mathematics for year 12 mathematical studies...HAESE HARRIS PUBLICATIONS& second edition mathematics for year 12 Robert Haese Sandra Haese Tom van Dulken Kim Harris Anthony Thompson

mathematicalstudies

mathematicalstudies

HAESE HARRIS PUBLICATIONS&

second editionsecond edition

mathematics for year 12

Robert Haese

Sandra Haese

Tom van Dulken

Kim Harris

Anthony Thompson

Mark Bruce

Michael Haese

SA_12STU-2magentacyan yellow black

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Y:\HAESE\SA_12STU-2ed\SA12STU-2_00\001SA12STU-2_00.CDR Friday, 10 November 2006 10:29:42 AM PETERDELL

MATHEMATICAL STUDIES (Second edition)MATHEMATICS FOR YEAR 12

This book is copyright

Copying for educational purposes

Disclaimer

Robert Haese B.Sc.Sandra Haese B.Sc.Tom van Dulken B.Sc.(Hons.), Ph.D.

Mark Bruce B.Ed.Michael Haese B.Sc.(Hons.), Ph.D.

Haese & Harris Publications3 Frank Collopy Court, Adelaide Airport SA 5950Telephone: (08) 8355 9444, Fax: (08) 8355 9471email:web:

National Library of Australia Card Number & ISBN 978-1-876543-58-7

© Haese & Harris Publications 2006

Published by Raksar Nominees Pty Ltd, 3 Frank Collopy Court, Adelaide Airport SA 5950

First Edition 2002Reprinted 2003Second Edition 2006

Cartoon artwork by John Martin. Artwork by Piotr Poturaj and David Purton.Cover design by Piotr Poturaj. Cover photography by Piotr Poturaj.Computer software by David Purton and Richard Milotti.Typeset in Australia by Susan Haese and Charlotte Sabel (Raksar Nominees).

Typeset in Times Roman 10 /11

Photo on p. 347 ©iStockphoto

. Except as permitted by the CopyrightAct (any fair dealing for the purposes ofprivate study, research, criticism or review), no part of this publication may be reproduced, stored in aretrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying,recording or otherwise, without the prior permission of the publisher. Enquiries to be made to Haese &Harris Publications.

: Where copies of part or the whole of the book are made underPart VB of the Copyright Act, the law requires that the educational institution or the body thatadministers it has given a remuneration notice to Copyright Agency Limited (CAL). For information,contact the CopyrightAgency Limited.

While every attempt has been made to trace and acknowledge copyright, the author and publishersapologise for any accidental infringement where copyright has proved untraceable. They would bepleased to come to a suitable agreement with the rightful owner.

: All the internet addresses (URL’s) given in this book were valid at the time of printing.While the authors and publisher regret any inconvenience that changes of address may cause readers,no responsibility for any such changes can be accepted by either the authors or the publisher.

Kim Harris B.Sc., Dip.Ed.Anthony Thompson B.Sc., Dip.T, Dip.Ed., Grad.Dip.Ed.Admin.

Reprinted 2007 , 2008

(with corrections)

(with corrections)

\Qw_ \Qw_

[email protected]


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mailto:[email protected]

http://www.haeseandharris.com.au/

FOREWORD

[email protected]

www.haeseandharris.com.au

This 2 edition of our established course in mathematics for Year 12 has been thoroughly

revised as a result of the recent changes to the . It

is not our intention to define the course, it is our interpretation of the concepts outlined in the

Statement and we encourage teachers and students to use other resources.

The main change in this new edition is the thorough overhaul of the two statistics chapters:

Chapter 7 now covers normal distributions and Chapter 8 covers binomial distributions only.

Other changes have been made to modelling in Chapter 2, the section ‘economic models’ has

been deleted from Chapter 4, a more detailed account of specific exponential functions is

included in Chapter 5, and Chapter 10 now includes transition matrices and other matrix

types. The book has been printed in full colour and is accompanied by a new and improved

version of our interactive Student CD.

The CD offers exciting possibilities for students and teachers. It contains links to

spreadsheets, graphing and geometry software, graphics calculator instructions, computer

demonstrations and simulations. Teachers will be able to demonstrate concepts quickly,

clearly and simply, and students have the opportunity to revisit the demonstrations and

experiment for themselves.

The book contains many problems from basic to advanced, to cater for a range of student

abilities and interests. While some of the exercises are designed simply to build skills, every

effort has been made to contextualise problems so that students can see everyday uses and

practical applications of the mathematics they are studying.

Emphasis has been placed on the gradual development of concepts with appropriate worked

examples. However, we have also provided extension material for those who look towards

further studies or applications of mathematics for their career choices. It is not our intention

that each chapter be worked through in full. Time constraints will not allow for this.

Consequently, teachers must select exercises carefully, according to the abilities and prior

knowledge of their students, in order to make the most efficient use of time and give as

thorough coverage of work as possible.

The extensive use of graphics calculators and computer packages throughout the book

enables student to realise the importance, application and appropriate use of technology. No

single aspect of technology has been favoured. It is as important that students work with a pen

and paper as it is that they use their calculator or graphics calculator, or use a spreadsheet or

graphing package on computer.

Instructions appropriate to each graphics calculator problem are on the CD. They are written

for Texas Instruments and Casio calculators, and can be printed from the CD.

We hope that this book, with the associated use of technology, will enhance the students’

understanding, knowledge and appreciation of mathematics.

We welcome your feedback. Email:

Web:

nd

Stage 2 Mathematics Curriculum Statement

RCH SHH

TvD KPH

AWT MFB PMH

The publishers would like to thank Michael Binkowski, David Martin, Carol Moule, and Paul

Urban for their assistance in editing this publication.


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mailto:[email protected]

http://www.haeseandharris.com.au/

page 4SA_12STU-2

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Y:\HAESE\SA_12STU-2ed\SA12STU-2_00\004SA12STU-2_00.CDR Friday, 3 November 2006 4:40:52 PM PETERDELL

TABLE OF CONTENTS 5

1 BACKGROUND KNOWLEDGE 9

2 FUNCTIONS AND INTRODUCTORY CALCULUS 29

3 DIFFERENTIAL CALCULUS 59

4 APPLICATIONS OF DIFFERENTIAL CALCULUS 99

5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 145

A Key concepts 10

B Constructing functions using geometry 21

C Review 27

A Functions 30

B Modelling from data 32

C Constructing exact models 37

D Basic theory of calculus 40

E When the rate of change is not constant 42

F Definite integrals 48

G Review 56

A The idea of a limit 62

B Derivatives at a given -value 65

C The derivative function 69

D Simple rules of differentiation 72

E Composite functions and the chain rule 76

F Product and quotient rules 80

G Implicit differentiation 84

H Tangents and normals 87

I The second derivative 93

J Review 95

A Rates of change 100

B Motion in a straight line 103

C Curve properties 110

D Optimisation 124

E Review 138

A Derivatives of exponential functions 148

B The natural logarithmic function 153

C Derivatives of logarithmic functions 158

D Exponential, surge and logistic modelling 162

E Applications of exponential and logarithmic functions 170

F Review 174

x

TABLE OF CONTENTS


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_00\005SA12STU-2_00.CDR Monday, 6 November 2006 12:25:30 PM PETERDELL

6 TABLE OF CONTENTS

6 INTEGRATION 179

7 STATISTICS 219

8 BINOMIAL DISTRIBUTIONS 281

9 SOLVING SYSTEMS OF LINEAR EQUATIONS 315

10 MATRICES 341

A Reviewing the definite integral 180

B The area function 183

C Antidifferentiation 184

D The Fundamental theorem of calculus 187

E Integration 190

F Linear motion 201

G Definite integrals 204

H Finding areas 206

I Further applications 211

J Review 214

A Key statistical concepts 220

B Describing data 224

C Normal distributions 228

D The standard normal distribution 234

E Finding quantiles ( -values) 241

F Investigating properties of normal distributions 244

G Distribution of sample means 245

H Hypothesis testing for a mean 257

I Confidence intervals for means 266

J Review 276

A Pascal’s triangle 282

B Assigning probabilities 285

C Normal approximation for binomial distributions 294

D Hypothesis testing for proportions 297

E Confidence intervals for proportions 301

F Review 310

A Solutions ‘satisfy’ equations 317

B Solving × systems of equations 322

C × systems with unique solutions 326

D Other × systems 330

E Further applications 334

F × and × systems 337

G Review 338

A Introduction 342

B Addition, subtraction and multiples of matrices 345

C Matrix multiplication 351

k

��

��

��

��


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_00\006SA12STU-2_00.CDR Monday, 6 November 2006 12:27:54 PM PETERDELL

D Transition matrices 364

E The inverse of a × matrix 376

F The inverse of a × matrix 381

G Determinants of matrices 385

H Review 390

��

��

ANSWERS 395

INDEX 447

TABLE OF CONTENTS 7


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The interactive CD is ideal for independent study.

Frequent use will nurture a deeper understanding of

Mathematics. Students can revisit concepts taught in

class and undertake their own revision and practice. The

CD also has the text of the book, allowing students to

leave the textbook at school and keep the CD at home.

Graphics calculators: instructions for using Texas

Instruments and Casio graphics calculators are also given

on the CD and can be printed. Click on the relevant icon

(TI or C) to access printable instructions.

Examples in the textbook are not always given for both types of calculator.

Where that occurs, click on the relevant icon to access the instructions for

the other type of calculator.

USING THE INTERACTIVE CD

The CD icons throughout the book denote active links to a range of interactive features

INTERACTIVE LINKS

DEMOSPREADSHEET

STATISTICS

PACKAGE

GRAPHING

PACKAGESIMULATION

Other icons used in this book:

HISTORICAL NOTE INVESTIGATION

DISCUSSION

TI

C

LAW

DETERMINER

AREA

FINDER

COMPUTER

DEMO

CALCULUS

GRAPHING PACKAGE MODELLING

APPENDIX INVESTIGATION

EXTRA

PROBLEMS

PRINTABLE

INVESTIGATION


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_00\008SA12STU-2_00.CDR Tuesday, 14 November 2006 2:32:55 PM PETERDELL

1

Contents:

Background knowledgeBackground knowledge

A

B

C

Key concepts

Constructing functions using geometry

Review


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_01\009SA12STU-2_01.CDR Thursday, 2 November 2006 9:09:23 AM PETERDELL

10 BACKGROUND KNOWLEDGE (Chapter 1)

The following is a summary of key definitions and formulae which is assumed knowledge

from Stage 1 Mathematics.

y = ax + b, a 6= 0 Linear

y = ax2 + bx + c, a 6= 0 Quadratic

y = ax3 + bx2 + cx + d, a 6= 0 Cubic

y = ax4 + bx3 + cx2 + dx + e, a 6= 0 Quartic

A linear function is commonly written as y = mx + c:

The slope of the line is m and the y-intercept is c.

If (x1; y1) and (x2; y2) are two points on the line,

then the slope m =y-step

x-step=

y2 ¡ y1x2 ¡ x1

.

KEY CONCEPTSA

POLYNOMIALS

LINEAR FUNCTIONS

y

x

y mx c��

yx

yz

xz xx

12 xxx ��

12 yyy ��

),( 22 yx

),( 11 yx

),0( c

slope is m

The graph of the linear function is a straight line, often just called a “line”.y mx c� � � �= +

This may be written as m =¢y

¢x, where ¢y = y2 ¡ y1 is the difference between the two

y values, and ¢x = x2 ¡ x1 is the corresponding difference between the two x values.

The capital Greek letter ¢ (Delta) is the initial letter of the word “difference”.

To find an equation of a straight line we need: ² the slope of the line

² a point on the line.


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_01\010SA12STU-2_01.CDR Wednesday, 25 October 2006 12:24:49 PM PETERDELL

BACKGROUND KNOWLEDGE (Chapter 1) 11

1 Find the equations of these lines:

a b c

d e f

a Find an equation of the straight line that passes through (1, 3) and (2, 5).

b Find an equation of the straight line with y-intercept 3 that is parallel to the

line in part a.

c Find where the line in part b meets the line 2y + 3x = ¡1:

d Sketch the graphs of parts b and c.

a The slope of the line though (1, 3) and (2, 5) is m =¢y

¢x=

5 ¡ 3

2 ¡ 1= 2.

If (x, y) is any point on the line theny ¡ 3

x¡ 1= 2 ) y ¡ 3 = 2x¡ 2

) y = 2x + 1

b The slope of the line is 2 and the y-intercept 3, so an equation of the

line is y = 2x + 3.

c Substituting y = 2x + 3 into 2y + 3x = ¡1 gives 2(2x + 3) + 3x = ¡1

4x + 6 + 3x = ¡1

7x = ¡7

x = ¡1d

EXERCISE 1A

y

x

3 slope��

y

x

3

2

y

x

4( )��,

y

x

( )��,

� � � ��x y

y

x

� �� x y

( )��,

y

x �

�

( )��,

Example 1

( )��,

( )��,

( ) ��,

y

x

3

y x��

y x�� y x When , ,

so, the two lines meet

at the point , .

x y� � � �

�

= 1 = 1

( 1 1)

¡

¡


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2 Sketch the graphs of each of the following lines, and find their equations:

a containing the point (1, ¡1) with slope of 2

b through the points (¡1, 3) and (1, ¡2)

c parallel to the line 3x + 4y = 4 with y-intercept ¡1

3 a Decide which of the following points lie on the straight line y = 2x¡ 1:

i (1, 1) ii (¡3, ¡5) iii (¡2, ¡5) iv (2, 5) v (3, 5)

b For each of the points (x, y) in part a that lie on the straight line y = 2x¡ 1,

calculatey ¡ 7

x¡ 4.

4 a Use the same axes to draw accurate graphs of y = 3 ¡ 2x and 3y + 2x = 4.

b Use your graphs to estimate the point where the two lines in a meet.

c Use algebra to find the exact point where the two graphs in a meet.

The graph of a quadratic function is called a

parabola. The point where the graph ‘turns’ is

called the vertex.

If the graph opens upward, the y-coordinate of

the vertex is the minimum. If the graph opens

downward, the y-coordinate of the vertex is the

maximum.

The vertical line that passes through the vertex is

called the axis of symmetry. All parabolas are

symmetrical about the axis of symmetry.

The value of y where the graph crosses the y-axis

is the y-intercept.

The values of x, (if they exist) where the graph crosses the x-axis should be called the

x-intercepts, but more commonly are called the zeros of the function.

For the quadratic y = ax2 + bx + c, a 6= 0:

QUADRATIC FUNCTIONS

x

y

vertex

axis

of

sym

met

ry

zero zero

y-intercept

parabola

minimum

² The coefficient of x2 (which is a) controls the degree of width of the

graph and whether it opens upwards or downwards.

I a > 0 whereas a < 0 produces

I If ¡1 < a < 1, a 6= 0 the graph is wider than y = x2:

If a < ¡1 or a > 1 the graph is narrower than y = x2:

² In the form y = a(x¡ ®)(x¡ ¯) the graph cuts the x-axis at ® and ¯.

² In the form y = a(x¡ ®)2 the graph touches the x-axis at ®:

² In the form y = a(x¡ h)2 + k the graph has vertex (h, k) and axis

of symmetry x = h.


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_01\012SA12STU-2_01.CDR Thursday, 2 November 2006 1:40:57 PM PETERDELL


Following is a list of expansion rules you should use:

² (a + b)(c + d) = ac + ad + bc + bd fsometimes called the FOIL ruleg² (a + b)(a¡ b) = a2 ¡ b2 fdifference of two squaresg² (a + b)2 = a2 + 2ab + b2

(a¡ b)2 = a2 ¡ 2ab + b2

)fperfect squaresg

5 Expand and simplify:

a ¡2x(5 ¡ x) b (2x + 1)(x + 4) c (3x¡ 4)(2x¡ 1)

d (x +p

7)(x¡p7) e ¡(x + 2)(x + 1) f ¡3(2x + 1)(1¡ 3x)

g (x + 6)(x¡ 6) h (x + 3)2 i (2x¡ 1)2

j (2x¡ 1)(2x + 1) k (4x + 5)2 l (x +p

3)(x¡p3)

m 1 ¡ 3x + (x + 2)2 n (2x + 1)2 + (x¡ 2)2 o 1 ¡ (x¡ 2)2

p (x + 2)(x2 + 3x¡ 4) q (x + 3)3 r (2x + 1)(x¡ 3)(x + 4)

A quadratic equation, with variable x, is an equation of the form

ax2 + bx + c = 0 where a 6= 0.

PRODUCT EXPANSION

Expression

Take out any

common factors

Recognise type

Difference of

two squares

a2 ¡ b2 = (a + b)(a¡ b)

Perfect squares

a2 + 2ab + b2 = (a + b)2

a2 ¡ 2ab + b2 = (a¡ b)2

Sum and product type

x2 + bx + c

x2 + bx+ c = (x + p)(x + q)where p + q = b and pq = c

Flow chart for factorising:

FACTORISATION OF QUADRATICS

QUADRATIC EQUATIONS

Sum and product type

ax2 + bx+ c, a 6= 0

² find ac

² find the factors of acwhich add to b

²

² complete the factorisation

if these factors are and ,replace by

p qbx px qx� � �+


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The roots (or solutions) of ax2 + bx + c = 0 are the values of x which satisfy the

equation (i.e., make it true).

To solve quadratic equations we can:

² factorise the quadratic and use the Null Factor law: “if ab = 0 then a = 0 or b = 0”

² complete the square

² use the quadratic formula

² use technology.

If ax2 + bx + c = 0, then x =¡b§

pb2 ¡ 4ac

2a.

¢

The quadratic formula becomes x =¡b§p

¢

2aif ¢ = b2 ¡ 4ac:

Notice that:

² if ¢ = 0, x =¡b

2ais the only solution (a double root)

² if ¢ > 0, i.e., ¢ is positive,p

¢ is a real number and so there are two distinct real

roots,¡b +

p¢

2aand

¡b¡p¢

2a.

For two real roots, ¢ > 0.

² if ¢ < 0, i.e., ¢ is negative,p

¢ involves the imaginary number i and so we have

no real roots.

Note: ² If a, b and c are rational and ¢ is a perfect square then the equation has two

rational roots which can be found by factorisation.

² Be aware that it is the sign of the discriminant which is significant.

Definition:pa is the non-negative number such that

pa£p

a = a:

Properties: ² pa is never negative, that is

pa > 0.

² pa is meaningful for real numbers only for a > 0.

² pab =

pa£p

b for a > 0 and b > 0.

²r

a

b=

papb

for a > 0 and b > 0.

THE QUADRATIC FORMULA

THE DISCRIMINANT

In the quadratic formula, which is under the square root sign is called the

. The symbol , is used to represent the discriminant.

b ac2�¡�4

¢discriminant delta

SURDS


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6 Factorise into linear factors:

a 5x2 + 20x b 7x¡ 2x2 c 2x2 ¡ 8

d x2 ¡ 7 e 4x2 ¡ 8 f x2 ¡ 2x + 1

g 2x2 + 4x + 2 h 3x2 + 12x + 12 i 2x2 + 5x¡ 12

j 3x2 ¡ 5x¡ 2 k 7x2 ¡ 9x + 2 l 6x2 ¡ x¡ 2

m 4x2 ¡ 4x¡ 3 n 10x2 ¡ x¡ 3 o 12x2 ¡ 16x¡ 3

7 Factorise into linear factors:

a (x + 5)2 + (x¡ 1)(x + 5) b 2(x + 1)(x¡ 3) ¡ (x + 1)2

c 3(x¡ 2)2 + 2(x¡ 2)(x + 4) d 5(x¡ 1)(x + 5) ¡ (x + 5)2

e (x + 3)2 ¡ 4 f 1 ¡ (2x + 1)2

g (2x + 3)2 ¡ (x¡ 4)2 h (x + 7)2 ¡ (1 ¡ 3x)2

8 Solve using factorisation:

a x2 = 5x + 6 b x2 = 2x c 5x¡ 10x2 = 0

d 2x2 ¡ 4 = 0 e x2 + 3x = 4 f 3x2 + 5x = 2

g 3x2 = 4x + 4 h 3x2 = 10x + 8 i 12x2 ¡ 11x = 15

9 Find the points of intersection of the following functions and check using a calculator:

a y = (x+1)2 and y = ¡x2 +x+4 b y = 6(x¡ 1

x) and y = 5

c y = 6x¡ 1 and y =10

x¡ 2

10 Use the quadratic formula to solve:

a x2 ¡ 4x¡ 3 = 0 b x2 ¡ 2x¡ 2 = 0 c x2 ¡ 2x + 2 = 0

d (2x¡ 1)2 = 5 ¡ 3x e x +1

x= 1 f 2x¡ 1

x= 2

QUADRATIC GRAPHS

Quadratic form, a 6= 0 Graph Facts

² y = a(x¡ ®)(x¡ ¯)®, ¯ are real

x-intercepts are ® and ¯

axis of symmetry is

x = ®+¯2

² y = a(x¡ ®)2

® is real

touches x-axis at ®vertex is (®, 0)

axis of symmetry is x = ®

� �

x =� �+

2

x = �

xV ( , 0)�


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11 For the quadratic y = 2x2 ¡ 4x + 1, find:

a the y-intercept b the equation of the axis of symmetry

c the coordinates of the vertex d the x-intercepts.

Hence, sketch the graph of the function.

Find the quadratic function

f(x) which has graph:

Since the x-intercepts are ¡3 and 1, f(x) = a(x + 3)(x¡ 1) where a 6= 0.

But f(0) = 6, ) a(3)(¡1) = 6) ¡3a = 6

) a = ¡2

) f(x) = ¡2(x + 3)(x¡ 1)

Quadratic form, a 6= 0 Graph Facts

² y = a(x¡ h)2 + k vertex is (h, k)

axis of symmetry is x = h

² y = ax2 + bx + c(general quadratic

form)

axis of symmetry is

x =¡b

2a

x-intercepts for ¢ > 0 are

¡b§p¢

2a

where ¢ = b2 ¡ 4ac

x h=

V ( , )h k

bp¢

x

x = b2a

p¢ �b

2a2a

QUADRATIC OPTIMISATION

For y = ax2 + bx + c:

² if a > 0, the minimum value of y occurs at x =¡b

2a

² if a < 0, the maximum value of y occurs at x =¡b

2a:

abx

2�

�

min. y

max. y

Example 2y

x

6

� �


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12

a b c

d e f

13

14 A doorway is to be parabolic in shape and 2 m

high. At a height of 1 m above ground level the

width of the opening is 1:6 m. How wide is the

door at floor level?

15 Find x if:

a (x¡ 3)(x2 + 2x¡ 2) = 0 b (x¡ 2)(x2 ¡ 4x¡ 6) = 0

y

x � �

�x

y

�

�

x

y

��

� �

x

y

�

��

x

y

�� x

y vertex ��

�

y

x

18

�

2 m

1 m

1.6 m

floor level

Find the intersection of the graphs of y = x2 + 2x + 2 and y ¡ x = 4

From y ¡ x = 4, we get y = 4 + x.

The graphs intersect if x2 + 2x + 2 = 4 + x

) x2 + x¡ 2 = 0

) (x + 2)(x¡ 1) = 0

) x = ¡2 or 1Using y = 4 + x, if x = ¡2 then y = 4 + (¡2) = 2

So, the graphs intersect at (¡2, 2) and (1, 5).

Example 3

Find the quadratic function which has graph:f x( )

Find given that the graph touches the

-axis at and ( , ) lies on the graph.

®x ® 1 8

and if thenx y := 1 = 4 + 1 = 5


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16 Find the intersection of the graphs of the following:

a y = x2 + 2x + 1 and y = 2x + 2

b y = x2 + 3x + 2 and y = x + 2

c y = 3x2 + 4x + 7 and y = x2 ¡ 4x + 1

17 Solve the following simultaneous equations:

a xy ¡ 3 = 3y + 1 and y ¡ x = 1

b xy = 16 and x + y = 5

c x2 + y2 = 16 and 2y + x = 3

The zeros of any polynomial are the values of x which make y have a value of zero.

These are clearly the x-intercepts of the graph of the polynomial.

Real zeros are x-intercepts, so a cubic can have:

² 3 real zeros, for example,

² 1 real and 2 imaginary zeros, for example,

18 Use technology to find the real zeros of:

a 2x3 ¡ 9x2 + 7x + 6 = 0 b x3 + 4x2 + 8x + 8 = 0

c 3x3 ¡ 7x2 ¡ 28x + 10 = 0 d 2x4 ¡ 2x3 ¡ 9x2 + 7x + 2 = 0

e x4 ¡ x2 = 2 f 4x4 ¡ 11x3 + 10x2 ¡ 11x + 6 = 0

Try to check yoursolutions using

technology.

CUBIC POLYNOMIALS

A cubic polynomial has form y = ax3 + bx2 + cx + d where a 6= 0and a, b, c and d are constants.

² If a > 0 the graph’s shape is or . If a < 0 it is or .

² For a cubic in the form y = a(x¡ ®)(x¡ ¯)(x¡ °) the graph has x-intercepts

®, ¯ and ° and the graph crosses over or cuts the x-axis at these points.

² For a cubic in the form y = a(x¡ ®)2(x¡ ¯) the graph touches the x-axis

at ® and cuts it at ¯.

² For a cubic in the form y = a(x2 + ¯x + °)(x¡ ®) where the discriminant of

the quadratic factor is < 0, the graph cuts the x-axis once only at ®.

x

THE ZEROS OF A POLYNOMIAL

xxor

1 repeatedzero


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If n is not a positive integer then an still has meaning, but we can no longer interpret it as a

product of n factors.

² am £ an = am+n ² am

an= am¡n ²

µa

b

¶n

=an

bn(b 6= 0)

² (am )n = amn ² (ab)n = anbn ² a0 = 1 for all a 6= 0

² a¡n and an are reciprocals, i.e., a¡n =1

anfor all a 6= 0:

² a1

2 =pa, a

1

3 = 3pa, a

1

n = npa, a

mn = n

pam

19 Write in the form xn :

apx b

1

xc 1 d e

px

x2f

20 Write as a sum or difference of terms:

ax + 1

x2b

x¡ 2

x3c

2x2 + x + 1px

dx2 ¡ 3x + 10

xpx

21 Solve for x:

a b c

The simplest exponential functions have form y = abx. They are all asymptotic to the

x-axis. The y-intercept is a:

To solve exponential equations we use a common base. If ax = an then x = n.

22 Solve for x: a 4x = 8 b 9x = 31¡x c 32x+1 = 13 d 52¡x = 1

In general, an = a£ a£ a£ a£ ::::£ a| {z }n factors

where n is a positive integer.

INDEX NOTATION

i.e., 8 = 23

LAWS OF INDICES

EXPONENTIAL FUNCTIONS

EXPONENTIAL EQUATIONS

y y y y

x x

x x

10

0

1

0

10

0

1

0

��

�

�

�

��

�

�

�

b

a

b

a

b

a

b

aa

a

a a

3px2 x2 £ 3

px2

(x + 3)¡3 = 8 (x + 2)¡ 1

3 = 2 (x¡ 2)¡ 1

2 =px¡ 2

base number

index, power or exponent


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23 Use algebra to solve these equations:

a 3 ¡ 4x = ¡7 b 3 ¡ 2x = x + 1 cx + 4

x¡ 4= 2

d2

x¡ 3=

3

x + 3e

2

x+ 2 = 3 f

2

x + 3¡ 2 = 4

gp

2 ¡ x = 3 hpx + 2 = ¡p

2 ¡ x i4p

2x + 3= 2

j x¡ 6 =px k

6

x2 ¡ 3= 1 l

1 ¡ x2

1 + x2= 2

3

If two objects of mass m1 and m2 are a distance d apart, then by Newton’s law of

gravity, the force F of attraction between the two objects is given by F =Gm1m2

d2,

where the constant G is known as the gravitational constant.

Find a formula for the distance d in terms of the other quantities.

F =Gm1m2

d2) Fd2 = Gm1m2 fmultiplying both sides by d2g

) d2 =Gm1m2

Ffdividing both sides by Fg

) d = §r

Gm1m2

F

) d =

rGm1m2

F

Use algebra to solve the equation x¡ 2 =px

x¡ 2 =px

) (x¡ 2)2 = x

) x2 ¡ 4x + 4 = x

) x2 ¡ 5x + 4 = 0

) (x¡ 1)(x¡ 4) = 0

) x = 1 or 4

Check: When x = 1, LHS = 1 ¡ 2 = ¡1 and RHS =p

1 = 1

So, x = 1 is not a solution.

When x = 4, LHS = 4 ¡ 2 = 2 and RHS =p

4 = 2

So, x = 4 is a solution.

Thus, x = 4 is the only solution.

When we square both sidesof an equation we may

introduce invalid solutions.Consequently we must checkeach solution in the original

equation.

Example 4

Example 5

f g�d is a distance, so is positive


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24 Transpose the following formulae to the variable indicated:

a F = ma2 to a b A = ¼r2 to r, where r > 0

c V = 43¼r

3 to r d C = 4x + ¼x to x

e V = x2l to x, where x > 0 f I =V

R1+

V

R2to V

g x2 + y2 = r2 to y, where y 6 0 h y =pr2 ¡ (x¡ 2)2, to x, for x > 2

i a =b + x2

b¡ x2to x, where x 6 0 j y =

rx¡ a

x + ato x

k A = ¼r2 + 2¼rl to l l1

f=

1

r¡ 1

sto s

We use surprisingly few formulae to connect known and unknown quantities.

The most commonly used connectors are:

² the theorem of Pythagoras

² formulae for perimeter, area and volume

² the ratio of sides of similar triangles

²² right angled triangle trigonometric ratios

² the sine rule and cosine rule for non-right angled triangles.

In any right angled triangle with legs of length a and b and

hypotenuse of length c, a2 + b2 = c2.

Rectangles

P = 2a + 2b

A = ab

TrianglesA = 1

2bh

Parallelograms

A = bh

Trapezia

A =

µa + b

2

¶h

CONSTRUCTING FUNCTIONSUSING GEOMETRY

B

THE THEOREM OF PYTHAGORAS

PERIMETER, AREA AND VOLUME

a

bc

a

b h h

b b

b

h

b

h

a

In calculus applications later in this course, it is necessary to involve geometrical figures and

to find relationships between their sides and angles.

coordinate geometry formulae


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Circles

C = 2¼r

A = ¼r2

Rectangular prisms

V = abc

A = 2(ab + bc + ca)

Cylinders

V = ¼r2h

S = 2¼r(r + h)

Cones

V = 13¼r

2h

If two triangles are similar then one triangle is an enlargement of the other, and vice versa.

If two triangles are equiangular then they are similar and their corresponding sides

are in the same ratio.

² slope =¢y

¢x=

y2 ¡ y1x2 ¡ x1

² distance =p

(¢x)2 + (¢y)2

² midpoint

µx1 + x2

2,y1 + y2

2

¶² y = mx + c is the equation of a line with slope m

The gable of a roof is illustrated.

Find, in terms of x:

a the length of beam AB

b the area A(x) of the gable ABC:

a y2 + x2 = 82 fPythagorasg) y2 = 64 ¡ x2

) y =p

64 ¡ x2 fas y > 0g) AB = 2

p64 ¡ x2 m

b A(x) = 12AB £ CM =

p64 ¡ x2 £ x

) A(x) = xp

64 ¡ x2 m2

r

ab

c

h

r

h

r

SIMILAR TRIANGLES

and AB

PQ=

BC

PR=

CA

QR:

A

B C� �

�Q

P

R

�

�

�

sides opposite angle

��

For example,for

COORDINATE GEOMETRY FORMULAE

Example 6

A BM

8 m

C

x m

y m

A BM

8 m

C

x m


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1 Amy’s rectangular paddock is 200 m by x m.

Find in terms of x only:

a the area of the paddock, A

b the perimeter of the paddock, P

c

d the measure of angle DBC.

2 Find formulae for the perimeter, P (x), and the area, A(x), for the following figures:

a b c

d e f

3 A 24 cm length of wire is cut into two pieces.

One piece is shaped into a square and the other

is used to form a circle.

a Find an expression for the total area, A(x),of the square and the circle.

b What is the significance of the values of

A(0) and A(24)?

The illustrated window has a perimeter

of 4:2 metres. The area of the glass is

A(x) m2. Find the A(x) function in terms

of the variable x only.

The radius of the semi-circle is x m.

Let the height of the rectangular part be y m.

The perimeter is 4:2 m.

) 2x + 2y + 12 (2¼x) = 4:2

) 2x + 2y + ¼x = 4:2

) 2y = 4:2 ¡ 2x¡ ¼x

) y = 2:1 ¡ x¡ ¼2x

EXERCISE 1B

A B

CD

�°

200 m

x m

x cm

2 cmx

rectangle

24 cm

x cm

cut

x cm4 cm

x cm

6 cm

cmx

7 cm

cmx

5 cmcmx

8 cm

the length of the fence BD which would divide

the paddock into two triangular paddocks

Example 7

2 mx

x my m

2 mx


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Now A(x) = area of rectangle + area of semi-circle

= 2x£ y + 12 (¼x2)

= 2x(2:1 ¡ x¡ ¼2x) + ¼

2x2

= 4:2x¡ 2x2 ¡ ¼x2 + ¼2x

2

= 4:2x¡ 2x2 ¡ ¼2x

2

i.e., A(x) = 4:2x¡ ¡2 + ¼2

¢x2 m2

4 Find the area function A(x) of:

a b

given that the perimeter is 20 cm given that the perimeter is 40 cm

c d

given that the perimeter is 50 cm given that the perimeter is 400 m.

x cm x cm

x m

x cm

Let the width of the chip be y cm.

As area = length £ width,

3:2 = x£ y

)3:2

x= y

Now P (x) = 2x + 2y

) P (x) = 2x + 2

µ3:2

x

¶P (x) = 2x +

6:4

xcm

A rectangular computer chip is to have an area of cm . If one side of it is cm

long, find the perimeter .

3 2( )

: xP x

2

Example 8

y cm

x cm3.2 cm2


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5 Find the perimeter P (x) of:

a b

given that the area is 1000 m2 given that the area is 400 cm2

c d

given that the area is 200 cm2 given that the area is 100 cm2.

A backyard garden plan is illustrated.

A fence of perimeter 50 m is to border it.

If A(x) is the total area of the garden:

a find A(x) in terms of x only

b find the possible values that x may have.

a As the perimeter is 50 m,

4x + 2y + 2(3x) = 50

) 2y + 10x = 50

) 2y = 50 ¡ 10x

) y = 25 ¡ 5x

Now h2 + (2x)2 = (3x)2 fPythagorasg) h2 = 9x2 ¡ 4x2

) h2 = 5x2

) h =p

5x fas h > 0gSo A(x) = area of rectangle + area of triangle

= 4x£ y + 12 (4x£ h)

= 4xy + 12(4x£p

5x)

= 4x(25 ¡ 5x) + 2p

5x2

= 100x + (2p

5 ¡ 20)x2 m2

b As 3x and 4x are positive, then x > 0

As y is positive, then 25 ¡ 5x > 0 ) 25 > 5x

) 5x < 25

) x < 5 So, 0 < x < 5:

x cm

2 cmxx m

2 cmx

Example 9

y m

y m

3 mx

3 mx

4 mx

3x

3x

h4x

2x

2x


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6 A paddock has the shape illustrated.

The perimeter is 400 m.

a Express y in terms of x.

b Hence find the area function A(x).

c Explain why 0 6 x 6 25:

7 For the following water containers, find the volume of water function V (x):

a b c

y m

5 mx

6 mx

A conical water tank has dimensions as shown.

x m is the radius of the cone of water.

Find V (x), the volume of water in the cone.

Let h m be the height of the water.

From the similar triangles,

h

x=

6

2and so h = 3x

Now V (x) = 13¼x

2h ffor a coneg= 1

3¼x2(3x)

i.e., V (x) = ¼x3 m3

x m

4 m

6 m

x m x m

h m

2 m2 m

6 m

Example 10

40 cm

2 m

x cm50 cm

2 m

4 m

x m

x m

6 m

8 m

8

Find the area function A(x) for

the area of the shaded rectangle.

A 8

B ,( )x y�C

6

x

y


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REVIEW SET 1B

REVIEWC

REVIEW SET 1A


1 Find the perimeter function, P (x), and the area function,

A(x), for the given triangle:

2 Alongside is a garden plan. The area is to be

80 m2.

Find the perimeter function P (r) in terms of

r only.

3 A right-circular cone of base radius 2 m and height

3 m is filled with water to a depth of x m.

a Find a formula for the volume of water

contained, in terms of x.

b How deep is the water if its volume is

1 m3? (Vcone = 13¼r

2h)

4 The perimeter of the figure is 200 cm. Express y in

terms of x and hence find the area function, A(x),for the figure.

What restrictions must be placed on the values of x?

5 Runner A heads east from point X at 2 pm and runs at a constant speed of 7 kmph.

Runner B also leaves X, but leaves at 2:12 pm and heads south at a constant speed

of 9 kmph.

If t is the number of hours after 2 pm, find an expression for their distance apart,

D(t) km, at time t hours.

6 A sphere of diameter 40 cm has a cylinder

of base radius r cm inscribed within it. Find

an expression for the volume of the cylinder,

V (r), in terms of r only.

1 The cross-section of a gutter has dimensions as

shown, with AB + BC + CD = 80 cm.

a Show that BC = 80 ¡ 2p

400 + x2 cm.

b Find A(x), the area of the cross-section.

x m

x cm

10 cm

13 cmx

10 cmx

y cm

r m

A

B C

D

20 cm

x cm


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2 1000 m of fencing is used to make 4 identical

horse paddocks. A river forms the fourth

boundary of each paddock.

a Find y in terms of x.

b Find A(x), the total area enclosed.

3 a Find A(x), the area of triangle ABC.

b Find V (x), the volume of the triangular

prism.

4 A person wishes to travel from a point B in the

desert to X on the road AC. If it is possible to

travel at 3 kmph in the desert and at 8 kmph

along the road, find a formula for the total time

taken to travel from B to X to C in terms of

x, the distance from A to X.

5 Infinitely many cylinders may be inscribed within a

right-circular cone of radius 5 cm and height 10 cm.

If the cylinder has radius r cm and height h cm,

express the volume V of the cylinder as a function

of: a r b h.

6 A sphere of radius 8 cm is inscribed within a right-

circular cone of radius r cm and height h cm.

Show that the volume of the cone V is given by

V (h) =64¼h2

3(h ¡ 16)cm3:

(Hint: Look for similar triangles.)

my

mx

river

A B

C

2 cmx8 cm

x cm10 cm

A C

B

5 km

X

10 km

x km

Q 10

P ,( )x y�R

5

x

y7 A(x) is the area function for

the shaded rectangle ORPQ.

Find A(x):


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_01\028SA12STU-2_01.CDR Monday, 6 November 2006 12:57:51 PM DAVID3

Contents: A

B

C

D

E

F

G

Functions

Modelling from data

Constructing exact models

Basic theory of calculus

When the rate of change is notconstant

Definite integrals

Review

2Functions and

introductory calculus


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Throughout this book we are concerned about relationships between numbers.

Often the information we want to know cannot be easily measured. For example:

² To work out how much fertiliser to use on a round lawn we need to know the area.

What we can easily measure is the radius.

From this we can find the area from the relationship A = ¼r2.

² In the 17th century, navigators knew the relationship between latitude and the height

above the horizon of the sun. The number of shipwrecks on the Western Australian

coast show that they had no satisfactory way of calculating their longitude. The

relationship between time and longitude led to the development of accurate clocks.

² When submerged, submarines can only measure acceleration with an accelerometer.

This information is used to calculate their velocity, which in turn is used to calculate

their position.

² When alcohol is consumed, it is possible to measure the concentration in the blood.

This information is used to estimate the concentration of alcohol in other organs such

as the brain.

A function is a special relationship between two quantities.

A function f with domain D and range R is a rule which assigns to each element

x of D exactly one element y of R:

To define a function we should: ² specify the domain D

² specify the range R

² specify the rule.

It is, however, common to only specify the rule relating x and y, as y = f(x) where f(x)is usually some formula.

Unless otherwise stated, the domain is understood to be the set of all values of x for which

the formula f(x) is defined.

Because of this, the words “formula” and “function”, although technically different, are often

used to mean the same thing.

FUNCTIONSA

Find the domain of the function y = f(x) where f(x) =px2 ¡ 2x:

As the range consists of real numbers only, the formula y =px2 ¡ 2x is

defined providing x2 ¡ 2x > 0

) x(x¡ 2) > 0

From the sign diagram this occurs if x 6 0 or x > 2:

So the domain of y =px2 ¡ 2x is the set of numbers x 6 0 or x > 2:

Example 1

� �

sign diagram

� �

30 FUNCTIONS AND INTRODUCTORY CALCULUS (Chapter 2)


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In this book it will be assumed that the range will consist of real numbers only.

Functions are written in many different ways, but we use only two of them in this book.

The second form is useful for evaluations. Instead of having to write, “Find y if x = 3” we

simply write “Find f(3)”.

1 Find the domain of these functions:

a f(x) =1

x + 1b f(x) =

x + 3

x(x + 2)c f(x) =

x

x2 ¡ 1

d y =px2 ¡ 4 e y =

p(3 ¡ x)(2 + x) f f(x) =

1p9 ¡ x2

2 Given that f(x) =1

x+ x find:

a f(3) b f(¡2) c f(p

2)

d f(t) e f(1

u) f f(

1

x)

3 If g(t) = 2t + 5t2, find:

a g(¡1) b g(x) c g(t2 + 1) d g(p

1 + y ¡ 1)

4 If y = x2 + x¡ 6, find the values of x for which y 6 0.

5 Graph the function y =x3 ¡ 8

x¡ 2on your calculator or graphing

package.

a What is the domain of this function?

b Calculate y for x = 1, 1:5, 1:8, 1:9, 1:99

c What do you think happens to y if x gets close to 2?

6

a What is the domain of this function?

b Give algebraic evidence to support your answer to a.

EXERCISE 2A

Given that f(x) = x2 + 2x + 3, find: a f(¡1) b f(u) c f(px)

a f(¡1)

= (¡1)2 + 2(¡1) + 3

= 1 ¡ 2 + 3

= 2

b f(u) = u2 + 2u + 3 c f(px)

= (px)2 + 2(

px) + 3

= x + 2px + 3

Note: In c, every x in the original formula is replaced bypx:

Example 2

GRAPHING

PACKAGE

, 1:999

For example: y =px2 ¡ 2x or f(x) =

px2 ¡ 2x both describe the same function.

Graph the function f(x) = x log10(x2 ¡ 16) on your calculator or graphing package.

FUNCTIONS AND INTRODUCTORY CALCULUS (Chapter 2) 31


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A mathematical model is a mathematical idealisation of a real problem. A good model is the

simplest one that retains the essential features of the problem.

If a model is too simple it may not make accurate enough predictions. If a model is too

complicated, it may be useless. For example, trying to take into account so much detail that

it will take a year of calculations to predict tomorrow’s weather.

Note: In higher mathematics, instead of writing exponential functions in base 2, 3 or 10,

say, we use base e where

e + 2:7182818::::::: , and is called exponential e.

The corresponding logarithms are called natural logarithms.

We write these, for example, as loge 6 or ln 6

In Chapter 5 we examine exponential e and natural logarithms in greater detail.

In almost every field of study models are used to make predictions. To construct suitable

models often requires the specialised knowledge of an expert such as a chemist who may be

concerned with the rates of chemical reactions. However, once a model has been supplied by

an expert, mathematics can be used to explore how well the model fits the information.

The models discussed in this book are functions that describe how the change in one quantity

is related to the change in another. The following is a list of the models with a description

of some of the most important features. Many of these functions will be discussed in more

detail in later chapters.

² linear

² quadratic y = at2 + bt + c

² power y = atm where a and m are constants.

For m > 0, the function passes through the origin and if a > 0, the function

increases without bound.

If m is not an integer, the graph may not be defined for t < 0.

MODELLING FROM DATAB

y

t

y

t

m>1

y

t

0< <1� �m

y

t

m<0

y mt c mc y

� � � �= + where is the slope of the line

and is the -intercept.

t

y

a��0

t

y

a��0

vertex


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² rational y =f(t)

g(t)where f and g are polynomials.

Many different shapes are possible and will

be discussed after an introduction to calculus

in later chapters.

The graph shown is of y =1 + 2t

1 + tand shows

some common features of rational functions.

This has a vertical asymptote t = ¡1(where the function is not defined), and a

horizontal asymptote y = 2.

² exponential

y = akt, with k > 0, or more commonly, y = aebt, where the number e is as

described previously. These graphs do not pass through the origin. They have a

horizontal asymptote y = 0, and for b > 0 (and a > 0) they grow without bound

at an ever increasing rate. For b < 0 the graph decays to the horizontal asymptote

y = 0.

Populations with unlimited growth are often modelled by exponential functions.

² logarithmic y = a + b ln t where a and b are constants.

Note: ln t is shorthand for loge t,i.e., ‘the logarithm of t in base e’.

The function is not defined for t 6 0.

t = 0 is a vertical asymptote.

² surge y = Ate¡bt, with b > 0.

The graph passes through the origin.

If A > 0, the graph rises rapidly to a maximum

value before decaying to the asymptote y = 0.

After taking medicine orally, the level of drugs

in the blood stream is often modelled by surge

functions.

² logistic P (t) =C

1 + Ae¡btwith C > 0, b > 0.

The graph grows steeply before reaching a point

of inflection after which it continues to grow at a

decreasing rate towards an asymptote.

Populations with limits to growth, such as re-

stricted food supplies, are often modelled by

logistic functions.

y

x

t��

y��

y

t

b<0a

y

t

b>0a

y

t

y

t

y

t

y C��


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By knowing the basic shapes of curves, and using simple arguments, one can often eliminate

many functions as possible models.

Consider the data shown in this table:

x 0 1 2 3 4 5 6 7 8 9 10

y ¡4:95 ¡0:03 13:11 26:04 43:09 64:45 80:81 97:04 131:45 165:88 195:98

a Decide which of the following functions is the best model for this set of data.

A The line y = 20:0x ¡ 26:2 B The quadratic y = 1:42x2 + 5:83x ¡ 4:95

C The power functiony = 4:21x1:66

D The exponential function y = 10:9e0:317x.

b Use each of the models to estimate the value of y if i x = 5:7 ii x = 20

Comment on the results.

a The graph shows the scatterplot of x and y.

A Since the graph has a definite curve, it is not

a straight line.

C Since the graph does not pass through the

origin (0, 0) it is not a power function. Note

that it is also negative for some values of x and

this also shows it cannot be a power function.

D Since the graph crosses the x-axis it cannot be an exponential function.

y = 1:42x2 + 5:83x ¡ 4:95.

The the graph of this quadratic superimposed on

the scatterplot shows how accurately this function

fits the data. From the four choices, the quadratic

function is the best model for the given data.

b For A, x = 5:7, y = 87:8 x = 20, y = 374

For B, x = 5:7, y = 74:4 x = 20, y = 670

For C , x = 5:7, y = 75:6 x = 20, y = 608

For D , x = 5:7, y = 66.4 x = 20, y = 6180

For x = 5:7 all models give roughly similar estimates. The estimate from the

linear function is too high; it is larger than the value at x = 6. The value from

the exponential function is a little on the low side; it is only just larger than the

value at x = 5. Both the quadratic and the power function give reasonable

results. For x = 20, the four models give very different estimates. The number

20 lies well outside the range of the data, and using any model to make

estimations can be very unreliable.

To see the instructions on how to plot these graphs on a calculator, click

on the icon.

Example 3

TI

C

Having eliminated the functions , , and the

only function left is the quadratic function

A C D


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_02\034SA12STU-2_02.CDR Friday, 8 December 2006 2:05:07 PM DAVID3


1 The scatterplot shows data from an experiment.

a Describe the main features of this graph and use

these to suggest a likely model.

b Explain, with brief reasons, why each of the

following is not a suitable model for the data.

A Polynomial B Power C Exponential D Logarithmic E Surge.

2 The following table gives the cost of tarpaulins of various sizes:

3 m £ 3 m $90 4 m £ 5 m $190 5 m £ 6 m $2803 m £ 4 m $120 4 m £ 6 m $225 5 m £ 8 m $3653 m £ 5 m $145 4 m £ 8 m $300 6 m £ 8 m $440

It is suspected that the cost is related to the area.

a Obtain a scatterplot of the cost against the area.

b What do you suspect the model to be?

c

d Use the given data to estimate the cost of a 5 m £ 10 m tarpaulin.

3 The rate of a chemical reaction in a certain plant depends on the number of frost-free

days experienced by the plant over a year, which, in turn, depends on altitude. The

higher the altitude, the greater the chance of frost. The following table shows the rate

of the chemical reaction R, as a function of the number of frost-free days, n.

Frost-free days 60 75 90 105 130 145

Rate of reaction (R) 44:6 42:1 39:4 37:0 34:1 31:2

a Produce a scatterplot of the data of R against n.

b Superimpose the graphs of the following functions on the scatterplot of part a:

i the linear function R = 53:5 ¡ 0:154n

ii the exponential function R = 57:1e¡0:0041n

c Which, if any, of the two functions in b seems to describe the data better?

d Use the two models above to estimate the value of R for:

i n = 100 ii n = 365:

e Explain briefly why the exponential function may be the better of the two models.

f Complete: “The higher the altitude, the ...... the rate of reaction.”

An is an estimation made with a model at a point that lies the range

of the data set. Interpolations are usually reliable.

An is an estimation made with a model at a point that lies the

range of the data set. Extrapolations are often unreliable, particularly if the point lies well

outside the data range.

interpolation within

extrapolation outside

EXERCISE 2B

A computer package may be useful to determine models quickly.LAW DETERMINER

Briefly give a theoretical reason why your selection of the model type is to be

expected.


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INVESTIGATION 1 FITTING MODELS TO DATA


4 When Alvin threw a baseball in a ‘longest throw’ competition, the height of the ball

above ground level at one second intervals was recorded as follows:

Time (seconds) 1 2 3 4 5 6

Height (m) 6:6 9:9 11:5 10:7 8:1 3:5

a Draw a scatterplot for the data.

b Briefly explain why none of the following functions are suitable models for the

data: A logarithmic B logistic C exponential.

c The following two models were suggested for the data:

i the quadratic y = ¡t2 + 6:42t + 1:17

ii the surge function y = 10:1te¡0:345t.

Use both models to predict the height when t = 0, t = 4:5 and t = 8. In each case

give a possible interpretation of the result.

d From your knowledge of throwing balls, which of the two models in c is the more

likely?

Are there any restrictions you might have to put on the values of the time t for

which the model can be used?

Data obtained from most experiments is subject to random fluctuations and is unlikely to

lie precisely on the graph of a simple function. A common way to construct models from

experiments is to construct a function whose graph, in some sense, best fits the data. Although

in this book we do not use this method, the following investigation is a reminder of how

technology may be used to construct models from experimental data.

Note:

The tabled results are: Pressure (hPa) 1015 2000 3000 4000 5000 6000

Volume (mL) 1000 505 330 255 205 170

1 Draw a scatterplot of the data for values of the volume V corresponding to the values

of pressure P .

2 Note that the exponential V = aebP , with b < 0, could be a suitable model for this

data. The aim of fitting this model to the data is to find the so-called parameters aand b so that the function ‘best fits’ the data.

Use technology to calculate the exponential that ‘best fits’ the data.

In , Robert Boyle carried out experiments on

compressing gases. He measured the pressure exerted

and the volume of gas each time. Today we can

repeat his experimentation using a blocked off

syringe. Heavy objects, like your maths book, are placed on the

plunger to apply pressure. (Each book of about kg produces a

pressure of around hPa on each cm .)

1660

110 2

The starting pressure is not zero. It is the atmospheric

pressure of the day of the experiment, hPa

(hectopascals).

1015

What to do:

rubberstopper

volu

me

pressure


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3 Superimpose the graph of the model on the scatterplot of the data.

4 There are a number of criteria that are used to assess how well the model fits the data.

² Residuals

Residuals are the difference between the predicted values of the model and the

actual data. Residuals are automatically calculated on the calculator.

² Coefficient of determination R2

R2 can be displayed on the calculator. R2 lies between 0 and 1 and measures

how much variation in the data can be explained by the model.

If R2 = 0:932 for example, then the model explains 93:2% of the variation in

the data. The other 6:8% is due to other factors, possible random fluctuations

in the data.

In the special case where the model is derived from a linear model, R2 = r2,

where r is the correlation coefficient.

Make a residual plot of the data. Note that the scale on the vertical axis is important

when considering residual plots. Note the value of R2.

5 Repeat steps 2 to 5 for the power function V = axP .

6 From your results decide which of the two models is a better fit for the data.

Often sufficient information is given so that we can construct an exact model of a situation.

If required, we can then use the model to obtain data. A graph could also be drawn connecting

the variables and other information can be obtained, for example, the maximum value of the

dependent variable,

CONSTRUCTING EXACT MODELSC

A duck farmer wishes to build a rectangular enclosure of area 100 m2. The farmer

must purchase wire netting for three of the sides as the fourth side is an existing

fence of another duck yard. Naturally the farmer wishes to minimise the length

(and therefore the cost) of the fencing required to complete the job.

a If the shorter sides are of length x m, show that the required length to be

purchased is L = 2x +100

x:

b Use your graphics calculator and/or computer graphing package to help you

sketch the graph of y = 2x +100

x:

c Find the minimum value of L and the corresponding value of x when this

occurs using technology.

d Sketch the optimum situation.

Example 4

i.e., model of situation data graph


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a If the other side is y m long, then

xy = 100 ) y =100

x:

Thus, L = 2x + y

) L = 2x +100

x:

b

c The coordinates of A are (7:07, 28:28) fto 2 dec. pl.gSo, the minimum value of the length required is 28:28 m when x is 7:07 m.

d When x = 7:07, y =100

7:07+ 14:14 ) shape is

1 An arena is to have a ground area of 1000 m2 and is to be

rectangular in shape. The owners wish to minimise the cost

of the fence required to enclose the arena.

a If x is the length of one side, find the length of

the other side.

b If P is the perimeter of the arena, show that P = 2x +2000

x.

c Use technology to help to sketch the graph of y = 2x +2000

x.

d Use technology to find the minimum length of fencing required and the correspond-

ing value of x.

e Sketch the optimum situation.

2 Anne has 20 m of fencing to form three sides of a

rectangular chicken pen.

a If the pen is y m by x m, show that the area

of the pen is given by A = x(20 ¡ 2x) m2.

b Use technology to sketch a graph of Aagainst x.

c What value of x will maximise the area of the enclosure? Sketch the optimum

solution.

EXERCISE 2C

100 m2

existing fence

x

y

14.14 m

7.07 m

15105

40

20A

x

L

min L

area = 1000 m2

TI

C

GRAPHING

PACKAGE

y m

x m

existing brick wall



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3 An industrial shed is to have a total floor space

of 600 m2 and is to be divided into 3 rectangular

rooms of equal size. The walls, internal and exter-

nal, will cost $60 per metre to build. Suppose we

have:


b Show that the total cost of the wall is given by C = 60(6x +800

x) dollars.

c Use technology to help you sketch the graph of y = 60(6x +800

x).

d Use technology to find the minimum cost of the walls and the corresponding value

of x.

e Show the optimum dimensions on a sketch.

4

a What is the inner length of the box?

b Explain why x2h = 100.

c Explain why the inner surface area of the box is given by A(x) = 4x2+600

xcm2.

d Use technology to help sketch the graph of y = 4x2 +600

x:

e Use technology to find the minimum inner surface area of the box and the corre-

sponding value of x.

f Draw a sketch of the optimum box shape.

5 Consider the manufacture of 1 L capacity tin cans where

the cost of the metal used to manufacture them is to be

minimised. This means that the surface area is to be as

small as possible but still must hold a litre.

a Explain why the height h, is given by h =1000

¼r2cm.

b Show that the total surface area A, is given by A = 2¼r2 +2000

rcm2:

c Use technology to help you sketch the graph of A against r.

d Use technology to find the value of r which makes A as small as possible.

e Draw a sketch of the dimensions of the can of smallest surface area.

f A manufacturer wishes to make cylindrical bins with an open top and each bin is

to have a capacity of 50 litres. What base radius and height would be chosen to

minimise the cost of material?

y m

x m

h cm

x cm

h cm

r cm

Radioactive waste is to be disposed of in

fully enclosed lead boxes of inner

volume cm . The base of the box

has dimensions in the ratio .

2002 : 1

3



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So far we have mainly used the function y = f(x) to calculate y given the value of x.

In this section we explore how functions can be used to answer the two questions:

² Given the rate of change of a quantity, how can you find the quantity?

For example, given the speed (rate of change of distance) of a car, how can you find

the distance the car has travelled?

² Given a quantity, how can you find its rate of change?

For example, given the distance a car has travelled, how can you find its speed?

The following example is very simple. You have probably answered similar questions without

even thinking when travelling in a car. It does, however, illustrate all the important features

of the theory of calculus developed in this course, and it pays to study the example carefully.

At time t = 0 hours a car is 200 km from Adelaide and travels at a constant speed

of v(t) = 80 km/hour away from Adelaide.

a Draw the graph the speed v(t) of the car against time t for t > 0:

b Calculate the distance s(t) the car is from Adelaide at time t > 0, and relate

this to the graph drawn in part a.

c Sketch the graph of s(t).

d The graph s(t) is a straight line. Use any two points on this line to find the

slope of the line. Give an interpretation of the slope and the s intercept of

this line.

a Since the car travels at a constant speed

v(t) = 80 km/hour, the graph is a

horizontal line.

Example 5

BASIC THEORY OF CALCULUSD

20

40

60

80

100

1 2 3 4

t (time in h)

v (speed in km/h)

t

Speed against time

b The distance travelled in t hours is speed £ time

= 80 km/h £ t hours

= 80t km

We can interpret this as the area = base £ height of the rectangular region, providing

we use the units of hour for the base and km/hour for the height.

Since the car was 200 km from Adelaide at time t = 0, the distance s(t) of the car

from Adelaide at time t is s(t) = 200 + 80t km:



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c

d The points P(1, 280) and Q(3, 440) lie on the line.

Using these points, the slope of the line is¢s

¢t=

(440 ¡ 280) km

(3 ¡ 1) hours

= 80 km/hour

The slope of the line is the speed of the car in km/hour.

The s intercept of the line is s(0) = 200 km, which is the distance of

the car from Adelaide at time t = 0 hours.

In each of the following questions pay particular attention to the units.

1 A car is travelling towards Adelaide at a constant speed of v(t) = 110 km/hour.

At 1 pm the car left Tailem Bend, 99 km from Adelaide.

a Draw the graph of the speed v in km/hour for 0 6 t 6 0:75

b Calculate the distance s(t) of the car from Adelaide at time 0 6 t 6 0:75, and relate

this to the graph drawn in a.

c Sketch the graph of s(t).

d The graph s(t) is a straight line. Use any two points on this line to find the slope

of the line. Give an interpretation of the slope and the s intercept of this line.

2 After the initial 2000 people have entered a sports stadium, a steady stream of

p(t) = 1500 persons per hour come through the gates. If t is the number of hours after

the initial intake:

a draw a graph of the number of people p(t) entering the stadium per hour for

0 6 t 6 2:

b Calculate the number of people P (t) inside the stadium at time t hours, and relate


c Sketch the graph of P (t).

d The graph P (t) is a straight line. Use any two points on this line to find the slope

of the line. Give an interpretation of the slope and the P intercept of this line.

3 Before it began to rain, a tank contained 500 litres of water.

At time t minutes after the rain started, water was flowing into the tank at a constant

rate of f(t) = 12 litres/minute.

a Draw a graph of the flow f(t) of water into the tank for t > 0.

EXERCISE 2D

1 2 3 4 t (time in h)

s (distance in km)

100

200

300

400

500

600

�� s km

�� t h

P ,� ��( )

Q ,� ��( )

Distance against time



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b Calculate the volume V (t) of water in the tank at t minutes after the rain started,

and relate this to the graph drawn in a.

c Sketch the graph of V (t).

d The graph of V (t) is a straight line. Use any two points on this line to find the

slope of the line. Give an interpretation of the slope and the V intercept of this line.

4 Drugs are generally cleared from the body ex-

ponentially. If, however, there is an overdose,

all enzymes involved in clearing the drug from

the body are saturated and fully working, and the

drug is cleared at a constant rate.

Measurable quantities of alcohol are cleared from

the body at a constant rate which depends on

a number of factors. A healthy 65 kg male

can be expected to clear alcohol at the constant

rate r(t) = 6:5 grams/hour:

Suppose such a male has just drunk 4 fluid ounces of whisky that loaded his body with

30 grams of alcohol:

a Draw the graph of the clearance r(t) for t the number of hours after drinking the

whisky.

b Calculate the amount A(t) of alcohol in the blood of this man for t > 0, and relate


c Sketch the graph of A(t) for t > 0.

d The graph of A(t) is a straight line. Use any two points on this line to find the

slope of the line. Give an interpretation of the slope and, in this case, the t intercept

of this line.

5 Let the function f(x) = c, where c is a constant.

a Sketch the graph of f(x) for x > 0:

b Calculate the area F (x) between the graph of f(x) and the x axis in the interval

from 0 to x.

c Sketch the graph of F (x) for x > 0:

d The graph of F (x) is a straight line. Use any two points on this line to find the

slope of the line. Relate the slope of the line to the function f(x):

By selecting suitable units for x and y = f(x), question 5 covers all the previous questions.

In Section D, we used the fact that the rate of change such as speed was constant, to calculate

the distance for any value of time t.

In this section we begin to develop ways of estimating quantities like the distance travelled

even though the rate of change is not constant.

We shall use the features developed in Section D, in particular that the area of a rectangle is

base £ height: Because the calculations are more involved, we shall only calculate distances

travelled for fixed time intervals.

WHEN THE RATE OF CHANGEIS NOT CONSTANT

E



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Consider the function f(x) = x2 + 1.

Consider the region between f(x) = x2 + 1,

the x-axis, and the vertical lines x = 1 and

x = 4.

We wish to estimate the area A of this region.

Let us divide the x-interval into equal parts,

each of length 1 unit.

The diagram alongside shows the creation of

upper rectangles, which are rectangles with

top edges at the maximum value of the curve

on that interval.

Notice that the original shaded area A is less

than the sum of the upper rectangular areas,

which we will denote AU .

Now, AU = 1 £ f(2) + 1 £ f(3) + 1 £ f(4) = 5 + 10 + 17 = 32 units2

and so the original shaded area A < 32 units2:

This diagram shows the creation of lower rect-

angles, which are rectangles with top edges at

the minimum value of the curve on that interval.

IfAL is the sum of the lower rectangular areas,

then

AL = 1 £ f(1) + 1 £ f(2) + 1 £ f(3) = 2 + 5 + 10 = 17 units2

and so the original shaded area A > 17 units2:

As AL < A < AU , the required area lies between 17 units2 and 32 units2.

If the interval 1 6 x 6 4 was divided into 6 equal intervals, each of length 12 , then

AU = 12f(11

2 ) + 12f(2) + 1

2f(212 ) + 1

2f(3) + 12f(31

2 ) + 12f(4)

= 12 (134 + 5 + 29

4 + 10 + 534 + 17)

= 27:875 units2

and AL= 12f(1) + 1

2f(112) + 1

2f(2) + 12f(21

2) + 12f(3) + 1

2f(312)

= 12(2 + 13

4 + 5 + 294 + 10 + 53

4 )

= 20:375 units2

When we create more subdivisions of the x-values, we narrow the lower and upper boundaries.

As the subdivisions become very small we will get a very accurate estimate for the area A.

UPPER AND LOWER RECTANGLES

4321

20

15

10

5

y

x1 25

10

17

1)( 2�� xxf

4321

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y

x1 25

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1)( 2�� xxf

4321

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1)( 2�� xxf


From this refinement we conclude that the required area lies between units and

units .

2

220 375

27 875:

:


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_02\043SA12STU-2_02.cdr Friday, 3 November 2006 1:59:54 PM PETERDELL

A car travels at the speed of v(t) = 100 ¡ 50

5t2 + 1km/hour away from a city.

The time t is the number of hours after the car leaves the city.

a Sketch a graph of the speed against time.

b Estimate the distance the car has travelled after 4 hours.

a

The graph shows that the function

v(t) = 100 ¡ 50

5t2 + 1is increasing.

b We start by dividing the t axis into four time periods.

Consider the interval 0 6 t 6 1.

The speed v(t) > 50 km/hour and so for that one hour the car will travel a

distance of more than 50km

hour£ 1 hour = 50 km.

Example 6

4321

100

50

v

t

4321

100

50

v

t

4321

100

50

v

t

4321

100

50

v

t

91.7 97.6 98.9 99.4 91.7 97.6 98.9 99.4

4321

100

50

v

t

This distance is represented by the shaded region below the graph in the

interval The following graphs show lower and upper rectangles.

The total of the sums of the areas of these give us under and over estimates

of the total distance travelled.

0 16 6t :

Notice that AU = 1 £ v(1) + 1 £ v(2) + 1 £ v(3) + 1 £ v(4)= 1 £ 91:7 + 1 £ 97:6 + 1 £ 98:9 + 1 £ 99:4+ 387:6

AL = 1 £ v(0) + 1 £ v(1) + 1 £ v(2) + 1 £ v(3)= 1 £ 50 + 1 £ 91:7 + 1 £ 97:6 + 1 £ 98:9 + 338:2

From which we conclude that: 338:2 km < total distance < 387:6 km

Notice that if we subdivide the time interval into 8 equal parts the rectangles are:

Now, the upper sum, AU + 379:5 and the lower sum, AL + 354:8

From which we conclude that: 354:8 km < total distance < 379:5 km



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From examples like the one above, we conclude that:

we can get better estimates of distance by dividing the time axis into smaller time intervals.

1

a Sketch a graph of the speed of the car for 0 6 t 6 1:

b What is the maximum, and what is the minimum speed of the car for 0 6 t 6 1?

c Show that the distance d of the car from Adelaide, 1 hour after starting its trip from

Port Wakefield, is less than 27 km from Adelaide.

d By dividing the time of travel into 2 half hour intervals, estimate the distance of the

car from Adelaide after 1 hour of travel.

e Improve the estimate you made in d by considering 4 time intervals of a quarter of

an hour each.

2 The marginal profit is the profit of each article that is sold. The marginal profit usually

increases as the number of articles sold increases.

Suppose that the marginal profit of selling the nth house is P(n) = 200n ¡ n2 dollars.

a Sketch a graph of the marginal profit for 0 6 n 6 100:

b Show that the total profit for selling 100 houses is less than $1 000 000:

c By considering the maximum and minimum marginal profit of the first 50 and the

second 50 houses sold, estimate the profit made for selling 100 houses.

d Improve the estimate you have made in d by considering four equal intervals.

e Suggest a way of finding the exact profit made by selling 100 houses.

3 The rate at which drugs are eliminated from the body is called the clearance rate. The

clearance rate depends on many individual factors as well as the amount of drugs present.

a Sketch the graph r(t) for 0 6 t 6 4:

b Show that 4 hours after the intake of 110 mg of caffeine, the amount Q(t) of

caffeine left in the body is between 10 and 70 mg:

c By considering the maximum and minimum clearance for each hour, estimate the

amount of caffeine left in the body 4 hours after an intake of 110 mg:

d Suggest a way of improving the accuracy of your estimate in c.

4 a Sketch the graph of the function f(x) =1

xfor 2 6 x 6 6:

b Find the maximum and minimum value of f(x) =1

xfor 2 6 x 6 6.

Use these values to show that the area between the graph of f and the x axis for

c Divide the interval 2 6 x 6 6 into 4 smaller intervals of equal length. By consider-

ing the smallest and the largest values of f on each of these smaller intervals, find

an estimate for the area between the graph of f and the x axis for 2 6 x 6 6:

d Suggest a way of improving the accuracy of your estimate in c.

EXERCISE 2E.1

At time t = 0 hours a car starts from Port Wakefield, a distance of 95 km from Adelaide,

and travels at a speed of v(t) = 50 + 50e¡t km/hour towards Adelaide.

Suppose that in a healthy adult the clearance rate r(t) of 110 mg of caffeine (about one

cup of coffee) is given by r(t) = 25e¡0:23t mg/hour, where t is the number of hours

after which the caffeine is taken.

2 6 x 6 6 lies between 23 and 2.



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y y

x x

(1, 1) (1, 1)

1 1

1 1y x� X y x� X

So, if A is the actual area then 0:219 < A < 0:469 .

a lower bound an upper bound

By subdividing the horizontal axis into small enough intervals we can, in theory, find estimates

of areas under curves which are as close as we want them to be to the actual value.

We illustrate this process by estimating the area between the graph of y = x2 and the x-axis

for 0 6 x 6 1.

This example is of historical interest. Archimedes (287 - 212 BC) found the exact area. In an

article that contains 24 propositions he developed the essential theory of what is now known

as integral calculus.

As you will see in a later chapter, calculating the exact area between the graph of y = x2

and the x-axis can now be done as a simple exercise in mental arithmetic.

Consider y = x2 and divide the interval 0 6 x 6 1 into 4 equal subintervals.

If AL represents the lower area sum and AU represents the upper area sum, then

AL = 14(0)2 + 1

4( 14)2 + 14(12 )2 + 1

4(34 )2 and AU = 14(14 )2 + 1

4 (12 )2 + 14 (34)2 + 1

4 (1)2

from which AL + 0:219 and AU + 0:469:

Here we used the fact that the function y = x2 is increasing. The minimum y-value of each

interval is at the left most point of the interval.

The following instructions show you how a calculator can be used to calculate these sums.

USING TECHNOLOGY


Step 1: Type the function Y1 = X2 in the function menu of the calculator.

Step 2: In List L1 enter the areas, using base £ height, of all the lower rectangles as

L1 = seq(1=4*Y1(X), X, 0, 1 ¡ 1=4, 1=4)

In this instruction, 1=4*Y 1(X) calculates the area of each rectangle using

the common base of 1=4 and height Y 1 defined in Step 1, for X values04 ;

14 ;

24 ;

34 at the left hand points of each of the 4 intervals.

Step 3: The instruction sum(L1) calculates the lower sum of the areas.

Step 4: To find the upper sum change the instruction in Step 2 to

L2 = seq(1=4*Y1(X), X, 1=4, 1, 1=4)

This calculates the area 1=4*Y1(X) = 14 £X2 of each rectangle at the right

hand points 14 ;

24 ;

34 ;

44 . The upper sum of the areas is found by sum(L2)

It was not until the time of and , some years later, that any further

progress was made.

Newton Leibniz 1900


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_02\046SA12STU-2_02.CDR Monday, 6 November 2006 9:22:54 AM PETERDELL

The number n of intervals can be easily adjusted in the above procedure. The following

diagrams show lower and upper rectangles for n subdivisions where n = 10, 25 and 50.

A summary of these results to 5 decimal places is given in the table below as well as the

average value of AL and AU :

n AL AU Average

4 0:218 75 0:468 75 0:343 75

10 0:285 00 0:385 00 0:335 00

25 0:313 60 0:353 60 0:333 60

50 0:323 40 0:343 40 0:333 40

The number of items you can store in lists on your calculator is probably

limited to 1000. Click on the icon to open an area finder on the computer

that can calculate upper and lower sums for larger numbers.

1 Use rectangles to find lower and upper sums for the area between the graph of the

function y = x2 and the x-axis for 0 6 x 6 1, using n = 10, 25, 50, 100, 500.

Give your answer to 5 decimal places.

As n gets larger, both AL and AU approach (or converge) to the same number, which is

a simple fraction. Can you recognise this fraction?

2 Use rectangles to find lower and upper sums for the area between the graphs of each of

the following functions and the x-axis for 0 6 x 6 1.

Use values of n = 5, 10, 50, 100, 500, 1000 and 10 000.

EXERCISE 2E.2

y

x

(1, 1)

1

n = 25

AL = 0.31 360�

y

x

(1, 1)

1

n = 25

AU = 0.35 360�

y

x

(1, 1)

1

n = 50

AL = 0.32 340�

y

x

(1, 1)

1

n = 50

AU = 0.34 340�

y

x

(1, 1)

1

n = 10AL = 0.28 500�

y

x

(1, 1)

1

n = 10AU = 0.38 500�

AREA

FINDER



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Give your answer to 5 decimal places in each case.

a i y = x3 ii y = x iii y = x1

2 iv y = x1

3

b For each of these, as n gets larger AL and AU converge to the same number, which

is a simple fraction. Can you recognise this fraction?

c On the basis of your answer to b, conjecture what the area between the graph of

y = xa and the x-axis for 0 6 x 6 1 might be for any number a > 0.

3 Consider the quarter circle of centre (0, 0)

and radius 2 units as illustrated.

Its area is 14 (full circle of radius 2)

= 14 £ ¼ £ 22

= ¼

a By calculating the areas of lower and upper rectangles for n = 10, 50, 100, 200,

1000, 10 000, find rational bounds for ¼.

b Archimedes found the famous approximation 31071 < ¼ < 31

7 .

For what value of n is your estimate for ¼ better than that of Archimedes?

We will now have a closer look at lower and upper rectangle sums for a function which is

above the x-axis on the interval [a, b], and is increasing.

Notice that the lower sum is

AL = f(x0)¢x + f(x1)¢x + f(x2)¢x + :::::: + f(xn¡2)¢x + f(xn¡1)¢x

=n¡1Pi=0

f(xi)¢x where ¢x =b¡ a

n:

AU = f(x1)¢x + f(x2)¢x + f(x3)¢x + :::::: + f(xn¡1)¢x + f(xn)¢x

=nP

i=1f(xi)¢x where ¢x =

b¡ a

n:

In these formulae, ¢ is the capital Greek letter delta. In this notation ¢x = xi ¡ xi¡1 is

the difference in the x-values of each small interval. Since all intervals have the same width,

¢x is the same for each interval.

DEFINITE INTEGRALSF

a bx1 x2 x3

x0

xn 2

xn 1 xn

y

x

y x= ƒ( )

....

a bx1 x2 x3

x0

xn 2

xn 1 xn

y

x

y x= ƒ( )

....

Likewise the isupper sum

2

2

x

y

24 xy ��



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From the work in the previous section you should have discovered the following:

² As n gets larger, as b¡ a is fixed, ¢x gets smaller and closer to 0.

² There exists a unique number A, say, such that for any value of nAL < A < AU and both AL and AU approach A as n gets very large.

² If f(x) > 0 on a 6 x 6 b, then A is the area between y = f(x), the x-axis and

the vertical lines x = a and x = b.

Notation:

We talk about n getting very large and write n ! 1.

n ! 1 could be read as n approaches infinity or n tends to infinity.

Using this notation, as n ! 1, AL ! A and AU ! A:

We define the unique number between all lower and upper sums as

Z b

a

f(x)dx and call

it “the definite integral of f(x) from a to b”,

i.e.,n¡1Pi=0

f(xi)¢x <

Z b

a

f(x)dx <nP

i=1f(xi)¢x where ¢x =

b¡ a

n:

More simply, AL <

Z b

a

f(x)dx < AU

and as n ! 1, AL !Z b

a

f(x) dx and AU !Z b

a

f(x) dx

The word integration means “to put together into a whole.”

An integral is the “whole” produced from integration. Here the word is used in the sense

that all areas f(xi)£¢xi of the thin rectangular strips are put together into one whole area.

The symbol

Zis called an integral sign.

stretched out letter s, but it is no longer part of the alphabet.

The notation

Z b

a

f(x) dx conveys the idea that we are summing the areas of all rectangles

in the interval x = a to x = b of height f (x) with widths ¢x shrunk to “infinitesimal”

size dx.

The ideas of “infinitesimals” are useful in loosely describing basic ideas in calculus, but are

not formally used in this book.

For us, the symbol

Z b

a

f(x) dx is a single number that lies between all upper sums and all

lower sums.

THE DEFINITE INTEGRAL


In the time of Newton and Leibniz it was the


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We can calculate this number as accurately as we need by making the number of subdivisions

n large enough. In a later chapter you will discover the remarkable fact that for many functions

the integral can easily be found exactly.

a Sketch the graph of y = x4 for 0 6 x 6 1.

Divide the interval 0 6 x 6 1 into 5 equal parts, and display the 5 upper and

lower rectangles.

b Use technology to calculate the lower and upper rectangle sums for n equal

subdivisions where n = 5, 10, 50, 100, 500. Give your answer to 4 dec. places.

c Use the information in b to find

Z 1

0

x4 dx to 2 significant figures.

a

b The upper and lower rectangular sums are displayed in the following table:

N AL AU

5 0:1133 0:3133

10 0:1533 0:2533

50 0:1901 0:2101

100 0:1950 0:2050

500 0:1990 0:2010

c When n = 500, AL + AU + 0:20, to 2 significant figures.

Since

Z 1

0

x4 dx is the number that lies between all upper and all lower sums,Z 1

0

x4 dx = 0:20, to 2 significant figures.

10.80.60.40.2

1

0.8

0.6

0.4

0.2

x

y

y x��

5 lower rectangles

10.80.60.40.2

1

0.8

0.6

0.4

0.2

x

y

y x��

5 upper rectangles

Example 7



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1 a Sketch the graph of y =px for 0 6 x 6 1.

Divide the interval into 5 equal parts and display the 5 upper and lower rectangles.

b Find the lower and upper rectangle sums for n = 5, 50, 100, 500.

c Use the information in b to find

Z 1

0

px dx to 2 significant figures.

2 a Sketch the graph of y =p

1 + x3 and the x-axis for 0 6 x 6 2.

b Find the lower and upper rectangle sums for n = 50, 100, 500.

c What is your best estimate for

Z 2

0

p1 + x3 dx?

3 a Sketch the region between the curve y =4

1 + x2and the x-axis on 0 6 x 6 1.

Divide the interval into 5 equal parts and display the 5 upper and lower rectangles.

b Find the lower and upper rectangle sums for n = 5, 50, 100 and 500.

c Give your best estimate for

Z 1

0

4

1 + x2dx and compare this answer with ¼.

Use graphical evidence and

known area facts to find:a

Z 2

0

(2x + 1) dx b

Z 1

0

p1 ¡ x2 dx

aR 2

0(2x + 1)dx

= shaded area

=¡1+52

¢£ 2

= 6

b As y =p

1 ¡ x2, then y2 = 1 ¡ x2 i.e., x2 + y2 = 1 which is the

equation of the unit circle. y =p

1 ¡ x2 is the upper half.R 1

0

p1 ¡ x2 dx

= shaded area

= 14(¼r2) where r = 1

= ¼4

EXERCISE 2F.1

If f(x) > 0 for all x on a 6 x 6 b thenZ b

a

f(x) dx is the shaded area.

x

y

a b

y x= ƒ( )

Example 8

x

y y x2 1� �

2

5

3

1

(2, 5)

1 1

1

y

y = ~`1``-̀!`� 2

x



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4 Use graphical evidence and known area facts to find:

a

Z 3

1

(1 + 4x)dx b

Z 2

¡1

(2 ¡ x) dx c

Z 2

¡2

p4 ¡ x2 dx

So far our arguments have been restricted to f(x) > 0.

But, what is

Z b

a

f(x)dx for f(x) 6 0 on the interval a 6 x 6 b?

Sincen¡1Pi=0

f(xi)¢x <

Z b

a

f(x) dx <nP

i=1

f(xi)¢x,

Z b

a

f(x) dx must be negative as clearly ¢x =b¡ a

nis always positive and f(xi)

values are always negative.

5 Find upper and lower bounds for

Z 1

0

(¡x2)dx using upper and lower sums when

n = 5 and 10:

6 Find upper and lower bounds for

Z 1

0

(x2 ¡ x)dx using upper and lower sums when

n = 200:

THE DEFINITE INTEGRAL WHEN f x( ) 06

Find upper and lower bounds for

Z 1

0

(x2 ¡ 1)dx using upper and lower sums

when n = 5.

a = x0 = 0 f(0) = ¡1

x1 = 0:2 f(0:2) = ¡0:96

x2 = 0:4 f(0:4) = ¡0:84

x3 = 0:6 f(0:6) = ¡0:64

x4 = 0:8 f(0:8) = ¡0:36

x5 = 1:0 f(1) = 0

and ¢x =1 ¡ 0

5= 1

5

4Pi=0

f(xi)¢x = 15 [f(0) + f(0:2) + f(0:4) + f(0:6) + f(0:8)] = ¡0:76

5Pi=1

f(xi)¢x = 15 [f(0:2) + f(0:4) + f(0:6) + f(0:8) + f(1)] = ¡0:56

So, ¡0:76 <

Z 1

0

(x2 ¡ 1)dx < ¡0:56

x

y

1

1

1

Example 9



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7 Consider the graph of f(x) = e¡x

a Sketch the graph of y = f(x) for 0 6 x 6 2:

b On the graph in a draw 8 upper rectangles and find the height of each one.

c Use b to find an upper bound for

Z 2

0

e¡xdx

d Find a lower bounds for

Z 2

0

e¡xdx using 8 lower rectangles.

e Use technology to find, correct to 4 significant figures, upper and lower bounds forZ 2

0

e¡xdx when n = 100.


If f(x) 6 0 for all x on a 6 x 6 b thenZ b

a

f(x) dx = ¡(the shaded area).

Use graphical evidence and known area facts to find:

a

Z 3

1

(2 ¡ 2x) dx b

Z 2

0

¡p4 ¡ x2 dx

a

Shaded area = 12 £ 2 £ 4

= 4

)

Z 3

1

(2 ¡ 2x)dx = ¡(the shaded area)

= ¡4

b As y = ¡p4 ¡ x2 , y2 = 4 ¡ x2 and so x2+ y2 = 4.

centre (0, 0) and radius 2. y = ¡p4 ¡ x2 is the lower half.

Shaded area = 14(¼r2)

= 14 £ ¼ £ 4

= ¼

)

Z 2

0

¡p4 ¡ x2dx = ¡(the shaded area)

= ¡¼

�

�

�

� �

( )� �,

�

�

x

y

�

�

y

x

Example 10

This is a circle with

y

xa b

y x��( )


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2

2

2

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y

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The following properties of the definite integral can be deduced by considering upper and

lower sums that approximate the integrals.

²Z b

a

[f(x) + g(x)]dx =

Z b

a

f(x)dx +

Z b

a

g(x)dx

²Z b

a

cf(x)dx = c

Z b

a

f(x)dx, c is any constant

In particular, if c = ¡1,

Z b

a

[¡f(x)] dx = ¡Z b

a

f(x) dx

²Z b

a

f(x) dx +

Z c

b

f(x) dx =

Z c

a

f(x)dx

These facts follow from the properties of sums. For example,P

c f(xi)¢x = cP

f(xi)¢x.

1 Given that

Z 1

0

xn dx =1

n + 1for integers n > 0, calculate these integrals:

a

Z 1

0

(1 + x + x2)dx b

Z 1

0

(3x2) dx c

Z 1

0

(¡x3) dx

d

Z 1

0

(2 ¡ 3x + 2x2) dx e

Z 1

0

(1 + x)2dx f

Z 1

0

(2 + x)(2 ¡ x) dx

2 The graph of y = f(x) is illustrated:

Evaluate the following integrals using area

interpretation:

a

Z 3

0

f(x) dx b

Z 7

3

f(x) dx

c

Z 4

2

f(x) dx d

Z 7

0

f(x) dx

PROPERTIES OF DEFINITE INTEGRALS

EXERCISE 2F.2

8 Use graphical evidence and known area facts to find:

a

Z 0

¡2

3xdx b

Z 4

1

(1 ¡ 2x) dx


Evaluate the following integrals using area

interpretation:

a

Z 3

1

f(x) dx b

Z 5

3

f(x) dx c

Z 7

5

f(x) dx

�

�

�

� �

y

x


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Evaluate the following using area interpretation:

a

Z 4

0

f(x) dx b

Z 6

4

f(x) dx

c

Z 8

6

f(x) dx d

Z 8

0

f(x) dx

4 Write as a single integral:

a

Z 4

2

f(x) dx +

Z 7

4

f(x) dx b

Z 3

1

g(x) dx +

Z 8

3

g(x) dx +

Z 9

8

g(x) dx

5 a If

Z 3

1

f(x) dx = 2 and

Z 6

1

f(x) dx = ¡3, find

Z 6

3

f(x) dx:

b If

Z 2

0

f(x) dx = 5,

Z 6

4

f(x) dx = ¡2 and

Z 6

0

f(x) dx = 7,

find

Z 4

2

f(x) dx:

2

2

2

4

y

x6 8

INVESTIGATION 2 ESTIMATING

The integral

Z 3

¡3

e¡

x2

2 dx is of considerable interest to statisticians.

1 Sketch the graph of y = e¡

x2

2 for :

2 Calculate the upper and lower rectangular sums of the function for the three intervals

0 6 x 6 1, 1 6 x 6 2 and 2 6 x 6 3 using n = 750 for each.

3 Combine the upper rectangular sums and the lower rectangular sums you found in 2

0 6 x 6 3 for n = 2250.

4 Use the fact that the function y = e¡

x2

2 is symmetric to find upper and lower

rectangular sums for for n = 2250.

5 Use your results of 3 and 4 to find an estimate for

Z 3

¡3

e¡

x2

2 dx.

How accurate is your estimate?

6 Compare your estimate of 5 withp

2¼.

What to do:

to obtain an upper and lower rectangular sum for

Z 3

¡3e

¡ x2

2 dx

In this investigation we shall use a calculator to estimate the value of this

integral using upper and lower rectangular sums for , which is a

number too large for a single list on most calculators.

n = 4500

¡3 6 x 6 0

¡3 6 x 6 3


TI

C


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_02\055SA12STU-2_02.CDR Thursday, 9 November 2006 3:03:52 PM DAVID3

REVIEW SET 2A

REVIEWG

x m

y m

x

y semi circle A

semi circle B

1 3 5

1 The population of a colony of koalas in Eastern Victoria has increased since it was

established in 1968. The following table shows the population numbers at five year

intervals:Year 1973 1978 1983 1988 1993 1998 2003

Number 12 20 32 47 77 113 181

Let t years be the time since 1968, and let the number of koalas be K.

a Produce the scatterplot of K against t.

b Explain, with brief reasons, which of the following is likely to be a suitable

model for the data:

A the linear function K = 5:27t ¡ 36:6

B the power function K = 1:01t1:36

C the exponential function K = 8:03e0:0892t

c Use the model you selected in b to estimate the likely colony size in

i 1990 ii 2008 iii 2020. Comment on these estimates.

2


b Find the total area A as a function of x.

c Use quadratic theory or a graphics calculator to find the dimensions of each

alpaca paddock when the total area is a maximum. Show your answer on a

diagram.

3 a Use your knowledge of geometry to calculate

Z 3

0

(1 + 2x) dx:

b If

Z 1

0

x2dx = 13 and

Z 3

1

3x2dx = 26 use

this and the information from a to calculate

Z 3

0

(x2 ¡ 2x ¡ 1)dx.

4 For the given function y = f(x), 0 6 x 6 6:

a Show that A has equation yA =p

4x ¡ x2:

b Show that B has equation

yB = ¡p10x ¡ x2 ¡ 24:

c FindR 4

0yAdx and

R 6

4yB dx:

d Hence, findR 6

0f(x) dx.

5 a Sketch the graph of y =px from x = 1 to x = 4.

b By using a lower sum and rectangular strips of width 0:5, estimate the area

between y =px, the x-axis, x = 1 and x = 4.

Susan has m of fencing to construct eight

alpaca paddocks which are to be rectangular and

of equal area as shown.

800



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Y:\HAESE\SA_12STU-2ed\SA12STU-2_02\056SA12STU-2_02.CDR Monday, 21 May 2007 10:14:31 AM DAVID3

REVIEW SET 2B

open

h cm

1 The flash unit of a camera stores charge on a capacitor. The charge is then rapidly

released when it flashes. The following table gives the charge remaining on the

capacitor (in ¹C, microcoulombs) at time t seconds.

Time (t sec) 0:01 0:02 0:03 0:04 0:05 0:06 0:07

Charge (Q ¹C) 74:6 69:7 64:7 60:5 56:4 52:6 48:9

a Draw a scatterplot of the data.

b Superimpose the graph of the linear function Q(t) = 78:1¡427t on the scatter

plot of a.

c Superimpose the graph of the exponential function Q(t) = 80:1e¡7:03t on the

scatter plot of a.

d Use both models to calculate the charge at i t = 0:035 ii t = 0:15

e Comment on the reliability of the answers obtained in d. Which of the two

models would be more reliable to use for extrapolating data?

2 Open cylindrical bins are to have a capacity of 100 litres

and the cost of the metal used to make them must be

a minimum, i.e., the surface area must be as small as

possible.

a Explain why h =105

¼r2cm.

b Show that the total outer surface area is given by:

A = ¼r2 +200 000

rcm2

c Use your graphics calculator to plot A against r and hence find the minimum

value of A and the value of r when it occurs.

d Draw the bin of optimum size. (It does not have to be of actual scale size but

must show the appropriate dimensions.)

3 a Find the domain of the function f(x) =

px2 + x¡ 6

x2 ¡ 1.

b Sketch the graph of f(x).

c Use technology to find the maximum value of f(x).

4 A culture is left to grow and a researcher is interested in how long it will take to reach

certain sizes. His recorded observations are:

Mass (m grams) 0:26 0:32 0:35 0:40 0:47 0:53

Time (t hours) 5:0 9:1 10:8 13:2 16:3 18:7

a Draw a scatterplot of the data. Which model seems to be the most appropriate

to fit the data?

b Find the model.

c How long should it take for the mass to reach: i 0:5 grams ii 0:65 grams?

d Which of the answers in c is more reliable? Why?



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REVIEW SET 2C

x

y

2 4 6

2

2

4

2y

x

1 Sketch the graph of y =1

xfor x > 0.

Find an estimate of

Z 5

1

1

xdx using 4 strips of width 1 and adding the areas of

four trapezia. Give your answer to 2 decimal places.

2 Joan needs to make a rectangular box with a square bottom. The box is to have a

volume of 8 cubic metres. The material to be used for the bottom costs $10 per m2,

and the material for the side costs $5 per m2. There is no top to this box.

Let the length of the square bottom be x m and the height be y m.

a Show that the cost of the box in dollars is C = 10x2 + 20xy.

b Use the fact that the volume is to be 8 cubic metres to show that, in terms of x,

the cost of the box in dollars is C(x) = 10(x2 +16

x).

c Sketch a graph of this function.

d Use technology to calculate the dimensions of the cheapest box Joan can make.

3 The function y = f(x) is graphed.

Find:

aR 4

0f(x)dx

bR 6

4f(x)dx

cR 6

0f(x)dx

4 The ellipse shown has equationx2

16+

y2

4= 1.

a Sketch the graph again and mark on it the area

represented byR 4

012

p16 ¡ x2 dx:

b Explain from the graph why we can say 8 <R 4

0

p16 ¡ x2 dx < 16.

5 A jet of water from a hose is videoed and a grid is placed behind the path.

The coordinates are given by: x 1:0 1:5 2:0 2:5 3:0 3:5 4:0

y 3:68 4:51 5:13 5:50 5:63 5:48 5:10

a Obtain a scatterplot of the data.

b Use your knowledge of the behaviour of water coming out of a hose to briefly

explain which of the following two functions is the better model for the data:

A the surge function y = 5:13x e¡0:342x

B the quadratic model y = ¡0:494x2 + 2:95x + 1:21c If x = 0 corresponds to the point where the water exits the nozzle, and y = 0 is

ground level, use the model you selected in b to calculate:

i how high the nozzle is above the ground

ii where the water hits the ground.



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3

Contents:

Differential calculusDifferential calculus

A

B

C

D

E

F

G

H

I

J

The idea of a limit

Derivatives at a given -value

The derivative function

Simple rules of differentiation

Composite functions and the chain rule

Product and quotient rules

Implicit differentiation

Tangents and normals

The second derivative

Review

x


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HISTORICAL NOTE

² slopes of tangents to curves at any point on the curve, and

² finding the rate of change in one variable with respect to another.

Note:

Calculus has applications in a wide variety of fields including engineering, biology, chemistry,

physics, economics and geography.

Consider trying to find the slope of the tangent to the curve y = x2 at the point (1, 1).

There are a few methods we could use to do this. Some are given below

Isaac Newton 1642 – 1727 Gottfried Leibniz 1646 – 1716

Differential Calculus

Sir Isaac Newton Gottfried Wilhelm Leibniz

is a branch of Mathematics which originated in the th

Century. and are credited with

the vital breakthrough in thinking necessary for the development of calculus.

Both mathematicians were attempting to find an algebraic method for solving

problems dealing with

17

Calculus is a Latin word meaning

pebble. Ancient Romans used

stones to count with.

INTRODUCTION

Let the tangent to y = x2 at (1, 1) have equation

y = mx + c.

Now y = x2 meets y = mx+ c where

x2 = mx + c and so x2 ¡mx¡ c = 0 ...... (1):

Because of the tangency, this quadratic equation has

a repeated root, x = 1 and so the quadratic equation

from which it comes is (x¡ 1)2 = 0.

This equation expanded is x2 ¡ 2x + 1 = 0 ...... (2).

Comparing equations (1) and (2) we see that

m = 2 and c = ¡1:

So, the slope of the tangent is 2 and the equation of the tangent is y = 2x¡ 1.

This technique can only be used when a quadratic equation results. Consequently it would

be unusable if we wanted, for example, to find the slope of the tangent to y = 2x at x = 0,

say.

A COORDINATE GEOMETRY METHOD

( )��,

x

y

y mx c��

y x�� X

,

60 DIFFERENTIAL CALCULUS (Chapter 3)


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Consider this table: x-coordinate y-coordinate slope of chord

2 4 4¡12¡1 = 3

1 = 3

1:5 2:25 2:25¡11:5¡1 = 1:25

0:5 = 2:5

1:1 1:21 1:21¡11:1¡1 = 0:21

0:1 = 2:1

1:01 1:0201

1:001 1:002 001 1:002 001¡11:001¡1 = 0:002 001

0:001 = 2:001

It is fairly clear that:

² the slope of the tangent at (1, 1) would be exactly 2

² the table method is tedious, but it does help to understand the ideas behind

finding slopes at a given point.

( (1 ) ).

Now the slope of chord MF is m =y-step

x-step=

(1 + h)2 ¡ 1

1 + h¡ 1=

1 + 2h + h2 ¡ 1

h

) m =2h + h2

h

=h(2 + h)

h

= 2 + h fif h 6= 0g

THE TABLE METHOD

(1, 1)

21

4

3

2

1

y

x

tangent

y x�� X

THE ALGEBRAIC METHOD

F (1, 1)

M (1+ ,(1+ ) )h h 2

y

x

y x= 2

chord

1:0201¡11:01¡1 = 0:0201

0:01 = 2:01

To illustrate the algebraic method we will consider the

curve y = x2 and the tangent at F(1, 1).

Let a moving point M have x-coordinate 1+h where

h is small.

) the y-coordinate of M is (1 + h)2 fas y = x2gM is 1 + h + h 2

1

1

Now as M approaches F, h approaches 0: Consequently, 2 + h approaches 2.

So, we conclude that the tangent at (1, 1) has slope 2.

Drawing a tangent to a curve at a given point in exactly the

correct position is extremely difficult.

Three different people may produce three different results. So

we need better methods for performing this procedure.

Consider the curve and the tangent at the point ( , ).

A table of values could be used to find the slope of the

tangent at ( , ). We consider a point not at ( , ), find the

slope to ( , ), and do the same for points closer and closer

to ( , ).

y x� �= 1 1

1 1 1 11 1

1 1

2

DIFFERENTIAL CALCULUS (Chapter 3) 61


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INVESTIGATION 1 THE SLOPE OF A TANGENT

Note:

THE IDEA OF A LIMITA

curve

tangent

chord

(secant)

Given a curve, how can we find the slope

of a tangent at any point on it?

For example, the point A(1, 1) lies on the

curve y = x2. What is the slope of the

tangent at A?

1 Suppose B lies on f(x) = x2 and B

has coordinates (x, x2).

a Show that the chord AB has slope

f(x) ¡ f(1)

x¡ 1or

x2 ¡ 1

x¡ 1.

b Copy and complete:

x Point B Slope of AB

5 (5, 25) 632

1:51:11:011:001

2 Comment on the slope of AB as x gets

closer to 1.

3 Repeat the process as x gets closer to 1,

but from the left of A.

4 Click on the icon to view a demonstration

of the process.

5 What do you suspect is the slope of the tangent at A?

What to do:

A (1, 1)

y

x

ƒ( ) =x x2

A (1, 1)

B ( , )x x2

y

x

�

�

A chord (secant) of a curve is a straight line

segment which joins any two points on the curve.

A tangent is a straight line which touches a curve

at a point.

We have already visited the concept of a limit in the previous

chapter, where we talked about the definite integral as the

unique number between the upper and lower sums.

We now investigate the slopes of chords (secants) from a fixed

point on a curve over successively smaller intervals.

DEMO



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The above investigation shows us that as x approaches 1, the slope of the chord approaches

the slope of the tangent at x = 1.

Notation: We use a horizontal arrow, !, to represent the word ‘approaches’ or the

phrase ‘tends towards’.

So, x ! 1 is read as ‘x approaches 1’ or ‘x tends to 1’.

In the investigation we noticed that the slope of AB approached a limiting value of 2 as xapproached 1, from either side of 1.

Consequently we can write, as x ! 1,x2 ¡ 1

x¡ 1! 2.

This idea is written simply as limx!1

x2 ¡ 1

x¡ 1= 2

and is read as:the limit as xapproaches 1

ofx2 ¡ 1

x¡ 1is 2

In general,

iff(x) ¡ f(a)

x¡ acan be made as close as we like to some real number L by making

x sufficiently close to a, we say thatf(x) ¡ f(a)

x¡ aapproaches a limit of L as x

approaches a and write

Fortunately we do not have to go through the graphical/table of values method (as illustrated

in the investigation) each time we wish to find the slope of a tangent.

Recall that the slope of AB =x2 ¡ 1

x¡ 1

) slope of AB =(x + 1)(x¡ 1)

x¡ 1= x + 1 provided that x 6= 1

Now as B approaches A, x ! 1 ) slope of AB ! 2 ...... (1)

From a geometric point of view:

as B moves towards A,

the slope of AB ! the slope of the tangent at A .... (2)

Thus, from (1) and (2), we conclude that as both limits

must be the same, the slope of the tangent at A is 2.

ALGEBRAIC/GEOMETRIC APPROACH

A

B1

B2

B3

B4

y x= X

limx!a

f(x)¡ f(a)

x¡ a= L



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The following are useful limit rules:

² limx!a

c = c c is a constant

² limx!a

c£ u(x) = c£ limx!a

u(x) c is a constant, u(x) is a function of x

² limx!a

[u(x) + v(x)] = limx!a

u(x) + limx!a

v(x) u(x) and v(x) are functions of x

² limx!a

[u(x)v(x)] =hlimx!a

u(x)i h

limx!a

v(x)i

u(x) and v(x) are functions of x

We make no attempt to prove these rules at this stage. However, all can be readily verified.

For example: as x ! 2, x2 ! 4 and 5x ! 10 and x2+5x ! 14 clearly verifies

the third rule.

Before proceeding to a more formal method we will reinforce the algebraic/geometric method

of finding slopes of tangents.

1 Use the algebraic/geometric method to find the slope of the tangent to:

a y = x2 at the point (3, 9) b y =1

xat the point where x = 2.

2 a Show that (x¡ a)(x2 + ax + a2) = x3 ¡ a3.

b Use the algebraic/geometric method and a to find the slope of the tangent to y = x3

at the point where x = 2.

3 a Show that

px¡p

a

x¡ a=

1px +

pa

.

b Hence, find the slope of the tangent to y =px at the point where x = 9.

LIMIT RULES

Use the algebraic/geometric method to find the slope of the tangent to

y = x2 at the point (2, 4).

Let B be (x, x2) ) slope of AB =x2 ¡ 4

x¡ 2

=(x + 2)(x¡ 2)

(x¡ 2)

= x + 2 provided that x 6= 2

) limx!2

(slope of AB) = 4 ...... (1)

But, as B ! A, i.e., x ! 2

limx!2

(slope of AB) = slope of tangent at A ...... (2)

) slope of tangent at A = 4 ffrom (1) and (2)g

A (2, 4)

B ( , )x x2

y x=2

Example 1

EXERCISE 3A



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Consider a general function y = f(x), a fixed point A(a, f(a)) and a variable point

B(x, f(x)).

The slope of chord AB =f(x) ¡ f(a)

x¡ a.

Now as B ! A, x ! a

and the slope of chord AB ! slope of

tangent at A

So, f 0(a) = limx!a

f(x) ¡ f(a)

x¡ a.

Thus

Note: ² The slope of the tangent at x = a is defined as the slope of the curve at the

point where x = a, and is the instantaneous rate of change in y with respect

to x at that point.

² Finding the slope using the limit method is said to be using first principles.

We are now at the stage where we can find slopes

of tangents at any point on a simple curve using a

limit method.

Notation: The slope of the tangent to a curve

y = f(x) at x = a is f 0(a),

read as ‘eff dashed a’.

DERIVATIVES AT A GIVEN VALUEx-B

a

y

x

y x= ƒ( )

tangent at a

point of contact

a x

y y x= ƒ( )

( )x, xƒ( )

ƒ( )x

ƒ( )a

ƒ( )x

A

B

tangent at A with slope ƒ'( )a

Find, from first principles, the slope of the tangent to:

a y = 2x2 + 3 at x = 2 b y = 3 ¡ x¡ x2 at x = ¡1

a Now f(2) = 2(2)2 + 3 = 11

and f 0(2) = limx!2

f(x) ¡ f(2)

x¡ 2

) f 0(2) = limx!2

2x2 + 3 ¡ 11

x¡ 2

= limx!2

2x2 ¡ 8

x¡ 2

= limx!2

2(x + 2)(x¡ 2)

x¡ 2

= 2 £ 4

= 8

b Now f(¡1) = 3 ¡ (¡1) ¡ (¡1)2 = 3

and f 0(¡1) = limx!¡1

f(x) ¡ f(¡1)

x¡ (¡1)

) f 0(¡1) = limx!¡1

3 ¡ x¡ x2 ¡ 3

x + 1

= limx!¡1

¡x¡ x2

x + 1

= limx!¡1

¡x(1 + x)

x + 1

= 1

1

11

1

Example 2

f 0(a) = limx!a

f(x) ¡ f(a)

x ¡ ais the slope of the tangent at x = a and is

called the derivative at x = a.



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1 Find, from first principles, the slope of the tangent to:

a f(x) = 1 ¡ x2 at x = 2 b f(x) = 2x2 + 5x at x = ¡1

c f(x) = 5 ¡ 2x2 at x = 3 d f(x) = 3x + 5 at x = ¡2

Find, from first principles, the derivative of:

a f(x) =9

xat x = 2 b f(x) =

2x¡ 1

x + 3at x = ¡1

a f 0(2) = limx!2

f(x) ¡ f(2)

x¡ 2

= limx!2

µ 9x¡ 9

2

x¡ 2

¶= lim

x!2

µ 9x¡ 9

2

x¡ 2

¶2x

2x

f2x is the LCD

of 9x

and 92g

= limx!2

18 ¡ 9x

2x(x¡ 2)

= limx!2

¡9(x¡ 2) 1

12x(x¡ 2)

= ¡94

b f 0(¡1) = limx!¡1

f(x) ¡ f(¡1)

x¡ (¡1)

= limx!¡1

Ã2x¡1x+3 + 3

2

x + 1

!

= limx!¡1

Ã2x¡1x+3 + 3

2

x + 1

!£ 2(x + 3)

2(x + 3)

= limx!¡1

2(2x¡ 1) + 3(x + 3)

2(x + 1)(x + 3)

= limx!¡1

4x¡ 2 + 3x + 9

2(x + 1)(x + 3)

= limx!¡1

7x + 7

2(x + 1)(x + 3)

= limx!¡1

7(x + 1)

2(x + 1)(1

1

x + 3)

= 72(2)

= 74

where f(¡1) =2(¡1) ¡ 1

(¡1) + 3

= ¡32

fas x 6= 2g

EXERCISE 3B

Example 3

Never ‘multiply out’the denominator.

There should alwaysbe cancelling of theoriginal divisor atthis step. Why?



