a mistake:

• do not underestimate yourself;

• be positive, say ‘I can’ and ‘I will’; or

• ask for help from your fellow students, work colleagues and, of course, yourtutor.

Let us now look at some of the techniques of algebra in practice.

1.3 Expansion

As we saw in section 1.1.3 the use of brackets (parentheses) in mathematics signifies thatthe contents of the bracket are to be considered one number. For example, 5× (3 + 4)indicates that it is the 3 + 4 that is multiplied by 5, not just the 3, as would be the casewere the brackets omitted (that is, 5× 3 + 4). You can show very easily that ‘5 lots of3 + 4’, is equivalent to ‘5 lots of 3 plus 5 lots of 4’. Writing this in mathematical notation,

5× (3 + 4) = 5× 3 + 5× 4 .

This is called the distributive law. Replacing the numbers with pronumerals (lettersstanding for any number) since this law applies to any three numbers we might choose,we write,

a(x + b) = ax + ab ,

or, since the order of multiplication does not matter (from the commutative law for mul-tiplication)

(x + b)a = xa + ba = ax + ab .

This process of multiplying by bracketed expressions is called expansion.

We can extend this to expand expressions such as (x + a)(x + b).

(x + a)(x + b) = (x + a)x + (x + a)b

= x× x + ax + xb + ab

= x2 + (a + b)x + ab .

For example,

(x + 3)(x + 4) = x2 + (3 + 4)x + 3× 4 = x2 + 7x + 12 ,

(x + 3)(x− 4) = x2 + (3 +−4)x + 3×−4 = x2 − x− 12 .

28 c© USQ, August 27, 2010

1.3. Expansion MAT1500

Some important examples worth remembering are,

(x + a)2 = (x + a)(x + a) = x2 + (a + a)x + aa = x2 + 2ax + a2 ,

and(x + a)(x− a) = x2 + (a− a)x + a×−a = x2 − a2 .

The process of expanding products of three brackets extends naturally from here, alwaysdo it in groups of two expressions.

(x + a)(x + b)(x + c) = [x2 + (a + b)x + ab](x + c)

= [x2 + (a + b)x + ab]x + [x2 + (a + b)x + ab]c

= x3 + (a + b)x2 + abx + cx2 + c(a + b)x + abc

= x3 + (a + b + c)x2 + (ab + bc + ca)x + abc .

For example,

(x + 1)(x− 2)(x + 3) = [x2 + (1− 2)x− 2](x + 3)

= [x2 − x− 2]x + [x2 − x− 2]3

= x3 − x2 − 2x + 3x2 − 3x− 6

= x3 + 2x2 − 5x− 6 .

A special case is

(x + a)3 = (x + a)(x + a)(x + a)

= x3 + (a + a + a)x2 + (a2 + a2 + a2)x + a3

= x3 + 3ax2 + 3a2x + a3 .

Take special note at this stage of what happens to the value of some of these expressionsfor particular values of x.

Look at the expansion (x + 3)(x − 4) = x2 − x − 12. When x takes the value −3, thebracket (x + 3) obviously becomes 0, so the product (x + 3)(x − 4) becomes (−3 +

3)(−3− 4) = 0×−7 = 0 also.

It may seem obvious, but just check this out with the expanded form:

x2 − x− 12 = (−3)2 − (−3)− 12

= 9 + 3− 12

= 0 .

This should also work for the other bracket in the original product. That is when x = 4,

c© USQ, August 27, 2010 29


the expanded form should also be 0. Let’s check it out

x2 − x− 12 = (4)2 − (4)− 12

= 16− 4− 12

= 0 .

What about for the example with 3 brackets in the product?

(x + 1)(x− 2)(x + 3) = x3 + 2x2 − 5x− 6 .

From our reasoning above the values x = −1, x = 2, and x = −3, should all make theexpression equal to 0. Let’s check this out.

When x = −1, then

x3 + 2x2 − 5x− 6 = (−1)3 + 2(−1)2 − 5(−1)− 6

= −1 + 2 + 5− 6

= 0 .

When x = 2, then

x3 + 2x2 − 5x− 6 = (2)3 + 2(2)2 − 5(2)− 6

= 8 + 8− 10− 6

= 0 .

When x = −3, then

x3 + 2x2 − 5x− 6 = (−3)3 + 2(−3)2 − 5(−3)− 6

= −27 + 18 + 15− 6

= 0 .

Note Two common errors you must avoid:

1. (x + a)2 is not x2 + a2.For example,

(1 + 2)2 = 32 = 9 .

12 + 22 = 1 + 4 = 5 .

Therefore (1 + 2)2 6= 12 + 22 .

2. (x + a)3 is not x3 + a3.

30 c© USQ, August 27, 2010


For example,

(1 + 2)3 = 33 = 27 .

13 + 23 = 1 + 8 = 9 .

Therefore (1 + 2)3 6= 13 + 23 .

Example 1.15: Expand the following expressions. Check your answer if possible by sub-stituting appropriate values.

(a) (x + 3)(x− 2)

(b) (2x− 1)(x + 2)

(c) (3a− 2b)(4a + 3b)

(d) (x− 2)(x2 − 3x + 4)

(e) (2x− 3y + z)(4x + 2y− z)

Solution:

(a)

(x + 3)(x− 2) = x(x− 2) + 3(x− 2)

= x2 − 2x + 3x− 6

= x2 + x− 6 .

Check: Try the values x = −3 and x = 2. When x = −3,

x2 + x− 6 = (−3)2 + (−3)− 6

= 9− 3− 6

= 0 ,

and when x = 2,

x2 + x− 6 = (2)2 + (2)− 6

= 4 + 2− 6

= 0 .

(b)

(2x− 1)(x + 2) = 2x(x + 2)− 1(x + 2)

= 2x2 + 4x− 1x− 1× 2

= 2x2 + 3x− 2 .

c© USQ, August 27, 2010 31


Check: Try the values12

, and x = −2. When x =12

,

2x2 + 3x− 2 = 2(

12

)2

+ 3(

12

)− 2

=12+

32− 2

= 0 ,

and when x = −2,

2x2 + 3x− 2 = 2(−2)2 + 3(−2)− 2

= 8− 6− 2

= 0 .

(c)

(3a− 2b)(4a + 3b) = 3a(4a + 3b)− 2b(4a + 3b)

= 12a2 + 9ab− 8ab− 6b2

= 12a2 + ab− 6b2 .

(d)

(x− 2)(x2 − 3x + 4) = x(x2 − 3x + 4)− 2(x2 − 3x + 4)

= x3 − 3x2 + 4x− 2x2 + 6x− 8

= x3 − 5x2 + 10x− 8

(e)

(2x− 3y + z)(4x + 2y− z)

= 2x(4x + 2y− z)− 3y(4x + 2y− z) + z(4x + 2y− z)

= 8x2 + 4xy− 2xz− 12xy− 6y2 + 3yz + 4xz + 2yz− z2

= 8x2 − 6y2 − z2 − 8xy + 2xz + 5yz .

32 c© USQ, August 27, 2010


Example 1.16: Expand (3y− 1)(2y + 2)(3− y).Solution:

(3y− 1)(2y + 2)(3− y) = (3y− 1)[2y(3− y) + 2(3− y)]

= (3y− 1)(6y− 2y2 + 6− 2y)

= (3y− 1)(6 + 4y− 2y2)

= 3y(6 + 4y− 2y2)− 1(6 + 4y− 2y2)

= 18y + 12y2 − 6y3 − 6− 4y + 2y2

= −6y3 + 14y2 + 14y− 6 .

Example 1.17: Expand the following expressions.

(a) (x2 + y2)(x− y)

(b) (2x− 1)(3x + 2)(x− 3)

Solution:

(a)

(x2 + y2)(x− y) = x2(x− y) + y2(x− y)

= x3 − x2y + xy2 − y3 .

(b)

(2x− 1)(3x + 2)(x− 3) = (2x− 1)[3x(x− 3) + 2(x− 3)]

= (2x− 1)(3x2 − 9x + 2x− 6)

= (2x− 1)(3x2 − 7x− 6)

= 2x(3x2 − 7x− 6)− 1(3x2 − 7x− 6)

= 6x3 − 14x2 − 12x− 3x2 + 7x + 6

= 6x3 − 17x2 − 5x + 6 .

Exercise 1.18: Expand the following

(a) (x + 2)(x + 1)

(b) (x− 2)(x + 1)

(c) (x− 2)(x− 1)

(d) (2x + 3)(x + 4)

(e) (3x− 1)(2x + 4)

(f) (2x− 5)(3x− 2)

(g) (x− 1)(x + 1)(x + 2)

(h) (2x− 1)(3x + 2)(4x + 1)

(i) (x2 − 3)(x + 2)

c© USQ, August 27, 2010 33


(j) (x + y)(x− y)

(k) (x + y)(x + y)

(l) (x2 + y2)(x− y)(2x + y)

Application: genetics Chromosomes are made up of genes. Genes in turn arelocated at a particular location or locus on the chromosome. Genes which occupythe same locus on the chromosome are called alleles. The genotype of an individualis its genetic constitution and contains the inherited characteristics determined bygenes. Assume that one locus has three alleles, A, B and C, and these three alleleshave frequencies of p, q and r, respectively in a population. If the populationis at Hardy-Weinberg equilibrium, what are the algebraic frequencies of the sixgenotypes in the population? The possible six genotypes are

A B CA AA BA CAB AB BB CBC AC BC CC

The Hardy-Weinberg equilibrium equation is given by

(p + q + r)2 = 1 ,

(p + q + r)(p + q + r) = 1 ,

p(p + q + r) + q(p + q + r) + r(p + q + r) = 1 ,

p2 + pq + pr + pq + q2 + qr + pr + qr + r2 = 1 ,

p2 + q2 + r2 + 2pq + 2pr + 2qr = 1 .

Since p is the frequency of A, q the frequency of B and r the frequency of C, thenthe algebraic frequencies of the six genotypes are:

AA = p2 ,

BB = q2 ,

CC = r2 ,

AB = 2pq ,

AC = 2pr ,

BC = 2qr .

Points to remember

• (x + a)(x + b) = x2 + (a + b)x + ab.

• (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2.

• (x + a)2 is not the same as x2 + a2.

34 c© USQ, August 27, 2010

1.4. Factorisation MAT1500

1.4 Factorisation

The reverse process of expansion is called factorisation.

Common factors

We can see that if two terms of an expression ax + ab contain the same factor (in this casea) it could have resulted from the expansion of a(x + b). In this case, a is called a commonfactor of ax and ab. In the same way we could say that 3 is a common factor of 18 and 33,since we could write 18 = 3× 6 , and 33 = 3× 11 .

For example, to factorise the expression 3x + 6 we note that 3 is a common factor of bothparts. We can then say 3x + 6 = 3(x + 2) .

Factorising quadratic expressions

Expansion and factorisation are actually reverse procedures. So let us examine some ofthe expansions to see how they can help us to factorise.

(p + 1)(p + 3) = p(p + 3) + 1(p + 3)

= p2 + 3p + 1p + 1× 3

= p2 + (3 + 1)p + 3× 1 .

6 6

3 + 1 = 4 3× 1 = 3

We have two numbers (3 and 1), which add to give 4 and multiply to give 3 .

So if we must factorise p2 + 4p + 3 , we try to find two numbers that add to give 4 andmultiply to give 3 .

Numbers to try, factors of 3 Sum Product

−3 and −1 −3 +−1 = −4 −3×−1 = 33 and 1 3 + 1 = 4 3× 1 = 3

√

The correct factors are 3 and 1 . So,

p2 + 4p + 3 = p2 + (3 + 1)p + 3× 1= p2 + 3p + p + 3 Group terms in pairs.

= p(p + 3) + (p + 3) Take out p as common factor from firsttwo terms.

= p(p + 3) + 1(p + 3) Take out 1 as common factor of next twoterms.

= (p + 3)(p + 1) . Take out (p + 3) as common factor.

c© USQ, August 27, 2010 35

MAT1500 1.4. Factorisation

Example 1.19: Factorise x2 + 7x− 30 .Solution:We need two numbers that multiply to give −30 and add to give 7 . Factors of −30are 3 and −10 , or −3 and 10 , or 2 and −15 or −2 and 15 plus others.

Numbers to try, Sum Productfactors of −30

3 and −10 3 +−10 = −7 3×−10 = −30−3 and 10 −3 + 10 = 7 −3× 10 = −30

√

2 and −15 2 +−15 = −13 2×−15 = −30−2 and 15 −2 + 15 = 13 −2× 15 = −30

The correct factors are −3 and 10, so

x2 + 7x− 30= x2 + (10− 3)x− 3× 10 Use factors to rewrite 7 and −30 .

= x2 + 10x− 3x +−3× 10 Remove brackets.

= x(x + 10)− 3x +−3× 10 Take out x as a common factor from thefirst two terms.

= x(x + 10)− 3(x + 10) Take out −3 from the next two terms. Inboth cases the remaining factor is thesame, x + 10 .

= (x + 10)(x− 3) . Take out x + 10 as a common factor.

Check your answer by expanding:

(x + 10)(x− 3) = x(x− 3) + 10(x− 3)

= x2 − 3x + 10x− 30

= x2 + 7x− 30 .

Therefore the final answer is x2 + 7x− 30 = (x + 10)(x− 3) .

As you become more proficient at factorisation you will not need to write out the table infull and will be able to do the guessing and checking in your head.

Hint When the going gets tough. . .

• re-read what you do know;

• move forward in the study material to see if it comes together;

• talk it out with somebody, like your study partner;

• put your questions to the discussion group;

• contact your tutor. There is no such thing as a stupid question.

. . . the tough get studying.

36 c© USQ, August 27, 2010


Now that you can factorise expressions in which the coefficient of the x2 term is one, weuse similar methods to factorise expressions where the coefficient is greater than 1.

Consider what you will have to do to factorise 6x2 + 7x + 2. In this instance we cannotfind the factors of the constant alone, but need to find the factors of the product of thecoefficient of x2 and the constant. That is, what are the factors of 6× 2 or 12?

Factors of 12 are 6 and 2, or −6 and −2, or 3 and 4 or −3 and −4 and others.

Following the pattern of before and the knowledge we have gained by expanding usingthe distributive law, we need to now look for two numbers which multiply to give 12 andadd to give 7.

Numbers to try, Sum Productfactors of 12

6 and 2 6 + 2 = 8 6× 2 = 12−6 and −2 −6 +−2 = −8 −6×−2 = 123 and 4 3 + 4 = 7 3× 4 = 12

√

−3 and −4 −3 +−4 = −7 −3×−4 = 12

Correct factors are 3 and 4.

6x2 + 7x + 2= 6x2 + (3 + 4)x + 2 Use factors to rewrite 7.

= 6x2 + 3x + 4x + 2 Remove brackets.

= 3x(2x + 1) + 2(2x + 1) Take out 3x as a common factor from thefirst two terms and 2 from the next twoterms. In each case the remaining factorshould be the same, 2x + 1.

= (2x + 1)(3x + 2) . Take out 2x + 1 as a common factor.


(2x + 1)(3x + 2) = 2x(3x + 2) + 1(3x + 2)

= 6x2 + 4x + 3x + 2

= 6x2 + 7x + 2 .

Therefore the final answer is 6x2 + 7x + 2 = (2x + 1)(3x + 2) .

Example 1.20: Factorise 10x2 + 17x + 3 .Solution:In this instance, again, we cannot find the factors of the constant alone but need tofind the factors of the product of the coefficient of x2 and the constant. That is, whatare the factors of 3× 10 or 30?

Factors of 30 are 3 and 10, or −3 and −10, or 15 and 2 or −15 and −2 and others.

Following the pattern of before and the knowledge we have gained by expanding

c© USQ, August 27, 2010 37


using the distributive law, we need to now look for two numbers which multiplyto give 30 and add to give 17.


3 and 10 3 + 10 = 13 3× 10 = 30−3 and −10 −3 +−10 = −13 −3×−10 = 3015 and 2 15 + 2 = 17 15× 2 = 30

√

−15 and −2 −15 +−2 = −17 −15×−2 = 30

The correct factors are 15 and 2.

10x2 + 17x + 3 = 10x2 + (15 + 2)x + 3 Use factors to rewrite 17.

= 10x2 + 15x + 2x + 3 Remove brackets.

= 5x(2x + 3) + 1(2x + 3) Take out 5x as a commonfactor from the first twoterms and 1 from the nexttwo terms. In each case theremaining factor should bethe same, 2x + 3.

= (2x + 3)(5x + 1) Take out 2x + 3 as acommon factor.


(2x + 3)(5x + 1) = 2x(5x + 1) + 3(5x + 1)

= 10x2 + 2x + 15x + 3

= 10x2 + 17x + 3 .

Therefore the final answer is 10x2 + 17x + 3 = (2x + 3)(5x + 1) .

Example 1.21: Factorise 3x2 − 10x + 3.Solution:We must find the factors of the product of the coefficient of x2 and the constant.That is, what are the factors of 3× 3 or 9.

Factors of 9 are 3 and 3, or −3 and −3, or 9 and 1, or −9 and −1.

Following the pattern of before and the knowledge we have gained by expandingusing the distributive law we need to now look for two numbers which multiply togive 9 and add to give −10.


3 and 3 3 + 3 = 6 3× 3 = 9−3 and −3 −3 +−3 = −6 −3×−3 = 99 and 1 9 + 1 = 10 9× 1 = 9−9 and −1 −9 +−1 = −10 −9×−1 = 9

√

38 c© USQ, August 27, 2010


Correct factors are −9 and −1.

3x2 − 10x + 3= 3x2 + (−9− 1)x + 3 Use factors to rewrite −10.

= 3x2 − 9x− x + 3 Remove brackets.

= 3x(x− 3)− 1(x− 3) Take out 3x as a common factor fromthe first two terms and −1 from the nexttwo terms. In each case the remainingfactor should be the same, x− 3.

= (x− 3)(3x− 1) . Take out x− 3 as a common factor.


(x− 3)(3x− 1) = x(3x− 1)− 3(3x− 1)

= 3x2 − x− 9x + 3

= 3x2 − 10x + 3 .

Therefore the final answer is 3x2 − 10x + 3 = (x− 3)(3x− 1) .

In practice you may find this is all you need to do.

Look at the expansion of expressions such as (x + a)(x + b).

Recall that(x + a)(x + b) = x2 + (a + b)x + ab .

Comparing this to x2 + 5x− 14, we see that we need to find a and b so that ab = −14 anda + b = 5 . By inspection, and a little trial and error, we find that the numbers −2 and 7satisfy this requirement. We can now see that the expression x2 + 5x− 14 results from anexpansion of (x− 2)(x + 7). That is, x2 + 5x− 14 can be factorised to (x− 2)(x + 7).

In practice, you should always be looking out for expressions which are perfect squares,and difference of squares. Referring back to section 1.2, you will recall that

x2 + 2ax + a2 = (x + a)2 .

Such an expression is called a perfect square. It is easily factorised as above. For example,the following expressions are perfect squares, with their factors shown below them:

c© USQ, August 27, 2010 39


x2 + 6x + 9 = x2 + 2× 3× x + 32

= (x + 3)2 .

x2 − 4x + 4 = x2 − 2× 2× x + 22

= (x− 2)2 .

16a2 + 24ab + 9b2 = (4a)2 + 2× 4a× 3b + (3b)2

= (4a + 3b)2 .

W2 − 2WL + L2 = W2 − 2× L×W + L2

= (W − L)2 .

Also notice from section 1.2 that

x2 − a2 = (x− a)(x + a) .

Such an expression is called a difference of squares and is also easily factorised.

x2 − 25 = x2 − 52

= (x− 5)(x + 5) .

a2 − 4b2 = a2 − (2b)2

= (a− 2b)(a + 2b) .

4y2 − 9z2 = (2y)2 − (3z)2

= (2y− 3z)(2y + 3z) .

Example 1.22: Factorise the following expressions.

(a) 2ab− 10a + 3b− 15

(b) y2 − 5y− 36

(c) 6x2 + 17x− 3

(d) 4x2 − 12x + 9

(e) 25x2 − 16

Solution:

(a) 2ab− 10a + 3b− 15 = 2a(b− 5) + 3(b− 5) = (b− 5)(2a + 3)

(b) We need two numbers whose product is −36 (so they will be of different sign)and whose sum is−5 (so the negative number will have the larger magnitude).

40 c© USQ, August 27, 2010


These numbers are −9 and 4. Thus

y2 − 5y− 36 = (y− 9)(y + 4) .

(c) Multiplying the coefficient of x2 and the constant term, 6×−3 = −18 . Factorsof −18 are 6 and −3; −6 and 3; 18 and −1; −18 and 1; 2 and −9; or −2 and 9.We need to now look for two numbers which multiply to give −18 and add togive 17.


6 and −3 6 +−3 = 3 6×−3 = −18−6 and 3 −6 + 3 = 3 −6× 3 = 1818 and −1 18 +−1 = 17 18×−1 = −18

√

−18 and 1 −18 + 1 = −17 −18× 1 = −18

Correct factors are 18 and −1.

6x2 + 17x− 3= 6x2 + (18 +−1)x− 3 Use factors to rewrite 17.

= 6x2 + 18x− x− 3= 6x(x + 3) +−1(x + 3) Take out 6x as a common factor from

the first two terms and −1 from thenext two terms. In each case theremaining factor should be the same,x + 3.

= (6x +−1)(x + 3) Take out x + 3 as a common factor.

= (6x− 1)(x + 3) .

(d) Recognise the form of a perfect square a2− 2ab + b2 where a = 2x , and b = 3 .

4x2 − 12x + 9 = (2x)2 − 2× 2x× 3 + 32

= (2x− 3)2 .

(e) Recognise the form of difference of squares a2 − b2 where a = 5x , and b = 4 .

25x2 − 16 = (5x)2 − 42

= (5x− 4)(5x + 4) .

If we choose a value for the pronumeral that makes either of the factors equal to 0, thatvalue will also make the original quadratic expression equal to zero. For example, inExample 1.19, we found that the factors of x2 + 7x − 30 are (x + 10) and (x − 3). Thevalue x = −10 would make the first factor 0, and the value x = 3 would make thesecond factor 0. Both of these values make the quadratic expression x2 + 7x− 30 = 0.

c© USQ, August 27, 2010 41

MAT1500 1.5. Factorising Cubics

Let’s verify: when x = −10, then

x2 + 7x− 30 = (−10)2 + 7(−10)− 30

= 100− 70− 30

= 0 ,

and when x = 3, then

x2 + 7x− 30 = (3)2 + 7(3)− 30

= 9 + 21− 30

= 0 .

Exercise 1.23: Factorise the following:

(a) x2 − x− 6

(b) x2 + 5x + 6

(c) x2 − 5x + 6

(d) x2 + x− 6

(e) 2x2 − 3x− 14

(f) 6x2 − 8x− 8

(g) a2b + ab2 − abc

(h) x2 − 13x− 48

(i) 4x2 − 8x + 3

(j) 8x2 + 22x− 6

(k) 4− 9x2

(l) 15x2 − 11x + 2

1.5 Factorising Cubics

A cubic expression in x is a polynomial of the form ax3 + bx2 + cx + d where a, b, c,and d are constants. For example 2x3 + 3x2 − x − 5. Cubic expressions result from theexpansion of product of three bracketed terms of the form (x + a)(x + b)(x + c) as insection Expansion. For example

(x + 1)(x− 2)(x + 3) = x3 + 2x2 − 5x− 6 .

In factorising cubics we want to reverse this process. The first step in finding one ofthe factors has to be one of ‘educated guess and check’. If you look back to Section 1.5Expansion, you will see that

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc .

42 c© USQ, August 27, 2010

1.5. Factorising Cubics MAT1500

See that the constant term abc is the product of the constant terms a, b, and c in thebrackets. So we can make a guess as to what at least one of those constants might be.Now to check. Remember in the previous section we noted that if (x + a) was one of thefactors, then x = −a would make the expanded form zero.

Let’s try this out. Suppose we want to find a factor of the cubic expression x3 + 10x2 +

31x + 30. Possible factors of 30 are 1, −1, 2, −2, 3, −3, 5, −5. We try substituting thesevalues into the expression until we find one that makes the expression zero.

x = 1 (1)3 + 10(1)2 + 31(1) + 30 = 72 .

x = −1 (−1)3 + 10(−1)2 + 31(−1) + 30 = 8 .

x = 2 23 + 10× 22 + 31× 2 + 30 = 140 .

x = −2 (−2)3 + 10(−2)2 + 31(−2) + 30 = 0 .√

So by trial and error we find that x = −2 makes the expression zero and therefore we candeduce that (x + 2) is a factor.

How do we now proceed to find the other factors? We can use a process very similar tothe long division process you would have learned in primary school. Start by setting upthe division process like this

x + 2)

x3 +10x2 +31x +30

where the expression to be divided is under the bracket, and the expression we are di-viding by is outside at the front. We then ask ourselves the question, what would (x + 2)need to be multiplied by to give the term with the highest power under the bracket (thatis the x3)? The answer of course is x2 which we put above the bracket directly over the10x2.

x2

x + 2)

x3 +10x2 +31x +30

We then multiply (x + 2) by x2 and put the answer directly under corresponding powersand subtract and bring down the next term.

x2

x + 2)

x3 +10x2 +31x +30

x3 +2x2

8x2 +31x

The process is now repeated. What do we need to multiply (x + 2) by to get 8x2? Clearly

c© USQ, August 27, 2010 43

MAT1500 1.5. Factorising Cubics

8x.x2 +8x

x + 2)

x3 +10x2 +31x +30

x3 +2x2

8x2 +31x8x2 +16x

15x +30

And repeating one last time:

x2 +8x +15

x + 2)

x3 +10x2 +31x +30

x3 +2x2

8x2 +31x8x2 +16x

15x +3015x +30

0

So the result of dividing x3 + 10x2 + 31x + 30 by (x + 2) is x2 + 8x2 + 15. This quadraticexpression can now be factorised as before (factors of 15 that add to give 8) giving x2 +

8x2 + 15 = (x + 3)(x + 5).

The factors of the cubic are therefore (x + 2)(x + 3)(x + 5).

Example 1.24: Factorise x3 − 6x2 + 5x + 12.

Solution:

Possible factors of 12: ±1, ±2, ±3, ±4, ±6.

Try: x = 1 13 − 6× 12 + 5× 1 + 12 = 12 .

Try: x = −1 (−1)3 − 6(−1)2 + 5(−1) + 12 = −1− 6− 5 + 12 = 0 .√

Therefore one factor must be (x + 1).

To find the other factors we need to divide x3 − 6x2 + 5x + 12 by (x + 1).

x2 −7x +12

x + 1)

x3 −6x2 +5x +12

x3 +x2

−7x2 +5x−7x2 −7x

12x +1212x +12

0

To find factors of the quadratic x2 − 7x + 12, find factors of 12 that sum to −7: −3

44 c© USQ, August 27, 2010

1.6. Algebraic fractions MAT1500

and −4.Therefore the factors of x2 − 7x + 12 are (x− 3)(x− 4) and the factors of the cubicexpression x3 − 6x2 + 5x + 12 are (x + 1)(x− 3)(x− 4).

Exercise 1.25: Factorise:

(a) x3 + 2x2 − 5x− 6

(b) x3 + 3x2 − 4x− 12

(c) x3 − 2x2 − 19x + 20

(d) x3 − 5x2 − 2x + 24

(e) x3 + 4x2 − 17x− 60

(f) x3 + 4x2 − x− 4

(g) x3 + 2x2 − 9x− 18

(h) x3 + 15x2 + 74x + 120.

Points to remember

• In an expression ax + ab, a is called a common factor.

• To factorise ax+ ab, take the common factor a outside the bracketsto give a(x + b).

• To factorise the quadratic expression x2 + bx + c find factors p,q of c that add to give b, replace b with p + q and find commonfactors twice.

• To factorise the quadratic expression ax2 + bx + c find factors p, qof ac that add to give b, and continue as above.

• Look out for perfect squares of the form x2 + 2ax + a2 which havefactors (x + a)2.

• Look out for differences of squares of the form x2− a2 which havefactors (x + a)(x− a).

1.6 Algebraic fractions

Algebraic fractions occur in many formulas you will come across in your university stud-ies. For example:

• In physics, when an object of mass, m, crashes into a stationary pendulum of mass,M, with a velocity of v, the pendulum will have a resultant velocity of

v =

(m

m + M

)v .

c© USQ, August 27, 2010 45

MAT1500 1.6. Algebraic fractions

• In ecology, the probability (p) that a bird originally banded with only one bandwould lose that band over the same period is

p =β1

2β2 + β1,

where β1 is the number of single banded birds and β2 the number of double bandedbirds.

• In chemistry, the pressure (P), volume (V) and temperature (T) of a gas are relatedby the equation

PVT

= constant .

• In electronics, the resistances of some types of circuits are related by the equation

1R

=1

R1+

1R2

.

Do not try to learn any of these formulas. They are just there as examples involvingalgebraic fractions.

So, when algebraic fractions occur in such profusion in business and science based stud-ies, how do we manipulate them?

You will already be familiar with operations involving fractions. Remember in algebra,the pronumerals stand for numbers. So, the same procedures hold for fractions that con-tain pronumerals as for ordinary numbers.

Addition and subtraction of fractions

Consider the process of adding fractions

xa+

yb

.

The first step is to find a way of expressing both as equivalent fractions with the samedenominator (remember the denominator is the bottom line, the numerator is the topline). That is we need to find a common denominator. In this case we can express bothas equivalent fractions with the denominator ab. This gives us

xa+

yb

=xa× b

b+

yb× a

a

=xbab

+yaab

=xb + ya

ab.

46 c© USQ, August 27, 2010


The process of subtracting fractions will be the same so that

xa− y

b=

xb− yaab

.

Multiplication or division of fractions

Again the same rules apply to multiplication and division of algebraic fractions as forordinary fractions. To multiply fractions you multiply the numerators and multiply thedenominators. So

xa× y

b=

xyab

.

To divide one fraction by another, we invert the dividing fraction and multiply.

xayb

=xa÷ y

b=

xa× b

y=

bxay

.

Example 1.26: Subtractx− 1x + 2

fromx

x + 1. Express your answer with no common factors.

Solution:

xx + 1

− x− 1x + 2

=x

(x + 1)× (x + 2)

(x + 2)− (x− 1)

(x + 2)× (x + 1)

(x + 1)Make (x + 1)(x + 2) the commondenominator by appropriatemultiplications of each term. Noteterms are bracketed to keep themtogether.

=x(x + 2)

(x + 1)(x + 2)− (x− 1)(x + 1)

(x + 1)(x + 2)

=x2 + 2x

(x + 1)(x + 2)− x2 − 1

(x + 1)(x + 2)Remove brackets from each numeratorby expansion.

=x2 + 2x− (x2 − 1)(x + 1)(x + 2)

Subtract numerators, put subtractednumerator in brackets so that entirenumerator is subtracted.

=x2 + 2x− x2 + 1(x + 1)(x + 2)

Remove brackets from numerator.

=2x + 1

(x + 1)(x + 2). Simplify numerator.

c© USQ, August 27, 2010 47

MAT1500 1.6. Algebraic fractions

Example 1.27: Simplify the expression below.

1s− 2

− s2(s− 2)

.

Solution:

1s− 2

− s2(s− 2)

=1

s− 2× 2

2− s

2(s− 2)Make 2(s− 2) the common denominator byappropriate multiplication of each term. Note because(s− 2) is common to both terms we only have toinclude it once in the common denominator.

=2

2(s− 2)− s

2(s− 2)Simplify each numerator.

=2− s

2(s− 2)Subtract numerators. Notice that 2− s is nearly thesame as s− 2.

=−1× (−2 + s)

2(s− 2)Take out -1 as a common factor of the numerator.

=−1(s− 2)2(s− 2)

Rewrite −2 + s as s− 2, and cancel s− 2 from thenumerator and denominator.

=−12

= −12

.

Example 1.28: Simplify the following expression.

xx− 3

+5

x + 2.

Solution:

xx− 3

+5

x + 2=

x(x + 2) + 5(x− 3)(x− 3)(x + 2)

=x2 + 2x + 5x− 15x2 + 2x− 3x− 6

=x2 + 7x− 15x2 − x− 6

.

48 c© USQ, August 27, 2010


Example 1.29: Simplify the following expression.

x + 2x− 2

− x− 2x + 2

.

Solution:

x + 2x− 2

− x− 2x + 2

=(x + 2)(x + 2)− (x− 2)(x− 2)

(x− 2)(x + 2)

=(x + 2)2 − (x− 2)2

(x− 2)(x + 2)

=(x2 + 4x + 4)− (x2 − 4x + 4)

x(x + 2)− 2(x + 2)

=x2 + 4x + 4− x2 + 4x− 4

x2 + 2x− 2x− 4

=8x

x2 − 4.

Exercise 1.30: Simplify the following expressions:

(a)x + 1x− 2

+x− 1x + 2

(b)x− 3x− 1

− 2x− 1x + 2

(c)3x− 1x + 2

− 2x + 1x− 3

(d)x

2x + 1+

x− 13x− 1

(e)4

x− 1+

x− 2x + 3

(f)2x− 3x + 2

+x− 1x− 3

Application: conservation biology The intrinsic rate of increase of a pop-ulation of deer is related to stock and yield at each level of harvesting by theequation

rm =KH1

K− N1=

KH2

K− N2,

where K is the population size at steady density, N the population size at twodifferent times, and H is the instantaneous rate of harvesting necessary to holdnumbers constant.

c© USQ, August 27, 2010 49

MAT1500 1.7. Equations

KH1

K− N1=

KH2

K− N2,

KH1

K− N1− KH2

K− N2= 0 ,

KH1

K− N1× K− N2

K− N2− KH2

K− N2× K− N1

K− N1= 0 ,

KH1(K− N2)− KH2(K− N1) = 0 ,

KH1K− KH1N2 − KH2K + KH2N1 = 0 ,

H1K− H1N2 − H2K + H2N1 = 0 ,

H1K− H2K = H1N2 − H2N1 ,

K(H1 − H2) = H1N2 − H2N1 ,

K =H1N2 − H2N1

H1 − H2.

K is now available to use in a range of other equations (Adapted from Caughley1977, p. 183) .

Points to remember

• To add or subtract fractions express both as equivalent fractionswith the same denominator then add or subtract numerators:

xa+

yb

=xbab

+yaba

=xb + ya

ab;

xa− y

b=

xbab− ya

ba=

xb− yaab

.

• To multiply fractions multiply the numerators and multiply thedenominators:

xa× y

b=

xyab

.

• To divide one fraction by another invert the dividing fraction andmultiply:

xayb=

xa÷ y

b=

xa× b

y=

bxay

.

1.7 Equations

Word Expressions

The great usefulness of mathematics lies in its ability to express real world situations inequations, which are then accessible to simple logic and an accumulated toolbox of tech-niques and experience that leads to a solution. To be useful, this solution then needs tobe translated back to the words of the real world situation. Translating between mathe-matical language and the language used in everyday life is an essential skill for anyone

50 c© USQ, August 27, 2010

1.7. Equations MAT1500

wishing to make mathematics useful.

Hint To translate from a real world situation, perhaps expressed in words, or evenjust as a mental concept, there are a number of useful practices.

• wherever possible draw a diagram, no matter how trivial it may seem;

• read the question aloud;

• take each phrase of the problem and convert it into equivalent mathematics;

• make logical/common sense connections between the information writtenabove.

For example, let us examine this problem.

I wish to paint my lounge room. It is 5 metres wide by 10 metres long and3 metres high. How much paint will I need to put two coats on the wallsgiven that the coverage using a roller is about 60 square metres per litre?

1. Draw a diagram.

2. Label the diagram with the information about the dimensions, and define variables.

How much paint will I need. . . . Since this is what we need to calculate, call it some-thing. Let us call it Q.

Q =? litres.

. . . the coverage using a roller is about 60 square metres per litre. . . . This piece of infor-mation also needs a name. Let us call it C for coverage.

C = 60 sq m/L.

3. From the units (square metres per litre) see that the coverage relates the quantity ofpaint to the area to be covered. Each litre of paint corresponds to C square metresof area. That area (call it A) is the coverage (C) times the quantity of paint (Q).

A = CQ sq m .

c© USQ, August 27, 2010 51


Area is also given by width multiplied by height. There are two pairs of walls to bepainted. Opposite walls have the same surface area.

Two walls have area 5× 3, the other two have area 10× 3, and these four walls needto be painted twice, so

A = [2× (5× 3) + 2× (10× 3)]× 2 = 180 sq m .

Solution of equations

To solve equations, you must develop the skill of manipulating algebraic expressions. Inparticular, you need to be able to rearrange an equation to make a particular variable thesubject, that is, by itself on the left hand side of the equation. A number of operations canbe used to rearrange equations. These include:

• interchanging the sides of the equation;

• adding or subtracting the same term from both sides;

• multiplying or dividing every term in the equation by the same quantity;

• changing sign of every term in the equation;

• taking the square roots of or raising to powers each side of the equation;

• taking reciprocals (inverting) of both sides of the equation; and

• taking the same function of both sides.

An equation is a statement that one quantity is the same as another. This statement re-mains true only if what we do to one of the quantities we also do to the other.

Note Whatever operation is performed on one side of an equation must also beperformed on every term of the other.

In the previous example involving painting a lounge room, we had found the equationto be

A = CQ ,

where we seek to determine a value for Q, the amount of paint required. To make Q thesubject, we initially interchange sides.

52 c© USQ, August 27, 2010


CQ = A ,

CQC

=AC

Divide both sides of the equation by C.

Q =AC

Cancel the C’s on the left hand side.

Q =18060

Substitute values for A and C.

= 3 litres.

Example 1.31: If an object is placed at a distance u from a spherical mirror of radius r, itsimage appears at a distance v from the mirror. The formula relating these quantitiesis:

1v+

1u=

2r

.

Make r the subject of the formula.

Solution:

1v+

1u

=2r

,

u + vuv

=2r

, Put LHS over a common denominator.

uvu + v

=r2

, Invert both sides.

Therefore r =2uv

u + v. Multiplying by 2.

Example 1.32: The oscillating time, t, of a pendulum of length L is given by:

t = 2π

√Lg

,

where g is the acceleration due to gravity. Find an expression for L.

c© USQ, August 27, 2010 53


Solution:

t = 2π

√Lg

,

t2π

=

√Lg

, Dividing by 2π.

(t

2π

)2

=Lg

Square both sides. ,

t2

4π2 =Lg

,

L =t2g4π2 . Multiply by g.

Example 1.33: Make p the subject of

k =

√f + pf − p

.

Solution:

k =

√f + pf − p

,

k2 =f + pf − p

, Square both sides.

k2( f − p) = f + p , Multiply by ( f − p).

k2 f − k2 p = f + p ,

−p + k2 f − k2 p = f , Subtract p from both sides.

−p− k2 p = f − k2 f , Subtract k2 f from both sides.

p + k2 p = − f + k2 f Change sign throughout (that is,multiply both sides by −1).

p(1 + k2) = − f + k2 f ,

p =− f + k2 f

1 + k2 . Divide by (1 + k2).

54 c© USQ, August 27, 2010


Example 1.34: Make y the subject of the following formulae.

(a) x =12y− 7

5

(b) x = 3− 7y

(c) x = 6a− 3y2

(d) x = 6a− 3b + 2a√

y + 1

Solution:

(a)

x =12y− 7

5,

5x = 12y− 7 ,

5x + 7 = 12y ,5x + 7

12= y ,

y =5x + 7

12.

(b)

x = 3− 7y

,

x− 3 = −7y

,

3− x =7y

,

13− x

=y7

,

73− x

= y ,

y =7

3− x.

(c)

x = 6a− 3y2 ,

x− 6a = −3y2 ,

6a− x = 3y2 ,6a− x

3= y2 ,

±√

6a− x3

= y ,

y = ±√

6a− x3

.

c© USQ, August 27, 2010 55


Note that both the positive and negative square root must be included.

(d)

x = 6a− 3b + 2a√

y + 1 ,

x− 6a + 3b = 2a√

y + 1 ,x− 6a + 3b

2a=

√y + 1 ,

(x− 6a + 3b

2a

)2

= y + 1 ,

y =

(x− 6a + 3b

2a

)2

− 1 .

Application: mechanical engineering A torque (forces tending to cause arotation) T is applied to a solid tapered shaft AB.

The angle of twist is given by

θ =7TL

12πGc4 .

To find the torque, we would rearrange the formula to make T the subject.

θ =7TL

12πGc4 ,

θ × 12πGc4 = 7TL ,θ × 12πGc4

7L= T ,

T =θ × 12πGc4

7L.

Exercise 1.35: Transform the following equations to make the stated variable the subject.

(a) s = rt ; r

(b) c = 2πr ; r

(c) e = mc2 ; m

56 c© USQ, August 27, 2010


(d) e = mc2 ; c

(e) F =Gm1m2

d; G

(f)1f=

1a+

1b

; f

(g) P =13

Nmv2 ; v

(h) V =1

2rL

√T

πd; T

(i) 3x− 4y− 12 = 0 ; y

(j) y =2x− 33x− 5

; x

1.7.1 Quadratic equations

Previously we looked at describing and factorising expressions which had the variableraised to the power two, that is, quadratic expressions. In reality many of these ex-pressions form part of an equation, the quadratic equation and has the general formax2 + bx + c = 0 . This type of equation occurs very commonly in a range of disciplines.

• In economics, the average cost equation for a particular production plant is C =

0.3Q2 − 3.3Q + 15.3 , where C is the average cost in dollars of producing each unitand Q is the quantity produced.

• In electrical engineering the thermal e.m.f. (E) corresponding to the temperature Tfor a copper-iron thermocouple is modelled by E = −0.0192T2 + 6.88T .

• In applied biology, the relationship C = 30t2 − 240t + 500 , predicts the number ofbacteria (C) in a swimming pool, measured as count per cm3, given the time (t) indays from treatment.

We can solve quadratic equations by a number of methods.

Factorisation

In general, a quadratic equation is of the form ax2 + bx + c = 0 , where a, b, and c areconstants. If the expression ax2 + bx + c can be factorised as (x + d)(x + e) , the solutionof the equation can be found as follows:

ax2 + bx + c = 0 ,

(x + d)(x + e) = 0 .

c© USQ, August 27, 2010 57


Since the product of two numbers can only be 0 if at least one of the numbers is 0, then

(x + d) = 0 or (x + e) = 0 .

That is,

x = −d or x = −e .

Example 1.36: Solve the quadratic equations below.

(a) x2 + 3x + 2 = 0

(b) x2 − 5x + 6 = 0

(c) x2 − x− 2 = 0

(d) 6x2 + 7x + 2 = 0

(e) x2 − 4 = 0

(f) x2 + 5x = 6

Solution:

(a)

x2 + 3x + 2 = 0 ,

(x + 1)(x + 2) = 0 ,

x = −1 or x = −2 .

(b)

x2 − 5x + 6 = 0 ,

(x− 3)(x− 2) = 0 ,

x = 3 or x = 2 .

(c)

x2 − x− 2 = 0 ,

(x− 2)(x + 1) = 0 ,

x = 2 or x = −1 .

58 c© USQ, August 27, 2010


(d)

6x2 + 7x + 2 = 0 ,

6x2 + 4x + 3x + 2 = 0 ,

2x(3x + 2) + (3x + 2) = 0 ,

(3x + 2)(2x + 1) = 0 ,

3x + 2 = 0 or 2x + 1 = 0 ,

x = −23

or x = −12

.

(e)

x2 − 4 = 0 ,

(x + 2)(x− 2) = 0 ,

x = −2 or x = 2 .

(f)

x2 + 5x = 6 ,

x2 + 5x− 6 = 0 ,

(x + 6)(x− 1) = 0 ,

x = −6 or x = 1 .

Quadratic formula

If the quadratic equation of the form ax2 + bx + c = 0 cannot be factorised (or if youcannot readily determine its factors), the solutions are given by the formula:

x =−b±

√b2 − 4ac

2a.

The + sign on the square root gives one solution and the − sign on the square root givesthe other. Thus,

x =−b +

√b2 − 4ac

2aor x =

−b−√

b2 − 4ac2a

.

This formula arises from a process called completing the square which you will meet inChapter 2.

The quadratic formula will obtain the solutions of any quadratic equation.

c© USQ, August 27, 2010 59


Example 1.37: Solve 3x2 − 15x + 17 = 9 .

Solution:

3x2 − 15x + 17− 9 = 0 ,

3x2 − 15x + 8 = 0 .

Comparing this with the standard form ax2 + bx + c we note that a = 3 , b = −15 ,and c = 8 .

x =−b±

√b2 − 4ac

2a

=−(−15)±

√(−15)2 − 4× 3× 8

2× 3

=15±

√225− 966

=15±

√129

6,

thus,

x =15 +

√129

6or x =

15−√

1296

,

x ≈ 4.39 or x ≈ 0.61 .

(Check that each value of x satisfies the original equation.)

Example 1.38: Solve 3 = x + 4x2 .

Solution:Rearranging: 4x2 + x− 3 = 0

Comparing this with the standard form ax2 + bx + c we note that a = 4 , b = 1 , andc = −3 .

60 c© USQ, August 27, 2010


x =−b±

√b2 − 4ac

2a

=−1±

√12 − 4× 4× (−3)

2× 4

=−1±

√49

8

=−1± 7

8,

thus,

x =68

or−88

= 0.75 or− 1 .

(Check that each value of x satisfies the original equation.)

Application: electrical engineering In electrical engineering the thermale.m.f. (E) corresponding to the temperature T for a copper-iron thermocoupleis modelled by E = −0.0192T2 + 6.88T . What would be the temperature of thethermocouple when the e.m.f. is 200 µV?

(Note that e.m.f. is electromotive force, the greatest potential difference that canbe generated by a particular source of electric current. It is measured in microvolts in this case.)

E = −0.0192T2 + 6.88T ,

200 = −0.0192T2 + 6.88T ,

0.0192T2 − 6.88T + 200 = 0 .

Here, a = 0.0192 , b = −6.88 , and c = 200 .

T =−b±

√b2 − 4ac

2a

=−(−6.88)±

√(−6.88)2 − 4× 0.0192× 200

2× 0.0192, so

T = 31.9 C or 326 C .

c© USQ, August 27, 2010 61


Discriminant — types of solutions

We have seen that the solutions of any quadratic equation ax2 + bx + c = 0 are given by

x =−b±

√b2 − 4ac

2a.

Quadratic equations do not always have two real solutions. The number of possible realsolutions depends on the quantity under the square root sign, b2 − 4ac .

• Two different solutions.

Provided b2 − 4ac is a positive number, there are two possible solutions

• No solution.

If however b2 − 4ac is a negative number, there are no solutions because the squareroot of a negative number is not defined.

• One solution.

If b2 − 4ac is zero, the formula gives us

x =−b±

√0

2a=−b2a

.

Thus there is only one solution.

The quantity, b2 − 4ac , is called the discriminant.

Example 1.39: Find the number of solutions for the following quadratics:

(a) 3x2 − 4x + 1 = 0

(b) x2 − 6x + 9 = 0

(c) 2x2 − 4x + 3 = 0

Solution:

(a) b2 − 4ac = 16− 4× 3× 1 = 4 . Thus there are 2 solutions.

(b) b2 − 4ac = 36− 4× 1× 9 = 0 . Thus there is 1 solution.

(c) b2 − 4ac = 16− 4× 2× 3 = −8 . Thus there are no real solutions.

Exercise 1.40: Solve the following quadratic equations where possible.

(a) x2 − 5x + 6 = 0

(b) 3x2 − 2x + 4 = 0

(c) x2 − 5x + 2 = 0

(d) 2x2 − 4x + 1 = 0

(e) y2 − 5y− 36 = 0

62 c© USQ, August 27, 2010


(f) 6t2 + 17t = 3

(g) x2 − 8x + 9 = 0

(h) 2x2 + 2x− 30 = 0

(i) x2 − 1.7x + 0.3 = 0

(j) x2 + 3.1x = 1.8

(k) 4x = x2 − 45

(l) m2 = 14m− 49

1.7.2 Simultaneous equations

Previously, we concerned ourselves with one unknown variable, but in practice there areoften two or more variables to contend with. For example:

A city bakery sold 1 500 bread rolls on Sunday, with sales receipts of $1 460. Plainrolls sold for 90 cents each while gourmet rolls sold for $1.45 each. How many of eachtype of roll were sold?

What information is given in this question?

• The total number of rolls sold was 1 500.

• The total money received from selling these rolls was $1 460.

• The price of each plain and gourmet roll.

What is the question asking?

• This question asks us how many of each type of roll so there are two variablesinvolved.

– The number of plain bread rolls, let us call it P.

– The number of gourmet rolls, let us call it G.

Is there a relationship between the variables?

• If the total number of rolls sold is 1 500 then number of plain rolls and number ofgourmet rolls must be 1 500. In algebraic language this is P + G = 1 500 .

• If the total receipts are $1 460 and plain rolls cost 90 cents each and gourmet rolls$1.45 each then the total receipts from plain rolls and the total receipts from gourmetrolls must be $1 460. In algebraic language this is 0.9P+ 1.45G = 1 460 . (Remember90 cents is $0.90)

c© USQ, August 27, 2010 63


We have generated two equations in two unknown variables.

P + G = 1 500 ,

0.9P + 1.45G = 1 460 .

Such equations are called simultaneous equations. You can have a pair with two unknownvariables or a set of three with three unknown variables. In fact, you can have as manyequations as you need as long as you have at least as many equations as you have un-known variables. Only then do you have a chance of finding values for each of the vari-ables if solutions exist. In this chapter we will focus only on two equations with twounknown variables.

Hint If you do not understand:

• go back and re-read the previous page;

• read ahead to discover anything new that might help;

• locate and review examples that explain things more clearly;

• take a break, have a cup of coffee, or come back to it when you have had agood night’s sleep;

• read the misunderstood section out loud;

• write out exactly what you do not understand and ask a study friend orcolleague about it; or

• contact your tutor.

Exercise 1.41: Set up equations for each of the following problems algebraically usingtwo variables. Remember to state what your variables represent. Do not solve at thisstage.

(a) A boot manufacturer produces two styles, Outback and Bush Walker. Fromprevious sales records it is known that 10 percent more of the Outback modelsell than the Bush Walker. A profit of $150 is made on each Outback sold, but$200 is made on each Bush Walker. If during the next year a profit of $730 000is planned, how many of each model must be sold?

(b) A woman invested part of her money at 5% and the rest at 8%. The incomefrom both investments resulted in a profit of $5 650. If she interchanged herinvestments, the profit would have been $4 750. How much did she have ineach investment?

(c) Flying with the wind it took a plane 1 hour and 30 minutes to travel 720 kilo-metres, while against the wind it took the plane 2 hours. Find the speed of thewind and the speed of the plane in still air.

(d) A mother is 4 times as old as her son. In six years time the mother will be3 times as old as her son. Find their present ages.

64 c© USQ, August 27, 2010


(e) The weekly wages of 6 carpenters and 8 apprentices amount to $5 160 and thewages of 8 carpenters and 6 apprentices amount to $5 760. Find the weeklywages of a carpenter and an apprentice.

(f) A box contains 40 coins made up of 10 cent coins and 20 cent coins. How manyof each are in the box if the value of the coins is $5.00?

Once you have mastered generating equations the next stage is to solve them. You mightalso have done this previously. There are a number of different ways that you can solve aset of simultaneous equations. We will look at two here and another method in chapter 5.All of the methods described are equally valid. It is up to you to choose the method thatis easiest for you to use.

Application: electrical engineering Kirchoff’s Laws are two laws relating toelectrical circuits. The current law states that the algebraic sum of the currentsflowing through all the wires of a network that meet at a point is zero. When it isapplied to a particular circuit the current (in Amperes) in two parts of the circuit,I1 and I2 can be represented by two equations.

2I1 + 5I2 = 4.5 .

5I1 + 3I2 = 6.5 .

When you have completed the following section, try and solve these equations forI1 and I2, you should get I1 = 1 A, I2 = 0.5 A.

Substitution

In this method, we substitute one variable from one equation into the other. The stepsinvolved are:

1. Using either of the equations, express one variable in terms of the other.

2. This expression is then substituted into the other equation to form an equation inone variable only.

3. Solve this equation to find the value of one of the variables.

4. Substitute the value of this variable into the equation formed in the first step to findthe value of the other variable.

5. Check your answer in both of the original equations.

Let us follow the steps through in a pair of simple equations.

2x + 5y = 6 . (1.1)

3x + 2y = −2 . (1.2)

c© USQ, August 27, 2010 65


Step 1: Express one variable in terms of the other.

Using Equation (1.1), rearrange to make x the subject.

2x + 5y = 6 ,

2x = 6− 5y ,

x =6− 5y

2.

Step 2 and 3: Substitute this into Equation (1.2) and solve for the single variable.

3x + 2y = −2 ,

3(

6− 5y2

)+ 2y = −2 ,

3(6− 5y) + 4y = −4 ,

18− 15y + 4y = −4 ,

−11y = −4− 18 ,

−11y = −22 ,

y = 2 .

Step 4: Substitute this value into the other equation. (You could also substitute it into theequation formed in Step 1 if you find this easier.)

2x + 5y = 6 ,

2x + 5× 2 = 6 ,

2x = 6− 10 ,

2x = −4 ,

x = −2 .

From our calculations our answers are x = −2 and y = 2 .

Step 5: Check the answer in both of the original equations.

Check: x = −2 and y = 2 .

Equation (1.1):

2x + 5y = 6 ,

LHS = 2×−2 + 5× 2

= 6

= RHS.

66 c© USQ, August 27, 2010


Equation (1.2):

3x + 2y = −2 ,

LHS = 3×−2 + 2× 2

= −2

= RHS.

Both values substitute correctly into both equations so the answer must be correct.

You should note that this is just one way of eliminating a variable, we could havejust as easily decided to eliminate y by multiplying one equation by 2 and the otherby 5 . . . the choice is yours.

Elimination

In this method we eliminate one variable and form one equation. The steps involved are:

1. Multiply one or both equations by constants so that one of the variables has thesame coefficient.

2. Subtract one equation from the other so that the variable with the same coefficientis eliminated.

3. Solve this equation to find the value of the variable.

4. Substitute the value of this variable into one of the equations to find the value ofthe other variable.

5. Check your answer in both of the original equations.

Let us look at the previous equations.

2x + 5y = 6 (1.1)3x + 2y = −2 (1.2)

Step 1: Multiply one or both equations by constants.

The coefficients of x are different in both equations but if we multiplied equation(1.1) by 3 and equation (1.2) by 2, the coefficient of x in both equations would be 6.Remember, however, that we cannot simply multiply the x term by a constant —we must multiply every term in the equation by it. Let us do it.

Multiply equation (1.1) by 3:

3× 2x + 3× 5y = 3× 6 ,

6x + 15y = 18 . (1.3)

c© USQ, August 27, 2010 67



2× 3x + 2× 2y = 2×−2 ,

6x + 4y = −4 (1.4)

Step 2: Subtract the equations to eliminate a variable.

6x + 15y = 186x + 4y = −4

+ 11y = 22 .

Thus the equation is11y = 22 .

Step 3: Solve the equation.

11y = 22 ,

y = 2 .

Step 4: Substitute to find the other variable.

2x + 5y = 6 ,

2x + 5× 2 = 6 ,

2x = 6− 10 ,

2x = −4 ,

x = −2 .

From our calculations the answers are x = −2 and y = 2 .

Step 5: Check the answers in both of the original equations.

Check: x = −2 , and y = 2 .

Equation (1.1):

2x + 5y = 6 ,

LHS = 2×−2 + 5× 2

= 6

= RHS.

Note that it is not absolutely essential to substitute into this equation to check be-cause we used it to originally find the value of the variable, but it is good practiceas a check on your arithmetic.

68 c© USQ, August 27, 2010


Equation (1.2):

3x + 2y = −2 ,

LHS = 3×−2 + 2× 2

= −2

= RHS.

Both values substitute correctly into both equations so the answer must be correct.

Example 1.42: Solve these equations simultaneously to find the values of x and y.

6x + 5y = 4 . (1.5)

8x + 3y = −2 . (1.6)

Solution:


3× 6x + 3× 5y = 3× 4 ,

18x + 15y = 12 . (1.7)


5× 8x + 5× 3y = 5×−2 ,

40x + 15y = −10 . (1.8)

Subtract equation (1.8) from equation (1.7):

18x + 15y = 1240x + 15y = −10

−22x + = 22

Equation is

−22x = 22 ,

x = −1 .

To find the value of y substitute x = −1 into equation (1.5).

6x + 5y = 4 ,

6(−1) + 5y = 4 ,

−6 + 5y = 4 ,

5y = 10 ,

y = 2 .

c© USQ, August 27, 2010 69


The solution is x = −1 and y = 2 .

Check:

In equation (1.5) LHS = 6(−1) + 5(2) = 4 = RHS.

In equation (1.6) LHS = 8(−1) + 3(2) = −2 = RHS.

Thus the solution is correct.

Example 1.43: Refer back to the example at the beginning of this section.

A city bakery sold 1 500 bread rolls on Sunday, with sales receipts of $1 460.Plain rolls sold for 90 cents each while gourmet rolls sold for $1.45 each. Howmany of each type of roll were sold?

Solution:

The two equations generated from these sentences are:

P + G = 1 500 , (1.9)

0.9P + 1.45G = 1 460 . (1.10)

Multiply equation (1.9) by 0.9.

