Oxford Mathematics: Applications and Interpretation Higher Level Course Companion [First Edition] 9780198427056

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O X

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MATHEMATICS: APPLICATIONS AND INTERPRETATION

H I G H E R

L E V E L

E N H A NC E D

Suzanne Doering

Tony Halsey

Panayiotis Economopoulos

Michael Or tman

Peter Gray

Nuriye Sirinoglu Singh

David Harris

Jennifer Chang Wathall

ON L IN E

/

M AT H E M ATI C S :

APPLICATIONS AND

INTERPRETATION

H I G H E R

E N H A NC E D

L E V E L

ON L IN E

C O U R S E

C O M PA N I O N

Paul Belcher

Peter Gray

Palmira Mariz Seiler

Jennifer Chang Wathall

Tony Halsey

Michael Or tman

Suzanne Doering

David Harris

Nuriye Sirinoglu Singh

Phil Duxbury

Lorraine Heinrichs

Nadia Stoyanova Kennedy

Panayiotis Economopoulos

Ed Kemp

Paula Waldman

Jane Forrest

Paul La Rondie

3

/

3 Great

Clarendon

Oxford

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Acknowledgements

The

author

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contributions

Paul

David

Blas

Economopoulos

Edinburgh

Tony

for

Julija

Markeviciute

Vilda

Markeviciute

Dane

Tom

Noon

Rogers

Daniel

Gray

Ioannadis

Kolic

Martin

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Peter

authors

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Fensom

Jane

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resources:

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Halsey

publisher

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p650:

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Panayiotis

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Course Companion denition

doing, they acquire in-depth knowledge and develop

understanding across a broad and balanced range The IB Diploma Programme Course Companions are of disciplines. designed to support students throughout their two-

Thinkers They exercise initiative in applying thinking year Diploma Programme. They will help students

skills critically and creatively to recognise and gain an understanding of what is expected from their

approach complex problems, and make reasoned, subject studies while presenting content in a way

ethical decisions. that illustrates the purpose and aims of the IB. They

reect the philosophy and approach of the IB and

Communicators They understand and express ideas

encourage a deep understanding of each subject by

and information condently and creatively in more

making connections to wider issues and providing

than one language and in a variety of modes of

opportunities for critical thinking.

communication. They work eectively and willingly

in collaboration with others. The books mirror the IB philosophy of viewing the

curriculum in terms of a whole-course approach and

Principled They act with integrity and honesty, with a

include support for international mindedness, the IB

strong sense of fairness, justice, and respect for the

learner prole and the IB Diploma Programme core

dignity of the individual, groups, and communities.

requirements, theory of knowledge, the extended

They take responsibility for their own actions and the

essay and creativity, activity, ser vice (CAS).

consequences that accompany them.

Open-minded They understand and appreciate their

IB mission statement own cultures and personal histories, and are open

The International Baccalaureate aims to develop to the perspectives, values, and traditions of other

inquiring, knowledgable and caring young people individuals and communities. They are accustomed to

who help to create a better and more peaceful world seeking and evaluating a range of points of view, and

through intercultural understanding and respect. are willing to grow from the experience.

To this end the IB works with schools, governments Caring They show empathy, compassion, and

and international organisations to develop respect towards the needs and feelings of others.

challenging programmes of international education They have a personal commitment to ser vice, and

and rigorous assessment. act to make a positive dierence to the lives of

These programmes encourage students across the

others and to the environment.

world to become active, compassionate, and lifelong Risk-takers They approach unfamiliar situations

learners who understand that other people, with their and uncertainty with courage and forethought, and

dierences, can also be right. have the independence of spirit to explore new roles,

ideas, and strategies. They are brave and ar ticulate

The IB learner prole in defending their beliefs.

The aim of all IB programmes is to develop Balanced They understand the importance of internationally minded people who, recognising their intellectual, physical, and emotional balance to common humanity and shared guardianship of the achieve personal well-being for themselves and planet, help to create a better and more peaceful others. world. IB learners strive to be:

Reective They give thoughtful consideration to their Inquirers They develop their natural curiosity. They own learning and experience. They are able to assess acquire the skills necessary to conduct inquiry and and understand their strengths and limitations in research and show independence in learning. They order to support their learning and professional actively enjoy learning and this love of learning will development. be sustained throughout their lives.

K nowledgeable They explore concepts, ideas, and

issues that have local and global signicance. In so

iii

/

Contents

Introduction vii

4 Modelling constant rates of change:

How to use your enhanced

linear functions and regressions 140

online course book ............................................... ix

1 Measuring space: accuracy and

geometry  2

1.1

Representing numbers exactly and

approximately .............................................. 4

1.2

Angles and triangles .................................. 13

1.3

Three dimensional geometry .................... 28

Chapter review

4.1

Functions .................................................. 142

4.2

Linear models ........................................... 155

4.3

Inverse functions ..................................... 168

4.4

Arithmetic sequences and series .......... 1 78

4.5

Linear regression ..................................... 190

Chapter review ...................................................198

Modelling and investigation activity ...............202

.................................................... 39

5 Quantifying uncer tainty:

Modelling and investigation activity ................. 42

probability 204

2 Representing and describing data:

5.1

Reecting on experiences in the world of

chance. First steps in the quantication

descriptive statistics4 4

of probabilities 2.1

Collecting and organizing data ................. 46

2.2

Statistical measures .................................. 51

2.3

Ways in which you can present data ....... 59

2.4

Bivariate data .............................................. 66

5.2

........................................ 206

Representing combined probabilities with

diagrams ................................................... 2 12

5.3

Representing combined probabilities with

diagrams and formulae

.......................... 2 16

Chapter review ..................................................... 75

5.4

Complete, concise and consistent

Modelling and investigation activity ................. 80

representations

....................................... 22 1

3 Dividing up space: coordinate geometry, Chapter review

Voronoi diagrams, vectors, lines

................................................. 226

 82 Modelling and investigation activity ...............230

3.1

Coordinate geometry in 2 and 3

6 Modelling relationships with dimensions

3.2

................................................ 84

The equation of a straight line in 2

dimensions

................................................ 86

3.3

Voronoi diagrams

....................................... 96

3.4

Displacement vectors

functions: power and polynomial

functions

 232

6.1

Quadratic models

............................. 104

6.2

Problems involving quadratics

3.5

The scalar and vector product. ............... 112

6.3

Cubic functions and models ................... 259

3.6

Vector equations of lines

6.4

Power functions, direct and inverse

Chapter review

........................ 118

.................................................. 130

.................................... 234

.............. 245

variation and models ...............................268

Modelling and investigation activity ............... 134

Chapter review

Paper 3 question and comments .................... 136

Modelling and investigation activity .............. 286

.................................................. 281

iv

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10 Analyzing rates of change: dierential

and logarithmic functions 288

calculus

7 .1

Geometric sequences and series ...........290

10.1

7 .2

Financial applications of geometric

10.2 Dierentiation: fur ther rules and

sequences and series ............................ 302

426

Limits and derivatives

........................... 428

techniques ............................................... 442

7 .3

Exponential functions and models ........ 310

10.3 Applications and higher derivatives

7 .4

Laws of exponents – laws of

Chapter review

7 .5

.................................................. 467

Modelling and investigation activity ............... 470

snoitcnuF

logarithms .................................................320

...... 455

Logistic models ........................................ 329

11 Approximating irregular spaces: Chapter review

.................................................. 332

integration and dierential Modelling and investigation activity

arbegla dna rebmuN

7 Modelling rates of change: exponential

..............336

equations  472

8 Modelling periodic phenomena: 11.1

trigonometric functions and complex

Indenite integrals and techniques of

8.1

Measuring angles .................................... 340

8.2

Sinusoidal models:

11.3

Applications of integration ..................... 498

f(x) = a sin (b (x − c)) + d .....................343

11.4

Dierential equations .............................. 515

8.3

Completing our number system ............. 352

11.5

Slope elds and dierential equations ..... 519

8.4

A geometrical interpretation of complex

Chapter review

8.5

...................................................356

..................................................526

Modelling and investigation activity .............. 530

Using complex numbers to understand

12 Modelling motion and change in t wo periodic models ........................................364

Chapter review

................................................. 368

Modelling and investigation activity ............... 370

and three dimensions  532

12.1

Vector quantities ...................................... 535

dna scitsitatS

numbers

integration ................................................487

dna y rtemoeG

11.2

y rtemonogirt

irregular regions....................................... 474

ytilibaborp

numbers 338

Finding approximate areas for

12.2 Motion with variable velocity ..................540

9 Modelling with matrices: storing and 12.3 Exact solutions of coupled dierential

analysing data  372 equations .................................................. 547

Introduction to matrices and matrix 12.4

operations

Approximate solutions to coupled

................................................ 374 linear equations ...................................... 556

9.2

Matrix multiplication and Proper ties

..... 377 Chapter review

9.3

..................................................563

suluclaC

9.1

Solving systems of equations using Modelling and investigation activity .............. 568

matrices ....................................................383

Transformations of the plane .................. 391

9.5

Representing systems ........................... 402

9.6

Representing steady state systems ..... 406

9.7

Eigenvalues and eigenvectors ................ 413

Chapter review

................................................. 422

Modelling and investigation activity ...............424

13 Representing multiple outcomes:

random variables and probability

distributions  570

13.1

Modelling random behaviour .................. 572

13.2 Modelling the number of successes in a

noitarolpxE

9.4

xed number of trials ............................. 580

v

/

13.3 Modelling the number of successes

in a xed inter val ................................. 587

13.4

15.5

Graph theory for weighted graphs: the

travelling salesman problem .................. 720

Modelling measurements that are

Chapter review

.................................................. 729

distributed randomly .......................... 593

Modelling and investigation activity ...............734

13.5 Mean and variance of transformed or

16 Exploration



736

combined random variables ............... 601

13.6 Distributions of combined random

variables

Chapter review

............................................... 604

Practice exam paper 1

 750

Practice exam paper 2

 756

Practice exam paper 3

 761

............................................... 6 16

Modelling and investigation activity ........... 620

Answers

 765

14 Testing for validity: Spearman’s,

Index

 849

hypothesis testing and χ2 test for

independence  622

14.1

Spearman’s rank correlation

Digital contents Coecient ..............................................625

14.2

Hypothesis testing for the binomial

Digital content over view probability, the Poisson mean and Click on this icon here to see a list of all

the product moment correlation the digital resources in your enhanced

coecient

14.3

.............................................629

Testing for the mean of a normal

online course book. To learn more

about the dierent digital resource

types included in each of the chapters

distribution ........................................... 638

and how to get the most out of your 2

14.4

χ

test for independence .................... 650 enhanced online course book, go to 2

14.5

χ

goodness-of-t test ......................... 657 page ix.

14.6

Choice, validity and interpretation

Syllabus coverage of tests ................................................... 670

This book covers all the content of

Chapter review

.............................................. 684 the Mathematics: applications and

Modelling and investigation activity ........... 688 interpretation HL course. Click on this

icon here for a document showing you

15 Optimizing complex net works: graph the syllabus statements covered in

theory

 690 each chapter.

15.1

Constructing graphs ............................ 692

15.2

Graph theory for unweighted

Practice exam papers

Click on this icon here for an additional

graphs ................................................... 699

15.3

set of practice exam papers.

Graph theory for weighted graphs: the

Worked solutions minimum spanning tree

.................... 708 Click on this icon here for worked

15.4

Graph theory for weighted graphs:

the

solutions for all the questions in the book.

Chinese postman problem................... 7 14

vi

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Introduction

The new IB diploma mathematics courses have been

own conceptual understandings and develop higher

designed to support the evolution in mathematics

levels of thinking as they relate facts, skills and

pedagogy and encourage teachers to develop

topics.

students’ conceptual understanding using the The DP mathematics courses identify twelve possible content and skills of mathematics, in order to fundamental concepts which relate to the ve promote deep learning. The new syllabus provides mathematical topic areas, and that teachers can use suggestions of conceptual understandings for to develop connections across the mathematics and teachers to use when designing unit plans and wider curriculum: overall, the goal is to foster more depth, as opposed

to breadth, of understanding of mathematics.

Approximation

Modelling

Representation

Change

Patterns

Space

Equivalence

Quantity

Systems

Generalization

Relationships

Validity

What is teaching for conceptual

understanding in mathematics?

Traditional mathematics learning has often focused

on rote memorization of facts and algorithms, with Each chapter explores two of these concepts, which little attention paid to understanding the underlying are reected in the chapter titles and also listed at concepts in mathematics. As a consequence, many the star t of the chapter. learners have not been exposed to the beauty and

creativity of mathematics which, inherently, is a

The DP syllabus states the essential understandings

network of interconnected conceptual relationships.

for each topic, and suggests some concept-specic

conceptual understandings relevant to the topic Teaching for conceptual understanding is a content. For this series of books, we have identied framework for learning mathematics that frames impor tant topical understandings that link to the factual content and skills; lower order thinking, these and underpin the syllabus, and created with disciplinary and non-disciplinary concepts investigations that enable students to develop and statements of conceptual understanding this understanding. These investigations, which promoting higher order thinking. Concepts represent are a key element of every chapter, include factual powerful, organizing ideas that are not locked in a and conceptual questions to prompt students to particular place, time or situation. In this model, the develop and ar ticulate these topical conceptual development of intellect is achieved by creating a understandings for themselves. synergy between the factual, lower levels of thinking

and the conceptual higher levels of thinking. Facts

A tenet of teaching for conceptual understanding

and skills are used as a foundation to build deep

in mathematics is that the teacher does not tell

conceptual understanding through inquiry.

the student what the topical understandings are

at any stage of the learning process, but provides The IB Approaches to Teaching and Learning investigations that guide students to discover these for (ATLs) include teaching focused on conceptual themselves. The teacher notes on the ebook provide understanding and using inquiry-based approaches. additional support for teachers new to this approach. These books provide a structured inquiry-based

approach in which learners can develop an

A concept-based mathematics framework gives

understanding of the purpose of what they are

students oppor tunities to think more deeply and

learning by asking the questions: why or how? Due

critically, and develop skills necessary for the 2 1st

to this sense of purpose, which is always situated

century and future success.

within a contex t, research shows that learners are

more motivated and suppor ted to construct their

Jennifer Chang Wathall

vii

/

In every chapter, investigations provide

inquiry activities and factual and conceptual

questions that enable students to construct

and communicate their own conceptual

understanding in their own words. The key

to concept-based teaching and learning, the

investigations allow students to develop a deep

conceptual understanding. Each investigation

has full suppor ting teacher notes on the

enhanced online course book.

Gives students the opportunity to reect on what they

have learned and deepen their understanding.

Every chapter starts with a question that students can begin to think

about from the start, and answer more fully as the chapter progresses.

The developing inquiry skills boxes prompt them to think of their

own inquiry topics and use the mathematics they are learning to

investigate them further .

The modelling and investigation activities are

open-ended activities that use mathematics

in a range of engaging contexts and to develop

students’ mathematical toolkit and build the

skills they need for the IA. They appear at the

end of each chapter .

The chapters in this book have been written to provide logical

progression through the content, but you may prefer to use

them in a dierent order, to match your own scheme of work.

The Mathematics: applications and interpretation Standard and

Higher Level books follow a similar chapter order, to make teaching

easier when you have SL and HL students in the same class.

Moreover, where possible, SL and HL chapters start with the same

inquiry questions, contain similar investigations and share some TOK and International-mindedness questions in the chapter reviews and mixed reviews – just as the are integrated into all the chapters. HL exams will include some of the same questions as the SL paper.

viii

/

How to use your enhanced online course book

Throughout the book you will nd the following icons. By clicking on these in your

enhanced online course book you can access the associated activity or document.

Prior learning

Clicking on the icon nex t to the ‘‘Before you star t” section in each chapter takes

you to one or more worksheets containing shor t explanations, examples and

practice exercises on topics that you should know before star ting, or links to other

chapters in the book to revise the prior learning you need.

Additional exercises

The icon by the last exercise at the end of each section of a chapter takes you to

additional exercises for more practice, with questions at the same diculty levels

as those in the book.

Animated worked examples

This icon leads you to an animated worked example,

explaining how the solution is derived step-by-step,

Click here for a transcript

of the audio track.

while also pointing out common errors and how to

avoid them.

Click on the icon on the page to launch

the animation. The animated worked

example will appear in a second screen.

Things to remember and

ex tra tips will appear here.

Click on the icon for the menu

Graphical display and then select your GDC model.

calculator suppor t

Supporting you to make the most of your TI-Nspire

CX, TI-84+ C Silver Edition or Casio fx-CG50 graphical

display calculator (GDC), this icon takes you to step-

by-step instructions for using technology to solve

specic examples in the book.

ix

/

Data sets

Access a spreadsheet containing a data set relevant to the tex t associated with

this icon.

Teacher notes

This icon appears at the beginning of each chapter and opens a set of

comprehensive teaching notes for the investigations, reection questions, TOK

items, and the modelling and investigation activities in the chapter.

Assessment opportunities

This Mathematics: applications and interpretation enhanced online course book

is designed to prepare you for your assessments by giving you a wide range of

practice. In addition to the activities you will nd in this book, fur ther practice and

suppor t are available on the enhanced online course book.

End of chapter tests and mixed review exercises

This icon appears twice in each chapter: rst, nex t to the “Chapter summary” section

and then nex t to the “Chapter review ” heading.

Click here for an end-of-chapter

summative assessment test,

designed to be completed in one hour.

Click here for the mixed review, a summative

assessment consisting of exercises and

exam-style questions, testing the topics you

have covered so far.

Each chapter in the printed book ends with a

“Chapter review ”, a summative assessment of the

facts and skills learned in the chapter, including

problem-solving and exam-style questions.

x

/

Exam-style questions

Plenty of exam practice questions, in Paper 1 (P1) or Paper 2 (P2)

style. Each question in this section has a mark scheme in the

worked solutions document found on the enhanced online course

book, which will help you see how marks are awarded.

The number of darker bars

shows the diculty of the

question (one dark bar = easy;

three dark bars = dicult).

Exam practice exercises provide exam style questions

for Papers 1, 2, and 3 on topics from all the preceding

chapters. Click on the icon for the exam practice

found

at the end of chapters 3, 7 , 11, and 15 in this book.

Introduction to Paper 3

The new HL exams will have three papers, and Paper 3 will have just two ex tended

response problem-solving questions. This introduction, on page 136, explains the

new Paper 3 format, and gives an example of a Paper 3 question, with notes and

guidance on how to interpret and answer it.

There are more Paper 3 questions in the exam practice exercises on the Enhanced

Online Course Book, at the end of Chapters 3, 7 , 11, and 15.

Answers and worked solutions

Answers to the book questions

Concise answer to all the questions in this book can be found on page 765.

Worked solutions

Worked solutions for all questions in the book can be accessed by clicking the icon

found on the Contents page or the rst page of the Answers section.

Answers and worked solutions for the

digital resources

Answers, worked solutions and mark schemes (where applicable) for the

additional exercises, end-of-chapter tests and mixed reviews are included with the

questions themselves.

1

/

Measuring space: accuracy

and geometry

1 In

this

chapter

measures

in

two

how

the

of

and

these

angles,

tools

to

aircrafts.

applications

assessment

In

approximate

distances,

are

used

nearby

to

in

will

dimensions.

distances

innavigation

and

you

three

distance

measure

Concepts

in

between

addition,

surveying,

and

You

and

will

measure

incartography

the

you

course

will

Quantity

Microconcepts

■

Rounding

■

Percentage error

■

Standard form

■

Operations with rational exponents

■

Right triangles: Sine, cosine, tangent ratios

■

Angles of elevation and depression

■

Bearings

■

Non-right triangles: Sine rule and cosine rule

■

Circles: Length of arc, area of sector

■

Volume and surface area of 3D gures

■

Angle between line and plane

to

and

of

ships

explore

architecture,

■

calculate

determine

landmarks,

Space

volumes

astronomyto

stars,

determine

areas

and

■

other

disaster

biology.

How does a sailor calculate the

distance to the coast?

How can the distance to a

nearby star be measured?

How would you decide which shape and

size make a cell more ecient in passing

materials in and out of its membrane?

How could you nd the height

of a mountain peak?

2

/

E

In

1856

India

Great

measured

known

as

The

in

Trigonometric

the

Napali

as

Chomolungma.

distance

then

between

measured

mountain

and

The

summit

two

points

height

two

points

at

of

Everest,

and

surveyors

angles

each

of

Mount

Sagarmatha

The

the

of

Survey

in

Tibetan

measured

sea

level

between

the

the

and

top

of

the

point.

Mount

Everest

is

labeled

E,

the A

are

33 km

ˆ E =

If

90°,

possible

A

and

B

are

roughly

at

sea

level

B

and

apart.

what

height

would

of

be

the

Mount

maximum

Developing inquiry skills

Everest?

Write down any similar inquiry questions you

Think

own

about

and

intuitive

Discuss

share

your

your

then

write

answer

to

answer

with

ideas

with

down

each

a

your

your

might ask if you were asked to nd the height

question.

friend

of tree, the distance between two towns or the

then

distance between two stars.

class.

What questions might you need to ask in these

•

Why

do

you

determine

•

Why

do

the

you

measure

think

it

took

elevation

think

angles,

so

of

long

Mount

surveyors

but

not

scenarios which dier from the scenario where

to

Everest?

prefer

lengths

of

you are estimating the height of Mount Everest?

Think about the questions in this opening problem

to

sides?

and answer any you can. As you work through the

chapter, you will gain mathematical knowledge

•

What

assumptions

are

made?

and skills that will help you to answer them all.

Before you start

Click here for help

You should know how to:

1

Round

to

signicant

gures

and

decimal

1

places.

eg

2

Write

the

number

a

2

decimal

b

3

signicant

a

80.43

Evaluate

b

80.426579

to:

places

gures.

and

rational

exponents.

2

each

a

2

decimal

b

3

signicant

i

0.6942

iii

77.984561

number

rounded

to:

places

ii

gures.

28.706

Evaluate

1

a

3

b

2

27

1

−4

3

down

3

1

eg

Write

80.4

integer

1

with this skills check

Skills check

2

=



,

16

=

16

=

4

4

3

3

Use

81

properties

Pythagoras’

of

triangles,

3

including

theorem.

In

each

length

of

of

these

side

right

triangles,

nd

the

x.

B

7 km

x km

eg

Find

the

length

of 9 km

AB

in

this

2

AB

triangle.

2

=

BC

2

–

AC

5 cm

13 km

2

AB

2

=

5

=

21

x km

5 km

2

–

2

2

AB

AB

=

21

=

4.58 cm

(3

s.f.)

2 cm

3

/

1

ME A S URING

1.1

S PA C E :

ACCURAC Y

AND

GE OM E T R Y

Representing numbers exactly

and approximately

In

mathematics

numbers

will

that

and

have

investigate

calculations

in

been

how

we

everyday

use

we

life,

measured

can

we

or

quantify

throughout

this

frequently

estimated.

encounter

In

uncertainty

this

in

section

the

you

numbers

and

course.

Investigation 1

Margaret Hamilton worked for NASA as the lead developer for Apollo

ight software. The photo here shows her in 1969, standing next to

the books of navigation software code that she and her team produced

for the Apollo mission that rst sent humans to the Moon.

1

Estimate the height of the books of code stacked together, as

shown in the image. What assumptions are you making?

2

Estimate the number of pages of code for the Apollo mission. How

would you go about making this estimation? What assumptions are

you making?

3

What is an estimate? What is estimation? How would

Factual

you go about estimating? How can comparing measures help you

estimate?

Conceptual

4

Recall

that

Why are estimations useful?

we

can

specify

that

can

be

accuracy

using

signicant

gures

(digits).

E X AM HINT The

digits

determined

accurately

are

called

signicant

Write your nal answers gures.

of

a

Thus,

gram

accuracy

a

mass

(4

scale

until

that

could

99.99 g,

signicant

only

could

register

only

up

to

measure

the

up

to

hundredths

4

gures

as exact values or

of

rounded to at least 3 s.f.,

gures).

unless otherwise In

your

IA

or

on

specic

exam

questions,

you

may

need

to

choose

instructed. Round only an

appropriate

degree

of

accuracy.

When

performing

operations

on

your nal answer and measurements,

round

answers

to

the

same

number

of

signicant

not any intermediate gures

as

example

the

least

accurate

measurement.

This

is

illustrated

in

the

calculations.

below.

Example 1

C

A

component

One

as

a

of

the

shown

Find

of

i

of

legs

in

the

an

aircraft

must

the

wing

measure

is

being

17 cm

designed

and

the

in

the

shape

hypotenuse

must

of

a

right

measure

triangle.

97.1 cm,

diagram.

height

of

the

triangle,

rounding

your

answer

to

the

given

degree

accuracy

To

4

d.p.

97.1 cm

(decimal

ii

places)

To

2

s.f.

2

b

Find

the

area

component

of

to

the

an

material

(in

appropriate

cm

degree

)

necessary

of

to

manufacture

the

accuracy.

B

A

c

Show

that

intermediate

rounding

to

2

s.f.

leads

to

an

inaccurate

answer.

17 cm

4

/

a

i

95.6003 cm

ii

96 cm

Use

Pythagoras

to

nd

2

If

the

question

accuracy,

1

b

A

of

the

 17

did



not

would

95.60026151

specify

round

to

a

3

level

s.f.:

of

95.6 cm.

1



bh



2

 17  95.60026151

Use

the

most

accurate

value

(store

or

copy/

2 paste

=

we

height

2

97.1

triangle:

the

arbegla dna rebmuN

1 .1

with

technology)

when

calculating.

812.602… You

could

also

write

down

your

work

2

=

810

cm

(2

1

s.f.) as

A



 17  95.6.

Do

not

use

this

to

2

calculate!

Round

area

accurate

to

2

s.f.

because

measurement,

the

17 cm,

least

has

2

s.f.

1

c

A



 17  96



816

This

shows

why

rounding

intermediate

2 answers

should

be

avoided

in

all

2

≠

810

(2

s.f.)

calculations.

2

cm

dna y rtemoeG

820cm

y rtemonogirt

=

Exercise 1A

1

A

restaurant

its

circular

want

as

the

the

is

tables

new

old

remodelling

with

tables

ones.

The

and

square

to

have

2

replacing

tables.

the

They

same

circumference

The

of

heights

nearest

area

88,

91,

tables

is

measured

to

be

4.1 m.

side

i

to

an

ii

to

3

length

of

the

appropriate

new

square

degree

of

are:

koalas,

81,

73,

measured

71,

80,

76,

to

84,

the

73,

75.

the

mean

(average)

height

of

the

koalas

Find to

the

10

the Find

circular

cm,

of

an

appropriate

degree

of

accuracy.

tables

accuracy

s.f.

Bounds and error

Suppose

to

the

you

nearest

anywhere

would

nd

in

round

the

0.1 g.

the

to

weight

Then

interval

the

of

a

bag

exact

541.45 g

≤

of

coffee

weight

w


100

n

2

n (20  (n  1 )  4)



2

Solve

with

technology

by

graphing

sum

larger

is

smaller

when

n

=

using

a

table.

Sum

n

The

or

9

54

10

80

11

110

than

100

when

n

=

10,

and

11.

Example 19

A

skydiver

second.

In

continues

jumping

each

until

from

succeeding

she

opens

a

Find

the

distance

b

Find

the

total

c

The

skydiver

Determine

a

that

distance

should

the

latest

plane

4200 m

second

her

she

she

open

time,

she

above

falls

7.8

the

ground

metres

falls

more

9.5

than

in

metres

the

in

the

previous

rst

one.

This

parachute.

falls

has

her

to

during

fallen

the

after

parachute

the

nearest

20th

20

at

second.

seconds.

or

before

second,

600 m

that

she

above

can

ground

open

her

level.

parachute.

186

/

4.4

A

second

falls

d

skydiver

metres

more

approximately

d

Find

e

Explain,

a

u

d

=

to

leaving

120

the

+

a

same

second

plane

than

the

also

falls

previous

9.5

metres

one.

After

in

5

the

rst

seconds

second,

she

has

but

fallen

metres.

nearest

with

9.5

each

the

tenth

reason,

(20

−

1)

×

of

a

which

7.8

=

metre.

skydiver

is

157.7 m

falling

The

faster.

distance

the

skydiver

has

travelled

after

20

each

rst

second

term

is

9.5

an

arithmetic

and

common

sequence

difference

with

7.8.

20 S

 (9 .5  157 .7) 



1672 m

Because

the

sequence

represents

distance

she

falls

each

distance

will

the

additional

20

2

second

be

with

the

sum

of

second,

these

her

total

second-by-

distances.

snoitcnuF

b

n

c

3700

 (2  9 .5

 (n

At

 1 )  7 .8) 

a

height

of

600 m

above

ground

level

2 the

skydiver

will

have

fallen

a

total

of

4200−600=3700 m y

The

sum

must

therefore

be

less

than

3700.

You

4000

use

(30.092, 3700)

this

to

set

up

an

inequality

and

solve

with

3000

technology:

2000

At

1000

30

seconds

31seconds

she

she

will

will

be

be

above

600 m;

at

below.

x 0 –10

n

≤

10

her

d

=

30

40

30.092

The

d

20

latest

the

parachute

skydiver

is

after

can

safely

open

30seconds.

For

7.3 m

the

9.5

but

You

second

the

know

skydiver

common

that

S

=

the

rst

difference

term

is

is

still

unknown.

120.

5

Substitute

sum

the

the

formula

known

and

unknown,

use

information

technology

into

to

the

solve

for

d:

5 120

 (2  9.5  ( 5  1 )d) 

2

d

e

The

rst

skydiver

distances

second

are

is

faster,

increasing

rather

than

by

as

=

7.25

=

7.3 m

to

the

nearest

tenth

of

a

metre.

her

7.8 m

each

7.3 m.

187

/

4

MODE LLING

C O N S TA N T

R AT E S

OF

C H A NGE :

LIN E A R

F UNCTIONS

AND

R E GR E SS ION S

Exercise 4N

1

Find

the

sum

arithmetic

of

the

series

rst

20

terms

of

6

the

6 +3+0−3−6−…

Montserrat

In

her

and

2

Find

the

sum

of

the

rst

30

multiples

of

rst

in

the

Consider

the

series

52

a

Find

the

number

b

Find

the

sum

+

of

62

+

72

+

…

+

training

training

second

than

462.

a

terms.

Every

the

her

rst

week

she

runs

training

the

week

previous

Determine

the

Montserrat of

for

week

race.

3 km,

she

runs

8. 3.5 km.

3

is

she

runs

0.5 km

more

one.

number

runs

in

of

her

kilometres

10th

training

terms.

week?

4

Janet

begins

farming

on

1

acre

of

land.

b Each

year,

as

her

business

land

to

farm.

grows,

she

Calculate

that more

She

buys

an

additional

in

the

third

year.

second

year

and

9

acres

Montserrat

in

7

Assuming

that

buys

year

the

amount

of

land

In

training

Exercise

continues

in

the

write

down

a

sum

of

22

total

land

she

owns

that

after

Write

c

Calculate

down

this

sum

in

ten

and

nd

sigma

S

10

d

Find

the

in

of

n

for

terms

of

interpret

the

and

Find

S

An

arithmetic

series

has

S

=

of

n

b

in

Write

down

an

question

rows

of

5,

seats

you

in

a

found

theatre

the

that

in

the

rst

row,

49

in

the

tenth

106

in

the

last

row.

Recall

that

the

of

seats

in

each

row

increases

by

a

amount.

the

total

number

of

seats

in

the

An

4

and

expression

for

architect

is

designing

another

theatre

context.

d

=

−3.

25

1

a

week.

=2000,

for

5

her

theatre.

which

meaning

by

n

n

and

run

notation.

n

value

have

years.

a S

will

represents

constant

b

4L

seats

number the

kilometres

same

row, pattern,

of

she

has each

number

the

number

a

total

5

15th acres

the

buys

S

a

rows,

and

in

university

a

a

rst

total

row

that

of

of

wants

at

16

least

to

have

6000

seats.

The

seats,

number

n

terms

of

of

n

seats

per

constant

b

Find

row

will

amount.

increase

Determine

by

a

the

least

S 10

possible

c

Find

the

smallest

n

for

total S




0

is

metres.

Find

the

maximum

height

that

the

shot-

the

symmetry

reaches.

and

b the

0.75x

horizontal

put coordinate

+

graph

a show

1.5

snoitcnuF

2

e

Write

down

the

equation

of

the

axis

of

range.

symmetry. 2

a

f(x)

=

−x

b

f(x)

=

−2x

,

for

−3

≤

x

≤

4

c

Find

the

positive

value

for

x

when

the

2

+

4x

−

3,

for

0

≤

x

≤

2 graph

1

what

2

c

f ( x)



x



4x

 5,

for

0

≤

x

≤

crosses

this

the

value

x-axis

and

represents

explain

in

context.

4

2

d

Find

1

this

2

d

f ( x)



x



4x

 5,

for

0

≤

x

≤

6

2

f(x)

95

=

≤

−5x

x

≤

+

1000x

−

49945,

y-intercept

value

and

represents

in

explain

what

context.

8

2

e

the

Ziyue

He

for

is

takes

vertical

105

the

a

goalkeeper

free

path

kick

of

the

in

from

ball

his

the

is

football

goal

team.

and

modelled

the

by

the

2

function

2

f

P(q)

=

75

−

0.000186q

,

where

P

is

is price

at

which

q

units

of

a

certain

be

sold

and

where

q

≤

Consider

the

curve

dened

−0.06x

+

1.2x,

where

f(x)

the

height

of

the

ball,

in

metres,

and

x

is

horizontal

distance

travelled

by

the

ball,

500

in

3

=

good

the can

f(x)

the

by

the

metres.

equation

a

Find

the

maximum

height

that

the

ball

2

y

=

0.4x

−

2x

−

8. reaches.

a

Find

the

coordinates

where

the

curve

b crosses

the

Write

down

the

equation

of

the

axis

of

x-axis. symmetry.

b

Find

the

coordinates

with

the

y-axis.

of

the

intercept

c

Find

the

these

c

Find

the

equation

of

the

axis

of

the

values

Find

the

point

The

of

intersection

of

with

the

what

represent.

vertical

cross-section

of

a

curve

given

by

park

is

modelled

this

ramp

in

a

by

the

function

2

f(x) curve

explain

curve.

skateboard

d

and

of

7 symmetry

x-intercepts

=

10.67

−

1.6x

+

0.0417x

,

where

x

is

the the

horizontal

distance

in

metres

from

the

3

equation

y

=

−5x

. start

4

Zander

the

is

ball

playing

and

the

a

game

height

of

of

baseball.

the

ball

is

He

hits

modelled

2

by

the

where

and

a

x

formula

y

>

is

0

Find

the

is

reaches.

=

−0.018x

height

the

the

y

of

the

horizontal

maximum

+

0.54x

ball,

in

distance

height

+

1.0,

metres,

in

that

metres.

the

ball

the

and

f(x)

ground

constructed

base

lies

around

is

in

in

vertical

metres.

such

below

the

the

the

a

The

way

level

ramp

that

of

the

above

has

part

been

of

its

ground

ramp.

a

Find

the

minimum

b

Find

the

x-intercepts

these

distance

values

depth

and

of

the

run.

explain

what

represent.

237

/

6

MODE LLING

8

Jin

throws

the

stone

a

R E L AT I O N S H I P S

stone

above

into

the

the

air.

ground

The

can

WITH

F UNCTIONS:

height

be

b

of

POWER

Use

modelled

the

plot

AND

table

the

P O LY N O M I A L

function

values

of

S

on

for

1

F UNCTIONS

the

GDC

≤

≤

n

to

8.

n 2

by

the

function

f(t)

=

1

+

7.25t

−

1.875t

,

c

where

t

is

the

time,

in

seconds,

that

Hence,

since

the

stone

was

thrown,

and

the

height

of

the

stone,

in

the

Find

the

maximum

possible

value

sequence.

Determine

the

greatest

value

of

n

for

metres. which

a

maximum

f(t)

d is

the

has of

passed

nd

height

that

the

the

sum

is

positive.

stone

11

The

equation

of

motion

when

throwing

reaches. something

b

Determine

how

long

it

takes

for

vertically

upwards

the

formula

h (t )



h



v

0

stone

to

land

on

the

is

given



t



0

by

2

1

the

g



t

,

2

ground. where

h

is

the

initial

height

above

the

0

9

A

bullet

is

red

from

the

top

of

a

cliff.

The ground,

path

of

the

bullet

may

be

modelled

by

v

is

the

initial

upward

speed

and

0

the

2

g

(=

9.81 m s

)

is

the

acceleration

due

to

2

equation

y

=

−0.0147x

+

2x

+

96

(x

≥

0), gravity.

where

x

is

the

horizontal

distance

from

roof foot

of

the

cliff

and

y

is

the

vertical

the

sea

which

meets

the

someone

standing

on

the

of

a

building

21 m

tall,

who

throws

a

distance ball

above

Consider

the

foot

of

vertically

upwards

with

an

initial

speed

the 1

of

15 m s

cliff.

a a

State

the

height

of

the

Sketch

ball

b

Find

the

maximum

height

reached

Find

the

of

Consider

three

a

distance

the

an

terms

Show

the

against

Determine

the

bullet

is

from

the

height

of

the

time.

the

maximum

height

that

the

cliff,

when

arithmetic

13,

that

9

the

arithmetic

and

it

hits

the

sequence

will

reach.

the

c foot

10

of

bullet. ball

c

graph

by

b the

the

cliff.

Determine

how

in

fall

long

the

ball

will

need

sea.

with

order

to

to

the

ground.

rst

5.

sum

of

the

sequence

rst

is

n

terms

given

by

of

the

2

quadratic

expression

S

=

15n

−

2n

,

n +

n

∈

ℤ

Problems involving quadratics

Example 3

A

a

rectangular

If

the

terms

mirror

length

of

of

has

the

perimeter

mirror

is

260 cm.

x cm,

nd

the

height

of

the

mirror

in

x x

cm

2

b

Find

c

Use

on

an

equation

your

the

d

Find

e

State

f

Find

g

State

GDC

y-axis

the

to

and

these

the

a

of

of

area

graph

length

two

equation

what

the

plot

coordinates

what

the

for

x

on

the

equation

of

of

x

the

mirror,

your

the

graph’s

of

the

A cm

equation

x-axis.

points

values

the

of

where

in

for

Choose

the

,

a

terms

the

area

suitable

graph

of

of

x

the

mirror,

domain

intersects

the

and

showing

area

A

range.

x-axis.

represent.

line

line

of

of

symmetry.

symmetry

represents

in

this

context.

238

/

6 .1

Let

the

height

of

the

rectangle

be

y cm.

Label

the

height

equation 2x

+

2y

=

b

+

Area

y

= 130

y

= 130 -

=

length

the

and

y

for

form

the

an

perimeter

A

=

xy

A

=

x( 130 -

rectangle.

x

×

height

Substitute

found

c

x

and

260

of x

in

y cm

in

y

=

part

130

−

x,

which

you

a.

x)

Use

y

your

GDC

to

locate

the

snoitcnuF

a

turning

4500

point.

A

reasonable

domain

would

be

4000

from

0

to

150,

and

range

from

0

to

3500

4500. 2

3000

f(x)

=

130x

–

x

2500

2000

1500

1000

500

x 0 10

20

30

40

50

d

The

x-intercepts

e

The

x-coordinates

and

lower

60

are

70

80

(0, 0)

90

100 110 120 130 140

and

(130, 0).

Use

your

GDC

to

nd

the

zeros

of

the

function.

must

limits

0

and

130

between

are

the

upper

which

the

value

of

x

lie.

You

f

The

line

of

symmetry

is

x

=

can

nd

by

nding

vertex

on

midpoint

g

This

is

the

area

of

value

the

of

x

which

the

line

of

symmetry

65

gives

the

largest

The

mirror.

the

axis

the

coordinates

your

of

of

vertex

the

two

or

the

in

by

nding

passes

graph,

this

the

the

x-intercepts.

symmetry

of

maximum

GDC

of

through

which

is

a

case.

Exercise 6B

1

A

rectangular

100

a

picture

frame

has

perimeter

e

Find

the

y-intercept.

f

Find

the

equation

cm.

The

width

of

expression,

of

the

the

in

frame

terms

of

is

x,

x

cm.

for

Find

the

the

axis

of

symmetry.

an

height

of

g

Find

h

Write

the

coordinates

of

the

vertex.

frame. down

the

maximum

area

of

the

2

b

Find

in

an

expression

terms

c

Sketch

d

Find

of

the

the

for

the

area,

A cm

, picture

frame,

picture

frame

and

the

dimensions

of

the

x.

graph

of

A

against

x.

that

give

this

maximum

area.

x-intercepts.

239

/

6

MODE LLING

2

A

rectangular

R E L AT I O N S H I P S

picture

frame

has

WITH

F UNCTIONS:

c

perimeter

Find

70 cm.

a

the

The

width

of

expression,

of

POWER

the

the

in

frame

terms

of

is

x,

x

cm.

for

Find

the

an

5

A

If

frame.

Find

an

expression

for

the

area

of

A

cm

jumping

,

in

terms

of

Sketch

varies

the

graph

with

x.

which

Use

a

shows

suitable

how

out

domain

will

fall

and

6

Marina

e

State

x-intercepts

of

the

graph

in

these

two

seconds

of

the

shape

the

of

which

than

water

a

880.

and

parabola.

maximum

after

it

started

of

the

water,

determine

when

values

of

back

in

the

water.

a

12

small

in

her

me tr e s

of

fe nci ng

p l a y g r ound

recta ng ul ar

for

10

to

he r

m

by

c

part

8 what

the

for

greater

out

reaches

bought

construct

the

follows

1.3

daughter Find

jumping

is

n

A

range.

d

terms

of

x

it

c

values

F UNCTIONS

the

2

frame,

at

P O LY N O M I A L

of

n

dolphin

height

b

to

is

motion

the

set

sum

dolphin

its

height

the

AND

m

backyard.

She

w a nts

to

de ci de

h ow

x to

place

the

fencing

in

o rd er

to

constr u c t

represent. the

f

Use

the

x-intercepts

to

nd

the

largest

sides

of

the

graph’s

line

of

symmetry,

what

this

tells

you

in

the

of

her

context

of

backya r d

adjacent

T h re e

are

e nclos e d

by

ho us e s

and

one

si d e

is

of connected

the

pl ay gr ound.

and walls

state

rectan g ul a r

equation

to

her

ho us e

thr oug h

a

sl id in g

problem. door.

3

A

company

produces

and

sells

books.

a

The

weekly

cost,

in

euros,

for

producing

is

€

0

1x



400

rst

analyse

x

2

books

The

design

was

playground



that

that

in

of

the

she

a

wanted

to

rectangular

middle

of

the

backyard. The

weekly

income

from

selling

x

books

is

2

€

a

 0

12 x

 30 x

Show

that



the

weekly

prot,

P(x),

can

be

2

written

b

Sketch

as

the

showing

and

c

the

d

what

the

tells





0

of

22 x

the

 30 x

prot

coordinates

of

 400

function

the

vertex

intercepts.

the

of

x-intercepts

the

equation

symmetry

this

x

graph

context

Find



the

axes

State

P

of

you

the

in

represent

in

problem.

of

the

graph,

the

axis

and

context

of

i

state

of

what

If

the

length

playground

the

other

side

of

is

in

one

x,

side

nd

terms

the

of

of

the

length

of

the

x.

problem.

ii

4

The

rst

four

sequence

terms

of

an

arithmetic

a

10

14

expression

in

for

terms

of

the

area

of

the

x.

are:

18

Show

an

playground

iii

6

Find

Using

area

that

the

sum

to

n

terms

can

be

iv

an

appropriate

function,

Hence

sketch

determine

the

domain

its

for

the

graph.

maximum

2

written

as

2n

+

4n

possible

area

that

the

playground

can

have.

b

If

the

sum

quadratic

to

n

terms

equation

to

is

880,

write

represent

a

this

information.

240

/

6 .1

b

The

to

second

analyse

design

was

playground

adjacent

that

with

wall

that

of

of

one

her

she

a

wanted

rectangular

side

being

the

backyard.

x

If

the

length

playground

other

ii

Find

side

an

If

the

length

playground

diagram,

side

ii

in

Find

terms

an

Using

area

is

nd

one

x

as

the

of

an

in

side

length

in

of

iii

the

the

Using

area

other

iv

for

terms

the

of

sketch

Hence

determine

possible

area

the

that

in

side

nd

terms

of

the

length

of

the

x

for

terms

of

the

of

area

of

sketch

determine

possible

the

appropriate

function,

the

area

of

the

x

area

that

the

the

domain

its

for

the

graph.

maximum

playground

can

have.

domain

for

v

the

State

the

graph.

the

dimensions

maximum

possible

iv

x,

expression

an

Hence

x

its

one

the

x

appropriate

function,

of

shown

expression

playground

iii

of

is

in

playground

i

of

snoitcnuF

i

area

that

from

produce

all

the

designs.

maximum

the

playground

can

have.

c

A

third

design

was

that

two

sides

corners

of

a

that

wanted

rectangular

being

of

she

her

on

the

to

analyse

playground

walls

of

one

with

of

the

backyard.

Restricted domains and the inverse

of a function

From

chapter

4

you

will

recall

the

following

facts

about

the

inverse

−1

function

f

:

-1

•

f

-1



f

(

x

)

=

f



f

of

y

=

(x)

=

x.

−1

•

The

y

•

=

To

f(x)

nd

and

This

to

graph

in

the

the

line

(x)

y

inverse

rearrange

section

f

will

to

=

can

be

by

reecting

the

graph

of

x.

function

make

explore

obtained

a

y

the

write

as

subject

particular

y

of

=

f(x),

the

interchange

x

and

y

expression.

requirement

for

inverse

functions

exist.

241

/

6

MODE LLING

R E L AT I O N S H I P S

WITH

F UNCTIONS:

POWER

AND

P O LY N O M I A L

F UNCTIONS

Investigation 1

h

A pyrotechnician has developed a model function for

200

the trajectory of the reworks that he launches:

2

h(t) = 180 − 5(t − 6)

180

. This is a function of height

against time. In this way he can predict the height

160

h of

the reworks, t seconds after launch.

140

He is planning a show with a large number of reworks.

120

To create the most amazing spectacle, he wants the

100

reworks to burst at dierent heights so that they do not 80

overlap, as shown on the graph. Knowing the desired 60

height of burst of each rework , he needs to work out the 40

corresponding time. What he has realized is that, instead

20

of knowing the input of the function, he now knows the

t

output and is looking for the input, which leads to a time-

0 1

2

3

4

5

6

7

8

9

10

11

12

consuming solution of quadratic equations each time.

So, he has decided to reverse the variables of the function and make the height

h the input and the time t the

output. To do this he needs to make t the subject of the equation.

After reaching the maximum height, the reworks will star t falling back to the ground. This means that they

will be at a cer tain desired height at two dierent times. It is required, though, that the reworks burst while

they are ascending.

1

How should the domain of the model function be restricted so that each height corresponds to a single time?

2

Rearrange the model function h(t) = 180 − 5

2

(t − 6)

so that it becomes a function of time t with

respect to height h

3

What do you notice? How can the restricted domain from question

1 help?

The function h(t) is an example of a many-to-one function: two (or more) dierent inputs map to the same

output. When each output corresponds to exactly one input, we say a function is

one-to-one. Here is an

example of each function type:

2

One-to-one: y = 3x −5

Many-to-one: y = x

y

y

8 12

4

8

x 0 4

–4

x

O –6

–4

–12

–2

2

4

6

–4

–16

4

You can create the inverse relation of any function by reecting it across the line

y = x, as you did for

inverse functions.

a

Sketch the graph of the inverse relation of each of the example functions above.

b

Use your graphs to explain why the one-to-one function’s inverse relation is a function, but the many-

to-one’s inverse relation is not a function.

5

Conceptual

Why is a function not always inver tible (inver tible means that its inverse function

exists), and when is it possible for a function to be inver tible?

242

/

6 .1

An

the

easy

way

to

horizontal

curve

more

test

line

than

whether

test.

once

This

then

or

not

test

an

a

function

says,

if

inverse

any

has

inverse

horizontal

function

If a function is many-to-one, we can sometimes

an

does

line

not

is

to

cuts

use

the

exist.

restrict its domain so that it is

one-to-one.

Example 4 y

2

the

function

f(x)

=

(x

−

2)

snoitcnuF

Consider

1

a

State

reasons

why

f

does

not

exist

if

the

domain

of

6

f

is

x

∈

ℝ

(0, 4)

b

The

domain

possible

is

value

now

of

k

restricted

such

that

to

x

the

≥

k.

Find

function

is

the

4

smallest

invertible.

2

(2, 0)

1

c

Restricting

f(x)

to

this

domain,

sketch

the

graph

of

f

(x).

x 0 –4

–2

2

4

6

8

10

–2

1

a

f

does

one.

not

For

exist

because

example,

f(0)

=

f

4

is

many-to-

and

f(4)

=

4.

Note

pass

line

that

the

a

many-to-one

horizontal

can

be

function

drawn

more

line

that

than

function

test :

A

intersects

will

not

horizontal

the

once.

y

6

(0, 4)

2

(2, 0) x 0 –4

–2

2

4

6

8

10

–2

b

{x | x

≥

The

2}

function

there

are

positive

the

is

two

symmetric

x-values

y-value.

function

around

corresponding

Eliminating

will

x=2,

remove

the

one

of

left

to

so

each

“half”

these

of

x-values.

1

c

The

graph

of

f

is

in

orange.

Sketch

the

function

line

will

y

be

=

x

to

see

where

the

reected.

y

16

Use

the

table

or

point-plotting

feature

of

14

your

graphing

technology

to

nd

several

12

coordinates

on

the

graph

of

f(x).

Invert

10

these

(5, 9)

8

to

nd

the

corresponding

coordinates

1

of

f

(x).

6

(1, 3) 4

(9, 5) (4, 4)

2

(0, 2)

(3, 1) x

–4

–2

0 (2, 0) 2

4

6

8

10

12

14

16

–2

–4

243

/

6

MODE LLING

R E L AT I O N S H I P S

WITH

F UNCTIONS:

POWER

AND

P O LY N O M I A L

F UNCTIONS

Exercise 6C

TOK 1

Restrict

the

domain

of

the

follow-

3

Consider

the

function

2

ing

quadratic

they

can

functions

have

an

so

inverse

that

f(x)

nd

their

inverse,

(x

+

5)

−

2.

How can you deal with

and

a

then

=

stating

Use

technology

to

sketch

the

its

using mathematics to

graph.

range.

cause harm, such as

−1

b

Explain

why

f

does

not

3

a

f(x)

=

x

+

plotting the course of a

4 exist

if

the

domain

of

f

is missile?

3

b

f(x)

=

(x

c

f(x)

=

2(x

+

d

f(x)

=

1

3(x

−

the ethical dilemma of

1)

−

2

x∈

3

c

3)

−

the

domain

3

−

Find

such

The

model

positive

that

the

2)

function

2

largest

function

linking

is

invertible.

the −1

d price

€p

of

a

can

of

soda

to

Sketch

by

the

Q(p)

quantity

model

=

175

−

Q,

is

€5

cost

and

cost

of

of

4

function

3.5p

restricted

Construct

function

the

producing

company

€875

in

one

has

making

a

can

xed

a

these

The

the

the

modelling

revenue

of

that

Find

for

the

the

of

two

Hence

nd

the

the

a

for

has

with

three

different

negative

and

one

gradient.

the

function

company.

b

The

function

is

continuous.

c

The

function

is

invertible

when

the

restricted

domain

to

x

on

≥

3,

the

is

but

is

not

domain

modelling

x function

the

gradients

invertible

c

meets

linear

function

modelling

costs

on

requirements:

pieces,

company .

b

(x)

domain.

piecewise

function

with

for

a

positive

Find

f

is

cans.

a

of

given

following The

graph

its this

demanded

the

the

prot

of

≥

k

for

k

smaller

than

3.

the

company.

Developing inquiry skills

Look back at the opening

problem for this chapter.

Oliver was trying to throw a

basketball through a hoop.

What type of function could

you use to model the path of

the basketball?

244

/

6.2

6.2

The

equation

with

the

Problems involving quadratics

two

will

be

of

linear

these

since

create

able

a

parameters

value

points,

of

to

two

(m

two

there

is

and

c)

has

single

line

equations

through

your

the

dening

parameters,

a

linear

solve

function

you

a

form

unique

need

going

with

general

the

through

two

line.

f(x)

To

=

mx

given

unknowns,

c,

determine

coordinates

two

+

of

two

points.

which

you

This

will

GDC.

2

equation

with

the

three

value

points,

This

will

be

a

quadratic

parameters

of

these

since

will

of

able

three

there

create

to

(a,

is

a

three

solve

b

function

and

c)

linear

you

parabola

equations

through

your

the

dening

parameters,

single

has

general

a

unique

now

going

with

need

f(x)

parabola.

the

through

three

form

=

ax

T o

+

given

unknowns,

+

c,

determine

coordinates

three

bx

of

three

snoitcnuF

Τhe

points.

which

you

GDC.

Example 5

A

student

respect

wants

to

to

model

the

path

of

a

rock

h

with

30

time.

V

25

He

is

standing

height

of

shown

at

20 m

in

the

the

and

top

of

throws

a

a

vertical

rock

cliff

at

upwards

a

as

diagram. 15

He

starts

the

counting

rock,

at

point

time

at

the

instant

he

throws

10

A. 5

t

He

measures

that

the

time

at

which

the

rock

is 0

again

He

at

eye

nally

The

level,

at

measures

simplest

point

that

function

B,

the

that

is

t

=

time

will

3.26 s.

at

which

model

the

the

rock

height

of

falls

the

into

rock

the

sea,

with

at

point

respect

to

P ,

is

time

t

is

=

4.8 s.

a

2

quadratic

of

Determine

There

are

Method

your

You

the

this

two

1:

form

h(t)

=

at

+

bt

+

c.

function.

methods

Using

you

can

Quadratic

use

to

do

this.

regression

on

GDC.

need

to

identify

three

clear

points

on

the

curve.

You

can

graph

Known

points

are

A(0, 20),

B(3.26, 20)

P(4.8, 0).

and

Use

At

the

as

the

A,

t

and

The

parabola

h(t)

=

is

given

by

the

equation

0

and

as

at

level,

GDC,

the

Then

as

three

they

information

same

your

any

long

=

ground

In

take

go

h

h

A,

regression.

20.

20.

put

the

in

P

=

t

List

on

the

exact.

At

At

h

Statistics

Here

are

from

where

values

to

=

points

the

B

question.

the

the

height

rock

is

is

at

0.

values

in

List

1

2.

Quadratic

you

see

the

values

of

the

2

−2.71t

+

8.82t

+

20 parameters

a,

b

and

c

Continued

on

next

page

245

/

6

MODE LLING

Method

solver

Curve

and

2:

on

Using

your

passes

P(4.8,

R E L AT I O N S H I P S

Simultaneous

WITH

F UNCTIONS:

AND

P O LY N O M I A L

F UNCTIONS

equation

GDC.

through

A(0,

20),

B(3.26,

20)

First,

0).

Substituting

POWER

as

these

points

into

the

given

function:

in

nd

the

Method

points

on

the

curve

1.

For

each

the

coordinates

general

of

three

the

three

of

equation

t

points,

and

of

a

h

substitute

into

the

parabola,

2

Using

point

A:

a(0)

+

b(0)

+

c

=

20

⇒

c

=

20 2

h

=

at

+

bt

+

c.

2

Using

point

B:

a(3.26)

+

b(3.26)

+

20

=

20 This

gives

you

three

simultaneous

2

⇒

a(3.26)

+

b(3.26)

=

0 equations

in

a,

b

and

c

2

Using

point

P:

a(4.8)

+

b(4.8)

+

20

=

0 Solve

these

equations

on

your

GDC

to

2

⇒

a(4.8)

+

b(4.8)

=

−20 give

the

solution.

2

h(t)

=

−2.71t

+

8.82t

+

20

Exercise 6D

Find

the

quadratic

functions

represented

by

y

3

6

the

following

graphs.

4

1

y

2 1

x x

0 –5

0

–4

–3

–2

–1

1

2

5

6

–2

–1

–4

–2 –6

–3

–4

–12

y

4

10 –7

9

8

2

y

7

6

6

5

4

2 3

0

x

2

1

–2

x 0 –1

1

2

3

4

–1

Find

approximate

represented

by

quadratic

the

functions

following

graphs.

246

/

6.2

y

5

a

In

a

free

kick,

the

player

sets

the

ball

at

20

the

point

where

they

were

fouled,

and

the

10

opponents

can

make

a

“wall”

at

a

distance

x

of

0

9.1

metres.

The

goal

at

which

the

player

–10

taking

the

free

kick

is

aiming,

has

a

height

–20

of

2.4

metres.

Assuming

that

the

player

–30

was

fouled

and

that

will

make

about

20

metres

from

the

goal

–40

the

tallest

football

players

who

y

b the

wall

have

a

height

of

about

8

2

metres,

we

are

faced

with

the

problem

6

of

creating

the

trajectory

to

describe

the

optimum

path

of

the

ball

towards

the

snoitcnuF

4

net.

2

x

y

5

4

B

3

C

2

x 0

6

The

graph

of

the

quadratic

1

function

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

2

f(x)

=

ax

+

bx

+

c

passess

through

point Assume

P(3,

the

modelled

a

Show

that

9a

+

3b

+

c

=

graph

Q(−2,

9)

b

Find

c

Solve

also

and

two

the

points

of

a,

equations

values

graph

for

of

of

this

a,

b

b

and

as

a

the

ball

can

be

parabola.

9

c

function

the

The

the

equation

just

over

wall

and

crossbar

c

to

and

the

passes

in

equations

system

the

through

−30).

more

the

Sketch

passess

R(5,

determine

d

of

−4. Find

The

path

−4).

of

the

arched

the

the

path

head

then

goal,

truss

following

of

of

passes

as

for

of

the

just

shown

an

the

ball

tallest

players

under

in

the

outdoor

which

the

diagram.

concert

has

shape:

clearly y

showing

7

A

points

mathematics

model

Louis

is

function

in

both

the

US

192 m

modelling

P ,

Q

and

student

for

the

state

wide

R.

wants

to

Gateway

of

and

create

Arch

Missouri.

high.

The

Find

a

a

in

St.

arch

suitable

function.

x

The

function

that

models

this

arch

is

known

2

to

be

the

f(x)

−0.12x

horizontal

corner

of

=

the

and

y

arch

of

is

the

concert,

the

1.92x

distance

the

rests

Because

+

from

height.

on

3m

wind

+

3,

where

the

The

expected

bottom

curved

vertical

x

is

left

part

sides.

during

the

O

arched

8

In

a

football

game,

if

a

player

is

the

penalty

area,

they

are

free

of

the

stage

down

keeping

the

vertical

to

by

bring

the

1 metre

section

at

3 m.

awarded Find

a

decided

fouled while

outside

part

organizers

the

new

function

that

will

model

the

kick. arched

part

of

the

stage.

247

/

6

MODE LLING

R E L AT I O N S H I P S

WITH

F UNCTIONS:

POWER

AND

P O LY N O M I A L

F UNCTIONS

Quadratic regression

In

order

the

data

the

•

to

choose

need

data

to

the

data

having

the

axis

the

•

being

of

data

a

more

curve

in

•

to

order

If

symmetry

data

to

vertex

of

the

a

As

describe

features.

or

a

parabola)

about

a

certain

a

certain

These

single

a

of

data,

include:

minimum

vertical

set

line

point

(corresponding

to

parabola)

rate

the

you

three

of

of

a

saw

change,

creating

a

curve

parabola).

more

quadratic

exactly

to

of

shape

have,

a

specic

variable

them.

create

have

a

you

model

you

the

model

maximum

symmetric

having

to

some

single

to

(corresponding

The

quadratic

have

(corresponding

•

a

certain

before,

you

you

are

about

need

at

the

least

choice

three

of

points

model.

points,

there

is

a

single

parabola

going

Internationalthrough

to

use.

them,

In

indeed

but

such

a

looking

not

case,

for

a

enough

you

evidence

need

parabola

to

in

that

know

order

in

to

it

is

the

advance

go

on

and

correct

that

nd

curve

you

mindedness

are

one. The

If

you

have

more

than

three

points

from

a

data

collection

process,

it

Sulba

is (around

almost

certain

that

they

will

not

lie

exactly

on

the

graph

of

a

function.

This

means

that

you

need

to

determine

500 BCE)

certain and

quadratic

Sutras

the

Bakhshali

the manuscripts

parabola

of

best

t

through

these

given

points.

Your

GDC

can

easily 300 AD),

and

quickly

perform

this

operation

through

quadratic

both

contained

algebraic difference

from

the

previous

example

is

that

now

you

need

whether

the

parabola

ts

the

data

well

or

not.

The

an

formula

to

for check

from

regression. India,

The

(around

solving

quadratic

measure

equations. for

this

is

the

calculates.

that

of

all

points

function

which

the

takes

determination

linear

You

the

It

coefcient

you

always

best

t

need

the

can

be

for

between

data

will

Chapters

keep

in

messy

determination ,

set

used

value

in

to

curve

in

its

met

values

of

lie

for

be

2

and

on

any

the

and

mind

0

which

your

1

where

1

the

curve.

The

curve.

same

as

If

used

the

would

to

GDC

indicate

coefcient

test

square

of

for

the

a

PMCC

4.

the

following

diagram

when

nding

data.

t c j

e

e R

real-world

a

Develop

problem

a

Test

model

tpeccA

Pose

the

model

Reect

on

and Extend

apply

the

model

Example 6

A

patient

amount

the

(C)

takes

of

the

specic

drug

medication.

is

a

measured

found

Time

in

drug

( t)

is

in

in

the

the

of

of

a

bloodstream

measured

milligrams

form

in

hours

pill.

of

the

and

the

medication

Data

is

collected

patient

the

found

as

soon

for

as

concentration

per

litre

of

the

he

of

has

the

taken

medication

blood.

t

0

1

2

3

4

5

6

7

8

9

10

C(t)

0

4.87

7.17

10.27

12.81

13.05

15.03

13.3

12.22

11.29

8.26

248

/

6.2

a

Plot

b

State

c

Use

d

Comment

e

Sketch

scatter

a

suitable

your

f

Hence,

g

Using

h

Use

GDC

on

the

original

fully

plot

of

the

type

to

of

given

function

determine

the

choice

model

data.

of

that

the

model

model

function

would

over

by

the

model

function

for

determining

scatter

plot

this

this

the

and

set

of

set

data

of

data.

coefcient

comment

points.

on

of

determination.

the

closeness

of

t

to

the

data.

determine

the

model

absorbed

the

model

the

time

function,

by

to

the

at

which

the

determine

medication

the

time

at

is

at

which

its

the

maximum

effect.

medication

will

have

been

patient.

determine

the

concentration

of

the

medication

after

24

snoitcnuF

a

hours?

C

a

16 G

14

H

F

E

I 12

J

D 10

K 8

C

6 B

4

2

A t 0

–2

b

There

is

strong

quadratic

there

data

evidence

behaviour.

seems

follows

to

a

be

that

There

some

parabolic

the

is

kind

a

of

data

depicts

maximum

symmetry

point,

and

the

shape.

2

c

C(t)

=

d

The

coefcient

−0.3741375t

of

+

4.56328438t

determination

+

0.12111888

is

2

R

=

0.98617438,

quadratic

e

which

shows

very

strong

association.

y

Sketching

the

graph

over

the

data

16

points

G

14

you

get

the

graph

shown.

H F

E

It

is

now

evident

that

the

I

12

regression

J

D

curve

ts

the

data

very

closely.

10

K

8 C

6 B

4

2

A

0 4

6

8

10

12

14

–2

Continued

on

next

page

249

/

6

MODE LLING

f

The

maximum

around

g

It

will

6.1

take

any

more

No,

as

the

amount

hours

about

medication

h

R E L AT I O N S H I P S

to

in

be

the

after

12.2

fully

of

it

WITH

medication

was

hours

taken

for

absorbed

F UNCTIONS:

occurs

by

the

the

not

Using

patient.

whole

and

POWER

of

be

AND

the

function

P O LY N O M I A L

graph

on

of

your

the

F UNCTIONS

model

GDC.

the

present

bloodstream.

value

would

be

negative.

C(24)


0

268

/

6.4

Investigation 4

HINT 1

Sketch the graphs of the following power functions.

Use a view window of 2

a

f(x) = x

d

f(x) = x

3

b

f(x) = x

e

f(x) = x

4

c

f(x) = x

f

f(x) = x

− 3 ≤ x ≤ 3 and 5

6

7

2

Can you group power functions in smaller categories?

3

How are they the same and how are they dierent from a parabola?

4

How are they the same and how are they dierent from a cubic graph?

5

How many maximum and minimum points can a power function have?

6

Are power functions symmetric? If yes, what type of symmetry do they

snoitcnuF

−6 ≤ y ≤ 6.

possess?

7

What happens to the graph of power functions as the value of the power

increases?

Conceptual

8

What are the distinguishing geometrical features of positive

and negative power functions?

The graphs of all even positive power functions have a shape similar to a

parabola and are symmetric about the

y-axis.

The graphs of all odd positive power functions have a shape similar to a cubic

function and are rotationally symmetric about the origin.

As the index of positive power functions increases, the graphs tend to

become atter around the origin

Example 11

A

designer

order

to

some

data

wants

be

able

to

to

points,

create

process

with

the

a

it

model

function

digitally.

centre

of

To

the

do

for

so,

a

bowl

she

bottom

of

put

the

that

the

she

sketched

sketch

bowl

over

being

the

A

L

G

K H

Point

origin.

M

F

the

marked

N

E

are

and

O

D

points

grid

in

P

C

data

hand

Q

B

The

a

by

I

J

following:

A

B

C

D

E

F

G

x

−16.5

−15.1

−14

−12.8

−11.3

−9.3

−7.5

−5.3

y

7

4.9

3.62

2.53

1.54

0.7

0.3

0.07

Point

H

I

0

0

J

K

L

M

N

O

P

Q

x

5.3

7.5

9.3

11.3

12.8

14

15.1

16.5

y

0.07

0.3

0.7

1.54

2.53

3.62

4.9

7

Continued

on

next

page

269

/

6

MODE LLING

a

Plot

She

the

rst

given

thought

b

Explain

c

Use

d

Find

e

Sketch

why

your

the

Not

being

quartic

f

Find

the

why

function

the

other

could

be

the

to

model

suitable

quadratic

over

POWER

AND

P O LY N O M I A L

F UNCTIONS

technology.

function

determination,

function

the

model

designer

a

model

and

to

plot

the

shape.

model

function

comment

scatter

function

to

might

quartic

of

have

function

determine

coefcient

and

on

this

for

your

shape.

this

set

of

data.

answer.

comment

on

the

closeness

of

t

to

the

she

created,

she

decided

to

determine

a

new

the

not

could

quartic

determination,

been

be

a

model

and

satised

suitable

the

model

alternative

function

comment

with

on

for

its

this

function

she

model.

set

of

data.

signicance

in

relation

to

the

model.

model

data

and

quartic

dimensions

j

of

or

quadratic

determine

with

the

GDC

the

original

the

a

GDC

F UNCTIONS:

function.

and

your

Sketch

Using

to

model

why

previous

i

using

quadratic

satised

created

h

a

your

WITH

data.

Explain

Use

of

on

coefcient

model

g

data

GDC

the

original

R E L AT I O N S H I P S

are

Determine

function

compare

model,

double

the

the

in

it

to

the

the

designer

size

equation

over

of

to

the

the

scatter

plot

previous

want

to

and

comment

on

the

closeness

of

t

to

the

model.

create

a

model

for

another

bowl

whose

original.

model

function

for

the

larger

bowl.

a

A

Q

P

B C

O N

D E

M F

L

G I

H

b

A

quadratic

model

to

this

have

their

formula

shape

vertical

constant

K J

would

since

the

symmetry

be

suitable

data

and

points

also

to

seem

because

of

concavity.

2

c

The

d

R

best

t

parabola

is:

f(x)

=

0.0263x

−

1.15

Using

quadratic

regression

in

the

GDC.

2

=

The

0.933

coefcient

model

of

function

determination

describes

the

shows

data

set

that

very

the

closely .

e

A

Q

P

B C

O N

D E

M F

L

G H

Although

close

point

to)

by

the

most

a

J

I

model

K

function

points,

signicant

it

still

goes

misses

through

the

(or

bottom

amount.

270

/

6.4

f

The

in

quadratic

describing

important

A

quartic

addition

The

had

shape

at

the

function

to

best

the

being

concavity,

g

model

it

t

also

of

is

the

be

a

of

close

function

=

0.0000946x

t,

to

most

bowl.

and

since

in

constant

the

origin.

is:

Using

4

f(x)

problems

the

the

better

symmetry

atter

quartic

major

bowl,

bottom

can

depicting

some

quartic

regression

in

the

GDC.

2

−

0.0000649x

+

0.000214

2

R

=

The

0.999

coefcient

perfect

than

t

the

and

of

is

determination

of

course

shows

signicantly

snoitcnuF

h

almost

better

quadratic.

i

A

Q

P

B C

O N

D E

M F

L

G H

The

the

model

data

bowl

in

g( x )



and

much



j

function

points

a

x

perfectly

describes

better

way

the

than

through

shape

the

of

all

the

quadratic.

4

4

0.0000942 x

 

goes



2 f



K

J

I



0.0000118 x

The

enlarged

bowl

is

double

the

size

 2

 both

in

the

x

and

in

the

y

directions.

Direct Variation

Two

quantities

that

n

y



be

ax

,

a, n  

directly

Because

is

are

related

by

a

power

law

of

the

form

+

said

proportional

there

required

are

in

is

only

order

one

to

to

to

be

vary

each

the

with

each

other,

or

to

other.

unknown

nd

directly

parameter

relationship,

just

one

assuming

pair

the

of

values

value

of

n

is

known.

Example 12

The

rate

of

the

spread

of

fungus

on

a

petri

disk

( x)

varies

directly

with

2

perimeter

3.2

of

the

area

covered

by

the

fungus

( p).

If

the

rate

is

2cm

s

the

square

of

the

1

when

the

perimeter

is

cm.

a

Find

the

equation

b

Find

the

perimeter

relating

the

rate

of

spread

2

when

the

rate

is

6 cm

to

the

perimeter.

1

s

Continued

on

next

page

271

/

6

MODE LLING

R E L AT I O N S H I P S

WITH

F UNCTIONS:

POWER

AND

P O LY N O M I A L

F UNCTIONS

2

a

x



The

ap

for

the

power

a  3.2

Substituting

2



equation

function.

2



2

general



a



0.195



x

the

values.

0.195p

2

b

6



0.195 p

2

p



30.72



p



5.54 cm

Power Regression HINT

The

power

regression

function

on

a

GDC

will

nd

the

best

t

curve

of

On most calculators

n

the

y

form



ax

,

a, n   .

power regression can

only be used if all

the data points are

greater than 0.

Exercise 6I

1

It

is

of

known

an

that

object

sometimes

the

falling

best

resistance

through

modelled

with

the

object’s

with

the

square

velocity

of

the

a

as

to

liquid

is

varying

and

e

motion

of

directly

sometimes

wishes

releases

By

a

to

test

weight

measuring

how

a

as

Emily’s

theories

tube

quickly

of

and

thick

it

falls,

The

that

when

for

on

sum

the

power

which

of

squares

function,

model

best

ts

data.

it

is

distance,

d

metres,

that

a

ball

rolls

so down

a

slope

of

time,

varies

directly

with

the

square

liquid. the

t

seconds,

it

has

been

rolling.

she In

calculates

residuals

the

velocity.

these

in

considering

comment

2

Emily

By

travelling

2

seconds

the

ball

rolls

9

metres.

at

a

Find

an

b

Find

how

c

Find

the

equation

connecting

d

and

t

1

2 cm s

a

the

Write

resistance

down

resistance

R

an

is

4.2 N.

equation

with

velocity

v

for

each

two

possible

test

the

models

further

she

collects

rolls

in

5

seconds.

time

it

takes

for

the

ball

to

roll

metres.

The

mass

of

a

uniform

sphere

varies

directly

the with

following

ball

models.

3 To

the

of 26.01

the

far

linking

the

cube

of

its

radius.

A

certain

sphere

data: has

radius

Find

v

3.2

4.0

R

7.1

11.1

the

3 cm

mass

and

of

mass

another

113.1 g.

sphere

with

radius

5 cm.

4 b

Find

the

value

of

R

predicted

by

each

Sketch

within the

models

for

v

=

3.2

and

v

=

the

By

considering

the

sum

of

the

of

the

following

functions,

the

specied

domains

and

hence

4.0. determine

c

graph

of

their

range.

square 5

a residuals

say

which

model

is

most

f(x)

=

−x

,

for

−3

≤

x

≤

3

,

for

−4

≤

x

≤

4

≤

x

likely 4

x

to

best

t

the

situation.

b

g(x)

= 4

d

Use

the

power

regression

function

on 12

c your

nd

GDC

and

another

all

three

possible

data

points

h(x)

=

2x

,

for

−2

≤

2

to

model.

272

/

6.4

5

Viola

is

conducting

demonstrate

radius

and

gathered

the

the

the

an

experiment

relation

volume

between

of

following

a

6

to

the

sphere.

According

physical

She

weight

data.

building

specications

measurements,

that

buildings

fourth

to

foundation

can

carry

power

of

the

their

maximum

columns

varies

and

of

directly

tall

with

the

perimeter.

r 1

2

3

4

5

6

7

8

In

(cm)

a

a

specic

column

sample

with

check,

a

perimeter

of

55 000 kg.

it

was

of

found

6 m

that

could

V 33.5

113

268

524

905

1440

2140

hold

)

a a

Plot

the

given

data

on

your

a

weight

Find

the

formula

perimeter

b

Use

your

function

GDC

to

model

determine

for

this

set

the

of

power

can

Comment

by

on

the

determining

choice

the

of

relates

the

of

the

column

to

the

weight

model

coefcient

Estimate

column

of

it

carry.

data.

b

c

that

GDC.

snoitcnuF

4.19 3

(cm

the

of

perimeter

weight

the

of

same

8 m

that

another

make

can

with

a

hold.

determination.

c

d

Sketch

the

scatter

plot

closeness

model

and

of

t

function

comment

to

the

over

on

the

should

the

original

Use

the

model

function

to

of

a

sphere

with

have

perimeter

in

order

to

that

be

a

column

able

to

hold

data.

determine

Find

radius

the

inverse

of

the

following

one-to-

the one

volume

the

200 000 kg?

7

e

Calculate

functions:

10 cm. 2

f

Comment

part

e

on

how

compares

the

with

value

the

found

actual

a

f(x)

=

x

b

f(x)

=

−x

c

f(x)

=

2x

in

value

for

x

≤

0

3

of 5

the

volume

found

using

of

a

the

sphere

of

radius

appropriate

10 cm

formula.

Inverse variation

Investigation 5

A mathematician drives along the same route to school every day.

1

How is the time that he takes to get to school aected by his driving speed?

2

One day when he was very early there was no trac so he traveled at twice

h is normal average speed.

How did the time taken to complete the journey change from normal?

3

On the way home he was caught in heavy trac so he only managed half his normal average speed. How

did the time taken to complete the journey change from normal?

How could you describe the variation between speed and time?

4

Factual

5

Conceptual

How would you describe real situations that model asymptotic behavior?

273

/

6

MODE LLING

R E L AT I O N S H I P S

WITH

F UNCTIONS:

POWER

AND

P O LY N O M I A L

F UNCTIONS

n

When

n

is

negative

and

x

>

0,

the

function

f(x)

=

ax

represents

inverse

variation.

Investigation 6

1

Sketch the graphs of the following inverse variation

1

a

f ( x)

1

b

=

f ( x)

2

f ( x)

=

3

4

x

1 f ( x)

1

c

=

x

d

functions:

x

1

e

=

f ( x)

1

=

5

f

6

x

f ( x)

=

x

7

x

2

How can you describe the shape of a inverse variation function?

3

Can you group inverse variation functions in smaller categories?

4

Are inverse variation functions symmetric? If yes, what type of symmetry

HINT

Use a view window of

−3 ≤

x

≤

3 and

−6 ≤ y ≤ 6 do they possess?

5

Describe what happens to the y-value of the function as

Factual

TOK a

x becomes very large

b

as x approaches 0 from the positive side

c

as x approaches 0 from the negative side.

Aliens might not be

able to speak an Ear th

language but would

6

What happens to the graph of inverse variation functions as the value of

they still describe the

the power becomes more negative?

equation of a straight

line in similar terms?

Conceptual

7

What are the distinguishing geometrical features of inverse Is mathematics a

variation functions? formal language?

International-

An asymptote is a line that a graph approaches but never crosses or touches.

mindedness

Asymptote comes n

If

a

quantity

y

varies

inversely

with

x

n

for

x

>

0

then

y

=

kx

or from the Greek word

k y

“ασύμπτωτη”,

=

In

these

situations

y

will

decrease

as

x

increases

and

vice

versa.

n

meaning “to not fall on

x

As

with

direct

parameter

variation

a

single

N

taken

point

is

needed

to

nd

the

value

of

top of each other ”.

the

k

Example 13

The

of

a

number

people

When

it

b

x

hours

who

three

takes

Given

of

to

it

are

available

people

build

takes

the

three

are

to

to

build

work

available

wall

when

hours

to

a

on

the

four

build

wall

varies

inversely

with

the

number

it.

wall

takes

people

the

are

wall,

two

hours

available

state

how

to

to

build.

work

many

Find

on

people

the

time

it.

worked

on

it.

274

/

6.4

For

k

a

N

inverse

variation,

the

variable

is

 1

x

1

written

as

or

x

which

is

a

power

x

k

N

=

2

when

x

=

3

function.

2 

so

3

k



6 Find

the

value

of

k

using

the

given

6

N



information. x

Substitute

6



the

4

So,

it

four

takes

people

1.5

hours

work

on

to

build

the

wall

when

equation

Now

value

of

k

you

found

into

nd

N

for

N.

when

x

=

4.

it.

Substitute

6

b

the

 1.5

snoitcnuF

N

N

=

3

into

the

equation

3  6

x

N 3x



 6

which

you

found

in

part

a

x

6

x





2

3

So,

two

Sometimes

the

context

people

the

were

terms

will

available

direct

make

it

or

clear

to

build

inverse

how

to

the

wall.

variation

form

the

might

not

required

be

used

but

equations.

Investigation 7

1

A local authority pays its workers depending on the number of hours that

they work each week . If the workers are paid €22 per hour, complete the

following table:

Number of hours

20

25

30

35

40

Pay (€)

Plot a graph of this information on your GDC.