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2 Find, from first principles, the derivative of:

a f(x) =4

xat x = 2 b f(x) = ¡3

xat x = ¡2

c f(x) =1

x2at x = 4 d f(x) =

4x

x¡ 3at x = 2

e f(x) =4x + 1

x¡ 2at x = 5 f f(x) =

3x

x2 + 1at x = ¡4

3 Find, from first principles, the instantaneous rate of change in:

apx at x = 4 b

px at x = 1

4

c2px

at x = 9 dpx¡ 6 at x = 10

An alternative formula for finding f 0(a) is

slope of AB =f(a + h) ¡ f(a)

h

Note that as B ! A, h ! 0

and f 0(a) = limh!0

(slope of AB)

which justifies the alternative formula.

x

y

A

Bƒ( + )a h

a a h+

ƒ( )a

y x= ƒ( )

h

Find, using first principles, the instantaneous rate of change in y =px at x = 9.

f(x) =px and f(9) =

p9 = 3

Now f 0(9) = limx!9

f(x) ¡ f(9)

x¡ 9

= limx!9

px¡ 3

x¡ 9

= limx!9

px¡ 3

1

1(px + 3)(

px¡ 3)

=1p

9 + 3

= 16

Example 4

f 0(a) = limh!0

f(a+ h)¡ f(a)

h

Treat as thedifference of two

squares.

x� �¡9



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4 Use the first principles formula f 0(a) = limh!0

f(a + h) ¡ f(a)

hto find:

a the slope of the tangent to f(x) = x2 + 3x¡ 4 at x = 3

b the slope of the tangent to f(x) = 5 ¡ 2x¡ 3x2 at x = ¡2

c the instantaneous rate of change in f(x) =1

2x¡ 1at x = ¡2

Use the first principles formula f 0(a) = limh!0

f(a+ h) ¡ f(a)

hto find:

a the slope of the tangent to f(x) = x2 + 2x at x = 5

b the instantaneous rate of change of f(x) =4

xat x = ¡3

a f 0(5) = limh!0

f(5 + h) ¡ f(5)

hwhere f(5) = 52 + 2(5) = 35

= limh!0

(5 + h)2 + 2(5 + h) ¡ 35

h

= limh!0

25 + 10h + h2 + 10 + 2h¡ 35

h

= limh!0

h2 + 12h

h

= limh!0

h(h + 12)

1

1

1

1

h

= 12

and so the slope of the tangent at x = 5 is 12.

b f 0(¡3) = limh!0

f(¡3 + h) ¡ f(¡3)

hwhere f(¡3) = 4

¡3 = ¡43

= limh!0

Ã4

¡3+h+ 4

3

h

!

= limh!0

Ã4

h¡3 + 43

h

!£ 3(h¡ 3)

3(h¡ 3)

= limh!0

12 + 4(h¡ 3)

3h(h¡ 3)

= limh!0

4h

3h(h¡ 3)

= ¡49

) the instantaneous rate of change in f(x) at x = ¡3 is ¡49 .

fas h 6= 0g

fas h 6= 0g

Example 5



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d the slope of the tangent to f(x) =1

x2at x = 3

e the instantaneous rate of change in f(x) =px at x = 4

f the instantaneous rate of change in f(x) =1px

at x = 1

5 Using f 0(a) = limh!a

f(a + h) ¡ f(a)

hfind:

a f 0(2) for f(x) = x3 b f 0(3) for f(x) = x4

Reminder: (a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

For a non-linear function with equation y = f(x),slopes of tangents at various points continually change.

Our task is to determine a slope function so that when

we replace x by a, say, we will be able to find the slope

of the tangent at x = a.

Consider a general function y = f(x) where A is (x, f(x)) and B is (x + h, f(x + h)).

The chord AB has slope =f(x + h) ¡ f(x)

x + h¡ x

=f(x + h) ¡ f(x)

h.

If we now let B move closer to A, the slope of

AB approaches the slope of the tangent at A.

So, the slope of the tangent at the variable point

(x, f(x)) is the limiting value of

f(x + h) ¡ f(x)

has h approaches 0.

Since this slope contains the variable x it is called a slope function.

DERIVATIVE FUNCTION

The slope function, also known as the derived function, or derivative function or

simply the derivative is defined as

THE DERIVATIVE FUNCTIONC

x

y y x= ƒ( )

x

y

A

Bƒ( + )x h

x x h+

ƒ( )x

y x= ƒ( )

h

f 0(x) = limh!0

f(x+ h) ¡ f(x)

h.



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INVESTIGATION 2 FINDING THE SLOPE OF FUNCTIONS

WITH TECHNOLOGY

[Note: limh!0

f(x + h) ¡ f(x)

his the shorthand way of writing

“ the� limiting value off(x + h) ¡ f(x)

has h gets as close as we like to zero.”]

1 By using a graphical argument only, explain why:

a for f(x) = c where c is a constant, f 0(x) = 0

b for f(x) = mx + c where m and c are constants, f 0(x) = m:

2 Consider f(x) = x2. Find f 0(x) for x = 1, 2, 3, 4, 5, 6 using technology.

Predict f 0(x) from your results.

3 Use technology and modelling techniques to find f 0(x) for:

a f(x) = x3 b f(x) = x4 c f(x) = x5

d f(x) =1

xe f(x) =

1

x2f f(x) =

px = x

1

2

4 Use the results of 3 to complete the following:

“if f(x) = xn, then f 0(x) = ::::::”

Unfortunately the way of finding slope functions by the method shown in the investigation is

insufficient for more complicated functions. Consequently, we need to use the slope function

definition, but even this method is limited to relatively simple functions.

What to do:

This investigation can be done by or by clicking on the icon

to open the The idea is to find slopes at various points on

simple curve in order to find and table -coordinates of points and the slopes of

the tangents at those points. From this table you should be able to predict or find

the slope function for the curve.

graphics calculator

demonstration. a

x

Find, from first principles, the slope function of f(x) = x2.

If f(x) = x2, f 0(x) = limh!0

f(x + h) ¡ f(x)

h

= limh!0

(x + h)2 ¡ x2

h

= limh!0

x2 + 2hx + h2 ¡ x2

h

= limh!0

h(2x + h)

h

= 2x

fas h 6= 0g

TI

C

GRAPHING

PACKAGE

Example 6



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1 Find, from first principles, the slope function of f(x) where f(x) is:

a x b 5 c x3 d x4

[Reminder: (a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4]

2 Find, from first principles, f 0(x) given that f(x) is:

a 2x + 5 b x2 ¡ 3x c x3 ¡ 2x2 + 3

3 Find, from first principles, the derivative of f(x) when f(x) is:

a1

x + 2b

1

2x¡ 1c

1

x2d

1

x3

Find, from first principles, f 0(x) if f(x) =1

x.

If f(x) =1

x, f 0(x) = lim

h!0

f(x + h) ¡ f(x)

h

= limh!0

"1

x+h¡ 1

x

h

#£ (x + h)x

(x + h)x

= limh!0

x¡ (x + h)

hx(x+ h)

= limh!0

¡h

hx(x + h)

= ¡ 11

¡1

x2fas h ! 0, x + h ! xg

fas h 6= 0g

Find, from first principles, the slope function of f(x) =px:

If f(x) =px, f 0(x) = lim

h!0

f(x + h) ¡ f(x)

h

= limh!0

px + h¡p

x

h

= limh!0

1

1

µpx + h¡p

x

h

¶µpx + h +

pxp

x + h +px

¶= lim

h!0

x + h¡ x

h(px + h +

px)

= limh!0

h

h(px + h +

px)

fas h 6= 0g

EXERCISE 3C

Example 7

Example 8



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=1p

x +px

=1

2px

4 Find, from first principles, the derivative of f(x) equal to:

apx + 2 b

1px

cp

2x + 1

5 Using the results of derivatives in this

exercise, copy and complete:

Use your table to predict a formula for

f 0(x) given that f(x) = xn where nis rational.

Function Derivative (in form kxn)

x

x2 2x = 2x1

x3

x4

x¡1

x¡2

x¡3

x1

2 12px

= 12x

¡

1

2

x¡

1

2

Differentiation is the process of finding the derivative (i.e., slope function).

Notation: If we are given a function f(x) then f 0(x) represents the derivative function.

However, if we are given y in terms of x then y0 ordy

dxare commonly

used to represent the derivative.

Note: ² dy

dxreads “dee y by dee x”, or “ the derivative of y with respect to x”.

² dy

dxis not a fraction.

² d(:::::)

dxreads “the derivative of (....) with respect to x.

From question 5 of the previous exercise you should have discovered that if f(x) = xn

then f 0(x) = nxn¡1:

Are there other rules like this one which can be used to differentiate more complicated

functions without having to resort to the tedious limit method? In the following investigation

we may discover some additional rules.

SIMPLE RULES OF DIFFERENTIATIOND



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INVESTIGATION 3 SIMPLE RULES OF DIFFERENTIATION

In this investigation we attempt to differentiate functions of the form cxn

where c is a constant, and functions which are a sum (or difference) of

terms of the form cxn.

1 Find, from first principles, the derivatives of:

a 4x2 b 2x3 c 5px

2 Compare your results with the derivatives of x2, x3 andpx obtained earlier.

Copy and complete: “If f(x) = cxn, then f 0(x) = ::::::”

3 Use first principles to find f 0(x) for:

a f(x) = x2+3x b f(x) = x3 ¡ 2x2

4 Use 3 to copy and complete: “If f(x) = u(x) + v(x) then f 0(x) = ::::::”

You should have discovered the following rules for differentiating functions.

f(x) f 0(x) Name of rule

c (a constant) 0 differentiating a constant

xn nxn¡1 differentiating xn

c u(x) cu0(x) constant times a function

u(x) + v(x) u0(x) + v0(x) sum rule

Each of these rules can be proved using the first principles definition of f 0(x).

The following proofs are worth examining.

² If f(x) = cu(x) where c is a constant then f 0(x) = cu0(x).

Proof: f 0(x) = limh!0

f(x + h) ¡ f(x)

h

= limh!0

cu(x + h) ¡ cu(x)

h

= limh!0

c

·u(x + h) ¡ u(x)

h

¸= c lim

h!0

u(x + h) ¡ u(x)

h

= c u0(x)

² If f(x) = u(x) + v(x) then f 0(x) = u0(x) + v0(x)

Proof: f 0(x) = limh!0

f(x + h) ¡ f(x)

h

= limh!0

µu(x + h) + v(x + h) ¡ [u(x) + v(x)]

h

¶

What to do:

Rules



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= limh!0

µu(x + h) ¡ u(x) + v(x + h) ¡ v(x)

h

¶= lim

h!0

u(x + h) ¡ u(x)

h+ lim

h!0

v(x + h) ¡ v(x)

h

= u0(x) + v0(x)

x

Find the slope function of f(x) = x2 ¡ 4

xand hence find the slope of the tangent

to the function at the point where x = 2.

f(x) = x2 ¡ 4= x2 ¡ 4x¡1

Using the rules we have now developed we can differentiate sums of powers of x.

For example, if f(x) = 3x4 + 2x3 ¡ 5x2 + 7x + 6 then

f 0(x) = 3(4x3) + 2(3x2) ¡ 5(2x) + 7(1) + 0

= 12x3 + 6x2 ¡ 10x + 7

Find f 0(x) for f(x) equal to: a 5x3 + 6x2 ¡ 3x + 2 b 7x¡ 4

x+

3

x3

a f(x) = 5x3 + 6x2 ¡ 3x + 2

) f 0(x) = 5(3x2) + 6(2x) ¡ 3(1) + 0

= 15x2 + 12x¡ 3

b f(x) = 7x¡ 4

x+

3

x3

= 7x¡ 4x¡1 + 3x¡3

) f 0(x) = 7(1)¡ 4(¡1x¡2) + 3(¡3x¡4)

= 7 + 4x¡2 ¡ 9x¡4

= 7 +4

x2¡ 9

x4

1 Find f 0(x) given that f(x) is:

a x3 b 2x3 c 7x2

d x2 + x e 4 ¡ 2x2 f x2 + 3x¡ 5

g x3 + 3x2 + 4x¡ 1 h 5x4 ¡ 6x2 i3x¡ 6

x

j2x¡ 3

x2k

x3 + 5

xl

x3 + x¡ 3

x

Example 9

EXERCISE 3D

Example 10

Each term is inthe form cxn.



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) f 0(x) = 2x¡ 4(¡1x¡2)

= 2x + 4x¡2

= 2x +4

x2

Substituting x = 2 into the slope function will give the slope of the tangent at

x = 2.

So, as f 0(2) = 4 + 1 = 5, the tangent has slope of 5.

Find the slope function of f(x) where f(x) is:

a 3px +

2

xb x2 ¡ 4p

xc 1 ¡ x

px

a f(x) = 3px +

2

x= 3x

1

2 + 2x¡1

) f 0(x) = 3(12x¡ 1

2 ) + 2(¡1x¡2)

= 32x

¡ 1

2 ¡ 2x¡2

=3

2px¡ 2

x2

b f(x) = x2 ¡ 4px

= x2 ¡ 4x¡ 1

2

) f 0(x) = 2x¡ 4(¡12x

¡ 3

2 )

= 2x + 2x¡3

2

= 2x +2

xpx

½x¡

3

2 =1

x3

2

=1

x1x1

2

¾c f(x) = 1 ¡ x

px = 1 ¡ x

3

2

) f 0(x) = 0 ¡ 32x

1

2

= ¡32

px

2 Find the slope of the tangent to:

a y = x2 at x = 2 b y =8

x2at x = 9

c y = 2x2 ¡ 3x + 7 at x = ¡1 d y =2x2 ¡ 5

xat x = 2

e y =x2 ¡ 4

x2at x = 4 f y =

x3 ¡ 4x¡ 8

x2at x = ¡1

Example 11



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3 Find the slope function of f(x) where f(x) is:

a 4px + x b 3

px c ¡ 2p

xd 2x¡p

x

e4px¡ 5 f 3x2 ¡ x

px g

5

x2px

h 2x¡ 3

xpx

4 a If y = 4x¡ 3

x, find

dy

dxand interpret its meaning.

b The position of a car moving along a straight road is given by S = 2t2+4t metres

where t is the time in seconds. FinddS

dtand interpret its meaning.

c The cost of producing and selling x toasters each week is given by

C = 1785 + 3x + 0:002x2 dollars. FinddC


Composite functions are functions like (x2 + 3x)4,p

2 ¡ 3x or1

x¡ x2:

These functions are made up of two simpler functions.

² y = (x2 + 3x)4 is y = u4 where u = x2 + 3x

² y =p

2 ¡ 3x is y =pu where u = 2 ¡ 3x

² y =1

x¡ x2is y =

1

uwhere u = x¡ x2

Notice that in the first example, if f(x) = x4 and g(x) = x2 + 3x then

f(g(x)) = f(x2 + 3x)

= (x2 + 3x)4

If y = 3x2 ¡ 4x, finddy


As y = 3x2 ¡ 4x,dy

dx= 6x¡ 4.

dy

dxis ² the slope function or derivative of y = 3x2 ¡ 4x from which

the slope at any point can be found

² the instantaneous rate of change in y as x changes.

COMPOSITE FUNCTIONS

AND THE CHAIN RULEE

All of these functions can be made up in this way where we compose function of function. Con-

sequently these functions are called

a a

, .composite functions

Example 12



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INVESTIGATION 4 DIFFERENTIATING COMPOSITES

a If f(x) = 3x2 and g(x) = 3x + 7, find f(g(x)).

b If f(g(x)) =p

3 ¡ x2, find f(x) and g(x).

a If f(x) = 3x2 and g(x) = 3x + 7 then

f(g(x)) = f(3x + 7) freplacing g(x) by 3x + 7g= 3(3x + 7)2 freplacing x in the f function by (3x + 7)g

b If f(g(x)) =p

3 ¡ x2 then f(x) =px and g(x) = 3 ¡ x2.

1 Find f(g(x)) if:

a f(x) = x2 and g(x) = 2x + 7 b f(x) = 2x + 7 and g(x) = x2

c f(x) =px and g(x) = 3 ¡ 4x d f(x) = 3 ¡ 4x and g(x) =

px

e f(x) =2

xand g(x) = x2 + 3 f f(x) = x2 + 3 and g(x) =

2

x

g f(x) = 2x and g(x) = 3x + 4 h f(x) = 3x + 4 and g(x) = 2x

2 Find f(x) and g(x) given that f(g(x)) is:

a (3x + 10)3 b1

2x + 4c

px2 ¡ 3x

d1p

5 ¡ 2xe (x2 + 5x¡ 1)4 f

10

(3x¡ x2)3

The purpose of this investigation is to gain insight into how we can

differentiate composite functions.

We might suspect that if y = (2x + 1)2 thendy

dx= 2(2x + 1)1 = 2(2x + 1)

based on our previous rule “if y = xn thendy

dx= nxn¡1”. But is this so?

What to do:

1 Consider y = (2x + 1)2. Expand the brackets and then finddy

dx. Is

dy

dx= 2(2x + 1)?

2 Consider y = (3x+ 1)2. Expand the brackets and then finddy

dx. Is

dy

dx= 2(3x+ 1)1?

3 Consider y = (ax + 1)2. Expand the brackets and finddy

dx: Is

dy

dx= 2(ax + 1)1?

DERIVATIVES OF COMPOSITE FUNCTIONS

Example 13

EXERCISE 3E.1



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4 If y = u2 where u is a function of x, what do you suspectdy

dxwill be equal to?

5 Consider y = (x2 + 3x)2. Expand it and finddy

dx.

Does your answer agree with your suspected rule in 4?

From the previous investigation you probably formulated the rule that:

If y = u2 then .dy

dx= 2u£ du

dx=

dy

du

du

dx

Now consider y = (2x + 1)3 which is really y = u3 where u = 2x + 1.

Expanding we have y = (2x + 1)3

= (2x)3 + 3(2x)21 + 3(2x)12 + 13 fbinomial expansiong= 8x3 + 12x2 + 6x + 1

)dy

dx= 24x2 + 24x + 6

= 6(4x2 + 4x + 1)

= 6(2x + 1)2

= 3(2x + 1)2 £ 2

= 3u2 £ du

dx

=dy

du

du

dx

From the investigation and from the above example we formulate the chain rule.

A non-examinable proof of this rule is included for completeness.

Proof: Consider y = f(u) where u = u(x).

For a small change of ¢x in x, there is a small change of u(x + h) ¡ u(x) = ¢u in uand a small change of ¢y in y.

x x + x

u + u

u

uu u x= ( )

ƒ( )x

x

u

u u + u

y + y

y

y

u

y u�ƒ( )

If y = f(u) where u = u(x) thendy

dx=

dy

du

du

dx.

If y = [f(x)]n thendy

dx= n[f(x)]n¡1f 0(x).



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) lim¢x!0

¢y

¢x= lim

¢u!0

¢y

¢u£ lim

¢x!0

¢u

¢xflimit ruleg

)dy

dx=

dy

du

du

dx

fIf in f 0(x) = limh!0

f(x + h) ¡ f(x)

h, we replace h by ¢x and f(x + h) ¡ f(x)

by ¢y we have f 0(x) =dy

dx= lim

¢x!0

¢y

¢x.g

1

a1

(2x¡ 1)2b

px2 ¡ 3x c

2p2 ¡ x2

d3px3 ¡ x2 e

4

(3 ¡ x)3f

10

x2 ¡ 3

Notice that thebrackets around

are essential.

Why?

� �x

Write in the form aun, clearly stating what u is:

Finddy

dxif: a y = (x2 ¡ 2x)4 b y =

4p1 ¡ 2x

a y = (x2 ¡ 2x)4

) y = u4 where u = x2 ¡ 2x

Nowdy

dx=

dy

du

du

dxfchain ruleg

= 4u3(2x¡ 2)

= 4(x2 ¡ 2x)3(2x¡ 2)

b y =4p

1 ¡ 2x

) y =4pu

where u = 1 ¡ 2x

i.e., y = 4u¡ 1

2 where u = 1 ¡ 2x

Nowdy

dx=

dy

du

du

dxfchain ruleg

= 4u¡ 3

2

= 4(1 ¡ 2x)¡3

2

= 4³¡1

2u¡ 3

2

´(¡2)

Example 14

EXERCISE 3E.2

Now¢y

¢x=

¢y

¢u£ ¢u

¢xffraction multiplicationg

Now as ¢x ! 0, ¢u ! 0 also.



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2 Find the slope functiondy

dxfor:

a y = (4x¡ 5)2 b y =1

5 ¡ 2xc y =

p3x¡ x2

d y = (1 ¡ 3x)4 e y = 6(5 ¡ x)3 f y = 3p

2x3 ¡ x2

g y =6

(5x¡ 4)2h y =

4

3x¡ x2i y = 2

µx2 ¡ 2

x

¶3


a y =p

1 ¡ x2 at x = 12 b y = (3x + 2)6 at x = ¡1

c y =1

(2x¡ 1)4at x = 1 d y = 6 £ 3

p1 ¡ 2x at x = 0

e y =4

x + 2px

at x = 4 f y =

µx +

1

x

¶3

at x = 1

4 If y = x3 then x = y1

3 .

a Finddy

dxand

dx

dyand hence show that

dy

dx£ dx

dy= 1.

b Explain whydy

dx£ dx

dy= 1 whenever these derivatives exist for any general

function y = f(x).

If f(x) = u(x) + v(x) then f 0(x) = u0(x) + v0(x):

That is, the derivative of a sum of two functions is the sum of the derivatives.

But, what if f(x) = u(x)v(x)? Is f 0(x) = u0(x)v0(x)?

That is, is the derivative of a product of two functions equal to the product of the derivatives

of the two functions?

The following example shows that this cannot be true:

If f(x) = xpx we could say f(x) = u(x)v(x) where u(x) = x and v(x) =

px.

Now f(x) = x3

2 ) f 0(x) = 32x

1

2 .

But u0(x)v0(x) = 1 £ 12x

¡ 1

2 = 12x

¡ 1

2 6= f 0(x)

THE PRODUCT RULE

If u(x) and v(x) are two functions of x and y = uv then

PRODUCT AND QUOTIENT RULESF

dy

dx=

du

dxv + u

dv

dxor y0 = u0(x)v(x) + u(x)v0(x).



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Consider the example f(x) = xpx again.

This is a product u(x)v(x) where u(x) = x and v(x) = x1

2

) u0(x) = 1 and v0(x) = 12x

¡ 1

2 .

According to the product rule f 0(x) = u0v + uv0

= 1£ x1

2 + x£ 12x

¡ 1

2 .

= x1

2 + 12x

1

2

= 32x

1

2 which is correct X

For completeness we now prove the product rule.

Proof: Let y = u(x)v(x) and consider the effect of a small change in x of ¢x.

Corresponding changes of ¢u in u, ¢v in v and ¢y in y occur and as y = uv,

Finddy

dxif: a y =

px(2x + 1)3 b y = x2(x2 ¡ 2x)4

a y =px(2x + 1)3 is the product of u = x

1

2 and v = (2x + 1)3

) u0 = 12x

¡ 1

2 and v0 = 3(2x + 1)2 £ 2

= 6(2x + 1)2

Nowdy

dx= u0v + uv0 fproduct ruleg= 1

2x¡ 1

2 (2x + 1)3 + x1

2 £ 6(2x + 1)2

= 12x

¡ 1

2 (2x + 1)3 + 6x1

2 (2x + 1)2

b y = x2(x2 ¡ 2x)4 is the product of u = x2 and v = (x2 ¡ 2x)4

) u0 = 2x and v0 = 4(x2 ¡ 2x)3(2x¡ 2)

Nowdy

dx= u0v + uv0 fproduct ruleg= 2x(x2 ¡ 2x)4 + x2 £ 4(x2 ¡ 2x)3(2x¡ 2)

= 2x(x2 ¡ 2x)4 + 4x2(x2 ¡ 2x)3(2x¡ 2)

y + ¢y = (u + ¢u)(v + ¢v)

) y + ¢y = uv + (¢u)v + u(¢v) + ¢u¢v

¢y = (¢u)v + u(¢v) + ¢u¢v

)¢y

¢x=

µ¢u

¢x

¶v + u

µ¢v

¢x

¶+

µ¢u

¢x

¶¢v fdividing each term by ¢xg

) lim¢x!0

¢y

¢x=

µlim

¢x!0

¢y

¢x

¶v + u

µlim

¢x!0

¢v

¢x

¶+ 0 fas ¢x ! 0, ¢v ! 0 alsog

)dy

dx=

du

dxv + u

dv

dx

Example 15



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1 Finddy

dxusing the product rule:

a y = x2(2x¡ 1) b y = 4x(2x + 1)3 c y = x2p

3 ¡ x

d y =px(x¡ 3)2 e y = 5x2(3x2 ¡ 1)2 f y =

px(x¡ x2)3


a y = x4(1 ¡ 2x)2 at x = ¡1 b y =px(x2 ¡ x + 1)2 at x = 4

c y = xp

1 ¡ 2x at x = ¡4 d y = x3p

5 ¡ x2 at x = 1

3 If y =px(3 ¡ x)2 show that

dy

dx=

(3 ¡ x)(3 ¡ 5x)

2px

.

Find the x-coordinates of all points on y =px(3 ¡ x)2 where the tangent is hori-

zontal.

THE QUOTIENT RULE

Expressions likex2 + 1

2x¡ 5,

px

1 ¡ 3xand

x3

(x¡ x2)4are called quotients.

Quotient functions have form Q(x) =u(x)

v(x).

Notice that u(x) = Q(x)v(x) and by the product rule,

u0(x) = Q0(x)v(x) + Q(x)v0(x)

) u0(x) ¡Q(x)v0(x) = Q0(x)v(x)

i.e., Q0(x)v(x) = u0(x) ¡ u(x)

v(x)v0(x)

) Q0(x)v(x) =u0(x)v(x) ¡ u(x)v0(x)

v(x)

) Q0(x) =u0(x)v(x) ¡ u(x)v0(x)

[v(x)]2and this formula is called the

quotient rule.

Use the quotient rule to finddy

dxif: a y =

1 + 3x

x2 + 1b y =

px

(1 ¡ 2x)2

or if y =u

vwhere u and v are functions of x then

dy

dx=

du

dxv ¡ u

dv

dxv2

.

EXERCISE 3F.1

Example 16

So, if Q (x) =u (x)

v (x)then Q0(x) =

u0(x)v(x)¡ u(x)v0(x)

[v(x)]2



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1 Use the quotient rule to finddy

dxif:

a y =1 + 3x

2 ¡ xb y =

x2

2x + 1c y =

x

x2 ¡ 3

d y =

px

1 ¡ 2xe y =

x2 ¡ 3

3x¡ x2f y =

xp1 ¡ 3x

Note: Most of the time, simplification ofdy

dxas in the above example is unnecessary,

especially if you want to find the slope of a tangent at a given point, because you

can substitute a value for x without simplifying.

a y =1 + 3x

x2 + 1is a quotient with u = 1 + 3x and v = x2 + 1

) u0 = 3 and v0 = 2x

Nowdy

dx=

u0v ¡ uv0

v2fquotient ruleg

=3(x2 + 1) ¡ (1 + 3x)2x

(x2 + 1)2

=3x2 + 3 ¡ 2x¡ 6x2

(x2 + 1)2

=3 ¡ 2x¡ 3x2

(x2 + 1)2

b y =

px

(1 ¡ 2x)2is a quotient where u = x

1

2 and v = (1 ¡ 2x)2

) u0 = 12x

¡ 1

2 and v0 = 2(1 ¡ 2x)1 £¡2

= ¡4(1¡ 2x)

Nowdy

dx=

u0v ¡ uv0

v2

=12x

¡ 1

2 (1 ¡ 2x)2 ¡ x1

2 £¡4(1¡ 2x)

(1 ¡ 2x)4

=12x

¡ 1

2 (1 ¡ 2x)2 + 4x1

2 (1¡ 2x)

(1 ¡ 2x)4

=

(1 ¡ 2x)

·1 ¡ 2x

2px

+ 4px

µ2px

2px

¶¸(1 ¡ 2x)4 3

=1 ¡ 2x + 8x

2px(1 ¡ 2x)3

=6x + 1

2px(1 ¡ 2x)3

flook forcommon factorsg

EXERCISE 3F.2



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IMPLICIT DIFFERENTIATIONG


a y =x

1 ¡ 2xat x = 1 b y =

x3

x2 + 1at x = ¡1

c y =

px

2x + 1at x = 4 d y =

x2

px2 + 5

at x = ¡2

3 a If y =2px

1 ¡ x, show that

dy

dx=

x + 1px(1 ¡ x)2

:

For what values of x isdy

dxi zero ii undefined?

b If y =x2 ¡ 3x + 1

x + 2, show that

dy

dx=

x2 + 4x¡ 7

(x + 2)2:

For what values of x isdy

dxi zero ii undefined?

For relations such as y3 + 3xy2 ¡ xy + 11 = 0 it is often difficult or impossible to make

y the subject of the formula.

Such relationships between x and y are called implicit relations.

To gain insight into how such relations can be differentiated we will examine a familiar case.

Consider the circle with centre (0, 0) and radius 2.

The equation of the circle is x2 + y2 = 4.

Suppose A(x, y) lies on the circle.

The radius OA has slope =y-step

x-step

=y ¡ 0

x¡ 0

=y

x

) the tangent at A has slope ¡x

yfthe negative reciprocalg

Thusdy

dx= ¡x

yfor all points (x, y) on the circle.

This result was achievable because of a circle property.

In general implicit relations do not have a simple means of findingdy

dxas in the case of a

circle.

Another way of findingdy

dxfor a circle is to split the relation into two parts.

A ( , )x y

y

x(0, 0)

2

tangent



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As x2 + y2 = 4, then y2 = 4 ¡ x2 and so y = §p4 ¡ x2.

Case 1:

y =p

4 ¡ x2 = (4 ¡ x2)1

2

)dy

dx= 1

2(4 ¡ x2)¡1

2 £ (¡2x)

= ¡xp4 ¡ x2

= ¡xy

Case 2:

y = ¡p4 ¡ x2 = ¡(4 ¡ x2)

1

2

)dy

dx= ¡1

2(4 ¡ x2)¡1

2 £ (¡2x)

= xp4 ¡ x2

= x¡y

the same result as in case 1.So,dy

dx= ¡x

y,

As stated before, the process of making y the subject is often difficult or impossible with

some implicit functions. So, is there a better way?

The process by which we differentiate implicit functions is called implicit differentiation.

We simply differentiate term-by-term across the equation.

For example, if x2 + y2 = 4 thend

dx(x2) +

d

dx(y2) =

d

dx(4)

) 2x + 2ydy

dx= 0 and so,

dy

dx= ¡x

y

Note:d

dx(y2) =

d

dy(y2) £ dy

dxfchain ruleg

= 2ydy

dxand

If y is a function of x find: ad

dx(y3) b

d

dx

µ1

y

¶c

d

dx(xy2)

ad

dx(y3) = 3y2

dy

dxb

d

dx

µ1

y

¶=

d

dx(y¡1)

= ¡y¡2 dy

dx

cd

dx(xy2) = 1 £ y2 + x£ 2y

dy

dxfproduct ruleg

= y2 + 2xydy

dx

IMPLICIT DIFFERENTIATION

y

x2

2 ~`4``-̀̀ ``x2

-~`4``-̀̀``x2

Example 17

d

dx(yn) = nyn¡1

dy

dx



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1 If y is a function of x, find:

ad

dx(2y) b

d

dx(¡3y) c

d

dx(y3) d

d

dx(1

y)

ed

dx(y4) f

d

dx(py) g

d

dx

µ1

y2

¶h

d

dx

µ1py

¶i

d

dx(xy) j

d

dx(x2y) k

d

dx(xy2) l

d

dx(x2y3)

2 Finddy

dxif:

a x2 + y2 = 25 b x2 + 3y2 = 9 c y2 ¡ x2 = 8

d x2 ¡ y3 = 10 e x2 + xy = 4 f x3 ¡ 2xy = 5

gy2

x+ y = 100 h y3 ¡ x

y2= x2 i 2x2 ¡ 5x2y2 = y3

Finddy

dxif: a x2 + y3 = 8 b x + x2y + y3 = 100

a x2 + y3 = 8

)d

dx(x2) +

d

dx(y3) =

d

dx(8)

i.e., 2x + 3y2dy

dx= 0

) 3y2dy

dx= ¡2x

)dy

dx=

¡2x

3y2

b x + x2y + y3 = 100

)d(x)

dx+

d

dx(x2y) +

d

dx(y3) =

d

dx(100)

fproduct ruleg

) (x2 + 3y2)dy

dx= ¡1 ¡ 2xy

)dy

dx=

¡1 ¡ 2xy

x2 + 3y2

i.e., 1 +

·2xy + x2 dy

dx

¸+ 3y2

dy

dx= 0

EXERCISE 3G

Example 18



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TANGENTS AND NORMALSH

A ( , ƒ( ))a a

x = a

y x= ƒ( )

point ofcontact

tangent

normal


a x + y3 = 4y at the point where y = 1

b x + y = 8xy at the point where x = 12

c y ¡ ¡xy2 = 6 at the point(s) where x = 1.

Consider a curve y = f(x).

If A is the point with x-coordinate a, then

the slope of the tangent at this point is f 0(a).

The equation of the tangent is

NORMALS

A normal to a curve is a line which is perpendicular to the tangent at the point

of contact.

Thus, the slope of a normal at x = a is ¡ 1f 0(a)

:

TANGENTS

Find the slope of the tangent to x2 + y3 = 5 at the point where x = 2.

First we finddy

dx.

d

dx(x2) +

d

dx(y3) =

d

dx(5)

i.e., 2x + 3y2dy

dx= 0

) 3y2dy

dx= ¡2x

)dy

dx=

¡2x

3y2

But when x = 2, 4 + y3 = 5

) y3 = 1

) y = 1

Consequentlydy

dx=

¡2(2)

3(1)2

= ¡43

So, the slope of the tangent at x = 2 is ¡43 .

Example 19

fequating slopesg

or

y ¡ f(a)

x ¡ a= f 0(a)

y ¡ f(a) = f 0(a)(x ¡ a)



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Note: ² If a tangent touches y = f(x) at (a, b) then it has equation

² Vertical and horizontal lines have equations x = k and y = c respectively.

Find the equation of the tangent to f(x) = x2 + 1 at the point where x = 1.

Since f(1) = 1 + 1 = 2, the point of

contact is (1, 2).

Now f 0(x) = 2x

) f 0(1) = 2

) the tangent has equation

y ¡ 2

x¡ 1= 2

i.e., y ¡ 2 = 2x¡ 2

or y = 2x

(1, 2)

y

x

ƒ( ) = +1x x2

1

Find the equation of the normal to y =8px


When x = 4, y = 8p4

= 82 = 4

) the point of contact is (4, 4).

Now as y = 8x¡ 1

2

dy

dx= ¡4x¡ 3

2

and when x = 4,dy

dx= ¡4 £ 4¡

3

2

= ¡12

) the normal at (4, 4) has slope 21 .

So, the equation of the normal is

y ¡ 4

x¡ 4= 2

i.e., y ¡ 4 = 2x¡ 8

i.e., y = 2x¡ 4.

(4, 4)

y

x

Example 20

Example 21

y ¡ b

x ¡ a= f 0(a) or y ¡ b = f 0(a)(x¡ a).



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3 a Find the equations of the horizontal tangents to y = 2x3 + 3x2 ¡ 12x + 1.

b Find all points of contact of horizontal tangents to the curve y = 2px +

1px

.

c Find k if the tangent to y = 2x3 + kx2 ¡ 3 at the point where x = 2 has slope 4.

4 a The tangent to the curve y = x2 + ax + b, where a and b are constants, is

2x + y = 6 at the point where x = 1. Find the values of a and b.

b The normal to the curve y = apx +

bpx

, where a and b are constants, has

equation 4x + y = 22 at the point where x = 4. Find the values of a and b.

d Find the equation of the tangent to y = 1 ¡ 3x + 12x2 ¡ 8x3 which is parallel

to the tangent at (1, 2).

1 Find the equation of the tangent to:

a y = x¡ 2x2 + 3 at x = 2 b y =px + 1 at x = 4

c y = x3 ¡ 5x at x = 1 d y =4px

at (1, 4)

e y =3

x¡ 1

x2at the point (¡1, ¡4). f y = 3x2 ¡ 1

xat x = ¡1

2 Find the equation of the normal to:

a y = x2 at the point (3, 9) b y = x3 ¡ 5x + 2 at x = ¡2

c y = 2px + 3 at x = 1 d y =

3px

at the point where x = 9

e y =5px¡p

x at the point (1, 4). f y = 8px¡ 1

x2at x = 1

Find the equations of any horizontal tangents to y = x3 ¡ 12x + 2.

Let f(x) = x3 ¡ 12x + 2

) f 0(x) = 3x2 ¡ 12

But f 0(x) = 0 for horizontal tangents and so 3x2 ¡ 12 = 0

) 3(x2 ¡ 4) = 0

) 3(x + 2)(x¡ 2) = 0

) x = ¡2 or 2

Now f(2) = 8 ¡ 24 + 2 = ¡14 and

f(¡2) = ¡8 + 24 + 2 = 18

i.e., points of contact are

(2, ¡14) and (¡2, 18)

) tangents are y = ¡14 and y = 18.

EXERCISE 3H

Example 22



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Find the equation of the tangent to y =p

10 ¡ 3x at the point where x = 3:

Let f(x) = (10 ¡ 3x)1

2

) f 0(x) = 12(10 ¡ 3x)¡

1

2 £ (¡3)

) f 0(3) = 12(1)¡

1

2 £ (¡3)

= ¡32

When x = 3, y =p

10 ¡ 9 = 1

) point of contact is (3, 1).

So, the tangent has equationy ¡ 1

x¡ 3= ¡3

2

i.e., 2y ¡ 2 = ¡3x + 9

or 3x + 2y = 11

5 Find the equation of the:

a tangent to y =p

2x + 1 at the point where x = 4

b tangent to y =1

2 ¡ xat the point where x = ¡1

c normal to y =1

(x2 + 1)2at the point (1, 1

4 )

d normal to y =1p

3 ¡ 2xat the point where x = ¡3

6 y = ap

1 ¡ bx where a and b are constants, has a tangent with equation 3x+y = 5 at

the point where x = ¡1. andFind ba .

Find the equation of the normal to y =x2 + 1

1 ¡ 2xat the point where x = 1

Let f(x) =x2 + 1

1 ¡ 2xf(1) = 2

¡1 = ¡2

) f 0(x) =2x(1 ¡ 2x) ¡ (x2 + 1)(¡2)

(1 ¡ 2x)2fquotient ruleg

) f 0(1) =2(¡1) ¡ (2)(¡2)

(¡1)2=

¡2 + 4

1= 2

The point of contact is (1, f(1)), i.e., (1, ¡2)

) the equation of the normal isy ¡¡2

x¡ 1= ¡1

2

i.e., 2y + 4 = ¡x + 1

i.e., x + 2y = ¡3

Example 23

Example 24



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7 Find the equation of:

a the tangent to f(x) =x

1 ¡ 3xat the point (¡1, ¡1

4 )

b the normal to f(x) =px(1 ¡ x)2 at the point where x = 4

c the tangent to f(x) =x2

1 ¡ xat the point (2, ¡4)

d the normal to f(x) =x2 ¡ 1

2x + 3at the point where x = ¡1.

8 Find the equation of the:

a tangent to x2 + y3 = 9 at the point (1, 2)

b tangent to 2x2 ¡ y2 = ¡7 at the point (¡1, 3)

c normal to x2 + xy + y2 = 3 at the point (2, ¡1)

d normal to x3 ¡ 2xy2 + y = ¡9 at the point (1, ¡2)

e tangent to x2 + xy ¡ 3y2 = 3 at the point (2, 1)

f normal to 2x2 ¡ xy2 = 6 at the point (¡1, 2).

9 Find the equation of the tangent(s) to y2¡3xy+x3 = 3 at the points where x = ¡1.

Find the coordinates of the point(s) where the tangent to y = x3 + x + 2 at (1, 4)

meets the curve again.

Find the equation of the tangent to x2 ¡ 3xy + y2 = 5 at (1, 4).

Differentiating x2 ¡ 3xy + y2 = 5 term-by-term we get:

) 2x¡·3(y) + 3x

dy

dx

¸+ 2y

dy

dx= 0

But when x = ¡1, y = 4

) 2 ¡·3(4) + 3

dy

dx

¸+ 2(4)

dy

dx= 0

) 2 ¡ 12 ¡ 3dy

dx+ 8

dy

dx= 0

) ¡10 + 5dy

dx= 0 and so

dy

dx= 10

5

) the tangent at (1, 4) has slope

) equation of tangent is

= 2

2

y ¡ 4

x¡ 1=

which simplifies to 14x¡ 11y = ¡58

2

Example 25

Example 26

It is not necessary to

make the

subject of theformula.

dy

dx



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10 a Find where the tangent to the curve y = x3, at the point where x = 2, meets the

curve again.

b Find where the tangent to the curve y = ¡x3 + 2x2 + 1, at the point where

x = ¡1, meets the curve again.

c Find where the tangent to the curve y = x2 ¡ 3

x, at x = 3, meets the curve

again.

d Find where the tangent to the curve y = x3+4

x, at the point where x = 1, meets

the curve again.

Find the equations of the tangents to y = x2 from the point (2, 3).

Let (a, a2) lie on f(x) = x2.

Now f 0(x) = 2x

) f 0(a) = 2a

) at (a, a2) the slope of the tangent is2a

1

) equation is 2ax¡ y = 2a(a) ¡ (a2)

i.e., 2ax¡ y = a2.

But this tangent passes through (2, 3).

) 2a(2) ¡ 3 = a2

) 4a¡ 3 = a2

( , )a a2

y

x

y x= 2

(2, 3)

f(x) = x3 + x + 2

) f 0(x) = 3x2 + 1

) f 0(1) = 3 + 1 = 4

) the tangent at (1, 4 4) has slope

Now y = 4x meets y = x3 + x + 2 where x3 + x + 2 = 4x

) x3 ¡ 3x + 2 = 0

and this cubic must have a repeated zero of x = 1 because of

) (x¡ 1)2(x + 2) = 0

x2 £ x = x3 2( 1)¡ £ 2 = 2

) x = 1 or ¡2 and when x = ¡2, y = (¡2)3 + (¡2) + 2 = ¡8

) tangent meets the curve again at (¡2, ¡8).

(1, 4)

( ) ��

the tangent

and therefore its equation isy ¡ 4

x¡ 1= 4

i.e., y ¡ 4 = 4x¡ 4

y = 4x

Example 27



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THE SECOND DERIVATIVEI

i.e., a2 ¡ 4a + 3 = 0

(a¡ 1)(a¡ 3) = 0

) a = 1 or 3

If a = 1, tangent equation is 2x¡ y = 1, with point of contact (1, 1).

If a = 3, tangent equation is 6x¡ y = 9, with point of contact (3, 9).

11 Find the equation of the tangent to y = x2 ¡ x + 9 at the point where x = a.

Hence, find the equations of the two tangents from (0, 0) to the curve. State the

coordinates of the points of contact.

Find the equations of the tangents to y = x3 from the point (¡2, 0).b

c Find the equation(s) of the normal(s) to y =px from the point (4, 0).

Note that: ² d2y

dx2=

d

dx

µdy

dx

¶² d2y

dx2reads “dee two y by dee x squared”.

Time of ride (t min) 0 2:5 5 7:5 10 12:5 15 17 19

Distance travelled (s m) 0 498 782 908 989 1096 1350 1792 2500

a

THE SECOND DERIVATIVE IN CONTEXT

t = 0 t = 5

t = 10

t = 17 t = 19t = 15

friend’s houseMichael’s place

The second derivative of a function f(x) is the derivative of f 0(x),i.e., the derivative of the first derivative.

Notation: We use f 00(x), or y00 ord2y

dx2

derivative.

to represent the second

Michael rides up hill and down the other side to his friend’ house. The dots on the graph show

Michael’s position at various times .

a s

t

The distance travelled by Michael from his place is given at various times

in the following table:

A cubic model seems to fit this data well with coefficient of determination r2 = 0:9992.

The model is s + 1:18t3 ¡ 30:47t2 + 284:52t¡ 16:08 metres.

Notice that the model gives s (0) = ¡16:08 m whereas the actual data gives s (0) = 0.

This sort of problem often occurs when modelling from data.

DEMO



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Nowds

dt+ 3:54t2 ¡ 60:94t + 284:52 metres/minute is the instantaneous rate of change in

displacement per unit of time, i.e., instantaneous velocity.

The instantaneous rate of change in velocity at any point in time is the acceleration of the

moving object and so,d

dt

µds

dt

¶=

d2s

dt2is the instantaneous acceleration,

i.e.,d2s

dt2= 7:08t¡ 60:94 metres/minute per minute.

Notice that, when t = 12, s + 1050 m

ds

dt+ 63 metres/minute

andd2s

dt2+ 24 metres/minute/minute

We will examine displacement, velocity and acceleration in greater detail in the next chapter.

1 Find f 00(x) given that:

a f(x) = 3x2 ¡ 6x + 2 b f(x) = 2x3 ¡ 3x2 ¡ x + 5

c f(x) =2px¡ 1 d f(x) =

2 ¡ 3x

x2

e f(x) = (1 ¡ 2x)3 f f(x) =x + 2

2x¡ 1

Find f 00(x) given that f(x) = x3 ¡ 3

x:

Now f(x) = x3 ¡ 3x¡1

) f 0(x) = 3x2 + 3x¡2

) f 00(x) = 6x¡ 6x¡3

= 6x¡ 6

x3

15105

2500

2000

1500

1000

500

y

x

y x x x��

Example 28

EXERCISE 3I



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REVIEW SET 3B

REVIEW SET 3A

2 Findd2y

dx2given that:

a y = x ¡ x3 b y = x2 ¡ 5

x2c y = 2 ¡ 3p

x

d y =4 ¡ x

xe y = (x2 ¡ 3x)3 f y = x2 ¡ x +

1

1 ¡ x

3 Find x when f 00(x) = 0 for:

a f(x) = 2x3 ¡ 6x2 + 5x + 1 b f(x) =x

x2 + 2.

1 Find the equation of the tangent to y = ¡2x2 at the point where x = ¡1.

2 Finddy

dxfor: a y = 3x2 ¡ x4 b y =

x3 ¡ x

x2

3 Find, from first principles, the derivative of f(x) = x2 + 2x.

4 Find the equation of the normal to y =1 ¡ 2x

x2at the point where x = 1.

5 Finddy

dxfor: a x2y + y3 = 3 b xy2 ¡ x2 = y:

6 Find where the tangent to y = 2x3 + 4x ¡ 1 at (1, 5) cuts the curve again.

7 The tangent to y =ax + bp

xat the point where x = 1 is 2x ¡ y = 1.

Find a and b.

8 Find, from first principles, the derivative of:

a f(x) = x2 + x at x = 2 b f(x) =px + 2 at x = 7.

9 Find a given that the tangent to y =4

(ax + 1)2at x = 0 passes through (1, 0).

1 Find, from first principles, the derivative of4

x2.

2 Find the equation of the normal to y =1px


3 Find, from first principles, the derivative of f(x) =1

2x + 3at x = 6.

4 Determine the derivative with respect to t of:

a M = (t2 + 3)4 b A =

pt + 5

t2

REVIEWJ



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REVIEW SET 3D

REVIEW SET 3C

5 Use the rules of differentiation to finddy

dxfor:

a y =4px¡ 3x b y = (x¡ 1

x)4 c y =

px2 ¡ 3x

6 Find the equation of the tangent to y =2px


7 Find the equation of the tangent to f(x) = x3 ¡ 5x at x = ¡1.

At what point(s) does this tangent meet the curve again?

8 Determine the equation of any horizontal tangents to the curve with equation

y = x3 ¡ 3x2 ¡ 9x + 2.

9 Differentiate with respect to x: a y3 +y

x= 2 b (xy)3 = y2 + 3x

1 Differentiate with respect to x:

a 5x¡ 3x¡1 b (3x2 + x)4 c (x2 + 1)(1 ¡ x2)3:

2 Find the equation of the normal to y =x + 1

x2 ¡ 2at the point where x = 1.

3 If f(x) = kg(x) where k is a constant, prove that f 0(x) = kg0(x).

4 Find the equation of the tangent to f(x) = ¡4x3 +3x at the point where x = a.

How many tangents can be drawn from the point A(0, 1) to the cubic?

Find the equation(s) of any such tangent(s).

5 Find, from first principles, the derivative of f(x) =x2 + 3

xat x = 1.

6 Find the equation of the normal to the curve with equation y3 +pxy = 3 at the

point (4, 1).

7 Find, from first principles, the derivative of f(x) =p

3x + 2.

8 Find all points on the curve y = 2x3 + 3x2 ¡ 10x + 3 where the slope of the

tangent is 2.

9 Find the equation of the tangent to f(x) = 2x3 ¡ 5x2 + 6 at the point (2, 2).

Determine the coordinates of the point where this tangent meets the curve again.

1 Differentiate with respect to x: a f(x) =(x + 3)3p

xb f(x) = x4

px2 + 3

2 Find f 00(x) for: a f(x) = 3x2 ¡ 1

xb f(x) =

px

3 Find, from first principles, the derivative of f(x) =x

x + 3at the point where x = ¡2.



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REVIEW SET 3E

4 y = 2x is a tangent to the curve y = x3 + ax + b at x = 1. Find a and b.

5 Find the equation of the normal to y =p

1 ¡ 4x at the point where x = ¡2.

6 Use first principles to find the derivative of f(x) =1px

.

7 Find the equation of the tangent to the curve x3 + xy + 3y2 = 12 at the point (0, 2).

8 Find, from first principles, the derivative ofx2

3 ¡ xat x = 2.

9 The tangent to y = x2p

1 ¡ x at x = ¡3 cuts the axes at A and B.

Determine the area of triangle OAB.


apx(1 ¡ x)2 b

p3x¡ x2 c

1

2 ¡ x2 Use first principles to find the derivative of f(x) = x2¡3 at the point where x = 3.

3 Find the equation of the normal to y =4px + 2


4 Differentiate with respect to x: a y3 ¡ xy2 = x3 b xy = x2 +py

5 The tangent to y = x3 + ax2 ¡ 4x + 3 at x = 1 is parallel to the line y = 3x.

Find the value of a and the equation of the tangent at x = 1.

Where does the tangent cut the curve again?

6 Find the equation of the tangent to y = x3 at the point where x = 2.

7 The curve f(x) = 2x3 +Ax+B has a tangent with slope 10 at the point (¡2, 33).

Find the values of A and B.

8 Determine the values of x for which f 00(x) = 0 where

f(x) = 2x4 ¡ 4x3 ¡ 9x2 + 4x + 7.

9 If the normal to f(x) =3x

1 + xat (2, 2) cuts the axes at B and C, determine the

length of BC.

10 The equation of a curve is x2 + 4xy + y2 = 3.

Finddy

dx, using implicit differentiation, and hence determine the equation of the

normal to the curve at the point (x1, y1).

Find the coordinates of each point on the curve for which the normal passes through

the origin.



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REVIEW SET 3G

REVIEW SET 3F

1 Differentiate with respect to x: a y = x3p

1 ¡ x2 b y =x2 ¡ 3xpx + 1

2 Find the equation of the normal to y =x + 1

x2 ¡ 2at the point where x = 1.

3 Find the equation of the tangent to y = 3 ¡ x + x2 ¡ 2x3 at the point where

x = ¡1 and hence determine where this tangent cuts the curve again.

4 Find the equation of the tangent to f(x) =px¡ 2(x + 1)2 at x = 3.

5 How many tangents can be drawn from (2, ¡4) to the curve y = x3 ¡ 4x?

Find the equation(s) of any such tangent(s).

6 Find, from first principles, the derivative of y =1 ¡ x

x2at x = 2.

7 Find the equation of the normal to the curve xy2+y3 = 3 at the point where y = ¡1.

8 Findd2y

dx2for: a y = 3x4 ¡ 2

xb y = x3 ¡ x +

1px

9 Find a and b if y =xp

1 ¡ xhas tangent 5x+by = a at the point where x = ¡3.

10 Find the equation of the normal(s) to y = x¡ x2 from the point (1, 0).

1 Finddy

dxfor: a y =

x2

3 ¡ 2xb y =

px(x2 ¡ x)3

2 Find the equation of the tangent to the curve with equation 2x3 + xy ¡ y2 = ¡31at the point (¡2, 3).

3 The curve f(x) = 3x3 +Ax2 +B has tangent with slope 0 at the point (¡2, 14).

Find A and B and hence f 00(¡2).

4 A cubic polynomial has equation y = 2x3 +3x2¡10x+3. Find all points on this

curve where the slope of the tangent is 2.

5 Find the equations of the tangents to y = x3 from the point (2, 0).

6 (3, ¡1) is a point on the curve x3 + xy2 + y = k.

a Find k. b Find the equation of the normal at the point (3, ¡1).

7 Find where the tangent to the curve y = 3x3 ¡ 11x2 + 8x+ 4 at x = 1 meets the

curve again.

8 Use first principles to find the derivative of y =x

x + 1at the point where x = 3.

9 The line joining A(2, 4) to B(0, 8) is a tangent to y =a

(x + 2)2. Find a.

10 Show that the curves whose equations are y =p

3x + 1 and y =p

5x¡ x2

have the same slope at their point of intersection. Find the equation of the common

tangent at this point.



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4

Contents:

Applications ofdifferential calculus

Applications ofdifferential calculus

A

B

C

D

E

Rates of change

Motion in a straight line

Curve properties

Optimisation (maxima and minima)

Review


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One application of differential calculus is the finding of equations of tangents and normals to

curves. There are many other uses, but in this course we consider only:

² rates of change

² motion on a straight line (displacement, velocity and acceleration)

² curve properties (monotonicity and concavity/convexity)

² optimisation (maxima and minima, local and global)

Earlier we discovered that:

if s(t) is a displacement function then s0(t) ords

dtis the instantaneous rate of

change in displacement with respect to time, which is of course the velocity function.

In general,dy

dxgives the rate of change in y with respect to x.

Note: If as x increases, y also increases, thendy

dxwill be positive, whereas

if, as x increases, y decreases, thendy

dxwill be negative.

RATES OF CHANGEA

According to a psychologist the ability of a person to understand spatial concepts

is given by A = 13

pt where t is the age in years, 5 6 t 6 18.

a Find the rate of improvement in ability to understand spatial concepts when the

person is i 9 years old ii 16 years old.

b Explain whydA

dt> 0 for 5 6 t 6 18 and comment on this result.

a A = 13

pt = 1

3 t1

2)

dA

dt= 1

6 t¡

1

2 =1

6pt

i When t = 9,dA

dt= 1

18

) rate of improvement is 118 units per year

for a 9 year old.

ii When t = 16,dA

dt= 1

24

) rate of improvement is 124 units per year for a 16 year old.

b Aspt is never negative,

1

6pt

is never negative, Note that the rate

of increase actually

slows down as tincreases.

i.e.,dA

dt> 0 for all t in 5 6 t 6 18.

This means that the ability to understand spatial

concepts increases with age. This is clearly shown by the graph.

Example 1

100 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 4)


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The height of , grown in ideal

conditions, is given by metres,

where is the number of years after the tree was

planted from an established juvenile tree.

pinus radiata

t

You are encouraged to use technology to graph the function for each question.

1 The quantity of a chemical which is responsible for ‘elasticity’ in human skin is given

by Q = 100 ¡ 10pt where t is the age of a person.

a Find Q at: i t = 0 ii t = 25 iii t = 100 years.

b At what rate is the quantity of the chemical changing at the ages of:

i 25 years ii 50 years?

c Show that the rate at which the skin loses the chemical is decreasing for all values

of t.

2

a How high is the tree at planting?

b

c Find the rate at which the tree is growing at

t = 0, 5 and 10 years.

d Show thatdH

dt> 0 for all t > 0. What is the significance of this result?

3 The resistance to the flow of electricity in a certain metal is given by

R = 20 + 110T + 1

200T2 where T is the temperature (in oC) of the metal.

a Find the resistance R, at temperatures of 0oC, 20oC and 40oC.

b Find the rate of change in the resistance at any temperature T .

c For what values of T does the resistance increase as the temperature increases?

4 The total cost of running a train is given by C(v) = 200v +10000

vdollars where v

is the average speed of the train in kmph.

a Find the total cost of running the train at i 20 kmph ii 40 kmph.

b Find the rate of change in the cost of running the train at speeds of:

i 10 kmph ii 30 kmph.

c At what speed will the cost be a minimum?

5 At a given temperature, the pressure p, and volume v, of gas in a balloon are such that

pv = c where c is a positive constant.

a Find a formula fordp

dv.

b Explain whydp

dv< 0 for all values of v and interpret this result.

EXERCISE 4A

Find the height of the tree at ,

and

t tt

= 4 = 8= 12 years.

H = 20 ¡ 9

t + 5

©iStockPhoto/Nicola Stratford

APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 4) 101


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6 Alongside is a land and sea profile where

the x-axis is sea level and y-values give

the height of the land or sea bed above (or

below) sea level and

y = 110x(x¡ 2)(x¡ 3) km.

a Find where the lake is located relative to the shore line of the sea.

b Finddy

dxand interpret its value when x = 1

2 and when x = 1 12 km.

c Find the deepest point of the lake and the depth at this point.

7 A salt crystal has sides of length x mm and is growing slowly in a salt solution. V is

the volume of the crystal (a cube).

a FinddV

dxand explain what it represents.

b FinddV

dxwhen x = 2. Interpret this result.

c Show thatdV

dx= 1

2A where A is the surface area of the crystal. This means

thatdV

dx/ A. Explain why this result seems reasonable.

8 A tank contains 50 000 litres of water. The tap is left fully on and all the water drains

from the tank in 80 minutes. The volume of water remaining in the tank after t minutes

is given by V = 50000µ1 ¡ t

80

¶2where 0 6 t 6 80.

a FinddV

dtand draw the graph of

dV

dtagainst t.

b At what time was the outflow fastest?

c Findd2V

dt2and interpret the fact that it is always constant and positive.

9

y

x

sea hill lake

A fish farm grows and harvests barramundi in a large

dam. The population P of fish at time t is of interest and

the rate of change in the populationdP

dtis modelled

bydP

dt= aP

µ µ1 ¡ P

b 100

¶ ¶¡ c

P where a, b

and c are known constants. a is the birth rate of the

barramundi, b is the maximum carrying capacity of the

dam and c is the percentage that is harvested.

a Explain why the fish population is stable whendP

dt= 0.

b If the birth rate is 6%, the maximum carrying capacity is 24 000 and the harvest

rate is 5%, find the stable population.

c If the harvest rate changes to 4%, what will the stable population increase to?



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INVESTIGATION 1 AVERAGE AND INSTANTANEOUS VELOCITY

s(t) is a displacement function and for any value of t it gives the displacement from O.

It is clear that if s(t) > 0, P is located to the right of O

if s(t) = 0, P is located at O

if s(t) < 0, P is located to the left of O.

Consider s(t) = t2 + 2t¡ 3 cm, then

s(0) = ¡3 cm, s(1) = 0 cm, s(2) = 5 cm, s(3) = 12 cm, s(4) = 21 cm.

Click on the demo icon to get a better idea of the motion.

Consider the displacement function s(t) = t2 + 2t¡ 3 cm for a particle P

moving along a straight line.

MOTION IN A STRAIGHT LINEB

P

s( )t

originO

MOTION GRAPHS

0 5 10 15 20 25t = 0 t = 1 t = 2 t = 3 t = 4

COMPUTER

DEMO

VELOCITY AND ACCELERATION

AVERAGE VELOCITY

What to do:

1 Find the displacement on the time intervals given in the table and copy

and complete the table.

Suppose an object P moves along a

straight line so that its position from

an origin, is given as some function of

time , i.e., where .

s

t s s t t= ( ) 0>

Fully animated, we not only get a good idea of the position of ut

also of what is happening with regard to velocity and acceleration.

P b

The average velocity of an object moving in a straight line in the time interval from

t = t1 to t = t2, is the ratio of the change in displacement to the time taken,

i.e., average velocity =s(t2)¡ s(t1)

t2 ¡ t1, where s(t) is the displacement function.

To appreciate the motion of we draw a .P motion graph



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Time interval Displacement Time taken Average velocity

t = 1 to t = 4 secs 21 cm 3 sec 7 cm s¡1

t = 1 to t = 3 secs

t = 1 to t = 2 secs

t = 1 to t = 1:1 secs

t = 1 to t = 1:01 secs

t = 1 to t = 1:001 secs

2 Find the average velocity on the time interval

t = 1 to t = 1 + h and then find: limh!0

s(1 + h) ¡ s(1)

h

3 Interpret your result from 2.

Examples like the one above lead us to conclude that:

If s(t) is a displacement function of an object moving in a straight line, then

v(t) = s0(t) = limh!0

s(t + h) ¡ s(t)

his the instantaneous velocity of the object

at time t.

A particle moves in a straight line with displacement from O given by

s(t) = 3t¡ t2 metres at time t seconds. Find:

a the average velocity in the time interval from t = 2 to t = 5 seconds

b the average velocity in the time interval from t = 2 to t = 2 + h seconds

c limh!0

s(2 + h) ¡ s(2)

hand comment on its significance.

a average velocity

=s(5) ¡ s(2)

5 ¡ 2ms¡1

=(15 ¡ 25) ¡ (6 ¡ 4)

3ms¡1

=¡10 ¡ 2

3ms¡1

= ¡4 ms¡1

b average velocity

=s(2 + h) ¡ s(2)

2 + h¡ 2

=3(2 + h) ¡ (2 + h)2 ¡ 2

h

=6 + 3h¡ 4 ¡ 4h¡ h2 ¡ 2

h

=¡h¡ h2

h

= ¡1 ¡ h ms¡1 as h 6= 0c limh!0

s(2 + h) ¡ s(2)

h= lim

h!0(¡1 ¡ h)

= ¡1 ms¡1

and this is the instantaneous velocity at time t = 2 seconds.

INSTANTANEOUS VELOCITY

Example 2



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1 A particle P moves in a straight line with a displacement function of s(t) = t2 + 3t¡ 2metres, where t > 0, t in seconds.

a Find the average velocity from t = 1 to t = 3 seconds.

b Find the average velocity from t = 1 to t = 1 + h seconds.

c Find the value of limh!0

s(1 + h) ¡ s(1)

hand comment on its significance.

d Find the average velocity from time t to time t + h seconds and interpret

limh!0

s(t + h) ¡ s(t)

h.

2 A particle P moves in a straight line with a displacement function of

s(t) = 5 ¡ 2t2 cm, where t > 0, t in seconds.

a Find the average velocity from t = 2 to t = 5 seconds.

b Find the average velocity from t = 2 to t = 2 + h seconds.


s(2 + h) ¡ s(2)

hand state the meaning of this value.

d Interpret limh!0

s(t + h) ¡ s(t)

h:

If a particle moves in a straight line with velocity function v(t), then the

3 A particle moves in a straight line with velocity function v(t) = 2pt + 3 cm s¡1,

where t > 0.

a Find the average acceleration from t = 1 to t = 4 seconds.

b Find the average acceleration from t = 1 to t = 1 + h seconds.


v(1 + h) ¡ v(1)

h. Interpret this value.

d Interpret limh!0

v(t + h) ¡ v(t)

h:

EXERCISE 4B.1

AVERAGE ACCELERATION

INSTANTANEOUS ACCELERATION

If an object moves in a straight line with velocity function then its

on the time interval from to is the ratio of its

to the time taken,

v tt t t t

( )= =

average

acceleration 1 2

change in velocity

i.e., average acceleration =v(t2) ¡ v(t1)

t2 ¡ t1.

the instantaneous acceleration at time t is a(t) = v0(t) = limh!0

v(t+ h) ¡ v(t)

h



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4 An object moves in a straight line with displacement function s(t), and velocity func-

tion v(t), t > 0. State the meaning of:

a limh!0

s(3 + h) ¡ s(3)

hb lim

h!0

v(5 + h) ¡ v(5)

h

c limh!0

s(t + h) ¡ s(t)

hd lim

h!0

v(t + h) ¡ v(t)

h

If a particle P, moves in a straight line and its position is given by the displacement func-

tion s(t), t > 0, then:

² the velocity of P, at time t, is given by

v(t) = s0(t) fthe derivative of the displacement functiong

² the acceleration of P, at time t, is given by

a(t) = v0(t) = s00(t) fthe derivative of the velocity functiong

We can use sign diagrams to interpret:

² where the particle is located relative to O

² the direction of motion and where a change of direction occurs

² when the particle’s velocity is increasing/decreasing.

SIGNS OF s(t):

s(t) Interpretation

= 0 P is at O

> 0 P is located to the right of O

< 0 P is located to the left of O

SIGNS OF v(t):

v(t) Interpretation

= 0 P is instantaneously at rest

> 0 P is moving to the right

< 0 P is moving to the left

Note: v(t) = limh!0

s(t + h) ¡ s(t)

h. If h > 0, so that t + h > t,

then v(t) > 0 implies that s(t + h) ¡ s(t) > 0 ) s(t + h) > s(t)

i.e.,

) P is moving to the right.

VELOCITY AND ACCELERATION FUNCTIONS

Note:

initial conditions

, and give us the position, velocity and acceleration of the particle at

time , and these are called the .

s v at

(0) (0) (0)= 0

SIGN INTERPRETATION

s( )t s +( )t h

Suppose a particle P moves in a straight line, with displacement function s(t) for locating the

particle relative to an origin O, and velocity function v(t) and acceleration function a(t).



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SIGNS OF a(t):a(t) Interpretation

> 0 velocity is increasing

< 0 velocity is decreasing

= 0 velocity may be a maximum or minimum

A useful table:Phrase used in a question t s v a

initial conditions 0

at the origin 0

stationary 0

reverses 0

maximum height 0

constant velocity 0

max. / min. velocity 0

As we have seen, velocities have size (magnitude) and sign (direction). The speed of a particle

is a measure of how fast it is travelling regardless of the direction of travel.

Thus the speed at any instant is the modulus of the particle’s velocity,

i.e., if S represents speed then S = jvj :The question arises: “How can we determine when the speed of a particle is increasing

or decreasing?”

² If the signs of v(t) and a(t) are the same, (i.e., both positive or both negative),

then the speed of P is increasing.

² If the signs of v(t) and a(t) are opposite, then the speed of P is decreasing.

We prove the first of these as follows:

Proof: Let S = jvj, be the speed of P at any instant

) S =

½v if v > 0

¡v if v < 0 fdefinition of modulusg

Case 1: If v > 0, S = v and )dS

dt=

dv

dt= a(t)

and if a(t) > 0,dS

dt> 0 which implies that S is increasing.

Case 2: If v < 0, S = ¡v and )dS

dt= ¡dv

dt= ¡a(t)

and if a(t) < 0,dS

dt> 0 which also implies that S is increasing.

Thus if v(t) and a(t) have the same sign, the speed of P is increasing.

SPEED

We employ a . This is:sign test



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INVESTIGATION 2 DISPLACEMENT VELOCITY

AND ACCELERATION GRAPHS

1 Click on the icon to see vertical projectile motion in a straight line. Observe first the

displacement along the line, then look at the velocity or rate of change in displacement.

2 Examine the three graphs ² displacement v time ² velocity v time

² acceleration v time

3 Pick from the menu or construct functions of your own choosing to investigate their

displacement, velocity and acceleration functions.

You are encouraged to use the motion demo in the questions of the following exercise.

A particle moves in a straight line with position, relative to some origin O, given by

s(t) = t3 ¡ 3t + 1 cm, where t is the time in seconds (t > 0).

a Find expressions for the particle’s velocity and acceleration, and draw sign

diagrams for each of them.

b Find the initial conditions and hence describe the motion at this instant.

c Describe the motion of the particle at t = 2 seconds.

d Find the position of the particle when changes in direction occur.

e Draw a motion diagram for the particle.

f For what time interval(s) is the particle’s speed increasing?

a Since s(t) = t3 ¡ 3t + 1 cm

) v(t) = 3t2 ¡ 3 cm s¡1 fas v(t) = s0(t)g= 3(t2 ¡ 1)

= 3(t + 1)(t¡ 1) which has sign diagram

Also a(t) = 6t cm s¡2 fas a(t) = v0(t)gwhich has sign diagram

b When t = 0, s(0) = 1 cm

v(0) = ¡3 cm s¡1

a(0) = 0 cm s¡2

) particle is 1 cm to the right of O, moving to the left at a speed of 3 cm s¡1.

fspeed = jvjgc When t = 2, s(2) = 8 ¡ 6 + 1 = 3 cm

v(2) = 12 ¡ 3 = 9 cm s¡1

a(2) = 12 cm s¡2

) particle is 3 cm right of O, moving to the right at a speed of 9 cm s¡1 and

the speed is increasing. fas a and v have the same signg

Note: t > 0) critical value

t = ¡1 is not

required.

In this investigation we examine the motion of a projectile

which is fired in a vertical direction under gravity. Other

functions of a different kind will be examined.What to do:

Example 3

t1

+

0

t+

0

COMPUTER

DEMO



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d Since v(t) changes sign when t = 1, a change of direction occurs at this

instant and s(1) = 1 ¡ 3 + 1 = ¡1) changes direction when t = 1 and is 1 cm left of O.

e

as t ! 1, s(t) ! 1 and v(t) ! 1f Speed is increasing when v(t) and a(t) have the same sign i.e., t > 1.

1 An object moves in a straight line with position given by s(t) = t2 ¡ 4t + 3 cm from

an origin O, t > 0, t in seconds.

a Find expressions for its velocity and acceleration at any instant and draw sign

diagrams for each function.

b Find the initial conditions and explain what is happening to the object at that instant.

c Describe the motion of the object at time t = 2 seconds.

d At what time(s) does the object reverse direction? Find the position of the object

at these instants.

e Draw a motion diagram of the object.

f For what time intervals is the speed of the object decreasing?

2 A stone is projected vertically upwards so that its position above ground

level after t seconds is given by s(t) = 98t¡ 4:9t2 metres, t > 0.

a Find the velocity and acceleration functions for the stone and draw

sign diagrams for each function.

b Find the initial position and velocity of the stone.

c Describe the stone’s motion at times t = 5 and t = 12 seconds.

d Find the maximum height reached by the stone.

e Find the time taken for the stone to hit the ground.

3 A particle moves in a straight line with displacement function

s(t) = 12t¡ 2t3 ¡ 1 centimetres, t > 0, t in seconds.

a Find velocity and acceleration functions for the particle’s motion.

b Find the initial conditions and interpret their meaning.

c Find the times and positions when the particle reverses direction.

d At what times is the particle’s: i speed increasing ii velocity increasing?

4 The position of a particle moving along the x-axis is given by x(t) = t3 ¡ 9t2 + 24tmetres, t > 0, t in seconds.

a Draw sign diagrams for the particle’s velocity and acceleration functions.

b Find the position of the particle at the times when it reverses direction, and hence

draw a motion diagram for the particle.

c At what times is the particle’s: i speed decreasing ii velocity decreasing?

d Find the total distance travelled by the particle in the first 5 seconds of motion.

Note:

on the

line

The motion

is actually

, not above it

as shown.0 1 1

EXERCISE 4B.2

s m( )t



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5 An experiment to determine the position of an object fired vertically upwards from the

earth’s surface was performed. From the results, a two dimensional graph of position

above the earth’s surface s(t) metres, against time t seconds, was plotted.

It was noted that the graph was parabolic.

Assuming a constant gravitational acceleration

g, show that if the initial velocity is v(0) then

a v(t) = v(0) + gt, and

b s(t) = v(0) £ t + 12gt

2.

(Hint: Assume s(t) = at2 + bt + c.)

6 For a particle moving in a straight line, prove that if the signs of v(t) and a(t) are

opposite at some time t, then the speed of the particle is decreasing.

7 Draw motion diagrams for a particle moving in a straight line with displacement functions

of:

a s(t) = (t¡1)3 +2 cm b s(t) = 4 ¡pt + 1 cm c s(t) = 40t +

10pt + 1

cm.

In each case find the initial conditions and discuss what happens to the particle as

t ! 1:

Recall that f 0(x) ordy

dxis the slope function of a curve. The derivative of a function is

For example, if f(x) =px then

f(x) = x1

2 and

f 0(x) = 12x

¡ 1

2 =1

2px

Substituting x = 14 , 1

2 , 1 and 4 gives:

f 0 ¡14

¢= 1, f 0 ¡1

2

¢= 1p

2, f 0(1) = 1

2 , f 0(4) = 14

i.e., the slopes are 1, 1p2

, 12 and 1

4 respectively.

Notice also that a tangent to the graph at any point, provided that x > 0, has a positive

slope.

This fact is also observed from f 0(x) =1

2px

aspx is never negative and x > 0.

Many functions are increasing for all x whereas others are decreasing for all x.

s( )t

t

CURVE PROPERTIESC

MONOTONICITY

1

1

y

x

m��Qw_

(1, 1)

another function which enables us to find the slope of a tangent to the curve at any point

on it.



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For example,

y = 2x is increasing for all x. y = 3¡x is decreasing for all x.

Notice that:

The majority of other functions have intervals where the function is increasing and intervals

where it is decreasing.

For example: y = x2

increasing for x > 0.

Note: x 6 0 is an interval of x values.

So is x > 0.

Some examples of intervals and their graphical representations are:

Algebraic form Means Alternative notation

x > 4 [4, 1)

x > 4 (4, 1)

x 6 2 (¡1, 2]

x < 2 (¡1, 2)

2 6 x 6 4 [2, 4]

2 6 x < 4 [2, 4)

�

y = 2xy

x

�

y

x

y = 3 x

increase in y

increase in x increase in x

decrease in y

y

x

@=!X

INTERVALS

4

4

2

2

2

2

4

4

is decreasing for andx 6 0

²²

for an an increase in produces an increase inincreasing function x y

for a an increase in produces a decrease in .decreasing function x y



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INCREASING / DECREASING INTERVALS

Definition

If S is an interval of real numbers and f(x) is defined for all x in S, then:

² f(x) is increasing on S , f 0(x) > 0 for all x in S, and

² f(x) is decreasing on S , f 0(x) 6 0 for all x in S.

Note: , is read as ‘if and only if’

1 Find intervals where f(x) is i increasing ii decreasing:

a b c

d e f

g h i

Sign diagrams for the derivative are extremely useful for determining intervals where a

function is increasing/decreasing.

yy = 3

x

y

x

( )��

( )��

y

x( )��

y

x

x = 4

( )��

(1 )��

y

x

(2 )��y

x

3

y

x

y

x

� ��

��

y

x � �

Find intervals where f(x) is:

a increasing

b decreasing.

a f(x) is increasing for x 6 ¡1 and for x > 2.

fsince tangents have slopes > 0 on these intervalsgb f(x) is decreasing for ¡1 6 x 6 2.

y x= ƒ( )( ) 1, 3

(2 ), 4

y

x

Example 4

EXERCISE 4C.1



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Consider the following examples:

² f(x) = x2 f 0(x) = 2x

which has sign diagram

So f(x) = x2 is decreasing for x 6 0

and increasing for x > 0.

² f(x) = ¡x2 f 0(x) = ¡2x


² f(x) = x3 f 0(x) = 3x2


² f(x) = x3 ¡ 3x + 4 f 0(x) = 3x2 ¡ 3

= 3(x2 ¡ 1)

= 3(x + 1)(x¡ 1)


y

x

y

x

y

x

� �

y

x

4

0

�

decreasing increasing

0

�

increasing decreasing

� �

� �

increasing increasingdecreasing

0

� �

increasing for all x

(never negative)

Find the intervals where the following functions are increasing/decreasing:

a f(x) = ¡x3 + 3x2 + 5 b f(x) = 3x4 ¡ 8x3 + 2

a f(x) = ¡x3 + 3x2 + 5

) f 0(x) = ¡3x2 + 6x

) f 0(x) = ¡3x(x¡ 2)


So, f(x) is decreasing for

x 6 0 and for x > 2 and

is increasing for 0 6 x 6 2.

0 2

�

DEMO

DEMO

DEMO

DEMO

Example 5



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1 Find intervals where f(x) is increasing/decreasing:

a f(x) = x2 b f(x) = ¡x3

c f(x) = 2x2 + 3x¡ 4 d f(x) =px

e f(x) =2px

f f(x) = x3 ¡ 6x2

g f(x) = ¡2x3 + 4x h f(x) = ¡4x3 + 15x2 + 18x + 3

i f(x) = 3x4 ¡ 16x3 + 24x2 ¡ 2 j f(x) = 2x3 + 9x2 + 6x¡ 7

k f(x) = x3 ¡ 6x2 + 3x¡ 1 l f(x) = x¡ 2px

Find intervals where f(x) = x4 ¡ 4x3 + 2x2 + 8x + 3 is increasing/decreasing.

f(x) = x4 ¡ 4x3 + 2x2 + 8x + 3

) f 0(x) = 4x3 ¡ 12x2 + 4x + 8

= 4(x3 ¡ 3x2 + x + 2)

So f 0(x) = 4(x¡ 2)(x2 + ax¡ 1) Coefficient of x2: ¡2 + a = ¡3

) a = ¡1

Coefficient of x: ¡2a¡ 1 = 1

) ¡2a = 2

) a = ¡1 X) f 0(x) = 4(x¡ 2)(x2 ¡ x¡ 1)

f 0(x) = 0 when x = 2 or1 §p1 ¡ 4(1)(¡1)

2

i.e., x = 2 or1 §p

5

2

b f(x) = 3x4 ¡ 8x3 + 2

) f 0(x) = 12x3 ¡ 24x2

= 12x2(x¡ 2)


So, f(x) is decreasing for

x 6 2 and is increasing

for x > 2.

0 2

�

EXERCISE 4C.2

Example 6



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2 Find intervals where y = f(x) is increasing/decreasing if:

a y = 3x4 ¡ 8x3 ¡ 6x2 + 24x + 11 b y = x4 ¡ 4x3 + 2x2 + 4x + 1

c y = ¡3x4 + 28x3 ¡ 84x2 + 72x¡ 7

Sign diagram is:

) f(x) is decreasing for x 6 1¡p5

2 and for 26 x 61+p5

2

and is increasing for 1¡p5

2 6 x 6 2and for x >1+p5

2 .

1 1¡ +p p5 5

2 22

� �

Consider f(x) =3x¡ 9

x2 ¡ x¡ 2.

a Show that f 0(x) =¡3(x¡ 5)(x¡ 1)

(x¡ 2)2(x + 1)2and draw its sign diagram.

b Hence, find intervals where y = f(x) is increasing/decreasing.

a f(x) =3x¡ 9

x2 ¡ x¡ 2

f 0(x) =3(x2 ¡ x¡ 2) ¡ (3x¡ 9)(2x¡ 1)

(x¡ 2)2(x + 1)2fquotient ruleg

=3x2 ¡ 3x¡ 6 ¡ [6x2 ¡ 21x + 9]

(x¡ 2)2(x + 1)2

=¡3x2 + 18x¡ 15

(x¡ 2)2(x + 1)2

=¡3(x2 ¡ 6x + 5)

(x¡ 2)2(x + 1)2

=¡3(x¡ 5)(x¡ 1)

(x¡ 2)2(x + 1)2which has sign diagram

b f(x) is increasing for 1 < x 6 2 and for 2 < x 6 5f(x) is decreasing for x < ¡1 and for ¡1 < x 6 1 and for x > 5.

Note: A screen dump of y = f(x) is:

- - + + -

-1 21 5

Example 7



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3 a Consider f(x) =4x

x2 + 1:

i Show that f 0(x) =¡4(x + 1)(x¡ 1)

(x2 + 1)2and draw its sign diagram.

ii Hence, find intervals where y = f(x) is increasing/decreasing.

b Consider f(x) =4x

(x¡ 1)2.

i Show that f 0(x) =¡4(x + 1)

(x¡ 1)3and draw its sign diagram.


c Consider f(x) =¡x2 + 4x¡ 7

x¡ 1.

i Show that f 0(x) =¡(x + 1)(x¡ 3)

(x¡ 1)2and draw its sign diagram.


d Consider f(x) =x2 ¡ 3x + 2

x2 + 3x + 2.

i Show that f 0(x) =6(x +

p2)(x¡p

2)

(x + 1)2(x + 2)2and draw its sign diagram.


4 Find intervals where f(x) is increasing/decreasing if:

a f(x) =x3

x2 ¡ 1b f(x) = x2 +

4

x¡ 1

MAXIMA/MINIMA

Consider the following graph which has a restricted domain of ¡5 6 x 6 6:

A is a global minimum as it is the minimum value of y and occurs at an endpoint of the

domain.

B is a local maximum as it is a turning point where the curve has shape and f 0(x) = 0at that point.

C is a local minimum as it is a turning point where the curve has shape and f 0(x) = 0at that point.

y

x

D(6, 18)

A( , ) � ��\Qw_\

B( , 4) �

C( , 4)�

2 �

y x�ƒ( )



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D is a global maximum as it is the maximum value of y and occurs at the endpoint of the

domain.

HORIZONTAL INFLECTIONS

It is not true that whenever we find a value of x where f 0(x) = 0 we have a local

maximum or minimum.

For example, f(x) = x3 has f 0(x) = 3x2

and f 0(x) = 0 when x = 0:

Notice that the x-axis is a tangent to the curve which

actually crosses over the curve at O(0, 0).

This tangent is horizontal and O(0, 0) is not a local

maximum or minimum.

It is called a horizontal inflection (or inflexion).

STATIONARY POINTS

A stationary point is a point where f 0(x) = 0. It could be a local maximum,

local minimum or a horizontal inflection.

Consider the following graph:

Its slope sign diagram is:

Summary:

y

x

��

y

x

horizontalinflection

local maximum

local minimum

��

� � �

localmaximum

localminimum


1

Note: horizontalFor local maxima/minima, tangents at these points are and thus have a

slope of , i.e., .��0 ( )=0f x\0

Stationary point Sign diagram of f 0(x) Shape of curve near x = anear x = a

local maximum

local minimum

horizontal inflection

a�

a �

a a� �

or

x = a

x = a

x = a x = aor



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Find and classify all stationary points of f(x) = x4 ¡ 4x3 ¡ 2.

f(x) = x4 ¡ 4x3 ¡ 2

) f 0(x) = 4x3 ¡ 12x2

= 4x2(x¡ 3), which has sign diagram:

So, we have a horizontal inflection at x = 0 and a local minimum at x = 3.

As f(0) = ¡2, the horizontal inflection is at (0, ¡2).

As f(3) = ¡29, the local minimum is at (3, ¡29).

��

Find and classify all stationary points of f(x) = x3 ¡ 3x2 ¡ 9x + 5:

f(x) = x3 ¡ 3x2 ¡ 9x + 5

) f 0(x) = 3x2 ¡ 6x¡ 9

= 3(x2 ¡ 2x¡ 3)

= 3(x¡ 3)(x + 1), which has sign diagram:

So, we have a local maximum at x = ¡1 and a local minimum at x = 3.

f(¡1) = (¡1)3 ¡ 3(¡1)2 ¡ 9(¡1) + 5

= ¡1 ¡ 3 + 9 + 5

= 10 ) local maximum at (¡1, 10)

f(3) = 33 ¡ 3 £ 32 ¡ 9 £ 3 + 5

= 27 ¡ 27 ¡ 27 + 5

= ¡22 ) local minimum at (3, ¡22)

��

��

Example 8

Example 9



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1 A, B and C are points where tangents are

horizontal.

a Classify points A, B and C.

b Draw a sign diagram for the slope

of f(x) for all x.

c State intervals where y = f(x) is:

i increasing

ii decreasing.

2 For each of the following functions, find and classify the stationary points and hence

sketch the function showing all important features.

a f(x) = x2 ¡ 2 b f(x) = x3 + 1

c f(x) = x3 ¡ 3x + 2 d f(x) = x4 ¡ 2x2

e f(x) = x3 ¡ 6x2 + 12x + 1 f f(x) =px + 2

g f(x) = x¡px h f(x) = x4 ¡ 6x2 + 8x¡ 3

i f(x) = 1 ¡ xpx j f(x) = x4 ¡ 2x2 ¡ 8

3 At what value of x does the quadratic function, f(x) = ax2 + bx + c, a 6= 0, have

a stationary point? Under what conditions is the stationary point a local maximum or a

local minimum?

4 f(x) = x3 + ax + b has a stationary point at (¡2, 3).

a Find the values of a and b.

b Find the position and nature of all stationary points.

5 A cubic polynomial P (x), touches the line with equation y = 9x+ 2 at the point

(0, 2) and has a stationary point at (¡1, ¡7). Find P (x).

Rational functions are functions of the form f(x) =g(x)

h(x)where g(x) and h(x) are

polynomials. One feature of a rational function is the presence of asymptotes.

Vertical asymptotes can be found by letting h(x) equal 0. Stationary points can be found

by letting f 0(x) equal 0.

yy ƒ x= ( )

x

C

B 2, 8( )

A 3, 11( )

RATIONAL FUNCTIONS

Consider f(x) =x2 ¡ 5x + 4

x2 + 5x + 4.

a Use technology to obtain a graph of the function.

b What are the function’s vertical asymptotes?

c What are the stationary points of the function?

EXERCISE 4C.3

Example 10



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6 For the following find i vertical asymptotes ii stationary points:

a f(x) =4x

x2 ¡ 4x¡ 5b f(x) =

3x¡ 3

(x + 2)2c f(x) =

x2 + 4

x¡ 2

d f(x) =x2 ¡ 3x¡ 5

x + 3e f(x) =

x2 ¡ x

x2 ¡ x¡ 6f f(x) =

3x2 ¡ x + 2

(x + 2)2

Use technology to graph and check for stationary points.

a b When x2 + 5x + 4 = 0, then

(x + 1)(x + 4) = 0

) x = ¡1 or ¡4

so, x = ¡1 and x = ¡4are vertical asymptotes.

c f 0(x) =(2x¡ 5)(x2 + 5x + 4) ¡ (x2 ¡ 5x + 4)(2x + 5)

(x2 + 5x + 4)2

=2x3 + 5x2 ¡ 17x¡ 20 ¡ [2x3 ¡ 5x2 ¡ 17x + 20]

(x + 1)2(x + 4)2

=10x2 ¡ 40

(x + 1)2(x + 4)2

=10(x + 2)(x¡ 2)

(x + 1)2(x + 4)2

f(¡2) = ¡9 and f(2) = ¡19

) we have a local maximum at (¡2, ¡9) and a local minimum at (2, ¡19 )

Do not forget to use technology to check these stationary points.

+++ --

� �

� �

Find the greatest and least value of x3 ¡ 6x2 + 5 on the interval ¡2 6 x 6 5.

First we graph y = x3 ¡ 6x2 + 5 on [¡2, 5].

The greatest value is clearly whendy

dx= 0

Nowdy

dx= 3x2 ¡ 12x

= 3x(x¡ 4)

= 0 when x = 0 or 4

So, the greatest value is f(0) = 5 when x = 0.

The least value is either f(¡2) or f(4), whichever is smaller.

Now f(¡2) = ¡27 and f(4) = ¡27

) least value is ¡27 when x = ¡2 and x = 4.

Example 11



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tangent slope = 0

point of inflection

y x= ƒ( )

point of inflection

y x= ƒ( )tangent

7 Find the greatest and least value of:

a x3 ¡ 12x¡ 2 for ¡3 6 x 6 5 b 4 ¡ 3x2 + x3 for ¡2 6 x 6 3

c x¡px for i 0 6 x 6 4 ii 1 6 x 6 9.

8 A manufacturing company makes door hinges. The cost function for making x hinges per

hour is C(x) = 0:0007x3¡0:1796x2+14:663x+160 dollars where 50 6 x 6 150.

The condition 50 6 x 6 150 applies as the company has a standing order filled by

producing 50 each hour, but knows that production of more than 150 an hour is useless

as they will not sell.

Find the minimum and maximum hourly costs and the production levels when each

occurs.

INFLECTIONS AND SHAPE TYPE

When a curve, or part of a curve, has shape we say that the shape is concave

(or concave downwards).

If a curve, or part of a curve, has shape we say that the shape is convex (or

concave upwards).

POINTS OF INFLECTION (INFLEXION)

A point of inflection is a point on a curve at which a change of shape occurs.

i.e.,

point ofinflection

point ofinflection

or

Notes: ² If the tangent at a point of inflection

is horizontal we say that we have a

horizontal stationary inflectionor .

For example,

²

For example,

² Notice that the tangent at the point of inflection (the inflecting tangent)

crosses the curve at that point.

² A Venn diagram summary of inflection points:

If the tangent at point of inflection is

not horizontal we say that we have

or

.

a

a

non-horizontal non-stationary in-

flection

DEMO

inflections( ) = 0f'' x

stationary points( ) = 0f' x stationary inflections



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Consider the concave curve:

Notice that as x increases for all

points on the curve the slope is

decreasing,

i.e., f 0(x) is decreasing,

) its derivative is negative,

i.e., f 00(x) < 0.

Likewise, if the curve is convex:

As the values of x increase for all

points on the curve the slope is

increasing,

i.e., f 0(x) is increasing,

) its derivative is positive,

i.e., f 00(x) > 0.

Consequently,

Observe that if f(x) = x4 then f 0(x) = 4x3 and f 00(x) = 12x2 and f 00(x) has sign

diagram

Although f 00(0) = 0 we do not have a point of inflection

at (0, 0) since the sign of f 00(x) does not change on

either side of x = 0. In fact the graph of f(x) = x4 is:

Summary:

m = 0

m = ��m = �

m = ��m = �

m = 0

m = �� m = �

m = �� m = �

y x�� 2

y x� 2

we have a point of inflection at x = a if f 00(a) = 0 and the sign of f 00(x) changes

on either side of x = a,

i.e., f 00(x) has sign diagram, in the vicinity of a, of either a a� �� or

0

� �

For a curve (or part curve) which is concave in

an interval S, for all x in S.

For a curve (or part curve) which is convex in an

interval S, for all x in S.

If f 00(x) has a sign change on either side of x = a, and f 00(a) = 0, then

² we have a horizontal inflection if f 0(a) = 0 also,

² we have a non-horizontal inflection if f 0(a) 6= 0.

concave

convex

( )�� ( )��

y

xlocal

minimum 0, )( 0

y x� 4

TEST FOR SHAPE


f 00(x) 6 0

f 00(x) > 0


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1 Find and classify, if they exist, all points of inflection of:

a f(x) = x2 + 3 b f(x) = 2 ¡ x3

c f(x) = x3 ¡ 6x2 + 9x + 1 d f(x) = x3 + 6x2 + 12x + 5

e f(x) = ¡3x4 ¡ 8x3 + 2 f f(x) = 3 ¡ 1px

Click on the demo icon to examine some standard functions for turning points,

points of inflection and intervals where the function is increasing, decreasing,

convex and concave.

For f(x) = x4 + 4x3 ¡ 16x + 3:

a find and classify all points where f 0(x) = 0

b find and classify all points of inflection

c find intervals where the function is increasing/decreasing

d find intervals where the function is convex/concave.

e Hence, sketch the graph showing all important features.

f(x) = x4 + 4x3 ¡ 16x + 3

a ) f 0(x) = 4x3 + 12x2 ¡ 16

= 4(x3 + 3x2 ¡ 4)

= 4(x ¡ 1)(x2 + 4x + 4)

= 4(x ¡ 1)(x + 2)2


) (¡2, 19) is a horizontal inflection ff(¡2) = 16 ¡ 32 + 32 + 3 = 19gand (1, ¡8) is a local minimum ff(1) = 1 + 4 ¡ 16 + 3 = ¡8g

��

��

�

Find and classify all points of inflection of f(x) = x4 ¡ 4x3 + 5.

f(x) = x4 ¡ 4x3 + 5

) f 0(x) = 4x3 ¡ 12x2

) f 00(x) = 12x2 ¡ 24x

= 12x(x ¡ 2) which has sign diagram

f 00(x) = 0 when x = 0 or 2

and since the signs of f 00(x) change about x = 0 and x = 2, we have

points of inflection at these two points.

Also f 0(0) = 0 and f 0(2) = 32 ¡ 48 6= 0

and f(0) = 5, f(2) = 16 ¡ 32 + 5 = ¡11

Thus (0, 5) is a horizontal inflection and (2, ¡11) is a non-horizontal inflection.

� �

0 2

DEMO

Example 12

EXERCISE 4C.4

Example 13



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Y:\HAESE\SA_12STU-2ed\SA12STU-2_04\123SA12STU-2_04.CDR Thursday, 9 November 2006 10:02:09 AM DAVID3

2 For each of the following functions:

i find and classify all points where f 0(x) = 0ii find and classify all points of inflection

iii find intervals where the function is increasing/decreasing

iv find intervals where the function is convex/concave.

v

a f(x) = x2 b f(x) = x3

c f(x) =px d f(x) = x3 ¡ 3x2 ¡ 24x + 1

e f(x) = 3x4 + 4x3 ¡ 2 f f(x) = x4 ¡ 6x2 + 8x + 1

g f(x) = x4 ¡ 4x2 + 3 h f(x) = 3 ¡ 4px

3 a Show that the x-coordinate of the point of inflection of a cubic polynomial is the

average of its 3 zeros.

(Hint: Consider f(x) = a(x¡ ®)(x¡ ¯)(x¡ °).)

b Show that, if f(x) = ax3 + bx2 + cx + d has two distinct turning points at

x = p and x = q then:

i b2 > 3ac ii the point of inflection occurs at x =p + q

2.

Consider the following problem.

OPTIMISATIOND

b f 00(x) = 12x2 + 24x

= 12x(x + 2) which has sign diagram:

) (¡2, 19) is a horizontal inflection falready discovered in agand (0, 3) is a non-horizontal inflection.

c f(x) is decreasing for x 6 1and increasing for x > 1.

d f(x) is concave for ¡2 6 x 6 0and convex for x 6 ¡2 and for x > 0.

ey

x

��

��

��

stationary inflection

non-stationary inflection

(-2' 19)

(0' 3)

local minimum(1' -8)

��

��

�

Sketch the graph showing important features.all

Many problems where we try to find the or value of variable can be

solved using differential calculus techniques. The solution is often referred to as the

solution.

maximum minimum

optimum

a



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An industrial shed is to have a total floor space of

600 m2 and is to be divided into 3 rectangular rooms

of equal size. The walls, internal and external, will

cost $60 per metre to build.

What dimensions should the shed have to minimise

the cost of the walls? We let one room be x m by ym as shown.

The total length of wall material is L where

L = 6x + 4y metres.

However we do know that the total area is 600 m2,

) 3x£ y = 600

and so y =200

xNote: x > 0 and y > 0

Knowing this relationship enables us to write L in terms of one variable (x in this case),

i.e., L = 6x + 4

µ200

x

¶m, i.e., L =

µ6x +

800

x

¶m

and at $60/metre, the total cost is C(x) = 60

µ6x +

800

x

¶dollars.

Clearly, C(x) is a minimum when C 0(x) = 0.

Now C(x) = 360x + 48000x¡1

) C0(x) = 360 ¡ 48 000x¡2

) C0(x) = 0 when 360 =48000

x2

i.e., x2 =48000

360+ 133:333

i.e., x + 11:547

Now when x + 11:547, y +200

11:547+ 17:321

and the minimum cost is

C(11:547) + 8313:84 dollars.

So, the floor design is:

and the minimum cost is $8313:84.

y m

x m

17.321 m

11.547 m

WARNING

The maximum/minimum value does not always occur when the first derivative is zero.

It is essential to also examine the values of the function at the end point(s) of the domainfor global maxima/minima, i.e., given you should also consider anda x b f a f b6 6 , .( ) ( )

The graph of C(x) is shown alongside.



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In the illustrated example,

the maximum value of y occurs at

x = b and the minimum value of yoccurs at x = p.

If one is trying to optimise a function f(x) and we find values of x such that f 0(x) = 0,

how do we know whether we have a maximum or a minimum solution? The following are

acceptable evidence.

If near x = a where f 0(a) = 0 the sign diagram is:

² we have a local maximum ² we have a local minimum.

If near x = a where f 0(a) = 0 and:

² d2y

dx2< 0 we have shape, i.e., a local maximum

² d2y

dx2> 0 we have shape, i.e., a local minimum.

If we have a graph of y = f(x) showing we have a local maximum and we

have a local minimum.

The following steps should be followed:

TESTING OPTIMAL SOLUTIONS

OPTIMISATION PROBLEM SOLVING METHOD

GRAPHICAL TEST

SECOND DERIVATIVE TEST

SIGN DIAGRAM TEST

dy

dx= 0

dy

dx= 0

x a= x p= x b=

y x= ƒ( )

a�

a�

Step 1:

Step 2:

Step 3:

Step 4:

Draw large, clear diagram of the situation. Sometimes more than one diagram is

necessary

Construct an equation with the variable to be or

as the subject of the formula in terms of convenient

say Also find what restrictions there may be on

Find the and find the value(s) of when it is

If there is restricted domain such as the maximum/minimum value of

the function may occur either when the derivative is zero or at or at

Show by the or the that you have

maximum or minimum situation.

a

.

(

) ,

. .

.

a ,

.

a

a

optimised maximised

minimised one variable

first derivative zer

sign diagram test second derivative test

xx

x

a x bx a x b

o

6 6

= =



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To illustrate the method we consider the following example.

A rectangular cake dish is made

by cutting out squares from the

corners of a 25 cm by 40 cm rect-

angle of tin-plate and folding the

metal to form the container.

What size squares must be cut out in order to produce the cake dish

of maximum volume?

Step 1:

Step 2: Now volume = length £ width £ depth

= (40 ¡ 2x)(25¡ 2x)x cm3

i.e., V = (40 ¡ 2x)(25x¡ 2x2) cm3

Notice that x > 0 and 25 ¡ 2x > 0

) x < 12:5

) 0 < x < 12:5

Step 3: NowdV

dx= ¡2(25x¡ 2x2) + (40¡ 2x)(25 ¡ 4x) fproduct ruleg= ¡50x + 4x2 + 1000 ¡ 50x¡ 160x + 8x2

= 12x2 ¡ 260x + 1000

= 4(3x2 ¡ 65x+ 250)

= 4(3x¡ 50)(x¡ 5)

)dV

dx= 0 when x = 50

3 = 1623 or x = 5

Step 4: Sign diagram test

dV

dxhas sign diagram:

or Second derivative test

d2V

dx2= 24x¡ 260 and at x = 5,

d2V

dx2= ¡140 which is < 0

) shape is and we have a local maximum.

So, the maximum volume is obtained when x = 5,

i.e., 5 cm squares are cut from the corners.

(40 2 ) cm x

(25 2 ) cm x

x cm

Let cm be the lengths of the sides

of the squares cut out.

x

�

��

�� We_�

� �

Example 14

DEMO



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Step 1:

Let OB = x cm

Step 2: In ¢OBC, (BC)2 + x2 = 102 fPythagorasg) BC =

p100 ¡ x2 as BC>0.

A = length £ width

) A = 2x£p100 ¡ x2

Step 3: As A = 2x(100¡ x2)1

2

dA

dx= 2(100¡ x2)

1

2 + 2x£ 12(100¡ x2)¡

1

2 £ (¡2x) fproduct ruleg

=2p

100 ¡ x2

1¡ 2x2

p100 ¡ x2

fon simplifyingg

=2(100¡ x2) ¡ 2x2

p100 ¡ x2

=200 ¡ 4x2p

100 ¡ x2

=4(50¡ x2)p

100 ¡ x2

=4(p

50 + x)(p

50 ¡ x)p100 ¡ x2

So,dA

dx= 0 when x = §p

50 fi.e., when the numerator is 0g

But x > 0 )dA

dx= 0 when x + 7:071 cm

Step 4: Sign diagram test

if x = 5 if x = 8

dA

dx+ 11:5

dA

dx+ ¡9:3

i.e., i.e.,> <0 0

So, we have a local maximum where

x + 7:07 cm

p100 ¡ x2 =

p50 + 7:07 cm

Note: The second derivative test is not used here as d2Adx2

� ��

� �

would be difficult to find.

14.14 cm

x cm

20 cm

O B

C

x cm

10 cm

The rectangle has area,

We draw one rectangle to

represent all possible cases.Notice thatas it is lengthand as itcannot exceedthe radius,

i.e.,

x >

x <

<x<

0

0

a

.

10

10

Infinitely many rectangles can be inscribedin semi-circle of diameter cm. Find theshape of the largest rectan

a 20gle which can be

inscribed.

DEMO

Example 15



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Find the most economical shape (minimum

surface area) for a box with a square base,

vertical sides and an open top, given that it

must contain 4 litres.

Step 1: Let the base lengths be x cm and the depth

be y cm. Now the volume

V = length £ width £ depth

) V = x2y

) 4000 = x2y .... (1) fas 1 litre ´ 1000 cm3g

Step 2: Now total surface area,

A = area of base + 4 (area of one side)

) A x( ) = x2 + 4xy

) A x( ) = x2 + 4x

µ4000

x2

¶fusing (1)g

) A x( ) = x2 + 16000x¡1 Notice:

x > 0 as x is a length.

Step 3: Thus A0(x) = 2x¡ 16 000x¡2

and A0(x) = 0 when 2x =16000

x2

i.e., 2x3 = 16000

x3 = 8000

x = 3p

8000

x = 20

Step 4: Sign diagram test or

if x = 10

A0(10) = 20 ¡ 16 000100

= 20 ¡ 160= ¡140

if x = 30

A0(30) = 60 ¡ 16 000900

+ 60 ¡ 17:8

+ 42:2

Second derivative test

A00(x) = 2 + 32000x¡3

= 2 +32 000

x3

which is always positive

as x3 > >0 0for all x .

Each of these tests establishes that minimum material is used to make the

container when x = 20, and y =4000

202= 10,

i.e., is the shape.

open

y cm

x cm

x cm

20 cm

10 cm

20 cm

��

��

Example 16



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EXERCISE 4D

h cm

x cm

Use calculus techniques in the following problems.

1 A small manufacturer can produce x fittings per day where 0 6 x 6 10 000. The costs

are: ² $1000 per day for the workers

² $2 per day per fitting

² $5000

xper day for running costs and maintenance.

How many fittings should be produced daily to minimise costs?

2 The cost function for producing x items is C(x) = 720 + 4x + 0:02x2 dollars and

likewise, the price per item is p(x) = 15¡0:002x dollars. Find the production level that

will maximise profits.

3 The total cost of producing x blankets per day is 14x

2+8x+20 dollars and each blanket

may be sold at (23 ¡ 12x) dollars.

How many blankets should be produced per day to maximise the total profit?

4 A manufacturer of DVD players has been selling 800 each week at $150 each. From a

market survey it is discovered that for each $5 reduction in price, they will sell an extra

40 DVD players each week.

a What is the price function?

b How large a reduction in price (rebate) should the manufacturer give a buyer so that

revenue will be maximised?

c If the weekly cost function is C(x) = 20 000 + 30x dollars, what rebate would

maximise the profit?

5 The cost of running a boat isv2

10dollars per hour where v is the speed of the boat.

All other costs amount to $62:50 per hour. Find the speed which will minimise the total

cost per kilometre.

6 A duck farmer wishes to build a rectangular enclosure of area 100 m2. The farmer must

purchase wire netting for three of the sides as the fourth side is an existing fence of

another duck yard. Naturally the farmer wishes to minimise the length (and therefore

the cost) of the fencing required to complete the job.

a If the shorter sides are of length x m, show that the required length of wire netting

to be purchased is L = 2x +100

x:

b Use technology to help you sketch the graph of y = 2x +100

x:

c Find the minimum value of L and the corresponding value of x when this occurs.

d Sketch the optimum situation with its dimensions.

7 Radioactive waste is to be disposed of in fully enclosed

lead boxes of inner volume 200 cm3. The base of the

box has dimensions in the ratio 2 : 1.

a What is the inner length of the box?

b Explain why x2h = 100.



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c Explain why the inner surface area of the box is given by A(x) = 4x2+600

xcm2.

d Use technology to help sketch the graph of y = 4x2 +600

x:

e Find the minimum inner surface area of the box and the corresponding value of x.

f Draw a sketch of the optimum box shape with dimensions shown.

8 Consider the manufacture of 1 L capacity tin cans where the

cost of the metal used to manufacture them is to be minimised.

This means that the surface area is to be as small as possible

but still must hold a litre.

a Explain why the height h, is given by h =1000

¼r2cm.

b Show that the total surface area A, is given by A = 2¼r2 +2000

rcm2:

c Use technology to help you sketch the graph of A against r.

d Find the value of r which makes A as small as possible.

e Draw a sketch of the dimensions of the can of smallest surface area.

9 Sam has sheets of metal which are 36 cm by 36 cm

square and wishes to use them. He cuts out identical

squares which are x cm by x cm from the corners of

each sheet. The remaining shape is then bent along

the dashed lines to form an open container.

a Show that the capacity of the container is given

by V (x) = x(36 ¡ 2x)2 cm3.

b What sized squares should be cut out to produce

the container of greatest capacity?

10 Infinitely many rectangles can be inscribed in a circle of

diameter 10 cm. One of these rectangles has maximum

area.

a Let ON = x cm and find the area of ABCD, in

terms of x only.

b Find the dimensions of ABCD when its area is a

maximum.

11 An athletics track consists of two ‘straights’ of length l m and

two semicircular ends of radius x m. The perimeter of the track

is to be 400 m.

a Show that l = 200 ¡ ¼x, and hence write down the

possible values that x may have.

b Show that the area inside the track is given by

A = 400x¡ ¼x2.

c What values of l and x produce the largest area inside

the track?

h cm

r cm

36 cm

36 cm

D

A

C

B

Nx cm

l m

x m



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12 A sector of radius 10 cm is bent to form a conical cup as shown.

Suppose the resulting cone has base radius r cm and height h cm.

a Show that in the sector, arc AC =µ¼

18:

b If r is the radius of the cone, explain why r =µ

36:

c If h is the height of the cone show that h =

q100 ¡ ¡ µ

36

¢2:

d If V (µ) is the cone’s capacity, find V (µ) in terms of µ only.

e Use technology to sketch the graph of V (µ):

f Find µ when V (µ) is a maximum.