0.9P + 0.9G = 1 350 . (1.11)

Subtract equation (1.11) from equation (1.10).

0.9P + 1.45G = 1 4600.9P + 0.9G = 1 350

0P + 0.55G = 110

Therefore the equation is

0.55G = 110 ,

G = 110÷ 0.55 ,

G = 200 .

To find the value of P substitute G = 200 into equation (1.9).

P + G = 1 500 ,

P + 200 = 1 500 ,

P = 1 300 .

Thus the solution is G = 200 and P = 1 300 .

Check:

In equation (1.9), LHS = 200 + 1 300 = 1 500 = RHS.

In equation (1.10), LHS = 0.9× 1 300 + 1.45× 200 = 1 460 = RHS.

70 c© USQ, August 27, 2010


So at the end of the day the shop had sold 200 gourmet rolls and 1 300 plain rolls.

Finally note that so far we have solved equations where multiples of the variables areonly added to or subtracted from each other. We call these linear equations. Situationsdo arise where the variables are related in other ways (non-linear equations) such as inthe example below.

Example 1.44: We may need to find the dimensions of a rectangle (perhaps a field or aroom) whose area is 6 square units and perimeter is 10 units. Let us translate thisinto mathematics.

Solution:

• Draw a diagram.

• Define the unknowns.

Dimensions: length = x, width = y.

x = ?, y = ?

• Area must be 6.xy = 6 .

• Perimeter must be 10.2x + 2y = 10 .

The resulting equations can only be solved by the substitution method.

xy = 6 . (1.12)

2x + 2y = 10 . (1.13)

c© USQ, August 27, 2010 71


The steps for solving this problem are:

(a) Rearrange one of the equations to express one of the variables in terms of the other.

From equation (1.12),

y =6x

. (1.14)

(b) Substitute this expression into the other equation to form an equation in one variableonly.

Substituting into equation (1.13),

2x + 2× 6x= 10 .

(c) Solve this equation and substitute back into the first one.

2x2 + 12 = 10x , Multiply through by x.

2x2 − 10x + 12 = 0 , Subtract 10x from both sides.

x2 − 5x + 6 = 0 , Divide through by 2.

(x− 2)(x− 3) = 0 , Factorise.

x = 2 , or x = 3 .

(d) Use this equation to find the first variable.

Substitute these values into equation (1.14).

y =62= 3 , or

y =63= 2 .

Thus the two solutions are (x = 2 , y = 3), or (x = 3 , y = 2).

(e) Check this solution.In equation (1.12) LHS=2× 3 = 6 =RHSIn equation (1.13) LHS=2× 2 + 2× 3 = 10 =RHS.As an exercise you should check the alternative solution x = 3 , and y = 2.

Exercise 1.45: Solve the following sets of simultaneous equations.

(a)

2x + y = 5 ,

x + 2y = 4 .

(b)

3x− y = 12 ,

x + y = 8 .

72 c© USQ, August 27, 2010


(c)

3x− 4y = 5 ,

5x− 12y = 3 .

(d)

x− y = 1 ,

2x + y = 8 .

(e)

x + y = 0 ,

2x− y = 3 .

(f)

2x− 8y = −1 ,

x− 4y = 1 .

Putting it all together! As we have visited Kirchhoff’s laws before, let us lookat another more complicated circuit and determine what the currents (I) in thiscircuit would be.

28− 6I1 − 5I2 = 0 ,

28− 6I1 − 10I3 = 0 ,

I1 − I2 − I3 = 0 .

While these equations appear to be more complicated we can solve them using the Elim-ination or Substitution methods. The difference here is that we have three variables. Thefollowing example shows how we can use both of these methods as long as we modifyour approach.

c© USQ, August 27, 2010 73


Example 1.46: Let us solve the equations from the Kirchoff example involving three vari-ables. Whilst initially daunting we will see to solve three equations that we onlyneed to modify our approach for two variables. In both methods of Substitutionand Elimination, we seek remove or eliminate one variable to obtain a system oftwo variables. Once obtained we can solve these as previously. Writing the equa-tions in the usual form we have

6I1 + 5I2 = 28 , (1.15)

6I1 + 10I3 = 28 , (1.16)

I1 − I2 − I3 = 0 . (1.17)

In this example we will look at using the Substitution Method first and then theElimination method.

Substitution

For the Substitution method we have the two new steps of Step A and Step B.

Step A: Express one variable in terms of the others. In this step we first find onevariable in terms of the other two. We rearrange Equation (1.17) to make I1 thesubject.

I1 − I2 − I3 = 0 ,

I1 = I2 + I3 .

Note we chose to use Equation (1.17) to find an expression for I1 since it hasthe simple coefficient of 1. We could easily have chosen to use this equation tosolve for I2 or I3 or one of the other two equations - the choice is yours.

Step B: We then substitute this into the other two equations (1.15) and (1.16) andsimplify. For Equation (1.15) we have

6I1 + 5I2 = 28 ,

6 (I2 + I3) + 5I2 = 28 ,

6I2 + 6I3 + 5I2 = 28 ,

11I2 + 6I3 = 28 , (1.18)

and for Equation (1.16)

6I1 + 10I3 = 28 ,

6 (I2 + I3) + 10I3 = 28 ,

6I2 + 6I3 + 10I2 = 28 ,

6I2 + 16I3 = 28 . (1.19)

74 c© USQ, August 27, 2010


This leaves us with the two equations which we know how to solve as beforeon page 65.

11I2 + 6I3 = 28 ,

6I2 + 16I3 = 28 .

Now we solve these equations following the steps for two equations from before.

Step 1: Express one variable in terms of the other.

Using Equation (1.18), rearrange to make I2 the subject.

11I2 + 6I3 = 28 ,

11I2 = 28− 6I3 ,

I2 =28− 6I3

11.

Step 2 and 3: Substitute this into Equation (1.19) and solve for the single variable.

6I2 + 16I3 = 28 ,

6(

28− 6I3

11

)+ 16I3 = 28 ,

16811− 36

11I3 + 16I3 = 28 ,

16811− 140

11I3 = 28 ,

−14011

I3 = 28− 2811

,

−14011

I3 = −14011

,

I3 = 1 .

Step 4: Substitute this value into the other equations in Step 1 and Step A. First intothe equation formed in Step 1 to find I2

I2 =28− 6I3

11,

=28− 6 (1)

11,

=2211

= 2 .

With I2 and I3 we can use the equation formed in Step A to find I1

I1 = I2 + I3 ,

= 2 + 1 ,

= 3 .

c© USQ, August 27, 2010 75


From our calculations our answers are I1 = 3, I2 = 2, and I3 = 1.

Step 5: Check the answer in all of the original equations.

Check: I1 = 3, I2 = 2, and I3 = 1.

Equation (1.15):

6I1 + 5I2 = 28 ,

LHS = 6× 3 + 5× 2

= 28

= RHS.

Equation (1.16):

6I1 + 10I3 = 28 ,

LHS = 6× 3 + 10× 1

= 28

= RHS.

Equation (1.17):

I1 − I2 − I3 = 0 ,

LHS = 3− 2− 1

= 0

= RHS.

Hence the three values substitute correctly into each equations so the answermust be correct.

Elimination

Now we instead use the Elimination Method to solve the equations Kirchoff exam-ple. Here we again introduce two new steps of Step A and Step B in the modifiedapproach. Here we first seek to eliminate one variable from each of the equations.We do this by considering the equations in pairs where the same equation is used ineach pair. We can then eliminate the same variable from each pair by multiplyingone or both of equations so the coefficient of the variable is the same. From before

76 c© USQ, August 27, 2010


we have the equations

6I1 + 5I2 = 28 , (1.20)

6I1 + 10I3 = 28 , (1.21)

I1 − I2 − I3 = 0 . (1.22)

Step A: Multiply one, two or all equations by constants.

The coefficients of I1 is the same in (1.20) and (1.21) but different for (1.22).

But if we multiplied equation (1.22) by 6 then the coefficient of I1 in all equa-tions would be 6. Remember, however, that we cannot simply multiply the I2

term by a constant — we must multiply every term in the equation by it. Letus do it.

Let us take our first pair as equations (1.20) and (1.22). Now multiply equation(1.22) by 6:

6× I1 − 6× I2 − 6× I3 = 6× 0 ,

6I1 − 6I2 − 6I3 = 0 . (1.23)

Step B: Subtract from two of the equations the third equation to eliminate a vari-able. Subtracting (1.23) from (1.20):

6I1 + 5I2 = 286I1 − 6I2 − 6I3 = 0

+ 11I2 + 6I3 = 28 .

This the same equation in (1.18)

Let us now consider our second pair as Equations (1.21) and (1.22) (note wehave used (1.22) twice). We have already multiplied (1.22) buy 6 to give (1.23)so there is no need to do this again. Subtracting (1.23) from (1.21):

6I1 + 10I3 = 286I1 − 6I2 − 6I3 = 0

+ 6I2 + 16I3 = 28 .

This the same equation in (1.19) This leaves us with the two equations

11I2 + 6I3 = 28 , (1.24)

6I2 + 16I3 = 28 . (1.25)

Now we can solve using the Elimination method as before. This is left for youas an exercise. You should find I2 = 2 and I3 = 1. With these values of we can

c© USQ, August 27, 2010 77


use equation (1.23) to find I1

I1 − I2 − I3 = 0 ,

I1 − 2− 1 = 0 ,

I1 = 0 + 2 + 1 ,

= 3 .

From our calculations our answers are I1 = 3, I2 = 2, and I3 = 1. Rememberto check your answers!

Note in the above we used Equation (1.23) twice to eliminate I1. However we couldhave as easily have chosen to use Equation (1.20) or Equation (1.21) to eliminateI1 from the other two remaining equations. We could have also chosen instead toeliminate I2, from each equation, by adding 10 times Equation (1.22) to Equation(1.21) or to eliminate I3 by adding 5 times Equation (1.22) to Equation (1.20). Thechoice is yours!

Note Solving three simultaneous equations using the Substitution method and Elimina-tion methods can easily become complicated and time-consuming. This is especially sowhen trying to decide which variable to remove at each step. In later study, you maylearn other methods which are more systematic in their approach.

78 c© USQ, August 27, 2010

1.8. Inequalities and absolute values MAT1500

Points to remember

• To translate from a real world situation expressed in words:

– draw a diagram;

– read the question aloud;

– convert each phrase individually into equivalent mathemat-ics; and

– make logical connections.

• Practise manipulating algebraic equations.

• A quadratic equation is of the form ax2 + bx + c = 0.

• Solve quadratic equations by

– factorisation; or

– formula.

• If a× b = 0, then a = 0 or b = 0.

• x =−b±

√b2 − 4ac

2a

• If discriminant b2 − 4ac is

> 0, there is two solutions;

= 0, there is one solution (two equal solutions); and

< 0, there are no real solutions.

• A set of simultaneous equations is where there are more than oneequation in more than one unknown.

• Simultaneous equations can be solved by substitution (p. 65) orelimination (p. 67).

1.8 Inequalities and absolute values

Inequalities are central to the definition of all limiting processes and are essential to en-gineers and applied mathematicians. When faced with a real world problem to translateto algebraic language, not all will result in simple expressions or equations.

Consider the following.

The total cost of landscaping a garden includes the cost of plants (call it P) and thecost of the soil and mulch (call it S). The owner declares that the total cost of the jobshould not be greater than $2 500. Express this relationship in algebraic form.

There is no way that this would be an equation. But rather this would be an inequation or

c© USQ, August 27, 2010 79

MAT1500 1.8. Inequalities and absolute values

an inequality. We would write it as P + S ≤ 2 500 , and say

The sum of the cost of plants and the cost of soil and mulch is less than or equal totwo thousand and five hundred dollars.

Recall from your previous experiences the meaning of the following signs:

< Less than≤ Less than or equal to> Greater than≥ Greater than or equal to

Simple versions of these inequations can be solved in ways similar to those used to solveequations. There are some differences, however, so look at the following examples forthe pitfalls.

Example 1.47: Find the value of x when 2x− 1 < 4 .

Solution:

2x− 1 < 4 ,2x− 1 + 1 < 4 + 1 , To remove −1 from the LHS add 1 to

both sides.2x < 5 ,

2x2

<52

, To isolate the x divide both sides by 2.

x <52

.

Note that our solution is not a single number as we would obtain from solving anequation. Rather, it is a range of numbers.

To check our solution we could try numbers either side of 52 in the original inequal-

ity.

Try a number less than 52 , say x = 2 .

Substitute this into the original inequality:

LHS = 2× 2− 1 = 3 ,

which indeed is less than 4 and the inequality is true.

Next try a number greater than 52 , say x = 3 .

Substitute this into the original inequality:

LHS = 2× 3− 1 = 5 ,

which is greater than 4. The inequality is not true for this value of x, which is whatour solution predicted.

80 c© USQ, August 27, 2010


This procedure does not ensure that we are correct, but it does give an opportunityto pick up some errors.

Example 1.48: Rearrange the following formula to make i the subject, j ≤ i2+ 4 .

Solution:

j ≤ i2+ 4 ,

j− 4 ≤ i2+ 4− 4 , To remove 4 from the RHS subtract 4 from both sides.

j− 4 ≤ i2

,

2× (j− 4) ≤ i2× 2 , To isolate the i multiply both sides by 2.

2(j− 4) ≤ i ,

i ≥ 2(j− 4) . When you switch i to the LHS of the inequation youmust change the direction of the inequality sign.

When you switch sides in an inequality you must reverse the sign.For example, 2 < 3 must become 3 > 2, otherwise it is not true.

Example 1.49: Solve the following inequation for x.

1− 2x < x + 2 .

Solution:

1− 2x < x + 2 ,1− 2x− x < x + 2− x , To group the x’s on the LHS subtract x from both sides.

1− 3x < 2 ,1− 3x− 1 < 2− 1 , To group the constants on the RHS subtract 1 from

both sides.−3x < 1 ,

−3x−3

>1−3

, To isolate x on the LHS divide both sides by −3.

x > −13

, When you divide an inequation by a negative numberyou must reverse the inequality sign.

When you divide or multiply by a negative number you must re-verse the inequality sign. For example, 2 < 3, but when multipliedon both sides by −1 it must become −2 > −3, otherwise it is nottrue.

c© USQ, August 27, 2010 81


One possible solution would be x = 0 , since 0 > − 13 .

Check this solution in the original inequation:

LHS = 1− 2(0) = 1 ,

RHS = 0 + 2 = 2 .

LHS < RHS, so solution is reasonable.

Try a value out of the solution range, say x = −1 , since −1 < − 13 .

Check this solution in the original inequation:

LHS = 1− 2(−1) = 3 ,

RHS = −1 + 2 = 1 .

So LHS is not less than RHS for this value of x, as our solution predicts.

Example 1.50: Find the values of x which satisfy the following double inequality.

−2 < 3x + 1 < 4 .

Solution:

−2 < 3x + 1 < 4 , We can rearrange all three parts of thisinequation at the same time.

−2− 1 < 3x + 1− 1 < 4− 1 , To remove the 1 from the centre partsubtract 1 from all three parts.

−3 < 3x < 3 ,

−33

<3x3

<33

, To isolate the x in the centre divide allthree parts by 3.

−1 < x < 1 .

This means that x lies between −1 and 1.

One possible solution is x = 0 , since it lies between −1 and 1.

Check this solution in the original inequation: The centre expression is 3(0) + 1 =

1 . This is between −2 and 4, so the solution is reasonable.

82 c© USQ, August 27, 2010


Exercise 1.51: Solve the following inequations for x.

(a)x3+ 2 < 5

(b)22 ≤ 5x− 3 ≤ 32

(c)3x5− 2x

3> −7

(d)3x− 2(2x− 7) ≤ 2(3 + x)− 4

(e)

−3 ≤ 2x− 13

< 3

Exercise 1.52: How are the solutions in (a) and (b) different?

(a) Solve 6− 3(2y− 3) = 4(2y− 2) for y.

(b) Solve 6− 3(2y− 3) ≤ 4(2y− 2) for y.

Exercise 1.53: Solve the following inequations:

(a) A mixed indoor cricket team must have 5 more women than men and the totalnumber of people must be at least 9 but not more than 15 (including reserves).What is the possible number of women on the team?

(b) The sum of two consecutive integers is less than 13. What positive values canthe smaller integer have?

(c) A food stall at a football ground sells on average 10 more sausage rolls thanpies. If each food stall must allow for at least 60 customers but not more than100, what is the possible number of sausage rolls that should be heated?

The real number line

Real numbers are made up of rational numbers (integers and fractions) and irrationalnumbers (those numbers which can be represented by a decimal that does not terminateor repeat, for example

√2, e and π).

These can be displayed on a line called the real number line, as shown below.

c© USQ, August 27, 2010 83


Interval notation

The inequality x > 2 is read as

‘x is (any real number) greater than two’.

Alternatively, we could use interval notation and write that

‘the range of x is (2, ∞)’.

This is also expressed as

‘x is in the interval (2, ∞)’, or

x ε (2, ∞).

Some valid values of x (that is, satisfying the inequality x > 2) are 2.1, 3, 125, 5, and soon.

The inequality y < 10 is read as

‘y is (any real number) less than 10’, or

‘the range of y is (−∞, 10)’.

Some valid values of y are 9.9, 8, 7.64, 2, −127.65, and so on.

If we allow y to take on the value of 10 as well as any real number below 10, we writey ≤ 10, or the range of y is (−∞, 10]. Note a square bracket indicates that the end pointof the interval is also a valid value of y. Infinite end points always have round brackets(because no number is ever equal to infinity). Similarly we can write x ≥ 2 as meaning xis greater than or equal to 2 or the range of x is [2, ∞).

Double inequalities further restrict the range of values that x can take.

For example, 1 ≤ x ≤ 3 is read as ‘x is greater than or equal to one but less than or equalto three’, or x ε [1, 3].

Other possibilities include:

1 < x ≤ 3 x is greater than 1 but less than or equal to 3, or x ε (1, 3].1 < x < 3 x is greater than 1 but less than 3, or x ε (1, 3).1 ≤ x < 3 x is greater than or equal to 1 but less than 3, or x ε [1, 3).

Note that inequalities like 1 > x > 3 do not make sense since they imply that 1 is greaterthan 3 which of course is not the case. Do not use inequalities like 2 > x < 4. Use twoseparate inequalities x < 2, x < 4. But if x is to satisfy x < 2, then it clearly must also beless than four, so the second inequality is superfluous, and all we need is the inequalityx < 2 or x ε (−∞, 2).

84 c© USQ, August 27, 2010


Example 1.54: Express the following inequalities in interval notation.

(a) x > −4

(b) x ≤ 32

(c) −10 < x ≤ −2.1

(d) x ≥ 4

(e) 4 ≤ x ≤ 8

Solution:

(a) (−4,+∞).

(b) (−∞, 32].

(c) (−10,−2.1].

(d) [4,+∞).

(e) [4, 8].

Application: applied mathematics Linear programming is one of the mostwidely used applications of mathematics with uses in business planning, industrialengineering, and social and physical sciences. It was developed during the secondworld war when the logistics associated with moving supplies and armies was enor-mous. At first calculations were long and tedious, but today with the advent ofsuper computers, it is now possible to deal with problems involving thousands ofvariables. Here is a very simple example.

A student wants to maximise an assignment grade. The student will get 5 pointsfor each page of originality (I), 4 for each page of quotations (Q), and 1 point foreach page of graphics (G). Using a linear programming technique, this problemcan be represented as a series of equations and inequations that can be solved bygraphing.

I + Q + G = 20 ,

I ≥ 3 ,

0.5Q + G ≤ 12 ,

Q ≥ 0 ,

G ≥ 0 ,

I ≥ 0 .

When this problem is solved, the student receives a maximum score of 78 basedon this ‘number of pages’ criteria.

c© USQ, August 27, 2010 85


Absolute value

In many instances, when you solve an equality, especially if it is a double inequality, youmay obtain a range of answers. The example −2 < 3x + 1 < 4 produces the answer−1 < x < 1.

We summarise this type of inequality by introducing a new notation called the absolutevalue, so that −1 < x < 1 can be written as |x| < 1.

Recall now where you might have seen absolute value in the past. We use absolute valuewhen we are interested in the magnitude or size of a number without regard to its sign.For example, if we return to the number line, we know that −2 and 2 are the same dis-tance from zero.

-

0 1 2 3 4−1−2−3−4∞−∞

So using absolute value notation we say that |2| = 2 and | − 2| = 2 .

If we introduce a variable into this type of notation and say that |x| = 2 , then we meanthat x has two possible values, x = 2 or x = −2 , which are both two units from zero. Youmight be familiar with this written as x = ±2 .

Extending this concept even further to inequalities, we can say that |x| < 2, so we knowthat x must lie somewhere between −2 and 2 on the number line, that is, −2 < x < 2.

Similarly, when we write |x| > 2, then we know that x must lie further than 2 on one sideof the number line and also further than −2 on the other side of the number line.

So in a general way,

When |x| < a then −a < x < a.

86 c© USQ, August 27, 2010


When |x| > a then x > a or x < −a.

Let us look at some examples which use this notation.

Example 1.55: Find the values of x which satisfy |x + 1| ≤ 2 .

Solution:

|x + 1| ≤ 2 , Replace absolute value notation with inequality.

−2 ≤ x + 1 ≤ 2 ,−2− 1 ≤ x + 1− 1 ≤ 2− 1 , Remove 1 from the centre part by subtracting 1

from all 3 parts.

−3 ≤ x ≤ 1 .

Check:

Two possible solutions are −3 and 1.

When x = −3,

LHS = | − 3 + 1|= | − 2|= 2

= RHS.

When x = 1,

LHS = |1 + 1|= |2|= 2

= RHS.

Example 1.56: Solve the following inequation for p, |3p− 4| > 5.Solution:This will produce two inequations.

3p− 4 > 5 , or 3p− 4 < −5 ,3p− 4 + 4 > 5 + 4 , 3p− 4 + 4 < −5 + 4 ,

3p > 9 , 3p < −1 ,3p3

>93

,3p3

<−13

,

p > 3 . or p < − 13 .

c© USQ, August 27, 2010 87


Thus the final answer is p > 3 or p < − 13 .

Check:

Two possible solutions are p = 4 and p = −1 .

When p = 4 ,

LHS = |3× 4− 4|= |8|= 8 > RHS.

When p = −1 ,

LHS = |3×−1− 4|= | − 7|= 7 > RHS.

Are you on schedule? Now is a good time to rethink your time scheduling.Think about the following:

• plan enough time for study;

• study at the same time every day;

• make use of the free hours during the day;

• plan study periods to follow class periods if you are an on campus student;

• space study periods and have short breaks;

• plan for weekly reviews;

• leave some unscheduled time for flexibility;

• do not forget to leave some time for fun.

Exercise 1.57: Solve the following inequalities.

(a) |x| ≤ 3

(b) |2x| > 8

(c) |x− 4| ≤ 1

(d) |8− 4x| > 0

(e) |3x− 4| < 8

(f) |2− (x + 1)| ≤ 3

(g) |x− 1| > 2

88 c© USQ, August 27, 2010

1.9. Review MAT1500

Points to remember

• When you switch sides in an inequality you must reverse the sign.

• When you divide or multiply by a negative number you mustreverse the inequality sign.

• ‘x is in the interval (a, b)’ implies x > a and x < b.

• ‘x is in the interval [a, b]’ implies x ≥ a and x ≤ b.

• ‘x ∈ (a, b)’ means ‘x is in the interval (a, b)’.

• Absolute value of a number is its distance from 0 irrespective ofits sign.

• |x| = a implies x = a or x = −a.

Congratulations, you have made it to the end of the first chapter in this course – you arewell on your way! Here are a number of things to check.

1. Have a close look at your action plan for study. Are you on schedule? Or do youneed to restructure your action plan or contact your tutor to discuss any delays orconcerns?

2. Make a summary of the important points in this chapter noting your strengths andweaknesses. Add any new words to your personal glossary. This will help withfuture revision and the open book exam.

3. Check your skill level by attempting the Post-test.

1.9 Review

an = a× a× a . . .× a︸︷︷︸n times

, an × am = an+m,an

am = an−m, (an)m = an×m = anm,

(ab)n = an × bn = anbn, a−n =1an , a

1n = n√

a, a0 = 1.

• In scientific notation a quantity is expressed as a number between 1 and 10 (calledthe mantissa), multiplied by a power of 10.

• (x + a)(x + b) = x2 + (a + b)x + ab.

• To factorise ax + ab, take the common factor a outside the brackets to give a(x + b).

• To factorise the quadratic expression ax2 + bx + c find factors p, q of ac that add togive b, replace b with p + q and find common factors twice.

• To add or subtract fractions express both as equivalent fractions with the same de-nominator then add or subtract numerators.

c© USQ, August 27, 2010 89

MAT1500 1.10. Post-test

• To multiply fractions multiply the numerators and multiply the denominators. Todivide one fraction by another invert the dividing fraction and multiply.

• To translate from a real world situation expressed in words: draw a diagram; readthe question aloud; convert each phrase individually into equivalent mathematics;and make logical connections.

• A quadratic equation is of the form ax2 + bx + c = 0:

– if discriminant b2 − 4ac > 0, then the equation has two solutions;

– if discriminant b2 − 4ac = 0, then the equation has one solution (two equalsolutions); or

– if discriminant b2 − 4ac < 0, then the equation has no real solutions.

• Simultaneous equations can be solved by substitution or elimination.

– ‘x is in the interval (a, b)’ implies x > a and x < b.

– ‘x is in the interval [a, b]’ implies x ≥ a and x ≤ b.

– ‘x ∈ (a, b)’ means ‘x is in the interval (a, b)’.

• Absolute value of a number is its distance from 0 irrespective of its sign.

1.10 Post-test

1. Simplify the following expression.

(9a2b6) 1

2

3 (a6b9)13

.

2. (a) Express 0.06425 in scientific notation.

(b) Express the sum of 0.14 Mg and 43.25 kg in grams.

3. (a) Expand x(x− 2)(3x− 1) .

(b) Expand (x + 3)2 .

4. (a) Factorise x2 − 5x− 6 .

(b) Factorise x2 − 9 .

5. Factorise 8x2 + 2x− 3 .

6. Simplify the following expression.

3x− 2

+2x

x + 3.

90 c© USQ, August 27, 2010

1.10. Post-test MAT1500

7. A Golden Rectangle is a very pleasing shape to the human eye. It has the property

that its length, L, is1 +√

52

times its width, W.

Write down an equation to represent this information.

An artist wants to use a canvas of these dimensions but needs to limit its area to1.7 m2. Write an equation using the same variables (L and W), to represent thisinformation.

Solve the equations to find the dimensions of the canvas.

8. (a) Solve the following quadratic equation using factorisation.

x2 − 7x + 6 = 0 .

(b) Find all solutions to x2 + 5x + 7 = 0 .

9. Solve the following set of simultaneous equations.

3x− y = 1 ,

x + y = 3 .

10. Simplify the inequality −3|x + 1| < −6 .

c© USQ, August 27, 2010 91

MAT1500 1.11. Solutions

1.11 Solutions

1.11.1 Answers to selected Exercises

1.4 (a) 332 .

(b) 12.

(c) 1.

(d) 175.

(e) 12.

(f) 847.

(g)8

a2b.

(h)4

a2b.

(i)1

3a16 b

43

.

(j)a8

b7 .

(k) 0.

(l) 1 075.

1.8 (a) 2.216× 108.

(b) 4.21× 10−7.

(c) 3.124 8× 10−1.

(d) 1.618 5× 105.

(e) 1.121 4× 10−1.

(f) 1.121 4× 103.

(g) 3.124 8× 107.

(h) 1.618 5× 10−3.

(i) 4.5× 102.

(j) 4.5× 10−10.

1.9 (a) 5.705 8× 103.

(b) 1.744× 103.

(c) 2.653 43× 1061.

(d) 1.2× 10−2.

(e) 3× 102.

(f) 1× 103.

92 c© USQ, August 27, 2010

1.11. Solutions MAT1500

1.11 (a) 240.8 kg.

(b) 16.8 km.

(c) 1.91 kL.

(d) 678 g.

(e) 68 mm.

(f) 620 mL.

(g) 16.425 g.

(h) 15 m.

(i) 26 kL.

(j) 625 pg.

(k) 6 pm.

(l) 24.9 L.

(m) 0.3 m2.

1.18 (a) x2 + 3x + 2 .

(b) x2 − x− 2 .

(c) x2 − 3x + 2 .

(d) 2x2 + 11x + 12 .

(e) 6x2 + 10x− 4 .

(f) 6x2 − 19x + 10 .

(g) x3 + 2x2 − x− 2 .

(h) 24x3 + 10x2 − 7x− 2 .

(i) x3 + 2x2 − 3x− 6 .

(j) x2 − y2 .

(k) x2 + 2xy + y2 .

(l) 2x4 − x3y + x2y2 − xy3 − y4 .

1.23 (a) (x− 3)(x + 2) .

(b) (x + 3)(x + 2) .

(c) (x− 3)(x− 2) .

(d) (x + 3)(x− 2) .

(e) (2x− 7)(x + 2) .

(f) 2(x− 2)(3x + 2) .

(g) ab(a + b− c) .

(h) (x− 16)(x + 3) .

(i) (2x− 3)(2x− 1) .

(j) 2(4x− 1)(x + 3) .

c© USQ, August 27, 2010 93


(k) (2− 3x)(2 + 3x) .

(l) (5x− 2)(3x− 1) .

1.25 (a) x3 + 2x2 − 5x− 6 = (x + 1)(x− 2)(x + 3).

(b) x3 + 3x2 − 4x− 12 = (x + 2)(x− 2)(x + 3).

(c) x3 − 2x2 − 19x + 20 = (x− 1)(x + 4)(x− 5).

(d) x3 − 5x2 − 2x + 24 = (x + 2)(x− 3)(x− 4).

(e) x3 + 4x2 − 17x− 60 = (x + 3)(x− 4)(x + 5).

(f) x3 + 4x2 − x− 4 = (x− 1)(x + 1)(x + 4).

(g) x3 + 2x2 − 9x− 18 = (x + 2)(x− 3)(x + 3).

(h) x3 + 15x2 + 74x + 120 = (x + 4)(x + 5)(x + 6).

1.30 (a)2x2 + 4x2 − 4

.

(b)−x2 + 2x− 7

x2 + x− 2.

(c)x2 − 15x + 1x2 − x− 6

.

(d)5x2 − 2x− 16x2 + x− 1

.

(e)x2 + x + 14x2 + 2x− 3

.

(f)3x2 − 8x + 7

x2 − x− 6.

1.35 (a) r =st

.

(b) r =c

2π.

(c) m =ec2 .

(d) c = ±√

em

.

(e) G =Fd

m1m2.

(f) f =ab

a + b.

(g) V = ±√

3PNm

.

(h) T = 4πr2L2V2d .

94 c© USQ, August 27, 2010


(i) y =3x− 12

4.

(j) x =5y− 33y− 2

.

1.40 (a) x = 2 or 3.

(b) No real solution.

(c) x =5±√

172

≈ 4.56 or 0.44 .

(d) x ≈ 1.71 or 0.29.

(e) y = −4 or 9.

(f) t = −3 or 16 .

(g) x ≈ 1.35 or 6.65.

(h) x ≈ −4.41 or 3.41.

(i) x = 0.2 or 1.5.

(j) x = −3.6 or 0.5.

(k) x = −5 or 9.

(l) m = 7.

1.41 (a) Let A be the number of Outback boots sold. Let B be the number of BushWalker boots sold.

150A + 200B = 730 000 ,

A = 1.1B .

(b) Let x be one amount of money invested. Let y be the other amount of moneyinvested.

0.05x + 0.08y = 5 650 ,

0.08x + 0.05y = 4 750 .

(c) Let x be the speed of the plane. Let y be the speed of the wind.

(Remember that speed =distance

time, so distance = speed × time.)

1.5(x + y) = 720 ,

2(x− y) = 720 .

(d) Let m be the mother’s age in years. Let s be the son’s age in years.

m + 6 = 3(s + 6) ,

m = 4s .

(Answer: m = 48, s = 12.)

c© USQ, August 27, 2010 95


(e) Let c be the weekly wage of a carpenter. Let a be the weekly wage of an ap-prentice.

6c + 8a = 5 160 ,

8c + 6a = 5 760 .

(Answer: c = 540, a = 240.)

(f) Let x be the number of ten cent coins. Let y be the number of twenty centcoins.

x + y = 40 ,

0.1x + 0.2y = 5 .

(Answer: x = 30 , y = 10.)

1.45 (a) x = 2 , y = 1 .

(b) x = 5 , y = 3 .

(c) x = 3 , y = 1 .

(d) x = 3 , y = 2 .

(e) x = 1 , y = −1 .

(f) No solution exists.

1.51 (a)

x3+ 2 < 5 ,

x3+ 2− 2 < 5− 2 ,

x3

< 3 ,

x3× 3 < 3× 3 ,

x < 9 .

(b)

22 ≤ 5x− 3 ≤ 32 ,

22 + 3 ≤ 5x− 3 + 3 ≤ 32 + 3 ,

25 ≤ 5x ≤ 35 ,255

≤ 5x5

≤ 355

,

5 ≤ x ≤ 7 .

96 c© USQ, August 27, 2010


(c)

3x5− 2x

3> −7 ,

15× 3x5− 15× 2x

3> 15×−7 ,

3× 3x− 5× 2x > −105 ,

9x− 10x > −105 ,

−x > −105 ,

−1×−x < −1×−105 ,

x < 105 .

(d)

3x− 2(2x− 7) ≤ 2(3 + x)− 4 ,

3x− 4x + 14 ≤ 6 + 2x− 4 ,

−x + 14 ≤ 2 + 2x ,

−x + x + 14 ≤ 2 + 2x + x ,

14 ≤ 2 + 3x ,

14− 2 ≤ 2− 2 + 3x ,

12 ≤ 3x ,123≤ 3x

3,

4 ≤ x ,

x ≥ 4 .

(e)

−3 ≤ 2x− 13

< 3 ,

−3× 3 ≤ 2x− 13× 3 < 3× 3 ,

−9 ≤ 2x− 1 < 9 ,

−9 + 1 ≤ 2x− 1 + 1 < 9 + 1 ,

−8 ≤ 2x < 10 ,−82

≤ 2x2

<102

,

−4 ≤ x < 5 .

c© USQ, August 27, 2010 97


1.52 (a) Solve 6− 3(2y− 3) = 4(2y− 2) .

6− 6y + 9 = 8y− 8 ,

15− 6y = 8y− 8 ,

15− 6y + 6y = 8y + 6y− 8 ,

15 = 14y− 8 ,

15 + 8 = 14y− 8 + 8 ,

23 = 14y ,2314

=14y14

,

y = 1914

.

(b) Solve 6− 3(2y− 3) ≤ 4(2y− 2) .

6− 6y + 9 ≤ 8y− 8 ,

15− 6y ≤ 8y− 8 ,

15− 6y + 6y ≤ 8y + 6y− 8 ,

15 ≤ 14y− 8 ,

15 + 8 ≤ 14y− 8 + 8 ,

23 ≤ 14y ,2314≤ 14y

14,

y ≥ 1914

.

The solution for (b) includes 1 914 and all values greater than this.

1.53 (a) Let w be the number of women on the team.

9 ≤ w− 5 + w ≤ 15 ,

9 ≤ 2w− 5 ≤ 15 ,

9 + 5 ≤ 2w− 5 + 5 ≤ 15 + 5 ,

14 ≤ 2w ≤ 20 ,142

≤ 2w2

≤ 202

,

7 ≤ w ≤ 10 .

Therefore, there may be 7, 8, 9 or 10 women on the team.

98 c© USQ, August 27, 2010


(b) Let x be the value of the smaller integer.

x + x + 1 < 13 ,

2x + 1 < 13 ,

2x + 1− 1 < 13− 1 ,

2x < 12 ,2x2

<122

,

x < 6 .

Therefore the small integer may have a value of 1, 2, 3, 4 or 5.

(c) Let s be the number of sausage rolls heated.

60 ≤ s + s− 10 ≤ 100 ,

60 ≤ 2s− 10 ≤ 100 ,

60 + 10 ≤ 2s− 10 + 10 ≤ 100 + 10 ,

70 ≤ 2s ≤ 110 ,702

≤ 2s2

≤ 1102

,

35 ≤ s ≤ 55 .

Therefore the possible number of sausage rolls that should be heated rangesfrom 35 to 55.

1.57 (a) |x| ≤ 3 :−3 ≤ x ≤ 3 .

(b) |2x| > 8 :2x > 8 , 2x < −8 ,

2x2

>82

,2x2

<−82

,

x > 4 . or x < −4 .

(c) |x− 4| ≤ 1 :

−1 ≤ x− 4 ≤ 1 ,

3 ≤ x ≤ 5 .

c© USQ, August 27, 2010 99


(d) |8− 4x| > 0 :

8− 4x > 0 , 8− 4x < 0 ,8− 4x + 4x > 0 + 4x , 8− 4x + 4x < 0 + 4x ,

8 > 4x , 8 < 4x ,

84

>4x4

84

<4x4

,

2 > x , 2 < x ,x < 2 . or x > 2 .

(e) |3x− 4| < 8 :

−8 < 3x− 4 < 8 ,

−8 + 4 < 3x− 4 + 4 < 8 + 4 ,

−4 < 3x < 12 ,−43

<3x3

<123

,

−43

< x < 4 .

(f) |2− (x + 1)| ≤ 3 :

−3 ≤ 2− (x + 1) ≤ 3 ,

−3 ≤ 2− x− 1 ≤ 3 ,

−3 ≤ 1− x ≤ 3 ,

−3− 1 ≤ 1− 1− x ≤ 3− 1 ,

−4 ≤ −x ≤ 2 ,

−4×−1 ≥ −x×−1 ≥ 2×−1 ,

4 ≥ x ≥ −2 ,

−2 ≤ x ≤ 4 .

(g) |x− 1| > 2 :

x− 1 > 2 , or x− 1 < −2 ,x− 1 + 1 > 2 + 1 , x− 1 + 1 < −2 + 1 ,

Therefore x > 3 . or x < −1 .

1.11.2 Post-test solutions

1.1a

.

2. (a) 6.425× 10−2 .

(b) 183 250 g.

100 c© USQ, August 27, 2010


3. (a) 3x3 − 7x2 + 2x .

(b) x2 + 6x + 9 .

4. (a) (x− 6)(x + 1) .

(b) (x + 3)(x− 3) .

5. (2x− 1)(4x + 3) .

6.2x2 − x + 9x2 + x− 6

.

7.

L =1 +√

52

w ,

And Lw = 1.7 .

Therefore L = 1.659 ,

w = 1.025 .

8. (a) x = 1 or x = 6 .

(b) No solutions.

9. x = 1 , y = 2 .

10. x < −3 or x > 1 .

c© USQ, August 27, 2010 101


102 c© USQ, August 27, 2010

Chapter 2

Functions and Graphing

Chapter contents2.1 The concept of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

2.1.1 What is a function? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

2.1.2 Functional notation and representation . . . . . . . . . . . . . . . . . . . . 105

2.1.3 Inverse function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

2.2 Linear functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

2.2.1 Slope-point equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

2.2.2 Point-point equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

2.2.3 Some special lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

2.2.4 Distance between two points . . . . . . . . . . . . . . . . . . . . . . . . . . 129

2.3 Quadratic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

2.3.1 Completing the square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

2.3.2 Quadratic Equation Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 138

2.3.3 Sketching parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

2.4 Rational functions — the rectangular hyperbola . . . . . . . . . . . . . . . . . 144

2.4.1 Graphing rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

2.5 Circle — not a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

2.6 When two functions meet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

2.7 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

2.8 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

2.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172



c© USQ, August 27, 2010 103

MAT1500

Objectives


• define the concepts of function, domain, range and inverse;

• recognise the equations and describe the general shapes of the graphs for straightlines, parabolas, hyperbolas and circles;

• find the equations of

– a straight line given either one point and its slope or two points, and

– a circle given its centre and radius;

• find the distance between two points in two dimensions;

• find the maximum or minimum point of a quadratic given its equation;

• find the vertical and horizontal asymptotes of a hyperbola given its equation;

• find the centre and radius of a circle given its equation;

• draw a graph of any mathematical relation by plotting selected points; and

• find an approximate solution of any mathematical equation by plotting graphs.

Introduction

Have you ever looked up the words function or graph in an English dictionary? Youmight have found the meaning of function as something that has to do with events, pur-pose, duty or action. While for graph you might find the words diagram, visual imageor plot. However, these definitions are not enough when studying mathematics. Just aspreviously we found that algebra was an essential tool, so the concept of a function andthe technique of representing them with graphs is also essential. For this reason mathe-maticians have built on everyday language to develop specific definitions for these twoimportant tools. In this chapter we will explore and develop these definitions and socontinue to develop your practical mathematical skills.

Hint Form a study group.

Working alone might sometimes be difficult. Find some students you get alongwith and form a study group. External students should contact USQAssist(see yourintroductory book for details) to find out names and phone numbers of studentsin your area who are keen to do this.

What you can get out of a study group:

• two or three brains are better than one. Students come from a range ofexperiences and backgrounds and often look at problems from different per-spectives – this helps;

104 c© USQ, August 27, 2010

2.1. The concept of a function MAT1500

• you support each other when things are not going well;

• working together helps you avoid procrastination.

2.1 The concept of a function

2.1.1 What is a function?

In mathematics a function is an extremely useful construct. A function is like a lookuptable, for example a price list in a shop. Think about how this works. The assistant needsto find the price of a particular item. The item name or some other identification is usedas an ‘input’, the output is the price for that particular item. It hardly seems necessaryto mention in this example, but for any item there will be only one price. However thisproperty is an essential in our concept of a function. A function relates an input valuefrom one set of entities to a single value from another set of entities. For example:

• The function ‘birth date’ links a person (from the set of all people) to a date (fromthe set of all dates).

• The function ‘rounding up’ links the actual price of your purchases (from the set ofall possible prices) to what you actually pay (from the set of prices that are multiplesof five cents).

• The function ‘area of a square’ links the length of a side (from the set of possiblemeasurements) to the square of that number (from the set of possible numbers).

These examples illustrate that a function can link the input to the output via a table ofvalues, a rule, or as is very frequently the case a formula. Whichever is the case, for eachinput value there is only one output value. A function must be single valued.

2.1.2 Functional notation and representation

A function can be represented in any way that defines the links between members ofthe input set, the domain, and members of the output set, the range. We could define aparticular function by a table of values such as:

c© USQ, August 27, 2010 105

MAT1500 2.1. The concept of a function

Table 2.1: Table of values for two related variables.

l 1 2 3 4 5 6A 1 4 9 16 25 36

where l is the input value, or independent variable,

and A is the output value, or dependent variable.

We could also define this information as a set of ordered pairs: (1, 1), (2, 4),(3, 9), (4, 16), (5, 25), (6, 36) . Note it is often useful to use a specialised function nota-tion to name a function as in the diagram below.

So in the above example we have A = f (l) , that is A is a function of l. It is often simplywritten A = A(l) . Returning to the function in Table 2.1, note that the domain for thisfunction is the set 1, 2, 3, 4, 5, 6 , and the range is 1, 4, 9, 16, 25, 36 .

We can represent this function visually by plotting the ordered pairs on a plane. A sys-tem invented by the French mathematician and philosopher, René Descartes (1596-1650)makes this possible. This system is called the Cartesian coordinate system. When ap-plied to a plane, the plane may be called the Cartesian plane or the xy-plane.

To construct a Cartesian coordinate system in a plane, two mutually perpendicular linesare drawn intersecting at a point called the origin (the zero point). Units of length, notnecessarily the same, are marked off on each of these perpendicular lines as in Figure 2.1.

106 c© USQ, August 27, 2010


Figure 2.1: The Cartesian coordinate system

The perpendicular lines (shown as solid lines in Figure 2.1) are coordinate lines or axes.The vertical coordinate line is the y-axis and the horizontal coordinate line is the x-axis.

Any point P in the plane now has two numbers associated with it. Its x-coordinate (callit x1) and its y-coordinate (call it y1) and these are written as an ordered pair, (x1, y1) andcalled the coordinates of P.

The first number in the ordered pair is the displacement in a horizontal direction, andthe second number is the displacement in a vertical direction from the origin. Thus (1, 1)is the point 1 unit to the right and 1 unit up from the origin. The point (2, 4) is a point2 units to the right and 4 units up from the origin and so on. If we continue to plot thepoints representing the function A we get:

Figure 2.2: Points representing the function A(l) .

c© USQ, August 27, 2010 107


The general point in the Cartesian plane has coordinates (x, y) and the origin O is (0, 0).

Functions are frequently defined by a formula. For example the function listed in thetable and depicted visually in figure 2.2, could be defined thus:

A(l) = l2 , l ε 1, 2, 3, 4, 5, 6 ,

meaning that the output value A is the square of the input value l, where l takes valuesfrom the set 1, 2, 3, 4, 5, 6.The domain and range can equally well be continuous sets, where all numbers in a rangeare included rather than the discrete sets we have been looking at in this example. If wetake the function A(l), area of a square of side l, in reality a more appropriate domainwould be l ε (0, ∞) (that is, l can be any real number greater than 0). In such a case, a tableof values cannot give a complete description of the function, and the function is usuallydefined by a formula.

Visually, we can illustrate the continuous function over a portion of the domain with agraph as follows:

In this case, the curve on the graph is made up of every point whose coordinates (l, A)

satisfy the function A = l2 . Of course if your graph is accurate enough, you can read offthe value of A(l) for any value of l.

Example 2.1: Plot on a Cartesian plane the graph of the functionf (x) = x3 .

Solution:

x −2 −1 0 1 2

f (x) −8 −1 0 1 8

108 c© USQ, August 27, 2010


-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

-8

-6

-4

-2

2

4

6

8

Example 2.2: Plot on a Cartesian plane the graph of the functionf (x) = x + 1 .

Solution:

We have a continuous domain (all values between −2 and +2), and we only tabu-late a few points.

x −2 −1 0 1 2

f (x) −1 0 1 2 3

c© USQ, August 27, 2010 109


Example 2.3: If f (a) = a2 − 2a + 1 , find f (−1), f (a2), f (a + 1), and f (a + h).

Solution:

f (−1) = (−1)2 − 2×−1 + 1

= 1 + 2 + 1

= 4 .

f (a2) = (a2)2 − 2(a2) + 1

= a4 − 2a2 + 1 .

f (a + 1) = (a + 1)2 − 2(a + 1) + 1

= a2 + 2a + 1− 2a− 2 + 1

= a2 .

f (a + h) = (a + h)2 − 2(a + h) + 1

= a2 + 2ah + h2 − 2a− 2h + 1 .

Example 2.4: If we are given a function T = f (c) , which is used to predict the timein minutes required to mow a lawn from the capacity (c) of the engine in cubiccentimetres, explain the meaning of the expressions f (c + 5) and f (c) + 5.

Solution:

The expressions both look similar because they both involve adding 5. However,in f (c + 5) we are increasing the capacity of the lawn mower engine by 5 cm3 andthen calculating the predicted time to mow the lawn. In contrast, f (c) + 5 involvesadding 5 minutes to the time taken to mow the lawn given a certain capacity c ofthe lawn mower.

Example 2.5: Two students were asked to evaluate k(x + h), for the function k(x) = x2 +

2x + 1 . Discuss the correctness of the solutions presented below.

Solution:

First student:k(x + h) = x2 + 2x + 1 + h .

Second student:

k(x + h) = (x + h)2 + 2(x + h) + 1 ,

= x2 + h2 + 2x + 2h + 1 .

The first student did not substitute into the function properly. Instead of replacingeach x by x + h he just added h to the end of the function.

110 c© USQ, August 27, 2010


The second student has made the substitution into the function correctly by re-placing every value of x by the value x + h, however an error has occurred be-cause he has expanded (x + h)2 incorrectly. (x + h)2 is actually (x + h)(x + h) =

x2 + xh + xh + h2 = x2 + 2xh + h2 . The correct solution should be:

k(x + h) = (x + h)2 + 2(x + h) + 1

= x2 + 2xh + h2 + 2x + 2h + 1 .

Example 2.6: If f (x) = x + 1 , and g(x) = x2− 1 , find the following functional combina-

tions: f (x) + g(x), f (x)− g(x), f (x)× g(x) andg(x)f (x)

.

Solution:

f (x) + g(x) = x + 1 + x2 − 1

= x + x2 , or

= x(x + 1) .

f (x)− g(x) = x + 1− (x2 − 1)

= x + 1− x2 + 1

= x− x2 + 2 , or

= −(x2 − x− 2)

= −(x− 2)(x + 1) .

f (x)× g(x) = (x + 1)(x2 − 1) , or

= (x + 1)(x− 1)(x + 1)

= (x + 1)2(x− 1) .

It is important not to confuse the brackets in the function notation with multiplication or

division. We cannot expand the left hand side to get f g and x2, because the expression is

not g times x but rather reads the value of the function at x.

g(x)f (x)

=x2 − 1x + 1

,

=(x + 1)(x− 1)

x + 1= x− 1 .

It is important not to confuse the brackets in the function notation with multiplication or

division. We cannot cancel out in the first line, because the expression is not g divided by f

but rather reads the value of the function at x in each case.

c© USQ, August 27, 2010 111


Example 2.7: When f (x) = x + 1 , and g(x) = x2 − 1 , find the value of f (g(x)).

This type of example is called a function of a function. Wherever you see x in the ffunction you replace it by the function g(x).

Solution:

f (x) = x + 1 , Insert the function g(x) into the f (x)function.

f (g(x)) = f (x2 − 1)= (x2 − 1) + 1 Evaluate the f (x) function. In this case

wherever you see x replace it withx2 − 1.

= x2 .

Example 2.8: If f (x) = x2 and g(x) =√

x , find the values of f (g(x)) and g( f (x)) andhence show that in this special case f (g(x)) = g( f (x)) = x .

Solution:

The left hand side (LHS) of the expression is

f (g(x)) = f (√

x)

= (√

x)2

= x .

The right hand side (RHS) of the expression is

g( f (x)) = g(x2)

=√

x2

= x .

As the left hand side equals the right hand side and both equal x, we have shownthat in this case

f (g(x)) = g( f (x)) = x .

Exercise 2.9: Write in words the meaning of the following mathematical expressions.

(a) f (12) = 3 .

(b) h(−2) = 12.75 .

Exercise 2.10: If p(t) = 3t2 − 2 find the value of the following.

(a) p(0.5)

(b) p(m + 2)

Exercise 2.11: If we coil a wire around a piece of metal and then put a current in the wire,the metal becomes magnetic. Assume that the function m = B(n) measures thestrength of the magnetic field produced when the coil of wire has n turns. Explainthe meaning of the following expressions.

112 c© USQ, August 27, 2010


(a) B(2n)

(b) B(n) + 2

Exercise 2.12: If p(m) = 2m− 3 and h(m) = 4m2 − 9 find the following functional com-binations.

(a) h(m)− p(m)

(b) p(m)× h(m)

(c)p(m)

h(m)

(d) p(h(m))

Exercise 2.13: If f (x) = 2x2 , and g(x) = x2 + 2 , evaluate f (g(x)) and g( f (x)).

Hint When making notes or summaries you could:

• use three columns, one for keywords, one for examples and one for explana-tions or rules;

• make a maths glossary, write formal definitions with explanations in your ownwords;

• make a mind or network map by putting the main idea at the centre of thepage with supporting points or details radiating out (you could do this withkey examples); or

• talk with other students about their summaries.

2.1.3 Inverse function

A function which reverses the effect of the original function, that is it takes the originaloutput value as its input, and produces the original input as its output, is called theinverse function.

For our example above, the inverse of A(l) (that is, l(A)) would be shown as a table:

Table 2.2: Table of values for inverse function of A(l).

A 1 4 9 16 25 36l 1 2 3 4 5 6

where A is the input value, and l is the output value.

Visually, it is represented by the graph below.

c© USQ, August 27, 2010 113


The inverse function expressed as a formula is

l(A) =√

A , A = 1, 4, 9, 16, 25, 36 .

In the continuous case, the function is expressed

l(A) =√

A , A ε (0, ∞) ,

and can be illustrated graphically by Figure 2.3. But when would we use the inversefunction? Think about this!

Have you ever wanted to reverse a function, for example calculate a Fahrenheit temper-ature from the Centigrade temperature instead of vice versa? We did something like thiswhen we changed the subject of a formula in Chapter 1. Finding the inverse function isanother perspective on this issue.

Let us consider the relationship between Centigrade and Fahrenheit temperatures. Wecould represent some of the temperatures by a set of ordered pairs.

114 c© USQ, August 27, 2010


Figure 2.3: Curve representing the function l(A) .

Centigrade (C) Fahrenheit (F)0 3210 5020 6830 8640 10450 12260 14070 15880 17690 194100 212

If Fahrenheit is a function ( f ) of Centigrade, for the first point we might say f (0) = 32 orfor the second point f (10) = 50 . If we wanted to go the other way and write centigradeas a function (g) of Fahrenheit, we would write the first point as g(32) = 0 , and thesecond point as g(50) = 10 .

Now because the second function g, has reversed or ‘undone’ the process involved in thefirst function f , it is the inverse function of f , and we use the notation g = f−1 .

f−1 is a special notation and is not the same as the index notation a−1 = 1a which we intro-

duced in Chapter 1. How do we tell which is which? The only way is by the context inwhich it is used. If you know f is the name of a function, f−1 means its inverse function.If on the other hand you know f represents a number, then f−1 means 1

f . So our set of

c© USQ, August 27, 2010 115


points would now become as shown below.Centigrade to Fahrenheit Fahrenheit to Centigrade

f (0) = 32 f−1(32) = 0f (10) = 50 f−1(50) = 10f (20) = 68 f−1(68) = 20f (30) = 86 f−1(86) = 30

......

f (100) = 212 f−1(212) = 100The original function The inverse function

The two functions are different but have the same meaning looked at from differentperspectives. We can continue this undoing process to find the formula for the Centi-grade/Fahrenheit relationship.

F = f (C) = 95 C + 32 , where F was the temperature in Fahrenheit and C the temperature

in Centigrade. If we want to undo the process we have to subtract 32 then multiply by 59 .

So if the original function is

f (C) =95

C + 32 = F ,then its inverse will be

f−1(F) =59(F− 32) = C .

Not all functions have an inverse. Remember at the very beginning we said that a func-tion had to be single valued. If a function connects more than one entity in the domain tothe same entity in the range, a relation that reverses that would not be singled valued.

In summary,

• if an inverse of the original function f (x) exists we call it f−1(x);

• the domain of the original function becomes the range of the inverse function;

• the range of the original function becomes the domain of the inverse function.

Application: business The manufacturing costs of producing x products isgiven by the function f (x) = 100 + 3x . The government has placed quotas onproduction and will allow the company to produce only 1 000 of each item. Whatis the formula for the inverse function and how would you interpret it?

The original function is f (x) = 100 + 3x , and valid only for the domain of 0≤ x ≤ 1 000. The restriction on the domain means that the range will also berestricted.

The value of the function at x = 1 000 is

f (1, 000) = 100 + 3× 1 000 = 3 100 ,

and at x = 0 the value of the function is

f (0) = 100 + 3× 0 = 100 .

116 c© USQ, August 27, 2010

2.2. Linear functions MAT1500

Thus the range is 100 ≤ f (x) ≤ 3 100.

The given function allows you to calculate the cost of producing a given numberof products.

To obtain the inverse function, we must reverse the processes acting on x in theoriginal function, by subtracting 100 and then dividing by 3, so

f−1(x) =x− 100

3.

The domain of the inverse function is also affected, as it is now equivalent to therange of the original function. The domain of the inverse function is 100 ≤ x ≤3 100 .

The value of the inverse function at x = 3 100 is

f−1(3 100) =3 100− 100

3= 1 000 .

Thus we see that the range of the inverse function is indeed the domain of theoriginal function.

The inverse function allows you to calculate the number of products produced ata given cost.

Points to remember

• A function relates an input value from one set of entities (the do-main) to a single value from another set of entities (the range).

• Functions must be single valued.

• The domain and range can be discrete or continuous sets.

• Functions may be described by one or more of a rule, formula,table of values, set of ordered pairs, graph.

• The Cartesian plane can be used to graphically represent somefunctions.

• A function which reverses the effect of the original function, thatis it takes the original output value as its input, and produces theoriginal input as its output, is called the inverse function.

2.2 Linear functions

A function in which the variable is only raised to a single power is called a linear function.That is a function described by an equation of the form y = f (x) = mx + c, where m andc are constants. The graph representing a linear function is a straight line (which is whyit is called a linear function). The equation expressing the relationship between the y

c© USQ, August 27, 2010 117

MAT1500 2.2. Linear functions

coordinate and the x coordinate for every point on the line (y = mx + c) is called theequation of the line. When we plot a few linear functions, we see that the constants mand c have very specific meanings.

Example 2.14:

y = x + 1 ,

m = 1 , c = 1 .

y = 2x + 1 , m = 2 , c = 1 .

118 c© USQ, August 27, 2010


y = −x + 2 , m = −1 , c = 2 .

From the above example, we see that c is the intercept on the y-axis, that is, the value ofy when x = 0 . On a graph, it is the point where the line crosses the y-axis.