Describe how a worker ’s pay varies with the number of hours worked.

2

The local authority has decided to put ar ticial grass tiles on a football eld.

If four people are available to lay the grass tiles, it takes them two hours to

complete the work .

Fill in this table showing the number of people available and the number of

hours it takes to complete the work .

Number of people

Number of hours

1

2

4

6

8

12

2

Plot a graph of this information on your GDC.

Factual

How do the number of hours to complete the work vary with the

number of men available?

3

Conceptual

For problems which involve direct and inverse variation,

how does understanding the physical problem help you to choose the

correct mathematical function to model the problem with?

275

/

6

MODE LLING

R E L AT I O N S H I P S

WITH

F UNCTIONS:

POWER

AND

P O LY N O M I A L

F UNCTIONS

Any value of x that makes the denominator of a function equal to zero is

excluded from the domain of the function. The ver tical line represented by

this x-value is called a ver tical asymptote on the graph of the function.

Investigation 8

1

Consider the function

f ( x)

=

x

1

Which value of x makes the denominator equal to zero?

2

Which ver tical line is the ver tical asymptote of the graph of

f(x)?

Now consider the function g(x) = f(x − 2)

3

Which value of x makes the denominator equal to zero?

4

Which ver tical line is the ver tical asymptote of the graph of

5

Give a full geometric description of the transformation represented

g(x)?

by g(x)

6

What do you notice? How does this transformation aect the ver tical

asymptote?

Now consider the function h(x) = f(x) + 3

7

To which value does the function tend as

x becomes innitely positive or

negative?

8

Which horizontal line is the horizontal asymptote of the graph of

h(x)?

9

Give a full geometric description of the transformation represented

by h(x)

10

What do you notice? How does this transformation aect the horizontal

asymptote?

Finally consider the function j(x) = f(x − h) + k

11

Give a full geometric description of the transformation represented

by

12

j(x)

Hence write down the equations of the asymptotes of

13

Conceptual

j(x)

How do you identify an asymptote?

1

The function

f ( x)

+ k has a ver tical asymptote

=

x

x = h and a

- h

horizontal asymptote y = k

276

/

6.4

Example 14

1 Consider

the

function

f ( x)

=

when

x

≠

0.

The

graph

of

y

=

f(x)

is

transformed

to

become

the

x

down

the

a

g(x)

b

h(x)

c

j(x)

a

functions.

=

=

equations

f(x)

=

f(x

f(x

g(x)

Give

−

−

+

of

a

the

description

asymptotes

of

of

the

the

transformations

transformed

they

1)

−

2

a

translation

asymptote

x

=

asymptote

represents

3

units

down.

The

vertical

a

y

=

x

=

will

be

c

g(x)

left

asymptote

represents

and

2

a

units

3

units

to

the

right.

The

vertical

y

=

1

units

to

the

units

The

down.

x

=

to

asymptote

3

units

asymptote

the

right.

will

vertical

translated

down.

will

The

be

translated

horizontal

remain

unaffected.

1

asymptote

units

to

will

the

asymptote

be

left.

will

The

be

translated

−1.

2 Horizontal

asymptote

0.

translation

asymptote

remain

3.

horizontal Vertical

horizontal

translated

asymptote Horizontal

The

will

−3.

translation

asymptote

asymptote

0.

3 Vertical

write

3)

Horizontal

g(x)

hence

function.

unaffected.

b

and

3

represents

Vertical

represent

snoitcnuF

following

y

=

units

down.

−2.

Exercise 6J

1

The

the

volume

V

of

a

p

of

the

pressure

gas

varies

gas.

inversely

The

pressure

with

4

is

The

density

varies

of

a

inversely

sphere

as

the

of

xed

cube

of

mass

its

radius.

3

20

Pa

when

its

volume

is

180 m

.

Find

the A

volume

when

the

pressure

is

90

sphere

A

group

The

of

children

number

child

of

receives

attend

pieces

varies

of

a

birthday

candy

inversely

c

that

with

party.

a

each

of

children

there

are

16

10

pieces

Find

number

of

receives

when

the

density

If

we

require

n

each

candy.

Find

100

kg/m

of

candy

Timothy

each

Sketch

within

the

graph

the

of

specied

determine

their

a

sphere

with

radius

a

sphere

to

have

a

density

are

20

the

,

nd

its

radius.

had

to

choose

a

topic

for

his

Exploration.

Being

a

child engineering

student,

he

decided

children. to

3

of

the

prospective there

of

one

Mathematical pieces

density

3

children,

of

a

the

5 receives

has

kg/m

of

When

10 cm

5 cm.

b

number

radius

3

500

2

with

Pa.

following

domains

and

functions,

hence

investigate

law

states

light

that

source

square

range.

of

validate

the

is

the

the

inverse-square

light

intensity

inversely

distance

physical

law.

from

a

proportional

from

law,

the

he

The

specic

to

source.

performed

the

To

an

1

a

f(x)

=

,

for

−3

≤

x

≤

3

x

≠

0. experiment

5

where

he

positioned

a

light

x source

at

various

distances

to

a

receiver

4

b

g(x)

=

,

for

−4

≤

x

≤

4

x

≠

measuring

0

its

intensity.

The

data

he

4

x

collected

appear

in

the

following

table. 1

c

h(x)

=

,

for

−2

≤

x

≤

2

x

≠

0

12

2x

277

/

6

MODE LLING

R E L AT I O N S H I P S

WITH

F UNCTIONS:

d 10

15

20

25

30

35

40

45

3232

1434

805

513

353

259

201

159

d (cm)

POWER

Find

and I (lu x)

e

a

Plot

b

Explain

this

data

on

your

the

plot

of

the

the

closeness why

function

set

of

an

could

inverse

be

6

A

Use

suitable

to

model

f

this

Use

GDC

model

for

company

selling

used

relating

The

of

their

the

data

are

Price (€)

a

Plot

b

Explain

c

Explain

d

Use

e

Find

f

Sketch

set

cars

in

to

the

the

of

the

original

on

the

data.

as

the

model

if

the

from

to

estimate

light

the

source

the

light

moves

to

receiver.

power

to

create

reference

number

mathematical

one

of

of

years

its

models

biggest

since

the

selling

car

for

the

models,

was

value

it

of

cars

collected

as

a

data

produced.

table:

2

3

4

5

6

7

8

9

10

21641

17890

14709

12613

12239

10223

10501

9744

8257

6772

data

why

on

a

why

the

your

linear

an

GDC

GDC.

function

inverse

to

model

of

would

variation

determine

coefcient

the

original

the

not

function

power

determination,

function

over

be

the

suitable

could

function

and

points

to

be

suitable

model

comment

and

model

for

on

this

to

comment

on

of

model

this

your

set

set

of

data.

this

set

of

data.

data.

answer.

the

closeness

of

t

to

the

data.

According

the

to

this

domain

(where

it

model,

of

is

the

nd

the

age

of

in

order

car

9

following

needed)

the

when

a

them

its

value

Describe

falls

the

below

€4000?

transformations

required

to

1 to

make

to

comment

the

data.

wants

Taking

price

shown

your

functions

this

and

over

1

this

Restrict

age.

selling

Age (years)

7

determine

function

function

g

to

answer.

function

points

t

determination,

your

model

of

your

intensity

data.

your

on

of

F UNCTIONS

variation

50 cm

c

P O LY N O M I A L

coefcient

comment

Sketch

GDC.

AND

convert

the

function

f ( x)

=

to

the

invertible. x

1

a

f ( x)

=

,

for

x

≠

2

0

4

function

g( x )

=

3 -

x x

2

b

f ( x)

b

=

,

for

x

≠

Determine

the

- 1

asymptotes

of

both

f(x)

0

3

and

x

1

c

f ( x)

=

c

-

,

for

x

≠

g(x).

State

the

transformations

described

in

0

2

part

x

the

8

Find

the

inverse

of

the

following

a

relate

to

the

asymptotes

of

relative

the

two

positions

of

functions

one-to2

one

functions.

10

The

graph

of

g

(

x

)

= 1 -

can

x

be

- 2

1

a

f(x)

=

,

for

x




0

to

go

back

to

f(x).

2

x

278

/

6.4

The

following

diagram

shows

a

a

demand

i

Explain

why

a

power

function

of

n

curve

for

shows

a

the

product

the

a

product.

price

for

and

percentage

would

buy

(p)

the

The

a

the

(N)

of

horizontal

company

vertical

the

product

could

axis

market

at

this

axis

the

sell

form

model

shows

a

who

N

for

of

kp

the

necessary

values

=

might

demand

restriction

be

a

good

curve,

on

the

stating

possible

n

price.

ii

State

one

reason

why

a

function

of

y

this

100

form

demand

Two

80

and

of

the

is

not

a

good

model

for

the

curve.

points

have

coordinates

(1, 57)

(4,13).

60

b

Use

these

points

to

nd

possible

values

40

for

k

and

n

20

x

snoitcnuF

11

0 1

2

3

4

5

Chapter summary

2

•

The standard form of a quadratic function is f(x) =

ax

+ bx + c

•

The fundamental (simplest) quadratic function is f(x)

•

The shape of the graph of all quadratic functions is called a parabola

2

=

x

and is symmetric about a ver tical line called the axis of symmetry,

going through its ver tex

•

Parabolas have constant concavity, either being constantly concave

upwards or constantly concave downwards.

2

•

The quadratic equation ax

-b

±

+ bx +

c = 0 has the solution

D 2

x 1

, where Δ = b

=

− 4ac

2

2a

•

In order for any function to have an inverse, its domain has to be

restricted in such a way as to make sure that it is a one-to-one

function.

•

Transformations of functions can be expressed in three dierent

ways. One is their mathematical notation, the second is the

geometric description and the third is the graphical

representation.

0 ❍

y = f(x) + k:

ver tical translation k units up, with vector

k

0 ❍

y = f(x) − k:

ver tical translation k units down, with vector

k

Continued

on

next

page

279

/

6

MODE LLING

❍

y = k

×

f(x):

f ( x) ❍

y

R E L AT I O N S H I P S

WITH

F UNCTIONS:



f (x ):

y = −f(x):

❍

y = f(x

+

F UNCTIONS

ver tical stretch with scale factor

k

❍

P O LY N O M I A L

1

 k

AND

ver tical stretch with scale factor k

1



POWER

k

ver tical reection in the x-axis

k):

horizontal translation k units left, with

 k 

vector



 0



❍

y =

f(x

−



k):

horizontal translation k units right, with

k

vector 0

1 ❍

y = f(kx):

horizontal stretch with scale factor k

x ❍

❍

y



f



k

1



y = f(−x):



f



 k

x

: horizontal stretch with scale factor k



horizontal reection in the y-axis

3

= ax

2

•

The standard form of a cubic function is f(x)

+ bx

+ cx + d

•

The fundamental (simplest) cubic function is f(x)

•

The graphs of all cubic functions are symmetric about a point called

3

=

x

the point of inexion

n

•

The standard form of a power function is f(x)

=

ax

•

The fundamental (simplest) power function is f(x) = x

•

The graph of all even direct variation functions have a shape similar

n

to a parabola and are symmetric about the y-axis. The graph of all odd

direct variation functions have a shape similar to a cubic function and

are symmetric about the origin.

•

The graph of all inverse variation functions have the coordinate axes

as asymptotes.

•

Any value of x that makes the denominator of a function equal to

zero is excluded from the domain of the function. The ver tical line

represented by this x-value is called a ver tical asymptote on the

graph of f(x)

•

If a function tends towards a constant value as the independent

variable (x) tends towards positive and/or negative innity, this

constant value represents a horizontal asymptote on the graph of the

function.

280

/

6

Developing inquiry skills

How many dierent trajectories would

there be if the ball were to follow a straight

line from the hands of the player to the

hoop?

How many dierent trajectories are there in the

case that the ball follows a curved line from the

snoitcnuF

hands of the player to the hoop, as in the drawing

here?

How would the initial angle at which the ball

leaves the hands of the player aect the

angle at which it arrives at the hoop?

How would this angle increase the chance

of the player making the basket?

Knowing the maximum height of the

ball, could we determine a unique

model for the trajectory it follows?

How does the height up to which the

ball goes increase the chance of the

player making the basket?

How can we determine the optimum

path for a basketball player to score a

basket?

What else would we need to know to

answer this question?

What assumptions have you made in

the process?

Click here for a mixed

Chapter review

1

Sketch

the

graph

review exercise

d

of:

your

2

a

f(x)

=

x

b

f(x)

=

x

Find

−

2

−

5x

the

x-intercepts

results

in

and

and

interpret

context.

2

+

4.

2

3

For

the

graph

of

the

function

f(x)

=

x

+

6x

−

7

nd:

2

The

a

perimeter

If

the

the

of

length

height

a

of

in

picture

the

terms

is

400 cm

picture

of

is

x

cm,

Find

the

c

an

expression

picture

Sketch

this

domain

in

graph

and

for

terms

the

coordinates

of

the

y-intercept

b

the

coordinates

of

the

x-intercepts

c

the

equation

the

d

the

coordinates

x.

2

b

a nd

of

the

area ,

A cm

x

using

a

,

of

axis

of

symmetry

of

of

the

vertex

suitable

range

281

/

6

MODE LLING

4

Anmol

throws

R E L AT I O N S H I P S

a

sto ne

in

the

ai r.

WITH

F UNCTIONS:

T he

9

The

POWER

line

2y

=

AND

x

+

−

5

P O LY N O M I A L

3

intersects

the

F UNCTIONS

curve

2

height

t

of

the

seconds

is

stone ,

h(t)

modell e d

metres,

by

the

at

time

f(x)

function

=

Find

x

+

the

2x

at

the

coordinates

points

of

A

A

and

and

B.

B

2

h(t)

=

− 2.2625 x

+

8.575 x

+

1.9

10

a

Find

the

y-intercept

and

explain

Sketch

what f(x)

this

graph

of

the

function

=

x

−

2x

−

3

for

the

domain

0

≤

x

≤

4

represents. Find

b

the

2

Find

the

maximum

height

of

the

the

range

of

y

=

f(x)

in

the

given

stone. domain.

c

Find

the

time

when

the

stone

lands

on

11 the

The

acceleration

proportional

since

5

For

i

the

the

following

functions,

coordinates

the

of

the

equation

the

b

rocket

is

inversely

of

the

coordinates

of

f(x)

f(x)

=

3(x

=

−

4(x

2)(x

axis

the

+

−

1)(x

take

to

off

the

square

root

of

the

time

t

−2

of

acceleration

symmetry

is

6.2 m s

when

−2

=

5.76 m s

vertex.

a

a

a

x-intercepts

t

iii

of

nd

Its

ii

a

ground.

Find

the

acceleration

of

the

rocket

as

a

4)

−

5)

b

function

of

Find

value

the

time.

of

t

at

which

the

−2

acceleration

6

Sketch

the

following

cubic

f(x)

=

2x

12

b

f(x)

=

(x

−

7

The

number

2004

until

1)(x

of

+

1)(x

lilies,

2020

can

N,

be

−

N(x)

0.5 m s

A

quadratic

curve

has

x-intercepts

at

(0,

0)

in

=

(6,

0),

and

the

vertex

is

at

( h,

8).

3)

a

pond

modelled

3

function

to

1

and −

equal

functions.

3

a

is

from

using

a

Find

the

value

b

Find

the

equation

of

h.

of

the

curve

in

the

2

the

form

y

=

ax

+

bx

+

c

2

−0.04x

+

0.9x

−

7x

+

70, n

13 where

x

is

the

number

of

years

after

The

curve

Sketch

the

graph

N(x)

−0.04x

for

b

0

≤

Find

x

≤

the

2

+

0.9x

−

7x

+

passes

through

the

points

and

(4,2).

Find

the

values

of

k

and

n:

of

lilies

a

without

b

using

using

a

GDC

70

20

number

kx

of

3

=

=

2004. (2,32)

a

y

after

5

the

power

regression

on

a

GDC.

years.

To do this without a GDC is beyond the

c

Find

the

number

d

Find

the

maximum

and

the

year

in

of

lilies

after

number

which

this

12

of

years.

scope of the syllabus. This is a good

extension exercise. If you are having

lilies

trouble, the worked solution found in the

occurs.

digital resources may help.

e

Find

the

the

year

minimum

in

which

number

this

of

lilies

and

occurs.

14 f

Find

when

there

are

60

lilies

in

The

dimensions

of

a

rectangular

the swimming

pool

are

shown

in

the

diagram.

pond.

8

The

number

directly

with

of

miles ,

the

m,

time,

t,

travelled

for

varies

which

the 7

train

a

If

has

the

been

train

hours,

nd

travels

a

100 miles

relationship

in

connecting

m

Find

the

after

2

number

of

miles

2

set

of

value(s)

that

the

x

should

take,

such

that

the

hours.

area Find

the

travelled

variable

c

+

s.

Determine

b

x

1.25

x

and

–

travelling.

how

long

it

takes

to

travel

of

the

swimming

pool

is

greater

300 than

16.

miles.

282

/

6

A

farmer

for

his

three

wants

sheep.

sides,

to

He

make

needs

with

the

a

rectangular

to

put

other

fencing

side

being

d

pen

i

Find

on

maximum

wall

as

shown

in

the

the

ball

reaches

height.

Find

the

maximum

height

Draw

the

graph

3

f(x)

A

18

=

The

to

of

the

−

x

−

x

+

determine

monthly

the

1,

for

−1

the

function

C

length

of

fencing

that

he

can

≤

a that

t

≤

the

length

of

side

AB

is

for

course

t

P(t)

is

can

the

Hence

total

an

expression

length

of

the

determine

area

of

the

aid

1

of

one

be

in

side

in

terms

of

year

(from

by

in

mm

January

the

2

0.11t

+

number

1.87t

of

the

−

6.91t

month,

+

45,

and

to

January.

the

the

an

b

BC

expression

of

for

or

the

length

of

the

Find

its

of

x)

which

maximum

part

AB

(ie

is

The

thrown

the

area

height,

ground

h

Find

e

after

metres,

t

upwards

of

seconds

the

is

and

during

lowest

which

the

precipitation

occur.

mean

of

the

two

precipitations

b

the

into

ball

given

month

during

which

the

mean

part

c

occurs.

take

value.

vertically

mm

the

Use

on ball

in

otherwise,

side

makes

months

the

from value

Oxford

July.

Find

c

graph,

in

the

pen.

a

precipitation

x

d determine

the

Oxford

modelled

corresponds

Find

in With

air.

=

the

greatest

A

and

x, in

determine

16

≤

m.

Given

c

x

use 0

b

≤

range.

precipitation

December)

where

a

1,

function

3

B

60

ball.

2

2x

during

is

the

D

hence

total

of

diagram.

17

The

its

a

ii

brick

when

snoitcnuF

15

your

what

results

the

from

position

a

of

–

d

the

to

comment

month

with

the

the

mean

rainfall

the

majority

tells

you

about

when

above

of

the

rain

falls

in

Oxford.

by

2

h

=

−t

+

6t

+

12.

19

a

Write

the

at

down

ground

the

the

of

instant

initial

the

ball

when

it

height

(that

is

is,

above

its

Sketch

within

height

Show

that

the

height

of

the

determine

graph

the

ball

of

the

specied

their

released).

following

domains

and

functions

hence

range.

4

a

b

the

f(x)

=

−x

,

for

−3

≤

x

≤

3

after 5

x one

second

is

17

metres.

b

g( x )

=

,

for

−4

≤

x

≤

4

4

c

At

a

later

time

the

ball

is

again

at

a 9

c height

i

of

Write

ii

20

The

the

h(x)

=

3x

,

for

−2

≤

x

≤

2

metres.

down

satisfy

17

17

an

when

equation

the

ball

is

that

at

a

t

must

height

of

metres.

Solve

the

average

equation

price

following

of

a

algebraically

litre

of

gasoline

in

France

in

each

year

from

2006

to

2017

is

given

in

table.

Year

Price (€)

2006

2007

2008

2009

2010

2011

2012

2013

2014

2015

2016

2017

29.66

29.39

28.84

28.57

28.51

28.62

28.69

29.52

29.79

30.16

30.31

30.96

a

Find

b

Comment

c

Estimate

a

cubic

model

on

the

the

for

this

data.

suitability

price

in

of

using

this

model

to

predict

future

gasoline

prices.

2021.

283

/

6

MODE LLING

21

Sketch

the

R E L AT I O N S H I P S

graph

of

the

WITH

F UNCTIONS:

following

24

P2:

POWER

AND

Consider

the

P O LY N O M I A L

quadratic

F UNCTIONS

function

2

functions,

and

within

hence

the

specied

determine

their

domains

f(x)

a

range.

f ( x)

=

(x

−

Write

3)

+

-

,

for

−3

≤

x

≤

3,

when

x

≠

7.

down

minimum

1

a

=

the

coordinates

point

of

this

of

the

quadratic.

0. (2

3

marks)

x

b

1

b

g( x )

=

,

for

−4

≤

x

≤

4,

when

x

≠

Find

the

largest

possible

domain

and

0. range

4

of

f(x)

which

include

the

point

x

(4, 8)

1

c

h ( x)

such

that

the

inverse

function

−1

=

,

for

−2

≤

x

≤

2,

when

x

≠

f

0.

(x)

exists.

(3

marks)

5

2x −1

c

Exam-style questions

Find

the

state

its

inverse

domain

function

and

f

(x),

range.

(5

22

P1:

The

height

(s

m)

of

an

object

freely

under

gravity

marks)

which 25

moves

and

is

given

P1:

The

cubic

by

function

3

2

- 105 x

x

(

+ 3000 x

)

2

s

=

c

+

ut

−

5t

,

where

c

m

is

the

initial

h

(

x

=

)

,

of

velocity

is

time

the

in

object,

the

the

u

m s

upward

object

has

is

the

direction,

been

releases

a

model

and

moving

t s

for.

models

the

ride,

metres,

in

rocket

from

1.7

m

above

the

ground.

initial

a

the

velocity

is

50

x

time

between

height

the

off

and

from

the

is

also

measured

the

returning

to

platform.

Hence

for

write

the

rocket

to

how

reach

the

the

coaster

rst

80

m

long

its

marks)

b

it

(2

horizontal

start

of

the

ride,

in

the

the

maximum

c

marks)

height

rocket.

Find

Find

Find

the

the

coordinates

the

of

the

local

point.

(3

coordinates

the

of

the

marks)

local

point.

set

roller

(3

marks)

of

values

coaster

is

of

x

for

above

which

50

m.

(2

friend

Paul

does

marks)

26

P1:

When

exactly

a

thing,

but

the

height

of

coin

is

well,

dropped

the

time

down

(t

into

seconds)

an

it

the takes

same

marks)

of

ancient Peter’s

and

metres.

(3 Calculate

its

takes

maximum

height.

c

of

strike

(2

down

roller

rocket

maximum

b

a

over

represents

minimum setting

of

m s

a Find

80

The

–1

rocket’s

≤

a

distance platform

x

initial

circuit. Peter

≤

1000

–1

height

0

the

coin

to

fall

a

distance

of

d

his 1

2

platform

is

1.8

metres

m.

a d

By

considering

the

difference

in

Find

displacement-time

the

two

the

answers

Paul’s

23

P2: a

model

rockets,

a,

parts

write

b

and

rocket.

Determine

for

to

graphs

which

the

the

set

of

values

c

quadratic

A

a

for

10

b

k

coin

long

it

by

takes

t

a

=

5d

coin

m.

the

Find

the

equation

is

splash

hits

marks)

of

how

modelled

to

(2

marks)

for

down

(4

be

the

fall two

can

dropped

is

heard

water

the

at

depth

into

8

well,

seconds

the

of

the

and

later

as

it

bottom.

the

well

waterline.

down

(2

to

marks)

2

x

+

kx

+

9

=

0,

x ∈



has If

i

one

repeated

a

frog

depth

b

ii

no

iii

two

For

real

k∈

roots

distinct

,

fall

roots.

determine

(8

how

many

roots

equation

has.

−x

+

kx

+

9

=

0,

(3

x∈



marks)

equal

that

catch

c

Find

well,

how

a

the

coin

at

a

if

the

he

time

it

takes

changes

into

to

a

prince.

far,

frog

waiting

to

depth,

handsome

marks)

2

the

can

root

should

he

handsome

below

wants

prince.

the

be

to

top

of

hanging

become

(2

the

on,

a

marks)

284

/

6

P1:

James

has

forgotten

abcd,

where

0–9.

To

written

which

help

will

he

(for

x

=

a

giving

the

example,

Find

then

the

two

passcode

d

4-digit

are

digits

remember

him

it

to

away

work

easily

password

x

if

x

=

a

+

ab

=

y

=

24

=

and

P2:

he

has

be.

an

archaeological

squares

are

mosaic.

It

part

it

system

out

others:

5

(2,

a

then

=

a

Find

with

a

dig

on

believed

curve.

the

A

and

ruin

passes

they

the

a

were

of

three

(1,

10),

of

a

quadratic

through

these

points.

fourth

of

red

115).

equation

that

small

coordinate

coordinates

(4,

the

three

that

introduced

27) and

Later

marks)

is

have

three

James’

found

red

is

curve

143.

(6

of

points

cd

xy

that

In

puzzle

two

=

4andb

possibilities

could

y

29

between

to

into

and

passcode

it,

mathematical

integers

45),

c,

allow

splits

2-digit

b,

him

down

without

If

a,

his

(4

red

coordinates

square

of

(5,

is

marks)

discovered

198).

2

28

P2:

Consider

the

function

f(x)

=

x

.

b

The

graph

vertically

g(x)

=

of

to

f(x)

the

function

dene

+

a

new

is

translated

the

Find

point

value

of

g(x)

of

k

such

passes

that

through

(3, 4).

graph

of

horizontally

h(x)

=

f(x

−

the

to

function

dene

a

is

four

the

Find

function

P2:

A

cable

of

l

such

that

of

h(x)

passes

through

height

bridge

graph

(3

a

of

the

function

is

new

function

m(x)

translated

=

f(x

Find

the

the

values

minimum

of

r

and

point

of

s

−

such

m(x)

is

(3

The

graph

formed

n(x)

d

=

af(x

Find

that

is

to

of

the

dene

−

the

the

(3, 4).

b)

+

function

a

new

is

of

does

a

marks)

cubic

pass

through

(5

suspension

it.

of

at

The

in

The

the

bridge

shape

by

a

all

marks)

of

of

has

the

quadratic

cable

each

cable

the

r)

how

is

the

10

m

two

a

cable

curve.

above

ends

of

the

is

2

m

above

the

middle

of

the

above

the

bridge.

far

bridge

the

cable

to

+

at

one

quarter

of

the

way

along

the

s bridge,

c

a.

part

marks)

is

dene

in

on

the

Find

The

lies

the

bridge (3, 4).

that

modelled

bridge.

point

found

points.

above

be

The

values

equation

horizontal

can

l).

the

graph

curve

point

marks)

translated

new

the

function

the

(2

Find

the

b

quadratic

new

(2

30

The

this

k

the

graph

if

function

c

a

Determine

snoitcnuF

27

from

one

end.

(8

marks)

that

(3, 4).

marks)

trans-

function

c

values

of

a, b

maximum

and

point

c

of

such

m(x)

(4

marks)

285

/

Approaches to learning: Thinking skills:

Hanging around!

Create, Generating, Planning, Producing

Exploration criteria: Presentation (A),

Personal engagement (C), Reection (D)

IB topic: Quadratic Modelling, Using

technology

ytivitca noitagitsevni dna gnilledoM Investigate

Hang a piece of rope or chain by its two ends. It must be free hanging

under its own weight. It doesn’t matter how long it is or how far apar t the

ends are.

What shape curve does the hanging chain resemble?

How could you test this?

Impor t the cur ve into a graphing

package

A

graphing

Take

a

What

photograph

do

Import

package

you

the

Carefully

need

image

follow

can

t

an

of

your

to

consider

into

the

a

equation

hanging

instructions

a

curve

to

a

photograph.

rope/chain.

when

graphing

of

taking

this

photo?

package.

for

the

graphing

package

you

are

using.

The

image

should

appear

in

the

graphing

screen.

286

/

6

Fit an equation to three points on the cur ve

Does

three

it

Would

In

matter

two

your

Now

points

the

chosen

Carefully

that

which

graphing

use

three

points

lie

on

three

be

the

points

curve.

you

select?

ytivitca noitagitsevni dna gnilledoM

Select

enough?

package,

graphing

enter

package

your

to

three

nd

the

points

best

t

as

x-

and

quadratic

y-coordinates.

model

to

your

points.

follow

the

instructions

for

the

graphing

package

you

are

using.

Test the t of your cur ve

Did

you

What

nd

reasons

The

shape

and

not

a

a

curve

are

that

a

which

there

that

ts

may

free-hanging

parabola

at

all.

the

This

shape

mean

chain

is

that

or

why

of

you

rope

you

your

image

did

not

makes

did

not

is

get

get

a

perfect

actually

a

International-

exactly?

perfect

a

t?

catenary

t.

mindedness

The word “catenary”

comes from the Latin

word for Research

the

difference

between

the

shape

of

a

catenary

and

a

“chain”.

parabola.

Did you know?

A football eld is often cur ved to allow

water to run o to the sides.

Ex tension

Explore one or more of the following – are they quadratic?

A well-known

landmark –

perhaps the

Sydney Harbour

The cur ve of

Bridge or the

arches at the

a banana.

bottom of the

Eiel Tower.

The cross section

of a football eld.

The path of a football when

Other objects that look like

kicked in the air – here you

a parabola – for example,

would need to be able to

the arch of a rainbow, water

use available soft ware to

coming from a fountain, the

trace the path of the ball

arc of a Satellite dish.

as it moves.

287

/

Modelling rates of change:

exponential and logarithmic

7 The

reduction

skateboard

under

can

all

in

Concepts

temperature,

ramp,

compound

be

functions

the

value

interest,

modelled

by

of

the

an

and

certain

curve

of

■

Change

■

Modelling

a

investment

radioactive

families

of

decay

Microconcepts

curves. ■

This

chapter

looks

at

the

characteristics

of

Common ratio, geometric sequences,

some

geometric series and innite geometric of

these

families

of

curves,

and

considers

how

sequences they

in

can

be

real-life

used

to

model

and

predict

outcomes ■

Convergence and divergence

■

Percentage, interest rate, compound

situations.

interest, compounding periods, present

value, future value, annuity and

How long will it take you to become amor tization

a millionaire if you receive US$1

■

Exponents, exponential growth/decay,

inverse functions, logarithms, exponential

the rst month, US$2 the second

functions and half-life

month, US$4 the third month,

US$8 the four th

■

Horizontal asymptotes

■

Logistic functions and log-log graphs

month and so on?

If the temperature of a cup of

tea reduces at a given rate,

will it ever reach 0°C?

How can you nd an equation to

model the slope of a skateboard

track? Or a ski slope?

Does the same amount

of money have the same

value today, yesterday

and tomorrow?

288

/

Look

back

at

1

Describe

2

Would

3

Describe

4

What

scale

5

the

on

Gapminder

how

the

scale

x-axis

how

would

What

the

the

type

reach

scale

the

the

on

on

p.

x-axis

45.

increases.

Developing inquiry

zero?

the

problem

y-axis

in

skills

increases.

using

the

same

Think about the questions in this opening

axes?

of

general

the

be

both

ever

on

graph

problem and answer any you can. As you

function

trend

of

could

the

you

data

use

to

work through the chapter, you will gain

model

mathematical knowledge and skills that will

points?

help you to answer them all.

Before you start Click here for help

You should know how to:

Skills check

1

1

Apply

the

2

eg

laws

3

x

×

of

exponents.

Find

the

=

a

of:

1

x

2

Find

value

5

x

a 2

with this skills check

specic

percentage

of

a

(x

 a







2

3

b

)

quantity.



b

c

x



x



6 eg

6%

of

24

=

 24

5

x

 1.44

d

e

(

100

x

)

3

x

3

Mappings

another.

ordered

eg

x

3

the

elements

Illustration

pairs,

given



of

the



by

tables,

=

2 × 3

one

means

of

diagrams

function

f(3)

of

+

f(x)

1

=

=

set

sets

and

2x

+

to

2

of

graphs.

1

when

means

that

number

a

3%

c

28%

3

from

the

set

domain

the

set

is

mapped

value

of

Similarly

the

b

24

of

Consider

onto

number

7

15%

of

72

150

the

function

f(x)

=

1

−

3x.

Copy

and

complete

the

following

table

on based

of

of:

of

a the

the

7

3

This

Find

on

f(x).

range.

using

more

x

x

1

2

3

4

y

3

5

7

9

x

values:

0

−1

1

2

y

Graphically:

b

Find

c

Graph

the

value

the

of

f(12).

function

f(x).

y

4

10

Find

D(4, 9)

9

the

value

of

10

the

following:

10

4

3

8

a

7

C(3, 7)

2i

i 1

 1

b

 i 5

2i

 1

c

i

i 1

6

B(2, 5)

5

4

3

A(1, 3)

2

1

f x 0 1

4

Sigma

2

3

4

5

6

notation.

n

S

 n



u



i

u

 u

1

 u

2

 u

3

  

u

4

n

i 1

5

2

i

eg

2

 1

2



2

2

 3

2



4

2

 5



55

i 1

289

/

7

MODE LLING

7.1

The

tennis

competing

champion

a

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

Geometric sequences and series

Australian

Slam

R AT E S

How

Open

Men’s

tournaments.

in

is

straight

Singles

In

this

competition

competition,

elimination

matches

is

part

128

until

of

the

players

the

nal,

Grand

start

where

the

declared.

many

matches

are

played

in

the

rst

three

rounds

of

the

tournament?

b

What

c

How

do

many

round)

d

How

you

of

course

of

matches

the

many

notice?

be

played

in

the

last

round

(the

nal

tournament?

matches

the

are

will

be

played

in

total

during

the

complete

tournament?

The nth term of a geometric sequence is given by the formula:

n

u

= u n

×

r

1

, r ≠ 1

1

Investigation 1

1

a

One common counting method is decimal counting. This uses

units, tens, hundreds, thousands, etc. ie multiples of 1, 10, 100, …

Determine the rule for going from one category to the next.

b

In a single elimination (or knockout) tournament, the number of

remaining par ticipants follows the pattern 64, 32, 16, …. Determine

the rule for going from one number to the next.

c

The value of a car depreciates because of age. A certain car has the

following value at the end of each year after purchase. €40 000, €32 000,

€25 600, … Determine the rule for going from one value to the next.

d

A pendulum consists of a hanging weight that can swing freely. When

moving from side to side, it gets displaced from its rest position. For a

cer tain pendulum the displacements at each end of its oscillation were

recorded to be 40, −20, 10, −5, …

Determine the rule for going from

one value to the next.

What do you notice in the pattern of the terms in all these examples?

Why aren’t these sequences arithmetic?

2

Complete the table for each geometric sequence.

1st term

2nd term

3rd term

a

1

10

100

b

64

32

16

c

40 000

32 000

25 600

d

40

−20

4th term

5th term

6th term

10

290

/

InternationalFor each sequence, determine whether it is increasing, decreasing

Factual

mindedness

or oscillating.

Conceptual

When do the terms of a geometric sequence increase? When

The Elements, Euclid’s

do they decrease? When do they oscillate?

book from 300BC,

contained geometric

3

a

To get from the rst term of a geometric sequence to the second you sequences and series.

need to multiply by the common ratio.

= u

u 2

× r = u 1

r

arbegla dna rebmuN

7. 1

1

How many terms after the rst term is the second term?

How does this relate to the expression above?

To get from the second term of a geometric sequence to the third you

snoitcnuF

b

2

=

need to multiply again by the common ratio. u

u

3

× r

=

u

2

r × r

=

u

1

r

1

How many terms after the rst term is the third term?

How does this relate to the expression above?

c

How many terms after the rst term is the eighth term?

How does this relate to an expression for the eighth term

in terms of

u 8

and the common ratio r?

the rst term u 1

d

How many terms after the fth term is the eighth term?

How does this relate to an expression for the eighth term

u

in terms of 8

the fth term u

and the common ratio r? 5

e

How many terms after the rst term is the

nth term of the sequence?

How does this relate to an expression for the

in terms of

nth term u n

the rst term u

and the common ratio r? 1

Conceptual

HINT

How can you nd the general term of a geometric

A sequence of sequence?

terms that has a

common ratio equal

to 1 (a sequence

A sequence of numbers in which each term can be found by multiplying the

of constant terms) preceding term by a common ratio, r, is called a geometric sequence.

is not a geometric

For a sequence to be geometric, r ≠ 1

sequence.

Example 1

For

each

of

specied

a

4,

the

term,

12,

36,

following

and

…,

u

an

,

geometric

expression

sequences,

for

the

b

…

nth

2,

nd

the

2.2,

2.42,

…,

−6,

18,

…,

u

,

d

…

2 ,

12 =

,

…,

3

u

a

u

u

8

×

r

4

×

3

=

u

…

to

determine

the

common

ratio

of

geometric

term

of

sequence

the

you

sequence

need

by

to

the

divide

preceding

8748 term.

1

= n

…

7

=

n

u

,

order

any 7

=

,

3

4

u

the

3

In =

of

7

3

r

value

4

,

6

a

the

5

1 2,

ratio,

term.

8

c

common

×

r

1

n

=

4

×

1

3

1

Continued

on

next

page

291

/

7

MODE LLING

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

-6

b

r

=

=

-3 Sometimes

when

the

term

you

are

trying

2

to u

=

u

5

×

r

×

=

18

×

(−3)

×

(−3)

=

=

×

1

n

r

=

2

×

term

very

close

to

the

ones

you

have,

instead

of

using

the

formula

1

(−3)

you

as

r

a

can

just

multiply

by

the

common

ratio

1

2

c

is

already

u

n

nd

162

3

n

u

r

many

times

as

is

necessary

until

you

2

=

= 1

reach

1

the

required

term.

2

5

u

=

u

6

×

5

r

=

n

u

=

u

n

×

1.1

=

3.22102

=

3.22

1

n

r

=

2

×

(3

s.f.)

1

1.1

2 

  r

×

1



d

2

1

 3





2



 1 

 

 3



1

64

6

u



u

7



6

r



2





1

3

3

n

1 u



u

n



1

2

n 1

n 1

r





2



1

3

3

Exercise 7A

1

1

For

each

one

of

the

following

d

geometric

u

4



and

u

1

= 4

27

2 sequences,

the

value

of

expression

a

5,

10,

nd

the

the

for

20,

common

specied

the

…,

nth

u

,

ratio,

term

and

e

an

u

=

−2,

u

1

has

term.

f

…

=

−1458

and

the

sequence

7

u

both

=

positive

13.5

and

u

3

and

=

negative

terms.

367.5

6

7

b

−3,

15,

−75,

…,

u

,

…

3

There

is

a

u

epidemic

in

Cozytown.

On

8

the

c

2,

6,

3

2,

…,

u

,

rst

day,

2

people

have

the

u.

On

the

…

6

second

2

3

d

,

1,

third

,

…,

u

,

day,

day,

10

50

people

people

have

have

the

the

u.

On

the

u.

…

4

3

2

a

Show

the

e

2,

20,

200,

…,

u

,

that

u

the

forms

number

the

rst

of

of

people

three

with

terms

in

a

…

10

geometric

2

For

each

one

of

sequences,

nd

expression

for

a

u

=

3

and

the

the

the

u

1

following

common

nth

=

geometric

ratio

and

an

b

Calculate

after

term.

c

24

Use

1

u

=

32

and

u

= 1

u



week

your

many

(7

people

have

the

u

days).

model

to

calculate

how

many

1,

u

=

81

will

have

the

u

after

1

year.

243

6

1

c

how

4

people

b

sequence.

Comment

and

all

terms

are

positive.

on

the

reasonableness

of

your

answer.

5

292

/

arbegla dna rebmuN

7. 1

Investigation 2

Censuses have taken place in the United Kingdom every ten years since

1801. Studies of the ndings on the population have shown that it is

increasing by 0.6% each year. The population in the last census at the

beginning of 2011 was found to be 63.2 million.

Assuming the population continues to increase at the same rate:

1

Explain why the population at the beginning of 2012 can be found by

multiplying 63.2 million by 1.006.

The population (in millions) at the beginning of 2017 can be given by

snoitcnuF

2

n

1.006

3

× 63.2. State the value of n

If a population increases at a rate p% per year write down an expression for

the population after n years (P

) in terms of the original population (P n

4

) 0

Explain how this formula is dierent from the usual formula for a geometric

series.

The population of Japan is decreasing on average by 0.2% per year. Its

population in 2018 is 127 million.

Assuming the rate of decrease remains the same:

5

Write down an expression for the population of Japan in 2025 and

calculate this value.

6

In which year would the population drop below 120 million?

7

If a population decreases at a rate p% per year write down an expression for

the population after n years (P

) in terms of the original population (P

n

8

Conceptual

)

0

What type of sequence models a situation with constant

percentage change?

9

Conceptual

How do geometric sequences model real-life situations?

Percentage change is a form of a geometric sequence.

When terms increase by p% from the preceding term, their common ratio

p

corresponds to

r

 1 

100

TOK

When terms decrease by p% from the preceding term, their common ratio

p

corresponds to

r

Why is proof impor tant

 1 

100

in mathematics?

Example 2

Costis

bought

a

car

for

€16 000.

The

a

Find

the

value

of

the

car

at

b

Find

the

value

of

the

car

after

c

Calculate

after

how

many

the

value

end

5

years

of

of

the

the

car

rst

depreciates

by

10%

each

year.

year.

years.

the

value

of

the

car

fall

below

half

its

original

value.

Continued

on

next

page

293

/

7

MODE LLING

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

10

a

10%

of

€16 000

is

 16000

=

€1600

100

V

=

16000

−

1600

=

€14400

1

100 - 10 or

r

=

=

and

0 .9

100

A V

=

16 000

×

=

€14 400

=

16 000 × 0.9

decrease

of

10%

is

found

by

multiplying

0.9

1

by

0.9.

5

b

V

=

€9447.84

5

c

Half

of

the

original

value

is

This

€8000

the

n

16 000 × 0.9

n

>

6.58

After

rst

7

⇒




300

>

19.1

raised

100 + The

n

power

is

to

7.

n

× r May

⇒

the

⇒

n

≥

20

⇒

n

=

20

common

ratio

is

r

2

5

=

= 1

⇒

025

100

min

after

20

users

months

will

the

exceed

November

of

number

300

of

million,

daily

ie

in

2020.

12

c

V

= May2019

V

× r

12

⇒

300

=

187 × r

From

May

number

100 + ⇒

r

=

1.040

⇒

p

=

2018

until

May

of

2019,

the

of

months

that

will

pass

is

12

and

p = 1

100

⇒

of

May2018

040

so

the

raised

power

is

to

which

the

common

ratio

is

12.

4.0%

The exponent n to which the common ratio r is raised will always be the

number of time periods that have elapsed since the growth or decay begins.

294

/

Exercise 7B

1

At

the

city

end

was

of

200

population

a

2016

000.

was

Assuming

follow

a

the

At

264

that

the

at

end

of

of

6

a

2018

the

these

the

of

year

sequence,

of

gures

nd

The

the

5%

Calculate

the

applied

their

diameter

each

number

in

of

the

times

that

following

the

cases

%

and

value.

of

sequoia

trees

increases

by

year.

2017.

a

b

is

the

determine

end

end

Determine

change

500.

geometric

population

population

arbegla dna rebmuN

7. 1

population

at

the

end

of

If

the

diameter

1.2 m,

nd

of

what

a

it

sequoia

will

be

tree

in

7

today

is

years.

2020.

c

Comment

on

whether

this

increase

will

One

end

Find

the

the

kilogram

of

2015.

the

end

of

tomatoes

Prices

cost

of

of

a

rise

at

costs

$2.20

2.65%

kilogram

of

per

at

the

c

year.

tomatoes

Petra

at

the

a

camper

camper

Find

van

for

€45

000.

The

end

the

of

was

annua l

1.2 m,

determine

at

metres,

the

nd

beginning

the

of

diameter

2010

at

the

2017.

van

decreases

in

world

value

of

the

value

camper

six

population

at

the

beginning

Each 2019

van

is

7.7

billion

people.

Assume

the

by rate

of

increase

throughout

the

at time

the

r s t

measurement.

1.2

annual

5%.

the

made

diameter

of

of

2019.

buys

the

If

end

of

year

diameter

7th

was

7

3

the

measurement

continue.

2

If

snoitcnuF

b

periods

considered

in

the

question

is

years. 1.72%.

4

Beau

spends

€15

000

buying

computer

a

materials

for

his

ofce.

Each

year

Find

the

world

population

after

one

the decade.

materials

depreciate

by

12%.

b

a

Find

the

value

of

the

materials

Determine

the

world

population

7

years

the

world

population

at

after ago.

three

years.

c

b

Find

how

many

years

it

takes

for

Determine

end

materials

to

be

worth

Find

the

common

ratio

representing

for

the

the

when

the

world

population

geometric doubled

compared

to

the

beginning

following of

percentage

2040.

Determine

has

sequence

of

€5000.

d

5

the

the

2019?

changes:

e

a

increase

c

becomes

by

12.5%

89%

b

decrease

d

increase

by

by

7.3%

Calculate

in

order

become

0.1%

the

for

10

annual

the

percentage

world

billion

in

increase

population

to

2090.

Investigation 3

Let S

denote the sum of the rst n terms of a geometric series, so n

5

2

S



5



u



i

u 1

 u

r

1

 u

r

3

 u

1

r

4

 u

1

r

1

i 1

1

Write down an expression for rS 5

2

Hence nd an expression in terms of u

and r for: 1

a

rS

− 5

b

S 5

S 5

n

u 1

The result obtained in par t b can be generalized to S

(r

- 1

)

= n

r

- 1

Continued

on

next

page

295

/

7

MODE LLING

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

n

u 1

3

Explain why the formula can also be written as S

(1

- r

) and why this form might be easier to use

= n

1 - r

when r < 1.

According to a story, Sissa ibn Dahir, invented the game

of chess for King Shihram to play. The king was so

pleased with the game that he asked Sissa what reward

he wanted for this great invention. Sissa said that he

would take this reward: the king should put one grain of

wheat on the rst square of a chessboard, two grains

of wheat on the second square, four grains on the third

square, eight grains on the four th square, and so on,

doubling the number of grains of wheat

4

Find how many grains of wheat the King would have

to give Sissa.

16

5

How does this compare with the annual world production of wheat consisting of approximately 1.2 ×10

grains?

The sum of the rst n terms of a geometric sequence is called a geometric

series and is given by the formula:

n n

u



1

S

 n



u

r

n

 1

u







1

1

 r



, r ≠ 1

k

r

k 1



 1

1  r

Example 4

The

rst

term

of

a

geometric

a

Find

the

sum

of

the

rst

b

Find

the

sum

of

the

next

sequence

10

terms

10

is

of

terms

equal

the

of

to

3

and

the

common

ratio

is

equal

to

2.

sequence.

the

sequence.

10

3 

a

S

2

 1







3069

10

2  1

20

3 

b

S

2

 1







3 145 725

The

20

sum

of

the

next

ten

terms

of

the

2  1 geometric

Thus

the

sum

of

the

next

10

terms

of

the

u

+

u

11

sequence

−

S

u

+

is

actually

...

+

u

13

.

the

You

sum

can

think

of

20

is

the

20

+ 12

this

S

series

=

3 145 725

−

3069

=

3 142 656

S

10

as

sum

=

u

20

S

sum

of

+

=

u

+

− 20

+

of

u

u

+

10

...

u

+

u

...

terms

minus

+

u

+ 11

...

+

u 20

u 10

+ 11

+ 10

+

u

20

terms.

3

= 10

+

rst

3

2

S

the

rst

2

1

S

the

u

1

10

⇒

the

u

+ 12

...

+

u 20

296

/

Example 5

in

the

The

day

a

students

campus.

money

is

15%

will

be

less

their

than

the

the

the

to

of

day

of

to

money

raise

was

money

of

the

full

purchase

raising

raise

the

€50.

in

order

required

The

to

install

money

money

that

of

hammocks

€300.

they

raise

on

each

subsequent

previous.

number

to

to

days

purchase

maximum

goal

10

rst

amount

money

the

decided

have

on

enough

enough

Find

school

They

the

Calculate

c

a

raised

Calculate

b

in

they

they

raise

in

total.

Comment

on

whether

this

would

need

to

fundraise

on

if

they

are

to

raise

hammocks.

daily

percentage

€300

in

10

to

hammocks.

days

the

expect

decrease

in

the

money

they

raise

if

they

are

to

reach

days.

10

50

a

1

 0.85





S

snoitcnuF

The

arbegla dna rebmuN

7. 1



€ 267.71

This

is

a

geometric

series

with

u

10

=

50

and

1

1  0.8 5

100 - 15 The

amount

will

not

be

enough

in

order

r

=

=

0.85

100 to

purchase

the

hammocks.

Use

the

sum

of

second

a

form

geometric

of

the

series

formula

as

r


1 or r < −1 (we can also write |r| > 1 )?

n 

2

Consider a piece of paper of area 1 m

the diagram.

which is divided as shown in

First one half is shaded, then one half of the remainder is

shaded, then one half again.

1

Let u

be the area shaded on the nth “shading”. Hence u

,

=

1

n

2

1 u

etc.

=

2

4

6

a

Find an expression for u

b

State the value of r for the sequence.

c

In the context of the diagram explain what is meant by

snoitcnuF

n

S n

d

From the diagram write down lim

S

n



n 

n

u 1

e

Use the formula S

(1

- r

) to justify your answer to 6d

= n

1 - r

f

Hence right down a formula for the limit as

n tends to innity for the sum of a geometric series giving

the condition on r for it to be valid.

This limit is referred to as the sum to innity of the series and is written as

S ∞

7

Use your formula to nd the sum to innity of the following geometric sequences.

a

8

1,

0.1, 0.01, 0.001, …

Conceptual

b

8, −4, 2, −1, …

When does an innite sum of a geometric series converge to a nite sum?

TOK

The sum to innity of a geometric sequence is:

u

How do mathematicians

1

S

, for |r| < 1





reconcile the fact that

1  r

some conclusions

conict with intuition?

Example 6

Sheldon

is

chemical

He

adds

carrying

to

a

out

collection

60 ml

to

the

an

of

rst

experiment

test

and

that

involve

adding

decreasing

amounts

of

a

solutions.

50 ml

to

the

second.

The

amounts

added

form

a

geometric

sequence.

a

Find

the

b

Find

which

c

Given

test

amount

added

solution

Sheldon

solutions

has

he

to

will

be

400 ml

has

the

fth

the

of

rst

the

included

solution.

in

to

have

chemical,

his

less

than

prove

that

10 ml

he

added.

will

have

enough

however

many

experiment.

Continued

on

next

page

299

/

7

MODE LLING

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

5

a

r

=

6

Let

the

amount

experiment

be

added

on

the

nth

u n

4

 5  60 



u

 

5

28.9 ml

 6





n−1

b

u


0 and a ≠ 1 a

Here a is called the base of the logarithm. Any positive number can be used

as a base for logarithms, but the only ones you need to know about for this

course are 10 and Euler ’s number e.

x as logx

You can write log 10

x as the natural logarithm lnx

You can write log e

Some examples follow.

Two fundamental exponential equations with their solutions are:

x

•

10

•

e

= c ⇒ x = logc

x

= c ⇒ x = lnc

and two fundamental logarithmic equations are:

c

•

logx = c ⇒ x = 10

•

lnx = c ⇒ x = e

c

On

your

GDC,

log

is 10

called

“LOG”

and

log

is

called

“LN”.

e

320

/

arbegla dna rebmuN

7. 4

Example 15

Find

the

exact

value

x

1

10

1

x

for

each

of

the

following

equations .

2x

=

x

of

=

2

5

log

e

=

3

12

log

x

=

4

3

3ln

5

x

By

=

7

the

result

denition

of

log

or

by

using

the

above.

1

2

2x



ln 12



x



ln 12

2

3

x

=

10

=

1000

7

7 3

4

ln

x





x



e

It

is

important

to

isolate

the

log

term

before

3 applying

the

rule.

snoitcnuF

3

Investigation 13

x

1

Consider the equations 10

x

x

= 1 and e

= 1

or even the general case a

= 1

Find the value of x

What is the solution in terms of logs?

x

2

Consider the equations 10

x

= 10 and e

x

= e or even the general case a

= a

Find the value of x

InternationalWhat is the solution in terms of logs?

x

3

Consider the equations 10

x

a

n

= 10

x

and e

mindedness

n

= e

or even the general case

n

= a

, where x is the unknown variable and n is a constant parameter.

Logarithms do not

have units but many

Find the value of x

measurements use

What is the solution in terms of logs? a log scale such as

4

Use your GDC to copy and complete the following table, giving your earthquakes, the pH

answers to three signicant gures. scale and human

hearing.

log

2

log

3

log

3

log

4

ln

5

ln

7

log

log

ln

What do you notice? What can you conjecture about

6

12

35

log

x + log a

5

y? a

Use your GDC to copy and complete the following table, giving your

answers to three signicant gures.

log

12

log

2

log

6

log

15

log

3

log

5

11

ln

11

ln

7

ln

7

What do you notice? What can you conjecture about

log

x − log a

Continued

y? a

on

next

page

321

/

7

MODE LLING

6

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

Use your GDC to copy and complete the following table, giving your

answers to three signicant gures.

2

log

(3

)

log

3

2log

3

1

ln

ln

2

log

2

2

2

n

What do you notice? What can you conjecture about

log

(x

)?

a

log

1 = 0, ln1 = 0 and in general, log

1 = 0 a

log

10 = 1,

ln e = 1 and in general, log

a = 1 a

n

log

10

n

= n,

n

ln e

n

=

and in general, log

a

= n

a

Laws of logarithms:

•

log

x + log a

y = log a

xy a

x

•

log

x

-

log

a

y

log

=

a

a

y

n

•

log

(x)

= nlog

a

x a

Example 16

Write

each

of

the

following

as

a

single

logarithm:

log x

a

3

log

b

x

c

2

log

x

+

log

y

d

log

x

−

2

log

y

2

e

f

−ln x

1

+

g

ln x

ln x

+

ln y

−

ln z

3

a

3

log x

=

log(x

)

1

1

log x

2

b

log x

=

2

=

log

(

x

)

=

log

x

2

2

c

2

d

log x

log

x

+

log

y

=

log(x

2

)

+

log

y

=

log(x

y)

x 2

 2 log

y

log x





log( y

) 

log 2

y

1 1

e

 ln

x

1 ln



x



ln( x

) 

ln

x

f

1

g

ln

+

ln x

=

ln e

+

ln x

=

ln(ex)

xy x



ln

y



ln z



ln( xy) 

ln z



ln

z

322

/

Exercise 7J

1

Find

the

value

of

each

of

the

following

c

3log x

e

−2ln x

g

ln x

+

2log y

d

log x

−

3log y

logarithms.

a

log

b

100

log

1

d

c

0.1

e

f

ln e

log



4

−

ln y

Find

the

−

log x

ln z

exact

piecewise

1

 

+

10

e

ln

2

2

ln

g

f

ln e



x

>

value

functions

of

a

are

if

the

following

continuous

arbegla dna rebmuN

7 7 . .4 4

for

0

e

a 2

Find

the

solution

to

the

f

fo ll owi ng



x





as

a

equations

logarithm

where

giving

the

f



x





x

=

b

10

10

=

x

10

e

e

g

e

x

=

38

d

e

f

e

x

100

3

=

1

x



2

2 ln ax

x



2

2

=

e

=

0.3

TOK

x

=

 4





c

 1

2

appropriate.

x

x

answer

b 10

 1

ax

2e

3 x

a

x



 exponential

 3

snoitcnuF

3

4 x

x

The phrase “exponential growth” is used

popularly to describe a number of phenomena.

3

Write

each

of

the

following

as

a

single

Do you think that using mathematical language

logarithm:

can distor t understanding?

log x

a

b

2log x

3

Investigation 14

x

1

a

Verify that if f(x) = log x and g(x) = 10

then f

g(x) = g °

b

f(x)

= x

°

Hence state the connection between the two functions

f and g

x

2

Use the result of 1 a to sketch the graphs y = 10

and y = log

x on the

same axes.

3

State the domain and range of f(x) = log x

4

a

Find the inverse of h(x) = e

b

Sketch the graph of y = h(x) and its inverse on the same axes.

x

+

2.

1

5

6

State the domain and range of h and h

Factual

What are the domain and range of the function

f(x) = log

x? a

7

Conceptual

What is the relationship between the logarithmic function

and the exponential function?

x

The fuctions y = 10

composition gives

and y = logx are inverse functions and so their

x

x

Similarly, the same thing holds for the functions

logx

10

y = e

and y = lnx

lnx

= x and e

= x

323

/

7

MODE LLING

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

Exercise 7K

1

Solve

the

following

using

logarithms.

exponential

Give

your

d

equations

answers

as

a

Determine

f(x)and

logarithm.

the

sketch

coordinate

inverse

it

on

axes,

function

the

same

indicating

of

set

any

of

possible

x

10 x

a

i

asymptotes

x

10

=

ii

3,

iii

= 15

e

=

and

intercepts

with

the

5

5

coordinate

axes.

x

10

x

5

x

b

i

3e

−

1

=

ii

14,

+ 3

=

Consider

the

function

f(x)

=

10

−

3.

5,

2

1

a

Find

b

Sketch

its

inverse

function

f

(x).

x

iii

2

3(e

−

1)

=

5

Solve

the

using

logarithms,

are

1

following

positive

real

exponential

given

numbers

x

a

3

a

that

all

and

equations

parameters

a

≠

c

=

Simplify

b

0

the

b e

the

x

=

c

2,

2 ×1 0

graph

show

asymptotes

1.

x

−

graph,

the

exact

c

following:

State

clearly

with

f

(x).

any

their

coordinates

intersections

= k

of

clearly

with

the

the

On

your

possible

equations

of

any

and

possible

coordinate

domain

and

axes.

range

of

1

the 3logx

a

lnx

b

10

inverse

function

f

(x).

lny

e x

6 2logx

c

logy

Consider

d

10

Given

the

function

f(x)

=

2e

−

6.

e

a 4

the

2lnx

functions

f(x)

=

ln(x

+

2)

Sketch

the

graph

of

f(x).

On

your

graph,

and show

clearly

any

possible

asymptotes

x

g(x)

=

2e with

a

Find(g ∘f )(x),

giving

your

answer

in

their

equations

coordinates

form

b

For

ax

the

+

the

exact

the

exact

of

any

possible

intersections

b

function

equation

and

the

of

any

f(x),

determine

possible

coordinates

of

the

b

Find

its

c

State

coordinate

axes.

the

asymptotes

any

with

1

inverse

function

f

(x).

and clearly

the

domain

and

range

of

possible 1

the intersections

c

Hence,

with

sketch

the

the

coordinate

curve

of

inverse

function

f

(x).

axes.

f(x).

Investigation 15

Kim recorded the battery remainder indication on her laptop during the course

of a day with respect to the time of the day and gathered the following data,

where t is the time in hours since she star ted the experiment and

B is the

battery left on her laptop.

t

B

1.6

3

4.4

6

6.8

9

10

12

14.8

16.4

96

79

65

52

47

34

30

23

15

12

She plotted the data and produced the following scatter diagram:

y

A 100

B

80 C

60

D E

40

F

G

I J

20 H

0 2

4

6

8

10

12

14

16

18

324

/

1

TOK

Find the PMCC for this data.

The scatter diagram she created did not have strong enough linear correlation

for her to conclude that the relation between her variables is linear.

mathematics enjoys

Her physics teacher suggested that in fact the relationship was more likely to

ct

be exponential of the form B = ae

of

“One reason why

and suggested she draw a semi-log graph

special esteem, above

all other sciences, is

that its propositions are

ln B against t

absolutely cer tain and

2

arbegla dna rebmuN

7 7 . .4 4

By taking the natural log of both sides of the equation, explain why a indisputable”–Alber t

graph of ln B against t for this function would be a straight line and give Einstein.

expressions for of the gradient and the

y-intercept of this line.

a

Copy and completed the table below being after all a product

of human thought

1.6

t

3

4.4

6

6.8

9

10

12

14.8

16.4

which is independent

ln B

snoitcnuF

How can mathematics,

3

of experience, be

appropriate to the

b

Create

a

scat ter

diagram

based

on

the

table

in

par t

a

with

t

on

the

objects of the real

horizontal

axis

and

ln

B

on

the

ver tical

axis.

This

is

called

a

semi-

world?

log

graph—one

varible

plot ted

against

the

logarithm

of

the

other

variable.

c

Find the PMCC for this data and compare your result with the one for a

linear model.

d

Use your values for the gradient and for the

y-intercept to estimate the

ct

parameters for the equation

B = ae

Her chemistry teacher suggested that in fact the relationship was more likely to be

c

of the form B

4

=

at

and suggested she draw a log-log graph of ln B against ln t

By taking the natural log of both sides of the equation, explain why a graph

of ln B against ln t

for this function would be a straight line and give an

expression for the gradient of the line and for the

5

a

Copy and complete the table below

ln

t

ln

B

b

y-intercept.

Create a scatter diagram based on the table in par t

a with lnt on the

horizontal axis and lnB on the ver tical axis. This is called a log-log

graph – the logarithm of one variable plotted against the logarithm of

the other variable.

c

Find the PMCC for this data and compare your result with the previous

two models.

d

Use your values for the gradient and for the

y-intercept to estimate the

c

parameters for the equation

6

B



at

By considering the value of B when t = 0 which of the two non-linear

models might you prefer irrespective of the values of the PMCC?

7

Conceptual

How does logarithmic scaling of graphs help you identify

the relationship between variables?

325

/

7

MODE LLING

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

b

A power function of the form y = ax

A graph of

ln

y

can be linearized as ln y = bln x + lna.

against ln x will have a gradient of b and a y-intercept of

ln a.

bx

An exponential function of the form y = ae

can be linearized as

ln y = bx + lna

A graph of

ln

y

against x will have a gradient of b and a y-intercept of ln a.

E X AM HINT

In most cases the parameters for the function can also be found by using

either power or exponential regression on your GDC.

But if instructed to

nd them using linearization techniques you must show sucient method

to demonstrate you have found them this way.

x

Some GDCs will give the exponential regression function in the form y = ab

.

r

This can be conver ted into base e by writing b = e

r

y = a (e

x

)

rx

= ae



r



and hence

ln b

xlnb

or y

=

ae

Investigation 16

a

b

Sketch the following data on a scatter diagram.

x

3

5

7

9

11

13

15

17

y

95

1071

9812

103 201

1 100 458

9 526 983

107 247 384

993 485 029

Write down the domain and the range of the data. What do you notice? What is the problem with viewing

this scatter diagram?

c

Complete the following table to include the logarithm of the

95

y

log

d

9812

103 201

1 100 458

9 526 983

107 247 384

993 485 029

y

Sketch the following data on a scatter diagram.

3

x

log

e

1071

y-values.

5

7

9

11

13

15

17

y

Write down the domain and the range of the new set of data. What do you notice? How was the problem

resolved?

326

/

7. 4

In

a

set

of

data,

measurements

a

have

been

of

manageable

found

and

scaling

recorded.

the

The

logarithm

new

set

of

(in

data

base

is

as

10)

of

the

follows.

A

B

C

D

E

F

G

H

I

J

log y

1.7

3.2

4.1

4.7

5.2

6

6.6

7.2

8

10.1

Find

the

value

State

c

Comment

Data

the

range

linking

income

on

the

(GNI)

y

for

measurement

B

and

measurement

G.

How

many

times

is

G

greater

GNI

advantages

of

expectancy

index

are

shown

set

in

of

using

data

a

log

countries

scaling

of

the

of

the

original

European

Union

data

to

their

gross

national

below.

CZE

CNK

EST

FIN

FRA

42 080

39 270

13 980

19 330

24 280

42 300

20 830

38 500

35 650

80.7

80.2

74.2

77.1

77.8

79.9

76.3

80.3

81.6

DEU

GRC

HUN

IRL

ITA

LVA

LTU

LUX

NLD

39 970

26 090

20 260

33 230

32 710

17 820

19 690

64 410

43 260

80.5

80.5

75

80.6

82

74.6

74.3

81.5

80.9

POL

PRT

ROU

SVK

SVN

ESP

SWE

GBR

20 480

24 480

15 140

22 230

26 960

31 660

42 200

35 940

76.5

80.2

74.5

76.1

79.9

82

81.7

80.5

Country

L.E.

the

this

HRV

L.E.

Calculate

life

in

BGR

Country

GNI

the

values

BEL

L.E.

GNI

of

AUT

Country

Let

of

B?

b

a

reasons

Value

than

2

for

snoitcnuF

1

arbegla dna rebmuN

Exercise 7L

the

log

value

of

of

both

GNI

be

sets

x

of

and

data

the

and

value

include

of

L.E.

your

be

results

in

an

enlarged

table.

y

b

It

3

is

thought

b

Find

c

Hence

d

Use

the

Frances

line

nd

the

that

of

the

best

t

estimates

power

suspects

relation

for

for

a

regression

that

the

ship

log

y

and

between

against

two

variables

is

of

the

form

y

=

ax

x

b

function

time

log

the

on

the

GDC

to

verify

your

values

of

a

and

b

taken

Total time (t

minutes)

to

answer

one

question

in

a

working on the multiple-choice

quiz

increases

with

5

20

30

40

50

2.4

2.8

3.4

4.0

5.1

the

quiz, q (minutes) total

time

(q

minutes)

that

she

has

spent

Time taken to working

on

the

quiz.

answer a question,

She

times

herself

over

different

periods

and

t (minutes)

collects

the

following

data.

327

/

7

MODE LLING

Frances

thinks

t

R AT E S

and

q

OF

are

C H A NGE :

related

by

E XPONENTIAL

an

b

AND

LO G A R I T H MI C

Write

down

a

linear

F UNCTIONS

equation

of

ln( s

−

c)

bt

equation

a

Find

of

the

the

given

values

in

the

b

Plot

c

Either

by

graph,

or

a

form

graph

function

q

of

=

ln

q

on

against

for

the

values

of

q

c

Plot

d

Find

of

ln

q

against

using

your

t.

d

and

ln

Hence,

time

values

the

from

linear

GDC,

nd

your

e

Use

taken

would

regression

the

values

or

ln( s

for

linear

nd

producing

−

c)

against

ln n

a

b

and

using

your

regression.

model

the

ln a

for

the

cost

of

shirts.

of

to

a

suitable

answer

model

for

the

Yenni

would

a

Use

model

take

to

like

to

produce

500

shirts

working

on

to

the

the

the

model

total

from

daily

part

e

to

cost.

questions.

estimate

Frances

day.

estimate

how

answer

long

a

it

quiz

for

Describe

the

dangers

estimate

the

cost

of

of

using

this

producing

model

500

a

to

day.

question

h

after

of

a

suggest

the

values

Hence

g

e

graph

graph

f

b

a

ln n

table.

measuring

by

ae

one

hour

i

Describe

how

you

could

enter

data

in into

your

GDC

to

nd

the

model

using

total. the

4

Yenni

owns

a

sma l l

fa ctor y

tha t

shirts.

She

wants

to

model

between

the

daily

total

cost

shirts

($ s)

and

the

Hence,

The

velocity

number

that

are

produced

different

believes

the

model

answer

to

part

e

(v)

of

a

falling

body

is

recorded

distances

( d)

from

the

point

at

it

was

launched.

Consideration

of

( n).

the She

your

of

which shirts

verify

of

at manufacturing

function.

the

5 relationship

regression

manu-

ii

factures

power

is

of

the

physics

involved

leads

to

the

following

form model

being

proposed:

b

s

=

an

+

c

v She

knows

shirts,

would

a

the

be

Find

Yenni

s

daily

even

cost

if

of

she

were

running

making

the

value

collects



a

d

 b

no

factory

$1000.

the

following

n

that,

Data

for

v

and

d

is

collected

and

shown

below.

of

data

c.

and

estimates

d

1

2

3

4

v

8.4

10.3

11.7

12.9

the

values

50

100

150

200

2000

2800

3400

3800

a

Plot

b

Estimate

a

graph

of

v

values

against

of

a

and

d

b.

328

/

7.5

The

Logistic models

progression

high

jump

1912

is

at

of

the

shown

the

world

beginning

on

the

record

of

scatter

each

for

men’s

year

Progression

since

of

Men's

High

Jump World

Record

by

Year

2.55

diagram. 2.45

a

How

could

you

describe

the

basic

features 2.35

the

progression

of

the

world

b

How

is

it

different

from

a

c

How

is

it

different

from

an

thgieH

of

record?

linear

graph?

arbegla dna rebmuN

7. 5

2.25

2.15

graph?

d

Where

2.05

else

do

you

encounter

this

shape? 1.95

1910

1930

1950

1970

1990

2010

Year

The

logistic

restriction

bacteria

in

function

on

a

the

petri

is

used

growth.

dish,

or

to

For

the

model

situations

example,

increase

where

population

in

height

of

on

a

there

an

is

snoitcnuF

exponential

a

island,

person

or

seedling.

TOK

Logistic

functions

have

a

horizontal

asymptote

at

f(x)

=

L Mathematics is all

around us in patterns,

shapes, time and L

The equation of a logistic function has the standard form

f ( x)

,



space.

kx

1  Ce

What does this tell you

where

L,

C

and

k

are

the

parameters

of

the

f u nction

a nd

e

is

Euler ’s

about mathematical

number.

knowledge?

Investigation 17

a

Sketch the graph of the following functions

10

i

f ( x)

20

ii



f ( x)

30

iii



x

f ( x)



x

1  e

x

1  e

1  e

What is the relation of the value of the numerator to the graph of the function?

HINT How does this link with graph transformations?

In a logistic equation How does the rate of change vary at dierent points on the graph? L

f(x) = At what point does the rate of change seem greatest?

, (

( 1  Ce

Give the equations of any asymptotes.

kx )

)

f(x) = L is an

asymptote to the b

Sketch the graph of the following functions:

curve and L is referred

10

i

f ( x)

10

ii



f ( x)

x

1  e

10

iii



f ( x)

to as the carrying



2x

1  e

3x

1  e

What is the relation of the parameter multiplying x to the graph of the function?

capacity for the model

being considered.

How does this link with graph transformations?

How does the rate of change vary at dierent points on the graph?

329

/

7

MODE LLING

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

HINT

Logistic functions have a varying rate of change. The most usual shape of a

logistic function star ts with the data being almost constant (close to zero rate

In some real-life

of change), then the data show a rapid increase and nally become almost

situations the change

constant again (close to zero rate of change again). Thus the data show two

will begin at a cer tain dierent horizontal asymptotes, one for small data values and one for large

time and so the lower data values.

horizontal asymptote

will not be par t of the

function.

Exercise 7M

1

The

function

people

that

globally

Internet

with

models

with

the

access

respect

to

percentage

to

time

c

of

Assuming

broadband

is

given

by

be

the

function

P(t )

ever



, 0

t

≥

be

the

graph

of

the

3

A

u

Write

down

the

range

of

values

function

and

that

able

to

the

state

Virgin

also

the

largest

Islands

will

accommodate.

epidemic

An

is

spreading

estimated

120

throughout

million

people

interpret

susceptible

to

this

particular

strain

and

it

of is

the

future,

will

function. are

b

the

model

5t

Europe. Sketch

this

0

1  10e

a

in

population

100 logistic

valid

that

predicted

that

eventually

all

of

them

will

their get

infected.

There

are

10 000

people

already

signicance. infected

c

Use

the

take

for

model

half

broadband

to

the

predict

people

how

to

long

have

it

will

access

to

the

when

number

infected

will

of

x

=

0

and

people

double

it

is

who

in

the

projected

will

have

next

two

that

been

weeks.

L

Internet. The

logistic

function

f ( x)



models kx

d

Calculate

the

will

access

20

2

Data

to

of

people

broadband

who

Internet

in

years.

for

Islands

the

have

percentage

1  Ce

the

measured

the

population

from

1950

following

to

logistic

of

2015

the

have

function

a

Virgin

model.

, 0

1 

years

where

t

is

the

number

b

Use

the

passed

since

1950

c

The

the

in

Estimate

population

of

the

is

L,

will

approximate

C

be

and

to

values

of

the

k

determine

infected

for

how

after

many

four

90%

of

considered

the

people

terminated

have

been

Use

the

model

to

determine

1950.

the

in

is

Virgin

population

of

the

long

it

will

be

until

the

infection

Virgin can

Islands

the

infection

how

b

x

weeks.

model

infected. Islands

where

weeks.

have

Estimate

infected

of

when

a

people

135t

4e

that

in

Determine

people



of

parameters

produced

107000 P(t )

number

be

considered

terminated.

2015.

330

/

arbegla dna rebmuN

7. 7 5

Chapter summary

•

A sequence of numbers in which each term can be found by multiplying the preceding term

by a common ratio, r, is called a geometric sequence. For a sequence to be geometric,

•

The nth term of a geometric sequence is given by the formula:

n

u

= u n

•

r ≠ 1

1

r

, r ≠ 1

1

Percentage change is a form of a geometric sequence.

100 +

•

When terms increase by p% from the preceding term, their common ratio corresponds to r

p

=

100 -

•

When terms decrease by p% from the preceding term, their common ratio corresponds to r

snoitcnuF

100

p

=

100

•

The sum of the rst n terms of a geometric sequence is called a geometric series and is given by

the formula:

n

u 1

S

n

r

 1

u





1

1

 r

 , r ≠ 1



n

r

•

 1

1  r

The sum to innity of a geometric sequence is:

u 1

, for |r| < 1



S 

1  r

•

If PV is the present value, FV the future value, r the interest rate, n the number of years and k the

compounding

frequency,

or

number

of

times

inter est

is

pa id

in

a

yea r

(ie

k

=

1

for

yearly,

k = 2 for half-yearly, k = 4 for quar terly and k = 12 for monthly, etc), then the general formula for

nding the future value is

kn

r

 FV





1 

PV 

 100k



•



When a constant investment of $P is made for a cer tain number of n periods always compounded

with the same interest rate, it is called an annuity.

•

All exponential functions have a horizontal asymptote

x

•

Exponential functions of the form f(x) = b

have the line y = 0 as a horizontal asymptote. Unlike

inverse propor tion functions, the asymptotic behavior occurs either as

x

→

∞ or as x

→ −∞,

but not on both.

x

•

The exponential function f(x) = ka

+ c has:

❍

The straight line y = c as a horizontal asymptote.

❍

It crosses the y-axis at the point (0, k + c)

❍

Is increasing for a > 1 and decreasing for 0 < a < 1

x

•

In general, the solution of the exponential equation a

= b is the logarithmic function x = log

b, a

where a, b > 0 and a ≠ 1

•

Here

a

is

called

logarithms,

number

•

•

but

the

base

the

only

of

the

ones

loga rithm.

you

need

to

Any

pos itive

k now

abou t

number

for

thi s

ca n

be

cou r se

used

a re

10

as

a

and

base

for

Euler ’s

e.

You can write log

x simply as log

x

10

You can write log

x simply as the natural logarithm lnx e

Continued

on

next

page

331

/

7

MODE LLING

•

R AT E S

OF

C H A NGE :

E XPONENTIAL

AND

LO G A R I T H MI C

F UNCTIONS

Two fundamental exponential equations with their solutions are:

❍

10x

❍

e

= c ⇒ x = log

c

x

= c ⇒ x = ln

c

and two fundamental logarithmic equations are:

c ❍

log

❍

ln

x = c ⇒ x = 10

c

x = c ⇒ x = e

Laws of logarithms:

•

log

x + log a

y = log a

xy a

x

•

log

x − log a

y = log a

a

y

n

•

log

(x

) = nlog

a

x a

x

The functions y = 10

and y = logx are inverse functions and so their composition gives x

x

Similarly, the same thing holds for the functions y = e

logx

lnx

= x and e

10

and y = lnx

= x

L

The equation of a logistic function has the standard form

f ( x)

, where L, C and k are the

 kx

1  Ce

parameters of the function and e is Euler ’s number. The logistic function is used in situations where

there is a restriction on the growth. L is referred to as the carrying capacity.

Developing inquiry skills

If you keep doubling a number, how long will it take before it becomes

more than 1 million times its original value?

Can something that grows indenitely never exceed a cer tain value?

What is the equation that models a skateboard quar ter pipe ramp?

Does the same amount of money have the same value today, yesterday

and tomorrow?

How can you decide on which is the best investment plan?

Will an ice cube left outside of the freezer or a boiling cup of water that

has been removed from heat ever reach room temperature?

Click here for a mixed

Chapter review

1

The

are

rst

2,

5,

four

terms

12.5,

of

a

review exercise

geometric

sequence

31.25.

2

Saul

bought

Every

year

decreased

a

Find

the

common

b

Find

the

8th

c

Find

the

sum

a

new

the

by

bicycle

value

12%.

of

for

the

Find

US$350.

bicycle

the

value

of

the

ratio.

bicycle

at

the

end

of

ve

years.

term.

of

the

rst

eight

terms.

332

/

7

Molly

the

bought

third

had

a

year,

increased

beginning

increased

of

to

at.

the

to

At

the

value

of

363000

the

fth

439230

beginning

Molly’s

euros.

year,

the

euros.

9

of

The

at

At

Assume

of

the

at

increases

by

a

constant

for

Find

6500

ve

how

are

to

euros

years

at

much

clear

for

a

2.5%

motorbike.

interest

Nathalie’s

the

per

monthly

loan.

the

10 value

is

payments

had

that

borrows

loan

annum.

the

value

Nathalie

A

population

of

rabbits

can

be

modelled

by

rate x

the each

is

a

function

f(x)

=

24 000

×

1.12

,

where

x

year.

Calculate

the

rate

of

increase

as

the

a percentage

of

the

value

of

the

time

in

years.

a

Find

the

number

of

rabbits

after

arbegla dna rebmuN

3

ve

house.

years. Calculate

how

many

originally

paid

Find

much

euros

Molly

b for

the

Calculate

how

population

c

how

beginning

of

the

the

at

ninth

is

worth

at

nite

geometric

will

take

for

the

of

rabbits

to

double.

year. The

can A

it

the

11

4

long

at.

snoitcnuF

b

sequence

has

terms

2,

temperature,

be

modelled

T°C,

by

of

the

a

cup

of

soup

eq ua ti on

6, −x

T(x) 18,

…,

=

time

a

21

+

74

×

(1.2)

,

where

x

is

the

118098.

Calculate

the

number

of

terms

in

in

minutes.

this

a

Find

the

initial

b

Find

the

temperature

c

Find

how

temperature.

sequence.

b

5

Find

The

of

the

school

sum

fees

ination.

of

the

increase

When

each

Barnaby

year

joined

by

the

the

soup

rate

fees

were

10

minutes.