13 Special boxes are constructed from lead. Each box is

to have an internal capacity of one cubic metre and the

base is to be square. The cost per square metre of lining

each of the 4 sides is twice the cost of lining the base.

a Show that y =1

x2.

b If the base costs $25 per m2 to line, show that the

total cost of lining the box is C(x) = 25(x2 + 8x¡1) dollars.

c What are the dimensions of the box costing least to line?

14 A retired mathematics teacher has a garden in which the paths are modelled by

y =100

x2(as shown), ¡20 6 x 6 20.

He plans a rose garden as shown by the shaded region.

a

b

c

15

becomes when edges AB and CB

are joined with tape.

��

10 cmA C

B

sector

10 cmjoin

top

y m

x m

x m

cut

24 cm

x cm

Colin bends sheet steel into square section

piping and circular section piping. A client

supplies him with cm wide sheets of

steel which must be cut into two pieces,

where one piece is bent into square-section

tubing and the other into circular tubing.

24

If OB units, find the dimensions of rectangle ABCD.= x

Show that as increases the area of the rectangle

decreases for all .

xx > 0

Show that the perimeter , of the rectangle is

given by for and hence find

the dimensions of the rectangle of least perimeter.

P

x>� �0P = 4x +200x2

y

y =100

x2

fence

20 20A B

CD

pathpath



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Two lamps are of intensities and candle-power respectively and are m apart. If

the intensity of illumination , at any point is directly proportional to the power of the

source and inversely proportional to the

square of the distance from the source, find

the darkest point on the line joining the two

lamps.

40 5 6I

However, the client insists that the sum of the cross-sectional areas is to be as small as

possible. Where could Colin cut the sheet so that the client’s wishes are fulfilled?

16 B is a row boat 5 km out at sea from A. AC

is a straight sandy beach, 6 km long. Peter

can row the boat at 8 kmph and run along

the beach at 17 kmph. Suppose Peter rows

directly from B to X, where X is some point

on AC and AX = x km.

a Explain why 0 6 x 6 6.

b If T (x) is the total time Peter takes to

row to X and then run along the beach

to C, show that T (x) =

px2 + 25

8+

6 ¡ x

17hrs.

c Find x whendT

dx= 0. What is the significance of this value of x? Prove your

statement.

17 A pipeline is to be placed so that it connects

point A to the river to point B.

A and B are two homesteads and X is the

pumphouse.

How far from M should point X be so that

the pipeline is as short as possible?

18

19

Sometimes the variable to be optimised is in the form of a single square root function.

“if A > 0, the optimum value of A(x) occurs at the same value of x as the optimum

value of [A(x)]2.”

5 km

A x km X

B

C

6 km

M

A

B

X N

2 km

1 km

river5 km

openOpen cylindrical bins are to contain litres. Find the

radius and the height of the bin made from the least

amount of material (i.e., minimise the surface area).

100

6 m

40 cp 5 cp

In these situations it is convenient to square the function and use the fact that



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An animal enclosure is a right-angled triangle with one

leg being a drain. The farmer has 300 m of fencing

available for the other two sides, AB and BC.

a Show that AC =p

90 000 ¡ 600x if AB = x m.

b Find the maximum area of the triangular enclosure.(Hint: If the area is A m2, find A2 in terms of x.

Notice that A is a maximum when A2 takes its maximum value.)

a (AC)2 + x2 = (300 ¡ x)2 fPythagorasg) (AC)2 = 90000 ¡ 600x + x2 ¡ x2

= 90000 ¡ 600x

) AC =p

90 000 ¡ 600x

b The area of triangle ABC is

A(x) = 12 (base £ altitude)

= 12 (AC £ x)

= 12x

p90 000 ¡ 600x

So [A(x)]2 =x2

4(90 000 ¡ 600x)

= 22 500x2 ¡ 150x3

)d

dx[A(x)]2 = 45000x¡ 450x2

= 450x(100 ¡ x)

with sign diagram:

So A(x) is maximised when x = 100

Amax = 12 (100)

p90 000 ¡ 60 000

+ 8660 m2

20 A right angled triangular pen is made from 24 m of

fencing used for sides AB and BC. Side AC is an

existing brick wall.

a If AB = x m, find D(x), the distance AC, in

terms of x.

b Findd[D(x)]2

dxand hence draw a sign diagram

for it.

c Find the smallest and the greatest value of D(x)and the design of the pen in each case.

Example 17

A

B

Cdrain

A

B

C

x m

(300 ) m x

��

� �

100 m200 m

D x( ) metres

B

C

Awall



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21 At 1:00 pm a ship A leaves port P, and sails in the direction 30oT at 12 kmph. Also, at

1:00 pm ship B is 100 km due East of P and is sailing at 8 kmph towards P. Suppose tis the number of hours after 1:00 pm.

a Show that the distance D(t) km, between the two ships is given by

D(t) =p

304t2 ¡ 2800t + 10000 km

b Find the minimum value of [D(t)]2 for all t > 0.

c At what time, to the nearest minute, are the ships closest?

22 AB is a 1 m high fence which is 2 m from

a vertical wall, RQ. An extension ladder PQ

is placed on the fence so that it touches the

ground at P and the wall at Q.

a If AP = x m, find QR in terms of x.

b If the ladder has length L(x) m show

that [L(x)]2 = (x + 2)2µ

1 +1

x2

¶.

c Show thatd[L(x)]2

dx= 0 only when x = 3

p2.

d Find, correct to the nearest centimetre, the shortest length of the extension ladder.

You must prove that this length is the shortest.

23 A strip of paper is 10 cm wide and much longer

than it is wide. The top left hand corner A is

pulled to the right hand edge. (It goes to A0:)The paper is pressed down to create a fold line

BC.

How should the paper be folded so that the fold

line is a minimum length?

Hint: Let AB = x cm.

24 A, B and C are computers. A printer P is networked

to each computer. Where should P be located so

that the total cable length AP + BP + CP is a

minimum?

1 m

2 m

Q

wall

R A P

B

10 cm

A

A'

B

C

A

P

B N4 m 5 m C

8 m

Sometimes the derivative finding is difficult and technology use is

recommended.

Use the or your to help

solve the following problems.

graphing package graphics calculator

GRAPHING

PACKAGE



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25 Three towns and their grid references are marked

on the diagram alongside. A pumping station is to

be located at P on the pipeline, to pump water to

the three towns. Grid units are kilometres.

Exactly where should P be located so that pipelines

to Aden, Bracken and Caville in total are as short

as possible?

Sometimes technology does not provide easy solutions to problems where optimisation is

required. This occurs when at least one quantity is unknown.

A square sheet of metal has squares cut

from its corners as shown.

What sized square should be cut out so

that when bent into an open box shape the

container holds maximum liquid?

Let x cm by x cm squares be cut out.

Volume = length £ width £ depth

= (a¡ 2x) £ (a¡ 2x) £ x

i.e., V (x) = x(a¡ 2x)2

Now V 0(x) = 1(a¡ 2x)2 + x£ 2(a¡ 2x)1 £ (¡2) fproduct ruleg= (a¡ 2x)[a¡ 2x¡ 4x]

= (a¡ 2x)(a¡ 6x)

and V 0(x) = 0 when x =a

2or

a

6

We notice that a¡ 2x > 0 i.e., a > 2x or x <a

2; So, 0 < x <

a

2

Thus x =a

6is the only value in 0 < x <

a

2with V 0(x) = 0

Second derivative test:

Now V 00(x) = ¡2(a¡ 6x) + (a¡ 2x)(¡6) fproduct ruleg= ¡2a + 12x¡ 6a + 12x

= 24x¡ 8a

) V 00(a

6) = 4a¡ 8a = ¡4a which is < 0

So, volume is maximised when x =a

6.

Conclusion:

When x =a

6, the resulting container has maximum capacity.

pipeline

y

x

8

(3, 11)Caville

(1, 2)Aden

(7, 3)Bracken

P

Example 18

a cm

a cm

x

( 2 ) cma x

concave



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26 Infinitely many lines can be drawn through the

fixed point (2a, a) where a > 0.

Find the position of point A on the x-axis so that

triangle AOB has minimum area.

27 The trailing cone of a guided long range torpedo

is to be conical with slant edge s cm where sis fixed, but unknown. The cone is hollow and

must contain maximum volume of fuel.

28 A company constructs rectangular seating plans

and arranges seats for pop concerts on AFL

grounds. All AFL grounds used are elliptical in

shape and the equation of the ellipse illustrated

isx2

a2+

y2

b2= 1 where a and b are the lengths

of its semi-major and minor axes as shown.

a Show that y =b

a

pa2 ¡ x2 for A as shown.

b Show that the seating area is given by A(x) =4bx

a

pa2 ¡ x2.

c Show that A0(x) = 0 when x =ap2

.

d Prove that the seating area is a maximum when x =ap2

.

e Given that the area of the ellipse is ¼ab, what percentage of the ground is occupied

by the seats in the optimum case?

29 Jeweller John has a given quantity of gold and wishes to cast it into a sphere and a

cube. These will be placed on a bracelet. To appeal to his customers he wants their total

surface area to be a maximum. Show that this will occur when the edge of the cube has

the same length as the sphere’s diameter.

Note: Vsphere = 43¼r

3, Asphere = 4¼r2, Vcube = s3, Acube = 6s2

Hint: a Show that s =¡k ¡ 4

3¼r3¢ 1

3 for some constant k.

b If A(r) is the area sum, find A(r) in terms of variable r only.

c Show that A0(r) = 8¼r

µ1 ¡ 2r

s

¶.

d Show that A(r) is maximised when the edge of the cube is equal to the

sphere’s diameter.

y

x

B

A

(2 , )a a

Find the ratio of en maximum fuel carrying capacity occurs.s r: wh

r cm

s cm

h cm

stage

seating

A( , )x y

y

b

b

aa

x



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REVIEW SET 4A

30

a Let MP = x km and hence show that

L(x) = PA + PB =pa2 + x2 +

pb2 + (c¡ x)2.

b Show that L0(x) =xp

a2 + x2¡ c¡ xp

b2 + (c¡ x)2.

c Find x in terms of a, b and c when L0(x) = 0, etc.

1

s(t) = 2t3 ¡ 9t2 + 12t¡ 5 cm, where t is the time in seconds (t > 0).

a

b Find the initial conditions.

c Describe the motion of the particle at time t = 2 seconds.

d Find the times and positions where the particle changes direction.

e Draw a diagram to illustrate the motion of P.

f Determine the time intervals when the particle’s speed is increasing.

2 The cost per hour of running a freight train is given by C(v) = 10v +90

vdollars

where v is the average speed of the train.

a Find the cost of running the train for:

i two hours at 15 kmph ii 5 hours at 24 kmph.

b Find the rate of change in the cost of running the train at speeds of:

i 10 kmph ii 6 kmph.

c At what speed will the cost be a minimum?

3 For the function f(x) = 2x3 ¡ 3x2 ¡ 36x + 7 :

a find and classify all stationary points and points of inflection

b find intervals where the function is increasing and decreasing

c find intervals where the function is convex or concave

d sketch the graph of y = f(x), showing all important features.

4 For f(x) =3x¡ 2

x + 3: a state the equation of the vertical asymptote

b find the axis intercepts

c find f 0(x) and draw a sign diagram for it

d find the position and nature of any stationary points.

A

B

b km

a kmPM N

c kmpipeline

A and B are two towns which are

km and km away from a

pipeline as shown. P is to be a

pumping station which supplies

water to A and B via straight

pipelines PA and PB.

a b

REVIEWE

Find expressions for the particle’s velocity and acceleration and draw sign


A particle P moves in a straight line with position relative to the origin O given by



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REVIEW SET 4B

5 Rectangle ABCD is inscribed within the parabola

y = k ¡ x2 and the x-axis, as shown.

a If OD = x, show that the rectangle ABCD

has area function A(x) = 2kx ¡ 2x3.

b If the area of ABCD is a maximum when

AD = 2p

3, find k.

6 A person walks towards a wall where a picture AB

which is 1 m high is hung 1 m above eye level.

How far from the wall should the person view the

picture so that µ, the angle of view, is a maximum?

Hint: Let angle BPC = ®o and PC = x m.

Find tan® and tan(µ + ®) and show that

tan µ =x

x2 + 2.

Note: µ is a maximum when tan µ is a maximum.

1 A rectangle has a fixed area of 500 m2,


b Finddy

dxand explain why

dy

dx< 0 for all values of x.

c Interpret your results of b.

2 The height of a tree at time t years after the tree was planted is given by:

H(t) = 6

µ1 ¡ 2

t + 3

¶metres, t > 0.

a How high was the tree when it was planted?

b Determine the height of the tree after t = 3, 6 and 9 years.

c Find the rate at which the tree is growing at t = 0, 3, 6 and 9 years.

d Show that H 0(t) > 0. What is the significance of this result?

e Sketch the graph of H(t) against t.

3 For the function f(x) = x3 + x2 + 2x ¡ 4 :

a state the y-intercept

b find the x-intercept(s), given that x = 1 is one of them

c find and classify any stationary points and points of inflection

d on a sketch of the cubic, show the features found in a, b and c.

4 A particle moves along the x-axis with position relative to origin O, given by

x(t) = 3t ¡ pt cm, where t is the time in seconds, t > 0.

a

b Find the initial conditions and hence describe the motion at that instant.

c Describe the motion of the particle at t = 9 seconds.

A

B C

D

x

y

y k x� 2

wall

A

B

C�°

�°

P eye level

1 m

1 m

Find expressions for the particle’s velocity and acceleration at any time and

draw sign diagrams for each function.

t

but its length m and breadth m may vary.y x� �



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REVIEW SET 4C

d Find the time(s) and position(s) when the particle reverses direction.

e Determine the time intervals when the particle’s speed is decreasing.

5 A manufacturer of open steel boxes has to make

one with a square base and a volume of 1 m3.

The steel costs $2 per square metre.

a If the base measures x m by x m and the

height is y m, find y in terms of x.

b Hence, show that the total cost of the steel

is C(x) = 2x2 +8

xdollars.

c Find the dimensions of the box costing the manufacturer least to make.

6

a If AC = 2x m, show that the area of triangle

ABC is A(x) = xp

2500 ¡ x2.

b Findd[A(x)2]

dxand hence find x when the

area is a maximum.

1 The weight of a fish t weeks after it is released from a breeding pond is given by

W (t) = 5000 ¡ 4900

t2 + 1grams, t > 0.

a How heavy is the fish at the time of release?

b How heavy is the fish after:

i 1 week ii 4 weeks iii 26 weeks?

c Find the rate at which the fish is growing at time:

i 0 days ii 10 days iii 20 days.

d Sketch the graph of W (t) against t.

2 Given that G =(t¡ 2)2

3+ 5 units where t is the time in seconds:

a find the values of t for which G is increasing

b for what values of t isdG

dt> 10 units per second?

3 For the function f(x) = x3 ¡ 4x2 + 4x :

a find all axis intercepts

b find and classify all stationary points and points of inflection

c sketch the graph of y = f(x) showing features from a and b.

y m

x m

open

B

50 m

A Criver

A triangular pen is enclosed by two fences AB and BC

each of length m, and the river is the third side.50



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REVIEW SET 4D

4 A particle moves in a straight line with position relative to O given by

s(t) = t3 ¡ 2t2 ¡ 4t + 5 metres, where t is the time in seconds, t > 0.

a

b For what values of t is the particle moving to the: i right ii left of O?

c Find any time and positions when the particle reverses direction.

d Find time intervals when the: i velocity is decreasing ii speed is decreasing.

5

of a rectangle with a semi-circle on one of its sides.

a Using the dimensions shown on the figure

show that y = 100 ¡ x ¡ ¼2x.

b Hence, find the area of the lawn A(x), in terms of x only.

c Find the dimensions of the lawn if it has maximum area.

6 The rigidity of a beam of fixed length is directly proportional

to the width and also to the cube of the depth.

a A log has diameter 1 m and if the width of the illustrated

rectangular-section beam is x m, show that the depth isp1 ¡ x2 m.

b Find the rigidity R(x) units of the beam, in

terms of x and the proportionality constant k.

c Show that [R(x)]2 = k2x2(1 ¡ x2)3.

d For what values of x isd[R(x)]2

dx= 0?

e

1 Consider the function, f(x) = xpx ¡ x.

a For what values of x does f(x) have meaning?

b Find the axis intercepts for y = f(x).

c Find the positon and nature of any stationary points and points of inflection.

d Sketch the graph of y = f(x) showing all special features found in a, b and c.

2 For the function f(x) =x2 ¡ 1

x2 + 1:

a find the axis intercepts.

b Why does f(x) have no vertical asymptotes?

c Find the position and nature of any stationary points.

d Show that y = f(x) has non-stationary inflections at x = §q

13 .

e Sketch the graph of y = f(x) showing all features found in a, b, c and d above.

Find expressions for the particle’s velocity and acceleration and draw sign

diagrams for them.

What are the dimensions of the beam of greatest rigidity which can be cut from

a log of diameter m?1

y m

2 mx

x m


A m fence is placed around a lawn which has the shape200


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REVIEW SET 4E

3 A particle P, moves in a straight line with position from O given by

s(t) = 15t ¡ 60

(t ¡ 1)2cm, where t is the time in seconds, t > 0.

a Find velocity and acceleration functions for P’s motion.

b Describe the motion of P at t = 3 secs.

c For what values of t is the particle’s speed increasing?

4

5 Find the maximum and minimum values of x3 ¡ 3x2 + 5 for ¡1 6 x 6 4.

6 A sheet of poster paper has total area of 1 m2.

No printing is to be done on the top and bottom

10 cm. Also, 5 cm on each side must be left.

a If the width of the sheet is x cm, find the

depth of the sheet in terms of x.

b Hence, show that the total area available

for printing is given by:

A(x) = (x ¡ 10)

µ10 000

x¡ 20

¶cm2.

c

1

If x suites are made per month and each suite is sold for

µ71 +

1800

x

¶dollars:

a how many suites should be made and sold each month to maximise profits

b what is the maximum profit?

2 f(x) = x3 + ax where a < 0 has a turning point when x =p

2.

a Find a.

b Find the position and nature of all stationary points of y = f(x).

c Sketch the graph of y = f(x).

24 cm

end view

10 cm

10 cm

5 cm5 cm

A manufacturer has a factory which is capable of producing up to bedroom suites

per month. The total cost of materials and labour needed to make the suites is

dollars, where . In addition there are fixed monthly costs of $ .

60

0 1200x >(50x+2x3

2 )


A rectangular gutter is formed by

bending a cm wide sheet of

metal as shown in the illustration.

24

Where must the bends be made in

order to maximise the water carried

by the gutter?

What dimensions should the poster have for

maximum printing area?


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_04\142SA12STU-2_04.CDR Wednesday, 8 November 2006 8:36:53 AM DAVID3

3 Consider f(x) =x¡ 2

x2 + x¡ 2.

a Determine the equations of any vertical asymptotes.

b Find the position and nature of its turning points.

c Find its axis intercepts.

d

e For what values of p doesx¡ 2

x2 + x¡ 2= p have two real distinct roots?

4 A particle moves in a straight line with position relative to O given by

s(t) = 2t¡ 4

tcm where t is the time in seconds, t > 0.

a Find velocity and acceleration functions for the particle’s motion and draw sign


b Describe the motion of the particle at time t = 1 second.

c Find the time(s) and position(s) when the particle reverses direction.

d Draw a diagram to illustrate the motion of the particle.

e Find the time intervals when the:

i velocity is increasing ii speed is increasing.

5 A machinist has a spherical ball of brass with

diameter 10 cm. The ball is placed in a lathe and

machined into a cylinder.

a If the cylinder has radius x cm, show that

the cylinder’s volume is given by

V (x) = ¼x2p

100 ¡ 4x2 cm3.

b Hence, find the dimensions of the cylinder

of largest volume which can be made.

6

7 Two roads AB and BC meet at right angles. A straight

pipeline LM is to be laid between the two roads with

the requirement that it must pass through point X.

a If PM = x km, find LQ in terms of x.

b Hence show that the length of the pipeline is

given by L(x) =px2 + 1

µ1 +

8

x

¶km

c Findd[L(x)2]

dxand hence find the shortest possible length for the pipeline.

Sketch the graph of the function showing all important features found in ,

and above.

a b

c

In an orange orchard there are trees per hectare and the average yield per tree is

oranges. For each additional tree planted per hectare the average yield per tree

falls by oranges. How many additional trees per hectare should be planted to

maximise the total crop?

40300

5

x cm

A

BC

P M

X8 km 1 kmQ

L



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REVIEW SET 4F (ALL OPTIMISATION)

1 A road has parabolic shape and in fact has

equation y = 2x2. O is the centre of the

city and the axes represent two other main

roads.

A is 1:75 km from O on the y-road. P(x, y)

represents a car travelling along the parabolic

road.

a Show that P’s distance from A is given

by S(x) =q

4x4 ¡ 6x2 + 4916 km.

b Find the position of P when it is nearest to A.

2 At noon, ship A is 45 km due south of ship B. Ship A sails north at 10 kmph and ship

B sails east at 7:5 kmph. At what time are they closest?

3

4 The manager of a picture theatre offers the following plan to schools:

² The manager will accept a minimum of 30 and a maximum of 200 students.

² The cost of a ticket is $5 per student and will decrease by 4 cents per student

for every person in excess of 30.

How many students attending will give the manager maximum revenue?

5 Infinitely many isosceles triangles can be inscribed in a

circle, one of which has maximum area.

a If ON = x cm and the circle’s radius is fixed at

r cm, show that the area of ¢ABC is given by

A(x) =pr2 ¡ x2(r + x)

b Hence, prove that the triangle of maximum area is

equilateral.

6 A square sheet of tin-plate is k cm by k cm and four

squares each with sides x cm are cut from its corners.

The remainder is bent into the shape of an open rectan-

gular container.

Show that the capacity of the container is maximised

when x =k

6.

A B

O

C

N

k cm

x cm

A B

C

x cm

x

y

A (0, 1.75)

P ( , )x y

y x2� 2

The sum of the lengths of AB and BC is cm.

What length must AB have if triangle ABC is to

have maximum area?

12



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5

Contents:

Exponential andlogarithmic functions

Exponential andlogarithmic functions

A

B

C

D

E

F

Derivatives of exponential functions

The natural logarithmic function

Derivatives of logarithmic functions

Exponential, surge and logisticmodelling

Applications of exponential andlogarithmic functions

Review


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INVESTIGATION 1 CONTINUOUS COMPOUND INTEREST

An = A0(1 + i)n where

An is the final amount

A0 is the initial amount

i is the interest rate per compounding period

n is the number of periods

(i.e., the number of times the interest is compounded)

We are to investigate the final value of an investment for various values of n and allow nto get extremely large.

What to do:

1 Suppose $1000 is invested for 4 years at a fixed rate of 6% p.a. Use your calculator

to find the final amount (sometimes called the maturing value) if the interest is paid:

a annually (once a year and so n = 4, i = 6% = 0:06)

b quarterly (four times a year and so n = 4 £ 4 = 16 and i = 6%4 = 0:015)

c monthly d daily e by the second f by the millisecond.

2 Comment on your answers obtained in 1.

3 If r is the percentage rate per year, t is the number of years, and N is the number of

interest payments per year, then i =r

Nand n = Nt.

This means that the growth formula becomes An = A0

³1 +

r

N

Ńt

:

a Show that An = A0

Ã1 +

1Nr

!N

r£rt

:

b Now letN

r= a. Show that An = A0

£¡1 + 1

a

¢a¤ rt.

4 For continuous compound growth, the number of interest payments per year N , gets

very large. Explain why a gets very large as N gets very large.

5

From the investigation we observed that:

“If interest is paid continuously (instantaneously) then the formula for calculating a

compounding amount An = A0(1 + i)n can be replaced by An = A0ert

where r is the percentage rate p.a. and t is the number of years.”

The simplest exponential functions are of the form f(x) = ax where a is any positive

constant, a 6= 1.

Recall that the formula for calculating the amount to which an investment

grows is given by:

146 EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 5)

Using lima!1

¡1 + 1

a

¢a= e explain why the compound growth formula becomes

An = A0ert.


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_05\146SA12STU-2_05.cdr Wednesday, 8 November 2006 8:37:30 AM DAVID3

INVESTIGATION 2 GRAPHING SIMPLE EXPONENTIALS

The exponential family has form f(x) = ax where a > 0, but a 6= 1.

For example, f(x) = 2x, f(x) = 3x, f(x) = 10x, f(x) = (1:2)x,

f(x) = (12 )x are members of this family.

Note:

1 Graph f(x) = ax where a = 1:2, 2, 3, 5 and 10. Comment on any similarities

and differences.

2 Repeat 1 but for a = 0:9, 0:7, 0:5 and 0:2. Comment on any similarities and

differences.

From the previous investigation you should have discovered that:

All members of the exponential family f(x) = ax have the properties that:

² their graphs pass through the point (0, 1)

² their graphs are asymptotic to the x-axis at one end

² their graphs are above the x-axis for all values of x

² their graphs are convex for all x

² their graphs are increasing for a > 1 and decreasing for 0 < a < 1.

For example,

( 2)

( )=( 2)

¡¡

x

x

has no meaning unless is an integer, and

is not an exponential function.

x

f x � �GRAPHING

PACKAGE

y y

x x

1 1

10 �� a1�a

2.0�y x

5.0�y x

8.0�y x2.1�y x

2�y x5�y x

What to do:

EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 5) 147


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INVESTIGATION 3 FINDING WHEN ANDa y = ax dy

dx= ax

Let us try to find the derivative of f(x) = ax from first principles.

Now f 0(x) = limh!0

f(x + h) ¡ f(x)

hffirst principles definitiong

= limh!0

ax+h ¡ ax

h

= limh!0

ax(ah ¡ 1)

h

= ax £µ

limh!0

ah ¡ 1

h

¶fas ax is independent of hg

But f 0(0) = limh!0

f(0 + h) ¡ f(0)

h

= limh!0

ah ¡ 1

h

) f 0(x) = axf 0(0)

At this stage we realise that if we can find a value of a such that f 0(0) = 1, then we have

found a function which is its own derivative.

In the following investigation we find an approximate value for a, so the slope of the tangent

to f(x) = ax at the point (0, 1) is 1.

Click on the icon to graph f(x) = ax and its tangent at

the point (0, 1).

1 Experiment with different values of a. First try a = 2 and a = 3. What is the slope

of the tangent at (0, 1) in each case?

2 Now try a = 2:5. Is the slope at (0, 1) closer to 1? Keep experimenting with values

of a until the slope of the tangent is as close to 1 as you can make it:

From Investigation 3 you should have discovered that

if a + 2:72 and f(x) = ax then f 0(x) = ax also.

To find this value of a more accurately we return to the algebraic approach.

We showed that if f(x) = ax then f 0(x) = axµ

limh!0

ah ¡ 1

h

¶:

DERIVATIVES OFEXPONENTIAL FUNCTIONS

A

What to do:

DEMO

y

x

xay �

slope is ƒ'(0)



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So if f 0(x) = ax we require limh!0

ah ¡ 1

h= 1:

Then, roughly speaking, we require

ah ¡ 1

h+ 1 for values of h which are close to 0

) ah + 1 + h for h close to 0.

Thus a1

n

+ 1 +1

nfor large values of n fh =

1

n! 0 as n ! 1g

) a +

µ1 +

1

n

¶n

for large values of n.

We now examine

µ1 +

1

n

¶n

as n ! 1. Notice:

n

µ1 +

1

n

¶n

10 2:593 742 460

102 2:704 813 829

103 2:716 923 932

104 2:718 145 927

105 2:718 268 237

106 2:718 280 469

107 2:718 281 693

108 2:718 281 815

109 2:718 281 827

1010 2:718 281 828

1011 2:718 281 828

this is the

value of

(1 + 0:1)10

In fact as n ! 1,

µ1 +

1

n

¶n

! 2:718 281 828 459 045 235 :::::: and this irrational

number is given the symbol e to represent it,

i.e.,

Thus,

e = 2:718 281 828 459 045 235 :::: and is called exponential e.

if f(x) = ex then f 0(x) = ex .

We have discussed y = ex near to the origin.

But, what happens to the graph for large positive

and negative x-values?

Notice thatdy

dx= ex = y: As x ! 1,

y = ex ! 1 very rapidly and alsody

dx! 1. ��

15

10

5

y

x

y e� x



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This means that the slope of the curve is very large for large x values.

For x large and negative, we write x ! ¡1.

As y = ex, y ! 0 and sody

dx! 0.

This means for large negative x, the graph becomes flatter.

Observe that ex > 0 for all x.

Functions such as e¡x, e2x+3, e¡x2

, x2e2x, etc need to be differentiated in problem

solving. How do we differentiate such functions?

Consider y = ef(x):

Now y = eu where u = f(x):

Butdy

dx=

dy

du

du

dxfchain ruleg

)dy

dx= eu

du

dx

= ef(x) £ f 0(x)

Summary: Function Derivative

ex ex

ef (x) ef (x)f 0(x)

Alternative notation: ex is sometimes written as exp (x):

For example, exp (1 ¡ x) = e1¡x.

Find the slope function for y equal to:

a 2ex + e¡3x b x2e¡x ce2x

x

a If y = 2ex + e¡3x, thendy

dx= 2ex + e¡3x(¡3)

= 2ex ¡ 3e¡3x

b If y = x2e¡x, thendy

dx= 2xe¡x + x2e¡x(¡1) fproduct ruleg= 2xe¡x ¡ x2e¡x

c If y =e2x

x, then

dy

dx=

e2x(2)x¡ e2x(1)

x2fquotient ruleg

=e2x(2x¡ 1)

x2

ef (x)THE DERIVATIVE OF

Example 1



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1 Find the slope function for f(x) equal to:

a e4x b ex + 3 c exp (¡2x) d ex

2

e 2e¡

x

2 f 1 ¡ 2e¡x g 4ex

2 ¡ 3e¡x hex + e¡x

2

i e¡x2

j exp (1

x) k 10(1 + e2x) l 20(1 ¡ e¡2x)

2 Use the product or quotient rules to find the derivative of:

a xex b x3e¡x cex

xd

x

ex

e x2e3x fexpx

gpxe¡x h

ex + 2

e¡x + 1

Find the slope function of f(x) equal to: a (ex ¡ 1)3 b1p

2e¡x + 1

a y = (ex ¡ 1)3 = u3 where u = ex ¡ 1

Nowdy

dx=

dy

du

du

dxfchain ruleg

= 3u2 du

dx

= 3(ex ¡ 1)2 £ ex

= 3ex(ex ¡ 1)2

b y = (2e¡x + 1)¡1

2 = u¡ 1

2 where u = 2e¡x + 1

Nowdy

dx=

dy

du

du

dx

= ¡12u

¡

3

2du

dx

= ¡12(2e¡x + 1)¡

3

2 £ 2e¡x(¡1)

= e¡x(2e¡x + 1)¡3

2

=1

ex(2e¡x + 1)3

2

3 Find the slope function of f(x) equal to:

a (ex + 2)4

d1

(1 ¡ e3x)2

b1

1 ¡ e¡x

e1p

1 ¡ e¡x

cpe2x + 10

f xp

1 ¡ 2e¡x

EXERCISE 5A

Example 2



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4 If y = Aekx, where A and k are constants:

a show that idy

dx= ky ii

d2y

dx2= k2y.

b Predict the connection betweendny

dxnand y. No proof is required.

5 If y = 2e3x + 5e4x, show thatd2y

dx2¡ 7

dy

dx+ 12y = 0:

Find the position and nature of any turning points of y = (x¡ 2)e¡x.

dy

dx= (1)e¡x + (x¡ 2)e¡x(¡1) fproduct ruleg= e¡x(1 ¡ (x¡ 2))

=3 ¡ x

exwhere ex is positive for all x.

So,dy

dx= 0 when x = 3:

The sign diagram ofdy

dxis:

) at x = 3 we have a maximum turning point.

But when x = 3, y = (1)e¡3 =1

e3

) the maximum turning point is (3,1

e3).

6 Find the position and nature of the turning point(s) of:

a y = xe¡x b y = x2ex c y =ex

xd y = e¡x(x + 2)

Find the slope of the tangent to 3y + y2ex = x3 at the point (0, ¡3).

For the implicit function 3y + y2ex = x3 we differentiate term by term.

)d

dx(3y) +

d

dx(y2ex) =

d

dx(x3)

i.e., 3dy

dx+ 2y

dy

dxex + y2ex| {z } = 3x2

product rule

But when x = 0, y = ¡3

) 3dy

dx+ 2(¡3)

dy

dxe0 + (¡3)2e0 = 0

Example 3

3

�

Example 4



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During Stage 1 mathematics we solved exponential equations of the form 2x = 30 by using

logarithms in base 10. Recall that we found the logarithm (in base 10) of both sides.

The laws of logarithms (in base 10) are: log(AB) = logA + logB

log

µA

B

¶= logA¡ logB

logAn = n logA

So, if 2x = 30 then log 2x = log 30

) x log 2 = log 30 flog law: log an = n log ag) x =

log 30

log 2

) x + 4:907

The question arises: “How do we find x if ex = 3 or ex = 10 say?”

It is possible to use base 10 logarithms, but instead we will use logarithms in base e, which

we call natural logarithms.

So, if ex = 3 we say “x is the natural logarithm of 3” and write

x = loge 3 or x = ln 3:

In general, if ex = a then x = lna and vice versa

i.e., ex = a , x = lna:

If y = lnx then x = ey:

So, the graph of y = lnx can be obtained from the graph of y = ex by swapping

(interchanging) the x and y coordinates.

) 3dy

dx¡ 6

dy

dx+ 9 = 0

) ¡3dy

dx= ¡9 and so

dy

dx= 3

So, the tangent at (0, ¡3) has slope 3.


a x¡ 2ey = y at the point where y = 1

b x2 + yex = y2 at the point(s) where x = 0:

THE NATURAL LOGARITHMIC FUNCTIONB

y = lnxTHE GRAPH OF



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For y = ex we have approximate values in the table of:

x ¡2 ¡1 0 1 2y 0:1353 0:3679 1 2:718 7:389

So, for y = lnx we have: x 0:1353 0:3679 1 2:718 7:389y ¡2 ¡1 0 1 2

Clearly the graph of y = ln x is the

reflection of the graph of y = ex in

the mirror line y = x.

Functions which are reflections in the mirror line y = x are called inverse functions.

y = ex and y = lnx are inverse functions.

From the definition ex = a , x = lna we observe that eln a = a

This means that

and also that

any positive real number can be written as a power of e,

ln en = n

The laws of logarithms in base e are identical to those for base 10.

These are: For a > 0, b > 0 ² ln(ab) = lna+ ln b

² ln

µa

b

¶= lna¡ ln b

² ln (an) = n lna Note: ln en = n

Proof of the first law:

If a > 0 and b > 0 then we can write a and b as a = eln a and b = eln b.

Likewise ab = eln(ab) :::::: (1)

But ab = eln aeln b = eln a + ln b::::::(2) findex lawg) ln(ab) = ln a + ln b fcomparing (1) and (2)g

The other two laws are likewise easily deduced.

Other useful results

If we substitute b = 1 into the first law, then ln(a£ 1) = ln a + ln 1

) ln a = ln a + ln 1

) ln 1 = 0

Consequently ln

µ1

a

¶= ln 1 ¡ lna fsecond log lawg= ¡ ln a fas ln 1 = 0g

xy ln�

ey x�

xy �

1

1



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i.e., ln 1 = 0 and ln

µ1

a

¶= ¡ lna

Notice also that if x = ln e then e = ex fex = a , x = lnag) x = 1

and so, ln e = 1

1 Write as a natural logarithmic equation:

a N = 50e2t b P = 8:69e¡0:0541t c S = a2e¡kt

2 Write in exponential form:

a lnD + 2:1 + 0:69t b lnG + ¡31:64 + 0:0173t

c lnP = ln g ¡ 2t d lnF = 2 lnx¡ 0:03t

[Hint: In c find lnP ¡ ln g:]

Find, without a calculator, the exact values of: a ln e3 b e3 ln 2

a ln e3

= 3

b e3 ln 2

= eln 23 fthird log lawg= eln 8 fas 23 = 8g= 8 fas eln a = ag

3 Without using a calculator, evaluate:

a ln e2 b lnpe c ln

µ1

e

¶d ln

µ1pe

¶e eln 3 f e2 ln 3 g e¡ ln 5 h e¡2 ln 2

a Write as a natural logarithmic equation: M = aekt

b Write lnN + 2:136 + 0:385t in exponential form.

a M = aekt

) lnM = ln(aekt) flogarithms of both sidesg) lnM = ln a + ln(ekt) fln ab = lna + ln bg) lnM = ln a + kt fas ln en = ng

b As lnN + 2:136 + 0:385t

then N + e2:136+0:385t

) N + e2:136 £ e0:385t

) N + 8:4655£ e0:385t

Example 5

EXERCISE 5B

Example 6



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4 Simplify to ln k, using the laws of logarithms:

a ln 5 + ln 6 b 3 ln 4 ¡ 2 ln 2 c 2 ln 5 d 2 + 3 ln 2

5 Write as a power of e: a 2 b 10 c a d ax

Solve for x: a ex = 7 b ex + 2 = 15e¡x

a ex = 7

) ln ex = ln7 ffind ln of both sidesg) x = ln7

b ex + 2 = 15e¡x

) ex(ex + 2) = 15e¡x £ ex fmultiply both sides by exg) e2x + 2ex = 15 fe0 = 1g

) e2x + 2ex ¡ 15 = 0

) (ex + 5)(ex ¡ 3) = 0

) ex = ¡5 or 3

) x = ln 3 fas ex = ¡5 is impossibleg

6 Solve for x:

a ex = 2 b ex = ¡2 c ex = 0

d e2x = 2ex e ex = e¡x f e2x ¡ 5ex + 6 = 0

g ex + 2 = 3e¡x h 1 + 12e¡x = ex i ex + e¡x = 3

7 Solve for x:

a lnx + ln(x + 2) = ln 8 b ln(x¡ 2) ¡ ln(x + 3) = ln 2

Find algebraically, the points of intersection of y = ex ¡ 3and y = 1 ¡ 3e¡x: Check your solution using technology.

The functions meet where ex ¡ 3 = 1 ¡ 3e¡x

) ex ¡ 4 + 3e¡x = 0

) e2x ¡ 4ex + 3 = 0 fmultiplying each term by exg) (ex ¡ 1)(ex ¡ 3) = 0

) ex = 1 or 3

) x = ln 1 or ln 3

) x = 0 or ln 3

when x = 0, y = e0 ¡ 3 = ¡2

when x = ln3, ex = 3 ) y = 3 ¡ 3 = 0

) functions meet at (0, ¡2) and at (ln 3, 0).

Example 7

Example 8TI

C

GRAPHING

PACKAGE



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8 Find algebraically the point(s) of intersection of:

a y = ex and y = e2x ¡ 6

b y = 2ex + 1 and y = 7 ¡ ex

c y = 3 ¡ ex and y = 5e¡x ¡ 3

Consider the function y = 2 ¡ e¡x:

a Find the x-intercept. b Find the y-intercept.

c Show algebraically that the function is increasing for all x.

d Show algebraically that the function is concave for all x.

e Use technology to help graph y = 2 ¡ e¡x:

f Explain why y = 2 is a horizontal asymptote.

a Any graph cuts the x-axis when y = 0, i.e., 0 = 2 ¡ e¡x

) e¡x = 2

) ¡x = ln2

) x = ¡ ln 2

) the x-intercept is ¡ ln 2 + ¡0:69

b The y-intercept occurs when x = 0,

i.e., y = 2 ¡ e0 = 2 ¡ 1 = 1.

cdy

dx= 0 ¡ e¡x(¡1) = e¡x =

1

ex

Now ex > 0 for all x, sody

dx> 0 for all x.

) the function is increasing for all x.

dd2y

dx2= e¡x(¡1) =

¡1

exwhich is < 0 for all x.

) the function is concave for all x.

e f As x ! 1, ex ! 1and e¡x ! 0

) 2 ¡ e¡x ! 2

i.e., y ! 2 (below)

9 The function f(x) = 3 ¡ ex cuts the x-axis at A and the y-axis at B.

a Find the coordinates of A and B.

b Show algebraically that the function is decreasing for all x.

c Find f 00(x) and hence explain why f(x) is concave for all x.

d Use technology to help graph y = 3 ¡ ex.

e Explain why y = 3 is a horizontal asymptote.

Use a graphicscalculator to check

your answers.

Example 9



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DERIVATIVES OFLOGARITHMIC FUNCTIONS

C

INVESTIGATION 4 THE DERIVATIVE OF ln�x

10 The function y = ex ¡ 3e¡x cuts the x-axis at P and the y-axis at Q.

a Determine the coordinates of P and Q.

b Prove that the function is increasing for all x.

c Show thatd2y

dx2= y. What can be deduced about the concavity/convexity of

the function above and below the x-axis?

d Use technology to help graph y = ex ¡ 3e¡x.

Show the features of a, b and c on the graph.

11 f(x) = ex ¡ 3 and g(x) = 3 ¡ 5e¡x.

a Find the x and y-intercepts of both functions.

b Discuss f(x) and g(x) as x ! 1 and as x ! ¡1.

c Find algebraically the point(s) of intersection of the functions.

d Sketch the graph of both functions on the same set of axes. Show all important

features on your graph.

If y = lnx, what is the slope functiondy

dx?

1 Click on the icon to see the graph of y = lnx.

A tangent is drawn to a point on the graph and the slope of this tangent is given.

The graph of the slope of the tangent is displayed as the point on the graph of

y = lnx is dragged.

2 What do you suspect the equation of the slope is?

3 Find the slope at x = 0:25, x = 0:5, x = 1, x = 2, x = 3, x = 4, x = 5:Do your results confirm your suspicion from 2?

From the investigation you should have observed that if y = lnx thendy

dx=

1

x.

Proof: As y = lnx then x = ey

Differentiating with respect to x, 1 = eydy

dxfby the chain ruleg

) 1 = xdy

dxfas ey = xg

)1

x=

dy

dx

What to do:

CALCULUS GRAPHING

PACKAGE



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Also, if y = ln f(x) then y = lnu where u = f(x).

Nowdy

dx=

dy

du

du

dx=

1

uf 0(x) =

f 0(x)

f(x)

Summary Function Derivative

lnx1

x

ln f(x)f 0(x)

f(x)

Find the slope function of: a y = ln(kx) where k is a constant

b y = ln(1 ¡ 3x) c y = x3 lnx

a If y = ln(kx), thendy

dx=

k

kx

f 0(x)

f(x)

=1

x

b If y = ln(1 ¡ 3x), thendy

dx=

¡3

1 ¡ 3x

f 0(x)

f(x)

=3

3x¡ 1

c If y = x3 lnx, thendy

dx= 3x2 lnx + x3

µ1

x

¶fproduct ruleg

= 3x2 lnx + x2

= x2(3 lnx + 1)

1 Find the slope function of:

a y = ln(7x) b y = ln(2x + 1) c y = ln(x¡ x2)

d y = 3 ¡ 2 lnx e y = x2 lnx f y =lnx

2x

g y = ex lnx h y = (lnx)2 i y =p

lnx

j y = e¡x lnx k y =px ln 2x l y =

2px

lnx

m y = 3 ¡ 4 ln(1 ¡ x) n y =px ln 4x o y = x ln(x2 + 1)

The use of the laws of logarithms can help us to differentiate some logarithmic functions

more easily.

Example 10

EXERCISE 5C



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Differentiate with respect to x:

a y = ln(xe¡x) b y = ln

·x2

(x + 2)(x¡ 3)

¸a If y = ln(xe¡x) then y = lnx + ln e¡x flog of a product lawg

) y = lnx¡ x fln ea = ag

Differentiating with respect to x, we getdy

dx=

1

x¡ 1

b If y = ln

·x2

(x + 2)(x¡ 3)

¸then y = lnx2 ¡ ln[(x + 2)(x¡ 3)]

= 2 lnx¡ [ln(x + 2) + ln(x¡ 3)]

= 2 lnx¡ ln(x + 2) ¡ ln(x¡ 3)

)dy

dx=

2

x¡ 1

x + 2¡ 1

x¡ 3

Note: Differentiating b using the ruledy

dx=

f 0(x)

f(x)is extremely tedious and difficult.

2 After using logarithmic laws to write in an appropriate form, differentiate with respect

to x:

a y = lnp

1 ¡ 2x b y = ln

µ1

2x + 3

¶c y = ln (ex

px)

d y = ln (xp

2 ¡ x) e y = ln

µx + 3

x¡ 1

¶f y = ln

µx2

3 ¡ x

¶

Example 11

a Show thatd

dx(ln y) =

1

y

dy

dx.

b Use a to finddy

dxif x ln y = y2 + 2x2:

ad

dx(ln y) =

d

dy(ln y) £ dy

dxfchain ruleg

=1

y

dy

dx

b Differentiating term by term x ln y = y2 + 2x2 we have

d

dx(x ln y) =

d

dx(y2) +

d

dx(2x2)

) 1 ln y + x

µ1

y

dy

dx

¶= 2y

dy

dx+ 4x

Example 12



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) ln y +x

y

dy

dx= 2y

dy

dx+ 4x

)

µx

y¡ 2y

¶dy

dx= 4x¡ ln y

)dy

dx=

4x¡ ln yx

y¡ 2y

(or4xy ¡ y ln y

x¡ 2y2)

3 Finddy

dxif: a x ln y = 5 b x2 ln y = y c ln y = xy2

Many complicated functions involving products, quotients or powers or combinations of these

can be differentiated more easily by first using logarithms and then using

d

dx(ln y)=

1

y

dy

dx.

4 Differentiate by first using logarithms:

a y = xex(2x + 1)2

d y =3x

x

b y = 2x

e y = xlnx

c y = 3¡x

f y =

px(x2 + x)3

e2x2

Finddy

dxif a y =

p1 ¡ x2

(2x + 1)4b y = xx

a If y =(1 ¡ x2)

1

2

(2x + 1)4then ln y = 1

2 ln(1 ¡ x2) ¡ 4 ln(2x + 1)

)1

y

dy

dx= 1

2

µ ¡2x

1 ¡ x2

¶¡ 4

µ2

2x + 1

¶)

dy

dx=

µ ¡x

1 ¡ x2¡ 8

2x + 1

¶y

=

p1 ¡ x2

(2x + 1)4

µ ¡x

1 ¡ x2¡ 8

2x + 1

¶b If y = xx then ln y = lnxx

) ln y = x lnx

)1

y

dy

dx= 1 lnx + x

µ1

x

¶)

dy

dx= (lnx + 1)y

)dy

dx= xx(lnx + 1)

Example 13



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5 By substituting eln 2 for 2 in y = 2x, finddy

dx.

6 Consider f(x) = ln(2x¡ 1) ¡ 3:

a Find the x-intercept.

b Can f(0) be found? What is the significance of this result?

c Find the slope of the tangent to the curve at x = 1.

d For what values of x does f(x) have meaning?

e Find f 00(x) and hence explain why f(x) is concave whenever f(x) has

meaning.

f Graph the function.

7 Consider f(x) = x lnx:

a For what values of x is f(x) defined?

b Show that the smallest value of x lnx is ¡1

e.

8 Prove thatlnx

x6

1

efor all x > 0.

(Hint: Let f(x) =lnx

xand find its greatest value.)

9 Consider the function f(x) = x¡ lnx:

Show that the graph of y = f(x) has a local minimum and that this is the only turning

point. Hence prove that lnx 6 x¡ 1 for all x > 0.

Many useful functions which model real world situations involve exponentials. Here are three

such functions:

I exponential growth and decay functions: y = aebt, t > 0 a and b are constants

b > 0 b < 0

Examples include the replication of bacteria, and radioactive decay.

Notice that ² the y-intercept is a

² if b > 0 the function is increasing

if b < 0 the function is decreasing

² the independent variable t is usually time, t > 0.

EXPONENTIAL, SURGEAND LOGISTIC MODELLING

D

y

t

a

y

t

a



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I surge functions: y = Ate¡bt, t > 0 A and b are positive constants

This model is used extensively in the study of medicinal doses where there is an initial

rapid increase to a maximum and then a slower decay to zero.

The independent variable t is usually time, t > 0.

I logistic functions: y =C

1 + Ae¡bt, t > 0 A, b and C are positive constants

The logistical model is useful for the growth of populations that are limited by

resources or predators.

The independent variable t is usually time, t > 0.

1 When a new pain killing injection is administered

the effect is modelled by E = 750te¡1:5t units,

where t > 0 is the time in hours after the injection

of the drug.

a Draw a sketch of the graph of E against t.

b What is the effect of the drug after

i 30 minutes ii 2 hours?

c When is the drug most effective?

d In the operating period, the effective level of the drug must be at least 100 units.

i When can the operation commence?

ii How long has the surgeon to complete the operation if no further injection is

possible?

e Find t at the point of inflection of the graph. What is the significance of this point?

2 a Prove that f(t) = Ate¡bt has

i a local maximum at t =1

bii a point of inflection at t =

2

b

b Use question 1 to check the facts obtained in a.

EXERCISE 5D.1

y

t

point of inflection

y C��

2,

ln

2

C

b

AC

1 A

C

�

µ ¶


y

t

point of inflection

2,

22be

A

b

µ ¶

b2

b1


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3 The velocity of a body after t seconds, t > 0, is given by v = 25te¡2t cm/s.

a Draw a sketch of the velocity function.

b Show that its acceleration at time t is 25(1 ¡ 2t)e¡2t cm/s2.

c When is the velocity increasing?

d Find the point of inflection of the velocity function. What is its significance?

e Find the time interval when the acceleration is increasing.

4 The number of ants in a colony after t months is modelled by A(t) =25 000

1 + 0:8e¡t:

a Draw a sketch of the A(t) function.

b What is the inital ant population?

c What is the ant population after 3 months?

d Is there a limit to the population size? If so what is it?

e At what time does the population size reach 24 500?

5 The number of bees in a hive after t months is modelled by B(t) =C

1 + 0:5e¡1:73t:

a What is the inital bee population?

b Find the percentage increase in the population after 1 month.

c Is there a limit to the population size? If so what is it?

d If after 2 months the bee population is 4500, what was the original population size?

e Find B0(t) and use it to explain why the population is increasing over time.

f Sketch the graph of the B(t) function.

6 Consider the logistic function f(t) =C

1 + Ae¡bt= C(1 + Ae¡bt)¡1.

a Show that f 0(t) = AbCe¡bt(1 + Ae¡bt)¡2.

b Show that f 00(t) = Ab2Ce¡bt

·Ae¡bt ¡ 1

(1 + Ae¡bt)3

¸.

c Hence, show that the logistic function has a point of inflection at

µlnA

b,C

2

¶.


7 For the following scatterplots, explain why the suggested model is not acceptable:

a b

A power model of the form

y = atn, a > 0, n > 1

A logistic model of the form

P =C

1 + Ae¡bt, A, b and C are > 0

y

t

P

t


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Given a set of data a scatterplot may be obtained using technology.

or

MODELLING FROM DATA

TI

C

GRAPHING

PACKAGE

y

t

y

t

² A simple growth or decay model, y = aebt, may apply if the scatterplot looks like:

c d

A surge model of the form

M = Ate¡bt, A and b > 0

An exponential model of the form

K = aebt, a > 0, b > 0

e f

A logistic model of the form

T =C

1 + Ae¡bt, A, b and C > 0

A cubic model of the form

D = at3 + bt2 + ct + d

g h

A surge model of the form

B = Ate¡bt, A and b > 0

An exponential model of the form

N = aebt, a > 0, b < 0

K

t

T

t

B

t

N

t

M

t

D

t


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y

t

y

t

y C��

Strawberry plants are sprayed with a pesticide-fertiliser mix. The data below gives

the average yield of strawberries per plant for various spray concentrations per litre

of water:Spray concentration (x mL) 0 2 4 5 8

Yield per plant (y strawberries) 8:0 11:5 16:7 20:0 34:1

a Generate a scatterplot of the data and suggest a suitable model.

b Find the model.

c Predict the average number of strawberries per plant if the spray concentration

is 3 mL per litre of water.

a

A simple exponential growth model is suggested.

b The model is y + 8:023 £ (1:199)x

or y + 8:023e0:1816x

which we check by plotting over the given data.

c When x = 3, y + 13:8 strawberries per plant.

Example 14

² A surge model, y = Ate¡bt, may apply if the scatterplot looks like:

In this case enter new lists for t andy

tand try a simple exponential growth

model.

² A logistic model, y =C

1 + Ae¡bt, may apply if the scatterplot looks like:

Your calculator should have a logistic

modelling facility.



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1 The mass B grams of bacteria present in a culture t days after it has been set aside is as

shown in the following table:

Time (t days) 1 2 3 4 5 6 7 8

Mass of bacteria (B grams) 1:58 2:04 2:73 3:48 4:38 5:32 6:78 8:14

a Obtain a scatterplot of the data and suggest a suitable model.

b Find the model and check how well it fits the data by plotting it over the scatterplot.

c According to the model, what was the original mass of the culture?

d Use the model to predict the mass of the culture after i 3:5 days ii 10 days.

e Why is your answer to d i likely to be accurate?

f What are the dangers in predicting the mass of the culture beyond 8 days?

2 When a cup of hot water was placed in a freezer the temperature of the water (T oC)was measured every 10 minutes and the results were displayed in a table.

Time (t min) 10 20 30 40 50 60 70 80

Temperature (T oC) 35:5 19:0 9:5 5:5 3:0 2:0 1:0 0:5

a Obtain a scatterplot of the data and suggest a suitable model.

b Find the model and check how well it fits the data.

c According to the model, what was the original temperature of the water?

d Use the model to predict the temperature of the water after i 15 min ii 90 min.

e Why is your answer to d i likely to be accurate?

f Is there a danger in predicting the temperature of the water beyond 80 minutes

assuming that there is no power failure, etc?

The effect of a pain killing injection t hours after it is given is shown in the table:

Time (t hours) 0:00 0:10 0:25 0:50 0:75 1:00 1:25 1:50 1:75 2:00

Effect (E units) 0 56 84 84 58 42 22 10 5 3

a Obtain a scatterplot of the data and suggest a possible model.

b Find the model and plot it over the scatterplot.

c When is the drug at its most effective stage?

d If an operation cannot commence before the effectiveness reaches 60 units:

i after what time can the operation commence

ii how long has the surgeon to complete the operation?

a

EXERCISE 5D.2

Example 15

A surge model is suggested.

So, we plot Et

values against tand use expotential modelling.

Note: (0; 0) must be ignored.



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b )E

t+ 804.5£ e¡3:147t

i.e., E + 804.5te¡3:147t

c The drug is most effective at the local maxi-

mum for E. This can be found by either:

finding t wheredE

dt= 0 from technology or using the fact that the surge

function is a maximum when

t =1

b+ 0:318 hours

i.e., about 19 min

d We need to find t such that E = 60

i.e., t + 0:103 hours or 0:721 hours from technology

i.e., t + 6 min or 43 min.

i The operation can start after 6 minutes.

ii It can last 43 ¡ 6 = 37 minutes.

3 The owner of a computer software shop advertises a new information game called

D’FACTO. Interest in the new product, from telephone calls and shop enquiries/purchases

is recorded on a weekly basis. The data is tabulated and shown below:

Time (t weeks) 0 1 2 3 4 5 6 7 8

Interest (I people) 0 59 68 60 50 37 26 17 11

a Obtain a scatterplot and suggest a possible model for the data.

b Find the model.

c Use the model to estimate the day of highest interest.

d Estimate the interest in week 9.

4 The following table gives the average pain relief effectiveness (APRE) of a 500 mg

aspirin tablet over a 12-hour period.

Time (t hours) 0 0:5 1 2 3 4 6 9 12

APRE (P units) 0 2:9 3:6 5:0 5:5 5:2 3:9 2:5 1:1

a Obtain a scatterplot of the data and suggest a possible model for the data.

b Find the model.

c What is the APRE at time i t = 5 ii t = 10?

d At what time does maximum pain relief happen?

e At what time does the APRE rate change from decreasing at an increasing rate to

decreasing at a decreasing rate?

f If the tablets produce acceptable pain relief when the APRE level is 2 or more units,

between what times does the tablet produce effective relief?

g At what time should a second tablet be taken to keep the dose at an acceptable

level?

GRAPHING

PACKAGE



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Bacteria are placed into a 2 litre carton of milk. After t hours there are B(t) grams

of bacteria present. The weight of bacteria present is recorded at two hour intervals.

t 2 4 6 8 10 12 14

B(t) 0:09 0:36 1:02 1:82 2:27 2:45 2:46

a

b Find the model.

c Estimate the weight of bacteria at time: i t = 7:5 h ii t = 16 h?

d Estimate the initial weight of bacteria added to the milk.

e At what rate is the weight of bacteria increasing after 6 hours?

f At what time does the rate of growth change from increasing at an increasing

rate to increasing at a decreasing rate?

a

A logistic model is suggested.

b B(t) +2:484

1 + 94:69e¡0:6960t

c i 1:64 grams ii 2:48 grams

d When t = 0, B(0) + 0:026 grams.

e We needdB

dtat t = 6 i.e.,

dB

dt+ 0:42 grams/hour.

f At t = 6:54 hours (whend2b

dt2= 0).

5 From past experience, in a small community eventually everyone hears a particular

rumour. A small group of people inadvertently start a rumour and the proportion of

people who have heard the rumour is recorded on an hourly basis during the day.

Hours (h) 0 1 2 3 4 5 6 7 8 9 10

Proportion

(P )0:02 0:04 0:09 0:18 0:33 0:54 0:72 0:86 0:91 0:96 0:98

a Obtain a scatterplot of the data and suggest a

possible model for the data.

b Find the model.

c Find the rate at which the rumour is spreading

after: i 2 hours ii 5 hours?

d

Example 16

MODELLING

After what time does the rumour spreading

rate change from increasing at an increasing

rate to increasing at a decreasing rate?

Draw a scatterplot of the data and suggest a possible model.



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6 A small island off Cape Yorke has an abundance of rodents which are the food of a

threatened species of brown snake. A few pairs of brown snakes were introduced to the

island which was previously snake free. Gulls and other birds on the island keep the

snake population in check. The population of snakes over many years has been recorded.

t years is the time since introduction.

Time (t years) 1 2 3 4 5 6 7 8

Population (S) 47 108 215 384 560 675 753 780

a Draw a scatterplot of the data and suggest a possible

model.

b Find the model.

c Estimate the population after 12 years.

d Is there a limit to the population size? If so, what is it?

e At what rate is the population increasing at:

i year 2 ii year 4 iii year 6?

The applications we consider here are: ² tangents and normals

² rates of change

² curve properties

² displacement, velocity and acceleration

² optimisation (maxima/minima)

APPLICATIONS OF EXPONENTIALAND LOGARITHMIC FUNCTIONS

E

Find the equation of the tangent to y = lnx at the point where y = ¡1.

When y = ¡1, lnx = ¡1

) x = e¡1 =1

e

) point of contact is (1

e, ¡1)

Now f(x) = lnx has derivative f 0(x) =1

x

) tangent has slope11e

= e

) tangent has equationy ¡¡1

x¡ 1e

= e,

i.e., y + 1 = e

µx¡ 1

e

¶i.e., y + 1 = ex¡ 1

i.e., y = ex¡ 2

1

1

�e

y

x

Example 17



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1 Find the equation of the tangent to y = e¡x at the point where x = 1.

2 Find the equation of the tangent to y = ln(2 ¡ x) at the point where x = ¡1.

3 The tangent at x = 1 to y = x2ex cuts the x and y-axes at A and B respectively.

Find the coordinates of A and B.

4 Find the equation of the normal to y = lnpx at the point where y = ¡1.

5 Find the equation of the tangent to y = ex at the point where x = a.

Hence, find the equation of the tangent to y = ex from the origin.

6 Consider f(x) = lnx:

a For what values of x is f(x) defined?

b Find the signs of f 0(x) and f 00(x) and comment on the geometrical significance

of each.

c Sketch the graph of f(x) = lnx and find the equation of the normal at the point

where y = 1.

7 Find, correct to 2 decimal places, the angle between the tangents to y = 3e¡x and

y = 2 + ex at their point of intersection.

8 The weight of bacteria in a culture is given by W = 200et

2 grams where t is the

time in hours.

a What is the weight of the culture at:

i t = 0 ii t = 30 minutes iii t = 112 hours?

b How long will it take for the weight to reach 1 kg?

c Find the rate of increase in the weight at time:

i t = 0 ii t = 2 hours.

d Sketch the graph of W against t.

9 The current flowing in an electrical circuit t seconds after it is switched off is given by

I = 50 £ e¡

t

10 amps.

a At what rate is the current changing at:

i t = 1 second ii t = 10 seconds?

b How long would it take for the current to fall to 1 amp?

10 A radioactive substance decays according to the formula W = 20e¡kt grams where

t is the time in hours.

a Find k given that the weight is 10 grams after 50 hours.

b Find the weight of radioactive substance present at:

i t = 0 hours ii t = 24 hours iii t = 1 week.

c How long will it take for the weight to reach 1 gram?

d Find the rate of radioactive decay at: i t = 100 hours ii t = 1000 hours.

e Show thatdW

dtis proportional to the weight of substance remaining.

EXERCISE 5E



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11 The temperature of a liquid after being placed in a refrigerator is given by

T = 5 + 95e¡kt degrees Celsius where k is a positive constant and t is the time in

minutes.

a Find k if the temperature of the liquid is 20oC after 15 minutes.

b What was the temperature of the liquid when it was first placed in the refrigerator?

c Show thatdT

dtis directly proportional to T ¡ 5.

d At what rate is the temperature changing at:

i t = 0 mins ii t = 10 mins iii t = 20 mins?

12 The height of a certain species of shrub t years after it is planted is given by

H(t) = 20 ln(3t + 2) + 30 cm, t > 0.

a How high was the shrub when it was planted?

b How long would it take for the shrub to reach a height of 1 m?

c At what rate is the shrub’s height changing:

i 3 years after being planted ii 10 years after being planted?

13 In the conversion of sugar solution to alcohol the chemical reaction obeys the law

A = s(1 ¡ e¡kt) where t > 0 is the number of hours after the reaction commences.

s is the original sugar concentration and A is the quantity of alcohol produced.

a Find A when t = 0.

b If s = 10, and after 3 hours A = 5, find k.

c Find the speed of the reaction at time 5 hours if s = 10.

d Show that the speed of the reaction is proportional to A¡ s.

14 Consider the function f(x) =ex

x.

a Does the graph of y = f(x) have any x or y-intercepts?

b Discuss f(x) as x ! 1 and as x ! ¡1:

c Find and classify any stationary points of y = f(x).

d Sketch the graph of y = f(x) showing all important features.

e Find the equation of the tangent to f(x) =ex

xat the point where x = ¡1:

15 Consider f(x) = x + e¡x.

a Find and classify any stationary points of y = f(x).

b Discuss f(x) as x ! ¡1:

c Explain why the graph of y = f(x) becomes more linear as x gets large and

positive.

d Find f 00(x) and determine its signs for all x.

e Sketch the graph of f(x) = x + e¡x.

f Deduce that e¡x > 1 ¡ x for all x.

16 The distribution function for experiments in radioactivity is f(t) = 1¡ e¡at where tis the time, t > 0 and a is a positive constant.

a Does f(t) have any stationary points? b Graph f(t) in the case where a = 1.

c Find the signs of f 0(t) and f 00(t) in the case where a = 1, and comment on

the significance of each of them.



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17 A body moves along the x-axis with displacement function x(t) = 100(2¡ e¡

t

10 ) cm

where t is the time in seconds, t > 0.

a Find the velocity and acceleration functions for the motion of the body.

b Find the initial position, velocity and acceleration of the body.

c Find the position, velocity and acceleration when t = 5 seconds.

d Find t when x(t) = 150 cm.

e Explain why the speed of the body is decreasing for all t and its velocity is always

decreasing.

18 A particle P moves in a straight line so that its displacement from the origin O is given

by s(t) = 100t + 200e¡

t

5 cm where t is the time in seconds, t > 0.

a Find the velocity and acceleration functions.

b Find the initial position, velocity and acceleration of P.

c Discuss the velocity of P as t ! 1.

d Sketch the graph of the velocity function.

e Find when the velocity of P is 80 cm per second.

19 A psychologist claims that the ability A(t) to memorise simple facts during infancy

years can be calculated using the formula A(t) = t ln t+ 1 where 0 < t 6 5, t being

the age of the child in years.

a At what age is the child’s memorising ability a minimum?

b Sketch the graph of A(t) against t.

20 The most common function used in statistics is the normal distribution function given

by f(x) =1p2¼

e¡

1

2x2

.

a Find the stationary points of the function and find intervals where the function is

increasing and decreasing.

b Find all points of inflection.

c Discuss f(x) as x ! 1 and as x ! ¡1.

d Sketch the graph of y = f(x) showing all important features.

21 Before making electric kettles, a manufacturer performs a cost control study.

They discover that to produce x kettles, the cost per kettle C(x) is given by

C(x) = 4 lnx +

µ30 ¡ x

10

¶2

hundred dollars

with a minimum production capacity per day of 10 kettles.

How many kettles should be manufactured to keep the cost per kettle a minimum?

22 Infinitely many rectangles which sit on the

x-axis can be inscribed under the curve

y = e¡x2

.

Determine the coordinates of C when rect-

angle ABCD has maximum area.

C

D

B

A

y

x

2xey �

�



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REVIEWF

REVIEW SET 5A

y

x

xy

axy

ln

2

�

�

1 b


23 The revenue generated when a manufacturer sells x torches per day is given by

R(x) + 1000 ln³1 +

x

400

´+ 600 dollars.

Each torch costs the manufacturer $1:50 to produce plus fixed costs of $300 per day, so

the total cost per day is given by C(x) = 1:5x + 300 dollars.

How many torches should be produced daily to maximise the profits made?

Hint: Profit = ¡ Cost, so P (x) = R (x) ¡ C (x).

24 A quadratic of the form y = ax2, a > 0, touches the logarithmic function

y = lnx.

a If the x-coordinate of the point of contact

is b, explain why ab2 = ln b and 2ab =1

b.

b Deduce that the point of contact is (pe, 1

2 ).

c What is the value of a?

d What is the equation of the common tangent?

25 A small population of wasps is observed. After t weeks the population is modelled by

P (t) =50 000

1 + 1000e¡0:5twasps, where 0 6 t 6 25.

Find when the wasp population is growing fastest.

26 f(t) = atebt2

has a maximum value of 1 when t = 2.

Find a and b given that they are constants.

1 Differentiate with respect to x: a y = ex3+2 b y =

ex

x2

2 Find the equation of the normal to y = e¡x2


3 Sketch the graphs of y = ex + 3 and y = 9 ¡ e¡x on the same set of axes.

Determine the coordinates of the points of intersection.

4 Consider the function f(x) =ex

x ¡ 1.

a Find the x and y-intercepts.

b For what values of x is f(x) defined?

c Find the signs of f 0(x) and f 00(x) and comment on the geometrical signifi-

cance of each.

d Sketch the graph of y = f(x) and find the equation of the tangent at the point

where x = 2.

Revenue


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y

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y e� xC

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B

REVIEW SET 5B


5 The height of a tree t years after it was planted is given by

H(t) = 60 + 40 ln(2t + 1) cm, t > 0.

a How high was the tree when it was planted?

b How long will it take for the tree to reach: i 150 cm ii 300 cm?

c At what rate is the tree’s height increasing after: i 2 years ii 20 years?

6 A particle P moves in a straight line such that its position is given by

s(t) = 80e¡t

10 ¡ 40t m where t is the time in seconds, t > 0:





e Find when the velocity is ¡44 metres per second.

7 Infinitely many rectangles can be inscribed under

the curve y = e¡x as shown. Determine the

coordinates of A when the rectangle OBAC has

maximum area.

8 The number of turtles which come on shore to lay their eggs on a Queensland beach

has been recorded as:

Year 1970 1975 1980 1985 1990 1995 2000

Number of turtles (T) 11 23 40 56 71 79 83

Let 1970 be represented by t = 0.

a Obtain a scatterplot for the data. What features of the plot indicate that a logistic

model may be appropriate?

b Find the logistic model.

c According to the logistic model, what is the limiting number of turtles likely to

come on shore in the future?

d Estimate the number of turtles likely to come on shore in: i 2005 ii 2015

e FinddT

dt. What is the significance of this function?

f Find t whend2T

dt2= 0. What is the significance of this value of t?

1 Differentiate with respect to x: a y = ln (x3 ¡ 3x) b y = ln

µx + 3

x2

¶2 Find where the tangent to y = ln (x2 + 3) at x = 0 cuts the x-axis.

3 Solve for x: a e2x = 3ex b e2x ¡ 7ex + 12 = 0


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REVIEW SET 5C


4 Consider the function f(x) = ex ¡ x:

a Find and classify any stationary points of y = f(x):b Discuss f(x) as x ! ¡1 and as x ! 1.

c Find f 00(x) and draw a sign diagram of it. State geometrical interpretations of

the signs of f 00(x).

d Sketch the graph of y = f(x).

e Deduce that ex > x + 1 for all x.

5 Finddy

dxby first taking natural logarithms of both sides:

a y = x2x b y =(x2 + 2)(x ¡ 3)

1 ¡ x3

6 A particle P moves in a straight line so that its displacement from the origin O is

given by s(t) = 250t ¡ 200e¡t

4 m where t is the time in seconds, t > 0.





e Find when the velocity is 260 m per second.

7 A shirt maker sells x shirts per day with revenue function

R(x) = 200 ln³1 +

x

100

´+ 1000 dollars.

The manufacturing costs are determined by the cost function

C(x) = (x¡100)2 +200 dollars. How many shirts should be sold daily to maximise

profits? What is the maximum daily profit?

8 A drug to reduce arthritic pain was tested among several sufferers and the average

effect was recorded in the following table:

Time (t hours) 0:0 0:5 1:0 1:5 2:0 3:0 4:0 6:0 8:0 12:0

Effect (E units) 0 11 18 23 26 27 25 18 12 4

t hours is the time after the administration of the drug.

a Obtain a scatterplot of the data and suggest a possible model.

b Find the model.

c At what time is the drug most effective?

d If the drug remains effective when the E-level is 12 units, at what time after

taking the first capsule should the next capsule be taken?


a f(x) = ln(ex + 3) b f(x) = ln

·(x + 2)3

x

¸


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y

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y ae� xB

P

A


2 Find the equation of the tangent to y = lnx at the point where x = a.

Hence, find the equation of the tangent to y = lnx from the origin.

3 A particle P moves in a straight line such that its position is given by

s(t) = 25t ¡ 10 ln t cm, t > 1, where t is the time in minutes.


b Find the position, velocity and acceleration when t = e minutes.

c Discuss the velocity as t ! 1.


e Find when the velocity of P is 12 cm per minute.

4 When producing x clocks per day a manufacturer determines that the total weekly

cost C(x) thousand dollars is given by C(x) = 10 lnx +³20 ¡ x

10

´2.

How many clocks per day should be produced to minimise the costs given that at

least 50 clocks per day must be made to fill fixed daily orders?

5 Solve for x: a 3ex ¡ 5 = ¡2e¡x b 2 lnx ¡ 3 ln

µ1

x

¶= 10

6 The graph of y = ae¡x for a > 0 is shown.

P lies on the graph and the rectangle OAPB is drawn.

As P moves along the curve, the rectangle constantly

changes shape. Find the x-coordinate of P when

rectangle OAPB has minimum perimeter.

7 For the function f(x) = x + lnx :

a find the values of x for which f(x) is defined

b find the signs of f 0(x) and f 00(x) and comment on the geometrical signifi-

cance of each

c sketch the graph of y = f(x) and find the equation of the normal at the point

where x = 1.

8 Consider f(x) =40

1 + e2¡x, 0 6 x 6 8

a Find i f(0) ii the limiting value of f(x) as x ! 1.

b Sketch the graph of y = f(x) on the given domain.

c Given that f(x) =C

1 + Ae¡bxhas a point of inflection which has a y-coordinate

ofC

2, find the coordinates of the point of inflection.

d Mark the point of inflection on your sketch in b.

9 In a country town a rumour is spread by two individuals and after t hours the number

N of people who have heard the rumour is given in the following table.

Time (t hours) 0 1 2 3 4 5 6 7

Number (N people) 2 9 27 72 133 194 217 230


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a Obtain a scatterplot of the data. Why do you think that a logistic model may

apply in this case?

b Find the logistic model.

c From the model, what is the likely population size of the town assuming they

eventually all hear the rumour?

d Estimate the number of people who have heard the rumour after 4:5 hours.

e Find the rate at which the rumour is spreading after 2 hours.

f Find the coordinates of the point of inflection. Interpret its meaning in the context

of the question.


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_05\178SA12STU-2_05.cdr Tuesday, 7 November 2006 5:07:05 PM DAVID3

6

Contents:

IntegrationIntegration

A

B

C

D

E

F

G

H

I

J

Reviewing the definite integral

The area function

Antidifferentiation

The Fundamental theorem of calculus

Integration

Linear motion

Definite integrals

Finding areas

Further applications

Review


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Recall from Chapter 2 that for a function

y = f(x) which is positive, continuous, and

increasing on the interval a 6 x 6 b, the area

under the curve can be approximated using

vertical rectangular strips.

If the strips go above the curve, the area can

be approximated by the upper sum

AU =nP

i=1

f(xi)¢x where ¢x is the strip

width.

If the strips remain below the curve, the area can be approximated by the lower sum

AL =n¡1Pi=0

f(xi)¢x:

We noticed that reducing the strip width ¢x improved the approximations.

So, AL 6 A 6 AU where AU and AL approach A as ¢x ! 0:

Now ¢x =b¡ a

n, so as ¢x ! 0, n ! 1:

Noticing that x0 = a and xn = b, we wrote

limn!1

nPi=0

f(xi)¢x =

Z b

a

f(x)dx

where the limit sum was called the definite integral.

Suppose a car travels at a constant positive velocity of 60 km/h for 15 minutes.

Since the velocity is always positive we can use speed instead of velocity. The speed is given

by v(t) = 60 km/h.

The distance travelled is 60 km/h £ 14 h = 15 km

However, when we graph speed against time, the graph

is a horizontal line.

Since average speed =distance travelled

time taken,

distance = speed £ time.

The distance travelled is therefore the shaded area.

Now suppose the speed decreases at a constant rate so that the car, initially travelling at

60 km/h, stops in 6 minutes. The speed is given by v(t) = 60 ¡ 600t km/h.

REVIEWING THE DEFINITE INTEGRALA

a bx1 x2 x3

x0

xn 2

xn 1 xn

y

x

y x= ƒ( )

....

DISTANCES FROM VELOCITY-TIME GRAPHS

60

Qr_ time ( hours)t

speed (km/h)v t( ) = 60

180 INTEGRATION (Chapter 6)


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Now average speed =distance travelled

time taken

so60 + 0

2=

distance110

) 30 £ 110 = distance i.e., distance = 3 km

But the triangle has area = 12 base £ altitude = 1

2 £ 110 £ 60 = 3

So, once again the shaded area gives us the distance travelled.

Using definite integral notation:

distance travelled =

Z 1

4

0

60 dt = 15 for the first example

and distance travelled =

Z 1

10

0

(60 ¡ 600t) dt = 3 for the second example.

These results suggest that:


Z t2

t1

v(t) dt, provided we do not change direction.

The area under the velocity-time graph gives distance travelled regardless of the shape of the

velocity function. However, when the graph is made up of straight line segments the distance

is usually easy to calculate.

The velocity-time graph for a train journey is

illustrated alongside. Find the total distance

travelled by the train.

Total distance travelled

= total area under the graph

= area A + area B + area C + area D + area E

= 12(0:1)50 + (0:2)50 +

¡50+30

2

¢(0:1) + (0:1)30

+12(0:1)30

= 2:5 + 10 + 4 + 3 + 1:5

= 21 km

When a velocity-time graph contains negative velocities, i.e., the object travels backwards

at some stage, the area above the t-axis gives positive displacements and the area below gives

negative displacements.

60

time

( hours)t

speed (km/h)

v t t( ) = 60 600

qA_p_

Example 1

0.20.1 0.3 0.4

30

0.5

v (km/h)

t (h)

0.6

60

0.2

50 30 3050

0.1 0.1 0.1 0.1

A B C D E

a bc

cba

��

�

2area

INTEGRATION (Chapter 6) 181


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A car travels along a straight road and its

velocity is shown in the graph alongside.

Find the final displacement of the car from

its starting point.

Displacement

= area A ¡ area B

=

µ0:2 + 0:05

2

¶60 ¡ 1

2 (0:1)20

= 7:5 ¡ 1

= 6:5

) it ends up 6:5 km from its starting

point (in the positive direction).

In general, for a velocity-time function v(t)where v(t) > 0 on t1 6 t 6 t2,


Z t2

t1

v(t)dt

1 A runner has velocity-time graph as

shown. Find the total distance travelled

by the runner.

2 A car travels along a straight road and its

velocity-time function is illustrated.

a What is the significance of the graph:

i above the t-axis

ii below the t-axis?

b Find the final displacement of the car.

3 A triathlete rides off from rest accelerating at a constant rate for 3 minutes until she

reaches 40 km/h. She then maintains a constant speed for 4 minutes until reaching a

section where she slows down at a constant rate to 30 km/h in one minute. She then

continues at this rate for 10 minutes before reducing her speed uniformly and is stationary

2 minutes later. After drawing a graph, find how far she has travelled.

Example 2

Note that the totaldistance travelled

would bearea A area B� ��

0.1

20

20

40

60

0.2

v (km/h)

0.3 0.4

t (h)

0.1

20

60

0.2

0.05

v (km/h)

t (h)

A

B

v

tt1 t2

EXERCISE 6A

5 10 15 20

2

4

6

8

velocity (m/s)

time(s)

0.20.1 0.3 0.4 20

20

40

60

0.5

t (h)

0.6 0.7

80velocity (km/h)



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INVESTIGATION 1 THE AREA FUNCTION

4 A household’s rate of consumption of

electricity in kWh (kilowatt hours) is

shown in the graph alongside over a

period of 12 hours from midnight.

Use the graph to estimate the total con-

sumption of electricity over this period.

5 Estimate the area bounded by the curve y =1

xand the x-axis from x = 1 to x = 3

by partitioning into 10 strips and using:

a upper rectangular sums b lower rectangular sums

We have seen that if f(x) > 0 on the interval [a, b]

then

Z b

a

f(x) dx is the area between the curve

y = f(x) and the x-axis from x = a to x = b.

Now consider the shaded region alongside, which is the

region between y = f(x) and the x-axis from x = ato x = t.

The area of the shaded region clearly depends on the

value of t, and we can hence represent it by an area

function A(t). In fact,

A(t) =

Z t

a

f(x) dx.

In the following investigation we will explore the form of the area function.

Consider the constant function f(x) = 5.

The corresponding area function is

A(t) =

Z t

a

5 dx

= shaded area in graph

= (t¡ a)5

= 5t¡ 5a

Notice that this can be written in the form A(t) = F (t) ¡ F (a) where F (t) = 5t or

equivalently, F (x) = 5x:

12108642

1.4

1.6

1.2

1

0.8

0.6

0.4

0.2time (hours)

kW

THE AREA FUNCTIONB

a b x

yy x= ƒ( )

a b xt

yy x= ƒ( )

a xt

y

y�5

t a�



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1 What is the derivative F 0(x) of the function F (x) = 5x? How does this relate to the

function f(x)?

2 Consider the simplest linear function f(x) = x.

The corresponding area function is

A(t) =

Z t

a

xdx


=

µt + a

2

¶(t¡ a)

a Can you write A(t) in the form F (t) ¡ F (a)?

b If so, what is the derivative F 0(x)? How does it relate to the function f(x)?

3 Consider f(x) = 2x + 3. The corresponding area function is

A(t) =

Z t

a

(2x + 3)dx


=

µ2t + 3 + 2a + 3

2

¶(t¡ a)

a Can you write A(t) in the form F (t) ¡ F (a)?

b If so, what is the derivative F 0(x)?How does it relate to the function f(x)?

4 Repeat the procedure in 2 and 3 for finding the area functions of

a f(x) = 12x + 3 b f(x) = 5 ¡ 2x

Do your results fit with your earlier observations?

5 If f(x) = 3x2+4x+5, predict what F (x) would be without performing the algebraic

procedure.

From the investigation you should have discovered that, for f(x) > 0,Z t

a

f(x) dx = F (t) ¡ F (a) where F 0(x) = f(x). F (x) is the antiderivative of f(x).

If F (x) is a function where F 0(x) = f(x) we say that:

² the derivative of F (x) is f(x) and

² the antiderivative of f(x) is F (x):

We have already seen the usefulness of derivatives in problem solving.

Antiderivatives also have a large number of useful applications.

What to do:

a xt

t

y y x�

t a

a xt

y y x� �2 3

2 3t�

2 3a�

t a

ANTIDIFFERENTIATIONC



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These include:

² finding areas where curved boundaries are involved

² finding volumes of revolution

² finding distances travelled from velocity functions

² finding hydrostatic pressure

² finding work done by a force

² finding centres of mass and moments of inertia

² solving problems in economics and biology

² solving problems in statistics.

In many problems in calculus we know the rate of change of one variable with respect to

another, i.e.,dy

dx, but we need to know y in terms of x.

For example:

² The slope function f 0(x) of a curve is 2x + 3, and the curve passes through the

origin. What is the function y = f(x)?

² The rate of change in temperature (in oC)dT

dt= 10e¡t where t is the time in

minutes, t > 0. What is the temperature function given that initially the

temperature was 11oC?

The process of finding y fromdy

dx(or f(x) from f 0(x)) is the reverse process of

differentiation and is called antidifferentiation.

Consider the following problem:

Ifdy

dx= x2, what is y in terms of x?

From our work on differentiation we know that y must involve x3 because when we differ-

entiate power functions the index reduces by 1.

If y = x3 thendy

dx= 3x2, so if we start with y = 1

3x3, then

dy

dx= x2:

However, if y = 13x

3 + 2,dy

dx= x2, if y = 1

3x3 + 100,

dy

dx= x2,

and if y = 13x

3 ¡ 7,dy

dx= x2:

So, there are in fact infinitely many such functions of the form y = 13x

3 + c where c is

an arbitrary constant.

However, if we ignore the arbitrary constant we can say that: 13x

3 is the antiderivative of

x2 as it is the simplest function which when differentiated gives x2.



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1 Find the antiderivative of:

a x4 by differentiating x5

b 6x2 + 4x by differentiating x3 + x2

c e3x+1 by differentiating e3x+1

dpx by differentiating x

px

e (2x + 1)3 by differentiating (2x + 1)4

2 Find y if: (Do not forget the integrating constant c.)

ady

dx= 6 b

dy

dx= 4x2 c

dy

dx= 5x¡ x2

ddy

dx=

1

x2e

dy

dx= e¡3x f

dy

dx= 4x3 + 3x2

Summary Function Antiderivative

k kx fk is a constantg

xnxn+1

n+ 1

ex ex

Find the antiderivative of: a x3 b e2x c1px

a We know that the derivative of x4 involves x3

i.e.,d

dx

¡x4¢

= 4x3, )d

dx

¡14x

4¢

= x3

So, the antiderivative of x3 is 14x

4:

b We know thatd

dx

¡e2x¢

= e2x £ 2

Henced

dx

¡12e

2x¢

= 12 £ e2x £ 2 = e2x

So, the antiderivative of e2x is 12e

2x:

c1px

= x¡ 1

2 Nowd

dx(x

1

2 ) = 12x

¡ 1

2

)d

dx(2x

1

2 ) = 2(12)x¡ 1

2 = x¡ 1

2

) the antiderivative of1px

is 2px:

Example 3

EXERCISE 6C



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Isaac Newton showed the link between differential calculus and the definite integral (the limit

of an area sum). This link is called the Fundamental theorem of calculus. The beauty of

this theorem is that it enables us to evaluate complicated summations.

We have already established that:

“if f(x) is a continuous positive function on

an interval [a, b] then the area under the curve

between x = a and x = b isZ b

a

f(x) dx ”.

i.e., A(t) =

Z t

a

f(x) dx.

A(t) is clearly an increasing function and

A(a) = 0 and A(b) =

Z b

a

f(x) dx ...... (1)

Now consider a narrow strip of the region between

x = t and x = t + h:

The area of this strip is A(t + h) ¡A(t).

Since the narrow strip is contained within two

rectangles then

area of smaller6 A(t + h) ¡A(t) 6

area of larger

rectangle rectangle

) hf(t) 6 A(t + h) ¡A(t) 6 hf(t + h)

) f(t) 6A(t + h) ¡A(t)

h6 f(t + h)

Now taking limits as h ! 0 gives

f(t) 6 A0(t) 6 f(t)

Consequently A0(t) = f(t)

So, the area function A(t) is an antiderivative of f(t), and so A(t) and F (t) may only

differ by a constant.

THE FUNDAMENTALTHEOREM OF CALCULUS

D

y

x

y x�ƒ( )

a b

y

x

y x�ƒ( )

a t b

A t( )

y

x

y x�ƒ( )

a

t t h�

b

y x�ƒ( )

t t h�

ƒ( )t ƒ( )t h�

x

enlarged strip

Consider a function which has antide-

rivative and an area function that is the

area from to ,

y f xF x A t

x a x t

� �

� � � �

= ( )( ) ( )

= =



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However A(a) = F (a) + c on letting t = a in (2)

) 0 = F (a) + c

) c = ¡F (a)

Thus

Z b

a

f(x) dx = F (b) ¡ F (a) and we have formally proven the result obtained in

Investigation 1.

We can now state the Fundamental theorem of calculus:

For a continuous function f(x) with antiderivative F (x),Z b

a

f(x)dx = F (b)¡ F (a).

Note: Considering a velocity-time function v(t) we know thatds

dt= v.

So, s(t) is the antiderivative of v(t) and by the Fundamental theorem of calculus,

Z t2

t1

v(t) dt = s(t2) ¡ s(t1) gives the displacement over the time interval [t1, t2].

The following properties of the definite integral can all be deduced from the Fundamental

theorem of calculus and some can be easily demonstrated graphically.

²Z a

a

f(x) dx = 0

²Z b

a

c dx = c(b¡ a) fc is a constantg

²Z b

a

f(x)dx +

Z c

b

f(x)dx =

Z c

a

f(x) dx

²Z a

b

f(x) dx = ¡Z b

a

f(x) dx

²Z b

a

c f(x)dx = c

Z b

a

f(x) dx

²Z b

a

[f(x) § g(x)]dx =

Z b

a

f(x) dx§Z b

a

g(x) dx

Hence, A(t) = F (t) + c ...... (2)

Now

Z b

a

f(x) dx = A(b) ffrom (1)g= F (b) + c ffrom (2)g



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Proof:

Z b

a

f(x) dx +

Z c

b

f(x) dx

= F (b) ¡ F (a) + F (c) ¡ F (b)

= F (c) ¡ F (a)

=

Z c

a

f(x) dx

1 Use the Fundamental theorem of calculus to show that:

a

Z a

a

f(x) dx = 0 and explain the result graphically

b

Z b

a

c dx = c(b¡ a), c is a constant

c

Z a

b

f(x) dx = ¡Z b

a

f(x) dx

d

Z b

a

c f(x)dx = c

Z b

a

f(x)dx, c is a constant

e

Z b

a

[f(x) + g(x)] dx =

Z b

a

f(x) dx +

Z b

a

g(x) dx

For example, consider

Z b

a

f(x)dx +

Z c

b

f(x) dx =

Z c

a

f(x) dx

EXERCISE 6D

y

x

y x�ƒ( )

A1 A2

a b cZ b

a

f(x) dx +

Z c

b

f(x) dx = A1 + A2 =

Z c

a

f(x) dxi.e.,

Use the Fundamental theorem of calculus to find the area:

a between the x-axis and y = x2 from x = 0 to x = 1

b between the x-axis and y =px from x = 1 to x = 9

a f(x) = x2 has antiderivative

F (x) =x3

3

So the area

=R 1

0x2 dx

= F (1) ¡ F (0)

= 13 ¡ 0

= 13 units2

1

y

x

2xy �

Example 4



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Earlier we showed that the antiderivative of x2 was 13x

3

i.e., if f(x) = x2 then F (x) = 13x

3:

We also showed that 13x

3 + c has derivative x2 for any constant c:

We say that “the integral of x2 is 13x

3 + c” and writeZx2dx = 1

3x3 + c

Note:

Zx2 dx can be read as “the integral of x2 with respect to x”

In general, if F 0(x) = f(x) then

Zf(x) dx = F (x) + c:

b f(x) =px = x

1

2 has antiderivative

F (x) =x

3

2

32

= 23x

px

So the area

=R 9

1x

1

2 dx

= F (9) ¡ F (1)

= 23 £ 27 ¡ 2

3 £ 1

= 1713 units2

2 Use the Fundamental theorem of calculus to find the area between the x-axis and:

a y = x3 from x = 0 to x = 1

b y = x3 from x = 1 to x = 2

c y = x2 + 3x + 2 from x = 1 to x = 3

d y =px from x = 0 to x = 2

e y = ex from x = 0 to x = 1:5

f y =1px

from x = 1 to x = 4

g y = x3 + 2x2 + 7x + 4 from x = 1 to x = 1:25

Check each answer using technology.

y

x1 9

xy �

INTEGRATIONE

TI

C



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Since integration or finding antiderivatives is the reverse process of differentiating we can

discover integrals by differentiation.

For example,

² if F (x) = x4, then F 0(x) = 4x3

)R

4x3dx = x4 + c

² if F (x) =px = x

1

2 , then F 0(x) = 12x

¡ 1

2 =1

2px

)

Z1

2pxdx =

px + c

The rules

and

Zk f(x) dx = k

Zf(x) dx, k is a constantZ

[f(x) + g(x)]dx =

Zf(x)dx +

Zg(x)dx may prove useful.

The first tells us that a constant c may be written before the integral sign. The second rule

tells us that the integral of a sum is the sum of the separate integrals. This rule enables us to

integrate term-by-term.

To prove the first of these rules we consider differentiating kF (x) where F 0(x) = f(x):

Nowd

dx(k F (x)) = k F 0(x) = k f(x)

)

Zk f(x) dx = k F (x)

= k

Zf(x)dx

If y = x4 + 2x3, finddy

dxand hence find

Z(2x3 + 3x2) dx:

If y = x4 + 2x3, thendy

dx= 4x3 + 6x2

)R

4x3 + 6x2 dx = x4 + 2x3 + c1

)R

2(2x3 + 3x2) dx = x4 + 2x3 + c1

) 2R

(2x3 + 3x2) dx = x4 + 2x3 + c1

)R

(2x3 + 3x2) dx = 12x

4 + x3 + c

DISCOVERING INTEGRALS

c1

c =c12

could be anyconstant so

could also be anyconstant.

Example 5



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1 If y = x7, finddy

dxand hence find

Rx6 dx:

2 If y = x3 + x2, finddy

dxand hence find

R(3x2 + 2x) dx:

3 If y = e2x+1, finddy

dxand hence find

Re2x+1 dx:

4 If y = (2x + 1)4, finddy

dxand hence find

R(2x + 1)3 dx:

5 If y = xpx, find

dy

dxand hence find

R pxdx:

6 If y =1px

, finddy

dxand hence find

Z1

xpxdx:

7 Prove the ruleR

[f(x) + g(x)] dx =Rf(x) dx +

Rg(x) dx:

Hint: Suppose F (x) is the antiderivative of f(x)

and G(x) is the antiderivative of g(x) then findd

dx[F (x) + G(x)]:

8 a Finddy

dxif y = (2x¡ 1)6 and hence find

R(2x¡ 1)5 dx:

b Finddy

dxif y =

p1 ¡ 4x and hence find

Z1p

1 ¡ 4xdx:

c Finddy

dxif y =

1p3x + 1

and hence find

Z1

(3x + 1)3

2

dx:

9 a If y = e1¡3x, finddy

dxand hence find

Re1¡3x dx:

b If y = ln(4x+1), finddy

dxand hence find

Z1

4x + 1dx for 4x+1 > 0.

10 a By consideringd

dx(ex¡x2

), findRex¡x2

(1 ¡ 2x)dx:

b By consideringd

dxln(5 ¡ 3x + x2), find

Z4x¡ 6

5 ¡ 3x + x2dx:

c By consideringd

dx(x2 ¡ 5x + 1)¡2, find

Z2x¡ 5

(x2 ¡ 5x + 1)3dx:

d By consideringd

dx(xex), find

Rxex dx:

e By consideringd

dx(2x), find

R2x dx: (Hint: 2x = (eln 2)x:)

f By consideringd

dx(x lnx), find

Rlnxdx:

EXERCISE 6E.1



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In earlier chapters we developed rules to help us differentiate functions more efficiently.

Following is a summary of these rules:

Function Derivative Name

c, a constant 0

mx + c, m and c are constants m

xn nxn¡1 power rule

cu(x) cu0(x)

u(x) + v(x) u0(x) + v0(x) sum rule

u(x)v(x) u0(x)v(x) + u(x)v0(x) product rule

u(x)

v(x)

u0(x)v(x) ¡ u(x)v0(x)

[v(x)]2quotient rule

y = f(u) where u = u(x)dy

dx=

dy

du

du

dxchain rule

ex ex

ef(x) ef(x)f 0(x)

lnx1

x

ln f(x)f 0(x)

f(x)

[f(x)]n n[f(x)]n¡1 f 0(x)

These rules or combinations of them can be used to differentiate almost all functions.

However, the task of finding antiderivatives is not so easy and cannot be contained by listing

a set of rules as we did above. In fact huge books of different types of functions and their

integrals have been written. Fortunately our course is restricted to a few special cases.

Notice thatd

dx(kx + c) = k )

Rk dx = kx + c

if n 6= ¡1,d

dx

µxn+1

n + 1+ c

¶=

(n + 1)xn

n + 1= xn )

Rxn dx =

xn+1

n + 1+ c

d

dx(ex + c) = ex )

Rex dx = ex + c

if x > 0,d

dx(lnx + c) =

1

x

if x < 0,d

dx(ln(¡x) + c) =

¡1

¡x=

1

x)

R 1

xdx = ln jxj + c

SIMPLE INTEGRALS



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SummaryFunction Integral

k kx + c fk is a constantg

xnxn+1

n+ 1+ c

ex ex + c

1

xln jxj+ c

c is always an arbitrary constant

called “the integrating constant”

or “the constant of integration”.

Find a

Z µ3x +

2

x

¶2

dx b

Z µx2 ¡ 2p

x

¶dx

a

Z µ3x +

2

x

¶2

dx

=

Z(3x)2 + 2(3x)

µ2

x

¶+

µ2

x

¶2

dx f(a + b)2 = a2 + 2ab + b2g

=

Z µ9x2 + 12 +

4

x2

¶dx

=

Z(9x2 + 12 + 4x¡2)dx

=9x3

3+ 12x +

4x¡1

¡1+ c

= 3x3 + 12x ¡ 4

x+ c

Find aR

(x3 ¡ 2x2 + 5)dx b

Z µ1

x3¡ p

x

¶dx

aR

(x3 ¡ 2x2 + 5)dx

=x4

4¡ 2x3

3+ 5x + c

b

Z µ1

x3¡ p

x

¶dx

=R

(x¡3 ¡ x1

2 ) dx

=x¡2

¡2¡ x

3

2

32

+ c

= ¡ 1

2x2¡ 2

3x3

2 + c

Example 6

Example 7

Notice that we expanded thebrackets and simplified to putthe function into a form where

integration could be done.



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b

Z µx2 ¡ 2p

x

¶dx

=

Z µx2

px¡ 2p

x

¶dx fsplitting into two fractions as

a¡ b

c=

a

c¡ b

cg

=

Z(x

3

2 ¡ 2x¡ 1

2 ) dx findex lawsg

=x

5

2

52

¡ 2x1

2

12

+ c

½Rxn dx =

xn+1

n + 1+ c

¾= 2

5x2px¡ 4

px + c fsimplifyingg

Note: There is no product or quotient rule for integration. Consequently we often have to

carry out multiplication or division before we integrate.

1 Find:

aR

(x4 ¡ x2 ¡ x + 2)dx b

Z µpx¡ 1p

x

¶dx c

Z2ex ¡ 1

x2dx

d

Zxpx¡ 1

xdx e

R(2x + 1)2 dx f

Zx2 + x¡ 3

xdx

g

Z2x¡ 1p

xdx h

Z1

xpx¡ 4

xdx i

R(x + 1)3 dx

2 Find y if:

ady

dx= 2x + 3 b

dy

dx= 3 ¡ 1

xc

dy

dx= (1 ¡ 2x)2

ddy

dx=

px¡ 2p

xe

dy

dx=

x2 + 2x¡ 5

x2f

dy

dx= (x + 2)3

3 Find f(x) if:

a f 0(x) = x3 ¡ 5x + 3 b f 0(x) = 2px(1 ¡ 3x) c f 0(x) = 3ex ¡ 4

x

The constant of integration can be found if we are given a point on the curve.

Find f(x) given that f 0(x) = x3 ¡ 2x2 + 3 and f(0) = 2.

If f 0(x) = x3 ¡ 2x2 + 3 then f(x) =R

(x3 ¡ 2x2 + 3) dx

i.e., f(x) =x4

4¡ 2x3

3+ 3x + c

But f(0) = 2

) 0 ¡ 0 + 0 + c = 2 and so c = 2

Thus f(x) =x4

4¡ 2x3

3+ 3x + 2

Example 8

EXERCISE 6E.2



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4 Find f(x) given that:

a f 0(x) = 2x ¡ 1 and f(0) = 3 b f 0(x) = 3x2 + 2x and f(2) = 5

c f 0(x) = ex +1px

and f(1) = 1 d f 0(x) = x ¡ 2px

and f(1) = 2.

5 Find f(x) given that:

a f 00(x) = 2x + 1, f 0(1) = 3 and f(2) = 7

b f 00(x) = 15px +

3px

, f 0(1) = 12 and f(0) = 5

c f 00(x) = 2x and the points (1, 0) and (0, 5) lie on the curve.

eax+b (ax+ b)n

Considerd

dx

¡eax+b

¢= a eax+b

)

Zeax+b dx =

1

aeax+b + c

If we are given the second derivative we need to integrate twice to find the function. This

creates two integrating constants and so we need two facts about the function in order to

find them.

Find f(x) given that f 00(x) = 12x2 ¡ 4, f 0(0) = ¡1 and f(1) = 4:

If f 00(x) = 12x2 ¡ 4

f 0(x) =12x3

3¡ 4x + c fintegrating with respect to xg

i.e., f 0(x) = 4x3 ¡ 4x + c

But f 0(0) = ¡1 ) 0 ¡ 0 + c = ¡1 and so c = ¡1

Thus f 0(x) = 4x3 ¡ 4x ¡ 1

) f(x) =4x4

4¡ 4x2

2¡ x + d fintegrating againg

i.e., f(x) = x4 ¡ 2x2 ¡ x + d

But f(1) = 4 ) 1 ¡ 2 ¡ 1 + d = 4 ) d = 6

Thus f(x) = x4 ¡ 2x2 ¡ x + 6

Example 9

INTEGRATING AND In theseexamples and

are constants.a

b



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Considerd

dx(ax + b)n+1 = a(n + 1)(ax + b)n

)

Z(ax+ b)n dx =

1

a

(ax+ b)n+1

n+ 1+ c, provided n 6= ¡1

Considerd

dx(ln(ax + b)) =

a

ax + bfor ax + b > 0

then

Z1

ax + bdx =

1

aln(ax + b) + c

In fact

Z1

ax+ bdx =

1

aln jax+ bj + c

Notice the presence of1

ain each result.

Find:

aR

(2x + 3)4 dx

b

Z1p

1 ¡ 2xdx

aR

(2x + 3)4 dx

= 12 £ (2x + 3)5

5+ c

= 110(2x + 3)5 + c

b

Z1p

1 ¡ 2xdx

=R

(1 ¡ 2x)¡

1

2 dx

= 1¡2 £ (1 ¡ 2x)

1

2

12

+ c

= ¡p1 ¡ 2x + c

1 Find:

aR

(2x + 5)3 dx b

Z1

(3 ¡ 2x)2dx c

Z4

(2x¡ 1)4dx

dR

(4x¡ 3)7 dx eR p

3x¡ 4 dx f

Z10p

1 ¡ 5x

gR

3(1 ¡ x)4 dx h

Z4p

3 ¡ 4xdx i

R3p

2x¡ 1 dx

2 a Ifdy

dx=

p2x¡ 7, find y = f(x) given that y = 11 when x = 8.

b A function f(x) has tangent-slope function4p

1 ¡ x, and passes through the

point (¡3, ¡11).

Find the point on the graph of the function y = f(x) with x-coordinate ¡8.

Example 10

EXERCISE 6E.3



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3 Find:

aR

3(2x¡ 1)2 dx bR

(x2 ¡ x)2 dx cR

(1 ¡ 3x)3 dx

dR

(1 ¡ x2)2 dx eR

4p

5 ¡ xdx fR

(x2 + 1)3 dx

Find aRe2x¡1 dx b

R(2e2x ¡ e¡3x) dx c

Z4

1 ¡ 2xdx

aRe2x¡1 dx

= 12e

2x¡1 + c

bR

(2e2x ¡ e¡3x) dx

= 2(12)e2x ¡ ( 1¡3)e¡3x + c

= e2x + 13e

¡3x + cc

Z4

1 ¡ 2xdx = 4

Z1

1 ¡ 2xdx

= 4³

1¡2

´ln j1 ¡ 2xj + c

= ¡2 ln j1 ¡ 2xj + c

4 Find:

aR ¡

2ex + 5e2x¢dx b

R ¡x2 ¡ 2e¡3x

¢dx c

Z ¡px + 4e2x ¡ e¡x

¢dx

d

Z1

2x¡ 1dx e

Z5

1 ¡ 3xdx f

Z µe¡x ¡ 4

2x + 1

¶dx

gR

(ex + e¡x)2 dx hR

(e¡x + 2)2 dx i

Z µx¡ 5

1 ¡ x

¶dx

5 Find y given that:

ady

dx= (1 ¡ ex)2 b

dy

dx= 1¡2x+

3

x + 2c

dy

dx= e¡2x +

4

2x¡ 1

6 To find

Z1

4xdx, Tracy’s answer was

Z1

4xdx = 1

4 ln j4xj + c

Nadine’s answer was

Z1

4xdx = 1

4

Z1

xdx = 1

4 ln jxj + c

Which of them has found the correct answer? Prove your statement.

7 a If f 0(x) = 2e¡2x and f(0) = 3, find f(x).

b If f 0(x) = 2x¡ 2

1 ¡ xand f(¡1) = 3, find f(x).

c If a curve has slope functionpx + 1

2e¡4x and passes through (1, 0), find the

equation of the function.

Example 11



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8 Show that3

x + 2¡ 1

x¡ 2=

2x¡ 8

x2 ¡ 4, and hence find

Z2x¡ 8

x2 ¡ 4dx:

9 Show that1

2x¡ 1¡ 1

2x + 1=

2

4x2 ¡ 1, and hence find

Z2

4x2 ¡ 1dx:

Zf(u)

du

dxdx

ConsiderR

(x2 + 3x)4(2x + 3)dx.

If we let u = x2 + 3x, thendu

dx= 2x + 3.

We can now write the integral as

Zu4 du

dxdx which is of the formZ

f(u)du

dxdx where f(u) = u4:

Zf(u)

du

dxdx =

Zf(u) du

Proof: Suppose F (u) is the antiderivative of f(u), i.e.,dF

dx= f(u).

Observe thatd

dx(F (u)) =

d

du(F (u))

du

dxfchain ruleg

=dF

du

du

dx

= f(u)du

dx

)Rf(u)

du

dxdx = F (u) + c =

Rf(u)du.

This theorem allows us to replacedu

dxdx by du.