We also see that m is the slope or gradient of the line, that is, the change in y for a unitchange in x, sometimes called the rate of change. The gradient puts a value on the steep-ness of straight line by comparing the change in height with the change in horizontaldistance, that is the increase in y divided by the increase in x.

c© USQ, August 27, 2010 119


Gradient (m) =change in height (increase in y)

change in horizontal distance (increase in x)

=∆y∆x

=y2 − y1

x2 − x1

=f (x2)− f (x1)

x2 − x1.

Note that ∆x is a mathematical abbreviation for the ‘the change in x’.

A line that slopes from the bottom left to top right will have a positive slope, while onethat slopes from top left to bottom right will have a negative slope.

Since m =y2 − y1

x2 − x1, if we know any two points on a straight line graph, we can use them

to calculate its gradient.

The equation y = mx + c is the slope-intercept form of the equation of a straight line.

All points (x, y) which satisfy the equation of the line (that is, all pairs of x and y whichmake the equation true) lie on the line.

Lines with the same slope are parallel lines. Lines with a point in common are calledconcurrent or intersecting lines. Since the point of intersection is on both lines, its co-ordinates simultaneously satisfy both linear equations represented by the two lines. Thecoordinates of the point of intersection are therefore the solution to the system of equa-tions representing the lines. Lines that are both parallel and concurrent are coincident,that is, their graphs lie on top of each other. There are infinite numbers of solutions to thesystem representing the equations of coincident lines as there are an infinite number ofpoints common to the lines.

Note that parallel lines never intersect, so they do not have a point in common. Thusthere is no solution to the system of equations representing the lines.

120 c© USQ, August 27, 2010


Note that intersecting lines have only one point in common. Hence there is a uniquesolution to the system of equations representing the lines.

Application: exercise physiology Heart volume is directly related to theamount of oxygen transported to the muscles. Experimental observations revealthe linear relationship, V = 133.3C + 300 , where V is the heart volume (in mL)and C is the maximal oxygen consumption (in litres/minute).

Example 2.15: Find the slopes and intercepts of the lines below and sketch their graphs.

(a) y = 2 + 3x

(b) y = 3− x

(c) x = 10− 2y

(d) 3x− 4y− 5 = 0

Solution:

(a) y = 2 + 3x

The y intercept is 2, and the slope is 3.

c© USQ, August 27, 2010 121


The easiest points to find to sketch the graph of a straight line are the pointswhere the graph cuts the axes. We know the y intercept is 2. Thus one of thesepoints is (0, 2).

The point where the graph cuts the x axis is where y = 0 .

When y = 0 ,

0 = 2 + 3x ,

−2 = 3x ,

x =−23

.

Thus the x intercept is − 23 , and (− 2

3 , 0) is another point on the line.

Note that we could have used any value of x, say x = 1 , and found the corre-sponding y value (that is, y = 2 + 3× 1 = 5, and thus (1, 5) is another pointon the graph).

(b) y = 3− x

The y intercept is 3, and the slope is −1.

From the y intercept, (0, 3) is on the line. Similarly, when y = 0 , 0 = 3− x ,thus x = 3 is the x intercept and (3, 0) lies on the line.

122 c© USQ, August 27, 2010


(c) x = 10− 2y Rearranging the equation to slope-intercept form,

y = 5− 12

x .

Thus the y intercept is 5, and the slope is − 12 .

Two points on the line are (0, 5) and (10, 0).

(d) 3x− 4y− 5 = 0 Rearranging the equation to slope-intercept form,

y = −54+

34

x .

Thus the y intercept is − 54 and the slope is 3

4 .

Two points on the line are (0, − 54 ) and ( 5

3 , 0).

c© USQ, August 27, 2010 123


2.2.1 Slope-point equation

If we know the slope, m, of a line and one point on the line, say (x1, y1), then the equationof the line may be found using the following procedure.

If (x2, y2) is any other point on the line, then the change in y in moving from (x1, y1) to(x2, y2) is y2 − y1, and the corresponding change in x is x2 − x1.

Thus the change in y for a unit change in x is

m =y2 − y1

x2 − x1=

riserun

.

But (x2, y2) is any other point on the line, so it may be replaced by (x, y) to obtain

m =y− y1

x− x1,

so y− y1 = m(x− x1) .

The equation for a line given the slope (m) and a point (x1, y1) is

y = y1 + m(x− x1) .

Example 2.16: Find the equations of the straight lines

(a) passing through (1, −1) with slope 1.

(b) passing through (2, 0) with slope − 12 .

(c) passing through (−3, 1) with slope 2.

124 c© USQ, August 27, 2010


Solution:

(a) Here x1 = 1 , y1 = −1 , and m = 1 , so the required equation is

y = −1 + 1(x− 1) ,

Therefore y = −2 + x .

(b) Here x1 = 2 , y1 = 0 , and m = − 12 , so the required equation is

y = 0− 12(x− 2) ,

Therefore y = 1− 12

x .

(c) Here x1 = −3 , y1 = 1 , and m = 2 , so the required equation is

y = 1 + 2(x− (−3)) ,

Therefore y = 7 + 2x .

Note that we should always check the answer by substituting x1 for x in the equa-tion. For example, in (c) put x = x1 = −3 and we find

y = 7 + 2(−3) = 1 = y1 ,

so we have the right equation.

Example 2.17: Find the equations of the lines parallel to x + 2y = 1 and passing through

(a) the point (2, 3).

(b) the origin.

Solution:

The slope of the line to be determined is the same as the slope of x + 2y = 1 , or inslope-intercept form,

y =12− 1

2x .

So the slope is − 12 .

(a) Here x1 = 2 , y1 = 3 , and m = − 12 , so

y = 3− 12(x− 2) ,

Therefore y = 4− 12

x .

(b) Here x1 = 0 , y1 = 0 , and m = − 12 , so

y = 0− 12(x− 0) ,

Therefore y = −12

x .

c© USQ, August 27, 2010 125


2.2.2 Point-point equation

We may need to find the equation of a straight line passing through two given points, say(x1, y1) and (x2, y2).

Let (x, y) be any other point on the line.

The slope, m, is given by

m =y− y1

x− x1or m =

y2 − y1

x2 − x1,

Thereforey− y1

x− x1=

y2 − y1

x2 − x1,

y− y1 =y2 − y1

x2 − x1(x− x1) .

The equation for a line given two points is

y = y1 +y2 − y1

x2 − x1(x− x1) .

Note this is the same as the slope-point equation except with

m =y2 − y1

x2 − x1.

Example 2.18: Find the equations of the lines passing through the points

(a) (1, 2) and (2, 1)

(b) (3, −4) and the origin

126 c© USQ, August 27, 2010


(c) (−2, 1) and (3,−1)

Solution:

(a) Here (x1, y1) = (1, 2) and (x2, y2) = (2, 1) , so

y = 2 +1− 22− 1

(x− 1)

= 2− (x− 1) ,

Therefore y = 3− x .

(b) Here (x1, y1) = (0, 0) and (x2, y2) = (3,−4) , so

y = 0 +−4− 03− 0

(x− 0) ,

Therefore y = −43

x .

(c) Here (x1, y1) = (−2, 1) and (x2, y2) = (3,−1) , so

y = 1 +−1− 1

3− (−2)(x− (−2))

= 1− 25(x + 2) ,

Therefore y =15− 2

5x .

Note you should check your substituting. Both the given points must satisfy theequation of the line.

Example 2.19: The distance a truck spring compresses is directly proportional to theload. If the spring is compressed 11.2 mm by the unloaded truck (3 200 kg) and18.9 mm by a combined mass of 5 400 kg of truck and load, find the equation relat-ing compression to mass and find the compression when the truck carries a load of1 500 kg.

Solution:

Let y be compression, and x be the mass. The line passes through (3 200, 11.2) and(5 400, 18.9).

y = 11.2 +18.9− 11.2

5 400− 3 200(x− 3 200)

= 11.2 +7.7

2 200(x− 3 200)

= 11.2 + 0.003 5(x− 3 200)

= 11.2 + 0.003 5x− 11.2

= 0.003 5x .

c© USQ, August 27, 2010 127


When x = 1 500 + 3 200 = 4 700 ,

y = 0.003 5× 4 700 = 16.45 mm .

The words ‘directly proportional’ in the question imply that the compression is aconstant multiple of the mass. We usually represent an ‘is proportional to’ relation-ship by the symbol ∝, for example y ∝ x. This equation infers that the relationshipbetween y and x may be represented by an equation of the form

y = mx .

This implies that the compression is 0 when the load is 0.

Once we realise this, only one point (other than (0, 0)) is necessary to find m, so letus use (3 200, 11.2), then

11.2 = 3 200×m ,

Therefore m =11.23 200

= 0.003 5 ,

so y = 0.003 5x .

2.2.3 Some special lines

1. Lines which pass through the origin (0, 0) have equations of the form

y = mx .

These lines have slope m and y intercept 0. For example,

y = 2x ,

y = −3x ,

y =12

x .

2. Lines which are parallel to the x-axes have equations of the form

y = c ,

where c is a constant. These lines have zero slope and y intercept c. For example,

y = 2 ,

y = −1 ,

y = −13

.

128 c© USQ, August 27, 2010


Note that y = 0 is the equation of the x-axis.

3. Lines which are parallel to the y-axis have equations of the form

x = k ,

where k is a constant. These lines have infinite slope and no y intercept. For exam-ple,

x = 1 ,

x = 2 ,

x =23

.

Note that x = 0 is the equation of the y-axis.

Example 2.20: Find the equation of the lines passing through the points

(a) (1, 2) and (−8, 2)

(b) (− 12 , 0) and (− 1

2 , 2)

Solution:

(a) We note that the y values are equal, that is, regardless of the value of x, y isalways 2. Thus the required line is

y = 2 .

(b) We note that here the x values are equal, that is, regardless of the value of y, xis always − 1

2 . Thus the required line is

x = −12

.

2.2.4 Distance between two points

The length of a straight line between two points (a, b) and (c, d) can be derived as follows.

c© USQ, August 27, 2010 129


L2 = (c− a)2 + (d− b)2 , [by Pythagoras]

Thus,

The distance between any two points (a, b) and (c, d) is

L =√(c− a)2 + (d− b)2 .

Application: physics Vectors are representations of quantities which have bothmagnitude and direction, for example, velocity is a vector quantity because it hasa magnitude of 30 km/hr (say) and a direction of 30N (say). A resultant vectorof a number of similar vectors is a single vector which would have the same effectas all the original vectors taken together. The magnitude of the resultant vectorcan be calculated using the distance formula giving a result of

R =√

R2x + R2

y .

Here Rx is the magnitude of the vector in the x direction and Ry the magnitudeof the vector in the y direction.

Example 2.21: Find the distance between the two points (−1, 4) and (−3, −2).

Solution:

Distance =√(−3 + 1)2 + (−2− 4)2

=√(−2)2 + (−6)2

=√

4 + 36

=√

40

≈ 6.325

130 c© USQ, August 27, 2010


Note that only the positive root is chosen as the distance must always be positive.

Example 2.22: An air traffic controller observes two aircraft on the radar screen. Oneis 7 km west and 3 km north and the other is 5 km west and 4 km north of thecontroller. Find the distance between the aircraft if they are at the same height.

Solution:

L =√(−5 + 7)2 + (4− 3)2

=√

4 + 1

=√

5

≈ 2.236 km.

Thus the aircraft are approximately 2.236 km from each other.

Example 2.23: If the first aircraft in the previous question is at a height of 3 km and theother at a height of 6 km, find the distance between them.

Solution:

Given that the horizontal distance between the two aircraft is√

5 km, and the ver-tical height between them is 3 km, we use Pythagoras’ Theorem to find the directdistance between the two aircraft.

We know that for a right-angled triangle with sides a and b and hypotenuse c,

c2 = a2 + b2 .

Clearly, we have a right-angled triangle in this situation, so we can use this formulato deduce the unknown distance, d,

d2 = 32 +(√

5)2

= 9 + 5

= 14 ,

d =√

14

≈ 3.742 km

c© USQ, August 27, 2010 131


Therefore there is a distance between the two aircraft of approximately 3.742 km.

Exercise 2.24: Plot the straight lines described by the following equations.

(a) y = 3x + 2

(b) y− 2 = 2(x− 3)

(c) y = 12 x

(d) y = 6

(e) y + 1 = −2(x− 4)

(f) y = 4x− 1

Exercise 2.25: Find the equations of the straight lines passing through the given points.

(a) (2, 6) and (4, 3)

(b) (1, 1) and (2, 4)

(c) (4, 8) and (2, 4)

(d) (11, 5) and (4, −2)

(e) (3, 2) and (5, −2)

(f) (1, 1) and (4, 10)

Exercise 2.26: Find the equations of the straight lines passing through the points belowwith the slopes, m:

(a) (2, 4) with m = 2

(b) (3, −2) with m = 12

(c) (1, 4) with m = 1

(d) (0, 0) with m = 3

(e) (1, −1) with m = −1

(f) (2, 1) with m = −2

Exercise 2.27: Find the equation of the straight line parallel to y = 3x− 2 passing through

(a) (2, 1)

(b) (1, 0)

(c) (0, 0)

(d) (−1, 3)

Exercise 2.28: Find the distance between the points in exercise 2.25 above.

Exercise 2.29: Find the distance between the points

(a) (4, −1) and (3, 4)

(b) (2, 0) and (0, 1)

132 c© USQ, August 27, 2010

2.3. Quadratic functions MAT1500

(c) (5, 1) and (1, −1)

Points to remember

• The equation defining a linear function is of the form y = mx + cwhere m is the gradient and c is the y-intercept of the straight linegraph representing the function.

• The gradient (or slope) m of the line is the change in y valuesdivided by the change in x values between two points on the line.

• Given the slope m and a point on the line (x1, y1), the equation ofthe line can be found from y = y1 + m(x− x1).

• Given two points on the line (x1, y1) and (x2, y2), the equation ofthe line can be found from

y = y1 +y2 − y1

x2 − x1(x− x1) .

• The distance between two points (a, b) and (c, d) is given by

L =√(c− a)2 + (d− b)2 .

2.3 Quadratic functions

Functions come in all shapes and forms. Some are easy to represent and interpret; othersare more difficult. Linear functions represent some of the easiest functions to understand.Other functions that are not linear, not too complex, and occur frequently in business andscience are quadratics.

The graph of a quadratic function is called a parabola (pronounced ‘pah/rab/oh/lah’).Let us quickly look and see what shape they are before we go into details. If you have agraphing calculator you may want to sketch them yourself first.

c© USQ, August 27, 2010 133

MAT1500 2.3. Quadratic functions

In economics the average cost equation for a particular production plant is C = f (Q) =

0.3Q2 − 3.3Q + 15.3 , where C is the average cost in dollars of producing each unit and Qis the quantity produced.

In electrical engineering, the thermal e.m.f . (E) corresponding to the temperature T for acopper-iron thermocouple is modelled by E = −0.019 2T2 + 6.88T.

In applied biology, the relationship C = f (t) = 30t2 − 240t + 500 is used to predict thenumber of bacteria C in a swimming pool, measured as count per cm3, given the time indays from treatment.

134 c© USQ, August 27, 2010


In physics, the height of a small object above the ground, thrown up with an initial ve-locity of 4.2 m/sec, can be predicted from the function,H = f (t) = −4.9t2 + 4.2t , where t is the amount of time in the air in seconds and H isthe height in metres above the ground.

Recall then that a quadratic function has a characteristic ‘satellite dish’ shape. It can befolded in half at its vertex which is called the turning point and thus the shape is said tobe symmetrical. The straight line which bisects the parabola through its turning point iscalled the axis of symmetry.

The general equation of this upward opening parabola is f (x) = ax2 + bx + c , where a ispositive.

c© USQ, August 27, 2010 135


The general equation of this downward opening parabola is f (x) = ax2 + bx + c , where ais negative.

Previously we have examined quadratic functions of the form y = f (x) = ax2 + bx + c .Let us have a look at another useful form of this function.

Consider the quadratic function

y = f (x) = 2x2 + 8x + 5 .

This may also be written in the form y = 2(x + 2)2 − 3 by using the technique of ‘com-pleting the square’ (Section 2.3.1).

In fact, any quadratic function can be written in the form

y = a(x− d)2 + e , (2.1)

where a, d and e are constants.

This function is symmetric about x = d as we see when we plot the graph of the function.The point (d, e) is the turning point of the parabola. If a is negative then the curve is saidto be downward opening or concave down. In this case, the turning point is a maximumvalue of the function. If a is positive the parabola is said to be upward opening or concaveupwards and the function will have a minimum turning point.

Application: civil engineering A single span arch is still a very commonstructure in today’s industry. Engineers commonly use a particular form of theequation of a parabola when dealing with a parabolic arch that is tied. The originof x and y are taken at one side of the arch rather than at the midpoint. Theequation would be:

y =

(4hL2

)x(L− x) ,

136 c© USQ, August 27, 2010


where L and h are detailed in the figure below.

Linear and quadratic functions are both examples of polynomial functions. Polynomialfunctions are of the form:

y(x) = anxn + an−1xn−1 + . . . + a1x1 + a0 ,

where the a’s represent constants, and n is called the order of the polynomial. So a linearfunction

y = mx + c

is a polynomial of order 1, and the quadratic function

y(x) = ax2 + bx + c

is a polynomial of order 2.

2.3.1 Completing the square

An algebraic technique called ‘completing the square’ is required to convert from theusual quadratic form to the one in equation 2.1. Consider the expansion of (x + p)2

where p is any constant,

(x + p)2 = x2 + 2px + p2 ,

Therefore (x + p)2 − p2 = x2 + 2px ,

that is, x2 + 2px = (x + p)2 − p2

Thus, if we want to replace an expression such as x2 + 8x by a squared term, then theconstant p is half of the coefficient of x. That is,

x2 + 2px = (x + p)2 − p2 ,

x2 + 8x = (x + 4)2 − 42

= (x + 4)2 − 16 .

c© USQ, August 27, 2010 137


Further examples are:

• x2 − 6x = (x− 3)2 − (−3)2 = (x− 3)2 − 9 .

• −2x2 + 4x = −2(x2 − 2x) = −2[(x− 1)2 − (−1)2] = −2(x− 1)2 + 2 .

Note: Make a special note of this technique now, we will return to it in the study of circleslater in this chapter.

2.3.2 Quadratic Equation Formula

The completing the square technique was used to derive the formula for finding the so-lution of quadratic equations. The general quadratic equation is

ax2 + bx + c = 0 .

Dividing through by a, the coefficient of x2 gives

x2 +ba

x +ca= 0 .

Applying the completing the square technique to the first two terms gives

x2 +ba

x =

(x +

b2a

)2

−(

b2a

)2

,

which when substituted into the quadratic equation above gives

(x +

b2a

)2

−(

b2a

)2

+ca= 0 .

Rearranging,

(x +

b2a

)2

=

(b

2a

)2

− ca

=b2

4a2 −ca

=b2 − 4ac

4a2 .

Take the square root of both sides and make x the subject

x +b

2a= ±

√b2 − 4ac

4a2

= ±√

b2 − 4ac2a

x = − b2a±√

b2 − 4ac2a

=−b±

√b2 − 4ac

2a,

138 c© USQ, August 27, 2010


which gives the formula:

x =−b±

√b2 − 4ac

2a.

2.3.3 Sketching parabolas

To sketch a parabola we must:

1. recognise that the equation is a quadratic function;

2. decide if the function is upward opening or downward opening by examining thecoefficent of the x2 term, a;

3. put the equation in the form y = a(x− d)2 + e to determine a, d and e;

4. mark the axis of symmetry, x = d , and the turning point (d, e) on a set of axes;

5. determine where the curve cuts the y-axis by putting x = 0 in equation 2.1 andmark it on the axes;

6. determine if the curve cuts the x-axis by putting y = 0 in equation 2.1 and solvingfor x. Mark these points, if any, on the axes; and

7. complete the sketch by joining the plotted points with a curved line and perhaps afew other points found by substituting a few selected values of x into the originalequation and finding the corresponding y values. Remember that the parabola issymmetrical about its axis of symmetry so only choose values of x on the left orright of the axis of symmetry.

Comment: The graph of y = ax2 + bx + c will cross the x-axis only if the quadratic equa-tion ax2 + bx + c = 0 has two solutions. Of course the two solutions are the points wherethe graph crosses the x-axis (y = 0 on the x-axis). The graph will just touch the x-axisif the quadratic equation has one solution, and will not cross the x-axis if there are nosolutions. (See Chapter 1.7.1 section on quadratic equations, particularly the role of thediscriminant.)

Example 2.30: Sketch the following parabolas.

c© USQ, August 27, 2010 139


(a) y = 2x2

(b) y = 2(x− 1)2 + 3

(c) y = 2(x + 1)2 − 3

(d) y = 1− 4x− 4x2

Solution:

(a) y = 2x2

i. Comparing y = 2x2 with equation (2.1), we have

y = 2(x− 0)2 + 0 .

Thus here a = 2, d = 0 and e = 0.

ii. The axis of symmetry is x = 0 (the y-axis). As a = 2 , which is greaterthan 0 the graph is concave up (or upward opening).

iii. Thus the turning point (0, 0) is a minimum point.

iv. When x = 0 , y = 0 ,⇒ (0, 0) is on the graph.

v. When y = 0 , x = 0 ,⇒ (0, 0) is on the graph.

vi. A few points help us sketch this parabola.

x 1 2 3

y = 2x2 2 8 18

(b) y = 2(x− 1)2 + 3

i. Comparing y = 2(x − 1)2 + 3 with equation (2.1), we have a = 2, d = 1and e = 3.

ii. The axis of symmetry is x = 1 . As a = 2 , which is greater than 0 thegraph is concave up.

iii. Thus the turning point (1, 3) is a minimum point.

140 c© USQ, August 27, 2010


iv. When x = 0 , y = 5 ,⇒ (0, 5) is on the graph.

v. The curve will not cut the x-axis as the minimum y value is 3. (Note: If wesubstitute y = 0 into the original equation we cannot find a solution.)

vi.x −1 −2

y = 2(x− 1)2 + 3 11 21

Note that the graph of y = 2(x− 1)2 + 3 is the same shape as the graph ofy = 2x2 , but the axis of symmetry has moved 1 unit to the right and theminimum point has moved 3 units up the axis of symmetry.

(c) y = 2(x + 1)2 − 3

As we know the graph of y = 2x2 , we can use its graph to determine the shapeof y = 2(x + 1)2 − 3 .

In this case the axis of symmetry will be moved 1 unit to the left (as we have2(x + 1)2 in the expression of y) and the minimum point will be moved 3 unitsdown the axis of symmetry (as we have−3 for e whereas previously we had 0).

c© USQ, August 27, 2010 141


(d) i.

y = 1− 4x− 4x2

= −4(x2 + x) + 1

= −4

((x +

12

)2

− 14

)+ 1

= −4(

x +12

)2

+ 2 .

So a = −4 , d = − 12 and e = 2 .

ii. The axis of symmetry is x = − 12 . As a = −4 which is less than 0, the

graph is concave down. The turning point (− 12 , 2) is a maximum point.

iii. When x = 0 , y = 1 ⇒ (0, 1) is on the graph.

iv. When y = 0 ,

−4(

x +12

)2

+ 2 = 0 ,

Therefore(

x +12

)2

=12

,

Therefore x +12

= ± 1√2

,

Therefore x = −12± 1√

2.

Thus x ≈ −1.2 , or x ≈ 0.2 .Note that the graph cuts the x axis in two places at (0.2, 0) and (−1.2, 0).

v. When x = 1 , y = 1− 4× 1− 4× 12 = −7 .Thus (1, −7) is also on the graph.

Exercise 2.31: Sketch the following quadratic functions showing maximum or minimumvalues.

(a) y = x2 + 4

(b) y = 2x2 − 3

142 c© USQ, August 27, 2010


(c) y = 3x2 + 2x + 4

(d) y = −2x2 + 3x + 9

(e) y = 3(x + 2)2

(f) y = 2x2 − x− 6

Application: physics In an experiment concerning momentum, two identicalballs were made to collide. The initial velocity of one is 0.747 m/s, while thesecond is −0.32 m/s. If the collision is perfectly elastic then the final velocity ofeach ball is determined by the following calculations.

Using concepts of conservation of momentum and energy, the equation involvingthe final velocity is determined.

0.747 = (0.32− v)2 + v2 ,

0.747 = 0.322 − 2× 0.32v + v2 + v2 ,

2v2 − 0.64v− 0.645 = 0 .

This equation could be solved by using the quadratic formula,

x =−b±

√b2 − 4ac

2a.

So the solution is:

v =0.64±

√(0.64)2 + 5.16

4= 0.16± 0.59

= 0.75 , or − 0.43 .

The final velocity is 0.75 m/s or −0.43 m/s.

In reality as physicists we would discard the first choice because it implies that theballs continue unchanged, that is, that no collision occurred.

This velocity equation could be solved by graphing a parabola of the form y =

2v2− 0.64v− 0.645 , and determining graphically where the curve intersected thehorizontal axis.

c© USQ, August 27, 2010 143

MAT1500 2.4. Rational functions — the rectangular hyperbola

Points to remember

• The equation defining a quadratic function is of the form y =

ax2 + bx + c.

• The equation defining a quadratic function can also be written inthe form y = a(x− d)2 + e.

• The graph representing a quadratic function is a parabola.

• In the two equations above,

– y = c is the y-intercept,

– x = d =−b2a

is the axis of symmetry,

– the turning point is (d, e),

– the sign of a determines if the graph is concave up or down.

2.4 Rational functions — the rectangular hyperbola

Rational functions are of the form

f (x) =Pn(x)Qm(x)

,

where Pn(x) and Qm(x) are polynomial functions of the order n and m respectively, andQ is not the zero function (means the function whose output is always zero). That is, afunction of the form

f (x) =axn + . . . + bx + cpxm + . . . + qx + r

.

For example, simple rational functions are

f (x) =x + 1

2x2 − 3x + 1,

orf (x) =

x + 1x− 2

.

In this section we will only consider the simplest of the non-trivial rational functions,those of the form:

f (x) =ax− dcx− b

.

But let’s start further back.

2.4.1 Graphing rational functions

In our investigation of functions so far, we have looked at functions that you have prob-ably seen in your past studies. Now let us look at something completely different. We allknow that when we drive anywhere, the speed we travel at is dependent on the distance

144 c© USQ, August 27, 2010

2.4. Rational functions — the rectangular hyperbola MAT1500

travelled and the time taken to get there. So if we were to travel 10 kilometres, we coulddetermine the average speed from the function,

S =10t

.

We say that the speed attained is indirectly related to the time taken. If we graph thisfunction by plotting points or using a graphing calculator over the domain t ≥ 0, we getthe following shape:

What is happening with this function? As you go faster and faster, you take less timeto travel the 10 kilometres. That is probably self evident. But what are the limits on thefunction? For example, when you drive, can you ever travel so fast that it will take youzero time? Probably not in this universe. On the other hand, will it take you so long toget there that you travel at zero speed (assuming that of course you actually leave)?

In this function when time gets very large (that is, t → ∞), speed will get extremelysmall but can never actually reach zero. We say that the limit of the function S as time tapproaches infinity is zero or in symbolic notation

limt→∞

S = 0 .

The values which the variables approach (but can never reach or touch) are called asymp-totes. In this function then, we have two asymptotes, t = 0 and S = 0 .

But there is much more to this function than we have discussed above. You might noticewhen you draw the graph that there appears to be another part of the graph in the 3rdquadrant. Let us investigate this in more detail by examining the more general function,

y =1x

,

for all real values of x. Consider this function and ask yourself the following questions:

• What is the shape of the graph?

• Where will it cut the vertical axis?

c© USQ, August 27, 2010 145


• Is it a continuous function?

• What happens when the variables become very large or small?

Sketch the graph of y =1x

for the domain of all real values of x by plotting some pointsor using a graphing calculator. Think about the characteristics of the graph as you do.

We were right. It certainly looks very different over the real domain from any graph wehave examined before, in that it comes in two parts and thus is not continuous every-where. Let us go through the questions above one at a time so we can get an understand-ing of the behaviour of this type of function.

• The graph is a curve in two parts in diagonally opposite quadrants of the plane. Itappears to have a discontinuity at x = 0 , when examined over the real domain.

• To find where this function cuts the vertical axis we would put x = 0 . If we do this,we get y = 1

0 . We know from previous work that this is one of the operations thatcannot be performed in arithmetic and so the expression is undefined. This meansthat the function will never cut the y-axis.

• To find where this function cuts the horizontal axis, we would put y = 0 . If wedo this in this function, we get 0 = 1

x . Again in the real number system we cannotfind a number such that when we divide 1 by it we will get zero. Thus the functioncannot take the value y = 0 . This means that the function will never cut the x-axis.

• From the graph, it appears that the function is not defined for all real numbers, buthas a break in the middle. This was confirmed when we attempted to evaluate thevertical and horizontal intercepts. The function is undefined for x = 0. Nor is itpossible to find a value of x such that the function is 0.

• If we investigate the behaviour of the graph as x → ±∞ , we find the functionapproaches zero. But how does it do this? Consider some points.

146 c© USQ, August 27, 2010


x y =1x

1 110 0.1100 0.011 000 0.00110 000 0.0001

As x gets very large (approaches positive infinity) the y value approaches zero fromthe positive direction, that is, from above the x-axis. We write this as

limx→∞

1x= 0+ .

x y =1x

−1 −1−10 −0.1−100 −0.01−1 000 −0.001−10 000 −0.0001

As x gets very small (approaches negative infinity) the y value approaches zerofrom the negative direction, that is, from below the x-axis. We write this as

limx→−∞

1x= 0− .

This behaviour means that y = 0 (the x-axis) must be a horizontal asymptote.

To investigate what happens when y gets very large or small, it is useful to inves-tigate what is happening around the vertical asymptote, that is, what happens as xapproaches zero.

Consider y becoming very large or small.

x y =1x

1 10.1 100.001 1 0000.000001 1 000 000

As x approaches 0 from the positive direction y becomes very large and approachespositive infinity. Write this as x → 0+ , y→ +∞ .

x y =1x

−1 −1−0.1 −10−0.001 −1 000−0.000001 −1 000 000

c© USQ, August 27, 2010 147


As x approaches 0 from the negative direction y becomes very small and approachesnegative infinity. Write this as x → 0− , y→ −∞ .

This behaviour assures us that x = 0 (the y-axis) must be a vertical asymptote.

Note that a short cut way to determine the position of the vertical asymptote is to thinkabout where the function is undefined. In this case, it will be undefined at x = 0, so thismust be the vertical asymptote.

Application: electrical engineering Ohm’s Law related the three quantitiesof voltage (V, in volts), current (I, in amps) and resistance (R, in ohms) in theequation

R =VI

.

When this equation is represented graphically over the domain I > 0, and rangeR > 0 when V = 1 , we get a graph as shown below.

Let us now generalise what we have learned about this type of function. A rectangularhyperbola represents a reciprocal relationship between y and x, that is

y =1x

.

However the same sort of behaviour would be exhibited by any relationship of the form

y− b =c

x− a.

Rearranging this we find the general form.

The general form of a rectangular hyperbola is

y =c

x− a+ b . (2.2)

When x approaches a, y tends to +∞ or −∞ depending on whether x approaches a fromthe positive or negative side. Under such conditions we say the curve has a vertical asymp-tote at x = a . As x tends to +∞, y approaches b, or−∞ and thus the curve has a horizontalasymptote at y = b .

148 c© USQ, August 27, 2010


The sketch below shows the important features of a graph of a hyperbola.

With a little manipulation,

y =c

x− a+ b =

bx + c− abx− a

=P(x)Q(x)

,

that is, clearly a rational function as defined at the beginning of this section. . .

Sketching hyperbolas

To sketch a hyperbola given the form in equation (2.2) we must

1. recognise from the equation that the curve will be a hyperbola shape and come intwo parts;

2. determine and plot the asymptotes x = a and y = b;

3. determine and plot the points where the curve cuts the y and x axes by putting x =

0 and y = 0 respectively, that is, the points(

0, b− ca

)and

(a− c

b, 0)

respectively;

4. find a few extra points by substituting some x-values and plot these; and

5. join all points with a smooth curve, bearing in mind that there are two distinct partsof the curve and they are mirror images of each other reflected in a line passingthrough (a, b) at an angle of either 135 or 45 to the horizontal asymptote.

Example 2.32: Sketch the graphs of:

(a) y =1x

c© USQ, August 27, 2010 149


(b) y =1

x− 2− 5

(c) y =3

1− x+ 2

Solution:

(a) i. Comparing y =1x=

1x− 0

+ 0 with equation (2.2) we have a = 0 , b = 0and c = 1 .Thus x = 0 is the vertical asymptote and y = 0 is the horizontal asymp-tote.The point where the asymptotes intersect is (a, b) = (0, 0) .

ii. The curve does not cut the x-axis as a− cb = 0− 1

0 , which cannot be deter-mined.The curve does not cut the y-axis as b− c

a = 0− 10 , which also cannot be

determined. (This is clearly the case because the axes are the asymptotes.)

iii.x 1 2 3 4 5

y = 1x 1 1

213

14

15

(b) i. Comparing y =1

x− 2− 5 with Equation (2.2) we have a = 2 , b = −5 ,

and c = 1 .Thus x = 2 is the vertical asymptote and y = −5 is the horizontal asymp-tote.The point where the asymptotes intersect is (2, −5).

ii. The curve cuts the x-axis at

x = a− cb

,

that is, at

x = 2− 1−5

= 215

,

therefore( 11

5 , 0)

is on the graph.

150 c© USQ, August 27, 2010


The curve cuts the y-axis at

y = b− ca

,

that is, at

y = −5− 12

= −512

,

therefore(0, − 11

2

)is on the graph.

iii.x 3 4

y = 1x−2 − 5 −4 − 9

2

(c) i. Comparing y =3

1− x+ 2 =

−3x− 1

+ 2 with equation (2.2) we have a = 1 ,b = 2 , and c = −3 .Thus x = 1 is the vertical asymptote and y = 2 is the horizontal asymp-tote.The point of intersection of the asymptotes is (1, 2).

ii. When x = 0 ,

y = b− ca

= 2− −31

= 5 ,

therefore (0, 5) is on the graph.When y = 0 ,

x = a− cb

= 1− −32

= 212

,

c© USQ, August 27, 2010 151


therefore ( 52 , 0) is on the graph.

iii.x 2 4

y = 31−x + 2 −1 1

Exercise 2.33: Sketch the following hyperbolas indicating vertical and horizontal asymp-totes.

(a) y =1

x− 2+ 3

(b) y =1

x + 3− 2

(c) y =2

x− 1+ 3

(d) y(x− 2) = 4

(e)1

y− 2= x− 3

(f) y = 1− 1x + 3

Exercise 2.34: Find the equation of the hyperbolas with the given asymptotes and pass-ing through the point supplied.

(a) Vertical asymptote is 3.

Horizontal asymptote is 2.

Passing through (1, 7).

(b) Vertical asymptote is −1.

Horizontal asymptote is −2.

Passing through (3, −4).

152 c© USQ, August 27, 2010

2.5. Circle — not a function MAT1500

(c) Vertical asymptote is 2.

Horizontal asymptote is −6.

Passing through (−3, 0).

Hint Organize your study:

Plan for 2 minutes

Act for 25 minutes

Survey for 3 minutes

Stop for a short break

Points to remember

• The general form of a rectangular hyperbola is

y =c

x− a+ b .

• The graph of the above equation has a vertical asymptote at x = aand a horizontal asymptote at y = b.

2.5 Circle — not a function

Circles occur in many branches of science and engineering from predicting the path ofan earthquake to determining how fluid flows in various situations. Although circles areuseful they are not functions. To see this look, at the graph in figure 2.4

Figure 2.4: Circles are not functions.

For many values of x, there are two corresponding values of y. That is, the relationshipis not single valued. However we could find a function which describes the upper semi-circle, and another function which describes the lower semicircle. With this limitation inmind we will press on with the relationship between x and y coordinates on a circle.

c© USQ, August 27, 2010 153

MAT1500 2.5. Circle — not a function

Figure 2.5: Representing a circle in Cartesian coordinates.

The relationship between x and y coordinates of points on a circle can be derived fromthe simple definition that every point on a circle is the same distance (the radius) from thecentre (Figure 2.5). The distance r from the centre (a, b) to any point on the circumference(x, y), is calculated from the formula for the distance between two points on a plane.

r =√(x− a)2 + (y− b)2

Thus the equation of a circle centred at (a, b) of radius r is given by:

(x− a)2 + (y− b)2 = r2 . (2.3)

In the special case of a circle centred at the origin, (2.3) becomes

x2 + y2 = r2 . (2.4)

When equation (2.3) is expanded we get an equation of the form

x2 − 2ax + y2 − 2by = c , (2.5)

where c = r2 − a2 − b2 .

154 c© USQ, August 27, 2010


Example 2.35: Find the position of the centre and the radius of the following circles.

(a)(x + 1)2 + (y− 2)2 = 5

(b)x2 + y2 − 4x + 2y + 4 = 0

Solution:

(a) Centre = (−1, 2), radius =√

5 = 2.2361 .

(b)

x2 + y2 − 4x + 2y + 4 = 0 ,

x2 − 4x + y2 + 2y + 4 = 0 ,

It is necessary to ‘complete the square’ (as in section 2.3.1) for x2 − 4x and fory2 + 2y .

[x2 − 4x + (−2)2 − (−2)2]+

[y2 + 2y + 12 − 12]+ 4 = 0 ,

[(x− 2)2 − (−2)2]+

[(y + 1)2 − 12]+ 4 = 0 ,

[(x− 2)2 − 4

]+[(y + 1)2 − 1

]+ 4 = 0 ,

(x− 2)2 + (y + 1)2 − 4− 1 + 4 = 0 ,

(x− 2)2 + (y + 1)2 − 1 = 0 ,

(x− 2)2 + (y + 1)2 = 1 .

Therefore the centre = (2, −1) and radius =√

1 = 1 .

Example 2.36: Find the points at which the circle

(x− 2)2 + (y− 1)2 = 5 ,

intersects the axes.

Solution:

The circle cuts the x-axis when y = 0 , that is,

(x− 2) + (−1)2 = 5 ,

(x− 2)2 + 1 = 5 ,

(x− 2)2 = 4 ,

x− 2 = ±2 ,

Therefore x = 4 or x = 0 .

Therefore the circle cuts the x-axis at x = 0 and x = 4 .

c© USQ, August 27, 2010 155

MAT1500 2.5. Circle — not a function

The circle cuts the y-axis when x = 0 , that is,

(−2)2 + (y− 1)2 = 5 ,

4 + (y− 1)2 = 5 ,

(y− 1)2 = 1 ,

y− 1 = ±1 ,

Therefore y = 2 or y = 0 .

Therefore the y-axis is cut at y = 0 and y = 2 .

Example 2.37: Find the equation of the circle which passes through the points (0, 0),(0, 3) and (1, 0).

Solution:

The general equation to a circle is (x− a)2 + (y− b)2 = r2 .

Given x = 0 and y = 0 ,

(−a)2 + (−b)2 = r2 ,

a2 + b2 = r2 .


(−a)2 + (3− b)2 = r2 ,

but we know that r2 = a2 + b2 , from above. Thus

a2 + (3− b)2 = a2 + b2 ,

(3− b)2 = b2 ,

9− 6b + b2 = b2 ,

9− 6b = 0 ,

9 = 6b ,

b =96= 1.5 .


(1− a)2 + (−b)2 = r2 ,

156 c© USQ, August 27, 2010


but we know that r2 = a2 + b2 , from above. Thus

(1− a)2 + b2 = a2 + b2 ,

(1− a)2 = a2 ,

1− 2a + a2 = a2 ,

1− 2a = 0 ,

1 = 2a ,

a = 0.5 .

Therefore the centre of the circle is (0.5, 1.5).

Now, r2 = a2 + b2 ,

Therefore r2 = 0.52 + 1.52

= 0.25 + 2.25

= 2.5 .

Thus the equation of the circle is

(x− 0.5)2 + (y− 1.5)2 = 2.5 .

Application: mechanical engineering Mohr’s Circle was once the leadingtool used to visualise relationships between normal and shear stresses, and toestimate the maximum stresses. Even today, Mohr’s Circle is still used by engineersall over the world. A chief benefit of Mohr’s Circle is that the principal stressesσ1 and σ2 and the maximum shear stress τmax are obtained immediately afterdrawing the circle. The circle is centred at the average stress value, and has aradius R equal to the maximum shear stress such that

y = ±√

R2 − (x− σavg)2 .

Plan stresses in two dimensions.

c© USQ, August 27, 2010 157

MAT1500 2.6. When two functions meet

Mohr’s Circle illustrates principal stresses and stress transformations viaa graphical format.

Exercise 2.38: For each of the following circles find (i) the position of the centre, (ii) theradius, and (iii) the points, if any, where it cuts the axes.

(a) (x− 3)2 + (y + 2)2 = 4

(b) (x + 1)2 + (y− 2)2 = 3

(c) x2 − 2x + y2 − 4y = 3

(d) x2 + x + y2 − 6y + 0.25 = 0

(e) x2 − 4x + y2 − 6y = 12

(f) x2 − 2x + y2 − 16y = 0

Points to remember

• The equation defining a circle is of the form

(x− a)2 + (y− b)2 = r2 .

• In the above equation (a, b) are the coordinates of the centre, andr is the radius.

• The above equation may also be of the form

x2 − 2ax + y2 − 2by = c .

2.6 When two functions meet

In Chapter 1, we looked at questions associated with equations with two unknown vari-ables. Let us revisit that section now and see the graphical approach to the solution. Theoriginal question in Chapter 1 was:

158 c© USQ, August 27, 2010

2.6. When two functions meet MAT1500

A city bakery sold 1 500 bread rolls on Sunday, with sales receipts of $1 460. Plainrolls sold for 90 cents each while gourmet rolls sold for $1.45 each. How many of eachtype of roll were sold?

Previously we solved the question by generating two equations and then solving themalgebraically. The equations were

P + G = 1 500 ,

and0.9P + 1.45G + 1 460 .

The final solution gave the number of gourmet rolls (G) as 200 and number of plain rolls(P) as 1 300.

However, each of the equations can be rewritten as a linear function and drawn on a setof coordinate axes. Rearranging the equations we will get:

P + G = 1 500 will become P = 1 500− G .

0.9P + 1.45G = 1 460 will become P =1 460− 1.45G

0.9.

Note we could have just as easily have made G the subject of the formula and graphed itas the dependent variable.

Graphing the two equations we obtain

So the point of intersection of two straight lines is the same as the solution to two simul-taneous equations with two unknowns. The graphical approach to solving simultaneousequations like the one above is often the first step in solving more complex questions inwhich the algebra is time consuming or impossible.

c© USQ, August 27, 2010 159


Example 2.39: Economics uses the graphical and algebraic approach in a type of analysiscalled break-even analysis. The break-even point for a business is the point at whichtotal revenue equals total costs. To determine the break-even point for a business acompany needs to consider two functions. For a specific company this might be:

The Cost function c(x) = 100 000 + 5x .

The Revenue function r(x) = 9x .

Solution:

In both of these equations, x is the quantity of goods. Note in break-even analysiswe are interested only in finding the value of the x-coordinate, that is, the quantityof goods that will result in the equality of the total revenue and total costs.

The x coordinate of the point of intersection or the break-even value of the quantityfrom the graph is 25 000 units. To test this algebraically substitute 25 000 into bothequations to ensure that it gives the same value for cost and revenue.

c(x) = 100 000 + 5x ,

c(25 000) = 100 000 + 5× 25 000

= 225 000 .

r(x) = 9x ,

r(25 000) = 9× 25 000 ,

= 225 000 .

So the break-even value of the quantity is 25 000 units.

Example 2.40: Solve the following systems of equations graphically (if it is possible).

(a)

2x + y = 5 ,

y = 7 + 2x .

160 c© USQ, August 27, 2010


(b)

x + y = 5 ,

x = 2− y .

Solution:

(a)2x + y = 5 , y = 7 + 2x .

Rearranging these two equations produces:

y = −2x + 5 , y = 7 + 2x .

The approximate solution from the graph is x ≈ −0.5 , y ≈ 6 .

Checking this approximate solution:

2(−1

2

)+ 6 = 5 ,

7 + 2(−1

2

)= 6 .

(b)x + y = 5 , x = 2− y .

Rearranging yieldsy = −x + 5 , y = −x + 2 .

c© USQ, August 27, 2010 161


Because the two lines are parallel to each other, the equations are inconsistentand there is no solution to the system.

Using this graphical method, we can obtain approximate solutions to linear equationswhich are not parallel or coincident. This same method can be extended to finding theapproximate points of intersection for non linear equations. But first let us practise doingsketches of unfamiliar non linear functions.

Example 2.41: Draw a graph of the function y = 3 − 1.5x , over the domain x = 0 tox = 3 .

Solution:

Pick some values of x and substitute into the function to develop a table of values.

x 0 0.5 1 1.5 2 2.5 3

y 2 1.775 1.5 1.163 0.75 0.244 −0.375

Remember that this is a curve function so you should connect the points with asmooth curve not a straight line.

162 c© USQ, August 27, 2010


Note that the y values are obtained by substituting the x values into the functiony = 3− 1.5x . For example, when x = 0.5 , y = 3− 1.50.5 = 3−

√1.5 = 3− 1.225 =

1.775 .

Example 2.42: Draw a graph of

y =2

x− 1+

3x− 2

.

Solution:

x y0 −3.50.5 −61 ±∞1.2 6.251.5 −21.8 −12.52 ±∞2.5 7.333 43.5 2.84 2.17

Note that we require extra points in the region x = 1 to x = 2 since the behaviourof the curve is unusual there.

Other points of interest include:

• when y = 0 ,2

x− 1+

3x− 2

= 0 , which solves to give x = 75 ;

• as x → +∞ , y→ 0 from above; and

• as x → −∞ , y→ 0 from below.

Remember that this is a curve function so you should connect the points with asmooth curve and not a straight line.

c© USQ, August 27, 2010 163


Example 2.43: Graph the function

f (x) =

x if − 1 ≤ x < 1 ,2 if x = 1 ,x + 2 if x > 1

.

Solution:

The domain of f is all the real numbers x ≥ 1. We use a filled circle • to indicatethat at x = 1 , the value of f is f (1) = 2 . We use an open circle to illustrate thatthe function does not assume the values 3 or 1 at x = 1 .

Hint Stressed about commitments?

Your study responsibilities will impact on your family, friends, work, and recre-ational life. Many times commitments will clash. What should you do?

• discuss your weekly and semester schedule with those involved;

• discuss the rewards and demands of study;

• arrive at a joint decision; then

• keep communicating.

Exercise 2.44: Draw graphs of the following functions over the ranges given.

(a) y =

(12

)x

, x = 0 to 4.

(b) y = 22x , x = −1 to 2.

(c) y = 0.5− 0.3x , x = 0 to 4.

(d) y =1

x− 2+√

x , x = 0 to 4.

(e) y = 3x−1.2 + 1 , x = 1 to 3.

164 c© USQ, August 27, 2010


Now that you can roughly sketch some unfamiliar non-linear functions let us use thisnew skill to find approximate solutions to some more complicated equations.

Application: biology Physiologists may be interested in the oxygen diffusionaround a capillary and develop a simple model that leads to an equation of theform

C(r) =Rr2

4K+ B1 log(r) + B2 .

Where R, K, B1, and B2 depend on the geometry, reaction rates, and other specificsof the problem. If some simple constants are included, the physiologist could findthe value of r by considering the following.

C(r) = 2r2 + 3 log r + 1 ,

2 = 2r2 + 3 log r + 1 ,

2r2 − 1 = −3 log r .

One way to solve this equation would be to approximate the solution graphicallyby drawing two graphs: y = 2r2 − 1 , y = −3 log r .

However in many situations, we cannot solve an equation or a system of equations byalgebraic means and we must find an approximate solution by other methods. Often it ispossible to get a rough approximation from carefully drawn graphs, just as we did whenwe had linear functions. Obviously we would not use the graphical method to obtainapproximate solutions to a system of equations if analytical methods are available whichwill yield the exact solution.

Consider the equation 2x − 1.5x = 0 . There is no algebraic method for solving thisequation. However, the equation can be rearranged as 2x = 1.5x . We can plot twosimple functions y1 = 2x and y2 = 1.5x on the same graph. At the point where thesegraphs intersect, the y-values must be the same, that is, y1 = y2 . Thus, at this point, wehave a solution of 2x = 1.5x . Then we have solved the equation 2x − 1.5x = 0 . Theaccuracy of this solution depends on the accuracy with which the graphs are plotted.

These curves must intersect between x = 0 and x = 1 , so we select values of x from 0 to1 and determine the corresponding values of y1 and y2.

c© USQ, August 27, 2010 165


x 0 0.2 0.4 0.6 0.8 1

y1 = 2x 0 0.4 0.8 1.2 1.6 2y2 = 1.5x 1 1.084 1.176 1.275 1.383 1.5

Thus the approximate solution is x = 0.65 .

Check that this solution satisfies (or at least nearly satisfies) the original equation.

Example 2.45: Solve the equation x2 − 3−√x = 0 for x.

Solution:

x2 − 3−√

x = 0

x2 − 3 =√

x ,

Let y1 = x2 − 3 , y2 =√

x .

x 0 0.5 1 1.5 2 2.5 3

y1 = x2 − 3 −3 −2.75 −2 −0.75 1 3.25 6y2 =

√x 0 0.707 1 1.225 1.414 1.581 1.732

166 c© USQ, August 27, 2010


Thus the solution is x ≈ 2.11 .

Example 2.46: Solve the equation1

x− 1+√

2x2 + 1− 5 = 0 for x.

Solution:

y1 =1

x− 1,

y2 = 5−√

2x2 + 1 .

(Recognise that y1 is a hyperbola.)

x y1 =1

x− 1y2 = 5−

√2x2 + 1

−4 −0.2 −0.74−3 −0.25 0.64−2 −0.33 2−1 −0.5 3.270 −1 41 ±∞ 3.272 1 23 0.5 0.644 0.33 −0.74

c© USQ, August 27, 2010 167


Note that there are three solutions.

These solutions are x ≈ −3.6, 1.4, 3.1.

Check that each solution satisfies the original equation.

The table below shows a number of possible equations that we may need to solve, andthe corresponding functions we must plot to do so.

Equation Functions to be plotted

x2 −√x = 0 y1 = x2 ,y2 =

√x .

x3 +1x= 0 y1 = x3 ,

y2 = −1x

.

2x+1 − x2 + 5x− 6 = 0 y1 = 2x+1 ,y2 = x2 − 5x + 6 .

x− 10x− 3

+ 4 y1 = x ,

y2 =10

x− 3− 4 .

Alternatively, y1 = x + 4 ,

y2 =10

x− 3.

168 c© USQ, August 27, 2010


Exercise 2.47: Use graphical methods to find approximate solutions to the followingequations:

(a) 0.5x = 0.3x .

(b) x2 + 3√

x = 2 .

(c) 3− x = x1.5 .

(d) 1 +1

x− 2=√

x2 + 1 .

Putting it all together! Surveyors use telescopes of one form or another. Thepurpose of a telescope in this area is to create for an observer a picture of thecross-wire’s position on the target with the greatest clarity and precision. Opticalfactors to be considered included resolving power, magnification, definition, eyedistance, size of pupil and field of view. Two of these are discussed below.

The ability of a lens to show detail is called resolving power. It is measured as thesmallest angular distance, expressed in seconds of arc, between two points just farenough apart to be distinguished as separate objects rather than a blurred one.The maximum resolving power that theoretically can be attained with a perfectlymade telescope depends entirely on the diameter of that part of the objective lensactually used (the effective aperture). The resolving power of an objective lens isindependent of magnification. It can be computed by the empirical formula:

R =140D

.

where R is the angle that can be resolved, in seconds, and D the diameter of thelens aperture in millimetres. For example if the objective lens of a certain telescopehas an aperture of 30 mm in diameter, its resolving power is about 4.7 seconds.The accepted standard for resolving power of a human eye is 60 seconds. Seediagram below:

1. The specification of telescope manufacturers could include a graph describingthe relationship between resolving power and diameter of the lens. Completea graph of this function.

2. What do you think the domain and range of the function should be?

3. What would be the resolving power of a 40 mm camera lens?

c© USQ, August 27, 2010 169

MAT1500 2.7. Review

4. Using a 40 mm lens how many times would the image have to be magnifiedso that it was resolvable by the human eye?

Points to remember

• Simultaneous equations can be solved graphically.

• The point where graphs intersect is the solution.

Well done! Another chapter complete! Here are a number of things to check.

• Have a close look at your action plan for study. Are you on schedule? Or do youneed to restructure your action plan or contact your tutor to discuss any delays orconcerns?

• Make a summary of the important points in this chapter noting your strengths andweaknesses. Add any new words to your personal glossary. This will help withfuture revision and the open book exam.

• Check your skill level by attempting the Post-test.

2.7 Review

• A function relates an input value from one set of entities (the domain) to a singlevalue from another set of entities (the range).

• A function which reverses the effect of the original function, that is it takes theoriginal output value as its input, and produces the original input as its output, iscalled the inverse function.

• The equation defining a linear function is of the form y = mx + c, where m is thegradient and c is the y-intercept of the straight line graph representing the function.

• Given the slope m and a point on the line (x1, y1), the equation of the line can befound from y = y1 + m(x− x1).

• Given two points on the line (x1, y1) and (x2, y2), the equation of the line can befound from

y = y1 +(y2 − y1)

(x2 − x1)(x− x1)

• The distance between two points (x1, y1) and (x2, y2) is given by

L =

√(x2 − x1)

2 + (y2 − y1)2

• The graph representing a quadratic function is a parabola.

170 c© USQ, August 27, 2010


• The equation defining a quadratic function can be of the form y = ax2 + bx + c ory = a(x− d)2 + e. In these equations y = c is the y-intercept, x = d = −b

2a is the axisof symmetry, the turning point is (d, e), the sign of a determines whether the graphhas a maximum or minimum.

• The equation defining a circle is of the form (x− a)2 + (y− b)2 = r2 where (a, b)are the coordinates of the centre, and r is the radius.

• The general form of a rectangular hyperbola is

y =c

x− a+ b .

• The graph of the above equation has a vertical asymptote at x = a and a horizontalasymptote at y = b.

• Simultaneous equations can be solved graphically, the point where the graphs in-tersect is the solution.

2.8 Post-test

1. Draw a graph of the following function from x = 1 to 4.

y =√

x− 1x

.

2. Find the slope and intercept of the line 4x − y = 3 . Sketch a graph of the linearfunction y(x).

3. Find the distance between the points (4, 3) and (2, 6).

4. Sketch the graph of the function y = (x− 3)2 + 1 .

5. What is the equation of the hyperbola with vertical asymptote at x = 3 , a horizontalasymptote at y = 2 , and passing through the point (2, 3)? Sketch a graph of thehyperbola.

6. Determine the centre and radius of x2 + y2 − 6x + 2y + 4 = 0 .

7. Indicate by a labelled sketch how you would graphically solve the equation (x −1)2 = 3 +

2x− 4

.

c© USQ, August 27, 2010 171


2.9 Solutions


2.9 (a) ‘The value of the function f at 12 is 3.’

(b) ‘The value of the function h at −2 is 12.75.’

2.10 (a) p(t) = 3t2 − 2

p(0.5) = 3× 0.52 − 2

= 3× 0.25− 2

= −1.25 .

(b) p(t) = 3t2 − 2

p(m + 2) = 3(m + 2)2 − 2

= 3(m2 + 4m + 4)− 2

= 3m2 + 12m + 12− 2

= 3m2 + 12m + 10 .

2.11 (a) This expression determines the strength of the magnetic field when there aretwice as many turns of wire.

(b) This expression gives the strength of the magnetic field when we leave theoriginal coil alone but add from an external source (for example a standardbar magnet) an extra two units of magnetic strength.

2.12 (a)

h(m)− p(m) = (4m2 − 9)− (2m− 3)

= 4m2 − 9− 2m + 3

= 4m2 − 2m− 6 .

(b)

p(m)× h(m) = (2m− 3)(4m2 − 9)

= 8m3 − 12m2 − 18m + 27 .

(c)

p(m)

h(m)=

2m− 34m2 − 9

= (2m− 3)

(2m− 3)(2m + 3)

=1

2m + 3.

172 c© USQ, August 27, 2010


(d)

p(h(m)) = p(4m2 − 9)

= 2(4m2 − 9)− 3

= 8m2 − 18− 3

= 8m2 − 21 .

2.13

f (g(x)) = 2 [g(x)]2

= 2[x2 + 2

]2, or

= 2[

x4 + 4x2 + 4]

.

g( f (x)) = [ f (x)]2 + 2

=[2x2]2

+ 2

= 4x4 + 2 .

2.24 Below are the two points that give the intercepts of each line and the axes. To drawthe lines, plot the points and join them with a ruler. An additional point is alsoincluded as a check.

(a) (0, 2), (− 23 , 0) and (3, 11).

(b) (0, −4), (2, 0) and (3, 2).

(c) (0, 0), (6, 3) and (8, 4). This line passes through the origin. Thus, the y inter-cept and x intercept are equal to 0.

(d) (0, 6), (4, 6) and (−2, 6). This is a horizontal line at y = 6 , thus there is no xintercept.

(e) (0, 7), ( 72 , 0) and (4,−1).

(f) (0, −1), ( 14 , 0) and (3, 11).

2.25 (a) y = 9− 32 x .

(b) y = 3x− 2 .

(c) y = 2x .

(d) y = x− 6 .

(e) y = 8− 2x .

(f) y = 3x− 2 .