UK£9500.

At

the

end

of

to

many

reach

minutes

it

takes

for

the

40°C.

school,

d

the

after

sequence.

Write

down

the

room

temperature.

the

x

rst

a

year

Find

the

the

second

The

rate

cost

of

of

ination

the

was

school

1.16%.

fees

for

12

the

An

a

year.

following

Write

year

the

rate

of

ination

was

b

Write

the

cost

of

the

school

fees

for

f(x)

equation

=

4

of

×

2.3

+

16.

the

the

coordinates

curve

cuts

the

of

the

point

y-axis.

the

13 third

the

the

is

asymptote.

down

where

Find

function

down

horizontal

1.14%.

b

exponential

The

spread

of

a

disease

can

be

modelled

by

year. x

the

in

6

Wei

invested

SGD

3000

(Singapore

equation

days

from

y

=

4

the

+

e

,

time

where

the

x

is

disease

the

was

time

rst

dollars)

detected. in

a

bank

that

compounded

paid

2.35%

interest

per

year

monthly.

a

Find

the

disease

a

Find

how

after

six

Find

the

SGD

5000

much

Wei

had

in

the

years.

b

number

in

the

of

years

before

he

Find

invested

JOD

The

the

a

bank

that

paid

4500

(Jordanian

has

JOD

interest

interest

quarterly.

5179.27

in

per

Silvia

is

the

After

six

bank.

years

an

at

4%

will.

per

monthly.

be

days

it

takes

for

infected.

metres,

each

of

a

runner

week

bean

according

to

the

function

h(t)

=

0.25

+

log(2t

−

0.6),

Find

where

t

is

the

time

in

weeks.

the

a

Find

the

height

of

the

plant

after

four

weeks.

annuity

The

of

annuity

annum

Find

in

increases

US$4000

in

Find

the

and

is

is

to

monthly

for

be

ve

paid

the

number

of

weeks

it

takes

height

of

for

her the

uncle’s

to

of

he

rate.

left

days.

annum

b 8

the

dinar)

t>0, compounded

number

people

height,

model in

with

bank.

plant Omar

seven

people

had

14

7

after

of

bank

25 000

b

number

plant

to

reach

a

2m.

years

out

payments.

333

/

7

MODE LLING

15

a

Find

the

R AT E S

value

of

OF

the

C H A NGE :

following

E XPONENTIAL

in

the

AND

LO G A R I T H MI C

were

rst

kept.

The

F UNCTIONS

rst

recorded

n

r

most

simplied

form:

( 5  2

height

was

2

record

tends

m.

As

t

increases,

the

) to

a

limit

of

3

m.

r 1

b

Find

the

value

of

the

following

in

a

the

Find

the

values

of

a

and

b.

2n

(4

marks)

r

most

simplied

form:

( 5  2

)

b

r 1

Calculate

the

record

height

years.

16

$900

year

is

invested

into

annum

total

the

an

at

the

account

earning

compounded

value

tenth

of

the

beginning

4%

annually.

investments

of

per

c

Find

the

the

h

end

20

year.

=

pet

parents

animal,

rock.

She

so

will

not

instead

keeps

it

in

let

she

her

her

has

P1:

a

the

of

age

(t

it,

but

sadly

this

wears

it

a

to

years)

can

piece

the

can

t

for

of

ratio

carbon

then

since

be

r

which

used

14

marks)

in

tree

living

the

was

estimate

Scientists

carbon

estimate

the

to

wood.

of

and

14

in

the

wood.

time

cut

down

-6000 using

strokes

of

(3

dating

They

pet

pocket

value

Carbon

measure

have

a

the

2.8.

sample

P1: Maria’s

marks)

of

Exam-style questions

17

(2

10

each

Find

at

after

the

model

t

=

ln

(r )

away. ln 2

Its

mass

is

given

by

the

equation

where

r




giving

your

answers

to

3

signicant

gures.

the

0, q >

0),

marks) α,

13





α

where

x

areas

Given

Give



0),

(8

marks)

degrees.

y

20

10

x 0 90

369

/

Approaches to learning: Critical thinking,

Making a Mandelbrot!

Communication,

Exploration criteria: Mathematical

cation

(B),

Personal

mathematics

engagement

communi-

(C),

Use

of

(E)

IB topic: Complex

numbers

ytivitca noitagitsevni dna gnilledoM Fractals

You may have heard about fractals.

This image is from the Mandelbrot set, one of the most famous examples

of a fractal:

This is not only a beautiful image in its own right. The Mandelbrot set as

a whole is an object of great interest to mathematicians. However, as yet

there have been no practical applications found!

This image appears to be very complicated, but is in fact created using a

remarkably simple rule.

Exploring an iterative equation

Consider

this

Z

)

iterative

equation:

2

=

(Z

n+1

+

c

n

2

Consider

a

value

of

c

=

0.5,

so

you

have

Z

=

(Z

n+1

Given

that

Z

=

0 ,

nd

the

value

of

nd

the

+

0.5.

Z

1

Now

) n

2

value

of

Z 3

Repeat

What

for

do

a

few

you

more

think

is

iterations.

happening

to

the

values

found

in

this

calculation?

A dierent iterative equation

2

Now

consider

a

value

of

c

=

−0.5,

so

you

have

Z

= n+1

Again

start

with

Z

=

(Z

)

+

0.5.

n

0

1

What

happens

this

time?

370

/

8

The Mandelbrot set

The

Mandelbrot

sequence

set

starting

consists

at

Z

=

0

of

all

does

those

not

values

escape

to

of

c

for

innity

which

(those

the

values

of

c

1

That

The

the

is

calculations

only

value

It

could

If

you

of

be

c

of

in

the

the

complex

pick

Consider

part

a

story

off

of

the

does

form

number

with

a

to

innity).

however.

function

complex

starting

zoom

ytivitca noitagitsevni dna gnilledoM

where

value

a

+

for

of

not

need

to

be

real.

bi.

c,

Z

is

it

=

0

1

+

in

+

the

0i,

Mandelbrot

rather

than

set

just

or

not?

“0”.

1

2

Consider,

for

example,

a

value

of

c

of

i,

so

you

have

Z

=

(Z

n+1

Find

)

+

(1

+

i).

n

Z 2

Repeat

to

nd

Z

,

Z

3

Does

this

Repeat

diverge

for

a

few

,

etc.

4

or

converge?

more

values

(Does

of

it

zoom

off

to

innity

or

not?)

HINT

c

One way to check is

Try,

for

a

=

c

example:

0.2

–

to plot the dierent

values of Z

0.7i

, 1

b

c

=

–0.25

+

Z

0.5i

, Z

Z

2

, 3

, etc, on an Argand 4

What

Try

happens

some

if

values

c

=

of

i?

diagram and to see

your

whether the points

own.

remain within a You

may

nd

a

GDC

will

help

with

these

calculations

as

it

can

be

used

boundary square, or for

complex

will

get

not

be

“big”

a

numbers.

able

lot

to

When

give

quicker

a

the

value.

than

number

You

may

gets

also

very

big,

notice

the

that

calculator

some

numbers

you could calculate

the modulus of

others.

each value and It

is

clearly

a

time-consuming

process

to

try

all

values!

see whether this is There

are

programs

that

can

be

used

to

calculate

the

output

after

several

increasing. iterations

of

c

The

belong

to

of

black)

all

you

the

Mandelbrot

diagram

(say

when

set

all

a

number

Mandelbrot

those

and

input

diagram

values

those

is

of

that

c

set

for

and

created

which

remain

c.

This

which

by

do

will

escape

bounded

which

values

don’t.

colouring

not

indicate

in

points

to

on

an

innity

another

Argand

in

colour

one

(say

colour

red).

Ex tension

•

Try to construct your own spreadsheet or write a code

that could do this calculation a number of times for

dierent chosen values of c

•

Explore what happens as you zoom in to the edges of

the Mandelbrot set.

•

What is the relationship between Julia sets and the

Mandelbrot set?

•

What is the connection to Chaos Theory?

•

What are Multibrot sets?

•

How could you nd the area or perimeter of the

Mandelbrot set?

371

/

Modelling with matrices:

storing and analysing data

9

Concepts

■

Systems

■

Modelling

Microconcepts

■

Matrix

■

Array

■

Order of matrix

■

Element

■

Dimensions of matrix

■

Eigenvalue

■

Eigenvector

■

Fractal

■

Determinant

■

Inverse

■

Identity

■

Linear system

■

Transformations

■

Iterative function systems

How can urban planners determine the

population of dierent cities if people

are constantly moving bet ween them?

How can athletes ensure

their diet provides all the

nutrients they need?

How do computers

simulate movement?

How do we send messages

securely?

372

/

Is

it

possible

The

Koch

found

When

ice

is

create

snowake

an

an

ice

ice

crystal

called

By

a

in

to

be

is

found

and

mathematical

carrying

out

stage

using

attempt

at

to

Koch

we

mathematical

describe

see

level

structures

describing

we

process

the

each

and

process

this

of

an

magnied

self-similarity

identifying

Each

is

snowake

the

formulae?

patterns

like

those

crystal.

crystal

can

a

the

can

of

that

versions

magnication.

that

display

smallest

create

smaller

the

piece

it

in

entire

This

are

this

of

property

called

fractals.

pattern

structure

the

with

simply

by

repeatedly.

snowake

is

described

by

the

iterative

process

shown.

We

begin

creating

•

by

an

dividing

equilateral

Write

down

could

use

square

line

shape

graphs

in

of

millimetre

segment

triangle

step-by-step

the

Construct

•

the

in

the

stages

graph

1

1

to

paper

2

0

of

using

the

the

the

into

third

describing

create

and

stage

middle

instructions

stage

in

of

the

Koch

in

stages

unit

0

Stage

1

Stage

2

segment.

2,

snowake

1

Stage

and

geometrically

shapes

scale

thirds

=

how

3

you

and

4.

on

30mm

y

like

the

one

shown.

Determine

the

exact

coordinates

of

each

1

vertex

Is

•

and

there

a

endpoint

pattern

points

that

points

on

you

in

for

the

could

each

gure.

values

use

to

of

the

coordinates

determine

the

of

each

coordinates

of

of

these

the 0

stage

3?

x

Save

your

work

as

you

will

need

the

results

to

check

your

answers 1

to

the

investigation

at

the

end

of

the

2

3

unit. y

1

Developing inquiry skills

0

Where else do fractals appear in the real world?

x

Can we use fractal formulae to model real-life situations? How might

1

2

3

these be useful?

Think about the questions in this opening problem and answer any

you can. As you work through the chapter, you will gain mathematical

knowledge and skills that will help you to answer them all.

Before you start Click here for help

You should know how to:

1

Identify

and

geometric

write

the

general

term

of

a

1

Find

a

formula

sequence.

The

general

term

for

the

nth

of

4,

−8,

16,

−32,

18,  12, 

8, 

…

is

u

=

4

×

Find

3

eg

The

sum

sum

 

(−2)

2

the

the

1

n

2

in

 ,...



n

term

16

 sequence:

eg

with this skills check

Skills check

of

of

8

geometric

terms

of

each

series.

the

series

Determine

3

the

sequence

Write

an

sum

in

of

the

question

expression

for

S

rst

15

terms

of

2.

for

the

series

n

12

+

6

+

3

+

1.5

+...

is

2

 1 

3

 1 

 1 

8

12

S

1

 0

5



6  6



 6 



23



9

2

 

 6









2









then

use

it

to

 2



8

1  0

5 nd

S 12

373

/

9

MODE LLING

9.1

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Introduction to matrices and

matrix operations

Computers

called

leads

an

you

store

array.

to

a

large

amounts

Understanding

study

of

of

data

how

matrices

and

using

data

is

a

programming

stored

matrix

and

structure

manipulated

operations.

InternationalA matrix A with dimensions m × n is a rectangular array of real (or complex)

numbers containing m rows and n columns where a

mindedness

refers to the i,j

element located in row i and column j. Such a matrix is said to have

English mathematician

order m × n.

and lawyer, James

Column

1

Column

2

Column

n

Sylvester, introduced

the term “matrix” in the

19th century and his

a

a 1, 1

...

a

1, 2

a

1, n

...

a 2, 1

row

friend, Ar thur Cayley,

a

2, 2

1

row

2, n

2

advanced the algebraic

A =

aspect of matrices.

a

...

a m, 1

a

m, 2

 1

m

3

4

2

1

5

0

7

3

4

2

6

row

n

m

 For

example,

matrix

P



has

dimensions

3

×

4

where

  

a

is

the

element

located

in

the

2nd

row

and

3rd

column

and

has

a

2,3

value

If

m

of

0.

n

(ie

=

then

we

3

the

say

Consider

turing

Note

that

number

that

the

3

matrix

×

P

5

different

is

a

of

P

contains

columns

square

matrix

C

products

showing

for

5

1



of

the

different

1

×

4

equal

matrix

Manufactuer

Product

is

3

=

to

12

the

order

cost

elements.

(in

number

of

rows)

n.

euros)

of

manufac-

manufacturers.

Manufactuer r

3.50

4.25

8 .2 5

10.50

2

Manufactuer

3.82

3

Manufactuer

4

Manufactuer

4 .10

3 .95

9 .45

7 .80

 C



Product





2

.75





 Product

5



3

10 .50

9 .75

8.95

11.00

10.80



For

example,

4

given

is

by

the

cost

element

of

producing

a

and

is

product

3

with

manufacturer

€11.00.

3,4

Matrices A and B are equal if and only if the matrices have the same

dimensions and their corresponding elements are equal.

374

/

9 .1

Three

models

of

manufacturing

elements

costs

(in

of

a

particular

processes

matrices

USD)

for

A

each

each

and

of

B

carried

three

125

35

275

40

the

models

shipping

at

each

factories.

and

of

two

factories.

1



55

and

B

+ 1,1

1is

b

=

425

50

210

35



the

325

total

1

Model

2

Model

3



3

65



that

 Model



 Model

indicates

300







a

Shipping

 2

 180

The

manufacturing







manufacturing

cost

of

1,1

$425.

Similarly,

adding

a

+

b

3,2

total

different

two

Manufacturing

Model



the

in

 Model



Model

out

undergo





Adding

television

Shipping

 A

of

represent

the

Manufacturing



brand

arbegla dna rebmuN

Matrix addition

shipping

cost

of

Model

3

is

=

120

and

indicates

that

3,2

$120.

TOK

3 0

3 0 0

5 0

4 2 5

8 5

2 7 5

4 0

2 1 0

3 5

4 8 5

7 5

1 8 0

5 5

3 2 5

6 5

5 0 5

1 2 0

mathematics, however

abstract, which may not

someday be applied

to phenomena of the

To add or subtract two or more matrices, they must be of the same order. You

real world.” – Nikolai

add or subtract corresponding elements.

Lobatchevsky

dna y rtemoeG

A + B =

“ There is no branch of

1 2 5

y rtemonogirt

Thus,

Where does the power

of mathematics come

The

zero

matrix

0

is

the

matrix

whose

entries

are

all

zero. from? Is it from its

0

For

2

×

2

0

matrices

is 0

0

0

0

0

0

0

the

zero

matrix,

for

2

×

3

matrices

it

is

0

as a language, from the

axiomatic proofs or from

.

For

any

ability to communicate

its abstract nature?

matrix

A,

A

+

0

=

A

and

A

–

0

=

A.

Scalar multiplication for matrices

Consider

the

shipping

and

110%

its

matrix

of

situation

where

manufacturing

old

cost.

representing

manufacturer

costs

Therefore,

the

new

by

10%.

=

35 ( 1 .10) 

is,

each

both

each

value

 137 .50

 275 ( 1 .10)

40( 1 .10)

  180( 1 .10) 

That

multiplying

 10 A

increases

new

by

its

cost

1.10

is

gives

a

costs.

( 1 .10)  125

1

A

 302

 







55 ( 1 .10)

50

44

HINT 198



38 .50

60

5



A scalar is a

quantity that is oneIn

this

way,

we

dene

scalar

multiplication

of

matrices.

dimensional. Some

examples of scalars

Given a matrix A and a real number k then kA is obtained by multiplying

every element of A

by k where k is referred to as a scalar.

are temperature and

weight.

375

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Example 1

 Find

a

and

b

if

2P

−

5Q

=

0,

P



1

2

2b 



 5

 3c



1

 5

0

4



 0 

 6c

0

 5

 



 1



4b

 0







 2  5a

1

a





 and

3c

Q



1

1

The

0

1

a

 

 

0 .4

zero

matrix

elements 0

are

is

a

matrix

all

of

whose

0.

0

0 

 

2b

 



1

 0

0



2 2  5a



0



a



If

two

matrices

are

equal

then

their

5 corresponding

elements

are

equal.

5 4b

 5



0





b

4

5 6c

 5



0



c



6

Exercise 9A

1

Given

3

that

Consider

W , 3

 3

 2

5 

and



 





2

4

6

,

B







 

1 

0

R

5

2

4

0

3









3

T,

U,

V ,

3

2

1

4

1

7



 S

7 













  9

4





=



5 8

S,

below.



 5



R,

,

4 



X

matrices



 A

the

6

2 

 and

C



,

write

down

the

 8

values

3

2 

 9 

 3  1

1

T

7

=





6

 of

a

,

a

1,2

2

The

,

b

2,3

,

c

3,1

matrix

Q

,

and

each

of

2,2

each

of

four

the

number

of

 2

units

products

that

are



manufacturers

where

by

amount

manufacturer

of

product

i





 1

1 

 4

X

=

0

340

575

420

100

200

375

425

8

7





2

3

2

1

 



 3



2 

 

145

W

j

produced

250

1



q

j.

 420

0



produced

 the

2





0

i

represents

7





by

 6

3,1

describes

three

12



c

V of

 U



Find

 6

4

each

of



the

following

if

possible.

If

not

 Q



possible,

explain

why.

Check

your

answers

 using



technology.

235



a a

Write

down

the

dimensions

b

Write

down

the

value

of

of

q

c

and

a

4

number

×

of

1

matrix

1

describe

d

meaning.

Find

products

representing

produced

−

W

c

4U

+

S

by

the

1 T

3

T

R

Q

2,4

its

b

3W



V

2

total

each

manufacturer.

376

/

9.2

Determine

the

points:

arbegla dna rebmuN

a

 32450



i

18725

2X



4

The

matrix

P



describes 

the

prices

1

24175

ii





(in

USD)

rate

is

of

four

7.5%

of

types

the

of

cars.

purchase

The

sales

iii

tax

Write

down

T

=

kP

−X

price.

b a

X

3

19250

where

T

Show

each

of

these

the

points

on

a

represents graph.

amount

of

tax

paid

for

each

car.

c b

The

matrix

C

represents

the

cost

Make

a

points each

car

including

equation

5

A

column

that

vector

tax.

describes

can

be

Write

C

in

a

as

a

of

2

P

×

6

Find

The

point

P

=

(−3, 6)

is

written

real

as

X





B



9.2



in

by

and

B



μ

set

of

kX

and

 10

λ

if

12 

 

.

of

8

4

 9 ,

where

A





12

 

 6

3

2

1 

the



6 

1

3

y rtemonogirt

form

values

 

 3 matrix

described

1 A

matrix.

about

matrix

terms

regarded

conjecture

of

Matrix multiplication and

properties

Consider

three

sold

a

matrix

types

at

a

of

café

F

whose

medium

on

each

Frapp é

day

drinks

of

the

represent

(frappé,

rst

Cappuccino

28



Sun

size

elements

week

Iced

32

the

number

cappuccino,

in

of

and

each

iced

dna y rtemoeG

the

of

coffee)

June.

coffee

16







34



51

8



Mon 

Tues





40

31

15

37

24

20

 W





Wed 

Thurs







75

47

29



Fri









38

29

19





Sat 



47



The

price

for

a

34

frappe

is

12

€4.00,

cappuccino



is

€3.50

and

iced

coffee

is

€2.50.

The

following

each

calculation

can

be

used

to

determine

the

revenue

R

for

day:

# frappes day

revenue

# cappuccinos ×

=

price

+

sold

Sunday

$264

=

28

# iced ×

price

+

sold

×

4.00

+

32

×

price

×

2.50

coee

×

3.50

+

16

377

/

9

MODE LLING

Similarly,

the

the

products

revenue

of

corresponding



28

WITH

 4 .00 

the

for

M AT R I C E S :

each

number

prices

for

of

each

day

each

34

on

 32 (3 .50)  16(2 .50) 



40

 4 .00 

4

00 

=



 31 ( 3 .50)  15 (2 .50)

 4 .00 



75  4 .00 



24(3 .50) 

20(2 .50)





 4 .00 

 34(3 . 5 0)  12 (2 .50)

A

more

way

of

week.

That

is,

50

282

537 

expressing

301



 337





compact

the

the



 47



of

and



29(3 .50)  19(2 .50) 



sold

306





 4 .00 

summing





 38

D ATA

264



4 7( 3 .50)  29(2 .50)



day

A N A LY S I N G







by

drink

each

334 



37

of

AND



 51 (3 .50)  8(2 .50)



R

type





obtained

drink





is

S TO R I N G

R

would

 4

be

to

represent

the

price

00

 of

each

drink

as

a

3

×

1

matrix

P



3

50

2

50

  

 28

32

16 











 34

51

8



Sunday’s

264

revenue

=

28

 4 .00 

 32

334.50

3 .50 

 4

40

31

15



 4



00 

306



 28

32

16 

 3

50

  so

that





WP



37

24



 3

20

50













 



75

47

29





38

29

19

47

34

12

 

2

282

264

 2



50







50



 

 

R

 2 .50 

00 

 

 16

537



















Friday’s

301

revenue

=

38

 4 .00 

 29

337

3 .50 

 4

 38

29

19 

 3

50



301

 2

If P = AB then each element of P (named as P

 

 

 2 .50 

00 

 

 19

50 

) is found by summing the i,j

products of the elements in row i of A with the elements in column j of B

 1

 For

example,

if

A



1

5

3

 



 1

4

2

and

B



3

4

2

1

then



2  

 1

 P



1

5

3 

 

 1

4

2



2



 3

4







 2

1



 

1 (1 )  5 (3 )  3 (2 )

1 (2 )  5 (4)  3 ( 1 )

 

 1 (1 ) 

4(3 ) 

2 (2)

1 (2) 

4(4) 

2 1 ( )

 22







19

 

 7

16

378

/

arbegla dna rebmuN

9.2

Investigation 1



Consider the matrices

A



1

2

2

0

3

1

 D

  1

3

2

,

B



1

3

 1

,

 

0

3

C



4

,

 

2

 2



, and

E



.

1









1

1



1

1

 2

 1





4

 



2



Factual

Find the products of AB,

AC,

BD,

DE, and BE.

Complete the table below to determine the dimensions

products AB, AC,

Product

BD,

(order) of the

DE, and BE

Dimensions of rst

Dimensions of

Dimensions of

matrix

second matrix

product

2 x 2

AB

2 x 3

AC

DE

BE

3

Factual

Based upon the results in question 1 explain how you could

dna y rtemoeG

y rtemonogirt

BD

determine the dimensions of the product of two matrices using the

dimensions of the two matrices being multiplied?

4

5

Explain why the products BA,

Conceptual

CB,

and

DB cannot be determined.

When does the product of two matrices exist?

TOK

How are mathematical

denitions dierent

If A has dimensions m × n and B has dimensions p × q

•

from denitions in other

then

the product AB is dened only if the number of columns in

areas of knowledge?

A is equal to

How are mathematical

the number of rows in B

denitions dierent

(that is, n = p) from proper ties,

•

and when the product does exist, the dimensions of the product is

m × q. axioms, or theorems?

Example 2

A

diet

research

participants

in

project

the

consists

survey

is

of

adults

given

by

the

and

 

both

sexes.

The

number

of

75

Children

180

 

of

matrix:

Adults

A

children



Male

 110

250



Female

Continued

on

next

page

379

/

9

MODE LLING

The

number

adult

is

of

given

WITH

daily

by

M AT R I C E S :

grams

the

of

S TO R I N G

protein,

fat,

and



D ATA

consumed

by

each

child

and

b

Explain

Carbohydrate

20

25

 Adult

 

AB.

carbohydrates

Fat

 15 B

Determine

A N A LY S I N G

matrix:

Protein

a

AND

 8

the

16

meaning

20

of



c

AB

Child

Explain

why

BA

does

not

exist.

3,2

75



a

AB



 

250

20

the

matrices.



75 (20)  1 80( 16)

110( 15) 

250( 16)

110(20) 

4380

5475 

6200

7750

 

AB

16

Multiply

 8

 

75 ( 15 )  180 (8)

 2565

b

25 

75(25)  180(20)

 



20

  110

 

180   15

250( 16)

110(25) 

250 (20)

 5650

=

110(25)

+



250(20)

=

7750

and

represents

the

total

Total

carbohydrates

2,3

consumed number

of

carbohydrates

consumed

by

females

in

the

110

c

BA

is

not

possible

since

the

number

of

columns

in

B

is

to

the

number

of

rows

in

the

adult

females

not is

equal

by

project.

110(25)

and

the

A total

carbohydrates

consumed

females

is

by

250

child

250(20).

Investigation 2

 2

Use the matrices

A



 1

,

 

1

3

4

B



and



1



 2

5

6

2

to answer each question. 4

3

= ba, for example 5 × 4 = 4 × 5 = 20.

Is matrix multiplication commutative?

a(b + c) = ab

+

ac and (b

+ c)a = ab + ac

Determine each of the following then describe your observations.

i

A(B + C)

ii

(B + C)A

iii

AB + AC

iv

BA + CA

Based upon your observations is matrix multiplication distributive? Is

correct to write

4

1

 

Multiplication of real numbers is distributive since

3



Multiplication of real numbers a and b is commutative since ab

Determine AB and BA.

2

C

(B

+

C)A = AB

+

A(B +

C) = AB + BC?

Is it

CA?

Multiplication of real numbers is associative which means that

associative? You can answer this by showing whether or not

a(bc) = (ab)c.

Is matrix multiplication

A(BC) = (AB)C



5

Consider the marix I =

1

0

0

1



 

a

6

Find AI and IA

Conceptual

b

Write down what you notice.

How do matrix and matrix multiplication work in terms of the commutative and

associative proper ties?

380

/

arbegla dna rebmuN

9.2

Proper ties of multiplication for matrices

For matrices A,

B, and C:

•

Non-commutative AB ≠ BA

•

Associative proper ty A(BC) = (AB)C

•

Distributive property A(B

+

C) = AB

+

AC and (B

+

C)A = BA +

CA

These proper ties only hold when the products are dened.

•

In A(B

•

In (B

+

+

C),

B

+

C is pre-multiplied by A

C)A,

B

+

C is post-multiplied by A

Multiplicative identity for matrices

If

any

A

In

is

real

any

other

identity

identity

number

square

words,

matrix,

any

then

real

and

I

square

the

is

the

matrix

answer

is

0



0

0

1



0









0

0



1

identity

is

the

The multiplicative identity of a square

 1

numbers

is

1

since

a

×

1

=

1

×

a

=

a

a.

matrix

if

for

pre-

matrix,

or

then

×

A

post-multiplied

original

I

=

by

I

×

A

=

A.

the

matrix.

n × n matrix A is given by

dna y rtemoeG

for

multiplicative

y rtemonogirt

The



 I

.

 n







Note that both A

× I

= A and I n

× A = A hold only in the case where A is a n

square matrix.

Unless needed for clarity, n is not normally written, and I is used alone to

denote the identity matrix, whatever the size.

Exercise 9B

1

Using

the

2

matrices

Given

2  1 A



2 



 2



3

 ,

B





2

4

1

0

 3

,

0 

 C

2

,

D

and











3



2

3

0

4

×

m

matrix,

determine

the

Q

is

values

a

of

m

n

if

PQ

is

a

4

×

3

matrix.

Find

a

matrix

that

has

the

effect

of

4

summing

the

entries

in

every

row

of

a

,   a

 3

2

b

c

e

f

g

h

i

a

 b

1





matrix,

a

1 







2

1



 1





is



3 1



n

P

1

 

×

that



 3

×

3

matrix.

That

is,

if

A



d

nd

 2



 E





 3





1

,

 

and

F

 









 

 2



3 

1 

4

,  

 5



each

of

the

c



 a

nd



2

following

matrices

if

possible.

matrix

B

such

that

AB

 



d



e



f



 If

it

is

a

(C

d

CF

not

+

possible,

F)A

state

b

DE

e

2A

the

reason.

c

−



. 





g

 h  i



ABC

3I 2

381

/

9

MODE LLING

WITH

M AT R I C E S :

3

2

1

1

5

6



4

Consider

the

matrix

A



S TO R I N G

A N A LY S I N G

New

 

AND



federal

Find

a

matrix

B

such

that

AB

b

Find

a

matrix

C

such

that

CA

c

Explain

=

facturer

A.

=

reduce

by

60%,

A

does

not

have

levels.

A.

The

A

manufacturer

products

P

,

P

1

plant

makes

,andP 2

L

,

L

1

gives

the

carbon

oxide

daily

,

each

types

each

of

,andL 3

during

its

daily

emissions

dioxide,

and

40%

to

reduce

and

of

its

of

nitric

carbon

oxide

current

takes

emissions

corrective

to

meet

the

daily

standards.

Let

matrix

are

the

.

(in

Matrix

kilograms)

dioxide,

the

A

the

and

nitric

The

daily

kilogram

nitric

dioxide

total

R

matrix

whose

number

released

cost

of

plants

given

is

that

L

nitric

by

L

N

USD)

carbon

and

for

of

each

measures.

(in

dioxide,

oxide

35

a

daily

such

1

20

be

corrective

matrix

of

manufacturing

sulfu ur

N

entries

kilograms

of

four

product.

monoxide



manu-

of

4

sulfur

carbon

P

its

manufacturer

pollutants

L

amount

produced

of

at

2

monoxide,

process

three

3

locations

the

identity.

a 5

that

a

minimum multiplicative

require

sulfur

20%,

measures why

laws

.

monoxide,

a

D ATA

of

Write

AR

oxide

L

nd

removing

at

the

N

each

sulfur

each

matrix

after

down

then

monoxide,

the

2

=

product

of

the

four

C:

L

3

4

15  10

1

8

12

8

2

4

3



carbon

monoxide

 A



P 2

18

28



12 C





 3

sulfur



dioxide



 P 3

45

72



32



 7

9

6

11

nitric



b

Find

oxide



AC

and

describe

the

meaning

of

its

entries.

Investigation 3

 2

Consider the matrix

A

=

3 

 

 4

7

 2 ,

B

=



0



 



0

3

0

 ,

C

=



2

 

 3

0

2

= A

×

A show that

A







2

Find B

2

, C

0

 

 0

1

 1





and

I

 2

0 

 

 0

1



(3)

 2

2





2

 2

Using the fact that A

D

2

2

1

 2 ,

.

2

4

7

2

, and (I

) 2

2

 a

b

 a

2

b

2

3

If

A



under what conditions is

 

A





 2

c

d 

c

? Does this same condition apply to all

2

d

n × n matrices? Explain.

2

3

4

k

+

4

Determine D

5

Without using technology write your own denition for the

, D

, and D

.

Write a formula for D

where k∈ℤ

.

0

zeroth power of a square matrix A

.

Explain

0

your reasoning. How does the GDC interpret

 a

A

and is this result consistent with your denition?

b k

If

A



then

 

A



A 

A   

A.

    c

d k

factors

of

A

k

 a

0 

 a

0

k

In the case where b = c = 0 then

A



 

 0

d









.

k



0

d

382

/

Exercise 9C

2

Complete

and

each

exercise

properties

check

your

you

using

learned

answers

using

the

in

4

denitions

this

section

In

algebra,

(a

+

2

then

 1 If

A



2

3



1

=

if

B



0

possible.

(T

+

+

properties

.

 

If

not

2ab

+

b

X)(T

−

X)

AB

c

A

1

explain

b

BA

d

B

5

(T

+

X)(T

(T

+

X)

Show

of

matrices.

−

X)

≠

questions

matrices

and

b

a,

to

(T

+

Explain

and

b∈ℝ.

X)

using

why

2

T

−

X

and

2

≠

2

T

+

2TX

that(SU)

matrix

6 following

2

−

+

X

why.

2

the

a

where

2

1

possible,

2

Use

=

Find

2

a

b)

2

a

2

each

−

2

2

3





b)(a

technology.

4 and

+

2

b)

Expand

1

(a

arbegla dna rebmuN

9.3

2

≠

S

2

U

.

Explain

why

using

multiplication.

Show

that

What

does

2(RT)

=

(2R)T

=

R(2T).

answer

this

suggest

about

scalar

2–6.

multiplication?

 R







3

7

3

2

1

4

1

3



 ,

S























5

4

7



5

 3

,



7

Consider

the

T



2



 

1

3

3

U



B



 4

5

2

and

2

1

5 .

 

1

3

,





5

 

2 

 ,



2

 2

1

A



 4



matrices





 0

a

Show

that

AB

=

BA

=

1

1



I 2

 10

b 1 W



2

3

2

1



 

 3

7

 ,

X





 

Determine

 6

4

8

Michael

Calculate

each

of

the

following

if

not

possible,

explain

b

SR

c

TR

WR

+

3

(S

Use

UR

−

+

9.3

R

conjectures

=

A

A

and

B

n

B

n

b

(kA)

n

=

k

n

A

d

X

W(S

−

you

agree

with

Michael’s

claims,

prove

U) If

you

disagree,

write

down

counter-

examples.

properties

=

following

why.

U)W

the

the

matrices

n

(AB)

them.

e

square

n

If

a

all

possible.

a If

makes

 for

2

(2AB)

2 

dna y rtemoeG

1 

y rtemonogirt

0



(U

+

of

I)R.

matrices

Verify

to

your

show

that

answer.

Solving systems of equations

using matrices

One

it

of

gives

the

you

equations.

expressed

For

many

an

a

matrix

 5y



to

of

express

containing

n

matrix

and

multiplication

solve

variables

 3 x

4s

 7y

 3t



equivalent

to

23 

 2z



0

and

n

of

that

linear

equations

can

be

5   x 

 

 

3



 4

7

 

3

 y

 35 



 

2  

 23

s 



0

 2s



2t

 3z



6

 

is

equivalent

to

  2

2

3

 

 t

 

 6

 











.

 6s



systems

is

equation.

 10 is

 

way

35 





system

applications

example,

 10 x



efcient

Any

as

useful

 t



z





2 

  6



1

1

z  

2

383

/

9

MODE LLING

In

general,

AX

a

n

=

B

×

1

an

where

you

as

2x

in

that

=

10,

solving

is

n

×

of

n

S TO R I N G

equations

matrix

the

can

be

containing

constants,

and

AND

A N A LY S I N G

expressed

the

X

is

in

the

coefcients,

an

n

×

1

D ATA

form

B

is

matrix

variables.

the

in

system,

=

B

where

the

an

interested

AX

M AT R I C E S :

system

containing

the

are

equations

notice

A

matrix

containing

Since

n × n

WITH

our

looks

you

matrix

the

values

goal

similar

would

equation

is

to

to

the

variables

determine

the

simply

by

of

more

divide

dividing

X.

At

familiar

both

both

that

rst

linear

sides

sides

solve

by

by

glance

you

equation

2.

the

all

such

However,

matrix

A

would

B not

lead

you

to

X

.



Unfortunately,

matrix

division

is

not

dened

in

A

the

same

answer

way

to

as

this

other

matrix

problem

operations.

utilizes

the

The

key

multiplicative

to

unlocking

inverse

the

property.

1 Recall

that

if

a

is

a

real

number

then

is

referred

to

as

the

a

1 multiplicative

inverse

of

a

since

1

a 

 1 and

a

2

3 and

-

-

3  

 are

multiplicative

inverses



since

For

example,

3

2  

 2

 1.

 a

a

 1 and

  2



 

International-

 3



mindedness 2  

 

3  



 1 .

  3



Note

that

there

are

two

conditions

for

which

a

number

 2

 

Matrix methods

 were used to solve

has

a

multiplicative

inverse.

First,

both

the

number

and

its

inverse simultaneous linear

must

have

a

product

equal

to

1

and

second,

this

product

must

satisfy equations in the

the

commutative

property

of

multiplication. second century BC

Similarly,

n

if

and

AB

=

I

×

n

matrices

BA

=

I

n

A and B

where

I

n

are

is

called

the

multiplicative

identity

matrix.

in China in the book

inverses

For

“Nine Chapters on

example

n

the Mathematical Ar t”  3 A



1

 

 2



Thus,

its



1

1 

1

the

 

 2

central

multiplicative

through

this

2

1 

  2

are

 

 

B

1

 3 AB

and

1

1



3

0



0

inverse?

for

a

BA



1

is,

The

2

×

 

given

a

2

1

  2

square

following

1   3

1

 and

question

process

written during the Han

since

Dynasty





inverses

3

 1 

multiplicative

3

 

matrix

 2

A

investigation

1

 1 



will

 

how

0

0

do

1

you

guide

nd

you

matrix.

Investigation 4

8



Determine the multiplicative inverse of

A



 w

i

Let

B



5

4

.



x

 



 

a

6 

 y

be the multiplicative inverse of A. Use the fact that

z

AB = I to show that



8w

 6y

 1





ii

5w



4 y



0





 and







8x

 6z



0



. 5 x



4z

 1 

Solve each system in i to nd the values of w, x, y, and z

Check your

answer by verifying that AB = I and BA = I.

384

/

 a

b

Consider that you are given a matrix

A



b

where a,

 

c

b,

c and d are

d

real numbers and you wish to determine the multiplicative inverse of

 w

Let

B



be the multiplicative inverse of A. y

z

aw

Use the fact that

AB = I to show that



by

 1





ii

A.

x

 

i

arbegla dna rebmuN

9.3

cw



dy



0

ax

 and







 bz



0

 cx



dz

 1



Solve each system in i to nd the values of w, x, y, and z in terms of a,

b,

c, and d.

Verify that

AB =

I and BA =

I.

The multiplicative inverse of a 2 × 2 matrix International

reveals

that

if

A



b then

 c



by:

the

inverse

of

A

mindedness

is

d Carl Gauss rst used

the term “determinant” 

1

d

b

1

A



where

 ad

 bc



c

ad

−

bc

≠

0

in 1801.

a

 a You

can

thus

determine

the

inverse

of

using



 

steps.

b 

c

d

the

following



ad

1

Subtract

product

the

of

product

the

of

the

numbers

on

values

the

on

main

the

dna y rtemoeG

given

4

y rtemonogirt

 a Investigation

minor

diagonal

from



bc

the

diagonal.



2

Interchange

sign

of

the

the

elements

numbers

on

on the

the

main

minor

diagonal

and

change

the



diagonal.

d



b

 ad

1

A d





b 

3

Divide

each

element

of

by



will

see

chapters

in

that

later

the

sections

value

( ad

of

−

−

ad

 bc

c

a

bc). 

c



You

(ad

 bc



a

this

bc)

 ad

chapter

has

as

well

geometric

as

in

future

signicance.

 bc

ad

 bc

E X AM HINT

For

this

Finding the reason,

this

value

is

given

a

name

and

is

dened

as

the

determinant

multiplicative of

matrix

A

and

is

written

as

det

A

or

|A|.

Restating

the

formula

inverse for square

−1

A

for

you

now

have:

1

d

b

c

a



matrices beyond a

1

A



where

 A



ad

−

bc

≠

0

2 × 2 requires using

technology, and you

1

In

the

case

singular

where

matrix

|A|

or

=

0,

A

does

not

exist

and

A

is

said

to

be

a

will need to know

non-invertible .

how to do this for

your exams.

385

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

E X AM HINT

Example 3 

Find

the

inverse

1

2

3

4

In examinations



of



you should be able



to demonstrate the

use the formula  4

1

2



 1



d

b

c

a

1

1



4







 3

2   3

1 

A



















4

1



 A



to compute the 2

2





1 Either

of

these

two

forms

can

be

inverse of a 2 × 2 2

3

1







1 .5

0 .5

used,

choose

whichever

is

most



matrix but are appropriate

for

the

context.

expected to use the

matrix utility using

technology to nd

the inverse of a

Example 4 3

2

2

2

3

6

1

1

 4

matrix that is larger

 Use

technology

to

determine

the

inverse

of

P



than a 2 × 2. .

  

1

Verify

that

PP

1

=

P

P

=

I 3

1

1

1

12

12

12







1



1

P





2

4

15

15



 3  1

11

7

   6



30

30



1

1

1

12

12

12



 4





2  



1

PP

3

2

3



1

2



4

 

3  6

1

1



15

0

0 

1

0







 



 1



  2



 0









 0



0

I 3

1

 1





15

11



7



 

 



 6

1

1

1

12

12

12









  4

1



1

30





PP

30



2

4

3

2 

 

 2



2

3

15

0 

1

0

 0













 

I 3

15





 6

1

1

 1

0

 

3

 



 1

11

 0

0

1 

7



 

 

You

are

 6

now

ready

the

beginning

AX

=

B

for

30

of

to

30



unlock

section

9.3,

the

solution

which

was

to

to

the

solve

problem

the

proposed

matrix

at

equation

X.

386

/

9.3



B

TOK 1

A

1

AX



A

1

B

 pre-multiplying

both sides by A

 Do you think that

1

I

X



A

X



A

multi plicative inverse property

B



B

 Multiplicative

n



one form of symbolic

1

representation is

identity pro perty 

preferable to another?

arbegla dna rebmuN

AX

Example 5

systems

of

equations

 10 x

 5y



35 



a



 7y



4s

rst

 3t

forming

 2z



0



 3 x

by



b

23 

2t

 3z



6



a



35





 10

 

 3 x

 7y



23

 t





z

 



5 



  7

2

 

 y



 1

7

  0

1

 

 x 



 

y



  3

7

0   x 

 





 y





 

AX

=

the

system

in

the

form

B

23

 

 7

1 



 



5

 70  15



 y



5 

 10

 3

7

10

1

 

Pre-multiplying

both

sides

by

A

23

  35 

  3



 35

 

1

 23

AA

=

I 2



 360 

1 

 55



Rewriting

1

5   x 

 10

 3



 35

1

 10





5   x 

3

equation.

 6s

 5y

matrix

 

2s



10 x

a

dna y rtemoeG

the

y rtemonogirt

Solve



 335

I 

X

=

X

2

 72 

  x 

 

 y

1

 11









67



calculation



Use

solution

to

the

system

is

x



4s

 3t

 2z



0

2s



 3z



6



2

 4

 t

3



z

2  

2

   

 6

1

1



 t

3













could

3

7

GDC

done

to

directly

obtain

the

on

a

GDC.

result.

y



11

0

 6

 











2

z  

been

23



s 

 

 2



35

67 and

11

6s





72

2t



5



have

b



10

 11 

The



 The





Continued

on

next

page

387

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

1

s 





 4







2













z 

2 





















t 

3

2

6

3

0

6

1

1

2

1 

  

 3





4







 3



 8



 



 3



The



solution

1 s





to

the

system

4 ,

t



8 ,

3

is

and

z





3

3

Applications to cryptography

Protecting

upon

a

sensitive

process

translating

it

electronic

called

into

an

data

information

encryption.

unrecognizable

such

Data

form

as

a

password

encryption

using

an

relies

protects

algorithm

data

called

by

a

cipher.

To

a

encrypt

number

your

to

characters

bank

each

and

account

lowercase

the

digits

0

password

and

“CharLie578Sam*!”

uppercase

through

9

as

letter

shown

along

with

you

the

assign

special

below.

a

b

c

d

e

f

g

h

i

j

k

l

m

n

o

p

q

r

s

t

u

v

w

x

y

z

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

A

B

C

D

E

F

G

H

I

J

K

L

M

N

O

P

Q

R

S

T

U

V

W

X

Y

Z

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

!

@

#

$

%

^

&

*

“

?

:

;

/

\

0

1

2

3

4

5

6

7

8

9

space

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

Using

table

above

converts

the

password

“CharLie578Sam*!”:

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h

a

r

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e

5

7

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75

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decide

password.

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upon

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the

case

length

you

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will

the

use

a

block

you

block

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will

three

use

to

which

divide

HINT

will

If the word length divide

the

word

into

ve

blocks

giving

the

following

3

×

5

matrix

M

was not a factor of 5

then the remaining

spaces could simply

be lled with zeros.

388

/

 29 



 18 





8

,

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 38



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1

 72

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encode

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 75 

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you

can

18

5

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message,

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9.3

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choose

any

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×

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74

matrix

7

2

1

0

3

1

3

4

2



for

our

 cipher.

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example,

if

you

choose

C



then

our

  

encrypted

message



E

is

  29

7

2

1

0

3

1

 E



CM

 220

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211

253

616

264

23

105

142

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125

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18

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80

125

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document

in

matrix

form

as



y rtemonogirt

 

To

decrypt

the

password

1

E

=

CM

⇔

C

requires

you

1

E

=

C

to

solve

E

=

CM

for

M:

1

CM

∴M

=

C

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dna y rtemoeG

.



1

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21

80

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616

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23

105

142

134

127

80

125

47

95



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

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18

5

75

13 

38

72

45

60

 8





 1

9

74

1

53





decrypted

29

8

1

C

h

a

95

253



the

47

21 1

 

So

125



7





616

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8

253

 23



211

18

r

password

38

9

L

i

5

is

72

e

5

74

75

7

8

45

S

1

a

13

60

m

53

*

!

Exercise 9D

1

Determine

the

multiplicative

inverse

of

each



3

1

4

2

 

1

d

b 

1

matrix

using

A



 det

A



 c

a

.

Verify

c





 0 .25

d 1



3 



0 .8 

 0 .75

0 .2



 

1

your

answer

 2

a

showing

4

3

5

that

 2

b

 

by

AA

=

2

4

I

3

 

1

0

389

/

9

MODE LLING

2

Determine

each

the

the

matrix

WITH

multiplicative

using

elements

M AT R I C E S :

as

inverse

technology

exact

S TO R I N G

values.

AND

5

of

A

expressing

Verify

A N A LY S I N G

manufacturer

different

your

D ATA

table

wishes

products

below

A,

shows

to

B,

the

produce

C,

and

D.

number

four

The

of

minutes

1

answer

by

showing

that

AA

=

I.

required

on

produce 2

1

4

1

2

0

1

1

2



1.2

2.5

3

0.8

1

4

1.5

 0.75

1

each

each

of

four

machines

to

product.



a

b

Machine

Product

Product

Product

Product

A

B

C

D

I

2

1

1

3

II

1

3

2

4

III

2

1

2

2

IV

3

4

1

2

  

2



3

1

  3

4

2



1



c

1

1

 3  The

5

maximum

amount

of

time

available

 0

1



for

3

Write

AX

each

machine

I,

II,

III,

and

IV

is

2



each

=

B

system

then

in

solve

the

the

form

system

of

using

240

minutes,

and

400

how

380

minutes

many

of

minutes,

280

respectively.

each

product

minutes,

Calculate

can

be

manu-

1

X

=

A

B



x

factured

 3z

y



30

available





a

5 x



3x 





2y



z

20



b







y

 7y

 12



 2 x

 5y



20

machine

of

its

6

a

Jonathan

makes

the

conjecture

that

1

(AB)

1

=

A

1

B

.

Show

that

Jonathan’s



 2a

 b

 c



4d



2

does

not

hold

using

the

  1



2

 4



5b



 3c



matrices

4

 4a

A











 3b



d

 1







a





2c

 8d

questions

equations

then

that

solve

to

4–6

0

5

create

system

all

 2

3



B







1



a

represents

the

dene



and

Michelle

does

system

each

using

claims

hold

using

that

the

1

( AB)

=

matrices

B

A

three

Sonya

different

1

A

and

B

situation

matrices.

i

Verify

ii

Choosing

that

Michelle’s

claim

is

correct

Be your

own

2

×

2

matrices

for

variables. 1

2016,

1

above.

of

A

In

3

75



b

4

1



c

sure

all

time.

conjecture

For

uses



 10

z



each





 2x

if

invested

index

a

total

funds

F

of

, F

1

$175,000

, and F

2

.

in

and

B

show

that

(AB)

1

iii

After

Prove

that

(AB)

1

=

B

1

=

B

1

A

.

1

A

using

the

1

fact

3

that

(AB)(AB)

=

I 2

one

year,

the

investments

on

each

each

of

of

was

these

investment

2.5%,

the

combined

year.

money

4.8%,

If

in

$181,615.

made

and

than

F

she

7

invested

Using

the

[James

,

of

her

collected

that

annual

respectively

invested

in

all

showed

average

3.5%

2

of

Data

investments

Sonya

F

value

twice

as

calculate

gains

during

much

the

amount

1

in

each

fund.

conversion

Joseph

contributions

table

Sylvester

to

matrix

on

page

386

(1814–1897)

theory,

–

decrypt

an

invariant

the

English

theory,

famous

quote

mathematician

number

theory,

by

James

who

Joseph

made

partition

Sylvester

fundamental

theory,

and

combinatorics.]

390

/

 224

288

84

192

279

75

38

158

103

96

83

220 

202

50

109

182

49

28

80

71

60

50

150

 Encrypted

arbegla dna rebmuN

9.4

 144

message: 





 83

163

49

89

174

44

15

79

51

51

47

131





 1

2

3

1

1

2

0

1

2

 Encryption

matrix:   

Develop

your

own

9.4

cipher

conversion

characters

giving

For

own

them

and

your

using

table

and

numbers.

encrypted

matrices

to

encryption

Check

your

message,

encrypt

matrix

work

by

encryption

a

by

sensitive

piece

assigning

having

key,

a

and

of

information.

different

friend

numbers

decrypt

conversion

your

to

Design

each

message

of

by

table.

Transformations of the plane

this

section

reection

and

you

need

to

enlargement,

be

as

familiar

with

illustrated

in

the

the

ideas

of

rotation,

example

below.

Example 6

Draw

the

image

of

the

triangle

shown

after

the

following

y

transformations:

a

rotation

b

reection

of

90˚

in

dna y rtemoeG

the

your

y rtemonogirt

8

3

counter

the

line

x

clockwise

=

(or

anticlockwise)

about

(0,

0)

2

0 1

c

enlargement

scale

factor

2

centre

(0,

0) x 0

d

a

stretch

parallel

a

to

the

x-axis,

scale

factor

The

y

1

2.

rotation

drawing

4

and

an

can

L

rotating

be

done

using

shape

from

the

the

L

shape

the

2

tracing

origin

to

required

3

4

paper

the

or

point

angle.

2

1

x 0 –3

–2

–1

1

2

3

4

–1

y

b

Reection

3

in

transformed

other

side

of

the

to

a

the

line

x

=

0

position

mirror

means

an

all

equal

points

are

distance

the

line

2

1

x 0 –4

–3

–2

–1

1

2

3

4

–1

Continued

on

next

page

391

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

c

An

AND

A N A LY S I N G

enlargement

scale

D ATA

factor

2

centre

(0,

0)

means

y

the

image

the

origin

of

each

point

moves

twice

as

far

from

4

as

before.

Their

new

positions

can

easily

3

2

be

calculated

by

2.

by

multiplying

all

their

coordinates

1

x 0 1

d

2

3

4

5

6

A

y

stretch

scale

horizontally)

4

by

2

and

factor

just

leaves

2

parallel

multiplies

the

all

to

the

the

y-coordinates

x-axis

(or

x-coordinates

unchanged.

3

Similarly

for

a

stretch

parallel

to

the

y-axis.

2

1

x 0 1

2

3

4

5

6

Investigation 5

Points in the plane can be represented by their position vectors. In the

y

 3

1

diagram below, for example, the position vector of

A is





C

1 2



The position vectors of the ver tices of the triangle ABC can be put in a single A

B

1

1

matrix

2

3

2

1

 

 1

x

 0 1

1

1

0

0

1

Find the product of

1

2

3

2

1

 

2

3

4

 1



Let the columns of the new matrix be the position vectors of the

1

0

0

1

image of triangle ABC under the

transformation represented by the matrix

2

On a copy of the diagram above draw the triangle ABC and its image after the transformation.

What is the transformation?

3

1

0

0

1

In the same way use the matrix

to nd the image of (1, 0) and (0, 1) under this transformation.

What do you notice about the image matrix?

4

Test your conjecture by considering the image of (1, 0) and (0, 1) under this transformation represented

 a

by the matrix

b 

 

 c

d



392

/

5

arbegla dna rebmuN

9.4

By considering the images of (1, 0) and (0, 1) suggest a matrix that represents an enlargement scale

factor 2, centre (0, 0).

Verify your answer by multiplying the points from the triangle above by this matrix and seeing if all the

coordinates are multiplied by two.

How can you use the points (1, 0) and (0, 1) to nd a transformation matrix?

All standard transformations of the xy-plane, including rotations, reections, stretches and enlargements

can be represented by 2 × 2 matrices provided the point (0, 0) is invariant. These are often referred to as

linear transformations.

 a

A matrix representing a linear transformation can be written

d)

the image of (0,





 

and (c,

c

b

d

where (a, b) is the image of

(1,

0)



1)

6

Find the matrix that represents a rotation of 90° clockwise about (0, 0).

7

Find the matrix that represents a stretch parallel to the

x-axis with a scale factor of 2 and the y-axis

invariant.

a

In the diagram below A and B are the images of (1, 0)

y

and (0, 1) under a counter clockwise rotation of θ B

about (0, 0). Use the diagram to show this rotation is

 cos 

represented by the matrix

θ

.

 

b

 sin 

sin 

cos 

A

dna y rtemoeG

8

y rtemonogirt

Using the same method some general formulae can be found.

What will be the matrix for a clockwise rotation of

magnitude θ? θ

c

Hence write down the matrix that represents a rotation

θ x 0

of 60° clockwise about (0, 0).

9

a

1

The line y = mx can be written as y = (tanα)x, where α

is the angle made with the x-axis.

y

In the diagram below A and B represent the images of (1, 0) and 1

(0, 1) respectively under a reection in the line

a

y = (tanα)x

Explain why the image of (1, 0) has coordinates

(cos2α, A

sin2α)

b

By nding

ˆ OBC in terms of α nd the image of (0, 1) under

the transformation.

α

c

Hence verify that the matrix that represents a reection in

α x

  cos2

sin 2 

C

0

1

the line y = (tanα)x is

 

d

By rst nding

sin 2 

cos2 

α, determine the matrix that represents a

reection in the line

y

=

3x

B

10

By considering the images of (1, 0) and (0, 1) nd the general

matrices for: –1

a

a one-way stretch, parallel to the x-axis, scale factor k

b

an enlargement scale factor k, centre (0, 0).

HINT

These formulae are all

in the formula book

393

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Example 7

a

Write

down

the

matrix

that

represents

a

rotation

of

a

horizontal

45°

anticlockwise

about

the

1 origin

given

that

cos 45 °



sin 45 °



2

b

Write

down

the

matrix

that

represents

stretch

of

scale

factor

2

c

by

a

horizontal

stretch

of

scale

factor

2

d

followed

by

a

horizontal

stretch

of

scale

factor

2

 1

1







2

2

a

This

 1

is

using

the

formula

for

an

for

a

1



anticlockwise

rotation.

 2

2







2

0

b This

is

using

the

formula

one-way

 0

1



stretch.

formula

 

 

2

c







 2

2

1





1

1

 

 



2

2









1 

 



point

is

rst

rotated

and

then

its

image

 

is

stretched.

1



 

  0

2

1













 2





2





 



 1

  0





1 

 2

5



5 



3



  0

book.

 The



 

2

the





 



in

2

 

  0

are





 

0

formulae

1





these

 1



Both

2

2









1

1

1

1



d  0 





 











 1

1

1 

2 

2

2

2





If a transformation represented by a matrix

A is followed by a transformation

represented by matrix B then the single matrix that represents both

transformations is BA

394

/

arbegla dna rebmuN

9.4

Example 8

to

the

(4,

2

×

2

that

will

transform

the

point

(2,

1)

to

T



  x 



and



c

 a 



Write

b   2

 





  c

d

 a 



 2a

 

1

that

 1  



b  









 b



1





 4

2a



b

 1

and

2c



d



d

 1

 

 3





down



 2c

4



describes



d

 3d





2

the

equation

transformation.

two

equations



a

 3b



4

and

c

 3d



2

and

×

2

systems

solve

each

of

system

9 using

elimination

or

technology.





T

7b



7

 b



1

and

a

 1



4 

9







2

1 Therefore,

 9





 c

matrix







the



 

  3b

and

−3)

4

 4 



  c

 a

(1,

y rtemonogirt

  x 

y

point

d

Develop



the

b

Therefore,

y

4)

 



(1,

9).

 a Let

matrix

7d



d

14



2

and

c



3



1 

  

 3

2

dna y rtemoeG

Find



Investigation 6

Consider the triangle ABC with ver tices A(0,

0), B(2,

2), C(−1, 5).

1

Show that ∆ABC is a right-angled triangle and nd its area.

2

Triangle ABC is enlarged by a factor of 3 to obtain triangle

A BC.

a

Use matrix multiplication to determine the coordinates of

b

Determine the area of triangle A′B′C′

A BC.

By what factor has the area

of triangle ABC changed? By what factor will the area change if it is

enlarged by a factor of k?

 1

3

Triangle ABC is transformed by

T



2

 

2

1

a

Find the coordinates of the image points of triangle

b

Show that the new triangle is a right-angled triangle and determine its

area. By what factor has the area of triangle

c

ABC

ABC changed under T?

Calculate det(T) and show that the area of triangle ABC enlarged by a

factor of |det(T)|

Continued

on

next

page

395

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND



4

Consider triangle ABC under the transformation

T



Use graphing software to show that area of

right triangle?

b

2

1

1

3

A′B′C′ =

42. Is ABC a

How do you know?

Show that this area is equal to |det(T)| × Area of triangle ABC.

 2

5

D ATA

 

a

A N A LY S I N G

Consider the transformation of triangle

ABC under

T



1

Does the

 

6

3

relationship |det(T)| × Area of triangle ABC hold for T? Explain why.

In the investigation above you discovered a very useful relationship between

the area of a gure’s pre-image and the area of its image under a linear trans-

formation T

Area

of

the

image = |det(T)| ×

Area

of

the

pre-image

Exercise 9E

y

1

Find

the

transformation

matrix

for:

8

a

A

b

An

reection

in

the

B

x-axis

=

(6, 7)

6

anticlockwise

origin

of

rotation

about

the

C

45˚

=

(10, 6)

4

A

=

(2, 4)



c

A

clockwise

rotation

2

of 2

x

d

A

reection

in

the

line

y

=

−x

0 –2

e

A

f

An

vertical

(0,

stretch

enlargement

of

scale

scale

factor

factor

2,

3

centre

a

0)

g

A

a

Write

2

Use

the

reection

in

the

line

y

=

matrix

image

reected

3x

triangle

2

down

the

transformation

an

enlargement

(0,

0).

scale

factor

4,

for

a

down

the

rotation

of

transformation

180°

about

(0,

rotation

of

180°

about

the

an

enlargement

of

0).

origin

scale

factor

b

Write

>

0

with

is

often

scale

referred

to

an

factor

as

Find

the

k

the

in

parts

a

and

b

two

and

x-axis

Sketch

the

to

A′B′C′

same

is

obtain

when

to

the

it

is

obtain

triangle

and

axis.

reected

across

the

A”B”C”.

a

single

of

matrix

P

equation

under

reections.

the

Show

the

to

nd

composition

of

Sketch

triangle

A”B”C”

same

axis.

that

the

if

triangle

x-axis

ABC

then

is

again

reected

across

the

matrices

x found

the

ABC

determine

enlargement

c

of

triangle

to

where

−k

product

on

image

across

c

14

followed

on k

12

matrix

both by

of

across

image

the A

10

centre

y-axis Write

8

multiplication

A′B′C′.

Triangle

b

6

matrix its

for

4

–2

axis

that

the

resulting

image

is

triangle

hence

ABC. explain

why

this

denition

is

justied.

d 3

Consider

vertices

 2 P



the

are

triangle

described

6

10

7

6

 

ABC

by

What

transformation

is

equivalent

to

whose

the

reection

in

the

x-axis

followed

reection

in

the

y-axis.

by

matrix



 4



396

/

9.4

a

Write

for

a

down

the

reection

transformation

in

the

line

y

=

9

matrix

Let

(tanα)x

Verify

that

a

reection

followed

by

be



reection

in

the

same

line

to

the

identity

2

2

1

a

Write

down

the

exact

of

45°

matrix

clockwise

that

.

Find:



the

coordinates

of

the

image

the

point

of

the

point

matrix.

R

for

about

4)

under

T

a

b rotation

such

is

P(−3,

5

transformation





a

a equivalent

linear

 

second

a

 1 T

b

T

arbegla dna rebmuN

4

the

coordinates

of

having

an

the image

of

(−12,

12).

1 origin,

given

that

cos 45 

sin 45 





10

Find

the

2

×

2

matrix

that

will

transform

2 the

point

(2,

1)

to

(1,

4)

and

the

point

8

b

Find

the

value

of

R

and

interpret

your

(1,

−3)

to

(4,

9).

result.

11

6

a

Find

the

area

of

the

triangle

A

triangle

at

(0, 5),

Use

result,

(4, 5)

and

the

area

of

is

rotated

(2, 7).

image

=

of

object

to

nd

the

area

|det

of

1),

(3,

1)

and

clockwise

about

(0,

0)

2

(T)|

then

reected

in

the

line

y

=

-x .

Let

R

× represent

area

(1,

 3)

and

b

vertices

with (3,

vertices

with

the

rotation

and

T

represent

the

the reection.

after

the

a

factor

vertical

2,

centre

ii

after

triangle:

stretch

followed

(0,

0),

by

scale

undergoing

with

an

enlargement,

factor

the

a

Write

b

Find

by

the

c

transformation

matrix

Hence

after 2

2

3

5

d

that

change

the

a

8

the

a

T



be

a

the

the

4

the

by

general

a

12

considering

matrix

Let

E

be

of

point

of

under

(−12,

P(−2,

that

of

5)

the

image

the

point

undergoes

R

is

a

such

of

the

having

two

a

Write

the

triangle

transformations.

or

scale

b

Find

down

the

series

otherwise

represented

representing

factor

describe

0.5,

the

by

RT

an

centre

(0,

0).

the

matrix

E

of

single

n

(0,

matrix

that

enlargements

represents

scale

factor

a

0.5,

0).

point

an

rotations

coordinates

of





R

followed

by

R

rotation

2

of

20° and

the

image

of

R

is

a

counterclockwise

rotation

of

P



cos 60

that

R

is

a



  x 

   

 y



 cos 20 



cos 40



 sin 20





sin 40

 cos 40



sin 20



 sin 40





cos 20

 



sin 40



cos 20

counterclockwise



 cos 40

rotation

of

α

1

cos(α

these

of

for:

that

counterclockwise

the

 sin 60

sin 60

show

image

2



Given

R

T



c

represents

that

 cos 60



that

12).

Determine

Show

the

matrix

enlargment

1

b

T



coordinates

40°.

and

Find:

12)

Given

R

T

sketch

a

1

a

matrix

transformation

centre .

matrices

rotation.

transformation

coordinates

image

The

for

object

b

linear

 2

P(14,

13

an

will

5 

 

b

of

by

nd

a

single

transformation

reection

 3

a

area

determinant

Let

T

neither

single

both

From



Prove

the





7

the

followed

3



represented

down

scale

dna y rtemoeG

i

of

y rtemonogirt

image





sin 20

and

R

 sin 40



sin 20



 cos 40

is

a

counterclockwise

sinθcosα

+

cosαsinθ



cos 20

rotation

of

x

 

 y



θ,

2

that

+

θ)

≡

cosαcosθ

−

sinαsinθ

and

sin(α

+

θ)

≡

397

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Ane transformations

Points

in

the

plane

(translations).

Afne

Ax

the

which

b

(4,

by

as

self-similar.

(3,

a

position

consists

matrix)

x

are

is

part

and

the

often

whole

A

using

2)

vector



 2

vectors

under

1

 











the

3

4

 



 

1

−1).

where

the

same)

point



transformations

for

have

transformation

+

the

3

(represented

form

transformed

3 



coordinates

An

be

example,

will



or

For

also

1

 translation

can

of

used

famous

is

a

linear

translation

point

object

itself

a

of

being

to

the

is

describe

same

the

and

acted

is

on.

or

(or

(self-similar),

example

transformation

of

the

Afne

create

objects

approximately

many

fractals

Barnsley

are

Fern.

Example 9

The

square

PQRS

transformation

1



 

 y

 2



 x 

  1





described

0



P( −2,

4),

Q(−6,

−3),

R(1,

−7),

S(5,

0)

undergoes

a

by

0





vertices



  x ’

with

 

 y

 3 



 2



 2 



 1  

2



a

By

multiplying

two

transformation

matrices

verify

that

0

is



equivalent

to

an

1 0

 

2



enlargement

b

Determine

c

Draw

 0

a

the

5

the

scale

0

  1

PQRS

0

5

 

 0

1



of

the

its

5

followed

by

vertices

image

on

a

of

the

reection

the

image

same

in

of

0

PQRS.

each

image.

1

point

The

in

full

turn

and

working

nd

for

the

5 point



y-axis.

axes.

its 0

the

Take

0

 



0.5

and

 0

0 

  0

factor

coordinates

square

 

of

(−2,

4)

is

shown.

 0





 x ’

b





 y ’

2 



 

1





0



1,

5, 0 





4



 3 





 2



 1 





 2



 3 





 2



 4 





3

2 

3

 7

2







 6, 











 6, 0.5 

 2.5,

 1.5 

 0.5, 2 

398

/

c

arbegla dna rebmuN

9.4

y

6

P

P’ 4

S’

2

Q’ b x 0

S

R’

–2

Q

–4

R

–8

Investigation 7

, 1

S

,

y

... where the

2

10

sides of each successive square is one-half of the previous

square, as shown. Each successive square is formed by a 8

transformation of the form AX + b where A is a 2 × 2 matrix

6

and b is a 2 × 1

, y

Let (x n

1

column vector.

) be a point in S n

1

and (x n

of this point in S

1

,

y

n

S

4

) be the image

dna y rtemoeG

, S 0

y rtemonogirt

Consider the series of squares S

0

n

such that

S

2

1

n

S 2

 x

 x

 n



n





A

1

x

1



y 







0

b.

2

6

4

8

10

12

14





n

1



By considering the points(0,

0),

(0,

8),

and (8,

0) in S

and their corresponding images in S 0

down and solve a system of equations to determine

2

16

y n

Using the coordinates of a point in the interior of

S

write 1

A and b

along with the coordinates of its image in

S

1

verify 2

your answer in question 1.

3

Determine the coordinates of the image of (6, 6) in

S 3

4

a

Apply the transformation to nd (x

, y 1

b

) in terms of (x 1

Apply the transformation again to nd

, y 0

(x

, y 2

) 0

) in terms of (x 2

, y 0

) 0



 



c

3

y



2

1  y

 

d





3



4

0

8

 

 8 

x 



Hence show that 

1

 x

0

8

Use the formula for a geometric series to nd an expression for

(x

, y n

5

) in terms of (x n

Use your conjecture to determine the coordinates of the image of (6, 2) in

, y 0

) 0

S 4

399

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Exercise 9F

1

The

points

andD(5,

A(−3,

8)

according

are

to

a

4),

B(2,

−3),

transformed

linear

C(0,

to

A ′,

ii

4),

B′,

transformation

C′,

matrix

D′

rotation

 4 X



0





 1



4

.





matrix

the



coordinates

of

Find:

line

the

image

B′,

C′,

the

of

the

point

P(−6,

Under

of

a

a

transformation

point

reection

in

the

y

=

3x

S

represents

a

stretch

of

factor

parallel

to

the

x-axis

and

a

stretch

(x,

y)

factor

4

parallel

to

the

y-axis.

Find

a

single

transformation

in

the

form

−3). of

2

a

whose

b is

represents

D′

coordinates

image

R

matrix

of

b

135°

points 2

A′,

clockwise

2

iv a

of

a

6

 

X

represents

dened

iii

by

Q

is

T

the

obtained

image

by

the

(x′,

y′)

Q

matrix

AX

+

b

that

followed

by

by

transforms

P

followed

a

point

by

R

(x,

y)

by

followed

S.

equation

c

  x 



 

 2 

y

1   x







Find

form .

  1

2

a

single

of

AX

the

b

when

a

point

y

 

image

+

of

the

point

the

c

the

image

with

of

the

(x,

y)

is

4



point

( −5,

by

the

vector

an

the

image

point

of

( a,

(3,

a)

by

6)

where

a∈ℝ

5

A

followed



8) 

b

in

Find

translated

a

transformation

2

R

sequence

of

equilateral

triangles

(T

)

are

n

d

the

point

with

an

image

of

( a,

a)

where

A

shape

below.

is

transformed

a

enlargement

a

rotation

of

by

a

by

factor

rst

of

3

undergoing

followed

90° anti-clockwise

then

by

T

is

formed

from

T

n

rotating

a∈ℝ.

3

shown

(0,

0)

(0,

0).

the

and

by n

triangle

60˚

enlarging

by

clockwise

a

factor

1

about

0.5

(centre

lastly y

 2 translated

by

the

vector



a

Determine

2

×

1

the

column

image

point

2

×

2

vector

(x,

y)

to

  x 



 3

2

.



1.5

matrix

b

that

its

T

and

maps

image

the

the

point

T

pre-

(x′,

0

y′)

x

 x  0

such

that

 

The

image



 y

T





point



(x′,

 y

y′)



b

–2

–1.5

–1

–0.5



is

0.5

1

–0.5

now

reected T

is

the

triangle

shown

with

two

vertices

at

0

across

the

origin

to

obtain

the

image

point (−2,

(

0)

and

The

b

Determine

the

2

×

2

matrix

A

and

0).

perimeter

×

1

column

vector

c

that

maps

point

(x,

y)

to

its

image

(

¢, y ¢ ¢ x¢ )

such

that

 

  y

Find

the

2

×

2

formed

from

triangles

is

the

distance

around

the

edge.

Find

the

perimeter

of

the

shape

shown

 x  

A



 

 y



made

c

up

of

triangles

T

to

T

0

3



b a

shape

point

a     x

the

the outside

pre-image

of

the these

2

4

(0,

¢, y ¢ ¢ x¢ )

transformation

matrices

Explain

why

the

maximum

perimeter

is

P, reached

once

T

is

added

to

the

diagram

5

Q,

R,

and

S

given

that and

i

matrix

P

represents

a

reection

in

nd

this

value.

the

x-axis

400

/

9.4

E

be

rotation

c

i

the

matrix

and

R

the

d

Write

e

Find

f

Use

down

Find

matrix

R

det(E).

the

area

of

all

T

ER

your

three

GDC

Hence

T

to

nd

the

to

vertices

nd

in

the

coordinates

of

T 8

0

iii

the

matrix.

Find

ii

enlargement

total

area

covered

by

T

0

6

arbegla dna rebmuN

Let

Developing inquiry skills

In the chapter opener, you found, using geometry, the exact values of the

coordinates of each ver tex and endpoint of stages 1 and 2 of the Koch

Snowake.

Stage 1

y

0

x

1

2

3

dna y rtemoeG

y rtemonogirt

1

You may have noticed that the basic building block of the Koch

Snowake is given by stage 1– that is, using a series of transformations

on stage 1 you can create each colored piece of the snowake in

stage 2.

Stage 2

red:

3

2

blue:

2

orange: √3

1

√3

√3

5

1

2 3

3

3

3

purple: 3

3

1

2

5

6

2

7

6

1

6

6

7 , 0

3

1

6

2 , 0

6

11

8 , 0

3

3

, 0 3

Describe the series of transformations that are necessary to map the

red piece in stage 1 to the red piece in stage 2. Write your answer in

the form of AX + b where A is a 2

vector, and X is a 2

of stage 1.

×

1

×

2 matrix, b is a 2

×

1 column

column vector of the ver tices and endpoints

Name this transformation T

.

1

By applying T

to each 1

ver tex and endpoint in stage 1, determine the image points of the red

piece in stage 2.

2

Repeat the steps in question 1 to determine the transformations of

the form AX +

b that maps the ver tices in stage 1 onto the blue,

orange, and purple pieces of stage 2. Name these transformations

T

, T 2

, and T 3

respectively. By applying these transformations 4

determine the coordinates of each ver tex and endpoint in each stage.

401

/

9

MODE LLING

9.5

Some

an

applications

probability

be

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Representing systems

outcome

the

WITH

genes

probability

depends

of

of

affected

of

our

our

by

on

the

inheriting

parents.

the

type

consider

previous

a

of

outcome.

particular

Whether

weather

or

sequences

genetic

not

from

For

today

the

of

events

example,

trait

sees

is

which

the

dependent

heavy

previous

in

on

rainfall

may

day.

Investigation 8

Sequences A and B are simulations of weather data over 21 days.

W represents the event “It was wet today” and D represents the event “It was dry today”.

A

D

D

D

D

W

D

W

W

D

D

W

D

W

D

W

W

W

D

D

D

D

B

W

D

D

W

W

D

W

W

W

W

W

W

W

D

D

D

W

D

W

W

W

You can investigate this data to see if there is any evidence of the weather tomorrow being dependent on the

weather today as follows.

In sequence A there are 8 wet days and 13 dry. You only consider the rst 20 days however since you do not

have information about the weather on the 22nd day. From the rst 20 days you have this data: {WD, WW,

WD, WD, WD, WW, WW, WD} to represent the days on which the weather was wet and then the weather on the

next day. From this set you can nd these experimental conditional probabilities:

3 P(Wet

tomorrow

Wet

5

and P(Dry

today) =

tomorrow

Wet

today) =

8

8

Show that for sequence B an estimate of the conditional probabilities are:

4 P(Wet

tomorrow

Dry

today) =

3

and P(Dry

tomorrow

Dry

today) =

7

7

One sequence was generated by a computer simulation programmed with these conditional probabilities:

P(Wet

tomorrow

P(Wet

|

Dry

tomorrow

|

today) = 0.4

Wet

and P(Dry

tomorrow

today) = 0.5 and P(Dry

|

Dry

tomorrow

|

today) = 0.6

Wet

and

today) = 0.5

The other sequence was simulated by ipping a coin. Discuss in a small group which method may have

generated each sequence

1

How did you decide?

2

Why is it dicult to be sure?

Conceptual

3

The

states

and

B

of

show

system,

you

this

the

How do you describe independent events?

system

system

cannot

are

“Dry”

changing

predict

the

and

state

“Wet”

from

outcome

of

and

one

a

the

day

given

sequences

to

trial

another.

with

A

In

this

absolute

TOK certainty,

the

but

system

in

you

can

use

any

given

probabilities

to

quantify

the

likely

state

of

trial.

How do “believing

that” and “believing in”

dier?

A Markov Chain is a system in which the probability of each event depends How does belief dier

only on the state of the previous event. from knowledge?

402

/

9.5

investigation

transition

8

in

the

probabilities

two

tree

in

diagrams

the

as

computer

below.

These

are

D

called

W

W

0.4

convenient

probabilities

is

in

way

this

D

0.5

D

to

0.5

summarise

transition

and

state

W

represent

these

0.5

diagram:

0.5

You

can

also

transition

to

the

The

represent

matrix

current

matrix

T

and

entries

transition

state

the

in

same

which

of

row

the

from

in

the

the

the

headings

current

column

system,

transition

the

information

the

next

matrix

state

to

etats

the

following

repeated

erutuF

T

=

investigation

multiplication

of

a

here

the

W

D

0.6

0.5

W

0.4

0.5

will

transition

as

0.6

0.4

the

state.

of

the

state.

state

D

you

D

refer

probability

future

W

a

(future)

show

the

in

headings

shown

Current

In

given

probabilities.

0.6

Another

simulation

arbegla dna rebmuN

represent

nd

out

how

to

interpret

dna y rtemoeG

in

could

y rtemonogirt

You

matrix.

Investigation 9

0.6

D

Coninuing with the situation described above You can use this tree diagram

to nd the probability that if is dry today, it is also dry two days from now by D

0.6

nding 0.4

W

D

0.5

0.4

D

0.6

×

0.6

+

0.4

×

0.5

=

0.56

and the probability that if it is dry today it will be wet two days from now by

W

nding

0.5

W

0.6

×

0.4

+

0.4

×

0.5

=

0.44

1

Apply a tree diagram to nd the probability that if is wet today, it is also wet two days from now.

2

Apply a tree diagram to nd the probability that if is

3

Given

 0 .6

wet today, it is dry two days from now.

0 .5  2

T



 

 0 .4

0 .5

,

Find T



2

4

How do you interpret the values in the matrix

T

?

2

5

Factual

What is more ecient to nd the probabilities in

T

- matrix multiplication or drawing the two

tree diagrams?

3

6

Factual

How do you interpret the values in the matrix

7

Factual

How can you interpret our results when multiplying by the transition matrix?

8

Factual

What does n represent in T

T

?

n

9

Conceptual

?

How can the probabilities of Markov chains be represented?

403

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

International

A transition matrix is a square matrix which summarizes a transition state

diagram. It describes the probabilities of moving from one state in a system

mindedness

to another state. Matrices play an

The sum of each column of the transition matrix is 1. impor tant role in the

The transition matrix is extremely useful when there are more than two states

projection of three-

in the system and when you wish to predict states of the system after several

dimensional images

time periods.

into a two-dimensional

screen, creating the

realistic seeming

motions in computer-

based applications.

Example 10

Dockless

Users

bicycle

can

Mathbike

By

company

unlock

divides

tracking

their

a

Mathbike

bicycle

the

city

bicycles

with

hires

their

into

three

with

GPS

bicycles

smartphone,

zones:

over

Inner

several

in

a

city

ride

(I),

it

through

to

Outer

weeks,

the

their

(O),

a

mobile

destination

and

Central

company

nds

phone

then

app.

lock

business

that

at

the

bicycle.

district

the

end

of

(C).

each

day:

•

50%

of

were

•

60%

left

of

10%

•

35%

the

bicycles

in

the

were

of

were

Zone

left

in

in

zone

C

remained

in

in

rented

zone

C.

rented

in

zone

in

zone

zone

I

O

remained

remained

in

C,

30%

were

left

in

Zone

I,

and

20%

zone

I,

O,

35%

30%

were

were

left

left

in

in

zone

zone

O,

I,

and

and

30%

C.

Show

this

information

in

a

transition

state

b

Show

this

information

in

a

transition

matrix.

c

Determine

a

in

zone

a

the

zone

O

bicycles

bicycles

left

rented

probability

that

after

three

diagram.

days,

a

The

0.60

O

bicycle

that

diagram

started

gives

the

relationships

put

in

the

right

a

in

C

quick

given

in

is

now

way

the

to

in

O.

check

question

that

are

places.