So, by letting u = x2 + 3x,du

dx= 2x + 3 and

Z(x2 + 3x)4(2x + 3) dx

=

Zu4 du

dxdx

=

Zu4 du

=u5

5+ c

= 15 (x2 + 3x)5 + c

INTEGRALS OF THE FORM

We can make integrals of this form much easier to find by applying the

:

substitution

theorem



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Use substitution to find:

aR p

x3 + 2x(3x2 + 2) dx b

Z3x2 + 2

x3 + 2xdx c

Rxe1¡x2

dx

aR p

x3 + 2x(3x2 + 2) dx =

Z pudu

dxdx fletting u = x3 + 2xg

=

Zu

1

2 du ftheoremg

=u

3

2

32

+ c

= 23(x3 + 2x)

3

2 + c fsubstituting u = x3 + 2xg

b

Z3x2 + 2

x3 + 2xdx =

Z1

x3 + 2x(3x2 + 2) dx

=

Z1

u

du

dxdx fletting u = x3 + 2xg

=

Z1

udu ftheoremg

= ln juj + c

= ln¯̄x3 + 2x

¯̄+ c

c

Zxe1¡x2

dx = ¡12

Z(¡2x)e1¡x2

dx

= ¡12

Zeuµ

du

dx

¶dx fletting u = 1 ¡ x2 )

du

dx= ¡2xg

= ¡12

Zeu du ftheoremg

= ¡12e

u + c

= ¡12e

1¡x2

+ c

1 Integrate with respect to x:

a 3x2(x3 + 1)4 b2xpx2 + 3

cpx3 + x(3x2 + 1)

d 4x3(2 + x4)3 e (x3 + 2x + 1)4(3x2 + 2) fx2

(3x3 ¡ 1)4

gx

(1 ¡ x2)5h

x + 2

(x2 + 4x¡ 3)2i x4(x + 1)4(2x + 1)

Example 12

EXERCISE 6E.4



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2 Find:

a

Z¡2e1¡2x dx

d

Zepx

pxdx

3 Find:

a

Z2x

x2 + 1dx

d

Z6x2 ¡ 2

x3 ¡ xdx

4 Find f(x) if f 0(x) is:

a x2(3 ¡ x3)2

d xe1¡x2

g4x + 3x2

x3 + 2x2 ¡ 1

b

Z2xex

2

dx

e

Z(2x¡ 1)ex¡x2

dx

b

Zx

2 ¡ x2dx

e

Z4x¡ 10

5x¡ x2dx

b3x

x2 ¡ 2

e1 ¡ 3x2

x3 ¡ x

h4

x lnx

c

Zx2ex

3+1 dx

f

Ze

x¡1

x

x2dx

c

Z2x¡ 3

x2 ¡ 3xdx

f

Z1 ¡ x2

x3 ¡ 3xdx

c xp

1 ¡ x2

f(lnx)3

x

i1

x(lnx)2

In this section we are concerned with motion in a straight line, i.e., linear motion.

Recall that for some displacement function s(t) the velocity function is v(t) = s0(t) and

that t > 0 in all situations.

From the displacement function we can determine total distance travelled in some time

interval a 6 t 6 b.

Consider the following example:

A particle moves in a straight line with velocity function v(t) = t¡ 3 cms¡1.

How far does it travel in the first 4 seconds of motion?

Notice that v(t) = s0(t) = t¡ 3 has sign diagram:

Since the velocity function changes sign at t = 3 seconds, the particle reverses direction at

this time.

Now s(t) =R

(t¡ 3) dt

) s(t) =t2

2¡ 3t + c cm and we are given no information to determine the value of c.

LINEAR MOTIONF

t3

�

0

So, given a we can determine the by integration.velocity function displacement function



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Except for the constant we have determined the displacement function.

The displacement in the first four seconds is s(4)¡s(0) = ¡4 cm. However, this is not the

total distance travelled because of the reversal of direction at t = 3 seconds.

We find the position of the particle at t = 0, t = 3 and t = 4.

s(0) = c, s(3) = c¡ 412 , s(4) = c¡ 4.

We can now draw a diagram of the motion:

Thus the total distance travelled is (412 + 1

2) cm = 5 cm.

This is clearly different to the displacement ¡4 cm.

Summary:

To find the total distance travelled given a velocity function v(t) = s0(t) on a 6 t 6 b:

² Draw a sign diagram for v(t) to help determine any changes in direction.

² Determine s(t) by integration, including an integrating constant c.

² Find s(a) and s(b). Also find s(t) at every time at which there is a direction reversal.

² Draw a motion diagram.

² Determine the total distance travelled from the motion diagram.

A particle P moves in a straight line with velocity function

v(t) = t2 ¡ 3t + 2 ms¡1. How far does P travel in the first 4 seconds of motion?

v(t) = s0(t) = t2 ¡ 3t + 2

= (t¡ 1)(t¡ 2)) sign diagram of v is:

Since the signs change, P reverses direction at t = 1 and t = 2 secs.

Now s(t) =R

(t2 ¡ 3t + 2) dt =t3

3¡ 3t2

2+ 2t + c

So, s(0) = c, s(1) = 13 ¡ 3

2 + 2 + c = c + 56

s(2) = 83 ¡ 6 + 4 + c = c + 2

3

s(4) = 643 ¡ 24 + 8 + c = c + 51

3

Motion diagram:

) total distance =¡c + 5

6 ¡ c¢

+¡c + 5

6 ¡ [c + 23 ]¢

+¡c + 51

3 ¡ [c + 23 ]¢

= 56 + 5

6 ¡ 23 + 51

3 ¡ 23

= 523 m

c �c �\Qw_ c

Example 13

t1 2

� �

0

c+\We_ c+\Ty_c+5\Qe_c



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The answer in Example 13 can be found graphically by considering the absolute value or

modulus function V (t) = abs (t2 ¡ 3t + 2) .

The v(t) = t2 ¡ 3t + 2 function The V (t) = abs (t2 ¡ 3t + 2) function has

graph:

Total distance travelled

=

Z 4

0

abs (t2 ¡ 3t + 2) dt

= 523

1 A particle has velocity function v(t) = 1 ¡ 2t cms¡1 as it moves in a straight line.

Find the total distance travelled in the first second of motion.

2 Particle P has velocity v(t) = t2 ¡ t¡ 2 cms¡1. Find the total distance travelled in

the first 3 seconds of motion.

3 A particle moves along the x-axis with velocity function x0(t) = 16t ¡ 4t3 units/s.

Find the total distance travelled in the time interval:

a 0 6 t 6 3 seconds b 1 6 t 6 3 seconds.

In the following problems we will also consider an acceleration function,

a(t) = v0(t) = s00(t).

4 The velocity of a particle travelling in a straight line is given by

v(t) = 50 ¡ 10e¡0:5t ms¡1, where t > 0, t in seconds.

a State the initial velocity of the particle.

b Find the velocity of the particle after 3 seconds.

c How long would it take for the particle’s velocity to

increase to 45 ms¡1?

d Discuss v(t) as t ! 1.

e Show that the particle’s acceleration is always positive.

f Draw the graph of v(t) against t.

g Find the total distance travelled by the particle in the

first 3 seconds of motion.

5 A train moves along a straight track with acceleration t10 ¡3 ms¡2. If the initial

velocity of the train is 45 ms¡1, determine the total distance travelled in the first minute.

6 Find the distance travelled in the first 4 seconds of motion, by a particle moving in a

straight line with initial velocity 3 ms¡1 and acceleration function 2t ¡ 4 ms¡2.

EXERCISE 6F

4

has graph:

2

1 2 4

2

1 2

Check youranswer

graphically.



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FindR 3

1(x2 + 2) dx

R 3

1(x2 + 2) dx

=

·x3

3+ 2x

¸ 31

=³

33

3 + 2(3)´¡³

13

3 + 2(1)´

= (9 + 6) ¡ (13 + 2)

= 15 ¡ 213

= 1223

Check:

fnInt (X2 + 2, X, 1, 3)

12:666 666 667

1 Evaluate the following and check with your graphics calculator:

a

Z 1

0

x3 dx b

Z 2

0

(x2 ¡ x) dx c

Z 1

0

ex dx

d

Z 4

1

µx¡ 3p

x

¶dx e

Z 9

4

x¡ 3px

dx f

Z 3

1

1

xdx

g

Z 1

¡1

(2x + 5)3 dx h

Z 6

2

1p2x¡ 3

dx i

Z 1

0

e1¡x dx

Example 14

EXERCISE 6G

If F (x) is the antiderivative of f(x) where f(x) is continuous on the interval a 6 x 6 bthen the definite integral of f(x) on this interval is

Z b

a

f(x) dx = F (b) ¡ F (a)

Note:

Z b

a

f(x) dx reads “the integral of f(x) with respect to x, fromx = a to x = b”

Notation: We write F (b) ¡ F (a) as [F (x)]ba:

DEFINITE INTEGRALSG

7 A body has initial velocity 20 ms¡1 as it moves in a straight line with acceleration

function 4e¡

t

20 ms¡2:

a Show that as t increases the body approaches a limiting velocity.

b Find the total distance travelled in the first 10 seconds of motion.



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= 12

Z 8

3

1

udu

= 12 [ln juj]83

= 12(ln 8 ¡ ln 3)

= 12 ln(83 )

b In

Z 1

0

6x

(x2 + 1)3dx, suppose we let u = x2 + 1

) du =du

dxdx = 2xdx and when x = 0, u = 1

when x = 1, u = 2

)

Z 1

0

6x

(x2 + 1)3dx =

Z 2

1

1

u3(3du)

= 3

Z 2

1

u¡3 du

= 3

·u¡2

¡2

¸ 21

= 3

µ2¡2

¡2¡ 1¡2

¡2

¶= 3

¡¡18 + 1

2

¢= 9

8

j

Z 2

1

(e¡x + 1)2 dx k

Z 6

5

1

2x¡ 1dx l

Z ¡2

¡1

µ1 ¡ 8

1 ¡ 3x

¶dx

Evaluate: a

Z 3

2

x

x2 ¡ 1dx b

Z 1

0

6x

(x2 + 1)3dx

a In

Z 3

2

x

x2 ¡ 1dx, Suppose we let u = x2 ¡ 1,

) du =du

dxdx = 2xdx and when x = 2, u = 22 ¡ 1 = 3

when x = 3, u = 32 ¡ 1 = 8

)

Z 3

2

x

x2 ¡ 1dx =

Z 8

3

1

u( 12 du)

Example 15



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INVESTIGATION 2AND AREAS

Z b

a

f(x) dx

Does

Z b

a

f(x) dx always give us an area?

1 Find

Z 1

0

x3 dx and

Z 1

¡1

x3 dx.

2

3 Find

Z 0

¡1

x3 dx and explain why the answer is negative.

4 Check that

Z 0

¡1

x3 dx +

Z 1

0

x3 dx =

Z 1

¡1

x3 dx:

We have already established that:

If f(x) is positive and continuous on the in-

terval a 6 x 6 b, then the area bounded

by y = f(x), the x-axis and the vertical

lines x = a and x = b is given byR b

af(x) dx:

What to do:

Explain why the first integral in gives an area whereas the second integral does

not. Graphical evidence is essential.

1

FINDING AREASH

y

xa b

y x�ƒ( )

2 Evaluate the following and check with your graphics calculator:

a

Z 2

1

x

(x2 + 2)2dx b

Z 1

0

x2ex3+1 dx c

Z 3

0

xpx2 + 16dx

d

Z 2

1

xe¡2x2

dx e

Z 3

2

x

2 ¡ x2dx f

Z 2

1

lnx

xdx

g

Z 1

0

1 ¡ 3x2

1 ¡ x3 + xdx h

Z 4

2

6x2 ¡ 4x + 4

x3 ¡ x2 + 2xdx i

Z 1

0

(x2 + 2x)n(x + 1)

[Careful!]

3 Show that3

x + 4¡ 2

x¡ 1=

x¡ 11

x2 + 3x¡ 4:

Hence show that

Z ¡1

¡2

x¡ 11

x2 + 3x¡ 4dx = 5 ln

¡32

¢:



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To check your results on a graphics calculator (e.g. TI-83),

press and enter y = x2 + 1: Then press

7. It will ask for the lower and upper limits,

so press 1 2 .

Alternatively, you can use the function (X2 + 1, X, 1, 2)

1 Find the area of the region bounded by:

a y = x2, the x-axis and x = 1

b y = x3, the x-axis, x = 1 and x = 2

c y = ex, the x-axis, the y-axis and x = 1

d the x-axis and the part of y = 6 + x ¡ x2 above the x-axis

e the axes and y =p

9 ¡ x

f y =1

x, the x-axis, x = 1 and x = 4

g y =1

x, the x-axis, x = ¡1 and x = ¡3

h y = 2 ¡ 1px

, the x-axis and x = 4

i y = ex + e¡x, the x-axis, x = ¡1 and x = 1:

Use technologyto check your

answers.

Y= GRAPH

2nd CALC

ENTER ENTER

EXERCISE 6H

Find the area of the region enclosed by y = x2 + 1, the x-axis,

x = 1 and x = 2.

Area =R 2

1(x2 + 1)dx

=

·x3

3+ x

¸ 21

=¡83 + 2

¢¡ ¡13 + 1¢

= 143 ¡ 4

3

= 313 units2

21

1

y

x

Example 16

fnInt



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2 Find the area bounded by:

a the x-axis and the curve y = x2 + x¡ 2

b the x-axis, y = e¡x ¡ 1 and x = 2

c the x-axis and the part of y = 3x2 ¡ 8x + 4 below the x-axis

d y = x3 ¡ 4x, the x-axis, x = 1 and x = 2.

Find the area of the region bounded by the x-axis and the curve y = x2 ¡ 2x.

The curve cuts the x-axis when y = 0

) x2 ¡ 2x = 0

) x(x¡ 2) = 0

) x = 0 or 2

i.e., x intercepts are 0 and 2.

Area =R 2

0[y1 ¡ y2] dx

=R 2

0[0 ¡ (x2 ¡ 2x)] dx

=R 2

0(2x¡ x2) dx

=

·x2 ¡ x3

3

¸ 20

=¡4 ¡ 8

3

¢¡ (0)

) the area is 43 units2:

2

y

x

y1� ��

xxy2 22��

Example 17

If two functions f(x) and g(x) intersect at

x = a and x = b and f(x) > g(x) for all

x in the interval a 6 x 6 b, then the area of

the shaded region between their points of in-

tersection is given byR b

a[f(x) ¡ g(x)] dx:

Note that this is the area between the curves regardless of the position of the x-axis.

Proof: If we translate each curve vertically through [0, k] until it is completely above

the x-axis, the area is preserved (i.e., does not change).

Area of shaded region

=R b

a[f(x) + k] dx¡ R b

a[g(x) + k] dx

=R b

a[f(x) ¡ g(x)] dx

AREA BETWEEN TWO FUNCTIONS

y x�ƒ( )

y g x� ( )

a b x

y

y g x k� �( )

y x k� �ƒ( )

a b x



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3 Find the area enclosed by the function y = f(x) and the x-axis for:

a f(x) = x3 ¡ 9x b f(x) = ¡x(x¡2)(x¡4) c f(x) = x4 ¡ 5x2 + 4:

Find the area of the region enclosed by y = x + 2 and y = x2 + x¡ 2.

y = x + 2 meets y = x2 + x¡ 2 where

x2 + x¡ 2 = x + 2

) x2 ¡ 4 = 0

) (x + 2)(x¡ 2) = 0

) x = §2

Area =R 2

¡2[(x + 2) ¡ (x2 + x¡ 2)]dx

=R 2

¡2(4 ¡ x2) dx

=

·4x¡ x3

3

¸ 2¡2

=¡8 ¡ 8

3

¢¡ ¡¡8 + 83

¢= 16 ¡ 16

3

= 1023 units2

��

�

�

y

x

2�� xy

22�� xxy

Example 19

Find the total area of the regions contained by y = f(x) and the x-axis for

f(x) = x3 + 2x2 ¡ 3x.

f(x) = x3 + 2x2 ¡ 3x

= x(x2 + 2x¡ 3)

= x(x¡ 1)(x + 3)

) y = f(x) cuts the x-axis at 0, 1, ¡3.

Total area

=R 0

¡3(x3 + 2x2 ¡ 3x) dx +

R 1

0[0 ¡ (x3 + 2x2 ¡ 3x)] dx

=R 0

¡3(x3 + 2x2 ¡ 3x) dx¡ R 1

0(x3 + 2x2 ¡ 3x) dx

=

·x4

4+

2x3

3¡ 3x2

2

¸ 0¡3

¡·x4

4+

2x3

3¡ 3x2

2

¸ 10

=¡0 ¡¡111

4

¢¡ ¡¡ 712 ¡ 0

¢= 115

6 units2

1 �

x

y

Notice the use of the modulus

function to shorten work.

Example 18



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7 a Explain why the total area shaded is not

equal toR 7

1f(x)dx:

b What is the total shaded area equal to in

terms of integrals?

8 The shaded area is 0:2 units2:

Find k, correct to 4 decimal places.

9 The shaded area is 1 unit2:

Find b, correct to 4 decimal places.

y

xb

xy �

1 k

1

y

x

xy

21

1

�

�

y

x

1 3 5 7

y x= ( )f

4 a Find the area of the region enclosed by y = x2 ¡ 2x and y = 3.

b Consider the graphs of y = x¡ 3 and y = x2 ¡ 3x.

i Sketch each graph on the same set of axes.

ii Find the coordinates of the points where the graphs meet. Check algebraically.

iii

c Determine the area of the region enclosed by y =px and y = x2.

d On the same set of axes, graph y = ex ¡ 1 and y = 2 ¡ 2e¡x, showing axis

intercepts and asymptotes.

Find algebraically, the points of intersection of y = ex ¡ 1 and y = 2¡ 2e¡x.

Find the area of the region enclosed by the two curves.

e Determine the area of the region bounded by y = 2ex, y = e2x and x = 0.

5 On the same set of axes, draw the graphs of the relations y = 2x and y2 = 4x.

Determine the area of the region enclosed by these relations.

6 Sketch the circle with equation x2 + y2 = 9.

a Explain why the ‘upper half’ of the circle has equation y =p

9 ¡ x2.

b Hence, determineR 3

0

p9 ¡ x2 dx without actually integrating the function.

c Check your answer using technology.

Find the area of the region enclosed by the two graphs.



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The marginal cost isdC

dx= 2:15 ¡ 0:02x + 0:000 36x2 $=urn

) C(x) =R

(2:15 ¡ 0:02x + 0:000 36x2) dx

= 2:15x¡ 0:02x2

2+ 0:000 36

x3

3+ c

= 2:15x¡ 0:01x2 + 0:000 12x3 + c

But C(0) = 185 ) c = 185

) C(x) = 2:15x¡ 0:01x2 + 0:000 12x3 + 185

C(100) = 2:15(100)¡ 0:01(100)2 + 0:000 12(100)3 + 185

= 420

) the total cost is $420:

1 The marginal cost per day of producing x gadgets is C 0(x) = 3:15+0:004x dollars per

gadget. What is the total cost of daily production of 800 gadgets given that the fixed

costs before production commences are $450 a day?

EXERCISE 6I

10 The shaded area is 2:4 units2:

Find k, correct to 4 decimal places.

11 The shaded area is 6a units2:

Find a.

y

x

ky �

xy �2

y

x a a

22�� xy

FURTHER APPLICATIONSI

dC

dx2:15 ¡ 0:02x + 0:000 36x2 dollars per urn provided that 0 6 x 6 120:

The set up costs before production starts are $ . Find the total cost of producing

urns per week.

185100

Example 20

The marginal cost of producing urns per week, , is given by:x



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A metal tube has an annulus cross-section as shown. The

outer radius is 4 cm and the inner radius is 2 cm. Within

the tube, water is maintained at a temperature of 100oC.

Within the metal the temperature drops off from inside

to outside according todT

dx= ¡10

xwhere x is the

distance from the central axis O, and 2 6 x 6 4:Find the temperature of the outer surface of the tube.

dT

dx=

¡10

x) T =

Z ¡10

xdx

) T = ¡10 ln jxj + c

But when x = 2, T = 100

) 100 = ¡10 ln 2 + c

) c = 100 + 10 ln 2

Thus T = ¡10 lnx + 100 + 10 ln 2

i.e., T = 100 + 10 ln¡2x

¢When x = 4, T = 100 + 10 ln

¡12

¢+ 93:07

) the outer surface temperature is 93:07oC.

Example 21

metal

x

waterat 100°C

tube cross-section

2 The marginal cost of producing x items is given by C0(x) = 10 ¡ 4px + 1

dollars

per item. If the fixed cost of production is $200 (i.e., the cost before any items are

produced), find the cost of producing 100 items.

3 Swiftflight Pty Ltd makes aeroplanes.

The initial cost of designing a new model and setting

up to manufacture them will be $275 million dollars.

The cost of manufacturing each additional plane is

modelled by 25x¡4x0:8+0:0024x2 million dollars,

where x is the number of aeroplanes made.

4 The marginal profitdP

dxfor producing x dinner

Find the total cost of manufacturing the first aeroplanes.20

plates per week is given by dollars

per plate. If no plates are made a loss of $

occurs each week.

15 0 03650

¡ : x

a

b

c

Find the profit function.

What is the maximum profit and when does

it occur?

What production levels enable a profit to

be made?



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h

x

metal

x

deflection y

x

5 The tube cross-section shown has inner radius of 3 cm and

outer radius 6 cm. Within the tube water is maintained at a

temperature of 100oC. Within the metal the temperature

falls off at a rate according todT

dx=

¡20

x0:63where x is the

distance from the central axis O and 3 6 x 6 6: Find the

temperature of the outer surface of the tube.

6

It is known thatd2y

dx2= ¡ 1

10(1 ¡ x)2.

a Find the equation for measuring the deflection from the horizontal at any point on

the beam.

(Hint: When x = 0, what are y anddy

dx?)

b Determine the greatest deflection of the beam.

7 A contractor digs roughly cylindrical wells to a depth

of h metres. He estimates that at the depth of x m, the

cost of digging is 12x

2 + 4 dollars per m3.

If a well is to have a radius r m, show that the total

cost is given by

C(h) = ¼r2µh3 + 24h

6

¶+ C0 dollars.

·Hint:

dC

dx=

dC

dV£ dV

dx

¸

8 A restaurant opens at 6 pm and closes at 12 midnight. The rate at which people enter

the restaurant over this period is modelled bydE

dt= 30te¡0:6t + 10 for 0 6 t 6 6.

dE

dtis measured in people per hour and t = 0 is 6 pm.

At 6:00 pm there are no people at the restaurant.

a Sketch the graph of y =dE

dt

b Find the rate at which people are entering the restaurant at 6 pm.

c Find the time at which the rate of entry is a maximum.

d Provide evidence which indicates that people continually enter during the 6-hour

period.

e

f Calculate the number entering between 7 pm and 9 pm.

g The rate at which people leave the restaurant isdL

dtwhere

dL

dt= 20te¡0:3t for

0 6 t 6 6.

i On the graph in a, sketch the graph of y =dL

dt.

ii Find the time when the rate of entry matches the rate of departure.

A thin horizontal cantilever of length metre has a

deflection of metres at a distance of m from the

fixed end.

1y x

On your graph in give a representation of the number entering between pm and

pm.

79

a



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REVIEW SET 6A

1 Integrate with respect to x: a4px

b3

1 ¡ 2xc xe1¡x2

2 Evaluate: a

Z ¡1

¡5

p1 ¡ 3xdx b

Z 1

0

4x2

(x3 + 2)3dx

REVIEWJ

h Given that

Z 6

0

µdE

dt¡ dL

dt

¶dt + 13, state the meaning of the value 13 in the

context of this problem.

9

E0(t) = 350te¡0:6t + 5 for 0 6 t 6 8.

The graph of y = E0(t) for 0 6 t 6 8 is shown

alongside.

a

b Using calculus, find the exact value of t when the rate at which the shoppers were

entering the store was a maximum

c Find the time when the rate of shoppers entering the store changes from decreasing

at an increasing rate to decreasing at a decreasing rate.

d i On a sketch of the graph given, provide representation of the number of shop-

ii Calculate the number of shoppers the model suggests entered the venue from

Time after 10 : 00 am (t hours) 0:5 1:5 2:5 3:5 4:5 5:5 6:5 7:5

Rate of shoppers leaving, L0(t) 21 41 70 103 130 147 157 161

e Fit a logistic function of the form L0(t) =C

1 + ae¡btfor 0:5 6 t 6 8 to the data

above and state its equation.

f Sketch the graph of y =L0(t) for 0:5 6 t 6 8 on your previous sketch.

g At what time is the rate at which shoppers were leaving the store greater than the

rate at which they were entering it?

h Given that

Z 8

0

E0(t)dt ¡Z 8

0.5

L0(t) dt + 146,

state the meaning of 146 in the context of the number of shoppers at the store.

A large furniture store opens at am

and closes at pm. The rate people per

hour at which shoppers entered the store over

the hours is modelled by the function

10:006 : 00 ( )

8

��

( )tE t0

� �=0

It was noticed that shoppers started leaving the store only after am. Data was

collected on the rate of departure from am to pm and is given in the table

below:

10 : 3010 : 30 5 : 00

� � ��

Find the rate at which shoppers were entering the store at am.10 : 00� � �

pers the model suggests entered the venue from am to pm� � � � � � �11 : 00 3 : 00

11 : 00 3 : 00� � � � � �am to pm

t

8642

200

150

100

50

y



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REVIEW SET 6B

3 By differentiating (3x2 + x)3, findR

(3x2 + x)2(6x + 1) dx:

4 A particle moves in a straight line with velocity given by v(t) = t2 ¡ 6t + 8 ms¡1

for t > 0.

a Draw a sign diagram for v(t).

b Explain exactly what is happening to the particle in the first 5 seconds of motion.

c After 5 seconds, how far is the particle from its original position?

d Find the total distance travelled in the first 5 seconds of motion.

5 Determine the area enclosed by the axes and y = 4ex ¡ 1.

6 A curve y = f(x) has f 00(x) = 18x + 10.

Find f(x) given that f(0) = ¡1 and f(1) = 13.

7 Find a given that the area of the region between y = ex

and the x-axis from x = 0 to x = a is 2 units2.

Hence determine b, given that the area of the region

between x = a and x = b is also 2 units2.

y e� x

a b

x

y

3 By differentiating y =px2 ¡ 4, find

Zxp

x2 ¡ 4dx:

4 A particle moves in a straight line with velocity v(t) = 2t¡ 3t2 ms¡1.

Find the distance travelled in the first second of motion.

5 Find the area of the region enclosed by y = x2 + 4x + 1 and y = 3x + 3.

6 The current I(t) amps, in a circuit falls off in accordance withdI

dt=

¡100

t2where

t is the time in seconds, provided that t > 0:2 seconds.

It is known that when t = 2, the current is 150 amps. Find a formula for the current

at any time (t > 0:2), and hence find:

a the current after 20 seconds b what happens to the current as t ! 1.

7 DetermineR 2

0

p4 ¡ x2 dx by considering the graph of y =

p4 ¡ x2 .

8 Is it true thatR 3

¡1f(x) dx represents the area

of the shaded region?

Explain your answer briefly.

1 Find: a

Z µ2e¡x ¡ 1

x+ 3

¶dx b

Z µpx¡ 1p

x

¶2

dx

2 Evaluate aR 2

1(x2 ¡ 1)2 dx b

R 2

1x(x2 ¡ 1)2 dx

x

y

� � �



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REVIEW SET 6C

5 Draw the graphs of y2 = x¡ 1 and y = x¡ 3.

a Determine the coordinates where the graphs meet.

b Determine the area of the enclosed region.

6 Determine k if the enclosed region has area

513 units2.

7 A function has slope function 2px +

apx

and passes through the points (0, 2) and

(1, 4). Find a and hence explain why the function y = f(x) has no stationary points.

8 Write¡2x

4 ¡ x2as

A

x + 2+

B

2 ¡ xand hence show that

Z 4

3

¡2x

4 ¡ x2dx = ln(125 ).

9 By appealing only to geometrical evidence,

explain why:Z 1

0

ex dx +

Z e

1

lnxdx = e:

y x�

y xln�

y e� x

x

y

1

1

x

y y x� 2

y k�

84x¡ 3

2x + 1can be written in the form A +

B

2x + 1.

a Find the values of A and B. b Hence find

Z 2

0

4x¡ 3

2x + 1dx:

9 Find a given that the shaded area is 4 units2.

Find the x-coordinate of A if OA divides the

shaded region into equal areas.

1 Find y if: ady

dx= (x2 ¡ 1)2 b

2 Evaluate: a

Z 0

¡2

4

2x¡ 1dx b

Z 1

0

10xp3x2 + 1

dx

3 By differentiatingp

3x2 + 1, find

Zxp

3x2 + 1dx.

4 O is a point on a straight line. A particle moving on this straight line has a velocity of

27 cms¡1 as it passes through O. Its acceleration t seconds later is 6t¡ 30 cms¡2.

Find the total distance (from O) that the particle has travelled when it momentarily comes

to rest for the second time.

x

y

y ax x( 2)�

A

2

dy

dx= 400 ¡ 20e

¡

x

2



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7 OABC is a rectangle and the two shaded

regions are equal in area. Find k.

8 Consider f(x) =3x¡ 5

(x¡ 2)2:

a State the axis intercepts.

b State the equation of any vertical asymptotes.

c Find the position and nature of any turning points.

d Using a graphics calculator to help, sketch the graph of y = f(x).

e Find constants A and B such that3x¡ 5

(x¡ 2)2=

A

x¡ 2+

B

(x¡ 2)2:

f Find the area of the region defined by y = f(x), the x-axis and the vertical

line x = ¡1:

9 Determine m and c if the enclosed

region has area 412 units2.

y x k� �2y

x

A B

C2

k

y mx c� �

y x x2 3� � �2

y

x 1

REVIEW SET 6D

1 Find f(x) if: a f 0(x) =p

3 ¡ 2x b f 0(x) = (ex ¡ e¡x)2

2 Evaluate: a

Z 2

1

(x + 1)2px

dx b

Z 4

1

e¡px

px

dx

3 If y =px e¡x, find

dy

dxand hence determine

Z 1

0

2 ¡ 4xpxex

dx.

4 A particle moves in a straight line with acceleration given by a(t) = 2t¡ 3 cms¡2.

When t = 3 its velocity is 2 cms¡1. Find:

a the velocity function

b the times when the particle reverses direction

c the total distance travelled by the particle in the first 3 seconds.

5 Determine the area enclosed by y =px and y = x3.

In what ratio does the straight line joining (0, 0) and (1, 1) divide the area of this region?

6

Z k

1

1

1 ¡ 3xdx = ¡ ln 4. Find k given that k > 1.



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REVIEW SET 6E

3dy

dx= a

p1 ¡ x for a given function.

The graph of the function passes through (0, 1) and (¡3, ¡27).

Find a and the equation of the tangent at (0, 1).

4 Find constants A and B given that1

x(x + 1)=

A

x+

B

x + 1.

Hence determine

Z1

x2 + xdx.

5 Findd

dx(lnx)2 and hence find

Zlnx

xdx:

6 Show that y = ¡3x + 2 touches y = x3 ¡ 6x at one point (A, say) and cuts it at

another point (B). Find the coordinates of A and B.

Sketch each graph on the same set of axes and find the area enclosed by the two graphs.

7 It can be shown that if y = f(x) is revolved

about the x-axis to form a solid between x = aand x = b, then the volume of the solid is given

by V =R b

a¼[f(x)]2 dx:

Use this formula to prove that the volume of the

solid of revolution when x2 + y2 = r2 revolves

about the x-axis is V = 43¼r

3.

8 The area of the region defined by y = x2 and y = mx is 43 units2. Find m.

x

y

a b

1 A boat travelling in a straight line has its engine turned off at time t = 0.

Its velocity in metres per second at time t seconds is then given by

v(t) =100

(t + 2)2ms¡1, t > 0.

a Find the initial velocity of the boat, and its velocity after 3 seconds.

b Discuss v(t) as t ! 1.

c Sketch the graph of v(t) against t.

d Find how long it takes for the boat to travel 30 metres.

e Find the acceleration of the boat at any time t.

f Show thatdv

dt= ¡kv

3

2 , and find the value of the constant k.

2 The graph of y = f(x) is illustrated below.

It is known thatR 3

0f(x) dx = 3.

a What can be deduced about the areas A and B?

b CanR 2

0f(x)dx = 2?

A

B

32

y

x

y f x( )�



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7

Contents:

StatisticsStatistics

A

B

C

D

E

F

G

H

I

J

Key statistical concepts

Describing data

Normal distributions

The standard normal distribution

Finding quantiles ( -values)

Investigating properties of normaldistributions

Distribution of sample means

Hypothesis testing for a mean

Confidence intervals for means

Review

k


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DISCUSSION SAMPLING

Words that are commonly used in Statistics:

² Population A collection of individuals about which we want to drawconclusions.

² Census The collection of information from the whole population.

² Sample A selection of information from a subset of the population.

² Data (singular datum) Information about individuals in a population.

² Parameter A numerical quantity measuring some aspect of a population.

² Statistic A quantity calculated from data gathered from a sample.It is usually used to estimate a population parameter.

² Distribution The pattern of variation of data.

A population generally consists of a large number of individuals. Because of expense and

time factors it is often only practical to select a sample rather than use the whole population.

A random sample is a sample where every individual has the same chance of being selected.

A sampling technique is biased if it tends to systematically select members of the population

with certain properties and not select those that do not have these properties. In other words

it favours some individuals above others.

INTRODUCTION

KEY STATISTICAL CONCEPTSA

In the following scenarios, can you suggest a likely population?

Can you think of any reasons the sampling techniques might be biased?

People in the local shopping centre on Saturday morning were askedhow many computers they have in their household.

After a program likely to be watched by older people, a televisionstation asked viewers to vote on the use of hand-held phones in cars.

A local paper advertised for volunteers to test the usefulness of fish oilin a diet.

²

²

²

The word was introduced into the English language by the

Scottish politician ( – ). He borrowed it

from Germany where, as he put it, it meant,

“

”.

The meaning he wished to give to the word was an

“

.”

You can still recognise the word “state” in statistics.

statistics

Sir John Sinclair 1754 1835

an inquiry for the purpose of ascertaining the political

strength of a country

inquiry into the state of a country, for the purpose of

ascertaining the quantum of happiness enjoyed by its

inhabitants, and the means of future improvement

RANDOM SAMPLES

220 STATISTICS (Chapter 7)


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Many sampling techniques have been developed to avoid bias. In this book it will be assumed

that any sample is a random, unbiased sample.

Descriptive statistics are concerned with collecting, summarising and describing the

characteristics of data.

With descriptive statistics we are only concerned with the data collected and make no effort

to generalise it to any other data, such as for the population.

In inferential statistics we select a random sample and we use the information from it

to make generalisations about the population from which the sample was taken.

Recall that:

a parameter is a numerical characteristic of a population and

a statistic is a numerical characteristic of a sample.

For example, when examining the mean age of people in retirement villages throughout

Australia, the mean age found would be a parameter. If we took a random sample of 300people from the population of all retirement village persons, then the mean age would be a

statistic.

Note:P

S

arameter

opulation

ample

tatistics

a What is the population size?

b What is the sample size?

c What population parameter is of interest to the business?

d What statistic is being used to estimate the parameter?

a The population is the number of blank CDs to be purchased and its size is

50 000.

b The sample size is 600:

c The population parameter being considered is the percentage of CDs which

are defective.

d The statistic being used is the percentage of CDs which are defective in

the sample. As 1:5% of 600 = 9, the business would make the purchase if

9 or less CDs in the sample were found to be defective.

A business is considering purchasing blank CDs to make CDs of their new text

books. It will make the purchase if no more than of the CDs are defective.

Because of the expense and time factors in testing all CDs the business decides

to test a random sample of for defects. They will then use the results of this sample

to estimate the percentage of defectives for the population to be purchased.

500001 5%50000

600

�

�:

Example 1

DESCRIPTIVE AND INFERENTIAL STATISTICS

EXAMPLES OF PARAMETERS AND STATISTICS

STATISTICS (Chapter 7) 221


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In this course the key application is to examine a random sample in order to make appropriate

statements or inferences about the population.

Generally speaking there are five steps to address in any inferential problem. They are:

THE PROCEDURE USED IN AN INFERENTIAL PROBLEM

Step 1: State the population we are interested in examining.

Step 2: Collect data from a random sample of sufficient size from the population.

Note: What is meant by sufficient size is covered in a later chapter.

Step 3: Examine the relevant information from the sample.

Step 4: Use the results of the sample analysis to make an inference about the

population.

Step 5: Give a measure of the reliability of the inference made.

For the CD purchase in Example 1 list the procedural steps for the inferential

problem.

Step 1: The population consists of all 50 000 CDs.

Step 2: To avoid unnecessary costs and wasting time we must first decide on the

sample size. 600 has been decided upon, so we collect 600 data values

at random. We record only whether the CD is defective or not.

Step 3: Find the percentage of defective CDs in the sample.

Step 4: The inference will be to provide an estimate of the percentage of defective

CDs for the whole population. For example, if 12 CDs are defective in

the sample our inference would be that approximately 12600 = 2% would

be defective in the population.

Step 5: The estimate from the sample is not likely to be equal to the exact

value for the population. Some indication of the possible error for the

estimate should therefore be given.

An example of such a statement as in Step 5 is:

If we had many shipments of 50 000 CDs and in each we found that 12 in a sample of

600 were defective, then in 95% of these shipments there would be between 440 and

1560 defective CDs.

This type of statement is usually condensed to:

We are 95% confident that about 440 to 1560 CDs are defective.

The main thrusts of this course are to:

² determine confidence intervals in which a certain population parameter should lie at

a particular level of confidence (commonly 90%, 95%, 99%)

² devise and use particular tests of hypotheses about population means

² determine what sample sizes should return a particular level of confidence in given

situations.

Example 2



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a In this city, bananas are cheaper than oranges.

If you buy a kilogram of each of the three different types of fruit from the one

store, you pay the same total amounts at stores A and D.

Of the four stores, the store with the most expensive apples also had the most

expensive oranges and bananas.

In general, store C has the

most expensive fruit.

Of the four stores, store C

has the most expensive

fruit. (Careful! What is the

population and what is the

sample?)

b

c

d

e

1 A new drug called Cobrasyl, a derivative of cobra

venom, is to be approved for the treatment of high

blood pressure in humans.

A research team treats 127 high blood pressure

patients with the drug and in 119 cases it reduces

their blood pressure to an acceptable level.

a What is the sample of interest?

b What is the population of interest?

2 In 2006, 800 computer workers throughout Australia were surveyed and asked a question.

The question was: “Is your main interest in developing software or in using already

developed software?” 83% said that developing software was their main interest.

a What is the population of interest?

b What is the parameter of interest?

c What statistic is used to estimate the parameter?

3

a

b What is the parameter of interest?

c

4 Last December Tina visited four super-

markets A, B, C and D on the same day.

She recorded the price per kilogram of

various fruits in the table opposite:

Determine whether the following state-

ments are descriptive or inferential:

Store Oranges Apples Bananas

A $2:35 $2:15 $1:70

B $2:45 $2:55 $2:00

C $2:50 $2:60 $2:10

D $2:25 $2:05 $1:90

EXERCISE 7A

What is the population the processor is

interested in?

A South Australian processor of seafood needs to

estimate the average weight of a prawn in a

catch. A sample of prawns was selected and

found to have an average weight of grams.

35253 8:

What statistic does the processor use to

estimate the parameter?



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This section will review the main concepts from Year 11 so that students will reacquaint

themselves with the terminology used in statistics.

A variable is a quantity that can have different values for different individuals in the

population.

Since variables are sometimes used to describe random processes, they are often called

random variables.

Variables are usually denoted by capital letters such as X. Individual values, called observa-

tions or outcomes, are denoted by lower case letters such as x.

We shall deal with two types of variables: categorical and quantitative.

A categorical or nominal variable can be described by a quality or characteristic that

is essentially non-numeric. Individuals are described by different categories.

DESCRIBING DATAB

Examples of categorical data are:

Variable Possible values

² X is the gender of a person x = male or female

² C is the type of motor car c = Holden, Ford, Toyota

² M is the membership of political party m = ALP, LIB, DEM

A quantitative or numerical variable takes numerical values.

There are essentially two different types of numerical variable.

A numerical discrete variable takes discrete number values only.

It is often a result of counting.

Examples of discrete variables are:


² X is the number of people in a household x = 1, 2, 3, 4 ::::::

² T is the mark out of 10 for a test t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Examples of continuous variables are:


² W is the weight of newborn babies w is likely to be in the interval from 0:5 kg

to 5 kg.

² X is the amount of water in a 500 litre

rain water tank

x is any volume between 0 and 500 litres.

A can take any numerical value in an interval.

A continuous variable is often a result of measuring.

numerical continuous variable



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In a sample of size n, the sample standard deviation, usually denoted by s, is:

s =

s(x1 ¡ x)2 + (x2 ¡ x)2 + :::::: + (xn ¡ x)2

n ¡ 1=

sP(xi ¡ x)2

n¡ 1

In a population of size n, the population standard deviation, usually denoted

by the Greek letter ¾ (sigma), is:

¾ =

s(x1 ¡ ¹)2 + (x2 ¡ ¹)2 + :::::: + (xn ¡ ¹)2

n=

sP(xi ¡ ¹)2

n

Since continuous variables take on values in intervals, they are also called interval variables.

The essential difference between a categorical and a quantitative variable is that we can do

arithmetic with quantitative variables, but not with categorical variables.

In this book we are mainly concerned with the mean and the standard deviation.

The mean of a sample of n numbers,

x1, x2, ......... , xn is: x =x1 + x2 + ::::::: + xn

n=

1

n

nPi=1

xi

The Greek letterP

(sigma) is used to denote the summation of numbers,

sonP

i=1xi = x1 + x2+ ::::::: +xn (read “the sum of all xi for i = 1 to n”).

The endpoints of the summation, i = 1 to n are sometimes omitted, so the mean can be

written as 1n

Pxi or even 1

n

Px.

The mean of a population is usually denoted by the Greek letter ¹ (mu), so ¹ = 1n

Px.

We can get a much clearer picture of a data set if, in addition to having a measure for the

centre, we also have an indication of how the data is spread.

For example, the mean weight of oranges from a particular orchard and the mean weight of

salt bagged by a machine may both be 500 grams, but the variation in the weights of oranges

is likely to be much greater than that of bags of salt. The data for oranges will therefore have

a greater spread.

The most commonly used measure of spread about the mean is the standard deviation.

The standard deviation of a sample is a little different from the standard deviation of a

population.

THE MEAN AND STANDARD DEVIATION (REVIEW)

The reason for this difference is rather technical and, at this stage we do not attempt to explain

the difference.

Statisticians know that the value of s, as calculated by the above formula, gives an unbiassed

estimate of the population standard deviation ¾.

Notice that for large n, the values of s and ¾ are virtually the same.



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The mean and standard deviation can also be calculated from frequency tables.

The frequency fi of a quantity xi is the number of times it occurs.

For a population of size n, the formulae for the mean and standard deviation become:

¹ =f1x1 + f2x2 + f3x3 + :::::: + fkxk

n

and ¾ =

r(x1 ¡ ¹)2f1 + (x2 ¡ ¹)2f2 + :::::: + (xk ¡ ¹)2fk

n

Notice that ¹ =

µf1n

¶x1 +

µf2n

¶x2 +

µf3n

¶x3 + :::::: +

µfkn

¶xk.

fin

is the proportion of xi in the population. For large values of n, the experimental

probability pi of randomly selecting xi from the population is taken to be pi =fin

.

So, using pi =fin

, ¹ = p1x1 + p2x2 + p3x3 + :::::: + pkxk =X

pixi :

Similarly for the population standard deviation:

¾ =

sµf1n

¶(x1 ¡ ¹)2 +

µf2n

¶(x2 ¡ ¹)2 + :::::: +

µfkn

¶(xk ¡ ¹)2

which leads to ¾ =qX

pi(xi ¡ ¹)2.

The probability table is: xi 0 1 2 3 4 5

pi 0:00 0:23 0:38 0:21 0:13 0:05

Now ¹ =X

pixi

= 0:23 £ 1 + 0:38 £ 2 + 0:21 £ 3 + 0:13 £ 4 + 0:05 £ 5

= 2:39

i.e., in the long run, the average number purchased per customer is 2:39

Also, ¾ =qX

pi(xi ¡ ¹)2

=q

0:23 £ (1 ¡ 2:39)2 + 0:38 £ (2 ¡ 2:39)2 + :::: + 0:05 £ (5 ¡ 2:39)2

+ 1:12

A magazine store claims of its customers purchase one magazine, purchase

two, purchase three, purchase four, and purchase five. Find the mean

and the standard deviation of , the number of magazines sold to a customer.

23% 38%21% 13% 5%

X

Example 3



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‘Cheap Car Insurance’ insures used cars valued at $6000 under these conditions.

A $6000 will be paid to the owner for total loss

B for damage between $3000 and $5999, $3500 will be paid

C for damage between $1500 and $2999, $1000 will be paid

D for damage less than $1500, nothing will be paid.

From statistical information the insurance company knows that in any year the

probabilities of A, B, C and D are 0:03, 0:12, 0:35 and 0:50 respectively.

If the company wishes to receive $80 more than its expected payout on each

policy, what should it charge for the policy?

Let X be the random variable of payouts, so the probability table is:

xi 0 1000 3500 6000

pi 0:50 0:35 0:12 0:03

The expected payout is the mean, ¹, and

¹ =P

pixi

= (0:50) £ 0 + (0:35) £ 1000 + (0:12) £ 3500 + (0:03) £ 6000

= 950

The company expects to pay out $950 on average in the long run, so it should

charge $950 + $80 = $1030:

1

xi 0 1 2 3 4 5 > 5

P (xi) 0:54 0:26 0:15 0:03 0:01 0:01 0:00

a What is the mean number of deaths per dozen crayfish?

b Find ¾, the standard deviation for the probability distribution.

2

Example 4

EXERCISE 7B

Australian crayfish is exported to Asian markets. The

buyers are prepared to pay high prices when the crayfish

arrive still alive. If is the number of deaths per dozen

crayfish, the probability function for is given by:

XX

A random variable X has probability function given by

P (x) = k(0:4)x(0:6)3¡x for x = 0, 1, 2, 3.

a Find P (x) for x = 0, 1, 2 and 3 and hence find k.

b Find the mean and standard deviation for the distribution.

3 An insurance policy covers a $20 000 sapphire ring against theft and loss. If it is stolen

the insurance company will pay the policy owner in full. If it is lost they will pay the

owner $8000. From past experience the insurance company knows that the probability

of theft is 0:0025 and of being lost is 0:03. How much should the company charge to

cover the ring if they want a $100 expected return?



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NORMAL DISTRIBUTIONSC

DISCUSSION THE EFFECT OF RANDOM FACTORS

4 Use technology to find the mean and standard deviation of the two samples, A and B,

of weights given in grams.

A 498:8 500:2 500:4 499:9 500:4 500:6 498:9 498:2 500:1 501:9500:8 498:6 499:7 498:6 499:0 498:8 499:1 500:7 500:7 501:3501:1 501:5 499:0 499:7 498:4 501:1 500:1 499:9 500:9 499:2

B 545:5 543:4 399:8 511:3 616:3 496:7 337:8 650:2 426:3 522:2664:0 415:1 416:0 425:4 419:9 503:7 427:8 474:2 459:9 390:5428:5 451:9 590:1 613:5 402:3 318:3 478:1 502:2 626:4 435:7

Which of the samples is the weights of bags of salt, and which is the weights of oranges?

5 Test marks out of 10 are recorded in the following frequency table:

Mark 0 1 2 3 4 5 6 7 8 9 10

Frequency 2 1 0 4 5 8 12 15 7 3 5

a Find the mean and standard deviation of these scores.

b Calculate the percentage difference between using the formulae for population

standard deviation and sample standard deviation.

6 Using ¾2 =P

pi(xi ¡ ¹)2 show that ¾2 =P

pix2i ¡ ¹2:

(Hint: ¾2 =P

pi(xi ¡ ¹)2 = p1(x1 ¡ ¹)2 + p2(x2 ¡ ¹)2 + :::::: + pn(xn ¡ ¹)2:

Expand ¾2 and regroup the terms.)

Many quantities reflect the combined effect of a large number of random factors.

For example:

²

²

² Consider at least three factors that affect each of the following:

a the weight of a newly born piglet

b the time to complete an assignment

c the mark achieved in an examination

d the number of goals scored in a netball match.

² For each of the above random variables, suggest why the distribution might be

a symmetric b bell shaped.

The next investigation explores the distribution of a quantity that is the combined result of

different factors.

The yield of a wheat plant is the combined result of many unpredictable factors such

as genes, rainfall, sunshine, and its position in the field where it was seeded.

The weight of a packet of sultanas is the sum of the weights of each individual

sultana, and it is unlikely a packet labelled as kg will weigh exactly kg.1 1



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INVESTIGATION 1 SOME PROPERTIES OF A NORMAL DISTRIBUTION

Stage What is happening Time

1 Cross the road in front of the school up to 1 minute

2 Walk to the shopping centre 5 § 2 minutes

3 Walk through the shopping centre 3 § 2 minutes

4 Cross a road up to 1 minute

5 Buy a loaf of bread up to 2 minutes

6 Talk with a friend up to 2 minutes

7 Walk the remaining distance home 2 § 1 minutes

Question: According to the table, what is the longest time it may take Les to walk

home? What is the shortest time?

If Les wanted to study the distribution of the time it takes to walk home, he could keep a

daily record, but the amount of data collected would be very small.

Les could also use the information given in the table and use a spreadsheet or a calculator

to simulate the time it takes to walk home.

The following instructions are set up for a spreadsheet, but the procedure will also work

on a calculator.

1 Open the spreadsheet “Normal distribution”.

A spreadsheet with the following headings will appear.

2 In each of the cells A2 to G2, under the headings ‘Stage 1’ to ‘Stage 7’, type in the

formulae shown in the table. Do not forget to start each formula with an = sign.

Note: rand() calculates a random number between 0 and 1.

Question: What does 5 + (4*rand( ) ¡ 2) calculate?

3 In cell N2, below the heading ‘Total time’, type in the formula =sum(A2:M2)

Question: What does this formula calculate?

4 Drag the formulae in cells A2 to N2 down to fill all cells A251 to N251. Pressing

the F9 function key will produce another random sample.

Consider the time it takes Les to walk home from school. We have broken

this into the following stages with the time it takes to complete each stage:

What to do:SPREADSHEET

The numbers in cell P2 under the heading ‘Mean’, and in cell Q2 under the heading

‘Standard Deviation’, are the mean and standard deviation of the numbers in cells N2to N251.

The number in cell R2 under the heading ‘No. within 1 st. dev.’ gives the number of



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values within 1 standard deviation of the mean. For example, if the mean x = 12:96and the standard deviation s = 1:82, then this cell gives the number of values that

lie between x ¡ s = 11:14 and x + s = 14:78 . Similarly, the numbers in cells

S2 and T2 give the number of values within 2 and 3 standard deviations of the mean

respectively.

If you are having difficulty setting up this spreadsheet, click on the tag ‘Normal 2’ to

open a finished version.

5 Calculate the proportion of data values within each interval. For example, if there are

169 values within 1 standard deviation of the mean, the proportion of values in the

interval = 169250 = 0:676 .

6 Copy and fill in the following table for 5 different samples. The entries of the first

line may not agree with your values.

Sample Mean Stdev x¡ s to x + s x¡ 2s to x + 2s x¡ 3s to x + 3sno. x s Count Propn. Count Propn. Count Propn.

1 12:96 1:82 169 0:676

2

3

4

5

What do you notice about the proportions of data in each of the intervals?

In the following we change the value of the factors and then add more factors.

7 Change the formulae in cells A2 to G2 as shown in the table.

8 Repeat steps 4 to 6.

9 Add the following formulae in cells H2 to M2:

10 Repeat steps 4 to 6.

The graph that appears is the

histogram of data in cells N to N .2 251



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From Investigation 1 you should have discovered that changing the number and values of

factors may change the mean and standard deviation, but leaves the following unchanged:

Note:

A smooth curve drawn through

the midpoints of each column

of the histogram would ideally

look like the graph displayed.

Note the points of inflection at

¹¡ ¾ and ¹ + ¾.

The above information is typical of a family of normal distributions. Curves with this shape

are known as normal curves. Because of their characteristic shape, they are also called

bell-shaped curves.

Variables which are the combined result of many random factors are often approximately

normal.

The normal variable X with mean ¹ and standard deviation ¾ is denoted by X » N(¹, ¾2).

34% 34%

13.5% 13.5%

2.35%0.15% 0.15%2.35%

� ��

¹¡¹ ¾ +¹ ¾

concave

convex convex

point of inflection point of inflection

² The shape of the histogram is symmetric about the mean.

² Approximately 68% of the data lies between 1 standard deviation below the mean

and 1 standard deviation above the mean.

² Approximately 95% of the data lies between 2 standard deviations below and 2standard deviations above the mean.

² Approximately 99.7% of the data lies between 3 standard deviations below and 3standard deviations above the mean.

It is a rare event for an outcome to be outside the standard deviation range between

and . In a sample of , you would only expect about cases.¡3 3 1000 3¾ ¾

For any distribution of data, whether it is a normal distribution or not, the function whose

smooth curve approximates the histogram of the data is called a probability density function

or pdf.

If the variable X is normally distributed, N(¹, ¾2), the probability density function is

f(x) =1

¾p

2¼e¡

12 (

x¡¹¾

)2 .

CONTINUOUS PROBABILITY DENSITY FUNCTIONS



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Probability density functions f have the following properties:

² f(x) > 0 for all values of x.

² The area between the graph of f and the horizontal axis is 1, since the total of all

probabilities is 1.

² The proportion of outcomes of the variable X between the values a and b is the

area between the graph of f and the horizontal axis for a 6 x 6 b.

Notice that: Pr(a 6 X 6 b) =

Z b

a

f(x)dx

For a continuous variable X, the probability X is exactly equal to a point a is zero.

For example, the probability an egg will weigh exactly 72:9 g is zero.

If you were to weigh an egg on scales that weigh to the nearest 0:1 g, a weight of 72:9 g

means the weight lies somewhere between 72:85 and 72:95 grams.

Presumably an egg has to weigh something, and it could be 72:9 grams, but you will never

know. No matter how accurate your scales are, you can only ever know the weight of an egg

within a range.

So, for a continuous variable we can only talk about the probability an event lies in an

interval.

Notice that:

if X is continuous, Pr(a 6 X 6 b), Pr(a < X 6 b), Pr(a 6 X < b)and Pr(a < X < b) all have the same value. Why?

This would not be correct if X was discrete.

87 95 103 111� � ��

��

��

34% 34%

13.5%

� � ��

The chest measurements of 18 year old male footballers are normally distributed with

a mean of 95 cm and a standard deviation of 8 cm.

a Find the percentage of randomly chosen footballers with chest measurements

between: i 87 cm and 103 cm ii 103 cm and 111 cm

b Find the probability of randomly choosing a footballer with a chest measurement

between 87 cm and 111 cm.

a i We need the percentage between

¹¡ ¾ and ¹ + ¾. This is 68%.

ii We need the percentage between

¹ + ¾ and ¹ + 2¾. This is 13:5%:

b The percentage between ¹¡ ¾ and

¹ + 2¾ is 68% + 13:5% = 81:5%:

So the probability is 0:815

For the distribution of chest measurements, the mean

cm and the standard deviation cm.¹ ¾� � � � � �=95 =8

Example 5



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1 What is the probability that a normally distributed value lies between:

a 1¾ below the mean and 1¾ above the mean

b the mean and the value 1¾ above the mean

c the mean and the value 2¾ below the mean

d the mean and the value 3¾ above the mean?

2 Suppose the heights of 16 year old male students are normally distributed with a mean

of 170 cm and a standard deviation of 8 cm. Find the percentage of male students whose

height is:

a between 162 cm and 170 cm b between 170 cm and 186 cm.

Find the probability that a student from this group has a height:

c between 178 cm and 186 cm d less than 162 cm

e less than 154 cm f greater than 162 cm.

3 The time T minutes it takes Charlotte to go to work is normally distributed with mean

50 minutes and standard deviation of 5 minutes. Every morning Charlotte leaves for

work at 8 am.

a If work starts at 9 am, what is the probability Charlotte will be late for work?

b If Charlotte works 250 days a year, how many times can she expect to be late?

4 Explain why each of the following variables might be normally distributed:

a the chest size of 18 year old Australian males

b the length of adult female sharks

c the protein content of each kilogram of corn grown in the same field.

5 A farmer has a flock of 237 crossbred lambs. The mean weight of the flock is 35 kg

with a standard deviation of 2 kg.

a Explain why the weights of the lambs might be normally distributed.

b If lambs between the weights of 33 to 39 kg are suitable for export, how many

lambs in this flock could the farmer expect to be able to export?

6 The weights of hens’ eggs are normally distributed with mean 65 grams and standard

deviation 6 grams.

a Determine the probability that a randomly selected egg has weight

i greater than 53 g ii less than 71 g iii between 59 g and 77 g.

b In one week the hens lay 1286 eggs. How many of these eggs are expected to be

i greater than 53 g ii less than 71 g iii between 59 g and 77 g.

7 The marks for a geography examination are normally distributed with mean 65 and

standard deviation 11.

a A geography student is chosen at random. Determine the probability that the student

scored i less than 76 marks ii between 43 and 76 marks.

b

c If 2582 students sit for the examination, how many of them would be expected to

score less than 32 marks?

EXERCISE 7C

If the top of students receive an A grade, what was the minimum mark

for an A?

16%



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For each value of ¹ and ¾ there is a different normal distribution N(¹, ¾2).

As illustrated by Investigation 1, all normal distributions have one important property in

common: the probability of an event occurring depends only on the number of standard

deviations the event is from the mean.

If x is an observation from a normal distribution with mean ¹ and standard deviation ¾,

the z-score of x is the number of standard deviations x is from the mean.

The diagram shows how

the z-score is related to

a normal curve.

�� x x x

THE STANDARD NORMAL DISTRIBUTIOND

34% 34%

13.5% 13.5%

2.35%0.15% 0.15%2.35%

Normal distribution curve

��

��

actual score

z-score

8 The weights of Jason’s oranges are normally distributed. 84% of the crop weigh more

than 152 grams and 16% weigh more than 200 grams.

a Find ¹ and ¾ for the crop

b What proportion of the oranges weigh between 152 grams and 224 grams?

9 The heights of 13 year old boys are normally distributed. 97:5% of them are above 131cm and 2:5% are above 179 cm.

a Find ¹ and ¾ for the height distribution

b A 13-year old boy is randomly chosen. What is the probability that his height lies

between 143 cm and 191 cm?

10 Using the same set of axes, quickly sketch the graphs of the density functions for each

of the following distributions:

a N(0, 32) b N(0, (0:5)2) c N(¡5, 12) d

11 Each of the following is a graph of a normal distribution with different vertical scales:

A B C

a Write down the mean ¹ for each of these distributions.

b Which of the distributions has standard deviation

i ¾ = 0:1 ii ¾ = 1 iii ¾ = 10 ?

c Which of the distributions has the largest spread?


N(3, 0:252).


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z-scores are particularly useful when comparing two measurements made using different ¹and ¾. But be careful! These comparisons will only be reasonable if both measurements are

approximately normal.

a i Sketch the graphs of the two distributions using the same scale for the

z-scores from ¡3 to +3.

ii Put the actual times/distances below each of the z-scores on the graphs.

iii Calculate the z-scores for John and Anne, and mark these on the graphs.

iv Shade the area under the respective graphs to represent performances that

were better than those of John and Anne.

b Of all the students who participated in these two events, what proportion would

have performed better than i John ii Anne?

c If 1000 students had participated in each of these two events, how many would

have performed better than i John ii Anne?

d Of the father and daughter, who had the better result?

a i/ii/iv

iii John’s time was 3:2 ¡ 3:4 = ¡0:2 minutes from the mean.

Since the standard deviation is 0:2 minutes, John ran the 800 metres in a

time of 1 standard deviation less than the mean.

The z-score of John’s performance is ¡1:

The distance Anne jumped was 5:1 ¡ 4:3 = 0:8 m above the mean.

Since the standard deviation is 0:4 metres, Anne jumped a distance of 2standard deviations above the mean.

The z-score of Anne’s performance is +2.

Example 6

The local school has kept records of all its athletics competitions. It was found that

the time, in minutes, to run the men’s metres was normally distributed as

N , . The women’s long jump, in metres, was normally distributed as

N , . In John won the metre race with a time of minutes. In

his daughter Anne came second in the long jump with a distance of m.

800(3 4 (0 2) )(4 3 (0 4) ) 1980 800 3 2

2006 5 1

: :: : :

:

2

2

34% 34%

13.5% 13.5%

2.35%0.15% 0.15%2.35%

John’s time

� � � � � � �actual time (min)

z-score��

��

34% 34%

13.5% 13.5%

2.35%0.15% 0.15%2.35%

Anne’s distance

� � � � � � �

actual distance (m)

z-score

better than John

better than Anne



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b i The proportion less than ¹¡ ¾ is 0:16, so 16% of all participants

performed better than John.

ii The proportion greater than ¹ + 2¾ is 0:025, so only 2:5% of all

participants performed better than Anne.

c i Of 1000 participants, 16% of 1000 = 160 were better than John.

ii 2:5% of 1000 = 25 were better than Anne; one of these happened

to be competing on the same day as Anne.

d Anne’s long jump was more outstanding than her father’s 800 metre race.

1 In a year 12 class, the marks for a Geography test marked out of 50 were normally

distributed with mean of 34 and standard deviation of 6. The marks for an English essay

out of 20 were normally distributed with a mean of 12 and standard deviation of 1:5 .

Val received a mark of 40 for her Geography and 15 for her English essay.

a Sketch the graphs of the two distributions below one another using the same scale

for the z-scores from ¡3 to +3.

Put the actual marks below each z-score on the graph.

b For which of the two subjects did Val receive the higher % mark?

c Calculate the z-score for each of Val’s results.

i Mark these z-scores on the two graphs.

ii Shade the region on the two graphs of scores which were better than Val’s.

d What proportion of the students performed better than Val in Geography, and what

proportion performed better than Val in English?

e If there were 32 students in the class, how many performed better than Val in

Geography and how many in English?

f In which of these two assessments did Val perform better?

2 Suppose that the weight W of bags of sugar filled by a machine are normally distributed

with mean ¹ = 504 grams and standard deviation ¾ = 2 grams.

A quality controller rejects any bags of sugar with weight less than 500 grams.

Across town, the weight A of bags of apples filled by an assistant in a green grocer shop

is normally distributed with mean weight 5 kilograms and standard deviation 500 grams.

Bags weighing less than 412 kg are rejected by a quality controller.

a Sketch the graphs of the two distributions below one another using the same scale

for the z-scores from ¡3 to +3.

Put the actual weights below each z-score on the graph.

b Calculate the z-score for each of the two quality controls, and shade in the regions

corresponding to the weights of bags that are rejected.

c Which of the two quality controllers is the more stringent, i.e., rejects the larger

proportion of bags?

EXERCISE 7D.1



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b Hua’s mark is ¡1:5 standard

deviations from the mean.

Since the standard deviation is

12, this is 12 £ (¡1:5) = ¡18marks from the mean.

Since the mean is 63, Hua’s

mark is 63 + (¡18) = 45.

3 Suppose the distribution of the diameter (in cm) of oranges from a tree is N(10, 22).

a Sketch a graph of the distribution that displays both the actual diameters as well as

the z-score along the horizontal axis.

b Find the z-score for each of the following diameters:

i 12 cm ii 9 cm iii 13 cm

c Oranges are to be dumped if their diameters have a z-score of less than ¡2.

What is the diameter of oranges that are to be dumped?

d If there are 120 oranges on the tree, how many will be dumped?

4 The volume of milk cartons filled by a machine is normally distributed with mean 504mL and standard deviation of 1:5 mL.

a What is the z-score of a carton containing 506 mL of milk?

b What is the volume of milk in a carton with a z-score of ¡1:5?

Hua’s mark

�

��

�

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�

��

�

��

�

��

�

��

�

��actual mark

z-score

If x is an observation from a normal distribution with mean ¹ and standard deviation ¾, the

z-score of x can be calculated from the formula z =x¡ ¹

¾.

If the variable X is normally distributed with mean ¹ and standard deviation ¾, then

Z =X ¡ ¹

¾is called the standard normal distribution.

The variable Z is the number of standard deviations X is from the mean.

Notice that, if x = ¹ then z = 0 and if x = ¹ + ¾ then z = 1.

Suppose examination scores are normally distributed with mean mark ¹ = 63 and

standard deviation of ¾ = 12 marks.

a What is the z-score for a mark of 80?

b If Hua’s z-score is ¡1:5, what is Hua’s actual score?

a A mark of 80 is 80 ¡ 63 = 17above the mean.

Since the standard deviation

is 12, this is 1712 = 1:42 standard

deviations above the mean.

So, the z-score is 1:42

score of 80

�

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�

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�

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�

��

�

��

�

��

�

��actual mark

z-score

Example 7

Hence, the mean of is and the standard deviation of is .Z Z0 1



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When working with normal distributions, you are advised to sketch a graph of the normal

distribution and shade in the areas of interest.

Use technology to illustrate and calculate:

a Pr(¡0:41 6 Z 6 0:67) b Pr(Z 6 1:5) c Pr(Z > 0:84)

a For a TI, Pr(a 6 Z 6 b)

can be calculated using normalcdf(a, b, 0, 1)

Pr(¡0:41 6 Z 6 0:67)

= normalcdf (¡0:41, 0:67, 0, 1)

+ 0:408

USING TECHNOLOGY TO FIND PROBABILITIES

TI

C

Example 9

0.41

0

0.67

The probability Z lies between ¡2 and 1 is the proportion of observations that lie

between 2 standard deviations to the left of the mean and 1 standard deviation to

the right of the mean. This is about 0:815 .

1 Subject Emma’s score ¹ ¾

English 12 10 1:1

Chinese 27 20 3:0

Geography 84 55 18

Biology 34 25 10

Mathematics 84 50 15

a Find the z-score for each of

Emma’s subjects.

b Arrange Emma’s subjects from

‘best’ to ‘worst’ in terms of the z-scores.

2 Calculate the following probabilities. In each case sketch the graph of the Z-distribution

shading in the region of interest.

a Pr(¡1 < Z < 1) b Pr(¡1 < Z < 3) c Pr(¡1 < Z < 0)

d Pr(Z < 2) e Pr(¡1 < Z) f Pr(Z > 1)

EXERCISE 7D.2

34% 34%

13.5%

� � � � � � �

z

Find the probability that the standard normal distribution Z lies between ¡2 and 1.

The graph of the Z-distribution is shown:

Example 8

The table shows Emma’s midyear exam

results. The exam results for each subject are

normally distributed with mean and

standard deviation shown in the table.

¹¾

So far we have only used integer -scores to calculate probabilities. By

refining the methods used in we can calculate probabilities for

other -scores. To see how to use your calculator to do this, click on the icon.

z

zInvestigation 1



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b Pr(Z 6 1:5)

= normalcdf(¡E99, 1:5, 0, 1)

+ 0:933

Note: ¡E99 is the largest negative

number on a calculator.

c Pr(Z > 0:84)

= normalcdf(0:84, E99, 0, 1)

+ 0:200

Note: E99 is the largest positive

number on a calculator.

1 If Z is the standard normal distribution, find the following probabilities.

In each case sketch the regions.

a Pr(¡0:86 6 Z 6 0:32) b Pr(¡2:3 6 Z 6 1:5) c Pr(Z 6 1:2)

d Pr(Z 6 ¡0:53) e Pr(Z > 1:3) f Pr(Z > ¡1:4)

g Pr(Z > 4)

With modern technology we can calculate probabilities for normal

distributions which have not been standardised. Click on the icon to

see how this is done.

1.50

0.840

EXERCISE 7D.3

TI

C

If X is N(10, 2:32), find these probabilities:

a Pr(8 6 X 6 11) b Pr(X 6 12) c Pr(X > 9). Illustrate.

a Pr(8 6 X 6 11)

= normalcdf(8, 11, 10, 2:3)

+ 0:476

b Pr(X 6 12)

= normalcdf(¡E99, 12, 10, 2:3)

+ 0:808

c Pr(X > 9)

= normalcdf(9, E99, 10, 2:3)

+ 0:668

1210

109

108 11

Example 10

Note:

When ¹ = 0 and ¾ = 1 we can simply use normalcdf (a, b)



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2 If the random variable X is N(70, 32), find these probabilities:

a Pr(60:6 < X 6 68:4) b Pr(X > 74) c Pr(X 6 68)

3 Suppose the variable X is normally distributed with mean ¹ = 58:3 and standard

deviation ¾ = 8:96 .

a Let the z-score of x = 50:6 be z1 and the z-score of x = 68:9 be z2.

i Calculate z1 and z2. ii Find Pr(z1 6 Z 6 z2)

b Find Pr(50:6 6 X 6 68:9) directly from your calculator.

c Compare the answers to a and b.

4 Suppose X is N(50, 52). Calculate Pr(a < X 6 51) for each of the following values

of a. Give your answers to 5 decimal places.

a a = 45 b a = 35 c a = 25 d a = 15 e a = 0

Compare the answers of a to e with Pr(X 6 51):

5 The height of 18 year old men is normally distributed with mean 182:3 cm and standard

deviation 9:6 cm. Find the probability that a randomly selected 18 year old man is:

a at least 180 cm tall b at most 190 cm tall c between 175 and 185 cm.

6 The weight of hens’ eggs is normally distributed with mean 42:3 g and standard deviation

5:9 g. Find the probability that a randomly selected egg is:

a at most 50 g b at least 45 g c between 35 g and 45 g.

7 The speed of cars passing the supermarket is normally distributed with mean 56:3 kmph

and standard deviation 7:4 kmph. Find the probability that a randomly selected car is

travelling at:

a between 60 and 75 kmph b at most 70 kmph c at least 60 kmph.

8 The lengths of metal bolts produced by a machine are found to be normally distributed

with a mean of 19:8 cm and a standard deviation of 0:3 cm. Find the probability that a

bolt selected at random from the machine will have a length between 19:7 and 20 cm.

9 The IQs of secondary school students from a particular area are believed to be normally

distributed with a mean of 103 and a standard deviation of 15:1. Find the probability

that a student will have an IQ:

a of at least 115 b that is less than 75 c between 95 and 105:

a player was: a at least 175 cm tall b between 170 cm and 190 cm.

If X is the height of a player then X is normally distributed with mean ¹ = 179and standard deviation ¾ = 7:

a We need to find

Pr(X > 175)

= normalcdf(175, E99, 179, 7)

+ 0:716

b We need to find

Pr(170 6 X 6 190)

= normalcdf(170, 190, 179, 7)

+ 0:843

In the heights of SANFL players was found to be normally distributed with

mean cm and standard deviation cm. Find the probability that in

1972179 7 1972

Example 11



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10 The average weekly earnings of the students at a local high school are found to be

approximately normally distributed with a mean of $40 and a standard deviation of $6:What proportion of students would you expect to earn:

a b

11 The lengths of Murray Cod caught in the River Murray are found to be normally

distributed with a mean of 41 cm and a standard deviation of 3:317 cm.

a Find the probability that a cod is at least 50 cm.

b What proportion of cod measure between 40 cm and 50 cm?

c In a sample of 200 cod, how many of them would you expect to be at least 45 cm?

Let X be the random variable of the length in mm of a snail shell.

Suppose that X is normally distributed with mean ¹ = 23:6and standard deviation ¾ = 3:1 mm. A snail farmer wants to

harvest some of his snails, but only those whose shell lengths

are amongst the longest 5%. The problem is to find k such that

Pr(X < k) = 95%.

When finding quantiles we are given a probability and are asked to calculate the corresponding

measurement. This is the inverse of finding probabilities, and we use the inverse normal

function.

Click on the icon to obtain instructions for using your calculator.

For the above example, the TI instruction is

k = invNorm(0:95, 23:6, 3:1) = 28:7

The instruction k = invNorm(0:95) will

assume that the mean ¹ = 0, and the

standard deviation ¾ = 1.

FINDING QUANTILES ( -VALUES)kE

TI

C

If Z has a standard normal distribution, find k if Pr(Z < k) = 0:73

Using a TI,

k = invNorm(0:73, 0, 1)+ 0:613

This means 73% of the values are expected to be less than 0:613

k��

73%

��

Example 12

The number is known as a , and in this case the quantile.k quantile 95%

k��.��

95%

X

between $ and $ per week30 50 at least $ per week?50

Z



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Let X denote the final examination result, so X » N(62, 132):

Pr(X > k) = 0:8

) Pr(X 6 k) = 0:2

) k = invNorm(0:2, 62, 13)

) k + 51:059

So, the minimum pass mark is 51.

A university professor determines that of this year’s History candidates should

pass the final examination. The examination results are expected to be normally

distributed with mean and standard deviation . Find the lowest score necessary

to pass the examination.

80%

62 13

Example 13

1 Z has a standard normal distribution. Illustrate with a sketch and find k if:

a Pr(Z 6 k) = 0:81 b Pr(Z 6 k) = 0:58

2 X » N(20, 32). Illustrate with a sketch and find k if:

a Pr(X 6 k) = 0:348 b Pr(X 6 k) = 0:878

c Pr(Z 6 k) = 0:17

c Pr(X 6 k) = 0:5

3 a Show that Pr(¡k 6 Z 6 k) = 2Pr(Z 6 k) ¡ 1:

b If Z is standard normally distributed, find k if:

i Pr(¡k 6 Z 6 k) = 0:238 ii Pr(¡k 6 Z 6 k) = 0:7004

4 The length of a fish species is normally

distributed with mean 35 cm and standard

deviation 8 cm. The fisheries department

has decided that the smallest 10% of the

fish are not to be harvested. What is size

of the smallest fish that can be harvested?

5 The length of screws produced by a machine is normally distributed with mean 75 mm

and standard deviation 0:1 mm. If a screw is too long it is automatically rejected. If 1%of screws are rejected, what is the length of the smallest screw to be rejected?

6 The average score for a Physics test was 46 and the standard deviation of the scores was

15. Assuming that the scores were normally distributed, the teacher decided to award

an A to the top 7% of the students in the class. What is the lowest score that a student

needed in order to achieve an A?

7 The volume of cool drink in a bottle filled by a machine is normally distributed with

mean 503 mL and standard deviation 0:5 mL. 1% of the bottles are rejected because they

are underfilled, and 2% are rejected because they are overfilled; otherwise they are kept

for retail. What range of volumes is in the bottles that are kept?

EXERCISE 7E

k ��

20%

X

We need to find

such that

k



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58.2 ��

15%

0.150.1

!z=20 � !x=29#z #x

Note: Z-scores are essential for finding unknown values of ¹ and/or ¾.

8 The arrival times of buses at a depot is normally distributed with standard deviation of

5 minutes. If 10% of the buses arrive before 3:45 pm, what is the mean arrival time of

buses at the depot?

9 The IQ of a population has a standard deviation of 15. In a school 20% of students have

an IQ larger than 125. What is the mean IQ of students in this school?

10 The distance an athlete can jump is normally distributed with mean 5:2 m. If 20% of

the jumps by this athlete are less than 5 m, what is the standard deviation?

11 The weekly income of a greengrocer is normally distributed with a mean of $6100. If

85% of the time the weekly income exceeds $6000, what is the standard deviation?

Find the mean and standard deviation of a normally distributed random variable Xif Pr(X 6 20) = 0:1 and Pr(X > 29) = 0:15

X » N(¹, ¾2) where we have to

find ¹ and ¾.

We start by finding z1 and z2 which

correspond to x1 = 20 and x2 = 29.

Now z1 =20 ¡ ¹

¾= invNorm(0:1) = ¡1:282 ) 20 ¡ ¹ = ¡1:282¾ .... (1)

and z2 =29 ¡ ¹

¾= invNorm(0:85) = 1:036 ) 29 ¡ ¹ = 1:036¾ ....... (2)

Solving these two equations gives ¹ + 25:0 and ¾ = 3:88

Let the mean weight of the population be ¹ g.

If X g denotes the weight of an adult scallop,

then X » N(¹, 5:92):

As we do not know ¹ we cannot use the

invNorm directly, but we can find the z-value.

Now Pr(X 6 58:2) = 0:15

) Pr(Z 658:2 ¡ ¹

5:9) = 0:15

)58:2 ¡ ¹

5:9= invNorm(0:15) = ¡1:0364

) 58:2 ¡ ¹ + ¡6:1

¹ + 64:3 So, the mean weight is 64:3 g.

An adult scallop population is known to have a standard deviation of g. If

of scallops weigh less than g, find the mean weight of the population.

5 9 15%58 2

::

Example 14

Example 15



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INVESTIGATION 2 THE GEOMETRIC SIGNIFICANCE OF AND¹ ¾

12 a Find the mean and the standard deviation of a normally distributed random variable

X, if Pr(X > 80) = 0:1 and Pr(X 6 30) = 0:15:

b In a Mathematics examination it was found that 10% of the students scored at least

80, and no more than 15% scored under 30. Assuming the scores are normally

distributed, what proportion of students scored more than 50?

13 The diameters of pistons manufactured by a company are normally distributed. Only

those pistons whose diameters lie between 3:994 and 4:006 cm are acceptable.

a Find the mean and the standard deviation of the distribution if 4% of the pistons

are rejected as being too small, and 5% are rejected as being too large.

b

In the previous section a number of assertions were made about the standard deviation. In

this section some of these assertions will be justified.

1 The normal probability density function is f(x) =1

¾p

2¼e¡

12 (

x¡¹¾

)2 .

Use technology to graph this function for a ¹ = 6, ¾ = 1 b ¹ = 6, ¾ = 2.

2 Show that the derivative of f(x) is f 0(x) = ¡x ¡ ¹

¾2f(x).

3 Use the result in 2 to show that f (x) has a maximum value at x = ¹.

4 Show that f 00(x) = ¡ 1

¾4(¾2 ¡ (x ¡ ¹)2) f(x) .

5 Use the result of 4 to find the points of inflection of f(x).

From Investigation 2 you

should have discovered that

the points of inflection occur

at x = ¹+¾ and x = ¹¡¾.

Consequently:

For a given normal curve the standard deviation is uniquely determined as the

horizontal distance from the vertical line x = ¹ to a point of inflection.

INVESTIGATING PROPERTIESOF NORMAL DISTRIBUTIONS

F

What to do:

x

��

� �

point of

inflection

point of

inflection


Determine the probability that the diameter of a randomly chosen piston lies

between . mm and . mm.3 997 4 003

GRAPHING

PACKAGE


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INVESTIGATION 3 CALCULATING PROBABILITIES

FROM NORMAL DISTRIBUTIONS

Suppose a dietician wants to know the mean

weight of thirteen year old Australian boys.

It is impractical to weigh each thirteen year

old boy in Australia, but the dietician could

find the mean weight of a randomly selected

sample of, say, 10 boys.

The mean weight of the sample of 10 boys

is a statistic that is then used to estimate the

population parameter.

Clearly the mean weight depends on the sam-

ple. If another health worker had selected a

different sample of 10 boys, it would be un-

likely that the two sample means would be

the same.

The statistic the sample weight is a new variable. Repeated sampling can be used to discover

how the variable sample weight is distributed. In particular we want to know how the mean

of the sample means and the standard deviation of the sample means is related to the parent

population of 13 year old boys.

The following investigation explores the relation between the statistic “sample mean” and the

parameter “population mean”.

SPREADSHEET

DISTRIBUTION OF SAMPLE MEANSG

To find probabilities from a normal distribution you need to be able to find

areas between the graph of f(x) = 1¾p2¼

e¡12 (

x¡¹¾

)2 and the x-axis.

A simple way to estimate these probabil-

ities is to approximate them with areas

of rectangles that fit snugly around the

curve.

The area beneath the smooth curve is

approximately equal to the sum of the

areas of the rectangles.

Use a spreadsheet to:

² calculate the area of each rectangle using area = base £ height

² add the areas of rectangles to find an approximate area below the curve.

Details of how to set up a spreadsheet can be found by clicking on

the icon.

What to do:



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INVESTIGATION 4 A SIMPLE RANDOM SAMPLER

Suppose a school has 216 thirteen year old boys.

Let the variable X be the weight in kg of the boys.

The table shows all the possible values of X in random order.

31:2 35:7 36:4 33:2 37:3 35:0 34:0 33:6 34:4 32:0 32:7 36:730:8 33:8 32:9 35:4 31:9 36:7 32:0 29:2 33:6 31:0 32:5 36:433:3 36:7 27:9 32:0 36:4 34:5 35:3 31:6 32:5 35:3 34:6 31:134:9 30:9 33:2 33:8 33:6 30:5 37:7 30:9 35:0 33:2 36:2 35:231:8 35:9 32:8 30:8 29:0 32:1 34:6 32:7 35:4 30:4 33:3 30:233:3 35:5 32:0 34:8 30:2 36:3 35:7 38:9 32:0 28:0 32:7 33:6

35:4 31:2 32:5 29:6 35:1 32:9 37:3 33:6 36:7 30:7 32:8 32:529:4 33:5 32:5 30:1 34:9 32:3 34:9 31:4 33:0 32:4 29:7 33:630:6 30:5 30:5 36:3 34:3 32:1 36:6 31:3 30:8 29:8 30:8 29:233:1 35:0 32:5 34:1 33:2 32:9 30:2 33:4 33:2 31:1 32:3 30:632:0 31:4 32:4 37:1 32:5 35:9 29:4 30:3 34:9 32:1 34:6 35:731:4 27:5 31:7 37:1 29:9 31:6 35:4 32:5 33:4 35:2 34:2 29:5

34:3 31:9 33:2 34:5 32:4 30:8 32:4 32:0 27:1 36:4 34:0 32:431:9 32:6 29:4 32:6 35:5 33:0 35:5 31:4 40:6 37:1 31:4 30:031:5 31:6 34:2 29:1 35:4 29:9 32:0 33:7 29:0 32:0 29:9 34:635:0 27:0 31:8 36:1 32:7 31:0 30:4 35:9 38:4 31:6 34:4 31:632:3 33:4 35:3 38:7 37:5 32:1 29:7 33:9 34:0 34:2 29:2 37:629:3 34:0 30:6 37:1 30:4 33:2 33:7 28:5 36:2 35:7 36:4 33:2

1 Select a sample of 10 boys from this population by:

a rolling a die to select one of the 6 blocks

b rolling the die again to select a row in the block

c rolling the die again to select a boy in the row

d count off 10 boys from left to right from the boy you selected.

If the 3 rolls of the die produced f3, 2, 4g, the boy selected has weight 30:1 kg.

The sample selected is presented in the first column of the table.

2 Copy and enter your data in the following table.

Number Sample 1 Sample 2 Sample 3 Sample 4 Sample 5

1 30:12 34:93 32:34 34:95 31:46 33:07 32:48 29:79 33:610 30:6

mean, x 32:3

1 2

3 4

5 6

What to do:



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INVESTIGATION 5 A COMPUTER BASED RANDOM SAMPLER

3 The last row in this table consists of 5 sample means.

The variable of sample means can be denoted by X10. The bar on the top indicates

it is a variable of means; the subscript 10 indicates that the means are of samples of

size ten.

The last row of your table is a sample of size 5 from the distribution of X10.

4 Combine your results with those of the other students of your class.

Draw a histogram of the sample means.

5 Calculate the mean and the standard deviation of the sample means.

6 Compare the mean and the standard deviation you found in 5 with the mean weight

33:1 kg and standard deviation 2:54 kg of the 216 boys.

From Investigation 4 you should have discovered that the sample means are close to the

population mean. The mean of the sample means should be particularly close to the population

mean.

You should also have noted that the standard deviation of the sample means is smaller than

the standard deviation of the population.

The following important investigation uses a computer to speed up sampling and obtain a

more accurate picture of how the standard deviation of the sample means is related to the

standard deviation of the population.

In this investigation it is important to distinguish between:

² The original population, sometimes referred to as the “parent population ”, with a

random variable X which has mean ¹ and standard deviation ¾.

In Investigation 4 the parent population consists of 216 thirteen year old boys.

The mean ¹ = 33:1 kg and standard deviation ¾ = 2:54 kg.

and

² The new population with variable Xn, consisting of all statistics of sample means.

The subscript n indicating the sample size is sometimes omitted and the variable

just written X.

A typical outcome of X is a sample mean ¹x =x1 + x2 + :::::: + xn

n

In Investigation 4 a typical outcome is the mean weight of 10 boys.

The investigation explores the shape of the distribution of the random variable X, its

mean ¹X

or ¹(X), and its standard deviation ¾X

or ¾(X).

We start by sampling from a population which has a normal distribution. The heights of

18 year old Australian males may be approximately normal.

In this investigation we examine the variation in sample means.

We examine samples taken from symmetric distributions as well as one that

is skewed.



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1 Click on the icon given alongside. This opens a worksheet named

Samples with a number of buttons. Click on each of these buttons

in turn.

2 Sample size: from which you can select the numbers n = 10, 20, 40, 80, 160.

Start with n = 10.

3 Find sample means: finds the means of each of two hundred different samples.

4 Analyse: lists the two hundred sample means.

It finds the standard deviation sX

and

draws a histogram of these sample means.

It also superimposes a normal probability density function.

Trial 1 Trial 2 Trial 3 Trial 4

n (sX

)2 (sX

)2 (sX

)2 (sX

)2

10

20

40

80

160

5 Make a copy of the table alongside.

Enter the value of (sX

)2 in the first

column next to n = 10.

6 Go back to the worksheet named

Samples and change the sample size

to 20. Repeat steps 3, 4, and 5.

Enter the value of (sX

)2 next to

n = 20 in the table.

7 Repeat for samples of size 40, 80and 160.

8 We wish to see how (sX

)2 is related to the standard deviation of the population.

However, (sX

)2 can vary quite a lot, so to spot the pattern more clearly you should

repeat the experiment another 3 times.

9 From your experiment, determine a relationship between the square of the sample

standard deviation (sX

)2 and the square of the population standard deviation.

10

11

What to do: STATISTICS

PACKAGE

STATISTICS

PACKAGE

STATISTICS

PACKAGE

Now click on the icon to sample data from a population with a

uniform distribution. These distributions are very commonly used

in computer games where, for example, cards have to be selected

at random. Complete an analysis of this data by repeating the

above procedure and recording all results.

Now click on the icon to sample data from a population with an

exponential distribution. These distributions are notoriously skew.

They are commonly used in modelling lifetimes, such as the

lifetime of light globes. Complete an analysis of this data by

repeating the above procedure and recording all results.


This output is shown on the worksheet named Analysis.

Note that the first graph on this worksheet is the graph of the probability density

function of the population, and that the axes differ from that of the other graphs.


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APPENDIX

From the investigation you should have discovered the following:

If X is a random variable with mean ¹ and standard deviation ¾ then the random

variable Xn of sample means of size n has:

² mean ¹X

= ¹, the same as the mean of the random variable X

² standard deviation ¾X

=¾pn

.

Furthermore, for large values of n, Xn is approximately normal.

² The histogram of the sample means becomes symmetric and starts

to take on a bell-like shape. For large values of n it becomes

approximately normal.

² The mean of the sample

means approximates the

population mean.

Individual points selected

from any distribution are

likely to come from either

side of the mean, and dif-

ferences are likely to av-

erage out.

² As the sample size increases, there is less variability.

² This diagram shows what happens if the sample size n increases.

The spread decreases since =¾pn

and ¹X

= ¹:

You should notice:

¾X

¾X

¾X

�

x x x x1 2 3, , ,..., n x x x x1 2 3, , ,..., n x x x x1 2 3, , ,..., n

Sample 1 x1 Sample 2 x2 Sample 3 x3

x1 x2 x3¹

X

¾X


In the the behaviour of the mean and the standard deviation are

explored algebraically. It is beyond the level of this course to show why

the distribution of the sample means is approximately normal.

Appendix

�


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1 A machine produces sheets of cardboard with mean thickness 3 mm and standard devi-

ation 0:12 mm. A quality controller checks the thickness of each sheet in 10 different

places. Let the random variable X be the thickness of the cardboard at any point, and

let the random variable X10 be the mean thickness of the 10 points.

a The quality controller records the following thicknesses in mm from a sample of

10 points: 3:02, 2:77, 3:08, 2:89, 3:21, 2:79, 2:97, 3:07, 2:94, 3:01: What is the

corresponding outcome of the random variable X10?

b If the quality controller records 10 outcomes of X as:

x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, what is the corresponding statistic of X10?

c What is the mean and standard deviation of X10?

2 Records show that a machine has been producing screws with mean length 75 mm and

standard deviation 0:5 mm. Screws are packaged in lots of 50. Let the random variable

X50 be the mean length of a screw in a packet.

Find the mean and standard deviation of X50.

The life expectancy , of a certain brand of AAA battery is known to have a

mean hours and standard deviation hours. The batteries are sold in

packets of . Let the random variable be the mean life expectancy of batteries in

a packet.

X¹ ¾ :

X� � � � �= 27 = 3 25

6 6

a

What is the corresponding outcome of the random variable X6?

b If the numbers of hours lasted by batteries in a packet of six were

x1, x2, x3, x4, x5, x6 what is the corresponding outcome of X6 ?

c What is the mean and standard deviation of X6?

a The outcomes of X6 are the means of the life expectancies of 6 batteries in

a packet. In this case the outcome of X6 is the statistic

x =25:3 + 21:6 + 27:75 + 22:25 + 35:5 + 28:5

6+ 26:8

b If the batteries in the packet lasted for x1, x2, x3, x4, x5, x6 hours, the

corresponding outcome of X6 is the statistic x =x1 + x2 + x3 + x4 + x5 + x6

6.

c The mean of X6 is the same as the mean of X, so ¹X6

= 27 hours.

Since the standard deviation of X is 3:25, the standard deviation of X6 is

¾X6

=¾p6

=3:25p

6+ 1:327

The batteries in a packet were tested and the number of hours they lasted

were: , , , , ,

625 3 21 6 27 75 22 25 35 5 28 5: : : : : :

Example 16

EXERCISE 7G.1



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3 The time it takes a train from Adelaide to Belair to complete its journey is known to

have a mean of 40 minutes and standard deviation of 3 minutes. An inspector times 8such trips. Let X8 be the mean travel time of a sample of 8 trips. Find the mean and

standard deviation of X8.

4 Suppose the probability a coin falls heads is p and the probability it falls tails is q = 1¡p.

Let the random variable X = 1 if it falls heads and X = 0 if it falls tails.

a Show that the mean of X is p.

b Show that the standard deviation of X isppq =

pp(1 ¡ p).

c Let Xn be the sample mean of n tosses of the coin.

i Find the mean and standard deviation of Xn.

ii Describe in words how Xn is related to the tosses of a coin.

In general, knowing the mean and standard deviation of a random variable X is insufficient

information to calculate probabilities. However, we are able to calculate probabilities in the

special case where X is normally distributed. Not only that, but if X is normally distributed,

the random variable Xn of sample means of size n is also normally distributed.

Example 17

Including yourself there are 12 persons in the line to be served.

To complete buying your ticket in less than 10 minutes the mean serving time per

person has to be less than10 £ 60

12= 50 seconds.

The time it takes to serve a customer at a railway station ticket booth is normally

distributed with mean seconds and standard deviation seconds. You only have

minutes to buy your ticket or you will miss your train. If there is a line of

people in front of you waiting to be served, what is the probability you will catch the

train?

T45 20

10 11

Example 18

Suppose the random variable X is normally distributed with mean 40 and standard

deviation 10. Let X20 be the sample means of size 20. Find:

a Pr(35 < X < 45) b Pr(35 < X20 < 45).

a Pr(35 < X < 45)

= normalcdf(35, 45, 40, 10)

+ 0:383

b The mean of X20 = mean of X = 40:

The standard deviation of X20 = 10p20

Pr(35 < X20 < 45)

= normalcdf(35, 45, 40, 10p20

)

= 0:975

Notice that about 38% of the individual outcomes are in the interval 35 < X < 45,

but almost all of the sample means lie in this interval.



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Y:\HAESE\SA_12STU-2ed\SA12STU-2_07\251SA12STU-2_07.CDR Monday, 29 October 2007 3:42:41 PM DAVID3

Let the random variable T 12 be the mean time to serve 12 persons.

Since T is normally distributed with mean 45 and standard deviation 20, T 12 is

normally distributed with mean 45 and standard deviation 20p12

.

Pr(T 12< 50)

= normalcdf(¡E99, 50, 45, 20p12

)

+ 0:807

5 Suppose the random variable X is normally distributed with mean 80 and standard

deviation 20. Let X10 be the sample means of size 10: Find:

a Pr(75 < X < 85) b Pr(75 < X10 < 85)

6 Let the random variable X be the IQ of 17 year old girls. Suppose X is normally

distributed with mean 105 and standard deviation 15.

a Find the probability that an individual 17 year old girl has an IQ of more than 110.

b Find the probability that the mean IQ of a class of twenty 17 year old girls is greater

than 110.

7 A manufacturer of chocolates produces chocolates of mean weight 20 g and standard

deviation 5 g. A box of 13 such chocolates is sold with the claim that the nett weight in

the box is 250 g. Assuming the weights are normally distributed:

a For what proportion of boxes is this claim correct?

b If the manufacturer decides to increase the number of chocolates to 15 per box, for

what proportion of boxes is the claim now true?

In the previous investigation, we also observed that the distribution of the sample means Xis approximately normal.

Note:

²

² In the special case where the population is normally distributed, the distribution X of

the sample means is always normal.

THE CENTRAL LIMIT THEOREM

The Central Limit Theorem

There is no simple answer as to how large should be before the central limit theorem

can be applied. It depends on many factors including how much accuracy is required. If

the population is very skew it may require a large sample size , whereas if the

population is symmetric a small sample size may be sufficient. As a rule of thumb,

is often used, but each case must be considered on its merits.

n

nn

n� �>30

So, the probability of catching the train is 0 807:

¹ and standard deviation ¾: For sufficiently large n, the distribution Xn of the sample means

of size n, is approximately normal with mean ¹X

= ¹ and standard deviation ¾X

=¾pn:

Suppose is a random variable which is not necessarily normally distributed, but has meanX



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The standard deviation ¾X

=¾pn

of the sample means X is a measure of the

variability of sample means, and is called the sampling error or the standard error.

Note:

² Unless the population is small, the population size is almost irrelevant.

²

For example, a sample size of 1000 gives a sampling error of ¾X

=¾p1000

+¾

32

whereas a sample of 4000, four times the size, only halves the sampling error.

Two histograms of samples, each of size , are shown below. One is from a

uniform distribution with mean and standard deviation . The other is from

the distribution of the sample means of size selected from the distribution .

Note that the scales are not the same in the two diagrams.

40010 5 77

36X :

X X36

a Which of the two histograms is from X36? Give reasons for your answer.

b From the diagram estimate Pr (X36 < 9).

c Find the approximate mean and standard deviation of X36.

d Use the histogram to estimate the probability X36 is one standard deviation

from the mean.

a The data in Histogram A is less spread out than that in Histogram B, and

appears clustered around 10. Histogram A is the histogram for the

distribution X36.

b To find Pr (X36 < 9) we count the numbers in all the bins before the

bin [9, 9:25), and use the fact that there are 400 in the sample. We get:

THE SAMPLING ERROR

The larger the value of , the smaller the sampling error. A sufficiently large sample

should give an accurate estimate of the mean. However, making the sample size too big

may be expensive and may not improve the reliability of the estimate by much.

n

Example 19

We are trying to estimate the using a . By only looking at a

small portion of the population, the sample mean is likely to be different from the population

mean.

population mean sample mean

Histogram A

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Histogram B

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Pr (X36 < 9) =15 + 15 + 12 + 3 + 2 + 3 + 2 + 1

400=

53

400+ 0:13

Your answer may vary a little depending on how well you can read the numbers

on the graph.

c The mean of X36 = mean of X = 10.

The standard deviation ¾X

+¾p36

=5:77

6= 0:962

d Pr(10 ¡ 0:96 < X < 10 + 0:96) = Pr(9:04 < X36 < 10:96)

+ Pr(9 < X36 < 11)

=30 + 27 + 39 + 44 + 45 + 42 + 31 + 30

400= 0:72

This crude estimate compares with 0:68 when using the normal approximation.

1 The IQ measurements of a population have mean 100 and standard deviation 15. Many

hundreds of random samples of size 36 are taken from the population and a relative

frequency histogram of the sample means is formed.

a What would we expect the mean of the samples to be?

b What would we expect the standard deviation of the samples to be?

c What would we expect the shape of the histogram to look like?

2 Two histograms of sample size 300 each are shown below. One is from a life expectancy

distribution X with mean 10 and standard deviation 10. The other is from the distribu-

tion X64 of the sample means of size 64 selected from the distribution X. Note that

the scales are not the same in the two diagrams.

a Which of the two histograms is from X64? Give reasons for your answer.

b From the diagram estimate Pr(X64 < 9).

c Find the approximate mean and standard deviation of X64 .

d Use the histogram to estimate the probability that X64 is one standard deviation from

the mean. How does this answer compare with using the normal approximation?

EXERCISE 7G.2

Histogram A

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40

60

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3 During a one week period in Sydney the mean price of an orange was 42:8 cents with

standard deviation 8:7 cents. Find the probability that the mean price per orange from a

case of 60 oranges was less than 45 cents.

4 The mean energy content of a fruit bar is 1067 kJ with standard deviation 61:7 kJ. Find

the probability that the mean energy content of a sample of 30 fruit bars is more than

1050 kJ/bar.

5 The mean sodium content of a box of cheese rings is 1183 mg with standard deviation

88:6 mg. Find the probability that the mean sodium content per box for a sample of 50boxes lies between 1150 mg and 1200 mg.

6 Customers at a clothing store are in the shop for a mean time of 18 minutes with standard

deviation 5:3 minutes. What is the probability that in a sample of 37 customers the mean

stay in the shop is between 17 and 20 minutes?

7 The mean contents of a can of cola is 382 mL, even though it says 375 mL on a can.

The statistician at the factory says that the standard deviation is steady at 16:2 mL. Find

the probability that a slab of three dozen cans has mean contents less than 375 mL per

can.

The age of men in Australia is distributed with mean 43 and standard deviation 8.

If a sample of 67 men is selected from the population of Australian men, what is

the probability the sample mean is:

a less than 42 b greater than 45 c between 40 and 45?

Let the random variable X be the mean age of samples of 67 Australian males.

Assuming n = 67 is sufficiently large for the Central Limit Theorem to apply,

X is approximately normal with mean 43 and standard deviation ¾X

= 8p67

.

a Pr(X < 42)

= normalcdf(¡E99, 42, 43, 8p67

)

+ 0:153

b Pr(X > 45)

= normalcdf(45, E99, 43, 8p67

)

+ 0:0204

c Pr(40 < X < 45)

= normalcdf(40, 45, 43, 8p67

)

+ 0:979

Example 20

43 45

43 45��

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INVESTIGATION 6 CHOCKBLOCKS

Chockblock produce mini chocolate

bars which vary a little in weight. The

machine used to make them produces

bars whose weights are normally

distributed with mean grams and standard

deviation grams. bars are then placed in a

packet for sale. Hundreds of thousands of packets

are produced each year with mean weight .

18 23 3 25

::

X

8 A sample of 375 people will be used to estimate

the mean number of hours that will be lost due

to sickness this year. Last year the standard de-

viation for the number of hours lost was 67 and

we will use this as the standard deviation this

year. What is the probability that the estimate is

9 A concerned union member wishes to estimate the hourly wage of shop assistants in

Adelaide. He decides to randomly survey 300 shop assistants to calculate the sample

mean. Assuming that the standard deviation is $1:27, find the probability that the estimate

of the population mean is in error by 10 cents or more.

1 What are the mean ¹X

and standard deviation ¾X

of X?

2 Printed on each packet is the nett weight of contents, 425 grams. What is the manu-

facturer claiming about the mean weight of each bar?

Let the random variable X be the mean of samples of 60. As the sample size is

larger than 30, we assume that X is normally distributed with mean ¹ and standard

deviation 8p60

.

We need to find Pr(¡2 < X ¡ ¹ < 2).

Now Pr(¡2 < X ¡ ¹ < 2) = Pr

µ ¡28p60

<X ¡ ¹

8p60

<28p60

¶= Pr

³¡p60

4 < Z <p604

´= normalcdf(¡

p60

4 ,p604 , 0, 1)

+ 0:947

A population is known to have a standard deviation of but has an unknown mean

. In order to estimate , the mean of a random sample of is found. Find the

probability that this estimate is out by less than .

860

2¹ ¹

Example 21

What to do:

in error by less than ten hours?



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3 What percentage of their packets will be rejected because they fail to meet the 425gram claim?

4 An additional bar is added to each packet with the nett weight claim retained at 425grams.

a What is the minimum acceptable claim now?

b What are the mean ¹X

and standard deviation ¾X

now?

c What percentage of these packets would we expect to reject?

Claims are often made about the population mean of some

quantities.

For example, it is claimed that the mean protein content of a

1 litre carton of milk is 39 grams. The truth of this claim can

only be known by measuring the protein content of every 1litre carton of milk, clearly an impossible task. It is, however,

possible to draw reasonable conclusions from measuring the

protein content of a random selection of cartons.

A statistical hypothesis is a statement about a population parameter. The parameter

could be a population mean or a proportion.

In this section we will test hypotheses concerning the mean ¹.

When a statement is made about a product, it is usually tested statistically before changes to

the product are made.

The alternative hypothesis denoted Ha is that the statistical evidence is sufficient to accept

the consumer’s claim, i.e., that the milk company’s statement is false.

So, we consider two hypotheses:

HYPOTHESIS TESTING FOR A MEANH

HYPOTHESIS ABOUT MEANS

²

²

a which is a statement of or . It is

assumed to be true until sufficient evidence is provided so that it is rejected.

an which is a statement that there or

which has to be established. Supporting evidence is necessary if it is to

be accepted.

null hypothesis

alternative hypothesis

H

H

0 no difference no change

is a difference

changea

For example, suppose a consumer makes the statement that the mean protein content in

litre cartons of milk is not grams. The milk company does not want to go to the

expense of changing packaging until it is statistically shown that the mean protein content is

indeed not grams. The company will start with the assumption that their claim is true,

and whatever tests the consumer did were just random fluctuations. This assumption or

statement of no change is called the and is usually denoted .

1 39

39

�

null hypothesis H0



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We want to test the claim that the mean protein content of 1 litre cartons of milk is 39 grams.

The null hypothesis is H0: ¹ = 39

The alternative hypothesis is Ha: ¹ 6= 39

Suppose we select a sample of 10 cartons of milk and find that for this sample the mean

protein content is ¹x = 38:4 grams.

Suppose it is known that the standard deviation of protein in 1 litre containers of milk is

¾ = 0:8 grams.

Let X be the protein content of a 1 litre container of milk, so according to the null hypothesis,

X » N(39, 0:82).

Let the random variable X be the mean protein content of a sample of 10 one litre cartons.

Hence X » N ¹,

µ ¾pn

¶2i.e., X » N 39 ,

0:8p10

2

.

HYPOTHESIS TESTING WHEN THE POPULATION IS NORMALLY

DISTRIBUTED

We need to determine the likelihood that this difference is

due to random fluctuation or chance, or whether it is

sufficient evidence to say the milk company’s statement is

incorrect.

Since the protein content of milk is a result of many

different factors, it is reasonable to assume that the protein

content of litre cartons of milk is normally distributed.1

µ ¶ µ ¶ ¶µWe use this to calculate the z-score of the observed value ¹x = 38:4 grams.

z =¹x¡ ¹¾pn

=38:4 ¡ 39

0:8p10

+ ¡2:37 So the number of standard deviations ¹x is

from the mean is ¡2:37 .

If the difference between the observed value of ¹x and the mean is due to chance alone, it

could just as likely have been 2:37 standard deviations to left or right of the mean. So, the

probability that X is 2:37 standard deviations or more either side of the mean is a measure

of how likely this is to occur.

Now Pr(Z 6 ¡2:37 or Z > 2:37) = 2 £ Pr(Z 6 ¡2:37) fsymmetryg= 2 £ normalcdf(¡E99, ¡2:37)

= 0:0178

so the probability of this event happening is small.

One of the problems with random processes is that differences can always be due to chance.

However, the practical solution is to reject the null hypothesis if the probability of the observed

or more extreme results occurring is small.

The probability ® at which we reject the null hypothesis is called the significance level of the

test. Common significance levels are ® = 0:05 or 5% and ® = 0:01 or 1%.

In the above example, Pr(Z 6 ¡2:37 or Z > 2:37) = 0:0178 . This is less than 0:05 so

we would reject the null hypothesis at the significance level of 0:05, but not at the significance

level of 0:01 .



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The procedure for testing a hypothesis is:

Step 1: State the null hypothesis H0: ¹ = ¹0

and the alternate hypothesis Ha: ¹ 6= ¹0.

Step 2: Select a significance level, usually 0:05 .

Unless otherwise stated, the level of 0:05 is used in

this book.

Step 3: From a sample, calculate the sample mean ¹x.

If the parent population is normally distributed with

mean ¹ and standard deviation ¾, then the random

variable X of sample means has the normal

distribution

is called the null distribution:

The null distribution is critical. It allows us to calcu-

late the probability of the observed or more extreme

events happening if the null hypothesis is true.

Step 4: Use the sample mean ¹x to find the test statistic

z =¹x¡ ¹¾pn

:

Step 5: Calculate the probability of all observations having

z-values more extreme than the test statistic z found

in Step 3.

Step 6: ² Reject the null hypothesis if the P-value is less

than the significance level decided on in Step 2.

The smaller the P-value is, the stronger the

evidence against the null hypothesis.

² If the P-value is larger than the significance

level decided on in Step 2, do not reject the

null hypothesis.

H0: ¹ = 39

Ha: ¹ 6= 39

X » N(39, 0:2532)

z = ¡2:37

P= Pr(Z 6 ¡2:37or Z > 2:37)

= 0:0178

The is the probability of all observations

having a -value more extreme than the test statistic.

P-value

z

N ¹,

µ ¾pn

¶2µ ¶.N ¹,

µ ¾pn

¶2µ ¶

Since we include the extreme outcomes either side

of the mean, we call this a . Only

two-sided tests are considered in this course.

two-sided -testZ

The name derives its name from this statistic.Z-test

Since P , we

do not reject the

null hypothesis at

the level.