2.26 (a) y = 2x .

(b) 2y = x− 7 .

(c) y = 3 + x .

(d) y = 3x .

c© USQ, August 27, 2010 173


(e) y = −x .

(f) y = 5− 2x .

2.27 (a) y = 3x− 5 .

(b) y = 3x− 3 .

(c) y = 3x .

(d) y = 3x + 6 .

2.28 (a)√

13 .

(b)√

10 .

(c)√

20 .

(d)√

98 .

(e)√

20 .

(f)√

90 .

2.29 (a)√

26 .

(b)√

5 .

(c)√

20 .

2.31 (a) Minimum at (0, 4).

(b) Minimum at (0, −3).

(c) y = 3(

x +13

)2

+113

.

Minimum at(−1

3,

113

).

(d) y = −2(

x− 34

)2

+818

.

Maximum at(

34

,818

).

(e) Minimum at (−2, 0).

(f) y = 2(

x− 14

)2

− 498

.

Minimum at(

14

, −498

).

2.33 (a) Vertical asymptote at x = 2 .

Horizontal asymptote at y = 3 .

When x = 3 , y = 4 .

174 c© USQ, August 27, 2010


(b) Vertical asymptote at x = −3 .

Horizontal asymptote at y = −2 .

(c) Vertical asymptote at x = 1 .


c© USQ, August 27, 2010 175


(d) Vertical asymptote at x = 2 .


(e) Vertical asymptote at x = 3 .


176 c© USQ, August 27, 2010


(f) Vertical asymptote at x = −3 .


2.34 (a)

y =−10x− 3

+ 2 .

(b)

y =−8

x + 1− 2 .

(c)

y =−30x− 2

− 6 .

2.38 (a) i. (3, −2) .

ii.√

4 = 2 .

c© USQ, August 27, 2010 177


iii. Touches at x = 3 .

(b) i. (−1, 2) .

ii.√

3 .

iii. y = 2 +√

2 and y = 2−√

2 .

(c) i. (1, 2) .

ii.√

8 .

iii. x = −1 and x = 3 , y = 2 +√

7 and y = 2−√

7 .

(d) i. (−0.5 , 3) .

ii. 3 .

iii. Touches at x = −0.5 .y = 3 +

√8.75 and y = 3−

√8.75 .

(e) i. (2, 3) .

ii. 5 .

iii. x = −2 and x = 6 , y = 3 +√

21 and y = 3−√

21 .

(f) i. (1, 8) .

ii.√

65 .

iii. x = 0 and x = 2 , y = 0 and y = 16 .

2.44 (a) Table of Values:

x y0 1

0.5 0.7071 0.5

1.5 0.3542 0.25

2.5 0.1773 0.125

3.5 0.0894 0.063

(b) Table of Values:

178 c© USQ, August 27, 2010


x y−1 0.25−0.5 0.5

0 10.5 21 4

1.5 82 16

(c) Table of Values:

x y0 −0.51 0.22 0.413 0.4734 0.4919

(d) Table of Values:

c© USQ, August 27, 2010 179


x y0 − 1

2

1 01.5 −0.7751.8 −3.6582.2 6.4832.5 3.5813 2.7324 2.5

(e) Table of Values:

x y1 4

1.5 2.8442 2.306

2.5 1.9993 1.803

2.47 (a) y1 = 0.5x , y2 = 0.3x .

x 0 0.2 0.4 0.6 0.8 1

y1 = 0.5x 0 0.1 0.2 0.3 0.4 0.5y2 = 0.3x 1 0.79 0.62 0.49 0.38 0.3

180 c© USQ, August 27, 2010


Thus the solution is approximately x = 0.78 .

(b) y1 = x2 , y2 = 2− 3√

x .

x 0 0.2 0.4 0.6 0.8 1

y1 0 0.04 0.16 0.36 0.64 1y2 2 0.658 0.103 −0.324 −0.683 −1


(c) y1 = 3− x , y2 = x1.5 .

x 1 1.2 1.4 1.6 1.8 2

y1 2 1.8 1.6 1.4 1.2 1y2 1 1.31 1.66 2.02 2.41 2.83


c© USQ, August 27, 2010 181


(d)

y1 = 1 +1

x− 2,

y2 =√

x2 + 1 .

x 2.2 2.4 2.6 2.8 3

y1 6 3.5 2.67 2.25 2y2 2.42 2.6 2.79 2.97 3.16



1. Graph of y =√

x− 1x

2. Slope = 4 , intercept = −3 .

182 c© USQ, August 27, 2010


3.√

13 .

4. Graph of y = (x− 3)2 + 1

5. Graph of y =−1

x− 3+ 2

-5 -4 -3 -2 -1 0 1 2 3 4 5

-6

-4

-2

2

4

6

6. Centre = (3, −1) , radius =√

6 .

c© USQ, August 27, 2010 183


7. Graphs of y = (x− 1)2 and y = 3 +2

x− 4

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

y = 3 +

2

x − 4

y = (x − 1)2

Solutions

184 c© USQ, August 27, 2010

Chapter 3

Exponential, logarithmic andtrigonometric functions

Chapter contents3.1 Exponential functions (ex, 10x and ax) . . . . . . . . . . . . . . . . . . . . . . . . 187

3.2 Logarithm as an inverse function . . . . . . . . . . . . . . . . . . . . . . . . . . 192

3.2.1 Properties of log functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

3.2.2 Working with logarithmic equations . . . . . . . . . . . . . . . . . . . . . . 203

3.2.3 Working with exponential equations . . . . . . . . . . . . . . . . . . . . . . 205

3.3 Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

3.4 Trigonometric identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

3.5 Triangle solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

3.6 Compound angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

3.7 Radian measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

3.8 Graphs involving trigonometric functions . . . . . . . . . . . . . . . . . . . . . 249

3.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

3.10 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

3.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258



Objectives


• demonstrate an understanding of the definition of logarithmic functions and expo-nential functions;

• use your calculator to evaluate log x, ln x, 10x and ex for any acceptable value of x;

• apply the logarithmic laws to simplify algebraic equations and formulae;

c© USQ, August 27, 2010 185

MAT1500

• define trigonometric ratios for any right-angled triangle;

• evaluate trigonometric functions of positive or negative angles of any size;

• evaluate positive or negative inverse trigonometric functions;

• solve problems involving trigonometric identities;

• apply the sine rule and cosine rule to find sides and angles for any triangle;

• find the area of any triangle;

• demonstrate an understanding of the meaning of radians and convert angles fromone system of units to the other;

• find the area of the sector of a circle; and

• know the general shapes of graphs of trigonometric, exponential and logarithmicfunctions.

Introduction

In the previous chapters, we looked at functions — both straight line and curved. Thereare other families of functions that form patterns besides these two. The families of func-tions we consider here, are exponential, logarithmic and trigonometric. Exponential andlogarithmic are relatively new functions being developed around the 1700’s, but the de-velopment of trigonometry has a long history. Like many things mathematical it is em-bedded in the practical, in this case surveying, astronomy or navigation. While it wasused as early as Babylonian times, and trigonometric tables were used in a form in Greektimes, it came into its own in the 15th century when trigonometry was related to otherareas of mathematics. Today these three families of functions are instrumental in thedevelopment of science and engineering.

Hint How to stop procrastination.

• Get rid of old excuses. Write down all your excuses on one side of a piece ofpaper. Start challenging the faulty reasoning behind each of the excuses.

• Give yourself a pep-talk.

“There is no time like the present."“The sooner I get it done, the sooner I can go out."“There is no such thing as perfectionism. It is an illusion that keepsme from doing what I have to do right now."“It is cheaper and less painful if I do it now rather than wait until itgets worse."

• Focus on the present and what positive steps you can take toward reachingyour goals.

186 c© USQ, August 27, 2010

3.1. Exponential functions (ex, 10x and ax) MAT1500

• Design clear goals — having goals too big can scare you away from starting.

• Set priorities — start at the top of the list and work your way down.

• Break large chapters into smaller more manageable sections.

• Get organised — have all your materials ready before you begin a task. Usea daily schedule.

• Commit yourself to doing the task and tell somebody else, like a study partner.

3.1 Exponential functions (ex, 10x and ax)

Have you ever considered how an epidemic grows, or how a fad or rumour spreads?Suppose we have a creature whose breeding patterns are such that its population doublesevery month. Beginning with a single pair, what will the population be at the end of theyear?

We can plot the growth as a table of values:

t months P t months P0 2 7 2561 4 8 5122 8 9 1 0243 16 10 2 0484 32 11 4 0965 64 12 8 1926 128

We call this type of growth exponential growth. The above table could also representthe number of ancestors you have from t generations ago. That is, 2 parents, 4 grandpar-ents, 8 great-grandparents, 16 great great-grandparents and so on, with 8 192 (great)11-grandparents.

We could express this function asP = P0 × 2t ,

where P0 is the starting population. We could plot the first few points of the function P =

2t as in Figure 3.1. The graph of the function y = 10x for the values x = 0, 1, 2, 3, 4, 5would differ only in the scale on the vertical axis. In fact, the same shape graph wouldbe drawn for any function of the form y = ax , where a is a positive number.

The exponential function is defined as y = ex , where e is the number 2.718 28. . . . Thenumber e is called the natural number. It is important in science because ex is the functionwhose rate of growth is exactly equal to its value. We will examine this more in Chapter6.

The curves representing y = ex and y = e−x are shown in Figures 3.2 and 3.3. Thesecurves are of particular importance in problems of growth and decay in many areassuch as biological populations, radioactivity, electrical charges, economics and statistics.

c© USQ, August 27, 2010 187

MAT1500 3.1. Exponential functions (ex, 10x and ax)

Figure 3.1: Exponential growth

Again the curves representing any function of the form y = ax would be the same shape.Note that both of these curves have a horizontal asymptote along the x axis.

Figure 3.2: Graph of y = ex .

Other examples would be:

• Population growth of rabbits where f is the number of rabbits and n the number ofseasons.

f (n) = 10n .

• The population growth of Australia, where P is the number of people in millionsand t is time in years.

P(t) = 19.5e0.016 3t .

• The time for a cup of coffee to cool, where C is the temperature in centigrade and t

188 c© USQ, August 27, 2010


Figure 3.3: Graph of y = e−x .

is time in minutes.C = 79.345e−0.016 6t .

• The amount of a drug in the body where a is the amount in milligrams and t is timesince ingestion in hours.

A = 25× 0.8t .

• The atmospheric pressure p in Pascals varies with the height (h) in kilometres abovesea level. p0 is the pressure at sea level.

p = p0e−0.15h .

• The charge q on a discharging capacitor is related to the time t, where Q0 is theinitial charge, R the resistance in the circuit and C the capacitance.

q = Q0et

CR .

Application: chemistry Some chemicals break down into different componentswhen in solution, so that the concentration of the original compound changes overtime. In 1864, Guildberg and Waage recognized that at a constant temperaturethe rate of this type of reaction followed an exponential decay curve. An exampleof this is the breakdown of di-nitrogen pentoxide into nitrogen oxide and oxygen.When this decomposition is graphed, we get the following figure.

c© USQ, August 27, 2010 189

MAT1500 3.1. Exponential functions (ex, 10x and ax)

The equation for this function is

C(t) = 0.87e−0.30t ,

where t is in hours and C(t) is concentration in moles per litre. The initial amountof chemical in the solution was 0.87 moles per litre.

Because the exponential function is a power of a number (the xth power of the numbere), its properties follow from the index laws. So:

exey = ex+y ,ex

ey = ex−y ,

eax = (ex)a .

Hint If you have trouble using your calculator, contact the Learning Centre orview the Foundation Mathematics website for access to calculator booklets whichhave instructions on what to do for the most common calculators.

Example 3.1: Use your calculator to find e1, e2, e3, e−2, e−3.

Solution:

e1 = 2.718 281 8 , (of course this is just e)

e2 = 7.389 056 099 ,

e3 = 20.085 536 92 ,

e−2 = 0.135 335 28 ,

e−3 = 0.049 787 07 .

190 c© USQ, August 27, 2010


Example 3.2: Use your calculator to evaluate y = 2.6× e−0.2x,when x = 2, x = 0.02, x = −0.1.

Solution:

When x = 2, y = 2.6× e(−0.2×2) = 2.6× e−0.4 ≈ 2.6× 0.670 = 1.74

When x = 0.02, y = 2.6× e(−0.2×0.02) = 2.6× e−0.004 ≈ 2.59(important to use brackets in these calculations).

When x = −0.1, y = 2.6× e(−0.2×−0.1) = 2.6× e0.02 ≈ 2.65

Example 3.3: Plot on the same set of axes the graphs of ex, 2ex, e2x, e−x.

Solution:

ex 2ex e2x e−x

−3 0.05 0.10 0.00 20.09−2 0.14 0.27 0.02 7.39−1 0.37 0.74 0.14 2.720 1.00 2.00 1.00 1.001 2.72 5.44 7.39 0.372 7.39 14.78 54.60 0.143 20.09 40.17 403.43 0.05

Hint You will sometimes see the exponential function expressed differently; thisespecially occurs when large expressions are involved or in computer packages. Inthese cases

e(

x2+2xx3−3

)= exp

(x2 + 2xx3 − 3

).

Warning! Do not confuse this notation with the button on your scientific calculatorlabelled exp. This button means that the number entered will be multiplied by 10to some power.

c© USQ, August 27, 2010 191

MAT1500 3.2. Logarithm as an inverse function

Points to remember

• The exponential function is defined as y = ex , where e is thenumber 2.718 28.

• The graph of y = ex is a curve that increases with an ever increas-ing slope, crossing the y axis at 1 and asymptotes to the negative xaxis. The graph of y = e−x is a curve decreasing with an ever de-creasing slope which asymptotes to the positive x axis and crossesthe y axis at 1.

• exey = ex+y;ex

ey = ex−y; eax = (ex)a.

3.2 Logarithm as an inverse function

The inverse function to any function of the form y = ax is a logarithmic function. It iswritten y = loga x where the subscript a is called the base of the logarithm.

If you have forgotten what an inverse function is, return to Chapter 2.

So the inverse function y = 10x , is y = log10 x and inverse function of the exponentialfunction y = ex is y = loge x . These two special cases have their own terminology. Thefunction loge x is called the natural logarithm and is written ln x (standing for ‘log nat-ural’) and the function log10 x is called the common logarithm and the subscript 10 isomitted, that is, log x. Schematically we can illustrate these functions as follows.

p = ln ep

Natural Logarithmy = ln x

Find the power of e that gives x.i.e. find y such that ey = x

ep

Exponential Functiony = ex

Raise e to the power of x.

192 c© USQ, August 27, 2010

3.2. Logarithm as an inverse function MAT1500

p = log 10p

Common Logarithmy = log x

Find the power of 10 that gives x.i.e. find y such that 10y = x

10p

10 to power of xy = 10x

Raise 10 to the power of x.

p = loga ap

Logarithm to the base ay = loga x

Find the power of a that gives x.i.e. find y such that ay = x

ap

a to power of xy = ax

Raise a to the power of x.

Because the exponential and natural logarithm functions are inverse functions we canwrite

eln x = x and ln ex = x .

Similarly,10log x = x and log 10x = x ,

(Check the correctness of these on your calculator using x = 2)

and in generalaloga x = x and loga ax = x .

The graph of the function y = ln x is shown in Figure 3.4. The graphs of y = log x and,in fact, y = loga x will be exactly the same shape but with different scales on the axes.

Hint Change the formula into something more simple.

The relationship between exponentials and logarithms is often hard to remember.Why not try this instead:

102 = 100 is the same as log10 100 = 2 .

If you cannot remember this then it is easy to check it on your calculator, rememberthe 10x key is always behind the log key.

c© USQ, August 27, 2010 193


Figure 3.4: Graph of y = ln x .

Note that the log of zero or a negative number does not exist. Try it on your calculator;it will give you a big E. This means that the domain of the logarithmic function is all realvalues greater than zero, that is, x > 0. In fact this function will have a vertical asymptotealong the y-axis.

You will observe that the curve flattens out for very large values of x. This is usefulbecause it allows us to reduce large, clumsy numbers to smaller, more convenient valuesby considering their logs rather than the numbers themselves.

Typical examples of the use of logarithms are given below.

Scale Function

Decibel scale for sound levels Noise level = 10 log(

II0

), where I is the sound inten-

sity and I0 a bench mark sound level.

Chemical acidity (pH) pH = − log(H+) , where H+ is the hydrogen ion con-centration in moles per litre.

Richter Scale for earthquakes M = log(

WW0

), where W0 is the strength of a bench-

mark earthquake and W the strength of the currectearthquake.

The effectiveness of any volt-age amplifier is dependenton the frequency of the volt-age signals.

The occurs in a region of frequencies called the mid-band. The voltage gain in the midband, called themidband gain, is given by the expression:

G = 20 log(

vo

vi

),

where G is the midband gain in decibels, vo is outputvoltage, vi the input voltage.

194 c© USQ, August 27, 2010


Many graphs are expressed in logs along one or both axes for this reason, you will noticethese as you progress further in your studies.

Example 3.4:

Plot the graph of y = 2 ln x

Solution:

x 0 0.1 0.5 1 2 3 4 5 6

y Not defined −4.6 −1.4 0 1.4 2.2 2.8 3.2 3.6

Example 3.5:

Plot the graph of y = 2 + log x

Solution:

x 0.1 1 2 3 4 5 6 7 8 9 10

y 1 2 2.30 2.48 2.60 2.70 2.78 2.85 2.90 2.95 3

Example 3.6:

Slick industries and C .A .R .T .E .L . are battling it out for control of the grease andpalm oil business. C .A .R .T .E .L . currently has 80% of the market, but Slick’s tough

c© USQ, August 27, 2010 195


new advertising campaign promises to change that. Thus industry experts forecastthe sales for the two companies to be the following functions of time(t) in yearsfrom now.

SS = 20× 1.3t , SC = 80− 10 log(t + 1) .

(a) Graph the two functions on the same set of axes.

(b) From the graph approximate the time it will take for Slick’s sales to catch upto C .A .R .T .E .L .’s.

Solution:

t 0 1 2 3 4 5

SS 20 26 33.8 43.94 57.12 74.26SC 80 76.99 75.23 73.98 73.01 72.22

Approximate time is 4.9 years from now.

Application: biology We use logarithmic functions to simplify graphs of func-tions. They are commonly used in biology to study allometry, which is concernedwith the comparative sizes of different parts of living organisms. For example, in aspecies of mammal, brain volume (V in cm3) varies with body weight (W, in kg).

196 c© USQ, August 27, 2010


W V31 36536 38038 38241 39542 39745 41047 41048 41550 42053 42755 43757 440

We could graph this directly on a graph or alternatively we could graph the dataon log-log paper.

This type of graph has both a horizontal and vertical scale which have been logged(specifically logged to the base 10).

V = 114W0.33 ,

log V = log(114W0.33) ,

log V = log 114 + log W0.33 ,

log V = log 114 + 0.33 log W .

(Note that the next section will explain these operations.)

Hint Do not underestimate your memory.

I bet you can remember:

• your birthday;

• what schools you went to;

• your phone number; and

c© USQ, August 27, 2010 197


• your pin number.

So your memory is fine, you just have to make it work for mathematics. Here aresome things to help:

• stop underestimating your memory;

• make your subject come alive to you;

• make use of mnemonics; and

• remember intelligence is not fixed, you can improve it.

Exercise 3.7: Find the natural logarithms of the following:

(a) 2

(b) 4.6

(c)12

(d) 18.4

(e) 0.34

Exercise 3.8: Find the numbers whose natural logs are given below:

(a) 2

(b) 1.4

(c) −1.6

(d) 0.4

(e) −4

Exercise 3.9: Draw graphs of the following functions:

(a) y = 2 log x

(b) y = 10 + 3 ln x

(c) y = 10 + 3 log x

(d) y = e−2x

(e) y = 2ex

(f) y = ln x2

Exercise 3.10: Use graphical methods to find rough solutions to the following equations:

(a) x2 − 4x = ln x

(b) 2e−x − 1x + 3

+ 4 = 0

198 c© USQ, August 27, 2010


3.2.1 Properties of log functions

There are a number of laws describing the behaviour of log functions (regardless of theirbase number). If a is any base.

1. Adding logarithms.

By definition x = aloga x, y = aloga y , and xy = aloga xy.

But nowxy = aloga xaloga y = aloga x+loga y , (index laws)

soaloga xy = aloga x+loga y .

Thus

loga xy = loga x + loga y .

For example,

log 6 = 0.778

log 2 + log 3 = 0.301 + 0.477

= 0.778

thus log 6 = log 2 + log 3 .

That is, the log of the product of two numbers is equal to the sum of the logs of eachof the numbers.

Try a few more on your calculator to convince yourself this is true.

2. Subtracting logarithms.

By definition x = aloga x, y = aloga y , andxy= aloga

xy .

But nowxy=

aloga x

aloga y = aloga x−loga y , (index laws)

soaloga

xy = aloga x−loga y ,

thus

logaxy= loga x− loga y .

c© USQ, August 27, 2010 199


For example,

log 4 = 0.602

log 8− log 2 = 0.903− 0.301

= 0.602

thus log 4 = log 8− log 2 .

That is, the log of the quotient of two numbers is the difference of the logs of each ofthe numbers.

Try a few more on your calculator.

3. Multiplying by a constant.

By definition x = aloga x , and xy = aloga xy.

Butxy =

(aloga x

)y= ay loga x , (index laws)

soaloga xy

= ay loga x ,

and thus,

loga xy = y loga x

For example,

log 32 = log 9

= 0.954

2 log 3 = 2× 0.477

= 0.954

thus log 32 = 2 log 3 .

That is, the log of a number raised to a power is the power multiplied by the log ofthe number. A special case of this law is

loga1x= loga x−1 = − loga x .

For example,

log12

= −0.301

− log 2 = −1× 0.301

thus log12

= − log 2 .

200 c© USQ, August 27, 2010


4. Special values.

Since a = a1 , and 1 = a0 , we see that

loga a = 1 ,

(for example log 10 = 1 , ln e = 1), and

loga 1 = 0 ,

(including log 1 = 0 , ln 1 = 0).

Note that

• loga x approaches −∞ as x approaches zero from the positive side.

• loga 0 is undefined and loga(−x) , (where −x is a negative number) is alsoundefined.

5. Change of base.

Often we have to determine a logarithm of a number with a base other than 10 or e.As these are the only bases provided on calculators, we can change the base so thatan expression involving only the base 10 or the base e is obtained.

For any N and b , N = blogb N

Take the natural log of both sides,

ln N = ln blogb N ,

= logb N × ln b , by properties of log in part 3. above.

Therefore logb N =ln Nln b

.

We could equally well have taken common logs (or indeed logs of any other base)of both sides above giving:

logb N =ln Nln b

=log Nlog b

=loga Nloga b

Example 3.11: Use the logarithmic laws to simplify the following:

(a) log5 25 + log5 0.04− 2 log5 5

(b) ln e2 − 2 ln e

c© USQ, August 27, 2010 201


(c)2 log x− 2 log(xy)

log x

Solution:

(a) log5 25 + log5 0.04− 2 log5 5 = log5 52 + log5 5−2 − 2 log5 5= 2 log5 5− 2 log5 5− 2 log5 5 = −2.

(b) ln e2 − 2 ln e = 2 ln e− 2 ln e = 0.

(c)2 log x− 2 log(xy)

log x=

2 log x− 2 (log x + log y)log x

=−2 log y

log x.

(note that this cannot be simplified further)

Example 3.12: Evaluate log2 5 using:

(a) ordinary logarithms

(b) natural logarithms

Solution:

(a) log2 5 =log10 5log10 2

=0.699 00.301 0

= 2.322.

(b) log2 5 =ln 5ln 2

=1.609 40.693 1

= 2.322.

Exercise 3.13: Find the logarithms (to base 10) of the following:

(a) 3× 10−9

(b) 6.7× 10−2

(c) 2.8× 10−5

(d) 9.2× 10−3

(e) 5.4× 10−13

(f) 0.003 2

(g) 0.15

(h) 5.7× 103

Exercise 3.14: The acidity of a solution is measured by the symbol pH, where pH is− logof the concentration of the hydrogen ion, i.e.

pH = − log[H+]

202 c© USQ, August 27, 2010


Find:

(a) the pH if[H+]

is

i. 10−3

ii. 1

iii. 2× 10−5

(b)[H+]

if the pH is

i. 5

ii. 3.4

iii. 0.49

Exercise 3.15: The average height of females (y) in metres is given approximately interms of age (x) by:

y = 0.61 + 0.796 log(x + 1) , for x ≤ 20

(a) Find the average height of an 8 year old girl.

(b) At what age is the average height 1.5 metres?

Exercise 3.16: Use the log laws to simplify these expressions (if possible):

(a) log3 81 + log319

(b) 2 log9 3− 4 log9 2

(c) ln 25 + 2 ln 0.2

(d) 2 log 3 + log 6− 2 log65

(e)log 6 + log 3

log 3(f) log 4× log 7

3.2.2 Working with logarithmic equations

In many instances you might be presented with a logarithmic equation that needs to besolved for a given unknown.

Example 3.17: Solve the following logarithmic equations for x:

(a) log3 x = log3 6 + log3 4

(b) log (x + 1) = 3

c© USQ, August 27, 2010 203


Solution:

(a)

log3 x = log3 6 + log3 4

log3 x = log3 (6× 4)

log3 x = log3 (24)

x = 24 .

(b)

log (x + 1) = 3

log (x + 1) = log 103 , since 3 = 3 log 10 = log 103

x + 1 = 103

x = 103 − 1

x = 999 .

Example 3.18: Rearrange the following equations to make the variable x the subject ofthe formula:

(a) y = 2 + ln x

(b) y = log (x− 2)− 3

Solution:

(a)

y = 2 + ln x

y− 2 = ln x

ln x = y− 2 , change to exponent form

x = ey−2 .

(b)

y = log (x− 2)− 3

y + 3 = log (x− 2) , change to exponent form

10y+3 = x− 2

10y+3 + 2 = x

x = 10y+3 + 2 .

Exercise 3.19: Determine the value of x in the following equations:

(a) 6.78 = ln( x

234

)

(b) 0.018 9 = ln(5x)

204 c© USQ, August 27, 2010


(c) 44 = 22 + 10 log (2x)

(d) 3 log (3x + 1)2 + 1 = 10

Exercise 3.20: Rearrange to make the given variable the subject of the formula

(a) N = 10 log(

I0.01

), I

(b) G = 20 log(

vv0

), v

(c) t =ln(

1 000P− 1)

−0.7, P

(d) t =ln (1− T)−k

, T

Exercise 3.21: The voltage, y, in volts across an inductor in an electrical circuit varieswith time, t in milliseconds according to the equation

t = −10 ln( v

200

).

Rewrite the equation to calculate v for any value of t and hence find the value of vwhen t = 0.01 ms.

Exercise 3.22: The atmospheric pressure, in Pascals, varies with height, h in kilometresabove sea level according to the equation

h =

ln(

pp0

)

−0.15

If p0 = 105 Pascals, rewrite the formula to make p the subject and hence calculatethe atmospheric pressure at 0.25 kilometres above sea level.

3.2.3 Working with exponential equations

On their own the exponential and logarithmic functions are powerful tools to help de-scribe and measure natural phenomena. However, the inverse relationship between thefunctions allows us to extend their application even further.

Solving exponential equations

Previously, you might have solved exponential equations by trial and error.

When we had to solve 3x = 81 , we had to recall that 34 = 81 and thus say that x = 4 .

However, we now can solve these types of equations using our knowledge of the rela-tionship between logarithms and exponentials, and the logarithmic properties.

c© USQ, August 27, 2010 205


To solve 3x = 81 , we do the following.

As we have two expressions or numbers that are equal, their logarithms must also beequal. Our next step is to take logarithms of both sides (we could pick any logarithm butusually we choose either base 10 or base e).

log 3x = log 81 .

Using the log property that log mn = n log m , we have

x log 3 = log 81 .

Dividing both sides by log 3,

x log 3log 3

=log 81log 3

,

x =log 81log 3

.

Use your calculator to evaluate the logarithms of these numbers and then divide, givingx = 4 .Check: When x = 4 , LHS = 34 = 81 = RHS.

Example 3.23: Solve the following equation for a, 72a = 2.73 .Solution:

72a = 2.73 ,ln 72a = ln 2.73 , Take natural log of both sides.

2a ln 7 = ln 2.73 , Use power property to bring 2a to thefront of ln 7.

2a =ln 2.73

ln 7, Divide both sides by ln 7.

a =ln 2.732 ln 7

, Divide both side by 2.

a ≈ 0.258 1 .

Check: When a ≈ 0.258 1 , LHS = 72×0.258 1 = 2.73 = RHS.

Example 3.24: What values of p satisfy the following exponential equation?

2× e−0.1p + 3 = 4 .

206 c© USQ, August 27, 2010


Solution:

2× e−0.1p + 3 = 4 ,2× e−0.1p = 1 , Subtract 3 from both sides.

e−0.1p =12

, Divide both sides by 2.

ln e−0.1p = ln12

, Take log to base e of both sides.

−0.1p ln e = ln12

, Use power property to bring −0.1p to the front of ln e.

−0.1p× 1 = ln12

, Recall that ln e = 1.

−0.1p = ln12

,

p =ln 1

2−0.1

, Divide both sides by −0.1.

p ≈ 6.931 5 .

Check: When p ≈ 6.931 5 , LHS = 2× e−0.1×6.931 5 + 3 ≈ 4 = RHS.

Example 3.25: Rearrange the following formula to make x the subject.

L = 2e−2x + 7 .

Solution:

L = 2× e−2x + 7 ,

L− 7 = 2× e−2x , Subtract 7 from both sides.

L− 72

= e−2x , Divide both sides by 2.

ln(

L− 72

)= ln e−2x , Take logs to base e of both sides.

ln(

L− 72

)= −2x ln e Use power property to bring −2x to the

front of the ln e.

−2x = ln(

L− 72

), Recall that ln e = 1 .

x = −12

ln(

L− 72

). Divide both sides by −2.

Example 3.26: Use logs to make y the subject of the following equations:

(a) x = (a + 2b)6y−1

(b) x =ab2y

cy

c© USQ, August 27, 2010 207


Solution:

(a)

x = (a + 2b)6y−1 ,

log x = log(a + 2b)6y−1 ,

log x = (6y− 1) log(a + 2b) ,

6y− 1 =log x

log(a + 2b),

6y =log x

log(a + 2b)+ 1 ,

y =log x

6 log(a + 2b)+

16

.

(b)

x =ab2y

cy ,

log x = log(

ab2y

cy

),

log x = log a + log b2y − log cy ,

log x = log a + 2y log b− y log c ,

log x− log a = y(2 log b− log c) ,

y =log x− log a

2 log b− log c

orlog( x

a

)

log b2 − log c

orlog( x

a

)

log(

b2

c

) .

Example 3.27: The rate at which a chemical is generated in a reaction is given by

x = 11.6× 2.71−2.4t ,

where x is the amount of chemical present at time t.

Solution:

A graph of x against t is very difficult to draw. Use logs to obtain a simpler graph.

208 c© USQ, August 27, 2010


This function is a straight line graph of log x against t.

x = 11.6× 2.71−2.4t ,x

11.6= 2.71−2.4t ,

log( x

11.6

)= log(2.71−2.4t) ,

log( x

11.6

)= (1− 2.4t) log 2.7 ,

log( x

11.6

)

log 2.7= (1− 2.4t) ,

2.4t = 1− log( x

11.6

)

log 2.7,

t =1

2.4× log

( x11.6

)

log 2.7− 1

2.4,

=log( x

11.6

)

1.035− 0.417 ,

=log x− log 11.6

1.035− 0.417 ,

=log x1.035

− log 11.61.035

− 0.417 ,

=1

1.035log x− 1.445 ,

= 0.966 log x− 1.445 .

This gives a straight line with log x plotted on the horizontal axis and t on the ver-tical axis. The gradient of this line is 0.966 and the vertical intercept is −1.445.

Example 3.28: Radioactive material is known to decay to inert material at a rate given bythe exponential equation:

x = x0e−kt ,

where x is the mass of radioactive material remaining at time t, x0 is the initial mass,k is the decay time, and t is the time elapsed.

Find the time required for 10 kilograms of radioactive material with a decay con-stant of 1.3 (years)−1 to decay to a mass of 6 kilograms.

Solution:

x = x0e−kt ,

x = 10e−1.3t ,x

10= e−1.3t .

Taking natural logs of each side gives

ln( x

10

)= ln e−1.3t

= −1.3t ln e

= −1.3t , since ln e = 1 .

c© USQ, August 27, 2010 209


Thus

t = − ln( x

10

)

1.3.

When x = 6 ,

t = − ln 0.61.3

≈ −−0.51081.3

≈ 0.393 years.

Therefore the sample will take approximately 0.393 years to decay to a mass of6 kilograms.

Example 3.29: The world’s population is known to increase at a rate given by

P = AeBt ,

where P is the population in millions at time t, t is the time elapsed in years, A isthe population at time zero, and B is a constant.

If the population was 2 500 million in 1945, and 4 000 million in 1970, find B andestimate the population in 1985 and 2000.

Solution:

Let time be zero in 1945. Then A = 2 500 , and P = 4 000 when t = (1970− 1945) =25 .

Therefore 4 000 = 2 500× e25B ,

e25B =4 0002 500

= 1.6 .

Taking the natural log of each side produces

ln(e25B) = ln 1.6 ,

25B ln e = ln 1.6 ,

25B = ln 1.6 ,

B =ln 1.6

25

=0.4725

= 0.018 80 .

Therefore P = 2 500e0.018 80t .

210 c© USQ, August 27, 2010


In 1985, when t = (1985− 1945) = 40 ,

P = 2 500e0.018 80×40

= 2 500e0.752

= 2 500× 2.121

= 5 303 million.

In 2000, when t = (2000− 1945) = 55 ,

P = 2 500e0.018 80×55

= 7 031 million.

Thus in 1985 we estimate a population of 5 303 million people, and in 2000 weestimate the population has grown to 7 031 million people.

Application: environmental science Radioactive decay is modelled by theequation N = N0e−kt , where N represents the mass of the substance, N0 theinitial mass of the substance and t the time. If a certain radioactive substancehas a half-life of 5 years and 20 grams of it was initially secured, how much ofthe substance would be left after 10 years? If the substance could only be safelymoved in batches of 0.1 g, when would the original 20 g be safe to move?

We know that we started with 20 g so N0 = 20 , and we know that in 5 years weonly have half of this (10 g), so N = 10 when t = 5 . Thus we use our model,

N = N0e−kt ,

10 = 20e−5k ,

0.5 = e−5k ,

ln(0.5) = ln(e−5k) ,

ln(0.5) = −5k ln e ,

ln(0.5) = −5k ,

k =ln 0.5−5

,

k ≈ 0.138 629 .

Now that we know the value of k we can determine the amount of the substanceleft after 10 years.

N = 20e−0.138 629×10

= 20e−1.386 29

≈ 5 .

There would be approximately 5 g left after 10 years. We can also determine how

c© USQ, August 27, 2010 211


long it would take to have decayed to 0.1 g of the substance.

0.1 = 20e−0.138 629t ,0.120

= e−0.138 629t ,

ln(

0.120

)= ln e−0.138 629t ,

ln(

0.120

)= −0.138 629t ln e ,

ln(

0.120

)= −0.138 629t ,

t =ln( 0.1

20

)

−0.138 629,

t ≈ 38.22 .

Thus it would be safe to transport the substance after 38.22 years.

Exercise 3.30: Solve the following for x:

(a) 3.7 = 3× 42x + 1

(b) 3ex = 10

(c) 4e1−2x + 7 = 10

(d) 43x−2 = 26x+1

Exercise 3.31: Rearrange the following formula to make x the subject:

(a) t = 11.6× 2.71−2.4x

(b) p = 2e−0.1x − 7

(c) 4e1−2x + 7 = 10y

(d) y− 103x−2 = 10

Exercise 3.32: If p = 10log10 x, show that p = x .

Exercise 3.33: Use logs to transform the equations below into straight lines:

(a) y = 1.4x

(b) y = 6× 1.2x

(c) y = 6× 0.6−2x

(d) y = 2× 33−2x

Exercise 3.34: The current, I, in amperes, in a circuit with an inductor in series with aresistor is given by the equation

I = 4(

1− e−kt)

.

If when t = 0.15 s, I = 3.10A, find the constant k in the equation.

212 c© USQ, August 27, 2010


Exercise 3.35: A certain population grows exponentially. The population grows from3 500 people to 6 245 people in 8 years.

(a) What was the population at the end of the first year?

(b) How long will it take for the original population to double?

Exercise 3.36: Radioactive strontium-90 is used in nuclear reactors and decays accordingto

A = Pe−0.024 8t ,

where P is the amount present at t = 0 and A is the amount left after t years. Findthe time required for 10 kilograms to decay to 5 kilograms.

Exercise 3.37: A team of archaeologists thinks they may have discovered a fossilised tool,but they want to determine whether the fossil is authentic before they report theirdiscovery. The radioactive substance, Carbon 14, has a half-life of 5 730 years. Bymeasuring the amount of C14 present in a fossil, scientists can estimate how old thefossil is. Analysis of the tool determines that 15% of the radioactive substance hasalready decayed. How old is the fossil?

Exercise 3.38: A ‘well known power station worker’ has been losing his hair due toradioactive decay ever since he began working at the Springfield Nuclear PowerPlant. After his first five years on the job, he had lost 60% of his hair. As of todayhe has lost 94% of his hair. How long has he worked at the power plant?

Exercise 3.39: In a newly created wild life preserve it is estimated that Species A willtriple every five years and Species B will quadruple every three years. If initiallythe preserve was stocked with forty-six individuals of Species A and twenty-fourindividuals of Species B, determine two formulas from which the number of indi-viduals can be determined over time and hence use these formulas to determinehow long it will be before the numbers of each species will be the same.

c© USQ, August 27, 2010 213

MAT1500 3.3. Trigonometric functions

Points to remember

• The inverse function of y = ax is a logarithmic function writteny = loga x where a is the base of the logarithm.

• loge x is called the natural log and is written ln x.

• log10 x is called the common log and is written log x.

• The graph of a logarithmic function y = loga x is only defined forx > 0, asymptotes to the negative y axis crosses the x axis at 1,and continues to increase with an ever decreasing slope.

• loga xy = loga x + loga y, logaxy= loga x− loga y, and

loga xy = y loga x.

• loga a = 1, loga 1 = 0.

• Change of base: logb N =log Nlog b

=ln Nln b

=loga Nloga b

.

• Exponential and logarithmic functions are inverse functions ofeach other.

3.3 Trigonometric functions

So far we have examined functions that either smoothly increase or decrease, or are sta-tionary, but there is one group of functions which is characterised by their repeated na-ture. These are the trigonometric functions, also called periodic or circular functions be-cause of the way they repeat, and due to their relationship to angles within a circle. Letus look into these functions further.

There are six trigonometric functions: sine θ, cosine θ, tangent θ, cosecant θ, secant θ andcotangent θ, where θ is a variable. We abbreviate these functions as: sin θ, cos θ, tan θ,cosec θ, sec θ and cot θ. Sometimes they are called trigonometric ratios since they can beexpressed in terms of the sides of a right-angled triangle.

214 c© USQ, August 27, 2010

3.3. Trigonometric functions MAT1500

In terms of this triangle,

sin θ =ac

,

cos θ =bc

,

tan θ =ab=

sin θ

cos θ,

cosec θ =1

sin θ=

ca

,

sec θ =1

cos θ=

cb

,

cot θ =1

tan θ=

ba

.

Application: physics A vector is a quantity that has both magnitude anddirection. Examples of vectors would be force and velocity. Consider a crocodileswimming across a river at an angle of 45 to the bank with a speed of 5 m/s.

The magnitude of the velocity vector of the crocodile would be 5 m/s and thedirection would be 45. Physicists, and biologists for that matter too, would beinterested in the vertical and horizontal components of this vector. We wouldcalculate these using trigonometric ratios.

c© USQ, August 27, 2010 215


Horizontal component would be 5 cos 45 and the vertical component would be5 sin 45. These would represent the current of the river and the forward speed ofthe crocodile respectively.

These trigonometric functions can be evaluated on any scientific calculator. Note thatmost machines work in ‘decimal’ degrees rather than degrees, minutes and secondswhich must be converted to decimal fractions of degrees before calculation, for exam-ple,

sin 24 12′ 2′′ = sin(

24 +1260

+2

60× 60

)= sin 24.200 556 ≈ 0.409 93 .

Your calculator can accept positive or negative angles of any size (a negative angle meansthat it is measured clockwise instead of anticlockwise). However, the trigonometric func-tions of angles outside the range of 0 to 90, that is, outside Quadrant I below, cannot beinterpreted by the triangle diagram above. In fact many of the results are negative.

These values maybe found by considering the cases which follow.

• 90 to 180 (angles of quadrant II).

(Recall that the Cartesian coordinate system divides the xy-plane into 4 quadrants.)

In the second quadrant b is negative and a is positive.

Thus,

sin θ = sin φ =ac

,

cos θ = − cos φ = −bc

,

tan θ = − tan φ = − ab

.

Note that in this quadrant,sin φ = sin θ ,

so in generalsin(180 − θ) = sin θ .

216 c© USQ, August 27, 2010


• 180 to 270 (angles of quadrant III).

In the third quadrant, both a and b are negative which means that sin θ and cos θ

are both negative and tan θ is positive. Note that since

tan φ =ab= tan θ ,

then in generaltan(180 + θ) = tan θ .

• 270 to 360 (angles of quadrant IV).

In this quadrant a is negative and b is positive, thus cos θ is positive and sin θ andtan θ are both negative.

Note that since herecos θ = cos φ ,

in generalcos(360 − θ) = cos θ .

c© USQ, August 27, 2010 217


These results may be summarised by the diagram below which gives the positive ratios ineach quadrant.

Hint Mnemonics are helpful tools to remember some facts. Here are a couple tohelp remember the trigonometric ratios. It is always better to invent your own.That way they are easier to remember.

All Stations To Central,

All Students Take Chemistry,

All Silver Tea Cups,

alternatively, starting in the 4th quadrant, you can obtain a word,

CAST.

Thus there are usually two angles between 0 and 360 which have the same trigonomet-ric ratio (except for sin θ = ±1 , and cos θ = ±1). For example, if we wish to find θ suchthat sin θ = 0.5 , then θ = 30 or 150, since sin θ = sin(180 − θ).

Given sin θ we find θ by using the SIN−1 key which lies behind the sin key on yourcalculator. The result is an angle between−90 and 90. From this first value of θ, we mayfind the other angle between 0 and 360 which also satisfies the trigonometric equation.

For example if sin θ = 0.5, the calculator shows that θ = 30.

The other angle must be found using our knowledge of trigonometric relationships. Weknow sin θ = sin(180 − θ) .

Thus if θ1 = 30 gives sin θ1 = 0.5 , then θ2 = (180 − 30) also gives sin θ2 = 0.5 .

So the two angles which satisfy the trigonometric equation are 30 and 150.

Similarly to find two angles between 0 and 360 such that sin θ = −0.5 , we find sin−1(−0.5) =−30 from the calculator.

Therefore sin(180− (−30)) = 0.5 , which means the other angle must be 210. So thetwo angles which satisfy sin θ = −0.5 are −30 and 210. To obtain angles between 0

and 360 we note that −30 is the same as 330.

218 c© USQ, August 27, 2010


Thus the two angles are 210 and 330.

Hint Trigonometric notation is not like the other algebraic notation.

You may see sin−1 x = arcsin x , where arcsin x is an older notation for the inversefunction, which you will find in numerous books. They both mean, in words, ‘theangle whose sine is x’.

However, sin−1 x (the inverse of sine) is not the same as (sin x)−1, in fact(sin x)−1 =

1sin x

(the sine function raised to a power).

To avoid this confusion, we generally use (sin x)−1 = cosec x instead of using thenegative index.

Warning! Positive index notation still holds the original meaning,

sin2 θ = (sin θ)2 = sin θ × sin θ .

The other trigonometric ratios also have inverses, that is, cos−1 x, tan−1 x, sec−1 x, cosec−1xand cot−1 x, where x is a number. These inverse trigonometric ratios have a unique value.

This implies that an equation like sin θ = 0.5 has two solutions for θ in the range 0 to360 (namely 30 and 150) but has only one value, 30.

In order to find the second angle when solving trigonometric equations, you must knowthe diagram showing the positive quadrants and the relationships;

sin(180 − θ) = sin θ ,

cos(360 − θ) = cos(−θ) = cos θ ,

tan(180 + θ) = tan θ .

The values of the trigonometric ratios for some angles commonly used are shown in thetable below.

The values of sin θ, cos θ and tan θ for angles 0 to 360 given in the table must be notedfor future reference.

c© USQ, August 27, 2010 219


θ sin θ cos θ tan θ sec θ csc θ cot θ

0 0 1 0 1 ND ND

3012

√3

21√3

2√3

2√

3

451√2

1√2

1√

2√

2 1

60√

32

12

√3 2

2√3

1√3

90 1 0 ND ND 1 0

180 0 −1 0 −1 ND ND

270 −1 0 ND ND −1 0

360 0 1 0 1 ND ND

Note that ND means ‘not defined’.

Two triangles which are used by many people to help them remember the trigonometricratios for angles of 30, 60 and 45 are given below.

Example 3.40: Evaluate the following expressions.

(a) cos 72 4′

(b) sec 105.37

(c) tan(−15 23′ 32′′)

(d) tan−1 0.3

(e) cos−1 0.3

(f) tan−1 2.7

(g) sin−1(−0.6)

220 c© USQ, August 27, 2010


(h) cot−1 2

Solution:Note that there are two systems for measuring angles: radians and degrees. Ensurethat your calculator is switched to degree mode.

(a)

72 4′ =

(72 +

460

)

= 72.066 7 ,

cos 72.066 7 ≈ 0.307 9 .

(b)

sec 105.37 =1

cos 105.37

=1

−0.265 1= −3.772 9 .

(c)

15 23′ 32′′ =

(15 +

2360

+32602

)

= (15 + 0.383 333 + 0.008 889)

= 15.392 222 ,

tan(−15.392 222) = −0.275 3 .

(d)tan−1 0.3 = 16.699 .

(Remember, tan−1 x is not the same as1

tan x.)

(e)cos−1 0.3 = 72.54 .

(f)tan−1 2.7 = 69.68 .

(g)sin−1(−0.6) = −36.87 .

Note that 360 may be added to this if a positive answer is preferred, so that360 − 36.87 = 323.13 . The angle −36.87 is the same angle as 323.13.

(h) If cot θ = 2 , then1

tan θ= 2 ,

sotan θ =

12= 0.5 .

c© USQ, August 27, 2010 221


Thus cot−1 2 is the same angle as tan−1 0.5 = 26.57.

Example 3.41: Find all angles in a 360 circle which satisfy the trigonometric equationsbelow.

(a) sin θ = 0.3

(b) tan θ = 1.3

(c) cos θ = 0.8

(d) sin θ = −0.6

(e) tan θ = −2.7

(f) cos θ = −0.2

Solution:

(a) One solution is θ = sin−1 0.3 = 17.46 and the other is(180 − θ) = 180 − 17.46 = 162.54 .

(Check on your calculator that sine of both these angles is 0.3.)

(b) Solutions are

θ = tan−1 1.3 and (180 + θ)

= 52.43 and (180 + 52.43)

= 52.43 and 232.43 .

(Check on calculator.)

(c)

θ = cos−1 0.8 and (360 − θ)

= 36.87 and (360 − 36.87)

= 36.87 and 323.13 .

(Check on calculator.)

(d)

θ = sin−1(−0.6) and (180 − θ)

= −36.87 and (180 − (−36.87))

= −36.87 and 218.37 .

(Note that −36.87 is equivalent to 360 − 36.87 = 323.13 .)

(e)

θ = tan−1(−2.7) and (180 + θ)

= −69.68 and (180 − 69.68)

= −69.68 and 110.32 .

222 c© USQ, August 27, 2010


(Note that −69.68 is equivalent to 360 − 69.28 = 290.32 .)

(f)

θ = cos−1(−0.2) and (360 − θ)

= 101.54 and (360 − 101.54)

= 101.54 and 258.46 .

Example 3.42: A surveyor on the roof of a building observes that the top of anotherbuilding below forms an angle of 10 32′ with the horizontal whereas the base ofthe building forms an angle of 31 17′ to the horizontal. If the building are 250 me-tres apart, find the height of the observed building.

Solution:

From the triangle ABC,tan 10 32′ =

x250

,

therefore

x = 250 tan 10 32′

= 250 tan 10.533 33

= 250× 0.185 9

= 46.49 m.

From the triangle ABD,

x + h250

= tan 31 17′ ,

x + h = 250 tan 31 17′ ,

46.49 + h = 250 tan 31.283 333 ,

46.49 + h = 250× 0.607 6 ,

46.49 + h = 151.90 m ,

Therefore h = 151.90− 46.49

= 105.41 m.

c© USQ, August 27, 2010 223


Thus the height of the observed building is approximately 105.41 m.

Example 3.43: The angle to the top of a tower is 21 above horizontal and from anotherpoint 150 m closer it is 43. Find the height of the tower.

Solution:

Let x be the height of the tower and y be the distance from the second point to thetower.

From the triangle DBC,

tan 43 =xy

,

y =x

tan 43,

=x

0.9325.

From the triangle ABC,

tan 21 =x

150 + y,

x = tan 21(150 + y) ,

Therefore x = tan 21(

150 +x

0.932 5

),

x = 0.383 9(

150 +x

0.932 5

),

x = 0.383 9× 150 +0.383 9x0.932 5

,

x = 57.580 + 0.411 6x ,

x(1− 0.411 6) = 57.580 ,

0.588 4x = 57.580 ,

x =57.5800.588 4

,

≈ 97.9 m.

Thus the height of the tower is approximately 97.9 m.

224 c© USQ, August 27, 2010


Example 3.44: Simplify

(a) cos A× tan A× cosec A

(b)sec A

cosec A

Solution:

(a)

cos A× tan A× cosec A = cos A× sin Acos A

× 1sin A

= 1 .

(b)

sec Acosec A

=1

cos A÷ 1

sin A

=sin Acos A

= tan A .

Exercise 3.45: Evaluate

(a) sin(−22 15′)

(b) cos 36.42

(c) tan(−120 36′)

(d) sin 223.4

(e) tan 96.7

(f) cos 192 33′ 22′′

Exercise 3.46: Find the following angles.

(a) sin−1 0.4

(b) cos−1 0.2

(c) tan−1 1.9

(d) sin−1(−0.25)

(e) cos−1(−0.51)

(f) tan−1(−2.6)

Exercise 3.47: Find all angles between 0 and 360 which satisfy the following equations:

(a) tan θ = 1.3

(b) sin θ = 0.63

(c) cos θ = 0.13

(d) sin θ = −0.28

c© USQ, August 27, 2010 225

MAT1500 3.4. Trigonometric identities

(e) tan θ = −1.6

(f) cos θ = −0.84

Exercise 3.48: From one point on the ground the angle to the top of a tower is 32 abovethe horizontal. From a point 200 m closer, the angle is 64. Find the height of thetower.

Points to remember

• For the triangle below:

sin θ =ac

; cos θ =bc

; tan θ =ab=

sin θ

cos θ

cosec θ =1

sin θ=

ca

; sec θ =1

cos θ=

cb

; cot θ =1

tan θ=

ba

.

• sin(180 − θ) = sin θ; cos(360 − θ) = cos(−θ) = cos θ;tan(180 + θ) = tan θ.

• The inverse function sin−1 θ gives the angle θ whose sine is x.The inverse function cos−1 θ gives the angle θ whose cosine is x.The inverse function tan−1 θ gives the angle θ whose tangent is x.

3.4 Trigonometric identities

Some important relationships between the trigonometric ratios can be established usingPythagoras’ Theorem.

sin θ =ac

,

cos θ =bc

.

226 c© USQ, August 27, 2010

3.4. Trigonometric identities MAT1500

Thus,

sin2 θ + cos2 θ =a2

c2 +b2

c2

=a2 + b2

c2

=c2

c2

= 1 .

sin2 θ + cos2 θ = 1 .

Note that sin2 θ means sin θ multiplied by sin θ, or sin2 θ = (sin θ)2 . It does not meansin(θ)2.

Now, sin2 θ + cos2 θ = 1 may be written as

sin2 θ = 1− cos2 θ , or

cos2 θ = 1− sin2 θ .

Dividing by cos2 θ givessin2 θ

cos2 θ+ 1 =

1cos2 θ

,

thus

tan2 θ + 1 = sec2 θ .

Dividing by sin2 θ gives

cot2 θ + 1 = cosec 2θ .

Example 3.49: Simplify the following.

(a)1

cos2 θ− tan2 θ

(b) (sin x + cos x)2 + (sin x− cos x)2

(c)tan θ√

1 + tan2 θ

Solution:

(a)1

cos2 θ− sin2 θ

cos2 θ=

1− sin2 θ

cos2 θ=

cos2 θ

cos2 θ= 1

c© USQ, August 27, 2010 227

MAT1500 3.4. Trigonometric identities

(b)

sin2 x + cos2 x + 2 sin x cos x + sin2 x + cos2 x− 2 sin x cos x

= 2 sin2 x + 2 cos2 x

= 2(sin2 x + cos2 x)

= 2× 1

= 2 .

(c)

tan θ√sec2 θ

=tan θ

sec θ

=sin θ/ cos θ

1/ cos θ= sin θ .

Example 3.50: Find all angles between 0 and 360 which satisfy

(a) 2 cos2 θ − sin θ − 1 = 0

(b) sec2 θ = 3 + tan θ

Solution:

(a)

2(1− sin2 θ)− sin θ − 1 = 0 ,

2− 2 sin2 θ − sin θ − 1 = 0 ,

−2 sin2 θ − sin θ + 1 = 0 ,

2 sin2 θ + sin θ − 1 = 0 ,

(2 sin θ − 1)(sin θ + 1) = 0 ,

Thus sin θ = 12 or sin θ = −1 .

sin θ =12⇒ θ = sin−1 0.5 ,

Therefore θ = 30 and (180 − θ) = 150 .

sin θ = −1 ⇒ θ = sin−1(−1) ,

Therefore θ = −90 ,

and −90 is the same as 360 − 90 = 270.

Thus solutions are θ = 30, 150, and 270.

228 c© USQ, August 27, 2010

3.5. Triangle solution MAT1500

(b)

sec2 θ = 3 + tan θ ,

1 + tan2 θ = 3 + tan θ ,

1 + tan2 θ − 3− tan θ = 0 ,

tan2 θ − tan θ − 2 = 0 ,

(tan θ − 2)(tan θ + 1) = 0 ,

therefore tan θ = 2 or tan θ = −1 .

tan θ = 2 ⇒ θ = tan−1 2 ,

Therefore θ = 63.43 and (180 + θ) = 243.43 .

tan θ = −1 ⇒ θ = tan−1(−1) ,

Therefore θ = −45 and (180 − 45) = 135 .

and −45 is the same as 360 − 45 = 315 .

Thus solutions are θ = 63.43, 135, 243.43 and 315.

Exercise 3.51: If tan A = 2 , find cos A using the trigonometric identities.

Exercise 3.52: If sin B = 0.4 , find cos B and cot B using the trigonometric identities.

Exercise 3.53: Prove the following:

tan θ + cot θ =1

sin θ cos θ.

Exercise 3.54: Find all angles between 0 and 360 which satisfy

(a) cos2 θ + 4 sin2 θ = 2

(b) 2 cos2 θ = 3 sin θ

(c) 2 sin2 θ − 9 cos θ + 3 = 0

Points to remember

• sin2 θ + cos2 θ = 1;

• tan2 θ + 1 = sec2 θ;

• cot2 θ + 1 = cosec 2θ

3.5 Triangle solution

One of the most common applications of trigonometry is to the problem of triangle ‘so-lutions’. In this problem, some sides and angles of a triangle (which is not necessarily

c© USQ, August 27, 2010 229

MAT1500 3.5. Triangle solution

right-angled) are known and we want to find the remaining sides and angles.

In the triangle ACD, CD = b sin A .

In the triangle BCD, CD = a sin B .

Therefore b sin A = a sin B ,b

sin B=

asin A

,

Similarly,a

sin A=

csin C

.

These are combined in the sine rule,

asin A

=b

sin B=

csin C

,

where a, b, and c are the sides and A, B, C are the opposite angles ofany triangle.

To use the sine rule, we need any 2 sides and an opposite angle or any 2 angles and anopposite side.

Example 3.55: Find all angles and sides of the triangles:

(a)

(b) AC = 11 , BC = 9 , 6 A = 50.

230 c© USQ, August 27, 2010


Solution:

(a)

asin 32

=12

sin 50,

Therefore a =12 sin 32

sin 50

=12× 0.529 9

0.766 0= 8.301 .

6 C = 180− 32− 50

= 98 ,

Thereforec

sin 98=

12sin 50

,

c =12 sin 98

sin 50

=12× 0.990 3

0.766 0= 15.512 .

(b)

11sin B

=9

sin 50,

sin B11

=sin 50

9,

sin B =11 sin 50

9,

=11× 0.766 0

9= 0.936 3 ,

Therefore B = sin−1 0.936 3 or (180 − B)

= 69.435 or (180 − 69.435)

= 69.435 or 110.565 .