0.10 0.35

0.30

0.20

0.30

0.35

C

I

0.50

0.30

b

Current

location

I

O

0.50

0.30

0.10

yad

dne

C

Be

eht

eht

ta

fo

C

careful

represent

and I

noitacoL

O

T

=

0.30

0.35

0.30

0.20

0.35

0.60

the

to

make

the

row

the

current

headings

column

state

the

of

headings

the

future

system

state.

404

/

c

The

most

efcient

method

is

to

store

arbegla dna rebmuN

9.5

the

3

matrix

T

0 . 5 0

0 . 3 0

0 . 1 0

0 . 5 0

0 . 3 0

0 . 1 0

0 . 3 0

0 . 3 5

0 . 3 0

0 . 3 0

0 . 3 5

0 . 3 0

0 . 2 0

0 . 3 5

0 . 6 0

0 . 2 0

0 . 3 5

0 . 6 0

0 . 3 0 7

0 . 2 8 0

0 . 2 4 3

0 . 3 1 6

0 . 3 1 6

0 . 3 1 6

0 . 3 7 7

0 . 4 0 5

0 . 4 4 1

in

a

GDC

as

T

and

nd

T

3 T

=

Read

The

probability

central

days

is

that

business

in

Reect

the

a

bicycle

district

outer

at

zone

starting

the

is

end

of

in

the

three

the

off

O

the

row,

conclusion

number

which

that

is

in

the

0.441

interprets

C

column

and

this

write

and

a

number.

0.441.

From what you have learned in this section, how can you best

represent a system involving probabilities?

For visualising a system, would you choose a tree diagram, a transition state

For calculating probabilities, would you choose a tree diagram, a transition

state diagram or a transition matrix?

dna y rtemoeG

y rtemonogirt

diagram or a transition matrix?

Exercise 9G

1

For

these

transition

corresponding

matrices,

transition

construct

a

the

diagrams:

Show

this

matrix

T

information

by

writing

in

a

down

transition

the

blank

entries.

a

Current

state

etats

X

Y

5

1

11

5

6

4

11

5

Current

location

C

P

X

Current

erutuF

b

=

etats

erutuF

T

Y

C

P

0.2

0.25

location

3

A

B

T

=

C

2

This

b

A

B

C

0.2

0

0.47

0.5

0.9

0

0.3

0.1

0.53

transition

state

that

from

3

diagram

shows

market

research

of

of

consumers

who

shop

T

and

person

to

further

more

realistic

hence

who

Ceko

Upon

buys

three

research,

model

nd

is

the

Popsi

weeks

it

to

is

probability

now

from

found

create

a

will

now.

that

a

third

state

the

the

to

represent

a

person

buying

neither

Ceko

buying

nor habits

a

change

N ndings

Calculate

weekly

Popsi.

The

transition

state

diagram

is

and

0.9

buy

either

example,

soft

the

drink

brand

probability

Ceko

that

a

or

Popsi.

person

For

buying N

Ceko

will

buy

Popsi

the

next

week

is

0.25.

0.25

0.05

0.04

0.25 0.75

C

P

0.8

0.26

0.2

405

/

9

MODE LLING

a

Complete

WITH

the

M AT R I C E S :

transition

Current

drink

C

S TO R I N G

5

matrix:

N

Two

elected

desahcrup

weeks

emit

after

knirD

txen

P

T

the

the

probability

that

a

person

prior

for

Ceko

nor

Popsi

now

will

A

in

coin

ve

weeks

begins

and

one

is

put

in

box

chosen

into

repeated

week

from

of

state

the

many

and

at

1

with

two

two

blue

random

other

State

the

in

from

then

the

candidate

to

weeks

each

one

state

The

game

to

box

of

and

number

of

to

the

show

State

matrix

to

transitioning

one

three

as

more

three

of

planning

to

vote

changed

for

their

candidate

mind

B

and

were

planning

to

vote

B

changed

their

mind

and

for

candidate

A.

before

the

election

a

nal

survey

showed

that

45 520

vote

candidate

people

were

planning

to

for

A

and

38 745

people

were

is planning

to

a

a

vote

for

candidate

B.

Write

matrix

number

n

of

equation

voters

weeks

that

describes

supporting

after

the

nal

each

survey.

3

Assuming

that

the

continue

trends

up

until

revealed

the

day

in

of

the

the

show determine

which

candidate

will

from Calculate

the

margin

of

victory.

blue

the

sophisticated

white

3W3B.

states.

in

the

to

the

transition

in

of

new

matrix

transitioning

another.

transition

situations

coins

another,

State

probabilities

state

Investigate

9.6

made

with

represented

3W2B

4.1%

another.

is

starting

other

that

box

win.

c

were

A

who

vote

election

from

showed

another.

process

2

transition

probabilities

one

who

to

those

survey Construct

by

several

white

b

b

over

times.

1

the

election

Statistics

approximately

candidate

candidate

a

taken

for

now.

the

State

ofce.

survey

the

running

buy

Five coins

to

decided

decided

game

a

are

buying

for

A

from

each

3.2% neither

4

B

government

candidates

vote

Popsi

and

0.04

N

Find

A

0.05

=

and

b

D ATA

candidates

gathered

0.65

C

A N A LY S I N G

an

purchased

P

AND

such

as

matrices

4W4B,

for

5W5B,

etc.

Representing steady state

systems

You

have

and

if

seen

in

section

9.5

that

if

T

is

an

m

×

m

transition

matrix

TOK

If we can nd solutions  a

a

11

a

12

....

a 1m

13

of higher dimensions,

 a

a 21



a

....

22

a 2m

23

can we reason that these

n

T





a

a

31

a

32

....

a

33

3m

spaces exist beyond our

 ...

....

....

....

....

sense perception?



 a 

a m1

a m2

.... m3

a mm

406

/

9.6

then

a

is

the

probability

that

the

system

transitions

from

state

j

to

n

state

i

after

behaves

n

in

stages.

different

n

sequence

u

r

The

sequence

ways

of

according

transition

to

T,

just

matrices

as

the

T

arbegla dna rebmuN

ij

+

,n

∈ ℤ

geometric

+

,

n

∈

ℤ

behaves

in

different

ways

according

to

the

1

value

of

r.

Investigation 10

Use technology to nd these powers of each transition matrix

2

T

b 

5

1

11

5

6

4

11

5

0.2

0

0.47

0.5

0.9

0





0.1

0.53

0

0.5

0

1

0

1











 0

0.5

0

0

0

0

0.8

0.3

1

0

0.1

0.3

0

1

0.1

0.4

0

0

0

d

2

Term





e

50

T

 0.3





20

T





c

4

T







3

T

dna y rtemoeG

a

T

T

y rtemonogirt

1



0.25

0

0.65

0

0

0.35

0.75

1

0

What patterns are there in these sequences of matrices?

Each sequence above is classied by one of these terms: absorbing,

regular, periodic

An absorbing state is one that when entered, cannot transition to

another.

3

What would a denition of the other terms be? Drawing

a transition state diagram may help you distinguish between the dierent

types.

4

5

Identify the appropriate term for each.

Conceptual

What happens to a regular Markov chain with transition

matrix T as n tends to innity?

407

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

+

A regular transition matrix T has the proper ty that there exists n ∈ ℤ

such

n

that all entries in T

are greater than zero. For high powers of

n a regular

transition matrix converges to a matrix in which all the columns have the

same values. In the rest of this chapter you only consider regular transition

matrices.

You

can

system

need

state

at

Ceko

this

after

to

which

use

property

many

consider

the

a

Popsi

make

periods.

population

beginning.

represents

or

time

to

the

each

For

predictions

In

in

practical

which

example,

probabilities

a

etats

126

erutuF

of

state

you

number

the

are

transition

of

a

often

in

either

matrix

between

buying

week:

T

survey

certain

transitioning

Current

A

the

applications

consider

of

about

=

consumers

The initial state vector S

location

C

P

C

0.75

0.2

P

0.25

0.8

showed

that

81

bought

Ceko

and

45

Popsi.

shows the initial state of the system. 0

Number

of

consumers

S

C

81

P

45

=

o

You

can

predict

the

market

share

after

 0 .75 S



TS

1



0

Hence

you

can

predict

these

126

these

customers

changed

will

consumers

from

have

will

Popsi.

changed

to

that

will

  0 .25

after

0 .8

one

purchase

have

Also,

initially

some

Popsi.

week

0 .2   81 

 

one

The

 

 45

week,

Ceko.

bought

by

 69 



nding

75

 

56

25

between

69

Remember

Ceko

customers

and

who

multiplication

Similarly,

all

the

you

subsequent

that

some

originally

S

=

TS

1

represents

and

70

some

will

of

of

have

bought

quanties

Ceko

and

0

transitions.

can

multiply

weeks,

and

by

look

T

for

successively

patterns

in

to

our

predict

the

states

in

predictions.

2

S

=

TS

2

= 1

T

S

.

Repeating

this

process

establishes

the

equation

0

n

S

=

T

n

S 0

408

/

Investigation 11

Using the Ceko and Popsi matrix and vector above, use technology to complete the table.

Algebra

n =

no. of

weeks

Vector representation

representation

Matrix representations of S

of S n

0

of S n

n

81

81

45

45

S 0

1

0.75

0.2

81

69.75

0.25

0.8

45

56.25

0.31

81

63.5625

0.69

45

62.4375

TS 0

0.2   69.75 

2

TS

=

T

S

1

 0



 0.75

  0.25

0.8

 

 56.25

0.2   63.5625 

3

3

TS

=

T

S

2

 0



  0.25

0.8

 

 62.4375

 0.6125 



 

0.3875

 0.536875 



 

60.1594

0.3705   81 

  0.463125

0.6295

 

 45

65.8406



4

10

20

dna y rtemoeG

 0.75 2

y rtemonogirt

1

arbegla dna rebmuN

9.6

49

2

Factual

3

Conceptual

What are the converging steady state patterns in your results?

Generalise these convergent patterns using the terms

long-term probability matrix and

steady state vector

4

What interpretations can you make of these patterns regarding the long-term market share of each

brand?

5

6

What assumptions are made?

Conceptual

How can you use powers of the transition matrix to make predictions?

If P is a square regular transition matrix, there is a unique vector

q such that

Pq = q. This q is called the steady-state vector for P.

Example 11

Dockless

bicycle

company

Mathbike

open

their

business

by

distributing

a

number

of

 110

 bicycles

in

a

city

according

to

the

initial

state

vector

S 0



80

.

  50 

Continued

on

next

page

409

/

9

MODE LLING

a

Calculate

the

transition

b

Find

c

Comment

the

state

WITH

likely

matrix

likely

on

vector

.

number

given

steady

how

S

M AT R I C E S :

of

in

AND

each

A N A LY S I N G

zone

of

the

D ATA

city

after

ve

days

using

the

10.

Mathbikes.

owners

the

bicycles

example

state

the

State

in

of

S TO R I N G

of

Mathbikes

limitations

of

this

should

reect

on

their

choice

of

the

initial

model.

0

5

a

 0.50

0.30

0.10 

0.35

0.30



 110 

 0.30





 65.746 





80 





 75.789





5

Find



S

=

T

S

5



 0.20

0.35





0.60



50

with

a

GDC

0

 98.465



There

would

be

approximately

66 Make

Mathbikes

98

in

in

Zone

zone

C,

76

in

Zone

I

sure

to

interpret

your

ndings.

and

O.

50

 0.50

b

0.30

0.10

110

0.35

0.30

80

65.263

Use

 0.30

your

GDC

to

investigate

higher

powers

75.789





of

the

transition

matrix

and

look

for

 0.20

0.35

0.60

50

98.947



convergence.

the So,

the

system

quickly

reaches

an

rate

65

bikes

in

C,

76

in

I

and

99

in

shows

that

Mathbike’s

of

good

but

the

guess,

needs

placing

of

50

bikes

to

they

the

bikes

in

C

would

and

sequences,

depends

on

the

are

given.

ten

matrix

is

reached

earlier

than

days.

original

zone

customers

convergence

state

after

Think distribution

you

steady

this,

This

geometric

O.

A

c

with

equilibrium values

at

of

As

I

was

have

better

110

in

critically

about

the

context.

a

Write

met

your

by

O

at

a

complete

sentence

that

presents

ndings.

the

start.

This

or

model

stolen

changing

due

to

from

does

bikes.

Nor

habits

weather

other

not

account

does

amongst

patterns

it

for

account

the

or

broken

for

customers

competition

companies.

Investigation 12

 0 .6

0 .2

n

1

By investigating the sequence T

show

that if

T





 1

, then the

 0 .4

0 .8

1



long-term probability matrix is

X



3

3

2

2

 3

3



 

2

Fill in the table with examples of your own initial state vectors and nd

values for the steady state population vector and the steady state

probability vector.

410

/

arbegla dna rebmuN

9.6

1

Total population

u



v

(the steady state

p

(p) represented

probability

S XS

by S

0

0

= v

vector)

0

63

 27 

81



18



 54



 568 

 

 123



56358

2

…

Then repeat with your own 2

3

×

2 regular transition matrix T

Does the steady state probability vector depend on the initial

Factual

What relationship exists between the steady state probability vector and

the long-term probability matrix?

 1

 0 .6

Let

T





0 .2

3

with long-term probability matrix

 

0 .4

1

X



0 .8

3

, and



2

2

 3

3



dna y rtemoeG

4

y rtemonogirt

state vector?



 u 1

let

u







be the steady state probability vector. Hence,

u 

2

 0 .6u

 0 .2u

1



2

u 1

 Tu



u



0 .4 u





2

u



5

 0 .8u

1



 u

1

u 2

 1

2

Explain why each of the three equations hold and show that the solution of

 1

 3

this system is

u







. 2

  3

6

With your own regular 2

× 2 matrix T, nd the long-term probability matrix

by repeated multiplication and then form three equations to solve

T u

= u

using technology.

7

8

9

Repeat with your own regular 3

Factual

×

3

matrix.

What do you notice?

Conceptual

How can you nd the steady state probabilities? Give two

dierent methods.

411

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Exercise 9H

1

Find

the

the

long-term

Markov

probability

processes

with

matrices

these

for

At

transition

the

customers

matrices:

25%

0.24

0

0.71

0.87

0.29

a

b

0.42

0.85

0.05

0.6

0.2

B

45%

provided

and

0.34

0.15

0.64

Find

the

30%

by

of

the

electricity

by

A,

C.

percentage

provided

after

b

0.1

by

are

time,

0.31

a 0.13

present

by

two

each

of

the

customers

electricity

provider

years.

Calculate

the

long-term

percentage

0.7 of

0.05

0.2

0.2

0.05

0.4

0.05

0.2

0.2

0.45

0.15

0.4

0.05

customers

electricity

who

by

B.

will

State

be

provided

your

assumptions.

c 2

c

By

comparing

the

matrices

T,

T

5

,

T

and

20

T

,

comment

that

2

A

regular

given

transition

Markov

matrix

process

is

T

in

a

dened

0 .1

the

the

likelihood

probabilities

long

remain

term.

as A

hire

four



in

on

car

owned

by

Mathcar

is

in

one

of

0 .08

 T

transition

constant

4  0 .81

the

critically

possible

states:

A:

functioning

normally,

 0 .09

0 .15

0 .5

.



B:

functioning

despite

needing

a

minor





 0 .1

0 .75

repair,

0 .42

C:

functioning

despite

needing

a

 major

repair,

D:

broken

down.

Mathcar

only

 u 1

takes

a

car

out

of

service

if

it

breaks

down.

 Let

u



u

be

the

steady

state

probability

2

 

The

transition

matrix

u 

vector

3

such

that

T u

= u.



0 .9

0

0

0 .96

0 .03

0 .85

0

0

0 .02

0 .05

0 .6

0

0 .05

0 .1

0 .4

0 .04



a

u

Find

by

solving

a

system

of

equations.

 F



represents 

b

Demonstrate

your

answer

for

part

the

a 

is

correct

by

probability

nding

the

long-term



matrix. probabilities

3

Three

electricity

compete

in

a

providers

market.

A

A,

B

survey

and

of

C

state

determined

the

probabilities

another

that

a

an

electricity

provider

in

Interpret

one

year

stay

with

that

provider

or

will

switch

next

year.

The

a

period

of

a

from

one

month.

probabilities

the

zeros

in

the

matrix

in

the

are

of

the

problem,

and

the

entry

in

fourth

row

fourth

column.

to

b another

after

transitioning

will the

either

car

person context

using

the

consumers

a has

to

of

Verify

C

that

F

is

a

regular

Markov

given

matrix. in

this

transition

matrix:

c

Mathcar

begins

500

cars

in

0

D.

business

with

a

eet

of

Current provider

A

B

C

in

state

Find

A,

how

25

in

many

B,

cars

0

in

are

C

in

and

each

Provider A

0.8

0.05

0.06

0.12

0.92

0.1

0.08

0.03

0.84

state

after

10

months.

after one

year

B

T



d

Comment

their

C

to

F

on

practices

which

their

how

so

Mathcar

that

improve

could

changes

the

are

change

made

availability

of

cars.

412

/

9.7

Eigenvalues and eigenvectors

TOK

Investigation 13

1



Consider the transformation

T



1

The multi-billion dollar

applied to the points on the line  2



eigenvector.

4

Google’s success derives

L

: y

arbegla dna rebmuN

9 .7

= 2x

1

in large part from its

i

Write down the coordinates of any ve points on

then show that images

L 1

PageRank algorithm,

of these ve points lie on the same line. which ranks the

Take a general point on the line with coordinates

(x, 2x) and nd its image.

importance of webpages

according to an

Based upon your observations in par t i write down the value of λ for which

ii

eigenvector of a weighted

1



x

1  







  2



4

 

2x

x

 



link matrix

.







What does this tell you about how T transforms

2x

How ethical is it to create

mathematics for nancial

the points on L

geometrically? 1

1

also maps the points on line L



: y

= x to image points that

2

2



4

are also on y = x.

HINT

Determine the scale factor of the stretch that describes

1   x 

1



this mapping by solving the equation

 

 

2

4

 

 x

When a transformation

 x







 

matrix T maps each

x

dna y rtemoeG



y rtemonogirt

1



T

iii

gain?

point on one line to a

point on a dierent

The

investigation

that

maps

stretch

by

points

a

above

on

a

λ.

factor

reveals

line

In

back

this

a

special

onto

section

the

you

type

of

same

will

linear

line

learn

transformation

and

how

is

described

this

idea

as

line we say that the

a

lines are invariant but

along

the points themselves

with

our

knowledge

of

matrix

algebra

will

allow

you

to

determine

are not. large

powers

study

of

of

matrices.

modelling

chapters

12

and

and

These

results

analyzing

lay

the

dynamical

foundation

systems

and

for

our

networks

in

14.

International-

mindedness

Eigenvalues and eigenvectors

Geometrically,

under

the

you

say

that

transformation

the

lines

y

=

2x

and

y

=

x

are

The prex “eigen-” is

invariant

German, meaning “to

T.

own” or “is unique to”

In

general,

you

are

interested

in

determining

the

equations

of indicating that that an

the

invariant

lines

and

their

associated

scale

factors,

λ,

for

a

given eigenvalue is a unique

transformation

matrix

A.

That

is,

you

wish

to

solve

the

equation value associated

 x  A

 

 y

 x 





with a unique vector or

 

expressed

more

simply

Ax

=

λx.

The

values

of

λ named the eigenvector.

y

Eigenvalues describe are

referred

to

as

the

eigenvalues

of

matrix

A.

Each

eigenvalue

the characteristics of

A

is

associated

with

a

particular

line

that

is

invariant

under

the

of a transformation transformation

A

whose

equation

is

described

by

the

eigenvector

x.

and for this reason

Consider

the

problem

of

determining

 1 of

the

transformation

matrix

A



the





2

3

and

eigenvectors

are also referred to as

characteristic values and

4 



eigenvalues

. eigenvectors referred to

 as characteristic vectors.

413

/

9

MODE LLING

You

would

like

WITH

to

nd

such

that

the

M AT R I C E S :

S TO R I N G

eigenvalues

λ

 1 eigenvectors

algebra

 1

2

3

2

3

 1

3

 x.



Proceeding

with

the

matrix





 x

x

  x

x





0

x



(add dition

property

of

equality)



4 



 2

3

  1

 I

4 

 2

0

(identity

property)

2



  3



 

 

4

0

 3 

1



x

0

(distributive

 



x



0

(matrix

the

matrix

can

be

4





 3  

2



equation

addition)



1   When

property)



 2

0  

 1





1  



2

x

4 



 









associated

yields:

x



 1



D ATA

4 

 

their

A N A LY S I N G

4 

 

and

AND

multiplied

by

has

an

inverse

then

both

sides

of

the



the

inverse

matrix

to

give

the

solution

International-

 0 x











.

If

looking

for

additional

solutions

they

must

occur

when

the

mindedness 0



inverse

does

singular

and

not

exist

which

therefore

its

happens

when

determinant

is

the

equal

matrix

to

(A

-

λI)

Belgian/Dutch

is

mathematician Simon

zero.

Stevin use vectors in his Thus: theoretical work on falling

1  

bodies and his treatise

4 

0 “Principles of the art of

3  

2

weighing” in the 16th

1  



3  



 8



0 century.

2



∴λ

=



−1

4

 5

λ

and

1



0

=

5

2

1   Now,

returning

to

the

original

equation

eigenvector

 2





1





1



  2

 4





5



2

4

 

4

 

The

4   x 

 y

eigenvectors

2

 

 













associated

 0



2x



4 y





0

4 y



the

to

nd

the

you:

1 equations

give

y











eigenvalue

x

2

0

  2y

0



both

 2x



0



 4 x 



with

4 y

gives



 0  



x

 3  

2

eigenvalue

2 x 

 0

  x 

y

each

 0  



  2

for



 

corresponding

4

0

both

equations

give

y



x



of

λ

=

−1

are

1

1 described

by

the

position

vectors

of

the

points

on

the

line

y

=

-

x

2

414

/

9 .7

arbegla dna rebmuN

t



 and

therefore

have

the

form

x

  

the

is,

there

are

eigenvalue

an

λ

of

innite

=

−1.

is

any

real

number.

2

number

One

t





That

where

t

1

of

eigenvectors

particular

associated

eigenvector

(when

t

with

=

2)

1

2

 is

x



.



1



Similarly,

there

are

an

innite

number

of

eigenvectors

1

associated

with

the

eigenvalue

of

λ

=

5

described

by

the

points

on

the

2

 t line

y

=

x

and

these

vectors

have

the

form

x

 2

where

 

t

is

any

real

t

1 One

particular

eigenvector

(when

t

=

1)

is

x





2



 a A



b

then the real number λ is called an eigenvalue of A if there

 

c

d

exists a nonzero vector x such that Ax = λx where

x is the associated

eigenvector with the eigenvalue λ.

dna y rtemoeG

If

1

y rtemonogirt

number.

HINT a

- 

b

The eigenvalues of A

=

0

The process for c

d

-



nding the steady referred to as the characteristic equation.

state vector of a

Markov chain, ie

solving Tu = u, is Matrix

algebra

provides

you

with

a

tool

for

modelling,

analyzing,

and

equivalent to nding predicting

(called

the

long-term

dynamical

behaviour

of

systems

that

change

over

time

the eigenvector for

systems).

an eigenvalue of 1.

Example 12  0 Find

the

eigenvalues

and

corresponding

eigenvectors

of

A



1



 3

2





1 

Solve

3

−λ

(−2

2 

−λ)

−3

=

0



0

This

can

using λ

=

−3

When

or

λ

be

expanded

or

solved

directly

technology.

1

=

−3

Continued

on

next

page

415

/

9

MODE LLING



  3



 





 

y 



 





3

1



3x



S TO R I N G

AND

A N A LY S I N G

Using

(A

−

λ

I)

D ATA

x

=

0

0





M AT R I C E S :



x

1

WITH

y







0

0

or

y

=

−3x

An

alternative

equation

 

  1

Similarly

when

λ =

1





x

1

0

is

 





 













3



x

+

y

=

0

or

y

=

y

 

the



 

 x





3



 

3

nd

x

1



0



 

 eigenvectors

to

0



3

2



 

y

 









y



x

 1 Possible

eigenvectors

are



therefore

 

In

3

the

rst

case

y-coordinate

−3

any

×

vector

the

which

has

x-coordinate

the

will

do.

 1

and





1



Exercise 9I

1

Determine

the

characteristic

equation

 1

of

c

2

each

matrix

in

the

form

λ

pλ

+

I

 2

+

q

=

0

then

its

0

a

A





c

C



b

B



1

 2

5







0

e

T





 3



4

0 .7

0 .6

0 .3

Consider the transition matrix

2

 

whether

or

not

λ

=

−1

eigenvalue

of

the

each

matrix.

1  a

1  b

where 0 ≤

Show

and

3

λ

that

determine

terms and

eigenvectors



 2

b

 2

3

1

is

an

eigenvalue

of

T

the

other

value

of

λ

in

of

a

and

Q





Determine

eigenvectors

of

T

in

for

terms

of

each

a

and

b

4

 

b.

of

eigenvalue 8 



=

8

b



b ≤ 1.

2

eigenvalues

 5

a ≤ 1 and 0 ≤

 

Find

b

an

a 3

a

 

C

0

5

1

1

Determine

a

3







T

3

3

 

2



eigenvalues.

2

3

P

1

 0 .4  4

0



d

 

determine

0

 5

3

Diagonalization and powers of a matrix

Investigation 14



Consider the matrix

A



.

 

1

3

5

6

2

Show that the eigenvalues of A are λ

= 1

 1  x 1





 

2



 and



x 2

−1 and λ

= 8 and that the corresponding eigenvectors are 2

1

 

1

416

/

 

0

 1

1

2

Let P be a matrix containing the eigenvectors

x

and x 1

of A and

D



2

. That is, P

 0



  1

and

D





1

 

2

arbegla dna rebmuN

9 .7

1

2

0

.

 0



8

a

Find AP and show that AP = PD

b

Explain using the denition of eigenvalues and eigenvectors why this will always be the case.

a

Use matrix algebra to show that if AP = PD

b

Use the result in par t a to show that A

c

How did you use the fact that matrix multiplication is associative?

d

How did you use proper ties of inverses?

1

3

then A = PDP

3

3

= PD

1

P

3

4

a

For the value of D given in question 2 nd the value of D

n

b

Can you generalize your result for D

3

5

a

Determine A

3

by nding PD

1

P

3

. Verify the result using the GDC to calculate

A

n

?

E X AM HINT  a

Let

A



b

containing real number entries and having distinct real

 

c

In an exam, matrices

d

dna y rtemoeG

Can you generalize your result for A

y rtemonogirt

b

will always have diseigenvalues λ

and λ 1

then you say that A is diagonalizable and can thus 2

tinct real eigenvalues.

1

where P = (X

be written in the form of A = PDP

X 1

) is the matrix of 2

0

  1

eigenvectors and

D



 0





2

International-

mindedness

Vectors developed quickly

in the rst two decades

Let A be a diagonalizable 2

×

2 matrix expressed in the form of of the 19th century with

  −1

A

=

n

then A

PDP

n

= PD

0

Danish-Norwegian Caspar

1

1

P

where P = (X

X 1

)

and

D



2

 0 

Wessel, Swiss Jean Robert

 2

Argand and German Carl

Friedrich Gauss.

Example 13

1

a

Find

the

diagonalization

of

A



2

 

 3

0

 4

4

b

Hence

c

Find

nd

an

expression

for

A

in

the

form

PD

1

P

.

4

an

expression

1  

for

A

as

a

product

of

3

matrices

with

no

exponents.

Eigenvalues

2

are

the

solutions

of

the

2

A



 I



0





0









 6



0 characteristic

3

 1



3

and



equation

|A

−

λI |

=

0



= 2 2

Continued

on

next

page

417

/

9

MODE LLING

1

2

 

WITH

1 x

 3

0



3x

1



M AT R I C E S :

2  x 



1





  3

0

 y

 

S TO R I N G

AND

A N A LY S I N G

There are an innite number

 x  

3







 y



D ATA



y

x of eigenvectors of the form



 t   1

x







1



 1





1

x

t



2t

 and



x 2



where

 3t



 t ∈ℝ

you arbitrarily choose

t = 1(for

simplicity .)

1

2

 

1 x

 3

0



2





2

2  x 



2



  3

0

 

 x  

 y

2



 

3 

 y

y





x

2







2

 2x



 x



 3





1 Using

P



2

c

b

a

 ad

1

d

1

Therefore,

 3



0

1  3



 bc



2

1

P



 

 1

3

,

D



 



,

 2

0



P



 5





 1

1

1

Therefore: The

can

be

taken

out

as

a

factor

5 

  1

1

2

 A



3

0

 

2 or

 

placed

inside

one

of

the

matrices,

 

5

1

3



3

1 

 

1

either

1

is

acceptable.

  0

 



   

3

2





2

0

  1

A

   



or

  3

0 .6

0 .4

0 .2

 0 .2

 

0

2

 

   

4

 

 1

4

b



A

3

2





3

1

 

2

1

3











0

 





0 .6

0 .4

0 .2

 0 .2

  81





2

  1



A

0

 

 4

c



0

 





0 .6

0 .4

0 .2

 0 .2

 

 

0

8

 

   

Return to transition matrices

In

by

section

9.5

you

considering

equation

This

Tu

section

=

high

u

diagonalization.

computers)

successive

An

nd

powers

of

the

u

the

how

Though

has

is

been

calculate

is

it

powers

state

is

of

matrices

a

a

can

longer

much

using

be

by

solving

obtained

for

than

system

the

vector.

process,

easier

this

dynamical

or

probability

results

initially

found

high

state

same

a

steady

transition

steady

the

this

the

once

you

by

using

(or

the

for

performing

multiplications.

additional

formula

to

show

form

to

how

where

will

diagonalised

saw

for

benet

the

state

is

of

that

the

it

also

system

gives

after

you

n

an

easy

way

to

nd

a

transitions.

418

/

arbegla dna rebmuN

9 .7

Investigation 15

Co n s i d e r

the

situation

where

people Neighbourhood A

move

in

a

between

par ticular

2015,

8%

of

two

city.

Each

people

neighbourhood

A

Neighbourhood

B

neighbourhoods

year

currently

move

to

since

living

in

neighbourhood 12%

B

and

12%

of

people

neighbourhood

B

currently

move

to

living

in

neighbourhood 8%

A.

Yo u

than

may

assume

between

exactly

the

balances

any

two

movement

other

neighborhoods

those

arriving

with

those

l e av i n g.

What will the population of each neighbourhood be in 2020 assuming the migration rates remain xed?

1

represents the probability of a person moving from

Let T be the transition matrix where T ji

neighbourhood i to neighbourhood j. Explain why the transition matrix is



 B

2

0 .12



0 .08

0 .88

The populations of neighbourhoods A and B are 24,500 and 45,200 respectively at the beginning of

 0 .92

2015. After 1 year the population can be expressed as the matrix

P 1



TP



0

0 .12   24500 

 

  0 .08

dna y rtemoeG

A  0 .92 T

B

y rtemonogirt

A

0 .88

 

 45200



What is the population of neighbourhood A after 1 year? Neighbourhood B?

 24500

Let

S





n

. The population after n

years is given by the expression T

S 0

0

 45200 

3

Find the eigenvalues and eigenvectors for

4

Diagonalize T and use the relationship T

T

n

n

= PD

P

1

to show that

n

 3

1

 1

  0 .2

0

0 .2

n

T



 

 

2

1

 

n

0

0

8

   

0 .4

0 .6

n

1

0

n

5

As n → ∞, what will

D

n



approach? What will T



approach?

n



6

0

0 .8

Hence nd the long-term population of the neighbourhoods.

The administration of both neighborhoods would like a formula to tell them how many residents are

expected to be in each community n

years after 2015.

7

Multiply out your expression found in question 3 to nd a suitable formula for each of the school boards.

8

Use your formula to write down the populations of neighborhood A in 2020.

419

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Exercise 9J

 2

1

Given

that

eigenvalues

of

R



1 are



2 3

2



company

S

keeps

of

its

customers

while

3

λ

=

−4

λ

and

1

=

−1

write

R

in

the

form

2

1

1

R

=

PDP

of

them

switch

to

company

R.

At

the

3  16

2

Consider

A



35

beginning

 

6

6500

a

Determine

the

eigenvectors

of

the

2005,

company

R

had

13

eigenvalues

of

A.

State

and

why

A

customers

while

company

S

had

5200

customers.

is

a

Write

down

a

transition

matrix

T

diagonalizable. representing

the

proportion

of

the

1

b

Express

A

in

the

form

of

A

=

PDP customers

moving

between

the

two

n

c

Find

of

a

general

expression

for

A

in

companies.

terms

n

b

Find

the

distribution

of

the

market

after

4

d

Use

your

Verify

on

expression

your

the

result

in

part

using

Consider

to

nd

matrix

A

two

.

years.

Describe

distribution

utility

from

the

the

change

in

this

2005.

1

GDC.

c  0 .4

3

the

c

matrix

T



Write

T

in

the

form

T



P P

0 .75 





.

 n



0 .6

0 .25

1



n

d

Show

that

T

 Express

b

Find

T

in

the

form

T

=

8  9p



1

a

n

 8  9p

17

PDP



n

n

9  9p

9  8p

4

T

using

your

answer

in

part

a.

Verify

5 where

p





.

4

 0 .4 your

result

by

nding

 

12

0 .75  using

 0 .6

0 .25

e

Hence,

of the

matrix

utility

on

the

Using

the

result

from

customers

part

b

long-term

behavior

Verify

of

Suppose

only

two

rival

companies,

R

manufacture

a

certain

the

number

buying

from

R

after

n

years.

your

formula

by

nding

how

customers

are

purchasing

from

product.

after

two

years.

and

g S,

for

T R

4

expression

determine many

the

an

GDC.

f c

nd



Each

Find

the

long-term

number

of

customers

year, buying

from

R.

1 company

R

keeps

of

its

customers

while

4 3 of

them

switch

to

company

S.

Each

year,

4

Chapter summary

•

To add or subtract two or more matrices, they must be of the same order. You add or subtract

corresponding elements.

•

Given a matrix A and a real number k then kA

is obtained by multiplying every element of

A by

k

where k is referred to as a scalar.

•

If P = AB then each element of P (named as P

) is found by summing the products of the elements in i,j

row i of A with the elements in column j of B.

420

/

•

If A has dimensions m × n and B has dimensions p × q

then

the product AB is dened only if the number of columns in

❍

arbegla dna rebmuN

9

A is equal to the number of rows in B

(that is, n = p)

and when the product does exist, the dimensions of the product is

❍

•

m × q

Proper ties of multiplication for matrices

For matrices A,

B, and C:

❍

Non-commutive AB ≠ BA

❍

Associative proper ty A(BC) =

❍

Distributive proper ty A(B

+

(AB)C

C) = AB

+

AC and (B

+

C)

A

=

BA

+

CA

These proper ties only hold when the products are dened.

 1

0



0

0

1



0









0

0



1





•

The multiplicative identity of a square

n

×

n matrix A is given by

I

.



Note that

n







I

= A and I n

× A

=

A hold only in the case where A is a square matrix.

n

k

•

A

If A is a square matrix then



A  A   

A

   

k

factors

of

.

A

k

 a

0 

 a

0

k

•

In the case where b = c = 0 then



A

 

•

 0

d





.

 k



0

d

dna y rtemoeG

×

y rtemonogirt

both A

The matrices which represent rotations, reections, enlargements and one way stretches are all given

in the IB formula book .

•

A Markov Chain is a system in which the probability of each event depends only on the state of the

previous event.

•

A transition matrix is a square matrix which summarizes a transition state diagram. It describes the

probabilities of moving from one state to another in a stochastic system.

❍

The sum of each column of the transition matrix is 1.

❍

The transition matrix is extremely useful when there are more than two states in the system and

when you wish to predict states of the system after several time periods.

+

•

A regular transition matrix T has the proper ty that there exists n∈ℤ

n

such that all entries in T

are

greater than zero.

•

The initial state vector S

shows the initial state of the system. 0

•

If P is a square regular transition matrix, there is a unique probability vector

q such that Pq = q. This q

is called the steady-state vector for P

 a

•

Let

A



b

containing real number entries and having distinct real eigenvalues

 

λ

and λ

1

c

then

2

d

1

you say that A is diagonalizable and can thus be written in the form of

A = PDP

where P

=

(x 1

 

x

) is

2

0

1

the matrix of eigenvectors and

D



.





0 

2

1

•

Let

A be a diagonalizable 2 × 2 matrix expressed in the form of A =

  n

A

n

= PD

P

0

PDP

then



1

1

where P = (x 1

x

)

2

and

D





 0



 2



421

/

9

MODE LLING

WITH

M AT R I C E S :

S TO R I N G

AND

A N A LY S I N G

D ATA

Click here for a mixed

Chapter review review exercise

1

 5



1

If

A







 ,

 2



B



 1

 1





 1



, C









3 





 2

4

 3 ,

D



1









 0

5

each

of

does

not.

a

2C

e

i

the

E





without

E

f

(C

CDA

j

C

−

using

the

matrix

B

2

utility

c

DF

g

EB

on

−



2

3 nd

 1

4

not

exist



 1

2

4 

the

GDC.

If

d

FE

h

C

2

+

F





1

 ,



A

3D

0





b

+

1

 2

3

following

1 

 ,

2 

2

it

does

state

why

it

1

D)

+

B

2I 2

2

Determine

the

values

 1  3a

2

of



2b



1

Solve

each

 2

equation

 2

a

3X



 

1



b

X

2

 2

Write

solve

each

the

c

Area

of

P′Q′R′S′

Determine

6

The

linear

  x 





P ′,

Q′,

R′,

=

k

×

y

the

value

1   x 







(x, y)

2

1

10

 5  2



of

of

PQRS.

k

to

  1

4

 

(x′, y′).

 y

 1  





maps

 2



the

points



Find

a

the

coordinates

of

the

image

b

the

coordinates

of

the

point

of

the

image

4 

 

system

is

as

2

6



c

a

matrix

equation

using

matrix

algebra

your

of

( −3, 5)

who

image

(−7, 0)



the

coordinates

answers

in

(if

exact

of

( a, 2a)

a∈ℝ

then

7

A

series

of

triangles

T

,

T

0

Write

Area

transformation

 3 



0 

 3I

system

possible).

of

S′

where

4

coordinates

9

5

7

1 

4

the

 

 





3



.

3

X

3

 5 X







3 

1



c

2



Determine



for

 



 

1 

1

b

that

6









3



a





such



4 

b

and

 8 



and

2b 



 1

a

,

T

1

,...

are

formed

2

form. by

rotating

each

consecutive

triangle

anti-

°

w

7y



4 x

2y

2y



clockwise

2z

by

b 

3 

 20

a 3x



3y

 4z



6 

a

θ

by

factor

of

then

enlarging

the

triangle

k.

w

 10

y

4w

 5y

 3



2z 4

6x



5y

3

 18

c 27  9 x



7

5y

2

T 1

T

T

0

2

5

Parallelogram

Q(−1, 7),

PQRS

R(8, 4),

with

T(5, −5)

vertices

is

P(−4,

translated

1

−2),

by

the

3

x 0 –1

3 vector

1

2

3

4

5

7

enlarged

by

a

factor

of

2

a

Given

that

the

series

of

 x reected

across

the

transformations

 x

 n

x-axis. can

be

described

by







A

Determine

a

single

transformation

of

n

1

n

1



y 

a

8

1 then

5

then

6

–1

y n





the

determine form

of

image

AX

+

b

P′Q′R′S′

that

maps

PQRS

to

its

i

the

2

ii

the

exact

×

2

matrix

A

coordinates

of

the

vertices

ofT 3

422

/

9

x

x n

0

−1

b

Given

that



C

a

Find

b

Hence

A

.

(2

marks)

y

y n

i

write

0

down

AC

1

ii

nd

C

, C 2

nd

the

the

matrix

C

such

that

=

B.

(3

marks)

,andC 3

4

13 iii

nd

C

sum

of

the

areas

P1:

Consider

the

following

system

of

of equations:

T

,T 0

,T 1

,... 2

5x

+

3z

=

23

x

−

2y

+

5z

=

23

3y

+

7z

=

122

Exam-style questions

3

8

P1:

Given

A

2

a

the

B



 

following

0

,



3

2

nd

 4 and



arbegla dna rebmuN

n

a

Write

the

system

in

the

form

AX

=

B,

 2

1



where

A,

X

and

B

are

matrices.

matrices:

2B

(2

(2

marks)

(2

marks)

marks) −1

c

AB

d

(A

−

B

(2

marks)

(2

marks)

(2

marks)

c

Hence

B)

x

9

P1:

Given

that

the

matrix

A

1

14

P1: At

a

3 

of

x

determine

the

possible

(5

1

P1:

P





3 





and

Q

2

1

The

2







 

screening

Wars

attendance

2,

one

cinema

had

of

750

people.

It

a

£8

for

adults,

£5

for

children,

£4

for

OAPs.

number

of

OAPs

attending

was

one-fth

of

the

total

The

amount

attendance.

3

1



night

marks)





1

3



3

2

2



opening

of





2

10

system

marks)

and



original

value(s)

charged x.



the

.

(3

particular

Galaxy

full

of

solve

A

is



4

singular,

matrix

equations.

2

+

the

dna y rtemoeG

A

Find

y rtemonogirt

b

b

2



2

2

total

tickets

was

received

in

entrance

£4860.

3

a

Find

the

matrix

(PQ)

.

(3

marks) Use

b

Hence,

nd

the

smallest

1

n

such

that

 PQ 

value

of

n

Show

that

the

that

determine

attended

(9

1

(2

a

children

to

screening.

15

P2:

of

method

matrix

A

P2:

Consider

the

matrix

A

the

the

marks)

marks)

4

1

3

6

a

Find

the

4

2

2

4



eigenvalues

and

 corresponding

eigenvectors

of

(7 satises

the

.

0

11

matrix

number

0



a

A.

marks)

equation

2

A

2

−

×

10A

2

+

21I

identity

=

0,

where

matrix.

I

is

(4

the

b

Hence

nd

a

matrix

P

and

a

matrix

−1

marks)

D

such

that

D

=

P

AP.

(2

marks)

n

c

3

b

Hence

express

A

in

the

Find

a

terms pA

+

qI.

general

expression

for

A

in

form

(3

of

n.

(5

marks)

marks)

4

c

Hence

rA

12

P1:

The

B



+

express

A

in

the

sI.

matrix

form

(3

A

1

0

1

0

1

0

1

0

1



3

2

1

0

1

2

1

1

3

marks)

and

423

/

Approaches to learning: Communication,

MarComm phones

Critical thinking

Exploration criteria: Mathematical

and Markov chains

communication (B), Use of mathematics (E)

IB topic: Markov chains, Matrices

ytivitca noitagitsevni dna gnilledoM

Example

A phone company, MarComm, has four package options,

A, B,

C and D.

A customer is initially allocated to package A for a year.

After this, every year, customers can either continue with

their present option or change to one of the other options.

The probabilities of each possible change are given in this

transition state diagram:

0.7

0.1

0.3

0.4

A

B

0.2 0.1 0.3 0.4

0.1

0.1

0.1

0.4

D

C

0.2

0.1

0.4

How would MarComm be able to determine the probabilities on the diagram?

Represent this information in a transition matrix, P, of the system.



If customers are all initially allocated to option A for the rst year, then write down an initial state vector,

x

, 0

w 



in the form

 x

 x







y 

 0





 z

 

The probabilities for the nex t state can then be found by multiplying

x

by the transition matrix P 0

Calculate the probabilities of being in each option after one year if the customer star ts with option

A

a 



 b

 

Write down these probabilities in the form

x



=

1

 c 





 d



How might MarComm nd this information useful?

Repeating this iteratively will give the probability of a customer being in any state in subsequent years.

Investigate the behaviour of the system over time.



 1





 

Let π be the stationary distribution of the system, where



2



 

 3



 



4



424

/

9

You

will

now

nd

the

stationary

distribution

using

three

different

methods.

Using technology 





19

and

19

This

a

will

Why

would

is

20

x

P

20

give

reasonable

(that

x

an

this

of

not

x

P

0

indication

degree

and

of

the

)

using

a

calculator

or

computer.

0

convergence

to

a

stationary

distribution

ytivitca noitagitsevni dna gnilledoM

 x

Calculate

(to

accuracy).

be

considered

a

sophisticated

approach

to

the

problem?

Solving a system of equations

This

method

multiplying

uses

by

the

the

fact

that

transition



once

the

matrix

stationary

has

no

effect

distribution,

⇒

Pπ

=

π,

is

π



 1









letting









you

can

solve

the

system

of

equations

this

verify

that

method

Does

method

using

this

Justify

the

feel

your









3

4







 

with

result



2

the

is

information

equivalent

to

in

the

the

example

answer

to

calculate

obtained

using









Use



P

3

 



 2

 ,

 1

 2

 by



 1



So,

reached,

π

the

4

 3







 

and



4



to

previous

technology.

more

sophisticated

than

the

previous

process

using

technology?

answer.

Using eigenvalues and eigenvectors

If

the

this

stationary

chapter

eigenvector

Calculate

the

π

the

eigenvectors

How

here

could

and

getting

How

will

process

scaled

you

from

you

four

Clearly

four

all

the

signicant

However,

Which

P

an

sum

from

that

of

the

Pπ

=

π,

then

eigenvalue

its

of

elements

example

1

is

and

as

you

with

have

seen

in

corresponding

1.

the

corresponding

that

you

the

understand

process

that

is

fully

what

taking

you

place

are

beyond

nding

just

the

mathematical

calculations

problem

to

answering

MarComm

π,

the

in

the

long

use

methods

the

so

that

problem

probabilities

of

a

the

is

logical?

customer

being

in

each

term.

this

information?

produce

the

same

approximate

vector

π

correct

to

gures.

which

method

the

company

three

have

the

of

such

answers?

give

options

could

that

understand

posing

of

How

must

is

technology).

organize

methods

π

demonstrate

correct

you

so

P

eigenvalues

These

the

matrix

(using

that

the

distribution

method

is

the

do

most

you

prefer?

sophisticated?

Justify

your

answer.

Ex tension

Does a system always reach a

steady state?

How quickly will the process

converge to the steady state?

425

/

Analyzing rates of change:

10 Calculus

change.

which

is

a

This

looks

changing

Change

is

mathematical

chapter

at

how

with

all

around

change

current

and

rates

reaction,

bacteria;

cost

and

the

changes

of

at

in

on

one

in

Change

■

Relationships

differentiation,

in

velocity,

biology,

rates

which

fashion.

these

■

studying

variable

Examples

we

the

include

we

a

is

Microconcepts

physics,

density,

growth

But

Concept of a limit

■

Derivatives of standard functions

■

Increasing and decreasing functions

■

Tangents and normals

■

Optimization

■

The chain rule

■

The product rule

■

The quotient rule

■

The second derivative

■

Kinematics

of

marginal

might

of

■

have

rumour

All

differentiation.

measuring

for

another.

sociology,

speed

tool

chemistry,

economics,

applications

about

in

and

in

to

position,

power;

prot;

measure

periodic

in

of

us.

Concepts

focus

quickly

respect

include

of

will

dierential calculus

want

to

spreads

these

how

can

do

or

be

we

go

changes?

If you are given a model for the number

of people who catch a disease, how

would you work out how quickly it is

spreading at a par ticular time?

How might you nd the

velocity of an asteroid if

you were just given the

equation for its position?

A company has a model that shows the

prot it could make for dierent levels

of output. How could it quickly nd the

output that would give maximum prot?

How do weather forecasters

predict the weather when the

changes seem so unpredictable? 426

/

A

rm

is

about

to

launch

a

new

product.

They

conduct

a

Percentage of the sample who survey

to

nd

the

optimum

price

at

which

it

should

be

Price ($x)

sold.

would buy at this price (d) They

The

obtained

rm

model

the

decides

these

results

that

results.

a

shown

in

demand

Two

the

table.

equation

models

are

is

needed

suggested:

a

5

68

10

55

15

43

20

33

25

22

30

14

35

8

40

4

to

power

a model

of

the

form

d

=

and

an

exponential

model

of

the

b

x

bx

form

d

=

ae

x

or

d

=

•

Plot

the

data

on

•

Explain

why

these

•

Find

•

The

•

best

business

If

information

The

out

demand

like

were

of

the

you?

e

might

each

would

How

of

they

available,

value

What

be

maximize

demand

demand”.

tell

for

to

maximum

gradient

=

models

information

the

“marginal

two

would

c

GDC.

equations

other

this

where

your

What

work

•

tting

b

a(c)

of

could

prot.

Developing inquiry skills

need?

is

might

you

What other information might you need to work

function?

called

the

you

models.

their

prot

curve

might

the

how

a

suitable.

out the prot?

the

What could the company do to persuade more

marginal

work

out

people to buy the product?

its

How might this aect the model? value

from

the

curve?

Think about the questions in this opening problem •

The

marketing

team

has

a

model

that

links

extra

and answer any you can. As you work through the demand

with

the

amount

of

money

spent

on

chapter, you will gain mathematical knowledge advertising.

How

can

this

be

incorporated

into

and skills that will help you to answer them all. your

model?

Before you start Click here for help

You should know how to:

Skills check

1

1

Find

eg

Find

5

1

c

=

=

or

2

equation

the

and

+

−9

use

y

y

−

the

a

line.

of

the

through

line

the

with

point

Find

=

5x

1

=

the

−

equation

perpendicular

gradient

passing

(2, 1).

to

through

5(x

of

−

2

indices.

following

with

a

single

exponent.

Write

line

y

point

=

2x

−

3

and

(3, 4).

the

following

with

negative

exponents.

1 5

3

2

2

x

2

b

x

=

=

3

2

c

x

x

=

x

x

1

a

2

x

x

maximum

and

using

your

minimum

values

of

x

Find

the

minimum

coordinates

3

point

of

y

=

x

value

of

GDC. 4

the

x

a

y

Find

c

7

3 function

3

b x

eg

the

line

1

=

Find

the

the

2)

5

3

of

9

x

a

the

c

laws

Write

of

equation

passing

5(2)

Use

eg

the

with this skills check

of

the

=

x

3

−

5x

2

−

4x

+

20,

0

≤

x

≤

2.

maximum

2

−

8x

+

15,

where

0

≤

x

≤

3.

y

10 (1.21, 8.2)

8

6

4

2

x

1

(1.21,

2

3

8.21)

427

/

10

A N A LY Z I N G

10.1

The

OF

C H A NGE :

DIFFERENTIAL

C A LC U L U S

Limits and derivatives

graph

increase

R AT E S

shows

with

how

the

the

number

prots

of

of

a

widgets

y

company

it

2500

sells.

(160, 2100)

2000

previous

for

that

each

the

new

rate

on

at

widget

linear

which

made

functions

the

prot

(prot

per

you

will

torP

know

work

)$(

From

increases

widget)

is

1500

1000

2100  500 

$ 10

per

widget. 500

160  0

x

This

is

equivalent

to

the

gradient

of

the

curve.

If

0

the

20

curve

had

a

steeper

gradient

the

prot

per

40

60

80

100

Number

would

prot

be

higher.

per

Hence,

widget

the

equivalent

This

is

rate

to

easy

investigation

might

be

the

be

change

gradient

calculate

will

gradient

would

of

the

to

If

less

140

160

180

200

steep

of

widgets

the

lower.

of

one

of

the

when

consider

were

120

widget

variable

respect

to

another

is

curve.

the

how

with

functions

the

are

gradient

at

linear.

a

point

The

on

following

a

curve

calculated.

Investigation 1

When a curve is increasing, we say that it has a positive gradient, and

y

when it is decreasing we say that it has a negative gradient. Consider

5

2

the curve y = x

B

shown in the diagram.

4

1

a

Give the range of x values for which the gradient of the curve is

i

negative

ii

positive

3

2

A 1

2

y

iii

=

x

x

equal to zero. 0 –3

–2

–1

1

2

3

–1

b

At which point is the gradient of the curve greater, A or B?

The tangent to a curve at a point is the straight line that just touches but does not y

cross the curve at that point. 3

2

2

a

On your GDC or other software, plot the curve

at the point (1,

y = x

and draw the tangent

2

1), as shown in the second diagram. 1

b

Zoom in to the point (1,

(1, 1)

2

1).

y

=

x

x

Factual

0

What do you notice about the gradient of the tangent and the

–1

gradient of the curve at the point of contact?

2

c

Use your GDC or online software to draw tangents to

y = x

at the points

listed below and in each case write down the gradient of the tangent.

x

0

−1

1

2

3

2

Gradient of tangent at x

2

d

Conjecture a function for the gradient of

y

=

x

2

at the point on the curve with coordinates

(x, x

).

428

/

10 . 1

2

To justify the conjecture above we will consider the gradient of a chord from the point

distance h fur ther along the x

(x, x

) to the point a

axis. This is shown in the diagrams below for dierent values of

y

h

y

2

(x

+

h, (x

+

h)

)

Tangent Tangent

2

(x,

x

) 2

(x,

x

)

h x

h x

y

y

Tangent Tangent

2

(x,

x

)

2

(x,

x

)

h h

x

x

3

a

What do you notice about the gradient of the chord as

b

What will happen as h approaches zero? (Note: we could write this as

h decreases?

h

→ 0

)

2

The answer to 3b can be used to work out an expression for the gradient of the curve at the point

(x, x

).

2

4

a

Explain why an expression for the gradient of the chord between the two points

(x, x

) and

2 2

(

(x

x

+ h)

-

x

2

+ h, (x + h)

)

is

h

b

What would happen if you let h → 0?

c

Expand and simplify your expression. What happens now if you let

d

Compare your answer with your answer to question

a

Now consider the curve y = x

h → 0?

3

5

. Draw tangents to the curve at the points listed below and in each case

write down the gradient of the tangent.

x

−2

0

−1

1

2

3

suluclaC

2d.

Gradient of tangent at x

3

3

b

Conjecture a function for the gradient of

c

Write down an expression for the gradient of the chord between

d

By rst expanding and simplifying your answer to par t

y = x

at the point on the curve with coordinates

(x,

3

(x,

x

) and (x + h, (x + h)

x

).

3

).

c nd an expression for the gradient of the

3

curve at (x,

6

Conceptual

x

). Is it the same as the answer you conjectured in par t

b?

How would you nd the gradient of the curve at a point using the gradient of a chord from

that point?

429

/

10

A N A LY Z I N G

R AT E S

OF

C H A NGE :

DIFFERENTIAL

The function that gives the gradient of the graph

C A LC U L U S

HINT

y = f(x) at the point x is

The notation lim ( A)  f  (x)

written as f ′(x) where



f

 h) 

(x

f ( x) 

h0

lim 



h0

h



is read as “the limit of



A as h tends to zero”.

The

gradient

function,

f ′(x),

is

referred

to

as

the

derivative

of

x.

E X AM HINT

You will not be asked

Reect

questions that require

What does the derivative of a function tell you?

you to use this

notation. However, the

ideas behind functions

Exercise 10A

approaching limits will

reappear at various 2

Use

the

7

formula

f(x)

=

2x

points in the course.

4

 f  ( x)



f

(x

 h) 

8

f (x )

f(x)

=

x

lim  h0

h



HINT

4

to

nd

(the

the

gradient

derivative)

for

(x

function

the

+

h)

4

=

x

3

+

4x

2

h

+

6x

2

h

3

+

4x

h

4

+

h

following 1

functions.

9

f ( x)



x

1

f(x)

=

2

4x

f(x)

=

−2

10

3

f(x)

=

3x

−

4

5

f(x)

=

Conjecture

an

expression

c

for

n

f ′(x)

given

that

f(x)

=

ax

,

a,

TOK

n ∈ℝ

5

f(x)

=

mx

6

f(x)

=

x

What value does the

knowledge of limits 2

−

3x

+

7 have?

Notice that the last

From the previous exercise, you may have conjectured the following

two follow the same

rules:

rule as the rst if you n

•

f(x) = ax

(n

1)

⇒ f ′(x) = anx

, n∈ (the power law) 0

write c as cx •

f(x) = c ⇒ f ′(x) = 0

and

mx

1

as mx

•

f(x) = mx ⇒ f ′(x) = m

In addition, if

h(x) = f(x) + g(x)

then h′(x) =

f ′(x) +

g′(x). Reect

What is the power law for dierentiation?

Example 1

Use

the

power

law

to

nd

the

derivative

of

each

of:

1

3 2

a

f(x)

=

3x

+

2x

+

7

b

f(x)

c

=

f ( x)



4

x



2

x

x

430

/

10 . 1

a

f ′(x)

b

f(x)

=

6x

+

Use

2

the

rules

above.

2

=

3x

To

use

the

the

power

expression

law

using

you

a

need

negative

to

rst

write

index.

6 3

f ( x )



6 x



 3

The

x

answer

exponent,

fraction

1

The f ( x)





as

a

sometimes

negative

writing

subsequent

work

it

as

a

easier.

rst

step

again

is

to

write

the

expression

x

1

fractional

and

negative

powers.

3





1 2



makes

left

2

4x

using

f  ( x)

but

be

1

2

c

can

2



2x

x

2

There

2

are

different

ways

to

write

the

nal

1

=

answer;

-

all

are

acceptable

in

an

exam

unless

3

x

you

x

are

told

otherwise.

Example 2

For

each

of

the

i

nd

f ′(x)

ii

nd

the

iii

sketch

iv

write

functions

gradient

the

graph

down

the

below:

of

the

curve

of

the

function

set

of

values

at

of

the

x

point

and

for

its

where

x

derivative

which

the

=

2

on

the

function

same

is

axes

increasing.

2

a

f(x)

=

2x

+

3x

−

5

2

b

f ( x)





x,

x

=

4x

≠

0

x

i

f ′(x)

ii

x

=

2

⇒

+

3

f ′(2)

=

11

Use

the

The

derivative

at

iii

that

laws

given

at

a

above.

point

gives

the

gradient

point.

suluclaC

a

y

15 2

f(x)

=

2x

+

3x

–

5

10

5

f ’(x)

=

4x

+

3

x 0

–10

iv

x

>

−0.75

This

can

be

found

by

either

nding

the

2

minimum

nding

point

where

of

f ′(x)

f(x)

>

=

2x

+

3x

−

5

or

by

0.

Continued

on

next

page

431

/

10

A N A LY Z I N G

R AT E S

OF

C H A NGE :

DIFFERENTIAL

C A LC U L U S

2 1

b

i



1

2x

⇒

f(x)

=

2x

+

x

x

-2 2

f ′(x)

=

−2x

+

1

=

+

1

1

=

2

x

ii

x

=

2

⇒

f ′(2)

=

−0.5

+

0.5

iii y

6

4

–2 f'(x)

=

+

1

2

2

x

x 0 –8

–6

–4

2 f(x)

=

+

x

x –4

–6

iv

x

>

1.41,

x




is

the

using

a

GDC

0.

What is the relationship between the gradient of a curve and the sign

of its derivative?

Example 3

In

economics,

approximated

A

company

the

by

marginal

the

cost

gradient

produces

is

of

motorbike

a

the

cost

cost

of

producing

one

more

unit.

It

can

be

curve.

helmets

and

the

daily

cost

function

can

be

modelled

as

3

C(x)

=

600

where

x

is

+

7x

the

a

Write

b

Find

an

c

Find

the

d

State

b

C′(x)

c

i

the

marginal

units

of

daily

expression

the

$600

0.0001x

number

down

a

−

of

cost

0

≤

helmets

cost

for

the

for

the

if

to

x

≤

150

produced

the

company

marginal

i

20

marginal

and

cost,

helmets

if

C

no

C′(x),

are

the

cost

in

helmets

of

US

are

produced.

producing

produced

ii

80

dollars.

x

helmets.

helmets

are

produced.

cost.

This

is

the

value

when

x

=

0.

2

ii

d

$

=

7

−

C′(20)

C′(80)

per

0.0003x

=

=

6.88

C’

represents

the

marginal

cost.

5.08

helmet

This

is

the

helmet

extra

cost

in

$

for

each

new

made.

432

/

10 . 1

Alternative notation International-

There

is

an

alternative

notation

for

the

derivative,

which

is

very

widely

mindedness used.

Maria Agnesi, an

dy 2

For

example,

if

y

=

2x

+

5

then

its

derivative

is

=

4 x 18th century, Italian

dx

mathematician,

The

notation

comes

from

the

fact

that

the

gradient

of

a

line

is published a text

difference

in

y on calculus and

difference

in

x

also studied

curves of the form One

useful

aspect

to

this

notation

is

that

it

indicates

clearly

the 2

3

variables

involved,

so

if

s

=

3t

+

4t

then

the

derivative

would

a

be y





2

written

a

2

ds as

=

9t

x +

4

dt

HINT Usually the prime notation, f ′(x), is used when dealing with functions and

There are no set dy

the fractional notation,

, is used when dealing with equations relating two

rules, however. dx

variables.

y′ = 2x − 1 and

df

= 3x are both dx

correct, though less

Example 4 frequently seen.

2

The

tangent

to

the

curve

gradient

of

Find

coordinates

the

y

=

2x

+

3x

of

at

the

point

A

has

a

A.

First 4 x

4

11.

dy 

−

nd

the

derivative.

 3

dx

+

3

=

11

4x

=

8

You

know

same

–

x

When

=

x

and

2

=

so,

the

the

for

derivative

of

the

derivative

is

the

tangent

to

11

x.

substitute

original

the

gradient

equate

solve

Now

2,

as

that

2

for

x

into

the

suluclaC

4x

equation.

2

y

=

So,

(2,

The

that

2(2)

the

+

3(2)

−

4

coordinates

=

10,

of

A

are

10),

following

they

gradient,

both

or

exercise

indicate

rate

of

contains

that

change,

the

of

a

mixture

of

both

derivative

of

a

the

notations.

function

gives

Remember

the

variable.

433

/

10

A N A LY Z I N G

R AT E S

OF

C H A NGE :

DIFFERENTIAL

C A LC U L U S

Exercise 10B

1

Find

the

derivative

functions

with

of

each

respect

to

of

x

the

and

6

following

nd

the

The

is

prot,

$P,

modelled

made

by

the

from

selling

c

cupcakes

function

2

gradient

a

y

=

of

the

curve

at

the

point

where

x

=

2.

P

=

−0.056c

6

5.6c

−

20.

dP

a b

y

c

f(x)

=

3x

d

f(x)

=

5x

e

f(x)

=

3x

f

f(x)

=

5x

=

+

Find

4x

dc

2

b

Find

2

−

3x

+

7x

−

3x

the

respect

c

=

20

rate

to

of

the

and

c

=

change

number

of

of

the

prot

cupcakes

with

when

60.

4

−

3

c 4

+

2x

−

Find

the

derivative

of

each

of

the

your

answers

for

part

b

The

distance

with

respect

to

x

and

of

a

bungee

jumper

below

his

following starting

functions

on

6

7 2

Comment

2

nd

point

is

modelled

by

the

function

the 2

f(t) gradient

of

the

curve

at

the

point

where

x

=

=

10t

−

5t

,

0

≤

t

≤

2,

where

t

is

the

time

1. in

seconds

from

the

moment

he

jumps.

3

a

f ( x)



b

x

y

=

2

3

x

+ 3

a

Find

b

State

c

Find

f ′(t).

the

quantity

represented

by

f ′(t).

6

2

c

y

2x

=

d

-

y

=

+

4 x

- 3

3

x

f ′(0.5)

and

f ′(1.5)

and

comment

on

x the

values

obtained.

7 4

e

y

=

+ 8x

2

- 6x

+

d

2

Find

f(2)

and

comment

on

the

validity

of

3

x

3

Find

the

the

derivative

of

each

of

the

following

8 functions,

and

nd

associated

curve

the

gradient

of

Points

t

=

A

the

and

3

f(x)

when

model.

=

x

B

lie

on

the

curve

and

the

gradient

2

+

x

+

2x

of

the

16. curve

at

both

A

and

B

is

equal

to

3.

8

a

s

=

4t

Find

-

the

coordinates

of

points

A

and

B.

t

9

The

outline

of

a

building

can

be

modelled

3

2

16

by

the

equation

h

=

2x

−

0.1x

where

h

4

b

v

=

4t

1

is

the

height

of

the

building

and

x

the

4

t horizontal

4

For

each

of

the

functions

i

nd

the

derivative

ii

nd

the

set

the

a

y

b

f(x)

c

g(x)

=

of

associated

(2x

−

1)(3x

+

below:

building,

An

values

of

function

x

is

for

which

increasing.

2x(x

+

3

5

The

=

x

area,

4x

−

A,

elevation

point

he

the

stands

from

can

his

see

height

at

the

A.

the

that

corner

diagram

The

position

on

of

in

one

angle

to

the

building

point

of

the

below.

of

highest

is

45°.

above

the

5)

2

+

shown

from

ground.

3

=

as

observer

Find

4)

distance

3x

of

a

−

9x

−

circle

h

8

of

radius

r

is

given

2

by

the

formula

a

Find

A

=

πr

dA

dr

b

Find

the

rate

of

change

of

the

area

with A

respect

to

the

radius

when

r

=

x

2.

20

434

/

10 . 1

Equations of tangents and normals

The

be

equation

found

for

using

the

the

tangent

gradient

to

to

a

curve

the

at

curve

a

given

and

the

point

can

easily

coordinates

of

the

point.

Example 5

2

Find

the

equation

of

the

tangent

to

the

curve

y

=

2x

+

4

x

at

the

point

where

x

=

4.

1

2

y

=

2

2x

+

4 x Write

can

dy

in

be

exponent

form

so

the

power

rule

used.

2

=

4 x

+

2x

dx

2 =

4 x

If

working

without

a

GDC,

it

is

easier

to

+ write

x

the

working

derivative

out

its

as

a

fraction

when

value.

dy When

x

=

= 17

4,

The

gradient

of

the

tangent

is

equal

to

dx the

gradient

of

the

curve

so

we

know

the

2

When

lies

x

on

=

4,

the

Equation

y

=

2

curve

of

the

×

4

and

+

4

on

tangent

×

the

2

=

40

so

(4,

40)

tangent.

tangent

we

need

−

40

=

17(x

−

=

17x

−

point

form

on

y

the

=

17x

+

c.

To

nd

c,

curve.

form

of

the

equation

is

4)

left y

the

point–gradient

often

or

a

of

is

The

y

is

the

like

easiest

this,

to

use.

though

it

Your

is

answer

usual

to

can

write

it

be

in

28.

gradient–intercept

The normal to a curve at a point is the line that

Normal

line

to

form.

cur ve

Internationalis perpendicular to the tangent to the curve at

mindedness that point.

The ancient Greeks used

the idea of limits in their y

=

f(x)

“Method of exhaustion”.

line

to

cur ve

suluclaC

Tangent

HINT

Example 6 If two lines are

Find

the

equation

of

the

normal

to

the

curve

for

equation

perpendicular then

3

f(x)

=

2x

+

3x

−

2

at

the

point

where

x

=

1.

the product of their

gradients is −1.

3

f(1)

=

2(1)

+

3(1)

−

2

=

3

Find

the

y-coordinate

of

the

point.

If the gradient of the

So

the

point

is

(1,

a

3).

tangent is

, then b

2

f ’(x)

=

6x

+

Next

3

nd

f ’(x).

the gradient of the

2

f ’(1)

=

6(1)

+

3

=

9

Find

f ’(1)

to

work

out

the

b

normal is gradient

of

the

tangent

line.

Continued

on

next

.

-

a

page

435

/

10

A N A LY Z I N G

The

gradient

line

is

of

R AT E S

the

OF

C H A NGE :

DIFFERENTIAL

C A LC U L U S

tangent

9.

Gradient

of

the

Gradient

of

1 normal

line

=

1

-

normal

=

9

gradient

of

tangent

1 Substitute 3

=

-

(1)

+

the

coordinates

(1, 3)

c

9

into

the

equation

of

the

normal.

1 c

=

3

9

The

equation

1 y

=

-

of

the

normal

is

1 x

In

+ 3

9

the

general

form,

this

is

9 x

+

9y

−

28

=

0.

Example 7

3

The

gradient

of

the

normal

to

the

graph

of

the

function

dened

by

f(x)

=

kx

−

2x

+

1

at

the

1 point

(1, b)

is

-

.

Find

the

values

of

k

and

b

4

2

f ′(x)

=

3kx

−

2

1 If

the

gradient

of

the

normal

is

,

-

then

4

the

gradient

of

the

tangent

is

4.

So,

2

f  ( 1 )



3k( 1 )

 2



3k



4

The

at

any



point.

is

the

Here

gradient

the

point

of

is

the

tangent

where

x

=

1.

6

So, k

derivative

you

put

x

=

1

into

the

derivative

and

2 equate

it

to

4

(which

is

the

gradient

of

the

3

f(1)

So,

=

b

2(1)

=

−

2(1)

+

1

=

tangent

1

To

1.

nd

the

at

b

that

you

original

point).

need

to

substitute

1

for

x

into

equation.

Exercise 10C

1

Find

the

graph

of

equation

the

of

the

function

tangent

dened

to

the

When

by

a

the

tangent

seesaw

to

the

is

log

level,

at

the

the

plank

point

of

where

wood

x

=

is

1.

2

f(x)

=

2x

−

4

at

the

point

where

x

=

3. By

2

Brian

part

makes

of

a

a

seesaw

log

and

the

log

a

for

plank

his

of

children

wood.

the

from

of

can

be

modelled

2x

for

by

f(x)

=

−x

+

0


0

•

if f is decreasing at x then f ′ (x) < 0

1

Consider the two curves shown below. Write down for which of the two curves

a

the gradient of the curve is increasing

b

the gradient of the curve is decreasing.

y

y

x

x

The rst curve is an example of a curve that is

concave up and the second an example of a curve that is

concave down

3

2

Consider the curve y = (x − 1)

a

Sketch the curve showing clearly the

b

Find

x-intercept.

dy

dx

c

Explain from your answer to par t b why the curve is always increasing.

The second derivative is obtained by dierentiating the rst derivative of a function. The notation is either

2

d

y

(said as “d two y by dx squared”) or f ″(x) . 2

dx

a

Find the second derivative of y = (x − 1)

b

State the values of x for which

2

d

2

y

i

d >

0

y

ii




0

f ’ (x)

y

f ’ ’ (x)

>




cos

x

in

others,

we

need

to

use

the

suluclaC

5.65685424 3

2

absolute A



sinx

 cosx

value.

dx

 0

A

≈

5.66(3

sf)

y

The area A

of the region bounded by

the curves y = f(x), y = g(x) and the

lines x = a and x = b, where f and g are

continuous, is given by a x

0

b

b

A





f



x





g



x



dx

a

499

/

11

A P P R O X I M AT I N G

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

AND

DIFFERENTIAL

E Q U AT I O N S

Finding the area bet ween a cur ve and the y-axis

If

the

equation

similar

If

the

to

is

nding

equation

make

y

given

the

the

is

in

area

given

as

the

form

x

=

between

a

curve

y

=

f(x)

then

f(y)

it

then

and

will

the

the

need

method

is

very

x-axis.

to

be

rearranged

to

subject.

The area A of the region bounded by the curve

and y = b, where h is continuous for all y in [a,

x = h(y), and the lines y = a

b], is given by

b

A



h ( y) dy

 a

Note

that,

if

we

are

absolute

value,

we

positive

have

but

trying

for

the

areas

nd

we

the

must

integral,

use

the

we

do

not

absolute

need

value

to

the

ensure

values.

Example 16

2

Find

the

area

of

the

region

bounded

by

the

curve

y

=

x

and

Sketch

y

the

the

y-axis,

graph

and

x ∈[−2,

shade

0].

the

required

area.

2

x 0 –2

Identify

x

=

0

⇒

y

=

0and

x

=

−2

⇒

y

=

4

Write

x

the

as

a

lower

and

function

upper

of

y

for

boundaries

the

given

of

y.

region.

2

y



x



x





y

Write

the

area

A

as

a

denite

integral.

4

You

A





y

can

use

your

GDC

or

integration

rules

dy



to

nd

the

value

of

the

integral.

0

A

≈

5.33(3

sf) 4

(|–√ Y|)dY

∫ θ

5.333333544

500

/

11 . 3

Exercise 11H

Find

the

area

of

the

a

shaded

d

regions.

y

y

1.2

f (x) x 1 0

0.6

0.4

x 0.2

x 0

–0.2

f (x)

–0.4

2

f(x)

2

f(x)

=

−2x

+

x

−

=

x

−

3x

−

4

1

g(x)

=

x

−

4

y

b e

y

x 0

g(x)

A

f (x)

x 0

f (x)

2

f(x)

=

x

3

f(x)

=

(x

−

3) g(x)

c

=

4

−

x

y y

f

suluclaC

1

0.5 x e

0

g(x)

f (x)

f (x)

f(x)

=

ln x

x 0

x

f(x)

=

e

−x

g(x)

=

e

501

/

11

2

A P P R O X I M AT I N G

Find

the

by

=

area

of

the

IRREGUL AR

shaded

region

S PA C E S :

I N T E G R AT I O N

bounded

a

Find

b

Write

AND

the

DIFFERENTIAL

values

of

a

and

E Q U AT I O N S

b.

3 y

,

the

y-axis

and

the

x

y

=

down

surface

1andy

=

the

equation

for

the

inner

lines

of

the

roof.

3. The

building

is

15

m

long.

y

c

5

Find

In

the

volume

economics,

the

quantity

willing

unit

to

a

of

concrete

supply

( q)

that

supply

for

a

a

in

function

the

roof.

relates

manufacturer

particular

is

price

per

( p).

3

The

diagram

below

shows

the

supply

curve

f (x)

1 p

=

ln

(q

+ 1)

2

p 1

x 0

1

A

3

Find

the

area

of

the

region

bounded

by 2

A

x

=

sin

y

=

2π

y,

the

y-axis

and

the

lines

y

=

0

and

q

4

A

building

is

in

the

shape

of

a

prism

with 0 a

cross-section

The

as

building

shown

has

a

in

the

concrete

diagram

roof

that

below.

is

1

m For

thick

when

measured

vertically.

The

a

particular

quantity,

the

area

of

the

outside region

above

the

curve

(A

)

gives

the

1

of

the

cross-section

of

the

roof

forms

a

curve producer

of

the

form

y

=

a sin

( bx )

when

placed

surplus

for

selling

that

quantity

on and

the

area

below

the

curve

(A

)

gives

2

a

coordinate

system

as

shown.

This

curve the

touches

the

ground

at

the

points

A(0,

0)

revenue

0)

and

the

highest

point

of

the

roof

is

of

the

the

producer.

The

whole

rectangle

is

the

actual

at producer

( 2.5, 4 )

by

and area

C(5,

needed

revenue.

A

quantity

q

=

a

is

a

for:

produced.

a

y

5

C

Find

an

expression

i

the

producer

ii

the

revenue

in

terms

of

surplus

needed.

4

b 1

If

the

surplus

is

at

least

50%

more

than

m

the

revenue

needed

by

the

producer,

3

nd

this

2

the

to

maximum

integer

value

of

a

for

occur.

1

1

m

A

B x

0

1

2

3

4

5

6

502

/

11 . 3

Developing inquiry skills

Look again at the opening problem. To work out the

y

area, we will now t two curves through the points

45

E

found previously: one curve through A, B, C, D, E 40

and F and one through A , J, I, H, G and F. F 35

D

Use cubic regression to nd an equation for each

ormido

30

curve and hence, an estimate for the area of San Brujo o

C

G

25

Cristobal.

s

Whit

y

B 20

H 15

A I

10 J

5

x 0 5

10

15

20

25

30

35

40

45

50

55

Solids of revolution

How

do

we

calculate

the

volumes

of

irregular

shapes?

TOK

We

have

formulae

to

calculate

the

volumes

of

regular

shapes,

for 1

example

a

cylinder

or

a

pyramid.

We

can

extend

our

knowledge

of

Consider f(x) =

, x

volumes

of

regular

shapes

and

integration

to

volumes

of

irregular

1 ≤ x ≤ ∞.

shapes. An innite area sweeps

Consider

f(x)

=

cos(0.5x

−

1)

+

2

in

the

interval

[0,

out a nite volume.

12].

How does this compare

y

to our intuition?

6

4

x 0 2

4

6

8

10

12

14

–2

When

would

you

get

rotate

a

3D

this

curve

shape,

like

through

a

2 πradians

around

the

x-axis,

suluclaC

–4

you

vase.

y

4

2

0

x

2

4

6

8

10

12

–2

–4

503

/

11

A P P R O X I M AT I N G

On

a

closer

sections

each

look

when

with

a

at

the

sliced

radius

IRREGUL AR

shape,

you

can

perpendicular

of

y

=

S PA C E S :

to

see

the

that

I N T E G R AT I O N

it

x-axis,

has

circular

centred

on

AND

DIFFERENTIAL

E Q U AT I O N S

cross-

the

x-axis,

f(x).

y

4

2

f(x1)

δ x

0

x

–2

–4

If

we

divide

(remember

in

the

the

shape

from

value

of

into

Section

x),

we

n

cylindrical

11.1

can

δx

that

is

approximate

discs,

used

the

each

to

of

indicate

total

δx

width

a

small

change

volume.

2

The

cross-sectional

area

of

the

disk

with

x-coordinate

x

is

A

i

=

π[f(x )]

i

i

2

using

the

Using

of

formula

the

for

formula

the

for

area

the

of

a

volume

circle

of

a

A

=

πr

cylinder,

each

disc

has

a

volume

approximately

2

V

=

δx.

π[f(x )]

i

i

Thus,

the

function

volume

y

=

f(x)

formed

through

by

the

revolution

2πradians

around

of

the

the

continuous

x-axis

is

approximately

E X AM HINT n 2

V







 f

x





i



 x .

This section has



i 1

given an indication

As

δx

→

0,

the

approximation

will

become

more

accurate.

of the origin of

n

the formula rather

2

Hence,

V



lim





 f 



x i



 x



than a proof of the

 x 0 i 1

formula. Knowledge Recall

from

Section

11.1

that

we

can

write

the

limit

with

the

integral

of the derivation is sign

and

δx

that

becomes

dx.

not required for the

Hence,

radians

the

formula

about

the

b

V

a

x-axis

volume

between

formed

x

=

a

by

and

rotating

x

=

b

y

=

f(x)

through

2π

examinations.

is

2





for





f



x

dx .



TOK

a

We

by

can

now

rotating

between

x

use

f(x)

=

0

=

the

formula

cos(0.5x

and

x

=

−

to

1)

nd

+

2

the

volume

through

2π

of

the

radians

solid

about

formed

the

x-axis

12:

You have been using

radians to measure

angles instead of degrees

in recent chapters.