� �> :

:

0 01

0 01

Since P we

reject the null

hypothesis at the

level.

� �< :

:

0 05

0 05

Milk cartons example



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When a null hypothesis is not rejected, the terms “retain” and “accept” are often used. This

does not mean that the null hypothesis is true, but rather that there is not enough evidence to

show it is not true.

Similarly, when rejecting the null hypothesis, it is often stated that the alternative hypothesis

is “accepted”. This does not mean that the alternative hypothesis is true. However, if the null

hypothesis is true, the outcome that led to rejecting it is a very unlikely one. The P-value

tells you just how unlikely.

Notice that we

use and not

for the test.

¾ sZ-


Note: If H0 is rejected,

² the direction of the difference is determined by the value of ¹x

² we still do not know how accurate the claim was.

1 A random variable X is normally distributed with a standard deviation ¾ = 4. It is

claimed that the mean of X is ¹ = 17.

a To test this claim a random sample of n = 50 was taken and the sample mean ¹xwas found to be 16.

i Write down the hypotheses H0 and Ha . ii Write down the null distribution.

Step 1: H0 : ¹ = 74 Ha : ¹ 6= 74

Step 2: Significance level is 0:05

Step 3: The sample mean, ¹x = 72

Let the random variable X be the sample means, so the null distribution

is X » N(¹,

µ¾pn

¶2

) i.e., X » N(74,

µ7p40

¶2

):

Step 4: The test statistic is z =¹x ¡ ¹¾pn

=72 ¡ 74

7p40

+ ¡1:81

Step 5: The P-value is P = Pr(Z 6 ¡1:81 or Z > 1:81)

= 2 £ Pr(Z 6 ¡1:81)

+ 0:0708

Step 6:

A Mathematics coaching school knows that the results for their final test are

normally distributed with population mean and standard deviation . A new

coaching technique which is cheaper to implement but reported to have the same

results is trialled by the school. In a trial of students it is found that the mean

score for the final test is with standard deviation . Is there sufficient

evidence at the level to conclude that the final test scores will be different?

74% 7%

4072% 6%

5%

Example 22

As P there is insufficient evidence to reject

the null hypothesis that the new coaching produces the

same results as the old technique. We thus accept that the

new technique has the same result as the old technique.

� � � �=0 0708 0 05: > :

EXERCISE 7H.1


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iii Calculate the test statistic.

iv Calculate the P-value.

v What conclusion is there at the 0:05 level?

b Suppose that a random sample of n = 70 was taken and ¹x = 16. What can you

now conclude at the 0:05 level?

2 A random variable X is normally distributed with a standard deviation ¾ = 6. A random

sample of 40 was taken and the sample mean was found to be ¹x = 61:4 .

Use this information to test the claim that the population mean of X is ¹ = 60.

.N ¹,

µ ¾pn

¶2µ ¶

The bottlers of Groutt claim that the mean volume of bottles is 503 mL.

To test this claim 10 bottles were selected.

The measurements are listed below to the nearest 0:1 mL:

502:5, 501:0, 501:5, 503:9, 498:7, 505:7, 504:6, 499:4, 501:8, 501:1

Test the claim made by the bottlers of Groutt at the 5% level if it is known that

the population standard deviation ¾ is 1:8 mL.

We need to test: the null hypothesis H0: ¹ = 503

against the alternative hypothesis Ha: ¹ 6= 503

Let X be the volume of each bottle of Groutt. As the bottling of liquids is subject to

many random fluctuations, it is reasonable to assume that X is normally distributed

with mean ¹ and standard deviation ¾.

Let X be the distribution of the sample means, so the null distribution of X is

From the null hypothesis we assume that ¹ = 503.

From the sample we find that ¹x = 502:02, so the test statistic

z =¹x ¡ ¹¾pn

+502:02 ¡ 503

1:8p10

+ ¡1:722

The P-value is P = Pr(Z 6 ¡ 1:722 or Z > 1:722)

= 2 £ Pr(Z 6 ¡1:722)

+ 0:0851

As P > 0:05 there is insufficient evidence to reject the claim that the volume

of bottles of Groutt is 503 mL,

i.e., we accept that the mean volume could be 503 mL.

Example 23



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TI

C

3 A market gardener claims that the carrots in his field have a mean weight of 50 grams.

Before buying the crop a buyer pulls 20 carrots at random. She finds that their individual

weights in grams are:

57:6 34:7 53:9 52:5 61:8 51:5 61:3 49:2 56:8 55:9

57:9 58:8 44:3 58:3 49:3 56:0 59:5 47:0 58:0 47:2

a Explain why it is reasonable that the distribution of carrots’ weights is normally

distributed.

b Test the claim made by the market gardener if it is known that the standard deviation

for the whole crop is 7:1 grams.

4 The length of screws produced by a machine is known to be normally distributed with

standard deviation ¾ = 0:08 cm.

The machine is supposed to produce screws with a mean length of ¹ = 2:00 cm.

A quality controller selects a random sample of 15 screws and finds that the mean length

of the 15 screws is ¹x = 2:04 cm with sample standard deviation of s = 0:09 cm.

Does this justify the need to adjust the machine?

To see how to do hypothesis testing using a calculator,

click on the appropriate icon.

In the examples we have seen so far, the variable X was normally distributed and so the

distribution of sample means X was normally distributed also. This may not be true if Xis not normally distributed. However, if the sample size n is sufficiently large, the Central

Limit Theorem tells us that X is approximately normally distributed with mean ¹ and standard

deviation¾pn

.

We can use this fact to test claims about population means.

Susan’s resting pulse rate has been 55 beats per minute

for many years with standard deviation ¾ = 2:6 bpm.

During a 5 day period she checks her resting pulse rate

8 times a day at regular intervals and finds that it has

mean 56:2.

Is there sufficient evidence, at a 5% level, to conclude

that Susan’s pulse rate has changed?

The null hypothesis is H0: ¹ = 55. The alternative hypothesis is Ha: ¹ 6= 55

The significance level ® = 0:05 .

The number in the sample is n = 5 £ 8 = 40 and the sample mean is ¹x = 56:2.

The population standard deviation ¾ = 2:6 .

HYPOTHESIS TESTING WHEN THE POPULATION IS NOT NECESSARILY

NORMALLY DISTRIBUTED

Example 24



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Let X be Susan’s resting pulse rate. We do not know how the random variable

X is distributed, but if we assume that n is large enough for the Central Limit

Theorem to apply then the null distribution for the sample means X is

approximately normally distributed with mean ¹ = 55 and standard

deviation¾pn

=2:6p40

= 0:411 .

Entering this information into the calculator gives a P-value of P = 0:003 51 . As

P = 0:003 51 < 0:05 there is evidence at the 0:05 level to reject the null hypothesis.

We accept the alternative hypothesis Ha that Susan’s pulse rate has changed.

1 Globe Industries make torch globes with standard deviation life time of ¾ = 9 hours. If

the globes last too long, people will have no need to buy new ones, but if they do not

last long enough, people will stop buying them. A quality controller is to ensure that

globes made by a machine have a mean life of 80 hours. The quality controller selects

a sample of 50 globes and finds that they have a mean life of 83 hours.

a What is the null hypothesis the quality controller is testing?

b Assuming that a sample of n = 50 is large enough for the Central Limit Theorem

to apply, what is the null distribution the quality controller will be using?

c Is there sufficient reason at the 5% level for the quality controller to adjust the

machine?

2 Let X be the outcome of the roll of a fair six-sided die. The mean outcome of such a

die is ¹ = 3:5 with standard deviation ¾ = 1:708. Jack thinks his die may not be

fair. To test this he rolls the die 100 times and finds that the mean of the 100 rolls is

3:2.

a What null hypothesis is Jack testing?

b Briefly explain why the outcomes of a roll of a fair die are not normally distributed.

c Assuming that a sample of size n = 100 is large enough for the Central Limit

Theorem to apply, what is the null distribution Jack should be using?

d Does Jack have enough evidence at the 5% level to claim the die is not fair?

e Jack’s sister Betty rolls the same die 200 times and finds that the mean of her sample

is also 3:2. Would Betty come to the same conclusion as Jack?

3 While peaches are being canned, 250 mg of preserva-

tive is supposed to be added by a dispensing device.

It is known that the standard deviation of preserva-

tive added is 7:3 mg.

To check the machine, the quality controller obtains

60 random samples of dispensed preservative and

finds that the mean preservative added was 242:6mg.

At a 5% level, is there sufficient evidence that the

machine is not dispensing a mean of 250 mg?

EXERCISE 7H.2



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0.025

1.960

RR of H0

0.025

1.96

RR of H0

4 In recent times the mean age for New Zealand women on their first wedding day is 23:6years with a standard deviation of 2:9 years. To determine if this differs from Australian

women, a survey of 32 women was carried out. It was found that the mean age was

24:3 years. Test whether there is a significant difference at a 5% level.

H0

To test the null hypothesis H0 : ¹ = ¹0 against the alternative hypothesis Ha : ¹ 6= ¹0

we have used the test statistic z =¹x¡ ¹¾pn

.

P = Pr(Z 6 ¡z orZ > z) < 0:05

i.e., 2 £ Pr(Z 6 ¡z) < 0:05

i.e., Pr(Z 6 ¡z) < 0:025 :

But invNorm(0:025) + ¡1:96, and so

we reject the null hypothesis at the 5%level if the test statistic

z 6 ¡1:96 or z > 1:96 .

The rejection region for the null hypothesis H0 is the set of values of the test

statistic for which the null hypothesis is rejected.

The 5% rejection region for the null hypothesis H0 : ¹ = ¹0 is the set

fz : z 6 ¡1:96 or z > 1:96g

Note that the calculator also calculates the test statistic z when using the 2-sided Z-test.

REJECTION REGION FOR THE NULL HYPOTHESIS

Example 25

Assuming that , our test at the significance level has been to reject the null

hypothesis if the P-value

z� �> 0 5%

H0 : ¹ = 13:45, Ha : ¹ 6= 13:45 We use s = 0:25 to estimate ¾ as n is large.

Assuming that the sample of size n = 389 is large enough for the Central Limit

Theorem to apply, we find the test statistic z =¹x¡ ¹¾pn

=13:30 ¡ 13:45

0:25p389

+ ¡11:8

Since z < ¡1:96 we reject the null hypothesis that there is no difference in the

price and accept the alternative hypothesis that the price has changed.

A liquor chain claims that the mean price of wine has not changed from what it was

months ago. Records show that months ago the mean price was $ for a

mL bottle. A random sample of prices of different bottles of wine is taken

from several stores and the mean price is $ and the standard deviation is $ .

Is there sufficient evidence at the level to reject the claim?

12 12 13 45750 389

13 30 0 255%

:

: :



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For questions 1 and 2, test the hypothesis using the rejection region for the null hypotheses.

In each case you may assume that the sample size n is large enough for the Central Limit

Theorem to apply.

1

2

To test the hypothesis H0 : ¹ = 40 against Ha : ¹ 6= 40, a random sample

of size 60 was taken and found to have mean ¹x and standard deviation s = 7.

For what values of ¹x will the null hypothesis be rejected at the 5% level? Assume

that the sample size is large enough for the Central Limit Theorem to apply.

The test statistic z =¹x¡ ¹¾pn

=¹x¡ 40

7p60

+¹x¡ 40

0:9037

The null hypothesis will be rejected if z 6 ¡1:96 or if z > 1:96

i.e., if¹x¡ 40

0:90376 ¡1:96 or if

¹x¡ 40

0:9037> 1:96

) ¹x 6 40 ¡ 1:96 £ 0:9037 or ¹x > 40 + 1:96 £ 0:9037

The null hypothesis will be rejected if ¹x 6 38:2 or ¹x > 41:8 .

EXERCISE 7H.3

Example 26

Quickshave produces disposable razorblades. They

claim that the mean number of shaves before a blade

has to be thrown away is 13. A researcher wishes to test

the claim and asks 30 men to supply data on how many

shaves they got from one of the Quickshave blades. The

researcher found that the mean of the sample was 12:8.

Use this information to test the manufacturer’s claim at

a 5% level if the population standard deviation ¾ is 1:6:

It is claimed that the mean disposable income of households in a country town is $50 per

week. To test this claim, 36 households were sampled and it was found that the mean

disposable income of the 36 families was $47. Use this to test the claim that the mean

disposable income is not $50 per week if the population standard deviation ¾ = $12.

3 To test the hypothesis H0 : ¹ = ¡23 against Ha : ¹ 6= ¡23, a random sample

of size 100 was taken and found to have mean ¹x.

For what values of ¹x will the null hypothesis be rejected at the 5% level? You may

assume that the sample size is large enough for the Central Limit Theorem to apply and

that the population standard deviation ¾ = 4.

4 The volume of soft drinks dispensed by a machine is normally distributed with standard

deviation 3 mL. A quality controller has to adjust the machine if the mean volume

dispensed is not 504 mL. To test the machine the quality controller finds the mean

volume ¹x of 20 randomly selected bottles every hour. For what values of ¹x should the

quality controller not adjust the machine?



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DISCUSSION

Does this mean that if you take a large enough sample, and have a measuring instrument that

can measure outcomes of X accurately enough, you can always reject the null hypothesis?

Compare the formal sentence, “There is a statistically significant difference between the

population mean ¹ and ¹0.” with what is commonly understood by, “There is a significant

difference between the population mean ¹ and ¹0.”

In this section we show how to use a sample mean x to calculate an interval in which we

expect the population mean ¹ to lie. As with all statistics, our estimate for x could by chance

be very far from ¹, and we can never be absolutely sure that ¹ lies within the interval. We

can, however, know how probable it is that ¹ lies in the interval.

A confidence interval estimate of a parameter (in this case the population mean ¹)

is an interval of values between two limits, together with a percentage indicating our

confidence that the parameter lies in that interval.

We now consider how a so-called 95% confidence interval is constructed.

We start by finding the number a for which the standard normal distribution Z has probability

Pr(¡a < Z < a) = 0:95 .

Pr(Z < ¡a) = 0:025

) ¡a = invNorm(0:025)

¡a = ¡1:95996

a + 1:96

So, Pr(¡1:96 < Z < 1:96) = 0:95

This means that:

In any normal distribution, 95% of the outcomes lie within 1:96 standard deviations

from the mean.

So, suppose the random variable X is normally distributed as N(¹, ¾2):

If X is the random variable of sample means of size n, then X » :

) 95% of all ¹x lie in the interval ¹¡ 1:96¾pn< ¹x < ¹ + 1:96

¾pn:

The null hypothesis assumes that the population mean is exactly

equal to . This is required to set up the null distribution needed to

calculate probabilities. However, if the variable that is being tested is

continuous, the probability that is exactly equal to is zero!

H ¹¹

X¹ ¹

0

0

0

CONFIDENCE INTERVALS FOR MEANSI

a a

0.95

0.0250.025

0

Because of the symmetry of the graph of the

normal distribution, the statement reduces to

N ¹,

µ ¾pn

¶2µ ¶



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In the diagram we have shown a few ¹xvalues in this interval as well as one that

is not in this interval.

Note that each of the ¹x is in the middle of a line segment. All of these segments have the

same length as the line segment from ¹ ¡ 1:96¾pn

to ¹ + 1:96¾pn

.

Since Pr(¡1:96 < Z < 1:96) = 0:95 we know Pr(¡1:96 <X ¡ ¹

¾pn

< 1:96) = 0:95 .

So for the outcome x within the confidence interval,

x ¡ ¹¾pn

< 1:96 andx ¡ ¹¾pn

> ¡1:96

) x ¡ ¹ < 1:96¾pn

and x ¡ ¹ > ¡1:96¾pn

) ¹ > x ¡ 1:96¾pn

and ¹ < x + 1:96¾pn

This says that if we were to take many samples of size n and calculate the sample mean ¹xfor each of these samples, then for about 95% of these sample means, the population mean

¹ would lie in the interval

x ¡ 1:96¾pn< ¹ < x + 1:96

¾pn:

Confidence intervals for different confidence levels can be constructed for the population ¹in a similar way. Remember that we cannot be absolutely sure that ¹ will lie within the

confidence interval, but we can be confident that 95% of the time it will be.

�

x1

x1

x3

x3

x2

x2

95%

x4

x4

So, the 95% confidence interval for ¹is from

x¡ 1:96¾pn

to x + 1:96¾pn: x–

n

�96.1n

�96.1

x–

n

�96.1� x–

n

�96.1�

lower limit upper limit

Notice that theinterval calculated

for does not`!vcontain .�



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INVESTIGATION 7 CONFIDENCE LEVELS AND INTERVALS

Note: Consider samples of different size but all with mean 10 and standard deviation 2.

The 95% confidence interval is 10 ¡ 1:960 £ 2pn

< ¹ < 10 +1:960 £ 2p

n.

For various values of n we have: n Confidence interval

20 9:123 < ¹ < 10:877

50 9:446 < ¹ < 10:554

100 9:608 < ¹ < 10:392

200 9:723 < ¹ < 10:277

We see that increasing the sample size produces confidence intervals of shorter width.

DEMO

9 9.5 10 10.5 11

��10

n = 20n = 50n = 100n = 200

To obtain a greater understanding of confidence levels and intervals, click

on the icon to visit a random sampler demonstration. This will

calculate confidence intervals at various levels of your

choice ( , , or ) and count the intervals

which include the population mean.

90% 95% 98% 99%

We are given that x = 84:6 and ¾ = 16:8.

The 95% confidence interval is: x¡ 1:96¾pn

< ¹ < x + 1:96¾pn

i.e., 84:6 ¡ 1:96 £ 16:8p60

< ¹ < 84:6 +1:96 £ 16:8p

60

) 80:3 < ¹ < 88:9

So, we are 95% confident that the population mean weight of yabbies lies

between 80:3 grams and 88:9 grams.

A sample of yabbies was taken from a dam. The sample mean weight of the

yabbies was grams. Find the confidence interval for the population mean if

the population standard deviation is grams.

6084 6 95%

16 8:

:

Example 27



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1 A random sample of n individuals is selected from a population with known standard

deviation 11. The sample mean is 81:6.

a Find a 95% confidence interval for ¹ if: i n = 36 ii n = 100.

b In changing n from 36 to 100, how does the width of the confidence interval change?

2 Neville works for a software company. He keeps records of the times customers have to

wait to receive telephone support for their software. During a six month period he logs

167 calls, and the mean waiting time is 8:7 minutes. Find a 95% confidence interval

for estimating the mean waiting time for all telephone customer calls for support if the

population standard deviation is 2:08 minutes.

3 A breakfast cereal manufacturer uses a machine to

deliver the cereal into plastic packets which then go

into cardboard boxes. The quality controller ran-

domly samples 75 packets and obtains a sample mean

of 513:8 grams. Construct a 95% confidence interval

in which the true population mean should lie if the

population standard deviation is 14:9 grams.

4 A sample of 42 patients from a drug rehabilitation program showed a mean length of

stay on the program of 38:2 days. Estimate with a 95% confidence interval the average

length of stay for all patients on the program if the population standard deviation is 4:7days.

The fat content (in grams) of 30 randomly selected pasties at the local bakery was

determined and recorded as:

15:1 14:8 13:7 15:6 15:1 16:1 16:6 17:4 16:1 13:917:5 15:7 16:2 16:6 15:1 12:9 17:4 16:5 13:2 14:017:2 17:3 16:1 16:5 16:7 16:8 17:2 17:6 17:3 14:7

From a calculator x = 15:90 and we are given ¾ = 1:35

The 95% confidence interval for ¹ is

x¡ 1:96¾pn< ¹ < x + 1:96

¾pn

) 15:90 ¡ 1:96 £ 1:35p30

< ¹ < 15:90 + 1:96 £ 1:35p30

) 15:4 < ¹ < 16:4

So, we are 95% confident that the mean fat content of all pasties produced lies

between 15:4 g and 16:4 g.

Determine a confidence interval for the mean fat content of all pasties made if

the population standard deviation is grams.

95%1 35:

Example 28

EXERCISE 7I.1



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TI

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84 53 66 61 80 75 67 74 59 56 81 68 74 6982 76 60 63 78 71 80 60 72 63 58 77 68 7263 71 67 76 54 72 64 70 70 61 82 68

A 95% confidence interval for a mean ¹ of a population was recorded as

8:5617 6 ¹ 6 9:4383. This estimate was based on a sample of size n = 60.

Use this information to calculate

a x, the sample mean

b ¾, the population standard deviation which was used to calculate the

confidence interval.

a The 95% confidence interval is x¡ 1:96¾pn< ¹ < x + 1:96

¾pn

So, x¡ 1:96¾pn

= 8:5617 and x + 1:96¾pn

= 9:4383

Adding these equations gives 2x = 8:5617 + 9:4383 = 18 and so x = 9.

b Substituting n = 60 and x = 9 into

x¡ 1:96¾pn

= 8:5617 gives 9 ¡ 1:96¾p60

+ 8:5617

) 1:96¾p60

+ 0:4383

) ¾ + 0:4383 £p

60

1:96+ 1:732

6 A 95% confidence interval for the mean ¹ of a population is based on a sample of

n = 50, and given by 3:5842 6 ¹ 6 4:4158. Find:

a x, the sample mean

b ¾, the population standard deviation which was used to calculate the confidence

interval.

7 A 95% confidence interval for the mean ¹ of a population is given by

19:685 6 ¹ 6 22:315. If the population standard deviation is ¾ = 6, what was the

sample size?

It is possible to obtain confidence intervals at any level of confidence

from graphics calculators. Click on the icon to see how to do this on

your calculator.

Example 29

a Determine the sample mean x and standard deviation s.

b Using s to estimate ¾, determine a 95% confidence interval that the company would

use to estimate the mean point score for the population of applicants.

To work out the credit limit of a prospective credit card holder, a company gives points

based on factors such as employment, income, home and car ownership, and general

credit history. A statistician working for the company randomly samples applicants

and determines the point total for each. These are:

40



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When designing an experiment in which we wish to estimate the population mean, the size

of the sample is an important consideration. Finding the sample size is a problem that can be

solved using the confidence interval.

Let us revisit Example 28 on the fat content of pasties. The question arises:

‘How large should a sample be if we wish to be 95% confident that the sample mean will

differ from the population mean by less than 0:3 grams?’

i.e., ¡0:3 < ¹¡ x < 0:3

Now the 95% confidence interval for ¹ is: x¡ 1:96¾pn

< ¹ < x + 1:96¾pn

Hence ¡1:96¾pn

< ¹¡ x < 1:96¾pn

and we need to find n when 1:96¾pn

= 0:3 .

So,pn =

1:96¾

0:3=

1:96 £ 1:35

0:3+ 8:82 and so n + 78.

Thus, a sample of 78 pasties should be taken.

1 A researcher wishes to estimate the mean weight of adult crayfish in South Australian

waters. She knows that the population standard deviation ¾ is 250:5 grams. How large

must a sample be so that she is 95% confident that the sample mean differs from the

population mean by less than 70 grams?

2 A porridge manufacturer samples 80 packets of porridge and finds that the sample stan-

dard deviation s, of the contents’ weight is 17:8 grams. If s is used to estimate the

population standard deviation ¾, how many packets must be sampled to be 95% confi-

dent that the sample mean differs from the population mean by less than 3 grams?

DETERMINING HOW LARGE A SAMPLE SHOULD BE

Now ¡1:96¾pn

< ¹¡ x < 1:96¾pn

so we need to find n such that 1:96¾pn

= 5 i.e.,1:96 £ 16:8p

n= 5

) n =

µ1:96 £ 16:8

5

¶2

+ 43:37

A sample of 44 yabbies should be taken.

Revisit the yabbies from the dam problem of . Suppose we wish to find

the sample size needed to be confident that the sample mean differs from the

population mean by less than grams. What sample size should be taken?

Example 27

95%5

Example 30

EXERCISE 7I.2



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xn

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3 Patients from an alcohol rehabilitation program participate for various lengths of time

with a standard deviation of 4:7 days. How many patients would have to be sampled to

be 95% confident that the sample mean number of days on the program differs from the

population mean by less than 1:8 days?

Consider the typical 95% confidence interval

shown in the diagram.

The width of this interval is w = 2 £ 1:96¾pn

.

In taking a sufficiently large sample size n we can make w as small as we like.

As w = 2 £ 1:96¾pn

,pn =

2 £ 1:96¾

wand so n =

µ2 £ 1:96¾

w

¶2

When we wish to estimate the population mean from a sample of size n at a 95%confidence level, the sample size is given by

n =

µ2£ 1:96¾

w

¶2

where ¾ is the population standard deviation

and w is the confidence interval width.

In Example 30, w = 2 £ 5 and ¾ + 16:8 : Thus, n =

µ2 £ 1:96 £ 16:8

10

¶2

+ 43:37, etc.

Since n is an integer, n = 44 would give a 95% confidence interval of width about 10 grams.

4 A population is known to have standard deviation ¾ = 34. Find the sample size n that

should be taken to find a 95% confidence interval for the population mean ¹ of width:

a w = 5 b w = 1 c w = 0:1

5 A manufacturer of bottled water knows that the machine dispenses water into 1 litre

bottles with a standard deviation of 2:3 mL. The machine needs to be checked regularly

to ensure it is still delivering the correct volume. How many bottles should a quality

controller be checking to find a 95% confidence interval of width:

a 2 mL b 1 mL c 0:5 mL?

6 a If the size n of a sample is doubled, by how much will the width of a 95% confidence

interval decrease?

b How much larger do you have to make a sample size to halve the width of a 95%confidence interval?

Confidence intervals provide an estimate for the size of the population mean ¹. They can

also be used to assess claims about population means. For example:

Suppose the volume V of fruit juice dispensed by a machine is normally distributed with

mean ¹ litres which can be adjusted, and standard deviation ¾ = 0:0015 litre (112 mL, about

14 of a teaspoon) which is fixed.

USING A CONFIDENCE INTERVAL FOR A CLAIM ABOUT ¹



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Suppose a manufacturer needs to fill cartons with 1 litre of fruit juice. To ensure that almost

all cartons contain at least 1 litre, the value of the mean ¹ is set at 1:005 litre.

Note that for sufficiently large n the null hypothesis will not be accepted at the 5% level.

For such values of n the difference is statistically significant at the 5% level even though the

difference of 0:01 mL (hardly a drop) is not significant as the word is commonly understood.

1 Suppose the time it takes Joan to run 100 metres is normally distributed with mean

¹ = 12:46 seconds and standard deviation 1 second. To improve her time Joan goes on

a training program. After the training program, Joan finds that the mean time from 12trial runs is now 11:62 seconds.

a Construct a 95% confidence interval for Joan’s mean assuming the standard deviation

has not changed.

b Use the result of part a to assess the claims:

i Joan’s time to run 100 metres has improved.

ii Joan is now better than Betty whose time for the 100 metres is 11:97 seconds.

a Construct a 95% confidence interval for the volume ¹ dispensed by the machine.

b Use the 95% confidence interval to assess the claim that the volume dispensed by

the machine has increased.

c Can we conclude that the volume of ¹ is now larger than 1005 mL?

a The confidence interval is 1003 6 ¹ 6 1011:

b Since 995 is less than all the values in the 95% confidence interval we can be

confident that the population mean has increased.

c Althouth the sample statistic 1007 mL is larger than 1005, the smallest number

in the 95% confidence interval for ¹ is 1003 mL. This means that ¹ could be as

small as 1003 mL, and there is not enough evidence to support the claim that

¹ > 1005 mL.

Suppose the volume of cool drinks dispensed into cartons by a machine is

normally distributed with mean which can be adjusted, and standard deviation

mL which is fixed. The value of is supposed to be mL, but the machine

operator notices that actually mL. The operator therefore adjusts the volume

dispensed by the machine. A quality controller tests cartons and finds that their

mean volume is mL.

V¹

¹¹

10 1005=995

251007

� ��

�

Example 31

Note: This question is closely related to testing the hypotheses ,

.

H ¹H ¹

0�

�

: = 1005: =1005

� � �� a 6

EXERCISE 7I.3

A quality controller takes a sample of n cartons and, with very accurate measurements, finds

that the sample mean v = 1:004 99 litres. We want to test the hypotheses H0 = 1:005,

Ha 6= 1:005 for various large values of n:



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50.0 56.1 57.2 58.3

CI

A buyer for a restaurant chain goes to a seafood wholesaler to inspect a large catch

of 50 000 prawns. She has instructions to buy the catch only if the prawns are heavy

enough. The buyer selects a sample of 60 prawns and finds that their mean weight is

57:2 grams. It is known that the population standard deviation ¾ is 4:2 grams.

a Find the 95% confidence interval for the population mean.

b The buyer claims she is 95% confident that no more than 10% of the prawns

weigh less than 50 grams. Use the confidence interval found in part a to justify

this claim. You may assume that the weights of prawns are normally distributed.

a Using technology, the 95% confidence interval for the population mean ¹ is

56:1 6 ¹ 6 58:3 .

b The smallest value in the 95% confidence int-

erval is 56:1, and so the buyer can be 95%confident that the population mean ¹ > 56:1 .

If W is the weight of prawns, then W » N(¹, ¾2).

If we use ¹ = 56:1 and ¾ = 4:2, then using technology Pr(W < 50) = 0:0732.

Hence 7:32%, or less than 10% of the prawns weigh less than 50 grams.

OTHER APPLICATIONS OF CONFIDENCE INTERVALS

Example 32

2 A complaint was made to a call centre that it took a mean time of 12 minutes before a

caller was put through to an operator. After changes were made, the call centre claimed

that the service had improved. To check this claim, a consumer group made 40 calls to

the centre. They found the mean waiting time was 8 minutes with a standard deviation

of 3 minutes. Assuming that 40 is large enough for the Central Limit Theorem to apply,

construct a 95% confidence interval for the mean waiting time ¹. Does the confidence

interval support the call centre’s claim? (Use s to estimate ¾.)

3 The distance D a golfer can hit a ball is randomly distributed with a mean ¹ = 115metres and standard deviation ¾ = 32 metres.

a After spending time with a professional the golfer measured the distance of 30drives. The results of the drives in metres were as follows:

133 153 110 93 142 135 62 150 127 112119 171 143 92 162 128 149 73 39 84138 152 163 174 152 141 129 87 118 149

Assuming that the sample of 30 is large enough for the Central Limit Theorem to

apply, calculate a 95% confidence interval for the mean distance ¹ the golfer can

now hit the ball. Does the confidence interval provide enough evidence to support

the claim the golfer has improved?

b The golfer decided to have another trial of 50 drives. Suppose the mean of the 50trials is the same as in part a.

i

ii Does the new information provide evidence that the golfer has improved?


Explain briefly why increasing the number of trials could make a difference

to a drive length.


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1 The manager of a golf club claimed that the income of most of its members was in excess

of $75 000 and thus its members could afford to pay increased annual subscriptions. To

justify this claim was not valid, the members sought the help of a statistician.

The statistician examined a random sample of 113 club members and found that the mean

income was $96 318. It is known that the standard deviation of the members’ incomes

is $14 268:

a Find the 95% confidence interval for the population mean income of all members.

b The statistician claimed that he was 95% certain that no more than 10% of the

members had a mean income of less than $75 000.

Assuming that the income of members is normally distributed, how could you justify

the statistician’s claim?

2 Fabtread manufacture motorcycle tyres. Under normal test conditions the stopping time

for motor cycles travelling at 60 km/h is 3:45 seconds with standard deviation 0:17seconds. Their production team has just designed and manufactured a new tyre tread.

They take 41 stopping time measurements with the new tyres and find the mean time is

3:03 seconds.

a Calculate a 95% confidence interval for the mean stopping time of the new tyres.

b The team claims that they are 95% certain that less than 15% of the stopping times

of their new tyres will exceed the 3:45 seconds of the old tyres.

Assuming that the stopping time is normally distributed, how could you justify the

team’s claim?

There are often good reasons to find confidence intervals other than those of 95%. In areas

like medicine, a researcher may want to have more certainty when making decisions and often

may prefer a confidence interval of 99%. In other areas where the outcomes of decisions are

not so important, people may be satisfied with 90% confidence intervals.

Your calculator can produce confidence intervals at any level.

1 The mean ¹ of a population is unknown, but its standard deviation is 10. In order to

estimate ¹ a random sample of size n = 35 was selected. The mean of the sample was

found to be 28:9.

a Find a 95% confidence interval for ¹. b Find a 99% confidence interval for ¹.

c In changing the confidence level from 95% to 99%, how does the width of the

confidence interval change?

2 If the P% confidence interval for ¹ is x¡ a

µ¾pn

¶< ¹ < x + a

µ¾pn

¶then

for P = 95, a = 1:960: Find a if P is: a 99 b 80 c 85 d 96.

3 The choice of the confidence level to be used is made by an experimenter. Why is it

that experimenters do not always choose confidence intervals of at least 99%?

EXERCISE 7I.4

EXTENSION TO CONFIDENCE INTERVALS OTHER THAN 95%

EXERCISE 7I.5



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REVIEW SET 7A

1 The arm lengths of 18 year old females are normally distributed with mean 64 cm

and standard deviation 4 cm.

a Find the percentage of 18 year old females whose arm lengths are:

i between 60 cm and 72 cm ii greater than 60 cm.

b Find the probability that if an 18 year old female is chosen at random, she will

have an arm length in the range 56 cm to 68 cm.

2 a If Z has a standard normal distribution, find k if Pr(Z 6 k) = 0:95 .

b If X » N(23, 2:62) find k if Pr(X < k) = 0:6 .

3 In a mathematics test out of 40 marks, the mean mark was 28:3 and the standard

deviation was 4:1. The marks were all integers and the minimum pass mark was set

at 24. Assuming marks were approximately normal, what proportion of the students:

a passed the test b scored more than 20 c scored between 25 and 35?

4 The weights of apples from an orchard are known to be normally distributed with

mean ¹ = 350 grams and standard deviation ¾ = 25 grams. The apples are packed

in boxes of 50 each.

a How many apples in a box would you expect to weigh more than 375 grams,

and how many less than 325 grams?

b In 500 boxes, how many apples would you expect to have a weight between 325and 375 grams?

5 To test the hypotheses H0: ¹ = 36 and Ha: ¹ 6= 36 a random sample of n = 20was selected. The outcomes are listed below:

38 22 43 21 36 44 20 49 36 3042 43 38 28 33 22 29 25 28 34

Use this information to test the null hypothesis at the 5% level if the population

standard deviation is 10 grams.

6 The standard deviation in the weight of cereal boxes is 23:6 grams. How many boxes

must be sampled from the population to be 95% confident that the sample mean differs

from the population mean by less than 4 grams?

7 A factory canning apricots uses a machine to deliver the fruit and syrup into cans. The

quality controller randomly samples 65 cans and finds that the mean mass of contents

is 828:2 grams.

a Construct a 95% confidence interval in which the true population mean should

lie if the population standard deviation is 16:3 grams.

b What should the sample size be to construct a confidence interval of half the

width of that in a?

8 a Kerry’s marks for an English essay and a Chemistry test were 26 out of 40 and

82% respectively.

i Explain briefly why the information given is not sufficient to determine

whether Kerry’s results are better in English than in Chemistry.

REVIEWJ



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REVIEW SET 7B

ii Suppose that the marks of all students in both the English essay and the

Chemistry test were normally distributed as N(22, 42) and N(75, 72) re-

spectively. Use this information to determine which of Kerry’s two marks

is better.

iii If there were 50 students sitting for the English essay, how many would

have scored more than Kerry?

b Les is to sit for five subjects in the final examination. Because of many different

factors that determine examination marks, the marks Les can expect in each exam

are normally distributed. Suppose that the mean ¹ and standard deviation ¾ = 2are the same for each exam.

If ¹ = 12 calculate the probability that Les will gain a total mark for the five

subjects of between 60 and 70.

c The value of the mean ¹ depends on the time t hours that Les studies. It is given

by ¹ = 16 ¡ 8=(t + 2).

i For how long must Les study to achieve a value of ¹ = 15?

ii Les’s total score for the five examinations was 65. Use this information to

test the hypotheses H0 : ¹ = 15 and Ha : ¹ 6= 15.

iii Use the total score of 65 to construct a 95% confidence interval for the mean

¹. Use this interval to estimate a range of times Les might have studied for

the examination.

1 Find the mean and standard deviation of these two samples of lengths given in cm:

A 170:1 169:4 169:5 170:4 169:8 170:5 170:0 170:0 170:3 170:8170:0 169:9 170:2 170:0 169:9 169:9 170:5 170:1 169:7 170:0

B 177 166 153 167 176 173 169 161 172 174170 162 178 174 179 171 148 184 178 175

Which of the above is a sample of heights of 15 year old boys, and which is a sample

of length of planks cut by a machine?

2 The contents of a certain brand of soft drink can is normally distributed with mean

377 mL and standard deviation 4:2 mL.

a Find the percentage of cans with contents:

i less than 368:6 mL ii between 372:8 mL and 389:6 mL

b Find the probability of randomly selecting a can with contents between 364:4mL and 381:2 mL.

3 The life of a Xenon battery is known to be normally distributed with a mean of

33:2 weeks and a standard deviation of 2:8 weeks.

a Find the probability that a randomly selected battery will last at least 35 weeks.

b For how many weeks can the manufacturer expect the batteries to last before 8%of them fail?

4 The length of steel rods produced by a machine is normally distributed with a standard

deviation of 3 mm. It is found that 2% of all rods are less than 25 mm long. Find

the mean length of rods produced by the machine.



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5 a If Z has a standard normal distribution, find a if Pr(Z 6 a) = 0:9 .

b If X » N(15:6, 22) find a if Pr(X < a) = 0:9 .

6 A manufacturer claims that his canned soup contains 135 mg of salt. To check this

claim a consumer tested 87 cans for salt content and found that the mean was 139:6mg. It is known that the population standard deviation is 22:8 mg. At a 5% level is

there sufficient evidence to reject the manufacturer’s claim?

7 To test the null hypothesis H0: ¹ = 2000 and Ha: ¹ 6= 2000, a random sample

of n = 75 was selected and found to have mean x = 1840.

a If the population standard deviation ¾ = 690, is there sufficient evidence to reject

the null hypothesis at the 5% level?

b For what values of the sample mean ¹x would you not reject the null hypothesis at

the 5% level?

8 A telephone call centre handles many calls each day. Let T be the time in minutes

taken to answer a call.

In 2006 the mean answering time for a call was ¹ = 4:3 minutes with standard

deviation ¾ = 1:2 minutes.

Let T be the mean time taken to answer a random sample of 100 calls.

a The two histograms below show the distribution of a sample of size 50 taken

from T . Note that the horizontal scale and the bin width are the same in both

histograms, but the vertical scales are different.

Identify the histogram that represents a sample from T .

Explain your answer.

b i Assuming that n = 100 is sufficiently large, explain why the distribution

of T is approximately normal with mean 4:3 minutes and standard deviation

0:12 minutes.

ii Calculate the probability Pr(T 6 4:35).

iii Hence calculate the probability that an operator in the call centre can be

occupied in answering 100 calls for less than seven and a quarter hours.

c As well as answering routine calls, the supervisor of the call centre also han-

dles unusual cases that are too complicated for other staff to handle. When the

supervisor was timed her mean time to answer 100 calls was T = 4:6 minutes.

i Use the statistic T = 4:6 minutes to test the hypothesis H0 : ¹ = 4:3 and

Ha : ¹ 6= 4:3, at level.5%ii The supervisor is asked to explain why she is taking too long to answer

questions. What reasons can the supervisor provide to claim that the Central

Limit Theorem does not apply to her?

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REVIEW SET 7C

1 Sketch the graph of X » N(3, 22).On the horizontal axis mark in the z-scores as well as their corresponding x values.

Calculate these probabilities: a Pr(¡1 6 X 6 1) b Pr(¡1 6 Z 6 1) .

2 Staplers are manufactured for $5:00 each and are sold for $20:00 each. The staplers

are guaranteed to last three years. The mean life is actually 3:42 years and the

standard deviation 0:4 years. If the life of these staplers is normally distributed, how

much profit would we expect from selling a batch of 2000 (with a maximum of one

replacement)?

3 The edible part of a batch of Coffin Bay oysters is normally distributed with mean

38:6 grams and standard deviation 6:3 grams. Given that the random variable X is

the mass of a Coffin Bay oyster, find:

a a if Pr(38:6 ¡ a 6 X 6 38:6 + a) = 0:6826 b b if Pr(X > b) = 0:8413.

4 King prawns are favourite items on the menu of Stirling Caterers. From past expe-

rience the manager knows that people on average eat 325 g of prawns with standard

deviation 86 g. The manager is to cater for a wedding of 80 guests and decides to

purchase 27:5 kg of prawns. What is the probability that the caterer will run out of

prawns?

5 For export purposes peaches must be neither too small nor too large. A grower claims

that the peaches in his orchard have a mean weight of 300 grams, just right for export.

A buyer knows that the population standard deviation is 30 grams, and he wants to

test the grower’s claim.

a What hypotheses should the buyer consider?

b Suppose the buyer selects a random sample of 100 peaches and finds that their

mean weight ¹x = 310 grams.

i What is the null distribution the buyer should use?

ii Calculate the test statistic z for this sample.

iii Does this sample support the grower’s claim at the 5% level?

6 Width (mm) Frequency

22 123 324 1725 4326 6827 4128 2429 3

a Find the sample mean.

b Determine a 95% confidence interval for the

population mean ¹.

7 Suppose the weight X of apricots is normally distributed with ¹ = 90 grams and

¾ = 10 grams.

a Calculate the proportion of apricots with weight less than 88 grams.

b In a box of 100 apricots, how many would you expect to weigh less than 88 g?

c The apricots are packaged into boxes of 100 each. What proportion of the boxes

will have apricots with a mean weight less than 88 g?

The average width of snail shells of a local species

needs to be estimated. It is known that the standard

deviation is . mm. Pauline takes a random sample of

snails and measures the width of each shell to the

nearest mm. The results are shown in the table

alongside.

��

�

1 4200



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d On each of the boxes of 100 apricots is printed that the nett weight is 8:8 kilo-

grams. In a shipment of 500 boxes, for how many is the weight less than 8:8kilograms?

8 The time T it takes Laura to travel to work is normally distributed with mean ¹minutes and standard deviation 10 minutes. Laura’s work starts at 9 o’clock in the

morning.

a Suppose ¹ = 40 minutes and Laura leaves for work at a quarter past eight in

the morning.

i What is the probability she will be late?

ii If there are 250 working days in a year, how often would Laura be expected

to be late to work in a year?

b

ii Suppose Laura found that for her sample of 10 days the mean time to travel

to work was T 10 = 35 minutes. Use this information to test the hypotheses

H0 : ¹ = 40 and Ha : ¹ 6= 40, at level.5%

iii Calculate the 95% confidence interval for ¹.

iv How large a sample should Laura take to obtain a 95% confidence interval

of width 2:48 minutes?

c

Laura does not know the value of ¹ and decides to keep a 10 day record of the

time it takes her to go to work. Let T 10 be the distribution of the mean time

over 10 days it takes Laura to go to work.

i Briefly describe the distribution T 10 in terms of the distribution T it takes

Laura to go to work.


After keeping records for a year consisting of working days, Laura found

that the mean travelling time to work was minutes. She wants to be

certain that she will be at work before o’clock at least of the time in the

following year. To the nearest minute, what is the latest time you would advise

Laura to leave home? Give reasons for your answer.

25031 52

9 90%: �

��

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8

Contents:

Binomial distributionsBinomial distributions

A

B

C

D

E

F

Pascal’s triangle

Assigning probabilities

Normal approximation for binomialdistributions

Hypothesis testing for proportions

Confidence intervals for proportions

Review


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In this chapter we explore binomial distributions. These arise from

situations where there are only two outcomes, such as an answer

being true or false, a child being a boy or a girl, or a coin showing

a head or a tail.

One of the main results we will find is that binomial distributions

can be approximated by normal distributions. This remarkable re-

sult was the first known special case of the Central Limit Theorem.

Isaac Newton (1642 - 1727) who was not known for praising his contemporaries, gave

someone who asked him a mathematical question the advice, “Go to Mr de Moivre, he knows

these things better than I.”

We will use the machinery of normal distributions to study binomial distributions. In particular

we will use the test hypotheses and construct approximate confidence intervals for proportions.

To describe a random variable X completely we need to

² specify precisely all possible outcomes of X

² assign probabilities to each outcome of X.

In this section we discuss the first of these tasks.

To analyse any random experiment it is important to start by

determining all possible outcomes that will be considered.

For example, when tossing a real coin the possible

outcomes are heads or tails fH, Tg.

All possible outcomes of tossing a coin twice are

displayed on the tree diagram opposite.

We can write them as fHH, HT, TH, TTg.

When tossing a coin four times the possible outcomes are cumbersome to put on a tree

diagram, but we can list them:

HHHH THHH HHTT HTTT TTTT

HTHH HTHT THTT

HHTH HTTH TTHT

HHHT TTHH TTTH

THTH

THHT

Notice that:

1 outcome is ‘4 heads’

4 outcomes are ‘3 heads’

6 outcomes are ‘2 heads’

4 outcomes are ‘1 head’

1 outcome is ‘0 heads’

INTRODUCTION

PASCAL’S TRIANGLEA

H

T

H

T

H

T

1st coin 2nd coin

HH

HT

TH

TT

It was discovered by the French-English mathematician

( - ). De Moivre made many contributions to

mathematics, including complex numbers.

Abraham

de Moivre 1667 1754� �

282 BINOMIAL DISTRIBUTIONS (Chapter 8)


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The number of possible outcomes in

tossing a coin 10 times is 210 or 1024:Constructing a tree diagram or listing

every outcome is practically impossible.

However, if we are only interested in

the number of heads, and not at which

tosses they appeared, then there are only

11 outcomes: 0 heads, 1 head, 2 heads,

...., 10 heads. In this case we can sim-

plify the tree diagram considerably.

The above diagram shows the history of tossing a coin a number of times. The lines sloping

to the right indicate a tail, those to the left a head. The number n indicates the number of

tosses. Along the horizontal line we record the frequency of the number of heads that could

occur after n tosses.

For example, the line for n = 4 reads from left to right: “4 heads occur once, 3 heads occur

4 times, 2 heads occur 6 times, 1 head occurs 4 times and no heads occurs once.”

The diagram also illustrates how we can calculate the next row from the previous one. For

example:

The only ways we can get 4 heads on the 7th toss are:

² if after the 6th toss we have 4 heads, then a tail will still give us 4 heads

² if after the 6th toss we have 3 heads, then another head will make 4.

But in the 6th toss, the frequency of 4 heads is 15, and of 3 heads is 20. It follows that the

frequency of 4 heads on the 7th toss is 15 + 20 = 35.

The above tree is known as Pascal’s triangle in honour of Blaise Pascal (1623 - 1662)

although the triangle was already known in China by Chia Hsien (about 1050) and by the

Persian mathematician Omar Khayyam (ca 1050 - 1120), better known for his poem in praise

of wine, The Rubaiyat.

The number of ways of getting r successes and n¡ r failures from n repetitions is usually

written as Cnr . These quantities are known as the binomial coefficients.

The C stands for Combinations. It is the combined or total number of r successes in n trials.

The following diagram shows Pascal’s triangle using the Cnr notation.

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

is the same as

C10 C1

1

C20 C2

1 C22

C30 C3

1 C32 C3

3

C40 C4

1 C42 C4

3 C44

C50 C5

1 C52 C5

3 C54 C5

5

C60 C6

1 C62 C6

3 C64 C6

5 C66

n��

n��

n��

n��

n��

n��

n��

( )�H

( )�H

( )�H

( )�H

( )�H

( )�H

(1 )H

(2 )H

(3 )H

(4 )H

(5 )H

(6 )H

H T

1 1

1 12

1 33 1

1 64 4 1

1 105 10 5 1

1 156 20 15 6 1

�H �H �H

�H �H

Cnr can also be calculated from the formula Cn

r =n!

r!(n¡ r)!, where n! is the product

of the first n positive integers.

BINOMIAL DISTRIBUTIONS (Chapter 8) 283


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The number of different ways of getting ‘ heads’ (and ‘tails’) when tossing coins

is .

6 19 25

C256

Notice that 4! = 4 £ 3 £ 2 £ 1.

You can see how this formula is derived by clicking on the icon.

You can also find Cnr from the combination key or on a scientific calculator.

When tossing 4 coins, the total number of possible outcomes which are ‘2 heads’ is C42 = 6.

To find C42 press: 4 2 or 4 PRB 3 then 2

We press 25 PRB 3 then 6 to get 177 100.

1 Display all possible outcomes of tossing a coin three times. How does this relate to

Pascal’s triangle?

2 a When tossing five coins, what are the possible outcomes? List all 32 of them.

b How many of the outcomes in a consist of:

i ‘5 heads’ ii ‘4 heads’ iii ‘3 heads’

iv ‘2 heads’ v ‘1 head’ vi ‘0 heads’

3 a What is the rule for finding the next row of Pascal’s triangle from the previous row?

b Predict the seventh row of Pascal’s triangle.

c Use your calculator to check the seventh row by finding C70 , C7

1 , C72 , ..... , C7

7 .

4 Draw Pascal’s Triangle to row 8. Use the triangle to find the number of ways of getting:

a 2 successes in 3 trials b 5 successes in 7 trials c 2 successes in 5 trials

d 1 success in 8 trials e 0 successes in 4 trials f 6 successes in 6 trials

Verify each using your calculator.

5 Use a calculator to find the number of ways of getting:

a 5 successes in 9 trials b 3 successes in 14 trials

c 1 success in 40 trials d 2 successes in 3 trials

e 10 successes in 40 trials f 20 successes in 40 trials

6 a When tossing 18 coins, in how many different ways can you get 10 heads and 8tails?

b Over her lifetime Jessie the cow had 23 calves, 14 male and 9 female. In how many

different ways could she have had her calves according to their sex?

(Note: MMFMFFFMFMMMFMFMMFMFMMM is one such way.)

7 (p + q)2 = p2 + 2pq + q2

a Find (p + q)3 using (p + q)(p + q)2. List the coefficients of each term in the

expansion.

b Likewise find the expansion of (p + q)4 and list its coefficients.

c Without using algebraic expansion, predict the expansions of:

i (p + q)5 ii (p + q)6

d If p + q = 1, what is the value of C40 p

4 + C41 p

3q + C42 p

2q2 +C43 pq

3 +C44 q

4?

nCrn rC

MATH ENTERn rC =

MATH ENTER

EXERCISE 8A

APPENDIX



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Probabilities may be assigned to events in a number of ways. For example:

²

² we can use symmetry to say the chances of a coin toss being a head or a tail are

both 12 , or a die showing a particular number being 1

6

²

However probabilities are assigned, they must satisfy the following rule:

If X is a random variable with sample space fx1, x2, x3, ...... , xng and corresponding

probabilities fp1, p2, p3, ...... , png then

² 0 6 pi 6 1 for all i = 1 to n

² p1 + p2 + p3 + :::::: + pn = 1.

The function P (xi) = pi is called the probability function of X.

We sometimes write P (xi) as Pr(X = xi) or P(X = xi).

Note that unlike the distribution of continuous variables, the end points of an interval do

matter for discrete variables.

The following table lists commonly used notation for probabilities.

Notation Statement

Pr(X = 3) the probability that X equals 3

Pr(X < 3) the probability that X is less than 3

Pr(X 6 3) the probability that X is at most 3, or no more than 3

Pr(X > 3) the probability that X is more than 3

Pr(X > 3) the probability that X is at least 3, or no less than 3

Pr(3 < X < 7) the probability that X is between 3 and 7

Pr(3 6 X 6 7) the probability that X is at least 3 but no more than 7

Pr(3 < X 6 7) the probability that X is more than 3 but no more than 7

Pr(3 6 X < 7) the probability that X is at least 3 but less than 7

For example, a coin does not have a memory.

If you toss a coin twice the outcome of the second toss is independent of what happened on

the first toss.

ASSIGNING PROBABILITIESB

we can conduct experiments where we perform trials many times over until a pattern

emerges

we can evaluate the form of tennis players to predict their chances in an upcoming

match.

INDEPENDENT EVENTS

Trials or events are if the outcome of one does not affect the outcome of

the others.

independent



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INVESTIGATION 1 SAMPLING SIMULATION

If you draw a marble from a bag that contains 1 blue,

1 red and 2 green marbles and replace the marble after

it has been drawn, there are again 1 blue, 1 red and 2green marbles in the bag. Your chance of drawing a

red marble is exactly the same in the next draw, and the

trials are independent.

If marbles are not replaced, the next selection depends

on which marble has already been selected. If you drew

a red marble in the first draw, there will be no more

red marbles in the bag and you cannot draw another red

marble in the next draw.

1 To simulate the results of tossing two coins,

set the bar to 50% and the sorter to show

Run the simulation 200 times and repeat this four times. Record each set of results.

2 A bag contains 7 blue and 3 red marbles and two marbles are randomly selected from

it, the first being replaced before the second is drawn.

The sorter should show and set the bar to 70%

as Pr(blue) = 710 = 0:7 = 70%:

3 From the bag of 7 blue and 3 red marbles, three marbles are randomly selected with

replacement.

Set the sorter to and the bar to 70%:

Run the simulation a large number of times to obtain experimental estimates of the

probabilities of getting:

a three blues b two blues c one blue d no blues.

What to do:

in

A B C D E

In this investigation balls are dropped into a sorting

device. When balls enter the ‘sorting’ chamber they hit

a metal rod and may go left or right with

. This movement continues as the balls

fall from one level of rods to the next. The balls finally

come to rest in collection bins at the bottom of the sorter.

Click on the icon to open the simulation. Notice that we

can use the sliding bar to alter the probabilities of balls

going to the left or right at each rod.

equal

likelihood

SIMULATION

Click on the icon to obtain ‘How much should I plant’ investigation.

Run the simulation a large number of times and use the results to estimate the

probabilities of getting: two blues one blue no blues.a b c

INVESTIGATION



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A binomial distribution describes the distribution of the number of successes that occur

in a sequence of n trials providing that:

² the number of trials n is fixed in advance

² the trials are independent

² each trial has exactly two possible outcomes, often called success and failure

² each trial has the same probability of success.

The two outcomes are not necessarily equally likely, so we suppose the probability of a

success is p and the probability of a failure is q.

The sum of the probabilities p + q = 1, so q = 1 ¡ p:

Suppose a spinner has three blue edges and one white edge.

On each occasion it is spun, the chance of finishing on blue is34 and on white is 1

4 .

If blue is a “success” and white is a “failure” then p = 34 and

q = 14 :

Let the binomial random variable X be the total number of successes in n trials.

The possible outcomes for the three spins, with their probabilities, are displayed on the tree

diagram:Event Number of blues Probability

BBB 3 (34)3

BBW 2 (34 )2(14)1

BWB 2 (34 )2(14)1

BWW 1 (34 )1(14)2

WBB 2 (34 )2(14)1

WBW 1 (34 )1(14)2

WWB 1 (34 )1(14)2

WWW 0 (14)3

Note: ² Pr(3 blues) = 1 (34 )3

Pr(2 blues) = 3 (34 )2(14 )1

Pr(1 blue) = 3 (34 )1(14 )2

Pr(0 blues) = 1 (14 )3

The coefficients 1 3 3 1 are

the third row of Pascal’s triangle.

² The results are also obtainable by expanding (p + q)3 = p3 + 3p2q + 3pq2 + q3

with p = 34 and q = 1

4 .

THE BINOMIAL PROBABILITY DISTRIBUTION

B

B

B

BB

B

B

W

W

W

W

W

W

W

Er_

Er_

Er_

Er_ Er_

Er_

Er_

Qr_

Qr_

Qr_

Qr_

Qr_

Qr_

Qr_

1 spinst

2 spinnd

3 spinrd

Consider twirling the spinner three times. Let the variable be the number of blue results

that could occur. The possible outcomes of are , , , or .

XX x = 0 1 2 3



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TI

C

² Since 1 = C30 , 3 = C3

1 , 3 = C32 , 1 = C3

3 we can write

Pr(x blues) = C3x (34 )x (14)3¡x for x = 0, 1, 2, 3.

In the case of n trials there are Cnx ways of selecting x blues and n ¡ x non-blues.

If p is the probability of selecting a blue from a single trial, then for n trials,

Pr(x blues) = Cnx px (1 ¡ p)n¡x for x = 0, 1, 2, 3, ...... , n.

Thus,

if the random variable X is the number of successes in n binomial trials and p is the

probability of a success in any one trial, then P(X = x) = Cnx px (1 ¡ p)n¡x

for x = 0, 1, 2, 3, ...... , n.

Note: ² Some events which are not strictly binomial can

still be modelled using a binomial distribution.

For example,

During quality control a small sample of n light

globes may be selected from tens of thousands

and tested without replacement. Since the globes

are not replaced the events are not independent,

but because the population is exceedingly large,

the binomial distribution still provides a good ap-

proximation.

²

THE GENERAL CASE

a three of the chocolates selected are Strawberry Delights

b at least three of the chocolates selected are Strawberry Delights.

Since n = 5,

X = 0, 1, 2, 3, 4 or 5 and p = 72% = 0:72 and q = 1 ¡ p = 0:28

X is distributed as Bin(5, 0:72).

a Pr(X = 3)

= C53 (0:72)3(0:28)2

+ 0:293

b Pr(X > 3)

= Pr(X = 3 or 4 or 5)

= C53(0:72)3(0:28)2 + C5

4 (0:72)4(0:28)1 + C55 (0:72)5

+ 0:862

In a bin of chocolates, are Strawberry Delights and the remainder are

Caramel Creams. Five chocolates are taken from the bin with the previous selection

replaced before the next chocolate is sampled. Find the probability that:

100 72

Example 1

A binomial variable is often specified in the form Bin , .( )n p

Bin , is a useful

notation. It indicates

that the distribution

is binomial and gives

the values of and .

( )n p

n p

You can use a calculator to calculate these probabilities.


Let denote the number of Strawberry Delights selected from the bin. Since there

is replacement before the next chocolate is selected, the distribution is binomial.

X �


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There are usually two binomial probability functions on calculators. For a TI 83 these are:

² the probability of exactly x successes in n trials is binompdf(n, p, x)

where pdf stands for probability density function.

² the probability X is at most x, i.e., Pr(X 6 x) is binomcdf(n, p, x)

where cdf stands for cumulative density function.

On a TI 83, the calculations for Example 1 are:

1 For which of these probability experiments does the binomial distribution apply? Justify

your answers.

a A coin is thrown 100 times. The variable is the number of heads.

b One hundred coins are each thrown once. Assume the probability of heads turning

up is the same for each coin. The variable is the number of heads.

c A box contains 5 blue marbles and 3 red marbles. I draw out 5 marbles, replacing

the marble each time. The variable is the number of red marbles drawn.

d A box contains 5 blue marbles and 3 red marbles. I draw out 5 marbles. I do

not replace the marbles that are drawn. The variable is the number of red marbles

drawn.

e A paper cup is thrown 100 times. For each throw it can land right-side up, upside-

down, or on its side. The variable is the position of the cup when it has landed.

f

g A large bin contains ten thousand bolts, 1% of which are faulty. I draw a sample

of 10 bolts from the bin. The variable is the number of faulty bolts.

2 If the births of boys and girls are assumed to be equally likely, calculate the probability

that in a family of six children:

a there are exactly 2 boys b all the children are boys

c there are at least 4 girls d most of the children are boys.

EXERCISE 8B.1

A union has members. of the members are in favour of a certain change

to their conditions of employment. A random sample of five members is chosen

and their opinions asked. The variable is the number of surveyed members that are

in favour of the change in conditions.

100 72%

a Pr(X = 3)

= binompdf(5, 0:72, 3)

+ 0:293

b Pr(X = 3 or more)

= Pr(X > 3)

= 1 ¡ Pr(X 6 2)

= 1 ¡ binomcdf(5, 0:72, 2)

+ 0:862

3 At an election 35% of the 21 million voters favoured the Red Party.

a Can the binomial distribution be applied?

b If 7 voters were selected at random, what is the probability that:

i exactly 3 voters favoured the Red Party

ii a majority of those selected favoured the Red Party

iii at most 3 voters favoured the Red Party?



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4 It is known that at 8 a.m. the sky is overcast on an average of two days out of every five.

If a week of the year is taken at random (i.e., 7 consecutive days), find the probability

of the sun shining at 8 a.m.:

a for the whole week b for the first 3 days only

c on any 5 days d on at least 4 days.

You can use your calculator to construct a probability spike graph for a binomial distribution,

say Bin(20, 12).

The basic instructions are: ² L1 = 0, 1, 2, ......, 20

² L2 = binompdf(20, 0:5, L1)

² scatter plot L1 against L2

5 Sketch the probability spike graph for each of the following distributions:

a Bin(10, 0:1) and Bin(10, 0:9)

b Bin(10, 0:3) and Bin(10, 0:7)

c Bin(10, 0:5)

6 A true-false test consists of 20 questions.

A student guesses answers at random.

a Draw the probability spike graph for this distribution.

b Find the probability that this student is correct in

i all 20 questions

ii exactly 10 questions

iii at most 10 questions

iv at least 15 questions.

7 Over a period of time it is found that 6% of the goods produced by a manufacturer are

defective. If a sample of 12 goods is selected at random, find the probability that there

will be:

a no defectives b at least one defective

c at most one defective d less than 4 defectives.

8 A machine produces items that have a 5% chance of being defective. Suppose a sample

of 25 is taken. Find the probability that:

a exactly 2 of these items are defective

b at least one is defective.

9 An infectious flu virus is spreading through a school.

The probability of a student having the flu next week

is 0:3.

a Calculate the probability that out of a class of 25students, 2 or more will have the flu next week.

b If more than 20% of the students are away with

the flu next week, a class test will have to be

cancelled. What is the probability that the test

will be cancelled?

c Draw the probability spike graph for this

distribution.

TI

C



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10 Over a long period of time, a target shooter is found to have a 94% success rate in

shooting a target. During a competition the target shooter fires 20 times at a target. Find

the probability that the target shooter is:

a successful on all 20 shots b successful on at least 18 shots.

11 If a fair coin is tossed 200 times, what is the probability that:

a between 90 and 110 heads turn up

b at least 100 but no more than 108 heads turn up

c more than 95 but no more than 105 heads turn up

d at least 99 but less than 110 heads turn up.

12

14 People found to have high blood pressure are started on

a course of tablets and have their blood pressure checked

at the end of 4 weeks. The drop in blood pressure over

the period is normally distributed with mean 5:9 units

and standard deviation 1:9 units.

a Find the proportion of people who show a drop of

more than 4 units.

b

13 Apples from a grower’s crop in 2006 were normally distributed with mean 173 grams

and standard deviation of 34 grams. Apples weighing less than 130 grams were too

small to sell.

a Find the proportion of apples from this crop which were too small to sell.

b Find the probability that in a picker’s basket of 100 apples, up to 10 apples were

too small to sell.

A fair die is rolled 100 times.

What is the probability of between 15 and 22 fives turning up?

Let X be the number of fives that turn up in n = 100 rolls.

Then X is Bin(100, 16).

We are asked to calculate Pr(15 < X < 22),

i.e., the probability that X = 16, 17, 18, 19, 20 or 21.

Now Pr(15 < X < 22)

= Pr(X 6 21) ¡ Pr(X 6 15)= binomcdf(100, 1

6 , 21) ¡ binomcdf(100, 16 , 15)

+ 0:512

Example 2

A new drug has probability of curing a patient within week. If patients are to

be treated using this drug, what is the probability that between and patients

(inclusive) will be cured within a week?

75% 1 3824 31

Eight people from the large population taking the

course of tablets are selected at random. What is the

probability that more than five of them will show a

drop in blood pressure of more than units?4

EXTRA

PROBLEMS



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INVESTIGATION 2 THE MEAN AND STANDARD DEVIATION

OF A BINOMIAL DISTRIBUTION

15 Large batches of digital displays for clocks are produced by Clockit Ltd. There is a

probability p a display is faulty. Each batch contains n of these displays. A quality

control measure is carried out by using the following sampling method: Four displays

are randomly selected from a batch. If there is at most one defective item in the sample,

the batch is accepted; otherwise the batch is rejected.

Define the random variable X to be the number of defective displays in the sample of

four.

a Show that Pr(accepting a batch) = (1 + 3p)(1 ¡ p)3, 0 6 p 6 1:

b Sketch a graph of Pr(accepting a batch) vs p.

c What value of p will ensure that there is a 95% chance of accepting a batch?

Similarly, if we roll a die with probability of 16 of a 4 turning up, then in 30 rolls we might

expect np = 30 £ 16 = 5 “4s” to turn up.

In general, if a binomial experiment is repeated n times and the proportion of successes

is p, we might expect np successes to occur. This suggests that the mean of the binomial

distribution is ¹ = np.

There is no such easy way to guess the standard deviation of a binomial distribution. However,

the following investigation indicates what it might be.

Let us calculate the mean and standard deviation

of the variable X » Bin(30, 0:25).

1 Enter L1 = 0, 1, 2, 3, ......, 30 .

2 Enter L2 = binompdf(30, 0:25, L1)

3 Draw the scatterplot of L1 against L2.

MEAN & STANDARD DEVIATION OF A BINOMIAL RANDOM VARIABLE

What to do:

In this investigation we shall use a calculator to calculate the mean and

standard deviation of a number of binomial distributions. A spreadsheet can

also be used to speed up the process and handle a larger number of examples.

4 The command 1¡Var Stats L1,L2 calculates

the descriptive statistics for the distribution.

This produces a relative frequency table

with the values of in L and their relative

frequencies (probabilities) in L .

X 1

2

¹

¾

¹

¾

CASIO

TI


Suppose we have a coin which has probability of 12 of falling heads. If we toss this coin

20 times we might expect it to fall heads half the time, i.e., we expect np = 20 £ 12 = 10

heads.


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_08\292SA12STU-2_08.CDR Wednesday, 8 November 2006 5:18:43 PM DAVID3

5 Copy and complete the following table.

n p = 0:1 p = 0:25 p = 0:5 p = 0:7

10

30¹ = 7:5

¾ + 2:3717:::::

50

6 Compare your values with the formulae ¹ = np and ¾ =pnpq =

pnp(1 ¡ p)

From this investigation you should conclude that in general:

If X is a random variable which is binomial with parameters n and p,

i.e., X » Bin(n, p), then the mean of X is ¹ = np and the standard deviation is

¾ =pnp(1 ¡ p).

A fair die is rolled twelve times and X is the number of sixes that could result.

Find the mean and standard deviation of the X-distribution.

This is a binomial distribution with n = 12 and p = 16 i.e., X is Bin(12, 1

6 ).

So, ¹ = np

= 12 £ 16

= 2

and ¾ =pnp(1 ¡ p)

=q

12 £ 16 £ 5

6

+ 1:291

This means that we expect a six to be rolled 2 times, with standard deviation 1:291 .

5% of a batch of batteries are defective. A random sample of 80 batteries is taken

with replacement. Find the mean and standard deviation of the number of defective

batteries in the sample.

This is a binomial sampling situation with n = 80, p = 5% = 120 :

If X is the random variable for the number of defectives then X is Bin(80, 120 ).

So, ¹ = np

= 80 £ 120

= 4

and ¾ =pnp(1 ¡ p)

=q

80 £ 120 £ 19

20

+ 1:949

This means that we expect a defective battery 4 times, with standard deviation 1:949.

Example 3

Example 4



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1 Suppose X is Bin(6, p). For each of the following cases:

i find the mean and standard deviation of the X-distribution

ii graph the distribution using a histogram

iii comment on the shape of the distribution

a p = 0:5 b p = 0:2 c p = 0:8

2 A coin is tossed 10 times and X is the number of heads which occur. Find the mean

and standard deviation of the X-distribution.

3 Bolts produced by a machine vary in quality. The probability that a given bolt is defective

is 0:04 . Random samples of 30 bolts are taken from the week’s production. If X is

the number of defective bolts in a sample, find the mean and standard deviation of the

X-distribution.

4 A city restaurant knows that 13% of reservations

are not honoured, i.e., the group does not come.

Suppose the restaurant receives five reservations.

Let X be the random variable of the number of

groups that do not come. Find the mean and stan-

dard deviation of the X-distribution.

5 Suppose X is Bin(3, p).

a Find P (0), P (1), P (2) and P (3) using

P (x) = C3x p

x q3¡x

and display your results in a table:

xi 0 1 2 3P (xi)

b If pi = P (xi), use ¹ =P

pixi to show that ¹ = 3p:

c Use ¾2 =P

x2i pi ¡ ¹2 to show that ¾ =

p3p(1 ¡ p):

Calculating binomial probabilities by hand is tedious and although with electronic technology

it is no longer a big problem, there are limits to the binomial probabilities that calculators can

compute. For example, your calculator might be able to calculate C10000050000 (12)100000 but

not C1000000500000 (12 )1000000 .

In this section we examine how binomial distributions can be approximated by normal dis-

tributions provided n is large. Not only will this allow us to calculate probabilities, but also

enable us to apply the theory of normal distributions to binomial distributions.

EXERCISE 8B.2

NORMAL APPROXIMATIONFOR BINOMIAL DISTRIBUTIONS

C



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INVESTIGATION 3 SHAPE OF BINOMIAL DISTRIBUTIONS

1 Calculate values of np and n(1 ¡ p) for every combination of n = 20, 30, 50and 100 and p = 0:1, 0:3, 0:5, 0:7 and 0:9 . Enter your results in the table:

n p np n(1 ¡ p) Description of the graph

20 0:1

20 0:3...

100 0:9

2

In this investigation we will examine the shape of binomial distributions

for various values of and .n p

What to do:

SIMULATION

10 15 20 25

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

x

P x( )

5 30

Use a calculator to draw the histograms of the binomial

distribution Bin , for each case. Complete the table, describing

each graph as either “skew” or “symmetric”.

Alternatively, click on the icon to do the same activity by computer.

20( )n p�

From the investigation you should have discovered that:

² providing both np and n(1 ¡ p) are bigger than about 10, the graphs are symmetric

about the mean np and are bell-shaped like a normal curve.

² as n increases the binomial distribution more closely resembles the normal distribution.

Consider the case of n = 30 and p = 12 :

The graph shows the binomial distribution

for these values of n and p.

The probabilities for x = 0 to 6 and 24 to

30 are virtually 0.

Since the only possible outcomes of Xare the integers from 0 to 30, we have

a scatterplot with values of P (x) for each

integer x.

Notice that this graph is similar in shape

to the bell-shaped normal distribution.

In this example the binomial distribution X has

mean ¹= np

= 30 £ 12

= 15

and standard deviation ¾ =pnp(1 ¡ p)

=q

30 £ 12 £ 1

2

+ 2:739

and so we might expect the normal variable X¤ » N(15, 2:7392) to be a good approximation

for X.



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To read probabilities from the graphs we need to be aware of the difference between the

discrete binomial variable and the continuous normal variable.

Let us consider trying to find Pr(X = 13) graphically using a normal approximation.

We already know that as X» Bin(30, 0:5) then Pr(X = 13) = binompdf(30, 12 , 13)

+ 0:1115

From the graph alongside,

Pr(X = 13) area of rectangle ABCD

+ AD £ DC

+ h£ 1

+ h

+ normalpdf(13, 15, 2:739)

+ 0:116

which is extremely close to 0:115

If the random variable X has a binomial distribution Bin(n, p) then X has an

approximate normal distribution with

mean ¹ = np and standard deviation ¾ =pnp(1 ¡ p):

The larger the value of n, the better the approximation.

As a rule of thumb, the approximation is good as long as both np > 10 and

n(1 ¡ p) > 10.

A fair coin is tossed 1 000 000 times. X is the number of heads which can result

and so X is Bin(1 000 000, 0:5).

a Check that np > 10 and n(1 ¡ p) > 10.

b Find the mean and standard deviation of X.

c Find the probability that exactly 500 000 heads result.

d Find the probability that between 499 800 and 500 200 (inclusive) heads result.

1 A recent fitness report claimed that only 56% of young Australians participate regularly

in vigorous sports. If a random sample of 40 young Australians is taken, find the

probability that 20 of them participate regularly in vigorous sports by:

a using the binomial distribution b using the normal approximation.

Example 5

EXERCISE 8C

��

A B

CD

12.5 13.5

X¤

+

h

a n = 1000 000 and p = 0:5 and so both np and n(1 ¡ p) are 500 000) np > 10 and n(1 ¡ p) > 10.

b ¹ = np = 500000 and ¾ =pnp(1 ¡ p) =

p1000 000 £ 0:5 £ 0:5 = 500

c Pr(X = 500000) + normalpdf(500 000, 500 000, 500)

+ 0:000 798

d If X¤ is the variable for the normal approximation then

Pr(499800 6 X 6 500 200) + Pr(499800 6 X¤ 6 500 200) + 0:3108



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In each of the following the normal approximation must be used.

Statisticians are often called upon to make decisions about proportions.

² Will the proportion of people who voted Liberal in the last federal election change

for the next one?

² Is the proportion of people smoking in 2006 different from that in 1990?

² Has a new medical drug increased the proportion of people cured?

Suppose that 247 students in a school of 839 support Port Power.

The proportion of students who support Port Power is the relative frequency 247839 .

The probability that a randomly selected student supports Port Power is 247839 + 0:294 .

It is feasible to ask every student in the school what team they support, but this is not possible

for every person in South Australia. A practical way of estimating such a proportion is to

select a random sample of say n persons and ask each of them which team they support.

Let x be the number of people from the sample who say they support Port Power.

The sample proportion p̂ =x

ncan be used to draw conclusions about the population pro-

portion p of all South Australians who support Port Power. The population proportion p can

be thought of as the probability of success in a binomial trial.

Before we can estimate how close the sample proportion is to the population proportion, we

must first understand how it is distributed.

Suppose we have a coin that has probability p of turning

up heads. Consider the population of all tosses of the coin.

Let the random variable X be the number of heads of the

coin in one toss. The probability distribution of X has

x 0 1

px 1 ¡ p p

mean ¹ =P

xpx

= 0(1 ¡ p) + 1(p)

= p

and standard deviation ¾ =pP

x2px ¡ ¹2

=p

02(1 ¡ p) + 12(p) ¡ p2

=pp ¡ p2 or

pp(1 ¡ p)

Taking a sample of size n from the population means we toss the coin n times.

HYPOTHESIS TESTING FOR PROPORTIONSD


2 If a fair coin is tossed 1 500 000 times, what is the probability it falls heads:

a on 750 000 occasions b at least 750500 times

c between 749 800 and 750 200 (inclusive) times?

3 If a fair die is rolled 1 200 000 times, what is the probability of throwing a ‘six’:

a on 200 000 occasions b at least 200 300 times?

4 A restaurant chain knows from past experience that 27% of its customers will order

a dessert after the main course. If during a year the restaurant chain has 1 175 072customers, determine the probability that at most 317 000 customers will order dessert.

5 A manufacturing company mass produces pens, 3% of which are faulty. If over the

course of a year, the company produces 1 400 000 pens, find the probability that the

number of faulty pens produced is

a at most 42 200 b between 41 800 and 42 100 (inclusive).


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Suppose a sample is fx1, x2, x3, ...... xng where xi =

½1 if the coin turns up heads

0 if it turns up tails.

Let the mean of this sample be bp =x1 + x2 + x3 + :::::: + xn

n=

x

n

where x is the number of heads in n tosses of the coin.

Note that p̂ is a mean of a sample of size n, and by the Central Limit Theorem:

for sufficiently large n, the distribution of the sample means is

approximately normal with

mean ¹bp = p and standard deviation ¾bp =¾pn

=

rp(1¡ p)

n

As in Section C, as a rule of thumb:

if both np > 10 and n(1 ¡ p) > 10, the distribution of p̂ is approximately normal.

The standard deviation ¾bp is called the standard error.

The standard error is a measure of the spread of the sample proportions p̂. The smaller the

standard error, the closer the sample proportions p̂ are crowded around the population propor-

tion p, and the more likely it is that an individual sample proportion is close to the population

proportion.

Using the fact that sample proportions are approximately normal we can now use the machin-

ery developed for normal distributions to test claims about proportions.

We follow these steps:

Step 1: State the null hypothesis and the alternative hypothesis.

Step 2: Select a significance level. The usual significance level is 5%.

Step 3: Calculate the sample mean.

Step 4: Calculate the test statistic z =p̂ ¡ p0

¾pn

.

In this case, from the null hypothesis it is assumed that p is known to be p0:In the case of proportions, we can use this to calculate the population

standard deviation as ¾ =pp0(1¡ p0) .

Step 5: Calculate the P-value.

Step 6: Draw a conclusion.

In the South Australian Election, of the voters voted for the Labor Party.

To see if this proportion has changed for the election, a pollster selects a

random sample of eligible voters. Of this sample, say they will vote for the

Labor Party in the next election. Does this support the claim at the level that the

proportion of voters voting for the Labor Party has changed since the election?

1997 35 2%2006

500 1985%1997

:

Example 6



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The calculations are identical to the calculations for testing a mean. The only difference is

that we do not have to estimate ¾ from a sample.

From the null hypothesis we know p0, and we can use this to find the popu-

lation standard deviation ¾ for a binomial distribution frompnp0(1 ¡ p0) .

Step 1: H0: p = 0:352 and Ha: p 6= 0:352

Step 2: The significance level is 5%:

Step 3: The sample mean is the proportion p̂ = 198500 = 0:396

Step 4: We use p0 = 0:352 and ¾ =pp0(1 ¡ p0) =

p0:352(1 ¡ 0:352):

The test statistic is z =p̂ ¡ p0

¾pn

=0:396 ¡ 0:352p0:352(1 ¡ 0:352)p

500

+ 2:06

Step 5: The P-value is P = Pr(Z 6 ¡2:06 or Z > 2:06)

= 2 £ Pr(Z 6 ¡2:06) = 0:0394

Step 6: Since P < 0:05 we reject the null hypothesis at the 5% level and accept

that there has been a change in the proportion of people who will vote for

the Labor Party in 2006 compared with those who voted for them in 1997.

These calculations can also be done on a calculator as a one variable -test for proportions.Z

TI

C

Use the significance level of 5% for each of the following questions:

1 To test if a coin was biased it was tossed 200 times and 91 heads appeared.

a State the null and alternative hypotheses that are to be tested.

b Calculate the test statistic z.

c Calculate the P-value.

d On this evidence, would you say the coin was biased?

2 To test if a six sided die was fair, it was rolled 150 times and 32 fours turned up.

a State the null and alternative hypothesis that are to be tested.


c Calculate the P-value.

d Does this support the claim that the die is biased?

3 A supplier of nuts claims that 25% of its nut mixes are pecans. A consumer did not

believe the claim and in a sample of 3187 nuts found that 746 were pecans. Does the

consumer’s evidence support the supplier’s claim?

4

EXERCISE 8D


Sixty migraine sufferers were asked to change from their old medication to a new one.

Thirty eight said the medication improved their condition; the others said it made their

condition worse. Test the null hypothesis at the 5% level that the new medication is no

better than the old.

(Hint: If the medication made no difference, the probability a person reports an im-

provement is the same as the probability a person becomes worse, i.e., H0: p = 12 .)


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5 A producer of instant scratch lottery tickets claims that 5% of its tickets win a prize. A

consumer group wants to test this claim.

a Explain why testing a random selection of 100 tickets may not be enough.

b In a randomly selected sample of 500 tickets, 19 were found to be winning tickets.

Does this support the manufacturer’s claim?

6 In 1996, 20:6% of people over the age of 16 said they expected to be better off in 5years’ time. To see if this proportion had changed in 2006, a pollster found that in a

random selection of 300 people, 66 said they expected to better off in 5 years’ time.

Does this support the hypothesis that people’s feelings about their future well-being have

changed?

7 A coin is tossed n times and the number of heads that appear is recorded. For each of

the following outcomes:

i calculate the proportion of heads that appear ii decide if the coin is biased.

a n = 100 tosses and 51 heads appear

b n = 1000 tosses and 510 heads appear

c n = 10000 tosses and 5100 heads appear.

For a normal distribution,

we reject the null

hypothesis at the level

if the test statistic

or if .

5%

1 96 1 96z< : z> :� � � � �¡

H0: p = 0:5 and Ha: p 6= 0:5

Let the number of heads that appear in the 100 tosses be x, so p̂ =x

100:

The test statistic is z =p̂ ¡ p0

¾pn

=p̂ ¡ p0pp0(1 ¡ p0)p

n

=

x

100¡ 0:5

0:05

and we reject the null hypothesis at the 5% level if P < 0:05

and thus z < ¡1:96 or z > 1:96 .

So,

x

100¡ 0:5

0:05< ¡1:96 or

x

100¡ 0:5

0:05> 1:96

)x

100< 0:402 or

x

100> 0:598

) x < 40:2 or x > 59:8

An experimenter wishes to test the claim that a coin is biased. The experimenter

decides to toss the coin times. What number of heads must appear for the

experimenter to accept that the coin is biased at the level?

1005%

Example 7

The experimenter will reject the null hypothesis and accept that the coin is biased

if less than 41 or more than 59 heads appear.

8 A television manufacturer wishes to test the claim that 20% of people living in a large

country town have digital television. To test this claim a random sample of 382 house-

holds was selected. How many householders will have to say they own a digital television

for the claim to be rejected?



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9 The theoretical chance of rolling a sum of seven with a pair of unbiased dice is 16 . In a

casino, dice are regularly tested. If the dice are to be tested by rolling 250 times, how

many times must a sum of seven appear before they are to be rejected at the 5% level

on the grounds that they are biased?

10

11 a Consider the hypotheses H0: p = p0 and Ha: p 6= p0:

Show that the null hypothesis will be rejected at the 5% level if the sample proportion

p̂ calculated from a sample of size n satisfies p̂ > p0 + 1:96

rp0(1 ¡ p0)

n.

b Suppose that p0 = 0:3 and the null hypothesis is rejected because the sample

proportion is found to be p̂ = 0:301 . What is the smallest sample size n?

In this section we show how an unknown population proportion can be estimated from a

sample proportion. As a sample proportion depends on the sample, we can never be sure

how close it is to the population proportion. However, we can obtain confidence intervals for

the proportions in the same way we did for means in Chapter 7. Our main tool is again the

Central Limit Theorem.

If p̂ is a proportion, then for sufficiently large n the distribution of proportions is

approximately normal with

mean ¹bp = p and standard deviation ¾bp =¾pn

=

rp(1¡ p)

n

We use this result to calculate 95% confidence intervals for proportions.

Since the actual population proportion p is unknown, we use bp to replace p.

The large sample 95% confidence interval for p is

bp¡ 1:96

rbp (1¡ bp)n

< p < bp+ 1:96

rbp (1¡ bp)n

A motor boat dealer claims that of his customers would recommend his boats to

their friends. A student wanted to test this claim and decided to ask the dealer’s

customers who were easily identified by the stickers on their boats.

If the student found that out of

questioned did recommend the dealer to

a friend, does this support the dealer’s

claim at the level?

Out of , how many customers would

have to deny they recommended the

dealer to a friend before the claim could

be rejected at the level?

85%

45 57

5%

57

5%

a

b

CONFIDENCE INTERVALSFOR PROPORTIONS

E



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A random sample of 200 South Australians showed that 76 supported the Adelaide

Crows AFL team.

a Find the sample proportion of Crows supporters.

b Find a 95% confidence interval for the proportion who support the Crows.

c Interpret your answer to b.

a The sample proportion of Crows supporters in SA is bp =x

n=

76

200= 0:38

We estimate that 38% of South Australians are Crows supporters.

Note: This estimate is called a point estimate.

b If p is the proportion of all South Australians who support the Crows,

then the 95% confidence interval for p is

bp¡ 1:96

rbp(1 ¡ bp)n

< p < bp + 1:96

rbp(1 ¡ bp)n

i.e., 0:38 ¡ 1:96

r0:38 £ 0:62

200< p < 0:38 + 1:96

r0:38 £ 0:62

200

) 0:313 < p < 0:447

c

Example 8

a Find a 95% confidence interval for each sample.

b Illustrate the limits. c Comment on the limits.

a Jason’s sampling: bp = 123300 = 0:41

and so his 95% confidence interval for the population proportion p is

bp¡ 1:96

rbp(1 ¡ bp)n

< p < bp + 1:96

rbp(1 ¡ bp)n

i.e., 0:41 ¡ 1:96

r0:41 £ 0:59

300< p < 0:41 + 1:96

r0:41 £ 0:59

300

) 0:354 < p < 0:466

Jason and Kelly wish to estimate the proportion of households that own at least one

dog. Jason sampled households and found that had at least one dog. Kelly

sampled households and found that had at least one dog.

300 123600 252

Example 9

We expect the population proportion to lie between and with

confidence. In other words, we are confident that the actual proportion of

Crows supporters in the whole SA population lies between and

p : :

: : :

0 313 0 447 95%95%

31 3% 44 7%

Kelly’s sampling: bp = 252600 = 0:42

and so her 95% confidence interval for the population proportion p is



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TI

C

0.3 0.4 0.5

Jason’s interval

Kelly’s interval

bp ¡ 1:96

rbp(1 ¡ bp)n

< p < bp + 1:96

rbp(1 ¡ bp)n

i.e., 0:42 ¡ 1:96

r0:42 £ 0:58

600< p < 0:42 + 1:96

r0:42 £ 0:58

600

) 0:381 < p < 0:460

b

c Jason estimates the actual population proportion to lie between 35:4% and 46:6%with 95% confidence whereas Kelly estimates the actual population proportion

to lie between 38:1% and 46:0%. Kelly’s larger sample has produced a more

precise proportion with a narrow interval.

We can use a graphics calculator to find these confidence intervals.

1 When 2839 Australians were randomly sampled,

1051 said they feared living close to overhead elec-

tricity power lines because of possible ‘increased

cancer risk’. Use the results of this survey to esti-

mate with a 95% confidence interval the proportion

of all Australians with this fear.

2 225 randomly selected elite sports people were asked the question “Should all team

players be tested for the HIV virus?”.

a Estimate with a 95% confidence interval the percentage of all team players in the

population who would say yes.

b Interpret your answer to a.

3 A die was rolled 420 times. A ‘six’ resulted on 86 occasions.

a Determine a 95% confidence interval to estimate the probability of rolling a ‘six’

with this die.

b Comment on the probable fairness of the die.

4 Jason expected that a coin he had was unfair. He tossed it 500 times and observed 267heads.

a Estimate with a 95% confidence interval the probability of getting a head when

tossing this coin.

b Comment on the probable fairness of the coin.

5 When a coin was tossed 100 times, 52 heads appeared. When the same coin was tossed

400 times, 213 heads appeared. In 1600 tosses of the same coin, 783 heads appeared.

a Construct a 95% confidence interval for the true population proportion of heads for

each of the three experiments.

EXERCISE 8E.1


209 responded “Yes”.


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b Draw a sketch of the three confidence intervals.

6 A bag contains some red marbles and some blue marbles. Amy drew out 15 marbles

with replacement and observed that 4 of these were red.

John drew out 60 marbles with replacement and observed that 16 were red.

a Find the two 95% confidence intervals for the true population proportion of red

marbles in the bag, and make a sketch of these intervals.

b Comment on the intervals from a.

7 The Transport Authority of Australia conducted a survey on motor vehicle accident

deaths. They found that 56 out of 173 drivers tested positive to having high levels of

drug or alcohol in their blood. Estimate with a 95% confidence level the true proportion

of drivers’ deaths in Australia where drivers had high levels of alcohol or drugs in their

blood.

8 The political party DNT asked 1000 persons in an

electorate how they would vote in an upcoming

election. 39% said they would vote for DNT. An

independent pollster found that 1360 persons out of

4000 surveyed said they would vote for DNT.

a Construct 95% confidence intervals for the

population proportion of people who will vote

for DNT using the two samples.

b Make a sketch of the two confidence intervals.

c Comment on the result of b.

9 A random sample of 2587 Australian adults was asked if they are better off now than

they were ten years ago. 1822 said that they were not.

a What proportion of the sample said that they were not better off now?

b Estimate with a 95% confidence interval the proportion of all Australian adults who

claim not to be better off now.

c In a town of 5629 adults, how many would you expect to be better off now?

The 95% confidence interval for a proportion p is given by 0:558 < p < 0:842. Find:

a the statistic bp on which this estimate is based b the number n of samples.

a The 95% confidence interval is bp¡ 1:96

rbp(1 ¡ bp)n

< p < bp + 1:96

rbp(1 ¡ bp)n

So, bp¡ 1:96

rbp(1 ¡ bp)n

= 0:558 and bp + 1:96

rbp(1 ¡ bp)n

= 0:842

Adding these equations gives 2bp = 0:558 + 0:842 = 1:4 and so bp = 0:7

b In the first equation, 0:7 ¡ 1:96

r0:7 £ 0:3

n= 0:558

which has solution n = 40

Example 10



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w

n

pppp

n

ppp

)ˆ1(ˆ96.1ˆ

)ˆ1(ˆ96.1ˆ

��

��

10 The 95% confidence interval for a proportion p is 0:304 6 p 6 0:496 .

a Find the statistic bp on which this estimate is based.

b Find the number n of samples used.

11 The 95% confidence interval for failures in a medical procedure is claimed to be

0:0923 6 p 6 0:1077.

Find the number of cases n used to obtain this confidence interval.

For the large sample 95% confidence interval

bp¡ 1:96

rbp (1 ¡ bp)n

< p < bp + 1:96

rbp (1 ¡ bp)n

,

the width w = 2 £ 1:96

rbp (1¡ bp)n

When we estimate a proportion it is desirable to make the width of the confidence interval as

small as possible. As for the confidence intervals for means, we can make the width smaller

by selecting a larger sample size n.

Before we could find the sample size for the confidence interval of a mean, we needed to

have some idea about the standard deviation ¾. In the case of proportions we need to know

something about the possible value of bp.

Case 1: We are given a preliminary value bp = p¤.

The width of the interval is w = 2 £ 1:96

rp¤(1 ¡ p¤)

n

)

µw

2 £ 1:96

¶2

=p¤(1 ¡ p¤)

n

) n =

µ2 £ 1:96

w

¶2

p¤(1 ¡ p¤)

So, for a 95% confidence interval of width w for a proportion, the sample

size n =

µ2£ 1:96

w

¶2

p¤(1¡ p¤)

CHOOSING THE SAMPLE SIZE



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2

3 An approximate 95% confidence interval of width 0:01 is to be found for a proportion.

a Find the sample size needed for each of the following values of p¤:

i p¤ = 0:1 ii p¤ = 0:2 iii p¤ = 0:3 iv p¤ = 0:4 v p¤ = 0:5

b Write down the sample size needed for

i p¤ = 0:6 ii p¤ = 0:7 iii p¤ = 0:8 iv p¤ = 0:9

4 Suppose we want a 95% confidence interval of width 0:0392 .

a Find a formula for the sample size n in terms of the preliminary probability p¤.

b Sketch a graph of n against p¤.

c For what value of p¤ is n biggest?

Case 2: We have no preliminary value for the proportion bp.

f(x) = x(1 ¡ x) = ¡x2 + x where 0 6 x 6 1 has a vertex at x =¡b

2a= 1

2 :

) bp(1 ¡ bp) has a maximum value of 14 which occurs when bp = 1

2 :

So, even if we have no information about bp, we can find a minimum value

for the sample size n by using p¤ = 0:5 in the formula

n =

µ2 £ 1:96

w

¶2

p¤(1 ¡ p¤).

Using the formula n =

µ2 £ 1:96

w

¶2

p¤(1 ¡ p¤) with p¤ = 0:8, w = 0:04

we get n =

µ2 £ 1:96

0:04

¶2

£ 0:8 £ 0:2 + 1537

So, about 1540 patients are required in the study.

A medical procedure is successful in of the patients treated. A new procedure is

being tried. If a confidence interval of width is needed to estimate the

probability of success of the new procedure, how many patients must be tested?

80%95% 4%

Example 11

EXERCISE 8E.2

A coaching method to teach people to type within amonth has a success rate of .

A new method is proposed which is claimed to have ahigher success rate.

How many people are needed to construct aconfidence interval of width for this new method?

70%

95%2%

The probability a bus is behind schedule is . A new bus is to be tried on the route to

see if the the arrival time can be improved. Assuming the new bus is less likely to fall

behind schedule, find the number of trial runs needed to find a confidence interval

of width for the probability the bus is behind schedule.

0 1

95%0 04

:

:



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An experimenter wishes to estimate, with a probability of 0:95, the proportion to

within 3% of mosquitos which carry a virus. How large must the sample be?

As we want the proportion to be within 3% of the mosquitos carrying the virus, the

confidence interval width is 6% (3% on either side of the population proportion).

Using the formula n =

µ2 £ 1:96

w

¶2

p¤(1 ¡ p¤) with p¤ = 0:5, w = 0:06

we get n =

µ2 £ 1:96

0:06

¶2

£ 0:5 £ 0:5 + 1067

So, a sample size of about 1070 is needed.

The reason p¤ = 0:5 is not always used is that if p is actually a long way away from 0:5,

it may result in the sample size n being unnecessarily large.

1 Publishers Karras Pty Ltd decides to survey 1500 of their readers to ask their opinion on

the new format and layout of their weekly magazine.

a If p¤ = 12 what is the width of the 95% confidence interval?

b If p¤ = 14 what is the width of the 95% confidence interval?

c How many people should they survey if they want a confidence interval of width

found in b but using the value p¤ = 12?

2 When 2750 voters were asked whether they felt the income tax rates were too high, 2106said ‘yes’.

a What is the 95% confidence interval for the population proportion of voters who

think the income tax rates are too high?

b If a pollster wants to find a confidence interval of half the width in a, how many

people should he sample if he uses: i p¤ = 21062750 ii p¤ = 1

2?

3 After the last frost, 189 apples were picked randomly

and 43 were found to be not fit for sale.

a Construct a 95% confidence interval for the

population proportion of unsaleable apples.

b How large a sample should be taken to estimate

the proportion of unsaleable apples to within 2%of the population proportion, using:

i p¤ = 43189 ii p¤ = 1

2?

Example 12

EXERCISE 8E.3



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A nutrition expert found that 43% of Victorian children ate insufficient fruit each day.

To check whether this figure was the same for South Australia, a university research

group sampled 625 children and found that 308 ate insufficient fruit each day.

a Find the 95% confidence interval for the proportion of South Australian

children eating insufficient fruit.

b How does the proportion of children not eating enough fruit in South Australia

compare with that of children in Victoria?

a For the sample, bp =x

n=

308

625= 0:4928

The 95% confidence interval for the population proportion p is:

bp¡ 1:96

rbp(1 ¡ bp)n

< p < bp + 1:96

rbp(1 ¡ bp)n

) 0:4928 ¡ 1:96

r0:4928 £ 0:5072

625< p < 0:4928 + 1:96

r0:4928 £ 0:5072

625

) 0:454 < p < 0:532

b As the proportion p = 0:43 of Victorian children is not in this interval, there is

evidence that the proportion of South Australian children who eat insufficient

fruit is different from those of Victorian children.

1 The manufacturer of Chocfruits claims that 90% of the one kilogram boxes have apricot

centres in more than half of the Chocfruits. To check this claim a consumer purchased

80 boxes at random and found the percentage of each box with apricot centres. She

found that 70 of the boxes had apricot centres in more than half of the Chocfruits.

a What proportion of the sample of boxes had more than half of the Chocfruits with

apricot centres?

b Estimate with a 95% confidence interval the proportion of all boxes produced by

the manufacturer which have more than half of the Chocfruits with apricot centres.

c Does the consumer’s data support the manufacturer’s claim?

2

a

b Does the Consumer Affairs data support the company’s claim?

ASSESSING CLAIMS WITH CONFIDENCE INTERVALS

Example 13

EXERCISE 8E.4

Growhair is the latest product of a pharmaceutical

company. The company claims that of users of the

product show significant hair gain after a period of four

months. To test the claim, Consumer Affairs randomly

sampled users and found that of them did show

significant hair gain.

43%

187 66

Construct a confidence interval for the

proportion of those who showed a significant gain.

95%



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3 An advertisement claimed that more than 20% of all households used Ongodo washing

powder. To test this claim a consumer group asked 100 households which washing

powder they used. Twelve households said they used Ongodo.

a Find the 95% confidence interval for the true population proportion of all households

that use Ongodo washing powder.

b Does the consumer group data support the manufacturer’s claim?’s

INVESTIGATION 5 CONFIDENCE INTERVALS OTHER THAN 95%

Confidence intervals of any level of confidence can easily be constructed

for proportions by adjusting the confidence level on your calculator. Other

common levels are and .90% 99%

An important question arises:

“Is it better to assess a claim by examining the confidence interval, or rather to use a single

proportion z-test?”

The reason for this question is that sometimes the two approaches give contradictory results.

This occurs because the z-test uses p0 in the hypothesis whereas the confidence interval uses p̂:

So, two different standard errors are used:

Example:

Answer using the z-test for a proportion

H0: p = 0:12 and Ha: p 6= 0:12

As P = 0:0312 < 0:05 there is

sufficient evidence at the 5% level

to reject the null hypothesis which

claims p = 0:12

Answer using a confidence interval

p̂ =x

n=

19

100= 0:19

So the 95% CI is 0:113 < p < 0:267and as 12% lies within the CI,

H0: p = 0:12 cannot be rejected.

It seems that a single proportion z-test for assessing a claim about a proportion should have

preference as it does not use an estimated standard deviation. It may be inappropriate to just

use a confidence interval to assess such a claim.

IMPORTANT NOTE ON CONFIDENCE INTERVALS

¾(p̂) =pp0(1 ¡ p0) for the z-test and ¾(p̂) =

pp̂(1 ¡ p̂) for the CI.

A report claims that the percentage of car crashes due to drivers being over the legal blood

alcohol level is 12%. From the police files, 19 out of 100 crashes were due to drink driving.

Comment on the report’s claim at a 95% level of confidence.



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REVIEWF

REVIEW SET 8A

1 A discrete random variable X has probability distribution function P where

P (x) = k¡34

¢x ¡14

¢3¡xwhere x = 0, 1, 2, 3 and k is a constant.

a Find k. b Find Pr(X > 1).

2 A manufacturer finds that 18% of the items produced from one of its assembly lines are

defective. During a floor inspection the manufacturer randomly selects ten items. Find

the probability that the manufacturer finds:

a one defective b two defective c at least two defective.

1 The manufacturer of Perfect Strike matches claim that 80% of their match boxes

contained 50 or more matches. To check this claim a consumer randomly chose 250boxes and counted the contents. The consumer found that 183 boxes contained 50 or

more matches.

a Find the 95% confidence interval for the proportion of match boxes in the pop-

ulation which contain 50 or more matches.

b Does the consumer’s data support the manufacturer’s claim?

c Repeat a and b with a 99% confidence interval.

2 A coin is tossed 100 times and 43 heads appear.

a Construct 90%, 95%, and 99% confidence intervals for the population proportion

of heads that appear for this coin.

b Sketch the confidence intervals.

c Is there evidence at any level that the coin is biased?

3 A quality controller tested a sample of 120 chocolates in a factory and found 18 were

underweight.

a Construct 90%, 95%, and 99% confidence intervals for the proportion of under-

weight chocolates produced.

b What sample size should be taken to get a confidence interval of width 0:05 for

each level of confidence? Use both p¤ = 18120 and p¤ = 1

2 .

4 The weight W kg of sugar in bags filled by a machine is normally distributed with

mean ¹ = 1:012 and standard deviation ¾ = 0:0055 .

A bag is considered underweight if it weighs less than 1 kg.

a What proportions of the bags are underweight?

b After an engineer adjusts the machine, a sample of 500 bags is weighed and 4are found to be underweight. On this basis the engineer claims the machine is

now working better.

i Construct a 90% confidence interval for the proportion of underweight bags

produced by the machine.

Does the 90% confidence interval support the engineer’s claim?

ii What would your answer be for a 99% confidence interval?

What to do:



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REVIEW SET 8B

d Is the phone company’s claim justified?

1 a List all possible outcomes for 4 tosses of a coin.

b Expand (H + T )4 and compare your answer with part a.

2 An x-ray has probability of 0:96 of showing a fracture in an arm. If four different x-rays

are taken of a particular fracture, find the probability that:

a all four show the fracture b the fracture does not show up

c at least three x-rays show the fracture d only one x-ray shows the fracture.

3 Let X be the weight in grams of bags of sugar filled by a machine.

Suppose that X » N(503, 22). Bags less than 500 grams are underweight.

a What proportion of bags are underweight?

b If a quality inspector randomly selects 20 bags, what is the probability that at most

2 bags are underweight?

4 Suppose X » Bin(20, 0:3).

a Draw the column graph for the distribution.

b Calculate the mean and standard deviation of X.

5 The probability that a 1 or a 6 turns up if a fair die is rolled is 13 . To test if a die was

fair, Joan rolled the die 100 times. In 60 rolls neither a 1 nor a 6 turned up.

a What hypotheses should Joan be considering?

b What is the test statistic Joan should calculate?

c What is the null distribution Joan should use?

d Does Joan have enough evidence at the 5% level to claim the die is biased?

6

7 a How large should a sample size be to find a 95% confidence interval of width 0:1for a population proportion p?

b It is known that the population proportion p is at least 0:8 . How large a sample

size is now required to find a 95% confidence interval for p?

c What is the sample size if it is known that p is at most 0:1?

8 A phone company claims that 90% of its customers are happy with the service they

provide. To test this claim, 345 of the company’s customers were surveyed, and 297were happy with the service.

a State the null and alternative hypotheses.


c Test the null hypothesis, using a 5% level of significance.

1758 105895%

Australians were randomly sampled and said they were in favour of

Australia becoming a republic. Use the results of this survey to estimate, using a

confidence interval, the proportion of all Australians who are in favour of Australia

becoming a republic.



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3 Suppose that the weight of apples is normally

distributed with mean 300 grams and standard

deviation 50 grams. If only apples with weights

between 250 and 350 grams are fit for sale:

a find the proportion of apples fit for sale.

b In a sample of 100 apples, what is the prob-

ability that at least 75 are fit for sale?

4 Suppose the probability of successfully treating a certain type of disease is 0:8 .

a How large a sample should be taken before using a normal distribution to estimate

probabilities?

b Let the random variable X be the number of successful treatments of 200 patients.

i Find Pr(X > 175), assuming the continuity correction.

ii What is the answer if you do not use the continuity correction?

5 In “two up”, two pennies are tossed. If the pennies are fair there is a probability of14 that two heads appear. To test if the pennies were fair, John tossed them together

150 times and observed that 2 heads appeared 41 times.

a What hypotheses should John consider?

b What is the test statistic John should calculate?

c What is the null distribution John should use?

d Does John have enough evidence at the 5% level to claim the pennies are not

fair?

6 The National Literacy Council gave 3500 adults a test consisting of 20 basic literacy

questions. The minimum pass mark was twelve correct responses and only 1348passed.

a Construct a 95% confidence interval for the proportion of adults able to pass the

basic literacy test.

b Does this support the claim that only 40% of adults are able to pass the basic

literacy test?

7 After a storm 317 flowers were picked and 87 were found to be not fit for sale.

a Construct a 95% confidence interval for the proportion of unsaleable flowers.

b How large a sample would need to be taken to estimate the proportion of un-

saleable flowers to within 2:5% with 95% confidence?

8 A manufacturer produces batteries. The batteries are deemed to be faulty if they last

less than 100 hours.

Let X denote the length of the battery’s life. It is found that X is normally distributed,

with a mean of 104:8 hours and a standard deviation of 2:6 hours.

a What proportion of the batteries will be faulty?

b A random sample of 20 batteries is selected from a huge batch. Let Y denote

the number of faulty batteries in the sample.

i What type of variable is Y ?

ii What is the probability that:

( )1 exactly one battery in the sample is faulty

( )2 at least three of the batteries are faulty?


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REVIEW SET 8C

6 A random sample of 213 SA high school students were asked if they were worried about

their future job prospects. 74 said that they were worried.

a What proportion of the sample were worried?

b Estimate with a 95% confidence interval the proportion of all SA high school stu-

dents who would be worried about their future job prospects.

7 It was found that 62% of households in a town participated in recycling. In an attempt

to increase this number, a publicity campaign was launched, informing residents of the

benefits of recycling. Six months later, to find out if the campaign was effective, 174households were surveyed, and 118 of those surveyed participated in recycling.

a State the null and alternative hypotheses.

b

c Use the confidence interval to test the null hypothesis.

d Is there sufficient evidence to conclude that the campaign was effective?

1 a Draw Pascal’s triangle down to row 5.

b Use Pascal’s triangle to find the number of ways of getting 3 successes in 5 trials.

c Calculate C50 + C5

1 + C52 + C5

3 + C54 + C5

5 .

2 Only 40% of young trees planted will survive the

first year. Adelaide Botanical Gardens buys five

young trees. Assuming independence, calculate

the probability that during the first year:

a exactly one tree will survive

b at most one tree will survive

c at least one tree will survive.

3 A random variable X has probability function P (x) = C4x (12 )x (12)4¡x

for x = 0, 1, 2, 3, 4.

a Find P (x) for x = 0, 1, 2, 3, 4. b Find ¹ and ¾ for this distribution.