Note that both angles are possible so the triangle above is only one represen-tation of the given information.

c© USQ, August 27, 2010 231


If B = 69.435, then

6 C = 180 − 50 − 69.435 = 60.565 .

csin 60.565

=9

sin 50,

Therefore c =9 sin 60.565

sin 50

=9× 0.870 9

0.766 0= 10.232 .

If B = 110.565 , then

6 C = 180 − 50 − 110.565 = 19.435 .

csin 19.435

=9

sin 50

Therefore c =9 sin 19.435

sin 50= 3.909 .

Cosine rule

232 c© USQ, August 27, 2010


In the triangle BCD,

a2 = (b− x)2 + h2 ,

= b2 − 2bx + x2 + h2 ,

= b2 − 2bx + c2 .

Since in triangle ADB, x2 + h2 = c2 .

But x = c cos A . Substituting gives the cosine rule.

a2 = b2 + c2 − 2bc cos A .

To use the cosine rule, we require any 2 sides and the included angle, or all 3 sides.

Example 3.56: Find all sides and angles of the triangle below.

Solution:

Note that the only angle given is labelled C in the diagram which does not agreewith the labelling in the cosine rule. We may either re-label the diagram to makethis angle A or adjust the letters in the equation to give:

c2 = a2 + b2 − 2ab cos C

= 22 + 32 + 2× 2× 3 cos 120

= 4 + 9− 12× (−0.5)

= 19 .

Therefore c =√

19

≈ 4.359 .

Note that only the positive square root is feasible for the length of a side of a trian-

c© USQ, August 27, 2010 233


gle.

4.359sin 120

=3

sin B,

Therefore sin B =3 sin 120

4.359= 0.596 ,

Therefore B = sin−1 0.596 or (180 − B)

= 36.585 or (180 − 36.585)

= 36.585 or 143.415 .

From the diagram, 143.415 is clearly impossible. Thus B = 36.585.

A = 180 − 120 − 36.585 = 23.415 .

Example 3.57: A weight hangs from the junction of two ropes 4.4 metres and 3.6 metreslong respectively. If the other ends of the ropes are attached to a beam at points5 metres apart, find the angle between the ropes.

Solution:

52 = 4.42 + 3.62 − 2× 4.4× 3.6 cos θ ,

25 = 19.36 + 12.96− 31.68 cos θ ,

25 = 32.32− 31.68 cos θ ,

25− 32.32 = −31.68 cos θ ,

−7.32 = −31.68 cos θ ,

cos θ =7.32

31.68,

cos θ = 0.231 1 ,

Therefore θ = cos−1(0.231 1) or (360 − θ)

= 76.64 or 283.36 .

Obviously 283.36 is impossible. Thus, the angle is 76.64.

Application: mechanical engineering Statics is the study of forces actingon bodies at rest or moving at uniform velocity. A common situation when studyingsuch forces on a body can be represented by a vector diagram as below. Two forcesare applied to a body and we have to determine the resultant force.

234 c© USQ, August 27, 2010


y

x

100 N

150 N

R

120

α

If the 100 N force is called P and the 150 N force is Q, then using the cosine rulewe get

R2 = P2 + Q2 − 2PQ cos θ ,

R2 = 1002 + 1502 − 2× 100× 150× cos 120.

R = 217.9 N.

If we want to find the angle, α, between the resultant and 100 N forces, then wewould use the sine rule.

Qsin α

=R

sin θ,

sin α =QR

sin θ ,

sin α =150

217.9sin 120 ,

α = 36.59 .

Triangle area

The area of any triangle ABC, may be found as follows:

Area =12× base× height =

12

b× BD .

ButBD

c= sin A ,

Area =12

bc sin A .

c© USQ, August 27, 2010 235


Thus, to find the area we require any 2 sides and the included angle.

Application: surveying In surveying a traverse survey consists of a series ofadjoining lines, the bearings and lengths which are determined in the field and/orby calculation. A closed traverse is often used to delineate property boundaries,hence the importance of determining the boundaries that will give a fixed areaabout a particular point. Consider the surveying situation below.

An area of 14 650 m2 must be cut off by a line PX to delineate the boundariesof a property of fixed area. If bearings and measurements indicate that the anglePBX is 50 and the length of BP is 201.5 m, then surveyors would use the area ofa triangle formula to determine the length of BX.

∆Area =12

bc sin A ,

14 650 =12× 201.5× BX× sin 50 ,

BX =2× 14 650

201.5× sin 50,

BX = 189.818 53 m.

Surveyors would now use this to determine the boundaries of the property.

Example 3.58: A geological exploration company takes out a mining lease of the shapebelow. Find its area.

236 c© USQ, August 27, 2010


Solution:In triangle CDE,

2802 = 4502 + 5002 − 2× 450× 500 cos θ1 ,

78 400 = 202 500 + 250 000− 450 000 cos θ1 ,

78 400 = 452 500− 450 000 cos θ1 ,

78 400− 452 500 = −450 000 cos θ1 ,

−374 100 = −450 000 cos θ1 ,

Therefore cos θ1 =374 100450 000

= 0.831 3 ,

Therefore θ1 = cos−1 0.831 3 or (360 − θ1)

= 33.764 or 326.24 .

From the diagram, θ1 must be 33.764.

Area of triangle CDE =12× 450× 500 sin 33.764

= 112 500× 0.555 8

= 62 525 m2.

In triangle CAE,

3202 = 2802 + 3002 − 2× 280× 300× cos θ2 ,

102 400 = 78 400 + 90 000− 168 000 cos θ2 ,

−66 000 = −168 000 cos θ2 ,

cos θ2 = 0.392 9 ,

θ2 = 66.868 .

Note the other answer of 293.13 is impossible.

Area =12× 300× 280 sin 66.868

= 42 000× 0.919 6

= 38 623 m2 .

c© USQ, August 27, 2010 237


In triangle BCA,

3202 = 3002 + 2002 − 2× 300× 200 cos θ3 ,

102 400 = 90 000 + 40 000− 120 000 cos θ3 ,

−27 600 = −120 000 cos θ3 ,

cos θ3 = 0.23 ,

θ3 = 76.703 .

Area =12× 300× 200 sin 76.703

= 29 196 m2

Therefore total area of lease = 62 525 + 38 623 + 29 196

= 130 344 m2.

Exercise 3.59: Find the remaining angles of the triangle ABC given that:

(a) b = 4 , c = 2 , A = 30

(b) b = 5 , c = 4 , A = 42

(c) a = 4 , b = 7 , C = 110

Exercise 3.60: Find the remaining side of the triangle ABC given that:

(a) c = 56 , a = 74 , C = 30 15′

(b) b = 15 , c = 14 , B = 57 22′

(c) b = 6 , c = 5 , B = 53 8′

Exercise 3.61: Find the angle A in the triangle ABC with the sides:

(a) a = 2 , b = 4 , c = 3

(b) a = 9 , b = 5 , c = 6

Exercise 3.62: Find all the remaining angles and sides of the triangle ABC where:

(a) a = 2 , b = 4 , C = 30

(b) a = 6 , c = 7 , B = 75 31′

Exercise 3.63: Find the areas of the triangles in exercise 3.59.

Exercise 3.64: Find the area of the quadrilateral ABCD given that AB = 12 , BC = 10 , CD= 14 , DA = 8 , and AC = 16 .

238 c© USQ, August 27, 2010

3.6. Compound angles MAT1500

Points to remember

• Sine rule:a

sin A=

bsin B

=c

sin Cwhere a, b, and c are the sides

and A, B, C are the opposite angles of any triangle.

• Cosine Rule: a2 = b2 + c2 − 2bc cos A where a, b, and c are thesides and A, B, C are the opposite angles of any triangle.

• Area: Area =12

bc sin A =12

ac sin B =12

ab sin C where a, b, and care the sides and A, B, C are the opposite angles of any triangle.

3.6 Compound angles

cos A =x1

1= x1 ,

sin A =y1

1= y1 ,

cos B =x2

1= x2 ,

sin B =y2

1= y2 ,

The coordinates of P are (cos A, sin A) and of Q are (cos B, sin B). The distance PQ isgiven by

PQ2 = (cos A− cos B)2 + (sin A− sin B)2

= (cos2 A + cos2 B− 2 cos A cos B)

+(sin2 A + sin2 B− 2 sin A sin B)

= 1− 2 cos A cos B + 1− 2 sin A sin B

= 2− 2 cos A cos B− 2 sin A sin B .

c© USQ, August 27, 2010 239

MAT1500 3.6. Compound angles

From the cosine rule in triangle POQ,

PQ2 = 12 + 12 − 2× 1× cos(A− B)

= 2− 2 cos(A− B) ,

Therefore 2− 2 cos A cos B− 2 sin A sin B = 2− 2 cos(A− B).

cos(A− B) = cos A cos B + sin A sin B .

If we write (−B) for B we get

cos(A + B) = cos A cos B− sin A sin B .

We derive a similar rule for the sine of a compound angle as follows:

sin(A + B) = cos(90− (A + B))

= cos((90− A)− B)

= cos(90− A) cos B + sin(90− A) sin B .

Now, since cos(90− A) = sin A , and sin(90− A) = cos A , we get

sin(A + B) = sin A cos B + cos A sin B .

Similarly,

sin(A− B) = sin A cos B− cos A sin B .

If we put B = A, we obtain the special results for double angles. From above,

sin(A + A) = sin A cos A + cos A sin A .

Thus,

sin 2A = 2 sin A cos A .

Similarly,cos(A + A) = cos A cos A− sin A sin A .

240 c© USQ, August 27, 2010


Thus,

cos 2A = cos2 A− sin2 A .

A summary of the results discussed to this point is given below.

Summary: definitions and basic identities

sin θ =yr=

1cosec θ

,

cos θ =xr=

1sec θ

,

tan θ =yx=

1cot θ

,

sin(−θ) = − sin θ ,

cos(−θ) = cos θ ,

sin2 θ + cos2 θ = 1 ,

sec2 θ = 1 + tan2 θ ,

cosec 2θ = 1 + cot2 θ ,

sin 2θ = 2 sin θ cos θ ,

cos 2θ = cos2 θ − sin2 θ ,

cos2 θ =1 + cos 2θ

2,

sin2 θ =1− cos 2θ

2,

sin(A± B) = sin A cos B± cos A sin B ,

cos(A± B) = cos A cos B∓ sin A sin B ,

sin(A− 90) = − cos A ,

cos(A− 90) = sin A ,

sin(A + 90) = cos A ,

cos(A + 90) = − sin A .

c© USQ, August 27, 2010 241

MAT1500 3.6. Compound angles

Summary: angles and sides of a triangle

Cosine rule:c2 = a2 + b2 − 2ab cos C .

Sine rule:a

sin A=

bsin B

=c

sin C.

Area of a triangle:

Area =12

bc sin A =12

ac sin B =12

ab sin C .

Example 3.65: Simplify the following.

(a)sin(A + B)cos A cos B

(b) √1 + cos 2θ

1− cos 2θ

(c)sin 4θ − cos 2θ

1− cos 4θ − sin 2θ.

Solution:

(a)

sin A cos B + cos A sin Bcos A cos B

=sin A cos Bcos A cos B

+cos A sin Bcos A cos B

=sin Acos A

+sin Bcos B

= tan A + tan B .

242 c© USQ, August 27, 2010


(b)

√1 + cos 2θ

1− cos 2θ=

√1 + cos2 θ − sin2 θ

1− (cos2 θ − sin2 θ)

=

√1− sin2 θ + cos2 θ

1− cos2 θ + sin2 θ

=

√cos2 θ + cos2 θ

sin2 θ + sin2 θ

=

√2 cos2 θ

2 sin2 θ

=

√cos2 θ

sin2 θ

=cos θ

sin θ= cot θ .

(c)

sin 4θ − cos 2θ

1− cos 4θ − sin 2θ=

2 sin 2θ cos 2θ − cos 2θ

1− (cos2 2θ − sin2 2θ)− sin 2θ

=cos 2θ(2 sin 2θ − 1)

1− cos2 2θ + sin2 2θ − sin 2θ

=cos 2θ(2 sin 2θ − 1)

sin2 2θ + sin2 2θ − sin 2θ

=cos 2θ(2 sin 2θ − 1)

2 sin2 2θ − sin 2θ

=cos 2θ(2 sin 2θ − 1)sin 2θ(2 sin 2θ − 1)

=cos 2θ

sin 2θ= cot 2θ .

Exercise 3.66: Write the following in terms of sin A, cos A, sin B and cos B.

(a) cos(A + 2B)

(b) sin(2A + B)

(c) cos(2B− A)

Exercise 3.67: Express the following as a single trigonometric ratio.

(a) cos 24 sin 35 − sin 24 cos 35

(b) cos2 45 − sin2 45

(c) cos 32 cos 58 − sin 32 sin 58

(d) sin 21 cos 46 + cos 21 sin 46

c© USQ, August 27, 2010 243

MAT1500 3.7. Radian measure

Exercise 3.68: Prove the identities:

(a)

cot A + cot B =sin(A + B)sin A sin B

(b)sin(A− B)cos(A + B)

=tan A− tan B

1− tan A tan B

Points to remember

For two angles A and B:

• sin(A + B) = sin A cos B + cos A sin B

• sin(A− B) = sin A cos B− cos A sin B

• cos(A + B) = cos A cos B− sin A sin B

• cos(A− B) = cos A cos B + sin A sin B

• sin 2A = 2 sin A cos A

• cos 2A = cos2A− sin2A

3.7 Radian measure

Radians and degrees are alternative means of measuring the size of an angle.

Consider a circle of radius r. The angle θ will be one radian if it corresponds to an arc ofthe circumference of length r. This means that

1 radian ≈ 57.295 8 degrees ,

1c ≈ 57.295 8 .

Note that a radian is really a ratio of arc length to radius although it corresponds to ameasure of angle. Thus it may be mixed with other measurements in an equation whereasdegree measurements cannot. Often the little c superscript to denote radians is omitted.If an angle does not have the degree sign you must assume it is in radians.

244 c© USQ, August 27, 2010

3.7. Radian measure MAT1500

Radians may be converted to degrees and vice versa by use of the relationship

π radians = 180 .

For instance, 1 is 1 one hundred and eightieth of π. So

60 = 60× π

180=

π

3=

3.141 593 . . .3

≈ 1.047 198c .

It is not necessary to convert radians to degrees before using a calculator since most ma-chines work directly in either system. Most scientific and engineering problems are ex-pressed in radians.

Make sure your calculator is in the correct mode. In future work we will use radians.

Example 3.69: Convert

(a) 217 to radians.

(b) 2.64 radians to degrees.

Solution:

(a)217× π

180= 3.787c .

(b)

2.64× 180π

= 151.26 .

Length of arc of a circle

The length of the arc of a circle may be deduced as follows.

Let θ be the angle subtended at the centre of a circle of radius r by an arc of length l.

Then θ =lr

radians (by definition of a radian).

c© USQ, August 27, 2010 245


l = rθ ,

where θ is the angle measured in radians, r the radius and l the arclength.

Example 3.70: A circular arc of radius 7.6 metres has a length 4.2 metres. Find the anglesubtended at the centre.

Solution:

4.2 = 7.6θ ,

θ =4.27.6

= 0.553c .

Area of a sector of a circle

We also use radians to derive a formula for the area of a sector of a circle as follows.

area of sectorarea of circle

=θ

2π,

Therefore area of sector =θ

2π·πr2 ,

area of a sector =12

r2θ ,

where θ is measured in radians.

(Note that this formula cannot be used if θ is measured in degrees.)

246 c© USQ, August 27, 2010

3.7. Radian measure MAT1500

Example 3.71: Find the area of the figure with the dimensions given below.

D

A

C

Bθ φ

AC = 3 m, AD = 2.5 m, AB = 5.2 m, CB = 5.2 m.

Solution:

In triangle DAC,

sin θ =1.52.5

= 0.6 ,

θ = sin−1 0.6 = 0.643 5c ,

Therefore 6 ADC = 2θ = 1.287c .

Thus the outer part of circle from A to C has an angle at the centre of (2π− 1.287) ra-dians.

Area of sector =12

r2(2π − 1.287)

=2.52

2(2π − 1.287)

= 3.125× 4.996

= 15.613 m2.

Area of triangle DAC =12× 2.52 sin 1.287c

= 3.125× 0.960 0

= 3 m2.

c© USQ, August 27, 2010 247


In triangle ABC,

sin φ =1.55.2

= 0.288 ,

Therefore φ = 0.292 6c.

Therefore Area of triangle ABC =12× (5.2)2 × sin(2× 0.292 6c)

= 13.52 sin 0.585 2c

= 7.468 m2.

Thus,

total area = 15.613 + 3 + 7.467

= 26.081 m2.

Exercise 3.72: Express the following angles in radians.

(a) 360

(b) 30

(c) 10

(d) 13715′12′′

Exercise 3.73: Express the following angles in degrees.

(a)πc

3(b) 1.674c

(c)5πc

4(d) 6.803c

Exercise 3.74: Find the arc length and area of the sector of the circle with radius and anglegiven below.

(a) r = 4 m, θ =π

3

c

(b) r = 5 m, θ = 2.5c

(c) r = 10 m, θ = 45

(d) r = 6 m, θ = 130

Hint Add the things to learn to your list and learn them. Revise your previouschapters. If you have been skipping these revisions, then you should do it now.You will be surprised at the amount you have forgotten. (You will be even moresurprised if you leave it until just before the exam!)

248 c© USQ, August 27, 2010

3.8. Graphs involving trigonometric functions MAT1500

Points to remember

• π radians = 180.

• Arc length of a circle: l = rθ where l is the arc length of a circle, ris the radius and θ is the angle measured in radians.

• Area of a sector of a circle: Area =12

r2θ where r is the radius andθ is the angle measured in radians.

3.8 Graphs involving trigonometric functions

In order to use graphical methods to solve equations involving trigonometric functions,it is necessary to know the graphs of the basic trigonometric functions.

Application: engineering Trigonometric functions are common in electricalengineering, for example:

• A sinusoidal alternating current can be represented byi = 10 sin 500t , where i is in mA and t in seconds.

• A sinusoidal alternating voltage across a component of a circuit can be rep-resented by V = 5.0 sin ωt , where V is in volts.

Using radian measure and your calculator, verify the following graphs of the trigonomet-ric functions.

• y = sin x .

c© USQ, August 27, 2010 249

MAT1500 3.8. Graphs involving trigonometric functions

• y = cos x .

• y = tan x .

Note that the x-axis is measured in radians. In most graphs involving trigonometric func-tions, other functions of x are also involved, for example y = x− sin x. In such situationsradians must be used since radians and trigonometric functions are all ratios (that is, theyare numbers without dimensions) and may be mixed. Degrees are not ratios and cannotbe combined with trigonometric functions.

Application: meteorology The change in the maximum and minimum tem-peratures around Toowoomba throughout the year varies according to a cosinefunction.

250 c© USQ, August 27, 2010


Example 3.75: Sketch the graph of y = sec x .

Solution:y = sec x =

1cos x

.

Therefore we must plot the reciprocal of the values for cos x.

x cos x sec x =1

cos xx cos x sec x =

1cos x

−π2 0 Undef. 5π

8 −0.383 −2.613

− 3π8 0.383 2.613 3π

4 −0.707 −1.41

−π4 0.707 1.41 7π

8 −0.924 −1.08

−π8 0.924 1.08 π −1 −1

0 1 1 9π8 −0.924 −1.08

π8 0.924 1.08 5π

4 −0.707 −1.41

π4 0.707 1.41 11π

8 −0.383 −2.613

3π8 0.383 2.613 3π

2 0 Undef.

π2 0 Undef.

c© USQ, August 27, 2010 251


Application: biological physics The spring-mass system is a basic physicsconcept that is often used in biological models such as those describing the be-haviour of the cochlea of the ear. A mass hanging from a spring oscillates aboutan equilibrium position (y = 0) and the maximum displacement will occur whenthe velocity is zero, that is, according to the equation

e−0.1t [3 cos(3t)− 0.1 sin(3t)] = 0 .

But since e0.1t can never be zero for any values of t, the equation is reduced to

3 cos(3t)− 0.1 sin(3t) = 0 .

To solve this equation we could graph the two equations,

y1 = 3 cos(3t) ,

y2 = 0.1 sin(3t) .

The approximate solution is 0.5 seconds.

252 c© USQ, August 27, 2010


Example 3.76: Find all solutions of x2 − 1− sin x = 0 .

Solution:

y1 = x2 − 1 ,

y2 = sin x .

The points of intersection of these graphs are the solutions.

Note that

• radians must be used; and

• we already know the general shapes of y1 and y2 (y1 is a parabola and y2 is thesine curve).

x y1 = x2 − 1 y2 = sin x

−π

21.47 −1

−3π

80.39 −0.92

−π

4−0.38 −0.71

−π

8−0.85 −0.38

0 −1 0

π

8−0.85 0.38

π

4−0.38 0.71

3π

80.39 0.92

π

21.47 1

Approximate solutions are x = 1.4 or x = −0.6

c© USQ, August 27, 2010 253


Example 3.77: Use graphical methods to solve the equation x− tan x = 0 .

Solution: One solution is obviously x = 0 but other solutions exist at approximately

x = ±3π

2(that is, x = ±4.5),

x = ±5π

2(that is, x = ±7.7).

More accurate answers can be obtained by using more accurate graphs.

This can be written as x = tan x .

Let y1 = x , and y2 = tan x .

The points of intersection of these graphs are the solutions.

Note that

• radians must be used here; and

• we already know the general shapes of y1 and y2, (y1 is a straight line and y2

is the tangent curve).

Exercise 3.78: Find all solutions of the following equations:

(a) x− sin x = 0

(b) x2 − 1− 2 sin x = 0

(c) x2 − 3x + 4 cos x = 0

Putting it all together! An electrical power cable hangs between two supports20 metres apart. Engineers model the shape of its suspension by the equation

y =410

cos( π

10x)+ 10 ,

where x is the distance in metres from the supports and y is the height in metresabove the ground. The graph of the function is depicted below.

254 c© USQ, August 27, 2010

3.9. Review MAT1500

1. How much does the cable sag?

2. Engineers need to know the slope of the cable where it attaches to eachpylon. Find the approximate value of this by finding the average slope of thecable one metre from the support.

3. What is the average slope of the cable one metre from the lowest point?

Points to remember

• Graphs of trigonometric functions are periodic. (i.e. they repeatthemselves)

Congratulations, you made it to the end of this chapter. Here are a number of things tocheck.

1. Have a close look at your action plan for study. Are you on schedule? Do youneed to restructure your action plan or contact your tutor to discuss any delays orconcerns?

2. Make a summary of the important points in this chapter noting your strengths andweaknesses. Add any new words to your personal glossary. This will help withfuture revision and the open book exam.


3.9 Review

• The exponential function is defined as y = ex, where e is the number 2.718 28.

• The graph of y = ex is a curve that increases with an ever increasing slope, crossingthe y axis at 1 and asymptotes to the negative x axis. The graph of y = e−x is a

c© USQ, August 27, 2010 255

MAT1500 3.9. Review

curve decreasing with an ever decreasing slope which asymptotes to the x axis andcrosses the y axis at 1.

• The inverse function of y = ax is a logarithmic function written y = loga x where ais the base of the logarithm.

• loge x is called the natural log and is written ln x.

• log10 x is called the common log and is written log x.

• The graph of a logarithmic function y = loga x is only defined for x > 0, asymptotesto the negative y axis crosses the x axis at 1, and continues to increase with an everdecreasing slope.

• Exponential and logarithmic functions are inverse functions of each other.

• For the triangle below:

sin θ =ac

; cos θ =bc

; tan θ =ab=

sin θ

cos θ

cosec θ =1

sin θ=

ca

; sec θ =1

cos θ=

cb

; cot θ =1

tan θ=

ba

.

• sin(180 − θ) = sin θ; cos(360 − θ) = cos(−θ) = cos θ; tan(180 + θ) = tan θ.

• The inverse function sin−1 θ gives the angle θ whose sine is x.The inverse function cos−1 θ gives the angle θ whose cosine is x.The inverse function tan−1 θ gives the angle θ whose tangent is x.

• sin2 θ + cos2 θ = 1.

• Sine rule:a

sin A=

bsin B

=c

sin Cwhere a, b, and c are the sides and A, B, C are the

opposite angles of any triangle.

• Cosine Rule: a2 = b2 + c2 − 2bc cos A where a, b, and c are the sides and A, B, C arethe opposite angles of any triangle.

• Area: Area =12

bc sin A =12

ac sin B =12

ab sin C where a, b, and c are the sides andA, B, C are the opposite angles of any triangle.

• π radians = 180.

• Arc length of a circle: l = rθ where l is the arc length of a circle, r is the radius andθ is the angle measured in radians.

• Area of a sector of a circle: Area =12

r2θ where r is the radius and θ is the anglemeasured in radians.

• Graphs of trigonometric functions are periodic. (i.e. they repeat themselves)

256 c© USQ, August 27, 2010


3.10 Post-test

1. Sketch on one set of axes the graphs of y = ex , y = e−x , and y = e2x.

2. Sketch on one set of axes the graphs of the functions y = ln x , y = − ln x , and y =12 ln x . What is it about these graphs that confirm these functions are the inversesof those in question 1?

3. If log x = a , and log y = b , express logy5

xin terms of a and b.

4. Simplify ln(e5x) .

5. A radioactive isotope decays according to the formula M(t) = M0e−kt , where M0

is the mass initially present, k is a constant, and t is the time in days. If its half-life(the time taken for half the mass to decay) is 14 days, how long would it take for100 g to decay to 15 g?

6. Sketch the graph of y = tan x from x = −2π to 2π. Draw a rough sketch only!

7. Find all angles between 0 and 360, which satisfy sin θ = 0.4 .

8. In the triangle below, find x.

Find the area of this triangle.

9. (a) Simplify√

1 + tan2 θ .

(b) Simplify sin A sin B + cos A cos B .

10. (a) Convert an angle of π4 radians to degrees.

(b) Find the perimeter of a sector of a circle of radius 2 metres which has a sub-tended angle at the centre of π

3 .

c© USQ, August 27, 2010 257


3.11 Solutions


3.7 (a) 0.6931

(b) 1.5261

(c) −0.6931

(d) 2.9124

(e) −1.0788

3.8 (a) 7.3891

(b) 4.0552

(c) 0.2019

(d) 1.4918

(e) 0.0183

3.9 (a) Graph of y = 2 log x

(b) Graph of y = 10 + 3 ln x

(c) Graph of y = 10 + 3 log x

258 c© USQ, August 27, 2010


(d) Graph of y = e−2x

(e) Graph of y = 2ex

(f) Graph of y = ln x2

c© USQ, August 27, 2010 259


3.10 (a) Graph of y = ln x and y = x2 − 4x, approximate solution is x = 0.3 or x = 4.3

(b) Graph of y = 2e−x and y =1

x + 3− 4, approximate solution is x = −3

3.13 Logarithms to base 10:

(a) −8.52

(b) −1.17

(c) −4.55

260 c© USQ, August 27, 2010


(d) −2.04

(e) −12.27

(f) −2.49

(g) −0.82

(h) 3.76

3.14 pH and hydrogen ion concentration

(a) the pH is

i. 3

ii. 0

iii. 4.7

(b)[H+]

is

i. 0.00001

ii. 0.0004

iii. 0.32

3.15 Height and age of females:

(a) 1.37 metres

(b) Approximately 12 years old, since

1.5 = 0.61 + 0.796 log(x + 1)1.5− 0.61

0.796= log(x + 1)

1.118 = log(x + 1)

101.118 = x + 1

x = 101.118 − 1 = 12.12

3.16 (a) log3 81 + log319= 4 log3 3 + −2 log3 3 = 2 log3 3 = 2

(b) 2 log9 3− 4 log9 2 = log9 32 − log9 24 = 1− log9 16

(c) ln 25 + 2 ln 0.2 = ln 25 + ln 0.22 = ln(25× 0.22) = ln 1 = 0

(d) 2 log 3 + log 6− 2 log(

65

)= log 32 + log 6− log

(65

)2

= log

(32 × 6÷

(65

)2)

= log(

752

)

(e)log 6 + log 3

log 3– not possible to simplify using the log laws.

(f) log 4× log 7 – not possible to simplify using the log laws.

c© USQ, August 27, 2010 261


3.19 (a)

6.78 = ln( x

234

)

e6.78 =x

234x = 234× e6.78

x ≈ 205 936

(b)

0.0189 = ln(5x)

e0.0189 = 5x

x =e0.0189

5≈ 0.204

(c)

44 = 22 + 10 log (2x)44− 22

10= log(2x)

102.2 = 2x

x =102.2

2≈ 79.2

(d)

3 log (3x + 1)2 + 1 = 10

log (3x + 1)2 =10− 1

3= 3

103 = (3x + 1)2

3x + 1 =√

103

x =

√103 − 1

3≈ 10.2

3.20 (a)

N = 10 log(

I0.01

)

N10

= log(

I0.01

)

10N10 =

I0.01

I = 0.01× 10N10

262 c© USQ, August 27, 2010


(b)

G = 20 log(

vv0

)

G20

= log(

vv0

)

10G20 =

vv0

v = v0 × 10G20

(c)

t =

ln(

1 000P− 1)

−0.7

−0.7t = ln(

1 000P− 1)

e−0.7t =1 000

P− 1

e−0.7t + 1 =1 000

P1

e−0.7t + 1=

P1 000

P =1 000

e−0.7t + 1

(d)

t =ln (1− T)−k

−kt = ln (1− T)

e−kt = 1− T

−T = e−kt − 1

T = 1− e−kt

3.21

t = −10 ln( v

200

)

t−10

= ln( v

200

)

v = 200e−t

10

When t = 0.01 , v = 200e−0.0110 ≈ 199.8 volts.

c© USQ, August 27, 2010 263


3.22

h =

ln(

pp0

)

−0.15

−0.15h = ln(

pp0

)

p = p0e−0.15h

When h = 0.25 , p = p0e−0.15×0.25 ≈ 0.96p0 = 96 319.44 Pascals.

3.30 (a)

3.7 = 3× 42x + 1

2.7 = 3× 42x

0.9 = 42x

ln 0.9 = ln 42x

ln 0.9 = 2x ln 4

x =ln 0.92 ln 4

≈ −0.038

(b)

3ex = 10

ex =103

ln ex = ln(

103

)

x ln e = ln(

103

), since ln e = 1

x = ln(

103

)≈ 1.204

(c)

4e1−2x + 7 = 10

4e1−2x = 3

e1−2x =34

ln e1−2x = ln(

34

)

(1− 2x) ln e = ln(

34

), since ln e = 1

1− 2x = ln(

34

)

−2x = ln(

34

)− 1

x =ln( 3

4

)− 1

−2≈ 0.644

264 c© USQ, August 27, 2010


(d)

43x−2 = 26x+1

ln (43x−2) = ln (26x+1)

(3x− 2) ln 4 = (x + 1) ln 26

3x ln 4− 2 ln 4 = x ln 26 + ln 26

3x ln 4− x ln 26 = ln 26 + 2 ln 4

x(3 ln 4− ln 26) = ln 26 + 2 ln 4

x =ln 26 + 2 ln 43 ln 4− ln 26

≈ 6.695

3.31 (a)

t = 11.6× 2.71−2.4x

t11.6

= 2.71−2.4x

ln(

t11.6

)= ln

(2.71−2.4x

)

ln(

t11.6

)= (1− 2.4x) ln 2.7

ln( t

11.6

)

ln 2.7= 1− 2.4x

−2.4x =ln( t

11.6

)

ln 2.7− 1

x =

ln ( t11.6 )

ln 2.7 − 1−2.4

, or

x =ln( t

11.6

)− ln 2.7

−2.4 ln 2.7

(b)

p = 2e−0.1x − 7

p + 7 = 2e−0.1x

lnp + 7

2= ln e−0.1x

lnp + 7

2= −0.1x ln e

−0.1x = lnp + 7

2

x =ln p+7

2−0.1

, or

x = −10 lnp + 7

2

c© USQ, August 27, 2010 265


(c)

4e1−2x + 7 = 10y

e1−2x =10y− 7

4

ln e1−2x = ln(

10y− 74

)

(1− 2x) = ln(

10y− 74

)

−2x = ln(

10y− 74

)− 1

x =ln(

10y−74

)− 1

−2, or

x =12

(1− ln

(10y− 7

4

))

(d)

y− 103x−2 = 10

y− 10 = 103x−2

log (y− 10) = log 103x−2

log (y− 10) = (3x− 2) log 10

3x− 2 = log (y− 10) , since log 10 = 1

x =log (y− 10) + 2

3

3.32

If p = 10log10 x

log10 p = log10 10log10 x

log10 p = log10 x log10 10

log10 p = log10 x , since log10 10 = 1

p = x , and thus 10log10 x = x

3.33 In all instances below, we are attempting to rewrite these equations in the form of astraight line y = mx + c.

(a)

y = 1.4x

log y = log 1.4x

log y = x log 1.4

A straight line with Y = log y and m = log 1.4.

266 c© USQ, August 27, 2010


(b)

y = 6× 1.2x

y6

= 1.2x

logy6

= log 1.2x

log y− log 6 = x log 1.2

log y = x log 1.2 + log 6

A straight line with Y = log y, m = log 1.2 and c = log 6.

(c)

y = 6× 0.6−2x

y6

= 0.6−2x

logy6

= log 0.6−2x

logy6

= log 0.6−2x

log y− log 6 = −2x log 0.6

log y = −2x log 0.6 + log 6

A straight line with Y = log y, m = −2 log 0.6 and c = log 6.

(d)

y = 2× 33−2x

y2

= 33−2x

logy2

= log 33−2x

log y− log 2 = (3− 2x) log 3

log y = −2x log 3 + (3 log 3 + log 2)

A straight line with Y = log y, m = −2 log 3 andc = 3 log 3 + log 2.

c© USQ, August 27, 2010 267


3.34 Substituting t = 0.15 s, I = 3.10A into I = 4(1− e−kt),

I = 4(

1− e−kt)

3.10 = 4(

1− e−k×0.15)

3.104

= 1− e−k×0.15

3.104− 1 = −e−k×0.15

(3.10

4− 1)×−1 = −e−k×0.15 ×−1

1− 3.104

= e−k×0.15

1− 3.104

= e−k×0.15

ln(

1− 3.104

)= ln e−k×0.15

ln(

1− 3.104

)= −k× 0.15

k =ln(1− 3.10

4

)

−0.15≈ 9.94

3.35 (a) Assume that the population follows the exponential growth formula of P =

P0ekt, where P is the size of the population and t is the time the populationhas been growing in years. When t = 0 , P = 3 500 , so 3 500 = P0ek×0 andP0 = 3 500. Using the formula P = 3 500ekt , when t = 8, P = 6 245

6 245 = 3 500ek×8

6 2453 500

= ek×8

ln6 2453 500

= ln ek×8

ln6 2453 500

= 8k

k =ln 6 245

3 5008≈ 0.0724

Using the formula P = 3 500e0.0724t, when t = 1, P = 3 500e0.0724×1 = 3 762.8.Population after one year is 3 762 people.

(b) Population when either P = 7 000 (or P = 2P0)

7 000 = 3 500e0.0724t

2 = e0.0724t

ln 2 = ln e0.0724t

0.0724t = ln 2

t =ln 2

0.0724≈ 9.58

Population will double in approximately 9.58 years.

268 c© USQ, August 27, 2010


3.36 When P = 10 and A = 5

5 = 10e−0.0248t

0.5 = e−0.0248t

ln 0.5 = ln e−0.0248t

ln 0.5 = −0.0248t

t =ln 0.5−0.0248

≈ 27.95

Time to decay is approximately 27.95 years.

3.37 Using the exponential decay formula A = A0e−kt, where A is amount of radiativematerial (g) and t is time in years, to find k when after 5 730 years A will be half ofA0, i.e. 0.5A0. Analysis of the tool determines that 15% of the radioactive substancehas already decayed, thus 85% is left. First of all we need to find k.

0.5A0 = A0e−k×5 730

0.5 = e−k×5 730

ln 0.5 = ln e−5 730k

ln 0.5 = −5 730k

k =ln 0.5−5 730

≈ 0.000121

Now with 85% of the radioactive substance left:

0.85A0 = A0e−0.000121t

0.85 = e−0.000121t

ln 0.85 = ln e−0.000121t

ln 0.85 = −0.000121t

t =ln 0.85−0.000121

≈ 1 343.5

The fossil is approximately 1 343.5 years old.

3.38 Let A = A0e−kt where A is the amount of hair and t is time in years.

A = A0e−kt

0.4A0 = A0e−k×5

0.4 = e−5k

ln 0.4 = ln e−5k

ln 0.4 = −5k

k =ln 0.4−5

≈ 0.1833

c© USQ, August 27, 2010 269


Now with k = 0.1833 and only 6% of his hair left:

A = A0e−kt

0.06A0 = A0e−0.1833t

0.06 = e−0.1833t

ln 0.06 = ln e−0.1833t

ln 0.06 = −0.1833t

t =ln 0.06−0.1833

≈ 15.35

He has worked at the power plant approximately 15.35 years.

3.39 For each species the exponential growth formula is P = P0ekt, where P is the numberof animals and t is the time in years. For Species A, when t = 0, PA = 46 and whent = 5, PA = 138. To find k:

138 = 46e5k

3 = e5k

k =ln 3

5≈ 0.22

Growth formula is PA = 46e0.22t

For Species B, when t = 0, PB = 24 and when t = 3, PB = 96. To find k:

96 = 24e3k

4 = e3k

k =ln 4

3≈ 0.46

Growth formula is PB = 24e0.46t

The number of each species will be the same when PA = PB

46e0.22t = 24e0.46t

e0.22t

e0.46t =2446

e0.22t−0.46t =1223

e−0.24t =1223

ln e−0.0.24t = ln1223

−0.24t = ln1223

t =ln 12

23−0.24

≈ 2.7

Populations will be equal in approximately 2.7 years.

3.45 (a) −0.3786.

270 c© USQ, August 27, 2010


(b) 0.8047.

(c) 1.6909.

(d) −0.6871.

(e) −8.5126.

(f) −0.9761.

3.46 (a) 23.578

(b) 78.46

(c) 62.24

(d) −14.48

(e) 120.66

(f) −68.96

3.47 (a) 52.43 and 232.43.

(b) 39.05 and 140.95.

(c) 82.53 and 277.47.

(d) 343.74 and 196.26.

(e) 302.01 and 122.01.

(f) 147.14 and 212.86.

3.48 179.76 m.

3.51

sec2 A = 1 + tan2 A ,

sec2 A = 5 ,

sec A = ±√

5 .

cos A =1

sec A,

= ± 1√5

,

= ±0.4472 .

c© USQ, August 27, 2010 271


3.52

cos2 B = 1− sin2 B ,

= 1− 0.42

= 0.84 ,

Therefore cos B = ±0.9165 .

cot B =cos Bsin B

= ±0.91650.4

= ±2.2913 .

3.53

tan θ + cot θ =sin θ

cos θ+

cos θ

sin θ

=sin2 θ + cos2 θ

cos θ sin θ

=1

sin θ cos θ.

3.54 (a) sin θ = ±0.5774 , therefore θ = 35.26, 144.74, 215.26, or 324.73.

(b) sin θ = 0.5 or −2. Now sin θ = −2 is impossible, thus θ = 30 or 150.

(c) cos θ = 0.5 or −5. Now cos θ = −5 is impossible, thus θ = 60 or 300.

3.59 (a) B = 126.21, C = 23.79.

(b) B = 85.14, C = 52.86.

(c) A = 24.19, B = 45.81.

3.60 (a) b = 22.14 or 105.71.

(b) a = 16.82 .

(c) a = 7.47 .

3.61 (a) A = 28.96.

(b) A = 109.47.

3.62 (a) A = 23.79, B = 126.21, c = 2.48 .

(b) A = 46.57, C = 57.91, b = 8.00 .

3.63 (a) 2.

(b) 6.691.

(c) 13.16.

3.64 115.92.

3.66 (a) cos A cos2 B− cos A sin2 B− 2 sin A sin B cos B.

272 c© USQ, August 27, 2010


(b) 2 sin A cos A cos B + cos2 A sin B− sin2 A sin B.

(c) 2 sin A sin B cos B + cos A cos2 B− cos A sin2 B.

3.67 (a) sin(35 − 24) = sin 11 .

(b) cos(2× 45) = cos 90 .

(c) cos(32 + 58) = cos 90 .

(d) sin(21 + 46) = sin 67 .

3.68 (a)

LHS = cot A + cot B

=cos Asin A

+cos Bsin B

=cos A sin B + sin A cos B

sin A sin B

=sin(A + B)sin A sin B

= RHS.

(b)

LHS =sin(A− B)cos(A + B)

=sin A cos Bcos A cos B − cos A sin B

cos A cos Bcos A cos Bcos A cos B − sin A sin B

cos A cos B

=tan A− tan B

1− tan A tan B= RHS.

Note in the last step we divided each term by cos A cos B.

3.72 (a) 2π = 6.283c .

(b)π

6= 0.524c.

(c)π

18= 0.1745c.

(d) 2.3955c.

3.73 (a) 60.

(b) 95.913.

(c) 225.

(d) 389.783.

c© USQ, August 27, 2010 273


3.74 (a)

Arclength = rθ

=4π

3= 4.189 m.

Area =12

r2θ

=8π

3= 8.378 m2.

(b)

Arclength = 12.5 m.

Area = 31.25 m.

(c)

45 =π

4= 0.7854c ,

Therefore Arclength = 7.854 m ,

Area = 39.27 m2.

(d)

130 = 2.269c ,

Therefore Arclength = 13.614 m ,

Area = 40.841 m2.

3.78 (a) x = sin x , therefore

y1 = x ,

y2 = sin x .

274 c© USQ, August 27, 2010


x y1 y2

−1 −1 −0.84−0.8 −0.8 −0.72−0.6 −0.6 −0.56−0.4 −0.4 −0.39−0.2 −0.2 −0.199

0 0 00.2 0.2 0.1990.4 0.4 0.390.6 0.6 0.560.8 0.8 0.721 1 0.84

Thus the only solution is x = 0 .

(b) x2 − 1 = 2 sin x , therefore

y1 = x2 − 1 ,

y2 = 2 sin x .

x y1 y2

−1.2 0.44 −1.86−0.8 −0.36 −1.43−0.4 −0.84 −0.78

0 −1 00.4 −0.84 0.780.8 −0.36 1.431.2 0.44 1.861.6 1.56 2.002 3 1.82

c© USQ, August 27, 2010 275


Thus solutions are roughly x = −0.4 and x = 1.7 .

(c) 4 cos x = 3x− x2 , therefore

y1 = 4 cos x ,

y2 = 3x− x2 .

x y1 y2

0 4 00.5 3.51 1.251 2.16 2

1.5 0.28 2.252 −1.66 2

2.5 −3.20 1.253 −3.96 0

3.5 −3.75 −1.754 −2.61 −4

276 c© USQ, August 27, 2010


Thus solutions are roughly x = 1.04 and x = 3.81 .


1. Graphs of y = ex , y = e−x , and y = e2x

2. Graphs of y = ln x , y = − ln x , and y = 12 ln x

This graph is the same shape as in question 1, but with the axes reversed.

3.

logy5

x= log y5 − log x

= 5 log y− log x

= 5b− a .

4. 5x .

c© USQ, August 27, 2010 277


5. First, determine k.

12

M0 = M0e−14k ,

k =ln( 1

2

)

−14.

Next we find t,

15 = 100e−kt ,

t =ln(0.15)−k

,

= 38.32 days.

6. Graph of y = tan x

7. 23 34′ 41′′, 156 25′ 19′′.

8. x =√

31, area =15√

32

.

9. (a) sec θ .

(b) cos(A− B) .

10. (a) 45.

(b)2π

3m.

278 c© USQ, August 27, 2010

Chapter 4

Vectors

Chapter contents4.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280

4.2 Displacement Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

4.3 Vectors in General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

4.4 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

4.5 Angle between vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

4.6 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

4.7 Post Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294

4.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295



Objectives


• identify the displacement vector for shifts of position within Cartesian space;

• identify physical quantities in 2 or 3 dimensions as vector or scalar;

• perform addition, subtraction and multiplication by a scalar for, and find the mag-nitude of, any 2 or 3 dimensional vector;

• find a unit vector in the direction of any 2 or 3 dimensional vector;

• use the i j k notation to describe vectors;

• find the scalar (dot) product of any pair of 2 or 3 dimensional vectors; and

• find the angle between any pair of 2 or 3 dimensional vectors.

c© USQ, August 27, 2010 279

MAT1500 4.1. Cartesian Coordinates

Introduction

Some aspects of vector quantities and their addition using the parallelogram rule appearto go back as far as Aristotle (384–322BC). Newton (1642–1727) deals with what we nowclassify as vector quantities such as force and velocity.

However the study of vectors and their interaction really didn’t until the early 19th Cen-tury. The concept of a pair of numbers being considered a single quantity and capableof representation geometrically on a two dimensional plane began with the developmentof the theory of complex numbers by European mathematicians such as Caspar Wessel(1745–1818), Jean Robert Argand (1768–1822), Carl Friedrich Gauss (1777–1855).

William Rowan Hamilton (1805–1865), Irish physicist, astronomer, and mathematician,who made important contributions to classical mechanics, optics, and algebra, tried toextend this idea of two dimensional numbers to three dimensions and developed theconcept of quaternions which contained two parts, a first number which he called thescalar part, and three other numbers which he called the vector part.

The development of the theory of vectors as we know them today began with the workof a little known German secondary school maths teacher, Hermann Grassmann (1809–1877). This work was extended by William Kingdon Clifford (1845–1879) who in his shortlifetime developed the ideas of scalar and vector products.

The theory was further systematised by the American university lecturer J. Willard Gibbs(1839–1903) whose notes for his students were printed privately but distributed widely tomathematicians in United States, Britain and Europe. These notes were later assembledand published as the first English language book on Vector Analysis by one of his lastgraduate students, Edwin B. Wilson (1879–1964).

Vectors are now widely used in Engineering, Mathematics and Physics where they arean essential tool for describing and understanding physical phenomena (see History ofVectors 2009).

4.1 Cartesian Coordinates

In Chapter 2 we introduced you to the concept of the Cartesian plane. You should re-visit that section now if you need to refresh your understanding of the concept. We nowextend this concept to three dimensions, by including a third axis at right angles to boththe x and y axes (Figure 4.1). We call this (naturally) the z axis.

You will notice in Figure 4.1 that the z axis could have pointed up or down and would stillhave been at right angles to the x–y plane. The convention is that just as in the x–y plane,the positive y direction is an anti-clockwise quarter turn from the positive x direction, inthe y–z plane, the positive z direction is an anti-clockwise quarter turn from the positivey direction.

280 c© USQ, August 27, 2010

4.1. Cartesian Coordinates MAT1500

Figure 4.1: Cartesian axes

Just as each point on the Cartesian plane has two coordinates which exactly defines itsposition, every point in the three dimensional space defined by these axes will have threecoordinates. We write these coordinates as three numbers (x1, y1, z1) corresponding to itsdistance from the origin O (x = 0, y = 0, z = 0) in the x, y, and z directions respectively(this is called an ordered triplet).

Example 4.1: What are the coordinates of a point in Cartesian space which is reached bytraveling 10 in the y direction, 14 in the z direction and then 5 in the x directionfrom the origin?

Solution:

(5, 10, 14)

Exercise 4.2: What are the coordinates of points in Cartesian space displaced from theorigin:

(a) 4 in the x direction, 2 in the y direction and 3 in the z direction;

(b) 13 in the z direction, 1 in the x direction and 4 in the y direction;

(c) 2 in the y direction, 3 in the z direction and 1 in the x direction;

(d) 5 in the x direction, 1 in the z direction and 2 in the y direction;

(e) 2 in the x direction, −1 in the y direction and −5 in the z direction;

(f) p in the x direction, q in the y direction and r in the z direction;

(g) sin t in the x direction, cos t in the y direction and t in the z direction;

(h) t2 in the x direction, 1− t2 in the y direction and k in the z direction;

(i) 1 in the x direction, 0 in the y direction and 0 in the z direction;

(j) 0 in the x direction, 1 in the y direction and 0 in the z direction;

c© USQ, August 27, 2010 281

MAT1500 4.2. Displacement Vectors

(k) 0 in the x direction, 0 in the y direction and 1 in the z direction;

(l) a1 in the x direction, a3 in the z direction and a2 in the y direction; and

(m) x0 in the x direction, y0 in the y direction and z0 in the z direction.

4.2 Displacement Vectors

The coordinates of the point P, (a1, a2, a3), indicate a displacement from the origin O of a1

in the x direction, a2 in the y direction and a3 in the z direction. The line from the origin

to the point P has a length OP =√(a1)2 + (a2)2 + (a3)2 (by Pythagoras’s theorem) and a

direction from O to P.

z

y

x

O(0, 0, 0)

a1

a2

a3

P(a1, a2, a3)

Figure 4.2: Position vector of a point in space.

The vector−→OP is said to be the position vector of the point P, and can be written (a1, a2, a3).

The numbers (a1, a2, a3) are called the components of the position vector−→OP. It represents

the displacement in moving from the origin O to the point P.

It is important to notice that the definition above of a vector does not specify a position.So the vector a = (a1, a2, a3) also represents any displacement of a1 in the x direction,a2 in the y direction and a3 in the z direction irrespective of the starting position. Thenumbers a1, a2, and a3 are the components of the vector a.

282 c© USQ, August 27, 2010

4.2. Displacement Vectors MAT1500

Figure 4.3: Vectors are not defined by their position.

In general vector quantities are expressed as a bold-faced letter in print, e.g. a, or inwritten material as a letter underlined or with an arrow above, e.g. a or −→a .

All of the above material also applies in 2 dimensions by simply deleting the third com-ponent. That is the displacement vector a in 2 dimensions representing a shift of a1 in thex direction, a2 in the y direction is a = (a1, a2).

Points to remember

• A vector is a quantity that has magnitude and direction.

• A scalar is a quantity that only has magnitude.

Example 4.3:

(a) What is the displacement vector in moving from the point P(1, 4, 2) to pointQ(3, 2, 3)?

(b) A displacement of (1,−1, 2) is applied at a point P(4, 5, 3). What are the coor-dinates of the resulting position Q?

Solution:

(a) The displacement vector moves from 1 to 3 (i.e. 2) in the x direction, 4 to 2 inthe y direction (i.e. −2), and 2 to 3 in the z direction (i.e. 1). The displacementis therefore (2,−2, 1).

(b) A displacement of 1 in the x direction from 4 brings the x value to 5. Similarlythe new y and z coordinates become 5+−1 = 4 and 3+ 2 = 5. The coordinatesof the resulting position are Q(5, 4, 5).

Exercise 4.4: What is the displacement vector in moving from the point:

(a) (1, 1, 1) to (2, 3, 4)?

(b) (1, 0, 2) to (0, 1, 1)?

(c) (1, 5, 2) to (3, 2, 2)?

(d) (6, 8, 7) to (0, 0, 0)?

c© USQ, August 27, 2010 283

MAT1500 4.3. Vectors in General

(e) (−1, 1, 0) to (0, 1, 1)?

(f) (4, 3) to (−1,−2)?

(g) (1, 0, 2) to (0, 1, 1)?

(h) (0, 0, 0) to (12, 13,−14)?

Exercise 4.5: What are the coordinates of the resulting position Q if

(a) a displacement of (2,−2,−3) is applied at a point P(−1,−3,−1)?

(b) a displacement of (6,−4, 0) is applied at a point P(−7, 8, 8)?

(c) a displacement of (3,−1,−9) is applied at a point P(−3, 5, 8)?

(d) a displacement of (5,−6,−8) is applied at a point P(0, 7, 2)?

(e) a displacement of (1,−6, 3) is applied at a point P(−3,−2, 5)?

(f) a displacement of (−6, 2) is applied at a point P(−2, 6)?

(g) a displacement of (−5,−2, 3) is applied at a point P(−7,−7, 9)?

(h) a displacement of (−8,−5) is applied at a point P(−5, 2)?

4.3 Vectors in General

Vectors are not confined to describing displacements. Any quantity which has magnitudeand direction can be considered a vector quantity. Examples of vectors are displacement,velocity, momentum, acceleration, electric field, magnetic field, force, angular velocity(spin), torque.

Scalar quantities are defined solely and completely by their magnitude and do not havea direction associated with them. Examples of scalar quantities are length, area, volume,mass, density, energy, temperature, speed (as distinct from velocity), electric potential.

Vectors are considered equal if they have the same magnitude and direction. This canonly happen if all the components of the vectors are equal. That is if we have two vectorsa = (a1, a2, a3) and b = (b1, b2, b3), then a = b if and only if a1 = b1, a2 = b2, and a3 = b3.

If two vectors are added (or subtracted), the corresponding components are added (orsubtracted). That is, if a = (a1, a2, a3) and b = (b1, b2, b3), a + b = (a1 + b1, a2 + b2, a3 +

b3), and a− b = (a1 − b1, a2 − b2, a3 − b3). Since the components of a and b are numbers(scalars), a + b = b + a, (i.e. addition is commutative), but a− b = −(b− a).

Geometrically a + b or a− b will be the vector which completes the triangle with vectorsa and b making the other two sides.

284 c© USQ, August 27, 2010

4.3. Vectors in General MAT1500

a

ba + b

a

ba− b

a

ba− b b

a

a + b

Figure 4.4: Geometric view of addition and subtraction of vectors.

Alternatively a + b or a− b will be the diagonals of the parallelogram formed where aand b make up two of the sides.

If a vector is multiplied by a scalar, each component of the vector is multiplied by thescalar. That is if a = (a1, a2, a3), then ka = (ka1, ka2, ka3).

The magnitude (or length) of a vector a is |a| =√(a1)2 + (a2)2 + (a3)2.

The magnitude of a vector ka is |k||a|. For example: multiplying a vector by a scalark ≥ 0 has the effect of multiplying its magnitude without changing its direction, whilemultiplying by a scalar k ≤ 0 has the effect of increasing its magnitude, while reversingthe direction.

Examples of vectors multiplied by scalars:Newton’s third law: force F (vector) is equal to mass m (scalar) timesacceleration a (vector).

F = ma .

In electromagnetism, force on a charged particle F is charge q timeselectric field E.

F = qE .

A vector of magnitude 1 is called a unit vector. A unit vector in the direction of thevector a can be found by dividing the vector by its own magnitude. That is, a =

a|a| . (The

carat symbol ‘∧ ’ above a vector indicates that it is a unit vector.)

Figure 4.5: Magnitude of vector multiplied by a scalar.

The above properties allow for a very convenient vector notation. We can define threespecial unit vectors i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) in the directions of the x, y, andz axes respectively.

c© USQ, August 27, 2010 285

MAT1500 4.3. Vectors in General

Then a vector a = (a1, a2, a3) can be written as

a = (a1, a2, a3)

= (a1, 0, 0) + (0, a2, 0) + (0, 0, a3)

= a1(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1)

= a1 i + a2 j + a3k

z

y

x

k

ji

1

1

1

Figure 4.6: i j k unit vectors.

Points to remember

• a = b if and only if a1 = b1, a2 = b2, and a3 = b3.

• a + b = (a1 + b1, a2 + b2, a3 + b3), and a − b = (a1 − b1, a2 −b2, a3 − b3).

• a + b = b + a, but a− b = −(b− a).

• The magnitude or length of a vector a is

|a| =√(a1)2 + (a2)2 + (a3)2 .

• A vector of magnitude 1 is called a unit vector.

• A unit vector a in the direction a is a =a|a| .

• i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1).

• a = (a1, a2, a3) = a1 i + a2 j + a3k

Example 4.6:

Two forces are acting on a particle, F1 = 8i− 20j+ 25k and F2 = 4i+ 2j+ 6k. Whatis the resultant (total) force acting on the particle? What is its magnitude? Find aunit vector in the direction of the resultant force.

286 c© USQ, August 27, 2010

4.3. Vectors in General MAT1500

Solution:

The resultant force is

F = F1 + F2

= (8 + 4)i + (−20 + 2)j + (25 + 6)k

= 12i− 18j + 31k .

The magnitude of the resultant forced is:

|F| =√

122 + (−18)2 + 312 =√

144 + 324 + 961 =√

1429 ≈ 37.8 .

The unit vector in the direction of the force is:

F =F|F|

=12i− 18j + 31k

37.8≈ 0.32i− 0.48j + 0.82k .

Exercise 4.7: (a) Identify the following quantities as either vector or scalar quantities:acceleration, angular velocity (spin), area, density, displacement, electric field,electric potential, energy, force, length, magnetic field, mass, momentum, speed,temperature, torque, velocity.

(b) Write the vector a = (2, 4,−3) in i j k notation.

(c) Convert the following pairs of vectors to i j k notation and then add them:

i. (1, 1, 1), (2, 3, 4)

ii. (1, 0, 2), (0, 1, 1)

iii. (1, 5, 2), (3, 2, 2)

iv. (6, 8, 7), (0, 0, 0)

v. (−1, 1, 0), (0, 1, 1)

vi. (4, 3, 2), (−1,−2,−1)

vii. (1, 0, 2), (0, 1, 1)

viii. (0, 0, 0), (12, 13,−14)

(d) Subtract the first vector from the second:

i. 2i− 2j− 3k, −i− 3j− k

ii. 6i− 4j, −7i + 8j + 8k

iii. 3i− j− 9k, −3i + 5j + 8k

iv. 5i− 6j− 8k, 7j + 2k

v. i− 6j + 3k, −3i− 2j + 5k

vi. −6i + 2j + 4k, −2i + 6j− 9k

vii. −5i− 2j + 3k, −7i− 7j + 9k

viii. −8i− 5j + 6k, −5i + 2j− k

c© USQ, August 27, 2010 287

MAT1500 4.4. Scalar Product

(e) The length of a vector a is 2, what is the length of the vector 3a?