12

2 3

V

  cos







 0. 5 x

 1



2

dx

 167. 27 unit

(2

dp)



Why has this change

0

been necessary? What To

nd

the

volume

formed

when

rotating

about

the

y-axis,

we

simply are its advantages?

interchange

x

and

y

504

/

11 . 3

Volumes of revolution

•

The volume formed when a continuous

function y

=

y

y

f(x) is rotated through

=

f (x)

2π radians about the x-axis is

b

b

y 2

V



2

  f ( x )

or V

dx











y



dx

x a

a

O

dx

y

•

The volume of the solid formed by

revolution of y

=

f(x) through

2π radians about the y-axis in the

interval y

=

c

to y

=

x

=

g (y)

d

d is

d

2

V



x

dy

x

dy

 c

c

x O

Example 17

2x f

(

x

)

a

the

a

4 ,

x

rotated

+

=

0

≤

x

≤

4.

Find

the

volume

of

revolution

formed

when

the

curve

f(x)

is

+ 1

through

2π

b

x-axis

radians

the

about

y-axis.

Sketch

y

the

graph

of

the

curve.

6

5

4

suluclaC

Let

f (x)

3

2

1

x 0 1

2

3

 2x 

0

V

≈



6

4 





5

2

4

V

4

dx  

Use

the

formula

and

evaluate

the

denite

 x

 1

 integral

on

the

GDC.

101(3 sf)

Continued

on

next

page

505

/

11

A P P R O X I M AT I N G

b

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

Sketch

y

4

the

AND

graph

DIFFERENTIAL

of

the

E Q U AT I O N S

curve.

f (x)

2.4

x 0

x

=

0

⇒

2x y

y



=

4

and

4

=

 y



x



x

x

4



⇒

y

=

2.4

Find

4

The

the

boundaries

formula

for

requires

x

y.

as

a

function

of

y,

so



y

 1



we

2

need

Use

2

the

to

rearrange

formula

and

the

formula.

evaluate

on

a

GDC.

4

  y V





≈

4  dy

 

2.4

V



  y

 2



9.93(3 sf)

Example 18

x

a

Sketch

b

Find

the

intersection

c

Find

the

volume

f(x)

rotated

a

=

cos x

through

andg(x)

of

2π

=

points

e

of

revolution

radians

on

the

the

two

formed

about

the

same

axes

curves

when

in

the

the

0

≤

x

given

region

≤

1.5.

interval.

enclosed

by

the

curves

f

and

g

is

x-axis.

Sketch

y

for

the

graphs

within

the

given

domain.

1.2

Use

(0, 1)

your

GDC

to

nd

the

intersection

points

1

remembering

to

use

at

least

3

sf.

0.8

0.6

0.4 (1.29, 0.275)

0.2 g(x)

f(x) x 0 0.5

b

1

Intersection

(1.29,

1.5

2

points

2.5

are

(0,

1)

and

0.275).

1.29

1.29 2

2 x

c

V







 cos x



0

dx







e







dx

x

If

we

the

two

≈

0.993(3 sf)

the

volume

formed

by

y

=

e

0

from

V

subtract

Use

the

volume

volume

formed

formed

by

by

the

y

=

cos

region

x,

we

get

between

the

curves.

your

GDC.

506

/

11 . 3

Exercises 11I

1

Each

on

of

the

the

following

domain

the

x-axis

by

the

generated

0

2π

≤

x

functions

≤

1

and

radians.

solids

of

are

rotated

Find

the

a

sketched

Write

about

volumes

down

volume

of

2π

integrals

obtained

by

that

represent

rotating

[AB]

the

by

radians:

revolution.

i

about

the

x-axis

ii

about

the

y-axis.

2

a

a(x)

=

b

x

b(x)

=

x

3

c

2

c(x)

The

=

x

d

x

same

parts

of

the

e(x)

=

e

functions

b

Calculate

are

rotated

about

integrals

and

conrm

your

in answers

question1

the

the

using

the

formula

for

the

y-axis 1 2

by

2π

radians.

Find

the

volumes

of

volume

the

of

a

V

cone:

 r



h.

3 generated

solids

of

revolution.

5

3

A

teardrop-shaped

Its

shape

is

based

earring

on

the

is

being

graph

Hany

an

designed.

and

Ahmed

American

are

trying

football.

of

to

Hany’s

model

model

is

2

s( x )

=

-4

+

=

4.5

72. 25 -

x

and

Ahmed’s

is

x 2

y

=

900 -

x

for

0

≤

x

≤

30

2

(where

f(x)

−

0.079x

100

a x

and

y

are

measured

in

mm),

which

Choose

the

estimate revolved

by

2π

radians

around

the

x-axis

model

and

hence

the

volume

of

the

American

as

football shown

best

is

in

cubic

units

using

the

below.

information

given

in

the

diagram.

F

(0, 4.5)

10 10 y

5 5

y

5

=

0

0

–5 –5 –10

E

0

=

=

(7.5, 0)

–10 10

0

10

20

30

x 20

40

–9 30

x

40 r

2

x



2

Find

900 



s

x

dx



 –5

100



b

Hence,

c

If



nd

the

volume

of

one

earring.

b

3

1 cm

gold

costs

used

4

The

for

line

where

of

A

gold

£30,

the

(0,

nd

the

19 g,

cost

and

of

1 g

the

11

of

4)

[AB]

and

B

is

=

placed

(3,

as

that

the

inches,

length

convert

of

your

the

football

answer

to

is

part a

into

gold

earring.

segment

=

weighs

Given

shown,

i

cubic

inches

ii

cubic

centimetres.

0).

6

The

following

regions

are

rotated

about

y

the

x-axis

by

2π

radians.

For

each

suluclaC

a



region:

5

i

Sketch

ii

Write

the

information

given.

A 4

3

down

quantify

iii

2

Hence,

the

the

integrals

volumes

calculate

the

that

required.

volumes.

2

a

The

region

bounded

by

x

=

0,

y

=

x

and

bounded

by

y

=

0,

y

=

ln(x)

1

x

y

=

e

.

B x 0

b 1

2

3

The

region

and

y

4

2

=

4

−

x

.

507

/

11

A P P R O X I M AT I N G

7

a

Use

technology

to

IRREGUL AR

sketch

the

S PA C E S :

functions

2

y

=

40 +

4

-

I N T E G R AT I O N

The

and

y

=

40 -

4

-

Sketch

the

the

region

shape

generated

bounded

by

y

by

and

the

x-axis

Write

the

its

8

In

be

the

O2

thought

of

Arena

as

a

in

segment

of

a

x

down

volume

diagram

with

[AD]

2π

an

of

with

length

a

radius

365

r

m

formed

by

a

chord

m.

about 1

radians.

integral

this

that

shape

quanties

and

hence,

nd

volume.

the

circle

by

of

rotating

y

2

c

can

of

E Q U AT I O N S

1

circle

b

DIFFERENTIAL

cross-section

London

2

x

2

AND

below,

radius

passes

r

A

and

through

C

is

the

the

the

centre

line

a

segment

middle

D

of

of

[BC].

B

Given

and

that

the

the

height

equation

of

a

of

circle

2

centre

at

(0,

0)

a

the

value

b

the

volume

is

x

the

2

+

y

O2

of

is

50 m

radius

r

with

2

=

r

,

nd:

r

A

a

Prove

that

ˆ AD B

is

of

r

of

the

O2.

90˚.

K inematics TOK

The

following

results

and

acceleration

for

an

object

with

displacement

s,

velocity

v Does the inclusion of

ds a

at

time

t

were

derived

in

Chapter

10:

v

=

and kinematics as core dt

2

dv

d

s mathematics reect

a

=

= 2

dt

dt

a par ticular cultural

heritage?

We

can

now

reverse

the

process

to

obtain

s



v dt

and

v





a dt .



Example 19

−1

A

stone

above

Take

is

the

the

thrown

vertically

upwards

with

an

initial

velocity

of

4.9

m s

from

a

point

2

m

ground.

ground

as

the

origin

of

the

coordinate

system

with

the

upward

direction

as

positive.

2

The

acceleration

due

to

a

Find

the

maximum

b

Find

the

speed

with

gravity

height

which

is

a

=

−9.8 m s

reached

it

hits

by

the

the

.

stone.

ground.

508

/

11 . 3

a

a

=

v



−9.8

9 .8 dt



9 .8t



Using

c

v



a dt .





When

⇒

s

v

=



t

=

0,

4.9

v

−

=

4.9

⇒

c

=

Using

4.9

the

boundary

condition

to

nd

c.

9.8t

Using

(4.9  9.8t )d t

s



v dt .





2

=

4.9t

When

t

−

4.9t

=

0,

s

+

=

2

c

⇒

t

=

2

2

⇒

s

=

4.9t

When

v

−

=

4.9t

0,

t

+

2

4

9

9

8

=

=

0

The

5

maximum

velocity

is

height

equal

to

will

occur

when

the

0.

Hence,

2

s

=

4.9

×

0.5

−

4.9

×

0.5

+

2

=

This

3.225

value

for

displacement

maximum

b

When

s

=

When

0,

2

4.9t

⇒

t

−

4.9t

=

+

1.31

2

=

=

4.9

−

is

put

equation

to

into

the

nd

the

height.

the

stone

hits

displacement

will

displacement

from

the

be

0

ground,

as

we

are

its

measuring

0

ground

level.

seconds

2

v

time

9.8

×

1.31

1

=

−7.95

The

m s

velocity

moving

in

is

negative

the

negative

as

the

stone

is

direction.

1

Speed

=

7.95

The

m s

the

question

asks

magnitude

of

for

the

the

speed,

which

is

velocity.

Example 20

A

particle

moves

so

that

its

velocity

at

time

t

is

given

by

the

equation

2

=

4t

sin(t

a

Given

b

Find

).

that

the

it

is

rst

initially

time

at

at

the

which

it

origin,

nd

returns

to

an

its

expression

starting

for

its

displacement

at

time

t.

suluclaC

v

point.

2

a

s





4t sin

t

 dt

“Initially”

implies

when

t

=

0.

2

=

−2cos(t

0,

+

s

c

When

t



0

 2 cos 0 



c



=

)

=

0

c

It

is

important

always 

be

not

equal

to

to

0

assume

just

that

because

c

will

initially

2

s

=

0.

2

s

=

2

−

2cos(t

)

Continued

on

next

page

509

/

11

A P P R O X I M AT I N G

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

AND

DIFFERENTIAL

E Q U AT I O N S

2

b

0



2  2 cos

t



2



cos

First

t

The

zeros

can

and

solving

be

or

found

by

from

drawing

a

rearranging

graph.

 1



positive

solution

is

when

2

Y

=

2–2cos(X

)

1

t

=

2.51

seconds

Zero

X

=

2.5066281

Y

=

0

Exercise 11J

1

a

Find

and

expression

for

displacement

the

of

a

3

velocity

particle

A

particle

moves

so

that

its

velocity

with

given

by

the

equation

v

(t )

=

(t )

t

for

k

>

0.

acceleration

of

2

3 a

time

t is

acceleration

at

kt

,

=

t

≥

0,

if

e

the

2

(t

particle

b

is

initially

Determine

particle

to

the

at

time

reach

a

+

1)

rest

it

at

will

the

take

distance

of

At

time

it

a

Determine

the

50 m

the

the

origin.

the

maximum

Find

the

set

of

values

of

k

for

which

the

origin.

particle

at

from

from

−1

A

1 cm

particle.

particle

2

is

origin.

b the

0,

time

t

moves

seconds

so

is

that

its

given

velocity

by

the

(m s

travel

at

least

5 cm

from

the

origin.

)

formula

will

4

a

Show

that

the

vertical

height

of

a

projectile

2sin3t

v

=

e

cos3t.

after

t

seconds,

if

it

is

acted

on

only

by

2

a

gravity

Find:

(acceleration

g

=

−9.8 m s

)

1 2

i

an

expression

the

for

the

acceleration

of

is

given

the

s

(t )

=

t

+

v

t

+

y

0

particle

time

gt

,

where

0

2

v

ii

by

at

which

the

is

its

initial

vertical

velocity

and

0

particle’s

its

y

is 0

initial

vertical

position.

−2

acceleration

is

rst

equal

to

−0.5 m s

b b

Find

an

expression

for

the

Hence

determine

maximum of

the

particle

at

time

(s)

from

its

initial

Write

c

t

down

the

exact

value

of

displacement

of

height

Determine,

for

the

reached

by

the

particle.

rocket

with

red

reasons,

with

an

whether

initial

vertical

the velocity

maximum

formula

position

a

c

a

displacement

of

310 m/s

from

10 m

above

the ground

level

is

capable

of

reaching

a

particle. target

at

a

height

of

4 km.

510

/

11 . 3

Area under a velocity–time graph

Investigation 6

The graph shows a velocity−time graph for the journey of an object.

v

The horizontal axis shows the time taken from the star t of the journey in

10

1

seconds and the ver tical axis shows the velocity of the object in m s 8

The equation for the velocity of the par ticle is

v = 10 − 2t,

0 ≤ t ≤ 8.

6

1

Explain why the par ticle has its maximum displacement when

2

a

t = 5.

4

Find an equation for the displacement ( s) of the par ticle, given that

2

s = 0 when t = 0.

b

t

Sketch the displacement graph for 0 ≤ t ≤ 8. 0

3

a

Find the maximum displacement of the par ticle.

b

Find the displacement at t = 8.

–2

–4

4

Hence, write down the total distance travelled by the particle for 0 ≤ t ≤ 8.

5

Find the area between the curve, the t

Conceptual

6

–6

axis, t = 0 and t = 8.

How do you nd the distance travelled by a par ticle?

8

7

Explain why this is not the same as

( 10  2t )d t .



0

b

b

∫

f (t)

and ∫

dt

a

f (t) dt • •

a

b

The

value

of

v



t 

dt



s

b 



s

a ,

which

is

the

change

in

a

displacement

Consider

by

v(t)

=

an

of

the

object

particle

moving

v

along

a

t

=

line

a

and

with

a

t

=

b

velocity

at

time

t

given

cos t.

ds Using

between

=

dv and

a

dt

=

,

we

obtain

dt

b

a

t 

dt



v

v

t 

dt



s



b 



v

a ,

the

net

change

in

the

velocity

of

the

particle

a



b 

 b

 a ,

the

net

change

in

the

displacement

(or

suluclaC

b

position)

a

of

the

particle.

The denite integral of a rate of change in a given interval calculates the net

change. This is a restatement of the fundamental theorem:

If f is a continuous function on a

≤

x

≤

b and F is any antiderivative of f then

b

f





x



dx



F

b 



F(a)

a

If the rate of change of a function F(x) is given by f(x) then the total change

b

in F between a and

b is

f





x



dx

a

511

/

11

A P P R O X I M AT I N G

As

v

is

the

rate

of

IRREGUL AR

change

of

s

then

S PA C E S :

the

total

I N T E G R AT I O N

distance

AND

DIFFERENTIAL

E Q U AT I O N S

travelled

b

between

t

=

a

and

t

=

b

is

v



t 

dt

a

There

are

various

applications

of

total

and

net

change

derived

from

TOK equations

equation

the

net

for

for

rates

a

rate,

change

modulus

of

beyond

the

denite

between

the

kinematics.

the

function

integral

two

will

Whenever

values

give

us

of

the

and

the

are

function

the

total

we

given

will

denite

an

give

us

integral

of

Why do we study

the

mathematics?

change. What’s the point?

For

example,

our

bank

calculate

we

if

we

account,

the

would

total

use

would

we

would

value

the

like

total

of

to

use

all

know

net

how

much

change,

transactions

but

into

money

if

we

and

is

left

would

out

of

the

in

like

Can we do without it?

to

account,

change.

Example 21

a

pipe.

During

periods

of

low

demand,

water

is

pumped

back

up

the

pipe

to

the

reservoir

above.

Let

the

leaves

the

volume

the

pipe

lake

water

during

during

dV

a

this

15

8 .2 sin



where

t

is

the

reservoir

period

period

is

is

be

V

through

given

by

and

the

the

assume

pipe.

that

The

following

rate

the

at

only

water

which

water

that

enters

ows

or

through

equation:





t

 5

 dt

in

24-hour



 

of

 12

12

measured

in



hours

after

midnight

and

V

is

measured

in

millions

of

litres.

dV

a

Sketch

the

curve

for

against

time

for

a

24-hour

period.

dt

b

By

in

c

calculating

the

By

reservoir

calculating

passed

d

i

an

Determine

volume

is

over

an

through

appropriate

V

24-hour

appropriate

the

a

a

pipe

formula

at

denite

in

a

for

integral,

nd

the

net

change

nd

the

total

in

the

volume

of

water

period.

denite

24-hour

the

integral,

amount

of

water

that

has

period.

volume

of

water

in

the

reservoir

at

time

t,

given

that

the

midnight.

0

ii

Find

the

value

of

V

when

t

=

24

and

use

this

For

dV

a

value

to

verify

most

of

your

the

answer

time

the

to

part

rate

b

is

dt

negative,

indicating

that

water

is

4

owing

out

of

the

reservoir.

2

t 0 16

18

20

22

24

–2

–4

–6

–8

–10

–12

–14

512

/

11 . 3

24

b

V













15

8 .2 sin

t



  5













dt

To

nd

the

net

change,

you

take

the

0

12





12

denite

integral

between

the

two

3

=

−120 m points.

24

c

V







15

8 .2 sin



t

  5





dt

To



nd

the

total

change,

you

take

0



12

12

 the

denite

integral

of

the

modulus

3

=

149.4 m

function



d

i

V











15

8 .2 sin

t

 5





12

12









dt

The

reverse

integrate

98

4







15 t

cos

When

=

0,

points.

chain

the

rule

is

needed

to

expression.

 5t



c

 12



t

two









the









between

V

=

12



V 0

98 ⇒

c



V

4





98

4









12

15

cos









0

V

 15 cos

t

ii

When

t

98 V



=

 5 t

12

15

 2

cos

 V



22. 15

0



 120  V  12



−120

+









=



 12

22. 15

24,

4



 0









V



22. 15

0



V 0

So

the

difference

is

−120

+

V

−

V

0

=

The

−120

difference

is

independent

of

V

in 0

0

the

same

know

the

denite

Both

are

way

+c

term

methods

acceptable

do

when

not

need

working

to

out

b

is

of

of

working

but

out

generally,

technology,

the

more

a

change

given

the

the

method

straightforward.

suluclaC

part

you

integrals.

availability

of

as

Exercise 11K

1

In

an

then

experiment,

heated.

temperature

The

of

a

liquid

rate

the

of

is

cooled

change

l i q uid

is

of

2

and

The

graphs

the

graphs

as

period

g iv en

of

of

show

four

the

particles

time.

The

=

t

4t

and

−

8

where

T(t)

the

t

is

the

time

temperature

in

in

y-axis

t

=

8,

is

velocity

For

each

how

than

b

the

many

it

degrees

was

total

that

time

(s)

and

the

(m s

).

graph,

nd

the

total

distance

nd: travelled

a

is

5-second

°C.

a

When

a

1

−

minutes

for

x-axis

2

T’(t)

velocity–time

at

t

=

change

higher

the

liquid

is

5

and

the

displacement

after

seconds.

0

in

temperature

over

period.

513

/

11

A P P R O X I M AT I N G

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

a i

Particle

ii

A

Particle

Find

AND

the

=

10

v(t)

=

10

−

b

travelled

by

the

after

5

ii

seconds

Calculate

the

time

10

seconds.

taken

for

the

particle

y

to 12

return

to

its

original

position.

12

5

A

suspension

Immediately

heavy

load,

bridge

after

the

is

the

being

designed.

bridge

vertical

sustains

movement

a

of

a

2

2

x

–1

x

2

–1

point

on

iv

C

Particle

5t

oscillation

=

10

−

10  4t ,

4t v(t )



is

modelled

with

a

−

0.08616

model,

sin

5t),

v(t)

=

5(0.65 )

where

time

is

D measured

v(t)

bridge

t

(cos

Particle

the

2

damped

iii

distance

2t

i

y

total

E Q U AT I O N S

B

particle v(t)

DIFFERENTIAL

0



t



in

seconds

and

v

is

measured

1

2 .5

in

cm s





4t

 10,

2 .5



t



5

a

Calculate

5

y

seconds

the

displacement

after

sustaining

of

a

the

point

heavy

load,

and

12

b

Find

the

point

at

total

this

distance

travelled

by

the

time.

y

6 2

Two

objects

A

and

B

are

launched

from

12

a

x

balloon

at

a

height

of

1000 m,

with t

–1

2 1

16

velocity

functions

v

(x)

 16  4 e

m s

a

t

2 1

16

and

x

–1

–12

v

( x)

 18  7e

m s

Identify

which

of

the

particles

a

changes

i

is

positive.

0

ii Use

technology

to

help

sketch

the

taken

Graph

direction.

3

≤

t

≤

both

graphs

for

each

of

downward

the

as

velocity

functions

for

20.

Calculate

after

how

many

seconds

the

velocity– objects

time

The

2

direction

b

.

b

are

rst

travelling

at

the

same

following velocity.

moving

bodies

distance

the

and

travelled

time

hence,

and

intervals

the

given.

nd

the

total

displacement

Time

is

in

b

on

Calculate

at

hours

this

how

far

above

the

ground

A

is

time.

1

and

velocity

in

km h

7





a

v(t)

=

cos t,

derivative

 2

=

3.1sin 2t

is

given

−

energy,

as

=



E (mJ),

of

a

sin

( 0.5t )

+ a.

It

is

dt

known

v(t)

the

dE particle





b

of



0,



The

3

1,

,

energy



that

the

change

in

the

particle’s

 between

t

=

0

and

t

=

π

is

4

mJ.

 6



2

 Find

the

value

of

a

2

c

v(t)

=

17

−

2t

,

[1,

4.1]

8

The

rate

of

change

in

the

area

( A)

covered

t

d

v(t)

=

1

−

e , [0,

by

2.39]

mould

on

a

petri

dish

after

t

days

is

given

dA rt

4

A

particle

moves

with

velocity

function

as

=

2e

where

r

is

t

=

the

rate

of

growth.

dt

4,



0



t

 5 Between

the

times

2

and

t

=

4,

the

area



2

4

 v(t )





covered

by

the

mould

increases

by

7.2

cm

.

2

4

(t



)  5

,

5



t

 10 Find

25

the

value

of

r.





1.6t

 16,

t

 10



where

time

distance

in

is

measured

in

seconds

and

metres.

514

/

11 . 4

9

The

rate

of

chang e

mosquitoes

( N ),

is

by

modelled

of



4

t

numb er

measured

the

in

b

of

By

10 0 0 0 s ,

t

much

mosquitoes

1

cos

how

does

the

population

of

increase:

e q uatio n

 

dN

the

where



t

is

i

the

rst

ii

the

second

year

year.

measured

 dt

6



in

months

(1

January).

2 The

from

the

beginning

of

a

year

actual

c

a

Find

in

which

population

11.4

of

month

each

mosquitoes

year

is

a

average

Find

the

for

increase

increase

the

the

divided

average

rst

in

two

population

by

increase

the

in

is

the

duration.

population

years.

minimum.

Dierential equations

International-

A dierential equation is an equation that contains a derivative.

mindedness

dy 2

For

example,

=

2x

French mathematician

+ 5.

dx

You

have

Jean d’Alember t’s

already

met

several

of

these

and

solved

them

by

integrating.

analysis of vibrating

strings using When

no

boundary

condition

is

given

then

the

solution

to

the dierential equations

equation

will

have

an

unknown

constant

(the

+c

term).

If

a

boundary plays an impor tant role

condition

is

given,

the

particular

solution

can

be

found. in modern theoretical

dy The

equation

above

is

written

in

the

physics.

form



f



x

,

but

often

the

dx

context

of

the

question

will

mean

that

the

differential

equation

is

of

dy the

form

=

f ( x , y).

These

equations

can

often

not

be

solved

directly,

dx

so

numerical

solutions

need

to

be

found.

dy One

situation

in

which

they

can

be

solved

is

when



f



x



g



y

.

dx

form

is

called

a

separable

1 It

can

be

written

=

with

respect

1

to

dx



(

x

)



can



f



x

be

shown

dy

sides

are

then

integrated

1 dx

that



two

dx

the

s.

This

reverse

is

dx

as

if

we

were

cancelling

 g

using

both

 dx



the

and

dx

1 It

f

dx

)

x:

 y

y

dy

 g

(

equation.

dy

as

g

differential

suluclaC

This



very

chain

y



dx

g

similar

to

the



y



process

used

when

integrating

rule.

1 Both

sides

of

the

dy

expression





f

 g



y



x



dx

can

now

be



integrated.

515

/

11

A P P R O X I M AT I N G

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

AND

DIFFERENTIAL

E Q U AT I O N S

dy In

practice,

many

people

often

go



from

f



x



g



y

to



dx

1 dy

g



y

this



f



x



dx

and

then

add

the

integral

operators.

Although



middle

stage

is

meaningless,

notationally

it

can

be

a

convenient

shortcut.

Example 22

dy 2

Find

the

solution

of

the

differential

equation

y

=

x

+ 1

given

that

y(0)

=

4.

dx

2

ydy



(x



 1 )d x

The



variables

have

been

separated

have

an

to

give

2

y dy

1

=

(x

+

1 2

3

y

=

x

2

+

x

+

c

We

do

not

need

to

integrating

3 constant,

If

the

in

a

the

y

1)dx.

=

⇒

4

8

when

=

x

=

0

c,

for

question

particular

answer

Substitute

both

asks

sides

for

form,

the

the

we

obtained

the

equation.

solution

may

from

given

of

need

the

values

to

to

be

given

rearrange

integration.

to

nd

c.

c

1

1 2

So

y

3

=

2

x

+

x

+ 8

3

Example 23

dy 2

Find

a

solution

in

the

form

y

=

f(x)

to

the

differential

equation

=

6x

y

given

that

y

=

2

dx

when

x

=

0.

1 2

dy



6x



dx



Separate

the

variables

and

integrate.

y

3

ln y

=

2x

+

c

2

2x

y

=

Ae Converting

when

so

the

The

that

full

y

=

⇒

2

2

when

=

x

=

0

=

e

a

it

is

often

c

=

Substitute

e

to

nd

coefcient

log

operator

done

in

a

is

single

2

2x

e

a

is

2

+ c

into

natural

process

2

2x

y

term

eliminating

common

step.

+c

2x

=

Ae

c

where

A

=

e

.

A

A

2

2x

So

y

=

2e

516

/

11 . 4

Such

the

an

exponential

growth

or

decay

model

of

a

from

a

variable

differential

is

equation

proportional

to

the

occurs

whenever

amount

present.

Example 24

The

rate

of

growth

of

mould

on

a

large

petri

dish

is

directly

proportional

to

the

amount

of

2

mould

present.

a

Find

an

b

Find

the

The

area

expression

covered

for

the

by

area

the

mould

covered

by

( A)

is

initially

the

mould

at

2

2 cm

and

time

t

after

2

days

it

is

9 cm

.

days.

2

value

of

t

at

which

the

mould

will

cover

20 cm

dA

a

=

kA If

two

variables

a

and

b

are

proportional

to

dt each

other,

then

a

=

kb.

1 dA



kdt



 A

ln

A

=

kt

+

c

kt

A

=

Be

A

=

2

c

where

B

=

e

0

⇒

There

so when

t

=

B

=

are

two

two

sets

of

unknowns

boundary

in

the

equation,

conditions

need

to

2

be

used.

kt

A

=

2e

A

=

9

when

t

=

2

2k

⇒

9

=

2e

This

equation

solved

using

and

the

the

log

one

laws

in

or

part

b

directly

can

on

be

a

9 2k



⇒

ln

k

=

0.752

=

3.06 days

GDC.

2

0.752t

So

A

20

=

=

2e

0.752t

b

2e

1 t

=

ln 10

0 .752

suluclaC

Exercise 11L

dy

1

Solve

4

=

xy ,

given

that

y

=

2

when

x

=

b

0.

Given

that

the

amount

initially

is

40 g

dx and

there

hours,

dy

is

nd

only

the

20 g

value

remaining

of

after

2

k

2

2

Solve

(

x

- 1

2 xy,

=

)

given

that

y(2)

=

9. 2

dx

4

a

Find

the

derivative

of

y

=

4

-

x

.

2

3

a

The

rate

of

substance

decay

is

of

a

b

radioactive

proportional

to

the

Verify

that

g

remaining.

Form

a

=

4

-

x

amount

satises

dy differential

M

y

equation

x =

in

M

and

show

that

this

dy

give

M

=

Ae

c where

A

and

k

Solve

x =

to

-

nd

the

general

are dx

positive

y

solves

kt

to

-

differential dx

equation

the

y

constants. solution

to

the

differential

equation.

517

/

11

5

A P P R O X I M AT I N G

Solve

the

following

IRREGUL AR

differential

equations.

I N T E G R AT I O N

9

dy x +

=

y



y

cos

b

e

makes

a

y

variety

of

the

following

of

given

nd

the

initial

dy

a

tracer.

of

It

2.8

is

days,

used

in

solution

that

including

of

blood

cell

components,

rare

cancers

etc.

satises In

the

early

19th

E.

Rutherford

century,

F .

Soddy

and

condition.

formula ,

methods,

differential

x =

dx

half-life

useful

diagnostic

labelling

diagnosing

the

a

short

 1 isotopic

equations,

it

a



dx

each

has

E Q U AT I O N S

x

2

For

DIFFERENTIAL

y

dx

6

AND

Indium-111

which

dy

a

S PA C E S :

y(1)

=

derived

the

radioactive

decay

as

2

2y dN =

dy

2x

 sin x



b

,

dx

- kN ,

y(0)

=

y’

tan x

=

radioactive

tan y,

y(2)

=

Newton’s

law

of

is

the

amount

of

a

cooling

material

that

and

depends

k

on

is

a

the

positive

radioactive

2 substance,

7

N

1

y

constant

c

where

dt

states

that

the

with

T

being

the

half-life

of

the

substance.

rate

ln 2 dT

a of

change

of



temperature,

,

of

an

Show

that

the

formula

for

k

is

k



object T

dt

is

proportional

its

own

to

the

temperature,

temperature,

T

(ie

difference

T(t),

the

and

b

between

the

If

after

ambient

temperature

of

its

N(0)

10

A

1

=

5 g,

nd

the

mass

remaining

day.

cylindrical

tank

of

radius

r

is

lled

with

a

2

water.

dT surroundings),

that

is

=

- k(T

- T

A

hole

of

area

A

m

is

made

in

the

).

a

bottom

dt

Suppose

coffee

room

a

b

you

with

a

where

have

the

most

quickly,

initial

of

time

at

which

justifying

differential

conditions

a

cup

75°C

temperature

the

a

poured

temperature

State

Write

just

is

the

of

in

equation

v

velocity

of

water

through

the

hole

is

m s

.

The

volume

of

water

in

the

tank

at

3

24°C.

time

cools

answer.

using

tank.

-1

a

coffee

your

The

of

the

t

seconds

Let

h

the

tank

a

be

the

height

after

Given

after

t

that

the

of

hole

is

water

made

is

V

remaining

m

in

seconds.

v

=

2 gh ,

show

that

given. dV =

c

Solve

the

differential

-A

2 gh .

equation. dt

d

Explain

from

the

equation

why

the dV

temperature

will

never

go

below

24°C.

b

i

Hence,

show

that

=

-k

V

where

k

dt is

8

The

is

to

number

observed

the

of

to

bacteria

grow

number

of

at

cells

a

in

a

liquid

rate

ii

proportional

present.

At

a

constant.

culture

State

and

the

the

value

of

k

in

terms

of

A,

r

g.

4

beginning

of

the

experiment

there

are

10

The

volume

of

the

cylinder

when

full

is

V

. 0

5

cells

and

after

3

hours

there

are

3

×

10

c

Find

how

long

it

will

take

the

cylinder

to

cells. empty

in

terms

of

k

and

V 0

a

Calculate

1

day

of

the

number

growth

if

this

of

bacteria

rate

of

after

growth

continues.

b

Determine

the

doubling

time

of

the

bacteria.

518

/

11 . 5

Investigation 7

In this investigation, you can assume that the number

i can be treated the

same as any other number when dierentiating or integrating.

dx

1

Solve



ix

given x = 1 when θ

= 0

d

Let x = cos θ + i sin θ

Verify that x = 1 when θ

2

= 0

dx

Show that



ix

d

iθ

Conceptual

3

What can you conclude about e

?

iθ

= cosθ

This is called Euler ’s formula: e

11.5

+

isinθ

Slope elds and dierential

International-

mindedness

equations Dierential equations

is

not

explicit

elds,

They

always

formula

are

are

nd

used

drawn

produced

and

possible

even

an

for

to

the

to

solve

unknown

represent

using

when

explicit

is

not

formula

function.

graphical

tangent

it

differential

lines

at

possible

for

the

equations

Slope

solutions

specic

to

solve

unknown

of

elds,

nd

or

They

an

rst became solvable

direction

differential

points.

the

and

equations.

can

differential

function.

often

be

equation

with the invention of

calculus by Newton

and Leibniz. In Chapter

2 of his 1736 work

Method of Fluxions,

Newton described three

types of dierential

equations.

Investigation 8

dy

Consider the dierential equation

y

y

- 3x

suluclaC

It

= 5

dx

2

The gradient of the tangent drawn to the curve

•

at (1, 1) will be −1

•

at (1,

4

y

3

1

2) will be

2

.



2

1

These points can be plotted on a grid and small line segments drawn x 0

1

with gradients of −1 and

, respectively, as seen in the diagram.



–1

2 –2

The process can be continued and is shown below for all the integer

values of x and y in the intervals −4 ≤ x ≤ 4 and −4

≤

–3

y ≤ 4. –4

Continued

on

next

page

519

/

11

A P P R O X I M AT I N G

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

AND

DIFFERENTIAL

E Q U AT I O N S

y

5

4

3

2

1

x 0 –5

–4

–3

–2

–1

1

2

3

4

5

–1

–2

–3

–4

–5

This type of diagram is called a slope eld.

The slope eld indicates the family of curves that satisfy the dierential equation. At each point the direction

of the curve is the same as the direction of the tangent. This means that given a star ting point (a boundary

condition), a solution curve can be drawn.

One way to think about this is that the slopes are like the current in a river and the curve follows the path a

cork might take if released in the river at that point.

The diagrams below show the solution curves that pass through

(−1,

0) and (−1,

y

1).

y

4

4

3

3

2

(–1, 1)

1

1

(–1, 0)

x

x

0

0

–1

–1

–2

–2

–3

–3

–4

–4

Cer tain features of the family of curves satisfying the dierential equation can be worked out from the

dy

y

- 3x

=

equation itself. For example, the maximum points will occur when dx

=

0.

2

So, they will lie on the line y = 3x

520

/

11 . 5

Exercise 11M

2

1

Consider

a

Find

of

the

the

the

differential

missing

tangents

equation

values

to

the

of

the

y’

=

x

+

y.

3

One

−3

integer

≤

y

≤

points

solution

for

−3

≤

x

≤

the

following

differential

equation

other

the

slope

y’

=

elds

cos(x

curves

3

is

for

differential

=

cos(x

−

0

1

2

3

−2

−3

−2

1

6

−1

−2

−1

2

7

−1

−3

6

1

−2

7

2

and

the

equation

y).

A

−2

y),

the

and

3.

−3

for

at

B

y

y

+

is

gradients

y’ the

of

y

x

x

−1

8

3

0

−1

0

3

8

0

9

4

1

0

1

4

9

1

10

5

2

1

2

11

6

3

2

3

12

7

4

3

a

i

For

write

b

Sketch

the

slope

eld

for

the

y’

=

x

cos(x

down

the

+

ii

y)

y’

equation

of

=

a

cos(x

line

−

y)

along

differential which

the

slope

eld

would

have

a

equation. gradient

c

Use

the

slope

eld

to

sketch

the

that

passes

through

(2,

State

By

evaluating

the

gradient

at

or

otherwise,

match

each

below

with

a

slope

y’

=

cos(x

TheGompertz

dy

dy =

x

-

b

2

=

dx

slope

eld

for

+

ii

y)

y’

=

cos(x

−

y)

equationis

a

model

that

used

to

describe

the

growth

of

certain

eld. populations.

a

the

differential is

equation

is

various

4 points,

which

−2).

i

x

+

In

the

1960s,

the

Gompertz

curve

to

t

the

growth

tumours.

AK

the

Laird

data

used

on

y

dx

cellular

of

population

A

tumour

is

a

growing

in

a

conned

availability

of

nutrients

dy 2

c

=

y

+

d

2

y’

=

x

2

+

space

y

where

the

is

dx limited.

i

ii IfT(t) y

is

the

size

of

a

tumour

then

under

y

dT this



model

dt

a

x

x

0

Sketch

6

a







.



slope

months,

 T T ln

0

≤

eld

t

≤

6.

2



for T(t)over

Take

0

≤

T

the

≤

rst

3.

0

b

On

your

shows

slope

the

eld,

growth

sketch

of

the

a

curve

tumour

that

suluclaC

2

zero.

solution

b curve

of

given

3

that

iii

iv

y

c

its

Write

volume

down

is

the

0.5 cm

when

maximum

t

=

0.

possible

size

y

of

d

a

tumour

according

Explain

why

suitable

for

this

to

model

modelling

this

model.

might

the

be

growth

of

a

tumour. x 0

x 0

521

/

11

A P P R O X I M AT I N G

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

AND

DIFFERENTIAL

E Q U AT I O N S

Euler ’s method for numerically solving dierential

equations

The

idea

used

in

slope

elds

can

also

be

applied

to

obtain

numerical

Internationalapproximations

of

differential

equations.

mindedness

dy Consider

the

differential

equation

=

f

(

x, y

)

dx

Let

(x

, n

y

1

) n

be

a

point

on

the

The Euler method is

curve

and

let

x

1

= n

x

+ n

h

(h

>

0).

named after Swiss

The

1

mathematician method

will

use

the

gradient

of

the

curve

at

(x

, n

y

1

) n

to

evaluate

1

Leonhard Euler y

,

an

estimate

of

the

y

coordinate

of

the

point

on

the

curve

with

x

n

(pronounced “oiler ”),

coordinate

x

. who proposed it in his

n

y

book Institutionum The

gradient

of

the

(x

,

f(x

tangent

at

the

point

Calculi Integralis in

n

y

1

)

n

is

1

,

n

y

1

).

n

The

diagram solution

1

cur ve

1768.

y

-

y

n

shows

that

it

is

n

1

also

.

(x

Hence,

, y

n

)

n

h

y



y

n

f



x

, y n

n

n

1





,

which

can

be

rearranged

h

h

y n–1

to

give

y

=

y

n

+ n

Generally,

hf(x

1

we

, n

are

y

1

given

). n

a

x

1

x

0

boundary

x

n–1

n

condition, y

which

can

be

written

as

(x

,

y

0

is

then

used

to

nd

the

).

The

formula

0

coordinates

(x

,

y

1

y

is

unlikely

to

be

on

the

curve,

but

if

). 1

h

is

1

sufciently

so

the

small

it

coordinates

values

to

obtain

should

can

(x

,

y

2

This

will

create

that

might

look

a

be

be

quite

used

as

close

the

and

starting

solution

cur ve

). 2

series

of

approximations

similar

to

the

red

lines

in

the

diagram.

This

iterative

differential

numerical

equations

is

approximation

named

after

of

Euler

as x

Euler’s

method

0

Example 25

Use

Euler’s

method

with

step

size

0.1

to

approximate

the

solution

to

the

initial

value

dy problem

=

xy

and

y(1)

=

1,

and

estimate

the

value

of

y(2).

dx

y

= n

y

+ n

hf(x

1

, n

y

1

) n

In

examinations,

down h

=

0.1,

f(x,

y)

=

it

is

important

to

write

1

the

formula

being

used,

as

method

xy marks

can

numerical

be

awarded

errors

in

even

the

if

there

are

calculations.

522

/

11 . 5

Hence,

There

is

though

y



y

n





0

1x

n 1

y



y n 1

1



0

n 1

put

1x n 1

to

sometimes

simplify

this

can

the

formula,

make

it

into

the

course

a

For

have

this

a

recommended

simple

way

to

for

this

calculate

an

formula.

example,

the

full

table

of

values

is

technology,

x

n

y n

≈

to

GDC.

calculators

iterative

y(2)

easier

 All

Using

need

y n 1

n

no

3.860891894

n

0

1

1

1

1.1

1.1

2

1.2

1.221

3

1.3

1.36752

4

1.4

1.5452976

5

1.5

1.761639264

6

1.6

2.025885154

7

1.7

2.350026778

8

1.8

2.74953133

9

1.9

3.24444697

10

2

3.860891894

Exercises 11N

Consider

dy

the

differential

4

equation

a

where

y

=

4

when

x

=

=

4.

Use Euler’s

method

with

step

size

0.1

to

approximate

value

of

y

when

x

=

Consider

the

differential

equation

y’

=

1

−

y

=

−1

when

x

=

Euler’s

method

an

=

=

=

1

when

x

=

1.

with

approximate

method

with

step

size

0.1

an

approximate

value

of

y

when

1.5.

Solve

the

step

size

0.05

value

of

y

the

differential

equation

and

hence,

error

in

your

approximation.

to

a

Consider

the

differential

equation

when

dy x

y

0.

5 nd

Euler’s

nd

nd

Use

where

y

b

where

- xy

4.5. x

2

equation

nd to

an

differential

dx

4 y

Use

the

dy

x =

dx

Consider

suluclaC

1

2 xy

0.15. =

where

y

=

3

when

x

=

1.

2

dx

3

Consider

the

differential

1 +

x

equation Use

Euler’s

method

with

step

size

0.1

to

2

dy

2x

+ 1

=

where

y

=

0

when

x

=

nd

1.

an

approximate

value

of

y

when

y

dx

xe

Use

Euler’s

x

method

with

step

size

0.025

an

approximate

value

of

y

1.3.

to

b

nd

=

Solve

the

=

equation

and

when hence,

x

differential

nd

the

absolute

error

in

your

1.1. approximation.

523

/

11

6

A P P R O X I M AT I N G

Let

the

population

meerkats

years,

can

in

the

be

units

rate

of

modelled

IRREGUL AR

size

of

10

of

be

change

by

the

a

colony

N.

of

At

the

S PA C E S :

of

time

I N T E G R AT I O N

b

rare

On

a

AND

copy

DIFFERENTIAL

of

the

slope

eld

E Q U AT I O N S

diagram,

show

t

i

the

line

ii

the

trajectory

N

=

a

+

bt

population

differential

of

the

population

if

at

equation t

=

0,

N

=

2.

dN =

0.5N

c

- t

Find

the

least

number

of

meerkats

in

dt the

a

Given

that

N

=

a

+

bt,

a,

b ∈ℝis

colony

at

population

solution

to

the

differential

equation

particular

initial

population,

nd

of

a

and

slope

eld

for

shown

that

will

ensure

the

not

become

extinct.

measuring

the

calculates

it

will

reach

a

b

the

differential

after

three

years

and

then

will

equation begin

is

does

environmentalist

maximum

The

0

the population

values

=

for An

a

t

a

to

decline.

below.

d

What

is

the

population

of

meerkats

at

N

this

time?

9

At

t

=

3,

they

decide

to

introduce

more

8

meerkats

from

another

colony.

7

e

If

the

model

remains

valid,

nd

the

least

6

number

of

meerkats

that

would

need

to

5

be

introduced

for

the

colony

to

increase

4

in

size

continually?

3

After

the

introduction

of

extra

meerkats,

2

the

size

of

the

population

is

120.

1

f

Use

Euler’s

method

with

a

step

size

of

t 0

0.1

to

estimate

the

population

one

year

–1

later.

Chapter summary

b

•

When f is a non-negative function for a ≤ x ≤ b,

f ( x )d x

gives



y

a

the area under the cur ve from x = a to x = b y

=

f(x)

b

•

If f(x) < 0 in an inter val [a,

b] then

f ( x ) dx





0.

a

•

If the area bounded by the cur ve y = f(x) and the lines

b

∫

f (x)dx

a

x = a and x = b

is required and if f(x) < 0 for any values in this

b

x

inter val then the area should be calculated using

f ( x ) dx



0

a

b

a

•

For a positive continuous function, the area

(A) between the graph of the function, the

x

axis and the

lines x = a and x = b can be approximated by the trapezoidal rule:

1 A



b h



y

 0

2

•



f



x



y

 n

2 (y 1

 ... 

y n

) where

h

- a

=

1

n

dx

is an indenite integral and the process of nding the indenite integral is called

integration.

524

/

11

d

•

F

If





x





f





x

 , then



f



x



dx



F



x





where c is an arbitrary constant.

c

dx

•

F(x) is the antiderivative of f(x).

•

Rules for integration can be derived from those for dierentiation:

The sum/dierence of functions:

❍





f ( x) 

Multiplying a function by a constant:

❍

af





g( x ) dx

x



dx





f ( x ) dx



a



f





g( x ) dx



x



dx

n 1

ax n

Power rule:

❍

ax

dx





c 

 n  1

sin

❍

x 

 cos x



c

cos x

❍





sin

x



c



1

x

dx

❍



ln

| x |

 c

❍

e





x

dx



e



c

x

1 ❍

dx

 cos

•



tan

x



c

2

x

In addition:

1

1 sin ax

❍

dx





cos ax



c



cos ax

❍

dx



sin ax



c

 a

a

1 ax

e

❍

ax

dx



e



c

 a

•

The reverse chain rule can be used whenever the integral consists of the product of a composite

function and a multiple of the derivative of the interior function. In these cases, the derivative will

cancel when the substitution is made.

•

The area A of the region bounded by the cur ves y = f(x), y = g(x) and the lines x = a

and x = b, where

f and g are continuous, is given by

b

A



f





x





g



x



dx

a

•

The area A of the region bounded by the cur ve

and y = b, where h is

b], is given by suluclaC

continuous for all y in [a,

x = h(y), and the lines y = a

b

A





h ( y) dy

a

•

The denite integral of a rate of change in a given inter val calculates the net change and the

denite integral of the modulus of a function gives the total change.

If f is a continuous function on [a, b] and F is any antiderivative of f

then

b

f





x



dx



F

b 



F(a)

a

Developing inquiry skills

Write down any fur ther inquiry questions you could ask and investigate how

you could nd the areas of irregular shapes and curved shapes.

525

/

11

A P P R O X I M AT I N G

IRREGUL AR

S PA C E S :

I N T E G R AT I O N

AND

DIFFERENTIAL

E Q U AT I O N S

Click here for a mixed

Chapter review

review exercise

1

The

graph

prototype

areas

90

of

shows

drone

how

the

designed

rainforest

velocity

to

changed

of

monitor

in

a

trial

3

a

A

large

prototype

tested.

lasting

solar-powered

Sensors

following

data

in

the

every

car

5

vehicle

record

is

the

seconds.

hours.

Time (s) Velocity

Velocity (m/s)

(km/h)

50

0

6.7

5

15.3

(30, 40)

40

30

10

26

20

15

27.3

20

29.4

25

32.1

30

35.5

10

time

(hours)

0 10

20

–10

(75, –20)

–20

–30

Apply

the

distance

a

Describe

the

motion

of

the

trapezoidal

the

car

has

rule

to

estimate

travelled

by

the

the

time

drone −1

its during

b

Write

the

the

entire

down

the

interval

acceleration

of

the

during

drone

is

which

4

Identify

can

negative.

to

c

Hence,

determine

slowing

velocity

reaches

35.5 m s

trial.

when

the

drone

be

be

which

found

of

the

following

analytically

found

using

sin(4 x

 3 )d x

and

integrals

which

need

technology.

is

down.

4

5 2

a

b



x cos( x

0

d

Find

e

Calculate

the

total

distance

travelled

in

the

2

trial.

5

the

position

of

the

drone

5 2

at

c

2

x

d

cos( x )d x



cos( x

end

of

the

)d x



2

the

)d x



2

trial.

4 4

2

The

graph

gives

information

about

10

e

f

(sin(4 x )  3)d x



dx



of

a

more

programmed

to

advanced

y

in

a

drone

straight

that

line

x

is

and

to

its

starting

point

once

it

has

g

4

x

10

 1

h

dx



0.7 x dx



4

5

5

up

its

enough

photographic

data

to

(x

 1 )

ll

memory.

x 10

i Velocity

8

5

0.7 x collected

 1

5 10

return

5

5

0

motion

0.7 x

the

0.7e dx



(m/s)

5

5

x

 1

10 2

f

(x)

=

10

–

(x

–

2)

, 0

≤

x

≤

4

5

8

A

population

of

midges

in

an

area

of

(4, 6)

western

Scotland

is

to

be

controlled

through

(6, 6)

6

sustainable

changes

to

the

environment.

4

The

rate

of

decrease

of

the

population

P

is

2 T ime

(10, 0)

(seconds)

proportional

any

0

–2

time

600 000

–4

t.

to

to

The

the

number

population

500 000

over

of

P

four

midges

at

decreases

from

years.

dP

(12, –3)

a

Write

down

the

relationship

between

dt

a

Calculate

the

distance

travelled

in

the and

rst

b

6

P .

seconds.

Determine

the

time

around

to

return

Hence,

calculate

to

when

the

the

drone

starting

b

Solve

c

Hence,

the

differential

takes

to

how

complete

its

long

for

P .

nd

the

time

taken

for

the

point

population

c

equation

turns

the

to

decrease

by

60%.

drone

journey.

526

/

11

6

The

diagram

shows

2

f ( x)

=

r

the

function

After

owner

2

-

x

,

0

≤

x

≤

r,

and

the

rainfall,

needs

to

the

empty

tank

the

is

full

tank

so

but

it

the

can

area be

beneath

heavy

cleaned.

it. The

tap

level

a

is

of

opened

the

Using

V

and

water

for

has

the

after

10

minutes

dropped

volume

of

to

the

64 cm.

water

in

the

(0, r)

tank

and

write

Apply

for

the

the

State

the

geometrical

shape

of

the

revolution

generated

by

the

x-axis

by

2π

rotating

Write

the

down

volume

an

of

integral

this

Hence,

show

that

that

rule

of

4p

dh

25

dt

a

.

and

the

cylinder

Hence,

formula

to

write

show

your

to

part

a

in

terms

of

h

and

t.

Predict

how

long

it

will

take

for

the

to

empty.

represents

9

solid.

the

equation

radians.

Xavier

public

c

water,

f(x)

tank

b

the

solid

c around

of

variables.

=

answer

of

depth

differential

chain

dt

a

the

volume

that

(r, 0)

a

these

dV 0

for

down

relating

b

h

volume

V

of

is

designing

space.

It

is

an

a

art

large

installation

water

tank

in

in

a

the

a shape

4

of

a

prism.

He

builds

a

scale

model

in

3

sphere

of

radius

r

is

V

 r



his

studio.

The

constant

cross-sectional

area

3 of

7

a

Consider

y’

=

2y(1

the

+

x)

differential

where

y

=

the

model

is

bounded

by

the

functions

equation

3

when

x

=

9

0.

x

f(x)

=

2

2

x

+

2

and

f ( x)

=

x

+

2

16

to

x

Euler’s

nd

=

method

with

approximate

0.1,

0.2

and

step

values

size

of

y

0.1

a

when

Xavier

the

0.3.

sketches

domain

calibrated

b

Solve

the

differential

equation

nd

the

absolute

errors

for

your

A

water

of

tank

in

the

shape

of

a

cylinder

Xavier

40 cm

and

height

100 cm

the

metres.

The

x-axis

Calculate

is

the

the

region

bounded

by

the

two

in

square

metres.

for

uses

his

graph

to

build

the

model

prism,

which

is

50 cm

deep.

Given

collects that

rainwater

2.

of of

radius

≤

approximations.

b

8

x

on

each

graphs of

in

≤

functions

and

area hence,

−2

both

the

model

is

1:40

scale,

calculate

the

recycling. volume

the

of

the

nearest

art

cubic

installation,

correct

to

metre.

Exam-style questions

2

10

P1:

a

Find

the

value

x

of

,



giving

2

0

x

your

3

answer

signicant

as

a



suluclaC

Use

4

decimal

gures.

correct

(2

to

marks)

x

b

Find

the

integral

dx



2

x



4

(3

c

Hence

2

nd

the

exact

value

marks)

of

x dx .



(2

marks)

2

0

x

Water

ows

the

tank

at

the

square

out

a

of

rate

root

a

tap

at

directly

of

the

the

bottom

of

proportional

to

depth

of

the



4

water.

527

/

11

A P P R O X I M AT I N G

11

P1:

Find

the

IRREGUL AR

following

indenite

S PA C E S :

I N T E G R AT I O N

12

integrals.

P1:

AND

Farmer

Davies

between

2

a

∫x

b

∫sin x

+

3x

+

1 dx

(2

marks)

(2

marks)

two

modelled +

cos(2x) dx

DIFFERENTIAL

owns

the

roads.

by

land

The

curve s

E Q U AT I O N S

enclosed

roads

w i th

can

be

eq ua ti ons

2

y x

c

=

x

−

8x +

23

and

y

=

x +

9

for

0

≤

x ≤

x

∫e

+

e

dx

(2

marks) 10,

where

x

and

y

each

represent

a

unit

6

d

∫(3x +1)

dx

(2

marks)

(2

marks)

of

100

m.

2

x

e



x

Find dx

the

area

of

Farmer

Davies’



(6

x

13

P1:

Alun,

boat

a

geographer,

and

Alun’s

land.

a

marked

results

distance

are

from

measures

piece

of

line

presented

the

shore,

the

in

and

depth

with

the

d

a

of

a

river

heavy

table

the

1 m

weight

below,

represents

at

at

where

depth

the

x

of

intervals

from

bank

using

a

bottom.

represents

the

the

marks)

river

the

(both

perpendicular

measured

in

metres).

x

0

1

2

3

4

5

6

7

8

9

10

11

12

d

0

6.71

10.4

12.15

12.8

12.95

12.96

12.95

12.8

12.15

10.4

6.71

0

a

Use

the

answer

trapezium

correct

to

rule

2

to

estimate

decimal

the

cross-sectional

area

of

the

river,

giving

places.

your

(3

marks)

(3

marks)

4

1296 -

It

is

later

discovered

that

the

equation

connecting

x

and

d

is

d

(

x

- 6)

=

.

100

b

Calculate

c

Find

d

Explain

the

the

true

value

percentage

why,

in

this

of

the

error

in

case,

the

cross-sectional

your

estimation

trapezium

rule

area

of

of

the

the

area

river.

you

made

underestimates

the

in

true

part

a.

(2

value.

marks)

(1

mark)

2 x

14

P1:

The

is

curve

rotated

y

=

e

between

through

2π

x

=

0

radians

and

x

about

=

1

the

16

P2:

A

differential

the

to

form

volume

a

of

solid

this

of

revolution.

solid.

(4

Find

P2:

The

acceleration

of

a

small

craft

is

given

by

=

xy ,

Use

of

rocket-

a

Find

an

expression

for

4e

the

–2

for

ms

the

time

craft,

t s,

from

v ms

given

as

a

the

function

Find

craft

rest.

an

started

(4

expression

displacement

function

craft

of

of

started

for

the

time

t s,

from

Write

the

down

craft

as

t

the

craft,

given

the

b

s

m,

that

very

Find

one

the

craft’s

minute

as

a

c

0

where

y(1)

to

y(2).

method

nd

For

an

with

a

=

2.

step

size

approximation

each

value

corresponding

of

value

x,

of

y

show

in

(5

Solve

the

differential

your

marks)

the

some

answer

in

function

equation,

the

form

f.

y

(7

Hence,

nd

the

exact

value

=

f(x)

marks)

marks)

velocity

of

Find

in

the

the

absolute

value

method.

of

Give

signicant

large.

of

(2

the

d

(1

d

Euler’s

giving

marks)

origin.

limiting

gets



the

(4

c

y

of

for

b

0,

working.

,

that



velocity

–1

of

x

0.25

the

a

by

marks)

−0.2t

propelled

dened

dx

a

15

is

dy 

x-axis

equation

marks)

percentage

y(2)

your

gures.

given

by

answer

error

Euler’s

to

(2

y(2).

2

marks)

mark)

di s p l a ce me nt

after

it

begins

to

(2

move.

marks)

528

/

11

2

17

P1:

a

Find

the

derivative

of

x

sin x.

(3

b

Hence

If

marks)

an

the

nd

If

equation

constant

an

is

an

value

equation

is

isocline,

of

not

the

an

state

gradient.

isocline,

2

∫4cos x

+

6x sin x

+

3x

cos x dx.

(2

P2: In

a

slope

curve

In

on

other

eld,

an

which

words,

isocline

the

an

is

dened

gradient

isocline

is

marks)

is

a

as

2

a



k,

with

k

a

curve

b

the

following

slope

eld

differential

2

x

x

ii

y

=

1

iii

y

=

x

iv

y

=

x

v

y

=

−x

of

to

the

the

−

8xy

=

0

+

1

(10

marks)

whether

2

x

the

solution

equation

2



y



, x

 0, y



0

has

any

y



 0, y

, x



2 xy

0

2 xy turning

a

y

differential

equation

dx

dx

+

Determine

2



not.

2

i

dy

dy

is

constant.

dx

Consider

it

constant.

dy where

why

State

which

equations

of

are

the

points.

Justify

of

the

answer.

(3

following

isoclines

your

marks)

above

Click here for fur ther

differential

equation. exam practice

suluclaC

18

justify

529

/

Approaches to learning: Research, Critical

In the footsteps of

thinking

Exploration criteria: Mathematical

Archimedes

communication (B), Personal engagement (C),

Use of mathematics (E)

ytivitca noitagitsevni dna gnilledoM

The area of a parabolic segment

A

parabolic

parabola

and

Consider

this

bounded

by

segment

a

is

a

region

IB topic: Integration, Proof, Coordinate geometry

bounded

by

a

line.

shaded

region

which

is

the

area

2

the

line

y

=

x

+

6

and

the

curve

y

=

x

:

y

12

10

B(3, 9)

8

6

A(–2, 4) 4

2

x 0 –6

–4

–2

2

4

6

–2

From

this

using

integration.

On

the

Point

C

points

What

chapter

diagram

is

A

and

are

Triangle

such

the

ABC

you

points

that

know

A( −2,

the

that

4)

x-value

you

and

of

C

can

B(3,

is

calculate

9)

are

halfway

the

shaded

area

marked.

between

the

x-values

of

B.

coordinates

is

of

constructed

point

as

C

on

the

curve?

shown:

y

12

10

B(3, 9)

8

6

A(–2, 4) 4

Archimedes showed

that the area of the

parabolic segment

C(0.5, 0.25)

4 x

is

of the area of

0 –6

–4

–2

2

4

6

3

triangle ABC –2

530

/

11

Calculate

What

the

the

area

methods

of

are

the

triangle

available

to

shown.

calculate

y

the

area

12

of

triangle?

10

Use

integration

to

calculate

the

area

between

the

two

ytivitca noitagitsevni dna gnilledoM

B(3, 9)

curves. 8

Hence

verify

parabolic

that

Archimedes’

result

is

correct

for

this

segment. 6

You

can

and

for

show

that

this

result

is

true

for

any

parabola

A(–2, 4)

any

Consider

starting

another

parabola

such

x-values

of

A

points

triangle

that

and

its

C,

A

by

and

on

choosing

x-value

similar

B

is

to

the

parabola.

point

halfway

D

on

between

4

the

the

before. D(–0.75, 0.5625)

C(0.5, 0.25) x

What

are

the

coordinates

of

point

0

D? –6

Calculate

the

area

of

triangle

–4

–2

2

4

6

ACD. –2

Similarly,

half-way

What

the

do

the

can

you

have

You

can

If

see

a

you

add

nd

and

E

such

that

its

x-value

is

B.

area

of

of

between

point

E?

triangle

the

BCE.

areas

of

the

new

triangles

and

original

triangle

ABC.

notice?

already

that

reasonable

improve

more

C

the

ratio

you

You

four

BC,

coordinates

calculate

Calculate

What

line

between

are

Hence

for

this

triangles

the

if

add

the

approximation

approximation

from

areas

you

of

sides

these

AD,

seven

areas

for

by

CD,

of

the

triangles

area

of

continuing

CE

and

triangles,

the

the

ABC,

ACD

parabolic

process

and

BCE,

segment.

and

forming

BE.

you

have

an

even

better

approximation.

How

could

the

approximation

be

improved?

Generalise the problem

Let

the

What

If

is

you

the

of

the

total

continued

would

By

area

have

the

summing

triangles,

area

triangle

be

of

the

next

adding

the

areas

exact

the

you

rst

area

areas

can

of

show

for

all

that

the

X

two,

of

four

an

and

innite

parabolic

eight

triangles

number

of

in

such

terms

of

triangles,

X?

you

segment.

the

they

Ex tension form

•

•

a

geometric

series.

What

is

the

common

What

is

the

rst

What

is

the

sum

What

has

ratio?

This task demonstrates the par t of the historical development of

the topic of limits which has led to the development of the concept of

term?

calculus.

•

•

this

of

the

shown?

series?

Look at another area of mathematics that you have studied on this

course so far or one that interests you looking for ward in the book.



What is the history of this par ticular area of mathematics?



How does it t into the development of the whole of mathematics?



How signicant is it?



Who are the main contributors to this branch of mathematics?

531

/

Modelling motion and change

in two and three dimensions

12 Change

to

a

in

a

changes

in

projectile,

horizontal

air

and

in

variable

another

for

example,

time

a

for

is

often

variable.

displacement,

the

Growth

single

Concepts

might

its

which

population,

on

The

height

depend

velocity

it

has

the

■

Change

■

Modelling

connected

of

Microconcepts

on

through

been

other

the ■

Vector quantities

■

Component of a vector in a given direction

travelling.

hand,

evaluated using scalar or vector product might

size

you

depend

of

not

another

develop

just

on

resources

competing

the

ideas

but

population.

from

earlier

on

the

How

chapters

■

Acceleration

■

Two-dimensional motion with variable

can

to

velocity predict

change

in

complex

systems

such

How much electricity can be

as

these?

■

Projectiles

■

Phase por traits

■

Euler method with three variables

■

Coupled systems

■

Eigenvalues and eigenvectors

■

Asymptotic behaviour

■

Second order dierential equations

generated by a solar panel?

How much will a building move

during an ear thquake?

How might the shark population

be aected by changes in the

sh population?

532

/

An

aircraft

research

aircraft

needs

station.

from

a

to

deliver

The

a

supply

package

height

of

will

150 m.

If

package

be

to

dropped

the

aircraft

a

polar

from

is

the

ying

at

1

180 km h

package

need

Make

•

how

a

far

to

from

be

guess

the

released

at

a

research

if

possible

air

station

resistance

answer

to

would

can

the

be

the

ignored?

question

above.

How

•

the

at

If

•

would

same

an

air

your

speed

angle

of

at

to

be

research

the

change

same

if

height

the

but

aircraft

was

still

had

ascending

30 °?

resistance

need

answer

cannot

released

be

ignored

further

from

would

or

the

closer

to

package

the

station?

Developing inquiry skills

Think about the questions in this opening problem and

answer any you can. As you work through the chapter, you

will gain mathematical knowledge and skills that will help

you to answer them all.

Before you start Click here for help

You should know how to:

1

Integrate

exponential

with this skills check

Skills check

and

trigonometric

1

Find:

functions. 1 3x

a

6e



eg

Find

dx



2x

2x

e

3e



2 sin 5 x d x



b 3

2

2x

2x

3e



2 sin 5 x d x



e



cos 5x



3 cos 6 x

 4 sin 2 x d x



c

 2

2

Find

of

and

two

use

the

5

scalar

and

vector

products

2

a

Find

the

 2



 Find

the

angle

between

i

+

2j

−

k

and

and

 1

−

If

×

θ

3 

3k

First

1

2



 

4

 3

2i

between

vectors.

eg

a

angle

take

2

is

+

2

the

×

the

0

scalar

+

(

angle

product

−1)

×

(

−3)

between

=

the

5

two

vectors

then

5 cos 

=

⇒

6

θ

=

55.5°

13

Continued

on

next

page

533

/

12

MODE LLING

b

Find

a

M OT I O N

vector

1

AND

C H A NGE

perpendicular

IN

TWO

AND

THREE

b

to

Find

a

DIMENSIONS

vector

2

2

0

and

3

1

and

vector

1

2



















1 

Find

eg

of

the











4

the

and

3

eigenvectors.

eigenvalues

and

eigenvectors

for

Find

the

eigenvalues

and

eigenvectors

for

3 





 1

4

First

solve

the

2

5

1

equation

3 

1

4

−

2



2  

(2

product

will

1



eigenvalues





vectors

 

3

Find

 2

vector

two

 6

0





the





2 

to

to

3

vectors.





3

perpendicular

parallel

two



2

3 2

be

to

2 2

A

perpendicular

0

 

λ)(4

−

λ)

5

=

−

3

=

0

2

λ

−

For

6λ

+

the

3

 1

+

3y

=

λ

x

=



3

0

⇒

 y

x

1,

5

above,

solve

0

 1, 

x

⇒

eigenvalues

1 

0

=





 0

−3y

3 Eigenvectors

are

parallel

to

1

1 Similarly

the

second

eigenvector

is

1

534

/

12 . 1

12.1

In

Vector quantities

Chapter

3

straight-line

In

the

other

19th

you

the

movement

century,

quantities.

usefully

met

divided

in

of

using

vectors

two-dimensional

physicists

They

into

idea

extended

realized

those

that

that

and

this

physical

needed

a

to

describe

position

and

three-dimensional

idea

of

vectors

quantities

direction,

as

to

could

well

space.

many

be

as

size,

to

E X AM HINT dene

them

completely,

and

those

that

did

not.

In exams you will not

For

example

an

object’s

mass

does

not

need

a

direction

but

its

velocity

1

does.

A

velocity

of

10

ms

need to understand

1

to

the

right

is

very

different

from

10

ms

the concepts of

to

the

left.

the quantities

Quantities

vectors

which

and

need

those

a

that

direction

do

not

to

were

dene

called

them

fully

were

classed

being used except

as

for those listed

scalars

in the syllabus Vector

quantities

include:

(displacement,

•

velocity

•

acceleration

•

force

velocity and

and

electric

elds

•

momentum.

acceleration). Other

quantities might

appear in questions

but the contex t will

make it clear how

The results obtained for displacement vectors in Chapter 3 apply equally well

the question is to be

dna y rtemoeG

magnetic

y rtemonogirt

•

to any vector quantity.

answered. In par ticular the single vector which will have the same eect as several

vectors acting together is called the resultant and is equal to the sum of all

the vectors.

Example 1



 2 force

is

given

by

the

vector

Newtons  

horizontal

A

particle

and

(A)

vertical

is

subjected

to

is

the

total

horizontal

2

What

is

the

total

vertical

3

Hence

experienced

by

as

the

a

second

force

 5

by

3

N

 



both

What

down

and

relative

to

2

axes.

1

write

(N)

a

these

force

force

it

column

particle

and

it

forces

as

suluclaC

A

shown.

experiences?

experiences?

vector

show

the

this

resultant

force

on

force

a A

diagram.

4

Find

the

the

magnitude

of

the

resultant

force

experienced

by

particle.

Continued

on

next

page

535

/

12

MODE LLING

M OT I O N

AND

C H A NGE

IN

TWO

AND

The

1

2  3



total

5  2

DIMENSIONS

horizontal

force

is

found

by

5N

adding

2

THREE



the

two

horizontal

components.

3N The

the

total

two

vertical

vertical

force

is

found

by

adding

forces.

5

3

N

The

resultant

two

vectors.

force

is

the

sum

of

the

 

3

individual

It

is

vectors

direction

the

on

by

other

As

can

also

no

be

be

equivalent

its

particle

seen

found

individual

so

to

the

direction

would

two

is

move

the

if

acted

forces.

from

the

diagram

geometrically

vectors

by

following

it

can

drawing

on

from

the

each

other.

2

4

5

2

 3



5

83

The

N

magnitude

sum

of

vectors

the

the

so

they

direction.

individual

and

the

resultant

magnitudes

unless

same

of

vectors

cancel

are

If

the

all

not,

act

each

of

not

the

individual

acting

parts

against

other

is

in

of

the

each

other

out.

Exercise 12A

1

Find

the

magnitude

following

angle

forces.

measured

positive

and

Give

direction

the

of

direction

counter-clockwise

4

the

as

an

from

the

x-axis.

(6i

b

+3j)N

object

a

a



the

following

with

lever

a

force

F

the

the

when

acting

displacement

by

State

a

equation

direction

comparison

4

b Write

(t ) produced

r

t

of

at

from



F

turning

the

the

end

an

of

object

is

 r

the

torque

in

N



2

torque

given

 3

a

The

velocities

as

Find

the

to

the

other

magnitude

of

two

the

vectors.

torque

column

 3

vectors:

 1

a

7ms

on

a

bearing

of

produced

045°

when

a

force

of

4

N

is

 

1

b

3

A

12ms

on

particle

a

bearing

experiences

of

the

0

330°.



following  2

accelerations

as

a

result

of

three

different

 acting

on

it

ms

 

displacement

1

m

from

the

 1 2

acting

a



 2 forces

at

2

,



ms

0

 4 

 2

object.  2 2

and

ms  

Find

0

the

resultant

magnitude

and

acceleration

direction

experienced

of

the

by

the

particle.

536

/

12 . 1

5

A

sledge

shown.

of

is

being

One

150 N

at

of

an

pulled

the

by

dogs

angle

of

is

two

dogs

pulling

10°

to

the

as

with

a

b

Find

the

resultant

c

Find

the

value

and

the

second

with

a

direction

force

of

D

angle

of

15°

to

the

direction

of

N

sledge

is

also

subject

to

a

of

70 N

as

of

D

if

the

sledge

is

in

the

intended

direction.

second

law

states

that

the

force

(F

N)

acting

on

a

body

will

resistance equal

to

the

product

shown.

of

its

mass

(kg)

and

2

the

(F

150

acceleration

=

(ms

)

of

the

body

ma).

N

d

70

D

travel.

be

force

of

at

resultant

The

terms

of

Newton’s an

in

force

heading

travel

force

Given

that

the

mass

of

the

sledge

and

N

occupant

10°

is

60

kg

nd

the

magnitude

15°

and

direction

of

the

sledge’s

acceleration.

D

along

the

perpendicular

a

Find

the

direction

to

the

expressions

column

of

of

vectors

travel

direction

for

the

of

and

travel.

three

forces

as

vectors.

Components of vector quantities in given directions

It

is

in

a

For

3i

often

given

4j

will

result

case

The

N

and

the

book

3 N

will

to

book

the

know

is

of

one

less

the

of

is

how

be

lift

lying

of

the

the

much

a

to

book

move

off

depends

book

table

in

base

which

is

the

on

the

straightforward

vector

our

on

acting

moving

weight

situation

along

a

acting

component

lie

if

then

be

in

to

of

a

vector

quantity

is

acting

direction.

example,

+

4 N

important

dna y rtemoeG

lie

components

y rtemonogirt

to

the

and

the

is

book

table.

the

subject

to

along

Whether

frictional

a

force

the

or

force

table

not

in

of

and

this

the

will

rst

second.

when

acting

we

in

a

are

trying

direction

to

nd

which

the

does

not

vectors.

International-

Investigation 1 mindedness Let θ be the angle between two non-parallel vectors

a and b. Think of the

Greek philosopher

direction of b as being along one of the axes.

suluclaC

Take

and mathematician,

1

Show that the component of a in the direction of b can be written as Aristotle, calculated the

| a | cosθ

where | a

| is the magnitude of a combined eect of two

2

Find a similar expression for the component of

a perpendicular to b within

Parallelogram law.

the plane formed by the two vectors.

3

Rewrite the expression for the component of

or more forces called the

a in the direction of b in

terms of the scalar product of a and b

4

Rewrite the expression for the component of

a perpendicular to the

direction of b in terms of the vector product of a and b

Continued

on

next

page

537

/

12

MODE LLING

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

 1



5

Hence nd the components of

2   1 

 3

 3



a

in the direction of



b

4

perpendicular to







 0

Conceptual

i

  0



6

 4



How do you nd the component of a vector

parallel to a vector b

ii

a acting

perpendicular to a vector b?

International-

mindedness

René Descar tes used

Given two vectors a and b the component of a in the direction of b is given (x,

a

y,

z) to represent

b points in space in the

by b

17th century.

The component of a

perpendicular to b, within the plane dened by the two

In the 19th century,

Ar thur Cayley reasoned a  b

that we might go fur ther b than three values.

Example 2

A

sailboat

is

travelling

on

a

bearing

of

045 °.

The

wind

is

blowing

with

a

velocity

of

1

40

km h

from

a

Write

b

Find

down

the

a

direction

the

of

velocity

component

of

of

the

200 °

the

wind

velocity

of

as

a

the

column

wind

in

vector.

the

same

direction

as

the

path

of

the

boat.

  40 cos 70



a 





 13

A 

40 sin 70 

7 







diagram

is

helpful

to

make

sure

that

the

 37.6



direction

of

the

vector

is

correct.

B

20°

40

70°

A

C

1

b

Direction

of

sailboat

=

Any

vector

in

the

required

direction

1 will

Required

give

product

component

the

is

same

divided

answer,

by

the

as

the

scalar

magnitude.

13 .7  1  37 .6  1 1



 2

1

36

3 km h

2

 1

538

/

12 . 1

Exercise 12B

A

1

a

Find

the

magnitudes

of

the

current

is

owing

with

speed

components

3 1

1 of

the

10 km h

vector

parallel

and

in

the

direction

of

perpen-

4

4

a

Write

the

velocity

of

the

current

as

a

1 column dicular

to

the

vector.

vector

2

b

Find

the

acting

component

in

the

of

direction

the

the

current

boat

is

travelling.

b

Find

the

magnitudes

of

the

components

4

2

of

the

computer

calculate

0

vector

A

parallel

off

and

a

at

knows

graphics

the

direction

surface

the

technician

of

(known

direction

of

the

as

the

needs

ray

ray

to

reected

tracing).

incident

ray

He

and

5

perpendicular,

in

the

plane

dened

the

direction

Let

the

of

the

normal

to

the

surface.

by

incident

ray

have

direction

b

, i

reected

ray

have

direction

b

and

y rtemonogirt

the

r

the

two

vectors,

to

the

2

vector

.

the

vector

1

Let

2

T wo

forces

are

pulling

a

truck

of

surface

mass

150

a

a

rail.

The

direction

of

the

rail

perpendicular

to

the

unit

n

be

a

vector

perpendicular

to

n

as

kg shown

along

be

in

the

diagram.

dna y rtemoeG

2

is

n

12

 16 

 5

and

the

two

forces

 8

are

N



and

a





a



0

2 



18

b

b

i

10

r

N.

5

a

Write

down

an

expression

for

b

in i

in Find

the

component

of

the

the

form

The

in

the

direction

acceleration

of

a

of

body

the

is

force

divided

by

the

mass

a

+

kn,

where

k

is

a

b

and

n

i

to

the

b

Find

of

an

expression

for

b

in

terms

of

b

r

and resultant

of

rail.

equal

= i

function force

b

resultant

i

k n

suluclaC

a

the

0 body.

c b

Find

the

the

rail,

that

acceleration

assuming

no

of

the

other

truck

forces

A

ray

with

direction

1

strikes

a

along

act

4

in

direction. 2 1

3

In

still

water

a

boat

can

travel

at

8 km h surface

on

full

power.

On

a

certain

day

the

boat

with

normal

vector

1

.

Find

is

2 1 travelling

in

the

direction

on

full

power.

1 the

direction

of

the

reected

ray.

539

/

12

MODE LLING

12.2

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

12 . 2

Motion with variable velocity

InternationalA

satellite

equation

is

of

in

its

a

spiral

path

descent

look

like?

back

to

What

the

earth.

features

What

would

it

would

need

to

the

have?

mindedness

Dierent ways of

representing vectors

appear around the world

such as row vectors

being shown as .

TOK

Do you think that

one form of symbolic

representation is

preferable to another?

In

Chapter

3

dimensions.

velocity

in

you

In

one

considered

Chapters

10

dimension.

motion

and

In

11

this

with

you

constant

covered

section

the

velocity

motion

two

in

with

concepts

two

variable

will

be

combined.

f 1

If the displacement of an object is given by

r

t 

then

 f 2



d

1

t 

2



df

t 

f 1



 2

dr v



dt



d

and 

r

dt





dt

2

a



 2

df



dt

2

2

d

f 2







dt

2





dt



Integrating the acceleration vector (by integrating each component) gives

the velocity vector, which can then be integrated to obtain the displacement

vector. In the case of integration there will be two unknown constants which

need to be found from the initial conditions.

International-

mindedness

Example 3

Vectors developed A

particle

has

velocity

v(t)

where

v

=

(4t

−

6)i

−

(3t)j quickly in the rst two

a

Find

the

speed

b

Find

the

time

of

the

particle

when

t

=

decades of the 19th

3.

century with Danish– at

which

the

direction

of

movement

of

the

Norwegian, Caspar particle

is

parallel

to

j

Wessel, Swiss, Jean

c

Show

that

the

acceleration

of

the

particle

is

constant. Rober t Argand, and

d

Given

an

that

the

expression

displacement

for

its

of

the

displacement

particle

at

time

t

when

t

=

0

is

4i

nd

German, Carl Friedrich

Gauss.

540

/

12 . 2

a

When

t

=

3,

v

Hence

speed

=

6i

−

9j

1

b

4t

−

6

=

0

⇒

=

t

10.8

=

m s

If

1.5

parallel

in

the

equal

i

to

j

the

direction

to

component

must

be

0.

dv

c

a





4i

 3 j

hence

There

constant

is

no

variable

in

the

dt acceleration

equation,

acceleration

is

Acceleration

quantity

left

as

a

of

13

the

question

ask

for

constant.

is

and

so

a

so

vector.

vector

should

If

an

be

answer

2

m s

the

were

required,

would

need

magnitude

of

to

the

acceleration.

TOK

r

 2t



6t





c 1

i





t



c

 

Because

j 2

2

there

are

two

 Why might it



components

to

the

vector be argued that

At

t

=

0,

r

=

4i

⇒

c

=

4,

1

c

=

two

0

different

constants

of

2

vector equations integration

are

required. are superior

 3

2

r



 2t

 6t

4





i

2





t

j

 

to Car tesian

 2

 equations?

dna y rtemoeG

d



2

y rtemonogirt

 3

2

Exercise 12C

1

The

position

vector

4t

is

given

as

r



of

a

particle

P

at

time

a

t

Find

the

velocity

of

P

at

time

t

2

,



where

r

is

4

measured

2

3t

The

 t

initial

displacement

of

P

is

.

 1 in

metres

and

t

in

seconds.

b a

Find

the

initial

displacement

b

Find

the

initial

velocity

c

Show

and

of

Find

the

speed

of

P .