4 In a poll conducted in 1997, 31% of the population of Australia said they were satisfied

with the world situation. To see if this had changed, another poll conducted in 2007found that out of 300 people interviewed 87 said they were satisfied. Does this provide

enough evidence at a 5% level that the proportion of people who were satisfied had

changed since 1997?

5 A football team has its players on a weights program in the off-season. Management

believes it will change the strength of its players. It was found that compared with the

last season’s performance, 20 out of 35 players improved their strength while 15 became

weaker. Test the null hypothesis at the 5% level that the program made no difference to

the strength of players.

(Hint: If the program made no difference, the probability a player became stronger is

the same as the probability a player became weaker, i.e., H0: p = 12 :)


Construct a confidence interval for the proportion of households who partici-

pate in recycling.

95%


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8 A community group wants to reduce the speed limits in residential areas to 40 km/hour.

They contact 70 households in the area, and 37 say they want the speed limit reduced.

a Construct a 95% confidence interval for the true population proportion of people

who want the speed limit reduced.

b Explain why a council might not be satisfied by the evidence from the 95% confi-

dence interval.

c The council insists that they need a survey which results in a 95% confidence

interval of at most 0:05 .

i Why do you think p¤ = 12 should be used in estimating the number required

to obtain a confidence interval of at most 0:05?

ii How many people should be surveyed to ensure a confidence interval of at

most 0:05?

d If a survey to satisfy the council’s requirement was carried out and the same pro-

portion of about 52:9% said they wanted the speed limit reduced, how should the

council react?

e Assuming it takes at least a quarter of an hour to interview a person in a household,

why do you think the community group would not carry out the size of the survey

required to satisfy the council?



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9

Contents:

Solving systems oflinear equations

Solving systems oflinear equations

A

B

C

D

E

F

G

Solutions ‘satisfy’ equations

Solving 2 × 2 systems of equations

3 × 3 systems with unique solutions

Other 3 × 3 systems

Further applications

4 × 4 and 5 × 5 systems

Review

� �

� �

� �

� � � �


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For example: A farmer has hens and sheep. In total there are 23 heads and 64 legs.

How many hens and how many sheep does the farmer have?

Recall a typical solution:

Suppose there are x hens and y sheep,

so x + y = 23 fas there are 23 headsgand 2x + 4y = 64 fas there are 64 legsg

We now need to find the solution which satisfies both of these equations simultaneously.

¡2x ¡ 2y = ¡462x + 4y = 64

) 2y = 18

and so y = 9

Consequently, x = 23 ¡ y = 14

Therefore, the farmer has 14 hens and 9 sheep.

For example:

½x + y = 23

2x + 4y = 64

is a 2 £ 2 system of linear equations

with unknowns x and y.

These are called linear equations because

their graphs are straight lines.

Suppose we plot the straight line graphs

for the hens and sheep example above.

The point of intersection (14, 9) satisfies

both equations simultaneously.

For example:

8<: 3x + 2y + 7z = 154x ¡ y + 2z = 102x + 3y ¡ z = ¡4

is a 3 £ 3 system of linear equations in

variables x, y and z.

However, these equations have no meaning in 2-dimensional coordinate geometry. They are

called linear equations because the powers of the variables are all 1. In other words there are

no squared terms, etc.

INTRODUCTION

Two-by-two 2 2or systems of linear equations are systems with two equations in two

unknowns.

� �£

30202323

2323

3232

10 14

25

20

15

10

5

y

x

(14' 9)6442 ��

23�� yx

yx

Three-by-three 3 3or systems of linear equations are systems with three equations in

three unknowns.

� �£

We could have dividedthe second equation by

2 and added the result

to the first equation.

Multiplying the first equation by and adding the second

equation to it will eliminate . We can then solve for .

¡2x y

Since Year we have been solving problems in which two unknowns need to be found from

two pieces of information given.

9

316 SOLVING SYSTEMS OF LINEAR EQUATIONS (Chapter 9)


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Likewise,

8>><>>:a + b + c + d = 22

a + b ¡ c + 2d = 29

a ¡ b + c ¡ 3d = 11

is a 3 £ 4 system of linear equations

in variables a, b, c and d.

In the hens and sheep problem x = 14 and y = 9 is the solution to½x + y = 23

2x + 4y = 64as

x + y = 14 + 9 = 23 X

and 2x + 4y = 2(14) + 4(9) = 28 + 36 = 64 X

We say that x = 14 and y = 9 satisfy the equations simultaneously.

Which of the following are solutions of:

8<:2x + 3y ¡ z = 5

x + 2y + 4z = 17

3x + 5y + 3z = 22

?

a x = 1, y = 4, z = 9 b x = 1, y = 2, z = 3 c x = ¡41, y = 29, z = 0

a If x = 1, y = 4, z = 9, then 2(1) + 3(4) ¡ (9) = 2 + 12 ¡ 9 = 5 X

(1) + 2(4) + 4(9) = 1 + 8 + 36 = 45 £As the values fail to satisfy the second equation, x = 1, y = 4, z = 9cannot be a solution.

b If x = 1, y = 2, z = 3, then 2(1) + 3(2) ¡ (3) = 2 + 6 ¡ 3 = 5 X

(1) + 2(2) + 4(3) = 1 + 4 + 12 = 17 X

3(1) + 5(2) + 3(3) = 3 + 10 + 9 = 22 X

As the values satisfy all three equations, x = 1, y = 2, z = 3 is a solution.

c If x = ¡41, y = 29, z = 0, then 2(¡41) + 3(29) ¡ (0) = 5 X

(¡41) + 2(29) + 4(0) = 17 X

3(¡41) + 5(29) + 3(0) = 22 X

) x = ¡41, y = 29, z = 0 is also a solution.

1 a Check that x = 3, y = 5 is a solution of

½3x + 2y = 197x ¡ 4y = 1

:

b x = 3, y = ¡2 is a solution of

½3x + 5y = a4x + 3y = b

: Find a and b.

c Show that x = t, y = 3 ¡ 2t is a solution of 2x + y = 3for all possible real values of t.

This means that 2x + y = 3 has infinitely many solutions.

Explain what this means and why it is so.

SOLUTIONS ‘SATISFY’ EQUATIONSA

Example 1

EXERCISE 9A.1

This system doesnot have a

unique solution.

SOLVING SYSTEMS OF LINEAR EQUATIONS (Chapter 9) 317


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2 Which of the following is a solution of

8<: x + 2y ¡ 3z = ¡9x¡ 2y + 4z = 162x + y ¡ 5z = ¡12

?

a x = ¡4, y = 2, z = 3 b x = 2, y = ¡1, z = 3

3 a Show that x = 2¡ t, y = 3 +2t, z = t is a solution of

½x + y ¡ z = 5

3x + 2y ¡ z = 12for all values of t.

b What is the solution when i t = 2 ii t = ¡1?

4 a Show that x =2 ¡ t

2, y =

5t¡ 16

2, z = t is a solution of8<: x + y ¡ 2z = ¡7

x¡ y + 3z = 93x¡ y + 4z = 11

for all possible values of t, i.e., that there are infinitely

many solutions.

b Show that x = s, y = ¡3 ¡ 5s, z = 2 ¡ 2s is a solution to the same set of

equations in a .

c Explain algebraically why the answers of the form x = s, y = ¡3¡5s, z = 2¡2s

are exactly the same as those generated by x =2 ¡ t

2, y =

5t¡ 16

2, z = t.

Consider the following problem:

The Rich River Invitational Golf Classic is held each year. Eight of the best professional

golfers are invited to play for prize money which this year totals 2:2 million dollars.

The rules for the distribution of the prize money are:

² all of the prize money must be distributed to the best five

players

² second receives two thirds of the winner’s amount

² third gets twice as much as fourth

² second gets the same amount as third and fourth combined

² the sum of first and fifth winnings equals the same amount

as the other three prizes combined.

How do we solve such a problem?

How would we solve similar problems where there are far more golfers

and lots more conditions (constraints) for the allocation of the prize money?

As mathematicians, we convert the given conditions into algebraic equations.

There are five players to distribute money to so we let their amounts be $a for first, $b for

second, $c for third, $d for fourth, and $e for fifth.

From the total prize money we find: a + b + c + d + e = 2:2 million

The other information yields: b = 23a, c = 2d, b = c + d and a + e = b + c + d

We hence have a 5 £ 5 system of linear equations, i.e., 5 equations in 5 unknowns.

PROBLEM SOLVING INVOLVING LINEAR EQUATIONS



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ACTIVITY

Discuss strategies to solve the golfers’ prize distribution equations.

Apply your strategies to try to solve the equations.

Would your strategy work if the equations were linear but more complicated?

Typical problems where linear equations arise are in:

² allocating resources in manufacturing

² distribution of prizes in sporting events or lotteries

² designing balanced diets

² fitting a set of data points to a polynomial model in economics,

for example a quadratic profit model or a cubic cost model

² balancing chemical equations

Let us consider the last of these: balancing chemical equations.

Although there are sometimes quicker methods, we could set up and solve a system of linear

equations. This may be the best method if the equation is very complicated.

Cement is used to make concrete. In the cement manufacturing process, one of the chemical

reactions involves calcium aluminosilicate (CaAl2Si2O8) combining with calcium carbonate

(CaCO3) at a high temperature to produce dicalcium silicate (Ca2SiO4), tricalcium aluminate

(Ca3(AlO3)2) and carbon dioxide (CO2).

A chemical equation can be written to show this concisely. If the equation is in balance, the

number of atoms of each element on the left hand side must equal the number of atoms on

the right hand side.

So, we seek numbers a, b, c, d and e such that

aCaAl2Si2O8 + bCaCO3 ! cCa2SiO4 + dCa3(AlO3)2 + eCO2

Equating the number of Ca atoms: a + b = 2c + 3d ..... (1)

Al atoms: 2a = 2d ..... (2)

Si atoms: 2a = c ..... (3)

O atoms: 8a + 3b = 4c + 6d + 2e ..... (4)

C atoms: b = e ..... (5)

Solving these five equations to get a, b, c, d and e will enable us to balance the chemical

equation.

A solution: From (2) and (3) d = a and c = 2a

Now in (1), a + b = 2(2a) + 3(a)

) a + b = 7a

) b = 6a and from (5), e = b = 6a also.

In (4), 8a + 3(6a) = 4(2a) + 6(a) + 2(6a) i.e., 26a = 26a X

) a : b : c : d : e = a : 6a : 2a : a : 6a = 1 : 6 : 2 : 1 : 6

Consequently the balanced equation is

CaAl2Si2O8 + 6CaCO3 ! 2Ca2SiO4 + Ca3(AlO3)2 + 6CO2



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In the following exercise we will be concerned with setting up systems of linear equations,

but not in their solution.

1 Seven cups and five plates cost a total of $90 whereas nine cups and eight plates cost a

total of $133.

a Clearly state the variables necessary to set up equations connecting them.

b What is the system of linear equations?

2 Sam buys a pad, a biro and a ruler for a total of $5:30. Jan buys two pads, two biros

and a ruler for a total of $8:35. Wei buys three pads, three biros and a ruler for a total of

$11:40. Clearly state the variables required to set up a system of linear equations, then

write down this system.

3 Because of overhunting and genetic problems, the population of brown bears in Canada

decreased considerably during the period from 1986 to 2006: Biologists attempted to

model the decrease using a quadratic model. They chose t = 1 to represent 1986,

t = 2 to represent 1996 and t = 3 to represent 2006. They estimated the number of

brown bears in hundreds to be P (1) = 38, P (2) = 32 and P (3) = 25.

a State clearly the quadratic model.

b Obtain a set of linear equations which when solved can be used to state the approx-

imate quadratic model for population size.

4 Jason breeds high quality Siberian Huskies. He

mixes his own food from three foods obtainable

in bulk. Manufacturer’s specifications are:

Food Units of vitamin/kg Calories/kg

A 428 214

B 256 605

C 179 713

Jason makes up his mixture in 5 kg batches, and the batches must contain 1400 units of

vitamins and 2650 calories.

a What variables should Jason use to set up his equations?

b What is the system of constraints Jason needs to solve?

5 The general equation of a circle is x2 + y2 + ax + by + c = 0.

Three points (1, 3), (5, 4) and (4, ¡1) lie on a particular circle. Find a set of linear

equations which when solved will enable us to find the exact equation of that circle.

6 A cost function for a carpet manufacturer is modelled by a cubic polynomial from the

following information:

The cost of making 2000 metres is $120 000, 4000 metres is $150 000,

7000 metres is $170 000, 10 000 metres is $250 000:

a If x is the number of 1000’s of metres made, write down the form of the cost model.

b State the linear constraints which need to be solved to find the actual cost model.

EXERCISE 9A.2



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7 Find linear equations which need to be solved to ‘balance’ these chemical equations:

a a NH3 + b O2 ! c NO + d H2O

b a Cu + b HNO3 ! c Cu(NO3)2 + d NO2 + e H2O

c a Cu + b HNO3 ! c Cu(NO3)2 + d NO + e H2O

There is no need to solve these equations unless you want to do so.

In past years, to solve 2 £ 2 systems of linear equations we have used the following:

² a graphical solution

² solution by substitution

² solution by eliminating one of the variables.

In solution by substitution we need one variable in terms of the other. We then substitute it

into the second equation. The system½y = 2x + 3

3x + 2y = 20is suitable for this method since we have y given in terms of x.

In solution by elimination we multiply one or both equations by non-zero constants so that

when the new equations are added either x or y will be eliminated.

Consider a traditional method for solving the 3£3 system

8<:x + 2y + z = 6 ...... (1)

x¡ y + 4z = ¡6 ...... (2)

2x + 3y ¡ z = 12 ...... (3)

Let z = t ) x + 2y = 6 ¡ t ...... (4)

and x¡ y = ¡6 ¡ 4t ...... (5)

) x + 2y = 6 ¡ t

2x¡ 2y = ¡12 ¡ 8t fmultiplying (5) by 2g) 3x = ¡6 ¡ 9t

) x = ¡2 ¡ 3t

So, in (2), ¡2 ¡ 3t¡ y + 4t = ¡6

) ¡y = ¡4 ¡ t

) y = 4 + t

Substituting x = ¡2 ¡ 3t, y = 4 + t and z = t into (3) gives:

2(¡2 ¡ 3t) + 3(4 + t) ¡ t = 12

) ¡4 ¡ 6t + 12 + 3t¡ t = 12

) ¡4t + 8 = 12

) ¡4t = 4) t = ¡1

and when t = ¡1, x = ¡2 ¡ 3(¡1) = 1

y = 4 + (¡1) = 3

z = ¡1

) x = 1, y = 3, z = ¡1 is the solution.

TRADITIONAL SOLUTIONS



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INVESTIGATION 1 SOLVING LINEAR EQUATIONS WITH

TRADITIONAL METHODS

1 Use the same approach as in the example above to solve:

a b8<: x + 2y + z = 6

x + 3y + 2z = 82x¡ y + 4z = 11

8<: x + y + z = 2x + 2y + 3z = 6

2x + 5y + 8z = 16

2 Try to solve the following using the method above:8>><>>:

a + b + c + d = 1a + 2b + c + 3d = 22a¡ b + c + 2d = 33a + b + 2c¡ d = 4

3

4 A typical allocation of resources in a manufacturing problem may be modelled by a

polynomial of degree 5. Explain why this would require a 6 £ 6 system of linear

equations to be solved.

The system of equations2x + y = ¡1

x¡ 3y = 17

is called a 2 £ 2 system because there are

2 equations in 2 unknowns.

In the method of ‘elimination’ used to solve these equations, we observe that the following

operations produce equations with the same solutions as the original pair.

² The equations can be interchanged without affecting the solutions.

For example, 2x + y = ¡1

x¡ 3y = 17

has the same solutions as x¡ 3y = 17

2x + y = ¡1

:

² An equation can be replaced by a non-zero multiple of itself.

For example, 2x + y = ¡1 could be replaced by ¡6x¡ 3y = 3

(obtained by multiplying each term by ¡3).

² Any equation can be replaced by itself plus (or minus) a multiple of another equation.

For example, E1: x¡ 3y = 17

E2: 2x + y = ¡1

becomes x¡ 3y = 17

7y = ¡35 if E2 ! E2 ¡ 2E1.

(E2 ! E2 ¡ 2E1 reads: equation 2 is replaced by equation 2 ¡ twice equation 1.)

What to do:

SOLVING 2 × 2 SYSTEMS OF EQUATIONS� �B

We will use these three legitimate operations in the method of solution described in the next

section.

So, what is wrong with this method? Actually there is nothing wrong with it, but we do not

use it. To find out why, complete the following investigation.

Using and explain why it would be desirable to find alternative methods of

solving higher order systems of linear equations.

1 2,



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Instead of writing 2x + y = ¡1x ¡ 3y = 17

we detach the coefficients and write the system in the

augmented matrix form

·2 1 ¡11 ¡3 17

¸.

We can then use elementary row operations equivalent to the three legitimate operations

with equations,

i.e., ² interchange rows

² replace any row by a non-zero multiple of itself

² replace any row by itself plus (or minus) a multiple of another row.

Interchanging rows is equivalent to writing the equations in a different order. It is often

desirable to have 1 in the top left hand corner.

So,

·2 1 ¡11 ¡3 17

¸becomes

·1 ¡3 172 1 ¡1

¸.

We now attempt to eliminate one of the variables in the second equation, i.e., obtain a 0 in

its place. To do this we replace R2 by R2 ¡ 2R1

So,

·1 ¡3 172 1 ¡1

¸becomes

·1 ¡3 170 7 ¡35

¸2 1 ¡1 Ã R2

¡2 6 ¡34 Ã ¡2R1

0 7 ¡35 adding

The second row of the matrix is really 7y = ¡35 and so y = ¡5.

Substituting y = ¡5 into the first equation, x ¡ 3(¡5) = 17 and so x = 2.

So, the solution is x = 2, y = ¡5:

We may not see the benefit of this method right now but we will certainly appreciate it when

solving 3 £ 3 or higher order systems.

AUGMENTED MATRICES

Use elementary row operations to solve:

½2x + 3y = 45x + 4y = 17

In augmented matrix form the system is:·2 3 45 4 17

¸

»·

2 3 40 7 ¡14

¸We can eliminate xif we replace

R2 by 5R1 ¡ 2R2 10 15 20¡10 ¡8 ¡34

0 7 ¡14

5R1

¡2R2

Re-introducing the variables we

have 7y = ¡14) y = ¡2

» is read as

“which has the

same solution as”

Example 2

f ¡ £ g(Row ) (Row )2 2 1



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intersecting parallel coincident

Substituting back into the first

equation we have 2x + 3(¡2) = 4

) 2x ¡ 6 = 4

) 2x = 10

) x = 5

So the solution is x = 5, y = ¡2:

a unique solution no solution one equation is an exact

multiple of the other,

) infinitely many solutions

1 Solve using elementary row operations:

a x ¡ 2y = 84x + y = 5

b 4x + 5y = 215x ¡ 3y = ¡20

c 3x + y = ¡102x + 5y = ¡24

2 By inspection, classify the following pairs of equations as either intersecting, parallel, or

coincident lines:

a x ¡ 3y = 23x + y = 8

b x + y = 73x + 3y = 1

c 4x ¡ y = 8y = 2

d x ¡ 2y = 42x ¡ 4y = 8

e 5x ¡ 11y = 26x + y = 8

f 3x ¡ 4y = 5¡3x + 4y = 2

3 Consider the equation pair

½x + 2y = 3

2x + 4y = 6.

a Explain why there are infinitely many solutions, giving geometric evidence.

b Explain why the second equation can be ignored when finding all solutions.

c Give all solutions in the form:

i x = t, y = :::::: ii y = s, x = ::::::

Don’t forget to checkyour solution by

substituting into theoriginal equations.

l1

l2

l1

l2

l1

l2

EXERCISE 9B

Type

Sketch

2x + 3y = 1

x ¡ 2y = 8

2x + 3y = 1

2x + 3y = 7

2x + 3y = 1

4x + 6y = 2

Example

Type of

solutions

Number of

solutionsone point of

intersection

no points of

intersection

infinitely many

points of intersection

In two dimensional geometry, where , and are constants, is the equation

of a straight line. If we graph two straight lines, then there are three different cases which

could occur. The lines could be:

ax by c a b c� � � �+ =



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4 a Use elementary row operations on the system2x + 3y = 52x + 3y = 11

to show that it

reduces to

·2 3 50 0 6

¸:

What does the second row indicate? What is the geometrical significance of your

result?

b Use elementary row operations on the system2x + 3y = 54x + 6y = 10

to show that it

reduces to

·2 3 50 0 0

¸: Explain geometrically.

5 a Use augmented matrices to show that3x¡ y = 2

6x¡ 2y = 4has infinitely many solutions

of the form x = t, y = 3t¡ 2.

b Discuss the solutions to3x¡ y = 2

6x¡ 2y = kwhere k can take any real value.

6 Consider

½3x¡ y = 8

6x¡ 2y = kwhere k is any real number.

a Use elementary row operations to reduce the system to:

·3 ¡1 80 0 ....

¸b For what value of k is there infinitely many solutions?

c What form do the infinite number of solutions have?

d When does the system have no solutions?

Find all solutions to

½x + 3y = 5

4x + 12y = kwhere k is a constant, by using elementary

row operations.

In augmented matrix form, the system is:·1 3 54 12 k

¸»·

1 3 50 0 k ¡ 20

¸R2 ! R2 ¡ 4R1

4 12 k¡4 ¡12 ¡20

0 0 k ¡ 20

R2

¡4R1

The second equation actually reads 0x + 0y = k ¡ 20

So, if k 6= 20 we have 0 = a non-zero number. This is absurd, so

no solution could exist.

If k = 20 we have 0 = 0.

This means that all solutions come from x + 3y = 5 alone.

Letting y = t, x = 5 ¡ 3t for all values of t

) there are infinitely many solutions of the form x = 5 ¡ 3t, y = t, t real.

Example 3



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7 Consider

½4x + 8y = 12x¡ ay = 11

a Use elementary row operations to reduce the system to:

·4 8 10 .... ....

¸b For what values of a does the system have a unique solution?

c Show that the unique solution is x =a + 88

4a + 16, y =

¡21

2a + 8

d What is the solution in all other cases?

8 Use elementary row operations to find the values of m when the system

½mx+ 2y = 62x + my = 6

has a unique solution.

a Find the unique solution.

b Discuss the solutions in the other two cases.

Click on the appropriate icon to obtain instructions on how to enter a

number array called an augmented matrix and then obtain the reduced

row-echelon form.

In the example

½2x + y = ¡1x¡ 3y = 17 ·

2 1 ¡11 ¡3 17

¸became

·1 0 20 1 ¡5

¸.

Solve using a calculator: a

½3x + 5y = 46x¡ y = ¡11

b

½0:83x + 1:72y = 13:761:65x¡ 2:77y = 3:49

A general 3 £ 3 system in variables x, y and z has form a1x + b1y + c1z = d1a2x + b2y + c2z = d2a3x + b3y + c3z = d3

where the coefficients of x, y and z are constants.24 a1 b1 c1 d1a2 b2 c2 d2a3 b3 c3 d3

35 is the system’s augmented matrix form. We need to reduce this

to echelon form

24 a b c d0 e f g0 0 h i

35 by using elementary row operations.

Notice the creation, where possible, of a triangle of zeros in the bottom left hand corner.

In this form we can easily solve the system. The last row is really hz = i.

USING A GRAPHICS CALCULATOR

TI

C

3×3 SYSTEMS WITH UNIQUE SOLUTIONS� �C

9

we could have found

reduced row-echelon form when



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² If h 6= 0 we can determine z =i

huniquely, and likewise y and x from the other

two rows. Thus we arrive at a unique solution.

² If h = 0 and i 6= 0, the last row reads 0 £ z = i where i 6= 0. This is absurd

so there is no solution and we say that the system is inconsistent.

²

1 Without using technology, solve:

a x + y + z = 62x + 4y + z = 52x + 3y + z = 6

b x + 4y + 11z = 7x + 6y + 17z = 9x + 4y + 8z = 4

c 2x¡ y + 3z = 172x¡ 2y ¡ 5z = 43x + 2y + 2z = 10

Solve the system

x + 3y ¡ z = 152x + y + z = 7x¡ y ¡ 2z = 0.

In augmented matrix form, the system is24 1 3 ¡1 152 1 1 71 ¡1 ¡2 0

35

»24 1 3 ¡1 15

0 ¡5 3 ¡230 ¡4 ¡1 ¡15

35 R2 ! R2 ¡ 2R1

R3 ! R3 ¡R1

»24 1 3 ¡1 15

0 ¡5 3 ¡230 0 ¡17 17

35R3 ! 5R3 ¡ 4R2

2 1 1 7¡2 ¡6 2 ¡30

0 ¡5 3 ¡23

R2

¡2R1

1 ¡1 ¡2 0¡1 ¡3 1 ¡15

0 ¡4 ¡1 ¡15

R3

¡R1

0 ¡20 ¡5 ¡750 20 ¡12 920 0 ¡17 17

5R3

¡4R2

The last row gives ¡17z = 17 ) z = ¡1

Thus in row 2, as ¡5y + 3z = ¡23

) ¡5y ¡ 3 = ¡23

) ¡5y = ¡20

) y = 4

and from row 1 x + 3y ¡ z = 15

) x + 12 + 1 = 15

) x = 2

A typical graphics

calculator solution:

Thus we have a unique solution x = 2, y = 4, z = ¡1.

Example 4

EXERCISE 9C

TI

C

If and , the last row is all zeros. Consequently, there are

solutions which can be written in terms of a . For example, by

letting we can write and in terms of , and the free parameter can

take any real value.

h i

z t x y t t

= 0 = 0

=

infinitely

many free parameter



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2 Use technology to check your answers to 1 a and 1 c.

3 Use technology to solve:

a x + 2y ¡ z = 23x¡ y + 3z = ¡23

7x + y ¡ 4z = 62

b 10x¡ y + 4z = ¡97x + 3y ¡ 5z = 89

13x¡ 17y + 23z = ¡309

c 1:3x + 2:7y ¡ 3:1z = 8:22:8x¡ 0:9y + 5:6z = 17:36:1x + 1:4y ¡ 3:2z = ¡0:6

Rent-a-car has three different makes of vehicles, P, Q and R, for hire. These cars

are located at yard A and yard B on either side of a city. Some cars are out (being

rented). In total they have 150 cars. At yard A they have 20% of P, 40% of Q and

30% of R which is 46 cars in total. At yard B they have 40% of P, 20% of Q and

50% of R which is 54 cars in total. How many of each car type do they have?

Suppose Rent-a-car has x of P, y of Q and z of R.

Since it has 150 cars in total, x + y + z = 150 ...... (1)

But yard A has 20% of P + 40% of Q + 30% of R and this is 46.

) 210x + 4

10y + 310z = 46

i.e., 2x + 4y + 3z = 460 ...... (2)

Yard B has 40% of P + 20% of Q + 50% of R and this is 54.

) 410x + 2

10y + 510z = 54

i.e., 4x + 2y + 5z = 540 ...... (3)

We need to solve (1), (2) and (3) simultaneously:

The augmented matrix is

24 1 1 1 1502 4 3 4604 2 5 540

35Using elementary row operations

»24 1 1 1 150

0 2 1 1600 ¡2 1 ¡60

35 R2 ! R2 ¡ 2R1

R3 ! R3 ¡ 4R1

»24 1 1 1 150

0 2 1 1600 0 2 100

35R3 ! R3 + R2

or using technology:

So, 2z = 100 ) z = 50

and 2y + 50 = 160

) 2y = 110

) y = 55

and x + y + z = 150

) x = 45

Thus Rent-a-car has 45 of P,

55 of Q and 50 of R.

Example 5



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4 a Find all solutions of the system of equations

8<: 2x + y + 3z = 903x + 2y + z = 81

5x + 2z = 104.

b

i State clearly what the variables x, y and z must represent if this situation is to

ii Northtown high school needs 4 cricket balls and 5 softballs, and wishes to

order as many netballs as they can afford. How many netballs will they be

able to purchase if there is a total of $315 to be spent?

5 Managers, clerks and labourers are paid according to an industry award.

Xenon employs 2 managers, 3 clerks and 8 labourers with a total salary bill of $352 000.

Xanda employs 1 manager, 5 clerks and 4 labourers with a total salary bill of $274 000.

Xylon employs 1 manager, 2 clerks and 11 labourers with a total salary bill of $351 000.

a If x, y and z represent the salaries (in thousands of dollars) for managers, clerks

and labourers respectively, show that the above information can be represented by

a system of three equations.

b Solve the above system of equations.

c Determine the total salary bill for Xulu company which employs 3 managers, 8clerks and 37 labourers.

6 Herbert and Agnes had plotted three points on the graph of a quadratic function. Un-

fortunately, they forgot the original function and were unable to plot any more points.

Given that the points were (1, ¡3), (3, ¡5) and (¡2, ¡15), can you help the two

poor students complete the table of values below?

x ¡3 ¡2 ¡1 0 1 2 3y ¡15 ¡3 ¡5

7 A mixed nut company uses cashews,

macadamias, and brazil nuts to make

three gourmet mixes. The table along-

side indicates the weight in hundreds

of grams of each kind of nut required

to make a kilogram of mix.

Mix A Mix B Mix C

Cashews 5 2 6Macadamias 3 4 1Brazil Nuts 2 4 3

If 1 kg of mix A costs $12:50 to produce, 1 kg of mix B costs $12:40, and 1 kg of mix

C costs $11:70, determine the cost per kilogram of each of the different kinds of nuts.

Hence, find the cost per kilogram to produce a mix containing 400 grams of cashews,

200 grams of macadamias, and 400 grams of brazil nuts.

8 Klondike High has 76 students at Matriculation level and these students are in classes P,

Q and R. There are p students in P, q in Q and r in R.

One-third of P, one-third of Q, and two-fifths of R study Chemistry.

One-half of P, two-thirds of Q, and one-fifth of R study Maths.

One-quarter of P, one-third of Q, and three-fifths of R study Geography.

Westfield school bought two cricket balls, one softball and three netballs for a total

cost of $ . Southvale school bought three cricket balls, two softballs and a netball

for a cost of $ . Eastside school bought five cricket balls and two netballs for $ .

9081 104

be described by the set of equations considered in .a



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Given that 27 students study Chemistry, 35 study Maths, and 30 study Geography:

a find a system of equations which contains this information, making sure that the

coefficients of p, q and r are integers.

b Solve for p, q and r.

c If the equation containing the Geography information is removed, solve the remain-

ing system. Give your answer in parametric form, i.e., in terms of a parameter such

as t.

9 Susan and James opened a new business in 2001. Their annual profit was $160 000 in

2004, $198 000 in 2005, and $240 000 in 2006. Based on the information from these

three years they believe that their annual profit can be predicted by the model

P (t) = at + b +c

t + 4dollars

where t is the number of years after 2004, i.e., t = 0 gives the 2004 profit.

a Determine the values of a, b and c which fit the profits for 2004, 2005 and 2006.

b If the profit in 2003 was $130 000, does this profit fit the model in a?

c Susan and James believe their profit will continue to grow according to this model.

Predict their profit in 2007 and 2009.

As with 2 £ 2 systems of linear equations, 3 £ 3 systems may have a unique solution where

a single value of each variable satisfies all three equations simultaneously. Alternatively they

could have ² no solutions or

² infinitely many solutions.

We will now consider examples which show each of these situations.

Solve the system

8<: x + 2y + z = 32x ¡ y + z = 8

3x ¡ 4y + z = 18:

In augmented matrix form, the system is:24 1 2 1 32 ¡1 1 83 ¡4 1 18

35

»24 1 2 1 3

0 ¡5 ¡1 20 ¡10 ¡2 9

35

»24 1 2 1 3

0 ¡5 ¡1 20 0 0 5

35R2 ! R2 ¡ 2R1

R3 ! R3 ¡ 3R1

R3 ! R3 ¡ 2R2

2 ¡1 1 8¡2 ¡4 ¡2 ¡6

0 ¡5 ¡1 2

3 ¡4 1 18¡3 ¡6 ¡3 ¡9

0 ¡10 ¡2 9

0 ¡10 ¡2 90 10 2 ¡40 0 0 5

OTHER 3 × 3 SYSTEMS� �D

Example 6



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The last line means 0x + 0y + 0z = 5 i.e., 0 = 5 which is absurd,

)

Using a graphics calculator:

Once again we have an absurd statement, 0 = 1,

Solve the system:

8<: 2x ¡ y + z = 5x + y ¡ z = 2

3x ¡ 3y + 3z = 8.

In augmented matrix form, the system is:24 1 1 ¡1 22 ¡1 1 53 ¡3 3 8

35»24 1 1 ¡1 2

0 ¡3 3 10 ¡6 6 2

35»24 1 1 ¡1 2

0 ¡3 3 10 0 0 0

35R2 ! R2 ¡ 2R1

R3 ! R3 ¡ 3R1

R3 ! R3 ¡ 2R2

2 ¡1 1 5¡2 ¡2 2 ¡4

0 ¡3 3 1

3 ¡3 3 8¡3 ¡3 3 ¡6

0 ¡6 6 2

0 ¡6 6 20 6 ¡6 ¡20 0 0 0

If we let z = t in row 2, ¡3y + 3t = 1¡3y = 1 ¡ 3t

) y =1 ¡ 3t

¡3

) y = ¡13 + t

Using row 1, x + (¡13 + t) ¡ t = 2

) x ¡ 13 = 2

) x = 73

) the solutions have form: x = 73 , y = ¡1

3 + t, z = t (t real)

1 Solve the following systems:

a x ¡ 2y + 5z = 12x ¡ 4y + 8z = 2

¡3x + 6y + 7z = ¡3

b x + 2y ¡ z = 43x + 2y + z = 7

5x + 2y + 3z = 11

We write the secondequation on the top lineof the augmented matrixso there is a 1 in the top

left corner.

Example 7

EXERCISE 9D

the system has .no solution

so there is .no solution

The row of zeros indicates .infinitely many solutions



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c 2x + 4y + z = 13x + 5y = 1

5x + 13y + 7z = 4

d 2x + 3y + 4z = 15x + 6y + 7z = 2

8x + 9y + 10z = 4

Consider the system

8<: x¡ 2y ¡ z = ¡12x + y + 3z = 13x + 8y + 9z = a where a can take any real value.

a Use elementary row operations to reduce the system to echelon form.

b When does the system have no solutions?

c When does the system have infinitely many solutions? What are the solutions?

a In augmented matrix form, the system is:24 1 ¡2 ¡1 ¡12 1 3 131 8 9 a

35

»24 1 ¡2 ¡1 ¡1

0 5 5 150 10 10 a + 1

35 R2 ! R2 ¡ 2R1

R3 ! R3 ¡R1

»24 1 ¡2 ¡1 ¡1

0 5 5 150 0 0 a¡ 29

35R3 ! R3 ¡ 2R2

2 1 3 13¡2 4 2 2

0 5 5 15

1 8 9 a¡1 2 1 1

0 10 10 a + 1

0 10 10 a + 10 ¡10 ¡10 ¡300 0 0 a¡ 29

b If a 6= 29 the last row reads zero = non-zero.

c If a = 29 the last row is all zeros.

Letting z = t, row 2 gives 5y + 5t = 15 ) y = 3 ¡ t

Using the first row, x¡ 2y ¡ z = ¡1

) x¡ 2(3 ¡ t) ¡ t = ¡1

) x¡ 6 + 2t¡ t = ¡1

) x = 5 ¡ t

Thus we have infinitely many solutions in the form:

x = 5 ¡ t, y = 3 ¡ t, z = t (t being real).

2 Write the system of equations x + 2y + z = 32x¡ y + 4z = 1x + 7y ¡ z = k in augmented matrix form.

a Use elementary row operations to reduce the system to

echelon form as shown:

24 ² ² ² ²0 ² ² ²0 0 ² ²

35b Show that the system has either no solutions or infinitely

many solutions and write down these solutions.

c Why does the system not have a unique solution?

Example 8

The system is and there are .inconsistent no solutions

This indicates .infinitely many solutions



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3 Consider the system of equations x + 2y ¡ 2z = 5x¡ y + 3z = ¡1

x¡ 7y + kz = ¡k.

a Reduce the system to echelon form.

b Show that for one particular value of k, the system has infinitely many solutions.

Find the solutions in this case.

c Show that a unique solution exists for any other value of k. Find the unique solution.

4 A system of equations is x + 3y + 3z = a¡ 12x¡ y + z = 7

3x¡ 5y + az = 16.

a Reduce the system to echelon form using elementary row operations.

b Show that if a = ¡1 the system has infinitely many solutions. Find their form.

c If a 6= ¡1, find the unique solution in terms of a.

5 Reduce the system of equations 2x + y ¡ z = 3mx¡ 2y + z = 1x + 2y + mz = ¡1

to a form in which the solutions may be determined for all real values of m.

a Show that the system has no solutions for one particular value of m (m = m1, say).

b Show that the system has infinitely many solutions for another value of m(m = m2, say).

c For what values of m does the system have a unique solution?

Show that the unique solution is x =7

m + 5, y =

3(m¡ 2)

m + 5, z =

¡7

m + 5.

6 Consider the system of equations x + 3y + kz = 2kx¡ 2y + 3z = k

4x¡ 3y + 10z = 5.

a Write the system in augmented matrix form and reduce it

by elementary row operations to the form:

24 1 3 k 20 ² ² ²0 0 ² ²

35b Show that for one particular value of k the system has

infinitely many solutions, and find the form of these

solutions.

c For what value(s) of k does the system have no solutions?



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½x + y + 2z = 22x + y ¡ z = 4

is a 2 £ 3 system of two linear equations in three unknowns and

It requires a further equation if a unique solution is to be obtained.

We call this an underspecified system because there are less equations than unknowns.

a Solve the system:

½x + y + 2z = 22x + y ¡ z = 4

b What can be deduced if the following equation is added to the system

i 3x¡ y ¡ 4z = 18 ii 3x + y ¡ 4z = 18?

a The augmented matrix is·1 1 2 22 1 ¡1 4

¸

»·

1 1 2 20 ¡1 ¡5 0

¸R2 ! R2 ¡ 2R1

2 1 ¡1 4¡2 ¡2 ¡4 ¡4

0 ¡1 ¡5 0

So, ¡y ¡ 5z = 0 and if z = t, y = ¡5t

From row 1, x + y + 2z = 2

) x¡ 5t + 2t = 2

) x = 2 + 3t.

The areinfinitely many solutions x = 2 + 3t, y = ¡5t, z = t for all real t.

b i If in addition 3x¡ y ¡ 4z = 18

then 3(2 + 3t) ¡ (¡5t) ¡ 4t = 18

) 6 + 9t + 5t¡ 4t = 18

) 10t = 12

) t = 1:2

When t = 1:2 x = 5:6, y = ¡6, z = 1:2.

ii If in addition 3x + y ¡ 4z = 18

then 3(2 + 3t) + (¡5t) ¡ 4t = 18

i.e., 6 = 18 which is absurd

)

1 Solve the following systems:

a 2x + y + z = 5x¡ y + z = 3

b 3x + y + 2z = 10x¡ 2y + z = ¡4

c x + 2y + z = 52x + 4y + 2z = 16

FURTHER APPLICATIONSE

Example 9

EXERCISE 9E

has .infinitely many solutions

we have the unique solution

no solution exists.



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2 Solve the systemx¡ 3y + z = 0

2x + y ¡ 2z = 0and hence solve

x¡ 3y + z = 02x + y ¡ 2z = 03x¡ y + z = 18.

3 Solve2x + 3y + z = 0x¡ y + 2z = 0

and hence solve

2x + 3y + z = 0x¡ y + 2z = 0ax + y ¡ z = 0

for all real numbers a.

4 An economist producing x thousand items attempts to model a profit function as a

quadratic model P (x) = ax2 + bx + c thousand dollars. She notices that producing

a Using the supplied information, show that

½a + b + c = 8

16a + 4b + c = 17

b Show that a = t, b = 3¡ 5t, c = 5 + 4t represents the possible solutions for

the system.

c If she discovers that the profit for producing 2500 items is $19 750, find the actual

profit function.

d What is the maximum profit to be made and what level of production is needed to

achieve it?

5 a Solve the system:8<: x + y ¡ z = 2x¡ y + 2z = 7

2x + 4y ¡ 5z = ¡1

b Now solve the system:8>><>>:x + y ¡ z = 2x¡ y + 2z = 7

2x + 4y ¡ 5z = ¡13x + y ¡ z = 8

6

A uniform beam is in perfect balance with a pivot point at its centre. Any mass hung

from a point on the right hand side causes the beam to rotate about the pivot point

(fulcrum) in a clockwise direction. Likewise, a mass hung on the balanced beam on the

left hand side causes it to move anticlockwise. Consider the following:

With masses w1, w2, w3 and w4 hung from the beam at distances d1, d2, d3 and d4 from

the pivot point,w1d1 + w2d2| {z } = w3d3 + w4d4| {z } .

on LHS on RHS

The three masses w1, w2 and w3 balance the beam when hung in the two positions

shown:

Position (1):

d cm d cm

d2 cm d3 cmd1 cm d4 cm

w1 w2 w3 w4

4 m 3 m 1 m

w1 w2 w3

1000 8000 4000 17 000items yields a profit of $ , and producing items yields a profit of $ .



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Y:\HAESE\SA_12STU-2ed\SA12STU-2_09\335SA12STU-2_09.CDR Tuesday, 14 November 2006 11:57:51 AM PETERDELL

INVESTIGATION 2 ELECTRICAL CIRCUITS

Position (2):

a Explain why 4w1 ¡ 3w2 ¡ 4w3 = 0.

b What equation results from the balance in position (2)?

c Solve simultaneously the two equations from a and b. How many solutions exist?

d If in addition we know that the total mass of the three weights is 105 kg, what are

the individual masses?

e If all four masses w1, w2, w3, w4 were hung from the beam to create balance, how

many positions are necessary to form equations so that:

i a ratio of w1 : w2 : w3 : w4 can be found

ii the exact values of the masses can be found?

7 The illustration shows a network of roads, each of which

is one-way. The nodes K, L and M are intersections.

The direction of flow is shown by the arrow heads. The

rate of flow is shown on the diagram by the pronumerals.

Their units are vehicles per minute.

The total rate of flow into an intersection is equal to the

total rate of flow from it. For example, a+ x = b+ y.

a By considering all three intersections, show that a

system of linear equations results, and this system

has augmented matrix

24 1 ¡1 0 b¡ a1 0 1 d¡ e0 1 1 c

35b If it was determined that a = 15, b = 32, c = 12 d = 40:

i find e when a solution exists ii find the solution when z = t.

c If the road from K to L is closed, find the rate of flow from K to M and from M

to L.

8 In this network of cables, data flows in the direction

shown by each arrowhead.

A, B and C are nodes. Data enters at B and C and exits

at A and C.

The rates of flow between the nodes are indicated by

the pronumerals x1, x2 and x3.

a If the total rate of flow into a node equals the total

rate from it, construct a system of linear equations to connect x1, x2 and x3.

b Use elementary row operations to reduce the system.

Show that a solution exists only if a + d = b + c.

c If x3 = t, solve the system and show that a valid solution exists if t 6 d¡ c.

3 m 2 m 2 m

w1 w2w3

Click on the icon to produce printable pages on a

further application of linear equations.

PRINTABLE

INVESTIGATION

a b

d ce

x y

z

M

K L

a

b cd

A

B

C

xz xx

xc

and



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The technique of solving systems of linear equations using elementary row operations can be

applied to systems of higher order. However we will use technology to solve problems where

the number of unknowns is 4 or 5.

1 East Beach Primary School bought 3 footballs, 2 netballs, 1 basketball and 2 softballs

for $283. West Lakes Primary School bought 2 footballs, 3 netballs, 2 basketballs and 1softball for $282. Southlands Primary School bought 1 football, 2 netballs, 3 basketballs

and 3 softballs for $289. Northlands Primary School bought 2 footballs, 1 netball, 3basketballs and 4 softballs for $313. Given that all five schools bought their sporting

gear from the same shop on the same day, what did Central City Primary pay in total

for 2 footballs, 4 netballs, 1 basketball and 3 softballs?

4 × 4 AND 5 × 5 SYSTEMS� � � �F

x2 + axy + by2 + cx + dy + e = 0.

The units along the axes are astronomical units where

1 astronomical unit + (1:50 £ 108) km. An astronomer observes the asteroid at

5 positions: (¡1:03, 2:164), (¡0:56, ¡1:868), (0:38, ¡1:668), (1:17, 4:876) and

(2:89, 1:019). Find the equation of the asteroid’s orbit.

Substituting the points into the equation gives:

¡2:229a + 4:683b¡ 1:03c + 2:164d + e = ¡1:061

1:046a + 3:489b¡ 0:56c¡ 1:868d + e = ¡0:3136

¡0:634a + 2:782b + 0:38c¡ 1:668d + e = ¡0:144

5:705a + 23:775b + 1:17c + 4:876d + e = ¡1:369

2:945a + 1:038b + 2:89c + 1:019d + e = ¡8:352

Using a graphics calculator the

solution is: a = ¡0:799

b = 0:538

c = ¡0:616

d = ¡1:065

e = ¡3:691

So, the elliptical orbit is:

x2 ¡ 0:799xy + 0:538y2 ¡ 0:616x¡ 1:065y ¡ 3:691 = 0

from which the position of the asteroid elsewhere in the orbit can be determined.

An asteroid orbits the sun in an elliptical path. If the sun is at the origin of a

Cartesian coordinate system the equation of the path is

Example 10

321 1

4

3

2

1

1

x

y

EXERCISE 9F



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REVIEW SET 9A

2 An economist for a manufacturing business observes that the most likely model for a

cost function when making x trailers each week is a cubic polynomial:

C(x) = ax3 + bx2 + cx + d dollars.

For making 10 trailers the cost is $457. For making 20 it is $1764. For 30 it is $2436and for 40 it is $3389.

Find the cost function and use it to estimate the cost per week for making 26 trailers.

3 An asteroid orbits the sun in an elliptical orbit with equation of the path given by

x2 + axy + by2 + cx+ dy + e = 0. It is observed at (¡1:57, ¡1:735), (¡0:83, 1:984),

(1:25, ¡0:878), (2:06, 5:453) and (3:41, 2:896).

a Find five linear equations in a, b, c, d and e.

b Use technology to solve the system.

c Plot the orbit of the asteriod on a set of axes.

d Find the position of the asteriod when y = 0.

4 It is suspected that the sum of the first n perfect squares is a cubic polynomial in n.

That is, 12 + 22 + 32 + 42 + :::::: + n2 = an3 + bn2 + cn + d. If n = 1, the left

hand side is 12 and the right hand side is a + b + c + d. So, a + b + c + d = 1.

a Find three other linear equations by substituting n = 2, 3 and 4.

b Solve for a, b, c and d.

c Hence find 12 + 22 + 32 + 42 + :::::: + 1002.

5 In 2004 Xenon’s profit was $75 187. In 2005 it was $83 843. In 2006 it was $98 491.

In 2007 it was $125 910. Based on the information from these four years, management

P (t) = at2 + bt + c +d

t + 2dollars, where t is the time after 2004 (t = 0 for 2004).

a Use the given data to find a, b, c and d.

b Use the model to predict the profit for 2008.

6 James believes that 1 £ 2 £ 3 + 2 £ 3 £ 4 + 3 £ 4 £ 5 + ::::: + n(n + 1)(n + 2)has a sum which is a quartic polynomial, an4 + bn3 + cn2 + dn + e:

a Using the same technique as in question 4, set up five linear equations in a, b, c, dand e.

b Solve the system.

c Find the sum of 1 £ 2 £ 3 + 2 £ 3 £ 4 + 3 £ 4 £ 5 + ::::: + 50 £ 51 £ 52.

1 When does the systemx + 4y = 2

kx + 3y = ¡6have a unique solution?

Comment on the solutions for the non-unique cases.

2 Solve the system 3x¡ y + 2z = 32x + 3y ¡ z = ¡3x¡ 2y + 3z = 2.

REVIEWG

GRAPHING

PACKAGE

believes that their annual profit could be predicted by the model:



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REVIEW SET 9B

3 The two points (¡2, 4) and (1, 3) lie on a circle with equation in the form

x2 + y2 + ax + by + c = 0.

a Find two equations in a, b and c and solve the system of equations.

b Explain why infinitely many solutions are obtained in a.

c If (2, 2) is also on the circle, find the equation of the circle.

4 2x + 3y ¡ 4z = 13x¡ y + 3z = ¡1

3x + 7y ¡ 11z = kSolve the system using elementary row operations, and

describe the solutions as k takes all real values.

5 Solve the system 3x + y ¡ z = 0x + y + 2z = 0:

6 Jason, Mary, Peter and Sue bought tickets for

the games and performances shown. These

are the number of tickets bought per person:

Play Concert AFL NBL

Jason 2 1 3 2Mary 2 2 1 4Peter 3 1 2 2Sue 1 4 2 3

a

were as follows: Jason $178,

Mary $206, Peter $197, Sue $237.

Write down a system of equations which will enable you to find the price of each

ticket type.

b Find the cost of each ticket type.

c If Jon wishes to purchase 4 play, 3 concert, 4 AFL and 1 NBL ticket, what will

be the total cost?

1 Solve the system of equations 2x + y + z = 84x¡ 7y + 3z = 103x¡ 2y ¡ z = 1.

2 Solve the systemx + 2y ¡ 3z = 3

6x + 3y + 2z = 4:

3 a Show that the system2x¡ 3y = 9mx¡ 7y = n

has augmented matrix after elementary

row operations of

·2 ¡3 9

14 ¡ 3m 0 63 ¡ 3n

¸b Under what conditions does the system have a unique solution?

4 Find the values of t for which the system of equations

x¡ y ¡ 2z = ¡3tx + y ¡ z = 3tx + 3y + tz = 13

does not

have a unique solution for x, y and z.

Show that no solution exists for one of these values of t. Find the solution set for the

other values of t.

5 A rock thrown upwards from the top of a cliff followed a path such that its distance

above sea level was given by s(t) = at2 + bt + c, where t is the time in seconds

after the rock was released. After 1 second the rock was 63 m above sea level, after

The total costs of all tickets per person



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REVIEW SET 9C

2 seconds 72 m, and after 7 seconds 27 m.

a Find a, b and c and hence an expression for s(t).b Find the height of the cliff.

c Find the time taken for the rock to reach sea level.

6 The profits of a business are

recorded in dollars as follows:2003 2004 2005 2006 2007

¡23 689 ¡3528 18 042 43 322 75 483Fit a quartic model to the data

using P = at4 + bt3 + ct2 + dt + e where t is the number of years since 2003.

a Write down a 5 £ 5 system of linear equations

b Solve the system for a, b, c, d and e.

c Use the model to predict the profits for 2008.

1 Solve the system of equationskx + 2y = 12x + ky = ¡2

as k takes all real values.

2 Find the solution set of the following:

a 2x + y ¡ z = 93x + 2y + 5z = 19x + y ¡ 3z = 1

b 2x + y ¡ z = 33x + 2y + z = 1

x¡ 3y = 5

3 The cost of producing x hundred bottles of correcting fluid per day is given by the

function C(x) = ax3 + bx2 + cx + d dollars where a, b, c and d are constants.

a If it costs $80 before any bottles are produced, find d.

b It costs $100 to produce 100 bottles, $148 to produce 200 bottles and $376 to

produce 400 bottles per day. Determine a, b and c.

4 Solve the system2x¡ 3y + z = 104x¡ 6y + kz = m

for all possible values of k and m.

5 Consider the system of equations x + 5y ¡ 6z = 2kx + y ¡ z = 35x¡ ky + 3z = 7 as k takes all real values.

a Show using elementary row

operations that the system

reduces to

24 1 5 ¡6 20 1 ¡ 5k 6k ¡ 1 3 ¡ 2k0 0 (k ¡ 2)(3k ¡ 2) ¡(k ¡ 2)(k + 18)

35b For what values of k does the system have a unique solution?

c For what value of k does the system have infinitely many solutions? Find the

solutions.

d For what value of k is the system inconsistent? How many solutions does the

system have in this case?

6 The following points lie on a curve with equation y = ax3 + bx2 + cx +d

x¡ 2:

A(0, 2), B(1, 9), C(3, 41), D(4, 118).

a Write down a system of linear equations in a, b, c and d.

b Solve the system. c If E(2:3, k) lies on the curve, find k.



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10

Contents:

MatricesMatrices

A

B

C

D

E

F

G

H

Introduction

Addition, subtraction and multiplesof matrices

Matrix multiplication

Transition matrices

The inverse of a 2 × 2 matrix

The inverse of a 3 × 3 matrix

Determinants of matrices

Review

� �

� �


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In the previous chapter we solved systems of linear equations using augmented matrices.

These matrices allowed us to represent the system as a rectangular array of numbers so that

we could carry out simple operations on them without the complications of the variables.

Spreadsheets displaying rectangular arrays of stock in hand numbers, costing, budgets, etc are

actually matrices, and they can be very large.

You have been using matrices for many years without realising it.

In general:

A matrix is a rectangular array of numbers arranged in rows and columns.

It is usual to put square or round brackets around a matrix.

Consider these two items of information:

We could write them in detailed matrix form as:

number

B 2J 1E 6C 1

and C T B

F 6 1 2U 9 2 3H 10 3 4

and if we can remember what makes up the rows and columns,

we could write them simply as:26642161

3775 and

24 6 1 29 2 310 3 4

35

INTRODUCTIONA

For example:

Goals Behinds Points

Crows 16 11 107Power 15 17 107

Ingredients Amount

sugar 1 tspn

flour 1 cup

milk 200 mL

salt 1 pinch

July 2001

M T W T F S S

1

2 3 4 5 6 7 8

9 10 11 12 13 14 15

16 17 18 19 20 21 22

23 24 25 26 27 28 29

30 31

Bread 2 loaves

Juice 1 carton

Eggs 6

Cheese 1

Furniture inventory

chairs tables beds

Flat 6 1 2Unit 9 2 3

House 10 3 4

Shopping list

342 MATRICES (Chapter 10)


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26642161

3775 has 4 rows and 1 column and we

say that this is a 4 £ 1 column matrix

or column vector.

24 6 1 29 2 310 3 4

35 has 3 rows and 3 columns and is

called a 3 £ 3 square matrix.

This element, 3, is in row 3, column 2.£3 0 ¡1 2

¤has 1 row and 4 columns and is called a 1 £ 4 row matrix

or row vector.

Note: ² An m£ n matrix has m rows and n columns.

rows columns

² m£ n specifies the order of a matrix.

Following are a few of many uses for the mathematics of matrices:

² Solving systems of equations in business, physics, engineering, etc.

² Linear programming where, for example, we may wish to optimise a linear

expression subject to linear constraints. For example, optimising profits of a

business.

² Business inventories involving stock control, cost, revenue and profit calculations.

Matrices form the basis of business computer software.

² Markov chains for predicting long term probabilities such as in weather.

² Strategies in games where we wish to maximise our chance of winning.

² Economic modelling where the input from various suppliers is needed to help a

business be successful.

² Graph (network) theory which is used in truck and airline route determination to

minimise distance travelled and therefore costs.

² Assignment problems where we have to direct resources in industrial situations in

the most cost effective way.

² Forestry and fisheries management where we need to select an appropriate

sustainable harvesting policy.

² Cubic spline interpolation which is used to construct fonts used in desktop

publishing.

Each font is stored in matrix form in the memory of a computer.

² Computer graphics, flight simulation, Computer Aided Tomography (CAT

scanning) and Magnetic Resonance Imaging (MRI), Fractals, Chaos, Genetics,

Cryptography (coding, code breaking, computer confidentiality), etc.

A matrix can be used to represent numbers of items to be purchased, prices of items to be

purchased, numbers of people involved in the construction of a building, etc.

USES OF MATRICES

MATRICES (Chapter 10) 343


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Lisa goes shopping at store A to buy 2 loaves of bread at $2:65 each, 3 litres of

milk at $1:55 per litre, and one 500 g tub of butter at $2:35.

Represent the quantities purchased in a row matrix and the costs in a column matrix.

The quantities matrix is£

2 3 1¤

bread milk butter

The costs matrix is

24 2:651:552:35

35 bread

milk

butter

Note: If Lisa goes to a different supermarket

(store B) and finds that the prices for the

same items are $2:25 for bread, $1:50 for

milk, and $2:20 for butter, then the costs

matrix to show prices from both stores is:

24 2:65 2:251:55 1:502:35 2:20

35 bread

milk

butter

store A store B

1 Write down the order of:

a £5 1 0 2

¤ b ·27

¸ c ·2 ¡11 3

¸ d24 1 2 3

2 0 45 1 0

352 A grocery list consists of 2 loaves of bread, 1 kg of butter, 6 eggs and 1 carton of cream.

The cost of each grocery item is $1:95, $2:35, $0:15 and $0:95 respectively.

a Construct a row matrix showing quantities.

b Construct a column matrix showing prices.

c What is the significance of (2 £ 1:95) + (1 £ 2:35) + (6 £ 0:15) + (1 £ 0:95)?

3

1000, 1500 and 1250 cans of each in week 1; 1500, 1000 and 1000 of each in week 2800, 2300 and 1300 cans of each in week 3; 1200 cans of each in week 4.

Construct a matrix to show February’s production levels.

4 Over Easter a baker produced the following food items: On

Friday he baked 40 dozen pies, 50 dozen pasties, 55 dozen

rolls and 40 dozen buns. On Saturday 25 dozen pies, 65 dozen

pasties, 30 dozen buns and 44 dozen rolls were made. On

Sunday 40 dozen pasties, 40 dozen rolls, 35 dozen of each of

pies and buns were made. On Monday the totals were 40 dozen

pasties, 50 dozen buns, and 35 dozen of each of pies and rolls.

Represent this information as a matrix.

Example 1

EXERCISE 10A

Ben’s Baked Beans factory produces cans of baked beans in sizes: g, g and

g. In February they produced respectively:

3 200 300500

� ��



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Before attempting to add and subtract matrices it is necessary to define what we mean by

matrix equality.

Two matrices are equal if they have exactly the same shape (order) and elements in

corresponding positions are equal.

For example, if

·a bc d

¸=

·w xy z

¸then a = w, b = x, c = y and d = z.

Notice that

·1 2 03 4 0

¸6=·

1 23 4

¸:

Sally has three stores (A, B and C). Her stock levels for dresses, skirts and blouses are given

by the matrix:Store

A B C

dresses

skirts

blouses

24 23 41 6828 39 7946 17 62

35Some newly ordered stock has just arrived. For

each store 20 dresses, 30 skirts and 50 blouses

must be added to stock levels.

Her stock order is given by the matrix

24 20 20 2030 30 3050 50 50

35Clearly the new levels are:

24 23 + 20 41 + 20 68 + 2028 + 30 39 + 30 79 + 3046 + 50 17 + 50 62 + 50

35or

24 23 41 6828 39 7946 17 62

35+

24 20 20 2030 30 3050 50 50

35 =

24 43 61 8858 69 10996 67 112

35So, to add two matrices they must be of the same order and then we simply

add corresponding elements.

If A =

·1 2 36 5 4

¸, B =

·2 1 60 3 5

¸and C =

·3 12 4

¸find:

a A + B b A + C

ADDITION, SUBTRACTIONAND MULTIPLES OF MATRICES

B

EQUALITY

ADDITION

Example 2



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a A + B =

·1 2 36 5 4

¸+

·2 1 60 3 5

¸=

·1 + 2 2 + 1 3 + 66 + 0 5 + 3 4 + 5

¸=

·3 3 96 8 9

¸

b A + C cannot be found

as A and C are not the

same sized matrices

i.e., they have different

orders.

If Sally’s stock levels were

24 29 51 1931 28 3240 17 29

35 and her sales matrix for the week is

24 15 12 620 16 1919 8 14

35 , what are her stock levels now?

It is obvious that we subtract corresponding elements.

That is

24 29 51 1931 28 3240 17 29

35¡24 15 12 6

20 16 1919 8 14

35 =

24 14 39 1311 12 1321 9 15

35So, to subtract matrices they must be of the same order and then we simply

subtract corresponding elements.

SUBTRACTION

If A =

24 3 4 82 1 01 4 7

35 and B =

24 2 0 63 0 45 2 3

35 find A ¡ B.

A ¡ B =

24 3 4 82 1 01 4 7

35¡24 2 0 6

3 0 45 2 3

35=

24 3 ¡ 2 4 ¡ 0 8 ¡ 62 ¡ 3 1 ¡ 0 0 ¡ 41 ¡ 5 4 ¡ 2 7 ¡ 3

35=

24 1 4 2¡1 1 ¡4¡4 2 4

35

Example 3



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In the pantry there are 6 cans of peaches, 4 cans of apricots and 8 cans of pears.

This information could be represented by the column vector C =

24 648

35 .

Doubling these cans in the pantry we would have

24 12816

35 which is C + C = 2C.

Notice that to get 2C from C we simply multiply all matrix elements by 2.

Likewise, trebling the fruit cans in the pantry gives

3C = C + C + C =

24 3 £ 63 £ 43 £ 8

35 =

24 181224

35and halving them gives

12C =

266412 £ 6

12 £ 4

12 £ 8

3775 =

24 324

35

In general,

If A is

·1 2 52 0 1

¸find a 3A b 1

2A

a 3A = 3

·1 2 52 0 1

¸=

·3 6 156 0 3

¸ b 12A = 1

2

·1 2 52 0 1

¸

=

"12 1 21

2

1 0 12

#

1 If B =

·6 1224 6

¸find: a 2B b 1

3B c 112B d ¡1

2B

2 If A =

·2 3 51 6 4

¸and B =

·1 2 11 2 3

¸find:

a A + B b A ¡ B c 2A + B d 3A ¡ B

MULTIPLES

Example 4

if a scalar is multiplied by a matrix the result is matrix

obtained by multiplying every element of by .

t tt

A A

A

EXERCISE 10B.1



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3 Frank’s order for hardware items is shown in matrix form as

H =

26646126030

3775hammers

screwdriver sets

packets of nails

packets of screws

Find the matrix if:

a Frank doubles his order

b Frank halves his order

c Frank increases his order by 50%:

4 Christine sells clothing made by four different companies which we will call A, B, C

and D. Her usual monthly order is:

A B C D

skirt

dress

evening

suit

266430 40 40 6050 40 30 7540 40 50 5010 20 20 15

3775Find her order, to the nearest whole number,

if:

a she increases her total order by 15%

b she decreases her total order by 15%.

5 A restaurant served 85 men, 92 women and 52 children on Friday night. On Saturday

night they served 102 men, 137 women and 49 children.

a Express this information in two column matrices.

b Use the matrices to find the totals of men, women and children served over the

Friday-Saturday period.

6 On Monday David bought

shares in five companies and

on Friday he sold them. The

details are:

Cost price per share Selling price per share

A $1:72 $1:79

B $27:85 $28:75

C $0:92 $1:33

D $2:53 $2:25

E $3:56 $3:51

a Write down David’s

i cost price column

matrix

ii selling price column matrix.

b What matrix operation is needed to find David’s profit/loss matrix?

c Find David’s profit/loss matrix.

7 During week days a video store finds that its average hirings are: 75 movies (VHS),

27 movies (DVD) and 102 video/computer games. On the weekends the average figures

are: 43 DVD movies, 136 VHS movies and 129 games.

a Represent the data using two column matrices. 24 35 VHS

DVD

games

b Find the sum of the matrices in a .

c What does the sum matrix of b represent?

8 a In November, Lou E Gee sold 23 fridges, 17 stoves and 31 microwave ovens and

his partner Rose A Lee sold 19 fridges, 29 stoves and 24 microwave ovens.

In December Lou’s sales were: 18 fridges, 7 stoves and 36 microwaves while

Rose’s sales were: 25 fridges, 13 stoves and 19 microwaves.

i Write their sales for November as a 3 £ 2 matrix.

ii Write their sales for December as a 3 £ 2 matrix.

iii Write their total sales for November and December as a 3 £ 2 matrix.



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b A builder builds a block of 12 identical flats. Each flat is to contain 1 table, 4chairs, 2 beds and 1 wardrobe.

If F =

26641421

3775 is the matrix representing the furniture in one flat,

what, in terms of F, is the matrix representing the furniture in all flats?

9 Find x and y if:

a b·

x x2

3 ¡1

¸=

·y 43 y + 1

¸ ·x yy x

¸=

· ¡y xx ¡y

¸

10 a If A =

·2 13 ¡1

¸and B =

· ¡1 22 3

¸find A + B and B + A.

b Explain why A + B = B + A for all 2 £ 2 matrices A and B.

11 a For A =

· ¡1 01 5

¸, B =

·3 4¡1 ¡2

¸and C =

·4 ¡1¡1 3

¸find

(A + B) + C and A + (B + C).

b Prove that, if A, B and C are any 2 £ 2 matrices then

(A + B) + C = A + (B + C).

[Hint: Let A =

·a bc d

¸, B =

·p qr s

¸and C =

·w xy z

¸, say.]

For real numbers, it is true that a + 0 = 0 + a = a for all values of a.

The question arises: “Is there a matrix O in which A + O = O + A = A for any

matrix A?”

Simple examples like:

·2 34 ¡1

¸+

·0 00 0

¸=

·2 34 ¡1

¸suggests that O consists of

all zeros.

A zero matrix is a matrix in which all elements are zero.

For example, the 2 £ 2 zero matrix is

·0 00 0

¸,

the 2 £ 3 zero matrix is

·0 0 00 0 0

¸:

Zero matrices have the property that:

If A is a matrix of any order and O is the corresponding zero matrix, then

A + O = O + A = A.

ZERO MATRICES



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The negative matrix A, denoted ¡A is actually ¡1A.

So, if A =

·3 ¡12 4

¸, then ¡A =

· ¡1 £ 3 ¡1 £¡1¡1 £ 2 ¡1 £ 4

¸=

· ¡3 1¡2 ¡4

¸Thus ¡A is obtained from A by simply reversing the signs of each element of A.

Notice that the addition of a matrix and its negative always produces a zero matrix. For

example, ·3 ¡12 4

¸+

· ¡3 1¡2 ¡4

¸=

·0 00 0

¸

Thus, in general, A + (¡A) = (¡A) + A = O.

Compare our discoveries about matrices so far with ordinary algebra.

Note: We will assume the matrices have the same order.

NEGATIVE MATRICES

MATRIX ALGEBRA FOR ADDITION

Ordinary algebra

² If a and b are real numbers then

a + b is also a real number.

² a + b = b + a

² (a + b) + c = a + (b + c)

² a + 0 = 0 + a = a

² a + (¡a) = (¡a) + a = 0

Matrix algebra

² If A and B are matrices then

A + B is also a matrix.

² A + B = B + A

² (A + B) + C = A + (B + C)

² A + O = O + A = A

² A + (¡A) = (¡A) + A = O

Explain why it is true that:

a if X + A = B then X = B ¡ A b if 3X = A then X = 13A

a if X + A = B

then X + A + (¡A) = B + (¡A)

) X + O = B ¡ A

i.e., X = B ¡ A

b if 3X = A

then 13 (3X) = 1

3A

) 1X = 13A

) X = 13A

Example 5



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Notice that the rules for addition (and subtraction) of matrices are identical to those of real

numbers but we must be careful with scalar multiplication in matrix equations.

1 Simplify:

a A + 2A b 3B ¡ 3B c C ¡ 2C

d ¡B + B e 2(A + B) f ¡(A + B)

g ¡(2A ¡ C) h 3A ¡ (B ¡ A) i A + 2B ¡ (A ¡ B)

2 Find X in terms of A, B and C if:

a X + B = A b B + X = C c 4B + X = 2C

d 2X = A e 3X = B f A ¡ X = B

g 12X = C h 2(X + A) = B i A ¡ 4X = C

3 a If M =

·1 23 6

¸, find X if 1

3X = M.

b If N =

·2 ¡13 5

¸, find X if 4X = N.

c If A =

·1 0¡1 2

¸, and B =

·1 4¡1 1

¸, find X if A ¡ 2X = 3B.

Suppose you go to a shop and purchase 3 soft drink cans,

4 chocolate bars and

2 icecreams

and the prices are soft drink cans

$1:30

chocolate bars

$0:90

ice creams

$1:20

Each of these can be represented using matrices,

i.e., A =

24 342

35 and B =£

1:30 0:90 1:20¤:

The total cost of the purchase is: $1:30 £ 3 + $0:90 £ 4 + $1:20 £ 2 = $9:90:

We can also determine this from the matrix multiplication

BA =£

1:30 0:90 1:20¤ 24 3

42

35= (1:30 £ 3) + (0:9 £ 4) + (1:20 £ 2)

= 3:90 + 3:60 + 2:40

= 9:90

EXERCISE 10B.2

MATRIX MULTIPLICATIONC



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Notice that we write the row matrix first and the column matrix second

and that£a b c

¤ 24 pqr

35 = ap + bq + cr:

1 Determine:

a b c£3 ¡1

¤· 54

¸ £1 3 2

¤24 517

35 £6 ¡1 2 3

¤266410¡14

3775

2 Show that the sum of w, x, y and z is given by£w x y z

¤26641111

3775 :

Represent the average of w, x, y and z in the same way.

3 Lucy buys 4 shirts, 3 skirts and 2 blouses costing $27, $35 and $39 respectively.

a Write down a quantities matrix Q and a price matrix P.

b Show how to use P and Q to determine the total cost.

4 In the interschool swimming carnival a first place is awarded 10 points, second place 6points, third place 3 points and fourth place 1 point. One school won 3 first places, 2seconds, 4 thirds and 2 fourths.

a Write down this information in terms of a points matrix P and a numbers matrix N.

b Show how to use P and N to determine the total number of points awarded to the

school.

Now consider more complicated matrix multiplication.

In Example 1 Lisa needed 2 loaves of bread, 3 litres

of milk and 1 tub of butter.

We represent this by the quantities matrix£2 3 1

¤.

To find the total cost Lisa needs to multiply the

number of items by their respective cost.

In Store A a loaf of bread is $2:65, a litre of milk is $1:55 and a tub of butter is $2:35, so

the total cost is2 £ $2:65 + 3 £ $1:55 + 1 £ $2:35 = $12:30

In Store B a loaf of bread is $2:25, a litre of milk is $1:50 and a tub of butter is $2:20, so

the total cost is2 £ $2:25 + 3 £ $1:50 + 1 £ $2:20 = $11:20

To do this using matrices, notice that:£2 3 1

¤ £24 2:65 2:25

1:55 1:502:35 2:20

35 =£

12:30 11:20¤

orders: 1 £ 3 3 £ 2 1 £ 2

EXERCISE 10C.1

�



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Now suppose Lisa’s friend Sam needs 1 bread, 2 milk and 2 butter.

The quantities matrix for both Lisa and Sam would be

·2 3 11 2 2

¸Lisa

Sam

bread milk butter

Lisa’s total cost at Store A is $12:30 and at Store B is $11:20

Sam’s total cost at Store A is 1 £ $2:65 + 2 £ $1:55 + 2 £ $2:35 = $10:45

Store B is 1 £ $2:25 + 2 £ $1:50 + 2 £ $2:20 = $9:65

We are now ready to give a formal definition of a matrix product.

The product of an m£ n matrix A with an n£ p matrix B is the m£ p matrix

AB. The element in the rth row and cth column of AB is the sum of the products of the

elements in the rth row of A with the corresponding elements in the cth column of B.

For example,

if A =

·a bc d

¸and B =

·p qr s

¸, then AB =

·ap + br aq + bscp + dr cq + ds

¸,

and if C =

·a b cd e f

¸2 £ 3

and D =

24 xyz

353 £ 1

, then CD =

·ax + by + czdx + ey + fz

¸.

2 £ 1

So, using matrices we require that

·2 3 11 2 2

¸£

24 2:65 2:251:55 1:502:35 2:20

35 =

·12:30 11:2010:45 9:65

¸2 £ 3 3 £ 2

the same

resultant matrix

row 1 column 1"

row 2 column 1"

row 1 column 2"

row 2 column 2"

MATRIX PRODUCTS

If A =£

1 3 5¤

, B =

24 247

35 and C =

24 1 02 31 4

35find: a AB b AC

a A is 1 £ 3 and B is 3 £ 1 ) AB is 1 £ 1

AB =£

1 3 5¤ 24 2

47

35 = [1 £ 2 + 3 £ 4 + 5 £ 7]

= [49]

�

Example 6



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�

b A is 1 £ 3 and C is 3 £ 2 ) AC is 1 £ 2

AC =£

1 3 5¤ 24 1 0

2 31 4

35 = [1 £ 1 + 3 £ 2 + 5 £ 1 1 £ 0 + 3 £ 3 + 5 £ 4]

= [12 29]

1 Explain why AB cannot be found for A =£

4 2 1¤

and B =

·1 2 10 1 0

¸.

2 If A is 2 £ n and B is m£ 3:

a Under what condition can we find AB?

b If AB can be found, what is its order?

c Why can BA never be found?

3 a For A =

·2 13 4

¸and B =

£5 6

¤, find BA.

b For A =£

2 0 3¤

and B =

24 142

35 find i AB ii BA.

4 Find: a b£1 2 1

¤ 24 2 3 10 1 01 0 2

35 24 1 0 ¡1¡1 1 00 ¡1 1

3524 234

355 At the Royal Show, tickets for the Ferris wheel are

$12:50 per adult and $9:50 per child. On the first

day of the show 2375 adults and 5156 children ride

this wheel. On the second day the figures are 2502adults and 3612 children.

a Write the costs matrix C as a 2 £ 1 matrix and

the numbers matrix N as a 2 £ 2 matrix.

b Find NC and interpret the resulting matrix.

c Find the total income for the two days.

6 You and your friend each go to your local hardware

stores, A and B respectively. You want to buy 1hammer, 1 screwdriver and 2 cans of white paint,

and your friend wants 1 hammer, 2 screwdrivers and

3 cans of white paint. The prices of these goods are:

Hammer Screwdriver Can of paint

Store A $7 $3 $19

Store B $6 $2 $22

a Write the requirements matrix R as a 3 £ 2 matrix.

EXERCISE 10C.2



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b Write the prices matrix P as a 2 £ 3 matrix.

c Find PR.

d What are your costs at store A and your friend’s costs at store B?

e Should you buy from store A or store B?

7 A greengrocer at the market sells 6 boxes of apples, 7 boxes of bananas and 9 boxes of

oranges. The next day, 5 boxes of apples, 8 boxes of bananas and 4 boxes of oranges

are sold. On the third day, 4 boxes of apples, 7 of bananas and 2 of oranges are sold.

The apples cost $18 a box, bananas $15 a box and oranges $13 a box over the 3-day

period.

Express this information in the form of two matrices and show how to use the matrices

to find the total cost of the fruit.

Click on the icon for your calculator to assist you to enter and

perform operations on matrices.

Click on the icon to obtain printable instructions on how to

use a speadsheet to perform operations with matrices.

1 Use technology to find:

a b24 13 12 4

11 12 87 9 7

35+

24 3 6 112 9 83 13 17

35 24 13 12 411 12 87 9 7

35¡24 3 6 11

2 9 83 13 17

35c d

22

24 1 0 6 8 92 7 4 5 08 2 4 4 6

352664

2 6 0 73 2 8 61 4 0 23 0 1 8

37752664

45611

3775Use technology to assist in solving the following problems:

2 For their holiday, Frank and Jean are planning to spend time at a popular tourist resort.

They will need accommodation at one of the local motels and they are not certain how

long they will stay. Their initial planning is for three nights and includes three breakfasts

and two dinners. They have gathered prices from three different motels.

The Bay View has rooms at $125 per night. A full breakfast costs $22 per person (and

therefore $44 for them both). An evening meal for two usually costs $75 including

drinks.

By contrast, ‘The Terrace’ has rooms at $150 per night, breakfast at $40 per double and

dinner costs on average $80.

USING A GRAPHICS CALCULATOR FOR MATRIX OPERATIONS

USING A SPREADSHEET FOR MATRIX OPERATIONS

TI

C

SPREADSHEET

EXERCISE 10C.3



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Things seem to be a little better at the Staunton Star Motel. Accommodation is $140 per

night, full breakfast (for two) is $40, while an evening meal for two usually costs $65.

a Write down a ‘numbers’ matrix as a 1 £ 3 row matrix.

b Write down a ‘prices’ matrix in 3 £ 3 form.

c Use matrix multiplication to establish total prices for each venue.

d Instead of the couple staying three nights, the alternative is to spend two nights.

In that event Frank and Jean decide on having breakfast just once and one evening

meal before moving on. Recalculate prices for each venue.

e Remake the ‘numbers’ matrix (2£3) so that it includes both scenarios. Recalculate

the product with the ‘prices’ matrix.

3 A bus company runs four tours. Tour A costs $125, Tour B costs $315, Tour C costs

$405, and Tour D costs $375. The numbers of clients they had over the summer period

are shown in the table below.

Tour A Tour B Tour C Tour D

NovemberDecemberJanuary

February

2664 50 42 18 6565 37 25 82120 29 23 7542 36 19 72

3775Use the information and matrix methods to find the total income for the tour company.

4

Mon Tues Wed Thurs Fri Sat

Beer 225 195 215 240 352 321Wine 75 62 50 92 80 97Spirits 62 54 55 72 102 112

Soft drinks 95 60 68 85 115 146

Write the information in a suitable matrix.

The cost price per drink averages as shown:

Cost price (in $) Beer Wine Spirits Soft drinks£1:95 2:10 1:45 0:95

¤The selling price for this data is:

Selling price (in $) Beer Wine Spirits Soft drinks£2:55 4:40 3:50 1:80

¤Use matrix methods to calculate the profit for the business for the week.

5 The Oregon Motel has three types of suites for guests.

Standard suites cost $125 per night. They have 20 suites.

Deluxe suites cost $195 per night. They have 15 suites.

Executive suites cost $225 per night. They have 5 suites.

The rooms which are occupied also have a maintenance cost:

A hotel mainly sells beer, wine, spirits and soft drinks. The number of these drinks sold

during a week is shown in the table below.



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Standard suites cost $85 per day to maintain.

Deluxe suites cost $120 per day to maintain.

Executive suites cost $130 per day to maintain.

The hotel has confirmed room bookings for the next week:

M T W Th F S Su

Standard

Deluxe

Executive

24 15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

35a The profit per day is given by

(income from room) £ (bookings per day)

¡ (maintenance cost per room) £ (bookings per day)

Create the matrices required to show how the profit per week can be found.

b How would the results alter if the hotel maintained (cleaned) all rooms every day?

Show calculations.

c Produce a profit per room matrix and show how a could be done with a single

matrix product.

In the following exercise we should discover the properties of 2 £ 2 matrix multiplication

which are like those of ordinary number multiplication, and those which are not.

1 For ordinary arithmetic 2 £ 3 = 3 £ 2 and in algebra ab = ba:

For matrices, is AB = BA always?

Hint: Try A =

·1 01 2

¸and B =

· ¡1 10 3

¸.

2 If A =

·a bc d

¸and O =

·0 00 0

¸find AO and OA.

3 For all real numbers a, b and c it is true that a(b + c) = ab + ac and this is known

as the distributive law.

a ‘Make up’ three 2 £ 2 matrices A, B and C and verify that

A(B + C) = AB + AC.

b Now let A =

·a bc d

¸, B =

·p qr s

¸and C =

·w xy z

¸and prove that in general A(B + C) = AB + AC.

c Use the matrices you ‘made up’ in a to verify that (AB)C = A(BC)

d As in b, prove that (AB)C = A(BC)

4 a If

·a bc d

¸ ·w xy z

¸=

·a bc d

¸i.e., AX = A,

deduce that w = z = 1 and x = y = 0.

SOME PROPERTIES OF MATRIX MULTIPLICATION

EXERCISE 10C.4



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b For any real number a, it is true that a £ 1 = 1 £ a = a.

Is there a matrix, I say, such that AI = IA = A for all 2 £ 2 matrices A?

5 Suppose A2 = AA, i.e., A multiplied by itself, and that A3 = AAA.

a Find A2 if A =

·2 13 ¡2

¸b Find A3 if A =

·5 ¡12 4

¸:

6 a If A =

24 1 23 45 6

35 try to find A2.

b When can A2 be found, i.e., under what conditions can

we square a matrix?

7 Show that if I =

·1 00 1

¸then I2 = I and I3 = I.

I =

·1 00 1

¸is the 2 £ 2 identity matrix, whereas

I =

24 1 0 00 1 00 0 1

35 is the 3 £ 3 identity matrix.

You should have discovered from the above exercise that:

Ordinary algebra Matrix algebra

²

² ab = ba for all a, b

² a0 = 0a = 0 for all a

² a(b + c) = ab + ac

(a + b)c = ac + bc

² (ab)c = a(bc)

² a £ 1 = 1 £ a = a

² an exists for all a > 0

²

² In general AB 6= BA.

² If O is a zero matrix then

AO = OA = O for all A.

² A(B + C) = AB + AC

(A + B)C = AC + BC

² (AB)C = A(BC)

² If I =

·1 00 1

¸then AI = IA = A

for all 2 £ 2 matrices A.

² An can be determined provided that A

is a square matrix and n is an integer.

If and are matrices that can be

multiplied then is also a matrix.

A B

AB

If and are real numbers then

so is .

a bab

12A is written not

There is no suchthing as divisionwith matrices.

A

2



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Expand and simplify where possible:

a (A + 2I)2 b (A ¡ B)2 fI is the identity matrixg

a (A + 2I)2

= (A + 2I)(A + 2I) fX2 = XX by definitiong= (A + 2I)A + (A + 2I)2I fA(B + C) = AB + ACg= A2 + 2IA + 2AI + 4I2 f(A + B)C = AC + BC again, twiceg= A2 + 2A + 2A + 4I fAI = IA = A and I2 = Ig= A2 + 4A + 4I

b (A ¡ B)2

= (A ¡ B)(A ¡ B) fX2 = XX by definitiong= (A ¡ B)A ¡ (A ¡ B)B

= A2 ¡ BA ¡ AB + B2

Note: b cannot be simplified further as in general AB 6= BA.

8 Given that all matrices are 2 £ 2 and I is the identity matrix, explain and simplify:

a A(A + I) b (B + 2I)B c A(A2 ¡ 2A + I)

d A(A2 + A ¡ 2I) e (A + B)(C + D) f (A + B)2

g (A + B)(A ¡ B) h (A + I)2 i (3I ¡ B)2

If A2 = 2A + 3I, find A3 and A4 in the form kA + lI, where k and lare scalars.

A2 = 2A + 3I

) A3 = A £ A2

= A(2A + 3I)

= 2A2 + 3AI

= 2(2A + 3I) + 3A

= 4A + 6I + 3A

= 7A + 6I

and A4 = A £ A3

= A(7A + 6I)

= 7A2 + 6AI

= 7(2A + 3I) + 6A

= 14A + 21I + 6A

= 20A + 21I

9 a If A2 = 2A ¡ I, find A3 and A4 in the form kA + lI, where k and l are

scalars.

b If B2 = 2I ¡ B, find B3, B4 and B5 in linear form.

c If C2 = 4C ¡ 3I, find C3 and C5 in linear form.

10 a If A2 = I, simplify:

i A(A + 2I) ii (A ¡ I)2 iii A(A + 3I)2

b If A3 = I, simplify A2(A + I)2.

Example 7

Example 8



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c If A2 = O, simplify:

i A(2A ¡ 3I) ii A(A + 2I)(A ¡ I) iii A(A + I)3

11 The result “if ab = 0 then a = 0 or b = 0” for real numbers does not have an

equivalent result for matrices.

a If A =

·1 00 0

¸and B =

·0 00 1

¸find AB.

This example provides us with evidence that

“If AB = O then A = O or B = O” is a false statement.

b If A =

"12

12

12

12

#determine A2.

c Comment on the following argument for a 2 £ 2 matrix A:

It is known that A2 = A, ) A2 ¡ A = O

) A(A ¡ I) = O

) A = O or A ¡ I = O

) A = O or I

d Find all 2 £ 2 matrices A for which A2 = A. [Hint: Let A =

·a bc d

¸.]

12 Give one example which shows that “if A2 = O then A = O” is a false statement.

Find constants a and b such that A2 = aA + bI for A equal to

·1 23 4

¸:

If A2 = aA + bI

then

·1 23 4

¸ ·1 23 4

¸= a

·1 23 4

¸+ b

·1 00 1

¸)

·1 + 6 2 + 83 + 12 6 + 16

¸=

·a 2a3a 4a

¸+

·b 00 b

¸·

7 1015 22

¸=

·a + b 2a3a 4a + b

¸Thus a + b = 7 and 2a = 10

) a = 5 and b = 2

Checking for consistency

3a = 3(5) = 15 X

4a + b = 4(5) + (2) = 22 X

13 Find constants a and b such that A2 = aA + bI for A equal to:

a b·

1 2¡1 2

¸ ·3 12 ¡2

¸

Example 9



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INVESTIGATION 1 DOMINANCE MATRICES

14 If A =

·1 2¡1 ¡3

¸, find constants p and q such that A2 = pA + qI.

a Hence write A3 in linear form rA + sI, where r and s are scalars.

b Also write A4 in linear form.

In round one matches of the tournament 1 plays 64, 2 plays 63, 3 plays 62, etc.

So, how are the original rankings determined?

We will consider simple ‘round robin’ events where each player plays every other player

in the school’s A golf team.

Suppose the players are A, B, C and D. In a round robin

event: B beat A, C and D; C beat A and D, A beat D.

The results can be displayed on a directed graph as

shown. A ! B means A defeated B.

Now, if we let 1 represent a win and 0 a non-win we can construct a dominance matrix.

On this occasion the dominance matrix is

loser

A B C D

winner

A

B

C

D

26640 0 0 11 0 1 11 0 0 10 0 0 0

3775 0’s are on the main diagonal as

players cannot play themselves.

We now add the elements in each row to create a dominance vector

26641320

3775A

B

C

DSo, clearly the rankings are B, C, A, D, as expected.

Note:2664

0 0 0 11 0 1 11 0 0 10 0 0 0

37752664

1111

3775 =

26641320

3775 can be used to find the dominance vector.

Now consider a more complicated event with directed graph, dominance matrix and vector:

The world’s best professional golfers

are ranked from to , the rankings

being continuously updated. A computer is

necessary to handle the addition and

multiplication of matrices.

The rankings may be used by tournament organisers to

‘seed’ the players who have entered a matchplay event.

Suppose that in a player knockout event, the highest

ranked player is seeded number , the next is number

and so on and the last is seeded .

10001 1000

1000 1000

641 264

£

D B

A

C



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E beat BE beat C

loser

A B C D E

winner

A

B

C

D

E

2666640 0 0 1 11 0 0 0 01 1 0 1 00 1 0 0 10 1 1 0 0

377775 = M and V =

26666421322

377775

The rankings are: C,

A

D

E

, B which is not really satisfactory as A, D and E are equally

ranked.

It is clear that we need to separate A, D and E and to do this we find M2.

Now M2 =

2666640 2 1 0 10 0 0 1 11 1 0 1 21 1 1 0 02 1 0 1 0

377775where the shaded element is the result of

£0 1 1 0 0

¤266664

01100

377775B beat A

C beat A

So, the 2 is a result of E beating B and B beating A

E beating C and C beating A

which are ‘second order’ influences of E over A.

The matrix M + 12M2 is often used to help sort equally ranked individuals.

The 12 is arbitrary, but second order influences are generally reduced (given less weight)

when added to the first order influences.

In this case M + 12M2 =

266666640 1 1

2 1 112

1 0 0 12

12

112 11

2 0 112 1

12 11

212 0 1

1 112 1 1

2 0

37777775 with dominance vector

266666644

2

512

312

4

37777775So, the rankings are now: C,

A

E, D, B.

To further split any equal rankings M + 12M2 + 1

3M3 could be used.

If we are still unable to separate A and E, we might consider

M + 12M2 + 1

3M3 + 14M4 or else toss a coin.

1 Alan, Bob, Colin and David play each other in a round robin

table tennis competition. The directed graph showing their

match results is drawn alongside.

a Find the dominance matrix M.

What to do:

A E

B

D

C

D B

A

C



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INVESTIGATION 2 NETWORK MATRICES

b Can the players be ranked using M only?

c Find M + 12M2 and use it to fully rank the players.

2 Rank the players in the following round robin tournaments:

a b

3 Rank the players in a team at your school according to their win/loss records in a

round robin competition.

to

A B C D

from

A

B

C

D

26640 1 0 00 0 1 01 1 0 10 1 1 0

37751 Construct network matrices for:

a b

2 Draw a network diagram which corresponds to the network matrix:

a b26640 0 1 01 0 0 11 0 0 00 1 1 0

3775266664

0 1 0 1 01 0 1 0 10 1 0 1 11 0 0 0 10 1 1 0 0

377775

P

QT

RS

A B

C

DE

F

Consider the roadways which connect the

four locations A, B, C and D. Single

arrowheads indicate travel in one

direction, i.e., one-way streets. Double

arrow heads indicate two-way travel, i.e., travel in

both directions.

Suppose represents a connection from one place to

another which is possible, whereas represents

impossible.

The corresponding is therefore

10

network matrix

A

B

CD

A

B

D C

B

C

DE

A

What to do:



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3 26640 1 1 01 0 0 01 1 0 00 1 0 0

3775 is the matrix which shows the existing roadways

between four towns A, B, C and D.

a If new routes are to be added so that every town is connected to every other

town, what matrix represents the final situation?

b What matrix represents the additional routes required?

c How are the three matrices connected?

4 Suppose we have three towns P, Q and R. The networks for the bus and train services

are:

a Draw a network diagram for the combined bus and train service using different

colours.

b Find the network matrix for the bus service (B) and the train service (T).

c Find matrix BT.

d Find the matrix of possible journeys of a bus ride followed by a train ride,

i.e., P Q R

P

Q

R

24 0

1

35 no combined trips go from P to R

by bus first and then by train

one combined trip goes from R and

back to R by bus then train.

e What do you notice about c and d?

f Find the matrices TB, BTB and B2. What trips do these matrices represent?

5 Repeat 4 for four towns A, B, C and D with networks for bus and train services:

We can use matrices to represent the state of many systems.

For example, if Adam has 7 marbles, Bianca has 4 marbles and Carly has 9 marbles, the

situation can be represented by the state matrix £7 4 9

¤Adam’s Bianca’s Carly’s

bus train

P Q

R

P Q

R

B

C

A

Dbus train

B

C

A

D

TRANSITION MATRICESD

Some systems may change over time. For example, the three children could compete to win

marbles from each other, the populations of nearby towns could fluctuate with time, and so on.



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The following problem involves the buying trends of icecream customers. We will analyse

the market on a weekly basis.

The proportions of a market that businesses have will be shown in a state matrix which we

denote S.

The original market distribution we call S0, the initial state matrix.

For example, the brand Ace makes icecream. If it currently has 63% of the market share,

then we can represent the system by the initial state matrix

S0 =£

0:63 0:37¤

Subscript 0 indicates Ace Other

the initial state after brands

0 weeks

A transition matrix is used to model how the system changes in time.

We usually denote the transition matrix by T.

By analysing the market, Ace determines the following information:

The transition matrix is

Ace Other

T =Ace

Other

·0:8 0:20:3 0:7

¸The entries tell us how the market varies from week to week, i.e., the transition from week

to week, but not how much is actually purchased.

Notice that the sum of each row is 1. This is because everyone who buys icecream either

buys Ace or another brand, and the total of the proportions of the market is 100% or 1.

We will now use elements of the initial state matrix S0 and the transition matrix T to

determine market shares in the future.

After one week, the proportion of people who will be buying Ace is 80% of the 63% who

currently buy Ace, plus 30% of the 37% who currently do not.

This is 0:8 £ 0:63 + 0:3 £ 0:37 = 0:615

The remainder will buy another brand, so this proportion is 0:385 :

Notice that S0T =£

0:63 0:37¤ · 0:8 0:2

0:3 0:7

¸=£

0:615 0:385¤:

This confirms that after 1 week the market share for Ace is 61:5% which is a decrease from

63%, and this is how we determine the state after one week using the transition matrix.

We label it S1 and we can use it to find the state matrix after 2 weeks, S2, the state matrix

after 3 weeks S3, and so on.

Buy

now

Buy next week

Ace Other

Ace 0:8 0:2

Other 0:3 0:7

Of those who buy Ace

brand this week, 80% will

buy Ace brand next week.

Of those who buy another


buy Ace brand next week.

Of those who buy Ace brand

this week, 20% will buy

another brand next week.

Of those who buy another


buy another brand next week.



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After two weeks, S2 = S1T =£

0:615 0:385¤ · 0:8 0:2

0:3 0:7

¸=£

0:6075 0:3925¤

so after two weeks, 60:75% will buy Ace and 39:25% will buy another brand.

After three weeks, S3 = S2T =£

0:6075 0:3925¤ · 0:8 0:2

0:3 0:7

¸=£

0:6038 0:3963¤

After four weeks, S4 = S3T =£

0:6038 0:3963¤ · 0:8 0:2

0:3 0:7

¸=£

0:6019 0:3982¤

After five weeks, S5 = S4T =£

0:6019 0:3982¤ · 0:8 0:2

0:3 0:7

¸=£

0:6010 0:3991¤

the market share for the market share for

Ace has decreased the other brands has

to almost 60% increased to almost 40%

Notice that there is very little change from S4 to S5. The proportions are converging to 0:6for Ace and 0:4 for the Other brands.

When there is very little change in the values of the state matrices from one state to the next

then we say we have reached steady state.

A steady state is indicated when Sn+1 + Sn for sufficiently large values of n.

Now observe the increasing powers of the transition matrix T.

Notice that for T =

·0:8 0:20:3 0:7

¸, T2 =

·0:70 0:300:45 0:55

¸,

T4 =

·0:625 0:3750:5625 0:4375

¸,

T8 +

·0:6016 0:39840:5977 0:4023

¸,

and T16 +

·0:6000 0:40000:6000 0:4000

¸.

Notice that the rows in T16 are identical to four decimal places, and that each row gives the

Most importantly, this will be the steady state of the system irrespective of what proportion

of the market Ace has originally.

Hence, we have two ways of finding the steady state of a system in transition:

² by continued multiplication of the state matrix Sn by the transition matrix T, until

there is very little difference between one state and the next, i.e., Sn+1 + Sn

² by determining Tn for large n.

To find the state matrix Sn for a particular value of n, notice that for the initial state of S0

and transition matrix T,

Values in column are

converging to

Values in column are

converging to

10 62

0 4

:

:

steady state proportions for Ace brand and the Other brands of icecream.



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S1 = S0T

S2 = S1T = S0TT = S0T2

S3 = S2T = S0T2T = S0T3

S4 = S3T = S0T3T = S0T4....... and so on

In general, the state of the system after n weeks will be Sn = S0Tn:

Note: Steady state can also be found using algebra.

If we suppose the steady state is S1 =£a b

¤, then since the state matrix does not

change from week to week, S1 = S1T.

)£a b

¤ · 0:8 0:20:3 0:7

¸=£a b

¤and so

£0:8a + 0:3b 0:2a + 0:7b

¤=£a b

¤Hence, 0:8a + 0:3b = a and 0:2a + 0:7b = b .

Looking at either one of these equations we find 2a = 3b and so b = 23a.

At the local high school, 30% of students who bring their lunch from home on one

day will buy their lunch at the canteen the next day. 50% of students who buy their

lunch at the canteen on one day will bring their lunch from home the next.

a Construct a transition matrix T for this situation.

On Monday, 400 students from the school were surveyed. 243 brought their lunch

from home, while the remainder bought their lunch at the canteen.

b Construct an initial state matrix S0 for this situation.

c How many students do you expect to buy their lunch from the canteen on

i Tuesday ii Friday?

d Find the steady state proportion of students who buy their lunch at the canteen.

a

Lunch

today

Lunch tomorrow

Home Canteen

Home 70% 30%

Canteen 50% 50%

So the transition matrix T =

·0:7 0:30:5 0:5

¸b The initial state matrix is S0 =

£243 157

¤c i The state matrix for Tuesday is

S1 = S0T =£

243 157¤ · 0:7 0:3

0:5 0:5

¸=£

248:6 151:4¤

Rounding to the nearest student, 151 students are expected to buy their

lunch at the canteen on Tuesday.

Example 10

home canteen

Since a + b = 1, a + 23a = 1 or 5

3a = 1. Hence, a = 35 = 0:6 and b = 0:4



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ii The state matrix for Friday is S4 = S0T4

Using a calculator,

S4 +£

249:99 150:01¤

d

The steady state proportion of students buying their lunch at the canteen

is 0:375 or 37:5%:

1 A country town has two fish-and-chip

shops: shop X, a well established shop, and

a new shop, Y. The following buying pat-

terns have been observed:This

week

Next week

Shop X Shop Y

Shop X 90% 10%

Shop Y 20% 80%

a Assuming the choice of shop is based entirely on the choice in the previous week,

construct a transition matrix T with elements in decimal form.

b State the exact meaning of i the 0:2 element of T ii the 0:8 element of T.

c Initially shop Y has no share of the market and so the initial state matrix S0 is£1 0

¤. Calculate S0 and give meaning to the elements in this matrix.

d Find shop Y’s market share after i two weeks ii five weeks.

e Calculate S19 and S20 correct to three decimal places, and hence indicate the steady

state proportions of market share for shops X and Y.

2 Two brands of cheese made from sheeps’

milk are available. The following buying

patterns have been recorded:

Brand bought

this time

Brand boughtnext time

Baaah Sheez

Baaah 0:7 0:3

Sheez 0:4 0:6

a State the meaning of the 0:4 in the

table.

b Construct a transition matrix T1.

c If the market shares for the first period are given by the row matrix£

0:8 0:2¤,

find T1

£0:8 0:2

¤and explain what it represents.

d Find Baaah’s market share for:

i the third buying period ii the sixth buying period.

e Use powers of T to find the steady state proportions of market share for brands

Baaah and Sheez.

The steady state

proportions can be

found by examining

as increases. Using

the calculator we

calculate and

T

T T

n

n

:10 20

150 students are expected to buy their

lunch at the canteen on Friday.

EXERCISE 10D

T



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f Baaah manufacturers embark on an advertising campaign to gain a greater portion of

the market. Their promotion produces new buying patterns which, given in matrix

form, are T2 =

·0:8 0:20:5 0:5

¸.

At the commencement of the campaign, the market shares were£

0:57 0:43¤.

What will be the new market shares after:

i one buying period ii two buying periods iii four buying periods?

g What will be the new steady state for market shares following the advertising period?

3 Two bus services, Clydes and Roos, operate between two cities. The following transition

matrix shows proportions of passengers who use these services on a monthly basis.

T = Now

Next month

C R

C

R

·0:84 0:160:21 0:79

¸a Explain the meaning of the entry i 0:84 ii 0:16

b If this month Clydes carry 425 passengers and Roos carry 716, how many passengers

will each bus service have: i next month ii the month after next?

c Find the steady state and describe what it means.

d If the transition matrix does not change, predict the monthly number of passengers

for each bus service.

e How reliable is the answer in d?

4 A group of 1000 smokers is attempting to quit. They

meet for a barbeque on the first Saturday of each month

to discuss their progress and encourage each other. In

January, 620 members of the group were still smoking,

while the rest of the group had gone the whole of the

previous month without.

Statistics indicate that in such situations some of the

smokers will give up in the next month, while some of

those who had gone without will return to the habit. The

proportions are described as follows:

next month

Smoking Not Smoking

this month Smoking 80% 20%

Not Smoking 10% 90%

a Show that 534 members are expected to have smoked between the January and

February meetings.

b Suppose S =£

620 380¤

and A =

·0:8 0:20:1 0:9

¸.

i Evaluate SA and hence estimate how many members will not have smoked

between the January and February meetings.

ii Evaluate SA2 and interpret the result.

c How many members do you expect to be smoking in January the following year?

d Suppose at a given time that s members of a group smoke and n do not.



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Let X =£s n

¤. i Find a relationship between s and n if X = XA.

ii What is the significance of this result?

5

State when they leave

Open Closed

State when Open 30% 70%

they arrive Closed 5% 95%

Suppose S =£

0:1 0:9¤

and A =

·0:3 0:70:05 0:95

¸a Explain the significance of S and A.

b Evaluate SA and hence determine the likelihood of the fridge being left open after

the next visitor.

c Evaluate (SA)A and interpret the result.

d Use SA10 to predict what proportion of the time the fridge will be left open in the

long term. Is Jess’ concern that the problem is getting worse justified?

e Suppose that at any given time, the probability of the fridge being open is x and

the probability of it being closed is y.

i Explain why x + y = 1.

ii Let X =£x y

¤. Determine x and y if X = XA.

iii Interpret the result in ii.

situation

this year

situation next year

very good usable very poor

very good 0:91 0:06 0:03

usable 0:30 0:65 0:05

very poor 0:02 0:28 0:70

a Write down the transition matrix T.

b Explain the meaning in the matrix of: i 0:91 ii 0:03 iii 0:28 .

c Find the matrix T2 and explain the meaning of the figure in:

i row 2, column 1 ii row 3, column 3.

d The initial land quality matrix is£

0 0:1 0:9¤. Explain what this means.

e What is the land quality after i one year ii two years iii four years?

f What is the steady state? Describe what it means.

An ambitious plan is to reverse the effect of salt damage on land adjacent to the

River Murray. Scientists examined techniques to improve land quality in the US and

believe that the same techniques will result in land improvement in South Australia.

Data was collected and examined from a South Australian property. The following

table compares the situation in the current year with the next year.

Example 11

Jess has noticed that some people keep leaving the fridge door at her work partly open.

She estimates that the door is currently left open of the time, and the

problem is getting worse. She has noticed that the state in which a person leaves the

fridge is dependent on the state in which they found it, as described in the table below:

10% she suspects



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a T =

24 0:91 0:06 0:030:30 0:65 0:050:02 0:28 0:70

35b i The 0:91 means that 91% of the land which is very good this year will be

very good next year.

ii The 0:03 means that 3% of the land which is very good this year will be

very poor next year.

iii The 0:28 means that 28% of the land which is very poor this year will be

usable next year.

c T2 =

24 0:91 0:06 0:030:30 0:65 0:050:02 0:28 0:70

352

=

24 0:8467 0:1020 0:05130:4690 0:4545 0:07650:1162 0:3792 0:5046

35i In row 2, column 1 we have 0:4690. This means that 46:90% of land

which is usable now will be very good in two years’ time.

ii In row 3, column 3 we have 0:5046. This means that 50:46% of land

which is very poor now will be very poor in two years’ time.

d£0 0:1 0:9

¤

e i£0 0:1 0:9

¤T =

£0 0:1 0:9

¤ 24 0:91 0:06 0:030:30 0:65 0:050:02 0:28 0:70

35=£0:048 0:317 0:635

¤i.e., after one year, 4:8% is very good, 31:7% is usable, 63:5% is very poor.

ii£0 0:1 0:9

¤T2 =

£0 0:1 0:9

¤ 24 0:8467 0:1020 0:05130:4690 0:4545 0:07650:1162 0:3792 0:5046

35=£0:1515 0:3867 0:4618

¤i.e., after two years, 15:15% is very good, 38:67% is usable, 46:18% is

very poor.

iii£0 0:1 0:9

¤T4 =

£0 0:1 0:9

¤ 24 0:7707 0:1522 0:07710:6192 0:2834 0:09740:3349 0:3755 0:2896

35=£0:3633 0:3663 0:2704

¤i.e., after four years, 36:33% is very good, 36:63% is usable, 27:04% is

very poor.

f Perhaps the easiest of the three methods for determining steady state for higher

than 2 £ 2 transition matrices is to find Tn for large n.

Using the calculator: T100 =

24 0:6952 0:2017 0:10310:6952 0:2017 0:10310:6952 0:2017 0:1031

35The steady state is that 69:52% of the land is very good, 20:17% is usable,

10:31% is very poor.

means that is very good, is usable and is very

poor this year.

0% 10% 90%



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6 A farmer hears of the success of soil restoration techniques and wishes to join the

program. Due to different soil types and conditions his transition matrix is

T =

VG U VP

VG

U

VP

24 0:8 0:1 0:10:2 0:7 0:10:0 0:3 0:7

35 VG = very good

U = usable

VP = very poor

a What is the meaning of:

i the number in row 1, column 3 ii row 3, column 2?

b Find T2 and give the meaning of the number in:


c The present land is all very poor. Write down the initial land quality matrix.

d What is the land quality after 3 years?

e The property next door is subject to the same transition matrix but consists of 21 ha

which is very good, 157 ha which is usable and 428 ha which is very poor. How

many hectares of each category will there be after 4 years?

f What is the steady state for both properties?

(Hint: Find Tn where n is large.)

7 An AFL club’s fitness coach has observed, over a long period, the following:

status

this

week

status next week

fully fit getting treatment cannot play

fully fit 0:88 0:06 0:06

getting treatment 0:75 0:17 0:08

cannot play 0:08 0:42 0:50

a Write down the transition matrix T for the fitness status of the club’s players.

b State the meaning of the number in:


c Find T2 and state the meaning of the number in row 2, column 1.

d If the club has 24 fully fit players, 6 getting treatment and 3 who cannot play due

to injury, how many would we expect in each category:

i next week ii in two weeks’ time iii in five weeks’ time?

e What is the steady state of fitness in the long term?