(f) Find 2a, 3b, and 2a− 3b when

i. a = 6i− 4j, b = −7i + 8j + 8k

ii. a = −6i + 2j + 4k, b = −2i + 6j− 9k

iii. a = −8i− 5j + 6k, b = −5i + 2j− k

(g) Find the unit vectors in the direction of the following vectors:

i. 2i− 2j− 3k

ii. 6i− 4j

iii. 3i− j− 9k

iv. −3i + 5j + 8k

v. 5i− 6j− 8k

vi. 7j + 2k

vii. i− 6j + 3k

viii. −3i− 2j + 5k

ix. −6i + 2j + 4k

x. −2i + 6j− 9k

(h) The current in a river flows with a velocity 0.8i− 1.4j metres per second. (Theunit vectors i and j are in the east and north directions). A boat moves rel-ative to the current with a velocity of 3.4i − 2.8j metres per second. What isthe boat’s velocity relative to the bank, assuming the current is parallel to thebank?

(i) Two particles collide and immediately combine to form a new particle. By thelaw of conservation of momentum, the momentum of the single particle afterthe collision must be equal to the total momentum of the two particles beforecollision. If the two original particles had momentum 4i− 2j and 2i + 4j, whatis the momentum of the combined particle after collision?

(j) An electric potential difference is applied to plates on either side of a chargedparticle creating an electric field at the position of the particle of E1 = 5i + 2j.Another set of plates creates an electric field at the same point of E2 = 4i− 10j.What is the resultant electric field E at the position of the particle? The force ona particle is given by the formula F = Eq where q is the charge on the particle.If the particle has a charge of 6, what is the force on the particle?

4.4 Scalar Product

In the previous section we saw that vectors could be multiplied by a scalar to give anothervector in the same direction but of a different length.

There are 2 other types of multiplication involving vectors, the scalar product which asthe name suggest results in a scalar quantity, and a vector product which produces avector result.

288 c© USQ, August 27, 2010

4.4. Scalar Product MAT1500

In this course we will only cover the scalar product. Vector products are left to a latercourse in maths.

The scalar product of two vectors a = (a1, a2, a3) and b = (b1, b2, b3) is defined as a ·b =

a1b1 + a2b2 + a3b3, that is the sum of the products of the corresponding components.Because a dot is used to indicate that it is a scalar product, it is often called a ‘dot product’.Be careful if you come across a× b. This indicates a vector product, often called a ‘crossproduct’.

It should be clear that since the components of the vectors a and b are numbers, so ab isthe same as ba, then a ·b = b · a.

Similarly, it is also easily verified that c · (a + b) = c · a + c ·b.

Note that i · i = 1× 1 + 0× 0 + 0× 0 = 1, and similarly j · j = k · k = 1. Also notice thati · j = 1× 0 + 0× 1 + 0× 0 = 0, and similarly i · k = j · k = 0.

We can also express the length of a vector in terms of the scalar product with itself, |a|2 =

(a1)2 + (a2)2 + (a3)2 = a · a.

Example of the use of the scalar product:The work done W by a force F in displacing a particle by r is given bythe equation

W = F · r

Points to remember

a ·b = a1b1 + a2b2 + a3b3

a ·b = b · a and c · (a + b) = c · a + c ·b

a · a = (a1)2 + (a2)

2 + (a3)2 = |a|2

Example 4.8:

(a) Find the scalar product of the vectors a = 2i− 3j + k, b = i− j + 2k

(b) Find the scalar product of a = 2i + 3j, b = i− 2j.

Solution:

(a) a ·b = 2× 1 +−3×−1 + 1× 2 = 2 + 3 + 2 = 7

(b) a ·b = 2× 1 + 3×−2 = 2− 6 = −4

c© USQ, August 27, 2010 289

MAT1500 4.5. Angle between vectors

Exercise 4.9: Find the scalar product of the following vectors.

(a) 2i− 2j− 3k, −i− 3j− k

(b) 6i− 4j, −7i + 8j + 8k

(c) 3i− j− 9k, −3i + 5j + 8k

(d) 5i− 6j− 8k, 7j + 2k

(e) i− 6j + 3k, −3i− 2j + 5k

(f) −6i + 2j + 4k, −2i + 6j− 9k

(g) −5i− 2j + 3k, −7i− 7j + 9k

(h) −8i− 5j + 6k, −5i + 2j− k

4.5 Angle between vectors

In this section we investigate how the angle between two vectors is related to the scalarproduct. You may need to refresh your understanding of the cosine rule from Chapter 3.

Figure 4.7: Angle between vectors.

In the triangle in Figure 4.7, the lengths of sides are

OA = |a| ,OB = |b| , and

AB = |a− b| .

By the cosine rule, in triangle OAB,

AB2 = OA2 + OB2 − 2OA×OB× cos θ .

Substituting the lengths above, this can be written

|a− b|2 = |a|2 + |b|2 − 2|a||b| cos θ .

290 c© USQ, August 27, 2010

4.5. Angle between vectors MAT1500

But if we consider that the length squared of a vector is the scalar product of itself, wecan also say that

|a− b|2 = (a− b) · (a− b)

= (a− b) · a− (a− b) ·b= a · a− a ·b− a ·b + b ·b= |a|2 + |b|2 − 2a ·b .

So|a|2 + |b|2 − 2a ·b = |a|2 + |b|2 − 2|a||b| cos θ .

That isa ·b = |a||b| cos θ .

This result allows us to work out the angle between two vectors, since we can now seethat

cos θ =a ·b|a||b| .

The other important result that comes from this is that since the cosine of a right angle iszero, (that is cos 90 = 0), the scalar product of vectors that are perpendicular (also calledorthogonal) is zero.

This could be written if a ⊥ b, then a ·b = 0.

This result can also be used to find the projection of one vector b onto another a.

From the diagram you can see that the projection of b onto a (the length on a of the sideof a right angled triangle of which b is the hypotenuse) is |b| cos θ. See Figure 4.8.

Figure 4.8: The projection of b onto a.

But |b| cos θ =a ·b|a| = a ·b. This value is also called the component of b in the direction

of a.

c© USQ, August 27, 2010 291

MAT1500 4.5. Angle between vectors

Points to remembera ·b = |a||b| cos θ ;

cos θ =a ·b|a||b| ;

If a ⊥ b, then a ·b = 0;

The component of b in the direction of a is |b| cos θ =a ·b|a| = a ·b.

Example 4.10:

(a) Find the angle between the vectors a = 2i− 3j + k, b = i− j + 2k.

(b) Find the angle between a = 2i + 3j, b = i− 2j.

(c) Which of the following vectors are perpendicular to i + 2j− k?

i. −i + 2j− 2k

ii. i + j + 3k

iii. 2i + 4j− 10k

Solution:

(a)

cos θ =a ·b|a||b| =

2 + 3 + 2√4 + 9 + 1

√1 + 1 + 4

=7√

14√

6≈ 0.763 763

θ = cos−1(0.763 763) ≈ 0.70 radians = 40.2 .

(b)

cos θ =a ·b|a||b| =

2− 6√4 + 9

√1 + 4

=−4√13√

5≈ −0.496 14

θ = cos−1(−0.496 14) ≈ 2.09 radians = 119.75 .

(c) i. (i + 2j− k) · (−i + 2j− 2k) = 1×−1+ 2× 2− 1×−2 = −1+ 4+ 2 = 5,therefore these vectors are not perpendicular.

ii. (i+ 2j− k) · (i+ j+ 3k) = 1× 1+ 2× 1− 1× 3 = 1+ 2− 3 = 0, thereforethese vectors are perpendicular.

iii. (i + 2j− k) · (2i + 4j− 10k) = 1× 2+ 2× 4− 1×−10 = 2+ 8+ 10 = 20,therefore these vectors are not perpendicular.

Exercise 4.11: (a) Find the angle (in degrees and radians) between the vectors

i. 2i− 2j− 3k, −7i + 8j + 8k

ii. 6i− 4j, −3i + 5j + 8k

iii. 3i− j− 9k, 7j + 2k

iv. 5i− 6j− 8k, −3i− 2j + 5k

v. i− 6j + 3k, −2i + 6j− 9k

vi. −6i + 2j + 4k, −7i− 7j + 9k

vii. −5i− 2j + 3k, −5i + 2j− k

292 c© USQ, August 27, 2010

4.6. Review MAT1500

viii. −8i− 5j + 6k, −i− 3j− k.

(b) Find a vector perpendicular to:

i. 2i− 2j− 3k

ii. 6i− 4j

iii. 3i− j− 9k

iv. −3i + 5j + 8k

v. 5i− 6j− 8k

vi. 7j + 2k

vii. i− 6j + 3k

viii. −3i− 2j + 5k

ix. −6i + 2j + 4k

x. −2i + 6j− 9k

(c) Find the component of b in the direction of a where:

i. b = 2i− 2j− 3k, a = i.

ii. b = 6i− 4j, a = j.

iii. b = 3i− j− 9k, a = k.

iv. b = 5i− 6j− 8k, a =1√2(j + k).

v. b = i− 6j + 3k, a =1√3(i + j + k).

vi. b = −6i + 2j + 4k, a =1√14

(i + 3j− 2k).

vii. b = −5i− 2j + 3k, a = −i.

viii. b = −8i− 5j + 6k, a = −k.

4.6 Review

• A vector is a quantity that has magnitude and direction.

• Vectors are considered equal if they have the same magnitude and direction.

• If a = (a1, a2, a3) and b = (b1, b2, b3), and i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1).

– a = a1 i + a2 j + a3k and b = b1 i + b2 j + b3k

– a = b if and only if a1 = b1, a2 = b2, and a3 = b3.

– a + b = (a1 + b1)i + (a2 + b2)j + (a3 + b3)k, anda− b = (a1 − b1)i + (a2 − b2)j + (a3 − b3)k

– ka = ka1 i + ka2 j + ka3k.

– The length (or magnitude) vector a is |a| =√(a1)2 + (a2)2 + (a3)2 = a · a.

– a ·b = a1 × b1 + a2 × b2 + a3 × b3.

– i · i = j · j = k · k = 1, and i · j = i · k = j · k = 0.

– a ·b = |a||b| cos θ and cos θ =a ·b|a||b| .

– The component of b in direction of a is a ·b.

c© USQ, August 27, 2010 293

MAT1500 4.7. Post Test

4.7 Post Test

1. What are the coordinates of a point in Cartesian space which is reached by travel-ling 4 in the y direction, 5 in the z direction and then 6 in the x direction from theorigin?

2. What is the displacement vector in moving from the point P(2, 1, 4) to point Q(3, 2, 3)?

3. A displacement of (2,−1, 1) is applied at a point P(1, 2, 1). What are the coordinatesof the resulting position Q ?

4. Identify the following quantities as either vector or scalar quantities: acceleration,angular velocity (spin), area, density, displacement, electric field, electric potential,energy, force, length, magnetic field, mass, momentum, speed, temperature, torque,velocity?

5. Two forces are acting on a particle, F1 = 4i− 15j + 12k and F2 = 2i + j + 3k.

(a) What is the resultant (total) force acting on the particle?

(b) What is its magnitude?

(c) Find a unit vector in the direction of the resultant force.

6. (a) Find the scalar product of a = 3i + j, b = i− 2j.

(b) Find the scalar product of the vectors a = i− 2j + 4k and b = i− j− k.

7. (a) Find the angle between a = 1i + 3j, and b = 3i− 2j.

(b) Find the angle between the vectors a = 3i− 2j + 2k, and b = i + 2j + k.

8. Which of the following vectors are perpendicular to 2i + j− k?

(a) −i− 2j + 2k

(b) 2i + 4j− 10k

(c) i + 2j + 4k

9. Find the component of b = i− j + 3k in the direction of1√3(i + j− k).

294 c© USQ, August 27, 2010


4.8 Solutions


4.2 (a) (4, 2, 3)

(b) (1, 4, 13) (Coordinates are always in x y z order)

(c) (1, 2, 3)

(d) (5, 2, 1)

(e) (2,−1,−5)

(f) (p, q, r)

(g) (sin t, cos t, t)

(h) (t2, 1− t2, k)

(i) (1, 0, 0)

(j) (0, 1, 0)

(k) (0, 0, 1)

(l) (a1, a2, a3)

(m) (x0, y0, z0)

4.4 (a) (2, 3, 4)− (1, 1, 1) = (1, 2, 3)

(b) (−1, 1,−1)

(c) (2,−3, 0)

(d) (−6,−8,−7)

(e) (1, 0, 1)

(f) (−1,−2)− (4, 3) = (−5,−5)

(g) (−1, 1,−1)

(h) (12, 13,−14)

4.5 (a) (−1,−3,−1) + (2,−2,−3) = (1,−5,−4)

(b) (−1, 4, 8)

(c) (0, 4,−1)

(d) (5, 1,−6)

(e) (−2,−8, 8)

(f) (−2, 6) + (−6, 2) = (−8, 8)

(g) (−12,−9, 12)

(h) (−13,−3)

c© USQ, August 27, 2010 295


4.7 (a)acceleration vectorangular velocity (spin) vectorarea scalardensity scalardisplacement vectorelectric field vectorelectric potential scalarenergy scalarforce vectorlength scalarmagnetic field vectormass scalarmomentum vectorspeed scalartemperature scalartorque vectorvelocity vector

(b) a = 2i + 4j− 3k

(c) i. i + j + k + 2i + 3j + 4k = 3i + 4j + 5k

ii. i + 2k + j + k = i + j + 3k

iii. i + 5j + 2k + 3i + 2j + 2k = 4i + 7j + 4k

iv. 6i + 8j + 7k + 0i + 0j + 0k = 6i + 8j + 7k

v. −i + j + j + k = −i + 2j + k

vi. 4i + 3j + 2k +−i +−2j +−k = 3i + j + k

vii. i + 2k + j + k = i + j + 3k

viii. 0i + 0j + 0k + 12i + 13j +−14k = 12i + 13j− 14k

(d) i. −i− 3j− k− (2i− 2j− 3k) = −i− 3j− k− 2i + 2j + 3k = −3i− j + 2k

ii. −7i + 8j + 8k− (6i− 4j) = (−7− 6)i + (8−−4)j + (8− 0)k = −13i +12j + 8k

iii. (−3i + 5j + 8k)− (3i− j− 9k) = (−3− 3)i + (5−−1)j + (8−−9)k =

−6i + 6j + 17k

iv. 7j + 2k − (5i − 6j − 8k) = (0 − 5)i + (7 − −6)j + (2 − −8)k = −5i +13j + 10k

v. −3i − 2j + 5k − (i − 6j + 3k) = (−3 − 1)i + (−2 − −6)j + (5 − 3)k =

−4i + 4j + 2k

vi. −2i + 6j− 9k− (−6i + 2j + 4k) = (−2−−6)i + (6− 2)j + (−9− 4)k =

4i + 4j− 13k

vii. −7i− 7j+ 9k− (−5i− 2j+ 3k) = (−7−−5)i+(−7−−2)j+(9− 3)k =

−2i− 5j + 6k

viii. −5i + 2j− k− (−8i− 5j + 6k) = (−5−−8)i + (2−−5)j + (−1− 6)k =

3i + 7j− 7k

296 c© USQ, August 27, 2010


(e) |a| = 2. |3a| = 3|a| = 3× 2 = 6.

(f)2a 3b 2a− 3b

a = 6i− 4j 12i− 8j 33i− 32j− 24kb = −7i + 8j + 8k −21i + 24j + 24ka = −6i + 2j + 4j −12i + 4j + 8k −6i− 14k + 35kb = −2i + 6j− 9k −6i + 18j− 27ka = −8i− 5j + 6k −16i− 10j + 12k −i− 16j + 15kb = −5i + 2j− k −15i + 6j− 3k

(g) i.1√

22 + 22 + 32(2i− 2j− 3k) =

1√4 + 4 + 9

(2i− 2j− 3k) =1√17

(2i− 2j− 3k)

ii.1√52

(6i− 4j)

iii.1√91

(3i− j− 9k)

iv.1√98

(−3i + 5j + 8k)

v.1√125

(5i + 6j− 8k) =1

5√

5(5i + 6j− 8k)

vi.1√53

(7j + 2k)

vii.1√46

(i− 6j + 3k)

viii.1√38

(−3i− 2j + 5k)

ix.1√56

(−6i + 2j + 4k) =1

2√

14(−6i + 2j + 4k) =

1√14

(−3i + j + 2k)

x.1√121

(−2i + 6j− 9k) =1

11(−2i + 6j− 9k)

(h)

Figure 4.9: Resultant velocity is flow velocity plus velocity relative to flow.

R = 0.8i− 1.4j + 3.4i− 2.8j = (0.8 + 3.4)i− (1.4 + 2.8)j = 4.2i− 4.2j

c© USQ, August 27, 2010 297


(i)

Momentum after collision = Momentum particle 1 before + Momentum particle 2 before

= 4i− 2j + 2i + 4j

= 4i + 2i + 4j− 2j

= 6i + 2j

(j) Total electric field

E = E1 + E2

= 5i + 2j + 4i− 10j

= 9i− 8j

Force on charged particle

F = Eq

= (9i− 8j)× 6

= 54i− 48j

4.9 (a) (2i− 2j− 3k) · (−i− 3j− k) = 2×−1+−2×−3+−3×−1 = −2+ 6+ 3 = 7

(b) (6i− 4j) · (−7i + 8j + 8k) = −42− 32 + 0 = −74

(c) (3i− j− 9k) · (−3i + 5j + 8k) = −86

(d) (5i− 6j− 8k) · (7j + 2k) = −58

(e) (i− 6j + 3k) · (−3i− 2j + 5k) = 24

(f) (−6i + 2j + 4k) · (−2i + 6j− 9k) = −12

(g) (−5i− 2j + 3k) · (−7i− 7j + 9k) = 76

(h) (−8i− 5j + 6k) · (−5i + 2j− k) = 24

4.11 (a) i.

cos θ =(2i− 2j− 3k) · (−7i + 8j + 8k)|2i− 2j− 3k|| − 7i + 8j + 8k|

≈ −5454.85435

= −0.9844 ,

θ = 2.96 radians = 169.87 .

ii. cos θ = −0.5323, θ = 2.13 radians = 122.16.

iii. cos θ = −0.3600, θ = 1.94 radians = 111.10.

iv. cos θ = −0.6239, θ = 2.24 radians = 128.60.

v. cos θ = −0.8712, θ = 2.63 radians = 150.60.

vi. cos θ = 0.6392, θ = 0.88 radians = 50.27.

vii. cos θ = 0.5331, θ = 1.01 radians = 57.78.

viii. cos θ = 0.4585, θ = 1.09 radians = 62.71.

(b) There are many possible correct answers, as long as the scalar (dot) product ofyour answer with the given vector is zero then you are correct.

298 c© USQ, August 27, 2010


i. 2i− 2j− 3k. Choose the i and j components of the required perpendicularvector to be 1 and 1. The k component then needs to be 0 for the dotproduct to be 0. Therefore one answer can be be i + j + 0k = i + j.

ii. 6i− 4j; For 2 dimensional vectors, one sure way is to interchange the twocomponents and then change the sign of one of them. In this case oneanswer can be: 4i + 6j.

iii. 3i − j − 9k ; You can also do a similar thing with 3 dimensional vectorsif an answer is not immediately clear. Make one of the components ofthe required vector 0, and the use the process above for the other twocomponents. In this case make the k component 0, then interchange the 3and −1, and change the −1 to +1. Hence i + 3j + 0k = i + 3j.

iv. −3i + 5j + 8k; −i + j− k.

v. 5i− 6j− 8k; 6i + 5j.

vi. 7j + 2k; 2j− 7k.

vii. i− 6j + 3k; 6i + j.

viii. −3i− 2j + 5k; i + j + k.

ix. −6i + 2j + 4k; i + j + k.

x. −2i + 6j− 9k; 6i + 2j.

(c) i. b = 2i− 2j− 3k, a = i; the component of b in the direction of a is a ·b = 2.

ii. b = 6i− 4j, a = j; a ·b = −4.

iii. b = 3i− j− 9k, a = k; a ·b = −9.

iv. b = 5i− 6j− 8k, a =1√2(j + k); a ·b =

−14√2

.

v. b = i− 6j + 3k, a =1√3(i + j + k); a ·b =

−2√3

.

vi. b = −6i + 2j + 4k, a =1√14

(i + 3j− 2k); a ·b =−8√

14.

vii. b = −5i− 2j + 3k, a = −i; a ·b = 5.

viii. b = −8i− 5j + 6k, a = −k; a ·b = −6.


1. (6, 4, 5).

2. (1, 1,−1).

3. (3, 1, 2).

4.

c© USQ, August 27, 2010 299


acceleration vectorangular velocity (spin) vectorarea scalardensity scalardisplacement vectorelectric field vectorelectric potential scalarenergy scalarforce vectorlength scalarmagnetic field vectormass scalarmomentum vectorspeed scalartemperature scalartorque vectorvelocity vector

5. (a)

F = F1 + F2

= 4i− 15j + 12k + 2i + j + 3k

= 6i− 14j + 15k

(b)

|F| = |6i− 14j + 15k|=

√62 + (−14)2 + (15)2

=√

457

≈ 21.38

(c)

F =F|F|

=6i− 14j + 15k

21.38

6. (a) 1.

(b) −1.

7. (a) cos θ ≈ −0.2631, θ ≈ 1.837 radians ≈ 105.26.

(b) cos θ ≈ 0.0990, θ ≈ 1.472 radians ≈ 84.32.

8. (a) (2i + j− k) · (−i− 2j + 2k) = −6, therefore not perpendicular.

300 c© USQ, August 27, 2010


(b) (2i + j− k) · (2i + 4j− 10k) = 18, therefore not perpendicular.

(c) (2i + j− k) · (i + 2j + 4k) = 0, therefore perpendicular.

9.a ·b = − 3√

3= −√

3

c© USQ, August 27, 2010 301

Chapter 5

Matrices

Chapter contents5.1 What is a matrix? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

5.2 Matrix addition and subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 309

5.3 Multiplication of a matrix by a scalar . . . . . . . . . . . . . . . . . . . . . . . . 313

5.4 Transpose of a matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

5.5 Multiplication of a matrix by a matrix . . . . . . . . . . . . . . . . . . . . . . . . 316

5.6 Some special matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

5.6.1 Identity matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

5.6.2 The inverse matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334

5.7 Linear equations in matrix notation . . . . . . . . . . . . . . . . . . . . . . . . . 340

5.8 Solution of linear equations using the inverse matrix . . . . . . . . . . . . . . 345

5.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354

5.10 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355

5.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357



Objectives


• explain the concept of a matrix;

• find the order and elements of any matrix;

• add and subtract matrices;

• multiply a matrix by a constant;

• find the transpose of a matrix;

c© USQ, August 27, 2010 303

MAT1500

• multiply matrices together;

• define zero and identity matrices;

• express a set of simultaneous equations in matrix form;

• find the inverse of a 2× 2 matrix; and

• use the inverse to solve sets of two or three simultaneous equations.

Introduction

This topic might sound new to you, but in fact the roots of matrices go back to the 2ndcentury B.C., with some traces as far back as 4th century B.C. The Babylonians studiedproblems which lead to simultaneous equations (see Chapter 1) which still survive onclay tablets today. For example:

There are two fields whose total area is 1 800 square yards. One produces grain ata rate of 2

3 of a bushel per square yard, while the other produces grain at the rate ofa bushel per square yard. If the total yield is 1 100 bushels, what is the size of eachfield?

The Chinese developed and used matrices in the Han Dynasty (200 to 100 B.C.). Todayof course matrices are used in a vast variety of ways in all of the many disciplines fromelectrical engineering, to business, to population biology to archaeology. But what is amatrix?

Hint Coping with something new:

• skim the materials first — get an idea of the major concepts and topics;

• circle or note in a summary book words you do not understand;

• re-read the material concentrating fully;

• stop at examples and go through step by step. If steps are skipped writethem in now;

• do activities when you come to them;

• remember maths is learnt by doing not just reading;

• if you get stuck talk with colleagues, fellow students or your tutor.

304 c© USQ, August 27, 2010

5.1. What is a matrix? MAT1500

5.1 What is a matrix?

In our everyday lives we often represent data in table form to make it meaningful andeasy to read, for example cricket scores, the milkman’s weekly order, or an inventory fora tyre retailer.

Often this information is tabulated in the form of a matrix. The word matrix originallycame from the Latin word of the same form meaning womb — that is, from which some-thing originates. In mathematics it means a table of rows and columns, but it has anextremely wide usage in science, engineering, mathematics, business and the arts.

For example, it is used in:

• business, especially in planning and production;

• studying vibrations in car engines;

• solving linear equations;

• developing quantum theory;

• studying population structure; and

• in many circumstances where there are rotations, reflections or other distortions ofgeometrical figures such as in the theory of building construction, in electricity andmagnetism and in aerodynamics.

Tables are an excellent way of displaying data but come in a range of forms so it is difficultto perform operations on them. With numbers we have a standard way of depictingthem, for example, the number ‘4’ (not usually IV, 104 or ||||) so that we can manipulatethese numbers more easily (for example add, subtract, multiply, square). This is the samewith tables. If we standardise them into matrices we now have a much more powerfultool to manipulate them.

Imagine you are a school teacher keeping a record of marks on tests in Geography (G)and History (H) of the students in your class.

You might have a table that looks like this:

Geometry HistoryAlison 9 8

Bob 5 7Chris 8 8Diane 6 9

Chances are, however, that in these technological times you would use a spreadsheet and

c© USQ, August 27, 2010 305

MAT1500 5.1. What is a matrix?

end up with an array of numbers like this:

G HAlison

BobChrisDiane

9 85 78 86 9

An obvious but important observation: if we used the same numbers but changed theirplaces, it would no longer be the same array.

A list of numbers like this where the position in the array is significant is an example of amatrix. Each number in the matrix is called an element of that matrix.

Alison’s marks, [9 8] are a row of the matrix.

The history marks of the class,

8789

are a column of the matrix.

It is conventional to use upper case variable names to represent matrices (the plural ofmatrix, pronounced ‘may/tri/sees’), so since these were the students’ marks for the firstterm, we could write:

T1 =

9 85 78 86 9

Because T1 has 4 rows and 2 columns, it is called a 4× 2 matrix (said “4 by 2"). We say ithas dimension or order 4× 2. Note that the number of rows is always first.

A matrix which has either one row, or one column, is also called a vector. Thus Alison’smarks, [9 8] are a 1× 2 matrix, or a row vector.

The class’s history marks

8789

are a 4× 1 matrix or column vector. You can see that our definition of a vector has beenextended. We can now have more than three elements (we call them components inchapter 4). Stay tuned there will be more developments.

If the marks in term 2 were exactly the same for each student as they got in term 1, wecould say that the matrix T1 of term 2 results was equal to the matrix T1 . That is, two ma-trices are said to be equal if and only if every element in one is equal to the corresponding

306 c© USQ, August 27, 2010

5.1. What is a matrix? MAT1500

element in the other.

Elements are labelled by the row and column that they are part of. Thus in matrix T1,the element in the second row, first column would be labelled t21, and the element in thefourth row, second column would be labelled t42. Notice that the row number comes firstin the subscript. So t21 = 5 , and t42 = 9 . For example in our matrix the elements wouldbe labelled as follows.

T1 =

t11 t12

t21 t22

t31 t32

t41 t42

=

9 85 78 86 9

Example 5.1:

(a) Write down the orders of the following matrices.

A =[

1 2 3]

B =

123

C =

[1 −1 00 1 −1

]

D =

1 0−1 1

0 1

E =

[1 −1 00 1 −1

]

F =

[0 1 −11 −1 0

]

G =

1 12 23 3

(b) Using the matrices above, find any pairs of matrices which are equal.

(c) Using the matrices above, find a13, c13, f21, d22, g21, f23.

Solution:

(a) (1× 3), (3× 1), (2× 3), (3× 2), (2× 3), (2× 3), (3× 2).

(b) C = E .

(c) a13 = 3 , c13 = 0 , f21 = 1 , d22 = 1 , g21 = 2 , f23 = 0 .

c© USQ, August 27, 2010 307

MAT1500 5.1. What is a matrix?

Application: business The matrix below represents the probability of changein wheat prices with differences in weather conditions.

Change in wheat prices

↑ ↓ none

Weatherconditions

drymoderate

damp

0.6 0.1 0.30.4 0.2 0.40.1 0.1 0.8

If a33 means there is an 80% chance that the wheat prices will not change whenthe conditions are damp, what do you think a21 means?

Exercise 5.2: Write down the orders of the following matrices.

A =

[3 1−2 0

]

B =

[4 6 80 −4 2

]

C =[

3 1 2]

D =

4 60 −48 2

E =

[0 0 00 0 0

]

F =

341

G =

4 60 −4−2 3

H =[

4]

Exercise 5.3: Using the matrices above (from exercise 5.2), find a11, b13, c12, d21, e33, f31,g22, h11.

308 c© USQ, August 27, 2010

5.2. Matrix addition and subtraction MAT1500

Application: anthropology Certain kinds of data collected by an anthropol-ogist can be formatted as a matrix. For example this matrix represents datacollected on sex, race, height, income, marital status and years of education inrespective columns on 9 individuals in respective rows. Data have been coded, forexample, male would have a code of 1 and female a code of 2.

1 1 70 50 1 121 2 72 100 2 202 1 55 250 1 161 2 65 20 2 162 1 60 10 3 161 1 68 30 1 122 5 66 25 2 92 4 66 43 1 211 1 69 67 1 18

Points to Remember

• A matrix is an array of numbers called elements.

• A matrix consists of rows and columns.

• A matrix of m rows and n columns has dimension (order) m× n.

• A matrix consisting of one column or one row is called a vector.

• Matrices are equal if all corresponding elements are equal.

5.2 Matrix addition and subtraction

Suppose the marks for term 2 are as follows:

Chris History 6 Geography 7Bob Geography 6 History 8Alison Geography 9 History 9Diane History 6 Geography 9

Write these as a matrix T2 suitable for comparison with the results T1.

You probably chose the matrix

T2 =

9 96 87 69 6

so that the positions in the matrix had the same meaning as in T1. For example, theelement in the 3rd row, 2nd column is the mark for Chris in History.

c© USQ, August 27, 2010 309

MAT1500 5.2. Matrix addition and subtraction

Most teachers would want to get total marks at the end of the second term for eachstudent in each subject. How would you create such a matrix?

Clearly,

Ttotal =

9 + 9 8 + 95 + 6 7 + 88 + 7 8 + 66 + 9 9 + 6

=

18 1711 1515 1415 15

that is, elements in corresponding positions are added together. We define this as matrixaddition, and write the operation as

Ttotal = T1 + T2 .

Alternatively, you may be interested as the teacher to look at the improvement of eachstudent in each subject from term 1 to term 2. An obvious measure of this would be thedifference between the term 2 mark and the term 1 mark for each student/subject. Inmatrix form we write,

Tdifference =

9− 9 9− 86− 5 8− 77− 8 6− 89− 6 6− 9

=

0 11 1−1 −2

3 −3

Thus the elements in T1 are subtracted from the corresponding elements in T2. We definethis operation as matrix subtraction and write it as

Tdiff = T2 − T1 .

It would of course be nonsense to add or subtract matrices which do not match, that is,which do not have the same number of columns and the same number of rows. So we saymatrix addition and matrix subtraction are defined only for matrices of the same order.

Example 5.4: Refer to the matrices in Example 5.1. Are the following sums defined? Ifso, find the sum.

(a) A + B

(b) C + F

(c) F + C

(d) F + G

(e) G + D

(f) C + E + F

310 c© USQ, August 27, 2010

5.2. Matrix addition and subtraction MAT1500

Solution:

(a) Undefined.

(b) [1 0 −11 0 −1

]

(c) Same as (b).

(d) Undefined.

(e)

2 11 33 4

(f) [2 −1 −11 1 −2

]

Example 5.5: Using the matrices of Example 5.1 find the following differences if theyexist.

(a) A− B

(b) C− F

(c) E− C

(d) F−G

(e) G−D

(f) C + E− F

Solution:

(a) Undefined.

(b) [1 −2 1−1 2 −1

]

(c) [0 0 00 0 0

]

(d) Undefined.

(e)

0 13 13 2

(f) [2 −3 1−1 3 −2

]

c© USQ, August 27, 2010 311

MAT1500 5.2. Matrix addition and subtraction

Application: biology A field biologist noted three types of birds in two differentfields. He defined a 2× 3 matrix:

Quail Pheasant BrushCaucal Turkey

Field 1Field 2

[14054

2545

3312

]

Six months later, the quail population had decreased. The matrix below showsthe number of birds that had disappeared from the fields.

[5 0 0

25 2 8

]

To find the population they would subtract the second matrix from the first matrix.[

140 25 3354 45 12

]−[

5 0 025 2 8

]

=

[140− 5 25− 0 33− 054− 25 45− 2 12− 8

]

=

[135 25 33

29 43 4

]

So now there are 135 quail in field 1; 29 quail in field 2 and so on. This type ofcalculation would be done routinely using spreadsheet software.

The zero matrix

A matrix, in which all the elements are zero, is called the zero matrix. This could bethought of as the results sheet for your class of geography an history students at thebeginning of the term when no results have been entered. If you add a zero matrix to anyother matrix of the same size, it leaves the other matrix unchanged — just like adding zeroto a number. If you subtract two identical matrices, you are left with the zero matrix.

Points to Remember

• To add (or subtract) matrices add (or subtract) corresponding el-ements

• Matrices can only be added or subtracted if the dimensions arethe same.

• A matrix, in which all the elements are zero, is called the zeromatrix.

312 c© USQ, August 27, 2010

5.3. Multiplication of a matrix by a scalar MAT1500

5.3 Multiplication of a matrix by a scalar

The matrices T1 and T2 represent the marks out of 10 at the end of term 1 and term 2. Thereporting policy of the school may require that the marks be given as percentages. Thusevery element in each of the matrices T1 and T2, must be multiplied by 10. Suppose wecall the matrix of the term 1 marks expressed as percentages, P1, then

P1 = 10T1 = 10

9 85 78 86 9

=

90 8050 7080 8060 90

We call this operation multiplication by a scalar (a scalar in matrix arithmetic is an or-dinary number, in this case ‘10’). Remember scalars from Chapter 4. Of course if wewanted the end of year marks expressed as percentages we would need to multiply thematrix Ttotal by 5. So that Ptotal = 5Ttotal .

Example 5.6: Find

(a)

4

[2 −21 3

]

(b)2[

3 −1 2]+ 4

[1 −2 0

]

(c) 3A− 2B , where

A =

[1 3 12 0 3

]B =

[−1 0 2−1 2 1

]

Solution:

(a) [8 −84 12

]

(b) [6 −2 4

]+[

4 −8 0]=[

10 −10 4]

(c)

[3 9 36 0 9

]−[−2 0 4−2 4 2

]

=

[3 9 36 0 9

]+

[2 0 −42 −4 −2

]

=

[5 9 −18 −4 7

]

c© USQ, August 27, 2010 313

MAT1500 5.4. Transpose of a matrix

Application: biology If, in a particular year, the population of birds in a studydepicted by the matrix below decreased by 20%, how would you calculate the newpopulation size of birds?

Quail Pheasant BrushCaucal Turkey

Field 1Field 2

[14054

2545

3312

]

Here we would multiply the matrix by 0.8 (the population decreased by 20%, sothe remaining bird population would be 80%)

0.8×[

140 25 3354 45 12

]=

[0.8× 140 0.8× 25 0.8× 330.8× 54 0.8× 45 0.8× 12

]

=

[112 20 2643 36 10

]

(rounded to the nearest bird).

So now there are 112 quail in field 1, 43 quail in field 2, and so on.

Point to Remember

• To multiply a matrix by a scalar, multiply every element of thematrix by that number.

5.4 Transpose of a matrix

Our marks matrix was,

G HAlison

BobChrisDiane

9 85 78 86 9

The same information is communicated by the matrix

Alison Bob Chris Diane

GH

[9 5 8 68 7 8 9

]

where we have interchanged the rows and columns. Note that the first row is now thefirst column, the second row is now the second column and so on.

This is called the transpose of the matrix. The transpose of a matrix is denoted AT.

314 c© USQ, August 27, 2010

5.4. Transpose of a matrix MAT1500

• The element aij in A is the same as the element aji in AT.

• The order of a transpose matrix is the reverse of the original matrix, that is, if theorder of A is (5× 3) the order of AT is (3× 5).

• A matrix A where A = AT is called a symmetric matrix. It must always be squareand elements are symmetrical about the diagonal from top left to bottom right.

Example 5.7: Find the transposes of the following.

(a)

A =

2 11 −23 1

(b)B =

[3 1 2 4

]

(c)

C =

1 3 43 2 74 7 1

Solution:

(a)

AT =

[2 1 31 −2 1

].

(b)

BT =

3124

(c) CT = C , C is symmetric.

Exercise 5.8: Given

A =

[3 1−2 0

]B =

[6 78 4

]C =

[1 −12 1

]

find:

(a) A + B

(b) B + C

(c) A− B + C

(d) 3A− 2B + C

(e) AT

c© USQ, August 27, 2010 315

MAT1500 5.5. Multiplication of a matrix by a matrix

(f) BT

Exercise 5.9: Find

(a)

3 1 −12 0 31 4 2

−

4 0 11 −1 00 1 3

+ 3

−1 −2 0−4 0 1

3 1 2

(b)

3

[−1 0 1

1 1 1

]+ 4

[1 2 3−1 1 0

]− 2

[0 1 23 −1 4

]

Exercise 5.10: Can a matrix of order (3× 4) be symmetric?

Hint Give yourself a do-able goal:

• divide your work into small, short-range subgoals;

• do not set a goal as vague and large as, ‘I am going to spend all weekendstudying non-stop!’ – you will only set yourself up for failure and discourage-ment;

• set your goal when you sit down to study but before you begin to work;

• set a goal that you can reach – you may, in fact, do more than your goal butset a reasonable goal even if it seems too easy;

• take the time block that you have scheduled for study and set a reachablestudy goal, (for example, finish one activity);

• tell somebody else about your goals;

• reward yourself when you achieve your goal.

Points to Remember

• The transpose of matrix A is the matrix found by interchangingrows and columns

• The transpose of a matrix is denoted AT.

• If matrix A = AT, the matrix is said to be symmetric.

5.5 Multiplication of a matrix by a matrix

We calculated before the total marks for each student in geography and history as being

Ttotal =

18 1711 1515 1415 15

316 c© USQ, August 27, 2010

5.5. Multiplication of a matrix by a matrix MAT1500

where each row contains a student’s marks and each column contains the whole class’smarks for geography and history respectively. The marks, if you remember, are out of amaximum 20.

In this complicated school that you teach in, geography has a 3 : 2 weighting over historyin determining a final percentage social science mark. That is, for each student, the socialscience percentage is 3× (Total geography mark) +2× (Total history mark). In matrixterms we have a new matrix S that represents the social science percentages of each stu-dent by a mark in the usual row. This matrix is found by multiplying each element in thefirst column by 3, and each element in the second column by 2, and then adding the tworesults.

If we define the weighting matrix as

W =

[32

]

we call the process just described as the product of Ttotal with W.

S = Ttotal ×W =

18 1711 1515 1415 15

×[

32

]

=

18× 3 + 17× 211× 3 + 15× 215× 3 + 14× 215× 3 + 15× 2

=

88637375

If the weighting matrix included an alternative scheme, with a ratio 4 : 1, as a secondcolumn, that is,

W =

[3 42 1

]

c© USQ, August 27, 2010 317


we write the result of the second weighting as a second column of S. That is,

S = Ttotal ×W =

18 1711 1515 1415 15

×[

3 42 1

]

=

18× 3 + 17× 2 18× 4 + 17× 111× 3 + 15× 2 11× 4 + 15× 115× 3 + 14× 2 15× 4 + 14× 115× 3 + 15× 2 15× 4 + 15× 1

=

88 8963 5973 7475 75

Can you see the pattern? The element in the first row, first column of S is the ‘product’of the first row of Ttotal and the first column of W. The element in the third row andsecond column of S is the ‘product’ of the third row of Ttotal and the second column ofW.

In general, the element in the nth row and mth column of S is the ‘product’ of the nth rowof Ttotal and the mth column of W. Here of course we are defining the term ‘product’ ofa row and column as being the result of multiplying each of the elements in the row bythe corresponding element in the column and summing the resultants. Do you recognisethis as the scalar product of vectors we defined in Section 4.4?

We call this process matrix multiplication.

If you reflect on this for a minute, and think about which rows and columns produce thebottom right hand element, you will recognise that the resultant matrix (in this case S),must have the same number of rows as the first matrix (Ttotal), and the same number ofcolumns as the second matrix (W).

Also reflect for a moment what would happen if the number of elements in the rows ofthe first matrix was not the same as the number of elements in the columns of the secondmatrix. The process of finding the product of a row with a column would not work. Forexample try multiplying

[2 1 3

]×[

11

]

We say this product is undefined.

Hint When the going gets tough. . .

• re-read what you do know;

• move forward in the study book to see if it comes together;

• talk it out with somebody, like your study partner;

318 c© USQ, August 27, 2010

5.5. Multiplication of a matrix by a matrix MAT1500

• put your questions to the Discussion group;

• contact your tutor — there is no such thing as a stupid question.

. . . the tough get studying.

To summarise, the product AB exists only when the number of columns of A equals the numberof rows of B.

As a further example, consider

A =

3 41 32 1

and B =

[2 10 3

]

There are six positions in the product matrix to be filled.

If Z = AB , then

z11 is obtained by forming the product of the 1st row of A and 1st column of B.

z11 = 3× 2 + 4× 0 = 6 + 0 = 6

z12 is obtained by forming the product of the 1st row of A and 2nd column of B.

z12 = 3× 1 + 4× 3 = 3 + 12 = 15

z21 is obtained by forming the product of the 2nd row of A and 1st column of B.

z21 = 1× 2 + 3× 0 = 2 + 0 = 2

z22 is obtained by forming the product of the 2nd row of A and the 2nd column of B.

z22 = 1× 1 + 3× 3 = 1 + 9 = 10

c© USQ, August 27, 2010 319


z31 is obtained by forming the product of the 3rd row of A and 1st column of B.

z31 = 2× 2 + 1× 0 = 4 + 0 = 4

z32 is obtained by forming the product of the 3rd row of A and the 2nd column of B.

z32 = 2× 1 + 1× 3 = 2 + 3 = 5

Thus, to find AB we multiply each column of B by the first row of A, then multiply eachcolumn by the second row of A, and so on.

Z = AB =

3 41 32 1

[

2 10 3

]

=

3× 2 + 4× 0 3× 1 + 4× 31× 2 + 3× 0 1× 1 + 3× 32× 2 + 1× 0 2× 1 + 1× 3

=

6 + 0 3 + 122 + 0 1 + 94 + 0 2 + 3

=

6 152 104 5

Note that the product BA is impossible. We cannot multiply a (2× 2) by a (3× 2). Ingeneral, if A is of order (m× n) and B is of order (n× p) then AB exists and is of order(m× p) but BA exists only if p = m.

To summarise: when we want to multiply two matrices together, we must do two thingsbefore we do the multiplication:

• See if the number of columns of the first matrix matches the number of rows of thesecond matrix.

• Decide on the order of the resulting matrix. The resulting matrix will have thenumber of rows in the first matrix and the number of columns in the second matrix.

If you write the two matrix orders side by side, you can see this more clearly.

Application: computer science In computer graphics, computer scientistsmanipulate objects which may include light sources, cameras, and models in theirscenes. Each of these is likely to be defined in its own coordinate system and thenplaced within the scene they are modelling. A necessary operation in computergraphics is to convert the coordinates of a point in one frame, to coordinates ofthe point in a second frame. These conversions can be defined via matrices and

320 c© USQ, August 27, 2010

5.6. Some special matrices MAT1500

the actual conversions carried out via matrix multiplication. A simplified examplewould be the following.

To scale the x-direction by a factor of 2 and the y-direction by a factor of 3, wewould have to multiply the coordinate matrix by a scaling matrix, that is,

[2 00 3

] [xy

]=

[2x + 0× y0× x + 3y

]=

[2x3y

]

Points to Remember

• The element in the nth row/mth column of the matrix product ABis the ‘scalar product’ of the nth row of A with the mth column ofB.

• For the matrix product AB to be defined, the number of columnsof matrix A must be equal the number of rows of column B.

• If the dimension of A is m× n and the dimensions of B are n× p,then the product AB is defined and has dimension m× p.

5.6 Some special matrices

We have looked at how to add, subtract and multiply matrices. We saw that in manycases this was similar to the ordinary decimal system with a few restrictions. Let uscontinue with this comparison. We cannot really divide by a matrix, but we can look atsome similar operations in the decimal number system related to division. These are theidentity and the inverse.

5.6.1 Identity matrices

Look at the following example in the real number system.

3× 1 = 3

The number 1 is called the identity element of multiplication, since when you multiply1 by any number it gives that number. (That is, x× 1 = x , or 1× x = x).

Similarly in matrix multiplication there is an identity matrix, such that if you multiply itby any matrix the resulting matrix is the same matrix. Look at the following examples:

[1 00 1

]×[

3 14 5

]=

[3 + 0 1 + 00 + 4 0 + 5

]=

[3 14 5

]

c© USQ, August 27, 2010 321

MAT1500 5.6. Some special matrices

which is the same matrix we started with.

1 0 00 1 00 0 1

×

1 0.7 1−5 3 −9

3 0 4

=

1 + 0 + 0 0.7 + 0 + 0 1 + 0 + 00 +−5 + 0 0 + 3 + 0 0 +−9 + 00 + 0 + 3 0 + 0 + 0 0 + 0 + 4

=

1 0.7 1−5 3 −9

3 0 4

Again, this is the same matrix we started with.

Look carefully at the matrix multiplication above. Can you answer the following ques-tions?

1. Does the first matrix always have to be the identity matrix?

2. Does the identity matrix have to have the same number of rows and columns?

3. Does the diagonal of ones have to run down from left to right with the remainingelements 0?

Let us have a closer look at these points.

1. If we multiplied an identity matrix by a matrix would we still get the original ma-trix? [

3 14 5

]×[

1 00 1

]=

[3 + 0 0 + 14 + 0 0 + 5

]=

[3 14 5

]

Yes! Pre- or post-multiplying by the identity matrix does not change a square ma-trix. Try it yourself. You will see that no matter what values we have in the originalmatrix, it will always work.

2. What if we had a matrix that was not square?

2 34 71 2

This is a 3 × 2 matrix. If we multiplied it by another matrix and we wanted theoriginal matrix, what size would it have to be?

It would have to be a 3× 3 matrix since a 3× 3 times a 3× 2 give a 3× 2 matrix.

1 0 00 1 00 0 1

2 34 71 2

=

2 34 71 2

322 c© USQ, August 27, 2010


But what if we post-multiplied by an identity matrix? Can you see a problem? Sincethe first matrix is a 3× 2 matrix we need the second matrix to have order 2× 2, sowe can get a 3× 2.

So would an identity matrix [1 00 1

]

work?

2 34 71 2

[

1 00 1

]=

2 + 0 0 + 34 + 0 0 + 71 + 0 0 + 2

Most of the matrices you will be dealing with in relation to the identity will besquare matrices, but as you can see you have to be careful when pre- and post-multiplying matrices.

Remember that an identity matrix is always a square matrix (that is, the number ofrows and columns are equal).

3. Try this multiplication:

0 0 10 1 01 0 0

2 34 71 2

Did you get the original matrix?

No? You should have

1 24 72 3

An identity matrix must always have the 1’s running from top left to bottom right.

Example 5.11: Find the following matrix products where they exist.

(a)[

1 2] [ 3

4

]

(b)

[3 4

]

123

(c)

[−1 3 2 5

]

2−3−4

7

c© USQ, August 27, 2010 323


(d)

[2 0 −3 0

]

050−1

Solution:

(a) [1× 3 + 2× 4 = 3 + 8] = [11] , notice the similarity with the scalar product(i + 2j) · (3i + 4j).

(b) Product not defined.

(c) [−2− 9− 8 + 35] = [16] .

(d) [0 + 0 + 0 + 0] = [0] = 0 .

Note that in matrix algebra, AB = 0 does not necessarily mean that either A or Bmust be a zero matrix.

Example 5.12: Find AB and BA when

A =

[1 −12 3

]and B =

[−2 1−1 4

]

Solution:

A is (2× 2) and B is (2× 2).

AB =

[1 −12 3

] [−2 1−1 4

]

=

[1×−2 +−1×−1 1× 1 +−1× 42×−2 + 3×−1 2× 1 + 3× 4

]

=

[−2 + 1 1− 4−4− 3 2 + 12

]

=

[−1 −3−7 14

]

324 c© USQ, August 27, 2010


BA is also defined and is of order (2× 2).

BA =

[−2 1−1 4

] [1 −12 3

]

=

[−2× 1 + 1× 2 −2×−1 + 1× 3−1× 1 + 4× 2 −1×−1 + 4× 3

]

=

[−2 + 2 2 + 3−1 + 8 1 + 12

]

=

[0 57 13

]

Note that although both products AB and BA exist, they are not equal. Thus ma-trices do not in general obey the commutative law of ordinary arithmetic, that is,xy 6= yx. The order in which we multiply matrices is therefore important. In theproduct AB we say that B is pre-multiplied by A. In the product BA we say that Bis post-multiplied by A.

Example 5.13: Find AB and BA when

A =[

1 2 3]

and B =

456

Solution:

A is (1× 3) and B is (3× 1).

Consider the product AB.

AB = [4 + 10 + 18] = [32]

that is, a single number.

Consider the product BA.

c© USQ, August 27, 2010 325


BA =

456

[

1 2 3]=

4 8 125 10 156 12 18

Note that not only are AB and BA different, but here they have different orders aswell.

Example 5.14: Find AI and IA when

A =

[a bc d

]and I =

[1 00 1

]

Solution:

AI =

[a bc d

] [1 00 1

]

=

[a× 1 + b× 0 a× 0 + b× 1c× 1 + d× 0 c× 0 + d× 1

]

=

[a bc d

]

= A

IA =

[1 00 1

] [a bc d

]

=

[1× a + 0× c 1× b + 0× d0× a + 1× c 0× b + 1× d

]

=

[a bc d

]

= A

Thus, in matrix algebra, pre- or post-multiplying by I does not change a matrix.

Example 5.15: Given the following matrices,

A =[

1 2 3]

B =

[1 −1 00 1 −1

]

C =

1 2 13 0 11 0 3

D =

3 11 −12 2

326 c© USQ, August 27, 2010


find these products if they exist.

(a) AB

(b) DB

(c) BTC

(d) AD

(e) ACD

(f) I2B

Solution:

(a) AB is undefined.

(b)

DB =

3 11 −12 2

[

1 −1 00 1 −1

]

=

3 + 0 −3 + 1 0− 11 + 0 −1− 1 0 + 12 + 0 −2 + 2 0− 2

=

3 −2 −11 −2 12 0 −2

(c)

BTC =

1 0−1 1

0 −1

1 2 13 0 11 0 3

and is undefined.

(d)

AD =[

1 2 3]

3 11 −12 2

=[

3 + 2 + 6 1− 2 + 6]

=[

11 5]

c© USQ, August 27, 2010 327


(e) Find AC first.

AC =[

1 2 3]

1 2 13 0 11 0 3

=[

1 + 6 + 3 2 + 0 + 0 1 + 2 + 9]

=[

10 2 12]

Thus

ACD =[

10 2 12]

3 11 −12 2

=[

30 + 2 + 24 10− 2 + 24]

=[

56 32]

(f)

I2B =

[1 00 1

] [1 −1 00 1 −1

]

=

[1 −1 00 1 −1

]

= B

as expected.

Example 5.16: The number of electrical appliances stored in the rooms of three motelsare given below:

Motel A: 10 television sets, 6 jugs and 5 irons.Motel B: 12 fans, 2 jugs and 14 heaters.Motel C: 14 fans, 8 televisions, 10 heaters and 6 irons.

(a) Construct a matrix R to represent the appliances of the three motels withcolumns representing televisions, fans, jugs, heaters and irons in that order.

(b) The power consumption matrix is

Power

(Watts)

P =

400100

1 0001 0001 500

TVfanjugheateriron

328 c© USQ, August 27, 2010


Find the column matrix B which is the maximum power consumption of eachmotel.

(c) Let U =[

1 1 1]

. Find S = UR and explain its meaning.

(d) Find UB and explain its meaning.

(e) Find SP and explain its meaning.

Solution:

(a) The rows of R represent motels and the columns represent appliances, that is,

TV Fan Jug Heater Iron

R =

10 0 6 0 50 12 2 14 08 14 0 10 6

ABC

In other words,R ≡ (motel× appliance)

(b) Similarly, P may be interpreted as:

P ≡ (appliance× power)

Thus, the product RP may be interpreted as (motel× power), that is, each rowis the power consumption for a motel.

RP = B =

4 000 + 0 + 6 000 + 0 + 7 5000 + 1 200 + 2 000 + 14 000 + 03 200 + 1 400 + 10 000 + 9 000

=

17 50017 20023 600

Thus, at most the appliances will consume

17 500 W in Motel A17 200 W in Motel B23 600 W in Motel C

(c)

S =[

1 1 1]

10 0 6 0 50 12 2 14 08 14 0 10 6

=[

18 26 8 24 11]

S represents the total numbers of each appliance across the three motels.

c© USQ, August 27, 2010 329


(d)

UB =[

1 1 1]

17 50017 20023 600

= 58 300

and this is the total power consumption across the three motels.

(e) SP = URP = UB , therefore it is the same as (d).

Example 5.17: A certain timber mill receives an order for flooring boards, weather boardsand framing timber. The number of units of each is given by the requirement matrixR:

R =[

100 50 75]

where r1 is the number of units of flooring boards, r2 is the number of units ofweather boards, and r3 is the number of units of framing timber.

Each type of timber requires time in the kiln for drying and time for milling. Therequired times in hours per unit of timber are shown in the process matrix P:

Drying Milling

P =

20 5040 600 30

Flooring boardWeather boardFraming timber

Find the total processing time required in each section for the current order.

Solution:

R is (units × type), P is (type × process), hence RP will be (units × process).

Check: R is (1× 3), and P is (3× 2), thus RP is (1× 2).

RP =[

100 50 75]

20 5040 600 30

=[

2 000 + 2 000 + 0 5 000 + 3 000 + 2 250]

=[

4 000 10 250]

That is, the total processing times are composed of 4 000 hours drying and 10 250 hoursmilling.

330 c© USQ, August 27, 2010


Example 5.18: A company manufacturing televisions, cars and refrigerators has the fol-lowing requirement matrix, R, to make one unit of each product.

TV Car Refrigerator

R =

1 5 20 6 11 4 00.5 1 1

CopperSteelGlassAluminium

(Note that R is (material × product).)

The sales were divided between local and export markets as given by the consump-tion matrix, C.

Local Export

C =

2 05 13 2

TVCarRefrigerator

(Note here that C is (product ×market).)

Find the quantities of raw materials required for:

(a) the local market.

(b) the export market.

Solution:

RC =

1 5 20 6 11 4 00.5 1 1

2 05 13 2

Local Export

=

33 933 822 49 3

CopperSteelGlassAluminium

Here RC is (materials ×market). (Check that the product RC is defined!)

The first column gives the materials required for the local market and the secondcolumn gives the materials required for the export market.

Example 5.19: A building contractor has orders for 5 Ranch-type houses, 7 Cape Codhouses and 12 Colonial-style houses. This information is represented by the matrixA.

A =[

5 7 12]

c© USQ, August 27, 2010 331


(where A is number × type).

The raw materials required to build each type of house are given by the matrix R:

Steel Wood Glass Paint Labour

R =

5 20 16 7 177 18 12 9 216 25 8 5 18

RanchCape CodColonial

(where R is type ×material).

The cost of purchase and transport to the building site of the materials is given bythe cost matrix Q:

Purchase Transport

Q =

15 4.58 25 31 0.5

10 0

SteelWoodGlassPaintLabour

(where Q is material × costs).

(a) Calculate the raw materials required.

(b) Find purchase and transport costs for each type of home.

(c) Find ARQ and explain what it represents.

(d) Let

E =

[11

]

Find ARQE and explain what it represents.

Solution:

(a)

AR =[

5 7 12]

5 20 16 7 177 18 12 9 216 25 8 5 18

Steel Wood Glass Paint Labour

=[

146 526 260 158 448]

AR is (number ×material) and gives the units of raw material needed.

332 c© USQ, August 27, 2010


(b)

RQ =

5 20 16 7 177 18 12 9 216 25 8 5 18

15 4.58 25 31 0.5

10 0

Purchase Transport

=

492 114528 108515 103.5

RQ is (type × cost) and gives the purchase and transport cost of materials foreach type of house.

(c)

ARQ =[

5 7 12]

492 114528 108515 103.5

Purchase Transport

=[

12 336 2 568]

ARQ is (number × cost) and gives the purchase and transport costs for thetotal order.

(d)

ARQE =[

12 336 2 568] [ 1

1

]

= 14 904

which is the total cost of the order.