4

The

velocity

magnetic that

the

acceleration

of

P

displacement

of

P

at

time

t

P .

of

eld

a

at

charged

time

t

is

particle

given

in

a

by

is 2

v and

nd

its

=

(2t

−

1)i

+

3t

j.

magnitude.

a

Find

the

time

at

which

the

particle

suluclaC

constant

is

1

2

A

particle

moves

with

velocity

v

(m s

) moving

given

by

v

=

(4t

−

1)i

+

(5

−

b Given

the

particle

is

initial

2i

−

j

displacement

from

the

of

nd

the

time

b

nd

displacement

origin

of

vector

j

Find

the

t

and

the

velocity

of

the

particle

when

=

2

hence

the

angle

its

direction

particle

with

the

vector

i

at The

initial

2i

3j.

displacement

of

the

particle

is

t

the

when

A

the

O,

distance

of

the

particle

from

t

particle

=

P

−

O

c

3

to

the

makes

a

parallel

2t)j

Find

an

expression

for

the

particle’s

the

distance

2.

moves

with

acceleration

displacement

at

time

t

Find

at

which

a

d

the

time

of

the

2 particle

2

where

a

m s



.

The

initial

velocity

from

the

origin

is

16.

of

1

2 1

P

is

m s

.

3

541

/

12

MODE LLING

5

During

the

rst

M OT I O N

few

AND

seconds

C H A NGE

after

IN

TWO

AND

a

take-off

THREE

Find

an

12 . 2

DIMENSIONS

expression

for

the

displacement

2

an

aircraft’s

acceleration

a

ms

at

time

t

of

A

6t seconds

(t

≥

0)

is

given

by

a

,



P

at

second

time

t

particle

acceleration

and

two

components

vertical

motion.

represent

When

t

=

of

.

a

constant

the

Its

initial

velocity

is

4

horizontal

1

has

0

2

the

Q

where

velocity

3 1

10

m s

parallel

to

the

vector

7 4

1

of

the

particle

is

m s

.

14

b

a

Find

an

time

t.

expression

for

the

velocity

at

c

b

Find

the

motion

time

of

the

at

which

aircraft

the

is

at

direction

45°

to

Find

an

time

t.

Given

of

expression

Q

particles

the

which

is

initially

collide

this

for

at

and

its

O

velocity

show

state

the

the

at

two

time

at

occurs.

horizontal.

7

A

boat

has

a

displacement

(km)

given

by

2

6

A

particle

P

has

velocity

vector

v

given

by

r

=

(4t)i

from 4t

v

the

+

(t

−

3t)j

beginning

where

of

its

t

is

time

in

hours

motion.

 2.5

.



The

initial

displacement

of

P Find

2

the

minimum

speed

of

the

boat

and

3t the

value

of

t

at

which

it

occurs.

2 from

an

origin

O

is

0

Projectiles

International-

Investigation 2 mindedness An object travelling under the force of gravity

Belgian/Dutch

alone is called a projectile. For example, a

mathematician Simon

baseball in ight would be a projectile.

Stevin used vectors in

The only acceleration experienced by a his theoretical work on

projectile is a downward acceleration of g falling bodies and his

2

due to gravity. This acceleration is

m s

treatise “Principles of

2

approximately equal to 9.81 m s

at the the ar t of weighing” in

surface of the Ear th. the 16th century.

1

Write down, in terms of g, the acce-

leration vector for a projectile, taking the

positive y-direction as “up”.

At t = 0 a projectile is launched with a

1

speed of

u m s

at an angle of

α to the

horizontal.

2

Find the initial velocity of the projectile as a column vector in terms of

u

and α

3

Use integration to nd the velocity vector for the projectile at time

t

542

/

12 . 2

 c



1

At t = 0 the par ticle has displacement





.

c 

4

2



Find an expression for the par ticle’s displacement at time

t

A par ticle is projected from the point (0, 0) on level ground.

5

Use your answer to question 3 to nd the time at which the par ticle

reaches its maximum height.

6

Use your answer to question 4 to nd an expression for this height.

7

If the initial speed of the par ticle is xed, which value of

α will result in

the par ticle maximizing the height reached?

8

a

Use your answer to question 4 to nd the time at which the particle hits

the ground.

b

Compare your answer to par t a with your answer to question 5. What

does this tell you about the motion of the par ticle?

The range of the par ticle is the total horizontal distance travelled until it

Find an expression for the range of the par ticle.

10

If the initial speed of the par ticle is xed which value of

α will result in the

par ticle maximizing its range?

1

A ball is projected from ground level with an initial speed of 15 m s

at an

dna y rtemoeG

9

y rtemonogirt

returns to the same height as that at which it began.

2

angle of 30°. The acceleration due to gravity is 9.81 m s

.

11

Find the horizontal distance travelled by the ball before it hits the ground.

12

Find the maximum height reached by the ball.

13

In a more realistic model what additional force will need to be taken into

account when predicting the ight of a ball?

14

Factual

15

Conceptual

What does a column vector describe in projectile motion?

How are vectors useful in describing projectile motion and

how do we nd the velocity equation?

2

In

the

following

1

questions

take

g

=

9.81m s

.

b

Find

an

expression

suluclaC

Exercise 12D

for

i

the

velocity

ii

the

displacement

0 and

0

are

unit

vectors

in

the

horizontal

1

of

and

vertical

directions

respectively.

All

the

object

t

seconds

after

it

has

been

external projected.

forces,

such

as

air

resistance,

may

be

ignored.

c

1

An

object

is

projected

with

an

Find

the

time

at

which

the

particle

initial strikes

the

distance

it

ground

has

and

the

travelled

horizontal

during

its

5 1

velocity

of



above

a

from

a

point

1.5 m

ight.

2

ground

Write

the

m s



level.

down

particle

the

acceleration

while

in

vector

for

ight.

543

/

12

MODE LLING

2

A

stone

cliff

is

50 m

M OT I O N

thrown

high

from

with

an

AND

the

top

initial

C H A NGE

of

a

IN

TWO

vertical

velocity

4

of

AND

An

THREE

object

the

is

12 . 2

DIMENSIONS

projected

horizontal

from

at

a

an

point

angle

at

of

30°

ground

to

level.

1

4i

+

a

2j

m s

Write

.

It

down

stone

at

the

time

t

velocity

seconds

vector

after

for

the

the

Find

been

the

the

ground

initial

again

speed

after

of

the

6.0

seconds.

object.

stone

5 has

hits

A

ball

is

thrown

from

a

height

of

1.5 m

of

45°

with

thrown. 1

a

b

Find

the

displacement

vector

for

speed

of

10

taking

the

point

of

projection

origin

of

the

coordinate

Find

the

time

at

which

the

point

stone

hits

Find

d

Find

at

its

when

the

base

distance

it

hits

the

of

from

the

the

An

particle

is

It

of

an

the

the

just

clears

a

wall

10 m

from

projection.

expression

for

the

displacement

ball

at

time

t

cliff.

base

of

the

b

Find

the

height

c

Find

the

horizontal

of

the

wall.

cliff distance

travelled

by

ground. the

3

to

the of

ground

angle

system.

a

c

an

as the

the

at

the horizontal.

stone,

m s

projected

from

ground

level

ball

before

it

hits

the

ground.

at 1

6 an

angle

of

30°

to

the

horizontal

and

with

A

particle

is

projected

at

a

speed

of

50

m s

a at

an

angle

of

30°

above

the

horizontal

1

speed

of

14.7

m s from

a

Write

as

b

a

the

initial

column

Find

an

velocity

of

the

a

point

a

Write

for

the

velocity

of

Find

the

b

Find

time

time

its

the

Find

an

at

which

the

the

greatest

Find

velocity

vector.

of

the

particle

point

of

projection

t

seconds

the

particle

is

projected.

Find

the

time

for

which

the

particle’s

height

height.

expression

for

the

the

ground

is

greater

than

22

m.

displacement Find

the

time

at

which

the

particle

hits

particle. the

e

initial

particle

d of

the

displacement

the

above

d

down

t

c attains

ground.

the

after

c

level

vector.

expression

at

above

particle

from particle

2 m

the

greatest

height

reached

by

ground

and

the

horizontal

distance

the travelled

during

the

ight.

particle.

f

Find

the

g

the

horizontal

particle

Find

the

particle

Fully

while

in

minimum

during

justify

its

your

distance

travelled

by

ight.

speed

attained

by

the

ight.

answer.

Exponential and trigonometric motion

Investigation 3

Consider a par ticle moving so that its position at time t is given by the equation

r cos t r

, where r and ω are constant values.



r sin t

1

By considering the distance of the par ticle from the origin at time

t, show

that the path followed by the par ticle is a circle centred on the origin and

state its radius.

544

/

12 . 2

2

a

b

TOK

Write down the initial position of the par ticle.

How do we relate a

Give the value of t when the par ticle rst returns to its initial position.

theory to the author?

3

a

Find the velocity of the par ticle at time

t Who developed vector

b

Show that the speed of the par ticle is constant and state its value.

c

Show that the velocity is always perpendicular to the par ticle’s

analysis, Josiah

Willard Gibbs or Oliver

Heaviside?

displacement vector.

d

Explain what this means geometrically about the direction of the

motion of the par ticle.

4

a

b

Find the acceleration of the par ticle at time

t

Show that the acceleration is always perpendicular to the velocity vector.

Let a be the acceleration vector of the par ticle.

2

v

c

Show that

a

.

=

r

Show that a = kr and state the value of k

How would you derive the velocity and acceleration equations

for a particle moving in a circle from the general equation for its displacement?

dna y rtemoeG

Conceptual

5

y rtemonogirt

d

Example 4 0.5t

10e A

particle

has

displacement

r

at

time

t

where

r

cos t .

 0.5t

sin t

10e



a

Find

the

position

of

the

particle

when

t



0,

3 ,

,

and

2 coordinate

Describe

the

c

Find

the

velocity

d

Find

the

magnitude

long-term

of

behaviour

the

particle

and

the

at

of

the

show

these

points

on

found

by

a

time

direction

particle.

t

of

the

velocity

a

when

The 0 ,

 2.08  ,

4.56

0 ,

 0

and

grid.

b

10

2π

2





=

0.

position

vectors

are

0.432 , substituting

 0

t

0.948

the

values

of

t

into

the

0 for

the

displacement.

suluclaC

equation

y

5

3

2

1

x 0 8

9

10

–2

Continued

on

next

page

545

/

12

MODE LLING

b

The

particle

M OT I O N

will

spiral

AND

C H A NGE

towards

the

IN

point

TWO

AND

(0, 0).

THREE

The

DIMENSIONS

velocity

is

differentiating

0 .5t



0 .5t

5e v

c



cos t

 10e

vector

using

found

the

the

by

displacement

chain

and

product

sin t

 0.5t

rules.

0.5t

 5e

sin t

 10e

cos t





0

5t

(cos t

5e

2 sin t )



  0



5t

(sin t

5e

2 cos t )





5 ( 1 

d

v

5

0) 



(0  5

Magnitude

positive

10

2 )

=

11.2,

direction

=

117°

to

the

x-axis.

Exercise 12E

1

A

particle

time

t

is

moves

given

such

that

its

velocity

b

at

Find

by

the

the

particle’s

origin

at

time

distance

from

t.

2t



v





2e

.

  

At

t

=

0

the

particle

is

at

Hence

d

Given

describe

the

path

of

the

particle.



2t

e



2

Find

an

that

a

=

kr

state

the

value

of

k



4 a

c (0, 0).

expression

for

the

A

particle

moves

such

that

its

displacement

particle’s

2t

acceleration

at

time

t.

e at

time

t

is

given

by

r

.

 2t

b

Find

an

expression

displacement

at

for

time

the

particle’s

3e

t.

a

2

A

particle

has

acceleration

at

time

t

given

by

a

an

particle

2t

10e

Find

b

Find

an

expression

at

time

for

the

velocity

of

the

for

the

acceleration

t.

expression

 1



of

the

particle

at

time

t.

2t

4e



2

c

The

particle

is

initially

at

rest

at

the

an

expression

for

its

displacement

A

a

=

kr

state

the

value

of

k

particle

moves

such

that

its

displacement

at t

time

that

origin.

5 Find

Given

seconds

after

its

motion

begins

is

given

by

t t

e

3

A

particle

moves

so

that

its

acceleration

is

r

cos 4t



2

 t

e 

sin 4t

 5

 8 cos 2 t

given

by

Given

the

the

expression

particle

has

a



initial





 8 sin 2 t







velocity

of

the

time

t

particle’s

b

Hence

c

Show

that

time

t

is

Find

the

describe

the

distance

the

path

speed

of

of

the

from

the

(2, 5)

at

particle.

particle

at

t



and

an

initial

displacement

of

4

:

Find

17e

.

0

d

a

Find

2

0 v

a

.

an

expression

displacement

at

for

time

the

particle’s

the

value

particle

is

of

t

at

which

reduced

by

the

speed

of

half.

t.

546

/

12 . 3

6

A

particle

speed

in

(0, 0).

moves

a

circle

Given

revolution

initially

at

is

anti-clockwise

of

that

6

radius

the

4

time

seconds

at

units

to

and

constant

centred

complete

the

on

one

particle

is

A

in

Find

the

a

(5,

particle

circle

4)

rst

of

with

radius

the

particle.

is

moving

4

same

units

anti-clockwise

centred

constant

Initially

it

is

at

on

speed

the

as

point

the

(5, 0).

(0, −4):

b

a

second

an

expression

particle

at

time

for

the

position

down

an

displacement

of

t.

Write

c

Find

the

expression

of

this

velocity

for

the

particle.

vector

at

time

t

Developing inquiry skills

Return to the opening problem for the chapter. The package will be

dropped from the aircraft from a height of 150 m. The aircraft has a speed

1

of 180 km h

. If air resistance can be ignored, nd how far from the

if the aircraft is ying horizontally

b

if the aircraft is ascending at an angle of 30°

12.3

dna y rtemoeG

a

y rtemonogirt

research station the package should be released:

Exact solutions of coupled

dierential equations

on

fungi,

a

tree.

fungus

will

and

is

in

will

spread

are

faster

it

and

growing

growing

encounters

benet

from

Y,

Y

the

X.

happen

as

the

two

fungi

out?

section

models

Y,

the

when

dominate

What

will

where

variable

of

X

but

nutrients

This

X

is

look

the

affected

at

simple

growth

by

suluclaC

Two

the

of

one

presence

another.

547

/

12

MODE LLING

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

A coupled system is one in which either the equation for the derivative of

x contains a function of y or the equation for the derivative of y contains a

function of x or both of these.

Investigation 4

Initially the investigation considers an uncoupled system.

dx

1

Solve the dierential equation



2x

given that x = 5 when t = 0.

dt

2

Solve the system of dierential equations

 x







 

 x 

 

 y

, given that  



 3y



 3 

 y

 2x  





when t = 0

 5

The vector x is often used to stand for the general vector of the variables, so

 x

in the example above x =

and x

 

will be the derivative of this vector with

y

respect to time.

For any system of equations that can be written in the form

x = Mx, and for

which the eigenvalues of M are distinct, the solution is

λ

x = Ae

λ

t

1

p

+ Be

t

2

p

1

where

2

A, B ∈ ℝ,

λ

λ

, 1

are the eigenvalues of M, and 2

p

and p 1

are the 2

corresponding eigenvectors.

3

a

Verify the system of equations in question

b

Find the eigenvalues and eigenvectors of

2 can be written as

M and hence show that your

λ

solution can be written in the form x = Ae

1

λ

t

p

+ Be 1

values of A and

Verify that x = Ae

2

t

p

, stating the 2

B

λ

4

x = Mx

λ

t

1

p

+ Be 1

2

t

p

is a solution to

x = Mx more generally

2

by dierentiating the right-hand side and showing that it is equal to

Mx

You will need to use the fact that from the denition of eigenvalues and

= λ

eigenvectors, Mp 1

p 1

= λ

and Mp 1

2

p 2

2

Two fungi, X and Y, are growing on a tree. X is the faster growing fungus but

when it encounters Y, Y will dominate and benet from the nutrients in X.

The area of tree covered by X is given as

x and that of Y by y. The growth of the

2

two fungi in cm

can be approximately modelled by the following system of

dierential equations where t is measured in weeks:

x



0.4 x

y



0.1 x

5

 0.2 y

 0.1 y

Explain why the signs of the dierent variables and the values of the

coecients indicate that these equations might be consistent with the

information given.

548

/

12 . 3

6

Show that the right-hand side of this coupled system can be written in the

 x 

form

Mx where M is a 2 × 2 matrix and

x



 

7

 y



Hence, nd expressions for the areas covered by X and by Y at time

2

given that X covers 10 cm

8

a

t,

2

and Y covers 6 cm

when t = 0

If this model continues to be a good approximation to growth what will be

the long-term ratio of the area covered by X to the area covered by Y?

b

9

Comment on why this model is unlikely to be valid as

Conceptual

t increases.

How are eigenvalues and eigenvectors useful when solving

a system of equations?

λ

The coupled linear system x

= Mx has solution x

= Ae

λ

t

1

p

+ Be

t

2

p

1

where A, B ∈

, 2

Example 5

a

Find

and

x

=

y

b

the

y

=

3x

=

x

solution

−2

-

-

to

when

t

the

=

following

system

of

differential

given

that

x

=

4

0.

4 y

2y

Determine

the

long-term

ratio

of

x

to

y

The



3 

equations,

dna y rtemoeG

y rtemonogirt

ℝ and are dependent on the initial conditions.

equations

can

be

written

as

4

a



1

0





2 











 



y 



 







x

  3



x

4

.

Hence

the

rst

stage

is

2

λ

λ

−

−

2

=

0

Corresponding

4

⇒

λ

=

2,

−1

eigenvectors

are

to

nd

1

x

4







Ae

eigenvalues

and

eigenvectors

for

3

4

1

2

t



y





Be

1



 1

The

the Putting

4



=



2,

initial



 

B

 A

=

y

The

of

formula

the

general

solution

is

given

in

book.

conditions





 

B

−4

4

1

2t



form

 B

 4 A 



 2

in

x

b

the

1

2t

A

y

 

1

Hence



2

1 and



1

suluclaC

⇒

t

2e

 4e

1

long-term

1

ratio

is

4:1.

As

t

increases

the

second

term

tends

to

0.

549

/

12

MODE LLING

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

Phase por trait

It

can

be

variables

useful

to

change

show

over

on

a

time.

diagram

Such

a

how

the

diagram

is

values

called

a

of

the

phase

two

portrait.

Investigation 5

A system of linear equations has the following solution:

x

1

1

t



y

1

5t

Ae



Be

2

a

1

Show that the equations of the lines through the origin which are in the direction of each of the

eigenvectors are y = 2x and y = −x

b

Choose an initial value for the system that lies on

y = 2x, for example (2, 4), and nd this par ticular

solution to the system of equations.

c

Choose an initial value for the system that lies on the line

y = −x and nd this par ticular solution to

the system of equations.

d

2

3

What do you notice?

A phase por trait indicates future trajectories for dierent initial values.

a

Draw a set of coordinate axes and on them draw

y = 2x and y = −x.

b

Indicate with arrows the direction of motion of any point initially on one of these lines.

For any initial values not on these lines, the values of both

a

A and B will be non-zero.

Conjecture from your general solution the long-term trajectory for systems which are not initially on

y = 2x or

b

y = −x

Conjecture the “history” of the trajectory by considering what happens as

t becomes increasingly

negative.

c

Find the par ticular solution for an initial value of (6, 6).

d

Enter the top and bottom lines of your solution into a GDC and use the table function to look at the

values of x and y as t increases. Set the step interval on your table to 0.1. Consider the values of

x

and y as t increases in both the positive and negative directions. Does this suppor t your conjectures

in par ts a and b?

4

e

Show (6, 6) and the trajectory through that point on your phase por trait.

f

Use technology to plot trajectories within each of the quadrants in the diagram drawn in question

3

Now consider another system of coupled dierential equations that has the following general solution:

1 

 x 



t



 

y





Ae

2



3t

 

 2





Be

 

 1



a

Write down the equations of the two lines through the origin that are parallel to the eigenvectors.

b

By considering the values of A and B for a particular point on either of these two lines, describe the

trajectory of any point initially on these lines.

c

For those trajectories beginning away from the two lines given in par t

trajectory approaches as t →

d

∞

a conjecture the direction the

. Verify your answer using your GDC or Geogebra.

Which direction do the majority of trajectories tend towards as

t → −∞?

550

/

12 . 3

e

As they approach this point write down the approximate gradient for the trajectories that are not

initially on either line through the origin parallel to the eigenvectors.

f

Sketch the phase por trait for the solution given.

g

How would your diagram change when drawing the phase por trait for the system with solution

1 

 x 

 

 y

3t

Ae









Conceptual

5

2



t



2



Be

?







1

Why are phase por traits a useful way of depicting the solutions to dierential

equations?

An

equilibrium

point

(or

equilibrium

solution)

is

a

point

at

which

TOK x ˙

=

If

0

all

and

y ˙

=

points

0.

An

close

equilibrium

equilibrium

to

point,

an

it

is

point

equilibrium

stable,

is

classed

point

otherwise

will

it

is

as

stable

move

or

unstable.

towards

the

“ There is no branch of

unstable.

mathematics, however

abstract, which may not In

both

the

examples

in

Investigation

5,

(0,

0)

is

an

equilibrium someday be applied

At

this

point



0

and



dt

0

so

if

a

particle

were

initially

at

to phenomena of the

dt real world.” – Nikolai

(0,

0)

it

would

stay

there. Lobatchevsky

Where does the power In

both

cases

(0,

0)

is

an

unstable

equilibrium

point

because

particles of mathematics come

that

begin

close

to

the

origin

will

generally

move

away

from

it

dna y rtemoeG

point.

dy

y rtemonogirt

dx

rather from? Is it from its

than

towards

the

origin. ability to communicate

In

the

rst

example

(0,

0)

is

a

saddle

as a language, from the

point

axiomatic proofs or from In

the

second

example

(0,

0)

is

a

source its abstract nature?

the

matrix

differential

through

give

and

the

as

When

the

t

equations

the

→

both

is

a

saddle

the

direction

for

the

long-term

increases

the

are

of

real

the

of

linear

eigenvalues,

eigenvectors

behaviour

positive

and

larger

eigenvalues

t

increases,

with

one

distinct

system

of

the

rst

the

of

order

lines

the

system

matrix

as

t

→

∞

all

towards

trajectories

the

eigenvalue.

direction

The

origin

move

of

is

away

the

an

from

eigenvector

unstable

point.

as

equilibrium

When

t

with

associated

two

coupled

in

eigenvalues

as

equilibrium

origin

has

a

−∞.

both

associated

the

origin

gradients

origin

When

representing

suluclaC

When

the

are

negative

and

more

all

towards

negative

trajectories

the

direction

eigenvalue.

The

move

of

the

towards

eigenvector

origin

is

a

stable

point.

eigenvalue

point.

the

origin

the

eigenvector

As

parallel

t

to

is

positive

increases

the

and

all

with

other

trajectories

eigenvectors

associated

the

the

move

positive

negative

not

on

towards

the

the

then

lines

the

origin

through

direction

of

eigenvalue.

551

/

12

MODE LLING

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

Exercise 12F

1

a

Find

i

the

x

=

y

=

x

=

y

=

solutions

2x

5x

+

to

the

following

systems

of

differential

2y

-

y

iv

x

=

−6,

y

=

22

when

t

=

0

b ii

x

+

equations.

x

=

y

=

Draw

a

x

+

3x

- 4 y

phase

x

=

1,

portrait

y

=

for

11

the

when

t

=

0

general

2y

solutions 3y

x

=

3,

y

=

1

when

t

=

to

case

give

lines x

=

-2 x

y

=

-x

+

the

coupled

differential

0 equations

iii

2y

obtained

the

in

part

Cartesian

through

the

a.

In

each

equations

origin

parallel

of

to

the

the

2y

eigenvectors.

2

Match

each

differential

- 5y

of

x

these

=

2,

y

phase

=

1

when

portraits

t

=

with

0

the

associated

solution

to

a

system

y

b

y

x

x

c

y

d

y

x

x

x

1

x

3

1

y

3

a

Find

the

system

general

of

x

1

solution

differential

1

of

the

following

b

equations:

5t



2x

=

x

Be

2

Sketch

the

system

of

1

phase

portrait

equations,

asymptotic =

for

making

this

clear

any

behaviour.

+ 3y

.

y

3 Ae

y

.

x

1

4t



D

Be

2

Be

3

2t



Ae

1 3t



Ae

y

1

t



B

Be

0



1 t

t



Ae

y

C

x

0

t



coupled

equations.

a

A

of

c +

Find

the

equation

of

the

particular

solution

4 y given

that

x

=

−9,

y

=

−1

when

t

=

0.

552

/

12 . 3

4

a

Find

the

system

all

general

of

nal

solution

differential

solutions

of

the

equations.

correct

to

3

a

following

Draw

Give

a

phase

solution

with

portrait

x, y

≥

for

this

general

0.

signicant The

system

of

equations

that

led

to

this

gures. solution

=

b

=

+

2x

y

-

x

=

0 .4 x

y

=

0 .1x

y

Sketch

the

system

of

phase

portrait

for

equations,

making

clear

the

area

the

equation

of

the

particular

5

a

By

given

that

x

considering

below

explain

the

0

=

1,

the

y

=

1

when

system

what

of

would

t

will

=

the

initial

no

=

fungus

longer

Write

down

that

will

is

extinct

=

ax

y

=

cx

conditions

are

x

=

+

i

=

Let

by

+

differential

replace

x

ii

0

give

y

the

the

=

considering

the

initial

times

be

x

areas

and

system

(0,

the

phase

portraits

from

part

conditions

0)

beyond

y

covered

where

x

the

Use

your

a,

or

otherwise,

these

point.

,

to

be

says

a

on

stable

that

the

phase

the

point

Fred’s

fungi

variables

x

X

and

x

point

to

4

be

the

and

y

eigenvalues

all

points

must

0.

0

Y

area

was

1

deduce

the

of

x : y

if

the

initial

were

such

that:

1

1 0




two

statement.

Investigation

two

y

portrait

proportions

equilibrium.

“Nearly

equilibrium

for

the

write

0

A

to

this

by

0

0

Fred

of

if

and

i



above

solution

0

conditions down

by

equations

0

general

for

long-term

In

tree

dy

solutions

6

that

0,

c By

on

on

0.

x

it

fungi

grow.

the

0

c

the

happen

fungi

b

of

equations

equations y

either

0.

and when

by

solu-

b tion

covered

behaviour. and

Find

0 .1y

any reaches

asymptotic

+

this If

c

- 0 .2 y

dna y rtemoeG

y

3x

y rtemonogirt

x

was

the

values

1

were

dependent

on

the

initial

suluclaC

conditions.

Systems with imaginary or complex eigenvalues

Sometimes

need

to

system

nd

in

portrait

the

Because

the

this

and

case,

complex

2

+

but

the

have

you

do

and

2

−

complex

they

eigenvalues.

eigenvectors

need

long-term

eigenvalues

equation

3i

will

corresponding

predict

characteristic

example

system

will

will

to

be

or

able

behaviour

come

always

a

to

of

from

be

the

You

solution

draw

the

do

the

to

not

the

phase

trajectories.

solutions

conjugate

to

the

pair,

for

3i.

553

/

12

MODE LLING

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

Investigation 6

1

a

Find the eigenvalues for the following system of equations:

x



2y

y



2 x

In a similar way to the case of real eigenvalues, the general solution to a system of equations with imaginary

 x  bit

eigenvalues ±

bi can be written in the form:

 

 y



Ae

bit

v



Be

v

1

, with b∈ 2



bit

b

State the value of b for the system given in par t a and write e

in the form cos(bt) + i sin(bt)

HINT

A, B and the elements of the two eigenvectors can be complex numbers

but values can be found in which both and x and y are real for all values of t.

bit

can be

This is beyond this course but you need to note that the fact e

written as cos(bt) + i sin(bt) means the solution will be periodic with a

2

period of

. b

c

For the system given in par t a nd the value of t which gives the rst positive time when the

coordinates (x, y) are again equal to their initial value.

y

y

For imaginary eigenvalues the phase por trait will consist of

concentric circles or ellipses centred on the origin.

x

2

a

A

b



y



3 x

2x



 4 y

B

y

x



y



2x

x



y

 2y

the real par t of the eigenvalue

and

3

x

Write down what you notice about

i

c

x

Find the eigenvalues for the systems below.

ii

the imaginary par t in both A and B

Explain why this is always the case.

Write down a condition on a for the trajectory to spiral towards the origin.

dy

at a point on the x-axis

To decide whether the spiral is clockwise or counter-clockwise you can evaluate dt dx

at a point on the y-axis.

or

dt

dy

b

For each of the systems in question 2 nd the values of

at the point (1, 0) and deduce what this

dt

means about the direction of motion as

4

t

increases.

c

Use the chain rule to also evaluate the gradient at (0, 1).

d

Hence sketch the phase por trait for the two systems given in question 2

Conceptual

How is

the motion of a par ticle dierent when the matrix for a coupled system has either

imaginary or complex eigenvalues?

554

/

12 . 3

The result from question 2 means that a solution can be written in the form



 x at







y 







e

bit



Ae

bit



v

Be

v

1

2



where a, b ∈ ℝ and the periodic factor is the

same as the one in question 1

The solution will be either a clockwise or counter-clockwise spiral with the

trajectories either moving towards or away from the origin as shown below.

y y

International-

x x

mindedness

“As long as algebra

and geometry have

been separated, their

If the eigenvalues are a ±

bi then the trajectories will:

y rtemonogirt

and their uses limited,

but when these two

sciences have been

•

form circles or ellipses with the origin as a centre if

•

spiral away from the origin if a > 0

a = 0 united, they have lent

each mutual force, and

•

spiral into the origin if a < 0.

dna y rtemoeG

progress has been slow

have marched together

dy

towards perfection.” –

for a point on the

The direction of movement can be found by evaluating

18th century French

dt dx

mathematician Joseph

for a point on the y-axis.

x-axis or

dt -Louis Lagrange

Exercise 12G

For

each

of

the

systems

2

below:

Its

i

Find

the

the

dy

ii

Find

displacement

and

the

gradient

of

the

Earth

with

the the

point

1,

the

Sketch

t,

measured

with

the

at

the

origin,

is

given

as

( x,

y)

coordinates

satellite’s

lying

motion.

The

in

the

plane

velocity

of

of

the

0 satellite

iii

time

curve

dt

at

at

eigenvalues.

suluclaC

1

phase

portrait

for

at

time

t

is

given

by

the

following

the system

of

equations:

system. dx 

2x



y



x



2y

dt

a

x

=

2x

y

=

x

b

- 5y

x

=

- x

- 3y

y

=

4 x

+ 3y

dy -

2y

dt

c

x

=

-0

2x

y

=

- x

-

x

=

3x

y

=

4 x

0

+

y

2y

d

x

=

y

=

x

+ 5y

-5 x

+

a

away

y

b e

Verify

that

from

Given

that

the

the

the

satellite’s

path

will

take

it

Earth.

satellite

passes

through

- 9y

the

point

(1, 1)

sketch

a

possible

future

- 3y trajectory

for

the

satellite.

555

/

12

MODE LLING

12.4

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

Approximate solutions to

coupled linear equations

Many

are

closely

for

to

ecosystems

connected,

limited

nd

resources.

population

species

can

always

come

These

through

Most

are

to

of

Chapter

possible

or

will

over

are

of

both

one

time?

addressed

coupled

non-

equations.

differential

and

cannot

methods

method

that

competing

where

equations

approximate

Euler

sizes

Fortunately,

numerical

it

dominate

of

linear

exactly.

Is

consideration

systems

species

often

together

differential

not

give

exist

kinds

linear

contain

be

there

that

you

solved

are

can

results,

which

equations

many

be

used

including

studied

to

the

HINT

in

11.

Though easy to apply,

the Euler method is not

dy For

the

equation

=

f

x, y

(

the

)

Euler

method

equations

are:

much used in practice

dx

due to its inaccuracies. x

=

x

n+1

+

h,

y

n

=

y

n+1

+

hf (x

n

,

y

n

),

where

h

is

the

step

size.

n

Other methods include

the improved Euler

method and the

Runge–Kutta method. For the coupled case in which

But these are not in the dx

dy 

f



1

x , y ,t



,



f



2

dt

x , y, t



the Euler method equations are:

Higher Level course.

dt

t

=

t

n+1

+

h

n

HINT x

=

x

n+1

+

hf

n

(x 1

,

y

n

,

t

n

) n

This is provided in the

y

=

y

n+1

+

hf

n

(x 2

,

y

n

,

t

n

)

formula book.

n

Example 6

a

Use

the

of

and

x

x

=

y

=

and

b

In

3x

x

x

y

-

-

=

Euler

method

when

t

=

4

given

5

y

=

the

−2

4



=

2e

when

exact

2

for

size

the

of

0.1

to

nd

following

the

system

approximate

of

values

differential

equations:

t

=

solution

0.

to

this

system

of

differential

equations

was

found

to

be

1

2t

y

t

step

2y

Example



and

a

4 y

x



1

with

t



 1

 4

e

. 



Find

the

percentage

error

in

your

answers

to

part

a

1

556

/

12 . 4

x

two

equations

=

+

x

n +1

0

1( 3x

n

= 1 .3 x

are:

-

4 y

n

n

)

There

Only

- 0 .4 y n

is

do



y

n 1



0

1 x

n



0 .1x



t

=

1,

x

2y

n

=

2,

=

x

more

48.1,

y

=

The

=

At

exact

t

=

1,

306,

values

x

it

y

=

an

is

=

=

2,

x

y

=

the

equation

are:

exam

be

need

but

to

the

errors

actual

13.3

list

intermediate

recurrence

=

436,

y

=

errors

this

are

calculated

1,

16%

it

is

increase

and

the

=

2,

30%

the

100.

clear

the

as

that

the

actual

the

Euler

values,

iteration

method

and

the

continues,

asymptotic

behaviour

of

the

17%

(y-value t

using

are:

approximation

At

used

error 

case,

though

=

relation

109

errors Percentage

solutions

amount

underestimates

t

next

given.

Percentage

In

At

the

76.2

from

57.6,

no

exact t

make

straightforward.

formula

At

will

equation.

11.0

should

b

feel

the

n

in t

you

simplify

 0 .8 y

There

At

if

to



n

n

At

so

need

n

stages

y

no

and

is

similar

approximately

to

the

four

exact

times

equation

x-value

as

30%

expected

from

the

exact

equation).

dna y rtemoeG

The

y rtemonogirt

a

Exercise 12H

1

Solve

each

of

the

equations

from

dx

question

2

b 1a

in

with

Exercise

a

step

12F

size

of

using

0.1

the

from

t

Euler

=

0

to

=

2x





2 xy

xy

dt

method

t



1.

dy 2

Use

your

results

to

sketch

the



y

trajectory dt

from

the

initial

point

and

compare

your

x answer

with

question

the

phase

portrait

drawn

=

1,

3

1b

Use

0.1 Comment

on

the

accuracy

of

your

in

particular

whether

the

ratio

=

1

when

t

=

0

the

to

Euler

nd

method

with

approximations

a

step

for

the

size

of

values

answers,

of and

y

for

x

and

y

when

t

=

0.5

for

the

following

of

systems. to

y-coordinates

expected

is

tending

towards

the

value.

suluclaC

x-

dx 2

a



2tx



3 x

 3y

dt

2

Use

to

x

the

nd

and

Euler

method

with

approximations

y

when

t

=

1

for

for

the

a

step

the

size

values

following

of

0.1

dy

of

2

 3ty

dt

systems.

x

=

−1,

y

=

2

when

t

=

0

dx

a



2 xy



x

dt dx

b

dy 

2y





2tx





3 x

y

dt

xy

dt dy 2

x

=

1,

y

=

1

when

t

=

0

 ty

dt

x

=

0,

y

=

1

when

t

=

0

557

/

12

MODE LLING

M OT I O N

AND

C H A NGE

IN

TWO

AND

THREE

DIMENSIONS

Predator–prey and other real-life models

Investigation 7

When a population X of size x increases at a steady rate α, propor tional to the

size of the population, we can write the dierential equation describing its rate

dx

of growth as



ax . This equation leads to exponential growth.

dt

Suppose now a dierent species Y , with population size y, is in competition with X.

The rate of increase of x is now aected by the size of y, such that as y

increases the rate of increase of x will diminish.

dx

The simplest equation that will do this is



a

 by



x

dt

As the population of Y (y) increases, the rate of growth of

depends on a

If Y

−

X

decreases as it

by

is competing with X for the same resources then the equation for the

growth in population of Y is likely to have a similar structure:

dy 

c



dx



y

dt

However, if species Y is a predator and species X is the prey then an equation

of the form

dy 

 cx



d



y might be more appropriate.

dt

HINT 1

Explain why this equation might be more appropriate than the previous

These types of

one for a predator–prey situation.

equations are often The populations, in thousands, of sh (x) and of sharks, in hundreds (y) at

referred to as the time t (years) are given by the equations:

Lotka–Volterra dx

dy 

3x

 3 xy

and

dt

2



xy

 2y

equations.

dt

Find the two equilibrium positions for the populations of sh and sharks.

Initially there are 3000 sh and 20 sharks.

3

Use the Euler method with a step size of 0.02 years to show that when

t =

0.02 there are approximately 3144 sh and 20.4 sharks.

Use either a spreadsheet or a graphing calculator to answer the questions below.

4

a

Plot the values of x and y on a set of axes for 0 ≤

t ≤ 3

Plot values

every 0.2 years.

b

Join the points to create a trajectory.

What does this suggest about the non-zero equilibrium point?

5

Explain from a consideration of the dierential equations why the population

of sharks continues to grow once the population of sh has star ted to fall.

6

If using a spreadsheet draw a graph of

x and y on the same axes, with t

on

the horizontal axis. Comment on the signicant features of the graph.

7

Conceptual

How can we develop a predator–prey model and what can

the model display?

558

/

12 . 4

Exercise 12I

1

In

the

predator–prey

following

of

the

prey.

equations

predators

All

model

let

and

Q

P

given

be

the

the

by

values

individuals

and

are

of

the

measured

of

time

is

of

the

the

fox

rabbit

population

population

is

a

maximum.

in Using

technology

verify

your

answers

to

measured parts

in

size

when

d

1000s

the

population

population

population

ii

the

b

and

c

and

describe

the

change

in

years. the

populations

over

one

cycle.

dP 



Q

2 P

dt

HINT

dQ

Because of the approximations involved 

1



P

Q

dt

the

two

equilibrium

points

for

becomes inaccurate and spirals out after

the

the rst cycle, rather than repeating the

populations.

b

Use

the

equations

happen

the

to

the

predators

to

explain

population

were

of

absent

what

the

(P

=

same cycle.

would

prey

if

3

0).

Two

species

grazing

c

Use

the

equations

to

explain

happen

to

the

the

population

of

if

the

prey

were

each

other.

absent

(Q

=

If

the

and

initial

of

the

method

population

predators

with

a

step

is

of

prey

2000

size

of

grass

in

X

and

Y,

are

competition

is

use

0.1

The

growth

of

their

can

be

given

by

the

following

0).

differential

d

same

animals,

the

populations predators

grazing

what

with would

on

of

equations:

2000

the

year

Euler

dx 

2



3



y

dna y rtemoeG

Find

y rtemonogirt

a

in the Euler method the trajectory

x



to dt

nd

the

population

of

each

after

1

year.

dy

Populations

of

rabbits

and

foxes

live

together



x



y

dt in

a

large

area

of

countryside. where

The

numbers

in

each

population

are

x

is

the

population

by

the

following

dx

Use

of x



2 xy

and

 1 .8 xy

dt

x

rabbits

foxes

of

Y,

the

measured

Euler

is

(in

(in

the

size

1000s)

0.1

for

of

the

and

y

population

is

to

nd

the

two

population

of

population

the

You

of

size

100s).

may

method

Find

for

the

the

non-zero

equilibrium

populations

of

is

less

in

y

is

the

hundreds.

with

a

=

there

0.75,

y

=

are

750

step

size

foxes

and

Use

rabbits.

rabbits

and

100

b

the

ini ti a l

va l ue s

of

is

the

equati o ns

the

populations

are

initially

of

s tate

fo x e s

increa s i ng

i

Write

into

the

differential

for

the

size

of

the

rabbit

the

fox

ra bb its

de cr e as i ng .

equations

is

a

ory




3,

explain >

If

y

d

the

DIMENSIONS

2

the

the

>

2

population

initial

the

of

Y

populations

population

of

will

decrease.

have

X

will

x




2

x

dt

dt

equation

is

4c =

Ae

t λ 1

+

Be

t λ 2

.

(11

marks)

2

quadratic

have

be

λ

two

and 1

equation

distinct

real

λ

+

roots,

bλ

let

+

c

=

0

these

will

roots

λ 2

dna y rtemoeG

The

y rtemonogirt

18

and

.



b 

dy



a  b

suluclaC

17

a

567

/

Approaches to learning: Communication,

Disease modelling

Critical thinking

Exploration criteria: Mathematical communica-

tion (B), Reection (D), Use of mathematics (E)

IB topic: Dierential equation (variables

ytivitca noitagitsevni dna gnilledoM

separable)

Hypothetical situation

Imagine a population threatened by an infectious disease.

Within the population there are two groups, those that are infected (I) and those that are healthy (H)

Assume that each year, the probability that a healthy person catches the disease is

c, and the probability that an

infected person recovers is r

Let x be the propor tion of the population that are infected and t be the number of years from the beginning of

recording.

Show that a dierential equation that will model the rate of change of the propor tion of the population aected

dx

= c(1

with respect to time is given by:

−

x) − rx

dt

Reect on the assumptions made in this hypothetical situation and how they may dier in a real-life situation

involving an infectious disease in a population.

By separating the variables nd the general solution of this dierential equation.

Using this general solution what would you expect to happen to the propor tion of people aected by the disease

over time? (ie what happens as t tends towards innity?)

568

/

12

Simulation

are

now

Assume

Use

that

these

going

c

=

to

0.3

values

simulate

and

of

c

r

and

=

r

a

specic

case

of

this

situation.

0.2.

to

simulate

the

proportion

of

infected

people

over

ytivitca noitagitsevni dna gnilledoM

You

time

To

(H

do

=

You

this,

7)

and

are

If

the

random

the

three

going

Consider

10

assume

to

the

If

number

is

person

a

they

(who

2

or

a

3

a

1

will

or

a

2

and

remain

a

10

of

of

off

to

the

year

disease

(I

of

the

healthy

in

year

person

will

in

each

numbers

the

which,

particular

happens

they

and

of

starts

list

Otherwise

size

with

what

from

1,

disease.

is

infected

simulate

numbers

catch

Otherwise

population

are

rst

number

the

a

from

is

is

to

healthy

remain

person

1

=

10

0,

seven

are

healthy

3).

people

0).

over

Generate

a

10

years.

series

of

10.

(c

=

0.3)

then

they

will

healthy.

infected

they

will

recover.

infected

Example:

Year

0

1

2

3

4

5

6

7

8

9

10

5

3

7

3

8

2

5

5

9

6

H

I

I

H

H

I

I

I

I

I

Random

number

Healthy or H infected?

Conduct

this

For

year

each

graph

What

Now

of

x

simulation

calculate

against

does

the

consider

10

the

times

value

for

of

each

x,

the

of

the

10

people

proportion

of

in

your

people

population.

infected.

Plot

a

time.

graph

the

suggest

will

differential

happen

equation

as

x

tends

to

innity?

again.

dx Rewrite

the

equation

for

but

use

the

specic

values

of

c

=

0.3

and

r

=

0.2

dt from

before.

Solve

this

differential

conditions

Sketch

What

How

the

of

t

0

graph

happens

well

=

does

as

equation

and

of

t

this

x

x

0.3

against

tends

t

=

to

with

by

to

separating

nd

the

the

value

variables

of

b

in

the

and

using

the

starting

equation.

t

innity?

your

data

in

your

simulation?

Ex tension

What happens to the solution if you vary c and r?

What happens to the solution if you vary the star ting

conditions of the propor tion of infected people in the

population?

569

/

Representing multiple outcomes:

random variables and probability

13

distributions

Concepts

Probability

of

events

enables

occurring

probabilities

below

us

can

in

be

predictions

the

to

and

evaluate

diverse

modelled

and

quantify

well

to

the

risk.

situations

help

informed

likelihood

you

■

Representation

■

Validity

The

described

Microconcepts

make

decisions.

■

Discrete random variables

■

Binomial distribution

■

Normal distribution

■

Probability distribution function

■

Continuous probability distribution function

■

Discrete probability distribution function

■

Probability distribution

■

Cumulative distribution function

■

Discrete and continuous data

■

Expected value

■

Variance

■

Poisson distribution

■

Parameters

■

Probability density functions

How many errors are likely in a

new translation of a book?

How can an

engineer quantify

the risk that an

unacceptable

number of micro

scopic aws are

present in an

aircraft wing?

How can an airline manage their booking

system so that overbooking the ight

maintains protability while treating

their customers fairly?

How likely is it that a sample

of sh is representative of

the entire population?

570

/

I

woke

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inspected

expect

than

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that

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average

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Do the results surprise 3.2

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Think about the questions elevator

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the

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and answer any you can. •

How

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this

tale

that

involve

probability?

As you work through the

•

What

kind

of

variables

are

involved?

chapter, you will gain

•

How

could

•

Write

you

represent

and

quantify

the

mathematical knowledge

variables?

and skills that will help you down

some

examples

where

chance

has

played

a

role

in

your

life.

to answer them all.

Before you start

Click here for help

You should know how to:

1

Find

eg

the

Find

mean

the

of

a

mean

with this skills check

Skills check

data

set.

1

Find

the

mean

value

of

x

of:

x

1

2

3

6

11

9

7

3

2

1

i

x

4

7

8

10

2

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5

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f f

i

i

 Mean

f

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125

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=







f

8.33

15

i

3

2

Use

technology

to

solve

2

eg

Use

technology

to

2

equations.

solve

x

Solve

the

equation

x

e

x

+

ln(x)

=

1

x

e

+

x

=

4

y

8

(3.65, 4)

(–1.23, 4)

f2(x)=4

2

f1(x)=x

x

•

e

+x

x

–1

–5

1

5

–1

x

=

−1.23

or

3.65

571

/

13

R E P R E S E N T ING

13.1

You

learned

event

and

apply

these

For

occur

the

The

in

is

real

in

of

determined

use

Let

X

the

by

to

how

to

combined

model

Chancer’s

on

a

random

that

different

following

the

Terminology

cafe

given

quantity

represent

of

5

quantify

events.

the

In

probability

this

behaviour

in

chapter

various

of

an

you

will

processes

life.

buses

another

We

Chapter

probability

percolators

number

time

in

concepts

example,

failing

OUTCOME S

Modelling random behaviour

have

that

M U LT I P L E

in

day

arrive

that

the

is

at

a

a

opening

quantity

bus

changes

stop

scenario

that

in

a

randomly.

the

number

changes

randomly.

5-minute

These

of

interval

quantities

of

are

processes.

terminology

number

of

and

failing

notation:

percolators

on

a

given

day.

Explanation

Discrete: X can be found by counting.

X is a discrete Random:

X is the result of a random process.

random variable.

Variable: X can take any value in the {0, 1, 2, 3, 4}.

A

rst

step

in

acquiring

knowledge

about

X

is

to

ll

in

a

table:

Number of failing 0

1

2

3

4

percolators (x)

P(X = x)

The

0

≤

ve

P(X

You

then

of

In

probabilities

=

will

This

P(X = 0)

row

x)

the

establishes

how

random

this

the

to

1)

1,

the

you

probabilities

the

P(X =

and

in

probability

entire

variable

example

determine

add

=

they

2)

P(X =

must

each

3)

P(X

=

4)

satisfy

1.

determine

shows

the

≤

must

P(X

in

have

probability

second

row

distribution

probability

its

the

of

1

is

of

in

X

distributed

section

since

13.2.

the

to

each

in

order

table

value

domain.

explored

data

distribution.

sets

You

for

can

patterns

also

explore

to

processes.

TOK

Investigation 1

“ Those who have

Daniel is playing a dice game in which he throws ve fair cubical dice until he knowledge, don’t

throws ve equal numbers. He tires of this game after throwing the dice 617 predict. Those who

times without success, and plans to carry out a simulation with a spreadsheet predict, don’t have

instead. knowledge.” – Lao Tzu.

He writes down the denition of a discrete random variable T:

“T is the number Why do you think

of trials taken until I throw ve equal numbers with ve fair cubical dice.” that people want to

1

What is the sample space for T ? believe that an outside

2

What is P(T = 1)?

3

How many trials would Daniel expect to carry out before he can expect to

inuence such as an

octopus or a groundhog

can predict the future?

have thrown ve equal numbers once?

572

/

13 . 1

You can test your answer with a spreadsheet as shown below:

a

Use a random number generator such as “ = RANDBETWEEN(1, 6)” in

each one of the rst ve columns of the rst row to simulate the ve dice.

b

Typing “ =IF(AND(A1=B1,B1=C1,C1=D1,D1=E1),1,0)” in the sixth

column makes the spreadsheet give the answer 1 if ve equal

numbers are thrown, 0 otherwise.

c

Copy and paste the rst row down to the 1500th row to simulate 1500

throws of the ve dice.

d

Typing “ =MATCH(1,G1:G1500,0)” will nd the number of the rst trial

of the 1500, if any, in which ve identical numbers were thrown. This is

the value of the random variable T. Pressing F9 will generate another

1500 trials.

Find P(T = 2), P(T = 3), P(T = 4), … Generalize to nd a function

P(T = t). This is called a probability distribution function

4

What is the general statement for f(t) = P(T = t)?

5

What should the values of f(t) add up to on its domain and why?

6

Find

f (t ) over all possible values of

Conceptual

How can a discrete probability distribution function be found?

Denition

A discrete probability distribution function

f(t) assigns to each value of the

random variable a corresponding probability.

f(t)

f(t) = P(T = t).

dna scitsitatS

7

T

ytilibaborp

f(t) =

is commonly referred to by the abbreviation “pdf ”.

The discrete cumulative distribution function

F(t ) assigns to each

value of the random variable its corresponding cumulative probability.

t

F(t )



P(T



t)





f (n) where a is the minimum value of the domain of

f(t)

n a

F(t) is commonly referred to by the abbreviation “cdf ”.

Example 1

A

fair

cubical

dened

a

b

as

dice

the

sum

Represent

the

i

of

a

table

Hence

i

P(S

nd

>

7)

and

of

a

fair

the

tetrahedral

numbers

probability

the

distribution

values

the

on

dice

ii

a

ii

P(S

bar

are

two

of

S

thrown.

The

discrete

random

variable

S

is

dice.

as:

chart.

probabilities:

is

at

most

5)

iii

P(S

≤

6

|

S

>

2).

Continued

on

next

page

573

/

13

R E P R E S E N T ING

M U LT I P L E

OUTCOME S

a

Draw

1

2

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

i

s

2

3

4

5

6

7

8

9

Read

10

1

1

1

1

1

1

1

1

24

12

8

6

6

6

8

12

24

ii

Probability

distribution

of

S

Don’t

the

P(S

sample

P(S = s)

sample

off

event 1

a

=

space

diagram.

probability

s)

space

forget

in

turn

of

from

each

the

diagram.

to

label

both

axes.

0.3

)s =

0.2

S( P

0.1

0

2

3

4

5

6

7

8

9

10

s

1

b

i

P(S

>

7)

=

1

8

P(S

≤

5)

=

12

24

1

=

1

+

1

ii

1

+

1

+

24

4

1

+

5

+

12

8

Take

=

6

5”

P(S

iii

P(S



S

6



) 2



care

to

interpret

“at

most

12

6  S



2 )

correctly.

Use

the

formula

for

conditional



P(S



probability

2 )

and

intersection

of

nd

the

the

two

sets.

13

P(3



S



6)



13

24 

P(S



 23

2 )

23

24

Example 2

The

probability

P(U

=

u)

=

a

Find

b

In

c

Hence

k(u

the

100

distribution

−

3)(8

value

trials,

of

−

k

u),

and

calculate

predict

the

of

a

discrete

u∈{4,

hence

the

me a n

5,

6,

random

expected

of

U

is

dened

by

7}.

represent

va lue

variable

the

value

U

after

of

a

probability

each

large

distribution

possible

numb e r

outcome

of

tr i al s.

of

of

U

U.

Inter pr et

y ou r

answer.

574

/

13 . 1

a

Represent

u

4

5

6

the

probability

distribution

in

a

7 table.

4k

P(U = u)

6k

6k

4k

1 4k

+

6k

+

6k

+

4k

=

20k

=



1

k



,

Use

the

fact

that

the

probabilities

must

add

20 hence

to

u

4

5

6

7

1

3

3

1

5

10

10

5

1

on

the

domain

of

the

pdf.

P(U = u)

b

The

expected

of

=

number

of

Use

occurrences

the

formula:

Expected

number

of

1 u

4

is

100 



20.

Similarly,

occurrences

the

of

A

=

nP(A).

5

expected

and

c

A

7

number

are

30,

grouped

expected

30

and

number

Frequency

the

occurrences

20

frequency

table

for

6

these

occurrences

is

6

7

20

30

30

20

value

5,

respectively.

5

mean

frequencies

of

of

4

u

Hence

of

of

U

with

Each

these

Use

number

the

is

formula

occurrences

for

the

is

a

frequency.

population

mean:

k

f

 20  4

of

 30  5  30  6 

20  7

x

i

i

k

550 i 1

is

not

a

the

probability

numb e r

Nevertheless,

value

100

You

can

briey

20  4

of

U

in

the

models

expected

in

the

a



where

n



n

of

Recall

function .



f i

i 1

the

measure

denition

of

central

of

the

mean

as

a

tendency.

ce ntra l

dat a

set

of

trials.

express

the

calculation

in

part

c

of

the

previous

example

more

as

 30  5  30  6 

20  7

20 

4 

100

Notice

d oma in

dis tr i b uti on

it

5.5

dna scitsitatS

5.5



100

ytilibaborp



100

that

variable

this

with

100

is

its

30  5 

the

sum

of

the

corresponding

product

30 

100

of

probability,

each

or

6 

20 

7 

100

value

u P( U



of

1 

4 

100

the

u).

3  5 

5

3  6 

10

1 

10

7 

5

random

This

leads

u

to

a

further

generalization

for

all

discrete

random

variables.

The expected value of a discrete random variable X is the mean score that

would be expected if X was repeated many times. It is calculated using the

formula: E (X ) 







xP(X



x)

x

575

/

13

R E P R E S E N T ING

M U LT I P L E

OUTCOME S

Investigation 2

Consider the distribution in worked example 1: “A fair cubical dice and a fair tetrahedral dice are thrown. The

discrete random variable S is dened as the sum of the numbers on the two dice.” Consider an experiment in

which 100 values of S are calculated in 100 trials.

Predict the average of these 100 values of

S values by:

a

deducing from the shape of the bar char t representation

b

application of the formula for the expected value of a discrete random variable

c

constructing a data set of 100 values of

formulae and dragging

1

2

S with a spreadsheet and nding its mean by entering these

A1, B1 and C1 down to the 100th row.

What are the strengths and weaknesses of each approach? Discuss.

Conceptual

What does the expected value of a discrete random variable predict about the outcomes

of a number of trials?

Application: Many countries organize national lotteries in which adults buy a ticket giving a chance to win

one of a range of cash prizes. Prots are often invested in “good causes”. For example, the UK National

Lottery has distributed over £37 billion to good causes including spor t, ar t and health projects since 1994.

It is possible to use the expected value formula to manage the prize structure of a lottery in order to maintain

protability.

Prize

Probability

Cash value per winner (£)

1

5 421 027

1st (Jackpot) 45 057

474

1

2nd

44 503 7 509 579

1

3rd

1 018 144

415

1

4th

84 2180

1

5th

25 97

1

6th

Free ticket 10.3

576

/

13 . 1

Looking at the cash prizes only for simplicity, we can nd the expected winnings as follows:

Expected cash winnings =

1

1

5 421 027 



45 057

44

1

503 

474

1

 1018 

7 509 579

1

 84 

144

415



25 

2180



0.430

97

This would appear to show that you would expect to make a prot (or a positive gain) playing this lottery!

However, the cost of a ticket is £2.00 so you expect a

loss (a negative gain) of £1.57. The attraction of the

game is based on the desire to win a large prize and/or contribute to good causes. But it should not be a

surprise that you would expect to make a loss on any one game.

If X is a discrete random variable that represents the gain of a player, and if

E(X) = 0, then the game is fair.

Example 3

Some

CAS

This

x

students

project.

have

Some

incomplete

a

meeting

of

the

design

decisions

probability

(prize in $)

to

a

made

distribution

dice

in

game

the

table

to

raise

meeting

are

funds

for

charity

as

part

of

a

lost.

remains:

1

2

4

6

7

11

1

1

40

4

8

P(X = x)

67 The

students

also

recall

that E (X )

=

.

Determine

an

b

expression

What

is

prot?

a

the

Let

by

the

a

for

P( X

would

missing

and

of

smallest

What

the

rest

the

=

table.

Given

the

probabilities

follow

cost

you

the

students

recommend

probabilities

be

could

as

the

set

cost

represented

b

for

of

playing

the

the

game

with

a

11 + b

= 1 -

1 -

40

1 -

4



2 

8

,

4a

 6b

 7 

4

40

a

+ b

also

67

8

=

down

in

order

to

predict

a

why?

unknown

and

writing

involving

them

quantities

down

is

a

true

problem-

strategy.

The

probabilities

must

add

to

1

and

the

for

the

expected

value

can

be

applied.



20

7 Hence

write

20

1 

model,

7 =

1

11

the

variable

formula

1

game

and

Representing

solving

a

linear

x).

statements

Then

a

dna scitsitatS

a

ytilibaborp

20

17 and

20

4a

+ 6b

=

Solve

the

system

of

simultaneous

10 equations.

1 So

a

=

3 and

5

b

=

Look

for

a

pattern

in

the

probability

20 distribution

the

y

=

general

mx

linear

+

c.

table

or

alternatively

equation

This

is

for

called

a

linear

apply

function

generalizing

to

a

model.

Continued

on

next

page

577

/

13

R E P R E S E N T ING

The

x

completed

M U LT I P L E

table

(prize in $)

is

OUTCOME S

:

1

2

4

6

7

11

10

8

6

5

40

40

40

40

40

P(X = x)

1 Hence

f ( x)

=

P(X

=

x)

=

( 12 -

x)

40

67

b

E (X )

=

=

$ 3. 35

so

charging

a

Apply

player

the

formula

for

the

expected

value

20 and

$3.35

would

charging

but

this

$3.36

could

charging

it

be

would

a

fair

would

easily

$4.00

game.

predict

give

would

predict

a

a

be

larger

reect

critically.

Therefore

a

small

loss.

more

prot,

Perhaps

practical

and

prot.

Designing a C A S project

Students

a

school

the

A

are

fair.

chance

fair

working

Adults

to

cubical

outcome

the

win

dice

to

are

cash

is

prizes

raise

money

invited

prizes.

thrown.

to

pay

The

To

for

to

rules

play

charity

play

are

the

a

as

in

a

CAS

simple

project

game

that

during

gives

follows:

game

once

costs

$5.

For

each

are:

Outcome

1

2

3

4

5

6

Prize ($)

3

3

3

4

5

6

The

expected

gain

1

1 3 

 3 

1  3 

6

6

a

Is

b

You

this

a

from

can

4 

generate

the

100

is

1

1

 5

6

game?

test

game

1 

6

fair

this



6 

6

Would



5 

4  5 

1

6

you

protability

expect

of

the

it

to

make

game

a

using

prot

a

for

the

charity?

spreadsheet

to

trials.

Enter the prizes in cells A1, B1, C1, D1, E1, F1 and “=RANDBETWEEN(1,6)”

cell

H1.

Then

type

in

cell

J1:

in

=“IF(H1=1,$A$1,IF(H1=2,$B$1,IF(H1=3,$C$1,

IF(H1=4,$D$1,IF(H1=5,$E$1,IF(H1=6,$F$1))))))”

Copy

nd

and

the

Work

in

drag

total

a

all

cells

winnings

small

group

c

Suggest

a

d

Suggest

changes

change

distribution

e

Suggest

of

to

and

on

to

to

the

changes

attractive

down

the

the

100th

total

these

the

the

to

design

game

game

gain

to

to

row.

in

tasks

make

make

it

it

Use

the

for

fair

fair

the

100

the

by

by

spreadsheet

to

trials.

CAS

project:

changing

changing

the

cost.

the

prizes.

to

the

game

that

make

it

protable

but

more

players.

578

/

13 . 1

Exercise 13A

1

Consider

table(s)

these

could

probability

three

not

tables.

represent

distribution

and

State

a

a

which

discrete

of

C.

why.

b

Find

E(C).

a 1

2

3

4

0.1

0.2

0.4

0.4

b

P(B = b)

5

T wo

fair

bered

tetrahedral

1,

2,

random

3,

4

are

variable

dice

with

thrown.

D

is

faces

The

dened

num-

discrete

as

the

product

b 4

3

1

0

−0.2

0.2

0.6

0.4

b

P(B = b)

of

a

the

two

numbers

Represent

D

in

a

the

thrown.

probability

distribution

of

table.

c 1

2

0.1

0.2

b

P(B = b)

3.104

4

0.3

b

The

probability

random

distribution

variable

A

is

of

dened

a

Marin

throws

$15

if

three

by

this


10 21

7

16

44

c

Combine

two

rows

so

that

all

expected

people values

Total

Test,

is

a

42

at

the

21

5%

signicance

connection

number

of

57

between

people

level,

time

walking

of

their

are

greater

than

5.

120

if

there

day

d

Find

e

Comment

the

p-value

on

for

your

this

data.

result.

and

dog.

TOK

Developing inquiry skills What does it mean if a

Let

the

height

heights

into

medium

Use

in

is

these

areas

height

Does

from

the

a

small,

4.5




p rob lem

w here

5%

of

area

suppor t

a

small

tab le

signif ica nce

the

be

is

h.

h

Divide

≤

the

data set passes one test

4.5 m,

but fails another?

5.0.

contingenc y

the

ind epend ent

conclusion

area

Justify

the

categories

A

of

of

greater

in

the

f or

the

level

which

it

w hether

is

hypothesis

height

than

hei g hts

the

grow ing.

that

those

the

from

trees

area

B.

answer.

656

/

14 . 5

2

14.5

χ

goodness-of-t test

The uniform and normal distributions

Investigation 6

Jiang wonders whether the die he was given is fair. He rolls it 300 times. His

results are shown in the table.

Number

Frequency

1

35

2

52

3

47

4

71

5

62

6

33

1

Write down the probability of throwing a 1 with a fair die.

2

If you throw a fair die 300 times, how many times would you expect to

throw a 1?

3

Write down the expected frequencies for throwing a fair die 300 times.

Since all the expected frequencies are the same, this is known as a

uniform

distribution

2

Factual

Is the formula for the χ

test suitable to test whether Jiang’s

The null hypothesis is H

ytilibaborp

results t this uniform distribution?

: Jiang’s die is fair. 0

5

Write down the alternative hypothesis.

6

Given that the critical value at the 5% signicance level for this test is

dna scitsitatS

4

2

( f

 o

2

2

11.07, use the formula for the χ

test,





f

) e



, to nd out if f e

Jiang’s results could be taken from a uniform distribution.

Normally you would solve this using your GDC, which may ask you to enter

the degrees of freedom.

2

7

Factual

What is the number of degrees of freedom in a

χ

goodness-of-

t test? (Consider in how many cells you have free choices when

completing the expected values table.)

8

Write down the number of degrees of freedom for this test.

9

Using your GDC verify the value for the test statistic found above and write

down the associated p-value.

10

11

What is your conclusion from this test?

Conceptual

2

What is the purpose of the χ

goodness-of-t test?

657

/

14

2

TESTING FOR VALIDITY: SPE ARMAN’S, HYPOTHESIS TESTING AND χ

These

types

measuring

data

for

a

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how

for

are

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called

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test

tables

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example

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with

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TEST FOR INDEPENDENCE

test,

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distribution.

2

In a χ

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ν

=

(n − 1)

Example 12

Marius

results

works

are

in

in

a

the

sh

table

shop.

measures

250

sh

before

selling

them.

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5

22

12 ≤ x < 15

71

15 ≤ x < 18

88

18 ≤ x < 21

52

21 ≤ x < 24

10

< 27

2

27 ≤ x

is

he

below.

x < 12

Marius

week

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lengths

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from

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table

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could

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of

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0.0814

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18 ≤ x < 21

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0.00383

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658

/

14 . 5

d

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e

Perform

the

missing

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be

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:

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not

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).

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685

/

14

2

TESTING FOR VALIDITY: SPE ARMAN’S, HYPOTHESIS TESTING AND χ

A

sample

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the

results

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TEST FOR INDEPENDENCE

the

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Up to 5 years old satisfaction

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686

/

14

13

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687

/

Approaches to learning: Collaboration,

Rank my maths!

Communication

Exploration criteria: Personal engagement (C),

Reection (D), Use of mathematics (E)

IB topic: Spearman’s rank , hypothesis test

ytivitca noitagitsevni dna gnilledoM The task

In this task you will be designing an experiment that will

compare the rankings given by dierent students in your class.

You will then use your results to determine

how similar they are and what agreement there is and then

to test the signicance of that relationship.

The experiment

Select

one

students

The

in

rank

Would

same,

What

the

a

the

set

you

of

would

what

is

the

or

the

to

you

to

from

the

subject

students

the

of

least

the

experimenter.

of

the

to

their

the

there

group

most

students

other

experiment.

whether

in

The

in

is

any

by

similarity

asking

them

favourite.

your

group

to

be

the

different?

predict

would

might

be

determine

other

circumstances

to

the

rankings

correlation

group,

be

completely

circumstances

your

going

of

group

will

selections

help

strongest

your

group

expect

what

in

tastes

similar

Under

In

your

experimenter

between

to

student

who

have

similar

tastes

similar

and

(where

be)?

might

they

might

the

be

rankings

be

under

different?

discuss:

What

could

you

use

What

other

ideas

the

can

Spearman’s

you

think

of

rank

that

method

this

could

to

be

compare?

used

to

test?

688

/

The

experimenter

other

members

should

of

the

prepare

group

to

their

own

compare

in

set

the

of

selections

area

they

for

have

ytivitca noitagitsevni dna gnilledoM

14

the

chosen

to

investigate.

They

should

correlation

make

will

experiment

be

they

experimenter

The

other

What

does

the

give

will

(n),

about

the

how

other

strong

students

they

with

think

regards

the

to

rank

the

doing.

will

students

favourite

prediction

between

are

The

least

a

their

rank

where

n

the

is

experimenter

set

of

selections

to

set

of

selections

from

the

number

need

to

be

of

the

group.

favourite

(1)

to

selections.

aware

of

when

providing

instructions?

The

experimenter

should

The

students

are

who

communicate

Why

is

this

with

record

doing

each

the

the

rankings

ranking

in

should

a

table.

not

collaborate

or

other.

important?

The results

Now

Do

nd

the

Are

Spearman

students

the

Discuss

been

the

results

as

a

a

what

group

accurate

Write

display

or

rank

strong

were

why

correlation

between

the

students.

correlations?

expected.

the

original

hypothesis

may

have

inaccurate.

conclusion

for

the

experiment

based

on

the

results

Ex tension you

have

found

so

far.

What other experiments or data

The statistical test collection exercises can you think of What

spearman’s

rank

value

would

mathematically

that will be suitable for a statistics-based be

considered

to

be

high?

exploration that will result in a hypothesis It

is

possible

to

test

the

signicance

of

a

relationship

test like the one in this task? between

the

two

rankings.

conducted

for

the

product

Conduct

hypothesis

The

test

moment

is

similar

to

correlation

the

one

coefcient.

Use the examples and questions in this

chapter to give you some inspiration and

you

a

have

test

for

the

Spearman’s

rank

value(s)

then design your experiment and how

found.

you will conduct it. What

are

conclusions

you

in

your

can

you

draw

from

this?

How

condent

results?

689

/

Optimizing complex

networks: graph theory

15 This

chapter

graphs)

can

between

circuits

also

to

parcels

be

used

metro

and

looks

nd

explores

how

optimal

in

a

to

a

for

in

can

in

electrical

network.

be

example

the

Representation

■

Systems

connections

social

algorithms

area

the

■

(called

components

within

routes,

given

diagrams

model

stations,

people

at

how

Concepts

Microconcepts

It ■

Simple, weighted and directed graphs

■

Minimum spanning tree

■

Adjacency and transition matrices

■

Eulerian circuits and trails

■

Hamiltonian paths and cycles

■

Graphs: directed, connected, complete;

implemented

for

shortest

delivering

possible

time.

degree of a ver tex, weight of an edge

■

Adjacency matrix, lengths of walks,

connectivity of a graph, transition matrix,

steady state probabilities

■

Spanning trees, minimum spanning trees,

Prim’s and Kruskal’s algorithms, Prim’s

algorithm from a table

■

Eulerian circuits and trails, leading to the

Chinese postman problem

■

Tables of least distances

■

Classical and practical travelling salesman

problems

How can the driver What is the quickest route of a snowplough bet ween t wo metro stations? ensure that they

clear every road in a

town after a blizzard

as eciently as

possible?

How can a

postman deliver

letters to every

house on their

How could an electrical engineer model

the connections on a circuit board?

route while

minimizing the

distance they

travel?

690

/

A

national

seven

of

park

has

viewpoints.

viewpoints

entrance

to

are

the

Viewpoints

bicycle

The

distances

given

park

A

trails

is

in

at

B

the

connecting

between

table

below.

The

A.

C

D

E

F

G

A

0

5

4.5

B

5

0

4

3.5

C

4.5

4

0

2

3.5

2

0

6

5.5

6

0

4.5

3

5.5

4.5

0

3.5

3

3.5

0

D

E

F

6

6

G

•

pairs

What

do

you

think

that

a

blank

cell

represents?

Developing inquiry skills •

Name

three

pieces

of

information

in

table.

Write down any similar inquiry questions you represented

the

could ask and investigate for your local park . What

•

Draw

a

diagram

to

represent

the

infor

information would you need to nd? mation

in

the

table.

Compare

with

Could you write similar inquiry questions about your others.

local supermarket or shopping mall? Or any other place?

•

How

could

you

use

your

diagram

to

Think about the questions in this opening problem nd

the

shortest

route

around

all

the

and answer any you can. As you work through the viewpoints?

chapter, you will gain mathematical knowledge and

skills that will help you to answer them all.

Before you start

Click here for help

You should know how to:

with this skills check

Skills check

1



1

0



1

Raise

a

matrix

to

an

integer

1

power.

If

A

2



1

1

0

2

nd





2

1

1

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4

eg

If

A

nd



1

A

0

2

a

4

b

A

A

 4

5 4

Using

a

GDC:

A







3 

 4 



 0.8

0. 3



2

Find

the

steady

state

vector

for

2

a

Let

P



be

  0.2

transition



matrix

which eg

Find

the

steady

state

vector

for

the

probability

a

 0 .9

to

state

Solve

x

=

+ 0 .6 y

=

0.4,

y

=

of

j

is

P

state

.

forming

the

+ 0 .4 y

along

a

system

Verify

steady

your

calculating

these

from

of

linear

equations

=

state

answe r

vector.

is

cor re ct

by

x

equations

either

By

nd

b

0 .9 x

moving

0 .4



Form

of

ji



0 .1x

in

0 .6

 P

matrix

the

i

0 .1

matrix

transition





transition

a

0. 7 

large

powers

of

P

y

with

x

+

y

=

1:

c

What

is

being

in

the

longterm

state

probability

of

j?

0.6

691

/

15

OP T IMI Z ING

15.1

For

like

the

NE TWORK S:

GR A P H

THEORY

Constructing graphs

opening

this.

C O M PL E X

This

is

problem

a

you

may

have

drawn

a

diagram

A

4

B

graph.

4 3.5

4.5

A graph is dened as set of ver tices and a set of edges. 2

D

6

E

C

A ver tex represents an object. An edge joins two ver tices.

4.5 5.5

3

6

In

of

the

the

In

graph

of

graph.

the

The

mathematical

and

in

the

the

trails

are

graph

positions

real

national

of

world.

A

park

the

are

the

vertices

F

graphs

vertices

graph

viewpoints

edges.

theory,

the

the

will

do

do

not

always

not

need

show

have

to

to

be

relate

which

drawn

to

their

vertices

to

3.5

scale

positions

are

G

TOK

directly

Have you heard of the connected.

Seven Bridges of The

graph

above

also

shows

the

distances

between

vertices.

A

graph

that Königsberg problem?

shows

values

like

these

“weight”

can

be

(called

weights)

is

called

a

weighted

graph. Königsberg is now

The

any

quantity,

such

as

cost,

time

or

distance. Kaliningrad in Russia.

The

map

of

the

unweighted

between

the

London

graph.

It

Underground

gives

no

only

the

stations,

is

a

famous

information

connections

example

regarding

between

the

of

an

Do all mathematical

distances

problems have a

them.

solution?

A walk is any sequence of ver tices passed through when moving

along the edges of the graph.

In

the

from

The

an

graph

A

to

of

the

park

ABC

and

ABDBC

are

both

walks

C.

information

adjacency

number

national

of

in

an

table.

direct

unweighted

In

an

graph

adjacency

connections

can

table

between

also

the

two

be

contained

entries

indicate

in

the

vertices.

Example 1

A The

table

small

shows

mountain

connection

a

Show

b

Write

a

the

connections

railway.

between

this

the

An

B

two

empty

cell

stations

indicates

on

no

on

possible

a

A

direct

B

1

C

1

graph.

walks

from

A

to

B

C

1

1

1

1

1

D

D.

To

D

a

stations.

information

down

between

draw

the

graph

1

begin

with

vertex

A.

A

From

the

rst

connected

to

row

B

in

and

the

C,

so

table,

add

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is

them

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the

graph. C

B

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and

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so

B

the

C.

you

to

second

The

rst

only

row,

of

need

B

is

connected

these

is

already

to

add

the

edge

to

A

included,

connecting

C.

C

692

/

15 . 1

The

B

third

row

shows

that

you

need

to

add

A

the

edge

The

CD.

nal

row

includes

no

extra

edges.

Use

it

D

to

check

your

graph

is

drawn

correctly.

C

b

eg

ABCD,

ACD

or

There

ABACD

you

are

can

each

innitely

pass

edge

as

many

through

many

answers

each

times

vertex

as

you

to

part

and

b

as

along

like.

Example 2

A shows

costs

in

dollars

of

B

C

D

4

7

6

A by

bus

Show

between

this

towns.

information

on

a

B

4

C

7

D

6

5

graph.

5

3

B 4

Because

also

7

show

Note

that

2

6

the

costs

them

there

6

3

E

A

E

travelling

are

on

are

2

given

the

in

the

table

you

must

graph.

many

different

ways

to

dna y rtemoeG

table

y rtemonogirt

The

draw

5

6

the

graph,

eg:

B

B

5

C

5

D 3

C

C

6

4 4

2

7

6

6

7

E

3

A A 3

E

E

2

6

6

2

D

D

7 A

C

3 6

6 5

4

D

2

E

B

Any

depiction

correct

Reect

is

ne

so

long

as

you

show

the

connections.

How do you construct a graph from information given in a table?

What information can be readily obtained from a graph that cannot easily be

seen in a table?

693

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Exercise 15A

1

State

which

information

of

the

given

graphs

in

the

represent

the

2

table.

Draw

direct

the

A

B

C

D

E

a

1

B

to

represent

connections

graphs

the

between

number

the

vertices

in

below.

B

b

S Q

1

1

of

F

a A

table

1

1 P

C

1

D

1

1

T

1

1 D R E

E

1

F

1

1

1

3

a A

Draw

the

B

the

weighted

graphs

represented

by

tables.

a A

E

F

A

D

B

15

C

12

D

10

B

C

D

15

12

10

E

13

9

C

b

12

B

13

E

16

9

12

16

F

C

A

b A

A

D

E

B

6

C

10

B

C

6

10

D

E

7

7

B

c

D

5

E A

5

C

E

4

Draw

tables

weighted

below.

column D

Each

the

represented

table

weight

is

read

of

the

to

decide

by

from

edge

the

row

from

A

to

to

F

B

d

so

graphs

F

is

the

B

10.

fact

same

as

You

will

that

B

A

to

need

to

B

is

not

how

necessarily

to

show

the

A.

a A C

B

C

A

10

7

B

E

D

A

8

D

C

e

D

8

12

D

C

F

B

b

A

A

B

C

D

10

A

B

10

6

E

C

D

7

8

11

694

/

15 . 1

unweighted

tables

below.

A

1

graphs

in

the

to

represent

intersection

the

of

b

A

A i

and

column

connection

j

indicates

from

i

to

that

there

is

D

1

1

1

B

j.

1

C B

C

a

a A

B

row

C

1

1

D

D A

B

1

1

1

1

1

C

D

6

A

1

salesman

to

needs

Shefeld.

below,

In

Write

distance

b

nder

the

in

miles

from

distances

an

estimate

Manchester

table

as

much

why

all

those

Shefeld

algorithm

distance

the

give

from

to

as

has

Manchester

been

between

the

Manchester

from

for

to

Shefeld

the

shortest

Nottingham

major

cities

Manchester

one

the

other

with

in

Nottingham

distance

via

Nottingham

programmed

to

to

to

the

data

England.

has

and

not

from

algorithm

town.

To

then

in

the

back

table

Unfortunately

been

nd

this

the

included.

Nottingham

will

do

and

the

to

Shefeld.

shortest

efciently

it

needs

a

graph

other

Hence

towns

possible.

towns

further

than

77

miles

from

Manchester

can

be

excluded

showing

all

remaining

routes

from

Manchester

to

Nottingham

which

state

and

an

estimate

return

to

for

the

shortest

distance

the

salesman

must

travel

to

visit

the

292

25

97

194

65

82

181

80

174

252

53

Southampton

128

74

164

107

229

217

77

208

319

190

160

65

17

195

76

164

37

240

34

72

160

38

128

146

39

130

208

141

94

175

124

242

231

71

222

332

184

173

78

Oxford

63

70

90

141

163

153

56

144

254

141

95

Nottingham

49

137

16

213

70

99

123

159

123

Norwich

161

209

139

282

174

217

112

185

Newcastle

204

288

161

364

94

155

274

131

81

162

59

238

41

34

184

111

114

123

170

191

198

90

161

81

237

73

Leeds

110

196

70

271

Exeter

164

75

203

Derby

40

127

Bristol

88

Liverpool

notpmahtuoS

dleehS

htuomst roP

drofxO

mahgnit toN

hciw roN

eltsacweN

retsehcnaM

nodnoL

loopreviL

sdeeL

retexE

ybreD

lotsirB

mahgnimriB

88

London

two

Shefeld.