8

The transition matrix for the fitness status of players is:

T =

24 0:86 0:12 0:020:68 0:24 0:080:00 0:32 0:68

35.

a How many would we expect in each group next week?

b What is the fitness status of the players:

i after two weeks ii in the long term?

A netball squad has players of which currently are fully fit, are getting

treatment and are unavailable due to injury.

40 30 55



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9 There are three islands Paua, Manu and

Chalk, off the northern coast of Queensland.

Each year there is considerable migration

between the islands. Over time an interested

party has observed the trends alongside.

this

year

next year

Paua Manu Chalk

Paua 75% 15% 10%

Manu 20% 60% 20%

Chalk 15% 20% 65%a Find the transition matrix T which shows

how the populations for the islands change from one year to the next.

b Find T2 and explain what the numbers in row 3 represent.

c Calculate T16 and explain what it represents.

d If in 2006 the population of the three islands is split so that 26% live on Paua, 39%live on Manu and 35% live on Chalk, what are the expected populations on each

island in: i 2007 ii 2021?

e What is the state of the islands’ population proportions in the long term?

10 a Solve the system of linear equations:

Give your answer in parametric form.

8x¡ 5y ¡ 10z = 0

4x¡ 5y = 0

y ¡ 2z = 0.

b A new 22 acre plantation of bamboo will be used to feed

the giant pandas at the Beijing zoo. The plantation is ready

for harvesting for the first time, so we call this time zero. To

provide the pandas with the correct diet, at the end of each

year 20% of the 1-year old bamboo, 50% of the 2-year old

bamboo, and all of the 3-year old bamboo will be harvested.

All harvested bamboo will be replanted with fresh shoots

so the area will yield new bamboo the following year.

Suppose M =£

22 0 0¤

and N =

24 0:2 0:8 00:5 0 0:51 0 0

35.

i Discuss the rows of matrix N.

ii Evaluate MN and MN2. Interpret the results.

iii How many acres of bamboo will there be in each age group 10 years after the

first harvest?

c Suppose that at any given time the area of bamboo at each age are given by

X =£x y z

¤.

i Explain why x + y + z = 22.

ii If X = XN, write down a system of equations connecting x, y and z.

iii Find x, y, and z such that X = XN. Interpret this situation.

11 a Consider the system of linear equations: 8a¡ 4b¡ 6c = 0

4a¡ 9b + 4c = 0

4a + 5b¡ 10c = 0,

i Use row operations to show that the system does not have a unique solution.

ii Give solutions to the system of linear equations in parametric form.

b Several different chemicals are used in horses to protect against worms. To ensure

the worms do not build up resistance to the chemicals, the chemical used in any

particular animal should be changed regularly. There are three brands currently on

the market: A, B, and C, with market shares 60%, 30%, and 10% respectively.



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It is estimated the next time horses are wormed,

the chemical used will change according to the

following table:

Next time

A B C

A 20% 40% 40%

This time B 40% 10% 50%

C 60% 40% 0%

i

Australia that are regularly wormed:

(1) How many horses are currently

being wormed using brand A?

(2) How many horses do you expect to be wormed using brand A next

time?

ii Suppose S =£

0:6 0:3 0:1¤

and T =

24 0:2 0:4 0:40:4 0:1 0:50:6 0:4 0

35.

(1) Evaluate ST and hence estimate how many horses will be wormed

with brand C next time.

(2) Evaluate ST2 and interpret the result.

iii Suppose at a given time that brands A, B, and C are being used in proportions

a, b and c respectively. Let X =£a b c

¤.

(1) Given an interpretation of the equation a + b + c = 1.

(2) If X = XT, write down a system of equations connecting a, b and c.

(3) Use part a to find a, b, and c such that X = XT.

(4) What does it mean when X = XT?

The following problems involve the modelling of population growth with time. They are

related to the transition matrices we have already seen because there is a matrix which

controls the change in population from one time to the next.

One difference you will notice is that the populations are usually expressed as column

matrices and we pre-multiply by the transition matrix.

Another difference is that, where previously the rows of the transition matrices added to 1,

the columns of these transition matrices do not. A consequence of this is that the steady state

populations will depend on the initial state.

12 Studies on a colony of penguins show that each year

80% of the female chicks die. At the end of the year

the surviving chicks become mature adults. Each year a

female adult will lay on average 4 eggs, each of which

will have a 50% chance of containing a female chick. A

female adult has a 60% chance of surviving to the next

year and laying once more. The population of chicks

and adults at any given time can be described by the

population matrix

P =

·ca

¸, and the change in the population year by year can be described by the

matrix T =

·0 2

0:2 0:6

¸.

MODELLING POPULATIONS

If there are currently in South100000�



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a Describe the significance of TnP where n is a positive integer.

b Calculate TP and T2P if P =

·250300

¸. Interpret their meaning.

c Calculate T5P and T10P. Predict what will happen to the population over time.

d A freak storm one year means that only a few chicks survive.

The population is left at P =

·20250

¸. Are the numbers likely to recover?

e Consider M =

·0 ab c

¸and P =

·xy

¸. If MP = P, show that ab = 1¡c.

Make a prediction about what would happen to the penguin population if a change

in environmental conditions changed the matrix T into

·0 2

0:2 0:7

¸.

Give evidence to support your answer.

13 Lengthy research into the rare pygmy butterfly has

revealed that after 24 hours, caterpillars will turn

into butterflies, and after 48 hours, the butterflies

are ready to mate. Any female butterflies that survive

this long will give birth and die after 72 hours.

It has also been determined that 34 of the female

caterpillars die before pupating into a butterfly, and23 of the adult female butterflies die before mating.

On average, those female butterflies that survive to mate produce 12 female caterpillar

offspring.

At a particular time there are 120 female caterpillars, 15 adolescent female butterflies,

and 8 adult female butterflies in a colony.

If A =

264 0 0 1214 0 0

0 13 0

375 and B =

24 120158

35:

a

b

c Explain why the numbers of females in each category after 48 hours are predicted

by A2B.

d According to the model, how many females will there be in each category after

i 48 ii 72 hours?

e Predict what will happen to the butterfly population in the future.

f It is known that in another colony of pygmy butterflies, the numbers of females in

each category remain constant.

i If the population is described by X =

24 xyz

35, explain why AX = X.

ii If there are 256 females in the colony, determine x, y and z, the numbers of

females in each category.

f

Explain how matrices and describe the system.A B

Calculate and hence predict the number of female caterpillars, adolescent

butterflies, and adult butterflies after hours.

AB

24



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In Chapter 9, Example 2 we solved

½2x + 3y = 45x + 4y = 17

to get x = 5, y = ¡2:

Notice that this system can be written as a matrix equation ·2 35 4

¸ ·xy

¸=

·417

¸:

The solution x = 5, y = ¡2 is easily checked as·2 35 4

¸ ·5

¡2

¸=

·2(5) + 3(¡2)5(5) + 4(¡2)

¸=

·417

¸X

Notice that these matrix equations have form AX = B where A is the matrix of coefficients,

X is the unknown column matrix and B is the matrix of constants.

The question arises: If AX = B, how can we find X using matrices only?

To answer this question, suppose there exists a matrix C such that CA = I.

If we pre-multiply each side of AX = B by C we get

CAX = CB

) IX = CB

and so X = CB

The matrix C such that CA = I does not always exist. However, if it does exist, we call it

the multiplication inverse of A and denote it C = A¡1.

In general, the multiplication inverse of A, if it exists, satisfies A¡1A = AA¡1 = I.

Notice that

·3 15 2

¸ ·2 ¡1

¡5 3

¸=

·1 00 1

¸= 1I

and that

·1 23 4

¸ ·4 ¡2

¡3 1

¸=

·¡2 00 ¡2

¸= ¡2

·1 00 1

¸= ¡2I

and that

·5 11

¡2 3

¸ ·3 ¡112 5

¸=

·37 00 37

¸= 37

·1 00 1

¸= 37I

In each case we

are multiplying·a bc d

¸by·

d ¡b¡c a

¸

These results suggest that

·a bc d

¸ ·d ¡b

¡c a

¸= kI for some scalar k.

On expanding this product

·ad ¡ bc 0

0 ad ¡ bc

¸= kI

i.e., (ad ¡ bc)I = kI

and so k = ad ¡ bc:

Consequently

·a bc d

¸£ 1

ad ¡ bc

·d ¡b

¡c a

¸=

·1 00 1

¸:

THE INVERSE OF A 2 × 2 MATRIX� �E

INVERSES OF 2 × 2 MATRICES� �



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So, if A =

·a bc d

¸then A¡1 =

1

ad ¡ bc

·d ¡b

¡c a

¸:

Notice that A¡1 exists provided ad ¡ bc 6= 0, otherwise1

ad ¡ bcwould be undefined.

If ad ¡ bc 6= 0, we say that A is invertible.

If A =

·2 35 4

¸, find A¡1 and hence solve

½2x + 3y = 45x + 4y = 17

:

Now in matrix form the system is:·2 35 4

¸ ·xy

¸=

·417

¸i.e., AX = B

) A¡1AX = A¡1B

) IX =1

2 £ 4 ¡ 3 £ 5

·4 ¡3

¡5 2

¸ ·417

¸) X = 1

¡7

· ¡3514

¸=

·5

¡2

¸and so x = 5, y = ¡2.

1 Find, if it exists, the inverse matrix of:

2

3

4 e

y y y

a b c d·

2 4¡1 5

¸ ·1 01 ¡1

¸ ·2 41 2

¸ ·1 00 1

¸e f g h

·3 5

¡6 ¡10

¸ · ¡1 24 7

¸ ·3 4

¡1 2

¸ · ¡1 ¡12 3

¸Perform the following products:

a b·

1 23 4

¸ ·xy

¸ ·2 31 ¡4

¸ ·ab

¸Convert into matrix equations:

a 3x ¡ y = 82x + 3y = 6

b 4x ¡ 3y = 113x + 2y = ¡5

c 3a ¡ b = 62a + 7b = ¡4

Use matrix algebra to solve equations a, b and c using AX = B and equations d,

and f using XA = B:

a 2x ¡ y = 6x + 3y = 14

b 5x ¡ 4y = 52x + 3y = ¡13

c x ¡ 2y = 75x + 3y = ¡2

d 3x + 5y = 42x ¡ = 11

e 4x ¡ 7y = 83x ¡ 5 = 0

f 7x + 11y = 1811x ¡ 7 = ¡11

Example 12

Both sides of the matrixequation are multiplied bythe inverse matrix in the

front or preposition. This iscalled premultiplication,

EXERCISE 10E.1



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5 a Show that if AX = B then X = A¡1B whereas if XA = B then X = BA¡1:

b Find X if:

i X

·1 25 ¡1

¸=

·14 ¡522 0

¸ii

·1 32 ¡1

¸X =

·1 ¡34 2

¸

6 Find A¡1 and state k when A¡1 exists if:

a b cA =

·k 1¡6 2

¸A =

·3 ¡10 k

¸A =

·k + 1 2

1 k

¸

7 a If A =

·1 0 2¡1 1 3

¸and B =

24 ¡1 2¡4 61 ¡1

35, find AB.

b Does your result in a imply that A and B are inverses? [Hint: Find BA.]

The above example illustrates that only square matrices can have inverses. Why?

8 If a matrix A is its own inverse, then A = A¡1.

For example, if A =

·0 ¡1¡1 0

¸then A¡1 =

1

¡1

·0 11 0

¸=

·0 ¡1¡1 0

¸= A.

a Show that if A = A¡1, then A2 = I.

b If

·a bb a

¸is its own inverse, show that there are exactly 4 matrices of this

form.

9 Given A =

·2 10 1

¸, B =

·1 2¡1 0

¸and C =

·0 31 2

¸, find X if AXB = C.

10 a If A =

·1 2¡1 0

¸find A¡1 and (A¡1)¡1.

b If A is any square matrix which has inverse A¡1, simplify (A¡1)¡1 (A¡1) and

(A¡1)(A¡1)¡1 by replacing A¡1 by B.

c What can be deduced from b?

Find A¡1 when A =

·4 k2 ¡1

¸and state k when A¡1 exists.

A¡1 =1

¡4 ¡ 2k

· ¡1 ¡k¡2 4

¸=

26641

2k + 4

k

2k + 4

2

2k + 4

¡4

2k + 4

3775So A¡1 exists provided that 2k + 4 6= 0, i.e., k 6= ¡2:

Example 13



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11 a If A =

·1 12 ¡1

¸and B =

·0 12 ¡3

¸find in simplest form:

i A¡1 ii B¡1 iii (AB)¡1

iv (BA)¡1 v A¡1B¡1 vi B¡1A¡1

b Choose any two invertible matrices and repeat question a.

c What do the results of a and b suggest?

d Simplify (AB)(B¡1A¡1) and (B¡1A¡1)(AB) given that A¡1 and B¡1 exist.

What can you conclude from this?

12 If k is a non-zero number and A¡1 exists, simplify (kA)(1

kA¡1) and (

1

kA¡1)(kA).

What conclusion follows from your results?

13 If X = AY and Y = BZ where A and B are invertible, find:

a X in terms of Z b Z in terms of X.

(Assume X, Y and Z are 2 £ 1 and A, B are 2 £ 2.)

If A2 = 2A + 3I, find A¡1 in linear form rA + sI, where r and s are scalars.

A2 = 2A + 3I

) A¡1A2 = A¡1(2A + 3I) fpremultiply both sides by A¡1g) A¡1AA = 2A¡1A + 3A¡1I

) IA = 2I + 3A¡1

) A ¡ 2I = 3A¡1

) A¡1 = 13 (A ¡ 2I) i.e., A¡1 = 1

3A ¡ 23 I

14 Find A¡1 in linear form given that

a A2 = 4A ¡ I b 5A = I ¡ A2 c 2I = 3A2 ¡ 4A

15 If A =

·3 2

¡2 ¡1

¸, write A2 in the form pA + qI where p and q are scalars.

Hence write A¡1 in the form rA + sI where r and s are scalars.

16 It is known that AB = A and BA = B where the matrices A and B are not necessarily

invertible. Prove that A2 = A. (Note: From AB = A,

you cannot deduce that B = I. Why?)

17 Under what condition is it true that AB = AC ) B = C?

18 If X = P¡1AP and A3 = I, prove that X3 = I.

19 If aA2 + bA + cI = O and X = P¡1AP,

prove that aX2 + bX + cI = O.

Example 14

)means

“implies that”



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If A =

·2 35 4

¸then A¡1 = 1

¡7

·4 ¡3¡5 2

¸or

" ¡47

37

57

¡27

#

It is so easy using the 2 £ 2 inverse matrix formula, so why do we need another method?

Have you thought about how to find inverses of 3 £ 3, 4 £ 4, 5 £ 5 etc. matrices?

A formula for even the 3 £ 3 inverse is extremely complicated and not worth pursuing.

So let us examine the following method:

Start with the augmented matrix£

A I¤

and use row operations to form the matrix I

on the left of the augmented matrix, i.e.,£

I A¡1¤:

Note that R2 ! 2R2 ¡ 5R1 reads ‘replace row 2 by 2£ row 2 ¡ 5 £ row 1’.£A I

¤i.e.,

·2 3 1 05 4 0 1

¸»·

2 3 1 00 ¡7 ¡5 2

¸R2 ! 2R2 ¡ 5R1

»"

1 32

12 0

0 1 57 ¡2

7

#R1 ! 1

2R1

R2 ! ¡17R2

»"

1 0 ¡47

37

0 1 57 ¡2

7

#R1 ! R1 ¡ 3

2R2

» £ I A¡1¤

10 8 0 2¡10 ¡15 ¡5 0

0 ¡7 ¡5 2

1 32

12 0

0 ¡32 ¡15

14614

1 0 ¡47

37

This method will also apply to matrices of higher order.

We can easily find A¡1 using technology.

1 Using the augmented matrix method as shown above, find the inverse of:

a bA =

·1 42 ¡1

¸A =

·3 ¡14 5

¸Check your answers

using technology.

2 Use technology to find the inverse of:

a bA =

·235 ¡176151 318

¸A =

·11:67 8:94¡6:72 3:65

¸

OTHER METHODS FOR FINDING 2 × 2 MATRIX INVERSES� �

EXERCISE 10E.2

TI

C



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In Example 4 of Chapter 9

we used augmented matrices

to show that

8<: x + 3y ¡ z = 152x + y + z = 7x¡ y ¡ 2z = 0

has solution x = 2, y = 4, z = ¡1:

The system has matrix equation

24 1 3 ¡12 1 11 ¡1 ¡2

3524 xyz

35 =

24 1570

35 ,

and the solution checks as24 1 3 ¡12 1 11 ¡1 ¡2

3524 24¡1

35 =

24 1(2) + 3(4) ¡ 1(¡1)2(2) + 1(4) + 1(¡1)1(2) ¡ 1(4) ¡ 2(¡1)

35 =

24 1570

35 X

1 Write as a single matrix:

a b24 1 1 2

1 3 ¡12 ¡1 4

3524 xyz

35 24 1 2 42 ¡1 13 2 ¡3

3524 abc

352 Write as a matrix equation:

a x¡ y ¡ z = 2x + y + 3z = 7

9x¡ y ¡ 3z = ¡1

b 2x + y ¡ z = 3y + 2z = 6

x¡ y + z = 13

c a + b¡ c = 7a¡ b + c = 6

2a + b¡ 3c = ¡2

3 Show that

24 1 ¡1 0¡1 0 10 2 ¡1

35 and

24 2 1 11 1 12 2 1

35 are inverses of each other.

4 If A =

24 2 0 31 5 21 ¡3 1

35 and B =

24 ¡11 9 15¡1 1 18 ¡6 ¡10

35 , find:

a AB b A¡1 in terms of B.

5 For A =

24 2 1 ¡1¡1 2 10 6 1

35 and B =

24 4 7 ¡3¡1 ¡2 16 12 ¡5

35 ,

calculate AB and hence solve the system of equations

4a + 7b¡ 3c = ¡8¡a¡ 2b + c = 3

6a + 12b¡ 5c = ¡15.

THE INVERSE OF A 3 × 3 MATRIX� �F

EXERCISE 10F.1



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6 For M =

24 5 3 ¡7¡1 ¡3 3¡3 ¡1 5

35 and N =

24 3 2 31 ¡1 22 1 3

35 ,

calculate MN and hence solve the system 3u + 2v + 3w = 18u¡ v + 2w = 6

2u + v + 3w = 16.

7 It is known that

24 1 0 00 a b0 c d

35 and

24 1 0 00 kd ¡kb0 ¡kc ka

35 are inverses.

a Find k:

b State the conditions under which the inverse exists.

To find the inverse of a 3 £ 3 matrix we can use elementary row operations.

For example, to find A¡1 when A =

24 1 2 42 0 13 ¡1 2

35 we start with

FINDING INVERSES OF 3 × 3 MATRICES� �

£A I

¤i.e.,

24 1 2 4 1 0 02 0 1 0 1 03 ¡1 2 0 0 1

35»24 1 2 4 1 0 0

0 ¡4 ¡7 ¡2 1 00 ¡7 ¡10 ¡3 0 1

35 R2 ! R2 ¡ 2R1

R3 ! R3 ¡ 3R1

»24 1 2 4 1 0 0

0 ¡4 ¡7 ¡2 1 00 0 9 2 ¡7 4

35R3 ! 4R3 ¡ 7R2

»

26641 2 4 1 0 0

0 1 74

12 ¡1

4 0

0 0 1 29 ¡7

949

3775 R2 ! ¡14R2

R3 ! 19R3

»

26641 0 1

2 0 12 0

0 1 74

12 ¡1

4 0

0 0 1 29 ¡7

949

3775R1 ! R1 ¡ 2R2

»

26641 0 0 ¡1

989 ¡2

9

0 1 0 19

109 ¡7

9

0 0 1 29 ¡7

949

3775R1 ! R1 ¡ 1

2R3

R2 ! R2 ¡ 74R3

I A¡1

2 0 1 0 1 0

¡2 ¡4 ¡8 ¡2 0 0

0 ¡4 ¡7 ¡2 1 0

3 ¡1 2 0 0 1

¡3 ¡6 ¡12 ¡3 0 0

0 ¡7 ¡10 ¡3 0 1

0 ¡28 ¡40 ¡12 0 4

0 28 49 14 ¡7 0

0 0 9 2 ¡7 4

1 2 4 1 0 0

0 ¡2 ¡7

2¡1

1

20

1 01

20

1

20

1 01

20

1

20

0 0 ¡1

2¡1

9

7

18¡2

9

1 0 0 ¡1

9

8

9¡2

9

0 17

4

1

2¡1

40

0 0 ¡7

4¡ 7

18

49

36¡7

9

0 1 01

9

10

9¡7

9



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Check:

AA¡1 =

24 1 2 42 0 13 ¡1 2

35£ 19

24 ¡1 8 ¡21 10 ¡72 ¡7 4

35 Using a calculator:

= 19

24 ¡1 + 2 + 8 8 + 20 ¡ 28 ¡2 ¡ 14 + 16¡2 + 0 + 2 16 + 0 ¡ 7 ¡4 + 0 + 4¡3 ¡ 1 + 4 24 ¡ 10 ¡ 14 ¡6 + 7 + 8

35= 1

9

24 9 0 00 9 00 0 9

35 = I X

1 Use the elementary row operations method to find A¡1 for:

a b

A =

24 3 2 31 ¡1 22 1 3

35 A =

24 2 0 31 5 21 ¡3 1

35 Check your answers

using technology.

2 Use technology to find B¡1 for:

a b

B =

24 13 43 ¡1116 9 27¡8 31 ¡13

35 B =

24 1:61 4:32 6:180:37 6:02 9:417:12 5:31 2:88

35

x¡ y ¡ z = 2Solve: x + y + 3z = 7

9x¡ y ¡ 3z = ¡1 using matrix methods and a graphics calculator.

In matrix form the system is:24 1 ¡1 ¡11 1 39 ¡1 ¡3

3524 xyz

35 =

24 27¡1

35 (i.e., AX = B)

)

24 xyz

35 =

24 1 ¡1 ¡11 1 39 ¡1 ¡3

35¡1 24 27¡1

35Into a calculator we enter A and B and calculate

[A]¡1

[B] i.e., x = 0:6, y = ¡5:3, z = 3:9

3 Use matrix methods and technology to solve:

a 3x + 2y ¡ z = 14x¡ y + 2z = ¡8

2x + 3y ¡ z = 13

b x¡ y ¡ 2z = 45x + y + 2z = ¡63x¡ 4y ¡ z = 17

c a + b¡ c + d = 8a¡ b + c + 3d = 1

2a¡ b + 2c¡ d = 114a + b + c + 3d = 0

EXERCISE 10F.2

Example 15



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INVESTIGATION 3 USING MATRICES IN CRYPTOGRAPHY

Messages are sent in code or cipher form. The method of converting text to ciphertext is

called enciphering and the reverse process is called deciphering.

The operations of matrix addition and multiplication can be used to create codes and the

coded messages are transmitted. Decoding using additive or multiplicative inverses is re-

quired by the receiver in order to read the message.

The letters of the

alphabet are first

assigned integer

values.

Notice that Z is

assigned 0.

A B C D E F G H I J K L M

1 2 3 4 5 6 7 8 9 10 11 12 13

N O P Q R S T U V W X Y Z

14 15 16 17 18 19 20 21 22 23 24 25 0

The coded form of the word SEND is therefore 19 5 14 4 which we could put in 2 £ 2

matrix form

·19 514 4

¸:

An encoding matrix of your choice could be added to this matrix. Suppose it is

·2 713 5

¸:

The matrix to be transmitted is then

·19 514 4

¸+

·2 713 5

¸=

·21 1227 9

¸

Now

·21 1227 9

¸becomes

·21 121 9

¸as any number not in the range 0 to 25 is

adjusted to be in it by adding or subtracting multiples of 26.

So,

·21 121 9

¸is sent as 21 12 1 9.

The message SEND MONEY PLEASE could be broken into groups of four letters and

each group is encoded.

SENDjMONEjYPLEjASEE Ã repeat the last letter to make group of 4.

This is a dummy letter.

For MONE the matrix required is

·13 1514 5

¸+

·2 713 5

¸=

·15 2227 10

¸i.e.,

·15 221 10

¸For YPLE the matrix required is

·25 1612 5

¸+

·2 713 5

¸=

·27 2325 10

¸i.e.,

·1 2325 10

¸For ASEE the matrix required is

·1 195 5

¸+

·2 713 5

¸=

·3 2618 10

¸i.e.,

·3 018 10

¸So the whole message is 21 12 1 9 15 22 1 10 1 23 25 10 3 0 18 10

Cryptography is the study of encoding and decoding messages.

Cryptography was first developed to send secret messages in written form.

However, today it is used to maintain privacy when information is being

transmitted via public communication services (by line or by satellite).



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The decoder requires the additive inverse matrix

· ¡2 ¡7¡13 ¡5

¸to decode the message.

1 Use the decoder matrix to check that the original message is obtained.

2 Use the code given to decode the message:

22 15 18 24 21 6 10 14 22 15 7 13 7 25 189 16 22 6 10 11 0 22 19 9 25 18 6 22 1114 19 9 12 5 23

3 Create your own matrix addition code. Encode a short message. Supply the decoding

matrix to a friend so that he/she can decode it.

4 Breaking codes where matrix multiplication is used is much more difficult.

A chosen encoder matrix is required. Suppose it is

·2 31 2

¸:

The word SEND is encoded as

·19 514 4

¸ ·2 31 2

¸=

·43 6732 50

¸which is converted to

·17 156 24

¸a What is the coded form of SEND MONEY PLEASE?

b What decoder matrix needs to be supplied to the receiver so that the message

can be read?

c Check by decoding the message.

d Create your own code using matrix multiplication using a matrix

·a bc d

¸where ad¡ bc = 1. Why?

e What are the problems in using a 2 £ 2 matrix when ad¡ bc 6= 1?

How can these problems be overcome?

5 Research Hill ciphers and explain how they differ from the methods given previously.

Recall that if A =

·a bc d

¸then A¡1 =

1

ad¡ bc

·d ¡b¡c a

¸

So, if A =

·2 35 4

¸then A¡1 = 1

8¡15

·4 ¡3¡5 2

¸= 1

¡7

·4 ¡3¡5 2

¸

But if A =

·1 22 4

¸then A¡1 = 1

4¡4

·4 ¡2¡2 1

¸which does not exist.

So the number ad¡ bc determines whether a 2 £ 2 matrix has an inverse or not.

What to do:

DETERMINANTS OF MATRICESGTHE 2 × 2 DETERMINANT� �

on removing multiples of .26



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For A =

·a bc d

¸,

the determinant is ad¡ bc, and is denoted by jAjor detA. A has an inverse if jAj 6= 0.

a If A =

·2 13 4

¸find jAj :

b Does

(2x + y = 4

3x + 4y = ¡1have a unique solution?

a jAj = 2(4) ¡ 1(3)

= 8 ¡ 3

= 5

b The system in matrix form is·2 13 4

¸ ·xy

¸=

·4¡1

¸Now as jAj = 5 6= 0, A¡1 exists

and so we can solve for x and y

) the system has a unique solution.

1 Find jAj for A equal to:

a b c d·

3 72 4

¸ · ¡1 31 ¡2

¸ ·0 00 0

¸ ·1 00 1

¸2 Find det B for B equal to:

a b c d·

3 ¡27 4

¸ ·3 00 2

¸ ·0 11 0

¸ ·a ¡a1 a

¸

3 a Consider the system

½2x¡ 3y = 84x¡ y = 11

i Write the equations in the form AX = B and find jAj.ii Does the system have a unique solution? If so, find it.

b Consider the system

½2x + ky = 84x¡ y = 11

i Write the system in the form AX = B and find jAj.ii For what value(s) of k does the system have a unique solution? Find the unique

solution.

iii Find k when the system does not have a unique solution. How many solutions

does it have in this case?

4 Find the following determinants for A =

·2 ¡1¡1 ¡1

¸a jAj b

¯̄A2¯̄

c j2Aj

5 Prove that, if A is any 2 £ 2 matrix and k is a constant, then jkAj = k2 jAj :

Example 16

EXERCISE 10G.1



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6 By letting A =

·a bc d

¸and B =

·w xy z

¸a find jAj and jBjb find AB and jABj, and hence

c show that jABj = jAj jBj for all 2 £ 2 matrices A and B.

7 A =

·1 23 4

¸and B =

· ¡1 20 1

¸.

a Using the results of 5 and 6 above and the calculated values of jAj and jBj, find:

i jAj ii j2Aj iii j¡Aj iv j¡3Bj v jABjb Check your answers without using the results of 5 and 6 above.

8 Use the results of questions 5 and 6 above to find jAj, given that:

a A2 = O b A2 = I c A2 = A.

9 If A is its own inverse and jAj = 1, show that A = kI where k is a scalar (real

number).

10 a If jAj 6= 0 where A is a square matrix of any order, prove that¯̄A¡1

¯̄=

1

jAj .

(Note: Do not use the formula for the inverse of a 2 £ 2 matrix.)

b If A is its own inverse, prove that jAj = §1.

11 If A =

·a bc d

¸, prove that A2 = (a + d)A ¡ jAjI.

Hence, deduce the inverse formula A¡1 =1

jAj·

d ¡b¡c a

¸.

If A and B are 2£2 matrices then: ² jABj = jAj jBj ² jkAj = k2 jAj ² ¯̄A¡1

¯̄=

1

jAj

On page 382 we found A¡1 when A =

24 1 2 42 0 13 ¡1 2

35 using an augmented matrix and

linear row operations.

24 1 2 4 1 0 02 0 1 0 1 03 ¡1 2 0 0 1

35 became

26641 0 0 ¡1

989 ¡2

9

0 1 0 19

109 ¡7

9

0 0 1 29 ¡7

949

3775This suggests that 9 or ¡9 is the determinant of A , i.e., jAj :

SUMMARY OF THE PROPERTIES OF 2 × 2 MATRICES� �

THE 3 × 3 DETERMINANT� �



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the determinant of A =

24 a1 b1 c1a2 b2 c2a3 b3 c3

35 as jAj = a1

¯̄̄̄b2 c2b3 c3

¯̄̄̄+ b1

¯̄̄̄c2 a2c3 a3

¯̄̄̄+ c1

¯̄̄̄a2 b2a3 b3

¯̄̄̄

Alternatively

Find jAj for A =

24 1 2 42 0 13 ¡1 2

35

jAj = 1

¯̄̄̄0 1¡1 2

¯̄̄̄+ 2

¯̄̄̄1 22 3

¯̄̄̄+ 4

¯̄̄̄2 03 ¡1

¯̄̄̄

= 1(0 ¡¡1) + 2(3 ¡ 4) + 4(¡2 ¡ 0)

= 1 ¡ 2 ¡ 8

= ¡9 fwhich checks with the earlier exampleg

same

same same

Example 17

Now if we use the same method on A =

24 a1 b1 c1a2 b2 c2a3 b3 c3

35 we would obtain an algebraic

form for jAj :

The result is jAj = a1(b2c3 ¡ b3c2) + b1(a3c2 ¡ a2c3) + c1(a2b3 ¡ a3b2) or its negative.

Consequently we define

Once again we observe that a 3 £ 3 system of linear equations in

matrix form AX = B will have a unique solution if jAj 6= 0:

Note: A graphics calculator or spreadsheet can be used

to find the value of a determinant.

1 Evaluate:

a¯̄̄̄¯̄ 2 3 0¡1 2 12 0 5

¯̄̄̄¯̄

d¯̄̄̄¯̄ 1 0 0

0 2 00 0 3

¯̄̄̄¯̄

g¯̄̄̄¯̄ 3 ¡1 ¡2

0 1 1¡1 ¡1 3

¯̄̄̄¯̄

b¯̄̄̄¯̄ ¡1 2 ¡3

1 0 0¡1 2 1

¯̄̄̄¯̄

e¯̄̄̄¯̄ 0 0 2

0 1 03 0 0

¯̄̄̄¯̄

h¯̄̄̄¯̄ 1 3 2¡1 2 12 6 4

¯̄̄̄¯̄

c¯̄̄̄¯̄ 2 1 3¡1 1 22 1 3

¯̄̄̄¯̄

f¯̄̄̄¯̄ 4 1 3¡1 0 2¡1 1 1

¯̄̄̄¯̄

i¯̄̄̄¯̄ 0 3 0

1 2 56 0 1

¯̄̄̄¯̄

EXERCISE 10G.2

jAj = a1

¯̄̄̄b2 c2b3 c3

¯̄̄̄¡ b1

¯̄̄̄a2a3

¯̄̄̄+ c1

¯̄̄̄a2 b2a3 b3

¯̄̄̄c2c3

DEMO



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2 Evaluate:

a b c¯̄̄̄¯̄ a 0 0

0 b 00 0 c

¯̄̄̄¯̄

¯̄̄̄¯̄ 0 x y¡x 0 z¡y ¡z 0

¯̄̄̄¯̄

¯̄̄̄¯̄ a b cb c ac a b

¯̄̄̄¯̄

3 For what values of k does

8<: x + 2y ¡ 3z = 52x¡ y ¡ z = 8

kx + y + 2z = 14have a unique solution?

4 For what values of k does

8<: 2x¡ y ¡ 4z = 83x¡ ky + z = 15x¡ y + kz = ¡2

have a unique solution?

5 Find k given that:

a b¯̄̄̄¯̄ 1 k 3k 1 ¡13 4 2

¯̄̄̄¯̄ = 7

¯̄̄̄¯̄ k 2 1

2 k 21 2 k

¯̄̄̄¯̄ = 0

6 Use technology to find the determinant of:

a b26641 2 3 12 0 1 23 1 4 01 2 0 5

3775266664

1 2 3 4 62 3 4 5 01 2 0 1 42 1 0 1 53 0 1 2 1

3777757 If Jan bought one orange, two apples, a pear, a cabbage and a lettuce the total cost would

be $6:30. Two oranges, one apple, two pears, one cabbage and one lettuce would cost a

total of $6:70. One orange, two apples, three pears, one cabbage and one lettuce would

cost a total of $7:70. Two oranges, two apples, one pear, one cabbage and three lettuces

would cost a total of $9:80. Three oranges, three apples, five pears, two cabbages and

two lettuces would cost a total of $10:90.

a Write this information in AX = B form where A is the quantities matrix, X is

the cost per item column matrix and B is the total costs column matrix.

b Explain why X cannot be found from the given information.

c If the last lot of information is deleted and in its place “three oranges, one apple,

two pears, two cabbages and one lettuce cost a total of $9:20” is substituted, can

the system be solved now, and if so, what is the solution?



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REVIEW SET 10A

1 If A =

24 1 2 32 5 7

¡2 ¡4 ¡5

35 and B =

24 3 ¡2 ¡1¡4 1 ¡12 0 1

35 ,

find AB and BA and

hence find A¡1 in

terms of B.

2 If A = 2A¡1:

a show that A2 = 2I

b

3 Find x if

¯̄̄̄¯̄ x 2 0

2 x + 1 ¡20 ¡2 x + 2

¯̄̄̄¯̄ = 0, given that x is real.

4 If A =

· ¡2 34 ¡1

¸, B =

· ¡7 99 ¡3

¸, C =

· ¡1 0 30 2 1

¸,

evaluate if possible:

a 2A ¡ 2B b AC c CB d D, given that DA = B.

5 Prove the property of inverses (AB)¡1 = B¡1A¡1 given that these inverses exist.

6 If A =

24 1 a 2a 1 12 ¡2 a + 2

35 , X =

24 xyz

35 and B =

24 518

35 , show using jAj that

AX = B has a unique solution provided a 6= 0 or ¡1:

a What solutions are there when a = ¡1?

b If a = 0, find the particular solution for which x + y + z = 0.

7 The 2 £ 2 matrix D has the property that D¡1 = kD where k is a real number.

a Find D2 in simplest form.

b Find det D.

c Write D(D ¡ 3I)(D + 4I) in simplest form.

8 Doctors classify people as underweight, normal weight, and

overweight. Several doctors pool data and the following ta-

ble shows the connection between the weight classifications

of people and their children.

current

generation

next generation

underweight normal overweight

underweight 0:43 0:47 0:10normal 0:18 0:56 0:26

overweight 0:09 0:55 0:36

a State the transition matrix T and calculate T2.

b What percentage of underweight people of the current generation are expected to

have: i normal weight children ii overweight grandchildren?

REVIEWH


simplify ( )( ), giving your answer in the form where

and are real numbers.

A I A I A I¡ + 3 +r s rs


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REVIEW SET 10B

d

e What is the steady state for each category?

1 Write 2x ¡ 3y = 117x + 8y = ¡4

in the form AX = B and hence solve for x and y using

matrix algebra.

2 a Under what conditions are the following true, assuming A and B are square matrices?

i AB = B ) A = I ii (A + B)2 = A2 + 2AB + B2

b If M =

·k 22 k

¸ ·k ¡ 1 ¡2¡3 k

¸has an inverse M¡1, what values can k have?

3 Find A¡1 given that A =

24 2 1 0¡1 1 11 3 ¡2

35 .

4 If A =

24 1 2 12 4 63 1 2

35 and B =

24 ¡1 2 ¡32 ¡1 43 4 1

35show by calculation that det (AB) = det A £ det B = 80.

5 It is known that the matrix

24 2 1 11 1 12 2 1

35 has inverse of the form

24 a b 0b 0 a0 2 b

35 .

a Find the values of a and b.

b Use the above to solve the system

2x + y + z = 1x + y + z = 62x + 2y + z = 5.

6 Solve for k if

¯̄̄̄¯̄ k 1 3

2k + 1 ¡3 20 k 2

¯̄̄̄¯̄ = 0.

7 If A =

·0 ¡2

¡1 1

¸, show that A satisfies the equation A2 = A + 2I.

Hence express A¡1 and A4 in terms of A and I.

8 The local Country Club has a weekly Seniors Night. The meal is a roast dinner offering

a choice of four different roast meats: beef, lamb, chicken or pork.

The choice of roast meat of 200 patrons is

recorded over several weeks. The initial choices

are shown alongside.

Beef Lamb Chicken Pork

56 45 39 60

Comment on the change in the proportion of normal weight people in each

generation.


c Currently the weight status of people is 15% underweight, 56% normal, and 29%overweight. Find how these proportions change:

i in the next generation ii in two generations.


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REVIEW SET 10C

night, 10% will change to beef, 20% will change to chicken and 10% will change to

pork. Of those who choose chicken on a given night, 30% will choose chicken on the

next night, 10% will change to beef, 40% will change to lamb and 20% will change

to pork. Of those who choose pork on a given night, 60% will choose pork on the

next night, 20% will change to beef, 10% will change to lamb and 10% will change to

chicken.

a Write a transition matrix T for this situation.

b Write an initial state row matrix S0.

c Assuming the number attending Seniors night remains constant at 200, determine

the number of different types of roast meat meals that will be served in

i the first week after the initial survey

ii the fifth week after the initial survey.

d Establish the steady state proportions of each type of roast meat meals that will be

served at Seniors Night.

e If 300 people attended Seniors Night, how many roast pork meals would the Country

Club expect to serve?

1 If 3A ¡ BX = C, find X in terms of A, B and C.

2 If 4x ¡ 3y = 82x + 3y = ¡1,

write the system of equations in the form AX = B and hence

solve for x and y using matrix algebra.

3 If A =

24 0 2 11 1 1

¡1 ¡2 ¡2

35 , find A2 and hence determine A¡1.

4 If B =

24 3 1 2¡1 ¡2 12 0 ¡1

35 , find B¡1.

5 Show that if A =

24 b + c c + a b + aa b c1 1 1

35 then det A = 0 for all a, b and c.

6 S is the set of all 3 £ 3 matrices of the form

24 p 0 q0 r 0q 0 p

35 .

If A =

24 a 0 b0 c 0b 0 a

35 and X =

24 x 0 y0 z 0y 0 x

35 , find AX and hence:

a show that AX is a member of set S.

b Find A¡1 by setting AX = I. State any constraints that must be placed on the

real numbers a, b and c in order that the inverse exists.


It is known that half the seniors who choose beef on a given night also choose beef on

the next night, while 20% change to lamb, 10% to chicken and the remainder change to

pork. Of those who choose lamb on a given night, 60% will choose lamb on the next


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REVIEW SET 10D

7 If A =

24 1 0 0¡1 1 0

1 ¡ 2k 1 k ¡ 1

35 and B =

24 1 1 k1 k 12 1 1

35 find AB.

a Hence find jAB j and determine all values of k when jAB j = 0.

b Write the system x + y + kz = 1x + ky + z = 12x + y + z = 1

in the form BX = C.

c By using AB, show that if k 6= 0 or 1,

the system has a unique solution ofx =

k ¡ 1

2k, y = z =

1

2k.

8 A legal firm has four city offices. A courier delivers

original documents between the offices with each de-

livery taking 10 minutes. Suppose the offices are A, B,

C and D. If the courier is at A, the chance that he will

be next at B is 50%, C 30% and D 20%. If the courier

is at B, the chance that he will be next at A is 40%,

C 10% and D 50%. If the courier is at C, the chances

are: A 20%, B 50% and D 30%. If the courier is at

D, the chances are: A 10%, B 30% and C 60%.

a Write this information in the form of a transition

matrix T.

b Find T2 and T3.

c If the courier is now at B, what is the chance that he is:

i at C in 10 minutes time ii at A in 20 minutes time

iii at D in 30 minutes time?

d Where is the courier most likely to be in:

i 20 minutes time ii 30 minutes time?

1 If A2 = 5A + 2I, find A3, A4, A5, A6 in the form rA + sI.

2 If P2 = P, find det P and show that (I + P)3 = I + 7P.

3 If A =

24 1 2 44 3 8¡4 ¡4 ¡9

35 calculate B = A + 4I and AB and hence deduce the

matrix A¡1.

4 If A =

24 1 2 22 1 22 2 1

35 show that A2 ¡ 4A ¡ 5I = O.

5 If M =

26643 ¡1 2 01 0 1 12 ¡1 0 31 4 1 4

3775 find M¡1 using your calculator.



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6 Given that some matrix A has the property that A2 = A¡1, show that:

a det A = 1 b (A + 2I)3 can be written in the form aA + bA¡1 + cI and

state the values of a, b and c.

7 A matrix A has the property that A2 = A ¡ I. Find expressions for An

for n = 3, 4, 5, ..., 8 in terms of A and I (i.e., in the form aA + bI). Hence:

a deduce simple expressions for A6n+3 and A6n+5

b express A¡1 in terms of A and I.

8 Prove that

¯̄̄̄¯̄ a + b c c

a b + c ab b c + a

¯̄̄̄¯̄ = 4abc.

9 If A =

24 ¡2 k 13 k + 2 2

k ¡ 1 1 1

35 , find the values of k for which jA j = 0.



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ANSWERS


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\395SA12STU-2_AN.CDR Thursday, 9 November 2006 12:06:36 PM PETERDELL

396 ANSWERS

1 a y = 4x+3 b 3x+2y = 6 c y = 3x+4d 2x ¡ y = ¡7 e x ¡ 2y = 8f lower line: x¡3y = 6, top line: x¡3y = ¡32

2 a y = 2x ¡ 3

b 5x+2y = 1

c 3x+4y = ¡4

3 a i Yes ii No iii Yes iv No v Yes

by ¡ 7

x ¡ 4= 2 in each case.

4 a

b approx

(1:2, 0:5)

c (1 14 , 1

2 )

5 a 2x2¡10x b 2x2+9x+4 c 6x2¡11x+4

d x2 ¡ 7 e ¡x2 ¡ 3x¡ 2 f 18x2 +3x¡ 3

g x2 ¡ 36 h x2 + 6x + 9 i 4x2 ¡ 4x + 1

j 4x2 ¡ 1 k 16x2 + 40x + 25 l x2 ¡ 3

m x2 + x + 5 n 5x2 + 5 o ¡x2 + 4x ¡ 3

p x3 + 5x2 + 2x ¡ 8 q x3 +9x2 + 27x +27

r 2x3 + 3x2 ¡ 23x ¡ 12

6 a 5x(x+4) b x(7¡2x) c 2(x+2)(x¡2)

d (x +p7)(x ¡

p7) e 4(x +

p2)(x ¡

p2)

f (x ¡ 1)2 g 2(x + 1)2 h 3(x + 2)2

i (2x ¡ 3)(x + 4) j (3x + 1)(x ¡ 2)

k (7x ¡ 2)(x ¡ 1) l (2x + 1)(3x ¡ 2)

m (2x + 1)(2x ¡ 3) n (5x ¡ 3)(2x + 1)

o (6x + 1)(2x ¡ 3)

7 a 2(x + 5)(x + 2) b (x + 1)(x ¡ 7)

c (x ¡ 2)(5x + 2) d 2(x + 5)(2x ¡ 5)

e (x + 5)(x + 1) f ¡4x(x + 1)

g (3x ¡ 1)(x + 7) h ¡4(x ¡ 4)(2x + 3)

8 a x = 6 or ¡1 b x = 0 or 2 c x = 0 or 12

d x = §p2 e x = 1 or ¡4 f x = ¡2 or 1

3

g x = 2 or ¡ 23

h x = 4 or ¡ 23

i x = ¡ 34

or 53

9 a (1, 4) and (¡ 32 , 1

4 ) b ( 32 , 5) and (¡23 , 5)

c (¡ 12

, ¡4) and ( 83

, 15)

10 a x = 2§p7 b x = 1§

p3 c x = 1§ i

d x =1§

p65

8e x =

1§ ip3

2

f x =1§

p3

2

11 a 1

b x = 1

c (1, ¡1)

d 1 + 1p2

,

1¡ 1p2

12 a f(x) = 2(x + 1)(x ¡ 4)

b f(x) = 2(x ¡ 2)2

c f(x) = ¡2(x + 2)(x ¡ 3)

d f(x) = 2(x + 1)(2x ¡ 5)

e f(x) = ¡2x2+7x f f(x) = ¡(x¡3)2+4

13 ® = 3 or 35 14 2:26 m

15 a x = 3, ¡1§p3 b x = 2, 2§ i

16 a (¡1, 0) and (1, 4) b (¡2, 0) and (0, 2)

c (¡1, 6) and (¡3, 22)

17 a x + ¡1:828, y + ¡0:8284 or

x + 3:828, y + 4:828

b no solutions

c x + ¡2:770, y + 2:885 or

x + 3:970, y + ¡0:4852

18 a x = ¡ 12

, 2, 3 b x = ¡2

c x + ¡2:317, 0:333, 4:317d x = ¡2, ¡0:225, 1, 2:225e x + ¡1:414, 1:414 f x = 3

4, 2

EXERCISE 1A

y

x

Ew_

( )� �,

�

y

x

( ) ��,

( )� �,

y

x-\Re_

�

x��

� �

�

�

�

�xy 23 ��

xy 423 ��

y

1

x

y

x = 1

(1, 1) 2

11�

2

11�

142 2�� xxy


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\396SA12STU-2_AN.CDR Monday, 21 May 2007 10:24:04 AM DAVID3

ANSWERS 397

19 a x1

2 b x¡1 c x0 d x2

3 e x¡ 3

2 f x8

3

20 a x¡1 + x¡2 b x¡2 ¡ 2x¡3

c 2x3

2 +x1

2 +x¡ 1

2 d x1

2 ¡ 3x¡ 1

2 +10x¡ 3

2

21 a x = ¡ 52

b x = ¡ 158

c x = 3

22 a x = 32 b x = 1

3 c x = ¡1 d x = 2

23 a x = 2 12

b x = 23

c x = 12 d x = 15

e x = 2 f x = ¡2 23 g x = ¡7

h no solution i x = 12

j x = 9

k x = §3 l x = § 1p5

24 a a = §

rF

mb r =

rA

¼c r = 3

r3V

4¼

d x =C

4 + ¼e x =

rV

lf V =

R1R2I

R1 + R2

g y = ¡p

r2 ¡ x2 h x = 2 +p

r2 ¡ y2

i x

rb(a¡ 1)

a + 1j x =

a(1 + y2)

1¡ y2

k l =A ¡ ¼r2

2¼rl s =

rf

f ¡ r

1 a A = (200x) m2 b P = (400 + 2x) m

c BD =p40 000 + x2 m d µ = tan¡1

³200

x

´2 a P (x) = 6x cm, A(x) = 2x2 cm2

b P (x) = (x + 4 +p16¡ x2) cm

A(x) =

µxp16¡ x2

2

¶cm2

c P (x) = (6 + x +p

x2 ¡ 36) cm

A(x) = 3p

x2 ¡ 36 cm2

d P (x) = (16 + 2p64¡ x2) cm

A(x) = xp64¡ x2 cm2

e P (x) = (2x + 2p

x2 ¡ 25) cm

A(x) = 5p

x2 ¡ 25 cm2

f P (x) = (14+2x) cm, A(x) = xp49¡ x2 cm2

3 a A(x) =

·³x

4

´2

+(24¡ x)2

4¼

¸cm2

b A(0) is the area of the circle when no square

is formed and A(24) is the area of the square

when no circle is formed (i.e., all of the wire

is used for one shape).

4 a A(x) = (10x ¡ x2) cm2

b A(x) = 20x ¡¡12+ ¼

8

¢x2 cm2

c A(x) = (25¡ x)p50x ¡ 625 cm2

d A(x) = ¡x2

¼+

40 000

¼m2

5 a P (x) = 2x +2000

xm

b P (x) = 2x +¼x

2+

400

xcm

c P (x) = 2x + 2

rx2 +

40 000

x2cm

d P (x) = 2x +

µ¡4x + 4

px2 + 50¼

¼

¶+¡¡2x + 2

px2 + 50¼

¢cm

6 a y = 200¡ 8x b A(x) = 1200x¡ 36x2 m2

c Hint: Lengths must be > 0.

7 a V (x) =8¼x3

9m3 b V (x) =

¼x3

48m3

c V (x) = 80x2 cm3

8 A(x) = ¡ 34x2 + 6x

1 P (x) = 20 + 2p100¡ x2 cm

A(x) = xp100¡ x2 cm2

2 P (r) = (2 + ¼

2)r +

80

rm

3 a V (x) =4¼x3

27m3 b 1:290 m deep

4 y = 100¡ 18x, A(x) = 1000x ¡ 120x2 cm2

0 < x 6 509

5 D(t) =p

130t2 ¡ 1625

t + 8125

km

6 V (r) = 2¼r2p400¡ r2 cm3

1 a Hint: Pythagoras’ theorem

b A(x) = 1600¡ 40p400 + x2 + 20x cm2

2 a y = 250¡5x

4b A(x) = 1000x ¡ 5x2 m2

3 a A(x) = 12x(p64¡ x2 +

p100¡ x2) cm2

b V (x) = x2(p64¡ x2 +

p100¡ x2) cm3

4 T (x) =

µpx2 + 25

3+

10¡ x

8

¶hours

5 a V = ¼r2(10¡ 2r) cm3

b V = ¼h(5¡h

2)2 cm3

6 Hint: Show thatr

8=

hp

h2 ¡ 16h.

7 A(x) = ¡ 12x2 + 5x

EXERCISE 1B

REVIEW SET 1A

REVIEW SET 1B

= ¡


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398 ANSWERS

1 a fx : x 6= ¡1g b fx : x 6= 0 or ¡2g

c fx : x 6= §1g d fx : x 6 ¡2 or x > 2g

e fx : ¡2 6 x 6 3g f fx : ¡3 < x < 3g

2 a 3 13 b ¡2 1

2 c3p2

2d

1

t+ t

e u +1

uf

1

x+ x

3 a 3 b 2x + 5x2 c 5t4 + 12t2 + 7

d 5y + 8¡ 8p1 + y

4 ¡3 6 x 6 2

5 a fx : x 6= 2g

b 7, 9:25, 10:84, 11:41, 11:9401, 11:994 001c y approaches 12

6 a fx : x < ¡4 or x > 4g

b Hint: ln a is defined only when a > 0

1 a The graph grows rapidly at first, then grows at

a decreasing rate as it heads towards an asymp-

tote. The logistic model is the most likely

model.

b A Polynomial functions do not have

asymptotes.

B Power models either pass through the

origin, or are asymptotic at x = 0:

C Increasing exponential functions grow at

an ever increasing rate.

D Logarithmic functions are undefined at

x = 0:

E Surge functions pass through the origin,

and head towards the asymptote y = 0as x increases.

2 a

b linear

c As the area increases, the cost appears to

increase at a constant rate,

d Cost = 8:92A + 11:55 A 5£ 10 tarpaulin

costs $458

3 a/b

c the linear function

d i linear: R = 38:1, exponential: R = 37:89

ii linear: R = ¡2:71, exponential: R = 12:79

e As n increases, we expect R to head towards

the asymptote R = 0, not become negative.

Hence, the exponential function may be better.

f “The higher the altitude, the faster the rate of

reaction.”

4 a

b A The logarithmic function does not change

direction.

B The logistic function grows slowly towards

an asymptote.

C The exponential function does not change

direction.

c i t = 0, y = 1:17 (the ball was 1:14 m above

ground level when thrown)

t = 4:5, y = 9:81 (after 4:5 sec the ball was

9:81 m above the ground.)

t = 8, y = 11:47 (after 8 sec the ball has

landed.)

ii t = 0, y = 0 (the ball was at ground level

when it was thrown.)

t = 4:5, y = 9:62 (after 4:5 sec the ball was

9:62 m above the ground.)

t = 8, y = 5:11 (after 8 sec the ball was

5:11 m above the ground)

d The quadratic model, 0 6 t 6 6:60

1 a1000

xm

EXERCISE 2A

EXERCISE 2B

EXERCISE 2C

�

��

��

��

��

��

� ��

A

C

n

14012010080604020

50

40

30

20

10

R n�� . .

neR 0041.01.57 �

�

�

�

�

�

�

��

��

� � � � � � �

H

t

¡


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ANSWERS 399

c

d 126:5 m

x = 31:6 m

e

2 b

c x = 5

3 a y =200

xc

d $8314, x = 11:55 m

e

4 a 2x cm

d

e 213:4 cm2, x = 4:217 cm

f

5 c A = 2¼r2 +2000

r

d r = 5:42 cm e

f r = 25:15h = 25:15

1 a

b s(t) = 99¡ 110t, 110t is the area of the

rectangle drawn in a.

c

d The slope is ¡110. The distance from Adelaide

is decreasing at a rate of 110 km per hour. The

s-intercept is 99, the distance from Adelaide at

t = 0.

2 a

b P (t) = 2000+1500t, 1500t is the area of the


EXERCISE 2D

31.6 m

10 m

5 m

�

��

��

��

��

� ��

y

x

�

��

��

��

��

��

��

� � � � � ��

A

x

8.434 cm4.217 cm

5.623 cm

�

��

��

��

� � ��

y

x

�

��

��

��

��

� � � � � ��

y

x

11.55 m

17.32 m

5.419 cm

10.84 cm

�

��

��

��

��

��

� � � � � ��

A

r

0.2 0.4 0.6 0.8

110

t

v

t

0.70.60.50.40.30.20.1

100

80

60

40

20

distance (km)

time (h)

1 2

1500

t

v

time (h)


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400 ANSWERS

c

d The slope is 1500. The number of people inside

the stadium increases at a rate of 1500 people

per hour. The P -intercept is 2000, the number

of people inside the stadium after the initial

intake.

3 a

b V (t) = 500 + 12t, 12t is the area of the


c

d The slope is 12. The volume of the water in the

tank increases at a rate of 12 litres per minute.

The V -intercept is 500, the volume of water

initially in the tank.

4 a

b A(t) = 30¡ 6:5t, 6:5t is the area of the


c

d The slope is ¡6:5. The amount of alcohol in the

blood decreases at a rate of 6:5 grams per hour.

5 a

b cx

c

d The slope of F (x) is c, the value of f(x).

1 a

b maximum speed is 100 km/h

minimum speed is 68:4 km/h

c Hint: V (t) > 68:4 km/h

d 4:84 km < distance from Adelaide < 20:64 km

e 9:28 km < distance from Adelaide < 17:18 km

2 a

b Hint: P (n) 6 10 000

c $375 000 < profit < $875 000

d $531 250 < profit < $781 250

e Use 100 intervals of 1 house each.

21

5000

4000

3000

2000

1000

P t( )

t

1 5

12

t

�( )t

time (m)

2 3 4

654321

600

500

400

300

200

100

V t( )

time (m)

1 5

6.5

t

r t( )

time (h)

2 3 4

4321

30

25

20

15

10

5

A t( )

time (h)

c

x

�( )x

x

F x( )

x

10.80.60.40.2

100

80

60

40

20

V t( )

time (h)

EXERCISE 2E.1

10080604020

10 000�

8000

6000

4000

2000

P n( )

n

The -intercept ist 4.61, the time in hours it takesto completely leave the man’s bloodstream.


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ANSWERS 401

3 a

b Hint: 10 mg 6 r(t) 6 25 mg

c 36:8 mg < amount < 51:9 mg

d Use more intervals of shorter length.

4 a

b maximum is 0:5, minimum is 0:167

c 0:95 < area < 1:28

d Use more intervals of shorter length.

1 n AL AU

10 0:285 00 0:385 0025 0:313 60 0:353 6050 0:323 40 0:343 40100 0:328 35 0:338 35500 0:332 33 0:334 33

converges

to 13

2 a i n AL AU

5 0:160 00 0:360 0010 0:202 50 0:302 5050 0:240 10 0:260 10100 0:245 03 0:255 03500 0:249 00 0:251 001000 0:249 50 0:250 5010 000 0:249 95 0:250 05

ii n AL AU

5 0:400 00 0:600 0010 0:450 00 0:550 0050 0:490 00 0:510 00100 0:495 00 0:505 00500 0:499 00 0:501 001000 0:499 50 0:500 5010 000 0:499 95 0:500 05

iii n AL AU

5 0:549 74 0:749 7410 0:610 51 0:710 5150 0:656 10 0:676 10100 0:661 46 0:671 46500 0:665 65 0:667 651000 0:666 16 0:667 1610 000 0:666 62 0:666 72

iv n AL AU

5 0:618 67 0:818 6710 0:687 40 0:787 4050 0:738 51 0:758 51100 0:744 41 0:754 41500 0:748 93 0:750 931000 0:749 47 0:750 4710 000 0:749 95 0:750 05

b i 14

ii 12

iii 23

iv 34

c area =1

a+ 1

3 a n Rational bounds for ¼

10 2:9045 < ¼ < 3:304550 3:0983 < ¼ < 3:1783100 3:1204 < ¼ < 3:1604200 3:1312 < ¼ < 3:15121000 3:1396 < ¼ < 3:143610 000 3:1414 < ¼ < 3:1418

b n = 10 000

1 a lower rectangles upper rectangles

b n AL AU

5 0:5497 0:759750 0:6561 0:6761100 0:6615 0:6715500 0:6656 0:6676

c

Z 1

0

px dx + 0:67

EXERCISE 2E.2

EXERCISE 2F.1

4321

25

20

15

10

5

r t( )

t (h)

654321

1

0.8

0.6

0.4

0.2

�( )x

x

10.80.60.40.2

1

0.8

0.6

0.4

0.2x

y

10.80.60.40.2

1

0.8

0.6

0.4

0.2x

yxy � xy �


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\401SA12STU-2_AN.CDR Tuesday, 7 November 2006 10:06:20 AM PETERDELL

402 ANSWERS

2 a

b n AL AU

50 3:2016 3:2816100 3:2214 3:2614500 3:2373 3:2453

c

Z 2

0

p1 + x3 dx + 3:24

3

a lower rectangles upper rectangles

b n AL AU

5 2:9349 3:334950 3:1215 3:1615100 3:1316 3:1516500 3:1396 3:1436

c

Z 1

0

4

1 + x2dx + 3:1416

4 a 18 b 4:5 c 2¼

5n = 5 ¡0:44 <

R 1

0(¡x2) dx < ¡0:24

n = 10 ¡0:385 <R 1

0(¡x2) dx < ¡0:285

6 ¡0:1679 <

Z 1

0

(x2 ¡ x) dx < ¡0:1654

(Since the graph is symmetric about x = 12 the

answer is found by using upper and lower sums

for

Z0

(x2 ¡ x) dx, then doubling the result.)

7 a/b

f(0) = 1, f(0:25) + 0:779, f(0:5) + 0:607f(0:75) + 0:472, f(1) + 0:368, f(1:25) + 0:287,

f(1:5) + 0:223, f(1:75) + 0:174, f(2) + 0:135

c 0:977 units2 d 0:761 units2

e AL + 0:8560 units2, AU + 0:8733 units2

8 a b ¡12 9 a 3 b ¡3 c ¡6

1 a 116

b 1 c ¡14

d 76

e 73

f 113

2 a 6:5 b ¡9 c 0 d ¡2:5

3 a 2¼ b ¡4 c ¼

2d 5¼

2¡ 4

4 a

Z 7

2

f(x) dx b

Z 9

1

g(x) dx 5 a ¡5 b 4

1 a

exponential

b The data points appear non-linear. The function

should not pass through the origin (there must

have been some koalas present when it was

established.) So, it is not a power function.

The exponential model appears to be suitable.

c i 57 ii 285 iii 830

The first estimate is found using interpolation

and so is likely to be accurate. The last two

estimates are by extrapolation, and are likely

to be inaccurate.

2 a y = 80¡ 1:2x

b A = 640x¡ 9:6x2

c 33 13£ 40 m

3 a 12 b ¡3

4 c

Z 4

0

yAdx = 2¼

Z 6

4

yB dx = ¡¼

2d 3

2¼

EXERCISE 2F.2

REVIEW SET 2A

x

10.80.60.40.2

4

3

2

1

y

x

10.80.60.40.2

4

3

2

1

y

x1

4y

21

4

xy

�

�

x

1.510.5

4

3

2

1

y

2

31 xy ��

x21.510.5

1y

xey �

�

�

�

��

�

� � �

K

t

33.3 m

40 m

1

2

¡6


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V:\BOOKS\SA_books\SA_12STU-2ed\SA12STU-2_AN\402SA12STU-2_AN.CDR Thursday, 14 August 2008 3:20:24 PM PETER

ANSWERS 403

5 a

b 4:41

1 a/b/c

d i linear: Q = 63:2, exponential: Q = 62:6

ii linear: Q = 14:05, exponential: Q = 27:9

e The estimates in i are more accurate as they

are interpolations, not extrapolations. The

exponential model would be more reliable

for extrapolation.

2 c

Min A = 9466 cm2

when r = 31:69 cm d

3 a x 6 ¡3 b

or x > 2

c 0:318

4 a

logarithmic

b t = 19:094 ln(m) + 30:776

c i 17:54 hrs ii 22:55 hrs

d i, as ii involves extrapolation

1

Area = 1:68 units2

2 c

d 2 m £ 2 m £ 2 m

3 a 2 + ¼ b ¡2 c ¼

4 a b

area ¢ <

Z 4

0

12

p16¡ x2 dx < area rectangle

) 4 <

Z 4

0

12

p16¡ x2 dx < 8

) 8 <

Z 4

0

p16¡ x2 dx < 16

5 a

b The water will follow a parabolic path.

) the quadratic model is better.

c i 1:21 ii 6:4

REVIEW SET 2B

REVIEW SET 2C

41

2

1

y

x

xy �

�

��

��

��

��

� ��

A

r

31.7 cm

31.7 cm

Q

�

��

��

��

��

� ��

ttetQ

ttQ

03.71.80)(4271.78)(

�

�

��

�

�

��

��

��

� ��

t

m

y

x

2

4

y

x

1

1

y = x1

y

x

�

�

�

�

�

�

� � � � � �

y

x

y

x

2

4

y

x � �

1

62

2

�

��

x

xxy


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Function Derivative Function Derivative

x 1 x¡2 ¡2x¡3

x2 2x1 x¡3 ¡3x¡4

x3 3x2 x1

212x¡ 1

2

x4 4x3 x¡ 1

2 ¡ 12x¡ 3

2

x¡1 ¡x¡2 xn nxn¡1

404 ANSWERS

1 a 6 b ¡ 14

2 b 12

3 a Hint: x¡a = (p

x+p

a)(p

x¡p

a) b 16

1 a ¡4 b 1 c ¡12 d 3

2 a ¡1 b 34

c ¡ 132

d ¡12 e ¡1

f ¡45289

3 a 14

b 1 c ¡ 127

d 14

4 a 9 b 10 c ¡ 225

d ¡ 227

e 14

f ¡ 12

5 a 12 b 108

1 a f 0(x) = 1 b f 0(x) = 0 c f 0(x) = 3x2

d f 0(x) = 4x3

2 a f 0(x) = 2 b f 0(x) = 2x ¡ 3

c f 0(x) = 3x2 ¡ 4x

3 a f 0(x) =¡1

(x + 2)2b f 0(x) =

¡2

(2x ¡ 1)2

c f 0(x) = ¡2

x3d f 0(x) = ¡

3

x4

4 a f 0(x) =1

2p

x + 2b f 0(x) = ¡

1

2xp

x

c f 0(x) =1

p2x + 1

5

1 a f 0(x) = 3x2 b f 0(x) = 6x2

c f 0(x) = 14x d f 0(x) = 2x + 1

e f 0(x) = ¡4x f f 0(x) = 2x + 3

g f 0(x) = 3x2+6x+4 h f 0(x) = 20x3¡12x

i f 0(x) =6

x2j f 0(x) = ¡

2

x2+

6

x3

k f 0(x) = 2x ¡5

x2l f 0(x) = 2x +

3

x2

2 a 4 b ¡ 16729

c ¡7 d 134

e 18

f ¡11

3 a f 0(x) =2p

x+ 1 b f 0(x) =

1

33p

x2

c f 0(x) =1

xp

xd f 0(x) = 2¡

1

2p

x

e f 0(x) = ¡2

xp

xf f 0(x) = 6x ¡ 3

2

px

g f 0(x) =¡25

2x3p

xh f 0(x) = 2 +

9

2x2p

x

4 ady

dx= 4 +

3

x2,

dy

dxis the slope function of

y = 4x ¡3

xfrom which the slope at any point

can be found.

bdS

dt= 4t + 4 metres per second,

dS

dtis the

instantaneous rate of change in position at the

time t, i.e., it is the velocity function.

cdC

dx= 3 + 0:004x dollars per toaster

dC

dxis the instantaneous rate of change in cost

as the number of toasters changes.

1 a f(g(x)) = (2x+7)2 b f(g(x)) = 2x2 +7

c f(g(x)) =p3¡ 4x d f(g(x)) = 3¡ 4

px

e f(g(x)) =2

x2 + 3f f(g(x)) =

4

x2+ 3

g f(g(x)) = 23x+4 h f(g(x)) = 3£ 2x + 4

2 a f(x) = x3, g(x) = 3x + 10

b f(x) =1

x, g(x) = 2x + 4

c f(x) =p

x, g(x) = x2 ¡ 3x

d f(x) =1p

x, g(x) = 5¡ 2x

e f(x) = x4, g(x) = x2 + 5x ¡ 1

f f(x) =10

x3, g(x) = 3x ¡ x2

1 a u¡2, u = 2x ¡ 1 b u1

2 , u = x2 ¡ 3x

c 2u¡ 1

2 , u = 2¡ x2 d u1

3 , u = x3 ¡ x2

e 4u¡3, u = 3¡ x f 10u¡1, u = x2 ¡ 3

2 ady

dx= 8(4x ¡ 5) b

dy

dx= 2(5¡ 2x)¡2

cdy

dx= 1

2 (3x ¡ x2)¡1

2 £ (3¡ 2x)

ddy

dx= ¡12(1¡ 3x)3 e

dy

dx= ¡18(5¡ x)2

EXERCISE 3A

EXERCISE 3B

EXERCISE 3C

EXERCISE 3D

EXERCISE 3E.1

EXERCISE 3E.2


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\404SA12STU-2_AN.CDR Wednesday, 8 November 2006 9:01:22 AM DAVID3

ANSWERS 405

fdy

dx= 1

3(2x3 ¡ x2)¡

2

3 £ (6x2 ¡ 2x)

gdy

dx= ¡60(5x¡ 4)¡3

hdy

dx= ¡4(3x¡ x2)¡2 £ (3¡ 2x)

idy

dx= 6(x2 ¡

2

x)2 £ (2x+

2

x2)

3 a ¡ 1p3

b ¡18 c ¡8 d ¡4 e ¡ 332

f 0

4 ady

dx= 3x2,

dx

dy= 1

3y¡

2

3

Hint: Substitute y = x3

bdy

dx£

dx

dy=

dy

dy= 1

1 ady

dx= 2x(2x¡ 1) + 2x2

bdy

dx= 4(2x+ 1)3 + 24x(2x+ 1)2

cdy

dx= 2x(3¡ x)

1

2 ¡ 12x2(3¡ x)¡

1

2

ddy

dx= 1

2x¡ 1

2 (x¡ 3)2 + 2px(x¡ 3)

edy

dx= 10x(3x2 ¡ 1)2 + 60x3(3x2 ¡ 1)

fdy

dx= 1

2x¡ 1

2 (x¡x2)3+3px(x¡x2)2(1¡2x)

2 a ¡48 b 406 14

c 133

d 112

3 x = 3 or 35

1 ady

dx=

3(2¡ x) + (1 + 3x)

(2¡ x)2

bdy

dx=

2x(2x+ 1)¡ 2x2

(2x+ 1)2

cdy

dx=

(x2 ¡ 3)¡ 2x2

(x2 ¡ 3)2

ddy

dx=

12x¡ 1

2 (1¡ 2x) + 2px

(1¡ 2x)2

edy

dx=

2x(3x¡ x2)¡ (x2 ¡ 3)(3¡ 2x)

(3x¡ x2)2

fdy

dx=

(1¡ 3x)1

2 + 32x(1¡ 3x)¡

1

2

1¡ 3x

2 a 1 b 1 c ¡ 7324

d ¡ 2827

3 a i never (note:dy

dxis undefined at x = ¡1)

ii x 6 0 and x = 1

b i x = ¡2§p11 ii x = ¡2

1 a 2dy

dxb ¡3

dy

dxc 3y2

dy

dxd ¡y¡2 dy

dx

e 4y3dy

dxf 1

2y¡

1

2

dy

dxg ¡2y¡3 dy

dx

h ¡ 12y¡

3

2

dy

dxi y + x

dy

dxj 2xy + x2 dy

dx

k y2 + 2xydy

dxl 2xy3 + 3x2y2

dy

dx

2 ady

dx= ¡

x

yb

dy

dx= ¡

x

3yc

dy

dx=

x

y

ddy

dx=

2x

3y2e

dy

dx=

¡2x¡ y

x

fdy

dx=

3x2 ¡ 2y

2xg

dy

dx=

y2

2xy + x2

hdy

dx=

2xy3 + y

3y5 + 2xi

dy

dx=

4x¡ 10xy2

3y2 + 10x2y

3 a 1 b ¡ 19

c 45

when y = 2, ¡ 95

when y = ¡3

1 a y = ¡7x+ 11 b 4y = x+ 8c y = ¡2x¡ 2 d y = ¡2x+ 6e y = ¡5x¡ 9 f y = ¡5x¡ 1

2 a 6y = ¡x+ 57 b 7y = ¡x+ 26c y = ¡x+ 6 d y = 18x¡ 161e 3y = x+ 11 f x+ 6y = 43

3 a y = 21 and y = ¡6 b ( 12

, 2p2)

c k = ¡5 d y = ¡3x+ 1

4 a a = ¡4, b = 7 b a = 2, b = 4

5 a 3y = x+5 b 9y = x+4 c y = 2x¡ 74

d y = ¡27x¡ 2423

6 a = 4, b = 3

7 a 16y = x¡ 3 b 57y = ¡4x+ 1042c y = ¡4 d 2y = x+ 1

8 a 6y = ¡x+ 13 b 3y = ¡2x+ 7c y = ¡1 d 5y = ¡9x¡1 e 4y = 5x¡6f 2y + x = 3

9 y = 1 and y = 3x¡ 1

10 a (¡4, ¡64) b (4, ¡31) c ( 13

, ¡ 809

)

d does not meet the curve again

EXERCISE 3F.1

EXERCISE 3F.2

EXERCISE 3G

EXERCISE 3H


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406 ANSWERS

11 a y = (2a¡ 1)x¡ a2 + 9y = 5x, contact at (3, 15)

y = ¡7x, contact at (¡3, 21)

b y = 0, y = 27x+ 54

c y = 0, y = ¡p14x+ 4

p14

1 a f 00(x) = 6 b f 00(x) = 12x¡ 6

c f 00(x) =3

2x5

2

d f 00(x) =12¡ 6x

x4

e f 00(x) = 24¡ 48x f f 00(x) =20

(2x¡ 1)3

2 ad2y

dx2= ¡6x b

d2y

dx2= 2¡

30

x4

cd2y

dx2= ¡ 9

4x¡ 5

2 dd2y

dx2=

8

x3

ed2y

dx2= 6(x2 ¡ 3x)(5x2 ¡ 15x+ 9)

fd2y

dx2= 2 +

2

(1¡ x)3

3 a x = 1 b x = 0, §p6

1 y = 4x+ 2

2 ady

dx= 6x¡ 4x3 b

dy

dx= 1 +

1

x2

3 f 0(x) = 2x+ 2 4 x = 1

5 ady

dx= ¡

2xy

x2 + 3y2b

dy

dx=

y2 ¡ 2x

1¡ 2xy

6 (¡2, ¡25) 7 a = 52

, b = ¡ 32

8 a f 0(2) = 5 b f 0(7) = 16 9 a = 1

2

1 f 0(x) = ¡8

x32 y = 16x¡ 127

2

3 f 0(6) = ¡ 2225

4 adM

dt= 8t(t2 + 3)3

bdA

dt=

12t(t+ 5)¡

1

2 ¡ 2(t+ 5)1

2

t3

5 ady

dx= ¡

2

xpx¡ 3

bdy

dx= 4

³x¡

1

x

´3 ³1 +

1

x2

ć

dy

dx= 1

2(x2 ¡ 3x)¡

1

2 (2x¡ 3)

6 8y = ¡x+ 12

7 y = ¡2x+ 2, meets again at (2, ¡2)

8 y = 7, y = ¡25

9 a 3y2dy

dx+

1

x

dy

dx¡

y

x2= 0

b 3x2y3 + x33y2dy

dx= 2y

dy

dx+ 3

1 a 5 + 3x¡2 b 4(3x2 + x)3(6x+ 1)

c 2x(1¡ x2)3 ¡ 6x(1¡ x2)2(x2 + 1)

2 5y = x¡ 11

3 Hint: Use first principles to differentiate

f(x) = kg(x).

4 one, y = 1 5 f(1) = ¡2 6 y = 20x¡ 79

7 f 0(x) =3

2p3x+ 2

8 (1, ¡2) and (¡2, 19)

9 y = 4x¡ 6, meets curve again at (¡ 32

, ¡12)

1 a f 0(x) =3(x+ 3)2

px¡ 1

2x¡ 1

2 (x+ 3)3

x

b f 0(x) = 4x3px2 + 3 + x5(x2 + 3)¡

1

2

2 a f 00(x) = 6¡2

x3b f 00(x) = ¡ 1

4x¡ 3

2

3 f 0(¡2) = 3 4 a = ¡1, b = 2

5 2y = 3x+ 12

6 f 0(x) = ¡1

2xpx

= ¡ 12x¡ 3

2

7 6y = ¡x+ 12 8 f 0(2) = 8

9 area = 3267152 units2

Hint: tangent is 4y = ¡57x¡ 99

1 a f 0(x) =1

2px(1¡ x)2 ¡ 2

px(1¡ x)

b f 0(x) = 12 (3x¡ x2)¡

1

2 £ (3¡ 2x)

c f 0(x) =1

(2¡ x)2

2 f 0(3) = 6 3 2y = 27x¡ 5593

4 a 3y2dy

dx¡ y2 ¡ 2xy

dy

dx= 3x2

b y + xdy

dx= 2x+

1

2py

dy

dx

5 a = 2 and the tangent is y = 3x¡ 1 which

meets the curve again at (¡4, ¡13)

6 y = 12x¡ 16 7 A = ¡14, B = 21

8 x = ¡ 12

, 32

EXERCISE 3I

REVIEW SET 3A

REVIEW SET 3B

REVIEW SET 3C

REVIEW SET 3D

REVIEW SET 3E


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ANSWERS 407

9 BC = 8p103

(Hint: normal is y = ¡3x+ 8)

10dy

dx= ¡

x+ 2y

2x+ y, ( 1p

2, 1p

2) and (¡ 1p

2, ¡ 1p

2)

Hint: normal at (x1, y1) is

y =

µ2x1 + y1

x1 + 2y1

¶x+

¡2y 2

1 ¡ 2x 21

¢

1 ady

dx= 3x2(1¡ x2)

1

2 ¡ x4(1¡ x2)¡1

2

bdy

dx=

(2x¡ 3)(x+ 1)1

2 ¡ 12(x2 ¡ 3x)(x+ 1)¡

1

2

x+ 1

2 5y = x¡ 11 3 y = ¡9x¡ 2, ( 52

, ¡ 492

)

4 y = 16x¡ 32

5 3: y = ¡x¡ 2

y = (8 + 6p3)x¡ 20¡ 12

p3

y = (8¡ 6p3)x¡ 20 + 12

p3

6 0 7 y = ¡5x+ 19

8 ad2y

dx2= 36x2 ¡

4

x3b

d2y

dx2= 6x+ 3

4x¡ 5

2

9 a = 9, b = ¡16 10 y = x¡ 1

1 ady

dx=

6x¡ 2x2

(3¡ 2x)2

bdy

dx= 1

2x¡ 1

2 (x2¡x)3+3x1

2 (x2¡x)2£(2x¡1)

2 8y = 27x+78 3 A = 9, B = 2, f 00(¡2) = ¡18

4 (¡2, 19), (1, ¡2)

5 y = 0 and y = 27x¡ 54

6 a k = 29 b 28y = ¡5x¡ 13

7 ( 53

, 23

) Hint: tangent is y = ¡5x+ 9

8 116 9 a = 64 10 4y = 3x+ 5

1 a i Q = 100 ii Q = 50 iii Q = 0

b i decreasing 1 unit per year

ii decreasing 1p2

units per year

c Hint: Consider the graph ofdQ

dtagainst t.

dQ

dt=

¡5pt< 0

for all t > 0

2 a 18:2 metres

b t = 4; 19 m, t = 8; 19:3 m, t = 12; 19:5 m

c t = 0: 0:36 m/year t = 5: 0:09m/year

t = 10: 0:04 m/year

d asdH

dt=

9

(t+ 5)2> 0, for all t > 0,

the tree is always growing,

anddH

dt! 0 as t increases

3 a 0oC; 20, 20oC; 24, 40oC; 32

bdR

dT= 1

10+

T

100

cdR

dT> 0 (increasing) for all T > ¡10

4 a i $4500 ii $8250

b i increase of $100 per kmph

ii increase of $188:89 per kmph

cdC

dt= 0 at v =

p50 i.e., 7:1 kmph

5 adp

dv= ¡

c

v2

b v2 > 0, c > 0 ) p0 < 0, v > 0

Pressure always decreases as volume increases.

6 a The near part of the lake is 2 km from the sea,

the furthest part is 3 km.

b x = 12 ;

dy

dx= 0:175, height of hill is

increasing as slope is positive

x = 1 12

;dy

dx= ¡0:225, height of hill is

decreasing as slope is negative

) top of the hill is between x = 12

and

x = 1 12 .

c 2:55 km from the sea, 63:1 m deep

7 a

the volume increases as the length of the sides

increase.

b

millimetre the sides increase the volume increases

by 12 mm3.

c For a small change in side length (¢x), the

increase in volume is approx. ¢x £ surface

area.

REVIEW SET 3F

REVIEW SET 3G

EXERCISE 4A

t

dQ

dt

dV

dx= 3x2 mm3/mm. This is the rate at which

dV

dx= 12 mm3/mm at x = 2. For every


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408 ANSWERS

8 adV

dt= ¡1250

³1¡

t

80

´

b at t = 0 (when the tap was first opened)

cd2V

dt2=

125

8This shows that the rate of change

of V is constantly increasing, i.e., the outflow

is decreasing at a constant rate.

9 a WhendP

dt= 0, the population is not changing

over time, i.e., it is stable.

b 4000 fish c 8000 fish

1 a v(t) = 2t ¡ 4, a(t) = 2

b

c

d At t = 2, s(2) = 1 cm to the left of the

e

f 0 6 t 6 2

2 a v(t) = 98¡ 9:8t, a(t) = ¡9:8

c t =p2, s(

p2) = 8

p2¡ 1 + 10:3

d i t >p2 ii never

4 a v(t) = 3t2 ¡ 18t + 24

a(t) = 6t ¡ 18

b x(2) = 20, x(4) = 16

c i 0 6 t 6 2 and 3 6 t 6 4

ii 0 6 t 6 3

d 28 cm

5 Hint: s0(t) = v(t) and s00(t) = a(t) = g

Show that a = 12g b = v(0) c = 0

EXERCISE 4B.1

EXERCISE 4B.2

t

dVdt

80

1250

s

v

a

(

(

(

t

t

t

):

):

):t

1 3

� �

0

t2

�

0

t�

0

s

v

a

(

(

(

t

t

t

):

):

):

t10

�

0

t�

0 t

0

0 1 3s

t2 4

� �

0

t3

�

0

160 20x

1 a 7 ms¡1 b (h + 5) ms¡1

c 5 ms¡1 = s0(1)

d av. velocity = (2t + h + 3) ms¡1,

limh!0

(2t + h + 3) = s0(t) ! 2t + 3 as h ! 0

2 a ¡14 cm s¡1 b (¡8¡ 2h) cm s¡1

c ¡8 cm s¡1 = s0(2)i.e., velocity = ¡8 cm s¡1 at t = 2.

d ¡4t = s0(t) = v(t)

3 a 23

cms¡2 b

µ2

p1 + h + 1

¶cms¡2

c 1 cms¡2 = v0(1)

d1p

tcms¡2 = v0(t) i.e., the instantaneous

acceleration at time t.

4 a velocity at t = 3b acceleration at a = 5c velocity at t, i.e., v(t)d acceleration at t, i.e., a(t)

The object is initially 3 cm to the right of the

origin and is moving to the left at 4 cm s¡1. It

is accelerating at 2 ms¡2 to the right.

The object is instantaneously stationary, 1 cm

to the left of the origin and is accelerating to

the right at 2 ms¡2.

b s(0) = 0 m above the ground

v(0) = 98 ms¡1 skyward

c t = 5 Stone is 367:5 m above the ground and

moving skyward at 49 ms¡1. Its speed is de-

creasing. t = 12 Stone is 470:4 m above the

ground and moving groundward at 19:6 ms¡1.

Its speed is increasing.

d 490 m e 20 seconds

3 a v(t) = 12¡ 6t2, a(t) = ¡12t

b s(0) = ¡1, v(0) = 12, a(0) = 0

Particle started 1 cm to the left of the origin and

was travelling to the right at a constant speed

of 12 cm s¡1.


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ANSWERS 409

6 see proof on page 107

but in case 1 consider a(t) < 0and in case 2 consider a(t) > 0

as t ! 1, s(t) ! 1, v(t) ! 40 ms¡1

1 a i x > 0 ii never

b i never ii ¡2 < x 6 3c i ¡2 6 x 6 0 ii 0 6 x 6 2d i x 6 2 ii x > 2e i never ii all real x

f i all real x ii never

g i 1 6 x 6 5 ii x 6 1, x > 5h i 2 6 x < 4, x > 4 ii x < 0, 0 < x 6 2i i x 6 0, 2 6 x 6 6 ii 0 6 x 6 2, x > 6

1 a increasing for x > 0, decreasing for x 6 0

b decreasing for all x

c increasing for x > ¡ 34

,

decreasing for x 6 ¡ 34

d increasing for x > 0, never decreasing

e decreasing for x > 0, never increasing

f increasing for x 6 0 and x > 4,

decreasing for 0 6 x 6 4

g increasing for ¡p

236 x 6

p23

,

decreasing for x 6 ¡p

23

, x >p

23

h decreasing for x 6 ¡ 12

, x > 3,

increasing for ¡ 126 x 6 3

i increasing for x > 0, decreasing for x 6 0

j increasing for x > ¡ 32+

p52

and x 6 ¡ 32¡

p5

2

decreasing for ¡ 32¡

p5

26 x 6 ¡ 3

2+

p5

2

k increasing for x 6 2¡p3, x > 2 +

p3

decreasing for 2¡p3 6 x 6 2 +

p3

l increasing for x > 1, decreasing for 0 6 x 6 1

2 a increasing for ¡1 6 x 6 1, x > 2

decreasing for x 6 ¡1, 1 6 x 6 2

b increasing for 1¡p2 6 x 6 1, x > 1 +

p2

decreasing for x 6 1¡p2, 1 6 x 6 1 +

p2

c increasing for x 6 2¡p2, 3 6 x 6 2 +

p2

decreasing for 2¡p2 6 x 6 3, x > 2 +

p2

3 a i

ii increasing for ¡1 6 x 6 1decreasing for x 6 ¡1, x > 1

b i

ii increasing for ¡1 6 x < 1decreasing for x 6 ¡1, x > 1

c i

ii increasing for ¡1 6 x < 1, 1 < x 6 3decreasing for x 6 ¡1, x > 3

d i

ii increasing for x < ¡2, ¡2 < x 6 ¡p2,

x >p2 and decreasing for ¡2 6 x < ¡1,

¡1 < x 6p2

4 a increasing for x >p3, x 6 ¡

p3

decreasing for ¡p3 6 x < ¡1, ¡1 < x < 1,

1 < x 6p3

b increasing for x > 2decreasing for x < 1, 1 < x 6 2

1 a A - local min B - local max

C - horizontal inflection

b

c i x 6 ¡2, x > 3 ii ¡2 6 x 6 3

2 a b

EXERCISE 4C.1

EXERCISE 4C.2

EXERCISE 4C.3

1 2s

stops for an instant

3s

10s

x 1 1

�

x 1 1

�

x 1 1 3

� �

~2̀ ~2̀x

2 1

��

x 2 0 3

� �

(0, 1)

x

ƒ( )x

1

horizontal

inflection

(0, 2)

x

ƒ( )x

~`2 ~`2

local min.

7 a s(0) = 1 cm, v(0) = 3 cm s¡1

a(0) = ¡6 cm s¡2

as t ! 1, s(t) ! 1, v(t) ! 1

b s(0) = 3 cm, v(0) = ¡ 12

cm s¡1

a(0) = 14

cm s¡2

as t ! 1, s(t) ! ¡1, v(t) ! 0

c s(0) = 10 cm, v(0) = 35 cm s¡1

a(0) = 7 12

cm s¡2


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\409SA12STU-2_AN.CDR Thursday, 9 November 2006 9:05:30 AM PETERDELL

410 ANSWERS

c d

e f

g h

i j

3 x = ¡b

2a, local min if a > 0, local max if a < 0

4 a a = ¡12, b = ¡13

b (¡2, 3) local max. (2, ¡29) local min

5 P (x) = ¡9x3 ¡ 9x2 + 9x + 2

6 a i x = ¡1, x = 5 ii no turning points

b i x = ¡2 ii (4, 14

) is a local max

c i x = 2

ii (2¡p8, 4¡ 2

p8) is a local max

(2 +p8, 4 + 2

p8) is a local min

d i x = ¡3

ii (¡3¡p13, ¡9¡ 2

p13) is a local max

(¡3 +p13, ¡9 + 2

p13) is a local min

e i x = 3, x = ¡2 ii ( 12 , 125 ) is a local max

f i x = ¡2 ii ( 613

, 2364

) is a local min

( 1, 4)

(1, 0)

2

x

ƒ( )x

local min.

local max.

2 ~`2 ~`2

( 1, 1) (1, 1)

(0, 0)

x

ƒ( )x

local min. local min.

localmax.

(2, 9)

1

x

ƒ( )x

horizontal

inflection 2

x

ƒ( )x

(no stationary points)

x

ƒ( )x

local min.

1

( , )Qr_ - Qr_( 2, 27)

(1, 0)

x

ƒ( )x

3 3

local min.

horizontal

inflection

(0, 1)

x

ƒ( )x

local max.

(1, 9) ( 1, 9)

(0, 8)

2 x

ƒ( )x


localmax.

2

x = 1 x = 5

x

ƒ( )x

( )2 8, 4 2 8 ~` ~`

( )2+ 8, 4+2 8~` ~`

x

ƒ( )x

local max.

local min.

x = 2

x

ƒ( )xlocal max.

Er_

(4, )Qr_

x = 2

x

local min.

x = 2

ƒ( )x

&qH_e_\' Wy_Er_\*Qw_

( ) 3+ , 9+2~`13 ~`13( ) 3 , 9 2~`13 ~`13

x

local max. local min.

ƒ( )x

Te_

x = 3

x = 3x = 2

(Qw_ wA_t_),

x

ƒ( )x

localmin.


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ANSWERS 411

7 a greatest value = 63 (at x = 5)

least value = ¡18 (at x = 2)

b greatest value = 4 (at x = 3 and x = 0)

least value = ¡16 (at x = ¡2)

c i greatest value = 2 (at x = 4)

least value = ¡ 14

(at x = 14

)

ii greatest value = 6 (at x = 9)

least value = 0 (at x = 1)

8 Maximum hourly cost = $680:95 when 150 hinges

are made per hour. Minimum hourly cost = $529:80when 104 hinges are made per hour.

1 a no inflection

b horizontal inflection at (0, 2)

c non-horizontal inflection at (2, 3)

d horizontal inflection at (¡2, ¡3)

e horizontal inflection at (0, 2)

non-horizontal inflection at (¡ 43

, 31027

)

f no inflection

2 a i local minimum at (0, 0)

ii no points of inflection

iii decreasing for x 6 0, increasing for x > 0

iv function is convex for all x

v

b i horizontal inflection at (0, 0)

ii horizontal inflection at (0, 0)

iii increasing for all real x

iv concave for x 6 0, convex for x > 0

v

c i f 0(x) 6= 0, no stationary points

ii no points of inflection

iii incr. for x > 0, vnever decreasing

iv concave for x > 0,

never convex

d i local maximum at (¡2, 29)

local minimum at (4, ¡79)

ii non-horizontal inflection at (1, ¡25)

iii increasing for x 6 ¡2, x > 4decreasing for ¡2 6 x 6 4

iv concave for x 6 1, convex for x > 1v

e i horizontal inflection at (0, ¡2)

local minimum at (¡1, ¡3)

ii horizontal inflection at (0, ¡2)

non-horizontal inflection at (¡ 23

, ¡ 7027

)

iii increasing for x > ¡1,

decreasing for x 6 ¡1

iv concave for ¡ 23 6 x 6 0

convex for x 6 ¡ 23

, x > 0

v

f i local minimum at (¡2, ¡23)

horizontal inflection at (1, 4)

ii horizontal inflection at (1, 4)

non-horizontal inflection at (¡1, ¡12)

iii increasing for x > ¡2,

decreasing for x 6 ¡2

iv concave for ¡1 6 x 6 1,

convex for x 6 ¡1, x > 1

v

g i local minimum at (¡p2, ¡1),

local maximum at (0, 3),

EXERCISE 4C.4

(0,0) x

ƒ( )x

local min.

(0,0) x

ƒ( )x

horizontal

inflection

x

ƒ( )x �

(1, 25)

(4, ) ��

( , ) ��

x

ƒ( )x

local min.

local max.

non-horizontal

inflection

80

� �

( 2, 23)

( 1, 12) (1, 4)

x

ƒ( )x

local min.


non-horizontalinflection

non-horizontal

inflection

(0, 2)

( 1, ) �

x

ƒ( )x

local min.

horizontal

inflection

1 1 2

&-\We_\'- Uw_Pu_\*


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412 ANSWERS

local minimum at (p2, ¡1)

ii non-horizontal inflection at (p

23

, 79

)

non-horizontal inflection at (¡p

23

, 79

)

iii increasing for ¡p2 6 x 6 0, x >

p2

decreasing for x 6 ¡p2, 0 6 x 6

p2

iv concave for ¡p

23 6 x 6

p23

convex for x 6 ¡p

23

, x >p

23

v

h i no stationary points ii no inflections

iii increasing for x > 0, never decreasing

iv concave for x > 0, never convex

v

3 a Hint: Show that f 00(x) = 2a(3x¡[®+¯+°])

and consider f 00(x) = 0

b i Hint: Show that if f(x) has two distinct

turning points then 0 = 3ax2 + 2bx + c

has two distinct solutions. Then consider the

discriminant.

ii Hint: Show that

3ax2+bx+c = k(x¡p)(x¡q) for some

k 6= 0 then equate the coefficients to show

p + q =2b

3a.

1 50 fittings 2 250 items 3 10 blankets

4 a P (x) = 250¡x

8, x > 800 b $25 c $10

5 25 kmph 6 b

c Lmin = 28:28 m,

x = 7:07 m

d

7 a 2x cm b V = 200 = 2x £ x £ h

c Hint: Show h =100

x2and substitute into the

surface area equation.

d

e SAmin = 213:4 cm2,

x = 4:22 cm

f

8 a recall that Vcylinder = ¼r2h and that

1 L = 1000 cm3

b recall that SAcylinder = 2¼r2 + 2¼rh

c

d A = 553 cm2,

r = 5:42 cm

e

9 b 6 cm £ 6 cm

10 a Area = 4xp25¡ x2 b 7:07 cm £ 7:07 cm

11 a 0 6 x 6 63:66c x = 63:66 m, l = 0 m (i.e., circular track)

12 a Hint: Show that AC = µ

360 £ 2¼ £ 10

b Hint: Show that 2¼r = AC

c Hint: Use the result from b and Pythagoras’

theorem.

d V (µ) = 13¼¡

µ

36

¢2q100¡

¡µ

36

¢2

EXERCISE 4D

(0, 3)

( ~̀ ) 2, 1

( ~̀ We_ Uo_ ) , (~` We_ Uo_ ),

(~` )2, 1

x

ƒ( )x


local max.

non-

horizontal

inflection

non-

horizontal

inflection

x

ƒ( )x �

Ql_Y_

33

y (m )2

x (m)

14.14 m

7.07 m

y (cm )2

x (cm)

450

10

8.43 cm4.22 cm

5.62 cm

A (cm )2

x (cm)

1500

15

10.84 cm

5.42 cm


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ANSWERS 413

e

f µ = 293:9o

13 b Hint: Show L = 25x2 + 200xy then use

the result from a.

c 1:59 m £ 1:59 m £ 0:397 m

14 a 2x units £100

x2units

b Hint: Show thatdA

dx= ¡

200

x2

c Pmin = 27:8 units, 9:28 units £ 4:64 units

15 13:44 cm from left (i.e., uses 13:44 cm for square

tubing)

16 a For x < 0 or x > 6, X is not on AC.

c x = 2:67 km This is the distance from A

to X which minimises the time taken to get

from B to C. (Proof: Use sign diagram or sec-

ond derivative test. Be sure to check the end

points.)

17 3:33 km

18 radius = 31:7 cm, height = 31:7 cm

(Note: 100 L = 0:1 m3)

19 4 m from the 40 cp globe

20 a D(x) =p

x2 + (24¡ x)2

bd[D(x)]2

dx= 4x ¡ 48

c Smallest D(x) = 17:0

Largest D(x) = 24,

which is not an accept-

able solution as can be

seen in the diagram.

21 a Hint: Use the cosine rule.

b 3553 km2

c 5:36 pm

22 a QR =³2 + x

x

´m

c Hint: Discard all solutions < 0 as x > 0.

d 416 cm

23 AB should be 7:5 cm

24 between A and N, 2:566 m from N

25 at grid reference (3:544, 8) 26 A = (4a, 0)

27p

32: 1 28 e 63:7%

29 b A(r) = 6(k ¡ 43¼r3)

2

3 + 4¼r2

30 c x =ac

a + b

c At t = 2, particle is 1 cm to the left of the

origin, is stationary and is accelerating towards

the origin.

d t = 1, s = 0 and t = 2, s = ¡1

e

f Speed is increasing for 1 6 t 6 1 12

and t > 2.

2 a i $312 ii $1218:75

b i $9:10 per kmph ii $7:50 per kmph

c 3 kmph

3 a local maximum at (¡2, 51)

local minimum at (3, ¡74)

non-horizontal inflection at ( 12 , ¡11:5)

b increasing for x 6 ¡2, x > 3decreasing for ¡2 6 x 6 3

c concave for x 6 12 , convex for x > 1

2

d

4 a x = ¡3

b y-intercept at y = ¡ 23 , x-intercept at x = 2

3

REVIEW SET 4A

V (cm )3

� (°)

500

360

12 �

0 24

12 m

17 m

24 m

t4.605

�

0

v(t):

a(t):

t1 2

� �

0

t1\Qw_

�

0

0 1 5s

x

y

local min (3, 74)

local max ( 2, 51) non-horizontalinflection

( , )\Qw_\ -37\Qw_\

1 a v(t) = (6t2 ¡ 18t + 12) cm s¡1

a(t) = (12t ¡ 18) cm s¡2

b s(0) = 5 cm to left of origin

v(0) = 12 cm s¡1 towards origin

a(0) = ¡18 cm s¡2 (reducing speed)


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414 ANSWERS

c f 0(x) =11

(x + 3)2

d There are no stationary points.

5 b k = 9 6 1:41 m

1 a y =500

x, x > 0

bdy

dx= ¡

500

x2as x2 > 0, ¡

500

x2< 0

c As the breadth of the rectangle increases, the

length decreases.

2 a 2 m

b H(3) = 4 m, H(6) = 4:67 m, H(9) = 5 m

c H0(0) = 1:33 m/year, H0(3) = 0:333 m/year,

H 0(6) = 0:148 m/year, H0(9) = 0:083 m/year

d As H0(t) > 0, the tree is always growing.

e

3 a y = ¡4 b x = 1 (only 1 intercept)

c no stationary points, non-horizontal inflection at

(¡ 13

, ¡ 12427

)

d

4 a v(t) = 3¡1

2p

t, a(t) =

1

4tp

t

v(t):

a(t):

b x(0) = 0, v(0) is undefined,

a(0) is undefined

d Changes direction at t = 136

, 0:083 cm to the

left of the origin.

e Particle’s speed is decreasing for 0 6 t 6 136

.

5 a y =1

x2c 1:26£ 1:26£ 0:630

6 a Hint: Draw a line from B to AC, bisecting the

angle ]ABC.

bd[A(x)2]

dx= 5000x ¡ 4x3

x = 35:4 (i.e., AC = 70:7 m)

1 a 100 g

b i 2250 g ii 4711:8 g iii 4992:8 g

c i 0 ii 1514 g/week iii 333:5 g/week

d

2 a t > 2 b t > 17

3 a y-intercept at y = 0,

x-intercept at x = 0 and x = 2

b local maximum at¡23

, 3227

¢, local minimum at

(2, 0), non-horizontal inflection at¡43

, 1627

¢c

4 a v(t) = 3t2 ¡ 4t ¡ 4, a(t) = 6t ¡ 4

v(t):

a(t):

b i 0 6 t < 1, t > 1+p21

2

ii 1 < t < 1+p21

2

c reverses direction at (2, ¡3)

d i velocity is decreasing for 0 6 t 6 23

ii speed is decreasing for 23 6 t 6 2

5 b A(x) = 200x ¡ 2x2 ¡ 12¼x2

c

REVIEW SET 4B

REVIEW SET 4C

6

2t (years)

H t( ) metres

x

y

1

4&- Qe_' -\Qs_Wj_R_*�

t�

0Ae_y_

t�

0

5000

100 t (weeks)

W t( ) (g)

t�

02

t�

0We_

local min(2, 0)

local max

,&\We_\ Ew_Wu_\*axis intercept

at (0, 0)


&\Re_\' Qw_Yu_\*

y

x

28.0 m

56.0 m

c Particle is 24 cm to the right of the origin and is

travelling to the right at 2:83 cm s¡1. Its speed

is increasing.


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ANSWERS 415

6 b R(x) = kx(1¡ x2)3

2

d x = 0, 12

, 1, however x = 12

is the only

sensible solution as x = 0 and 1 result in

a beam of width or depth = 0.

e 12

m £p32

m

1 a x > 0

b y-intercept at y = 0, x-intercepts at x = 0, 1

c local minimum at ( 49

, ¡427

), no points of

inflection

d

2 a y-intercept at y = ¡1x-intercepts at x = 1, x = ¡1

b x2 + 1 > 0 for all real x

(i.e., denominator is never 0)

c local minimum at (0, ¡1)

e

c Speed is increasing for 0 6 t < 1

4 6 cm from the ends

5 maximum = 21 (at x = 4)

minimum = 1 (at x = 2 and ¡1)

6 a depth =10 000

xcm c 80:7 cm £ 161:4 cm

1 a 49 suites b $943 per month

2 a a = ¡6

b local maximum at (¡p2, 4

p2),

local minimum at (p2, ¡4

p2)

c

3 a x = ¡2, x = 1

b local min. at (0, 1), local max. at (4, 19

)

c y-intercept at y = 1, x-intercept at x = 2

d

e p < 0, 0 < p < 19

or p > 1

4 a v(t) = 2 +4

t2, a(t) = ¡

8

t3

c Particle never changes direction.

d

e i velocity is never increasing

ii speed is never increasing

5 a Hint: Use Pythagoras to find h as a function

of x and then substitute into the equation for

the volume of a cylinder.

b radius = 4:08 cm, height = 5:77 cm

6 10 additional trees

7 a LQ =8

xkm b Hint: Show that

(length of pipe)2 = (LQ + 1)2 + (8 + x)2

then simplify.

c 11:2 km (when x = 2 km)

1 b³p

32

, 32

´2 2:53 pm 3 4 cm

4 77 or 78 (both give revenue of $240:24)

5 a Hint: Draw in construction lines OA and OC

to find the base length and height respectively.

REVIEW SET 4D

REVIEW SET 4E REVIEW SET 4F

v(t): a(t):

1

axisintercept

(0,0)

local min

&Ro_\'\-\Fw_u_*

x

y

non-horizontalinflection at

non-horizontalinflection at

local min(0, 1)

�21

3

1 ,& *��21

3

1 ,& *

x

y

local min

&~`2\' -4~̀2\*

local max

&-~`2\' 4~̀2\*

~`6-~`6

x

y

t�

0

t

0

s

local max

&4' Qo_*

xx-intercept

2�

local min(0, 1)

x

f x( )

x 2� x 1�

3 a v(t) = 15 +120

(t ¡ 1)3cm s¡1

a(t) = ¡360

(t ¡ 1)4cm s¡2

b The particle is 30 cm to the right of the origin

and is travelling to the right at 30 cm s¡1. It is

slowing down at 22:5 cm s¡2.

b The particle is 2 cm to the left of the origin and

is moving to the right at 6 cm s¡1. Its speed is

decreasing at a rate of 8 cm s¡2.


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416 ANSWERS

b Hint: Show that x =r

2and use ¢OAN and

CNB to show that all sides have lengthp3r.

6 Hint: Show that V = x(k ¡ 2x)2

1 a f 0(x) = 4e4x b f 0(x) = ex

c f 0(x) = ¡2e¡2x d f 0(x) = 12e

x

2

e f 0(x) = ¡e¡x

2 f f 0(x) = 2e¡x

g f 0(x) = 2ex

2 + 3e¡x h f 0(x) =ex ¡ e¡x

2

i f 0(x) = ¡2xe¡x2

j f 0(x) = e1

x £¡1

x2

k f 0(x) = 20e2x l f 0(x) = 40e¡2x

2 ad

dx[xex] = ex + xex

bd

dx[x3e¡x] = 3x2e¡x ¡ x3e¡x

cd

dx

hex

x

i=

xex ¡ ex

x2d

d

dx

hx

ex

i=

1¡ x

ex

ed

dx[x2e3x] = 2xe3x + 3x2e3x

fd

dx

·¡expx

¸=

xex ¡ 12ex

xpx

gd

dx[pxe¡x] = 1

2x¡ 1

2 e¡x ¡ x1

2 e¡x

hd

dx

hex + 2

e¡x + 1

i=

ex + 2 + 2e¡x

(e¡x + 1)2

3 a f 0(x) = 4ex(ex+2)3 b f 0(x) =¡e¡x

(1¡ e¡x)2

c f 0(x) =e2x

pe2x + 10

d f 0(x) =6e3x

(1¡ e3x)3

e f 0(x) = ¡e¡x

2

¡1¡ e¡x

¢¡ 3

2

f f 0(x) =1¡ 2e¡x + xe¡x

p1¡ 2e¡x

4 bdny

dxn= kny

5 Hint: Finddy

dxand

d2y

dx2and substitute into the

equation.