Exercise 5.20: Find, if they exist, the following products.

(a)

1 −3 22 1 −34 −3 −1

1 4 1 02 1 1 11 −2 1 2

(b)

123

[−1 2 1

]

c© USQ, August 27, 2010 333


(c)

[−1 2 1

]

1 4 1 02 1 1 11 −2 1 2

(d)

1 4 1 02 1 1 11 −2 1 2

123

Exercise 5.21: A consumer buys 2 dozen eggs, 6 apples, 4 oranges and 5 lemons.

(a) Express the purchases as a row matrix R.

(b) If eggs cost $2.12 per dozen, apples cost 20 cents each, oranges cost 22 centseach and lemons cost 30 cents each, express the price per unit of each item asa column matrix P.

(c) Obtain the total amount spent.

Exercise 5.22: Sue, Mary and Paul purchase the following items:

Sue: 12 apples, 8 lemons, 6 pears.Mary: 2 lemons, 1 dozen bananas, 18 oranges.Paul: 2 dozen bananas, 6 apples, 9 oranges, 4 pears.

The cost per item is 18 cents for an apple, 22 cents for a banana, 16 cents for a lemon,20 cents for an orange and 16 cents for a pear.

Find matrices to express:

(a) the purchases for each person;

(b) the cost for each person; and

(c) the number of the purchases of each type of fruit.

5.6.2 The inverse matrix

Let us revisit the real number system. Have a look at the equation:

3× 13= 1

Any number multiplied by its reciprocal is equal to 1. In mathematics terms, we call 13

the multiplicative inverse of 3. You have come across this idea of inverses in arithmetic,in functions, and now you see it is also an important part of matrices.

An inverse matrix, when multiplied by the original matrix gives the identity matrix.

Let us have a look at some examples.

334 c© USQ, August 27, 2010


Example 5.23: Is

A =

17

47

−17

37

the inverse of

B =

[3 −41 1

]?

To show that the matrix is the inverse, we multiply them together. If the result isthe identity matrix, then it is the inverse.

17

47

−17

37

[3 −41 1

]=

37+

47−4

7+

47

−37+

37

47+

37

=

[1 00 1

]

or we could have multiplied them the other way:

[3 −41 1

]

17

47

−17

37

=

37+

47

127− 12

7

17− 1

747+

37

=

[1 00 1

]

So A is the inverse of B. Since they are square matrices it does not matter whetherwe pre- or post-multiply the inverse.

Thus[

3 −41 1

]and

17

47

−17

37

are inverses of each other.

In summary, this means that if A is a matrix and A−1 is its inverse, then

AA−1 = I = A−1A

c© USQ, August 27, 2010 335


Inverse of a 2× 2 matrix

The inverse of a (2× 2) matrix

A =

[a bc d

]

is given by

A−1 =1

ad− bc

[d −b−c a

]

This result can be proved as follows:

AA−1 =

[a bc d

]· 1

ad− bc

[d −b−c a

]

=1

ad− bc

[ad− bc −ab + abcd− cd −bc + ad

]

=

[1 00 1

]

= I2

Similarly, A−1A = I2 .

Note that the inverse exists only if ad− bc 6= 0 .

ad− bc is called the determinant of A and is usually denoted by Det A, ∆, or

∣∣∣∣∣a bc d

∣∣∣∣∣ .

So for the inverse of A to exist, it is essential that the determinant of A be non-zero, sincewe are dividing by this number.

Example 5.24: Find inverses (if they exist) of the following matrices.

(a) [2 00 3

]

(b) [0 20 3

]

(c) [−5 1

3 2

]

(d) [6 43 2

]

336 c© USQ, August 27, 2010


Solution:

(a)

16− 0

[3 00 2

]=

12

0

013

(b)1

0− 0

[3 −20 0

]

and since 10 is undefined, no inverse exists.

(c)

1−10− 3

[2 −1−3 −5

]=

− 213

113

313

513

(d)1

12− 12

[2 −4−3 6

]

therefore no inverse exists.

Note: Always check your answer by multiplying it with the original matrix. Theresult should be the identity matrix, I.

Application: cryptology There are many ways to encrypt a message. One wayis to use a matrix and its inverse. First develop an invertible matrix such as

A =

−1 5 −1−2 11 7

1 −5 2

Now get your message and code it simply using numbers for letters. Put thesenumbers in matrix form. Let us call it B. Then calculate AB and send this. Atthe other end they will need to know the inverse of A so that they can uncodeAB using A−1(AB) = B .

Exercise 5.25: Find the inverses of these matrices.

(a) [1 −4−1 3

]

(b) [2 31 −4

]

c© USQ, August 27, 2010 337


(c) [1 23 4

]

(d) [4 68 6

]

Exercise 5.26: Show that the matrices below are inverses of each other.

A =

[2 15 3

]B =

[3 −1−5 2

]

Exercise 5.27: Without determining the inverse of the matrices below, decide whether ornot it is possible to find the inverse matrix.

(a) [−1 7−4 3

]

(b) [8 44 2

]

How to find the inverse of matrices larger than 2 × 2 (for interestonly) There are many ways to find the inverse of a matrix. One method is calledrow reduction or Gaussian elimination. In this method, a given matrix A is changedto the identity matrix I by a systematic series of row operations. At the same timethe identity matrix I is changed to the inverse of A by identical row operations.

338 c© USQ, August 27, 2010


For example,

1 2 3 1 0 02 5 3 0 1 01 0 8 0 0 1

1 2 3 1 0 00 1 −3 −2 1 00 −2 5 −1 0 1

We added −2 times the first

row to the second row, and−1 times the first row to thethird.

1 2 3 1 0 00 1 −3 −2 1 00 0 −1 −5 2 1

We added 2 times the

second row to the third.

1 2 3 1 0 00 1 −3 −2 1 00 0 1 5 −2 −1

We multiplied the third row

by −1.

1 2 0 −14 6 30 1 0 13 −5 −30 0 1 5 −2 −1

We added 3 times the third

row to the second and −3times the third row to thefirst.

1 0 0 −40 16 90 1 0 13 −5 −30 0 1 5 −2 −1

We added −2 times the

second row to the first.

Therefore A−1 =

−40 16 9

13 −5 −35 −2 −1

Thus we have found the inverse. Today we are lucky in that most of this type ofcalculation can be done by computer or on a graphics calculator. Do not learnthis.

c© USQ, August 27, 2010 339

MAT1500 5.7. Linear equations in matrix notation

Points to Remember

• An identity matrix is a square matrix with ‘1’s down the diagonaland ‘0’s elsewhere.

• If I is an identity matrix of the appropriate order for the productto be defined, then IA = A and AI = A.

• AB is usually not equal to BA.

• A−1 is the inverse of A if AA−1 = I = A−1A.

• The inverse of a (2 × 2) matrix

[a bc d

]is given by

A−1 =1

ad− bc

[d −b−c a

].

5.7 Linear equations in matrix notation

Suppose we want to produce a model to describe and forecast economic behaviour. Thereare many variables involved in the modelling process. Economists look at such things asaggregate demand, consumption, investment, supply and demand and may obtain a setof equations such as the ones below:

1Y− 1C− 1I + 0R = G

−bY + 1C + 0I + 0R = a

−dY + 0C + 1I − eR = 0

1Y + 0C + 0I +gf× R =

1f×M

With your knowledge of solving simultaneous equations in Chapter 1, this would be avery difficult process. With the help of matrices, the solution of 4 equations with 4 un-knowns (or even more) becomes much simpler.

Let us think about the systems of linear equations that we studied in Chapter 1. They canbe expressed in terms of matrix notation.

Consider the system of equations:

2x− 5y + 3z = 8

−x + 2y + 5z = 3

x + 6y− 2z = −1

We could express this as an equality between two vectors representing the left sides and

340 c© USQ, August 27, 2010

5.7. Linear equations in matrix notation MAT1500

right sides of the equations.

2x− 5y + 3z−x + 2y + 5zx + 6y− 2z

=

83−1

However the left hand side is in the form of the product of two matrices,

2 −5 3−1 2 5

1 6 −2

and

xyz

So we can write the matrix equation as

2 −5 3−1 2 5

1 6 −2

xyz

=

83−1

This is of the formAX = B

where

A =

2 −5 3−1 2 5

1 6 −2

X =

xyz

and B =

83−1

Example 5.28: Find A, X and B for the following systems of linear equations.

(a)

x− 3y = 7

3x− 5y = 6

(b)

2x− y + 5 = 0

x + 2y− 7 = 0

(c)

3u− 5v = 2

−u + 2v = −5

c© USQ, August 27, 2010 341


(d)

3x + 2y− z = 10

3x− 3z = 5

3y + 5z = 6

(e)

3p− 2q− r = 1

2p− 5q + 2r = 4

(f)

3x + 2y = 7

y− 5z = 8

4x + 7z = −2

(g)

u + v = w

2v + 3w = u

u + v + w = 1

Solution:

(a)

A =

[1 −33 −5

]X =

[xy

]B =

[76

]

(b)

A =

[2 −11 2

]X =

[xy

]B =

[−5

7

]

(c)

A =

[3 −5−1 2

]X =

[uv

]B =

[2−5

]

(d)

A =

3 2 −13 0 −30 3 5

X =

xyz

B =

1056

(e)

A =

[3 −2 −12 −5 2

]X =

pqr

B =

[14

]

342 c© USQ, August 27, 2010

5.7. Linear equations in matrix notation MAT1500

(f)

A =

3 2 00 1 −54 0 7

X =

xyz

B =

78−2

(g)

A =

1 1 −1−1 2 3

1 1 1

X =

uvw

B =

001

Example 5.29: The tolls across Pollution River Bridge are 50 cents for cars, $1 for trucksand $2 for buses. A total of 5 000 vehicles crossed the bridge on a particular dayyielding $3 100 in tolls and there were three times as many trucks as buses. Write amatrix equation for this problem.

Solution:

Let x be the number of cars, y the number of trucks and z be the number of buses.Then,

x + y + z = 5 000 (total number of vehicles)x2+ y + 2z = 3 100 (total tolls)

y = 3z (three times as many trucks as buses)

Thus we have a set of three simultaneous equations.

x + y + z = 5 00012

x + y + 2z = 3 100

y− 3z = 0

Expressing this set of equations as the matrix equation AX = B , we find,

A =

1 1 112 1 20 1 −3

X =

xyz

B =

5 0003 100

0

Exercise 5.30: Express the following systems of equations in the matrix form AX = B .

(a)

2x− y = 3

x + 2y = 4

(b)

x + y + z = 10

2x− y− z = 5

c© USQ, August 27, 2010 343


(c)

x− y + z = −2

2x + z = 1

x− 5y− 2z = 3

(d)

4x− 2y + z + w = −13

−x− z− w = 2

y + 3z + 2w = 0

2x + 5y− z = 20

Exercise 5.31: An investor has $100 000 to invest between stocks in uranium and oil. Ura-nium has a 6% dividend rate and oil has a 9% dividend rate. If the investor re-quires a $7 000 annual dividend, find a set of equations whose solution will givethe amounts to be invested. Express the equations as a matrix equation by definingappropriate matrices.

Exercise 5.32: When braking from high speed, the velocity of a vehicle t seconds afterapplying the brakes is given by

v = a + bt2 + c2t

where a, b and c are constants.

Under ideal test conditions, a car travelling at 30 ms−1 decelerates to 25 ms−1 after1 second and to 16 ms−1 after 2 seconds from the start of braking. Set up equationsto find a, b and c.

Application: engineering technology Kirchhoff’s Current Law (KCL) statesthat the algebraic sum of currents in an electrical circuit flowing towards a nodeis zero. Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of voltagesaround an electrical circuit is zero. In a particular circuit this can produce a seriesof equations. For example:

I1 + I2 + I3 = 0

(2× 103)I1 + (4× 103)I3 = 2

(1× 103)I2 + (4× 103)I3 = 6

344 c© USQ, August 27, 2010

5.8. Solution of linear equations using the inverse matrix MAT1500

This can be written in matrix notation as RI = V, where

R =

1 1 12× 103 0 4× 103

0 1× 103 4× 103

I =

I1

I2

I3

V =

026

so that

1 1 12× 103 0 4× 103

0 1× 103 4× 103

I1

I2

I3

=

026

5.8 Solution of linear equations using the inverse matrix

Before we try to solve these matrix equations let us recap one important point whensolving ordinary equations.

If we had an equation 3y = 21 , to find a value for y we would multiply both sides by 13 .

So we have,

13× 3y =

13× 21

1× y = 7

In this case the 13 is called the multiplicative inverse. We use the same process with matrix

equations. However, while in ordinary arithmetic it does not matter in which order weplace the factors 3, 1

3 and y, in matrix algebra the position of the matrices is important.

So if we had a set of equations we would first convert them into matrix form.

3x + 4y = 10

2x + 5y = 2

Change this into a matrix equation:

[3 42 5

] [xy

]=

[102

]

We want to get the unknown matrix

[xy

]by itself.

Just as in ordinary solving of algebraic equations, we must multiply both sides by the

c© USQ, August 27, 2010 345

MAT1500 5.8. Solution of linear equations using the inverse matrix

multiplicative inverse which in this case is

57−4

7

−27

37

Therefore

57−4

7

−27

37

[3 42 5

] [xy

]=

57−4

7

−27

37

[102

]

The first two matrices multiply to give the identity matrix (check this carefully yourself).We then multiply the last two matrices to get the solution to our equation. Remember wehave to pre-multiply by the inverse on both the left and right hand sides of the matrixequation, otherwise we will not get the correct answer.

[1 00 1

] [xy

]=

507− 8

7

−207

+67

[xy

]=

427

−147

[xy

]=

[6−2

]

So x = 6 and y = −2 .

Check in both equations.

LHS = 3× 6 + 4×−2

= 10

= RHS.

LHS = 2× 6 + 5×−2

= 2

= RHS.

Let us have a look at the steps involved in the following examples.

Example 5.33: Using matrices solve the simultaneous equations

x = 2y

2x = y + 4

346 c© USQ, August 27, 2010


Solution:

Step 1: Rearrange into a standard form.

x = 2y ⇒ 1x− 2y = 0

2x = y + 4 ⇒ 2x− 1y = 4

Step 2: Put new equations into matrix form.

[1 −22 −1

] [xy

]=

[04

]

Step 3: Find the inverse of the matrix on the LHS.

A−1 =1

ad− bc

[d −b−c a

]

=1

−1 + 4

[−1 2−2 1

]

=

−13

23

−23

13

Step 4: Pre-multiply both sides by the inverse matrix.

−13

23

−23

13

[1 −22 −1

] [xy

]=

−13

23

−23

13

[04

]

[1 00 1

] [xy

]=

83

43

[xy

]=

83

43

Step 5: State the solution.

Therefore x = 83 and y = 4

3 .

c© USQ, August 27, 2010 347


Step 6: Check in the original equations.

x = 2y ⇒ RHS = 2× 43=

83= LHS.

2x = y + 4 ⇒ LHS = 2× 83=

163

RHS =43+ 4 =

163

= LHS.

Example 5.34: Given that

A =

[2 4−3 1

]and B =

[−5 −3

6 4

]

solve the matrix equation A = BX . Note that the inverses of A and B are

A−1 =

114−2

7

314

17

and B−1 =

−2 −32

352

Solution:

To solve A = BX , we must have X by itself, so we must pre-multiply both sides bythe inverse of B.

B−1BX = B−1A

X = B−1A

X =

−2 −32

352

[2 4−3 1

]

=

−4 +92−8− 3

2

6− 152

12 +52

X =

12−19

2

−32

292

or we could write it as12×[

1 −19−3 29

]

348 c© USQ, August 27, 2010


Example 5.35: Solve the following simultaneous equations:

2x + 3y− 2z = 5

2x + y + z = 11

3x + 2y− 3z = 0

given that the inverse of

2 3 −22 1 13 2 −3

is

− 13

13

13

35 0 − 2

5

115

13 − 4

15

Solution:

In matrix form the above equation becomes:

2 3 −22 1 13 2 −3

xyz

=

5110

−13

13

13

35

0 −25

115

13− 4

15

2 3 −22 1 13 2 −3

xyz

=

−13

13

13

35

0 −25

115

13− 4

15

5110

=

−53+

113

+ 0

3 + 0 + 0

13+

113

+ 0

=

234

So the solution to the given simultaneous equations is x = 2 , y = 3 , and z = 4 .

You should check these solutions in each of the 3 original equations.

In practice, this method of solving systems of linear equations is only used in cases where:

1. the inverse matrix is already known; or

2. many systems of equations with the same left hand sides are solved for differentright hand sides.

c© USQ, August 27, 2010 349


In all other cases there would be less work involved in solving the equations directly.

Exercise 5.36: Write the following simultaneous equations in matrix form and then solvefor x and y using matrix multiplication.

2x + y = 5

x + 2y = 4

Exercise 5.37: Solve the simultaneous equations

x + 2y + 3z = 2

2x + 5y + 3z = 6

x + 8z = 1

using matrices given that the inverse of

1 2 32 5 31 0 8

is

−40 16 9

13 −5 −35 −2 −1

Exercise 5.38: If

C =

[−8 4−16 4

]and M =

[7 34 2

]

and the inverses of C and M are

C−1 =

18−1

8

12−1

4

and M−1 =

1 −32

−272

solve for P in the equation CP = M .

Exercise 5.39: Solve the following systems of equations:

(a)

2x− y = 7

−x + 2y = −5

(b)

x + 2y = 12

4x− 3y = 4

350 c© USQ, August 27, 2010


(c)

x + y + z = 1

2x− y + 4z = −2

−x + 3y− 3z = −7

(d)

−x + 2y + 3z = 10

3x− 2y + z = −16

−2x− y− 4z = 5

(e)

x− y + z = 3

2x + 3y− 4z = 9

−x + 2y− z = 0

(f)

x + y + z = 6

3x + 4y + 2z = 17

−x + 2y + z = 6

Here are some inverses that you may need:

Matrix Inverse

1 1 12 −1 4−1 3 −3

92−3 −5

2

−1 1 1

−52

232

1 −1 12 3 −4−1 2 −1

56

16

16

1 0 1

76−1

656

[1 24 −3

]

311

211

411− 1

11

c© USQ, August 27, 2010 351


Matrix Inverse

1 1 13 4 2−1 2 1

015−2

5

−125

15

2 −35

15

[2 −1−1 2

]

23

13

13

23

−1 2 3

3 −2 1−2 −1 −4

−0.9 −0.5 −0.8−1 −1 −10.7 0.5 0.4

Points to Remember

• A system of linear equations

a1x + a2y = b1

a3x + a3y = b2

can be written in the form AX = B where A =

[a1 a2

a3 a4

],

X =

[xy

]and B =

[b1

b2

](and similarly for 3× 3 systems).

• If a system of linear equations is written in the form AX = B, thenthe solution is X = A−1B.

Putting it all together! Vertical curves for roads are often designed to have aparabolic cross-section as in the diagram below. The equation of these curves isthe parabolic form y = ax2 + bx + c .

In laying out these curves surveyors use four pieces of information to determinethe coefficients a, b and c.

• the elevation above the datum line at A where the curve begins;

• the gradient (G1) of the road at A;

352 c© USQ, August 27, 2010


• the gradient (G2) of the road at B where the curve ends; and

• the horizontal length (L) of the curve from A to B

Using this information surveyors develop the following formulas to find a, b and cin the general parabolic equation:

a =G2 − G1

2Lb = G1

c = elevation at A,

so that the equation of the parabola becomes

y =

[G2 − G1

2L

]x2 + G1x + A

A road with a vertical curve of horizontal length 500 m passes through the threepoints given below where x is measured from A and y is the height above thedatum line.

x = 100 m, y = 320 m,

x = 200 m, y = 322 m,

x = 400 m, y = 321 m.

1. Using the formula for y given above, develop three equations in the threeunknowns (G1, G2 and A)

2. If the inverse of the matrix

90 10 1160 40 1240 160 1

is

−0.02 0.025 −0.0050.013 −0.025 0.01172.667 −2 0.333

find using matrices

(a) elevation at A,(b) the first gradient G1,(c) the second gradient G2,

3. Using your knowledge of parabolas find the maximum elevation of the curve.

Hints Preparing for the exam now!

One more chapters to go! Now is a good time to start preparing for the OpenBook Exam. You might have a few questions.

• What is an open book exam?An open book exam is one where you may take any written materials intothe exam room, including but not restricted to text books, your own notes(either written or typed) and copies of web pages. Scientific calculators are

c© USQ, August 27, 2010 353

MAT1500 5.9. Review

also permitted. The most fatal mistake you could make is to assume thatyou would not have to know the materials or study for an open book exam!

• How do I prepare for an open book exam?You must prepare as if you were sitting a closed book exam. If you have notstarted already then start today.

• What do I have to know?Use the objectives at the beginning of each chapter to help you determinewhat is important.

• Any other suggestions?Make a summary. Your summary should be designed to suit your needs.

More on this later. . .

Well done! One more chapter to go! Here are a number of things to check.

1. Have a close look at your action plan for study. Are you on schedule? Do youneed to restructure your action plan or contact your tutor to discuss any delays orconcerns?


5.9 Review

• A matrix is an array of numbers called elements.

• A matrix consists of rows and columns.

• A matrix of m rows and n columns has dimension (order) m× n.

• A matrix consisting of one column or one row is called a vector.

• Matrices are equal if all corresponding elements are equal.

• To add (or subtract) matrices add (or subtract) corresponding elements

• Matrices can only be added or subtracted if the dimensions are the same.

• A matrix, in which all the elements are zero, is called the zero matrix.

• To multiply a matrix by a scalar, multiply every element of the matrix by that num-ber.

• The transpose of matrix A is the matrix found by interchanging rows and columns.

• The transpose of a matrix is denoted AT.

• If matrix A = AT, the matrix is said to be symmetric.

354 c© USQ, August 27, 2010


• The element in the nth row/mth column of the matrix product AB is the ‘scalarproduct’ of the nth row of A with the mth column of B.

• For the matrix product AB to be defined, the number of columns of matrix A mustbe equal the number of rows of column B.

• If the dimension of A is m× n and the dimensions of B are n× p, then the productAB is defined and has dimension m× p.

• An identity matrix is a square matrix with ‘1’s down the diagonal and ‘0’s else-where.

• If I is an identity matrix of the appropriate order for the product to be defined, thenIA = A and AI = A.

• AB is usually not equal to BA.

• A−1 is the inverse of A if AA−1 = I = A−1A.

• The inverse of a (2× 2) matrix A =

[a bc d

]is given by is given by A−1 =

1ad− bc

[d −b−c a

].

• A system of linear equations

a1x + a2y = b1

a3x + a4y = b2

can be written in the form AX = B where A =

[a1 a2

a3 a4

], X =

[xy

]and B =

[b1

b2

]

(and similarly for 3× 3 systems).

• If a system of linear equations is written in the form AX = B, then the solution isX = A−1B.

5.10 Post-test

In Questions 1 to 4 below, let

A =

[1 −3 20 1 4

]

B =

[1 23 4

]

C =

[1 03 1

]

If any of the matrix expressions are undefined, write UD as your answer.

1. Evaluate AC.

c© USQ, August 27, 2010 355


2. Evaluate 2AT.

3. Evaluate B + C.

4. Evaluate CB.

5. Express the simultaneous equations

3x− 2y = 5

x− y = −4

as a single matrix equation.

6. Write down the inverse of [6 42 3

]

7. Let A and B be square matrices of the same size, let I be the identity matrix of thesame size as A and B, and let A have an inverse.

Which (if any) of the matrix equations below are necessarily true? (Name all of theequations that are necessarily true.)

(a) IA = A

(b) A−1A = I

(c) AB = BA

(d) AT = A

8. A toll bridge charges $0.40 for cars, $1.00 for trucks, and $1.40 for buses. Fourthousand vehicles crossed the bridge on a particular day yielding $2 200.00 in tolls.There were twice as many trucks as buses. If x, y, z are the number of cars, trucksand buses respectively, write down a set of 3 equations in x, y, z that describe theabove information and hence write this system as a single matrix equation (that is,as AX = B where A is a 3× 3 and B is a 3× 1 matrix).

9. In the matrix equation AX = B ,

A =

2 −2 31 2 −11 1 1

and B =

374

Find X, given that the inverse of A is

17

3 5 −4−2 −1 5−1 −4 6

356 c© USQ, August 27, 2010


5.11 Solutions


5.2 A is (2× 2), B is (2× 3), C is (1× 3), D is (3× 2), E is (2× 3), F is (3× 1), G is(3× 2), H is (1× 1).

5.3 a11 = 3 , b13 = 8 , c12 = 1 , d21 = 0 , e33 does not exist, f31 = 1 , g22 = −4 , h11 = 4 .

5.8 (a) [9 86 4

]

(b) [7 6

10 5

]

(c) [−2 −7−8 −3

]

(d) [−2 −12−20 −7

]

(e) [3 −21 0

]

(f) [6 87 4

]

5.9 (a) −4 −5 −2−11 1 6

10 6 5

(b) [1 6 11−7 9 −5

]

5.10 No, since a symmetric matrix must be square.

5.20 (a) −3 −3 0 1

1 15 0 −5−3 15 0 −5

c© USQ, August 27, 2010 357


(b) −1 2 1−2 4 2−3 6 3

(c) [4 −4 2 4

]

(d) Product is not defined.

5.21 (a)R =

[2 6 4 5

]

(b)

P =

212202230

(c) RP = [782] , or $7.82.

5.22 (a)

apples bananas lemons oranges pears

A =

12 0 8 0 60 12 2 18 06 24 0 9 4

(b)

B =

12 0 8 0 60 12 2 18 06 24 0 9 4

1822162016

=

440656880

Sue spent $4.40, Mary spent $6.56 and Paul spent $8.80.

(c)

[1 1 1

]

12 0 8 0 60 12 2 18 06 24 0 9 4

=

[18 36 10 27 10

]

5.25 (a) [−3 −4−1 −1

]

358 c© USQ, August 27, 2010


(b)

411

311

111− 2

11

(c)

−2 1

32−1

2

(d)

−14

14

13−1

6

5.26

AB =

[2 15 3

] [3 −1−5 2

]

=

[6− 5 −2 + 2

15− 15 −5 + 6

]

=

[1 00 1

]

5.27 (a) Yes, because determinant of the matrix is −1× 3− 7×−4 = 25 .

(b) No, because determinant of the matrix is 8× 2− 4× 4 = 0 .

5.30 (a) [2 −11 2

] [xy

]=

[34

]

(b)[

1 1 12 −1 −1

]

xyz

=

[105

]

(c)

1 −1 12 0 11 −5 −2

xyz

=

−2

13

(d)

4 −2 1 1−1 0 −1 −1

0 1 3 22 5 −1 0

xyzw

=

−1320

20

c© USQ, August 27, 2010 359


5.31 Let x be the amount invested in uranium and y be the amount invested in oil.

x + y = 100 000

0.06× x + 0.09× y = 7 000

Let

A =

[1 1

0.06 0.09

]X =

[xy

]B =

[100 0007 000

]

Then AX = B .

5.32

v = 30 , when t = 0, Therefore 30 = a + cv = 25 , when t = 1, Therefore 25 = a + b + 2cv = 16 , when t = 2, Therefore 16 = a + 4b + 4c

Therefore

1 0 11 1 21 4 4

abc

=

302516

5.36

2x + y = 5

x + 2y = 4

In matrix form this is [2 11 2

] [xy

]=

[54

]

Next we find the inverse of the coefficient matrix,

[2 11 2

]−1

=13

[2 −1−1 2

]

Multiplying both sides by the inverse,

[xy

]=

13

[2 −1−1 2

] [54

]

=

[21

]

Thus the solution is x = 2 , y = 1 .

Check this in original equations.

360 c© USQ, August 27, 2010


5.37

xyz

=

−40 16 9

13 −5 −35 −2 −1

261

=

25−7−3

so the solution is x = 25 , y = −7 , z = −3 .

Check this in the original equations.

5.38

CP = M

P = C−1M

=

18−1

8

12−1

4

[7 34 2

]

=18

[1 −14 −2

] [7 34 2

]

=18

[3 1

20 8

]

Check in the original equation by showing CP = M .

5.39 (a) x = 3 , y = −1 .

(b) x = 4 , y = 4 .

(c) x = 28 , y = −10 , z = −17 .

(d) x = −5 , y = 1 , z = 1 .

(e) x = 4 , y = 3 , z = 2 .

(f) x = 1 , y = 2 , z = 3 .


1. UD – undefined.

2.

2 0−6 2

4 8

c© USQ, August 27, 2010 361


3. [2 26 5

]

4. [1 26 10

]

5. [3 −21 −1

] [xy

]=

[5−4

]

6.110

[3 −4−2 6

]

7. (a) and (b).

8.

1 1 10.4 1 1.4

0 1 −2

xyz

=

4 0002 200

0

9.

41−1

362 c© USQ, August 27, 2010

Chapter 6

Calculus – Differentiation

Chapter contents6.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364

6.2 Rate of change - the problem of the curve . . . . . . . . . . . . . . . . . . . . . 366

6.3 Instantaneous rate of change - the derivative . . . . . . . . . . . . . . . . . . . . 369

6.4 Finding the derivative algebraically . . . . . . . . . . . . . . . . . . . . . . . . . 378

6.5 Derivatives of compound functions . . . . . . . . . . . . . . . . . . . . . . . . . 386

6.6 Application: maxima and minima . . . . . . . . . . . . . . . . . . . . . . . . . . 392

6.7 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

6.8 Post-test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

6.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405



Objectives


• explain the relationship between rate of change and the derivative;

• find the derivatives of basic functions; and

• find the local maximum and minimum values (if any) of a function.

Introduction to calculus

Calculus is the mathematics of motion and change. The calculus we use today is thecontribution of many people. Its roots can be traced to classical Greek geometry, but itsinvention is chiefly the work of the astronomers, mathematicians, and physicists of theseventeenth century. Its invention is thought to be one of the greatest of modern mathe-matics and has revolutionised all of the science and mathematically based disciplines. At

c© USQ, August 27, 2010 363

MAT1500 6.1. Overview

the time, however, it created an enormous scandal as its two inventors (Sir Isaac Newtonand Gottfried Leibnitz) battled for the right to be called first inventor (see boxed historylater in the chapter).

Today, we use calculus to investigate how things change, whether they be growing babiesor dwindling profits; predicting the orbits of satellites; designing navigation systems andradar systems; exploring problems of space travel; testing theories about ocean currents;or determining the safety of a new invention. It is applied increasingly to solve problemsin biology, business, economics, linguistics, medicine, political science and psychology.Calculus is also the gateway to many fields of higher mathematics.

Hint Do not believe the bad press!

You might have heard the many stories that circulate about calculus and howdifficult it is. Think about this. Look how far you have come and what you haveachieved. Calculus will use all these skills. For a light hearted approach to calculusand to understand what all the fuss is about, have a look at the video Off on aTangent (available from USQ Library).

6.1 Overview

In calculus, we look at rates of change and summation of quantities, how we can repre-sent them, and how we can calculate them.

Let us think about what we already know about rates of change. We know that if wewant to compare values of a single variable we might use ratios or percentages, but if wewant to compare between two variables we use a rate of change or more briefly a rate.You have come across many different rates during your life, in either your work or study.Here are some you might be familiar with. Recall that rates always have units.

Speed metres per secondkilometres per hourrevolutions per seconddegrees centigrade per minute

Volumetric flow rate cubic metres per secondMass flow rate kilograms per secondEnergy content kilojoules per gramAcceleration kilometres per hour per hourDensity grams per cubic centimetrePressure dynes per centimetre squared (pascals)Power joules per secondDiffusion square metres per secondConcentration grams per cubic centimetreLatent heat kilojoules per kilogram

Speed is an example of a rate of change with which we are all familiar. Speed is the rate

364 c© USQ, August 27, 2010

6.1. Overview MAT1500

at which distance is covered, or in other words, the rate at which distance from a fixedpoint is changing with time.

If we look at the question of speed from another perspective, this would involve sum-mation. ‘If we know the speed of a car at any given instant, how much distance has itcovered in a given period of time?’. In essence this is a process of adding all the smallpieces of distance covered in each second (say) of time elapsed over the full time period.

Let us look at some graphs representing this. Figure 6.1 is a graph of distance (let us callit s) covered over a time (let us call it t).

Figure 6.1: Probe’s speed over time

Speed is the change of distance, ∆s, divided by the elapsed time, ∆t. As you know fromChapter 2, this quantity is also the slope or gradient of the line in Figure 6.1.

(Note that ∆ is a Greek letter used to indicate a change, so ∆s is the change in distance,∆t is the change in time.)

Let us imagine that the graph represents the distance an interplanetary probe travels overa time t. Since the graph is a straight line it will have a constant slope meaning that thespeed that the interplanetary probe travels is constant.

When we now graph speed (called, v) over time for the same interplanetary probe (Figure6.2) we get a straight line parallel to the horizontal axis. Such a line has a slope or gradientof zero, which means that the speed is not changing over time (it is a constant).

The distance covered by our interplanetary probe travelling at constant speed v1 for agiven time, say t1, will be its speed multiplied by the time elapsed, v1 × t1. This corre-sponds in Figure 6.2 to the shaded area underneath the graph between t = 0 , and t = t1 .This means that the slope of a graph and the area under the graph have significant mean-ing in the context of constant speed.

c© USQ, August 27, 2010 365

MAT1500 6.2. Rate of change - the problem of the curve

Figure 6.2: Probe’s speed over time

In this course we will focus on differentiation, the calculus of rates of change. The studyof integration will be left to later courses in Mathematics.

6.2 Rate of change - the problem of the curve

It is quite easy to determine the gradient or rate of change of a function if it is a linearfunction, because linear functions always have a constant gradient or rate of change.Curved functions are not so easy, because the rate of change of one variable with respectto the other is always changing. However, we can approximate the process by findingthe gradient between two points of interest on the graph. We call this the average rate of

366 c© USQ, August 27, 2010

6.2. Rate of change - the problem of the curve MAT1500

change of the curve.

Look at Figure 6.3. We approximate the average rate of change of the curve between thepoints x = −5 and x = −4 , by finding the gradient of the straight line connecting thesetwo points. We have done this in a previous chapter. This type of line segment whichintersects the curve in two places is called a secant.

Figure 6.3: Illustration of average rage of change.

The average rate of change for the curve shown between the points (−5, 0) and (−4, 210)is

m =∆ f∆x

=f (x2)− f (x1)

x2 − x1

=210− 0−4−−5

= 210 .

Between −5 and −4 the curve changes at a rate of 210 units of y for each unit of x.

This means that when we know any two points on a curve, we can use them to calculatethe average rate of change of the function, by calculating the gradient of the straight lineconnecting the points (gradient of the secant).

But what if we chose x coordinates of −4.5 and −3.7? (Figure 6.4)

c© USQ, August 27, 2010 367

MAT1500 6.2. Rate of change - the problem of the curve

Figure 6.4: Illustration of average rate of change.

Here we would calculate that the average rate of change is zero. But as you can see fromthe graph there is actually quite a lot happening to the function and its rate of changebetween −4.5 and −3.7. In this situation and many others it is not enough to find theaverage rate of change, we must find what is happening exactly at a particular instant (ata point on the graph).For example:

• If you were gathering data on the fuel consumed by a rocket, the instant the rate offuel consumption changed would tell you that the rocket was boosting its speed.

• A manufacturer might monitor production costs over time so that they would knowthe instant there is a variation in the rate of change of cost with time. They couldrelate this instant with machinery or human error.

• Human population biologists might want to monitor the rate of change of popula-tion size so that predictions could be made for town planning for a particular year.

• Pattern makers know that there are a range of rectangles with different dimensionsthat can be made with the same perimeter. The rectangular dimensions that pro-duce the maximum area can be found by determining at what specific width (orlength) the rate of change of area with width is zero (at a turning point).

We will return to more of these examples later in the chapter. But it appears that we needa more accurate idea of the rate of change of a function. We often need to know the rateof change at a particular instant (point on the function), that is, the instantaneous rate ofchange.

368 c© USQ, August 27, 2010

6.3. Instantaneous rate of change - the derivative MAT1500

A bit of history

(For interest only) Who invented calculus?

In the mid 1660’s, about one thousand years after the Greeks first thought of theconcepts, Isaac Newton (an Englishman) is said to have developed his method offluxions, which today we call calculus. But Newton is described as a suspicious,neurotic and tortured personality and only distributed his discovery to a few col-leagues. He did not publish it. Ten years later Gottfried Leibnitz (a German)made virtually the same discoveries. Letters passed between Newton and Leibnitz,but still Newton did not publish. Leibnitz was the first to publish on differentialcalculus in 1684. He did not acknowledge Newton’s work.

Sounds like something out of the Sunday papers, doesn’t it? But it does notend there. Charge and countercharge followed. A commission held by the RoyalSociety in 1713 supported Newton’s claims. Not surprising perhaps as by this timeNewton was the President of the Royal Society. But still accusations continued.

Newton is often thought to be the greatest genius of all time making discoveriesin mathematics, physics, astronomy and numerous other sciences. Leibnitz wasalso extremely talented and made numerous discoveries in mathematics from hisearly years. Dunham (1993) has said that

The mutual denunciations of two of the greatest mathematicians of alltime make a sad chapter in European intellectual history. That indi-viduals of such genius descended to petty and outrageous mudslingingdoes not bode well for those of us with more modest intellects.

Perhaps it just makes them appear more human.

You might be interested to know that today many people incorrectly associateonly Newton’s name with calculus, but it was actually Leibnitz who first coinedthe word calculus and whose mathematical notation we use today.

6.3 Instantaneous rate of change - the derivative

As indicated before in many scientific, engineering and business situations, we are con-cerned with the instantaneous rate of change of one quantity with respect to another, forexample the rate of conversion during a chemical reaction (moles/sec); the populationgrowth rates (number of bacteria per day); the rate of change of voltage (volts/sec); thespeed of objects (km/h), and so on.

The process by which we compute such rates of change is called differentiation. Thequantity we find is called the derivative.

How do we do this? Let us do some investigating.

Suppose the graph of the distance travelled versus time is as shown in Figure 6.5. (Here

c© USQ, August 27, 2010 369

MAT1500 6.3. Instantaneous rate of change - the derivative

we have used the function

s =−(t− 3)3 + 27(t− 3) + 54

50,

to draw the graph and calculate the table of values in Table 6.1)

Figure 6.5: Instantaneous rate of change

Clearly the speed, that is the slope, is not constant over the graph. It is much greater in themiddle of the time period than towards the end where for a short time the distance doesnot change with time (zero speed), or perhaps even reduces slightly (negative speed).

How could we get the speed of the vehicle at time t = 5 say, from the graph?

The graph is curved so it does not have a constant slope that we can measure.

Table 6.1: Distance for s =−(t− 3)3 + 27(t− 3) + 54

50from t = 0 to 7.

t s0 01 0.162 0.563 1.084 1.65 26 2.167 1.96

Look at the region of the graph around t = 5 . We have drawn straight lines joining thepoints on the curve at t = 4 and 5, and the points at t = 5 and 6.

370 c© USQ, August 27, 2010


When we calculate the slope of the line from t = 5 to t = 6 we get

∆s∆t

=2.16− 2

6− 5= 0.16 .

When we calculate the slope of the line from t = 4 to t = 5 we get

∆s∆t

=2− 1.65− 4

= 0.4 .

In actual fact, what we are finding here is the average rate of change of distance betweenthese times, or in other words the average speed over these time intervals (see Chapter2). Clearly neither of these could be thought of as representing the ‘slope’ of the curve att = 5 because they are so different.

See what happens when we ‘zoom in’ around the region of t = 5 . The graph becomesthat shown in Figure 6.6 and a corresponding table of values would be as in Table 6.2.

Figure 6.6: Instantaneous rate of change.


50, from t = 4.7 to 5.4.

t s4.7 1.899 744.8 1.935 364.9 1.968 825 25.1 2.028 785.2 2.055 045.3 2.078 665.4 2.099 52

c© USQ, August 27, 2010 371


The curve now looks much closer to a straight line, doesn’t it? Look at the straight (dot-ted) lines joining the points on the curve from t = 4.9 to t = 5.0 and from t = 5.0 tot = 5.1 . They are virtually indistinguishable from the curve itself.

When we calculate the slope of these lines as before, we find that the slope from t = 4.9to 5.0 is

∆s∆t

=2− 1.968 82

5− 4.9= 0.311 8 ,

and the slope from t = 5.0 to 5.1 is

∆s∆t

=2.028 78− 2

5.1− 5.0= 0.287 8 .

Notice that these two slopes (0.311 8 and 0.287 8) are much closer in value than for thewider intervals we looked at before. Zoom in further and look at Figure 6.7 and table ofvalues for t = 4.97 to t = 5.03 , shown in Table 6.3.

Figure 6.7: Instantaneous rate of change.


50, from t = 4.97 to 5.04.

t s4.97 1.990 894.98 1.993 954.99 1.996 995 25.01 2.002 995.02 2.005 955.03 2.008 895.04 2.011 81

Notice how the curve is indistinguishable from a straight line; we cannot distinguish the

372 c© USQ, August 27, 2010


lines joining the reference points between t = 4.99 and t = 5 and between t = 5 andt = 5.01 from the ‘curve’ itself.

When we use the table of values again to calculate the slopes of these lines we find thatbetween t = 4.99 and t = 5 ,

∆s∆t

=2− 1.996 988

5− 4.99= 0.301 2 ,

and between t = 5 and t = 5.01

∆s∆t

=2.002 988− 2

5.01− 5= 0.298 8

Did you notice that as our ∆t values became smaller and smaller, the values of the slopeof the two lines, ∆s

∆t , converges on the value 0.3?

Try it out for yourself. Remember that the function we used to draw the graph andcalculate the table of values is given by

s =−(t− 3)3 + 27(t− 3) + 54

50

We would expect that as ∆t became even smaller the slope of the curve would even moreclosely approach 0.3. This value is called the limit of ∆s

∆t at t = 5 as ∆t approaches zero.The mathematical notation for this statement is

lim∆t→0

∆s∆t

∣∣∣∣t=5

This quantity is defined as the derivative of s with respect to t at t = 5 . We use thenotation ds

dt or s′(t) to refer to the derivative. The process of finding the derivative iscalled differentiation.

When we draw a straight line on the graph through the point on the curve where t = 5with the slope equal to the derivative we have just found, 0.3, it would be the tangent tothe curve at that point, as shown in Figure 6.8.

In fact, the value we have found represents the slope of the tangent to the curve (seeChapter 2) at t = 5 . It is often also referred to as the instantaneous rate of change of swith respect to t.

c© USQ, August 27, 2010 373


Figure 6.8: Tangent slope and derivative at t = 5.

Example 6.1: Create tables of values and plot graphs to compute the derivative of y = x2

at x = 1, 2 and 3. (Use Graphmatica, another computer program, or a graphingcalculator if you wish). Can you spot a pattern?

Solution:

The graphs below are produced directly from Graphmatica (Ver.1.6N c©1999 kSoft.Inc.).

x y−0.5 0.250 00.5 0.251.0 1.01.5 2.252.0 4.02.5 6.253.0 9.0

374 c© USQ, August 27, 2010


Now, zoom in around x = 1 ; the resulting graph should look similar to the follow-ing graph.

x y0.997 0.994 0090.998 0.996 0040.999 0.998 0011.0 1.01.001 1.002 0011.002 1.004 0041.003 1.006 009

From the table of values calculate the difference between each x value, ∆x = 0.001.

c© USQ, August 27, 2010 375


The slope of the graph between x = 0.999 and x = 1 is

∆y∆x

=1− 0.998 001

0.001= 1.999 .

The slope of the graph between x = 1 and x = 1.001 is

∆y∆x

=1.002 001− 1

0.001= 2.001 .

Thus the slope of the graph at x = 1 approaches 2 as ∆x approaches 0.

Next, zoom in around x = 2 .

x y1.997 3.988 0091.998 3.992 0041.999 3.996 0012.0 4.02.001 4.004 0012.002 4.008 004

This table of values has ∆x = 0.001 .


∆y∆x

=4− 3.996 001

0.001= 3.999 .


∆y∆x

=4.004 001− 4

0.001= 4.001 .


376 c© USQ, August 27, 2010


Finally, zoom in on x = 3 .

x y2.997 8.982 0092.998 8.988 0042.999 8.994 0013.0 9.03.001 9.006 0013.002 9.012 0043.003 9.018 009

This table of values has ∆x = 0.001 .


∆y∆x

=9− 8.994 001

0.001= 5.999 .


∆y∆x

=9.006 001− 9

0.001= 6.001 .


We see the pattern here that the slope at x appears to always be 2x.

c© USQ, August 27, 2010 377

MAT1500 6.4. Finding the derivative algebraically

6.4 Finding the derivative algebraically

In example 6.1 we use tables of numerical values and graphs to determine the derivativeof the function y = x2 at points where x = 1, 2 and 3. We now investigate how we canfind an algebraic formula for the derivative of the function f (x) = x2 , which allows usto calculate the instantaneous rate of change (or derivative) of the function at any point.

To calculate the derivative we have determined the slope over smaller and smaller incre-ments, and found a limiting value. In the previous examples we have called this smallchange ∆x or ∆t, depending on the variable used. In this discussion let us call the smallchange h. Algebraically the slope between two points on the curve f (x) at x = x0 andx = x0 + h is

Change in fChange in x

=f (x0 + h)− f (x0)

h.

The derivative is the limiting value of this slope for vanishingly small h. That is,

d fdx

∣∣∣∣x=x0

= limh→0

f (x0 + h)− f (x0)

h.

For f (x) = x2 , f (x0) = x20 and f (x0 + h) = (x0 + h)2 , so the slope between the two

points on the curve is

(x0 + h)2 − x20

h=

x20 + 2x0h + h2 − x2

0h

=2x0h + h2

h= 2x0 + h .

378 c© USQ, August 27, 2010

6.4. Finding the derivative algebraically MAT1500

As h gets smaller = 2x0+h

. . . and smaller = 2x0+h

. . . and smaller = 2x0+h

its limiting value is = 2x0

Thus as h becomes vanishingly smaller and smaller the limiting value of the slope is 2x0.This agrees with our observations in Example 6.1, where the limiting value when x = 1was 2.

In general we dispense with x = x0 and write: when f (x) = x2 ,d fdx

= 2x .

What we have actually done here is to find a formula for the slope of the straight linewhich touches the function at one point. This straight line is called the tangent to thecurve and its slope mimics the slope of the curve at the point of interest.

A note on notation

Many different notations are associated with finding the derivative. A small change in avariable could be called ∆x or ∆y or h.

Functions can be described using a single symbol such as y = x2 or using function nota-tion f (x) = x2.

If the function is called just y or g or h, the derivative would be calleddydx

,dgdx

,dhdx

respec-tively, if we differentiated with respect to x.

If the function notation such as f (x) is used, the derivative is called:

d fdx

ddx

f (x) f ′(x) f

(The last one in this group is usually reserved for the derivative with respect to time i.e.f = d

dt f (t).)

c© USQ, August 27, 2010 379


Example 6.2: The distance (in metres) a car moves from a fixed point is a function of time(in seconds) given by s = f (t). Explain the meaning of f ′(2) = 5 including whatunits you would expect the value 5 to have.

Solution:

If the original function is s = f (t), f ′(2) = 5 is the first derivative of that functionwith respect to t and means that at t = 2 seconds the car is moving at an instanta-neous rate of change (or velocity) of 5 metres per second.

Example 6.3: Water flows through a pipe at a constant rate of 3 m3 per second. Interpretthis rate as a derivative of some function. What are units of the variables in thatfunction?

Solution:

Water flowing through a pipe at a constant rate of 3 m3 per second represents a rateof change of volume with respect to time, since m3 is a cubic measure of volumeand seconds represents time. If we let volume be V in m3 and t be time in seconds,

we would saydVdt

= 3 or V ′(t) = 3.

Exercise 6.4: The population of Australia since 1900 is given by the function P = f (n),where P is measured in thousands and n is in years since 1900. What are the unitsfor f ′(n)? Explain the meaning of the three situations:

(a) f ′(105) = 10

(b) f ′(105) = −2

(c) f ′(105) = 0

Exercise 6.5: The current (I) in amperes in an electrical circuit varies with the time t inseconds according to the function I = g(t). Explain the meaning of g′(0.01) = 108.

Exercise 6.6: The volume (V, in m3) in a reservoir varies with time (t in minutes) aftera valve is opened and is given by the function V = h(t). Explain the meaning ofdVdt

∣∣∣∣t=2

= −112 , include in your explanation the units for this quantity and the

sign.

Exercise 6.7: The e.m.f., E in millivolts, produced by a thermocouple is given by the func-tion E = f (T), where T is the temperature in degrees Celsius. If the rate of changeof e.m.f with respect to temperature is 4.8 mV/ C, write this using mathematicalnotation for the derivative.

Exercise 6.8: For weightlifters the relationship between cardiac output (volume of bloodin litres) and the workload (w) in kilogram metres (kg.m) for 1 minute of weightlift-ing is given by the function C = f (w). Explain the meaning of f ′(300) = 0.02,include the units for 0.02.

380 c© USQ, August 27, 2010


Example 6.9: Use algebraic methods to find a formula for the derivative of the functiony = x2 − 1 .

Solution:

The slope between two points at x0 and x0 + h is


=f (x0 + h)− f (x0)

h

=

[(x0 + h)2 − 1

]− (x2

0 − 1)h

=x2

0 + 2x0h + h2 − 1− x20 + 1

h

=2x0h + h2

h= 2x0 + h .

The limiting value as h→ 0 is again 2x0.

That is, for y = x2 − 1 ,dydx

= 2x also.

Example 6.10: Use algebraic methods to find a formula for the derivative of the function

y =1x

.

Solution:



=f (x0 + h)− f (x0)

h

=

1x0 + h

− 1x0

h

=1h

(1

x0 + h− 1

x0

)

=1h

(x0 − (x0 + h)

x0(x0 + h)

)

=1h

( −hx2

0 + x0h

)

=−1

x20 + x0h

.

Thus the limiting value as h→ 0 is − 1x2

0.

That is, for y =1x

,dydx

= − 1x2 = −x−2 .

Example 6.11: Use algebraic methods to find a formula for the derivative of the function

y =1x2 .

Solution:


c© USQ, August 27, 2010 381



=f (x0 + h)− f (x0)

h

=1

(x0+h)2 − 1(x0)2

h=

1h

(1

(x0 + h)2 −1

(x0)2

)

=1h

((x0)2 − (x0 + h)2

(x0 + h)2(x0)2

) (x0)2 − (x0 + h)2 can be factorised asdifference of two squares (Chapter 1)to

=1h

( −h(2x0 + h)(x0 + h)2(x0)2

)(x0)2 − (x0 + h)2)

= (x0 − (x0 + h))(x0 + (x0 + h))

=−(2x0 + h)

(x0 + h)2(x0)2 = −h(2x0 + h)

The limiting value as h→ 0 is−2x0

(x0)2(x0)2 =−2(x0)3 .

Thus for y =1x2 ,

dydx

=−2x3 = −2x−3 .

Can you spot a pattern in these results?

Function: x2 x−1 x−2

Derivative: 2x −x−2 −2x−3

Could you spot it? The power in the function becomes the coefficient of the derivative,and the power in the derivative is one less than the power in the function. That is iff (x) = xn, then

dydx

= nxn−1 .

Obviously, finding the derivative in this manner is a rather long and difficult process.However there is an easier way. Since we are always dealing with basic functions, orcombinations of them, we find the derivatives of these basic functions using the abovemethod, and simply use these results to differentiate more complicated functions. Table6.4 gives the derivatives of some of these basic functions.

382 c© USQ, August 27, 2010


Table 6.4: Derivatives of selected basic functions

Function Derivative

constant 0

x 1

xn nxn−1 where n is any number

eax aeax where a is a constant

ax ax ln a where a is a constant

ln x1x

log x1

x ln 10

sin(ax) a cos(ax) where a is a constant

cos(ax) −a sin(ax) where a is a constant

tan(ax) a sec2(ax) where a is a constant

Note: For trigonometric functions (including sin x, cos x or tan x) always perform calcula-tions using radians not degrees. It would be a good idea to change your calculator to theradian mode now!

Points to remember

• The average rate of change of a function is the slope of the straightline joining two points on the function.

• The derivative is the instantaneous rate of change of a function ata point.

• The process of finding the derivative is called differentiation.

• The derivative at a point on a curve is the slope of the tangent tothe curve at that point.

• The derivative of a function f (x) is defined asdydx

(also f ′(x))

= lim∆x→0

∆ f∆x

.

• We can find the derivative of a function at any point graphically,numerically, algebraically or by using a formula.

c© USQ, August 27, 2010 383


Example 6.12: Find the derivative of y = x10

Solution:

If y = x10 ,dydx

= 10× x10−1 = 10x9

Example 6.13: If f (x) =1x2 , find f ′(x).

Solution:

If f (x) =1x2 , first write the function in the form xn, so the rule for differentiating

can be applied.

f (x) =1x2 = x−2 , f ′(x) = −2x−2−1 = −2x−3 or

−2x3

Example 6.14: If g = cos(2t), what isdgdt

.

Solution:

If g = cos(2t),dgdt

= 2×− sin(2t) = −2 sin(2t)

Exercise 6.15: Find the derivative of the following functions with respect to x:

(a) y = x

(b) y = 3

(c) y = x−32

(d) y =√

x

(e) y =1√x

(f) y = π

(g) y = x5

(h) y = ln x

(i) y = e2x

(j) y = x3.1

(k) y = x52

(l) y = sin 3x

(m) y = cos 2x

(n) y = e3.1x

(o) y = 3x

(p) y = log x

(q) y = x12

(r) y = e

(s) y = tan 5x

(t) y =√

x3

384 c© USQ, August 27, 2010


Exercise 6.16: Differentiate the following functions with respect to the given indepen-dent variable.

(a) f (x) = x8

(b) f (t) = e2t

(c) g(p) = sin(3p)

(d) T(x) = ln x

(e) Q(t) = e−t

(f) z(x) = log x

(g) s(t) = t5

(h) P(t) = ln t

(i) T(t) = cos(

2π

365t)

(j) A(r) = r2

(k) E(q) =1q2

(l) V(r) =1r2

(m) y(x) = sin (2πx)

Hint Coping with all these rules.

If you have not done so already:

• make a list of the formulas;

• identify what the variables and symbols mean;

• make sure you know what each one is for; and

• include a short example next to each formula.

Application: physics Capacitance in an alternating current (AC) circuit.

A capacitor is essentially a pair of parallel plates, separated by an insulator. Whena potential difference (that is, a voltage) is applied to either side of the capacitor,charge builds up on the plates. When this voltage is removed the plates discharge.See the diagram below:

c© USQ, August 27, 2010 385

MAT1500 6.5. Derivatives of compound functions

The amount of charge that collects on the plates depends on the voltage that hasbeen applied and the capacitance of the plates (C). That is: q = Cv, where q isthe charge, C the capacitance, measured in farads (F) and v the voltage in volts.When the capacitor is charging or discharging the rate at which this happens isgiven by:

dqdt

= Cdvdt

,

(which we obtained by differentiating both sides of the above equation).

In electronics, the rate at which the charge varies is defined as the current i inamperes. That is:

i =dqdt

= Cdvdt

.

6.5 Derivatives of compound functions

Unfortunately life is never simple. Most functions that we need to deal with in real lifeare more complicated than this. For example the decay of a sample of a radioactiveisotope is given by the function Q(t) = Q0e−kt (remember Chapter 3). This is one of ourbasic functions multiplied by a constant Q0. The interference pattern of two interactingwaves (perhaps in optics, or water) would involve in some way the function f (x) =

sin kx + cos hx , that is, the sum of two of our basic functions. Let us see how we coulduse the known derivatives of the basic functions to find the derivative of the compoundfunctions.

Derivative of the sum of functions

The derivative is a slope — the change in the function divided by the change in thevariable. Where a function p(x) is the sum of two functions f (x) and g(x), the change inp(x) will be the sum of the changes in f (x) and g(x).

Take a simple example, p = x + x2 , so that f = x and g = x2. Look at a small section ofa table of values for this function.

x x = 1 x = 2 Change in x = 1f (x) = x f (1) = 1 f (2) = 2 Change in f = 1

g(x) = x2 g(1) = 1 g(2) = 4 Change in g = 3p(x) = f + g = x + x2 p(1) = 2 p(2) = 6 Change in p = 4

equals change in f + change in g

Thus

Change in pChange in x

=Change in f + change in g

Change in x

=Change in fChange in x

+Change in gChange in x

.

386 c© USQ, August 27, 2010

6.5. Derivatives of compound functions MAT1500

Taking the limiting value as the change in x becomes very small makes no difference tothis. Consequently we obtain the following result:

If f (x) and g(x) are two functions of x, then the derivative of their sumis the sum of their derivatives, or

ddx

[ f (x) + g(x)] =d

dx[ f (x)] +

ddx

[g(x)] = f ′(x) + g′(x) .

For example,

ddx(x3 + x− 7

)=

ddx

(x3) +d

dx(x) +

ddx

(−7)

= 3x3−1 + 1 + 0

= 3x2 + 1 .

Derivative of a constant times a function

Similar to the above, the [change in k× f (x)] is k× [change in f (x)]. Try this out yourselffor the function p = 5x2 (k = 5, f = x2).