217

Manchester

via

town.

129

Por tsmouth

from

go

York

Sheeld

to

search.

Draw

one

d

the

travel

distances

on

from

Explain

the

c

to

to

route

down

trying

reduce

A

showing

information

a

1

dna y rtemoeG

Draw

y rtemonogirt

5

239

260

695

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

In a connected graph it is possible to construct a walk between any two

ver tices.

A subgraph of a graph G consists entirely of ver tices and edges that are also

in G

The

graph

from

Exercise

15A

question

3b

is

an

example

A 6

of

an

unconnected

B

graph.

E

It

is

unconnected

because

no

walk

exists

from

vertex

10

A

7

5

to

vertex

D,

for

example.

C

D

It

is,

however,

made

up

of

two

connected

subgraphs.

In a directed graph all the edges are assigned a specic direction.

A directed graph is connected if a walk can be constructed in at least one

direction between any two ver tices. A directed graph is

strongly connected if

it is possible to construct a walk in either direction between any two ver tices.

The

graphs

The

graph

is

Exercise

drawn

sometimes

It

from

this

possible

to

is

in

15A

question

referred

go

from

to

A

to

question

4a

as

is

a

for

are

directed

connected,

weakly

D,

4

graphs.

directed

A

but

7

not

from

D

to

8

graph

possible

drawn

to

go

in

in

question

both

4b

directions

is

strongly

between

connected.

any

two

It

C

12

is

vertices.

10

A

The

context

often

makes

it

clear

whether

a

directed

or

8

A.

D

The

B

graph;

connected.

example,

10

B

undirected 10

graph

is

required.

For

example,

a

bus

may

go

from

town

A

to 7

11

town

B

but

not

make

the

reverse

D

Reect

6

journey.

C

8

When is it appropriate to use directed or undirected

graphs?

Except

in

those

will

be

taken

will

be

explicitly

to

cases

where

mean

an

extra

clarity

undirected

is

needed,

graph.

If

the

the

graph

word

is

graph

directed

it

mentioned.

The degree (or order) of a ver tex in a graph is the number of edges with that

ver tex as an end point.

E

A

vertex

whose

degree

is

an

even

number

is

called

an

A

even

vertex

or

is

said

to

have

even

degree

F

The

A

2,

degree

B

3,

C

of

2,

each

D

4,

vertex

E

3,

F

in

the

graph

shown

is:

D

2.

B

C

696

/

15 . 1

The

of

is

in-degree

edges

the

In

with

that

number

the

graph

outdegree

of

of

a

vertex

edges

shown

of

vertex

A

in

as

an

with

has

a

directed

end

that

an

graph

point.

vertex

indegree

as

of

The

a

1

is

the

A

B

D

C

number

out-degree

starting

and

point.

an

2.

Investigation 1

Five people shake hands before a meeting as indicated in the table.

A

B

A

1

1

1

1

E

1

E

1

1

1

C

D

D

1

1

1

1

Show this information in a graph with each edge representing a

handshake between two people.

b

From the graph write down the total number of handshakes that take

place before the meeting.

2

dna y rtemoeG

B

a

C

y rtemonogirt

1

If possible, draw a graph consisting only of ver tices with the degrees

listed. If it is not possible, explain why not.

a

3

1, 2, 2, 2, 3

b

1, 2, 2, 2

c

2, 2, 3, 3

d

2, 2, 2, 4

From your graphs in questions 1 and 2, can you nd a link between the

degrees of the ver tices in a graph and the number of edges? If necessary

draw more graphs and record the degrees of the ver tices and the number

of edges.

4

Using your result from question 3, explain why it is not possible to construct

a graph with a degree sequence of 3, 2, 4, 1, 5, 2, 3, 3.

5

The link conjectured in question 3 is called the handshaking theorem.

Explain why with reference to your answer to question

6

1

Can you nd a similar theorem linking the degrees of the ver tices in a

directed graph and the number of edges?

Exercise 15B

1

A

group

which

the

of

can

ve

the

an

the

it

share

shown

as

in

the

several

a

graph,

vertices.

connections

following

whether

or

be

people

represent

For

people

would

undirected

best

to

An

edge

represents:

with

The

between

connections

be

attributes

a

lives

b

is

the

c

is

a

friend

d

is

a

follower

in

the

same

block

as

edges sister

of

them.

of

discuss

draw

a

directed

on

Twitter

of.

graph.

697

/

15

2

OP T IMI Z ING

a

Draw

a

table

contained

directed

in

C O M PL E X

to

show

each

of

NE TWORK S:

the

the

GR A P H

5

information

following

THEORY

The

the

table

shows

ferries

on

part

Lake

of

the

timetable

Starnberg

in

for

Germany.

graphs.

Stop (village)

i

ii

1

Starnberg

8

3

9.35

12.00

14.30

9

B

C

2

A

C

D

5

Possenhofen

10.07

14.44

Tutzing

10.31

15.06

Ammerland

12.44

15.22

13.01

15.39

13.23

15.58

13.48

16.23

14.08

16.43

14.34

17.09

14.50

17.25

7

Bernried

4

10.49

D

Ambach

b

State

whether

connected,

each

of

strongly

the

graphs

connected

or

is

Seeshaupt

11.08

Ambach

11.32

not

connected.

Bernried

3

a

i

Draw

an

unweighted

containing

the

the

following

A

information

graph

given

in

table.

B

A

directed

C

1

B

Ammerland

11.51

Tutzing

12.08

D

Possenhofen

1

Starnberg

12.28

1

Starnberg

C

1

D

1

1

1

1

Possenhofen

ii

For

and

b

i

each

the

Draw

a

vertex

state

the

indegree

outdegree.

weighted

directed

graph

Ammerland

containing

the

represented

A

by

the

B

A

Tutzing

information

following

C

table.

D

6

Bernried

4

Ambach

B

6

C

5

7

Seeshaupt

D

ii

For

and

4

Without

by

each

the

vertex

and

the

state

the

indegree

a

the

the

table

graph,

write

outdegree

of

down

each

the

b

B

State

connections

on

a

one

to

piece

see

D

E

1

1

Give

two

1

d B

of

graph.

information

the

the

graph

that

than

on

is

the

From

in

your

of

one

graph

towns

that

direction

give

four

are

directly

only.

possible

1 routes

C

1

D

E

on

pairs

connected

A

directed

between

timetable.

below.

C

the

easier

vertex

c A

Show

towns

outdegree.

drawing

indegree

given

7

1

1

1

1

1

do

e

not

Use

the

in

from

call

the

at

any

timetable

routes

one

Starnberg

in

part

to

town

to

d

Ambach

more

nd

than

how

could

be

which

once.

many

of

completed

day.

698

/

15 . 2

Developing inquiry skills

In the opening problem, you drew a graph to represent the

information in the table. Check that your graph is a weighted

graph that contains all of the information in the table.

Is the graph connected?

State the degree of each ver tex.

Often,

in

we

the

interested

of

of

such

electronic

friends

on

the

only

the

connections

between

vertices

and

not

edges.

situations

circuit

social

in

might

board

or

be

in

a

a

map

large

of

a

metro

lestorage

system,

system,

connections

or

a

network

media.

y rtemonogirt

of

an

are

weights

Examples

on

Graph theory for unweighted graphs

This

information

will

normally

be

shown

in

an

unweighted

dna y rtemoeG

15.2

graph.

Two ver tices are said to be adjacent if they are directly connected by an edge.

TOK

of an adjacency matrix A is the number of direct connections

An element A

To what extent can shared

ij

between ver tex i and ver tex j knowledge be distorted

and misleading?

Unlike

in

matrix,

rather

if

a

table,

there

than

is

when

no

leaving

giving

information

connection

the

entry

a

0

has

to

about

be

connections

put

in

as

a

place

in

a

holder

empty.

Example 3

Find

the

a

adjacency

matrices

for

the

two

graphs

shown.

b

B

B

A A

D

C

D

C

Continued

on

next

page

699

/

15

OP T IMI Z ING

a

A

B

C

0

1

1

0

B  1

0

1

0

1

1

0

1

0

0

1

0

A

B

0

1

C O M PL E X

NE TWORK S:

GR A P H

As

D

THEORY

the

graph

symmetric

 A

is

in

undirected

the

diagonal

the

matrix

from

top

is

left

to

 bottom

right

(the

leading

diagonal).

 C 

D

 

b

C

D

An

 A

1

adjacency



movement

B  0

matrix

indicates

possible

0

0

0

0

1

1

0

0

0

0

1

0

opposite

to

from

the

row

to

column;

convention

for

this

is

the

transition

 C 

D

matrices.

 

An edge from a ver tex back to itself is called a

loop.

If there are multiple edges between two ver tices the graph is a multigraph.

A simple graph is one with no loops or multiple edges.

Reect

What can you deduce about the adjacency matrix of a simple graph

from the above denition?

Example 4

Find

the

adjacency

a

matrices

for

the

two

graphs

shown.

b

B

B

A

A

D

D

C

C

a

A

A

B

C

D

 0

1

2

0

In

a

an

2

in

undirected

an

graph

adjacency

a

loop

matrix

(as

is

shown

going

as

both

 B

1

0

1

0

ways

along

it

is

possible).



C



2

1

0

1

 D



b

0

0

1

A

B

C

2

D In

A

 0

1

1

this

graph

the

loop

is

shown

as

a

1

in

the

0 adjacency

matrix

as

the

edge

is

directed.

 B

0

0

0

0

2

0

0

0

0

0



C



 D



1

1

700

/

15 . 2

Exercise 15C

1

Find

the

adjacency

following

matrices

for

4

the

graphs.

The

following

relationships

where

a

b

A

1

in

between

the

a

entry

matrix

group

in

row

shows

of

i

the

people

and

column

A

j

indicates

person

B

D

a

adjacency

that

i

knows

the

name

of

j.

B

D

person

A

B

A  1

C

D

E

1

0

0

1

1

1

1

0

0

0

1

1

1

1

0

0

1

1

1

0

0

0

1

1

 C

B

C



c

A

d

B

C

A



 D  B

 E

E

C

E



F

a

C

i

Explain

each

D

why

entry

the

on

matrix

the

has

leading

a

1

in

diagonal.

D

ii the

features

of

an

adjacency

Explain

to

which

tell

you

that

a

graph

is

Explain

matrix

a

the

how

to

you

can

use

the

be

does

not

have

Without

drawing

the

graph

nd:

adjacency

i

how

ii

the

many

names

A

knows

nd:

degree

undirected

of

each

vertex

in

number

the

indegree

ii

the

outdegree

people

who

know

the

an of

iii

who

C

iv

whose

knows

vertex

in

a

directed

the

name

number

each

of

graph

i

of

graph

symmetric.

name

b

the

simple.

b

3

why

matrix

dna y rtemoeG

State

y rtemonogirt

2

of

is

most

names

known

people

in

by

the

the

largest

group.

graph.

c

State

not

whether

know

know

A ’s

there

name

anyone

who

is

anyone

and

also

knows

who

does

A ’s

does

not

name.

Investigation 2

1

For this graph nd the number of walks of the given length between the

two given ver tices. Write down each of the walks. B

A

a

length 2 between A and C

b

length 2 between C and C

c

length 3 between A and C

d

length 3 between D and D

D

C

2

a

Find the adjacency matrix M for the graph shown.

2

3

and M

b

Find M

c

Conjecture a link between the powers of an adjacency matrix and the

lengths of walks between ver tices

Continued

on

next

page

701

/

15

OP T IMI Z ING

3

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Draw a directed graph of your own and verify that the conjecture in

question 2c still holds.

4

How do you nd the number of walks of length

Factual

n between two

ver tices in a graph?

2

5

Explain how the matrices M, M

3

and M

can be combined to give a matrix

which shows the number of walks of length 3 or less between the ver tices

of the graph.

6

What can be calculated from the powers of an adjacency

Factual

matrix?

Factual

7

How can you use the powers of an adjacency matrix to nd the

minimum length of path between two ver tices?

Conceptual

8

How does the adjacency matrix allow you to analyse the

paths between two points on a graph?

Let M be the adjacency matrix of a graph. The number of walks of length

n

n

from ver tex i to ver tex j is the entry in the ith row and the jth column of M

The numbers of walks of length r or less between any two ver tices are given

2

where S

by the matrix S r

r

= M + M

+ ... + M

r

Exercise 15D

1

The

adjacency

Exercise

15C

matr i ce s

ques ti o n

for

1

the

are

gr a phs

d

in

A

shown

B

C

D

E

F

0

0

1

0

 A

0

1

0

0

1

0

0

0

0

0

1

1

0

0

0

1

0

0

0

0

1

0

0

1

0

1

0

0

0

0

0

0



below.

B





a

A

B

C

A

b

D

B

C

D

C 

A

 0

1

0

A

1

 0

1

0

0

D



  B



 1

0

1

B

1



0

0

0

1

E

0

F







 C



0

1

0

1



C



0

1

0

  D





 1

1

1

0

D 



1

0

1

0

Use

these

walks

c

A

B

C

D

length

3

to

nd

the

number

of

between:

E

i

A  0

of

matrices

1

0

0

1

1

0

1

1

1

0

1

0

1

0

0

1

1

0

1

1

1

0

1

0

In

A

each

and

case

ii

A

use

the

C

graphs

and

in

D.

Exercise

15C

 B

question

1

to

verify

your

result

by

listing

all



C



the

possible

walks.

 D

2

By

considering

an

appropriate

power

of

a



 E



matrix,

verify

question

4

your

part

c

answer

to

Exercise

15C

i

702

/

15 . 2

Use

the

adjacency

answer

the

1

0

0

1

0

1

0

1

0

1

 B

1

eeriT

1

 0

muR

D

giallaM

C

ggiE

B

yabeltsaC

A

to

questions.

A

a

M

yromreboT

following

matrices

eladsiobhcoL

3



C



1

 D



b

0

A

B

A  0

0

C

1

D

0

Castlebay

0

0

1

0

0

1

1

Eigg

0

0

0

0

1

0

0

Lochboisdale

1

0

0

0

0

0

1

Mallaig

0

0

0

0

1

0

0

Rum

0

1

0

1

0

0

0

Tiree

1

0

0

0

0

0

1

Tobermory

1

0

1

0

0

1

0

E

0

0

 B

1

0

1

0

0



C



0

1

0

1

1

 D

0

0

1

0

1



 0

0

1

1

0

a 2

i

Find

Find

and

Write

down

length

iii

Find

less

2

the

the

number

between

number

between

C

C

and

of

and

of

walks

A.

walks

b

of

length

diameter

of

a

3

By

to

is

the

length

of

vertices

from

less

shortest

(so

any

than

all

the

other

or

walk

vertices

vertex

equal

between

to

in

the

a

can

be

walk

any

diameter

c

extra

of

powers

of

the

adjacency

nd

the

diameters

of

each

your

ferry

route

The

following

hence

a

Use

the

adjacency

between

the

pair

of

is

is

adjacency

possible

ferry

now

new

state

the

directed

15B

adjacency

matrix

shows

to

get

system.

added

between

adjacency

two

ports

matrix

that

are

graph

question

you

5

to

drew

write

for

down

matrix

for

the

ferry

on

Lake

Starnberg.

By

taking

some

ports

in

off

the

coast

of

in

both

directions

a

suitable

power

of

the

the matrix

nd

all

the

pairs

of

the that

are

not

connected

directly

or

Scotland. at

most

one

stop,

stating

clearly

the

between direction

each

the

it

the

graph.

with

go

ports

of

answer.

Islands

ferries

in

matrix

towns

The

Castlebay

apart.

adjacency

connections

Shetland

travel

Lochboisdale.

down

furthest

the

b 4

to

powers

using

connections Justify

the

which

Tiree

and

Write

an to

can

trips.

nd

from

and

length

5

the

you

two

reached

of

ferry

Exercise Use

routes

the

graph).

iv

of

Lochboisdale

considering

Mallaig maximum

from

matrix

or

A.

graph

get

three

of

An The

number

M to

ii

the

3

M

dna y rtemoeG



y rtemonogirt

E

of

the

connection.

ports.

703

/

15

OP T IMI Z ING

Reect

C O M PL E X

NE TWORK S:

GR A P H

THEORY

TOK

How might you work out the maximum number of steps

necessary to move between any two ver tices on a graph? Matrices are used in

computer graphics for three-

How can we use powers of the adjacency matrix to determine whether or

dimensional modelling.

not a graph is connected?

How can this be used in real-

life situations in other areas

of knowledge?

Transition matrix

A

random

chosen

In

a

walk

on

randomly

graph

Markov

with

a

graph

from

a

is

a

those

nite

walk

in

which

the

vertex

moved

to

is

available.

number

of

vertices

a

random

walk

is

a

nite

chain.

HINT A

transition

matrix

can

be

constructed

for

both

directed

and

undirected

See Chapter 9 for

graphs.

an introduction to The

probability

of

moving

from

one

vertex

to

any

of

the

adjacent

vertices

Markov chains. is

dened

graph

As

i

to

and

be

the

discussed

to

vertex

j

the

reciprocal

reciprocal

in

earlier

will

be

of

of

the

the

outdegree

chapters,

the

entry

degree

in

the

the

of

in

the

a

directed

probability

ith

vertex

of

column

in

an

graph.

moving

and

undirected

from

jth row

of

vertex

the

matrix.

The direction of movement in a transition matrix is

E X AM HINT

opposite to that in an

adjacency matrix.

Within the IB

syllabus questions

will only be set in The

steady

state

probabilities

indicate

the

proportion

of

time

that

which the graph is would

be

spent

at

each

vertex

if

a

random

walk

was

undertaken

for

a

connected. In the long

period.

The

transition

time

between

the

vertices

is

ignored.

case of a directed

The

Google

PageRank

algorithm

was

developed

by

Larry

Page

and

graph, it will be Segei

Brin

in

1996

while

working

at

Stanford

University.

They

wanted

strongly connected. to

be

most

The

able

to

likely

rank

to

algorithm

edges

in

a

be

web

pages

useful

they

directed

in

come

Internet

towards

developed

graph.

an

The

the

considered

steady

search

top

of

links

state

so

the

from

that

the

ones

list.

web

probabilities

pages

as

calculated

the

E X AM HINT

from

If a walk around a the

transition

matrix

indicate

which

site

would

be

visited

most

often

graph is equally if

someone

were

randomly

clicking

on

links

in

web

pages.

This

site

likely to take any of would

come

top

of

the

list

in

a

search.

the edges leading

The

justication

for

this

ranking

is

that

the

sites

most

likely

to

be

from a ver tex, it is

visited

will

either

have

lots

of

links

to

them,

or

be

linked

to

by

sites

called a random with

lots

of

links

going

to

them.

In

either

case

this

is

likely

to

reect

walk their

relative

importance.

704

/

15 . 2

A

Example 5

In

this

graph

vertices

A

to

F

represent

web

pages

and

the

edges B

F

indicate

For

links

example,

links

to

B,

D

between

page

and

A

the

has

pages.

links

from

both

C

and

F

and

contains

F . D

a

Construct

this

transition

matrix

for

a

random

walk

around

graph.

b

Find

c

Hence

a

the

the

The

steady

rank

the

transition

state

probabilities

vertices

matrix

in

order

for

of

the

network.

importance.

is

For

it

1

example,

to

vertices

A

B,

has

D

three

and

F ,

edges

and

so

connecting

each

of

these

1

1 0

0

0

0

2

has

a

probability

of

,

which

is

shown

in

the

2

3

1

1

1

rst

0

0

3

2

column

of

the

matrix.

0

3

Because

the

question

does

not

specify

which

0

0

0

0

method

to

use,

you

can

just

look

at

high

y rtemonogirt

0

3

powers

of

the

transition

matrix

using

your

1 1

0

0

1

0

0

0

0

GDC.

0

3

You

1 0

to

2

1 0

ensure

to

0

0

gure.

The

steady

the

would

be

solve

a

system

of

six

linear

equations

or



The

given



0

326

0

076

0

152





 F

are

109





probabilities

matrix.

 0



state

the

207 









ranking

would

be

D,

B,

F ,

A,

C,

E.

The

vertex

placed

Reect

alternative

third

130



c

the



 0

C

E

An

in

 0



D

change

enough

vector

 A

B

no

large

3

diagonalize

by

is

powers

0

to

b

consider

there

signicant

1

3

need

dna y rtemoeG

1

at

with

the

the

top

of

highest

the

probability

is

list.

How do you construct a transition matrix for a given graph?

Why might a random walk indicate the most impor tant sites on a graph?

How can the steady state probabilities be used to rank lists?

705

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Exercise 15E

1

For

each

of

transition

state

the

two

matrix,

graphs

P,

probabilities

and

below

hence

nd

the

3

the

steady

The

following

websites.

by:

one

of

there

i

considering

ii

solving

a

a

high

system

powers

of

of

linear

The

the

is

diagram

a

arrows

pages

link

to

shows

indicate

another;

from

four

website

a

link

for

D

to

from

example,

website

C.

P

A

B

C

D

equations.

b

A

A

B B

D D

a

Write

C

down

represent

C

a

the

transition

random

walk

matrix

to

around

this

graph.

2

In

a

an

experiment

robot

is

put

in

on

a

articial

maze

intelligence

which

has

b

six

Show

to rooms

labelled

A

to

F .

Each

of

the

C

that

using

gaps

in

the

wall

connecting

it

not

exactly

possible

three

to

link

links,

from

and

B

write

two

other

connections

that

are

also

to

not adjacent

is

rooms

down has

it

possible

using

exactly

three

links.

rooms.

c

E

By

solving

four

linear

equations

nd:

A

F

i

the

steady

ii

the

proportion

state

probabilities

of

time

that

a

person

D

following

random

links

will

be

on

B

site

B.

C

d

a

Draw

the

maze

as

a

graph,

vertices

representing

and

edges

the

between

the

with

the

different

representing

the

order

rooms

a

gaps

robot

is

search

moves

randomly

and

once

in

the

equally

likely

to

leave

through

the

the

down

Within

a

large

created

a

transition

matrix

for

the

i

from

robot

begins

probability

Find

it

through

the

likely

from

A

state

one

small

to

two

be

in

is

room

again

four

of

rooms

visited

A,

what

in

room

the

gaps?

which

by

the

are

A

is

to

to

Determine

the

might

the

be

one

probability

the

listed

with

by

the

rst.

network

links

can

person’s

page

to

group

of

people

share

links

the

robot

be

in

page

pages

below.

it

is

of

For

possible

Belle,

example,

to

link

Charles

and

after Antoine

Belle

Charles

Dawn

Emil

Frances

Belle

Antoine

Antoine

Emil

Antoine

Dawn

Charles

Charles

Belle

Frances

Dawn

most

robot.

percentage

will

table

Emil.

of

Dawn

time George

that

the

the

Emil

ii

giving

social

Antoine’s

directly

passing

d

engine,

sites

down

gaps.

Write

If

the

write

any

maze.

c

which

steady

according

b

information

a

another. of

in

this

rooms.

be room

on

largest

4 The

Based

each

of

is

also

part

of

the

network

and

these friends

with

to

the

all

those

listed.

He

decides

rooms. visit

pages

of

each

of

them

in

the

706

/

15 . 2

following

page

he

will

choosing

links

way:

on

having

then

move

randomly

his

chosen

to

from

current

a

another

the

5

starting

list

page

of

a

by

Explain

for

possible

this

the

why

the

graph

starting

longterm

will

not

be

probabilities

independent

of

position.

page. C

a

Show

a

the

information

directed

from

the

table

in

graph. B

b

From

the

possible

friends’

graph

for

explain

George

pages

in

to

the

why

visit

rst

it

is

each

six

D

not

of

the

pages

he F

if

he

begins

on

Antoine’s

page.

b c

Determine

whether

it

is

possible

to

Write

down

random each

of

the

friends’

pages

in

the

rst

give

if

a

he

begins

possible

on

another

order;

if

not,

page.

say

If

Construct

a

transition

matrix

why

for

not.

begins

by

visiting

Antoine’s

walk

at

Find

the

probability

he

is

back

c

If

X

is

the

page

after

visiting

end

at

vertex

A

if

the

walk

to

of

reach

steps

A

it

would

from

B

nd

take

E( X).

page.

d

Find

e

By

the

transition

evaluating

matrix

high

matrix

for

powers

nd

the

of

the

graph.

the

longterm

on

ve

of

being

at

vertex

E

if

you

further began

at

vertex

C.

pages.

f

Determine

the

pages

Charles,

g

George

web

in

probability

the

Dawn,

order

Emil,

continues

pages

time.

the

Find

in

the

which

he

visited

Antoine,

it

B.

number

probability Antoine’s

would

vertex

transition

e

a

your

graph.

George

that

so,

for

d

probability

six

began visits

the

visit

dna y rtemoeG

at

y rtemonogirt

looks

Belle,

Antoine.

moving

same

page

through

way

he

for

is

a

the

long

likely

to

visit:

i

the

most

ii

the

least.

Developing inquiry skills

In the opening problem, what is the minimum

number of trips needed to travel between

any two of the viewpoints?

A walker got lost in the park and was randomly

travelling along trails.

After a long period of

time a rescuer went out to try and nd him.

Given that the rescuer decides to wait at one

of the viewpoints, which one should they

choose if they want to maximize the chance

that the walker will pass through that

viewpoint rst.

707

/

15

OP T IMI Z ING

15.3

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Graph theory for weighted

graphs: the minimum spanning

tree

A cycle is a walk that begins and ends at the same

A B

ver tex and has no other repeated ver tices.

H

In the graph shown ABDGHA is a cycle.

C

D

G E

F

TOK

A tree is a connected graph which has no cycles, such as

the one shown on the right. Is imagination more

A spanning tree for a graph is a subgraph which is a tree and impor tant than

contains all the ver tices in the graph. knowledge?

The

two

trees

B

below

C

4

are

both

spanning

trees

D

4

for

B

5

the

given

C

4

graph.

B

D

4

3

A

E

3

E

A

E

3 5

5

4

4

H

9

8

3

D

5

9

8 A

C

6

7

G

Let

the

graph

the

weighted

the

H

new

costs

houses

(in

built

€1000s)

4

4

F

represent

edges

4

on

of

6

an

7

G

extensive

connecting

F

H

estate

them

to

G

F

and

mains

electricity.

If

connected

with

The

of

the

second

problem

least

the

would

edge

still

Reect

the

rst

would

needs

which

is

tree

be

to

the

total

cost

would

be

€33 000,

and

€38 000.

be

solved

normally

is

nding

referred

to

the

as

spanning

the

tree

minimum

tree

minimum

cycle

it

that

weight,

spanning

The

using

be

cost

with

solution

cannot

contain

greatest

weight

could

any

be

cycles.

removed

If

there

and

the

were

a

houses

connected.

What would be the least-weight way to connect all the ver tices in a

graph?

708

/

15 . 3

You

need

Kruskal’s

to

learn

and

two

Prim’s

algorithms

to

nd

the

minimum

spanning

tree,

algorithms.

Kruskal’s algorithm

1

Find the edge of least weight anywhere in the graph. If there are two or

more edges with the same weight any may be chosen.

2

Add the edge of least weight that has not already been selected and does

not form a cycle with the previously selected edges.

3

Repeat the second stage until all the ver tices are connected.

Example 6 B

Use

Kruskal’s

algorithm

to

nd

the

minimum

cost

C

4

D

4

of 5

connecting

all

the

houses

on

the

estate.

The

costs

3

of

9

8 A

3

E

3 5

each

pair

of

houses

is

shown

on

the 4

4

graph.

The

weights

are

the

costs

in

€1000

of

H

connecting

the

6

7

G

F

houses.

1

B

Find

in

the

the

with

edge

graph.

the

of

If

same

least

there

weight

are

weight

anywhere

multiple

any

may

edges

be

chosen.

3

In

3

this

can

example

be

any

of

the

edges

of

dna y rtemoeG

weighted

y rtemonogirt

connecting

weight

selected.

H

2 B

Add

the

edge

of

least

weight

that

has

not

D

already

a

cycle

been

with

selected

the

and

does

previously

not

form

selected

edge.

3

Add

another

of

the

edges

of

weight

3.

F

H

3 B

Repeat

the

second

stage

until

all

the

D

vertices

are

connected.

3

The

A

3

4

B

C

4

edge

edges

with

be

as

EF

AH,

F

H

next

added

will

be

the

The

fourth

third

of

the

E

3

All

D

4

BC

the

weight

this

or

Neither

of

would

CD

edges

3.

is

of

the

create

a

one

cycle

cannot

and

so

added.

weight

edges

of

4

are

added

weight

5

next.

can

be

3

added A

3

as

they

both

form

cycles,

so

GH

of

E

3

weight

6

is

added.

All

the

vertices

are

now

4

connected.

H

The

6

minimum

houses

to

4

4

+

3

+

F

G

cost

for

electricity

+

4

+

3

+

3

connecting

will

+

6

all

the

be

=

27,

ie

€27 000

709

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Prim’s algorithm

1

Select any ver tex and add the edge of least weight adjacent to it.

2

Add the edge of least weight that is incident to the tree formed in rst step

and does not connect to a ver tex already in the tree.

3

Repeat this process until all the ver tices have been added.

Example 7

Use

all

Prim’s

the

algorithm

houses

on

to

the

nd

estate.

the

The

minimum

costs

of

cost

of

connecting

connecting

B

C

4

D

4

each 5

pair

of

houses

is

shown

on

the

weighted

graph.

The

3

weights 9

8 A

are

the

costs

in

€1000

of

connecting

the

3

E

3

houses.

5

4

4

H

1

A

Select

least

4

For

H

any

vertex

weight

example,

connected

if

to

6

and

adjacent

vertex

A

7

G

add

to

is

the

F

edge

of

it.

selected

it

is

H.

B

2

Add

the

incident A

edge

to

of

the

least

tree

weight

formed

that

in

is

the

rst

3

step

and

does

not

connect

to

a

vertex

4

already

in

the

tree.

H

If

B

C

4

vertices

formed

D

4

two

can

be

have

adjacent

the

same

to

the

tree

weight

being

then

either

used.

3

In A

3

this

example

3

4

Repeat

have H

6

minimum

houses

to

cost

for

electricity

connecting

will

all

the

+

3

+

4

+

4

+

3

this

been

+

3

+

6

tree

process

edge

would

be

HB.

until

all

the

vertices

added.

formed

spanning

tree

will

for

be

the

a

minimum

graph.

be

The

4

next

F

G

The

The

the

E

3

=

27,

ie

€27 000

the

order

tree

in

are

which

AH,

the

HB,

edges

BC,

All minimum spanning trees will have the same weight but may not consist of

CD,

were

DE,

added

DF ,

to

HG.

E X AM HINT

exactly the same edges.

Unless told which

algorithm to use,

you can use either For

very

large

graphs

Prim’s

algorithm

is

usually

quicker

than

Kruskal’s

Prim’s or Kruskal’s as

there

is

no

need

to

check

whether

adding

an

edge

will

form

a

cycle.

algorithm in an It

is

possible

to

use

Prim’s

algorithm

directly

from

a

table

without

exam. needing

to

draw

the

graph.

710

/

15 . 3

Example 8

minimum

spanning

tree

for

the

A

Begin

with

vertex

graph

represented

C

D

E

A

0

40

45

25

10

B

40

0

25

15

30

C

45

25

0

20

35

D

25

15

20

0

15

E

10

30

35

15

0

A.

Cross

B

the

B

are

A

by

C

D

out

the

added

to

table

columns

the

tree

below.

of

vertices

and

as

indicate

they

at

the

E end

of

have

these

the

been

rows

the

added.

rows

to

nd

order

You

the

can

in

which

then

entry

they

look

with

along

least

B

weight.

These

entries

should

be

circled.

C

D

A

B

C

D

E

dna y rtemoeG

the

y rtemonogirt

Find

B

C

E

It

is

often

easiest

to

draw

the

tree

at

each

10

stage

of

the

algorithm.

15

A

The

D

sum

weight

A

B

C

D

E

A

B

C

D

E

of

of

the

the

circled

entries

minimum

will

spanning

give

the

tree.

C

Continued

on

next

page

711

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

10 E

A

15

D

15

B

20

C

0

+

15

+

Reect

15

+

20

=

60

Why is Prim’s a better algorithm than Kruskal’s for a large graph or

when the information is given in a table?

How should the algorithm be changed if extra restrictions are added, for

example if there must be a direct connection between A and B?

Exercise 15F

1

Use

Prim’s

spanning

algorithm

tree

beginning

for

with

the

to

nd

the

following

vertex

2

minimum

graphs,

Use

Kruskal’s

minimum

A.

in

are

clearly

the

order

in

which

the

spanning

question

edges State

are

algorithm

1.

List

to

tree

the

nd

for

a

the

order

in

graphs

which

the

selected.

edges

selected.

3

Determine

be

in

a

Justify

the

number

spanning

your

tree

answer

of

edges

there

containing

through

a

v

will

vertices

consideration

E X AM HINT of

the

application

of

Prim’s

algorithm.

It is likely an exam question

4

The

following

graph

shows

the

ofces

of

a

will require this as evidence

nance

company.

The

management

wants

you have performed the to

connect

all

the

ofces

with

extremely

algorithm correctly. fast

internet

$1000s)

B

a

b

9

are

cables.

The

indicated

costs

on

the

of

doing

graph

so

(in

below.

C

10 8

B

B

A

A 15

11

10

10 7

13

9

17 18

11 12

C

G

F

A

8

C

D

20

12 9 10

8

F

8

13 7

5

D

6

9

E

6

14

E

E

a c

10

F

D 9

A

6

B

d

Use

Kruskal’s

algorithm

to

nd

the

8 A

minimum

cost

for

connecting

all

the

10

B

C

ofces. 7 4

3

6

9 7

are

7

11

5

List

the

order

in

which

the

edges

6 F

connected.

D

F

C

G 5

b

6

12 8

i

State

algorithm

E

and E

how

you

would

adapt

the

10

5

D

to

if

be

it

is

essential

for

ofces

B

connected.

D 7

ii

Find

the

minimum

cost

in

this

situation.

712

/

15 . 3

5

Use

of

Prim’s

the

algorithm

minimum

following

to

nd

spanning

the

tree

a

weight

for

Find

the

to

tables.

the

minimum

connect

all

the

corresponding

b

a A

B

C

D

E

A

0

40

25

15

20

B

40

0

25

45

30

C

25

25

0

15

35

D

15

45

15

0

20

E

20

30

35

20

0

It

is

decided

directly

of

7

a

i

pipe

houses

spanning

that

A

and

connected.

that

Use

length

will

Prim’s

minimum

graph

be

of

pipe

and

needed

draw

the

tree.

B

Find

must

the

be

extra

length

required.

algorithm

spanning

represented

to

nd

tree

by

for

the

the

the

table

shown.

B

C

D

E

A

0

4

7

9

12

B

4

0

8

5

3

C

7

8

0

10

5

D

9

5

10

0

6

E

12

3

5

6

0

ii

State

the

A

A

c

Comment

difcult

Prim’s

on

to

why

use

it

would

Kruskal’s

algorithm

when

be

more

algorithm

the

than

weight

of

the

B

C

D

E

10

7

12

9

8

5

B

10

C

7

8

D

12

5

E

9

6

The

given

a

following

between

a

in

mains

9

5

A

B

table

shows

needing

source

C

at

the

to

be

connected

7

6

6

5

3

D

G

7

H

6

b

E

F

23

11

10

15

B

20

0

9

16

21

15

10

C

17

9

0

22

16

10

12

A

ninth

vertex,

to

the

tree.

to

vertex

D

23

16

22

0

19

13

18

E

11

21

16

19

0

16

12

F

10

15

10

13

16

0

25

nd

i

the

the

to

ii

25

A

I,

4

4

needs

the

and

graph

tree

is

9

to

I

be

is

added

connected

A

is

5

vertex

your

of

the

B.

tree

By

from

new

part

a

minimum

if:

of

and

weight

to

of

weight

weight

A

the

to

12

In

consideration

spanning

18

3

G

17

12

6

to

A.

20

10

H

distances

0

15

G

information

A

G

tree.

table.

homes

water

F

9

F is

spanning

dna y rtemoeG

A

y rtemonogirt

b

to

of

and

the

B

the

to

B

edge

is

I

connecting

I

10

edge

is

connecting

6.

0

Developing inquiry skills

Look back to the opening question about the national

park .

Extreme weather causes extensive damage to all of the

trails. The park rangers decide to close some trails and

repair the others to keep them open. All viewpoints still

need to be accessible, and repair costs need to be kept

to a minimum. Which trails should they keep open?

What assumptions can you make about how the cost

of the repairs relates to the length of a trail?

713

/

15

OP T IMI Z ING

15.4

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Graph theory for weighted graphs:

the Chinese postman problem

International-

Investigation 3 mindedness

1

Find four dierent ways to draw this graph without taking your

A

The Chinese postman

pen o the paper. Begin from at least two dierent vertices. problem was rst

2

a

What do you notice about the degree of each ver tex?

b

What do you notice about the star ting ver tex and

F

B

posed by the Chinese

mathematician Kwan

Mei-Ko in 1962.

nishing ver tex? E

c

Conceptual

i

C

From the above results conjecture a

sucient condition for being able to traverse all the

D

edges in a graph exactly once, ending at the ver tex

at which you began.

ii

Verify your conjecture on a few graphs of your choosing.

iii

Explain in your own words why your conjecture might also be a

necessary condition.

3

Is it possible to draw this graph without taking your pen o

A

the paper? If so, nd at least two ways this can be done,

star ting from at least two dierent ver tices.

4

a

E

B

D

C

What do you notice about the star ting and nishing

ver tices and the degrees of the ver tices in the graph?

b

Conceptual

i

Conjecture necessary conditions for

being able to draw a graph without taking your pen

o the paper, ending at a dierent ver tex from the

one at which you began.

ii

Draw some graphs to verify your conjecture.

iii

Justify your conjecture.

A trail is a walk that repeats no edges. A circuit is a trail that star ts and

nishes at the same ver tex.

An Eulerian trail is a trail that traverses all the edges in a graph, and an

Eulerian circuit is a circuit that traverses all the edges in a graph.

The

•

results

For

a

even

•

For

a

from

graph

It

can

have

will

to

3

can

therefore

have

an

Eulerian

circuit

have

an

Eulerian

trail

all

be

the

summarized

vertices

as

must

follows:

have

degree.

graph

vertices,

the

Investigation

to

and

the

trail

must

begin

at

there

one

of

must

these

be

exactly

vertices

two

and

odd

end

at

other.

also

an

be

proved

Eulerian

have

an

that

circuit

Eulerian

every

and

graph

every

whose

graph

vertices

with

are

exactly

all

two

even

odd

will

vertices

trail.

714

/

15 . 4

Investigation 4

After a fall of snow the maintenance depar tment in a school needs to clear all

the paths before the students get to school. A graph of the school is shown.

The weights on the graph show the time it would take to clear the paths. The

time it takes to walk back along any cleared paths is one quar ter of the time

taken to clear them.

Art

and

Technology

12 13

Gym Cafeteria

11

18

14

20

9

16

Middle Senior

school school 17

10

y rtemonogirt

8

17 14

Junior

school

16 Estate

ofce 15 9

dna y rtemoeG

11

Theatre

16 8

Reception

Main 13 entrance 12

Car

park

The maintenance depar tment needs to begin and end their clearing at the

estate oce.

1

What would be the minimum time to clear all the paths if none of the paths

needed to be repeated?

2

By considering the degrees of the ver tices explain why it is not

possible to clear all the paths without having to walk back along some

of them.

3

The path between two buildings will need to be repeated; say which

buildings these are.

4

In which order would you recommend the maintenance depar tment clear

the paths if they are to take the least possible time?

5

Given that there is an easy alternative to taking the path from the estate

oce to the gym (go via the theatre instead), it is decided that the direct path

does not need to be cleared. How much time would be saved by not clearing

this path?

715

/

15

OP T IMI Z ING

The

Chinese

around

a

slightly

above

by

Another

the

as

context

Clearly

if

Eulerian

the

to

as

to

example

the

of

time

along

to

return

a

to

problem

walk

is

to

the

the

Chinese

walk

is

all

nd

that

the

GR A P H

of

starting

a

the

on

weight

The

problem,

cleared

postman

streets

least

vertex.

postman

along

of

route

THEORY

paths.

seeking

their

modied

round,

to

nd

repeating

possible.

graph

circuit

and

reduced

for

way

distance

a

an

NE TWORK S:

problem

graph

was

having

shortest

little

postman

weighted

question

C O M PL E X

and

consists

the

of

least

just

even

weight

vertices,

will

simply

you

be

can

the

nd

sum

an

of

TOK

the What is most important in

weights

of

all

the

edges. becoming an intelligent

Recall

from

the

handshaking

odd

number

of

in

graph

two.

a

an

is

Eulerian

edges.

least

One

You

odd

trail

then

If

vertices,

there

from

must

theorem

so

are

one

the

return

to

to

the

it

is

minimum

exactly

vertex

that

two

the

odd

other

starting

not

possible

number

of

vertices,

that

vertex

to

odd

you

goes

can

the

an

human being: nature or

vertices

along

using

have

nurture?

nd

all

the

route

of

weight.

way

vertices

to

on

vertices,

think

the

and

about

repeated

hence

this

is

to

route.

there

is

an

add

This

a

second

creates

Eulerian

a

edge

graph

between

with

just

the

even

circuit.

Example 9

A

Solve

the

Chinese

postman

problem

(nd

a

route

of

least

weight) 4

for

this

graph,

weight

for

this

beginning

and

ending

at

A,

and

nd

the

total

B

5

route.

6 4 F

5 C 8

7

E

4

5 D

There

are

two

odd

vertices,

B

and

E.

Begin

by

vertices.

considering

In

this

the

example

degrees

there

of

are

all

two

the

odd

A

4

vertices.

B

5

By

inspection

nd

the

route

of

least

weight

6 4

4

F

between

these

two

vertices.

5 C

Here

8

it

is

BCE.

Add

extra

edges

to

your

7

7

E

graph

4

Use

5

to

this

show

this

second

route.

graph

to

nd

an

Eulerian

D

circuit

A

possible

postman

solution

problem

to

is

the

Chinese

ABCDEFBCECFA.

There

beginning

are

normally

many

and

ending

possible

relatively

easy

at

routes

to

nd

A.

and

it

is

one

by

inspection.

The

total

weight

is

48

+

11

=

59

The

of

weight

all

the

the

of

this

edges

weights

of

in

the

route

the

will

be

original

repeated

the

weight

graph

plus

edges.

716

/

15 . 4

Reect

How do the conditions for the existence of an Eulerian circuit or trail

help in solving the Chinese postman problem?

Exercise 15G

1

Solve

the

graphs

edges

the

below,

that

route

a

Chinese

postman

starting

are

at

repeated

problem

vertex

and

the

A.

for

State

weight

i

the

the

it

of

the

could

minimum

take

complete

taken.

would

ii

8

A

Find

If

the

for

his

need

length

the

be

postman

and

to

to

which

walked

has

time

postman

round,

to

of

streets

along

walk

twice.

down

10

B

both

sides

of

every

street

explain

C

9

why

6

6

no

extra

streets

will

need

to

be

7 F

covered.

7 D

11 G

Give

5

12

a

possible

route

he

could

take.

10

3

The

following

graph

represents

bus

routes

c

A 20

costs 5

F

towns,

and

the

weights

are

the

A

of

the

journeys

in

dollars.

A

tourist

4

B

35

staying F

town

A

wishes

to

travel

along

7

all 45

at

B

20

40

the

routes

and

return

to

A

as

cheaply

as

C 7

50

6

possible. 9 30

E 45

E

C

D

A

6

4

B

C

5

D

4

5

5

6

3

2

a

Explain

why

the

following

graph

3

has 5 H

an

Eulerian

trail

but

not

an

6

G

E

D

Eulerian

circuit.

F

B

C

17

16

13 14

9

A

y rtemonogirt

between

b

dna y rtemoeG

E

Find

b

A

the

new

minimum

direct

between

12

23

a

bus

towns

G

possible

route

and

is

E.

total

cost.

added

Explain

why

D

the

addition

of

this

route

will

result

in

14 12

a

G

b

The

the

times

for

the

a

A.

to

in

which

If

he

as

and

graph

along

the

begins

it

does

down

not

for

write

for

down

travelling

the

new

all

the

routes

amount.

a

streets

and

takes

cost

represent

minutes

walks

time

he

E

the

walk

second

time,

on

taken

round,

vertex

14

F

weights

postman

his

16

lower

ends

any

one

need

on

at

street

third

to

of

deliver

letters.

717

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Investigation 5

1

The graph on the right has four odd ver tices. List these ver tices.

A

To solve the Chinese postman problem for this graph you will need to connect the

odd ver tices in pairs and nd the route of least weight between them. You should

2

then select the lowest of these as your solution.

2

G

B 6

3

2

List

the

three

pairings

of

the

fou r

ver tices

a nd

s tate

the

wei g ht

of

the

least3 H

weight

route

connect ing

the

ver tices

in

ea ch

of

t he

4

pairs .

7 C

3

3

Show, on a copy of the graph, the routes connecting the pairs which have the F

1

2

least total weight . Hence solve the Chinese postman problem, beginning and

ending at ver tex A .

1 D

3

4

Find the total weight of the route of least weight that will traverse all the edges in

E

the graph at least once.

The requirement to star t and nish at the same ver tex is now removed.

5

Use the weights listed in question 2 to nd where you should now choose to star t and nish.

6

How many extra possible routes do we need to consider when the graph has four ver tices of

Factual

odd degree?

Conceptual

7

How does an understanding of an Eulerian trail help determine how the algorithm should

be adapted if the trail does not have to begin and end at the same ver tex?

What does the solution to the Chinese postman problem represent?

Exercise 15H

1

For

each

of

the

following

graphs

list

b

the

A

10

three

odd

pairings

degree,

that

and

connect

nd

the

two

vertices

minimum

of

weight

G

11

of

the

walk

between

the

vertices

in

12

each

9

pair.

F

10

Hence

nd

which

edges

need

to

be

repeated B

in

the

solution

to

the

Chinese

11

postman

7

8

C

problem.

13

a

A

8

E

6

3 9 D

H

B

4

9

2

8

4

6

5

3

below

shows

the

lengths

of

roads

in

a

factory

complex.

C

Each

evening

all

the

a

security

guard

walks

along

2

2

5

7

of

vertex

6

F

graph

connecting

I G

The

4

roads

and

returns

to

his

ofce

at

A.

D

6

E

718

/

15 . 4

A

A

40

B

38

30 20 G

50

35 I

B

88 56 26

60

23

H

60 80

43 29 D

C 35

C

H 40

20 32

40

40

80 30

50 F

G

61 F

E

D

a

E

a

Find

of

b

A

a

his

route

walk

friend

of

the

and

offers

vertices

another.

be

which

security

to

its

drop

and

State

chosen

to

to

which

has

to

the

walk

length

him

two

off

him

at

one

up

vertices

the

b

The

graph

below

shows

the

roads

H

State

to

be

taken

by

a

postman

back

letters.

The

weights

the

lengths

the

postman

of

the

roads

in

on

the

deliveries

at

vertex

to

start

metres,

and

to

route.

pick

of

A.

decrease

him

his

up

from

round

Explain

and

why

this

the

length

he

would

where

would

you

advise

the

to

be

collected

in

order

to

nish

the

length

he

needs

to

walk,

edges

nd

the

total

length

of

the

repeated

and

roads needs

this

has

when

and are

of

he

in

that

minimize delivering

distance

length

end

to

take

answer.

to

the

could

walk.

postman need

at

not

to

your

he

the

the

offers

him

have

walk.

state

justify

would

the

route

minimize

friend

take

should

c 3

to

vertex

at

distance

A

possible

and

Fully

length.

pick

minimize

guard

a

order

minimizes

state

Find

in

this

dna y rtemoeG

40

y rtemonogirt

30

case.

his

A.

E X AM HINT

Developing inquiry skills

Look back to the opening question about the national park .

In an exam question

there can only be 0,

2 or 4 ver tices of odd What is the minimum distance you would need to cycle to travel along

all the trails? Find one way of cycling along all the trails. Do you need to

degree.

cycle along any trails more than once?

719

/

15

OP T IMI Z ING

15.5

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Graph theory for weighted

graphs: the travelling salesman

problem

A table of least distances (weights) shows the length of the shor test route

between each pair of ver tices in a graph.

Example 10

This

graph

weights

shows

are

the

the

times

direct

of

bus

the

connections

journeys

in

between

minutes

and

four

the

towns.

buses

The

travel

A

both

45

ways

along

the

50

routes.

110

Construct

between

a

table

each

of

pair

least

of

distances

towns.

to

Assume

show

that

the

there

shortest

is

no

journey

need

to

C

times

change

B

buses 60

on

any

of

the

70

journeys.

D

From

A

B

C

D

0

50

45

105

the

routes

A

direct

B

50

C

45

D

105

from

shortest

route

Because

the

row

B

C

0

50

45

to

B

and

and

rst

is

see

to

but

be

graph

the

a

will

can

C

are

from

via

A

route

the

shortest

just

to

D

the

the

C.

undirected

column

direct

that

can

will

both

be

be

the

rst

completed.

quicker

than

D a

A

A

we

connections,

Normally

A

graph

route

that

passes

through

other

vertices,

105 but

B

50

0

95

70

C

45

95

0

60

D

105

70

60

0

it

is

it

95

the

route

on

important

shorter

takes

In

is

to

the

check.

from

B

to

In

C

this

via

example

A,

which

minutes.

context

is

go

to

over

it

might

poor

be

roads

that

or

has

the

direct

more

stops

way.

A path is a walk which does not pass through any ver tex more than once. A

TOK

cycle is a walk that begins and ends at the same ver tex, but otherwise does Hamiltonian paths and

not pass through any ver tex more than once. cycles are named after

A Hamiltonian path or cycle is a path or cycle which passes through all the

the 19th-century Irish

ver tices in a graph.

mathematician William

A complete graph is one in which every ver tex is directly connected to every

Rowan Hamilton.

other ver tex.

720

/

15 . 5

The

diagram

on

the

right

shows

a

complete

graph

with

ve

vertices.

Investigation 6

There is just one Hamiltonian cycle in a complete graph with

B

three ver tices, ABC.

This is because a cycle which passes through the ver tices in

exactly the reverse order of another cycle is regarded as the

same cycle. For example, CBA would be considered the same

A

C

as ABC.

Any cycle with the ver tices in the same order but beginning at a dierent

ver tex is also regarded as the same cycle. For example, BCA is the same

as ABC.

Therefore, when counting cycles you can always star t from the same ver tex.

How many dierent Hamiltonian cycles are there in a complete graph with:

a

2

4

b

5

c

6

ver tices?

Write down a formula for the number of Hamiltonian cycles in a

Factual

y rtemonogirt

complete graph with n ver tices.

80

3

There are about 10

atoms in the universe. How many ver tices would

there need to be in a complete graph for the number of Hamiltonian cycles

to exceed the number of atoms in the universe?

dna y rtemoeG

1

Before doing the calculation guess what you think the answer might be,

and then compare this guess with the calculated value.

Investigation 7

A

$130

$180

$160 $115

E

$200

B

$150 $120

$125 $190

C D

$170

In the diagram the ver tices represent towns and the edges represent the costs

of ying between those towns. A salesman wishes to visit all the towns and

return to his star ting point at A .

1

What is the minimum number of ights he would need to take?

2

Explain why the total cost of visiting all the towns would be greater than

$640.

Continued

on

next

page

721

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

3

Find the weight of any Hamiltonian cycle beginning and ending at A .

4

Use your answers to questions 2 and 3 to give a lower and an upper bound

for the cost of visiting all the cities.

5

In order to try and nd a cheaper route the salesman decides that he will

take the cheapest ongoing ight from each airpor t that takes him to a town

he has not already been to and then back to A . Find the cost of the route if

he follows this method beginning the algorithm at A.

6

Investigate the cost of visiting all the towns if he begins the algorithm at a

dierent town.

7

Which of the routes you found would you recommend he take, beginning

at town A?

8

Can you nd a cheaper route? How likely do you think it is to be the

cheapest? Justify your answer.

The

classical

Hamiltonian

The

cycle

practical

through

vertex.

each

In

context

visit

a

From

TSP

case

for

number

possible

of

cycle

the

a

the

graph,

graph

and

problem

in

a

of

starting

TSP

not

to

least

and

be

would

return

(TSP)

complete

walk

need

practical

towns

length

it

guarantee

a

One

of

weight

nd

that

you

will

you

individually

Unfortunately,

which

to

investigation

considerable

Instead

least

is

the

salesman

is

to

weighted

weight

nd

graph.

that

nishing

at

the

passes

the

same

complete.

be

his

a

salesman

starting

needing

point

to

in

the

shortest

nding

the

weight

time.

the

each

of

vertex

this

One

of

travelling

has

procedure

a

to

been

see

realized

which

shown

you

nd

range

for

to

have

is

that

shortest

would

take

a

time.

route

need

give

of

will

have

upper

of

doing

is

there

found

and

values

so

that

in

is

no

indeed

lower

which

illustrated

is

other

the

way

to

shortest.

bounds

the

simple

for

solution

the

must

solution

lie.

below.

Finding an upper bound

A 11 B

The

graph

shows

the

connections

between

seven

cities.

5

10

A

driver

needs

to

make

deliveries

to

all

of

these

11

cities,

9 F

12 5

beginning

and

ending

at

city

A.

The

weights

are

the

driving 6 C

times

between

the

cities,

measured

in

G

hours.

10

E

13

7

21

Any

to

route

the

that

starting

passes

point

through

would

be

all

an

the

vertices

upper

and

bound

returns

for

the D

solution

greater

For

be

It

the

weight

example,

an

is

to

upper

unlikely

TSP

as

than

by

the

this

shortest

ABFCDGEA

namely

though

cannot

have

a

route.

inspection

bound,

route

that

11

this

is

+

10

the

+

is

12

best

a

+

cycle

7

+

13

possible

so

+

its

6

+

upper

length

11

=

would

70

hours.

bound.

722

/

15 . 5

The best upper bound is the upper bound with the smallest value.

A

method

that

neighbour

The

1

the

Using

and

has

your

not

to

is

two

not

give

a

better

upper

bound

is

the

nearest

table

been

the

(or

around

already

stages:

complete,

showing

move

have

has

graph

(weights)

2

likely

algorithm .

algorithm

If

is

been

visited,

shortest

the

the

create

graph,

in

of

going

the

route

least

between

graph)

always

shortest

table

route

equivalent

included

the

a

each

choose

to

cycle.

back

distances

the

the

of

all

vertices.

starting

nearest

Once

to

a

pair

the

vertex

vertex

that

vertices

starting

point

is

taken.

E X AM HINT

Completing a table of least distances conver ts a practical TSP into a classical

In exams if there is TSP because it ensures that the graph is complete and so a Hamiltonian cycle

more than one route

route of least weight

The

the

algorithm

triangle

adjacent

should

only

inequality

vertices

is

be

used

(which

always

the

on

means

a

complete

the

shortest

direct

route).

graph

route

This

that

between

will

between them is

satises

always

found by inspection.

two

be

If there is a direct

the

case

when

working

with

a

graph

or

table

that

shows

the

dna y rtemoeG

between ver tices the

y rtemonogirt

exists.

least

route between two distances.

vertices this is often

the shortest route, but

it does not have to be.

Example 11

This

graph

shows

the

connections

between

seven

cities.

A

A

driver

11 B

needs

to

make

deliveries

to

all

of

these

cities,

beginning

and

5

ending

at

city

A.

The

weights

are

the

driving

times

between

the 10

cities,

measured

in

11

hours.

9 F

12 5

a

The

table

of

least

distances

is

given

below.

Find

the

values

of

a, 6 C

b

and

c. G

10

E

A

B

C

D

E

F

G

13

7

21

A

0

11

a

23

11

5

10

B

11

0

9

16

b

10

15 D

C

a

9

0

7

16

12

10

D

23

16

7

0

c

18

13

E

11

b

16

c

0

11

6

F

5

10

12

18

11

0

5

G

10

15

10

13

6

5

0

Continued

on

next

page

723

/

15

OP T IMI Z ING

b

Use

the

length

c

If

the

C O M PL E X

nearest

of

the

driver

neighbour

driver’s

were

to

NE TWORK S:

algorithm,

GR A P H

THEORY

beginning

at

town

A,

to

nd

an

upper

bound

for

the

journey.

follow

this

route,

which

towns

would

he

pass

through

more

than

once?

a

a

=

17,

b

=

21,

c

=

The

19

values

graph.

It

is

routes

as

shortest.

longer

it

is

of

Note

is

b B

C

D

E

F

by

to

check

obvious

example,

along

the

inspection

the

the

three

is

edges

the

alternative

not

route

two

from

always

from

outside

through

the

B

to

edges

the

E

is

than

centre

graph.

also

not

that

the

the

direct

using

shortest

route

from

D

to

E

route.

the

algorithm

on

a

table

of

G least

A

For

most

going

When

A

taken

important

the

along

the

are

weights

it

is

a

good

idea

to

cross

off

the

a columns

B

use

b

the

of

the

rows

vertices

to

nd

already

the

visited

nearest

and

to

neighbours.

C

D

c

E

16

b

c

F

G

The

nal

solution

is:

So

beginning

These AFGECDBA

with

+

5

+

6

+

16

+

7

+

16

+

11

=

time

c

The

an

the

upper

driver

actual

would

AFGEGCDCBA

would

be

bound

route

for

take

would

and

passed

the

so

the

is

length

66

of

hours.

be

towns

through

F

columns

row

G

twice.

and

C

weight,

After

is

shortest

are

used

return

from

Though

no

some

the

pass

which

reaching

of

if

the

this

distance

crossed

to

is

out

as

is

to

F .

shown

nd

the

next

edge

of

vertex

are

shortest

other

the

[FG].

nal

vertices

through

included

Reect

the

66 least

Hence

A

weight and

5

two

at

vertex

to

repeated

towns.

What algorithms can you use to nd an upper bound for the TSP for a

must

A.

routes

driver’s

you

in

the

between

These

route

has

towns

need

to

table,

be

to

be

given.

TOK

par ticular graph? How long would it take a

computer to test all of the

Hamiltonian cycles in a

complete weighted graph

with just 30 vertices?

724

/

15 . 5

Exercise 15I

Keep

they

your

will

answers

be

to

needed

this

for

exercise

Exercise

available

2

as

The

15J.

nearest

be

1

For

a

each

of

the

Construct

b

Hence

a

use

algorithm

following

table

the

to

of

a

neighbour

done

on

the

least

algorithm

complete

triangle

inequality.

the

algorithm

comment

neighbour

cycle

graph

and

on

directly

your

route

that

would

A

also

the

original

should

graph

In

need

to

be

to

b

5

which

each

the

the

total

more

than

one

satises

of

route

apply

4

the

is

6

7

4

6

D

route.

possible

and

B

C

5

4

If

always

B

taken

graph.

weight

case

graph

A

State

the

3

C

around

why

ndings.

500

the

illustrate

weights.

Hamiltonian

complete

a

the

a

around

diagrams

graphs:

nearest

nd

following

3

give D

the

upper

bound

for

both

routes

and 5

select

the

“better”

a

upper

bound.

A

b

7

6

B

6

F

3

E

C B

3

y rtemonogirt

4 6 4

5 A

F

9

C

D

8

3

6

5 8

10 E

7

E

c

d

B

6

A

D

B

50

A

dna y rtemoeG

7

9

15 6

4

4 10 D

5 25 15

E E

C 3

20

2

D

5

C

4

F

Finding a lower bound

For

at

a

graph

least

the

v

An

edges,

weight

weights.

with

of

In

so

the

most

alternative

The

v

procedure

is

is

vertices

one

edge

cases

to

as

a

lower

with

this

use

bound

least

is

the

solution

not

to

the

would

weight

likely

deleted

or

to

TSP

will

clearly

the

be

vertex

a

need

be

sum

good

v

of

to

traverse

multiplied

the

lower

ν

by

lowest

bound.

algorithm .

HINT

follows.

This algorithm should 1

Choose

a

vertex

and

remove

it

and

all

the

edges

incident

to

it

from

also be used only the

graph.

on a complete graph 2

Find

the

minimum

spanning

tree

for

the

remaining

subgraph.

that satises the

3

A

lower

bound

combined

is

the

weight

of

weight

the

two

of

the

edges

minimum

of

least

spanning

weight

tree

removed

plus

in

the

step

1.

triangle inequality.

This will always

be the case when 4

The

process

The

best

can

then

be

repeated

by

removing

a

different

initial

vertex.

working with a graph lower

bound

(the

one

with

the

largest

weight)

is

then

taken.

or table showing least The

example

below

includes

an

explanation

of

why

this

algorithm

distances. gives

a

lower

bound.

725

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Example 12

D

Find

a

lower

bound

by

deleting

vertex

A

from

the

graph.

9

C

7

4 6

E

8 4 B

5

F

6

A

With

vertex

spanning

A

tree

deleted,

the

minimum

Explanation

of

why

this

will

give

a

lower

bound:

is The

actual

solution

to

the

TSP

is

D D 9

C

9 C 7 7

4 6

4

E

6

E

8 8 4 B B F F

5

This

has

weight

6

25. A

When

A

solution

the

tree.

two

have

deleted

weights

5

edges

and

of

least

A

weight

is

to

In

deleted

the

of

the

this

this

to

part

of

example

it

using

combined

A

in

remaining

will

weight

The

incident

the

TSP

reconnected

weight.

4.

to

weight

equal

The

is

the

clearly

will

the

is

be

part

be

a

of

tree,

greater

minimum

the

and

than

so

or

spanning

28.

the

two

weight

solution

edges

of

of

the

the

of

least

two

TSP

edges

will

be

D

C

9

greater

than

or

equal

to

the

combined

weight

of

7

these

two

edges.

In

this

example

they

add

to

11.

4 6

E

8 4 B

5

F

6

A

Lower

bound

is

25

+

5

+

4

=

The

34

total

weight

algorithm

will

the

weight

of

the

example

of

edges

therefore

the

is

28

produced

be

solution

+

11

=

less

of

using

than

the

or

TSP ,

the

equal

which

to

in

37.

HINT

If the weights are given in a Reect

How do the algorithms used in the travelling salesman problem help

table then Prim’s algorithm address the fact that it is often not possible to test all the possible routes?

can be used to nd the How

ca n

you

make

sure

that

the

nearest

neighbour

algorithm

and

the

minimum spanning tree, deleted

ver tex

algorithm

will

give

the

upper

and

lower

bounds

for

a

having removed the rows p r a c t i ca l

TS P ?

and columns of the deleted

vertex.

726

/

15 . 5

TOK

Could we ever reach a point where everything impor tant in a mathematical sense is

known?

What does it mean to say that the travelling salesman problem is “NP hard”.

Is there any limit to our mathematical knowledge?

Exercise 15J

each

of

produced

15I,

i

for

a

lower

bound

vertex

A

higher

different

use

the

15I

plus

weights

given

using

and

lower

in

A

B

C

D

E

F

G

A

0

11

17

23

11

5

10

B

11

0

9

16

21

10

15

C

17

9

0

7

16

12

10

D

23

16

7

0

19

18

13

E

11

21

16

19

0

11

6

F

5

10

12

18

11

0

5

G

10

15

10

13

6

5

0

Exercise

the

deleted

deleting

bound

by

deleting

a

vertex

upper

bounds

least

1:

algorithm

a

of

graphs

vertex

nd

iii

tables

the

question

nd

ii

the

your

for

bound

calculated

answer

the

to

solution

part

of

ii

the

in

to

Exercise

give

TSP .

4

2

Answer

relate

the

to

following

the

graph

questions

in

which

complete

minutes

to

graph

walk

below

shows

between

ve

the

times

shops

in

a

city.

below.

A

The

C

B

40

dna y rtemoeG

For

y rtemonogirt

1

5

4

10

15

25

30

B

D 3

8

8 10 20

8

3

5

40

D

a

By

deleting

vertex

E

6

C

on

the

graph,

use A

E 5

the

deleted

lower

b

bound

Complete

c

a

this

graph.

Use

the

nd

vertex

an

for

table

algorithm

the

of

to

nd

a

a

TSP .

least

distances

Use

the

nd

for

an

time

nearest

upper

neighbour

bound

for

algorithm

the

State

what

reason

for

you

notice

and

b

and

ending

Use

the

In

Example

the

time

10

an

it

up p e r

woul d

between

distances

Prim’s

for

the

se v e n

vertex

is

alg orithm

above

A.

all

of

the

the

length

shops,

of

starting

A.

that

21

vertex

minutes

for

the

bound

wa s

algorithm

is

a

lower

to

bound

time

to

walk

By

consideration

to

all

the

shops.

ta ke

for

a

ci ti e s.

dr ive r

T he

sho wn.

to

nd

pro b l e m,

Use

a

by

the

l owe r

rs t

of

the

edges

adjacent

f ou n d C

and

ta bl e

tabl e

to

E

explain

why

it

is

not

to to

nd

a

cycle

with

length

22

of or

least

at

deleted

possible travel

to

for

to

this.

to for

bound

algorithm

the

c 3

walk

neighbour

to

TSP .

explain

upper

to

verify

d

nearest

less.

and

b o u nd

de let in g

d

The

shopper

decides

they

and

nish

shop

but

before

State

returning

how

might

at

be

the

A

to

A

adapted

to

their

has

nearest

still

to

need

last

be

allow

for

start

visit

to

neighbour

to

shop

B.

algorithm

this.

727

/

15

OP T IMI Z ING

C O M PL E X

NE TWORK S:

GR A P H

THEORY

Chapter summary

•

A graph is a set of ver tices connected by edges.

•

When raised to the power n an adjacency matrix gives the number of walks of length n between

two ver tices.

•

The transition matrix for a graph gives the probability of moving from one ver tex to another if all

edges from the ver tex are equally likely to be taken.

•

The steady states of a transition matrix give the propor tion of time that would be spent at each

ver tex during a random walk around the graph.

•

Kruskal’s and Prim’s algorithms are used to nd the minimum spanning tree for a graph.

•

The Chinese postman problem is to nd the walk of least weight that goes along every edge at least

once.

•

In the Chinese postman problem for a graph with two odd ver tices, the shor test route between the

two odd ver tices is found and indicated using multiple edges. The solution to the Chinese postman

problem will be an Eulerian circuit around the graph formed.

•

With four odd ver tices the shor test total route between any two pairs of the odd ver tices is found

and indicated using multiple edges. The solution to the Chinese postman problem will be an

Eulerian circuit around the graph formed.

•

The classical travelling salesman problem is to nd the Hamiltonian cycle of least weight in a

weighted complete graph.

•

The practical travelling salesman problem is to nd the route of least weight around a graph which

visits all the ver tices at least once and returns to the star ting ver tex.

•

A practical travelling salesman problem should be conver ted to the classical problem by completion

of a table of least distances where necessary.

•

The nearest neighbour algorithm is used to nd an upper bound for the travelling salesman

problem.

•

The deleted ver tex algorithm is used to nd a lower bound for the travelling salesman problem.

Developing inquiry skills

Look back to the opening questions

about the national park . What would

be the shor test route for a visitor

wanting to visit all of the viewpoints

and return to A. Approximately how

long would their journey be?

728

/

15

Click here for a mixed

Chapter review

1

a

Explain

why

Eulerian

b

Write

the

review exercise

graph

shown

has

b

a

State

trail.

down

of

a

possible

B

Eulerian

trail.

c

the

your

whether

A

Kruskal’s

algorithm

spanning

State

added

part

the

vertices

b

state

has:

ii

an

Eulerian

circuit

iii

an

Eulerian

trail.

to

nd

Hamiltonian

cycle

tree

for

your

the

order

in

and

the

the

By

considering

of

the

the

M

state

it

is

possible

to

move

between

vertices

of

G

in

two

or

fewer

steps.

tree. your

answer.

C

8

5 4

5

of

edges

Justify

B

powers

graph

which

weight

answers.

the

all

are

G

to

a

whether

shown.

of

E

d

minimum

answer

i

Justify

Use

each

D

F

2

of

G

Using

C

degree

a

Use

Prim’s

algorithm

to

nd

the

6 3 F

minimum

5

State 9

the

Chinese

weighted

State

the

graph

which

total

postman

shown,

edges

weight

the

graph

by

the

table

below.

the

weight

the

B

problem

beginning

need

of

of

the

minimum

D

spanning

Solve

for

8

E

3

tree

to

be

for

at

the

vertex

repeated

A

B

C

D

E

F

G

A

0

20

10

30

15

11

15

B

20

0

8

22

27

13

9

C

10

8

0

24

16

10

12

D

30

22

24

0

29

13

18

E

15

27

16

29

0

16

14

F

11

13

10

13

16

0

25

G

15

9

12

18

14

25

0

A.

and

route.

C

30

tree.

dna y rtemoeG

represented

4

2

6

spanning

2

G

y rtemonogirt

7 A

22

45

35

A 32

50

47

F

40

25

30

Another 38

E

is

4

Consider

the

represents

adjacency

the

graph

A

vertex

H

matrix

M

which

G.

connected

by

an

edge

weight

B

C

D

1

1

2

b

E

to

of

A

now

by

an

weight

added

to

the

graph.

H

edge

12

and

of

weight

to

C

by

9,

an

to

B

edge

of

11.

Find

a

lower

salesman A  0

is

D

bound

problem

for

for

the

this

travelling

extended

0  graph.

 B

 1

0

2

0

0





6 C



1

2

0

1

0

1

0

1

 D

0





 E

This

State

i

iii

the

0

graph

G

1

be

the

edges

of

a

graph

with

A,

following

B,

C,

D

and

E

are

given

in

the

table.

complete

0



done

without

A

B

C

D

E

A

0

16

10

25

15

B

16

0

8

22

27

C

10

8

0

24

16

D

25

22

24

0

19

E

15

27

16

19

0

G.

is:

ii

directed

your

0

should

whether

Justify

of

 0

question

drawing

a



weights

vertices

 2

The



iv

simple

a

tree.

answers.

729

/

15

OP T IMI Z ING

a

i

State

C O M PL E X

whether

represented

or

by

NE TWORK S:

not

the

the

table

GR A P H

8

graph

THEORY

Consider

the

weighted

graph

below.

is B C

3

complete.

1

ii

Draw

the

graph

represented

by

4

the

2

G

3

4

table.

D

A 3

iii

Use

the

nearest

neighbour

2 1

2 F

algorithm,

beginning

at

vertex

A, 2

to

nd

an

upper

travelling

bound

salesman

for

the

E

problem.

a

b

i

Use

Kruskal’s

minimum

algorithm

spanning

to

tree

nd

for

the

obtained

by

A

from

the

weight

of

this

Find

a

graph

and

Hence

nd

a

of

walk

The

graph

G

is

lower

bound

salesman

shown

of

odd

and

ending

at

total

weight,

which

every

edge

at

least

once,

and

tree.

for

the

weight

of

this

walk.

the justify

your

answer.

problem.

c

7

starting

minimum

Fully travelling

vertices

state

state

ii

four

removing

includes the

the

the

A, vertex

down

degree.

b subgraph

Write

Find

that

below.

if

it

the

minimum

includes

is

nish

State

no

at

a

every

longer

the

weight

at

necessary

same

possible

edge

of

a

walk

least

to

once

start

and

vertex.

starting

and

nishing

A

vertex

9 D

Let

G

be

for

the

this

walk.

weighted

graph

below.

B

A

B

14

5

8

C

C

a

State

whether

G

is: 4

i

ii

directed

7

simple

10

E

iii

strongly

Justify

your

a

answers.

Explain

why

Hamiltonian

b

For

each

vertex

indegree

and

write

the

down

i

State

b

outdegree.

whether

G

has

an

Conjecture

for

a

a

necessary

directed

Eulerian

graph

to

Give

Does

G

an

down

a

condition

have

d

Use

to

an

the

nd

Eulerian

trail?

If

Conjecture

a

a

necessary

directed

graph

to

d

Construct

e

Find

the

a

transition

on

Hamiltonian

path

table

of

least

neighbour

graph

least

distances

G.

algorithm

cycle

on

represented

distances

for

by

which

the

the

begins

at

vertex

A.

If

this

cycle

was

taken

around

G,

state

condition

have

number

of

times

the

cycle

passes

an vertex

C.

trail.

steady

comment

a

Hamiltonian

ends

through Eulerian

of

trail.

the for

a

so

e iv

a

of

a

nearest

complete

possible

have

cycle.

example

Construct

and write

an

c

circuit.

have

not

G.

table

iii

does

Eulerian

circuit.

ii

G

the

in

c

D

connected.

state

your

matrix

for

G.

probabilities

and

results.

730

/

15

13

P1:

Consider

the

network.

Exam-style questions B

A

10

P1:

Consider

the

graph

below.

A

D C

B C

E

F

E

a

Write

down

a

starting

at

Explain

why

Hamiltonian

the

point

cycle,

F .

(2

marks)

D

b

construct

F

how

circuit

you

exists

know

in

the

a

Write

down

one

c

Eulerian

above

such

Eulerian

possible

circuit

(2

Suggest

which

removed

graph.

mark)

in

edge

order

could

that

impossible

construct

cycle

any

from

a

point.

it

(2

be

Hamiltonian

Justify

Prim’s

algorithm

(starting

at

(2

A)

a

minimum

the

State

spanning

marks)

P1:

From

the

graph,

write

down

all

the

odd

to

vertices. determine

your

marks)

14 Use

marks)

be

answer.

P1:

this

Eulerian

circuit.

11

to

from

graph.

(1

b

not

tree

(1

mark)

for A

graph.

clearly

in

which

order

1

2

you

are

adding

the

edges.

(3

marks)

3 B 4

6

C

F

dna y rtemoeG

State

is

y rtemonogirt

a

a

it

A 3

B

2

5

7

3

6

6

3

4

6

F

C 5

4

E

D

6

E 4 D

a

Determine

spanning

the

weight

of

the

starts

minimum

tree.

(2

Hence

P1:

The

on

by

following

a

computer

electronic

any

graph

two

pounds)

shows

network,

cable.

The

computers

(in

is

the

respective

given

arc

of

by

the

B

be

cost

of

15

P2:

shortest

nishes

at

through

State

A

network

following

connecting

number

a

also

route

vertex

every

its

that

A,

edge

weight.

at

(6

least

marks)

points

connected

hundreds

is

represented

by

the

table.

of

on

A

B

C

D

E

A

0

24

17

18

21

B

24

0

31

26

20

C

17

31

0

19

13

D

18

26

19

0

25

E

21

20

13

25

0

each

graph.

D

11

15

to

seven

and

travelling

marks)

once.

12

nd

14

16 24 E

8 G

A

9 12

17

13

a

C

the

F

b a

By

using

Kruskal’s

Draw

algorithm,

a

weighted

minimum

graph,

added

b

Draw

and

stating

the

the

spanning

the

order

tree

in

arcs.

Using

minimum

Kruskal’s

hence

connecting

nd

the

the

least

cost

computers.

you

marks)

tree

for

(4

algorithm,

nd

marks)

a

spanning

tree

for

the

graph.

the

which

spanning

(3

nd

for

(2

representing

network.

minimum the

graph

State

add

c

clearly

the

the

order

in

which

edges.

Hence

nd

minimum

the

you

(2

weight

spanning

of

tree.

marks)

the

(2

marks)

marks)

731

/

15

16

OP T IMI Z ING

P2:

The

C O M PL E X

following

four

islands

shipping

directed

served

by

NE TWORK S:

graph

a

GR A P H

represents

18

THEORY

P2:

particular

The

by

graph

letters

shows

A,

B,

seven

C,

D,

schools,

E,

F ,

denoted

G.

company. G

3

2

B

C

A

A 4

3

5

B

D

12 10

5

5

D

C

6

a

Construct

for

the

the

adjacency

shipping

matrix

routes.

(2

marks)

A

3

b

i

Find

ii

Hence

the

matrix

school

and

M

is

the

number

it

is

possible

to

travel

inspector

returning

➝

C

in

exactly

three

visit

to

from

each

his

school

of

A,

the

starting

point.

weights

of

each

edge

denote

the

from

travelling A

to

starts

of

The ways

E

required

schools, determine

7

F

M

times

for

the

inspector.

journeys.

(3

a

marks)

Construct

a

table

of

least

distances.

3

(3

marks)

i

c

i

Find the matrix

, where S

S





3

3

M

b

By

using

the

nearest

neighbour

i 1

algorithm,

ii

Hence

nd

the

number

of

ways

bound is

possible

to

travel

from

B



C

determine

a

best

of

the

upper

it

for

the

length

journey.

in

(3 fewer

than

four

c iii

List

all

such

possible

State

the

actual

route

taken

inspector.

The

shipping

company

from

question

to

introduce

an

express

P2:

Consider

A



D

and

vice

marks)

the

graph

below,

where

A

to

E

form

a

network

of

route

directed from

(2

8

vertices decides

the

marks)

19 P2:

by

ways.

(8

17

marks)

journeys.

routes

as

shown.

versa.

A

B

B

A

C

C

E

D

D

a

Construct

a

transition

matrix

T

for

the

a new

set

of

shipping

Construct

that

a

ship

starting

its

for

this

port

B

is

just

as

likely

If

“random

a

stationed

at

any

of

the

three

20

at

per

at

the

steady

P2:

Consider

state

probabilities

shipping

vertex

hour

each

your

determine

(to

the

nearest

vertex.

the

average

minute)

(4

following

the

marks)

network,

possible

a

mouse

trapped

routes

available

in

a

maze.

marks)

D

B

Using

A,

network.

(2

d

undertaken

for for

the

was

marks)

illustrating

Find

walk”

journeys.

(3

c

marks)

ports spent

following

(3

to time

be

graph.

journey starting

from

matrix

marks)

b Show

transition

routes.

(3

b

the

answer

to

part

c,

determine

E

A C

the

to

best

base

your

port

its

for

the

shipping

headquarters,

answer.

company

justifying

(2

marks)

732

/

15

a

Construct

this

the

adjacency

matrix

network.

(2

M

for

22

P2:

Consider

the

following

graph.

marks) B

7

b

i

By

evaluating

that

there

is

the

only

matrix

one

M

,

show

journey

15

in

13 20

this

network

that

can

be

done

19

in

17

exactly

seven

moves.

C

A

ii

Describe

the

the

journey ,

vertices

listing

visited.