6 a local maximum at (1, e¡1)

b local max. at (¡2, 4e¡2), local min. at (0, 0)

c local minimum at (1, e)

d local maximum at (¡1, e)

7 a slope =1

1 + 2e(+ 0:1554)

b at (0, 1), slope = 1 at (0, 0), slope = 0

1 a lnN = ln 50 + 2tb lnP = ln 8:69¡ 0:0541tc lnS = 2 ln a¡ kt

2 a D + 8:166£ e0:69t

b G + 1:815£ 10¡14 £ e0:0173t

c P = ge¡2t d F = x2e¡0:03t

3 a 2 b 12

c ¡1 d ¡ 12

e 3 f 9

g 15

h 14

4 a ln 30 b ln 16 c ln 25 d ln 23e2

5 a eln 2 b eln 10 c elna d ex ln a

6 a x = ln 2 b no real solutions

c no real solutions d x = ln 2 e x = 0f x = ln 2 or ln 3 g x = 0 h x = ln 4

i x = ln³

3+p5

2

ór ln

³3¡

p5

2

´7 a x = 2 (note that x > 0)

b no solutions exist

8 a (ln 3, 3) b (ln 2, 5)

c (0, 2) and (ln 5, ¡2)

9 a A = (ln 3, 0) B = (0, 2)

b f 0(x) = ¡ex which is < 0 for all x

c f 00(x) = ¡ex which is < 0 for all x,

so f(x) is concave

d

e as x ! ¡1, ex ! 0, so f(x) ! 3

10 a P = ( 12ln 3, 0) Q = (0, ¡2)

b f 0(x) = ex + 3e¡x > 0 for all x

c f(x) is concave below the x-axis and convex

above the x-axis

d

11 a f(x): x-intercept is at x = ln 3y-intercept is at y = ¡2

g(x): x-intercept is at x = ln¡53

¢y-intercept is at y = ¡2

EXERCISE 5A

EXERCISE 5B

ln 32

y 3�

x

y

2 &\Qw_\ *ln3, 0


x

y


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ANSWERS 417

b f(x): as x ! 1, f(x) ! 1as x ! ¡1, f(x) ! ¡3 (above)

g(x): as x ! 1, g(x) ! 3 (below)

as x ! ¡1, g(x) ! ¡1

c intersect at (0, ¡2) and (ln 5, 2)

d

1 ady

dx=

1

xb

dy

dx=

2

2x + 1c

dy

dx=

1¡ 2x

x ¡ x2

ddy

dx= ¡

2

xe

dy

dx= 2x lnx + x

fdy

dx=

1¡ lnx

2x2g

dy

dx= ex lnx +

ex

x

hdy

dx=

2 lnx

xi

dy

dx=

1

2xplnx

jdy

dx=

e¡x

x¡ e¡x lnx

kdy

dx=

ln 2x

2p

x+

1p

xl

dy

dx=

lnx ¡ 2p

x(lnx)2

mdy

dx=

4

(1¡ x)n

dy

dx=

ln 4x

2p

x+

1p

x

ody

dx= ln(x2 + 1) +

2x2

x2 + 1

2 ady

dx=

¡1

1¡ 2xb

dy

dx=

¡2

2x + 3

cdy

dx= 1 +

1

2xd

dy

dx=

1

x¡

1

2(2¡ x)

edy

dx=

1

x + 3¡

1

x ¡ 1f

dy

dx=

2

x+

1

3¡ x

3 ady

dx=

¡y ln y

xb

dy

dx=

2xy ln y

y ¡ x2

cdy

dx=

y3

1¡ 2xy2

4 ady

dx=³1

x+ 1 +

4

2x + 1

´¡xex(2x + 1)2

¢= (2x2 + 7x + 1) (ex(2x + 1))

bdy

dx= 2x ln 2 c

dy

dx= 3¡x(¡ ln 3)

ddy

dx=³ln 3¡

1

x

´3x

x

edy

dx=

2 lnx

x£ xlnx = 2 lnx £ xlnx¡1

fdy

dx=³

1

2x+

6x + 3

x2 + x¡ 4x

´µpx(x2 + x)3

e2x2

¶5

dy

dx= 2x ln 2

6 a x =e3 + 1

2(+ 10:54)

b no, hence there is no y-intercept

c slope = 2 d x > 12

e f 00(x) =¡4

(2x ¡ 1)2< 0 for all x > 1

2, so

f(x) is concave

f

7 a x > 0

b Hint: Find when f 0(x) = 0 and show this

point to be a local minimum. You must also

consider f(x) as x ! 0 and x ! 1.

8 Hint: Show that as x ! 0, f(x) ! ¡1, and

x ! 1, f(x) ! 0.

9 Hint: Show that f(x) > 1 for all x > 0.

1 a

b i 177:1 units ii 74:7 units

c When t + 0:667 hours = 40 min

d i 0:1728 h + 10 min ii 91 min

e t = 43

or 80 min. This is when the effective-

ness changes from decreasing at an increasing

rate to decreasing at a decreasing rate.

2 a i Show that f 0(t) = Ae¡bt(1¡ bt)

ii Show that f 00(t) = Abe¡bt(bt ¡ 2)

b Local max. in c is at t = 23

Point of inflection in e is at t = 43

EXERCISE 5C

EXERCISE 5D.1

y

xln 3

ln Te_

(0, 2)

(ln 5, 2)

y 3�

y 3�

y g x( )�

y f x( )�

x

x = Qw_f x( )

213�e

t654321

200

150

100

50

E

¡


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75

50

50

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95

100

100 0 05 5

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25

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75

50

50

95

95

100

100


418 ANSWERS

3 a

c 0 6 t 6 12

d Point of inflection is at (1, 3:38). This is

when the velocity changes from decreasing

at an increasing rate to decreasing at a

decreasing rate.

4 a

b + 13 900 ants c + 24 000 ants

d Yes, 25 000 ants e after 3:67 months

5 a2C

3bees b 37:8% increase

c Yes, C bees d 3047 bees

e B0(t) =0:865C

e1:73t(1 + 0:5e¡1:73t)2

and so B0(t) > 0 for all t > 0) B(t) is increasing over time.

f

7 a All power functions of this form pass through

(0, 0):

b For the logistic function with horizontal asymp-

tote y = C, the y-coordinate of the point of

inflection is y = C

2. For this function it appears

to be about C

3:

c If a is the x-coordinate of the local maximum

of a surge function, its point of inflection is at

x = 2a. This is not the case for the graph given.

d No exponential function of this form passes

through (0, 0):

e No part of a logistic function can be negative.

f No cubic function can have 3 turning points.

g Surge functions pass through (0, 0):h No exponential function of this form is negative.

1 a/b

suggests the model is exponential

B + 1:306e0:2347t grams

c approximately 1:31 grams

d i 2:97 g ii 13:65 g

e We know that the data for 3 and 4 days fits

the model well, and so it is likely that growth

patterns between those times are accurately pre-

dicted by the model.

f Any prediction beyond 8 days fails to take into

account any potential limits on the growth. It

is unlikely that the bacteria would grow expo-

nentially indefinitely.

2 a/b

suggests that the model is exponential

T + 61:32e¡0:0593t oC

c 61:32oC d i 25:19oC ii 0:29oC

e As we have data for 10 and 20 minutes that

fits the model, predictions in this interval (i.e.,

15 minutes) are likely to be accurate.

f No. Assuming no power failure, soon after the

80 minute measurement, the water would have

reached and stayed at 0oC. The model predicts

this result with good accuracy.

3 a


b I + 101:0te¡0:5312t people

c Day 13 d 8 people

EXERCISE 5D.2

t

4321

5

4

3

2

1

v

t

108642

25000

20000

15000

10000

5000

A

+��

t

32.521.510.5

5000

4000

3000

2000

1000

B

C��

3047

2

4

6

8

1 2 3 4 5 6 7 8 9

t (days)

B (grams)

10

20

30

20 40 60 80

T (°C)

t (mins)

20

40

60

80

2 4 6 8 t

I


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50

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100

100 0 05 5

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75

50

50

95

95

100

100


ANSWERS 419

4 a


b P + 5:399te¡0:3404t units

c i 4:92 units ii 1:79 units

d after approximately 2:94 hours

e after approximately 5:88 hours

f for 0:43 6 t 6 9:55

g At 9:55 hours

5 a

A logistice model is suggested.

b P +

³0:9877

1 + 54:53e¡0:8333h

óf the community

c i 0:066 of the community per hour

ii 0:204 of the community per hour

d after approximately 4:80 hours

6 a

logistic because of its shape

b S +

³804:6

1 + 39:80e¡0:8980t

´snakes

c 804 snakes d yes - 805 snakes

e i 82:51 snakes per year

ii 180:3 snakes per year

iii 94:10 snakes per year

1 y = ¡1

ex +

2

e2 3y = ¡x + 3 ln 3¡ 1

3 A is¡23

, 0¢

, B is (0, ¡ 2e)

4 y = ¡2

e2x +

2

e4¡ 1

5 y = eax + ea(1¡ a) so y = ex is the tangent

to y = ex from the origin

6 a x > 0

b f 0(x) > 0 for all x > 0, so f(x) is al-

ways increasing. Its slope is always positive.

f 00(x) < 0 for all x > 0, so f(x) is con-

cave for all x > 0:

c

normal has equation f(x) = ¡ex + 1 + e2

7 63:43o

8 a i 200 grams ii 256:8 grams iii 423:4

b 3 hours 13 minutes

c i 100 grams per hour

ii 271:8 grams per hour

d

9 a i decreasing by 4:524 amps per second

ii decreasing by 1:839 amps per second

b 39:1 seconds

10 a k = 150

ln 2 (+ 0:0139)

b i 20 g ii 14:34 g iii 1:948 g

c 9 days and 6 minutes (216:1 hours)

d i ¡0:0693 grams per hour

ii ¡2:644£ 10¡7 grams per hour

e Hint: You should finddW

dt= ¡ 1

50ln 2£ 20e¡

1

50ln 2t

11 a k = 115

ln¡9515

¢(+ 0:1231) b 100oC

d i decreasing by 11:69oC per minute

ii decreasing by 3:415oC per minute

iii decreasing by 0:998oC per minute

12 a 43:86 cm b 10:37 years

c i growing by 5:45 cm per year

ii growing by 1:88 cm per year

13 a A = 0 b k =ln 2

3(+ 0:2310)

c 0:7278 units of alcohol produced per hour

1

2

3

4

5

6

5 10 15

t (h)

P

0.2

0.4

0.6

0.8

1

2 4 6 8 10

P

h (hours)

200

400

600

800

2 4 6 8

S

t (years)

t (hours)

W (grams)

200

�( )x

( )e,��

x

EXERCISE 5E


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50

50

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100

100 0 05 5

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25

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75

50

50

95

95

100

100


420 ANSWERS

14 a f(x) does not have any x or y-intercepts

b as x ! 1, f(x) ! 1, as x ! ¡1,

f(x) ! 0 (negative)

c local minimum at (1, e)

d

e ey = ¡2x ¡ 3

15 a local minimum at (0, 1)

b as x ! ¡1, f(x) ! +1

c e¡x ! 0 so f(x) ! x

d f 00(x) = e¡x > 0 for all x

e

f Hint: Use your result from a.

16 a No, because f 0(t) > 0 for all x for any

value a.

b

c f 0(t) > 0 and f 00(t) < 0 for all t, so the

curve is always concave and f(t) is always

increasing.

d

e after 3:466 seconds

19 a at 4:41 months old

b

20 a There is a local maximum at

µ0,

1

¼p2

¶.

f(x) is increasing for all x 6 0 and

decreasing for all x > 0.

b Inflections at

µ¡1,

1

¼p2e

¶and

µ1,

1

¼p2e

¶c as x ! 1, f(x) ! 0 (positive) as

x ! ¡1, f(x) ! 0 (positive)

d

21 20 kettles 22 C =³

1p2

, e(¡1

2)´

23 267 torches

24 a Hint: They must have the same y-coordinate

at x = 6 and the same slope.

c a =1

2ed y = e¡

1

2 x ¡ 12

25 after 13:8 weeks 26 a =

pe

2, b = ¡ 1

8

1 ady

dx= 3x2ex

3+2 bdy

dx=

xex ¡ 2ex

x3

2 y =e

2x +

1

e¡

e

2

REVIEW SET 5A

x

f x( )local min

(1, )e

vertical asymptote0x �

x

f x( )

local min(0, 1)

t

f t( )y 1�

t (years)

A t( )

minimum(e , 0.6321) 1

(5, 5ln 5 1)�

t (sec)

v t( ) (cm s )� 1

60

y�100

x

f x( )


e2

1,1#& *


�e2

1,1#& *

local max

2

1,0#& *

17 a v(t) = 10e¡t

10 cm s¡1

a(t) = ¡e¡t

10 cm s¡2

b x(0) = 100 cm v(0) = 10 cm s¡1

a(0) = ¡1 cm s¡2

c x(5) = 139:3 cm v(5) = 6:065 cm s¡1

a(5) = ¡0:6065 cm s¡2

d 6:931 seconds

e As v(t) and a(t) are opposite sign, speed

is decreasing. Because a(t) < 0, velocity is

decreasing also.

18 a v(t) = 100¡ 40e¡t

5 cm s¡1

a(t) = 8e¡t

5 cm s¡2

b s(0) = 200 cm on positive side of origin

v(0) = 60 cm s¡1 a(0) = 8 cm s¡2

c as t ! 1, v(t) ! 100 ms¡1 (below)


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ANSWERS 421

3

4 a y-intercept at y = ¡1 no x-intercept

b f(x) is defined for all x 6= 1

c f 0(x) < 0 for x < 1 and 1 < x 6 2 and

f 0(x) > 0 for x > 2, f 00(x) > 0 for x > 1and f 00(x) < 0 for x < 1.

So the slope of the curve is negative for all

defined values of x 6 2 and positive for all

x > 2. The curve is concave for x 6 1 and

convex for x > 1.

d tangent is y = e2

5 a 60 cm b i 4:244 years ii 201:2 years

c i 16 cm per year ii 1:951 cm per year

6 a v(t) = ¡8e¡t

10 ¡ 40 ms¡1 t > 0

a(t) = 45e¡

t

10 ms¡2 t > 0

b s(0) = 80 m v(0) = ¡48 ms¡1

a(0) = 0:8 ms¡2

c as t ! 1, v(t) ! ¡40 ms¡1 (below)

d

e t = 6:931 seconds

7 A(1, e¡1)

8 a

because of its shape and as the curve seems to

approach an upper limit as x ! 1, this sug-

gests a logistic model would be appropriate.

b T +86:53

1 + 6:473e¡0:1681tc 87 turtles

d i 85 turtles ii 86 turtles

edT

dt=

94:15e¡0:1681t

(1 + 6:473e¡0:1681t)2> 0 for all t,

so the number of turtles is increasing

fd2T

dt2= 0 when t = 11:11

This is when the rate at which the number of

turtles is increasing changes from increasing at

an increasing rate to increasing at a decreasing

rate.

1 ady

dx=

3x2 ¡ 3

x3 ¡ 3xb

dy

dx=

1

x + 3¡

2

x

2 It does not.

3 a x = ln 3 b x = ln 4 or ln 3

4 a local minimum at (0, 1)

b As x ! 1, f(x) ! 1,

as x ! ¡1, f(x) ! ¡x (above)

c f 00(x) = ex Thus f(x) is convex for all x.

d

5 ady

dx= 2x + ln 2 x 2x

bdy

dx=

2x(x ¡ 3)

1¡ x3+

x2 + 2

1¡ x3+3x2(x2 + 2)(x ¡ 3)

(1¡ x3)2

6 a v(t) = 250 + 50e¡t

4 ms¡1

a(t) = ¡12:5e¡t

4 ms¡2

b v(0) = 300 ms¡1 s(0) = ¡200 m

a(0) = ¡12:5 ms¡2

c As t ! 1, v(t) ! 250 ms¡1 from above

d

e t = 4 ln 5 = 6:438 s

7 100 or 101 shirts, $938:63 profit

x

y

y 3�

y 9�

xey �� 9

xey �� 3

�� 226,223& (ln ) *

�� 226,223& (ln ) *

y e� 2

x 1�

1��

xex

y

x

y

20

40

60

80

5 10 15 20 25 30

t (years)

T

t

v t( ) ms 1

48

v t( )� ��

REVIEW SET 5B

642

300

280

260

t (s)

v (ms ) 1

21 1 2

6

4

2

x

yy f x( )�

y x�

11


0 05 5

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25

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25

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75

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50

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95

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100


422 ANSWERS

8 a

b A surge model is suggested

E = (26:51te¡0:3628t) units

c after 2 h 45 min d after 7 h 19 min

1 aex

ex + 3b

3x ¡ (x + 2)

x(x + 2)

2 y =x

a+ ln a ¡ 1 y =

x

e

d

e t = 1013

min

4 197 5 a x = ln 23

or x = 0 b x = e2

6 x = ln a

7 a x > 0

b f 0(x) > 0 for all x > 0 ) f(x) is increas-

ing for all x > 0f 00(x) < 0 for all x > 0 ) f(x) is concave

for all x > 0

c

normal is x + 2y = 3

8 a i 4:77 ii f(x) ! 40

b/d

c (2, 20)

9 a

the shape of the graph and as t becomes large,

N approaches a limiting value, so a logistic

model is approximate.

b N +

³234:4

1 + 78:05e¡1:169t

´people

c 234 people

d 167 people

e 28:3 people per hour

f (3:73, 117:2) The rumour is spreading fastest

after 3:73 hours.

1 110 m

2 a i travelling forwards

ii travelling backwards (or in the opposite

direction)

b 8 km from starting point (on positive side)

3 9:75 km

4 approx. 8:85 kWh

5 a 1:168 square units b 1:035 square units

1 ad

dx(x5) = 5x4

) antiderivative of x4 = 15x5

bd

dx(x3 + x2) = 3x2 + 2x

) antiderivative of 6x2 + 4x = 2x3 + 2x2

cd

dx(e3x+1) = 3e3x+1

10

20

30

2 4 6 8 10 12

E (units)

t (hours)

1284

25

15

5

v t( )

t

REVIEW SET 5C

EXERCISE 6A

EXERCISE 6C642 2

6

4

2

2

y x�ƒ( )

y x� ��

50

100

150

200

1 2 3 4 5 6 7

N (people)

t (hours)

10

2 4 6 8 10 12 14 16 18 20

20

30

40velocity (km/h)

t (mins)

3 a v(t) = 25¡10

tcm min¡1, a(t) =

10

t2cm min¡2

b s(e) = 25e¡10, v(e) = 25¡10

e, a(e) =

10

e2

c As t ! 1, v(t) ! 25 cm min¡1 from below

x

8642

50

40

30

20

10

y

point of inflection

y��

4.77


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\422SA12STU-2_AN.CDR Monday, 21 May 2007 12:02:26 PM DAVID3

ANSWERS 423

) antiderivative of e3x+1 = 13e3x+1

dd

dx(xp

x) = 32

px

) antiderivative ofp

x = 23x

px

ed

dx(2x + 1)4 = 8(2x + 1)3

) antiderivative of (2x+1)3 = 18(2x+1)4

2 a y = 6x + c b y = 43x3 + c

c y = 52x2 ¡ 1

3x3 + c d y = ¡

1

x+ c

e y = ¡ 13e¡3x + c f y = x4 + x3 + c

2 a 14

units2 b 3 34

units2 c 24 23

units2

d 4p2

3 units2 e 3:482 units2 f 2 units2

g 3:965 units2

1dy

dx= 7x6;

Rx6 dx = 1

7x7 + c

2dy

dx= 3x2 + 2x;

R(3x2 + 2x) dx = x3 + x2 + c

3dy

dx= 2e2x+1;

Re2x+1 dx = 1

2e2x+1 + c

4dy

dx= 8(2x + 1)3;R

(2x + 1)3 dx = 18(2x + 1)4 + c

5dy

dx= 3

2

px;

R px dx = 2

3xp

x + c

6dy

dx= ¡

1

2xp

x;

Z1

xp

xdx = ¡

2p

x+ c

8 ady

dx= 12(2x ¡ 1)5;R

(2x ¡ 1)5 dx = 112(2x ¡ 1)6 + c

bdy

dx=

¡2p1¡ 4x

;Z1

p1¡ 4x

dx = ¡ 12

p1¡ 4x + c

cdy

dx= ¡

3

2(3x + 1)3

2

;Z1

(3x + 1)3

2

dx =¡2

3p3x + 1

+ c

9 ady

dx= ¡3e1¡3x;

Re1¡3x dx = ¡ 1

3e1¡3x + c

bdy

dx=

4

4x + 1;Z

1

4x + 1dx = 1

4 ln(4x+1)+c (4x+1 > 0)

10 a ex¡x2

+ c

b 2 ln(5¡ 3x + x2) + c (5¡ 3x + x2 > 0)

c ¡ 12(x2 ¡ 5x + 1)¡2 + c d xex ¡ ex + c

e1

ln 22x + c f x lnx ¡ x + c

1 a 15x5¡ 1

3x3¡ 1

2x2+2x+c b 2

3xp

x¡2p

x+c

c 2ex +1

x+ c d 2

5x

5

2 ¡ ln jxj+ c

e 43x3 +2x2 + x + c f 1

2x2 + x ¡ 3 ln jxj+ c

g 43x

3

2 ¡ 2p

x + c h ¡2p

x¡ 4 ln jxj+ c

i 14x4 + x3 + 3

2x2 + c

2 a y = x2 + 3x + c b y = 3x ¡ ln jxj+ c

c y = x¡2x2+ 43x3+c d y = 2

3x

3

2 ¡4p

x+c

e y = x + 2 ln jxj+5

x+ c

f y = 14x4 + 2x3 + 6x2 + 8x + c

3 a f(x) = 14x4 ¡ 5

2x2 + 3x + c

b f(x) = 43x

3

2 ¡ 125

x5

2 + c

c f(x) = 3ex ¡ 4 ln jxj+ c

4 a f(x) = x2 ¡ x + 3 b f(x) = x3 + x2 ¡ 7

c f(x) = ex + 2p

x ¡ 1¡ e

d f(x) = 12x2 ¡ 4

px + 11

2

5 a f(x) = 13x3 + 1

2x2 + x + 1

3

b f(x) = 4x5

2 + 4x3

2 ¡ 4x + 5

c f(x) = 13x3 ¡ 16

3x + 5

1 a 18(2x + 5)4 + c b

1

2(3¡ 2x)+ c

c¡2

3(2x ¡ 1)3+ c d 1

32(4x ¡ 3)8 + c

e 29(3x ¡ 4)

3

2 + c f ¡4p1¡ 5x + c

g ¡ 35(1¡ x)5 + c h ¡2

p3¡ 4x + c

i 38(2x ¡ 1)

4

3 + c

2 a y = 13(2x ¡ 7)

3

2 + 2 b (¡8, ¡19)

3 a 12(2x ¡ 1)3 + c b 1

5x5 ¡ 1

2x4 + 1

3x3 + c

c ¡ 112(1¡ 3x)4 + c d x ¡ 2

3x3 + 1

5x5 + c

e ¡ 83 (5¡ x)

3

2 + c f 17x7 + 3

5x5 + x3 + x+ c

4 a 2ex + 52e2x + c b 1

3x3 + 2

3e¡3x + c

c 23x

3

2 + 2e2x + e¡x + c d 12ln j2x ¡ 1j+ c

e ¡ 53 ln j1¡ 3xj+c f ¡e¡x¡2 ln j2x + 1j+c

EXERCISE 6D

EXERCISE 6E.1

EXERCISE 6E.2

EXERCISE 6E.3

+ x


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Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\423SA12STU-2_AN.CDR Monday, 29 October 2007 3:51:23 PM DAVID3

424 ANSWERS

g 12e2x + 2x ¡ 1

2e¡2x + c

h ¡ 12e¡2x ¡ 4e¡x + 4x + c

i 12x2 + 5 ln j1¡ xj+ c

5 a y = x ¡ 2ex + 12e2x + c

b y = x ¡ x2 + 3 ln jx + 2j+ c

c y = ¡ 12e¡2x + 2 ln j2x ¡ 1j+ c

6 Both are correct. Recall that:

d

dx(ln jAxj) =

d

dx(ln jAj + ln jxj) =

1

x

7 a f(x) = ¡e¡2x + 4

b f(x) = x2 + 2 ln j1¡ xj+ 2¡ 2 ln 2

c f(x) = 23x

3

2 ¡ 18e¡4x + 1

8e¡4 ¡ 2

3

8

Z2x ¡ 8

x2 ¡ 4dx = 3 ln jx + 2j ¡ ln jx ¡ 2j+ c

9

Z2

4x2 ¡ 1dx = 1

2ln j2x ¡ 1j ¡ 1

2ln j2x + 1j+ c

1 aR3x2(x3 + 1)4 dx = 1

5(x3 + 1)5 + c

b

Z2x

px2 + 3

dx = 2p

x2 + 3 + c

cR p

x3 + x(3x2 + 1) dx = 23(x3 + x)

3

2 + c

dR4x3(2 + x4)3 dx = 1

4(2 + x4)4 + c

eR(x3 + 2x + 1)4(3x2 + 2) dx

= 15(x3 + 2x + 1)5 + c

f

Zx2

(3x3 ¡ 1)4dx = ¡

1

27(3x3 ¡ 1)3+ c

g

Zx

(1¡ x2)5dx =

1

8(1¡ x2)4+ c

h

Zx + 2

(x2 + 4x ¡ 3)2dx = ¡

1

2(x2 + 4x ¡ 3)+c

iR

x4(x + 1)4(2x + 1) dx = 15(x2 + x)5 + c

2 a e1¡2x + c b ex2

+ c c 13ex

3+1 + c

d 2epx + c e ¡ex¡x

2

+ c f e1¡1

x + c

3 a ln¯̄x2 + 1

¯̄+ c b ¡ 1

2ln¯̄2¡ x2

¯̄+ c

c ln¯̄x2 ¡ 3x

¯̄+ c d 2 ln

¯̄x3 ¡ x

¯̄+ c

e ¡2 ln¯̄5x ¡ x2

¯̄+ c f ¡ 1

3ln¯̄x3 ¡ 3x

¯̄+ c

4 a f(x) = ¡ 19(3¡ x3)3 + c

b f(x) = 32ln¯̄x2 ¡ 2

¯̄+ c

c f(x) = ¡ 13(1¡ x2)

3

2 + c

d f(x) = ¡ 12e1¡x

2

+ c

e f(x) = ¡ ln¯̄x3 ¡ x

¯̄+ c

f f(x) = 14(lnx)4 + c

g f(x) = ln¯̄x3 + 2x2 ¡ 1

¯̄+ c

h f(x) = 4 ln jlnxj+ c i f(x) =¡1

lnx+ c

1 12

cm 2 5 16

cm 3 a 41 units b 34 units

d as t ! 1, v(t) ! 50

e a(t) = 5e¡0:5t and as ex > 0 for all x,

a(t) > 0 for all t.

f

g 134:5 m

5 900 m 6 4 m

b 370:4 m

1 a 14

b 23

c e ¡ 1 (+ 1:718) d 1 12

e 6 23

f ln 3 (+ 1:099) g 290 h 2

i e ¡ 1 (+ 1:718) j 1:524

k 12ln¡119

¢(+ 0:1003)

l ¡1 + 83 ln

¡74

¢(+ 0:4923)

2 a 112

b 1:557 c 20 13

d 0:0337

e 12ln¡27

¢(+ ¡0:6264) f 1

2(ln 2)2 (+ 0:2402)

g 0 h 2 ln 7 (+ 3:892) i3n+1

2n + 2, n 6= ¡1

3 Hint: lnA ¡ lnB =

1 a 13

units2 b 3 34

units2

c e ¡ 1 (+ 1:718) units2 d 20 56

units2

e 18 units2 f ln 4 (+ 1:386) units2

g ln 3 (+ 1:099) units2 h 4 12

units2

i 2e ¡2

e(+ 4:701) units2

2 a 4 12 units2 b 1 + e¡2 (+ 1:135) units2

EXERCISE 6F

EXERCISE 6E.4

EXERCISE 6G

EXERCISE 6H

40

50

v t( ) (ms ) �

t(sec)

1050)( 5.0��

� tetv

4 a 40 ms¡1 b 47:77 ms¡1 c 1:386 seconds

7 a Show that v(t) = 100¡ 80e¡1

20t ms¡1 and

as t ! 1, v(t) ! 100 ms¡1

ln³

A

B

´


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ANSWERS 425

c 1 527

units2 d 2 14

units2

3 a 40 12

units2 b 8 units2 c 8 units2

4 a 10 23

units2

b i, ii

iii 1 13

units2

c 13

units2

d

enclosed area = 3 ln 2¡ 2 (+ 0:0794) units2

e 12

units2

5

enclosed area = 13 units2

6

b 9¼4

units2 (+ 7:07 units2)

7 a

Z 5

3

f(x) dx = ¡(area enclosed between

x = 3 and x = 5)

b

Z 3

1

f(x) dx ¡

Z 5

3

f(x) dx +

Z 7

5

f(x) dx

8 k + 1:7377 9 b + 1:3104 10 k + 2:3489

11 a =p3

1 $4250 2 $1127:60 3 $4793:2 million dollars

4 a P (x) = 15x ¡ 0:015x2 ¡ 650 dollars

b maximum profit is $3100, when 500 plates are

made

c 46 6 x 6 954 plates (you can’t produce part of

a plate)

5 76:27 C

6 a y = ¡ 1120

(1¡ x)4 ¡x

30+ 1

120

b 2:5 cm (at x = 1 m)

7 Extra hint:dC

dV= 1

2x2 + 4 and

dV

dx= ¼r2

8 a/g

b 10 people/hour c 7:40 pm

ddE

dt> 0 for all t

e Shaded region in a

f + 55 people g ii + 9:06 pm

h

9 a 5 shoppers/hour b t = 53

c 1:20 pm

d i

ii + 680 shoppers

e L0(t) =165

1 + 10:2e¡0:8t

f Shown on the graph g + 2:12 pm

h At 6:00 pm there are still 146 shoppers at the

store.

EXERCISE 6I

y = 1

y = 2

x

y1��

xey

22 ��xey

(1, 2)

(3, 0)

(0, 3)

x

y

32�� xxy

3�� xy

y

x3

3

3

3

922�� yx

y

x

42� xy

2� xy

At midnight there are people at the

restaurant.

13

t8642

250

200

150

100

50

y

y L t� '( )

y E t� '( )

t

654321

30

20

10

y

dt

dLy �

dt

dEy �

o


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426 ANSWERS

1 a

Z4p

xdx = 8

px + c

b

Z3

1¡ 2xdx = ¡ 3

2ln j1¡ 2xj+ c

cR

xe1¡x2

dx = ¡ 12e1¡x

2

+ c

2 a 12 49

b 554

3dy

dx=

xp

x2 ¡ 4;

Zx

px2 ¡ 4

dx =p

x2 ¡ 4+c

4 29:6 cm 5 4:5 units2

6 I(t) =100

t+ 100

a 105 amps b as t ! 1, I ! 100

7 ¼ units2

8 no -R 3

1f(x) dx = ¡ (area bounded by curve

between x = 1 and 3)

1 a ¡2e¡x ¡ ln jxj+ 3x + c

b 12x2 ¡ 2x + ln jxj+ c

2 a 2 815

b 4 12

3d

dx(3x2 + x)3 = 3(3x2 + x)2(6x + 1)R

(3x2 + x)2(6x + 1) dx = 13(3x2 + x)3 + c

4 a v(t):

b The particle moves in the positive direction ini-

tially, then at t = 2, 6 23 from its starting

point, it changes direction. It changes direc-

tion again at t = 4, 5 13

from its starting point,

and at t = 5, it is 6 23

m from its starting point.

c 6 23

m d 9 13

m

5 3¡ ln 4 (+ 1:614) units2

6 f(x) = 3x3 + 5x2 + 6x ¡ 1

7 a = ln 3, b = ln 5

8 a A = 2, B = ¡5

b

Z 2

0

4x ¡ 3

2x + 1dx = 4¡ 5

2ln 5 (+ ¡0:0236)

9 a = ¡3 A has x-coordinate3p4

1 a y = 15x5 ¡ 2

3x3 + x + c

b y = 400x + 40e¡x

2 + c

2 a ¡2 ln 5 (+ ¡3:219) b 103

3d

dx[p3x2 + 1] =

3xp3x2 + 1

)

Zx

p3x2 + 1

dx = 13

p3x2 + 1 + c

4 269 cm

5

a (2, ¡1) and (5, 2) b 4:5 units2

6 k = 3p16

7 a = 13 slope > 0 for all x in the domain

8¡2x

4¡ x2=

1

x + 2¡

1

2¡ x

9 Hint: Show that the areas represented by the

integrals can be arranged to form a 1£ e unit

rectangle.

1 a f(x) = ¡ 13(3¡ 2x)

3

2 + c

b f(x) = 12e2x ¡ 2x ¡ 1

2e¡2x + c

2 a 5:129 b 0:4651

3dy

dx=

1¡ 2x

2exp

x

)

Z 1

0

2¡ 4xp

xexdx =

4

e(+ 1:472)

4 a v(t) = t2 ¡ 3t + 2b t = 1 and t = 2 seconds c 1 5

6cm

5 Area = 512 units2

area above line : area below line = 2 : 3

6 k = 43 7 k = 1 13

8 a y-intercept is ¡1 14 , x-intercept is 1 2

3

b x = 2

c local minimum at (1 13

, ¡2 14

)

d

e A = 3, B = 1 f area = 3:925 units2

t(seconds)2 4

� �

0

(2, 1)

3

(5, 2)

31

y

x

2 1�� xy3�� xy

REVIEW SET 6A

REVIEW SET 6B

REVIEW SET 6C

REVIEW SET 6D

x = 2

(0, 1 ) \Qr_

(1 , 0)\We_

(1 , 2 )\Qe_ \Qr_

x

local min.

y-intercept

x-intercept

2)2(

53)(�

��

x

xxf


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ANSWERS 427

9 m = 1, c = 1

1 a v(0) = 25 ms¡1, v(3) = 4 ms¡1

b as t ! 1, v(t) ! 0

c

d 3 seconds

e a(t) =¡200

(t + 2)3, t > 0 f k = 1

5

2 a A = B + 3

b no, that would require area B to be ¡1, which

is impossible

3 a = 6, y = 6x + 1

4 A = 1, B = ¡1,Z1

x2 + xdx = ln jxj ¡ ln jx + 1j+ c

5d

dx[lnx]2 =

2 lnx

x

Zlnx

xdx = 1

2(lnx)2 + c

6 A = (¡1, 5) B = (2, ¡4)

enclosed area = 6 34

7 Hint: Show that f(x) =p

r2 ¡ x2

8 m = 2 or ¡2

1 a The sample is the 127 high blood pressure

patients.

b The population is all people with high blood

pressure.

2 a The population is all computer workers in

Australia.

b

c 83%

3 a All prawns in a catch.

b The average weight of prawns. c 53:8 grams

4 a Inferential b Descriptive c Descriptive

d Inferential e Inferential

1 a ¹ = 0:74 b ¾ + 0:9962

2 a x 0 1 2 3P (x) 0:216k 0:144k 0:096k 0:064k

k = 1:923 b ¹ + 1:015, ¾ + 1:045

3 $390

4 For sample A, x + 500 g and s + 1:02 g

For sample B, x + 483 g and s + 90:6 g

We expect bags of salt to have less variation and

therefore smaller standard deviation.

So, sample A is the salt and sample B the oranges.

5 a x + 6:18 and s + 2:251

b As ¾ + 2:233, the difference is 0:018 and

% difference + 0:803%

1 a 0:68 b 0:34 c 0:475 d 0:4985

2 a 34% b 47:5% c 0:135 d 0:16e 0:025 f 0:84

3 a 0:025 b 6 times

4 a Chest size depends on many factors including

parents’ genes, diet and exercise.

b The length depends on many factors including

parents’ genes and environment.

c Protein content depends on many factors in-

cluding genes, amount of sunshine, rain, avail-

ability of nutrients and soil.

5 a The weight depends on many factors including

genetic makeup and nutrition.

b 193 lambs

6 a i 0:975 ii 0:84 iii 0:815b i 1254 eggs ii 1080 eggs iii 1048 eggs

7 a i 0:84 ii 0:815 b 76 marks c 4 students

8 a ¹ = 176 g, ¾ = 24 g b 81:5%

9 a ¹ = 155 cm, ¾ = 12 cm b 0:8385

10

11 a For A, ¹ = ¡2 For B, ¹ = 0For C, ¹ = 1

b i A ii C iii B c B

REVIEW SET 6E

EXERCISE 7A

EXERCISE 7B

EXERCISE 7C

v t( ) m/s

t(sec)

2)2(

100)(�

�t

tv

25

2

x

y

~`6 ~`6

A

B

We_

xxy 63��

xy 23 ��

The parameter is the percentage of computer work-

ers who are interested in developing software. ��

2

1.5

1

0.5

a bc d


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428 ANSWERS

1 a/c ii

b Geog. (80%) is higher than Eng. essay (75%).

c i For Geography, z-score = 1For English essay, z-score = 2

ii Shaded on the above graphs.

d 16% in Geog. and 2:5% in Eng. essay

e About 5 performed better in Geog. and 1 in

Eng. essay.

f English essay

2 a

b For sugar, z-score = ¡2For apples, z-score = ¡1

c For apples

3 a

b i z = 1 ii z = ¡12

iii z = 1 12

c 6 cm d 3 oranges

4 a z-score = 1 13

b 501:8 mL

1 a z-scores: Geography + 1:61English + 1:82 Biology = 0:9Chinese + 2:33 Maths + 2:27

b Chinese, Maths, English, Geography, Biology

2 a b

0:68 0:839

c d

0:34 0:975

e f

0:84 0:16

1 a b

0:431 0:922

c d

0:885 0:298

e f

0:0968 0:919

g

3:17£ 10¡5

2 a 0:296 b 0:0912 c 0:252

EXERCISE 7D.1

EXERCISE 7D.2

EXERCISE 7D.3

� � � � � � �actual

z-score

��

� � � � � � �actual

z-score

��

Val

Val

Geography

English essay

� � � � � � �actual

z-score

��

� � � � � � �actual

z-score

��

Sugar

rejected

Apples

rejected

(g)

(kg)

� � � � � � �actual

z-score

� � � ��

Orange

dumped

� �

� � � � � �

� �

� � � �

� ��

� ��

� ��

� �


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100

Y:\HAESE\SA_12STU-2ed\SA12STU-2_AN\428SA12STU-2_AN.CDR Friday, 10 November 2006 9:18:18 AM PETERDELL

ANSWERS 429

3 a i z1 + ¡0:8594, z2 + 1:1830 ii 0:6865

b 0:6865

c The same if calculations are to 4 dec. places.

4 a 0:420 60 b 0:577 91 c 0:579 26d 0:579 26e 0:579 26 Pr(X 6 51) = 0:579 26 fto 5 d.p.g

5 a 0:595 b 0:789 c 0:387

6 a 0:904 b 0:324 c 0:568

7 a 0:303 b 0:968 c 0:309

8 0:378 9 a 0:213 b 0:0318 c 0:255

10 a 0:904 b 0:0478

11 a 0:003 33 b 0:615 c 23 cod

1 a b

k + 0:878 k + 0:202

c

k + ¡0:954

2 a b

k + 18:8 k + 23:5

c

k = 20

3 a Use

and Pr(¡k 6 z 6 k)= Pr(z 6 k)¡ Pr(z 6 ¡k) etc.

b i k + 0:303 ii k + 1:037

4 24:7 cm 5 75:2 mm

6 Lowest score for an A is 68:

7 between 501:8 mL and 504:0 mL 8 3:51:24 pm

9 112:4 10 0:238 m 11 $96:50

12 a ¹ + 52:4, ¾ + 21:6

b ¹ + 52:4, ¾ + 21:6, 54:4%

13 a ¹ + 4:000 cm, ¾ + 0:003 53 cm b 0:603

1 a 2:975 bx1 + x2 + x3 + :::::: + x10

10

c mean = 3 mm, stand. dev. = 0:0379 mm.

2 mean = 75 mm, standard deviation = 0:0707 mm.

3 mean = 40 min, standard deviation = 1:06 min.

4 a xi 1 0

pi p 1¡ p

Use ¹ =P

pixi

b

c i mean of Xn = p

standard deviation of Xn =

rp(1¡ p)

n

ii Xn is the proportion of heads in n tosses

of a coin.

5 a 0:197 b 0:571 6 a 0:369 b 0:0680

7 a 71:0% b 99:5%

1 a 100 b 2:5 c a normal curve

2 a B - symmetric and has much smaller spread.

b 0:217

c mean = 10, standard deviation = 1:25

d We need Pr(8:75 < X64 < 11:25) + 68:3%Normal distribution 68%, so the approximation

is good.

3 0:975 4 0:934 5 0:908 6 0:864

7 0:004 76 8 0:996 9 0:173

1 a i H0: ¹ = 17 Ha: ¹ 6= 17

ii X » N

µ17,

³4p50

´2¶

iii ¡1:76

iv P + 0:0771

v As P-value > 0:05 there is insufficient

evidence to reject the null hypothesis.

b X » N

µ17,

³4p70

´2¶

and P + 0:0365

which is < 0:05. There is now sufficient


2 The P-value is + 0:1400 and so we do not reject

the null hypothesis.

3 a Since the growth of carrots depends on many

factors such as genetic makeup and environ-

ment, it is reasonable to assume the weight of

carrots is distributed normally.

EXERCISE 7E

EXERCISE 7G.1

EXERCISE 7G.2

EXERCISE 7H.1

��k

� k � k

�k

��k �� k

� k k

Use ¾ =pP

pi(xi ¡ ¹)2


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430 ANSWERS

b For the sample, x = 53:58.

P + 0:0243 which is < 0:05

So, we reject the null hypothesis at a 5% level.

4 P + 0:0528 which is > 0:05:

So, we do not reject the hypothesis that

¹ = 2:00 cm at the 5% level. This does not justify

having to adjust the machine.

1 a H0: ¹ = 80 Ha: ¹ 6= 80

b X » N

µ80,

³9p50

´2¶

c P + 0:0184 which is < 0:05:

So, we reject the null hypothesis at a 5% level.

Consequently the machine should be adjusted.

2 a H0: ¹ = 3:5 Ha: ¹ 6= 3:5

b All the outcomes are equally likely to occur,

i.e., a uniform distribution.

c X » N

³3:5,

¡1:70810

¢2´d P + 0:0790 which is > 0:05, so there is not

enough evidence at a 5% level to reject H0.

e This time P + 0:0130 which is < 0:05, so

there is enough evidence at a 5% level to reject

H0.

3 H0: ¹ = 250 mg Ha: ¹ 6= 250 mg

P + 4:13£10¡15 which is < 0:05 and so there is

strong evidence at the 5% level to reject H0. Ac-

cept that the machine is not dispensing the preser-

vative at a mean of 250 mg.

4 H0 is ¹ = 23:6 years, Ha is ¹ 6= 23:6 years

P + 0:172 which is > 0:05 and so there is not

enough evidence at the 5% level to reject the null

hypothesis.

1 The test statistic is z + ¡0:685 which lies within

¡1:96 6 z 6 1:96. So, we do not reject the null

hypothesis at the 5% level.

2 The test statistic is z = ¡1:5 which lies within

¡1:96 6 z 6 1:96. So, we do not reject the null

hypothesis at the 5% level.

3 H0 will be rejected if x6¡23:78 or x>¡22:22

4 H0 will not be rejected if 502:7 6 x 6 505:3

1 a i 78:0 < ¹ < 85:2 ii 79:4 < ¹ < 83:8

b The width decreases for larger n.

2 8:38 6 ¹ 6 9:02 3 510:4 g 6 ¹ 6 517:2 g

4 36:8 days 6 ¹ 6 39:6 days

5 a x + 69:1 points, s + 8:21 points

b 66:6 6 ¹ 6 71:6

6 a 4 b 1:5 7 n = 80

1 50 crayfish 2 136 packets 3 27 patients

4 a 711 b 17 800 c 1 780 000

5 a 21 b 82 c 326

6 a w is divided byp2 b n is 4 times as large.

1 a 11:05 6 ¹ 6 12:19

b i 12:46 seconds is slower than all the values

in the 95% confidence interval, so there is

evidence that Joan has improved.

ii ¹ could be as high as 12:19, so there is

not enough evidence that Joan is better than

Betty.

2 7:07 6 ¹ 6 8:93, there is evidence that the

service has improved.

3 a 114:5 6 ¹ < 137:5 . As 115 is inside the

95% confidence interval for ¹ , there is not

enough evidence that the golfer has improved.

b i Increasing the number of trials decreases

the width of the confidence interval and

increases the accuracy of the estimate of ¹:

ii 117:1 6 ¹ 6 134:9, there is now evi-

dence the golfer has improved.

1 a 93 700 6 ¹ 6 98 900

b

The statistician is 95% confident that

93 700 < ¹ < 98 900.

X » N(93 700, 14 2682)

Pr(X < 75 000) + 0:095

So, about 9:5% have income < 75 000justifying the claim.

2 a 2:978 < ¹ < 3:082

b

The team is 95% confident that the stopping

time lies in 2:978 < ¹ < 3:082.

Using the extreme of 3:082 and assuming that

the time T » N(3:082, 0:172)

Pr(T > 3:45) + 0:0152 which is 1:52% and

much less than 15%.

EXERCISE 7H.2

EXERCISE 7H.3

EXERCISE 7I.1

EXERCISE 7I.2

EXERCISE 7I.3

EXERCISE 7I.4

75 000� 93 700� 98 900�

95% CI

2.978 3.082 3.45

95% CI


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ANSWERS 431

1 a 25:6 6 ¹ 6 32:2 b 24:5 6 ¹ 6 33:3

c The 99% CI is wider.

2 a 2:576 b 1:282 c 1:440 d 2:054

3 To select a CI of a given width requires a larger

sample for a 99% CI than for a 95% CI.

1 a i 81:9% ii 84:1% b 0:819

2 a k + 1:645 b k + 23:66

3 a 85:3% b 96:3% c 63:0%

4 a 8 apples > 375 g, 8 apples < 325 g.

b about 17 000 apples

5 x + 33:05, P + 0:187 which is > 0:05:

So, there is insufficient evidence to reject the null

hypothesis.

6 134 packets

7 a 95% CI is 824:2 < ¹ < 832:2

b width = 8, so 12

width = 4, n = 255

8 a i The relative difficulty of each test is not known.

ii z-score for English = 1, z-score for

Chemistry = 1 ) Kerry’s performance

is the same in both tests.

iii 8 students b 0:487

c i 6 hours

ii x = 13, P-value + 0:0253 < 0:05) we reject the null hypothesis.

iii 11:2 6 N 6 14:8, Les may have studied

between 0 and 4:42 hours.

1 A has x + 170:05 and s + 0:338

B has x + 170:35 and s + 8:89

We expect the plank length to have less variability

and hence standard deviation.

So, sample A is the plank length data and B is the

heights data.

2 a 2:28% b 84:0% c 0:840

3 a 0:260 b k-value + 29:27 so the manufac-

turer can expect 8% of batteries to

fail after 29 weeks, 2 days.

4 ¹ + 31:2 5 a a + 1:2816 b a + 18:16

6 H0 : ¹ = 135 g, Ha : ¹ 6= 135 g

P + 0:0599 which is > 0:05:

So, there is insufficient evidence at a 5% level to

reject the manufacturer’s claim.

7 a P + 0:0446 which is < 0:05

So, sufficient evidence exists at a 5% level to

reject the null hypothesis.

b

8 a Histogram A, as it is more symmetric and has

a smaller spread than Histogram B.

b i Using the Central Limit Theorem,

T »N

µ4:3,

³1:2p100

´2¶= N(4:3, (0:12)2)

ii 0:662 iii 0:662

c i P-value + 0:0124 < 0:05 ) we reject

the null hypothesis.

ii The supervisor handles unusual calls, so

the sample is biased, not random.

1

a 0:136 b 0:683

2 Pr(X 6 3) + 0:1469, Profit = $28 530

3 a a + 6:3 grams b b + 32:3 grams

4 + 0:0256 or about 2 12% of the time

5 a H0 : ¹ = 300 g, Ha : ¹ 6= 300 g

b i X » N(300, 32) ii z + 3:33

iii

fz : z 6 ¡1:96 or z > 1:96g.

So, we reject the null hypothesis that the

mean weight is 300 g. The test does not

support the grower’s claim.

6 a x 26:04 b 95% CI is 25:85 6 ¹ < 26:23

7 a 42:1% b about 42 of them

c about 2:3% d about 11 of them

8 a i 0:309 ii about 77 days

b i If T » N(¹, ¾2), T10 »

ii P-value + 0:114 > 0:05 ) insufficient


c Laura is 95% confident that 30:286¹6 32:76

Using T »N(32:76, 102), Pr(T 6 45:58)=0:9

) Laura should leave at 8:14 am.

1

EXERCISE 7I.5

REVIEW SET 7A

REVIEW SET 7B

REVIEW SET 7C

EXERCISE 8A

HTHTHTHT

H

T

H

T

H

T

HHHHHTHTHHTTTHHTHTTTHTTT

� � � � � � �actual

z-score � � � � � � �

)2,3( 2X »N

x

For 1844 < x < 2156

3:33 lies inside the 5% rejection region

=

N

µ¹,

³¾p10

´2¶

iii 28:8 6 ¹ 6 41:2 iv 250


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432 ANSWERS

There is 1 outcome of 3 heads

3 outcomes of 2 heads

3 outcomes of 1 head

1 outcome of 0 heads.

2 a HHHHH HTHHT HHTTT TTHTH

HHHHT THHHT HTHTT TTTHH

HHHTH HHTTH THHTT HTTTT

HHTHH HTHTH HTTHT THTTT

HTHHH THHTH THTHT TTHTT

THHHH HTTHH TTHHT TTTHT

HHHTT THTHH HTTTH TTTTH

HHTHT TTHHH THTTH TTTTT

b i 1 ii 5 iii 10 iv 10 v 5 vi 1

3 a The rule is to add the two terms directly above

the new row.

b 1 7 21 35 35 21 7 1

4

a C32 = 3 b C7

5 = 21 c C52 = 10

d C81 = 8 e C4

0 = 1 f C66 = 1

5 a C95 = 126 b C14

3 = 364 c C401 = 40

d C32 = 3 e C40

10 = 847 660 528

f C4020 + 1:378r £ 1011

6 a C1810 = 43 758 b C23

14 = 817 190

7 a p3 + 3p2q + 3pq2 + q3 1 3 3 1

b p4 + 4p3q + 6p2q2 + 4pq3 + q4 1 4 6 4 1

c i p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5

ii p6 + 6p5q + 15p4q2 + 20p3q3 + 15p2q4

+6pq5 + q6

d 1

1 a Binomial as trials are independent and for each

trial the probability of occurrence is the same.

b binomial c binomial

d Not binomial and the trials are not independent.

e Not binomial as there are not 2 possible out-

comes at each trial.

f Not binomial and the trials are not independent.

g Not binomial as the trials are not independent.

The probability of a success is not constant

at each trial. Note: As the number of bolts

is huge, a binomial model could be used to

approximate the situation.

2 a 0:234 b 0:0156 c 0:344 d 0:344

3 a Even though these outcomes are not strictly bi-

nomial, as n is vast, the binomial model gives

an excellent approximation.

b i 0:268 ii 0:200 iii 0:800

4 a 0:0280 b 0:005 53 c 0:2613 d 0:7102

5 a

c

6 a

b i 9:54£ 10¡7 ii 0:176 iii 0:588iv 0:0207

7 a 0:476 b 0:524 c 0:840 d 0:996

8 a 0:2305 b 0:7226

9 a 0:9984 b 0:8065

c

EXERCISE 8B.1

1 2 3 4 5 6 7 8 9 10 11 12

0.1

0.2

0.3)5.0,10(Bin

n

Pr

1 11 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 11 6 15 20 15 6 1

1 7 21 35 35 21 7 11 8 28 56 70 56 28 8 1

0.1

0.2

0.3

0.4

1 2 3 4 5 6 7 8 9 10 11 12

n

Pr )1.0,10(Bin )9.0,10(Bin

0.1

0.2

0.3Pr )3.0,10(Bin

)7.0,10(Bin

1 2 3 4 5 6 7 8 9 10 11 12

n

n� ��

)5.0,20(Bin

��

��

Pr

n� ��

)3.0,25(Bin

��

��

��

Pr

b


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ANSWERS 433

10 a 0:2901 b 0:8850

11 a 0:821 b 0:414 c 0:519 d 0:494

12 0:837 13 a 10:3% b 0:544

14 a 84:13%

15 a Pr(accepting a batch)

= Pr(x = 0) + Pr(x = 1)

= C40 p

0(1¡ p)4 + C41 p

1(1¡ p)3

= (1¡ p)4 + 4p(1¡ p)3

= (1¡ p)3[1¡ p+ 4p]

= (1¡ p)3(1 + 3p)

b

c Where Pr(accepting a batch) = 0:95,

p + 0:0976 fuse a graphics calculatorg

1 a i ¹ = 3, ¾ = 1:2247

xi 0 1 2 3P (xi) 0:0156 0:0938 0:2344 0:3125

xi 4 5 6P (xi) 0:2344 0:0938 0:0156

ii

iii The distribution is bell-shaped.

b i ¹ = 1:2, ¾ = 0:9798

xi 0 1 2 3P (xi) 0:2621 0:3932 0:2458 0:0819

xi 4 5 6P (xi) 0:0154 0:0015 0:0001

ii

iii The distribution is positively skewed.

c i ¹ = 4:8, ¾ = 0:9798

xi 0 1 2 3P (xi) 0:0001 0:0015 0:0154 0:0819

xi 4 5 6P (xi) 0:2458 0:3932 0:2621

ii

iii This distribution is negatively skewed and

is the exact reflection of b.

2 ¹ = 5, ¾ = 1:5811

3 ¹ = 1:2, ¾ = 1:0733 4 ¹ = 0:65, ¾ = 0:7520

5 a xi 0 1 2 3

P (xi) (1¡ p)3 3p(1¡ p)2 3p2(1¡ p) p3

1 a 0:0938 b ¹ = 22:4, ¾ = 3:139, 0:0949

2 a 0:000 651 b 0:207 c 0:256

3 a 0:000 977 b 0:231

4 0:288 5 a 0:839 b 0:529

1 a H0: p = 0:5, Ha: p 6= 0:5

b z = ¡1:27 c P = 0:203

d Since p > 0:05 there is not enough evidence

to reject H0 at the 5% level. So, we do not

accept that the coin is biased.

2 a H0: p = 16

, Ha: p 6= 16

b z = 1:53 c P = 0:125

d Since p > 0:05, the p-value does not give

support the die being biased.

3 H0: p = 0:25, Ha: p 6= 0:25P = 0:0379 which is < 0:05

So, we reject the null hypothesis. The evidence does

not support the supplier’s claim.

EXERCISE 8B.2

EXERCISE 8C

EXERCISE 8D

Pr

p

1

1

0.1

0.2

0.3

0 1 2 3 4 5 6

probability

x

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6

probability

x

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6

probability

x

4 H0: p = 12 , Ha: p 6= 1

2 , P = 0:0389 < 0:05

So, we reject the null hypothesis. The new medi-

cation is better than the old.

5 a n = 100 and p = 0:05 so np = 5 which

is not > 10 and so the normal approximation

may not be good enough.

b H0: p = 0:05, Ha: p 6= 0:05, P = 0:218which is > 0:05

So, there is not enough evidence to reject the

manufacturer’s claim.

6 H0: p = 0:206, Ha: p 6= 0:206, P = 0:549which is > 0:05

b 0:880


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434 ANSWERS

There is not enough information to reject the

hypothesis that people’s feelings had not changed.

7 a i bp = 0:51

ii P = 0:841 which is > 0:05

So, there is not enough evidence to say

the coin is biased.

b i bp = 0:51

ii P = 0:527 which is > 0:05

So, there is not enough evidence to say

the coin is biased.

c i bp = 0:51

ii P = 0:0455 which is < 0:05

So, there is evidence at a 5% level that

the coin is biased.

8 The number would have to be 6 61 or > 92for the claim to be rejected.

9 The number would have to be 6 30 or > 54for the claim to be rejected.

10 a H0: p = 0:85, Ha: p 6= 0:85P = 0:201 which is > 0:05

So there is not enough evidence to reject the

dealer’s claim at a 5% level.

b The number would have to be 6 3 or > 14for the claim to be rejected.

11 b 806 737

1 0:352 < p < 0:388

2 a 0:895 < p < 0:962

b We can be 95% confident that the true popu-

lation proportion of all players who would say

yes lies between 89:5% and 96:2%.

3 a 0:166 < p < 0:243

b For a fair die p = 0:167 which just lies inside

the 95% CI.

4 a 0:490 < p < 0:578

b For a fair coin p = 0:5 and this lies within

the 95% CI. So, there is no reason to believe

the coin is not fair.

5 a i 0:422 < p < 0:618ii 0:484 < p < 0:581iii 0:465 < p < 0:514

b

6 a Amy: 0:0429 < p < 0:4905John: 0:1548 < p < 0:3786

b John’s CI is much shorter than that which Amy

obtained. This is expected as John’s sample is

much larger.

7 0:254 < p < 0:393

8 a Party: 0:360 < p < 0:420Poll: 0:325 < p < 0:355

b

c The two confidence intervals do not overlap. It

would suggest that polling by party may have

been biased.

9 a bp = 0:704 i.e., 70:4% said they were not

better off now.

b 0:687 < p < 0:722

c Between 1566 and 1764

10 a bp = 0:4 b n = 100

11 bp = 0:1, n = 5830

1 8070 people 2 864 trial runs

3 a i 13 800 ii 24 600 iii 32 300iv 36 900 v 38 400

b i 36 900 ii 32 300 iii 24 600 iv 13 800

4 a n = 10 000p¤(1¡ p¤)

b

c n is largest when p¤ = 12

1 a w = 0:0506 b w = 0:0438 c 2000

2 a 0:750 < p < 0:782b i n = 11 000 ii n = 15300

3 a 0:168 < p < 0:287 b i + 1690 ii + 2400

EXERCISE 8E.1

EXERCISE 8E.2

EXERCISE 8E.30.4 0.45 0.5 0.55 0.6

0.422 0.618

0.484 0.581

0.465 0.514

i

ii

iii

0.3

0.360 0.420

0.325 0.355

0.35 0.450.4

Party

Poll

0

0.0429 0.4905

0.1548 0.3786

0.1 0.2 0.3 0.4 0.5

Amy

John

p

2500

Qw_ 1

n


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ANSWERS 435

1 a bp = 78= 0:875 b 0:803 < p < 0:947

c The 90% claimed by the manufacturer lies

within the CI and so we cannot dismiss the

manufacturer’s claim. We note that the per-

centage could be as low as 80%.

2 a 0:284 < p < 0:421

b The company’s claim of 43% lies outside the

CI and so the Consumer Affairs data does not

support the company’s claim.

3 a 0:0563 < p < 0:1837

b The claim of 20% (or 0:2) lies outside the CI.

So, the consumer group’s data does not support

the manufacturer’s claim.

1 a x 0 1 2 3 k = 1:6

P (x)k

64

3k

64

9k

64

27k

64

b Pr(X > 1) = 1¡ Pr(X = 0) = 0:975

2 p = 0:18 a 0:302 b 0:298 c 0:561

3 a 6:68% b 0:854

4 a

b u = 6, ¾ + 2:05

5 a H0: p = 13

, Ha: p 6= 13

b z + 1:414

c N

µ13

,

³p2

30

´2¶

d P 6= 0:157 which is > 0:05 so Joan would not

reject the null hypothesis. There is insufficient

evidence to conclude that the die is biased.

6 bp + 0:602 and the 95% CI for p is

0:579 < p < 0:625 which contains bp.

7 a n = 384 b n = 246 c n = 138

8 a H0: p = 0:9, Ha: p 6= 0:9 b z + ¡2:42

c P-value + 0:0154 < 0:05 ) reject the null

hypothesis.

d The phone company’s claim is not justified.

1 a HHHH

HHHT HHTH HTHH THHH

HHTT HTHT HTTH TTHH THTH THHT

TTTH TTHT THTT HTTT

TTTT

b (H + T )4

= H4 + 4H3T + 6H2T 2 + 4HT 3 + T 4

There is 1 of all Heads

4 of 3 Heads and 1 Tail

6 of 2 Heads and 2 Tails

4 of 1 Head and 3 Tails

1 of all Tails.

2 a 0:849 b 2:56£ 10¡6 c 0:991d 2:46£ 10¡4

3 a 68:3% b 0:0884

4 a n > 50 b 0:005 18

5 a H0: p = 14

, Ha: p 6= 14

b z + 0:660

c N

µ14

,

³1

4p50

´2¶

d P = 0:509 which is > 0:05. We do not have

enough evidence to claim the pennies are unfair.

6 a 0:369 < p < 0:401

b As 40% lies within the CI, the claim is sup-

ported.

7 a 0:225 < p < 0:324 b n = 1224

8 a 3:24%

b i virtually binomial

ii (1) 0:347 (2) 0:0257

1 a 1 11 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 1

b 10 ways c 32

2 a 0:259 b 0:337 c 0:922

3 a x 0 1 2 3 4

P (x) 116

416

616

416

116

b ¹ = 2, ¾ = 1

4 H0: p = 0:31, Ha: p 6= 0:31

P + 0:454 which is > 0:05 so there is insuf-

ficient evidence to reject the null hypothesis at

a 5% level. The proportion who were satisfied

appears unchanged.

5 H0: p = 12

, Ha: p 6= 12

. p + 0:398 which is

> 0:05, so there is insufficient evidence at a 5%level that strength has improved.

6 a 34:7% b 0:283 < p < 0:411

7 a H0: p = 0:62, Ha: p 6= 0:62

b 0:609 < p < 0:748

c 0:62 is inside the 95% confidence interval, so

we do not reject the null hypothesis.

EXERCISE 8E.4

REVIEW SET 8A

REVIEW SET 8B

REVIEW SET 8C

Pr

n5 10 15

0.2

0.1


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100

100 0 05 5

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95

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100


436 ANSWERS

d There is not sufficient evidence to conclude the

campaign was effective.

8 a 0:412 < p < 0:646

b The confidence interval is too large.

c i To maximise the sample size needed.

ii n + 1540

d Accept that a majority of the residents want the

reduced speed limit.

e It would take 385 hours which is time consum-

ing and therefore costly.

1 a It is a solution. b a = ¡1, b = 6

c 2x + y = 3 is a straight line, so there are

infinitely many values for x and y which will

satisfy the equation.

2 b x = 2, y = ¡1, z = 3 is a solution.

3 b i x = 0, y = 7, z = 2ii x = 3, y = 1, z = ¡1

4 c Hint: Let s =2¡ t

2and compare the two forms.

1 a c = the cost per cup in dollars, p = the cost

per plate in dollars

b 7c + 5p = 90, 9c + 8p = 133

2 p = the cost per pad in $s

b = the cost per biro in $s

r = the cost per ruler in $s

p + b + r = 5:32p + 2b + r = 8:353p + 3b + r = 11:4

3 a P (t) = at2 + bt + c, a 6= 0 where P (t)is the number of brown bears in hundreds and

t is the number of decades after 1976 (t = 1is one decade after 1976).

b P (1) = a + b + c = 38P (2) = 4a + 2b + c = 32P (3) = 9a + 3b + c = 25

4 a A = the number of kilograms of food A used in

his mixture, B = the number of kilograms of

food B used in his mixture, C = the number

of kilograms of food C used in his mixture

b A + B + C = 5428A + 256B + 179C = 1400214A + 605B + 713C = 2650

5 a + 3b + c = ¡105a + 4b + c = ¡414a ¡ b + c = ¡17

6 a C(x) = ax3 + bx2 + cx + d, a 6= 0 where

C(x) = the cost of making the carpet in 1000’s

of dollars and x = the number of 1000’s of

metres of carpet made.

b 8a + 4b + 2c + d = 12064a + 16b + 4c + d = 150

343a + 49b + 7c + d = 1701000a + 100b + 10c + d = 250

7 a N: a = c, H: 3a = 2d, O: 2b = c + d

b Cu: a = c, H: b = 2e, N: b = 2c + d,

O: 3b = 6c + 2d + e

c Cu: a = c, H: b = 2e, N: b = 2c + d,

O: 3b = 6c + d + e

Balanced equations are:

a 4NH3 + 5O2 ! 4NO + 6H2O

b Cu + 4HNO3 ! Cu(NO3)2 + 2NO2 + 2H2O

c 3Cu + 8HNO3 ! 3Cu(NO3)2 + 2NO + 4H2O

1 a x = 2, y = ¡3 b x = ¡1, y = 5c x = ¡2, y = ¡4

2 a intersecting b parallel c intersecting

d coincident e intersecting f parallel

3 a The lines representing the two equations are

coincident.

b It does not add any new information.

c i x = t, y =3¡ t

2ii y = s, x = 3¡ 2s

4 a The second line indicates that 0x + 0y = 6,

which is impossible, ) no solutions. The

lines must be parallel.

b The lines are coincident. Infinite solutions.

5 b Infinitely many solutions of the form x = t,

y = 3t ¡ 2 if k = 4 (lines coincide), no

solutions if k 6= 4 (lines parallel).

6 a ..... = k ¡ 16 b k = 16c x = t, y = 3t ¡ 8, t 2 R d when k 6= 16

7 a»

·4 8 10 ¡2a ¡ 8 21

¸b for all

a 6= ¡4

d There are no solutions if a = 4.

8 a x =6

m + 2, y =

6

m + 2, when m 6= §2

b When m = ¡2 there are no solutions.

When m = 2 there are infinitely many

solutions of the form x = t, y = 3¡ t.

9 a x = ¡ 1711

, y = 1911

b x + 8:59, y + 3:86

1 a x = 2, y = ¡1, z = 5b x = 4, y = ¡2, z = 1c x = 4, y = ¡3, z = 2

3 a x = 4, y = 6, z = ¡7b x = 3, y = 11, z = ¡7c x + 0:33, y + 7:65, z + 4:16

EXERCISE 9A.1

EXERCISE 9A.2

EXERCISE 9B

EXERCISE 9C

¡


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ANSWERS 437

4 a x = 14, y = 11, z = 17

b i x represents the cost per cricket ball in dol-

lars, y represents the cost per softball in

dollars, z represents the cost per netball in

dollars

ii 12 netballs

5 a 2x+ 3y + 8z = 352x+ 5y + 4z = 274

x+ 2y + 11z = 351

b x = 42, y = 28, z = 23 c $1 201 000

6 x ¡3 ¡2 ¡1 0 1 2 3y ¡23 ¡15 ¡9 ¡5 ¡3 ¡3 ¡5

7 $11:80 per kg

8 a 5p+ 5q + 6r = 40515p+ 20q + 6r = 1050

15p+ 20q + 36r = 1800

b p = 24, q = 27, r = 25c p = 114¡ 18t, q = 12t¡ 33, r = 5t

9 a a = 50 000, b = 100 000, c = 240 000b yes c 2007 + $284 000, 2009 + $377 000

1 a x = 1 + 2t, y = t, z = 0, t real

b no solutions

c x =¡1 + 5t

2, y =

1¡ 3t

2, z = t

d no solutions

2"

1 2 1 32 ¡1 4 11 7 ¡1 k

#a"

1 2 1 30 5 ¡2 50 0 0 k ¡ 8

#b If k 6= 8 there are no solutions. If k = 8

there are infinitely many solutions of the form

x =5¡ 9t

5, y =

5 + 2t

5, z = t (t is real).

c No solution because the system reduces to

2 equations in 3 unknowns for k = 8.

3 a"

1 2 ¡2 50 3 ¡5 60 0 k ¡ 13 ¡k + 13

#b If k = 13 there are infinitely many solutions of

the form x =3¡ 4t

3, y =

6 + 5t

3, z = t

(t is real).

c For k 6= 13, 3rd element in bottom row 6= 0and so there is a unique solution:

x = 73

, y = 13

, z = ¡1.

4 a"

1 3 3 a¡ 10 7 5 2a¡ 90 0 a+ 1 a+ 1

#b If a = ¡1, the bottom row is all zeros, so

the system will have infinitely many solutions

of the form x =19¡ 6t

7, y =

¡11¡ 5t

7z = t (t is real).

c If a 6= ¡1, x =a+ 14

7, y =

2a¡ 14

7, z = 1

5"

2 1 ¡1 30 4 +m ¡2¡m 3m¡ 20 0 (m+ 5)(m+ 1) ¡7(m+ 1)

#a If m = ¡5, the bottom row reads 0 = 28,

which is impossible, so there are no solutions.

b If m = ¡1, the bottom row is all zeros, so

there are infinitely many solutions.

c Unique solution if m 6= ¡5 or ¡1:

x =7

m+ 5, y =

3 (m¡ 2)

m+ 5, z =

¡7

m+ 5

6 a"

1 3 k 20 3k + 2 k2 ¡ 3 k

0 0 (3k + 25)(k ¡ 1) 6(k ¡ 1)

#b When k = 1, system has solutions of the form

x =7¡ 11t

5, y =

2t+ 1

5, z = t (t is real).

c k = ¡ 253

1 a x =8¡ 2t

3, y =

t¡ 1

3, z = t (t is real)

b x =16¡ 5t

7, y =

22 + t


c no solutions

2 x = 5t, y = 4t, z = 7t (t is real)

) x = 5, y = 4, z = 7

3 x = ¡7t, y = 3t, z = 5t (t is real)

if t = 0, i.e., a 6= ¡ 27

, x = 0, y = 0, z = 0

if t 6= 0, i.e., a = ¡ 27

, x = ¡7t, y = 3t,z = 5t, (t is real)

4 c P (x) = ¡ 299x2 + 172

9x¡ 71

9

d $20 448, when 2966 items are made

5 a x =9¡ t

2, y =

3t¡ 5

2, z = t, (t is real)

b x = 3, y = 2, z = 3

6 b 3w1 ¡ 2w3 ¡ 4w2 = 0

c There are infinite solutions of the form

w1 = 10t, w2 = 4t, w3 = 7t (t is real).

d w1 = 50 kg, w2 = 20 kg, w3 = 35 kg

EXERCISE 9D

EXERCISE 9E


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438 ANSWERS

e i 3 ii 4

7 a a+ x = b+ y

x+ z + e = d

y + z = c

so x¡ y = b¡ a

x+ z = d¡ e

y + z = c

Hence the given augmented matrix.

b i e = 11 ii x = 29¡ t, y = 12¡ t, z = t

t is any non-negative integer.

c z = 0 ) x = 29 and y = 12Rate K to M is 29 vehicles/min.

Rate M to L is 12 vehicles/min.

8 a x1 ¡ x2 = a

x1 + x3 = b

x2 + x3 = d¡ c

b"

1 0 1 b

0 1 1 d¡ c

0 0 0 a+ d¡ b¡ c

#If a+ d¡ b¡ c 6= 0, no solution exists and if

a+ d¡ b¡ c = 0, infinitely many solutions

exist, i.e., solutions if a+ d = b+ c:

c x1 = b ¡ t, x2 = d ¡ c ¡ t, x3 = t, t > 0solutions are non-negative if t 6 d¡ c

1 $367

2 C(x) = 0:152 67x3¡12:335x2+393:88x¡2410:0C(26) + $2185

3 a 2:724a+ 3:010b¡ 1:57c¡ 1:735d+ e = ¡2:465¡1:647a+ 3:936b¡ 0:83c+ 1:984d+ e = ¡0:6889¡1:098a+ 0:7709b+ 1:25c¡ 0:878d+ e = ¡1:563

11:23a+ 29:74b+ 2:06c+ 5:453d+ e = ¡4:2449:875a+ 8:387b+ 3:41c+ 2:896d+ e = ¡11:63

b a + ¡0:9349b + 0:4603c + ¡0:3807d + ¡0:6600e + ¡3:047

c

d (1:946, 0) or (¡1:566, 0)

4 a 8a+ 4b+ 2c+ d = 527a+ 9b+ 3c+ d = 1464a+ 16b+ 4c+ d = 30

b a = 13

, b = 12

,

c = 16

, d = 0

c 131003 + 1

21002 + 1

6100 = 338 350

5 a a + 8645:2, b + ¡22 585, c + 142 977,

d + ¡135 580

b Profit for 2008 + $168 364

6 a a+ b+ c+ d+ e = 616a+ 8b+ 4c+ 2d+ e = 3081a+ 27b+ 9c+ 3d+ e = 90256a+ 64b+ 16c+ 4d+ e = 210625a+ 125b+ 25c+ 5d+ e = 420

b a = 14

, b = 32

, c = 114

, d = 32

, e = 0

c 1 756 950

1 unique solution if k 6= 34

, no solution if k = 34

2 x = 1, y = ¡2, z = ¡1

3 a ¡2a+ 4b+ c = ¡20a+ 3b+ c = ¡10) a = 2¡ t, b = ¡4¡ 3t, c = 10t (t is real)

b 3 unknowns but only 2 pieces of information.

c x2 + y2 + 4x+ 2y ¡ 20 = 0

4 When k 6= 27, there are no solutions.

When k = 27, there are infinite solutions of the

form x = 2¡ t, y = 2t+ 3, z = t (t is real).

5 x = 3t, y = ¡7t, z = 2t (t is real)

6 a 2p+ c+ 3a+ 2n = 1782p+ 2c+ a+ 4n = 2063p+ c+ 2a+ 2n = 197p+ 4c+ 2a+ 3n = 237 where

p = cost of a ticket to the play in $s

c = cost of a ticket to the concert in $s

a = cost of a ticket to the AFL match in $s

n = cost of a ticket to the NBL match in $s

b a ticket to the play costs $35a ticket to the concert costs $32a ticket to the AFL game costs $16a ticket to the NBL game costs $14 c $314

1 x = 2, y = 1, z = 3

2 x =¡13t¡ 1

9, y =

20t+ 14


3 b when m 6= 143

4 When t = 3 there are no solutions.

When t = 2 there are infinite solutions of the

form x = 1 + s, y = 4¡ s, z = s (s is real).

For t 6= 2 or 3 there is the unique solution

x =3(t¡ 5)

t¡ 3, y =

6t¡ 8

t¡ 3, z =

¡8

t¡ 3

5 a a = ¡3, b = 18, c = 48

) S(t) = ¡3t2 + 18t+ 48

b 48 m c 8 seconds

6 a e = ¡23 689a+ b+ c+ d+ e = ¡3528

16a+ 8b+ 4c+ 2d+ e = 18 04281a+ 27b+ 9c+ 3d+ e = 43 322

256a+ 64b+ 16c+ 4d+ e = 75 483

b a = 36:25, b = 166, c = ¡47:25,

d = 20 006, e = ¡23 689

c From the model, profit in 2008 will be $118 566.

EXERCISE 9F

REVIEW SET 9A

REVIEW SET 9B

��

y

�

�

�

x


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


ANSWERS 439

1 If k = §2 there are no solutions.

If k 6= §2, x =k + 4

k2 ¡ 4, y =

¡2¡ 2k

k2 ¡ 4

2 a x = 6, y = ¡2, z = 1

b x = 32

, y = ¡ 76

, z = ¡ 76

3 a d = 80 b a = 2, b = 8, c = 10

4 If k = 2 and m 6= 20 there are no solutions.

If k = 2 and m = 20 there are infinite

solutions of the form x =10 + 3s¡ t

2, y = s,

z = t, s, t are real.

If k 6= 2, z =m¡ 20

k ¡ 2, y = s,

x = 12

³10 + 3s¡

m¡ 20

k ¡ 2

´5 b k 6= 2 or 2

3

c k = 2, solutions are x =13¡ t

9, y =

11t+ 1

9,

z = t, t is real

d When k = 23

, the system is inconsistent and

so has no solutions.

6 a ¡ 12d = 2

a+ b+ c¡ d = 9

27a+ 9b+ 3c+ d = 41

64a+ 16b+ 4c+ 12d = 118

b a = 103

, b = ¡ 253

,

c = 10, d = ¡4

c k = 6:14

1 a 1£ 4 b 2£ 1 c 2£ 2 d 3£ 3

2 a£

2 1 6 1¤

b

264 1:952:350:150:95

375c total cost of groceries

3 264 1000 1500 12501500 1000 1000800 2300 13001200 1200 1200

3754 264 40 50 55 40

25 65 44 3035 40 40 3535 40 35 50

375

1 a b·

12 2448 12

¸ ·2 48 2

¸c d·

12

1

2 12

¸ ·¡3 ¡6¡12 ¡3

¸2 a b

·3 5 62 8 7

¸ ·1 1 40 4 1

¸c d

·5 8 113 14 11

¸ ·5 7 142 16 9

¸

3 a b c264 122412060

375264 3

63015

375264 9

189045

3754 a A B C D

skirt

dress

evening

suit

264 35 46 46 6958 46 35 8646 46 58 5812 23 23 17

375b A B C D

skirt

dress

evening

suit

264 26 34 34 5143 34 26 6434 34 43 439 17 17 13

3755 a bFriday Saturday"

859252

# "10213749

# "187229101

#

6 a i ii266641:7227:850:922:533:56

3777526664

1:7928:751:332:253:51

37775b subtract cost

price from

selling price

c 266640:070:900:41¡0:28¡0:05

377757 a "

7527102

#Ã VHS

Ã DVD

Ã games

"13643129

#Ã VHS

Ã DVD

Ã games

b "21170231

#Ã VHS

Ã DVD

Ã games

c total weekly

average hirings

8 a i Lou Rose"23 1917 2931 24

#fridge

stove

microwave

ii Lou Rose"18 257 1336 19

#fridge

stove

microwave

iii Lou Rose"41 4424 4267 43

#fridge

stove

microwave

b 12F

REVIEW SET 9C

EXERCISE 10A

EXERCISE 10B.1


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


440 ANSWERS

9 a x = ¡2, y = ¡2 b x = 0, y = 0

10 aA + B =

·1 35 2

¸B + A =

·1 35 2

¸11 a Both (A + B) + C and A + (B + C) are·

6 3¡1 6

¸

1 a 3A b O c ¡C d O e 2A + 2B

f ¡A ¡ B g ¡2A + C h 4A ¡ B i 3B

2 a X = A ¡ B b X = C ¡ B c X = 2C ¡ 4B

d X = 12A e X = 1

3B f X = A ¡ B

g X = 2C h X = 12

B ¡ A i X = 14

(A ¡ C)

3 a bX =

·3 69 18

¸X =

·12

¡ 14

34

54

¸c

X =

·¡1 ¡61 ¡1

2

¸

1 a [11]

b [22]

c [16]

2 £w x y z

¤ 266414141414

37753 a P =

£27 35 39

¤Q =

"432

#

b total cost =£

27 35 39¤" 4

32

#= $291

4 a P =£

10 6 3 1¤

N =

264 3242

375b total points =

£10 6 3 1

¤264 3242

375= 56 points

1

2 a m = n b 2£ 3c

3 a b i ii£28 29

¤ £8¤ "

2 0 38 0 124 0 6

#

4 a£

3 5 3¤

b

"¡211

#5 a

C =

·12:59:5

¸N =

·2375 51562502 3612

¸b

·78 669:565 589

¸income from adult rides

and children’s rides

c $144 258:50

6 a bR =

"1 11 22 3

#P =

·7 3 196 2 22

¸c

·48 7052 76

¸d My costs at store A are $48,

my friend’s costs at store B

are $76.e store A

7F =

"6 7 95 8 44 7 2

#C =

"181513

#FC =

"330262203

#total cost = $795

1 a b"

16 18 1513 21 1610 22 24

# "10 6 ¡79 3 04 ¡4 ¡10

#c

"22 0 132 176 19844 154 88 110 0176 44 88 88 132

#d264 115

13646106

3752 a b£

3 3 2¤ "

125 150 14044 40 4075 80 65

#c d

£657 730 670

¤ £369 420 385

¤e

·657 730 670369 420 385

¸3 $224 660 4 $3398:10

5 a £125 195 225

¤ "15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

#

¡£85 120 130

¤ "15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

#=£1185 800 1350 970 845 1130 845

¤gives the profit for each day. Profit = $7125.

b £125 195 225

¤ "15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

#

¡£85 120 130

¤ "20 20 20 20 20 20 2015 15 15 15 15 15 155 5 5 5 5 5 5

#

EXERCISE 10B.2

EXERCISE 10C.1

EXERCISE 10C.2

EXERCISE 10C.3

Number of columns in does not equal number of

rows in .

A

B

B Ahas columns, has rows3 2


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


ANSWERS 441

=£¡820 ¡1840 ¡455 ¡1485 ¡1725 ¡920 ¡1785

¤gives the profit for each day. In this case there is a

total loss of $9030.

c¡£

125 195 225¤¡£85 120 130

¤ ¢£

"15 12 13 11 14 16 84 3 6 2 0 4 73 1 4 4 3 2 0

#

1AB =

·¡1 1¡1 7

¸BA =

·0 23 6

¸AB 6= BA

2 AO = OA = O 4 bI =

·1 00 1

¸5 a b

·7 00 7

¸ ·97 ¡59118 38

¸6 a A2 does not exist

b when A is a square matrix

8 a A2 + A b B2 + 2B c A3 ¡ 2A2 + A

d A3 + A2 ¡ 2A e AC + AD + BC + BD

f A2 + AB + BA + B2

g A2 ¡ AB + BA ¡ B2 h A2 + 2A + I

i 9I ¡ 6B + B2

9 a A3 = 3A ¡ 2I A4 = 4A ¡ 3I

b B3 = 3B ¡ 2I B4 = 6I ¡ 5B

B5 = 11B ¡ 10I

c C3 = 13C ¡ 12I C5 = 121C ¡ 120I

10 a i I + 2A ii 2I ¡ 2A iii 10A + 6I

b A2 + A + 2I

c i ¡3A ii ¡2A iii A

11 a bAB =

·0 00 0

¸A2 =

·12

12

12

12

¸c false as A(A ¡ I) = O

; A = O or A¡ I = O

d ·0 00 0

¸,

·1 00 1

¸,

"a b

a¡ a2

b1¡ a

#,

b 6= 0

12 For example,

A =

·0 10 0

¸, gives A2 =

·0 00 0

¸13 a a = 3, b = ¡4 b a = 1, b = 8

14 p = ¡2, q = 1

a A3 = 5A ¡ 2I b A4 = ¡12A + 5I

1 aT =

·0:9 0:10:2 0:8

¸b i 20% of those who chose Shop Y in the pre-

vious week will buy from Shop X in the

next week.

ii 80% of those who chose Shop Y in the pre-

vious week will again buy from Shop Y in

the next week.

c£1 0

¤T =

£0:9 0:1

¤After week one Shop X will have 90% of the

market and Shop Y will have 10% of the market.

d i 17% ii 27:7%

e S19 =£0:667 0:333

¤S20 =

£0:667 0:333

¤Steady state proportions are Shop X has 2

3,

Shop Y has 13:

2 a 40% of Sheez customers this week will buy

Baaah brand milk next week.

b T1 =

·0:7 0:30:4 0:6

¸c£0:8 0:2

¤T1 =

£0:64 0:36

¤This represents the market shares of the two

brands for the second buying period (64% for

Baaah, 36% for Sheez).

d i 59:2% ii 57:2%

e Baaah has 57:14% of the market and Sheez has

42:86%:

f i 67:1% ii 70:1% iii 71:3%

g 71:4% Baaah, 28:6% Sheez

3 a i 84% of passengers who use Clydes this

month will use Clydes next month.

ii 16% of passengers who use Clydes this

month will use Roos next month.

b i Clydes: 507 people, Roos: 634 people

ii Clydes: 559 people, Roos: 582 people

c£0:568 0:432

¤In the long term, Clydes’ market share will

stabilise at 56:8%.

d Clydes: 648 passengers, Roos: 493 passengers.

e It will not be exact as consumer behaviour can-

not be predicted with certainty, but while market

trends continue, it should be relatively accurate.

4 a 620£ 0:8 + 380£ 0:1 = 534 smokers

b i SA =£534 466

¤, so 466 will not have

smoked between the Jan-Feb meetings.

EXERCISE 10C.4

EXERCISE 10D


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


442 ANSWERS

ii SA2 =£473:8 526:2

¤. We expect about

474 smokers and 526 non-smokers at the

March meeting.

c SA12 =£337:3 662:7

¤. So, we expect 337

smokers and 663 non-smokers.

d i n = 2sii In the long term there will be twice as many

non-smokers as smokers.

5 a S describes the current state of the system. The

door is currently open for 10% of the time and

closed 90% of the time.

A is the transition matrix which predicts how the

system will change each time someone visits.

b SA =£0:075 0:925

¤, 7:5% chance.

c SA2 =£0:06875 0:93125

¤, 6:9% chance.

d SA10 =£

115

1415

¤In the long term the door is left open 1

15th of

the time + 6:7%. Jess’ concern seems unjustified.

e i At any time the fridge is either open or

closed. Hence, x+ y = 1.

ii x = 115

, y = 1415

iii This confirms that the fridge door, in the

long term, will be left open 115

th of the

time and closed for 1415

th of it.

6 a i 10% of the soil which is very good this year

will be very poor next year.

ii 30% of the soil which is very poor this year

will be usable next year.

b

T2 =

"0:66 0:18 0:160:30 0:54 0:160:06 0:42 0:52

#i 16% of the soil that is usable this year will

be very poor in two years time.

ii 42% of the soil that is very poor this year

will be usable in two years time.

c£

0 0 1¤

d 13% is very good,

46% is usable,

41% is very poor.

e 152:7 ha will be very good, 265:9 ha will be

usable, 187:3 ha will be very poor.

f 37:5% will be very good, 37:5% will be us-

able, 25% will be very poor. (Steady state is

identical for both properties as they have the

same transition matrix.)

7 a

T =

"0:88 0:06 0:060:75 0:17 0:080:08 0:42 0:50

#

b i 88% of players who are fully fit this week will

be fully fit again next week.

ii 42% of players who cannot play this week

will be getting treatment next week.

c

T2 =

"0:824 0:088 0:0880:794 0:108 0:0990:425 0:286 0:288

#79:4% of players getting treatment this week will

be fully fit the week after next.

d i 26 players fully fit, 4 players receiving treat-

ment, 3 players cannot play

ii 26 players fully fit, 4 players receiving treat-

ment, 4 players cannot play. This result is

obviously not accurate as you cannot have

more players than you started with. This is

an error caused by rounding and demonstrates

that the result is only an approximation.

iii 26 players fully fit, 4 players receiving treat-

ment, 4 players cannot play

e 77:6% of players fully fit, 11:2% of players re-

ceiving treatment, 11:1% of players cannot play

8 a 29 fully fit players, 6 players receiving treatment,

4 players unable to play

b i 29 fully fit, 6 receiving treatment, 4 unavail-

able

ii 30 fully fit, 6 receiving treatment, 3 unavail-

able

9 a

T =

"0:75 0:15 0:100:20 0:60 0:200:15 0:20 0:65

#b

T2 =

"0:608 0:223 0:1700:300 0:430 0:2700:250 0:273 0:478

#Row 3 indicates the migration patterns of the

residents of Chalk over the next 2 years. i.e., in 2years time, 47:8% of Chalk’s current population

will still live in Chalk, 27:3% will move to Manu

and 25:0% will move to Paua.

c

T16 =

"0:412 0:299 0:2890:412 0:299 0:2890:412 0:299 0:289

#T16 represents the migration transition matrix

for the next 16 years.

d i 32:6% live on Paua, 34:3% live on Manu,

33:2% live on Chalk

ii 41:2% live on Paua, 29:9% live on Manu,

28:9% live on Chalk

e 41:2% live on Paua, 29:9% live on Manu,

28:9% live on Chalk


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


ANSWERS 443

10 a x = 52t, y = 2t, z = t, t in R.

b i N describes how the areas of each age of

bamboo vary each year.

In row 1, 20% of 1-year old bamboo is har-

vested and the area replanted. 80% is left

to grow into 2-year old bamboo.

In row 2, 50% of 2-year old bamboo is har-

vested and the area replanted. 50% is left

to grow into 3-year old bamboo:

In row 3, all the 3-year old bamboo is har-

vested.

ii MN =£4:4 17:6 0

¤After 1 year there are 4:4 acres of 1-year

old bamboo and 17:6 acres of 2-year

bamboo.

MN2 =£9:68 3:52 8:80

¤After 2 years there are 9:68 acres of 1-year

old, 3:52 acres of 2-year old and 8:80 acres

of 3-year old bamboo.

iii MN10 =£9:91 8:00 4:10

¤indicating 9:9 acres of 1-year old,

8:0 acres of 2-year old and

4:1 acres of 3-year old bamboo.

c i The total number of acres is 22.

) x + y + z = 22

ii x = 0:2x + 0:5y + z, y = 0:8x, z = 0:5y

iii x = 10, y = 8, z = 4: In time we

expect a steady state of 10 acres of

1-year old, 8 acres of 2-year old and 4acres of 3-year old.

11 a i The original system becomes24 1 ¡12

¡ 34

0

0 1 ¡1 00 0 0 0

35 The row of 0s

indicates a non-

unique solution.

ii a = 54t, b = t, c = t, t in R.

b i (1) 60 000 horses (2) 30 000 horses

ii (1) ST =£0:3 0:31 0:39

¤39 000 horses

(2) ST2 =£0:418 0:307 0:275

¤41 800 with A, 30 700 with B,

27 500 with C

iii (1) A, B and C represent 100% of the

market and so a + b + c = 1.

(2) 0:2a + 0:4b + 0:6c = a

0:4a + 0:1b + 0:4c = b

0:4a + 0:5b = c

(3) a = 513

, b = 413

, c = 413

(4) In the long term the market shares will

stabilise in the ratio A : B : C = 5 : 4 : 4.

12 a T describes the change in population year by

year. TP is the population after 1 year.

T2P is the population after 2 years.

TnP is the population after n years.

b TP =

·600230

¸So, after 1 year we have

600 chicks and 230 adults.

T2P =

·460258

¸So, after 2 years we have

460 chicks and 258 adults.

c T5P =

·502:6249:5

¸, T10P =

·499:97250:01

¸suggesting the stabilising of numbers at 500chicks and 250 adults.

d T10P =

·362:8181:4

¸, T20P =

·362:9181:4

¸suggesting that the population stabilises at a

lower level so the numbers are not likely to

recover.

f For the new T, ab = 1¡ c found in e no

longer holds. The population no longer

reaches a stable state.

T10P =

·968:6519:6

¸, T20P =

·19571050

¸indicate rapid increase in population.

13 a B describes the present state with the number

of female types in each category.

A describes how the system evolves from one

day to the next.14

of caterpillars survive to become adolescents.

13

of adolescents survive to become adults.

Each adult produces an average of 12 caterpillars.

b AB =

"96305

# After 24 hours there are:

96 caterpillars

30 adolescents

5 adults

c AB is the state of the system after 24 hours.

A2B is the state of the system after 48 hours.

d i A2B =

"602410

#Ã caterpillars

Ã adolescents

Ã adults

ii A3B =

"120158

#Ã caterpillars

Ã adolescents

Ã adults

e From d ii we notice that a return to original

levels has occurred.

This suggests a 3 day cycle.

f i If the numbers remain constant each day and

the numbers after one day are given by AX

then AX = X.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


444 ANSWERS

ii

24 0 0 1214

0 0

0 13

0

35" x

y

z

#=

"x

y

z

#) z = t, x = 12t and y = 3tBut x+ y + z = 256 and so t = 16) we have 192, 48 and 16 of each type.

1 a b114

·5 ¡41 2

¸ ·1 01 ¡1

¸c d

does not exist

·1 00 1

¸e f

does not exist ¡ 115

·7 ¡2¡4 ¡1

¸g h1

10

·2 ¡41 3

¸ ·¡3 ¡12 1

¸2 a b

·x+ 2y3x+ 4y

¸ ·2a+ 3ba¡ 4b

¸3 a

·3 ¡12 3

¸ ·x

y

¸=

·86

¸b

·4 ¡33 2

¸ ·x

y

¸=

·11¡5

¸c

·3 ¡12 7

¸ ·a

b

¸=

·6¡4

¸4 a x = 32

7 , y = 227 b x = ¡ 37

23 , y = ¡ 7523

c x = 1713

, y = ¡ 3713

d x = 5913

, y = ¡ 2513

e x = ¡40, y = ¡24 f x = 134

, y = 5534

5 b iX =

·¡1 32 4

¸ii

X =

·137

37

¡ 27

¡ 87

¸6 a

A¡1 =1

2k + 6

·2 ¡16 k

¸, k 6= ¡3

bA¡1 =

1

3k

·k 10 3

¸, k 6= 0

cA¡1 =

1

(k + 2)(k ¡ 1)

·k ¡2¡1 k + 1

¸,

k 6= ¡2 or 1

7 aAB =

·1 00 1

¸b A and B are not

inverses since

AB 6= BA.

8 b·1 00 1

¸,

·¡1 00 ¡1

¸,

·0 11 0

¸,

·0 ¡1¡1 0

¸

9X =

·14

34

1 0

¸10 a

A¡1 =

·0 ¡112

12

¸, (A¡1)¡1 =

·1 2¡1 0

¸b (A¡1)¡1(A¡1) = (A¡1)(A¡1)¡1 = I

c A¡1 and (A¡1)¡1 are inverses

11 a i ii· 1

313

23

¡ 13

¸ · 32

12

1 0

¸iii iv

· 56

13

13

13

¸ · 56

16

23

13

¸v vi

· 56

16

23

13

¸ · 56

13

13

13

¸c (AB)¡1 = B¡1A¡1 and (BA)¡1 = A¡1B¡1

d (AB)(B¡1A¡1) = (B¡1A¡1)(AB) = I

AB and B¡1A¡1 are inverses

12 (kA)³1

kA¡1

´=³1

kA¡1

´(kA) = I

kA and1

kA¡1 are inverses

13 a X = ABZ b Z = B¡1A¡1X

14 a A¡1 = 4I ¡ A b A¡1 = 5I + A

c A¡1 = 32

A ¡ 2I

15 A2 = 2A ¡ I 16 A¡1 = 2I ¡ A

17 If A¡1 exists, i.e., jAj 6= 0.

1 a b·

19

49

29

¡ 19

¸ ·519

119

¡ 419

319

¸2 a

·0:003 139 0:001 737¡0:001 491 0:002 320

¸b

·0:035 55 ¡0:087 070:065 45 0:113 66

¸

1 a b"

x+ y + 2zx+ 3y ¡ z

2x¡ y + 4z

# "a+ 2b+ 4c2a¡ b+ c

3a+ 2b¡ 3c

#

2 a"

1 ¡1 ¡11 1 39 ¡1 ¡3

#"x

y

z

#=

"27¡1

#

EXERCISE 10E.1

EXERCISE 10E.2

EXERCISE 10F.1


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50

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100

100 0 05 5

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95

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100


ANSWERS 445

b"

2 1 ¡10 1 21 ¡1 1

#"x

y

z

#=

"3613

#c

"1 1 ¡11 ¡1 12 1 ¡3

#"a

b

c

#=

"76¡2

#4 a b

"2 0 00 2 00 0 2

#A¡1 = 1

2B

5 AB = I, a = 2, b = ¡1, c = 3

6 MN = 4I, u = ¡1, v = 3, w = 5

7 a k =1

ad¡ bcb ad¡ bc 6= 0

1 a b24 5

434

¡ 74

¡ 14 ¡ 3

434

¡ 34

¡ 14

54

35 24 ¡ 112

92

152

¡ 12

12

12

4 ¡3 ¡5

352 a

"0:050 23 ¡0:011 48 ¡0:066 34

4:212£ 10¡4 0:013 53 0:027 75¡0:029 90 0:039 33 0:030 06

#b"

1:596 ¡0:9964 ¡0:1686¡3:224 1:925 0:62912:000 ¡1:086 ¡0:3958

#

3 a x = 2310

, y = 1310

, z = ¡ 92

b x = ¡ 13

, y = ¡ 9521

, z = 221

c a = 1018

, b = ¡ 31116

, c = ¡ 1518

, d = ¡ 6516

1 a ¡2 b ¡1 c 0 d 1

2 a 26 b 6 c ¡1 d a2 + a

3 a i·

2 ¡34 ¡1

¸ ·x

y

¸=

·811

¸,

jAj = 10

ii Yes, x = 2:5, y = ¡1

b i·

2 k

4 ¡1

¸ ·x

y

¸=

·811

¸,

jAj = ¡2¡ 4k

ii k 6= ¡ 12

x =8 + 11k

2 + 4ky =

5

1 + 2k

iii k = ¡ 12

no solutions

4 a ¡3 b 9 c ¡12

6 a jAj = ad¡ bc jBj = wz ¡ xy

bAB =

·aw + by ax+ bz

cw + dy cx+ dz

¸jABj = (ad¡ bc)(wz ¡ xy)

7 a i ¡2 ii ¡8 iii ¡2 iv ¡9 v 2

8 a 0 b §1 c 0 or 1

1 a 41 b ¡8 c 0 d 6 e ¡6 f ¡12g 11 h 0 i 87

2 a abc b 0 c 3abc¡ a3 ¡ b3 ¡ c3

3 k 6= ¡3 4 for all values of k except 12

or ¡9

5 a k = 52

or 2 b k = 1 or¡1§

p33

26 a 16 b ¡34

7 a26664

1 2 1 1 12 1 2 1 11 2 3 1 12 2 1 1 33 3 5 2 2

3777526664

o

a

p

c

l

37775 =

266646:36:77:79:810:9

37775b jAj = 0

c oranges 50 cents, apples 80 cents, pears 70 cents,

cabbages $2:00, lettuces $1:50

1 AB = I, BA = I, A¡1 = B

2 b 2A ¡ I

3 x = ¡1, 2 or ¡4

4 a b·

10 ¡12¡10 4

¸ ·2 6 ¡3¡4 ¡2 11

¸c d

not possible

·2:9 ¡0:3¡0:3 2:1

¸5 Hint: Show that (AB)(B¡1A¡1) = I

and (B¡1A¡1)(AB) = I.

6 a inconsistent, no solution

b x = ¡1, y = ¡2, z = 3

7 a D2 =1

kI b detD = jDj = §

1

k

c D

³1

k¡ 12

´+

1

kI

8 a

T =

"0:43 0:47 0:100:18 0:56 0:260:09 0:55 0:36

#

T2 =

"0:279 0:520 0:2010:202 0:541 0:2570:170 0:548 0:282

#

EXERCISE 10F.2

EXERCISE 10G.1

EXERCISE 10G.2

REVIEW SET 10A


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95

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100


446 ANSWERS

b i 47% ii 20:1%

c i 19:1% will be underweight, 54:4% will be

of normal weight, 26:5% will be overweight

ii 20:4% will be underweight, 54:0% will be

of normal weight, 25:6% will be overweight

d The proportion of people in the ‘normal weight’

category is slowly decreasing.

e 21:0% underweight, 53:9% normal weight,

25:2% overweight

1·2 ¡37 8

¸ ·x

y

¸=

·11¡4

¸, x = 76

37 , y = ¡ 8537

2 a i jBj 6= 0

ii AB = BA (i.e., they commute)

b k is any real number except 3, §2.

3

A¡1 =

24 511

¡211

¡111

111

411

211

411

511

¡311

355 a a = 1, b = ¡1 b x = ¡5, y = 4, z = 7

6 k = ¡ 14

or 2

7 A¡1 = 12 (A ¡ I) A4 = 5A + 6I

8 a

T =

264 0:5 0:2 0:1 0:20:1 0:6 0:2 0:10:1 0:4 0:3 0:20:2 0:1 0:1 0:6

375b S0 =

£56 45 39 60

¤c i

£48 60 32 60

¤B L C P

ii£43 68 33 56

¤B L C P

d£0:2127 0:3433 0:1679 0:2761

¤i.e., 21:3% beef, 34:3% lamb, 16:8% chicken,

27:6% pork

e 83 roast pork meals

1 X = B¡1(3A ¡ C) 2 x = 76

, y = ¡ 109

3 A2 = I, ) A¡1 = A

4

B¡1 =

24 215

115

13

115

¡ 715

¡ 13

415

215

¡ 13

356 a

AX =

"ax+ by 0 ay + bx

0 cz 0ay + bx 0 ax+ by

#

b

A¡1 =

266664a

a2 ¡ b20

¡b

a2 ¡ b2

01

c0

¡b

a2 ¡ b20

a

a2 ¡ b2

3777757

AB =

"1 1 k

0 k ¡ 1 1¡ k

0 0 ¡2k(k ¡ 1)

#a detAB = ¡2k(k ¡ 1)2, k = 0 or 1

b "1 1 k

1 k 12 1 1

#"x

y

z

#=

"111

#

8 a

T =

264 0 0:5 0:3 0:20:4 0 0:1 0:50:2 0:5 0 0:30:1 0:3 0:6 0

375b

T2 =

264 0:28 0:21 0:17 0:340:07 0:40 0:42 0:110:23 0:19 0:29 0:290:24 0:35 0:06 0:35

375

T3 =

264 0:152 0:327 0:309 0:2120:255 0:278 0:127 0:3400:163 0:374 0:262 0:2280:187 0:255 0:317 0:241

375c i 10% ii 7% iii 34% d i C ii D

1 A3 = 27A + 10I, A4 = 145A + 54I

A5 = 779A + 290I, A6 = 4185A + 1558I

2 jPj = 0 or 13

B =

"5 2 44 7 8¡4 ¡4 ¡5

#AB =

"¡3 0 00 ¡3 00 0 ¡3

#= ¡3I

A¡1 =

24 ¡ 53

¡ 23

¡ 43

¡ 43

¡ 73

¡ 83

43

43

53

35 = ¡ 13

B

5

M¡1 =

266435

¡ 75

15

15

625

¡ 1925

¡325

725

¡ 725

4325

¡925

¡425

¡ 825

1725

425

¡ 125

37756 b a = 12, b = 6, c = 9

7 a A6n+3 = ¡I A6n+5 = I ¡ A

b A¡1 = I ¡ A

9 k = 4§p11

REVIEW SET 10B

REVIEW SET 10C

REVIEW SET 10D


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INDEX 447

INDEX equal matrices

exponential decay function

exponential function

exponential growth function

extrapolation

first principles

frequency

function

global maximum

global minimum

horizontal inflection

identity matrix

implicit differentiation

implicit relations

increasing function

independent events

index laws

inferential statistics

initial condition

instantaneous acceleration

instantaneous velocity

integrating constant

interpolation

intersecting lines

inverse function

limit rules

linear function

local maximum

local minimum

logarithmic function

logistic function

lower area sum

lower bound

matrix

mean

midpoint of line

motion graph

multiplication inverse

natural logarithm

negative matrix

network matrix

non-horizontal inflection

normal

345

162

19, 33

162

35

65

226

30

117

116

117

358

85

84

111

285

19

221

106

105

104

194

35

324

154

64

10

116

116

33

33, 163

48, 180

46

342

225

22

103

376

153

350

363

121

87

alternative hypothesis

antiderivative

area under curve

augmented matrix form

average acceleration

average velocity

axis of symmetry

bell-shaped curve

bias

binomial coefficient

binomial distribution

categorical variable

census

central limit theorem

chain rule

chord

coefficient of determination

coincident lines

column matrix

column vector

composite function

concave

confidence interval

continuous variable

convex

cubic polynomial

data

decreasing function

definite integral

derivative

descriptive statistics

determinant

directed graph

discriminant

displacement function

distribution

domain

dominance matrix

elementary row operations

257

184

180

323

105

103

12

231

220

283

287

224

220

252

78

62

37

324

343

343

76

121

222, 267

224

121

18

220

111

49, 204

65

221

386

361

14

103

220

30

361

323


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25

75

75

50

50

95

95

100

100 0 05 5

25

25

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75

50

50

95

95

100

100

Y:\HAESE\SA_12STU-2ed\SA12STU-2_DX\447SA12STU-2_DX.CDR Friday, 10 November 2006 10:26:31 AM PETERDELL

448 INDEX

significance level

similar triangles

slope function

slope of line

slope of tangent

square matrix

standard error

standard normal distribution

state matrix

stationary point

statistic

steady state

sum rule

surd

surge function

tangent

test statistic

transition matrix

two-sided -test

unique solution

upper area sum

upper bound

vertex

-intercept

-intercept

zero matrix

-score

-test

259

22

69

22

62

343

253, 298

237

364

117

220

366

73

14

33, 163

62

259

365

Z 259

324

48, 180

46

12

x 12

y 12

349

z 234

Z 259

normal distribution

null distribution

Null Factor law

null hypothesis

numerical discrete variable

optimum solution

order of matrix

parabola

parallel lines

parameter

Pascal’s triangle

point of inflection

population

population standard deviation

power function

probability density function

probability function

product rule

proportion

-value

Pythagoras’ theorem

quadratic formula

quadratic function

quantile

quantitative variable

quartic function

quotient rule

random sample

random variable

rate of change

rational function

rejection region

residual

row matrix

row vector

row-echelon form

sample

sample standard deviation

sampling error

second derivative

second derivative test

sign diagram

sign diagram test

231

259

14

257

224

124

343

12

324

220

283

121, 244

220

225

32

231

285

80

226

P 259

21

14

12

241

224

10

82

220

224

100

33, 119

264

37

343

343

326

220

225

253

93

126

106

126


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25

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75

50

50

95

95

100

100 0 05 5

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25

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75

50

50

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95

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100

Y:\HAESE\SA_12STU-2ed\SA12STU-2_DX\448SA12STU-2_DX.CDR Friday, 10 November 2006 10:26:42 AM PETERDELL

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mathematics for year 12 mathematical studies...HAESE HARRIS PUBLICATIONS& second edition mathematics for year 12 Robert Haese Sandra Haese Tom van Dulken Kim Harris Anthony Thompson