The derivative of a constant multiplied by a function is the constantmultiplied by the derivative of the function.

ddx

[k f (x)] = kd

dx[ f (x)] = k f ′(x) .

For example,

ddx

(5x4) = 5× ddx

(x4) = 5× 4x4−1 = 20x3 .

Example 6.17: Differentiate the following with respect to the variable on the RHS.

(a) y = x3 − 4x2 + 4

(b) y = 4 ln x + 2√

x

(c) y = 2ex − 4√x+

tan x3

(d) t =log u

3−√

5u

(e) v =1

2m2 + 3 cos m−m

(f) y = sin 2θ + 2 cos θ

c© USQ, August 27, 2010 387


Solution:

(a) y = x3 − 4x2 + 4dydx

= 3x2 − 4× 2x + 0 = 3x2 − 8x

(b) y = 4 ln x + 2√

x = 4 ln x + 2x12

dydx

= 4× 1x+ 2× 1

2× x

12−1 =

4x+ x−

12 , or

4x+

1√x

(c)

y = 2ex − 4√x+

tan x3

,

y = 2ex − 4x−12 +

13

tan x ,

dydx

= 2ex − 4×−12

x−32 +

13

sec2 x

= 2ex +2

x32+

sec2 x3

, or

= 2ex +2

x√

x+

sec2 x3

(d)

t =13

log u−√

5×√

u ,

t =13

log u−√

5× u12 ,

t =13× ln u

ln 10−√

5× u12 ,

dtdu

=13× 1

u ln 10−√

5× 12

u−12

=1

3u ln 10−√

52√

u

(e)

v =12

m−2 + 3 cos m−m ,

dvdm

=12×−2m−3 + 3×− sin m− 1

= − 1m3 − 3 sin m− 1

(f)

y = sin 2θ + 2 cos θ

dydθ

= 2 cos(2θ)− 2 sin(θ)

388 c© USQ, August 27, 2010


Example 6.18: Find the slope of the tangent to the following curve at the point x = 4 .

y =x

32

4− 3√

x + 8 .

Solution:

The slope of the tangent to a curve at any point is given bydydx

.

dydx

=14× 3

2x

32−1 − 3× 1

2x

12−1 + 0

=14× 3

2x

12 − 3× 1

2√

x+ 0

=3√

x8− 3

2√

x.

At x = 4 ,

dydx

=3√

48− 3

2√

4

=34− 3

4= 0 .

(The slope of zero means that the tangent to the curve is parallel to the x-axis at thispoint.)

Exercise 6.19: State in your own words, what is a derivative?

Exercise 6.20: Write down the derivatives of the following functions with respect to x.

(a) x7

(b) 5x

(c)x3

(d) 0.06x

(e)14

x5

(f) 15x4 + 7

(g)2x6

3(h) 1.5x3

(i) (4x)2

(j) 17.3

(k) 3x2

(l) πx2

(m) 9x13

(n) 2 sin 4x

c© USQ, August 27, 2010 389


(o) 7x−3

(p) 4 ln x

(q) 3e3x

(r) 4 tan 0.5x

(s)ax

ln a

(t)13

sinπ

3x

(u) 4√

x

Exercise 6.21: If V = 43 πr3 , find

dVdr

.

Exercise 6.22: If P =20V2 , find

dPdV

.

Exercise 6.23: Differentiate the following functions with respect to x:

(a) 6x2 + 5x

(b) 3x2 + x + 1

(c) 4x4 + 3x2 − x

(d)12

x2 +17

x +14

(e)5x+ 4√

x

(f) 7 +4√x+ 2 ln x

(g) 3ex + sin x

(h) 4 tan x− cos x

(i) x(5− x + 3x2) + 2 ln x (Hint: expand brackets first)

(j) 3√

x− x√

x

(k) 5x2 + 6x− 7

(l) −x4 + 2x2 + 6

(m)x3 + 2x + 1

x(n) 2x− 5

√x

(o) 5ex − sin 2x

(p)4x4 +

3x3 +

2x2

(q) 4 sin 4x + 4 cos 4x

(r)14

(x8 − x4

)

(s) x− ln x

(t) 2√

x− 2√x

390 c© USQ, August 27, 2010


Exercise 6.24: Find the slope of the tangent to the curve y = x + tan x at the followingpoints.

(a) x = 0

(b) x = 3

Exercise 6.25: If f (x) = x3 + 2x , find

(a) f ′(x)

(b) f ′(0)

(c) f ′(2)

Points to remember

• If f (x) and g(x) are two functions of x, then the derivative of theirsum is the sum of their derivatives, or

ddx

[ f (x) + g(x)] =d

dx[ f (x)] +

ddx

[g(x)] = f ′(x) + g′(x)

• The derivative of a constant multiplied by a function is the con-stant multiplied by the derivative of the function.

ddx

[k f (x)] = kd

dx[ f (x)] = k f ′(x)

Hint How to translate from words to symbols:

• Read the problem twice.

• Read the problem out aloud to yourself, if possible.

• Ask yourself four questions:

– What is the problem asking me?– What facts are given in the problem?– Are there any special conditions?– Is any information irrelevant?

• Draw a diagram or picture.

• Break the problem down into parts.

• Define the variables.

• Look for connections between variables.

• Write the connection out in words, then write them out as an algebraicexpression.

• Talk to somebody else about it.

c© USQ, August 27, 2010 391

MAT1500 6.6. Application: maxima and minima

6.6 Application: maxima and minima

Think about the maximum height a ball reaches as it is thrown straight upwards by ajubilant sportsman. At the moment of its maximum value, it has just ceased movingupwards, and is just about to begin its descent. At that moment it is stationary. That is,its velocity equals zero. We could also say it is at its turning point (as in Chapter 2). If wewere to graph its height versus time, we would see something similar to that shown inthe following figure.

Notice that the instantaneous slope at its peak (that is, the slope of the tangent) is hor-izontal, so that it has zero slope. The slope of this graph, and velocity, are of coursemanifestations of the derivative of height versus time, so we can say that height is a max-

imum whendhdt

= 0 . We could think up other examples, and draw graphs similar to thatabove, where we can see that where a function has a minimum value, its derivative willalso be zero.

392 c© USQ, August 27, 2010

6.6. Application: maxima and minima MAT1500

The point A is called a local maximum since the y value at A is greater than its valueimmediately either side of A, and B is called a local minimum, since its y value is lowerthan those on either side. (It is important to remember that these points are not necessar-ily the absolute maximum or minimum values of the function as you can clearly see fromlooking at the graph.) At both of these points, the tangent to the curve is horizontal andhence its slope is zero. Thus

dydx

= 0 at A and B.

So we can find all the local maximum and minimum values of a function by differentiat-

ing and solving fordydx

= 0 .(Of course the maximum and minimum values are the y values. The x values are thelocations of these maxima and minima.)

The question is, when we have solved the equationdydx

= 0 and found the points whichare either maxima or minima, how do we know which of them are maxima and which ofthem are minima?

The maximum and minimum points, or ‘turning points’, can be distinguished by two al-ternative methods. Normally Method A is used except where the differentiation becomesdifficult.

Method A - Second derivative test

1. Differentiate y to finddydx

.

2. Find values of x which satisfydydx

= 0 .

3. Differentiatedydx

with respect to x to find the second derivative.

This is written asd2ydx2 or f ′′(x) and measures the rate of change of the slope of the

curve.

Look at the change of slope in the figure below as we traverse a maximum point inthe direction of x increasing.

c© USQ, August 27, 2010 393


See how the slope goes from positive, through zero to negative. That is, it is de-creasing as x increases, so the rate of change of slope, that is, the second derivativewill be negative.

Convince yourself with the same argument that the second derivative will be posi-tive at a minimum point.

4. Substitute each of the x values found in Step 2 above intod2ydx2 .

(a) Ifd2ydx2 is positive, a minimum occurs at that value.

(b) Ifd2ydx2 is negative, a maximum occurs at that value.

(c) Ifd2ydx2 = 0 , then use Method B, or roughly sketch the curve.

5. Find the values of y by substituting into the original function.

Method B - Adjacent slopes test

1. Differentiate y to finddydx

.

2. Find values of x which satisfydydx

= 0 .

3. Substitute values just before and just after each x value intodydx

.

(a) Ifdydx

is positive just before and negative just after, then the point is a maximum.

(b) Ifdydx

is negative just before and positive just after, then the point is a minimum.

4. Find the values of y by substituting into the original function.

There is another type of point for whichdydx

= 0 , which is neither a maximum nor aminimum. See the following figure.

394 c© USQ, August 27, 2010


These are called points of inflection and they may be distinguished from maxima and

minima by the fact thatd2ydx2 = 0 and the slope of the curve does not change sign on each

side of the point.

or

Note thatd2ydx2 = 0 does not by itself indicate that the point is a point of inflection.

Example 6.26: Find all maximum and minimum values for the functions:

(a) y = 2x3 − 8x2 + 10x + 4

(b) y = 2x3 + 6x2 − 18x + 4

(c) y = x3 − 3x2 + 3x + 12

Solution:

(a) y = 2x3 − 8x2 + 10x + 4dydx

= 6x2 − 16x + 10 .

Turning points occur whendydx

= 0 .

Therefore 6x2 − 16x + 10 = 0 ,

x =16±

√162 − 4× 6× 10

12

=16±

√16

12

= 1 or53

.

To determine whether these points are maxima or minima, findd2ydx2 , (that is,

differentiatedydx

).

d2ydx

= 12x− 16 .

c© USQ, August 27, 2010 395


When x = 1 ,d2ydx2 = 12− 16 = −4 .

Since this is negative, there is a maximum at x = 1 .

When x = 53 ,

d2ydx2 = 12× 5

3− 16 = 4 .

Since this is positive there is a minimum at x = 53 .

When x = 1 , y = 2× 13 − 8× 12 + 10× 1 + 4 = 8 .

When x = 53 , y = 2× 125

27 − 8× 259 + 10× 5

3 + 4 = 7.704 .

Therefore the function has a maximum value of 8 at x = 1 , and a minimumvalue of 7.704 at x = 5

3 .

Note: that we used Method A to determine the nature of the turning points,but Method B could have been used as follows:

At a point just below x = 1 , say x = 0.5 ,

dydx

= 6× 0.52 − 16× 0.5 + 10 = 3.5 .

At a point just above x = 1 say x = 1.5 ,

dydx

= −0.5 .

Since the slope has changed from + to −, the point is a maximum.

At a point just below x = 53 , say x = 1.5,

dydx

= −0.5 .

(Note we use the point found earlier.)

At a point just above x = 53 , say x = 2 ,

dydx

= 2 .

Since the slope has changed from − to +, the point is a minimum.

Note: that when selecting points above or below the turning point, take carenot to choose a point on the far side of another turning point, as this may leadto error in your conclusions.

396 c© USQ, August 27, 2010


(b) y = 2x3 + 6x2 − 18x + 4dydx

= 6x2 + 12x− 18 .


= 0 .

Therefore 6x2 + 12x− 18 = 0

6(x2 + 2x− 3) = 0

6(x + 3)(x− 1) = 0 .

thus x = 1 or −3.

d2ydx2 = 12x + 12 .

When x = 1 ,d2ydx2 = 24 ,

therefore minimum point at x = 1 .

When x = −3 ,d2ydx2 = −24 .

therefore maximum point at x = −3 .

Minimum value is y = 2× 13 + 6× 12 − 18× 1 + 4 = −6 .

Maximum value is y = 2× (−3)3 + 6× (−3)2 − 18×−3 + 4 = 58 .

(c) y = x3 − 3x2 + 3x + 12dydx

= 3x2 − 6x + 3 .


= 0 .

Therefore 3x2 − 6x + 3 = 0

3(x2 − 2x + 1) = 0

3(x− 1)2 = 0 .

thus x = 1.d2ydx2 = 6x− 6 .

When x = 1 ,d2ydx2 = 0 , so use Method B.

Below x = 1 , say x = 0 ,dydx

= 3 .

Above x = 1 , say x = 2 ,dydx

= 3× 22 − 6× 2 + 3 = 3 .

The slope is positive on both sides of the point. Thus x = 1 is a point ofinflection and there are no maxima or minima.

c© USQ, August 27, 2010 397


Example 6.27: The flow of air, F, through the trachea or ‘windpipe’ of radius r cm isgiven by

F = b(ar4 − r5) cm3s−1 ,

where a and b are positive constants.

During coughing, the radius is reduced to increase the velocity of the air. The ve-locity is given by

v =FA

,

where A is the cross-sectional area. Find the radii which give maximum velocityand maximum flow.

Solution:A = πr2 ,

therefore

v =b(ar4 − r5)

πr2

=bπ(ar2 − r3) .

Thereforedvdr

=bπ(2ar− 3r2) ,

which is zero when 2ar− 3r2 = 0 . That is, r(2a− 3r) = 0 , so r = 0 , or r =2a3

.

d2vdr2 =

bπ(2a− 6r) .

When r = 0 =d2vdr2 =

2abπ

, which is always positive since a and b are positiveconstants.

Therefore the minimum velocity occurs when r = 0 cm (windpipe is closed).

When r =2a3

,

d2vdr2 =

bπ(2a− 4a)

=−2ab

π,

which is always negative.

Therefore the maximum velocity occurs at r =2a3

cm.

To find maximum flow,

dFdr

= b(4ar3 − 5r4)

= br3(4a− 5r) ,

398 c© USQ, August 27, 2010


and this is zero when r = 0 or4a5

.

d2Fdr2 = b(12ar2 − 20r3) .

When r = 0 ,d2Fdr2 = 0 .

On sketching the curve, this is a minimum point.

When r =4a5

,

d2Fdr2 = b

(12a

16a2

25− 20× 64a3

125

)= − ba3

125× 320 ,

which is always negative.

Therefore the maximum flow occurs at r =4a5

.

Example 6.28: A church congregation has commissioned a lead-light window manufac-turer to make replica Norman windows for their new church. Norman windowsare modelled as a rectangle surmounted by a semicircle. The length of framing forthe heavy outside of a window is limited to 12 metres. Find the dimensions of sucha window so that the maximum amount of light will be allowed into the buildingthat has the window fitted.

Solution: First step is to draw a diagram of the situation.

What do you know?

Let L be the length of the rectangle and r the radius of the semicircle as indicated.Perimeter is 2r + 2L + 1

2 × 2πr , but perimeter is 12 m, so

2r + 2L +12× 2πr = 12

L =12− 2r− πr

2.

What do we need to find?

c© USQ, August 27, 2010 399


A function for the area of the window so that we can use it to find the maximumarea of the window: A = 1

2 πr2 + 2r× L .

Substituting in the expression for L to get the function in terms of one variable only:

A =12

πr2 + 2r×(

12− 2r− πr2

)

= 12r− 12

πr2 − 2r2 .

What do we need to do?

Use the first derivative of the area function to find the maximum value:

A =12

πr2 + 2r×(

12− 2r− πr2

),

SodAdr

= 12− πr− 4r .

The maximum occurs whendAdr

= 0 ,

Therefore 12− πr− 4r = 0

r =12

π + 4≈ 1.68 .

To check that this is a maximum, find the 2nd derivative:d2 Adr2 = −π − 4 , which is negative and so we have a maximum. Thus the dimen-

sions of the window which allows the maximum amount of light through are over-

all width,24

π + 4metres (≈ 3.36m) and overall height,

24π + 4

metres (≈ 3.36m),

since the radius of the semicircle is12

π + 4metres (≈ 1.68m) and the length of the

rectangle is12

π + 4metres (≈ 1.68m).

Application: electrical engineering An electrical voltage, V, in volts, de-pends on the time, t, in seconds represented by the function, V = 30 cos 500t +15 sin 500t . To determine the maximum value of the voltage then an engineerwould differentiate the given function and find the values of t that made thederivative equal to zero.

V = 30 cos 500t + 15 sin 500t ,dVdt

= 30× 500×− sin 500t + 15× 500× cos 500t

= 7 500 cos 500t− 15 000 sin 500t .

For a maximum dVdt

= 0 .

7 500 cos 500t− 15 000 sin 500t = 0 ,

7 500 cos 500t = 15 000 sin 500t ,sin 500tcos 500t

=7 50015 000

,

400 c© USQ, August 27, 2010


(assuming that cos 500t 6= 0 )

tan 500t = 0.5 ,

500t ≈ 0.4636 ,

t ≈ 0.0009 ,

(considering solutions in the first quadrant only).

The engineer would check that this was actually a maximum value by findingthe second derivative, then once this is confirmed, would substitute back intothe original equation to find the maximum voltage. In this instance it would beapproximately 33.5 V.

Exercise 6.29: Find the maximum and/or minimum values of the following functions ofx.

(a) y = x2 − 4x + 3

(b) y = 3 + 2x− x2

(c) y = 2x2 − 6x + 9

(d) y = x3 − 12x

(e) y = 2− 9x + 6x2 − x3

(f) y = x4

Exercise 6.30: A contractor wishes to fence in a rectangular enclosure of area 128 m2. Oneside is formed by an existing brick wall. Find the minimum length of fencing forthe other three sides.

Exercise 6.31: Weekly revenue for the latest best-seller, is given by

R = −p3 + 33p2 + 9p ,

where R is the revenue per week and the price is p dollars. What price should thecompany charge to obtain the largest revenue, and what is the largest revenue?

Exercise 6.32: The bending moment, M, of a beam supported at one end at a distance ofx from the support is given by

M =12

wLx− 12

wx2 ,

where L is the length of the beam and w is the uniform load per unit length. Assum-ing that L and w are given constants, find the point on the beam where the momentis greatest.

c© USQ, August 27, 2010 401


Exercise 6.33: The power transmitted by a belt drive is proportional to

Tv− wv3

g,

where v is speed of the belt, and T is the tension on the driving side, w is the weightper unit length of the belt and g, the acceleration due to gravity; all T, w and g areconstant. Find the speed at which the transmitted power is a maximum.

Exercise 6.34: A cylindrical metal container is open at the top. It must be designed tohave a capacity of 1078 cm3. In order to galvanise the container the manufacturerneeds to know what dimensions of the container would minimise surface area.What are these dimensions?

Exercise 6.35: A settling pond is to be constructed to hold 324 m3 of waste. The pond hasa square base and four vertical sides, all made of concrete and a square top madeof steel. If the steel costs twice as much per unit area as the concrete, determine thedimensions of the pond that will minimise the cost of construction.

That brings us to the end of this chapter covering differentiation. You have worked hardto reach here, and most probably have had to absorb some new concepts along the way.Go back over the chapter and make sure you have a solid grasp of what a derivative is,how to calculate it, how it relates to rates of change and the slope of a graph.

Points to remember

• The derivative of a function at local maximum and minimum val-ues is 0.

• The maximum and minimum points can be distinguished by theSecond Derivative test or the Adjacent Slopes test.

• In the second derivative test if d2 fdx2 = f ′′(x) > 0 the turning point

is a minimum.

• In the second derivative test if d2 fdx2 = f ′′(x) < 0 the turning point

is a maximum.

Putting it all together! Designing buildings is not always a matter of trial anderror. Consider this situation. You will need to use concepts from differentialcalculus as well as some basic trigonometry to investigate the solution to this.

One corridor 1.2 metres wide meets another twice the width at a right angle.What is the maximum length of a piece of furniture or ladder which can be carriedhorizontally around the corner? How narrow do you think corridors could be?What implications does this have for building design?

402 c© USQ, August 27, 2010

6.7. Review MAT1500

Hint Keep calm, do not panic!

As this is your final chapter, it is a good time to take stock on how much you haveachieved. Look back on your maths journal or just flick through your work booksand reflect upon:

• how far you have come;

• how you have overcome many things you thought difficult at first; and

• the strategies you have used that have been successful (note these for later).Use these positive thoughts to push you that last step through the last chapterand onto the exam.

Well done! The last chapter is completed. Check your skill level by attempting the Post-test.

6.7 Review

• The derivative is the instantaneous rate of change of a function at a point.

• The process of finding the derivative is called differentiation.

• The derivative at a point on a curve is the slope of the tangent to the curve at thatpoint.

• The derivative of a function f (t) is defined as d fdx (also f ′(x)) = lim∆t→0

∆ f∆x .

• Make sure you can use Table 6.4.

• ddx [ f (x) + g(x)] = d

dx [ f (x)] + ddx [g(x)] = f ′(x) + g′(x).

• ddx [k f (x)] = k d

dx [ f (x)] = k f ′(x).

• The derivative of a function at local maximum and minimum values is 0.

• The maximum and minimum points can be distinguished by the Second Derivativetest or the Adjacent Slopes test.

• At a local minimum d2 fdx2 = f ′′(x) > 0.

• At a local maximum d2 fdx2 = f ′′(x) < 0.

6.8 Post-test

1. On the graph mark where the derivative is

(a) < 0

(b) > 0

c© USQ, August 27, 2010 403


(c) = 0

2. If V is the volume of a weather balloon and r is its radius describe in a short sentencethe meaning of

dVdr

.

For the following five questions, differentiate the given expression.

3. sin x + ln x

4. 5e−x + 2 cos 3x

5. The height of a projectile shot vertically in the air at time t = 0 with an initialvelocity of 10 m/s is given by h = 10t− 4.9t2 .

What is the time when the projectile reaches its maximum height?

404 c© USQ, August 27, 2010


6.9 Solutions


6.4 (a) f ′(105) = 10, means that at exactly 105 years after 1900, the population is chang-ing at a rate of 10 000 people per year. The sign is positive so the population isincreasing at that rate.

(b) f ′(105) = −2, means that at exactly 105 years after 1900, the population ischanging at a rate of 2 000 people per year. The sign is negative so the popula-tion is decreasing at that rate.

(c) f ′(105) = 0, means that at exactly 105 years after 1900, the rate of change ofthe population is zero, so the population in neither growing nor declining.

6.5 g′(0.01) = 108, means that at 0.01 seconds the current in the electrical circuit is chang-ing at a rate of 108 amperes per second. The sign is positive so the current is increas-ing.

6.6dVdt

∣∣∣∣t=2

= −112 , means that at 2 minutes after opening the valve the rate of change

of volume with respect to time is 112 m3 per minute. The sign is negative so thismeans that the volume in the reservoir is decreasing. This value could be calledflow rate.

6.7 f ′(T) = 4.8

6.8 f ′(300) = 0.02, means that at a workload of 300 kilogram.metres the rate of changeof cardiac output with respect to workload is 0.02 litres per kilogram.metre.

6.15 (a)dydx

= 1

(b)dydx

= 0, since 3 is a constant.

(c)dydx

= −32× x−

32−1 = −3

2x−

52 , or − 3

2x52

(d) y =√

x = x12 ,

dydx

=12× x

12−1 =

12× x−

12 , or

1

2x12

, or1

2√

x

(e) y =1√x= x−

12 ,

dydx

= −12× x−

12−1 = −1

2× x−

32 , or − 1

2x32

,

or − 12x√

x

(f)dydx

= 0, since π is a constant.

(g)dydx

= 5x5−1 = 5x4

(h)dydx

=1x

(i)dydx

= 2e2x

(j)dydx

= 3.1x3.1−1 = 3.1x2.1

c© USQ, August 27, 2010 405


(k)dydx

=52

x52−1 =

52

x32

(l)dydx

= 3 cos 3x

(m)dydx

= −2 sin 2x

(n)dydx

= 3.1e3.1x

(o)dydx

= 3x ln 3

(p)dydx

=1

x ln 10

(q)dydx

=12

x−12

(r)dydx

= 0 (e is a constant.)

(s)dydx

= 5 sec2 5x

(t) y =(

x3) 12 = x

32 ,

dydx

=32

x12 =

32√

x

6.16 (a) f ′(x) = 8× x8−1 = 8x7

(b) f ′(t) = 2e2t

(c) g′(p) = 3 cos(3p)

(d) T′(x) =1x

(e) Q′(t) = −e−t

(f) z(x) =ln xln 10

, z′(x) =1

x ln 10(g) s′(t) = 5t4

(h) P′(t) =1t

(i) T′(t) = − 2π

365sin

2π

365t ,

(2π

365is just a constant.

)

(j) A′(r) = 2r

(k) E(q) =1q2 = q−2 , E′(q) = −2q−2−1 = −2q−3 = −2

q3

(l) V ′(r) = − 2r3 , (as in (k).)

(m) y′(x) = 2π cos 2πx

6.20 (a) 7x6.

(b) 5.

(c)13

.

(d) 0.06.

(e)54

x4.

(f) 60x3.

406 c© USQ, August 27, 2010


(g) 4x5.

(h) 4.5x2.

(i) 32x , (note that (4x)2 = 16x2).

(j) 0.

(k) 6x.

(l) 2πx.

(m) 3x−23 .

(n) 8 cos 4x.

(o) −21x−4.

(p)4x

.

(q) 9e3x.

(r) 2 sec2 0.5x.

(s) ax.

(t)π

9cos

π

3x.

(u)2√x

, (note that√

x = x12 ).

6.21 4πr2 .

6.22 P = 20V−2 , sodPdV

= −40V−3 .

6.23 (a) 12x + 5 .

(b) 6x + 1 .

(c) 16x3 + 6x− 1 .

(d) x +17

.

(e) − 5x2 +

2√x

.

(f) −2x−32 +

2x

.

(g) 3ex + cos x .

(h) 4 sec2 x + sin x .

(i) Expand first to get 5x− x2 + 3x3 + 2 ln x .

Thus f ′(x) = 5− 2x + 9x2 +2x

.

(j) Write this first as 3x12 − x

32 , thus

f ′(x) =32

x−12 − 3

2x

12 .

(k) 10x + 6.

(l) −4x3 + 4x.

(m) 2x− 1x2 ,(

Note:x3 + 2x + 1

x= x2 + 2 +

1x

).

c© USQ, August 27, 2010 407


(n) 2− 52√

x.

(o) 5ex − 2 cos 2x.

(p) −16x5 −

9x4 −

4x3 .

(q) 16 cos 4x− 16 sin 4x.

(r) 2x7 − x3.

(s) 1− 1x

.

(t)1√x+

1x√

x,(

Note: 2√

x− 2√x= 2x

12 − 2x−

12

).

6.24 (a)dydx

= 1 + sec2 x .

At x = 0 ,

dydx

= 1 + sec2 0

= 1 +1

cos2 0= 1 + 1

= 2 .

(b) At x = 3 ,

dydx

= 1 + sec2 3

= 1 +1

cos2 3= 1 + 1.0203

= 2.0203 .

(Note in both of these exercises radians must be used.)

6.25 (a) f ′(x) =d f (x)

dx= 3x2 + 2 .

(b) f ′(0) means the value of f ′(x) when x = 0 , which is 0 + 2 = 2 .

(c) f ′(2) = 3× 22 + 2 = 14 .

6.29 (a)dydx

= 2x− 4 = 0 , when x = 2 ,

d2ydx2 = 2 .

Therefore the minimum value of y = −1 , at x = 2 .

(b) Maximum value of y = 4 , at x = 1 .

(c) Minimum value of y = 4.5 , at x = 1.5 .

(d) Minimum value of y = −16 , at x = 2 .

Maximum value of y = 16 , at x = −2 .

408 c© USQ, August 27, 2010


(e) Minimum value of y = −2 , at x = 1 .

Maximum value of y = 2 , at x = 3 .

(f) Minimum value of y = 0 , at x = 0 .

(Note thatd2ydx2 can be zero at a maximum or minimum point.)

6.30

Area = xy

= 128 ,

Therefore y =128

x.

Length of fence = L = 2x + y = 2x +128

x.

dLdx

= 2− 128x2 ,

and this is zero when x = 8 .d2Ldx2 =

256x3 .

When x = 8 ,d2ydx2 is positive.

Minimum of y =128

x= 16 , and occurs at x = 8 .

Minimum length of fence = 2× 8 + 16 = 32 m.

6.31 WhendRdp

= 0 ,−3p2 + 66p + 9 = 0 .

p =−66±

√662 − 4×−3× 92×−3

≈ 22.14 ,−0.136 .

Because the problem is interested in the price of the best seller only 22.14 is valid.To determine if this is a maximum or a minimum we use the second derivative test.d2Rdp2 = −6p + 66 . When p = 22.14 ,

d2Rdp2 = −66.84 , hence p ≈ 22.14 gives a max-

imum value. To determine the largest revenue for the best seller, substitute 22.14into the original equation.

When p = 22.14 , R = −(22.14)3 + 33× (22.14)2 + 9× 22.14 = 5 522.61 The pricethat the company should charge for the best seller is $22.14 , which would producea total revenue of $5 522.61 per week.

c© USQ, August 27, 2010 409


6.32 The maximum value will occur when the first derivative is zero.

M =12

wLx− 12

wx2 ,

ThereforedMdx

=12

wL− 2× 12

wx

= 0 , to find a maximum ,

Therefore x =

12

wL

w

=12

L .

To determine if this is a maximum or minimum value consider the 2nd derivative,d2Mdx2 = −w .

Since w is a constant weight and is positive, then the second derivative must benegative and the value will be a maximum stationary point. Thus the point alongthe beam where the greatest moment occurs is half way along the length of thebeam.

6.33

P = k×(

Tv− wv3

g

).

The maximum power will occur whendPdv

= 0 .

P = k×(

Tv− wv3

g

),

= kTv− kwv3

g

= kT × v− kwg× v3

ThereforedPdv

= kT − 3× kwg

v2


kT = 3× kwg

v2

v2 = kT × g3kw

v2 =gT3w

v = ±√

gT3w

.

To determine if these are maximum or minimum values consider the 2nd deriva-tive.

dPdv

= kT − 3× kwg

v2

Therefored2Pdv2 = −6× kw

gv .

410 c© USQ, August 27, 2010


If v =

√gT3w

,d2Pdv2 = −6kw

g×√

gT3w

< 0 , a maximum.

If v = −√

gT3w

,d2Pdv2 = −6kw

g×−

√gT3w

> 0 , a minimum.

Thus the speed at which the transmitted power is a maximum is

√gT3w

.

6.34 A cylindrical metal container is open at the top. It must be designed to have acapacity of 1078 cm3. In order to galvanise the tank the manufacturer needs to knowwhat dimensions of the tank would minimise surface area. Let r be the radius of thetank and h be its height. To determine the relationship between height and radius,we use what is known about the volume of a cylinder.

V = πr2h = 1078

h =1078πr2 .

The formula for the surface area (A) of a topless cylinder,

A = πr2 + 2πrh ,

but h =1078πr2 ,

Therefore A = πr2 + 2πr1078πr2

A = πr2 +2156

r.

(Note: This area would be multiplied by 2 if both the inside and outside of thecontainer were to be galvanised, but the resulting dimensions would be the same.)

The minimum surface area will occur whendAdr

= 0 .

A = πr2 +2156

r= πr2 + 2156r−1 ,

ThereforedAdr

= 2πr +−1× 2156r−2

= 2πr +−1× 2156r2


Therefore 2πr +−1× 2156r2 = 0

2πr3 − 2156r3 = 0

2πr3 − 2156 = 0

r3 =21562π≈ 343.14

r ≈ 7.0

c© USQ, August 27, 2010 411


To determine if this is a maximum or minimum value consider the 2nd derivative.

dAdr

= 2πr +−1× 2156r−2

Therefored2Adr2 = 2π − 2156× 2× r−3

= 2π +4312

r3 .

If r ≈ 7 ,d2Adr2 > 0 , hence it is a minimum.

The dimensions of the container would be a diameter of 14 cm, and height of 7 cm.

6.35 Let s be the length of the side of the square base and h be its height. To determinethe relationship between height and side, we use what is known about the volumeof a rectangular prism.

V = s2h = 324

h =324s2 .

The formula for the surface area (A) of the prism, A = s2 + 4sh + s2 .

If the cost of the steel top is $2k per sq metre, then the cost of the concrete partswill be $k per sq metre. The formula for the cost (C) of the pond will be C =

k(s2 + 4sh) + 2k(s2) .

Substitute the expression for h into the equation to write it in terms of s only.

C = k(s2 + 4sh

)+ 2k

(s2)

= k(

s2 + 4s(

324s2

))+ 2k

(s2)

= k(

s2 +1296

s

)+ 2k

(s2) .

412 c© USQ, August 27, 2010


The minimum cost will occur whendCds

= 0 .

C = k(

s2 +1296

s

)+ 2k

(s2)

= 3ks2 +1296k

s,

ThereforedCds

= 6ks− 1296ks2

= 0 , to find a minimum ,

Therefore 6ks− 1296ks2 = 0

6ks3 − 1296ks2 = 0

6ks3 − 1296k = 0

6k(s3 − 216) = 0

s3 = 216

s = 6 .

To determine if this is a maximum or minimum value consider the 2nd derivative.

dCds

= 6ks− 1296ks2

Therefored2Cds2 = 6k +

2592ks3 .

If s = 6 ,d2Cds2 > 0 , hence it is a minimum. The dimensions of the pond would

be length of side of square base of 6 m, and height of 9 m.


1.

c© USQ, August 27, 2010 413


2.dVdr

is the rate at which volume increases per unit increase in radius.

3. cos x +1x

.

4. −5e−x − 6 sin 3x .

5. t = 1.02 sec.

414 c© USQ, August 27, 2010

Appendix A

Technical communication

Appendix contentsA.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

A.2 Writing Mathematics and numerical data . . . . . . . . . . . . . . . . . . . . . . 416

A.2.1 Whole numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416

A.2.2 Decimals and fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418

A.2.3 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

A.2.4 Mathematical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421

A.3 Tables and illustrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

A.3.1 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

A.3.2 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428

A.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431

A.1 Introduction

Successfully communicating complex technical, scientific and mathematical material is askill that only comes with practice and patience. Whether you are writing an assignment,a scientific paper, a technical report or a user guide, it is an essential skill. This appendix isdesigned as a user guide to help improve and master the complexities of writing technicalmaterial. As such, it does not cover grammar, and English usage1, but rather focuses onthe presentation of mathematical and numerical data in reports etc. As you read throughthe appendix you will find hints on good practice when presenting such material. If youconsciously implement these hints in your normal writing style, you find that after awhileyou will start automatically implementing these ideas whenever you write material.

1if you are interested in this you should read Strunk & White (1979) and Fowler (1996)

c© USQ, August 27, 2010 415

MAT1500 A.2. Writing Mathematics and numerical data

A.2 Writing Mathematics and numerical data

This section covers how to present whole numbers, decimals, fractions, units and mathe-matical equations in assignments and technical documents. Key concepts and exceptionsin this section are highlighted in bold. Examples are also provided to give you a betterfeel for the concepts and exceptions.

A.2.1 Whole numbers

Numbers below 10 are written in words. Most scientific writing guides also recommendthat 0 should be written as zero. For example:

• five cars;

• one command centre; and

• zero faulty items.

Exceptions to this are when numbers are used with: units of measure, age, time, dates,page numbers, percentages, money, ratios and proportions. You would write:

• 2 metres;

• 9 second intervals;

• 2 pm;

• page 3;

• $6; and

• 49:1.

Write all numbers as numerals when two or more numbers are used in the same sectionof writing. Writing numbers in the same format makes it easier for the reader to comparethe numbers.

A full irrigation system contains 10 metres of pipe, 2 water pumps and12 water storage tanks.

If however, none of the numbers in the section are greater than 9, then you shouldwrite them all as words.

A full irrigation system contains nine metres of pipe, two water pumpsand six water storage tanks.

416 c© USQ, August 27, 2010

A.2. Writing Mathematics and numerical data MAT1500

Numbers large larger than 999 should be written with a space to indicate thousands ofunits. For example you should write:

• 1 001;

• 100 000 001; and

• 234 123;

This makes the numbers easier to compare and understand at a glance. Note prior to1974 the standard in Australia was to use a comma to separate thousands of units (Stylemanual for Authors, Editors and Printers 1995). This change in Australia occurred on therecommendation of the Metric conversion board. The use of spaces is now endorsed bythe Standards Association of Australia (AS1000-1979).

If all numbers in a phrase are written as numerals the reader can become confused. Con-sider, 12 21-ohm resistors. The reader could assume you mean 1 221 ohm resistors, orthat you want twelve 21-ohm resistors. To make your meaning clear you should writeone of the two numbers in word form e.g. twelve 21-ohm resistors. This way the readerknows that you want twelve 21-ohm resistors.

Examples

Word form Preferred form45 4 000 component generators forty-five 4 000 component generators.three seven-person groups three 7-person groupsnine eight-colour images nine 8-colour images

There are a number of ways to write large numbers. For example:

• 310 000 000;

• 310 million;

• 3.1× 108; and

• 310× 106.

The format you choose should be:

1. consistent throughout the document;

2. in a format easily understood by your audience; and

3. suitable for any comparisons you wish to make between the numbers.

Normally most audiences find it easier to think in terms of 100 million or 25 000 insteadof 108 or 2.5× 104. However, if you want to compare orders of magnitude of numbers itis easier to compare 5.1× 103 and 5.1× 108, rather than having to count zeros.

The terms “billion”, “trillion” and “quadrillion” denote quantities shown in Table A.1.

c© USQ, August 27, 2010 417


Table A.1: Billion, trillion and quadrillion.

Term Numbermillion 1× 106

billion 1× 109

trillion 1× 1012

quadrillion 1× 1015

Note: this change in 1998 (Style manual for Authors, Editors and Printers 2002), and bringsAustralia into line with business practices used International (AS ISO 1000:1998). Mostscientists usually avoid the terms billion, trillion and quadrillion, as these are not consistentinternational; preferring to write large numbers in scientific notation.

A.2.2 Decimals and fractions

Use the singular, rather than the plural to describe a fraction less than one. For exam-ple, 0.5 kilometres may seem correct, but since 0.5 is less than one we have the singular,not plural. You would not say “half a kilometres”. The correct term is 0.5 kilometre.

Examples:

• 0.9 kg;

• 0.44 m3; and

• 0.25 cup of coffee.

Decimals and fractions should be written as numerals and not in words. In numericform decimals and fractions are more concise and readable.

Examples:

Word form Preferred numeral formzero point six seven 0.67six point seven zero eight one 6.7081three-sevenths 3/7four-fifths 4/5

When written inline a fraction should be constructed with a slash mark (/). For exam-ple: 1/3; 2/3; 1/2; and 7/10.

In Australia a full stop (not a raised point) should be used to represent the decimalpoint. Note: in Europe it is common to use a comma as a decimal marker, this is notfavoured in English speaking countries, and therefore should not be used (Style manualfor Authors, Editors and Printers 1995).

A zero should always be placed before the decimal point in numbers less than one.That is .34 should always be written 0.34. This ensures consistency when such numbersare mixed with numbers greater than one. Examples:

418 c© USQ, August 27, 2010


• 225.022;

• 0.34;

• 10.23; and

• 0.607 8.

A.2.3 Units

Units of measurement must be consistent within your entire document. Do not changeunits within a document. For example, don’t use kilometres in one place, and metres inanother place for the same variable. This forces the reader to convert between the units.It frustrates the reader and is prone to creating errors. In Australia, we use the SI units(see Tables A.2, and A.3). All other necessary units can be derived from the SI base units(Table A.2). Where a derived unit has a special name and symbol (Table A.4), these shouldbe used instead of the derived units.

Table A.2: SI base units

Quantity Name Symbollength metre mmass gram gtime second selectric current ampere Athermodynamic temperature kelvin Kamount of substance mole molluminous intensity candela cd

Table A.3: Examples of SI derived units

Derived quantity Name Symbolarea square metre m2

volume cubic metre m3

speed, velocity metre per second m/sacceleration metre per second squared m/s2

wave number reciprocal metre m−1

mass density gram per cubic metre g/m3

specific volume cubic metre per gram m3/gcurrent density ampere per square metre A/m2

magnetic field strength ampere per metre A/mamount-of-substance concentration mole per cubic metre mol/m3

luminance candela per square metre cd/m2

c© USQ, August 27, 2010 419


Table A.5: Prefixes for SI units

Prefix Symbol Valueexa E 1018

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

hecto h 102

deka da 101

deci d 10−1

centi c 10−2

milli m 10−3

micro µ 10−6

nano n 10−9

pico p 10−12

femto f 10−15

atto a 10−18

Table A.4: Units with special names and symbols

Name Symbol Expression inSI derived units

radian rad m m−1

degree (π

180

)rad

hertz Hz s−1

newton N m kg s−2

pascal Pa N m−2

joule J N mwatt W J s−1

coulomb C A svolt V W A−1

farad F C V−1

ohm Ω V A−1

litre L 10−3m3

degree Celsius C The Celsius temperature (tc) isgiven by tc = tk − 273.14, where tkis the temperature in kelvins.

Names of units and prefixes (Table A.5), when spelt out in full are expressed in lower-case letters (except at the beginning of a sentence). The exception to this is the capitalC in degree Celsius. Example: 10 kilometres per hour and 10 degrees Celcius.

Unit symbols are expressed in lower-case except for litre (i.e millilitre (mL), millimetre(mm)), units named after people (i.e. hertz (Hz), pascals (Pa), newton (N), watt (W) etc.)and symbols containing exa, peta, giga and mega (i.e. gigawatt (GW), gigametre (Gm)).

The term “per” should only be used when the unit is expressed in words, whereas“/” should only be used with symbols. For example kilometres per hour or km/h not

420 c© USQ, August 27, 2010


kilometres per h or km/hour.

Symbols for units should not be set in italics. The only exceptions are ohms which isrepresented by the greek letter Omega (Ω) and the prefix “micro”, which is the greekletter mu (µ).

A thin space should included between numerical value and unit names and symbols(e.g. 27 m not 27m). Exceptions to this are the symbols for degrees of arc (), minute (′),and second(′′) (e.g. note the difference between 10, and 30 C) .

Note we have not dealt with dates, times, roman numerals and currencies etc. Rulesgoverning how these should be typeset to Australian Government Standard are given inthe Style manual for Authors, Editors and Printers (1995).

A.2.4 Mathematical equations

Letters used as variables should be typed in italics (A, B, C, x, y, z) unless they repre-sented vector/matrix quantities, in which case they should be typeset in bold (A,B,C,x,y,z).It is essential that the same font be used for the same variable throughout the document.

Equations should be centred on a separate line in documents unless they are short andsimple.

Exampleax2 + bx + c = 0 . (A.1)

Equations appearing on a separate line should be numbered in the order that theyappear in the document. This makes it easy to reference a given equation later in thedocument. Numbering all equations provides the document with a look of consistency,which makes it easier for the reader to follow and find specific references.

Exampleax2 + bx + c . (A.2)

The general form of a quadratic is given in Eqn. A.2.

Short equations may be placed on a separate line or inline, whichever you prefer.

The current in the wire was calcu-lated using E = IR, where E is thecurrent, I is the electric potential,and R is the resistance.

The current in the wire was calcu-lated using

E = IR, (A.3)

where E is the current, I is the elec-tric potential, and R is the resis-tance.

c© USQ, August 27, 2010 421


All equal signs, fraction lines, multiplication, addition and minus signs should behorizontally aligned.

Examples

∆T =4π

KTT0

. (A.4)

I = e−kt × (sin ωt + cos ωt) +π

4. (A.5)

dTdy

= kT + C . (A.6)

y− y0

x− x0=

y1 − y0

x1 − x0. (A.7)

The equal signs should be aligned for a series of connected equations.

Example

z = r (cos θ + i sin θ) (A.8)

= reiθ . (A.9)

Note in this case it is acceptable to not number all the equations in the block. For exam-ple it is perfectly acceptable to follow the example below which just references the keyequation in an equation block.

z = r (cos θ + i sin θ)

= reiθ . (A.10)

Also as you would not write 2 = 3, you should not write a mathematic expression thatis not correct. For example do not write something like the following:

z =12− 6

6(A.11)

= 1 (A.12)

= 0.16 . (A.13)

The above statement can not be true as 1 is not equal to 0.16. In this case, the authormeant the following.

The standard z score is

z =12− 6

6(A.14)

= 1 .

This gives the area in the tail of the standard normal of 0.16.

The hierarchy of brackets is normally parentheses, square brackets, and than braces [()].According to the Style manual for Authors, Editors and Printers (1995) this is stylistic ratherthan obligatory. However, you should be aware that in certain mathematical contexts

422 c© USQ, August 27, 2010

A.3. Tables and illustrations MAT1500

brackets have special meaning. For example [1, 2] denotes a line segment from 1 to 2inclusive, while (1, 2) denotes an ordered pair.

A.3 Tables and illustrations

Diagrams, pictures, graphs2 and tables can often save you a large amount of words in atechnical document, as they can illuminate for the reader key findings. To clearly presentfindings tables and illustrations need to be clear, uncluttered and efficient. While tablesand illustrations share common characteristics in achieving these goals, we will deal withthem separately to illustrate subtle differences between well presented tables and figures.

A.3.1 Tables

Well presented tables have the following properties:

1. A caption that summarised the information presented in the table.

2. Clear and concise row and column headings as required.

3. Contents that are well organised, so it is easy for readers to find what they need.

4. The table is introduced in the text before it appears, and appears close to the textwhere it is first mentioned.

These properties of good tables, will be illustrated in following sections by consideringgood and bad examples.

Headings

Consider Table A.6. Reading the information in the table we can gather that informationis referring to unemployment. However, Table A.6 as it stands does not provide thereader with sufficient data to be able to interpret the numerical information. For example,it is not clear what the difference between 663 399 and 624 400 is? Are they differentmonths, years etc?

Table A.6: No column headings.

Unemployed persons 663 300 624 400 585 982 545 558 539 472Unemployment rate 6.7 6.2 5.8 5.3 5.1

The obvious way to fix this problem is provide a clear and concise headings for each col-umn and row of the table. Addition headings have been added to Table A.7. In Table A.7we now see that the data relates to the years 2002–2006. However, we still don’t have

2Diagrams, pictures, and graphs in documents are commonly referred to as illustrations and normallylabelled as figures in a technical document.

c© USQ, August 27, 2010 423

MAT1500 A.3. Tables and illustrations

information about the precise data or whether these relate to Australia, UK, US, or NewSouth Wales. Information to clarify these questions should be presented in the captionwhich is discussed in the following section.

Table A.7: Column headings.

Years2002 2003 2004 2005 2006


Captions

A caption should summarise the information presented in the table. If we consider Ta-ble A.7, the reader would have a number of questions about the data which the captionshould answer. Some of these questions would be:

• What country or region do these figures relate to?

• Are these figures yearly, quarterly averages, or something else?

Without this information, it is impossible for the reader to successfully understand thedata being presented. A suggested caption for Table A.7 is shown in Table A.8. Noticethat this caption tells the reader that the data is for Australia and that the yearly figuresare just for the June quarters over the period 2002–2006.

Table A.8: Australia unemployment figures for the June quarters 2002–2006.

Years2002 2003 2004 2005 2006


All tables must have a caption. The normal technical writing practice is for the captionto be above a table, as the reader needs to know what the figures in the table representbefore they can accurately interpret the information being presented.

Units

While Table A.8 is a significant improvement over Table A.6 it still can be improved.How? Consider the numbers in Table A.8, at present the reader has to guess that 663 300refers to the number of people and that 6.7 is a percentage of the population. If wedon’t formally tell the reader what units these numbers represent, misunderstandingsand errors can occur. Adding this information (Table A.9), makes it much harder forthese misunderstands and errors to occur.

424 c© USQ, August 27, 2010


Table A.9: Australia unemployment figures for the June quarters 2002–2006.

Years2002 2003 2004 2005 2006

Unemployed persons (number) 663 300 624 400 585 982 545 558 539 472Unemployment rate (%) 6.7 6.2 5.8 5.3 5.1

There are several ways that units can be included in tables. The most common ways are:

1. add units to each cell;

2. in table footnotes; or

3. in headings within the table.

While each method is acceptable there are a couple of issues with 1 and 2, that you shouldbe aware of. Adding units in each cell does provide unambiguous data, however it doesthis at the expense of introducing additional clutter into the table. An example of thisadditional clutter is shown in Table A.10. Compare this to Table A.11 where the unitshave been included in the headings.

Table A.10: Characteristics of selected fruit trees (Units in headings).

Species Height Trunk diameter Leaf length Fruit lengthBlack Walnut 21.4–27.4 m 0.6–0.9 m 304–610 mm 38–51 mmPawpaw 12.2 m 0.3–0.5 m 153–305 mm 102 mmRed Mulberry 7.62–12.2 m 0.3–0.5 m 76-127 mm 25 mm


Species Height (m) Trunk diameter (m) Leaf length (mm) Fruit length (mm)Black Walnut 21.4–27.4 0.6–0.9 304–610 38–51Pawpaw 12.2 0.3–0.5 153–305 102Red Mulberry 7.62–12.2 0.3–0.5 76–127 25

Like adding units in each cell, adding the units as footnotes does clarify the data, howeverit can be difficult for the reader to find this unit information, especially in a large table.The reader must move their eyes and search each time they want to find the informationon units, than move their eyes back to the data. If you have a lot of different units ina table this can become a frustration for the reader. An example of using footnotes isprovided in Table A.12. Compare how hard it is for you to find the units in Table A.12,with Table A.11 where the units are in the column headings. Consequently, unless youhave a very strong reason for doing so, the best option is to add units to row/columnheadings as appropriate.

c© USQ, August 27, 2010 425



Species Heighta Trunk diametera Leaf length b Fruit lengthb

Black Walnut 21.4–27.4 0.6–0.9 304–610 38–51Pawpaw 12.2 0.3–0.5 153–305 102Red Mulberry 7.62–12.2 0.3–0.5 76–127 25

aIn metresbIn millimetres

Ruled tables

In tables vertical and horizontal rules (lines) can be used to separate information. Whencreating tables you should use rules sparingly, as too many rules can clutter the table andmake it hard to read. However, used correctly rules can help guide the reader in inter-preting the information presented. Compare Table A.13 with Table A.14, and Table A.15with Table A.16. Most people find that Tables A.14 and A.16 are easier to read. This is dueto the eye not being over guided (or distracted) by unnecessary lines within the table.

Table A.13: Characteristics of selected fruit trees (Heavy use of rules).


Table A.14: Characteristics of selected fruit trees (Light use of rules).


Table A.15: Australia unemployment figures for the June quarter 2002–2006 (Rules heavyuse).

Years2002 2003 2004 2005 2006


426 c© USQ, August 27, 2010


Table A.16: Australia unemployment figures for the June quarter 2002–2006 (Rules lightuse).

Years2002 2003 2004 2005 2006


Note in Tables A.14 and A.16 the absence of vertical rules. In both cases white space andalignment has been sufficient to reduce the need to vertical rules. In addition, to this theonly other rules are used to a) separate the column headings; and b) indicate the end ofthe table. The only other feature to note relating to rules in Table A.16, is the use of theshort rule under Years. This short line is enough guidance for the reader to clearly seethat 2002–2006 refer to specific years.

Referencing

To be able to reference tables easily in text, each table must be numbered in order ofappearance. The reference number is normally added to the caption as shown in Ta-ble A.16. That is, the reference number is included in the caption following the wordTable. When referencing the table in text the word Table is followed by the unique refer-ence number of the required table (e.g. Table A.16).

Note all tables should be introduced in the text before they appear for the benefit ofthe readers. Consequently, no table should appear without a textual reference. Unlessspecified otherwise a table should appear immediately after it is first mentioned in thetext. Why does this have to occur? The reader maybe able to interpret the data from thetable you present, however the reader cannot put it in the context of your argument un-less you explain how the tabulated information fits into your argument. This contextualinformation has to occur before the reader looks at your table, otherwise the reader mayassume this data is going to be used in a different fashion.

The location of the table should not have the reader a flipping large number of pages tofind it. If this happens, the readers often forget the context in which the table was goingto be used, and more importantly get frustrated reading your document (after all you didput this information in writing for someone to read?).

Hints for word processing tables

1. Use thin lines for rules. Thick lines just look clumsy and make the table look clut-tered.

2. Make sure you use the correct tools for creating tables within the word processor(Do Not Use Tabs to create tables). Using the table option will give you additionaloptions in terms of formatting, rules, and alignment.

3. The fonts in tables should be the same throughout the document. Do not have atable with a font smaller than 8pt, as it is to hard to read.

c© USQ, August 27, 2010 427


A.3.2 Figures

Good figures like tables have the following properties:

1. A caption that summarises the information presented;

2. The contents are well organised, so it is easy for readers to interpret the information;

3. Legends, labels etc. are clear and concise;

4. Figures are introduced in the text and appear close to where they are first men-tioned; and

5. The style of figure is appropriate to the information being conveyed.

As there is a wide variety of types of figures (from pie charts to medical diagrams) wewill not try to cover the key points of every type. This will be done by using a simplebar chart to illustrate the key points. Additional information on including other types offigures can be found in books, such as Rosenberg (2005), which can be found in the USQLibrary. It is also worth mentioning while discussing other types of charts that 3D versionof bar and pie charts do not add any additional information over their 2D counterparts.However, 3D charts are harder to read accurately, and thus can be a source of errors.

Figure A.1: Break down of registered vehicles in Australia by fuel type over the period2002–2006.

Captions

Like tables sufficient background information should be provided in the caption to allowthe reader to understand the data being presented. An example of a good caption isshown in Figure A.1. This caption tells the reader that the information presented relates

428 c© USQ, August 27, 2010


to registered vehicles in Australia and that the information for the period 2002–2006 willbe broken down by fuel type.

Note a minor change from tables, is that the caption for a figure is normally locatedbelow the figure (not above as was the case for tables). This is due to the eye beingdistracted by discontinuities, and hence distracted directly by the figure when it occurs.Thus, when looking for the caption, since we read from the top to the bottom of the page,the reader will automatically look for the caption below the figure, as this is where thetext continues.

Labels, and legends

Compare Figure A.2 with Figure A.1, you will notice that the axes labels and legend aremissing from Figure A.2. What implications does this have for Figure A.2? Things youshould notice about Figure A.2 is that:

1. While it has a vertical scale, there is no indication of units. Thus, it is not possibleto tell what the height of the bars physically represent.

2. The scale on the horizontal axis just is 2002–2006. The reader can infer these relateto years, but it is not clear. It is quite possible that these numbers could relate totime, fuel price or some other quantity. Anything that the reader has to infer isdangerous and should be avoided.

The issues outlined above are fixed in Figure A.1 by adding labels with appropriate unitsto the horizontal and vertical axes.

Figure A.2: Break down of registered vehicles in Australia by fuel type over the period2002–2006 (no labels or legend) .

Adding the labels to the axes as suggested gives Figure A.3. While this is an improvementit is still not as good as Figure A.1. Why? In Figure A.3 there is no legend. Consequently,

c© USQ, August 27, 2010 429


in Figure A.3 there is no way to identify what each column represents. Figure A.1 fixesthis issue by adding a legend which clearly indicates what each column represents.

Figure A.3: Break down of registered vehicles in Australia by fuel type over the period2002–2006 (no legend).

Note these concepts apply to any type of figure. That is, any figure must have sufficientlabels (which include units as appropriate) to allow the reader to interpret the informa-tion within the figure using just these labels and the caption. If the reader, cannot do thiswith your figure, than there is no benefit in including it in the document.

Referencing

Figures are referenced in documents in a similar fashion to Tables, except the word Figureis used instead of Table (e.g. Figure A.2). It is also important to remember that figures liketables should not appear without a corresponding textual reference to place it in contextfor the reader.

Colour blindness

A significant percentage of people are colour-blind. Thus, the good technical writer whileloving the power of colour coding, understands that colour without due considerationcan frustrate a large proportion of readers. Most colour-blind people do not see the worldin black and white. In fact, most people are red–green colour-blind, this means that theysee red and green (and the shades in between) as the same colour. Nearly everyonewho is colour-blind can distinguish red from blue. Hence the default choice in manygraphing packages is blue and red. However, often it is necessary to choose a morecomplex colouring that has more than two colours. There are two approaches to thisproblem. One is to use shading instead of colour to make the distinction required, as done

430 c© USQ, August 27, 2010

A.4. Summary MAT1500

in Figure A.1. The alternative is to go to a site like http://colorschemedesigner.com

and select an appropriate colour scheme which will meet your needs, and allow peoplewith different levels of colour-blindness to distinguish the data.

A.4 Summary

At this stage you should be aware of two important points, namely that a good technicaldocument has the following properties:

1. it is consistent in terminology throughout; and

2. it presents information in a way that does not confuse or frustrate the reader.

All the stylistic and rules mentioned in this appendix really just reflect the conventionsused to achieve these propoerties in a technical document. If you use these conventionsyou are more likely to produce a document that gets the information across without frus-trating the reader.

c© USQ, August 27, 2010 431

http://colorschemedesigner.com

http://colorschemedesigner.com

References

Caughley, G. (1977), Analysis of Vertebrate Populations, Wiley. 50

Dunham, W. (1993), The Mathematical Universe, Wiley New York. 369

Fowler, H. (1996), Modern english usage, 3rd edn, Oxford. 415

History of Vectors (2009).URL: http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html 280

Rosenberg, B. (2005), Spring into technical writing for Engineers and Scientists, Addison Wes-ley. 428

Sherman, W. (1985), Changing values in the worksplace, in W. Ainsworth & Q. Willis,eds, ‘Australian Organizational Behaviour: Readings’, 2nd edn, Macmillan. 3

Strunk, W. & White, E. (1979), The elements of style, 3rd edn, Macmillan. 415

Style manual for Authors, Editors and Printers (1995), 5th edn, AGPS Press. 417, 418, 421,422

Style manual for Authors, Editors and Printers (2002), 6th edn, Wiley. 418

c© USQ, August 27, 2010 433

http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html

Mat1500 Sb Colour

Documents

Transcript of Mat1500 Sb Colour