(in

(6

20

order)

16

marks)

21

The

diagram

pathways

park,

shows

joining

with

the

a

network

seven

weights

of

points

14

cycle

in

a

indicating

the 18

E

time

of

each

journey

Nasson

aims

to

pathway

possible

returns

at

to,

cycle

least

time.

(in

along

once

He

point

in

starts

D

minutes).

every

the

a

shortest

from,

and

A.

By

using

the

nearest

algorithm,

and

determine

an

travelling

State

starting

upper

salesman

clearly

neighbour

the

at

vertex

bound

for

A,

the

problem.

order

in

which

10

you

b

By

are

taking

deleting

the

vertices.

vertex

A

and

(3

marks)

using

the

B

deleted

6

a

1

lower

vertex

algorithm,

bound

for

the

obtain

travelling

7

salesman

5

D

2

c

Show

same

8

problem.

that

by

lower

(4

deleting

bound

vertex

will

be

marks)

B,

the

dna y rtemoeG

P2:

y rtemonogirt

21

found.

5

(4

marks)

5

2

2

d

Hence

the

write

down

minimum

an

length

inequality

( l)

F

a

State

which

need

to

the

total

along

time

a

tour.

(2

pathways

cycle

of

that

Nasson

twice,

his

will

and

journey

nd

e

Write

down

a

tour

satisfying

inequality.

for

marks)

this

(1

mark)

will

Click here for fur ther

take.

(7

marks) exam practice

b

c

Hence,

nd

a

suitable

Nasson

to

take.

Instead

of

his

decides

to

start

needs

to

amount

other

to

at

point

every

time,

plan,

but

F .

path

can

marks)

Nasson

He

in

still

the

nish

least

at

any

point.

Determine

aim

of

for

(2

original

cycle

route

at

nish,

which

and

point

justify

Nasson

your

should

answer.

(4

marks)

733

/

Approaches to learning: Communication,

Choosing the right

Research, Reection, Creative thinking

Exploration criteria: Presentation (A),

path

Mathematical communication (B), Personal

engagement (C), Reection (D)

IB topic: Graph theory

ytivitca noitagitsevni dna gnilledoM

Minimum spanning trees

How many dierent spanning trees can you nd for

this graph?

What is the minimum length spanning tree?

50

A

B

30

25

35 25

40 E

D

C

25 20

35

45

25

40

W

60 F

Hamiltonian paths and c ycles (the

travelling salesman)

Find the minimum length Hamiltonian cycle on this

graph:

B

6

7

3

5

5

C

A E

4

9

8

D

3

Eulerian trails (The Chinese postman)

Find the minimum length Eulerian circuit on this graph. 5

5

4

5

4

What if you did not need to return to the star ting ver tex? 3

3

4 5

6

4

5

3

What real-life situations could you use these to solve?

In

•

•

•

order

clearly

solve

dene

consider

identify

or

•

to

the

be

able

to

reallife

what

it

method

what

collect

a

the

the

is

or

problem,

that

vertices

a

and

to

you

edges

required

representation

must:

want

algorithm

information

draw

you

you

of

of

to

nd

will

the

put

the

follow

graph

on

this

graph,

to

make

represent

this

possible

and

nd,

record

graph

perhaps

using

technology.

734

/

15

•

•

might

reect

also:

on

the

accuracy

assumptions

or

collect

and

them

consider

out

so

what

and

reliability

simplications

nally

it

answer

extensions

you

has

the

of

the

been

values

you

necessary

to

have

make

found

in

and

order

any

to

question

could

solve

based

on

what

you

have

ytivitca noitagitsevni dna gnilledoM

You

found

far.

Your task

Choose one of these problems and think of a situation that is relevant to you that you

could use this process to solve.

Based on this, try to give a brief answer to each of these questions:

•

What real-life issue are you going to try to address?

•

What is the aim of this exploration?

•

What personal reason do you have for wanting to do this exploration?

•

What data will you need to collect or nd?

•

What research will you need to do for this exploration?

•

What sources of information will you use?

•

What denitions will you need to give?

•

What possi ble representations will you need to include?

•

How will you draw these?

•

What technology will you require?

•

What assumptions or simplications have you needed to make in this exploration?

•

What possible ex tensions could there be to your exploration?

•

Are there any fur ther information/comments that may be relevant with regards

to your exploration?

Ex tension

There are also other types of problems in Graph theory

Domination:

that have not been covered in this chapter. Here are two Find the domination number (the size of a examples on ver tex colouring and domination. smallest dominating set which is a set of ver tices

What real-life situations could you use these to solve?

of the graph such that all ver tices not in the set

are adjacent to a ver tex in the set) of this graph:

Ver tex colouring: a

b

Find the chromatic number (the smallest number of colours c

needed to colour the ver tices so that no two adjacent d f

ver tices are the same color) of this graph:

g e

h

i

735

/

Exploration

16 All

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interests

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particular,

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that

exploration.

solving

exploration.

area

area

your

opportunity

exploration

exploration

do

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to

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(your

an

of

your

teacher

exploration

your

IB

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will

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set

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a

you

their

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deadline

receive

a

for

grade

Diploma.

736

/

How the exploration is marked

After

you

have

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teacher

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mark

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IA tip

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to

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external

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This

in black and white, external

moderation

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students’

of

IA

ensures

that

all

teachers

in

all

schools

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same

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To

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teacher’s

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moderator’s

then

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your

mark

will

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teacher’s

mark

within

marks

three

2

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stay

the

your

of

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information such as

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your

is lost. Make sure

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your exploration is If

the

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mark

is

more

than

2

marks

higher

or

lower

than

uploaded in colour your

teacher’s

mark,

the

external

moderator

will

mark

the

remaining

if it contains colour explorations

in

the

sample.

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may

increase

the

mark

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teacher

diagrams. marked

too

leniently.

reason

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for

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change

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the

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mark

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teacher

school

marked

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too

the

marks.

IA tip

Make sure that

Internal assessment criteria

all the pages are

Your

exploration

given

below.

The

will

IB

be

assessed

external

by

your

moderator

teacher,

will

use

against

the

the

same

criteria

assessment

criteria.

uploaded. It is

almost impossible to

mark an exploration

with pages missing. The

nal

criterion.

nal

The

mark

The

mark

for

criteria

for

each

exploration

maximum

possible

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cover

ve

is

nal

the

mark

applications

areas,

A

to

sum

is

and

of

the

20.

scores

This

is

for

20%

interpretation

each

of

your

Higher

level.

E

IA tip

Criterion A

Presentation

Criterion B

Mathematical communication

Criterion C

Personal engagement

Make sure you

understand these

criteria. Check your

exploration against

Criterion D

Reection

Criterion E

Use of mathematics

the criteria frequently

as you write it.

Criterion A: Presentation

This

your

criterion

assesses

the

organization,

coherence

and

conciseness

of

exploration.

Achievement level

Descriptor

The exploration does not reach the standard described by 0 the descriptors below.

1

The exploration has some coherence or some organization.

The exploration has some coherence and shows some 2 organization.

3

The exploration is coherent and well organized.

4

The exploration is coherent, well organized and concise.

737

/

16

E X P L O R AT I O N

To get a good mark for Criterion A: Presentation

•

A

well

a

❍

organized

rationale

exploration

which

includes

has:

an

explanation

of

why

you

chose

this

topic

an

❍

introduction

in

which

you

discuss

the

context

of

the

exploration

a

❍

statement

clearly

a

❍

•

A

of

the

aim

of

the

exploration,

which

should

be

identiable

conclusion.

coherent

❍

is

❍

should

❍

includes

exploration:

logically

developed

“read

well”

any

and

and

graphs,

easy

to

express

tables

and

follow

ideas

clearly

diagrams

IA tip

where

they

are

For more on needed—not

attached

as

appendices

to

the

document.

citing references,

•

A

concise

exploration:

academic honesty ❍

focuses

on

❍

achieves

❍

explains

the

aim

and

avoids

irrelevancies

and malpractice, see the

aim

you

stated

at

the

beginning

pages 741 – 742.

•

References

be

all

stages

must

considered

be

in

the

cited

academic

exploration

where

clearly

appropriate.

and

concisely.

Failure

to

do

so

could

malpractice.

Criterion B: Mathematical communication

This

criterion

•

use

•

dene

•

use

assesses

appropriate

how

IA tip

you:

mathematical

language

(notation,

symbols,

terminology)

Only include forms

of representation key

terms,

where

required

that are relevant multiple

forms

of

mathematical

representation,

such

as

formulas,

to the topic, for diagrams,

tables,

charts,

graphs

and

models,

where

appropriate.

example, don’t draw

a bar char t and pie Achievement level

Descriptor char t for the same

The exploration does not reach the standard described by data. If you include

0 the descriptors below.

a mathematical

There is some relevant mathematical communication

process or diagram

which is par tially appropriate.

without using or

1

commenting on it,

The exploration contains some relevant, appropriate 2

then it is irrelevant.

mathematical communication.

The mathematical communication is relevant, appropriate 3 and is mostly consistent.

The mathematical communication is relevant, appropriate 4 and consistent throughout.

IA tip

To get a good mark for Criterion B: Mathematical Use technology to

enhance the

communication

development of

•

Use

appropriate

mathematical

language

and

representation

when

the exploration—

communicating

mathematical

ideas,

reasoning

and

ndings.

for example, by

•

Choose

and

use

graphic

display

spreadsheets,

appropriate,

appropriate

calculators,

databases,

to

enhance

mathematical

screenshots,

drawing

and

and

ICT

tools

mathematical

word-processing

mathematical

such

as

reducing laborious

software,

software,

and repetitive

as

calculations.

communication.

738

/

16

•

Dene

•

Express

•

Label

•

Set

•

Dene

key

terms

that

you

use.

IA tip results

scales

to

and

an

appropriate

axes

clearly

in

degree

of

accuracy.

Students often

graphs.

copy their GDC out

proofs

clearly

and

logically.

display, which variables.

makes it unlikely

•

Do

not

use

calculator

or

computer

notation.

they will reach the

higher levels in this

Criterion C: Personal engagement criterion. You need

This

criterion

assesses

how

you

engage

with

the

exploration

and

make

to express results in

it

your

own.

proper mathematical

notation, for example,

Achievement level

Descriptor x

use 2

and not 2^x

The exploration does not reach the standard described use × not *

0 by the descriptors below.

use 0.028 and not 1

There is evidence of some personal engagement. 2.8E-2

2

There is evidence of signicant personal engagement.

3

There is evidence of outstanding personal engagement.

To get a good mark for Criterion C: Personal engagement

•

Choose

this

a

topic

makes

it

•

Find

•

Demonstrate

a

topic

for

your

easier

that

to

exploration

display

interests

that

personal

you

and

ask

you

are

interested

in,

as

engagement

yourself

“What

if…?”

IA tip

practices

personal

from

the

engagement

mathematician’s

❍

Creating

mathematical

❍

Designing

❍

Running

experiments

❍

Running

simulations.

❍

Thinking

and

❍

Thinking

creatively.

❍

Addressing

❍

Presenting

❍

Asking

and

models

implementing

to

❍

Considering

❍

Exploring

of

these

skills

and

Just showing

real-life

personal interest

situations.

in a topic is not

surveys.

enough to gain

data.

this criterion.

independently.

your own voice and

interests.

ideas

making

in

your

conjectures

own

and

demonstrate your

way.

own experience with

investigating

the mathematics in

ideas.

historical

unfamiliar

You

need to write in

mathematical

mathematical

some

toolkit.

for

collect

personal

questions,

using

the top marks in

working

your

by

and

global

the topic.

perspectives.

mathematics.

Criterion D: Reection

This

criterion

assesseshow

you

review,

analyse

and

evaluate

your

Achievement level

tnemssessa lanretnI

exploration.

Descriptor

The exploration does not reach the standard described 0 by the descriptors below.

1

There is evidence of limited reection.

2

There is evidence of meaningful reection.

3

There is substantial evidence of critical reection.

739

/

16

E X P L O R AT I O N

To get a good mark for Criterion D: Reection IA tip •

Include

reection

in

the

conclusion

to

the

exploration,

but

also

Discussing your throughout

the

exploration.

Ask

yourself

“What

next?”

results without

•

Show

reection

in

your

exploration

by:

analysing them is not ❍

discussing

the

❍

considering

❍

stating

❍

making

❍

considering

❍

explaining

implications

of

your

results

meaningful or critical the

signicance

of

your

ndings

and

results

reection. You need possible

limitations

and/or

extensions

to

your

results

to do more than just links

to

different

elds

and/or

areas

of

mathematics

describe your results. the

limitations

of

the

methods

you

have

used

Do they lead to why

you

chose

this

method

rather

than

another.

further exploration?

Criterion E: Use of mathematics

This

criterion

assesses

how

you

use

mathematics

Achievement level

in

your

exploration.

Descriptor

The exploration does not reach the standard described by 0 the descriptors below.

1

Some relevant mathematics is used.

Some relevant mathematics is used. The mathematics

2

explored is par tially correct. Some knowledge and

understanding are demonstrated.

Relevant mathematics commensurate with the level of the

3

course is used. The mathematics explored is correct. Good

knowledge and understanding are demonstrated.

Relevant mathematics commensurate with the level of the

4

course is used. The mathematics explored is correct. Good

knowledge and understanding are demonstrated.

Relevant mathematics commensurate with the level of the

course is used. The mathematics explored is correct and 5 demonstrates sophistication or rigour. Thorough knowledge

and understanding are demonstrated.

Relevant mathematics commensurate with the level of

the course is used. The mathematics explored is precise 6

IA tip

and demonstrates sophistication and rigour. Thorough

Make sure the

knowledge and understanding are demonstrated.

mathematics in your

exploration is not

To get a good mark for Criterion E: Use of mathematics

only based on the

prior learning for the

•

Produce

work

that

is

commensurate

with

the

level

of

the

course

you

syllabus. When you are

studying.

the

syllabus,

The

mathematics

you

explore

should

either

be

part

of

are deciding on a at

a

similar

level

or

beyond.

topic, consider what

•

If

the

level

of

mathematics

is

not

commensurate

with

the

level

of

the

mathematics will be course

you

can

only

get

a

maximum

of

two

marks

for

this

criterion.

involved and whether

•

Only

use

mathematics

relevant

to

the

topic

of

your

exploration.

Do

it is commensurate not

just

do

mathematics

for

the

sake

of

it.

with the level of the

•

Demonstrate

your

that

you

fully

understand

the

mathematics

used

in

course.

exploration.

740

/

16

Justify

❍

(do

•

not

just

you

use

using

❍

Apply

mathematics

❍

Apply

problem-solving

❍

Recognize

and

justify

all

Demonstrate

particular

mathematical

technique

conclusions.

in

different

is

contexts

techniques

explain

mathematics

at

a

it).

Generalize

Precise

and

are

❍

accuracy

•

why

patterns

error-free

where

where

where

and

appropriate.

appropriate.

appropriate.

uses

an

appropriate

level

of

times.

IA tip

sophistication

of

mathematics

in

your

exploration. Each step in the

Show

❍

that

you

understand

and

can

use

challenging

mathematical mathematical

concepts development of the

Show

❍

that

you

can

extend

the

applications

of

mathematics exploration needs to

beyond

those

you

learned

in

the

classroom be clearly explained

❍

look

at

a

❍

identify

problem

from

underlying

different

structures

mathematical

to

link

different

perspectives

areas

so that a peer can

follow your working

of

without stopping to

mathematics.

think .

•

Demonstrate

rigour

mathematical

by

using

arguments

clear

and

logic

and

language

in

your

calculations.

Academic honesty

This

is

very

Academic

class,

to

important

Honesty

make

consequences

According

“We

and

act

to

with

with

sure

of

all

Policy

that

your

which

you

IB

Learner

integrity

for

and

the

you

Your

should

school

be

what

will

given

to

have

an

discuss

malpractice

is

in

and

the

malpractice.

Prole

honesty,

dignity

work.

understand

committing

the

respect

in

for

with

and

a

rights

Integrity:

strong

of

sense

people

of

fairness

everywhere.

and

We

justice,

IA tip

take Reference any

responsibility

for

our

actions

and

their

consequences.” photographs you use

Academic

Honesty

means:

in your exploration,

•

that

your

work

is

authentic

•

that

your

work

is

your

•

that

you

conduct

•

that

any

work

including to decorate

the front page.

•

is

•

can

work

taken

from

properly

another

property

during

source

examinations

is

properly

cited.

work:

based

draw

yourself

intellectual

on

on

the

your

work

acknowledged

in

•

must

use

own

•

must

acknowledge

your

own

of

original

others,

footnotes

and

language

all

but

this

must

be

fully

bibliography

and

sources

ideas

expression

fully

and

appropriately

in

tnemssessa lanretnI

Authentic

own

a

bibliography.

741

/

16

E X P L O R AT I O N

Malpractice IA tip

The

IB

Organization

denes

malpractice

as

“behaviour

that

results

Plagiarism detection in,

or

may

result

in,

the

candidate

in

one

or

any

other

candidate

gaining

an

software identies unfair

advantage

or

more

assessment

components.”

text copied from

Malpractice

online sources. The

includes:

probability that a

•

plagiarism—copying

from

others’

work,

published

or

otherwise,

16-word phrase whether

intentional

or

not,

without

the

proper

acknowledgement

match is “just a

•

collusion—working

together

with

at

least

one

other

person

in

order

1

coincidence” is

1

12

10

to

gain

write

•

an

undue

your

duplication

any

other

taking

(this

includes

having

someone

else

exploration)

assessment

•

advantage

of

work—presenting

the

same

work

for

different

components

behaviour

unauthorized

examination

that

gains

materials

materials,

an

unfair

into

disruptive

an

advantage

examination

behaviour

such

as

room,

during

stealing

IA tip

examinations,

Discussing falsifying

CAS

records

or

impersonation.

individual

exploration

proposals with your

Collaboration and collusion

peers or in class

It

is

important

to

understand

the

distinction

between

collaboration

before submission

(which

is

allowed)

and

collusion

(which

is

not).

is collaboration.

Individually

Collaboration collecting data and

In

several

subjects,

including

mathematics,

you

will

be

expected

to

then pooling it to

participate

in

group

work.

It

is

important

in

everyday

life

that

you

create a large data

are

able

to

work

well

in

a

group

situation.

Working

in

a

group

entails

set is collaboration.

talking

to

others

and

sharing

ideas.

Every

member

of

the

group

is

If you use this

expected

to

participate

equally

and

it

is

expected

that

all

members

of

data for your own

the

group

will

benet

from

this

collaboration.

However,

the

end

result

calculations and

must

be

your

own

work,

even

if

it

is

based

on

the

same

data

as

the

rest

write your own

of

your

group.

exploration, that is

collaboration. If you

Collusion

This

is

when

deceive

with

write the exploration

two

others.

someone

friend

to

copy

or

more

Collusion

else

and

your

people

is

a

work

type

presenting

of

together

plagiarism.

the

work

as

to

as a group, that is

intentionally

This

your

could

own

be

or

working

allowing

collusion.

a

work.

References and acknowledging IA tip

sources

Be consistent and

use the same style The

IB

does

not

tell

you

which

style

of

referencing

you

should

of referencing use—this

is

left

to

your

school.

throughout your

exploration.

1

Words

&

matched

P .,

Ideas.

text

March

09,

is

The

Turnitin

likely

to

be

Blog.

Top

completely

15

misconceptions

coincidental

or

about

common

Turnitin.

Misconception

knowledge

(posted

by

11:

Katie

2010).

742

/

16

The

main

reasons

for

citing

references

are:

IA tip •

to

acknowledge

•

to

allow

the

work

of

others

Cite references to your

teacher

and

moderator

to

check

your

sources.

others’ work even if To

refer

to

someone

else’s

work:

you paraphrase or

•

include

a

brief

reference

exploration—either

the

appropriate

as

to

the

part

of

source

the

in

the

main

exploration

or

body

as

a

of

rewrite the original

your

footnote

on

text.

page

You do not need

•

include

a

full

reference

in

your

to cite references

bibliography.

to formulas taken The

bibliography

should

include

a

list

with

full

details

of

all

the

from mathematics sources

that

you

have

used.

textbooks.

Choosing a topic IA tip You

need

to

choose

a

topic

that

interests

you,

as

then

you

will

enjoy

You must include working

on

the

exploration

and

you

will

be

able

to

demonstrate

a brief reference personal

engagement

by

using

your

own

voice

and

demonstrating

in the exploration your

own

experience.

as well as in the Discuss

the

topic

you

choose

with

your

teacher

and

your

peers

before

bibliography. It is you

put

too

much

time

and

effort

into

developing

the

exploration.

not sucient just to Remember

that

the

work

does

not

need

to

go

beyond

the

level

of

include a reference the

course

which

you

are

taking,

but

you

can

choose

a

topic

that

is

in the bibliography. outside

the

choosing

syllabus

topics

and

that

are

is

at

too

a

commensurate

ambitious,

or

level.

below

You

the

should

level

of

avoid

your

course.

IA tip

These

questions

may

help

you

to

nd

a

topic

for

your

exploration:

Your exploration

•

What

areas

of

the

syllabus

are

you

enjoying

•

What

areas

of

the

syllabus

are

you

performing

•

Would

to

on

should contain a

most?

substantial amount best

in?

of mathematics you

prefer

work

purely

analytical

work

or

on

at the level of modelling

problems?

your course, and

•

Have

you

discovered,

mathematics

to

•

look

What

hope

•

through

courses,

areas

of

reading

or

talking

mathematics

that

to

peers

might

be

on

other

interesting

into?

be descriptive.

Although the history

mathematics

to

should not just

is

important

for

the

career

that

you

eventually

of mathematics can

be very interesting

follow?

it is not a good What

are

your

special

interests

or

hobbies?

Where

can

mathematics

exploration topic.

One

and

on

applied

way

of

create

a

mind

design,

for

744

on

area?

topic

map.

This

is

to

can

mathematics

how

explorations

of

there

your

a

shows

depletion

page

either

of

pages

suggestions

this

choosing

applications

following

On

in

fossil

is

own

an

or

lead

to

the

such

and

to

wor ki ng

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when

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B(0.5).

760

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Paper 3

Time

allowed:

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a

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a

H

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favourite

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a

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38

miles,

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The

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the

miles

10

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39

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15

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77

distribution

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town

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9

more

than

77

Manchester

miles

will

B

necessarily

be

longer.

b

Chapter 15

c C

B

7

Shefeld

Skills check 38

10

a

3

1

1

retsehcnaM

1

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er

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A

3

1

4 E

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3

D

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9

22

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C

12

d 2

a

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7

x

y

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y

=

=

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to

give

x

38

+

59

+

16

+

39

=

152

with

=

0 .6

8

miles

8

and

0 .4

Exercise 15B

A

b

As

an

10

B

example 1 0.6004

0.5994

0.3996

0.4006

10

P

 C

8

and

d

are

c

are

undirected,

b

and

directed.

D

2

c

a

b

a

i

0.4

A 6

7

B

C

D

11

Exercise 15A A

1

10

1

All

except

c B

B

1

1

C

1

1

A 10

D

1

842

/

b

ii

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is

easier

towns A

B

C

to

are

see

which

connected,

2

and

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adjacency

only

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zeros

matrix

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consists

ones

and

D

the

routes

between

the

has

zeros

on

the

diagonal.

A towns 3 B

8

C

9

c

7

i

Strongly

ii

Connected

strongly

a

i

Tutzing

and

Starnberg

and

Ammerland

of

entries

row

or

that

vertex

connected

but

Ammerland,

b

i

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not

column

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Ammerland,

ii

of

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row

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headed

by

elements

Bernreid,

Starnberg

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4

a

i

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of

by

elements

headed

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own

Ammerland,

ii

Bernreid,

Seeshaupt,

headed

in

the

that

by

in

the

that

vertex

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is

to

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knows

their

name

possible

know

the

someone

for

someone

name

else

of

without

C

D

them e A

either

vertex

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B

Sum

column

Starnberg

connected

A

Starnberg,

Starnberg

ii

in

5 d

3

and

Sum

4

D

b

Starnberg

a

in-degree

2

All,

except

knowing

their

Starnberg

out name. Ammerland,

degree

2;

B

Bernreid, 2-out

Tutzing,

in-degree

degree

1;

C

Seeshaupt,

b

i

3

ii

3

in-

Ambach degree

2

out-degree

in-degree

2

3;

D

out-degree

iii

iv

E

2 c

Exercise 15C

b

C

D

6

i

A

B

1

a A

B

C

D

Exercise 15D 6

A  0

1

0

1

1

0

1

1

0

1

0

1

1

1

1

0

A

B

C

1

0

0

0

0

1

1

 4

5

B

a

i

2

ABDA,

ADBA

ii

5

CBAD,

CDBD,



7

C





CDCD,

 D

D

C 7



b ii

A

in-degree

degree

2;

B

1

out-

in-degree

2

A  0

degree

1

1;

C

in-

B

out-degree

2;

D C

in-degree

2

out-degree









0

1

0

0

1

0

1

0



1



i

1

ABD

ii

c

i

2

ABEA,

AEBA

ii

5

CBED,

CBCD,

CDBD,



 D

b

 0

CBCD

0

D

 out-degree

CDAD,

d

i

0

CDCD,

CDED

ii

1

CCCD



2

4

A

in-degree

1

out-degree

3;

B

in-degree

2

out-degree

1;

C

in-degree

2

out-degree

2;

D

in-degree

2

out-degree

2;

2

DA

entry

in

M

is

0

c A

A  0

B

C

D

E

1

0

0

1

2

1

1

1

1

2

1

1

1

1

3

0

1

1

0

1

2

3

4

1

3

2

4

1

4

4

2

3

1

1

3

0

2

3

 B

1

0

1

1

a

i

M



1

 E

in-degree

3

out-degree

2 C

5



0

1

0

1

0

1

0



a D



1

0

1

Starnberg

 E



1

1

0

1

0

3

M

d

A

B

C

D

1

0

0

E



F

Possenhofen

A  0

1

0

Ammerland

 B

0

0

1

0

0

0 ii



Tutzing

C



0

0

1

1

0

0

0

0

1



Ambach

D



 4

0

1

0

0

6

2 

4

6

2



Bernreid

 5

 E

5



1

0

0

1

0

1

iii

S







6

3

 Seeshaupt



6

6

5

4

2

2

4

1



 F



0

0

0

0

0

0





 

 

843

/

ANS WERS

iv

S

contains

0s

but

S

1

b

does

Starnberg

to

Bernreid 1

2

1

0

not,

so

diameter

is

2

Starnberg

to

Possenhofen b

0

Seeshaupt

0

1

0

to

0

1

1

1

0

3

1

1

1

1

2

4

0

Possenhofen 2

3

1

Seeshaupt

0

0

to

2

Ambach

1

Seeshaupt

2

M



to

Starnberg

Seeshaupt 1

0

1

1

2

1

to

0

1

Starnberg

Ambach

to

Possenhofen

1

0

2

2

0

1

1 

0

4

1

1



3

2

0

0











0

4

2

4

4





1

4

2

3

1

1

4

3

2

 

1 0



iii

1



and

S

a

0s

1

1

2

3

4

0.222

d

i

does

not,

so

but

3.

5

b

Castlebay,

1

1

1

2

3

2

3

a

T

Lochboisdale

and

E

ii

18.75%

0

0.5

0.5

0

1

0

0.5

0.5

0

0

0

0.5

0

0.5

0

0

=



b

0.3 

ii 

Tobermory

0.2





 57

2

56

20

10

47

3

6

4

8

2



43

28

8

36

56



1

4

28

11

14

0.75

0.125

0.375

0.625

0.25

0

0.125

0.125

0

0.375

0.25

0.125



1

1 0

2



 20

0.125

3

T





6

0.25

2



56

0.5

58 

 2

 0.3

0 





c



0

 0.2 

and



3

3

diameter



5

4

1

3



S

1

0

a



B

0

3

is

c

2

S

1 0

3

1

contains

0

i

1

S

1

2

0

0

iv

1 0

0

2

1

2

1

3



1

ii

3

4

0

 1

1

Exercise 15E

3

M

1

1

2

 2

0

0

1

 0

0

4

1

to

0

b

Possenhofen

1

Ambach

0

1

0

3

0

0

3

i 1

0

18

20

2

2

Hence

not

B

in

possible

to

go

from

1 0



1

0

to

C

3

steps

or

from

A

to

A

2



 10

8

8

14

7

4

b

10



i

and



 47

2

36

18

4

32

0

47 

0



2

56

20

10

47

57

D.

 c

 58

to

0

2



A

1

i

A

number

of

B

two

of

to

the

get

0

0



trips

between

ports

is

5.

S

,

 19

C

4

2 ,



D 19

 19 needed

8



1

2

largest

 ,

 19



0

the

5



any

has



6

8



4

ii 

a

zero

element

at

Eigg-Tiree

 13



intersection,

so

this

is

19  0.462

the



4



d

B,

A,

D,

C

ii route

that

takes

ve



trips.

 13



5

a



0

1

0

1

0

0

 1



0

0

1

0

0

0



4

a

A

B

0.0769



 13



1

0.308

0.154 

2

0







 C

M



1

1

0

1

1

0

0

0

0

1

0

0

1

0

 13 

2

0

0

1

0

0

0

a

D

1 A

E

0

0

0

1

0

0

1

0

0

0

0

1

1

0

F

D

Town

order

is

Starnberg, B

Possenhofen,

b

It

would

only

be

possible

if

Tutzing, George

ended

on

Frances’

C

Ammerland,

Bernried, page,

Ambach,

otherwise

he

would

Seeshaupt

844

/

need

to

twice.

visit

This

Dawn’s

is

not

page

7

Exercise 15F

a

i

A

possible 7

1 as

you

cannot

pass

a

Minimum

the

other

pages

is

35, C

through

AF , all

weight

and

FE,

FD,

DC,

AB

E

end 8

at

Dawn’s

page

b

without

Minimum

weight

is

67,

F

5

B

repeating

some

AB,

pages.

AF ,

FE,

ED,

DC,

3

DG

5

c

Yes,

if

you

begin

on

c

Frances’

Minimum

weight

is

G

23, D

page;

Frances,

Dawn,

Emil,

AF ,

FC,

CB,

CD,

4

FE 6

Antoine,

Bella

and

H

Charles d

Minimum

weight

is

39,

FC,

CD

ii 1

1

2

3

AB,

1

0

0

1

(or

1

2 0

3

0

BG,

a

FC,

CD,

Minimum

b

BG)

weight

is

35,

DC,

FA,

1

3

0

b

0

Minimum

weight

is

ii

41

a

A

route

is

67,

2

ABFCFEFGBCDEGA, CD,

d

1 0

47

AB

1 0

i

Exercise 15G FD,

0

38

EF ,

0

3

1

FE,

0

2

0

BF ,

AB

(or

GD),

GD

1

0

repeated

0

(or

1

3

AB),

AF ,

FE,

2

weight c

1

Minimum

edges

CF

and

EF ,

ED

weight

is

23,

CD),

CD

103

1 0

0

0

3

b

0

BC,

AF ,

FC

(or

A

route

is

2

ABCDEFBFDBA, (or

FC),

repeated

FE

1 0

0

0

0

edges

0

d

Minimum

weight

is

39,

AB

and

BF ,

weight

EF ,

2

340 CD

(or

FB),

FB,

(or

CD), c

85

e



CF

0.131

(or

BG),

AB,

A

route

is

BG ABCDEFBECEFA,

648

3

You

begin

with

a

single repeated

1

f

1 

1 

1 

1

vertex.



Every

time

you

2

3

2

2

new

vertex

to

the

tree

Most

likely

EF

CB

and

BF)

weight

66

you 2

i

and

72

a

g

CE

add (or

3

edges

1



to

visit

Dawn’s

also

add

an

edge.

You

need

vertices

and

so

a

There

odd page

add

v

1

are

two

vertices

of

to degree

the

30

ii

Least

likely

to

visit

spanning

Belle’s

tree

will

have

v

1

b

i

160 +

= 170

3 edges.

page

minutes,

5

a

Because

two

states

4

are

a

AE,

CD,

ED

(or

CB),

ii

absorbing

states

probability

of

them

starting

b

of

will

and

entering

depend

(or

the

each

on

b

i

the

position.

1

c

ED),

AF;

Begin

by

B

D

and

$380,

algorithm

apply

from

because

the

ii

BD,

be

shown

by

having

this

and

DE,

FE

could

EA,

an

AF;

on

each

graph

edges

vertex,

every

would

even

a

two

hence

vertex

DC,

this

between

point.

3

and

000

connecting

and

GF

CB

have

degree.

1 1

0

0

0

Any

$400 000

0

route

which

2

traverses 5

a

75

b

17

each

edge

of

1

the 0

0

0

0

graph

twice

e.g.

1

3

ABCDEFGABGBFBCF

1 0

0

0

c 0

Because

it

is

difcult

to

CECDEFGA

0

check

3

whether

adding

an

3

a

$59

b

The

d

1 0

0

edge 0

0

will

form

a

cycle.

0

cost

will

go

down

as

3

6

a

63

all

vertices

now

have

even

1 A

0

0

0

1

0

10

F

degree

and

so

no

routes

D

3

11

13

need

be

repeated.

There

10

1

1

1

0

E

0

2

3

is

0

an

additional

cost

of

$7

C

3

10

9 B

but

a

saving

of

$9.

Total

G

cost e

is

$57.

0.125

b

10

845

/

ANS WERS

b

Exercise 15H

1

a

HB

6

4

BF

FD

7;

5;

HF

6

repeat

BD

5;

edges

HD

HB,

b

2

a

8

BE

21

BC

and

FE

BF

7;

B

C

D

E

F

0

6

7

12

10

4

13

CE

repeat

B

6

0

3

8

11

8

IC

to

70;

110

ID

110

DF

70;

repeat

DG,

route

C

7

3

0

5

8

5

D

12

8

5

0

7

10

E

10

11

8

7

0

6

F

4

8

5

10

6

0

GC;

Deleting

650

bound

iii

is

b

170

Either

a

lower

Odd

I

=

solution

29

ii

Deleting

and

a actual

route

AB

73;

to

or

I

and

are

38

A,

DH

B,

49;

AH

35

BD

repeat

AB,

possible

TSP

and

39

or

E

will

bound

of

C

will

give

weight =

lower

bound

of

31

iii

37

Solution

to

the

between

30

or

TSP

31

is

and

37

i

F

A

B

C

D

E

F

A

0

6

6

5

4

9

B

6

0

4

6

9

8

i

22

ii

Deleting

D B

will

give

a

AD bound

of

23,

and

88;

DG

route

D

deleting

C

and

6

4

0

2

5

is

D

5

6

2

0

C

gives

a

lower

4 bound

A

30

lower

deleting

need

the

is

length

C

a

lower

BH

to

gives

820

vertices

H;

32

neighbour

c

GH.

gives

between

30,

c

43

D

of

i

AIHFHIBHGFE

+

The

give Nearest

AFCBCDEFA,

and

ii

Possible

DGCDGCBA,

a

30

IH,

AFCBDEA,

3

i

edges

algorithm

b

a

13;

ii

HF ,

1

FE

DC

90;

need

11;

CF

100

CF

A

IF

BC

IF

Exercise 15J

DI A

and

i

3

of

24

6 iii

Solution

to

the

between

23

TSP

is

ABCDGABDGFDEFHGHA.

Weight

is

526

+

38

+

49

E

4

9

5

3

0

5

F

9

8

4

6

5

0

or

24

and

27

=

d

i

85

ii

Deleting

613 m.

b

He

would

have

to

repeat

ii

BD

Hamiltonian

bound which

he

is

has

and

88 m.

had

DH

to

repeat

which

AEDCBFA

Previously

total

routes

AB

30

87 m

or

As

he

needs

to

return

to

AEDCFCBA

best

place

to

be

30

or

27,

iii

27,

27

a

lower

of

95

upper

Solution

to

the

between

95

TSP

and

is

110

is 2

better

gives

AEDCBFA,

A

the the

or

AEDCBCFEA

weights c

D

cycle

a

Lower

bound

is

115

bound

picked b

up

is

need

at

to

length

will

B

he

repeat

of

be

so

will

DH,

repeated

d

only

so

a

A

B

C

D

E

A

0

40

45

25

10

B

40

0

25

15

30

C

45

25

0

20

35

D

25

15

20

0

15

E

10

30

35

15

0

49 m.

i

A

B

C

D

E

0

9

14

8

10

ii A

Hamiltonian

AEDBCA, B

9

0

7

A

B

C

D

E

A

0

35

20

25

10

B

35

0

25

35

15

C

20

25

0

30

10

D

25

35

30

0

20

E

10

15

10

20

0

roads

Exercise 15I

1

i

4

c

AECEBEDA

d

Lower

cycle

bound;

7 weight

7

0

6

8

4

6

0

a

Because

10

ii

7

Nearest

9

it

does

not

follow

ADEBCA,

3

0

triangle

upper

inequality

bound

the

NNA

is

the

solution

the

produced

far

higher

direct

by

ADEDBCDA,

is

not

to

the

route

weight

=

is

39

b

Once

E

it

is

the

not

adjacent

always

the

on

the

3

52

another

not

to

vertex

already

not

hold

graph)

4

a

reaches ADBCEA

been

upper

directly =

23

min

that

b has

does

triangle

hours

bound

reach

(the

TSP .

algorithm

possible

route

inequality

than

gives

actual

between

vertices

neighbour

algorithm

shortest

3 the

E

because

9 2

D

lower

110 route

14

than

route

AEDBCDEA,

C

105

Deleting

A

or

E

gives

a

passed.

lower

bound

of

21

minutes

846

/

c

The

of

9

minimum

the

edges

and

the

weight

adjacent

minimum

of

to

d

two

C

No,

of

is

there

length

between

weight

is

1

B

no

or

walk

iv

length

and

E.

All

2

The

vertices

must

to

have

degree

equal

except

for

has

out-degree

one

in-

out-degree

vertex

which

2

of

the

E

is

two

11.

Five

edges

This

edges

adjacent

adds

are

up

to

needed

matrix

to

20.

for

entries

and

as

the

smallest

+

for

M

has

these

zero

values.

is

an

more

a

than

another 5

cycle

M

a

Total

weight

its

one

in-degree

vertex

which

and

has

65

3

an

in-degree

one

more

than

C

then

the

must

be

solution

at

least

to

the

TSP

its

23.

A

d

d

Start

at

A

and

move

to

the

vertex

already

been

which

has

B

C

0

0

0.5 

0

0

0.5

A  0.5

F

nearest

out-degree.

D



G

not



B

0.5



 visited

but

not



C

B.

When

all

other

visited

go

to

to

to

go

A.

An

B

rst

to

alternative

and



0

  0



0

1

0



B

b then

0



E

D been

1



vertices D

have

0

then

65

+

9

+

11

=

85

is

use

0.25

6

a

i

The

graph

is

complete. 0.25

the

NNA

as

usual.

The

e

route

In

a

random

walk

ii 0.25

would

then

be

the

reverse

of

B

0.25

the

one

obtained.

16

approximately

equal

of

spent

amounts

8 22

Chapter review

time

will

be

at

each

10

1

a

There

are

C

F

two

odd

vertex.

C

A

vertices,

27

8 and

24

15

b

e.g.

a

C,

D,

b

CD

F ,

G

16

and

FG

1

+

4

=

5;

CF

CDECBEFBAF and

DG

5

+

4

=

9;

CG

and

25 B

C

DF

is

19

E

5

+

G

iii

74

i

Weight

c b E

8;

possible

walk

ABCDCEDGEFEGBFA;

weight

F

=

D

A

2

4

is

28:

is

32

begin

at

F

and

end

at

G,

43

D

or

begin

at

G

and

end

at

F

8

Order

of

CD(EF),

total

edge

CG,

weight

selection:

BF ,

FG(AB),

C

B

EF(CD),

9

a

The

cycle

through

AB(FG);

would

vertex

have

C

to

twice

pass

to

16

return

21 E

to

the

starting

point.

D

19

3

Possible

FCA.

Repeat

weight

4

a

route

b

ABCDEAEDFEB-

AE

and

ED.

ii

Total

472

7

i

No,

it

is

not

ii

No,

it

contains

a

No,

ii

multiple

edge

example,

between

A

there

and

is

ABCDE

c

Yes,

No,

iii

for

example

as

the

edges

are

A

B

C

D

E

A

-

13

5

12

9

B

13

-

8

22

12

C

5

8

-

7

16

D

12

15

7

-

10

E

9

12

16

10

-

directed.

symmetric

edges

iii

i

68

For

as

Yes,

it

as

circuit

no

the

E

contains

it

a

contains

that

loop.

a

includes

all

vertices.

b iv

No,

for

it

contains

example

a

circuit,

A

B

C

D

In-degree

2

2

1

1

Out-degree

2

1

1

2

BCB

10

b

A4,

c

i

B3,

No,

it

C4,

D4,

has

a

d

ACDEBA

a

Every

b

One

e

vertex

3

is

of

even

order.

E1

vertex

possibility

is

of ACEFDEBDCBA

degree

1

c

i

No

11 ii

No,

not

all

vertices

are

ii

even

For

each

degree

vertex

must

the

equal

a

AF ,

AB,

BC,

b

Algorithm

DF ,

DE;

17

in-

attempts

to

nd

the

the

‘optimum

choice’

at

each

out-degree iii

Yes,

it

has

2

odd

vertices stage iii

Yes,

DAABCDB

847

/

ANS WERS

12

a

EG,

EF ,

b

£6700

BD,

AC,

CE,

DG

ii

iii

B

D

11

b

4

B

→

C,

B

→

A

B

→

C

B

→

D

→

→

B

→

C,

→

B

→

A

=

4 min,

C

=

13 min,

E

=

11 min

 0

C

1 1

0

8

13 min,

1

1

0

0 

0

0

0

0



2

20

a

M





0

1

1

0

1

0

0

0

0

0

1

1

0

1

0

0





0



2 T

0



1

9



0 3

G

13

=



a 1

12

D

19 min,

 17

A

=

C,

14

E

B





 1



0 3

C

a

b

E.g.



2

F

1

1

2

3

0

13



1

0

0

6

8

7

3

2

5

4

3

2

2

5

9

5

5

1

5

6

3

3

3

9

8

8

3

3

FEDBAC

For

a

Eulerian

circuit,

1

all

1

must

be

even,

3

7

b 12

vertices

1

4

6

5

1

1

8

4

2

1

1

1

3

12

4

6

8

5

1

1

1

24

4

6

4

i

M



8

and

0

there

are

two

odd

vertices 3

T

b here

c

BF ,

(B

EF ,

and

AB

then

leave

1,

no

so



E).

or

a

AC:

This

vertex

of

Hamiltonian

would

degree

Therefore

journey

cycle

there

the

is

is

required

from

only

C

to

one

E

since

element

of

0.222

would

be

possible.

‘1’

here.

0.333

c 14

Weight

of

route

is

48.

ii

CBABACDE

0.222

One

possible

route

is 21

0.222

a

AB,

DF;

Weight

of

AFACABFEBCDEDA route d

15

After

a

of

a

large

time,

a

=

64

amount

ship

is

b

more

One

possible

route

is

B

likely

port

to

B,

be

based

hence

it

at

ACBABDCGFDFEDA

would c

24

If

Nasson

starts

at

F ,

the

31

likely 20

make

the

best

26

only place

for

the

possible

17

that

need

to

be

repeated

will

headquarters.

C

A

routes

company’s

either

be

AB

(=

6),

BD

13

18

a

18

A

B

C

D

E

F

G

A

0

4

5

10

9

9

3

B

4

0

3

8

5

5

5

C

5

3

0

5

8

8

2

(=

BC

+

(=

5).

so

this

CD

BD

=

is

3)

the

or

AD

shortest,

21 19

should

Therefore,

starts 25

E

CE,

AC,

AD,

BE

c

68

D

10

8

5

0

6

13

7

E

6

5

8

6

0

7

10

22

0

1

0

0

1

0

1

1

a

a

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Order

F

9

5

8

13

7

0

0

1

0

0

0

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1

0

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5

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7

By

0

2

1

0

2

1

2

2

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10

b

AGCBFEDA

c

AGCBFEDCGA



10

0

0

2

1

0

0

is

Upper

81.

A,

Kruskal

gives

for

CD,

3

 1

 2 1

0

1

1

19

By

the

remainder

CE;

weight

bound

deleting

=

B,

=

as

43,

75

Kruskal

gives

 MST



1

1

1 

2

3

2 

CD,

0

for

CE,

the

remainder

CA;

weight

as

47.

a Lower 

1 0



2

bound

is

therefore



1 0

0 

3

47

+

d

75

≤

e

E.g.

(13

+

15)

=

75.

1 

 1

1

3

2

1







3

4

3

 3 

i

ABCDEA.

0

BC,

1 0





0

c

is

deleting

MST

3

ii

nish



c

i

should

10

lower

b

he

Nasson

A.

bound

b 16

F ,

given

repeated.

D

at

b

at

be

0



1

3

2

1

1

1

2

1



L

≤

81



0 2

0

0 

tour

for

original

upper

 bound:

ABCDEA



 



 2

1

  2



1 0

2

1



0

 

848

/

Index

2D

see

two

dimensional

3D

see

three

dimensional

approximate

to

coupled

equations

acceleration

integration

459–62

508–10

dimensional

motion

account

540–7

balance

accuracy

afne

692–5

699

see

area

25–7

intersections

circles

also

629–83

sine

20–1

Cartesian

positive

22–3,

116–18

surface

bounded

area

by

curves

applications

488–9

of

approximation

358–9

269

functions

causation–correlation

relationship

248

bar

charts

67,

59,

573–4,

cdf

base

of

base

vectors

logarithms

bearings

bell

see

195,

cumulative

distribution

curves

320

cell

sampling

central

central

chain

arcs

of

25

limit

611–13,

distributions

581–5

diagrams

angles,

circles

47,

48–50

of

96–103

104

594–9

function

boundaries

Voronoi

104

19–20,

binomial

474–86

numbers

626

biased

498–503

form,

complex

power

585

integration

cases,

integration

343

25–7,

differential

equations;

median

symmetry

quadratic

anti-derivatives

hypothesis

mean;

functions

triangles

transformations

ambiguous

of

also

differentiation;

also

even

341–3

398–401

rule

axis

radians

of

see

of

51–2

see

derivatives

455–9

tendency)

234,

circle

second

275,

(measures

central

340–3

tables,

alternative

of

circles

sectors

vertices

averages

spaces

in

functions

276–8

area

graphs

networks

adjacent

of

lengths

699–703

adjacency

arcs

183

matrices,

network

556–61

472–531

angles

4–7

adjacency

irregular

power

differential

approximation

differentiation

three

solutions

theorem

638–9

rule

442–4,

448,

2

amortization

307–8

between

curves

amplitude

a

between

curve

of

function

periodic

347

axis

angles

bearings

between

planes

19–20,

lines

104

and

34

axis

and

474–86,

between

498–9

x

495–6

curve

and

χ

tests

661–5,

denition

expected

Poisson

500

link

495–7

binomial

492

chance

values

and

see

change,

585

probability

differential

equations

distribution

comparison

denite/indenite

integrals

582

variance

y

667

588

systems

of

547–61

characteristic

probability,

coupled

equation,

eigenvalues

415

2

of

depression

elevation

and

14–15

measurement

radians

in

340–3,

non-right

345

triangles

sectors

of

13–17

circles

25–7

vectors

annuities

121

307

495–6,

488–90,

data

boundary

66–72

conditions,

differential

movement/curves

sequences

equations

arrays

approximately

linear

complex

357,

358

sequences

see

and

lower

and

190–6

approximate

representations

4–11

series

184–8

658,

666

parameters

666–7

657–69

independence

650–6

whisker

Poisson

60–1

calculus

426–531

differential

515–24,

equations

547–61

derivatives

and

behavior

exponential

functions

geometric

496,

459–62,

sequences

298–300

rates

theorem

binomial

of

applications

508–10

change

applications

and

chords

26,

462–4

661–5

normal

distributions

657–61

429

circles

in

radians

340–3

arcs/sectors

511

kinematic

uniform

angles

426–54

fundamental

312

of

freedom

estimating

for

bounds

diagrams

rst

374

asymptotic

652,

of

distributions

numbers

arithmetic

tests

goodness-of-t

bounds

box

plane,

178–84

188–9

numbers

bivariate

upper

356–63

arithmetic

approximately

situations

diagrams

χ

degrees

515

482–5

arguments,

511

arithmetic

testing

362–3

antiderivatives,

integration

596–9

trapezoidal (trapezium)

Argand

104,

density

functions

Argand

hypothesis

631–3

476–82

rule

triangles

integrals

probability

17–25

right

denite

chords

unit

714,

25–7

circles

circuits,

and

343–5

network

graphs

716–17

849

/

INDE X

cis

form,

complex

numbers

classical

358–9

problem

248

and

organizing

variables

46–50

random

602–3,

216–17,

29–30,

conjugates

connected

604–13

32–3,

of

graphs,

event

probability

212–25

difference,

arithmetic

sequences

179

rate

of

ratio,

sequences

geometric

291–8

change

156

content

solution

or

for

439–40

validity,

type

critical

II

(d.p.)

4–5

455

area

553–5

550–5

bounded

of

type

indenite

errors

versus

633,

634,

link

total

511

(order)

vertices

677

integral

495–7

degree

I/

by

476–81

change

639–43

298–300

integrals

curves

tests

testing

values

sequences

293,

denite

regions

probability

tests

673

549

net

630–4,

307–8

derivatives

decreasing

complex

validity,

hypothesis

differentiation

428,

673

critical

constraints,

to

portraits

criterion

optimization

common

phase

places

function

291,

eigenvalues

270,

551

equations

imaginary

696

concavity

constant

points

547–55

general

decimal

306,

decreasing/increasing

solutions

linear

complex

271

combined

debt/loans

equations

equilibrium

34

354

networks

constant

548

547–61

exact

numbers

data

combinations,

common

probability

cones

determination

collecting

events,

218

of

denition

differential

conditional

723

coefcient

derivatives

455–6

travelling

salesman

722,

second

716,

of

696–7,

714,

719

degrees,

radian

conversion

340–1

2

complementary

probability

events,

216,

218,

223

contexts

of

functions

142–3

graphs,

networks

complex

data

725

eigenvalues

553–5

continuous

piecewise

functions

continuous

complex

46,

57–8

720–3,

numbers

density

353–66

166

diagrams

356–63

probability

functions

continuous

random

Cartesian

354–6

form

358–9

Euler/exponential

593,

continuous

change

rate

594–9

of

functions

511–12

matrix

applications

388–9

cubic

functions

cubic

regression

function

582–3,

259–67

263–5

distribution

(cdf)

573,

tests

of

freedom,

652,

deleted

658,

vertex

frequency

666

algorithm

725–6

demand

function

164

dependent

events

dependent

variables

integration

Poisson

probability

218

66,

derivatives

anti-derivatives

63–5

cumulative

χ

144–5

597

cumulative

variables

arithmetic

degrees

cumulative

594–9

Argand

653

cryptography,

continuous

complete

639,

function

differential

and

488–9

equations

515–24

591

cumulative

probability

of

equations

433

x

form

360

convenience

geometric

49

interpretations

356–63

polar/cis

form

364–6

powers

models

364–5

361–2

353–4

component

356–62

form

of

104

composition

of

functions

compound

interest

310

computer

402–3

concave

also

635–7

parametric

function

graphs

263,

270–1,

280

264,

triangles

plane

and

23–4

362–3

tangents

derivatives

344–5,

waves

348–52,

sketching

708,

438,

network

720–2,

cylinders

random

572,

593

systems

approximate

solutions

non-linear

equations

556–61

kinematics

and

limits

notation

28,

457

graphs

723

30,

31

430,

433,

433

459–62

428–42

430,

433

notation

430,

439

of

change

applications

462–4

second

364–5

notation

functions

rates

455–6

cycles,

derivatives

prime

sine/cosine

non-right

the

derivatives

455–9

trigonometric

functions

descriptive

449–51

statistics

44–81

outcomes,

discrete

to

second

product

rule,

in

435–6

625–9

452–4

433

of

431–2

Argand

rank

correlation

rst

gradients

normals

coefcient

coupled

system

lnx

fractional

movement

(PMCC)

coefcient

variables

curves

a

and

426–54

428–9,

moment

Spearman’s

counting

simulations

of

403

66–70

194–5

coefcient

cosine

302–8,

data

moment

174–7

213,

product

see

state

nding

correlation

planes

vectors

correlation

625–9,

formulae

e

589

curves

correlation

quadratic

complex

geometry

coefcients

functions

periodic

coordinate

bivariate

358–9

function

current

84–95

modulus-argument/

periodic

sampling

data

collection

46–50,

determinants

of

matrices

385–6

670–1

data

encryption

388–9

data

mining

data

representation/

671

description

44–81

determination

coefcient

248

diagonalization

matrices

of

416–19

850

/

differential

calculus

edges

426–71

e

differential

equations

515–24

systems

elds

519–23

rule

442–4,

minima

430,

law

433

444–6

446–8

442–8

direction,

108–9,

(slope)

elds

functions

159–60,

functions

data

discrete

probability

46,

51–6

functions

random

discriminant

quadratic

593

of

displacement

353

vectors

104–12

distance

and

distribution

speed

of

populations

643–5,

distribution

sample

domain

122

647,

of

638–

666

the

mean

608–12

restrictions

236,

241–4

cubic

functions

quadratic

235–6,

259

151–3

functions

241–4

decimal

places

form

5–8

versus

I/type

577,

10

data

57

parameters

data

638–40,

of

of

standard

643–5,

666

620–1

complex

numbers

functions

from

diagrams

interest

curves

264,

to

316

data

270–1,

287

goodness-of-t

fractals

370,

398,

tests

functions

433

142–54

Euler’s

formula

360

denition

Euler’s

method

domain

(e)

314,

452–4

vertices,

graphs

696,

radian

and

range

151–3

network

714,

716

values

and

logarithmic

coupled

differential

equations

547–55

inputs

values

binomial

distributions

585

discrete

and

sequences

290–310

series

296–300

events,

space

216,

sample

217

linear

form,

functions

157

gradients

curves

428–9,

nding

and

of

430

local

maximum

minimum

points

linear

431–2

437–8

equations

inverse

linear

outputs

functions

functions

to

and

curves

155–6

normals

435

graphical

representations

inverse

202–3

functions

logarithmic

171

functions

325–6

power

functions

quadratic

functions

variables

sine

and

345,

functions

155–67

between

143–4,

cosine

sketching

144–7

solving

and

235

differential

equations

147–9

functions

165–6

solving

of

curves

347–50

drawing

non-functions

notation

linear

146–7

piecewise

random

geometric

relationship

144–5

and

functions

234–5

168–77

to

quadratic

of

269–71

288–337

solutions

expected

174–5

145–6

exponential

number

401

notation,

composition

equations

descriptions,

tangents

734

differential

89

86–90

716–17,

for

equation

line

functions

183–4,

314,

derivatives

circuits/trails

geometric

derivatives

714,

320,

sites

99–101

248,

of

straight

87–92,

fractional

360

a

form

gradient-intercept

657–65

(exponential)

Euler’s

197

chance

calculations

599

normal

geometric

71,

Voronoi

tting

curves

253–7

320–8

point

fair

transformation

9–10

302–10,

63–5

sample

326

315–19

in

578,

nance,

cumulative

from

316

487–8

on

676–9

4–5,

314,

311

games

farthest

670

II

frequency

exact

511

chance,

also

given

families

random

exact

functions

see

164

340

domains

d.p.

price

error

even

496,

of

distributions

of

551,

522–3

629,

exponents

fair

Eulerian

formula

312

linearization

outcomes

points

see

functions

extrapolation

form,

572–92,

theorem

578

general

numbers

207

Euler

variables

of

391–2

of

Gaussian

360

logarithms

likely

629,

573

40,

413–18,

population

discrete

e

games

577,

complex

modelling

548–54

from

573

distribution

points

80–1

(Euler)

investments

551

continuous

distribution

calculus

310–19

base

estimation

cumulative

discrete

complex

stable/unstable

equally

form,

numbers

553–5

type

271–4

discrete

or

systematic

variation

207,

experimentation

exponential

measurement

519–23

direct

413–18

systems

equilibrium

537–9

directional

rate

558

vectors

119–21,

coupled

equilibrium

696

fundamental

probabilities

exponential

enlargement

graphs,

networks

interest

eigenvectors

439–40

rule

430,

489

experimental

208–9

equilibrium

430

rule

quotient

and

rules

optimization

product

448

437–40

integration

directed

320

imaginary

maxima

notation

314,

548–55

differentiation

nding

number)

eigenvalues

slope/directional

rules

(Euler’s

28–9

303

547–61

power

polyhedra

effective

coupled

chain

of

519–23

equations

single

variables

149–50

x

e

452–4

edges

of

variables

network

692–727

graphs

Poisson

591

575

distribution

power

and

polynomial

functions

232–87

trigonometric

338–71

transformation

quadratic

of

functions

253–7

851

/

INDE X

graph

theory

networks

Chinese

of

i

690–735

postman

problem

714–19

constructing

graphs

692–8

spanning

708–13

travelling

720–7

unweighted

upper

function

identity

line

identity

matrices

images

699–707

and

bounds

salesman

problem

solutions

722–6

weighted

graphs

692,

720–2,

histograms

593–4,

734

62,

(i)

parts

of

numbers

for

356

sequences

of

probability

algorithm

diagrams

lines

697

form

normal

simple

functions

graphs,

arithmetic

178

linearization,

exponential

323

power

241

linear

variation,

314,

157

sequences

power

274–8

functions

190–6,

transformations

393

random

310

variables

601–2

functions

316

linear

transformations

plus

interest

326

636–8

matrices

interest

and

regression

linear

investments

vertex

variation

165–6

598–9

functions

155–67

156

piecewise

260

exponential

a

functions

linear

302–8,

487–98

118–22

gradient-intercept

functions

quadratics

245

159–60

120–1

cumulative

integrals

equation

denition

sets,

of

compound

of

53–4,

216–19

intersections

inverse

Voronoi

linear

range

logarithms

293

in-degree

testing

197

direct

intersection

derivatives

quadratic

vectors

63–4

cubics

455

indenite

329–30

interquartile

equations

relationship

168–77

increasing/decreasing

428,

310

183–4

predictions

inverse

96

asymptotes

hypothesis

553–5

302–8,

interpolation,

90–2,

incremental

605

horizontal

312,

59,

723,

147

complex

number

increasing

cycles/

381

353–6

291,

paths

or

linear

compound

60,

eigenvalues

354,

interest

simple

391–2

functions,

708–27

Hamiltonian

171

transformations

complex

travelling

177

functions

imaginary

lower

of

of

imaginary

graphs

number

identity

imaginary

salesman

problem

692,

imaginary

image

minimum

trees

see

183–4

translations,

afne

transformations

2

622–98

independence,

binomial

probability

650–69

irregular

spaces,

398

approximations

independent

631–3

test

χ

events

218,

lines

472–531

intersections

90–1

2

χ

goodness-of-t

test

657–69

608–10

iterative

independent

variables

numerical

midpoint

approximations,

of

segments

line

84–6

2

χ

test

for

66,

independence

innity,

650–6

and

interpretation

statistical

of

tests

choosing

signicance

676

670–1

80–1

distributions

638–50

634–5

moment

coefcient

to

relationship

truth

of

hypothesis

sample

635–7

null-

681

selection

670–1

Spearman’s

correlation

coefcient

type

I/type

676–9

denite

indenite

integrals

II

errors

also

see

velocity

and

acceleration

Kruskal’s

algorithm

498–515

integrals

link

lateral

and

least

508–10

cones

length

arcs

process

488–9

likelihood

reverse

chain

limits

rule

492

rules

vectors

of

motion

volume

intercept

linear

155–7

see

of

variable

in

three

of

540–7

solids

of

503–7

parameter,

functions

636–8

25–7

probability

428–42

geometric

of

notation

298–300

430

combinations,

random

variables

602–3

linear

coupled

exact

systems,

solutions

549–51

best

also

t

68,

see

tting

69,

curves

data

natural

logarithms

loans/debts

local

306,

maximum

minimum

259–61,

or

points

(logarithms

10)

307–8

437–8

base

320

logarithmic

sequences

linear

lnx

logx

derivatives

489–94

regression

191–2,

of

pyramids

33

squares

line

kinematics

of

line

190

see

area

straight

equations

lines

709,

also

to

applications

93–4,

97–103

integration

revolution

625–9

459–62,

710

dimensions

rank

bisectors

508–11

495–7

correlation

p-value

calculus

vector

see

93

perpendicular

see

integrals

and

perpendicular

sequences

306

state

parallel

522–3

denite/indenite

distributions

product

ination

equations

kinematics

integrals;

experimentation

Poisson

of

408–9

collection

normal

limits

298–300

initial

670–83

data

differential

geometric

choice/validity

levels

144–5

functions

320–8

logarithm

laws

logarithms

320,

322–3

(natural/lnx)

452–4

logistic

functions

log-log

graphs

long-term

matrix

329–30

325

probability

409

852

/

loops,

network

graphs

700

lower

quadratic

functions

power

234–6

limit,

integration

476–7

second

quartile

lower

and

53

upper

bounds

approximation

bounded

mean

derivatives

by

of

area

curves

474–6

51,

process

596,

57

non-linear

problems

systems

models

343–52,

matrices

17–18,

364–5

normal

coupled

556–61

non-right

sinusoidal

634–5

rate

385

245–51

distribution

590–1,

functions

real-world

638–50

Poisson

non-invertible

234–58

distributions

interest

303

160–3

quadratic

54–5,

normal

nominal

268–79

456–7

lower

functions

triangles

23–4

distributions

2

travelling

salesman

problem

722–6

uncertainty

probability

density

functions

in

5–7

combined

variables

magnitude

of

vectors

measure

108–9

diagrams/

mapping

see

relationships

144–5

Markov

415,

of

and

random

601–3

424–5,

mathematical

402,

5–7,

transformation

quadratic

continuous

notation,

of

253–7

matrices

variables

measures

functions

arithmetic

measuring

eigenvalues

median

and

eigenvectors

matrices

60,

of

minimum

multiplication

space

segments

413–18

of

trees

377–82,

394

dispersion

in

345

2–43

63,

64

line

84–6

(scalar)

375–7

modal

708–13

of

identity

381

mode

57

inverse

385–6

network

powers

of

593–614

functions

discrete

701–2

products

random

representing

381

random

samples

for

643–5,

647–9

networks

events

two-tailed

394

identity

z-test

of

for

mean

381

multiplicative

tests

647–9

of

377–82,

multiplicative

648–9

population

variance

362–3

inverse

of

n

exclusive

vectors

curves

to

123

435–6

innity,

geometric

218

641–3

638–40

to

tending

tests

population

normalising

normals

385–6

320,

logarithms

(lnx)

sequences

nearest

functions

315–19

nearest

nth

term

null

neighbor

arithmetic

travelling

numbers

form

problem

sets

neighbor

interpolation,

179

hypothesis

in

629–83

standard

9–10

number

diagrams

of

sequences

452–4

722–5

572–92

exponential

systems

t-test

the

596

two-sample

salesman

variables

379,

standard

deviation

and

700

algorithm,

259–67

416–17,

and

paired

plane

multigraphs,

natural

continuous

cubic

699–705

in

testing

298–300

variables

graphs

mean

626

51

modelling

multiplicative

Argand

mutually

data

velocity

tests

638–50

511–12

behavior

movement

matrices

class

hypothesis

for

acceleration

matrices

statistics

continuous

multiplicative

see

594–9

657–61

function,

integral

matrices

spanning

misleading

a

multiplication

61–2

multiplication

of

random

goodness-of-t

357

monotonic

52

645–6

variables

complex

606–11

population

mean

numbers

change

motion

340–3,

51,

midpoint

random

angles

measuring

375–83

modulus

total

593

of

radians

372–425

of

denite

measurements,

704

for

form

666

intervals

continuous

numbers

670

Chains

combinations

systems

358–60

error

657–61,

condence

complex

modulus

tests

χ

modulus-argument

of

median

measurement

to

402–12

(averages)

mean;

matrices

represent

central

tendency

mapping

596

transformed

measurement

using

system,

number

353–4

Voronoi

observed

98

frequencies

2

402–12

steady

tting

state

systems

406–12

transition

rst

the

matrices

406–9,

418–19,

maxima

of

392–401

403–4,

and

704–5

minima

derivatives

437–8

local

248,

264,

to

data

270–1,

287

transformations

plane

curves

linear

points

437–8

optimization

439–40

142–54

functions

155–67

logistic

329–30

motion

and

change

exponents

negative

integrals

net

change,

change

nets,

2D

3D

rate

68

11

481–2

338–71

functions

odd

test

χ

vertices,

graphs

633,

function

objects

31–2

651

network

714,

716,

tests

719

630,

639

one-to-one

representations

versus

many-to-one

functions

242–3

optimization

theory

unweighted

692,

table,

one-tailed

of

networks

graph

phenomena

232–87

negative

of

532–69

polynomial

correlation

511–12

functions

periodic

turning

259–61,

functions

negative

690–735

graphs

708–27

graphs

constraints

and

439–40,

470–1

699–707

weighted

differentiation

692,

graph

theory

networks

of

690–735

853

/

INDE X

order

of

matrices

order

of

vertices

714,

716,

oscillating

374–5

696–7,

719

Poisson

291

algorithm

710–13,

587–92

binomial

sequences

Prim’s

distributions

distribution

comparison

principle

726

axis

curves

588

domain

236,

of

periodic

restrictions

241–4

graphical

347

representations

2

out-degree

of

a

vertex

χ

goodness-of-t

661–5

697

outliers

53,

60,

combinations

629

parabolas

271,

234,

479,

parallel

of

lines

polynomial

93,

674

and

94

graphs

correlation

coefcient

69–71,

352,

(PMCC)

194–5,

625–9,

percentage

248,

in

sequences

percentage

error

percentiles

from

cumulative

6–7

frequency

106–7

correlation

68

268–79

variation

271–4

variation

amplitude

and

period

power

326

law/rule

phase

shift

periodic

348,

365

phenomena

modelling

338–71

sinusoidal

models

343–52,

numbers

364–6

perpendicular

93–4,

components

numbers

of

of

vectors

practical

perpendicular

lines

93,

435

489–91

722,

272

matrices

701–2

travelling

systems

723

379,

348,

365

product-

form

equations

of

455–6

using

of

89

inexion

261,

prime

160,

193–5,

601–2

mean

distribution

coefcient

248,

635–7

444–6

matrices

608–10

random

walks,

graphs

range

52,

range

of

of

60

functions

151–3,

rates

653–4,

29–30,

235–6

change

288–337

functions

511–12

derivative

681

32–3

network

704–5

continuous

636–7,

of

combinations

140–203,

631–3,

quadratic

162

673

59

430,

quadratic

formula,

applications

462–4

differential

roots

gradients

353–4

functions

calculus

of

net

symmetry

234,

248

433,

parameters

from

245–58

versus

also

rational

real

511–12

functions

change

see

determination

428–9

integration

linear

234–58

axis

notation,

derivatives

439

sample

hypothesis

complex

data

and

426–71

behavior

presenting

mean

combinations

linear

42–3

pyramids

variables

transformations/

415

models

validity

coefcient

point-gradient

402–4,

670

variance

572

381

testing

580–1

correlation

1

and

rule,

p-values,

47,

601–13

zero

194–5,

639–42,

probability

random

moment

68–9

regression

408–9,

functions

165–6

see

data

194–5

periodic

functions

proof

bivariate

linear

550–5

shift,

piecewise

coupled

least

of

see

sampling

changing

moment

products

670

probability

number

differentiation

problem

events/

successes

625–9,

product

580–1

random

580–7

correlation

352,

48

errors

interval

equals

69–71,

predictions

portraits,

product

670

random

random

(PMCC/Pearsons’)

361–2

salesman

538

sum

complex

416–17,

perpendicular

271

59

linear

of

at

46,

behavior,

prediction

48,

successes

xed

values

data

340–3

random

573

xed

number

regression

powers

97–103

of

trials

of

theorem

random

of

powers

bisectors

a

204–31

sampling

processes

638–9

discrete

59

of

questionnaires

radians

588

limit

612,

46,

continuous

distribution

430–1

power

complex

central

data

quantitative

quota

distribution

a

integration

364–5

570–621

587–92

proportional

qualitative

random

in

quantities

multiple

variables

differentiation

directly

248–51

uncertainty

density

581–5,

in

245,

regression

quantication

415

variables

derivatives/

347

of

402–4,

functions,

number

linearization

functions

states

representing

binomial

functions

704–5

functions

274–8

periodic

points

tests

vectors

inverse

quadratic

graphs

probability

245–50

251–7

594–9

647–8

power

212–24

408–9,

random

643–5

direct

63–5

using

from

638–40,

situations

218

probability

parameters

629,

positive

293

laws

real-world

transformations

outcomes

647

samples

modelling

events

systems

non-

t-tests

estimation

position

combined

predicting

234–67

versus

234–5

204–31

network

functions

models

pooled

590

complex

two-sample

637

change

geometric

variance

31

212–25

28–9

population

product

testing

358–9

tests

pooled

moment

line

form,

polyhedra

network

Pearson’s

PMCC

and

checking

720

phase

polar

28–9,

diagrams

numbers

538

forms,

parallel

of

604–6

634–5

mean

530–1

reliability

paths,

269,

components

vectors

parallel

248,

prisms

probability

hypothesis

phase

test

155–6

total

512

acceleration

exponents

numbers

353,

11

593

854

/

real

parts

of

numbers

real

value

complex

354,

356

of

domain

of

modelling

random

variables

572–92

processes

248

reasonable

functions

rectangular

151–2

256,

398

cubic

functions

linear

quadratic

regular

636–8

functions

245,

272

functions

248–51

related

of

change

positions,

674

fallacy

outcomes

570–621

representing

numbers

approximations

standard

form

representing

4–11

391–2

67,

69

9–10

systems,

of

circles

25–7,

of

models

height,

stretches

391–2

strongly

cones

99–101

pyramids

33

differential

see

696,

graphs,

networks

probabilities

206

of

square

to

elds,

three

geometric

solids

coupled

sequences

function

surface

area

of

164

solids

11,

31–3

functions

248,

261,

269

systematic

errors

systematic

sampling

670

47,

78

rank

systems,

coefcient

velocity

122,

represented

matrices

systems

and

of

299–300

234,

551

Spearman’s

residuals

innity

symmetry,

systems

696

708

supply

equations

points,

47

connected

subgraphs,

sum

(directional)

speed

sampling

191–2

Voronoi

96,

118–22

stratied

sum

364–5

diagrams

slant

385

form

of

by

402–12

linear

equations

383–91

459–60

398,

401

see

also

velocity

and

acceleration

differential

spheres

515

square

178–89,

29,

30,

matrices

33

t-test

374,

equilibrium

signicant

gures

643–5,

of

points

tangents

deviation

52,

distances

network

720,

to

428–9,

standard

647–9

least

graphs

551

212–24

table

643–4,

676

(weights),

stable/unstable

184–8

t-distribution

646,

381

probability

see

matrices

625–9

290–309

s.f.

singular

form

subjective

correlation

derivatives

sequences

sets,

triangles

17–18

source

340–3

series

20–1

120

89

directed

cases

form

gradient

networks

dimensional

535

equations

multiple

364–5

ambiguous

solids

vector

diagrams

345,

519–23

375–7

factors,

separable

580–1

representing

of

curves

rule

slope

self-similarity

representativeness

347–50,

and

products

versus

sectors

reliability/repeatability

tests

207–25

455–9

124–5

(cosine)

sites

multiplication

scatter

sine

343–52,

214

investments

(dot)

second

vectors

212,

217

see

sampling

equations

87–92

parametric

vector

sinusoidal

space

matrices

scalar

random

simultaneous

limit

638–9

transformations

462–4

of

scalar

scale

408

rates

relative

612,

quantities

transition

matrices

central

112–14

190–6,

power

size,

183–4,

point

non-right

theorem

scalar

263–5

47–50,

670–1

savings

interest

simple

sine

probabilities

regression

combined

selection

givens

simple

246

distributions

diagrams

255,

666

608–10

sample

form

254,

393,

see

643–5,

distribution,

sample

form

Cartesian

391,

of

gradient-intercept

form

47

629,

mean

sample

domain

range

reection

sample

networks

302

626–9

parameters

normal

modelling

between

population

638–40,

discrete

graphs,

700

data

estimating

236

simple

551

variables

quadratic

functions

and

sample

306

context

160–3,

points

correlation

investments

real-world

saddle

723–4

curves

435

technology

(GDC’s

etc)

2

matrices

rescaling

402–12

vectors

residuals,

123

linear

regression

restricted

Sicherman

sigma

dice

domains,

214

54–5,

57,

standard

notation,

arithmetic

190–2

212,

states

series

of

585,

form

596

9–10

signicance

statistical

levels

test

for

independence

653–4

systems

cubic

402–12

184–5

χ

functions

259–60

measures

2

functions

resultant

151–2

vectors

χ

110,

reverse

chain

revolution

rule

solids

right

pyramids

right

triangles

roots

492

503–7

13–15,

independence

650,

rotation

parabolas

391

rotational

symmetry

269

testing

in

type

errors

I/

676

moment

4–5

network

probabilites,

gures

637

(s.f.)

437

the

area

probability

curve

478

under

density

596–8

formulae

523

704–5

systems

logarithms

320

multiplicative

steady-state

vector

straight

equations

line

general

nding

iterative

graphs

406–12

636,

integrals

derivatives

a

steady-state

correlation

coefcient

629–83

descriptive

steady-state

II

signicant

tests

44–81

630–1

product

479

statistical

statistics,

hypothesis

type

denite

52–9

importance

33

346

of

261,

for

653

535

343,

test

form

89

409

of

matrices

numbers

form

in

inverse

385–6

standard

10–11

855

/

INDE X

power

functions

270–1,

product

tree

272

probability

moment

trees,

correlation

coefcient

quadratic

and

635

terms

234–5,

addition

diagrams

96

178

checking

reliability

theoretical

28–36,

dimensional

two

area

84–6

solids

31–3

dimensional

area

using

28–30,

dimensional

ratios

111–12,

TSP

integral

of

function

waste

problem

trails,

denite

modulus

t-test

of

511–12

two

99–101

716–17

trajectories,

portraits

phase

of

550

complex

the

numbers

plane

quadratic

361

391–401

functions

type

4,

variables

604–13

matrices

406–9,

234–5,

study

unit

71,

nets

dimensional

tests

647,

I/type

II

from

linear

velocity

193–5

tests

concavity,

263,

52,

53,

integration

cubic

508–11

dimensions

velocity

57

equations

122–4

distributions

distributions

velocity–time

Venn

647

graph

diagrams,

probability

distribution

216–18,

vertical

variables

density

277,

and

601–3

of

lines

221

functions

line

products

vector

quantities

function

or

function

146

115

for

non-

or

minimum

535–9

of

quadratics

network

addition

276,

vertices

vectors

of

test

maximum

vector

213,

278

vertical

random

212,

asymptotes,

power

596

equations

104–6

234–5

graphs

692–727

triangles

polyhedra

volume

systems

of

28–9

solids

28–30,

503–7

Voronoi

direction

676–9

459–62

540–7

264

548–9

errors

acceleration

511

coupled

678

equations

and

three

116–18

630,

123–4

differentiation

670,

590–1

area

31–2

106–7

vectors

velocity

118–22

geometry

of

122–4

functions

vector

law

addition

within

combined

437–8

unbiased

estimator

population

403–

418–19,

704–5

transition

of

540–7

119–21,

diagrams

96–103,

134–5

537–8

601–2,

transition

trends

variable

vector

dimensions

triangle

671

transformed

251–7

random

data

functions

647–9

dimensional

641–2,

mining

probability

dimensional

two-tailed

data

Poisson

trig

449–51

points

objects

transformations

of

of

problem

643–5,

three

622–89

643–5,

86–95

two

validity

normal

radians

coordinate

graphs

three

585

13–27

travelling

turning

53

111–12,

binomial

13–15

259–61,

dump

network

714,

in

salesman

change,

toxic

see

quartile

variance

19–20

functions

540–7

a

13–15

17–23

derivatives

503–7

total

23–4

340–3

volume

vectors

rule

triangulation

31–2

three

17–18,

angled

angles

steady-state

673–4

23–4

systems

409

statistical

rule

of

408–9

integration

regression

vectors

trigonometry

nets

states

bounds

predictions

22–3

sine

lower

197

area

right

limit,

range

radians

non-right

dimensional

surface

vector

116–18

207–11

three

of

106–7

in

cosine

upper

see

207–9

13–25

angles

674

probabilities

geometry

upper

341–2

sequences

test-retest,

three

law

triangles

upper

bounds

476–7

experimental

triangle

640

in

graphs

probabilities

643

z-test

network

trials,

functions

models

Voronoi

and

222

708–13

248

t-test

upper

diagrams,

403

variance

uncertainty

networks

uniform

displacement

eigenvectors

104–12

4–7,

204–31

graphs,

walks,

413–18,

548–54

643–5

unconnected

probabilities

of

692,

waves,

forces

535–6

initial

state

distributions,

696,

see

vector

weighted

graphs

magnitude

graphs

701–8

also

sinusoidal

models

408–9

696

network

and

network

692,

Chinese

708–27

postman

2

transition

state

diagrams

translations

plus

linear

transformations

398

trapezoidal

(trapezium)

salesman

problem

(TSP)

union

of

test

720–7

probability

216–19

notations

position

unit

circles

unit

vectors

343–5

123–4

data

unweighted

graphs

direction

108–9

normalising

sets,

univariate

482–4

travelling

goodness-of-t

657–61

403–4

rule

χ

66

network

692,

699–707

relative

problem

123

minimum

104

vectors

trees

106–7

spanning

708–13

travelling

positions

714–19

salesman

problem

720–7

124–5

rescaling

123

resultant

vectors

535

y

intercepts

155–7,

235

110,

z-test

638–40

856

/

MATHEMATICS: APPLICATIONS AND INTERPRETATION

HIGHER LEVEL

Written by experienced practitioners and developed in cooperation with the IB, this concept-

Authors

based print and enhanced online course book pack offers the most comprehensive suppor t

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practice questions and papers, and a full chapter suppor ting the new mathematical

Jennifer Chang Wathall exploration (IA)

➜

Suppor t to develop a mathematical toolkit, as required by the syllabus, with modelling

and investigation activities presented in each chapter, including prompts for reflection,

and suggestions for fur ther study

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investigation activities

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Complete alignment with the IB philosophy, challenging learners with activities to

FOR FIRST ASSESSMENT target ATL skills, 'international-mindedness' features and topical Theory of Knowledge

IN 2021 links woven into each chapter

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